Lie metabelian restricted universal enveloping algebras - Springer Link

Arch. Math. 84 (2005) 398–405 0003–889X/05/050398–08 DOI 10.1007/s00013-005-1315-0 © Birkh¨auser Verlag, Basel, 2005

Archiv der Mathematik

Lie metabelian restricted universal enveloping algebras By Salvatore Siciliano and Ernesto Spinelli

Abstract. We characterize the restricted Lie algebras L whose restricted universal enveloping algebra u(L) is Lie metabelian. Moreover, we show that the last condition is equivalent to u(L) being strongly Lie metabelian.

1. Introduction. Let R be an associative algebra with a unit over a field F . R can be regarded as a Lie algebra via the Lie commutator [x, y] = xy − yx for every x, y ∈ R. The Lie derived series of R is defined by induction as follows: δ [0] (R) = R and δ [n+1] (R) is the additive subgroup generated by all the Lie commutators [x, y] with x, y ∈ δ [n] (R) (the term δ [1] (R) is also denoted by R ). On the other hand, the strong Lie derived series of R is defined by setting δ (0) (R) = R and δ (n+1) (R) as the associative ideal generated by all Lie commutators [r, s], where r, s ∈ δ (n) (R). R is said to be Lie solvable (resp. strongly Lie solvable) if δ [n] (R) = 0 (δ (n) (R) = 0) for some n. In particular, R is said to be Lie metabelian (resp. strongly Lie metabelian) if δ [2] (R) = 0 (δ (2) (R) = 0). Clearly, strong Lie solvability implies Lie solvability of R, but the converse is in general not true. Let u(L) be the restricted universal enveloping algebra of a restricted Lie algebra L with p-map [p] over a field F of characteristic p > 0. In [2], D. Riley and A. Shalev investigated the Lie structure of u(L) by giving sufficient and necessary conditions in order that u(L) is Lie solvable under the assumption that p > 2. It should be noted that, while the Lie nilpotency index of u(L) can be computed using rather satisfactory methods (cf. [3]), it seems that there are no results about Lie solvable restricted universal enveloping algebras and their derived lengths. The aim of this paper is to characterize the restricted Lie algebras L whose restricted universal enveloping algebra is Lie metabelian by proving the following Theorem 1. Let L be a nonabelian restricted Lie algebra over a field F of characteristic p > 0. Then the following three conditions are equivalent: Mathematics Subject Classification (2000): 16S30, 17B50, 17B60. The authors were partially supported by the National Research Project “Group Theory and Application”.

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1) u(L) is strongly Lie metabelian; 2) u(L) is Lie metabelian; 3) p and L satisfy one of the following conditions: (i) p = 3, dimF L = 1, L is central and L [p] = 0; (ii) p = 2, dimF L = 2, L is central and L [p] = 0; (iii) p = 2, dimF L = 1 and L [p] = 0 where L [p] = {x [p] |x ∈ L }. As remarked in the previously mentioned paper [3], there exists a strong analogy between the Lie theory of modular group rings and the Lie theory of restricted universal enveloping algebras. Actually, many results on the former are shared by the latter. Surprisingly, Theorem 1 does not continue such an analogy. In fact, in [1] F. Levin and G. Rosenberger proved that a nonabelian group ring F G is Lie metabelian if and only if either p = 3 and the commutator subgroup G is central of order 3 or p = 2 and G is central elementary abelian of an order dividing 4. In particular, the condition that FG is Lie metabelian forces G to be nilpotent (also, FG is forced to be Lie nilpotent of class at most 3: see Lemma 6 of [1]). On the contrary, the statement of our main result shows that if u(L) is Lie metabelian then the Lie algebra L need not be nilpotent. The rest of the paper covers the proof of Theorem 1. In the first section we state the necessary conditions in order that u(L) is Lie metabelian. In the final section we complete the characterization by studying the second term δ (2) (u(L)) of the strong Lie derived series of u(L). 2. Necessary conditions for Lie metabelianity. In this section we prove the most difficult part of the theorem by means of several intermediate results. Throughout, L denotes a nonabelian restricted Lie algebra over a field of positive characteristic. A powerful tool in the proofs will be the well-known PBW Theorem (for restricted Lie algebras): we refer the reader to [4, Chapter 2, Theorem 5.1]. The following assertion is stated in [1]: Lemma 1. Let R be a Lie metabelian ring and a, b ∈ R such that [a, b] = 0. Then the following identities hold: 1) [[r, s], a]t[b, d] = 0 and [b, d]t[[r, s], a] = 0 for any r, s, t, d ∈ R; 2) [a, t][b, r][s, d] = 0 for any r, s, t, d ∈ R provided [r, s] = 0; 3) [a, r][b, s][t, d] = 0 for any r, s, t, d ∈ R provided [r, s] = 0. The Lie metabelianity of u(L) requires a strong limitation on the characteristic of the ground field. In fact, the following holds: Lemma 2. Let L be a restricted Lie algebra over a field F of characteristic p > 0. If u(L) is Lie metabelian, then p 3 and L [p] = 0.

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P r o o f. Let x and y be two noncommutating elements of L. According to Lemma 1(3), in u(L) we have that (1)

[x, y]3 = 0.

In view of the PBW Theorem, we have necessarily p 3. If p = 3, the relation (1) gives immediately [x, y][p] = 0. Let now p = 2 and suppose that [x, y]2 = 0. Let B be a basis for L containing [x, y]. Then there are x1 , x2 , . . . , xn ∈ B and λ0 , λ1 , . . . , λn ∈ F such that (2)

[x, y][p] =

n

λi x i ,

i=0

where we have put x0 = [x, y]. By (1) and (2), in u(L) we have that 0 = [x, y][x, y][p] =

n

λ i x0 xi =

n i=0

i=0

λ0 λi x i +

n

λi x0 xi .

i=1

The PBW Theorem yields λ0 = λ1 = . . . = λn = 0, hence also in this case we have that [x, y][p] = 0. m Finally, let z ∈ L . Then z = αi [vi , wi ] for some α1 , α2 , . . . αm ∈ F and v1 , w1 , . . . , i=1

vm , wm ∈ L. Since L is abelian, the p-map [p] acts p-semilinearly on it. Consequently, [p] m m p [p] z = αi [vi , wi ][p] = 0 αi [vi , wi ] = i=1

and the statement is proved.

i=1

Note that, when u(L) is Lie metabelian, by the previous result, L trivially turns out to be a restricted Lie subalgebra of L. We proceed by considering, separately, the p = 2 and p = 3 cases. Proposition 1. Let L be a restricted Lie algebra over a field F of characteristic 3. If u(L) is Lie metabelian, then L is central and dimF L = 1. P r o o f. It is enough to prove that [[x, y], z] = 0 for any x, y, z ∈ L. Suppose [x, y] = 0, being the claim trivial when x and y commute. According to Lemma 1(1), we have that [[x, z], y][x, y] = 0 and [[z, y], x][x, y] = 0. By the Jacobi identity, it follows that (3)

[[x, y], z][x, y] = 0.

Consequently, the PBW Theorem forces [[x, y], z] = k[x, y] for some k ∈ F ; hence the relation (3) becomes k[x, y]2 = 0. Again, by the PBW Theorem, we have necessarily k = 0, and the first part of the proof follows.

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Finally, suppose by contradiction that dimF L > 1. Then there exist a, b, c, d ∈ L such that [a, b] and [c, d] are linearly independent. By Lemma 1(3) we have that [a, b]2 [c, d] = 0, contrary to the PBW Theorem.

When char F = 2, the Lie metabelianity of u(L) does not imply the centrality of L . Such a phenomenon appears in the following result: Lemma 3. Let L be a restricted Lie algebra over a field F of characteristic 2. If u(L) is Lie metabelian, then either L is central or L is nonnilpotent and dimF [L , L] = 1. P r o o f. Assume that L is not central and let x, y, z ∈ L such that [[x, y], z] = 0. As in the proof of Proposition 1, by Lemma 1(1) we have that (4)

[[x, y], z][x, y] = 0.

It is obvious that [x, y] = 0, so that the PBW Theorem forces (5)

[[x, y], z] = k[x, y]

for some k ∈ F , k = 0. As a consequence, L cannot be nilpotent. It remains to show that dimF [L , L] = 1. Suppose, if possible, the contrary. Then there exist a, b, c ∈ L such that [[a, b], c] and [[x, y], z] are linearly independent. For what already observed, one must have (6)

[[a, b], c] = h[a, b]

for some h ∈ F , h = 0. By Lemma 1(1), we have h[a, b][c, z] = [[a, b]c][c, z] = 0. By the PBW Theorem, it follows that [c, z] is proportional to [a, b]. In a similar fashion, we get that [c, z] is also proportional to [x, y]. Therefore, in view of (5) and (6), the linear independence of [[a, b], c] and [[x, y], z] forces [c, z] = 0. According to Lemma 1(1), it follows that [[a, b], c][z, [x, y]] = 0. Such a relation contradicts the PBW Theorem, and the proof is complete.

Let R be an associative algebra. Then, for every a, b, c ∈ A, one has: (7)

[a, b][c, d] + [a, d][c, b] = [[ac, b], d] − [[a, b], d]c − a[[c, b], d].

The proof of such an identity can be easily obtained by a straightforward computation.

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Proposition 2. Let L be a restricted Lie algebra over a field F of characteristic 2. If u(L) is Lie metabelian and L is central, then dimF L 2. P r o o f. Assume, if possible, dimF L > 2. Then there exist a, b, c, d, e, f ∈ L such that [a, b], [c, d] and [e, f ] are linearly independent. We claim that, for every x ∈ {a, b}, y ∈ {c, d} and z ∈ {e, f }, the elements [x, y] and [x, z] are linearly dependent. In order to show this fact, suppose the contrary. Denote by V the vector space generated by [x, y] and [x, z]. As u(L) is Lie metabelian, by Lemma 1(2) the following identities hold: [x, y][x, z][c, d] = 0; [x, y][x, z][e, f ] = 0. By the PBW Theorem, these relations imply that [c, d] and [e, f ] are in V . Because of centrality of L , (7) and Lemma 1(1) yield [x, y][x, z][a, b] = [[xz, y], x][a, b] = 0 and so, by the PBW Theorem, we have also that [a, b] ∈ V . As dimF V = 2, this is a contradiction. A similar argument shows also that [x, y] and [y, z] are linearly dependent and, again, the same property holds for [x, z] and [y, z]. In particular, we have that either [a, c] = 0 or [a, e] = 0 or [a, c] and [a, e] are nonzero and [a, c] = λ[a, e] for some λ ∈ F . Let us consider these cases. Suppose first [a, c] = 0. By (7) and Lemma 1(1) we have that (8)

[a, b][c, d][e, f ] = [a, b][f, c][e, d].

Moreover, we have proved above that [b, e] and [d, e] are linearly dependent. Now, if [b, e] = 0, by Lemma 1(2) we have that [a, b][c, d][e, f ] = 0 which is impossible for the PBW Theorem. On the other hand, if [b, e] = 0, by linear dependence, we have that [d, e] = η[b, e] for some η ∈ F . By (8) and Lemma 1(2), it follows again [a, b][c, d][e, f ] = 0 which is impossible. The [a, e] = 0 case is analogous to the [a, c] = 0 case. Finally, assume that [a, c] and [a, e] are nonzero and [a, e] = λ[a, c] for some λ ∈ F . Then, Lemma 1(2) yields [a, c][a, b][c, d] = 0; [a, e][a, b][e, f ] = 0. By the PBW Theorem, these relations force [a, c] ∈ span[a, b], [c, d]F = V1 and [a, e] ∈ span[a, b], [e, f ]F = V2 . As [a, e] and [a, c] are proportional, it is obvious

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that [a, e] ∈ V1 . Furthermore, we have previously proved that [a, e] and [c, e] are linearly dependent. Now, if [c, e] = 0, proceeding as in the [a, c] = 0 case, we obtain [a, b][c, d][e, f ] = 0 which is not possible. On the other hand, if [c, e] = 0, then there exists µ ∈ F such that [a, e] = µ[c, e]. By Lemma 1(2) it follows that [a, e][c, d][e, f ] = µ[c, e][c, d][e, f ] = 0 and so, by the PBW Theorem, we have [a, e] ∈ span[c, d], [e, f ]F = V3 . In conclusion, we have proved that [a, e] ∈ V1 ∩ V2 ∩ V3 = {0}. This contradicts our assumption, and the proof is complete.

Proposition 3. Let L be a restricted Lie algebra over a field F of characteristic 2. If u(L) is Lie metabelian and L is not central, then dimF L = 1. P r o o f. Assume, if possible, dimF L > 1. Let a, b, c ∈ L such that [[a, b], c] = 0. By (4) we have that [[a, b], c][a, b] = 0. Therefore, the PBW Theorem implies that [[a, b], c] = k[a, b] for some k ∈ F , k = 0. Put x = [a, b] and y = k −1 c. In such a way, we have [x, y] = x. In view of Lemma 3 one has dimF [L, L ] = 1, therefore we have necessarily (9)

[L, L ] = F x.

By assumption, there are v, w ∈ L such that x and z = [v, w] are linearly independent. According to Lemma 1(1) we have (10)

[x, v]z = [[x, y], v][v, w] = 0.

On the other hand, by (9) it follows that [[x, y], v] = hx for some h ∈ F . The PBW Theorem implies that xz = 0, so that for (10) we have h = 0 and then [x, v] = 0. In a similar fashion, from [x, w]z = [[x, y], w][v, w] = 0 we see that [x, w] = 0. In particular, the Jacobi identity yields [x, z] = [x, [v, w]] = [[x, v], w] + [v, [x, w]] = 0. Moreover, by Lemma 1(1), it follows from [[x, y], y][v, w] = xz = 0

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that y and w do not commute. By the PBW Theorem, the relation x[y, w] = [[x, y], y][y, w] = 0 forces [y, w] = µx for some µ ∈ F , µ = 0. By standard calculations we have that [yv, w] = [y, w]v + y[v, w] = µxv + yz. Since u(L) is Lie metabelian, the combination of the above relations yields 0 = [[x, y], [yv, w]] = [x, µxv + yz] = µ[x, x]v + µx[x, v] + [x, y]z + y[x, z] = xz. As x and z are linearly independent, such a relation contradicts the PBW Theorem and the claim is proved. 3. Proof of the main theorem. Let L be a restricted Lie algebra over a field of characteristic p > 0. We denote by ω(L) the augmentation ideal of u(L), that is, the associative ideal generated by L in u(L). It is well known that, for every restricted ideal I of L, the kernel of the canonical map φ : u(L) −→ u(L/I ) is given by ω(I )u(L). Suppose now that u(L) is Lie metabelian. In this case, we have seen in the previous section that L is a restricted ideal of L. Moreover, as u(L/L ) is abelian, it follows that (11)

[u(L), u(L)]u(L) = δ (1) (u(L)) ω(L )u(L).

We are now in position to prove our main result: P r o o f o f T h e o r e m 1. Clearly, 1) implies 2). On the other hand, it was proved in Section 2 that 2) implies 3). Thus, in order to complete the proof, it remains only to show that 3) implies 1). In view of (11), by standard calculation we obtain δ (2) (u(L)) [ω(L )u(L), ω(L )u(L)]u(L) = ω(L )[u(L), ω(L )u(L)]u(L) + [ω(L ), ω(L )u(L)]u(L) = ω(L )2 [u(L), u(L)]u(L) + ω(L )[u(L), ω(L )]u(L) + [ω(L ), ω(L )]u(L). Since ω(L ) is abelian, by using (11), it follows that (12)

δ (2) (u(L)) ω(L )3 u(L) + ω(L )[u(L), ω(L )]u(L).

We proceed by considering separately each case.

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Suppose first that the condition (i) of the statement holds. Then we have ω(L ) = F x ⊕ F x 2 for some x ∈ L . As L [p] = 0, it follows that ω(L )3 = 0. Moreover, the centrality of L in L forces [u(L), ω(L )] = 0 and so, by (12), we conclude that δ (2) (u(L)) = 0. Assume now the validity of (ii). In this case we have ω(L ) = F x ⊕ F y ⊕ F xy for some x, y ∈ L . Since L [p] = 0, we obtain that ω(L )3 = 0. Because of the centrality of L , the claim follows at once by (12). Finally, assume that the condition (iii) holds. Then there exists x ∈ L such that ω(L ) = F x and so, as L [p] = 0, we have that ω(L )2 = 0. By (11) and (12), it follows that δ (2) (u(L)) ω(L )[u(L), ω(L )]u(L) ω(L )δ (1) (u(L)) ω(L )2 u(L) = 0 and the proof is complete.

We conclude this section by providing the simplest example of a nonnilpotent restricted Lie algebra L such that u(L) is Lie metabelian. E x a m p l e. Let L be the nonabelian 2-dimensional Lie algebra over a field F of characteristic p = 2. Then there is a basis {x, y} for L such that [x, y] = y. Consider L as a restricted Lie algebra via the p-map [p] defined by the conditions: x [p] = x;

y [p] = 0.

By Theorem 1 we have that u(L) is Lie metabelian while L is not nilpotent. References [1] F. Levin and G. Rosenberger, Lie metabelian group rings, Group and Semigroup Rings, 153–161, Amsterdam 1986. [2] D. M. Riley and A. Shalev, The Lie structure of enveloping algebras. J. Algebra 162, 46–61 (1993). [3] D. M. Riley and A. Shalev, Restricted Lie algebras and their envelopes. Canad. J. Math. 47, 146–164 (1995). [4] H. Strade and R. Farnsteiner, Modular Lie algebras and their representations. New York 1988. Received: 1 October 2004 S. Siciliano and E. Spinelli Dipartimento di Matematica “E. De Giorgi” Universit`a degli Studi di Lecce Via Provinciale Lecce-Arnesano I-73100-Lecce Italy [email protected] [email protected]