Representation Theory Of Finite Dimensional Lie Algebras

1.3. NILPOTENT AND SOLVABLE LIE ALGEBRAS 13 Proposition 1.3.5. Let V 6 0 be a finite dimensional k-vector space, and g gl(V) a Lie subalgebra, such that g contains only nilpotent elements of Endk (V). Then: 1. There exists v V, v 6 0, such that gv {0}. 2. There is a chain of k-vector spaces {0} V1 · · · Vd V where dim(Vi ) i, such that g(Vi ) Vi 1 . 0 3. There is a basis of V such that g { . . . }, where elements of 0 0 g are viewed as matrices in this basis. (In particular g is nilpotent). Proof. We know that if x g is nilpotent, then ad(x) Endk (gl(V)) is nilpotent because n X n n ad(x) (y) ( 1)n j [xj , y]xn j j j 0 1. We proceed by induction on dim(g). If dim(g) 0, it’s clear. Choose l g a maximal Lie subalgebra. Consider ad : g g, ad : g/l g/l. Since ad(l) consists of nilpotent elements, there exists x 6 0 g/l such that ad(l)(x) 0, by induction. Hence, there exists x 6 0 g l such that [l, x] l, thus k · x l is a Lie subalgebra of g, hence is equal to g by maximality of l. Now consider W {w V : l · v 0}. By induction hypothesis, W 6 {0}. Claim. xW W because if y l, v W, yxv xyv [y, x]v 0. Since x is nilpotent, there exists v W such that xv 0, hence g·v 0 because g k · x l. 2. Consider the inclusion p : g gl(V). The image consists of nilpotent endomorphisms. Claim. There exists a flag V0 V1 · · · Vd V, where dim(Vi ) i, such that p(g)(Vi ) Vi 1 .

14 CHAPTER 1. INTRODUCTION TO LIE ALGEBRAS Induction on dim(V): If dim(V) 1, this is clear. Now let dim(V) 1. There exists v V, v 6 0 such that p(g)v 0. Let V1 : k · v, V0 V/V1 , consider can : V V/V1 . Let Vi : can 1 (Vi 1 ) and proceed by induction. 1.4 g-modules, basic definitions Theorem 1.4.1. (Engel) Let g be a finite dimensionalLie algebra. Then, g is nilpotent if and only if all its elements are ad-nilpotent. Definition 1.4.2. A representation of a k-Lie algebra g is a k-vector space V with a Lie algebra homomorphism ρ : g gl(V). A representation is called irreducible or simple if the only sub vector spaces W of V such that ρ(W) V are {0} or V. Remark 1.4.3. A representation ρ : g gl(V) is also called a g-module: Alternativel, a g-module is a k-linear map g V V (x, v) 7 x · v such that x · (y · v) y · (x · v) [x, y] · v. Remark 1.4.4. The representation is called indecomposable if there are no sub vector spaces W, W0 such that V W W0 , ρ(g). Clear: For g-modules, irreducible implies indecomposable, however the other implication doesn’t hold! Theorem 1.4.5. (Lie, abstract form) Every irreducible representation of a finite dimensional complex solvable Lie algebra is 1-dimensional. Theorem 1.4.6. (Lie, concrete form) Let g gl(V) be a solvable linear Lie algebra, dim(V) , V complex vector space. Then, there exists v V, v 6 0, such that g · v Cv(i.e. v is a common eigen vector for all the elements in g). Remark 1.4.7. This is wrong for a general field k. For example, sl(2, k) for char(k) 2 is solvable, the 2-dimensional standard representation/vector representation sl(2, k) gl(k 2 ) is irreducible but not one dimensional.

1.4. G-MODULES, BASIC DEFINITIONS 15 Proof of Engel. The first implication holds by Proposition 1.3.4. If, for all x g, ad(x) is nilpotent, then, again by Proposition 1.3.4, g is nilpotent because ad(g) g/Z(g) by isomorphism theorem. �—————————————— Lecture 3, 11/04/2011. Theorem 1.4.8. (Lie) Let g be a finite dimensional sub Lie algebra of gl(V) for V some finite dimensional vector space. Then there exists v V, v 6 0 such that g · v Cv. Definition 1.4.9. Let V be a k-vector space and and g gl(V) a Lie subalgebra. A linear map λ : g k is called a weight for g if Vλ : {v V : xv λ(x)v x g} 6 {0} and then Vλ is called a weight space. Lemma 1.4.10. (Invariance Lemma) If char(k) 0, (e.g k C as in the Theorem) and g gl(V) is a Lie subalgebra, where V is a finite dimensional vector space, ICg, λ : I k a weight for I. Then Vλ is g-stable, i.e. gVλ Vλ (i.e. xv Vλ x g, v Vλ ). (Proof: Excercise) Proof of Lie’s Theorem by induction on dim(g). dim(g) 1 Clear! dim(g) 1 If g is solvable, then [g, g] g. Choose a vector subspace U od codimension 1 in g, so g U Cz for some z g (as vector space). So, U is an ideal of g. Since Lie subalgebras and ideals inherit solvability we get that if U is solvable, then by induction hypothesis, there exists w V such that Uw Cw. Let λ : U C be the corresponding weight. By the Invariance Lemma (1.4.10), Vλ 6 0 is g-stable, in particular z-stable, i.e. zVλ Vλ . Hence, there exists an eigenvector 0 6 v Vλ for z. Let µ be the corresponding eigenvalue. Then v is the vector we are looking for: x g U Cz, x u βz, u U, β C. Then, xv (u βz)v λ(u)v βµv (λ(u) βµ)v

16 CHAPTER 1. INTRODUCTION TO LIE ALGEBRAS Remark 1.4.11. One could weaken the assumption for the field to be C by passing to a subfield which contains at least all of the occurring eigenvalues. However, char(k) 0 is required. Remark 1.4.12. Let ϕ : g1 g2 a Lie algebra morphism. Then g is solvable if and only if ker(ϕ) and im(ϕ) are both solvable. In particular, for any Lie algebra g: a. I, J C g solvable ideals, then I J is again a solvable ideal because I J/J I/I J and it follows since then I J/J and J are solvable. In particular, there is always a unique maximal proper solvable ideal. b. Let ICg. Take a 2-dimensional Lie algebra with basis x, y and [x, y] x solvable, not nilpotent. Then I : span(x) is a 1-dimensional ideal (in particular nilpotent), and the quotient g/I is also 1-dimensional nilpotent but g is not. Corollary 1.4.13. A Lie algebra g is solvable if and only if [g, g] is nilpotent. Proof. The quotient g/[g, g] is abelian, hence solvable, and [g, g] is nilpotent, hence solvable. Thus g is solvable. If g is solvable, then ad(g) gl(g) is solvable. By Lie’s Theorem, ··· . . . . ad(g) . . . 0 ··· After some good choice of basis in g, hence 0 · · · . . . . [ad(g), ad(g)] . . . 0 ··· 0 and therefore a nilpotent Lie subalgebra. Now, ker(ad [g,g] ) Z([g, g]), hence, we have a short exact sequence 0 ker(ad) [g, g] ad([g, g]) 0

1.5. TESTING SOLVABILITY AND SEMISIMPLICITY 17 Since ker(ad) is an ideal contained in Z(g), by Proposition 1.3.4, [g, g] is nilpotent. Definition 1.4.14. Let g be a k-Lie algebra. The maximal solvable ideal (which exists by the previous Remark) is called the radical of g and denoted rad(g). The Lie algebra g is called semi-simple if rad(g) {0}. 1.5 Testing solvability and semisimplicity Aim. ”Explicit” criterion for solvability/ semi-simplicity. Motivation: V finite dimensional complex vector space, g gl(V) a solvable Lie subalgebra. Then, · · · . g . (Lie’s Theorem) 0 So: tr(xy) 0 x g, y [g, g]. We develop traces to find the criterion we want. Definition 1.5.1. Let g be a finite dimensional k-Lie algebra. The Killing form is the bilinear form K(x, y) : Tr(ad(x) ad(y)) Properties of Killing form: symmetric K(x, y) K(y, x) (clear!) (1.3) invariance K([x, y], z) K(x, [y, z])(”associativity”) (1.4) To see this, note that for matrices X, Y, Z we have: [X, Y]Z XYZ Y(XZ) X[Y, Z] XYZ (XZ)Y In particular:

18 CHAPTER 1. INTRODUCTION TO LIE ALGEBRAS a. rad(g) {x g : K(x, y) 0 y g} C g (use invariance) b. I C g, I : {x g : K(x, y) 0 y I} is again an ideal (use invariance). Remark 1.5.2. The condition 1.4 is called ”invariance” because K corresponds to an element K (g g) ( dual vector space of g g) which is invariant under the natural action of g. Here invariant means sent to zero under the action of g. This makes sense by the following: Let G be an affine algebraic group, and V a representation. Then it is also a representation of g Lie(G). Now, v V is said to be G-invariant in the sense that gv v for all g G g-invariant in the sense that xv 0 for all x g. Theorem 1.5.3 (Cartan’s irreducibility criterion). Let k be a field, char(k) 0. Let g be a finite dimensional k-Lie algebra. g solvable K(X, Y) 0 x g.y [g, g] Proof. First we need a Lemma: Lemma 1.5.4. (Cartan criterion gl(V)) Let V be a complex vector space, finite dimensional, and g gl(V) Lie subalgebra. Then g solvable Tr(xy) 0 x g, y [g, g]. Assuming the lemma, we can deduce the theorem (k C): K(x, y) 0 x g, y [g, g] Tr(ad(x) ad(y)) ad(g) solvable. Now, we have a short exact sequence of Lie algebras: 0 Z(g) g ad g 0 So, g is solvable by the above. Theorem 1.5.5 (Characterization of semi-simple Lie algebras). For g a finite dimensional, complex Lie algebra. Then the following are equivalent: 1 g is isomorphic to a direct sum of simple Lie algebras.

1.5. TESTING SOLVABILITY AND SEMISIMPLICITY 19 2 g is a direct sum of its simple ideals (viewed as Lie algebras.) 3 The Killing form K is non-degenerate. 4 g has no non-zero abelian ideals. 5 g has no non-zero solvable ideals. �—————————————— Lecture 4, 13/04/2011. Proof of Theorem 1.5.5. First note that the Cartan criterion says that if Kg 0, then g is solvable. 4 5 Every abelian ideal is solvable. On the other hand, if I C g is solvable, then the last non-zero step in the derived series is an abelian ideal. Claim. Let g be a finite dimensional Lie algebra (over arbitrary k), and I C g an ideal. Then KI Kg I I where KI , Kg are the killing forms of I, g respectively. Proof. Assume U g is a vector subspace, and ϕ : g g is a linear map. Then Tr(ϕ) Tr(ϕ U ). Now take U I and ϕ ad(x) ad(y) for some x, y I. Then ad(x) ad(y)(z) I. 5 3 First, rad(K) {x g : K(x, y) 0 x g}. Now, since K is invariant, then rad(K) C g(x rad(K), z g K([x, z], y) K([x, [z, y]] 0)), hence by the Claim, Krad(K) Krad(K) rad(K) 0. By Cartan’s criterion, rad(K) is solvable. Since by assumption there are no nonzero solvable ideals, then rad(K) 0, so K is non-degenerate. 3 4 Assume {0} 6 I C g is an abelian ideal, then (ad(x) ad(y))2 0 x g, y I, i.e. ad(y) ad(x) ad(y)(z) [y, [x, [y, z]]] [I, I] 0. This means ad(x) ad(y) is nilpotent, so Tr(ad(x) ad(y)) 0 for x g, y I. Hence K is degenerate, a contradiction! 2 1 4 Clear! 5 2 Let I C g. Then I : {x g : K(x, y) 0 y I} is an ideal of g.(x I , z g, K([x, z], y) K(x, [z, y]) 0) By Cartan’s criterion, I I is a solvable ideal, hence zero. On the other hand, K is non-degenerate (because 5 3), so g I I , and hence every

20 CHAPTER 1. INTRODUCTION TO LIE ALGEBRAS ideal of I or I is an ideal of g. By induction on dimension, I, I satisfy 2. Hence g is a direct sum of simple ideals. It is left to show that every simple ideal occurs: Assume that J g is a simple ideal. Then, r M [J, Ii ] J [J, J] [J, g] i 1 Lr where g i 1 Ii is our decomposition into a sum of simple ideals. This means that, since the Ii are simple, and J is also simple, J [J, Ii ] Ii for some i. Hence every simple ideal occurs in the decomposition. Example 1.5.6. The Killing form for sl(2, k) is given by the following ma 0 0 4 0 1 1 0 0 0 trix 0 8 0 , in the basis e ,h ,f . 0 0 0 1 1 0 4 0 0 So, sl(2, k) is semisimple if and only if char(k) 6 2. Remark 1.5.7. One can show that, for sl(n, C), K(x, y) 2n tr(xy) and for gl(n, C): K(a, b) 2n tr(ab) 2 tr(a) tr(b) 1.6 Jordan Decomposition and Proof of Cartan’s Criterion Lemma 1.6.1. Let V be a finite dimensional C-vector space, and x : V V a C-linear endomorphism. Then there exist unique xs , xn C-linear endomorphisms such that xs is diagonalizable and xn is nilpotent, x xs xn and xs xn xn xs . Moreover, every subspace of V stabilized by x is stabilized by xs and xn .

1.6. JORDAN DECOMPOSITION ANDPROOF OF CARTAN’S CRITERION21 Example 1.6.2. i. λ 1 λ 0 0 λ 0 λ 0 1 0 0 x xs xn ii. 1 2 1 0 0 3 0 3 0 2 0 0 x xs xn The latter is not a Jordan decomposition because xn and xs don’t commute. Q Proof. Let a1 , · · · , ak be the distinct eigenvalues of x, so x(t) ki 1 (t ai )mi for some mi N. Let V(x, ai ) : ker(x ai )mi be the generalized eigenspace of x corresponding to ai . Then, V k M V(x, ai ). (1.5) i 1 Hence, by the Chinese remainder Theorem, there exists P(t) C[t] a polynomial satisfying: P(t) ai (mod(t ai )mi ); 1 i r P(t) 0 mod(t) Define xs : P(x) xn : Q(x) where Q(t) t P(t) C[t]. Now, xs and xn commute: in fact they commute with whatever x commutes with. By construction, xs ai V(x,ai ) 0, hence xs is diagonalizable with basis given by the decomposition 1.5 as above, so xn x xs must be nilpotent. This proves existence.

22 CHAPTER 1. INTRODUCTION TO LIE ALGEBRAS To prove uniqueness, assume there exist two such decompositions of x, namely x xs xn and x x0s xn0 . We then have: xs x0s x0n xn (1.6) The right hand side of (1.6) is nilpotent and the left hand side is diagonalizable, hence both sides must be zero, and uniqueness is proven. 1.6.1 Properties 1) Functoriality: Assume the following diagram commutes in the category of finite dimensional C-vector spaces. V f - W y x ? V ? f W Then the two following diagrams commute: V f - xs /xn W ys /yn ? V ? f W Proof. f x y f implies that f (V(x, λ)) V(y, λ). Then go through the definition of xs , xn . 2) Im(xs ) im(x). (This holds since xs P(x), and P(0) 0. This means xs (v) x r(x)(v) for some r(t) C[t]). Lemma 1.6.3. If x xs xn is the Jordan decomposition of x, then ad(xs ) ad(xn ) is the Jordan decomposition of ad(x). Remark 1.6.4. This lemma is crucial for defining the ”abstract” Jordan decomposition in a complex finite dimensional Lie algebra g.

1.7. THEOREMS OF LEVI AND MALCEV 23 Proof. We have [ad(xs ), ad(xn )] ad([xn , xs ]) 0, so ad(xs ), ad(xn ) commute. Further, ad(xn ) is nilpotent because xn is nilpotent. To see that ad(xn ) is diagonalizable, choose an eigenbasis v1 , · · · , vk of V for xs , with eigenvalues xs (vi ) λi vi for all vi Vi . Then Eij is an eigenvector for ad(xs ) with eigenvalue λi λj . (Notation: Eij is the ij-matrix unit , i.e. (Eij )kl δik δjl ) Lemma 1.6.5. Let V be a finite dimensional C-vector space. Let A B End(V) be vector subspaces, and let T : {z End(V) : ad(z)(B) A}. Then, if Tr(xz) 0 for all z T, then x is nilpotent. (Proof later!) Proof of Cartan’s Criterion (Lemma 1.5.4). Assume Tr(xy) 0 for all y g gl(V), x [g, g]. It is enough to show that [g, g] is nilpotent. We use Lemma 1.6.5 in the following way: Take x [g, g]. We show Pn that ( T r)(xy) 0 for all y gl(V) such that [y, g] [g, g]. Write x i 1 [xi , yi ]; xi , yi g. Take y gl(V) as above. Then: Tr(xy) n X Tr([xi , yi ]y) i 1 1.7 n X Tr(xi [yi , y]) 0 (by hypothesis.) i 1 Theorems of Levi and Malcev Aim. Let g be a finite dimensional Lie algebra. Then rad(g) l g as vector spaces for some semisimple Lie algebra l, and g rad(g) n l as Lie algebras. Then l is called the Levi compliment of g. It is unique up to some inner automorphism. Definition 1.7.1. Let g be a k-Lie algebra. We define Der(g) : {Derivations of g} {δ Endk (g) : g g : δ([a, b]) [δ(a), b] [a, δ(b)]} It is a Lie subalgebra of Endk (V). Example 1.7.2. Let x g. Then ad(x) : g g is a derivation. (Jacobi identity!)

24 CHAPTER 1. INTRODUCTION TO LIE ALGEBRAS Remark 1.7.3. ad(g) : {ad(x) : x g} C Der(g) Definition 1.7.4. Assume x g such that ad(x) is nilpotent, and char(k) 0. Then exp(ad(x)) : X (ad x)k k 0 k! Aut(g) : {f Endk (g); f invertible } An automorphism f Aut(g) is called inner, if it is contained in the subgroup generated by the exp(ad x), x g as above. Proposition 1.7.5. Let g be a finite dimensional semi-simple Lie algebra over C. Then Der(g) ad(g) Proof. Since g is semi-simple, Z(g) 0. Thus g ad(g) Z(g) Lie algebras. Let I C Der(g) an ideal of Lie algebras. Consider form KI KI I . In particular, I I 0 so [I, I ] 0. Now, hence we have for x, y g: ad(g) as the Killing let δ I , 0 [δ, ad(x)](y) δ([x, y]) [x, δ(y)] [δx, y] Hence ad(δ(x)) 0 for all x g, hence δ 0. This means there are no proper ideals, hence ad(g) Der(g) as Lie algebras. Definition 1.7.6. Let g be a finite dimensional k-Lie algebra. If g I l as vector spaces, where I is an ideal and l is a Lie subalgebra, then g is called the semi direct product of I and l, and is denoted by g Inl If such a decomposition exists, then α :l Der(I) x 7 ad(x) I (check!)

1.7. THEOREMS OF LEVI AND MALCEV 25 Conversely, assume that we are given α : l Der(h) a Lie algebra homomorphism for some k-Lie algebras h, l. Then g : h l becomes a Lie algebra, via: [(x, y), (x0 , y 0 )] (α(y)(x0 ) α(y 0 )x [x, x0 ], [y, y 0 ]) For α 0 this is just the direct sum of Lie algebras. Lemma 1.7.7. Let g be a finite dimensional Lie algebra. Then g/ rad(g) is semi-simple. Proof. We show that rad(g/ rad(g)) is zero. Let I C g/ rad(g) be a solvable ideal. Consider can :g g/ rad(g). Set J : can 1 (I) C g, an ideal containing rad(g). We have the following short exact sequence: 0 - rad(g) - J - J/ rad(g) - 0 Hence, J must be solvable, hence J rad(g), so J rad(g), so there is no non-trivial solvable ideal. Theorem of Levi 1.7.8. Let g be a finite dimensional Lie algebra. If α : g l is a Lie algebra homomorphism, surjective, and l is semi-simple, then there exists β : l g, a Lie algebra homomorphism such that α β Id l . Corollary 1.7.9. Any finite dimensional Lie algebra g is a semi-direct product of it’s radical and a semi-simple Lie algebra; more precisely, g rad(g) n g/ rad(g) where can α : g g/ rad(g). Proof of Corollary 1.7.9. The map can : g g/ rad(g) is a surjective Lie algebra homomorphism. Therefore there exists a Lie algebra homomorphism β : g/ rad(g) g as in Levi’s Theorem, hence β(l) g is a Lie subalgebra, and rad(g) β(l) g as vector spaces.

26 CHAPTER 1. INTRODUCTION TO LIE ALGEBRAS Theorem of Malcev 1.7.10. For two Levi complements l and l0 of g, there exists some x [g, rad(g)], ad(x) nilpoten

The Lie algebra g 1 g 2 is called the direct sum of g 1 and g 2. De nition 1.1.2. Given g 1;g 2 k-Lie algebras, a morphism f : g 1!g 2 of k-Lie algebras is a k-linear map such that f([x;y]) [f(x);f(y)]. Remarks. id: g !g is a Lie algebra homomorphism. f: g 1!g 2;g: g 2!g 3 Lie algebra homomorphisms, then g f: g 1! g 2 is a Lie algebra .