Basic superranks for varieties of algebras
aa r X i v : . [ m a t h . R A ] J a n Basic superranks for varieties of algebras
Alexey Kuz’min ∗ , Ivan Shestakov † Abstract
We introduce the notion of basic superrank for varieties of algebras generalizingthe notion of basic rank. First we consider a number of varieties of nearly asso-ciative algebras over a field of characteristic 0 that have infinite basic ranks andcalculate their basic superranks which turns out to be finite. Namely we prove thatthe variety of alternative metabelian (solvable of index 2) algebras possesses the twobasic superranks (1 ,
1) and (0 , ,
2) and (1 , r, s ) = (0 ,
0) of nonnegative integers we provide a varietythat has the unique basic superrank ( r, s ). Finally, we construct some examples ofnearly associative varieties that do not possess finite basic superranks.
Key words: alternative algebra, Jordan algebra, Malcev algebra, metabelianalgebra, Grassmann algebra, superalgebra, variety of algebras, basic rank of variety,basic superrank of variety, basic spectrum of variety.
MSC 2010:
Dedicated to Efim Isaakovich Zelmanov,on the occasion of his 60th birthday
Introduction
Throughout the paper, all algebras are considered over a field of characteristic 0. Let V be a variety of algebras and V r be a subvariety of V generated by the free V -algebra ofrank r . Then one can consider the chain V ⊆ V ⊆ · · · ⊆ V r ⊆ · · · ⊆ V , where V = S r V r . If this chain stabilizes, then the minimal number r with the property V r = V is called the basic rank of the variety V and is denoted by r b ( V ) (see [13]).Otherwise, we say that V has the infinite basic rank r b ( V ) = ℵ .Let us recall the main results on the basic ranks of the varieties of associative (Assoc ), Lie (Lie ), alternative (Alt),
Malcev (Malc ), and some other algebras . It was first shown ∗ The first author is supported by the FAPESP 2010/51880–2 and the PNPD/CAPES/UFRN–PPgMAE † The second author is supported by the FAPESP 2014/09310–5 and the CNPq 303916/2014–1
1y A. I. Mal’cev [13] that r b (Assoc) = 2. A. I. Shirshov [28] proved that r b (Lie) = 2and r b (SJord) = 2, where SJord is the variety generated by all special Jordan algebras.In 1958, A. I. Shirshov posed a problem on finding basic ranks for alternative and someother varieties of nearly associative algebras [3, Problem 1.159]. In 1977, the secondauthor proved that r b (Alt) = r b (Malc) = ℵ [18, 28]. The similar fact for the variety ofalgebras of type ( − ,
1) was established by S. V. Pchelintsev [14]. Note that the basicranks of the varieties of Jordan and right alternative algebras are still unknown.A proper subvariety of associative algebras can be of infinite basic rank as well. Forinstance, so is the variety Var G generated by the Grassmann algebra G on infinitenumber of generators, or the variety defined by the identity [ x, y ] n = 0 , n > V of algebras is called Spechtian or is said to have the Specht property if every algebra of V possesses a finite basis for itsidentities. The Kemer Theorem states the Specht property of the variety of associativealgebras. This result is obtained by certain reduction to the case of graded identitiesof finite dimensional superalgebras. Namely it is proved that the ideal of identitiesof arbitrary associative algebra coincides with the ideal of identities of the Grassmannenvelope of some finite dimensional superalgebra.This result suggests a generalization of the notion of basic rank. Namely we shall saythat a variety V has a finite basic superrank if it can be generated by the Grassmannenvelope of some finitely generated superalgebra. Then Kemer’s result implies that everyvariety of associative algebras has a finite basic superrank. This means that the notionof basic superrank is a more refined one then that of basic rank, and we can distinguishvarieties of infinite basic rank by their superranks.The notion of basic superrank is the main subject of our paper. First we considera number of varieties of nearly associative algebras over a field of characteristic 0 thathave infinite basic ranks and calculate their basic superranks which turns out to be finite.Namely we prove that the variety of alternative metabelian (solvable of index 2) algebraspossesses the two basic superranks (1 ,
1) and (0 , ,
2) and (1 , r, s ) = (0 ,
0) of nonnegative integers we provide a varietythat has the unique basic superrank ( r, s ). Finally, we construct some examples of nearlyassociative varieties of algebras that do not possess finite basic superranks.
Let A = A + A be a superalgebra ( Z -graded algebra) with the even part A and the odd part A , i. e. A i A j ⊆ A i + j (mod2) for i, j ∈ { , } ; G be the Grassmann algebra on a countable set of anticommuting generators { e , e , . . . | e i e j = − e j e i } with the nat-ural Z -grading (G and G are spanned by the words of even and, respectively, oddlength on { e i } ). The Grassmann envelope
G ( A ) of a superalgebra A is the subalgebraG ⊗ A + G ⊗ A of the tensor product G ⊗ A .It is well known that G ( A ) satisfies a multilinear polynomial identity f = 0 if andonly if A satisfies the graded identity ˜ f = 0 called the superization of f = 0 . Here,2 f denotes the so-called superpolynomial corresponding to f and we say that A satisfiesthe superidentity ˜ f = 0. The detailed descriptions of the process of constructing ofsuperpolynomials (the superizing process ) can be found in [19, 20, 24, 27]. Roughlyspeaking, one should apply the so-called Koszul rule (or
Kaplansky rule ): one shouldintroduce the sign ( − ij always when a variable of parity i passes through a variable ofparity j .Let V be a variety of algebras defined by a system S of multilinear identities. Recallthat A is said to be a V -superalgebra if its Grassmann envelope lies in V , i. e. if A satisfies the system ˜ S of all superidentities corresponding to the defining identities of V .Thus one can consider the set ˜ V of all V -superalgebras as a supervariety defined bythe system ˜ S . It is clear that ˜ V can be generated by the free V -superalgebra on acountable set of even and a countable set of odd generators. Let V r,s be the supervarietygenerated by the free V -superalgebra on r even and s odd generators for ( r, s ) = (0 , V , = (cid:8) { } (cid:9) . By L ( V ) denote the lattice V , ⊆ V , ⊆ · · · ⊆ V r, ⊆ · · · p ∩ p ∩ p ∩V , ⊆ V , ⊆ V , ⊆ · · · ⊆ V r, ⊆ · · · p ∩ p ∩ p ∩ p ∩ ... ... ... ... p ∩ p ∩ p ∩ p ∩V ,s ⊆ V ,s ⊆ V ,s ⊆ · · · ⊆ V r,s ⊆ · · · p ∩ p ∩ p ∩ p ∩ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . It is clear that L ( V ) is partially well-ordered with the relation of inclusion. A pair ( r, s )is called the basic superrank of the variety V if V r,s is a minimal element of L ( V ) withthe property V r,s = ˜ V . If V r,s = ˜ V for all r, s , we say that the basic superrank of V isinfinite. The set of all possible finite basic superranks of V is called the basic spectrumof V and is denoted by sp b ( V ). For instance, one can easily check that sp b (Assoc) =sp b (Lie) = { (2 , , (1 , , (0 , } . It is not hard to prove that if V possesses at least onefinite basic superrank ( r, s ), then sp b ( V ) is a finite set of at most r + s + 1 elements.The first examples of varieties of infinite basic superrank were constructed by M. V. Za-itsev [25, 26].Let us fix some notations. While writing down nonassociative monomials we usethe symbol · instead of parentheses to indicate the correct order of multiplication. Forinstance, we write xy · z instead of ( xy ) z and x · yz instead of x ( yz ). By[ x, y ] = xy − yx and x ◦ y = xy + yx we denote, respectively, the commutator and the Jordan product of the elements x, y .The notations( x, y, z ) = xy · z − x · yz and J ( x, y, z ) = xy · z + yz · x + zx · y are used, respectively, for the associator and the Jacobian of the elements x, y, z .3ecall that the varieties
Alt , Jord, and Malc of alternative, Jordan, and
Malcevalgebras are defined by the following pairs of identities:Alt : ( x, x, y ) = 0 , ( x, y, y ) = 0; (1)Jord : [ x, y ] = 0 , (cid:0) x , y, x (cid:1) = 0; (2)Malc : x ◦ y = 0 , J ( x, y, z ) x = J ( x, y, xz ) . (3)By V (2) we denote a subvariety of all metabelian (solvable of index at most ) algebrasof a given variety V , i. e. V (2) is distinguished in V by the identity xy · zt = 0 . (4)By NAlt we denote a subvariety of all nilalgebras in Alt of index at most 3, i. e. NAltis distinguished in Alt by the identity x = 0 . (5)Let us formulate the results of the paper. First, in Section 2, we describe the inclusionsin the lattices L (cid:0) Alt (2) (cid:1) and L (cid:0) NAlt (2) (cid:1) and calculate the basic spectrums of thesevarieties. Namely, we prove the following theorems.
Theorem 1. sp b (cid:0) NAlt (2) (cid:1) = (cid:8) (0 , (cid:9) and all inclusions NAlt (2) n ⊂ NAlt (2) n +1 are strict. Theorem 2.
For V = Alt (2) we have V , ⊂ V , ⊂ · · · ⊂ V r, ⊂ · · ·∩ ∩ ∩V , ⊂ V , = V , = · · · = V r, = · · · pp pp pp pp V , ⊂ V , = V , = · · · = V r, = · · ·∩ pp pp pp V , = V , = V , = · · · = V r, = · · · pp pp pp pp V , = V , = V , = · · · = V r, = · · · pp pp pp pp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Corollary 2.1. sp b (cid:0) Alt (2) (cid:1) = (cid:8) (1 , , (0 , (cid:9) . In Sections 3 and 4, we describe the inclusions in the lattices L (cid:0) Jord (2) (cid:1) , L (cid:0) Malc (2) (cid:1) and calculate the basic spectrums sp b (cid:0) Jord (2) (cid:1) , sp b (cid:0) Malc (2) (cid:1) , respectively.
Theorem 3.
For V = Jord (2) we have V , ⊂ V , ⊂ · · · ⊂ V r, ⊂ · · ·∩ ∩ ∩V , ⊂ V , ⊂ V , ⊂ · · · ⊂ V r, ⊂ · · ·∩ ∩ ∩ ∩V , = V , = V , = · · · = V r, = · · · pp pp pp pp V , = V , = V , = · · · = V r, = · · · pp pp pp pp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . orollary 3.1. sp b (cid:0) Jord (2) (cid:1) = (cid:8) (0 , (cid:9) . Theorem 4.
For V = Malc (2) we have V , ⊂ V , ⊂ · · · ⊂ V r, ⊂ · · ·∩ ∩ ∩V , ⊂ V , = V , = · · · = V r, = · · ·∩ pp pp pp V , ⊂ V , = V , = · · · = V r, = · · ·∩ pp pp pp ... ... ... . . . ... ∩ pp pp pp V ,s ⊂ V ,s = V ,s = · · · = V r,s = · · ·∩ pp pp pp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Corollary 4.1. sp b (cid:0) Malc (2) (cid:1) = (cid:8) (1 , (cid:9) . Further, let M be the variety of all metabelian algebras. In Section 5, we describethe inclusions in the lattice L ( M ) and provide an example of variety of unique arbitrarygiven finite basic superrank. Theorem 5.
The inclusion M r ′ ,s ′ ⊆ M r,s holds only if r ′ r and s ′ s . The equality M r ′ ,s ′ = M r,s takes place only if r ′ = r and s ′ = s . Corollary 5.1.
The basic superrank of M is infinite. Corollary 5.2.
For an arbitrary pare ( r, s ) = (0 , of nonnegative integers, the varietyof algebras generated by the Grassmann envelope of the free M r,s -superalgebra has theunique basic superrank ( r, s ) . In Section 6, we provide some examples of varieties of nearly associative algebras ofinfinite basic superrank. For ε = ±
1, by V h ε i denote a subvariety of M distinguished bythe identities ( x, y, z ) = ε ( x, z, y ) , (6) (cid:10) h x, y i ε , z (cid:11) ε = 0 , (7)where h x, y i ε = xy − εyx . Thus, V h +1 i is the variety of metabelian right symmetricalgebras that are Lie-nilpotent of index at most V h− i is the variety of metabelianright alternative algebras that are Jordan-nilpotent of index at most . Theorem 6.
The basic superrank of V h ε i is infinite. Finally, in Section 7, we suggest some open problems dealing with the introducednotions of basic superrank and basic spectrum for varieties of algebras.5 ommon notations
Throughout the paper, we use the following notations. By [ r ] we denote the integerpart of number r ; R x and L x are operators of right and left multiplication , respectively, by an element x ; T x is a common notation for R x and L x ; T ∗ x = ( L x , if T x = R x ,R x , if T x = L x ; X = { x , x , . . . } is a countable set ; X n = { x , x , . . . , x n } , n ∈ N ; F is a field ofcharacteristic char F = 0; F V [ Y ] is a free algebra of variety V on a set Y of freegenerators over F ; F ( s ) V [ Z ] is a free V -superalgebra on a set Z of free even and oddgenerators over F ; P n ( V ) is a subspace of F V [ X n ] of all multilinear polynomials ofdegree n >
2; ( f ) T is a T -ideal of algebra F V [ X ] generated by the given polynomial f ;S n is the symmetric group on the set , , . . . , n ; A n is the alternating subgroup of S n ;C n is the subgroup of S n generated by the cycle (1 2 . . . n ) ; | σ | is a parity of permutation σ ∈ S n , i. e. | σ | = ( , if σ is even,1 , if σ is odd . In order to avoid complicated formulas while writing down the elements of the space P n ( V ) we omit the indices of variables at the operator symbols R, L and assume themto be arranged in the ascending order. For example, the notation ( x x ) LR means themonomial ( x x ) L x R x R x . Throughout this section, we set V = Alt (2) and N = NAlt (2) . V -algebras Recall that by the Artin Theorem [28, Chap. 2.3] every two-generated alternative algebrais an associative one. It is also well known that every alternative algebra satisfies thecentral Moufang identity x · yz · x = xy · zx. Thus by virtue of metability (4) in the free algebra F V [ X ] , we have x · yz · x = 0 . Moreover, combining alternativity (1) with (4), we get x ( x · yz ) = 0 , ( yz · x ) x = 0 , ( x, zt, y ) = ( zt · y ) x = x ( y · zt ) . Therefore the relations T x T ∗ x = T x T x = 0 , [ L x , R y ] = R y R x = L y L x hold for the operators of multiplication acting on ( F V [ X ]) . Using these relations, onecan prove the following 6 emma 2.1 ([7, 15]) . The free algebra F V [ X ] is a linear span of the monomials of theform ( x i x i ) T x i R x i . . . R x in , which are skew-symmetric with respect to their variables x i , x i , . . . , x i n . Note that in view of non-nilpotency of F V [ X ] (see [5, 19]), Lemma 2.1 implies imme-diately the infiniteness of the basic rank of V .Further, by metability of F V [ X ], the Artin Theorem yields the following Proposition 2.1.
The algebra F V [ X ] satisfies the identities x T y = 0 , (8)( xy ) T x R y = 0 . (9)Combining Lemmas 2.1 with identities (8), (9), it is not hard to prove the following Lemma 2.2. ( F V [ X n ]) n +2 = 0 . Let us set ϕ ( x , x , x ) = X σ ∈ S (cid:0) x σ (1) x σ (2) (cid:1) x σ (3) . Lemma 2.3.
The free metabelian superalgebra on one odd generator satisfies the superi-dentity ˜ ϕ ( x , x , x ) = 0 .Proof. Let A be the free metabelian superalgebra on one odd generator. Consider thevalue ˜ ϕ ( x , x , x ) for arbitrary homogeneous elements x , x , x ∈ A . By metabilityof A , we may assume that at least two of the elements x , x , x are generators of A .But in this case, taking into account that A has only one odd generator, we obtain thatthe linear combination ˜ ϕ ( x , x , x ) contains with every its monomial αw ( α = ±
1) themonomial − αw . Hence, ˜ ϕ ( x , x , x ) = 0 in A . Lemma 2.4.
The intersection I = ( x ) T ∩ (cid:16)P ∞ n =3 P n ( V ) (cid:17) is spanned over F by theelement ϕ ( x , x , x ) .Proof. It is clear that ϕ ( x , x , x ) ∈ I as a linearization of x . Furthermore, by (8),the element x and all its linearizations lie in the annihilator of F V [ X ]. On the otherhand, Lemma 2.1 and identity (4) yield ϕ ( w, x , x ) = 0 for every element w ∈ F V [ X ] .Therefore every element of I is proportional to ϕ ( x , x , x ). Lemma 2.5.
The free V -superalgebra on two odd generators is an N -superalgebra.Proof. Let A be the free V -superalgebra on two odd generators. By Lemma 2.4, itsuffices to check ˜ ϕ ( x , x , x ) = 0 assuming that x , x , x are generators of A . Butin this case, at least two of the elements x , x , x coincide. Hence, with the similararguments as in Lemma 2.3, one can prove that ˜ ϕ ( x , x , x ) = 0.7 .3 Auxiliary N -superalgebra on one odd generator Let U = F · x be a superalgebra generated by an odd element x such that x = 0.Consider a Z -graded space M = M + M over F such that M i = F [ ε ] · a i , i = 0 , , where F [ ε ] is an algebraic extension of F with a primitive 3-th root of 1, i. e. such an ele-ment ε that ε + ε +1 = 0. It is clear that if ε ∈ F , then M is a 2-dimensional space over F , otherwise, a 4-dimensional one. We define on M a structure of an U -superbimodulesuch that the action of the element x is given by the equalities a i · x = a − i , x · a i = ( i + ε ) a − i , i = 0 , . Consider the superalgebra A = U ∔ M with the Z -grading A = A + A , A = M , A = U + M and the multiplication( u + m ) ( u + m ) = u m + m u , u , u ∈ U, m , m ∈ M. This superalgebra is called the split null extension of U by M (see [23, 28]). It isknown [19, 23] that A is an alternative superalgebra. Lemma 2.6. A is an N -superalgebra generated by one odd element.Proof. Let us show that A can be generated by the element y = a + x . First we have y = ( a + x ) = a · x + x · a = (2 + ε ) a . Hence for ε ∈ F , we get a = αy , a = αy · y, where α = 1 − ε . Otherwise, we calculate y · y = (2 + ε ) a · x = (2 + ε ) a ,y · y = x · (2 + ε ) a = (2 ε + ε ) a = ( ε − a . Considering the difference of the obtained equalities, we have3 a = y · y − y · y . Therefore to express any element of M with y over F it suffices to use the relations a = a · y, εa = y · a , εa = εa · y. To conclude the proof note that A is metabelian by definition. Hence by Lemma 2.3,it satisfies identity (5). 8 .4 Additive basis of P n ( N ) Definition 2.1.
The basis words of P n ( N ) are the polynomials of the following types:1) ( x x ) T R n − ,
2) ( x ◦ x i ) T R n − , i = 2 , . . . , n. Lemma 2.7.
The space P n ( N ) is spanned by its basis words.Proof. Let u be an arbitrary monomial of P n ( N ). By Lemma 2.1 we may assume that u has the form u = ( x i x j ) T R n − , i, j ∈ { , . . . , n } . Consider the linear span I n of the polynomials of P n ( N ) of the form( x i ◦ x j ) T R n − . Using the linearization X σ ∈ C (cid:0) x σ (1) ◦ x σ (2) (cid:1) x σ (3) = 0of identity (5) it is not hard to show that I n is the linear span of basis words of type 2).On the other hand, the linearizations x ( T y ◦ T z ) = ( y ◦ z ) T ∗ x of identities (1) imply that the monomial u is skew-symmetric with respect to all itsvariables modulo I n . Lemma 2.8.
Any nontrivial linear combination of basis words of P n ( N ) is not an iden-tity of G ( A ) .Proof. Consider an arbitrary linear combination f n = g n + h n of basis words of P n ( N ),where g n = α ( x x ) R n − + α ′ ( x x ) LR n − ,h n = n X i =2 (cid:0) β i ( x ◦ x i ) R n − + β ′ i ( x ◦ x i ) LR n − (cid:1) , for some scalars α, α ′ , β j , β ′ j ∈ F . Suppose that A satisfies the superidentity ˜ f n = 0.Then let us show that all the coefficients in f n are zero.Fix i > f n = 0 the substitution x i = a , x j = x for all j = i . Then we will get β i (1 + ε ) a R n − − β ′ i (1 + ε ) a R n − = 0 , which gives β i = (1 + ε ) β ′ i for all i >
2. If we substitute x i = a instead of a , we willsimilarly get β i = − εβ ′ i , which implies that β i = β ′ i = 0 for all i >
2. Therefore, h n hasa form h n = β ( x ◦ x ) R n − + β ′ ( x ◦ x ) LR n − , β, β ′ ∈ F. x i = a , i = 1 , x j = x for all j = i we will get the two equalities αa R n − − α ′ (1 + ε ) a R n − + H = 0 ,ε ( αa R n − − α ′ (1 + ε ) a R n − ) + H = 0 , where H = ˜ h n ( a , x, . . . , x ) = ˜ h n ( x, a , . . . , x ). This implies that α = (1 + ε ) α ′ and H = 0. Similarly, making the substitutions x i = a , i = 1 , x j = x for all j = i we get α = − εα ′ and H = 0, where H = ˜ h n ( a , x, . . . , x ) = − ˜ h n ( x, a , . . . , x ). Thus, α = α ′ = 0. Finally, we have H = β (1 + ε ) a R n − − β ′ (1 + ε ) a R n − = 0 ,H = − βεa R n − − β ′ ε a R n − = 0 , which implies that β = β ′ = 0. Therefore all the scalars in f n are zero.By Lemmas 2.6–2.8, we obtain N , = ˜ N . F N [ X n ] Lemma 2.9.
The index of nilpotency of F N [ X n ] is equal to n + 2 .Proof. By Lemma 2.2, the index of nilpotency of F N [ X n ] is not more than n + 2. Letus provide a nonzero element of F N [ X n ] of degree n + 1. Consider the monomial w = x R x . . . R x n . Note that the linearization of w can be written as a basis word of type 2) of P n +1 ( N ).Hence by Lemmas 2.6 and 2.8, w = 0 in F N [ X n ].Lemma 2.9 implies the strict inclusions N n ⊂ N n +1 for n ∈ N .Theorem 1 is proved. V -superalgebras Let A be the superalgebra defined in Section 2.3. Consider two superalgebras B = B + B and B ′ = B ′ + B ′ defined by the following conditions:1. B = A + F · e, B = A + F · ex + F · xe + F · exe, B ′ = A + F · yx + F · xz, B ′ = A + F · y + F · z + F · yxz ;2. A is a subalgebra of B and B ′ ;3. all nonzero products of basis elements of B and B ′ , in the case when at least oneof the factors doesn’t lie in A , are the following: e · x = ex, x · e = xe, ex · e = e · xe = exe ; y · x = yx, x · z = xz, yx · z = y · xz = yxz. emma 2.10. B is a V -superalgebra generated by one even and one odd elements; B ′ is a V -superalgebra generated by three odd elements.Proof. It is clear that B can be generated by the elements e, x and B ′ can be generatedby the elements x, y, z . Thus by virtue of Lemma 2.6, it remains to notice that if someproduct ρ = 0 of three basis elements of B or B ′ contains a factor not lying in A , then ρ is associative and lies in the annihilator of the corresponding algebra B or B ′ . P n ( V ) Definition 2.2.
The basis words of P n ( V ) are all basis words of P n ( N ) and the poly-nomial ϕ ( x , x , x ).It follows easily from Lemmas 2.4 and 2.7 that the space P n ( V ) is spanned by itsbasis words. Lemma 2.11.
Any nontrivial linear combination of basis words of P n ( V ) is not anidentity of either G ( B ) or G ( B ′ ) .Proof. By definition of G ( B ) and G ( B ′ ), taking into account Lemma 2.8, it suffices tocheck that ˜ ϕ ( x , x , x ) takes a nonzero value on some elements of B and B ′ . Indeed,˜ ϕ ( e, x, e ) = 2 ( e · x ) · e = 2 exe = 0; ˜ ϕ ( y, x, z ) = ( y · x ) · z = yxz = 0 . By Lemmas 2.10 and 2.11, we have V , = V , = ˜ V . Combining Lemmas 2.5 and 2.11with Theorem 1, we obtain V , ⊆ ˜ N = V , ⊂ ˜ V . V n ⊂ V n +1 Finally, to complete the proof of Theorem 2, it suffices to verify the strictness of allinclusions in the first row of the lattice L ( V ) . In fact, it follows from Lemmas 2.4and 2.9 that the index of nilpotency of F V [ X n ] for n > n + 2. Therefore, V n = V n +1 for all n >
2. It remains to notice that the variety V is commutative and V is not.Theorem 2 is proved. Throughout this section, we set V = Jord (2) .11 .1 Additive basis of the free V -algebra It is known [4, 9, 15, 22] that an additive basis of the free algebra F V [ X ] can be formedby the following monomials ( x k x i ) R x j R x i R x j . . . R x it R ′ x jt , where k > i < i < · · · < i t , j < j < · · · < j t , and the symbol ′ means that the operator R x jt is absent when the degree of monomialis even. In what follows, we use this basis with no comments. F V [ X n ] Proposition 3.1.
The algebra F V [ X ] satisfies the identities x R y R x = 0 , (10)( zt ) R x R y R x = 0 , (11)( zx ) R y R x R t R z = 0 . (12) Proof.
First we stress that (10) is a direct consequence of metability and (2). Further,taking into account commutativity, we write the partial linearization of (10) in the form2( zx ) R y R x + x R y R z = 0 . (13)Then by setting z := zt in (13), in view of metability, we get (11). Finally, multiply-ing (13) by R t R z and applying (11), we obtain (12). Lemma 3.1.
The algebra F V [ X n ] is nilpotent of index n + 2 .Proof. Identities (10)–(12) imply immediately that ( F V [ X n ]) n +2 = 0. Therefore it re-mains to note that the element x R x R x R x . . . R x n of the additive basis of F V [ X ] is a nonzero element of ( F V [ X n ]) n +1 .Lemma 3.1 yields that all inclusions V n ⊂ V n +1 are strict and, consequently, the basicrank of V is infinite. V First note that superizing (11), one can prove the following
Proposition 3.2.
Let F ( s ) V [ Z ] be a free V -superalgebra on an arbitrary set Z of evenand odd generators. Then the operators of multiplication acting on (cid:0) F ( s ) V [ Z ] (cid:1) satisfy therelation R x R y R z = ( − | x || y | + | x || z | + | y || z | +1 R z R y R x , (14) where x, y, z ∈ Z and | x | denotes the parity of x . U = U + U be the superalgebra U = { } , U = F · x + F · y with null multiplication and M = M + M be the vector space M = F · a, M = F · v. Consider a split null extension A = U ∔ M with a supercommutative multiplication suchthat all nonzero products of the basis elements of A , up to the order of factors, are thefollowing: a · x = v, v · y = a. The supercommutativity rule means that even elements commute with all elements ofsuperalgebra but products of two odd elements are anticommutative.
Lemma 3.2. A is a V -superalgebra generated by two odd elements.Proof. By definition, A is metabelian, supercommutative, and can be generated by theelements v + x and y . It remains to check that A is a Jordan superalgebra. Note thatby construction of A it suffices to verify that relation (14) holds in A . But it followstrivially from the definition of multiplication in A . Lemma 3.3.
The variety V is generated by G ( A ) .Proof. In view of Lemma 3.2 it suffices to prove that G ( A ) doesn’t satisfy any nontrivialidentity in V . Consider an arbitrary linear combination of basis monomials of P n ( V ) : f n = X I α I ( x k x i ) R x j R x i R x j . . . R x it R ′ x jt , where t = (cid:2) n (cid:3) and I runs all possible sets k, i , . . . , i t , j , . . . , j t of indices such that k > i < i < · · · < i t , j < j < · · · < j t . Suppose that A satisfies the superidentity ˜ f n = 0. Let us show that all the scalars in f n are zero. We fix I = { k, i , . . . , i t , j , . . . , j t } and make the substitution x k = a, x i = · · · = x i t = x, x j = · · · = x j t = y. Then it is not hard to see that ˜ f n turns out to be proportional with the coefficient ± α I to the element ( a · x ) R y R x · · · = ( v, if n is even ,a, if n is oddthat is nonzero in A . Therefore, α I = 0.Lemma 3.3 implies that V , = ˜ V . Thus, to complete the proof of Theorem 3, itremains to show that the chain V , ⊂ V , ⊂ V , ⊂ · · · ⊂ V r, ⊂ · · · ⊂ ˜ V ascends strictly and all the inclusions V n, ⊂ V n, are also strict. One can also note that A is a special Jordan superalgebra isomorphic to a subalgebra of the matrixsuperalgebra M (+)2 , (see [19]) with the generators e − e + e − e and e + e + e + e + e + e . .4 Strictness of the inclusions V n, ⊂ V n, and V n − , ⊂ V n, Let U ( n ) = U ( n )0 + U ( n )1 be the superalgebra U ( n )0 = n X i =1 F · e i , U ( n )1 = F · y with null multiplication and A ( n ) = A ( n )0 + A ( n )1 be an associative superalgebra with theunit generated by the even elements , e , . . . , e n and one odd element y with thedefining relations e i e j = 0 , y = , e i ye j = − e j ye i . Consider a split null extension B ( n ) = U ( n ) ∔ A ( n ) with a supercommutative multiplicationinduced by the actions · e i = 0 , a · e i = ae i , = a ∈ A ( n ) , b · y = by , b ∈ A ( n ) . Lemma 3.4. B ( n ) is a V -superalgebra generated by n even and one odd elements.Proof. By definition, B ( n ) is metabelian and supercommutative. Moreover, it is not hardto see that B ( n ) can be generated by the elements + e , e , . . . , e n , y . Thus it remainsto prove that B ( n ) is a Jordan superalgebra. Actually by definition of multiplication in B ( n ) , taking into account that (cid:0) B ( n ) (cid:1) ⊆ A ( n ) , it suffices to verify that relation (14) holdsfor the operators of the form R u R u R u acting on A ( n ) for u i ∈ { e , . . . , e n , y } . Indeed,besides the trivial case u = u = y , we check that R e i R e j annihilates A ( n ) : a R e i R e j = ae i e j = 0; R y acts on A ( n ) identically: a R y = ay = a ;and the action of R e i R y R e j on A ( n ) is skew-symmetric with respect to e i , e j : a R e i R y R e j = ae i ye j = − ae j ye i = − a R e j R y R e i . By Lemmas 3.1 and 3.4, in view of non-nilpotency of B ( n ) , we obtain V n, ⊂ V n, .Further, let us denote by f n = f n ( a, b, x , . . . , x n ) the polynomial f n = ( ab ) ( R x ◦ R x ) . . . (cid:0) R x n − ◦ R x n (cid:1) , where R x ◦ R y = R x R y + R y R x . Lemma 3.5.
The free V -superalgebra on n − even and one odd generators satisfiesthe superidentity ˜ f n = 0 . roof. Let A n − , be the free V -superalgebra on n − f n on some homogeneous elements ˜ a, ˜ b, ˜ x , . . . , ˜ x n ∈ A n − , . In viewof metability, we may assume that all the elements of the set S = { ˜ x , . . . , ˜ x n } aregenerators of A n − , . Thus by definition of f n , it is clear that ˜ f = 0 in A , .Let us prove by induction on n that ˜ f n = 0 in A n − , . For n >
2, by e , . . . , e n − wedenote the even generators of A n − , and y denotes its odd generator. For a monomial w = (˜ a ˜ b ) R ˜ x R ˜ x . . . R ˜ x n we set S ( w ) = { ˜ x , ˜ x , . . . , ˜ x n − } and ¯ S ( w ) = S \ S ( w ). Assumethat w = 0 in A n − , . Then it follows from (14) that every e i can be presented onlyonce in each set S ( w ) and ¯ S ( w ). Thus, ˜ f n can be nonzero only if every e i is included in S not more than twice. On the other hand, if some e i is included in S only once, thenone can represent ˜ f n in the form˜ f n = ± ˜ f n − (cid:0) ˜ a, ˜ b, ˜ x , . . . , ˜ x j − , ˜ x j +1 , . . . , ˜ x n − (cid:1) (cid:0) R ˜ x j − ◦ R ˜ x j (cid:1) , where e i ∈ { ˜ x j − , ˜ x j } . In this case, by inductive hypothesis, we have ˜ f n = 0. Therefore,it suffices to consider the case when every e i is included in S twice exactly and w =0. This assumption yields that the sets S ( w ) and ¯ S ( w ) consist of the same elements e , . . . , e n − , y . Consequently, except of w , there is only one more nonzero monomial w ′ in the linear combination ˜ f n of the form w ′ = (˜ a ˜ b ) R ˜ x R ˜ x R ˜ x R ˜ x . . . R ˜ x n R ˜ x n − . Hence we have ˜ f n = w + ( − | ˜ x || ˜ x | + | ˜ x || ˜ x | + ··· + | ˜ x n − || ˜ x n | w ′ = w + w ′ . By virtue of (14), taking into account the equalities S ( w ′ ) = ¯ S ( w ) and ¯ S ( w ′ ) = S ( w ), itis not hard to see that w ′ is proportional to w . Thus it remains to prove that the coef-ficient of this proportionality is equal to −
1. Let σ be the permutation that transforms¯ S ( w ) into S ( w ). It is clear that one can transform w ′ into w acting by σ on S ( w ′ ) andby σ − on ¯ S ( w ′ ). While that, a scalar ± σ , but will depend only on number of transpositions made by theodd elements. Taking into account that σ ( y ) = y , it is not hard to understand that sucha transposition will be only one. Therefore, w ′ = − w and, consequently, ˜ f n = 0.By Lemmas 3.4 and 3.5, to prove the strictness of inclusions V n − , ⊂ V n, it sufficesto verify that ˜ f n takes a nonzero value in B ( n ) . Indeed,˜ f n ( , y, e , y, e , y, . . . , e n , y ) = y ( R e ◦ R y ) ( R e ◦ R y ) . . . ( R e n ◦ R y ) == ( ye · y + · e ) ( R e ◦ R y ) . . . ( R e n ◦ R y ) == ye y ( R e ◦ R y ) . . . ( R e n ◦ R y ) = · · · = ye ye y . . . e n y = 0 . Theorem 3 is proved.
Remark 3.1.
Note that Theorem 3 gives a more detailed description of the inclusions inthe lattice L ( V ) than one needs to deduce the uniqueness of the basic superrank (0 , V . Actually proving that sp b ( V ) = (cid:8) (0 , (cid:9) we could restrict with establishing theequality V , = ˜ V and the strict inclusions V r, ⊂ ˜ V for r = 0 , , , . . . Namely, in viewof Lemmas 3.2 and 3.5, it is enough to check that the superpolynomial ˜ f n doesn’t vanishon some elements of the superalgebra A . 15 Malcev algebras
Throughout this section, we set V = Malc (2) . It’s well-known [17] that an anticommutative algebra over a field of characteristic distinctfrom 2 is a Malcev one if and only if it satisfies the Sagle identity X σ ∈ C (cid:0) x σ (1) x σ (2) (cid:1) R x σ (3) R x σ (4) = ( x x ) ( x x ) . By virtue of metability, the Sagle identity gets the form X σ ∈ C (cid:0) x σ (1) x σ (2) (cid:1) R x σ (3) R x σ (4) = 0 . (15)For x = w ∈ (cid:0) F V [ X ] (cid:1) , identity (15) implies wR x R x R x = wR x R x R x . (16)Moreover, combining (15) with anticommutativity and taking into account thatchar F = 2, we obtain ( xy ) [ R x , R y ] = 0 . (17)Finally, applying (16) and (17), we have( xy ) ρR x ηR y = ( xy ) ρR y ηR x , (18)for any operator words ρ, η . V -superalgebra Let U = U + U be the superalgebra U = F · e, U = F · y with null multiplication and M = M + M be the vector space M = F · a, M = F · v + F · w. Consider a split null extension A = U ∔ M with a superanticommutative multiplicationsuch that all nonzero products of the basis elements of A , up to the order of factors, arethe following: a · y = v, v · y = a, w · e = w. The superanticommutativity rule means that even elements anticommute with all ele-ments of superalgebra and odd elements commute to each other.
Lemma 4.1. A is a V -superalgebra on one even and one odd generators. roof. By definition, A is metabelian and superanticommutative. Moreover, it is nothard to see that A can be generated by the even element a + e and by the odd element w + y . It remains to check that A is a Malcev superalgebra. By construction of A , itsuffices to verify that the superization of (16) holds in A , i. e. that the action of anoperator R z R z R z on M , for homogeneous z i ∈ U , satisfies the relation R z R z R z = ( − | z || z | + | z || z | R z R z R z . Actually it is not hard to see that this relation holds trivially in both possible nonzerocases z = z = z = y and z = z = z = e . P n ( V ) By virtue of anticommutativity, we consider the space P n ( V ) only for n >
3. Followingthe fixed above notations, we write down the monomials of P n ( V ) omitting some uniquelyrestored indices that are assumed to be arranged in the ascending order. Definition 4.1.
The basis words of P n ( V ) are the polynomials of the following types:1) ( x x ) R n − , ( x x ) R n − , ( x x ) R n − ,
2) ( x x i ) R n − , i = 4 , . . . , n,
3) ( x x ) ( R x ◦ R x ) R n − , ( x x ) ( R x ◦ R x ) R n − , ( x x ) ( R x ◦ R x ) R n − ,
4) ( x x i ) ( R x ◦ R x ) R n − , i = 4 , . . . , n, where R x ◦ R y = R x R y + R y R x . Lemma 4.2. P n ( V ) is spanned by its basis words.Proof. In view of anticommutativity, it is clear that P ( V ) is spanned by its basis wordsof type 1).Let I n be the linear span of basis words of P n ( V ) for n >
4. Then using (15) andanticommutativity, we have( x x ) R x R x = − ( x x ) R x R x − ( x x ) R x R x − ( x x ) R x R x =( x x ) ( R x ◦ R x ) − ( x x ) R x R x − ( x x ) R x R x ++ ( x x ) ( R x ◦ R x ) − ( x x ) R x R x ≡ I ) . Similarly, it is not hard to check that( x x ) R x R x , ( x x ) R x R x , ( x x ) R x R x ∈ I . Thus, I = P ( V ) .Further, let u be an arbitrary monomial of P n ( V ) for n >
5. Then applying (16)one can order the indices of variables of u as follows: u = ( x i x i ) R x i . . . R x in , i < i , i < · · · < i n . Thus, similarly to the case n = 4, combining (15) with anticommutativity and (16), onecan obtain u ∈ I n . 17 emma 4.3. Any nontrivial linear combination of basis words of P n ( V ) is not an iden-tity of G ( A ) .Proof. Consider first the case n = 3. We set f = α ( x x ) x + β ( x x ) x + γ ( x x ) x , α, β, γ ∈ F and assume that ˜ f = 0 in A . Then by the substitution x = a , x = x = y , we have˜ f = α ( a · y ) · y − γ ( y · a ) · y = ( α + γ ) a = 0 . Similarly, after two more substitutions x = a , x = x = y and x = a , x = x = y ,we get the system α + γ = 0 ,α + β = 0 ,β + γ = 0 . The unique solution of this system in F is α = β = γ = 0.Now consider a linear combination f n = g n + h n + p n + q n of basis words of P n ( V ) for n >
4, where g n = α ( x x ) R n − + α ( x x ) R n − + α ( x x ) R n − ,h n = n X i =4 α i ( x x i ) R n − ,p n = β ( x x ) ( R x ◦ R x ) R n − + β ( x x ) ( R x ◦ R x ) R n − + β ( x x ) ( R x ◦ R x ) R n − ,q n = n X i =4 β i ( x x i ) ( R x ◦ R x ) R n − , α i , β i ∈ F. Suppose that A satisfies the superidentity ˜ f n = 0. Then let us show that all the coeffi-cients in f n are zero.First we fix i > f n = 0 the substitution x i = a, x j = y for all j = i .Then we get α i ( y · a ) R n − y = ( − α i v, if n is even , − α i a, if n is odd , whence, α i = 0 for i > f n = g n + p n + q n . Similarly, for i >
4, by the substitution x i = w, x j = e for all j = i , one can prove that β i = 0. Thus, f n = g n + p n . Further, for i = 1 , ,
3, by the substitution x i = a, x j = y for all j = i , similarly to thecase n = 3, we obtain α i = 0. Consequently, f n = p n . Finally, for i = 1 , ,
3, by the substitution x i = w, x j = e for all j = i , we get β i = 0.18y Lemmas 4.1–4.3, we have V , = ˜ V . Thus, to complete the proof of Theorem 4, itremains to show that the chains V ⊂ V ⊂ · · · ⊂ V r ⊂ · · · ⊂ V , V , ⊂ V , ⊂ · · · ⊂ V ,s ⊂ · · · ⊂ ˜ V ascend strictly. V n ⊂ V n +1 Let G n be the Grassmann algebra with the unit on the set e , . . . , e n of anticommutinggenerators and U n be the algebra on the set e , . . . , e n of generators with null multipli-cation. Consider a split null extension A n = U n − ∔ G n − with an anticommutativemultiplication induced by the actions( e i . . . e j ) · e k = e i . . . e j e k . Lemma 4.4. A n is a V n -algebra.Proof. By definition, A n is metabelian and anticommutative. Moreover, it is not hard tosee that A n can be generated by the elements , e , . . . , e n − . Thus it remains to checkthat A n satisfies (15). Indeed, for arbitrary w ∈ G n , we have( w · e i ) R e j R e k + ( e k · w ) R e i R e j = w e i e j e k − w e k e i e j = 0 . Further, let us denote by f n = f n ( x , . . . , x n ) the polynomial f n = X σ ∈ S n ( − | σ | (cid:0) x σ (1) x σ (2) (cid:1) R x σ (3) . . . R x σ ( n ) . Lemma 4.5.
The free algebra F V [ X n ] satisfies the identity f n +1 = 0 .Proof. By definition, the value of f n is zero when values of two of its variables coincide.Hence, by virtue of metability, it remains to verify that f n +1 takes zero value after asubstitution x n +1 = w , where w = ( x i x j ) τ for some operator word τ . Indeed, in viewof anticommutativity and metability, we have f n +1 ( x , . . . , x n , w ) = ( − n wχ n , χ n = X σ ∈ S n ( − | σ | R x σ (1) . . . R x σ ( n ) . Thus, on one hand, the action of χ n is skew-symmetric with respect to any pair of itsvariables by definition. On the other hand, by virtue of (18), χ n acts at w symmetricallyw.r.t. x i , x j . Therefore, this action is zero.In view of Lemmas 4.4 and 4.5, to prove the strict inclusions V n ⊂ V n +1 it suffices tocheck that f n +1 takes a nonzero value in A n +1 . Indeed, f n +1 ( , e , . . . , e n ) = 2 X σ ∈ S n ( − | σ | R e σ (1) . . . R e σ ( n ) = 2 n ! e . . . e n = 0 . Remark 4.1.
V. T. Filippov [6] proved the strict inclusion Malc n ⊂ Malc n +1 for all n = 3 and suggested the hypothesis Malc = Malc . This hypothesis is still known as adifficult open problem. 19 .5 Strictness of inclusions V ,n ⊂ V ,n +1 Proposition 4.1.
The free V -superalgebra F ( s ) V [ Y ] on an arbitrary set Y of odd gener-ators satisfies the relations wR x R y R z = wR z R x R y , (19)( xy ) ρR x ηR y = ( xy ) ρR y ηR x , (20) x ρR x ηR y = x ρR y ηR x , (21) where x, y, z ∈ Y , w ∈ (cid:0) F ( s ) V [ Y ] (cid:1) , and ρ, η are arbitrary operator words of the form R y i . . . R y j , y i , . . . , y j ∈ Y .Proof. Superizing (15) and taking into account metability and superanticommutativity,we have wR x R y R z + ( − | w | wL z R x R y = wR x R y R z − wR z R x R y = 0 , ( xy ) R x R y − ( yx ) R y R x + ( xy ) R x R y − ( yx ) R y R x = 2( xy ) [ R x , R y ] = 0 ,x R x R y − x R y R x + ( xy ) R x − ( yx ) R x = x [ R x , R y ] = 0 . Thus, (19) is proved and to conclude the proof of (20), (21) it remains to combine (19)with the last two obtained relations.Further, let us denote by g n = g n ( a, b, x , . . . , x n ) the polynomial g n = ( ab ) X σ ∈ S n R x σ (1) . . . R x σ ( n ) . Lemma 4.6.
The free V -superalgebra F ( s ) V [ Y n ] on a finite set Y n = { y , . . . , y n } of oddgenerators satisfies the superidentity ˜ g n = 0 .Proof. By definition, the value of ˜ g n is zero when values of at least two of its variables x , . . . , x n coincide. Hence, by virtue of metability, it suffices to consider the case when˜ g n takes a value ˜ g n = ( ab ) χ n , χ n = X σ ∈ S n ( − | σ | R y σ (1) . . . R y σ ( n ) for some a, b ∈ F ( s ) V [ Y n ] . Thus, on one hand, the action of χ n is skew-symmetric withrespect to any pair of its variables by definition. On the other hand, by virtue of (20)and (21), χ n acts at ab symmetrically with respect to some pair of its variables. There-fore, this action is zero.To conclude the proof of strict inclusions V ,n ⊂ V ,n +1 let us construct a V -superalgebraon a set of n + 1 odd generators that does not satisfy the superidentity ˜ g n = 0.Let G ( n ) be the Grassmann algebra with the unit on the set e , . . . , e n of anticom-muting generators and ¯G ( n ) be the Grassmann algebra without unit on the set ¯ e , . . . , ¯ e n of anticommuting generators. For an element w ∈ G ( n ) of the form w = e i . . . e i k =
20e use the notations | w | = k , ¯ w = ¯ e i . . . ¯ e i k , and set | | = 0, ¯ = 0. For i = 0 ,
1, byG ( n ) i and ¯G ( n ) i we denote, respectively, the subspaces of G ( n ) and ¯G ( n ) spanned by all thewords w and ¯ w such that | w | ≡ i (mod 2). Consider the direct some M ( n ) = G ( n ) + ¯G ( n ) of vector spaces, where we set M ( n )0 = G ( n )0 + ¯G ( n )1 to be an even component of M ( n ) and M ( n )1 = G ( n )1 + ¯G ( n )0 to be its odd component.Let U ( n ) be the superalgebra on the set y , . . . , y n of odd generators with null multipli-cation. Consider a split null extension A ( n ) = U ( n ) ∔ M ( n ) with a superanticommutativemultiplication induced by the actions w · y k = we k , ¯ w · y k = ¯ w ¯ e k , w ∈ G ( n ) . By virtue of skew-symmetry of the elements of G ( n ) and ¯G ( n ) with respect to theirgenerators it is not hard to prove the following Lemma 4.7. A ( n ) is a V -superalgebra. Further, we consider an extension ¯ A ( n +1) = F · x + A ( n ) of the superalgebra A ( n ) suchthat ¯ A ( n +1)0 = A ( n )0 , ¯ A ( n +1)1 = F · x + A ( n )1 , and all nonzero products of basis elements with x are the following: x = , x · y i = y i · x = ¯ e i , x · ¯ w = ( − | w | ¯ w · x = 12 w, ¯ w ∈ ¯G ( n ) . Lemma 4.8. ¯ A ( n +1) is a V -superalgebra generated by n + 1 odd element.Proof. By Lemma 4.7, taking into account skew-symmetry of the elements of G ( n ) and¯G ( n ) with respect to their generators, it suffices to verify that the superization of (15)holds in the following cases: x R y i R y j − ( x · y i ) R y j R x − ( y j · x ) R x R y i = ( · y i ) · y j − (¯ e i · y j ) · x − (¯ e j · x ) · y i == e i · y j − ¯ e i ¯ e j · x + 12 e j · y i = e i e j − e i e j + 12 e j e i = 0 , ( x · y i ) R x R y j − ( y i · x ) R y j R x + ( x · y j ) R x R y i − ( y j · x ) R y i R x == (¯ e i · x ) · y j − (¯ e i · y j ) · x + (¯ e j · x ) · y i − (¯ e j · y i ) · x == − e i · y j − ¯ e i ¯ e j · x − e j · y i − ¯ e j ¯ e i · x = − e i e j − e i e j − e j e i − e j e i = 0 , ( x · ¯ w ) R y i R y j − ( − | w | ( ¯ w · y i ) R y j R x == 12 ( w · y i ) · y j − ( − | w | ¯ w ¯ e i ¯ e j · x = 12 we i e j − we i e j = 0 ,
21 ¯ w · x ) R y i R y j − ( − | w | ( y j · ¯ w ) R x R y i = ( − | w | w · y i ) · y j − ( ¯ w ¯ e j · x ) · y i == ( − | w | we i e j − ( − | w | +1 we j · y i = ( − | w | we i e j + we j e i ) = 0 , ( ¯ w · y i ) R x R y j − ( − | w | ( y j · ¯ w ) R y i R x = ( ¯ w ¯ e i · x ) · y j − ( ¯ w ¯ e j · y i ) · x == ( − | w | +1 we i · y j − ¯ w ¯ e j ¯ e i · x = − ( − | w | we i e j + we j e i ) = 0 . In view of Lemmas 4.6–4.8, to prove the strict inclusions V ,n ⊂ V ,n +1 it remains tocheck that ˜ g n takes a nonzero value in ¯ A ( n +1) . Indeed,˜ g n ( x, x, y , . . . , y n ) = X σ ∈ S n ( − | σ | R y σ (1) . . . R y σ ( n ) = n ! e . . . e n = 0 . Theorem 4 is proved.
Remark 4.2.
We stress that the description of the inclusions in the lattice L ( V ) ob-tained in Theorem 4 is more detailed than one needs to deduce the uniqueness of the basicsuperrank (1 ,
1) for V . Actually proving that sp b ( V ) = (cid:8) (1 , (cid:9) we could restrict withestablishing the equality V , = ˜ V and the strict inclusions V n ⊂ V , V ,n ⊂ ˜ V . Namely,in view of Lemmas 4.1, 4.5, and 4.6, it is enough to check that the superpolynomials ˜ f n ˜ g n don’t vanish on some elements of the superalgebra A . Remark 4.3.
It is not hard to check that in fact the superalgebra ¯ A ( n +1) is a Lie su-peralgebra. The variety Lie (2) of metabelian Lie algebras has a finite basic rank: it isgenerated by a 2–dimensional non-abelian algebra and hence r b (cid:0) Lie (2) (cid:1) = 2. Since weare interested in metabelian varieties of infinite basic rank, it was not planed to considerthe variety Lie (2) . But in view of the results of this section we notice that the lattices L (cid:0) Lie (2) (cid:1) and L (cid:0) Malc (2) (cid:1) have a quite similar structure in spite of the difference in theirinitial chains. Indeed, one can easily prove that Lie (2) also possesses the basic super-rank (1 , (2)0 ,n ⊂ Lie (2)0 ,n +1 are provided by the factthat the free metabelian Lie superalgebra on n odd generators is nilpotent of index n + 2.Thus the lattice L ( V ), for V = Lie (2) , has the form V , ⊂ V , = · · · = V r, = · · ·∩ pp pp V , ⊂ V , = V , = · · · = V r, = · · ·∩ pp pp pp V , ⊂ V , = V , = · · · = V r, = · · ·∩ pp pp pp V , ⊂ V , = V , = · · · = V r, = · · ·∩ pp pp pp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . In particular, sp b (cid:0) Lie (2) (cid:1) = (cid:8) (2 , , (1 , (cid:9) .22 Metabelian algebras
Let d be a Young diagram of order n and τ be a permutation of S n . Following [1,Chap. 3], by τ d we denote the Young table obtained by filing in the diagram d with thenumbers τ (1) , . . . , τ ( n ) in the order from the top down and from left to right. The groupS n acts naturally on the set of all tables corresponding to d : σ ( τ d ) = ( στ ) d . By C τd we denote the column stabilizer of the table τ d , i. e. the subgroup of S n consisting ofall permutations preserving the set of symbols in each column of τ d . Similarly, one candefine the row stabilizer R τd of τ d .By P n we denote the linear span of all associative multilinear words of degree n onthe set X n of variables. For every Young table τ d of order n ′ n we define the followingendomorphisms: ϕ τd , ψ τd : P n
7→ P n ,ϕ τd ( w ) = X σ ∈ C τd X ρ ∈ R στd ( − | σ | w (cid:0) x ρσ (1) , . . . , x ρσ ( n ′ ) , x n ′ +1 , . . . , x n (cid:1) ,ψ τd ( w ) = X ρ ∈ R τd X σ ∈ C ρτd ( − | σ | w (cid:0) x σρ (1) , . . . , x σρ ( n ′ ) , x n ′ +1 , . . . , x n (cid:1) . For example, if τ d = 1 23 4 and w = x x x x , then ϕ τd ( w ) can be calculated asfollows: ϕ τd ( w ) = X ρ ∈ R τd x ρ (1) x ρ (2) x ρ (3) x ρ (4) − X ρ ∈ R (13) τd x ρ (3) x ρ (2) x ρ (1) x ρ (4) −− X ρ ∈ R (24) τd x ρ (1) x ρ (4) x ρ (3) x ρ (2) + X ρ ∈ R (13)(24) τd x ρ (3) x ρ (4) x ρ (1) x ρ (2) and, taking into account that(13) τ d = 3 21 4 , (24) τ d = 1 43 2 , (13)(24) τ d = 3 41 2 , we get ϕ τd ( w ) = ( x ◦ x )( x ◦ x ) − ( x ◦ x )( x ◦ x ) −− ( x ◦ x )( x ◦ x ) + ( x ◦ x )( x ◦ x ) == ( x ◦ x ) ◦ ( x ◦ x ) − ( x ◦ x ) ◦ ( x ◦ x ) . Similarly, for the same table τ d and w ′ = x x x x , we have ψ τd ( w ′ ) = X σ ∈ C τd ( − | σ | x σ (1) x σ (3) x σ (2) x σ (4) + X σ ∈ C (12) τd ( − | σ | x σ (2) x σ (3) x σ (1) x σ (4) ++ X σ ∈ C (34) τd ( − | σ | x σ (1) x σ (4) x σ (2) x σ (3) + X σ ∈ C (12)(34) τd ( − | σ | x σ (2) x σ (4) x σ (1) x σ (3) τ d = 2 13 4 , (34) τ d = 1 24 3 , (12)(34) τ d = 2 14 3 , we obtain ψ τd ( w ′ ) = [ x , x ][ x , x ] + [ x , x ][ x , x ] + [ x , x ][ x , x ] + [ x , x ][ x , x ] == [ x , x ] ◦ [ x , x ] + [ x , x ] ◦ [ x , x ] . We stress that, by definition, the polynomial ϕ τd ( w ) is skew-symmetric with respectto each pair of its variables whose indices stand at the same column of τ d . Similarly, ψ τd ( w ) is symmetric with respect to each pair of its variables whose indices stand at thesame row of τ d .In what follows, the introduced endomorphisms are considered for rectangular Youngtables only. The notation k × m means that a table has k rows and m columns.By A r,s denote the free associative superalgebra on the set Y r,s of r even and s oddgenerators . Lemma 5.1.
Let ξ : P n A r,s be a homomorphism induced by a mapping ξ : X n Y r,s ,where n > ( r + 1)( rs + s + 1) . Then for every rectangular ( r + 1) × ( rs + s + 1) table τ d and every word w ∈ P n , the superization of the polynomial ξ ( ϕ τd ( w )) is equal to zero.Proof. By definition of the polynomial f = ϕ τd ( w ), we have f = P σ ∈ C τd ( − | σ | f ′ σ ,where f ′ σ = X ρ ∈ R στd w (cid:0) x ρσ (1) , . . . , x ρσ ( n ′ ) , x n ′ +1 , . . . , x n (cid:1) , n ′ = ( r + 1)( rs + s + 1) . Let us substitute the element i by ξ ( x i ) in each cell of the table τ d . The obtained tablewe denote by ( τ d ) ξ .First, in view of skew-symmetry of f w.r.t. each pair of its variables whose indicesstand at the same column of τ d , we note that g ξ ( f ) can be non-zero only if none ofthe columns of ( τ d ) ξ contains the same even generator of A r,s twice. Consequently, thewhole number of even elements in ( τ d ) ξ is not more then r ( rs + s + 1).Further, if g ξ ( f ) = 0, then we may assume that ] ξ ( f ′ σ ) = 0 at least for one σ ∈ C τd .In this case, we stress that f ′ σ is symmetric w.r.t. each pair of its variables whose indicesstand at the same row of στ d . Hence, none of the rows of ( στ d ) ξ contains the same oddelement twice. Consequently, the whole number of odd elements in ( τ d ) ξ is not morethen ( r + 1) s .Therefore, the number of elements in ( τ d ) ξ is not more then r ( rs + s + 1) + ( r + 1) s = ( r + 1)( rs + s + 1) − . The obtained contradiction completes the proof.By the similar arguments one can prove the following
Lemma 5.2.
Let ξ : P n A r,s be a homomorphism induced by a mapping ξ : X n Y r,s ,where n > ( rs + r + 1)( s + 1) . Then for every rectangular ( rs + r + 1) × ( s + 1) table τ d and every word w ∈ P n , the superization of ξ ( ψ τd ( w )) is equal to zero. .2 Inclusions in the lattice L ( M ) Lemma 5.3.
For all naturals r, s , we have M r * M r − ,s .Proof. Let f k = f k ( u, v, x , . . . , x kr ) be the polynomial f k = ( uv ) X σ ∈ C τd X ρ ∈ R στd ( − | σ | R x ρσ (1) . . . R x ρσ ( kr ) , where τ d = 1 r + 1 . . . ( k − r + 12 r + 2 . . . ( k − r + 2... ... . . . ... r r . . . kr . We stress that applying Lemma 5.1, it is not hard to prove that ˜ f k = 0 is a superidentityin M r − ,s for k = rs + 1. Thus to prove the Lemma it suffices to construct some M r -algebra A ( r ) such that f k takes nonzero values in A ( r ) for all k .Consider the algebra U ( r ) = r X i =1 F · e i with null multiplication and the vector space M ( r ) = X n ∈ Z r F · a n . Let A ( r ) = U ( r ) ∔ M ( r ) be the split null extension with the multiplication induced by thefollowing actions: a n · e i = ( a n +1 (mod r ) , n ≡ i (mod r ) , , n / ≡ i (mod r ) . By definition, A ( r ) is a metabelian algebra generated by the elements a ¯0 + e r , e , . . . , e r − .Consider the mapping ξ : X kr
7→ { e , . . . , e r } defined by the r × k table( τ d ) ξ = e e . . . e e e . . . e ... ... . . . ... e r e r . . . e r , i. e. ξ ( x i ) is the element of the ( τ d ) ξ standing in the cell corresponding to the index i in τ d . To conclude the proof it remains to verify that f k takes a nonzero value in A ( r ) .Indeed, f k ( a ¯0 , e r , ξ ( x ) , . . . , ξ ( x kr )) = k ! r a ¯1 = 0 . Lemma 5.4.
For all naturals r, s , we have M ,s * M r,s − . roof. Let f k = f k ( u, v, x , . . . , x ks ) be the polynomial f k = ( uv ) X ρ ∈ R τd X σ ∈ C ρτd ( − | σ | R x σρ (1) . . . R x σρ ( ks ) , where τ d = 1 2 . . . ss + 1 s + 2 . . . s ... ... . . . ...( k − s + 1 ( k − s + 2 . . . ks . We stress that applying Lemma 5.2, it is not hard to prove that ˜ f k = 0 is a superidentityin M r,s − for k = rs + 1. Thus to prove the Lemma it suffices to construct some M ,s -superalgebra A ( s ) such that ˜ f k takes nonzero values in A ( s ) for all k .Let U ( s ) = U ( s )0 + U ( s )1 be the superalgebra U ( s )0 = { } , U ( s )1 = s X i =1 F · y i with null multiplication and M ( s ) = M ( s )0 + M ( s )1 be the vector space M ( s )0 = X n ∈ Z s F · a n , M ( s )1 = X n ∈ Z s F · a n +1 . Consider a split null extension A ( s ) = U ( s ) ∔ M ( s ) with a multiplication induced by thefollowing actions: a n · y i = ( a n +1 (mod 2 s ) , n ≡ i (mod s ) , , n / ≡ i (mod s ) . One can easily check that A ( s ) is a metabelian superalgebra generated by the odd ele-ments a ¯1 + y , y , . . . , y s .Let us consider the mapping ξ : X ks
7→ { y , . . . , y s } defined by the k × s table( τ d ) ξ = y y . . . y s y y . . . y s ... ... . . . ... y y . . . y s , i. e. ξ ( x i ) is the element of the ( τ d ) ξ standing in the cell corresponding to the index i in τ d . To conclude the proof it remains to verify that ˜ f k takes a nonzero value in A ( s ) .Indeed, 1 k ! s ˜ f k ( a ¯0 , y s , ξ ( x ) , . . . , ξ ( x ks )) = ( a ¯1 , if k is even ,a s +1 , if k is odd . M r ′ ,s * M r,s ′ and M r,s ′ * M r ′ ,s for all nonnegativeintegers r ′ < r , s ′ < s . Moreover, the inclusions M r ′ ,s ⊂ M r,s and M r,s ′ ⊂ M r,s arestrict.Theorem 5 is proved.Corollary 5.1 is an immediate consequence of Theorem 5. Let us deduce Corollary 5.2. For an arbitrary pare ( r, s ) = (0 ,
0) of nonnegative integers,consider the free M r,s -superalgebra A r,s . Let N = Var G ( A r,s ) be the variety generatedby the Grassmann envelope of A r,s . Then, by definition, we have ˜ N = M r,s . It remainsto prove that N doesn’t possess any other basic superrank ( r ′ , s ′ ) such that at least oneof the inequalities r ′ < r , s ′ < s holds. Indeed, we stress that Theorem 5 states thestrict inclusion ( M r,s ∩ M r ′ ,s ′ ) ⊂ M r,s . At the same time, it is clear that N r ′ ,s ′ = M r,s ∩ M r ′ ,s ′ . Thus, N r ′ ,s ′ ⊂ ˜ N , i. e.( r ′ , s ′ ) / ∈ sp b ( N ). Therefore, sp b ( N ) = { ( r, s ) } . The variety of right alternative algebras is a well-known source of examples of nonfinitelybased varieties [2, 8, 12, 16] over a field of characteristic zero. The similar results canbe obtained for the variety of right symmetric algebras. In this section, we constructthe varieties of right alternative and right symmetric algebras having no finite basicsuperrank.Recall that by V h ε i ( ε = ±
1) we denote the subvariety of M distinguished by iden-tities (6) and (7). Let us prove Theorem 6. V h ε i -superalgebra Let U = U + U be the superalgebra U = { } , U = ∞ X i =1 F · y i with null multiplication and M = M + M be the vector space M = ∞ X i =1 F · a i , M = ∞ X i =1 F · w i . Consider a split null extension A h ε i = U ∔ M such that all nonzero products of the basiselements of A h ε i are the following: y i · a i = ε a i · y i = w i +1 , y i · w i = a i . emma 6.1. A h ε i is a V h ε i -superalgebra on a countable set of odd generators.Proof. By definition, A h ε i is metabelian and can be generated by the elements w , y , y , . . . .Moreover, it is not hard to check that the only nonzero associators on the basis elementsof A h ε i are the following:( y i , w i , y i ) = ( y i · w i ) · y i = a i · y i = εw i +1 , ( y i , y i , w i ) = − y i · ( y i · w i ) = − y i · a i = − w i +1 , ( y i +1 , a i , y i ) = − y i +1 · ( a i · y i ) = − y i +1 · ( εw i +1 ) = − εa i +1 , ( y i +1 , y i , a i ) = − y i +1 · ( y i · a i ) = − y i +1 · w i +1 = − a i +1 . Thus one can see that the superization of (6) holds in A h ε i . Finally, it remains toprove that A h ε i satisfies the superization of (7). Actually it suffices to notice that h M , U i ε = 0. V h ε i Let f k,n = f k,n ( u, v, x , z , x , z , . . . , x kn − , z kn − , x kn ) be the polynomial f k,n = ( uv ) X ρ ∈ R τd X σ ∈ C ρτd ( − | σ | L x σρ (1) L z L x σρ (2) L z . . . L x σρ ( kn − L z kn − L x σρ ( kn ) , where τ d = 1 2 . . . nn + 1 n + 2 . . . n ... ... . . . ...( k − n + 1 ( k − n + 2 . . . kn . By Lemma 5.2, the superpolynomial ˜ f k,n is a superidentity of M r,s for k = rs + r + 1and n = s + 1. Therefore in view of Lemma 6.1, to prove the strictness of inclusions V h ε i r,s ⊂ ˜ V h ε i it suffices to verify that ˜ f k,n takes a nonzero value in A h ε i . Indeed, let usdenote by λ k,n = λ k,n (cid:0) L x , L z , L x , L z , . . . , L x kn − , L z kn − , L x kn (cid:1) the superpolynomial on operators of left multiplication such that ˜ f k,n = ( uv ) λ k,n . Thenby the substitution u = y , v = w , x i = y i , and z i = y i +1 for i = 1 , . . . , kn , we have˜ f k,n = a λ k,n (cid:0) L y , L y , L y , L y , . . . , L y kn − , L y kn , L y kn (cid:1) = w L y L y . . . L y kn = w kn +1 = 0 . Theorem 6 is proved.
1. Is it true that for every pare of nonnegative integers ( r, s ) = (0 ,
0) there is a variety V of associative algebras that has the unique basic superrank ( r, s )?28. What condition should satisfy a set of n > Acknowledgments
The article was carried out at the Department of Mathematics and Statistics of theUniversity of S˜ao Paulo (IME-USP), Brazil. The authors are very thankful to the IME-USP for the kind hospitality and the creative atmosphere. For the first author, this paperis a part of the postdoc project supported by the FAPESP 2010/51880–2 (2011–2014)and the PNPD/CAPES/UFRN–PPgMAE (2015). The second author is supported bythe FAPESP 2014/09310–5 and the CNPq 303916/2014–1.
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