Journal of Lie Theory Volume 13 (2003) 1{6 c 2003 Heldermann Verlag On the Multiplier of a Lie Algebra Bill Yankosky∗ Communicated by J. D. Lawson Abstract. Under fairly general conditions, we extend the 5-sequence of cohomology for nilpotent Lie algebras a step further. We then derive some consequences of the construction. Keywords: 5-sequence, multiplier, second cohomology group Mathematics Subject Classification: 17B30, 17B56 1. Introduction We begin with the following definitions from [2]. Definition 1.1. A pair of Lie algebras (K; M) is a defining pair for L if 1:L ∼= K=M 2:M Z(K) K2 ⊂ \ Definition 1.2. (K,M) is a maximal defining pair if the dimension of K is maximal. For these maximal defining pairs, K is called a cover for L and M , is called the multiplier for L and is denoted M=L. In her dissertation on the Lie algebra version of the Schur multiplier, Peggy 2 Batten showed M(L) ∼= H (L; F ) and extended, under certain conditions, the 5-sequence of cohomology for Lie algebras a step further. In particular, if L is a Lie algebra over a field F; H is an ideal in L and F is a trivial L-module, the 5-sequence is: 0 Hom (L=H; F ) Inf Hom(L; F ) Res −! −! −! Hom(H; F ) Tra H2 (L=H; F ) Inf H2(L; F ) −! −! As is known and is shown in [1], H2(L; F ) is the multiplier of L. When H is central and L is nilpotent, the exact sequence extends by δ : H2(L; F ) L=L2 H . This ! ⊗ ∗ This paper was a part of the author's dissertation, directed by Dr. Ernest Stitzinger at North Carolina State University ISSN 0949{5932 / $2.50 c Heldermann Verlag 2 Yankosky result has been used in [5] in further investigations of the multiplier. Using group results [8] as a guide, another possible extension should exist when L is nilpotent 2 n j of class n and δ : H (L; F ) L= n 1 L where j and L are the j th terms in the upper and lower central! seriesZ − of⊗L. ShowingZ that this is the case is one of the objectives of this paper. In fact, we show a Theorem which contains both of the results as consequences. We then obtain a number of corollaries which are in the form of inequalities. Finally, we record an example which shows that several of the inequalities are the best possible. Further discussions of multipliers are found in [1], [2], [3], [5] and [7]. 2. The Main Result Theorem 2.1. Let L be a nilpotent Lie algebra over a field F . Let A and B be ideals of L with L2 A and B (L). If f(A; B) = 0 for all f 2(L; F ), then there exists a homomorphism⊆ ⊆δ, Z defined below, such that 2 Z H2(L=B; F ) Inf H2(L; F ) δ Bil(L=A; B : F ) −! −! is exact. 2 Proof. Let x L; y B and f 0 (L; F ). Define f 00 Bil(L=A; B : F ) by 2 2 2 Z 2 f 00 (x + A; y) = f 0(x; y). Since f 0(A; B) = 0, it follows that f 00 is well defined. Let 2 δ0 : (L; F ) Bil(L=A; B : F ) be defined by δ0(f 0) = f 00 . Z ! 2 If f 0 B (L; F ), then there exists a linear function g : L F such that 2 ! f 0(x; y) = g([x; y]). For y B , f 00 (x + A; y) = f 0(x; y) = g([x; y]) = 0. − 2 2 − Hence, δ0(f 0) = 0. Therefore, δ0 induces δ : H (L; F ) Bil(L=A; B : F ) by 2 ! δ(f 0 + B (L; F )) = f 00 . 2 We now show exactness. For f (L=B; F ), let f 0(x; y) = f(x + B; y + B). Recall that Inf:H2 (L=B; F ) H2(L; Z F ) is given by Inf(f + B2(L=B; F )) = 2 ! 2 2 f 0 + B (L; F ) and is induced by I : (L=B; F ) (L; F ) where I(f) = f 0 . Z !Z For x L; y B , it follows that f(x + B; y + B) = f 0(x; y) = f 00(x + A; y) 22 2 2 where f (L=B; F ) and f 0 and f 00 are induced. Now δ(Inf(f +B (L=B; F ))) = 22 Z δ(f + B (L; F )) = 0 with f 00(x + A; y) = f(x + B; y + B) = 0. Hence, f 00 = 0 and Image(Inf) Ker(δ). ⊆ 2 2 Now let f 0 (L; F ) be such that f 0 + B (L; F ) Ker(δ). Then 2 2 Z 2 δ(f 0 + B (L; F )) = f 00 = 0. Let x L; y B . Then 0 = f 00(x + A; y) = f 0(x; y). 2 2 By anti-symmetry, it follows that f 0(y; x) = 0. Set g(x + B; y + B) = f 0(x; y). In order to see that g is well defined, let x + B = x1 + B and y + B = y1 + B . Then x1 = x + b; y1 = y + c for b; c B . Then 2 g(x1 + B; y1 + B) = f 0(x1; y1) = f 0(x + b; y + c) = f 0(x; y) + f 0(b; y) + f 0(x; c) + f 0(b; c) = f 0(x; y) = g(x + B; y + B): Hence, g is well defined. Furthermore, g satisfies the cocycle condition since 2 f 0 does and g(x + B; y + B) = f 0(x; y). Therefore, g (L=B; F ). By the 2 2 Z 2 definition of I , I(g) = f 0 and, therefore, Inf(g + B (L=B; F )) = f 0 + B (L; F ). Thus, Ker(δ) Image(Inf) and the sequence is exact. ⊆ Since Bil(L=A; B : F ) L=A B , the extension can be written as ' ⊗ H2(L; F ) δ L=A B with exactness at H2(L; F ). −! ⊗ Yankosky 3 3. Applications If we let A = L2 and B = Z Z(L), then Theorem 2.1 provides an alternate way to prove Theorem 4.1 on p.⊆ 45 in [1]. Proposition 3.1. Let L be nilpotent and (L). Then Z ⊆ Z H2(L= ;F ) Inf H2(L; F ) δ L=L2 Z −! −! ⊗ Z is exact. Proof. All that needs to be verified is that f(L2; ) = 0 for all f 2(L; F ). But f(L2; ) = f([L; ];L) + f([ ;L];L) = 0. Z 2 Z Z Z Z It should be noted that Proposition 3.1 was used to show [5], Proposition 2. Corollary 3.2. If L is nilpotent, finite dimensional with (L) L2 and dim = 1, then Z ⊆ Z \ Z dim H2(L; F ) + 1 dim H2(L= ;F ) + dim(L=L2) ≤ Z The next result is an analogue of a theorem of Vermani's in group theory [8]. Theorem 3.3. Let L be nilpotent of class n. Then 2 n Inf 2 δ n H (L=L ;F ) H (L; F ) L= n 1 L −! −! Z − ⊗ is exact. n 2 Proof. Since L =Z(L) and L n 1 , it remains to verify the cocycle condition in Theorem⊆ 2.1. ⊆ Z − s First, note that [L ; i] i s . This holds for s = 1 and is shown generally by induction on s. Z ⊆ Z − s To obtain that f( s 1;L ) = 0 for all s, we use induction. For s = Z − 1; f( 0;L) = 0. Suppose that the result holds for s = k. Now, Z k+1 k f( k;L ) = f([L; L ]; k) Z Z k k = f([L ; k];L) + f([ k;L];L ) Z Z k = f( 0;L) + f( k 1;L ) Z Z − = 0 Therefore, Theorem 3.3 follows from Theorem 2.1. 4 Yankosky Corollary 3.4. Let L be nilpotent of class n and finite dimensional. Then 2 2 n n n dim H (L; F ) dim H (L=L ;F ) + dim L dim(L= n 1) dim L : ≤ Z − − Proof. Set q = dim Hom(Ln;F ); r = dim H2(L=Ln;F ); s = dim H2(L; F ) and n t = dim(L= n 1 L ) where the terms are from the extended 5-sequence: Z − ⊗ Res n Tra 2 n Inf 2 δ n Hom(L; F ) Hom(L ;F ) H (L=L ;F ) H (L; F ) L= n 1 L −! −! −! −! Z − ⊗ Since F n = 0 for all n 2, for any f Hom(L; F ); f : Ln F n = 0. Thus, Res(f) = 0 = Ker(Tra).≥ Hence, q = dim(Image(Tra))2 = dim(Ker(Inf)).! Then r q = dim H2(L=Ln;F ) dim(Ker(Inf)) = dim(Image(Inf)) s. − − ≤ Also, dim H2(L; F ) dim(Ker(δ)) = dim(Image(δ)) t. − ≤ Therefore, s dim(Ker(δ)) t. Since r q = dim(Image(Inf)) = dim(Ker(δ)), it follows that s− (r q) t.≤ Hence, − − − ≤ dim H2(L; F ) dim H2(L=Ln;F ) ≤ n n + dim(L= n 1 L ) dim(Hom(L ;F )) Z − ⊗ − 2 n n n = dim H (L=L ;F ) + dim(L= n 1) dim L dim L : Z − − The next corollary is a Lie algebra analogue of a result of Gasch¨utz, Neub¨userand Yen [4]. Corollary 3.5. Let L be nilpotent and finite dimensional. Then dim H2(L; F ) dim H2(L=L2;F ) + dim L2[dim L=Z(L) dim(L=Z(L))2 1] : ≤ − − Proof.