LECTURE NOTES TO HUMPHREYS INTRODUCTION TO LIE ALGEBRAS AND REPRESENTATION THEORY

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1 LECTURE NOTES TO HUMPHREYS INTRODUCTION TO LIE ALGEBRAS AND REPRESENTATION THEORY ZHENGYAO WU Abstract. Reference: [Hum78]. Since our course is called Finite Dimensional Algebra, we assume that a vector space has finite dimension unless otherwise specified. Contents 1. February 8th Introduction to Lie algebra :00-8:45 [Hum78, 1] :55-9:40 [Hum78,.1,.] :00-10:45 [Hum78,.3, 3.1] 10. March 7th Solvable and nilpotent Lie algebras :00-8:45 [Hum78, 3.] :55-9:40 [Hum78, 3.3] :00-10:45 [Hum78, 4.1] March 14th Lie theorem, Jordan decomposition, Cartan s criterion and Killing form :00-8:45 [Hum78, 4.1] :55-9:40 [Hum78, 4.] :50-10:35 [Hum78, 4.3] :45-11:30 [Hum78, 5.1] 8 4. March 1st Semisimple decomposition, Lie modules, Casimir element and Weyl s theorem :00-8:45 [Hum78, 5., 5.3] :55-9:40 [Hum78, 6.1] :50-10:35 [Hum78, 6.] :45-11:30 [Hum78, 6.3] March 8th Jodan decomposition, sl(, F )-modules, maximal toral subalgebras and its centralizer :00-8:45 [Hum78, 6.4] :55-9:40 [Hum78, ] :50-10:35 [Hum78, 8.1] :45-11:30 [Hum78, ] April 11th Root systems and examples, pairs of roots :00-8:45 [Hum78, 8.4, 8.5] :55-9:40 [Hum78, 9.1] :50-10:35 [Hum78, 9.] :45-11:30 [Hum78, 9.3] April 18th Simple roots and the Weyl group 63 1

2 ZHENGYAO WU :00-8:45 [Hum78, 10.1] :55-9:40 [Hum78, 10.] :50-10:35 [Hum78, 10.3] :45-11:30 [Hum78, 10.3] April 5th Irreducible root systems, two root lengths and Cartan matrix :00-8:45 [Hum78, 10.3] :55-9:40 [Hum78, 10.4] :50-10:35 [Hum78, 10.4] :45-11:30 [Hum78, 11.1] May nd Coxeter graphs, Dynkin diagrams and types A, B :00-8:45 [Hum78, 11.] :55-9:40 [Hum78, 11.3] :50-10:35 [Hum78, 11.4] :45-11:30 [Hum78, 1.1] May 9th Types C,D,E,F,G, automorphisms and weights :00-8:45 [Hum78, 1.1] :55-9:40 [Hum78, 1.1] :50-10:35 [Hum78, 1.] :45-11:30 [Hum78, 13.1] May 16th Saturated set, universal enveloping algebra and Poincaré- Birkhoff-Witt theorem :00-8:45 [Hum78, 13., 13.3] :55-9:40 [Hum78, 13.4, 17.1] :50-10:35 [Hum78, 17.] :45-11:30 [Hum78, 17.4] May 3th Standard cyclic modules and condictions for finite dimension :00-8:45 [Hum78, 17.3] :55-9:40 [Hum78, 0.1, 0.] :50-10:35 [Hum78, 0.3, 1.1] :45-11:30 [Hum78, 1.] June 6th Applications of Freudenthal s formula June 13th Applications of Kostant s formula and Steinberg s formula 15 References 133

3 LECTURE NOTES 3 1. February 8th Introduction to Lie algebra :00-8:45 [Hum78, 1]. Sophus Lie ( ) established the theory of Lie groups and Lie algebras in late 1880s in Oslo, Norway Definition Let F be a field. Let L be a vector space over F. We say that L is a Lie algebra if there exists a map L L L, (x, y) [x, y] such that (L1) [x, y] is bilinear over F. (L) [x, x] = 0 for all x L. (L3) Jacobi identity [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 for all x, y, z L Lemma If char(f ), then (L ) [x, y] = [y, x] for all x, y L. Proof. By Definition 1.1.1(L1), By Definition 1.1.1(L), Hence (L ). [x + y, x + y] = [x, x] + [x, y] + [y, x] + [y, y]. 0 = 0 + [x, y] + [y, x] Definition Let L, L be two Lie algebras. An isomorphism of vector spaces f : L L is an isomorphism (of Lie algebras) if f([x, y]) = [f(x), f(y)] Example Let V be a vector space over F. Let End F (V ) be the space of endomorphisms (linear transformations) of V. For all x, y End F (V ), define [x, y] = x y y x. Then gl(v ) = (End F (V ), [, ]) is a Lie algebra, called the general linear algebra. Proof. (L1) Since all endomorphisms are linear over F, [x, y] is bilinear. (L) [x, x] = x x x x = 0. (L3) [x, [y, z]]+[y, [z, x]]+[z, [x, y]] = x (y z z y) (y z z y) x+y (z x x z) (z x x z) y+z (x y y x) (x y y x) z = x y z x z y y z x+z y x+y z x y x z z x y+x z y+z x y z y x x y z+y x z = Notation Let H, K be subsets of L. We write [H, K] = Span{[x, y], x H, y K}. By Definition 1.1.1, [H, K] = [K, H]. We write [x, K] = [{x}, K], for all x L Definition A vector subspace K of L is a subalgebra if [K, K] K. By Definition 1.1.1, K is a Lie algebra Example Subalgebras of gl(v ) are called Linear Lie algebras. If dim F (V ) = n, then we write gl(n, F ) = gl(v ). which is the Lie algebra of all n n matrices such that [e ij, e kl ] = δ jk e il δ li e kj where e ij is the n n matrix whose (i, j)-entry is 1 and all other entries are 0; δ ij = 1 if i = j, otherwise δ ij = 0.

4 4 ZHENGYAO WU Definition Classical algebras are the following proper Lie subalgebras of gl(v ) of type A l, B l, C l, D l, l Example Type A l (l 1): special unitary algebras sl(v ) = {x gl(v ) Tr(x) = 0}, dim F (V ) = l + 1 sl(l + 1, F ) = {x gl(l + 1, F ) Tr(x) = 0}. A l has standard basis {e ii e i+1,i+1 1 i l} {e ij 1 i j l + 1}. dim(a l ) = l + l Example Type B l (l 1): orthogonal algebras of odd degree Let f be a nondegenerate symmertic bilinear form on V whose matrix is s. o(v ) = {x gl F (V ) f(x(v), w) = f(v, x(w)), v, w V }, dim F (V ) = l + 1 o(l + 1, F ) = {x gl(l + 1, F ) sx = x t s}, s = I l. 0 I l 0 o(l + 1, F ) = 0 b 1 b b t m n gl(l + 1, F ) n t = n, p t = p b t 1 p m t. B l has standard basis {e i+1,i+1 e l+i+1,l+i+1 1 i l} {e 1,l+i+1 e i+1,1 1 i l} {e 1,i+1 e l+i+1,1 1 i l} {e i+1,j+1 e l+j+1,l+i+1 1 i j l} {e i+1,l+j+1 e j+1,l+i+1 1 i < j l} {e l+i+1,j+1 e l+j+1,i+1 1 j < i l}. dim(b l ) = l + l Example Type C l (l 1): symplectic algebras Let f be a nondegenerate skew-symmertic form on V whose matrix is s. sp(v ) = {x gl F (V ) f(x(v), w) = f(v, x(w)), ( v, w) V }, dim F (V ) = l. 0 sp(l, F ) = {x gl(l, F ) sx = x t Il s}, s =. I l 0 {( ) } m n sp(l, F ) = p m t gl(l, F ) n t = n, p t = p. C l has standard basis {e i,i e l+i,l+i 1 i l} {e ij e l+j,l+i 1 i j l} {e i,l+i 1 i l} {e i,l+j + e j,l+i 1 i < j l} {e l+i,i 1 i l} {e l+i,j + e l+j,i 1 i < j l}. dim(c l ) = l + l Example Type D l (l 1): orthogonal algebras of even degree: Let f be a nondegenerate symmertic bilinear form on V whose matrix is s. o(v ) = {x gl(v ) f(x(v), w) = f(v, x(w)), v, w V }, dim F (V ) = l.

5 LECTURE NOTES 5 ( ) 0 o(l, F ) = {x gl(l, F ) sx = x t Il s}, s =. {( ) I l 0 } m n o(l, F ) = p m t gl(l, F ) n t = n, p t = p. D l has standard basis {e ii e l+i,l+i 1 i l} {e ij e l+j,l+i 1 i j l} {e i,l+j e j,l+i 1 i < j l} {e l+i,j e l+j,i 1 j < i l}. dim(d l ) = l l Theorem Ado-Iwasawa: Every Lie algebra is isomorphic to a linear Lie algebra. Proof. Omit Example Let t(n, F ) be the set of upper triangular matrices. Let n(n, F ) be the set of strictly upper triangular matrices. Let d(n, F ) be the set of diagonal triangular matrices. They are linear Lie algebras Lemma (1) t(n, F ) = d(n, F ) + n(n, F ) as vector space direct sum. () [d(n, F ), n(n, F )] = n(n, F ). (3) [t(n, F ), t(n, F )] = n(n, F ) Proof. See Exercise Definition A derivation of an F -algebra A is a linear map δ : A A such that δ(ab) = aδ(b) + δ(a)b for all a, b A. Let Der(A) be the set of all derivations of A Lemma Der(A) gl(a). Proof. See Exercise Definition Let L be a Lie algebra. Let ad : L Der(L) be the adjoint representation of L such that ad(x)(y) = [x, y] for all x, y L Lemma ad(l) Der(L). Proof. We need to show that ad(x)([y, z]) = [ad(x)(y), z] + [y, ad(x)(z)]. It follows from the Definition 1.1.1(L)(L3) that [x, [y, z]] = [z, [x, y]] [y, [z, x]] = [[x, y], z] + [y, [x, z]] Definition Any derivation of the form ad(x) is called an inner, all other derivations are called outer.

6 6 ZHENGYAO WU Example Let ad L : L gl(l) be the map such that ad L (x)(y) = [x, y] for all x, y L. Let K be a Lie subalgebra of L. Then ad K (x) and ad L (x) are different in general. For x d(n, F ) gl(n, F ). ad d(n,f ) (x) = 0 and ad gl(n,f ) (x) Definition A Lie algebra L is abelian if [x, y] = 0 for all x, y L. Proof. (a) follows from Definition 1.1.1(L). (a) follows from Definition 1.1.1(L3) Example Let L be a finite dimensional vector space over F. Then with [x, y] = 0 for all x, y L, L is an abelian Lie algebra Example Every Lie algebra L of dimension 1 is abelian, since Definition 1.1.1(L) Example A two dimensional Lie algebras is either abelian or isomorphic to F x + F y such that [x, y] = x. Proof. Suppose L is not abelian and L = F a+f b. Then [L, L] = F x, for x = [a, b]. Take z L\F x. Then [x, z] = cx for some c F. Let y = c 1 z. Then L = F x+f y and [x, y] = x Definition (1) Let L be a Lie algebra over F with basis x 1, x,..., x n such that [x i, x j ] = n a k ij x k. {a k ij F 1 i, j, k n} are called structure constants of L. k= Lemma (a) a k ii = 0 = ak ij + ak ji for all 1 i, j, k n. n (b) (a k ij am kl + ak jl am ki + ak li am kj ) = 0 for all 1 i, j, l, m n. k=1

7 LECTURE NOTES :55-9:40 [Hum78,.1,.] Definition A subspace I of a Lie algebra L is an ideal if [I, L] I. 1.. Example 0 and L are ideals of L Example The center Z(L) = {z L [z, L] = 0} of L is an ideal of L, since [Z(L), L] = 0 Z(L) Example The derived algebra [L, L] = Span{[x, y] x, y L} of L is an ideal of L, since [[L, L], L] [L, L] Lemma L is abelian iff [L, L] = Lemma If L is a classical algebra, then L = [L, L]. Proof. Exercise Lemma If I and J are ideals of L, then I + J = {x + y x I, y J} is an ideal of L. Proof. [I + J, L] = [I, L] + [J, L] I + J Lemma If I and J are ideals of L, then [I, J] = { [x i, y i ] x i I, y i J} is an ideal of L. Proof. By Jacobi identity, [[I, J], L] [[J, L], I] + [[L, I], J] [I, J] Definition We call L simple if. (1) L has no ideals except for 0 and L; () [L, L] Lemma If L is simple, then Z(L) = 0. Proof. By Example 1..3 and Definition 1..9(1), Z(L) = 0 or L. If Z(L) = L, then [L, L] = 0, a contradiction to Definition 1..9() Lemma If L is simple, then L = [L, L]. By Defini- Proof. By Example 1..4 and Definition 1..9(1), [L, L] = 0 or L. tion 1..9(), [L, L] 0, hence [L, L] = L Example If char(f ), then sl(, F ) is simple. Proof. A standard basis for L = sl(, F ) is ( ) ( ) x =, y =, h = ( ) Let I 0 be an ideal of L. We show that I = L. Suppose 0 ax + by + ch I. (1) If a 0, then [y, ax + by + ch] = ah + cy I. Then [y, ah + cy] = ay I and hence y I.

8 8 ZHENGYAO WU () If b 0, then [x, ax + by + ch] = bh cx I. Then [x, bh cx] = bx I and hence x I. (3) If a = b = 0, then c 0. Then ax + by + ch = ch I and hence h I. Since [x, y] = h, [h, x] = x, [h, y] = y and I contains one of {x, y, h}, then I = L. Hence L is simple Definition If I is an ideal of L, then the quotient space L/I with is called the quotient algebra. [x + I, y + I] = [x, y] + I Proof. Suppose x 1 + I = x + I and y 1 + I = y + I, we need to show that [x 1, y 1 ] + I = [x, y ] + I. In fact, x x 1 I, y y 1 I = [x, y ] [x 1, y 1 ] = [x x 1, y ] + [x 1, y y 1 ] I Definition The normalizer of a subspace K of L is N L (K) = {x L [x, K] K}. If K = N L (K), then K is called self-normalizing Lemma N L (K) is a subalgebra of L. Proof. Suppose [x, K] K and [y, K] K. By Jacobi identity [[x, y], K] [[y, K], x] + [[K, x], y] [K, x] + [K, y] K Lemma Let K be a subalgebra of L. Then N L (K) is the largest subalgebra of L with K as an ideal. Proof. If A is another subalgebra of L such that [K, A] K. By Definition 1..14, A N L (K) Definition The centralizer of a subset X of L is Lemma C L (X) is a subalgebra of L. C L (X) = {x L [x, X] = 0} Proof. Identify {0} and 0. Suppose [x, X] = 0 and [y, X] = 0. By Jacobi identity Lemma C L (L) = Z(L). [[x, y], X] [[y, X], x] + [[X, x], y] = [0, x] + [0, y] = 0

9 LECTURE NOTES 9 Proof. It follows from Example 1..3 and Definition Definition (1) A linear transformation φ : L L of Lie algebras over F is a homomorphism if φ([x, y]) = [φ(x), φ(y)] for all x, y L. () A homomorphism φ is a monomorphism if ker(φ) = 0. (3) A homomorphism φ is an epimorphism if im(φ) = L. (4) A homomorphism φ is an isomorphism if it is both a monomorphism and an epimorphism Lemma ker(φ) is an ideal of L. Proof. If φ(x) = 0, then φ([x, L]) = [φ(x), φ(l)] = [0, φ(l)] = 0. Hence [x, L] ker(φ). 1.. Lemma φ(l) is a subalgebra of L. Proof. It follows from Definition 1..0(1) Lemma There is a 1-1 correspondence between homomorphisms from L and ideals of L. Given a homomorphism φ, ker(φ) is an ideal of L; Given an ideal I of L, there exists a homomorphism L L/I, x x + I. Proof. Given I, we have ker(l L/I) = I. Conversely, given φ, f : L/ ker(φ) φ(l), f(x + ker(φ)) = φ(x) is an isomorphism. By definition, f is epimorphism. Also, f(x + ker(φ)) = 0 iff x ker(φ) iff x + ker(φ) = 0 L/ ker(φ) Proposition If I ker(φ) is an ideal of L, then there exists a unique homomorphism ψ : L/I L such that the following diagram commute L φ L π ψ L/I Proof. Define ψ(x + I) = φ(x). If x 1 + I = x + I, then x x 1 I ker(φ). Then ψ(x 1 + I) = φ(x 1 ) = φ(x ) = ψ(x + I) Proposition If I J are ideals of L, then J/I is an ideal of L/I and (L/I)/(J/I) L/J. Proof. Since [J/I, L/I] = [J, L]/I J/I, J/I is an ideal of L/I. Define φ : L/I L/J such that φ(x + I) = x + J. It is well-defined because I J. Also, ker(φ) = J/I and φ(l/i) = L/J. The claim follows from Lemma Proposition If I and J are ideals of L, then (I + J)/J I/(I J). Proof. Define φ : I (I + J)/J such that φ(x) = x + J. Also, ker(φ) = I J and φ(i) = I + J/J. The claim follows from Lemma 1..3.

10 10 ZHENGYAO WU :00-10:45 [Hum78,.3, 3.1] Definition A representation of L is a homomorphism φ : L gl(v ) for some vector space V over F. We call φ faithful if it is one-to-one. Here V is allowed to have arbitrary dimension Lemma The adjoint representation ad : L gl(l), x ad(x) = [x, ] is a linear representation. Proof. We need to show that [ad(x), ad(y)] = ad([x, y]). It follows from the Definition 1.1.1(L)(L3) that [x, [y, z]] [y, [x, z]] = [x, [y, z]] + [y, [z, x]] = [z, [x, y]] = [[x, y], z] Notation Let λ x be the left multiplication by x. Let ρ x be the right multiplication by x. Then ad(x) = λ x ρ x Lemma ker(ad) = Z(L). Proof. Since ad(x) = 0 iff [x, L] = Lemma If L is simple, then ad is a monomorphism. Any simple Lie algebra is isomorphic to a linear Lie algebra. Proof. Suppose L is simple. By Lemma 1..10, Z(L) = 0. By Lemma 1.3.4, ker(ad) = 0. Then L ad(l), which is linear by Lemma Definition An automorphism of L is an isomorphism L L. Let Aut(L) be the group of automorphisms of L Example If L gl(v ) and g GL(V ) such that glg 1 = L (e.g. L = gl(v ) or sl(v )), then Int(g) Aut(L). Here Int(g) : L L, x gxg Example Suppose char(f ) = 0 and δ Der(L) such that δ k = 0 for some k > 0. Then (1) exp(δ(x)) exp(δ(y)) = exp(δ(xy)). () exp(δ) Aut(L). Proof. (1) By the Leibniz rule, Then δ n (xy) = n i=0 δ n n n! (xy) = i=0 ( ) n δ i (x) δ n i (y) i ( δ i (x) i! ) ( δ n i ) (y) (n i)! exp(δ(x)) exp(δ(y)) = exp(δ(xy)).

11 LECTURE NOTES 11 () In fact, for exp(δ) = 1 η, its inverse is 1 + η + η + + η k 1, where η = k 1 δ i k 1 δ i 1 = δ and η k = 0. i! i! Example Suppose char(f ) = 0 and x L such that ad(x) k = 0 for some k > 0. Then exp(ad(x)) = 1 + ad(x) + (ad(x)) + + (ad(x))k 1 Aut(L).! (k 1)! Definition An automorphism of the form exp(ad(x)) is inner. The subgroup of Aut(L) generated by exp(ad(x)), x L is denoted by Int(L), its elements are called inner automorphisms Lemma Int(L) is a normal subgroup of Aut(L). Proof. If φ Aut(L) and x L, then (φ ad(x) φ 1 )(y) = φ([x, φ 1 (y)]) = [φ(x), y] = ad(φ(x))(y). Hence φ ad(x) φ 1 = ad(φ(x)). Therefore φ exp(ad(x)) φ 1 = exp(ad(φ(x))) Int(L) Example Let L = sl(, F ). Let σ = exp(ad(x)) exp(ad( y)) ( exp(ad(x)) ) Int(L). Then 0 1 σ = Int(s) where s = exp(x) exp( y) exp(x) =. 1 0 Proof. By ( Exercise ) 1.10, σ(x, ( y, h) = )( y, x, h) Since x =, exp(x) =. ( 0 0 ) ( 0 1 ) Since y =, exp( y) = ( ) ( ) ( ) ( ) Then s = = Also, s 1 = s. ( ) ( ) ( ) ( ) sxs = = = y, ( ) ( ) ( ) ( ) sys = = = x, ( ) ( ) ( ) ( ) shs = = = h Hence σ = Int(s) Lemma Suppose char(f ) = 0. If L gl(v ) is a Lie algebra, and x L is nilpotent, then for all y L. (exp(x))y(exp(x)) 1 = exp(ad(x))(y)

12 1 ZHENGYAO WU Proof. Since ad(x) = λ x ρ x, exp(ad(x)) = exp(λ x ρ x ) = exp(λ x ) exp(ρ x ) 1 = λ exp(x) ρ 1 exp(x) Definition The derived series of ideals of a Lie algebra L is L (0) = L, L (1) = [L, L],, L (i) = [L (i 1), L (i 1) ], We call L solvable if L (n) = 0 for some n Example Abelian Lie algebras are solvable. Proof. If L is abelian, then L (1) = Example Simple Lie algebras are nonsolvable. Proof. If L is simple, then L = L (1) = L () = 0. Hence L is not solvable Remark Define j i = l(e ij ). Suppose i < j, a < b, i b, we have [e ij, e ab ] = δ ja e ib. If j = a, then l(e ib ) = b i = (b a) + (j i) = l(e ab ) + l(e ij ) min{l(e ab ), l(e ij )} Example t(n, F ) is solvable. Proof. Let L = t(n, F ). By Exercise 1.5, L = n(n, F ) d(n, F ) and [d(n, F ), n(n, F )] n(n, F ). Then L (1) = n(n, F ). Suppose i < j, a < b, i b. We have [e ij, e ab ] = δ ja e ib. If j = a, by Remark , l(e ib ) = l(e ab ) + l(e ij ) min{l(e ab ), l(e ij )}. Since L (1) is spanned by e ij such that l(e ij ) 1, L (k) is spanned by e ij such that l(e ij ) k 1. Hence L (m) = 0 whenever m 1 > n 1 because n 1 is the largest possible value for l(e ij ) Proposition If L is solvable, then its subalgebras are solvable. Proof. Let K be a subalgebra of L. If K (n 1) L (n 1), then K (n) = [K (n 1), K (n 1) ] [L (n 1), L (n 1) ] = L (n) Hence K (n) L (n) for all n. If L (n) = 0, then K (n) = Proposition If L is solvable, then its homomorphic images are solvable. Proof. Let φ : L M be an epimorphism. If φ(l (n 1) ) = M (n 1), then φ(l (n) ) = φ([l (n 1), L (n 1) ]) = [φ(l (n 1) ), φ(l (n 1) )]) = [M (n 1), M (n 1) ] = M (n) Hence φ(l (n) ) = M (n) for all n. If L (n) = 0, then M (n) = Proposition Let I be an ideal of L. If I and L/I are solvable, then L is solvable.

13 LECTURE NOTES 13 Proof. If (L/I) (n) = 0, then L (n) I. Since an ideal I of L is also a subalgebra of L, by Proposition , L (n) is solvable with derived series L (n) L (n+1) L (n+). Hence L is solvable.

14 14 ZHENGYAO WU. March 7th Solvable and nilpotent Lie algebras.1. 8:00-8:45 [Hum78, 3.]..1.1 Proposition If I, J are solvable ideals of L, then I + J is solvable. Proof. Since I is solvable, by Proposition 1.3.0, its homomorphic image I/(I J) is solvable. By Proposition 1..6, (I + J)/J I/(I J) is solvable. Since J and (I + J)/J are solvable, by Proposition 1.3.1, I + J is solvable..1. Lemma There exists a unique maximal solvable ideal of any Lie algebra. Proof. Existence. Let L be a Lie algebra. Since 0 is a solvable ideal of L, there exists a maximal solvable ideal S of L. Uniqueness. If I is a solvable ideal of L, then by Proposition.1.1, S + I is a solvable ideal of L. By the maximality of S, S = S + I and hence I S..1.3 Definition The unique maximal solvable ideal of L is called the radical Rad(L)..1.4 Definition If Rad(L) = 0, then L is called semisimple..1.5 Example A simple algebra is semisimple. Proof. Let L be a simple Lie algebra. By Definition 1..9, Rad(L) = 0 or Rad(L) = L. By Example , L is nonsolvable, then Rad(L) L. Hence Rad(L) = Example 0 is semisimple, since Rad(0) = Lemma For all Lie algebra L, L/ Rad(L) is semisimple. Proof. Let R be the ideal of L containing a solvable ideal Rad(L) of R such that R/ Rad(L) = Rad(L/ Rad(L)). Since Rad(L) and Rad(L/ Rad(L)) are solvable, by Proposition 1.3.1, R is solvable. By the maximality of Rad(L), R = Rad(L). Then Rad(L/ Rad(L)) = R/ Rad(L) = 0. Hence L/ Rad(L) is semisimple..1.8 Definition The descending (lower) central series of ideals of a Lie algebra L is L 0 = L, L 1 = [L, L],, L i = [L, L i 1 ], A Lie algebra L is nilpotent if L n = 0 for some n, or equivalently, ad(x 1 ) ad(x ) ad(x n )(y) = 0 for all x i, y L..1.9 Example Any abelian algebra L is nilpotent, since L 1 = 0. The abelian algebra d(n, F ) is nilpotent Lemma Nilpotent algebras are solvable.

15 LECTURE NOTES 15 Proof. Let L be a Lie algebra. We show that L (n) L n for all n. We use induction. L (1) = [L, L] = L 1. If L (n) L n, then L (n+1) = [L (n), L (n) ] [L, L (n) ] [L, L n ] = L n+1. Since L is nilpotent, L n = 0 for some n. Then L (n) = 0 and hence L is solvable Example t(n, F ) is solvable but not nilpotent. Proof. Let L = t(n, F ) By Example , t(n, F ) is solvable and L 1 = n(n, F ) 0. By Exercise 1.5, L = [t(n, F ), n(n, F )] = [d(n, F ) + n(n, F ), n(n, F )] = [d(n, F ), n(n, F )] + [n(n, F ), n(n, F )] = n(n, F ) + [n(n, F ), n(n, F )] = n(n, F ) = L 1. Similarly, L 1 = L = L 3 =, i.e. t(n, F ) is not nilpotent..1.1 Example n(n, F ) is nilpotent. Proof. Let M = n(n, F ). Similar to Example , M = M 0 is spanned by e ij such that l(e ij ) 1. Suppose i < j, a < b, i b, We have [e ij, e ab ] = δ ja e ib. If j = a, by Remark , l(e ib ) = l(e ab ) + l(e ij ) > max{l(e ab ), l(e ij )}. If e ij M and l(e ab ) m, then l(e ib ) m + 1. By induction, M m is spanned by e ij such that l(e ij ) m + 1. We have M n 1 = 0 since n 1 is the largest possible value for l(e ij ). Hence M is nilpotent Proposition If L is nilpotent, then its subalgebras are nilpotent. Proof. Suppose K is a subalgebra of L. If K n 1 L n 1, then K n = [K, K n 1 ] [L, L n 1 ] = L n. Hence K n L n for all n. If L n = 0, then K n = Proposition If L is nilpotent, then its homomorphic images are nilpotent. Proof. Suppose φ : L K is an epimorphism. If φ(l n 1 ) = K n 1, then φ(l n ) = φ([l, L n 1 ]) = [φ(l), φ(l n 1 )] = [K, K n 1 ] = K n Hence φ(l n ) = K n for all n. If L n = 0, then K n = Proposition If L/Z(L) is nilpotent, then so is L. Proof. If (L/Z(L)) n = 0, then L n Z(L). Hence L n+1 = [L, L n ] [L, Z(L)] = Proposition If L is nilpotent and nonzero, then Z(L) 0. Proof. Suppose L n 0 and [L, L n ] = L n+1 = 0. Then Z(L) L n 0.

16 16 ZHENGYAO WU.1.17 Definition For x L, x is ad-nilpotent if ad(x) is a nilpotent endomorphism Lemma If L is nilpotent, then all elements of L are ad-nilpotent. Proof. By Definition.1.8, ad(x 1 ) ad(x ) ad(x n )(y) = 0 for all x i, y L for some n. Let x = x 1 = = x n. We have (ad(x)) n = Lemma If x gl(v ) is a nilpotent endomorphism, then ad(x) is also nilpotent. Proof. Suppose x n = 0. Then λ n x = 0 and ρ n x = 0. By Notation 1.3.3, n ( ) n (ad(x)) n = (λ x ρ x ) n = (λ x ) i ( ρ x ) n i = 0. i.1.0 Example I n gl(n, F ) is ad-nilpotent but not nilpotent, since ad(i n ) = 0 and In m all m. i=0 = I n for

17 LECTURE NOTES :55-9:40 [Hum78, 3.3]...1 Lemma If f gl(v ) is nilpotent, then there exists 0 v V such that f(v) = 0. Proof. Suppose f n 0 and f n+1 = 0. Then there exists 0 w V such that f n (w) 0. Let v = f n (w). Then v 0 and f(v) = 0... Lemma Let K be a maximal proper subalgebra of L and an ideal of L. Then L = K + F z for some z L \ K. Proof. If dim(l/k) > 1, then a one-dimensional subalgebra of L/K corresponds a subalgebra of L strictly containing K, a contradiction. Then dim(l/k) = 1. Suppose L = K + F z for some z L \ K...3 Lemma Let L be a Lie subalgebra of gl(v ), where V is finite dimensional. Let K be a maximal proper nonzero subalgebra of L. If L consists of nilpotent endomorphisms and V 0, then K is an ideal of L. Proof. By Lemma.1.19, K acts on L via ad and elements of ad(k) are nilpotent. Since [K, K] K, K also acts on L/K. Since dim(l/k) < dim(l), by induction, there exists K x + K L/K such that ad(k)(x + K) = 0. Then x K and [K, x] K. Hence K N L (K). By maximality of K, N L (K) = L, i.e. K is an ideal of L...4 Theorem Let L be a Lie subalgebra of gl(v ), where V is finite dimensional. If L consists of nilpotent endomorphisms and V 0, then there exists nonzero v V such that x(v) = 0 for all x L. Proof. We prove by induction. If dim(l) = 1, then L is generated by some f gl(v ). By Lemma..1, there exists 0 v V such that f(v) = 0. Then x(v) = 0 for all x L. Let K be a maximal proper nonzero subalgebra of L. By Lemma..3, K is an ideal of L. By Lemma.., L = K + F z for some z L \ K. Let W = {w V y(w) = 0, y K} By induction, W 0. For all w W and y K, we have [x, y] K and hence y(x(w)) = x(y(w)) [x, y](w) = x(0) 0 = 0. Then x(w ) W for all x L. As F z gl(w ), by Lemma..1, there exists 0 v W such that z(v) = 0. Then x(v) = (y + cz)(v) = y(v) + cz(v) = 0 for all x L where x = y + cz, y K and c F...5 Theorem Engel: If all elements of L are ad-nilpotent, then L is nilpotent. Proof. Since ad(l) gl(l) consists of nilpotent endomorphisms, by Theorem..4, there exists 0 v L such that [L, v] = 0, i.e. Z(L) 0. Then all elements of L/Z(L) are ad-nilpotent and dim(l/z(l)) < dim(l). By induction, L/Z(L) is nilpotent. By Proposition.1.15, L is nilpotent.

18 18 ZHENGYAO WU..6 Definition If dim F (V ) <, a flag in V is a chain of subspaces 0 = V 0 V 1 V n = V, dim F (V i ) = i If x End(V ), we say that x stablizes (or leaves invariant) this flag if xv i V i for all i.

19 LECTURE NOTES :00-10:45 [Hum78, 4.1]..3.1 Corollary Suppose L is nilpotent, x L gl(v ), there exists a flag (V i ) in V, stable under L, with xv i V i 1 for all i. In other words, there exists a basis of V relative to which the matrices of L are all in n(n, F ). Proof. By Theorem..4, there exists v V such that x(v) = 0 for all x L. Since the action of L on V/(F v) is by nilpotent endomorphisms, by induction, there exists a flag (W i ) 0 i n 1 in V/(F v) such that x(w i ) W i 1. Let π : V V/(F v) be the quotient map. Let V i = π 1 (W i 1 ) for all 1 i n. Then and V 0 = 0, V 1 = F v, V = π 1 (W 1 ),..., V n = π 1 (W n 1 ) = π 1 (V/(F v)) = V x(v i ) = π 1 (x(w i 1 )) π 1 (W i ) V i Lemma Let L be nilpotent. Let K be an ideal of L. If K 0, then K Z(L) 0. In particular, Z(L) 0. Proof. Since L acts on K via ad, by Theorem..4, there exists 0 v K such that [L, v] = 0. Hence v K Z(L). For the rest of today, let F be an algebraically closed field of characteristic Lemma Let L be a finite dimensional solvable Lie algebra. There exists an ideal K of L of codimension 1. Proof. Since L is solvable and dim(l) > 0, by definition, [L, L] L. Let K be a codimension 1 subspace of L/[L, L]. Since L/[L, L] is abelian, K is an ideal of L/[L, L]. Let π : L L/[L, L] be the quotient map. Then K = π 1 (K ) is an ideal of L of codimension dim(l) dim(k) = (dim(l) dim([l, L])) (dim(k) dim([l, L])) = dim(l/[l, L]) dim(k/[l, L]) = dim(l/[l, L]) dim(k ) = Lemma Let V be a finite dimensional vector space over F. Let L be a solvable subalgebra of gl(v ). Let K be an ideal of L of codimension 1. Let λ : K F be a linear function. Let W = {w V y(w) = λ(y)w, y K}. If W 0, then x(w ) W for all x L. Proof. (1) Fix w W. Let n be the smallest integer such that w, x(w),..., x n (w) are linearly dependent. Let W 0 = 0; W i = Span{w, x(w), x (w)..., x i 1 (w)}, 1 i n. Then dim(w n ) = n, W n = W n+1 = W n+ =, x(w i 1 ) W i and x(w n ) W n. Next, we show that for all y K, y(x i (w)) λ(y)x i (w)(mod W i ), i 0

20 0 ZHENGYAO WU The case i = 0 follows from the definition of W. By induction, for some w W i 1. Then y(x i 1 (w)) = λ(y)x i 1 (w) + w y(x i (w)) = x(y(x i 1 (w))) [x, y](x i 1 (w)) = x(λ(y)x i 1 (w) + w ) λ([x, y])x i 1 (w) = λ(y)x i (w) + x(w ) λ([x, y])x i 1 (w) where x(w ) λ([x, y])x i 1 (w) W i. Therefore, y(w i ) W i and Tr Wn (y) = nλ(y). Since [x, y] K, nλ([x, y]) = Tr Wn ([x, y]) = Tr Wn (x y) Tr Wn (y x) = 0. We have λ([x, y]) = 0 since char(f ) = 0. () For all x L, y K and w W, we have [x, y] K and y(x(w)) = x(y(w)) [x, y](w) = x(λ(y)w) λ([x, y])(w) = λ(y)x(w) λ([x, y])(w) = λ(y)x(w). Therefore x(w) W.

21 LECTURE NOTES 1 3. March 14th Lie theorem, Jordan decomposition, Cartan s criterion and Killing form :00-8:45 [Hum78, 4.1] Theorem Let L be a solvable subalgebra of gl(v ). If V 0, then V contains a common eigenvector for all endomorphisms in L. Proof. Use induction on dim(l). If dim(l) = 0, then L = 0. Every nonzero vector of V is an eigenvector of 0. If dim(l) > 0, by Lemma.3.3, there exists a codimension one ideal K of L. Since K is a subalgebra of the solvable algebra L, by Proposition , K is solvable. By induction, there exists a common eigenvector 0 v V for all endomorphisms in K, and there exists a linear function λ : K F such that y(v) = λ(y)v for all y K, i.e. W 0 as in Lemma.3.4, where λ(y) are eigenvalues of v. Let W be the set of common eigenvectors of K, W 0. By Lemma.., L = K + F z. Since F is algebraically closed, by Lemma.3.4, z(w ) W, there exists an eigenvector v 0 W of z. Hence λ can be extend to L and v 0 is a common eigenvector of L Corollary Lie s theorem. Let L be a solvable subalgebra of gl(v ). dim F (V ) = n <. Then L stabilizes some flag in V. In other words, the matrices of L to a suitable basis of V are in t(n, F ). Proof. By Theorem 3.1.1, there exists a common eigenvector v of L. Then L acts on V/(F v), by induction, there exists a flag (W i ) 0 i n 1 in V/(F v) such that x(w i ) W i for all x L. Let π : V V/(F v) be the quotient map. Let V i = π 1 (W i 1 ) for all 1 i n. Similar to Corollary.3.1, (V i ) 1 i n is a flag and x(v i ) = π 1 (x(w i 1 )) π 1 (W i 1 ) = V i Corollary Let L be solvable. Then there exists a chain of ideals of L such that dim(l i ) = i. 0 = L 0 L 1 L n = L Proof. Let ad : L gl(l) be the adjoint representation of L. Since ad(l) is the homomorphic image of the solvable algebra L, by Proposition 1.3.0, ad(l) is solvable. By Corollary 3.1., there exists a flag of subspaces L i of L of dim(l i ) = i such that ad(l)(l i ) = [L, L i ] L i. Then L i is an ideal of L for all 0 i n Corollary Let L be solvable. If x [L, L] then ad L (x) is nilpotent. In particular, [L, L] is nilpotent. Proof. By Corollary 3.1.3, there exists a flag of ideals (L i ) 0 i n. Let {x 1, x,..., x n } be a basis of L such that L i = Span{x 1,..., x i }. Then matrices of ad(l) are in t(n, F ). Then matrices of [ad(l), ad(l)] = ad([l, L]) are in [t(n, F ), t(n, F )] =

22 ZHENGYAO WU n(n, F ), which is nilpotent by Example.1.1. Therefore ad L (x) is nilpotent for all x [L, L]. Then ad [L,L] (x) is nilpotent for all x [L, L]. By Theorem..5, [L, L] is nilpotent Definition Let F be a field of arbitrary characteristic. Let V be a finite dimensional vector space over F. We call x End F (V ) semisimple if the roots of its minimal polynomial over F are distinct Lemma Let F be an algebraically closed and let V be a finite dimensional vector space over F for the rest of this subsection. The x gl(v ) is semisimple iff x is diagonalizable. Proof. The minimal polynomial of the n a n a Jordan block a a a a is (X a) na. Since x is similar to its Jordan canonical form, x is semisimple iff n a = 1 for all eigenvalues a of x iff the Jordan canonical form of x is diagonal Lemma If x is semisimple and x(w ) W for some W V, then x W is semisimple. Proof. Let m(t ; x) be the minimal polynomial of x. By Definition 3.1.5, roots of m(t ; x) are distinct. Since m(t ; x W ) is a factor of m(t ; x), roots of m(t ; x W ) are distinct. By Definition again, x W is semisimple Lemma Commuting endomorphisms of finite dimensional vector spaces can be simultaneously diagonalized. Proof. We treat endomorphism as matrices. Suppose x, y gl(v ) are diagonalizable and xy = yx. We need to show that there exists z gl(v ) such that zxz 1 and zyz 1 are diagonal. Let E λ = {v V y(v) = λv} be eigenspaces of y. Since y(x(v)) = x(y(v)) = x(λv) = λx(v), x(e λ ) E λ for all eigenvalues λ of y. By Lemma 3.1.7, x Eλ is diagonalizable, there exists a basis of E λ and z λ gl(e λ ) (where row vectors of z λ are basis vectors) such that z λ (x Eλ )z 1 λ is diagonal for all λ. Since the basis vectors of E λ are eigenvectors of y, z λ (y Eλ )z 1 λ is diagonal for all λ because row vectors of z λ are basis vectors. Let z = z λ. Then zxz 1, zyz 1 λ are diagonal. For more commuting endomorphisms, use induction and a similar argument Lemma The sum or difference of two commuting endomorphisms are semisimple.

23 LECTURE NOTES 3 Proof. We treat endomorphisms as matrices. Let x, y be two semisimple endomorphisms of gl(v ) such that xy = yx. By Lemma 3.1.6, x, y are diagonalizable. By Lemma 3.1.8, there exists z gl(v ) such that zxz 1 = x, zyz 1 = y are diagonal. Hence z(x ± y)z 1 = x ± y are diagonal. By Lemma again, x ± y are semisimple Lemma For all x End F (V ), if the characteristic polynomial of x is there exists a polynomial p(t ) F [T ] such that p(t ) a i (mod(t a i ) mi ), p(t ) 0(mod T ) k (T a i ) mi, then Proof. If there exists a i = 0, then the last congruence is redundant. Otherwise polynomials (T a i ) mi, 1 i k or T are pairwise coprime. By Chinese Remainder Theorem, such p(t ) exists.

24 4 ZHENGYAO WU 3.. 8:55-9:40 [Hum78, 4.] Proposition Jordan-Chevalley decomposition. For all x End F (V ), there exists unique x s, x n End F (V ) such that (1) x s, x n are polynomials in x whose constant term is 0; () x = x s + s n ; (3) x s x n = x n x s ; if xy = yx, then x s y = yx s and x n y = yx n. (4) x s is semisimple; (5) x n is nilpotent; (6) If A B V and x(b) A, then x s (B) A and x n (B) A. (7) The decomposition is unique. Proof. Take p(t ) from Lemma (1) Let x s = p(x) and x n = x p(x). The fact that p(0) = 0 follows from p(t ) 0(mod T ); Also 0 p(0) = 0. () x = p(x) + (x p(x)) = x s + x n. (3) Because of x s and x n are polynomials of x. (4) If W is a subspace of V such that x(w ) W, then x s (W ) W and x n (W ) W. Let V i = ker(x a i 1) mi. Then x(v i ) = (x a i 1)(V i ) + a i V i V i and hence x s (V i ) V i. Since p(t ) a i (mod(t a i ) mi ), (x s a i 1) Vi = 0. Hence x s acts on V i diagonally by multiplying a i. Since V = k V i, x s is semisimple. (5) Since x n = x x s, (T p(t )) mi 0(mod(T a i ) mi ). Hence ((x n ) Vi ) mi = 0. Therefore x m n = 0 where m > m i for all 1 i k, x n is nilpotent. (6) follows from p(0) = 0. (7) If x = x s + x n = x s + x n are two decompositions, then, by Lemma and binomial theorem respectively, x s x s = x n x n is both semisimple and nilpotent, hence is Definition We call x = x s + x n the additive Jordan-Chevalley decomposition of x if x s is semisimple, x n is nilpotent and x s x n = x n x s We call x s the semisimple part of x and x n the nilpotent part of x Lemma Consider the the adjoint representation of gl(v ), dim F (V ) <. (1) If x gl(v ) is nilpotent, then ad gl(v ) (x) gl(gl(v )) is nilpotent. () If x gl(v ) is semisimple, then ad gl(v ) (x) gl(gl(v )) is semisimple. Proof. (1) was proved in Lemma () Suppose that relative to a basis of V, x = diag(a 1,..., a n ). Then [x, e ij ] = (a i a j )e ij, i.e. the matrix of ad(x) is diagonal relative to the basis {e ij 1 i, j n} of gl(v ) Lemma Let x End(V ), (dim(v ) < ). If x = x s + x n is the Jordan decomposition, then ad(x) = ad(x s ) + ad(x n ) is the Jordan decomposition of ad(x) in End F (End F (V )). Proof. By Lemma 3..3, ad(x s ) is semisimple, ad(x n ) is nilpotent. Also [ad(x s ), ad(x n )] = ad([x s, x n ]) = ad(0) = 0. By Proposition 3..1(7), ad(x s ) = (ad(x)) s and ad(x n ) = (ad(x)) n.

25 LECTURE NOTES Lemma Let A be a finite dimensional F -algebra, then Der(A) contains the semisimple and nilpotent parts (in gl(a)) of all its elements. Proof. Suppose x Der(A), we need to show that x s, x n Der(A). Let A a = {y A (x a 1) ka (y) = 0, k a }. By Lemma and Proposition 3..1(1), x s acts on A a by multiplication of a. For all y A a and z A b, we have x s (y)z + yx s (z) = ayz + ybz = (a + b)yz. On the other hand, since (x (a + b) 1)(yz) = (x a 1)(y)z + y(x b 1)(z), we have k a+k b (x (a + b) 1) ka+k b (yz) = (x a 1) ka+kb i (y) (x b 1) i (z), we obtain A a A b A a+b. Then yz A a+b, x s (yz) = (a + b)yz. i=0 Together x s (yz) = x s (y)z + yx s (z) for all y A a and z A b. Since A = A a, where a runs through all eigenvalues of x, we have x s Der(A). Since x n = x x s, x n Der(A) Lemma If [L, L] is nilpotent, then L is solvable. Proof. By Lemma.1.10, [L, L] is solvable. By Example , L/[L, L] is solvable. By Proposition 1.3.1, L is solvable.

26 6 ZHENGYAO WU :50-10:35 [Hum78, 4.3] Lemma Lagrange interpolation. Suppose a 1,..., a n, b 1,..., b n F. If a i a j for all 1 i < j n, then there exists a polynomial r(t ) F [T ] such that r(a i ) = b i. Proof. n n r(t ) = j=1,j i x a j b i a i a j 3.3. Lemma Let A B be two subspaces of gl(v ), dim F (V ) <. Set M = {y gl(v ) [y, B] A}. Suppose x M satisfies Tr(xy) = 0 for all y M. Then x is nilpotent. Proof. Let x = x s + x n be the Jordan decomposition of x. Take a basis of V such that x s = diag(a 1,..., a n ) gl(v ). Let E = Span Q {a i 1 i n}. If we can show that any linear function f : E Q is 0, then Hom(E, Q) = 0. Since char(f ) = 0, Q F. Then E = 0, x s = 0 and hence x = x n is nilpotent. Let y = diag(f(a 1 ),..., f(a n )). Then [x s, e ij ] = (a i a j )e ij and [y, e ij ] = (f(a i ) f(a j ))e ij. By Lemma 3.3.1, there exists a polynomial r(t ) F [T ] such that r(0) = 0 and r(a i a j ) = f(a i ) f(a j ). Here r is well-defined because f is linear. Hence r(ad(x s )) = ad(y). By Lemma 3..4, ad(x s ) = (ad(x)) s is a polynomial of ad(x) without constant term. Then ad(y) is also a polynomial of ad(x) without constant term. Since x M, ad(x)(b) = [x, B] A, we have ad(y)(b) A. Then 0 = Tr(xy) = n a i f(a i ). Apply f on both sides, we have n f(a i ) = 0 and hence f(a i ) = 0 for all 1 i n. Therefore f = Lemma Tr([x, y]z) = Tr(x[y, z]) for all x, y, z gl(v ). Proof. Tr([x, y]z) = Tr(x[y, z]) Tr(xyz yxz) = Tr(xyz xzy) Tr(xyz) Tr(y(xz)) = Tr(xyz) Tr((xz)y) Tr(y(xz)) = Tr((xz)y) The last step follows from Tr(AB) = Tr(BA) for all A, B F n n Theorem Cartan s criterion. Let L be a subalgebra of gl(v ), dim F (V ) <. Suppose that Tr(xy) = 0 for all x [L, L], y L, then L is solvable.

27 LECTURE NOTES 7 Proof. Let M = {c gl(v ) [c, L] [L, L]}. Then L M. For all a, b L, we have [b, c] [L, L] and Tr([a, b]c) = Tr(a[b, c]) by Lemma As {[a, b] a, b L} are generators of [L, L], Tr(xc) = 0 for all x [L, L], c M. By Lemma 3.3., x is nilpotent for all x [L, L]. By Lemma.1.19, ad(x) is nilpotent for all x [L, L]. By Theorem..5, [L, L] is nilpotent. By Lemma 3..6, L is solvable Corollary Let L be a Lie algebra such that Tr(ad(x) ad(y)) = 0 for all x [L, L], y L. Then L is solvable. Proof. By Theorem 3.3.4, ad(l) is solvable. By Lemma 1.3.4, ker(ad) = Z(L). By Lemma 1..3, L/ ker(ad) ad(l). Then L/Z(L) is solvable. By Example , Z(L) is solvable. By Proposition 1.3.1, both Z(L) and L/Z(L) are solvable implies that L is solvable.

28 8 ZHENGYAO WU :45-11:30 [Hum78, 5.1] Definition The Killing form of a Lie algebra L is κ(x, y) = Tr(ad(x) ad(y)), x, y L 3.4. Lemma κ is an associative, symmetric bilinear form over L. Proof. Associativity follows from Lemma Lemma Let W be a linear subspace of V. Tr(φ) = Tr(φ W ). If φ End(V ) such that φ(v ) W, then Proof. Let e 1,..., e m be a basis of W. Extend it to a basis e 1,..., e m, e m+1,..., e n of V. Then φ(e i ) = m a ij e j, a ij = 0 for m + 1 j n. Hence Tr(φ) = n a ii = m a ii = Tr(φ W ). j= Lemma Let I be an ideal of L. Then κ I = κ I I. Proof. For x, y I, (ad(x) ad(y))(l) I, by Lemma 3.4.3, we have κ(x, y) = Tr(ad(x) ad(y)) = Tr((ad(x) ad(y)) I ) = Tr(ad I (x) ad I (y)) = κ I (x, y) Definition The radical of a symmetric bilinear form β of L is Lemma The radical of κ is an ideal of L. S = {x L β(x, y) = 0, y L}. Proof. Let S be the radical of κ. For all x S, by Lemma 3.4. hence [x, y] S for all y L. κ([x, y], z) = κ(x, [y, z]) = 0, y, z L Definition A bilinear form β : V V F is nondegenerate if its radical S is Lemma Let x 1,..., x n be a basis of V. Then a bilinear form β is nondegenerate iff det((β(x i, x j )) 1 i,j n ) 0. Proof. S = ker( β) where β : V V, β(x)(y) = β(x, y). β is nondegenerate iff β is invertible iff its matrix (β(x i, x j )) 1 i,j n is nonsingular Example The Killing form of sl(, F ) is nondegenerate.

29 LECTURE NOTES 9 Proof. Take the standard basis {x, h, y} as column vectors. We have ad(x) = 0 0 1, ad(h) = 0 0 0, ad(y) = Then κ has matrix whose determinant is It follows from Lemma that κ is nondegenerate Lemma A Lie algebra is not semisimple iff it has at least one nonzero abelian ideal. Proof. If L has a nonzero abelian ideal, the ideal is solvable, then Rad(L) 0, a contradiction to Definition.1.4. Suppose L has no nonzero abelian ideals. If L is not semisimple, then Rad(L) 0. Since Rad(L) is solvable, by Definition , the last nonzero term in the derived series of Rad(L) is abelian. By Exercise 3.1, this term is a nonzero abelian ideal of L, a contradiction Lemma Let S be the radical of κ. Then S Rad(L) but the reverse inclusion need not hold. Proof. By Definition 3.4.5, for all x S, 0 = κ(x, y) = Tr(ad(x) ad(y)) for all y L. By Cartan s criterion Corollary 3.3.5, S is solvable. By Lemma 3.4.6, S is an ideal of L. Hence S Rad(L). For the reverse, see Exercise Lemma Let S be the radical of κ. Every abelian ideal I of L is included in S. Proof. For all x I and y L, (ad(x) ad(y))(l) = [x, [y, L]] [x, L] I. Then (ad(x) ad(y)) (L) (ad(x) ad(y))(i) = [x, [y, I]] [I, I] = 0. Then ad(x) ad(y) is nilpotent and hence κ(x, y) = Tr(ad(x) ad(y)) = 0. Therefore I S Theorem A Lie algebra L is semisimple iff its Killing form κ is nondegenerate. Proof. Let S be the radical of κ. If L is semisimple, by Definition.1.4, Rad(L) = 0. By Lemma , S = 0. By Definition 3.4.7, κ is nondegenerate. Conversely, if κ is nondegenerate, by Definition 3.4.7, S = 0. By Lemma 3.4.1, all abelian ideals of L are 0. By Lemma , L is semisimple. See also Exercise Definition A Lie algebra is the direct sum of ideals if L = I I t as direct sum of subspaces. We write L = I 1 I t. In this case, [I i, I j ] I i I j = Definition Let I be an ideal of L. Define I = {x L κ(x, I) = 0}.

30 30 ZHENGYAO WU Lemma I is an ideal of L. Proof. For all x I and y, z L, by Lemma 3.4., Then [x, y] I, I is an ideal of L Lemma If L is semisimple, then L = I I. κ([x, y], z) = κ(x, [y, z]) = 0 Proof. Since L is semisimple, I L, x κ(x, ) is injective. Since L is finite dimensional, by duality, L L I, x κ(, x) is surjective. Then L I + I I + I and dim(i ) = dim(ker(l I )) = dim(l) dim(i ) = dim(l) dim(i). Let J = I I. Then κ(x, y) = 0 for all x, y J. Then κ(x, y) = 0 for all x [J, J], y J. By Cartan s criterion Corollary 3.3.5, J is a solvable ideal of L, then J Rad(L) Since L is semisimple, Rad(L) = 0, then J = 0. Therefore L = I I.

31 LECTURE NOTES March 1st Semisimple decomposition, Lie modules, Casimir element and Weyl s theorem :00-8:45 [Hum78, 5., 5.3] Theorem Let L be semisimple. Then there exists simple ideals L 1,, L t of L such that L = L 1 L t. Proof. If L has no nonzero proper ideal, then L is simple, t = 1 and L 1 = L. If L 1 is a minimal nonzero proper ideal of L, then Rad(L 1 ) Rad(L) = 0 and hence L 1 is semisimple. Hence L 1 is simple by minimality and L = L 1 L 1. Similarly, L 1 is semisimple. By induction L 1 is a direct sum of simple ideals of L 1, which are also ideals of L Theorem Every simple ideal of L coincides one of L i. The Killing form of L i is κ Li L i. Proof. Let I be a nonzero simple ideal of L. Since [I, L] is an ideal of I and I is simple, [I, L] = 0 or [I, L] = I. Since Z(L) = 0, [I, L] = I. On the other hand, I = [I, L] = [I, L 1 ] [I, L t ]. So I = [I, L i ] for some i and [I, L j ] = 0 for all j i. Then I = [I, L i ] is an ideal of L i. Since L i is simple, we have I = L i. The last assertion follows from Lemma Corollary If L is semisimple, then L = [L, L]. Proof. Suppose L = L 1 L t where L i are simple ideals of L. By Lemma 1..11, [L i, L i ] = L i. By Definition , [L i, L j ] = 0 for all i j. Then [ t ] t t t [L, L] = L i, L i = [L i, L i ] = L i = L Corollary All ideals of a semisimple Lie algebra L are semisimple. Proof. Let I be an ideal of L. By Lemma and Lemma , det(κ I ) det(κ I ) = det(κ). Since L is semisimple, by Theorem , κ is nondegenerate. By the above equality, κ I is nondegenerate. By Theorem again, I is semisimple Corollary All homomorphic images of a semisimple Lie algebra L are semisimple. Proof. Let φ : L M be an epimorphism. If I is a simple ideal of L, then φ(i) is a simple ideal of M. Otherwise there exists a nonzero proper ideal J of φ(i) such that φ 1 (J) is a nonzero proper ideal of I. By Theorem 4.1.1, L = L 1 L t where L i are simple. Then M = φ(l) = φ(l 1 ) φ(l t ) is a direct sum of simple ideals Corollary Each ideal of L is a direct sum of certain simple ideals of L.

32 3 ZHENGYAO WU Proof. By Corollary 4.1.4, an ideal I of L is semisimple. Theorem Lemma [δ, ad(x)] = ad(δ(x)) for all x L, δ Der(L). Proof. For all y L, The rest follows from [δ, ad(x)](y) = δ(ad(x)(y)) ad(x)(δ(y)) = δ([x, y]) [x, δ(y)] = [δ(x), y] = ad(δ(x))(y) Proposition If L is semisimple, then L is linear. Proof. Since L is semisimple, Rad(L) = 0. Since Z(L) is solvable, Z(L) Rad(L). Then Z(L) = 0. By Lemma 1.3.4, ker(ad L ) = 0. Therefore L ad(l) gl(l) Theorem If L is semisimple, then ad(l) = Der(L) i.e. every derivation of L is inner. Proof. By Proposition 4.1.8, L ad(l) is semisimple, by Theorem , ad(l) has a nondegenerate Killing form κ ad(l). By Lemma 4.1.7, [Der(L), ad(l)] ad(l), i.e. ad(l) is an ideal of Der(L). By Lemma 3.4.4, κ ad(l) = κ Der(L) ad(l). Let ad(l) be the complement ad(l) in Der(L) relative to κ Der(L). Since κ ad(l) is nondegenerate, ad(l) ad(l) = 0 and hence [ad(l), ad(l) ] = 0. Now we show that ad(l) = 0. If δ ad(l), then by Lemma 4.1.7, ad(δ(x)) = [δ, ad(x)] = 0 for all x L. Then δ(x) = 0 for all x L. Hence δ = 0. Therefore ad(l) = 0, Der(L) = ad(l) Proposition If L is semisimple, then for all x L, there exists unique x s, x n L such that x = x s + x n, [x s, x n ] = 0, x s is ad-semisimple (i.e. ad(x s ) is semisimple), x n is ad-nilpotent (i.e. ad(x n ) is nilpotent). We call it the abstract Jordan decomposition of x in L. Proof. Let ad(x) = (ad(x)) s + (ad(x)) n be the Jordan decomposition of ad(x) in gl(l). Since ad(x) Der(L), by Lemma 3..5, ad(x) s, ad(x) n Der(L). By Theorem 4.1.9, ad(l) = Der(L) and hence ad(x) s, ad(x) n ad(l). Since L is semisimple, by Proposition 4.1.8, L ad(l). Then there exists unique x s, x n gl(l) such that ad(x s ) = (ad(x)) s and ad(x n ) = (ad(x)) n Remark If L = sl(v ), then the abstract and the usual Jordan decomposition coincide. Proof. Let x = x s + x n be the usual Jordan decomposition in gl(v ). Since x n is nilpotent, Tr(x n ) = 0, we have x n L and x s = x x n L. By Lemma 3..3(1), ad gl(v ) (x s ) is semisimple. Since ad(x s )(L) [L, L] L, by Lemma 3.1.7, ad L (x s ) is semisimple. By Lemma 3..3(), ad gl(v ) (x n ) is nilpotent. Then its restriction ad L (x n ) is nilpotent. Also [x n, x s ] = 0. By the uniqueness, x = x s + x n is the abstract Jordan decomposition of x.

33 LECTURE NOTES :55-9:40 [Hum78, 6.1] Definition Let L be a Lie algebra. Let V be a vector space with an operation L V V, (x, v) x v such that (M1) (ax + by) v = a(x v) + b(y v), (M) x (av + bw) = a(x v) + b(x w), (M3) [x, y] v = x (y v) y (x v), for all x, y L, v, w V and a, b F. Then we call V an L-module. 4.. Lemma x v = φ(x)(v), x L, v V. (1) Given a representation φ : L gl(v ), this defines an L-module. () Given an L-module V, this defines a representation φ : L gl(v ). Proof. (M1) iff φ is linear. (M) iff φ(x) is linear for all x L. (M3) iff φ is a homomorphism of Lie algebras Definition A homomorphism of L-modules is a map φ : V W such that φ(x v) = x φ(v) for all x L and v V Lemma Let φ : V W be a homomorphism of L-modules. Then ker(φ) is an L-module. Proof. If x L and v ker(φ), then φ(x v) = x φ(v) = x 0 = 0 and hence x v ker(φ). (M1-3) of ker(φ) are induced by (M1-3) of V Example L is an L-module via ad. (M1,) follows from the fact that [, ] is bilinear. (M3) follows from the Jacobi identity Definition If a homomorphism of L-modules φ : V W is an isomorphism of vector spaces, then we call φ an isomorphism of L-modules. Or we say that V and W afford equivalent representations of L Definition An L-module V is irreducible if it has precisely two L-submodules 0 and V Example 0 is not an irreducible L-module since it only has one submodule Example A one-dimensional vector space on which L acts trivially is an irreducible L-module Definition An L-module is called completely reducible if V is a direct sum of irreducible L-submodules Lemma An L-module is completely reducible iff each L-submodule W of V has a complement W, an L-submodule such that V = W W Example A simple algebra L is an irreducible L-module, since L-submodules equivalent to ideals Example A semisimple algebra L is a completely reducible L-module, since Theorem

34 34 ZHENGYAO WU Remark Given a representation φ : L gl(v ), the associative algebra with 1 generated by φ(l) has the same invariant subspaces as L. Then all the usual results for modules over associative rings hold for L as well, such as Jordan-Hölder theorem Lemma Schur. Let φ : L gl(v ) be irreducible. Then the only endomorphism of V commuting with all φ(x), x L are the scalars. Proof. Suppose ψ gl(v ) such that φ(x) ψ = ψ φ(x) for all x L. Since F is algebraically closed, ψ has eigenvalues. Let a be an eigenvalue of ψ. Then ψ a 1 gl(v ). Since eigenvectors of ψ are nonzero, ker(ψ a 1) 0. Since V is irreducible, ker(ψ a 1) = V and hence ψ a 1 = 0. Therefore ψ = a Example If W and W are L-modules, then W W is an L-module with x (w, w ) = (x w, x w ) for all x L, w W and w W Example Let V be an L-module. Then V is an L-module by (x f)(v) = f(x v) for all x L, v V and f V. (M1,) follows from the fact that f is linear. (M3) ([x, y] f)(v) = f([x, y] v) = f(x (y v) y (x v)) = f(x (y v))+f(y (x v)) = (x f)(y v)+(y f)(x v) = (y (x f))(v)+(x (y f))(v) = ((x (y f)) (y (x f)))(v) Example Let V and W be L-modules. Then V F W is a L-module by x (v w) = x v w + v x w. (M1,) follows from the fact that is bilinear. (M3) [x, y] (v w) = ([x, y] v) w + v ([x, y] w) = (x (y v) y (x v)) w + v (x (y w) y (x w)) = (x (y v)) w + v (x (y w)) (y (x v)) w v (y (x w)) = (x (y v)) w + (y v) (x w) + (x v) (y w) + v (x (y w)) (y (x v)) w (x v) (y w) (y v) (x w) v (y (x w)) = x ((y v) w) + x (v (y w)) y ((x v) w) y (v (x w)) = x (y (v w)) y (x (v w)) Example Let V and W be L-modules. Then Hom F (V, W ) is a L-module by (x f)(v) = x f(v) f(x v) for all x L, f Hom(V, W ) and v V. (M1,) follows from the fact that is bilinear.