On speciality of Bernstein Jordan algebras ^{∗}

M. Concepción López-Díaz Departamento de Matemáticas, Universidad de Oviedo, Oviedo, Spain

Ivan P. Shestakov Sobolev Institute of Mathematics, Novosibirsk, Russia

Sergei R. Sverchkov Novosibirsk State University, Novosibirsk, Russia

Dedicated to Professor Holger Petersson on occasion of his 60-th anniversary

Abstract The problem of speciality for Bernstein Jordan algebras is considered. Two

examples of exceptional Bernstein Jordan algebras with special nucleus are constructed. The examples correspond to two different types of exceptional Jordan algebras: the first one is i-exceptional, and the second one is i-special.

1 Introduction

In this work we study two problems on speciality of Bernstein Jordan algebras, which are related. The first more general problem is

Whether any Bernstein Jordan algebra is special? The second one is more specific: Let A′ be a Bernstein Jordan algebra and A be its nucleus. Then, if A is a special

Jordan algebra, would A′ be special as well?

Remind that a Jordan algebra J is called special if it admits an isomorphic embedding into an associative algebra with respect to a symmetrized multiplication a ∘ b = 1/2(ab + ba); otherwise it is called exceptional.

∗The second and third authors were supported by The Ministry of General and Professional Education of Russia (The Fundamental Mathematical Research Grant). The second author acknowledges also the financial support by the FAPESP, Proc. 98/2162-6.

We construct two examples of exceptional Bernstein Jordan algebras. Therefore, the first problem has a negative answer. Moreover, both our examples have special nucleus, so the second problem has a negative answer as well. These two examples correspond to two different types of exceptional Jordan algebras: the first one does not satisfy the Glennie identity which is valid in all special algebras, and so is i-exceptional; the second one is a homomorphic image of a special algebra, and so is i-special.

There are several reasons that motivate our study of these problems. The first one is that all the known examples of Bernstein Jordan algebras were special; so it was natural to ask whether it is always the case? If the answer were positive, the class of Bernstein Jordan algebras would “lie into associative algebras” and so could be studied by means of the well-developed associative theory.

Another reason is that the second problem relates closely to the problem posed by H. Petersson and A. M. Slin’ko [2] about the minimal dimension of exceptional Jordan algebras. A natural way to solve this problem is to try to describe Jordan algebras of low dimension, going up step by step and looking when an exceptional algebra would first appear. In this process, the Jordan algebras of the form J′ = F e + J appear quite often, where e is an idempotent and J is an ideal of J′. Now, if we already know that J is special, we would like to know whether J′ is special. Our second problem is just a special case of this question, with J′ being a Jordan Bernstein algebra and J its nucleus.

Furthermore, in the study of identities of an “idempotent extension” A′ = F e + A, an interesting notion of a so called Peirce derivative arises. Here by Peirce derivatives of an identity f in a given variety of algebras M we mean the “Peirce graded” identities obtained from f by replacing one or more of its arguments by an idempotent e and the others by arbitrary “Peirce homogeneous” elements from Peirce subspaces, according to the Peirce decomposition for algebras in M. Since Peirce subspaces are invariant under multiplication by e, Peirce derivatives, as a rule, have smaller degrees than the original identity. For example, the nontrivial Peirce derivatives of the commutativity xy − yx = 0 in the class of associative algebras are

x_{10} = 0, x_{01} = 0;

and those of the associativity (xy)z − x(yz) = 0 in the class of alternative algebras are

x_{10}y_{10} = 0, x_{01}y_{01} = 0.

Now, assume we have an algebra A′ ∈ M which is an idempotent extension of a subalgebra A; then A′ satisfies an identity f if and only if A satisfies both the identity f and all the Peirce derivatives in M of all partial linearizations of f, with respect to the Peirce decomposition induced by e. So, if we are looking for a certain “minimal” example of an algebra in M that does not satisfy an identity f, we could try an idempotent extension A′ = F e + A, where A satisfies f but does not satisfy some of its Peirce derivatives. For example, if A = F x_{10}, with x_{10}^{2} = 0, then A′ = F e + A is a minimal example of a noncommutative associative algebra; if A = F x_{10} + F y_{10} + F z_{01} with the only nonzero products x_{10}y_{10} = −y_{10}x_{10} = z_{01}, then A′ = F e+A is a minimal example of an alternative nonassociative algebra.

Our purpose was to look for nontrivial Peirce derivatives of the Glennie s-identity G_{8} and then to use them for construction of an i-exceptional idempotent extension of a special Jordan algebra. We did not describe all the Peirce derivatives of G_{8} in general case because of the problems with the expressions containing (J_{\frac{1}{2}} )^{2}e, where we could not cancel e and reduce a degree. So we restrict ourself to the case when J_{1} = 0 (and hence (J_{\frac{1}{2}} )^{2} ⊆ J_{0}). This includes the case of Jordan Bernstein algebras which are of the form J′ = F e + J, with J = J_{1/2} + J_{0}.

In the case of Jordan Bernstein algebras the Glennie identity has a unique nontrivial Peirce derivative of degree 6 described in Proposition 4.1. Using this derivative, we construct our first example of an i-exceptional idempotent extension of a special Jordan algebra.

The next question that then arises is whether a special Jordan algebra could have an exceptional idempotent extension of another type, that is, an i-special extension. Our second example answers this question in positive. In this example, as well as in the prove of speciality of the nucleus in the first example, we use the methods developed in [8] and investigate so called Cohn elements in a free special Jordan algebra. Observe that, up to our knowledge, the dimension 13 for this example is the least dimension for the known examples of exceptional Jordan algebras.

Moreover, the last example lies in the Grassmann envelope of a 7-dimensional Jordan superalgebra which is automatically exceptional. This raises a question about the minimal dimension of exceptional Jordan superalgebras, which we leave open.

Throughout the paper, F denotes a field of characteristic ≠ 2.

2 Bernstein algebras and their nucleus

In this section, we will briefly recall the definitions and some known facts concerning Bernstein algebras and their nucleus (see, for example, [3, 5, 10]).

A Bernstein algebra is a commutative algebra over a field F with a nonzero algebra homomorphism σ : B ⟶ F satisfying the identity

(x^{2})^{2} = σ(x)^{2}x^{2}.

Any Bernstein algebra B has an idempotent e ≠ 0, and with respect to e the algebra B has a Peirce decomposition:

B = F e ⊕ U_{e} ⊕ W_{e},

where U_{e} = \{ x ∈ B | ex = \frac{1}{2}x \}, W_{e} = \{ x ∈ B | ex = 0 \}, and N = U_{e} ⊕ W_{e} = ker σ. The ideal N = U_{e} ⊕ W_{e} is called a nucleus of a Bernstein algebra B. Furthermore, the Peirce components satisfy the relations

U_{e}^{2} ⊆ W_{e}, W_{e}^{2} ⊆ U_{e}, U_{e}W_{e} ⊆ U_{e},
u^{3} = 0, u(uw) = 0, uw^{2} = 0, (u^{2})^{2} = 0, (uw)^{2} = 0,

for any u ∈ U_{e}, w ∈ W_{e}.

The first examples of Bernstein algebras were Bernstein-Jordan algebras, that is, they satisfied also the Jordan identity

(x^{2}y)x = x^{2}(yx).

The Bernstein-Jordan algebras (BJ-algebras for short) play an important role in the theory of Bernstein algebras (see, for example, [5, 9, 10]).

Proposition 2.1 The following conditions for a Bernstein algebra B are equivalent: 1) B is a Jordan algebra.

2) W_{e}^{2} = 0 and uw · w = 0 for all u ∈ U_{e}, w ∈ W_{e} for any idempotent 0 ≠ e ∈ B.

3) The identity x^{3} = σ(x)x^{2} holds in B.

One of the difficulties in the study of Bernstein algebras is caused by the fact that they do not form a variety of algebras. In [4], a definition of a (U, W)-graded Bernstein algebra was introduced, that allows to study instead of Bernstein algebras their graded nucleus, which do form a (graded) variety.

Definition 2.1 Let A = U + W be a commutative algebra over a field F which is a direct sum of vector subspaces U and W such that:

U^{2} ⊆ W, W^{2} ⊆ U and UW ⊆ U

Then, A is called a (U, W)-graded Bernstein algebra or simply a (U, W)-Bernstein algebra if A satisfies the identities:

u^{3} = 0, u(uw) = 0, uw^{2} = 0, (u^{2})^{2} = 0, (uw)^{2} = 0,

for any u ∈ U, w ∈ W.

Observe that a (U, W)-Bernstein algebra A = U + W is graded by the commutative groupoid S = {u, w |u^{2} = w, w^{2} = u, uw = wu = u}, if we set A_{u} = U, A_{w} = W. This justifies the term graded Bernstein algebra.

Proposition 2.2 [4] (i) For any Bernstein algebra B and any its idempotent e ≠ 0 the nucleus N(B) forms a (U_{e}, W_{e})-Bernstein algebra. (ii) Let A = U + W be a commutative algebra over a field F which is a direct sum of subspaces U and W. Consider the vector space direct sum A(e) = F e + A with a multiplication defined by the conditions:

e^{2} = e, eu = \frac{1}{2}u, ew = 0 for any u ∈ U, w ∈ W,

and A being a subalgebra of A(e). Furthermore, let the map σ : A(e) ⟶ F be defined by σ(αe + u + w) = α. Then, A is a (U, W)-Bernstein algebra if and only if B = A(e) is a Bernstein algebra, in which case the nucleus N(B) = A and U_{e}(B) = U, W_{e}(B) = W.

3 Free (U, W)-graded BJ-algebras

It is clear that if A(e) is a Jordan algebra, then so is A. The following example shows that the contrary is not true.

Example: Let A = F u + F w be a (F u, F w)-Bernstein algebra with the multiplication: u^{2} = uw = 0 and w^{2} = u. Then A is a Jordan algebra but A(e) = F e + F u + F w is not Jordan, because W^{2} = (F w)^{2} ≠ 0.

For this reason, we give the next definition.

Definition 3.1 Let A = U + W be a (U, W)-Bernstein algebra. The algebra A is called a (U, W)-graded Bernstein-Jordan algebra (or (U, W)-graded BJ-algebra for short) if A(e) is a Jordan algebra.

It is easy to see that a (U, W)-graded BJ-algebra is just a nucleus of a BJ-algebra with a fixed idempotent and corresponding Peirce decomposition.

Proposition 3.1 Let A = U + W be a (U, W)-Bernstein algebra. The following conditions are equivalent:

1) A is a (U, W)-graded BJ-algebra.

2) W^{2} = 0 and uw · w = 0 for all u ∈ U and w ∈ W.

3) W^{2} = 0 and the identity x^{3} = 0 holds in A.

Proof. The implications 1) ⇒ 2) and 2) ⇒ 3) follow from Propositions 2.2 and 2.1. Let us prove that 3) ⇒ 1). Let A = U + W be a (U, W)-Bernstein algebra such that W^{2} = 0 and x^{3} = 0 holds in A. Consider the Bernstein algebra A(e) = F e + U + W given by Proposition 2.2. We claim that A(e) satisfies a^{3} = σ(a)a^{2} for all a ∈ A(e). In fact, if a is in A(e) then a = αe + u + w for some u ∈ U, w ∈ W, α ∈ F, and we have

a^{3} = (αe + (u + w))^{3} = α^{3}e + α^{2}[(u + w)e^{2} + 2(e(u + w))e]
+α[(u + w)^{2}e + 2(e(u + w))(u + w)] + (u + w)^{3}
= α^{3}e + α^{2}u + α(2uw + u^{2}) = α(αe + u + w)^{2} = σ(a)a^{2}.

Now, by Proposition 2.2, A(e) is a Jordan algebra and A is a (U, W)-graded BJ-algebra. •

Corollary 3.1 A Jordan algebra J is a (U, W)-graded BJ-algebra if and only if J has a Z_{2}-grading J = J_{0} + J_{1} (with U = J_{1} and W = J_{0}) such that J_{0}^{2} = 0 and the identity x^{3} = 0 holds in J.

It follows from Proposition 3.1 that (U, W)-graded BJ-algebras form a variety of graded algebras; thus, for any graded set X = X_{u} ∪ X_{w} of generators there exists a free (U, W)-graded BJ-algebra BJ[X_{u}, X_{w}]. Using the previous corollary, one get easily the following construction for these free algebras.

Proposition 3.2 Let N = ComNil_{3}[X_{u} ∪ X_{w}] be a free algebra in the variety of commutative algebras defined by the identity x^{3} = 0, on the set of generators X = X_{u} ∪ X_{w}, with a Z_{2}-grading induced by the following Z_{2}-grading of the generators: X¯0 = X_{w}, X¯1 = X_{u}. Let, furthermore, I be a (graded) ideal of N generated by N 2 ¯0. Then the quotient algebra \overline{N} = N/I is a (U, W)-graded BJ-algebra (with U = \overline{N} ¯1, W = \overline{N} ¯0), isomorphic to BJ[X_{u}, X_{w}].

Proof. Notice first that any commutative algebra A that satisfies the identity x^{3} = 0 is Jordan. In fact, in this case for any a, b ∈ A we have

a^{2}b + 2(ba)a = 0;

hence

a^{2}(ba) = −2(ba · a)a = (a^{2}b)a.

Now, the algebra \overline{N} = \overline{N} ¯0 + \overline{N} ¯1 is a Jordan Z_{2}-graded algebra with (\overline{N} ¯0)^{2} = 0. By Corollary 3.1 it is a (U, W)-graded BJ-algebra. Without loss of generality, we may also assume that \overline{N} is generated by the set X = X_{u} ∪ X_{w}.

Let B be a (U, W)-graded BJ-algebra and τ : X ⟶ B be a mapping that preserves the grading. Since N is a free commutative nil of index 3 algebra on the set of generators X = X_{u} ∪ X_{w}, there exists a unique homomorphism of Z_{2}-graded algebras τ̃ : N ⟶ B which extends τ. Furthermore, we have τ̃(N 2 ¯0) ⊆ W^{2} = 0; hence N 2 ¯0 ⊆ ker τ̃ and I ⊆ ker τ̃. Therefore, the homomorphism τ̃ induces a homomorphism of the algebra \overline{N} = N/I to B which extends τ. This proves that \overline{N} is a free algebra. •

Remark that in [3], S.González and C.Martínez give a different construction for nucleus of free Bernstein Jordan algebras.

With Propositions 3.1 and 3.2 at hand, one can easily build a base and a multiplication table for certain free (U, W)-graded BJ-algebras.

Lemma 3.1 The following set forms a base of the free algebra N[2, 2] = BJ[u, u′; w, w′]:

Degree  U  W
1  u, u′  w, w′
2  uw, uw′, u′w, u′w′  u^{2}, u^{′2}, uu′
3  uww′, u′ww′, u^{2}u′, u^{′2}u  uwu′, uw′u′
4  u^{2}u′w, u^{2}u′w′, u^{′2}uw, u^{′2}uw′  uww′u′
5  u^{′2}uww′, u^{2}u′ww′
6  u^{′2}uww′u

Here the associative word abc … d means the left-normed product (… ((ab)c) …)d. For example, u^{′2}uww′ = (((u′u′)u)w)w′.

Proof. Let us first prove that the elements given in the table span the algebra N[2, 2]. Most of the cases follow easily from the conditions U^{2} ⊆ W, UW ⊆ U, W^{2} = 0, and x^{3} = 0 for all x in the algebra. For example, we have

uw′ · w = −uw · w′ − ww′ · u = −uww′,
u′w · u = −uw · u′ − uu′ · w = −uwu′,
uw · u′w′ = −(uw · w′)u′ − (uw · u′)w′ = −uww′u′,
u′w · uw′ = −(u′w · w′)u = (u′w′ · w)u = −u′w′ · uw,
uw · uw′ = −uw′ · uw = 0,
uw · uu′ = −(uu′ · u)w = 1/2u^{2}u′w,
uww′ · u = −(uw · u)w′ − uw · uw′ = 0,
uww′ · w = −(uw · w)w′ = 1/2(uw^{2})w′ = 0,
uww′ · uw′ = −(uww′ · u)w′ ∈ U^{2}W = 0.
uww′ · u′w = −(uww′ · u′)w ∈ U^{2}W = 0,
uww′ · uu′ = −(uw · uu′)w′ = −1/2u^{2}u′ww′,
u^{2}u′ · u′w ∈ W u′ · W u′ = 0,
uww′u′ · u = −uww′ · uu′ = (uw · uu′)w′ = 1/2u^{2}u′ww′,
uww′u′ · u′ = −1/2uww′ · u^{′2} = −1/2u^{′2}uww′,
uww′ · u^{′2}u = −(uww′ · u^{′2})u = −u^{′2}uww′u,
uww′ · uwu′ = −(uww′ · uw)u′ − (uww′ · u′) · uw =
= 1/2((uw)^{2}w′)u′ + (uw · u′w′) · uw = −1/2(uw)^{2} · u′w′ = 0.

The less evident case is the following one:

u^{2}u′ww′ · u′ = (u′ww′ · u^{2})u′ = −u′ww′ · u^{2}u′ = 2u′ww′ · (uu′ · u′)
= −2(u′ww′ · uu′)u′ = u^{′2}uww′ · u = u^{′2}uww′u.

In order to prove that the given elements are linearly independent, we construct the vector space A with the base consisting of these elements and define a multiplication on it, using the expressions for the products obtained in the first phase of the proof. Clearly, the resulted algebra A is commutative and has a natural (U, W)-grading. Moreover, using MATHEMATICA computer program, we check that A satisfies the identities which define (U, W)-graded BJ-algebras. Since A is generated by u, u′ and w, w′, it should be a homomorphic image of the algebra N[2, 2]. But dim N[2, 2] ≤ dim A; hence A is isomorphic to N[2, 2] = BJ[u, u′; w, w]. •

Lemma 3.2 The subalgebra of the N[2, 2] generated by the elements u, w_{1} = w, w_{2} = w′, w_{3} = u^{′2} is isomorphic to the free algebra N[1, 3] = BJ[u; w_{1}, w_{2}, w_{3}] and has the following base:

Degree  U  W
1  u  w_{1}, w_{2}, w_{3}
2  uw_{1}, uw_{2}, uw_{3}  u^{2}
3  uw_{1}w_{2}, uw_{1}w_{3}, uw_{2}w_{3}
4  uw_{1}w_{2}w_{3}
5  uw_{1}w_{2}w_{3}u

Proof. Notice first that the elements given in the table are linearly independent in N[2, 2] and form a base of the subalgebra generated by u, w_{1}, w_{2}, w_{3}. On the other hand, it is easy to see that the free algebra N[1, 3] is spanned by the products of the types given in the table. Therefore, N[1, 3] is isomorphic to the indicated subalgebra. •

4 i-exceptional BJ-algebra with special nucleus

All the known s-identities, that is, nontrivial identities which are satisfied by all special Jordan algebras, follow from the so-called Glennie s-identity:

G_{8}(x, y, z) = 2zU_{xy,zU_{x}U_{y}} − 2zU_{xy,zU_{y}U_{x}} + (xy)U_{z}U_{y}U_{x} − (xy)U_{z}U_{x}U_{y},

where yU_{x,z} = {x, y, z} = xy · z + zy · x − y · xz and U_{x} = U_{x,x}.

The following proposition gives a necessary and sufficient condition for a BJ-algebra to satisfy the Glennie identity.

Proposition 4.1 Any (U, W)-graded BJ-algebra A satisfies the Glennie identity. Moreover, the extended BJ-algebra B = A(e) satisfies the Glennie identity if and only if the nucleus A satisfies the graded identity

[(u^{′2}u · w)\overline{w}]\overline{u} = 0,  (1)

for all u, u′, \overline{u} ∈ U, w, \overline{w} ∈ W.

Proof. To prove the proposition, we consider all the possible substitutions of the element e and homogeneous elements from A into the complete linearization of the Glennie identity. The results from [1] are used essentially in these computations. The element (1) appears only in the cases G_{8}(e, u, w; e, \overline{u}, \overline{w}; u′, u′) and G_{8}(e, u, w; u′, u′, \overline{w}; e, \overline{u}); all the other substitutions are equal to zero. Let us consider one of these cases:

\frac{1}{2}G_{8}(e, u, w; e, \overline{u}, \overline{w}; u′, u′) = 2u′U_{e,u′U_{u,w}U_{\overline{u},\overline{w}}} + 2u′U_{\frac{1}{2} \overline{u},u′U_{u,w}U_{e,\overline{w}}} + 2u′U_{\frac{1}{2} u,u′U_{e,w}U_{\overline{u},\overline{w}}}

+ 2u′U_{u\overline{u},u′U_{e,w}U_{e,\overline{w}}} + 2u′U_{u\overline{w},u′U_{e,w}U_{e,\overline{u}}} + 2u′U_{w\overline{u},u′U_{e,u}U_{e,\overline{w}}} + 2u′U_{w\overline{w},u′U_{e,u}U_{e,\overline{u}}} − 2u′U_{e,u′U_{\overline{u},\overline{w}}U_{u,w}} − 2u′U_{\frac{1}{2} \overline{u},u′U_{e,\overline{w}}U_{e,w}} − 2u′U_{\frac{1}{2} u,u′U_{\overline{u},\overline{w}}U_{e,w}} − 2u′U_{u\overline{u},u′U_{e,\overline{w}}U_{e,w}} − 2u′U_{w\overline{w},u′U_{e,\overline{u}}U_{e,w}} − 2u′U_{w\overline{u},u′U_{e,\overline{w}}U_{e,u}} − 2u′U_{w\overline{w},u′U_{e,\overline{u}}U_{e,u}} + eU_{u′}U_{\overline{u},\overline{w}}U_{u,w}

+ \frac{1}{2}\overline{u}U_{u′}U_{e,\overline{w}}U_{u,w} + \frac{1}{2}uU′_{u}U_{\overline{u},\overline{w}}U_{e,w} + u\overline{w}U_{u′}U_{e,\overline{u}}U_{e,w} + u\overline{u}U_{u′}U_{e,\overline{w}}U_{e,w}

+ w\overline{u}U′_{u}U_{e,\overline{w}}U_{e,u} + w\overline{w}U_{u′}U_{e,\overline{u}}U_{e,u} − eU_{u′}U_{u,w}U_{\overline{u},\overline{w}} − \frac{1}{2}\overline{u}U′_{u}U_{u,w}U_{e,\overline{w}} − \frac{1}{2}uU_{u′}U_{e,w}U_{\overline{u},\overline{w}}

− u\overline{w}U_{u′}U_{e,w}U_{e,\overline{u}} − u\overline{u}U′_{u}U_{e,w}U_{e,\overline{w}} − w\overline{u}U_{u′}U_{e,u}U_{e,\overline{w}} − w\overline{w}U_{u′}U_{e,u}U_{e,\overline{u}}.

By the definition of the operator U_{x,y}, we have

\frac{1}{4}G_{8}(e, u, w; e, \overline{u}, \overline{w}; u′, u′) = \{u, u′, \{\overline{u}, u′w, \overline{w}\}\} + 2\{u\overline{u}, u′, u′w.\overline{w}\}

− \{\overline{u}, u′, \{u, u′w, \overline{w}\}\} − 2\{u\overline{u}, u′, u′\overline{w}.w\} + \{u, \{\overline{u}, u^{′2}, \overline{w}\}, w\}

+ \frac{1}{2}\{u, \overline{w}.\{u′, \overline{u}, u′\}, w\} − \{\overline{u}, \{u, u^{′2}, w\}, \overline{w}\} − \frac{1}{2}\{\overline{u}, w.\{u′, u, u′\}, \overline{w}\}.

Extending the triple products, we get

\frac{1}{8}G_{8}(e, u, w; e, \overline{u}, \overline{w}; u′, u′) = 4[([u′w.\overline{w}]\overline{u})u′]u + 4[u\overline{u}.u′][u′w.\overline{w}] − 4[([u′\overline{w}.w]u)u′]\overline{u}
−4[u\overline{u}.u′][u′\overline{w}.w] + 2[([u^{′2}\overline{u}]\overline{w})w]u − 2[([u^{′2}u]w)\overline{w}]\overline{u}.

Till this moment, the result obtained is valid for every Jordan algebra J = F e+J_{\frac{1}{2}} +J_{0}, with J_{1} = F e. But B is a Bernstein Jordan algebra and A is a (U, W)-graded BJ-algebra. Using these facts and the calculations like those made in the proof of Lemma 3.1, we have finally

G_{8}(e, u, w; e, \overline{u}, \overline{w}; u′, u′) = 16[(u^{′2}u · w)\overline{w}]\overline{u}.

The other cases are analyzed in the same way. •

Using the terminology from the introduction, we can reformulate the Proposition in the following way:

Corollary 4.1 The only nontrivial Peirce derivative of the Glennie identity in the class of Bernstein Jordan algebras is the identity

[(a_{1/2}^{2}b_{1/2})a_{0}]b_{0} · c_{1/2} = 0.

Notice that in [1] it is proved that the left side of this identity lies in the annulator AnnA of a BJ-algebra A.

Our first main result is

Theorem 4.1 The free algebra N[2, 2] is a special Jordan algebra, but its idempotent extension N[2, 2](e) does not satisfy the Glennie identity, and so is an i-exceptional BJalgebra (of dimension 26).

Proof. Consider the nontrivial Peirce derivative of the Glennie identity from Proposition 4.1:

\frac{1}{16}G_{8}(e, u, w; e, u, w′; u′, u′) = [(u^{′2}u · w)w′]u.

Calculating the right part in the algebra N[2, 2], we get the basic element u^{′2}uww′u. This proves that the algebra N[2, 2](e) does not satisfy the Glennie identity.

Now, our objective is to prove that the Jordan algebra A = N[2, 2] is special. Observe that this algebra is homogeneous of type [2,2,1,1] relatively to the generators u, u′, w, w′, which means that it has no nonzero homogeneous components of multidegrees (n_{1}, n_{2}, m_{1}, m_{2}), with n_{i} > 2 or m_{j} > 1 for some i, j ∈ {1, 2}. In particular, A is nilpotent of degree 7. By Glennie’s theorem (see [11]), A verifies all the s-identities and so A = SJ[u, u′, w, w′]/I, where SJ[u, u′, w, w′] is a free special Jordan algebra and I is the ideal of relations of A. Notice that this ideal is generated by homogeneous elements.

We will use the next result in the proof of the theorem.

Cohn’s Criterion. (See [11, ?]). Let J be a special Jordan algebra, S = S(J) be the universal associative enveloping algebra of J, I be an ideal of J, and Î be the ideal of S generated by I. Then, the quotient algebra J/I is special if and only if I = Î ∩ J.

Suppose A is not special. By Cohn’s criterion, applied to J = SJ[u, u′, w, w′] and S = Ass[u, u′, w, w′], a free associative algebra, we have

Î ∩ SJ[u, u′, w, w′] ≠ I,

so there exists a nontrivial Cohn element for the ideal I, that is, an element k ∈ Î ∩ SJ[u, u′, w, w′] such that k ∉ I. Since I is homogeneous, we can assume that k is homogeneous. Futhermore, since A^{7} = 0 and k is not in I, we have deg k ≤ 6. On the other hand, it is known that there are no nontrivial Cohn elements of degree less then 5 (see [8]). So we have two possibilities: deg k = 5 or deg k = 6.

Case 1. deg k = 5.

Since k ∈ Î, we can write

k = ∑_{i} f_{i}(a_{i}, u, u′, w, w′),

where a_{i} = a_{i}(u, u′, w, w′) are certain (homogeneous) generators of I, and f_{i}(x_{1}, …, x_{5}) are associative homogeneous polynomials, with deg_{x_{1}}(f_{i}) = 1 for all i. Moreover, since k ∈ SJ[u, u′, w, w′], we have k = k^{∗} for the “reversal” involution ∗ of the algebra Ass[u, u′, w, w′], which is identical on u, u′, w, w′ (see [11]), and whence

k = \frac{1}{2}(k + k^{∗}) = ∑_{i}\{a_{i}z_{2}....z_{t}\},

where {x_{1}x_{2}...x_{t}} = x_{1}x_{2}...x_{t} + x_{t}...x_{2}x_{1} and z_{i} ∈ {u, u′, w, w′}, for all i ∈ {2, ..., t}. Observe that a_{i} ∈ SJ[u, u′, w, w′]; in particular, a_{i}^{∗} = a_{i}.

Now, if deg a_{i} > 2 for some i, the condition deg k = 5 implies that the corresponding summands of k are of the kind {a_{i}z_{2}....z_{t}}, t ≤ 3. By [11, Theorem 3.3], these elements are Jordan polynomials on a_{i}, z_{2}, z_{3} and hence belong to I. Thus, we can assume that deg a_{i} = 2 for all i. (It is clear that I does not contain elements of degree 1).

Observe that any Jordan polynomial of degree more than 1 on w or w′, or of degree more than 2 on u or u′ is in I. Thus, we can assume, without loss of generality, that deg_{w} k ≤ 1, deg_{w′} k ≤ 1, and deg_{u} k ≤ 2, deg_{u′} k ≤ 2. Under these conditions, the only element a_{i} in k of degree 2 may be w ∘ w′, and we therefore have

k = ∑_{i}{(w ∘ w′)z_{2}...z_{t_{i}}},

with z_{j} ∈ {u, u′}. By [11, Theorem 3.3] again, k = j(w ∘w′, u, u′) for a Jordan polynomial j(x, y, z), and k is in I. This finishes the case deg k = 5.

Case 2. deg k = 6.

Similarly to case 1, we have that 2 ≤ deg a_{i} ≤ 3 for all i. If deg a_{i} = 2, then, as in case 1, a_{i} = w ∘ w′ and the corresponding summand of k lies in I; so we can assume that deg a_{i} = 3 for all i.

Now, the generators of I of degree three are:

u^{2} ∘ w, u^{2} ∘ w′, u^{′2} ∘ w, u^{′2} ∘ w′,  (2)
(u ∘ u′) ∘ w, (u ∘ u′) ∘ w′,  (3)
n_{3}(u, u, w), n_{3}(u, u, w′), n_{3}(u′, u′, w), n_{3}(u′, u′, w′),  (4)
n_{3}(u, u′, w), n_{3}(u, u′, w′),  (5)
n_{3}(u, w, w′), n_{3}(u′, w, , w′),  (6)

where n_{3}(x, y, z) = (x ∘ y) ∘ z + (y ∘ z) ∘ x + (z ∘ x) ∘ y.

If a_{i} = u^{2}∘w, then the correspondent summand in k has the form k_{1} = {(u^{2}∘w)z_{1}z_{2}z_{3}}, with z_{i} ∈ {u′, w′}. Therefore, by [11], k_{1} = j(u^{2} ∘ w, u′, w′) ∈ I. The same argument shows that all the summands of k with a_{i} of type (2),(4), or (6), lie in I. Thus, we have to consider only the following summands in k:

k_{2} = {((u ∘ u′) ∘ w′)wuu′}, k_{3} = {((u ∘ u′) ∘ w)w′uu′},
k_{4} = {n_{3}(u, u′, w)w′uu′}, k_{5} = {n_{3}(u, u′, w′)wuu′}.

Note (see [7]) that {x_{1}x_{2}x_{3}x_{4}} is a skew-symmetric function module SJ[X] and

\{x_{1}x_{2}x_{3}x_{4}\} ≡ _{(mod SJ[X])} \frac{1}{2}[x_{1}, x_{2}] ∘ [x_{3}, x_{4}].  (7)

Therefore, we have

k_{2} = \{((u ∘ u′) ∘ w′)wuu′\} ≡ _{(mod I)} \frac{1}{2}[(u ∘ u′) ∘ w′, w] ∘ [u, u′]
= −\frac{1}{2}[(u ∘ u′) ∘ w, w′] ∘ [u, u′] + \frac{1}{2}[u ∘ u′, w ∘ w′] ∘ [u, u′].

Since w ∘ w′ ∈ I, we have by (7) again,

[u ∘ u′, w ∘ w′] ∘ [u, u′] ≡ _{(mod I)} − [u ∘ u′, u] ∘ [w ∘ w′, u′]
= −\frac{1}{2}[u′, u^{2}] ∘ [w ∘ w′, u′] = j(w ∘ w′, u, u′) ∈ I.

Thus,

k_{2} ≡ _{(mod I)} − \frac{1}{2}[(u ∘ u′) ∘ w, w′] ∘ [u, u′] ≡ _{(mod I)} − k_{3}.

Similarly, k_{4} ≡ _{(mod I)} [n_{3}(u, u′, w), w′] ∘ [u, u′] = [n_{3}(u, u′, w), w′ ∘ [u, u′]] − [n_{3}(u, u′, w), [u, u′]] ∘ w′. Since [[x, y], z] is a Jordan polynomial on x, y, z, the second summand lies in I. Consequently,

k_{4} ≡ _{(mod I)} [n_{3}(u, u′, w), w′ ∘ [u, u′]]
= n_{3}([u, w′ ∘ [u, u′]], u′, w) + n_{3}(u, [u′, w′ ∘ [u, u′]], w) + n_{3}(u, u′, [w, w′ ∘ [u, u′]]).

Observe that

[u, w′ ∘ [u, u′]] = [u, [w′ ∘ u, u′]] − [u, u ∘ [w′, u′]]
= [u, [w′ ∘ u, u′]] − \frac{1}{2}[u^{2}, [w′, u′]] ∈ SJ[u, u′, w′].

Therefore,

n_{3}([u, w′ ∘ [u, u′]], u′, w) ∈ n_{3}(SJ[u, u′, w′], u′, w) ⊆ I,

since A is nil of index 3. Similarly, n_{3}(u, [u′, w′ ∘ [u, u′]], w) ∈ I, and so k_{4} ≡ _{(mod I)} n_{3}(u, u′, [w, w′ ∘ [u, u′]]) = n_{3}(u, u′, [w, w′] ∘ [u, u′]) + n_{3}(u, u′, [w, [u, u′]] ∘ w′). The second summand lies in n_{3}(u, u′, SJ[w, w′, u, u′]) ⊆ I, and so we obtain finally

k_{4} ≡ _{(mod I)} n_{3}(u, u′, [w, w′] ∘ [u, u′]) ≡ _{(mod I)} − k_{5}.

Thus, we can assume that our Cohn element k is of the kind

k = α[(u ∘ u′) ∘ w, w′] ∘ [u, u′] + β[n_{3}(u, u′, w), w′] ∘ [u, u′].

Consider the operators of partial linearization Δ′_{u} and Δ′_{u′} (see [11, chapter 1]). Since k ∈ SJ[u, u′, w, w′], the partial linearization kΔ′_{u}Δ′_{u′} ∈ SJ[u, u′, w, w′] as well. Therefore, α[w, w′] ∘ [u, u′] + 3β[w, w′] ∘ [u, u′] ≡ _{(mod SJ[X])} 2(α + 3β){w w′ u u′} ∈ SJ[u, u′, w, w′].

Since {w w′ u u′} is not a Jordan polynomial (see [11]), we have α + 3β = 0 and α = −3β. Consequently,

k = β[−3(u ∘ u′) ∘ w + n_{3}(u, u′, w), w′] ∘ [u, u′] ∈ SJ[u, u′, w, w′].

Consider the Clifford algebra Cl_{n} = Cl[e_{1}, ..., e_{n}], with e_{i}^{2} = 1, e_{i} ∘ e_{j} = 0, and the Jordan algebra of bilinear form J = F · 1 + F · e_{1} + ⋯ + F · e_{n}, embedded naturally in Cl_{n}. Since k is a Jordan element, all its values in Cl_{n} with arguments from J should lie in J. Now, put u = e_{1} + e_{2}, u′ = e_{2}, w = e_{3}, w′ = e_{4}; then

k = β(−3e_{3} + e_{3}) ∘ 2e_{1}e_{2} = −4βe_{1}e_{2}e_{3}.

Therefore, β = 0 and k = 0, proving the theorem. •

Observe that one doesn’t need proposition 4.1 in all its generality in order to get the i-exceptionality of N[2, 2](e); it suffices just to calculate any one of the elements G_{8}(e, u, w; e, \overline{u}, \overline{w}; u′, u′) or G_{8}(e, u, w; u′, u′, \overline{w}; e, \overline{u}), according to the multiplication table of N[2, 2].

Notice also that using proposition 4.1, one can prove that the algebra N[2, 2] is in fact a (U, W)-graded BJ-algebra A of a minimal possible dimension for which the extended Bernstein Jordan algebra B = A(e) does not satisfy the Glennie identity.

5 Exceptional i-special BJ-algebra with special nucleus

In this section, we will deal mainly with the free algebra N[1, 3]. We first observe that its idempotent extension B = N[1, 3](e) is a special Jordan algebra. In fact, consider the 3 × 3 matrix algebra U = M_{3}(G) over the Grassman algebra G on the generators g_{1}, g_{2}, g_{3}, …; then the elements e = e_{33}, u = e_{13} + e_{32}, w_{i} = g_{i}(e_{11} − e_{22}), i = 1, 2, 3, generate a subalgebra in the symmetrized Jordan algebra M_{3}(G)^{+} isomorphic to B.

Let I = ⟨u^{2}⟩ be the ideal of B generated by u^{2}, and \overline{B} = B/I. It is clear that I = F · u^{2}, so the algebra \overline{N} = N[1, 3]/I has the following base:

Degree  U  W
1  u  w_{1}, w_{2}, w_{3}
2  uw_{1}, uw_{2}, uw_{3}
3  uw_{1}w_{2}, uw_{1}w_{3}, uw_{2}w_{3}
4  uw_{1}w_{2}w_{3}
5  uw_{1}w_{2}w_{3}u

Theorem 5.1 The algebra \overline{B} is an exceptional 13-dimensional i-special BJ-algebra with a special nucleus \overline{N}.

Proof. In fact, \overline{B} is a homomorphic image of a special algebra B, and so is i-special. Assume that it is special, that is, there exists an associative algebra C such that \overline{B} ⊆ C^{+}. It is easy to see that ew_{i} = w_{i}e = 0 in the algebra C. Therefore, we have in C the equality

16((((u ∘ w_{1}) ∘ w_{2}) ∘ w_{3}) ∘ e) ∘ u = (euw_{1}w_{2}w_{3} + w_{3}w_{2}w_{1}ue) ∘ u = 2((e ∘ u)w_{1}w_{2}w_{3} + w_{3}w_{2}w_{1}(u ∘ e)) ∘ u = (uw_{1}w_{2}w_{3} + w_{3}w_{2}w_{1}u) ∘ u

= [u, w_{1}w_{2}w_{3}] ∘ u = \frac{1}{2}[u^{2}, w_{1}w_{2}w_{3}] = 0.

On the other hand, the left part is equal in \overline{B} to 8uw_{1}w_{2}w_{3}u ≠ 0. The contradiction obtained proves that \overline{B} is not special.

Finally, let us check that there are no Cohn elements for the ideal K of relations of the algebra \overline{N} = N[1, 3]/I in the free special Jordan algebra SJ[u, w_{1}, w_{2}, w_{3}]. Since (\overline{N})^{6} = 0, any nontrivial Cohn element k should have degree 5; moreover, since the algebra \overline{N} is homogeneous of type [2,1,1,1], the only possibilities for such an element are k = {u^{2}w_{1}w_{2}w_{3}}, k′ = {(w_{i} ∘ w_{j})w_{k}uu}. The second element evidently lies in K, while the element k is not Jordan; for instance, in the Clifford algebra Cl_{n} we have for w_{i} = e_{i}, i = 1, 2, 3, and u = 1 + e_{4},

{(1 + e_{4})^{2}e_{1}e_{2}e_{3}} = −2e_{1}e_{2}e_{3}e_{4} ∉ J.

Therefore, the algebra \overline{N} is special. •

Remark. Notice that the algebra N lies in the Grassmann envelope G(A) of a 7dimensional BJ-superalgebra A (see [4] for a definition), with the base

Degree U_{0}  U_{1} W_{0} W_{1}
1  a  w
2  aw  a^{2}
3  aww
4  awww
5  awwwa

It is easy to check that the superalgebra A(e) is isomorphic (as an ordinary superalgebra) to the subsuperalgebra generated by the elements e = e_{33}, a = e_{13} + e_{32}, w = (e_{11}−e_{22})u in the Jordan matrix superalgebra (M_{3}[u])^{+}, where M_{3}[u] = M_{3}(F)+M_{3}(F)u, with u an odd central element such that u^{2} = 1 (see, for example, [6]). In particular, the superalgebra A(e) is Jordan and special. Moreover, the subspace F a^{2} forms an ideal in A(e), and it is easy to see that the algebra \overline{B} lies in the Grassmann envelope of the superalgebra \overline{A(e)} = A(e)/F a^{2}. Since \overline{B} is not special, neither is the superalgebra \overline{A(e)}. Thus, we have a 7-dimensional exceptional Jordan superalgebra. Clearly, it is i-special.

We conclude with two open problems.

Problem 1 What is the minimal dimension of exceptional (respectively, i-exceptional) Jordan superalgebra?

Problem 2 Is it true that any Bernstein Jordan algebra has a special nucleus?

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