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ISSN: 1736-4337
Journal of Generalized Lie Theory and Applications
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2-Dimensional Algebras Application to Jordan,G-Associative and Hom-Associative Algebras

Remm E* and Goze M

Universite de Haute Alsace, LMIA, 4 rue des Freres Lumiere, 68093 Mulhouse, France

*Corresponding Author:
Elisabeth Remm
Universite de Haute Alsace
LMIA, 4 rue des Freres Lumiere
68093 Mulhouse, France
Tel:
03 89 33 66 52
Fax: 03 89 33 66 53
E-mail: [email protected]

Received Date: July 18, 2017; Accepted Date: July 26, 2017; Published Date: July 30, 2017

Citation: Remm E, Goze M (2017) 2-Dimensional Algebras Application to Jordan, G-Associative and Hom-Associative Algebras. J Generalized Lie Theory Appl 11: 278. doi: 10.4172/1736-4337.1000278

Copyright: © 2017 Remm E, et al. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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Abstract

We classify, up to isomorphism, the 2-dimensional algebras over a field . We focuse also on the case of characteristic 2, identifying the matrices of GL(2, 2) with the elements of the symmetric group Σ3. The classification is then given by the study of the orbits of this group on a 3-dimensional plane, viewed as a Fano plane. As applications, we establish classifications of Jordan algebras, algebras of Lie type or Hom-Associative algebras.

Keywords

2-Dimensional algebras; Classification; Hom-associative algebras

Introduction

An algebra Equation over a field Equation is Equation-vector space equipped with a product which corresponds to a bilinear map on Equation with values in Equation. For a given dimension, one of the basic problems is the determination up to linear isomorphism of all these algebras. Sub classes of algebras where widely studied. These subclasses where often obtained setting a quadratic relation on μ. Among other examples of such classes are Lie algebras (in this case μ is skewsymmetric and satisfies Jacodi identity), associative algebras, Lie-admissibles algebras, Pre-Lie algebras in particular. In all these examples, classifications where established in a general frame work, that is, with no other hypothesis on these classes and only in very small dimensions. For example for Lie algebras, we know the general classifications up to the dimension 6. In bigger dimension we impose additional algebraic properties if we hope to continue this classification. For example simple Lie algebras are fully classified since the work of Killing and Cartan, in any dimension. Unfortunately it is more and less the only solved case. If we consider complexe nilpotent Lie algebras, the classification is known only up to the dimension 7. It is the same for the associatives algebras. If we are only interested in general algebras, the only known cases are the dimension 2 and 3. It is true that the problem is equivalent to the classification of tensors of type (2,1) on a finite dimensional vector space. We are then facing to a basic multilinear algebra problem which is subject to a lack of informations on the tensors.

Here we reconsider this problem from the beginning, that is in dimension 2. This work is certainly not the first one of the subject. There is for example the work of Petersson. Our approach is not similar. We are not fully interested by the classification up to isomorphism but by the determination of subclasses, minimal in a certain sense, which are invariant up to isomorphism. The motivation comes from the constatation of what happen in greater dimensions for nilpotent Lie algebras for example In this case, the classification is established in dimension 7 but quasi unusable in its present forme. This means that if we have a precise example of nilpotent Lie algebra of this dimension, it is long and fastidious to recognize it in the given list because most of the time it is not adapted to the invariants used to established the classification. Moreover the length of the list can be puzzling. In greater dimensions, the number of isomorphy classes, the need to write invariant parametrized families seems to be an unrealistic goal. Hence the idea to reduce the classification problem to a determination of invariant classes. This is the aim of this work. However we will established the link with Petersson’s work. Our approach is quite basic. In characteristic different from 2, we decompose a tensor μ as a skewsymmetric and symmetric one. Since the skewsymmetric case is elementary, we classify those which are symmetric modulo the automorphism group of the associated skeysymmetric law. In characteristic 2, the problem is equivalent to the determination of the orbits of the Fano plane modulo the symmetric group. Finally, we use these results to describe or find again certain classes of algebras whose a direct approach is rather difficult. In particular, we determine the 2-dimensional Jordan algebras and we find again the results of ref. [1], the G-associative algebras and the Hom-associative algebras.

We have begun the study of the determination of general algebras in ref. [2] which was specially an introduction to a more precise work developed in this paper but with the same idea to describe "minimal" families invariant by isomorphism rather than a precise list for which the use is difficult. Recently, we were acquainted with the work of Pertersson, based on an Kaplansky result which permits to describe all the algebras from some unital algebras and to give isomorphism criteria. We try in this paper to look our description in a Petersson point of view. We note also a recent work, on the same subject of H. Ahmed, U. Bekbaev and I. Rakhimov [3].

Generalities

Let Equation be a field whose characteristic will be precise later. An algebra over a field Equation is a Equation -vector space V with a multiplication given by a bilinear map

μ:V × VV.

We denote by I=(V,μ) a Equation-algebra structure on V with multiplication μ. Throughout this paper we fix the vector space V. Since we are interested by the 2-dimensional case we could assume that Equation. Two Equation-algebras A=(V,μ) and A′=(V,μ′) are isomorphic if there is a linear isomorphism,

f:VV

such as;

f(μ(X,Y))=μ′(f(X), f(Y)),

for all X,YV. The classification of 2-dimensional Equation-algebras is then equivalent to the classification of bilinear maps on Equation with values in V. Let {e1,e2} be a fixed basis of V. A general bilinear map μ has the following expression:

Equation

and it is defined by 8 parameters. Let f be a linear isomorphism of V. In the given basis, its matrix M is non degenerate. If we put:

Equation

Then,

Equation

with Δ=adbc≠0. The isomorphic multiplication.

Equation

Satisfies,

Equation

With,

Equation (1)

These formulae describe an action of the linear group Equation on Equation parameterized by the structure constants (αi, βi), i=1,2,3,4 and the problem of classification consists in describing an element of each orbit.

Algebras Over a Field of Characteristic Different from 2

We assume in this section that Equation. We consider the bilinear map μa and μs given by:

Equation

for all X,YV. The multiplication μa is skew-symmetric and it is a Lie multiplication (any skew-symmetric bilinear application in Equation is a Lie bracket). It is isomorphic to one of the following:

Equation

In fact, if μa is not trivial, thus Equation. If α≠0, we consider the change of basis:

Equation

We have Equation.

If α=0, thus β≠0 and we take:

Equation

This gives Equation. In any case, if μa≠0, then it is isomorphic to μ1a

Case Equation

An automorphism of the Lie algebra Equation is a linear isomorphismEquation such that:

Equation

for every X,YA. The set of automorphisms of this Lie algebra is denoted by Equation.

Lemma 1: We have:

Equation

Proof. In fact, assume that Equation is the matrix of the automorphism f in the given basis Equation. Then,

Equation

and,

Equation

Then,

c=0, a=ad.

But detM=ad≠0 so a=ad implies that d=1 This gives the lemma.

Let μ be a general multiplication of 2-dimensional Equation-algebra such that μa is isomorphic to Equation. It is isomorphic to a the bilinear map (always denoted by μ) whose structural constants are given by:

Equation

The classification, up to isomorphism, of the Lie algebras (V,μ) such that μa is isomorphic to Equation is equivalent to the classification up an isomorphism belonging to Equation of the abelian algebras isomorphic to:

Equation

In this case (1) is reduced to:

Equation (2)

1. Assume that β1≠0.

• Suppose that Equation is algebraically closed and consider the isomorphism Equation. The isomorphic algebra is such thatEquation. We deduce that in this case μs is isomorphic to:

Equation

Then μ is isomorphic to:

Equation

with Equation.

• If Equation is not algebraically closed (for example if Equation is a finite field), let Equation be the multiplicative subgroup of elements a2 with Equation. In this case μ is isomorphic to a Lie bracket belonging to the 4 parameters family:

with Equation andEquation. For example, if Equation, then λ∈{−1,1}.

2. Assume β1=0, β2≠0. In this case (1) is reduced to:

Equation (3)

and taking b=−β4/2β2 and Equation, we see that μs is isomorphic to:

Equation

We obtain the following multiplication, Equation being algebraically closed or not:

Equation

3. Assume now that Equation. In this case (1) is reduced to:

Equation (4)

and taking b=−α2/α1 and Equation, we obtain Equation and Equation. In this case, μ is isomorphic to:

Equation

4. Assume now that Equation. In this case, considering Equation, the Lie bracket μ is isomorphic to:

Equation

5. Assume now that Equation. The Lie bracket μ is isomorphic to:

Equation

6. If Equation, then μ is isomorphic to Equation with β4=2α2

Theorem 2: Any 2-dimensional non commutative algebras isomorphic to one of the following algebras:

• If Equation is algebraically closed:

Equation

with Equation

• If Equation is not algebraically closed:

Equation

Equation

Let us make the link with the results of Petersson [4]. The main idea of this work is to construct algebras from unital algebra. Recall that an algebra A=(V,μ) is called unital if there exists 1∈V such that μ(1, X)=μ(X,1)=X for any XV for any XV.

Lemma 3: If μa is not trivial, then A is not unital.

Proof. Assume that there exists 1 satisfying μ(1, X)=μ(X,1)=X, then:

Equation

for any XV. Then μa(1,X)=0 for any X and 1 is in the center of Aa=(V, μa). But if μa is not trivial, the center of Aa is reduce to {0}. The algebra A cannot be unital.

The algebra A=(V,μ) is called regular if there exists U,TV such that the linear applications:

Equation

are linear isomorphisms. From ref. [5], for any regular algebra A=(V,μ) there exist a unique, up an isomorphism, unital algebra B=(V,μu) and two linear isomorphisms f, g of V such that:

Equation

for any X, YV. The algebra B is called the unital heart of A. To compare Theorem 2 with the Petersson results, we have to determine the regular algebras. Let us consider the first family. The application LU is not regular for any U if and only if its determinant is identically null that is:

Equation

Likewise RT is not regular for any T if and only if its determinant is identically null that is:

Equation

We deduce that any algebra Equation is regular except the algebras given by:

Equation

Let us note that Equation is left-singular but right-regular and Equation is right-singular and left-regular. An algebra which is left and right singular is called bi-singular. We can summarize the results in the following array:

1. Equation regular except Equation and Equation.

2. Equation is left-singular and right-regular,

3. Equation is right-singular and left-regular,

4. Equation is regular,

5. Equation is regular except Equation,

6. Equation is bisingular.

7. Equation is regular exceptEquation

8. Equation is bisingular,

9. Equation is left-singular and right-regular as soon as β4≠0,

10. Equation is left-regular and right-singular as soon as β4≠0,

11. Equation is regular except for α2=0, 1 or −1,

12.Equation is bisingular,

13. Equation is left-singular and right-regular as soon as β4≠0,

14. Equation is left-regular and right-singular as soon as β4≠0,

We deduce.

Proposition 4: We consider the following algebras,

1. Equation withEquation

2. Equation

3. Equation withEquation

4. Equation with α2 ≠ 0,1,−1.

For anyone of these algebras A, there exists an unital Equation algebra Equation and linear endomorphisms fA, gA such that the multiplication of A is given by:

Equation

This unital algebra BA is called the unital heart of A. Since BA is unital, then [5] it is an etale algebra, that is Equation where Equation is the algebraic closure of A, or BA is isomorphic to the dual algebra defined by Equation. To find this heart algebra we use the Kaplansky’s Trick. If A is regular, we consider U and V such that LU and RV are non singular and Equation. The multiplication μu of the heart B is Equation and the identity of B is Equation.

1. Let be Equation. If α2≠1 or −1 then Equation and Equation are not singular. In fact,

Equation

Thus,

Equation

Then the identity element of BA is e2 and,

Equation

and BA is etale. If α2=−1, then we can take U=e2 and T=e1 as soon as Equation. If not we take U=e1+e2 and T=e1. We have the same calcul for α2=1.

2. Let be Equation. This algebra is regular. If α1≠0, then Equation and Equation are not singular and BA is etale.

Case Equation

The multiplicatio μ n is symmetric. The group of automorphisms of μa is Equation. Moreover the multiplication writes:

Equation

We assume that there exists two independent idempotent vectors. If e1 and e2 are these vectors, then:

Equation

We obtain the following algebras:

Equation

Remark that if any element is idempotent, thus Equation. In fact:

Equation

In the general case, if ae1+be2 is an idempotent with ab≠0, then a and b satisfy the system:

Equation

If 4α2β2=1, then the system has solutions as soon as Equation. In this case we obtain the multiplication Equation and for any a, the vectors ae1+(1−a)e2 are idempotent. If Equation, the vector:

Equation

is an idempotent and the only idempotents are e1, e2 and v. The changes of basis {e1,v} or {e2, v} do not simplify the number of independent parameters.

We assume that there exists only one idempotent vector. If e1 is this vector, thus μ(e1, e1)=e1. If we consider a vector v=xe1+ye2 such that μ(v,v)=v, then x and y have to satisfy:

Equation (5)

If we assume that y≠0, the second equation gives as soon as β2≠0, Equation and thus:

Equation (6)

Let us consider a change of basis which preserves e1 that is,

Equation (7)

with d≠0. Since in this new basis we have Equation, we can find b such that Equation. Then we can assume that β4=0.

If moreover α2≠0, taking Equation, we obtainEquation and we have the algebra:

Equation

Equation (6) simplifies as:

Equation (8)

If we assume that Equation is algebraically closed, then this equation has in general two roots. It has no root if β2=0 which is excluded. Then to have only one idempotent, 0 must be the only root which is equivalent to α4=0 and β2=1/2. We obtain the following algebra:

Equation

If Equation is not algebraically closed, then we have no idempotent other than 0 if α4=0 and β2=1/2 and we obtain the previous algebra μ7 or if Equation is irreducible in Equation. We obtain:

Equation

with Equation irreducible in Equation (so α4≠0).

If α2=0 and if Equation is algebraically closed, we consider in the change of basis (7) defined above, b=0 and Equation if α4≠0:

Equation

There exits only one idempotent if and only if β2=1/2. We obtain the following algebra:

Equation

If α2=α4=0, we have only one idempotent if and only if 2β2≠1. We obtain:

Equation

Assume Equation not algebraically closed and α2=0. If the equation d2α4 has a root in Equation, we find μ8. If not, let Equation such that Equation In this case we have only one idempotent if and only if (2β2=1) or Equation. We obtain:

Equation

and,

Equation

Assume now that β2=0. Then (5) implies y2β4=y. If β4=0, then y=0 and we have:

Equation

The change of basis Equation. We obtain:

Equation

if α2≠0. Assume now that α2=0 and α4≠0. If Equation is algebraically close, we obtain:

Equation

No vector is idempotent. If there exists v with μ(v,v)≠0, thus we can consider that μ(e1,e1)=e2 that is,

Equation

1. If α4=0, that is Equation, then the vectorEquation is idempotent as soon as β4≠0. Then the hypothesis implies β4=0. Let be v=xe1+ye2. The equation μ(v,v)=v is equivalent to:

Equation

that is,

Equation

If α2=0, then x=y=0, and no elements are idempotent. We obtain the algebras, corresponding to β2≠0 or β2=0

Equation

Equation

If α2≠0 and Equation then x satisfies the equation:

Equation (9)

If Equation is algebraically closed, such equation admits a non trivial solution. This is not compatible with our hypothesis. Assume that Equation is not algebraically closed. If β2≠0, the change of basis Equation and Equation permits to consider β2=1 and the (9) becomes,

Equation

This equation has a non solution if Equation whereEquation. We obtain the algebras:

Equation

and,

Equation

2. If α4≠0 the vector v=xe1+ye2 is idempotent if and only if:

Equation

Then Equation. Let us note that 1−22≠0 because 1−22=0 implies y2α4=0 that is y=0 and in this case x=0 and v=0 . We deduce that y is a root of the equation:

Equation

that is:

Equation

If Equation is algebraically closed, this equation admits always a solution except if:

Equation

Then Equation. We note that β2=0 implies, if the characteristic of Equation is not 3, α2=α4=0. From hypothesis, we can assume that β2≠0 and the change of basis Equation which preserves the condition e1e1=e2 changes β2 in 2 and we can take β2=3. Then Equation , then α2=−2 and α4=4, β4=8 and we obtain the algebra:

Equation

Let us note that if the characteristic of Equation is 3, then α4β2=0 and β2=0. This gives α2(α2+β4)=0 and Equation. Since α2=0 implies α4=0 and 424=α2+β4=0 we obtain β4=2α2 and Equation. By a change of basis we can take α2=1 and we obtain the algebra:

Equation

which correspond to μ15 in characteristic 3.

If Equation is not algebraically closed, we have to consider all the algebras for which the polynomial:

Equation (10)

has no root this is equivalent to say that PA is irreducible. If we consider the coefficient of y3, that is Equation , it is equal to the discriminant of the determinant of the endomorphism LV, that is q3(A)=Disc(det(LV)). We deduce:

Proposition 5: The algebra A is regular if and only if PA(y) is strictly of degree 3.

It remains to examine the case μ(v,v)=0 for any v. That is:

Equation

If α2β2≠0 we can find some idempotents. In all the others cases, we have no idempotent. We obtain:

Equation

Theorem 6: Any commutative 2-dimensional algebra over an algebraically closed field is isomorphic to one of the following:

Equation

If Equation is not algebraically closed, we have also the following algebras where Equation

Equation

Let us examine the property of regularity for these algebras. Since they are commutative, the left and right regularity are equivalent notions. Computing directly the determinant of the operator Equation we deduce in the case Equation algebraically closed:

1. The algebras Equation are regular,

2. Equation is regular if β2≠0,

3. The algebras Equation and Equation are bisingular.

Algebras Over a Field of Characteristic 2

Let Equation be a field of characteristic 2. Assume that Equation. If A is a 2-dimensional Equation-algebra and if {e1, e2} is a basis of A, then the values of the different products belong to {e1, e2, e1+e2}. If f is an isomorphism of A, it is represented in the basis {e1, e2} by one of the following matrices:

Equation

Each of these matrices corresponds to a permutation of the finite set {e1, e2, e3=e1+e2}. If fact we have the correspondance:

Equation

where ij is the transposition between i and j and c the cycle {231}. In fact, the matrix M2 corresponds to the linear transformation f2(e1)=e2, f(e2)=e1 and in the set (e1, e2, e3) we have the transformation whose image is (e1, e2, e3) that is the transposition τ12. The matrix M3 corresponds to the linear transformation f2(e1)=e1+e2, f(e2)=e2 which corresponds to the permutation (e3, e2, e1) that is τ13. For all other matrices we have similar results. We deduce:

Theorem 7: There is a one-to-one correspondance between the change of Equation-basis in Aand the group Σ3.

If we want to classify all these products of A, we have to consider all the possible results of these products and to determine the orbits of the action of Σ3. More precisely the product μ(ei,ej) is in values in the set (e1, e2, e3=e1+e2). If we write μ(ei,ej)=ae1+be2+ce3, thus the matrix (a, b, c) is one of the following:

Equation

Let us consider the following sequence:

Equation

As Equation, if Equation and Equation then Equation with the relations:

Equation

for i, j, k all different and non zero. Thus the four first terms of this sequence determine all the other terms. More precisely, such a sequence writes:

Equation

Consequence: We have 44=256 sequences, each of these sequences corresponds to a 2-dimensional Equation-algebra.

Let us denote by S the set of these sequences. We have an action of Σ3 on S: if σ∈Σ3 and sS, thus s′=σs is the sequence:

Equation

with Equation when Equation and Rk≠0. If Rk=0, then Equation. The classification of the 2-dimensional Equation-algebras corresponds to the determination of the orbits of this action. Recall that the subgroups of Σ3 are Equation.

1. The isotropy subgroup is Σ3. In this case we have the following sequence (we write only the 4 first terms which determine the algebras:

Equation

Recall that Equation meansEquation means Equation and so on.

2. The isotropy subgroup is Equation We have only one orbit:

Equation

3. The isotropy subgroup is of order 2.

Equation

4. The isotropy subgroup is trivial. In this case any orbit contains 6 elements. As there are 256−46=46=210 elements having Σ3 as isotropy group, we deduce that we have 35 distinguished non isomorphic classes.

Conclusion

We have 52 classes of non isomorphic algebras of dimension 2 on the field F2.

Applications : 2-dimensional G-associative and Jordan algebras

G-associative commutative algebras

The notion of G-associativity has been defined in ref. [4]. Let G be a subgroup of the symmetric group Σ3. An algebra whose multiplication is denoted by μ is G-associative if we have:

Equation

where ε(σ) is the signum of the permutation . Since we assume that μ is commutative, all these notions are trivial or coincide with the simple associativity. Now, if the algebra is of dimension 2, then the associativity is completely determined by the identities:

Equation

We deduce that the only associative commutative 2-dimensional algebras are:

μ6 for (α2, β2)∈{(0,1),(1,0),(0,0)},

μ9 for β2=0 or 1,

μ12, μ16, μ17 .

• if Equation for β2=1 and λ=−1.

We find again the classical list [6].

G-associative noncommutative algebras

Let us consider now the noncommutative case. From Theorem 2, the multiplication μ is isomorphic to some μi,i=1,…,5 (we consider here that Equation is algabraically closed). Let be the associator of μ, that is Equation and μ is associative if and only if =0. The examination of this list allows to find the classification of the 2-dimensional noncommutative associative algebras: these algebras are isomorphic to one of the following:

Equation

Now, for any nonassociative algebra, we examine the Gi-associativity. Note that all these algebras are Lie-admissible, that is Σ3-associative. We focuse essentially on the G2-associativity, G2={Id, τ12}, because we deduce immediately the affine structures on the associated Lie algebra μa. Then we compute for any algebra Equation and Equation. We deduce that Equation is G2- associative if and only if β2=α4=0 and α2=−1, β4=−4. The algebrasμ2 and μ3 are never G2-associative, Equation is G2-associative for α2=−1 or (β4=α2−1). Likewise, Equation is G2-associative for α2=−1 or α2=1.

Proposition 8: Any 2-dimensional noncommutative G2-associative algebra is isomorphic to one of the following:

1. Equation or Equation, that is μ is associative,

2. Equation

3. Equation

4. Equation

5. Equation

6. Equation

Jordan algebras

In a Jordan algebra, the multiplication μ satisfies:

Equation

for all v,w. We assume in this section that Equation is algebraically closed and that the Jordan algebra are of dimension 2. Thus the multiplication μ is isomorphic to μi for i=11,…,16. To simplify the notation, we will write vw in place of μ(v,w). If v is an idempotent, thus v2=v and the Jordan identity gives:

v(vw)=v(vw)

for any w, that is, this identity is always satisfied.

Lemma 9: If v1 and v2 are idempotent vectors, thus:

Equation

for any w.

Proof. In the Jordan identity, we replace v by v1+v2. We obtain:

Equation

Since v1 and (v2) are idempotent, this equation reduces:

Equation

Proposition 10: If v1 and v2 are idempotent vectors such that v1v2 and v1+v2 are independent, thus the Jordan algebra is associative.

Proof. Let x and y be two vectors of the algebra. Thus, by hypothesis, Equation. Thus:

Equation

and,

x( yw) = y (xw)

By commutativity we obtain:

x( yw) = x(wy) = y (xw) = (xw) y

this proves that the algebra is associative.

If μ is given by,

Equation

the Jordan algebra admits two idempotents e1 and e2. Since Equation, the vectors e1e2 and e1+e2 are independent if and only if α2β2. In this case the algebra can be associative and we obtain the following associative Jordan algebra corresponding to:

1. α2=1, β2=0

2. α2=0, β2=1

These Jordan algebras are isomorphic. This gives the following Jordan algebra:

Equation

If e1e2 and e1+e2 are dependent, that is e1e2=λ(e1+e2), then λ=−1 or ½ or 0. If e1e2=0, the product is not a Jordan product. If λ=−1 the product is never a Jordan product. If ë = ½, we obtain the following Jordan algebra,

Equation

μ is given by:

Equation

This product is a Jordan product if β2=1 or 0. We obtain:

Equation

If μ=μ11 we have also a Jordan structure,

Equation

μ=0, we have the trivial Jordan algebra.

• If Equation is not algebraically closed, we consider,

Equation

We obtain a Jordan structure:

Equation

We find the list established in ref. [1].

2-dimensional Hom-algebra

The notion of Hom-algebra was introduced to generalized form of Hom-Lie algebra which appeared naturally when we are interested by the notion of q-derivation on the Witt algebra. In dimension 2, this notion is equivalent to the classical notion of Lie algebra. In dimension 3, we have shown that any skew-symmetric algebra is a Hom-Lie algebra. Then our interest concerns Hom-associative algebra [7,8], that is algebra A=(V,μ) such that there exists f∈End(V) satisfying the Hom- Ass identity:

Equation

for any X, Y, ZV. Using previous notations, we consider the algebras A(Id, f) and its opposite A(f, Id). Their multiplication law are respectively defined by:

Equation

and the Hom-Ass identity can be written:

Equation

Assume now that the algebra A is regular. In this case, assuming that the field is algebraically closed, there exists an unital algebra whose product is denoted XY and two endomorphisms u and v of V such that:

μ ( X ,Y ) = u( X ) ⋅ v(Y )

Then,

Equation

Then the Hom-Ass identity becomes:

Equation

Maybe, it is better to look the Hom-Ass identity from the previous list. Assume that A is non commutative.

1. Equation, let f be an endomorphism of V satisfying the Hom-Ass identity. To simplify notations we write XY for μ(X,Y) and [X,Y] for μa(X,Y). We have in particular:

Equation

We deduce f(e1)=ae2. Likewise we have [e2e2,f(e2)]=0 and f(e2)=k(α4e1+β4e1). Other identities give :

(a) Equation implies a=0 or e2e2=0.

(b) If a=0, then (e1,e2)f(e2)(e1e1)=0 implies f(e2)e2=0 and (e1,e1) f(e2)−f(e1)(e1e2)=0 implies e2f(e2)=0. Then [e2, f(e2)]=0 and f(e2)=ke2. This gives 0=f(e2)e2=be2e2 that is f=0 or e2e2=0. But we have seen that Equation, then in all the cases, f=0.

(c) If a≠0, then e2e2=0 and f(e2)=0. We deduce that (e1e2)f(e1)−f(e1) (e2e1)=0 implies α2=β2=0. Thus (e2e1)f(e1)−f(e2)(e1e1)=−a(e1e2)=−ae1=0 and a=0.

We deduce that the algebra Equation is not a Hom-associative algebra.

2. Equation. With similar simple computation we can look that also this algebra is not a Hom-Ass algebra.

3. Equation. In this case also, if we compute Equation, we obtain f(e1)=k1e1. Also we have Equation and f(e1)=0. We deduce e1f(e2)=0 and f(e2)e1=0 and f(e2)=0. Thus f=0 and A3 is not a Homassociative algebra.

4. Equation. If β4≠0, then the Hom-Ass condition implies α2=1 or −1. We obtain the following Hom-Ass algebras:

Equation

In each of these two cases, f is a diagonal endomorphism. These algebras are for β4≠2 or −2, not associative.

5.Equation. If α2=0, any linear endomorphism with values in Equation satisfies the Hom-Ass identity. Then the following algebra is Homassociative:

Equation

Assume now that α2≠0. If á2 ≠ ±1 , then any endomorphism satisfying the Hom-Ass identity is trivial. If α2=1 or −1, we have non trivial solution and the following algebras are Hom-associative algebras:

Equation

with Equation in the first case and Equation in the second case.

Then we have the list of noncommutative Hom-associative algebras. The commutative case can be established in the same way. In this case the Hom-Ass identity is reduced to:

Equation

Then f is in the kernel of the linear system whose matrix is:

Equation

Then A is a Hom-associative algebra if and only if H(A)=det(HAA)=0. We deduce that the set of 2-dimensional commutative Hom-associative algebra can be provided with an algebraic hypersurface embedded in the affine variety Equation. From Theorem 6, when Equation is algebraically closed, we obtain:

1. Equation. It is equal to 0 for α2=0 or β2=0 or α2=1−β2 or Equation orEquation

2. Equation and A7 is not a Hom-associative algebra.

3. Equation and A8 is not a Hom-associative algebra.

4. H(Ai)=0 for i = 9,10,11,12,13,14,15,16,17 and A9 , A10 , A11, A12 , A13, A14 , A15 , A16 , A17 are a Hom-associative algebras.

References

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