On dimension of the Schur multiplier of nilpotent Lie algebras

Cent. Eur. J. Math. 9() 0 57-64 DOI: 0.478/s533-00-0079-3 Central European Journal of Mathematics On dimension of the Schur multiplier of nilpotent Lie algebras Research Article Peyman Niroomand School of Mathematics and Computer Science, Damghan University, Damghan, Iran Received 8 April 00; accepted 9 October 00 Abstract: Let L be an n-dimensional non-abelian nilpotent Lie algebra and s(l) = (n )(n ) + dim M(L) where M(L) is the Schur multiplier of L. In [Niroomand P., Russo F., A note on the Schur multiplier of a nilpotent Lie algebra, Comm. Algebra (in press)] it has been shown that s(l) 0 and the structure of all nilpotent Lie algebras has been determined when s(l) = 0. In the present paper, we will characterize all finite dimensional nilpotent Lie algebras with s(l) =,. MSC: 7B30, 7B60, 7B99 Keywords: Schur multiplier Nilpotent Lie algebras Versita Sp. z o.o.. Introduction For a given group G, the notion of the Schur multiplier M(G) arose from the work of Schur on projective representations of groups, as the second cohomology group with coefficients in C. For a group G, it can be shown by Hopf s formula that M(G) = R [F; F]/[R; F] where 0 R F G 0 is a free presentation of G. Many results for the Schur multiplier of various classes of groups have been obtained by several authors. Among famous classes of groups some classes took special attention. The most considered is the class of finite p-groups because of brilliant results on the Schur multiplier. One of the most important results for the Schur multiplier was presented by Green [8]. He showed that for every finite p-group G of order p n, M(G) = p n(n ) t(g) for some nonnegative t(g). Green s result suggests an interesting problem: Can we classify finite p-groups by t(g)? Berkovich in [4] showed that t(g) = 0 if and only if G is an elementary abelian p-group. He could also classify all p-groups with t(g) =. In 998, Zhou [8] considered the case t(g) = and determined the structure of p-groups with the desired property. Ellis in [6] showed how an upper E-mail: niroomand@du.ac.ir, p niroomand@yahoo.com 57

On dimension of the Schur multiplier of nilpotent Lie algebras bound of Gaschütz et al. [7] can help to shorten the procedure of finding the structure of p-groups with t(g) = 0,,, and he solved the problem for t(g) = 3. The author in [4] improved the upper bound of Jones []. By handling this result, one can obtain the structure of G when t(g) =,, 3 by a quite different and shorter way to that of [4, 6, 8]. Furthermore, the author succeeded in extending the result to t(g) = 4, 5. Analogous to the Schur multiplier of group, the Schur multiplier of Lie algebra, M(L), can be defined as M(L) = R [F, F]/[R, F] where L = F/R and F is a free Lie algebra (see [ 3, 5, 9, 0, 3, 6, 7] for more information). A result of Moneyhun [3], similar to the result of Green, shows that for a Lie algebra L of dimension n we have dim M(L) = n(n ) t(l), where t(l) 0. Batten et al. in [] succeeded in obtaining results similar to that of Berkovich and Zhou for nilpotent Lie algebras of finite dimension. In [0] and later in [9] with a different method to [], the cases t(l) = 3, 4, 5, 6 and t(l) = 7, 8 were considered, respectively. As before mentioned, a new upper bound for the dimension of the Schur multiplier of a Lie algebra of finite dimension may simplify the problem of determining a Lie algebra in terms of t(l). The author in his joint paper [5] presented an upper bound for the dimension of the Schur multiplier of a non-abelian nilpotent Lie algebra as follows dim M(L) = (n )(n ) + s(l) where s(l) 0, and classified the structure of L when s(l) = 0. Note that, if L is abelian, then s(l) might be negative. It seems classifying nilpotent Lie algebras by s(l) causes classification of those Lie algebras in terms of t(l). In the present paper we intend to classify all nilpotent Lie algebras with s(l) =,.. Some notations and known results In this section we give some notations and known results without proofs which will be used throughout the paper. Let L be a finite dimensional Lie algebra. Then the following notation will be used throughout: [, ] Lie bracket, [x, y] commutator of x and y, [I, J] span of the commutators of elements of I and J, Z(L) L center of L (the set of elements x L such that [x, y] = 0 for all y L), commutator subalgebra of L (L = [L, L]), H(m) Heisenberg algebra of dimension m+ (a Lie algebra such that L = Z(L) and dim L = ), A(n) abelian Lie algebra of dimension n, L 3 Z (L)/Z(L) [L, L], Z(L/Z(L)), L(3, 4,, 4) central extension of A() by H() and A() L, L(4, 5,, 4) Lie algebra with basis {x, y, z, c, r} and multiplication [x, y] = z, [x, z] = [y, z] = [z, c] = [y, c] = 0, [x, c] = r. Note that the two latest notations were introduced in [0]. Theorem. ([5, Theorem 3.]). Let L be an n-dimensional nilpotent Lie algebra and dim L = m. Then dim M(L) (n + m )(n m ) +. Moreover, if m =, then the equality holds if and only if L = H() A(n 3). 58

P. Niroomand Lemma. ([3, Example 3] and [3, Theorem 4]). Let H(m) be Heisenberg algebra of dimension m +. Then (i) dim M(H()) =, (ii) dim M(H(m)) = m m for all m. Corollary.3 ([5, Corollary.3]). Let L be a finite dimensional Lie algebra, K a central ideal of L and H = L/K. Then dim M(L) + dim (L K) dim M(H) + dim M(K) + dim (H/H K). Theorem.4 ([, Theorem ]). Let A and B be finite dimensional Lie algebras. Then dim M(A B) = dim M(A) + dim M(B) + dim ( A/A B/B ). We say that a nilpotent Lie algebra L has t(l) = m or s(l) = k, when its Schur multiplier has dimension n(n ) m (n )(n ) + k, respectively. or 3. Characterizing nilpotent Lie algebras with s(l) = In this section, first we characterize all finite dimensional Lie algebras with s(l) =, and then we illustrate how this idea can simplify determining the structure of all finite dimensional Lie algebras when t(l) =,, 3, 4. The following lemma is an upstanding result of [3, Lemma 3]. Lemma 3.. There is no n-dimensional abelian Lie algebra with s(l) =. Theorem 3.. Let L be a nilpotent Lie algebra and dim L = k. Then there is no nilpotent Lie algebra with s(l) = when k 3. In the case k =, the same result holds when dim Z(L) 3 or dim Z(L) = and Z(L) L. Proof. Suppose that k 3 and that s(l) =. By virtue of Theorem. we have what is impossible. (n )(n ) = dim M(L) (n + )(n 4) +, In the case k = and dim Z(L) 3, by choosing a -dimensional central ideal I such that I L = 0, Theorem. implies that dim M(L/I) (n )(n 3). Using Corollary.3 and by assumption, (n )(n ) = dim M(L) dim M(L/I) + dim ( I ( L/(I L ) )) (n )(n 3) + (n 3) = n(n 3), which is a contradiction. The case dim Z(L) = and Z(L) L is obtained similarly. The following lemma is a part of proof of [5, Theorem 3.]. 59

On dimension of the Schur multiplier of nilpotent Lie algebras Lemma 3.3. Let L be an n-dimensional Lie algebra and dim L =. Then for some m L = H(m) A(n m ). Proof. Let H/L be the complement of Z(L)/L in L/L. Thus, L = H + Z(L) and L = H. One can check that Z(H) = L, and hence H = H(m) for some m. Since L Z(L), let A(n m ) be the complement of L in Z(L). It is seen that L = H(m) A(n m ), as required. Lemma 3.4. There is no n-dimensional nilpotent Lie algebra with dim L = and s(l) =. Proof. It is obtained by virtue of Lemmas. and 3.3, and Theorem.4. The extended result of the following lemma is proved in [, Theorem 3..6 and Proposition 3..] for groups. Lemma 3.5. Let L be a finite dimensional nilpotent Lie algebra such that dim L = and dim Z(L) =. Then dim L 3 + dim M(L) dim M(L/L 3 ) + dim ( L/Z (L) L 3). Proof. Let L = F/R where F is a free Lie algebra. Since L/L 3 = F/R (F/R) 3 = F/(R + F 3 ), we have M(L/L 3 ) = ( ( (R + F 3 ) F ) /[R + F 3, F] (R F ) F 3) /[R, F] =. [R + F 3, F]/[R, F] On the other hand, ( (R F ) F 3) /[R, F] Therefore R F /[R, F] = ( (R F ) F 3) /R F = (R + F 3 )/R = L 3. dim L 3 + dim M(L) dim M(L/L 3 ) + dim ([R + F 3, F]/[R, F]). Since L is a nilpotent Lie algebra we have L 3 L, and hence L 3 = Z(L) by the fact that dim L =. Thus the rest of proof is obtained by a suitable epimorphism θ : L/Z (L) L 3 [R + F 3, F]/[R, F], as required. Lemma 3.6. There is no nilpotent Lie algebra of dimension n, n 6, with s(l) = and dim Z(L) =. Proof. We may assume that dim L = by invoking Lemmas 3. and 3.4, and Theorem 3.. Lemma 3.3 implies that L/Z(L) = H(m) A(n m ). In the case m, the result is obtained by using Corollary.3 when K = Z(L) and Theorem.. Thus L/Z(L) = H() A(n 4), and hence dim Z (L) = n. Using Theorem.4 and Lemma 3.5, we have + dim M(L) (n )(n 3) + 3, and hence dim M(L) (n )(n ). The result follows. Theorem 3.7. There is no nilpotent Lie algebra of dimension n, n 6, with s(l) =. 60

P. Niroomand Proof. Owing to Lemmas 3., 3.4 and 3.6, and Theorem 3., we may assume that Z(L) = L and dim L =. It is known that L/I = H(m) A(n m ) for any -dimensional central ideal I, using Lemma 3.3. Suppose that for two distinct central ideals of Z(L) denoted as I i, i =,, we have m =. Therefore by using the method involved in the proof of Lemma 3.3 we have L/I i = T i /I i S i /I i, where S i /I i L /I i = Z(L/I i ), T i /I i = H() and S i /I i = A(n 4) for i =,. But from L S i = Z(L) S i I i it follows that S S Z(L) I I = 0, and hence S S = 0 because they are ideals of the nilpotent Lie algebra L. Since dim S i = n 3 we have L = S S and n = 6. Therefore dim M(L) = dim M(S ) + dim M(S ) + dim ( S /S S /S ) 8, as required. Now assume that there exists a central ideal I such that L/I = H(m) A(n m ), m. By invoking Theorem.4 and Lemma., dim M(L/I) = (n )(n 4) which implies that dim M(L) (n )(n ) because of Corollary.3. The result follows. Lemma 3.8. Let L be a non-abelian nilpotent 4-dimensional Lie algebra and dim L =. Then L = L(3, 4,, 4) and dim M(L) =. Proof. Since dim Z(L) =, we have L/Z(L) = H() and so L = L(3, 4,, 4). Assume that L has a basis {x, y, z, r} with [x, y] = z, [x, z] = r, [y, z] = 0 and r Z(L). Some calculations analogous to those from [0, page 3530] imply dim M(L) =. Theorem 3.9. L is an n-dimensional nilpotent Lie algebra with s(l) = if and only if L = L(4, 5,, 4). Proof. Lemmas 3. and 3.4 and Theorems 3. and 3.7 imply that n 5 and Z(L) = L is of dimension. On the other hand, by using Lemma 3.8, there is no 4-dimensional Lie algebra with s(l) =. Hence n = 5, and for each central ideal I, we have L/I = H() A(). Thus L = L(4, 5,, 4). We use these results to give a short proof for the following theorem. Theorem 3.0. Let L be an n-dimensional nilpotent Lie algebra. Then (i) t(l) = if and only if L = H(); (ii) t(l) = if and only if L = H() A(); (iii) t(l) = 3 if and only if L = H() A(); (iv) t(l) = 4 if and only if L = H() A(3), L = L(3, 4,, 4) or L = L(4, 5,, 4). Proof. (i). When t(l) =, Theorem. implies that s(l) = 0, n = 3 and L = H(). (ii). The proof is similar to (i). (iii). It is obtained by Theorem. and Lemma 3.8. (iv). By using Theorem., we have n(n ) 4 = dim M(L) (n )(n ) +. Hence n = 6 and L = H() A(3) or dim M(L) (n )(n ) and n 5. When n = 5, Theorem 3.9 implies that L = L(4, 5,, 4). Thus, one may assume that dim L = 4. Now Lemma 3.4 and Theorem 3. imply that dim L =. Thus L = L(3, 4,, 4) by using Lemma 3.8, as required. 6

On dimension of the Schur multiplier of nilpotent Lie algebras 4. Characterizing nilpotent Lie algebras with s(l) = This section is devoted to characterizing the structure of all finite dimensional nilpotent Lie algebras with s(l) =. Lemma 4.. There is no n-dimensional nilpotent Lie algebra with the property s(l) = and dim L 3. Proof. Taking k = 3 in Theorem., we have (n )(n ) = dim M(L) (n + )(n 4) +, which is a contradiction. First we describe the structure of all nilpotent Lie algebras with s(l) = and dim L =. Lemma 4.. There is no nilpotent Lie algebra of dimension n, n 5, with s(l) = and dim L = when L Z(L) or dim Z(L) 4. Proof. In the case L Z(L), suppose first that dim Z(L) =. If n = 5, one can check that there is no Lie algebra satisfying the assumption by a similar method to that involved in the proof of Lemma 3.8. For n 6, the result is directly obtained by using Lemma 3.5 (see also Lemma 3.6). Now assume that dim Z(L). We may suppose that there is a central ideal I such that I L = 0. Since dim Z(L/I) = and dim (L/I) =, we have dim M(L/I) (n )(n 3) because of Theorems. and 3.9. Hence by using Corollary.3, we have dim (M(L)) dim (M(L/I)) + dim (( L/(L I) ) I ) (n )(n 3) + n 3 = (n )(n ), as required. The case dim Z(L) 4 can be obtained by a similar method. Theorem 4.3. Let L be an n-dimensional nilpotent Lie algebra with s(l) = and dim L =. Then L = L(3, 4,, 4) or L = L(4, 5,, 4) A(). Proof. In the case n = 4, L = L(3, 4,, 4). Therefore we may assume that n 5. By using Lemma 4., dim Z(L) 3 and L Z(L). First suppose that dim Z(L) =. In the case n = 5, there is no nilpotent Lie algebra satisfying s(l) =, because for any -dimensional central ideal I we have L/I = H() A(). Hence L = L(4, 5,, 4), and so s(l) = by Theorem 3.9. For all n 6, the proof of Lemma 3.6 shows that dim M(L) (n )(n ), as required. Now assume that dim Z(L) = 3. By taking K, the complement of L in Z(L), and using Theorem., we have dim M(L/K) (n )(n 3). On the other hand, Corollary.3 and the assumption imply that (n )(n ) = dim M(L) dim M(L/K) + dim (( L/(L K) ) K ) = dim M(L/K) + n 3, which implies that dim M(L) = (n )(n 3), hence L/K = L(4, 5,, 4) by Theorem 3.9, and so L = L(4, 5,, 4) K. For classifying the structure of all Lie algebras with s(l) =, the only case left to be verified is the case dim L =. 6

P. Niroomand Theorem 4.4. Let L be an n-dimensional nilpotent Lie algebra with dim L = and s(l) =. Then L = H(m) A(n m ) and m. Proof. Since dim L =, Lemma 3.3 implies that L = H(m) A(n m ). Using Lemma. and Theorem.4, for all m dim ( M(H(m) A(n m )) ) = (m m ) + (n m )(n m ) + (n m ) m = (n )(n ). A similar technique shows that dim ( M(H() A(n 3)) ) = (n )(n ) +, as required. We summarize the results as follows. Theorem 4.5. Let L be an n-dimensional nilpotent Lie algebra. Then s(l) = if and only if L is isomorphic to one of the following Lie algebras: (i) L(3, 4,, 4), (ii) L(4, 5,, 4) A(), (iii) H(m) A(n m ), m. Acknowledgements The author would like to thank the referees for making useful suggestions and improving Theorem 3.7. References [] Batten P., Multipliers and Covers of Lie Algebras, PhD thesis, North Carolina State University, 993 [] Batten P., Moneyhun K., Stitzinger E., On characterizing nilpotent Lie algebras by their multipliers, Comm. Algebra, 996, 4(4), 439 4330 [3] Batten P., Stitzinger E., On covers of Lie algebras, Comm. Algebra, 996, 4(4), 430 437 [4] Berkovich Ya.G., On the order of the commutator subgroups and the Schur multiplier of a finite p-group, J. Algebra, 99, 44(), 69 7 [5] Bosko L.R., On Schur multipliers of Lie algebras and groups of maximal class, Internat. J. Algebra Comput., 00, 0(6), 807 8 [6] Ellis G., On the Schur multiplier of p-groups, Comm. Algebra, 999, 7(9), 473 477 [7] Gaschütz W., Neubüser J., Yen T., Über den Multiplikator von p-gruppen, Math. Z., 967, 00(), 93 96 [8] Green J.A., On the number of automorphisms of a finite group, Proc. Roy. Soc. London Ser. A., 956, 37, 574 58 [9] Hardy P., On characterizing nilpotent Lie algebras by their multipliers III, Comm. Algebra, 005, 33(), 405 40 [0] Hardy P., Stitzinger E., On characterizing nilpotent Lie algebras by their multipliers t(l) = 3, 4, 5, 6, Comm. Algebra, 998, 6(), 357 3539 [] Jones M.R., Multiplicators of p-groups, Math. Z., 97, 7(), 65 66 [] Karpilovsky G., The Schur Multiplier, London Math. Soc. Monogr. (N.S.),, The Clarendon Press, Oxford University Press, New York, 987 [3] Moneyhun K., Isoclinisms in Lie algebras, Algebras Groups Geom., 994, (), 9 [4] Niroomand P., On the order of Schur multiplier of non-abelian p-groups, J. Algebra, 009, 3(), 4479 448 63

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