On abelian subalgebras and ideals of maximal dimension in supersolvable Lie algebras

arXiv:1110.2389v3 [math.RA] 5 Jun 2013 ON ABELIAN SUBALGEBRAS AND IDEALS OF MAXIMAL DIMENSION IN SUPERSOLVABLE LIE ALGEBRAS Manuel Ceballos 1 Departmento de Geometria y Topologia, Universidad de Sevilla Apartado 1160, 41080, Seville, Spain and David A. Towers Department of Mathematics, Lancaster University Lancaster LA1 4YF, England Abstract In this paper, the main objective is to compare the abelian subalgebras and ideals of maximal dimension for finite-dimensional supersolvable Lie algebras. We characterise the maximal abelian subalgebras of solvable Lie algebras and study solvable Lie algebras containing an abelian subalgebra of codimension 2. Finally, we prove that nilpotent Lie algebras with an abelian subalgebra of codimension 3 contain an abelian ideal with the same dimension, provided that the characteristic of the underlying field is not two. Throughout the paper, we also give several examples to clarify some results. Mathematics Subject Classification 2010: 17B05, 17B20, 17B30, 17B50. Key Words and Phrases: Lie algebras, abelian subalgebra, abelian ideal, solvable, supersolvable, nilpotent. 1 Introduction Nowadays, there exists an extensive body of research of Lie Theory due to its own importance from a theoretical point of view and also due to its 1Supported by MTM2010-19336 and FEDER 1 applications to other fields like Engineering, Physics and Applied Mathematics. However, some aspects of Lie algebras remain unknown. Indeed, the classification of nilpotent and solvable Lie algebras is still an open problem, although the classification of certain other types of Lie algebras (like semi-simple and simple ones) were already obtained in 1890, at least over the complex field. In order to make progress on these and other problems, the need for studying different properties of Lie algebras arises. For example, conditions on the lattice of subalgebras of a Lie algebra often lead to information about the Lie algebra itself. Studying abelian subalgebras and ideals of a finite-dimensional Lie algebra constitutes the main goal of this paper. Throughout Lwill denote a finite-dimensional Lie algebra over a field F. The assumptions on Fwill be specified in each result. Algebra direct sums will be denoted by ⊕, whereas vector space direct sums will be denoted by ˙ +. We consider the following invariants of L: α(L) = max{dim(A)|Ais an abelian subalgebra of L}, β(L) = max{dim(B)|Bis an abelian ideal of L}. Both invariants are important for many reasons. For example, they are very useful for the study of Lie algebra contractions and degenerations. There is a large literature, in particular for low-dimensional Lie algebras, see [9, 6, 13, 15, 8], and the references given therein. The first author dealing with the invariant α(g) was Schur [14], who studied in 1905 the abelian subalgebras of maximal dimension contained in the Lie algebra of n×nsquare matrices. Schur proved that the maximum number of linearly independent commuting n×nmatrices over an algebraically closed field is hn2 4i+ 1, which is the maximal dimension of abelian ideals of Borel subalgebras in the general linear Lie algebra gl(n) ( where [x] denotes the integer part of a real number x). Let us note that this result was obtained only over an algebraically closed field such as the complex number field. Almost forty years later, in 1944, Jacobson [10] gave a simpler proof of Schur’s results, extending them from algebraically closed fields to arbitrary fields. This fact allowed several authors to gain insight into the abelian subalgebras of maximal dimension of many different types of Lie algebras. More specifically, for semisimple Lie algebras sthe invariant α(s) has been completely determined by Malcev [12]. Since there are no abelian ideals in s, we have β(s) = 0. The value of αfor simple Lie algebras is reproduced in table 1. In this paper, we will study several properties of 2 these invariants and compare them for supersolvable, solvable and nilpotent Lie algebras. Table 1: The invariant αfor simple Lie algebras sdim(s)α(s) An, n ≥1n(n+ 2) ⌊(n+1 2)2⌋ B321 5 Bn, n ≥4n(2n+ 1) n(n−1) 2+ 1 Cn, n ≥2n(2n+ 1) n(n+1) 2 Dn, n ≥4n(2n−1) n(n−1) 2 G214 3 F452 9 E678 16 E7133 27 E8248 36 We shall call Lsupersolvable if there is a chain 0 = L0⊂L1⊂... ⊂ Ln−1⊂Ln=L, where Liis an i-dimensional ideal of L. The ideals L(k)of the derived series are defined by L(0) =L, L(k+1) = [L(k), L(k)] for k≥0; we also write L2for L(1) and L3for [L2, L]. It is well known that every supersolvable Lie algebra is also solvable. Moreover, these classes coincide over an algebraically closed field of characteristic zero (Lie’s theorem). There are, however, examples of solvable Lie algebras over algebraically closed field of non-zero characteristic which are not supersovable (see for instance [11, page 53] or [3]). The Frattini ideal of L,φ(L), is the largest ideal of L contained in all maximal subalgebras of L. We will denote the centre of L by Z(L) = {x∈L: [x, y] = 0,∀y∈L}and the centralizer of a subalgebra Aof Lby CL(A) = {x∈L: [x, A] = 0}. Given a subalgebra Aof L, the core of A, denoted by AL, is the largest ideal of Lcontained in A. The abelian socle of L, AsocL, is the sum of the minimal abelian ideals of L. The structure of this paper is as follows. In section 2 we give some bounds for the invariants αand β. In section 3, we consider the classes of supersolvable, solvable and nilpotent Lie algebras Lwith α(L) = n−1 or n−2. In particular, we characterise n-dimensional solvable Lie algebras Lfor which α(L) = n−2 and prove that every supersolvable Lie algebra, L, of dimension nwith α(L) = n−2 also satisfies β(L) = n−2. In the final section we show the αand βinvariants also coincide for nilpotent Lie 3 algebras Lwith α(L) = n−3, provided that Fhas characteristic different from two. We also give an example to show that the restriction on Fis necessary. 2 Some bounds on α(L)and β(L) We shall call a Lmetabelian if L2is abelian. First we have a bound on β(L) for certain metabelian Lie algebras. Proposition 2.1 Let Lbe a metabelian Lie algebra of dimension n, and suppose that dim L2=k. Then dim(L/CL(L2)) ≤[k2/4] + 1. If, further, L splits over L2then β(L)≥n−[k2/4] −1. Proof. Let ad : L→Der L2be defined by ad x(y) = [y, x] for all y∈L2. Then ad is a homomorphism with kernel CL(L2). It follows that L/CL(L2)∼ = Dwhere Dis an abelian subalgebra of Der L2∼ =gl(n, F ). It follows from Schur’s Theorem on commuting matrices (see [10]) that dim(L/CL(L2)≤ [k2/4] + 1. Now suppose that L=L2⊕Bwhere Bis an abelian subalgebra of L. Then CL(L2) = L2⊕B∩CL(L2) which is an abelian ideal of L. We call Lcompletely solvable if L2is nilpotent. Over a field of characteristic zero, every solvable Lie algebra is completely solvable. Next we note that if Lis completely solvable, has an abelian nilradical (so is metabelian) and the underlying field is perfect then α(L) and β(L) are easily identified. If ¯ Fis the algebraic closure of Fwe put ¯ S=S⊗F¯ Ffor every subalgebra Sof L. Lemma 2.2 α(¯ L)≥α(L),β(¯ L)≥β(L). Lemma 2.3 Let Lbe any solvable Lie algebra with nilradical N. Then CL(N)⊆N Proof. Suppose that CL(N)6⊆ N. Then there is a non-trivial abelian ideal A/(N∩CL(N) of L/(N∩CL(N) inside CL(N)/(N∩CL(N). But now A3⊆ [A, N] = 0, so Ais a nilpotent ideal of L. It follows that A⊆N∩CL(N), a contradiction.  Theorem 2.4 If Fis a perfect field and Lis a completely solvable Lie algebra with abelian nilradical Nthen α(L) = β(L) = dim N. 4 Proof. It is clear that Nis the unique maximal abelian ideal of L. Let Abe an abelian subalgebra of Lof maximal dimension. If N⊆A, then A⊆CL(N) = N, by Lemma 2.3, so N=Aand the result is clear, so suppose that N6⊆ Aand put U=N+A. Consider first the case where Fis algebraically closed and φ(L) = 0. Pick any a∈Aand put C=N+Fa. Then φ(C) = 0, by [18, Theorem 2.5], so N⊆N(C) = Asoc(C) by [17, Theorem 7.4], and Nis completely reducible as an Fa-module. Write N=⊕k i=1Ni, where Niis an irreducible Fa-module for 1 ≤i≤k. Then the minimal polynomial of the restriction of ad ato Niis irreducible for each i, and so {(ada)|N:a∈A}is a set of commuting diagonalizable operators. Thus N= AsocL=Fn1+. . . Fnr, where Fniis a minimal ideal of Lfor 1 ≤i≤r. If ni∈A, then CU(ni) = U; if ni/∈Athen dim CU(ni)≥dim U−1, since U/CU(ni)∼ =D, where Dis a subalgebra of Der Fni. But N=CL(N)⊇CU(N) = r \ i=1 CU(Fni), so dim N≥dim U−(r−dim(N∩A)) = dim A, and the result holds in this case. So suppose now that φ(L) is not necessarily trivial. Then N/φ(L) is the nilradical of L/φ(L), by [17, Theorem 6.1], and so L/φ(L) has abelian nilradical. Also (A+φ(L))/φ(L) is an abelian subalgebra of L/φ(L). It follows from the above that dim A+φ(L) φ(L)≤dim N φ(L),whence dim A≤dim N. Finally consider the case where Fis not necessarily algebraically closed. Since Fis perfect, N(L) = N(¯ L), by [4, page 42]. Hence β(L)≤α(L)≤α(¯ L) = β(¯ L) = dim N(¯ L) = dim N(L) = dim N(L) = β(L) by Lemma 2.2 and the above.  We obtain bounds for supersolvable Lie algebras by following a development similar to [16, Lemma 2]. Lemma 2.5 Let Lbe a supersolvable Lie algebra and let Abe a maximal abelian ideal of L. Then CL(A) = A. 5 Proof. We have that CL(A) is an ideal of L. Suppose that CL(A)6=A. Let B/A be a minimal ideal of L/A with B⊂CL(A). Then, for some b∈B, B=A+Fb, which is an abelian ideal of L, contradicting the maximality of A. The result follows.  Proposition 2.6 Let Lbe a supersolvable Lie algebra, Aany maximal abelian ideal of L. Suppose dim A=k. Then L/A is isomorphic to a Lie algebra of k×klower triangular matrices. Proof. Let ad : L→Der Abe defined by ad x(y) = [y, x]. Then ad is a homomorphism with kernel CL(A) = A, by Lemma 2.5. Since Lis supersolvable there is a flag of ideals 0 = A0⊂A1⊂... ⊂Ak=Aof L. Choose a basis e1,...,ekfor Awith ei∈Ai. With respect to this basis the action of Lon Ais represented by k×klower triangular matrices, since [Ai, L]⊆Aifor each 0 ≤i≤k. Corollary 2.7 Let Lbe a supersolvable Lie algebra with a maximal abelian ideal Aof dimension k. Then dim L≤k(k+3) 2and Lhas derived length at most k+ 1. Proof. The Lie algebra of k×klower triangular matrices has dimension k(k+1) 2and derived length k. Corollary 2.8 Let Lbe a supersolvable Lie algebra of dimension n. Then β(L)≥√8n+ 9 −3 2. Corollary 2.9 Let Lbe a non-abelian solvable Lie algebra of dimension n over an algebraically closed field of characteristic zero. Then √8n+ 9 −3 2≤α(L)≤n−1. Proof. Simply use Corollary 2.8 and [7, Proposition 2.5].  3 Supersolvable Lie algebras with α(L) = n−1or n−2 Proposition 3.1 Let Abe an abelian subalgebra of a Lie algebra L. Suppose that K=A+Fe1is a subalgebra of L, and that there is an x∈Lsuch that [x, K]⊆K, but [x, A]6⊆ A. Then either Kis abelian or K2is one dimensional and Z(K)has codimension at most one in A. 6 Proof. Let e2,...,ekbe a basis for Asuch that e1= [x, e2], say. Let [x, ej] = Pk i=1 αjieifor 1 ≤j≤k. Then [e2,[x, ej]] = αj1[e2, e1], so, for 2≤j≤k, 0 = [x, [e2, ej]] = −[e2,[ej, x]] −[ej,[x, e2]] = αj1[e2, e1]−[ej, e1]. Hence [e1, ej] = αj1[e1, e2]. It follows that K2=F[e1, e2]. Put vj=αj1e2− ejfor 3 ≤j≤k. Then v3,...,vk∈Z(K)∩A. The above result deals with the case where an abelian subalgebra of maximal dimension has codimension one in an ideal of L. Corollary 3.2 Let Lbe a supersolvable Lie algebra and let Abe an abelian subalgebra of maximal dimension in L. If A⊂Kwhere Kis an ideal of L and Ahas codimension 1in K, then α(L) = β(L). Proof. If Ais an ideal of Lthen the result is clear, so suppose that it is not an ideal of L. With the same notation as in Proposition 3.1 the hypotheses of that result are satisfied. Then v3,...,vk∈Z(K); in fact, the maximality of Agives Z(K) = F v3+···+Fvk. Let B/(Fv3+···+Fvk) be a chief factor of Lwith B⊂K. Then Bis an abelian ideal of Lwith the same dimension as A. The result follows.  Next we consider the situation where Lhas a maximal subalgebra that is abelian: first when Lis any non-abelian Lie algebra and Fis algebraically closed, and then when Lis solvable but Fis arbitrary. Proposition 3.3 Let Lbe a non-abelian Lie algebra of dimension nover an algebraically closed field Fof any characteristic. Then Lhas a maximal subalgebra Mthat is abelian if and only if L=A˙ +Ff for some f∈gl(V), where f6≡ 0,Ais abelian and [f, v] = −[v, f] = f(v). In particular, α(L) = β(L) = n−1. Proof. Suppose first that Lhas a maximal subalgebra Mthat is abelian. If Mis an ideal of Lwe have finished. So suppose that Mis self-idealising, in which case L2is one-dimensional and φ(L) = 0, by [19, Proposition 3.2]. Write L2=Fb and note that AsocL=L2⊕Z(L). Now L= AsocL˙ +C, where Cis abelian, by [17, Theorem 7.4]. For each c∈Cwe have that [c, b] = λ(c)b, for some λ(c)∈F. Since C∩Z(L) = 0, Cmust be onedimensional and the result follows. The converse is clear.  The following is a generalisation of [19, Proposition 3.1] 7 Proposition 3.4 Let Lbe a solvable Lie algebra. Then Lhas a maximal subalgebra Mthat is abelian if and only if either (i) Lhas an abelian ideal of codimension one in L; or (ii) L(2) =φ(L) = Z(L),L2/L(2) is a chief factor of L, and Lsplits over L2. Proof. By [19, Proposition 3.1] it suffices to show that if Lhas a maximal subalgebra Mthat is not an ideal then L2is nilpotent. By maximality, M is self-idealising, and from solvability there is a k≥1 such that L(k)6⊆ M but L(k+1) ⊆M. Then L=M+L(k), whence L2⊆L(k). It follows that L(2) ⊆Mand we have M⊆CL(L(2)). By maximality and the fact that M is self-idealising, we get CL(L(2)) = L, which yields that L2is nilpotent.  Next we characterise solvable Lie algebras Lwhose biggest abelian subalgebras have codimension two in L. The following proof relies on [7, Propositions 3.1 and 5.1] which are only stated for Lie algebras over fields of characteristic zero. However, it is easy to see that this assunption is not used in their proofs, and that the results are, in fact, valid over an arbitrary field. Theorem 3.5 Let Lbe a solvable Lie algebra of dimension nwith α(L) = n−2, and let Abe an abelian subalgebra of dimension n−2. Then one of the following occurs: (i) β(L) = n−2; (ii) L=L2˙ +B, where Bis an abelian subalgebra of L,L2is the threedimensional Heisenberg algebra, L(2) =φ(L) = Z(L)and L2/Z(L)is a two-dimensional chief factor of L(in which case β(L)≤n−3); (iii) Ahas codimension one in the nilradical, N, of L, which itself has codimension one in L. Moreover, N2is one dimensional, Z(N)is an abelian ideal of maximal dimension and β(L) = n−3. Proof. Let Abe a maximal abelian subalgebra of Lof dimension n−2 and suppose that (i) doesn’t hold. (a) Suppose first that Ais a maximal subalgebra of L. Then Lis as in Proposition 3.4(ii) and Ais a Cartan subalgebra of L. Let L=A˙ +L1 be the Fitting decomposition of Lrelative to A. Then L1⊆L2and dim L1= 2. Let L1=Fx +Fy. 8 If [x, y] = 0 then L1is an ideal of Land L/L1is abelian, so L2⊆ L1⊆L2. This yields that Lis metabelian and L2is a two dimensional minimal ideal over which Lsplits. It follows from Proposition 2.1 that β(L) = n−2, a contradiction. If [x, y]6= 0, then L2=F[x, y] + L1and F[x, y]⊆L(2) =Z(L), so F[x, y] = Z(L) and we have case (ii). Moreover, if Cis a maximal abelian ideal of L, then Z(L)⊆Cand L26⊆ C. It follows that C∩L2=Z(L). If dim C=n−2, then dim(L2+C) = dim L2+ dim C−dim C∩L2= 3 + n−2−1 = n, so L=L2+C. But then L2=Z(L), a contradiction. Hence, β(L)≤n−3. (b) So suppose that Ais not a maximal subalgebra of L. Then A⊂M⊂L, where dim M=n−1. Moreover, there is such a subalgebra Aof M which is an ideal of M, by [7, Proposition 3.1]. Suppose first that A does not act nilpotently on L. Then the Fitting decomposition of L relative to Ais L=M˙ +L1, and L1is a one-dimensional ideal of L. Put B=A˙ +L1, which is an ideal of L. Then CB(L1) has codimension one in Band so is an abelian ideal of codimension two in L. It follows that β(L) = n−2, a contradiction. Finally, suppose that Ais an ideal of Mand that Aacts nilpotently on L. Then there is a k≥0 such that L(ad A)k6⊆ Mbut L(ad A)k+1 ⊆ M. Let x∈L(ad A)k\M, so L=M˙ +Fx. Suppose first that Mis not an ideal of L. Then the core of M,MLhas codimension one in M, by [1, Theorem 3.1 and 3.2]. If A=MLthen we have case (i), so suppose that A6=MLand M=A+ML. Then [A, x]⊆Mwhich implies that [L, A]⊆Mand [L, M] = [L, ML] + [L, A]⊆M; that is, Mis an ideal of L. Let Nbe the nilradical of L. If N⊆Athen A⊆CL(N)⊆N, so N=Aand we have case (i) again. If A⊂Nthen N=Lor we can assume that N=M. If A6⊆ Nand N6⊆ Athen either A+N=L, in which case Lis nilpotent, or we can assume that A+N=M, in which case Mis a nilpotent ideal of Land so M=N. If Lis nilpotent, then we have case (i), by [7, Proposition 5.1].If not, then we have case (iii) by Proposition 3.1.  Corollary 3.6 Let Lbe a supersolvable Lie algebra of dimension nwith α(L) = n−2. Then β(L) = n−2. 9 [18] D. A. Towers and V. R. Varea, ‘Elementary Lie Algebras and Lie A-algebras’, J. Algebra 312 (2007), 891–901. [19] D. A. Towers, ‘The index complex of a maximal subalgebra of a Lie algebra’, Proc. Edin. Math. Soc. 54 (2011), 531–542. 16