Automorphisms and superalgebra structures on the Grassmann algebra
aa r X i v : . [ m a t h . R A ] S e p Automorphisms and superalgebra structures onthe Grassmann algebra
Alan de Araújo Guimarães ∗ , Plamen Koshlukov † Department of Mathematics, State University of Campinas651 Sergio Buarque de Holanda, 13083-859 Campinas, SP, Brazil ‡ e-mails: [email protected] , [email protected] Abstract
Let F be a field of characteristic zero and let E be the Grassmann al-gebra of an infinite dimensional F -vector space L . In this paper we studythe superalgebra structures (that is the Z -gradings) that the algebra E admits. By using the duality between superalgebras and automorphismsof order we prove that in many cases the Z -graded polynomial iden-tities for such structures coincide with the Z -graded polynomial identi-ties of the "typical" cases E ∞ , E k ∗ and E k where the vector space L ishomogeneous. Recall that these cases were completely described by DiVincenzo and Da Silva in [13]. Moreover we exhibit a wide range of non-homogeneous Z -gradings on E that are Z -isomorphic to E ∞ , E k ∗ and E k . In particular we construct a Z -grading on E with only one homoge-neous generator in L which is Z -isomorphic to the natural Z -grading on E , here denoted by E can . Let F be a field of characteristic zero. If L is a vector space over F withbasis e , e , . . . , the infinite dimensional Grassmann algebra E of L over F has a basis consisting of and all monomials e i e i · · · e i k where i < i < · · · < i k , k ≥ . The multiplication in E is induced by the rule e i e j = − e j e i for all i and j ; in particular e i = 0 for each i . The Grassmann algebra isthe most natural example of a superalgebra , it is widely used in various partsof Mathematics and also in Theoretical Physics. The Grassmann algebra E is one of the most important algebras satisfying a polynomial identity, alsoknown as PI-algebras. Its polynomial identities were described by Latyshev[24], and later on, in great detail, by Krakowski and Regev [23]. It was observedby P. M. Cohn that the Grassmann algebra satisfies no standard identities.Recall that the standard polynomial s n is the alternating sum of all monomialsobtained by permuting the variables in x x · · · x n . In fact the true importance ∗ Supported by PhD Grant from CNPq, Brazil, and by CNPq grant No. 421129/2018-2 † Partially supported by FAPESP grant No. 2014/09310-5 and by CNPq grant No.304632/2015-5. ‡ A. A. Guimarães’ current address: Department of Mathematics, Federal University of RioGrande do Norte, 59078-970 Natal, RN, Brazil
1f the Grassmann algebra in the area of PI algebras was revealed through thetheory developed by A. Kemer in the mid eighties. Kemer proved that everyassociative PI-algebra over a field of characteristic zero is PI-equivalent (thatis it satisfies the same polynomial identities) to the Grassmann envelope of afinite dimensional associative superalgebra, see [20, 21]. In Kemer’s structuretheory for the ideals of identities of associative algebras the natural Z -gradingon E was used. Here we point out that this superalgebra structure on E isdefined by E can = E (0) ⊕ E (1) . Here E (0) is the vector subspace of E spannedby 1 and all monomials of even length and E (1) is spanned by all monomials ofodd length. It is immediate to see that E (0) is the centre of E while E (1) is the"anticommuting" part of E . Recall here that if A = A ⊕ A is a Z -gradedalgebra then its Grassmann envelope is defined as E ( A ) = A ⊗ E (0) ⊕ A ⊗ E (1) .Clearly the Grassmann algebra admits various other gradings. A naturaland interesting problem in this direciton is to investigate the structure of groupgradings on the Grassmann algebra and the corresponding graded polynomialidentities.Let [ x , x ] = x x − x x be the commutator of x and x . We define [ x , x , x ] = [[ x , x ] , x ] and so on for more variables. In other words when noadditional brackets are written we assume the longer commutators left normed.As mentioned above, Latyshev, and Krakowski and Regev proved that the triplecommutator [ x , x , x ] forms a basis for the ordinary polynomial identities of E (see [24, 23]). Krakowski and Regev also computed the codimension sequenceof the ideal of identities of E . The interested reader could find more detailsconcerning the polynomial identities of the Grassmann algebra in [17] and thereferences therein.Group gradings on algebras and the corresponding graded polynomial iden-tities have been extensively studied in PI theory during the last three decades.The graded polynomial identities satisfied by an algebra are “easier” to describethan the ordinary ones. Here we recall that the polynomial identities of an asso-ciative algebra A are known in very few instances. These include the Grassmannalgebra E (over any field), the × matrix algebra [28, 15] (in characteristic0), [22] (over infinite fields of characteristic different from 2), and [26] (over afinite field); the upper triangular matrix algebras [25, 9], and the algebra E ⊗ E [27] in characteristic 0.On the other hand the gradings on matrix algebras are known [6]. Thegraded identities of these algebras are well understood, see for example [30,31, 4, 5] for the natural gradings on the n × n matrices. The gradings andthe corresponding graded identities on the upper triangular matrices are alsowell known [29, 10]. The graded identities for natural gradings on classes ofimportant algebras also have been described, see for example [11, 12], and also[1], as well as the references in these three papers.The Grassmann algebra admits the natural grading by the cyclic group Z of order two. Its structure from the point of view of the PI theory is well knownand easy to deduce, see for example [18]. In recent years a substantial numberof papers has presented results on gradings and their graded identities for theGrassmann algebra. In all of them, two conditions have been imposed, namely: • The grading group G is finite. • All generators e , e , . . . , e n , . . . of E are homogeneous. This means thatthe basis of the vector space L is homogeneous.2hen a grading on E satisfies the latter condition it is called homogeneousgrading . In [13] the authors studied all homogeneous superalgebra structuresdefined on the Grassmann algebra. These were denoted as E k , E k ∗ and E ∞ (we give, for the readers’ convenience, their definitions in the next section of thepaper). Their Z -graded polynomial identities were also described in [13]. Inthis context the following question arises naturally: • Is the list { E k , E k ∗ , E ∞ } of the Grassmann superalgebras complete?Clearly one has to seek an answer to the above question up to graded iso-morphisms.In order to look for a possible structure of a superalgebra on the Grassmannalgebra (which does not fit in any one of the three cases above) it is necessaryto study Z -gradings on E without the hypothesis of homogeneity.We refer the reader to [14, 8] and the references therein for various resultsconcerning group grading on the Grassmann algebra. In [14] the authors ap-proach the problem of grading on E by a cyclic group of prime order, and in [8]the author addresses the case of finite abelian groups.In the present paper we shall study general structures of superalgebras onthe Grassmann algebra, without supposing the condition of homogeneity ofthe vector space L . To this end we use the duality between Z -grading andautomorphisms of order on E . This duality is well known. It relies on thefact that if G is a finite abelian group then G is isomorphic to its dual groupassuming that the field is large enough. As we are interested in gradings bythe group Z we need no further assumptions on the base field (apart from itscharacteristic being different from 2). Thus if ϕ ∈ Aut ( A ) is an automorphismof an algebra A of order two, that is ϕ = 1 then one has a Z -grading on A given by A = A ,ϕ ⊕ A ,ϕ . Here A ,ϕ and A ,ϕ are the eigenspaces in A associated to eigenvalues 1 and − of the linear transformation ϕ . Reciprocallyto each Z -grading on A one associates an automorphism of A of order 2 asfollows. If A = A ⊕ A is the Z -grading the automorphism ϕ is defined by ϕ ( a + a ) = a − a for every a i ∈ A i , i = 0 , 1. We shall need this duality inthe form of a duality between group gradings and group actions, see for example[18] for a discussion in the general case.Let us fix a basis β = { e , e , . . . , e n , . . . } of the vector space L and anautomorphism ϕ ∈ Aut ( E ) such that ϕ = 1 , we consider the set I β = { n ∈ N | ϕ ( e n ) = ± e n } . There exist the following four possibilities:1. I β = N .2. I β = N is infinite.3. I β is finite and non empty.Given a basis β of L it is possible that I β = ∅ but I β ′ = ∅ for some otherbasis β ′ of L , see Example 2. Hence the fourth possibility that we shall consideris the following.4. I γ = ∅ for every basis γ of the vector space L .3he first possibility, when I β = N , corresponds to the homogeneous case. Asalready mentioned, it was completely described by Di Vincenzo and Da Silva intheir paper [13].Therefore in order to construct non-homogeneous Grassmann superalgebrasone is led to deal with automorphisms ϕ on E satisfying either one of the con-ditions 2, 3 or 4. In this paper we shall exhibit structures of types 2 and 3.We shall also prove that these structures are in fact equivalent to homogeneoussuperalgebras. In particular we shall construct one kind of a Grassmann super-algebra where only one generator among the e i is homogeneous. Neverthelessit will turn out that such a superalgebra is equivalent to the one coming fromthe natural grading E can .Furthermore we shall prove that the fourth structure does not exist in quitemany cases. We will not provide examples of Grassmann superalgebras of type4 since we could not find any. In fact we have some ground to conjecture thatthe fourth case does not happen at all. But we have not been able to prove ityet. Let F be a field and let A be a unitary associative F -algebra. We say that A is a Z -graded algebra (or superalgebra) whenever A = A ⊕ A where A , A are F -subspaces of A satisfying A i A j ⊂ A i + j for i , j ∈ Z . The vectorsubspace A i is called the i -homogeneous component of A . If a ∈ A i we say that a is homogeneous and is of (homogeneous) degree k a k = i . A vector subspace(subalgebra, ideal) W ⊂ A is homogeneous if W = ( W ∩ A ) ⊕ ( W ∩ A ) .Here we point out that we use "freely" the terms superalgebra and Z - graded algebra as synonymous although this is an abuse of terminology. Inthe associative case they are indeed synonymous while in the nonassociativesetting they are not. Indeed, a Lie or a Jordan superalgebra is not, as a rule,a Lie or a Jordan algebra. The correct setting in the general case should beas follows. Let A = A ⊕ A be a Z -graded algebra and let V be a varietyof algebras (not necessarily associative). Then A is a V -superalgebra whenever A ⊗ E (0) ⊕ A ⊗ E (1) is an algebra belonging to V . (We draw the readers’attention that one does not require A ∈ V .) Since we shall deal with associativealgebras only such a distinction is not relevant for our purposes, and we are notgoing to make any difference between superalgebras and Z -graded algebras.If A , B are superalgebras, a homomorphism f : A → B is a Z - graded ho-momorphism if f ( A i ) ⊂ B i for all i ∈ Z . When there exists a Z -gradedisomorphism between A and B we say that A and B are Z -isomorphic or equivalent .One defines a free object in the class of superalgebras by considering the free F -algebra over the disjoint union of two countable sets of variables, denoted by Y and Z . We assume further that the elements of Y are of degree zero andthe elements of Z are of degree . This algebra is denoted by F h Y ∪ Z i . Itseven part is the vector space spanned by all monomials whose degree countingonly the elements of Z , is an even integer. The remaining monomials span theodd component. It is straightforward that F h Y ∪ Z i is a free algebra in thesense that for every superalgebra A and for every map ϕ : Y ∪ Z → A such that ϕ ( Y ) ⊆ A and ϕ ( Z ) ⊆ A there exists unique homomorphism of Z -graded4lgebras F h Y ∪ Z i → A that extends ϕ .We say that the polynomial f ( y , . . . , y l , z , . . . , z m ) ∈ F h Y ∪ Z i is a Z - graded polynomial identity for a superalgebra A if f ( a , . . . , a l , b , . . . , b m ) = 0 for all substitutions such that k a i k = 0 and k b j k = 1 . The set T ( A ) of all Z -graded polynomial identity of A is a homogeneous ideal of F h Y ∪ Z i . It iscalled the T -ideal of A , and denoted by T ( A ) .The description of all structures of a superalgebra on a given algebra is animportant task in Ring theory, and in particular in PI theory. As commentedabove, in [13] the authors considered Z -grading on the Grassmann algebra E over a field of characteristic zero such that the vector space L is homogeneousin the grading. There exist three structures, namely: k e i k k = ( , if i = 1 , . . . , k , otherwise , k e i k k ∗ = ( , if i = 1 , . . . , k , otherwise , and k e i k ∞ = ( , if i is even , otherwise . These provide us with three superalgebras, E k , E k ∗ and E ∞ , respectively. The Z -graded polynomial identities of E k , E ∞ , and E k ∗ were also described in [13].These structures were studied over an infinite field of positive characteristic p > in [7] and also over finite fields, see [19]. From now on we shall call E k , E ∞ and E k ∗ the homogeneous Grassmann superalgebras . Consider A an associative F -algebra. There exists a natural duality between Z -gradings and automorphisms of order 2 on A . The duality, as commentedabove, is defined as follows.If ϕ ∈ Aut ( A ) is such that ϕ = 1 then A ϕ = A ,ϕ ⊕ A ,ϕ where thehomogeneous components are the eigenspaces corresponding to the eigenvalues and − of ϕ , respectively. The decomposition in a direct sum of the eigenspacesexists since F is of characteristic different from 2.The general facts about duality between gradings and actions of groups canbe found, for example, in [16, Chapter 3] and also in the paper [18]. We observethat the homogeneous Z -gradings on E correspond to the automorphisms on E satisfying ϕ ( e i ) = ± e i . If ϕ ∈ Aut ( E ) with ϕ = 1 we observe that e i = ( e i + ϕ ( e i )) / e i − ϕ ( e i )) / , for i ∈ N . Setting a i = e i + ϕ ( e i ) / we have • ϕ ( e i ) = − e i + 2 a i , • ϕ ( a i ) = a i , that is, a i is of degree zero in the Z -grading E ϕ , and5 ϕ ( e i − a i ) = − ( e i − a i ) , that is, e i − a i is of degree in the Z -grading E ϕ . Definition 1
Assume ϕ ∈ Aut ( E ) is of order two and let E can = E (0) ⊕ E (1) be the natural Z -grading on E . We say that ϕ is of canonical type if1. ϕ ( E (0) ) = E (0) ;2. ϕ ( E (1) ) = E (1) . Concerning the canonical automorphisms it is easy to check that • The superalgebras E ∞ , E k ∗ and E k correspond to automorphisms ofcanonical type. • If ϕ is an automorphism of order on E we have that ϕ is of canonicaltype if only if a i ∈ E (1) for all i ∈ N .Let us fix a basis β = { e , e , . . . , e n , . . . } of the vector space L and an au-tomorphism ϕ ∈ Aut ( E ) such that ϕ = 1 . Then ϕ , as a linear transformation,has eigenvalues 1 and − only, and moreover, there exists a basis of the vectorspace E consisting of eigenvectors. (It is well known from the elementary Linearalgebra that this fact does not depend on the dimension of the vector space aslong as the characteristic of F is different from 2.) Then E = E (1) ⊕ E ( − where E ( t ) is the eigenspace for the eigenvalue t of the linear transformation ϕ .One considers the intersections L ( t ) = L ∩ E ( t ) , t = ± . Changing the basis β , if necessary, one may assume that L ( t ) is the span of β ∩ L ( t ) . Clearly thischange of basis gives rise to a homogeneous automorphism of E and we can takethe composition of it and then ϕ . We shall assume that such a change of basishas been done.Denote I β = { n ∈ N | ϕ ( e n ) = ± e n } . We shall distinguish the following fourpossibilities:1. I β = N .2. I β = N is infinite.3. I β is finite and non empty.4. I γ = ∅ for every linear basis γ of L .The automorphisms of type 1 have been completely described by Di Vincenzoand Da Silva [13]. Therefore we shall focus on the remaining three cases.We shall call these automorphisms (and also the corresponding Z -gradings), automorphisms ( Z -grading) of type 1, 2, 3, and 4, respectively. In this section we shall describe automorphisms of type 2. We start with thefollowing proposition.
Proposition 1
Let ϕ ∈ Aut ( E ) be an automorphism such that ϕ = 1 . Supposethat ϕ is an automorphism of type 2, that is, I β = N is infinite. Then ϕ is ofcanonical type. roof Let E ϕ = E ,ϕ ⊕ E ,ϕ be the superalgebra structure on E induced by ϕ , we denote J = { j ∈ N | j / ∈ I β } . For the sake of simplicity we shall write I = I β .For each j ∈ J we have ϕ ( e j ) = ± e j . Therefore e j is not homogeneous inthe grading E ϕ = E ,ϕ ⊕ E ,ϕ . Thus we can write e j = b j + c j , with b j ∈ E ,ϕ and c j ∈ E ,ϕ . Moreover we have that each summand b j and c j is non zero.Since ϕ ( c j ) = − c j we have ϕ ( e j − b j ) = − e j + b i and this implies the equality ϕ ( e j ) = − e j + 2 b j . Next we prove that each b j isa linear combination of monomials of odd length. To this end we write b j = b ej + b oj . Here b ej is the linear combination of the monomials of even length in b j and b oj is the one consisting of all monomials of odd length in b j .Choose now t ∈ I such that e t does not belong to the support of the sum-mands of b ej (this is always possible since the set I is infinite). As e j e t + e t e j = 0 we obtain that ϕ ( e j ) ϕ ( e t ) + ϕ ( e t ) ϕ ( e j ) = 0 . By ϕ ( e t ) = ± e t it follows that ( − e j + 2 b j ) e t + e t ( − e j + 2 b j ) = 0 which in turn implies b j e t + e t b j = 0 . Hence ( b ej + b oj ) e t + e t ( b ej + b oj ) = 0 . It follows that e t b ej = 0 and b ej = 0 . Thereforethe element b j is a linear combination of monomials of odd length for all j ∈ J ,and the proof follows.Fix a basis β of L and let ϕ be an automorphism of type 2. We define thefollowing sets of indices: • I + = { i ∈ I | ϕ ( e i ) = e i } , • I − = { i ∈ I | ϕ ( e i ) = − e i } , and • J = { j ∈ N | j / ∈ I } . Since ϕ is of type 2 the set I = I + ∪ I − is infinite. Therefore we have to takeinto account the following five cases:S1. Both | I + | and | I − | are infinite.S2. | I + | is infinite but | I − | and | J | are finite.73. | I + | is infinite, | I − | is finite and | J | is infinite.S4. | I + | is finite, | I − | is infinite and | J | is finite.S5. | I + | is finite, | I − | is infinite and | J | is also infinite.Using the notation introduced above we have the following proposition. Proposition 2
The T -ideal of the graded identities of a Grassmann superal-gebra of Case S1 coincides with T ( E ∞ ) . Proof
Suppose that ϕ is as in case S1, then the sets I + and I − are infinite. Let B be the subalgebra of E ϕ generated by and e n , for every n ∈ I + ∪ I − . In thiscase B ⊂ E ϕ is a homogeneous subalgebra and B ≃ E ∞ (isomorphism of Z -graded algebras). Hence h [ x , x , x ] i T = T ( E ∞ ) ⊃ T ( E ϕ ) where α ( x i ) = 0 or α ( x i ) = 1 , for all i = 1 , 2, 3. Therefore it follows T ( E ϕ ) = T ( E ∞ ) .Suppose now E is equipped with a grading as in Case S2. Thus up toreordering the basis of L the action of ϕ on the generators is given by ϕ ( e n ) = − e n , if ≤ n ≤ k − e n + 2 a n , if k + 1 ≤ n ≤ k + te n , if n > k + t . Here we write, as above, a j = ( e j + ϕ ( e j )) / . We have that • E ,ϕ ⊃ V = span F { , e n , a k +1 , . . . , a k + t | n > k + t } . • E ,ϕ ⊃ V = span F { e , . . . , e k , ( e k +1 − a k +1 ) , . . . , ( e k + t − a k + t ) } . Remark 1
More precisely, the component E ,ϕ is generated by all products ofelements in V and V with an even number of factors in V . The component E ,ϕ is generated by these products with an odd number of factors in V . Let B be the subalgebra of E ϕ generated by the elements e , . . . , e k , e n , forevery n > k + t . Then B is homogeneous and E k ∗ ≃ B ⊂ E ϕ . Moreover, each translation ( e k +1 − a k +1 ) , . . . , ( e k + t − a k + t ) has zero squarebecause the automorphism ϕ is canonical. Thus it follows immediately that z z . . . z k + t +1 is a Z -graded polynomial identity for E ϕ . Therefore we obtain T ( E k ∗ ) ⊃ T ( E ϕ ) ⊃ T ( E ( k + t ) ∗ ) . By imposing an additional condition we shall describe the T -ideal T ( E ϕ ) .Beforehand we state a theorem due to Anisimov’s that will be important forour goals. Theorem 3 (Anisimov, [3])
Let ϕ be an automorphism of order 2 of E , andlet ϕ be of canonical type. Suppose that at least one of the following two condi-tions holds: . dim L − = l < ∞ and Y l +1 j =1 ( ϕ ( e i j ) − e i j ) = 0 , for any l + 1 generators e i , . . . , e i l +1 .2. dim L = l < ∞ and Y l +1 j =1 ( ϕ ( e i j ) + e i j ) = 0 , for any l + 1 generators e i ,. . . , e i l +1 .Then one has that T ( E ϕ ) = T ( E ϕ l ) . By using Anisimov’s theorem 3 and an additional we can describe the Z -graded polynomial identities for the Grassmann superalgebras in the Case S2. Theorem 4
Let ϕ be an automorphism of E of type S2 and T ( E ϕ ) its T -idealof Z -graded polynomial identities. If each element a k +1 , . . . , a k + t is a linearcombination of monomials of length ≥ then T ( E ϕ ) = T ( E ϕ l ) . Proof
By hypothesis we know that ϕ is of type S2 and therefore I − and J arefinite sets. Let us assume | I − | + | J | = l. By Proposition 1 we have that ϕ is of the canonical type and dim L − = l < ∞ ,and moreover Y l +1 j =1 ( ϕ ( e i j ) − e i j ) = 0 , for any choice of l + 1 generators e i , . . . , e i l +1 . By Theorem 3 it follows that T ( E ϕ ) = T ( E ϕ l ) . Theorem 5
Let ϕ be an automorphism of E of type S4 and T ( E ϕ ) its T -idealof Z -graded polynomial identities. If each element a k +1 , . . . , a k + t is a linearcombination of monomials of length ≥ then T ( E ϕ ) = T ( E ϕ l ) . Proof
The proof repeats verbatim the one of Theorem 4.In the remaining cases we can not guarantee the equality but at least one ofthe inclusions still holds.
Theorem 6
Let ϕ be an automorphism of type 2 of E . The following state-ments hold:1. If ϕ satisfies S3 then T ( E ϕ ) ⊆ T ( E k ∗ ) .2. If ϕ satisfies S5 then T ( E ϕ ) ⊆ T ( E k ) . Proof
The proof is similar to that of Theorem 4 and therefore we omit it.These results show that the Z -graded polynomials identities of many su-peralgebras of type 2 coincide with the identities of the homogeneous case. Inother words one cannot distinguish these superalgebras by means of their gradedpolynomial identities. In this subsection we construct certain Z -gradings of type 2. We shall provethat in many “typical” cases the superalgebras of type 2 are Z -isomorphic tohomogeneous superalgebras. In what follows we present a method to constructsuch structures.Choose an infinite set I ⊂ N , I = N , and define an action ϕ ( e i ) = ± e i whenever i ∈ I . As before we form the sets:9 I + = { i ∈ I | ϕ ( e i ) = e i } ; • I − = { i ∈ I | ϕ ( e i ) = − e i } ; • J = { j ∈ N | j / ∈ I } .Note that I = I + ∪ I − , a disjoint union. For each j ∈ J , we construct theelements d j ∈ E in the following way:1. The element d j is a linear combination of monomials of odd length. All monomials that occur in d j are products of generators whose indicesbelong to I .3. All monomials that occur in d j have an even numbers of factors in I − .We extend the action of ϕ on all e i as follows. ϕ ( e i ) = e i , if i ∈ I + − e i , if i ∈ I − − e i + 2 d i , if i ∈ J Hence • Condition (1) implies that ϕ can be extended to an endomorphism of E ,and • Conditions (2) and (3) imply that ϕ ( d j ) = d j for every j ∈ J .We claim that ϕ is an automorphism of order 2. Indeed, if i ∈ I we notethat ϕ ( e i ) = e i . If j ∈ J we have ϕ ( e j ) = ϕ ( − e j + 2 d j ) = − ( − e j + 2 d j ) + 2 d j = e j . This ensures that ϕ is an automorphism of order 2 of E . Let E ϕ = E ,ϕ ⊕ E ,ϕ be the superalgebra defined by ϕ . Its homogeneous components are • E ,ϕ ⊃ span F { e i , d j | i ∈ I + , j ∈ J } and • E ,ϕ ⊃ span F { e i , ( e j − d j ) | i ∈ I − , j ∈ J } .We recall that the precise description for these components was done inRemark 1. We call the previous method for obtaining a Z -grading on E the method . For each automorphism ϕ constructed by means of the method , itis straightforward to deduce that that the superalgebra E ϕ is not homogeneous.Moreover if J is infinite then there exist infinitely many generators of E whichare not homogeneous.In the following proposition we show that the construction from method ,while yielding many non homogeneous generators of E , produces superalgebraswhich do not differ significantly from the homogeneous ones. Proposition 7
Let ϕ ∈ Aut ( E ) be an automorphism of order 2 constructed bythe method . Then there exists a Z -graded isomorphism between E ϕ and someof the superalgebras E ∞ , E k ∗ or E k , for some k . roof Let E ϕ = E ,ϕ ⊕ E ,ϕ be a Z -grading on E induced by some automor-phism ϕ constructed by method .We define E = E (0) ⊕ E (1) the homogeneous Z -grading induced by thefollowing grading on the generators of E : k e n k = 0 , if n ∈ I + , k e m k = 1 , otherwise.Clearly such a grading produces one of the superalgebras E ∞ , E k ∗ or E k . Nowlet f ϕ : E (0) ⊕ E (1) −→ E ϕ = E ,ϕ ⊕ E ,ϕ be defined by f ϕ ( e i ) = ( e i , if i ∈ Ie i − d i , if i ∈ J .
Since the images of f ϕ satisfy the relation x x + x x = 0 we can extend f ϕ toa homomorphism of E . By the definition of f ϕ we can see that it preserves thedegrees of the generators, it is invertible and its inverse g ϕ : E ϕ −→ E (0) ⊕ E (1) is of the form g ϕ ( e i ) = ( e i , if i ∈ Ie i + d i , if i ∈ J .
Therefore E ϕ is isomorphic to one of the superalgebras E ∞ , E k ∗ or E k . Example 1
Let I = I + = { , , , . . . } and define the automorphism ϕ by itsaction on the generators of E : ϕ ( e i ) = ( e i , if i = 1 − e + 2 e e e , if i = 1 . In this case the grading is • E ,ϕ ⊃ span F { e , e , e , . . . } . • E ,ϕ ⊃ span F { e − e e e } .Consider E endowed with the Z -grading produced by the method presented inthe previous proposition. In other words, k e k = 1 and k e i k = 0 for i = 1 (it isexactly the superalgebra E ∗ ). Hence the action f ϕ : E → E ϕ , given by f ϕ ( e i ) = ( e i , if i = 1 e − e e e , if i = 1 , can be extended to a Z -graded isomorphism between E ∗ and E ϕ . Consequently, T ( E ϕ ) = h [ x , x , x ] , z z i T . On the other hand we can define another action as follows: ψ ( e i ) = ( e i , if i = 1 − e + 2( e + e + e e e + e e e e e ) , if i = 1 . Note that ψ yields the superalgebra E ,ψ ⊃ span F { e , e , e , . . . } . • E ,ψ ⊃ span F { e − ( e + e + e e e + e e e e e ) } .By the previous proposition we obtain that T ( E ϕ ) = T ( E ψ ) = T ( E ∗ ) . Z -gradings on E In this subsection we describe a particular way of constructing Z -gradings on E . We call these gradings triangular Z -gradings . For each N ∈ N , we shallconstruct a subgroup of Aut ( E ) of order N which has the property that all itselements induce Grassmann superalgebras.Recall that we denote by E can = E (0) ⊕ E (1) the natural Z -grading of E ; if k ∈ N , we denote: E ( e , . . . , e k ) = the span of all monomials without factors in { e , . . . , e k } . Definition 2
For every n ∈ N , the automorphism T n : E → E defined by T n ( e j ) = ( e j , if j = n, − e j + 2 P j , if j = n, where P n ∈ E ( e , . . . , e n ) ∩ E (1) , is called a triangular automorphism of index n . Remark 2
It is easy to prove that each triangular automorphism • is of order 2, • can be constructed by the method , • induces a Grassmann superalgebra that is Z -isomorphic to E ∗ . Take N ∈ N and let T , . . . , T N be triangular automorphisms such that P j ∈ E ( e , . . . , e N ) ∩ E (1) for every j = 1 , . . . , N .Denote by τ N the subgroup of Aut ( E ) generated by T , . . . , T N , that is τ N = h T , . . . , T N i . Proposition 8
Let τ N be the subgroup of Aut ( E ) constructed above. Then wehave that:1. τ N is an abelian group.2. τ N is of order N .3. Each element of τ N induces a Grassmann superalgebra.4. If ϕ ∈ τ N and ϕ = T j ◦ · · · ◦ T j s , with j i = j l , for i = l , then E ϕ is Z -isomorphic to E s ∗ . roof We start with the proof of statement 1. Let u , v ∈ { , . . . , N } with u < v . If j = u , j = v , then clearly T u ◦ T v ( e j ) = T v ◦ T u ( e j ) = e j . Moreover T u ◦ T v ( e u ) = T u ( e u ) = − e u + 2 P u , T v ◦ T u ( e u ) = T v ( − e u + 2 P u ) = − e u + 2 P u hence T u ◦ T v ( e u ) = T v ◦ T u ( e u ) . In the same way we compute that T u ◦ T v ( e v ) = T v ◦ T u ( e v ) , proving that τ N is an abelian group. Statement 2 is an immediateconsequence of statement 1.Now we prove statement 3. If ϕ ∈ τ N , it is enough to show that ϕ = 1 . Let ϕ = T j ◦ · · · ◦ T j s . Obviously ϕ ( e n ) = e n whenever n / ∈ { j , . . . , j s } . On the other hand if n = j t for some t = 1 , . . . , s , it is easy to see that: ϕ ( e j t ) = − e j t + 2 P j t and we get ϕ ( e j t ) = e j t .In order to prove Statement 4 one applies the argument used in the case ofmethod 1. We focus our attention on automorphisms of type 3. Some of the statementsare similar to results of the previous section. We start with the following propo-sition.
Proposition 9
Let ϕ ∈ Aut ( E ) be an automorphism of order 2 and supposethat ϕ is of type 3, that is, I β = { , . . . , k } where k ≥ . Then one of thefollowing two possibilities holds: • either ϕ is of the canonical type, or • ϕ is defined by ϕ ( e n ) = ( ± e n , if ≤ n ≤ k − e n + 2 e · · · e k V n + 2 W n , if n > k . Here the set { V n } n>k consists of nonzero elements, its elements are of the sameparity as that of k , each V n has no summands with factors among e , . . . , e k , and the set { W n } n>k is formed by elements of odd length. Moreover thefollowing relations hold: e · · · e k ( V p e r + V r e p ) = 2 e · · · e k ( V p W r + V r W p ) , for all r , p > k (1) Proof
As in the previous arguments we can write, for n > k , ϕ ( e n ) = − e n + 2 a n where each a n = ( e n + ϕ ( e n )) / is invariant under ϕ , that is ϕ ( a n ) = a n . Onceagain we denote a n = a en + a on where a en and a on are the even and the odd13art of a n , respectively. For each t ∈ { , . . . , k } and for each n > k , we have e t e n + e n e t = 0 . Therefore ϕ ( e t ) ϕ ( e n ) + ϕ ( e n ) ϕ ( e t ) = 0 and thus e t a en = 0 .Hence either a en = 0 or each monomial in the part a en has e t as a factor. Inboth cases, since t = 1 , . . . , k we obtain a en = e · · · e k V n where either V n = 0 or the monomials in V n have no factors among e , . . . , e k ,and all of them are of even length if k is even, or of odd length, whenever k isodd.Now we consider n > k and define W n = a on . In this case ϕ ( e n ) = − e n + 2 e · · · e k V n + 2 W n where each W n is a linear combination of odd monomials and V n is a linearcombination of even or odd monomials.If p , r > k , we have ϕ ( e p ) ϕ ( e r ) + ϕ ( e r ) ϕ ( e p ) = 0 and then ( − e p + 2 e · · · e k V p + 2 W p )( − e r + 2 e · · · e k V r + 2 W r )+( − e r + 2 e · · · e k V r + 2 W r )( − e p + 2 e · · · e k V p + 2 W p ) = 0 . Hence − e p e · · · e k V r − e · · · e k V p e r + 4 e · · · e k V p W r + 4 W p e · · · e k V r − e r e . . . e k V p − e · · · e k V r e p + 4 e · · · e k V r W p + 4 W r e · · · e k V p = 0 and in this way − e p e · · · e k V r − e r e · · · e k V p + 8 e · · · e k V p W r + 8 e · · · e k V r W p = 0 . All these computations yield the relation: e · · · e k ( V p e r + V r e p ) = 2 e · · · e k ( V p W r + V r W p ) , for all p , r > k. We have to examine separately the following two cases:Case 1. If V n = 0 , for all n > k , then clearly ϕ is of canonical type.Case 2. Otherwise we have that ϕ is defined by ϕ ( e n ) = ( ± e n , if ≤ n ≤ k − e n + 2 e · · · e k V n + 2 W n , if n > k where { V n } n>k and { W n } n>k both satisfy the conditions stated in theproposition. Furthermore the following relations hold e · · · e k ( V p e r + V r e p ) = 2 e · · · e k ( V p W r + V r W p ) . Therefore the proof is now complete. 14 .1 Concrete superalgebras of type 3
Now we exhibit a method that provides us with automorphisms of E of type 3.We shall see that they need not be necessarily of canonical type.Let t , k be integers such that t is odd and k ∈ { , , . . . } . We shall denoteby I the set I = { , . . . , k, k + 1 , . . . , k + t } ⊂ N . For each i ∈ I we define theaction ϕ ( e i ) = ± e i so that I + = { , . . . , k } and I − = { k + 1 , . . . , k + t } . For n > k + t we put ϕ ( e n ) = − e n + 2 e · · · e k e k +1 · · · e k + t e n . Hence the action of ϕ on the generators is given by ϕ ( e n ) = e n , if ≤ n ≤ k − e n , if k + 1 ≤ n ≤ k + t − e n + 2 e · · · e k e k +1 · · · e k + t e n , otherwise . Now we note that • ϕ preserves the relations e i e j + e j e i = 0 and so it can be extended to ahomomorphism of E . • For n > k + tϕ ( e n ) = ϕ ( − e n + 2 e · · · e k e k +1 · · · e k + t e n )= − ( − e n + 2 e · · · e k e k +1 · · · e k + t e n )+2 e · · · e k ( − t e k +1 · · · e k + t ( − e n + 2 e · · · e k e k +1 · · · e k + t e n ) . Therefore ϕ ( e n ) = e n .Thus ϕ is an automorphism of order 2 on E . The induced superalgebra E ϕ = E ,ϕ ⊕ E ,ϕ has its homogeneous components described by • E ,ϕ ⊃ span F { e , . . . , e k } , • E ,ϕ ⊃ span F { e k +1 , . . . , e k + t , ( e n − e · · · e k e k +1 · · · e k + t e n ) | n > k + t } . We call the method just described the method for obtaining Grassmann super-algebras. Whenever k is an even integer, the automorphism ϕ is not of canonicaltype. Nevertheless, as we shall see below, the superalgebras obtained in this waywill be isomorphic to homogeneous ones. Proposition 10
Let E ϕ be a Z -grading obtained according to method . Then E ϕ and E k are isomorphic as Z -graded algebras. Proof
Let f ϕ : E k −→ E ϕ be the function defined on the generators by f ϕ ( e n ) = e n , if ≤ n ≤ ke n , if k + 1 ≤ n ≤ k + te n − e · · · e k e k +1 · · · e k + t e n , otherwise .
15t is easy to check that f ϕ preserves the relations e i e j + e j e i = 0 and thus wecan extend f ϕ to an endomorphism of the algebra E . Moreover f ϕ ( e · · · e k e k +1 · · · e k + t e n ) = e · · · e k e k +1 · · · e k + t f ϕ ( e n )= e · · · e k e k +1 · · · e k + t ( e n − e · · · e k e k +1 · · · e k + t e n )= e · · · e k e k +1 · · · e k + t e n . Now we consider the homomorphism g ϕ : E ϕ → E k defined by g ϕ ( e n ) = e n , if ≤ n ≤ ke n , if k + 1 ≤ n ≤ k + te n + e · · · e k e k +1 · · · e k + t e n , otherwise . By using the same argument as above we obtain g ϕ ( e · · · e k e k +1 · · · e k + t e n ) = e · · · e k e k +1 · · · e k + t e n . Therefore if n > k + t we conclude that f ϕ ◦ g ϕ ( e n ) = f ϕ ( e n + e · · · e k e k +1 · · · e k + t e n )= ( e n − e · · · e k e k +1 · · · e k + t e n ) + e · · · e k e k +1 · · · e k + t e n = e n . We also obtain that g ϕ ◦ f ϕ ( e n ) = e n , and thus we prove that f ϕ is an automor-phism of E . By construction the automorphism f ϕ is Z -graded. Therefore thesuperalgebras E ϕ and E k are Z -isomorphic, and our proof is complete. Proposition 11
Let ϕ : E → E be the homomorphism defined by its action onthe generators of E : ϕ ( e n ) = ( − e , if n = 1 − e n + 2 e e n , if n > . Then ϕ is an automorphism of order 2, and E ϕ and E can are Z -isomorphic. Proof
The proof follows from the construction of method . Remark 3
Let E ϕ = E ,ϕ ⊕ E ,ϕ be the superalgebra defined in Proposition 11.It is immediate that its homogeneous components are as follows:1. E ,ϕ ⊃ span F { e , ( e n − e e n ) | n > } ,2. E ,ϕ = Z ( E ) .In this case just one of the generators, namely e , of E is homogeneous. More-over the automorphism ϕ is not of canonical type. However Proposition 11guarantees that E ϕ is isomorphic to the natural Z -grading E can . Gradings and automorphisms of type 4
In this section we consider automorphisms of type . Recall that these areautomorphisms such that for every basis β of the underlying vector space L ,one has I β = ∅ . Here I β are the eigenvectors of ϕ belonging to β . (We recallthat one may change the basis of L so that ϕ ( e n ) = ± e n + a linear combinationof monomials of higher degree.) We shall prove that such structures do not existin quite many cases. Proposition 12
There do not exist automorphisms of type 4 defined by ϕ ( e n ) = − e n + b n where the b n ∈ E are nonzero monomials for every n ∈ N , possibly multipliedby some (nonzero) scalars. Proof
Let us suppose that for every n ∈ N , one has ϕ ( e n ) = − e n + b n where b n = e n · · · e n kn = 0 is a monomial such that ϕ ( b n ) = b n for every n ∈ N .Since ϕ is of type 4, it follows that b n is of length k n = | b n | > , for all n ∈ N . Furthermore e n · · · e n kn = ϕ ( b n ) = ( − e n + b n ) · · · ( − e n kn + b n kn ) . Hence one obtains immediately that e n · · · e n kn = ( − k n e n · · · e n kn + X | m n | >k n m n . It follows that k n is an even integer and therefore b n ∈ Z ( E ) .On the other hand, due to the equality ( − e n + b n ) = 0 , we conclude that e n must be a factor of b n . We may assume that e n occurs in the first positionof the monomial b n . Rearranging the indices and with certain abuse of notationwe can write ϕ ( e n ) = − e n + e n e n · · · e n kn where k n is an odd integer.Consequently by direct computation we obtain ϕ ( e n ) ϕ ( e m ) = ( − e n + e n e n · · · e n kn )( − e m + e m e m · · · e m km )= e n e m − e n e m e m · · · e m km − e n e n · · · e n kn e m +2 e n e n · · · e n kn e m e m · · · e m km , and analogously ϕ ( e m ) ϕ ( e n ) = ( − e m + e m e m · · · e m km )( − e n + e n e n · · · e n kn )= e m e n − e m e n e n · · · e n kn − e m e m · · · e m km e n +2 e m e m · · · e m km e n e n · · · e n kn . Then the equality ϕ ( e n ) ϕ ( e m ) + ϕ ( e m ) ϕ ( e n ) = 0 implies − e n e m e m · · · e m km − e n e n · · · e n kn e m + 2 e n e n · · · e n kn e m e m · · · e m km − e m e n e n · · · e n kn − e m e m · · · e m km e n + 2 e m e m · · · e m km e n e n · · · e n kn = 0 . − e n e m e m · · · e m km − e m e n e n · · · e n kn + 2 e m e m · · · e m km e n e n · · · e n kn = 0 . Let us fix the integer m and choose n / ∈ { m , . . . , m k } . Then we conclude that e n · · · e n kn = e m · · · e m k for every n / ∈ { m , . . . , m k } .In this case we have ϕ ( e m ) = − e m + e m P , . . . , ϕ ( e m k ) = − e m k + e m k Q ,and ϕ ( e n ) = − e n + e n e m · · · e m k , for every n / ∈ { , . . . , k } (here we denoteby P , . . . , Q some monomials in E ).Now we fix m = m . Choosing n / ∈ { m , . . . , m k } with e n / ∈ supp ( P ) , andusing the same idea as in the case m = 1 , we obtain e n · · · e n kn = P. Therefore we conclude that e n · · · e n kn satisfies the equalities P = e n · · · e n kn = e m · · · e m k . Recall that m is fixed and there exist infinitely many n as above. Thus weconclude that ϕ ( e m ) = − e m + e m P = − e m + e m ( e m · · · e m k ) = − e m ,which is a contradiction. The proposition is proved.Another way of constructing automorphisms of type 4 could be by translationby a constant. These are automorphisms defined by ϕ ( e n ) = − e n + 2 P where P ∈ E is constant and contains some summand of length ≥ . However, as weshall see in the next proposition, this also turns out to be impossible. Proposition 13
Let ϕ be an automorphism of type 4 defined by the followingaction on the generators of E : ϕ ( e n ) = − e n + 2 a n , a n = ( e n + ϕ ( e n )) / . Then the set T = { a n | n ∈ N } is linearly independent over F . Proof
Let k ∈ N be a positive integer, a i , . . . , a i k ∈ T and λ , . . . , λ k ∈ F and assume that λ a i + · · · + λ k a i k = 0 . Then we have ϕ ( λ e i + · · · + λ k e i k ) = λ ( − e i + 2 a i ) + · · · + λ k ( − e i k + 2 a i k )= − ( λ e i + · · · + λ k e i k ) + 2( λ a i + · · · + λ k a i k ) . In this way we obtain ϕ ( λ e i + · · · + λ k e i k ) = − ( λ e i + · · · + λ k e i k ) . If the linear combination λ e i + · · · + λ k e i k does not vanish we can choose abasis γ of the vector space L containing the vector λ e i + · · · + λ k e i k (say as afirst vector of the basis). But this is a contradiction because ϕ is of type 4.Therefore λ e i + · · · + λ k e i k = 0 and λ = · · · = λ k = 0 . This proves thatthe set T is linearly independent. 18 orollary 14 There does not exist automorphism of type 4 which is a transla-tion by a constant.
Proof
The corollary follows immediately from the above proposition.The results we have obtained in this section show that in many situationsthe fourth structure of a Grassmann superalgebra cannot exist. We believe wehave grounds to conjecture that this is true in the general case.
Conjecture 1
Let E = ε ⊕ ε be an arbitrary Z -grading on the Grassmannalgebra. Then there exists a basis of the underlying vector space L that containsat least one homogeneous generator of E = ε ⊕ ε . We shall describe another situation where one cannot obtain an automor-phism of type 4, thus giving an extra strength to our conjecture. Recall thatif ϕ : E → E is an automorphism of order n where n is a positive integer, thelinearization ϕ ℓ of ϕ is defined as follows. Assume ϕ ( e i ) = u i + v i where u i ∈ L and v i is a linear combination of monomials of degrees ≥ . Then ϕ ℓ ( e i ) = u i ,and we extend then ϕ ℓ from the e i to a homomorphism of E . Anisimov proved(see for example [2, 3]) that if ϕ is an automorphism the so is ϕ ℓ . Moreover if ϕ is of order 2 then ϕ ℓ is also of order 2.We assume ϕ is an automorphism of type 4, and choose a basis { e i } of L such that ϕ ℓ ( e i ) = ± e i for every i (this is always possible since ϕ ℓ is also oforder 2).We shall prove that ϕ ℓ cannot be the identity map on E . If it were then ϕ ℓ ( e i ) = e i for each i . We write ϕ ( e i ) = e i + v i where v i is a linear combinationof monomials of lengths ≥ , and let s be the least length of a monomial thatappears with non-zero coefficient in some of the { v i | i ∈ N } . Take an index j such that ϕ ( e j ) = e j + βe j · · · e j s + w where β = 0 is in F and w is a linearcombination of monomials of lengths ≥ s . Then we have ϕ ( e j ) = e j and thiscan be written as e j = ϕ ( e j ) = ϕ ( e j + βe j · · · e j s + w )= ( e j + βe j · · · e j s + w ) + βϕ ( e j ) · · · ϕ ( e j s ) + ϕ ( w ) . Clearly ϕ ( e j ) · · · ϕ ( e j s ) = e j · · · e j s + some terms of lengths ≥ s all of themdifferent from e j · · · e j s .Therefore we obtain e j = e j + 2 βe j · · · e j s + terms of lengths ≥ s , all of themdifferent from e j · · · e j s . Hence β = 0 , a contradiction. In this way we haveproved the following theorem. Theorem 15
There is no automorphism ϕ of type 4 such that ϕ ℓ is the identityon E . Corollary 16 If ϕ is an automorphism of type 4 then the eigenspace of ϕ ℓ in L corresponding to the eigenvalue − is non-zero. The following example is also related to our conjecture.
Example 2
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