On Simple Completely Reducible Binary-Lie Superalgebras over sl₂(F)

In this article, we prove that if B is a simple binary-Lie superalgebra whose even part is isomorphic to sl₂(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 sl₂(F) not completely reducible.

• Binary-Lie
• Super-Algebras

1. Introduction

All algebras mentioned in this article are algebras over a fixed arbitrary field of characteristic zero.

A superalgebra is a -graded algebra, i.e., an algebra such that for every

. The elements of are called homogeneous. Given a homogeneous element x we de-

fine if, If is homogeneous we say that is even if, and that is odd if. A superalgebra is said to be anti-commutative if

Let us remember that for any anti-commutative algebra we define the Jacobian by the formula, and a Lie algebra is an anti-commutative alge- bra whose Jacobian is the null function (see [1] or [2] for properties of Lie algebras). The super-analog of the Jacobian, i.e. the analog of the Jacobian for anti-commutative superalgebras is the function defined by

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

for every homogeneous. A superalgebra is a Lie superalgebra if and only if for every, (see [3] for information about Lie superalgebras).

Lie algebras are a particular case of Malcev algebras (see [4] - [7] for definition and properties of Malcev algebras). Analogously Lie superalgebras are a particular case of Malcev superalgebras, (see [8] - [10] for infor- mation about Malcev superalgebras).

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 [11] , Section 2, p. 142). Gainov proved that an anti-commutative algebra was binary Lie if and only if

or equivalently

for every. If we define the function

we have that the identity is the complete linearization of (3), so it is satisfied in every binary- Lie algebra (see also [12] -[18] for information about binary-Lie algebras).

In consequence, we say that an anti-commutative superalgebra is a binary-Lie superalgebra, if it satisfies for every homogeneous in, where is the super-analog of the function, i.e.

A subset of a superalgebra is said to be a super-ideal of, if and only if is an ideal of and where for every. We say that a superalgebra is simple if its unique super- ideals are and the superalgebra itself. Simple Lie superalgebras have been classified by V. G. Kac in [3] and simple Malcev superalgebras have been classified by I. P. Shestakov in [19] .

For every superalgebra, the space is an algebra and is a module over. If is a Lie superalgebra then is a Lie algebra and is a Lie module. The same is true for binary-Lie superalgebras, i.e., for any binary-Lie superalgebra the algebra is a binary-Lie algebra and is a binary-Lie mo- dule over.

As usual we call the Lie algebra consisting in all two by two matrices with coficients in and null trace. This algebra is a simple Lie algebra of dimension three, moreover if is algebraically closed, is the unique 3-dimensional simple Lie algebra over. Our aim is to characterize binary-Lie superalgebras whose even part is isomorphic to and whose odd part is a completely reducible module over the even part. In particular we want to prove the following theorem.

Main Theorem. Let be a simple binary-Lie superalgebra, such that is isomorphic to and is a completely reducible module over. Then is a Lie superalgebra.

In Section 2, we explain some basics facts about binary-Lie superalgebras and irreducible binary-Lie modules over We also give an example of a non-completely reducible binary-Lie module over. In Section 3, we prove that, under the conditions described in the last parragraph the odd part is a Lie module over the even part. Finally in Section 4, we prove the main theorem.

2. Modules and Superalgebras

According to our main purpose, we must pay attention to the theory of modules over. We know that has a basis whose products are given by, ,. If is an irreducible Lie module over of dimension, then has a basis whose products are defined by

We call this module, the irreducible Lie module of type. Besides those modules, there is a non-Lie, binary- Lie module over (in fact it is Malcev) called the irreducible module of type (see Id. (5) in [16] , Section 1, p. 245). This module have a basis with products given by:

The following result of A. N. Grishkov (see Lemma 3 in [16] , Section 1, p. 247) implies that there is no other irreducible -module:

Let be a binary-Lie module over Then has a Lie sub-module such that can be decomposed as the direct sum of -modules of type.

We conclude that if an irreducible module over is not Lie it has to be isomorphic to the irreducible binary-Lie module of type.

Remark 1. For every let be the adjoint operator in, i.e.,. Since and are nilpotent operators, both have 0 as its only eigenvalue. Therefore, if satisfies that or for some, then. The set of eigenvalues of is. Therefore, if satisfies that for some, then.

Remark 2. For every we have that implies for some, also implies for some, and finally implies for some.

We notice that not every binary Lie module over is completely reducible as we can see in the following example:

Example 1. Let be the -module where the products are given by

We observe that is an irreducible Lie module of type 0 and the quotient is an irreducible mo- dule of type. Let be a non trivial sub-module of, then we have that for some scalars and at least one of them is different from zero. We have that and. Thus, if or, then. If, then with. We conclude that in any case therefore is a sub-module of, but is irreducible. This implies that either or. We conclude that is the only irreducible sub-module of which is not completely reducible.

Therefore, it only remains to prove that the split null extension is a binary-Lie algebra. To do that we define by:

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 the ordered basis, where and hence the expression

is checked to be identically 0.

In what follows is a binary-Lie superalgebra whose even part is and whose odd part is a completely reducible module over, Under such conditions we have that

for some pair, where is a Lie irreducible module of type for every, and is a module of type for every.

We finish this section with the following lemma.

Lemma 1. Let be a simple binary-Lie superalgebra with a direct decomposition as in (9).

Then for any.

Proof. Assume for some. Without loss of generality we can suppose that. It follows that, where. We conclude. Define

We have that is a sub-module of over. Next define and note that.

We can easily see that, and. It follows that, whence is a

super-ideal of with and. It is not zero because, and it is not since. We obtain a contradiction with the simplicity of. □

3. Sub-Modules of Type

The aim of this section is to prove that none element of type can be found in the decomposition given by (9). This implies that is a Lie module over. Since is a Lie algebra, is zero for every. So, if we set even and odd in (5), we get

On the other hand, if we set, and in (5) we get

Lemma 2. Let be a binary-Lie superalgebra with a decomposition given by (9). Let be the basis of that satisfies (7). Then, there are three functions such that,

for every, we have, and

.

Remark 3. Observe that, since odd elements commute, the functions and are always symmetric while might fail to be.

Proof. Using (10), a straightforward computation gives

Therefore, Remark 2 implies that for some. In the same way we obtain

,

and

.

It follows that, for some, and, for some. □

Lemma 3. Let be a binary-Lie superalgebra with a decomposition given by (9). Let for some, be a sub-module of type. Then.

Proof. Let be the basis of satisfying (7), while we write, and for simplicity. It suffices to prove that. Using (11), straightforward computations give both

,

and

.

It follows that. Assuming this result, and using (11) again, we obtain

,

we conclude as claimed. □

Lemma 4. Let be a simple binary-Lie superalgebra with decomposition given by (9). Let be and for some pair with, and, two sub-modules of type. Then.

Proof. If the result follows from lemma 3. Fix a pair with. Denote by and

the basis of and respectively satisfying (7). Because of remark 3, it suffices to prove that

. Using (11) and Lemma 3, straightforward computations give us the following results:

It follows that. Using this last result and (11) we get that

and

We conclude as claimed. □

Lemma 5. Let be a superalgebra with a direct decomposition given by (9). Let for some be a Lie sub-module of type and let for some, be a sub-module of type. Then.

Proof. Call, , the elements of the basis of satisfying (6) and, the elements of the basis of satisfying (7). Let. Using (10) we obtain, whence it follows

for every. On the other hand, setting, , , in (10), we obtain

. So (12) is satisfied by every. Computing the left hand side of (12), we find that, for every, identity holds. At this point, a simple in- duction proves that

for every (the case is trivial). Now using (10), we obtain

.

Since, by Lemma 1, we have that. Thus Remark 1 implies that, and (13) implies that for every.

Let be. Using (10) again we conclude

and since it follows that

for every. On the other hand, setting, , , in (10), we obtain . So (14) is satisfied by every. Computing the left side of (14) when, we have

Form here a simple induction proves

for every, where

Now using (10) we obtain. Since we have, whence remark 1 implies that, and therefore (16) implies for every. □

Now we can prove the following:

Theorem 1. Let be a simple binary-Lie superalgebra, such that is and has a decomposition given by (9). Then, i.e there is no sub-module of type in the decomposition. In other words is a Lie module over.

Proof. Let be a space of type in the direct decomposition given by (9). Thanks to Lemma 4 we have for every. As a consequence of Lemma 5 we get for every. Thus, and since, we conclude that is a super-ideal of with and Since and is simple, necessarily.

We have prove that the decomposition of given by (9) reduces to

where is a Lie irreducible sub-module of type, for every. Therefore is a Lie module over. □

4. The Main Theorem

In this section we are going to prove that is a Lie superalgebra. We need two previous lemmas.

Lemma 6. Let and be two sub-modules of type and m respectively in the decomposition given by (17). Then, which clearly implies that.

Proof. Since is a Lie module if and we have that, whence (11) becomes

Without loss of generality, we assume that. Let w be an arbitrary element of. Set, , , and in (18). We have

Therefore, if satisfy then. In particular, if is odd, then for every and. Assume that is an even integer (zero included). If we set, , , and in (18) we get

If are two indices such that and, then

. Thus Equation (18) implies

.

for every and every. On the other hand (19) implies

for every. Observe that implies. Then for every .

Next, set, , , and in (18). It follows that

Since by lemma (1), necessarily. We conclude that for every. □

Lemma 7. Let, be two submodules of type and respectively in the decom-

position given by (17). Then, which clearly implies that.

Proof. Let be and the basis of and respectively satisfying (6). It is enough to prove that every pair of indices satisfies

. For simplicity we use the notation,

and. We call -matrix to the by matrix with entrances in whose entrence is (called the -coefficients). Similarly we define the -matrix and the -matrix. We need to prove that these three matrices are the null matrix. Table 1 shows us some identities obtained using (10) by evaluating in different 4-tuples in.

We claim that the eight identities of Table 1 suffice to prove that the -coefficient and the -coefficient are zero, with the only possible exception of when.

To explain how this implication works we notice that identity (1) of Table 1 implies that the -matrix is zero if is odd and has the form in Figure 1 if is even. We also see that identities (4), (5), (7) and (8) imply that the -matrix is as in Figure 2, note that, in every position where neither (4) nor (5) implies that -coefficient is zero, either (7) or (8) does.

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 and are two matrices of the same size we put both in a double matrix diagram as in the left side of Figure 3.

Table 1

Some identities involving H-coefficients and X-coefficients implied by (10)

		Identity	Restriction
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)

Figure 1

The H-matrix with m − n even

Figure 2

The X-matrix with m and n arbitrary

Figure 3

Left: Example of a 2 × 2 double matrix. Right: Digram of identity (6) in Table 1

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 Table 1 is encoded in Figure 4, and information from identity (6) of Table 1 is encoded in the right side of Figure 3. If a coefficient of the -matrix is in the last column or in the last row it is zero as we see in Figure 2. Otherwise we can see in Figure 4 that we can apply identity (2) or (3) in Table 1 and conclude that it is zero except in the following situations.

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

2) Identity (2) cannot be applied and the -coefficient that is pointed by the empty triangle in the right side of Figure 4 corresponds to a bullet in Figure 1. This occurs only when and.

3) Identity (3) cannot be applied and the -coefficient that is pointed by the empty triangle in the left side of Figure 4 corresponds to a bullet in Figure 1. This occurs only when and.

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

call them the exceptional -coefficients.

As we see in Figure 2, if any of this three coefficients is in the last row or the last column, then it is zero. Otherwise this -coefficient can be put in the right side of Figure 3 with a non-exceptional -coefficient in

the other side of the arrow. Therefore, it is zero (notice that and can be put in the left side of Figure 3 in each tip of the arrow, but if one them is not in the last row or column, then either or is available).

We have proved that the -matrix is the matrix zero. Now we can see that every -coefficient different than can be put either in the left side or in the right side of Figure 4 with an -coefficient on the other side of the arrow. We conclude that, for every.

Now we have to look at Table 2. Information from identities (1) and (2) is encoded in Figure 5. We can see that the -coefficients are 0 except in three cases.

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

2) Identity (1) cannot be applied, and the -coefficient pointed by the triangle in the right side of Figure 5 is. This occurs only for when.

Figure 4

Diagram of identities (2) and (3) in Table 1 with their restrictions

Figure 5

Diagram of identities (1) and (2) in Table 2

Table 2

Identities involving the H-coefficients and A-coefficients implied by (10) and the nullity of the X-matrix

		Identity	Restriction
(1)
(2)
(3)
(4)

3) Identity (2) cannot be applied, and the -coefficient pointed by the triangle in the left side of Figure 5 is. This occurs only for when.

To prove that we set in identity (3) of Table 2. This always can be done since. The second and the third cases occur only if, and in this case, it is enough to prove that just one among, , is zero. In this case the desired result can be proved using identity (3) of Table 2 with, but we are able do that provided. Thus, the only coefficients we have not proved yet to be zero, are when. This is the reason why we included identity (4) in Table 2. At this point we have proved that every -coefficient is zero, every -coefficient is zero and every -coefficient is zero. This means that

for every pair finishing thus the proof of the lemma. □

Proof of the main theorem: We need to prove that for every w, u, v homogeneous. Since is a Lie algebra whenever these three elements are even. Theorem 1 implies that the identity holds whenever two of these elements are even and the other one is odd. So we only have to prove that when at least two of this three elements are odd. Thanks to the simmetries described in identity (1) it suffices to prove that when u and v are odd. Because of the decomposition given by (17) we only have to prove that the identity holds when for some and for some, but this follows from Lemma 7 if w is even, and from Lemma 6 if w is odd. This proves that is a Lie superalgebra. □

Funding

Supported by FONDECYT process 2010/11100092.

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