# Nilpotent Lie and Leibniz Algebras

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2 Communications in Algebra, 42: , 2014 Copyright Taylor & Francis Group, LLC ISSN: print/ online DOI: / NILPOTENT LIE AND LEIBNIZ ALGEBRAS Chelsie Batten Ray 1, Alexander Combs 1, Nicole Gin 1, Allison Hedges 1, J. T. Hird 2, and Laurie Zack 3 1 Department of Mathematics, North Carolina State University, Raleigh, North Carolina, USA 2 Department of Mathematics, West Virginia University, Institute of Technology, Montgomery, West Virginia, USA 3 Department of Mathematics and Computer Science, High Point University, High Point, North Carolina, USA We extend results on finite dimensional nilpotent Lie algebras to Leibniz algebras and counterexamples to others are found. One generator algebras are used in these examples and are investigated further. 1. INTRODUCTION Results on nilpotent Lie algebras have been extended to Leibniz algebras by various authors. We focus on necessary and sufficient conditions for determining whether a Leibniz algebra A is nilpotent. Previous results on this topic include: Engel s Theorem ([2], [6], [8], [13]), the nilpotency of algebras which admit a prime period automorphism without nonzero fixed points, and the equivalence of (a) nilpotency, (b) the normalizer condition, (c) the right normalizer condition, (d) that all maximal subalgebras are ideals, and (e) that all maximal subalgebras are right ideals (which follows from material in [5]). In this work we show that a Leibniz algebra A is nilpotent exactly when it has a nilpotent ideal N such that A/N 2 is nilpotent, when A satisfies condition k (as defined in [11]), and when A is an S algebra (using a generalized version of the definition in [10]). We also show several nilpotency conditions which fail to extend from Lie to Leibniz algebras, using cyclic Leibniz algebras as counterexamples. We study these algebras in Section 4, characterizing maximal subalgebras, Frattini subalgebras, Cartan subalgebras and various ideals of these algebras in terms of annihilating polynomials for left multiplication by the generator. Finally, we consider Leibniz algebras whose center or right center is one dimensional and extend non-embedding results from Lie theory ([9], [14]). There are results in [1] for when the left center is one dimensional. In this work we consider only finite dimensional Leibniz algebras. Received July 16, 2012; Revised July 25, Communicated by K. Misra. Address correspondence to Chelsie Batten, Department of Mathematics, North Carolina State University, Raleigh, NC 27695, USA;

3 NILPOTENT LIE AND LEIBNIZ ALGEBRAS PRELIMINARIES Let A be the Frattini subalgebra of the Leibniz algebra A, which is the intersection of all maximal subalgebras of A. As in Lie theory, A is an ideal when the algebra is of characteristic 0 [7], but not generally, even if the algebra is solvable [7], which is counter to the case for solvable Lie algebras. We will consider left Leibniz algebras, following Barnes [5]. Hence a Leibniz algebra is an algebra that satisfies the identity x yz = xy z + y xz. We consider only finite dimensional algebras over a field. Let A be a Leibniz algebra. The center of A will be denoted by Z A and R A will be the right center, a A Aa= 0. We let A 2 = AA and define the lower central series by A j+1 = AA j. It is known that A j is the space of all linear combination of products of j elements no matter how associated. Thus it is often sufficient to consider only left-normed products of elements. A is nilpotent of class t if A t+1 = 0 but A t 0. When A has class t, then A t Z A. 3. EXTENDING LIE NILPOTENCY PROPERTIES In this section we extend several properties equivalent to nilpotency in Lie theory to Leibniz algebras. We show two properties that hold for Leibniz algebras as in the Lie algebra case, one property which fails to extend, and one which holds for Leibniz algebras when we generalize a definition from Lie theory. Theorem 3.1. Let A be a Leibniz algebra and N be a nilpotent ideal of A. Then A is nilpotent if and only if A/N 2 is nilpotent. The proof of this theorem is identical to the Lie algebra case, shown in [10], and we omit it here. The following corollary is a consequence of the proof of Theorem 3.1. Corollary 3.2. Let A be a Leibniz algebra and N be an ideal of A. Suppose that N is nilpotent of class c and A/N 2 is nilpotent of class d + 1. Then A is nilpotent of class at most ( ) c+1 2 d c 2. We say that a Leibniz algebra A satisfies condition k if the only subalgebra K of A with the property K + A 2 is K = A. Theorem 3.3. condition k. Let A be a Leibniz algebra. Then A is nilpotent if and only if A satisfies The proof of this result is the same as in the Lie algebra case [11], and follows from the result of [5] mentioned in the introduction. The following two examples are of concepts that when applied to Leibniz algebras are no longer equivalent to nilpotency. Let S be a subset of the Lie algebra A. The normal closure, S A,ofS is the smallest ideal of A that contains S. A Lie algebra, A, is nilpotent if and only if there is exactly one non-zero nilpotent subalgebra whose normal closure is A [11] (there is a requirement that the dimension of A is large compared to the cardinality of the field). This result fails for Leibniz algebras when the normal closure is defined as above.

4 2406 BATTEN RAY ET AL. Example 3.4. Let A be a Leibniz algebra with basis a a 2, and aa 2 = a 2. H = spn a a 2 is a nilpotent subalgebra and, since Ha is not contained in H, H A = A. H is unique with respect to this property. For any proper subalgebra is of the form J = spn a + a 2, and J is a subalgebra if and only if = 1. Therefore, although A is not nilpotent, the nilpotent subalgebra whose normal closure is A is unique. A Lie algebra is an S algebra if each non-abelian subalgebra H has dim H/H 2 2. A Lie algebra is an S algebra if and only if it is nilpotent [10]. This result does not extend directly to Leibniz algebras. Example 3.5. Let A be a Leibniz algebra with basis a a 2, and aa 2 = 0. Then A is nilpotent, however dim A/A 2 = 1. We can alter the definition of an S algebra to obtain a property equivalent to nilpotency which restricts to the original definition in the Lie algebra case. Define a Leibniz algebra to be an S algebra if every non-abelian subalgebra H has either dim H/H 2 2orH is nilpotent and generated by one element. Theorem 3.6. A Leibniz algebra is an S algebra if and only if it is nilpotent. Lemma 3.7. Let A be a non-abelian nilpotent Leibniz algebra. Then either dim A/A 2 2 or A is generated by one element. Proof. Since A is nilpotent, A 2 = A, the Frattini subalgebra of A. Clearly, dim A/A 2 0 since A is nilpotent. If dim A/A 2 = 1, then A is generated by one element. Otherwise, dim A/A 2 2. Lemma 3.8. If A is not nilpotent but all proper subalgebras of A are nilpotent, then dim A/A 2 1. Proof. Suppose that dim A/A 2 2. Then there exist distinct maximal subalgebras, M and N which contain A 2. Hence M and N are ideals and A = M + N is nilpotent, a contradiction. Proof of Theorem 3.6. If A is nilpotent, then every subalgebra is nilpotent, so A is an S algebra by Lemma 3.7. Conversely, suppose that there exists an S algebra that is not nilpotent. Let A be one of smallest dimension. All proper subalgebras of A are S algebras, hence are nilpotent. Thus dim A/A 2 1 by Lemma 3.8. Since A is an S algebra, it is generated by one element and is nilpotent, a contradiction. 4. CYCLIC LEIBNIZ ALGEBRA In the last section, we found that Leibniz algebras generated by one element provide counterexamples to the extension of several results from Lie to Leibniz algebras. It would seem to be of interest to find properties of these algebras. In this section, we study them in their own right. Let A be a cyclic Leibniz algebra generated by a, and let L a denote left multiplication on A by a. Let a a 2 a n be a basis for A and aa n = 1 a + +

5 NILPOTENT LIE AND LEIBNIZ ALGEBRAS 2407 n a n. The Leibniz identity on a, a 2, and a shows that 1 = 0. Thus A 2 has basis a 2 a n. Let T be the matrix for L a with respect to a a 2 a n. T is the companion matrix for p x = x n n x n 1 2 x = p 1 x n 1 ps x n s, where the p j are the distinct irreducible factors of p x. We will continue using this notation throughout this section. We will show the following theorem. Theorem 4.1. Let A be a cyclic Leibniz algebra generated by a, and notation as in the last paragraph. Then A = b A q L a b = 0 where q x = p 1 x n 1 1 p s x ns 1. Proof. Let A = W 1 W s be the associated primary decomposition of A with respect to L a. Then W j = b A p j L a n j b = 0. Here p x is also the minimal polynomial for L a on A, and therefore, each W j is of the form 0 U j 1 U j nj = W j, where U j i = b A p j L a i b = 0, each U j i+1 /U j i is irreducible under the induced action of L a, and dim U j i = ideg p j x. Since x is a factor of p x, we let p 1 x = x. For j 2, W j A 2 and for i n 1, U 1 i A 2. A 2 is abelian, and left multiplication by b A 2 has L b = 0onA. Hence W j W k = 0 for 1 j k s W j W 1 = 0 for 2 j W 1 W j W j for 1 j s U 1 n1 1 W j = 0 for 2 j Hence each U j i except U 1 n1 = W 1 is an ideal in A. W 1 is generally not a right ideal. Let M j = W 1 U j nj 1 W s. Since dim A/A 2 = 1, A 2 = U 1 n1 1 W 2 W s, and M 1 is a maximal subalgebra of A. We show that M j, j 2, is a maximal subalgebra of A. Since a = b + c, where b W 1 and c A 2, L a = L b.it follows that any subalgebra that contains W 1 is L a invariant. If M is a subalgebra of A that contains M j properly, then M U j nj contains U j nj 1 properly. Since U j nj /U j nj 1 is irreducible in A/U j nj 1, M U j nj = U j nj and M = A. Thus each M j is maximal in A and A M j. Let M be a maximal subalgebra of A. IfM = A 2, then M = b g L a b = 0 where g x = p x /x, som = M 1. Suppose that M A 2. Then A = M + A 2. Hence a = m + c, m M, and c A 2, and L a = L m. Hence M is invariant under L a. Thus the minimum polynomial g x for L a on M divides p x. If there is a polynomial h x properly between g x and p x, then the space H annihilated by h L a is properly between M and A, is invariant under left multiplications by a and by any element in A 2, hence by any element in A so it is a subalgebra. Since the minimum polynomial and characteristic polynomial for L a on A are equal, the same is true of invariant subspaces. Hence dim H = deg h x and H is properly between M and A, a contradiction. Hence M is the space annihilated by g x = p x /p j x for some j and M = M j. Therefore, A = s j=1 M j = b A q L a b = 0 where q x = p 1 x n 1 1 p s x ns 1. As a special case, we obtain Corollary 3 of [12]. Corollary 4.2. A = 0 if and only if p x is the product of distinct prime factors.

6 2408 BATTEN RAY ET AL. Corollary 4.3. The maximal subalgebras of A are precisely the null spaces of r j L a, where r j x = p x /p j x for j = 1 s. Now let A 0 and A 1 be the Fitting null and one components of L a acting on A. Since L a is a derivation of A, A 0 is a subalgebra of A. L a acts nilpotently on A 0 and L b = 0 when b A 2. Therefore, for each c A, L c is nilpotent on A 0, and A 0 is nilpotent by Engel s theorem. Let a = b + c, where b A 0 and c A 1. Then L a = L b since A 1 A 2 yields that L c = 0. Then ba 1 = aa 1 = A 1. For any nonzero x A 1, bx is nonzero in A 1, and x is not in the normalizer of A 0. Hence A 1 N A A 0 = 0 and A 0 = N A A 0. Hence A 0 is a Cartan subalgebra of A. Conversely, let C be a Cartan subalgebra of A and c C. Then c = d + e, d A 0, and e A 1. Since A 1 A 2, A 1 A 1 = A 1 A 0 = 0, A 0 A 1 = A 1, and A 0 A 0 A 0. Therefore, A 1 is an abelian subalgebra of A. Now 0 = L n c c = Ln d c = Ln d d + e = L n d e = Ln c e, where we used that ea = 0 and d A 0 which is nilpotent. Since L c is non-singular on A 1, e = 0 and c = d. Hence C A 0. Since C is a Cartan subalgebra and A 0 is nilpotent, C = A 0, and A 0 is the unique Cartan subalgebra of A. Theorem 4.4. A has a unique Cartan subalgebra. It is the Fitting null component of L a acting on A. Using these same ideas, the following corollary can be shown. Corollary 4.5. The minimal ideals of A are precisely I j = b A p j L a b = 0 for j>1 and, if n 1 > 1, I 1 = b A p 1 L a b = 0 Corollary 4.6. Asoc A = b A u L a b = 0, where u x = p 2 x p s x if n 1 = 1 and u x = p 1 x p s x otherwise. Corollary 4.7. The unique maximal ideal of A is M 1 = b A t L a b = 0, where t x = p x /p 1 x. 5. NON-EMBEDDING Non-abelian Lie algebras with one dimensional centers cannot be embedded as certain ideals in the derived algebra of any nilpotent Lie algebra [9], [14]. In this section, we extend these results to Leibniz algebras. Let A be a Leibniz algebra. Define the upper central series as usual; that is, let Z 1 A = z A za= Az = 0 and inductively, Z j+1 A = z A Azand za Z j A. IfA is an ideal in a Leibniz algebra N, then the terms in the upper central series of A are ideals in N. Suppose that A is nilpotent of dimension greater than one and dim Z 1 A is 1. We will show that A cannot be any N i i 2, for any nilpotent Leibniz algebra N. This is an extension of the Lie algebra result in [9]. Suppose to the contrary that A = N i, where N is nilpotent of class t, and let z be a basis for Z 1 A. For n N, nz = nz z and zn = zn z. If one of these coefficients is not 0, then N is not nilpotent. Hence, Z 1 A Z 1 N. Since N t is an ideal in A, N t Z 1 A. Then, since dim Z 1 A = 1 N t = Z 1 A. Our initial assumptions guarantee that N t 1 A. Hence there exists a y N t 1 A, y Z 1 A, such that yu = yu z and uy = uy z for all u N. Let w also be in N. Then y uw = yu w + u yw = yu zw +

7 NILPOTENT LIE AND LEIBNIZ ALGEBRAS 2409 u yw z = 0. Similarly uw y = 0. Since A N i, it follows that y is in the center of A. Since y and z are linearly independent, this is a contradiction. Hence, we have the following theorem. Theorem 5.1. Let A be a nilpotent non-abelian Leibniz algebra with one-dimensional center. Then A cannot be any N i i 2 for any nilpotent Leibniz algebra N. Let A be a Leibniz algebra. Define R 1 A = r A Ar= 0 and, inductively, R j+1 A = r A Ar R j A. The R j A are left ideals of A. Let B be an ideal in A, and let L be the homomorphism from A into the Lie algebra of derivations of B given by L a = L a, left multiplication of B by a. Let E B A be the image of L. E B A is a Lie algebra, and E B B is an ideal in E B A. For any left ideal, C, of A that is contained in B, let E B A C = E E B A E C = 0. E B A C is an ideal in E B A, and E B A /E B A C is isomorphic to E C A. Theorem 5.2. Let B be a nilpotent Leibniz algebra with dim R 1 B = 1 and dim B 2. Then B is not an ideal of any Leibniz algebra A in which B A. Proof. Suppose that B is an ideal in A that contradicts the theorem. Then E B B = L B L A L A = E B A. Let z 1 z 2 z k be a basis for R 2 B and z k be a basis for R 1 B. Let be the restriction map from E B A to E R 2 B A. Since E B B + E B A R 2 B /E B A R 2 B = E B B / E B A R 2 B E B B = E B B /E B B R 2 B = E R 2 B B, it follows that E R 2 B B = E B B E B A E R 2 B A. Now we show that E R 2 B B is not contained in E R 2 B A by showing that E R 2 B B is complemented by a subalgebra in E R 2 B A. For i = 1 k, let e i z j = ij z k for j = i k, where ij is the Kronecker delta. Let S = spn e 1 e k 1. We claim that S = E R 2 B B. Since BR 2 B R 1 B, it follows that E R 2 B B S. To show equality, we show dim E R 2 B B = k 1 = dim S. For x B, L x induces a linear functional on R 2 B ; that is, E R 2 B B is contained in the dual of R 2 B. Hence dim E R 2 B B = dim R 2 B dim R 2 B B, where R 2 B B = z R 2 B L x z = 0 for all x B = R 1 B. Hence dim E R 2 B B = k 1 = dim S, and S = E R 2 B B. We now show that S is complemented in E R 2 B A. Let M = E E R 2 B A E z i = k 1 j=1 ijz j ij F i = 1 k 1. M is a subalgebra of E R 2 B A and M S = 0. We claim that M + S = E R 2 B A. Let E E R 2 B A. Then E z i = k 1 j=1 ijz j + ik z k for i = 1 k 1 and E z k = k z k. Now E = E k 1 i=1 i ke i + k 1 i=1 i ke i M + S. Therefore, E R 2 B A = M + S. Suppose that M = 0. Then E R 2 B A = E R 2 B B, which contradicts E R 2 B B E R 2 B A. Thus, M 0. Hence, S is complemented in E R 2 B A, which contradicts S E R 2 B A. This contradiction establishes the result. ACKNOWLEDGMENTS This work was completed as part of a Research Experience for Graduate Students at North Carolina State University, supported by the National Science

8 2410 BATTEN RAY ET AL. Foundation. The authors would like to thank Professor E. L. Stitzinger for his guidance and support. The work of the first three authors was supported by NSF grant DMS REFERENCES [1] Albeverio, S., Ayupov, A., Omirov, B. (2005). On nilpotent and simple Leibniz algebras. Comm. Algebra 33: [2] Ayupov, S. A., Omirov, B. A. (1998). On Leibniz algebras. Algebra and Operator Theory. Proceedings of the Colloquium in Tashkent, Kluwer. [3] Barnes, D., Lattices of subalgebras of Leibniz algebras. Comm. Algebra To appear. arxiv: [4] Barnes, D. On Levi s theorem for Leibniz algebras. Bull. Austral. Math. Soc. To appear. arxiv: [5] Barnes, D. (2011). Some theorems on Leibniz algebras. Comm. Algebra 39: [6] Barnes, D. (2012). On Engel s theorem for Leibniz algebras. Comm. Algebra 40: [7] Batten, C., Bosko-Dunbar, L., Hedges, A., Hird, J. T., Stagg, K., Stitzinger, E. A Frattini theory for Leibniz algebras. Comm. Algebra To appear. arxiv: [8] Bosko, L., Hedges, A., Hird, J. T., Schwartz, N., Stagg, K. (2011). Jacobson s refinement of Engel s theorem for Leibniz algebras. Involve. 3: [9] Chao, C.-Y. (1967). A nonimbedding theorem of nilpotent Lie algebras. Pacific J. Math. (22) 2: [10] Chao, C.-Y. (1968). Some characterizations of nilpotent Lie algebras. Math Z. 103: [11] Chao, C.-Y., Stitzinger, E. (1976). On nilpotent Lie algebras. Arch. Math. 27: [12] Hedges, A., Stitzinger, E. More on the Frattini subalgebra of a Leibniz algebra. arxiv: [13] Patsourakos, A. (2007). On nilpotent properties of Leibniz algebras. Comm. Algebra. 35: [14] Stitzinger, E. (1970). On the Frattini subalgebra of a Lie algebra. J. London Math. Soc. 2:

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