Lie-algebras-with-triangular-decompositions.pdf

P'

E'

Since jS is injective, so is /3', and furthermore A — coker /3 = coker j8' by Lemma 1. This allows us to uniquely define a': E' ^ A so that E:

0

B

-^ E

-^ A

► 0

I ^ E '^ A

► 0

4 0

B'

One now has to check that this procedure maps equivalent extensions into equivalent extensions. Hence we have a map
E x t(^ , J3').

Furthermore (i) if B ^ B ' ^ B", then (^V) *=
1.12

75

Extensions of Modules

In precisely the same way if il/: A' ^ A is a. morphism then starting from the extension (¿ ), we can use pull-backs to get 0 ---- >B

>0 1

0 ---- - ^ E ' ^

A ’ ---------^0

and hence derive iff*:Ext{A,B)

Ext{A',B)

with

Lemma 2

Let ¿’e ExtiA, B), and let ij/: A' ^ A and
Following [Str] we prefer to write
and

Lemma 2 follows directly from the easily proved next lemma: Lemma 3

Suppose that we have the commutative diagram E:

0

E':

0 ---- » B'

A' ---- >0

B

1 >E'

Then < p ^ ( E ) = r { E ’).

*1

so that

76

Lie Algebras

Proof o f Lemma 2. Apply Lemma 3 to the top and bottom rows of 0 ---- » B

E' *A’ ---- »0 I Pull-back I

0 ---- > B

E --------- *A ---- »0 Push-out I

1

E:

1 1 0 ---- > 5 ' --------- > E"

>A ---- >0

Using this we can now define an abelian group structure on Ext(A, B), Let E:

Q --> B -^E ^A ^Q

and E^\

Q^ B

jEj

—^ A —>0

be representatives of two classes ^ and obvious maps,

of Ext(yl, B). Then, with the

E e E^: 0 ^ B e B ^ E ® E i - ^ A e A - ^ 0 is a representative of a class in Ext(^ 0^4, B 0 J?). Let V: B ® B ^ B and A: ^4 0 ^ be the maps + ¿2 a (a, a), respectively. We define ^

= V (^ 0

(see notation following the statement of Lemma 2). We leave it to the reader to prove that this is indeed an associative operation and that the split extension 0-^B ^A ® B -^A -^0 is the identity element. The inverse of the class (E ) of E:

O ^B ^E ^A -^0

is (¿ ( —id)) obtained from the morphism —id: A ^ A, a ^ —a. This can be

1.12

Extensions of Modules

77

easily shown: E + E ( - i d ) = V (£ © £ ( - id ) ) A = V{E © £ )(id , -id )A = (V (£ © £ ) ) a , where K: A ^ A ® A given by a •-» (a, —d). Now the construction of V(£ © £ ) is shown as the top two rows of E e E

B ®B

A ®A

Push-out A

B

V(£ © E):

A ®A

E2 (id,0)

0

proj

B ®A

V A

0

In the lower part of this diagram, the part that needs explaining is the map ¡jl: a - ^ ^ 2* Given a ^ A , we let e ^ E be any lift of a, and define fi(a) = X(ie, —e)). This is independent of the choice of e because V is addition of components. Now the bottom right-hand square is a pull-back because ()8, i n d u c e s an isomorphism between ker(proj) —B and ker a — B (use the dual of Lemma 1). But the pull-back of E2

A ®A Ta A

defines (V(E 0 E))K, so this extension is split. We now come to the main object of study. Consider an exact sequence of modules in ^ : B: Let A be any module in Section 3.1].

0

B ^ B' ^ B" ^ 0. Then we have the usual exact sequence [Ja 2,

0 ^ H o m (^ ,B ) ^ H o m ( ^ ,£ ')

Ho m{A,B").

From the above we have a sequence of maps E xt(yl,B ) - ! ^ E x t( ^ ,B ') ^ E x t { A , B " ) .

78

Lie Algebras

We construct a connecting homomorphism 5:

H o m (^,B ") ^ Ext(y4,B)

by 5 (/) = (Bf). (The top two rows of the diagram below illustrate this.) Proposition 4 The sequence 0 ^ H o m ( A , B ) ^ H o m (^ ,B ') ^ H om (/4,B") - ^ E x t( ^ ,B ) - i^ E x t( /l,B ') - ^ E x t ( ^ , S " ) is exact. Proof. We content ourselves with showing that the sequence of Proposition 4 is a null sequence. The rest appears in [Str] or may be taken as an exercise. First observe that = (il/(p)^ = 0. Now let / g Hom(>l, B"), and con­ sider E = 8f. The definition of 8f is given by the top two rows of the diagram. B'

B .

1

E:

0

B

hB'

y

B" Pull-back

4 E Ufj)

(id,0)

B' ® A

proj

1^ A 1 A

The bottom two rows clearly form a commutative diagram. By Lemma 1, since induces an isomorphism coker

5 = 0. In the same way let

B' — > J?" «A

1.12

Extensions of Modules

79

for some g, and let us see that 8f = 0. Looking at the diagram B'

> l>

(»>.g)| 0 ---- * B ----- *B e A (id,0)

B"

^0

i' proj

we see from the dual of Lemma 1 that the right-hand square is a pull-back. But this is the definition of 8f, and hence 3 f = 0. □ Corollary 1 Let 0 ^ B ^ B' ^ B" 0 be exact. Then Ext(A, B') = 0 if both Ext(v4, B) = 0 and Ext(^, B") = 0. □ Corollary 2 Let B = Bf^ Z) Bf^_i D • • z) Bi ^ B q = (0) be a filtration o f B. Then E x t(^,

= 0

/ = ! , . . . , / : = > Ext(y4, B) = 0.

Proof Use the exact sequences 0

B j_ ^

to prove inductively that Ext(A,

Bj B f)

B j/ B j _ ^

-> 0

= 0, ; = 1 ,2 ,... .

Along similar lines given B\

0 ^ B ^ B '^ B " -^ 0

and A as above, we have the exact sequence 0

Hom(5", A) ^ H om (5', A ) ^ Hom(B, A)

and also E x t(5 " ,^ ) ^ E x t ( B ', ^ ) - ! ^ E x t( B ,^ ) . If the connecting homomorphism 5:

H om (S, A ) ^ Ext(B", A )

80

Líe Algebras

is now defined by d( f) = (fB), then we have Proposition 4' The sequence 0 ^ Hom(B", A )

Hom(B', A )

Ext(fi", A) ^

A) ^

Hom(B, A) Ext(B, A )

is exact. Corollaries analogous to those following Proposition 4 can then be ob­ tained. EXERCISES 1.1 Let F be a finite-dimensional vector space over a field /3: F X F ^ IKbe a bilinear form. (a) Prove that

and let

9 = g(/3) — [a G End(V)\ß{ax,y) + ß ( x , a y ) = 0 for all jc, y e F} is a Lie subalgebra of gI(F). In the rest of this exercise we look at this important construction of Lie algebras more carefully. An endomorphism a e g(jß) is said to be an invariant transformation of ß. The Lie algebra g(j8) is the “linearized” version of the group Giß) of isometries of F relative to ß; that is, Giß) '= [a e GLiV)\ßiax, ay) = ßix, y) for all x , y ^ F}. Henceforth assume that ß is nondegenerate; that is, for x g F, ßix, F ) = (0) <=>X = 0 <=> ßiV, x) = (0). (b) Let p: F ^ F* be the linear map defined by {p(x),y) = ß (y ,x )

(c)

forall ac,y G F.

Show that p is an isomorphism. Show that for each a e End(F) there exists a unique element e End(F) such that piax, y) = /3(x, a*y), and show that the mapping a ^ a"^ of End(F) into itself is linear, bijective, and satisfies iab)* = Similarly define by pi*ax, y) = Pix, ay), and show that *ia*) = a = i*a)* for all a e End(F).

Exercises

81

(d) For each a e End(F) let a‘ e End(F*) be the transpose map defined by ( a ‘f , x ) = ( f , a x } .

Show that a* = p * ° a' o p

for some p e GL(V) and hence that tr a = tr a* (e)

for all a e End( F ) .

Show that for a e End(F), a e g(/3) «»a + a* = 0<=>a +*a = 0.

Conclude that g(/3) c (f) Let B be some basis of V, and relative to this basis write the vectors of V as column vectors with entries in K. Let J be the matrix of B relative to B so that B(x, y) = x^Jy

for all x , y e V.

Show that in matrix terms A* = and g(j8) = {A ^ qI„(IK)| JA = -A^J). (g) Show that the symmetric bilinear form k: Q(B) X 9(/3) ^ K k { a , b ) = tTy(ab)

is nondegenerate. 1.2 Let L be a finite-dimensional vector space over IK with basis B = (Cj,. . . , e„). (a) Define /3 to be the symmetric bilinear form on V whose matrbc relative to B is I„ in X n identity matrix). The Lie algebra g(/3) of Exercise 1.1 is called §o(n), or go(n,IK) in this case (special orthogonal Lie algebra). Show that a e §o(n) iff a is skew-sym­ metric as a matrix. Determine dim §o(n). (b) Suppose that n = 2m, and define /3 to be a skew-symmetric bilinear form whose matrix B is

82

Líe Algebras

(c)

This time g()8) is denoted by ^p(m) (symplectic Lie algebra). Determine dim §p(m). Let F be a vector space of finite dimension n over C, and let <*, • >: F x F - ^ C b e a nondegenerate hermitian form. Prove that the unitary Lie algebra U{n) := [a e qX{V)\{ax,y) + { x, a y) = 0 for all x , y e V] is a Lie algebra over R (but not over C!) and that the special unitary Lie algebra §U(n) := [a G Vi{n)\\x a = 0}

is an ideal of codimension 1 in U(«). Prove that relative to an orthonormal basis of F, a g U(n) <=> ú: is skew-hermitian. Find dim \X{n) and dim §U(n). (d) Show that the Killing form of is negative definite. Show that ^U(n) is a simple (real) Lie algebra and that its complexification (Section 1.1) is §I(n,C). 1.3 (a)

Let a and b be Lie algebras, and suppose that we are given a homomorphism of Lie algebras b ^ from b into Der(a). Define [ • , * ] : a e b x a 0 b - > a 0 b b y [a + b,

(b)

+ b'] = [a, a'] + b¿,(a') - 5^.(iz) + [b, b']

for all a,a' ^ a and b, b' g b. Show that this gives the space a 0 b a Lie algebra structure and that a is an ideal of this algebra. In particular show that for a Lie algebra a and any subalgebra b of Der a, a 0 b has the natural structure of a Lie algebra. The holomorph of a is the Lie algebra a 0 Der(a) constructed in this way.

1.4 Let ^ = 0 be an associative algebra which is graded by the abelian group Q. Show that 1 g A^. 1.5 (a)

Let A be an algebra over K, and suppose that IK is algebraically closed. Let / g Der(.^4), and suppose that for each a g IK we define :=

G y 4 l(/- a Y a = Ofor somes r

g

Z+).

Suppose that A = (this always happens if dim^^ ^ < oo). Prove that this decomposition affords a grading of A by (C, +). [Hint: ( / - (a + p)Xab) = ( / - aXd)b + a{f - /3Xb).]

(b) Find a similar statement for / e A ut(^). (c) Let g be a finite-dimensional complex Lie algebra. Let 6 e Aut(g) be of finite-order N. Set ^ and for each 0 < ; < iV define g-' = {ac e glflAc = ^^x). Show that g = ®j=o 9^ constitutes a grading of g by Z / N Z . (d) Let 0 be the automorphism of gIsiC) defined by = £,+i,y+i (indices mods) where is the ( i,;) matrix unit of gljfC). Determine explicitly the spaces g 13(C)-', j g Z /3Z for 0. 1.6 Let F be a vector space over IK with basis Show that by defining

{ v / ^l k g

Z}. Let a, /3

g

K.

= {k + a + I3{n + \))v„ +k J we obtain a representation of the Witt algebra SB on V. We denote V with this SB-module structure by F„^. Determine the conditions on a and ¡3 under which ^ is irreducible. 1.7 Let g be a finite-dimensional simple Lie algebra over K. Let g X g ^ IK be the Killing form of g. Let /B(g) be the IK-space of invariant bilinear forms on g. (a) Show that the elements of IB( q) are symmetric (use g = D(g)) and that they are either trivial or nondegenerate. (b) Let Z := {A G End,^(g)|A ad(x) = ad(x)A for all x g g} (the centroid of g). Show that Z is a field extension of IK and that g has a natural Lie algebra structure over Z (henceforth denoted 9z)(c) Show that the following are equivalent: CS 1: Z = J<. CS 2: The IK-Lie algebra IK®||^g is simple (IK = algebraic closure of IK). If g satisfies either of the two equivalent conditions CS 1 or CS 2, it is called central simple. (d) For A G Z let k^(A): g X g IK be defined by k^(A)(x, y) = K,^(x, A(y)). Show that / : A •-> K|,^(A) is a IK-linear isomorphism of Z onto IB(q). (e) Let g be a finite-dimensional simple Lie algebra over an alge­ braically closed field IK. Then I B ( q) = K k .

Líe Algebras

84

1.8 Let 1^ c g be Lie algebras. Let 3 = 3g(l^) == {x e g |[j:,^] = 0}, It = iig(]^) ■■= [x ^ g|[jc, 1^] cl^}. These are the centralizer and nonnalizer of in g, respectively. (a) Prove that 3 an n are subalgebras and that 3 is an ideal of n. (b) Show that if the representation of i) on g: p: f| ^ qK q) given by X >-» adg(jc) is semisimple, then = 1^ ® 3g('^)1.9 Let F be a IK space and A e V*. Let A: S(F) IK denote the extension of A to an algebra homomorphism. Show that kerA = Z S ( V ) { v - k ( v ) l ) . u^V 1.10 Let g be a Lie algebra and U(g) its universal enveloping algebra. Let A^: U(g) -> U(g) be given by A^^: u ^ xu - ux. Let x, / e g c U(g). For all e Z+ show that L « ; ( A ^ ) r - '( a d / y ( x ) , j=i where a,(AT) = ( - i y

N—j+\ f J.J. E k=l \ ■'

j

1.11 Let g be a Lie algebra, g ¥= (0). Show that g admits modules that are not semisimple. 1.12 Let G be a group and p: G -* GL(V) a representation of G on a K-space V. (a) Show that there exists a unique representation p I:

G

G L (F ® • • • ® F )

(n-times), « > 1,

satisfying p

U

x

) ( V

i

e> ■ ■ ■

<8>

v

„)

=

p ( x ) ( u i) ® • • • ® p (x )v „.

Let pqI G K be the trivial representation. Show that we obtain a representation pT:= Q pj;-. G G L ( r ( y ) = 0

n>0

^

n>0

r '* ( F ) ) .

'

85

Exercises

(b)

Show that determines a natural representation of G on S(F) an also on each component S"(F) so that p^(xXf x • • ‘ p(x)vi ■ p(x)v„. Show that p^ determines a natural representation p^ of G on A(F) and also on each component A"(F) so that =

(c)

• •

Л ■■■ AV„) = p ( x) { Vi ) A ■■■ A p{ x) ( v „) .

(d)

Assume that G is finite and that dim^^ F = / is ^nite. Given ^ G, let WiCac), . . . , щ(х) be its I eigenvalues in K. Verify the following identities in IK[[i]]:

X

П (1 i=l

= L ( - 1 ) " Ч р „^(д:)) г". n=0

1.13 Let a be an ideal of the Lie algebra g. Show that U(g)a = all(g) is an ideal of U(g) and U (g /a) = U(g)/U(g)a. 1.14 Let g be a finite-dimensional Lie algebra. Prove that U(g) is left (right) noetherian. [Hint: Use the associated graded algebra gr(U) and the noetherian property of 5(g).] 1.15 Let be a (not necessarily commutative) ring with no zero divisors, (a) Prove that the following conditions are equivalent ORE 1: Aa r\A b (0) for all a, b e A \ {0}. ORE 2: There exists a division ring D and a ring embedding A ^ D such that for all d e Z) \ {0} we have d = x~^y for some x ,y ^ A. ORE 3: For every left .,4-module M the set = {m e M\am = 0 for some

a e ^ \ {0}}

is a submodule of M. A ring satisfying the above equivalent conditions is called an Ore domain [see JA II]. (b)

Let {b„)„ez ^ nonnegative real numbers such that Iim sup„_Jb„)'/" < 1. If we define a„ := E"=ib„ show that lim su p „^ ia„)‘/" < 1.

Lie Algebras

86

(c)

Let A be an associative algebra with no zero divisors. Assume A admits a filtration of subspaces A = \J7=o-^i IK =. ^0 <=•••• AiAj c Ai+j for all i, j e N. dimjj Ai < 00 for all i e N.

Defineb„ == d i m S h o w that if limsup„ < 1, be the Virasoro then A is an Ore domain. Let S3 and § l2 algebra and affine —§ l2 respectively. Show that U(9S) and U(§l2(IK)) are Ore domains. (d) Prove that if g is a finite-dimensional Lie algebra, then U(g) is an Ore domain. (e) Let g = ® “=ig" be a graded Lie algebra such that dim,^ g" < oo so for all n e Z+. Suppose that limsup„_j[dim g")*/" < 1. Show that U(g) is an Ore domain. 1.16 Let R be an Ore domain, and let S c i? be a subring. (a) Show that if R is free as a right S-module, then S is an Ore domain. (b) Let a c g be Lie algebras. Show that if U(g) is an Ore domain so is t/(a). 1.17 (See J. -P. Serre, Lie Algebras and Lie Groups, Benjamin, New York, 1965.) Let Y be a nonempty set. Let A ( X ) and ^(36) denote the free associative and free Lie algebra on X. Recall the augmentation ideal A+ (X ) = ®„^ qA(X)". Define a linear map

by d( jii,. . . ,

[■^i> • • • >■^n] := a d ^ i ••• adx„_i(A:„)

for all eX Let ad: § (X ) ^ gI(g(X )) be the adjoint representation. It extends uniquely to an associative algebra homomorphism ad: A { X ) ^ E n d g (Y ). (a) (b)

Prove that for all u ^ A i X ) , v e ^ ( X ) + , d(,uv) = 2id(u)d(v). Prove that dlg(Ar) is the derivation of g (X ) for which d\^^xr is scalar multiplication by n (Dynkin’s theorem).

Exercises

(c)

87

Let X, y ^ X, X ¥=y, and let z := Iog(exp X exp y) m+1 ( - 1) = E E — m m= l P+<J>0 and let z = Ez„, where z„ g g(AT)". Prove the following explicit formula for the terms z„ of the Campbell-Baker-HausdorflF formula: ~

"

^

^P,q’>

p + q= n

where

Pl+ ••• +Pm=P

(-!)"■ +' (a d x ,r(a d y i)« ' • • • (ad x)^"-(y) m P i ' - - - q x '--“ P j

- +q^_^=q-\

Pi-^Qi>\ (ad X i/'(ad y,)^' • • • (ad y / ”- ( x ) P l + • • • + P m - l = P - ‘^

m

P l'-Q l'-

^1+ ••• +l

- --

Q m -l'-

1.18 Let Z be a set, g(AT) the free Lie algebra on X over K, and let M be an g(A!’)-module. Let d: X ^ M he. any map. Show that there exists a unique extension of d to a linear mapping d: S(A') M satisfying d{[a,b]) = ad (b ) - bd(a)

for all

In particular show that any mapping d: X a derivation of §(Ar).

a ,b ^^{X ).

g(A') extends uniquely to

1.19 Compute all automorphisms of the Witt and Virasoro algebras. 1.20 Let g be a perfect Lie algebra, and let § be its universal covering algebra. Prove that every derivation (automorphism) of g lifts uniquely to a derivation (automorphism) of §. 1.21 Prove all the results quoted without proof in Section 1.12. [Hint: See Lang, Algebra, Exercise p. 175.] 1.22 The following exercise, based on Bourbaki, covers the fundamental facts about solvable and nilpotent Lie algebras.

Lie Algebras

88

Let g be a Lie algebra over an arbitrary field IK. Recall that Dg := [g, g] is an ideal of g. Define decreasing sequence of ideals D®(g) d D^(g) D • • • and ^ T f\g ) 3 • • ♦ inductively as follows: (derived series)

D°( q) == g D"^*(9) = [D"(9)>D"(g)]

(central series)

for all n > 0

^ °(g ) = 9 ^ " ^ ‘(9) = [9, ^"(9.)]

for all n > 0.

(a)

Show that all the terms in these series are ideals of g that are stable under every derivation of g. (b) Let /: g i) be a Lie algebra epimorphism. Show that = 3% ii) and that /(-^"(g)) = Conclude that 3 g is the smallest ideal a of g making g /a abelian. Henceforth in this exercise we assume all Lie algebras to be finite dimensional. (c ) For a Lie algebra g show that the following conditions are equivalent N1: = (0) for some n ^ N2: There exists a decreasing sequence of ideals 9o ^ 9 ^ 9i ^ • ** ^ ^k = (0) such that [g, g j c g.^^ for all 0 < / < A:. N3: There exists a decreasing sequence of ideals g^ 3 • • • d g^ as in N2 and such that dim,,^ 9i/9/+i = 1 for all 0 < f < A:. N4: There exists N ^ N such that ad jCj ® ° ad = 0 for all 9. (d)

(e)

A Lie algebra satisfying the equivalent conditions of (c) is said to be nilpotent. Show that subalgebras, quotients, and central exten­ sions of nilpotent Lie algebras are nilpotent. Let K be a (K-space, and let x e Endj^(F). Consider ad x:gi(V) gl(V) by ad x : y [ x , y] — a : ° y — y © x. Show that for all A/^ e Z . a d : c ''( y ) - L ( ^ ) ( - 1)' X N —i oyoJC.i / =0 ^ ^

(f)

Conclude that if x is nilpotent so is ad x. Let V ¥= (0) be a K-space, and let g be a finite-dimensional subalgebra of gI(L) consisting of nilpotent elements of Endj^(L). Show that there exists i; e F, # 0 such that g • i; = 0. (Assume that the results holds for dimension < n '= dim g. If is a subalgebra of g of dimension m < n, use (e) and induction

Exercises

89

to conclude that some nonzero element of g/f) is annihilated by all of adf). Conclude that g has an n - 1-dimensional ideal ]^. If ;c e g \]^ , show that U •= {v ^ V\ii • u = 0} is a nonzero subspace of V that is A:-stable.) (g) (Engel’s theorem). Show that for g to be nilpotent it is necessary and sufficient that ad ;c e gl(g) be nilpotent for all x e g. [Use (f) to show that g has nontrivial center c. Now use induction on dim(g/c) and (d).] (h) For a Lie algebra g show that the following conditions are equivalent: SI: D"(g) = (0) for some n ^ S2: There exists a decreasing sequence of ideals g = go 3 g^ d • • • D g^ = (0) such that g,/g,+i is commutative for all 0 < / < A:. Such a Lie algebra is said to be solvable. Note that every nilpo­ tent Lie algebra is solvable. (i) Show that subalgebras and quotients of solvable Lie algebras are solvable. (j) If a is an ideal of g and both a and g /a are solvable (as Lie algebras), show that g is solvable. (k) Let g ¥= (0) be solvable. Show that there exists an ideal of g of codimension 1. 1.23 Let g be a finite-dimensional Lie algebra. Let ^ ( 9 ) == n

«=0

(See Exercise 1.22.) Show that for an ideal a of g, g /a is nilpotent if and only if a D T^"(g).

Chapter Two

Lie Algebras Admitting Triangular Decompositions Agathon: But it was you who proved that death doesn’t exist. Hey, listen—Fve proved a lot of things. That’s how I pay my rent. Theories and little observations. A puckish remark now and then. Occasional maxims. It beats picking olives, but let’s not get carried away... So—what else is new? A g a t h o n : Oh, I ran into Isosceles. He has a great idea for a new triangle. —Adapted from W. Allen’s “My Apology”

A lle n :

The Lie algebra gI„([K) can be decomposed in a rather obvious way into the sum of three subalgebras: the strictly upper triangular matrices, the strictly lower triangular matrices, and the diagonal matrices. This “triangular decom­ position” is a basic ansatz that appears in all the split semisimple Lie algebras, the Kac-Moody algebras, and the Virasoro algebra. A considerable amount of the basic theory of these various types of Lie algebras depends upon nothing but this triangular decomposition, and hence is common. In this chapter we develop the triangular decomposition theory. The technical definition that we adopt here in Section 2.1 was inspired by the work of Rocha-Caridi and Wallach [R-CW]. At the foundation of the representation theory of Lie algebras with a triangular decomposition are the highest weight modules, and particularly the universal highest weight mod­ ules or Verma modules of Sections 2.2 and 2.3. Section 2.4 considers the special case of the three-dimensional simple Lie algebra §I2(IK). It is hard to overemphasize the importance of the represen­ tation theory of this Lie algebra in the study of semisimple and Kac-Moody Lie algebras (also in physics!). The material of this section is used extensively in Chapters 4, 6, and 7. 90

2.1

Triangular and Weight Space Decompositions

91

A basic tool in representation theory is the notion of formal characters. These characters are essentially formal power series that assemble into a single expression all the information about the dimensions of the weight spaces of a given representation. The fact that oftentimes characters can be given by unexpected and remarkably beautiful formulas (see Section 6.4) is one of their fascinations. In Section 2.5 we introduce characters, derive the character formula for Verma modules, and relate it to the theory of parti­ tions. In Section 2.6 we introduce the module category & of Bernstein-GelfandGelfand (BGG). This category provides a natural setting for highest weight modules. All the modules in 0 have characters, and we can even make a ring out of them. There are various series, particularly a weak form of composi­ tion series, that are available for modules in 0 , and these series play an important technical role in the development. Section 2.7 deals with various notions of radical ideals for Lie algebras with triangular decompositions. Many of these ideals vanish in important cases, although there are difficult, and even unanswered, aspects to this which we will come in Section 4.5. The structure of Verma modules has turned out to be a rich and subtle subject, in both the Kac-Moody and Virasoro settings. One of the tools by which we may understand this is the Shapovalov form and Shapovalov determinant. We introduce these objects in Section 2.8 and then use them in Section 2.9 to develop the Jantzen filtration, a device that allows one to study the effect on the structure of Verma modules as the highest weight (the single characterizing parameter) is allowed to vary. The BGG duality of Section 2.10 introduces a new collection of categories of modules, and in particular a class of projective modules that allow us to see a remarkable duality between projective and Verma modules, on the one hand, and Verma and irreducible modules, on the other. This result is used in Section 2.11 to give a general result about the possible embeddings of Verma modules into Verma modules. In Section 6.7, where we specialize to the Kac-Moody case, we get definitive results on this. The results of Sections 2.9, 2.10, and 2.11 are not used until Section 6.6 and not at all in Chapter 7. For this reason they might well be skipped on first reading. 2.7

TmANGlJLAR AND WEIGHT SPACE DECOMPOSITIONS

Let ^ be an abelian Lie algebra, and let ^ qI(M ) be a representation of in a IK-space M. Let g 1^* (the dual space of 1^). A nonzero vector i; G M is called a weight vector of weight (relative to and tt) if for all A G

7r(h)v = <¡>{h)v.

Líe Algebras Admitting Triangular Decompositions

92

Recall from Section 1.8 that U(l^) may be identified with the symmetric algebra 5(^) of 1& and that any representation 17 of lifts uniquely to a representation tt of U(]^). Suppose that u is a weight vector for i) of weight . The linear mapping <^: 1^ ^ IK extends uniquely to a homomorphism : S(f|) ^ K. If hi, /12, . . . ,

are any elements of

then

••• *1*) = E «i<^(*i)‘'<^(*2)'" ••• where i = (/i, /2, . . . , ( 1)

^

tt( u) u

For each e

and a-^ e K, It follows that

= {u)v

for all w G 5(]^), i; G M.

define the -weight space

by

= {y G M \7r(h)v = (f>(h)u, for all /z e

.

Clearly is an 1^-submodule of M. Let 'f) be an abelian Lie algebra, and let ir: ij qI(M ) be a representa­ tion of We say that the 1^-module M admits a weight space decomposition if M = This sum is necessarily direct (see Proposition 1). The (f) for which ¥= (0) are called the weights of M relative to and tt. The set of all weights of M will be denoted by P(M), For each e 1^*, dim is called the multiplicity of in M and is denoted by mult^(<^). Evidently mult^( 1 if and only if is a weight. Proposition 1 Let ^ be an abelian Lie algebra, and let M be an f)-module admitting a weight space decomposition. Then (i) the sum is direct, (ii) every ^-submodule N of M admits a weight space decomposition, and moreover n N for 12// e f)*. Proof Let be a submodule of M. We have n A/^ for all e ]^* by definition. Let x ^ N and write x = where x"^ e and R is some finite set of distinct weights of M . l f R = {}, then x e O N = AT^. We show by induction on card R that each x"^ ^ Suppose that card R > 1, and let and (f>2 ^ R, (l)^ ¥= 02- Choose /z g so that i(/z) # 2(h)x =

52

s r \[4>2)

- 2Íh))x'‘‘ :N.

2.1

Triangular and Weight Space Decompositions

93

By the induction hypothesis we may assume that each component (f>2(h))x'^ e In particular so that ^ Applying the induction assumption again, it follows that each x"^ belongs to N and hence that N admits a weight space decomposition. This estab­ lishes part (ii). Now part (i) follows from the same argument when x is taken to be 0. □ Example 1 (^I„(IK)) Let Then has a basis ...,

be the set of all diagonal matrices in ^I„([K). where

0. *0 hi. '=

(1 in the ( k , A:) -position).

-1

*•0 / Let M = K", and let tt be the natural representation of (Example 1.6.4). Then

on M

0

(1 in the A:th position)

lo is a weight vector for ij with weight

< f > k { h j) =

k

0 1 [-1

given by

(

if /:

l

if k = j ,

=

+ 1,

if A: = ; + 1; j

=

Furthermore and M = weight space decomposition.

-

1.

showing that M admits a

Example 2 (The Virasoro Algebra) Let 93 = E“ = _oolKL„ 0 (Kc be the Virasoro algebra, and let = IKLq ® IKc. Think of S as an f)-module through the adjoint representation restricted by Then [L q, L„] =

94

Lie Algebras Admitting Triangular Decompositions

shows that L„ is a weight vector of weight A„, where A„ e 1^* is defined for all Me Z by An(^o) =«> A„(i) = 0. Since A„ = nAj, we have iB"'" = 93"'^' = KL„ if n # 0 and 93° = f). This shows that 93 admits a weight space decomposition relative to ^ Proposition 2

Let Q be a Lie algebra, and let ^ Der %be a representation o f an abelian Lie algebra as a Lie algebra o f derivations on g. (i) Suppose that g admits a weight space decomposition 9 = 0

9“

with respect to under the representation tt. Then (ia) for all a,l3 ^ f)*, [9“, 9^] c 9“"^^ (thus g is graded, as a Lie algebra, by the Z-span of its weights)-, (ib) U(g) admits a weight space decomposition {relative to the repre­ sentation o f on U(g) extending that o f b, on % {Proposition 1.8.3)), and for all a, /3 e U (g )“U (g)'’ c U ( g ) “ ^ ^ Thus U(g) is graded, as an associative algebra, by the Z-span of the weights o f g. Moreover this grading is the grading o f U(g) inherited from that of g in part (i) {Section 1.8, Remark 2). (ii) Suppose that g is generated as a Lie algebra {see Section 1.10) by a set of weight vectors Xj, j e J, with Tr{h)xj = 4>j{h)xj for each j and for all h e ]^. Then g admits a weight space decomposition g = ©g“ with Xj e Q'I’J and each weight a e LZj. In particular the results o f part {ia) apply to g and U(g). Proof Identify g as a Lie subalgebra of U(g) in the usual way. We know that every element of U(g) is a sum of products x^X2 . . . x ^ , where each x, e g. In view of the assumptions on g (in either part (i) or (11)), we can assume that each X; lies in a weight space, say, g“'. Then TT{h){xiX2...Xk)

=

ix^...TT{h)xj...Xk

= («1

+ak){h)x^X2...Xk.

2.1

Triangular and Weight Space Decompositions

95

Part (ii) and part (ib) clearly follow from this (for (ii) invoke Proposition 1 with g taken as an l^-submodule of U(g). Since for all x, y e g, [x, y] = - y;c we may deduce (ia) from (ib). □ Let g be a Lie algebra over a field K. A triangular decomposition of g consists of an abelian subalgebra (0) and two subalgebras g+ and g_ such that TDl: TD2:

g = g_© f| 0 g^; g+^ (0), [1^, g+] c g+, and g+ admits a weight space decomposi­ tion relative to f) (under the, adjoint representation) with weights a =5^ 0 lying in a free additive semigroup 0 + ^ ^*5 TD3: there exists an anti-involution a (i.e., antiautomorphism of period 2) on g such that
mj

The triangular decomposition is said to be regular if in addition the weight spaces of g+ in TD(2) are finite dimensional. Notice that the condition 0 ^ 0 + is equivalent to ( 2)

C + n (- ! 2 + ) = 0

and also that (3)

o-(g_) = g+.

Remark 1 We will usually refer to a triangular decomposition of a Lie algebra g as a 4-tuple (^, g+, 0+ , ir), where f),g+, 0+, and < t are as above. It will be understood that g_== cr(g+) and that g = g_0 ^ 0 g+. Sometimes we simply write g = (^, g+, 0+, c7)t o describe this situation. Let g be a Lie algebra admitting a triangular decomposition, and assume that the above notation is in effect. Let x e g+, a 0 0^_. Then [h,ax] = [crh, ax] = a[x, h] = —a{h)ax for all A 0 from which we see that a x e g l“. It follows that g_ admits a weight space decomposition relative to

96

Lie Algebras Admitting Triangular Decompositions

with weights in —0+; (4)

g_“ =
for all a e Q_^.;

and g admits a weight space decomposition relative to S+ (5)

with

if a e e + , if - a e if a = 0, otherwise.

Let Q be the group generated by We know that Q = (which is a lattice in 1^* whenever J is finite). We call Q the root lattice (regardless of whether or not J is finite). We call the a ^ Q for which g" ¥= (0) the roots of g (relative to 1^) and the corresponding spaces g“ the root spaces. We define the height function h tiQ ^Z by

ht\

^j^j] = H

This terminology, due originally to W. Killing in this work on finite­ dimensional simple Lie algebras, derives from the fact that for each h the quantities a{h) are the characteristic roots of ad h (which is obviously a diagonalizable transformation). There is some nonuniformity in whether or not 0 is to be considered as a root. Classically in the case of semisingle Lie algebras it is not, but in the more general context here it seems more natural to do so (as we have just done). For each root a, dim g“ is called the multiplicity of a and the set of all roots is called the root system of g rela­ tive to ^ ' = (t), Q^, a) and is denoted most commonly by A. Define A+= A n 0 + and A_= - A+. Then TD2 and (4) and (5) give us that A = A+u{0} U A_. We call A+ (resp. A_) the set of positive (resp. negative) roots of A. We call the diagonal subalgebra of the decomposition and g+ (resp. g_) the upper (resp. lower) triangular subalgebra of the decomposition. Remark 2 \i ^ = (1^, g+, 0+, a) is a triangular decomposition of g, then it is immediate that g _ , - 0 + , cr) is another. We call this the triangular decomposition opposite to S^. Example 3 (g = ^I„(K), n > 1). Let g+ (resp. g_) be the subalgebra of strictly upper (resp. strictly lower) triangular matrices. Let 1^" be the subalge­ bra of diagonal matrices in gI„(K), and let Ij = 1^" n §I„((K). B o th 't)" and are abelian subalgebras of gI„(IK). The triangular decomposition that we are looking for is ^I„(IK) = g_0 f) e g+.

2.1

Triangular and Weight Space Decompositions

97

Transposition X ^ serves as the anti-involution, pointwise fixing and interchanging g_ and To determine the roots, we introduce the matrix units 1 < /, j < n: Eij is the matrix with a 1 in the (i,;)-entry and O’s elsewhere. Let Cj,. . . , e be defined by

and let a^,. . . ,

Since

be defined by

+ • • • +£„ vanishes on 1^, it is easy to see that { (E « ie,)|Ja,- s K, £ a ,. = o} =

and hence that

a„_i is a basis of 1^*. For h e 1^*, [h,

= /t£,, - £,,£ = (s, - e ,)(;i)£ ,,.

For i < j this gives = (a , + ••• +aj_i)(h)Eij. Thus E^j e is a weight vector with weight a, + • • • +ay_i. Set Q+-= {Em,a,|m, e l\j} \{0}. Then 9+=

®

g: = 0

i<;

KE,j

is the required weight space decomposition. Of course Ej^ = Ejj has weight -{ai + • • • -\-aj_i) for all i < j\ and g_=

© i>j

KE¿j

is the weight space decomposition of g_. Thus ^I„(IK) admits a regular triangular decomposition. It follows that gI„(IK) also has a triangular decom­ position with the same g+ and g_ but with replaced by %. These two examples are the origin of the terminology. Example 4 (The Virasoro algebra) Referring to Example 2, set g+ = ^ ® Let Aj e 1^* be the weight of Lj. Then with Q+ = we verify TD l, TD2, and TD4 of our definition. Let a be the linear transformation defined by aLj = L_y, j ^ Z, = 0.

98

Lie Algebras Admitting Triangular Decompositions

This provides S3 with a regular triangular decomposition. Of course it is easy to modify this for the Witt algebra. Example 5 (The affine algebra ^ l2(IK) of type

See the Appendix.

Example 6 (Heisenberg algebras extended by derivations) Let a = a_ 0 Kc 0 a+ be a Heisenberg algebra so that and a_ are abelian subalge­ bras that are nondegenerately paired by the associated skew-symmetric bilinear from «/r. Suppose that there are bases {a^\ and {a}}, ; g J, of and a_ such that [ay, a\] = 5y^c (this is always possible if a is finite dimensional). Define cr by cr{aj)=a'j,

y e J,

cr(c) = c. Then axioms TDl and TD3 of the definition of triangular decompositions are fulfilled with = Kc. However, is the centre of a and hence cannot produce the root spaces required by TD2. Just as in Example 5 this problem is easily cured by enlarging 1^^. For each i e J and integer g Z+, we let d be the linear operator on a defined by ¿(a,) = d¿a¿, d{a\) = -d^a\, d\i^^ = 0. Then d is di derivation of a. We let if •= if^ ^ Kd = Kc ® Kd and make f i : = a 0 l K d = a _ 0 ] ^ 0 a + into a Lie algebra containing a as a subalgebra by defining (Exercise 1.3) [d, x] = d{x)

for all X G a ,

[d,d] = 0. Then, if we define a g 1^* by a(c) = 0, a(d) = 1, and set = E^.=^IKa„ k a we see that a has a weight space decomposition with a+= i< — ka ij = a . After extending cr to a by a(d) = d, « - = ®;t>0 a (]^, a+, Z^_a, cr) is a triangular decomposition of a with root lattice Q = Za. The problem that was raised in these last two examples, namely that of extending the algebra in order to produce more eigenspaces, is one that we will see again in our study of affine Lie algebras. Remark 3 Later on in the text (Section 2.9) we will find it useful to have the notion of triangular decomposition for Lie algebras over rings. Suppose that ^ is a commutative ring of characteristic 0 (i.e., Z is a subring of AX Let fj be an abelian Lie algebra over A that is free as an .^4-module (Remark 1 of Section 1.1). Set r

= H o m ^ (^ ,^ ).

2.1

99

Triangular and Weight Space Decompositions

Suppose that M is a free ^-module that is also an 1^-module. We define the weight space /x e 1^*, just as above: := {y e M\h • V = ii{h)v for all h

,

M has a weight space decomposition if M = The sum is necessar­ ily direct. Let g be a Lie algebra over A that is free as an ^4-module. A triangular decomposition of g consists of an abelian subalgebra f) and two subalgebras g+ and g_ of g (all free as ^4-modules) for which TD1-TD4 hold. Let g = (]^, g+, (2+, a) be a Lie algebra over A with triangular decomposi­ tion. Let jB be a commutative ring containing A, and let g == B 0^ g be the Lie algebra over B obtained by extension of the base ring from A io B (see Section 1.1). Let 'i) = B and g+= B 0^ g+. These are subalgebras of g that are free as S-modules. Every ^ induces an ^-bilinear map : i5 X by At: (6,/ï)

biJi{h),

and hence an ^-linear map (also denoted ¡lI) /1: B 0^ f) ^ 5 . One verifies that ]I is jB-linear, and hence there exists an (injective) group homomorphism "ril* ^ 5 * . Along similar lines one shows that there exists a unique antiinvolution â of g satisfying a{b

0jc)

= b

0

ot( a:).

Then is a triangular decomposition of g which is said to be obtained from (1^, g+, Q^, a) by extension of the base ring from A to B, The most obvious examples are when g is defined over the field IK and IK' is some extension field of IK. Example 7 (Complexification of a real triangular decomposition) Let g be a Lie algebra over the real numbers IR. Assume (1^, g+, Ô+, a ) is a triangular decomposition of g. Let gc '= C 0ff^ g be the complexification of g. Then evidently g^ = (gc)-® ® (9c)+ ^be obvious notation) and, with as the C-linear extension of a to g^, (^ c?(9 c)+ j Ô+? a triangular decom­ position of g^ (see the preceding remark). However, it is convenient for

100

Lie Algebras Admitting Triangular Decompositions

Other purposes to extend a to a semilinear map so that cr(ojc) = acr{x) for all e C and for all x e g. We denote this extension, which is unique, simply by cr, and we call the 4-tuple the hermitian complexifícation of(l^,g+,(2+>i^)- The real subalgebra 5 of g^ is recovered as the set of fixed points of a- in Conversely, suppose that we are given a complex Lie algebra m, an abelian subalgebra f, and a semilinear anti-involution cr on m that stabilizes f. Let be the set of fixed points of a in f. Then if is a real form of f since, for any jc e f, we have x = l/2 ( x + o-jc Kix + aix)). The 4-tuple (f, m+, Q^, a ) is called a (regular) hermitian trian­ gular decomposition if it is the complexification of a (regular) triangular decomposition (^,g+,j2+,o-)of some real form g of m. Note that Q^czi^* and ]^* can be thought of as the real subspace of f* consisting of those functions that are real valued on i). We will see in Chapter 4 that all the affine algebras, and indeed the far wider class of contragredient Lie algebras, possess regular triangular decom­ positions that are fundamental to their study. Proposition 3

Let q be a Lie algebra admitting a triangular decomposition (^, g+, Ô+, (r), and suppose that 9=

© 9"

aGA

is the corresponding root space decomposition. Let be an automorphism o f g stabilizing Then (¡) maps root spaces onto root spaces. More precisely (f|) = Í), cind if (f>*: ]^* -> Í)* is defined by *(/) = f°(f>~^ for all f ^ 1^*, then 0(g") = g"^*“ for all a ^ A. In particular * determines by restriction a bijectiue mapping o f A onto itself. Proof Let CKe A, jc e g", and H

which shows that

e g° = 1^. Thus

Then

Now

and hence (x) e Together these show that (f> maps root spaces into, and hence onto, roots spaces in the desired way. It follows that <^*A = A. □ The group of automorphisms of g that stabilize f) will be denoted by Aut(g, ij). Let Aut(A) := { e AutK(i|*)l<^(A) = A}.

2.1

Triangular and Weight Space Decompositions

101

Proposition 2.3 gives rise to an exact sequence \

K ^ A ut(g, ]^)

Aut(A).

Let 6: Q ^ be a group homomorphism (from an additive into a multi­ plicative group). There exists a unique e Aut(g, 1^) such that /elg“ = scalar multiplication by ^(a) . We define KQ ={f,\e^H om {Q ,K ^)}. Clearly K q

c

K,

Proposition 4 [B el]

Let % be a Lie algebra admitting a triangular decomposition (i), Q+,Q+,(t ). Suppose that dim < oo and dim < dim g+. Let D q c Der(g) be the deriva­ tions of g stabilizing each root space o f g. Then Der g = D q + ad g and D q n ad g = ad 1^. Proof Let 5 be any derivation of g mapping ^ into 'i). Then for any root a and any e g“ we have for all e a{h)8x'^ = which we write as (6)

(ad/i - a ( h ) l ) 8 x ^ = ~ [ 8 h , x ^ ] e g“.

If 8x°" were to have some component in g^, /3 ^ a, then with /t e such that p(h) a (h \ we have a contradiction to (6). Thus 6g“ c g“, and hence 8 e D q. It is obvious that D q n ad g = ad 1^. Let dim ^ = n, and let ..., ^ A+ be arbitrary. We can find a basis {hy, . . . , /i„) of f) such that a^ih^) = 1, i = l , . . . , n . In fact proceeding inductively, if linearly independent h^,...,hf^_^ have already been chosen so that afhi ) = l , i = 1 , k - 1, then set •= {h e ^ a f h ) = 0}. Since cannot be written as the union of two proper subspaces, we can choose X

+ •••

UH^}.

Then hi^ := x / a j f x ) is as desired. Now let basis of g+ taken so that each Xj^ lies in some root space g“^, e A+, A: e S. Let {h,^}f^^^ be a subset of such that ^k(^k) = 1

for

k ^ S.

102

Lie Algebras Admitting Triangular Decompositions

By the above we can assume that spans 1^. Evidently {crAc^l^es is a basis of g_. Let d e Der g, and define bf^j e K and h\ e by

Since dim dO)) < «>, and b/^j are zero except for a finite set of values of j (independent of k). In particular ajj and bjj are zero except for finitely many j. From o = d[h^,h^] = [dh^,h^] + [h^,dh^] we have

- b^jaj(h^) + b^¡aj(h^) = 0

for all p, q, j.

With p = j, ^ pp^ pV^p )

^qp^

^pp^pi^q)

^qp’

Let ------ E«ppO-JCp + 'LbppXp e g. P

Then for all

P

^ S, [^^>3^] “

p P

^J^pp^p{j^q)^p P

= Hagpcrxp + T^bgpXp P

P

= d h ^ ~ h',, so we have (d + a d y ) ( / i J = A ' , e ^ . Then d' •= d proposition.

ad y: if

if, since the {h^} span 1^, so d' e D q, providing the □

As an example one may easily show that every derivation of the Virasoro algebra is inner (Exercise 2.24).

2.2

Highest Weight Modules

103

2.2 HIGHEST WEIGHT MODULES Let 9 be a Lie algebra over K admitting a triangular decomposition (1^, 9+> Q+, and write g = g_0 0 g+. Let tt be a representation of g on a space M. We know from Section 1.8 that ir lifts to a representation of U(g) on M. If y is a vector in M, then 7r{n(q))v

{7t(

u

)

v

\u

is a submodule of M [since it is closed by the action of 7r(g)], and indeed it is the smallest submodule of M containing v. We call it the submodule of M generated by v, A nonzero vector z; e M is called a highest weight vector for g (relative to the triangular decomposition) if (i) z; is a weight vector relative to the action of 1^. (ii) tt{ x ) v = 0 for all X g g+. A g-module M is said to be a highest weight module if M contains a highest weight vector u that generates it. The weight A g ]^* of the highest weight vector v is called the highest weight of M. We call the pair (A, z;) a highest weight pair of M. The representation of g on M afforded by M as a g-module is called a highest weight representation. Remark I We will see shortly (Remark 2) that highest weight vectors of a highest weight module are unique up to nonzero scalar factors. In particular all highest weight vectors have the same weight. In addition all the other weights will be seen to be “lower” than A (Proposition 2.1). These two facts justify the above terminology. Highest weight modules occur very frequently in the sequel. Although in detail there are many differences between highest weight modules for differ­ ent Lie algebras and many subtleties to be understood, the broad features are quite uniform. In the physics literatures as well as in many mathematical papers, highest weight vectors are called vacuum or null vectors. Let M be a highest weight module affording the representation tt and with highest weight pair (A, z;). Using the Poincaré-Birkhoff-Witt theorem, we have U(g) = U(g_)U(]^)U(g+). Thus M = 7r(U(g))z; = 7T(U(g_))7r(U(^))7r(U(gJ)z; = 7r(U(g_))7T(U(^))z; = 7r(U(g_))z;,

Lie Algebras Admitting Triangular Decompositions

104

where the last equality is a consequence of the fact that is a simultaneous eigenvector for 7t(U(]^)) = [see 2.1(1)]. According to Proposition 2.1.2 and 2.1(4), g and U(g) admit weight space decompositions with weights in Q and are also graded by Q. All the weights of g_ lie in -Q+, and those of ll(g_) in -(2 +^(0)Let P be a subgroup of 1^*. A (2 +-fan in P is any set of the form A I 0 + - {A} U {A -/3li3 e 0+}, where A e P. We call A the source of the fan. Fans are useful objects for keeping track of the weights of representation that we will study. Note that Q-U{0} = 0 i Q ^ Proposition 1 Let q be a Lie algebra admitting a triangular decomposition (ij, Q+,Q+,cr), and let Q denote its root lattice. Let M be a highest weight module with highest weight pair (X,v) and ajfording the representation tt. Then (i)

(ii)

(iii)

M admits a weight space decomposition relative to where all the weights lie in the fan A i Q^. Furthermore = '7r(U(g_)““)i;/or all a ^ !2+^{0}. In particular = Ko. the weight space decomposition o f part (i) defines a grading o f M by ZA + Q and, for each a ^ ¡2? 9“ tind U(g)“ act as homogeneous operators o f degree a on M. if the decomposition o f q i s regular, then all o f the weight spaces o f M are finite dimensional.

Proof By the preceding remarks, M is spanned by vectors of the form v{u)v, where u e U(g_)“", a e (2+Li{0}. We have (1)

Tr{h)7r{u)v

=

\'Tr{h),'Tr{u)\v

+

=

—a { h ) 7 r { u ) v - \ - \ { h ) T r { u ) v

'Tr{u)7r{h)v

= (A —a){h)Tr{u)v, which shows that M is spanned by weight vectors and hence has a weight space decomposition. In addition it shows that = 7r(U (g_)"“)o and that all weight spaces of M are of this form. (Since modules are by definition graded only by groups, we use the group ZA + (2 for the grading even though it is clear that all the weights lie in the coset A + Q.) That U(g)“ and g“ act homogeneously is clear from (1).

2.2

105

Highest Weight Modules

Finally, suppose that the decomposition of g is regular. Now U(g_) has all its weights in O iQ ^, and except for U(g_)® (= K.l), each weight space U(g_)”^ is spanned by products x ^ .. . Xf^ of weight vectors with weights -Pi, e - Q ^ and p = • -\-p¿. Since there are only finitely many ways to decompose p ^ Q+ into parts that are also in Q+, and since dim g l" = dim g““ < oo for all a e Q_^, we see that dim U(g_)“^ < oo. Thus = 7r(U(g_)“^)¿; is also finite dimensional, and this proves part (iii). □ Remark 2 The use of the terminology highest weight is now apparent. A is a weight and all other weights are “lower” because they are of the form A - a, a e (2_^. Proposition 1 shows that A and IK¿; are unique. Indeed, if w is a highest weight vector for M of highest weight ¡jl, then \ —a = (fjL — P) — a for some a, P ^ (2 + U{0}=>a+)S = 0 => a = p = 0 => fi =A. Then dim = 1 => Kw = Ki;. Example 1 (The Natural representation of §I„([K)). We use the notation established in Examples 1 and 3 in Section 2.1 with M = K" and tt the natural representation of ^I„(IK) on M. Then M = is a weight space decomposition of M. Since E^je^ = 0 for all i < j, g+Ccj) = 0, so is a highest weight vector. Since for / = 1 ,..., n, M is generated by ej, and hence M is a highest weight module. The weights are 2 = 0i All the weight Oil, 3 = i - a 2 , . . . , (f>n = i - Oil - ' spaces are one dimensional. Remark 3 It is a fact that if K is algebraically closed, all finite-dimensional irreducible representations of ^I„(1K) are highest weight modules. Example 2 (Heisenberg algebras) We return to the Heisenberg algebras of operators described in Section 1.4. Thus a is the linear span of the operators /(xy), d/dxj, and 1 acting on 5 = Let j j be any map of J into Z+. Let a_= ElK/(jCy), a+= T.Kd/dxj. hct the degree operator D be defined on S by ...,

for all monomials in S. Thus Xj is given degree derivation of 5 (see Example 1.5.1), and [d, K ^ j ) ] f =

. . . , x "Jnk k

= («1/, + • • • + n j.

Let

= - h j f + Xjdf - Xjdf =

d

•= —D. Then d is a

-¡l(xj)f

for / e 5 shows that [d,l(xj)]= —jl(xj). Similarly [d,d/dxj] = jd/dxj. Thus d be­ haves like the extension element of Example 6 of 2.1 and fi — Kd 0 a is a Lie algebra of operators on S with triangular decomposition a _ 0 0 a+, 0 K l.

106

Líe Algebras Admitting Triangular Decompositions

Consider 1 ^ 5 . Evidently where A e 1^* is defined by

1 = 0 and h • 1 = \(h)l for all h

A(l) = 1,

X(d) = 0.

1, we see that 5 is a highest Since also <* weight module. It is quite easy to see that it is irreducible, but we leave the proof of this until Section 2.3. Let a e 1^* be defined by a ( l) = 0,

oi(d) = 1.

Then ..., is a weight vector of weight A + ••• It follows that the weights of 5 lie in A - Na. Except for = K ^ 1, the weight spaces need not be finite dimensional unless J is finite. Proposition 2 Highest weight modules are indecomposable. Proof. Let M be a highest weight module with highest weight pair (A, y). Suppose that M = ® M2, where and M2 are submodules of M. Then =

Mf e

forces either = (0) or M^ = (0). Say, M^ = (0). Then v : M^ so that M = U(g) • V c Mj, and hence M2 = (0). Remark 4 In general highest weight modules are far from being irre­ ducible. We will see this explicitly in Section 2.4. Remark 5 Analogous to highest weight modules we may define lowest weight modules M. In this case M is generated by a lowest weight vector (G_* = 0, • i; c K v ), and the weights of M lie in upward fans of the form At!2+- Lowest weight modules for g with triangular decomposition Q+, 0-) are evidently highest weight modules for g with the opposite triangular decomposition (see Section 2.1, Remark 2).

2.3

VERMA M O D U LES

In this section we show how to actually construct highest weight modules for Lie algebras with triangular decompositions. In fact the modules that we construct are universal in the sense that all other highest weight modules occur as homomorphic images of them. They carry natural (contragredient) bilinear forms (see Section 2.8), which are used later in the book to gain insight into questions of submodules and irreducibility.

2.3

Yerma Modules

107

We begin with a Lie algebra g admitting a triangular decomposition The root lattice will be denoted as usual by Q. Unless explicitly assumed, the decomposition need not be regular. Let M be a highest weight module with highest weight vector v and highest weight A. In this section we will put more emphasis on modules than representations. We will not usually specify the representation explicitly but will simply write jc • w to denote the effect of the operator defined by jc e g on an element w of M. Suppose that g_ has a basis of root vectors indexed by some totally ordered set J. For each ; e J let e —Q^ denote the root corresponding to Xj, that is, Xj e g^>. Then, by Proposition 2.2.1, each of the vectors Y^k V, (1) Z+, is a weight vector of M of where A: > 0, ‘ ’ * > h, ^ weight A + + ••• and collectively they span M. We are going to show how, starting with any A e ]^*, we can construct a highest weight module M(A), as above, for which the vectors (1) form a basis of M(A). We will also see that such a module is unique up to isomorphism. Let A e A highest weight module M(A) for g with highest weight pair (A, i;+) is called a Verma module if given any highest weight module M with highest weight pair (A, v) (the same highest weight A), there exists a g-module homomorphism r¡: M(A) ^ M such that v, A Verma module is universal in the sense that every highest weight module with the same highest weight is a homomorphic image of it. It is clear that the homomorphism rj in the above definition is unique [for the pair (A, i;)] and surjective: for if v, then for all u e U(g_), u * u • v;on the other hand, M = U(g_) *v. Also M(A) is itself unique up to isomor­ phism by the standard type of argument (see Section 1.2). The existence of M(A) will be shown in two different ways. The first is by the method of inducing modules. The second, which is more prosaic, is much more explicit and gives one a better feeling for what the module looks like. It also takes more work. Proposition 1 Let (]^, g+, ¡2+, a ) be a triangular decomposition o f g. Then, for all A e 1^*, g has a Verma module M(X) o f highest weight A. Proof 1. Let b be the subalgebra 0 g+ of g. Let Kv+ be a one-dimen­ sional vector space and make it into a b-module by defining g+* v += 0, h • Vj^= \{h)u^

for all /r e ]^.

Líe Algebras Admitting Triangular Decompositions

108

Now IKd+ is a left U(b)-module and U(b) is a subring of U(g)- Hence we may form the induced module obtained from by extension of the base ring from U(b) and U(s) (see Section 1.1, Remark 1) M(A) == U(g) ®u(6)IKt;+. Clearly, M(A) is a left U(g)-module, and hence a g-module. Consider 1® A/(A). For X e g+, X • (1 ® Ü+) = 1 ®

0»

and for h /i • (1 ® Ü+) = 1 ® /1 • v + = 1 ® A(/i)i;+. Also U(gXl ® y+) = U(g) ®u(b)IKi’+= Af(A). Combining these, we see that M (\) is a highest weight module with highest weight vector 1 ® of weight A. Let M' be any highest weight module with highest weight vector v' and highest weight A. The mapping / : U(g) X f ( u , av^) = au ■v', is bilinear and U(b)-balanced: Namely for w s 11(b), f(uw,av+) = au • w • v' = f { u , w ■av^). Hence there exists an induced linear mapping /: U ( 9 )

+

satisfying u (S> au • v'. This is a U(Q)-niodule map, and it satisfies 1 0 i; + L?' as desired (see [BA2] for more on tensor products). □ Before starting on the second proof we isolate the following fact which we will use several times in the sequel. Lemma 2 Let A e 1^*. Then (2)

U(g) = U (g _ ) e |u ( g ) g + + £ U(9)((* "

2.3

Verma Modules

109

The reader may wonder about the notation ~ which the sum is incredibly redundant. Evidently the sum may be replaced by a sum over any basis of f). Proof. The mapping A:

IK extends uniquely to a homomorphism A: 5 (^ )

whose kernel is

IK

~ A(/z)l) (Exercise 1.9). Thus 5 (i|) = K 0

L 5 (^)(/г - A (/i)l).

Let b_!= b ® 9_, and recall that (PBW theorem. Corollary 2), U(b_) is a free U(g_)-module admitting any IK-basis of U(b) as a U(g_)-basis. Now the triangular decomposition and the Poincaré-Birkhoff-Witt theorem gives us U(g) = U (g _ ) U ( b )U ( g J = U (g _ )U (b )(K l® U (g J g + ) = U (g_)U (b) ® U (g)g^ = U (g_)(K ® E U ( i ) ( A - A ( / i ) l ) ) ® U (g)g^ (because U(b) = 5 (b )) = U (9_) ® E U (b_)(/i - A (/i)l) ® U (g)g^ Asf) (because U(b_) is a free U (g_) module). Now 9+(/i —A(/i)l) c U(g)g^ since g^. is an b-module under the adjoint representation, and thus U (g)(/i - A (/i)l) c U (b_)(/i - A (/i)l) + U (g)g^.

□

We now give the second proof of Proposition 1. Proof 2. Let p :U (g ) ^ U ( g _ ) be the projection determined by the direct sum decomposition (2) of U(g).

lio

Lie Algebras Admitting Triangular Decompositions

The kernel of p is clearly a left ideal of U(9>- Define an action of U(9) U(g_) by y-u= p(yu)

for all Me U (g_), y e U (g).

This makes U(g_) into a U(g)-module: Let y', y e U(g), u e U(g_), and write yu = piyu) + r, where p(.r) = 0. Then y' • (y • m) =p{ y'p {yu )) = p {y '{ yu - r)) = p(y'yu) = (y 'y ) • u, since y'r is in the kernel of p. For u', u e U(g_), u' ' u = p{u'u) = m' m, and therefor the action of 11(9_) on itself is left multiplication. Thus U(9-) is generated by 1 as a U(g)-module. For all y G U(g+), y • 1 = p ( y ) = 0, and for all

e ]^, /z • 1 = p{h) = p(A (/i)l + h - A(/z)l) = A(/z)l.

This shows that U(g_) has been made into a highest weight module with highest weight vector 1 and highest weight A. If M' is any highest weight module with highest weight pair (A, v ' \ then the kernel of p annihilates v'. Thus the mapping U(g_) ^ M', u• a g-module map showing that U(g_) (as a g-module) is a Verma module.

□ Remark 1 It is clear from Proposition 1 (especially the second proof) that, as U(g_)-modules, M(A) = U(g_) and M(A)^”“ = U(g_)““, where the ac­ tion on U(g_) is left multiplication. We see that as U(g_)-modules, all the M(A)’s look the same, namely like U(g_). It is the actions of and U(g+) that are different. Since U(g_) is an domain (Proposition 1.8.5), we have the important fact that a Verma module M = M(A) is a torsion-free U(g_)-module; that is, for u e U(g_), m ^ M, (3)

w • m = 0 <=> M= 0 or

m = 0.

Any homomorphism of M(A) that is not injective annihilates some nontrivial

2.3

Yerma Modules

111

element u • v, u ^ U(g_) \ {0}. Therefore we arrive at the following impor­ tant conclusion: Proposition 3 Let M be a highest weight ^.-module. For M to be a Verma module it is necessary and sufficient that M be torsion free as a \liQ_)-module [i.e., (3) holds], O Remark 2 If M is a Verma module, then as a U(g_)-module it is monogenic ( = cyclic) and torsion free, and hence free. Corollary Any submodule of a Verma module that is itself a highest weight module is a Verma module. From the left U(g)-module mapping U(g) ^ M ( A ) , u ^ u • u ,. we conclude that M(A) = U ( g ) //( A ) ,

(4)

where /(A) is the annihilator of in U(g). From (3) and the second proof, we see that (5)

/(A)

=U(9)g^+ ElI(s)(^-A(/i)l).

Often (4) and (5) are used to define Verma modules. The next result shows us that irreducible highest weight modules exist and are unique. Proposition 4 Let (^,g+,!2+jOr)

^ triangular decomposition o f g, and fei A e 1^*.

(i) I f M ( \ ) is the Verma module o f highest weight A, then (ia) every proper submodule N o f M ( \ ) has a weight space decomposi­ tion relative to and the weights lie in A —0+j (ib) M(A) has a unique maximal proper submodule MA); (ic) L(A) := M(A)/MA) is irreducible, and up to isomorphism, L(A) is the unique irreducible highest weight module o f g with highest weight A. (ii) Any highest weight module M with highest weight A contains a unique maximal proper submodule N. Furthermore M /N = L(A).

112

Lie Algebras Admitting Triangular Decompositions

Proof, (ia) If A/" is a submodule of M then in particular N is an 1^-submodule of M. By Proposition 2.1.1, N has a weight space decomposition with weight spaces n M If A is a weight of N, then = «1;+ and then N = M, since generates Af as a g-module. Thus, if N is proper, all the weights of lie in A — (ib) Using (ia), we see that the sum A^(A) of all the proper submodules of M is still proper (since none of them can contribute to M^). This establishes (ib). (ic) L(A) = M(X)/N(X) is clearly a highest weight module with highest weight A, and since A^(A) is maximal, it is irreducible. Let L' be any irreducible highest weight module with highest weight A. Then, since M(A) is a Verma module, there is a nonzero homomorphism 77: M(A) with M{XY for all weights (j) of M(A). We have already seen that rj is surjective, and therefore its kernel is a maximal proper submodule of M(A). Thus ri(N(\)) = 0 by part (ii), and M(A)/iV(A) = L '. (ii) From the universal property of M(A), there is a surjective homomor­ phism M(A) M. This takes weight spaces to weight spaces from which it is clear that N ( \) maps to a proper submodule N of M. Then N is maximal, and M / N = M{k)/N{k) = L(A). □ Example 1 (Verma modules for Heisenberg algebras) Following Example 2.1.6, assume that a = a _ 0 lK c 0 a + is a Heisenberg algebra with an anti-involution a and bases {fly}, (cr(fly)} (; e J) of and a_ such that [fly, cr(fl^)] = 8jj^c j, A: e J. We extend a to & = a _ 0 0 a^_, where = IKc 0 Kd and where [ d , fly] = fly, [d,a(aj)] = —o -(fly ) so that fi acquires a triangular decomposition with root lattice Q = Za(a e 1^* given by a(c) = 0 and a(d) = 1). Let A G ]^*, and let M(A) be the Verma module for fi with highest weight pair (A, ¿;+). Then M(A) = is the corresponding weight space decomposition of M(A). We have M(A) = U(a_) ♦v^A = U(a) • and the annihilator .^4 of v, in U(fi) is U (a )a ^ + Vi{a){c - A(c)) + Vi{á)(d - k ( d ) ) . This is the content of (5) for this example. Thus M(A) = Vi(&)/A.

2.3

Yerma Modules

113

If we restrict the representation to a, then the annihilator ^ of U(a) is quite easily seen to be

in

U (a)a^+U (a)(c-A (c)), and M(A) = U(a)//4. Since we end up with the same representation space [i.e., M(A)], whether or not we adjoin d to a, it is worthwhile to comment on its role. The element c e is central in a (and in a) and consequently acts as scalar multiplication by A(c) on all of M(A). Indeed for u e U(a), c -( m - i;+) = w *c *i; += A(c)w • i;+. Thus c is unable to decompose M(A) into its characteristic form as a one-dimensional “highest” weight space and a sum of weight spaces of lower weights. The element d does not have this defect, and as we saw already, leads to the decomposition 0M(A)^“'*'^. It is primarily because we want these kinds of weight space decompositions that we have formulated the definition of triangular decomposition as we did. Of course û = û _ 0 lK c 0 û + is not a triangular decomposition [it violates axiom TD2]. We return to our discussion of M(A) as an d-module. We propose now to show the following: (7)

If

A(c) # 0,

then

M(A)

is irreducible.

We introduce some terminology: A multi-index is an element n e i^l'^of finite support. If , «i j are the nonzero values of n. A: > 0, . . . , ^ J, then we write

a" |n|

for the operator <2"^* • • * for the operator for and n! for

on M(A); on M(A);

In this notation a basis for M(A)^“ '^“ is the set of elements +, |n| = r. We have the following inductive calculations in U(a): (8)

[a,.,o-(a^)”] =

V,

n e Z+,

m

(9)

=5,7«

m ,n r=l

114

Lie Algebras Admitting Triangular Decompositions

Thus, if m > « > 0, flX fly )" •

[a^,(r(aj)"]v^ = 8¡jncaf’-^a(ajY~^ • v+ =

• Ü+

_ j 8¿jn\c'^ • \0

by (9)

by induction

if m = «, in m > n .

It follows that for multi-indices m and n with |m| = |n|, ( 10)

if m

0 \ n!c'"‘ •

n (then for at least one j, nij > rij)

if m = n.

It is now easy to see that M(A) is irreducible. Let r e Z+, and let X = Ze^a'!. ■

e„ e K .

Then for any multi-index m with |m| = /*,

ÍZ™ •X =

•v ^ = m!e^A(c)*^ •v + G: N ( X ) ^ = (0).

Thus each ^„ = 0 [if A(c) # 0], and hence x = 0. As for M(0), (10) shows that • u += 0 whenever |m| = |n| > 1, and hence is a submodule of M(0). Obviously this submodule is maximal, and hence is MO). We conclude that L(A) =

(M(A) ([

i f A ( c ) ^ 0, if A(c) = 0.

Example 2 (The Virasoro algebra of Example 2.1.4). The diagonal subalge­ bra of the Virasoro algebra is two-dimensional with basis L q and 0. Thus any linear functional A on is determined by the two values /i :=A(L q)

and

c — A(0).

We denote the corresponding Verma module by M = M{h, c) and a highest weight vector by Since 0 is in the centre of S3, we have 0 * x * i ; + = x * 0* u^= cx ‘ for all X e U(S3). Thus 0 acts as scalar multiplication by c on

2.3

115

Yerma Modules

M. With Ai e ]^* defined by A^CLo) = 1 and A^Cc) = 0, we have M = Using the Remark 1 the Poincare-Birkhoff-Witt theorem applied to the ordered basis L _ i , L _ 2, . . . of SS_ we find that a basis of is the set and

V^\n^
= n |.

The problem of determining the maximal submodule N{h, c) of M(/i, c) is quite complex (see the references in Section 2.8). Remark 3 There is obviously a lowest weight analogue of Verma modules. Verma modules for g with the triangular decomposition are lowest weight Verma modules for g with the triangular decomposition and vice versa. Proposition 5 Let g have triangular decomposition (f), g+, o-) and let = be its opposite triangular decomposition {see Remark 2.1.2). Let ( tt, M ) be any representation o f g. We define a pair ( tt^, M^) consisting of a K-space and a mapping g ^ gl(M ^) as follows: • as a vector space, • for X ^ V Ei

= M, 7t ' ^ ( x )

v

=

-7 r(a (x ))v .

Then (i) (tt®", M^) is a representation o f g; (ii) ((7r^)^(M -)-) = (7r,M); (iii) = M~^ for all /1 e ]^* Furthermore M is an ^-weight module if and only if is an ^-weight module’, (iv) N (zM is a (¿-submodule relative to ir if and only if •= N

=

(p,h)v,

—Tr{o’h)v = ( p , h } v , <=>

7r(h)v

=

(

—

p,h}v.

e 1^, for all

e M

116

Lie Algebras Admitting Triangular Decompositions

Again (iv) is trivial, and for (v), (A, y+) is a highest weight pair for Af(A) <=> U+ G M^, v + ¥= 0, Tr(Q^)v + = 0,

M = ir(U (g))i;+ , « D+e

0

tt'^{Q-) v +=

0,

M‘^ = 7T(U(g))£;^, <=><—A,i; + >

a highest weight pair for M.

□

Now (vi) follows from (iv) and (v).

2.4

^l2(K)-THEORY

We recall the Lie algebra ^I2(1K) consisting of 2 x 2 matrices of trace 0 with entries in K. The representation theory of this Lie algebra is of the utmost importance for the theory of semisimple Lie algebra. A basis {E, H, F} of §I2(IK) satisfying the relations [H ,E ]= 2E ,

[H ,F ]= -2F ,

[E,F]= H ,

is called an § 12-triplet. We recall that ^I2(IK) = K f 0 Kh 0 Ke, where

/=(;

2 ).

- ( i

-M '

«=(2

D-

As we saw earlier, {e, h , f ) is an § 12-triplet. Evidently if we set §_;= [K/,

:= Kh,

§ += Ke,

and define an anti-involution a on ^l2([K) by transposition, then we have a regular triangular decomposition ^I2(1K) = ^_0

0 ^+

with Q+= Z+a, where a(h) = 2 (Section 2.1, Example 3). Let A 0 f)*, and consider the ^I2(1K)-Verma module M(A) of highest weight A. Fix a highest weight vector of M(A). Then by Propositions 2.2.1

2.4

117

ёКСЮ-ТЬеогу

and 2.3.3. = K /" • +^ (0),

for all

n ^ N.

It follows from Proposition 2.1.1 that any submodule R of M(A) is either (0) or has the form for some Hq > 0. In the latter case e -/"0 • y + = 0.

(1)

Conversely, if Hq is a positive integer for which (1) holds, then we see that /'^0 • is a highest weight vector and U(g) *u += U(^_)/''o • +c M(A) is a proper submodule of M(A). We conclude that the (unique) maximal proper submodule MA) of Af(A) is generated by the highest weight vector /"0 • ¿;+, where is the least positive integer satisfying (1). Lemma 1 Let \(h) = a e K. Then for all n ^ N, (i) /z • /" • u^= (a - 2n)f^ • (ii) e • /" • z; + = n{a — n + 1 )/" “ ^ *z;+, where we make the convention that f~^ ‘ v += 0. Proof (i) follows from the fact that / " • +e M(A)'^“"". We prove (ii) by induction on n, it being clear if n = 0. For n > 1 we have e

• ü+= [ e , / ] • / " ^ - v ^ + f - e - f n - l • V, = (a — 2(n — 1) ) / " ^ • y + + (/i — 1)(л — n

2) / " ^

□

= n(a —n + 1) / ” ^ v +. Proposition 2

Let M(A) be the Verma module for §12(Ю with highest weight pair (A,z;+). Then (i) M(A) is irreducible if and only if \(h) ^ 141; (ii) if A(A) e 141, then N ( \ ) is a Verma module with generator and highest weight A —(A(/z) + l)a. The irreducible module L(A) is of finite dimension X(h) H- 1 and is the only nontrivial quotient module of M(A). Proof Let a = A(/z). After the preparatory remarks and Lemma 1, we see that M(A) is reducible if and only if По(л —«0 + 1) = 0

for some n^ e Z^_.

118

Lie Algebras Admitting Triangular Decompositions

This is possible if and only if a e I4j and п^ = a This shows (i) and also shows that equation (1) has precisely one solution whenever a e N. In that case M(A) has only one nontrivial submodule, namely ЖА), and MA) is generated as a highest weight module by ' v Thus MA) has highest weight A —(a + l)a. Since (A —(я + \)aXh) = —л —2 e Z _ , it follows from (i) that the Verma module M(A —(a + l)a ) is irreducible, and so MA) is a Verma module, namely MA) = M(A - (a H- l)a) (Proposition 2.3.3). If M(A) ^ L(A) is the natural homomorphism [with kernel MA)], then {/^ • Ü+10 < 7 < й} is a basis for L(A). □ The finite-dimensional irreducible highest weight modules L(A) are pre­ cisely those for which A(A) = a e N. We sometimes denote L(A) by L(a) if \(h) = a. In the physics literature these modules are often denoted simply (a) or (a) [and also unfortunately by their dimensions (a + 1)]. If v+ denotes a highest weight vector of L(A), then we have the basis (2)

v^,f ■

and corresponding weights A,A —a,A —2 a , . . . , A — aa = —A. The values of the weights at h form the symmetric string (3)

a,a - 2,a - 4 , . , , , - a .

Trivial though it may seem, this symmetry will eventually lead us (in Chapters 4, 5, and 6) to a complete theory of symmetry in root and weight systems of Kac-Moody algebras. If we set у _=/"* • i;+, then, up to scalar multiples, the vectors in (2) can be written e"* '

'v

v_,

which displays L(A) from the point of view of a lowest weight module. Another way to look at the symmetry is to interchange the roles of e and / in the triangular decomposition (and then to replace й = [e,/ ] by —h = [/, e]). Then L(A) is a highest weight module with highest weight vector v_ and highest weight -A ( - A (- A ) = д). One of the central results of the classical theory of semisimple Lie algebras is Weyl’s theorem, which states that their finite-dimensional repre­ sentation are completely reducible. Eventually we will prove a version of this theorem (set in a much wider context). However, we will find it useful to have the result for §12(Ю much earlier. The proof that we present here is in essence the same as the more general one that will appear in Chapter 6. We will make use of the concept of primitive vectors, a subject that is more fully developed in Section 2.6.

2.4

SljOlO-Theory

119

Let M be an §l2(lK)-module admitting a weight space decomposition relative to IK/z. A weight vector v e is called a primitive vector if there is an §l2(lK)-submodule N oi M such that u ^ N and such that e • v ^ N, For instance, highest weight vectors are primitive [with N taken as (0)]. Proposition 3 Let K / 0 K/z 0 Ke be the triangular decomposition o f §I2(1K) described above, and let M be a finite-dimensional ^l2(K)-module admitting a weight space decomposition M = 0M ^ (p 0 1^*) relative to f). Then M= © L„ ¿=1 where each is an irreducible ^i2(K)-module isomorphic to L(A^) for some = /¿a, I f e N. The decomposition is unique to the extent that the collection (Aj,... , \jf) of highest weights is unique up to rearrangement. To prove this result, we introduce the element (4)

C = |/z2 + e/ + /e

of U = U(§l2(IK)). The next lemma says that this element lies in the center 3(U) of U. In general, elements of 3(U) for semisimple g are called Casimir operators. However, there is always a quadratic Casimir operator like (4) which is essentially unique and is usually called the Casimir operator. We will never have occasion to use any other. (In Section 4.5 we will introduce quadratic Casimir operators in a far wider context.) Lemma 4 C lies in the center of U. Proof It suffices to show that [e, C] = [/, C] = 0, since e and / generate §I2(IK) and hence U. We have [e,^h^ + ef + fe] = \{ [e ,h ]h + h{e,h]} + e { e , f ] + [ e , f ] e = —eh — he The case of / is similar

eh

he = 0. □

Suppose that F is a highest weight module for §I2(1K) (relative to a given triangular decomposition), and let be a highest vector of weight A. Let

120

Líe Algebras Admitting Triangular Decompositions

a = A(/z) e IK. Then C •f

f

++ e •/ • +

=

+ /e ) • f +

= { W + «)í'+Also for all MG U, C • M•

u ' C ' v^= [\a?’ +

• v^.

Since i;+ generates F, (5)

C V=

+ a)v

for all f g F.

Proof (of Proposition 3). For some algebraic extension IK' of IK all the eigenvalues of C acting on M lie in IK' [if IK is algebraically closed, then IK' = IK]. Let M' = IK' M. Then M' is a module for the Lie algebra IK' ^ ^I2(1K) = §I2(IK'), and C can be interpreted as an element of 301'), where U' == U(§l2(IK')). From M = we have M' = 0(IK' ^ M ^ \ Each space M'^ •= IK' 0^^ is a weight space for 1^' — IK' % 1^, from which it follows that we have produced a weight space decomposition of M' relative to f)'. Let t be an eigenvalue of C with corresponding generalized eigenspace M' := {i; G M '|(C - i)^ • = 0 for some A: > 0} (see Section 7.1). Then Af' is an §l2(lK')-submodule for M' since C g 301')- In particular it is 1^'invariant and hence decomposes into weight spaces M[^ '= M[ n M'^. If VG then e ' V E: (where we are identifying §I2(W inside §I2(IK') via X 1 0 x). In view of the finite dimensionality of M' over IK', there can only be finitely many nontrivial weight spaces, and hence there is a weight ¡i and a weight vector u g \ {0} such that e • ¿; = 0. Now is a highest weight vector, and in particular a primitive vector of M'f. Let us show that for any primitive vector w of weight v in Af/, we have ( 6)

K/t)

and

j v { h f + v(h) = t.

Let AT be a submodule of M'„ with w ^ N, e ■w ^ N. Then on w + N ^ M't/N, C acts as multiplication by \v(,hY + v{h), and hence C • w = (\v{h)^ + v{h)^w mod N. Since (C —i)* • *v = 0 for some A: > 0, we obtain \ v i h y + vih) = t. Since v

2.4

121

gl2(IK)-Theory

is a highest weight for the finite-dimensional quotient module M¡/N, v{h) by Proposition 2. This establishes (6). Now (6) shows that there is only one primitive weight in M/, namely above, and that a — e N. In particular t e K. Thus all the eigenvalues of C lie in K, and the extension of K to K' is unnecessary. Let u \ , . . . , be a basis for M^. If e • 0, then + /:a is a highest weight for some A: > 0, contradicting our observation above about primitive weights. Thus for each i, L,(/¿) == U • is an irreducible ^l2(IK)-module isomorphic to Liix). It follows that n (Ey^,U • = (0) (since iv{) and that R = ® L j ( f i ) CM,. ; =i Consider the quotient module M^/R. It is a finite-dimensional §l2(W-module with a weight space decomposition on which C acts as multiplication by t. The argument above shows that if it is not (0), then it has a highest weight v. Then V is 2i primitive weight of M ,, and hence it is equal to ¡ by (6). But { M J R Y = M>t/R^ = (0). Thus R = M„ and jl

(V)

© Lj(jx).

y=l

From M = (the sum taken over all the eigenvalues of C) and (7), we have the decomposition of M into irreducible submodules, as required. For the uniqueness property we consider the composition series k

k

M = © Ly D © Ly D y=l y= 2

э L ^ э (0)

(i.e., each submodule is a maximal submodule of the previous one). By the Jordan-Holder theorem (which is a very general fact whose scope lies far beyond Lie theory) the factors

j=i + \

are unique (including repetitions) up to the order in which they appear. [Ja2, 3.3]. □ Remark 1 It is quite impossible to expect any stronger uniqueness in the decomposition of a module into irreducible submodules. For example, the module M — L 0 L also decomposes nicely as L + 0 L_, where L = =

122

Lie Algebras Admitting Triangular Decompositions

{(i^, ± ^ L). However, one does have uniqueness of the isotypical com­ ponents (the Af, in the preceding proof) [Ja2, Thm. 3.11]. CoroUary 1 Under the hypotheses of Proposition 3, the Casimir operator C acts as a semisimple transformation on M with eigenvalues of the form \a^ + a, a The eigenspace of M is the sum o f all the irreducible submodules isomorphic to L{a) [the L(a)-isotypical component]. Example 1 Consider ^I2(1K) c gl2(lK). The adjoint representation, adgi^^^^^, restricted to ^I2(1K) determines a representation of ^Í2([K) on gl2(IK). We do not need to invoke Proposition 3 to conclude that gÍ2(W is completely reducible: clearly gl2(lK) = IKl e ^I2(1K). As a preview of what is to follow below, we explicitly describe the corresponding representation of 5L2(IK). Let A" g gl2(IK). Then ad = L x — R x ’>where L x and R x are, respectively, left and right matrix multipli­ cation by X. If X is nilpotent, then exp ad A" = exp(L;^ —R x) = exp L;^(exp — R x) = exp L ; ^ ( e x p s i n c e L x and R x commute [in End(gÍ2(IK))]. In other words, for all Y e gl2(IK), (exp ad X \ Y ) = exp(AOexp(y)exp(AT)“ ^ As we will see, this shows that the representation of SL2QÍ) on gl2(W is p i - X ) : Y ^ x Y x - ^ for all x e SL2(K). The Lie group associated with the Lie algebra ^l2(K) is the special linear group SL2ÍK)

ad-bc = 1 .

-{l(:

Any of the finite-dimensional §l2(IK)-modules that we have been discussing above carries a natural compatible 5L2(lK)-module structure. We begin with a noteworthy fact about the representations of §I2(1K). Let denote the space of 2 X 1 column vectors with coefficients in K, and let ir be the natural representation of §I2(IK) on it (i.e., acting as matrix multiplication). Then in and let fact K ^ s L(l) in our notation above. Let M= ^ j j , i ; = | ^ j i rm Then S is graded by 5 = 5([K^) be the symmetric algebra on the space Z, 5 = following Section 1.2, 5'^ =

© Ku^vL i +j = n

Each X e §I2(IK) lifts to a derivation on S, say, X (Proposition 1.1.5), and it is immediately clear that [X,Y] = [X,Y] [both are derivations extending [AT,y] e §I2(IK)] showing that we obtain a representation on S. Since X(S") c 5", we obtain a representation of §I2(IK) on each 5", n = 0,1,2,... .

2.4

§ 12(1K) -Theory

123

Proposition 5 Let n ^

Then ( tt^, 5 ”) is an irreducible ^l2ÍK)-module isomorphic to L{n).

Proof. The case « = 0 being obvious, we can assume that n > 0. Let {e, h, /} be the standard basis for §I2(IK). Then for all X e §I2(1K), = nu^ ^tt{ X ) u . Since

(s

m^i i )

and

we have 7T„(e)w" = 0, 7T^{h)u’^ = nw". Thus is a highest weight vector of weight n for §I2(1K), and the result follows from part ii of Proposition 2 and the fact that dim 5" = « + 1.

□ Now we extend the action of SL2(K) to 5(IK^) in an analogous fashion. Again, let 77 denote the natural representation of 5L2([K) on Each X e SL2(K) determines a linear mapping on ® ••• 0 (« factors) by i;j 0 • • • 0 7r(jc)i;i 0 • • • 0 v(x)u^, and hence on the tensor algebra T(IK^) by linear extension [where on T([K^)° = K, the action is trivial]. Thus 5L2(IK) acts as a group of automorphisms of r([K^) (as an associative algebra). Since it stabilizes the ideal J generated by the set of vectors y 0 w - w 0 y , y,w e we have a group action 77 of SL2(K) on 5(1K^) = T(K^)/J (see Section 1.2) with tt( x )( v ^ • • • v^) = irix)v^ • • * 77(jc)i;„. Let ^ Then 7T^ is our desired representation of SL2(K). Using the isomorphism 5" —L(n), we have an action of 5L2(IK) on L(n). Next we study in what sense t7„ is compatible with t7„ [see (8) below]. Let X e §I2(IK) be a nilpotent matrix; that is, X ^ = 0 for some /: > 0. In fact = 0, fc > 0 => (K^ d ATIK^ d AT^IK^ = (0) the containments being strict, so X ^ = 0. Relative to some basis {u^, U2) of IK^, A", as a linear transformation, has the matrix form ( ^ ^ |, whence exp AT = 1 + X + ( : : ) ' ATV2! + • • • is similar to “j, which shows that exp X ' 5L,

124

Líe Algebras Admitting Triangular Decompositions

Lemma 6 For all nilpotent X e §I2([K) and for all n ^ N, and

( 8)

is nilpotent on L(n),

expir„(A') = ■j7„exp(X).

Proof. We may take S" as our model of L(n). For Mj, . . . , m„ e ATe SI^OK),

and

n

^ n ( ^ ) ( “ i • • • « « ) = L «1 • • • ATwy • • • u„, J=i

and one sees inductively that 7T„(Xf(u, ■■■ U„)=

E . , *,+ ••• +*„=* '^1---- '^n-

•••

It follows at once that if X is nilpotent, so is tt„(X), and that

expv „(X )(u i • • • « „ ) = E — k _

— ( “ i ••• u„) _

X ’^^

X^-

k A:i+ • • +A:„=A: ^1* = exp X{u f) • • • exp X(u^) = 'n-„{expX)(ui ■■■ u„).

□

We say that an §l2(K)-moduIe (p, F ) is integrable relative to the basis {e,h, f ) if V is decomposable as a direct sum of IK/i-weight spaces and pe and p f are locally nilpotent on V, Proposition 7 An integrable ^l2(K)-module (p, F ) can be written as a direct sum o f finitedimensional irreducible ^l2(K)-submodules, In particular (p, F ) is completely reducible. For every nilpotent element X e §I2(IK), p{X) is locally nilpotent. The space V admits a unique SL2(K)-representation p with the compatibility condition (8)

exp p { X ) = p(exp X )

for all nilpotent X e §I2([K).

2.4

êl2(K)-Theory

125

Proof. If f e K is a [K/z-weight vector, then the êl2(lK)-submodule generated by V is ¿,;>0

(by the PBW theorem), and this is finite dimensional because of the local nilpotence of pe and p /. Thus W is completely reducible by Proposition 3 and it follows that F is a sum, hence direct sum, of finite-dimensional irreducible submodules of êl2(K) (Proposition 1.6.2). Using Lemma 6, every nilpotent X e êl2(lK) is locally nilpotent in its action on V. Furthermore the Lie algebra action êl2(IK) on each irreducible submodule W of F “integrates” to an SL2(K) action, as we described above, and the compatibility condition (8) holds on W. Piecing the representations together in the direct sum, we have a representation p of 5L2(IK) on F which is compatible with p. The uniqueness of p follows from the fact to be shown immediately below that

{(i

o)L ““ ((!

2)},.K

form a set of generators of the group SL2(IK).

□

We now gather together some useful facts about SL2(IK). Define B ==

a 0

b a-*

H -=

t 0

0

\b ^ K

0 - t -1

AT:=//U

t e

It is easy to see that each of these is a subgroup of 5L2(IK), that H
(o

!)(-!-■

?)(o

;)■ (-?-'

shows that the set i(A - { ( i

#\/l

n\

;)■!(:

Î)

te K

o)

126

Líe Algebras Admitting Triangular Decompositions

generates the element n{t) and hence also the matrices

( 10 )

h( t) ■■=n ( t ) n ( l ) ' = 1^

t^K ^.

j

Given an arbitrary matrix A =

* e 5L2(IK), we want to reduce it to 1

by left multiplications by elements of U. Applying n(l) if necessary, we can assume that a 0. After applying

1 —ca~^

0 1

we can assume that A =

a 0

b a-^

Then /i(a-i)y4 = | j

and

= 1.

We have proved the first part of Proposition 8 (i) U generates 5L2(IK). (ii) SL2(K) is generated by B KJ N\ and H = B C\ N, (iii) 5L2(1K) =

^

(Bruhat decomposition).

Proof. It remains to prove parts (ii) and (iii). From (9) we see that U lies in the subgroup of 5L2(IK) generated by B and N, whence the first part of (ii). Obviously B n N H. Set

=(-?

i)-

Notice that s is a representative for the nontrivial coset of H in N. After part (ii) we will prove part (iii) if we show that B U BsB is closed by right

2.4

SijiW -Theory

127

multiplications by elements of N. Since N = H U Hs, this reduces to showing that sBs (Z B U BsB. In showing this, we may replace each occurrence of 5 by any element that we wish from the coset Hs = sH of H in N. Using (9), we see that

1 -r-i

0 G BsB 1

for all t e

Thus for all t e Let a typical element of B be written as

[l

/-)(J

S i G K.

!)•

Then

^( o

which is what we wanted.

□

Parts (ii) and (iii) of Proposition 8 express a basic ansatz, which we will see fully developed in Section 6.3. An interesting application of Proposition 8 is to use it to prove the following. Proposition 9 The only proper normal subgroups o f SL2(K) are

( ( i ;)}

-

(

m

:

:)}■

The same proof {due to J. Tits) may be used to prove that many other groups are ''essentially"’ simple [773]. Proof Suppose that K # 5L2(K) and that AT is a normal subgroup of SL2(K). Consider the subgroup BK. Since it is a union of .B-double cosets, BK = B, or BK = SL2{K). The latter implies that SL2(K)/K = B K / K = B/B n K, However, is a solvable group; hence also B / B OK , whereas 5L2(IK) is its own derived group, and the same goes for SL2(K)/K. These are

128

Lie Algebras Admitting Triangular Decompositions

contradictory, so we have BK = B\ that is, B Z) K. However. /

0 1

l\la 0 /\0

feW

0

-1 \ 0/

\ —b

0 a

. A trivial calculation with shows - ( : : ) □ that fl = +1. Thus K has the required form. and hence K c

( : ; - )

Corollary The action of 5L2(1K) on 5" is faithful if n is odd and has kernel {±1) if n is even. Proof By Proposition 9 the kernel is either trivial or is {± 1}. Since

the result is clear.

□

2.5 CHARACTERS Suppose that M is a vector space which is graded by an abelian group P and for which the degree subspaces are finite dimensional. In many situations we can, or would like to, make useful numerical statements about all of these subspaces at once. For instance, we might like to know the dimensions {dim or we might have a homogeneous linear operator T of degree 0, and we might be interested in the traces (tr(r|A/a)}^gp (of which the dimensions are the special case for T = id^). As a simple example consider 5(K), F = K x 0 lKy, asa Z-graded module under the total grading. Then 5(K) == IK[x, y], and the degree subspace S%V) of elements of degree n admits the monomials

as a basis. Thus dim S%V) = n + 1. This information is neatly stored in the formal power series (1)

E (« +1)9"/1 = 0

This series goes by various names: the Poincare series of (the Z-graded module) 5(K) or the generating function (for the dimensions of the degree

2.5

129

Characters

subspaces) of 5(K). We often call it the character of 5(K), for reasons to be explained later. The use of the word “function” in this context is not to be interpreted to mean that (1) is some real or complex valued function of q. Of course such an interpretation might be useful (in which case questions of convergence appear), but our only intention at the moment is to consider (1) as an element of the formal power series ring K[[^]]. Part of this section is devoted to a basic introduction to power series. For the moment it suffices to say that (1) is an invertible element in the ring K[[q]] and that its inverse is (1 - qY. The ring structure of K[[q]] is manifestly useful when we go on to consider the graded algebra K[x, y, z]. Regarding [K[x:, y] and K[z] as subalgebras of K[x, y, z], the space K[x, y, z T of elements of degree n in K[x, y, z] is E K [x ,y M zr"^ / =0 and (2)

dim(IK[x, y, z ] ”) = ^ dim(lK[x, y]')dim(lK[z]" i= 0

Clearly the generating function of K[z] is E7=o^' = (1 —^) \ and hence by (1) and (2), the generating function of K[x, y, z] is -3

n=0

n=0

Induction on dim V quickly shows that the generating function of SiV ) for dim F = m is (3)

(1

It is not hard to expand this as

(!-«)■” -

£ j" '

We now establish a suitable generalization of this for treating jP-graded spaces. Let P be an additive abelian group. We let e{P) = [e(a)\a e P}

130

Lie Algebras Admitting Triangular Decompositions

be a multiplicative copy of P; that is, we define e(a)e(P) '■=e{a + P ) ,

a,P ^P .

The elements e(a) are called formal exponentials. Given a commutative ring R, we define P[P] = Re(a) to be the free module over R with basis e(P). Then R[P] becomes an associative i?-algebra with 1 — e(0) as the identity element if we define a multiplica­ tion by (4)

H

a^P

a„e(a) • E /3e/>

= E ( E a +^ =y

'

R[P] is called the group algebra of P with coefficients in R. If M is a finite-dimensional P-graded space, then M has only finitely many nontrivial degree spaces, and we can define the generating function or character of M by ch(M ) = i ; dim M“e (a ) e Z [P ]. Unfortunately, most of our graded spaces are not so simple, and we have to extend Z[P] to express their characters. To this effect let us assume the following. (5) There exists a nonempty family {yy}y of elements of P indexed by a set J such that the group Q •= Ey^jZyy is freely generated (as an abelian group, by the {yy}yej* Under this assumption we let denote the semigroup of Q generated by the 7y’s. A typical example of this situation is Q+= \ {0} in Q = Z^. Assuming (5) is in effect, we define sl Q +- fan in P to be a set of the form U {A - ^ 1^ e (2 +}, where A e P (Section 2.2). We call A the source of the fan. We introduce a partial ordering < on P by setting / x
2.5

Characters

131

we define the support of / by su p p (/) == {a e Pla„

0}.

Notice that R[P] = { / ^ R[PT\supp(f) is finite}. We cannot supply R[PT with a composition law by simply extending (4) because meaningless infinite sums of elements of R will arise on the right-hand side. Consider, however, R [ P i Q ^ ] ■■= { /=

a ) e jR[P]"^|supp(/)lies in a finite number of ¡2+-fans|,

and J.0+] == { / = £ a „ e ( a ) |s u p p ( / ) lies in A J. g + j , where A is an arbitrary element of P. Both R[PIQ+] and i?[0i<2+] become associative algebras under (4). Indeed let / = La^eia), and g = Lb^eip) lie in R[P j Q+] with supp(/) c (J (a, i(2 + ), 1=1

supp(g) c U j=i

Let 5 ^ P, and consider writing 8 = a p, a e supp(/), p e supp(g). Then 8 = pj - e ^ (a^ + Pj)i Q^, where s is the sum of two elements of C+U{0}. By the assumption on <2+ we see that there are only finitely many ways of writing 8 in such a manner. Thus

a +(B=8 is a well-defined sum and (4) makes sense. Clearly R[0i Q+l is a subalgebra of R [ P i Q j . For each A e F we define ^[■Pi!2 + ] ( A ) ~ { E « a ^ ( “ ) ^ R[P i Q +] \ ^a = Ofor all a rel="nofollow"> a). Let { /J/ei

^ family of elements of R[P iQ+], and write fi=

£ « « ,« (« )•

a^P

132

Líe Algebras Admitting Triangular Decompositions

We say that the above family is summable if the following two conditions hold. (51) There exist , A„ in P such that supp(/,) c (J J=i(Ay i Q_^,) for all i e I. (52) For all A in F there exists only a finite number of i in I for which /; i R[P i Q ^ \,y Proposition 1 Let R[P I Q

be as above.

(i) Two elements x and y o f R[P iQ ^ ] are equal if and only if x = y mod R[P i !2+](a) ^ (ii) If (//)/ei ^ ^ summable family o f elements o f R[P IQ+] with f = there exists a unique element f = of >10 +] that a^ = ^i^ts sum has only finitely many nonzero terms). Proof (i) is straightforward. (ii) Let a ^ P. Then =>f^ ^ R[P>IQ+](«). Axiom S2 then shows that the sum E^ei^a, ¡s finite. Consider / == e R[PT • By SI we see that supp(/) lies inside the union of a finite number of fans. Thus f^R [P iQ ^l □ The element f ^ R[P i Q+] defined in Part (ii) of Proposition 1 is de­ noted by E, ei//Now suppose that M is a P-graded vector space with finite-dimensional degree subspaces such that all nontrivial degrees [degrees a for which M“ (0)] lie in a finite union of 0+-fans. Then we define the character of M by (6)

ch(M) = £ dimM“e( a ) e Z [ P i 0 + ] . a^P

All important case of this is the situation where M is a highest weight module of highest weight A for a Lie algebra g admitting a regular triangular decomposition (f), g+, Q+, a). Then M is graded by the group P = ZA + 0, where 0 is the root lattice, so that (5) holds for this choice of 0 , and all nontrivial weights of this weight space decomposition lie in the 0+-fan A i 0+. Also by Proposition 2.2.1 all the weight spaces are finite dimensional because of the regularity of the triangular decomposition. The character (6) is

2.5

Characters

133

thus defined. Likewise we may define (7)

ch(g_) =

Y.

dim g “e ( - a ) e Z[Oi g +]-

It is primarily because of this type of example that it is more convenient to base fans and the definition of IK[0i (2+1 on - Q + rather than Q+. The use of the word character in this connection has its origin in the complex representation theory of compact Lie groups. In this situation is a finite-dimensional abelian Lie algebra over R, and there is associated with each h ^ if an operator on the complex space M whose action on the weight space M" is multiplication by [whereas h acts as multiplication by a(h)]. The operators form a compact abelian group T, If M has finite dimension then tr^ = E d i m M “e^“^^\ The mapping tr^(e^) is a character : T ^ C (in the sense of group theory). We have a formal version of this with e(a) replacing the function e" : -> We can try to define ch(M) as a function on i) through h

Y dim

cc(h)

This of course raises very interesting questions of convergence and the rate of growth of the dimensions of the weight spaces, which we do not treat in this book. The ring R[0i 2+] is easily identified as a ring of formal power series. Let the basis of Q+ given in (5). Each a ^ Q+ has a unique coordinate representation in terms of the y/s: ( 8)

ri:y: + JrJi

Jk'Jk'

The coordinates n = (itj) are elements of (mappings of J into [^) for which ftj = 0 for all but a finite subset of J. Let Xj •= e(-yj). Then for a with coordinates n as in (8) e { - a ) =x;/i... which we denote simply as x". Then

where the sums run over all coordinates of elements of 2 +u{0}. As we have already mentioned, these are formal sums involving (in general) infinitely many nonzero coefficients. The most common situation is with J = {1,..., m} in which case each monomial x" and /?[0J,2+] is the well-known ring of formal power series in m commuting variables, R[[xi,. . . , x^]]. We will also have

134

Líe Algebras Admitting Triangular Decompositions

occasion to use the case J = In both cases the exponential and monomial notations are useful, and we work with them interchangeably. Generally, when we are using the monomial notation, we will denote R[0i Q+]hy i?[[jc]] or The first question that we settle is which elements of R[[x]] are units (i.e., have multiplicative inverses). For m = (rrij) ^ (2 +*-^{0} we define k l := Recall the partial ordering < defined earlier. For m, n ^ ¡2 +^(0}? m < n if and only if

for all i.

Let R[[;c]]" be the R-span of the elements X*" for which In particular

lm| = n.

= R.\.

Proposition 2 For / = ^ ^[[^]] to be a unit, it is necessary and sufficient that üfi a(0, . . . , 0) be a unit o f R, I f f is a unit, then its inverse g = Ebm*™ ^ recursively defined through

(9)

L

aA-

r + s = m, s < m

Proof Writing fg = l leads directly to (9). If a^ is a unit, these equations admit a unique solution which is found inductively on |m|. If a^ is not a unit, then there is no solution to a^bQ = 1, and hence / is not invertible. □ A particularly important instance of this is the well-known (10)

(1 - x ) ~ ^ = l + x + x ^ +

.

We have already pointed out the expansion {i-x)

- £ (

„

)* ’ .

We will have a number of occasions to write down product expansions (11)

n (i+ /,).

2.5

Characters

135

where /¿^R[[x]\ and / / = (i.e., = 0). These products make sense if we assume that for each n e Q + there is a finite subset D(n) of J such that = 0

whenever j ^ D(n)

and

m < n.

Indeed in that case only finitely many /, can contribute to the coefficient of a given monomial x". More precisely the coefficient of x" in n (1 +//)

is independent of S provided that S d D(n). This coefficient is then the coefficient of x" in (11). Consider the rather natural looking product ( 12 )

n ( i + « 0 ^ »<[[«]]•' y=l

This has an expansion L Poin)q", n= 0

and Po(n) has a simple and beautiful combinatorial interpretation. The coefficient Po(n) of was already established by the time we expanded

fl(i +^0-

y=l

Since expanding this product amounts to forming simple products by choos­ ing either the 1 or the from each factor, we see that p^in) is the number of sequences 1 < 7i < 72 < **• < Jr

such that E-=i 7/ = n, that is, the number of partitions of n into distinct parts. In the same way if m^, m2, . . . belong to N and we write (13)

n ( i + «0 ^= L r ( n ) g " , ;=i

then r(n) can be interpreted as being the number of partitions of n into ^The use of q rather than x^, or simply x, is to conform with the standard notation used for partition functions in the literature and elsewhere in the book.

136

Lie Algebras Admitting Triangular Decompositions

distinct parts where the number j may be considered available in different colours. For example,

y=l

generates the partitions of n into distinct parts where the odd numbers are of two colours but then even numbers of just one colour. Suppose that we distinguish the two types of odd numbers by plain and bold fonts. Then, for instance, the coefficient of is 6, with the actual partitions being 4,31,31,31,31,211. The above partitions are restricted; this means that we can use each part only a limited number of times. The product (14)

0(1-90 ; =i

= E p («)9"

enumerates the unrestricted partition function: p(n) is the number of se­ quences j\ <J 2 < • • • <Jr,

with

= n. To see why, expand each factor of (14) using (10) to get

{l + g + q^ + ■■■)[! + q^- + ( q ^ f +

+q^ + { q ^ f +

Now interpret the /cth factor (1 +
1

1 part of size k 2 parts of size k <->

2

Then any given monomial

is interpreted as the partition 1 . . . 1 2 . . . 2 . ..A:...A: Si

$2

Sf^

ofn =

,

2.5

137

Characters

and the entire expansion has the interpretation as the generating function of the unrestricted partition function. To interpret (15)

n ( i -« 0 ; =i

>

m,

we imagine that each integer j is available in rrij “colours.” This is the generating function for the number of partitions of n into parts where we may use each positive integer any number of times and also with any combination of colours. If rrij = 0, then j cannot be used in any partition (i.e., j is not available in any colour). For more on partitions, see [HW] and [An]. We come now to the main results of this section. Proposition 3

Let n be a Lie algebra graded by an abelian group P for which the following conditions hold, (i) All the nontrivial degrees lie in —Q+ where Q+ is a semigroup o f P satisfying (5); GO For all a e •= dim n ““ is finite. Let U(n) be the universal enveloping algebra o f n with the inherited P-grading {Section 1.8). Then ch(U(n))=

n

(1

Proof. Let [yj}jej be the given basis for <2+- Let i;*., A: e K, be an ordered basis of n consisting of homogeneous elements, say, g n “**, 8 ^ ^ Q+Then according to the Poincare-Birkhoff-Witt theorem, the products (16)

ki< ...< k„

form a basis for U(n). On the other hand, the expansion of (17)

r i ( l - e ( - 8, ) )

-1

A:eK

enumerates all the products (18)

e ( - 8 , y “' . . . e { - 8 , y \

k,< ...
138

Lie Algebras Admitting Triangular Decompositions

The coefficient of e ( - 8 ) in (17) is thus the number of products (18), with

which in turn is the number of products (16) that lie in U(n)“^. Since every a appears times among the 5y’s, the proposition follows. □ Example 1 (Abelian Lie algebras) As a simple example we take a finite­ dimensional vector space a with a basis and consider it as an abelian Lie algebra. By Proposition 1.8.1 U(a) is naturally isomorphic to the symmetric algebra 5(a). If we set Q = Z, Q+= Z+, a~^ = a (all elements have degree —1), and q = e( —1), then Proposition 3 gives ch (5 (a))= (l-« )

-N

which is precisely what we had deduced before in (3). If instead we set Q = Z", ¡2+= \ {0}, = (0,..., 1,.. ., 0) (1 in the ith place), and = KXi, then we have a completely different grading, and

ch(5(a)) =¿=n(l-K-a,))“‘. 1 An immediate consequence of Proposition 3 and Remark 1 of Section 2.2 is the character formula for Verma modules: Proposition 4 Let q be a Lie algebra admitting a regular triangular decomposition. Write 9= ^ ® 9+, and let Q denote the root lattice. Let ch(g_) =

L " !« « (-« ) ^ Z[04, e + ] . a^Q^.

Then (i) ch(U(g.)) = - K -a))-"*« e Z[01 (ii) for each A e 1^* the character o f the corresponding Verma module M(A) is given by ch(M(A)) = e(A) n where P = Z \

Q,

( l - e ( - a ) ) " ' ”“ e Z [ P i < 2 j

2.5

139

Characters

Just as with (15) we can give the product in Part (i) of Proposition 4 a combinatorial interpretation: ch(U(g_))=

E K{ p ) e { - I 3 ) , j8eQ+u{0}

where K(p) is the number of partitions of ¡3 into parts a ^ Q+ and 1. each a ^ Q+ occurs in different colours, 2. each a ^ Q+ of each colour may be used any number of times. This gives rise to a function K:Q (defined to be 0 outside Q ^) which is called the Kostant partition function [Kol]. Example 2 (§I2(K)) We have §I2(IK) = IK/ 0 Kh © Ke graded by Q = l a c ]^* with / of degree —a. For A © 1^*, M(A) = 'L n > o ^ f * where u+ is 3. highest weight vector, and by Proposition 4, ch(M(A)) = e(A)(l - e ( —a))~^ =

^ ( A ) + ^ ( A — Of) + ^ ( A — 2o f ) + * * * .

Example 3 (The Virasoro algebra) We use the triangular decomposition of S3 given in Example 2.2.4 with SS_= ^ ® where Ai(L q) = 1 and A/c) = 0. Let q = e(-Ai). Then 1KL„ = and ch(SS_) = n= \

Let (19)

e ]^*. Then

ch(M(/x)) = e(/i,) n (1 /2

=1

\

and we see from (14) that dimAf(/i,)'^ ' ' ^ = p { n ) , where p{n) is the partition function. This agrees with the explicit basis constructed in Example 2.3.2.

140

Líe Algebras Admitting Triangular Decompositions

The final example is an application to the theory of free Lie algebras. Suppose that card X = m < oo. According to the discussion in Section 1.10, there are two natural gradings of S (^ ), one by Z in which every element of X has degree 1 and one by ß = 2 '”- The next result gives formulas for the dimensions of the degree subspaces with respect to each of these two gradings. We do not need these formulas in the sequel, and they can safely be omitted. Proposition 5 [Witt] Let X be a set o fm elements^ and let S(AT) be the free Lie algebra on X over K. Let pi be the Mobius inversion function. Then (i) for all n

^ r\n

(ii) for all m-tuples, n = (n^,. . . , n^) of nonnegative integers

where r|n means that there exists a positive integer k such that nj = krj for all j = 1, . . . , m, and in that case n / r — k. Proof Both parts are proved in essentially the same way using a result from the exercises. We give the proof of the second one. We refer readers unfamiliar with the Mobius function to [Jal] and [Bo2, Appendix]. For each m-tuple n let = dim^^ By Proposition 3 the character of the universal enveloping algebra of ^ ( X ) is n ( l - x " ) ~ ‘' " e K [ [ x ] ] , n

where x = {xj,. . . , x^}. (Do not confuse the commuting variables Xj,. . . , x„ with the elements of X ) On the other hand, we know that the free associative algebra A ( X ) is the universal enveloping algebra of S(X ), and its generating function is easily seen to be (1 - (x i + ••• +x„,)) \ Thus ( 20)

n ( l - x " ) " " = l - ( X i +

--+xJ.

2.6

141

The Category 0

By Exercise 2.16, the dimensions d„ are given in terms of the set 5(r) of partitions of r into other m-tuples and certain multinomials C(p) for each partition p. Since the only coefficients on the right-hand side of (20) are limited to terms of degree 0 or 1, the multinomials C(p) are 0 unless p is the partition r = r i ( l , 0 , . . . , 0 ) -I-r2(0, l , . . . , 0) 4- • •• -t-r„(0,0,...,l) in which case it is (/•i + ••• \

1• ***' m•

1 ''i +

□

Formula (ii) follows immediately. Example 4 ( ^ ( ^ ) , card(Ji) = 2) n 1

2 3 4 5

dim /x(l)2^ = 2 (1/2){ m(1)2" (1/3){ m(1)2^ (l/4){/x(l)2^ (l/5){/x(l)2^

dim

+ m(2)2'} = 1/2(4 - 2) = 1 + m(3)2'} = 1/3(8 - 2) = 2 + m(2)22 + jtx(4)2'} = (1/4)(16 - 4) = 3 + A^(5)2'} = (1/5X32 - 2) = 6

= (1 /6){m(1)6!/3!3!+ m(3)2!/1!1!} = 3

Note There is a procedure due to P. Hall [see Bo2] that produces an explicit basis for each degree subspace of a free Lie algebra. 2.6 THE CATEGORY & Throughout this section g will denote a Lie algebra admitting a triangular decomposition (1^, 0+? which we fix once and for all. Our aim is to introduce a category <^(g, [BGGl] whose objects might loosely be described as consisting of those g-modules whose characters lie in Z[f)* iQ+]. In fact the last part of this section is devoted to putting a ring structure on the isomorphism classes of modules in ^ to form the representation ring of which is then related to Z[f)* J, ¡2+] via the character map. Unfortunately, g itself, under the adjoint representation, does not appear in & in general. With this exception ^ and its a opposite category contain all the representations of g that we will need in this book, including in particular highest weight representations. The first half of this section introduces the (local) concept of composition series of modules in the

142

Lie Algebras Admitting Triangular Decompositions

notion of the multiplicity of an irreducible submodule of a module M in category and more on primitive vectors which were briefly used in Sec­ tion 2.4. The category is deflned as follows: (i) Objects of 3^). Those g-modules M having the following prop­ erties (ia) M admits a weight space decomposition M=

©

with respect to the action of f); (ib) the set P(M) of weights of M, that is,

is included inside a finite number of fans of ij* iQ+; (ic) for all e P(M), dim^^iM^) < oo. (ii) Morphisms of S^). All g-module homomorphisms between the modules in Obj(<^). We often write M e «^) to mean that M is an object of The category of g-modules is the category <^(g, We are already familiar with highest weight modules. Provided that its weight spaces are finite dimensional, such a module lies in For instance, if the triangular decomposition is regular, then for all A e 1^* the Verma module M(A) and the irreducible module L(A) are in Remark 1 (i) It is useful to note that if 0 + is countable, then every module in & is of countable dimension. (ii) We recall (from Section 2.5) that a partial ordering > was defined in by declaring that for all A, /u, e 1^*, A > / t < = > ^ t e A i ¡2 +* We also introduce for each A e 1^* the depth function

defined for all /8 e ¡2 _^u{0} by ¿ , ( A - / 3 ) := ht(/3).

2.6

The Category 0

143

Lemma 1 Let F = U i 0+) union o f a finite number o f fans in 1^*. Let S be a nonempty subset o f F, Then (i) if p ^ S, the set {A e 5|A > p) is finite; (ii) 5 admits maximal elements. Proof (i) Let p ^ S and ; e {1, . . . , n}. By definition of <2+-fans, it is immediate that if p ^ Ay i Q+, then there is no element ^ e Ay i with ^ > P, and if /1 e Ay I Q+, then there is only a finite number of elements i e A; I (2 +, with ^ > p . It is clear therefore that the set in question is finite. Part (ii) follows from Part (i). □ Let M e ^ (g , and let M = sition. We recall the formal character of M ch(M) =

be its weight space decompo­

E dim M^ e( p) ^ Z [ V iQ+]^ /Ltei)*

If M = M', then the isomorphism must map weight spaces isomorphically onto weight spaces so ch(M) = ch(M'). Proposition 2 Let M and N e ^ (g , ^ ) , and let P{M) and P{N) be their corresponding sets of weights. (i) I f both A and B belong to

is an exact sequence o f ^-modules, then Moreover, for a// e

0^

->

^ 0

is an exact sequence o f K-spaces and ch(M) = c h ( ^ ) + ch(B). (ii) Every submodule and every homomorphic image o f M belongs to ^ (g , ^ ) . (iii) The direct sum M 0 AT belongs to ^ (g , Moreover P{M ^ N ) = P(M) U P( N) and ch(M ® N) = ch(M) + ch(A^). (iv) The tensor product M ^ N belongs to ^ (g , Moreover P{M 0 N ) c P{M) + P{N) and ch(M ® N ) = ch(M)ch(iV).

144

Lie Algebras Admitting Triangular Decompositions

Proof. Let M = Identify A with a submodule of M to conclude (Proposition 2.1.1) that A admits a weight space decomposition A = where

e

A'",

= M'^ n A. It is clear then that A e B = M/A=

© M V 0 A -'s usi,* Ae6*

3^). Also 0

M>^/A>^,

and hence B=

® B>^,

where B'^ = M'^/A'^ for all fi e i)*. This shows that B of (i) follows easily. (ii) The proof is obvious from Part (i). (iii) For every ju. e 1^*,

,

The rest

(M ® AT)'‘ = M'" ® JV**, and thus M®N=

0

Evidently P( M ® N ) = P(M) u P(N), and the weight spaces are finite dimensional. Thus M ® AT e (^, and Part (iii) follows. (iv) Let {y,}, g j and ek basis of M and N, respectively, consisting of weight vectors, say, e M^ ‘ and e M*'* for all i e I and it e K. The set {d, ® Wfcl/ei, *eK is a basis of the space M ® AT and these are weight vectors: Indeed for h /i • ( y,- ® H'fc) = h ■Vi ® =

iJLi(h)Vi ®

+ Vi (8) h • +

Vi ® A * ( h ) w ^

= {Pi + ^k)(h)Vi ® w ^ . . This shows that M ® N admits a basis consisting of weight vectors and hence a weight space decomposition. Moreover, if y e 1^*, then (M ® AT)'*' has as a basis the set {y, ® consisting of those i and k such that /r, + A;t = y. In other words. (1)

( M <8>N y =

© (M>^(8i N^). fl+\=y

2.6

The Categoiy 0

145

Thus F(M <S) c P(M ) + P(A/^), and this is contained inside the union of finitely many fans. Finally, from (1) we get ch(M (S>N) = ch(M)ch(N). □ Let M be a g-module. An element v ^ M is called a primitive vector if 1 y is a weight vector relative to 1^; 2. there exists a submodule N of M such that u ^ N and

- i; c N.

If this situation arises and A the weight of v then we also say that A above is a primitive weight of M {relative to N). Every highest weight vector of a g-module is a primitive vector. Lemma 3 L e t M ^ & {% ,Sr\ (i) For a weight vector v o f M o f weight /x, the following are equivalent: (ia) V is primitive, Gb) V ^ U(g) • V, (ic) there exists a submodule N o f M such that N c U(g) *v and U(g) • v/N = L{p), (ii) For a primitive weight vector o f weight ¡jl to exist in M, it is necessary and sufficient that L{ jjl) be a subquotient o f M [i.e., that there exist submodules X and Y o f M such that X :d Y and X / Y = Lip)]. Proof (ia) => (ib) Let v g be primitive. Then there exists a submodule M' of M such that v ^ M' but g^* i; c Af'. We have v ^ M ' = Vi{%) - M ' D U ( g ) -g^-i; proving Part (ib). (ib) => (ic) Suppose that v ^ U(g) • g+* == M'. Then M' is a submodule of M and g+* v c M'. This shows that v is primitive. Let M *= U( q) *v/M' , M is a highest weight module generated by the coset v + M'. According to PartjGO of Proposition 2.3.4, M Ims a unique maximal submodule N and M/ N = L{p). The preimage N of N in U(g) *v gives us U(g) • v / N = L{p), proving Part (ic). (ic) => (ia) The submodule N of Part (ic) serves to show that v is primitive. (ii) After Part (i) it remains only to show that if M has a subquotient X / Y = L{p), then M has a primitive vector of weight p. But any preimage ¿;+ in of a generator D+e L { p Y of L{p) under the map X X/Y Lip) is primitive: T, g+* + c y. LI

Líe Algebras Admitting Triangular Decompositions

Proposition 4 Let M and N be objects of ^ ( 9, ^ ) , and let f : M -^ N be an epimorphism. Suppose that v ^ M is a highest weight vector o f weight A and that w ^ N is a primitive vector of weight p. Then (i) there exists a primitive vector x ^ M o f weight p such that f ( x ) = w; 00 if f(v) ^ 0, then f (v) is a highest weight vector o f N o f weight A. Proof 0) Let N ' be a submodule of N with w ^ N ' but w c N', Since / maps onto N^, we can find x ^ with f ( x ) = w. Set M' = f ~ \ N ' l Then X ^ M' and g+- jc c A/'. (ii) The proof is straightforward. □ Remark 2 It is generally not true that the homomorphic image of a primitive vector is a primitive vector. Proposition 5 Let M e ^ ( 9, Then

be a nonzero module, and let P(M) be its set o f weights.

(i) every maximal element o f P(M) (with respect to > ) is a highest weight', in particular highest weight vectors, and hence primitive vec­ tors, for M exist. (ii) The set of primitive vectors o f M generate M as a q_-module (and a fortiori as a q-module). Proof (i) Since M ¥= (0), we have P(M) # 0 , and hence P(M) admits maximal elements by Lemma 1. Let p be such an element, and let v e M^, V ¥= 0. Then 9+* V c ^ which shows that i; is a highest weight vector. (ii) To see the primitive vectors generate M as a 9 .-module, we consider the space

L

v= V

U(g_)i;cM .

primitive

Clearly V is an 1^-module and hence admits a weight space decomposition (being included in M). Suppose M ^ V , and choose p e P(M) maximal such that (tV .lix^ \ V, then x is not primitive, and hence X e U(9) • 9+- ;c by Lemma 3. Now U(9+) • 9^- X c c K by the maximality of p. But then x e U ( 9 _ ) - U ( ^ ) *1 1 (9 ^ - 9 ^;c

c

U ( 9 . ) - U( f t ) - K c F

contrary to the choice of x. This shows that V = M.

□

2.6

The Category 0

147

Our next two results characterize irreducible and completely reducible objects of Proposition 6 Let M ^

^ ) , (M ^ (0)). Then the following conditions are equivalent

(i) M is irreducible, (ii) M = L(A) for some A e 1^*. Proof Assume that M is irreducible. Let v e be a highest weight vector (Proposition 5). Then M = U(g)i;, so M is a highest weight module and hence a homomorphic image of M(A). Since M is irreducible, the kernel of this map is a maximal submodule of M(A) and it equals MA) (Proposition 2.3.4). Thus M = L(A) as desired. □ Proposition 7 Let M Ei ^ (g , ^ ) , (M =5^ (0)). The following conditions are equivalent, (i) M is completely reducible [in ^ (g , (ii) M is generated by its highest weight vectors, and every highest weight submodule of M is irreducible. Proof, (i) =►(ii) Suppose that M is the direct sum of a family of irreducible modules of <^(g, *^). By our last result M = © L( A ,) ¿el for some family {Aj^gj c Evidently M is generated by its highest weight vectors. Let be a highest weight vector of weight A, and write t; = E i;, where v¡ e L(A,) for all i e I. Then g^.- ü = 0 =» g^.- v¡ = 0 for all i => v¡ L(.X¡y‘ for each i. Now for all /t e we have E a(/ i ) ü, = \ { h ) v = h • V = /e l

/e l

148

Lie Algebras Admitting Triangular Decompositions

which shows that f = E A, = A

It is easy to see then that the submodule U(g)i; generated by v is isomorphic to L(A) and hence is irreducible. (ii) => (i) By assumption M is a sum of irreducible submodules. Then M is the direct sum of irreducible modules by Proposition 1.4.2. □ Corollary 1 Let M ^ be a module with the following property: I f A and p are primitive weights o f M, then \ > p ^ k = p. Then M is completely reducible. Proof Let be a highest weight vector of M, say, of weight A. If U(g)i; is not irreducible, then any primitive weight /a of a proper submodule of U(g)i; (such p exists by Proposition 5) will satisfy k > p, which is not possible. Thus U(g)¿; is irreducible. Let M' := with the sum being taken over all highest weight vectors of M. If M / M ' (0), we can consider a primitive vectors of M, say, x e M^, such that x ^ M' (Proposition 4 applied to the canonical map M M/M' ), Now g+- a : # (0) (for otherwise x e M'), and hence the space U(g+)g+* x contains some highest weight vector, say, of weight A'. Then X > p, contrary to our hypothesis. □ We now introduce composition series. It admits a composition following local version

the concept of local, highest weight, and Verma can be shown that not every module in ^ (g , series (see Exercise 2.13). We have, however, the of composition series.

Proposition 8 Let M e ^ (g , and A e if*. Then there exists a finite sequence M q, , ,.,Mf^ of modules in ^ (g , S^) and a subset I c {1,..., A:} with the following properties, (LCSl) (LCS2)

M = M¡^z> D • • • D Mo = (0). I f i G I, then there exists A¿ e 1^* such that A¿ > A and M,/ M, _i= L( A, )

(LCS3)

I f i ^ I, then (M ,/M ,_i)^ = (0) for all p > k.

2.6

The Category 0

149

In words, all weight spaces o f M for weights equal or above A are involved in irreducible quotients o f the series. Proof Define á(M, A) = dim(M^). (This sum is finite by Lemma 1). We reason by induction on d{M, A). If d{M, A) = 0, we simply set = M, Mq = (0), and 1 = 0 . Assume that d(M, A) > 0 and that the result holds for all modules N e <^(g, 5") for which d(N, A) < d(M, A). Let fi e P(M) be a maximal element satisfying p > \ ( jjl exists by Lemma 1), and let v e y ¥= 0. Then i; is a highest weight vector (part (i) of Proposition 5). The highest weight module V = U(g)i; contains a (unique) maximal proper submodule U (Proposition 2.3.4) and V / U = Lip). We observe that d{M/V,X)
since dim(F^) = 1,

and that d(U, A) < d{M, A),

since dim([/^) = 0.

By induction hypothesis both M / V and U have sequences of submodules with the desired properties. We now consider the inverse image in M of the sequence of M / V under the canonical map M M / V , say, M = M^z> ♦ D Mq = F, and the sequence of U, say, U = U^z:> • • • ^ Uq = (0). Since Mo/Up = Lifi) (see above), it then follows that the sequence M = M _D

••• z^ M o ^ U ^ z>

••• D i / o = ( 0 )

is as desired. A sequence of the type appearing in Proposition 8 is called a local composition series of M at A. The factors for some /x > A are called proper. The remaining ones are called extraneous. Remark 3 We make the simple observation that if we insert a submodule N between Mj and Mj_i, in the local composition series M = Mf^^

• •• d Mo ={0}

relative to A, then the resulting series M

= M f,:D

••• M y

D

••• d M o = {0 }

is also a local composition series at A and has the same proper factors, including multiplicities, as the original series. Indeed there are only two

Líe Algebras Admitting Triangular Decompositions

150

cases: 1.

is proper, in which case N = Mj or N = Mj_^, and one of N/Mj_^ and Mj / N is proper and one extraneous [actually (0)]. 2. is extraneous, in which case both Mj / N and N/Mj_-^ are extraneous. Two local composition series for M, say, M = d • • • =) Mq = (0) and M = Ni ^ • • • z>N q = (0) are equivalent if k = I, and there is a permutation i ^ i' of {1, . . . , k) such that for all i, Proposition 9 Let M e (^(g, and fei A, /x e 1^*. Any two local composition seríes {M¿} and {M/} o f M at A and /x, respectively, have common refinements that are equivalent to a series {M") that is local at both A and jjl. Furthermore for ^ > A {resp. ^ > /x) the number proper factors of type L ( 0 in {M¿} and [M¡'} (resp. {M/} and {M"}) are the same. Proof Using the Schreier refinement theorem [Ja2] find a refinement {M-} equivalent to refinements of both {M¿} and {M/} (as modules). According to the last remark, this series will be local at both A and p and the number of proper factors will not change. □ Let M e ^ (g , 5^) and p e 1^*. Let A g f)* be arbitrary with p > X, and let {M¡\ be any local composition series of M at A. We claim that the number of proper factors of type L(/x) in this series is independent of the choice of A and the particular series {Mj that we have chosen. Indeed if {M/} is a local composition series at some A' with p > A', then {MJ and {M/} have a common refinement (M/'} as in Proposition 9. Then the number of proper factors of type L(p) is the same in (MJ and (M"} and also in (M/} and (M/'}, hence in (M j and (M/}. □ Thus we can make the following definition. For M e (^(g, and for /x e we define the multiplicity of L(p) in M, denoted by [M : L(p)] to be the number of proper factors of type L(/x) in any local composition series of M at /x. We will see another interpretation of these numbers in Proposition 12. Let M e ^ (g, filtration

A highest weight series (HWS) for M is an increasing

(0) = Mo c Ml c M2

The Category 0

2.6

151

of submodules of M so that (HWSl).

U M =M;

i=0

(HWS2). If M,_i M¡, then with highest weight /i.,.

is a highest weight module, say,

Let {A/,} be a HWS for M. If 9 e 1^* is any weight, then dim M ’’ < 00, and the sequence {dim is increasing to dim M ’’. Thus (HWS3). For each e f)* there is a A: e Z+ such that for all i > k, M f = M ‘< ‘. If the highest weight series satisfies •••

•••>

we say that it is a finite highest weight series of length equal to n. Proposition 10 [G L, DGK]

Suppose that the root lattice Q is finitely generated (.equivalently the index set J of the triangular decomposition S ' is finite). Let M e Then M has a HWS, (M,}7=o- Furthermore the series can be constructed so that (HWS4).

> fij => i < j, where

Proof Since M e

there exist P { M) c Ai i

is the highest weight of e

such that

• • • U A^ i e +-

If A^ i (2+n Ay i (2 + =5^ 0 then there exists A with A i A¿ i <2+^ Ay i ¡2+- Thus we can assume that the union of the fans is disjoint. Then for any e P ( M \ ¡JL = - p for some unique / e {1,..., A:} and /3 e (2 +’^{0}Let d()L¿) := = ht(p) (Remark 2.6.1(ii)). Clearly from the assumption on QAP' ^ P(M)\d(/jL) = N} is finite for each N ^ N. Let fi^ G P(M) be chosen with d(fji^) minimal. Then is a maximal weight of M relative to > . Thus, if ¿;i e \ {0}, is a highest weight vector, and generates a highest weight module M^. Now repeat the argument with M/M^ to get a submodule M2 of M, M2 ^ Mj for which M2/ M 1 is a highest weight module. In this way we obtain a filtration (0) := M q (zMi d M2 c • • • . Since the value of d(fi¿) is increasing and can only remain at one value for finitely many i, we see that for any weight

152

Lie Algebras Admitting Triangular Decompositions

= • • • . Thus UM, = M. The con­ □

For each M e O bji^iq, y ) ) we have its associated character ch(M) e i Q A s we have pointed out before, M = M' => ch M = ch M'. Let S = S( q, be the set whose elements are the isomorphism classes of g-modules in We denote the equivalence class of M by M. Evidently ch determines a mapping, also denoted ch ch: satisfying ch(M), where M ^ M. We let R = R(q, denote the free abelian group on the set 5. A typical element of R has the form

(finite sum). R can be made into a commutative ring by defining multiplica­ tion by bilinear extension of M-N= This is evidently well defined. The character map extends to a homomor­ phism ch : R ^ i Q+], We denote its kernel by K, and set R = R(ij, ^ )8 } :R /K , We call R the representation ring of g relative to Factoring ch through K, we have the character map ch:R(q,^)

iG+].

For M e O bj(^) we denote its image in R( q, y ) by 0^ A B ^ C 0 is an exact sequence of modules then ch(B —^ —C) = 0 by Proposi­ tion 2 and hence [B] = [A] + [C] in R. In particular for modules M and N in [M © AT] = [M] + [TV]. We also have [A/® TV] = [TV/][TV] by the definition of the multiplication in R.

2.6

The Category t

153

Remark 4 It is useful to note that any element Z oi R can be written in the form [Z+] - [Z_], where Z+ and Z _ e For if Z = then we simply set Z^_==

0

(Af 0 • • * © Af)

(^[Af] summands),

Z_:=

0

(M ®

(-Crjv^j summands).

®M)

Proposition 11 Let {c;^}^ e 6* t>e a family o f elements o f Z. (0 The family ch ¿(A));^eE)* o f Z[if* | <2+1 is summable if and only if there exists a finite union F o f fans in Z[k*iQ^] such that [X e 0) c F . (ii) For every element f if Z['i)* IQ+] there exists a unique family o f integers {C;^}Aefi* such that the family {c^ o f zli)* J, 0 +] is summable, and f = ch(L(A)). Proof (i) For each ju. e h* let = c^ch

L { f i)

=

c J

Y .

-

a)].

If the family (/^)^ ^ f,* is summable, then by definition (see (SI) of Section 2.5) {supp/^lju e h*} c F , where F is a finite union of fans. Now since supp 0, and hence {p, e f)*|c^ Conversely, assume that {/x e holds, since

= 1, we see that ¡x e

0} c F . # 0} c F. Then clearly condition (SI)

{supp 4 I m e

c

F.

Fix A e ]^*. Then for /x e 1^* we have ffjL ^ Z[]^* i 2

+ ] ( a)

^

^ 0 and

=> jjL ^ F

and

for some ¡jl — a > \ jJL > \ fJL > \

154

Líe Algebras Admitting Triangular Decompositions

and this set is finite by Lemma 1. This establishes (S2) of Section 2.5 and hence the summability of the family in question by Proposition 2.5.1. (ii) Suppose that / = La^eifi) e Z[l^* i Q+], We wish to write / in the form Ec^ ch(L(^t)) as in Part (i). We may assume, as in Proposition 10, that all the weights in supp(/) lie inside a finite disjoint union of fans. Then there is no loss in assuming that all the weights appearing in supp(/) lie in one fan be the depth of fi (see A i G + . For each ¡x ^ A i Q + let difx) == Remark 1). We compute the coefficients by induction on the depth. Suppose that / has been written in the form /=

L c^ch(L(M)) + / . dilxXd

for some d > 0, where contain nontrivial terms only for weights v with d{v) > d (we begin at / = / q). Let denote the set of weights of depth d in A i (2+, and for each i e I let denote the coefficient of c()ll^) in /¿. By part (i) the set {c^. ch L(/x^)}^^/ is summable. Setting /¿+1 ■■=fa - E,^,c^.ch(L(/u,,)), we have / = E d if j¿ )< d- ¥ 1 c c h L ( ^ ) +/ d+i. where contains nontrivial terms only for weights v with d{v) > d + 1. The entire family {c^ ch L(^t)} is easily seen to be summable, proving the existence of part (ii). Finally, if the uniqueness were to fail, then E^ g f ^a L(A) = 0 for some union F of finitely many fans. Let 5 — {A e Flc;^ = 0}. If S 0, then it has a maximal element Aq and

CA„chL(Ao)-----E C;^chL(A) A # Aq

gives a contradiction. Indeed the coefficient of e(Ao) is on the right.

on the left and 0 □

Remark 5 In Proposition 11 we have used the characters of the irreducible modules in as a “Z-basis” for Z[l^* i 0+]. However, the only facts that we have used about ch(L(A)) are 1. ch L(A) involves e(/¿) only if /a e A J, 2. the coefficient of e(A) is 1. Thus Proposition 11 holds, for instance, if we replace each ch(L(A)) by the character ch(M(A)) of the corresponding Verma module.

2.6

155

The Category &

Proposition 12 Let M e

5^). Then chM=

12 [ M : L(iJi)]ch L(ix), /xei)*

and this is the unique way o f writing ch M as an integral combination o f the characters o f irreducible modules from Proof The uniqueness follows from part (ii) of Proposition 11. We prove that for each A e 1^* chM =

L [ M :L ( /i ) ] ch L ( /x ) fji>\

mod Z [ ^ i !2 +](a)

(see Proposition 2.5.1). For if {M¡)¡^o „ is a local composition series of M at A, then for each /a > A, [M : L(/i)] is the number of proper factors of type L(/i) in {M,}. We have c h M = ¿ ch ¿= 0

^

[ M :L ( M ) ] c h L ( M ) m o d K [ n e + ] ( A ) -

□

/11 ^ A

Let F e A Verma composition series (VCS) (of length r) for M is a descending sequence of modules of V F = M o = ) M i3 •••

DM, = (0)

so that M¿_i/M^ is a Verma module, / = 1,2,..., r. Such an object need not exist for a given module V. However, if it does, then c h F = E chM¿_i/M¿, and hence by Remark 5 the Verma factors M^_i/M^ are, up to permutation, uniquely determined by V and are independent of the particular VCS chosen. The number of times that M(/x) occurs in a VCS of M is denoted by [M:M(p)l Proposition 13 Let p

be fixed. There exists a unique group homomorphism

156

Lie Algebras Admitting Triangular Decompositions

satisfying

Proof Let

: R(g,

-» Z be the unique group homomorphism satisfying

We show that this maps factors through the defining kernel K
x = E c ,M „ 1= 1

where N

E c,ch(M ,.) = 0. ¿=1 Thus in Z[]^* i 2+] we have by Proposition 12, N

0=

E [ M, : L ( n ) ] c h L ( f i )

¿=1

= E ÍE^,[M,:L(M)]]chL(M), ^ e f ) * \/ = l

and hence for all

/

e 1^* we obtain N

£ c, [ M , : L ( m)] = 0 ¿=1 by Proposition 11. Thus N

¿= 1 =

L

c

, [ M , : L (

m

)]

= 0 .

/=1 We have therefore an induced group homomorphism

as desired.

2.6

The Category ^

157

Corollary 1 (i)

If 0 ^ I,

^ M ^ M /N -» 0 is an exact sequence o f modules from then for each ¡jl e 1^*, [ M : L ( m )] = [AT:L( m )] +

(ii) For M j, . . . ,

[ M / N : L { t i ) \ .

e i^(g, [Ml Ф

0 M , : L ( m)] = Е [ М : Ц м ) ] . k =\

Proposition 14 ch: i?(g,

-» Z[if* 1 0+] ^ an isomorphism o f rings.

Proof ch is injective by the definition of i?. Let / e I[i>* j, g+l. By Proposi­ tion 11 we can write / = Ec^ ch(L(A)). Let M+:= 0 c^L(A), «A>0 M _ ~

Both of these lie in

© c*<0

— c ^ L ( X ) .

and ch([M j-[M _ ])= /.

□

Let M e ^ (g , ^ ) . Then M has a weight space decomposition M=

© M^.

Recall that we have defined the restricted dual space M^es of M by Mres == { / ^ M*K/, M^> = (0)

for all but a finite number of

Proposition 15 Let

:= M^es

a K-space. The map g X M^ -> M^

e 1^*}.

Lie Algebras Admitting Triangular Decompositions

158

denoted by {x,f) ^ x - f , and given by {x ' f , m ) = gives

for all jc e g, / g

m ^ M,

a Q-module structure.

Proof. The map in question is clearly bilinear. Moreover, if jc, y g g and m ^ M ,^ e have i [ x , y ] ■f , m } = < /, [o-(y),
■a { x ) ■m) - ( f , a { x ) ■

= { x - y • f , m ) - ( y ■X ■f , m ) , so [^>y] - f = x ■y ■f - y ■X ■f . That AC• / e

is clear.

The g-module

□

is called the restricted dual module of M.

Proposition 16 Let M G ^ (g , (i)

and let

be its restricted dual. Then

admits a weight space decomposition relative to

=e with

(ii) e ^ (g , ^ y , (iii) ch = ch M; (iv) (v) M is an exact contravariant functor o f i^(g, (vi) L(A)^ = L(A) for all A e 1^*.

into itself;

Proof (i) Given / € we can define the map =/ ° ^ is the natural projection M that annihilates Af** for all ju.

where A. By the

2.6

159

The Category <

definition of M^, f ■= where 5 is some finite set (depending on /) . Let us see that each is a weight vector of weight A. Indeed, given m ^ M and writing m =

£

m ^,

e M>^,

we have, for all /i g f), {h ■f^ ,m ) = {f¡^,a{h) ■m) = {f ^ , h • m> =

< /

a>

= k{h){f)^,m) = k{h){f)^,m).

This shows that f¡^ g Now part (i) follows. Evidently - (M^)* as K-spaces, and hence dim,^ M'^ = dimi^ M^, since all the spaces in question are finite dimensional. This establishes parts (ii) and (iii). (iv) Write = M for convenience. If m g M, let m g M be de­ fined by ñi{f) = { f , m)

ioxdWf^Mg.

The resulting linear map ~ :M ^M is clearly injective. Moreover c

for all A G f)*

so that ~ is surjective because of the finite dimensionality of the weight spaces and part (iii). Finally, ~ is a g-module map. Indeed, if a: g g, w G M, and / G Mg, we have X ■m ( f ) = { f , x ■m) =
, F : AT*

Given a g-module homomorphism M*

160

Lie Algebras Admitting Triangular Decompositions

by

for all / e N* and m ^ M (thus is the transpose of F). It is clear that if / vanishes on N^, then F^f vanishes on M^. We therefore get an induced map F 'N

M

We verify that this is a g-module homomorphism F :N

M .

In fact for all x ^ q, {F„{x ■f ) , m) = ( x ■f , F{ m) ) = ( f , cr {x) ■F ( m ) ) = ( f , F { a { x ) ■m)> = ( F^ f , a ( x ) ■m) = {x • F„f , m). If G : AT ^ P is another morphism in

then

( G oF) ^ = F„ oG^, showing that

is a contravariant functor of ^ (g , into itself. The exactness is left as an exercise. (vi) Let ¿;+ be a highest weight vector of L(A), and let v% be the linear functional = 1, = 0 if ^ X. Then by part (i), u% e (L(A)^)^, and for all x e g" , jc • y* e (L(A)^)'^'^“ = (0). This shows that i;* generates a highest weight module with highest weight A. Since ch L(A)^ = ch L(A), we conclude from Proposition 12 that L(A)^ —L(A). □ Caution

is not the same thing as 2.7

defined in Proposition 2.3.5.

THE RADICAL

In this section we introduce four different ideals of a Lie algebra g with a triangular decomposition (i), g+, Q+, a). Each of them has a right to be

2.7

The Radical

161

called a radical of g. However, in general they are not all equal. They are 1. t y ( g ) == the set of all elements of g that annihilate every irreducible module in 2. m ^(g) == the largest ideal of g intersecting trivially; 3. n(g) == the sum of all the residually nilpotent ideals of g; 4. rad(g) := n the intersection being over all triangular decom­ positions of g. We discuss these in turn. We find that always r ^ = m ^ , and in the case of primary interest to us, that is when E , e j 9~“' + h + E , e j 9“' generates g as a Lie algebra, r^ ( g ) = m ^ (g ) = rad(g) and Z(g) ffi rad(g) = n(gX We make use of these results in Chapter 4. Throughout this section g will denote a Lie algebra over IK admitting a triangular decomposition, and we will let ■^= (i),Q+,Q+,o’) denote one such decomposition. To describe this situation, we will henceforth say that (g, is a triangular pair. Proposition 1 Let (g, y ) be as above, and write g = g _ ® ® g+. Let Z(g) and D(g) denote the centre and derived algebra o f g respectively. Then (0 Every ideal o f q is graded by Q; (ii) Z(g) = {A e l^|a(/i) = 0 for all a e A}; (iii) D(g) = g_®([g, g] n fi) ® g+. Proof (i) Let a c g be an ideal. Then [1^, a] c a and therefore a is a submodule of g under the adjoint action of 1^. Now part (i) follows from Proposition 2.1.1. (ii) By part (i) and element belongs to the centre of g if and only if all of its homogeneous components do. If x e g“, a e then a(,h) # 0 for some h Thus [h, x] = a(h)x # 0, and hence x i Z(g). Similarly g_n Z(g) = (0). It follows that Z(g) c fi, and part (ii) is then clear. (iii) As in part (ii) we see that g_® g+c D(g) and hence that part (iii) holds.

□ Let (m,), s i be any family of ideals of g such that m, n = (0). Then the ideal Em, of g, being graded, satisfies (Em,) n = (0). We conclude that there exists a unique largest ideal of g intersecting t) trivially. Thus m ^(g) exists.

Lie Algebras Admitting Triangular Decompositions

162

Proposition 2 Let (g,

be a triangular pair as above. Then r

= m ^(g).

Proof. Let L be an irreducible module in ^ (g , ^ ) . Then L s L(A) for some A e 1^* (Proposition 2.6.2). The annihilator of L is clearly an ideal of g, and hence also is an ideal of g. Suppose e r ^ n ]&. Then for any A for which A(/i) # 0, we have h • L(A) # (0) contradicting the fact that h e r ^ . Thus r ^ c xn^. Conversely, let L(A) be an irreducible module of (s, "^X and let i )+ g be a generator forit. Let V ~ v^. Since m ^ n = (0) and since is graded, we have V=

ti+c g_- v^+ g+- v^= g_- v+
Now g+- V = 9+-

Ü+C ( m ^ + m ^g + ) • v^= V.

Thus U(g)-FcU(g_)U(^)-Fc

E L(A)^
which shows that U(g) • F is a proper submodule of L(A) and hence (0). This proves that m^* u^= (0). Let N '= {u Ei L(A)|m^* v = (0)}. Then N is a submodule of L(A) since (9 •

^

g] • N + 9 •

N = (0),

Since v +E N, it follows that N = L(A) and hence that m ^ annihilates L(A). Since Awas arbitrary, this shows that m ^ c as desired. □ The ideal r^ ( g ) (or simply if g is understood from the context) described in the last proposition is called the (Wedderburn-Jacobson) radical of g with respect to We will see in Proposition 5 that for a large class of Lie algebras admitting triangular decomposition, the radical does not depend upon the choice of triangular decomposition Define r 5r : = r ^ n g + Then

and

r^:= r^ng_.

2.7

The Radical

Since cr(r^r) is an ideal of g intersecting

163

trivially it is easy to conclude that

= r^ and that

Example 1 (Extended Heisenberg algebra a) The notation is that of Example 2.6, where we extended a Heisenberg algebra by a derivation in order to obtain a triangular decomposition. Recall that S = IK[^y]ye j can be made into a highest weight module of type (1, A) for any A e ]^* where if = Kc ^ Kd. Moreover these modules are irreducible whenever A(c) 0. (Example 2.3.2). Now Kxj^) ■■ ^ 0 so that r^ (fi)_ = (0). It follows that r^(a)^.= cr(r^(a)_) = (Ó) and hence that a is radical free with respect to the triangular decomposition in question. Example 2 Let g = Ka © Kh © Kb be a three-dimensional Lie algebra with bracket defined by [A, a] = —2a, [h, b] = 2b, and [a, b] = 0. It is easy to see that g admits a triangular decomposition with root lattice Q = Za where a e (K/z)* is defined by a(h) = 2. The two-dimensional space r = Ka © Kb is an ideal of g intersecting Kh trivially, and hence r is the radical of g relative to the triangular decomposition in question. Proposition 3 Let g and ^ be as above, and suppose that r ^ ( g ) = 0. Then for g_ we have

g+ and

[ z+ ,s _ ] = (0) <»z+= 0, [z_, g+] = (0) « z_= 0. Proof Suppose that [z+, g_] = (0), g+. Without loss of generality we may assume that z+ is homogeneous. By assumption U(g_) z+= Kz+, where the action is given by restriction of the adjoint representation. Then U(g) • z^ = U(g J U ( ^ ) U ( g _ ) • z , c g^, which gives an ideal of g (i.e., the one generated by z+) intersecting trivially. Thus z+= 0. The second assertion follows by applying a to the first. □ Our next result deals with the problem of how Lie algebras with triangular decomposition behave under extension of the base field. (For Lie algebras obtained by extension of the base field see Remark 1.1.1).

164

Lie Algebras Admitting Triangular Decompositions

Proposition 4 Let q be a Lie algebra over IK admitting a triangular decomposition (^5 9+ rel="nofollow"> ¡2+? ^)* Let IK' be a field containing IK. Let g' = IK' the Lie algebra over IK' obtained from g by extension o f the base field from IK to IK', and let = (IK' IK' ®(k9+, Q+> I ® cr) be the corresponding triangular decomposition of g' {see Remark 3, Section 2.1). Then

Proof. For convenience let us set i)' = IK' and g'^. = IK' <S>|,^g+. It is immediate that IK' c To establish the reverse inclusion, it suffices to show that r^ ,(g ')‘^c IK' (for we can then apply 1 ® a). Let be a basis of IK' over IK. Let a e g+j and consider a nonzero element z e r^ /(g ')“. Write z = T, k^<8>x^, A eA

where x “ e g“. Suppose that e g^',. . . , y, e g^' is a sequence of ele­ ments where /Sj • • • -1-/3, = —a. Then ad(l ® y , ) . . . ad(l ® y,Xz) e r^Xg') n £)' = (0), and 0=

® adyi...ady,(ji:J'). AeA

We conclude that ad ... ad = 0. It follows that for each A e A , the ideal generated by in g intersects trivially and hence that each x “ er ^ ( g ) '^ . Thus z e IK'<8>j^r^(g)‘^ as desired. □ Remark 1 Let g, y , r ^ , and

be as above. Define 9 == 9 / r ^ ,

and let : g ^ g be the canonical map. Then g = g_e fi e g+, where g+— 9+ A 5r and §_== g _ A 5^ (we are identifying with a subalge­ bra of both g and g). It is clear that a induces an antiinvolution of g since o-(r^) = r ^ , as we just saw. As long as 5+=?^= (0), we obtain an induced

2.7

165

The Radical

triangular decom position

of g with root spaces given by 9“ =

g“

for all a e g .

Finally, observe that if a is an ideal of g such that d n = (0), then the preimage a of d in g also intersects ij trivially (because the restriction of “ to if is injective). Thus a c r We conclude that = (0).

( 1)

We next consider the problem of the independence of r^r from We begin with some general facts about Lie algebras. If A is an ideal of an arbitrary Lie algebra I, we inductively define a decreasing series A =A^ A^ ^ • • • of ideals of I as follows: A^ :=A, A'^^^ := [ A, A^ ] . The ideal A above is said to be residually nilpotent if n ^ " = (o). n>l

We define n(I) to be the sum of all residually nilpotent ideals of I. Observe that n(I) itself need not be residually nilpotent. Returning now to our Lie algebra g we define the (Wedderburn-Jacobson) radical rad(g) of g by rad(g) = the intersection being taken over all triangular decompositions of g. We say that g is radical free if rad(g) = (0). Notice that g/rad(g) is always radical free (see Exercise 2.18). Proposition 5 Let (g, ^ ) be a triangular pair as above. Assume that (CTl) The subspace algebra.

5

^ 9 generates % as a Lie

166

Lie Algebras Admitting Triangular Decompositions

Then (0 r ^ , r Jr, and are residually nilpotent ideals o f n (ii) r ^ c for any other triangular decomposition o f g, in particular = ^ 5^(9) = rad(g); (iii) n(g) n Í) = Z(g); (iv) ^ induces a natural triangular decomposition on g/Z(g), and we have rad(g/Z(g)) = it (g/Z(g)), Z(g) e rad(g) = it(g); (v) n(g) is residually nilpotent] (vi) I f S ci)* spans f)*, then rad(g) = fl a s 5 ann L(A). Proof (i) Let U(g) • r j- be the ideal of g generated by r j- (here U(g) acts by the adjoint action). We must show that U (9) • r ; ^ c r ^ . Suppose not. Then, since U(g) • r j- c r ^ , there exists some element z e r ^ n g“, a e Q+, such that [ x i [ x2,...)[-'"A;>-2] ' ’ ' ] ] ^

{0}

for some elements Xj,. . . , g g. From condition (CTl) we can assume that e g * “ >. for some g J. If we choose k above minimal, then X j g g “ “ A, and

Xi

w ■■= [x2,

[ x 3 , . . . , [ x * , z ] ]

G

But w is an element of n = (0), contradicting the fact that [xj, w] # 0. It follows that r jr is an ideal of g. Moreover ( 2)

(rj.)

c

©

g“,

ht(a)>n

which shows that r J- is residually nilpotent. The analogous results for follow by applying a. To show that is residually nilpotent, first observe that [t;^, rjr] c = (0) and that hence r ^ c ( r » " + (rj-)".

2.7

The Radical

167

Combining this with (2), it follows that r^ c

e aeA

g“.

\ht{a)\ >n

Thus 0 = (0), and is residually nilpotent. (ii) Let 3^' = be any triangular decomposition of g. We claim that c that is, Pi 1^' = (0). For suppose not. Let h 0 belong to this intersection. If a'(h) = 0 for all roots a' of (g, then h e Z(g), and hence ^ripart (i) of Proposition 1). Thus h e r ^ n = (0), a contradiction. Therefore let a' be a root of (g, such that a'(h) 0, and let X ^ X ¥= 0. Since (ad h ^ x = a ' ( h y x , we conclude that x e for all contradicting the fact that is residusually nilpotent. Thus x ^ c z x ^ r , and the rest of (ii) follows from Proposi­ tion 2 and the definition of rad(g). (iii) Let a be a residually nilpotent ideal of g. If e a n and a(h) ¥= 0 for some a e A, then (ad/l)"g- = g ^ and hence g" c a ” for all n, which contradicts the fact that a is residually nilpotent. It follows that a n c Z(g). Let a^,. . . , be residually nilpotent ideals of g. (Uj + • • • + a^) n ]^ - (Ui + • • • +a^)^ = a? + • • • +a^ (3)

£ a , ni)cZ(g) /=1

by the above. Now let h e n(g) Pi 1^. Then by definition h belongs to a finite sum of residually nilpotent ideals of g, so e Z(g) by (3). This shows that n(g) Pi c Z(g). The reverse inclusion is obvious. (iv) We can assume that for each i e J, g“' ¥= ( 0 ) . Then from part (ii) of Proposition 1 , o : ,lz (g ) = 0 for each i e J, so we have a natural interpretation of j2 as a set of linear functions on íi/ Z ( q). Furthermore {«/l.ej remains linearly independent in this setting. Let tt : g 9/Z (g) denote the natural mapping. Then 'Tr(g) = '7r(g_) 0 7t(]^) 0 7r(g+) [note that 7r(g+) ¥= ( 0 ) ] ad­ mits a weight space decomposition relative to 7r(fi), and a factors through tt. Thus we have an induced triangular decomposition Evidently Z(7r(g)) = (0 ), and (CTl) holds for 7r(g).

168

Lie Algebras Admitting Triangular Decompositions

By Part (i), tyÍTTÍg)) c nCirCg)). To establish the reverse inclusion notice that Tr(ii) n n(Tr(g)) = (0) by Part (iii). Thus rad(g/Z(g)) = n(g/Z(g)). Let a be a residually nilpotent ideal of g. For notational convenience let denote the map v . W contend that 5 is a residually nilpotent ideal of g. (Note that this is not “obviously” the case.) Set 21 = Write 21 = ®2l“ relative to the induced triangular decomposition ^ of g/Z (g) (see Proposition 5). Suppose that ic g 21“ \ {0} for some a 9^ 0. For each n = 1,2,... there exists x„ G (a")“ such that x„ = x. Since ~ restricted to g“ is injective, x„= x„=- X ^ g“ for all m, n. Then x g = (0). Since any ideal of g lying in 5 must be (0), 21“ = (0) and 21 = 21 n ^ c Z(g/Z(g)) = (0). This finishes the proof of our claim. Thus, if a is any residually nilpotent ideal of g, then 3 c n(g/Z(g)) = rad(g/Z(g)). Combining this with the fact that the preimage of rad(g/Z(g)) in g is (clearly) rad(g) ® Z(g), it follows that a c rad(g) ® Z(g). Thus n(g) c rad(g) ® Z(g). The reverse inclusion fol­ lows from (ii) and (iii). This fibnishes the proof of Part (iv). Finally, n(g) is a residually nilpotent ideal. Indeed n(g)" = ( r ( g ) ® Z ( g ) ) " = r(g)", and r(g) is residually nilpotent. (vi) ann(L(A) := [x G q\x • L(A)) = (0)}, so by definition rad g c n as5 L(A) =' I. As for the reverse inclusion, note that / is an ideal of g; hence if I
THE SHAPOVALOV FORM

The problem of describing the submodules of a Verma module is in general very difficult. For instance, it is not at all obvious whether M(A) is irreducible and how irreducibility might depend on A. In this regard the construction of the Shapovalov form is extremely useful. This is a so-called contragredient bilinear form on U(g) with values in 5(1^) which is then transferred to Verma modules M(A). In principle it provides information about the weight spaces MA)'^"“ of the maximal submodule N( \ ) of M(A), although in practice a lot more work is required before any specific information is forthcoming (see Sections 6.6 and 6.7). We assume that we have a Lie algebra g admitting a triangular decompo­ sition g_ 0 0 g+, root lattice Q, and anti-involution cr. This is the first time the anti-involution is going to play a significant role.

2.8

The Shapovalov Form

Using the Poincare-Birkhoif-Witt theorem, we may write

(1)

U (g)

= U(^) e {g_U(9) + U(9)9+}.

Note that which interchanges 9_H(g) and U(g)9+ and pointwise fixes UO^). Both components on the right-hand side of (1) are fi-modules under the adjoint representation. We let q : U % ) ^ U(^) be the projection defined by (1) (so that q annihilates the summand in braces). It is convenient to use Proposi­ tion 1.8.2 to identify U(fi) with the symmetric algebra S(^) so that q: U(g) S(l^). We define the Shapovalov form f : U ( g ) x U (9 ) by F { x , y ) = q{
(i) (ii) (iii) (iv)

F is symmetric. For all u , x , y ^ 11(9), F(ux, y) = Fix, a(u)y). For a , P ^ Q and a ¥=p, U(g)“ ± U(9)^ relative to F. H I, 1) = 1.

Proof, ii) Let X, y e U(g). Writing 9+, and applying
X

M

V

of a g-module M into some space V is called contr^redient (relative to tr) if

170

Lie Algebras Admitting Triangular Decompositions

for all x , y Ei M and for all w e g B{ux, y) = B{x,(T{u)y). The Shapovalov form F is contragredient relative to the anti-involution of the triangular decomposition. Remark 1 More common in physical applications are the hermitian contra­ gredient forms. For these we need a complex Lie algebra g with a semilinear anti-involution cr. Then, if M is a g-module, a hermitian contragredient form on M is a mapping

which is linear in the second variable and antilinear in the first—that is, {ax + y\z) = a{x\z) + and {z\ax + y> = a{z\x) + —such that {u jc|y> = <jc|o-(m), y> for all x, y e M and for all w e g. If the hermitian contragredient from < • | • > is positive definite, then we call it a unitary contragredient form and say that the representation of g on M is unitary (relative to < • I * » . We will usually construct hermitian contragredient forms by beginning with a real form f of g with an anti-involution a and a f-module M with a contragredient form ( | •) on M. Then the complexification of g acts on the complexification of M. Furthermore a extends uniquely to a semilin­ ear anti-involution cr on gc» (*!*) extends uniquely to a hermitian contragredient form < • | • > on M^. We have {x

iy\u + iv) = {x\u)

i{x\v) — i(y\u) +

for x, y, m, y e M.

One checks immediately that {g • x\y) = for all g e f and for all jc, y e M^. Then for g, e I and x , y ^ M^, we have ( ( g + ih) *jr|y> = - i(h -xly) = {x\(Tg *y> - i{x\ah *y> = {x\{(Tg - iah) *y) = { x \a { g + ih) *y), which shows the contragredient. Also < • | • ) is unitary if and only if (*| •) is positive definite. Let us return to the discussion of the Shapovalov form. Any A e 1^* extends uniquely to a homomorphism /- > /( A ) of S(ij) into K. Since in particular the Shapovalov form takes values in we may define, for each Ae a [K-valued symmetric bilinear form f (A) on U(g) by F (A )(x ,y ) = ( F (x ,y ))(A ). Now if y e U(g)g+, then for all x e U(g), F(x, y) = q((r(x)y) = 0 (since

2.8

171

The Shapovalov Form

cr(jc)y e 11(9)9+) and in particular F(A)(x, y) = 0 for all x e U(9)- Also if y G Xli^Xh — A(]^)X then it is a straightforward exercise to show that F(x, y) = f(h —A(/i)) for some / e 5(1^) and hence again F(A)(x, y) = 0. It follows that the left ideal 1(A) [see (2.3(5))] of U(9) is in the radical of F(A). If we recall that M(A) == U(g)/I(A) and use 1 e U(9) to determine the highest weight vector ¿;+ of M(A), then we obtain an induced blinear form F^: M(A) XM(A)

K

satisfying F;,(x • i; + ,y • u^) = F (A )(x ,y )

for all x ,y e U (g).

Remark 2 The symmetry between + and - in (1) is deceptive. The definition of the Shapovalov form given here is appropriate for highest but not lowest weight modules. Proposition 2 (0 (ii) (iii) (iv)

F^ is symmetric and contragredient. Relative to F^, unless a = /3. The maximal submodule is the radical o f F^. Writing M(A) = lKi; ++ have for all x, y e U(9) that F^(x • v^, y • v^) is the component of o-(x) *y • i^+ in

Proof Parts (i) and (ii) are straightforward. To move part (iii), let R be the radical of F;^. F is a submodule by the contragredience of F and is proper since F^(i^+, 1;+) = 1. Thus R c MA). Conversely, let w e MA). Then for all jc e U(9), Fj ,(x -v^ ,w ) =F^(v +,o-(x) -w ) = 0 by part (ii), since cr(x)w e N(Á) c This shows that MA) c R. To move part (iv), write a-(x)y = F(x, y) + r as in the proof of Proposi­ tion 1, and apply it to i > □ We have established that M(A) is irreducible if and only if is nonde­ generate. However, we can be more precise than this. Let y ^ Q+ U{0}, and define FL as the restriction of F to U(q_)"'’': F |^ :U ( g _ ) " ^ x U ( g _ ) '

S (^).

172

Lie Algebras Admitting Triangular Decompositions

Proposition 3

F|y(A) is singular if and only if N ( \Y Proof, We have

# (0). by

u

U ' V.

and F|y(A)(Mi,«2) = F ( « i ,M2)(A)

- v + , U 2 - v ^) .

For «2 e U (g_)' «2 is in the radical of

( 2)

F|,y(A)

<=> U2 *v ^ ±

relative to Fj^,

<=> «2 *^ + -*- ^ (A )

relative to

<=> «2 *^ + ^

□

Now suppose that g admits a regular triangular decomposition. Then for any y G ¡2+ U{0} we may find a basis, say, « i,. . . , u„, of U(g_)~'^ (See the proof of part Proposition 2.2.1 (iii)). Consider the n X n matrix (3)

Sh^(A) =

From (2) we have the following: Corollary

I f g admits a regular triangular decomposition and Sh^ is defined by (3), then (i)

the dimension of N(XY~^ equals the dimension o f the null space of Sh/A ), (ii) M(A) is irreducible if and only if det(Sh^(A)) ^ 0 for all y ^ U{0}. Amazingly det(Sh^) can be explicitly computed for some of the algebras that are most interesting to us, for instance, affine algebras and Virasoro algebras (see Section 6.6 for the invariant Kac-Moody case). The result is that det(Sh^) factors into linear factors in 5(1^). The knowledge of these factors obviously provides direct information about MA).

2.8

173

The Shapovalov Form

Example 1 (Heisenberg algebras) We continue directly from Example 2.3.1. In the notation established there, for |m| = |nl > 1 we have (4)

F ^{a 'lv^,a 'iv^) = F j^v^,a'la'Lv^)

=

„m!A(c)'"’'

[see equation 2.3(8)].

If A(c) ^ 0, we see that the basis of M(A) is an orthogo­ nal basis. If A(c) = 0, then every basis element |m| > 1 is in the radical of Thus we see once again that M(A) is irreducible if and only if A(c) ^ 0 and MA) = if A(c) = 0. If is finite dimensional, then so is M(A)^“^“ for each r > 0 and det(Sh,_,„(A)) = n

m!A(c)^

|m| = r

If IK = R and A(c) > 0, then (4) shows that the form is positive definite. If we complexify a and M(A), then according to Remark 1 above, the resulting representation is unitary. Example 2 (The Virasoro algebra) We use the notation of Example 2.3.2. The diagonal algebra is spanned by L q and c, and M = M{h, c) denotes the Verma module with highest weight A: A(Lq) = /z, A(c) = c. Let denote a highest weight vector of M. The subspace M" has as a basis the set of elements • K+, where < • • • < rij^ and «j + • • • = n. We compute out the value of F^ on this basis for the values n = 0,1,2. This requires some simple calculations: L iL _i • V

(

2L q + L _ iL i) •

= —Ihv^ _\ F _j *

“ 3 L jL _j

F _1^ 2.^ —\ *^ +

= 6hu^ F^F^F_2 •

6hv^

L iL iL _ iL _ i • v += (8/z^ —4h)u + L2L_2 ■u+= { - 4 h + (i)c)ü + .

174

Lie Algebras Admitting Triangular Decompositions

The resulting matrices for « = 0,1,2 with respect to the bases {v^}, {L_ • i; +}, and {L _2 *u+, • z;+} are

(1),

i-2h),

—Ah + (^)c

6h

6h

8h^ - Ah

In the case that K = R, we may, as before, produce a hermitian form on the complexification of M(A). The problem of deciding for which pairs h,c the resulting representation M(A)c is unitary is much more involved and inter­ esting than first appears. (See [GO] for more on this.). Example 3 ( ê l2(IK)) For A e and a highest weight vector of the Verma module M(A), a basis for M(A)^“"“ is / " • The computation

^ a( / " •

■i'+) = «! n

k= l

((A + p)(h) - k ) ,

is left as an exercise (Exercise 2.19). We have written this using the function p e p(h) = 1, because it appears this way in the generalization (Section 6.6). We observe that the formula shows us in another way that is singular (hence M(A) is reducible) if and only if \(h) g I^. Proposition 4 [Ka] Let q be a Lie algebra over C with triangular decomposition ^ = (^j 9+j Q+j cr). Suppose that the basis Yi o f Q is finite, and let a 0 for all a e n . Then any unitary q-module M in category 0 is completely reducible as an a-module. Proof Let M be a unitary module. Define F: R>o by ♦= ^o)- For each c e C let (resp. g^) be the sum of the a-eigenspaces of /iq on M (resp. ad /iq s) for which F(a) = c. Since the weights of M lie in a finite number of fans and F takes a discrete set of positive real values on ¡2+, it is clear that for all c e C, dim < oo. Note that g^ = ^ and g^ • c for all c, d e C.

2.9

Jantzen Filtrations

175

Decompose a into ad/iQ-^ig^nspaces: a = Define (resp. a_, resp. Oq) as Ea^^ summed over all c such that /(c ) > 0 (resp. /(c ) < 0, resp. /(c) = 0). Let a := a + C/iq. Then a = a_ + (ao + C/iq) + a+

and

dg := Qq + C/iq c ]^.

Let < • I • > be the contragredient form relative to which M is unitary. We begin by showing that M is completely reducible as an á-module. Indeed, if N is an d-submodule, then N = where C\ N. Since < • | • > is positive definite on M"" and dim < oo, has an orthogonal supplement in M^: = N"" ± P"". Thus N M = N ® ^ , and N ^ is an d-submodule because < ♦| • > is contragredient. This shows the complete reducibility. Now suppose that AT c M is an irreducible fi-module. We prove that it is an irreducible a-module. For if not, then N ¥= (0), and we let x = Ex^, be a vector chosen with the following properties: 1. x e N \ { 0 ) , U(a) - jc #iV, 2. cdLiá{c\x^ # 0} is minimal among all vectors x satisfying property 1. Note that the weights of M lie in finitely many fans and hence the values of /«A ,/zo» form a discrete set bounded above as A runs through the weights of M. Thus among the vectors jc = E jc^' satisfying properties 1 and 2 we may choose one for which 3. m3x{ f(c)\x^ ¥= 0} is maximal among all vectors x satisfying properties 1 and 2. Let X satisfy properties 1, 2, and 3, and let x = Ex^, with x^ satisfying the maximizing condition of property 3. Now from the choice of x, - x^ = 0 for all d > 0; that is, a+= 0 and hence a +* = 0 for all c. Thus for each c with x^ ¥= 0, N = U(a) • ^ 0. This can only be true = U(a_) • U(ao) • U(a^) • c E ^ ^,.A^^, for one value of c, and hence x = for some c. Therefore x is a /zro-eigenvector, It follows that U(a) • x = U(a) X = N, contradicting the choice □ of X . 2.9 JAN TZEN FILTRATIONS Jantzen filtrations provide a technical tool for deeper study of Verma modules. The results that we establish here are not used until we discuss the Shapovalov determinant formula in Section 6.5, and the reader can (and probably should) omit this paragraph until it is needed. This section is a natural continuation of Section 2.8, and we use the notation established there without further comment.

176

Líe Algebras Admitting Triangular Decompositions

Let A e 1^*, and let M(A) be the corresponding Verma module. Let ze and imagine that we make a “small perturbation” A^ = A + iz, i e K, in A. Presumably M(A^) is closely related to M(A). Indeed, if we identify M(A^) as a subspace of U(q_), these Verma modules can all be compared directly. In doing this, we find that for w e U(g_), the Shapovalov bilinear form F^(v,w) is a polynomial function in t. Loosely speaking is defined to be the subset of vectors u of M(A) for which F^lu, M(A)) c t^K[t] where t is now a variable. This gives rise to a filtration M( A) = Mq ^

D M2 D *• •

of g-submodules of M(A), a Jantzen filtration [see [Jz], Chapter 5]. We use this filtration to give us a detailed view of the way in which the singularities of arise, the main result being Theorem 4. Preliminary to even writing down the statement of Theorem 4, we require an algebraic apparatus to formalize the heuristic discussion above. Let g have a regular triangular decomposition (1^, g+2+,o-), and let K[i] be the polynomial ring in one indeterminate over K. Set 9 == IK[i] ®k 9Then g is a free (K[i]-module of rank equal to dim^^g. There exists a unique IK[i]-bilinear mapping [*, •]: 9 X g ^ g extending the Lie multiplication of g: [ p ( i) ® x , q ( t ) ®y] = p ( t ) q ( t ) ® [j;,y ] for all p ( t) ; q{t) e IK[r]; x, y e g. This makes g into a Lie algebra over K[t], It is also a Lie algebra over IKwith g as a subalgebra through x 1 ® jc. (See Section 1.1 for Lie algebras over rings. We will use the full generality of the Poincaré-Birkhoff-Witt theorem (see Remark 4 in Section 1.8) in the arguments to follow.) Now g has a triangular decomposition as a lK[i]-algebra in an obvious way, as we saw in Remark 2.1.3. We define §

H[t]

Í)

and 9+== ^ [t]

9,

2.9

Jantzen Filtrations

We identify 1^* with a subset of

177

— Honi|,^[,j(§, K[t]) by linear extension of

A(p ( t ) (S>h) = p ( t ) \ { h )

for all p ( t ) e K[t]; /i e 1^; A e f)*.

We make this identification without comment hereafter. Extend cr to <7 on § by linear extension: a(p(t)

= p ( i ) ®
Then our triangular decomposition is

where Q+<^

5*. For a e Q, r == K [i] ® 9“.

We set about constructing a suitable “Verma module” for g. Let U(§) be the universal enveloping algebra of g [viewed as IK[r]-module (Remark 4 Section 1.8)], and let A, z e 1^* be fixed in our discussion. Set A = A + rz e 5*. Define /(A) to be the left ideal of U(g) generated by g^. and all elements of the form h — Ae i { k ) = { U , h - \{h)\h

i/(g )-le ff

Then M = M(A) := U ( § ) / / is a §-module and is our desired Verma module (see 2.3(5)). For each /1 e let

M4 :=

! ^ v ^ M \ h - v = p ,(h )v

for all ft e §}.

Using Remark 3 of Section 2.1 and the arguments at the beginning of Section 2.2 we see that M=

© aeG+U{0}

178

Lie Algebras Admitting Triangular Decompositions

and that = IK[í]í)^.,

where i+== 1 + /,

M‘' ““ = U (g_) “

via M• (1 + / ) <-^ M-

This uses the Poincaré-Birkhoíf-Witt theorem (over IK[f]). In addition we will need the Verma modules (over g), A/(A + az), for various a e IK, which are defined as U(g)//(A + az), where /(A + az) = (g+, /z — (A + azXh) \h G ^>u(3)-ieft [see 2.3(5)]. Let a G K, and let £^;K [t] ^ K, p ( t ) *^p(«)> be the evaluation map. Since U(§) = »<[^1 «k « (s )’ ^ e have a K-linear homomorphism of associative algebras e„:U (g) - ^ U (9 ) defined by £a- P ( 0 ® ^ We evidently have e j9 = idg

(9 identified inside g ),

^0( 9 +) = 9±>

kei£a = (t - n)U (9)We recall the decomposition of 2.8(1): (1)

U(g) = U(^) ® (9 -fi(9 ) + ^ ( 9) 9+)

and the projection map ^ :U (g )-.U (^ )-W determined by this decomposition. In the same way we have (2)

U(g) = U ( § ) ® (9-U (§)

2.9

Jantzen Filtrations

179

and a projection i':U ( § ) ^ U ( § ) = 5 ( § ) . These projections will enter the picture in a little while when we consider the Shapavalov form. We next assemble the properties of that we need. Given / e 5(1^) and A e 1^*, we sometimes use the notation /[A] to refer to its value /(A) at A. We use similar notation for / e 5(§). This may avoid some confusion in the thicket of brackets that occasionally appears. Lemma 1 (i) ë ^ ° â = <7° s^. (ii) For all Ü G U(§), (iii) éa(/(Â)) c /(A + az).

= (qej^idxx + az],

Proof. (i) The proof is obvious. (ii) Write M= «1 + «2, where ii, lies in the ith direct summand in (2). Then q(ii) = Mj, qs^,(u) = ^(e^(iJi)). It is therefore enough to prove the result for u = Ui, where has the simple form p(t) ® f, pU) e IK[i], / G 5(f|). Then e<,(«(“ i)[-^]) =

+ tz])

= p ( ^ ) f [ ^ + «z], while (i?M “ i))[A + (iii) Since

= ( « ( P ( « ) / ) ) [ ^ + «2 ] = p ( a ) /[ A + az].

is a homomorphism, it suffices to check that 8+) ^ 9+j

which we already know, and that e^{h - A(A)) = e^(h) - (A + az)(e^(A )). It suffices to check the latter for h = p(t) ® h, p(t) e K[t], h

180

Lie Algebras Admitting Triangular Decompositions

Then ^a{pi0 =

i { p { t ) ® h)) - e ^ (p (0 ((A + tz){h)))

= e^(h) - p { a ) { \ + az)(h) = B^(h) - (A + az)s^{h) as required.

□

After part (iii) of Lemma 1 we have from

a natural mapping

= U ( g ) //( A ) ^ U ( g ) / / ( A + az) =M (A + a z) . In the sequel M(A) and M(A + az) are denoted by M and M, respectively. Lemma 2 (i)

i/i is a (U(§), U(g))-m<2/7 in the sense that for all v ^ M and for all X e U (§ ), • v) =

(ii) ker

= it — a)M.

Proof, (i) This is clear since (ii) M

M

is a homomorphism.

=

U (g )//(A )

=

\. U (g )//(A )

The result now follows from ker £„ - it

-

U (g_) i^a

^ a)U(§).

« (9 -) □

Now recall that the Shapovalov form on M(A + az) is obtained by defining F-. U(g) X U(9) ^ through F iu, v) = q i a iu ) y \ evaluating

2.9

181

Jantzen Filtrations

F(u, i;) at A + az and observing that the ideal /(A + az) is in the radical of this form. Thus we have an induced form on U(g)//(A + az) — M. Similarly we define F: U(g) X U(§) ^ S(§) through F(u, u) = q(a(u)v) and, after evaluation at A, we induce a lK[i]-valued bilinear form on M = U(§)//(A). Lemma 3 For all a ^ K and all x, y ^ M,

Proof. Let u,v ^ U(g_) correspond to x and y. Then £aFx{x,y) = e^(i((r(M )i;)[A]) =

+ az] A+ a z

by parts (i) and (ii) of Lemma 1.

□

We now make an important hypothesis (NDG)

F^ is nondegenerate i.e.

M(A)) = 0 <=>jc = 0.

For instance, suppose that to e 1^* can be chosen so that F^ in M(o)) is nondegenerate. Then for any A e f)* setting z = co - A, we have F^+z nondegenerate. Now let A = A + tz, and suppose that F^ (x, M) = 0 for some X Ei M. If X ¥= 0, then we can suppose that (t — 1) does not divide x [otherwise, divide out (t - 1)]. But then il/^ix) ¥= 0, and this contradicts Lemma 3. Thus, in this case (NDG) holds for A. For each A: = 0 ,1 ,2 ... define M,:= { i e M |( i - a ) * |F x ( i , M ) } . In other words, i e <=> Fj^x, y) is a multiple of (t - a)* for all y ^ M. Since is contragredient, it is obvious that is a §-submodule of M. Thus we have the filtration (JFl)

M = Mq ^

z>M2 ^ ■■■

of M. Since F^ is nondegenerate, n “ =o^* = (0)- We define M* =

A: = 0,1,2, • • • .

182

Líe Algebras Admitting Triangular Decompositions

By Lemma 2, M* is a g-module, so we have the Jantzen filtration (JF2)

M(A + az) = Mo =) Ml D M2 3 • • •

of M(A + az). Furthermore This last result is not entirely obvious. We prove it in Theorem 4. The case a = 0 gives us a filtration of M(A). This completes the formalization of the heuristic discussion that opened this paragraph. If y e <2+ U{0}, then we have corresponding filtrations of weight spaces M^ where Mf-^ = M* nM^-^ = {jc eM^-T'|(i -a)*lFj;(jc,M^-^)} which is mapped by

onto

This last sequence stabilizes at (0) after a finite number of steps because M^^az-y ¡5 a finite-dimensional K-space for each y. Let y e <2+^{0}- We know that M^~'^ is a free IK[i]-module of finite rank (since U(g_) is a free lK[i]-module by PBW and K[t] is a principal ideal domain). Let {x^,. . . , be a basis for it. We are interested in the Shapo­ valov determinant det(Sh,,) = det(fx(x,,ii),. .) e K [i]. This is unique up to nonzero factors from IK. Let f i t ) e IK[i]\(0). The number o rd ,_ ^ /(0 is defined as the unique nonnegative integer k so that (i - a)^\f(tX but (t does not divide fit). Theorem 4 [Jan] Suppose that Fj^ is nondegenerate. Then (0 For all y ^ Q + n{0}, ord,_„(det Sh^(A)) = E^=i dim (ii) ch M^. = E^eQ^n,o, ord,_^(detSh.^(A)c(A + az - y); (iii) Mj is the maximal submodule N(X + az) o f M = M(A + az); (iv) n “ =o^* = (0).

< oo;

2.9

183

Jantzen Filtrations

The theorem is a direct consequence of the following general result that is also due to Jantzen. Proposition 5 Let A be a principal ideal domain and let p be a prime ideal o f A. Let (K be the field A / p . Suppose that X is a free A-module o f finite type, and set X •= X / p X with natural mapping ijj: X ^ X. Suppose that F: X X X A is a nondegen­ erate symmetric bilinear form, and let det(F) be the determinant o f the matrix {F{Xi, Xj)) for some basis {Jcj of X over A (note that det(F) is unique up to a unit of A). Let X^ = { x ^ X \ F { x , X ) and set

Then ordp(det(F)) = Y.

X*

k= l

(where ordp has the obvious meaning). Proof B. Let X ^ = Hom^(X, A) he the -.4-dual of X. Each x ^ X induces an element F ( jc, • ) of Let Y (zX^ be the submodule of all functionals induced by X in this way. Since the mapping x F(x, • ) is injective, ranky = rank X = rank X^. Using the well-known basis theorem [Jal] for modules over principal ideal domains, we can find a basis {e?,. . . , of X^ so that {a^e^,. . . , a^e^ is a basis for Y for some suitable elements a ^ ^ A \ {0}. Choose fi e X with F(f^, • ) = a^e^, and let {e^,. . . , be the basis of X dual to the basis {e^,. . . , We have

Since {/)), is a basis for X (because {a,e,}, is a basis of Y), there is a matrix S = (Sij) e GL„(A) with /, = £y_iS,yey, i = Thus d e tF = d e t ( F ( / ,,/,) ) = d e t|F (i;s ,.;C p /* jj = ( a ^... a^)Aet S.

184

Lie Algebras Admitting Triangular Decompositions

Since det 5 is a unit of ^4, we conclude that n

ordp(det F) = Y j ^rdp k=l

Let X = T.bjfj e X, Then Jc e

<=> ordp F[ej, x) > k <=> ordp ajbj > k

for ; = 1 ,..., n, for y = 1 ,..., n,

<=> ordp bj > k - ordp aj. Now feyCmod p) = 0 unless ordp bj = 0. Thus X¡^ = € A / p ) = Eord„a )•/'(/;)> where J(,k) ■■={j e { 1 ,..., n}\k - ord„ a. < 0}. Thus ' diniK Xk = card(/ lordp aj > A:}, and hence L dim^ k=l

= L ordp ÜJ = ordp(det F ). j=l

Proof o f Theorem 4. (i) This follows from Proposition 5 with ^ = IK[r], X = M ^~f, p = U - a)IK[r], and so on (ii) This is an immediate consequence of part (i) (iii) Let X e M, and let jc e M be any preimage of x. Then using Proposition 2.8.2 and Lemma 3, X e AT(A + az) <=i>

M) = 0

<=>Jc e Ml jc e Mj. (iv) From part i, fl

= (0) for all y e Q+ U{0}.

□

2.10 BERNSTEIN-GEUFAND-GELTAND DUALITY The origin of this section is a paper by Bernstein, GelTand, and GelTand [BGG2]. Their theory was worked out for finite-dimensional split semisimple

2.10

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185

Lie algebras. The generalization to the infinite-dimensional case was carried out by Rocha-Caridi and Wallach [R-CW], and this section follows their paper. The motivation for [BGG2] was a certain duality principle that had appeared in the category of finite-dimensional ^-modules for certain finite-dimensional associative algebras A over a field IK. Suppose that {Lj,. .. , L„} are the irreducible ^-modules (up to isomorphism). For each L, there is a unique indecomposable projective module for which Hom^(/^, L^) ^ 0. Let = [¡¿: Lj] be the number of occurrences of Lj in a composition series of The matrix C = is called the Cartan matrix of A (no relation to the Cartan matrices that appear in Chapter 3 and play such a fundamental role in all of the rest of the book!). For certain A,C = D^D for some integral matrix D (not necessarily square). In particu­ lar C is symmetric and positive semidefinite. One natural way to interpret this would be to find a collection . . . , M„} of “intermediate” modules that occurs as the factors in certain descending series for each /¿. If we denote by [/^: M^] the number of occurrences of in li in such a series and by [M^ : Lj] the number of occurrences Lj in then c^y = [li : L y ] = : L y ] . Thus, if it were true that the “duality principle” [/^: M^] = held, then C = D^D for D = (d^tyX where == : L y ]. In this section we construct various subcategories of the category for which just such a duality principle holds, the and Ly being the Verma modules and irreducible modules in this subcategory, respectively. Our main task is to construct the indecomposable projective modules for which the Verma modules are to play the intermediate role. In Section 2.11 we will use this duality principle to prove a basic result about embeddings of Verma modules into Verma modules. Let g be a Lie algebra over IK with regular triangular decomposition (f), g+, Q+, cr). Fix A e ]^* once and for all. We define if(A) to be the full subcategory of <^(g, T) whose objects are those modules M of ^ for which M=

© tei)*

and = (0) if fjL ^ A i Q^. We define Fif(A) to be the full subcategory of if(A) whose modules are finitely generated g-modules. Note that if M is a module of F-^iA) with the finite set of generators Xi = E x f, i = 1 ,..., AT, where x f g then {xf'l/ = 1 ,..., n, e 1^*} is also a finite set of generators. Thus modules in F-^ have finite homogeneous generating sets.

186

Lie Algebras Admitting Triangular Decompositions

Proposition 1 (i)

T^(A) is closed under taking submodules, forming finite direct sums, and forming quotient modules. In particular i^(A) is an abelian category. (ii) Fif(A) is closed under forming finite direct sums and forming quotient modules. (iii) For all fjL E: M(fji) and L(fi) are objects in Fif{A) and -é'(A) and the modules L(/i), At g A J, ¡2+? precisely, up to isomorphism, the irreducible modules in if{A). Proof. The only part that is not immediately obvious is the statement about the irreducibles in A), and for that we use Proposition 2.6.6. □ Since A is fixed, we will feel free to abbreviate if(A) and F ^(A ) to F-é" respectively.

and

Lemma 2 Let M be a module in F€. Then any proper submodule N o f M in -é" lies in a maximal proper submodule o f M in -é. Proof Let jCj,. . . , be a set of generators of M, and let N be any proper submodule of M in Choose k as large as possible so that the module generated by {N, . . . , x¡^} is not all of M. Let ^ be the set of all submodules K oí M ijm -é") such that Nj^
¡ jl

and P^, P2 as defined above,

P^ is finite, a G P , P^Q+\J{0}=>a+P 2

G P2 .

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187

Proof, (i) See Lemma 2.6.1. (ii) If on the contrary a + /3 e then /x + o :H -)S e A l(2 + , and hence also /i + a e A l Q ^ . This contradicts a e P2^ Consider the decomposition 1 1 ( 9 © U ( 9 j “ e © U(gj". a^Pi ol^P2 Then from Lemma 3 we obtain Lemma 4

is an ideal o f U(g+).

□

For /1 e A 1 ¡2+ we provide U(g+) with the unique g+0 1^-moduIe struc­ ture satisfying (jc + /z) • M= for all

e

+ a , h )u + XU /z e ]^, w e U (g+)“ ; a e (2+U{0}.

(This can be thought as the restriction to g+0 with lowest weight ^t). Let

of the Verma module of g

IV= W ( pl) : = U ( g J / © U ( g j \ a^Pz Now from the fact that 0^eP2^^^+^'' U(9+) we see that it is stable under the action of g+0 which we have just described. Lemma 5

W is a g+0 ^-module with the action x ' и = x и for all x e g+0 f) and и e U(g+X ^here “ :ll(g+) ^ Wis the natural quotient map. □ P(ju) is defined to be the induced g-module P ( p ) = P ( p ; A ) := и ( д ) 0 ц , ^ е ^ W( p ) .

Lemma 6

Let {iV/Xe / basis {1 ®

basis o f W{fx). Then P(/x) is a free left Vi(g_)-module with Furthermore, if for each i, w¿ is a weight vector o f weight %

Líe Algebras Admitting Triangular Decompositions

188

for 'i), then 7/ = M + ßi for some unique weight % in P(ix),

: P^, and in this case 1

w¿ is of

Proof By Corollary 2 of the PBW theorem U(g) is a free right U(g+0 module with any basis, say, {ujij^j of U(g_) as a basis. Thus P(/x) is the linear span of the vectors {uj ® w^}j and these vectors are a IK-basis for P(fi). Then the vectors {1 <S>w¡\¿ certainly generate Pifi) as a left U(g_)module. Moreover, if £¿^¿{1 w¡\ = 0 for some e U(g_), then writing z, = LjCijUj, Cij e K, we obtain 0 = LijCijiuj ® w¿X whence each c^j = 0. This proves the freeness. The statement about weights is obvious. □ Proposition 7 Pip) e F i f i M Proof W as a vector space. Since P^ is finite and, for all a e ¡2^, dim U(g^_)" is finite, we see that dim W is finite. Thus, by Lemma 6, Pip) is finitely generated as a g-module. Choosing a basis of W consisting of weight vectors, we see from Lemma 6 that Pip) is a weight module and that all of its weights lie in UaepjiM Л- a ) l Q ^ ( z К I Q ^ . Furthermore, since U(g_) has finite-dimensional weight spaces, it follows from Lemma 6 that Pip) does too. Thus Pip) ^ -^(A). □ Proposition 8 For every M e -^(Л), Hom^iPip), M) - Hom^(IK^, M). Explicitly the iso­ morphism is given by

Ф:Ф(Л)(г;о) =Л(1 0 1) for all A e Нот ^iPip), M), and its inverse is given by linear extension of 4 ^ (/)(x ® y) =д: у -/(i^o) for all f e Horn j(IK^, M). Proof. Let us check that Ф(А) is an ^-module map from IK^ ^ M. For all Ae h ■Ф (^)(У о) = h A ( i ® T) = A { h ■(1 ® 1)) = A{h ® 1) =A{1 ® /г - I ) = A(1 ® = Ф(А){({1,1г}оо) = Ф ( А ) ( к ■Vq), where we have used the fact that 1 e U(g+)°. Now consider First we should prove that it is well defined, namely that X • у ■fivo) is independent of the choice of у e U(g+) with у y. But, if

2.10

B ern stein -G el’fand -G el’fand D u ality

1»9

also y ' ^ y , then y ' = y + u , where u e e„ei>2^(9+)“- However, V q ^ and / e Hom^iK^, M) - /(t;«) e U(g^)“ • fiv^) e = (0) f^r all a Ei P2 since M ^ -^(A). Thus u ' fiv^) = 0. It is now clear that ^ exists and is linear. Let z e g. Then ^ (/)(^ ■

® ^)) = ■ ^ ( / ) ( ( ^ ) ® y) = ( z x ) y - / ( V o ) = z - x - y -f{Vo) =z •

®y ),

showing that ^ ( / ) ^ Homg(P(/i), M). Finally, we have < I > ( ^ ( / ) ) ( ^ o ) = 'P ( / ) ( l ® i ) = / ( ^ o ) and ^((^))(a;: ® y) = jc y • (^)(i;o) = X • y • A(1 <8> 1) = .4 (;r y

(l® I))

= A ( x ® y ), showing that and ^ are inverses of one another. Proposition 9 P(li) is a projective module in the category ^(A). Proof. Consider a diagram Pitx) N

^ M -

0

where all the modules and maps are in ^(A ). Then, in the notation of Proposition 8, we have

f/ / N ^

/

IK. 4>(A)

M ^

0

for which we can obviously find f e HomgOK , A/^), making the diagram

Lie Algebras Admitting Triangular Decompositions

190

commute. Then by Proposition 8, 'ir(f) ^ prove that g = A. But g o ^ ir(/)(jc ® y )

N). We want to

= g { x - y -fiVo)) = x - y - gi f i vo) ) = X ■y • ^ { A) V q =

'P (< I> (^ ))(JC ®

y)

= A { x ® y).

□

Proposition 10 Every object of F-^(A) is the image o f a projective module of the form 0/L i P(A^) for some A, 0 Proof Let M be a module in Fif(A), and let be a finite generating set consisting of weight vectors, say, m^ g From /¿: M^‘, 1 •-> as ]^-modules, we have ^(/¿) ^ Homg(F(A^), M). Then adding, we obtain a surjective map M. i= l

But ef=iP(A^) is projective and lies in Fif(A).

□

An important method for proving results in ^(A ) is the use of Verma composition series (VCS). The definition and basic properties appear in Section 2.6. The next few results deal with Verma composition series for certain modules in i^(A). If M has a VCS, then /(M ) denotes the length of one (and hence any) such series of M. Proposition 11 (i)

P(/x) has a Verma composition series. (ii) For a l l V 0 A i (2 + , [ P ( ijl): dim H om g(P(^), M(v)) (iii) [Pip): M(v)] = 0 unless < v < A.

M(v)] = dim

=

jjl

Proof Let {wJ/Li be a basis for W •= Wip), which is chosen so that e for some /3^ e P^ and so that > Pj => i > j. Let Wf^ := k = 1,2,... ,n, and define + i == (0). Thus W = W^z>W2 D • • • => Z) + i = (0). Observe that g+IT; d since e and Pi + a > p^ for all a e A^_. Since is also a weight module, is a (g+0 ]^)-module.

2.10

Bernstein-Gel’fand-GeFfand Duality

Inside P iti) = U(g)

191

define M ,:= U (g_) • (1 ® = U(g) -(1 ® IT,).

Then P{li) = M i d Mz Since the elements {1 ® w^} from a basis of P()ti) as a U(g_)-module, it is obvious that is a free U(g_) module of rank 1. Since 1 ® has weight ßk + f i , we see that = M(ß^ + ¡jl) (Proposition 2.3.3). This proves part (i). Since the multiplicities [P{^í)■.M(.v)] are invariants of the first equality of part (ii) follows from [P()u,):M(v)] = card{*lj3;t + ¡i = v) = dim For the second part of (ii) we have for e A | Ö+. dimHomg(F()u.), M (v)) = dimHom^(lK^, Af(v)) = dim M (v)^ = K(^v — ¡x) = dim U(g+)*^ ^ , where K is the Kostant partition function (Section 2.5). We get zero here unless V — IX a ^ Q+ U{0}. Then ix + a = v < h. v — ¡x = a ^ P^. Thus dim U(g+)’^-'‘ = dim (iii) In the proof of part (i) the only Verma factors M{v) occurring have V = !x + ßi^iox some jß^ e P^. □ Proposition 12 For M ,N ^ T^(A), M ® N has a VCS if and only if M and N have VCSs. Furthermore, if M, N and M ® N have VCSs, then for all weights fx, [M ® N, Miix)] = [M, Miix)] + [N, M(ix)l We prove Proposition 12 using the following lemma: Lemma 13 Suppose that M has a VCS, and suppose that fx ^ M is a maximal weight in M. Let 0 V M ' ^ , and set M ' ■= U(g) • v+. Then M ' = Mip ) and M / M ' has a VCS with l i M / M ’) = /(M ) - 1. Proof M' is a highest weight module with highest weight pair (ix,v+). Let M = M q Z) • • • D M; = (0) be a VCS (of length /) with M¡/M¡+l = M(ju,,). We prove the result by induction on /. If / = 1, M ' = M q = M q/M^, and we are done. Suppose that / > 1. If then by the induction assumption applied to Afj we obtain M' - Mip), and A /j/M ' has a VCS

Líe Algebras Admitting Triangular Decompositions

192

with KM^/M') < Then M / M ' has a VCS by adjoining the top term M /M j and KM /M') = 1 + KM^/M') = 1 + l(M^) - 1 = /(M) - 1. We are left with the case / > 1, u +^ Let M' ^ M/M^ be the canonical map. Then ¥= 0, and by the maximality of fi we obtain fi = ijlq and M /M^ - M(/i) = U(g) • D+. From the universal property of Verma modules we obtain a homomorphism (p : M ( f i ) M ' with v+. Thus the exact sequence 0 ^ Ml ^ M ^ M /M i == M(fi)

0

is split, and

M Now M' == M(fi) and M /M '

Ml 0 M'. has a VCS with /(Mi) = /(M ) - 1.

□

Proof of Proposition 12. If M = Mq D Ml D • • • D M^ = (0) and N = N q D Vi D • • • D = (0) are VCSs, then M e V = M o 0 A /^ D M i0 V D ••• D M^ 0 V = Vo D Vi D • • • D = (0) is a VCS for M 0 V. Conversely, let / be the length of a VCS for M 0 V. If / = 1, then evidently M = (0) or N = (0) since Verma modules are indecomposable (Proposition 2.2.2). Thus we may assume that / > 1. Let be a maximal weight of M 0 V, and for definiteness suppose that M^ ¥= (0). Let 0 # i;+e M^, and let M' = U(g) • u^. By the last lemma, M' == and (M e N ) / M ' — M / M ' 0 N has a VCS of shorter length than M. By induction on / we may assume that M /M ' and N have VCSs, whence also M and N. Now the statement on multiplicities follows from the first part of the proof. □ The next sequence of results aims at proving that every projective module in F ^(A ) has a VCS. Proposition 14 Every module in F-^ is a finite sum o f indecomposable modules, which are also in F-^. Proof. Any M e Fi^ has a finite set of generators consisting of weight vectors. Let G •= {¿;i,. . . , be such a set of generators, and let v^ 0 M^‘. Define o-(G) == E dim M^. We define o-(M) — min^ o-(G) as G runs over all such sets of generators of M. We prove this result by induction on o-(M). If cr(M) = 1, then M = U(g)i; for some v e M^ with dim M^ = 1. If M = Ml 0 M2 for some submodules, then M^ = M f 0 M^ => M f = (0), or

2.10

Bernstein-GeFfand-GeFfand Duality

193

= (0). Say, = (0). Then v e M f and M = Thus M is indecom­ posable. Suppose that 1. If M is indecomposable, there is nothing to prove. Suppose that M = 0 M2, where and M2 are nonzero submod­ ules. Let G be a generating set of weight vectors of M with a{G) = cr(M). For each g e G write g = gj + g2, g, e M,.. Then G^ := {gjg ^ G) ^ {0} generates M^, i = 1,2. From = Mf 0 for each ¡jl we see that < cr(Gi) < <j(G) = a(M), Thus and M2 have finite decomposi­ tions into indécomposables, hence also M. If M = where each Ij is indecomposable then clearly M e => Ij e F if for each ;. □ Proposition 15 (i) For all p. Ei K iQ ^ , P(fi) is a finite sum of indecomposable modules. (ii) A module M in Fi^(A) is projective if and only if it is a direct summand of 0/LiF(A^) for some A^,. . . , 0 1^*. (hi) Every projective module in Fi^(A) has a VCS. Proof, (i) Use Propositions 7 and 14. (ii) Let M e F # be projective. By Proposition 10 we have Aj,. . . , A^ e 1^* so that there exists an exact sequence

© F(A ^) ^ M - ^ 0 .

Since M is projective the sequence splits, and hence M is a direct summand of 0F(A¿). Conversely, any direct summand of a projective module is projective. (hi) Use Propositions 11 and 12. □ We now move toward the connection between indecomposable projectives in F ^ and irreducible modules in F if. Proposition 16 Each projective indecomposable object in Fif(A) has exactly one maximal proper submodule {in i^(A)). Proof. The existence of maximal submodules is given by Lemma 2. Now let I be finitely generated, projective, and indecomposable, and suppose that Mj and M2 are distinct maximal proper submodules of /. Then M^ + M2 = I.

194

Líe Algebras Admitting Triangular Decompositions

Form the direct sum of

and M2. Then we have Mj 0 M2

0,

where m ^) = + .m2 and i/i is known to exist so that cpij/ = id^ by the projectivity of I. Let p^: M^ 0 M2 ^ M^ be the natural projection, and define f¿ = / -^ c I, Then /1 + /2 = id/. Indeed for jc e 7, (/j + f2)(x) = (Pi + P2^° = (p° il/(x) = jc. Thus in particular /1 ° /2 = /2 ° /1 in the ring Endg(7). Let 7 be a finite sum of weight spaces 7^ so that J generates 7 (since 7 is in Fif). Fix 1 = 1 or 2. Since is a g-module map, //(7^) c 7^ and hence /.(7) c 7. Now we make a slight variation on Fitting’s lemma [Ja2]. Since dim 7 < 00, the descending chain / d / , ( / ) d / 2 ( / ) d ••• stabilizes at some A/j, so we have f^>U) = = • • • . Since / = U(g) • J, we have /■^'(/) = U(g)/^^'(/) = U(g)_^^''^*(/) = = • • • . Fix any weight fjb, and consider ■= Then by Fitting’s lemma [Ja2]

where s Z i n == n k= l

and « r ( 0 ) == U /t=i

= 0}.

Now by what we just proved = g^'C/**) = and hence = /^^‘( /) which is a submodule. We also see that if x = Ex^ e ®g~”(0), then for each nonzero x^ there exists ^ with /^ '‘Xj^ = g^*‘x^ = 0, and hence for k ■= max{A:^}, /¿*x = 0. Thus, in the obvious notation, x e / ~ “(0). Conversely, x = Ex^ e / r “(0) x^ G g ; “(0) for all M. Thus / " “(O) = e g ; “(0). In summary then, /= =

e { g Z i n ® 8Z^(0)) At © /r(0 ).

2.10

Bernstein-GeFfand-Gerfand Duality

195

This decomposes I into two submodules. Since I is indecomposable, one of them is (0). But /;.(/) c M, # /, and hence f ^ ‘i n * 1. Thus = (0). Set N = max{iV,, Then finally, since /, and / , commute, we have the contradiction id, = id2^ = ( / i + / 2 f ” = 0. This proves the uniqueness.

□

Proposition 17 Let L be an irreducible object in -^(A). Then, up to isomorphism, there is a unique finitely generated projective indecomposable module in i f that has L as a quotient. Proof (Existence): Since L is irreducible it lies in F-f. By Proposition 10, L is the homomorphic image of a finitely generated projective module P of -f. By Proposition 14, P = ®/Li/y where each /y is indecomposable. Since L is irreducible, L is the homomorphic image of one of the Ifs. (Uniqueness): Suppose that and I2 are finitely generated projective inde­ composable modules and that 7t¿: L are surjective homomorphisms. From

// / A ^1

TT2 0

we obtain / , making 771° / = 772- Now ker 77 ^ is the unique maximal proper submodule of (by Proposition 15). It follows that / is surjective, for if not, by Lemma 2, / ( / 2) lies in a maximal proper submodule of I, and hence lies in ker TTp But then 772 = ttj ° / = 0, which is not the case. Since 0 ker / ^ /2 ^ /1 0 is exact and is projective, there exists a homomorphism g: 12 splitting this sequence. Then 12 = k e r /e g /^ . Since ¡2 is indecomposable, ker / = 0. Then I 2 - h^ Propositions 16 and 17 establish a 1-1 correspondence between irre­ ducible objects in -f(K) and finitely generated indecomposable projective modules in -f(K). However, we already know (Proposition 1) the irreducible objects in -f(hj), namely the modules L(p), e A 1 (2+. Let / 1 < A . We define K / jl) to be the unique (up to isomorphism) finitely generated indecomposable projective module in i f ( A ) having L(/i) as an irreducible quotient.

196

Lie Algebras Admitting Triangular Decompositions

Every finitely generated indecomposable projective module in -^(A) has the form /(/x), /jl < A, Proposition 18 Let M G. t^(A). Then for all

¡jl

^ A iQ+,

[M:L{ii)] = dimHomg(/()Lx),M). In particular dim Horng(/()Lx), M) = 0 if

= (0).

Proof We begin with the simple observation that if I, M q, a category, I is projective, and M q Z) M^, then

are modules in

H om (/,M o/M i) = H o m (/,M o )/H o m (/,M i). Indeed every map / : / ^ Mq induces another, namely / : / -> M q/M^ and / = 0 iff / ( / ) c Mj. Thus there is an injective map A:Hom(/, M o)/H om (/, M^) Furthermore, if M q/M^, then since / I -> M q for which g = g, thus proving that It follows th at dim H o m (/, M q) dim Hom(/, Mf), Extending this by induction, we see that then

H om (/, M q/M^). is projective, there is a map g: A is surjective. = dim H o m (/, M q/ M ^ ) + if M = Mq 3

( 1) d im H o m (/,M )= E dim Hom (/, /=1

z> • • d M^,

+ dim Hom (/,M ;t)-

In our present case let M = Mq Z) Z) • • • d Mi^ = (0) be a local com­ position series at fi (Section 2.6). For each i either Mi_^/Mi — ¡Hi > f i o r M f ^ _ ^ / M i has no weights r > /x. In the former case X (2)

VÍ = (0)

if A

AX,,

2.10

Berastein-Gerfand-Gerfand Duality

197

Since

1. I(fji) has only one maximal proper submodule, call it 2. = Ufi), 3. Endg(L(/x)) - K.

Rifi),

In the latter case let us show that Н отд(/(д), = (0). Let / e Н отд(/(д), Since Li/i) is a quotient of I(fi) and {M i-i/MiY = (0), we conclude that kei f Rifi), = I( ijl)/R(/ jl)). Now Lemma 2 and Proposition 16 combine to show that / = 0. Since [M: L(/x)] is the number of times L(/i) occurs as a quotient in a local composition series at fi, Proposition 18 is proved. □ Proposition 19 Lei Д e Л 1 2+. Then ®

(finite sum),

where = dimHomg(P(jLi),L(i/)). We have

m j^v) = 0

fi < v < A,

Proof. By Proposition 14, P(jLt) is a finite sum of indécomposables. Of course each of these is finitely generated since P(/i) is. As we have seen, every finitely generated projective indecomposable module is an I(v), i; < Л. Thus

for some finite set of p and integers m jiv). Furthermore dim Horn g( P ( Д ), L(iü)) = dimHomg( ф mJ^p)I{p),L((o)^ = Lm^i/)dimHomg(/(i/),L(û>))

from 2. But from Proposition 8, dim Н о т д ( Р ( д ) , L((o)) = dimHom^CK^, L((o)) = dim L{o)Y. Thus mjfii) = 1 and mjfo)) = 0 unless (i < (o < A. □

198

Lie Algebras Admitting Triangular Decompositions

Theorem 20 (BGG duality) For all ii,v Ei A i Q+, [M{n.)-.L{v)] = [ / ( v) : M ( m)] in the category i^(A). Proof. From Proposition 18, it will suffice to show that [ l( v ) : M { fi ) ] = dim Horn g(/(i/),A i(iu,)). Let /Ji,v E A i Q^. Using Proposition 19, write (3)

P{v) = © m^{(o)I{(o),m^{a)) E I^.

We have (4)

[ P ( i') : M ( m)] = 'LrnXo>)[l((o) : M(v)]

by Proposition 12. From (3), Homg(P(v),M(M)) = Em,(a>)Hom3(/(a>),M(»^)) so from part (ii) of Proposition 11, (5)

[P(v): M {fi )] = E "i.(i^ )d im H o m g (/(a )),M (i/)).

As ¡jL,v vary we may consider (4) and (5) as systems of linear equations with the same matrix of coefficients (m^(ci))). Since mj^v) = 1 for all v and m^((o) = 0 unless i/ < co < A, we see that the matrix (m^(co)) is unipotent relative to any ordering of A i Q ^ that respects < . In particular the matrix is invertible, and we obtain at once that for all (o,v E A i Q^, dimHomg(/(io)

= [l((o):M(v)]

as required.

□

Corollary For any VCS o f Kfi), / ( m) =/o=>/l=5 ••• =>/„-!=>/„ = (0),

2.10

Bernstein-Gerfand-Gel’fand Duality

199

we have Io/I,^M (ß), Ij/Ij+i = M(vj),

vj> fi, j = 1 , 2 , . . . , n - 1.

Proof, [/(/x): Miß)] = [Miß) : Liß)] = 1. Thus in any VCS for I(ß) exactly one subquotient is isomorphic to Miß). From li ß ) Miß )

Liß )

0

we have a map ilr. l i ß ) M iß ) and iJ/Hiß)'^) ¥= (0). Thus iffHiß)) contains a generator of Miß), so if/ is surjective. Consider /j. Since it lies in the unique maximal proper submodule of l i ß ) (i.e., /j c ker tt), we have if/Hi) c Miß). Thus if/Ui) c N iß ) and I q/ I i -* M i ß ) / N i ß ) = Liß), which shows that /o //i = Miß). □ We may interpret the duality theorem in the way suggested in the introduction to this section. Set ==

Liv)]

and '■= [ ^ i f J - ) '■Liv)] Consider the matrices C = c^„- =[l i ß) - H^) ]= =

D =

for all

V e A i Ö+.

We have

E [/(cu):M (ir>)][M (a>):L(i^)] /1, V<(0

E [Mico):Liß)][Mi fJL,V<0)

Thus C =

and in particular C is symmetric and positive semidefinite.

200

Lie Algebras Admitting Triangular Decompositions

2.11

EMBEDDINGS OF VERMA MODULES

Throughout this section g is a Lie algebra with a fixed regular triangular decomposition and & is the category Let A e ]^* and consider the Verma module M(A). It is a very interesting problem (going back to Verma’s original work on these modules) to decide when another Verma module M(/¿) appears as a submodule of M(A). In the case of an invariant Kac-Moody algebra g, we will obtain an exact answer in Section 6.7 to this problem for a large range of values of A (all A when dim g is finite). If M ( ijl) is a submodule of M(A), then clearly Homg(M(/¿), M(A)) ¥=(0). Conversely, if e Horn ^ ( M ( ijlX (p ^ 0, then ( p ( M ( f i ) ) - M(/x) by Proposition 2.3.3, and hence M(/x) is in effect a submodule of M(A). Thus we are looking for conditions under which Homg(M(/¿), M(A)) ¥= (0). The main result of the section is Theorem 1 (Ku [Kuly Neidhardt [Ne]) Let

f)*. Then [M(A): L(/x)] ^ 0 <=> Homg(M (/x),M (A)) ^ 0.

The hard direction is “ => .” Our proof is taken from Neidhardt [Ne] and relies on the BGG duality theorem (Proposition 2.10.20). Proposition 2 Let )LL e and let N be a ^-module in category & that has no weights A with A > jLt. Then Extg(M(^i), N ) = (0). Proof. Consider any extension (1)

0^

^ P ^ M( f i ) - ^ 0 .

Let P^ be chosen so that fjL, (pv^ is 2lhighest weight vector of P. Since M(¡i) is a Verma module, there is a unique homomor­ phism (/í : M( jjl) -> P with {¡/((pu^) = u^. Clearly

/x. Thus M{iD and L(A) lie in the category ^(A). Consider the indecomposable

2.11

201

Embeddings of Yerma Modules

projective g-module /(/x) in ^(A). We are going to prove that > 0 and hence by BGG duality that [M(A): L(/x)] > 0. Let

[ K

i j l)

:

M(A)]

/ ( m) = / o ^ / i => ••• 3 4 _ i D/„ = (0) be a VCS for /(/x). We have 4 / 4 +1 for some weights /x^, where /xq = fi. (See Corollary of the BGG theorem, Proposition 2.10.2). Let 0 -> L (A )

0

be a nontrivial extension. From the natural map tt : I(fi) = 4 ^ M(/x) with kernel 4 from the projectivity of /(/x), we have /( m) T_T/ / /

0 ^

L(A)

P ^

M (fi)

0

with (p o TT = 77.

Now (p o 77(4 ) = '^■(4 ) = (0), so ^ ( 4 ) ^ ker 9 = L(A). In fact 77(4 ) = L(A), for otherwise ^ ( 4 ) = (0). Therefore 77 induces ¥: 7( m) / / i(= V A = M (n )) ^ P, and then splits the sequence contrary to it being nontrivial. Since L ( A ) = 7 7 ( 4 ) ^ D • ‘ • D 7 7 ( 4 - 1 ) ^ ^ ^ ( 4 ) ^ (OX there is a A : such that ^ ( 4 ) = L ( \ \ 'ñr(4+i) = (0) and hence a surjective map M(jjLf^) ^ 4 /4 + 1 ^ ^(>^)- This can only mean that /x^ = A, and hence [/(/x): M(A)] > 0, as we wanted. □ Proposition 4

Let M E: 0 be a highest weight module o f highest weight A. I f Extg(M(jLx), M ) (0), then Á. > jjb, and for some v e ij* with X > v > 11 we have [M(X):L(p)] > 0 and [M(v): L(fji)] > 0. Proof We have A > ju by Proposition 1. By assumption there is an exact sequence 0 -^N

M

L(X)^0,

where N is some maximal submodule of M. Every weight cp of N satisfies X >
202

Lie Algebras Admitting Triangular Decompositions

Using Corollary 1 of Proposition 1.12.4 we have Extg(M(iU.), L(A)) # (0), or Extg(M()u.), N ) ¥= (0). In the first case we obtain [M(A): L(^)] > 0 by Proposition 3, and hence we obtain our desired result with v = k. Suppose that Extg(M(/i), N ) # (0). Let ( 2)

N = N ^ z> N ,^ ••• 3AT„_i DA^„ = (0)

be a local composition series for N dX yi. By Corollary 2 of Proposition 1.12.4, Extg(M(^t), ¥= (0) for some ;. Using Proposition 2 = L{v) for some v. Then by Proposition 3, [M{v) : L()it)] > 0. It remains to see that [M(A): L(z/)] > 0. But is a subquotient of M(A) (since M is a quotient of M(A)) and[A/^:L(i/)] > 0. The result follows from the Corollary 1 of Proposi­ tion 2.6.13. □ Proof o f Theorem 1, (<=) By assumption there is a nontrivial g-module homomorphism 0 at once. (=>) The proof is for all pairs A, m,(A > m)? by induction on h t(\ — ¡i). If h t{\ - m) = 0, then \ = ¡JL, and the result is trivial. Suppose that A > m- Lot MA) be the maximal submodule of M(A). Since [M(A): L( m)] > 0 and A > Mj we see by refining M(A) d MA) d (0) that jjl

(3)

[N (k) :

>0.

Let

(0) = A^o c ATj c iVj c be a HWS of MA) (Proposition 2.6.10). Each is a highest weight module, say, of highest weight i = 1 ,2 ,... . Evidently (4)

A > )u.,

From

: L(/a,)] > 0 we obtain

(5)

[M A ) : L (n ,)] > 0,

for all i.

[M(A) : L ( m,)] > 0.

By part (iii) of HWS of Section 2.6, there is a A: such that = N'^. It follows at once that [M A )/M ^LC^u.)] = 0, and hence from (3), [M •

2.11

Embeddings of Yerma Modules

203

Lifi)] > 0. Thus for some j < k , [Nj/N j _ i : L(ju,)] > 0, (6)

and

[M(/ty) : L(/ii)] > 0

My ^

Case 1. fjL Ф fjLj. Then к > ¡Xj> ji. Using the induction assumption, Ф

(0),

n o m ^ {M ifij),M (X )) Ф

(0),

and so combining, we obtain Homg(M(/x), M(A)) Ф (0) as required. Case 2. 11 = fij. Then Nj/Nj_^ is a nonzero homomorphic image of Mill), and clearly Нотд(М (д), ^ (0). Let 0 < / < ; be chosen minimal so that Нот J^Mifi), N ik )/N¿) Ф (0). If i = 0, then Homg(M(/i), MA)) Ф (0). Consequently Нот^iM ifi), M(A)) Ф (0), and we are done. If i > 0, then from the exact sequence 0^N,/N,_, ^Nik)/N,_, -^Nik)/N,-^0 we have the exact sequence (Proposition 1.12.4),

Since the first term is (0) and the second is not, neither is the third; that is, Extg(Ai(/i), Ni/N i_i) ^ (0). Applying Proposition 4, there is a weight V e ]^* with X > ( i j > V > f x s o that : L (v)] > 0 and

[ M {

p

)

: L (/i)] > 0.

The induction assumption gives us Homg(A/(v), ¥= (0) and Homg(Ai(ju,), M{v)) ¥= (0). Furthermore from (5) and the induction assumption Homg(Af(ju,,), Af(A))) # (0). Combining all these, we obtain the desired result. □

2(M

Lie Algebras Admitting Triangular Decompositions

2 .12

D EC O M PO SITIO N OF M O D U LES IN CATEGORY Û

According to Section 2.11 the embedding of Verma modules Mijx) ^ M(A) is possible if and only if [M(A): Lifi)] > 0. A central issue is then when is [M iX ):U fi)] > 0? This question can only be answered adequately by spe­ cializing the class of Lie algebras g under discussion. This is precisely what we do in Chapter 6, where g is assumed to be an invariant Kac-Moody algebra. However, the relation on 1^* defined by A /i if and only if [M (A):L( m>] > 0 does lead to a useful equivalence relation on certain subsets (characteristic subsets) of 1^*. The main result of this section shows how this notion leads to a natural decomposition of any module M lying in Û into submodules, each of which lies in a subcategory of ^ governed by the equivalence relation. The material is derived from [DGK] with some simplifications from Section 2.11. These results are not used until Section 6.8 and are not necessary for any other part of this book. We will assume throughout that g is a Lie algebra with fixed regular triangular decomposition <5^= (^, g+, Ô+? ^ind that Q is finitely gener­ ated. A nonempty subset 5 of 1^* will be called characteristic if it has the following property Ae 5

and

[M(A) : L(/¿)] > 0 =>

e 5,

or equivalently, A e5 = > ch M (A )= for some family sets:

Y. c^ch L{/jl) fJL^S

in N. The following are examples of characteristic

1. Í)*. 2. P := {A eri< A ,a,-^ > 3. A 1 (2+ for A e

Z for all i G J}.

For A, we write Á > ¡jl if and only if [M(A): Li/i)] > 0. Let 5 be a characteristic subset of 1^*. We define an equivalence relation = '^s ^ as the transitive closure of in 5; that is, A /x <=►, there exists Aq, Aj, . . . , A„ e 5 such that Aq = A, A^ = ¡jl, and either A¿_j -< A¿ or A¿ A,_i for all i < i < n. For y e 5 we will let f denote its equivalence class in S. Each equivalence class of S is itself a characteristic set. Let 5 be a characteristic set. A module M in <^(g, 5^) is said to be of type 5 if [M :L (A )] > 0 => A e 5.

2.12

Decom position o f M odules in Category

205

&

Equivalently M is of type A if c h M = £ c ;,c h L ( A ) . Ae5

Let S c ]^* be a characteristic set. We define a category as follows:

= ^ 5(9,

1. The objects of 0^ are the objects M of ^ of type 5 2. The morphisms of 0^ are all the g-module homomorphisms between the objects of 0^, It is clear that direct sums, submodules, and quotient modules (see Proposition 2.6.13) of modules in 0^ are still in 0^. In the sequel we will generally have a single characteristic set 5 in mind (this will he K ' in Section 6.8). The primary interest will be in the equiva­ lence classes of • For A e 5 its equivalence class A defines a subcategory which for simplicity we sometimes denote by 0^ (so ^ if A /a). The following is obvious: Proposition 1

Let 5 be a characteristic set, and let A e 5. Then (i) M(A), L(A) e (ii) every highest weight module with highest weight A is in 0^\ (iii) E ^ 0 x ^ 0^. Proposition 2

Let S be a characteristic set, and let S2 ^ S be subsets that are unions of equivalence classes. Suppose that 5^ Pi 52 = 0 . Then for E ^ 0^ , F ^ 0^^ we have (i) E n F = (0X (ii) H om g(£,F) = (0). Proof (0 Evidently [E n F : L(p)] > 0 => [E : L(fi)] > 0 and [F : L(p)] > 0. Hence fi ^ Si n S2, contradiction. (ii) If e Homg(E, F), then
206

Lie Algebras Admitting Triangular Decompositions

Proposition 3

Let S be a characteristic set, and let A and fx be in S. Suppose that E and F are highest weight modules in &(,%, o f highest weights A and fi, respectively. If A and ¡X are not equivalent, then E x t(F ,£ ) = (0).

Proof. Consider first the case F = M(fx). By Proposition 2.11.4, if Ext{M(.fx), E) ¥= (0), then \ > fx and there is a ^ e such that [M(A): L{v)] > 0 and [M{v): Lip,)] > 0. Thus Á ~ v ~ p , contradiction. Thus ExtiM ip), E) = (0). Now in the general case we have an exact sequence of g-modules 0

K ^

M (p) - ^ F ^ O

and hence an exact sequence (Proposition 1.12.4') H om g(/¡:,£) ^ E x t ( F , £ ) ^ Ext(M (jit),£ ) . But K is of type ¡X, and E is of type Á. Thus by Proposition 2, Homg(£, E) = (0). Since ExtiM ip), E) = (0) by what we have just proved, we obtain Ext(F, E) = (0). □ Corollary

Suppose that Á , p ^ S , are not equivalent. Assume E g F g are modules each of which has a finite highest weight series iSection 2.6). Then ExtiF, E) = (0). Proof. Let He ) and /(F) denote the lengths of the given HWS of E and F. If HE) + /(F) < 2, then the result is immediate from Proposition 3. In general we reason by induction on /(F ) + /(F). For example, if (0) c F i c

c £

= F

and n > 2, then from 0 ^ F „ _ i - ^ F ^ £ /£ „ _ ! - > 0 we have the exact sequence E x t(F ,F „_ i) ^ E x t(F ,F ) ^ E x t(F ,F /F „ _ i),

2.12

Decomposition of Modules in Category 0

207

and the outside terms are (0) by the induction hypothesis. Thus also Ext(F, = (0). The case for /(F) > 2 is similar. □ Theorem 4 Let S under

be a characteristic set, and let S denote its set o f equivalence classes . Let M be a %-module in 0^, Then M decomposes as a sum M=

where lies in

Furthermore is the unique maximal submodule o fM that In particular this decomposition o f M is unique.

Proof Let (0) = Mo c Ml c M2 c • • ‘ be a HWS for M (Proposition 2.6.10). Thus each nontrivial module M„/M„_i is a highest weight module, say, of highest weight A„, n = 1 ,2 ,... . We are going to construct for each equivalence class ft e 5 a module having a HWS (0) = M o o C M i o C such that ® M „n,

(1)

( 2)

i(0)

if e ft, if A„ ^ ft.

We will then have M = © ^ ^ required. We construct all the M„ ft g 5, at once by induction on n. For n = 1 we set Ml = Ml, Mi ^ = (0) if ft Ai. Suppose that n > 1 and by induction we have constructed M„_i ft G 5, satisfying (1) and (2) for n — 1. We have M„/M„_i ~\ . Thus ^ we define (3)

if ft ^ A„.

To define M„ (4)

we consider the exact sequence

0 ^ M„_i/M„_i,;^^

0.

208

Lie Algebras Admitting Triangular Decompositions

From the induction assumption

SO

Ext(M„/M„_i,

=

e

Ext(M „/M „_„

= (0)

by the Corollary to Proposition 3. Thus (4) splits, and hence there is a submodule E of such that

and

Define M„ ■= E. Since Af„_i ^ and £ G . We" check (1).

are in

, so also

M„ = £ + M„_, =£ +

® 0 n#A„

^ =£+

But

£

n 0 ^ ^ A „ ^ n , ii

= (0)

0 M„ n#A„ since £

M„ = £ ®

0

g

>

by(3).

Thus

M„_n= 0

Condition (2) is obvious. This completes the induction step. Define — U “ =iAf„ Clearly e and we have our required decomposition of M. As for the second part of the Theorem, let be the sum of all submodules K oi M such that K e Then e and z) M^, But ^ n ^ n ' "" (0) Proposition 2, and hence = M^, □

209

Exercises

EXERCISES 2.1 Let H be an abelian group and tt : H //. Let e Hom(//, K^). We define

G LiV) a representation of

= {u ^ M \7 r(h ){u ) = (f>{h)v for all h ^ H } . Show that is a submodule of V and prove the “group version’’ of Proposition 2.1. Describe also the “group version” of Example 1 of 2.1 for the group H = {A ^ SL(n, K)|>1} is diagonal. 2.2 (a)

Let (^ ,g + ,2 + ,< j)b e a triangular decomposition of g. Show that any subalgebra § of g satisfying ^ and c § also possesses a triangular decomposition. (b) Let g¿, i = 1 ,..., r, be Lie algebras with triangular decomposi­ tions ^ g^+, o}). Prove that the direct product gj X • • • X g^ of the Lie algebras g¿ also has a triangular decompo­ sition. (c) Let (]^, g+, (2+, a) be a triangular decomposition of g and let a be an ideal of g satisfying a n = (0), ga = a, g+
2.3 Let g = §I„(IK) be considered with the triangular decomposition of Example 3 of 2.1, and let M = IK" be the g-module with the weight space decomposition given in Example 1 of 2.1. Show by direct compu­ tation that these two gradings are compatible. 2.4 Classify (up to isomorphism) all three-dimensional Lie algebras admit­ ting a triangular decomposition. 2.5 [Be2] Let g be a Lie algebra with triangular decomposition (Q+j G+j 9 ^ 9 be any automorphism of order 2 such that i/f(g“) = g“" for all a e g . The objective of this exercise is to obtain some information about the Lie subalgebra § of all fixed points of í/í. Of course § = {jc + il/(x)\x G. g}. Let / : 0 -» Z be a group homomorphism satisfying f(a ) e Z+ for all a e 0+. Set 9" =

L

9“,

f ( a) = n

thereby obtaining a Z-grading on g for which g+= E„>o9", ft = 9^* Let {e^li e K}, where each e g"', n- e Z+, be a set of genera­ tors of g+ as a Lie algebra. Define == ^ i E: K.

210

Lie Algebras Admitting Triangular Decompositions

Also define the subspaces

and

n = - 1 ,0 ,1 ,... inductively

by 9-1 = (0),

9 o = 5,

9„=

E

9^

0<j : ^n

§_i = (0),

= [x + il/(x)\x e 1^} = i),

§„ = linear spcin of

and all products

... ,e,^],

where M; + • •• +n¡ < n . Evidently 9 = U "=_i9„, § => (a) Prove by induction on N that {x + tlf{x)\x G g^jm od §^_i and [«,V • • • ’ (b) Prove that § = U (c) Let

• • •’

= h v •••

^N-V

and that {e¡\i g K} generates §.

gr(§)+=«e= 1§„/§„_! be the graded Lie algebra associated with the filtration ^ c • • • of Prove that 9+— under the mapping x ^ x + il/(x)mod for all jc e g" , « = 1 ,2 ,... . 2.6 Show that every highest weight submodule of a Verma module is a Verma module. 2.7 Let (a ) and <6>, a, ft e N, denote the ^^(W-modules of dimensions a 1 and & + 1, respectively. Prove that (a ) ® (b ) - (a

b) ^ (a + b - 2)

e ••• e . 2.8 Let B c 5L2(IK) be defined as in Proposition 2.4.8. Let = {X^\X e B}. Prove that 5L2(IK) = B^B U J (Birkhoff decomposi­ tion) and that this union is disjoint.

Exercises

2.9

211

Let M be any finite-dimensional § l2(lK)-module admitting a weight space decomposition M = where M"* = {x e M\h • x = ¿zx}. Prove that for all a ^ Z, (a) dim = dim M “"*, (b) the sequence {dim is monotonically increasing until a = 0 and monotonically decreasing after 0.

2.10 Let Af(A) be a Verma modules of the category T). Show that for M(A) to be isomorphic to M(A)^, it is necessary and sufficient that M(A) be irreducible. The results of Exercises 2.11 and 2.12 are due to D.-N. Verma. 2.11 Let n be a finite dimensional Lie algebra graded by a free abelian group Q — Z^ whose degrees lie in —Q+= —(N^\(0}). Let ch U (n )=

n

(l-e (-a ))'

L K{
be the character of U(n) (see Proposition 2.5.3). (a) Prove that the coefficients K( 1 and any
tion Q+, (t ), and let M(A) be a Verma module for g. (a) Show that any two Verma submodules M ( / jl) and M ( p ) of M(A) must have a nontrivial intersection. (There is a counting argu­ ment using Exercise 2.11.) (b) Prove that M(A) is Noetherian (see Exercise 1.14). In the remaining exercises we will assume that every Verma module has a (finite) composition series. This is not in general true (see Exercise 2.13). However, it is true for split semisimple finite-dimensional Lie algebras (see Exercise 4.7). (c) Prove that every Verma module M(A) has a unique irreducible highest weight submodule M (v) (v depending on A) and that it is a submodule of every highest weight submodule of M(A). (d) Prove that for all A, /x e 1^*, dimHomu(g)(M(ju), M(A)) < 1. (Hint: Let M and M' be the image of two distinct embeddings of M ( ijl) in M(A). Then M(v) of (c) is in both. Write the generators of M(v), M, M' in terms of the generator í;+ of M(A), and use the fact that U(g_) is a domain.)

Lie Algebras Admitting Triangular Decompositions

212

2.13 Give an example of a triangular pair (g, (with dim g < oo) and a module M e ^ ( g , for which M does not admit a (finite) composi­ tion series. 2.14 Let be an abelian Lie algebra and V an l^-module admitting a weight space decomposition. Let V* be the dual space of V viewed as a ]^-module. (a) If / e for some A e ]^*, show that f(V ^ ) = (0) whenever JJL

(b)

(K^)* —(K*)"'^ as K-spaces for all A e 1^*.

2.15 Let M e <^(g, ^ ) . Let N+ and N_ be l^-modules admitting weight space decompositions: ©

N i ,

N _ =

©

N Z \

where all weight spaces are finite dimensional. Let if/: N _

be a vector space isomorphism satisfying {¡/(h • n) = —h • il/(n) for a\\ h phism

and n e N_. Show that there exists a unique monomor­

satisfying f{ m ® n) = f(>lf{n))(m ) for all / G Uom^(N+, M “^), n e N_, m ^ M. Moreover for all fi the above linear map composed with the isomorphism of Exercise 2.14(b) determines an isomorphism ~ :Hom(A^+,M‘^)'' s 2.16 [BM] Let Q = ®/=iZa, be a free abelian group. Set Q+'= (Ei=iNa,) \ {0}, and consider the power series ring Q[0T Q+\ which is the set of all formal sums e Q. Suppose that (s^, ^2,. - is a sequence (possibly finite) of distinct elements of Q+ arranged in order of increasing height, and suppose that for each 5^ there is associated a sign e(5,) = ± 1. For each sequence (n) = (n^, « 2?• •) of

Exercises

213

nonnegative integers, all but a finite number of the n, being 0 and for all A e (2+ U{0} define X(n) = 2 ( ( n ) ) = 2 « , ,

B {n) = B {(n )) -

(X n,)!

sgn(n) = sgn((«)) = r ie ( s ,) " ', S(A) =

{(n)|Xn,i,-

= A}.

For A, a e write X\a if a = rk for some r g Z+. In this case A /a := 1/r. Let m: ¡2+ ^ N, a be any function, and suppose that for some choice of s = ^2, . . . ) as above: n

( 1 - € ( « ) ) '" “ = 1 - £ £ ( 5 ,)e(s,). i

(This situation occurs for several types of graded Lie algebras n = ®n “, where = dim n “ (see Proposition 2.5.5 and the Corollary to Theo­ rem 6.4.1).) The objective is to give a closed formula for the integers m „.

(a) Let X := 1 - E,e(s,)e(s,). Show that e{ka ) -log X =

£

m,

a, k

(b) Apply the operator I E:= £ e ( a , ) i= l

d de{ai)

to obtain £(X )

(c)

= XC(A)e(A),

where C(A) — E„|AW„/ii(a). By computing -£ (X )/X in a different way, show that C(A) = X e(i,)/ii(5,)

£ ( n ) s 5 ( A - i ,)

sgn(n)B (n).

214

Lie Algebras Admitting Triangular Decompositions

(d)

Show that the right-hand side in (c) can be rewritten to give C (A )=Ä i(A )

E

sgn(n)

•

(«)e5(A )

(e)

Use Möbius inversion (see Bourbaki, Ch. 1-3, App. Ch. 2) to prove that

"ior

E Ala

2.17

^

(ri£i(i/) )

“ (n)e5(A )

pr

,

1 1 '^/-

(a)

Show that the homomorphic image of a residually nilpotent Lie algebra need not be residually nilpotent. (b) Show that the sum of a finite number residually nilpotent ideals of a Lie algebra is residually nilpotent. (c) Show that the sum of nilpotent ideals of a Lie algebra need not be residually nilpotent.

2.18 Let (g, be a Lie algebra with triangular decomposition. Show that g/rad(g) is radical free. 2.19 Prove that the Shapovolov form for §I2(1K) is given by ^ a( / ” *

*^+) =

n

k= l

(('^ + P )(^) “

for

n e N.

(See Section 2.8 Example 3). 2.20 Determine all the characteristic sets 5 c 1^* and their equivalence classes in the case of §I2(1K). 2.21 Prove that for the Lie algebra §I2(IK) the decomposition M =

® Mj,

Ae5 of Theorem 2.12.4 is a decomposition into eigenspaces of the Casimir operator. 2.22 Show that for t e C , a b one has M(a) ^ M(b) as ^12(C) modules if and only if i) e N and a = —b — 2. (See Section 2.4 for notation.) 2.23 Let I = ]^ 0 h be a Lie algebra consisting of an abelian subalgebra and an ideal b that decomposes as a l^-module under the adjoint representation into weight spaces b = where G +c 1^* is a free additive semigroup.

Exercises

215

Let t)' be a copy of b as a Lie algebra, and let o-: b b' be the anti-isomorphism between them that associates to each i; e b, the element - u , as seen in b'. We make b' an 1^-module by defining h • cr(v) = —o-[h, u], (a) Prove that f' == f) 0 b' has the structure of a Lie algebra compat­ ible with this action of on b'. (b) Let g := b' 0 0 b, and extend a to an endomorphism of g by defining (rli) = id ij,c7 ^ = l. Prove that g may be given the struc­ ture of a Lie algebra with subalgebras b and b' by defining [b,b'] = [b',b] = (0),

(c)

and show that (1^, b, ¡2+? is a triangular decomposition of 9. Let g be any Lie algebra with triangular decomposition 0), ^+,Q+,o-)- Set 5g = g_0 0 g+ as a vector space with multiplication defined as in g except that now [g+, g_] = (0) = [ 9 - , 9+ ].

Prove that ^g is a Lie algebra with triangular decomposition s ^ = (ij, Q+,Q+,(t ), We call sq the suspension of g. A Lie algebra g, with triangular decomposition is called suspended if [g+, g_] = 0. (d) Let (g, be a suspended Lie algebra. Prove that ^ t (Q) = i^ r(9 ) = n (g ) = rad(g) = g++ g_. (e)

Let (g, <5^) be a suspended Lie algebra, and let M(A) be a Verma module. Show that every subspace fx e A J, (2^., is a submodule of M(A). Show that all the irreducibles modules in <^(g, D are one dimensional.

2.24 Show that every derivation of the Vivasoro algebra is inner.

Chapter Three

Lattices and Finite Root Systems Symbols are to mathematics as wooden pieces are to chess —L. Wittgenstein

Any Lie algebra g with triangular decomposition У has associated with it a root system Д c 1^* so that g = There is no a priori reason to expect Д to have a particularly rich structure of its own. But, as Example 1.1.8 shows, it can happen. In fact Д is remarkably interesting for all the finite-dimensional semisimple Lie algebras over C. The root systems that arise from such semisimple Lie algebras are simply called finite (reduced) root systems. Over time it has been realized that finite root systems appear in other contexts that often have no apparent Lie theoretical connection. Thus there has developed an axiomatic approach to root systems (exemplified by Bourbaki [B03]), free from Lie algebras, which allows them to stand on their own. We adopt this approach here, prefacing it by a section on lattices and concluding with sections that show how to construct Lie algebras out of lattices, in certain cases. The investigation of root systems was begun at the end of the 19th century by W. Killing and E. Cartan in their study of complex semisimple Lie algebras g. Both realized that to each nonzero root there is an associated involution that stabilizes Д. Later on, influenced by the work of H. Weyl [Wy], the group W generated by all these involutions (now called the Weyl group) assumed a major role in understanding and describing the representa­ tion theory of g. These Weyl groups are finite groups generated by reflec­ tions (in euclidean space) and as such fall into H. S. M. Coxeter’s classifica­ tion of reflection groups [Cxi], A simplified version of the classification of simple and semisimple Lie algebras making use of root systems was achieved by B. L. van der Waerden [vdW]. 216

3.1

Lattices

217

3.7 LATTICES A geometric lattice is a pair (L ,( | *)) consisting of a free Z-module L of finite rank together with a symmetric bilinear form ( I ) : L X L ^ Z. Normally the bilinear form is understood from the context, and we simply speak of geometric lattices L. By assumption, L has a basis Cj,. . . , over Z: L = Z^i 0 • • • e Ze„. If we extend the scalars to IKby forming V - = K ^ i L and extend the bilinear form to a K-valued bilinear form on V in the only way possible, then V becomes a quadratic space by considering the quadratic form associated to ( I ) (see below). It is useful to be able to view a geometric lattice as a subgroup of a real or rational space (take IK = R or IK = Q above). The spaces R Q ®jL are called the realization and rationalization of L, respectively. Notice that L is a discrete subgroup of R The principal example of a geometric lattice is 0 • • • 0 Zs„

in R" = R^i 0 • • • 0 Re„,

where {e^,...,e„} is the standard orthonormal basis of R" as a euclidean space. This lattice is denoted by . The simplicity of this example belies the subtlety of the subject. Let L be a geometric lattice, and consider the mapping q:x of L into |Z . Then q is the quadratic form on L associated with ( I ). The quantity (x|jc) is called the norm or square norm of x. Of course (-I*) can be recovered from its quadratic form in the usual way: (jcly) = q { x + y ) - q(,x) - q{y) . The concept of isomorphism between geometric lattices is defined in the obvious way: We say that (L,(-10) and ( L ,(•!•)') are isomorphic if there is an isomorphism f : L - ^ L ' of groups such that (x|y) = (/(x )|/(y ))' for all x,y ^ L. If q and q' are the corresponding quadratic forms, then this last condition can be replaced by q(x) = q'(f(x)) for all x e L. The concept of automorphism of a geometric lattice is defined in the obvious fashion. If L is a geometric lattice, then its group of automorphisms is denoted by Aut(L). We list below a series of invariants for geometric lattices—that is, quanti­ ties attached to a lattice that depend only on its isomorphism class. 1. Rank, Two free Z-modules are isomorphic if and only if they have the same rank. In particular the rank, rank(L), of a geometric lattice L is an invariant. 2. Determinant. Let e = be a Z-basis of the geometric lat­ tice (L ,(1 )), and consider the n X n integral matrix = (B^ ), where

218

Lattices and Finite Root Systems

Bij '= If e' = {e\,. . . , is another basis for L and A is the matrix corresponding to the change of basis; e\ = Ae^ = then we have (< l< ) =

E a =l

E

Aue,

1=1 Ij

k,l

where A^ is the transpose of A. Thus B^,=A^B^A, Since det(y4) e Z and A has an integral inverse, det(^4) = ±1, from which d et(5 ,0 = d e t(B J. The common value of this determinant, which is independent of the choice of basis, is called the determinant of L and is denoted by det(L). It is an invariant of L. The bilinear form (*|*) is nondegenerate or nonsingular if and only if det(L) 0. In the case that det(L) 0, we say that L is nondegener­ ate or nonsingular. Otherwise, it is said to be singular. 3. Signature, Suppose that L is a nonsingular geometric lattice and that IK c R. Then by extending (•!•) to V •= we may apply Sylvester’s theorem (Jacobson [Ja2, Ch. 6] and write V as an orthogonal sum 1. V~ of a positive definite space and negative definite space V~. The dimensions and m " of these subspaces are invariants of (L,(*|*)) and are indepen­ dent of the choice of IK. Their difference sgn(L) = m'^ — m~ is called the signature of L. L is positive definite (resp. negative definite) if sgn(L) = rank(L) [resp. sgn(L) = -rank(L)]. It is indefinite if it is neither positive definite nor negative definite. 4. Type, We say that the geometric lattice L is even or of type II if the square norm of x is even for all x. Otherwise, L is said to be odd or of type I. If L is a free Z-module of rank n and L is a subgroup (Z-submodule) of L, then L is a free Z-module of rank n' < n, [Ja2, Ch. 3]. It follows that if L has a geometric structure (-I ), then L together with the restriction of ( I ) to L X L is itself a geometric lattice. We describe this situation by saying that L is a (geometric) sublattice of L, We say that L is a fiill sublattice of L in case rank(L') = rank(L). If L is a full sublattice of L, then the index [L : L] of L in L; that is, the number of cosets of L in L is finite. There is a simple way to compute this index: Let ..., be a Z-basis for L and / i , . . . , a Z-basis for L , Then the two bases are related by the relations matrix («¿y) defined through fj

52 ¿=1

j = l,,,,,n.

3.1

Lattices

219

It is a standard fact [Ja2, Ch. 3] that by suitably replacing the bases {«1, and { /,,...,/„ } , we can arrange to have a relations matrix (a'y), / / = E a'ije'i, ¿=1 where (a' ,) is of the form

d'y

and the d, e Z^. satisfy ¿¡\dj if i < j. Then L / L = Z /d ^ l e • • • ® Z/d„Z, and therefore [L : L'] = d, • • • d„ = det(a'y). Since det(a'y) = |det(a,y)|, (1)

[ L :L '] = |det(a,y)|.

The reader will observe that in obtaining (1) we use only the facts that L', L are free abelian groups of the same rank n and that L' c L. Now using the bilinear form, det(L') = dct{{f!\fj)) = det(d,(e;.|e;.)dy) = dl ■■■ dl det((e'|c})) = dl • • •

det(L ).

Thus (2)

det(L') = [L :L ']^ d e t(L ).

5. Dual lattices. Let L be a free Z-module of rank n. We denote by L° the dual module of L. By definition, L° is the Z-module of all homomorphisms of L into Z. Let e = [e^,. . . , be a basis of L, and define ef e L° for all 1 < i < n by (Cj, e f ) = Sij

for all 1 < j < n.

Clearly e* — { e f , . . e f } is a basis of L°, which shows that L° is free of the same rank as L. The basis e* is said to be dual to e. Suppose now that L is a geometric lattice. For each jc e L we define G L° by A:®(y) = (3:|y) for all y e L. The mapping x ^ of L into L°

220

Lattices and Finite Root Systems

is a homomorphism of modules and is injective if and only if L is nonsingu­ lar. If this is the case, we can identify L with a submodule of L°. Since L and L° have the same rank, [L®: L] is finite. This index is called the index of duality of L and can be computed as follows: By the definition of e* we have

y=i and hence from (1) [L»:L] = |det(L)|.

(3)

Especially important is the case |det(L)| = 1.

= L, which happens if and only if

A geometric lattice L is said to be unimodular if det(L) = ± 1. From the preceding remarks it is clear that Proposition 1 The following conditions are equivalent: (i) L is unimodular, (ii) L° = L {under the above identification). (iii) The mapping x •-> jc° is an isomorphism o f L onto V . Unimodular or not, the bilinear form ( I ) of a nonsingular geometric lattice L extends to a nonsingular bilinear form on the extension V = of L. One has a canonical identification of V and V* through x x®, where x°(y) = (x|y) just as before. Then L® can be identified with the following subgroup of V: = [v ^ V\{v\x) ^ Z f o r a ll x e L } . 6. Direct sums. If {L¿,{'\‘)¿), / = 1 ,..., r, are geometric lattices, then we may form a new geometric lattice by considering the Z-module L = © L, ¿=1 with symmetric bilinear form (• | ♦) defined by

¿= 1

E y, = E (^ily,), i=l I /=1

for all X,, y, e L,-.

3.1

Lattices

221

Evidently rank(L) = Erank(L,), sgn(L) = E sgn(L^), det(L) = ndet(L^), L is even if and only if each is even, L is positive definite if and only if each

is positive definite.

It is easy to identify L° with One identifies with the Z-valued linear functionals of L that vanish on Then the linear mapping x° of L L° takes into L® for each i. We have [L “ :L ] = n m - L , ] . ¿= 1 The lattice (L,(*|*)) thus obtained is called the orthogonal sum of the family (L^,(-I*)/), i = 1 ,..., r. To describe this situation, we write L = JL L,. /=1 Example 1 (Matrix representation of lattices) Let B = be any n X n symmetric integral matrix. Let L be a free Z-module with basis ..., and define (*1*) on L X L by {e^\ej) = B^j. This makes L into geometric lattice that is nonsingular if and only if det(5) # 0 and even if and only if B^^ is even for all i. The isomorphism problem amounts to determining when symmetric integral matrices B' and B are related by invertible integral matrices A so that B' = BA. This prob­ lem is far more difficult (and interesting) than first appears, and in one form or another has a long history in number theory. For more on the arithmetic and geometric properties of lattices see [Se2] and [CS]. Example 2 (Some unimodular type I lattices). Let m, e and consider with a standard basis the bilinear form defined by the matrix m

-1 -1

and

Lattices and Finite Root Systems

222

We denote the lattice together with this bilinear form by This is an odd geometric lattice of rank m + «, signature m - n, and determinant ( - D". For convenience we denote by and by I". It turns out that every indefinite odd unimodular geometric lattice is of the form ^ 0 If L is any geometric lattice and we let L q == {x e L\(x\x) = 0m od2}, then it is obvious that L q is a subgroup of L and that 2L — {2x\x e L} c Lq. Thus L q is an even full sublattice of L. If L is even, then of course L q = L. If, however, L is odd, then [L: L q] = 2, and detiLg) = 4det(L). To see this, let X q ^ L with (jcqIjco) odd. Then, if y e L with (y|y) odd, y - X q E: L q. So we see that L = L q \ J ( L q + X q). The value of det(Lo) follows immediately from (2). The simplest example of this procedure for constructing even lattices is the construction of the lattices. Example 3 (Lattices of type D/) Beginning with I/", we construct L := {x: 0 I/‘l(jc|jc) = 0mod2}. Here L is an even lattice of determinant 4. L is called the lattice of type Di. The terminology comes from Lie theory and E. Cartan’s classification of simple Lie groups and Lie algebras. In Section 3.7 we will construct the Lie algebra of type Di directly out of this lattice. If £ i,. . . , £/ is the standard orthonormal basis of I'l' and / > 2, then the / elements •“ ^/-1 form a basis of L. Each of these Z elements has norm 2. Norm 2 elements play an important part in the theory of lattices, particularly as they appear in relation to Lie theory. An element x = e l/" has norm 2 if and only if X is of the form ±e- ± Sj with i ¥=j . There are 2(/^ - /) of these elements. By definition they all lie in our lattice L. The matrix ((«¿1«^)) determined by the basis above is 2 -1

0 B =

-1

2 -1

-1

2

0 -1 -1

2 -1 0 0

-1 2 -1 -1

0 -1 2 0

-1 0 2)

3.1

Lattices

223

Finally, let L° be the dual of L. Recall that [L°: L] = |det(L)| = 4. Thus /L is isomorphic either to Z/4Z or to Z/2Z X Z/2Z. Let co = e Q Then (ajo)) e Z for all / so that o> e L°. If / is odd, it is clear that po) ^ L =>p = 0 (mod 4). Thus lP

^ 1

77"" 4Z

if / is odd.

On the other hand, if / is even, then lo) e L. It follows that if L °/L - Z/4Z, then there exists y e L° such that 2y = o) mod L. But this is impossible since ^ L°. Thus L °/L - Z/2Z X Z/2Z

if / is even.

Example 4 (Lattices of type Ai) Fix / e Z+, and form the lattice I/+i with the standard orthonormal basis £i,..., £/+i. Define //+1 L := I X; n,e,- E « i = 0 • L is a subgroup of

and it is easy to see that

defined by

a: := 6: - 6: form a basis of L as a Z-module. Thus L is a sublattice of rank /. The definition of L guarantees that L is even and, since it is a subgroup of it is positive definite and is called the lattice of type Ai. Relative to the basis «1, .. ., the matrix of L is

-1

-1 2

-1

B = -1

2 -1

Let z •=

+ *• •

^ I /+1? and define

-1 2

224

Lattices and Finite Root Systems

One checks by direct calculation that (cojay) =

from which we have

0 ••• 0 From the matrix of

one easily sees that

det(y4/) = 2det(y4/_i) - det(y4/_2),

with d et(^ i) = 2, d e t(^ 2^ = Thus [L°: L] = det(^/) = / + 1 and the in­ dex of duality of is / + 1. Since (o^ has order / + 1 modulo L, we have lV l

-

Example 5 (The hyperbolic plane) The simplest example of an indefinite even unimodular lattice is the so-called (by E. Artin) hyperbolic plane U, Here U = Ze^ Ф Ze2, and the matrix defining ( I ) is (?

:) ■

This lattice has rank 2, determinant - 1 , signature 0, and is even and indefinite. It is a useful building block in the classification of indefinite unimodular lattices [Se2]. Up to now we have not seen any examples of even unimodular positive definite lattices. In fact such objects must have rank divisible by 8 (not obvious). The next example gives a method for constructing one for each rank of the form Sk. Example 6 (The lattices Fix a positive integer k, and set n = Ak. Let L = subgroup of Q " with the standard inner product. Let 0 mod 2}, let v ^ Q”, and let r =

, and view this as a L

q

=

{x

e

L |( x lx ) =

:= ZV + L q.

From Example 3 we see that is a group halfway between the lattice D,Ak and its dual D®*. Let r ' be the subgroup of Q comprised of elements satisfying for all 1 < i, j < n. (4)

2?,- s Z,

(5)

Qi -

Qj

e Z,

n

( 6)

^ = 0 mod 2. ¿= 1

3.1

225

Lattices

We show that T = F : (r c F): Clearly y e F (since n = 4A:). Suppose now that x = E7=i«,e, ^ L. Then X e L q <=» T,nf = 0m od2 <=> = 0mod2. Thus L q = I E «¿e,; n, G ZiEn, = 0mod2 ki = 1

(7)

In particular L q c F , and hence F = l u + L q F'. (F c F): Suppose that x = ^ F', Qi e Q. If some e Z, then e Z for all i [because of (5)] and hence x ^ L q [because of (7) and (6)]. If no Qi e Z, then there exist integers ..., such that X = $21(1 + 2m^)Si = i; + [because of (4)]. On the other hand, because of (3.6) n

$2 i ( l + 2 '” /) = 2A: + $2'^/ = 0mod2. /=1 Hence

e L q by (7), and therefore x e F.

This finishes the proof that F = F'. Moreover it shows that every element of F is congruent either to 0 or to i; modulo L q, Thus [F: L q] = 2. Let us show that F is a geometric lattice. We have Ak {v\v) = — = f c e Z . 4 If X e L q, X = En^e^, n, e Z, then En, = 0mod2, and hence (ylx) = i E ^ i =\ Thus ( I ) takes integral values on F x F. Notice also that for x g L q we have {v -\r x\v + x ) = (¿;|i;)mod2 and hence F is even if and only if k is even, or equivalently, if and only if n = 0 mod 8. We know that F is positive definite (being included in Q ® /^). Finally, recall that [L: L q] = 2 = [F: L q]. Thus 1 = det(L) = 4det(Lo) = det(F). This proves Proposition 2 For each positive integer k the geometric lattice and unimodular.

is positive definite^ even^ □

226

Lattices and Finite Root Systems

Example 7 (The lattices Ei, I = 6 ,1 ,8,9,10) Let be the free Z-module of rank 10 with basis a_i, a^, ag. Consider the following diagram (this is a Coxeter-Dynkin diagram; see Section 3.4). 8

?

-o -o -o

-o -o

(

O - 0 - 0 - 0 ^lo) “ 1 0 1 2 3 4 5 6 7 We give £’io ^ geometric lattice structure by defining (-I*) E^q x E^10 Z as follows: For all - 1 < i, j < 8, (aja^) = (a,lay) = joined (ajay) =

2, - 1 if / ^ y and the /th and ;th node of the diagram above are by an edge, 0 otherwise.

We define geometric sublattices E¿, i = 6, 1, 8,9 of E^^ as follows: 8

£ 5 = 0 Za,-, ¿=3

8

^7 = 0 Za^, /=2

8

F^g = 0 Za^, i=\

8

£ 9 = 0 Za^. i =0

We could (and later will) attach to each of these four lattices a diagram that carries the information of the corresponding bilinear form. This is done by removing an appropriate number of nodes from the original £^0 diagram in the obvious way. We will later see that the lattices £g ,. . . , E^q are intimately related to Lie theory. Corresponding to each of E^, £ 7 , and £ g , there is an exceptional finite-dimensional simple Lie algebra. £9 corresponds to the affine KacMoody Lie algebra E\. The lattice £^0 corresponds to a hyperbolic KacMoody Lie algebra that is of much interest to physicists. We study these lattices in a number of steps ( - 1 . . . 3). [1] It is convenient to start with £g. Let . . . , eg be an orthonormal basis of I^, and define ..., ^ by Pi = ^i+i - ^/+2. ^ 7 = i ( £ l + eg) -

1 < / < 6, K «2 +

• • • +£7).

^8 = e? + eg. It is trivial to verify that usual inner product of Q

• Pj = (ajay) for all 1 < /,; < 8 (here • is the Ig), and hence £g = ©f^iZjS,. Using the j8/s,

3.1

Lattices

227

we are now going to show that ^8 —Tg (see the previous example for notation). It is clear that eZ/S^cTg. Conversely, suppose that x = (jCj,... , Xg) belongs to Fg. By using Pj if necessary, we can eliminate jCj. The resulting vector lies in L q [see (4)] and has = 0. By means of j8g we can assume that the sum of its coordinates Ef=2^/ = 0*We then have (0,X2, . . . , Jig) =X2(s2 - £3) + (^2 + ^3)(^3 - ^4) + • = X2)3i + (X2+X^)P2 + • • •

+ (^2 + **• +Xj){Sj + {x2

+ •••

+Xj)l3^

£g)

e eZ/3,-.

We conclude that is an even unimodular positive definite lattice. [0] To determine the structure of Eg, we consider the element a = qiq + 2ai + 3a2 + 4a 3 H- 50:4 + 6a^ + 4a^ + 2aj + 3ag. The coefficients of a have considerable significance as we will have reason to see later in Section 3.5. For now we note that they have the remarkable property that each n¿ is half the sum of the n/s for the nodes j that are adjacent to i in the diagram —for instance, «5 = 6 = |(5 + 4 + 3) = ^(«4 + «5 + n^). Writing this as 2n,. = (a ,|a ,> ,. = - 52 («,1«;)«; leads to

(a,la) =Ia,[52 j =0, i =0,...,8. In particular (a\a) = (Lw/aja) = 0, so a is isotropic. Since {a, a^} is a basis of Eg it follows that Eg is singular; that is, det(£9) = 0. We also observe that since E^ is positive definite. Eg is positive semidefinite. [-1 ] Consider the basis { - a _ i - a, a, a^,. . . , ag} of E^q. The matrix of Eiq with respect to this basis is given by 0

/M« 0 1

1 0

where M,g is the matrix for Eg Thus E jo = E^ ± U, where [/ is a hyperbolic plane and detCEjo) = det(£g)det(t/) = - 1 , showing that E^q is unimodular. E^q is even, and its signature is sgn(Eg) + sgn(t/) = 8. An example of an element of E^q of negative norm is given by a _ j + 2a. Summarizing, E^q is a rank 10 (indefi­ nite) even unimodular lattice of signature 8. [2] Ej, being a sublattice of Eg, is positive definite and even. To compute its determinant, we write a above in the form a = «0 +

228

Lattices and Finite Root Systems

Let L '= Z2aj 0 Z«2 0 * * * 0 Zag. Then {)3, «2» • • • ?^8^ ^ basis of L and [^gi L] = 2. Writing p = a - aQ immediately gives (a ,1)8) = 0, for 2 < / < 8, and ()S|)8) = 2. The matrix of L in this basis is therefore M,

o\ 0

y0

•••0

2/

where M7 is the matrix for E-^ and hence det(L) = 2 det(£7). On the other hand, det(L) = [£^gi L p det(jEg). Thus det(L) = 4, and therefore det(^7> = 2. [3] As above, is positive definite and even. Let )8 be as in [2] and write )3 = 2«! + )8'. Set L := Z3q'2 0 Zo:3 0 • • • 0 Zag. Then {)8', « 3, . . . , ag} is a basis of L', (a ,1)8') = 0 for 3 < i < 8, and ()8'|)8') = 6. Thus det(L') = 6 det(£g). On the other hand, [E^: L ] = 3 and det(L') = [E^: L Y so that det(L') = 18. Combining these two, we conclude that deti^^) = 3[4,5,6,7 ,8] The reader who is finding this game amusing may continue by removing nodes 3 ,4 ,5 ,6,7 in succession, obtaining lattices that might be called ^ 5, J&4, JE3, E2, El with determinants 4,5,6,4,2,. In fact it is easy to identify these with the lattices of type D5, .^4,^42 ± A i , A i ±A^, and A^, respectively. (These games, 24 in all, can be found in Exercise 3.6.) If L is a geometric lattice, we define for all n e Z, L(n )

: = { a e L | ( q :| q : ) =

n)

.

Proposition 3 Suppose that L is a definite geometric lattice. Then (i) each set L{n) is finite, (ii) Aut(L) is a finite group. Proof. Replacing (*| ) by - ( I*), if necessary, we may assume that (*1) is positive definite. Then L(n) = 0 if n < 0, so we consider L(n) for n > 0. There are two ways to see that L(n) is finite, each with its own merits 1. Let £ = be the realization of L. Then E is a euclidean space, L is a discrete subgroup of E ^ d for each n > 0, S{n) •= {x e E\(x\x) = /1} is a sphere of radius yfn . Since S(n) is compact and L is closed discrete, L(n) = S(n)C\ L is finite. 2. (*1*) extends to a positive definite bilinear form in V = Q Lti e = {ci,. . . , e^} be an orthogonal basis of V, and let ^ Q>o be such that (eje^) = q^. Each element of L can be written as a rational linear

3.2

Finite Root Systems

229

combination of the e/s. Since L is finitely generated, we conclude that there exists an integer N such that all elements of L can be written as integral linear combinations of e\ , e ', where e- — N for all 1 < i < m. Define

L

© Ze'. i= l

If x' = e L', then {x'\x') = N~^Lnjq^. Hence for each n e Z there exist only a finite number of x' e L' such that (x'\x') < n. Since L c L, we conclude that each of the sets L{n) is finite. To prove that Aut(L) is finite, let {ej,. . . , be a basis of L. Define e Z >0 by Qi '= and set 5 == U Then 5 is a finite set that generates L and is stable under Aut(L). Two automorphisms of L are equal if their actions on S coincide (because S spans L). But there are only finitely many ways in which Aut(L) can act on S (because S is finite). Thus Aut(L) is finite. □

3,2 FINITE ROOT SYSTEMS In this section we introduce the beautiful combinatorial objects called finite root systems. Historically these first appeared in the analysis of finite-dimen­ sional simple and semisimple Lie algebras. Since then they have been used also in the theory of lattices, the representation theory of associative alge­ bras, and the theory of singularities, just to mention a few instances. The simplest example of a finite root system is the set of norm 2 vectors of a positive definite geometric lattice L. Usually root systems appear in the theory of Lie algebras during the classification of split simple and semisimple Lie algebras. These algebras are decomposed into what amounts to a triangular decomposition, and the corresponding root systems are finite root systems in the special sense defined below. The classification problem reduces to classifying finite root systems. Our approach is the reverse. We begin by defining finite root systems, determining their basic properties in Sections 3.2 and 3.3, and using them to construct simple and semisimple Lie algebras in Section 3.7. The notion of a base for a finite root system, which is the main subject in Section 3.3, plays an important part in the theory. In the first place, bases allow us to classify finite root systems. Equally important for our purposes, they allow us to write down a canonical presentation for the corresponding Lie algebras, which in turn is the starting point for generalizing the finite-dimensional semisimple Lie

230

Lattices and Finite Root Systems

algebras to infinite-dimensional counterparts. The abstract theory of root systems was initiated by Witt [Wil]. We begin with a few generalities about reflections. Let L be a vector space over a field IK of characteristic 0. An endomorphism 5 of F is called a reflection (of K) if 1. = 1 and 5 ^ 1 , 2. s pointwise fixes a hyperplane (= subspace of codimension 1) H of V. If V is finite dimensional, it follows from this definition that the minimal polynomial of 5 is — 1 and that s admits - 1 as an eigenvalue of multiplicity 1 and 1 as an eigenvalue of multiplicity dim F - 1. In particular the minimal polynomial of s has no multiple roots, and all of its eigenvalues lie in IK. As a consequence 5 is a diagonalizable automorphism of F (see Section 7.1). Furthermore dct(s) = - 1 . If a e F is an eigenvector of s with eigenvalue - 1 (i.e., if sa = - a \ then V=Ka®H,

(1)

where H = {x ^ V\s(x) = x}. We call H the hyperplane of fixed points of s. Given an element a e F, a ¥= 0, we say that an endomorphism 5 of F is a reflection in a if 5 is a reflection of F and sa = —a. Notice that there are as many reflections in a as there are hyperplanes H of V supplementary to a [i.e., satisfying (1)]. In particular this number is infinite whenever dimtt,(F) > 1. Let 5 be a reflection in a, and decompose F as in (1). Let F* be the dual space of F, and let F X F* -> IK be the natural pairing. Define «V

^

by

(a, a ^ ) = 2, = ( 0),

and consider the endomorphism ( 2)

of F given by >

X - (x,a^)a.

Then it is evident that s = Conversely any decomposition (1) of F into a hyperplane H, and a one-dimensional supplement IKa gives rise to a reflection in a by using (2). Note When using this notation, it is inevitable that we have to make a choice whether to define (., .> from F X F* to IK or from F* X F to K. We have chosen the former, but later, when F is finite dimensional and we talk about reflections on F and F* at the same time, we will identify F** with V

3.2

231

Finite Root Systems

in the usual way and write reflection on V* in the form x* (a, x*)a^.

x* —

Proposition 1 Let V be finite dimensional, and let and S2 be reflections of V in the same a. If Si S2, then S1S2 is not a semisimple transformation o f V. Proof Let Hi and H 2 be the fixed point hyperplanes of tively, and set Í = Clearly

and ^2, respec­

1. det(0 = 1, 2. t pointwise fixes Ka 0 (//1 fl //2)If t is semisimple then t becomes diagonalizable by a suitable extension of the base field K (Section 7.1). It is clear then that conditions 1 and 2 imply that t = 1 and hence that Si = S2□ Proposition 2 Let ^ be a subset of V, and let Aut(A) be the group o f automorphisms of V stabilizing A. Suppose that A is finite and that A spans V. Then (i) Aut(A) is a finite group, (ii) For each a e A there exists at most one reflection s o f V in a such that s e Aut(A). Proof (i) If two automorphisms of V coincide on A, then they are identical (since A spans V). Thus Aut(A) is finite whenever A is. (ii) Let a e A, and suppose that Si and ^2 are distinct reflections in a that stabilize A. Then t = S1S2 ^ Aut(A) is an automorphism of V of finite order [by (i)] and hence semisimple by Maschke’s theorem [Ja2]. Now (ii) follows from our previous proposition. □ Suppose now that V carries a nondegenerate symmetric bilinear form ( 1 ) and that 5 is a reflection of F in a which is also an isometry of (*|*). Then, if we decompose F = Ka 0 / / as in (3.8), we have for all x e / / (a\x) = ( 5( a)U (x )) = ( - a l x ) , so (a|x) = 0. Thus Ka = H ^ , and we have (3)

V=Ka±H,

It follows that the restriction of ( I*) to both Ka and H is nondegenerate. In

232

Lattices and Finite Root Systems

particular (a\a) ¥= 0. Conversely, if a e F is nonisotropic, then the restric­ tion of ( I ) to / / := IKa-^ is nondegenerate and then the reflection in a given by the orthogonal decomposition (3) is an isometry. This is the unique isometry of V that is also a reflection in a. It is called the orthogonal reflection in a and is denoted by If one uses (• | •) above to identify V with F* (so that x e F < ^ ( | A : ) e F * ) , then it is clear that la (4)

(a|a)

In other words, (5)

s^(x) = x

2(x\a)a (ala)

Let F be a flnite dimensional vector space over a field K of characteristic 0. Assume that dimB^(F) > 1. A subset A of F is called a finite root system in F if the following conditions hold: RSI: A is finite, 0 ^ A, and A spans F. RS2: For each a e A there exists a '' e F* such that ( a , a ' ' ) = 2 and such that the reflection of F defined by ^ ^

-

(v,a-}a

stabilizes A. RS3: <j8, a " ) ^ Z for all a and p in A. The reflection whose existence is assumed in RS2 is necessarily unique (by part (ii) of Proposition 2). In particular a" is uniquely defined for each a, and this makes RS3 meaningful. In the sequel we will write instead of The dimension / of the space F spanned by A is called the rank of the root system. We refer to F as the ambient space of A and sometimes write (A, F ) to emphasize this. The elements of A are called roots. (This terminol­ ogy will be made clear when root systems are linked to Lie algebras.) For a e A,a'' is called the coroot associated with a. The set of coroots is denoted by A". The elements of GL(F) that stabilize A are called automorphisms of the root system A. The group of all these automorphisms is denoted by Aut(A). Axiom RSI together with part (i) of Proposition 2 imply that Aut(A) is a finite group. By assumption, if a e A, then e Aut(A). Thus {r^;|a e A} generates a subgroup W = IF(A) of Aut(A). This group is called the Weyl group of the root system A and is of fundamental importance in Lie theory.

3.2

Finite Root Systems

233

Since rj^a) = —a for all a e A, it follows that A = —A and that —1 e Aut(A). Whether or not - 1 is in depends on the root system (see the ^3,^3 Examples). Notice that if Aa e A for some A e (K^, then is a reflection in a stabilizing A. By the above observation In particular Ta ^ ' ' - a a e A. Let V and V' be IK spaces and A and A' root systems in V and V', respectively. We say that A and A are isomorphic if there exists a linear isomorphism F ^ F ' such that (A) = A. We write A = A to describe this situation. Note that if o- e Aut(A), then e Aut(A) and that the map or cr(f>~^ is a group isomorphism from Aut(A) onto Aut(A). More­ over, if a e A, then r^(f>~^ is a reflection of F ' in <^(a) stabilizing A, and therefore (6)

4>rA ‘ = r.
for all a e A.

This shows that under the above isomorphism IF(A) is mapped onto 1F(A). Proposition 3 Let is he a root system in F. Then IF(A) is a normal subgroup o f Aut(A). Proof If cr G Aut(A), then cr g GL(F), and we can view a as an isomor­ phism of A into itself. Our present result then follows from (6). □ As our first examples of root systems we have the following: Proposition 4 Let L be a positive definite geometric lattice o f rank I with rationalization F'. Suppose that the set L(2) of norm 2 vectors in L is nonempty and spans a subspace V of V'. Then L(2) is a finite root system in V and its reflections are the orthogonal reflections r^ o f V in the elements o f L(2). Proof By Proposition 3.1.3 L(2) is finite, and hence RSI holds. For each a e L(2) the orthogonal reflection r^ in a is 2(x\ a)a r ^ - . x - ^ x ----- , I , = (a|a)

- (jc|a)a.

Evidently r^ is an automorphism of L and hence stabilizes L(2); therefore RS2 holds. Identifying F* and F by ( 1 ) , we see from (4) that = a. Thus for G L(2), = (j8|a) ^ Z, and RS3 holds. □ Example 1 Figure 3.1 illustrates the root system A formed by the vectors L(2) of the lattice of type A^. The roots are the vectors from the centre 0 to the vertices of the cube-octahedron. The edges join vectors a, p

234

Lattices and Finite Root Systems

A, 0-0-0

B3 o — o = ^ o Figure 1.

for which a./3 = 1. The planes through which the reflections take place pass through the diagonals of opposite faces of the surrounding cube. The Weyl group is isomorphic to the symmetric group 54. In fact, if T denotes the set of four pairs of opposite triangular faces of the cube-octahedron then each of the six reflections generated by root pairs ± a acts as a distinct transposition on T, thereby giving all of the transpositions. If we adjoin to A the vectors from 0 to the centres of the square faces, we obtain another root system A' (later we will see it is of type B^) with 12 + 6 = 18 roots. The three reflections determined by these new roots evidently lie in Aut(A). However, their product is the mapping - 1 (central inversion), which acts trivially as a permutation on T and is not in W(A). On the other hand, the product tt of any two of them is a 180° rotation about the line L that is the intersection of their reflecting hyperplanes. The same line L is the intersection of two orthogonal reflecting hyperplanes of A so that tt e IF(A). Thus W{A) = {±1} X IT(A), [PT(A'):IT(A)] = 2. In fact we have W{A) = Aut(A') = Aut(A). The pair of graphs labelled

and ^3 are explained in Section 3.4. The

3.2

Finite Root Systems

235

reasons for changing the labelling of the roots in passing from -^3 to become clearer when we discuss bases in Section 3.3

will

Example 2 {Ai) (See Example 3.1.4) The set of roots L(2) is precisely the set of vectors of the form - Sj with i j, and the orthogonal reflection r in Si - Sj is the linear automorphism of V defined by

In other words, r is the linear automorphism defined by the transposition Si <-> Sj of the basis {e^,. . . , ei+J. It follows that W = 5^+^. Although we have defined root systems for spaces over arbitrary fields of characteristic 0, it is more convenient to develop the theory for spaces over the rational numbers Q. At the end of the next section we will see that it is easy to reduce the general situation to the rational case. However, until further notice, V denotes a rational vector space of dimension / and A c F a finite root system. Proposition 5 Let tx he a root system on V. Then there exists a positive definite symmetric bilinear form (*|*) on V that is invariant under Aut(A). I f V and F* are identified via such a form, then a" = 2a/ {a\ a) for all a e A. Proof Let (.,.) be any positive definite symmetric bilinear form on F. Define (•!•): F X F ^ Q by (^ly ) =

L

forall

e F.

o-e Aut(A)

It is clear that this form has the required properties. The statement concern­ ing a = 2a/{a\ a) now follows from (4). □ Proposition 6 Let ^ C.V be a root system. Then (0 A" is a root system in F*; (ii) PF(A) = W(A^) and Aut(A) = AutCA""), the isomorphisms being de­ fined by inverse transposition (see below); all a ^ A and A = ^ (under the canonical identifica­ (iii) q; = Qjv V tion o f V and F**). Proof Let (*1*) be a positive definite form on F that is invariant under Aut(A). Using (*|*) to identify F and F*, we have (as above) a" = 2a/(a\a). This shows that A" c F* satisfies RSI.

236

Lattices and Finite Root Systems

For each a e Aut(A) define
<£;,(r*(i;*)> =

for all y e F, r* e F*.

(
^ (y ),i3 '') = |(7 ‘(y)!

2(vla(/3)) (^1^)

2P

\

W )

I

2(v\o-(/3)) ( ( r(^ )k (/3 ))

= (v,{aiP)y}, and hence ( 8)

a*{p'^) = ( a { ^ ) y .

It follows that O'* stabilizes A"; in particular r* stabilizes A'" for each a e A. Now for i; e F and v* e F*, ( v , r : { v * ) ) = (r^(v), v*y = (v - ( v , a ^ } a , v * y = (u,v* - ( a , v * ) a - ) , which shows that r * : i;*

i;* - < a ,i;* > a''.

Thus r* is a reflection in a" and = a (see the note on pairings in the previous section). This proves RS2 and part (iii). Finally, for all a ' ' , ^ A'",

which proves RS3. The mapping o- >-> o-* is evidently a homomorphism of Aut(A) into Aut(A") [resp. W(A) into W(A" )]. It is seen to be injective from o-** = a and surjective by reversing the roles of A and A". □ The root system A" is called the root system dual to A. One should be aware that the mapping A ^ A" defined by a a '' is not the restriction of a linear map in general. In fact it is obtained by inversion with respect to a suitable (/ - l)-sphere.

3.2

Finite Root Systems

237

Example 3 The root systems dual to ^3 and are illus­ trated in Figure 3.1. Notice that for a e A(y43), a" = 2a/ ( a\ a) = a, whereas for a e A(^3) \ a" = 2a/ (a\ a) = 2a. The root system AiB^Y is of type C3. We leave it as an exercise to verify that AiB^) and A(^3)" are not isomorphic. The reader may wonder about the role of ( •| •), which seems to be quite arbitrary and yet quite significant. In fact it is essentially unique. We leave the precise sense of the uniqueness to Propositions 3.4.7 and 3.4.8. Some indication of this already appears in the following discussion. Let (*|*) be a symmetric bilinear form on V as constructed in Proposition 5. Then

(9)

2iß\a) ( a la )

for a,/3 e A.

Recall, on the other hand, that in the euclidean space U between a and is given by the expression

cos(6) =

the angle 6

(i3|a) [(a |a )(/3 lj8 )]

1/2

Hence ( 10 )

( ß , a ^ ) ( a , ß" ) = 4cos^(ö).

The left-hand side of (10) is the product of two integers, whereas the right-hand side is in absolute value < 4. Thus the possible values of 6 are severely restricted. We collect below all information that follows from the above. There are two main cases to consider: Case 1. a and ß are proportional. Suppose that ß = Aa, A e Q. Since both ( ß , a " ) and ( a , ß " ) are integers, it follows from (9) that both 2 A and 2A“ ^ are integers. We conclude that A e {± ± 1, ± 2}. These are the only possible multiples of a root. [In the sequel we will say that a root a g A is indivisible if ^ A. If a root system consists only of indivisible roots we will say that it is reduced. Otherwise, we say that it is nonreduced. We denote the set of reduced roots of A by Case 2. a and ß not proportional. The only possible solutions of (10) (up to interchanging a and ß) are given by Table 3.1.

238

Lattices and Finite Root Systems

Table 3.1

0

0

1

1

1

2 3

1 -1

-1

-1

-2 -3

-1

0 IT/ 2 7t/3 7t/4 ir/b 27t/3 3-77/4 57t/6

(a|a)/(j8li3) Undefined 1

2 3 1

2 3

Order of 2 3 4 6 3 4 6

We determine the order of as follows: Both act on the and space = Qa © Qp, and on this space the transformation has the prescribed order. There are two simple ways, one geometric and one alge­ braic, to see this. Embedding ^ into real 2-space U one can use the fact that the product of two reflections is the rotation through twice the angle between their reflecting hyperplanes. A completely algebraic way is to observe that the matrix for relative to the basis [a, of ^ is

\

-1

)

and its characteristic polynomial + (2 - 4cos^0)A + 1 has roots Now let and be the hyperplanes of V consisting of fixed points of and respectively. Let V' = Then L = K' ® Since acts like the identity on K', it has the prescribed order as a transformation on V, The rest of the information contained in the table is clear. Here are a few more facts about root systems. Proposition 7

is a reduced root system. Proposition 8

Let a, p ^ A be nonproportional roots. I f (a|j8) > 0, then a — p is a root. If (a\p) ^ Oj theft ot “1“ ^ is Cl foot. Proof. If (a|)3) > 0, then <j8, a '' > > 0 and > 0. Since their product is less than 4, at least one of these two quantities equals 1. If a = )3 - a e A, and hence a - /3 e A. If = 1, then r^a = a - ¡3 ^ A. The second assertion follows from the first by replacing )3 by —/3. □

3.3

Bases for Finite Root Systems

239

Let i= , r, be K spaces, and for each i let be a root system in Let V = and identify each L; with a subspace of V. Each can then be thought as a subset of V. We let A == U [=iA^. Let us show that A is a root system in V. It is clear that A spans V. Let a e A^ c L; c V, and let a" e K* be defined by means of the canonical identification F* = Then is a reflection on V stabilizing A (it acts like in Vi and like the identity on Vj, j i). Also = (0), and (A^,a ") (z I . Thus RS2 and RS3 hold for A. We denote the reflection in a by both for a e A^ and for a e A. The root system A in F constructed above is called the direct sum of the root systems A^. To describe this situation, we write A = Ai V

V A.

Let W d e ^ te the Weyl group of A and WK the Weyl group of A^. The subgroup of W generated by the r^, a e A^, is evidently isomorphic to (the restriction map w ^ w\y. being the isomorphism). S i n c e a n d commute if a g A^, j8 e Ay with i ^ j \ we conclude tjiat W = W j X • • • X Wj , - Wi X • • • XW^. There is no harm in identifying and and we do so freely in the sequel. A root system A is called decomposable if there are root systems Aj, A2 such that A = Ai V A2. Otherwise, A is called indecomposable. Any root system can be decomposed uniquely into indecomposable root systems (see Section 3.3). Proposition 9 A root system is decomposable if and only if there exist nonempty subsets A^, A2 of A such that (i) A = Ai UA 2, (ii) (aljS) = 0 for all a e A^ and j8 e A2. Proof (=>) The proof is clear from the definition. («=) Let Fi and F2 be the span of A^ and A2, respectively. It is trivial to verify that Aj and A2 are root systems in F^ and F2, respectively, and that ASA1VA2. □ 3.3 BASES FOR FINITE ROOT SYSTEMS Let A be a finite root system with ambient space F. It is evident that we can choose a basis of F from A. However, it is possible to do this in a very special way that naturally divides A into positive and negative subsets A+

Lattices and Finite Root Systems

240

and A_ with A_= -A+. The key result (Proposition 5) is that any two of these special bases are translates of each other by the Weyl group W. Thus the structure of a base is an invariant of W. This leads immediately to the definition of the Cartan matruc of a finite root system and the introduction of the Coxeter-Dynkin diagrams. Let A be a root system in a I -space V. A subset n = [a^,. . . , a:/} is a base for A if Bl: n is a basis for K as a vector space, B2: every root can be expressed (necessarily uniquely) as an integral linear combination of the a^, where all the coefficients are either nonnegative or all are nonpositive (loosely, all the coefficients have the same sign). We now return to the case K = Q. Throughout this section (-I*) denotes an Aut(A)-invariant positive definite symmetric bilinear form on V (see Proposition 3.2.5). Proposition 1 Every finite root system has a base. Proof We start by putting a total ordering > on the rational space V that makes V into a totally ordered vector space. By definition this means that > is a partial ordering on V and that 1. for all v,w ^ V either u > w,w > v, or w = u, the possibilities being mutually exclusive; 2. if V > w and X e F, then v + x > w + x; 3. if V > w and c G Q+, then cu > cw. For instance, if we choose a basis . . . , Ui), we may define the lexico­ graphical ordering by declaring that for two unequal elements, 'La¿v¿ > Lb^Vi if and only if at the smallest i for which a^ b^ we have a, >b, . ^ In any case let V+= {z; e V\v > 0} so that V = U{0}U V- (U denotes disjoint union). Let A+:=Ari]^+ and A_— Afll^-. A root a e A+ is called simple if a cannot be written as ¡ 3 y, where j8, y e A + . Let n = {«1, . . . , a^} c A+ be the set of simple roots. We reason in steps: 1. I f a, P ^ Yl, a ¥=p, then (a|j8) < 0 and (a, ) < 0: We have ex­ cluded the possibility that a and jS are proportional since then, say, P = 2a = a a and p is not simple. If (a\p) > 0, then a — p and p - a dire roots (Proposition 3.2.8). One of these, say, p - a, belongs to A+, and then p = (p — a) a shows that p is not simple.

33

Bases for Finite Root Systems

241

2. The elements o f O are linearly independent : Suppose that = 0, where not all the c, are 0. Let / ‘^= {i|c, > 0} and / “ = {i\c^ < 0}. Then both and I~ are proper subsets of k} (otherwise, 0 would belong to either or V~). Let x = Then x ^ 0, and X= c^a^. Thus 0 < (x|x) = ^ step 1, which is a contradiction. 3. Yl is a base o f A: If y e A+ is not simple, then y = a + )3, where a, p ^ A^. Then y —a = )3 > 0, so y > a, and similarly y > p. It follows by induction on > that every element of A^. is a sum of simple roots. Since A_= —A+, B2 holds; and since A spans F, O is a basis of V, □ Example 1 Let L be a positive definite geometric lattice for which L(2) 0. Consider the root system L(2) (Proposition 3.2.4). Suppose that 11 c L(2) has the properties 1. The Z-span of Yl contains L(2); 2. (a|j8) < 0 for all a, P ^ Yl, a Then n is a base of L(2). For suppose that y e L(2), y = E ^ e n '^ a ^ ’ e Z. If the are of mixed signs, we may write y = y++ y_, where y+— y_:= y - y+. Then (y|y) = (y+|y+) + 2(y+|y_) + (y_ly_). Since (y J y +) e 2Z+ and the assumption (2) gives (y+|y_) > 0, we have (y|y) > 4, which is impossible. The elements of II are linearly independent by the argument of step 2 of the last proposition and span the same rational space as L(2). Referring back to Section 3.1, this shows that the a-bases given there for the lattices L of types Ai, j g are actually bases for the corresponding root systems L(2). Proposition 2 Let ^ d V be a root system, and let S cz A be any nonempty subset. Let V' be the subspace of V spanned by the elements of S, and set A = A fl F'. Then (i) A is a root system in F', (ii) any base of A can be extended to a base o f A. Proof If a e A, then a" can naturally be viewed as an element of F'* by restriction. Then for all a, e A, r , ) 8 = ) 8 - < ) 8 , a - > a e A n F ' = A,

and it is clear then that A is a root system in F'.

242

Lattices and Finite Root Systems

Now let {«1, . . . , be a base of A', and consider a basis for V of the form {v^, q:^}. Then the base of A obtained from the lexicographic ordering of V determined by this basis contains « i , . . . , □ A subset 2 of a root system A is called a positive system of roots if S = A n 1^+ for V+'-= {v ^ V\v > 0}, where > is some total ordering on V. We have seen that the simple roots of a positive system of roots 2 form a base n in X. In fact FI is the only base of A lying in X. To see this, we introduce the height function ht := h t^ : A

(11) by

htl £ m „a\ =

E aen

Notice that /3 e A is in if and only if ht(jS) > 0. Let O ' be another base of A inside X. Then ht(a') > 1 for all a' e O'. Each a e O is expressible as a = where the m^, e Z are all of the same sign. In fact the m^> e N for otherwise a e A_. Now 1 = ht(a) = Xm^.ht(o:') =>« = «' for some a' e O'. Thus O c O ' and, in view of card(n) = dim(F) = card(n'), we have Yl = O'. Every base Yl can be used to determine a lexicographical ordering of V and then a positive system in which it lies as the simple roots. We call A+= A+(n) the system of positive roots corresponding to O. Clearly A ,(n) =

e A

/3=

E

rn^a,m ^

>

o}.

Summarizing, we obtain Proposition 3. Proposition 3 (i)

Every positive system o f roots contains exactly one base. This estab­ lishes a 1-1 correspondence between positive systems o f roots and bases. (ii) The elements of any base are indivisible. (iii) I f {«1, is a base, then (a^, aj" > < 0 for all i ¥=j. □ Proposition 4 Let Y\ be a base o f A. Then (i) { r j a G n} generates W as a group-, (ii) A^^^ = WY\ == {wa\w ^ W , a ^ O}.

3.3

Bases for Finite Root Systems

243

Proof, Let n = {aj,. . . , aj}. Let Wq be the subgroup of W generated by Tj,..., A*/ (for convenience, we denote by r^). We begin by proving ( 12 )

= W ^n.

We use induction on the height function ht = htp^ relative to Yl to prove that := A^®^ n A+c W^Yl, Let /3 e If ht(iS) = 1, then j3 e n . Suppose that ht(j8) > 1. If (jSla,) < 0 for all i, then writing )3 = e N, we have 0 < ( ^ 1^ ) = Lnii{l3\ai) < 0,

which is impossible. Thus (p\ a , ) > 0 for some i. Then (13) r,.(/3) = 13 - {p, a,'' >a, = Y,

+ (m, - )a, e A"®'*.

j * i

If rrij = 0 for all i i then ^ = m,a,. Since /3 is indivisible, /3 = a„ contrary to ht(/3) > 1. Thus some Wy ¥= 0, and hence nij > 0. It follows that r,(/3) e A+'’. Since = 2()3|a,)/(o:,|Q:,) > 0, ht(r,(/3)) < ht(/3), and we may assume by induction that /-¿(jS) e WqTI. Write r,(^) = for some w'^ in Wq. Then /3 = WqU^. with Wq = /-¿Wq e Wq. In addition by (6), -1 The conclusion of all this is that c IFoIl and that e Wq for all j3 G A™**. Since whenever /3 and Aj8 e A, we see that g for all -y G A. Thus Wq = IT, proving part (i). Finally, r,(a,) = —a,-, showing that - n c ITon, so IT o n -----ITon and A^f“ = -A'®“ c W q U . This proves that ^r®d (- ]YqTI. The reverse inclusion is obvious. This finishes part (ii). □ If n is a base for A+ and a e O, then Q a fl A+ is one of {a} or {a, 2a}. In either case we have the following simple facts:

(14)

1. stabilizes A + \(Q a fl A+) 2. r ,A ^ = - ( Q a n A ^ ) U ( A A ( Q a n A ^ ) .

The point is that for any root )3, /3 and rj^p) differ only by a multiple of a. It follows that if )8 e A+ and j8 ^ Q a f l A+, then the expression for in terms of elements of II still contains nontrivial (hence positive) contributions from elements 7 ^ n \ { a } [see (13)]. Thus r^(j3) e A +\ Q a fl A+ also. Simple as these remarks are, they play an important role in understanding the structure of W (see Chapter 5).

244

Lattices and Finite Root Systems

We have the following key result: Propositions. (Conjugacy o f bases) Let (A, V) be a root system. (i) I f li is a base o f A and w then iv(n) is also a base. (ii) Any two bases o f A are related in this way. Proof (i) The proof is obvious. To prove part (ii), let and II2 be two bases of A with corresponding positive systems A+ and A+. Let A_ := -A+ for / = 1,2. It suffices to prove that there is w ^ W with w A \ = A+ (see Proposition 3). If A + n A^_ = 0 , then A+ c A+, whereupon they are equal and hence III = Il2 by (i) of Proposition 3. Suppose that card(A+ n A^_) = n > 0. Then III ^ ^ since, otherwise, 111 c A+ and Hi = II2. Choose a e III n A^_, and consider r^A \. It differs from A+ only in that —a (and possibly —2a) replaces a (and 2a). But a e A^_, so - a e A+. Thus card(r^A^^ n A^_) < n. Since r^A \ is the system of positive roots for r^IIi, we are done by induction on card(A+ n A^_). □ This proposition says that the Weyl group is transitive on the set of bases of A. In fact it is simply transitive on the set of bases; we prove this in Chapter 5. Proposition 6 Let tsbe a reduced root system in V and let A'' c F* be its dual root system. Let Yl be a subset of A. For U to be a base o f A, it is necessary and sufficient that ••= [a"\a ^ A) be a base o f A". Moreover, if A^ and A \ denote the positive systems o f roots defined by II and II'', respectively, then a e A+ if and only if ^ A+. In other words, {A ^Y = A+. Proof Let identify V and F* via (• ). Fix an ordered basis of F*, and use its lexicographic order to define a system of positive roots A+ and a base n '' := {aj',. . . , a I } of A ''. The above choice of ordered basis of F* = F defines a positive system of roots A+ of A. Since = 2 a / ( a , a), we conclude that for all a e A, (15)

a e A+<=> a '' e A \.

We claim that II — {ai, .. .,ajf is a base of A. Since II c A+, we must show that a^ = p + y, with and 7 in A+, is not possible. Now if a^ = + 7

33

Bases for Finite Root Systems

245

with P and y in A+, then = ''a,«/ = SOthat either r^,p e A_ or r^y e A_. Assume that y is similar). Then {r^ p Y e A l, and hence by (8)

= r%p^ =

e A_ (the case of

^ a: .

By (14) (applied to A" and the base II"), it follows that P" = a ," , whence P= Then y = 0, which is absurd. This shows that II is a base of A. The rest of the proof is easy. □ Up to now most of our results have been proved about root systems under the restriction IK = Q. In what follows we show how the general situation is dealt with. Let F be a vector space over a field IK of characteristic 0. By restriction of the base field to Q c IK, we can think of F as a rational space, which we denote by F(Q). Notice that dimQ(F(Q)) = dim„^(F)dimQ(IK). A (rational) subspace U of F(Q) is called a rational form or a rational structure of F if the canonical K-linear mapping K 0Q i/

F

is an isomorphism. (This map is the unique linear map satisfying k ^ u ^ ku for all A: e IK and m e i/). Equivalently C/ is a rational form of F if 1/ has a (rational) basis that is also a basis for the K-space F. In any case notice that dimQ(i/) = dim„^(F). Example 2 Let

^ ^ be a basis of F over IK. Then i / :=

© QU^

AeA

is a rational form of F. Next we show how the concept of rational form allows us to understand the relationship between root systems (A, F ) for IK-spaces and rational root systems. Proposition 7 Let (A, V) be a finite root system where the ambient space Vis a K-space. Then the rational span Vq o f L in V is a rational form o f F, and (A, Fq) is a finite root system. Furthermore for each a ^ ^ the reflection r^ o f (A, Vq ) is the

246

Lattices and Finite Root Systems

restriction to Vq o f the corresponding reflection o f (A, F), and the coroot Vq is the restriction to Vq of the corresponding coroot of

of

Proof Let Vq be the rational span of A in V. Let a e A. Clearly a" (F q) c Q and hence a" induces an element also denoted by a " , of Fq . It is now easy to see that (A, Vq ) is a root system. It remains to show that Vq is a rational form of F. Let II = {a^, be a base of (A, Vq ), hence a Q-basis of Vq . Then II spans F (over K) since A does. Suppose, if possible, that Y, CiUi = 0, /=1

c,- e K,

where some c^ ¥= 0. Then for each jS e II, Ec, = 0, ¿=1 which shows that the matrix A = « a ,, a p ) defined by II is singular (where A is viewed as a matrix with coefficients in IK). However, we know that A is nonsingular when viewed as a matrix with coefficients in Q. Since nonsingu­ larity remains unchanged by extension of the base field, we conclude that the elements of II are linearly independent over IK and hence that II is a IK-basis of F as desired. □ Remark 1 The above proposition makes it clear that all our results about the structure of rational root systems generalize to arbitrary root systems. For example, any base of (A, Fq) is a base of (A, F ) (showing the existence of bases). Similarly any base of (A, F ) is a base of (A, F q ) , and hence any two such bases are conjugate under W. Let n = {«1, . . . , O'/} be a base of A. The Cartan matrix of A (with respect to n ) is the / X / integral matrk A = where Aij ■■= = («yl“ /) ’ Interchanging the ordering of and changes A by simultaneous inter­ change of rows i and j and columns i and j. Combining this with the IF-conjugacy of bases, we see that A is uniquely determined by A up to simultaneous row/column permutations.

3.4

Graphs and Coxeter-Dynkin Diagrams

247

We note the following properties of A: CMl CM2 CM3 CM4

A¡¿ = 2 for all i. Aij G - N if / ¥=j.

There is a positive diagonal matrix D (D = diagidj,. . . , d j, > 0) such that AD is symmetric. CMS: AD in CM4 is positive definite. For CM4 and CM5 simply observe that A diag

, (“ ii«i)

(«/I«/)

= ((a,l«y)).

In Section 3.4 we use CM1-CM3 as the definition of a “generalized” Cartan matrix. We will also see that CMl-CMS characterize the Cartan matrices of finite root systems. Notice that CMS implies that A is nonsingular.

3.4

GRAPHS AND COXETER-DYNKIN DIAGRAMS

We have many occasions to use square matrices of integers in the sequel, notably matrices like the matrices A satisfying the conditions of CM in Section 3.3. Representing these by graphs is an extremely economical and useful idea. In this brief section we show how we wish to attach a graph to a matrix and review the basic definitions and facts about graphs that we will need. There are no proofs in this section. The reader may refer to [Bg] or provide them as an (easy) exercise. Let N* = U {oo}. Let J be a nonempty set. A matrix M = is said to be combinatorially symmetric if for all i and j in J, CSl: Mij G N*, CS2: CS3: M,, = 0. Example 1 Let F be a IK-space, and let ^ family of automor­ phisms of V, For all i and j in J, define M^j as follows: ^

I order of s^Sj

if s^Sj is of finite order, otherwise,

M,- = 0. Then M =

is combinatorially symmetric.

if i ^ ;,

248

Lattices and Finite Root Systems

Let M = be a combinatorially symmetric matrix. We may represent M by a labeled graph T = (V,E,M) with vertices (or nodes) V = {vj\j ^ J] and directed edges (or arrows) E = e J X J, i ¥=j, M^j 0}, which are assumed to be marked by the quantities (M^y). Note that if M^j = 0, then there is no edge e¿J. By definition, then also will be missing. M is called the incidence matrix of F. We often think of as being M^j arrows from i to j [or of type (i, j)] and refer to them as such. Example 2

(0 M= 2 lo r;

2 0 0 00 1 0 o V2 0

J = {1,2,3} ^

0 .

There are several simplifying conventions that we will use for our graphs: 1. If M,y = 1 and Mji > 1, we omit the edge e,y. 2. If Mij = Mji # 0, we replace the edges and eJ¿ by a single undi­ rected edge labeled by their common value. 3. If M,, is small (< 4 usually), then we may elect to replace a labeled edge by a corresponding multiple edge. We also use this convention in 2. 2

Example 3 0 ^ 0

becomes O

by convention 3. O ^

2

O by convention 1, and then O => 0

O becomes 0 — 0 by convention 2, and then

O — O by convention 3. M of Example 2 now appears as

0 = 0 ^ o. Example 4

0—

0 1

0

o

3.4

Graphs and Coxeter-Dynkin Diagrams

249

corresponds to fo 1 0 io

1 0 3 1

0 1 0 0

o' 1 0 oj

In the sequel all graphs are assumed to be reduced by these conventions. We assume that the reader is familiar with the standard notions of graph theory. For definiteness we repeat the definitions that we will adopt for this book together with a few well-known facts. Let r = (V, E, M), F = (K', M'), where M and M' are indexed by J and J', respectively. 1. r and F' are isomorphic if there exists a bijection a: J ^ J' so that ^a(i),«(;) = ^ i j for all (j, y) e J X J. From this we obtain the concept of an automorphism of F. 2. F is a subgraph of F' if there is an injection a: J J' so that K(0,a(;) = for all i , j e J. 3. F is disconnected if J is the disjoint union of two nonempty sets Jj and J2 so that = 0= for all Jz) ^ Ji X Jz4. If F is not disconnected it is connected. 5. A path (of length n) in F is a sequence (/ q, ^1, • • •, of elements of J so that for each k = ¥= 0. The path is said to connect V: ^0 to V ih 6. F is pathwise connected if every two vertices are connected by some path. 7. Pathwise connected and connected are equivalent. 8. A cycle or loop in a graph is a path (/ q, ¿1, . . . , of length n > 3 for which ¿Q, are all distinct and 9. A forest is a graph with no cycles. A tree is a connected forest. 10. Two vertices (or nodes) are adjacent if there is an edge between them. 11. The valence of a vertex is the number (possibly infinite) of vertices adjacent to it. 12. Any finite tree with more than one vertex has at least two vertices of valence 1. 13. A connected graph F is bipartite if J can be partitioned into two nonempty subsets Ji and J2 so that M^j = 0 whenever /, j e or 14. Any finite tree with at least two nodes is bipartite.

250

Lattices and Finite Root Systems

Let J be a nonempty set, and let A = be a matrix with integer coefficients. We say that ^ is a Cartan matrix (some people call it a generalized Cartan matrix) if CMl: CM2: CM3:

A,.= 2 for all / g A¿j< 0 for all i ¥= A¿j= 0 <=>Aj¿ = 0

J, j in J, for all i

^ j E:

J.

Example 5 Let II = {a^,. . . , a j be a base of a finite root system A. Define A¿j for all 1 < i, j < I by A¡j = . Then ^ is a Cartan matrix. Let A he a Cartan matrix. Then M = 2I-A (where / is the J X J identity matrix) is a combinatorially symmetric matrix and defines a graph r = r(A) ^ (V,E,M). r is called the Coxeter-Dynkin diagram of A. Example 6

A =

2 -1 0 0 l:

-1 2 -1 0

0 -2 2 -1

O' 0 -1 2.

O — o =?• O — O

Example 7 Consider the finite root systems L(2) of types D,, Ai, E^, E^, Eg. (Example 3.3.1). Referring back to Section 3.1, we see the Cartan matrices of D, and A i displayed. The corresponding Coxeter-Dynkin diagrams are T{Ai):

0 -0 -

------0 - 0 O

r(A ):

0 —0 —

—0 —0 ^ O

For the E series we have already used graphs to define the matrices. The

3.4

Graphs and Coxeter-Dynkin Diagrams

251

Coxeter-Dynkin diagrams are 0 1 0 —0 —0 —0 —0

^ £ 7 ):

r(£ 8 ) :

0 1 0 —0 —0 —0 —0 —0 0 1 0 —0 —0 —0 —0 —0 - 0

The Cartan matrix A is indecomposable if T (^ ) is connected, and it is decomposable otherwise. To say A is decomposable is equivalent to saying that A can be reindexed (by simultaneous row/column permutation) so that it assumes the form 0 0

^2

where A^ and A 2 are (necessarily) Cartan matrices. A Cartan matrix A = i.Aij)i is symmetrizable if there exist nonzero rational numbers e„ i e J, so that for all i , j e J or equivalently e,- '-<4,7 = ej

for all i , j e J.

This concept is of considerable importance in the sequel so it is good to have an easy way to recognize a symmetrizable matrix. Proposition 1 The Cartan matrix A is symmetrizable if for each cycle (/ q, • • •, A:

i = A: :

in r(>l),

A: : .

Since \Aij\ = is the number of arrows (edges) joining i to this condition can be restated as the products of the numbers of arrows in one direction around any cycle equals the product of the number of arrows in the other direction.

252

Lattices and Finite Root Systems

We now return to the case of finite root systems to illustrate and apply some of the concepts just introduced. Example S Bases for the root systems of types A 2, and along with their Coxeter-Dynkin diagrams are shown in Figure 3.1. We will see shortly that to each diagram there is (up to isomorphism) only one reduced root system. Let us clarify the relationship between root systems and Coxeter-Dynkin diagrams. Given a root system A and a base II of A, once we index the elements of II, we can write down a Cartan matrix. From the matrix we derive an incidence matrix and corresponding diagram. Reindexing the elements of II leads to the “same” diagram with the nodes similarly reindexed. Choosing a different base II' of A, we find a w e IF such that wU = n '. Since (w(a)\w(p)) = (a|j8), II and vvll lead to the same matrix A and then to the same diagram. Now we show that the diagram r(A) determines A to within isomorphism. Proposition 2 I f A and A' are two reduced root systems such that F(A) —F(A'), then A ~ A'. Proof We may assume that F(A) = F(A'). Index the / nodes of the diagram with the numbers 1,..., / in any way. This gives rise to Cartan matrices A and A' for A and A. Let II = { a ^ , a n d II' = {a[ , . . . , a]} be the corresponding bases. Let V and V' be the ambient spaces of A and A'. There is a unique linear isomorphism : F ^ F ' such that (a¿) = a', i = 1,. . ., /. Let r¿ and r¡ be the reflections in a¿ and a', respectively. Then

Since Aji = A'ji, it follows immediately that r[ = r¿(l>-1 Hence we have an isomorphism W ^ W' with <^(w) = ) in the category of group representations; that is, the diagram W

X

♦1 W' commutes.

V

---- >

[* X

V

---- V

3.4

Graphs and Coxeter-Dynkin Diagrams

253

In particular, <^(A) = <^(W^II) = ^(W)(l>(Ji) = W'U' = A' (Use Proposi­ tion 3.3.4) which shows that (f>is an isomorphism. □ Remark 1 It would be more accurate to require that r(A) ^ r(A) (in the sense of graph theory) rather that F(A) = F(A') in the last proposition. It is easy to make this minor formal modification if needed. Example 9 (Rank 1 and rank 2 root systems) Suppose that A is a root system of rank 1. Then A has a base II = {a}, and the Coxeter-Dynkin diagram is a single node O and the ambient space is V= Ka. The Weyl group W is generated by and rj^a) = - a . Thus ^red = w H = {a, —a}. If A then ± 2 a are also roots (Section 3.2, case 1) and A = {a,2a, —a, —2a) (said to be of type If A = then A is said to be of type Suppose next that A is a root system of rank 2 and that II = {a^, «2) is a base. Then A ^2 '= («i, «2" ) ^21 '= ( ^ 2^^ 1" ) ^^e nonpositive integers. Reindexing, if necessary, so that |^2il ^ fhe possibilities listed in Table 3.2 (see Section 3.2, case 2) Reduced root systems for each of these diagrams are easily constructed. They are shown below in real euclidean 2-space (as far as a sheet of paper can represent euclidean 2-space!) The reader can easily construct these: For instance, for A 2 one has the Cartan matrix is generated by

| ^ ^j and its Weyl group W

and ^2- where ^1( ^ 1) =

'*1(^2) = «1 + 0^2.

''2(^1) = Oi^+ a2,

''2(«2) = “ «2-

Thus .

-« 1 <— ^ «1 (aj + a2)

- («1 + «2) —ao <— > an determines IF • II = Table 3.2 ^12

0 -1 -1 -1

'■ 21

0 -1 -2 -3

Coxeter-Dynkin diagram

0 0 0 —- 0 0 <= 0 0 - ^ 0

Name Ai VA Ai Cl Gi

254

Lattices and Finite Root Systems

vAi

+ 3«2

Figure 3.2. Reduced finite root systems of rank 2

If A is not reduced, then 2 a is a root for some indivisible root a. Using W, we obtain 2a¿ as a root for / = 1 or 2. Then (2a^)'" = and we must have <j8, \af ) e Z for all )8 g A. This happens only in and in C2 with / = 1. Up to isomorphism all the possibilities are shown in Figures 3.2 and 3.3. Proposition 3 Let s be the reflection in a vector a given by a decomposition IKa ® H of a vector space V. I f M is a subspace o f V invariant by s, then either IKa a M or M(zH, Proof. If there exists x g M \ / / , then sx —x = ca ^ Ka \ {0}, so IKa c M. □ Proposition 4 The following are equivalent for a root system A in a K-space V: (i) A is indecomposable. (ii) r(A) is connected. (hi) V is an irreducible W-space.

3.4

255

Graphs and Coxeter-Dynkin Diagrams

2ao «2

, «2

«1 «1

2«!

2(

A, V (BC\

iB C \ V (,BC\

(B0 2

Figure 3.3. Nonreduced finite root systems of rank 2

Proof, (ii) => (i) Suppose that A is decomposable: A = Aj V A2. Any bases III and II2 of Ai and A2 combine to form a base II = Hi U Il2 of A. Since for a e III, jS e ^2? = <)3, a '' > = 0, the diagram F(A) is composed of the two disconnected subdiagrams r(Ai) and F(A2). (iii) => (ii) If r(A) decomposes into two disconnected subdiagrams Fi and F2, then any base II of A decomposes nontrivially as IIi U II2 with = 0, whenever a e 111, p e II2. Then the linear spans Fi and F2 of III II2 are both W invariant, so V is reducible. (i) => (iii) Suppose that M is a nontrivial IT-invariant subspace of V. Using Proposition 3.4.3 we partition A as Ai U A2, with Ai = ( a e Ala e M},

A2= ( a e A|<M,a-> = (0)). If A2 = 0 , then A c M, so M = F. If Ai = 0 , then {M\a^ > = (0) for all a e A, so M = (0). By assumption neither of these happens. Thus Ai # 0 , A2 ^ 0 . By definition (p, a '' > = 0 for jS e Ai, a e A2. Thus A = Ai V A2. □

256

Lattices and Finite Root Systems

Proposition 5 Let V be a K-space and A a root system in V, Define a bilinear form К on Vby K{x,y)=

Y, аеД

Then K is nondegenerate, symmetric, and invariant under Aut(A). Moreover the restriction of K to the rational span Vq of A is positive definite. Proof K is clearly symmetric, and its restriction to Vq is positive definite since c Z . It follows that this restriction is nondegenerate and hence that K itself is nondegenerate (since A spans V). The invariance of K follows from the fact that = = (;c,(c7-(a))" > for all X El Vq and for all a Aut(A) [see Proposition 3.2.6 and equation (3.2.8). □ The bilinear form K constructed above is called the natural bilinear form of A. It is deeply connected with the Killing form of certain semisimple Lie algebras (see Exercise 6.4). In the previous paragraph we indicated that invariant bilinear forms are essentially unique. We now set about giving precise meaning to this. Proposition 6 Let Vbe a K-space, and A be an indecomposable root system in V. Let B be a bilinear form on V that is invariant under the Weyl group W o f A. Suppose that B Q. Then (i) B is nondegenerate, symmetric, and invariant under Aut(A). (ii) I f B' is any bilinear form on V that is W-invariant, then B' = kB for some fc G IK. Proof Consider the left radical of B, namely {x e V\ B(x, y) = 0 for all y e F}. Since B is IT-invariant, this is a IT-stable subspace of V, which is also proper (since B ^ 0). By Proposition 4 the left radical is (0). Similarly the right radical of B is trivial, and hence B is nondegenerate. Let B' be as in part (ii) of Proposition 6. Since V is finite dimensional, there exists / e Endj^(F) such that B\x,y)=B{f{x),y). Let us show that / commutes with IT. Indeed, if vv e IT, then using the fact

3.4

Graphs and Coxeter-Dynkin Diagrams

257

that B and B' are PT-invariant, we obtain B{fw{x),y) =B'{w{x),y) = B ' { x , w - \ y ) ) = B{f{x),w-\y))=B{wf{x),y). Thus Bdf w - wf)(x), y) = 0 for all x and y in V, and hence Jw = wf since B is nondegenerate. Now let a e A, and consider e W. Write V = Ka O H, where H is the hyperplane of V consisting of fixed points of r^. Since fr^ = r^f, we have f { K a ) = / ( l - r J ( F ) = (1 - r J / ( F ) c Ka, and therefore there exists A: e K such that /( a ) = ka. Consider now V' — ker(/-Al). Then V' is IT-invariant (because / commutes with W) and V ^ (0) (because K a c K'). By Proposition 4, V' = V, and therefore B' = kB. This establishes part (ii). Now part (i) follows from part (ii) together with Proposition 6. □ Proposition 7 Suppose that A is decomposable and that B is a Wdnvariant bilinear form on V, Let V ”^iA^ be the decomposition of A into indecomposable factors. Let Bi denote the restriction o f B to where is the ambient space o f A T h e n B(Vi,Vj) = (0) for i ¥=j and B = Furthermore, if is the natural bilinear form of A^, then B = for unique c^s in K. Proposition 8 Let A be an indecomposable root system in V. Let (*|*) be a non-degenerate W-invariant symmetric bilinear form on V. Then for a and P in A we have (ala) = (P\p) ^ W a = Wp. Proof is obvious since ( I ) is PT-invariant by definition. =» We may assume that K = Q and (• | •) is positive definite (Proposition 3.3.7 and Proposition 6). Since Wa spans V, we may replace a by a suitable wa and assume that (a\p) 0. Then 2(a\p)

and by replacing p by - p if necessary, we may assume that (p, a " ) > 0. From (a, p" ) ( p , a ^ ) < 4 we are left with two possibilities: 1. {a, p^ ) = 2 in which case a = p (see Section 3.2), 2. (a, p " ) = (p, a - > = 1 in which case r^r^r^p = r^r^ip - a) = r j , - p —(a —p)) = a. □

258

Lattices and Finite Root Systems

3 .S

CLASSIFICATIO N OF CARTAN M ATRICES A N D FINITE R O O T SYSTEM S

Let A = (Ajj) be an (/ + 1) X (/ + 1) Cartan matrix. (For convenience we will number the rows and columns of A from 0 to /). We think of A as an endomorphism of via A \ \ ^ \A for all X = ( xq, Xj, . . . , Xi) e An element n e of A of eigenvalue 0.

is called a null root of

In other words n = ( wq,

if n is a (left) eigenvector

satisfies

(1)

n A = 0,

(2)

n ^ 0,

(3)

Hi > 0

for all 0 < / < /.

A Cartan matrix is said to be affine if it is indecomposable and admits a null root. If A is affine then det(^) = 0. Example 1 Let ^ = possibilities are ' ^ - ' ^ ■’ - ( - 2

h t sl 2

i )

X

2 affine matrix. Then ab = 4. The

”

-< )

(we may assume that a < b). Examples of null roots are given by (1,1) for A^l^ and (1, 2) for A^^^. Our next objective is to classify all affine matrices. As a consequence of this classification we will obtain a classification for all finite root systems. (In this section we establish only that finite indecomposable reduced root sys­ tems lie in a certain list. In Section 5.3 we will complete the classification by showing that all these root systems actually exist.) Our approach follows [BMW]. Let A = (^¿y), 0 < /, j < /, be an affine matrix, and let n = (riQ, be a null root. Then =0 /=0

for all 0 < y < /,

C lassification o f Cartan M atrices and Finite Root System s 259

3.5

and hence for all 0 < 7 < /,

(4)

Y , \ ^i j \ ni = 2tij. 1=0

i*j

Proposition 1 Let A be an (/ + 1) X (/ + 1) affine matrix, and let n be a null root o f A. Then (i) all the coefficients o f n are positive and the 0-eigenspace o f A is Un. In particular, if n' is a null root o f A, then n' = qn for some q ^ (ii) There exists a unique null root { n ^ , n f ) o f A satisfying (5)

(6) Proof. Write {0,

ni ^

for all 0 < i < I,

g.c.d. {«0, •••,«;} = 1( 6) = I U J, where I = {i|n, > 0}, J =

= 0).

Because of (4) we see that Aij = 0 whenever / e l and j e J. Since A is indecomposable, either I or J is empty. Hence J = 0 , since n 0. Thus > 0 for all /. Let m = (mo,. . . , m^) be a left 0-eigenvector of A. We can choose A e R such that n^ — Am, > 0 for all 0 < / < /, n, - Am, = 0 for some /. Then n - Am is a null root. By the above argument n, —Am, = 0 for all /. Thus m = An. This proves part (i). As for part (ii), since A has rational coefficients and 0 e Q is an eigenvalue of ^ has a rational 0-eigenvector. By part (i) it is a multiple of n. Thus we may assume first that n e then by clearing denominators that n e and finally, removing common factors, that n is as in part (ii). Uniqueness follows from part (i). □ The null root n = (« q, . . . , n^) of A described in part (ii) of last proposi­ tion will be called the null root of A, We call n, the mark of the / node of the Coxeter-Dynkin diagram of A.

260

Lattices and Finite Root Systems

Example 2 Let 2 -1 0

-1 2 -1

0 -3 2

The Coxeter-Dynkin diagram of A is

00 0^0 1 2 -

and n = (1,2,3) is the null root, since with its marksisgiven by 1

= 0. The Coxeter-Dynkin diagram 2

3

0 0^0 -

0

1

2

Next weclassify allCoxeter-Dynkin diagrams arising from affine matrices A, Let r be the Coxeter-Dynkin diagram of A. The nodes are indexed from 0 to /, and we let be the mark of the i node. To each (/,;)-arrow (an arrow from the i to the j node with i j) we attach a weight n^/rij. Since there are ;)-arrows, and we have from (4) that for all 0 < j < I, (7)

n: L \Au\- = 2 ¿= 0 i^j

We conclude that 1. The sum of all the weights o f the arrows arriving at any node is 2. If there is a (y, /)-arrow there is also an (/, ;)-arrow and its weight is n^/n^ (the reciprocal of nj nj ) . Together with property 1 this implies property 2. 2. The weight w o f any arrow satisfies \ < w < 2, and hence we have property 3, 3. There are at most four arrows arriving at any node. Using properties 1 and 2, it follows that property 4 holds. 4. There exists no weight w satisfying f < w < 2, and hence by taking reciprocals, we have property 5. 5. There exists no weight w satisfying \ < w < \ . Because of property 5 it is impossible to complete a sum of weights lying between f and f to 2. Thus we have property 6. 6. There exists no weight w satisfying j < vr < f , and hence taking recipro­ cals results in property 7. 7. There exists no weight w satisfying \ < w < \ .

3.5

Classification of Cartan Matrices and Finite Root Systems

261

By using the seven properties above, we investigate the diagrams that may arise according to the number of arrows arriving to a node. Four Arrows Arrive at a Node. Suppose that 4 arrows arrive at a node i. Each of these arrows must have weight If one of these arrows is of type (;, /), then there is at least one arrow of type (/, j) and each of these is of weight 2. Hence there is a unique arrow of type (/, j) (of weight 2), and hence no other arrow may arrive at j. The only possibilities are (each diagram is given with its marks). 1

1

o

0

1 0 ^ 0

0 0 -0 -

1

|2

1

|2

0

0

1 1

1

1

2

1

1

G®

0 =>0 ^ 1 1

Bcf>

2

1

0 0^0 -

2

o » 0 -

Each of these diagrams is complete in the sense that it satisfies property 1. (See below for some remarks on the labeling convention.) Three Arrows Arrive at a Node. Let the weights of these arrows he w^ < W2 <Wy Recall that w^ > ^. Suppose that w^ = ^. Then W2 w^ = f - Thus W2 < I, and hence either W2 = or We see that the only possibilities are (w^,W2,w^) = {^ , ^ , 1) ,

or

or

Suppose that vwj = f . Then W2 + w^ = j. Thus W2 < f , and hence w^ W2 = W3 = f . We have {W^,W2,W3) = Suppose w^ > f . Then W2 + w^ < j , and hence vi^2 < f which violates w^ < W2. Having determined all possible weights we study each case separately

262

Lattices and Finite Root Systems

Case (j,

1). We start with

‘Q\

2

2

1

2

2

or 0 ^ 0 - 0

p - O

O In either case all nodes except the extreme right-hand node are full (sum of arriving weights = 2). These diagrams can be extended to include an arbitrary number, say, /, of nodes of the form 1

Q\

2

2

2

/ 0 - 0 ---------------O

1

"

2

2

0 ^ 0 - 0 -

2

-o

0 1

and then completed to one of the following diagrams (with 1 + 1 nodes)

.

0

a /O

0-0---- o

o

1

1

1

2

2

2

2

0 - 0 ----------- o « o 0 1

1

BCP>:

22

2

2

0 ^ 0 - 0 ----------0 ^ 0 1

1

2

2

0^0-0---- o'\ 1

cp>:

0

2

^

0

2

- 0

O

o

2

---------------- 0 ^

1

0

3.5

Classification of Caitan Matrices and Finite Root Systems

263

Case ( |, f , f ). We start with

0—0—0 3

4

o—o<=o

3

In either case the node with a mark 2 is full. In the first diagram the nodes with mark 3 have one arrow of weight | arriving at them. There must be one more arrow of weight f arriving at them. This leads to

o-o-o-o-o 2

3

4

3

2

and, by a similar argument, completes to

0 0 0 6 0 0-0 -

-

1

-

2

-

3

-

4

3

2

1

The same type of reasoning shows that our other diagram leads to O -O ^ O -O -O 2

4

3

2

1

Case (j, f , f ). We start with

o-o-o 5

6

4

The node with a 3 is full, and the diagram can only be completed to 3

0 0 0 0 0 6 0-0 -

-

1

2

-

-

-

3

4

5

2

3

2

-

6

4

2

Case ( f , f , f ). We start with 2

0 - 0<=0

Q 0 0-0 -

2

3

2

3

2

0^ 0

264

Lattices and Finite Root Systems

and these complete in unique ways to 1

?

£:

0 0 6 0 -0 -

-

-

O -O -O ^ O -O

6 6-0 ^

A t Most Two Arrows Arrive at Every Node. We start with one of the following 1

.1

1

o-3!o

A^P:

1

^

1

1

o ^ o = ^ o

0-0 ----- 0-0 1 1

1 1

or

0^0-0---o1 1 1 1 The first two diagrams are complete, while the other two complete to one of 1

.o

0-0 ------- 0-0 1 1

1 1

or o « = o - o -------------- 0 ^ 0

1

1

1

1 1

-

A t Most One Arrow Arrives at Every Node. Since the diagram is connected, the only possibility is Q —Q , which is not affine. No Arrows Arrive at Every Node. Since the diagram is connected, the only possibility is Q , which is not affine.

3.5

Classification of Caitan Matrices and Finite Root Systems

265

Proposition 2

Let A be an (/ + 1) X (/ + 1) affine Cartan matrix. Then its Coxeter-Dynkin diagram T is of the form where is in the following list {all diagrams have / + 1 nodes and are given with their marks): In particular affine matrices are symmetrizable. 1

,o 0-0 ---------1

/> 2

0-0

-

1

1

1

1

Q

\,

0 0 -

o^^

2

^^ 3 2

^

1

CP:

--------- 0 ^ 0

—

0 ^ 0 - 0 - - ------ 0 - 0 < ^ 0 1

2

2

2

2

^^ 2

1 1

1

/O O i” :

^

0 - 0 ------- ---------

0 - 0 ,

' i “ 1

1

E<‘>;

o-o-O-o-o 1

E y '- -

2

2

1

0 0 0 6 0 0-0 -

1

^

3

-

2

-

3

-

4

-

3

2

1

5 01 02 03 04 05 66 04 - 02 " ^

-

-

-

-

01 0 03 0 -02 2 4 -

-

= ^

-

-

266

Lattices and Finite Root Systems

G<»: r:

0 - 0 ^ 0 1

2

3

0

O

1

1

Q ^ 0 - 0 ---------0 ^ 0 2

2

2

^^ 3 1

1

0^ 0-0 1

fiC p :

1

------------------------------------------------------------------

1

0

-

1

0 ^ 0 - 0 - ---------0 ^ 0 1

2

2

2

0=^0

1

1

^^2

' ^ 2 2

o-o-o^o-o 1

2

3

2

B C P>=^P: G p:

0 0^0 -

1

2

1

Remark 1 The meaning of the number a e {1,2,3} appearing in be found in [Mo3]. An alternate nomenclature is used in [Ka5].

can

Remark 2 Suppose that A is an affine Cartan matrix. Then is clearly a Cartan matrix. The Coxeter-Dynkin diagram of A ^ is obtained by reversing the arrows on the diagram of A (since we are interchanging the roles of Aij and Aji), By inspection in Proposition 2 it follows that A ^ is also affine (or see Proposition 3.6.5). For example, has matrix 2-1 O] ^ = ^-3 2 - i j and diagram Q ^ Q _ Q , while A'^ has diagram Q

__Q

and is affine of type G^^\

We now begin the classification of the Coxeter-Dynkin diagrams of all finite root systems. Let A = (A^jX 1 < i, j < I, be the Cartan matrix of a finite indecomposable reduced root system A arising from a base II = ( a j,. . . , of A. Thus ^ ij

{

}•

3.5

Let

Classification of Cartan Matrices and Finite Root Systems

267

e A be chosen so that the height of 4> is maximal. That is, ht(o;) <

(We will see that and hence


for all Of e A.

is unique.) Then for all

( (f), a / ) > 0

e n we have ht

< ht

(f>,

for all 1 < i < 1.

If we choose a positive definite PT-invariant bilinear form as in Proposition 3.4.6, we have > 0,

1
Define for all 1 < i < I,

^ 0/ = -

I.. ^ (a ,la ,)

= - ( « ; )

and 2 (g ,l< /> )

^io — We can then define an (/ + 1) X (/ + 1) Cartan matrix A by A^j = {A^j) for all 0 < i, j < I where A qq == 2. Lemma 3 A is an affine Cartan matrix. Proof. That ^ is a Cartan matrix follows from construction. We show that A is indecomposable. Indeed the 0 node of the Coxeter-Dynkin diagram f of is connected to some other node for otherwise (<^|<^) = 0. With the 0 node removed, the diagram left is connected (since A is indecomposable). Thus f is connected, and hence A is indecomposable. To establish the lemma, we must show that A admits a null root. Write /
-(f>.

Clearly n^ > 0. Let Wq == 1, and let n == (nQ, n^,..., n^). Then for all 0 < ; < /

268

Lattices and Finite Root Systems

we have

1=0

= A qj + ^ /=1 = 0.

(« > ;)

( “ yl“ ;)

This shows that n A = 0 and hence that A is affine.

□

Theorem 4 (i) Up to isomorphism there exists a 1-1 correspondence between finite indecomposable reduced root systems and the diagrams listed below. This correspondence is given by attaching to a root system its Coxeter-Dynkin diagram with respect to any base,

0-0-0----------O 1

Br.

2

2

2

0 - 0 — 0 ^ 0 2

2

2

Q: 0 - 0 —

1

0<=0

^^2 1

Or-

1 2 0 - 0 ----------------

2/ o

/> 4

0 1

1

2

^

2

1

o-o-o-o-o z.

1

2

3

?|4.

3

2

2

3

4

5

s |6

4

2

3

4

2

2

3

£7: o-o-o-o-o-o

f.^2-

2

o-o-o-o-o-o-o o-o=o-o 0 ^ 0

3.5

Classification of Cartan Matrices and Finite Root Systems

269

(ii) If A is an indecomposable reduced finite root system, H = • • • aj) a base o f A, and T is its Coxeter-Dynkin diagram with respect to H, then A has a unique root (f>whose height with respect to H is maximal. Moreover, if

4> = H riiOLi, i= l

then the n^ are the marks o f Y as they appear in the graphs o f part (/) of the Theorem, and for all a E: ^ we have a < r^a for all 1 < f < /. That is, r^a = a - ka^, where A: > 0, or equivalently (a, a ^ } > 0 for all 1 < / < /, or equivalently (a|a^) > 0 for all 1 < i < 1. 1. If a = is dominant, then m^ > 0 for all i: First, 0 < (a |a ) = Urnfala^) together with (a|a^) > 0 for all i im­ plies that some > 0, and hence a e A+. Write {1, ...,/} = I U J, where I = {i\m^ > 0} and J = {i\m^ = 0}. Since (a, a ^ ) > 0, we see that = 0 for all i e I and j e J. Since A is indecomposable, we conclude that J is empty. 2. If a = and p = are both dominant then with respect to the Q^-partial ordering (see Section 2.5) either a > ¡3 or p > a: From argument 1 we have (a\p) > 0. Thus {a, = 2(alj8)/()3|/3) > 0, and hence a - p e A (Proposition 3.2.8). Suppose > Pi for some i. Then a — p e A forces a - p e A+ c (2+. Thus mi > Pi for all i. 3. Let a E A be of maximal height with respect to H. Then a is dominant. Moreover (a |a ) > ip\p) for all roots P: That a is dominant is clear. Suppose that A has roots of unequal lengths. If e A is of length different than that of a, we must show

Lattices and Finite Root Systems

270

that (a\a) > (p\p). By replacing p by wp for some vv e we may assume that p is dominant. Indeed, if p is not dominant, then (p,a^^) < 0 for some i and r^p > p. Repeating this argument if necessary, we obtain at last a dominant root. The argument of Step 2 shows that a - j8 e A^. Thus («1«) - (P\P) = ( a + P\a - P) = {a\a —)8) + (P\a — p) > 0 (the last inequality since a and p are dominant and a — p ^ Q+), Finally, (a\a) > (p\p), since a and p are supposed to have different lengths. 4. There exists a unique root (f> = of maximal height. Moreover i\(t>) > {p\p) and 4> > P for all e A: The first two assertions follow from arguments 2 and 3. Let j8 e A. Using the same argument as in 3, we conclude that there is a dominant root y with y > p. Now (j) > y > p hy 2. The root (f> above is called the highest root of A (with respect to II). We reserve the notation ..., for its coefficients. A root is called long if it has the same square length as (f>. 5.

If n^ = 1 for some 0 < f < / , then is long: Note that Uq = —(!> is long by definition. If f > 0 and if is not a long root, let i = Iq, ¿1, . . . , t)e a minimal set of distinct indices in { !,...,/} so that = («/J«/,) = ^ each 1 < j < k, A: , ¥= 0. Thus from the Table 3.1,

0 - 0 -----------0 - ^ 0 ^0 ^k-l with m > 1. Then r,^0 n, = 1.

r,^k-la:^k = ma:^0 + ma, +

H-m^

+ a, < S contradicts

Proof of Theorem (except for existence). Let A be an indecomposable finite root system, and let II = {a^,...,« /} be a base of A. Suppose that the Cartan matrix of n (hence of A) is A. Let (f> = be the highest root of A

3.5

Classification of Caitan Matrices and Finite Root Systems

271

with respect to II. Using Lemma 3, form the affine matrix A, The CoxeterDynkin diagram t of A is the Coxeter-Dynkin diagram of A extended by one node, the extended node being determined by —(f>. The marks are riQ= 1, /Ip , rii and — has the same square length as some i = 1, . . . , /). 5. f has marks = 1 only over nodes corresponding to long roots (by argument 5 above). Regarding claim 3, it is useful to observe that whenever we have the appearance of two nodes like Q s ^ Q , Q ’ O Coxeter-Dynkin diagram F or F, then, naming the corresponding roots a, ¡3, we have 2(a\p)/(p\p) = —4, —3, or —2 and 2(p\a)/(a\a) = —1, whence (a\a)/(p\p) = 4, 3, or 2, respectively. Thus the node has multiple arrows into it only if there are roots of longer square length than a¿. Now looking at the classification of affine matrices in Proposition 2, we observe that claims 1-5 eliminate all the diagrams except those of the form Moreover, up to a diagram automorphism these diagrams have a unique node with mark = 1. The diagram Xi is obtained by removing such a node. □ Remark 3 The number of nodes whose mark = 1 in is actually the index of connection [P: Q], where P and Q are the weight and root lattices of the root system of type Afp Simply by staring at the diagrams in Theorem 4, we obtain Corollary Let A be an indecomposable reduced finite root system with Coxeter-Dynkin diagram F. Then (i) there are at most two different root lengths in A; (ii) there is at most one multiple arrow in F; (iii) there is at most one node that is connected to more than two nodes, and such a node is connected to exactly three nodes; (iv) the corresponding Cartan matrix is symmetrizable.

272

Lattices and Finite Root Systems

3.6

THE PERRON-FROBENIUS THEOREM AND ITS CONSEQUENCES

On the set oi m X n real matrices we define two partial orderings > and ^ as follows: Given m X n matrices C = (c,,) and D = C >D C^ D

Cij > dij for all

i
either C = D or 1
and

and

1 < j < n,

> d^j for all

1 < ; < n.

As usual we write C > D (respectively C > D) io denote that C ^ D and C > D (respectively C D), We say that C is positive (resp. nonnegative) if C > 0 (resp. C > 0, where 0 denotes the zero matrix of appropriate size. If C in a square matrix, we say that C is primitive if 0 for some A: e Z+ C is called semiprimitive if e l C is primitive for all e e Throughout this section will denote the set of all column vectors X = (jCi,. . . , such that x > 0. The main result of this section is the following:

Theorem 1 Let C be a semiprimitive nonnegative n eigenvalue r satisfying

X

n matrix. Then C has a real

(i) r > 0. Furthermore r > Q if C is primitive or if n > 1; (ii) there exist positive left and right eigenvectors for r, each unique up to positive scalar factors; (iii) r > 1^1 for all other eigenvalues s of C and the multiplicity o f r is 1; (iv) if C > D > 0 and d is an eigenvalue of D, then \d\ < r and \d\ = r if and only if D = C. The eigenvalue r is called the Perron-Frobenius eigenvalue of A. Example 1 Let T be a finite labeled graph, and let C = C(T) be the incidence matrix of T. Thus the vertices of T are labeled ! , . . . , « in some way and is defined as the number of arrows from vertex i to vertex j for i # j and c¿¿ = 0 for all i. Certainly C > 0. It is straightforward to see that C is semiprimitive if and only if T is connected. Indeed, if « = 1, this is trivial. Suppose that n > 1, C is semiprimitive and (e / + C)^ >- 0. Then for all > 0, where the sum runs over all . . . , j ^[i,n] i, j, Ec;.,., • c;.Ill2

3.6

The Perron-Frobenius Theorem and Its Consequences

273

and dpq '= (s i + C)p^. Observe that the inequality itself is independent of s (> 0). Suppose that i ¥=j. Take any summand > 0, and omit all factors that are diagonal entries (Cp^X The remaining product is still positive, and by definition of C defines a path from vertex i to vertex j. Conversely, suppose that we have a path of length m = m(i,j) from i to Then we obtain a product ‘ j > 0, and we see that (C'”),y > 0. Then, for all e Z+ and e > 0, ( ( e / + includes the summand • ** c, fS — e > 0 (k factors of eX and hence we see > 0. Now if r is connected taking M == max{m(/,y)}, it is easy to see that (el + C )^ 0. Thus C is semiprimitive. In particular, if ^ is a Cartan matrix, then C '= 2 —A is its incidence graph, and C is semiprimitive if and only if the graph is connected (i.e., A is indecomposable). Later on we will prove the remarkable fact that the Perron-Frobenius eigenvalue of C(T) is 2 if and only if r is the Coxeter-Dynkin diagram of an affine Cartan matrix. Our proof of Theorem 1 follows [Sn]. We begin with a sequence of lemmas. We assume first that C is primitive and that m has been chosen so that > 0. Lemma 2 If a> Q and Cx = ax for some x e Proof If X ¥= 0, then a’^x = C ^x

q,

then either x ^ Q or x = Q.

0, so jc

0.

□

Lemma 3 Let C j,. .. , C„ be the rows o f C, and define

by I \ R ( x ) = m i n i ----->

I

i

T forallx = ( x ^ , . . . , x ^ ) .

(IfXj = 0, CjX/Xj is understood to be +oo.) Then R assumes a maximum value r at some point x e q\ {0}. Proof Let M = max^ y{c,y}. Then for x e R > o \ R(x)Xj < CjX =

^

for each j = 1, . . . , n.

k Adding, we obtain R{ x) Y,Xj < nM(l,Xi^),

274

Lattices and Finite Root Systems

whence

< nM. Thus R is bounded above. Define r '=

sup

Evidently 00 > r > 0. Since each of the functions CjX/Xj is continuous where it is defined, R is continuous. Furthermore R(x) = R{ax) for all a e and hence r = sup x^S where S ’= [R>o ^ [x\x^x = 1}. The latter being compact, R assumes a maximum on S, hence on R>o\№}. This proves the existence of x as required. □ Henceforth r will be the quantity defined in Lemma 3. Lemma 4 If X e

\ {0}

if rx < Cc, then rx = Cx and x

0.

Proof By assumption (C - r)x > 0. If this is not zero, then C(C'”x) - r{C^x) = C^{C - r ) ( x ) > 0. Setting x' := C"*x, we have rx' < Cx' and hence rx'j < Cjx' for all ;, contrary to the definition of r. Thus (C - r)x = 0 and x 0 by Lemma 2. □ Proof (of Theorem 1). We begin by assuming that C is primitive. By the definition of X (Lemma 3), rx < Cx and hence by Lemma 3.4, rx = Cx,

X

0.

Suppose that Cx = sx for some x e C" \ {0}, ^ e C. Then for each j =

( 1)

\sxj\ = |C^-x| < Y,Cji\Xi\ (=

+ooii Xj = 0).

3.6

The Perron-Frobenius Theorem and Its Consequences

275

Set

Then \s\ < R { x ) < r. Now if \s\ = r, then from (1) rx < Cx so by Lemma 4 rx = Cx, Set

jc >- 0.

Then

Since the Xy e C \ {0} and the > 0, we see from this equality that jCj,..., all have the same argument, say, 6. Thus Xj = e'^\xj\, j = 1, . . . , and X = e^^x. Hence Cx = sx => Cx = sx, which proves that s is real and finally that s = r. Suppose that x, x' are nonzero eigenvectors for r. We can assume from the argument just given that x, x' > 0 and choose a e R minimal so that some component of x — ax' equals 0. Then x — ax' > 0, and by Lemma 2, X - ax' = 0. This proves that x and x' are linearly dependent (hence multiples of Jc) and the multiplicity of r is one. Replacing C by C^, we obtain a positive left eigenvector for C with Perron-Frobenius eigenvalue r'. Thus all eigenvalues s of satisfy \s\ < r'. Since C and have the same eigenvalues, r = r'. This concludes the proof of parts (0, (ii), and (iii) in the primitive case. For part (iii) suppose that D < C and that D > 0. Suppose that y is a nonzero eigenvector for D with eigenvalue d. Then using (1), we get Idly < D y < C y , where y = ( ly j,. . . , ly„l)^. Let 3c^ be a positive left eigenvector for C with eigenvalue r. Then |dl;c^y <

x^Dy < x^Cy = rx^y.

276 SO

Lattices and Finite Root Systems

\d\ < r. If \d\ = r, then ry < Cy, and by Lemma 4, ry = Cy,

y > 0.

Thus ry < Dy < Cy = ry. It follows that Dy = Cy, and since all entires of y are positive, D = C. Now suppose that C is semiprimitive. Then for all e > 0, C(e) ■= el + C is primitive. Now C(e) has the same eigenvectors as C and has all its eigenvalues shifted by e. Thus, if x is the right Perron-Frobenius eigenvector of C(e) for some, hence all, e > 0 and the corresponding eigenvalue is r^, we have C ( s ) x = ex + Cx = r^x so Cx = rx. where r ‘= r^ - e is independent of e. Obviously r > 0, since > 0. If n > 1, then some row, say, Q, of C is not 0, and 'LJ^iC^Xj = rxj together with X > 0 gives r > 0. The remaining statements, parts (ii)-(iv), are now obvious. For part (iv) we use C(e) > D(e) > 0 for all e > 0. □ Let A be an indecomposable Cartan matrix, and write A = 21 — P. Then, as we saw in Example 1, P > 0 and P is semiprimitive. We let r = r{A) be defined as the Perron-Frobenius eigenvalue of P. A is called of finite (resp. affine, indefinite) type according as r{ A) < 2

(resp.= 2, > 2).

The apparent conflict in the use of the word affine is resolved in the following proposition. Proposition 5 [Vi] Let A be indecomposable Cartan matrix indexed by J = { ! ,..., /}. One of the three mutually exclusive cases occurs: (i) Finite. A is o f finite type <=>

e

<=>

e IR^\ {0},

<=>3 x ^

\ {0},

jc > 0 with Ax < 0, X > 0 with Ax > 0, Ax with Ax > 0.

0,

3.6

111

The Perron-Frobenius Theorem and Its Consequences

(ii) Affine. A is o f affine type <=>

\ {0},

<=>

G -+

a:

> 0 with Ax = 0,

with Ax = 0.

(iii) Indefinite. A is o f indefinite type <=» 3 j : e

«=>

R ^ \ {0},

G

\ {0},

<=> 3 x G Z+

a:

> 0

with Ax

<

0, Ax

0,

jc > 0 with Ax > 0,

with Ax < 0.

Moreover, A and A ^ fall under the same case. Proof Write A = 21 - P, and let r > 0 be the Perron-Frobenius eigenvalue of P. Let u and u be associated left and right Perron-Frobenius eigenvectors for P. Thus w 0, i; 0 and uP = ru, Pv = ru. If r = 2, then uP = 2u => uA = 0 A is affine. Conversely, if A is affine with null root n, then nP = 2n and 2n • i; = nPi; = rn ' v. Since n • i; > 0, r = 2. Thus A

affine

r{ A ) = 2.

The rest of the affine case is already done in Proposition 3.5.1. Henceforth assume that A is not affine so that r # 2. Suppose that AG R^ \ {0}, a : > 0, and Ax < 0. Then A x < 0 = > P x > 2 x = > uPx > 2wc => rwc > 2wc => r > 2. Thus A is indefinite. In precisely the same way, A G R^ \ {0}, A > 0, and Ax > Q =>A is finite. This proves all the implica­ tions “ < = 3 a ... Suppose that A is indefinite. Then the set F :=

{a G

WjPx

>

2

a

}

is a nonempty open cone (note that u ^ V) and hence there exists a point AG K n Q +. Scaling A , we can assume that a g Z +. Then ^ 0. In precisely the same way, A is finite => 3 a g Z+ with Ax > 0. This proves all the implications “ => 3 a ... The remaining implications now follow auto­ matically. □ Remark 1 Proposition 5 is a version of Vinberg’s lemma [Vi]. We have phrased it in terms of Cartan matrices in order to make it fit most easily into the material to follow. Vinberg uses some properties of dual convex cone theory to establish his lemma, although he points out that the proof can be carried out by the Perron-Frobenius theory, as we have done.

278

Lattices and Finite Root Systems

We have already classified the Cartan matrices of affine type. We now show that the Cartan matrices of finite type are precisely those whose diagrams appear in Theorem 3.5.4. Let A = (Aij) and A' = (A]j) be Cartan matrices of dimensions / X / and r X /', respectively. We say that A is inferior to A if there is an injective map

such that IA'a \ < I

I

for 1 < i, j < V.

Thus A is inferior to A if the Coxeter-Dynkin diagram of A can be obtained from that of A by removing nodes and arrows. Proposition 6 Let A be an indecomposable Cartan matrix with Coxeter-Dynkin diagram T. (i) (ii)

I f r appears in the list o f Theorem 3.5.4, then A is inferior to some affine matrix, I f r does not appear in the list o f Theorem 3.5.4, then there is an affine matrix inferior to A,

Proof (i) The proof is clear by the way in which we obtained the list of Theorem 3.5.4. (ii) Suppose that T is not in the list of Theorem 3.5.4. If T contains a cycle then some A^l^, A: > 2, is inferior to F. If F contains Q Q with ab > 4, then either

Q^ O

O^ O

inferior to F. If F contains a node of

valence > 4 (i.e., a node with 4 or more adjacent nodes) or two nodes of valence 3, then is inferior for some k. If there are two multiple edges (i.e., O # 0> !>’ then O = 0 - 0 --- 0 - 0 = O some orientation of the arrows is inferior. If there is a single multiple edge and a node of valence 3, then

o =o-o-o-o;

,o 'O

(again with some orientation) is inferior. If there is a single multiple edge and every node is of valence < 3, then Q — Q — 0 ^ 0 — inferior since Bi, Cl, and are excluded by hypothesis. Finally, we are reduced to

3.6

The Perron-Frobenius Theorem and Its Consequences

279

the case of only single edges, and the hypothesis leads us to conclude that one of or is inferior. □ Let A' = 21 — P \ A = 21 — P be two indecomposable Cartan matrices of dimensions /' X /' and / X /, respectively. Then A' is inferior to A if and only if there is an injective map *: /'} ---- > such that F/y < for all 1 < i, j < /'. Then /' < /, and we may assume, if we wish, that /* = i for each i. If I' < I , then extend P' to an / X / matrix P' by filling the new entries with zeros. Then 0 < P' < P, and the spectrum (= the set of eigenvalues) of P' differs from that of P' only by the possible addition of zero. By part iii of Theorem 1 all eigenvalues d of P' satisfy d < r(A) (:= Perron-Frobenius eigenvalue of P), and equality occurs if and only if ?' = P. It follows that KA'X the Perron-Frobenius eigenvalue of P \ satisfies r(A') < r ( A ) with equality if and only if ^ ' = ^4. Thus Proposition 7 Let A and A be indecomposable Cartan matrices. Then A inferior to A => r(A) < r(A) with equality if and only if A = A. Putting Propositions 6 and 7 together, we have Proposition 8 The Cartan matrices o f finite type are precisely those whose diagrams are listed in Theorem 3.5.4. Proposition 9 Let A be a symmetrizable indecomposable Cartan matrix, and suppose that A is symmetrized by e = diagie^,. . . , e^}, where each > 0. Then (i) Ae is positive definite <=>A is o f finite type, (ii) A s is positive semidefinite A is o f affine type, (iii) Ae is indefinite A is o f indefinite type. Furthermore the eigenvalues o f A are all real. Proof Let sY'^ be a square root of s^, and let s^^'^ -= diag{sj/^,. . . , sY^}. Then the equations A^Sj = A -^s^ show that is symmetric, and of course A is similar to Furthermore for y e U^, y^(As)y = so A s is positive definite, positive semidefinite, or indefinite as s'^'^^Ae^^^ is. But the latter is determined by the eigenvalues

280

Lattices and Finite Root Systems

of being all positive, all nonnegative, or not all nonnegative, respectively. However, the spectrum S p e c i e o f is real and is equal to S pecif) = Spec(2 —P). Since the eigenvalues of P are maximized by the Perron-Frobenius eigenvalue r, Spec(2 —P) is minimized by 2 —r. The result follows from the definition of finite type, affine type, indefinite type. □ The approach to Cartan matrices through the use of the Perron-Frobenius theory came to our attention through the work of N. Iwahori [Iw]. The theory of the spectra of graphs is quite extensive; see, for instance, [CDS].

3 .7

CONSTRUCTING L IE ALG E BR A S F RO M LATTICES

In this section we show how one can construct a Lie algebra from any positive definite, even geometric, lattice L by using the elements of norm 2 in L. The construction produces a free Z-module of finite rank (a finitely generated torsion-free abelian group) of the form 9^ = L ®

( 1)

© e L (2 )

together with a multiplication [•, •]: 9 l X 9^^ ^ 9z. that satisfies the two Lie identities: skew-symmetry and the Jacobi identity. In as much as our scalars are integers rather than elements of some field, is a Lie algebra over Z, sometimes called a Lie ring. This Lie ring (see Section 1.1) will be graded by L with as the elements of degree a and L itself as the elements of degree 0. Of course should it happen that L(2) = 0 , none of this would be of any interest. Henceforth we will assume that L(2) 0. One easily obtains L-graded Lie algebras Qo< over a field IK(even if IKis of characteristic 0) by extension of the base ring (see Section 1.1) 9iK ==

= ( 11^

®

© e L (2 )

The Lie algebras obtained in this way from the lattices of types ^4/, Z)/, £5, Ej, and £’g are the Lie algebras of the same name, and they provide us with about one-half of the so-called finite-dimensional split simple Lie algebras. Our approach follows [FLM]. We make the following observation: Suppose that ( L , ( | • )) is a positive definite geometric lattice. Then for a, /3 e L(2) precisely one of the following

3.7

C onstructing Lie Algebras from Lattices

281

holds: a

(2)

/3 ^

U

a

p ^

and

a

p = 0 and

+

{0} and

(a|/3) > 0, («1/3) = - 1 , (a|/3) = - 2 .

by the Schwarz inequality. |( a |^ ) |< ( ( a k ) ( / 3 |/ 3 ) ) ‘/^ = 2 with equality if and only if a and p are linearly dependent, and (2) follows at once. From the point of view of defining the multiplication on the main task is to define the products [X^, X^]. Of these the most interesting is the case when (a\^) = —1. Then [ X ^ ,X , ] = The choice of sign is a very interesting problem which ultimately is related to making a suitable central extension of L by Z/2Z. The details of this are left to Section 3.8, where we study such central extensions more carefully. For now it suffices to assume that there is a biadditive mapping s: L X L satisfying (3)

e (a , a ) = | ( a |a ) (mod2)

fo ra lla e L .

and hence also (4)

e(a, p) + e(/3, a) = (a|j8) (m od2)

for all a, p ^ L

We then define [X^, X^] = ( — for (a\p) = —1. Notice how skew-symmetry follows form (4). The last part of the section is devoted to determining the basic properties of like showing that admits a triangular decomposition with an anti-involution o) interchanging X^ and - X _ ^ for all a e L(2). As we know L(2) is a finite root system in its Q-span. Once a base {a^,. . . , a/} of this root system is chosen, we use it to determine a set of Lie algebra generators of g^^ and a set of relations, which, as we will see in Chapter 4, actually gives a presentation of g^. Since L(2) is a root system, we call the elements of L(2) roots. Define g^ by (1).

282

Lattices and Finite Root Systems

Define [•, • ] on

as follows:

[L,L] = (0), [a, Xp] =

= - [Xp, a]

0

(5)

for all a e L, j8 e L(2);

if(«l/3)>0, if ( a |/ 3 ) = - l , if( a |/3 ) = - 2 .

Xjg] = I -a

for all a,/3 e L(2),

Proposition 1 with the bracket defined by bilinear extension o f (5), is a Lie ring. Proof. First we look at the skew-symmetry. The only nontrivial case is when (a\fi) = —1. Then, as we noted above,

Note that (a|j8) = - 2 happens precisely when j8 = - a in view of the fact that (• I ■) is positive definite. The Jacobi identity involves looking at a number of cases. The hard part is checking that (6 )

[x.,[x,.x,]\

+

-0

when a, j8, y e L(2) and a + /3 + y ^ f^(2). There are three subcases: (1) two pairs of roots add to 0, (2) exactly one pair of roots adds to 0, (3) no pair of roots adds to 0. 1. Suppose that a + /3 = 0 = ^ + 'y- Then a = y and (6) reads

which trivially holds. 2. Suppose that a + /3 = 0. Then /3 + y = - a + y, and certainly at most one of a + y and - a + y may be a root [we need (±a:|y) = -1 ]. If neither is a root, then by (2) either (a\y) = 0 or a = +y. Since we are in case 2, the

3.7

Constructing Lie Algebras from Lattices

latter is impossible, and hence (a\y) = 0. Also (p\y) =

283

so clearly (6) reads

[X ^,[X ^,X .J]= 0. We may assume then that exactly one of + a + y is a root, say, a + y (the other case is similar, since then j8 + y is a root). Now the left-hand side of (6) reads [ x _ ,,[ x ^ ,x ,]] + [X ^,[X ,,X _J]

= 0, since e(a,a) = ^(ala) mod 2 = 1 mod 2 and (a|y) = ~ 13. If none of a + ^ , ^ + y , y + a are roots then (6) is trivially true. Suppose then that /3 -I- y is a root. Then (a|/3) -t- (a|y ) = («1^ + y) = —1, since a -t- /3 -I- y e L(2). Because we are in case 3, (a\P) > —1, (a|y ) > —1. The only possible solution is for («1)3) = 0 (resp. —1) (“ ly) = ~ 1 (resp. 0). Assume the first case. We have then (a|/3) = 0 , ( a |y ) = - 1 ,

and

(^ ly ) = “ 1-

The left-hand side of (6) reads (7)

{( -

^

Adding e(a, ^ + y) -I- e(/3, y) = e(a, /3) -f e(a, y) + + e(y, a) = sip, y) -I- e(/3, a ) + e(y, a), we get e ( a , p ) + £(j8, a ) -I- e ( a ,y ) -I- c(y>“ ) = ((ali3) -I- (a ly )) mod2 = lm od2, which shows that (7) is 0. The other case is similar. Define a bijective linear mapping 9l

0L

y

+

284

Lattices and Finite Root Systems

by =

for all a g L (2) ,

« li = id. It is easy to see that a> is an anti-involution of g^: Obviously it has order 2, and we have [
= [a,X_^]

= - ( a \ P ) X _ p = (a\l3) 0 , if(a|/3 ) = - 1 , if(a |j8) = - 2. =
—1 0

for all a e L(2), /3 g L,

if a + ¡3 = 0, otherwise.

Proposition 2

The symmetric bilinear form on 9^ is even, nondegenerate, and invariant. It admits co as an isometry. Relative to {'V) we have the orthogonal decomposi­ tion g^ = L ±

±

(ZAT, e Z 2 f _ J .

{a, - a )

Each submodule 0 Z X _^ is a hyperbolic plane. The determinant of (-I*) on is ( - 1)^ det(L), where k = \ card(L(2)X Proof. The orthogonal decomposition is obvious, and it is clear that 0 ZZ_^ is a hyperbolic plane (Section 3.1). The nondegeneracy follows imme­ diately, and the statement about determinants is left as an exercise. The

3.7

Constructing Lie Algebras from Lattices

285

invariance condition has to be checked on a basis made up of the and a basis of L. There are four cases. Case (;), j = 0,1,2,3, accordingly as there are 0, 1, 2, or 3 basis elements of L in the expression ( [ a , 6 ]|c) + ( è |[ a ,c ] ) . We treat only the case (0), which is also the most involved. We suppose that we are looking at (8)

We have to show that it is 0. If a + j8 + y 0, this is automatic. Suppose that a + /3 + y = 0. Then - y = a + /3 e L(2), and (8) becomes

(9)

- ( - 1)

eioc,

/3)

( - 1)

e(a,y)

However, £(oc,p) + e ( a ,y ) = e ( a ,j8) + e{a, - a ) + s (a , -/3 ) = e(a, —a) = s(a, a ) = 1 (mod 2), which shows that (9) reduces to 0. Finally,

= (x jx ,), from which we can see that co is an isometry.

□

Using the results of Section 3.3, we choose a system of positive roots L(2)+ and a base c L(2)+. Then every element of L(2)+ is a nonnegative integral linear combination of a ¡ and L(2) = L(2) + U - L(2) ^ ,

L(2) + n - L(2) + = 0 .

Let 9 - '“ ®aeL(2)+ ^ ^ -a “ ^^9+)- Then are obviously Z-subalgebras of and

and g_

= g_0 L 0 g+. The spaces ZX^ are root spaces in the usual sense that they are eigenspaces for ad L (strictly speaking we should be talking about Z-modules rather than spaces). The corresponding eigenfunctions are the functionals á: h ^ {a\h). We usually identify á and a. With this convention we have a triangular decomposition (g+, L, (2+, co) of g^ with root lattice Q = ©/^jZa¿. (Here

Lattices and Finite Root Systems

286

we are working with a Lie algebras over Z rather than over some field; see Section 2.1.) Note that [K

9^ = IK 0^ 9_©

K

L ®K

is a triangular decomposition of K 0^ 9 l * In the case where L is generated (as an abelian group) by L(2), we have Q = L. This can easily be arranged by replacing L by the Z-span of its norm 2 vectors from the outset. Proposition 3 Suppose that L is an even positive-definite geometric lattice. Let {a^,... ,a¡\ be a base of the finite root system L(2). Let

be the corresponding triangular decomposition o f the associated Lie algebra Then (i)

= 1, . . . , /} generates g+ and = 1, . . . , /} generates g_ as Lie algebras over Z; (ii) ifL(2) generates L (as an abelian group), then {X^,, X _ ^fi = 1 ,..., /} generate as a Lie algebra over Z; (hi) if L = L^ L L2 (see Section 3.1), then 9^ = 9^^ X 9^^^. Proof (i) It suffices to prove the first statement. Let A+ be the set of positive roots of L(2) relative to the base {a^,. . . , and let m be the subalgebra of 9^ generated by X ^^,. . . , X^^. Let a = Xc^a^ e A+ be a root of height k > 1 (see Section 3.3 for the definition of height). By a standard argument (Proposition 3.3.4) we have that (alay) > 1 for some 1 < j < I and hence that (alay) = 1 by the Schwarz inequality. It follows that a — Oj = rjU ^ L (2). Since ht(a - af) = k - 1 > 0, a - ay e A+, we may assume inductively that ^ Then

shows that g m. It follows that m = 9+. (ii) Use part (i) and the relations [X^,, X_^] = -a^. (iii) If L = Lj _L L 2, then L(2) = L^d) V ¿>2(2). The subalgebra 9 ^; i = 1,2 of 9^ spanned by L^ and [X^\a g L f l ) ) is evidently isomorphic to g^., and it is clear that 9^ = 9 1 © 92 and [9^, 92] = (0). □

3.7

Constructing Lie Algebras from Lattices

287

Theorem 4 Let L be an even positive definite geometric lattice. Let L(2) be the set of elements of norm 2 in L, and let L be the sublattice o f L spanned by L(2). Let K be a field o f characteristic 0, and consider the Lie algebras g= K § = (K 02 where and are the Z~Lie algebras o f the lattices L and L . Let 5 be the centre of g. Then (i) g = 3 X (ii) 3 is the subspace o f K L orthogonal to K ^ respect to the invariant bilinear form o f given by Proposition 2. Moreover dim(3) = rank(L) - rank(L'),

and

[g ,g ] =

(iii) ^ is semisimple. I f L(2) = Aj V • • • V is the decomposition of L(2) into indecomposable root systems and L^ is the sublattice o f L generated by A^, then = IK 9/, is a simple Lie algebra and ^ X *** X In particular g is simple if and only if the root system L(2) is indecomposable and spans L. Proof Let I) = K

L. Then g = i) ©

© KX^ aeL(2)

is a triangular decomposition of g. Let X=a

a:

e

3, and write

^ x “, aeL(2)

where a and a:“ e By the way the multiplication in g is defined it is clear that all the ;c“ = 0 and that (ala) = 0 for all a e L(2). In other words, 3 c L' , where L'^ := (jc © f)l(A:|y) = 0 for all y © L'}. Conversely, it is clear that L' c 3, so we have 3 = L' - rank(L') = rank(L) —rank(L'). Moreover g = 3 ® (K 02 L') © and since [3, §] = (0), we have g = 3 X

and dim(3) = dim()^)

= 3®

288

Lattices and Finite Root Systems

From Proposition 3, ^ X ••• X Suppose that we know that each is simple. Then, since 3 = (0) ^ L = L', g is simple if and only if L = L' and L(2) is indecomposable. To finish the proof of the theorem, we prove that each is simple. For this we may assume that L is generated by L(2), that L(2) is indecompos­ able, and that g = § = Let a be a nonzero ideal in We begin by showing that a n L # 0. As above, let ]^ = IK 02 Then a is an 1^-module by the adjoint representation and hence is a sum of 1^-weight spaces (Proposition 2.1.1). The weight spaces of § are f) and the one-dimensional spaces KX^, a ^ L(2). If e a, then [X_^, X^] = a e a. Thus in all cases a n L ¥= 0. Let {«1, . . . , a/} be a base of L(2), and let 0 9^= g a n L. Then for at least one ¿, (j8|a^) 0, and we have [X^,,p] = -(a^\IB)X^, e a. Thus suc­ cessively X^,, X^], and X_^_ = a-]/2 lie in a. The argu­ ment can be repeated to deduce that X^,, aj, X_^. are in a if (aj\a^) 4^ 0, that is, if «y and are joined by an edge in the Coxeter-Dynkin diagram. Since L(2) is indecomposable, the Coxeter-Dynkin diagram is connected (Proposition 3.4.4) and hence all the X lie in a. By Proposition 3, a = ^. □

Remark 1 We have shown that the Lie algebras of type Ai, Di, F7, and £g over fields of characteristic 0 are simple. It is not hard to consider­ ably relax the restriction on the characteristic. In the light of Proposition 3, we have a set of Lie algebra generators of which arise from a choice of a base for the underlying root system L(2). There is obviously something rather special about this since we know that bases of L(2) are all conjugate by the Weyl group W, In the next section, where we look at the way in which the sign problem is solved, we will also see that W lifts (in a slightly subtle way) to a group of automorphisms of g^. These automorphisms take a given set of generators into some other sets {±X^^^J, w ^ W, showing that all the generating sets obtained in choosing a base of L(2) are essentially alike. This means that any relations that we find amongst the are essentially invariants of the algebra (relative to L). Let L be an even positive definite geometric lattice which is generated by L(2). Let {«1, . . . , a/} be a base of L(2), and let A = (A^j) be the Cartan matrix of L(2): A^j = (ajay). Set

( 10 )

fi

a/’

^

1, . . . ,/.

Then we easily have (11)

[hi,ej] =Aji€j, [ h i , f j ]

=

[ci, fj] =

- A j i f j ,

for all I, y e ( 1, . . . ,

.

3.8

Central Extensions of Lattices

289

Now consider i * j. If = (ay|a,) = 0, then a, + i A and = 0. If = - 1, then a^ + aj e A, but 2a, + aj i A, so [e,,[c,, e^]] = 0. Since these are the only possibilities, we obtain ( 12 )

(ade,.)

= 0

for all i

j,

(a d /,)-^ " " V , = 0

for all i

j.

and similarly

The relations (11) and (12) provide the correct starting point for constructing a far more general class of Lie algebras. This is the subject of Chapter 4. The apparently unnatural ordering of the indices which appears in (11) and (12) is chosen to conform with this more general setting. Of course it is immaterial here, since A^j = Aj^.

3.8

CENTRAL EXTENSIONS OF LATTICES

Let L be an abelian group, and let be a central extension of L by a (necessarily abelian) group A. By definition this means that we have an exact sequence of groups ( 1)

0

0

and that i{A) lies in the center of . It is convenient to adopt additive notation for L and A and multiplicative notation for (which will not be abelian in general). Let (f>: L ^ be a section of L in L " , that is, an arbitrary mapping satisfying tt • (¡>= id^. For convenience we assume that <^(0) = 1. Then every element of L" is uniquely expressible as for some a ^ L and a E: A. When the maps (j>and i are understood, we will sometimes, for convenience, write (a, a) instead of Given a central extension L" of L by yl, as above, and a section of (f) of L in L ^, we define a mapping e: L X L

>A

by

[This definition makes sense, since i is injective and (a + p)~^(f>(a)(f>(p) e ker(7r) = i{A).] If our central extension is fixed, then the map s depends on the choice of section (f>. Later in this section we will discuss this dependence

290

Lattices and Finite Root Systems

in more detail. Notice that

( a , a ) ( ^ , b ) = (a)i(a)(p)i(b) = (a)4>(P)i(a)i(b) =
( a , a ) ( ^ , b ) = {a + p , e { a , p ) + a + b),

and therefore s allows us to understand the structure of L ^ in terms of that of A and L. It is a straightforward consequence of the associative law that s satisfies the condition (3) e(a -h p , y ) + e (a ,

= e ( a ,)8 + y)

s(P,y)

for all a, p , y G L.

A map e: L X L ^ A satisfying condition (3) is called a 2-cocycle on L with values in A. One knows (see [Ja2], Chapter 6 [Ro]) that any such map can be used to make the set L X ^ into a group by using (2) as multiplica­ tion; moreover the resulting group L " is then a central extension of L by A. A simple way to obtain a 2-cocycle is to take any biadditive mapping s: L X L -> A. Then e{a p , y ) = e(a, y) + y) and e(a, /3 + y) = s(a, p) + s(a, y) so that (3) trivially holds, and one can easily see how the above construction makes into a central extension of L by We are interested in the case A = Z/2Z. For our purposes it is enough to assume that e is bilinear (this is in fact not a restriction). Then e(2a, j8) = 0 = e(a, 20) for all a, p ^ L, so s determines a bilinear form e: L / 2 L

X

L/2L —

Conversely, any bilinear form s on L / 2 L lifts uniquely to a Z-bilinear mapping on L with values in Z/2Z. At this point we introduce quadratic forms over the field F2 == Z/2Z. This theory was originally worked out by Chevalley [Chi]. The principal thing to be aware of here is that the usual relationship between quadratic forms and symmetric bilinear forms is no longer valid. Instead we have a relationship of quadratic forms to alternating forms and arbitrary bilinear forms to quadratic forms. Let K be a vector space over F2. A mapping q:V

F.

is called a quadratic form on V if the associated mapping b : V x V ---- > F,

3.8

Central Extensions of Lattices

291

defined by (^)

+

—

—^ (y )

is bilinear. It is evidently symmetric. Furthermore 0 = ¿ ( 0, 0) = q(0) + g (0) + ^ (0) = > q{0) = 0, whence b(x, x) = qix) + qix) = 0 for all x, and hence b is alternating. The set Q(V) of all quadratic forms on V is also a vector space over F2 if we use pointwise addition of functions. Let B(V) denote the vector space of all bilinear maps on V, and let denote the subspace of alternating bilinear forms b on V. Obviously the bilinear forms in B%V) are symmetric. By definition we have a linear map II : 0 ( F ) ~ ^ B \ V ) ,

which attaches to a quadratic form its associated bilinear form. Notice that II (q) = 0 if and only if ^ is a linear map (linear maps are quadratic forms). A quadratic form is entirely determined by knowing the values q(xi) and b{Xi, Xj) on a basis {x,}, s j of F. In fact this follows simply from the identity q( x + y ) = q ( x ) + q{y) + b { x , y ) . Given any bilinear form s on F, we may define a quadratic form q^ by = e{x,x) thus establishing a linear map □: B ( V ) ---- > 0 ( F ) with kernel precisely B%V). The associated bilinear form of q^ is b^, where be(x, y) = s{x, y) + e{y, x).

(5)

In the next result we put this backwards: Given a quadratic form q, find a bilinear form e such that q = q^Proposition I

(i) 0 ^ B%V)

B(V)

0 (F )

0 is an exact sequence-,

(ii) if dim(F) = «<<», then dim B%V)

|

Proof Let q G 0 (F ). Let {x,|i e J} be a basis for F over F2, where J is some well-ordered set. We define s by establishing values for £(x„ X j ) for all

292

Lattices and Finite Root Systems

/, j e J. Choose six^, Xj), i < j\ arbitrarily and define e(xj, Xi) = b(Xi, Xj) + s(Xi, Xj)

for / <j,

where b is the bilinear form associated with q. This is forced by (5). Finally, define e(x,,jc,) =q(Xi). Consider the quadratic form defined by e. It agrees with q on the x^ and its associated bilinear form, given by (5), agrees with b on the basis {;cJ. Thus q^ = q, and q ( x) = s{x, x)

for all x ^ V,

Obviously, if card(J) = n, then there are 2( 2) choices for the bilinear form s.

We now turn to the problem that was left unresolved in Section 3.7: to determine a cocycle for a geometric lattice (L,(*| • )) that satisfies equations (3.7.3) and (3.7.4). Proposition 2 Let L be an even geometric lattice. Then there exists a central extension L" of L by Z/2Z with cocycle s : L X L ^ Z/2Z satisfying (i) e(a, j8) + s(p, a) = (a|j8) mod 2 (ii) s(a,a) = I (a |a ) mod 2 for all a, p ^ L. Proof Let L = L /2 L , and let : L L be the natural homomorphism. L is a vector space over F2 = Z/2Z. Define q :L

F.

by q{a) = ^(a\a) mod2. This is a quadratic form with associated bilinear form b ( a , p ) = (a\p) mod2. By the last proposition there is a bilinear form e: L X L q(a ) = s(a, a)

for all a ^ L,

F2 such that

3.8

293

Central Extensions of Lattices

and we know that (6)

e (a , P) + e()8, a ) = ¿ ( a , jS) = (a|/3) mod 2, and

(7)

e ( a , a ) = j( a \ a ) mod2.

Finally define e: L X L ^ Z/2Z by s(a, p) = e(a, $),

□

Remark 1 In order to avoid a lot of cuml^rsome notation we will often omit the overbars in symbols like q{a), b(a, p) writing simply q(a), b(a, p), and so on instead. Remark 2 The proof of Proposition 2 can be used, with only minor modification, to show that for every quadratic form q on L / 2 L there is a 2-cocycle e on L with values in Z/2Z such that e(a, a) = q{a) for all a. The second cohomology group H \ L / 2 L , Z / 2 Z ) of L / 2 L and the space Q of quadratic forms in L /2L are isomorphic as groups. The construction of L" leaves open the question of uniqueness, since there are many choices for a cocycle satisfying condition (7). To establish the fact that L" is unique up to isomorphism, we need Proposition 3. Proposition 3 Let V be a vector space over F2, and let b be any alternating bilinear form on V, Let B = {x^\i E: J } be a basis of V , and for each i e J let c, e F 2 be arbitrary. Then there is a unique quadratic form q on V such that (i) the bilinear form associated to q is b (ii) q(xi) = Ci for all i e J. In particular we have 0 --------

'B(V)

Q(y)

where the kernel of I1 is F* as noted above. Proof We can assume that J is totally ordered. If q exists, it must satisfy (8)

q{x + y ) = q{x) + q{y) + b { x , y ) .

As a consequence, if {w^,. . . , have (9)

is any subset of the basis B, then we must

«(«iMi + ••• +a„M„) =

“y) i<j

for all a-^,,,,,a^

e

F2.

294

Lattices and Finite Root Systems

Use (9) and the conditions qix^) = c, to define q on V. It is straightfor­ ward to show that it is a quadratic form satisfying (8). □ Now let us suppose that s and e' are two cocycles on L satisfying the hypotheses set down in Proposition 2. Let the bilinear mapping 6 on L be defined by 6( a , )8) = e (a , jS) + e '(a , )8)

for all a , )8 e L.

Then & (a ,a )

= e(a ,a )

e'(a,a)

= |( a |a ) + ^(ci\a) = 0 mod2. Thus b is alternating and by Proposition 3 there is a quadratic form u on L satisfying u(a + jS) + u(a) + u(p ) = b(a, p)

for all a, ¡3 ^ L.

Write this in the form ( 10)

s'{a,P) + u(a + )8) = s ( a , ^ ) + u{a) + w(jS).

Use s to define the group structure on L ^ = L X (Z/2Z); that is ( a , a ) ( p , b ) = {a + p , e { a

p) + a + ¿).

Define (!>': L by (f)'(a) = ( a , u { a ) ) . This section gives rise to another cocycle s" of L. Assuming that the extension group Z/2Z is identified as a subgroup of L "" in the obvious way, then s" is defined through

3.8

Central Extensions of Lattices

295

But

= ( a + ^ , s { a , p ) + u{a) + u(l3)) = ( a + ^ , s '( a , )8) + w(a + /3)) = '(a + p ) s ' ( a , p ) . Thus e" = s'. In other words, e' is a cocycle on L arising from some other section of L in L ^. Alternatively, the group extensions and L'^deter­ mined by e and s' are isomorphic: We use the set L X (Z/2Z) and the corresponding group laws given by (2). The mapping 0 : L" L'^defined by (a, a) ^ (a, u(a) + a) is an isomorphism, and we have the commutative diagram

( 11)

Proposition 4 Let L be an even geometric lattice. Up to isomorphism o f the type given by (11); the extension o f L by Z/2Z given in Proposition 2 is unique. □ Let L = ® j^ j Z a j be a free abelian group, and let 0 ---- > Z/2Z

L"

L ---- > 0

be a central extension of L by Z/2Z. Fix a section of L in L ", and let e: L X L ^ Z/2Z be the corresponding cocycle. Define z ^ L " by z *•= /(1) where 1 •= l(mod2) g Z/2Z. Then z^ = 1, and if we adopt the notation e" for (f>{a\ every element of L'" can uniquely be written in the form e“z", where a ^ L and n g Z/2Z. The multiplication of is now given by [see (2)]

( 12 )

_ ^oc+fi^eia,P)+n+m

Let J? be a commutative ring, and let /?[L"] be the group algebra of L" over R. Thus /?[L^] is a free i?-module admitting the set {e“z"|a g L, n G Z/2Z} as a basis and has multiplication given by bilinear extension of

296

Lattices and Finite Root Systems

(12). Consider next the free submodule j of i?[L"] defined by ;■= © a^L

+ e“z)

(the sum is direct since the terms e" + by assumption). By (12) we have ( e ^

+

e ^ z ) e ^ z ^

are linearly independent over R

_|_ ^ a + / 3 ^ e ( a , / 3 ) + H - n

=

= e"^^(l + z ) =

+ e^-^^z e j,

showing that j is a right ideal of A similar argument shows that j is also a left ideal (recall that z is in the center of L ") and hence that j is two sided. Because 1 + z = + e^z, it is immediate that j is generated by the single central element 1 + z. Consider the .R-module homomorphism f:R [L ^]^R [L ] of i?[L"] into the group algebra /?[L] of L (Section 2.5) defined by / : e“z"

e^(-iy.

This map is surjective, and its kernel is precisely j. Hence R[L^]/j == R[L] as i?-modules. The multiplication of the quotient algebra R[ L^]/j induces, via this isomorphism, a new multiplication on R[L]. We denote the resulting algebra by R[LY , and call it the twisted group algebra of L (by the cocycle e; see Borcherds [Bel]). Explicitly R[LY is a free /^-module admitting the family {e“|a e L} as a basis and has multiplication satisfying the identity (13) Evidently f?[L] (with its usual algebra structure) is not in general isomor­ phic to since R[L]' need not be commutative. Let L" be the central extension of L with a cocycle e determined by Proposition 2. Let e Aut(L). We will show that there is an automorphism " of L " (as a group) that stabilizes the central subgroup Z/2Z in L^ and that induces on L. We begin by considering the cocycle on L defined by £*^(«,/3) = e((a),{P))

for all a,f3 ^ L .

3.8

Central Extensions of Lattices

297

Since e * ( a , ^ ) + e'^{p,a) = e{(^)) + = ({a)\ on L by b (a,/3 ) = s { a ,P ) + £"^(«,/3), just as we did above, with b(a, a) = 0 mod 2 for all a. Let u be a quadratic form satisfying (14)

u{a + /3) + « (a ) + m( j8) = b ( a , p ) .

Define '' [in the notation of (12)] by (15)

^>"(£“7") =

We have '{e“z ‘’e'^z‘’) = = ^(a) + {p)^uia) + u ( ^ ) + b ( a , ^ ) + e ( a , P ) + a + b ^

g<^(a)+a>(P)^«(a) +u(£) +£*<“•

_

^ 4 > ( a ) ^ u ( a ) + a^((3)^u(P) + b

= ^{z )= z. Thus ^ is an automorphism of L" fixing z. Evidently ^ lifts on L. Lemma 5 (i) Every lifting o f an automorphism ^ and iff^ are liftings o f (j> and if/ with quadratic forms u and v, then is a lifting o f il/(f> with quadratic form u + v.

298

Lattices and Finite Root Systems

Proof. Begin with the case ф = 1. Let ф" be any lifting of ф. Then

for some Л: L F2. From (12) we see that Л is Z-linear. Now recall that linear maps are quadratic forms, and with и replaced by Л and b replaced by e Hequation (14) is satisfied. This proves part (i) when ф = 1. Now suppose that ф ^ and ф^ are as in part (ii). Then ф^ф"{е^г^) = _

^фф(а)^иф(а) + и(а) + а ^

which is what we wanted to show. Finally to finish part (i) in general, let ф" be any lifting of e Aut(L). Let ф^ be 2i lifting of ф given by a quadratic form u. Then (o^ = ф^ lifts 1 and so is given through some linear map Л, as above. Then ф" = ф"о)^ is given through the quadratic form Л + w by part (ii). □ Define Auto(L") to be the group of all automorphisms of L" that fix z. Each ф^ ^ Auto(L) determines an automorphism of L, and we have an exact sequence (16)

1 ---- --------- > A uto(L^) ---- > A ut(L) ---- ^ 1,

where К is the subgroup of elements of Auto(L^) inducing 1 on L. Proposition 6 к = (L/2L)* {the dual space of L /2 L ) via the map Л •-> Л" for all Z-linear maps Л: L ---- > F2, where A^(c“z^) =

Proof The mappings A" are indeed automorphisms of fixing z, and the proof of Lemma 5 shows that every lifting of 1 e Aut(L) is of this form. From part ii of the lemma, A^ = (/i + A)" for all /x, A e L. □ To obtain results about the automorphisms of the corresponding Lie algebra 9^, we apply the above results on automorphisms to the special case when L is a positive-definite geometric lattice. We will assume that L " is the extension of L given by a cocycle e satisfying the thesis of Proposition 2. We return to the ordered pair notation (a, a) for elements of L '".

3.8

Central Extensions of Lattices

299

Let (l>^ ^ Auto(L^), and let denote the automorphism of L induced by (¡>^, We know that for some quadratic form u, (f>"(a,a) = {(l>(a),u{a) + a) where, for & = e + u satisfies (14). Let = L 0 ®aeL(2)^'^a the Z-Lie algebra of the lattice L, as constructed above, and define an endomor­ phism of g^^ through

= Let us verify that ^ G L, then

a e L (2 ).

is an automorphism of g^. Indeed, if a e L(2) and

= ( - ! ) “<“>( while if a, ^ G L(2) then fO

i f ( a |^ ) > 0 ,

-<^(o)

if («1/3) - - 2 ,

and

iO

if{(a)\(p))>0,

if(<^(a)|«^(/3)) = - 2 . We see that "([X^, X^]) = ["(Xj,"iX^)] by taking into consideration the following facts: .

( a |/3 )

=

(< ^(a)l< f>(i3))

and

(a + P) = (f>(a) +

300

Lattices and Finite Root Systems

• From (14), u(a + /3) + e(a,IB) =

u (a )

+ m( j8) +

b ( a ,P )

e(a,/3)

+

= u{a) + m(/3) + e(a,l3) +

+ e(a,/3)

= uia) + u(p) + ei<j>(a),HP))-

♦ if (a |^ ) = —2 then j8 = —a and hence (16)

m( “ )

Suppose next that <(>'' and

+ m()S) = 0. e Auto(L^), and write

4>"(a,a) = { ^ ( a ) , u { a ) + a), i/r"(a,a) = (tl /(a),v(a) + a).

Then by Lemma 5 il>"4>"(a,a) =

, {u + v
Therefore for all a e A, (u+v4>Xoi) ^

(«)•

On the other hand,

= (-l)

(u+vXa)

X.

Thus we have established the existence of a natural homomorphism 4> of Auto(L^) into the group Aut(g^, L) of automorphisms of g stabilizing L (see Section 2.1). Proposition 7

If L(2) generates L, then AutoCL"") - Aut(g^, L) under the mapping i>. Proof If e Auto(L") determines the identity automorphism of g^, then (f> induces 1 on L, and hence the quadratic form determined by is a linear

3.8

301

Central Extensions of Lattices

mapping A by Proposition 6 and (16). Since A van­ ishes on L(2) and hence on L. From (15) we see that (f>" = 1 and hence that is injective. Now let e Aut(g^, L). Then by restriction r¡^ determines a (group) automorphism of L. By IK-linear extension of ry" we obtain an automor­ phism of IK ® stabilizing IK <8>L. Extend (*| * ) to a IK-bilinear form on K 0 L. Recall that K 0 g^ has a triangular decomposition with root spaces K0 corresponding to the roots a: h ^ (a\h) (see Section 3.7). By Proposition 2.1.3, 7]^ determines an endomorhism * of (IK 0 L)*, which permutes the roots. Using ( 1 ) to identify IK 0 L and (IK 0 L)*, we may view *(a) where

e IK^ (actually

foT 3.11 a e L(2),

e {1, —1}). Now we have

(Q^ I^ )^ /3 ^ < /> *( /3 ) “

'*7^ ([<^5 ^ / 3 ] )

~

['*7^ ( ^ ) j

= a^(*(^))AT^*(^), from which (17)

(a\p) = ((a)\*{p))

for all

L(2).

On the other hand, the equation V - { [ X , , X p ] ) = [r,-X^,v-Xp] together with the definition of multiplication on g^ leads directly to (18)

(a\p) = (^(a)\cl>*(p))

fo ra lla ,i8 g L (2).

Since L(2) generates L, (17) and (18) give = 0

and

(i> eA u t(L ).

Let denote both a lifting of 4> to Auto(L") and the corresponding automorphism of g stabilizing L. Then ¡1" :=" “ ^77" e Aut(g^, L) and pointwise fixes L. Let { aj,. . . , a„} be a base for L(2). Then = c¿X^,, where c¿= ±1. Let A: L F2 be the unique linear mapping defined by Ci = ( - 1)'^^"'^ Using Proposition 6, we may lift A to A" ^ K and replace is surjective. □

302

Lattices and Finite Root Systems

Remark 3 As an important application of these last two propositions we consider the Weyl group W of L. We assume that L is positive definite and that L(2) generates L. Then, since W c Aut(L), by (16) and Proposition 6, W has a preimage W" in Auto(T"), which is a 2*^^"*^^-fold cover of W. According to Proposition 7, W" may be identified with a group of automor­ phisms of stabilizing L. If w" e and w" ^ w then (19)

W^\l = W,

and for all roots a, ( 20)

= ±X ^^.

The group W is sometimes called the Demazure-Tits group. For more on its structure, see [MPS].

EXERCISES 3.1 Let 6i,

be the standard basis for /-dimensional euclidean space

(a)

Show that {±e, ± Sj\i j} U { ± e j is a root system of type and that {±e^ ± Sj\i # /} U {±2eJ is a root system of type Q (see Sections 3.1,3.3,3.4). (b) Show that the Weyl groups of these root systems are isomorphic to Si tx (Z/2Zy. (Interpret this as all permutations and sign changes of the e/s.) (c) Consider the root system {±Si ± 6j\l < i, j <1, i ^ j) of type D/ (Sections 3.1,3.3). ( i) Prove that the long roots of Bi and the short roots of C / form root systems of type (ii) Prove that the Weyl group of Di is isomorphic to 5/ K (Z/2Zy~^, (Interpret this as all the permutations and even number of sign changes of the e/s.) (d) Show that the set {±£, ± £y|l ^ i, j < 4, i ¥=j} U {^{ ±£i ± £2 ± ^3 ± «4} u (± £ ,|l < i < 4}}

(e)

is a root system of type F^. Show that both the sets of short and long roots of form a root systems of type D^. Let A be a root system of type D^, Show, by looking at the Coxeter Dynkin diagram, that [Aut(A): W(A)] = 6. Show that Aut(A) is isomorphic to the Weyl group of type F4.

Exercises

303

3.2 Compute the determinants of the root lattices of types Bi, Ci, and G2. 3.3 Let A be a finite indecomposable reduced root system. Let II = {ai,. . . , a/} be a basis of A. Show that there exists a unique short root in A of maximal height (with respect to II). Show that if = then p is dominant, and hence all c, > 0. p is called the highest short root of A. Compute this root for A of type G2, F4, Bi, and C/. 3.4 Let A be a finite indecomposable reduced root system of rank / on a K-space V. Let II = { aj,. . . , a j be a base of A. Let A^ be the dual root system and II [a^ , . . . , a /} the base of A^ dual to II. Recall the root and coroot lattices Ô == E 2 a = 0 Za, c V, ae A

^= 1

Q ''= E 2 aeA (a)

a

I © Za^cz V * . ^'=1

Let P^:= [y e F*|<jt:,y> e Z for all x ^ Q ) and P := [ x ^ K|<x,y> e Z for all y e

c K.

Show that P (resp. is a lattice in F* (resp. V) (called the weight and coweight lattice respectively). (b) Show that Q P and that the group P /Q is finite (the order of this group is called the index of connection of A. The cosets (o Q, (0 ^ P, are called the congruence classes). (c) Let (o^,. . . , (Oi ^ P be defined by a ^ ) = 8^j. (The co/s are called fundamental weights. They form a basis of P dual to the basis of (2^.) Given a, e II, show that Oii where A = (-^¿P is the Cartan matrix of A. (d) Show that the index of connection of A equals det(^). (e) Let W be the Weyl group of A. Show that W stabilizes P, thereby acting as automorphisms of P. Show that W acts trivially on P/Q,

304

Lattices and Finite Root Systems

(f)

Compute P / Q and the congruence classes for all indecomposable reduced root systems A.

3.5 Let A c K be a reduced finite root system of rank /, and let II = be a basis of A. Let Q and P be the root and weight lattices. (See Exercise 3.4.) (a) Let Z[P] be the integral group algebra of P. Show that Z[F] has no zero divisors. (b) Let W be the Weyl group of A. Then W acts as a group of automorphisms of Z[P] via w' • E n^e{ii) = ^ n^e{wii.). fX^P /i, sP

(c)

Let Z[P]^ := {jc e Z[P]\wx = x for all w e W}. Let K' and K denote the field of fractions of Z[P] and Z[P]^, respectively. Show that Z[P] is the integral closure of Z[P]^ in K' and that K' is a Galois extension of K with Galois group canonically isomorphic to W. Fix iV e Z+, and let ^ e C. Fix z* e 0 and define ring homomorphisms

by

and by restriction f:Z {P T Let p' = k e r/', p = k e r/, k' = field of fractions of Z[P]/p', and k = field of fractions of Z [F ]^/p. Show that k' is a Galois extension of k with Galois group canonically isomorphic to Wd/ W j , where Wo = { w ^ W \ w V ' = X>'), Wj = [w e W\wx = X mod p' for all x e Z [F ]} . Let i := {/i e

z*) = 0 mod N). Show that

IV^ = {w ^ W\w\ = i}, W[ = [w ^ W\wii. = fi mod i for all

/j,

e P }.

Exercises

305

3.6 Consider a connected affine Dynkin diagram F with nodes / e { 0 ,..., and marks rii. We play a “game” on F as follows: Step 0: Pick a node (starting node). Step 1: Remove the last node j you picked, and let Fy be the diagram left over (y = if you are starting). Step 2: Compute the determinant of the root lattice with diagram Tj. If Dj = rij, you may continue playing the game. Otherwise, stop: You’ve lost the game. Step 3: Pick a node connected to the last node you removed. If you cannot do this, you have finished playing a winning game. Otherwise, return to Step 1 and keep playing. Show that in the only diagram all of whose games (24 of them) are winning games. 3.7 (J. Milnor and D. Husemoller, Symmetric Bilinear Forms, SpringerVerlag, 1973). In this exercise we construct the famous Leech lattice: the unique positive-definite even unimodular lattice of dimension 24 with no lattice points of square length 2. Let F2 = {0,1} be the field with two elements. For N define §0 := (1 < i < N\Si = 0} = {1 < i < N \S i = 1}. Let V = (i^i)o^i^io = (1, h 1,0,1,1,0,1,0,0,0) e 11 square matrix A = (a^j) with entries in F2 by aij := i^(/+y-2)modll

for all 1 <

Define an 11 X

j < 11.

The 11 rows of this matrix we will denote by a^ ,. . . , Let |5| denote the cardinality of an arbitrary set S, (a) Show that \a] n a]\ = 3 for all 1 < i < j < 11. (b) Fix S c { 1 ,..., 11}. Let s(5) = s = ^ F2^ where by con­ vention s = 0 if 5 = 0 . Show that |s^| = 0 mod 2 and that for all j € S, |s^ n aj\ - |s^ n aj\ = 0, ± 4 if |5| is odd and = ±2, ± 6 if |5| is even.

306

Lattices and Finite Root Systems

(c)

Let B = (bn) be the 12 X 12 square matrix defined by 0 1 B ■■= 1

1 1

• •

1

A

1 Show that any two distinct rows of B are orthogonal and that B^ = I. (d) Let C be 12 X 24 matrix obtained by gluing together I and B

C ■■=

/1 0

1

0 0 B 1

,0

t

and let 7 1 , , 712 ^ be the rows of C. Show that I7/I = 8,12 for all 1 < I < 12. Show that any two distinct rows of C are orthogonal. (e) Let S c F|‘* be the span of the rows of C. Show that (i) (1 ,1 ,...,1 ) g 5, (ii) Is^l = 0 mod 4 for all s G S, (iii) |s^| > 8 for all s G 5 \ {0}, (iv) dim,:^ 5 = 12. (f) Let L q c be a lattice with an orthogonal basis (b,)i^is24 satisfying b, • b, = i- Define two subsets E and I of L q as follows. 24

^

{L U= i • i/ ^ 2Z,

such that for all 1 < / < 24,

24

•

= 0 mods, i=\ mod 2 , .. ., 1^24 niod 2) e S.

/ = I 51 tib}j such that for all 1 < / < 24, •

^ 2Z + 1, 24

• 51 ^ - 4 mods, /=1 • { j( t i + 1) m od2,. . . , ^(^24 + 1) mod2) e 5.

307

Exercises

We define the Leech lattice L by L = £ U / c L q.

Show that if jc = ^ L, then 'Ltf = 0 mod 16. Use this to show that L is an even integral lattice. (h) Show that [L qI L] = 2^^. (i) Let

(g)

M = {(^¿) e

= 0 mod 2 for all 1 < / < 24},

and let 0=

e

= OmodS^.

Show that [Mo: E] = 2^\ (j) Use (h) and (i) to show that L is unimodular. (k) Show that L has no vectors of square length 2. 3.8 Follow all the constructions of Section 3.7 to construct explicitly the Lie algebra of the root lattice of type Follow Section 3.8 to explicitly determine the cocycle needed for your construction. 3.9 Consider the root system of type Di A = [ ± s ^ ± sj\l < i, j < l , i ^ ;} and its base n

{a^,. . . ,

where —^2? • • • ?^ /- i ^ ^/-i ~ ^ ^i-i (a) Show that the automorphism cr: -> W which fixes 1
308

Lattices and Finite Root Systems

(d)

(e)

Show that the fixed points in of o- form a subalgebra I and that f is linearly spanned by the elements of S, the elements a = d, and the elements + X^, a ¥=a. For each define 6 ^ = 2 5 /(5 |5 ). Let a , /3,y e A , and suppose that a ^ a, ^ y = y. (0 Show that {X^ + X^, a d, -X _ ^ - X_^} is an ^l2-triplet. (ii) Show that [a + d,X^] = ( fla ^ )Z ^ , [y,\x,+x^] =

[a + d , X p + X^] =

+ x^).

(iii) Let fi

^ / —1 ^ /-1

^ a i_ i

=

^ /-1

^

f l —l

1, . . . , /

2,

a / _ i —a/ J

+

Prove that e^,. . . , ei_^, h^,. . . , hi_^, f ^ , . . . , fi_^ satisfy the relations (11) and (12) of Section 3.7 for the Cartan matrix

of the base [d^,. . . , a/_i) of 2. (f) Determine a triangular decomposition of I. (g) Let K be a field, and let = IK Show that the elements 10 1 0 h^, 1 0 /^ derived from (e) (iii) generate as a Lie algebra. 3.10 (This exercise will be used in the exercises of Ch. 6). Let T be a graph that is a forest with vertices /}. Suppose that G is a group and that

is a mapping with the property that whenever i and j are not joined, /(/) and / ( ; ) commute. We wish to show that for all permutations tt of { ! ,..., /}, := /(7 t(1)) *• • f(7r(l)) lies in the same conjugacy class. (a) Consider / points (also labeled 1 ,..., /) placed uniformly in some order on a circle. Allowing only the rule that adjacent points i and j on the circle may be interchanged if / and j are not joined by an edge of F, show that (up to cyclic permutation) any ordering of the points 1 ,..., / may be obtained from any other. (b) Use (a) to prove that the conjugacy class of is independent of 77. 3.11 Prove Proposition 3.4.1

Chapter Four

Contragredient Lie Algebras T h e o p in io n s e e m s to h a v e g o t a b r o a d t h a t in a f e w y e a r s a l l th e g r e a t p h y s ic a l c o n s t a n t s w i l l h a v e b e e n a p p r o x i m a t e l y e s t im a t e d , a n d

th a t th e o n ly o c c u p a t io n

w h ic h w il l t h e n b e l e f t t o m e n o f s c i e n c e w i l l b e t o c a r r y o n t h e s e m e a s u r e m e n t s to a n o th e r p la c e

of

d e c im a ls . . . .

But

we

have

no

r ig h t

to

t h in k

th u s

of

th e

u n s e a r c h a b le r ic h e s o f c r e a t io n , o r th e u n t r ie d f e r t ilit y o f th o s e f r e s h m in d s in t o w h ic h t h e s e r i c h e s w i l l c o n t i n u e t o b e p o u r e d .

—James Clerk Maxwell, 1871, quoted from Abraham Pais (Oxford: Clarendon, 1991).

N ie ls

B o h r 's

T im e s ,

A contragredient Lie Algebra is, roughly speaking, a Lie algebra g, with a triangular decomposition S^, that is generated by the diagonal algebra if and subalgebras ^ l 2\ j ^ J, each isomorphic to êl2(!K). The remaining chapters, beginning with this one, are devoted to the general structure of contragredi­ ent Lie Algebras and constitute the core of the book. Section 4.1 introduces the contragredient algebras and concentrates largely on the necessary and sufficient conditions for the subalgebras t>e “integrable” to actions of 5L2([K) as an automorphism of g. The concept of the Weyl group IV of a contragredient algebra arises naturally in this context, although a full discussion on the structure of W is the subject of Chapter 5. Every contragredient algebra has an associated structure matrix A = {A^jX i, j, e J that plays a similar role to the Cartan matrix of Chapter 3. Using a relaxed version of the conditions CM of Section 3.3, namely CM1-CM3, we prove the basic result that a radical free contragredient algebra is integrable if and only if its structure matrix is a Cartan matrix. Section 4.2 is devoted to the existence of contragredient algebras with a given structure matrix A. The construction proceeds by first constructing a suitable diagonal subalgebra i) (a realization of and then by constructing a universal contragredient algebra u around which has A as its structure matrix. If y4 is a Cartan matrix, then u/rad(u) is integrable, thereby proving the existence of integrable contragredient algebras. There are a number of things about embeddings, field extensions, radicals, and decomposability

309

Contragredient Lie Algebras

310

which are rather obvious but nonetheless need to be written down. These make up Section 4.3. A very important class of contragredient algebras consists of the invariant (or symmetrizable) contragredient algebras, those carrying a (proper) invari­ ant bilinear form. Such a form can exist only if the structure matrix of g is symmetrizable (a generalization of symmetric). After establishing this, the remainder of Section 4.4 is devoted to showing the converse of this result. This is a crucial fact since most of the subsequent development depends significantly on the existence of such forms. For instance, in this section we have the construction of Kac’s generalization of the (quadratic) Casimir operator. In reality this is a whole family of operators one for each module in the category ^ (g , ^ ) , with the property that and inter­ twine g-module maps from M to N, The construction of these operators is possible (as far as is known) only if g is radical free and invariant. The Casimir-Kac operator is used at critical points throughout the entire theory, mainly as a tool to keep track of the type of subquotient modules that can occur in modules from ^ (g , *^). Our first application of these operators appears in Section 4.6, where we prove the Gabber-Kac theorem. This gives information about the generators of rad(u), where u is the universal contragredient algebra of a symmetrizable structure matrix. An important consequence is that one obtains an explicit presentation for each integrable invariant contragredient algebra g (always assuming symmetrizability) and in fact concludes that integrable is equivalent to radical free. A slight modification of the Casimir-Kac operator gives us an operator F acting on g itself. In Section 4.7 we define F and use it to prove the remarkable fact that there exists a contragredient bilinear form on g, obtainable by a slight twist of the invariant bilinear form, that is positive definite on the nonzero root spaces of an integrable invariant contragredient algebra g over R. This leads to the existence of a hermitian contragredient form on g^, which is likewise positive definite on nonzero root spaces.

4.1

CONTRAGREDIENT L IE ALGEBRAS

Let g be a Lie algebra over a field K, We say that g is contragredient triangular or simply contragredient if there exists a triangular decomposition * ^ = (g + ,^ ,í2 + ,cr)o f g satisfying the following: CTl: If {«ylyej ^ is the given basis of according to the axiom of triangular decomposition TD4 of Section 2.1, then the Lie algebra 5/^^^ generated by g“^ and g““^ is isomorphic to ^12(1K). CT2: The elements of g“>, g"“^ j e J, and generate g (as a Lie algebra). CT3: The sum Eye 9“^] is direct.

4.1

311

Contragredíent Lie Algebras

If we wish to indicate the triangular decomposition relative to which g is contragredient, we refer to the contragredient Lie algebra or contragredient pair (g, S^). Let j G J. Using CTl and cr it is clear that dim g“> = dim g”“^ = 1. Let Ej G g“>, Fj G g~“/ be any nonzero elements. Then Hj ~ [Ej, Fj] ^ 0 (by CTl) and [Hj, Ej] = kjEj for some kj # 0. With Cj ■■=2 k J % , a/== I kJ^Hj, fj •■=Fj, we have an il2-triple for §1^^: ( 1)

[«;

2 e y , [tty , / y ]

2 / y , [^ y ,/ y ]

C ij •

There is no reason why aej should be /y, but we can easily alter a to have this happen. Indeed for each j e J, let crej = tjfj, tj e Then there exists an unique automorphism 0 of g such that Oej = iyey, = t-% e\if = id^. (see Section 2.1, Remark 2). Of course 0g“ = g" for all roots a. Set r = da. Then T is an antiautomorphism of g satisfying r\t) = id^. Furthermore T6j = 0iy/y = /y, r/y = = 6j. Thus by modifying the triangular decom­ position ^ by replacing a by r, we can assume at the outset that cr(ey) = fjJ e J. Note that = id from CT2. Henceforth we assume that for each j e J, Cy e g«^/y e g-->,

a /e i|

satisfy (1) and ( X l C j ^ f j .

We call such a choice of

Cp

a j, and /y a display of (g,

The display {cy, a j , /y}ye j is finite if J is finite. Remark 1 Let {cy, a / , /y}ye j and {e'p , //lye j Then there exists a family {^ylyej such that e '= t e

two displays of (g, S^).

f- =

Since a{e'j) = / / we obtain iy = ±1 for all j. In particular a / is unique, since it is the only element in [g“>, g““^] that has g“^ as a 2-eigenspace. We leave it as a simple exercise to show that if every element of IK has a square root then displays can always be found without altering a.

312

Contragredíent Lie Algebras

We define yej Remark 2 The assumption CT3 is extremely mild. In fact it is not used here at all. We retain it to avoid a certain awkwardness in the presentation, but it is straightforward to adjust the results appropriately in the case that the set {a/ly e J} is linearly dependent. In Chapter 5, when we consider general root systems, dependence relations are an important consideration. Proposition 1 Let (g, T) be contragredient, and let [ej, a j , Then

be a display.

(i) for all i e J, dim g“' = 1; (ii) g+ is generated by {ej\j e J} and g_ is generated by {fj\j e J} as Lie algebras; (iii) for all a ^ L \ {0}, dim g" is finite. Proof, (ii) Let g'+ denote the subalgebra of g generated by the e,, i e J, and g'_ the subalgebra of g generated by the /,, i e J. It suffices to show that 9' == g'_+ ^ + g'+ is a subalgebra of g. First, ad /i(g') c g' for all g is clear. Next it is easy to see that ad

, e ,J e ]^ + g'^.

for all A: > 1,

and hence ad /y(g'+) ^ il + 9'+the same way ad ey(g'_) c + g'_. This proves part (ii). Since [e,|,. . . , e,^] lies in gt^“o and g“' is spanned by e„ it is clear that dimgt^“' is finite. Using cr we complete part (iii). □ We next attach to (g,

a matrix A = (/1,^), An

by defining

= ( a ,- ,« / ) .

The matrix A is called the structure matrix corresponding to (g,T). The condition eJ = 2ei shows that An = 2

for all/.

Example 1 Let g be one of the Lie algebras of type A , £>, and E given in Section 3.7. Each of these is based on an even lattice L in which L(2) generates L and forms a finite root system. For each we have a base

4.1

n = {«1,. . . ,

Contragredient Lie Algebras

313

relative to which L(2) = L(2)+U L(2)_. We have g=

©

0 g“ = 0 g"“ © Í) © 0 g“, asU2) asi,(2)+ a^U2)^

the latter being a triangular decomposition of g. The spaces 9"“' + [9"''^ 9“^] + 9"^

j = 1, • • •, ^

are isomorphic to ^ Í 2ÍK) and collectively generate g as a Lie algebra. It follows that g is contragredient. The elements i^y>^y»/y}y=i,.defined in (3.7.10) form a display and the corresponding structure matrbc is the Cartan matrix A of L(2). Example 2 (§I2(IK)) (see the Appendix iT¡(lK) = < /o,/i> e $ ® <eo,ei> determines a triangular decomposition of ^l2(IK). The subalgebras = Kf¡ + Kh¡ + Ke¡,

j = 0,1,

are isomorphic to §I2(IK), from which we see that §I2(IK) is contragredient. The corresponding matrix is (-^

1

)'

We now define a very important group W (called the Weyl group) of linear automorphisms of As we will see, the Weyl group is closely related to the group Aut(g, ]^) of elementary automorphisms of g that stabilize if. Initially we use W to get a better idea of what the root system A looks like. Let j e J. Define 0: r by rji p ^ P - <j8, a ^ ) a j

for all P

and

by rjy : h ^ h — (aj, h ) a j

for all /2 e

It is easy to see that r,(ay) = - a ^ and that Vj pointwise fixes the hyperplane = {j8 e ]^*|<j8, = 0} of Thus rj is a reflection of f)* in aj

314

Contragredíent Líe Algebras

(Section 3.2). Similarly ry is a reflection of ^ in a / . In particular r / = id^*,

r / ^ = id^,

rj^G L ir),

r/e G L (^ ).

Notice that 0(

^ u

’

Thus rj stabilizes Q stabilizes

A^j ^ Z

for all / g J,

A

for all / g J.

e Z

The group W c GL(ij*) generated by those r, for which ad Cj is a locally nilpotent transformation of g is called the Weyl group of (g, T): W = (rj\j G J, ad Cj is locally nilpotent). Similarly we define the dual Weyl group GL{^),

W ^= {r^ I ;

G

J, ad

is locally nilpotent).

If none of the ad ej are locally nilpotent, then by convention W = {1} and W ^= {!}, where 1 denotes the identity map of the appropriate space. Remark 3 Both W and depend on (g, but not on the chosen display. The reason for asking that ad ej be nilpotent will become apparent as we move along. In the end we are going to be interested in the case when ad 6j is locally nilpotent for every j g J. Readers seeing this for the first time might like to assume this from the outset. Proposition 2 (i) W ^ and W are isomorphic groups, an explicit isomorphism being given by r,, j G J. (ii) I f these two groups are identified by the isomorphism in part (/), then (a , h) = { w{ a ),w { h) )

for all a

G

1^*,

G

1^, and w E:W.

4.1

Proof. Let ^ e 3.2.6). We have

315

Contragredient Lie Algebras

and /i e

Recall the inverse transpose (Proposition

= { ^ ,h ~ { a j,h )a f) = {l3 ~ { l 3 ,a ^ ) a j , h ) = { r j( ^ ) ,h )

fo r a ll/3 G f)*,

G f).

This shows that (ry^)* = rj and also shows that (2)

{ ^ , r / ( h ) ) = { r j ( p ) ,h )

fo ra ll^

Now it follows that fr/

V

Jk

J

= r,j \ . .. r ,Jk

and hence that (3)

(■ )* :W ''^W

is a (surjective) group homomorphism. Now consider the inverse transpose r* of rj, y

^^

Then for all t e 1^** and for all /3 e 1^*, (4)

{ P , r * i t ) ) ={ rj(P),t) ={l3,t-{aj,t)a^).

Here we have written the pairing of 1^* and 1^** as < • , • > : f|* X

^

IK

and have identified a f g 1^ as an element of 1^** via the canonical identifi­ cation of 1^ in 1^**. Thus (4) reads rr If. = r / , and we have a homomorphism (5)

W ^ W '^

Contragredíent Líe Algebras

316

given by VT

w* 1^

which is evidently the inverse of the homomorphism (3). (ii) This follows immediately from part (i) and equation (2). Proposition 3 Let t ^ K^, and let j e J. Assume that ad ej is a locally nilpotent endomor­ phism of g. Then (i) n^U) '= expad(iey)expad( —i~ypexpad(iey) is an {elementary) auto­ morphism of g stabilizing f); (ii) ny(i)li, = r /. Proof Since ad e^ is locally nilpotent so is ad aiej) = ad fj. Thus nj(t) is an automorphism of g (Proposition 1.5.3). Let Hj = { h ^ f i \ ( a j , h y = 0}. Since (oij, a j ) = 2,

and since [h, ej] = 0 = [/i, fj]

for all h e Hj,

we have \h ¡ =

1-

On the other hand, 0 (« /) =

ad(tey)exp ad( - 1~‘/ y ) ( a / - 2ie^)

= exp ad(ie^) ( - « / - 2iey) = —a¡ This shows that n^it) stabilizes 1^. Moreover let with W e H:. Then

e 1^, and write h = ka^ + h!

nj{t){h) = k a j h ! = h — { a j , h ) a j = r]'{h).

4.1

317

Contragredíent Lie Algebras

Proposition 4 Let w

and write w = rj . . . r j . Let

...,

e

be arbitrary. Then

(0 n = n{w) •= is an elementary automorphism o f g such that ng“ = g^“ for all a e A. Moreover n\\^ = w under the identification o f W with W ^\ (ii) ivA = A and dim g“ = dim g"^" for all o: e A; (iii) for j ^ J with ad Cj locally nilpotent and for all w El W, the elements of g"^“^ are diddocally nilpotent. Proof (i) We see that n stabilizes by Proposition 3, and hence by Propositions 2.1.3 and 2 above we obtain ng“ = g'^/i o*« = with n |^ = w, as desired. (ii) This follows from part (i). (iii) «(g“0 = g'^“^ and n is an automorphism. Since g“>= Ke^, g"^"> = IK/ie , and part (iii) is immediate. □ Of course the n{w) of Proposition 4 is far from unique for a given w. We come back to this point in detail in Chapter 6. Notice, however, that if n{w) and ri{w) are as above, then n{w\n'{w))~^ e K, where K c Aut(g, 1^) is as in the discussion following Proposition 2.1.3. Let (g, ^ ) be contragredient. A root « e A is said to be real if a = wa^ for some w EiW and some ; e J for which ad e^ is locally nilpotent. If a root is not real, then it is said to be imaginary. Set '’^A := (a e A I a is real}, '^A := A V"A, '■"A^:= A+n""A, '■"A_:= A_n''^A, ^■'”A^:= A+n^'”A, "'”A_:= A.n'^^A, n =={«;!;• e j } ,

n'^== { a / l / e J},

:= {«y I y G J, ad €j is locally nilpotent}, ‘11'^:=

I y G J, ad Cj is locally nilpotent}.

318

Contragredient Lie Algebras

Thus "A = w r W . We have

g= © 9“ ae A

0

9“ ©

0 g"" 0 ae"A_

0

0 g"" ® 0 9"* ae"^A+ aG''”A+

The root spaces g", a are relatively easy to understand. This is not so for the imaginary root spaces. Proposition 5 Let (g, r ) be contragredient, and let Wbe its Weyl group. Then with the above notation we have (i) (ii) (hi) (iv) (v) (vi)

Q a n A = {a, 0, —a) for all a e'^^A; dim g“ = 1 for all a e'^^A; ^"A = -""A ««¿"'"A = -"'"A; Wstabilizes '^^A/'”A^_, and ""A^= -""A_ and"'”A^= -'^A _; ^"n = nn^"A.

Proof (i) Let a e^'^A, and suppose that na ^ A for some n e Q^.. By assumption there exists aj and w ^ W such that w{na) = naj. Since W{A) = A, it follows that naj e A. Assume that nuj e A+c Q_^. Then by TD4 of Section 2.1, n e Z+. The space g"“> is generated by all expres­ sions of the form [ej^,...,ej^] with 0 J, A: 0 Z^_, and «y. + • • • +ay^ = noj (Proposition 1). It follows that k = n and that j) = j for all 1 < / < A:. Thus g"“>= (0) unless n = 1. If naj e A_ or if n e Q_, we reason in the same fashion. This proves (i). (ii) We know that dim g“>= 1 for all ; e J (Proposition 1). Since dim g“ = dim g"^“ for all a 0 A and w ^ W (Proposition 4) (ii) follows. (hi) Let a e'^^A. Choose w ^ W and aj e'^^II such that a = waj. Then - a = wrjOj e'^^A, showing that '^^A = -'^^A and hence that "'"A = - " ”A (since A = -A ).

4.1

Contragredient Lie Algebras

(iv) We have W C ^ ) Let us show that (6)

if aj

and hence that

319

='"’A, since M A ) = A.

^ n ,th e n rX A A { a;} ) = AA{«y}.

Indeed, since for all n e Q^, naj ^ A+\{ay}, we can establish (6) by reason­ ing as in (3.3.14). Now we want to show that +) Let a and let aj e '’^n. Then rja e'^'A and by (4.6) Vja e A+. Thus Vja e A^_n^'”A. It follows that ITa c (A +n^'”A) c^'^A^ which is what we want. Finally, WCA_) = 1^(-^‘'”A+) = -1TC'”A+) = -^■'”A^ = ''”A_. (v) The proof is obvious. (vi) Clear from part (iii) of Proposition 4. □ Example 3 {A, D, E) This continues Example 1. We have A = L(2) U {0}. Since A is finite, the set jo: + kaj I A: e Zj n A is finite for all a e A, ; e J. Thus ad and ad /y are nilpotent. Thus the Weyl group W (in the current context) is

and from the definition of each rj and Proposition 3.3.4, it is clear that W is precisely the Weyl group of the finite root system L(2). The same proposition shows that L(2) = ITIl ='^^A, and hence that "'"A = {0}. The action of W on = LK a^ = (]^*)* is precisely the transpose action discussed in Proposition 3.2.6. The role of the dual root system A^= {Wa^ | ; = 1 ,...,/} is made apparent below. Fix ay Then ad ey and ad/y are locally nilpotent, and we have already had occasion to see, exp(ad te^) and exp(ad i/y), i e IK are welldefined automorphisms of g. By Proposition 2.4.7 the representation == ^I2(IK) Ql(g) given by :X

ad X

gives rise to a representation ^<^>:5L2(IK) ^ A ut(g), satisfying ^ exp(adtó^),

(o

(]

?)

320

Contragredient Lie Algebras

We can interpret this as “integrating” the representation of representation. It is interesting to see that under 7t^^\ 0 - t -1

" ( ( ;

= nj(t)

to a group

:i(;

:il

fo ra lli e K ''

(see Proposition 2.4.7). We denote the group 7r^^^5L2(lK)) by that is,

or SL^{\

:= (exp ad tej,exp ad tfj U e 1K>. According to the way in which g decomposes into irreducible §I^2^^‘niodules under ad, S L f - 5L2(K) or - SL2(K)/{t ± 1} (see Proposition 2.4.9). The group generated by those exp ad te^ and exp ad tf¿, i e K, / e J for which ad and ad are locally nilpotent is called the adjoint group of g and is denoted by ^ad- There is much more to come on <Jad in Chapters 6 and 7. Proposition 6 Let (g, ^ ) be contragredient as above, and let a aj be such U that a = waj. Then

Let w

and

(i) the space := g“ 0 [g“,g "] 0 g “ is a three-dimensional subalgebra o f g isomorphic to ^I2(1K). (ii) The elements of and g~“ are locally ad-nilpotent, and the group SL^2 ^ := (exp ad e, exp ad / | e e g", / e g~“> is a subgroup o f isomorphic to SL2(K) or 5L2(IK)/ {± 1}. (iii) Either of the following properties characterizes the same unique ele­ ment ]^: (iiia) belongs to an 2-triplet of in other words, there exist ^a’fa ^ [ a ^ ,e „ ] = 2 e „ ,

[a'",/« ] = - 2 / ^ ,

[e^,fj= a'^;

(iiib) a ^ = wa^ {under the identification o f W with W ^). Proof By Proposition 4.1.4 there exists n = n{w) «9^ = 9*^^

for all /3 e A.

such that

4.1

Contragredient Lie Algebras

321

Thus 0 [g«y, g-«y] 0 g"«y)

which establishes part (i). Moreover, since n(g“>) = g", it follows that g" (and hence g““) consists of locally ad-nilpotent elements. This shows that is a well-defined subgroup of Aut(g). It is easy to verify that n(expad

= expad(njc)

wherever n e Aut(g) and x e g is locally ad-nilpotent. Thus = SL^^\ whence we have part (ii). From part (i) it follows that {ncj, n a j , n/y} is an ^l2-triplet for Since n\^ = w, the existence of having properties (iiia) and (iiib) follows by setting a^:= As for the uniqueness of we recall that g", g““, Ka ^ are the 1^-eigenspaces of 5/^"^ and that the space [g“, g““] has a unique element that can be extended to an ^ 12-triplet of ^ Remark 4 Let a and let w e IF and be such that a = waj. Let be the (unique) element defined in part (iii) of Proposition 6. Then a ^ = wa^, and this identity holds independently of the choice of w and aj above. We call the coroot of a. The set of coroots is denoted by Notice that from the definition it follows immediately that for all a e ""A and w e IF, {waY=w{a^) and that '’^A^c 2 ^ From [a^, e^] = 2e^ we clearly have = 2. Proposition 7 Let a e '^^A, and let

be its coroot. Then

(i) the reflection r^ E GL(i)*) deflned by

belongs to W: (ii) I f w ^ W and aj

are such that a = waj, then r^ = wrjW~^.

In particular r . = r,;

322

Contragredient Líe Algebras

(iii)

under the identification o f W with h

we have

= r^v, where for all

r^w :h ^ h — {a, h )a ^ . Proof Let w ^ W and Uj way. If e ]^*, we have

be such that a = waj. By Proposition 6,

wrjW~^p = w(^w~^fi - (w~^P,ay)aj^

= )8 - ( p , w a ^ ) a

(by part (ii) of Proposition 2)

= fi - <)8,a^>a which shows that r^ = wrjw~^. This establishes part (ii), and also part (i), since wrjw~^ e W. Finally, part (iii) follows from Proposition 2 (ii). □ We continue to assume that g is a contragredient Lie algebra. Proposition 8 Let A be the structure matrix o f g. Fix j e J. (i)

ad Cy is locally nilpotent if and only if for all i e J, / ^ ;, we have (ia) ^ , y e Z < o , (ib) dfj := (ad Cy)"^'^^e¿ = 0. (ii) I f ad Cy is locally nilpotent, then for all i ^ j, Aij = 0<=^ [e„ey] = 0, ^ , y = 0 = > ^ y , = 0.

(O f course, if ad e^ is also locally nilpotent, we obtain A^j = 0 - ^ y , = 0.) (iii) Properties (i) and (ii) hold if ej and c, are replaced by f andfi. Proof (i) Suppose that ad e^ is locally nilpotent. Since for each i e J, rja^ = —A^jUj e A, and since no root can have ‘'mixed signs,” ^¿y ^ whenever i ^ j. Furthermore + (1 “ ^ i j ) ^ j ) = «<• - «y Í A which shows that

4.1

Contragredient Lie Algebras

323

Conversely suppose that for all i ¥=j, e Z < q and (ad = 0. Note that (ad = 0 for all i and (ad ejYh = 0 for all /z e 1^. Now set a = |jc G g I(ad

= 0 for some rij{x) e l\l|.

It is straightforward to prove that a is a subalgebra of g. Since a contains a set of generators of g (see CT2) we have a = g. (ii) Suppose that ad ej is locally nilpotent and that A¿j = 0. Then (ad ej)e¿ = 0 by part (ib), and we have ^ji^J = [«."'>«;■] =

= 0

by the Jacobi identity. Thus Aj¿ = 0. This last argument works equally well to show that [e¿, ej] = 0 =>A¿j = 0, thus finishing the proof of part (ii). □ A contragredient Lie algebra (g, y ) is called integrable (relative to if for some (hence any) display {ej, a^, ad 6j (hence, using the involution (7, also ad fj) is locally nilpotentfor every j g J. Example 4 The Lie algebras A, D, E and §I2(IK) are integrable.For an integrable contragredient Lie algebra the Weyl group W is generated by the set of all rj, j G J. Proposition 9 The structure matrix o f an integrable contragredient Lie algebra g satisfies three of the conditions CM o f Section 3.3: CMl: All = 2 for all i g J; CM2: Aij G —N for all z, j CM3: A^j = 0 <=>Aji = 0.

g

J, z

;;

Using the notation established above, we have the relations

R l:

i[ar,f,] = -AjJj, for all i, j G J. ■■= ( ade¡ )

R2:

I d~j ■■= (ad f¡)

= 0,

=0

for all i # j.

324

Contragredíent Lie Algebras

A matrix A ^ satisfying CMl-3 is called a Caitan matrix. In the literature the expression “generalized Cartan matrix” is often used instead. We refer to a Cartan matrix satisfying CM l-5 of 3.3 as a Cartan matrix of finite type. The relations R1 and R2 are familiar from Section 3.7. However, the situation here is far more general. In Section 4.2 and subse­ quent sections we will show that a contragredient Lie algebra g can be constructed whose structure matrix is any prescribed matrix A for which CMl holds. If A satisfies CM l-3, then g can be constructed to be inte­ grable. In the exercises we outline a generalization of integrable contragredi­ ent Lie algebras that relaxes CMl. The terminology “integrable” comes from the fact that each representa­ tion of j e J, in g (given by the adjoint action) can be integrated to a group representation : 5L2(K) ^ A ut(g), as we have seen above. A contragredient Lie algebra (g, is called a Kac-Moody algebra if it is integrable. We say that g is of finite type if A is of finite type and of affine type if A is of affine type. Otherwise, it is said to be of índefiníte type. These Lie algebras were introduced in [Kal, Mol]. Let (g, be contragredient. We recall (Proposition 2.7.5) that the radical r = rad(g) of g can be characterized as the maximal ideal of g intersecting trivially. Since [e,, /¿] = e \ {0}, we see that r n

( 0 »
Let ~ : g g /r denote the natural mapping. By the Remark 1 of Section 2.7, g /r has a triangular decomposition ^ = (1^, g+, Q^, cr), where we have identified and and written a for the anti-involution induced by cr on g. Furthermore the display {e¿, , /J /e j rise to a display {é¿, a'^, of g, showing that g /r is contragredient. If ad is locally nilpotent, so are ad and ad /^. Thus we have a natural homomorphism ^:<^ad(s) ^ Gad(g) given by exp ad

exp ad

,

exp ad í/¿

exp ad í/¿,

4.1

Contragredient Lie Algebras

325

whenever ad (and hence ad f¿) is locally nilpotent. Note, however, that it can easily happen that ad é¿ is locally nilpotent, although ad is not; in particular 6 need not be surjective. Likewise we evidently have a homomor­ phism of Weyl groups, W{Q) ^ W(Q), which need not be surjective. However, this map is injective^ since W( q) can be viewed as a group of linear maps on Í) and i) and can be identified. Thus we can simply identify W( q) as a subgroup of W^(§). Similar considera­ tions apply to the algebra § •= 9 /t if t is any ideal of 9 such that t e r . More precisely Proposition 10 Let (g, be a contragredient Lie algebra, and let § = g /t, where t is an ideal of g contained in rad{%) such that o-(t) = t. Then the natural mapping “ : g ^ g is injective on th^subspace ®^ ® Identify­ ing 5 and ^j_we have that ^ = (^, g+, cr) is a triangular decomposition of g and (g, is contragredient. Furthermore (g, and (g, have the same structure matrix, and g is integrable if g is integrable. We have ^^A(g) c^^A(g),

PL(g) c iy(g)

with equalities if g is integrable. Proof. After the remarks above it only remains to prove the statements of the last sentence. If a e'^^A(g), then a = waj for some «y e '‘^n. Under the identification of IF(g) inside W{%), the same equality holds relative to g, so a e '’^A(g). If g is integrable, then = < r/ | ; e J> = W{%) and ^^A(g) = =^^A(g). □ Proposition 11 Let ( g , r ) be contragredient with structure matrix Л = g = g/rad(g), and let ~ ^ q be the canonical map. Then

y^j* Define

(i) I f A¿: e Z_ or if A¿: =Aj¿ = О for some i Ф j, then (adey) and d¿j = ( a d / y ) belong to rad(g); (ii) For ad éy {and hence ad /y) to be locally nilpotent, it is necessary and sufficient that for all i ¥=j, A¿j e Z Aj¿ = 0; (iii) g is integrable if and only if A is a Cartan matrix. Proof, (i) Suppose that A^j e Z < q for some i ¥-j. Consider the §1^2^^-sub­ module M of g (under the adjoint representation) generated by /¿. We have [ej,f] = 0 and [af,f¿] = - A ^ j f . By part (ii) of Proposition 2.4.2 (with f

Contragredíent Lie Algebras

326

replacing z;+, - a , replacing A, etc.), either d~j = (ad = 0 or M is a Verma module and dj] generates a maximal proper submodule. In either case ad ej{d¡j) = 0. Also (a d e ,)(a d /,)"^ '^ " 7 , = (

a

d

= 0.

Indeed, If A¿j < 0, this is clear, and if A¿j = 0, then also Aj¿ = 0 and [fj, h¿] = Aj¿f¿ = 0. Finally, it is obvious that ad ( a d = 0 if k ^ i, j. Thus d'j is a highest weight vector for g in the adjoint represen­ tation. We know then that the module M' that it generates is ad(U(g_))(ad 9_, and hence M' is an ideal of g whose intersec­ tion with is trivial. Thus M' c rad g, and we conclude that d~ e rad g. Since rad g is cr-invariant, d~^j e rad g proving part (i). (ii) If ad éj is locally nilpotent, then the A¿j are as desired by Proposition 8. (Recall that g and g have the same structure matrix, see Proposition 10). Conversely if ^4^y e Z A^ - = 0 => Aj¿ = 0, then (ad ij)

- A a + 1

e¿ = 0

for all i

j

by part (i). Thus ad Cj is locally nilpotent (part (i) of Proposition 8). (iii) The proof follows from part (ii). Proposition 12 Let (g, r ) be a contragredient algebra, and let Dg = g_e

=

Then

e g^..

Proof. For any a e A\{0}, [1^, g“] = g“, which together with [e„/;] = shows that Dg = g_+(^ n Dg) + g+D g_+ + g^. It remains to show that n Dg c . To prove this, we need only to show that [g_ , 9 + ] n ^ c Q ^, which we can reduce to looking at a single product [ j:, y], where X e g““ and y e g", a e A+. We use induction on ht a. If ht a = 1, then a = a,, X = c/„ y = de^ for some c, d e i e J, and [x, y] e Oth­ erwise, ht a > 1, and since g_ is generated by the /y, j e J, we may assume that X has the simple form [/ y, z], z e g“^, ¡3 e A+, ht = ht a — 1. Then [jc, y] = [[/;, z], y] = [fjiz, y]] - [z,[fj, y]]. Now [z, y] € g“', and [fj, y] e g^, so the proof is finished by the induction. □

4.1

Contragredíent Lie Algebras

327

Corollary ¡f ^ = Qk ^ ihen g is perfect.

□

Remark 6 The decomposition Dg = g_0 !2k ® 9+ (or more precisely (g+, Q^, o-log)) is not in general a triangular decom­ position. The problem is that it is possible (and always happens in affine algebras) that there are roots 5 0 for which g^] = 0. Thus is not in general large enough to serve as a diagonal algebra. This phenomenon, which can occur only if the structure matrix is singular, is largely responsible for the necessity in Section 4.2 of defining a realization of the matrix as a preliminary to constructing contragredient algebras with a given structure matrix. Let (g, Let a

be a Kac-Moody algebra with root system A. and let

0 A. The a-root string through ^ is 5(j8,a) = A n (/3 + Za).

The next result is the fundamental fact about root strings. Proposition 13 Let ( q, be a Kac-Moody algebra. Let a 0 '^^A. Then for any /3 e A there are nonnegative integers d, u so that 5(/3, a) is an unbroken sequence RSI

p - d a , . . . , p , . . . , p + ua,

and furthermore RS2

d - u = .

Proof Since a g“ + [g“, g““] + g”“ = ^1^2^ is isomorphic to ^l2(IK). Furthermore g affords an integrable representation of (Proposition 6). Let N ‘= E^ez9^^^“- Then N is an integrable submodule, and its weight system P(N) is precisely S(p, a). From Proposition 2.4.7, N decomposes into finite-dimensional irreducible submodules, each of which has the form K= where dim = 1 and ad is scalar multiplica­ tion by 5 - 2/, / = 0 , .. ., 5. Since ad a is multiplication by <j3, + 2k, we have K S —2Í

C

+

0 , o : ' ' » /2 ] - i } a

328

Contragredíent Líe Algebras

This implies that s = <)8, (mod 2). Now we use the Freudenthal mid­ point argument [FdV]. The midpoint of the set of roots ¡3 + {^(5 — (¡3, a'^)) - ¿}a as i runs from 0 to 5 is 1/ (

2

\ is-(p,a^y ^ l« + ^ + ( ^ )“ —

= /3 - -a, which is independent of s. Actually /n is a root of the string only if <13, a^> e 2Z, but the point is that all the irreducible submodules K of N share the same midpoint and they all involve fi or fi j a according to the parity of . □ We finish with a prescription to inductively compute the entire set of positive roots of a Kac-Moody Lie algebra: Proposition 14 Let ( q,

be a Kac-Moody algebra with root system A. Let

Q+{n) = (a e ¡2+ lht(a) = «}, A+(n) = A+n Q ^{n), n ^ l . {For n < 0, Q^{n) = A+(n) = 0.) Then (i) 0 ^ ( l ) = { a j / e J} = A^(l); (ii) for a e Q^{n), n > 1, we have a e A+(n) if and only if one of the following two conditions hold: (iia) there is an / e J such that (a, a f ) > 0 and r¿a s A+(n - ( a , a f ) ) , (iib) for all i e J, (a, < 0 and for some / G J, a —a¿ G (n - 1). The roots that occur in (iib) are imaginary, and a root a occurring in (iia) is imaginary if and only if the corresponding root r^a is imaginary.

4.1

Contragredient Líe Algebras

329

Proof, (i) Clear. (ii) Let a e Q^(n); n > 1. First, we show the conditions are necessary. Assume that a e A+(/t). If condition (iia) fails to be the case, then from (4.6) we see that 0 and hence by Proposition 13, a e A. Finally if the root a occurring in condition (iia) is imaginary, then so is r^a [Proposition 5(iv)]. If a occurs in condition (iib), then to show that a e'^'A, we must anticipate Proposition 5.2.6. Indeed, if a then In particular a ^ = ^ 0? and hence ( a , a ' ' ) = E y e j c / a ,a / > < 0, contrary to
rel="nofollow"> = 2.

□

Corollary Let t be any ideal contained in rad(g) such that or(t) = t. Then A (g/t) = A(g) {see also Proposition 10). Proof The algorithm of the proposition determines the same subset of 2 ^ for g and g /t. □ We finish this section with a description of the Lie algebra of derivations of a contragredient Lie algebra g. Any linear mapping e : g/D g Z(g) lifts to a linear map e : g Z(g) g annihilating Dg, and s is clearly a derivation. Identify the space of these derivations with Hom|,^(g/Dg,Z(g)) Proposition 15 Let Q be a finitely displayed contragredient Lie algebra with dim Then D erg = H om K (g/D g,Z (g)) + a d g . In particular, if Q is perfect or Z(g) = (0), then Der g = ad g.

< dim g+.

330

Contragredient Lie Algebras

Proof. Using Proposition 2.1.4 it will suffice to show that a derivation 5 e Dq belongs to Honi(|^(g/Dg,Z(g)) + ad g. Let a : / , j be a finite display. Then 8a^= d([ej, fj]) c K a / and 28ej = Cj]) = [5 a /, ej] + [a /, Scj] shows that 5 a / = 0. It follows that if 8ej = XjCj, then 8fj = -Ay/y for some Ay ^ IK. Choose /i e so that aj(h) = Ay for each j. Then 8 — aid h kills Dg. Also for any h' ^ ij and any x e Dg, we have 0 = (5 — ad h)[h', x] = [(5 - ad h)h', x] showing that (5 - ad h)(h') centralizes Dg. Since it lies in 1^, it also centralizes ij, hence g. Thus (5 - ad /z)g c (6 - ad AXi) + Dg) c Zg, and finally 5 e Hom(g/Dg, Zg) + ad g. □ 4.2

REALIZATIONS OF CONTRAGREDIENT LIE ALGEBRAS

In Section 4.1 we saw how every contragredient Lie algebra gives rise to a structure matrix. In this section we reverse the process and show how to construct a contragredient Lie algebra with a given structure matrix A. When A is ai Cartan matrix, this leads to the construction of a Kac-Moody algebra with structure matrix A. As a preliminary step it is necessary to construct from A a space 1^, which will eventually be the diagonal subalgebra of our Lie algebra. This involves the idea of a realization of A. Let J be a nonempty set, and let A = (A¿J), ¿,j e J, be a matrix with coefficients in IK(more properly A is an element of I K By a realization of A over IK, we understand a triple /? = ( 5 , n , n ^ ) consisting of a vector space over IK together with two subsets II = {a, | i e J} c ij* (the dual space of 1&) and II^= { a/ | / g J} c such that Rl: each of the sets II and II ^ consists of linearly independent elements; and if we denote the natural pairing of 1^* and l^by<*,->:l^*xf)-^IK, then R2: = A,j for all /, j g J. The realization is finite dimensional (of dimension dim 1^) if 1^, and hence is finite dimensional. Example 1 Consider A = ^

a,b, ^ K . We distinguish two cases:

1. ab ¥= A. We may take to be a two-dimensional IK-space with basis n ^ = {aj^, a /} and then define II = {a^, a2) c if* in the only possible way, namely (a¿, a /> =-^4¿y; 1 < /, j < 2. That H is linearly indepen­ dent follows from the fact that det(^) = A — ab 0.

4.2

Realizations of Contragredient Lie Algebras

331

2. ab = 4. Introduce a third symbol d, and define = Let n

Ka^,

[a^, «2 }, and II = {a^, «2} c 1^* be defined by a /> =Aij,

1 < i, j < 2,

( a i , d ) = i. Then R '= (]^, n , n ^ ) is a realization of A. Example 2 Let J = Z, and define A e

by

An = 2, A^j= Aij = 0, Let ]^ = ®

and

| / - ; | = 1,

otherwise.

and define

e 1^* for all i e Z by

=A^ j , By setting n = {a, I / ^ Z} and 11^ = yl.

i , j e Z. | i e Z}, we obtain a realization of

Proposition 1 Suppose that card(J) = / is finite. Let 1 = I -\- corank(y4). Given nonzero elements , S/ o f IK, we claim that there exists an I X l-matrix A with the following properties : mini: min2: min3: min4:

Aij = Aij for all 1 < i, j < 1. det(^) 0. A^j = AjiS^ for a l l \ < i < l < j < l andA^j = Aji for all I < i, j < L If A is a Cartan matrix and all belong to then A is a Cartan matrix.

Proof. To see that A exists, first observe that if corank(^) = 0, then A = A, and we are done. Otherwise, we construct A in corank(^)-steps as follows: Consider an (/ + 1) x (Z + l)-matrix A^ of the form

^1 =

SiOi 2

Contragredient Lie Algebras

332

Let S := { ( c j , e = 0 for all j}. This is a nontrivial subspace o f Thus, if H — {(a^ . . . , aj) | Ej^iCjCjaj = 0 for all (c^,..., Cj) e S}, then H is a proper subspace of Let R be the row space o f A. Then R is a proper subspace o f IK^ Hence there exists (a^,. . . , a|) e ZL\{R U H}. By construction the matrix defined by this choice o f a^ satisfies corank{Af) = corank(A) — 1. Indeed its first I rows have row rank = rank(A) + 1, and the last row is evidently independent of these first I rows. We now repeat this argument for A ^ with respect to e\ , , . ., where e\ = and = 1- After corankiAysteps we obtain a matrix A satisfying min(l)-(3). Suppose that the hypotheses of min(4) holds. Then A is a Cartan matrix provided that each Eia^ e Z. Thus must be chosen to cancel the denominator o f which is easily arranged. □ Remark 1 Condition min(l) states that ^ is a submatrix of A. Condition min(2) will be used to explicitly construct the realization of A. Condition min (4) allows us to assume that ^ is a Cartan matrix whenever A is. (We just make a suitable choice of Condition minO) is important later on when we define symmetrizable matrices (Section 4.4). It says that if A is symmetrizable by (e^,. . . , then A is symmetrizable by (e^,. . . , 1, . . . , 1). Proposition 2 Let A G

be a matrix. Then

(i) a realization of A exists', (ii) if card(J) = I is finite, then ij can be taken to be o f dimension I + corank(^), and this is the smallest dimension possible. Proof (i) Define to be a C-space admitting the set {af, basis. For each i e J, let =f|* be defined by

| / e J} as a

=Aij, Then I I : = { a J / e J } is clearly linearly independent, and with II {af \i ^ J}, R — (]^, n , n ^ ) is a realization of A. (ii) Using Proposition 1, let ^ be a Cartan matrix which is 1 X 1, where / = / + corank A, which contains A as the submatrix (A¿J)l^¿ J^l, and which satisfies d e t (^ ) ¥= 0. Let f) be an /-dimensional (K-space. Fix a basis a^, . . . , ai of Define « i , . . . , ^ if* by a / > = A^j. Then the a¿ are linearly independent because d et(^) ¥= 0. It is clear that if II — {a^,..., a/} and { a^ , . . . , af}, then (]&, II, 11^) is a realization of A of dimension I Finally, let R = 0),YI,U^) he any finite-dimensional realization of A. Choose bases B and B* of f) and 1^* so that they begin with a f , . . . , a / and ai , . . . , a i , respectively. Since the matrix M = has

4.2

Realizations of Contragredient Lie Algebras

333

linearly independent rows, in particular its first / rows are linearly indepen­ dent. Now, if S is any k X k matrix of rank s and S' is any (k) X (k 1)matrix of the form 5' = [5 I *], then S' has rank < 5 + 1. Applying this to the submatrix A which occupies the top left corner of M, we see that for its first / rows to be linearly independent, the number of columns must be at least / + (/ —rank(^)) = / + corank(^). Thus / + corank(>l) is the mini­ mum dimension for if in any realization of ^4. □ If J is finite and R •= ( / z , n , n ^ ) is a realization of A ^ with diml^ = / + corank(y4) then we say that R is a minimal realization of A. Let A and = (f), 11,11^) be as above. We construct a realization R^ of the transpose of A. Set R^ = ( f ) *, n" ',n ) where II^c]^

]^** under the canonical identification of f) inside 1^**. Let < • , • > : f)* X f)** ^

K

be the canonical pairing. (Note that this pairing is written with the factors in the reverse order that one might expect. We do this so that we may think of the new < * , • > as being an extension of the original pairing under the canonical identification of i) in 1^**), Then (aj,ar)=A,=Ajj. Since both n ^ c f)** and II c 1^* are linearly independent, it follows that is indeed a realization of A^. We say that is the realization dual to R. We now begin to construct Lie algebras out of matrices. Let A ^ be a matrix and R = (1^, I I ,11'^) be a realization of A. View as an abelian Lie algebra, and let X = {e^, | / e J} be a set of symbols. We let g = be the free Lie algebra on X and define f

^*g

to be the free product of f) and g. Let R be the subset of f , consisting of the following elements for all ¿,j e J and e 1^:

(1) [h,fj] + (aj , h) f j . The first Lie algebra that is of interest for us is then defined by

334

Contragredient Lie Algebras

where I(R) is the ideal of f generated by R. We call u(A, R) the universal algebra of the realization R of A. Remark 2 Let {h^ I i e 1} be any basis of Then u (^ , /?) can be described as the Lie algebra generated by the set ej, fj | i e / , ; g J} subject to the relations [ e j , f J - 8j^a]',[hi,ej] - ( a ^ , [ / i „ / ^ ] + (aj,hi)fj, where i , k ^ I and j, m g J. We now look more closely at the Lie algebra u = n(A, R), For conve­ nience the notation for the canonical map f ^ u will be suppressed when­ ever no confusion is possible. Thus we will use expressions of the form “ Cy G u ” where, strictly speaking, this is to be understood as meaning €j H- I(R) G u. For the time being, we will denote the image of fj under the canonical map by that is, + I(R) c u. Later we will see that is actually an injection, and we will be able to identify i) and Via the adjoint representation acts on u. Combining this fact with the canonical map ^ 1^^, we see that u has a natural 1^-module structure, and under this action h • Cj = a j ( h ) e j , h • f j = —a j ( h ) f j , and h • if ^ = (ff) for all /z G ]^. In other words, each ej and fj and all the elements of are weight vectors of the l^-module u. Their weights are aj, -Uj and 0, respectively. Since u is generated by these elements it follows from Proposition 2.1.2(ii), that u admits a weight space decomposition relative to 1^. The weights all lie in the group Q := 0

l a j c i)=*

and u = 0 u “, a^Q where u" := (jc e u I[h, x] = ( a, h ) x for all h We define

e+==(V;e0j

/

\{o},

Q- = -Q ^,

£tnd the corresponding subalgebras u+=

0

u “.

We also define, in the usual way, the height function h t : 0 ^ Z, ht: £ c ,a ,.

L c, .

g

4.2

Realizations of Contragredient Lie Algebras

335

Finally, set Qk

•= Ф

the K-span of П in 1^*.

In the same way we define QX, QX, and giK inside 1&. Our intention now is to show that u = u _ 0 ^ ® u + and that u admits a regular triangular decomposition. We begin by establishing the following proposition: Proposition 3 There exists a unique anti-involution a o f n satisfying cr(cy) = fj, o-(fj) = and c r I = 1. Proof The uniqueness is clear since u is generated by together with the e/s and f/s. The map e^ -e^ extends to an automorphism of the free Lie algebra Since is abelian, the map x —jc is an automor­ phism of ]^. By Proposition 1.10.4(ii) there exists an automorphism (o of f = ]^ * g satisfying (o(e^) = - f , caifi) = and
e I(R).

Hence (o induces an automorphism (also denoted) co of u. It follows that (T '= -0) is our desired anti-involution. □ Proposition 4 Let the notation be as above. Then (i)

is generated by {ej I y e J} and n_ is generated by {fj I ; ^ J} (as Lie algebras); (ii) u = u_® ® u+. Proof The argument is precisely the same as that in Proposition 4.1.1.

□

Proposition 5 Let the notation be as above. Then (i) The sum YKf j + + ElKcy is direct and the restriction to it o f the canonical map f u й injective; (ii) and u_ are free Lie algebras, freely generated by {ej I ; ^ J} and {fj I j ^ J}) respectively.

Contragredient Líe Algebras

336

Proof, Let A be the free associative algebra on the set X = {xj \ j e J}. Let A e ]^* be arbitrary. We proceed to show that A can be made into a highest weight u-module with highest weight A and highest weight vector 1 ^ A, Define a representation of ]&on rel="nofollow">1 by h ' 1=

h-Xj^

Jk

for a l l ; i , . . . , ^ G J,/i el^. Next we make A into an g-module as follows. Since S is freely generated by {ej, fj I j e J}, it suffices to define the action of each Cj and fj in A. (Recall the correspondence between ^-modules and left U(g)-modules and the fact that U(3) is the free associative algebra on {ej,fj\j e J}. See Proposition 1.10.2). We define f j ' X = XjX

for all y e J, jc e y4.

The action of the e/s is defined recursively on each monomial Xj^... Xj^ as follows: ey • 1 = 0, Xf

Jk

=

X:(e:

Jl^ J

• JC,

Jl

. . .

X: ) Jk'

+ djj^X - aj^

---- ''Jk'

[The action of e, is defined in such a way as to be compatible with = fjl ■ ^ifj = fi^ + We have given A both an and an g module structure. Thus there is a unique f = * ^-module structure for A compatible with the above (part (ii) of Proposition 1.10.4). Straightforward computation shows that the ideal I{R) of f lies in the annhilator of this module and hence that A inherits a u := f//(jR)-module structure. For instance, consider /y] [e,,/;] • 1 = C,- •/; • 1 - fj • = C,. • Xj

•1

= Xj{e^ ■1) + 5,/A,a>'>l = 5,/A,a,y>l = 5,,ar- 1. Similarly [e,, fj\

■Xj ^... Xj^

= e, •

...

x¡^

-

x ¡ ■ e,

•

...

x¡^

= 5„
Jk

4.2

Now let

Realizations of Contragredient Lie Algebras

337

e ]^ \ {0}. Then for the u-module A we have (h + I { R ) ) • 1 = /i • 1 = l.

Hence h + I(R) = 0 (in u) only if (A, /z> = 0. Since we are free to choose A arbitrarily, we may assume that ^ 0. It follows that +/ (/? ) ^ 0. In other words, the canonical map i) ^ is injective. The representation of u_ on is just the left regular representation of A on itself; that is, each fj acts on A as left multiplication by Xj, This gives us a homomorphism U(u_) A with fj Xj. Since A is free, the inverse map with Xj ^ fj exists, and U(u_) —A. Thus the Lie subalgebras of Lie (U(u_)) and Lie {A) generated by the fj and Xj, respectively, are also isomorphic. But the latter is free on the [xj] (Proposition 1.10.1). Thus u_ is freely generated by the set [fj | j e J}. Finally, a interchanges u_ and u+, and hence part (ii) is proved. From part (ii) of Proposition 4 and the freeness of u the sum ElK/y + + LKcj is direct. □ Remark 3 The u-module A described above is evidently the Verma module of type (1, A) for u. Remark 4 Part (i) of Proposition 5 allows us to identify in u. We assume this identification henceforth.

with its image

Proposition 6 Let the notation be as above. Then (i) for all a ^ Q \ {0}, dim u “ < oo; (ii) for all i e J, dim u “' = dim u " “' = 1; (iii) if J is countable and ^ is of countable dimension, then u is of countable dimension. Proof. We know that u “ = (0) unless a e ± 0 + or a = 0. Take a e Q^. By part (i) of Proposition 4, u “ is spanned by the products ej^ ...]], where + ••• = a, of which there are only a finite number. For the special case a = u “' is spanned by e^. Using a, parts (i) and (ii) now follow. Part (iii) is immediate from part (i). □ Putting everything together, we have Proposition 7 Let u = Vi{A, R) be the Lie algebra defined by a matrix A and a realization R. In the notation established above. (i) ,5^ := (]^, u+, 0+, O') is a regular triangular decomposition o f the Lie algebra u; (ii) if An # 0 for all i e J, then (u, ^ ) is contragredient.

Contragredient Lie Algebras

338

Remark 5 The structure matrix S of the contragredient algebra (u, *9") above need not equal A, In fact, assuming all ¥= 0, S = A if and and only if all ¿, Aii = 2. Example 3 Let A = {d) be a 1 X 1 matrix, a 0. Then with — Kh and a e ]^* defined by (a , h) = a, we obtain a realization R = (^,{o:},{M) of A and u = u(A, R) = K f 0 K/z 0 Ke with [e, f ] = h, [h, e] = ae, [h, /] = -af. It is easy to replace / , h, and e by suitable multiples (see Section 4.1) and obtain the standard §l2(IK)-triplet [e,f]=h,

[h,e] = 2e,

[A,/] = - 2 / ,

from which we conclude that u = ^I2(1K). The structure matrix of u is (2) and of course any minimal realization R' = (]^',{a},{a^}) of this matrix (i.e., (a, a ^ } = 2) yields u((2), R') = n(A, R). Let (g, be contragredient. Since later we will work extensively with such algebras, we will set once and for all the notation to be employed in this context. •^ = (^, Q+, Q+, o-) n =

Triangular decomposition Fundamental roots

n '-

Fundamental coroots

< 2 = 0 Zaj jsj

Root lattice

0 Za/

Coroot lattice

;s j <2k = 0

IK«; c

= 0 yeje += 0 yej

0 _= - Q ^ The structure matrix of ( g , ^ )

< -.-> :rx 5 ^ K J /jr}yej

The natural paring A display of (g ,

(in particular

cr(ej) = fj for all j e j ) Consider the triple R == (1^, II, 11^) arising from (g, 3^). It is immediate that R is a realization of the structure matrix ^ of (g, (note that we

4.2

Realizations of Contragredient Lie Algebras

339

need CT3 for this). We call R the natural realization of A in the context of (g, ^ ) . Let us now construct the universal Lie algebra u = \i(A, R) as above. Set

and f=

Then u = f//(i^),

where I(R) is the ideal of f generated by the elements of the set R defined in (1). Recall (Proposition 5(i)) that

(2 )

I 0 K/,.j ® ^ e I ©

cf

can be identified with a subspace of u. Accordingly we henceforth identify with a subalgebra of both u and g and think of ej and fj as elements of both of these algebras. (It will always be clear to which algebra we are making reference.) As we have seen above the algebra u admits <5^ as a triangular decompo­ sition, where = (f |,u^ ,G^ ,o-). Moreover (u, ^ ) is contragredient (Proposition 7(ii)). Because of the way in which f is defined (i.e., as a free product), there exists a (unique) Lie algebra epimorphism

satisfying

Hfj) =fj’ (¡/(h) = h

for all

e ]^, 7 G J.

Since i/r(/(i?)) = (0), there exists an induced epimorphism q. This allows us to realize g as a homomorphic image of u. We refer to

340

Contragredient Lie Algebras

(u, Ty, if/) above as the universal covering of (g, «^). Evidently (3)

^(u^)=g«

for all a e g .

Proposition 8

Let ( g , Then

be contragredient, and let

(u ,

, i/r) be its universal covering.

(i) ker((/r) c rad(u), (ii) (/i(rad(u)) = rad(g). Proof. In view of Proposition 2.7.5, we have t ^ ( u ) = m^^(u) = rad(u) and = my^Cg) = rad(g). Since is an isomorphism, it follows that ker(i/i) n ]^ = (0), whence we have part (i). Similarly (/r(rad(u)) is an ideal of g which intersects trivially, while i/i”Krad(g)) is an ideal of u which intersects trivially. This establishes part (ii). □ Let A G

and let R he a realization of A. Define Q= q ( A , R ) := u (^ ,i^ )/ra d (u ).

By (2.7(i)), rad(g) = (0). We call g (^ , /?) the radical free algebra of the pair (A, R). Note that if A^^ ¥= 0 for all i e J, then g is contragredient, and in this case we call q(A, R) the radical free contragredient algebra of the pair (A,R\ We have seen that any contragredient algebra (e, with the natural realization R has a universal covering (u, i/r). If furthermore e is radical free, then by Proposition 8 ker if/ = rad(u), and it follows at once that e is canonically isomorphic to g (^ , jR). Proposition 9

If e is a radical free contragredient algebra with structure matrix A and realization R, then there is a canonical isomorphism e g (^ , i?). Our interest is primarily in the case that ^ is a Cartan matrix. In that case = u (^ ,/? )/ra d (u ) is integrable, and hence, a Kac-Moody algebra, by Proposition 4.1.11. However, there may be intermediate Kac-Moody algebras, between n(A, R) and g(v4, R). q( A , R )

Proposition 10

Let A be a Cartan matrix with realization R, and let n{A, R) be the corre­ sponding contragredient algebra. Let J{R) be the ideal o f vl{A, R) generated by

4.2

Realizations of Contragredient Lie Algebras

341

the elements \dtj = {&áe¡)

i*j

1^.7 =

i* i

(R2)

Then R) •= u(yl, R)/ J{R) is a Kac-Moody Lie algebra. Furthermore for any Kac-Moody algebra % with realization R, there exist surjective homomorphisms q\ A

, R ) ^ q ^ q( A , R )

{with the obvious effects on their displays) under which ra d g ^ ( ^ , i? )

rad(g) ^ r a d g ( ^ , i ? ) = (0).

Proof. That g^(^, R) is integrable follows from Propositions 4.1.8 and 4.1.10. The existence of the map q^{A, i?) g follows from factoring the covering homomorphism u{A, R) ^ % through q^(A, R) by Proposition 4.1.9. The rest follows from Proposition 8. □ This proposition is a natural complement to Proposition 4.1.9. In the sequel (Section 4.6) we will find that for the cases of most interest Q^(A, R) = %{A, R). Example 1 A =

bring the notation in line with our usual

notation for §Í2(IK) in the Appendix on A^^\ we will take II = {«(„«1}, n'^= a^} with <«,-,«/> =A¡j and set h = IKif +

+ IKa^.

Using R = {§, n , n ''} , we define u(A, R). It consists of two free Lie algebras ^ +~

u _ —( / o , / i ) ,

and the abelian subalgebra h: u

=

< /o> /i>

®

< eo,C i> .

342

Contragredient Lie Algebras

The algebra u is contragredient and simple considerations with free Lie algebras show that its root system is A(u) = A(u)_u{0}

U

A(u) + ,

where A(u) + = [ka^ 4- ma^ | f c , m > 0 , A: + m > 0 , and fc(resp. m) = 1 if m (resp. k) = 0} A ( u ) _ = “ A(u) + . The root space multiplicities (dim u “, a e A(u)) can be computed by Witt’s formula (Proposition 2.5.5). Since ^l2(IK) is contragredient with structure matrix A there is homomor­ phism i/i : u

êl2(IK)

with e, ar^h„

I = 0,1 (the fact that corresponding elements u and § 12(11^) have the same name should not cause confusion). This example continues in the Appendix on

4.3

(see Section A.3.)

EMBEDDINGS, FIELD EXTENSIONS, AND DECOMPOSABILITY

Let J c J', and let A g be a submatrix of A g Suppose that R = (1^, n , n'^) and R' = (^', II', 11'^) are realizations of A and A , respec­ tively, where II = II'^= n' = and 11'''= ej-. An embedding of i? in i?' consists of an ordered pair 17 = (•>7^, 77^), where 77^ (resp. is an injective linear map (resp.

-» 1^')

4.3

Embeddings, Field Extensions, and Decomposability

343

such that (i) 77^:11^11' with 7]^a^) = a- for all i e J; (ii) n'"" with for all i e J; (iii) for all a e f)*, for all f| e i|, t7^(/z)>' = (a, h); where ( • , • ) pairings.

X ij

and <•, •>' : f)'* X i)'

are the canonical

Note that since <«;, = A]j = A^j = (a^, a j ) for all /,; e J the con­ ditions (i) and (ii) are compatible with condition (iii). At the lattice level there exists induced Z-linear maps Vl ‘ Q ^ Q' • Note that for ¡L e

( 1)

^

V

l

( Q )

if and only if supp(/i') c J.

Remark 1 The most obvious and usual example of an embedding occurs when one takes f) to be any subspace of 1^' containing such that {a-lygj, if restricted to functions on 1^, is a linearly independent set of 1^* (e.g., 'tj = 1^'). Define re s: 1^'* -> 1^* to be the restriction map arising in a natural way from the inclusion c 1^'. Then with II — [a\ I / ^ J} and n^:= {a'^ I i e J}, we see that /? = (1^, II, n ^ ) is a realization of A. Fur­ thermore, with the maps inclusion map, ^ any linear map that lifts the elements of 1^* to linear functions on ij' (i.e., res • rjj^ = id) and such that T7^(res(a^)) = for all i e J, we have that 17 == (17^, is an embedding of R in R \ We call R a natural realization of A in the context of R \ Remark 2 We have assumed in our definition that J c J' and that ^ is a submatrix of A'. It is rather easy to extend the definition to an embedding (p: J -> J', and matrices A and A' indexed by J X J and J' x J', respectively, satisfying = ^ ij for all i , j e J. Results involving 17 have obvious generalizations to this case. Proposition 1 With the notation as above let rj = ( v l ^Vr ^ tm embedding o f the realization R of A in the realization R! o f A . Then there is a canonical embedding {also denoted 77) 77

: u := n ( A , R)

u' := n { A , R')

344

Contragredient Lie Algebras

such that (0

7j(]^) = ■njiiif)

C

1^';

(ii) T7(e,) = e'i and rjifi) = for all i e J; (iii) T7U" = for all a ^ Q \ {0}; (iv) a'(rix) = riicrx) for allx e u, where a and a' are the anti-involutions on u and u', respectively. Proof Let X ■■=[ej, fj}j ^ j and X' ~ {e'j, fJ}j ^ j. and define f and f' to be the free products ii * and 1^' * that are used to define u and u', respectively. We define a Lie algebra homomorphism i/i: f ^ f'

i

by = Vr , il>(e¡)=e’¡, , ¡ , { f ) = f ¡ .

for all i e J,

and observe that the relations R involved in the definition of u are annihi­ lated by ip. For example, - «0«") = [<’ fj] -

0-

Thus ip factors through the ideal /(R) to give 77: u u'. Since 17(6^) = e) and and are freely generated by the {e^} and {e'¿}, respectively, we see that 7] embeds u+ into u+. The same applies to u_ and u'_. All the statements of the Proposition are now obvious except perhaps for the equality in part (iii). For that, observe first that for all a ^ Q \ {0} we have u “ = (0) = unless a ^ Q+. Assume then that a ^ (the Q_ case is similar). Then u" is spanned by products [Cj^,. . . , CjJ with j) in J and aj^ + • • • +ay^ = a e Q+. Next ^rvLM jg spanned by products [e'y,. . . , e'j^] with in J' and a'j^ + • • • = i7^(a). By (1) {j\ , . . . , ^} c J, and hence each of these products lies in t7(u"). □

Remark 3 The most common situation in which the last proposition arises is the one in which A is some submatrix, say, of some matrix ij-'l i,j < kl A = that has some realization R! = (f)', II', 11'^). For the real­ ization R one may take (f)', II, 11^), where II — [a\ , . . . , a'j^} c II' and — . . . , a '/ ) c (see Remark 1). Then u = u(yl, R) embeds into u' = Vi{A, R ) by ^ fi ^ fi, a\^ so that u may be identified with a subalgebra of u' in the most obvious way imaginable.

43

Embeddings, Field Extensions, and Decomposability

345

Another common and equally obvious embedding occurs when A, as above, is nonsingular (even though A' may be singular). Then we may use the realization R = ((2k >n , n'^), where 2 ^ == 0 <=i * '' Example 1 Let be a realization of A g and let 7 g J be fixed with Ajj # 0. Let = (Ka/,{a^},{a/}). Then, as above, is a realiza­ tion of ( A j j ), and we have the canonical embedding u((Ajj), R^^'>) -» u(^4, R). As we see in Example 4.2.3, u ( ( > l ^ . p , = gljClK). Now u((y4^.p, is isomorphic to the subalgebra of \i(A, R) generated by Cj, fj, a / , which we know also to be isomorphic to SljiK). Proposition 2 Let Tj be an embedding o f a realization R in realization R’. Let 17: u(A, R) -» \i(A, R') be the corresponding embedding o f Lie algebras. Then (i) 77(rad(u)) = rad(u') n tj( u ); (ii) the induced homomorphism q(A, R) g(A', R') is injective-, (iii) 77^ A = A' n t]j(Q), where A and A' denote the root systems o f q(A, R) andgiA', R'Xrespectively.Moreoverforalla G A \ {0},77(g(y4, i?)“) = g(A', (iv) if g(A, R) and g(A', R') are Kac-Moody algebras, then -nr.C^A) =^^A' n 7,^(2), 77^(""A) =""A' n 77^(2). Proof Let J and J' denote the indexing sets for A and A', respectively. Let a' ■■=Ttifa) G 2'+. where a g A+, and let x g u “. Assume x g rad(u). To show that y ~ t](x ) g rad(u'), we must verify that given y^ g g ' " ' ..., ^ 2'+ and -I- • • • +iS'* = a', then Vk ^ S -Pi where

[yi. • • •, y*. y] = 0. Since supp(a') = U f=i supp(j8-) it follows from (1) that supp(j8p c J for all 1 < i < A:. Thus = t7^(/3,) for some /3, G 2+ , and hence y, = t7^(x,) for some x, g u “^' (part (iii) of Proposition 1). Thus 0 = 77(0) = T7([x i,. . . , X*, x]) = [yj,. . . , y*., y]. In the same way 77(rad u)“ c radu' if a G A_. On the other hand, if x i rad(u), then it is evident that y € rad(u'). Thus 77(rad(u)“) = rad(u')’’i-(“) n t7( u ). Now parts (i) and (ii) follow at once. It is clear from (ii) and Proposition 1 that TjjfA) c A' n rjjfQ). For the reverse inclusion it suffices to take a' g A'+n r]j(Q). Then a' g -q^fQ^), and so a' = ■ +a-^ for some i j , . . . , G J. Since a' g A'+, for some permutation of i j , . . . , say, the trivial one, [e^, ■■■, e^] * 0. Thus [e,y...,e,^] # 0 and 6' ■= T7^(a,^ -!-••• +a,^) ^ t7^(A). Furthermore, since

346

Contragredient Lie Algebras

q(A', R')“' is spanned by the products of the form [e\^,. . . , eJJ, riiqiA, i?)“) = q(A', R')“'. This proves part (iii). According to Proposition 4.1.14, for n e Z+, n > 1,

(2)

a e" A +( n ) => 3; e J,

such that (a , a j ) > 0,

a - { a, a' j ) aj

- < a ,a / > ) .

Similarly (3)

e"A '+(n) ^ 3; e J', rtL*^ -

such that <7,^«, < > > 0,

e''^A'+(n - <17^«, a}'")).

Now supp(i7^(a)) e {a -1 i e J}, and hence a!^) > 0 is possible only if j e J. Thus in (3), J' can be replaced by J. Now part (iv) follows by induction on n, starting from the obvious fact that ry^C^Ad)) = t7^(A(1)) = A'(l) n 71¿ Q ) =^"A'(1) n 71¿ Q l □ Remark 4 If we are in the cases described in Remark 3, then part (iii) of Proposition 2 assumes the simple form (4)

A = A' n E

Remark 5 In this proposition q(A, R) and q(A', R') can be replaced by and q \A ',R ').

q\ A , R )

We now look more closely at derived algebras. Let g = q(A, R) be the radical free Lie algebra defined by a matrix A and realization R. The proof of Proposition 4.1.12 (which is for contragredient algebras) carries over directly to show that (5)

Dg = g_0

0 g+.

Here we will give an entirely different description of Dg that shows that it is dependent only on A (not on R). Let Л be arbitrary, and let b be the Lie algebra defined by the generators e^, (/ e J), together with the relations

(R')

43

Embeddings, Field Extensions, and Decomposability

347

where h¡ ■= [e,, /,]. We begin by noting that [ht,hj] = [ hi , [ e j J j ] \ = [Aj^ejJj] + [cj, - A j J j ] = 0 ,

which shows that a == is an abelian subalgebra of b. We can grade b by (the set of mappings of J into Z). Let be defined by «/(;) = ^¿j, and set deg

= a^,

deg/, = -a,., deg hi = 0. Then the relations (R') are homogeneous, which suffices to determine the grading on b. We have need only of the subgroup Q' of consisting of functions a with finite support. Let Q'+ = {En,o:, e Q' \ > 0} \ {0}, and let Q'_ = Then with b+:= we have by a trivial adaptation of Proposition 4.1.1, b = b_0 a © b , and b +=

e J>,

b_= . Evidently there is a surjective homomorphism (/) = / , and hence with (f>(hi) = a / , since the corresponding elements in u satisfy the relations (R'). Since u+ is freely generated by 1/ e J}, :b+ —^ u+. Likewise (j>: b_ —^ u_. Since { a/ | / g J} c c u are linearly independent, we have (f>: a , and we conclude that is a Lie algebra isomorphism. Furthermore there is an isomorphism : Q' —^ Q with a, and Define (R'')

r := maximal homogeneous ideal of b intersecting a trivially.

Since r is graded, (x) = n u “) from which it is clear that (j>(x) is invariant by ad 1^. Thus <^(r) is an ideal of u (not just Du), and hence (/)(r) c rad(u). On the other hand, rad(u) c D(u), and <^"^rad(u)) is a graded ideal of u intersecting a trivially. Thus r = <^“ Krad(u)), and we have

Contragredient Lie Algebras

348

finally b /r = Dg. This gives the alternative description of Dg by (R') and (R"). Proposition 3 Let A e

be a matrix, and let R b e a realization o f A,

(i) Dg(yl, R) —b /r , where b is the Lie algebra defined by the presenta­ tion (RO and r is the ideal defined by (R"). In particular Dq(A,R) depends only A. (ii) Dg w perfect if and only if A has no row consisting entirely of zeros. (iii) I f A is a Cartan matrix then Dg^(^, R) (see Proposition 4.2.10) has the presentation with generators e^, /¿, h^, i e J, and the relations

(R l) [e¿,fj]

=

8¿j hj

for all

i j ;

i(adc,)'^'^"^e, = 0, (R2) \

fj)

V/ = 0

for

ii^j.

Proof (i) The proof has already been established. (ii) and (iii) The proofs are left as exercises.

□

Corollary Let (t, and (c', ^ ' ) be radical free contragredient Lie algebras admitting the same structure matrix. Then De = Dc'. Proposition 4 Let A e and let R be a realization o f A. Let g = g(yl, R). Then the centers of g and Dg are given by Zg = (/i e

I

h) = 0 for all i e j},

ZDg = {A e G k K«/, A> = 0 for alii e j).

43

Embeddings, Field Extensions, and Decomposability

349

Proof, Let z be central in g or Dg. Then writing z = ^ 9“? see immediately that each z" is central. Proposition 2.7.3 shows that z “ = 0 if a e A+U A_. Thus z e ]^, and the result is immediate. □ Let (g, be contragredient, and let {e^,ay,f^} be a display of g and R = {]^,n, n^} the natural realization of the structure matrix A of g. The situation when J is finite and i? is a minimal realization is the most common and most important. Assuming that J is finite, we put together a number of conditions which are equivalent to R being a minimal realization. Write ^ = i2iK ® ^ for some subspace t of i). Then each f e e t induces a linear functional on <2iK by a , and we obtain a linear mapping i \ i ^ Q\IK-

Proposition 5 Let ( g , r ) be contragredient with natural realization R and structure matrix A= Assume that J is finite. Then the following are equivalent. (i) (ii) (iii) (iv)

R is a minimal realization o f A. Zg c Dg. Zgce„^. For every supplement t o f in 1^, the corresponding map i : t is injective.

Qt

Proof. Since Zg c and Dg n = 0ii< >it is clear that parts (ii) and (iii) are equivalent. (i) => (iii) We have dim finite by Proposition 4.2.2. Suppose that z e ( ^ \ G k ) Zg. Then {ai , z } = 0 for all i e J, and II induces a srt II of line^ly independent functionals on 'ii/Kz. The realization R •= {i)/Kz, n , n'^(mod Kz)) contradicts the assumption of (i). in f). Then (iii) (iv) Let t be a supplement of ker t = {z e 11 (a,, x) = 0 for all / G j}

= t n Z g c t n ( 2 K = (0)(iv) ^ (i) Let t be a supplement of in f). Since i : t -» <2k is injective and dim 0 k is finite, dim (| = dim + dim t is finite. Thus through the pairing <• , • >, induces all of ^**; in particular ii induces all the functionals of 0K- Thus we have the exact sequence Zg

^ ^ 0K

0.

350

Contragredient Lie Algebras

Now, by assumption, t n Zg = (0). Since t was arbitrary, Zg c Q'^. Thus Zg = {/z e |
is minimally realized if

MRl: it has a finite display (i.e., J is finite) MR2: any one of the conditions i-iv of Proposition 5 hold. Proposition 6 If (ind (g', ^ ' ) are minimally realized radical free contragredient Lie algebras with displays S ' = a / , j and S) = , fj)j^j tmd with the same structure matrix A, then g and g' are isomorphic by a (noncanonical) isomorphism