Introduction To Lie Algebras

Workshop on Group Theory: Classification of Reductive Algebraic Groups Indian Statistical Institute, Bangalore Centre December 18, 2006 - January 5, 2007

Introduction to Lie Algebras Amber Habib Mathematical Sciences Foundation, Delhi

Abstract An introduction to Lie algebras. This material sets the stage for the study of semisimple and reductive Lie algebras via their root systems. In turn, those Lie algebras are the means of understanding reductive groups.

Contents 1 Lecture 1 – Basic Definitions and Examples

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2 Lecture 2 – Ideals, Quotients & Homomorphisms

6

3 Lecture 3 – Solvable and Nilpotent Lie Algebras

9

4 Lecture 4 – Theorems of Engel and Lie

13

5 Lecture 5 – Jordan Decomposition

19

1

Lecture 1 – Basic Definitions and Examples

For our basic example we consider the vector space L(V ) of linear operators on a vector space V (over a field F). Besides the vector space operations of addition and scaling, this has another natural operation: composition of linear operators. This operation is not commutative: in general, f ◦ g 6= g ◦ f . We can try to capture the amount of non-commutativity by defining [f, g] = f ◦ g − g ◦ f.

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1 LECTURE 1 – BASIC DEFINITIONS AND EXAMPLES

2

Now we have a new operation, [· , ·] : L(V ) × L(V ) → L(V ). First, we easily see it is bilinear: [αf + βg, α′f ′ + β ′ g ′ ] = αα′ [f, f ′ ] + αβ ′ [f, g ′] + βα′ [g, f ′] + ββ ′ [g, g ′], where α, β, α′, β ′ ∈ F. Next, it is not commutative. In fact, we have [f, g] = −[g, f ] and [f, f ] = 0. Finally, let us consider associativity: [f, [g, h]] − [[f, g], h] = f [g, h] − [g, h]f − [f, g]h + h[f, g] = f gh − f hg − ghf + hgf − f gh + gf h + hf g − hgf = g[f, h] − [f, h]g = [g, [f, h]]. So the bracket is not associative either. However, the last calculation can be rewritten in a form which is quite useful: [f, [g, h]] + [g, [h, f ]] + [h, [f, g]] = 0. Note the cyclic pattern. The properties listed above lead to the following abstract notion: Definition 1.1 A Lie algebra g is a vector space (over a field F) with a bilinear operation g × g → g called the bracket or commutator, and denoted (X, Y ) 7→ [X, Y ], such that: 1. [X, X] = 0

∀X, Y ∈ g.

2. [X, [Y, Z]] + [Z, [X, Y ]] + [Y, [Z, X]] = 0

∀X, Y, Z ∈ g.

The first property of the bracket is called anti-commutativity while the second is the Jacobi identity. Exercise 1.2 Show that in a Lie algebra, [X, Y ] = −[Y, X]. Lie algebras can be studied for their own sake, but our interest in them arises out of their applications to the study of certain groups. Roughly, to each such group we will assign a Lie algebra which will contain local information about this group. Its job will be to convert problems about group structure to problems in linear algebra.

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Exercise 1.3 Let A be an associative algebra over F. Define [a, b] = ab − ba for a, b ∈ A. Show that this bracket makes A a Lie algebra. Example 1.4 Consider L(V ) with the bracket [f, g] = f ◦ g − g ◦ f . We have seen that it becomes a Lie algebra, and we shall call this Lie algebra the general Lie algebra and denote it by gl(V ).  Definition 1.5 Let g be a Lie algebra. We have the following definitions. 1. The Lie algebra g is abelian if the bracket is trivial: [X, Y ] ≡ 0. 2. A subset h ⊂ g is a Lie subalgebra of g if it is a vector subspace and is closed under the bracket operation. 3. A subset h ⊂ g is an ideal of g if it is a vector subspace and H ∈ h, X ∈ g implies [H, X] ∈ h. 4. If h is another Lie algebra, then ϕ : g → h is a Lie algebra homomorphism if it is linear and preserves the bracket: ϕ[X, Y ] = [ϕX, ϕY ] ∀X, Y ∈ g. 5. A Lie algebra h is isomorphic to g if there is a bijective Lie algebra homomorphism ϕ : g → h. Then ϕ is called an isomorphism. (Note that ϕ−1 is then an isomorphism from h to g.) 6. Let V be a vector space. A Lie algebra homomorphism g → gl(V ) is called a representation of g in V . Exercise 1.6 Classify the one and two dimensional Lie algebras up to isomorphism. Exercise 1.7 Let ϕ : g → h be a Lie algebra homomorphism. Show that im ϕ is a Lie subalgebra of h and ker ϕ is an ideal in g. Example 1.8 If the vector space V has a basis of size n, it becomes identified with Fn and L(V ) with M(n, F) - the n×n matrices with entries in F. Under this identification, composition becomes matrix multiplication and so the bracket is now defined by [A, B] = AB − BA. M(n, F) with this bracket is denoted by gl(n, F). Clearly gl(V ) and gl(n, F) are isomorphic. 

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Example 1.9 With the Lie algebra gl(n, F) in hand, we obtain others by considering various familiar subspaces: 1. sl(n, F) = {X ∈ gl(n, F) : Trace(X) = 0.}. (Special Linear Algebra) 2. skew(n, F) = {X ∈ gl(n, F) : X + X t = 0.}. 3. t(n, F) = {X ∈ gl(n, F) : X is upper triangular}. 4. n(n, F) = {X ∈ gl(n, F) : X is strictly upper triangular}. 5. d(n, F) = {X ∈ gl(n, F) : X is diagonal}.  Exercise 1.10 Which of the above Lie algebras depend on the choice of basis, and to what extent? Since sl(n, F) is independent of the choice of basis, we can denote it by sl(V ). Definition 1.11 A Lie algebra is called linear if it is a Lie subalgebra of gl(n, F). Example 1.12 We shall describe a machine for generating many linear Lie algebras. Let V = Fn and J ∈ M(n, F). Then define gJ := {X ∈ gl(n, F) : JX + X t J = 0}. It is easily verified that gJ is a vector subspace and also closed under bracket, hence it is a Lie subalgebra of gl(n, F). For example, J = I gives gI = o(n, F).  Exercise 1.13 Show that if J and K are orthogonally similar, then gJ and gK are isomorphic. Example 1.14 Let us consider various choices of J. (Note: The explicit descriptions of the Lie algebras below involve the assumption that char(F) 6= 2.) 1. Let n = 2p and consider J=



0 I −I 0



,

1 LECTURE 1 – BASIC DEFINITIONS AND EXAMPLES

5

where I is the p × p identity matrix. Then    X Y t t gJ = : X, Y, Z ∈ M(p, F), Y = Y , Z = Z Z −X t is called the symplectic algebra and denoted by sp(n, F). 2. Let n = 2p and consider K=



0 I I 0



,

where I is the p × p identity matrix. Then    X Y t t gK = : X, Y, Z ∈ M(p, F), Y + Y = Z + Z = 0 Z −X t is called the orthogonal algebra and denoted by o(n, F) = o(2p, F). 3. Let n = 2p + 1 and consider 

 1 0 0 L =  0 0 I , 0 I 0 where I is the p × p identity matrix. Then     0 −bt −ct  p b, c ∈ F , X, Y, Z ∈ M(p, F), Y : gL =  c X Y + Y t = Z + Zt = 0   b Z −X t

is also called the orthogonal algebra and denoted o(n, F) = o(2p + 1, F).  Exercise 1.15 Show that o(n, F) is isomorphic to skew(n, F), provided that F is algebraically closed. Exercise 1.16 Consider g = R3 with the vector cross-product [X, Y ] := X × Y. Verify g is a Lie algebra. Show it is isomorphic to o(3, R).

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Exercise 1.17 Let Eij ∈ M(n, F) be defined as having all entries equal 0, except that the (i, j) one equals 1. Show that [Eij , Ekl ] = δjk Eil − δli Ekj . Exercise 1.18 Show that dim(sp(n, F)) = 12 n(n+1), dim(o(n, F)) = 12 n(n− 1). Exercise 1.19 Prove the isomorphisms sl(2, F) ∼ = o(3, F) ∼ = sp(2, F), assuming char(F) 6= 2.

2

Lecture 2 – Ideals, Quotients & Homomorphisms

Let g be a Lie algebra. Given A, B ⊂ g we define their bracket by [A, B] = span{[X, Y ] : X ∈ A, y ∈ B}. Thus, a subspace h ⊂ g is a Lie subalgebra if [h, h] ⊂ h. It is an ideal if [h, g] ⊂ h. Exercise 2.1 Let h1 , h2 be ideals of g. Then so are h1 ∩ h2 , h1 + h2 and [h1 , h2 ]. Definition 2.2 The derived algebra of g is the ideal [g, g]. It is also called the commutator ideal. Let h be an ideal in g. Then the quotient vector space g/h becomes a Lie algebra under the bracket [X + h, Y + h] := [X, Y ] + h. Exercise 2.3 Let h be an ideal in g. The quotient map π : g → g/h, X 7→ X + h is a Lie algebra homomorphism. One has the usual results about ideals and quotients: Exercise 2.4 Let ϕ : g → h be a Lie algebra homomorphism. Then g/ ker ϕ is isomorphic to im ϕ.

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Exercise 2.5 Let ϕ : g → h be a Lie algebra homomorphism, and j an ideal of g contained in ker ϕ. Then there is a unique homomorphism ψ : g/j → h such that the following diagram commutes: g

ψ

-

?

g/j

ϕ

π

-

h

Exercise 2.6 If j, k are ideals of g such that j ⊂ k, then k/j is an ideal in g/f j and g/j is naturally isomorphic to g/k. k/j Exercise 2.7 If j, k are ideals of g, then j+k k is naturally isomorphic to . j j∩k Example 2.8 As the Workshop proceeds, we shall see that the Lie algebra g = sl(2, F) has a special role to play. Let us look at its structure in detail. A natural choice of basis for g is:       0 1 0 0 1 0 X= , Y = , H= . 0 0 1 0 0 −1 Since the bracket is bilinear, we only have to understand the bracket relations between these basis elements. They turn out to have a simple form: [H, X] = 2X,

[H, Y ] = −2Y,

[X, Y ] = H.

In fact, consider the linear g → g map defined by Z 7→ [H, Z]. The map is diagonalizable and the basis elements are its eigenvectors! This suggests that it would be useful to study the bracket via the linear maps it induces. An implication of these calculations is that [g, g] = g if char(F) 6= 2.



Definition 2.9 Let g be a Lie algebra. For any X ∈ g, define a linear map ad(X) : g → g by ad(X)Y = [X, Y ]. This map is called the adjoint of X.

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Exercise 2.10 Show that ad(X) = ad(Y ) implies [X, Y ] = 0. Is the converse true? Exercise 2.11 Show that ad(X)[Y, Z] = [ad(X)Y, Z] + [Y, ad(X)Z]. Definition 2.12 A linear map D : g → g is a derivation if D[X, Y ] = [DX, Y ] + [X, DY ],

∀X, Y ∈ g.

The collection of all derivations of g is denoted by Der(g). Exercise 2.13 Show that any derivation D of g satisfies the Leibniz rule: n

D [X, Y ] =

n X

n

Ci [D i X, Dn−i Y ].

i=0

Remark The binomial coefficients are defined for arbitrary F recursively by n

C0 = n Cn = 1 and

n

Ci = n−1 Ci−1 + n−1 Ci for 0 < i < n.

Exercise 2.14 Show that Der(g) is a Lie subalgebra of gl(g). Exercise 2.15 Show that the adjoint map ad : g → gl(g) is a representation of g in g. Note that the image of the adjoint representation lies in Der(g). Members of this image are called inner derivations. Derivations which are not inner are called outer. Exercise 2.16 g/[g, g] is an abelian Lie algebra. Exercise 2.17 Show that t(n, F) = d(n, F) + n(n, F) (vector space direct sum). Also: [d(n, F), n(n, F)] = n(n, F) [t(n, F), t(n, F)] = n(n, F) Exercise 2.18 Let X ∈ gl(n, F) be diagonalizable with eigenvalues a1 , . . . , an . Then ad(X) is diagonalizable with eigenvalues ai − aj (1 ≤ i, j ≤ n).

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Exercise 2.19 Show that the center of gl(n, F) equals s(n, F), which consists of the scalar matrices. In addition, char(F) ∤ n

=⇒

char(F) 6= 2 2 6= char(F) ∤ n

=⇒ =⇒

s(n, F) + sl(n, F) = gl(n, F) (vector space direct sum), [sl(n, F), sl(n, F)] = sl(n, F), [gl(n, F), gl(n, F)] = sl(n, F).

Exercise 2.20 If char(F) ∤ n then the center of sl(n, F) is 0. Else, it is s(n, F). Exercise 2.21 If char(F) 6= 2 then the only non-trivial ideals in gl(2, F) are sl(2, F) and s(2, F). Exercise 2.22 Let X ∈ g and D ∈ Der(g). Then [D, ad(X)] = ad(DX). (Hence the inner derivations form an ideal in Der(g).) Exercise 2.23 Show that sl(2, F) has no non-trivial ideals if char(F) 6= 2.

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Lecture 3 – Solvable and Nilpotent Lie Algebras

Definition 3.1 The center of g is defined by Z(g) = {X ∈ g : [X, Y ] = 0 ∀Y ∈ g} = ker(ad). Z(g) is an ideal of g. If Z(g) = g, then g is abelian. If Z(g) = {0}, then ad is one-one and so g is isomorphic to the linear Lie algebra ad(g). Definition 3.2 A Lie algebra g is called simple if it has no ideals except itself and 0, and [g, g] 6= 0. (The last requirement exactly excludes the onedimensional Lie algebra!) For example, sl(2, F) is simple if char(F) 6= 2. A simple Lie algebra must have zero center and hence the adjoint representation makes it isomorphic to a linear Lie algebra. Definition 3.3 An automorphism of g is an isomorphism with itself. The collection of all automorphisms of g is denoted Aut(g).

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Example 3.4 Let g = gl(V ) or sl(V ). Let A ∈ L(V ) be invertible. Then the map X 7→ AXA−1 is an automorphism of g.  Until further notice, assume char(F) = 0. Suppose X ∈ L(V ) is nilpotent: X M = 0. Then we can define its exponential: exp(X) =

N −1 X n=0

1 n X . n!

Note that exp(X) ∈ L(V ). Exercise 3.5 Suppose X, Y ∈ L(V ) are nilpotent and commute. Then exp(X + Y ) = exp(X) exp(Y ). In particular, exp(X) has inverse exp(−X). Suppose X ∈ gl(V ) is nilpotent: X M = 0. Let l(X) denote left multiplication by X and r(X) denote right multiplication by X. Then l(X) and r(X) are commuting nilpotent maps. In fact l(X)M = r(X)M = 0. Also, ad(X) = l(X) − r(X). Therefore ad(X)

2M

= (l(X) − r(X))

2M

=

2M X

2M

Cn l(X)n r(X)2M −n = 0.

n=0

Thus ad(X) is nilpotent. So we can define exp(ad(X)) : g → g and it is a linear isomorphism. In fact, exp(ad(X))Y

= exp(l(X) − r(X))Y = exp(l(X)) exp(−r(X))Y = exp(l(X))Y exp(−X) = exp(X)Y exp(−X)

It is easy to see from this that exp(ad(X)) ∈ Aut(gl(V )).

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More generally, suppose D is a nilpotent derivation of g. Then exp(D) is defined and is a linear isomorphism of g. If D M = 0, we calculate: [exp(D)X, exp(D)Y ] =

M M X Dk X Dl  X, Y k! l! k=0 l=0

M X M X Dk Dl  = [ X, Y k! l!

= =

k=0 l=0 2M X n X

[

D n−i  Di X, Y i! (n − i)!

n=0 i=0 2M n X X n=0

1 n!

n

Ci [D i X, Dn−i Y

i=0

2M X 1 n = D [X, Y ] n! n=0



(Leibniz Rule)

= exp(D)[X, Y ] Thus, exp(D) ∈ Aut(g). In particular, if X ∈ g such that ad(X) is nilpotent, then exp(ad(X)) ∈ Aut(g). Definition 3.6 Int(g) denotes the subgroup of Aut(g) generated by automorphisms of the form exp(ad(X)), where X ∈ g and ad(X) is nilpotent. Members of Int(g) are called inner automorphisms. Exercise 3.7 Int(g) is a normal subgroup of Aut(g): If ϕ ∈ Aut(g) and X ∈ g then ϕ exp(ad(X))ϕ−1 = exp(ad(ϕX)) We now remove the assumption that char(F) = 0. So far, we have developed the basic theory of Lie algebras without any restriction on their dimension (though our examples have been finite dimensional). From here on, we shall assume the Lie algebras to be finite dimensional (though certain infinite dimensional ones will temporarily appear later in this Workshop). And if g ⊂ gl(V ) is linear, we assume V is finite dimensional. We shall now start exploring the structure of a Lie algebra via its ideals. On the one extreme, we have simple Lie algebras such as sl(2, F) which have no non-trivial ideals. On the other, are the abelian ones in which every subspace is an ideal. In between are the algebras in the following example.

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Example 3.8 Consider t = t(n, F). Its commutator ideal is [t, t] = n(n, F), which is non-trivial. Every superspace of n(n, F) is clearly an ideal in t. However, d(n, F) is not an ideal. Now let us consider n = n(n, F). Its commutator  0 0 ∗    .. ..  . .  1 n := [n, n] =  .  ..     0

ideal is  ∗     ∗   .  0    0

Moreover, n1 is also an ideal of t: [n1 , t] = n1 . We can repeat these calculations using matrices where the non-zero entries keep shifting more and more towards the top-right corner.  To bring order to these observations, we set up two series of nested ideals: 1. Derived Series: Define g(0) = g, g(1) = [g, g], g(2) = [g(1) , g(1) ], . . . , and in general g(i+1) = [g(i) , g(i) ].

2. Lower (or Descending) Central Series: Define g0 = g, g1 = [g, g], g2 = [g, g1 ], . . . , and in general gi+1 = [g, gi ]. Exercise 3.9 Prove that each g(i) , gi is an ideal in g. (Hence each series is descending.) Exercise 3.10 Let h be an ideal in g. Show that each h(i) , hi is an ideal in g. Definition 3.11 A Lie algebra g is solvable if g(i) = 0 for some i. It is nilpotent if gi = 0 for some i. Clearly gi ⊃ g(i) and so nilpotent Lie algebras are also solvable. Exercise 3.12 Show t(n, F) is solvable but not nilpotent. On the other hand, n(n, F) is nilpotent. Exercise 3.13 If g is solvable or nilpotent, then so is every subalgebra or homomorphic image of g.

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Exercise 3.14 Let char(F) = 0. Consider sl(2, F) with the standard basis (X, Y, H). Consider the inner automorphism defined by σ = exp(adX) exp(−adY ) exp(adX). Show that σ has the following action: H 7→ −H,

X 7→ −Y,

Y 7→ −X.

Further, σ is the same as conjugating by   0 1 s= . −1 0 Exercise 3.15 If dim(g) = 3, g is either simple or solvable. Exercise 3.16 If g is nilpotent and non-zero then Z(g) 6= 0. Exercise 3.17 The Lie algebra g is semisimple iff it has no non-zero abelian ideals. Exercise 3.18 If char(F) = 2 then sl(2, F) is nilpotent.

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Lecture 4 – Theorems of Engel and Lie

Exercise 4.1 A Lie algebra g is solvable if and only if it has a sequence of Lie subalgebras g = g0 ⊃ g1 ⊃ · · · ⊃ gk = 0, such that each gi+1 is an ideal in gi and gi /gi+1 is abelian. Exercise 4.2 Let h be an ideal of g. Then g is solvable if and only if h and g/h are solvable. Exercise 4.3 If h1 , h2 are solvable ideals of g, so is h1 + h2 . These results don’t hold for nilpotent Lie algebras: Example 4.4 Let g be the non-abelian two dimensional Lie algebra. It has a basis {X, Y } such that [X, Y ] = X. Now, let h = FX and j = FY . Then h is an ideal and h, g/h are nilpotent because they are one dimensional. However g is not nilpotent: gi = FX ∀i > 0. Similarly, h and j are nilpotent but g = h + j is not.



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However, there is the following result. Exercise 4.5 If g/Z(g) is nilpotent, so is g. (Hint: (g/Z(g))i = gi /Z(g).) Exercise 4.6 A Lie algebra g is solvable or nilpotent iff ad(g) is so. Exercise 4.7 Every Lie algebra g has a unique maximal solvable ideal. Definition 4.8 The unique maximal solvable ideal of the Lie algebra g is called its radical and is denoted Rad(g). Definition 4.9 If Rad(g) = 0, g is called semisimple. A semisimple Lie algebra must have zero center, hence its adjoint representation is injective. Simple Lie algebras are semisimple. So is the 0 algebra. Exercise 4.10 For any Lie algebra g, g/Rad(g) is semisimple. Let us turn to nilpotent Lie algebras. The condition gn = 0 means that for any X1 , . . . , Xn+1 ∈ g we have [X1 , [X2 , . . . [Xn , Xn+1 ] . . . ]] = 0,

or ad(X1 )ad(X2 ) · · · ad(Xn ) = 0

In particular: ad(X)n = 0 ∀X ∈ g. Definition 4.11 If ad(X) is nilpotent, we say X is ad-nilpotent. We have just observed that if g is nilpotent, then each element of g is adnilpotent. Amazingly, the converse is also true. Theorem 4.12 Let g ⊂ gl(V ) be a linear Lie algebra, V 6= 0. If each X ∈ g is nilpotent, then ∃v ∈ V such that v 6= 0 and Xv = 0 ∀X ∈ g. Proof. We proceed by induction on dim(g). When dim(g) = 0, the result is easy. Now consider an arbitrary g satisfying the hypotheses of the theorem. We first show g has an ideal h of co-dimension one. Choose h to be any maximal proper subalgebra of g. For any H ∈ h, ad(H) preserves h, hence we can define ad : h → gl(g/h),

ad(H)(X + h) = ad(H)X + h.

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Since each ad(H) is nilpotent, so is each ad(H). Applying the induction hypothesis to ad(h), we find a non-zero Y + h such that ad(H)(Y + h) = h for every H ∈ h. Then Y ∈ / h and ad(H)Y ∈ h for each H ∈ h. This shows h + FY is a subalgebra, properly containing h. Hence we must have g = h + FY . So h has co dimension one, and is also an ideal. Now apply the induction hypothesis to h: ∃v ∈ V , v 6= 0, such that Hv = 0 for every H ∈ h. Let W = {v ∈ V : Hv = 0 ∀H ∈ h}. Also, fix Y ∈ / h: then g = h + FY . We have to find a non-zero v ∈ W such that Y v = 0. For any w ∈ W and H ∈ h, HY w = [H, Y ]w + Y Hw = 0, and so Y w ∈ W . Therefore Y acts on W . Since the action must be nilpotent, there is a non-zero v ∈ W such that Y v = 0.  Corollary 4.13 Let g ⊂ gl(V ) be a linear Lie algebra. If each X ∈ g is nilpotent, then there is a basis of V such that each member of g is represented by a strictly upper triangular matrix. Corollary 4.14 (Engel’s Theorem) Let g be a Lie algebra such that each element is ad-nilpotent. Then g is nilpotent. In fact, the entire family of results 4.12-4.14 is generally grouped under the name of Engel’s Theorem. Exercise 4.15 Let g be a nilpotent Lie algebra and h a non-zero ideal in g. Then h ∩ Z(g) 6= 0. Now we shall start exploring the structure of Lie algebras via eigenvectors and eigenvalues, hence: We assume that the underlying field F is algebraically closed. We have seen that prime characteristic creates various exceptions to general patterns, and so we also assume that char(F) = 0. Example 4.16 Let g be the non-abelian two dimensional Lie algebra, with basis {X, Y } such that [X, Y ] = X. Then it is solvable: g(1) = [g, g] = FX,

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g(2) = [g(1) , g(1) ] = 0. Consider its adjoint representation:   0 1 ad(X) = 0 0   −1 0 ad(Y ) = 0 0 Hence for a general element T = aX + bY ∈ g,   −b a ad(aX + bY ) = . 0 0 Clearly, there is no non-zero vector Z such that ad(T )Z = 0 for every T ∈ g. However X is at least a common eigenvector for each ad(T ): ad(aX + bY )X = −bX.  Theorem 4.17 Let g ⊂ gl(V ) be a solvable Lie algebra. If V 6= 0 then it contains a common eigenvector for all the elements of g. Proof. We proceed by induction on dim(g), broadly following the scheme used for Engel’s Theorem. If dim(g) = 0, the claim is trivial. For a general g we first locate an ideal of co dimension one. Since g is solvable, [g, g] 6= g. Hence we can choose some vector subspace k ⊂ g such that it has co dimension one and contains [g, g]. But then, [k, g] ⊂ [g, g] ⊂ k, hence k is an ideal. By the induction hypothesis, there is a common eigenvector v ∈ V for each element of k: Kv = λ(K)v, ∀K ∈ k. It is easy to see that λ : k → F is linear. Now define: W = {w ∈ V : Kw = λ(K)w,

∀K ∈ k}.

Since v ∈ W , W is a non-zero subspace of V . We shall show g preserves W . Let X ∈ g and w ∈ W . Then for any K ∈ k, KXw = [K, X]w + XKw = λ([K, X])w + λ(K)Xw.

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To show Xw ∈ W we have to prove that λ([K, X]) = 0. Let w ∈ W be non-zero and consider the sequence {w, Xw, X 2w . . . }. Let n be largest such that Wn = {w, Xw, . . . , X n w} is linearly independent. We have: KX i w = KXX i−1 w = [K, X]X i−1 w + XKX i−1 w = λ([K, X])X i−1 w + λ(K)X i w. Hence the action of K on Wn is given by an upper triangular matrix whose diagonal entries are all λ(K). Applying this to the action of [K, X], we find that its trace is Tr [K, X]|Wn = (n + 1)λ([K, X]). By our choice of n, we also have XWn ⊂ Wn , and so [K, X]|Wn = [K|Wn , X|Wn ]. But then Tr [K, X]|Wn = 0, and so λ([K, X]) = 0 (Since char(F) = 0). So we have shown that g preserves W . Let g = k+FZ. Since F is algebraically closed, Z has an eigenvector v ∈ W . Then v is clearly a common eigenvector for g.  Corollary 4.18 (Lie’s Theorem) Let g ⊂ gl(V ) be a solvable Lie algebra. Then V has a basis such that the matrix of each X ∈ g is upper triangular. Proof. Proceed by induction on n = dim(V ). The n = 0 case is trivial. Assume n > 0. We have a common eigenvector v for the g action on V . Let V1 = V /Fv. Define π : g → gl(V1 ) by π(X)(w + Fv) = Xw + Fv. Since π(g) is solvable, by the induction hypothesis, there is a basis {v1 + Fv, . . . , vn + Fv} of V1 such that each π(g) is upper triangular. Then {v, v1, . . . , vn } is a basis of V which makes g upper triangular.



Exercise 4.19 Let g be solvable and π : g → gl(V ) a representation of g. Then V has a basis in which each π(X) is upper triangular. Exercise 4.20 Let g be solvable. Then there are ideals hi of g such that 0 = h0 ⊂ h1 ⊂ h2 ⊂ · · · ⊂ hn = g,

and

dim(hi ) = i.

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Exercise 4.21 Let g be solvable. Then ad(X) is nilpotent for each X ∈ [g, g]. Exercise 4.22 A Lie algebra g is solvable if and only if [g, g] is nilpotent. Exercise 4.23 The sum of two nilpotent ideals is nilpotent. Hence each g has a maximal nilpotent ideal. Exercise 4.24 Let X, Y ∈ L(V ) commute. Let their Jordan decompositions be Xs +Xn and Ys +Yn respectively. Then the Jordan decomposition of X +Y is (Xs + Ys ) + (Xn + Yn ). Exercise 4.25 Show the previous result can fail if X, Y do not commute. Exercise 4.26 Let g ⊂ gl(V ) be solvable. Show that Tr (XY ) = 0 for all X ∈ [g, g] and Y ∈ g. Exercise 4.27 Let char(F) = 2 and let g ⊂ gl(2, F) be the span of     0 1 0 0 X= , Y = . 1 0 0 1 Show g is a solvable Lie algebra but its elements have no common eigenvector in F2 . Exercise 4.28 Let g be as in the previous exercise. Consider the vector space direct sum h = g ⊕ F2 and define a bracket on it by [X ⊕ x, Y ⊕ y] = [X, Y ] ⊕ (Xy − Y x). Show that h is a solvable Lie algebra but [h, h] is not nilpotent. Exercise 4.29 If g is a real Lie algebra, its complexification is the complex vector space gC = g⊗R C, with bracket defined by [X ⊗w, Y ⊗z] = [X, Y ]⊗wz. Verify gC is a Lie algebra over C. Exercise 4.30 Let g be a real Lie algebra. Show it is solvable if and only if gC is. Exercise 4.31 If g is a solvable real Lie algebra, then [g, g] is nilpotent.

5 LECTURE 5 – JORDAN DECOMPOSITION

5

19

Lecture 5 – Jordan Decomposition

Jordan Decomposition We keep in place the requirement that F is algebraically closed, but allow arbitrary characteristic. Diagonalizable linear maps over an algebraically closed field are also called semisimple. We have the following standard result from linear algebra: Theorem 5.1 (Jordan Decomposition) Let V be a finite dimensional vector space over an algebraically closed field F. Then any X ∈ L(V ) has a unique decomposition X = S + N where S is semisimple, N is nilpotent and [S, N] = 0. Moreover, 1. There exist polynomials p, q without constant term such that p(X) = S and q(X) = N. 2. For any eigenvalue λ of X, define its generalized eigenspace by Vλ = {v ∈ V : (X − λI)k v = 0 for some k}. Then V is the direct sum of the Vλ ’s and S acts on Vλ by λ. S and N are called (respectively) the semisimple and nilpotent parts of X. Exercise 5.2 Let X = S + N be the Jordan decomposition of X ∈ L(V ). Then 1. If M ∈ L(V ) commutes with X, it commutes with S and N. 2. If A ⊂ B ⊂ V are subspaces such that X(B) ⊂ A, then S(B), N(B) ⊂ A. Exercise 5.3 Let X ∈ g = gl(V ) have Jordan decomposition X = S + N. Then ad(X) ∈ gl(g) has Jordan decomposition ad(X) = ad(S) + ad(N). Theorem 5.4 Let g be a Lie algebra. Then Der(g) contains the semisimple and nilpotent parts of all its elements.

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Proof. Let D ∈ Der(g) with Jordan decomposition D = S + N. We have to show S ∈ Der(g). Consider the generalized eigenspaces of D: gλ = {X ∈ g : (D − λI)k X = 0 for some k}. To see that S is a derivation, it is enough to apply it to brackets of the form [X, Y ] where X ∈ gλ and Y ∈ gµ . We first show that [gλ , gµ ] ⊂ gλ+µ . Note that (D − λ − µ)[X, Y ] = [DX, Y ] + [Y, DX] − λ[X, Y ] − µ[X, Y ] = [(D − λ)X, Y ] + [X, (D − µ)Y ]. This is easily generalized by induction to (D − λ − µ)n [X, Y ] =

n X

n

Ci [(D − λ)i X, (D − µ)n−iY ].

i=0

Let (D − λI)k X = (D − µI)k Y = 0. Then it follows that (D − (λ + µ)I)2k [X, Y ] = 0. Therefore, for X ∈ gλ and Y ∈ gµ , we have S[X, Y ] = (λ + µ)[X, Y ] = [λX, Y ] + [X, µY ] = [SX, Y ] + [X, SY ]. The direct sum decomposition g = ⊕gλ now implies that S ∈ Der(g).



Cartan’s Criterion We assume that F is algebraically closed and that char(F) = 0. For a linear map X ∈ L(V ) the following criterion for solvability is quite easy to obtain: Let A ⊂ B ⊂ V be subspaces, and define M = {T ∈ L(V ) : T (B) ⊂ A}. Suppose X ∈ M satisfies Tr (XT ) = 0 ∀T ∈ M. Then X is nilpotent. We have a version of this for the adjoint action. Lemma 5.5 Let A ⊂ B be subspaces of L(V ). Define M = {X ∈ gl(V ) : [X, B] ⊂ A}. Suppose X ∈ M satisfies Tr (XY ) = 0 for all Y ∈ M. Then X is nilpotent.

21

5 LECTURE 5 – JORDAN DECOMPOSITION

Proof. Let X have Jordan decomposition S + N. Fix a basis of V in which S has a diagonal matrix:   a1   .. [S] =  . . an Since char(F) = 0 we have the rationals Q ⊂ F. Treat F as a vector space over Q and let E be the span (over Q) of a1 , . . . , an . We have to show E = 0 (as that will give S = 0). We will show E ∗ = 0. Let f : E → Q be linear. Define Y ∈ gl(V ) by   f (a1 )   .. [Y ] =  . . f (an ) Then ad(S) has eigenvalues ai −aj and ad(Y ) has eigenvalues f (ai )−f (aj ), in both cases corresponding to the eigenvectors Eij . By Lagrange interpolation there is a polynomial r(t) ∈ F[t] such that r(ai − aj ) = f (ai ) − f (aj ),

∀i, j.

It follows that r(ad(S)) = ad(Y ). Note that r has zero constant term. We know ad(S) is a polynomial in ad(X) without constant term, hence ad(Y ) is itself a polynomial in ad(X) without constant term. Since ad(X) maps B into P A, so does ad(Y ). Hence ad(Y ) ∈ P M. Therefore Tr (XY ) = 0, which gives i ai f (ai ) = 0. Applying f , we get i f (ai )2 = 0, and hence f (ai ) = 0 ∀i.  Exercise 5.6 If X, Y, Z ∈ L(V ) then Tr ([X, Y ]Z) = Tr (X[Y, Z]). Recall the earlier exercise that g solvable implies Tr (ad(X)ad(Y )) = 0 for every X ∈ [g, g] and Y ∈ g. We shall establish the converse. Theorem 5.7 (Cartan’s Criterion for Solvability) Let g ⊂ gl(V ) be a Lie algebra such that Tr (XY ) = 0 for every X ∈ [g, g] and Y ∈ g. Then g is solvable. Proof. Choose A = [g, g], B = g, and define M = {T ∈ gl(V ) : [T, B] ⊂ A}.

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Clearly, [g, g] ⊂ g ⊂ M. Now let X, Y ∈ g and T ∈ M. Then Tr ([X, Y ]T ) = Tr (X[Y, T ]) = Tr ([Y, T ]X). By definition of M, [T, Y ] ∈ B, hence by the hypothesis of the theorem Tr ([Y, T ]X) = 0. So we have obtained that Tr ([X, Y ]T ) = 0 for every T ∈ M. It follows that Tr (ZT ) = 0 for every Z ∈ [g, g] and T ∈ M. The earlier Lemma therefore implies that Z is nilpotent for each Z ∈ [g, g]. Therefore [g, g] is nilpotent and g is solvable.



Exercise 5.8 Let g be a Lie algebra such that Tr (ad(X)ad(Y )) = 0 for every X ∈ [g, g] and Y ∈ g. Then g is solvable.

Killing Form Definition 5.9 Let g be a Lie algebra. Its Killing form κ : g × g → F is defined by κ(X, Y ) = Tr (ad(X)ad(Y ). Exercise 5.10 The Killing form of g is bilinear, symmetric and associative: κ([X, Y ], Z) = κ(X, [Y, Z]). Definition 5.11 Given a bilinear symmetric form β : V ×V → F, its radical is defined to be S = {x ∈ V : β(x, y) = 0 ∀y ∈ V }. If S = 0 we call the form non-degenerate. Exercise 5.12 The radical of the Killing form of g is an ideal of g. Exercise 5.13 Let h be an ideal of g. Let κ be the Killing form of g and κh the Killing form of h. Then κh = κ|h×h. For the remaining results we again restrict to a field F which is algebraically closed with char(F) = 0. First, note that the Cartan Criterion corollary and its converse can now be expressed as: g is solvable iff the radical of its Killing form contains [g, g]. Exercise 5.14 The radical of the Killing form of g is solvable. (Hence the radical of the Killing Form is in Rad(g).)

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Theorem 5.15 (Criterion for Semisimplicity) A Lie algebra is semisimple if and only if its Killing form is non-degenerate. Proof. Let S be the radical of the Killing form κ of g. If g is semisimple, S ⊂ Rad(g) = 0. Now, suppose Rad(g) 6= 0. Then the last non-zero term h in its derived series is an abelian ideal of g. If X ∈ h and Y ∈ g then ad(X)ad(Y ) maps g → h and (ad(X)ad(Y ))2 maps g → [h, h] = 0. Therefore ad(X)ad(Y ) is nilpotent and so κ(X, Y ) = 0. This shows h ⊂ S 6= 0.  Remark The first half of the proof uses Cartan’s Criterion and so needs F algebraically closed and char(F) = 0. The second half works for any F and shows that for any g every abelian ideal is in S. Exercise 5.16 Let g = F. Give representations of g in which: 1. Every element of g acts semisimply. 2. Every element of g acts nilpotently. 3. No element of g acts semisimply or nilpotently. Nor are the semisimple or nilpotent parts of the images in the image of the representation. Exercise 5.17 If g is nilpotent, its Killing form is identically 0. Exercise 5.18 The radical of a Lie algebra need not equal the radical of its Killing form. Exercise 5.19 Compute the Killing form of sl(2, F).