The geometric classification of nilpotent algebras
aa r X i v : . [ m a t h . R A ] F e b The geometric classification of nilpotent algebras Ivan Kaygorodov a , Mykola Khrypchenko b & Samuel A. Lopes c a CMCC, Universidade Federal do ABC, Santo Andr´e, Brazil b Departamento de Matem´atica, Universidade Federal de Santa Catarina, Florian´opolis, Brazil c CMUP, Faculdade de Ciˆencias, Universidade do Porto, Rua do Campo Alegre 687, 4169-007 Porto, PortugalE-mail addresses:Ivan Kaygorodov ([email protected])Mykola Khrypchenko ([email protected])Samuel A. Lopes ([email protected])
Abstract : We give a geometric classification of n -dimensional nilpotent, commutative nilpotent andanticommutative nilpotent algebras. We prove that the corresponding geometric varieties are irre-ducible, find their dimensions and describe explicit generic families of algebras which define each ofthese varieties. We show some applications of these results in the study of the length of anticommuta-tive algebras. Keywords : Nilpotent algebra, commutative algebra, anticommutative algebra, irreducible compo-nents, geometric classification, degeneration, length function.
MSC2010 : 17A30, 17D99, 14L30. I
NTRODUCTION
The geometry of varieties of algebras defined by polynomial identities has been an active area ofinterest and research since the work of Nijenhuis–Richardson [25] and Gabriel [10] in the 1960’s and1970’s. The relationship between geometric features of the variety (such as irreducibility, dimension,smoothness) and the algebraic properties of its points brings novel geometric insight into the structureof the variety, its generic points and degenerations.Given algebras A and B in the same variety, we write A → B and say that A degenerates to B , or that A is a deformation of B , if B is in the Zariski closure of the orbit of A (under the base-change action of the general linear group). The study of degenerations of algebras is very rich andclosely related to deformation theory, in the sense of Gerstenhaber [11]. Degenerations have alsobeen used to study a level of complexity of an algebra [12, 21]. There are many results concerning The authors thank Alexander Guterman for some constructive discussions about the length of algebras. The workwas partially supported by CNPq 404649/2018-1, 302980/2019-9; RFBR 20-01-00030 and by CMUP, which is fi-nanced by national funds through FCT—Fundac¸ ˜ao para a Ciˆencia e a Tecnologia, I.P., under the project with referenceUIDB/00144/2020. degenerations of algebras of small dimensions in a variety defined by a set of identities (see, forexample, [1, 3, 8, 13, 14, 16–22, 22] and references therein). An interesting question is to study thoseproperties which are preserved under degenerations. Recently, Chouhy [5] proved that in the case offinite-dimensional associative algebras, the N -Koszul property is one such property.Concerning Lie algebras, Grunewald–O’Halloran [14] calculated the degenerations for the varietyof -dimensional nilpotent Lie algebras while in [3], Burde and Steinhoff constructed the graphs ofdegenerations for the varieties of - and -dimensional Lie algebras and in [8] Fern´andez Ouaridi,Kaygorodov, Khrypchenko and Volkov described the full graphs of degenerations of small dimen-sional nilpotent algebras.One of the main problems of the geometric classification of a variety of algebras is a descriptionof its irreducible components. In [10], Gabriel described the irreducible components of the varietyof -dimensional unital associative algebras and the variety of -dimensional unital associative al-gebras was classified algebraically and geometrically by Mazzola [24]. Later, Cibils [6] consideredrigid associative algebras with -step nilpotent radical. Goze and Ancoch´ea-Berm ´udez proved thatthe varieties of - and -dimensional nilpotent Lie algebras are reducible [13]. All irreducible com-ponents of -step nilpotent, commutative nilpotent and anticommutative nilpotent algebras have beendescribed in [16, 27].In many cases, the irreducible components of the variety are determined by the rigid algebras,i.e. algebras whose orbit closure is an irreducible component. It is worth mentioning that this isnot always the case and Flanigan had shown that the variety of -dimensional nilpotent associativealgebras has an irreducible component which does not contain any rigid algebras—it is instead definedby the closure of a union of a one-parameter family of algebras [9]. Here, we will encounter similarsituations. Our main results are based on Theorems 14, 21, 35 and [8, 22]. We are summarizing thembelow. Theorem A.
For any n ≥ , the variety of all n -dimensional nilpotent algebras is irreducible and hasdimension n ( n − n +1)3 .Moreover, we show that the family R n for n ≥ given in Definition 10 is generic in the varietyof n -dimensional nilpotent algebras and inductively give an algorithmic procedure to obtain any n -dimensional nilpotent algebra as a degeneration from R n . The case of n = 2 follows from [22]. Theorem B.
For any n ≥ , the variety of all n -dimensional commutative nilpotent algebras isirreducible and has dimension n ( n − n +4)6 .As above, we show that the family S n for n ≥ given in Definition 15 is generic in the varietyof n -dimensional commutative nilpotent algebras and inductively give an algorithmic procedure toobtain any n -dimensional nilpotent commutative algebra as a degeneration from S n . The cases of n = 2 and n = 3 follow from [8, 22]. Theorem C.
For any n ≥ , the variety of all n -dimensional anticommutative nilpotent algebras isirreducible and has dimension ( n − n +2 n +3)6 . We show also that the family T n for n ≥ given in Definition 33 is generic in the variety of n -dimensional anticommutative nilpotent algebras and inductively give an algorithmic procedure toobtain any n -dimensional nilpotent anticommutative algebra as a degeneration from T n . The cases of n = 2 , , , follow from [8, 22].The notion of length for nonassociative algebras has been recently introduced in [15], generalizingthe corresponding notion for associative algebras. Using the above result, we show in Section 5 (cf.Corollary 39) that the length of an arbitrary (i.e. not necessarily nilpotent) n -dimensional anticommu-tative algebra is bounded above by the n th Fibonacci number, and prove that our bound is sharp.1. V
ARIETIES OF ALGEBRAS , CENTRAL EXTENSIONS AND NILPOTENT ALGEBRAS
Throughout this paper, we work over the field C of complex numbers and, unless otherwise noted,all vector spaces, linear maps and tensor products will be taken over C . The identity matrix is denotedby I and the matrix unit corresponding to the row i and the column j is E ij . For a subset X of a givenvector space, the linear span of X is denoted by h X i .1.1. Central extensions and the method of Skjelbred and Sund.
An algebra is a vector spaceendowed with a bilinear multiplication. Formally, it is a pair A = ( V , µ ) , where V is a vector spaceand µ ∈ Hom( V ⊗ V , V ) , which gives the algebra law. The annihilator of A is Ann A = { x ∈ A | x A + A x = 0 } . For the purposes of this paper, we just need to consider -dimensional centralextensions. We will give here an overview of the algebraic classification method of Skjelbred andSund [30].A bilinear map θ : A × A −→ C determines the -dimensional central extension A θ = A ⊕ C withthe product ( x + v ) · θ ( y + w ) = xy + θ ( x, y ) , for all x, y ∈ A and v, w ∈ C . We let Z ( A , C ) be thevector space of all bilinear maps A × A −→ C , which we refer to as -cocycles (of A with values in C ). Then, the space of -coboundaries is B ( A , C ) = { δf | f ∈ A ∗ } ⊆ Z ( A , C ) , where δf = f ◦ µ ,so that δf ( x, y ) = f ( xy ) , for all x, y ∈ A . The second cohomology of A with values in C is thequotient space H ( A , C ) = Z ( A , C ) (cid:14) B ( A , C ) , which is well known to parametrize equivalenceclasses of -dimensional central extensions of A .Once a basis ( e i ) ni =1 of A is fixed, define the bilinear maps ∆ ij = e ∗ i × e ∗ j , for ≤ i, j ≤ n ,so that ∆ ij ( e k , e ℓ ) = δ ik δ jℓ , for all ≤ i, j, k, ℓ ≤ n . Then (∆ ij ) ≤ i,j ≤ n is a basis of Z ( A , C ) . Theautomorphism group Aut A acts on Z ( A , C ) via φ ( θ ) = θ ◦ ( φ × φ ) , for φ ∈ Aut A and θ ∈ Z ( A , C ) and the action induces a well-defined one on H ( A , C ) . If A is the matrix of θ and M is the matrix of φ , then φ ( θ ) has matrix M T AM .It is easy to see that Ann A θ = (Ann A ∩ Ann θ ) ⊕ C , where Ann θ = { x ∈ A | θ ( x, A ) + θ ( A , x ) = 0 } . Thus, given algebras A , A ′ and respective -cocycles θ , θ ′ such that Ann A ∩ Ann θ =0 = Ann A ′ ∩ Ann θ ′ , then A θ ≃ A ′ θ ′ implies that A ≃ A θ / Ann A θ ≃ A ′ θ ′ / Ann A ′ θ ′ ≃ A ′ .Therefore, it remains to determine the precise conditions on θ, θ ′ ∈ Z ( A , C ) for A θ and A θ ′ to beisomorphic, under the assumption that Ann A ∩ Ann θ = 0 = Ann A ∩ Ann θ ′ . This is given by thefollowing result. Lemma 1.
Let A be an algebra and θ, θ ′ be -cocycles, represented by the nonzero cohomologyclasses [ θ ] , [ θ ′ ] in H ( A , C ) . Suppose that Ann A ∩ Ann θ = 0 = Ann A ∩ Ann θ ′ . Then A θ isisomorphic to A θ ′ if and only if the orbits of [ θ ] and [ θ ′ ] under the action of Aut A span the samevector space. Varieties of algebras.
Given an n -dimensional vector space V , an algebra structure on V (oran n -dimensional algebra law) is naturally seen as an element of Hom( V ⊗ V , V ) ∼ = V ∗ ⊗ V ∗ ⊗ V ,a vector space of dimension n . Once we fix a basis e , . . . , e n of V , this space can be identified with C n and given the structure of an affine variety whose coordinate ring if the polynomial ring in thevariables (cid:0) c ki,j (cid:1) ni,j,k =1 . Accordingly, a subset of Hom( V ⊗ V , V ) is Zariski-closed if it can be definedby a set of polynomial equations in the variables (cid:0) c ki,j (cid:1) i,j,k . For simplicity, we identify points in thevariety with the corresponding maximal ideals of the coordinate ring so, if no confusion arises, wealso think of the c ki,j as scalars.Henceforth, having fixed the vector space V and its basis ( e i ) ni =1 , we will identify points (cid:0) c ki,j (cid:1) ni,j,k =1 in C n with n -dimensional algebras via their structure constants relative to the basis ( e i ) ni =1 . Con-cretely, any µ ∈ Hom( V ⊗ V , V ) is determined by the n structure constants c ki,j ∈ C such that µ ( e i ⊗ e j ) = P nk =1 c ki,j e k . Example 2.
The following polynomial identities define well-known varieties of n -dimensional alge-bras:(1) Anticommutative algebras c kij + c kji = 0 , ≤ i, j, k ≤ n. (2) Associative algebras n X ℓ =1 c ℓij c mℓk − c miℓ c ℓjk = 0 , ≤ i, j, k, m ≤ n. (3) Lie algebras n X ℓ =1 c ℓij c mkℓ + c ℓjk c miℓ + c ℓki c mjℓ = 0 and c kij + c kji = 0 , ≤ i, j, k, m ≤ n. Nilpotent algebras.
Recall that an algebra N is nilpotent if N k = 0 , for some k ≥ , where N = N , N k +1 = k X i =1 N i N k +1 − i , k ≥ . The smallest (if any) positive integer k satisfying N k = 0 is called the nilpotency index of N . It iseasy to see that the n -dimensional algebras N with N k = 0 form a closed set (as elements of C n ) andthence so do all nilpotent n -dimensional algebras. Lemma 3.
Let N be an algebra of dimension n . Then N is nilpotent if and only if it is isomorphic toan algebra M whose structure constants ( γ ki,j ) ni,j,k =1 satisfy γ ki,j = 0 , ∀ k ≤ max { i, j } . (1) Proof.
Indeed, (1) holds for the algebra with zero multiplication. Moreover, if an algebra A is acentral extension of another algebra B by some vector space and the structure constants of B relativeto some basis satisfy (1), then completing this basis of B to a basis of A we see that the correspondingstructure constants of A will also satisfy (1). It remains to note that each finite-dimensional nilpotentalgebra can be obtained, up to isomorphism, from an algebra with zero multiplication by consecutivelyapplying the central extension procedure.Conversely, if the structure constants of an algebra M satisfy (1), then all the products in M oflength n are zero, so M is nilpotent. (cid:3) Definition 4.
Let n ≥ . We denote by Nil n the variety of all nilpotent algebra structures (on V , withrespect to the basis ( e i ) ni =1 ) and by Nil γn its subvariety determined by the system of equations (1).Recall that the general linear group GL n ( C ) acts on the space of algebra structures on V viabase-change and the orbits parametrize the isomorphism classes of algebras. Given an algebra A , itsorbit under this action will be denoted by O ( A ) . Lemma 3 can thus be restated simply as Nil n =GL n ( C ) · Nil γn . Proposition 5.
Let n ≥ . The variety Nil n of all nilpotent n -dimensional algebras and its subvari-eties of commutative and anticommutative nilpotent algebras are irreducible.Proof. The coordinate ring O n of the variety Nil γn is the complex polynomial ring in the variables (cid:8) c ki,j | ≤ i, j ≤ n, k > max { i, j } (cid:9) . Thus, O n being a domain, the variety Nil γn is irreducible. If ℓ n is the transcendence degree of O n , then ℓ n +1 = ℓ n + n , so the dimension of Nil γn is ℓ n = n ( n − n − .By Lemma 3, Nil n = GL n ( C ) · Nil γn . Since GL n ( C ) is a connected (i.e. irreducible) algebraic groupand the product of irreducible varieties is irreducible, it follows that GL n ( C ) × Nil γn is irreducible.Hence, so is the continuous image Nil n = GL n ( C ) · Nil γn .A similar argument holds for the varieties of n -dimensional commutative and anticommutativenilpotent algebras. (cid:3) Definition 6.
Let V be an irreducible variety of algebras and R ⊆ V be a family of algebras. Thefamily R is said to be generic in V , if S A ∈ R O ( A ) = V . For an algebra B ∈ V , we also write R → B as shorthand for B ∈ S A ∈ R O ( A ) .2. T HE GEOMETRIC CLASSIFICATION OF NILPOTENT ALGEBRAS
In this section we find a generic family of algebras in the variety
Nil n of all nilpotent n -dimensionalalgebras and use it to compute the dimension of the variety. Recall that Nil γn is the subvariety ofalgebras satisfying (1) and that ( e i ) ni =1 is our fixed basis. Lemma 7.
Let A ∈ Nil γn . Then, for all x ∈ C , the linear endomorphism ϕ defined by ϕ ( e ) = e + xe n , ϕ ( e i ) = e i , ≤ i ≤ n , is an automorphism of A . Proof.
Obviously, ϕ ( e i ) ϕ ( e j ) = e i e j = ϕ ( e i e j ) for all ≤ i, j ≤ n , in view of (1). Observe that e n ∈ Ann A , so ϕ ( e ) ϕ ( e i ) = ( e + xe n ) e i = e e i = ϕ ( e e i ) and similarly ϕ ( e i ) ϕ ( e ) = ϕ ( e i e ) ,for all ≤ i ≤ n , by (1). Finally, ϕ ( e ) = ( e + xe n ) = e = ϕ ( e ) , again by (1). Thus, ϕ isan endomorphism of A . It is clearly invertible, with ϕ − ( e ) = e − xe n and ϕ − ( e i ) = e i , for all ≤ i ≤ n . (cid:3) Lemma 8.
Let A ∈ Nil γn +1 such that Ann A = h e n +1 i , e e n = 0 and e i = e i +1 , for all ≤ i ≤ n .Suppose that Aut ( A / h e n +1 i ) = { I + xE n | x ∈ C } in the basis ( e i + h e n +1 i ) ni =1 . Then Aut A = { I + xE n +1 , | x ∈ C } in the basis ( e i ) n +1 i =1 .Proof. Let ϕ ∈ Aut A . Then ϕ (Ann A ) = Ann A , so ϕ induces ˜ ϕ ∈ Aut ( A / h e n +1 i ) defined by ˜ ϕ ( e i + h e n +1 i ) = ϕ ( e i ) + h e n +1 i , ≤ i ≤ n . We know that the matrix of ˜ ϕ in ( e i + h e n +1 i ) ni =1 is ofthe form I + xE n for some x ∈ C . Since ϕ (Ann A ) = Ann A , we have ϕ ( e n +1 ) = ye n +1 for some y ∈ C . Then, the matrix of ϕ in ( e i ) n +1 i =1 is of the form I + xE n + ( y − E n +1 ,n +1 + n X j =1 a n +1 ,j E n +1 ,j for some { a n +1 ,j } nj =1 ⊆ C . In particular, ϕ ( e n ) = e n + a n +1 ,n e n +1 . Since e n = e n +1 , we have e n +1 = ( e n + a n +1 ,n e n +1 ) = ϕ ( e n ) = ϕ ( e n ) = ϕ ( e n +1 ) = ye n +1 . Hence, y = 1 . Now, ϕ ( e ) = e + xe n + a n +1 , e n +1 and e e n = 0 imply ϕ ( e e n ) = ϕ ( e ) ϕ ( e n ) = ( e + xe n + a n +1 , e n +1 )( e n + a n +1 ,n e n +1 ) = xe n +1 , so x = 0 . Therefore, ϕ ( e ) = e + a n +1 , e n +1 . Finally, using ϕ ( e i ) = e i + a n +1 ,i e n +1 and e i = e i +1 for all ≤ i ≤ n − we get e i +1 = ( e i + a n +1 ,i e n +1 ) = ϕ ( e i ) = ϕ ( e i ) = ϕ ( e i +1 ) = e i +1 + a n +1 ,i +1 e n +1 . Therefore, a n +1 ,i = 0 for all ≤ i ≤ n . Thus, the matrix of ϕ relative to the basis ( e i ) n +1 i =1 has theform I + a n +1 , E n +1 , .Conversely, the linear map ϕ defined by ϕ ( e ) = e + xe n +1 , ϕ ( e i ) = e i , ≤ i ≤ n + 1 , is anautomorphism of A , by Lemma 7. (cid:3) Lemma 9.
Let n ≥ and A ∈ Nil γn such that Ann A = h e n i and e i = e i +1 , for all ≤ i ≤ n − .Suppose that Aut ( A ) = { I + xE n | x ∈ C } in the basis ( e i ) ni =1 . Then, there is a parametric familyof pairwise non-isomorphic -dimensional central extensions B of A with basis ( e i ) n +1 i =1 , extendingthe basis of A , such that Ann B = h e n +1 i , e e n = 0 , e i = e i +1 , for all ≤ i ≤ n , and the structureconstants c n +1 ij of B in this basis are arbitrary independent complex parameters for all ≤ i = j ≤ n , ( i, j ) = (1 , n ) .Proof. We have B ( A ) = h ∆ i ⊕ n − M m =2 D ∆ mm + X ≤ i = j ≤ m c m +1 ij ∆ ij E , so H ( A ) = h [∆ ij ] | ≤ i = j ≤ n i ⊕ h [∆ nn ] i .Let φ = I + xE n ∈ Aut ( A ) and θ = P ≤ i = j ≤ n α ij [∆ ij ] + α nn [∆ nn ] ∈ B ( A ) . Consider thecorresponding matrix A = P ≤ i = j ≤ n α ij E ij + α nn E nn . Then φ T Aφ = ( I + xE n ) X ≤ i = j ≤ n α ij E ij + α nn E nn ! ( I + xE n )= A + x n X i =1 α ni E i + x n X i =1 α in E i + x α nn E = x ( α n + α n + xα nn ) E + n X i =2 ( α i + xα ni ) E i + n X i =2 ( α i + xα in ) E i + X ≤ i = j ≤ n α ij E ij + α nn E nn . (2)So, φ · θ = P ≤ i = j ≤ n α ∗ ij [∆ ij ] + α ∗ nn [∆ nn ] , where α ∗ i = α i + xα ni , α ∗ i = α i + xα in , for all ≤ i ≤ n , α ∗ ij = α ij , for all ≤ i = j ≤ n , and α ∗ nn = α nn .If α nn = 0 , then choosing x = − α n α − nn , we obtain the family of representatives of distinct orbits D X ≤ i = j ≤ n, ( i,j ) =(1 ,n ) c n +1 ij [∆ ij ] + [∆ nn ] E c n +1 ij ∈ C . It determines the desired family of algebras B . (cid:3) Definition 10.
Let n ≥ . Denote by R n the family of nilpotent algebras with basis ( e i ) ni =1 satis-fying (1), such that e i = e i +1 , for all ≤ i ≤ n − , c = 1 , c i +11 i = 0 , for all ≤ i ≤ n − ,and with the remaining structure constants c kij being arbitrary independent complex parameters, forall ≤ i = j ≤ n , k > max { i, j } .Notice that, given A ∈ R n +1 with n ≥ , since by (1) we have e n +1 ∈ Ann A , it follows that h e n +1 i is an ideal of A and A / h e n +1 i can be seen naturally as an element of R n , relative to the ordered basis ( e i ) ni =1 , where e i = e i + h e n +1 i . This property will be important in arguments by induction, as theone which follows. Lemma 11.
Let A ∈ R n with n ≥ . Then the following hold:(a) Ann A = h e n i .(b) Aut A = { I + xE n | x ∈ C } , relative to the basis ( e i ) ni =1 .Proof. We prove both statements simultaneously by induction on n ≥ . Indeed, R consists of asingle point, which as an algebra is defined by the following multiplication (as usual, only nonzeroproducts of the basis elements are shown): A : e e = e , e e = e , e e = e . It is easy to see by direct inspection that
Ann A = h e i and Aut A = { I + xE | x ∈ C } in ( e i ) i =1 . Now let A ∈ R n +1 . Then, viewing A / h e n +1 i as an algebra from R n , as explained above, theinduction hypothesis implies that Ann A / h e n +1 i = h e n i . Let v = P n +1 i =1 λ i e i ∈ Ann A . Since e n +1 ∈ Ann A , we can assume that λ n +1 = 0 and our goal is to show that λ i = 0 for all i . Since v = P ni =1 λ i e i ∈ Ann A / h e n +1 i , we deduce that λ i = 0 for all i ≤ n − . Thus, λ n e n ∈ Ann A , so λ n e n e n = λ n e n +1 and λ n = 0 , proving our claim that Ann A = h e n +1 i .The induction hypothesis also gives that Aut ( A / h e n +1 i ) = { I + xE n | x ∈ C } in ( e i ) ni =1 . Then,by Lemma 8 , Aut A = { I + xE n +1 , | x ∈ C } in the basis ( e i ) n +1 i =1 . (cid:3) Proposition 12.
We have dim (cid:16)S A ∈ R n O ( A ) (cid:17) = n ( n − n +1)3 .Proof. For any A ∈ R n , we know that dim Aut A = 1 , by Lemma 11, and thus dim O ( A ) =dim GL n ( C ) − n − . Moreover, the algebras in R n are pairwise non-isomorphic, by Lemma 9,so the corresponding orbits are disjoint.Let p n = dim R n , the number of independent parameters of the family R n . We calculate p n byinduction on n . We have p = 0 and p n +1 = p n + n ( n − − , for all n ≥ . Therefore, p n = ( n − n − . Thus, dim (cid:16)S A ∈ R n O ( A ) (cid:17) = n − p n = n ( n − n +1)3 . (cid:3) Before our main result of this section, we need a technical observation on the inverse of a lowertriangular matrix.
Lemma 13.
Let A = ( a ij ) ni,j =1 be an invertible lower triangular matrix of size n and A − = ( a ′ ij ) ni,j =1 .Then for all i > j we have a ′ ij = − a − ii a − jj a ij − a − jj P i − k = j +1 a ′ ik a kj . In particular, for all i ≥ j , a ′ ij isuniquely determined by a pq with i ≥ p ≥ q ≥ j .Proof. If i = j , then a ′ ij = a − ij . Otherwise, P ik = j a ′ ik a kj = 0 , whence a ′ ij = − a − jj P ik = j +1 a ′ ik a kj = − a − ii a − jj a ij − a − jj P i − k = j +1 a ′ ik a kj . The second statement follows by backward induction on j with i fixed. (cid:3) Now we can state and prove our main result about the variety
Nil n of nilpotent algebras. Noticethat the proof specifically gives an algorithmic construction of a degeneration R n → N from thefamily R n to any given n -dimensional nilpotent algebra N . Theorem 14.
For any n ≥ , the family R n is generic in Nil n . In particular, dim( Nil n ) = n ( n − n +1)3 .Proof. Given N ∈ Nil n , we will prove that R n → N . Recall that ( e i ) ni =1 is our fixed basis ofthe underlying vector space. Without loss of generality, we may assume that N ∈ Nil γn . Indeed,by Lemma 3, N is isomorphic to an algebra M whose structure constants satisfy (1) in some basis ( f i ) ni =1 . Take g ∈ GL n ( C ) such that g ( f i ) = e i , for all ≤ i ≤ n . Then the structure constants of gM in ( e i ) ni =1 are those of M in ( f i ) ni =1 , and thus satisfy (1). So, we may replace N by gM , if necessary.We will thus assume that N ∈ Nil γn and prove by induction on n that there is a parametric basis E i ( t ) = P nj = i a ji ( t ) e j , with a ji ( t ) ∈ C ( t ) , ≤ i ≤ j ≤ n , and a choice of structure constants c kij ( t ) ∈ C ( t ) , satisfying the conditions of Definition 10, with ≤ i = j ≤ n, k > max { i, j } , ( i, j, k ) = (2 , , , ( i, j, k ) = (1 , k − , k ) , (3) giving a degeneration of N from R n .The case n = 3 is proved in [8]. Let N ∈ Nil γn +1 . It follows that e n +1 ∈ Ann N , so h e n +1 i is an ideal of N and N/ h e n +1 i is seen as an element of Nil γn via the identification of e i + h e n +1 i with e i , ≤ i ≤ n . By the induction hypothesis, there is a parametric basis E i ( t ) = P nj = i a ji ( t ) e j , ≤ i ≤ n , and a choice of parameters c kij ( t ) satisfying the conditions of Definition 10 and determininga degeneration of N/ h e n +1 i from R n . Observe that the degeneration does not depend on a n ( t ) because e n ∈ Ann R for all R ∈ R n . More generally, since any such R satisfies (1), we have E i ( t ) E j ( t ) = n X p = i,q = j a pi ( t ) a qj ( t ) e p e q = n X p = i,q = j a pi ( t ) a qj ( t ) X r> max { p,q } c rpq ( t ) e r = n X r =2 r − X p = i,q = j c rpq ( t ) a pi ( t ) a qj ( t ) ! e r . (4)We see that a n ( t ) cannot appear among the a pi ( t ) or a qj ( t ) above. Moreover, each e r from the sum (4)belongs to h e , . . . , e n i = h E ( t ) , . . . , E n ( t ) i , so its coordinates in the basis ( E i ( t )) ni =1 do not dependon a i ( t ) , ≤ i ≤ n .We are going to redefine a n ( t ) and choose a n +1 ,i ( t ) , ≤ i ≤ n + 1 , with a n +1 ,n +1 ( t ) = 0 , and c n +1 ij ( t ) , ≤ i = j ≤ n , ( i, j ) = (1 , n ) , such that ˜ E i ( t ) := E i ( t ) + a n +1 ,i ( t ) e n +1 , ≤ i ≤ n , ˜ E n +1 ( t ) := a n +1 ,n +1 ( t ) e n +1 is a parametric basis giving a degeneration of N from R n +1 . Denote by A ( t ) the lower triangular matrix ( a ij ( t )) n +1 i,j =1 whose ( n + 1) -st row consists of unknown parameterswhich will be defined below and let A − ( t ) = ( a ′ ij ( t )) n +1 i,j =1 be its formal inverse. Observe that theupper left ( n × n ) -block of A − ( t ) is the inverse of the upper left ( n × n ) -block of A ( t ) and thusdoes not depend on the choice of a n +1 ,i ( t ) , ≤ i ≤ n + 1 . Since the coordinates of e i in the basis ( ˜ E j ( t )) n +1 j =1 are given by the i -th column of A − ( t ) , for all ≤ i ≤ n + 1 , we can further develop (4)to get ˜ E i ( t ) ˜ E j ( t ) = n +1 X k =2 k X r =2 a ′ kr ( t ) r − X p = i,q = j c rpq ( t ) a pi ( t ) a qj ( t ) ! ˜ E k ( t ) , (5)for all ≤ i, j ≤ n + 1 . Notice that these new structure constants satisfy (1).Let γ kij be the structure constants of N in ( e i ) n +1 i =1 . Thence, to construct the degeneration R n +1 → N ,we need that lim t → k X r =2 a ′ kr ( t ) r − X p = i,q = j c rpq ( t ) a pi ( t ) a qj ( t ) ! = γ kij , ≤ i, j < k ≤ n + 1 , (6)be satisfied. Observe that (6) holds for all ≤ i, j < k ≤ n by the choice of ( E i ( t )) ni =1 and ( c kij ( t )) with (3), because γ kij is the corresponding structure constant of N/ h e n +1 i for such ( i, j, k ) . Thus,it remains to consider k = n + 1 , which we do below by appropriately defining a n ( t ) , a n +1 ,i ( t ) , ≤ i ≤ n + 1 , and c n +1 ij ( t ) , ≤ i = j ≤ n , ( i, j ) = (1 , n ) (so that the conditions of Definition 10hold).We will proceed in n steps, from k = 0 to k = n − . At the end of Step k we will have defined c n +1 p,q ( t ) for all p, q ≥ n − k and a n +1 ,r ( t ) for all r ≥ n + 1 − k . We will also have established theconvergence lim t → n +1 X r =2 a ′ n +1 ,r ( t ) r − X p = i,q = j c rpq ( t ) a pi ( t ) a qj ( t ) ! = γ n +1 ij (7)for all n ≥ i, j ≥ n − k . Step . Since we must have c n +1 nn ( t ) = 1 , it remains to define a n +1 ,n +1 ( t ) . The left-hand side of (7)for i = j = n becomes lim t → ( a nn ( t ) a n +1 ,n +1 ( t ) − ) , so we set a n +1 ,n +1 ( t ) := ( ( γ n +1 nn ) − a nn ( t ) , if γ n +1 nn = 0 , t − a nn ( t ) , if γ n +1 nn = 0 .By definition, (7) holds for i = j = n . Notice also that a n ( t ) does not occur in the formula above. Step k . Let ≤ k < n − and assume that Step k − has been successfully completed and that a n ( t ) has not been used to define any new coefficients.Suppose first that n ≥ i > j = n − k . We will define c n +1 ij ( t ) and establish (7) in this case. Thecoefficient of c n +1 ij ( t ) on the left-hand side of (7) equals a n +1 ,n +1 ( t ) − a ii ( t ) a jj ( t ) which has alreadybeen defined and is non-zero. We thus put c n +1 ij ( t ) := a n +1 ,n +1 ( t ) a ii ( t ) a jj ( t ) γ n +1 ij − n +1 X r =2 a ′ n +1 ,r ( t ) r − X ′ p = i,q = j c rpq ( t ) a pi ( t ) a qj ( t ) ! , (8)where the primed sum is over all ( p, q ) such that ( p, q, r ) = ( i, j, n + 1) . Note that on the right-handside of (8) we must have n − k < i ≤ r − , so r ≥ n − k + 2 . Thence, by Step k − and Lemma 13,all the terms of the form a ′ n +1 ,r ( t ) on the right-hand side of (8) have already been defined. The sameholds for all remaining terms except those of the form c n +1 pj ( t ) with p > i . Thus, (8) is a recurrenceformula which defines c n +1 ij ( t ) in terms of c n +1 pj ( t ) with p > i . So, starting recursively with c n +1 nj ( t ) ,we can define all of the terms c n +1 p,n − k ( t ) , with p > n − k and by doing so we force the convergence (7)for all i > n − k and j = n − k . Similarly, we can define all the terms c n +1 n − k,q ( t ) , with q > n − k recursively, making sure that (7) holds for i = n − k and all j > n − k . This will work as beforebecause we are assuming that k < n − so ( n − k, q ) = (1 , n ) . Moreover, also by that assumptionon k , the coefficient a n ( t ) has not been used in (8) to define c n +1 ij ( t ) , as i, j ≥ n − k ≥ . Hence,given that c n +1 n − k,n − k ( t ) = 0 , all c n +1 p,q ( t ) with p, q ≥ n − k are defined and (7) holds for all i, j ≥ n − k ,except in the case i = j = n − k , which will be analyzed next. Now we will define a n +1 ,n +1 − k ( t ) so that (7) holds for i = j = n − k . Assume thus that i = j = n − k . Using Lemma 13 we have n +1 X r =2 r − X p,q = i a ′ n +1 ,r ( t ) c rpq ( t ) a pi ( t ) a qi ( t ) = a ′ n +1 ,i +1 ( t ) a ii ( t ) + n +1 X r = i +2 a ′ n +1 ,r ( t ) r − X p,q = i c rpq ( t ) a pi ( t ) a qi ( t )= − a n +1 ,n +1 ( t ) − a i +1 ,i +1 ( t ) − a ii ( t ) a n +1 ,i +1 ( t ) − a i +1 ,i +1 ( t ) − a ii ( t ) n X s = i +2 a ′ n +1 ,s ( t ) a s,i +1 ( t )+ n +1 X r = i +2 a ′ n +1 ,r ( t ) r − X p,q = i c rpq ( t ) a pi ( t ) a qi ( t ) . Hence, we put a n +1 ,i +1 ( t ) := − a n +1 ,n +1 ( t ) a i +1 ,i +1 ( t ) a ii ( t ) γ n +1 ii − n +1 X r = i +2 a ′ n +1 ,r ( t ) r − X p,q = i c rpq ( t ) a pi ( t ) a qi ( t ) ! − a n +1 ,n +1 ( t ) n X s = i +2 a ′ n +1 ,s ( t ) a s,i +1 ( t ) , (9)where the right-hand side defines a n +1 ,n − k +1 ( t ) in terms of a n +1 ,r ( t ) with r ≥ n − k + 2 (alreadydefined in the previous steps) and c n +1 p,q ( t ) with p, q ≥ n − k (defined above). Also, (7) holds for i = j = n − k and a n ( t ) does not occur in the definition (9) above. This step is thus finished. Step n − . When we reach this final step, all c n +1 p,q ( t ) with p, q ≥ and all a n +1 ,r ( t ) with r ≥ have been defined without using the coefficient a n ( t ) and (7) has been shown to hold for all i, j ≥ .Hence, as a n ( t ) also has no role in (4) nor on the invertibility of A ( t ) , we can redefine it at this point.We will do it so as to guarantee that (7) holds for ( i, j ) = (1 , n ) . This is necessary because we arebound to having c n +11 n ( t ) = 0 , so we cannot force (7) in case ( i, j ) = (1 , n ) by choosing c n +11 n ( t ) as weplease.Suppose thus that ( i, j ) = (1 , n ) . We have n +1 X r =2 r − X p = i,q = j a ′ n +1 ,r ( t ) c rpq ( t ) a pi ( t ) a qj ( t ) = a n +1 ,n +1 ( t ) − a nn ( t ) n X p =2 c n +1 pn ( t ) a p ( t ) , in which the coefficient of a n ( t ) is a n +1 ,n +1 ( t ) − a nn ( t ) = 0 . Hence we set a n ( t ) := a n +1 ,n +1 ( t ) γ n +11 n a nn ( t ) − n − X p =2 c n +1 pn ( t ) a p ( t ) , the right-hand side of which has already been defined and does not involve a n ( t ) . Now we can proceed as in the previous (generic) step with k = n − , defining c n +1 p ( t ) for p ≥ and then c n +11 q ( t ) for n − ≥ q ≥ and finally a n +1 , ( t ) , ensuring that (7) holds in the remainingcases. The coefficient a n +1 , ( t ) is unrestrained and can be chosen arbitrarily (which agrees with ourprevious observations).This finishes the construction and the proof. (cid:3)
3. T
HE GEOMETRIC CLASSIFICATION OF COMMUTATIVE NILPOTENT ALGEBRAS
In this section, we consider the variety of commutative nilpotent n -dimensional algebras. Ourmethods will be analogous to those of Section 2. Definition 15.
Let n ≥ . Denote by S n the family of commutative algebras in Nil γn such that e i = e i +1 for all ≤ i ≤ n − , c = 1 , c = 0 and c i +11 i = 0 for all ≤ i ≤ n − . The remainingstructure constants c kij are arbitrary, subject only to (1) and the commutativity constraint.As with the algebras R n from Definition 10, if A ∈ S n +1 , for some n ≥ , then A / h e n +1 i is seennaturally as an element of S n , relative to the ordered basis ( e i + h e n +1 i ) ni =1 . Example 16.
Let n = 4 and α ∈ C . Define the commutative algebra A α by the multiplication (10) A α : e = e , e e = αe , e e = 0 , e = e , e e = e , e = e , where e ∈ Ann A α . Then S consists of the algebras A α with α ∈ C ∗ . Considering the new basis f = e − αe , f = e + α e , f = e and f = e , we see that A α is isomorphic to the algebra C ( − α ) defined in [8].Recall that in [8, Thm. 5] it was shown that the family C ( α ) , with α ∈ C , is generic in thevariety of -dimensional nilpotent commutative algebras. Since A ∈ S α ∈ C ∗ O ( A α ) , it follows thatthe family S is also generic in the variety of -dimensional commutative nilpotent algebras. As we will see next, our restriction in Definition 15 that c = 0 ensures that the algebras in S have a sufficiently small automorphism group. Lemma 17.
Let A ∈ S n . Then the following hold:(a) Ann A = h e n i .(b) Aut A = { I + xE n | x ∈ C } , relative to the basis ( e i ) ni =1 .Proof. The proof is essentially the same as that of Lemma 11, since Lemmas 7 and 8 will still applyto the algebras in S n . We just need to verify the base cases for (a) and (b). Assume thus that n = 4 .Then S is described in Example 16; more specifically, it consists of the algebras A α with α = 0 andmultiplication given by (10), commutativity and the fact that e ∈ Ann A α .Let v = P i =1 λ i e i ∈ Ann A α . As e ∈ Ann A α , we can assume that λ = 0 . Then ve = λ e + λ αe , so λ = 0 ; ve = λ e + λ e , so λ = λ = 0 .So indeed Ann A α = h e i , for every α . Now, for (b), Lemma 7 guarantees that
Aut A α ⊇ { I + xE n | x ∈ C } . Conversely, let ϕ ∈ Aut A α , with matrix φ = ( a ij ) ≤ i,j ≤ relative to the ordered basis ( e i ) i =1 . Since Ann A α = h e i , itfollows that ϕ ( e ) = ze , for some z ∈ C ∗ . So ϕ induces an automorphism ϕ : A α / h e i −→ A α / h e i and, relative to the basis e i = e i + h e i , i = 1 , , , the nonzero products among basis vectors are just e i = e i +1 , for i = 1 , . It is easy to see (cf. [8, 3.1.1], where this -dimensional algebra is denotedby C ) that the matrix of ϕ is of the form x x y x , for x, y ∈ C with x = 0 . Thus, we conclude that φ = x x y x a a a z . Applying ϕ to the relation e e = 0 yields ϕ ( e ) ϕ ( e ) = ( xe + ye + a e )( x e + a e ) = yx e , so y = 0 . Similarly, using e = e we deduce that a = 0 ; then e = e gives a = 0 and by e e = e we get z = x . So it remains to show that x = 1 , which we do by using e = e and e e = αe , with α = 0 . The former relation implies that x = z = x , so x = 1 , and then αe = αx e = ϕ ( αe ) = ( xe + a e )( x e ) = x αe = xαe . As α = 0 , we can deduce from the above that x = 1 . (cid:3) Remark 18.
The proof of Lemma 17 shows also that, although
Ann A = h e i , the automorphismgroup of A is slightly larger: Aut A = (cid:26)(cid:18) ± x (cid:19) | x ∈ C (cid:27) . Since we are now working in a variety of commutative algebras, we need to slightly adapt themethod described in Subsection 1.1. Specifically, for a commutative n -dimensional algebra A ,let Z C ( A , C ) be the subspace of Z ( A , C ) consisting of the -cocycles θ : A × A −→ C suchthat θ ( x, y ) = θ ( y, x ) , for all x, y ∈ A . Then B ( A , C ) ⊆ Z C ( A , C ) and we set H C ( A , C ) =Z C ( A , C ) (cid:14) B ( A , C ) . This is a subspace of H ( A , C ) and A θ is commutative if and only if θ ∈ Z C ( A , C ) . We define ∆ cij = ∆ ij + ∆ ji and ∆ cii = ∆ ii , for ≤ i = j ≤ n , so that (cid:8) ∆ cij | ≤ i ≤ j ≤ n (cid:9) is a basis of Z C ( A , C ) .We have the following analogue of Lemma 9. Lemma 19.
Let n ≥ and A ∈ Nil γn be commutative such that Ann A = h e n i and e i = e i +1 forall ≤ i ≤ n − . Suppose that Aut ( A ) = { I + xE n | x ∈ C } in the basis ( e i ) ni =1 . Then there isa parametric family of pairwise non-isomorphic commutative -dimensional central extensions B of A with basis ( e i ) n +1 i =1 , extending the basis of A , such that Ann B = h e n +1 i , e e n = 0 , e i = e i +1 for all ≤ i ≤ n , and the structure constants c n +1 ij of B in this basis are arbitrary independent complexparameters for all ≤ i < j ≤ n , ( i, j ) = (1 , n ) .Proof. The proof is identical to that of Lemma 9, essentially replacing ∆ ij by ∆ cij and i = j by i < j .For example, H C ( A ) = h (cid:2) ∆ cij (cid:3) | ≤ i < j ≤ n i ⊕ h [∆ cnn ] i . (cid:3) Proposition 20.
Let n ≥ . Then dim (cid:16)S A ∈ S n O ( A ) (cid:17) = n ( n − n +4)6 .Proof. This proof is just an adaptation of the proof of Proposition 12. Since dim Aut A = 1 , forevery A ∈ S n , we have dim O ( A ) = n − . Moreover, we have observed in Example 16 that S = { A α | α = 0 } and, by [8, 3.1.3], A α ≃ A α ′ if and only if α ′ = ± α . Thus, the isomorphismclasses in S form a -parameter family and the isomorphism classes in S n are obtained by iterated -dimensional central extensions of this family, as shown in Lemma 19.Let q n be the number of independent parameters of the family S n . We have q = 1 and q n +1 = q n + n ( n − − , for all n ≥ . Therefore, q n = n ( n +1)( n − + 1 . Thus, dim (cid:16)S A ∈ S n O ( A ) (cid:17) = n − q n = n ( n − n +4)6 . (cid:3) Theorem 21.
For any n ≥ , the family S n is generic in the variety of all n -dimensional commutativenilpotent algebras. In particular, that variety has dimension n ( n − n +4)6 .Proof. The proof is identical to the proof of Theorem 14, the homologous result for the variety of all n -dimensional nilpotent algebras. Indeed, the base step for n = 4 is given in Example 16 and for theinductive step we just need to observe that if c rij = c rji and γ rij = γ rji for all ≤ i, j, r ≤ n + 1 , then(7) holds for the pair ( i, j ) if and only if it holds for ( j, i ) . (cid:3)
4. T
HE GEOMETRIC CLASSIFICATION OF ANTICOMMUTATIVE NILPOTENT ALGEBRAS
In this section, we consider the variety of anticommutative nilpotent n -dimensional algebras. Ourmethods will be analogous to those of Sections 2 and 3 but there will be some additional technicaldifficulties coming from larger automorphism groups in lower dimensions.To shorten the coming statements, we make the following auxiliary definition. Definition 22.
Let n ≥ . Denote by T ′ n the family of anticommutative algebras in Nil γn such that e i e i +1 = e i +2 for all ≤ i ≤ n − . The remaining structure constants c kij are arbitrary, subject onlyto (1) and the anticommutativity constraint. Lemma 23.
Let A ∈ T ′ n , with n ≥ . Then the following hold:(a) Ann A = h e n i .(b) For all α, β ∈ C , the linear map ϕ ( e ) = e + αe n , ϕ ( e ) = e + βe n , ϕ ( e i ) = e i , ≤ i ≤ n , isan automorphism of A .Proof. The first statement follows easily by induction, as in the proof of Lemma 11, and the secondstatement follows just as in the proof of Lemma 7, using the anticommutativity of A . (cid:3) Our immediate goal is to prove the converse of the second part of Lemma 23, for sufficiently large n and given a few extra conditions on the structure constants c kij . We will do this over a series oflemmas, providing just the key steps in the proofs. Lemma 24.
Let A ∈ T ′ with c = 0 . Then, relative to the basis ( e i ) i =1 , we have Aut A = ( x a y a a xy a a − a y xy ! | a ij ∈ C , x, y ∈ C ∗ ) . Proof.
Let ϕ ∈ Aut A and assume that, relative to ( e i ) i =1 , the matrix of ϕ is ( a ij ) ≤ i,j ≤ . We knowthat Ann A = h e i , so ϕ ( e ) = a e , with a = 0 . Moreover, A / h e i can be seen naturally as anelement of T ′ and ϕ induces an automorphism of A / h e i with matrix ( a ij ) ≤ i,j ≤ relative to the basis ( e i + h e i ) i =1 . The reasoning above then gives a = a = 0 and a = 0 .Next we apply ϕ to the identity e e = 0 , which follows from c = 0 and (1), to get a e + a e + a e + a e )( a e + a e ) = a a e . Thus, a = 0 . Below we list, for each identity in A , the corresponding relations we obtain when weapply ϕ , as above. I DENTITY RELATION e e = 0 a = 0 e e = e a = a a , a = − a a e e = e a = a a These show the direct inclusion in the statement. Since the listed relations comprise all relations in A , the reverse inclusion follows as well. (cid:3) Next, we look at the n = 5 case. Lemma 25.
Let A ∈ T ′ such that c = c = c = 0 . Then, relative to the basis ( e i ) i =1 , we have Aut A = x a y xy a a xy a a a x y | a ij ∈ C , x, y ∈ C ∗ , a = c x (1 − y ) , a = xy ( a c − a ) . Proof.
Let ϕ ∈ Aut A . Using, as before, the fact that Ann A = h e i and Lemma 24, we concludethat the matrix of ϕ relative to the standard basis is of the form x a y a a xy a a − a y xy a a a a a ! , with x, y, a = 0 . The remaining relations follow, as in the proof of Lemma 24, by applying ϕ to theidentities in A . We summarize these below. I DENTITY RELATION e e = 0 a = 0 e e = 0 a = 0 e e = e a = x y e e = e a = 0 e e = c e a = c x (1 − y ) e e = e a = xy ( a c − a ) Thus, the direct inclusion in the statement follows and the reverse follows as well since we have usedall the identities in A . (cid:3) Lemma 26.
Let A ∈ T ′ such that c = c = c = c = c = c = c = c = 0 and c c = 0 .Then, relative to the basis ( e i ) i =1 , we have Aut A = x x x x a a x | a , a ∈ C , x ∈ C ∗ . Proof.
The proof follows the same pattern as before. So, if ϕ ∈ Aut A , then the matrix of ϕ relativeto the standard basis is of the form x a y xy a a xy a a a x y a a a a a a , with x, y ∈ C ∗ , a = c x (1 − y ) and a = xy ( a c − a ) . Now we apply ϕ to the identities in A and the result is summarized in what follows.I DENTITY RELATION e e = e a = x y e e = c e c ( a − x y ) = 0 e e = 0 a = 0 e e = 0 a = 0 Therefore, as c = 0 and x, y = 0 , we get y = 1 and a = x . Notice also that the above isconsistent with our previously deduced relation a = c x (1 − y ) , and we also get a = xa c .Proceeding as before, we obtain the following additional relations.I DENTITY RELATION e e = 0 a = 0 e e = 0 a = 0 e e = e a = 0 e e = e a = 0 e e = c e a = 0 e e = e a = xa c Therefore, as xc = 0 , we deduce that a = 0 . But we had a = xa c and c = 0 , so a = 0 .The proof is thus complete. (cid:3) Lemma 27.
Let A ∈ T ′ such that c = c = c = c = c = c = c = c = 0 and c c c = 0 . Then, relative to the basis ( e i ) i =1 , we have Aut A = { I + a E + a E | a , a ∈ C } . Proof.
The proof follows the ongoing pattern. So, if ϕ ∈ Aut A , then the principal × submatrix ofthe matrix of ϕ relative to the standard basis of A is of the form given in the statement of Lemma 26and ϕ ( e ) = a e . We proceed as before listing the relations deduced from each of the identities in A . I DENTITY RELATION e e = e a = x e e = c e ( x − a ) c = 0 Since xc = 0 , we deduce from the above that x = 1 .I DENTITY RELATION e e = e a = 0 e e = e a = 0 e e = c e a = 0 e e = e a = 0 e e = c e a = 0 e e = e a = 0 Therefore, ϕ is of the desired form, which proves the direct inclusion in the statement. For the reverseinclusion, thanks to Lemma 23, we needn’t check the remaining identities as any linear map withmatrix of the form I + a E + a E , relative to the standard basis, is an automorphism of A . (cid:3) Example 28.
Let n = 6 and take A ∈ T ′ such that c = c = c = c = c = c = c = c =0 and c c = 0 . Then A depends just on the two nonzero parameters α = c and β = c and wedenote this algebra by A ( α, β ) . Define a new basis for A ( α, β ) as follows: E = βe , E = − e − αe , E = βe , E = β e , E = β e , E = β e . Then, we obtain E E = ( βe )( − e − αe ) = βe e = βe = E , E E = ( βe )( βe ) = β e = E , and similar computations show that the multiplication in this basis is given by: E E = E , E E = E , E E = 0 , E E = 0 , E E = 0 , E E = 0 , E E = 0 , E E = − α/β E , E E = 0 , E E = E , E E = E , E E = 0 , E E = E , E E = 0 , E E = 0 . It follows that A ( α, β ) ≃ A ( − α/β ) , where the family A ( γ ) , for γ ∈ C ∗ , was defined in [18, Thm.1] and shown to be generic in the variety of -dimensional complex nilpotent anticommutative alge-bras in [18, Thm. 2]. Moreover, since the algebras { A ( γ ) } γ ∈ C ∗ are pairwise non-isomorphic, itfollows that A ( α, β ) ≃ A ( α ′ , β ′ ) if and only if αβ ′ = α ′ β . Thus, A ( α, β ) ≃ A ( α/β , and wecan assume without loss of generality that β = c = 1 . We conclude that the algebras { A ( α, } α ∈ C ∗ are pairwise non-isomorphic and generic in the variety of -dimensional complex nilpotent anticom-mutative algebras. The results above, along with Example 28, motivate the following auxiliary definition.
Definition 29.
Let n ≥ . Denote by ˆ T n the family of those algebras in T ′ n such that c = c = c = c = c = c = c = c = 0 , c = 1 and c c = 0 (in case n = 6 , the latter conditionshould be replaced with c = 0 ).We can finally prove that the algebras in ˆ T n , with n ≥ , have the smallest possible automorphismgroup among n -dimensional nilpotent anticommutative algebras. Proposition 30.
Let n ≥ and suppose that A ∈ ˆ T n . Then Aut A = { I + a n E n + a n E n | a n , a n ∈ C } , relative to the basis ( e i ) ni =1 .Proof. The proof is by induction on n ≥ and the base step has been settled in Lemma 27. So supposethat the result holds for all algebras in ˆ T n and let us take A ∈ ˆ T n +1 , with n ≥ , and ϕ ∈ Aut A .Then, since Ann A = h e n +1 i we conclude that ϕ ( e n +1 ) = a n +1 ,n +1 e n +1 with a n +1 ,n +1 ∈ C ∗ . Inparticular, ϕ induces an automorphism of A / h e n +1 i . We can see A / h e n +1 i as an element of ˆ T n viathe basis ( e i + h e n +1 i ) ni =1 and the induction hypothesis implies that ϕ ( e ) = e + a n e n + a n +1 , e n +1 ,ϕ ( e ) = e + a n e n + a n +1 , e n +1 and ϕ ( e i ) = e i + a n +1 ,i e n +1 , for all ≤ i ≤ n .So it remains to show that a n +1 ,n +1 = 1 and a n = a n = a n +1 ,i = 0 , for all ≤ i ≤ n . As before,we have the following table.I DENTITY RELATION e n − e n = e n +1 a n +1 ,n +1 = 1 e i e i +1 = e i +2 , for ≤ i ≤ n − a n +1 ,i +2 = 0 , for ≤ i ≤ n − The last relation shows that a n +1 ,i = 0 for all ≤ i ≤ n . Using these we get the following additionalrelations, which conclude the proof.I DENTITY RELATION e e n − = c n ,n − e n + c n +11 ,n − e n +1 a n = 0 e e n − = c n ,n − e n + c n +12 ,n − e n +1 a n = 0 e e = e a n +1 , = 0 e e = e a n +1 , = 0 (cid:3) In our next step we consider central extensions, as in Lemmas 9 and 19. Since we are now work-ing in the variety of anticommutative algebras, as in the previous section, we accordingly adaptthe method described in Subsection 1.1. So, for an anticommutative n -dimensional algebra A , Z AC ( A , C ) is the subspace of Z ( A , C ) consisting of the -cocycles θ : A × A −→ C such that θ ( x, y ) = − θ ( y, x ) , for all x, y ∈ A . We set H AC ( A , C ) = Z AC ( A , C ) (cid:14) B ( A , C ) and it followsthat A θ is anticommutative if and only if θ ∈ Z AC ( A , C ) . We also define ∆ aij = ∆ ij − ∆ ji , for ≤ i, j ≤ n , so that (cid:8) ∆ aij | ≤ i < j ≤ n (cid:9) is a basis of Z AC ( A , C ) . Lemma 31.
Let n ≥ and A ∈ Nil γn be anticommutative such that Ann A = h e n i and e i e i +1 = e i +2 ,for all ≤ i ≤ n − . Suppose that Aut ( A ) = { I + xE n + yE n | x, y ∈ C } in the basis ( e i ) ni =1 .Then there is a parametric family of pairwise non-isomorphic, anticommutative -dimensional centralextensions B of A with basis ( e i ) n +1 i =1 , extending the basis of A , such that Ann B = h e n +1 i , c n +11 ,n − =0 = c n +12 ,n − , e i e i +1 = e i +2 for all ≤ i ≤ n − , and the structure constants c n +1 ij of B in this basis arearbitrary independent complex parameters for all ≤ i ≤ j − ≤ n − , ( i, j ) = (1 , n − , (2 , n − .Proof. The proof is similar to that of Lemma 9, but there are a few differences, also related to the factthat the automorphism group of A is larger. Thus, we will just highlight the differences. We have B ( A ) = n − M m =2 D ∆ am − ,m + X i Let A ∈ ˆ T with c = 0 = c . Then A is isomorphic to a unique algebra A ′ ∈ ˆ T with d = 0 = d and d = 1 , where the d kij are the structure constants of A ′ .Proof. Notice first that the hypotheses on A imply that there are exactly parameters of freedom,namely: c , c , c , c , c , c , c , c , c , with c c = 0 . Let x = c and consider the new basis E = xe , E = e , E = xe , E = xe , E = x e , E = x e , E = x e . Let d kij be the structure constants of A relative to this basis. Then, it is straightforward to see thatthese structure constants satisfy all of the conditions determined by ˆ T , along with d = 0 = d .For example, E E = x e e = x ( c e + c e ) = x e + x c e = E + c x − E , so d = 1 . Moreover, E E = x e e = x c e = x e = E , and thus d = 1 . This proves the existence part of the statement.For the uniqueness, with A ′ as in the statement, suppose also that c = 1 and φ : A −→ A ′ is an isomorphism. Then, φ induces an isomorphism φ on the quotient algebras by their respectiveannihilators. By Example 28, it follows that c = d and φ is thus an automorphism. We can thenceapply Lemma 26 to get the matrix of φ and then lift it to φ . From this point it is a simple matter to usethe multiplication tables of A and A ′ and the matrix of φ to deduce that the respective free parametersin A and A ′ are equal. (cid:3) At last we define the family T n , which will be shown to be generic in the variety of n -dimensionalanticommutative algebras. Definition 33. Let n ≥ (in case n = 6 , the condition c = 1 is to be ignored). Denote by T n thefamily of those algebras in ˆ T n such that c = 1 and c i +21 i = 0 = c i +22 i , for all i ≥ . In other words, T n is the family of n -dimensional complex anticommutative algebras whose structure constants ( c kij ) i,j,k relative to the basis ( e i ) ni =1 satisfy (1) and such that: • e i e i +1 = e i +2 for all ≤ i ≤ n − ; • c i +21 i = 0 = c i +22 i , for all ≤ i ≤ n − ; • c = c = c = c = c = c = 0 ; • c = 0 ; • c = c = 1 .The remaining structure constants c kij are arbitrary, subject only to the anticommutativity constraint. Proposition 34. Let n ≥ . Then dim (cid:16)S T ∈ T n O ( T ) (cid:17) = ( n − n +2 n +3)6 .Proof. In case n = 6 , the statement follows from Example 28 and [18, Thm. 2]. So assume that n ≥ . The remainder of the proof is again an adaptation of the proof of Proposition 12. By Proposition 30, for every A ∈ T n with n ≥ , we have dim Aut A = 2 , so dim O ( A ) = n − . Moreover, by Lemma 32, different choices of structure constants in T give rise to distinctisomorphism classes. Thus, as seen in the proof of Lemma 32, the isomorphism classes in T form an -parameter family and the isomorphism classes in T n are obtained by iterated -dimensional centralextensions of this family, as shown in Lemma 31.Let r n be the number of independent parameters of the family T n . We have r = 8 and r n +1 = r n + (cid:0) n − (cid:1) − , for all n ≥ . Therefore, r n = ( n − n +1)( n − . Thus, dim (cid:16)S T ∈ T n O ( T ) (cid:17) = n − r n = ( n − n +2 n +3)6 . (cid:3) Finally, we show that the family T n is generic and determine the dimension of the variety of com-plex n -dimensional anticommutative nilpotent algebras. Theorem 35. For any n ≥ , the family T n is generic in the variety of n -dimensional anticommutativenilpotent algebras. In particular, that variety has dimension ( n − n +2 n +3)6 .Proof. The proof is similar to that of Theorem 14.For arbitrary N ∈ Nil γn we will prove by induction on n that there is a parametric basis E i ( t ) = P nj = i a ji ( t ) e j , with a ji ( t ) ∈ C ( t ) , ≤ i ≤ j ≤ n , and a choice of structure constants c kij ( t ) ∈ C ( t ) ,satisfying the conditions of Definition 33 and giving a degeneration of N from T n .The base case n = 6 has already been proved in Example 28 and [18]. Denote by γ kij the structureconstants of N in ( e i ) n +1 i =1 . For the inductive step from n to n + 1 , as in the proof of Theorem 14,it suffices to establish the convergence (7) by the appropriate choice of c n +1 ij ( t ) and a n +1 ,i ( t ) . When n + 1 = 7 , we replace c = 1 by the more general condition c = 0 , which is permitted in viewof Lemma 32. We may also redefine a n ( t ) and a n ( t ) , as no degeneration from T n depends on thesecoefficients.We will proceed in n − steps, from k = 1 to k = n − . At the end of Step k we will have defined c n +1 p,q ( t ) for all q > p ≥ n − k and a n +1 ,r ( t ) for all r ≥ n − k + 2 . We will also have obtained (7) forall n ≥ j > i ≥ n − k . Step . Since we must have c n +1 n − ,n ( t ) = 1 , it remains to define a n +1 ,n +1 ( t ) . The left-hand sideof (7) for i = n − and j = n becomes lim t → ( a n − ,n − ( t ) a nn ( t ) a n +1 ,n +1 ( t ) − ) , so we set a n +1 ,n +1 ( t ) := ( ( γ n +1 n − ,n ) − a n − ,n − ( t ) a nn ( t ) , if γ n +1 n − ,n = 0 , t − a n − ,n − ( t ) a nn ( t ) , if γ n +1 n − ,n = 0 .By definition, (7) holds for i = n − and j = n . Notice also that a n ( t ) , a n ( t ) do not occur in theformula above. Step k . Let ≤ k < n − and assume that Step k − has been successfully completed and that a n ( t ) , a n ( t ) have not been used to define any new coefficients.Suppose first that n ≥ j > i = n − k and j = i + 1 . We will define c n +1 ij ( t ) and establish (7) in thiscase. The coefficient of c n +1 ij ( t ) on the left-hand side of (7) equals a n +1 ,n +1 ( t ) − a ii ( t ) a jj ( t ) which has already been defined and is non-zero. We thus put c n +1 ij ( t ) := a n +1 ,n +1 ( t ) a ii ( t ) a jj ( t ) γ n +1 ij − n +1 P r =2 a ′ n +1 ,r ( t ) r − X ′ p = i,q = j c rpq ( t ) a pi ( t ) a qj ( t ) ! , if this is non-zero , ta n +1 ,n +1 ( t ) a ii ( t ) a jj ( t ) , otherwise , (11)where the primed sum is over all ( p, q ) such that ( p, q, r ) = ( i, j, n + 1) . Note that on the right-handside of (11) we must have n − k + 1 < j ≤ r − , so r ≥ n − k + 3 . Thence, by Step k − andLemma 13, all the terms of the form a ′ n +1 ,r ( t ) on the right-hand side of (11) have already been defined.The same holds for all remaining terms except those of the form c n +1 iq ( t ) with q > j . Thus, (11) isa recurrence formula which defines c n +1 ij ( t ) in terms of c n +1 iq ( t ) with q > j . So, starting recursivelywith c n +1 in ( t ) , we can define all of the terms c n +1 n − k,q ( t ) , with q > n − k + 1 and by doing so we forcethe convergence (7) for all j > n − k + 1 and i = n − k . This will work because we are assumingthat k < n − so ( n − k, q ) = (1 , n − , (2 , n − . Moreover, also by that assumption on k , thecoefficients a n ( t ) , a n ( t ) have not been used in (11) to define c n +1 ij ( t ) , as j > i = n − k ≥ . Hence,given that c n +1 n − k,n − k +1 ( t ) = 0 ( k > ), all c n +1 p,q ( t ) with q > p ≥ n − k are defined and (7) holds for all j > i ≥ n − k except if i = n − k and j = n − k + 1 .Next, we will define a n +1 ,n +2 − k ( t ) so that (7) holds for i = n − k and j = n − k + 1 . Assume thusthat i = n − k and j = n − k + 1 . Using Lemma 13 we have n +1 X r =2 r − X p = i,q = i +1 a ′ n +1 ,r ( t ) c rpq ( t ) a pi ( t ) a q,i +1 ( t ) = a ′ n +1 ,i +2 ( t ) a ii ( t ) a i +1 ,i +1 ( t ) + n +1 X r = i +3 a ′ n +1 ,r ( t ) r − X p = i,q = i +1 c rpq ( t ) a pi ( t ) a q,i +1 ( t )= − a n +1 ,n +1 ( t ) − a i +2 ,i +2 ( t ) − a ii ( t ) a i +1 ,i +1 ( t ) a n +1 ,i +2 ( t ) − a i +2 ,i +2 ( t ) − a ii ( t ) a i +1 ,i +1 ( t ) n X s = i +3 a ′ n +1 ,s ( t ) a s,i +2 ( t )+ n +1 X r = i +3 a ′ n +1 ,r ( t ) r − X p = i,q = i +1 c rpq ( t ) a pi ( t ) a q,i +1 ( t ) . Hence, we put a n +1 ,i +2 ( t ) := − a n +1 ,n +1 ( t ) a i +2 ,i +2 ( t ) a ii ( t ) a i +1 ,i +1 ( t ) γ n +1 i,i +1 − n +1 X r = i +3 a ′ n +1 ,r ( t ) r − X p = i,q = i +1 c rpq ( t ) a pi ( t ) a q,i +1 ( t ) ! − a n +1 ,n +1 ( t ) n X s = i +3 a ′ n +1 ,s ( t ) a s,i +2 ( t ) , (12)where the right-hand side defines a n +1 ,n − k +2 ( t ) in terms of a n +1 ,r ( t ) with r ≥ n − k + 3 (alreadydefined in the previous steps) and c n +1 p,q ( t ) with q > p ≥ n − k (defined above). Also, (7) holds for i = n − k and j = n − k + 1 and a n ( t ) , a n ( t ) do not occur in the definition (12) above. This step isthus finished. Step n − . When we reach this step, all c n +1 p,q ( t ) with q > p ≥ and all a n +1 ,r ( t ) with r ≥ have been defined without using the coefficients a n ( t ) , a n ( t ) and (7) has been shown to hold for all j > i ≥ .Consider first the case ( i, j ) = (2 , n ) . Then, from the right-hand side of (7), we get n +1 X r =2 a ′ n +1 ,r ( t ) r − X p = i,q = j c rpq ( t ) a pi ( t ) a qj ( t ) = a ′ n +1 ,n +1 ( t ) a nn ( t ) n X p =2 c n +1 pn ( t ) a p ( t )= a n +1 ,n +1 ( t ) − a nn ( t ) n − X p =2 c n +1 pn ( t ) a p ( t ) , so we put c n +12 n ( t ) := a ( t ) − a n +1 ,n +1 ( t ) a nn ( t ) − γ n +12 n − a ( t ) − n − X p =3 c n +1 pn ( t ) a p ( t ) , in which the right-hand side has already been defined in the previous steps and does not depend on a n ( t ) or a n ( t ) .Now let ( i, j ) = (2 , n − . We will redefine a n ( t ) at this point. This is necessary because weare bound to having c n +12 ,n − ( t ) = 0 , so we cannot force (7) in the case ( i, j ) = (2 , n − by choosing c n +12 ,n − ( t ) as we please. We have n +1 X r =2 r − X p = i,q = j a ′ n +1 ,r ( t ) c rpq ( t ) a pi ( t ) a qj ( t ) = a n +1 ,n +1 ( t ) − a n − ,n − ( t ) n X p =3 c n +1 p,n − ( t ) a p ( t )+ a n +1 ,n +1 ( t ) − a n,n − ( t ) n − X p =2 c n +1 p,n ( t ) a p ( t )+ a ′ n +1 ,n ( t ) a n − ,n − ( t ) n − X p =2 c np,n − ( t ) a p ( t ) , in which a ′ n +1 ,n ( t ) = − a − n +1 ,n +1 ( t ) a − nn ( t ) a n +1 ,n ( t ) has already been defined and the coefficient of a n ( t ) is − a n +1 ,n +1 ( t ) − a n − ,n − ( t ) = 0 . Hence we set a n ( t ) := − a n +1 ,n +1 ( t ) γ n +12 ,n − a n − ,n − ( t ) + n − X p =3 c n +1 p,n − ( t ) a p ( t ) + a n,n − ( t ) a n − ,n − ( t ) n − X p =2 c n +1 p,n ( t ) a p ( t ) − a n +1 ,n ( t ) a nn ( t ) n − X p =2 c np,n − ( t ) a p ( t ) , the right-hand side of which has already been defined and does not involve a n ( t ) or a n ( t ) . Now we can proceed as in the previous (generic) step with k = n − defining c n +12 q ( t ) for n − ≥ q ≥ and finally a n +1 , ( t ) , ensuring that (7) holds in the remaining cases. Step n − . This step is totally analogous to the previous one. We define c n +11 n ( t ) , then redefine a n ( t ) and after that find c n +11 q ( t ) for n − ≥ q ≥ and finally a n +1 , ( t ) , ensuring that (7) holdsin the remaining cases. The coefficients a n +1 , ( t ) and a n +1 , ( t ) are unrestrained and can be chosenarbitrarily (which agrees with our previous observations). (cid:3) 5. C OROLLARIES , AN OPEN QUESTION AND A CONJECTURE Nilpotency index. Recall from Subsection 1.3 that the nilpotency index of a nilpotent algebra A is the smallest positive k such that A k = 0 . The nilpotency index of an n -dimensional nilpotentalgebra is not greater than n − + 1 but, in general, this upper bound can be attained. On the otherhand, for n -dimensional nilpotent Lie algebras, it is known that the upper bound for the nilpotencyindex is n − , while for n -dimensional nilpotent associative, Leibniz and Zinbiel algebras, the upperbound on the nilpotency index is n .Thanks to our Theorem 35, we know that each n -dimensional nilpotent anticommutative algebra isa degeneration from the generic family T n . This implies the following result. Corollary 36. The nilpotency index of an n -dimensional nilpotent anticommutative algebra is at most F n + 1 , where F n is the n th Fibonacci number. This bound is sharp and it is attained by the algebrasfrom the generic family T n given in Theorem 35 (see Definition 33). Length of algebras. Let A be a finite-dimensional algebra and S be a finite subset of A . Wedefine the length function of S as follows (see [15]). Any product (with any choice of bracketing) of afinite number of elements of S is a word in S , the number of letters (i.e. elements of S ) in the productbeing its length. For i ≥ , the set of all words in S having length less than or equal to i is denotedby S i . Then set L i ( S ) = h S i i , the linear span of S i , and L ( S ) = ∞ S i =1 L i ( S ) . Definition 37. Assume that S is a finite generating set for the finite-dimensional algebra A . Then thelength of S is defined as l ( S ) = min { i ≥ | L i ( S ) = A } . The length of A is l ( A ) = sup { l ( S ) | S ⊆ A finite and L ( S ) = A } . The length of the associative algebra of matrices of size was first discussed in [28] in the contextof the mechanics of isotropic continua. The more general problem for the algebra of matrices of size n was posed in [26] but is still open (recently, some interesting new results about this problem aregiven by Shitov [29]). The known upper bounds for the length of the matrix algebra of size n are ingeneral nonlinear in n .For our main corollary, we need the following key lemma. Lemma 38. Let A be an n -dimensional anticommutative algebra of length k . Then there is an n -dimensional nilpotent anticommutative algebra with nilpotency index k + 1 . Proof. Let S = { a , . . . , a t } be a generating set of A such that l ( A ) = k = l ( S ) . Our idea forthe construction of an n -dimensional nilpotent anticommutative algebra ( B , ⋆ ) with nilpotency index k + 1 is based on a reduction of the multiplication of A to a nilpotent case. The reduction of themultiplication of A is given in the following k + 1 steps. Step . We consider an algebra B with the same underlying vector space as A . Step . Fix a complement K for L ( S ) in L ( S ) , so that L ( S ) = K ⊕ L ( S ) . For each pair a, b of elements from L ( S ) , the product ab can be written as ab = ℓ + ℓ ∗ , where ℓ ∈ K and ℓ ∗ ∈ L ( S ) .Then set a ⋆ b = ℓ . Step R . Let ≤ R ≤ k . Fix a complement K R for L R − ( S ) in L R ( S ) , so that L R ( S ) = K R ⊕ L R − ( S ) . For each pair a, b of elements a ∈ L R − q ( S ) \ L R − q − ( S ) and b ∈ L q ( S ) \ L q − ( S ) ,where ≤ q < R , the product ab can be written as ab = ℓ + ℓ ∗ , where ℓ ∈ K R and ℓ ∗ ∈ L R − ( S ) .Then set a ⋆ b = ℓ . Step k + . The remaining multiplications are zero.By construction, ( B , ⋆ ) is an n -dimensional nilpotent anticommutative algebra of length k andnilpotency index k + 1 . (cid:3) Combining Corollary 36 and Lemma 38 gives the following corollary. Corollary 39. The length of an n -dimensional anticommutative algebra is bounded above by F n , the n th Fibonacci number. This bound is sharp and it is reached by the algebras from the generic family T n given in Theorem 35 (see Definition 33). Open question and conjecture. The study of n -ary algebras is an interesting topic which hasseen good developments recently. There are many different generalizations to the n -ary case of thecommutative and anticommutative properties. Let us give a more general definition below. Definition 40. Let N be an n -ary algebra with multiplication [ · , . . . , · ] . Given two disjoint subsets A and C of { , . . . , n } , we say that N is an ( A, C ) -commutative n -ary algebra if the following holds:(1) the multiplication [ x , . . . , x n ] is anticommutative on elements indexed by A ,(2) the multiplication [ x , . . . , x n ] is commutative on elements indexed by C .The main examples of ( A, C ) -commutative n -ary algebras are: • Lie triple systems and Tortkara triple systems ( ( { , } , ∅ ) -commutative -ary algebras), • anti-Jordan triple systems [7] and algebraic N = 6 3 -algebras [4] ( ( { , } , ∅ ) -commutative -ary algebras), • algebraic N = 5 3 -algebras [4] ( ( ∅ , { , } ) -commutative -ary algebras), • Jordan quadruple systems [2] ( ( ∅ , { , } ) -commutative -ary algebras and also ( ∅ , { , } ) -commutative -ary algebras), • commutative (resp. anticommutative) n -ary algebras ( ( ∅ , { , . . . , n } ) -commutative (resp. ( { , . . . , n } , ∅ ) -commutative) n -ary algebras). The special case of ( { , . . . , a } , { n − c + 1 , . . . , n } ) -commutative n -ary algebras we will called ( a, c ) -commutative n -ary algebras.The geometric study of varieties of n -ary algebras defined by a family of polynomial identites hasbeen started in [23]. Hence, we have an obvious open question. Open question. It is clear that the variety of k -dimensional nilpotent ( A, C ) -commutative n -aryalgebras is irreducible. What is the geometric dimension of this variety?In order to formulate our conjecture concerning a bound on the length of k -dimensional ( A, C ) -commutative n -ary algebras we need to introduce the N -generated Fibonacci numbers. Definition 41. Let N = p a · · · p a r r be the prime decomposition of N , where p r denotes the r th primenumber. We can define the N -generated Fibonacci number F N ( n ) recursively as F N ( n ) = a F N ( n − 1) + a F N ( n − 2) + · · · + a r F N ( n − r ) ,where F N ( n ) = 1 if n ≤ r . Conjecture. Let N be a k -dimensional ( A, C ) -commutative n -ary algebra. Then the length of N isat most F n − a +1 · ··· p a ( k ) , where a = | A | , if | A | > and a = 1 , if | A | = 0 . Remark 42. If the conjecture is true, then the bound is sharp and it gives the sharp bound for thenilpotency index of k -dimensional nilpotent ( a, c ) -commutative n -ary algebras. In the case of k -dimensional ( a, c ) -commutative n -ary algebras, it is confirmed by the following n -ary algebra N with the multiplication given by [ e j − a +1 , . . . , e j − , e j , . . . , e j ] = e j +1 , a ≤ j ≤ k − . For arbitrary ( A, C ) -commutative n -ary algebras, an extremal example can be obtained by a similarway using a suitable permutation of the indices in the above multiplication. 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