_{1}

^{*}

In this article, we prove that if B is a simple binary-Lie superalgebra whose even part is isomorphic to sl_{2}(F) and whose odd part is a completely reducible binary-Lie-module over the even part, then B is a Lie superalgebra. We introduce also a binary-Lie module over which is sl_{2}(F) not completely reducible.

All algebras mentioned in this article are algebras over a fixed arbitrary field

A superalgebra is a

fine

Let us remember that for any anti-commutative algebra we define the Jacobian

Since the Jacobian is an 3-linear alternating function, its super-analog

for every

Lie algebras are a particular case of Malcev algebras (see [

An algebra is called binary-Lie if every pair of elements generates a Lie algebra. This class of algebras con- tains properly the class of Malcev algebras and it was characterized by A. T. Gainov (see Id. (2) in [

or equivalently

for every

we have that the identity

In consequence, we say that an anti-commutative superalgebra

A subset

For every superalgebra

As usual we call

Main Theorem. Let

In Section 2, we explain some basics facts about binary-Lie superalgebras and irreducible binary-Lie modules over

According to our main purpose, we must pay attention to the theory of modules over

We call this module, the irreducible Lie module of type

The following result of A. N. Grishkov (see Lemma 3 in [

Let

We conclude that if an irreducible module over

Remark 1. For every

Remark 2. For every

We notice that not every binary Lie module over

Example 1. Let

We observe that

Therefore, it only remains to prove that the split null extension

Note that the vector on the right hand side is written as a column to fit the equation in one line. This function gives the coordinates of the product

In what follows

for some pair

We finish this section with the following lemma.

Lemma 1. Let

Then

Proof. Assume

We have that

We can easily see that

super-ideal of

The aim of this section is to prove that none element of type

On the other hand, if we set

Lemma 2. Let

for every

Remark 3. Observe that, since odd elements commute, the functions

Proof. Using (10), a straightforward computation gives

Therefore, Remark 2 implies that

and

It follows that

Lemma 3. Let

Proof. Let

and

It follows that

we conclude

Lemma 4. Let

Proof. If

It follows that

and

We conclude

Lemma 5. Let

Proof. Call

for every

for every

Since,

Let be

for every

Form here a simple induction proves

for every

Now using (10) we obtain

Now we can prove the following:

Theorem 1. Let

Proof. Let

We have prove that the decomposition of

where

In this section we are going to prove that

Lemma 6. Let

Proof. Since

Without loss of generality, we assume that

Therefore, if

If

for every

for every

Next, set

Since

Lemma 7. Let

position given by (17). Then

Proof. Let be

We claim that the eight identities of

To explain how this implication works we notice that identity (1) of

We introduce now a diagram notation to keep track of the information involved by the other identities. First we write down two matrices in the same diagram as follows: If

Identity | Restriction | ||
---|---|---|---|

(1) | |||

(2) | |||

(3) | |||

(4) | |||

(5) | |||

(6) | |||

(7) | |||

(8) |

In those diagrams we draw an arrow from a coefficient to another one if the nullity of the second one can be obtained from the nullity of the first one. We use a full triangle on the tip of the arrow if the implication works without any restriction and an empty triangle on the tip of the arrow if the implication depend of some restriction explained in the legend of the figure. With this notation, the information from identities (2) and (3) of

1) Neither identity (2) nor identity (3) can be applied. This occurs only when

2) Identity (2) cannot be applied and the

3) Identity (3) cannot be applied and the

Therefore, the only coefficients that might be different from zero are

call them the exceptional

As we see in

the other side of the arrow. Therefore, it is zero (notice that

We have proved that the

Now we have to look at

1) Neither identity (1) nor identity (2) can be applied. This occurs only for

2) Identity (1) cannot be applied, and the

Identity | Restriction | ||
---|---|---|---|

(1) | |||

(2) | |||

(3) | |||

(4) |

3) Identity (2) cannot be applied, and the

To prove that

for every pair

Proof of the main theorem: We need to prove that

Supported by FONDECYT process 2010/11100092.