212

Lie Groups Notes By Jeffrey S. Meyer Based On Lectures Given By Gopal Prasad Contents Introduction

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Part 1. Introduction to the Objects of Study 1. Topological Groups 2. Group Actions 2.1. Abstract Group Actions 2.2. Continuous Group Actions 2.3. Linear Group Actions 2.4. Groups Arising from H 3. Compact Topological Groups 4. Smooth Manifolds 4.1. Smooth Manifolds and Tangent Spaces 4.2. Smooth Maps and Vector Fields 4.3. Integrating Vector Fields 4.4. Appendix: Smooth Manifolds From The Perspective of Algebraic Geometry 5. Lie Groups and Lie Algebras 6. The Exponential Map

3 4 7 7 8 10 13 14 17 17 19 21 21 22 26

Part 2. Classification of Lie Groups 7. Abelian Lie Groups 8. Cartan’s Theorem 8.1. Historical Results Leading Up to the Theorem 8.2. Statement and Proof of Cartan’s Theorem 8.3. Consequences of Cartan’s Theorem 9. Simply Connected Lie Groups 9.1. Covering Spaces in Algebraic Topology 9.2. Covering Spaces of Lie Groups 10. Adjoint Representations 11. Lie Subgroups and Quotient Groups 11.1. Applications to SOn and SUn 12. Nilpotent and Solvable Lie Groups 12.1. Nilpotent Lie Groups

28 29 32 32 35 37 40 40 40 44 48 55 56 57

1

2

12.2. Solvable Lie Groups 12.3. Appendix: Levi Decomposition of Lie Algebras and Semidirect Sums 13. Compact Lie Groups 13.1. Invariant Inner Products on the Lie Algebra of a Compact Lie group 13.2. Invariant Riemannian Metrics on a Compact Lie Group 13.3. Root Systems of Compact Lie Groups 13.4. The Tori of Compact Lie Groups 13.5. Compact Semisimple Lie Groups 14. Noncompact Semisimple Lie Groups 14.1. Compact Forms of a Semisimple Lie Algebra 14.2. Cartan Decomposition of a Semisimple Lie Algebra 14.3. Cartan Decomposition of a Semisimple Lie Group 14.4. Iwasawa Decomposition of a Semisimple Lie Algebra 14.5. Iwasawa Decomposition of a Semisimple Lie Group 15. Appendix

62 66 67 67 69 70 72 75 78 78 80 85 90 93 95

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Introduction What follows are course notes for a Lie groups course taught by Prof. Gopal Prasad in the Winter of 2008. These notes were LATEXed and edited by Jeff Meyer. Part 1. Introduction to the Objects of Study

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1. Topological Groups Definition 1.1. A topological group G is a group with a Hausdorff topology such that the maps f : G → G and g : G × G → G given by f (x) = x−1 and g(x, y) = xy are continuous. Example 1.2. Topological groups include familiar spaces like (Rn , +) and (Cn , +). Example 1.3. There are also less familiar topological groups, such that Qp , the field of p-adic numbers. Example 1.4. Many important topological groups are matrix groups, such as GLn (R), On , SOn , GLn (C), Un , and SUn . Definition 1.5. A subset E ⊆ G is said to be symmetric if whenever x ∈ E, we have x−1 ∈ E. Remark 1.6. Let G be a topological group, and U a neighborhood of the identity e. Then U ∩ U −1 is a symmetric neighborhood of e. Thus, by shrinking neighborhoods, we may assume without loss of generality that neighborhoods of e are symmetric. Remark 1.7. Any neighborhood of an x ∈ G is of the form xU for some neighborhood U of e. Proposition 1.8. Let U be a neighborhood of e, and n a positive integer. Then there is a neighborhood V of e such that V n = V · · · V ⊆ U . Proof. By induction, the multiplication map ϕ : G × · · · × G → G is continuous. Therefore ϕ−1 (U ) is a neighborhood of (e, . . . , e). We may then choose neighborhoods V1 , . . . , Vn of e such that V1 × · · · × Vn ⊆ ϕ−1 (U ). Setting V = V1 ∩ · · · ∩ Vn gives the proposition.  Definition 1.9. Let X a topological space, and Y ⊆ X. We say that Y is locally closed in X if for all y ∈ Y there is a neighborhood Ny of y in X such that Ny ∩ Y is closed in Ny . Proposition 1.10. Let X be a topological space with Y ⊆ X a subspace. Then Y is locally closed if and only if Y = O ∩ C for some open set O ⊆ X and some closed set C ⊆ X. Proof. (=⇒): For each y ∈ Y , let Ny be an open neighborhood of y such that Ny ∩ Y is closedSin Ny . Then Uy = Ny \Y is open in Ny for eachS y, and hence open in X. In particular, U = y Uy is open in X, and U is contained in O = y Ny . But Y = O \ U , so Y is closed in O. (⇐=): Clear.  Example 1.11. Trivially we see that both closed subsets and open subsets of X are locally closed. Furthermore, any locally compact subset of a Hausdorff space X is locally closed. Lemma 1.12. Let G be a topological group and H ⊆ G a subgroup. Then H ⊆ G is a subgroup.

5

Proof. Suppose that x, y ∈ H, and let U be a neighborhood of xy −1 . Then by the continuity of the multiplication map there exist neighborhoods V of x and W of y −1 such that V W ⊆ U . Since x, y ∈ H, one has V ∩ H 6= ∅ and W −1 ∩ H 6= ∅. Let g ∈ V ∩ H and h ∈ W −1 ∩ H.  Then gh−1 ∈ H ∩ V W ⊆ H ∩ U . Therefore xy −1 ∈ H, completing the proof. Lemma 1.13. Let G be a topological group with H ⊆ G a subgroup. If H is open, then H is closed. Proof. Since H is open, G. Using coset representatives g, F each translate Hg is open for g ∈ F we have that G = g Hg, so that in particular H = G \ g6=1 Hg. Then since each Hg is open, H is closed.  Corollary 1.14. If G is a connected topological group, then any open subgroup H ⊆ G must be G. Corollary 1.15. A locally closed subgroup of a topological group is closed. Proof. Let H ≤ G be locally closed. By replacing G with H, we may without loss of generality assume that H is dense in G. By Proposition 1.10, there exists an open set O ⊆ G and a closed set C ⊆ G such that H = O ∩ C. As H is dense in G and C contains H, we can conclude that C = G, so that H = O. The result then follows from Lemma 1.13.  Lemma 1.16. Let G be a topological group and let N ⊆ G be a subset with nonempty interior. Then H = hN i is an open subgroup of G. Proof. Let O be a nonempty open subset of N , and let x ∈ O. Then for all h ∈ H, hx−1 O is an open neighborhood of h contained in H. Thus H is open.  Corollary 1.17. A connected topological group G is generated by any subset with a nonempty interior. In particular, any neighborhood of e generates G. Definition 1.18. Let G be a topological group and H ≤ G. Let π : G → G/H be the projection map. The quotient topology on G/H is the finest topology such that π is continuous. In particular, a subset E ⊆ G/H is open ⇐⇒ π −1 (E) is open in G. Lemma 1.19. Let G be a topological group, and H ≤ G. Then π : G → G/H is an open map. S Proof. Suppose that U ⊆ G is open. Then so is hU for all h ∈ H, so that h∈H hU = HU is open in G. But π −1 (π(U )) = U H, so it must be that π(U ) is open.  Proposition 1.20. Let G be a topological group and let H ⊆ G be a subgroup. Then G/H is Hausdorff if and only if H is closed. Proof. (=⇒): Since G/H is Hausdorff, one point sets are closed. In particular, {e·H} = {H} is closed, so that π −1 ({H}) = H is closed in G.

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(⇐=): Let f : G × G → G be the map (x, y) 7→ x−1 y. Then f is continuous, so f −1 (H) is closed in G × G. But f −1 (H) = {(x, y) | xH = yH}. Thus if xH 6= yH in G/H, (x, y) ∈ / f −1 (H), and thus there exist open neighborhoods U and V of x and y respectively such that (U × V ) ∩ f −1 (H) = ∅. By Proposition 1.20, π(U ) and π(V ) are open neighborhoods of xH and yH, respectively. Furthermore, it is easily seen that the condition (U × V ) ∩ f −1 (H) = ∅ forces π(U ) ∩ π(V ) = ∅. 

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2. Group Actions 2.1. Abstract Group Actions. Definition 2.1. Let G be a group and X a set. A group action of G on X is a map µ:G×X →X satisfying the following two conditions (1) (Associativity) For all g, h ∈ G, and x ∈ X, µ(g, µ(h, x)) = µ(gh, x). (2) (Identity) For all x ∈ X, µ(e, x) = x Remark 2.2. We denote a group action of G on X by G y X and when the map µ is understood, denote µ(g, x) by gx, in which case the above conditions become (1) (Associativity) For all g, h ∈ G, and x ∈ X, g(hx) = (gh)x. (2) (Identity) For all x ∈ X, ex = x Recall that set automorphisms are simply permutations, and as such Aut(X) = SX is a group. A group action µ : G × X → X induces a set map ψ : G → Aut(X)

g 7→ µ(g, −)

Proposition 2.3. The map ψ is a group homomorphism. Definition 2.4. Let G y X be a group action. The action is said to be (1) faithful if gx = x for all x ∈ X implies g = e. (2) fixed point free if gx = x for any x ∈ X implies that g = e Definition 2.5. The kernel of an action G y X is the set {g ∈ G | gx = x for all x ∈ X}. Observe that the kernel of an action is precisely ker(ψ). Then an action if faithful if and only if the kernel of the action is trivial if and only if ker(ψ) = {e}. We often wish to study a group G via its action on some set X. In some sense, we cannot see the effect of elements in the kernel of the action, so we prefer to work with faithful representations. Example 2.6. Let G be a group. Then left translation G y G is a fixed point free action. Definition 2.7. Suppose that G y X and let x ∈ X. Then the stabilizer of x (or isotropy subgroup at x) is the set Gx := {g ∈ G | gx = x} Definition 2.8. Let G y X be a group action and let x ∈ X. Then the set Ox := {y ∈ X | There exists a g ∈ G such that gx = y} is called the orbit of x under the action of G.

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2.2. Continuous Group Actions. Definition 2.9. Let G be a topological group and X is a topological space. A continuous action of G on X is a group action µ : G × X → X which is continuous. If G and X are differentiable manifolds, then the action is said to be smooth if µ is smooth. Example 2.10. If H ≤ G is a closed subgroup, then G/H is Hausdorff and the left multiplication action µ : G × G/H → G/H

(g, kH) 7→ gkH

is continuous. We show why this is true. Let π : G → G/H be the projection map, and then define the map p : G × G → G × G/H

(g, k) 7→ (g, π(k)) = (g, kH)

This is a quotient map, so by the universal property of quotient maps, µ is continuous if and only if µ◦p is continuous. Hence it suffices to show that µ◦p is continuous. If f : G×G → G is the group multiplication map, we observe that the following diagram commutes p G × G -- G × G/H f

µ ?

G

π --

?

G/H

(g, k)

p-

(g, kH) µ

f ?

gk

π - ? gkH

So µ ◦ p = π ◦ f , the composition of two continuous maps, and thus µ ◦ p continuous. Lemma 2.11. If G y X is a continuous action, then Gx is a closed subgroup of G for each x ∈ X. Proof. Suppose µ : G × X → X is the action. Since X is Hausdorff, one point sets are closed, so µ−1 (x) is closed. In the product topology G × {x} ⊆ G × X is closed, so that G × {x} ∩ µ−1 (x) is closed in G × X. Clearly this intersection is simply Gx × {x}, so that Gx is closed in G.  Lemma 2.12. Let G y X be a continuous action, and x ∈ X. Consider the natural map ϕ : G/Gx → Ox given by gGx 7→ gx. This is a continuous bijection. Proof. That ϕ is a bijection is a result of elementary group theory, so we only prove continuity. The projection π : G → G/Gx is a quotient map, so by the universal property of quotients

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ϕ is continuous if and only if ϕ ◦ π is continuous. π--

G

G/Gx

ϕ

ϕ

◦ π -

?

Ox

Now ϕ ◦ π is the map G → Ox given by g 7→ gx, which is continuous by hypothesis.



If G where compact, then ϕ would be a closed map, and hence a homeomorphism. We can in fact conclude that ϕ is a homeomorphism under weaker conditions: Theorem 2.13. (Baire) Let X be a locally compact Hausdorff space or a complete metric space. Let {An }n∈N be any S countable collection of closed sets in X each of whose interior is empty. Then the union n∈N An also has empty interior. Theorem 2.14. Suppose G is a locally compact second countable topological group, and Ox is locally compact (or a complete metric space). Then the orbit map ϕ : G/Gx → Ox is a homeomorphism. This is an immediate corollary of the following proposition: Proposition 2.15. Let G act continuously and transitively on X. Assume G is locally compact and second countable, and that X is locally compact (or a complete metric space). Then for any open set U ⊆ G and x ∈ X, U x is open in X. Proof. Let K ⊆ U be a compact subset with nonempty interior. Such a K exists since G is locally S compact. Because G is second countable, there exists a sequence gn ∈ G such that G = n gn K. Therefore [ X = Gx = gn Kx. n

Now each gn Kx is compact, and hence closed. By the Baire category theorem, not all of the gn Kx can have nonempty interior. By translation, Kx has nonempty interior. Let g ∈ U , so that g −1 U is a neighborhood of e. Since G is locally compact, we have by Proposition 1.8 that there is a symmetric neighborhood V ⊆ g −1 U of e such that V · V ⊆ g −1 U , with V compact. Thus gV is a compact subset of U such that gV · V ⊆ U . As shown above, V x has nonempty interior. Suppose hx ∈ int(V x), for h ∈ V . Then gh−1 hx = gx ∈ int(gh−1 V x). But g −1 hV ⊆ U , so that gx ∈ int(U x). But g was arbitrary, so U x is open. 

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2.3. Linear Group Actions. In the following section V shall be a finite dimensional vector space over a topological field F (which will usually be R or C). Definition 2.16. Let G be a group and let V be a vector space. A linear action of G on V is an action µ : G × V → V such that for each g ∈ G, the map µg : V → V

v → µ(g, v)

is linear. 2 We topologize GL(V ) by choosing a basis for V and identifying GL(V ) ∼ = GLn (F) ∼ = Fn .

Lemma 2.17. The topology on GL(V ) obtained in this way is independent of the basis chosen. 2

2

Proof. The map Fn → Fn induced via conjugation by a fixed matrix in GLn (F) is a homeomorphism.  Proposition 2.18. Let G be a topological group and let V be a finite dimensional F-vector space. Then the continuity of a linear action G y V is equivalent to the continuity of the induced homomorphism G → GL(V ). Proposition 2.19. Given a linear action G y V , we get a natural action G y V ∗ , given by (g · v ∗ )(w) = v ∗ (g −1 w). This action is continuous. Definition 2.20. Let G be a group and let V be a vector space. If G y V is a linear action, V is said to be a G-module. Remark 2.21. If V and W are G-modules, then (1) V ⊕ W is a G-module under the action g(v, w) = (gv, gw). (2) V V ⊗ W is a G-module under the action g(v ⊗ w) = gv ⊗ gw. (3) n V is a G-module under the action g(v1 ∧ · · · ∧ vn ) = gv1 ∧ · · · ∧ gvn . Definition 2.22. A G-module V is irreducible if there are no vector subspaces W ⊆ V such that GW = W other than 0 and V . Definition 2.23. A G-module V is completely reducible if for any G-submodule W ⊆ V , there exists a complimentary G-submodule W 0 ⊆ V such that V = W ⊕ W 0 . Definition 2.24. Let G be a group and let V and W be G-modules. A G-module homomorphism f : V → W is a linear map that commutes with the action of G, i.e., f (gx) = gf (x). Lemma 2.25 (Schur). Given two irreducible G-modules V and W , a G-homomorphism V → W is either 0 or an isomorphism.

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Proof. Let f : V → W be a G-module homomorphism. Then f (V ) and ker f are Gsubmodules. Since f (V ) is a G-submodule and W is irreducible, f (V ) = 0 or W . If f (V ) = 0, then f ≡ 0. Suppose then that f (V ) = W 6= 0, so f is surjective. Since f 6≡ 0, ker f 6= V . But then since V is irreducible, ker f = 0, so f is injective.  Corollary 2.26. Given an irreducible G-module V , EndG (V ) is a division algebra. Proposition 2.27. Over C, the only finite dimensional division algebra is C itself. Proof. Suppose that D is a finite dimensional division algebra over C. Let d ∈ D, and define `d : D → D by `d (x) = dx. Then `d is a linear map D → D. Since C is algebraically closed, `d must have an eigenvalue λ. Thus there is an non-zero x ∈ D such that `d (x) = λx = dx. In particular, (d − λ)x = 0, which is only possible in a division algebra if d − λ = 0.  Corollary 2.28. If M is an irreducible G-module over C, then EndG (M ) ∼ = C, i.e., all G-module endomorphisms are just multiplication by scalars. Corollary 2.29. More generally, if M1 and M2 are irreducible G-modules over C, and f, f 0 : M1 → M2 are G-module isomorphisms, then f 0 = λf for some λ ∈ C. Proof. Just consider f −1 ◦ f 0 ∈ EndG (M1 ) ∼ = C.



Definition 2.30. Let M be a G-module. A bilinear form (−, −) : M × M → F is said to be G-invariant if for all x, y ∈ M and g ∈ G, (gx, gy) = (x, y). The form is called symmetric if (x, y) = (y, x) for all x, y ∈ M , and called antisymmetric of (x, y) = −(y, x) for all x, y ∈ M . Remark 2.31. Any (G-invariant) bilinear form (−, −) : M × M → F can be written as the sum of a symmetric and a (G-invariant) antisymmetric bilinear form: 1 1 [(x, y) + (y, x)] + [(x, y) − (y, x)] . 2 2 Proposition 2.32. There is a natural correspondence between bilinear forms on M and linear maps M → M ∗ . Under this correspondence, G-invariant bilinear forms correspond to G-module homomorphisms. (x, y) =

Proof. Suppose that (−, −) is a bilinear form on M . For each x ∈ M , we get a map fx ∈ M ∗ given by fx (y) = (x, y). Conversely, given linear map f : M → M ∗ , the map M × M → F given by (x, y) = f (x)(y) is a bilinear form. This establishes the correspondence. Suppose then that (−, −) is a G-invariant bilinear form on M , and let f : M → M ∗ be the induced linear map x 7→ fx . Then f (gx)(y) = fgx (y) = (gx, y) = (x, g −1 y) = fx (g −1 y) = f (x)(g −1 y) = [gf (x)](y).

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Thus f (gx) = gf (x), and f is a G-module homomorphism. Conversely, suppose that one is given a G-module homomorphism f : M → M ∗ , and let (−, −) be the induced bilinear form. Then (gx, gy) = f (gx)(gy) = [g −1 f (gx)](y) = f (g −1 gx)(y) = f (x)(y) = (x, y).  Proposition 2.33. If M is an irreducible G-module if and only if M ∗ is irreducible. Proof. First note that M is naturally isomorphic to M ∗∗ as G-modules via the map x 7→ δx , where δx (f ) = f (x). Thus we need only prove one direction of the statement. Suppose that M is reducible, with N ⊆ M a non-zero proper G-submodule. Then we have the exact sequence 0 −→ N −→ M. Taking adjoints gives an exact sequence M ∗ −→ N ∗ −→ 0. The kernel of the first map in this sequence must be a non-zero proper G-submodule of M ∗ , so M ∗ is reducible.  Corollary 2.34. Suppose that M is an irreducible G-module over C. Then there is a Ginvariant bilinear form on M if and only if M ∼ = M ∗ as G-modules. In this case, all bilinear forms are the same up to scalar multiple. Definition 2.35. A representation M of G is said to be self-dual if M ∼ = M ∗. Corollary 2.36. Suppose M is an irreducible G-module over C. Then M cannot have both a non-zero symmetric G-invariant bilinear form and a non-zero antisymmetric G-invariant bilinear form. Proof. This is clear, for any two G-invariant bilinear forms must be scalar multiples.



Theorem 2.37. Over R, there are only 3 associative division algebras, namely R, C, and H, where H is Hamilton’s Quaternions: H = R ⊕ Ri ⊕ Rj ⊕ Rk, i2 = j 2 = k 2 = −1, ij = k = −ji. In particular, if M is an irreducible G-module over R, then EndG (M ) ∼ = R, C or H. Proposition 2.38. We have EndG (M ) ∼ = R ⇐⇒ C ⊗R M is irreducible as a G-module over C. Proof.



Proposition 2.39. We have EndG (M ) ∼ = C if M arises from an irreducible complex representation. Proof.



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Proposition 2.40. We have EndG (M ) ∼ = H ⇐⇒ C ⊗R M = M1 ⊕ M2 , where M1 , M2 are ∼ G-modules over C with M1 = M2 . Proof.



2.4. Groups Arising from H. Definition 2.41. If x = a0 + a1 i + a2 j + a3 k ∈ H, we define the conjugate of x to be x = a0 − a1 i − a2 j − a3 k. There is a norm on H, define by kxk2 = xx = a20 + a21 + a22 + a23 . The norm is multiplicative: kxyk = kxkkyk. Remark 2.42. Because the norm is multiplicative, {x ∈ H | kxk = 1} is a subgroup of H× . Topologically, this subgroup is homeomorphic to S3 , so we will denote this subgroup by S3 . There is a linear action S3 × S3 y H given by (x, y) · z = xzy −1 . The kernel of this action is {(1, 1), (−1, −1)}. Via three homomorphisms ϕ1 , ϕ2 , ϕ3 : S3 → S3 × S3 , we get linear actions S3 y H. These homomorphisms are ϕ1 (x) = (x, 1),

ϕ2 (x) = (1, x),

ϕ3 (x) = (x, x),

and result in the actions x · z = xz,

x · z = zx−1 ,

x · z = xzx−1

respectively. Each of the ϕi are injective, and thus the kernel of each of these actions is ϕ−1 i ({(1, 1), (−1, −1)}). In particular, the first two actions have trivial kernel, and the third has kernel {±1}. The first two actions result in irreducible representations of S3 , as, for instance, they act transitively on the sphere S3 ⊆ H. In particular, the original action must also be irreducible. The last action, on the other hand, fixes R and the space of pure quaternions H0 = Ri ⊕ Rj ⊕ Rk, and is thus a reducible representation. Now each of these representations preserves the norm, so from each of these actions we get homomorphisms S3 → O(H) ∼ = O4 . As S3 is connected, this (continuous) homomorphism must map into the connected component of the identity, i.e., we actually get homomorphisms S3 → SO4 . In the case of the last (conjugation) action, looking at the reduced representation of S3 on H0 gives a homomorphism S3 → SO(H0 ) ∼ = SO3 . We will see later from topological considerations that this map must be surjective. Hence by the first isomorphism theorem, S3 /{±1} ∼ = SO3 . As one might expect from this SO3 ∼ = RP 3 . Note that if we view C as the subspace R ⊕ Ri ⊆ H, the the left multiplication action is C-linear, and hence induces a homomorphism S3 → U2 . This will turn out to be an isomorphism S3 → SU2 .

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3. Compact Topological Groups Definition 3.1. A Borel measure µ is said to be regular if and only if (i) µ(K) < ∞ for all compact sets K. (ii) Every Borel set E satisfies µ(E) = inf{µ(U ) | E ⊆ U where U is open.} (iii) Every Borel set E satisfies µ(E) = sup{µ(K) | K ⊆ E, where K is compact.} Definition 3.2. Let G a topological group. A Haar measure on G is a regular Borel measure µ such that µ(U ) > 0 for all nonempty open sets U and such that µ is left invariant, i.e., satisfies µ(g · E) = µ(E) for all Borel sets E. Theorem 3.3. Suppose G is a locally compact topological group. Then there is a Haar measure µ on G. Furthermore, this measure is unique up to a positive multiple. Remark 3.4. Given a left invariant measure µ, one naturally gets a right invariant measure, defined by ν(E) = µ(E −1 ). Definition 3.5. A group G is called unimodular if µ = ν. Proposition 3.6. If G is a compact topological group, then G is unimodular. Proof. Let µ be a Haar measure on G. For each g ∈ G, define the measure µg by µg (E) = µ(Eg). Certainly µg is still left invariant, and hence a Haar measure on G. Thus there is a real positive number ∆(g) such that µg = ∆(g)µ. Since G is compact, µ(G) < ∞, and thus µg (G) = µ(G · g) = µ(G) = ∆(g)µ(G) implies that ∆(g) = 1 for each g. In particular, µ(E) = µ(Eg) for each E.



Definition 3.7. The function ∆ : G → R+ defined in the proof of Proposition 3.6 is called the modular function of G. As is evident from the proof, ∆ ≡ 1 ⇐⇒ G is unimodular. Proposition 3.8. ∆ is a continuous homomorphism. Proof.



Proposition 3.9. Suppose G is a compact topological group, and V is a finite dimensional real (complex) G-module. Then there is a positive definite symmetric (Hermitian) inner product on V which is G-invariant. Proof. We do the complex case; the real case is similar. Fix a positive definite Hermitian inner product (−, −) on V . Define h−, −i : V × V → C by Z hv, wi = (gv, gw) dµ(g), G

15

where µ is the unique Haar measure on G such that µ(G) = 1. First note that the function g 7→ (gv, gw) is continuous, since the action G y V is a continuous action. In particular, the function is measurable. Since G is compact, it is bounded, and hence this integral exists and is finite. All of the Hermitian inner product properties of h−, −i are inherited from those of (−, −). Thus we only must show h−, −i is G-invariant. But Z Z Z −1 (gv, gw) dµ(gh ) = (gv, gw) dµ(g) = hv, wi hhv, hwi = (ghv, ghw) dµ(g) = G

Gh

since µ is also right invariant.

G



Corollary 3.10. Let G be a compact topological group. Then any finite dimensional representation of G (over R or C) is completely reducible. Proof. Let V be a finite dimensional G-module. By proposition 3.9, V carries a G-invariant positive definite symmetric (Hermitian) inner product h−, −i. If V is irreducible, we are done, so assume otherwise. Let W ⊆ V be a nonzero proper submodule. Then W ⊥ = {v ∈ V | hv, W i = 0} is a submodule of V , such that V = W ⊕ W ⊥ as G-modules.



Remark 3.11. Note that this fails when G is not compact. For instance, consider the group    1 t ∼ t∈R R= 0 1 acting on R2 . This action fixes the subspace generated by (1, 0), and thus is reducible. However, this is the only proper irreducible subspace:        1 t a a + tb a = =λ ⇐⇒ b = 0. 0 1 b b b Lemma 3.12. Let A be a (not necessarily topological) abelian group, and V an irreducible A-module over an algebraically closed field k. Then dim V = 1. Proof. This is clear if A = 0, so assume otherwise. Let a ∈ A be any non-zero element. Since k is algebraically closed, there is a nonzero element v0 ∈ V such that av0 = λv0 . Let W = {v ∈ V | av = λv}. Since v0 ∈ W , W 6= 0. For any element b ∈ A and v ∈ W , we have that a(bv) = bav = bλv = λbv, so that bv ∈ W . In particular, bW ⊆ W . From this we have that AW ⊆ W , so that W is a submodule of V . Since V is irreducible, this implies V = W . This shows in particular that for all a ∈ A, the action of a on W is that of scalar multiplication, and hence every 1-dimensional subspace of V is A-stable. Since V is irreducible, this can only happen if dim V = 1. 

16

Corollary 3.13. If A is a compact abelian group, then every representation of A over C is a direct sum of 1-dimensional representations. Suppose that A is a compact abelian group, and V is a representation of A over C. Then we have shown that V = V1 ⊕ · · · ⊕ Vn , where the Vi are 1-dimensional representations of A. Choose a basis {vi } of V such that vi ∈ Vi for each i. Then for each a ∈ A, we have avi = λi (a)vi , where λi (a) ∈ C× . The maps a 7→ λi (a) are then homomorphisms A → C× . Definition 3.14. A character of a group G is a (continuous) homomorphism χ : G → C× Remark 3.15. The set of characters χ : G → C× is in bijective correspondence with isomorphism classes of 1-dimensional representations of G over C, the correspondence being that if V is a 1-dimensional representation of G, then g · v = χ(g)v for some character χ. Remark 3.16. If G is a compact group and χ is a character of G, then χ(G) is a compact subgroup of C. Then only compact subgroups of C× are contained in S1 , for if ζ ∈ / S1 , then n n ζ → ∞ or ζ → 0 as n → ∞. Suppose that A and V are as above, with basis {vi }. Then with respect to this basis, the action of a ∈ A on V is given by the diagonal matrix   λ1 (a) ... , a 7→  λn (a) where, as shown, |λi (a)| = 1. We have thus completely characterized complex representations of compact abelian groups: Theorem 3.17. If A is a compact abelian group and V an n-dimensional complex A-module, then with respect to some basis, the element a ∈ A acts on V via the diagonal matrix   λ1 (a) .. ,  . λn (a) where each λi is a character of A. Definition 3.18. A linear transformation T ∈ GL(V ) is semisimple if the roots of its minimal polynomial are all distinct. Corollary 3.19. Let G be a compact topological group and ρ : G → GL(V ) a finite dimensional complex representation. Then for all g ∈ G, ρ(g) is semisimple. Proof. Let g ∈ G. Then hgi is an abelian compact group. We may then apply Theorem 3.17 directly. 

17

4. Smooth Manifolds 4.1. Smooth Manifolds and Tangent Spaces. Definition 4.1. A topological manifold is a Hausdorff second countable topological space that is locally Euclidean. Definition 4.2. Let M be a topological n-manifold. Let {Ui }i∈I be an open cover such that for each i ∈ I, there is a homeomorphism ϕi : Ui → Rn satisfying the compatibility condition that if Ui ∩ Uj 6= ∅, the map (ϕj ◦ ϕ−1 i )|ϕi (Ui ∩Uj ) : ϕi (Ui ∩ Uj ) → ϕj (Ui ∩ Uj ) is a smooth map. The collection A = {Ui , ϕi }i∈I is called an atlas on M , and the collection (M, A) is a smooth manifold. Definition 4.3. Let M be a smooth manifold, and let x ∈ M . Define the set Rx := {(f, U ) | x ∈ U ⊂ M is open and f : U → R is a smooth function}/ ∼ where (f, U ) ∼ (g, V ) is there exists an open set W ⊆ U ∩ V containing x such that f |W = g|W . We call an equivalence class of such elements a germ of functions at x, and we denote such an equivalence class f = [(f, U )]. Proposition 4.4. Let M be a smooth manifold with x ∈ M . Define mx := {(f ∈ Rx | f (x) = 0} Then (Rx , mx ) is a local ring. Given a curve c at x ∈ M , we can take derivatives of a germ f ∈ Rx as follows d d [f ◦ c]|t=0 := [f ◦ c]|t=0 dt dt where f is a representative of f. Exercise 4.5. This derivative is independent of the choice of representative. Definition 4.6. Let M be a smooth manifold with x ∈ M . Two curves c1 and c2 are tangential at x if for all germs f ∈ Rx d d [f ◦ c1 ]|t=0 = [f ◦ c2 ]|t=0 dt dt Lemma 4.7. Let M be a smooth manifold with x ∈ M . Then being tangential is an equivalence relation on the set of smooth curves through x. Definition 4.8. Let M be a smooth manifold with x ∈ M , The the tangent space to M at x is the set Tx M := {c : (−0 , 00 ) → M | c(0) = x and c is smooth}/ ∼ where two curves are equivalent if they are tangential.

18

Definition 4.9. Let M be a smooth manifold with x ∈ M . A derivation is an R-linear map D : Rx → R satisfying the Leibnitz condition D(fg) = D(f )g(x) + f (x)D(g) For a fixed curve c, we may define a map δc : Rx → R

f 7→

d [f ◦ c]|t=0 dt

where f is a representative of the germ f Proposition 4.10. δc is a derivation which is independent of the choice of curve and germ representatives. Remark 4.11. We now make the following observations (1) If f ∈ Rx , but f ∈ / mx , then f is invertible and (Rx , mx ) is a local ring. (2) As a vector space, Rx = mx ⊕ R (3) For any curve c, δc (m2x ) = 0 = δc (R) Exercise 4.12. Show that any linear map satisfying the 3rd property listed above is a derivation. We then have the following commutative diagram of R-linear maps α -

mx /m2x

R

6

δ|

m

x

q mx from which we conclude that we have bijections {Derivations}

←→

(mx /m2x )∗

←→

Tx M

Definition 4.13. Let M be a smooth manifold. The tangent bundle is the set [ T M := Tx M x∈M

The cotangent bundle is the set T ∗ M :=

[ x∈M

Tx∗ M

19

Locally, Tx M has a natural basis

∂ , ∂ , . . . , ∂x∂n ∂x1 ∂x1

and Tx∗ M has a natural basis dx1 , dx2 , . . . , dxn

Proposition 4.14. Let M be a smooth manifold of dimension m. Then T M is a smooth manifold of dimension 2m. Proposition 4.15. Let M, N be smooth manifolds. Then T(x,y) M × N = Tx M × Ty N . We may also define the bundles [

T m,n M :=

Txm,n M =

x∈M i ^

[

Tx M ⊗m ⊗ Tx∗ M ⊗n

x∈M ∗

T M :=

i [ ^

Tx∗ M

x∈M

Sections into this last bundle gives differential i-forms. P fj1 ,j2 ,...,ji dxj1 ∧ dxj2 ∧ · · · ∧ dxji .

Locally, they are of the form

4.2. Smooth Maps and Vector Fields. Let M and N be smooth manifolds with charts {Ui , ϕi } and {Vj , ψj } respectively. A continuous map f : M → N . Without loss of generality, we may refine the charts on M and reindex to assure that f |Ui maps into Vi . Then f is smooth if for all i f |Ui - Vi Ui ψi ∼ =

∼ = ϕi

the composition ψi ◦ f |Ui

? ψi ◦ f |Ui ◦ ϕ−1 i Rm ◦ ϕ−1 is smooth. i

?

-

Rn

A smooth map f : M → N induces a linear map dfx : Tx M → Tf (x) N in the following way. Fix a curve c going through x. Then given ϕ ∈ Rf (x) , define d [ϕ ◦ f ◦ c]|t=0 dt Definition 4.16. Let M and N be smooth manifolds. Let x ∈ M be a fixed point. (1) f is an immersion at x if dfx : Tx M → Tf (x) N is injective. (2) f is an immersion if dfx is an immersion at every point of M . (3) f is a submersion at x if dfx : Tx M → Tf (x) N is surjective. (4) f is an submersion if dfx is an submersion at every point of M . dfx (δc )(ϕ) := δc (ϕ ◦ f ) =

20

Theorem 4.17. (Inverse Function Theorem) Let f : M → N be a smooth map of smooth manifolds with x ∈ M . If dfx is an isomorphism, then f is a local diffeomorphism. Definition 4.18. Let M be a smooth manifold. A vector field is a map X : M → T M such that π ◦ X = IdM . A vector field is said to be continuous (resp. smooth, analytic) if the map X is continuous (resp. smooth, analytic). Let X(M ) be the set of analytic vector fields on M . Let C ω (M ) be the set of analytic functions on M . If X ∈ X(M ) and x ∈ M , then Xx := X(x) ∈ Tx M , and hence may be viewed as a derivation on Rx . For a fixed C ω (M ), the values of Xx (f ) smoothly varies Pf ∈ ∂ as we vary x ∈ M . In fact locally X = ai ∂xi , and hence X is smooth if and only if the ai are smooth. Hence we have a natural map X(M ) × C ω (M ) → C ω (M ) (X, f ) 7→ X(f ) where X(f )(x) := Xx (f ). Definition 4.19. Let A be an R-algebra. Then a map D : A → A is a derivation (of R-algebras) if D is R-linear and D satisfies the Leibnitz property D(ab) = D(a)b + aD(b). Proposition 4.20. Let M be a smooth manifold and let X ∈ X(M ). Then X : C ω (M ) → C ω (M )

f 7→ X(f )

is a derivation (of R-algebras). Exercise 4.21. Let M be a smooth manifold and let D and D0 be derivations on C ω (M ). Show that DD0 is not a derivation, but that [D, D0 ] := DD0 − D0 D is a derivation. Definition 4.22. Let M be a smooth manifold and let D and D0 be derivations on C ω (M ). The derivation [D, D0 ] is called the Poisson Bracket of D and D0 Given two vector fields X, Y ∈ X(M ), we define [X, Y ] to be the vector field associated to the Poisson bracket of the derivations associated to X and Y. Definition 4.23. Let k be a field and let g be a k-vector space. Let [−, −] : g × g → g be a map satisfying the following three properties (1) (Bilinearity) For all x, y, z ∈ g and a ∈ k, [ax + y, z] = a[x, z] + [y, z] (2) (Skew symmetry) For all x, y ∈ g, [x, y] = −[y, x] (3) (Jacobi Identity) For all x, y, z ∈ g, [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 The pair (g, [−, −]) is a Lie algebra. Theorem 4.24. Let M be a smooth manifold. Let [−, −] be the Poisson bracket of vector fields. Then (X(M ), [−, −]) is a real Lie algebra.

21

4.3. Integrating Vector Fields. Given a curve c : (−, ) → M , we may define a vector field along c in the following way. At each point c(t0 ) ∈ Im(c), dtd [c(t)]|t=t0 ∈ Tc(t0 ) M gives a tangent vector yielding a vector field along the curve. However, we can go the other direction constructing a curve from a vector field. Definition 4.25. Let M be a smooth manifold and let X be a smooth vector field on M . An integral curve of X through a point x ∈ M is a curve c : (−, ) → M such that c(0) = x and the tangent vectors at each point of the curve equals the value of the vector field at that point. Theorem 4.26. Let M be a smooth manifold and let X be a smooth vector field on M . Let x ∈ M be a fixed point. Then there exists a unique integral curve of X going through x. Proof. Let ϕ : U → Rn be a chart around x such that ϕ(x)P= 0. We may push forward via ϕ so it suffices to show this result for Rn . In this case X = ni=1 fi ∂x∂ i . We then wish to find a curve c : (−, ) → Rn such that c(0) = 0 and where it is defined Xc(t) = c0 . In particular we want c(t) = (c1 (t), c2 (t), . . . , cn (t)) such that fi (c(t)) = c0i (t) for all 1 ≤ i ≤ n. By the theory of ordinary differential equations a unique solution exists.  Proposition 4.27. If a family of vector fields is parametrized by a compact set, we can find simultaneous solutions. 4.4. Appendix: Smooth Manifolds From The Perspective of Algebraic Geometry. Given a topological space, X, with topology τ we may view τ as a category with morphism given by inclusions. We then have to following structure Definition 4.28. Let X be a topological space, with topology τ . A sheaf of rings on M is a contravarient functor F : τ → Rings satisfying the following two compatibility conditions (1) Let U ∈ τ and a, b ∈ F(U ). If for all V ∈ τ such that iV : V ,→ U , FiV (a) = FiV (b), then a = b. S (2) Let Uk ∈ τ let U = k U . If for each k there exists ak ∈ F(Uk ) such that for all j 6= k the inclusion ij∩k,k : Uj ∩ Uk ,→ Uk gives Fij∩k,k (ak ) = Fij∩k,j (aj ), then there exists a ∈ F(U ) such that FiUk (a) = ak Proposition 4.29. Let M be a topological manifold. A sheaf of smooth function on M completely determines the smooth structure on M .

22

5. Lie Groups and Lie Algebras Definition 5.1. A Lie group G is a locally compact topological group which is an analytic µ  manifold such that the group structure maps G × G → − G and G → − G are analytic. Example 5.2. (Rn , +), (Cn , +), (GLn , ·) Let G be a Lie group. For each g ∈ G, we have a left translation map Lg : G → G

h 7→ gh

As multiplication in G is analytic, this map is analytic. For a fixed h ∈ G, we then have a map dh Lg : Th G → Tgh G (Note: We often drop the h from dh Lg and write dLg when it will not cause confusion.) Let X(G) be the set of analytic vector fields of G. Let C ω (G) be the set of analytic functions on G. Now both of these sets admit natural actions by left translation as follows: G × C ω (G) → C ω (G);

(g, f ) 7→ Lg f

where the function is defined point-wise to be (Lg f )(h) := f (g −1 h) Similarly G × X(G) → X(G);

(g, X) 7→ Lg X

where the vector field is defined pointwise to be (Lg X)h := (dLg )(Xg−1 h ) Definition 5.3. A left invariant vector field on G is a vector field X on G such that Xgh = dLg (Xh ) Remark 5.4. Due to the homogeneity of Lie groups, this definition is equivalent to insisting that a vector field satisfy Xg = dLg (Xe ) for all g ∈ G. Definition 5.5. A left invariant derivation is a derivation on C ω (X) whose corresponding vector field is left invariant. It then follows the derivation associated with a left invariant vector-field commutes with the actions of left translation, i.e. (Lg X)f = X(Lg f )

23

Exercise 5.6. Let D be a left invariant derivation on G, let f ∈ C ω (G), and let g ∈ G. Investigate the function Df . You should observe that D and Lg commute in the sense that Lg (Df ) = D(Lg f ) Exercise 5.7. Let D and D0 be left invariant derivations. Show that [D, D0 ] is left invariant. Therefore the set of left invariant vector fields, denoted L(G) (or g) form a Lie subalgebra of X(G). This Lie algebra is called the Lie algebra of the Lie group G. Remark 5.8. The R-vector space X(M ) is infinite dimensional. Proposition 5.9. dimR L(G) = dim G = dimR Te (G). Proof. Since Xg = (de Lg )(Xe ), it immediately follows that dim L(G) ≤ dimR Te (G) = dim G. It now suffices to show that given Xe ∈ Te (G), the vector field X defined by Xg = (de Lg )(Xe ) is analytic. Given an analytic function f in a neighborhood of e, we may define a function Ff : G × G → G;

Ff (g, x) := f (g −1 x) = (Lg f )(x)

which by the analytic properties of multiplication in G, in analytic in both g and x. But now we see Xg (f ) = ((de Lg )(Xe ))(f ) = Xe (Lg f ) = Xe (Ff (g, −)) and it follows that X is analytic.



Definition 5.10. A homomorphism of Lie groups is a set map which is both a homomorphism of groups and analytic. Definition 5.11. A Lie subgroup of G is a subgroup which is an immersed submanifold. Remark 5.12. Let H be a Lie subgroup of G. Then the topology of H may be finer than that induced by the subspace topology. Example 5.13. Consider a line ` through the origin of the plane R2 with an irrational slope. Then the image of ` in the quotient T 2 = R2 /Z2 is a dense immersion. Furthermore, it inherits a group structure from the line, and hence is a Lie subgroup of T 2 which is not embedded as a manifold. Definition 5.14. An embedded Lie subgroup of G is a Lie subgroup of G whose manifold topology coincides with the induced topology. Proposition 5.15. If H is an embedded Lie subgroup of G, then H is closed in G. Proof. By assumption H is an embedded submanifold, and hence locally compact. Then H is locally closed, but any locally closed topological group is closed. 

24

Definition 5.16. Let π : M → N be a smooth map of smooth manifolds. X ∈ X(M ) and Y ∈ X(N ). Then X is π-related to Y if (dx π)(Xx ) = Yπ(x) for all x ∈ M . Exercise 5.17. Show X and Y are π-related iff for functions f on an open subset U ⊂ N , (Y f )(πx) = X(f ◦ π)(x) for all x ∈ π −1 (U ). Exercise 5.18. Given vector fields X and X 0 on M which are π-related to Y and Y 0 on N , show [X, X 0 ] is π-related to [Y, Y 0 ]. Suppose f : (G, e) → (G0 , e0 ) is a Lie group homomorphism. Identifying L(G) with Te G, the differential map de f : Te G → Te0 G0 yields a map L(G) → L(G0 ), which we shall also denote df Proposition 5.19. Let f : G → G0 be a Lie group homomorphism of Lie groups. Then df : L(G) → L(G0 ) is Lie algebra homomorphism. Proof. df is already a linear map, so it suffices to show that it commutes with the Poisson bracket. Pick Xe ∈ Te G and denote Ye0 := (de f )(Xe ). These each extend uniquely to left invariant vector fields X ∈ L(G) and Y ∈ L(G0 ) respectively. These are f -related. Pick another X 0 ∈ L(G) and let Y 0 ∈ L(G0 ) be the unique left invariant vector field on G0 which is f -related to X 0 . Then by the exercise, [X, X 0 ] is f -related to [Y, Y 0 ]. But this precisely means that the differential commutes with the Poisson bracket.  Proposition 5.20. Let G be a Lie group with H ⊂ G a Lie subgroup. Then L(H) is a Lie subalgebra of the Lie algebra L(G). ˜ → G is a local isomorphism of Lie groups if it is a Definition 5.21. A map π : G homomorphism of groups and a local diffeomorphism of manifolds. ˜ → G be a local isomorphism of Lie groups. If G is connected, Proposition 5.22. Let π : G π is surjective. Proof. As π is a local diffeomorphism, its image is open. The fact that π is a homomorphism implies that Im(π) is an open subgroup of G. But open subgroups are closed, and hence G connected implies Im(π) = G.  ˜ → G is a local isomorphism of Lie groups, then L(G) ˜ and L(G) Corollary 5.23. If π : G are isomorphic as Lie algebras. Theorem 5.24. A simply connected Lie group is uniquely determined by its Lie algebra.
Theorem 5.25. The Lie algebra of GLn is Mn , [−, −] where [A, B] = AB − BA.

25

Proof. As GLn = Mn \ Det−1 (0), it is open in Mn . Hence L(GLn ) ∼ = T1 (GLn ) = T1 (Mn ) ∼ = Mn . We now must compute the bracket.  Definition 5.26. A smooth manifold of dimension n is parallelizable if there exists n linearly independent nowhere vanishing sections of M into T M . Theorem 5.27. A Lie group is parallelizable. Remark 5.28. S 1 , S 3 , S 7 are the only parallelizable spheres. Corollary 5.29. S 2 cannot be made into a Lie group.

26

6. The Exponential Map Definition 6.1. Let G be a Lie group. An exponential map is an analytic map exp : L(G) → G satisfying the following three conditions: (1) exp(0) = e (2) d0 (exp) = Id (3) For each X ∈ L(G), the map R→G

t 7→ exp(tX)

is an analytic homomorphism such that

d [exp(tX)]|t=0 dt

=X

Theorem 6.2. (Existence) Let G be a Lie group. Then there exists an exponential map. Proof. By the results of a prior section, for a fixed X ∈ L(G), there exists a unique integral curve γX : (−0 , 00 ) → G for small 0 , 00 > 0 such that γX (0) = e. We shall now extend γX to all of R in the following way. For a fixed s ∈ (−0 , 00 ), we may now define the map fs : (−0 + s, 00 + s) → G

fs (t) = γX (s)γX (t − s)

Observe that fs (s) = γX (s)γX (0) = γX (s). Hence fs is an integral curve that agrees with γX at a point. By the uniqueness of integral curves, they must be the same on their intersection, and hence there exists an integral curve γ˜X : (−0 , 00 ) ∪ (−0 + s, 00 + s) → G  0 0 Taking s = min 2 , 2 , and iterating this process, we extend γX to all of R. Moreover by construction γX (s)γX (t − s) = γ(t) for all s, t ∈ R, and hence γX is an analytic homomorphism. Define the map exp : L(G) → G

exp(X) := γX (1)

It is now clear that exp satisfies the 1st and 3rd conditions above. We now must show that exp is analytic and that d0 (exp) = Id. There exists a neighborhood U ⊂ L(G) of 0 such that there exists a map f : U × (−, ) → G which is analytic and f (X, t) = γX (t) for t ∈ (−, ). Now given X0 ∈ L(G), we want to show that exp is analytic around X0 . Choose a small neighborhood around X0 such that there exists an integer M such that M1 V ⊂ U . Choose a positive integer N ∈ N such that 1 < . Then for X ∈ U , N  M  M N  M N 1 1 1 1 M exp(X) = exp X = γ 1 X (1) = γ 1 X =f X, M M M N M N This last term is analytic, and hence we conclude that exp is analytic on all of L(G).



27

Example 6.3. When G = GLn , the map A 7→ eA :=

Mn → GLn

∞ X 1 n A n! 0

satisfies the above criteria. Proposition 6.4. Let f : G → G0 be an analytic homomorphism. Then f (exp(X)) = exp(df (X)), i.e. the following diagram commute. df L(G) L(G0 ) exp

exp

? f - ? G G0 Corollary 6.5. Let H be a Lie subgroup of G. Then expH = expG |L(H) .

Remark 6.6. Let U ⊂ G be a locally Euclidian neighborhood and let V ⊂ U be symmetric. Typically the best charts are given by exp. Then the multiplication on G yields a map µ

V ×V → − U

µ(x, y) = xy

Identifying V with its image in Rn , the above product yields n absolutely convergent power series (F1 , F2 , . . . , Fn ) in 2n variable: (x1 , x2 , . . . , xn , y1 , y2 , . . . , yn ) satisfying Fi (x, y) = xi if y = 0 and Fi (x, y) = yi if x = 0. So Fj (x, y) = xj + yj + higher order terms For Lie groups, this information also encodes the inverse map. However, in general, one needs n additional power series in n variable. The above data is called formal group laws. We now record the following properties about exp in GLn .

28

Part 2. Classification of Lie Groups

29

7. Abelian Lie Groups Proposition 7.1. Let A, B ∈ Mn such that AB = BA. Then eA+B = eA eB . Proposition 7.2. Let G be a connected abelian Lie group. Then exp : L(G) → G is an analytic homomorphism. Proof. We wish to show that for any X, Y ∈ L(G), exp(X)exp(Y ) = exp(X + Y ). Consider the map η:R→G t 7→ γX (t)γY (t) Observe that as G is abelian η(s + t) = γX (s + t)γY (s + t) = γX (s)γX (t)γY (s)γY (t) = γX (s)γY (s)γX (t)γY (t) = η(s)η(t) and hence η is an analytic homomorphism. Furthermore, by formal group law we have   dη =X +Y dt t=0 By the uniqueness of integral curves, η(t) = γX+Y (t). Hence exp(X)exp(Y ) = η(1) = γX+Y (1) = exp(X + Y )  As exp : L(G) → G is a local diffeomorphism, it is an open map, so if G is connected, exp is surjective. Hence exp is a covering map, so ker(exp) is discrete. As a Lie group, L(G) ∼ = Rn for some n ∈ N. So we conclude that any connected abelian Lie group G is isomorphic to Rn /Γ for some discrete subgroup Γ ⊂ Rn . Exercise 7.3. Show that a discrete subgroup of a topological group is closed. Proposition 7.4. Let Γ be a discrete subgroup of Rn . Then there exists a basis v1 , . . . , vn of Rn and an integer 1 ≤ m ≤ n such that Γ = Zv1 ⊕ · · · ⊕ Zvm . Proof. As we may replace Rn by the subspaces spanned by Γ, we may assume that Γ spans Rn . As such, there exists a basis w1 , . . . , wn of Rn contained in Γ. Define Λ := Zw1 ⊕ · · · ⊕ Zwn ⊂ Γ By construction Λ is a full lattice of Rn , hence Rn /Λ ∼ = Tn . Since Γ is discrete in Rn , n n Γ/Λ is discrete in T , which by the compactness of T implies that Γ/Λ is finite. Hence Γ is a finitely generated abelian group. As Γ is a subgroup of Rn which is torsion free, the fundimental theorem of finitely generated abelian groups, implies that Γ is free on s generators, v1 , . . . , vs ∈ Rn . Clearly s ≥ n otherwise Γ could not span Rn . Suppose s > n. Then the first n form a basis for Rn . Hence there exists a change of basis matrix A over R such that Av = w. Furthermore, as these bothPlie in a discrete subgroup, it follows that A n+1 must in fact lie over Q. Given any Pnexpression 1 ai vi = 0 over Z where an+1 6= 0 we may apply A to get a new expression 1 bi wi + an+1 vn+1 = 0 over Z. As the image of vn+1 in Γ/Λ is of finite order, it follows that vn+1 must be a Z-linear combination of v1 . . . , vn or an+1 = 0 In either case, this contradicts our assumption about vn+1 , and we conclude n = s. 

30

Corollary 7.5. Rv1 ⊕ · · · ⊕ Rvn ∼ Rn /Γ ∼ = (R/Z)m × Rn−m = Tn × Rn−m = Zv1 ⊕ · · · ⊕ Zvm Corollary 7.6. A connected abelian Lie group of dimension n is homeomorphic to Tm ×Rn−m for some 0 ≤ m ≤ n. Corollary 7.7. A simply connected abelian Lie group is isomorphic to Rn . Corollary 7.8. (The Fundamental Theorem of Algebra) Proof. Suppose the p ∈ C[x] is an irreducible polynomial of degree d ≥ 2. Then F :=
C[x]/ p(x) is a field extension of C of degree n. Then F× ⊂ F ∼ = Cn ∼ = R2n F × is a Lie group since multiplication is given by 2n polynomials. A basic result in algebraic topology says that if M is a manifold and K ⊂ M is a sub-manifold of codimension r ≥ 3 then π1 (M ) = π1 (M \ K). As F has a vector space structure of dimension at least 4 and as 0 has dimension 0, 0 = π1 (F ) = π1 (F × ). Hence F × is a simply connected Lie group of dimension 2n, and by the above corollary F × ∼ = R2n . But F × contains infinitely many elements of finite order, namely e2πi/m ∈ C× ⊂ F × but R2n has only one element of finite order. Then these spaces could not possibly isomorphic, hence leading to a contradiction. Thus no such p can exists and hence C is algebraically closed.  Proposition 7.9. Let f : Rm → Rn be a continuous homomorphism of Lie groups. Then f is R-linear. χ

Definition 7.10. A character on G χ is a continuous homomorphism G → − R/Z Let Char(G) denote the set of all characters on G. Define a multiplication on Char(G) point-wise via (χ1 χ2 )(g) = χ1 (g)χ2 (g). Proposition 7.11. Char(G) is an abelian group. Theorem 7.12. Char(Tn ) ∼ = Zn Proposition 7.13. Suppose G and H are topological groups and f : G → H a continuous ˜ pG ) and (H, ˜ pH ) be their respective universal covers. Then the homomorphism. Let (G, continuous lift of f making the below diagram commute which sends eG˜ 7→ eH˜ is a continuous homomorphism. f˜ - ˜ ˜ G H pG

pH ?

G

f

? -

H

31

Theorem 7.14. (Kronecker) If v1 , . . . vn ∈ R are such that {1, v1 , . . . , vn } are linearly independent over Q. Then the image of (v1 , . . . , vn ) in Rn /Zn generate a dense cyclic subgroup. Proof. Let H be the closure of the cyclic group generated by π(v1 , . . . , vn ) in Tn . We wish to show that H = Tn , so suppose not. Then K := π −1 (H) ⊂ Rn is a closed proper subgroup. Observe that Zn = Ze1 ⊕ Ze2 ⊕ · · · ⊕ Zen ⊆ K. As Rn /K is abelian, we conclude Rn /K ∼ = Rk ×Tl for some k, l ∈ N. But Rn /K ∼ = Tn /H and hence is compact, so Rn /K ∼ = Tk . n Let χ ∈ Char(R /K) be a nontrivial character. By the above lifting property, there exists a continuous homomorphism f : Rn → R such that f (0) = 0 making the following diagram commute. f Rn R p

π

? χ - ? R /K S1 We make the following two observations. First, since p(ei ) = 0, we conclude ai := f (ei ) ∈ Z for all 1 ≤ i ≤ n. As χ is nontrivial, at least on of the ai are nonzero. Hence ! n n X X f vi ei = ai vi n

? ?

i=1

i=1

is a nontrivial Z-linear Pn as ei ∈ K for all 1 ≤ i ≤ n, Pn combination of the vi . Secondly, P n i=1 v1 ei ) ∈ Z. Hence i=1 vi ei ) = 0, and hence a0 := f ( i=1 vi ei ∈ K, so p ( n X ai vi = a0 i=1

which implies that {1, v1 , v2 , . . . , vn } were in fact linearly dependent over Q.  Definition 7.15. Let G be a topological group. G is monogenic if there exists a g ∈ G such that G = hgi. Corollary 7.16. A torus is monogenic.

32

8. Cartan’s Theorem 8.1. Historical Results Leading Up to the Theorem. Theorem 8.1. Given a closed subgroup H < Rn , there exists a decomposition of subspaces A, B, C such that (1) Rn = A ⊕ B ⊕ C (2) A ⊂ H (3) H = A ⊕ [(B ⊕ C) ∩ H] (4) (B ⊕ C)∩ H is a full lattice in B Proof. We may replace Rn by the subspace generated by H, and by doing so set C = 0. Let A be the unique largest R-subspace contained in H. Let B be the supplement of A, whence conditions conditions 1-3 are immediately satisfied. It now suffices to show that B ∩ H is a full lattice in B. By assumption, H generates Rn and hence B ∩ H generates B, and is thus full. Suppose that B ∩ H is not discrete. Then there exists some point h which is a limit point in B of points {bi }. By translating, we may take this point to be 0. Let vi ∈ T0 (Rn ) be the directional vectors defined by the {bi }. As the {bi } converge to 0, none are zero, and hence define an infinite sequence of points ||vvii || on S n−1 . As S n−1 is a compact metric space, it is sequentially compact, hence there is a finite subsequence of ||vvii || that converges to a value v on the sphere (hence v 6= 0). By restricting the sequence to this subsequence, we may say ||vvii || 7→ v. Let b be the point on the sphere defined by the vector v. For any a ∈ R a = Nn + Sn where Nn is an integer and 0 < Sn < 1. As vn 7→ 0, it follows that vn Sn 7→ 0, ||vn || hence vn a n→∞ av ←−−− a = vn = vn (Nn + Sn ) = vn Nn + vn Sn ||vn || ||vn || So Nn vn 7→ av, and as Nn vn ∈ B ∩ H for all n ∈ N, it follows that av is a limit point of B ∩ H, and hence H, for all a ∈ R. But by assumption, H is closed, and as such, contains its limit points, thus Rv ⊂ H. As v ∈ / A, we see Rv ⊕ A ⊂ H, which contradicts the maximally of A. Thus B ∩ H must be discrete, and hence a lattice.  Definition 8.2. A map f : M m → N n of smooth manifolds if a map of constant rank if dx f has the same rank r for all x ∈ M . Remark 8.3. A map of constant rank has a simple local description. Locally, in euclidean neighborhoods, f = i ◦ p where p is a projection and i is an inclusion. p

i

(x1 , x2 , . . . , xm ) → − (x1 , x2 , . . . , xr , 0, 0, . . . , 0) → − (x1 , x2 , . . . , xr , 0, 0, . . . , 0) Example 8.4. (1) Immersions are when r = m (i.e. just inclusion) (2) Submersions are when r = n (i.e. just projection)

33

Proposition 8.5. Let G and G0 be Lie groups and let f : G → G0 be a analytic homomorphism. Then f is a map of constant rank. Proposition 8.6. Let M and N be manifolds and let f : M → N be an analytic open map of constant rank. Then f is a submersion. Lemma 8.7. Suppose we have the following diagram of smooth manifolds g -

N

-

M

⊂

i

f ?

P Where g is smooth, f is continuous, and i is a smooth immersion. Then f is smooth. Proof. Let U ⊂ M , V ⊂ N and W ⊂ P be local charts such that g|U : U → V , f |U : U → W , and i|W : W → V . Then in local coordinates g = (g1 , g2 , . . . , gn ) where each gi is smooth, f = (f1 , f2 , . . . , fp ) where each fi is continuous, and i is the canonical inclusion. The fact that the above diagram commutes implies that fi = gi for 1 ≤ i ≤ p. Hence on U f is smooth. As the choice of the above neighborhoods was arbitrary, we conclude that f is smooth when restricted to all charts, and hence smooth.  Corollary 8.8. Suppose M is a subspace of a differentiable manifold N. Then there is at most 1 differentiable structure on M such that the inclusion i : M ,→ N is an immersion. Proof. Suppose there exists two smooth structures on M such that the inclusion of M into N in an immersion. Let us denote these two manifolds as M1 and M2 respectively. Denote by f the identity set map M1 → M2 . As M1 and M2 are the same topologically, f is a homeomorphism. We then have the following commutative diagrams -

N

M2

i2 -

N

-

i1 -

M1

⊂

⊂

i2

i1

f −1

f ?

?

M2

M1

By the above lemma, we conclude that identity map and its inverse are smooth. Thus f is a diffeomorphism, and hence M1 and M2 have the same smooth structures. 

34

Lemma 8.9. Suppose we have the following diagram of smooth manifolds π -M N f g -

?

P Where g is smooth, f is continuous, and π is a smooth submersion. Then f is smooth. Corollary 8.10. Suppose M is a manifold and N is a topological space, and π : M → N is continuous. Then there is at most 1 differentiable structure on N such that the π : M → N is a submersion. Proof. Exercise



Definition 8.11. A topological group G does not contain a small subgroup if there exists a neighborhood U of the identity e such that U does not contain any nontrivial subgroups. Theorem 8.12. A real Lie group does not contain any small subgroups. Remark 8.13. The converse to this statement is also true, but it is rather difficult to prove. In fact it was 5th among Hilbert’s problems. Proof. Let G be a Lie group. Recall that exp : L(G) → G is a local diffeomorphism around 0. Thus we may pick  > 0 such that exp is a local diffeomorphism on the open ball of radius  around 0, which we shall denote B. Furthermore, let B 0 be the open ball of radius  around 0. Let U := exp(B 0 ). We shall show that this U does the job. Suppose there exists 2 a nontrivial subgroup H < U . Then there exists e 6= h ∈ H. Let x := exp−1 (h) ∈ B 0 . As B 0 is of finite radius, and as x by assumption has positive norm, there exists a minimal natural number n ∈ N such that nx ∈ / B 0 . Furthermore, as ||x|| < 12 , it follows that n||x|| < 1 and as such nx ∈ B. As exp is a homomorphism, exp(B \ B 0 ) 3 exp(nx) = [exp(x)]n = hn ∈ H ⊂ U Thus there exists y ∈ B \ B 0 such that exp(y) = hn . But as H is a group, for all m ∈ N, hm ∈ H, hence hn ∈ H, so there exists z ∈ B 0 such that exp(z) = hn . Hence by construction y 6= z but exp(y) = exp(z), so exp is not injective on B which contradicts our choice of B. We conclude the no such subgroup H can exist.  Remark 8.14. Qp contains a strictly decreasing chain of groups Zp ⊃ pZp ⊃ p2 Zp ⊃ p3 Zp ⊃ · · · The p-adic norm says that these groups become arbitrarily small, and hence we conclude that Qp is not a real Lie group.

35

8.2. Statement and Proof of Cartan’s Theorem. Theorem 8.15. (E. Cartan) A closed subgroup of a real Lie group is an embedded Lie subgroup. Remark 8.16. These types of results rely of the fact that a continuous Lie group homomorphism Rm → Rn is R-linear. This was proven by using the fact that Q is dense in R. Hence many of the same results, such as continuous homomorphism implies linear hold for Qp but fail for C. As such there is an analogous result to Cartan’s theorem for the p-adics, but not for C. [Consider the R as a closed subgroup of C, it is certainly not a complex Lie group.] Lemma 8.17. Let G be a Lie group with H ⊂ G a closed subgroup. Then the set h = {X ∈ L(G) | exp(tX) ∈ H for all r ∈ R} is a vector subspace of L(G) Proof. h is clearly closed under scalar multiplication. Therefore it suffices to check that it is closed under summation. Note that given X, Y ∈ h, X + Y ∈ h ⇔ exp(t(X + Y )) ∈ H for all t ∈ R By formal group law exp(sX) exp(tY ) = exp(sX + tY + O(st)) so for a fixed N ∈ N     N    2 N t t t t exp X exp Y = exp (X + Y ) + O N N N N2   2  t = exp t(X + Y ) + O N As N → ∞ for a fixed t, we get the right hand term becomes exp(t(X + Y )). As H is closed, and as the choice of t ∈ R was arbitrary, we conclude that exp(t(X + Y )) ∈ H for all t ∈ R.  Proof of Cartan’s Theorem. By the homogeneity of Lie groups, and the lemmas of the previous section, to show that H is an embedded submanifold it suffices to show that it is embedded in a neighborhood of e. This means we must find a Euclidean neighborhood U × V ⊂ G of e where each U and V is diffeomorphic to Euclidean space such that (U × V ) ∩ H = U Let h be defined as in the above lemma. Let W ⊂ L(G) be the complimentary vector space, i.e. L(G) = h ⊕ W . Then the map h⊕W →G

X + Y 7→ exp(X) exp(Y )

36

is a local diffeomorphism around (0, 0). We will prove that there is a small neighborhood of (0, 0) of the form U0 × V0 ⊆ h ⊕ W such that exp(U0 ) exp(V0 ) ∩ H = exp(U0 ) Since we are free to readjust the size of U0 , it suffices to prove that exp(U0 ) exp(V0 ) ∩ H ⊂ exp(U0 ). Suppose this is not the case. Then there exists a sequence {Ui } and {Vi } of neighborhoods of 0 in h and W respectively where each decreases to zero, but exp(Ui ) exp(Vi )∩H * exp(Ui ) for all i ∈ N. Then there exists sequences of vectors Xi ∈ Ui and nonzero vectors Yi ∈ Vi such that exp(Xi ) exp(Yi ) ∈ H but exp(Xi ) exp(Yi ) ∈ / exp(Ui ). Note that as exp(Xi ) ∈ H and exp(Xi ) exp(Yi ) ∈ H, then exp(Yi ) ∈ H. Viewing W as Euclidean space with the standard metric, ||YYii || is sequence which lies in the unit sphere, which is sequentially compact. Hence there exist a subsequence of {Yi } and an element Z on the unit sphere of W such that Yi i→∞ −−−→ Z ∈ W ||Yi || Which then implies that, for a fixed t ∈ R,   t i→∞ Yi −−−→ exp(tZ) exp ||Yi || Let

t ||Yi ||

i→∞

= ni + ai , for 0 ≤ ai < 1, ni ∈ Z. Then ai Yi −−−→ 0, so   t i→∞ exp Yi = exp((ni + ai )Yi ) −−−→ exp(ni Yi ) = (exp(Yi ))ni ∈ H ||Yi ||

As this holds for all t ∈ R, Z ∈ h, we conclude that Z ∈ h, but h ∩ W = ∅, hence such sequences cannot exist. We conclude that H is an embedded submanifold. We now must show that the multiplication µH in H is analytic. As H is embedded, the multiplication of G when restricted to H is continuous. As following diagram commutes, the lemmas of the prior section imply that µH is analytic. H ×H

µH -

H ∩

∩

immersion

immersion ?

G×G

µG analytic

? -

G 

37

8.3. Consequences of Cartan’s Theorem. Theorem 8.18. Let f : G → G0 be a continuous homomorphism of real Lie groups. Then f is analytic. Proof. Let π1 and π2 be the canonical projections of G × G0 onto the first and second coordinates respectively. Let H := Graph(f ) ⊂ G × G0 . Then the restrictions π1 |H and π2 |H are both continuous homomorphism. We know that π1 |H is a bijection and hence f factors as (π1 |H )−1H G f

π2 |H - ?

G0 Saying that f is continuous is equivalent to saying that H is closed in G × G0 . Saying that f is a homomorphism is equivalent to saying that H is a subgroup of G × G0 . By Cartan’s theorem, H is an embedded Lie subgroup of G × G0 . It immediately follows that π1 |H and π2 |H are analytic. As π1 |H is a homomorphism, it is a map of constant rank. Furthermore, as π1 |H is an open map, it is a submersion at every point. As G and H are the same dimension manifold, we conclude that π1 |H is also an immersion at every point. By the inverse function theorem, π1 |H is a local diffeomorphism at every point, and hence (π1 |H )−1 is analytic.  Theorem 8.19. Let G be a Lie group and let h ⊂ L(G) be a Lie subalgebra. Then there exists a connected immersed Lie subgroup H < G such that L(H) = h. Definition 8.20. Let M be a smooth manifold. An m-dimensional distribution if a smooth association to each point x ∈ M an m-dimensional subspace Vx ⊂ Tx M . In otherwords, it is a smooth section of M into the Grassman bundle Grm,n (T M ). Question: Can one find a submanifold N through a given point x0 ∈ M such that Ty N = Vy for all y ∈ N ? Frobenius gives us necessary and sufficient conditions. Definition 8.21. A distribution V is involutive if given two vector fields X, Y such that for all x ∈ M , Xx , Yx ∈ Vx , then [X, Y ]x ∈ Vx for all x ∈ M . Theorem 8.22. (Frobenius)Let M be a smooth manifold with a smooth m-dimensional distribution V . Then V is involutive if and only if for each point x ∈ M there exists a unique maximal immersed submanifold N containing x such that T N = V . Corollary 8.23. Let (G, e) be a Lie group with a smooth m-dimensional distribution V . If V is involutive there exists a maximal immersed submanifold H containing e such that TH = V .

38

Lemma 8.24. Let G be a Lie group and let h ⊂ L(G) be a Lie subalgebra. Then h defines an involutive distribution. Proof. Observe that canonically L(G) ∼ = Te G and hence h is canonically isomorphic to a subspace Ve ⊂ Te G. Furthermore, by the homogeneity of Lie groups, such a vector subspace Ve defines a distribution by left translation. Now fix a basis {H1 , H2 , . . . , Hn } of h. Then given two X, Y ∈ h, there exists analytic functions fi , gi such that n n X X X= fi (x)Hi and Y = gj (x)Hj 1

1

Then " [X, Y ] =

n X

n X

fi (x)Hi ,

1

=

X

=

X

# gj (x)Hj

1

[fi (x)Hi , gj (x)Hj ]

i,j



fi (x)Hi gj (x) Hj − gj Hj fi (x) Hi + fi (x)gj (x)[Hi , Hk ]

i,j

All of whose terms lie in h, hence h is involutive.



Proof. of theorem. Having proved the lemma, the Frobenius theorem implies that there exists a maximal connected immersed submanifold H through the identity e for the distribution defined by the Lie subalgebra h. Pick h ∈ H. Then h−1 H is an immersed submanifold of G containing the identity e. By uniqueness, H = h−1 H. Then hH is an immersed submanifold of G containing the identity e. By uniqueness, H = hH. Hence H is closed under inverses and multiplication, and hence H is a subgroup of G. To show that the multiplication of H is analytic, the lemmas in the above section in conjunction with the below commuting diagram imply that it suffices to check that the multiplication is continuous. µH - H H ×H ∩

immersion

∩

immersion

? µG -G analytic We must consider the possibility that multiplication restricted to H is not continuous because H is only immersed in G, and hence may have a finer topology than that induced by being a subspace of G. Let V ⊂ H be a connected open neighborhood of the identity. While we may not be able to find an open set U ⊂ G such that H ∩ U = V , we can find an open U such ?

G×G

39

that the connected component of H ∩ U containing the identity is V . Now µ−1 G (U ) is open in −1 −1 −1 G×G and as µ−1 (H ∩U ) = µ (H)∩µ (U ) it follows that µ (H ∩U ) is open in H ×H. As G G G G −1 −1 V is a connected in G, µG (V ) is a union of connected components of µG (G ∩ U ) mapping onto V . Hence µ−1 G (V ) is open in H × H. By the commutivity of the above diagram, it −1 follows that µH (V ) is open in H × H. By the homogeneity of H, the inverse image of any open set is open. We concluded that µH is continuous, and hence analytic.  Theorem 8.25. (Ado) Any finite dimensional Lie algebra g can be embedded in End(V ) = Mn Corollary 8.26. Given g, there exists a Lie group with Lie algebra g. Proof. Combine Ado with theorem 9.19.



40

9. Simply Connected Lie Groups 9.1. Covering Spaces in Algebraic Topology. Definition 9.1. Let X be a topological space. A covering map (or covering space) is a ˜ π) where X ˜ is a topological space and π : X ˜ → X is a surjective continuous map pair (X, satisfying the condition that there exists a discrete set C and an open cover {Ui } of X such that there exists a homeomorphism ϕ making the below diagram commute of each i.

π

π1

ϕ π −1 (Ui ) ∼- Ui × C =



-

Ui ˜ → X be a a covering map. Let Y Theorem 9.2. Let X be a topological space, and π : X be a path-connected, locally path-connected space and let f : Y → X be a continuous map. ˜ if and only if f∗ (π1 (Y )) ⊂ π∗ (π1 (X)). ˜ Then f lifts to a map f˜ : Y → X

f˜

-

˜ X π f

Y

? -

X

Theorem 9.3. Let X be a path connected, locally path connected, semilocally simply connected topological space. Then X admits a simply connected cover. 9.2. Covering Spaces of Lie Groups. Theorem 9.4. Let G and G0 be Lie groups such that G is simply connected, and let f : L(G) → L(G0 ) be a Lie algebra homomorphism. Then there exists a lift f˜ : G → G0 such that df˜ = f . G 6

exp L(G)

f˜ - 0 G 6

exp f L(G0 )

41

Proof. Let h = Graph(f ) ⊂ L(G)×L(G0 ) = L(G×G0 ). It is clear that h is a vector subspace, and it is not hard to see that it is in fact a Lie subalgebra. By the existence theorem of the above section, there exists an immersed Lie subgroup H ⊂ G × G0 such that L(H) = h. Let π1 and π2 be the canonical projections of G × G0 onto the first and second coordinates respectively. Then dπ1 and dπ2 are the canonical projections of L(G) × L(G0 ) onto the first and second coordinates respectively. It follows that dπ1 |h is an isomorphism of Lie algebras, and hence f factors as (dπ1 |h )−1 ◦ dπ2 |h Furthermore, as dπ1 |h is an isomorphism, π1 |H is a local diffeomorphism, and hence an open map. As G is connected, π1 |H is surjective, and hence a covering map. As G is simply connected, the uniqueness of universal covers implies that π1 |H is a homeomorphism. G

-

f˜

6

1 −

π2

|H

) |H 1 (π

6

G0

-

H 6

expG

expH

expG0

-

(d π1

|h 2 dπ

|h ) −

1

h

f

-

L(G)

-

L(G0 )

As expG is a submersion, and (dπ1 |h )−1 ◦ expH is analytic, the lemmas of the above section implies that (π1 |H )−1 is analytic. Define f˜ := (π1 |H )−1 ◦ π2 |H  Exercise 9.5. Construct a counter-example to the above theorem for when G is not simply connected. Corollary 9.6. Let G and G0 be simply connected Lie groups such that L(G) and L(G0 ) are isomorphic as Lie algebras. Then G and G0 are isomorphic as Lie groups. Lemma 9.7. Let G and G0 be Lie groups where G is connected, and let f1 , f2 : G → G0 be two analytic homomorphisms such that df1 = df2 . Then f1 = f2 .

42

Proof. By the commutivity of the following diagram, f1 (exp(X)) = f2 (exp(X)) for all X ∈ L(G). f1 - 0 -G G f2 6 6 exp

exp

df1 = df2 - L(G0 ) L(G) Now {exp(X)| X ∈ L(G)} contains a neighborhood of e in G and as G is connected, any such neighborhood generates G. Therefore f1 = f2 on a generating set, and hence they are equal everywhere.  ˜ → G be a covering map. Then G ˜ is a Theorem 9.8. Let G be a Lie group and let π : G Lie group and π is an analytic homomorphism. Proof.



˜ → G be a topological covering Theorem 9.9. Let (G, e) be a connected Lie group. Let π : G ˜ e˜) is a Lie group and π is an analytic homomorphism. space. Let e˜ ∈ π −1 (e). Then (G, Proof. ˜×G ˜ G

f˜ - ˜ G

π×π

π ?

G×G

f

? -

G

 ˜ Theorem 9.10. Let G be a Lie group. Then there exists a simply connected Lie group G ˜ an a covering homomorphism π : G → G. Proof.



Definition 9.11. A Lie algebra is abelian or commutative, if it has trivial bracket, (i.e. [X, Y ] = 0 for all vector fields X, Y ) Example 9.12. The Lie Algebra of Rn On Rn , the left invariant vector fields are spanned by { ∂x∂ 1 , ∂x∂ 2 , . . . , ∂x∂n }. Furthermore, for C 2 functions, and hence all analytic functions, ∂ 2f ∂ 2f = ∂xj ∂xi ∂xi ∂xj

43

Hence L(Rn ) = Rn = span{ ∂x∂ i } with [−, −] = 0. Theorem 9.13. A connected Lie group is abelian iff its Lie algebra is abelian. Proof. First suppose the G is an abelian Lie group. We have already shown that this imples that G ∼ = Rm × Tn−m as Lie groups. Hence G has Rn as its universal cover. Thus ˜ L(G) = L(G) = L(Rn ) = Rn hence its bracket it trivial. Now we prove the converse. Suppose that G is a connected Lie group whose Lie algebra L(G) is abelian. Then L(G) ∼ = Rn with trivial bracket. Hence the identity map integrates makeing the below diagram commute. f Rn G 6

exp

6

exp

Id L(Rn ) L(G) By construction f is a local diffeomorphism, hence it is an open map. G is connected implies that f is then surjective, hence G is abelian.  Definition 9.14. Let G and G0 be connected Lie groups. They are isogenous if L(G) ∼ = 0 L(G ).

44

10. Adjoint Representations Let G be a Lie group. Pick g ∈ G. Then we may define the following Lie group automorphism ig : G → G x 7→ gxg −1 Observe that g is central iff ig = IdG . We then have the induced differential Ad(g) := dig : L(G) → L(G) Varying g yields a map Ad : G → Aut(L(G)) ⊂ GL(L(G)) This map is called the adjoint representation. The image Ad(G) is called the adjoint group of G. Exercise 10.1. Show Ad is a group homomorphism. Exercise 10.2. Show Ad is analytic without using Ado’s theorem. Proposition 10.3. G a connected Lie group. Then g ∈ G is central iff Ad(g) = IdL(G) Proof. Suppose that g ∈ G is central. Then ig is the identity map. Hence Adg = dig is the identity map. Now suppose that Adg = IdL(G) . Pick some X ∈ L(G). Then g(exp(X))g −1 = ig (exp(X) = exp(Adg X) = exp(X) As the choice of X was arbitrary, we conclude that g commutes with the exp(X) for all X ∈ L(G). As G is connected, exp is an open map, and hence its image generates G. We conclude that g commutes with all elements of G, and is thus central.  Corollary 10.4. Ad(G) ∼ = G/C(G) Proposition 10.5. G a connected Lie group. Then Ad is analytic. Proof. Outline: (1) Compute the adjoint representation of GLn (2) Use Ado’s theorem to induce structure on G. Step 1: P Xn n Consider the curve in GLn defined γ(t) := exp(tX) = ∞ 0 n! t where X ∈ Mn . Then the image of the curve is ! ∞ ∞ n X X (gX n g −1 ) n X −1 n −1 t g = t ig (γ(t)) = g exp(tX) g = g n! n! n=0 n=0

45

Hence "∞ # d X gX n g −1 n Ad(g) = dig = t dt n=0 n!

" =

∞ X gX n g −1 n=1

t=0

(n − 1)!

# = gXg −1

tn−1 t=0

It is clear that conjugation is analytic (in fact algebraic), hence Ad is analytic for GLn Step 2: It may not be possible to embed G → GLn , however, by Ado’s theorem, L(G) ,→ Mn , hence ˜ → GLn where G ˜ is a simply connected cover we may lift this inclusion to an analytic map G of G. π ˜ - GLn G  G 6

6

exp

exp L(G) 

∼ =

6

exp

˜ ⊂ L(G)

Mn

-

Then we have the following commutative diagram where the top row is analytic, the bottom row are analytic immersions,, and by step 1, the right vertical map is analytic. As we know Ad to be continuous, we may then conclude the middle, and then subsequently the left vertical arrows are analytic. G  Ad

π

˜ G Ad

-

GLn

Ad

? ? ∼ = ˜ ⊂ - GL(Mn ) GL(L(G))  GL(L(G)) ?

Hence Ad is analytic for an arbitrary Lie group G.



Lemma 10.6. G a connected topological group. N C G a normal discrete subgroup. Then N < C(G). Proof. Consider the map x 7→ xnx−1 for n ∈ N . As N is normal, the image lies in N . As G is connected, the image of this map lies in a connected component of N . As N is discrete, it follows that ImG is one point, and as the identity is in the image we conclude ImG = e. The result immediately follows.  ˜ and Remark 10.7. As Ad kills precisely central elements, the above lemma implies that G, G ImG ⊂ GLn will have the same image in GL(Mn ). Since the restriction of an analytic map

46

is analytic, letting G0 = ImG, we actually have the following commutative diagram. π ˜ - G0 G  G Ad

Ad

Ad

? ? = ˜ = - GL(L(G0 )) GL(L(G)) - GL(L(G)) ?

Since Ad is analytic we may compute its derivative: ad := d(Ad) : L(G) → End(L(G)) Theorem 10.8. ad(X)(Y)=[X,Y] Proof. Adexp(tX) Y = exp(tX)Y exp(−tX)     1 2 2 1 2 2 = I + tX + t X + · · · Y I − tX + t X − · · · 2 2 2 = Y + t(XY − Y X) + O(t ) = Y + t[X, Y ] + O(t2 ) Hence ad(X)(Y ) =


d d Adexp(tX) Y t=0 = Y + t[X, Y ] + O(t2 ) t=0 = [X, Y ] dt dt 

Exercise 10.9. Prove without using Ado’s theorem. Proof. Outline: (1) Compute ad of GLn (2) Use Ado’s theorem to induce structure on G. Step 1: P Xn n Consider the curve in GLn defined γ(t) := exp(tX) = ∞ 0 n! t where X ∈ Mn . Note that P∞ −1 n Xn n γ(t) = 0 (−1) n! t . Then the image of the curve is "∞ # "∞ # n X Xn X X Adγ(t) (Y ) = γ(t)Y γ(t)−1 = tn Y (−1)n tn = Y + (XY − Y X)t + O(t2 ) n! n! 0 0 Hence

i dh 2 ad(X)(Y ) = Y + (XY − Y X)t + O(t ) = XY − Y X = [X, Y ] dt t=0

47

So for GLn , we have the desired result. Step 2: Follows from the the following commutative diagram i G G ∩ ∩ g

?

GLn

? -

GLn 

48

11. Lie Subgroups and Quotient Groups Proposition 11.1. Let f : G → G0 be a homomorphism of Lie groups where G is connected. Then f (G) is a Lie subgroup of G0 such that it Lie algebra L(f (G)) = df (L(G)). Proof. The image h := df (L(G)) is a Lie subalgebra of L(G0 ). Then there exists a connected Lie subgroup H of G0 with Lie algebra h. To prove this result, it suffices to show that f (G) = H. The set expG (L(G)) generates G, so f (expG (L(G))) generates f (G). But f (expG (L(G))) = expG0 (df (L(G))) = expG0 (h) which generates H. Hence f (G) = H.  Proposition 11.2. Let G be a Lie group and H ⊂ G a Lie subgroup. Then L(H) = {X ∈ L(G) | exp(tX) ∈ H for all t ∈ R} Proof. Let h = {X ∈ L(G) | exp(tX) ∈ H for all t ∈ R}. Clearly L(H) ⊂ h. Suppose X ∈ h.  Definition 11.3. Let (A, [−, −]) be a Lie algebra. A Lie subalgebra B ⊂ A is an ideal if for all a ∈ A and b ∈ B, [a, b] ∈ B. Proposition 11.4. Let G be a connected Lie group and N ⊂ G a connected Lie subgroup. Then N is normal in G if and only if L(N ) is an ideal in L(G). Proof. First suppose that N ⊂ G is normal. Then for any g ∈ G, ig (N ) ⊆ N . Now the following diagram commutes G 6

Ad

-

Aut(L(G)) 6

exp L(G)

exp ad -

Der(L(G))

so for any g ∈ G, Adg (L(N )) ⊆ L(N). Hence for  any X ∈ L(G) and Y ∈ L(N ), Adexp(tX) (Y ) ∈ L(N ) for all t ∈ R. So dtd Adexp(tX) (Y ) ∈ L(N ) for all t ∈ R. But Adexp(tX) (Y ) = exp(tad(X))Y
= I + tad(X) + O(t2 ) Y = Y + tad(X)Y + O(t2 ) = Y + t[X, Y ] + O(t2 ) Hence

d dt



 Adexp(tX) (Y ) |t=0 = [X, Y ]. The conclusion then follows.

Now suppose that L(N ) ⊂ L(G) is an ideal. We wish to show that ig (N ) ⊂ N for all g ∈ G. To do so, it suffices to prove this for generating sets, and hence, for all X ∈ L(G),

49

t ∈ R, and Y ∈ L(N ) we wish to show iexp(tX) (exp(Y )) ∈ N Now the following diagram commutes iexp(tX)

N

-

G

6

6

exp

exp Adexp(tX)

L(N )

-

L(G)

so iexp(tX) (exp(Y )) = exp(Adexp(tX) Y ). But
Adexp(tX) Y = exp(tadX) Y   t2 2 t3 3 = I + tad(X) + ad (X) + ad (X) + · · · Y 2! 3! 2  t3 h  i t = Y + t[X, Y ] + X, [X, Y ] + X, X, [X, Y ] + · · · 2! 3! By assumption L(N ) is an ideal, so every term in the above sum lies in L(H). We conclude that Adexp(tX) Y ∈ L(H) and hence exp(Adexp(tX) Y ) ∈ H. It then follows that N is normal.  Theorem 11.5. Let G be a simply connected Lie group and N ⊂ G a connected normal Lie subgroup. Then N is closed. Proof. By the above proposition, as N is normal in G, L(N ) is normal in L(G). Then L(G)/L(N ) is a Lie algebra. Let G0 be a connected Lie group with Lie algebra L(G)/L(N ). Let π : L(G) → L(G)/L(N ) be the quotient map. As G is simply connected, π lifts to a map f : G → G0 such that df = π making the following diagram commute. G

f -

6

6

exp L(G)

G0 exp

π-

L(G)/L(N )

As ker(f ) = f −1 ({1}), it is closed in G and hence by Cartan, ker(f ) ⊂ G is an embedded Lie subgroup. Let H be the connected component of ker(f ) containing the identity. As the

50

connected components of a space are closed, it follows that H is closed in G. Furthermore it follows that L(H) = L(ker(f )) and L(ker(f )) = {X ∈ L(G) | exp(tX) ∈ ker(f ) for all t ∈ R} = {X ∈ L(G) | f (exp(tX)) = 0 for all t ∈ R} = {X ∈ L(G) | exp(tπ(X)) = 0 for all t ∈ R} = {X ∈ L(G) | π(X) = 0} = ker(π) = L(N ) So N and H are connected Lie subgroups of G with the same Lie algebra, which by uniqueness implies they are equal. Hence N is closed.  Remark 11.6. For notational reasons, given a group G, and elements a, b ∈ G, we shall denote the commutator (a, b). Recall the following universal property of commutators. Proposition 11.7. Let G be a group and H, K ⊂ G subgroups and N ⊂ G be a normal subgroup. Suppose that for all h ∈ H, k ∈ K h and k commute modulo N . Then (H, K) ⊂ N . Exercise 11.8. Let A be a Lie algebra with two Lie algebra ideals I and J. (1) Show that [I, J] is the linear span of [a, b] for a ∈ I and b ∈ J. (2) Show that [I, J] is an ideal. (Hint: Use the Jacobi Identity) Lemma 11.9. Let V be a real vector space with basis {v1 , v2 , . . . , vn }. Then the set of vectors {v1 + O(s), v2 + O(s), . . . , vn + O(s)} is a basis for V for sufficiently small s ∈ R. Proof.



Lemma 11.10. Let G be a Lie group and X, Y ∈ L(G). Then (exp(sX), exp(tY )) = exp(st[X, Y ] + stO(1)) Proof. If either s or t are 0, then the left hand term must be e. So we observe that there cannot be a constant term. This imples that the power series expansion on the right hand

51

side is divisible by st. To complete the proof, we must determine the degree two term.
exp(sX), exp(tY ) = exp(sX)exp(tY )exp(−sX)exp(−tY )
= exp Adexp(sX) tY exp(−tY )  
= exp t exp s ad(X) (Y ) exp(−tY )  
= exp t I + s ad(X) + O(s2 ) (Y ) exp(−tY )
= exp tY + st [X, Y ] + O(s2 )t exp(−tY )
= exp st [X, Y ] + stO(1)  Exercise 11.11. Let G = GLn . Using the concrete definition of exp in this case, explicitly perform the above computation. Theorem 11.12. Let G be a connected Lie group and M, N ⊂ G connected normal Lie subgroups. Then (M, N ) is a connected normal Lie subgroup with Lie algebra [L(M ), L(N )]. π ˜→ ˜ = L(G). Let M ˜,N ˜ ⊆G ˜ Proof. Let G − G be the simply connected cover of G. Then L(G) be the unique connected Lie subgroups with Lie algebras L(M ) and L(N ) respectively. As ˜ and N ˜ are normal in G. ˜ By the above these Lie subalgebras are ideals, we conclude that M theorem, they are then closed. As commuting is a closed condition, we then conclude that ˜,N ˜ ) is a closed normal embedded Lie subgroup of G. ˜ Furthermore it is clear that (M ˜,N ˜) (M is generated by the image of the map

˜ ×N ˜ →G ˜ M

(m, n) 7→ mnm−1 n−1

˜,N ˜ ) has a Since the continuous image of a connected set is connected, we conclude that (M ˜,N ˜ ) is connected. connected generating set, and hence (M ˜ with Lie algebra [L(M ), L(N )]. As its Let C˜ be the unique connected Lie subgroup of G ˜ ˜ Then G/ ˜ C˜ is a Lie group with Lie Lie algebra is an ideal, C is normal, and hence closed in G. algebra L(G)/[L(M ), L(N )]. Given X ∈ L(M ) and Y ∈ L(N ), exp(sX) and exp(tY ) com˜ To see this consider the image of X and Y in L(G)/[L(M ), L(N )]. Theremute modulo G. ˜ C˜ corresponding to this Lie subalgebra in L(G)/[L(M ), L(N )] is fore the Lie subgroup of G/ ˜ By the universal property of the abelian. Hence hexp(sX), exp(tY )i is abelian modulo C. ˜,N ˜ ) ⊆ C. ˜ commutator subgroup, we then conclude that (M

52

Now fix Xi ∈ L(M ), Yi ∈ L(N ) such that {[Xi , Yi ]}1≤i≤n is a basis for [L(M ), L(N )]. For a fixed and s ∈ R, consider the map φs : Rn → C˜ (t1 , t2 , . . . , tn ) 7→

n Y


exp(sXi ), exp(ti Yi )

i=1

˜ Observe that the image of φs actually lies in (M ˜,N ˜ ). By This is a smooth map into C.
the above lemma, exp(sXi ), exp(ti Yi ) = exp(sti [Xi , Yi ] + stO(1)). So its derivative at (t1 , t2 , . . . , tn ) = 0 yields d0 φs : T0 Rn → Te˜C˜ d 7→ s[Xi , Yi ] + O(s2 ) = s ([Xi , Yi ] + O(s)) dti Small perturbations to a basis do not effect whether it is a basis. So {[Xi , Yi ] + O(s)}1≤i≤n is a basis for small s, and hence d0 φs is onto. But by dimensional considerations, d0 φs must then be an isomorphism. By the inverse function theorem, φs is then a diffeomorphism onto ˜ Hence the generating set of C˜ lies in (M ˜,N ˜ ), from a neighborhood of the identity in C. ˜ ˜ ˜ which we conclude that C ⊆ (M , N ). ˜,N ˜ ), and From the results of the preceding two paragraphs, we conclude that C˜ = (M ˜,N ˜ ) is a connected normal Lie subgroup of G ˜ with Lie algebra [L(M ), L(N )]. Now hence (M
˜,N ˜ ) = (M, N ), and as π is an analytic homomorphism, it π is a homomorphism, so π (M follows that (M, N ) is a Lie subgroup of G. Clearly it is normal, and is generated by the image of the map M ×N →G

(m, n) 7→ mnm−1 n−1

˜,N ˜ ) is an embedded submanifold of G ˜ and from which we conclude it is connected. As (M π is a local diffeomorphism, we conclude that π|(M˜ ,N˜ ) is a local diffeomorphism and thus

˜,N ˜ ) = L (M, N ) . The theorem then follows. L (M  Remark 11.13. This theorem fails if either or both M and N fail to be connected or normal. Definition 11.14. Let G be a Lie group and C a closed Lie subgroup. Then the map π : G → G/C is a locally trivial C-bundle if for each x ∈ G, there exists a neighborhood Ux containing x such that there exists an analytic isomorphism ϕ making the below diagram

53

commute. ϕ π −1 (Ux ) ∼- Ux × C =

-



π

π

Ux Theorem 11.15. Let G be a Lie group and C ⊂ G a closed subgroup. Then G/C is an analytic manifold such that the natural map π : G → G/C is a submersion. Moreover, π : G → G/C is a locally trivial principle C-bundle. Proof. As C is closed we may push forward the analytic atlas on G over to G/C via π. In so doing, it is clear that π becomes a submersion. We now wish to show that this is a locally trivial bundle. By the homogeneity of Lie groups, it suffices to find a locally trivial neighborhood over the identity. Let us write L(G) = V ⊕ L(C) as vector spaces. We know that the map V ⊕ L(C) → G (X, Y ) 7→ exp(X)exp(Y ) is a local diffeomorphism of a neighborhood U of (0,0) to a neighborhood of 1G . Note that as Y ∈ L(C), exp(Y ) ∈ C and hence the image of exp(X)exp(Y ) in G/C is simply π(exp(X)). As π is an open map and exp is a local diffeomorphism, restricting our domain to a small enough open neighborhood V0 ⊂ U containing (0, 0), we can ensure that π ◦ exp is an open map. Furthermore, on this domain π ◦ exp is injective, as we shall now show. Suppose that there exists X, X 0 ∈ V0 such that π(exp(X)) = π(exp(X 0 )). π(exp(X)) = π(exp(X 0 ))

⇐⇒

exp(X)c = exp(X 0 ) for some c ∈ C

⇐⇒

c = exp(−X)exp(X 0 ) = exp(Y ) for some Y ∈ L(C)

⇐⇒

exp(X)exp(Y ) = exp(X 0 )exp(0)

X = X 0 and Y = 0 Hence there is a diffeomorphism π −1 (π(exp(V0 ))) ∼ = exp(V0 )C ∼ = exp(V0 ) × C in a canonical way such that the following diagram commutes ⇐⇒

ϕπ(exp(V0 )) × C ∼ =

-



π

π1

π −1 (π(exp(V0 )))

exp(V0 ) 

54

A basic result from algebraic topology states than whenever we have a Serre fibration F → E → B, we have the following long exact sequence of higher homotopy groups. · · · → πi (F ) → πi (E) → πi (B) → πi−1 (F ) → · · · → π0 (F ) → π0 (E) → π0 (B) → 0 In particulal, a principle C-bundle over an analytic manifold is such a fibration, hence we have the following corollary Corollary 11.16. Let G be a Lie group and C ⊂ G a closed subgroup. Then we have the following long exact sequence of higher homotopy groups. · · · → πi (C) → πi (G) → πi (G/C) → πi−1 (C) → · · · → π0 (C) → π0 (G) → π0 (G/C) → 0 Corollary 11.17. Let G be a Lie group and C ⊂ G a closed subgroup. Suppose that both C and G/C are connected. Then G is connected.

55

11.1. Applications to SOn and SUn . This long exact sequence is a powerful computational tool as is seen in the following application. Recall that for a topological space (1) X is connected ⇔ π0 (X) = {∗}. (2) X is simply connected ⇔ π1 (X) = {∗}. Proposition 11.18. L(SOn ) = {A ∈ Mn | At = −A} Remark 11.19. We can such matrices skew symmetric Proof. We have previously shown that X ∈ L(SOn ) ⇔ exp(sX) ∈ SOn for all s ∈ R. But exp(s(−X)) = exp(−sX) = exp(sX)−1 = exp(sX)t = exp(sX t ) As exp is a local diffeomorphism, we conclude that −X = X t . Corollary 11.20. dim SOn =



n(n−1) 2

Corollary 11.21. The natural map π : SU2 → SO3 is a surjection. Proposition 11.22. (1) SOn and SUn are connected. (2) SUn is simply connected for n ≥ 2 (3) π1 (SOn ) = Z/2Z Proof. By induction on n.



56

12. Nilpotent and Solvable Lie Groups For any abstract group G we may recursively define the two following series of normal subgroups. The Central series: C1 (G) = (G, G)

Cn (G) = (G, Cn−1 (G))

The Derived series: D1 (G) = (G, G)

Dn (G) = (Dn−1 (G), Dn−1 (G))

An abstract group is said to be nilpotent if the Cental series terminated in a finite number of steps (i.e. Cm (G) = {1} for m >> 0). An abstract group is said to be solvable if the derived series terminated in a finite number of steps (i.e. Dm (G) = {1} for m >> 0). Clearly Dn (G) ⊂ Cn (G) for all n ∈ N, hence it follows that we have the following hierarchy of groups: abelian ⊂ nilpotent ⊂ solvable In our pursuit to understand the structure of Lie groups, we have already in some sense completely classified all abelian Lie groups. We are now wish to understand the next most simple types of Lie groups: Definition 12.1. A Lie group is nilpotent (resp. solvable) if it is nilpotent (resp. solvable) as an abstract group. We may similarly define for any Lie algebra A: The Central series: C1 (A) = [A, A]

Cn (A) = [A, Cn−1 (A)]

The Derived series: D1 (A) = [A, A]

Dn (A) = [Dn−1 (A), Dn−1 (A)]

A Lie algebra is said to be nilpotent if Cm (G) = {0} for m >> 0 and solvable if Dm (A) = {0} for m >> 0. Proposition 12.2. Let G be a connected Lie group. Then L(Cn (G)) = Cn (L(G))

and

L(Dn (G)) = Dn (L(G))

Proof. This follows from theorem 11.12 of the previous section.



Corollary 12.3. Let G be a connected Lie group. Then G is nilpotent (resp. solvable) if and only if L(G) is nilpotent (resp. solvable). From Lie algebra theory we have Proposition 12.4. Let A be a solvable Lie algebra. Then [A, A] is a nilpotent ideal.

57

Corollary 12.5. Let G be a connected solvable Lie group. Then (G, G) is a nilpotent normal subgroup. 12.1. Nilpotent Lie Groups. From linear algebra we are already familiar with the following two classes of matrices. Upper triangular nilpotent n × n matrices:  0 ∗ ∗ ···      0 0 ∗ ···    0 0 0 · · · Nn :=   ... ... ... . . .          0 0 0 ··· 0 0 0 ··· Upper triangular unipotent n × n  1    0     0 Un :=   ...          0 0

∗ ∗ ∗ .. .

  ∗     ∗     ∗  ∈ M ..  n  .      0 ∗   0 0

matrices: ∗ 1 0 .. .

∗ ∗ 1 .. .

  ··· ∗ ∗     · · · ∗ ∗     · · · ∗ ∗  . . .. ..  ∈ GLn  . . .     0 0 · · · 1 ∗   0 0 ··· 0 1

It turns out that these provide a concrete example of a nilpotent Lie group. Proposition 12.6. Un is a nilpotent Lie group with nilpotent Lie algebra Nn . Proof.



Remark 12.7. As a topological space, Un ∼ = Rn(n−1)/2 and hence is contractible. In particular, it is simply connected From Lie algebra theory, we have a strengthening of Ado’s theorem. Theorem 12.8. Let A be a nilpotent Lie algebra. Then there exists a faithful representation of A into Mn for some n such that the image lies in Nn . Recall from calculus that for −1 < x < 1 1 1 1 (−1)n+1 n ln(1 + x) = x − x2 + x3 − x4 + · · · + x + ··· 2 3 4 n so for 0 < x < 2 1 1 1 (−1)n+1 2 3 4 ln(x) = (x − 1) − (x − 1) + (x − 1) − (x − 1) + · · · + (x − 1)n + · · · 2 3 4 n

58

Much as we did for the exponential map, we would like to define a log map for matrices, but unfortunately such a series need not converge. However, for a unipotent matrix, A, then A − I is nilpotent, and as such the series terminates and hence converges. Thus we may define the map log : Un → Nn ∞ X (−1)n+1 A 7→ log(A) := (A − I)n n n=1 Proposition 12.9. If N is a connected nilpotent Lie group, then exp : L(N ) → N is surjective. Proof. We prove this result by induction on the length of the central series {Ni }. Let i be the smallest index such that Ni+1 = {e}. Suppose that i = 0. Then N is abelian and we know N = Rm × Tn for some m, n ∈ N. We then have the following commutative diagram in which we know the bottom the arrows to be surjective. exp L(Rm × Tn ) - Rm × Tn 6 6

6 6

p

dp

exp L(Rm × Rn ) -- Rm × Rn We conclude that the top arrow is surjective and hence this holds for i = 0. Now suppose the hypothesis holds for i − 1. By construction, Ni is a nontrivial central subgroup of N . Let C = Ni . Then C is a connected closed central subgroup of N . As C is central it is an abelian subgroup of N , and hence exp |L(C) : L(C) → C is surjective. Furthermore N/C is a nilpotent group with central series of smaller length than that of N . By the inductive hypothesis, exp : L(N )/L(C) → N/C is surjective. We then have the following commutative diagram of exact sequences di dπ - L(C) - 0 0 L(N ) -- L(N )/L(C) exp ? ?

1

-

C

exp

exp ?

i

? -

N

π -- ? N/C

-

1

Pick n ∈ N . By surjectivity there exists an X ∈ L(N ) such that exp(dπ(X)) = π(n). As exp(dπ(X)) = π(exp(X)) and π is an analytic homomorphism we have π([exp(X)]−1 n) = 1N/C . Thus [exp(X)]−1 n ∈ kerπ = C. By surjectivity, there exists a Z ∈ L(C) such that exp(Z) = [exp(X)]−1 n. Thus n = exp(X) exp(Z). As exp(Z) is central in G, Z is central in

59

L(G). So X and Z commute and hence exp(X) exp(Z) = exp(X + Z). We conclude that X + Z maps onto n, thus proving the inductive step.  Exercise 12.10. Show that SL2 R has elements that are not squares. Conclude that exp : L(SL2 R) → SL2 R is not surjective. Proposition 12.11. The maps exp : Nn → Un and log : Un → Nn are inverses of oneanother. Proposition 12.12. Let H ⊂ Un be a connected Lie subgroup. Then exp|L(H) and log|H are inverse maps. Proof. We already know that exp and log are inverses and that exp|L(H) : L(H) → H is surjective. It suffices to show that the image of log|H lies in L(H). Given h ∈ H, log(h) ∈ Nn , but h = exp(log(h)) since these are inverse maps. As exp is onto there exists an X ∈ L(H) such that exp(X) = h. Hence log(h) = X ∈ L(H)  Corollary 12.13. A connected Lie subgroup H of Un is closed and as a manifold diffeomorphic to L(H). Moreover H is contractible as a topological space and hence simply connected. Theorem 12.14. A simply connected nilpotent Lie group is isomorphic to a subgroup of Un for some n. Proof. As G is a nilpotent Lie group, L(G) is a nilpotent Lie algebra. There then exists an n ∈ N for which there is an embedding i : L(G) → Nn . As G is simply connected, this lifted to a map f : G → Un making the following diagram commute such that df = i f G Un 6 6

exp

6

exp

i L(G) ⊂ Nn As G is connected, f (G) is a connected Lie subgroup of Un . By the above corollary, f (G) is simply connected. By construction f (G) has Lie algebra L(G) so f : G → f (G) is a local isomorphism Thus f is open hence surjective, and thus a covering map. But both G and f (G) are simply connected, hence f is an isomorphism.  Corollary 12.15. Let N a be simply connected nilpotent Lie group and H ⊂ N be a connected subgroup. Then H is closed and as a manifold diffeomorphic to Euclidean space. Proof. H ,→ N ,→ Un



Lemma 12.16. Let G be a simply connected Lie group and let H ⊂ G be a connected normal subgroup. Then G/H is simply connected.

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Proof. We have a fibration H → G → G/H. This induces a long exact sequence of homotopy groups which yields · · · → π1 (G) → π1 (G/H) → π0 (H) → · · · But both the left and right term are zero by assumption, so by exactness, the middle term is zero.  Proposition 12.17. Let N be a simply connected nilpotent Lie group and let C ⊂ N be a compact Lie subgroup. Then C is trivial. Proof. Let {Ni } be the central series. Then the Ni are closed connected normal subgroups of N . Let Nk be the last nontrivial term of the series. Then the image of C in N/Nk is compact. Furthermore, by the above lemma, N/Nk is simply connected, so we may induct on the length of the central series. First suppose k = 0. Then N is abelian. By the classification of simply connected abelian Lie groups, N ∼ = Rm for some m. For any x ∈ Rm −{0}, Zx is unbounded, m and hence R has no nontrivial compact subgroups. Thus the hypothesis holds for k = 0 Suppose the hypothesis holds for all k < n. Now N/Nn is a simply connected nilpotent group with central series of smaller length than that of N . By the inductive hypothesis, π(C) ⊂ N/Nn is trivial. Hence C ⊂ Nn , which is central. As N is simply connected and N is normal, it is closed and hence a Lie subgroup. As Nn is connected, the above corollary implies Nn contractible. As Nn is abelian, we conclude Nn is isomorphic to Rk as Lie groups for some k ∈ N. Thus C is a compact Lie subgroup group of Rk which we showed above implies C is trivial, hence proving the inductive step.  Proposition 12.18. Let N be a simply connected nilpotent Lie group. Then the center Z(N ) is connected. Proof. It suffices to show that if c ∈ N is central, then log(c) ∈ L(N ) is central. This is because the analytic homomorphism t 7→ exp(t log(c)) yields a one-parameter subgroup (i.e. a path) connecting 1and c. Denote X := log(c). Now for all g ∈ G exp(Adg X) = g(exp(X))g −1 = gcg −1 = c = exp(X) As exp is a diffeomorphism, we conclude that Adg X = X for all g ∈ G



Remark 12.19. This need not be true for non-nilpotent groups. For example Z(SL2 R) = {−I, I} is discrete. Also note that there center is always closed as commutation is a closed operation. Proposition 12.20. Let G be a connected nilpotent Lie group. Then any compact subgroup of G is central. π ˜→ ˜ ,→ Un . Proof. Let G − G be the simply connected cover of G. We then have an embedding G ˜ Let π(˜ ˜ by By the below exercise Adg˜ is unipotent for all g˜ ∈ G. g ) = g. As L(G) = L(G),

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the commutativity of the below diagram it follows that Adg˜ = Adg ˜ G

Ad -

˜ Aut(L(G)) ||

π ?

Ad -

?

Aut(L(G)) G Hence Adg is unipotent for all g ∈ G. On the other hand, if C ⊂ G is a compact Lie subgroup, then Adg is semisimple for all g ∈ C. So Adg must be 1 for all g ∈ C, and hence C is central.  Exercise 12.21. Show that if g ∈ GLn is unipotent (resp. semisimple), then Adg is unipotent (resp. semisimple). Proof.



Theorem 12.22. (Engel) Let g be a Lie algebra such that ad(X) is a nilpotent endomorphism for all X ∈ g. Then g is a nilpotent Lie algebra.

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12.2. Solvable Lie Groups. Proposition 12.23. Let G be a simply connected solvable Lie group. Then (G, G) is a connected normal Lie subgroup of G with Lie algebra [L(G), L(G)]. Proof. We know from Lie’s theorem in Lie algebra theory that [L(G), L(G)] is a nilpotent Lie algebra, hence (G, G) is a nilpotent Lie group.  Exercise 12.24. Show that π2 (SOn ) and π2 (SUn ) are trivial. Theorem 12.25. (E. Cartan) Let G be a Lie group. Then π2 (G) is trivial. Proposition 12.26. Let G be a simply connected Lie group and let N ⊂ G be a connected normal Lie subgroup. Then N is simply connected. Proof. As G is simply connected, it follows that N is closed, and hence N → G → G/N is a fibration of Lie groups. The induced long exact sequence of higher homotopy groups yields · · · → π2 (G/N ) → π1 (N ) → π1 (G) → · · · But both the left and right term are zero by assumption, so by exactness, the middle term is zero and the conclusion follows.  Remark 12.27. Normal is a necessary element of the hypothesis. Consider the case of SUn , for n ≥ 2. The subgroup of diagonal matrices in SUn is isomorphic as a Lie group to Qn−1 S 1 , which is connected but normal. It is clear that this subgroup is very much not 1 simply connected. Proposition 12.28. Let G be a simply connected solvable Lie group and let C ⊂ N be a compact Lie subgroup. Then C is trivial. Theorem 12.29. Let G be a connected compact solvable Lie group. Then G is abelian and hence a torus. We shall now present a method for constructing new Lie groups. We recall the construction of a semidirect product from group theory. Given two groups G and H, and a group homomorphism ϕ : H → Aut(G), then we may endow the set G×H with the group structure given by (g, h)(g 0 , h0 ) = (g ϕh (g 0 ), hh0 ) Now in the context of Lie groups, we already know that G × H is a smooth manifold so the above multiplication is analytic as long as the map ϕ : H → Aut(G) is an analytic homomorphism of Lie groups. Proposition 12.30. Let G be a simply connected Lie group. Then Aut(G) is a Lie group.

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Proof. We know that for any real vector space V , GL(V ) is a Lie group. Now Aut(L(G)) is an abstract subgroup of GL(L(G)). Furthermore, as preserving the bracket is a closed condition, it is a closed subgroup. By Cartan’s theorem, it is an embedded Lie subgroup. By the lifting property of simply connected Lie groups, there is a canonical isomorphism Aut(L(G)) ∼  = Aut(G), which endows Aut(G) with a Lie group structure. ˜ Show Aut(G) ⊂ Exercise 12.31. Let G be a Lie group with simply connected cover G. ˜ is a Lie subgroup. Aut(G) Exercise 12.32. Let N and H be Lie groups and let ϕ : H → Aut(N ) be an analytic homomorphism. Compute L(N oϕ H). Definition 12.33. Let g be a Lie algebra. A derivation of g is an R-linear map D : g → g such that D[X, Y ] = [DX, Y ] + [X, DY ] Then the set of derivations of g is denoted Der(g). Proposition 12.34. Let g be a Lie algebra. Then Der(g) is a Lie algebra with bracket [D1 , D2 ] := D1 D2 = D2 D1 . Exercise 12.35. Let g be a Lie algebra. Show that the Lie algebra of Aut(g) is Der(g). Theorem 12.36. Let G be a simply connected Lie group and let N ⊂ G be a connected normal Lie subgroup. Then as a manifold, G is diffeomorphic to N × G/N Proof of the Semidirect Product Structure Theorem. Let G be a simply connected Lie group with Lie algebra g. Let N have Lie algebra n. Let r ⊂ g/n be the radical ideal of g/n and let r be its’ pullback ideal in g. Then the quotient map . g/n → (g/n) (r/n) := s . yields a semisimple Lie algebra. As (g/n) (r/n) ∼ = g/r, it follows that we have an exact sequence of Lie algebras 0→r→g→s→0 By Levi decomposition, this splits, and hence g = r + ⊃ϕ s. Let R and S be the connected Lie subgroups of G corresponding to r and s respectively. It follows that G = RS and as r ∩ s = 0, R ∩ S is discrete. Furthermore, as r is an ideal in g, r ∩ s is an ideal in s, and thus R∩S is normal in S. By the second isomorphism theorem, it follows that RS/R ∼ = S/(R∩S), and hence S is a covering space of G/R. As r is an ideal, R is normal, and hence by the long exact sequence of higher homotopy, G/R is simply connected. By the uniqueness of simply

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connected covers, we conclude that S is simply connected and that R ∩ S = e. As such we may lift ϕ as follows f S Aut(R) 6

6

exp

exp

- Der(r) df = ϕ It then follows that G = R of S. Furthermore, topologically, G ∼ = R × S.

s

Now observe that by construction, N ⊆ R, so it now suffices to show that topologically R∼ = N × R/N . By assumption, r/n is solvable and hence we have the derived series r/n = d0 ) d1 ) d2 ) · · · ) dm ) dm+1 = n/n Which lifts to a filtration of ideals of r r = d0 ) d1 ) d2 ) · · · ) dm ) dm+1 = n Which lifts to a filtration of Lie subgroups of R R = D0 ) D1 ) D2 ) · · · ) Dm ) Dm+1 = N Hence it suffices to show that topologically Di ∼ = Di−1 × Di /Di−1 . Then pick some X1 ∈ di \ di−1 . Then we have di ⊃ di−1 + RX1 ) di−1 We have a natural map d1 : RX1 → Der(di−1 ) X1 7→ [−, X1 ] This map gives the vector space di−1 + RX1 a Lie algebra structure, namely di−1 + ⊃d1 RX1 . As R is simply connected, we lift to a map ϕ1 R Aut(di−1 ) 6

6

exp RX1

exp -

Der(di−1 )

As Di−1 is a connected normal Lie subgroup of Di , it is simply connected. Hence Aut(di−1 ) = Aut(Di−1 ), so we may view ϕ1 : R → Aut(Di−1 )). We may construct the semidirect product Di−1 oϕ1 R. As the Lie algebra of a semidirect product is a semidirect sum of Lie algebras,

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we conclude that L(Di−1 oϕ1 R) = di−1 + ⊃d1 RX1 . We may now iterate this process, and as di is finite dimensional, it terminates, yielding a maximal filtration di = di−1 + RX1 + RX2 + · · · RXn ) · · · ) di−1 + RX1 + RX2 ) di−1 + RX1 ) di−1 We have just inductively constructed a simply connected Lie group  
Di−1 oϕ1 R oϕ2 R oϕ3 · · · oϕn+1 R with Lie algebra di . By the uniqueness of simply connected Lie groups, we then have an isomorphism of Lie groups:  
Di ∼ = Di−1 oϕ1 R oϕ2 R oϕ3 · · · oϕn+1 R Now iterating this process for 0 ≤ i ≤ m + 1 we conclude that topologically R ∼ = N × RN for some N ∈ N, and hence G ∼  = S × RN × N ∼ = G/N × N . Corollary 12.37. Let G be a simply connected Lie group and let N ⊂ G be a connected normal Lie subgroup. Then for all n ∈ N, πn (G) = πn (N ) × πn (G/N ) Proposition 12.38. Let G be a simply connected solvable Lie group. Then G is contractible. Proof. As G is simply connected, (G, G) is a connected closed normal Lie subgroup. Hence the projection map G → G/(G, G) is a fibration, inducing the following long exact sequence of higher homotopy groups. · · · → πi+1 (G/(G, G)) → πi ((G, G)) → πi (G) → πi (G/(G, G)) → πi−1 ((G, G)) → · · · As G/(G, G) is the abelianization of G, it is a connected abelian Lie group. By assumption π1 (G) = {1} = π0 ((G, G)) so by the long exact sequence, G/(G, G) is simply connected. By the classification of abelian Lie groups G/(G, G) ∼ = RN for some N ∈ N. Hence G/(G, G) is contractible and πi (G/(G, G) = {1} for all i ∈ N. By the long exact sequence of homotopy groups, πi (G) ∼ for all i ∈ N = πi ((G, G)) We conclude that (G, G) is simply connected. As G is solvable, (G, G) is nilpotent. As simply connected nilpotent groups are contractible, we conclude that πi (G) = {1} for all i ∈ N. As manifolds have the homotopy type of a CW-complex, it then follows from Whitehead’s theorem that G is contractible. 

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12.3. Appendix: Levi Decomposition of Lie Algebras and Semidirect Sums. Proposition 12.39. Let A be a Lie algebra and let I, J ⊂ A be solvable ideals. Then I + J is solvable. Corollary 12.40. Let A be a Lie algebra. Then there exists a unique maximal solvable ideal R ⊂ A. Definition 12.41. Let A be a Lie algebra. The unique maximal solvable ideal of A is called the radical of A. It is denoted Rad(A). Definition 12.42. Let A be a Lie algebra and let I ⊂ A be an ideal. I is semisimple if Rad(I) = 0. As there is a semidirect product construction for groups, there is a semidirect sum construction for Lie algebras. Let A and B be Lie algebras, and let ϕ : B → Der(A) be a Lie algebra homomorphism. Then we may endow the vector space A × B with a Lie algebra structure via the bracket  
(a, b), (a0 , b0 ) = [a, a0 ] + ϕ(b)a0 − ϕ(b0 )a, [b, b0 ] We denote the resulting Lie algebra A + ⊃ϕ B. This is equivalent to saying that we have an exact sequence of Lie algebras 0→A→A+ ⊃ϕ B → B → 0 Theorem 12.43. (Levi Decomposition) Let A be a Lie algebra. Then there exists a semisimple ideal S ⊂ A such that A = Rad(A) + ⊃S Definition 12.44. The Lie subalgebra S in the above semidirect sum is called the Levi subalgebra Proposition 12.45. Let G and H be Lie groups and let ϕ : H → Aut(G). Then L(G oϕ H) = L(G) + ⊃dϕ L(H)

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13. Compact Lie Groups 13.1. Invariant Inner Products on the Lie Algebra of a Compact Lie group. Proposition 13.1. The connected components of a Lie group are open. Proposition 13.2. Let G be a compact Lie group. Then G has finitely many connected components. Definition 13.3. Let (V, h−, −i) be an inner product space and let T : V → V be a linear map. The inner product is T-invarient if hT v, T wi = hv, wi for all v, w ∈ V . Proposition 13.4. Let G be a compact Lie group. Then L(G) carries an Ad(G)-invariant positive definite inner product. Proof. As L(G) is a vector space over R, it naturally comes with a positive definite inner product, (−, −). Then for X, Y ∈ L(G), define Z hX, Y i := (Adg X, Adg Y )dg G

This is clearly a positive definite symmetric bilinear form. Fix h ∈ G. Observe Adg Adh = Adgh Z hAdh X, Adh Y i = (Adg Adh X, Adg Adh Y )dg ZG = (Adgh X, Adgh Y )dg ZG = (Adg0 X, Adg0 Y )dg 0 = hX, Y i G

 Proposition 13.5. Let G be a compact Lie group. Let h−, −i be an Ad(G)-invariant positive definite inner product on L(G). Then h[Z, X], Y i + hX, [Z, Y ]i = 0 Proof. This is a direct computation. Fix X, Y, Z ∈ L(G). Then hX, Y i = hAdexp(tZ) X, Adexp(tZ) Y i = hexp(tadZ )X, exp(tadZ )Y i = h(I + tadZ + O(t2 ))X, (I + tadZ + O(t2 ))Y i = hX + tadZ X + O(t2 )X, Y + tadZ Y + O(t2 )Y i = hX + t[Z, X] + O(t2 )X, Y + t[Z, Y ] + O(t2 )Y i = hX, Y i + th[Z, X], Y i + thX, [Z, Y ]i + O(t2 )

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So th[Z, X], Y i + thX, [Z, Y ]i + O(t2 ) = 0 Evaluating the derivative of this expression at t = 0 yields h[Z, X], Y i + hX, [Z, Y ]i = 0  Definition 13.6. Let (g, [−, −]) be a Lie algebra with inner product h−, −i. The inner product is invariant if for all X, Y, Z ∈ g h[Z, X], Y i + hX, [Z, Y ]i = 0 Exercise 13.7. Let G be a connected compact Lie group. Let h−, −i be an invariant positive definite inner product on L(G). Show that h−, −i is Ad(G)-invariant. Theorem 13.8. Let G be a connected compact Lie group with Lie algebra g. Let h−, −i be a positive definite inner product on g. Then the following are equivalent (1) h−, −i is Ad(G)-invariant. (2) h−, −i is invariant. (3) ad(Z) is skew-symmetric for all Z ∈ g. (4) If n = dim G then the image of ad : g → End(g) = gln in fact lies in son Remark 13.9. Basic linear algebra implies that the eigenvalues of skew-symmetric transformations over R are purely imaginary. Also, as adZ Z = [Z, Z] = 0, we have zero is always an eigenvalue of adZ for all Z ∈ L(G). Let g be a Lie algebra with an invariant inner product h−, −i. For a subset s ⊂ g, we may define s⊥ := {v ∈ g | hv, si = 0 for all s ∈ s} Exercise 13.10. Let g be a Lie algebra with an invariant inner product and let I ⊂ g be an ideal. Then I ⊥ is an ideal such that g = I ⊕ I ⊥ as Lie algebras. Proposition 13.11. Let g be a finite dimensional Lie algebraL with an invariant inner product. Then there exists minimal ideals I1 , I2 , . . . such that g = n In Lemma 13.12. Let g be a Lie algebra and let I, J ⊂ g be ideals. Then [I, J] ∈ I ∩ J. Proposition 13.13. Let g be a finite dimensional Lie algebra with an invariant inner product. Then 1-dimensional ideal are central. Proof. Let I ⊂ g be a 1-dimensional ideal. Then I is generated by a single vector v, hence [I, I] = 0. Then, by the above exercise. [g, I] = [I ⊕ I ⊥ , I] = [I, I] + [I ⊥ , I] = [I ⊥ , I] ⊂ I ∩ I ⊥ = 0 

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Proposition 13.14. Let g be a finite dimensional Lie algebra with an invariant inner product. Let I ⊂ g be a minimal ideal of dimension greater than 1. Then I is semisimple. Proof. Let c ⊂ g be the span of all one dimensional ideals. Then c itself is an ideal. So g = L c ⊕ c⊥ . By the above proposition, there exists minimal ideals I1 , I2 , . . . such that ⊥ c = n In . By construction, the dimension of all such minimal ideals is greater than 1. It follows that I must be one of these. As [I, I] ⊂ I, minimality implies either [I, I] = 0 or [I, I] = I. Suppose the former. Then by the same computation as in the above proposition, [g, I] = 0 hence I is central. But central ideals are abelian, hence every 1 dimensional subspace is an ideal. This contradicts our assumptions about I. Hence it must be the case that [I, I] = I. The conclusion follows.  Corollary 13.15. Let g be a finite dimensional Lie algebra with an invariant inner product. Let c ⊂ g be the center. Then there exists a semisimple ideal s ⊆ g such that g = c ⊕ s. Corollary 13.16. Let G be a connected compact solvable Lie group. Then G is abelian. Proof. As G is compact, L(G) is a finite dimensional Lie algebra with an invariant inner product. As G is solvable, L(G) contains no semi-simple ideals. So in the above decomposition, S = 0, hence L(G) = C, which is abelian. Then G is abelian.  13.2. Invariant Riemannian Metrics on a Compact Lie Group. In this brief aside, we collect important result from differential geometry without giving proofs. Take any positive definite inner product on Te (G). We shall us this and left translation to get a left invariant Riemannian metric. By a similar process we can also construct a right invariant inner product. But not necessarily simultaneously left and right invariant, which we shall call bi-invariant. Observe that d(Rg−1 Lg ) = d(ig ) = Adg Theorem 13.17. Let G be a compact Lie group. Then G admits a bi-invarient Riemannian metric. Theorem 13.18. Let G be a Lie group with a bi-invarient metric. Then geodesics through the identity are of the form t 7→ exp(tX) for some X ∈ L(G). Theorem 13.19. Let M be a complete Riemannian manifold. Then any two points are joined by a geodesic. Theorem 13.20. Let G be a compact connected Lie group. Then the exponential map is surjective. Corollary 13.21. Let G be a compact connected Lie group and let n ∈ N. Then G have nth roots.

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13.3. Root Systems of Compact Lie Groups. In this section we take a brief detour from the study of compact Lie groups to introduce some important results from Lie algebra theory. We wish to study real Lie groups, and hence real Lie algebras, but complex Lie algebras are often easier to study. Here we shall introduce techniques that allow us to glean information about real Lie algebras from associated complex Lie algebras. Given a real Lie algebra (g, [−, −]R ) we may construct a complex Lie algebra in the following way. Define the complex vector space gC := C ⊗R g Observe that gC = {X + iY | X, Y ∈ g} √ where i = −1. This is then made into a complex Lie algebra with bracket [−, −]C defined via

[X + iY, X 0 + iY 0 ]C := [X, X 0 ]R − [Y, Y 0 ]R + i [X, Y 0 ]R + [X 0 , Y ]R As the usage should be clear from context, the R and C subscripts will be dropped. Definition 13.22. Let g be the Lie algebra of a compact Lie group. A Lie subalgebra h ⊂ g is a Cartan subalgebra if it is a maximal abelian subalgebra. Let be g a Lie algebra with h ⊂ g a Cartan subalgebra. Given a linear map α : hC → C, define gαC := {X ∈ gC | [H, X] = α(H)X for all H ∈ hC } Definition 13.23. Let g be a Lie algebra and let h ⊆ g be a Cartan subalgebra. A linear map α ∈ h∗C is called a root of g with respect to h if α 6= 0 and gαC 6= 0. The collection of all roots, Φ(g, h) is called the root system of g with respect to h. It follows that Φ ⊂ h∗C is a finite set. Proposition 13.24. Let g be the Lie algebra of a compact Lie group let h ⊆ g be a Cartan subalgebra. Let Φ be the set of roots of g with respect to h. Then Φ is an abstract root system. Let 0 denote the zero map hC → C. Then observe that g0C = {X ∈ gC | [H, X] = 0 for all H ∈ hC } which is precisely the centralizer ofL hC in gC . Furthermore, as ad(hC ) is diagonalizable, gC is a sum of these spaces, i.e. gC = α gαC . In fact we have the stronger result. Lemma 13.25. Let g be a Lie algebra and let h be a Cartan subalgebra. Then g0C = hC . Theorem 13.26. Let g be a Lie algebra and let h be a Cartan subalgebra. Then ! M α gC = hC ⊕ gC α∈Φ

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Definition 13.27. Let g be the Lie algebra of a compact Lie group let h ⊆ g be a Cartan subalgebra. Let Φ be the set of roots of g with respect to h. (1) An element v ∈ h is regular if α(v) 6= 0 for all α ∈ Φ. (2) An element v ∈ h is singular α(v) 6= 0 for some α ∈ Φ. Definition 13.28. Let g be the Lie algebra of a compact Lie group let h ⊆ g be a Cartan subalgebra. Let Φ be the setSof roots of g with respect to h. For α ∈ Φ let Hα = ker(α). A connected component of h \ α∈Φ Hα is called a Weyl chamber. Let G be a compact Lie group and let T ⊆ G be a maximal torus. The adjoint representation restricted to T gives a representation of a compact abelian Lie group. Complexifying, this representation is then completely reducible with irreducible components of dimension 1. Definition 13.29. Let G be a compact Lie group and let T ⊆ G be a maximal torus. An analytic homomorphism α : T → S 1 is called a root of G with respect to T if α is a nonzero character that is induced by the adjoint representation restricted to T . The collection of all roots, Φ(G, T ) is called the root system of G with respect to T . Proposition 13.30. Let G be a compact Lie group with maximal torus T ⊆ G. Then there is a natural bijection between the Lie group root system Φ(G, T ) and its corresponding Lie algebra root system Φ(L(G), L(T )). Definition 13.31. Let G be a Lie group and T ⊂ G a maximal torus. Let Φ be the set of roots of G with respect to T . (1) An element t ∈ T is regular if α(t) 6= 1 for all α ∈ Φ. (2) An element g ∈ G is regular if it is a conjugate of a regular element of T . (3) An element g ∈ G is singular if it is not regular.

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13.4. The Tori of Compact Lie Groups. Definition 13.32. Let G be a Lie group. A compact connected abelian Lie subgroup of G is called a torus. Definition 13.33. Let G be a compact Lie group. Any abelian Lie subalgebra of L(G) is called a toral subalgebra. Let G be a compact Lie group. L(G) contains such subalgebras since, for any X ∈ L(G), RX is abelian. Definition 13.34. Let g be a Lie algebra with h ⊂ g a Lie subalgebra. h is a Cartan subalgebra if it is a maximal abelian subalgebra. Proposition 13.35. Let G be a compact connected Lie group. Then G contains nontrivial tori. Proof. If dim G = 0, this result is trivial, so suppose dim G > 0. Then any one-dimensional subspace of L(G) is an abelian subalgebra. Let a ⊆ L(G) be a nonzero abelian subalgebra. Let A ⊂ G be the connected Lie subgroup with Lie algebra a. It follows that A is abelian and nontrivial. Then A is a closed, and hence compact, Lie subgroup. Furthermore, as commuting is a closed condition, A is abelian, and hence A is a torus.  Remark 13.36. (1) Let G be a compact connect Lie group and let h ⊂ L(G) be a Cartan subalgebra. Let H ⊂ G be a connected Lie subgroup with Lie algebra h. Then H is abelian, and by maximality, closed. Hence H is a torus. (2) We know L(G) = c ⊕ [L(G), L(G)]. Let C be the connected Lie subgroup of G with Lie algebra c. It follows that C is central in G, and hence so too is C. Thus L(C) is central in L(G). We conclude that L(C) = c = L(C), and hence C is closed. (3) We now have a bijection {Cartan subalgebras of L(G)} ↔ {Maximal Tori of G} Theorem 13.37. Let G be a compact Lie group and let t, t0 ⊂L(G) be maximal toral subalgebras. Then there exists a g ∈ G such that Adg t0 = t (i.e. any two maximal toral subalgebra are conjugate to eachother under the adjoint action of G). Proof. Let X ∈ t and X 0 ∈ t0 be regular. Consider the map f :G→R

g 7→ hAdg X − X 0 , Adg X − X 0 i

By applying Adg0 we may assume that f is minimized at g = 1. So for a fixed Y ∈ L(G), df 0 0 f (t) = hAdexp(tY ) X − X , Adexp(tY ) X − X i such that dt t=0

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Hence f (t) = hAdexp(tY ) X − X 0 , Adexp(tY ) X − X 0 i = hX − X 0 + t[X, Y ] + O(t2 ), X − X 0 + t[X, Y ] + O(t2 )i = hX − X 0 , X − X 0 i + 2th[X, Y ], X − X 0 i + O(t2 ) df = 2h[X, Y ], X − X 0 i + O(t) dt 0 = h[X, Y ], X − X 0 i = hY, [X, X − X 0 ]i = hY, [X, X 0 ]i As this is true, independent of the choice of Y , we conclude that [X, X 0 ] = 0. Ranging X we conclude that t ⊂ Zg (X 0 ) By maximality, we get equality.  Theorem 13.38. Let G be a compact Lie group and let T, T 0 ⊂ G be maximal tori. Then there exists a g ∈ G such that ig T 0 = T (i.e. any two maximal tori are conjugate to eachother under the conjugation action of G). Proof. By the above theorem, there exists a g ∈ G such that Adg (L(T )) = L(T 0 ). Hence ig (T ) = T 0 .  Corollary 13.39. Let G be a compact Lie subgroup. Then any two maximal toral subalgebras of L(G) have the same dimension. Proposition 13.40. Let G be a Lie group and T ⊂ G a maximal torus. Let t ∈ T be regular. Then ZG (t) = T . Proposition 13.41. Suppose the G is a compact Lie group with T ⊂ G a maximal torus. Then G = G◦ NG (T ). Proof. Suppose the G is a disconnected compact Lie group. Let G◦ be the identity component, which is open, and hence a closed normal subgroup of G. Let g ∈ G and let T ⊂ G be a maximal torus. Then T and gT g −1 a re both maximal tori contained in G◦ . By the above theorem, there exists a h ∈ G◦ such that h−1 gT g −1 h = T . Then h−1 g ∈ NG (T ). So G = G◦ NG (T ).  Proposition 13.42. Let G be a Lie group and H ⊂ G a connected Lie subgroup. Then NG (H) is closed in G. Proof. An element g ∈ G normalizes H if and only iff Adg (h) = h. Thus NG (H) = {g ∈ G | Adg (h) = h}.  Proposition 13.43. Let G be a connected Lie group and let T ⊂ G be a torus which is normalized by G. Then T is central in G.

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Proof. Let Un ⊂ T be the set of elements of T of order n. Observe that for each n ∈ N, Un is finite. As conjugation preserves order, and as T is normal in G, we conclude that conjugation maps each Un into itself. So the orbit of an order n element lies entirely in Un . But an orbit of a point is the continuous image of G, which is connected. As Un is discrete, we conclude that for all n ∈ N and for all x ∈ Un , gxg −1 = g for all g ∈ G. Let U := ∪n∈N Un . Hence U is central in G. But being central is a closed condition, and by Kronecker’s theorem, U is dense in T. Hence U = T is central in G.  Lemma 13.44. Let G be a compact connected Lie group, let T ⊂ G be a torus and let g ∈ ZG (T ). Then there exists a torus T 0 containing T ∪ {g}. Proof. Let H ⊂ G be the closure of subgroup generated by the set T ∪ {g}. As being central is a closed condition, this group is abelian, and as such the connected component containing the identity H ◦ is a torus. As T ⊂ H and T is connected, T ⊂ H ◦ . Let u ∈ H ◦ be a generator. As H is compact it has finitely many connected components, hence [H, H ◦ ] = n < ∞. Thus g n ∈ H ◦ and hence g −n u ∈ H ◦ . As H ◦ is abelian, it has nth roots, and as such, there exists a v ∈ H ◦ such that v n = g −n u. As G is connected and compact, the exponential map is surjective. Hence there exists an X ∈ L(G) such that exp(X) = gv. Let T 0 be the torus generated by the exponential curve exp(tX). Observe that u = g n v n = (gv)n = exp(nX) ∈ T 0 . Hence T ⊂ H ◦ ⊂ T 0 . Furthermore, v ∈ H ◦ implies v ∈ T 0 . But by construction, gv ∈ T 0 and hence g ∈ T 0 .  Theorem 13.45. Let G be a compact connected Lie group and let T ⊂ G be a torus. Then ZG (T ) is connected. Proof. As tori are abelian, any torus containing T must lie in ZG (T ). However, by the above lemma, given any element g ∈ ZG (T ), there is a torus containing T ∪ {g}. Hence we have [ ZG (T ) = T0 T 0 torus T ⊂T 0

Now each torus is connected, and as every torus contains the identity, their pairwise intersection in nontrivial. From point-set topology, it follows that their union is connected.  Corollary 13.46. Let G be a compact connected Lie group and let T ⊂ G be a torus. Then NG (T )◦ = ZG (T ). Proof. Clearly ZG (T ) ⊂ NG (T ). By the above theorem the former is connected, and hence ZG (T ) ⊂ NG (T )◦ . By the above proposition, NG (T ) is closed and hence NG (T )◦ is a connected Lie group. As every element of NG (T )◦ normalizes T , by the above proposition, T is central in NG (T )◦ . Hence NG (T )◦ ⊂ ZG (T ). 

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13.5. Compact Semisimple Lie Groups. Theorem 13.47. (H. Weyl) Let G be a compact connected semisimple Lie group. Then π1 (G) is finite. Proof. If G is simply connected, we are done, so suppose not. As G is a topological group, π1 (G) is abelian. G is connected so the Hurewicz isomorphism gives π1 (G) = π1 (G)Ab ∼ = H1 (G; Z). As G is a compact manifold its homology groups are finitely generated, in particular H1 (G; Z). By the fundamental theorem of finitely generated abelian groups, H1 (G; Z) ∼ = Zn ⊕ {f inite group}. So it suffices to show that n = 0. The universal coefficients theorem for cohomology H k (G; A) ∼ = HomAb (Hk (G; Z), A) ⊕ Ext1 (Hk−1 (G; Z), A) Z

gives for real coefficients H 1 (G; R) ∼ = HomAb (H1 (G; Z), R) ∼ = HomAb (Zn ⊕ {finite group}, R) ∼ = Rn By DeRham’s theorem, for a manifold, singular cohomology with real coefficients is isomorphism to DeRham cohomology. As G is a Lie group, the collection of bi-invarient differential forms H∗ (G) form a subcomplex of the DeRham complex Ω∗ (G) with trivial differentials and with the same cohomology. Thus H1 (G) ∼ = H 1 (G; R). But by the below exercise G G Hk (G) = (Λi (g∗ )) . Hence H1 (G) = (Λ1 g∗ ) = (g∗ )G ⊆ c where c is the center of g. As G is compact, L(G) = c + [L(G), L(G)]. As G is semisimple, L(G) has trivial center, so H1 (G) = 0 and the conclusion follows.  Corollary 13.48. Let G be a compact connected semisimple Lie group with simply connected ˜ Then G ˜ is compact. cover G. Definition 13.49. Let g be a Lie algebra. The Killing form is the map κ:g×g→R

κ(X, Y ) := T r(ad(X)ad(Y ))

Proposition 13.50. Let g be a Lie algebra. The Killing form is a symmetric bilinear form. Theorem 13.51. (Cartan) Let g be a Lie algebra. Then the Killing form is non-degenerate if and only if g is semisimple. Definition 13.52. Let g be a semisimple Lie algebra. Then g is of compact type if the Killing form κ is negative definite. Lemma 13.53. Let g be a Lie algebra. Then ker(ad) = z(g). Theorem 13.54. Let g be a semisimple Lie algebra. Then ad(g) = Der(g). Corollary 13.55. Let g be a semisimple Lie algebra. Then ad : g → Der(g) is a Lie algebra isomorphism.

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Theorem 13.56. Let G be a connected Lie group with Lie algebra g. Then G is compact semisimple if and only if g is of compact type. Proof. First suppose that G is compact semisimple. As G is compact ad(X) is skew symmetric, so its eigenvalues are purely imaginary. Hence κ(X, X) = T r(ad(X)2 ) ≤ 0 Furthermore, as g is semisimple, it has trivial center, and hence if X 6= 0, it has at least one nonzero eigenvalue. Thus for X 6= 0, κ(X, X) < 0. Now suppose that κ is negative definite. As this implies that κ is non-degenerate, it immediately follows that G is semisimple. Now the collection of Killing form preserving linear maps  O(g, κ) := T ∈ GL(g) κ(X, Y ) = κ(T X, T Y ) for all X, Y, ∈ g is clearly compact. Given any T ∈ Aut(g) and X, Y, Z ∈ g, then ad(T X)ad(T Y )Z = [T X, [T Y, Z]] = T [X, T −1 [T Y, Z]] = T [X, [Y, T −1 Z]] = T ad(X)ad(Y )T −1 Z Hence κ(T X, T Y ) = T r(ad(T X)ad(T Y )) = T r(T ad(X)ad(Y )T −1 ) = T r(ad(X)ad(Y )) = κ(X, Y ) As preserving the Lie bracket is a closed condition, Aut(g) is closed in O(g, κ). Hence Aut(g) is compact. As g is semisimple, it has trivial center, and hence the center Z(G)of G must be discrete. We have seen that the kernel of Ad : G → Ad(G) is Z(G), and we conclude that this is a covering map. Thus g = L(Ad(G)). By the above structure theorem of semisimple Lie algebras, we know ad : g → Der(g) = L(Aut(g)) is an isomorphism. Thus L(Ad(G)) = g = ad(g) = Der(g) = L(Aut(g)), and as G is connected, G = Aut(g)◦ . G

-

- Aut(g) ⊂ - O(g, κ) Ad(G) ⊂ closed closed

As connected components are closed, we conclude that Ad(G) is compact. As Ad(G) is ˜ of Ad(G) is compact. By the semisimple, by Weyl’s theorem, the simply connected cover G ˜ universal property of simply connected covers, G covers G, and hence G is compact.  Theorem 13.57. A complex semisimple Lie algebra is uniquely determined up to isomorphism by its root system. Definition 13.58. Let h be a complex Lie algebra. A real Lie algebra g whose complexification gC equals h is called a real form corresponding to h. Remark 13.59. It may be the case that many different real Lie algebras have the same complexification. For example sl(2)C = su(2)C but clearly sl(2) 6= su(2). However we do have the following compactness condition.

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Theorem 13.60. Given a complex semisimple Lie algebra h, there exists a unique real Lie algebra g of compact type such that gC = h. Bringing all of the above results together, we then have proved the following classification theorem. Theorem 13.61. A simply connected semisimple Lie group whose Lie algebra is of compact type is completely determined by its root system.

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14. Noncompact Semisimple Lie Groups Let G be any noncompact Lie group with Lie algebra g. Let r ⊂ g be the radical ideal. Then the quotient map g → g/r := s yields a semisimple Lie algebra. By Levi decomposition, this splits, and hence g = r + ⊃s Let R (resp. S) be the connected Lie group of G with Lie algebra r (resp. s). As r is a maximal ideal, it follows that R is closed and normal in G. Thus G = SR. However, this may not be a semidirect product, as S ∩ R need not be trivial. However, as s ∩ r = 0 and L(S ∩ R) = s ∩ r we conclude that S ∩ R is discrete. Furthermore, s ∩ r is an ideal in s, and hence S ∩R is normal in S. As S is semisimple, we then conclude that S ∩R is central. Thus the study of any noncompact Lie group boils down to understanding solvable Lie groups, semisimple Lie groups, and discrete central subgroups of semisimple Lie groups. As we already understand solvable Lie groups, we turn out attention toward semisimple Lie groups. 14.1. Compact Forms of a Semisimple Lie Algebra. Definition 14.1. Let g semisimple Lie algebra. An R-subalgebra u of gC is an R-form if the induced map u ⊗R C → gC is an isomorphism. An R-subalgebra u of gC is a compact R-form if it is a form on which the Killing form κ is negative definite. Proposition 14.2. Let G be a noncompact, semisimple Lie group with Lie algebra g. Let u ⊂ gC be an R-form. Then u is semisimple. Theorem 14.3. (Weyl) Let G be a noncompact, semisimple Lie group with Lie algebra g. Then gC contains a compact form. Proof. Fix a maximal toral subalgebra tC ⊂ gC . Let Φ be the set of roots. We may choose a Chevalley basis ∆ := {Xα ∈ gαC } in such a way that [Xα , Xβ ] = Nα,β Xα,β √ for some Nα,β ∈ Z where α, β, α + β ∈ Φ Let i = −1. Define the subalgebra D E u := i[Xβ , X−β ], Xα − X−α , i(Xα + X−α ) β ∈ ∆, α ∈ Φ+ It is now an exercise to show that u is a compact R-form.



Example 14.4. Let G = SLn , in which case g = sln (R). Then gC = sln (C). Now u := sun ⊂ sln (C) is a compact form. Letting Xα be elementary matrices, it is clear that the process described in the above proof generates sun from these Xα . We have proven the existence of a compact for semisimple Lie algebras. However, this u need not be unique, as is shown by the following proposition.

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Proposition 14.5. Fix a real semisimple Lie algebra g. Let u be a compact form of g. Let GC is a compact Lie group with Lie algebra gC and adjoint group GC . Then (1) GC = Aut(gC )◦ . In particular GC is nontrivial. (2) For any g ∈ GC , gu is a compact form of g. Before we proceed with decomposing semisimple Lie groups, we shall take an aside to use the machinery of compact forms and compact Lie groups to prove a classical result from Lie algebra theory. Exercise 14.6. An R-linear action of a Lie algebra g on a real vector space V induces a canonical C-linear action of gC on VC . Exercise 14.7. Let g be a real Lie algebra and let the real vector space V be a g-module. Suppose VC is completely reducible as a gC -module. Then V is completely reducible as a g-module. Proposition 14.8. Let g be a real semisimple Lie algebra and let the real vector space V be a g-module. Then VC is completely reducible as a gC -module. Proof. Let u ⊂ gC be a compact form. As u spans gC over C, a C-subspace is u-stable if and only if it is stable under gC . So we need to only show that the u-module VC is completely reducible. As the Killing form of u is negative definite, by Weyl’s theorem, any connected Lie group with Lie algebra u is compact. Let U be the simply connect Lie group with Lie algebra u. Since U is simply connected, we now have a representation of a compact Lie group, and we know implies complete reducibilty. - GL(VC ) U 6

6

exp u

exp -

End(VC ) 

Corollary 14.9. Let g be a real semisimple Lie algebra and let the real vector space V be a g-module. Then V is completely reducible as a g-module.

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14.2. Cartan Decomposition of a Semisimple Lie Algebra. Let g be a semisimple Lie algebra and fix a compact form u in gC . Define the maps σ : C ⊗R g → gC

(z ⊗ X) 7→ zX

τ : C ⊗ R u → gC (z ⊗ X) 7→ zX Observe that σ and τ are conjugate linear Lie algebra automorphisms of order two. Now σ and τ may not commute, but στ is a complex linear automorphism of gC with the following properties (1) (στ )−1 = τ σ (2) σ(στ )σ −1 = τ σ = (στ )−1 (3) τ (στ )τ −1 = τ σ = (στ )−1 (4) στ preserves the Killing form Exercise 14.10. The Killing form is preserved by any Lie algebra automorphism. For τ , its +1 eigenspace is u and its −1 eigenspace is iu. Observe that the iu is a subspace, not a subalgebra, of gC and the Killing form of gC restricted to the subspace iu is positive definite. For σ, its +1 eigenspace is g and its −1 eigenspace is ig. Proposition 14.11. στ = τ σ if and only if the +1 eigenspace of σ, g, is stable under the action of τ . Suppose now that we have chosen u such that σ and τ commute. Then g is stable under the action of τ , hence we have the vector space decomposition g = (g ∩ u) ⊕ (g ∩ iu) := k ⊕ p This is called the Cartan decomposition of g. Unfortunately we do not know if such a u exists. We shall spend the rest of the sections finding such a u. Define the map hτ : gC × gC → C (X, Y ) 7→ −κ(X, τ (Y )) where κ is the Killing form on gC . Proposition 14.12. hτ is a positive definite Hermitian form on gC . Proof. hτ is clearly Hermitian. For any Z ∈ gC , there exists X, Y ∈ g such that Z = X + iY . So hτ (Z, Z) = −κ(X + iY, τ (X + iY )) = −κ(X + iY, X − iY ) = −κ(X, X) − iκ(Y, X) + κ(X, Y ) − κ(Y, Y ) = −κ(X, X) − κ(Y, Y )

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As the Killing form is negative definite on g, the conclusion then follows. 
Definition 14.13. Let V, (−, −) be a complex inner product space and let T : V → V be an endomorphism. T is Hermitian if for all v, w ∈ V (T v, w) = (v, T w)
Proposition 14.14. Let V, (−, −) be a complex inner product space and let T : V → V be a Hermitian endomorphism. Then T is semisimple with all real eigenvalues. Lemma 14.15. στ is Hermitian with respect to hτ . Proof. hτ (στ X, Y ) = −κ(στ X, τ Y ) = −κ(X, (στ )−1 τ Y ) = −κ(X, (τ σ)(τ Y )) = −κ(X, τ (στ Y )) = hτ (X, στ Y )  Corollary 14.16. στ is semisimple with all real eigenvalues. Let ϕ := (στ )2 . Then ϕ is Hermitian with positive real eigenvalue. Observe σϕσ −1 = σ(στ στ )σ −1 = σστ (σ −1 σ)στ σ −1 = σ(στ )σ −1 σ(στ )σ −1 = (τ σ)(τ σ) = (στ )−2 = ϕ−1 Similarly ϕ has the following properties (1) σϕσ −1 = ϕ−1 (2) τ ϕτ −1 = ϕ−1
Definition 14.17. Let V, (−, −) be a complex inner product space and let T ∈ GLC (V ) be a Hermitian transformation with positive real eigenvalues. Let Λ be the set of eigenvalue of T and let Vλ be the eigenspace associated to λ ∈ Λ. Let Eλ : V → Vλ be the canonical projection. For a fixed λ ∈ Λ, define the transformation Tλt : Vλ → Vλ

v 7→ λt v

82

Then define the T t =

P

λ∈Λ

Tλt ◦ Eλ

Proposition 14.18. Let V be a complex vector space and let T ∈ GLC (V ) be a Hermitian transformation with positive real eigenvalues. If S ∈ GLC (V ) such that ST = T S, then ST t = T t S for all t ∈ R. Now consider the map ϕt τ ϕ−t . This is the involution given by the compact form ϕt u. 1

1

1

1

Proposition 14.19. Let σ, τ, ϕ be defined as above. Then ϕ 4 τ ϕ− 4 σ = σϕ 4 τ ϕ− 4 Proof. This proof is a result of several algebraic manipulations using only the following four properties: Given any s ∈ R: (1) σ = σ −1 (2) τ = τ −1 (3) σϕs = ϕ−s σ for any s ∈ R (4) τ ϕs = ϕ−s τ for any s ∈ R Observe that for a fixed t ∈ R   
−1  t −t t −t τ σ ϕ τϕ σ = τ (σϕ )τ (ϕ σ)   = τ (ϕ−t σ)τ (σϕt ) = (τ ϕ−t )στ σϕt = (ϕt τ )στ σϕt = ϕt (τ σ)2 ϕt = ϕt ϕ−1 ϕt = ϕ2t−1 Now suppose that we have picked t ∈ R such that σϕt τ ϕ−t σ −1 = ϕt τ ϕ−t . Then 
−1  t −t τ σ ϕ τϕ σ = τ (ϕt τ ϕ−t ) = (τ ϕt )τ ϕ−t = (ϕ−t τ )τ ϕ−t = ϕ−t ϕ−t = ϕ−2t Hence t satisfies the above condition if and only if ϕ2t−1 = ϕ−2t , or ϕ4t−1 = Id, from which we conclude that the above condition is satisfied when t = 41 .  Theorem 14.20. There exists a compact form u0 of gC such that g is invariant under the involution given by u0 .

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Proof. By Weyl’s theorem, gC admits a compact form u. Then, using the above terminology, let u0 := (στ )1/2 u, and the conclusion follows by the above theorem.  From here on out, let u be a compact form of gC under which whose involution g is invariant. Let k:g∩u p : g ∩ iu Then as a vector space g = k ⊕ p. Then observe that we may also decompose u as u = (u ∩ g) ⊕ (u ∩ ig) = (u ∩ g) ⊕ i(iu ∩ g) = k ⊕ ip Hence given a Cartan decomposition k ⊕ p of g, k ⊕ ip is a compact form of gC . Theorem 14.21. Let g be a Lie algebra with Cartan decomposition g = k ⊕ p. Then (1) [p, p] ⊆ k (2) k is a Lie subalgebra of g (3) The Killing form on g restricted to k is negative definite. (4) The Killing form on g restricted to p is positive definite. Let θ := τ |g . This is called the Cartan involution associated with the Cartan decomposition g = k ⊕ p. Observe that for X ∈ k and Y ∈ p, θ(X + Y ) = θ(X) + θ(Y ) = X − Y . Let κ be the Killing form on g. As θ is a Lie group automorphism κ(θ(X), θ(Y )) = κ(X, Y ) But by the above observations about θ κ(θ(X), θ(Y )) = κ(X, −Y ) = −κ(X, Y ) We conclude that κ(X, Y ) = 0. Hence, with respect to the Killing form, k ⊥ p, and p = k⊥ Define the map B :g×g→R (X, Y ) 7→ −κ(X, θ(Y )) Proposition 14.22. B is a positive definite form in g. Lemma 14.23. With respect to the positive definite form B, ad(X) is skew symmetric for X ∈ k and is symmetric for X ∈ p Proof. We will prove this result for when X ∈ k, and the other case follows similarly. B(ad(X)Y, Z) = −κ([X, Y ], θ(Z)) = κ([Y, X], θ(Z)) = κ(Y, [X, θ(Z)]) = κ(Y, θ([X, Z])) = κ(Y, θ(ad(X)(Z))) = −B(Y, ad(X)(Z))

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 Proposition 14.24. For Y ∈ p, ad(Y ) is semisimple with real eigenvalues. that ad(k) consists of skew symmetric matrices. Meanwhile exp(ad(p)) consists of positive definite transformations. Observe that for X, Y ∈ p, [X, Y ] ∈ k so B|p is positive definite.

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14.3. Cartan Decomposition of a Semisimple Lie Group. Recall SLn = {A ∈ GLn | det A = 1} SOn = {A ∈ SLn | At = A−1 } and define Pn = {A ∈ SLn | A = At and A is positive definite} The following is a classical result decomposing SLn into SOn and Pn . Theorem 14.25. (1) SOn is a maximal compact subgroup of SLn . Furthermore it is the unique maximal compact Lie subgroup up to conjugation. (i.e. If C ⊂ SLn a compact Lie subgroup, then there exists g ∈ SLn such that gCg −1 ⊆ SOn . 1 (2) Pn ∼ = R 2 n(n+1)−1 as smooth manifolds. (3) SLn = SOn Pn and SLn ∼ = SOn × Pn as smooth manifolds Proof. (1) Given a compact subgroup C of SLn , we can find a positive definite inner product on Rn which is C invariant. AnyPpositive definite quadratic form on Rn is equivalent to the standard inner product x · y := ni=1 xi yi , the stabilizer of which is SOn . We conclude that a conjugate of C is contained in SOn . As conjugation is analytic and preserves dimension and number of connected components, it follows from (1) that SOn is maximal. (2)Consider the smooth maps exp : {symmetric matrices} → {positive definite symmetric matrices}

X 7→ exp(X)

log : {positive definite symmetric matrices} → {symmetric matrices} X 7→ log(X) Then are inverses, and hence these two are diffeomorphic manifolds. However, we know the 1 first to be canonically diffeomorphic to R 2 n(n+1)−1 and the result then follows. (3) The map µ : SOn × Pn → SLn (X, Y ) 7→ XY is a smooth. We wish to find an inverse. In other words, given g ∈ SLn , we wish to find X ∈ On and Y ∈ Pn such that g = XY . We shall construct them explicitly. If such X and Y exist, then g t = (XY )t = Y t X t = Y X −1 The positive definite matrix g t g = Y X −1 XY = Y 2 1 1 Hence we may take the square root yielding Y = (g t g) 2 . So then X = gY −1 = g(g t g)− 2 . We must now show that indeed such an X is orthogonal. Observe that 1

1

X t X = (g t g)− 2 g t g(g t g)− 2 = 1

86

Now define the smooth map η : SLn → SOn × Pn

  1 1 g 7→ g(g t g)− 2 , (g t g) 2

If immediately follows that η = µ−1 and hence η is a diffeomorphism.



SLn is the prototype for a semisimple Lie group. The above result suggests we may decompose an arbitrary semisimple Lie group in a similar way. ˜ be the unique connected, simply connected Definition 14.26. Let g be a Lie algebra let G ˜ Lie group with Lie algebra g. The adjoint group of g is Ad(G). There are numerous benefits to studying Ad(G) over G, in particular Ad(G) is a matrix group. Observe that the center of a semisimple Lie group is discrete, and as such is a covering space of its adjoint group. Thus to study semisimple Lie groups, it suffices to study adjoint groups of semisimple Lie algebra together with covering space theory. Lemma 14.27. The adjoint group of a Lie algebra is connected. Proposition 14.28. Let G be the adjoint group of a semisimple Lie algebra g. Then G = Aut(g)◦ . In particular G is an embedded matrix group. Proof. As g is semisimple, ad is a Lie algebra isomorphism. Thus L(G) = g = ad(g) = Der(g) = L(Aut(g)), and as G is connected, G = Aut(g)◦ . As connected components are closed, we conclude that G is closed in Aut(g), and hence is embedded in GL(g). As preserving the Lie bracket is a closed condition, Aut(g) is closed in GL(g). G⊂

closed -

closed - GL(g) Aut(g) ⊂ 

Proposition 14.29. Let G be the adjoint group of a semisimple Lie algebra g with Cartan decomposition g = k ⊕ p and Cartan involution θ. Define K := {g ∈ G | Adg θ = θAdg }. Then K is a compact Lie subgroup of G with Lie algebra L(K) = k. Proof. Define H := {T ∈ GL(g) | T θ = θT }. Commuting with θ is a closed condition in GL(g) so H is closed in GL(g). Clearly K = G∩H, and by the above proposition G is closed in GL(g), so K is closed in both G and GL(g). By Cartan’s theorem, K is an embedded Lie subgroup of both G and GL(g).

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Let C ⊆ G be the unique connected Lie subgroup with Lie algebra k. Pick k ∈ k and X = Xk + Xp ∈ g where Xk ∈ k and Xp ∈ p. Observe that θ commutes with ad(k) (θ ◦ ad(k))X = θ([k, Xk + Xp ]) = θ [k, Xk ] +θ [k, Xp ] | {z } | {z } ∈k

∈p

= [k, Xk ] − [k, Xp ] = [k, Xk − Xp ] = [k, θ(Xk + Xp )] = (ad(k) ◦ θ)X P 1 So we see that Ad(exp(k)) = exp(ad(k)) = (ad(k))n commutes with θ. By construction n! ◦ exp(k) ∈ C, and we conclude that C ⊆ K and hence k ⊆ L(K). Now pick X ∈ L(K). Then Ad(exp(X))θ = θAd(exp(X)), so  
−1 θ(X) = Ad(exp(X)) ◦ θ ◦ Ad(exp(X)) (X)  
−1 = exp(ad(X)) ◦ θ ◦ exp(ad(X)) (X)
= exp(ad(−X)) ◦ θ ◦ exp(ad(X)) (X)
= exp(ad(−X)) ◦ θ (exp(ad(X))(X)  ∞ 
X 1 n = exp(ad(−X)) ◦ θ (ad(X)) (X) n! n=0 = (exp(ad(−X)) ◦ θ)(X) = exp(ad(−X))(θ(X)) = exp(ad(−X))(X)  ∞  X 1 n = (ad(−X)) (X) n! n=0 =X So X lies in the +1 eigenspace of θ, which is precisely k. Thus L(K) ⊆ k and hence L(K) = k. As B(−, −) := −κ(−, θ(−)) is a positive definite form on g, we may consider the corresponding orthogonal group O(g, B) := {T ∈ AutVect (g) | B(T X, T Y ) = B(X, Y ) for all X, Y ∈ g}

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We know such spaces to be compact. For g ∈ K, B(Ad(g)X, Ad(g)Y ) = −κ(Ad(g)X, θ(Ad(g)Y )) = −κ(Ad(g)X, Ad(g)θ(Y )) = −κ(X, θ(Y )) = B(X, Y ) so K ⊆ O(g, B). As K is closed in GL(g), it then follows that K is compact.  Definition 14.30. Let M be a connected Riemannian manifold. M is a symmetric space if for each p ∈ M , there exists an isometry jp : M → M such that jp (p) = p and (djp )p = −Idp . Proposition 14.31. Let M be a symmetric space. Then M is geodesically complete. Theorem 14.32. Let G be the adjoint group of a semisimple Lie algebra g with Cartan decomposition g = k ⊕ p and Cartan involution θ. Let K = {g ∈ G | Adg θ = θAdg }. Define S := G/K. Let B(−, −) := −κ(−, θ(−)). (1) For every gK ∈ S, TgK S ∼ = p. (2) (S, B|p ) is a Riemannian manifold. (3) S is a symmetric space of non-positive curvature. (4) The geodesics of S through the identity coset eK are of the form π(exp(tX)) for X ∈ p. Proof. As K is closed in G, the natural projection π : G → G/K is a submersion. Hence dπe : Te G → TeK G/K is surjective. Observe that for X ∈ Te G, d d [π(exp(tX))]|t=0 = [exp(tX)K]|t=0 dt dt As k is a Lie subalgebra, it immediately follows that k ⊆ ker(dπe ). Furthermore, if dπe (X) = 0, then there exists a t0 > 0 such that for all t < t0 , exp(tX) ∈ K, hence X ∈ k.  dπe (X) =

Corollary 14.33. Let G be as above, Then the exponential map exp : p → G/K is surjective and hence G = K(exp p). Theorem 14.34. (Cartan’s Fixed Point Theorem) Let S be a symmetric space and let C ⊆ Isom(S) be a compact Lie subgroup. Then C has a fixed point in S (i.e. there exists y ∈ S such that Cy = y). Proof. Fix x ∈ S. Then Cx ⊂ S is compact. Let H be the intersection of all convex subsets containing Cx and let d be the metric on H given by geodesic distance. Define f :H→R

f (y) := max d(y, z) z∈H

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As H is compact, the image of f is compact and hence there exists y ∈ H such that this function is minimized. We wish to show that this point is unique. Suppose there exists two such points y1 and y2 . Then there exists a geodesic γ ⊂ H connecting these two points. Let m denote the midpoint of this geodesic. For a fixed z ∈ H, let θ be the angle at m between γ and the geodesic connecting m and z. As S has non-positive curvature, for i = 1, 2 1 (d(yi , z))2 ≥ d(y1 , y2 )2 + d(m, z)2 + {positive number} 4 2 2 So d(m, z) ≤ d(yi , z) , hence m is a strictly better center than y1 or y2 , contradicting our assumption. As isometries send geodesics to geodesics, CH is a convex set containing Cx, and hence H ⊆ CH. Clearly no points in CH \ H could satisfy the above minimizing property, and hence we conclude that y must be fixed by C.  Theorem 14.35. Let G be the adjoint group of a semisimple Lie algebra g with Cartan decomposition g = k ⊕ p and Cartan involution θ. Let K ⊆ G be the subgroup of G consisting of elements which commute with θ. Then K a maximal compact Lie subgroup of G with Lie algebra L(K) = k. Furthermore, K is unique up to conjugation. Proof. Now if C ⊆ G is a compact subgroup, then C fixes a point gK, so CgK = gK which implies that gCg −1 K = K hence gCg −1 ⊆ K.  Theorem 14.36. (Cartan Decomposition) Let G be the adjoint group of a semisimple Lie algebra. (1) Let K ⊆ G be a maximal compact Lie subgroup.
(2) Let P ⊆ G be the manifold exp {X ∈ L(G) | κ(X, L(K)) = 0} . Then G = KP and G ∼ = K × P as manifolds.

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14.4. Iwasawa Decomposition of a Semisimple Lie Algebra. Definition 14.37. Let g be a semisimple Lie algebra. g is simple if it does not contain a proper nonzero ideal. Example 14.38. sln , so(p, q), so(n), su(n), sp(2n) Proposition 14.39. Let g be a simple Lie algebra with Cartan decomposition g = k ⊕ p. Then p is an irreducible k-module and k = [p, p]. Proof. Suppose that p0 ⊆ p is a nontrivial irreducible k-module. Let p00 = p0⊥ (in p) with respect to the Killing form. So then p00 is a k-submodule, and p = p0 ⊕ p00 . Now define the Lie subalgebra g0 := [p0 , p0 ] ⊕ p0 We shall now show that g0 is an ideal (i.e. [g0 , g] ⊆ g0 ). Now h i 0 0 0 0 0 00 [g , g] = [p , p ] ⊕ p , k ⊕ p ⊕ p ] h i h i = [p0 , p0 ], k ⊕ [p0 , p0 ], p ⊕ [p0 , k] ⊕ [p0 , p0 ] ⊕ [p0 , p00 ]   It immediately follows from the Cartan decomposition that [p0 , p0 ], p = 0. Let Y 0 ∈ p0 , Y 00 ∈ p00 and X ∈ k. Then [Y 0 , Y 00 ] ∈ k, so, as the Killing form is invariant on k, κ(X, [Y 0 , Y 00 ]) = κ([X, Y 0 ], Y 00 ) but as the first term lies in p0 and the second in p00 , the above expression is 0. As this conclusion is independent of the choice of X we conclude by the non-degeneracy of the Killing form that [Y 0 , Y 00 ] = 0. Hence [p0 , p00 ] = 0. Thus h i 0 0 0 [g , g] = [p , p ], k ⊕ [p0 , k] ⊕ [p0 , p0 ] By assumption [p0 , k] ⊆ p0 . Furthermore, given Y1 , Y2 ∈ p0 and X ∈ k, the Jacobi identity gives       [Y1 , Y2 ], X = Y1 , [X, Y2 ] + Y2 , [X, Y1 ] ∈ p0 hence [p0 , p0 ] ⊆ p0 . We conclude that g0 is a nonzero ideal. As g is simple it follows that that g0 = g. As [p0 , p0 ] ∩ p = 0 we conclude that that p0 = p and [p, p] = k.  Definition 14.40. Let g be a simple Lie algebra with Cartan decomposition g = k ⊕ p. A maximal abelian Lie subalgebra a ⊆ p is called a Cartan subspace Recall that for Y ∈ p, ad(Y ) is semisimple with real eigenvalues. Given a linear map α : a → R, define gα := {X ∈ g | [H, X] = α(H)X for all H ∈ a}

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Definition 14.41. Let g be a semisimple Lie algebra and let a be a Cartan subspace. A linear map α ∈ a∗ is called a root of g with respect to a if α 6= 0 and gα 6= 0. The collection of all roots, Φ is called the root system of g with respect to a. It follows that Φ ⊂ a∗ is a finite set. Proposition 14.42. Let g be a semisimple Lie algebra with Cartan decomposition g = k ⊕ p and let a ⊆ p be a Cartan subspace. Let Φ be the set of roots of g with respect to a. Then Φ is an abstract root system. Let 0 denote the zero map a → R. Then observe that g0 = {X ∈ gC | [H, X] = 0 for all H ∈ a} which is precisely the centralizer of a in g, z(a). Theorem 14.43. Let g be a semisimple Lie algebra with Cartan decomposition g = k ⊕ p and let a ⊆ p be a Cartan subspace. Let Φ be the set of roots of g with respect to a. Then ! M α g g = z(a) ⊕ α∈Φ

Remark 14.44. Unlike the root space domposition of gC , it may be the case that a ( z(a). Theorem 14.45. (Cartan) Let g be a semisimple Lie algebra with Cartan decomposition g = k ⊕ p. Then any two Cartan subspaces in p are conjugate. Proof.



Corollary 14.46. Up to conjugation, there is a unique Cartan subspace of g. Corollary 14.47. All Cartan subspaces have the same dimension. Lemma 14.48. θ(gα ) = g−α Let Φ be the root system of g with respect to a. We may give Φ an ordering as follows. Pick some regular H ∈ a. Then we say that α ∈ Φ is positive if α(H) > 0 and we say that α ∈ Φ is negative if α(H) < 0. Additionally we may say that α < β if β − α is positive, however it is often the case that β −α is not a root. As such we usually just consider whether a root is positive or negative. Observe that any two values in the the same Weyl chamber give the same ordering on Φ. As such, these ordering are parametrized by the Weyl chambers. Now define n+ := n :=

X

gα

α∈Φ α>0 −

n :=

X

gα

α∈Φ α<0

Lemma 14.49. Suppose that α, β ∈ Φ and α + β is a root.

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(1) If α, β ∈ Φ are positive, then α + β is positive (2) [gα , gβ ] ⊆ gα+β Proof.



Proposition 14.50. n ⊆ g is a Lie subalgebra. Furthermore it is nilpotent. Proof. By Engel’s theorem 12.22.



Lemma 14.51. θ(n) = n− Theorem 14.52. (Iwasawa Decomposition) Let g be a semisimple Lie algebra with Cartan decomposition g = k ⊕ p. Let a ⊆ p be a Cartan subspace and let n be the positive root spaces of g with respect to a. Then g = k ⊕ a ⊕ n as vector spaces. Proof. Theorem 14.43 can be restated g = z(a) ⊕ n ⊕ n− Given a positive root α and an element X α ∈ gα θ(X α + θ(X α )) = X α + θ(X α ) As k is the +1 eigenspace of θ we conclude that X α + θ(X α ) ∈ k. Hence g−α = θ(gα ) ⊆ k + n As this result is independent of the choice of α, it follows that n ⊕ n− ⊆ k + n. Furthermore z(a) = z(a) ∩ k ⊕ z(a) ∩ p = z(a) ∩ k ⊕ a ⊆ k + a We conclude that g=k+a+n Suppose there exists k ∈ k, a ∈ a, n ∈ n such that k + a + n = 0 Applying the Cartan involution then yields k − a + θ(n) = 0 where θ(n) ∈ n− . Taking the difference yields 0 = 2a + n − θ(n) where 2a ∈ z(a), n ∈ n and θ(n) ∈ n− . By theorem 14.43 we conclude each of these terms are zero and hence k = a = n = 0. Thus g = k ⊕ a ⊕ n as vector spaces.  Lemma 14.53. a normalizes n.

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14.5. Iwasawa Decomposition of a Semisimple Lie Group. Definition 14.54. Let G be a semisimple Lie group. G is simple if its Lie algebra is simple. Proposition 14.55. Let G be a semisimple Lie group. G is simple if and only if it does not contain a proper nonzero normal subgroup. Example 14.56. SLn , SO(p, q), SO(n), SU (n), Sp(2n) Exercise 14.57. Let g be a simple Lie algebra with Cartan decomposition g = k ⊕ p where p is nontrivial. Show that (1) dim(z(k)) ≤ 1. (2) p ⊗R C is the direct sum of one or two irreducible nontrivial k-submodules depending on whether the center of k is trivial or 1-dimensional. Theorem 14.58. (Harish-Chandra) Let g be a simple Lie algebra with Cartan decomposition g = k ⊕ p where p is nontrivial. Let G be the adjoint group of g and let S = G/K be the symmetric space of G. Then G carries a G-invariant complex analytic structure if and only √ if dim(z(k)) = 1. This complex structure is unique up to multiplication by −1. Definition 14.59. Let G be the adjoint group of a semisimple Lie algebra g and let K ⊆ G be a maximal compact subgroup. The R-rank of g (or the rank on the symmetric space G/K) is the dimension of a Cartan subspace a ⊆ g. Proposition 14.60. Let G be the adjoint group of a semisimple Lie algebra, and let N ⊆ G be a connected Lie subgroup whose Lie algebra is the positive root space of L(G) with respect to some Cartan subspace. Then N is unipotent and closed. Lemma 14.61. Let G be a connected Lie group and let M1 , M2 ⊆ G be Lie subgroups such that L(M1 ) + L(M2 ) = L(G). Then G = M1 M2 . Proof. Consider the action of M1 × M2 on G given by (M1 × M2 ) × G → G (m1 , m2 ) · g := m1 gm−1 2 Let Og be the orbit of g ∈ G under this action. A simple computation shows that for any point h ∈ Og , Th Og = g. By the implicit function theorem, the orbits of this action are open. As G is connected, each orbit is closed. Hence G = Oe , the orbit of the identity.  Exercise 14.62. Show that the above lemma is false if instead we have M1 , M2 , . . . , Mn ⊆ G Lie subgroups such that L(M1 ) + L(M2 ) + · · · + L(Mn ) = L(G) and n > 2 Exercise 14.63. Let G be a Lie group and let H ⊆ G be a Lie subgroup. Show that NG (L) is a Lie group with Lie algebra L(NG (L)) = NL(G) (L(N )).

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Theorem 14.64. (Iwasawa Decomposition) Let G be the adjoint group of a semisimple Lie algebra. (1) Let K ⊆ G be a maximal compact Lie subgroup. (2) Let A ⊆ G be a connected Lie subgroup with Lie algebra a Cartan subspace in L(K)⊥ . (3) Let N ⊆ G be a connected Lie subgroup with Lie algebra the positive root spaces of L(G) with respect to L(A). Then G = KAN as groups and G ∼ = K × A × N as manifolds. Proof of Iwasawa Decomposition. Let L(G) = g, L(K) = k, L(A) = a, L(N ) = n. It follows from the Iwasawa decomposition of Lie algebras that g=k⊕a⊕n As a normalizes gα for each root α, a normalizes n, and hence A normalizes N . Thus AN ⊆ G is a subgroup. As A and N are closed in G, so too is AN , and hence by Cartan’s theorem AN is a Lie subgroup. Consider the map η : A × N → AN

(a, n) 7→ an

Since A consists of semisimple transformations and N consists of nilpotent transformations, A ∩ N = {1}. Hence η is a diffeomorphism. Now by the above lemma it follows that G = KAN . To finish the proof of the theorem, it suffices to show that K ∩ AN = {1}. But this follows from considering eigenvalues. 

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15. Appendix Lemma 15.1. Theorem 15.2. Let G be a Lie group. Then π2 (G) is trivial. Proof. As π2 only sees the identity component of G, it we may assume that G is connected. ˜ be the simply connected Lie group covering G. Then πi (G) ˜ = πi (G) for all i > 1, in Let G particular, for i = 2. As such it suffices to assume that G is simply connected. In this case, we have previously shown in the proof of theorem 12.36 that G = R o S where R is solvable and S is semisimple. Observe that both R and S are simply connected. A simply connected solvable Lie group is contractible. Hence G is diffeomorphic to R × S which is homotopy equivalent to S. So π2 (G) = π2 (S), and without loss of generality, we may assume that G is semisimple. Let T ⊂ G a maximal torus and let Φ be the set of roots of G with respect to T . Then a fixed root α ∈ Φ gives a character T → S 1 which we may lift as follows. α ˜ Rn R

? α - ? T S1 By the commutivity of the above diagram, it follows that α ˜ (Zn ) ⊂ Z. Define the map

ϕ : G × T sing → Gsing

(g, t) 7→ gtg −1

It is clear that the this map is smooth and surjective. Hence for any x ∈ Gsing dim(G) + dim(T sing ) ≥ dim(ϕ−1 (x)) + dim(Gsing ) so codim(Gsing ) ≥ dim(ϕ−1 (x)) − dim(T sing ) Fix t ∈ T and α ∈ Φ such that α(t) = 1. Define Tα := {t ∈ T | α(t) = 1}◦ As this is closed and connected, it is a torus. Let Gα := ZG (Tα ). As commuting is a closed condition, Gα is a closed subgroup. As o Xn β g := L(G) = L(T ) + (gC + g−β ) ∩ g C β∈Φ

it follows that

n o −α α L(Gα ) = L(T ) + (gC + gC ) ∩ g

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And as such dim(Gα ) = dimR (L(Gα )) = dim T + 2  Exercise 15.3. G compact. Then (G, G)- is compact. Exercise 15.4. G compact. T ⊆ G maximal torus. Then G/T is simply connected.