3-dimensional algebras. Part 1. Skew-symmetric case

TL;DR

This paper classifies all 3-dimensional skew-symmetric algebras over fields with characteristic not 2 and explores related Hom-Lie algebra structures, including some results on 4-dimensional cases.

Contribution

It provides a complete classification of 3-dimensional skew-symmetric algebras and characterizes the subvariety of 3-dimensional Hom-Lie algebras, extending to 4-dimensional cases.

Findings

Complete classification of 3D skew-symmetric algebras

Identification of the subvariety of 3D Hom-Lie algebras

Initial results on 4-dimensional cases

Abstract

An algebra is called skew-symmetric if its multiplication operation is a skew-symmetric bilinear application. We determine all these algebras in dimension $3$ over a field of characteristic different from $2$. As an application, we determine the subvariety of $3$-dimensional Hom-Lie algebras. For this type of algebras, we study also the dimension $4$.

Equations

$$\mu :V\times V\to V.$$

$$\mu (X,Y)=-\mu (Y,X).$$

$$f:V\to V$$

$$f(\mu (X,Y))={\mu }^{\prime }(f(X),f(Y))$$

$$\left\{\begin{array}[]{l}\mu(e_{1},e_{2})=\alpha_{1}e_{1}+\beta_{1}e_{2}+\gamma_{1}e_{3},\\ \mu(e_{1},e_{3})=\alpha_{2}e_{1}+\beta_{2}e_{2}+\gamma_{2}e_{3},\\ \mu(e_{2},e_{3})=\alpha_{3}e_{1}+\beta_{3}e_{2}+\gamma_{3}e_{3},\\ \end{array}\right.$$

$$(g,\mu )\in GL(V)\times ss\mathcal{A}l{g}_3\to {\mu }_g\in ss\mathcal{A}l{g}_3$$

$${\mu }_g(X,Y)=g^{-1}\mu (g(X),g(Y))$$

$${\delta }_{\mu }f(X,Y)=\mu (f(X),Y)+\mu (X,f(Y))-f(\mu (X,Y))$$

$$v_f=(f_{1,1},f_{2,1},f_{3,1},f_{1,2},\cdots ,f_{2,3},f_{3,3}).$$

$$\delta f(e_i,e_j)=\mu (f(e_i),e_j)+\mu (e_i,f(e_j))-f(\mu (e_i,e_j)$$

$$\left(\begin{array}[]{l}\delta f(e_{1},e_{2})\\ \delta f(e_{1},e_{3})\\ \delta f(e_{2},e_{3})\\ \end{array}\right)$$

$$M_{\mu }\cdot {}^t{v}_f.$$

$$\mu \to M_{\mu }$$

$$M_{\mu }=\left(\begin{array}{ccccccccc}0& 0& -{\alpha }_3& -{\beta }_1& {\alpha }_1& {\alpha }_2& -{\gamma }_1& 0& 0\\ {\beta }_1& -{\alpha }_1& -{\beta }_3& 0& 0& {\beta }_2& 0& -{\gamma }_1& 0\\ {\gamma }_1& 0& -{\gamma }_3-{\alpha }_1& 0& {\gamma }_1& {\gamma }_2-{\beta }_1& 0& 0& -{\gamma }_1\\ 0& {\alpha }_3& 0& -{\beta }_2& 0& 0& -{\gamma }_2& {\alpha }_1& {\alpha }_2\\ {\beta }_2& -{\alpha }_2+{\beta }_3& 0& 0& -{\beta }_2& 0& 0& -{\gamma }_2+{\beta }_1& {\beta }_2\\ {\gamma }_2& {\gamma }_3& -{\alpha }_2& 0& 0& -{\beta }_2& 0& {\gamma }_1& 0\\ -{\alpha }_3& 0& 0& {\alpha }_2-{\beta }_3& {\alpha }_3& 0& -{\alpha }_1-{\gamma }_3& 0& {\alpha }_3\\ 0& -{\alpha }_3& 0& {\beta }_2& 0& 0& -{\beta }_1& -{\gamma }_3& {\beta }_3\\ 0& 0& -{\alpha }_3& {\gamma }_2& {\gamma }_3& -{\beta }_3& -{\gamma }_1& 0& 0\end{array}\right)$$

$$f=\left(\begin{array}{ccc}0& 0& 0\\ 0& b_2& 0\\ 0& 0& -b_2\end{array}\right)$$

$$\begin{array}[]{l}(-\alpha_{1}\alpha_{2}\beta_{1}+\alpha_{1}^{2}\beta_{2}-\alpha_{2}^{2}\gamma_{1}+\beta_{3}^{2}\gamma_{1}+\alpha_{1}\alpha_{2}\gamma_{2}-\beta_{1}\beta_{3}\gamma_{3}+\beta_{3}\gamma_{2}\gamma_{3}-\beta_{2}\gamma_{3}^{2},\\ \alpha_{2}\beta_{1}^{2}+\alpha_{1}\beta_{1}\beta_{2}-2\beta_{2}\gamma_{1}(\alpha_{2}+\beta_{3})+\alpha_{2}\beta_{1}\gamma_{2}+\alpha_{1}\beta_{2}\gamma_{2}+\beta_{1}\beta_{3}\gamma_{2}-\beta_{3}\gamma_{2}^{2}+\beta_{1}\beta_{2}\gamma_{3}+\beta_{2}\gamma_{2}\gamma_{3},\\ 2\beta_{2}\gamma_{1}(\alpha_{1}-\gamma_{3})-\alpha_{2}\beta_{1}\gamma_{1}+\beta_{1}\beta_{3}\gamma_{1}-\alpha_{1}\beta_{1}\gamma_{2}-\alpha_{2}\gamma_{1}\gamma_{2}+\beta_{3}\gamma_{1}\gamma_{2}+\alpha_{1}\gamma_{2}^{2}-\beta_{1}^{2}\gamma_{3}+\beta_{1}\gamma_{2}\gamma_{3},\\ \alpha_{1}\alpha_{3}\beta_{1}-\alpha_{1}^{2}\beta_{3}+2\alpha_{3}\gamma_{1}(\alpha_{2}+\beta_{3})-\alpha_{1}\alpha_{3}\gamma_{2}-\alpha_{1}\alpha_{2}\gamma_{3}-\alpha_{3}\beta_{1}\gamma_{3}-\alpha_{1}\beta_{3}\gamma_{3}+\alpha_{3}\gamma_{2}\gamma_{3}-\alpha_{2}\gamma_{3}^{2},\\ \alpha_{3}\beta_{1}^{2}-\alpha_{1}\beta_{1}\beta_{3}-\alpha_{2}^{2}\gamma_{1}+\beta_{3}^{2}\gamma_{1}+\alpha_{1}\alpha_{2}\gamma_{2}-\alpha_{3}\gamma_{2}^{2}-\beta_{1}\beta_{3}\gamma_{3}+\alpha_{2}\gamma_{2}\gamma_{3},\\ \alpha_{1}\beta_{3}\gamma_{1}-\alpha_{1}\alpha_{2}\gamma_{1}-2\alpha_{3}\gamma_{1}(\beta_{1}+\gamma_{2})+\alpha_{1}^{2}\gamma_{2}+\alpha_{1}\beta_{1}\gamma_{3}+\alpha_{2}\gamma_{1}\gamma_{3}-\beta_{3}\gamma_{1}\gamma_{3}+\alpha_{1}\gamma_{2}\gamma_{3}+\beta_{1}\gamma_{3}^{2},\\ \alpha_{2}\alpha_{3}(\beta_{1}-\gamma_{2})-2\alpha_{3}\beta_{2}(\alpha_{1}-\gamma_{3})+\alpha_{1}\alpha_{2}\beta_{3}+\alpha_{3}\beta_{1}\beta_{3}-\alpha_{1}\beta_{3}^{2}-\alpha_{3}\beta_{3}\gamma_{2}+\alpha_{2}\gamma_{3}(\alpha_{2}-\beta_{3}),\\ \alpha_{1}\alpha_{2}\beta_{2}-\alpha_{2}^{2}\beta_{1}+\alpha_{2}\beta_{1}\beta_{3}+\alpha_{1}\beta_{2}\beta_{3}-2\alpha_{3}\beta_{2}(\gamma_{2}-\beta_{1})+\alpha_{2}\beta_{3}\gamma_{2}-\beta_{3}^{2}\gamma_{2}+\alpha_{2}\beta_{2}\gamma_{3}+\beta_{2}\beta_{3}\gamma_{3},\\ \alpha_{1}\alpha_{2}\beta_{1}+\alpha_{3}\beta_{1}^{2}-\alpha_{1}^{2}\beta_{2}-\alpha_{1}\beta_{1}\beta_{3}-\alpha_{3}\gamma_{2}^{2}+\alpha_{2}\gamma_{2}\gamma_{3}-\beta_{3}\gamma_{2}\gamma_{3}+\beta_{2}\gamma_{3}^{2}).\end{array}$$

$$\left(\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right)$$

$$v_f={}^t(a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3).$$

$$L_{e_1}=\left(\begin{array}{ccc}0& {\alpha }_1& {\alpha }_2\\ 0& {\beta }_1& {\beta }_2\\ 0& {\gamma }_1& {\gamma }_2\end{array}\right)$$

$$\left\{\begin{array}[]{l}-\alpha_{1}\beta_{3}+\alpha_{3}\beta_{1}-\gamma_{3}\alpha_{2}+\alpha_{3}\gamma_{2}=0,\\ \beta_{2}\alpha_{1}-\beta_{1}\alpha_{2}-\gamma_{3}\beta_{2}+\beta_{3}\gamma_{2}=0,\\ \gamma_{2}\alpha_{1}+\gamma_{3}\beta_{1}-\beta_{3}\gamma_{1}-\gamma_{1}\alpha_{2}=0\\ \end{array}\right.$$

$$M_{\mu }=\left(\begin{array}{ccccccccc}0& 0& 0& -{\beta }_1& {\alpha }_1& 0& -{\gamma }_1& 0& 0\\ {\beta }_1& -{\alpha }_1& 0& 0& 0& 0& 0& -{\gamma }_1& 0\\ {\gamma }_1& 0& -{\alpha }_1& 0& {\gamma }_1& -{\beta }_1& 0& 0& -{\gamma }_1\\ 0& 0& 0& 0& 0& 0& 0& {\alpha }_1& 0\\ 0& 0& 0& 0& 0& 0& 0& {\beta }_1& 0\\ 0& 0& 0& 0& 0& 0& 0& {\gamma }_1& 0\\ 0& 0& 0& 0& 0& 0& -{\alpha }_1& 0& 0\\ 0& 0& 0& 0& 0& 0& -{\beta }_1& 0& 0\\ 0& 0& 0& 0& 0& 0& -{\gamma }_1& 0& 0\end{array}\right)$$

$$\left\{\begin{array}[]{l}\mu(e_{1},e_{2})=e_{2},\\ \mu(e_{1},e_{3})=\gamma_{2}e_{3},\\ \mu(e_{2},e_{3})=e_{1}.\\ \end{array}\right.$$

$$\left(\begin{array}{ccc}0& -a_3& a_2\\ -a_2& b_2& 0\\ a_3& 0& -b_2\end{array}\right)$$

$$\left(\begin{array}{ccc}0& 0& 0\\ 0& b_2& c_2\\ 0& b_3& -b_2\end{array}\right)$$

$$\left\{\begin{array}[]{l}\mu(e_{1},e_{2})=e_{1}+e_{4},\\ \mu(e_{1},e_{3})=e_{2}+e_{3},\\ \mu(e_{1},e_{4})=e_{3}+e_{4},\\ \mu(e_{2},e_{3})=e_{1}+e_{3},\\ \mu(e_{2},e_{4})=e_{2},\\ \mu(e_{3},e_{4})=e_{2}+e_{3},\\ \end{array}\right.$$

$$M_{\mu }=\left(\begin{array}{cccc}0& 0& -\beta & \alpha \\ \beta & -\alpha & 0& 0\end{array}\right)$$

$$\dim \mathcal{O}(\mu )\le 8.$$

$$\left\{\begin{array}[]{l}\mu(e_{1},e_{2})=e_{3},\\ \mu(e_{1},e_{3})=\beta_{2}e_{2}+\gamma_{2}e_{3},\\ \mu(e_{2},e_{3})=\alpha_{3}e_{1}+\beta_{3}e_{2}+\gamma_{3}e_{3},\end{array}\right.$$

$${\mathcal{C}}^0(\mu )=A,{\mathcal{C}}^1(\mu )=\mu (A,A),{\mathcal{C}}^k(\mu )=\mu ({\mathcal{C}}^{k-1}(\mu ),A),k\ge 2.$$

$$\left\{\begin{array}[]{l }\mu(e_{i},e_{3})=0,\\ \mu(e_{1},e_{2})=\gamma_{1}e_{3},\end{array}\right.$$

Full text

Université de Haute Alsace, 4 rue des Frères Lumière, F68093 Muhouse.

Key Words: $3$-dimensional skew-symmetric algebras. Hom-Lie algebras Variety.

2010 Mathematics Subject Classification: 17A01. 15A72

INTRODUCTION. An algebra over a field $\mathbb{K}$ is a $\mathbb{K}$-vector space equipped with a bilinear product. The multiplication operation in an algebra may or may not be associative, leading to the notions of associative algebras and nonassociative algebras. If we denote by $\mu$ this multiplication, we do not assume that $\mu$ satisfies some quadratic, ternary, or $n$-ary relations. When we consider only the finite dimensional framework, that is when the $\mathbb{K}$-vector space is of finite dimension, one of the first natural problem which then arise is the determination of all these algebras. For example, the classification up to isomorphism of algebras for a given dimension seems interesting and it looks at first easy to solve. It is strange that this problem was solved only in dimension $2$ (see the works of Goze-Remm, or Peterssen or Bekbaev ([1, 3, 9]). For special classes of algebras, this classification work was carried on for greater dimensions, but always relatively small. For example, associative algebras are classified up to dimension $6$, Lie algebras up to dimension $7$ ([4]). Of course, the classification is complete for some special classes, for example the simple Lie algebras, the simple associative algebras. But for the general case, we have not many informations, probably because we know very few invariants in multilinear algebra. Furthermore, when the classification exists (always with the fear that this classification could be incomplete), it is often difficult to use. Consider for example the classification of the complex nilpotent Lie algebras of dimension $7$. We can think that this list is complete. There are in this case $5$ one-parameter families and more than one hundred of algebras (this number depends of the authors and it is not an invariant). Then it is often difficult to recognize among this list a given Lie algebra especially when if it is not written in a basis respecting the invariants used to obtain the classification. And it is not convenient to determine subclasses, for example to determine the contact nilpotent Lie algebras. To test each algebra of this list is more than boring. That is why, as we presented it in previous works, we prefer to determine classes invariant by isomophism and minimal in a sense that we specify.

We are interested in all this work in the $3$-dimensional algebras over a field $\mathbb{K}$ with skew-symmetric multiplication. We suppose in this work that $\mathbb{K}$ is an arbitrary field of characteristic different to $2$. The plane of this work is the following: first of all we study the automorphism group of a $3$-dimensional algebra. In particular, we characterize among all these algebras those which are Lie algebras by studying the dimension of the automorphism group. In the second part, we classify the nilpotent skew-symmetric algebras by showing that all these algebras are Lie algebras (but it is not the case in greater dimension). Next we classify the solvable skew-symmetric algebras and, to end this classification, the non-solvable case. The last section is essentially dedicated to the study of Hom-Lie algebras, which are a particular class of skew-symmetric algebras. We show that any skew-symmetric algebra of dimension $3$ is Hom-Lie. This is no longer true in dimension $4$ where we study the algebraic variety of Hom-Lie algebras.

1. The automorphism group of a $3$-dimensional skew-symmetric algebra

1.1. Generalities

An algebra over a field $\mathbb{K}$ is a $\mathbb{K}$-vector space $V$ with a multiplication operation given by a bilinear map

[TABLE]

Such a multiplication is a linear tensor on $V$ of type $(2,1)$, that is $2$-times contravariant and $1$-time covariant. In all this paper, the vector space $V$ is finite-dimensional and fixed. We can for example consider $V=\mathbb{K}^{n}$. We denote by $A=(V,\mu)$ a $\mathbb{K}$-algebra with multiplication $\mu$. We also assume in this work that the field $\mathbb{K}$ is of characteristic zero and that the multiplication $\mu$ is skew-symmetric, that is

[TABLE]

In this case, we shall say that $A$ is a skew-symmetric algebra, and we shorten the name by writing ss-algebra. Two $\mathbb{K}$-algebras $A=(V,\mu)$ and $A^{\prime}=(V,\mu^{\prime})$ are isomorphic if there is a linear isomorphism

[TABLE]

such that

[TABLE]

for all $X,Y\in V$ and we denote by $Aut(A)$ or $Aut(\mu)$ the group of automorphisms of the algebra $A=(V,\mu)$.

Assume now that $\dim V=3$. Let $\{e_{1},e_{2},e_{3}\}$ be a fixed basis of $V$. A general skew-symmetric bilinear map $\mu$ has the following expression

[TABLE]

and the set of these applications is a $9$-dimensional vector space $ss\mathcal{A}lg_{3}$ parametrized by the structure constants $\alpha_{i},\beta_{i},\gamma_{i}$, $i=1,2,3$, identifying $\mu$ with its structure constants. The linear group $GL(V)$ acts on $ss\mathcal{A}lg_{3}$:

[TABLE]

with

[TABLE]

for any $X,Y\in V.$ Let $\mathcal{O}(\mu)$ be the orbit of $\mu$ corresponding to this action. It is a nonsingular algebraic subvariety of $ss\mathcal{A}lg_{3}$ (see for example [7]). To compute the dimension of the subvariety $\mathcal{O}(\mu)$, it is sufficient to compute the dimension of its tangent space at the point $\mu$. Since $ss\mathcal{A}lg_{3}$ is linear, its tangent space $T_{\mu}(ss\mathcal{A}lg_{3})$ at $\mu$ is identified to itself and it is constituted by skew-symmetric bilinear map $\psi\in ss\mathcal{A}lg_{3}$. The tangent space $T_{\mu}(\mathcal{O}(\mu))$ at the point $\mu$ to the orbit is a linear subspace of $T_{\mu}(ss\mathcal{A}lg_{3})=ss\mathcal{A}lg_{3}$ whose elements are the skew-symmetric applications of type $\psi=\delta_{\mu}f$ with $f\in End(V)$ and

[TABLE]

for any $X,Y\in V$. Recall also that $f$ is a derivation of the algebra $A=(V,\mu)$ is $\delta_{\mu}(f)=0$. The set $Der(\mu)$ of derivations of $A$ is a Lie algebra which is the Lie algebra of the algebraic group $Aut(\mu)$.

1.2. The matrix $M_{\mu}$ associated to $\mu$

To compute the dimension of $T_{\mu}(\mathcal{O}(\mu))$, we shall consider the following matrix $M_{\mu}$ associated to $\mu$, which will give a linear representation of $\mu$. Recall that we have fixed a basis $\{e_{i}\}$ a basis of $V$ and identified the skew-symmetric applications on $V$ with its structure constants related to this basis. Let $f\in gl(V)$ and we denote also by $f$ the matrix of $f$ related to the fixed basis. If $f=(f_{ij})$, we consider the vector $v(f)$ of $\mathbb{K}^{9}$:

[TABLE]

We have

[TABLE]

for $1\leq i<j\leq 3$ and we still denote $\delta f(e_{i},e_{j})$ the column matrix o its components in the basis $\{e_{1},\cdots,e_{3}\}$. The vector

[TABLE]

is then a matrix in $\mathcal{M}_{9,1}.$ It corresponds to a matrix product:

[TABLE]

This matrix $M_{\mu}$ is a square matrix of order $9$ and the map

[TABLE]

gives a linear representation of $\mu$.

If we consider the structure constants of $\mu$ related to the basis $\{e_{i}\}$ given in (1), then $M_{\mu}$ is equal to

[TABLE]

Proposition 1**.**

For any $\mu$, the matrix $M_{\mu}$ is singular. Moreover, there exists $\mu$ such that ${\rm Rank}(M_{\mu})=8.$

*Proof. *The first part is a direct computation of the determinant. We can see also that [math] is always an eigenvalue. Now we consider $\mu$ given by $(\beta_{1}=1,\gamma_{2}=2,\alpha_{3}=1)$. For this algebra the kernel of $M_{\mu}$ is of dimension $1$ and constituted of matrices

[TABLE]

and ${\rm Rank}(M_{\mu})=8.$ For a generic algebra, that is without algebraic conditions between the structure constants, the kernel of $M_{\mu}$ is a $1$-dimensional vector space generated by the vector

[TABLE]

Corollary 2**.**

For any $3$-dimensional ss-algebra $A=(V,\mu)$ over $\mathbb{K}$, the dimension of the automorphism group $Aut(\mu)$ is greater than or equal to $1$.

*Proof. *In fact, $Aut(\mu)$ is an algebraic group whose Lie algebra is isomorphic to $Der(\mu)$, the Lie algebra of derivations of $\mu$, that is the subspace of $gl(V)$ whose element $f$ satisfy $\delta(f)=0$ or equivalently the vector ${}^{t}v_{f}$ is in the kernel of $M_{\mu}$. Since the rank of this matrix is smaller that $8$, its kernel is bigger than $1$.

Proposition 3**.**

The algebra $A=(V,\mu)$ is a Lie algebra if and only if there exists a non zero vector $X$ in $V$ such that the linear endomorphism $L_{X}:Y\rightarrow\mu(X,Y)$ is in the kernel of $M_{\mu}$.

*Proof. *If $f\in End(A)$ then, considering the basis $\{e_{i}\}$, we identify $f$ with the matrix

[TABLE]

and we associate it with the vector

[TABLE]

If we take $X=e_{1}$, then

[TABLE]

and $M_{\mu}(v_{L_{e_{1}}})=0$ is equivalent to

[TABLE]

which corresponds to Jacobi’s conditions.

Proposition 4**.**

If the ss-algebra $A=(V,\mu)$ is a Lie algebra, then the rank of $M_{\mu}$ is smaller than or equal to $6$.

*Proof. *In fact, since $A=(V,\mu)$ is a Lie algebra, the endomorphisms $L_{X}(Y)=\mu(X,Y)$ are derivations and the vectors $v_{L_{X}}$ are in the kernel of $M_{\mu}$. If $L_{e_{1}},L_{e_{2}},L_{e_{3}}$ are linearly independent, then $\ker M_{\mu}\geq 3$ and the rank of $M_{\mu}$ is smaller than $6$. If these vectors are linearly dependent, we have a non trivial linear combination between these vectors and without loss of generality, we can assume that $L_{e_{1}}=0$. But $v_{L_{e_{1}}}=(0,0,0,\alpha_{2},\beta_{2},\gamma_{2},\alpha_{3},\beta_{3},\gamma_{3})$ and $v_{L_{e_{1}}}=0$ implies that

[TABLE]

which rank is less or equal to $3$.

The converse of this proposition is not true. We can find non Lie algebras whose associated matrix $M_{\mu}$ is of rank $6$. For example we consider the algebra

[TABLE]

This algebra is a Lie algebra if and only if $\gamma_{2}=-1.$ In this case the rank is $6$ and the kernel is composed of the endomorphisms

[TABLE]

If $\gamma_{2}=1$, the algebra is not a Lie algebra but the rank is also $6$. In this case the kernel is composed of endomorphisms

[TABLE]

While this last algebra is not a Lie algebra, we can embedded this algebra in a larger class of algebras such as algebras of Lie type (see [6]) or Hom-Lie algebras. We shall see that in the last section.

1.3. What happen in dimension $4$?

If $A=(V_{4},\mu)$ is a $4$-dimensional skew-symmetric $\mathbb{K}$-algebra, then $M_{\mu}$ is a non square matrix of order $24\times 16$. It is not too much complicated to write this matrix, but we do not make here. If we identified $A$ with $\mu$ and $\mu$ with its structure constant related with a fixed basis $\{e_{1},e_{2},e_{3},e_{4}\}$ of $V$, the set of $4$-dimensional ss-algebras is a $24$-dimensional vector space denoted $ss\mathcal{A}lg_{4}$.

Proposition 5**.**

There exists a Zariski-open set in the affine variety $ss\mathcal{A}lg_{4}$ constituted of ss-algebras $(V,\mu)$ whose automorphism group $Aut(\mu)$ is of dimension [math].

*Proof. *In fact, for a generic algebra of $ss\mathcal{A}lg_{4}$, that is without relations between the structure constants, the corresponding matrix $M_{\mu}$ is of rank $16$. Then, any derivation is trivial and the algebraic group $Aut(\mu)$ is of dimension [math]. For example, for the following algebra

[TABLE]

satisfies this property.

Let us note that $ss\mathcal{A}lg_{4}$ is an affine algebraic variety isomorphic to $\mathbb{K}^{24}$ and the Zariski open set constituted of algebras $A=(V_{4},\mu)$ whose automorphism group $Aut(\mu)$ is of dimension [math] is a finite union of algebraic components where each one is the complementary of an hypersurface, then they are algebraic subvarieties of $ss\mathcal{A}lg_{4}$.

Remark: Case of Dimension $2$. Before studying the dimension $3$, we have naturally studied the $2$-dimensional case in [GRKhukhro]. In this case, if $\mu(e_{1},e_{2})=\alpha e_{1}+\beta e_{2}$, then

[TABLE]

Its rank is $2$ as soon as $\alpha\neq 0$ or $\beta\neq 0$. In this case $\ker(M_{\mu})$ is of dimension $2$. This implies that the group of automorphisms of $A=(V,\mu)$ is an algebraic group of dimension $2$. We have also noted that any ss-algebra of dimension $2$ is a Lie algebra.

1.4. Deformations and rigidity

A deformation of $\mu$, in the Gerstenhaber’sense, is given by a tensor $(2,1)$ in $\mathbb{K}[t]\otimes ss\mathcal{A}lg_{3}.$ Grosso-modo, the notion of deformation permits to describe a neighborhood of $\mu$ in $ss\mathcal{A}lg_{3}$. There exists a cohomological approach of the notion of deformations. In our situation, the complex of cohomology is well known. The space of $p$-cochains is the space of $p$-linear skew symmetric applications, the first space of cohomology corresponds to the Kernel of $M_{\mu}$, and the second space is a factor space isomorphic to $T_{\mu}(ss\mathcal{A}lg_{3})/ImM_{\mu}$. This complex is well described in the context of operads. We denote by $\mathcal{S}ign$ the quadratic operad encoding the category of skew-symmetric algebras. Recall its construction (see also (see [8]). Let $\mathbb{K}[\Sigma_{2}]$ be the group algebra of the symmetric group of degree 2. Considered as a $\Sigma_{2}$-module we have that $\mathbb{K}[\Sigma_{2}]=1\!\!1_{2}\oplus sgn_{2}$ where $1\!\!1$ is the one-dimensional representation and $sgn_{2}$ the one-dimensional signum representation. We then consider $\Gamma(sgn_{2})$ the free operad generated by a skew symmetric operation. For this operad, we have in particular $\Gamma(sgn_{2})(0)=0$, $\Gamma(sgn_{2})(1)\simeq\mathbb{K},$ $\dim\Gamma(sgn_{2})(2)=1$ because $\Gamma(sgn_{2})(2)$ is the $\mathbb{K}[\Sigma_{2}]$-module generated by $\{x_{1}x_{2}=-x_{2}x_{1}\}$, $\dim\Gamma(sgn_{2})(3)=3$, $\Gamma(sgn_{2})(3)$ is the $\mathbb{K}[\Sigma_{2}]$-module generated by \{(x_{1}x_{2})x_{3}),(x_{2}x_{3})x_{1},(x_{3}x_{1})x_{2})\},$$\dim\Gamma(sgn_{2})(4)=15 because $\Gamma(sgn_{2})(2)$ is the $\mathbb{K}[\Sigma_{2}]$-module generated by

\begin{array}[]{l}\{(x_{1}x_{2})(x_{3}x_{4}),(x_{1}x_{3})(x_{2}x_{4}),(x_{1}x_{4})(x_{2}x_{3}),((x_{1}x_{2})x_{3})x_{4},((x_{1}x_{2})x_{4})x_{3},((x_{1}x_{3})x_{2})x_{4},\\ ((x_{1}x_{3})x_{4})x_{2},((x_{1}x_{4})x_{2})x_{3},((x_{1}x_{4})x_{3})x_{2},((x_{2}x_{3})x_{1})x_{4},((x_{2}x_{3})x_{4})x_{1},((x_{2}x_{4})x_{1})x_{3},\\ ((x_{2}x_{4})x_{3})x_{1},((x_{3}x_{4})x_{1})x_{2},((x_{3}x_{4})x_{2})x_{1}\}.\end{array}

This operad is a Koszul operad, then the cohomology which parametrizes the deformations is the operadic cohomology.

Recall also the notion of rigidity which traduces the fact that any deformations of $\mu$ is isomorphic to $\mu$.

Definition 6**.**

A skew-symmetric algebra $A=(V,\mu)$ is rigid if the orbit $\mathcal{O}(\mu)$ is Zariski-open in the algebraic variety $ss\mathcal{A}lg_{3}$.

This topological notion can be repalced by an algebraic condition. If we denote by $H^{*}_{S}(\mu,\mu)$ the complex of deformations, since the variety $\mathcal{A}lg_{3}$ is an affine reduced variety, an element $\mu\in\mathcal{A}lg_{3}$ is rigid if and only if $H^{2}_{S}(\mu,\mu)=0$. We deduce of the previous results

Proposition 7**.**

For any $\mu\in ss\mathcal{A}lg_{3}$, we have

[TABLE]

In particular none of $3$-dimensional ss-algebra is rigid in $ss\mathcal{A}lg_{3}$.

*Proof. *Since $ss\mathcal{A}lg_{3}$ is a linear plane and then a reduced algebraic variety, and since $\dim ss\mathcal{A}lg_{3}-\dim\mathcal{O}(\mu)\geq 1$, none of these algebras is rigid in $ss\mathcal{A}lg_{3}$.

This notion of rigidity which concerns an element can be extended to parametrized families of algebras.

Definition 8**.**

Let $F_{t_{1},\cdots t_{k}}$ be a family of algebras of $ss\mathcal{A}lg_{3}$ parametrized by $t_{1},\cdots,t_{3}.$ This family is rigid if its orbit by the action of the group $GL(V)$ is Zariski open in $ss\mathcal{A}lg_{3}.$

If $\mathcal{O}(F_{t_{1},\cdots t_{k}})$ is this orbit, its rigidity implies that $\overline{\mathcal{O}(F_{t_{1},\cdots t_{k}})}$ (the closure in the Zariski sense) is an algebraic component of $ss\mathcal{A}lg_{3}.$ But $ss\mathcal{A}lg_{3}$ is connected so if $F_{t_{1},\cdots t_{k}}$ is rigid, $\overline{\mathcal{O}(F_{t_{1},\cdots t_{k}})}=\mathcal{A}lg_{3}.$

Proposition 9**.**

The family $F_{\beta_{2},\gamma_{2},\alpha_{3},\beta_{3},\gamma_{3}}$ whose elements are the algebras (7):

[TABLE]

is rigid in $ss\mathcal{A}lg_{3}$.

In fact any deformation of any element of this family is isomorphic to an element of this same family.

2. Nilpotent case

Let $(A,\mu)$ be a $\mathbb{K}$-algebra. We consider the descending central series

[TABLE]

The algebra $A$ is nilpotent if there exists $k$ such that $\mathcal{C}^{k}(\mu)=0.$ The smallest $k$ such that $\mathcal{C}^{k}(\mu)=0$ is the nilindex. For a $3$-dimensional nilpotent algebra $(A,\mu)$, we have only the following sequences:

1. (1)

$A\supset\mathcal{C}^{1}(\mu)=0,$ 2. (2)

$A\supset\mathcal{C}^{1}(\mu)\supset\mathcal{C}^{2}(\mu)=0,$ 3. (3)

$A\supset\mathcal{C}^{1}(\mu)\supset\mathcal{C}^{2}(\mu)\supset\mathcal{C}^{3}(\mu)=0.$

It is obvious that the last term $\mathcal{C}^{k}(\mu)$ is contained in the center. Thus the first case corresponds to abelian case, the last is impossible because it would imply that $\dim\mathcal{C}^{1}(\mu)=2$. Then it remains $A\supset\mathcal{C}^{1}(\mu)\supset\mathcal{C}^{2}(\mu)=0.$ We consider an adapted basis $\{e_{1},e_{2},e_{3}\}$. It satisfies

[TABLE]

with $\gamma_{1}\neq 0.$ We deduce

Proposition 10**.**

Any $3$-dimensional nilpotent algebra is a Lie algebra. It is isomorphic to the abelian 3-dimensional Lie algebra or to the Heisenberg algebra $\mathfrak{h}_{3}$.

Let us note that in the non abelian case the rank of $M_{\mu}$ is equal to $3$.

Remark: What happen in dimension greater than or equal to $4$? We have seen that in dimension $3,$ any nilpotent algebra is a Lie algebra. This property doesn’t extend to greater dimension. Let us consider for example the filiform case, that is nilpotent algebra such that

[TABLE]

with $\dim A/\mathcal{C}^{1}(\mu)=2$ and $\dim\mathcal{C}^{k}(\mu)/\mathcal{C}^{k+1}(\mu)=1,\ i=1,\cdots,k-1.$ If in dimension less than or equal to $4$ these algebras are also Lie algebras, in dimension 5 (or greater) we can find filiform non Lie algebras. Consider the family

[TABLE]

Such an algebra is not a Lie algebra as soon as $a-c\neq 0$ or $d\neq 0$. But this algebra is provided with a Hom-Lie algebra structure (see Section 5).

3. Solvable case

Let $(A,\mu)$ a $\mathbb{K}$ algebra. We consider the derived series:

[TABLE]

The algebra $A$ is solvable if there exists $k$ such that $\mathcal{D}^{k}(\mu)=0.$ The smallest $k$ such that $\mathcal{D}^{k}(\mu)=0$ is the solvindex. For a 3-dimensional solvable algebra $(A,\mu),$ we have only the following sequences:

1. (1)

$A\supset\mathcal{D}^{1}(\mu)=0,$ 2. (2)

$A\supset\mathcal{D}^{1}(\mu)\supset\mathcal{D}^{2}(\mu)=0,$ 3. (3)

$A\supset\mathcal{D}^{1}(\mu)\supset\mathcal{D}^{2}(\mu)\supset\mathcal{D}^{3}(\mu)=0.$

In the first case $A$ is abelian. In the second case, if $\dim A/\mathcal{D}^{1}(\mu)=2$, we consider an adapted basis $\{e_{1},e_{2},e_{3}\}$ and the hypothesis imply that

[TABLE]

and, not to consider again the nilpotent case, we consider $a_{13}\neq 0$ or $a_{23}\neq 0$. This algebra is a solvable Lie algebra for any $a_{ij}$ and the rank of $M_{\mu}=5$. In particular the automorphism group is of dimension $4$. Let us note that any algebras of (4) is isomorphic to

[TABLE]

and in this case, the identity component of this group is constituted of matrices

[TABLE]

with $yt\neq 0.$

Assume now that $\dim A/\mathcal{D}^{1}(\mu)=1$ and consider an adapted basis $\{e_{1},e_{2},e_{3}\}$. This means that $\{e_{2},e_{3}\}$ is a basis of $\mathcal{D}^{1}(\mu)$. By hypothesis $\mu(e_{2},e_{3})=0$. Then $\mu$ satisfies

[TABLE]

Such algebra is also a Lie algebra. Recall that the classification, up to isomorphism, of $3$-dimensional solvable Lie algebras is given in the book of Jacobson ([Ja]). The rank of $M_{\mu}$ is smaller or equal to $5$. For example, it is equal to $5$ for

[TABLE]

and equal to $3$ for

[TABLE]

Assume now that the descending sequence is filiform, that is $A\supset\mathcal{D}^{1}(\mu)\supset\mathcal{D}^{2}(\mu)\supset\mathcal{D}^{3}(\mu)=0$. Let $\{e_{1},e_{2},e_{3}\}$ be an adapted basis to this flag. We have

[TABLE]

with $\gamma_{3}\neq 0$ and $\beta_{1}\neq 0$ or $\beta_{2}\neq 0$. Such algebra is never a Lie algebra. Let us note that, since $\gamma_{3}\neq 0$, giving $e^{\prime}_{2}=\gamma_{3}^{-1}e_{3}$, we can consider that $\gamma_{3}=1$.

Proposition 11**.**

Any $3$-dimensional solvable algebra is a Lie algebra or is isomorphic to the non-Lie algebra given by

[TABLE]

with $\beta_{1}\neq 0$ or $\beta_{2}\neq 0.$

We have

[TABLE]

as soon as $\beta_{2}\neq 0$ and the kernel is generated by

[TABLE]

If $\beta_{2}=0$, then ${\rm Rank}(M_{\mu})=7$ and the kernel is generated by

[TABLE]

It would be interesting to present this parameter $\beta_{2}$ as an invariant. The following proposition gives an answer.

Proposition 12**.**

The Killing symmetric form $K_{\mu}(X,Y)=tr(L_{X}\circ L_{Y})$ where $L_{U}(V)=\mu(U,V)$ of an algebra (6) is nondegenerate if and only $\beta_{2}\neq 0.$

*Proof. *The matrix of $K_{\mu}$ in the basis $\{e_{1},e_{2},e_{3}\}$ is

[TABLE]

and its determinant is equal $-\beta_{2}^{2}.$ Let us note that $(A,\mu)$ is not a simple algebra, $I=\mathbb{K}\{e_{2},e_{3}\}$ is an ideal of $A$, but its Killing form is nondegenerate as soon as $\beta_{2}\neq 0.$ Let us note also that $K_{\mu}$ is, in this case, a not invariant pseudo scalar product.

4. Non-Solvable case

If $(A,\mu)$ is not solvable, there exists $k$ with

[TABLE]

We have the following possibilities:

1. (1)

$A=\mathcal{D}^{1}(\mu),$ 2. (2)

$A\supset\mathcal{D}^{1}(\mu)=\mathcal{D}^{2}(\mu)\neq 0,$ 3. (3)

$A\supset\mathcal{D}^{1}(\mu)\supset\mathcal{D}^{2}(\mu)=\mathcal{D}^{3}(\mu)\neq 0$.

The sequence $A\supset\mathcal{D}^{1}(\mu)\supset\mathcal{D}^{2}(\mu)=\mathcal{D}^{3}(\mu)\neq 0$ is impossible because this implies that $\dim\mathcal{D}^{2}(\mu)=1$ and $\mu(\mathcal{D}^{2}(\mu),\mathcal{D}^{2}(\mu))\neq 0$, this is impossible since $\mu$ is skew-symmetric. Similarly with the sequence $A\supset\mathcal{D}^{1}(\mu)=\mathcal{D}^{2}(\mu)\neq 0.$ Thus, if $(A,\mu)$ is not solvable, we have

[TABLE]

that is

[TABLE]

with

[TABLE]

Lemma 13**.**

There exists $X,Y$ independent in $A$ such that $X,Y,\mu(X,Y)$ are also independent.

*Proof. *Assume that this property is not true. In any basis $\{e_{1},e_{2},e_{3}\}$ we must have

[TABLE]

with

[TABLE]

Let $e^{\prime}_{2}=ae_{2}+be_{3}$. Then $rank\{e_{1},e^{\prime}_{2},\mu(e_{1},e^{\prime}_{2})\}=2$ implies $\beta_{1}=\gamma_{2}.$ Likewise $rank\{e_{2},e^{\prime}_{1}=ae_{1}+be_{3},\mu(e^{\prime}_{1},e_{2})\}=2$ implies $-\gamma_{3}=\alpha_{1}$ and $rank\{e_{3},e^{\prime}_{2}=ae_{1}+be_{2},\mu(e_{3},e^{\prime}_{2})\}=2$ implies $\beta_{3}=\alpha_{2}.$ We deduce

[TABLE]

This gives a contradiction.

From this lemma we can find a basis $\{e_{1},e_{2},e_{3}\}$ of $A$ such that

[TABLE]

and $\alpha_{2}\beta_{3}-\alpha_{3}\beta_{2}\neq 0.$ We shall show that this family can be reduced to a family with only $5$ parameters. This condition implies that $\alpha_{2}$ or $\alpha_{3}$ is non zero. We can assume that $\alpha_{3}\neq 0$. In fact, if $\alpha_{3}=0$ and $\alpha_{2}\neq 0$ we have our result. If $\alpha_{2}\neq 0$, then the change of basis $e^{\prime}_{1}=e_{1}-\alpha_{2}/\alpha_{3}e_{2}$ and $e^{\prime}_{i}=e_{i}$ for $i=2,3$ permits to consider $\alpha_{2}=0$. We deduce:

Proposition 14**.**

Let $(A,\mu)$ a $3$-dimensional nonsolvable algebra. Then this algebra is isomorphic to the algebra

[TABLE]

with $\alpha_{3}\beta_{2}\neq 0,$ or

[TABLE]

with $\alpha_{2}\beta_{3}\neq 0$.

The algebras (7) are Lie algebras if and only if $\beta_{3}=\gamma_{2}=\gamma_{3}=0$, that is if we have

[TABLE]

with $\alpha_{3}\beta_{2}\neq 0$ and for the algebras (8) if and only if $\gamma_{2}=\gamma_{3}=0,\beta_{3}=-\alpha_{2}$, that is if we have

[TABLE]

with $\alpha_{2}\neq 0.$

For any generic algebra $\mu$ belonging to the family (7), the rank of $M_{\mu}$ is $8$, the kernel being generated by the vector

$(-\beta_{3}^{2}-\beta_{3}\gamma_{2}\gamma_{3}+\beta_{2}\gamma_{3}^{2},2\beta_{2}\beta_{3}+\beta_{3}\gamma_{2}^{2}-\beta_{2}\gamma_{2}\gamma_{3},-\beta_{3}\gamma_{2}+2\beta_{2}\gamma_{3},-2\alpha_{3}\beta_{3}-\alpha_{3}\gamma_{2}\gamma_{3},\beta_{3}^{2}-\alpha_{3}\gamma_{2}^{2},$ $2\alpha_{3}\gamma_{2}+\beta_{3}\gamma_{3},\alpha_{3}\beta_{3}\gamma_{2}-2\alpha_{3}\beta_{2}\gamma_{3},2\alpha_{3}\beta_{2}\gamma_{2}+\beta_{3}^{2}\gamma_{2}-\beta_{2}\beta_{3}\gamma_{3},\alpha_{3}\gamma_{2}^{2}+\beta_{3}\gamma_{2}\gamma_{3}-\beta_{2}\gamma_{3}^{2})$

and of rank $6$ is case of Lie algebra, that is if $(\gamma_{2},\beta_{3},\gamma_{3})=(0,0,0).$

We have a similar result for the family (8), but in this case the kernel is generated by

$(\beta_{3}\gamma_{2}\gamma_{3}-\beta_{2}\gamma_{3}^{2},-\beta_{3}\gamma_{2}^{2}+\beta_{2}\gamma_{2}\gamma_{3},0,\alpha_{2}\gamma_{3}^{2},-\alpha_{2}\gamma_{2}\gamma_{3},0,\alpha_{2}^{2}\gamma_{3}-\alpha_{2}\beta_{3}\gamma_{3},\alpha_{2}\beta_{3}\gamma_{2}-\beta_{3}^{2}\gamma_{2}+\alpha_{2}\beta_{2}\gamma_{3}+\beta_{2}\beta_{3}\gamma_{3},\alpha_{2}\gamma_{2}\gamma_{3}+\beta_{3}\gamma_{2}\gamma_{3}-\beta_{2}\gamma_{3}^{2}.)$

Proposition 15**.**

Any $3$-dimensional nonsolvable algebra is a simple algebra and belongs to the family (7) or (8). For all these algebras, the automorphism group is of dimension $1$ except if the algebra is a Lie algebra, in this case it is of dimension $3$

5. Applications: Hom-Lie algebras

5.1. Hom-Lie algebras [5]

The notion of Hom-Lie algebras was introduced by Hartwig, Larsson and Silvestrov in [5]. Their principal motivation concerns deformation of the Witt algebra. This Lie algebra is the complexification of the Lie algebra of polynomial vector fields on a circle. A basis for the Witt algebra is given by the vector fields

[TABLE]

for any $n\in\mathbb{Z}.$ The Lie bracket is given by

[TABLE]

The Witt algebra is also viewed as the Lie algebra of derivations of the ring $\mathbb{C}[z,z^{-1}]$. Recall that a derivation on an algebra with product denoted by $ab$ is a linear operator satisfying $D(ab)=D(a)b+aD(b)$. The Lie bracket of two derivations $D$ and $D^{\prime}$ is $[D,D^{\prime}]=D\circ D^{\prime}-D^{\prime}\circ D.$ We can also define a new class of linear operators generalizing derivations, the Jackson derivate, given by

[TABLE]

It is clear that $D_{q}$ is a linear operator, but its behavior on the product is quite different as the classical derivative:

[TABLE]

The authors interpret this relation by putting

[TABLE]

where $\sigma$ is given by $\sigma(f)(z)=f(qz)$ for any $f\in\mathbb{C}[z,z^{-1}]$. Starting from (9) and for a given $\sigma$, one defines a new space of ”derivations” on $C[z,z^{-1}]$ constituted of linear operator $D$ satisfying this relation. With the classical bracket we obtain is a new type of algebra so called $\sigma$-deformations of the Witt algebra. This new approach leads naturally to considerer the space of $\sigma$-derivations, that is, linear operators satisfying (9), to provide it with the multiplication associated with a bracket. This new algebra is not a Lie algebra because the bracket doesn’t satisfies the Jacobi conditions. The authors shows that this bracket satisfies a ”generalized Jacobi condition”. They have called this new class of algebras the class of Hom-Lie algebras. This notion, introduced in [5], since made the object of numerous studies and was also generalized (see [MDL]).

We denote by $\mathcal{H}Lie_{n}$ the subset $ss\mathcal{A}lg_{n}$ whose elements are $n$-dimensional Hom-Lie algebras. We have seen that $ss\mathcal{A}lg_{n}$ is an affine variety isomorphic to $K^{n^{2}(n-1)/2}$. We shall study in particular the case $n=3$ and $n=4$, proving that in dimension $3$ any skew-symmetric algebra is a Hom-Lie algebra and in dimension $4$, $\mathcal{H}Lie_{4}$ is an algebraic hypersurface in $ss\mathcal{A}lg_{4}$. We end this work by the determination of binary quadratic operads whose associated algebras are Hom-Lie.

Definition 16**.**

A Hom-Lie algebra structure on the vector space $V$ is a triple $A=(V,\mu,\alpha)$ consisting of a skew-bilinear map $\mu:V\times V\rightarrow V$ and a linear space homomorphism $f:V\rightarrow V$ satisfying the Hom-Jacobi identity

[TABLE]

for all $x,y,z$ in $A$, where $\circlearrowleft_{x,y,z}$ denotes summation over the cyclic permutations on $x,y,z$.

For example, a Hom-Lie algebra whose endomorphism $f$ is the identity is a Lie algebra. We deduce, since any $2$-dimensional skew-symmetric algebra (the multiplication $\mu$ is a skew-symmetric bilinear map) is a Lie algebra, that any $2$-dimensional algebra is a Hom-Lie algebra.

In the following section we are interested by the determination of all Hom-Lie algebras for small dimensions.

5.2. Hom-Lie algebras of dimension $3$

Any $3$-dimensional skew-symmetric $\mathbb{K}$-algebra $A=(V,\mu)$ is defined by its structure constants $\{\alpha_{i},\beta_{i},\gamma_{i}\}_{i=1,2,3}$ with respect to a given basis $\{e_{1},e_{2},e_{3}\}:$

[TABLE]

Let $f$ be an element of $gl(3,\mathbb{K})$ and consider its matrix in the same basis $\{e_{1},e_{2},e_{3}\}$

[TABLE]

We then define the vector

[TABLE]

For such an algebra we associate the following matrix, $HL_{\mu}$, belonging to $\mathcal{M}(3,9)$ the space of matrices of order $(3\times 9)$ and given by

[TABLE]

where

[TABLE]

Using the notation $ij.k$ in place of $\mu(\mu(e_{i},e_{i}),e_{k})$, we have

[TABLE]

with

[TABLE]

[TABLE]

[TABLE]

Theorem 17**.**

Any skew-symmetric $3$-dimensional algebra is a Hom-Lie algebra.

*Proof. *Consider a basis $\{e_{1},e_{2},e_{3}\}$ of the algebra $(A,\mu)$ and let $(\alpha_{i},\beta_{i},\gamma_{i})$, $i=1,2,3$ be its structure constants defined previously. Let $f$ be in $gl(3,\mathbb{K})$ and let us consider the associated vector

[TABLE]

The endomorphism $f\in gl(3,\mathbb{K})$ satisfies the Hom-Jacobi condition if and only if its corresponding vector $v_{f}=\ ^{t}(a_{1},a_{2},a_{3},b_{1},b_{2},b_{3},c_{1},c_{2},c_{3})$ is in the kernel of the matrix $HL_{\mu}.$ But this matrix belongs to $\mathcal{M}(3,9)$ and represents a linear morphism

[TABLE]

From the rank theorem we have

[TABLE]

Then this kernel is always non trivial and for any algebra $\mu$, there exists a non trivial element in the kernel. Then this algebra always admits a non trivial Hom-Lie structure.

5.3. Classification of Hom-Lie algebras of dimension $3$

In the previous section we have determinate the $3$-dimensional skew-symmetric $\mathbb{K}$-algebra. Since any Hom-Lie algebra is skew-symmetric, we deduce the classification of Hom-Lie algebras. Moreover, for a given skew-symmetric algebra, we can calculate the endomorphisms $f$ associated with the Hom-Lie-Jacobi condition solving the linear system:

[TABLE]

For example, the identity map whose associated vector is

[TABLE]

is in the kernel of $HL_{\mu}$ if and only if $\mu$ satisfies

[TABLE]

that is if it is a Lie algebra. We deduce

Proposition 18**.**

If the Hom-Lie algebras $A=(V,\mu)$ and $A^{\prime}=(V,\mu^{\prime})$ are isomorphic, then the kernels of the associated matrices $HL_{\mu}$ and $HL_{\mu^{\prime}}$ are isomorphic.

5.4. Dimension $4$

Let $A=(V,\mu)$ a $4$-dimensional skew-symmetric $\mathbb{K}$-algebra. Let us choose a basis $\{e_{1},e_{2},e_{3},e_{4}\}$ of $V$ and let us consider the corresponding constants structure of $\mu$:

[TABLE]

This algebra is a Hom-Lie algebra if there exists a linear endomorphism $f$ satisfying the Hom-Lie Jacobi equations. The endomorphism $f$ is represented in the basis $\{e_{1},e_{2},e_{3},e_{4}\}$ by a square matrix $(a_{i,j}$ of order $4$. As in dimension $3$, we consider the vecteur $v_{f}\in\mathbb{K}^{16}$

[TABLE]

Then $f$ satisfies the Hom-Lie conditions if and only if $v_{f}$ is solution on the linear system

[TABLE]

where $HL_{\mu}$ is the square matrix of order $16$:

[TABLE]

and $ij.k$ is the matrix

[TABLE]

The kernel of $HL_{\mu}$ is not trivial if and only if the rank of $HL_{\mu}$ is smaller or equal to $15$. We deduce that $A=(V,\mu)$ is provided with a Hom-Lie structure if the structure constant $C_{i,j}^{k}$ satisfy the homogeneous polynomial equation of degree $16$:

[TABLE]

Proposition 19**.**

The set $\mathcal{HL}_{4}$ of $4$-dimensional $\mathbb{K}$-Hom-Lie algebras is provided with a structure of algebraic variety embedded in $\mathbb{K}^{24}$.

For any $A=(V,\mu)\in\mathcal{HL}_{4}$ we consider the vector space $\ker HL_{\mu}$. We thus define a singular vector bundle $K(\mathcal{HL}_{4})$ whose fiber over $A=(V,\mu)$ is $\ker HL_{\mu}$. This fiber corresponds to the set of Hom-Lie structure which can be defined on a given $4$-dimensional algebra.

Remarks.

1. (1)

In dimension $3$, $\mathcal{HL}_{3}$ is the affine variety $ss\mathcal{A}lg_{3}$ which is isomorphic to the affine space $\mathbb{K}^{9}$. In this case, the fibers of $K(\mathcal{HL}_{3})$ are vector spaces of dimension greater or equal to $6$. 2. (2)

In dimension $4$, we are confronted with the resolution of the equation of degree $16$ with $24$ variables $\det(HL_{\mu})=0.$ If $\mathbb{K}$ is algebraically closed, we can simplify this problem because the endomorphism $f$ admits a general reduced form

[TABLE]

In this case, we have to consider only the reduced matrix $HL^{\prime}_{\mu}$ of order $16\times 7$:

[TABLE]

It remains to verify that there exist at least one point of $ss\mathcal{A}lg_{4}$ which not belong to $\mathcal{HL}_{4}$. Let us consider, for example, the following algebra:

[TABLE]

The associated matrix is

[TABLE]

and its determinant is not zero. This algebra is not Hom-Lie. As consequence we deduce that there exists an open set in $ss\mathcal{A}lg_{4}$ whose elements are not Hom-Lie algebras.

Proposition 20**.**

In the affine plane $ss\mathcal{A}lg_{4}$, there is a Zariski open set whose elements are $4$-dimensional algebras without Hom-Lie structure.

6. Algebras of Lie type of dimension $3$

Since the multiplication $\mu$ is skew-symmetric, any quadratic relation concerning $\mu$ can be reduced to a relation of type

[TABLE]

where $h_{1},h_{2},h_{3}$ are functions $h_{i}:V^{3}\rightarrow V$. In particular, we have the class of algebras of Lie type ([6]) given by the relation

[TABLE]

with $a\neq 0$. For example, In the first section, we have considered the non Lie algebra

[TABLE]

This algebra satisfies the identity

[TABLE]

where $b$ is a trilinear form satisfying $b(e_{1},e_{2},e_{3})=-b(e_{3},e_{1},e_{2})=1.$ It is an algebra of Lie type. An important study of the automorphism group of these algebras is presented in ([6]). In this example, $\dim Aut(\mu)=3$ and the identity component is the group of matrices

[TABLE]

with $ad-bc=1$. It is isomorphic to $SO(2)$. For all the non Lie algebras determined in the above sections, let us determine the algebras of Lie type.

1. (1)

Algebras (6). Such algebras are of Lie type if $\beta_{2}=\beta_{3}=0$, that is if it is also a Lie algebra. 2. (2)

Algebras (7). Such algebras are never algebras of Lie type. 3. (3)

Algebras (8). Such an algebra is a Lie algebra and only if it is a Lie algebra ($\gamma_{2},\beta_{3},\gamma_{3})=(0,0,0))$.Assume $\alpha_{2}=1,\beta_{2}=0$. We have a Lie algebra structure if and only if $(\gamma_{2},\beta_{3},\gamma_{3})=(0,-1,0).$ We have a complex structure of algebra of Lie type, which is not a Lie algebra if and only if $\gamma_{2}=\gamma_{3}=0$ and $\beta_{3}$ is a complex root of $-1$ (we assume here that $\mathbb{K}=\mathbb{C})$ that is $\beta_{3}=-j$ or $-j^{2}$. In this case the relation is

[TABLE]

or

[TABLE]

In this case $M_{\mu}$ is of rank $8$ and the automorphism group is of dimension $1$ and generated bu the automorphism

[TABLE]