The free alternative superalgebra on one oddgenerator

Ivan Shestakov∗

Instituto de Matematica e Estatıstica,Universidade de Sao Paulo, Sao Paulo, Brasil

e-mail: shestak@ime.usp.brand

Sobolev Institute of Mathematics, Novosibirsk, Russia

andNatalia Zhukavets†

Faculty of Electrical Engineering,Czech Technical University in Prague, Czech Republic

e-mail: natalia@math.feld.cvut.cz

Received 17.05.2006Revised 13.10.2006

Dedicated to Professor B. I. Plotkin on his 80th anniversary

AbstractA base of the free alternative superalgebra on one odd generator is

constructed. As a corollary, a base of the alternative Grassmann algebrais given. We also find a new element of degree 5 from the radical of thefree alternative algebra of countable rank.

Keywords : alternative algebra, superalgebra, free algebra, radical.Mathematics Subject Classification 2000: 17D05, 17A50, 17A70

∗Partially supported by CNPq (grant 304991/2006-6) and FAPESP (grants 05/60337-2,05/60142-7).†Partially supported by the research plan of the Ministry of Education MSM 6840770010,

Czech Republic.

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1 Introduction

One of the most important and difficult problems in the theory of non-associative algebras is the construction of effective bases of free algebras. Hereby an effective base we mean an explicit base given in terms of a free generatingset and accompanied by a multiplication table for basic elements or by a processof reduction of any element to a linear combination of basic words in a finitenumber of steps. There are not many classes of algebras where such bases areknown: free non-associative, free (anti)commutative and free Lie algebras are themost well-known examples besides polynomials and free associative algebras.

Normally, the construction of a base consists in two steps. First, one has toconstruct a “pre-base”, that is, a set that spans the algebra, and then to prove thatthe elements of the pre-base are linearly independent. The linear independencemay be proved by induction on degree of homogeneous basic elements or bydefining formally a multiplication on the pre-base and then proving that theobtained algebra satisfies the desired axioms. The two methods were effectivelyused by A. I. Shirshov [14, 15, 16] who constructed a series of bases of free Liealgebras and bases of free (anti)commutative algebras.

In [1, problem 1.160], A. I. Shirshov formulated the problem of constructionof a base of the free alternative algebra. In fact, in his solution of the Kuroshproblem for alternative PI-algebras, he constructed and used a pre-base (so calledr2-words, see [19, Theorem 5.5]) of this algebra. This pre-base was sufficient forhis purpose, but it was not a base. Note that, contrary to the classes mentionedabove, the free alternative algebra contains non-trivial nilpotent elements and zerodivisors [19]; moreover, the free alternative ring has elements of finite additiveorder [5]. This makes the problem of the base more complicated, and it seemsnatural to consider first some special cases.

For every variety of algebras V , one can consider the corresponding V-Grass-mann algebra (see [11]), which is isomorphic as a vector space to the subspace ofall skew-symmetric elements of the free V-algebra. Thus, it seems interesting toconstruct a base for this subspace. Due to [9, 17], the problem is reduced to thefree V-superalgebra on one odd generator, which is easer to deal with.

If A is an alternative algebra, then the algebra A− obtained from A by re-placing the product xy with the commutator [x, y] = xy− yx is a Malcev algebra(see [4]). In [10] a base of the free Malcev superalgebra M = Malc[∅;x] gener-ated by an odd element was constructed. It was proved later in [12, 13] (see also[20]) that this superalgebra is isomorphic to the corresponding free special Malcev

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superalgebra, that is, the smallest subspace of the free alternative superalgebraA = Alt[∅;x] which contains x and is closed with respect to the super-commutator[a, b]s = ab− (−1)abba.

In the proof of this result, a pre-base of A was obtained. However, it remainedan open question whether this set forms a base of A.

Here we prove that the pre-base found in [12] is in fact a base of A. Conse-quently, a base of the alternative Grassmann algebra is also obtained.

As corollaries, we find a new element of degree 5 from the radical of the freealternative algebra and give a new proof of the fact that the square of the radicalis not zero.

The paper is organized as follows. In Section 2 we recall some known results onthe structure of A and start to compute products of the pre-base elements. Thecomputations continue in Section 3. In Section 4 we prove that the pre-base isin fact a base of A, by providing exact formulas for an alternative multiplication(motivated by results of Sections 2-3) on the space formally generated by theelements of the pre-base. Section 5 is devoted to applications.

In the following, all the (super)algebras will be over a field F of characteristicdifferent from 2 and 3.

2 Multiplication in the superalgebra A(first properties)

Recall that an algebra A is called alternative if it satisfies the identities

(x, x, y) = 0 (left alternativity),

(x, y, y) = 0 (right alternativity),

where (x, y, z) = (xy)z − x(yz) is the associator of x, y and z.We will consider alternative superalgebras. Recall that, in general, a superal-

gebra means a Z2-graded algebra, that is an algebra A which may be written as adirect sum of subspaces A = A0 +A1 subject to the relation AiAj ⊆ Ai+j (mod 2).The subspaces A0 and A1 are called the even and the odd parts of the superalge-bra A; and so are called the elements from A0 and from A1, respectively. Belowall the elements are assumed to be homogeneous, that is, either even or odd, andfor an element x ∈ Ai, i ∈ {0, 1}, the symbol x = i means its parity.

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A superalgebra A = A0 +A1 is called an alternative superalgebra if it satisfiesthe following super-identities:

(x, y, z) + (−1)xy(y, x, z) = 0 (left super-alternativity),

(x, y, z) + (−1)yz(x, z, y) = 0 (right super-alternativity).

Note that the super-identities are obtained from the linearized identities of leftand right alternativity

(x, y, z) + (y, x, z) = 0,

(x, y, z) + (x, z, y) = 0,

by the so called “superization rule” which says that whenever two odd variablesare transposed a negative sign is introduced.

Denote by [x, y]s = xy− (−1)xyyx the super-commutator of the homogeneouselements x, y, and by x ◦s y = xy + (−1)xyyx their super-Jordan product. Whenat least one of the elements x, y is even, we will often omit the index s and writesimply [x, y] and x ◦ y. Note that sometimes it is useful to express a product ofgraded elements as a sum of super-commutator and super-Jordan product:

2xy = x ◦s y + [x, y]s. (1)

We will use the following identities (and their superizations) valid in alterna-tive algebras (respectively, superalgebras) [19]:

[yx, z] = y[x, z] + [y, z]x+ 3(y, x, z), (2)

[x ◦ y, z] = x ◦ [y, z] + y ◦ [x, z], (3)

[xm, y] =m−1∑i=0

xi[x, y]xm−1−i, (4)

(x, xy, z) = (x, y, z)x, (5)

(x, yx, z) = x(x, y, z), (6)

(x ◦ y, z, t) = x ◦ (y, z, t) + y ◦ (x, z, t), (7)

(xm, y, z) =m−1∑i=0

xi(x, y, z)xm−1−i. (8)

Observe that all these identities are valid in alternative superalgebras if the ele-ment x is even.

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Moreover, note that for any σ ∈ Sym(3)

(xσ(1), xσ(2), xσ(3)) = sign(σ) signodd(σ)(x1, x2, x3),

where signodd(σ) is the sign of the permutation afforded by σ on the odd xi.

Let A = Alt[∅;x] be the free alternative superalgebra on one odd generator x.Define by induction

x[1] = x, x[i+1] = [x[i], x]s, i > 0,

and denotet = x[2], z[k] = [x[k], t], u[k] = x[k] ◦s x[3], k > 1.

The following proposition summarizes some previous results from [12] and [13]on the structure of A.

Proposition 2.1 (i) For any k > 0,

z[4k−1] = z[4k−2] = u[4k−1] = 0, u[2] = 0, u[4k+2] = −tz[4k+1], (9)

(ii) For any i, j > 1,

[x[i], x[j]]s = −(−1)c(j)z[i+j−2], (10)

[z[i], x[j−1]]s = [z[i], z[j]]s = 0,

z[i] ◦s z[j] = x[i+1] ◦s z[j] = 0,

x[i+1] ◦s x[j+1] = −(−1)c(j)(u[i+j−1] − δjtz[i+j−2]), (11)

where c(j) = j(j−1)2

, δj = 1 + (−1)j.

(iii) For any i, j, k > 1,

(x, x, x) = 12x[3], (12)

(x[i], x, x) = 13x[i+2] − 1

6z[i], (13)

(x[i], x[j], x) = 13(−1)c(j+1)z[i+j−1], (14)

(x[i], x[j], x[k]) = 0, (15)

and any associator on x, x[i], z[j] vanishes if at least one entry is z[j].

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(iv) The superalgebra A is spanned by the elements

tmxσ, m+ σ ≥ 1, tm(x[k+2]xσ),

tm((x[4k+ε]x[3])xσ), tm(z[4k+ε]xσ),

where k > 0, m ≥ 0; ε, σ ∈ {0, 1}.

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Observe that the function c(i) satisfies the equality

c(i+ j) = c(i) + c(j) + ij; (16)

in particular, c(i+ 1) = c(i) + i.

Corollary 2.2 For any i, j > 2,

z[i]z[j] = x[i]z[j] = z[j]x[i] = 0, (17)

x[i]x[j] = 12(−1)c(j+1)(u[i+j−3] + δjtz

[i+j−4] − (−1)jz[i+j−2]), (18)

u[i]x[j] = 0. (19)

Proof. Identities (17) and (18) follow easily by (1) from the second part of theproposition. Prove (19) first for j = 3. By (18), we have for i > 2

x[i]x[3] = 12(u[i] + z[i+1]), (20)

hence

u[i]x[3] (20)= (2x[i]x[3] − z[i+1])x[3] (15),(17)

= 2x[i](x[3])2

= x[i][x[3], x[3]]s(10)= x[i]z[4] (17)

= 0.

If j > 3, then similarly

u[i]x[j] = 2x[i](x[3]x[j]) = 2x[i]((−1)jx[j]x[3] + [x[3], x[j]]s)(10),(15)

= 2(−1)j(x[i]x[j])x[3] ± 2x[i]z[j+1] (17),(18)= ±u[i+j−3]x[3] = 0.

2

In the sequence, we will also need the following symmetric version of (20):

x[3]x[i] = (−1)ix[i]x[3] − (−1)i[x[i], x[3]]s = (−1)i 12(u[i] − z[i+1]). (21)

The next corollary follows immediately from (20).

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Corollary 2.3 The superalgebra A is spanned by the elements

tmxσ, m+ σ ≥ 1, tm(x[k+2]xσ),

tm(u[4k+ε]xσ), tm(z[4k+ε]xσ), (22)

where k > 0, m ≥ 0; ε, σ ∈ {0, 1}.

Our objective is to prove that the pre-base (22) is in fact a base of the su-peralgebra A. In this and the next section, we will express products of elementsfrom (22) as linear combinations of pre-base elements. Then we will use theseexpressions to define multiplication on the vector space formally generated by(22) and will prove that the obtained superalgebra is alternative.

Actually, we will not use the fact that (22) is a pre-base of A until Section 4.Moreover, the formulas obtained in Sections 2 and 3 provide an alternative proofof this result. We specify this later in Remark 3.5.

We will need the following lemma.

Lemma 2.4 Let B be an alternative superalgebra generated by a set X, and letN(B) be the nucleus (associative center) of B. If an element a ∈ B0∪B1 satisfiesthe conditions

(a,X,X) = (a,X,XX) = (a,XX,XX) = 0, (23)

then a ∈ N(B). Moreover, if a ∈ N(B) and aX ⊆ N(B), then the ideal idB〈a〉is contained in N(B).

Proof. A standard passage to the Grassmann envelope G(B) easily implies (see,for instance, [11]) that it suffices to prove the lemma only for the case when Bis an alternative algebra. Let a ∈ B satisfies (23) and let u, v be monomials inX of degrees deg(u) = n and deg(v) = m. We prove by induction on n + mthat (a, u, v) = 0. If n > 3 then by [19, Exercise 5.5.1, Proposition 5.3] we haveu =

∑i αiui◦u′i where αi ∈ F and ui, u

′i are nonempty monomials in X of degrees

deg(ui) + deg(u′i) = n, and we have by the induction assumption

(a, u, v) =∑i

αi(a, ui ◦ u′i, v)(7)=∑i

αi(ui ◦ (a, u′i, v) + u′i ◦ (a, ui, v)) = 0.

Thus we may assume that n,m ≤ 3. Furthermore, if u = u1u2 then by linearizedidentity (5) and the induction assumption we have

(a, u1u2, v) = −(a, vu2, u1) + (a, u2, v)u1 + (a, u2, u1)v = (a, u1, vu2).

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Thus we may assume that n = 1,m ≤ 3. Finally, by the previous equalities wehave for x, y, z, t ∈ X

(a, t, (xy)z) = −(a, (xy)z, t) = −(a, xy, tz) = 0,

(a, t, x(yz)) = −(a, t, (yz)x) + (a, t, x ◦ (yz)) = 0,

which proves the first part of the lemma.Assume now that aX ⊆ N(B) and let u be a monomial in X of degree n > 1,

then we have

(au,X,XX) = (u, aX,XX) = 0,

(au,XX,XX) = (au,X, (XX)X) = (u, aX, (XX)X) = 0,

which proves that au ∈ N(B). Note that [a,B] ⊆ N(B) (see [19, Corollary 1 ofLemma 7.3]), hence X(au) ⊆ [X, au] + (au)X ⊆ N(B) + a(uX) ⊆ N(B).

Therefore, idB〈a〉 ⊆ N(B). 2

Proposition 2.5 Let U = idA〈u[k], z[k] | k > 2〉 and D = idA〈x[k] | k > 2〉, then

(i) U ⊆ N(A);

(ii) UD = DU = 0;

(iii) [U, t] = 0.

Proof. (i) For k > 2, by super-(7),

(u[k], x, x) = x[k] ◦s (x[3], x, x) + (−1)kx[3] ◦s (x[k], x, x)(13)= 1

3(x[k] ◦s (x[5] − 1

2z[3]) + (−1)kx[3] ◦s (x[k+2] − 1

2z[k]))

(17)= 1

3(x[k] ◦s x[5] + (−1)kx[3] ◦s x[k+2])

(11)= 1

3(−u[k+2] + u[k+2]) = 0,

(u[k], t, x) = x[k] ◦s (x[3], t, x) + (−1)kx[3] ◦s (x[k], t, x)(14)= −1

3(x[k] ◦s z[4] + (−1)kx[3] ◦s z[k+1]) = 0,

(u[k], t, t) = 0.

Since A is generated by x, Lemma 2.4 implies that u[k] ∈ N(A). It follows fromProposition 2.1 (iii) that z[k] ∈ N(A) as well.

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Furthermore, for any v, w ∈ A, we have by [19, Corollary 2 of Lemma 7.3]

(u[k]x, v, w) = u[k] (x, v, w),

(z[k]x, v, w) = z[k] (x, v, w),

and

u[k](x, x, x)(12)= 1

2u[k]x[3] (19)

= 0,

u[k](x, t, x)(13)= −1

3u[k]x[4] (19)

= 0.

Similarly,

z[k](x, x, x) = z[k](x, t, x) = 0.

As before, this proves that

u[k] x, z[k] x ∈ N(A),

and Lemma 2.4 finishes the proof of (i).

(ii) Observe that D coincides with the ideal of A generated by all the associ-ators (u, v, w) where u, v, w ∈ A. Now (ii) follows from [19, Proposition 8.10].

(iii) Since z[i], u[j] ∈ N(A), an arbitrary element from U may be written as alinear combination of the elements of type az[i]b, au[j]b, where a, b ∈ A and oneor both elements a, b may not appear. It is easy to see that [t,A] ⊆ D (since thequotient superalgebra A/D is commutative). Now by (2) and (i), (ii) we have

[az[i]b, t] = a[z[i], t]b = 0,

and

[au[j]b, t] = a[u[j], t]b = a[x[j] ◦s x[3], t]b(3)= a(x[j] ◦s [x[3], t] + [x[j], t] ◦s x[3])b

= az[j] ◦s x[3]b ∈ UD +DU = 0.

Thus [U, t] = 0. This proves the proposition. 2

Since now on, we will write the products

tmu[k]xσ, tmz[k]xσ,

without parenthesis.

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Corollary 2.6 The elements tmu[k]xσ, tmz[k]xσ annihilate all elements of pre-base (22) except the elements of first type.

Proof. It suffices to note that for i > 2 we have x[i], u[i], z[i] ∈ D. 2

Lemma 2.7 Let v = x[k]xε, w = x[m]xτ , where k,m > 2, ε, τ ∈ {0, 1}. Then

(v, w, t) = [v, t]w = 0.

Proof. We prove the lemma by induction on ε + τ . The base of induction forε = τ = 0 follows from (15) and (17). Assume that τ = 1, then by induction

2(v, x[m]x, t)(1)= (v, [x[m], x]s, t) + (v, x[m] ◦s x, t)= (v, x[m+1], t) + (v, x[m] ◦s x, t)(7)= (v, x[m], t) ◦s x± x[m] ◦s (v, x, t) = ±x[m] ◦s (v, x, t).

If ε = 0 then we are again done by (17). Otherwise, by superized (5), we have

(x[k]x, x, t) = (xx, x[k], t) + (−1)k(x, x[k], t)x+ (x, x, t)x[k]

= 12(t, x[k], t)− 1

3z[k+1]x+ 1

3x[4]x[k] (24)

= −13z[k+1]x+ 1

3x[4]x[k] ∈ U,

and Proposition 2.5 (ii) implies the first equality of the lemma.For the second equality, it suffices to prove that [v, t] ∈ U . It is obvious when

ε = 0, and for ε = 1 we have

[x[k]x, t](2)= x[k][x, t] + [x[k], t]x+ 3(x[k], x, t) ∈ U,

finishing the proof. 2

Proposition 2.8 For any m,n ≥ 0, k > 2, ε, σ ∈ {0, 1}, n+ σ ≥ 1, we have

(tmu[k]xε)(tnxσ) =

{tm+nu[k]xε+σ, ε+ σ < 2,12tm+n+1u[k], ε+ σ = 2,

(tmz[k]xε)(tnxσ) =

{tm+nz[k]xε+σ, ε+ σ < 2,12tm+n+1z[k], ε+ σ = 2,

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(tnx)(tmu[k]xε) = (−1)k+1

{tm+n(u[k]x− 2tz[k]), ε = 0,

tm+n+1(12u[k] − 2z[k]x), ε = 1,

(tnx)(tmz[k]xε) = (−1)k

{tm+nz[k]x, ε = 0,12tm+n+1z[k], ε = 1.

Proof. By Proposition 2.5 and Corollary 2.6, we may forget about parenthesesand evaluate all the products as if they were in an associative superalgebra.

The first two equalities of the proposition are implied easily by the condition[U, t] = 0.

Furthermore, since [tm, x] ∈ D, by Proposition 2.5 (ii), we have

[tm, x]u[k] = [tm, x]z[k] = 0.

Finally,

[u[k], x]s = [x[k] ◦s x[3], x]s = x[k] ◦s [x[3], x]s − [x[k], x]s ◦s x[3]

= x[k] ◦s x[4] − x[k+1] ◦s x[3]

= u[k+1] + 2tz[k] − u[k+1] = 2tz[k],

[z[k], x]s = 0,

which implies the last two equalities. 2

3 Multiplication in the superalgebra A(continuation)

Now we will compute products of the pre-base elements of the first two types.For k > 2, we have

xx = 12t,

xx[k] = (−1)kx[k]x+ [x, x[k]]s = (−1)k(x[k]x− x[k+1]),

xt = tx+ [x, t] = tx− x[3],

x[k]t = tx[k] + z[k].

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Moreover, for m > 1

[x, tm] =m−1∑i=0

ti[x, t]tm−1−i = −m−1∑i=0

tix[3]tm−1−i = −mtm−1x[3],

[x[k], tm] =m−1∑i=0

ti[x[k], t]tm−1−i = mtm−1z[k], k > 1,

6(tm, x, x) = 6m−1∑i=0

ti(t, x, x)tm−1−i = 2m−1∑i=0

tix[4]tm−1−i

= 2mtm−1x[4] + 2m−1∑i=0

ti[x[4], tm−1−i]

= 2mtm−1x[4] + 2tm−2(m−1∑i=0

(m− 1− i)) z[4]

= 2mtm−1x[4] +m(m− 1) tm−2z[4].

In the computations below, we will use the obtained expressions for xx, xt,xx[k], x[k]t, [x, tm], [x[k], tm], (tm, x, x) without special references.

Proposition 3.1 For m,n ≥ 0, σ ∈ {0, 1}, n+ σ ≥ 1

tm(tnxσ) = tm+nxσ,

(tmx)tn = tm+nx− ntm+n−1x[3],

(tmx)(tnx) = 12tm+n+1 − ntm+n−1(x[3]x) + m+2n

3tm+n−1x[4]

+m(m+2n−1)6

tm+n−2z[4].

Proof. The first formula is obvious. To verify the next two formulas we will needthe following associators for m,n > 0, j > 2

6(x[j], tm, x)(8)= 6

m−1∑i=0

ti(x[j], t, x)tm−1−i (14)= −2mtm−1z[j+1],

6(tmx, tn, x) = −6(tn, tmx, x)(5)= −6(tn, x, x)tm

= −2ntn−1x[4]tm − n(n− 1) tm+n−2z[4]

= −2ntm+n−1x[4] − 2ntn−1[x[4], tm]− n(n− 1) tm+n−2z[4]

= −2ntm+n−1x[4] − (2mn+ n(n− 1)) tm+n−2z[4].

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Now

(tmx)tn = tm(xtn) = tm(tnx− ntn−1x[3]) = tm+nx− ntm+n−1x[3],

x(tx) = (xt)x− (x, t, x) = (tx− x[3])x+ (t, x, x) = 12t2 − x[3]x+ 2

3x[4],

x(tnx) = (xtn)x− (x, tn, x) = (tnx− ntn−1x[3])x+ (tn, x, x)

= 12tn+1 + 2(tn, x, x)− ntn−1(x[3]x)− n(tn−1, x[3], x)

= 12tn+1 − ntn−1(x[3]x) + 2

3ntn−1x[4],

(tmx)x = tm(xx) + (tm, x, x) = 12tm+1 + m

3tm−1x[4] + m(m−1)

6tm−2z[4],

(tmx)(tnx) = ((tmx)tn)x− (tmx, tn, x) = (tm+nx− ntm+n−1x[3])x− (tmx, tn, x)

= 12tm+n+1 + (tm+n, x, x)− ntm+n−1(x[3]x)− n(tm+n−1, x[3], x)− (tmx, tn, x)

= 12tm+n+1 + m+n

3tm+n−1x[4] + (m+n)(m+n−1)

6tm+n−2z[4] − ntm+n−1(x[3]x)

− n(m+n−1)3

tm+n−2z[4] + 13ntm+n−1x[4] + 2mn+n(n−1)

6tm+n−2z[4]

= 12tm+n+1 − ntm+n−1(x[3]x) + m+2n

3tm+n−1x[4] + m(m+2n−1)

6tm+n−2z[4]. 2

Consider now for k > 2 the products

(x[k]x)x = 12x[k]t+ (x[k], x, x) = 1

2(tx[k] + z[k]) + 1

3x[k+2] − 1

6z[k]

= 12tx[k] + 1

3x[k+2] + 1

3z[k], (25)

x(x[k]x) = (xx[k])x− (x, x[k], x)

= (−1)k(x[k]x− x[k+1])x+ (−1)k(13x[k+2] − 1

6z[k])

= (−1)k(12tx[k] + 2

3x[k+2] − x[k+1]x+ 1

6z[k]). (26)

Proposition 3.2 For m,n ≥ 0, k > 2, ε ∈ {0, 1}

tm (tn(x[k]xε)) = tm+n(x[k]xε),

(tm x)(tnx[k]) = (−1)k(tm+n((x[k]x)− x[k+1])− tm+n−1(n

2u[k] + 2m+n

6z[k+1])

),

(tm x)(tn(x[k]x)) = (−1)k(

12tm+n+1x[k] + tm+n(2

3x[k+2] − x[k+1]x

+ 2m+4n+16

z[k])− tm+n−1(n2u[k]x− m+2n

6u[k+1] +2m+n

6z[k+1]x− m

6z[k+2])

).

13

Proof. The first formula is obvious. For the next two products, in view of theknown formula for xx[k] and (26), we may assume that m+ n > 0. We will needthe products

2x[4]x[k] = x[4] ◦s x[k] + [x[4], x[k]]s = x[k] ◦s x[4] − [x[k], x[4]]s

= u[k+1] + 2tz[k] + z[k+2],

x[3](x[k]x) = (x[3]x[k])x− (x[3], x[k], x)(21)= 1

2(−1)k(u[k]x− z[k+1]x) + (−1)k(x[k], x[3], x)

= (−1)k(12u[k]x− 1

2z[k+1]x+ 1

3z[k+2]),

and the associators

(tm, x, x[k]) =m−1∑i=0

ti(t, x, x[k])tm−1−i = (−1)km−1∑i=0

ti(x[k], t, x)tm−1−i

= m3

(−1)k+1tm−1z[k+1],

(t, x, x[k]x) = (−1)k(x[k]x, x, t)(24)= 1

3(−1)k+1z[k+1]x+ (−1)kx[4]x[k]

= 16(−1)k(u[k+1] + 2tz[k] − 2z[k+1]x+ z[k+2]),

(tm, x, x[k]x) = m6

(−1)ktm−1(u[k+1] + 2tz[k] − 2z[k+1]x+ z[k+2]).

Now for n = 0 we have

(tmx)x[k] = tm(xx[k]) + (tm, x, x[k])

= (−1)k(tm(x[k]x− x[k+1])− m3tm−1z[k+1]),

(tmx)(x[k]x) = tm(x(x[k]x)) + (tm, x, x[k]x)(26)= (−1)ktm(1

2tx[k] + 2

3x[k+2] − x[k+1]x+ 1

6z[k])

+m6

(−1)ktm−1(u[k+1] + 2tz[k] − 2z[k+1]x+ z[k+2])

= (−1)k(12tm+1x[k] + tm(2

3x[k+2] − x[k+1]x+ 2m+1

6z[k])

+m6tm−1(u[k+1] − 2z[k+1]x+ z[k+2])),

14

and for n > 0, by the previous formulas,

(tmx)(tnx[k]) = ((tmx)tn)x[k] − (tmx, tn, x[k])

=(tm+nx− ntm+n−1x[3])x[k] − (x, tn, x[k])tm

=tm+n(xx[k]) + (tm+n, x, x[k])− ntm+n−1(x[3]x[k]) + (tn, x, x[k])tm

=(−1)ktm+n(x[k]x− x[k+1])− (−1)k m+n3tm+n−1z[k+1]

− n2(−1)ktm+n−1(u[k] − z[k+1])− n

3(−1)ktm+n−1z[k+1]

=(−1)k(tm+n(x[k]x− x[k+1])− tm+n−1(2m+n

6z[k+1] + n

2u[k])

),

(tmx)(tn(x[k]x)) = ((tmx)tn)(x[k]x)− (tmx, tn, x[k]x)

=(tm+nx− ntm+n−1x[3])(x[k]x)− (x, tn, x[k]x)tm

=(−1)k(12tm+n+1x[k] + tm+n(2

3x[k+2] − x[k+1]x+ 2m+2n+1

6z[k])

+ m+n6tm+n−1(u[k+1] − 2z[k+1]x+ z[k+2]))− ntm+n−1(x[3](x[k]x))

+ n6(−1)ktn−1(u[k+1] + 2tz[k] − 2z[k+1]x+ z[k+2])tm

=(−1)k(12tm+n+1x[k] + tm+n(2

3x[k+2] − x[k+1]x+ 2m+4n+1

6z[k])

+ m+2n6tm+n−1(u[k+1] − 2z[k+1]x+ z[k+2]))

− ntm+n−1(12(u[k]x− z[k+1]x) + 1

3z[k+2]))

=(−1)k(

12tm+n+1x[k] − tm+n(x[k+1]x− 2

3x[k+2] − 2m+4n+1

6z[k])

− tm+n−1(n2u[k]x− m+2n

6u[k+1] +2m+n

6z[k+1]x− m

6z[k+2])

).

2

Proposition 3.3 For m,n ≥ 0, k > 2

(tnx[k])tm = tm+nx[k] +mtm+n−1z[k],

(tnx[k])(tmx) = tm+n(x[k]x) + tm+n−1(mz[k]x+ 2m+n3z[k+1]),

(tn(x[k]x))tm = tm+n(x[k]x)−mtm+n−1(12u[k] − z[k]x− 1

2z[k+1]),

(tn(x[k]x))(tmx) = 12tm+n+1x[k] + tm+n(1

3x[k+2] + 7m+2n+2

6z[k])

−tm+n−1(m2u[k]x− 2m+n

6u[k+1] + m+2n

6z[k+1]x

−2m+n6z[k+2]).

15

Proof. To verify the first and the third formulas use

(tn(x[k]xε))tm = tm+n(x[k]xε) + tn[x[k]xε, tm],

and

[x[k]x, tm] = x[k][x, tm] + [x[k], tm]x+ 3(x[k], x, tm)

= −mx[k](tm−1x[3]) +m(tm−1z[k])x+mtm−1z[k+1]

= mtm−1(−x[k]x[3] + z[k]x+ z[k+1])

= mtm−1(−12(u[k] + z[k+1]) + z[k]x+ z[k+1])

= 12mtm−1(−u[k] + 2z[k]x+ z[k+1]).

Compute now for m,n > 0 the super-commutator

[tmx,tnx[k]]s(2)= tm[x, tnx[k]]s + (−1)k[tm, tnx[k]]sx+ 3(tm, x, tnx[k])

(2),(5)= tm(tn[x, x[k]]s + [x, tn]sx

[k] − 3(x, tn, x[k]))

+ (−1)k(tn[tm, x[k]]s)x+ 3(tm, x, x[k])tn

=(−1)k+1tm+nx[k+1] − ntm+n−1(x[3]x[k])− (−1)kntm+n−1z[k+1]

− (−1)km(tm+n−1z[k])x− (−1)kmtm+n−1z[k+1]

(21)= (−1)k+1(tm+nx[k+1] + n

2tm+n−1(u[k] − z[k+1]) + ntm+n−1z[k+1]

+mtm+n−1(z[k]x) +mtm+n−1z[k+1])

=(−1)k+1(tm+nx[k+1] + tm+n−1(n

2u[k] + 2m+n

2z[k+1] +mz[k]x)

).

Therefore,

(tnx[k])(tmx) = (−1)k((tmx)(tnx[k])− [tmx, tnx[k]]s)

= tm+n(x[k]x− x[k+1])− tm+n−1(2m+n6z[k+1] + n

2u[k])

+ tm+nx[k+1] + tm+n−1(n2u[k] + 2m+n

2z[k+1] +mz[k]x)

= tm+n(x[k]x) + tm+n−1(mz[k]x+ 2m+n3z[k+1]),

16

(tn(x[k]x))(tmx) = (tn(x[k]x)tm)x− (tn(x[k]x), tm, x)

=(tn+m(x[k]x))x+ (tn[x[k]x, tm])x− (x[k]x, tm, x)tn

=tn+m((x[k]x)x) + (tn+m, x[k]x, x)− (x[k]x, tm, x)tn

+ 12m(tn+m−1(−u[k] + z[k+1] + 2z[k]x))x

(25)= tn+m(1

2tx[k] + 1

3x[k+2] + 1

3z[k])

+ 2m+n6tm+n−1(u[k+1] + 2tz[k] − 2z[k+1]x+ z[k+2])

+ 12mtn+m−1(−u[k]x+ z[k+1]x+ z[k]t)

=12tm+n+1x[k] + tm+n(1

3x[k+2] + 7m+2n+2

6z[k])

− tm+n−1(m2u[k]x− 2m+n

6u[k+1] + m+2n

6z[k+1]x− 2m+n

6z[k+2]).

2

Proposition 3.4 For m,n ≥ 0, i, j > 2

(tmx[i])(tnx[j]) = 12(−1)c(j+1)tm+n(u[i+j−3] + δjtz

[i+j−4] − (−1)jz[i+j−2]),

(tmx[i])(tn(x[j]x)) = 12(−1)c(j+1)tm+n(u[i+j−3]x+ δjt(z

[i+j−4]x)

− (−1)jz[i+j−2]x− 23z[i+j−1]),

(tm(x[i] x))(tnx[j]) = 12(−1)c(j)tm+n(u[i+j−3]x+ (−1)ju[i+j−2]

+ δjt(z[i+j−4]x) + (−1)jδj−1tz

[i+j−3] − (−1)jz[i+j−2]x− 13z[i+j−1]),

(tm(x[i] x))(tn(x[j]x)) = 12(−1)c(j)tm+n(1

2tu[i+j−3] + (−1)ju[i+j−2]x

− 13u[i+j−1] + 1

2δjt

2z[i+j−4] − δj−1t(z[i+j−3]x)− (5

6δj − 1

2)tz[i+j−2]

+ 13z[i+j−1]x− 1

3(−1)jz[i+j]).

Proof. First compute the products for m = n = 0:

x[i] (x[j]x) = (x[i]x[j])x− (x[i], x[j], x)

= 12(−1)c(j+1)(u[i+j−3] + δjtz

[i+j−4] − (−1)jz[i+j−2])x− 13(−1)c(j+1)z[i+j−1]

= 12(−1)c(j+1)(u[i+j−3]x+ δjt(z

[i+j−4]x)− (−1)jz[i+j−2]x− 23z[i+j−1]),

17

(x[i] x)x[j] = x[i](xx[j]) + (x[i], x, x[j])

= (−1)jx[i](x[j]x− x[j+1])− (−1)j(x[i], x[j], x)

= 12(−1)c(j)(u[i+j−3]x+ δjt(z

[i+j−4]x)− (−1)jz[i+j−2]x− 23z[i+j−1])

+ 12(−1)c(j)+j(u[i+j−2] + δj−1tz

[i+j−3] + (−1)jz[i+j−1])− 13(−1)c(j)z[i+j−1]

= 12(−1)c(j)(u[i+j−3]x+ (−1)ju[i+j−2] + δjt(z

[i+j−4]x) + (−1)jδj−1tz[i+j−3]

− (−1)jz[i+j−2]x− 13z[i+j−1]).

Since

2(x[i] x, x[j], x) = (x[i] ◦s x+ x[i+1], x[j], x)

=x[i] ◦s (x, x[j], x) + (−1)ix ◦s (x[i], x[j], x) + 13(−1)c(j+1)z[i+j]

= (−1)j+1x[i] ◦s (13x[j+2] − 1

6z[j]) + 1

3(−1)i+c(j+1)x ◦s z[i+j−1] + 1

3(−1)c(j+1)z[i+j]

= 13(−1)j+1x[i] ◦s x[j+2] + 1

3(−1)j−1+c(j+1)z[i+j−1] ◦s x+ 1

3(−1)c(j+1)z[i+j]

= 13(−1)c(j)(u[i+j−1] + δjtz

[i+j−2] − 2z[i+j−1]x+ (−1)jz[i+j]),

we have

(x[i] x)(x[j]x) = ((x[i]x)x[j])x− (x[i]x, x[j], x)

= 12(−1)c(j)(u[i+j−3]x+ (−1)ju[i+j−2] + δjt(z

[i+j−4]x)

+ (−1)jδj−1tz[i+j−3] − (−1)jz[i+j−2]x− 1

3z[i+j−1])x

− 16(−1)c(j)(u[i+j−1] + δjtz

[i+j−2] − 2z[i+j−1]x+ (−1)jz[i+j])

= 12(−1)c(j)(1

2tu[i+j−3] + (−1)ju[i+j−2]x− 1

3u[i+j−1] + 1

2δjt

2z[i+j−4]

− δj−1t(z[i+j−3]x)− (5

6δj − 1

2)tz[i+j−2] + 1

3z[i+j−1]x− 1

3(−1)jz[i+j]).

Now note that for v = x[i]xε and w = x[j]xσ we have by Lemma 2.7

(tmv)w = tm(vw),

v(tnw) = (vtm)w = (tmv)w − [tm, v]w = tm(vw),

(tmv)(tnw) = tm(v(tnw)) = tm+n(vw).

2

18

Remark 3.5 Propositions 2.8, 3.1 − 3.4, and Corollary 2.2 give an alternativeproof of the fact that the set (22) is a pre-base of A.

Proof. In fact, we have not used this result in our proofs. On the other hand,the refereed results imply that the subspace B of A spanned by elements (22) isclosed under multiplication and is hence a subalgebra of A. Since x generates Aand x ∈ B, we have B = A. 2

4 The pre-base is a base

In order to prove that the set (22) forms a base of the superalgebra A, wewill need the following example. Consider the vector space V over F with thebase

BV = {v1(m1, ε1), vi(mi, ki, εi), i = 2, 3, 4 },

where mi ≥ 0, ki > 1 are integers, k3, k4 ∈ {4n, 4n+ 1}, εi ∈ {0, 1},m1 + ε1 ≥ 1,and define on it a multiplication ∗.

First of all, the elements v3 and v4 annihilate all vj except v1. Their productswith the elements of type v1 are defined by

vi(m, k, ε) ∗ v1(n, σ) = (1− εσ2

)vi(m+ n+ εσ, k, ε⊕ σ),

v1(n, σ) ∗ vi(m, k, ε) = (−1)(k+i)σ(1− εσ2

)vi(m+ n+ εσ, k, ε⊕ σ)

+(−1)kσδi−1v4(m+ n+ 1, k, ε),

where ε⊕ σ = ε+ σ (mod 2), and δi = 1 + (−1)i. Note that δi−1 = 1− (−1)i.Thus we need to define products only for the elements v1, v2. To simplify the

formulas for multiplication, we define the elements v3, v4 for arbitrary values ofthe second argument, by setting

vi(m, 4n+ 3, ε) = 0, i = 3, 4,

v4(m, 4n+ 2, ε) = 0,

v3(m, 4n+ 2, ε) = −v4(m+ 1, 4n+ 1, ε).

19

We define

v1(m, ε) ∗ v1(n, σ) = (1− εσ2

)v1(m+ n+ εσ, ε⊕ σ)− εnv2(m+ n− 1, 3, σ)

+ εσ3

(m+ 2n)v2(m+ n− 1, 4, 0) + εσ6m(m+ 2n− 1)v4(m+ n− 2, 4, 0),

v1(m, ε) ∗ v2(n, k, σ) = (−1)kε((1− εσ2

)v2(m+ n+ εσ, k, ε⊕ σ)

− εv2(m+ n, k + 1, σ) + 2εσ3v2(m+ n, k + 2, 0)

− ε2nv3(m+ n− 1, k, σ) + εσ

6(m+ 2n)v3(m+ n− 1, k + 1, 0)

+ εσ6

(2m+ 4n+ 1)v4(m+ n, k, 0)− ε6(2m+ n)v4(m+ n− 1, k + 1, σ)

+ εσ6mv4(m+ n− 1, k + 2, 0)),

v2(m, k, ε) ∗ v1(n, σ) = (1− εσ2

)v2(m+ n+ εσ, k, ε⊕ σ)

+ εσ3v2(m+ n, k + 2, 0)− ε

2nv3(m+ n− 1, k, σ)

+ εσ6

(m+ 2n)v3(m+ n− 1, k + 1, 0)

+ (n+ εσ6

(2m+ n+ 2))v4(m+ n− 1 + εσ, k, ε⊕ σ)

+ (σ3(m+ 2n) + ε

2n− εσ

3(2m+ 4n))v4(m+ n− 1, k + 1, εσ)

+ εσ6

(m+ 2n)v4(m+ n− 1, k + 2, 0),

v2(m, k, ε) ∗ v2(n, j, σ) = 12(−1)c(j+1)+εj((1− εσ

2)v3(m+ n+ εσ, k + j − 3, ε⊕ σ)

+ (−1)jεv3(m+ n, k + j − 2, σ)− εσ3v3(m+ n, k + j − 1, 0)

+ (1− εσ2

)δjv4(m+ n+ 1 + εσ, k + j − 4, ε⊕ σ)

+ (−1)jεδj−1v4(m+ n+ 1, k + j − 3, σ)

− (−1)j(1− εσ2

+εσδj

3)v4(m+ n+ εσ, k + j − 2, ε⊕ σ)

− 13(2σ + ε− 4εσ)v4(m+ n, k + j − 1, εσ)− (−1)j εσ

3v4(m+ n, k + j, 0)).

Furthermore, define on V a Z2-grading by putting

V0 = vect〈v1(m, 0), vi(m, k, ε) | i+ k + ε = even〉,V1 = vect〈v1(m, 1), vi(m, k, ε) | i+ k + ε = odd〉.

Proposition 4.1 With the products and grading defined above, V is an alterna-tive superalgebra.

Proof. We have to verify that the identities of left and right super-alternativityare valid in V . It follows easily from the multiplication table that non-zeroassociators may appear only for the arguments v1, v1, vi, i ∈ {1, 2, 3, 4}, and forv1, v2, v2.

20

Note that for ε, σ, e ∈ {0, 1} we have ε⊕ σ = ε+ σ − 2εσ and

εσ + (ε⊕ σ)e = ε(σ ⊕ ε) + σe. (27)

Denote α = m+ n+ s. For i = 3, 4 we have

(vi(s, k, e), v1(m, ε), v1(n, σ))

= (1− ε1ε2

)vi(s+m+ eε, k, e⊕ ε) ∗ v1(n, σ)

− vi(s, k, e) ∗ (1− εσ2

)v1(m+ n+ εσ, ε⊕ σ)

= (1− ε1ε2

)(1− (e⊕ε)σ2

)vi(α + eε+ (e⊕ ε)σ, k, e⊕ ε⊕ σ)

− (1− εσ2

)(1− e(ε⊕σ)2

)vi(α + εσ + e(ε⊕ σ), k, e⊕ ε⊕ σ)(27)= 0.

Similarly,

(v1(m, ε), vi(s, k, e), v1(n, σ))

= (−1)(k+i)ε(1− εe2

)vi(m+ s+ εe, k, ε⊕ e) ∗ v1(n, σ)

+ (−1)kδi−1εv4(m+ s+ 1, k, e) ∗ v1(n, σ)

− v1(m, ε) ∗ (1− eσ2

)vi(s+ n+ eσ, k, e⊕ σ)

= (−1)(k+i)ε(1− εe2

)(1− (ε⊕e)σ2

)vi(α + εe+ (ε⊕ e)σ, k, ε⊕ e⊕ σ)

+ (−1)kδi−1ε(1− eσ2

)v4(α + 1 + eσ, k, e⊕ σ)

− (1− eσ2

)(−1)(k+i)ε(1− ε(e⊕σ)2

)vi(α + eσ + ε(e⊕ σ), k, ε⊕ e⊕ σ)

− (1− eσ2

)(−1)kδi−1εv4(α + eσ + 1, k, e⊕ σ)(27)= 0,

(v1(m, ε), v1(n, σ), vi(s, k, e))

= (1− εσ2

)v1(m+ n+ εσ, ε⊕ σ) ∗ vi(s, k, e)− v1(m, ε) ∗

((−1)(k+i)σ(1− eσ

2)vi(s+ n+ eσ, k, e⊕ σ)

+(−1)kδi−1σv4(s+ n+ 1, k, e))

= (1− εσ2

)(

(−1)(k+i)(ε⊕σ)(1− e(ε⊕σ)2

)vi(α + εσ + (ε⊕ σ)e, k, e⊕ ε⊕ σ)

+(−1)kδi−1(ε⊕ σ)v4(α + εσ + 1, k, e))− (−1)(k+i)σ(1− eσ

2)×(

(−1)(k+i)ε(1− (e⊕σ)ε2

)vi(α + eσ + (e⊕ σ)ε, k, e⊕ σ ⊕ ε)

+(−1)kδi−1εv4(α + eσ + 1, k, e⊕ σ))

− (−1)kδi−1σ(−1)kε(1− εe2

)v4(α + εe+ 1, k, ε⊕ e)

21

= (−1)kδi−1

((1− εσ

2)(ε⊕ σ)v4(α + εσ + 1, k, e)

− (−1)(k+i)σ(1− eσ2

)εv4(α + eσ + 1, k, e⊕ σ)

−(−1)kε(1− εe2

)σv4(α + εe+ 1, k, ε⊕ e)).

The last expression is obviously zero for i = 4. Moreover,

(v1(m, 0), v1(n, σ), v3(s, k, e)) =2(−1)k (σv4(α + 1, k, e)− σv4(α + 1, k, e)) = 0,

(v1(m, ε), v1(n, 0), v3(s, k, e)) =2(−1)k (εv4(α + 1, k, e)− εv4(α + 1, k, e)) = 0,

(v1(m, 1), v1(n, 1), v3(s, k, e)) =2(−1)k((−1)k(1− e

2)v4(α + e+ 1, k, e⊕ 1)

−(−1)k(1− e2)v4(α + e+ 1, k, 1⊕ e)

)= 0.

Therefore, we have to consider only the associators on v1, vi, vj, i, j ∈ {1, 2}.Compute first the associators (v1(m, ε), v1(n, σ), v1(s, e)). We have

(v1(m, 1), v1(n, 1), v1(s, 1))

= 12v2(α, 3, 0) + 1

3(m+ n+ s)v2(α− 1, 4, 1)

− 16(mn+ms+ ns)v3(α− 2, 4, 0)

+ 16(m2 + n2 + s2 + 4(mn+ms+ ns)− (m+ n+ s))v4(α− 2, 4, 1)

+ 19(m2 + n2 + s2 + 5

2(mn+ms+ ns)− (m+ n+ s))v4(α− 2, 5, 0),

which yields the super-alternativity identities for ε = σ = e = 1. Furthermore,

(v1(m, 0), v1(n, σ), v1(s, e))

= σe3mv2(α− 1, 4, 0) + σe

6m(m+ 2n+ 2s− 1)v4(α− 2, 4, 0),

(v1(m, ε), v1(n, 0), v1(s, e))

=− εe3nv2(α− 1, 4, 0)− εe

6n(2m+ n+ 2s− 1)v4(α− 2, 4, 0),

(v1(m, ε), v1(n, σ), v1(s, 0))

=− (ε⊕ σ)sv2(α− 1, 3, 0)− εnv2(α− 1, 3, σ)

+ ε(n+ s)v2(α− 1, 3, σ) + σs(−1)εv2(α− 1, 3, ε)

+ εσ3sv2(α− 1, 4, 0) + εσ

6s(2m+ 2n+ s− 1)v4(α− 2, 4, 0).

These associators are zero for ε+ σ + e ≤ 1, and

(v1(m, 1), v1(n, 1), v1(s, 0))

= 13sv2(α− 1, 4, 0) + 1

6s(2m+ 2n+ s− 1)v4(α− 2, 4, 0).

22

Now it is easy to see that the identities of super-alternativity are satisfied whenε+ σ + e = 2, which finishes the case v1, v1, v1.

Consider next v2, v1, v1. The direct calculations give

(v1(n, 0), v1(s, 0), v2(m, k, ε)) = (v1(n, 0), v2(m, k, ε), v1(s, 0))

= (v2(m, k, ε), v1(n, 0), v1(s, 0)) = 0,

(v1(s, 0), v1(n, 1), v2(m, k, ε)) = −(v1(n, 1), v1(s, 0), v2(m, k, ε))

= (v1(n, 1), v2(m, k, ε), v1(s, 0)) = (−1)ks(ε6v3(α− 1, k + 1, 0)

+ ε3v4(α, k, 0)− 1

3v4(α− 1, k + 1, ε) + ε

6v4(α− 1, k + 2, 0)

),

(v1(n, 0), v2(m, k, ε), v1(s, 1)) = −(v2(m, k, ε), v1(n, 0), v1(s, 1))

= (v2(m, k, ε), v1(s, 1), v1(n, 0)) = n(ε6v3(α− 1, k + 1, 0)

+ ε3v4(α, k, 0) + (−1)ε 1

3v4(α− 1, k + 1, ε) + ε

6v4(α− 1, k + 2, 0)

),

(v1(n, 1), v1(s, 1), v2(m, k, 0) = −(−1)k(v1(n, 1), v2(m, k, 0), v1(s, 1))

= (v2(m, k, 0), v1(n, 1), v1(s, 1)) = 13v2(α, k + 2, 0)

+ m6v3(α− 1, k + 1, 0) + 1

3(m− 1

2)v4(α, k, 0)

+ 13(n+ s)v4(α− 1, k + 1, 1) + 1

3(1

2m+ n+ s)v4(α− 1, k + 2, 0),

(v1(n, 1), v1(s, 1), v2(m, k, 1)) = (−1)k(v1(n, 1), v2(m, k, 1), v1(s, 1))

= (v2(m, k, 1), v1(n, 1), v1(s, 1)) = 13v2(α, k + 2, 1)

+ 14v3(α, k, 0) + 1

6(m+ n+ s)v3(α− 1, k + 1, 1)

− 16(n+ s)v3(α− 1, k + 2, 0) + 1

3(m+ n+ s− 1

2)v4(α, k, 1)

− 16(n+ s+ 1

2)v4(α, k + 1, 0) + 1

2(1

3m+ n+ s)v4(α− 1, k + 2, 1)

+ 16(2

3m+ n+ s)v4(α− 1, k + 3, 0),

and the identities of super-alternativity are true in this case.Denote β = k + j and consider v1, v2, v2:

(v2(m, k, ε), v1(s, 0), v2(n, j, σ)) = (v1(s, 0), v2(m, k, ε), v2(n, j, σ))

= (v2(m, k, ε), v2(n, j, σ), v1(s, 0)) = 0,

(v2(m, k, 0), v1(s, 1), v2(n, j, 0)) = −(−1)k(v1(s, 1), v2(m, k, 0), v2(n, j, 0))

= −(−1)j(v2(m, k, 0), v2(n, j, 0), v1(s, 1)) = −13(−1)c(j)v4(α, β − 1, 0),

(v2(m, k, 0), v1(s, 1), v2(n, j, 1)) = −(−1)k(v1(s, 1), v2(m, k, 0), v2(n, j, 1))

= (−1)j(v2(m, k, 0), v2(n, j, 1), v1(s, 1)) = (−1)c(j)(

16v3(α, β − 1, 0)

+16δjv4(α + 1, β − 2, 0)− 1

3v4(α, β − 1, 1)− 1

6(−1)jv4(α, β, 0)

),

23

(v2(m, k, 1), v1(s, 1), v2(n, j, 0)) = (−1)k(v1(s, 1), v2(m, k, 1), v2(n, j, 0))

= −(−1)j(v2(m, k, 1), v2(n, j, 0), v1(s, 1)) = (−1)c(j−1)(

16v3(α, β − 1, 0)

16δjv4(α + 1, β − 2, 0)− 1

3v4(α, β − 1, 1) + 1

6(−1)jv4(α, β, 0)

),

(v2(m, k, 1), v1(s, 1), v2(n, j, 1)) = (−1)k(v1(s, 1), v2(m, k, 1), v2(n, j, 1))

= (−1)j(v2(m, k, 1), v2(n, j, 1), v1(s, 1)) = (−1)c(j)(

16v3(α, β, 0)

+ 16v4(α + 1, β − 1, 0)− 1

3v4(α, β, 1) − 1

18(−1)jv4(α, β + 1, 0)

).

It is easy to see that the equalities

(v2(m, k, ε), v1(s, 1), v2(n, j, σ)) + (−1)k+ε(v1(s, 1), v2(m, k, ε), v2(n, j, σ)) = 0,

(v2(m, k, ε), v1(s, 1), v2(n, j, σ)) + (−1)j+σ(v2(m, k, ε), v2(n, j, σ), v1(s, 1)) = 0

hold. To show that

(v2(m, k, ε), v2(n, j, σ), v1(s, 1))+

(−1)(j+σ)(k+ε)(v2(n, j, σ), v2(m, k, ε), v1(s, 1)) = 0,

write

(v2(m, k, 0), v2(n, j, 0), v1(s, 1)) + (−1)jk(v2(n, j, 0), v2(m, k, 0), v1(s, 1))

= 13

((−1)c(j+1) + (−1)c(k+1)+jk

)v4(α, β − 1, 0), (28)

(v2(m, k, 0), v2(n, j, 1), v1(s, 1)) + (−1)(j+1)k(v2(n, j, 1), v2(m, k, 0), v1(s, 1))

= 16

((−1)c(j)+j + (−1)c(k)+(j+1)k

)v3(α, β − 1, 0)

+ 16

((−1)c(j)δj + (−1)c(k)+(j+1)kδk

)v4(α + 1, β − 2, 0)

− 13

((−1)c(j)+j + (−1)c(k)+(j+1)k

)v4(α, β − 1, 1)

− 16

((−1)c(j) − (−1)c(k)+jk

)v4(α, β, 0), (29)

(v2(m, k, 1), v2(n, j, 1), v1(s, 1)) + (−1)(j+1)(k+1)(v2(n, j, 1), v2(m, k, 1), v1(s, 1))

= 16

((−1)c(j)+j + (−1)c(k)+k+(j+1)(k+1)

)v3(α, β, 0)

+ 16

((−1)c(j)+j + (−1)c(k)+k+(j+1)(k+1)

)v4(α + 1, β − 1, 0)

− 13

((−1)c(j)+j + (−1)c(k)+k+(j+1)(k+1)

)v4(α, β, 1)

− 118

((−1)c(j) + (−1)c(k)+(j+1)(k+1)

)v4(α, β + 1, 0). (30)

24

In view of (16),

(−1)c(k) =

−(−1)c(j)+kj, k + j − 2 = 4n, 4n+ 1;

(−1)c(j)+(k+1)(j+1), k + j − 1 = 4n, 4n+ 1;

(−1)c(j)+kj, k + j = 4n, 4n+ 1;

−(−1)c(j)+(k+1)(j+1), k + j + 1 = 4n, 4n+ 1.

The element v4(α, β − 1, 0) is non-zero only for β − 1 = 4n, 4n + 1, but in thiscase

(−1)c(k+1)+jk = (−1)c(k)+k+jk = (−1)c(j)+(k+1)(j+1)+k+jk

= (−1)c(j)+j+1 = −(−1)c(j),

hence the value of (28) is always zero.Consider (29). If β−1 = 4n, 4n+1 then the coefficients in front of v3(α, β−1, 0)

and v4(α, β − 1, 1) are zero. If β = 4n, 4n + 1 then (−1)c(j) − (−1)c(k)+jk = 0 aswell. For β − 2 = 4n, 4n+ 1 we obtain

(−1)c(j)δj + (−1)c(k)+(j+1)kδk = (−1)c(j)δj − (−1)c(j)+kδk

= (−1)c(j)(δj − δk) = (−1)c(j)((−1)j − (−1)k),

and this is zero when both k and j are even or odd, i.e. β = 4n+ 2. If β = 4n+ 3then v3(α, β − 1, 0) = −v4(α + 1, β − 2, 0) and (−1)c(j)+j + (−1)c(k)+(j+1)k =(−1)c(j)+j − (−1)c(j)+k. Hence the value of (29) is always zero as well.

Using the same algorithm, we prove that the value of (30) is also zero.Finally, the equality

(v1(s, 1), v2(m, k, ε), v2(n, j, σ))+

(−1)(j+σ)(k+ε)(v1(s, 1), v2(n, j, σ), v2(m, k, ε)) = 0

follows easily from the previous one since

(v1(s, 1), v2(m, k, ε), v2(n, j, σ)) = (−1)j+k+ε+σ(v2(m, k, ε), v2(n, j, σ), v1(s, 1)).

2

Now we can prove our main result.

Theorem 4.2 The pre-base (22) is a base of the superalgebra A.

25

Proof. Denote v = v1(0, 1), then it is easy to see that the superalgebra V isgenerated by v. Since V is alternative and v ∈ V1, we may consider the surjectivehomomorphism ϕ : A → V , defined by the condition ϕ(x) = v. We have

ϕ(tmxσ) = v1(m,σ),

ϕ(tm(x[k]xσ)) = v2(m, k, σ),

ϕ(tmu[k]xσ) = v3(m, k, σ),

ϕ(tmz[k]xσ) = v4(m, k, σ).

Since the elements vi are linearly independent, so are their pre-images in (22). 2

5 Applications

Following [10, 11], we apply here the results obtained in the previous sectionsto get some new information on the structure of the free alternative algebra andto construct a base of the alternative Grassmann algebra. For simplicity, allalgebras and superalgebras are considered here over a field F of characteristic 0,though some of the obtained results are valid without this assumption.

Let Alt [T ] = Alt [T ; ∅] be the free alternative algebra on a set of even gen-erators T = {t1, t2, . . . , tn, . . .}. Let, furthermore, f = f(x) be a homogeneousnon-associative polynomial of degree n on one variable x. It may be written in theform f(x) = f(x, x, . . . , x), for a certain multilinear polynomial f(t1, t2, . . . , tn).Define the skew-symmetric polynomial Skew f = Skew f (t1, t2, . . . , tn) as fol-lows:

Skew f (t1, t2, . . . , tn) =∑

σ∈Sym(n)

sign(σ)f(tσ(1), tσ(2), . . . , tσ(n)).

Then Skew : A → Alt [T ] is a linear map which maps isomorphically the ho-mogeneous component A[n] of degree n of A to the subspace Skew(Alt[Tn]) ofmultilinear skew-symmetric elements on Tn = {t1, t2, . . . , tn} of Alt [T ].

Therefore, we have the following

Theorem 5.1 The elements

Skew f(ti1 , ti2 , . . . , tik),

where f = f(x) runs through the set (22), k = deg(f), i1 < i2 < · · · < ik, forma base of the space Skew(Alt[T ]) of skew-symmetric elements of Alt [T ].

26

In view of [12, Corollary 4.4], we have also

Corollary 5.2 Let d(n) denotes the dimension of the subspace Skew(Alt[Tn]).Then d(1) = d(2) = 1, d(3) = 2, and for n > 3

d(n) = 2(n− 3) + 12(1 + (−1)c(n+1)).

It was proved in [11] that the elements Skew z[k](t1, . . . , tk+2), k ∈ {4n, 4n+1},k > 4, are non-zero skew-symmetric central functions in Alt [T ]. It would beinteresting to find all skew-symmetric central and nuclear functions for alternativealgebras. Evidently, they all should be of the type Skew f , where f ∈ Z(A) andf ∈ N(A) for central and nuclear functions, respectively.

Therefore, we have to describe first the nucleus N(A) and the (super)centerZ(A) of the superalgebra A.

Proposition 5.3 The nucleus N(A) and the center Z(A) are given by

N(A) = U = idA〈u[k], z[k] | k > 2〉,Z(A) = vectF 〈tmz[k], tm(2z[k]x− u[k]) | m ≥ 0, k > 2〉.

Proof. The proof follows easily from the multiplication table of A. 2

Now, it seems natural to formulate the two problems:

• Describe the elements n ∈ N(A) for which the corresponding skew-sym-metric function Skew n is a nuclear function, that is, takes values in thenucleus N(Alt[T ]);

• Describe the elements z ∈ Z(A) for which the corresponding skew-symmetricfunction Skew z is a central function in Alt[T ].

Note that not every element in Z(A) produces central or nuclear function.For example, z[4] ∈ Z(A) but Skew z[4] is neither a central nor a nuclear functionin the algebra of octonions O.

Besides the centers, the radical R = Rad(Alt[T ]) of the free alternative algebrais an important and intriguing object. It is known that R coincides with the set ofall nilpotent elements of Alt[T ] and is equal to the intersection of the associatorideal D(Alt[T ]) and the ideal T (O) of identities of the algebra of octonions O[7, 19].

It was proved in [8, 18] that if the characteristic of the ground field is zero orthe set of generators T is finite then R is nilpotent, but no estimate for the index

27

of nilpotence is known. It is only known that R2 6= 0 [21]. Moreover, no exactset of generators of R as a T -ideal (or ideal of identities) is known.

Till now, the smallest degree for known elements from the radical R was 6. Forexample, such is famous “Kleinfeld’s element” k = ([t1, t2]2, t3, t4) or the element[[t1, t2] ◦ (t3, t4, t5), t6].

The proposition below gives some new information on the structure of R.

Proposition 5.4 Denote h = [x, x]s(x, x, x). The element

Skew h =∑

σ∈Sym(5)

sign(σ)[tσ(1), tσ(2)](tσ(3), tσ(4), tσ(5))

lies in the radical Rad(Alt[T ]) and it is non-zero if |T | ≥ 5.

Proof. It is clear that Skew h ∈ D(Alt[T ]). On the other hand,

Skew h = Skew (12tx[3]) = Skew (t ◦ x[3])

= Skew (t ◦ [t, x]) = Skew ([t2, x]),

which implies easily that Skew h is an identity for the algebra O. Thus Skew h ∈Rad(Alt[T ]) and Skew h 6= 0 in Alt [Tn] for n ≥ 5 since tx[3] 6= 0 in A. 2

It is known that Rad(Alt[T3]) = 0 (see [3]), hence Skew h is identically zeroin Alt([T3]). It would be interesting to know whether Skew h is zero in Alt([T4]).

Observe that there are no elements of degree 4 in Rad(Alt[T ]), since thealgebra O does not satisfy any non-trivial identity of degree 4 (see [6]).

Proposition 5.5 (Rad(Alt[Tn]))2 6= 0 for n > 9.

Proof. Consider again the element Skew h ∈ Rad(Alt[T ]). It suffices to showthat Skew h(t1, . . . , t5)Skew h(t6, . . . , t10) 6= 0. In fact, we have h2 = 1

4t2z[4],

which is not zero in A. Therefore, Skew (h2) 6= 0 in Alt[T ]. But

(5!)2Skew (h2)(t1, . . . , t10) =∑σ∈Sym(10)

sign(σ)Skew h(tσ(1), . . . , tσ(5))Skew h(tσ(6), . . . , tσ(10)),

which implies the needed result. 2

Now, recall the definition of the Grassmann algebra in a variety of algebras V .Our definition differs from the one given in [11] but one can easily check that thetwo definitions are equivalent.

28

Let G = G0 + G1 be the (ordinary) Grassmann superalgebra on odd gener-ators e1, e2, . . . , en, . . ., subject to the relations e2

i = 0, eiej = −ejei, with thecanonical Z2-grading: G0 is spanned by the empty word and the products of evenlength, and G1 is spanned by the products of odd length. It is well known thata superalgebra A = A0 + A1 is a V-superalgebra if and only if its Grassmannenvelope G(A) = G0 ⊗ A0 +G1 ⊗ A1 belongs to V .

Consider the free V-superalgebra V [∅;x] on one odd generator x, then itsGrassmann envelope G(V [∅;x]) belongs to V . The subalgebra of G(V [∅;x]) gen-erated by the elements

e1 ⊗ x, e2 ⊗ x, . . . , en ⊗ x, . . . ,

is called the V- Grassmann algebra and is denoted by G(V).The following proposition is evident.

Proposition 5.6 The V-Grassmann algebra G(V) has a base of the form:

eµ ⊗ v, |µ| = deg(v),

where v runs a (monomial) base of the superalgebra V [∅;x], µ = {i1, . . . , im},i1 < i2 < · · · < im, |µ| = m, eµ = ei1ei2 · · · eim ∈ G. 2

A base of the alternative Grassmann algebra G(Alt) is now given by base (22)of the superalgebra A.

Note that the alternative Grassmann algebra was first considered by G. V. Do-rofeev [2] who constructed its base up to degree 6 and proved that the elementSkew(t2, x, x) is non-zero in the free alternative algebra Alt[T6] but is an identityin every 3-generated alternative algebra.

References

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