Lie Algebra Homology and Cohomology

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1 Lie Algebra Homology and Cohomology Shen-Ning Tung November 26, 2013 Abstract In this project we give an application of derived functor. Starting with a Lie algebra g over the field k, we pass to the universal enveloping algebra U(g) and define homology H n (g, M) and cohomology groups H n (g, M) for every (left) g-module M, by regarding M as a U(g)- module. In the last section, we prove Whitehead s theorem: If g is a finite dimensional semisimple Lie algebra over field of characteristic 0 and M is a nontrivial irreducible module then H i (g, M) = H i (g, M) = 0, and then deduce Weyl s theorem and Whitehead s lemmas as corollaries. 1 Chain complexes Definition 1 A chain complex C of R-modules is a family {C n } n Z of R- modules, together with R-modules maps d = d n : C n C n 1 such that each composite d d : C n C n 2 is zero. The maps d n are called differentials of C. The n-th homology groups of C is the quotient H n (C) = Ker(d n )/Im(d n+1 ). Definition 2 A chain complex map f : C D is a family of R-module homomorphisms f n : C n D n commuting with d in the sense that f n 1 d n = d n 1 f n That is, such that the following diagram commutes d C n+1 d C n d C n 1 d f f d D n+1 d D n d D n 1 d f By the diagram above, we have Ker(d n ) Ker(d n) and Im(d n+1 ) Im(d n+1). Hence f induces a map H n (f) : H n (C) H n (D). Similarly, we can define cochain complex and cohomology by dualizing. 1

2 f Proposition 1 Let 0 A B C 0 be an exact sequence of chain complexes. Then there are natural maps n : H n (C) H n 1 (A), called connecting homomorphisms, such that g g Hn+1 (C) H n (A) f H n (B) g H n (C) H n 1 (A) f is an exact sequence. Similarly, if 0 A f Ḃ g Ċ 0 be an exact sequence of cochain complexes, there are natural maps n : H n (Ċ) Hn+1 ( A), called connecting homomorphisms, and a long exact sequence g H n 1 (Ċ) H n (Ȧ) f H n (Ḃ) g H n (Ċ) H n+1 ( A) f. Definition of connecting homomorphism B n+1 g n+1 C n A n f n Bn d n+1 g n Cn d n+1 0 d n 0 A n 1 f n 1 Bn d n d n g n 1 Cn 0 d n 1 d n 1 0 A n 2 f n 2 B n 2 Let c H n (C), and let c Ker(d n) represent c. Choose b B n such that g n (b) = c. Let b = d n(b). Then g n 1 (b ) = 0. Therefore, there exists a A n 1 such that b = f n 1 (a). Since (d n 1f n 1 )(a) = (d n 1d n)(b) = 0, it follows that (f n 2 d n 1 )(a) = 0 i.e. a Ker(d n 1 ). Let ā be the canonical image of a in H n 1 (A). It is easily seen that ā does not depend on the choice of c and b. We define n : H n (C) H n 1 (A) by n ( c) = ā. Clearly, n is an R-homomorphism. Similarly, we can define the connecting homomorphism for cochain complex. 2 δ-funtors and Derived functors Definition 3 A (covariant) homological (resp. cohomological) δ-functor between abelian category A and B is a collection of additive functors T n : A B (resp. T n : A B) for n 0, together with morphisms (resp. δ n : T (C) T (A) δ n : T (C) T (A)) 2

3 defined for each short exact sequence 0 A B C 0 in A. With two condition imposed: 1. For each short exact sequence as above, there is a long exact sequence... T n+1 (C) δ T n (A) T n (B) T n (C) δ T n 1 (C)... (... T n 1 (C) δ T n (A) T n (B) T n (C) δ T n+1 (C)...) In particular, T 0 is right exact and T 0 is left exact. 2. For each morphism of short exact sequence from 0 A B C 0 to 0 A B C 0, the δ s give a commutative diagram T n (C ) δ T n 1 (A ) resp. T n (C ) δ T n+1 (A ) T n (C) δ T n 1 (A) T n (C) δ T n+1 (A) Definition 4 A morphism S T of δ-functors is a system of natural transformations S n T n (resp. S n T n ) that commute with δ. A homological δ-functor T is universal if, given any other δ-functor S and a natural transformation f 0 : S 0 T 0, there exists a unique morphism {f n : S n T n } of δ-functors that extends f 0. A cohomological δ-functor T is universal if, given S and f 0 : S 0 T 0, there exists a unique morphism {f n : S n T n } of δ-functors extending f 0. Definition 5 We say an abelian category A has enough projectives if for every object A of A there is a surjection P A with P projective. Similarly, we say a category A has enough injectives if for every object A of A there is a injection A I with I injective. Definition 6 Let M be an object of A. A projective resolution of M is a pair (P, ε), where P is a left complex with each P i projective together with a map ε : P 0 M so that the augmented complex is exact. P n d n P n 1 P 0 ε M 0 Fact The category mod R has enough projectives. 3

4 Definition 7 Let F : A B be a right exact functor between two abelian categories. If A has enough projectives, we can construct the left derived functors L i F (i 1) of F as follows. If A is an object of A, choose (once and for all) a projective resolution P A and define We always have L 0 F (A) = F (A). Facts L i F (A) = H i (F (P )) 1. The object L i F (A) of B are well defined up to natural isomorphism. That is, if Q A is a second projective resolusion, then there is a canonical isomorphism: L i F (A) = H i (F (P )) H i F (Q). 2. If A is projective, then L i F (A) = 0 for i If f : A A is any map in A, there is a natural map L i F (f) : L i F (A ) L i F (A) for each i. 4. The derived functors L F form a homological δ-functor. 5. Assume that A has enough projectives. Then for any right exact functor F : A B, the derived functors L n F form a universal δ-functor. Definition 8 Let F : A B be a left exact functor between two abelian categories. If A has enough injectives, we can construct the right derived functors R i F (i 0) of F as follows. If A is an object of A, choose an injective resolution A I and define R i F (A) = H i (F (I)) Note that since 0 F (A) F (I 0 ) F (I 1 ) is exact, we always have R 0 F (A) = F (A). Since F also defines a right exact functor F op : A op B op, and A op has enough projectives, we can construct the left derived functors L i F op as well. Since I becomes a projective resolution of A in A op, we see that R i F (A) = (L i F op ) op (A) Therefore all the results about right exact functors apply to left exact functors. 4

5 3 Ext Functors and Tor Functors Definition 9 (Tor functors) Let B be a left R-module, so that T (A) = A R B is a right exact functor from mod R to Ab. We define the abelian groups In particular, T or R 0 (A, B) = A R B. T or R n (A, B) = (L n T )(A) Definition 10 (Ext functors) For each R-module A, the functor F (B) = Hom R (A, B) is left exact. Its right derived functors are called the Ext groups: Ext i R(A, B) = R i Hom R (A, )(B) In particular, Ext 0 R (A, B) is Hom(A, B). Recall that these groups are computed by finding a projective(resp. injective) resolution P A(resp. I B) and taking the homology(resp. cohomology) of P R B(resp. Hom(A, I)). Remark. 1. T or and Ext satisfy all the facts above. In particular, they are independent of choice of resolutions and are universal δ-funtors. 2. L n (A R )(B) = L n ( R B)(A) = T or R n (A, B) for all n. This means that we can alsp compute T or R n (A, B) by finding projective resolution of B and taking the homology. 3. R n (Hom R (A, )) = R n (Hom R (, B)) = Ext n R (A, B) for all n. This means that we can also compute Ext n R by finding projective resolution of A and taking the cohomology. 4 Lie algebra Homology and Cohomology Let g a Lie Algebra over field k and M a left g-module, we have 1. The invariant submodule M g of a g-module M, M g = {m M xm = 0 x g} Considering k as a trivial g-module, we have M g = Homg (k, M). 2. The coinvariant M g of a g-module M, M g = M/gM. 5

6 Definition 11 Let M be a g-module. we write H (g, M) or H Lie (g, M) for the left derived funtors L ( g )(M) of g and call them the homology groups of g with coefficients in M. By definition, H 0 (g, M) = M g. Similarly, we write H (g, M) or HLie (g, M) for the right derived funtors R ( g )(M) of g and call them the cohomology groups of g with coefficients in M. By definition, H 0 (g, M) = M g. Example 1 Let g be a free k-module on basis {e 1,...e n }, made into an abelian Lie algebra with zero bracket. Thus g-module is just k-module with n commuting endomorphisms {e 1,..., e n }. It follows that g-module is just R-module where R = k[e 1,...e n ]. If k is trivial g-module viewed as R-module with e i acting as 0, then M g = k R M and M g = Hom R (k, M). Therefore we have H Lie (g, M) = T or R (k, M) and H Lie(g, M) = Ext R(k, M). Example 2 If f is free Lie algebra on set X, then an f-module is a k-module M with an arbitrary set {e x : x X} of endormorphisms. Hence, if view k as trivial f-module, then M f = k R M and M f = Hom R (k, M), where R is the free associative k-algebra k{x}. Therefore, H Lie (f, M) = T or R (k, M) and H Lie(f, M) = Ext R(k, M). Proposition 2 The ideal J = Xk{X} of a free ring k{x} is free as right k{x}-module with basis the set X. Hence 0 J k{x} k 0 is a free resolution of k as a right k{x}-module. Proof. As a free k-module, k{x} has basis words W of the set X, and J is a free k-module on basis W {1}. Every elements of W {1} is of the form wx with w W and x X, so {xw : x X, w W } is also basis for J as k-module. For each x X the k-span xk{x} is isomorphic to k{x}, and J is a direct sum of the xk{x}, both as k-modules and as right k{x}-modules. That is, J is a free right k{x}-module with basis X. Corollary 1 If f is the free Lie algebra on X, then Hn Lie (f, M) = HLie n (f, M) = 0 and for all n 2 and all f-modules M. Moreover, H0 Lie (f, M) = HLie 0 (f, M) = k while H1 Lie (f, k) = HLie 1 (f, k) = x X k. Proof. Using the free resolution of k in Proposition 2, H Lie (f, M) is the homology of the complex 0 J f M f f M 0 and HLie (f, M) is the homology of the complex 0 Hom f (f, M) Hom f (J, M) 0, hence for n 2 the homology groups and cohomology groups are zero. For M = k, the differentials are 0. Furthermore, J f k = x X k = Hom f (J, k) and f f k = k = Hom f (f, k). we get the result for i = 0, 1. 6

7 5 Universal Enveloping Algebra Definition 12 If M is any k-module, the tensor algebra is defined by T (M) = n=0m n, where M n denotes M... M, the tensor product of n copies of M over k. Writing i : M T (M) for the inclusion, T (M) is generated by i(m) as k-algebra. Definition 13 If g is a Lie algebra over k, the universal enveloping algebra U(g) is the quotient of T (g) by the two sided ideal generated by {i([x, y]) i(x)i(y) i(y)i(x) : x, y g}. For every associated algebra A, there is a natural isomorphism Hom Lie (g, Lie(A)) = Hom k alg (U(g), A) where Lie(A) is the Lie algebra with underlying set A with Lie bracket defined by [a, b] = ab ba. Given a g-module M and a monomial x 1 x n in U(g) (x i g), define (x 1 x n )m = x 1 (x 2 ( (x n m) )), m M makes M into a U(g)-module. Conversely, if M is a U(g)-module and x g, define xm = i(x)m (m M) makes M into a g-module. Theorem 1 If g is a Lie algebra, every left g-module is naturally a left U(g)- module. The category g-mod is naturally isomorphic to the category U g-mod of left U(g)-modules. Proof. Let M be a k-module and write E = End k (M) for the k-algebra of all k-module endormorphisms of M. Take A = E in the above identification, we have Hom Lie (g, Lie(E)) = Hom k alg (U(g), E) A g-module is a k-module M together with a Lie algebra map g Lie(E). On the other hand, a U(g)-module M a k-module together with an associative algebra map U(g) End k (M), so the theorem follows. Example 3 There is a k-algebra homomorphism ε : U(g) k, sending i(g) to zero, called the augmentation. We define the augmentation ideal J to be the kernal of ε, which is an ideal of U(g) generated by i(g) as left module. Therefore J is a U(g)-module and k = U(g)/J = U(g) g. 7

8 Corollary 2 Let M be a g-module. Then H Lie (g, M) = T or Ug (k, M) and HLie(g, M) = Ext U g (k, M). Proof. To show any two derived funtors are isomorphic, we only need to show the underlying functors are isomorphic. Observe that k U(g) M = U(g)/J U(g) M = M/JM = M/gM = M g Hom U(g) (k, M) = Hom g (k, M) = M g Example 4 The map i : g U(g) maps [g, g] to J 2, hence induced an isomorphism i : g ab J/J 2, where g ab = g/[g, g]. Example 5 Let M and N be left g-modules. Then Hom k (M, N) is a g-module by xf(m) = xf(m) f(xm), x g m M. This gives us a natural isomorphism Hom g (M, N) = Hom k (M, N) g. Extend this we get a natural isomorphism of δ funtors: Ext U(g)(M, N) = Hom Lie(g, Hom k (M, M)). 6 H 1 and H 1 We have the exact sequence of g-module 0 J U ( g) k 0. If M is a g-module, applying T or U (g) (, M) yields H n (g, M) = T orn U(g) (k, M) = T or U(g) n 1 (J, M) for n 2 and the exact sequence 0 H 1 (g, M) J U(g) M M M g 0. ( ) Theorem 2 For any Lie algebra g, H 1 (g, k) = g ab. Proof. Putting M = k in ( ) yields the exact sequence 0 H 1 (g, k) J U(g) k k k g 0 where J U(g) k = J U(g) (U(g)/J) = J/J 2 = g ab Corollary 3 If M is any trivial g-module, H 1 (g, M) = g ab k M. 8

9 Proof. Since M = M g, ( ) yields H 1 (g, M) = J U(g) M = (J U(g) k) k M = (J U(g) U(g)/J) k M = J/J 2 k M = g ab k M. Definition 14 If M is a g-module, a derivation from g into M is a k-linear map D : g M such that D([x, y]) = x(dy) y(dx). The set of all such derivations is denoted Der(g, M); it is a k-submodule of Hom k (g, M). Note that if g = M, then Der(g, g) is the derivation algebra Der(g). If M is a trivial g-module, then Der(g, M) = Hom k (g ab, M). Example 6 If m M, define D m (x) = xm x g is a derivation, the D m is called an inner derivations of g into M, they form a k-subalgebra Der Inn (g, M) of Der(g, M). If ϕ : J M is a g-homomorphism, let D ϕ : g M be defined by D ϕ (x) = ϕ(i(x)). This is a derivation since D ϕ ([x, y]) = ϕ(i(x)i(y) i(y)i(x)) = xϕ(i(y)) yϕ(i(x)). Lemma 1 The map ϕ D ϕ is a natural isomorphism of k-modules: Hom g (J, M) = Der(g, M). Proof. The map ϕ D ϕ defines a natural homomorphism, so it suffices to show that it is an isomorphism. Note that U(g) k g U(g)g = J is onto with kernal generated by the terms (u [xy] ux y + uy x) with u U(g) and x, y g. Given a derivation D : g M, consider the map f : U(g) k g M f(u x) = u(d(x)) Since D is a derivation, f(u [xy] ux y + uy x) = 0 for all u, x and y. Therefore f induced a map ϕ : g M, which is a left g-module map. Since D ϕ (x) = ϕ(i(x)) = f(1 x) = Dx lifts D to a element in Hom g (J, M). On the other hand, given D = D h for some h Hom g (J, M), we have ϕ(ux) = u(dx) = uh(x) = h(ux) for all u U(g), x g. Hence ϕ = h as map from J = U(g)g to M. Theorem 3 H 1 (g, M) = Der(g, M)/Der Inn (g, M). Proof. If ϕ : J M extends to a g-map U(g) M sending 1 to m M, then D ϕ (x) = ϕ(x 1) = xm = D m (x) Hence D ϕ is an inner derivation. This shows that the image of M Hom g (J, M) = Der(g, M) is a submodule of inner derivations, as desired. Corollary 4 If M is trivial g-module H 1 (g, M) = Der(g, M) = Hom Lie (g, M) = Hom k (g ab, M). 9

10 7 The Chevalley-Eilenberg complex Let g be a Lie algebra over k. Denote p g the p-th exterior product of g, which is generated by monomials x 1... x n with x i g. Define chain complex V p (g) = U(g) k p g, which is free as a left U g -module. Since 0 (g) = k and 1 = g, so V 0 = U(g) and V 1 = U(g) k g. We define ε : V 0 (g) = U(g) k i(g) 0 to be the augmentation and d 1 : V 1 (g) = U(g) g V 0 (g) = U(g) to be the product map d(u x) = ux whose image is the ideal J. We have the exact sequence d V 1 (g) V 0 (g) ε k 0 Definition 15 For p 2, let d : V p (g) V p 1 (g) be given by d(u x 1... x p ) = θ 1 + θ 2, where u U(g) x i g and θ 1 = p ( 1) i+1 ux i x 1... ˆx i... x p i=1 θ 2 = i<j ( 1) i+j u [x i, x j ] x 1... ˆx i... ˆx j... x p V (g) with this differential is called Chevalley-Eisenberg complex. If p = 2, then d(u x y) = ux y uy x u [xy]. Theorem 4 V (g) ε k is a projective resolution of the g-module k. Corollary 5 (Chevalley-Eilenberg) If M is a right g-module, then the homology modules H (g, M) are the homology of the chain complex M Ug V (g) = M Ug U g k g = M k g. If M is a left g-module, then the cohomology modules H (g, M) are the cohomolofy of the cochain complex Hom g (V (g), M) = Hom g (U g g, M) = Hom k ( g, M). In this complex, an n-cochain f : n g M is just an alternating k-multilinear function f(x 1,..., x n ) of n variables in g taking values in M. The coboundary of such an n-cochain is the n+1 cochain df(x 1,..., x n+1 ) = ( 1) n x i f(x 1,..., ˆx i,..., x n+1 ) + ( 1) i+j f([x i, x j ], x 1,..., ˆx i,..., ˆx j,..., x n+1 ). 10

11 We thus constructed cochain complex (C (g, M), d) where C (g, M) = Hom k (g, M). Denote by Z p (g, M) = {f C p (g, M) dc = 0} the space of p-cocycles, and by B p (g, M) = {f C p (g, M) f C p 1 (g, M) such that c = dc } the space of p-coboundaries. Then H p Lie (g, M) = Zp (g, M)/B p (g, M). In particular, we have Z 1 (g, M) = {c C 1 (g, M) c([x, y]) = x c(y) y c(x)} = Der(g, M) B 1 (g, M) = {c C 1 (g, M) c(x) = x m m M} = Der Inn (g, M) Reinterpret H 0 (g, M) and H 1 (g, M) H 0 (g, M) = Z 0 (g, M) = {m M dm = 0} = {m M x m = 0 x g} H 1 (g, M) = Z 1 (g, M)/B 1 (g, M) = Der(g, M)/Der Inn (g, M). Example 7 Consider sl(2, C) =< x, y, h >, where ( ) ( ) ( x =, y =, h = with bracket relation [x, y] = h, [h, x] = 2x, [h, y] = 2y. This shows easily that H 1 (sl(2, C), sl(2, C)) = 0, i.e. all derivations are inner, and that H 1 (sl(2, C), k) = 0. Application If g is a n-dimensional over k, then H i (g, M) = H i (g, M) = 0 i > n. ( i g = 0) ) 8 H 2 and Extensions Definition 16 An extension of Lie algebra of g by M is a short exact sequence 0 M i e 1 π g 0 in which M is an abelian Lie algebra. Indeed M is a g-module via x m = [ g, m] in e where g lift of g in e, x g m M. 11

12 Give g-module M, say two extensions are equivalent if ϕ : e 1 e 2 isomorphism such that the diagram commutes 0 M i e 1 π g 0 0 M i e 2 π g 0 Denote Ext(g, M) the set of equivalent classes of extensions. Classification Theorem Let M be a g-module. The set Ext(g, M) of equivalent classes extensions of g by M is in 1-1 correspondence with HLie 2 (g, M). Sketch of proof. In one direction, associate to a given 2-cocycle c the extension e c = M g(as vector space) with the bracket [(a, x), (b, y)] = (x b y a + c(x, y), [x, y]). ϕ In other direction, choose a linear section s([x, y]) [s(x), s(y)] of π. default of s being a Lie algebra morphism gives a cocycle: The c(x, y) = s([x, y]) [s(x), s(y)] x, y g As Ker(π) = Im(i), c takes values in i(m) = M 9 H 3 and Crossed Modules Definition 17 A crossed module of Lie algebras is a homomorphism of Lie algebra µ : m n together with an action of n on m by derivations, denoted by m n m, such that for all m, m m and for all n n (a) µ(n m) = [n, µ(m)] (b) µ m = [m, m ] The requirements (a) and (b) imply that the cokernel of µ, denote by g, is a Lie algebra, that the kernel of µ, denote by M, is a central ideal, and that g acts on M. Therefore a crossed module corresponds to a 4 term exact sequence of Lie algebras 0 M m µ n g 0. Fact. H 3 (g, M) is isomorphic to the set of equivalent classes of crossed modules with cokernal g and kernal M. 12

13 10 Semisimple Lie Algebras Let g be finite dimensional Lie algebra over k where char k=0. Proposition 3 If g is finite dimensional and semisimple, then g = [g, g]. Consequently, H 1 (g, k) = H 1 (g, k) = 0. Proof. Since g = g 1... g k where g i simple Lie algebra and [g i, g i ] = g i (If [g i, g i ] = 0, then g i Rad(g) = 0), hence [g, g] = g. This gives us g ab = 0. On the other hand, we saw in Corollary 2 and Corollary 4 that H 1 (g, k) = g ab and H 1 (g, k) = Hom k (g ab, k). The Casimir operator Let g be a semisimple Lie algebra and let M a m-dim g-module, if h be the image of the structure map ρ : g Lie(End k (M)) = gl m (k) then ρ = h ker(ρ) h gl m (k). By Cartan s criterion, tr is non-degenerate on h. Choose e 1,..., e r basis of h, then there is a dual basis e 1,..., e r dual basis of h with respect to tr. Then the element c M := e i e i U g is called the Casimir operator for M, which is independent of choice of basis. Recall c M has the following properties: 1. If x g and [e i, x] = c ij e j, then [x, e j ] = c ij e j. 2. c M center(u(g)) and c M J. 3. Image of c M in End k (M) is r m id M, where r=dim(h). Theorem 5 Let g be a semisimple over field of characteristic 0. If M is a simple g-module M k, then H i Lie(g, M) = H Lie i (g, M) = 0 i. Proof. Let C be the center of U(g). We saw in Corollary 2 that H (g, M) = T or U(g) (k, M) and H (g, M) = Ext U(g) (k, M) are naturally C-modules. Multiplication by c C is induced by c : k k as well as c : M M. Since the Casmier element c M acts by 0 on k (as c M J) and by the invertible scalar r/m on M, we must have 0 = r/m on H (g, M) and H (g, M). This can only happen if they are zero. Corollary 6 (Whitehead s first lemma) Let g be a semisimple Lie algebra over field of characteristic 0. If M is any finite dimensional g-module, then H 1 Lie (g, M) = 0. That is, Der(g, M) = Der Inn(g, M). 13

14 Proof. Induction on dim(m). If M is simple, then either M = k and H 1 (g, k) = g/[g, g] = 0 or else M k and H (g, M) = 0 by the theorem. Otherwise, M contains a proper submodule L. By induction, H 1 (g, L) = H 1 (g, M/L) = 0. Since 0 L M M/L 0 induces exact sequence of cohomology,...h 1 (g, L) H 1 (g, M) H 1 (g, L/M)... We have H 1 (g, M) = 0. By Theorem 3, H 1 (g, M) = Der(g, M)/Der inn (g, M). Theorem 6 (Weyl s Theorem) Let g be a semisimple Lie algebra over field of characteristic 0. Then every finite dimensional g-module M is completely reducible i.e. direct sum of g-module. Proof. Suppose M is not a direct sum of simple modules. Since dim(m) is finite, M contains a submodule M 1 minimal with respect to this property. Clearly M 1 is not simple, so it contains a proper g-submodule M 0. By induction, both M 0 and M 2 = M 1 /M 0 are direct sums of simple g-modules but M 1 is not, so the extension of M 2 by M 0 is not split. Hence it is represented by a nonzero element of Ext 1 U(g)(M 2, M 0 ) = H 1 Lie(g, Hom k (M 2, M 0 )) by Example 5 and this contradicts Whitehead s first lemma. Corollary 7 (Whitehead s Second lemma) Let g be a semisimple Lie algebra over field of characteristic 0. If M is any finite dimensional g-module, then HLie 2 (g, M) = 0. Proof. Since H commutes with direct sums, and M is a direct sum of simple g-modules by Weyl s theorem, we may assume that M is simple. If M k, then H 2 Lie (g, M) = 0 by Theorem 5, so it suffices to show that H2 (g, k) = 0. That is, every Lie algebra extension 0 k e π g 0 splits. We claim that e can be made into a g-module in such a way that π is a g-map. To see this, let x be any lift of x g to e and define x e to be [ x, e] for e e. This is independent of choice of x because k is in the center of e. This defines a g-module, and by construction π(x e) = [x, π(e)]. By Weyl s theorem, e and g splits as g-modules, and there is a g-module homomorphism τ : g e splitting π such that e = k g as a g-module. If we choose x = τ(x), then it is clear that τ is a Lie algebra homomorphism and that e = k g as a Lie algebra. This proves that H 2 (g, k) = 0. 14

15 Theorem 7 (Levi-Malcev) If g is a finite dimensional Lie algebra over a field of characteristic zero, then there is a semisimple Lie subalgebra call of g(called a Levi factor of g) such that g is isomorphic to the semidirect product g = (radg) L. Proof. Since g/rad(g) is semisimple, it suffices to show that the following Lie algebra extension splits. 0 rad(g) g g/rad(g) 0 If rad(g) is abelian, then this extension is classified by H 2 (g/radg, radg), which vanishes by Whitehead s second lemma, so every extension splits. If rad(g) is not abelian, we proceed by induction on the derived length of rad(g). Let r denote the ideal [g, g] of g. Since rad(g/r) = rad(g)/r is abelian, the extension 0 (radg)/r g/r g/(radg) 0 splits. Hence there is an ideal h of g containing r such that g/r = (radg)/r h/r and h/r = g/(radg). Now rad(h) = rad(g) h = r, and r has a smaller derived length than radg. By induction there is a Lie subalgebra L of h such that h = r L and L = h/r = g/(radg). Then L is our desired Levi factor of g. References [1] C. Weibel: An introduction to Homological Algebra [2] P.J. Hilton and U. Stammbach: A Course in Homological Algebra [3] H. Cartan and S. Eilenberg: Homological Algebra [4] F. Wagemann: Introduction to Lie algebra cohomology with a view towards BRST cohomology [5] S. Raghavan, R. Balwant Singh and R. Sridharan: Homological methods in Commutative Algebra 15