POISSON PI ALGEBRAS 1. Introduction

Home , Poisson algebra

TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 359, Number 10, October 2007, Pages 4669–4694 S 0002-9947(07)04008-1 Article electronically published on May 1, 2007

POISSON PI ALGEBRAS

S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

Abstract. We study Poisson algebras satisfying polynomial identities. In particular, such algebras satisfy “customary” identities (Farkas, 1998, 1999) Our main result is that the growth of the corresponding codimensions of a Pois- son algebra with a nontrivial identity is exponential, with an integer exponent. We apply this result to prove that the tensor product of Poisson PI algebras is a PI-algebra. We also determine the growth of the Poisson-Grassmann algebra and of the Hamiltonian algebras H2k.

1. Introduction: Poisson algebras Throughout the paper K denotes a field of characteristic zero. A vector space A is called a Poisson algebra provided that, beside addition, it has two K-bilinear operations which are related by derivation. First, with respect to multiplication, A is a commutative associative algebra with unit; denote the multiplication by a·b (or ab), where a, b ∈ A. Second, A is a Lie algebra; traditionally here the Lie operation is denoted by the Poisson brackets {a, b},wherea, b ∈ A. It is also assumed that these two operations are connected by the Leibnitz rule {a · b, c} = a ·{b, c} + b ·{a, c}, a,b,c∈ A. Poisson algebras arise naturally in different areas of algebra and topology. Let us give two examples. First, consider the commutative and associative polynomial ring K[X, Y ]. It is turned into a Poisson algebra if we introduce the Poisson bracket as follows: ∂f ∂g ∂f ∂g {f,g} = − ,f,g∈ K[X, Y ]. ∂X ∂Y ∂Y ∂X This is the Hamiltonian algebra H2; see Section 5. Second, given a Lie algebra L, there is a standard way of constructing a corresponding Poisson algebra P = P (L) as follows. As a vector space P coin- cides with S(L), the symmetric algebra of L.NotethatS(L)isidentifiedwith the commutative polynomial ring K[v1,v2,...], where v1,v2,... is a linear basis of L over the field K. Now define the Poisson brackets {, } on P as follows: {vi,vj} =[vi,vj]inL, and extend by linearity and by derivations to all of P . For example, {vi · vj ,vk} = vi ·{vj,vk} + vj ·{vi,vk}, etc. In this way we construct thefreePoissonalgebraonthesetX; see Section 2. We remark that P = P (L) can also be defined via the universal enveloping algebra U(L); see [5]. A Poisson algebra A is PI (i.e. satisfies a polynomial identity) if there exists a non zero polynomial in the free Poisson algebra which vanishes under any substitution

Received by the editors August 23, 2004 and, in revised form, February 22, 2005. 2000 Mathematics Subject Classiﬁcation. Primary 17B63, 17B01, 16P90, 16R10. This research was partially supported by Grant RFBR-04-01-00739.

c 2007 American Mathematical Society 4669

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4670 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

in A. Note that many Poisson algebras are PI. For example, any Poisson algebra which is ﬁnitely generated as an associative algebra is PI. The goal of this paper is to study such PI algebras. Similar to the associative and to the Lie case, our main tools here are combinatorial, mainly applying the representations of the symmetric group Sn. We therefore assume that the characteristic of the base ﬁeld K is zero. Also, some special type of polynomials, customary, play a major role here; see below. Some of the main features of the combinatorial PI theory in the associative case are: P1. The exponential bound on the codimensions, and the integrality of that exponent. P2. The cocharacters lie in a hook. Note that these properties no longer hold in the Lie PI case. The main results of this paper are that the Poisson PI algebras do satisfy these properties P1 and P2, as well as their consequences. Let x1,x2,... be associative and noncommutative variables. The polynomial σ sn = sn(x1,...,xn)= (−1) xσ(1) ···xσ(n)

σ∈Sn is called the standard polynomial (of degree n), where (−1)σ is the sign of σ.It plays an important role in the theory of associative PI algebras. For example, the celebrated Amitsur-Levitsky Theorem says that Mk(R), the k × k matrices over a commutative ring R, satisﬁes the identity s2k ≡ 0. Furthermore, by a well-known Amitsur’s theorem, any associative PI algebra satisﬁes some ’s power of some k-th standard polynomial (sk) ≡ 0. In the theory of Poisson PI algebras, the analogue of the standard polynomial is σ (−1) {xσ(1),xσ(2)}···{xσ(2n−1),xσ(2n)},

σ∈S2n called a standard customary polynomial, where { , } are the Poisson brackets. { }···{ } The products xi1 ,xi2 xi2n−1 ,xi2n are called customary monomials, and their linear combinations – customary polynomial. These polynomials where introduced by D. Farkas [8], [9]. In this paper we study in detail the customary identities satisﬁed by a Poisson PI algebra. This is done by studying the intersection of the ideal of the identities of a given Poisson PI algebra - with the (free) space of the multilinear customary polynomials in 2n variables. In the free case, this space of multilinear customary polynomials of degree 2n has an over–exponential growth, more or less like n!Our main result asserts that when we mod-out the identities, the growth of the customary codimensions of a Poisson PI algebra is exponential – and with an integer exponent (Theorem 8.1). The exponential bound is a “Poisson” analogue of the corresponding result for associative PI algebras [17]. The integrality of the exponential growth is a “Poisson” analogue of the corresponding result of A. Giambruno and M. Zaicev on the integrality of the exponent for associative algebras [10]. In Section 9 we apply Theorem 8.1 to prove that the tensor product of PI Poisson algebras is again a PI-algebra. In Section 5 we study in detail the customary identities of the Poisson algebras H2k. As Lie algebras, these algebras are one-dimensional central extensions of the

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POISSON PI ALGEBRAS 4671

simple inﬁnite dimensional Lie algebras of the Hamiltonian series H2k. We refer to H2k as Hamiltonian Poisson algebras as well. In Section 6 we construct the Grassmann-Poisson algebra G and study in detail all its polynomial identities. This algebra G is an analogue of the associative Grassmann algebra. The authors are grateful to A.Guterman for attracting their attention to this subject.

2. Customary polynomials The main objective of this section is the study of the customary polynomial identities. These polynomials and their importance in the Poisson PI theory are discussed below. Recall from Section 1 the construction of the Poisson algebra P = P (L)ofa Lie algebra L. In the special case when L = L(X) is the free Lie algebra on the set of variables X, this yields P (L(X)) = F (X), the free Poisson algebra on the variables X [18]. Let us describe its basis. We consider the Hall basis family ∞ R(X)= n=1 Rn(X)ofL(X), where the elements of Rn are Lie elements of degree n in X [2]. In particular, R1 = {xi | i ∈ I}, R2 = {[xi,xj] | i

where all Xi, i =1,...,n, appear exactly once, so in particular αi,βij,γv ∈{0, 1}. Also consider the subspace Q2n ⊆ P2n spanned by the elements { }·{ }···{ } xα1 ,xα2 xα3 ,xα4 xα2n−1 ,xα2n .

Following D. Farkas [8], we call the elements of Q2n customary polynomials. The following spanning set is a canonical basis for Q2n:

(1) Q2n = {xτ(1),xτ(2)}·{xτ(3),xτ(4)}···{xτ(2n−1),xτ(2n)}|τ ∈ S2n, τ(1) <τ(2),τ(3) <τ(4),...,τ(2n − 1) <τ(2n),

τ(1) <τ(3) < ···<τ(2n − 1)K .

Denote by T2n the set of the permutations of S2n that satisfy the above conditions. Note that |T2n| is equal to the number of partitions of the set {1, 2,...,2n} into pairs; this number will be denoted by h(2n)=|T2n|.Itiswellknownthat (2n)! |T | = =1· 3 · 5 ···(2n − 1). 2n 2n · n!

Moreover, T2n can be chosen as a transversal of the left cosets of the wreath-product Z2 Sn in S2n.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4672 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

In addition, we deﬁne Rn ⊆ Pn to be the subspace spanned by products of customary monomials and singletons: { }·{ }···{ }· ··· xα1 ,xα2 xα3 ,xα4 xα2m−1 ,xα2m xα2m+1 xαn . We call such elements extended customary polynomials. The following spanning set is a basis of Rn:

(2) Rn = {xτ(1),xτ(2)}·{xτ(3),xτ(4)}···{xτ(2m−1),xτ(2m)}·xτ(2m+1) ···xτ(n) |

τ ∈ Sn, 0 ≤ 2m ≤ n, τ(1) <τ(2),τ(3) <τ(4),...,τ(2m − 1) <τ(2m), τ(1) <τ(3) < ···<τ(2m − 1),

τ(2m +1)<τ(2m +2)< ···<τ(n)K.

Note that the inclusion Rn ⊂ Pn is strict; for example, {{x1,x2},x3} belongs to P3 but not to R3. The deﬁnition of a Poisson PI algebra is standard, the identities being elements ofthefreePoissonalgebraF (X) of countable rank. The collection of Poisson algebras satisfying the same set of identities is called a variety. We assume that the basic facts about varieties of PI algebras are known to the reader; see e.g. [7] and [2]. The importance of customary polynomials is explained by the following fact, which was discovered by D. Farkas. Theorem 2.1 ([8]). Suppose that V is a nontrivial variety of Poisson algebras. Then it satisﬁes a nontrivial customary identity: ατ {xτ(1),xτ(2)}·{xτ(3),xτ(4)}···{xτ(2n−1),xτ(2n)}≡0.

τ∈T2n 3. Codimensions

Denote X = {xi | i ∈ N}.LetV be a variety of Poisson algebras, the polynomial identities being some elements of F (X). Denote by F (V,X) the relatively free algebra of V generated by X. This induces the natural epimorphism π : F (X) → F (V,X). Denote by Q2n(V,X), Rn(V,X), and Pn(V,X) the natural images of Q2n, Rn,andPn, respectively. In what follows, the standard generating set X may be omitted and we simply write, for example, Q2n(V), etc. This yields the following three sequences of codimensions:

• q2m(V)=dimQ2m(V) (customary codimensions), • rn(V)=dimRn(V) (extended customary codimensions), • pn(V)=dimPn(V) (regular codimensions). Deﬁne the upper and lower exponents for these sequences as follows: Q 2n Q 2n Exp V = lim sup q2n(V), Exp V = lim inf q2n(V), →∞ n→∞ n R Exp V = lim sup n r (V), ExpR V = lim inf n r (V), n →∞ n n→∞ n n n Exp V = lim sup pn(V), Exp V = lim inf pn(V). n→∞ n→∞ These limits always exist (of course they might be inﬁnite). If the upper and the lower limits coincide, we use the notations ExpQ V,ExpR V, and Exp V,the

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POISSON PI ALGEBRAS 4673

corresponding exponents of the variety V. By an exponent of a Poisson algebra A denote the corresponding exponent of the variety V =var(A) generated by this algebra.

Lemma 3.1. Let V be a variety of Poisson algebras. Then the sequences q2m(V) and rn(V) are related by the following formula: n r (V)= · q (V),n∈ N. n 2m 2m 0≤2m≤n

Proof. Fix Xn = {x1,...,xn}.Let2m ≤ n,andletY = Y2m ⊆ Xn be a subset of cardinality |Y2m| =2m.Denoteby{v2m,α(Y ) | 1 ≤ α ≤ q2m(V)} a basis of V \ { } ··· Q2m( ,Y). Also, if X Y = xi1 ,...,xin−2m ,thendenoteZY = xi1 xin−2m . We prove that the elements in the set

(3) {v2m,α(Y ) · ZY | Y ⊆ X, |Y | =2m, 0 ≤ 2m ≤ n, 1 ≤ α ≤ q2m(V)}

form a basis of Rn(V,Xn). It is obvious that these elements span Rn(V,Xn), so we prove they are linearly independent. Assume, by way of contradiction, that (4) β2m,α,Y (v2m,α(Y ) · ZY )=0, 2m,α,Y

and that some β2m,α,Y =0.Letm0 be minimal with β2m0,α,Y¯ =0,forsome Y¯ = Y2m . Substituting xi → 1 in (4) for all the variables in Xn \ Y¯ ,weobtain 0 β v (Y¯ )=0, 2m0,α,Y 2m0,α α a contradiction.

Next, deﬁne the following two exponential generating functions: ∞ q (V) CQ(V,z)= 2m z2m; (2m)! m=0 ∞ r (V) CR(V,z)= n zn. n! n=0 Simple manipulations with these series imply: Corollary 3.1. CR(V,z)=CQ(V,z) · ez. The following formula is the “Poisson analogue” of the relations between the proper and the ordinary codimensions of associative PI algebras [3], [7], and [15]. Corollary 3.2. Assume that ExpQ(V) exists. Then ExpR(V) = ExpQ(V)+1.

Q Proof. Let Exp (V)=. Then by Lemma 3.1 it is suﬃcient to show that n 2m 1 n 0≤2m≤n 2m behaves asymptotically like 2 ( +1) . This clearly follows since >0and n 1 · 2m = · ((1 − )n +(1+)n) . 2m 2 0≤2m≤n

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4674 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

4. Cocharacters and the action of the symmetric group Following D. Farkas [8], [9], we deﬁne the customary standard polynomial as follows. Let n =2m. Recall from Section 2 the deﬁnition of the set T2n ⊆ S2n. Then the customary standard polynomial is τ St2m =St2m(x1,...,x2m)= (−1) {xτ(1),xτ(2)}···{xτ(2m−1),xτ(2m)}.

τ∈T2m

Clearly, St2m(x1,...,x2m) is customary. Recall the notation h(2m)=|T2m| = (2m)! m . The “full” standard polynomial is deﬁned as m!2 π St2m(x1,...,x2m)= (−1) {xπ(1),xπ(2)}···{xπ(2m−1),xπ(2m)}.

π∈S2m It is easy to see that (2m)! St (x ,...,x )= St (x ,...,x )=2mm!St (x ,...,x ). 2m 1 2m h(2m) 2m 1 2m 2m 1 2m Lemma 4.1. Let n =2m be even. Then

St2m+2(x1,...,x2m+2)=St2m(x1,...,x2m) ·{x2m+1,x2m+2} 2m − St2m(x1,...,xj−1,x2m+1,xj+1,...,x2m) ·{xj,x2m+2}. j=1 Proof. Here is a hint of the proof. Note that the number of customary monomials

{xπ(1),xπ(2)}···{xπ(2m+1),xπ(2m+2)} on both sides is the same - with no cancellations on either side. The minus signs on the right are caused by switching xj and x2m+1.

By deﬁnition, St2 = {x1,x2}. An example of the previous lemma is

St4 = {x1,x2}·{x3,x4}−{x3,x2}·{x1,x4}−{x1,x3}·{x2,x4}. In the odd case n =2m+1, the customary standard polynomial is an “extended” polynomial, deﬁned as 2m+1 j+1 St2m+1 =St2m+1(x1,...,x2m+1)= (−1) St2m(x1,...,xˆj,...,x2m+1) · xj. j=1 The “full” customary standard polynomial is π St2m+1(x1,...,x2m+1)= (−1) {xπ(1),xπ(2)}···{xπ(2m−1),xπ(2m)}

π∈S2m+1

· xπ(2m+1) m =2 m!St2m+1(x1,...,x2m+1).

Cocharacters. Let σ ∈ Sn. Then the mapping σ(xi)=xσ(i) is extended to an automorphism of the free Poisson algebra. The spaces Pn(V)andRn(V)areSn- modules, and similarly Q2m(V)isanS2m-module. This deﬁnes the corresponding P V R V Q V sequences of cocharacters χn ( ), χn ( ), and χ2m( ), which are characters of the corresponding symmetric groups.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POISSON PI ALGEBRAS 4675

Let χn be an Sn character and let λ n be a partition (or a Young diagram). As usual, we say that λ appears in χn if the corresponding irreducible character χλ λ (also denoted χ in the literature) appears in χn with a nonzero multiplicity. Let λ n and consider the largest subdiagram µ ⊂ λ with even columns (i.e. with columns of even length). Then µ can be calculated as follows. Let λ be the t conjugate partition of λ,andwriteλ =(2n1 + 1,...,2ns + s, 1 )where i ∈{0, 1} and ns > 0. Then µ =(2n1,...,2ns). Clearly, µ 2m where m = n1 + ···+ ns. Denote by T = Tλ the standard Young tableau constructed as follows. Fill in the numbers 1, 2,...,2m consecutively into the first column of µ, into the second column of µ, etc., then insert the numbers 2m+1,...,ninto the boxes corresponding to 1,..., s; finally, fill in the remaining boxes in the first row. For example let λ =(5, 3, 2). Then Tλ is 1 3 5 9 10 2 4 6 7 8

As usual, given λ n and a tableau T of shape λ,leteT = RT CT denote the corresponding semi-idempotent in KSn.HereRT is the sum of the row permutations of T , while CT is the signed sum of the column permutations of T . Return now to the above specific tableau Tλ, and define the following corresponding multilinear element of the free Poisson algebra: ˜ fλ = RT CT ({x1,x2}···{x2m−1,x2m}·x2m+1 ···xn). Next, define the polynomial

f =St ···St , λ λ1 λs+t

where s + t = λ1 is the length of the first row of the diagram λ and s is the number of columns of length ≥ 2. Note that fλ is not multilinear, but up to a nonzero scalar it is obtained from f˜λ as follows: substitute one variable for the variables of the first row of Tλ, substitute a second variable for the variables of the second ˜ row, etc. Let us show that fλ is a nonzero element of the free Poisson algebra. We use the Hamiltonian algebras (see their definition in Section 5 below). Let X1,...,Xm,Y1,...,Ym ∈ H2m, substitute x1 = X1,x2 = Y1,..., x2m−1 = Xm, x2m = Ym, and finally substitute 1 for the rest of the variables. Since {Xi,Yj } = δi,j , it follows that the result of this substitution in fλ is equal to 1. ˜ Also define gλ = CT fλ. Clearly, gλ is skew–symmetric with respect to the sets of variables whose indices are written in each column of λ.Itisknownthat RT CT RT CT = γRT CT where γ = 0, hence RT · gλ = 0, so conclude that gλ =0. We say that λ lies in the hook H(i, j)ifλi+1 ≤ j. We say that this hook has leg of width j and arm of width i.WecallH(i, 0) a horizontal strip. Lemma 4.2. Let A be a Poisson algebra which, as an associative algebra, is generated by k elements. Then the customary and the extended customary cocharacters lie in a horizontal strip: Q χ2n(A)= mλχλ; λ2n; λ even; λ ≤k i 1 R χn (A)= m¯ λχλ. ≤ λ n; λ1 k+1

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4676 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

Proof. Let A be generated by a1,...,ak. Consider ﬁrst the customary cocharacter Q χ2n(A). Suppose mλ =0forsomeλ with λ1 >k. Consider a simple submodule (of KSn) that corresponds to λ. We may assume that such a submodule is generated

by an element f that corresponds to CTλ RTλ CTλ for some tableau Tλ. Such an element has a set of alternating variables, with = λ1 >k.

Let f be alternating, say, on the ﬁrst variables. Then clearly, for any ai1 ,ai2 ,... ∈{ } a1,...,ak , f(ai1 ,ai2 ,...) = 0 (there must be a repetition among ai1 ,ai2 ,...,a as >k). Since f is multilinear, we can restrict ourselves to substitutions xi → x¯i which are products of the generators a1,...,ak. Now, customary polynomials are derivations in each variable: f(ab,...)=af(b,...)+bf(a,...), etc. It follows that with such substitutions, · f(¯x1, x¯2,...)= w(i) f(ai1 ,ai2 ,...)=0, (i)

Q where w(i) are some monomials in the ajs. This completes the proof for χ2n(A). R Consider now χn (A). A set of at most k skew–symmetric letters in brackets can be extended by at most one letter in singletons because they commute. Now apply the above same arguments.

5. Hamiltonian algebras An important role in the theory of infinite dimensional Lie algebras is played by the (infinite dimensional) simple Lie algebras. Elie Cartan introduced four series of such algebras. These are: the general series Wn (Witt algebras); the special series Sn (not to be confused with the symmetric group!); the Hamiltonian series H2n; and the contact series K2n+1,wheren ∈ N. V. Kac proved that these algebras together with the Kac-Moody algebras and the Virasoro algebra, are the only infinite dimensional graded simple Lie algebras of polynomial growth (as algebras) over an algebraically closed field of characteristic zero [11]. Actually, he imposed some technical assumptions that were later removed by O. Mathieu [13]. The algebras H2m play an important role in later sections, and hence are de- scribed below. For more detailed discussions see [11], [13]. The derivations of the commutative polynomial ring Bn = K[X1,...,Xn] form a Lie algebra with respect to the commutator. This Lie algebra Wn =DerBn is called the Witt algebra. Any n ∂ element a ∈ Wn is of the form a = fi ,wherefi ∈ Bn. The Hamiltonian i=1 ∂Xi Lie algebra H2m is defined as the subalgebra of W2m, spanned as follows: m ∂f ∂ ∂f ∂ H2m = − + f ∈ K[X1,...,X2m] . ∂Xm+i ∂Xi ∂Xi ∂Xm+i i=1 K

Consider the polynomial ring H2m = K[X1,...,Xm,Y1,...,Ym], set {Xi,Yj } = δi,j and extend this bracket by the derivation rule. The result is a Poisson algebra with respect to the natural multiplication, where the Poisson brackets are computed as follows: m ∂f ∂g ∂f ∂g {f,g} = − ,f,g∈ H . ∂X ∂Y ∂Y ∂X 2m i=1 i i i i

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POISSON PI ALGEBRAS 4677

Identify the variables Y1,...,Ym with Xm+1,...,X2m, then one checks that the following mapping Φ is an epimorphism of Lie algebras: H2m → H2m m ∂f ∂ ∂f ∂ Φ:f → − + ,f∈ H . ∂X ∂X ∂X ∂X 2m i=1 m+i i i m+i This homomorphism has a one-dimensional kernel, consisting of the constant polynomials. Thus, as a Lie algebra, H2m is a one-dimensional central extension of the simple Hamiltonian Lie algebra H2m. We shall refer also to the Poisson algebra H2m as the Hamiltonian algebra. The following theorem describes the module structure of the customary and of the extended customary polynomials for the free Poisson algebra. Similar methods were used by V. Drensky to study varieties of representations of Lie algebras which are nilpotent of class 2 [6]. The structure here is similar: all the simple Sn–modules appear, and with multiplicity one. { | ∈ N} Theorem 5.1. LetF = F (X), X = xi i , be the free Poisson algebra. Then R ∈ N (1) χn (F )= χλ,forn ; λn Q ∈ N (2) χ2n(F )= χλ,forn . λ 2n; λi even

Proof. Fix the following notations: Rn = Rn(F ), Q2n = Q2n(F ), rn =dimRn, and q2n =dimQ2n; see Section 3. Recall that Rn(F ) has the canonical basis given by (2). Let g(n) denote the number of involutions in Sn. An element in (2) corresponds to the involution (in cycle notation) (τ(1),τ(2)) ···(τ(2m − 1),τ(2m)), and the assumptions on the orders in (2) imply that this correspondence is a bijec- tion. Thus r = g(n). On the other hand, it is well-known that n (5) g(n)= dim χλ λn (see e.g. [21, proof of Proposition 2]; it also follows from the identity on Schur functions [12, Problem 1.5.4]). We next show that every partition λ n does appear in the module Rn.Fix λ n and let λ =(m1,...,mk) be its conjugate, where k = λ1. Recall that · ··· fλ =Stm1 Stm2 Stmk .

We need to show that fλ = 0 in the free Poisson algebra, which is not that obvious. We prove it by showing that fλ(X1,Y1,...) = 0 for a certain substitution in H2s, the Hamiltonian algebra, where 2s ≥ m1. The following element B(m) ∈ H2s has the following nonzero value, since {X ,Y } = δ : i j i,j St (X ,Y ,...,X ,Y )=1,m=2k, (6) B(m)= m 1 1 k k Stm(X1,Y1,...,Xk,Yk,Xk+1)=Xk+1,m=2k +1. Thus

(7) fλ(X1,Y1,...)=B(m1)B(m2) ···B(mk) =0 .

Hence Rn contains a simple submodule corresponding to λ. Applying (5), conclude that for all λ n, Rn contains a corresponding simple module Vλ,andwith multiplicity one. This proves (1).

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4678 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

To prove the second claim, recall that Q2n ⊂ R2n. The same computations (6) and (7) also prove that all λ 2n with even (length) columns appear in Q2n.Since Q2n ⊂ R2n, these appear with multiplicity 1. If λ 2n has a column of odd length, then the monomials in fλ contain at least one singleton, hence fλ ∈ Q2n.This proves the second claim.

The above theorem clearly implies

Corollary 5.1. Let V be a variety of PI Poisson algebras; then R V ∈ N ∈{ } (1) χn ( )= mλχλ,forn ,andallmλ 0, 1 . Moreover, let λn · ··· fλ =Stm1 Stm2 Stmk ; V then mλ =0if and only if fλ is an identity of . Q V ∈ N ∈{ } (2) χ2n( )= m¯ λχλ,forn ,andallm¯ λ 0, 1 . Moreover, λ 2n; λi even · ··· there is a similar statement about fλ =Stm1 Stm2 Stmk ,wherethemi’s are even.

Theorem 5.1 has some additional implications. From formula (1) it follows that q2n is equal to the number of all the partitions of {1, 2,...,2n} into pairs, i.e. q2n = h(2n) in the previous notations. We obtain the well-known fact (e.g. [12, Problem 1.5.5 or 7.(2.4)]): h(2n)= dim χλ. λ 2n, λi even Corollary 5.2. Consider the Hamiltonian algebra H .Then 2k Q (1) χ2n(H2k)= χλ; λ2n, λ ≤2k, λ even 1 i R (2) χn (H2k)= χλ; ≤ λ n, λ1 2k+1 Q (3) Exp (H2k)=2k; R (4) Exp (H2k)=2k +1. Proof. It is well known that claim (1) implies claim (3) and that claim (2) implies claim (4). We therefore prove claims (1), (2) only. By Lemma 4.2, all the diagrams for Q2n(H2k) are in the horizontal strip of height 2k, and all the diagrams of Rn(H2k) are in the horizontal strip of height 2k + 1. On the other hand, the above computations show that the respective diagrams are present in Q2n(H2k) and Rn(H2k) (for the biggest columns in Rn(H2k), of length 2k + 1, we change the last substitution in (6) into x2k+1 =1).

Corollary 5.3. Consider the Hamiltonian algebra H .Then 2 1 2n (1) q2n(H2)= n+1 n is the n-th Catalan number; (2) rn(H2) is the Motzkin number.

Proof. By the previous corollary, Q2n(H2) consists of one diagram λ =(n, n)and (2n)! 1 2n by the hook formula dim χλ = n!(n+1)! = n+1 n . For the second claim see e.g. [19, Problem 7.16.b].

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POISSON PI ALGEBRAS 4679

6. The Poisson–Grassmann algebra Let Λ denote the (ordinary) associative Grassmann algebra, generated by the countable-dimensional vector space V = e1,e2,...K . The multiplication in this algebra is denoted by ∧, i.e. let a, b ∈ Λ; then denote the product by a∧b.Tomake this algebra commutative, define a new multiplication · as follows: let a, b ∈ Λ; then 1 a · b = (a ∧ b + b ∧ a). 2 This new multiplication is clearly bilinear, and 1 ∈ Λ remains a unit with respect to this multiplication. The associativity a · (b · c)=(a · b) · c is easily verified. (Note: given any noncommutative algebra, one can try to “make” it commutative by introducing a similar new · multiplication. Then that · is associative if and only if the algebra satisfies the identity [a, [b, c]] = 0.) Define also the Poisson brackets on Λ as {a, b} =[a, b]=a ∧ b − b ∧ a;thenit is easy to verify the Leibnitz rule: {a · b, c} = a ·{b, c} + b ·{a, c}.ThusΛisa Poisson algebra with respect to these two new operations. We call it the Poisson– Grassmann algebra and denote it by G. Assume K is of characteristic zero. It is well known that all the associative polynomial identities of Λ follow from the identity [[x, y],z] ≡ 0. Note that {x, y} is central in Λ, hence {x, y}·z =[x, y]∧z. It follows that in evaluating the polynomial Stn(x1,...,xn)inG, we can essentially evaluate it in Λ. Thus for example,

π (−1) {eπ(1),eπ(2)}···{eπ(2m−1),eπ(2m)} π =(−1) [eπ(1),eπ(2)] ∧···∧[eπ(2m−1),eπ(2m)] π m =(−1) 2 eπ(1) ∧ eπ(2) ∧···∧eπ(2m−1) ∧ eπ(2m) m =2 e1 ∧ e2 ∧···∧e2m.

Note also that {1+e1,e2} = {e1,e2}, hence

St2m(1 + e1,e2,...,e2m)=St2m(e1,e2,...,e2m). This and similar calculations yield that for any k and r

r m (8) St2m(1 + e1,e2,...,e2m) · (1 + e1) =2 |T2m| e1 ∧···∧e2m; r St2m+1(1 + e1,e2,...,e2m+1) · (1 + e1) = αe1 ∧···∧e2m+1 m (9) +2 |T2m| e2 ∧···∧e2m+1. Theorem 6.1. Let G be the Poisson-Grassmann algebra. Then Q ∈ N (1) χ2n(G)=χ(12n) and q2n(G)=1, n ; n − R − n 1 ∈ N (2) χn (G)= χ(r,1n r ) and pn(G)=rn(G)=2 , n ; r=1 (3) ExpQ G =1; (4) ExpR G =ExpG =2; (5) all the identities of G follow from {{x, y},z}≡0.

Proof. Since G satisﬁes the identity {{x, y},z}≡0, it follows that Pn(G)=Rn(G). We show next that all the partitions in the hook H(1, 1) appear in Rn(G). Let λ = − r−1 n r − · − (r, 1 ) n;thenf(r,1n r ) =Stn−r+1 x1 ,andf(r,1n r )(1+e1,e2,...,en−r+1) =

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4680 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

0 by (8) and (9). Therefore n n − n 1 n−1 (10) p (G)=r (G) ≥ dim χ n−r = =2 . n n (r,1 ) r − 1 r=1 r=1 Let U be the variety of Poisson algebras deﬁned by the identical relation {{x, y}, z}≡0. Consider its consequence obtained as a result of the “customarization” process [8]:

0 ≡{{x1 · x2,y},z}−x1 ·{{x2,y},z}−{{x1,y},z}·x2

= {x1,y}·{x2,z} + {x1,z}·{x2,y}.

This identity allows one to change letters on the second places in Q2n(U). Recall that we can also freely interchange brackets and letters inside brackets. Therefore, the space Q2n(U) is spanned by one monomial. Hence q2n(G) ≤ q2n(U) ≤ 1 for all n ≥ 1. By Lemma 3.1, n n (11) r (G) ≤ r (U)= q (U) ≤ =2n−1. n n 2m 2m 2m 0≤2m≤n 0≤2m≤n Now the arguments above, along with (10) and (11), yield claims (1)–(4). To prove claim (5) recall that we obtained the upper bound (11) relying on the identity {{x, y},z}≡0 only. Therefore, all the identities of G follow from this single identity.

Theorem 6.2. Let V be a variety of Poisson algebras such that G ∈V/ . Then the R V Q V customary characters χn ( ) and χ2n( ) lie in a horizontal strip. Proof. By assumption G ∈V/ . Hence, there exists a diagram λ =(r, 1n−r)fromthe space Pn(G)=Rn(G) such that the respective identity is satisfied in V.Sincethe multiplicities of the extended customary polynomials in the free Poisson algebra V · r−1 ≡ are equal to one, we conclude that satisfies Stn−r+1 x1 0. Assume first that n − r + 1 is odd; then the standard identity contains singletons. Substitute x1 =1 and conclude that Stn−r ≡ 0. Assume next that n − r + 1 is even. Consider the result of the following partial linearization: Substitute x1 + y1 for x1; then take the homogeneous component of degree r − 1iny1. In this component we set y1 =1 and obtain the standard identity Stn−r+1 ≡ 0. Let 2m be the degree of the standard identity in both cases. By Corollary 5.1 this implies that all the diagrams of Rn(V) lie in the horizontal strip of height 2m +1.

7. Varieties of customary almost polynomial growth

The goal of this section is to show the critical role played by H2 and G (Theo- rem 7.3) in studying the customary growth of Poisson algebras. Let V be a variety of Poisson algebras and let H be a Poisson algebra. We write H/∈R V if there exists an extended customary identity of V which is not satisﬁed in H.WesaythatV is of almost polynomial growth if it has exponential growth of pn(V), but any proper subvariety U⊂Vhas a polynomial growth of pn(U). We also write U⊂R V to denote that U⊂V, but these two varieties diﬀer by some nontrivial extended customary identity. Similarly, we say that V is of customary

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POISSON PI ALGEBRAS 4681

almost polynomial growth if rn(V) is of exponential growth, but for any U⊂R V the sequence of rn(U) is of polynomial growth. We now study the variety var H2. In the proof of the next two theorems we use the following notation. Let f(x1,...,xm,y1,...) be a polynomial; then we use the bar on the variables x1,...,xm to denote the skew–symmetrization in these variables: σ f(¯x1,...,x¯m,y1,...) (−1) f(xσ(1),...,xσ(m),y1,...).

σ∈Sm

Theorem 7.1. Consider the variety V generated by H2.Then Q (1) χ2n(H2)=χ(n,n); R (2) χn (H2)= χλ; ≤ λ n, λ1 3 Q (3) Exp (H2)=2; R (4) Exp (H2)=3; (5) all the customary identities of H2 follow from the identity St4 ≡ 0. Proof. Claims (1)–(4) are partial cases of Corollary 5.2, so we prove claim (5). Denote by U the variety deﬁned by the identity St4 ≡ 0. Consider the submodule U ≤ corresponding to λ that is contained in Rn( ). We shall prove that λ1 3. Recall that V =varH2 ⊆U, and we are going to prove that Rn(U)andRn(V)havethe same characters. This would imply that all the customary identities of H2 follow from St4 ≡ 0. Here is a technical proof of claim (5). Clearly, 1 St = {x ,x }·{x ,x }+{x ,x }·{x ,x }+{x ,x }·{x ,x } = {x¯ , x¯ }·{x¯ ,x }. 4 1 2 3 4 2 3 1 4 3 1 2 4 2 1 2 3 4 Thus, three skew–symmetric (i.e. alternating) variables inside two brackets yield zero. Four skew–symmetric letters inside three brackets also yield zero because then, at least three of these skew–symmetric letters enter two brackets. Decompose the following element with respect to position of x4 in it:

0 ≡{x¯1,x5}·{x¯2,x6}·{x¯3, x¯4}

= {x¯1,x5}·{x¯2,x6}·{x¯3,x4}−{x¯1,x5}·{x¯2,x6}·{x4, x¯3}

+ {x¯1,x5}·{x4,x6}·{x¯2, x¯3}−{x4,x5}·{x¯1,x6}·{x¯2, x¯3}. On the right, the ﬁrst two terms are equal, while the other two terms equal zero by the above remark. Therefore, U satisﬁes the following identity:

{y1, x¯1}·{y2, x¯2}·{y3, x¯3}≡0. Thus, any customary element with three skew–symmetric letters is equal to zero. Since singletons commute, an extended customary element with four skew–symmetric letters is also zero. Hence, the cocharacter of Q2n(U) has at most two rows and Rn(U) has at most three rows. The result now follows. Corollary 7.1. Consider the two varieties of Poisson algebras: U which is de- ﬁned by St4,andvar(H2). They are diﬀerent, but both satisfy the same customary identities. Proof. Namely, let us prove that U has no “pure Lie” identical relations at all, by latter we mean identical relations given by Lie polynomials.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4682 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

≥ Indeed, let us consider the free Poisson algebra F = F (X). Let k 0; denote (k) k { } ∈ by F the ideal of F generated by all elements i=1 ai,bi ,whereai,bi F .We observe that F (k) are verbal ideals, i.e. they are stable under all endomorphisms of F . Indeed, let φ : F → F be an endomorphism. Consider free generating elements x, y ∈ X. Suppose that φ(x)=α + f, φ(y)=β + g,whereα, β ∈ K and f,g are of degree at least one with respect to X.Then{φ(x),φ(y)} = {f,g}.Moreover,we observe that the number of brackets in monomials is not decreasing after φ. Hence, F (k) are verbal ideals. All embeddings F (k+1) ⊂ F (k), k ≥ 0, are proper; see the (2) construction of F (X) in Section 2. Remark that St4 ∈ F . By our observation, (2) all consequences of St4 belong to F , hence U has no “pure Lie” identities at all, i.e. homogeneous identities given by elements from F (1) \ F (2). On the other hand, the Lie algebra H2 is a central extension of the Lie algebra H2 ⊂ W2, where the Witt Lie algebra W2 is a PI Lie algebra with exponential growth of codimensions [2]. More precisely (see Chapter 6, Section 43 in [16]) each inﬁnite dimensional simple Lie algebra from the four Cartan series (in particular the algebra H2) satisﬁes the standard Lie identity of some degree m +1 π (−1) [...[x0,xπ(1)],xπ(2)],...],xπ(m)] ≡ 0.

π∈Sm

It is clear that the Poisson algebra H2 satisﬁes the same “pure Lie” Poisson identity. Finally, by Theorem 7.1.5, U and var(H2) have the same customary identities.

Theorem 7.2. Let V be a variety of Poisson algebras such that H2 ∈/R V (see the m beginning of this section). Then V satisﬁes the customary identity {x1,x2} ≡ 0 for some m.

Proof. By Corollary 5.2, Rn(H2) consists of all the diagrams with three rows, the multiplicities being one. Then Rn(V) does not contain a certain diagram λ of the form λ =3p2q1r. By standard methods this implies the following identity of V that corresponds to λ: p · q · r { } p ·{ }q · r ≡ St3 St2 St1 =( x¯1, x¯2 x¯3) x¯1, x¯2 x1 0.

Here, x1,x2 and x3 alternate in the ﬁrst factor, and x1 and x2 alternate in the second factor. Since {x2,x1} = −{x1,x2}, the alternation in the second factor can be removed, namely, the variety V satisﬁes the identity { }· p ·{ }q · r ≡ ( x¯1, x¯2 x¯3) x1,x2 x1 0.

Substitute x3 = 1. Since in a Poisson algebra {1,x} = 0, it follows that V satisfies { }p ·{ }q · r { }t · r ≡ x1,x2 x1,x2 x1 = x1,x2 x1 0, V { }t · r−1 ≡ where t = p + q. We show that satisfies x1,x2 x1 0. Repeating that argument clearly completes the proof. Substitute x1 + y1 for x1 and take the homogeneous component of degree 1 in y1.SinceK is infinite (char(K)=0),that component is also an identity of V.Now, t t {x1 + y1,x2} = {x1,x2} + {y1,x2}·f(x1,x2)+ highertermsiny1,

for some polynomial f(x1,x2). Similarly r r · r−1 (x1 + y1) = x1 + ry1 x1 + higher terms in y1.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POISSON PI ALGEBRAS 4683

Thus, after the substitution x1 + y1 for x1, that component of degree 1 in y1 in the above identity is { }t · · r−1 { }· · r r x1,x2 y1 x1 + y1,x2 f(x1,x2) x1. V { }t · r−1 ≡ Substitute y1 = 1 to deduce that satisﬁes x1,x2 x1 0. The proof is complete.

The following is the main result of this section.

Theorem 7.3. Let V be a variety of Poisson algebras such that H2 ∈/R V, G ∈V/ . Then

(1) rn(V) is of polynomial growth; (2) there exists N0 such that q2n(V)=0for all n ≥ N0. R V Proof. By Theorem 6.2, the character χn ( ) lies in a horizontal strip of some height t k. Also, by Theorem 7.2, we have the identity {x1,x2} ≡ 0forsomet.Consider a diagram λ =(λ1,...,λk)thatappearsinRn(V), and show that there is a bound on λ2,...,λk. Since the multiplicities Rn(V) are bounded by 1, the hook–formula then implies a polynomial bound on rn(V). Also, the columns of the diagrams in Q2n(V) are of even length, hence the sizes of the diagrams in Q2n(V) are bounded. Write λ = knk (k − 1)nk−1 ...2n2 1n1 . The identity corresponding to λ is of the form n − nk · k 1 ··· n2 · n1 Stk Stk−1 St2 St1 .

Consider i ∈{k, k − 1,...,2} and ﬁnd a bound on ni. Case 1: i =2m.Then

n (12) (St )ni =(St (x ,...,x )) i 2m 2m 1 2m ni τ = (−1) {xτ(1),xτ(2)}···{xτ(2m−1),xτ(2m)} .

τ∈T2m { } 2m 2m The number of different pairs xi1 ,xi2 is 2 . Assume ni >N= t 2 ,wheret t satisfies {x1,x2} ≡ 0. Decompose (12) and consider one of the resulting monomials. Then at least { }t ≥ one of the above pairs is raised to a sufficiently large power xi1 ,xi2 , t t. Hence, (12) is equal to zero. Case 2: i =2m +1.Here ⎛ ⎞ ni 2m+1 ni ⎝ j ⎠ (St2m+1) = (−1) St2m(x1,...,xˆj,...,x2m+1) · xj , j=1 where the hat denotes the omitted argument. The same arguments prove that this power is zero for sufficiently large ni. The proof is complete.

This theorem easily yields the following two corollaries. Corollary 7.2. (1) The variety generated by G is of almost polynomial growth. (2) The variety generated by H2 is of customary almost polynomial growth.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4684 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

n−1 Proof. 1). By Theorem 6.1(2) pn(G)=2 , hence is exponential. By Theo- rem 7.1(2), H2 ∈ var(G). If V⊂var(G) is a proper subvariety, then G ∈V as well as H2 ∈V,andbyTheorem7.3 pn(V)=rn(V) is polynomially bounded. (Recall that Pn(G)=Rn(G); see Section 6.) 2). A similar proof, now applying Theorem 6.1(1), Theorem 7.1(1) and Theo- rem 7.3.

Corollary 7.3. Let V be a variety of customary almost polynomial growth. Then either the customary identities of V coincide with the customary identities of var G, or they coincide with those of var H2. Proof. Case 1. Assume that every extended customary identity of V is also an identity of H2;thenrn(var H2) ≤ rn(V) for all n ∈ N. Recall that var H2 is of exponential customary growth. By the assumption on V, rn(var H2)=rn(V) hence V and var H2 have the same extended customary identities, and we are done. Case 2. Assume that every extended customary identity of V is also an identity of G. By the same argument conclude that V and var(G) have the same extended customary identities. Case 3. Assume neither Case 1 nor Case 2, namely H2 ∈/R V and G ∈/R V. Then by Theorem 7.3 rn(V) is of polynomial customary growth, contradicting the assumption that rn(V) is of exponential growth.

We say that a variety V has an intermediate customary growth if the sequence rn(V) lies properly between polynomial and exponential growth. Corollary 7.4. There are no Poisson varieties of intermediate customary growth.

Proof. Suppose that U is a variety of intermediate customary growth. Since H2 and G have exponential customary growth, we have H2 ∈/R U and G ∈/R U.By Theorem 7.3, U is of polynomial customary growth.

8. Exponential growth

Let F = F (X), X = {xi|i ∈ N}. InSection2wesawthatdim(Q2n(F )) = | | (2n)! T2n = n!2n . Then the Stirling formula implies that (2n)! n!2n (13) q (F )=dimQ (F )=|T | = ≈ √ ,n→∞. 2n 2n 2n n!2n πn

Thus, the spaces Q2n(F ) of the customary polynomials have an over–exponential growth. We show that the customary growth is exponentially bounded if we impose a nontrivial identity. This is is the Poisson–analogue of the result which asserts that the codimension growth of a nontrivial variety of associative algebras is at most exponential [17]. We use a technique similar to those in [21]. The following is the main result of the paper. Theorem 8.1. Let V be a nontrivial variety of Poisson algebras. Then ExpQ V – the exponent of the customary growth – exists, and moreover, it is an integer. The proof consists of several steps. Throughout this section we assume that V is a Poisson variety with a nontrivial identity. Let n be even, let λ n and assume λ appears in Qn(V); then it appears in Qn(V) with multiplicity 1. Denote

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POISSON PI ALGEBRAS 4685

β β 1 2 βs ··· λ =(α1 ,α2 ,...,αs ), where α1 >α2 > >αs > 0. We also assume that all αi are even. The corresponding polynomial is

β1 · β2 ··· βs fλ =Stα1 Stα2 Stαs .

By Corollary 5.1 we can assume that our variety satisﬁes fλ ≡ 0. Let φj be the (i) substitution of z for xj and φj the substitution of xi for xj, while other letters remain invariant. (4) (4) For example, φ2 (St2)=φ2 (St2(x1,x2)) = St2(x1,x4). Consider some j ∈{1,...α1}; it corresponds to the j–th row in the diagram λ. Make the partial linearization of fλ with respect to xj by ﬁrst substituting xj → xj + z, and then taking the homogeneous part of degree one in z.Denotethe (j) result of this process by fλ ; it depends of course on (the row) j. For the rows j =1,...,αs we have s φ (St ) f (j) = f · β j αi . λ λ i St i=1 αi

For the rows j = αs +1,...,αs−1 we have − s 1 φ (St ) f (j) = f · β j αi , λ λ i St i=1 αi

and so on. Finally, for the last rows j = α2 +1,...,α1 we obtain φ (St ) (j) · j α1 fλ = fλ β1 . Stα1 We remark that the fractions above are introduced in order to simplify the notations. Our goal is to prove that we can glue two boxes to λ in some special corner; namely, if that extended diagram is µ,thenalsofµ ≡ 0 is an identity of the algebra. Note that since αs−1 >αs and both are even, αs−1 ≥ αs +2. Proposition 8.1. Let V satisfy fλ ≡ 0;thenV also satisﬁes fµ ≡ 0,whereµ = − β1 βs 1 βs−1 (α1 ,...,αs−1 ,αs +2,αs ). Proof. Consider

αs ·{ }− (j) ·{ } (14) βsfλ z,t fλ xj,t j=1 V which is an identity of . Here we substitute z = xαs+1 and t = xαs+2.Note (αs+1) − that φj (Stαk )=0fork =1,...,s 1, since after the substitution the letter

xαs+1 appears twice inside the standard polynomial – which is alternating. Thus, (αs+1) βs the action of φj can be restricted to the last factor Stαs only. In other words,

when z = xαs+1 − (j) ·{ } β1 ··· βs−1 · βs 1 · (αs+1) ·{ } fλ xj,xαs+2 = βsStα1 Stαs−1 Stαs (φj (Stαs )) xj,xαs+2 . By (14), this implies that V satisﬁes the following identity:

αs − β1 ··· βs−1 · βs 1· ·{ }− (αs+1) ·{ } ≡ βsStα1 Stαs−1 Stαs (Stαs xαs+1,xαs+2 φj (Stαs ) xj,xαs+2 ) 0. j=1

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4686 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

By Lemma 4.1, the expression in the parentheses equals Stαs+2.Thus,weobtained the desired identity fµ ≡ 0.

Assume the variety V satisﬁes the identity fλ ≡ 0, where the last row of λ is of length q and the last column is of length 2r. The next proposition shows that we can glue to λ an arbitrary “leg” of width q and an arbitrary “arm” of width 2r, and the resulting diagram µ will yield the identity fµ ≡ 0 for the variety V.

2r 2l Proposition 8.2. Let λ =(p ,λ2r+1,...,λ2(r+s),q ) be a partition of 2n with columns of even length, and

2l µ =(p + α1,...,p+ α2r,λ2r+1,...,λ2(r+s),q ,µ2(r+s+l)+1,µ2(r+s+l)+2,...) a partition of 2N, also with columns of even length. Then the customary identity fλ ≡ 0 implies fµ ≡ 0. Proof. As a ﬁrst step we glue the leg, namely, denote

2r 2l ν =(p ,λ2r+1,...,λ2(r+s),q ,µ2(r+s+l)+1,µ2(r+s+l)+2,...),

a partition of 2M. We then show that the identity fν ≡ 0 follows from the identity fλ ≡ 0. Let T = Tλ be the standard Young tableau constructed as follows. Fill in the numbers 1, 2,... consecutively into the ﬁrst column, into the second column, etc. As usual, for a tableau T of shape λ,leteT = RT CT denote the corresponding semi-idempotent in KSn.SinceRT CT RT CT = γRT CT ,whereγ = 0, conclude that CT RT CT isanonzeroelementofKSn. As in Section 4, deﬁne the following corresponding multilinear element of the free Poisson algebra: ˜ fλ = RT CT ({x1,x2}···{x2n−1,x2n}).

Also define gλ = CT f˜λ. Clearly, this element is skew–symmetric with respect to the sets of variables whose indices are written in each column of Tλ.Notethat gλ ∈ Q2n,soKSngλ ⊆ Q2n,namelygλ generates the irreducible module for λ in Q2n.Considergλ = gλ(x1,y1,...,xn,yn) (i.e. with a different set of variables). Set m = r + s + l.Then2m is the length of the first q columns of the diagram λ.Byconstructiongλ is skew–symmetric in each of the next q sets of variables Ai = {x(i−1)m+1,y(i−1)m+1,...,xim,yim},wherei =1,...,q.Moreovergλ is skew– | | symmetric in each of the sets of variables Ai, i = q +1,...,p.Notethat Ai = λi is the length of the i–th columns, and each of the columns consists of the next variables. Let τ 2(M − n) be the “leg”–partition which is being attached to λ,namely τ =(µ2(r+s+l)+1,µ2(r+s+l)+2,...). Construct the multilinear element

gτ = gτ (xn+1,yn+1,...,xM−n,yM−n)

in a similar way. Thus gτ is skew–symmetric in each of the sets of variables Bi = {x ··· ,y ··· ,...,x ··· ,y ··· },i=1,...,q.In n+τ1+ +τi−1+1 n+τ1+ +τi−1+1 n+τ1+ +τi n+τ1+ +τi particular, B1 = {xn+1,yn+1,...,xn+τ ,y } and is not empty. Note that some 1 n+τ1 Bj might be empty. Construct gλ · gτ , and in this product alternate each of the q sets of variables Ci = Ai ∪ Bi, i =1,...,q. Note that this alternation corresponds to the ﬁrst q

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POISSON PI ALGEBRAS 4687

columns of ν. Formally this is done as follows: Given the set Ci = Ai ∪ Bi,let S(Ci) be the symmetric group on this set. Denote Cλ,µ = S(C1) ×···×S(Cq), − − π Cλ,µ = ( 1) π, π∈Cλ,µ − · then consider Cλ,µ(gλ gτ ). Examine the decomposition into irreducibles of the left module generated by this element. First, the decomposition into irreducibles of the left module generated by gλ · gτ is given by the Littlewood–Richardson (L–R) rule: the diagrams of these modules must contain λ, τ.Letη be such a diagram. ≤| | | | ≤| | | | | | By L–R η1 C1 ;ifη1 = C1 ,thenη2 C2 ;ifη1 = C1 and η2 = C2 , ≤| | ∪ then η3 C3 , etc. Because of the alternations of the variables of C1 = A1 B1 − · | | in Cλ,µ(gλ gτ ), the ﬁrst column of η must be of length at least C1 . Therefore | | | | | | ≤ η1 = C1 . A similar argument then shows that η2 = C2 ,etc.Thusηi = Ci , 1 ≤ − · i q. This implies that η = ν, i.e. the left module generated by Cλ,µ(gλ gτ ) corresponds to the partition ν. By applying H2M we now show that that module is nonzero in the free Poisson algebra. Rewrite the element gλ as the sum of monomials (namely products of {xi,xj},i< j, {xi,yj} or {yi,yj},i

{x1,y1}···{xm,ym}· ··· ·{xm(q−1)+1,ym(q−1)+1}···{xmq,ymq}

·{xmq+1,ymq+1}···{xn,yn}.

Since we substitute xi → Xi,yi → Yi and {Xi ,Yj} = δi,j in H2M ,theresultafter that substitution will be zero for all the other monomials. The structure of gλ implies that

gλ = α{x1,y1}···{xm,ym}· ··· ·{xm(q−1)+1,ym(q−1)+1}···{xmq,ymq}·g˜λ + hλ,

where α =0,˜ gλ = {xmq+1,ymq+1}···{xn,yn}, and the result of the above substitution in hλ is equal to 0. The same arguments give us that

g = β{x ,y }···{x ,y }· ··· τ n+1 n+1 n+τ1 n+τ1 ·{x ··· ,y ··· }···{x ··· ,y ··· } + h n+τ1+ +τq−1+1 n+τ1+ +τq−1+1 n+τ1+ +τq n+τ1+ +τq τ

where β = 0 and the result of our substitution in hτ is equal to 0. Note that for this partition τ, the element gτ is withoutg ˜τ because the number of columns is less than or equal to q. Thus, the resulting element is

g = γ{x ,y }···{x ,y }·{x ,y }···{x ,y }· ··· ν 1 1 m m n+1 n+1 n+τ1 n+τ1

·{xm(q−1)+1,ym(q−1)+1}···{xmq,ymq} ·{x ··· ,y ··· }···{x ··· ,y ··· }·g˜ + h , n+τ1+ +τq−1+1 n+τ1+ +τq−1+1 n+τ1+ +τq n+τ1+ +τq λ ν

where γ = 0. So the result is nonzero and the element gν generates a nonzero irreducible module in Q2M that corresponds to ν. Since the multiplicities there are one, this element generates the same module as that of fν . Clearly the identity gλ ≡ 0 implies that gλgτ ≡ 0 hence also that gν ≡ 0. Since gν and fν generate the same module, it follows that fν ≡ 0. This shows that fν ≡ 0 follows from fλ ≡ 0. Finally, we glue the arm τ to the diagram ν.Say,thearmisτ =(α1,...,α2r). Note that fµ = fν · fτ and conclude that fµ ≡ 0 follows from fλ ≡ 0 as well. Q V R V Proposition 8.3. The cocharacters χ2n( ) and χn ( ) lie in a hook.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4688 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

Proof. Assume V satisfies a customary identity fλ ≡ 0, where λ has columns of even length. First, apply Proposition 8.1 successively to deduce that V satisfies fµ ≡ 0whereµ is any rectangle with even-length columns that contains λ.This 2n1 2n2 2nk ··· is done as follows. Let λ =(α1 ,α2 ,...,αk )whereα1 >α2 > >αk > 0. Thus the 2n1 + 1–th and the 2n1 + 2–th rows of λ are of equal length α2,which is shorter than the 2n1–th row (whose length is α1). By Proposition 8.1 we can glue one box to the 2n1 + 1–th row and one box to the 2n1 + 2–th row of λ,and the resulted diagram still corresponds to an identity of V. Now repeat this process until rows 2n1 +1 and 2n1 +2 also become of length α1. Then continue this process 2m ··· until obtaining a rectangle µ =(α1 ), where m = n1 + +nk. By Proposition 8.1 V satisfies fµ ≡ 0. Furthermore, apply this process – of adding two vertical boxes – to this rectangle to obtain any rectangle µ of even height ≥ 2m and length ≥ α1, and V satisfies fµ ≡ 0. Let η be a diagram with even columns which contains λ. There is a maximal rectangle µ which is contained in η and which contains λ (there might be several such rectangles). Clearly, η is obtained from such µ by gluing an arm and a leg. Thus, applying Proposition 8.2, deduce that V satisfies fη ≡ 0. Together with Theorem 5.1 Q V this implies that the cocharacters χ2n( ) lie inside the hook H(2m, α1). For the tail consisting of singletons we use commutativity and conclude that R V Q V ≤ χn ( ) can be obtained from χ2m( ), 2m n, only by gluing a horizontal strip (in Q V sense of [12]) to the respective λ’s in χ2m( ).

Proposition 8.4. q2n(V),andrn(V) are exponentially bounded. R V Proof. This follows from Proposition 8.3. Let q2n( ) lie in the hook H(i, j). The Q asymptotic calculations in Section 7 in [4] imply that Exp V≤i+j. By Lemma 3.1 R it follows that Exp V≤i + j +1.

Q Proposition 8.5. Let k be an integer. Suppose that ExpQ V

then pn(t) ≤ 2s. Deﬁne

(16) p(t)= max pn(t). n even ≥ ≥ ≥ Since λt λt+1 and λt λt+1, hence p(t) p(t + 1). Therefore the following limit exists: (17) p = lim p(t). t→∞

This means that there exists t0 such that for any t ≥ t0 there exists nt and λ nt V which appears in Qnt ( ), and λt + λt = p. Moreover let n be arbitrary and let V ≤ µ n be any partition appearing in Qn( ); then by deﬁnition, µt0 + µt0 p.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POISSON PI ALGEBRAS 4689

Therefore µ lies in the union of the t0 × t0 square and H(i, j)–forsomei + j = p. 1 By the hook formula it follows that as n tends to infinity, (dim χµ) n ≤ p.Sincethe number of partitions of n in H(i, j) is polynomially bounded (≤ ni+j), it follows that Exp V≤p. First, suppose that p ≤ k − 1; then it follows that Exp V≤p ≤ k − 1andweare done. Second, assume p ≥ k and derive a contradiction. By the very definition (17), there exist infinitely many values t such that p(t)=p. Next, in (16), there exist infinitely many n’s with pn(t)=p. For each such t and n there exists λ n V appearing in Qn( ) such that λt + λt = p. Note that there are only finitely many pairs (a, b) of nonnegative integers such that a+b = p. We conclude that there exist integers i0,j0 ≥ 0 such that i0 + j0 = p and there exist infinitely many pairs (λ, t) such that λt = i0 and λt = j0. Hence we obtain p for infinitely many diagrams that lie in the union of the square t0 × t0 and the hook H(i0,j0). We consider the intersections of these diagrams with the t0 × t0 square. There are only finitely many diagrams contained in the t0 × t0 square, hence there exists a diagram µ which is the intersection of the t0 ×t0 square with infinitely | | ≤ 2 many such diagrams that yield the number p.Denoteµ = n0,wheren0 t0. Fix some T ≥ t0. We can find an irreducible character χλ appearing in Qn(V), and V t>T, such that λT = λt = i0 and λT = λt = j0.Sinceλ appears in Qn( ), by T Proposition 8.2 mλT = 0. Consider the diagram λ : except for its intersection with the t0 ×t0 square it has a rectangular arm of width i0, and a rectangular leg of T awidthj0, both being of length T .Thenumberofboxesinλ is n = n0 +(T −t0)p. By the hook formula

T γ n0+(T −t0)p (18) dim λ ≈ C0n p ,T→∞.

This asymptotic is valid for inﬁnitely many values of type n = n0 +(T −t0)p,where n0, t0 and p are ﬁxed while T is arbitrary such that T ≥ t0.Thesenumbersforma subsequence of the arithmetic progression ni = m0 + ip, i =0, 1, 2,.... Since these numbers satisfy (18), we obtain

1 n (19) lim inf (qn (V)) i ≥ p. i→∞ i

Let us consider all the even numbers between such ni and ni −p;say,weconsider numbers of type ni − 2. From the structure of the customary polynomials (1) it follows that n Qn(V,x1,...,xn)= {x1,xj}·Wj(x2,...,xˆj,...,xn), j=2

where Wj are some subspaces that are homomorphic images of Qn−2(V). This obviously implies that

qn(V) (20) q − (V) ≥ . n 2 n − 1 From (19) and (20) it follows that

1 n −2 lim inf (qn −2(V)) i ≥ p. i→∞ i From this we conclude that Exp V≥p ≥ k, a contradiction to the assumption of the proposition.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4690 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

The proof of Theorem 8.1 now follows from Proposition 8.5. We now deduce a few corollaries from Theorem 8.1. Corollary 8.1. Let A be a Poisson algebra with a nontrivial identity. Suppose that A is ﬁnitely generated as an associative algebra. Then ExpQ A is even and ExpR A is odd.

Proof. Consider ﬁrst Q2n(A). By Lemma 4.2, the diagrams that appear in Q2n(A) lie in a horizontal strip. In the proof of Theorem 8.1 all the diagrams are of even columns. In that proof it was shown that the cocharacters contain tableaux which – asymptotically – are rectangular. Hence ExpQ A is even. Now ExpR A is odd by Corollary 3.2. From the proof of Theorem 8.1 also follows Corollary 8.2. Let V be a nontrivial variety of Poisson algebras. Then the cus- R V Q V tomary characters χn ( ) and χ2n( ) lie in a hook. Corollary 8.3. Let V be a nontrivial variety of Poisson algebras. Then V satisﬁes a nontrivial customary identity of the special type απ{x1,yπ(1)}···{xm,yπ(m)}≡0,απ ∈ K.

π∈Sm

Proof. The number of the above monomials {x1,yπ(1)}···{xm,yπ(m)}, π ∈ Sm,is m!, while the customary codimensions are exponentially bounded.

The following is an analogue of the well-known Amitsur’s theorem. Corollary 8.4. Let V be a nontrivial variety of Poisson algebras. Then V satisﬁes k a power of a standard identity (St2m) ≡ 0, for some integers m, k.

Proof. By Corollary 8.2 there exists a hook H(i, j) such that all the characters of Q V × 2m χ2n( ) lie in this hook. Let 2m>iand k>j; then the 2m k rectangle λ = k is not in H(i, j), hence the corresponding customary polynomial fλ is an identity k for V. By Corollary 5.1, this identity is fλ =(St2m) ≡ 0.

9. Tensor products of Poisson algebras Recall the notion of the tensor product of Poisson algebras (see e.g. [20]). Let A and B be two Poisson algebras. We consider the tensor product of the associative algebras A ⊗K B and equip it with the Poisson bracket via

{a1 ⊗ b1,a2 ⊗ b2} = {a1,a2}⊗b1 · b2 + a1 · a2 ⊗{b1,b2},a1,a2 ∈ A, b1,b2 ∈ B.

It is easy to verify that A ⊗ B is a Poisson algebra. Moreover, with 1 = 1A ∈ A, 1 ⊗ B and B are isomorphic as Poisson algebras, so we can write B ⊆ A ⊗ B (here we use the fact that {1,a} =0foralla ∈ A). It can be shown, for example, that ∼ ⊗···⊗ (21) H2m = H 2 H2 . m times The main result of this section is the following theorem. It is an analogue of the respective result of the third author for associative PI-algebras [17]. Theorem 9.1. Let A and B be Poisson PI algebras; then A ⊗ B is PI.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POISSON PI ALGEBRAS 4691

The proof is given below. First, we construct extended customary monomials of special type. Denote n = {1,...,n}.Wealsodenotei = {i1,...,ik}⊆n,andj = n\i is its complement. Given such i = {i1,...,ik}⊆n,withj = n\i = {j1,...,j}, let Mi(x)=Mi(x1,...,x2n) be the following extended customary monomial: k { }· · (22) Mi(x)= x2ir −1,x2ir (x2js−1 x2js ) r=1 s=1 c (Poisson brackets, then parenthesis) and let Mi (x)=Mj(x) denote the corresponding “complementary” monomial: k c · · { } Mi (x)= (x2ir −1 x2ir ) x2js−1,x2js r=1 s=1

(parenthesis, then Poisson brackets). Since each subset i = {i1,...,ik}⊆n deter- mines such a monomial, in total there are 2n such extended customary monomials Mi(x)inx1,...,x2n. The deﬁnition of the Poisson bracket in the tensor product yields, by induction, the following lemma.

Lemma 9.1. Let A and B be Poisson algebras and let a1,...,a2n ∈ A, b1,...,b2n ∈ B.Then

{a ⊗ b ,a ⊗ b }···{a − ⊗ b − ,a ⊗ b } 1 1 2 2 2n 1 2n 1 2n 2n ⊗ c = Mi(a1,...,a2n) Mi (b1,...,b2n). i ⊆ n

If η ∈ S2n,then

{a ⊗ b ,a ⊗ b }···{a − ⊗ b − ,a ⊗ b } η(1) η(1) η(2) η(2) η(2n 1) η(2n 1) η(2n) η(2n) ⊗ c = Mi(aη(1),...,aη(2n)) Mi (bη(1),...,bη(2n)). i ⊆ n Proof of Theorem 9.1. Let g(x)=g(x1,...,x2n)= γη{xη(1),xη(2)}···{xη(2n−1),xη(2n)}

η∈T2n

be a general multi-linear customary polynomial, with the coeﬃcients γη considered as unknowns at the moment. By multi-linearity, g(x)isanidentityofA ⊗ B if it vanishes for any substitution xu = au ⊗ bu ∈ A ⊗ B,foru =1,...,2n.Withsuch a substitution, by Lemma 9.1

g(a ⊗ b)=g(a ⊗ b ,...,a ⊗ b ) 1 1 2n 2n · ⊗ c = γη Mi(aη(1),...,aη(2n)) Mi (bη(1),...,bη(2n)) ∈ ⊆ η T2n i n · ⊗ c = γη Mi(aη) Mi (bη).

η∈T2n i ⊆ n Fix some i ⊆ n and consider the first products equation in (22): these are customary elements. By definition of customary codimensions, there exist q2|i|(A) customary elements such that the first products are expressed via these elements.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4692 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

We multiply by the singletons of the second product of (22). Thus, there exist extended customary polynomials Ki,s(x1,...,x2n), 1 ≤ s ≤ q2|i|(A), and coeﬃcients αi,s,η with η ∈ T2n, such that for any a1,...,a2n ∈ A,

q2|i|(A) Mi(aη)= αi,s,ηKi,s(aη(1),...,aη(2n)). s=1 c c \ Similarly, for Mi (bη), let i = n i. There exist polynomials Lic,t(x1,...,x2n), 1 ≤ t ≤ q2|ic|(B), and coeﬃcients βic,t,η with η ∈ T2n, such that for any b1,...,b2n ∈ B,

q2|ic|(B) c Mi (bη)= βic,t,ηLic,t(bη(1),...,bη(2n)). t=1 Thus, ⊗ · ⊗ c g(a b)= γη Mi(aη) Mi (bη) ∈ ⊆ η T2n i n ⎛ ⎞ q2|i|(A) ⎝ ⎠ = γη · αi,s,ηKi,s(aη(1),...,aη(2n)) η∈T i ⊆ n s=1 ⎛ 2n ⎞ q2|ic|(B) ⎝ ⎠ ⊗ βic,t,ηLic,t(bη(1),...,bη(2n)) t=1 ⎡ ⎤ q2|i|(A) q2|ic|(B) ⎣ ⎦ = αi,s,η · βic,t,η · γη · Ki,s(aη) ⊗ Lic,t(bη).

i ⊆ n s=1 t=1 η∈T2n

It follows that g(x)=0isanidentityofA⊗B if for all i ⊆ n,all1≤ s ≤ q2|i|(A), and all 1 ≤ t ≤ q2|ic|(B), αi,s,η · βic,t,η · γη =0.

η∈T2n c This is a system of linear equations in |T2n| unknowns γη.Seta = |i| and b = |i |; then a + b = n. The number of equations is given by n (23) q (A) · q (B). a 2a 2b a+b=n

By Theorem 8.1, the customary codimensions q2n(A)andq2n(B) are exponentially bounded. Hence, the number of equations is exponentially bounded. On the other hand, |T2n| grows over–exponentially; see equation (13). If n is large enough, this system has a nontrivial solution, namely a nontrivial g(x) which is an identity of A ⊗ B.

Corollary 9.1. Let A and B be two Poisson PI algebras. Then we have the following estimate on the customary exponent of A ⊗ B:

ExpQ(A ⊗ B) ≤ ExpQ A +ExpQ B.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POISSON PI ALGEBRAS 4693

2a 2b Proof. For simplicity, suppose that q2a(A) ≤ λ and q2b(B) ≤ µ for all a, b. Remark that the number of equations (23) yields the upper bound on the customary codimension growth of A ⊗ B.Weget n n q (A ⊗ B) ≤ q (A) · q (B) ≤ λ2aµ2b 2n a 2a 2b a a+b=n a+b=n 2n 2n ≤ λ2aµ2b ≤ λiµj =(λ + µ)2n. 2a i a+b=n i+j=2n

We conclude by noting that this estimate is exact for Hamiltonian algebras; see Corollary 5.2 and (21).

References

[1] Andrews G.E., The theory of partitions, Addison-Wesley, 1976. MR0557013 (58:27738) [2] Bahturin Yu. A., Identical relations in Lie algebras. VNU Science Press, Utrecht, 1987. MR0886063 (88f:17032) [3] Bahturin Yu., Mishchenko S., Regev A., On the Lie and associative codimensions growth. Comm. Algebra 27 (1999), no. 10, 4901–4908. MR1709214 (2000m:16034) [4] Berele A., Regev A., Hook Young diagrams with applications to combinatorics and to representations of Lie superalgebras, Adv. Math. 64 (1987), no. 2, 118–175. MR0884183 (88i:20006) [5] Dixmier J., Enveloping algebras. AMS, Rhode Island, 1996. MR1393197 (97c:17010) [6] Drensky V., Identities of representations of nilpotent Lie algebras. Comm. Algebra 25 (1997), no. 7, 2115–2127. MR1451682 (98d:17015) [7] Drensky V., Free algebras and PI-algebras. Graduate course in algebra. Springer-Verlag Sin- gapore, Singapore, 2000. MR1712064 (2000j:16002) [8] Farkas D. R., Poisson polynomial identities. Comm. Algebra 26 (1998), no. 2, 401–416. MR1603345 (98m:16029) [9] Farkas D. R., Poisson polynomial identities. II. Arch. Math. (Basel) 72 (1999), no. 4, 252–260. MR1678045 (2000d:16034) [10] Giambruno A., Zaicev M., Exponential codimension growth of PI algebras: an exact estimate. Adv. Math. 142 (1999), no. 2, 221–243. MR1680198 (2000a:16048) [11] Kac V. G., Simple irreducible graded Lie algebras of finite growth. Izv. Akad. Nauk SSSR Ser. Mat. 32 (1968) 1323–1367. MR0259961 (41:4590) [12] Macdonald I. G., Symmetric functions and Hall polynomials, 2-nd ed., Oxford: Clarendon Press, 1995. MR1354144 (96h:05207) [13] Mathieu O., Classification of simple graded Lie algebras of finite growth. Invent. Math. 108 (1992), no. 3, 455–519. MR1163236 (93h:17069) [14] Mischchenko S.P., Regev, A., Zaicev M., Integrality of exponents of some abelian-by- nilpotent varieties of Lie algebras. Comm. Algebra, 28 (2000), no. 9, 4105–4130. MR1772005 (2001g:17019) [15] Petrogradsky V.M., On complexity functions for T -ideals of associative algebras. Mat. Za- metki 68 (2000), no. 6, 887–897; translation in Math. Notes 68 (2000), no. 5-6, 751–759. MR1835188 (2002c:16031) [16] Razmyslov Yu. P., Identities of Algebras and Their representations, A.M.S Translations of Math Monographs Vol 138 (1994). MR1291603 (95i:16022) [17] Regev A. Existence of identities in A ⊗ B. Israel J. Math. 11 (1972), 131–152. MR0314893 (47:3442) [18] Shestakov I.P., Quantization of Poisson superalgebras and speciality of Jordan Poisson superalgebras. Algebra i Logika. 32 (1993), no. 5, 571–584; English translation: Algebra and Logic, 32 (1993), no. 5, 309–317. MR1287006 (95c:17034) [19] Stanley R. P., Enumerative combinatorics, Vol. 2. Cambridge University Press, Cambridge, 1999. MR1676282 (2000k:05026)

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4694 S. P. MISHCHENKO, V. M. PETROGRADSKY, AND A. REGEV

[20] Tarasov A.A., On uniqueness of raising of maximal commutative subalgebras in Poisson algebras into universal enveloping algebra. Mat. Sbornik. 194 (2003), No. 7, 155-160. MR2020383 (2004k:17023) [21] Volichenko I. B., Varieties of Lie algebras with identity [[X1,X2,X3], [X4,X5,X6]] = 0 over a ﬁeld of characteristic zero. Sibirsk. Mat. Zh. 25 (1984), no. 3, 40–54. MR0746940 (85k:17016)

Faculty of Mathematics, Ulyanovsk State University, Leo Tolstoy 42, Ulyanovsk, 432970 Russia E-mail address: [email protected] E-mail address: [email protected] Faculty of Mathematics, Ulyanovsk State University, Leo Tolstoy 42, Ulyanovsk, 432970 Russia E-mail address: [email protected] Department of Theoretical Mathematics, Weizmann Institute of Science, Rehovot, Israel E-mail address: [email protected]

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use