The geometric classification of nilpotent algebras

We give a geometric classification of $n$-dimensional nilpotent, commutative nilpotent and anticommutative nilpotent algebras. We prove that the corresponding geometric varieties are irreducible, find their dimensions and describe explicit generic families of algebras which define each of these varieties. We show some applications of these results in the study of the length of anticommutative algebras.


INTRODUCTION
The geometry of varieties of algebras defined by polynomial identities has been an active area of interest and research since the work of Nijenhuis-Richardson [25] and Gabriel [10] in the 1960's and 1970'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 structure of 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 basechange action of the general linear group). The study of degenerations of algebras is very rich and closely related to deformation theory, in the sense of Gerstenhaber [11]. Degenerations have also been used to study a level of complexity of an algebra [12,21]. There are many results concerning We show also that the family T n for n ≥ 6 given in Definition 33 is generic in the variety of n-dimensional anticommutative nilpotent algebras and inductively give an algorithmic procedure to obtain any n-dimensional nilpotent anticommutative algebra as a degeneration from T n . The cases of n = 2, 3, 4, 5 follow from [8,22].
The notion of length for nonassociative algebras has been recently introduced in [15], generalizing the 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 anticommutative algebra is bounded above by the n th Fibonacci number, and prove that our bound is sharp.

VARIETIES 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 denoted by I and the matrix unit corresponding to the row i and the column j is E ij . For a subset X of a given vector space, the linear span of X is denoted by X . Skjelbred and Sund. An algebra is a vector space endowed with a bilinear multiplication. Formally, it is a pair A = (V, µ), where V is a vector space and µ ∈ Hom(V ⊗ V, V), which gives the algebra law. The annihilator of A is Ann A = {x ∈ A | xA + Ax = 0}. For the purposes of this paper, we just need to consider 1-dimensional central extensions. We will give here an overview of the algebraic classification method of Skjelbred and Sund [30].

Central extensions and the method of
A bilinear map θ : A × A −→ C determines the 1-dimensional central extension A θ = A ⊕ C with the product (x + v) · θ (y + w) = xy + θ(x, y), for all x, y ∈ A and v, w ∈ C. We let Z 2 (A, C) be the vector space of all bilinear maps A × A −→ C, which we refer to as 2-cocycles (of A with values in C). Then, the space of 2-coboundaries is B 2 (A, C) = {δf | f ∈ A * } ⊆ Z 2 (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 the quotient space H 2 (A, C) = Z 2 (A, C) B 2 (A, C), which is well known to parametrize equivalence classes of 1-dimensional central extensions of A.
Once a basis (e i ) n i=1 of A is fixed, define the bilinear maps ∆ ij = e * i × e * j , for 1 ≤ i, j ≤ n, so that ∆ ij (e k , e ℓ ) = δ ik δ jℓ , for all 1 ≤ i, j, k, ℓ ≤ n. Then (∆ ij ) 1≤i,j≤n is a basis of Z 2 (A, C). The automorphism group Aut A acts on Z 2 (A, C) via φ(θ) = θ•(φ×φ), for φ ∈ Aut A and θ ∈ Z 2 (A, C) and the action induces a well-defined one on H 2 (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 2-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 2 (A, C) for A θ and A θ ′ to be isomorphic, under the assumption that Ann A ∩ Ann θ = 0 = Ann A ∩ Ann θ ′ . This is given by the following result. Lemma 1. Let A be an algebra and θ, θ ′ be 2-cocycles, represented by the nonzero cohomology classes [θ], [θ ′ ] in H 2 (A, C). Suppose that Ann A ∩ Ann θ = 0 = Ann A ∩ Ann θ ′ . Then A θ is isomorphic to A θ ′ if and only if the orbits of [θ] and [θ ′ ] under the action of Aut A span the same vector space.
1.2. Varieties of algebras. Given an n-dimensional vector space V, an algebra structure on V (or an n-dimensional algebra law) is naturally seen as an element of Hom(V ⊗ V, V) ∼ = V * ⊗ V * ⊗ V, a vector space of dimension n 3 . Once we fix a basis e 1 , . . . , e n of V, this space can be identified with C n 3 and given the structure of an affine variety whose coordinate ring if the polynomial ring in the variables c k i,j n i,j,k=1 . Accordingly, a subset of Hom(V ⊗ V, V) is Zariski-closed if it can be defined by a set of polynomial equations in the variables c k i,j i,j,k . For simplicity, we identify points in the variety with the corresponding maximal ideals of the coordinate ring so, if no confusion arises, we also think of the c k i,j as scalars. Henceforth, having fixed the vector space V and its basis (e i ) n i=1 , we will identify points c k i,j n i,j,k=1 in C n 3 with n-dimensional algebras via their structure constants relative to the basis (e i ) n i=1 . Concretely, any µ ∈ Hom(V ⊗ V, V) is determined by the n 3 structure constants c k i,j ∈ C such that µ(e i ⊗ e j ) = n k=1 c k i,j e k . Example 2. The following polynomial identities define well-known varieties of n-dimensional algebras: (1) Anticommutative algebras (3) Lie algebras n ℓ=1 c ℓ ij c m kℓ + c ℓ jk c m iℓ + c ℓ ki c m jℓ = 0 and c k ij + c k ji = 0, 1 ≤ i, j, k, m ≤ n.

Nilpotent algebras. Recall that an algebra
The smallest (if any) positive integer k satisfying N k = 0 is called the nilpotency index of N. It is easy to see that the n-dimensional algebras N with N k = 0 form a closed set (as elements of C n 3 ) and thence 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 to an algebra M whose structure constants (γ k i,j ) n i,j,k=1 satisfy γ k i,j = 0, ∀k ≤ max{i, j}.
Proof. Indeed, (1) holds for the algebra with zero multiplication. Moreover, if an algebra A is a central extension of another algebra B by some vector space and the structure constants of B relative to some basis satisfy (1), then completing this basis of B to a basis of A we see that the corresponding structure constants of A will also satisfy (1). It remains to note that each finite-dimensional nilpotent algebra can be obtained, up to isomorphism, from an algebra with zero multiplication by consecutively applying the central extension procedure.
Conversely, if the structure constants of an algebra M satisfy (1), then all the products in M of length 2 n are zero, so M is nilpotent.
Definition 4. Let n ≥ 1. We denote by Nil n the variety of all nilpotent algebra structures (on V, with respect to the basis (e i ) n i=1 ) and by Nil γ n 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 via base-change and the orbits parametrize the isomorphism classes of algebras. Given an algebra A, its orbit under this action will be denoted by O(A). Lemma 3 can thus be restated simply as Nil n = GL n (C) · Nil γ n . Proposition 5. Let n ≥ 1. The variety Nil n of all nilpotent n-dimensional algebras and its subvarieties of commutative and anticommutative nilpotent algebras are irreducible.
Proof. The coordinate ring O n of the variety Nil γ n is the complex polynomial ring in the variables c k i,j | 1 ≤ i, j ≤ n, k > max{i, j} . Thus, O n being a domain, the variety Nil γ n is irreducible. If ℓ n is the transcendence degree of O n , then ℓ n+1 = ℓ n + n 2 , so the dimension of Nil γ n is ℓ n = n(n−1)(2n−1)

6
. By Lemma 3, Nil n = GL n (C)·Nil γ n . Since GL n (C) is a connected (i.e. irreducible) algebraic group and the product of irreducible varieties is irreducible, it follows that GL n (C) × Nil γ n is irreducible. Hence, so is the continuous image Nil n = GL n (C) · Nil γ n . A similar argument holds for the varieties of n-dimensional commutative and anticommutative nilpotent algebras. Definition 6. Let V be an irreducible variety of algebras and R ⊆ V be a family of algebras. The family R is said to be generic in V, if A∈R O(A) = V. For an algebra B ∈ V, we also write R → B as shorthand for B ∈ A∈R O(A).

THE GEOMETRIC CLASSIFICATION OF NILPOTENT ALGEBRAS
In this section we find a generic family of algebras in the variety Nil n of all nilpotent n-dimensional algebras and use it to compute the dimension of the variety. Recall that Nil γ n is the subvariety of algebras satisfying (1) and that (e i ) n i=1 is our fixed basis. Lemma 7. Let A ∈ Nil γ n . Then, for all x ∈ C, the linear endomorphism ϕ defined by ϕ(e 1 ) = e 1 + xe n , ϕ(e i ) = e i , 2 ≤ i ≤ n, is an automorphism of A.
Lemma 9. Let n ≥ 3 and A ∈ Nil γ n such that Ann A = e n and e 2 i = e i+1 , for all 1 ≤ i ≤ n − 1.
i=1 , extending the basis of A, such that Ann B = e n+1 , e 1 e n = 0, e 2 i = e i+1 , for all 1 ≤ i ≤ n, and the structure constants c n+1 ij of B in this basis are arbitrary independent complex parameters for all 1 ≤ i = j ≤ n, (i, j) = (1, n).

Proof. We have
Consider the corresponding matrix A = 1≤i =j≤n α ij E ij + α nn E nn . Then nn , we obtain the family of representatives of distinct orbits It determines the desired family of algebras B.
Definition 10. Let n ≥ 3. Denote by R n the family of nilpotent algebras with basis (e i ) n i=1 satisfying (1), such that e 2 i = e i+1 , for all 1 ≤ i ≤ n − 1, c 3 21 = 1, c i+1 1i = 0, for all 2 ≤ i ≤ n − 1, and with the remaining structure constants c k ij being arbitrary independent complex parameters, for all 1 ≤ i = j ≤ n, k > max{i, j}.
Notice that, given A ∈ R n+1 with n ≥ 3, since by (1) we have e n+1 ∈ Ann A, it follows that e n+1 is an ideal of A and A/ e n+1 can be seen naturally as an element of R n , relative to the ordered basis (e i ) n i=1 , where e i = e i + e n+1 . This property will be important in arguments by induction, as the one which follows.
Lemma 11. Let A ∈ R n with n ≥ 3. Then the following hold: We prove both statements simultaneously by induction on n ≥ 3. Indeed, R 3 consists of a single point, which as an algebra is defined by the following multiplication (as usual, only nonzero products of the basis elements are shown): A : e 1 e 1 = e 2 , e 2 e 1 = e 3 , e 2 e 2 = e 3 .
It is easy to see by direct inspection that Ann A = e 3 and Aut Now let A ∈ R n+1 . Then, viewing A/ e n+1 as an algebra from R n , as explained above, the induction hypothesis implies that Ann A/ e n+1 = e n . Let v = 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 = n i=1 λ i e i ∈ Ann A/ e n+1 , we deduce that λ i = 0 for all i ≤ n − 1. Thus, λ n e n ∈ Ann A, so 0 = λ n e n e n = λ n e n+1 and λ n = 0, proving our claim that Ann A = e n+1 .
The induction hypothesis also gives that 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) − 1 = n 2 − 1. 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 by induction on n. We have p 3 = 0 and p n+1 = p n + n(n − 1) − 1, for all n ≥ 3. Therefore, Before our main result of this section, we need a technical observation on the inverse of a lower triangular matrix.

j=1 be an invertible lower triangular matrix of size n and
The second statement follows by backward induction on j with i fixed. Now we can state and prove our main result about the variety Nil n of nilpotent algebras. Notice that the proof specifically gives an algorithmic construction of a degeneration R n → N from the family R n to any given n-dimensional nilpotent algebra N.
Theorem 14. For any n ≥ 3, the family R n is generic in Nil n . In particular, dim(Nil n ) = n(n−1)(n+1) Proof. Given N ∈ Nil n , we will prove that R n → N. Recall that (e i ) n i=1 is our fixed basis of the underlying vector space. Without loss of generality, we may assume that N ∈ Nil γ n . Indeed, by Lemma 3, N is isomorphic to an algebra M whose structure constants satisfy (1) in some basis , and thus satisfy (1). So, we may replace N by gM, if necessary. We will thus assume that N ∈ Nil γ n and prove by induction on n that there is a parametric basis E i (t) = n j=i a ji (t)e j , with a ji (t) ∈ C(t), 1 ≤ i ≤ j ≤ n, and a choice of structure constants c k ij (t) ∈ C(t), satisfying the conditions of Definition 10, with giving a degeneration of N from R n .
The case n = 3 is proved in [8]. Let N ∈ Nil γ n+1 . It follows that e n+1 ∈ Ann N, so e n+1 is an ideal of N and N/ e n+1 is seen as an element of Nil γ n via the identification of e i + e n+1 with e i , 1 ≤ i ≤ n. By the induction hypothesis, there is a parametric basis E i (t) = n j=i a ji (t)e j , 1 ≤ i ≤ n, and a choice of parameters c k ij (t) satisfying the conditions of Definition 10 and determining a degeneration of N/ e n+1 from R n . Observe that the degeneration does not depend on a n1 (t) because e n ∈ Ann R for all R ∈ R n . More generally, since any such R satisfies (1), we have We see that a n1 (t) cannot appear among the a pi (t) or a qj (t) above. Moreover, each e r from the sum (4) belongs to e 2 , . . . , e n = E 2 (t), . . . , E n (t) , so its coordinates in the basis We are going to redefine a n1 (t) and choose a 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 parameters which will be defined below and let A −1 (t) = (a ′ ij (t)) n+1 i,j=1 be its formal inverse. Observe that the upper left (n × n)-block of A −1 (t) is the inverse of the upper left (n × n)-block of A(t) and thus does not depend on the choice of a n+1,i (t), 1 ≤ i ≤ n + 1. Since the coordinates of e i in the basis (Ẽ j (t)) n+1 j=1 are given by the i-th column of A −1 (t), for all 1 ≤ i ≤ n + 1, we can further develop (4) to getẼ for all 1 ≤ i, j ≤ n + 1. Notice that these new structure constants satisfy (1).
be satisfied. Observe that (6) holds for all 1 ≤ i, j < k ≤ n by the choice of (E i (t)) n i=1 and (c k ij (t)) with (3), because γ k ij is the corresponding structure constant of N/ e n+1 for such (i, j, k). Thus, it remains to consider k = n + 1, which we do below by appropriately defining a n1 (t), a n+1,i (t), 1 ≤ i ≤ n + 1, and c n+1 ij (t), 1 ≤ i = j ≤ n, (i, j) = (1, n) (so that the conditions of Definition 10 hold).
We will proceed in n steps, from k = 0 to k = n − 1. 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 the convergence for all n ≥ i, j ≥ n − k.
Step 0. 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) By definition, (7) holds for i = j = n. Notice also that a n1 (t) does not occur in the formula above.
Step k. Let 1 ≤ k < n − 1 and assume that Step k − 1 has been successfully completed and that a n1 (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. The coefficient of c n+1 ij (t) on the left-hand side of (7) equals a n+1,n+1 (t) −1 a ii (t)a jj (t) which has already been defined and is non-zero. We thus put where the primed sum is over all (p, q) such that (p, q, r) = (i, j, n + 1). Note that on the right-hand side of (8) we must have n − k < i ≤ r − 1, so r ≥ n − k + 2. Thence, by Step k − 1 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 same holds for all remaining terms except those of the form c n+1 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 before because we are assuming that k < n − 1 so (n − k, q) = (1, n). Moreover, also by that assumption on k, the coefficient a n1 (t) has not been used in (8) 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 Hence, we put 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 (already defined 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 n1 (t) does not occur in the definition (9) above. This step is thus finished.
Step n − 1. When we reach this final step, all c n+1 p,q (t) with p, q ≥ 2 and all a n+1,r (t) with r ≥ 3 have been defined without using the coefficient a n1 (t) and (7) has been shown to hold for all i, j ≥ 2. Hence, as a n1 (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 are bound to having c n+1 1n (t) = 0, so we cannot force (7) in case (i, j) = (1, n) by choosing c n+1 1n (t) as we please.
Suppose thus that (i, j) = (1, n). We have the right-hand side of which has already been defined and does not involve a n1 (t).
Now we can proceed as in the previous (generic) step with k = n − 1, defining c n+1 p1 (t) for p ≥ 2 and then c n+1 1q (t) for n − 1 ≥ q ≥ 2 and finally a n+1,2 (t), ensuring that (7) holds in the remaining cases. The coefficient a n+1,1 (t) is unrestrained and can be chosen arbitrarily (which agrees with our previous observations).
This finishes the construction and the proof.

THE GEOMETRIC CLASSIFICATION OF COMMUTATIVE NILPOTENT ALGEBRAS
In this section, we consider the variety of commutative nilpotent n-dimensional algebras. Our methods will be analogous to those of Section 2. The remaining structure constants c k ij 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 ≥ 4, then A/ e n+1 is seen naturally as an element of S n , relative to the ordered basis (e i + e n+1 ) n i=1 . Example 16. Let n = 4 and α ∈ C. Define the commutative algebra A α by the multiplication we see that A α is isomorphic to the algebra C 19 (−α) defined in [8].
Recall that in [8,Thm. 5] it was shown that the family C 19 (α), with α ∈ C, is generic in the variety of 4-dimensional nilpotent commutative algebras. Since A 0 ∈ α∈C * O(A α ), it follows that the family S 4 is also generic in the variety of 4-dimensional commutative nilpotent algebras.
As we will see next, our restriction in Definition 15 that c 4 12 = 0 ensures that the algebras in S 4 have a sufficiently small automorphism group.
Lemma 17. Let A ∈ S n . Then the following hold: The proof is essentially the same as that of Lemma 11, since Lemmas 7 and 8 will still apply to the algebras in S n . We just need to verify the base cases for (a) and (b). Assume thus that n = 4. Then S 4 is described in Example 16; more specifically, it consists of the algebras A α with α = 0 and multiplication given by (10), commutativity and the fact that e 4 ∈ Ann A α .
So indeed Ann A α = e 4 , for every α.

Now, for (b), Lemma 7 guarantees that Aut
Conversely, let ϕ ∈ Aut A α , with matrix φ = (a ij ) 1≤i,j≤4 relative to the ordered basis (e i ) 4 i=1 . Since Ann A α = e 4 , it follows that ϕ(e 4 ) = ze 4 , for some z ∈ C * . So ϕ induces an automorphism ϕ : A α / e 4 −→ A α / e 4 and, relative to the basis e i = e i + e 4 , i = 1, 2, 3, the nonzero products among basis vectors are just e i 2 = e i+1 , for i = 1, 2. It is easy to see (cf. As α = 0, we can deduce from the above that x = 1. Since we are now working in a variety of commutative algebras, we need to slightly adapt the method described in Subsection 1.1. Specifically, for a commutative n-dimensional algebra A, let Z 2 C (A, C) be the subspace of Z 2 (A, C) consisting of the 2-cocycles θ : . This is a subspace of H 2 (A, C) and A θ is commutative if and only if . We have the following analogue of Lemma 9. Proof. The proof is identical to that of Lemma 9, essentially replacing ∆ ij by ∆ c ij and i = j by i < j. Let q n be the number of independent parameters of the family S n . We have q 4 = 1 and q n+1 = q n + n(n−1) 2 − 1, for all n ≥ 4. Therefore, q n = n(n+1)(n−4) Theorem 21. For any n ≥ 4, the family S n is generic in the variety of all n-dimensional commutative nilpotent algebras. In particular, that variety has dimension n(n−1)(n+4) 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 the inductive step we just need to observe that if c r ij = c r ji and γ r ij = γ r ji for all 1 ≤ i, j, r ≤ n + 1, then (7) holds for the pair (i, j) if and only if it holds for (j, i).

THE GEOMETRIC CLASSIFICATION OF ANTICOMMUTATIVE NILPOTENT ALGEBRAS
In this section, we consider the variety of anticommutative nilpotent n-dimensional algebras. Our methods will be analogous to those of Sections 2 and 3 but there will be some additional technical difficulties coming from larger automorphism groups in lower dimensions.
To shorten the coming statements, we make the following auxiliary definition.
Definition 22. Let n ≥ 3. Denote by T ′ n the family of anticommutative algebras in Nil γ n such that e i e i+1 = e i+2 for all 1 ≤ i ≤ n − 2. The remaining structure constants c k ij are arbitrary, subject only to (1) and the anticommutativity constraint.
Proof. The first statement follows easily by induction, as in the proof of Lemma 11, and the second statement follows just as in the proof of Lemma 7, using the anticommutativity of A.
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 k ij . We will do this over a series of lemmas, providing just the key steps in the proofs. Proof. Let ϕ ∈ Aut A and assume that, relative to (e i ) 4 i=1 , the matrix of ϕ is (a ij ) 1≤i,j≤4 . We know that Ann A = e 4 , so ϕ(e 4 ) = a 44 e 4 , with a 44 = 0. Moreover, A/ e 4 can be seen naturally as an element of T ′ 3 and ϕ induces an automorphism of A/ e 4 with matrix (a ij ) 1≤i,j≤3 relative to the basis (e i + e 4 ) 3 i=1 . The reasoning above then gives a 13 = a 23 = 0 and a 33 = 0. Next we apply ϕ to the identity e 1 e 3 = 0, which follows from c 4 13 = 0 and (1), to get 0 = (a 11 e 1 + a 21 e 2 + a 31 e 3 + a 41 e 4 )(a 33 e 3 + a 43 e 4 ) = a 21 a 33 e 4 .
Thus, a 21 = 0. Below we list, for each identity in A, the corresponding relations we obtain when we apply ϕ, as above. Proof. Let ϕ ∈ Aut A. Using, as before, the fact that Ann A = e 5 and Lemma 24, we conclude that the matrix of ϕ relative to the standard basis is of the form Therefore, as c 6 35 = 0 and x, y = 0, we get y = 1 and a 66 = x 3 . Notice also that the above is consistent with our previously deduced relation a 41 = c 5 13 x(1 − y 2 ), and we also get a 54 = xa 12 c 5 13 . Proceeding as before, we obtain the following additional relations. Proof. The proof follows the ongoing pattern. So, if ϕ ∈ Aut A, then the principal 6 × 6 submatrix of the matrix of ϕ relative to the standard basis of A is of the form given in the statement of Lemma 26 and ϕ(e 7 ) = a 77 e 7 . We proceed as before listing the relations deduced from each of the identities in A.
The results above, along with Example 28, motivate the following auxiliary definition. = 0 (in case n = 6, the latter condition should be replaced with c 5 13 = 0). We can finally prove that the algebras inT n , with n ≥ 7, have the smallest possible automorphism group among n-dimensional nilpotent anticommutative algebras.
Proposition 30. Let n ≥ 7 and suppose that A ∈T n . Then Aut A = {I + a n1 E n1 + a n2 E n2 | a n1 , a n2 ∈ C}, relative to the basis (e i ) n i=1 . Proof. The proof is by induction on n ≥ 7 and the base step has been settled in Lemma 27. So suppose that the result holds for all algebras inT n and let us take A ∈T n+1 , with n ≥ 7, and ϕ ∈ Aut A. Then, since Ann A = e n+1 we conclude that ϕ(e n+1 ) = a n+1,n+1 e n+1 with a n+1,n+1 ∈ C * . In particular, ϕ induces an automorphism of A/ e n+1 . We can see A/ e n+1 as an element ofT n via the basis (e i + e n+1 ) n i=1 and the induction hypothesis implies that ϕ(e 1 ) = e 1 + a n1 e n + a n+1,1 e n+1 , ϕ(e 2 ) = e 2 + a n2 e n + a n+1,2 e n+1 and ϕ(e i ) = e i + a n+1,i e n+1 , for all 3 ≤ i ≤ n.
So it remains to show that a n+1,n+1 = 1 and a n1 = a n2 = a n+1,i = 0, for all 3 ≤ i ≤ n. As before, we have the following table.
IDENTITY RELATION e n−1 e n = e n+1 a n+1,n+1 = 1 The last relation shows that a n+1,i = 0 for all 5 ≤ i ≤ n. Using these we get the following additional relations, which conclude the proof.
We need yet another restriction on the structure constants of the algebras inT 7 to ensure that different parameter choices give different isomorphism classes.
Let d k ij be the structure constants of A relative to this basis. Then, it is straightforward to see that these structure constants satisfy all of the conditions determined byT 7 , along with d 7 15 = 0 = d 7 25 . For example, This proves the existence part of the statement. For the uniqueness, with A ′ as in the statement, suppose also that c 7 46 = 1 and φ : A −→ A ′ is an isomorphism. Then, φ induces an isomorphism φ on the quotient algebras by their respective annihilators. By Example 28, it follows that c 5 13 = d 5 13 and φ is thus an automorphism. We can thence apply Lemma 26 to get the matrix of φ and then lift it to φ. From this point it is a simple matter to use the multiplication tables of A and A ′ and the matrix of φ to deduce that the respective free parameters in A and A ′ are equal.
At last we define the family T n , which will be shown to be generic in the variety of n-dimensional anticommutative algebras.
Definition 33. Let n ≥ 6 (in case n = 6, the condition c 7 46 = 1 is to be ignored). Denote by T n the family of those algebras inT n such that c 7 46 = 1 and c i+2 1i = 0 = c i+2 2i , for all i ≥ 4. In other words, T n is the family of n-dimensional complex anticommutative algebras whose structure constants (c k ij ) i,j,k relative to the basis (e i ) n i=1 satisfy (1) and such that: Proof. In case n = 6, the statement follows from Example 28 and [18,Thm. 2]. So assume that n ≥ 7. 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 ≥ 7, we have dim Aut A = 2, so dim O(A) = n 2 − 2. Moreover, by Lemma 32, different choices of structure constants in T 7 give rise to distinct isomorphism classes. Thus, as seen in the proof of Lemma 32, the isomorphism classes in T 7 form an 8-parameter family and the isomorphism classes in T n are obtained by iterated 1-dimensional central extensions of this family, as shown in Lemma 31.
Finally, we show that the family T n is generic and determine the dimension of the variety of complex n-dimensional anticommutative nilpotent algebras.
Theorem 35. For any n ≥ 6, the family T n is generic in the variety of n-dimensional anticommutative nilpotent algebras. In particular, that variety has dimension (n−2)(n 2 +2n+3) 6 .
Proof. The proof is similar to that of Theorem 14. For arbitrary N ∈ Nil γ n we will prove by induction on n that there is a parametric basis E i (t) = n j=i a ji (t)e j , with a ji (t) ∈ C(t), 1 ≤ i ≤ j ≤ n, and a choice of structure constants c k ij (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 k ij the structure constants 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 7 46 = 1 by the more general condition c 7 46 = 0, which is permitted in view of Lemma 32. We may also redefine a n1 (t) and a n2 (t), as no degeneration from T n depends on these coefficients.
We will proceed in n − 1 steps, from k = 1 to k = n − 1. 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) for all n ≥ j > i ≥ n − k.
By definition, (7) holds for i = n − 1 and j = n. Notice also that a n1 (t), a n2 (t) do not occur in the formula above.
Step k. Let 2 ≤ k < n − 2 and assume that Step k − 1 has been successfully completed and that a n1 (t), a n2 (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 this case. The coefficient of c n+1 ij (t) on the left-hand side of (7) equals a n+1,n+1 (t) −1 a ii (t)a jj (t) which has already been defined and is non-zero. We thus put where the primed sum is over all (p, q) such that (p, q, r) = (i, j, n + 1). Note that on the right-hand side of (11) we must have n − k + 1 < j ≤ r − 1, so r ≥ n − k + 3. Thence, by Step k − 1 and Lemma 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) is a recurrence formula which defines c n+1 ij (t) in terms of c n+1 iq (t) with q > j. So, starting recursively with 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 force the convergence (7) for all j > n − k + 1 and i = n − k. This will work because we are assuming that k < n − 2 so (n − k, q) = (1, n − 1), (2, n − 1). Moreover, also by that assumption on k, the coefficients a n1 (t), a n2 (t) have not been used in (11) to define c n+1 ij (t), as j > i = n − k ≥ 3. Hence, given that c n+1 n−k,n−k+1 (t) = 0 (k > 1), 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 thus that i = n − k and j = n − k + 1. Using Lemma 13 we have c r pq (t)a pi (t)a q,i+1 (t).
Hence, we put 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 (already defined 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 n1 (t), a n2 (t) do not occur in the definition (12) above. This step is thus finished.
Step n − 2. When we reach this step, all c n+1 p,q (t) with q > p ≥ 3 and all a n+1,r (t) with r ≥ 5 have been defined without using the coefficients a n1 (t), a n2 (t) and (7) has been shown to hold for all j > i ≥ 3.
Now we can proceed as in the previous (generic) step with k = n − 2 defining c n+1 2q (t) for n − 2 ≥ q ≥ 3 and finally a n+1,4 (t), ensuring that (7) holds in the remaining cases.
Step n − 1. This step is totally analogous to the previous one. We define c n+1 1n (t), then redefine a n1 (t) and after that find c n+1 1q (t) for n − 2 ≥ q ≥ 2 and finally a n+1,3 (t), ensuring that (7) holds in the remaining cases. The coefficients a n+1,1 (t) and a n+1,2 (t) are unrestrained and can be chosen arbitrarily (which agrees with our previous observations). 5. COROLLARIES, AN OPEN QUESTION AND A CONJECTURE 5.1. 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 nilpotent algebra is not greater than 2 n−1 + 1 but, in general, this upper bound can be attained. On the other hand, for n-dimensional nilpotent Lie algebras, it is known that the upper bound for the nilpotency index is n − 1, while for n-dimensional nilpotent associative, Leibniz and Zinbiel algebras, the upper bound on the nilpotency index is n.
Thanks to our Theorem 35, we know that each n-dimensional nilpotent anticommutative algebra is a 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 algebras from the generic family T n given in Theorem 35 (see Definition 33).

5.2.
Length of algebras. Let A be a finite-dimensional algebra and S be a finite subset of A. We define the length function of S as follows (see [15]). Any product (with any choice of bracketing) of a finite number of elements of S is a word in S, the number of letters (i.e. elements of S) in the product being its length. For i ≥ 1, the set of all words in S having length less than or equal to i is denoted by S i . Then set L i (S) = S i , the linear span of S i , and L(S) = ∞ i=1 L i (S).
Definition 37. Assume that S is a finite generating set for the finite-dimensional algebra A. Then the length of S is defined as l(S) = min{i ≥ 1 | 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 3 was first discussed in [28] in the context of 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 are given by Shitov [29]). The known upper bounds for the length of the matrix algebra of size n are in general 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 ndimensional nilpotent anticommutative algebra with nilpotency index k + 1.
The special case of ({1, . . . , 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 has been 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-ary algebras 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 1 1 · · · p ar r be the prime decomposition of N, where p r denotes the r th prime number. We can define the N-generated Fibonacci number F N (n) recursively as F N (n) = a 1 F N (n − 1) + a 2 F N (n − 2) + · · · + a r F N (n − r), where F N (n) = 1 if n ≤ r.
Remark 42. If the conjecture is true, then the bound is sharp and it gives the sharp bound for the nilpotency index of k-dimensional nilpotent (a, c)-commutative n-ary algebras. In the case of kdimensional (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−1 , e j , . . . , e j ] = e j+1 , a ≤ j ≤ k − 1. For arbitrary (A, C)-commutative n-ary algebras, an extremal example can be obtained by a similar way using a suitable permutation of the indices in the above multiplication.