Deformation Quantization of Nonassociative Algebras

Abstract

We investigate formal deformations of certain classes of nonassociative algebras including classes of $\mathbb{K}\left[{\Sigma }_3\right]$-associative algebras, Lie-admissible algebras and anti-associative algebras. In a process which is similar to Poisson algebra for the associative case, we identify for each type of algebras $(A,\mu )$ a type of algebras $(A,\mu ,\psi )$ such that formal deformations of $(A,\mu )$ appear as quantizations of $(A,\mu ,\psi )$. The process of polarization/depolarization associates to each nonassociative algebra a couple of algebras which products are respectively commutative and skew-symmetric and it is linked with the algebra obtained from the formal deformation. The anti-associative case is developed with a link with the Jacobi–Jordan algebras.

1. Introduction

In this work, $\mathbb{K}$ is a field of characteristic 0. By a $\mathbb{K}$-algebra, we mean a $\mathbb{K}$-vector space A with a bilinear map $\mu$ and we denote this algebra by $(A,\mu )$. We assume that $\mu$ satisfies a quadratic relation denoted $\mu •\mu =0.$ For example, for the associative case, we have $\mu •\mu =\mu \circ (\mu \otimes Id-Id\otimes \mu )$. The set of n-dimensional algebras satisfying a quadratic relation $\mu •\mu =0$ is an algebraic variety ${\mathcal{V}}_n$ over $\mathbb{K}$, and the classical notion of formal deformation enables a description of neighborhoods of any point of this variety (for a topology adapted to the structure of the algebraic variety). A naive definition of a formal deformation of a point $\mu \in {\mathcal{V}}_n$ is a formal series ${\mu }_t=\mu +{\sum }_{k\ge 1}{t}^k{\phi }_k$ considered as a bilinear map on the $\mathbb{K}\left[\right[t\left]\right]$-algebra $A\left[\right[t\left]\right]=A\otimes \mathbb{K}\left[\right[t\left]\right]$, where the maps ${\phi }_k$ are bilinear maps on A which satisfy quadratic relations resulting from the formal identity

$${\mu }_t•{\mu }_t=0.$$

In particular, we have $\mu •\mu =0$ in degree 0, $\mu •{\phi }_1+{\phi }_1•\mu =0$ in degree 1, and so on. Formal deformations are mainly used for the local study of ${\mathcal{V}}_n$. For example, a point of this variety with only isomorphic deformations is topologically rigid, that is, its orbit is open under the natural action of the linear group. But there are other applications of these deformations. If we consider a formal deformation of a given point $\mu$, it determines new algebra multiplications that are related to the original one. In fact, the linear term ${\phi }_1$ of ${\mu }_t$ is also a multiplication on A whose quadratic relation is a consequence of the degree 1 relation $\mu •{\phi }_1+{\phi }_1•\mu =0$. A fundamental consequence is deformation quantization theory introduced in [1]. In a simplified way, if we consider a formal associative deformation ${\mu }_t={\mu }_0+t{\phi }_1+\cdots$ of a commutative associative algebra $(A,{\mu }_0)$, the first term ${\phi }_1$ is a cocycle for the Hochschild cohomology associated with $(A,{\mu }_0)$ and is also a Lie-admissible multiplication whose associated Lie bracket ${\psi }_1$ satisfies the Leibniz identity with the initial commutative associative multiplication. Then, this formal deformation naturally determines a Poisson algebra $(A,{\mu }_0,{\psi }_1)$, and the algebra $(A\left[\left[t\right]\right],{\mu }_t)$ is a deformation quantization of the Poisson algebra $(A,{\mu }_0,{\psi }_1)$. In [2], we have enlarged this classical notion by considering not formal associative deformations but weakly associative formal deformations by considering the associative algebra ${\mu }_0$ as a weakly associative algebra. In this case, $(A,{\mu }_0,{\psi }_1)$ is still a Poisson algebra.

One of the aims of this work is to extend this construction for nonassociative formal deformations of commutative associative algebras. The nonassociative algebra world is very wide [3]. A description of the algebraic varieties associated with nonassociative laws can be found in [4]. In the present paper, we focus on a class of nonassociative algebras whose quadratic defining relation has symmetric properties linked with the symmetric group and previously studied in [5]. These algebras are called v-algebras, where v is a vector of $\mathbb{K}\left[{\Sigma }_3\right]$ the group-algebra over $\mathbb{K}$ associated with the symmetric group ${\Sigma }_3$. In this context, for a vector v given, any associative is also a v-algebra and we can naturally consider a v-formal deformation of $\mu$. For example, an algebra $(A,\mu )$ is a $(Id+c+c^2)$-algebra if its associator ${\mathcal{A}}_{\mu }$ satisfies the relation

$${\mathcal{A}}_{\mu }(x_1,x_2,x_3)+{\mathcal{A}}_{\mu }(x_2,x_3,x_1)+{\mathcal{A}}_{\mu }(x_3,x_1,x_2)=0$$

(these algebras are called $G_5$-algebras in [5] or $A_3$-associative algebras in [6]). It is clear that any associative algebra is also a $(Id+c+c^2)$-algebra. Thus, we can consider formal deformation ${\mu }_t$ of the associative multiplication ${\mu }_0$ but assume that ${\mu }_t$ is a $(Id+c+c^2)$-algebra.

This leads to generalizing the notion of Poisson algebras. Recall that a Poisson algebra is an algebra $(A,•,\{,\left\}\right)$ with a commutative associative multiplication • and a Lie bracket $\{,\}$ tied up by the Leibniz identity

$${\mathcal{L}}_{•,\{,\}}(x,y,z)=\{x,y•z\}-y•\{x,z\}-\{x,y\}•z=0.$$

In this paper, we introduce the notion of v-Poisson algebras, where v is a vector of $\mathbb{K}\left[{\Sigma }_3\right]$. The axioms of v-Poisson algebras are those of Poisson algebras weakening the Leibniz identity using the vector v. For example, if $v=(Id-{\tau }_{12})$, a $(Id-{\tau }_{12})$-Poisson algebra corresponds to the v-Leibniz rule

$${\mathcal{L}}_{•,\{,\}}(x,y,z)-{\mathcal{L}}_{•,\{,\}}(y,x,z)=0.$$

To obtain quantization deformation of a v-Poisson algebra, we consider v-formal deformation of a commutative associative product. For example, from a $G_5$-formal deformation

$${\mu }_t={\mu }_0+t{\phi }_1+\cdots$$

of a commutative associative product ${\mu }_0$, we obtain the algebra $(A,{\mu }_0,{\psi }_1)$, which is not a Poisson algebra but a v-Poisson algebra.

A useful trick to understand the properties of the algebra obtained by formal deformation or deform a given algebra in a good class is to use the polarization/depolarization process introduced in [7] in the case of Poisson algebras. Considering a nonassociative multiplication, this process consists of looking at the properties of the symmetric and skew-symmetric bilinear applications that are attached to it. We develop in Section 6 the polarization/depolarization process for the algebras studied in the first sections. A similar study on the link between polarization and deformations have been performed in [6]. For example, the polarization/depolarization process applied to a $G_5$-associative algebra $(A,\mu )$ gives a triple $(A,\rho ,\psi )$, where $\rho$ and $\psi$ are the commutative and anti-commutative multiplication associated with $\mu$, which is a nonassociative $(Id+c+c^2)$-Poisson algebra (by nonassociative Poisson algebras, we relax the associativity of the commutative multiplication $\rho$).

The $\mathbb{K}{\left[{\Sigma }_3\right]}^2$-algebra case, which is a generalization of the v-algebra case, including Leibniz algebras, is investigated in Section 5, and we study $\mathbb{K}{\left[{\Sigma }_3\right]}^2$-formal deformations of these algebras. The polarization/depolarization process is also developed in Section 7. A particular look is given to the anti-associative case, that is, related to the relation $\left(xy\right)z+x\left(yz\right)=0$. We recall in Section 5.2 that the corresponding operad is non-Kozsul, the description of the “natural cohomology” and the cohomology of the minimal model which parametrizes the deformations [8]. The deformation quantization process concerns in this case skew-symmetric anti-associative algebras, which are related to anti-Poisson algebras, which are defined in Theorem 3 and where the Lie Poisson bracket is replaced by a Jacobi–Jordan product, also called a mock-Lie product (see [9,10]). So, we obtain Jacobi–Jordan algebras by polarization of anti-associative algebras and we study the corresponding operads and describe free Jacobi–Jordan algebras with a small number of generators.

This paper also gives a generalization of the Leibniz identity in a graded version (see Equation (12)), which gives the usual Leibniz identity for $(\rho ,\psi )$ a couple of (commutative-skew-symmetric) multiplications but also Jacobi identity for $(\psi ,\psi )$ with a skew-symmetric multiplication $\psi$. If we consider it for $(\rho ,\rho )$ with a commutative multiplication $\rho$, we then obtain the Jacobi–Jordan identity. We also obtain for $(\psi ,\rho )$ a couple of (skew-symmetric-commutative) multiplications an identity appearing in the anti-associative algebra case.

2. $\mathbb{K}\left[{\mathbf{\Sigma }}_3\right]$-Associative Algebras

Let ${\Sigma }_3=\{Id,{\tau }_{12},{\tau }_{13},{\tau }_{23},c,c^2\}$ be the symmetric group of degree 3, where c is the cycle $\left(231\right)$ and ${\tau }_{ij}$ is the transposition between i and j. The product $\sigma {\sigma }^{\prime }$ corresponds to the composition $\sigma \circ {\sigma }^{\prime }$. Let $\mathbb{K}\left[{\Sigma }_3\right]$ be the group algebra of ${\Sigma }_3$. It is provided with an associative algebra structure and with a ${\Sigma }_3$-module structure. The left-action of ${\Sigma }_3$ on $\mathbb{K}\left[{\Sigma }_3\right]$ is given by

$$(\sigma \in {\Sigma }_3,v=\sum a_i{\sigma }_i\in \mathbb{K}\left[{\Sigma }_3\right])\to \sum a_i\sigma {\sigma }_i.$$

For any $v\in \mathbb{K}\left[{\Sigma }_3\right],$ the corresponding orbit is denoted by ${\mathcal{O}}_l\left(v\right)=\{v,{\tau }_{12}v,{\tau }_{13}v,{\tau }_{23}v,cv,c^{2}v\}$ or simply $\mathcal{O}\left(v\right)$ and $F_v=\mathrm{Span}\left(\mathcal{O}\left(v\right)\right)$ is the $\mathbb{K}$-linear subspace of $\mathbb{K}\left[{\Sigma }_3\right]$ generated by $\mathcal{O}\left(v\right)$. It is also a ${\Sigma }_3$-module.

Some notations:

(1)

We call the canonical basis of $\mathbb{K}\left[{\Sigma }_3\right]$ the ordered family $\{Id,{\tau }_{12},{\tau }_{13},{\tau }_{23},c,c^2\}$, and $(a_1,a_2,a_3,a_4,a_5,a_6)$ are the coordinates of the vector $v=a_{1}Id+a_2{\tau }_{12}+a_3{\tau }_{13}+a_4{\tau }_{23}+a_{5}c+a_6{c}^2$ in the canonical basis. We denote $M_v$ the matrix composed of the column component vectors of the family $(v,{\tau }_{12}v,{\tau }_{13}v,{\tau }_{23}v,cv,c^{2}v)$ in the canonical basis:

$$M_v=\left(\begin{array}{cccccc}{a}_1& a_2& a_3& a_4& a_6& a_5\\ a_2& a_1& a_5& a_6& a_4& a_3\\ a_3& a_6& a_1& a_5& a_2& a_4\\ a_4& a_5& a_6& a_1& a_3& a_2\\ a_5& a_4& a_2& a_3& a_1& a_6\\ a_6& a_3& a_4& a_2& a_5& a_1\end{array}\right)$$

(2)

Let A be a $\mathbb{K}$-vector space. The symmetric monoidal structure on the category of vector spaces naturally turns $A^{\otimes 3}$ into a representation of ${\Sigma }_3$. We denote this representation by

$${\Phi }^A:{\Sigma }_3\to Aut\left(A^{\otimes 3}\right)\subset End\left(A^{\otimes 3}\right).$$

The universal property of the group algebra allows to extend this representation to

$${\Phi }^A:\mathbb{K}\left[{\Sigma }_3\right]\to End\left(A^{\otimes 3}\right).$$

Thus, if $X,Y,Z$ are three vectors of A, and if we denote ${\Phi }_{\sigma }^A$ instead of ${\Phi }^A\left(\sigma \right)$, we have:

$$\begin{array}{ccccccc}& \hfill (X,Y,Z)& \hfill (Y,X,Z)& \hfill (Z,Y,X)& \hfill (X,Z,Y)& \hfill (Y,Z,X)& \hfill (Z,X,Y)\\ {\Phi }_{Id}^A\hfill & \hfill (X,Y,Z)& \hfill (Y,X,Z)& \hfill (Z,Y,X)& \hfill (X,Z,Y)& \hfill (Y,Z,X)& \hfill (Z,X,Y)\\ {\Phi }_{{\tau }_{12}}^A\hfill & \hfill (Y,X,Z)& \hfill (X,Y,Z)& \hfill (Y,Z,X)& \hfill (Z,X,Y)& \hfill (Z,Y,X)& \hfill (X,Z,Y)\\ {\Phi }_{{\tau }_{13}}^A\hfill & \hfill (Z,Y,X)& \hfill (Z,X,Y)& \hfill (X,Y,Z)& \hfill (Y,Z,X)& \hfill (X,Z,Y)& \hfill (Y,X,Z)\\ {\Phi }_{{\tau }_{23}}^A\hfill & \hfill (X,Z,Y)& \hfill (Y,Z,X)& \hfill (Z,X,Y)& \hfill (X,Y,Z)& \hfill (Y,X,Z)& \hfill (Z,Y,X)\\ {\Phi }_c^A\hfill & \hfill (Y,Z,X)& \hfill (X,Z,Y)& \hfill (Y,X,Z)& \hfill (Z,Y,X)& \hfill (Z,X,Y)& \hfill (X,Y,Z)\\ {\Phi }_{c^2}^A\hfill & \hfill (Z,X,Y)& \hfill (Z,Y,X)& \hfill (X,Z,Y)& \hfill (Y,X,Z)& \hfill (X,Y,Z)& \hfill (Y,Z,X)\end{array}$$

For any $w={\sum }_{i=1}^6{w}_i{\sigma }_i\in \mathbb{K}\left[{\Sigma }_3\right]$,

$${\Phi }_w^A=\sum _{i=1}^6{w}_i{\Phi }_{{\sigma }_i}^A.$$

In particular, for any $v,w\in \mathbb{K}\left[{\Sigma }_3\right]$,

$${\Phi }_w^A\circ {\Phi }_v^A={\Phi }_{vw}^A.$$

Definition 1. 

Consider a nonzero vector $v\in \mathbb{K}\left[{\Sigma }_3\right]$. An algebra $(A,\mu )$ is a v-associative algebra or simply a v-algebra if

$${\mathcal{A}}_{\mu }\circ {\Phi }_v^A=0,$$

where ${\mathcal{A}}_{\mu }$ is the associator of μ, that is, ${\mathcal{A}}_{\mu }(X,Y,Z)=\mu (\mu (X,Y),Z)-\mu (X,\mu (Y,Z)).$

A v-algebra $(A,\mu )$ is also a $v^{\prime }$-algebra if $v^{\prime }\in F_v=\mathrm{Span}\left(\mathcal{O}\left(v\right)\right)$. But for any $v^{\prime }\in F_v$ such that $dim{F}_{v^{\prime }}<dim{F}_v$, the $v^{\prime }$-associativity does not imply the v-associativity. For example, if $v=Id-{\tau }_{12}+c$, the vector $v^{\prime }=Id+{\tau }_{13}$ is in $F_v$. But $dim{F}_{v^{\prime }}=3$, this space is generated by the vectors $Id+{\tau }_{13},{\tau }_{12}+c^2,{\tau }_{13}+c$ and $Id-{\tau }_{12}+c\notin F_{v^{\prime }}$. The $(Id+{\tau }_{13})$-associativity does not imply the $(Id-{\tau }_{12}+c)$-associativity. Of course, v-algebras are the same in that $\sigma \left(v\right)$-algebras for any $\sigma \in {\Sigma }_3$, as we trivially have that $F_v=F_{\sigma \left(v\right)}$ and more generally the class of v-algebras coincides with the class of $v^{\prime }$-algebras if and only if $F_v=F_{v^{\prime }}$.

We obtain from the ${\Sigma }_3$-module structure of $F_v$ that it decomposes in a direct sum of $F_{v_i}$ associated with the irreducible representations of ${\Sigma }_3$. There exist two particular vectors in $\mathbb{K}\left[{\Sigma }_3\right]$ denoted here by $v_{Lad}$ and $v_{3Pa}$ corresponding to the only one-dimensional irreducible signum and trivial representations:

$$v_{Lad}=\sum _{\sigma \in {\Sigma }_3}\epsilon \left(\sigma \right)\sigma \mathrm{and}{v}_{3Pa}=\sum _{\sigma \in {\Sigma }_3}\sigma ,$$

where $\epsilon \left(\sigma \right)$ is the signature of the permutation $\sigma$. The vectors $v_{Lad}$ and $v_{3Pa}$ are the unique vectors v such that $F_v$ is one-dimensional up to a scalar factor.

Proposition 1. 

An algebra $(A,\mu )$ is

1. 

Lie-admissible if and only if it is $v_{Lad}$-associative,

2. 

3-power-associative if and only if it is $v_{3Pa}$-associative.

Proof. 

See [5]. □

The classes of Lie-admissible algebras and power-associative algebras have been introduced by Albert in [11]. An algebra is called Lie-admissible if the skew-symmetric bilinear map $\psi$ related to $\mu$ is a Lie bracket. This is equivalent to write ${\mathcal{A}}_{\mu }\circ {\Phi }_{v_{Lad}}^A=0$. Recently, results on the structure of certain classes of Lie-admissible algebras have been published in [12,13]. An algebra is said to be power-associative if every subalgebra generated by one element is associative. Over a field of characteristic 0, an algebra is power-associative if it satisfies ${\mathcal{A}}_{\mu }(x,x,x)={\mathcal{A}}_{\mu }(x^2,x,x)=0$ for any $x\in A$. An algebra is said to be 3-power-associative if it satisfies ${\mathcal{A}}_{\mu }(x,x,x)=0$ for any $x\in A$. This last condition is equivalent, by linearization, to ${\mathcal{A}}_{\mu }\circ {\Phi }_{v_{3Pa}}^A=0.$

Remark 1. 

If $(A,\mu )$ is 3-power-associative, then ${\mathcal{A}}_{\mu }\circ {\Phi }_{v_{3Pa}}^A=0$, which implies ${\mathcal{A}}_{\mu }(x^2,x,x)+{\mathcal{A}}_{\mu }(x,x^2,x)+{\mathcal{A}}_{\mu }(x,x,x^2)=0.$ In fact

$${\mathcal{A}}_{\mu }\circ {\Phi }_{v_{3Pa}}^A(x_1,x_1,x_3)=2({\mathcal{A}}_{\mu }(x_1,x_1,x_3)+{\mathcal{A}}_{\mu }(x_1,x_3,x_1)+{\mathcal{A}}_{\mu }(x_3,x_1,x_1)=0$$

for any $x_1,x_3$. But we also have ${\mathcal{A}}_{\mu }(x^2,x,x)-{\mathcal{A}}_{\mu }(x,x^2,x)+{\mathcal{A}}_{\mu }(x,x,x^2)=0.$ In fact, since $x{x}^2=x^{2}x$, then

$${\mathcal{A}}_{\mu }(x^2,x,x)-{\mathcal{A}}_{\mu }(x,x^2,x)+{\mathcal{A}}_{\mu }(x,x,x^2)=x^2\left(x^2\right)-\left(x^{2}x\right)x-x\left(x^{2}x\right)+\left(x{x}^2\right)x+x\left(x{x}^2\right)-x^2{x}^2=0.$$

We deduce ${\mathcal{A}}_{\mu }(x,x^2,x)=0$ and ${\mathcal{A}}_{\mu }(x^2,x,x)+{\mathcal{A}}_{\mu }(x,x,x^2)=0.$ Then, a 3-power-associative algebra is power-associative if and only if ${\mathcal{A}}_{\mu }(x^2,x,x)-{\mathcal{A}}_{\mu }(x,x,x^2)=0.$ So, a sufficient condition for a 3-power-associative to be power-associative is ${\mathcal{A}}_{\mu }\circ {\Phi }_{Id-{\tau }_{13}}=0.$

There is a third irreducible representation of the group ${\Sigma }_3$, the first two being associated with the vectors $v_{Lad}$ and $v_{3Pa}$. It is a representation of degree 2. It will be used later when we give the classification of v-algebras, which are Lie-admissible or 3-power-associative algebras using the rank of v.

3. Formal Deformations of $\mathit{v}$-Algebras

3.1. Generalities

Let $(A,{\mu }_0)$ be a v-algebra where $v=\sum a_i{\sigma }_i\in \mathbb{K}\left[{\Sigma }_3\right]$. Let ${\phi }_1$ and ${\phi }_2$ be two bilinear maps on A. We define the trilinear maps on A

$${\phi }_1•{\phi }_2={\phi }_1\circ ({\phi }_2\otimes Id)-{\phi }_1\circ (Id\otimes {\phi }_2)$$

and

$${\phi }_1{•}_v{\phi }_2=({\phi }_1\circ ({\phi }_2\otimes Id)-{\phi }_1\circ (Id\otimes {\phi }_2))\circ {\Phi }_v=({\phi }_1•{\phi }_2)\circ {\Phi }_v.$$

Let $(A,{\mu }_0)$ be a v-algebra. A v-formal deformation of $(A,{\mu }_0)$ is given by a family of bilinear maps on A

$$\{{\phi }_j:A\otimes A\to A,j\in \mathbb{N}\}$$

with ${\phi }_0={\mu }_0$ and satisfying

$$\sum _{i+j=k,i,j\ge 0}{\phi }_i{•}_v{\phi }_j=0,k\ge 0.$$

(1)

If we denote by $\mathbb{K}\left[\right[t\left]\right]$ the algebra of formal series with one indeterminate t, this definition is equivalent to consider on the space $A\left[\right[t\left]\right]=A\otimes \mathbb{K}\left[\right[t\left]\right]$ (recall that A is of finite dimensional) of formal series with coefficients in A a structure of $\mathbb{K}\left[\right[t\left]\right]$-v-associative algebra such that the canonical map $A\left[\right[t\left]\right]/tA\left[\right[t\left]\right]\to A$ is an isomorphism of v-algebras. It is useful to write

$${\mu }_t={\mu }_0+t{\phi }_1+t^2{\phi }_2+\cdots$$

Equation (1) implies at the order $k=0$ that ${\mu }_0$ is v-associative. The order $k=1$ writes

$${\mu }_0{•}_v{\phi }_1+{\phi }_1{•}_v{\mu }_0=0.$$

To be consistent with the conventional cohomological approaches to deformations, we will denote by ${\delta }_{v,{\mu }_0}^2\phi$ the trilinear map

$${\delta }_{v,{\mu }_0}^2\phi ={\mu }_0{•}_v\phi +\phi {•}_v{\mu }_0.$$

In fact, we know that a cohomological complex which parametrizes formal deformations of algebras over a quadratic operad exists, and ${\delta }_{v,{\mu }_0}^2$ corresponds to the second coboundary operator. For example, if $v=Id$, then $(A,{\mu }_0)$ is associative and ${\delta }_{Id,{\mu }_0}^2$ is the coboundary operator associated with the Hochschild complex of A classically denoted by ${\delta }_{H,{\mu }_0}^2$ and we have

$${\delta }_{H,{\mu }_0}^2\phi (x,y,z)=-x\phi (y,z)+\phi (xy,z)-\phi (x,yz)+\phi (x,y)z=({\mu }_0•\phi +\phi •{\mu }_0)(x,y,z)$$

where, to simplify the notations, $xy$ means ${\mu }_0(x,y)$. Then

$${\delta }_{v,{\mu }_0}^2\phi ={\delta }_{H,{\mu }_0}^2\phi \circ {\Phi }_v.$$

Coming back to Equation (1), we obtain

$$\begin{array}{cc}\mathrm{order}0:\hfill & {\mathcal{A}}_{{\mu }_0}\circ {\Phi }_v=0,\hfill \\ \mathrm{order}1:\hfill & {\delta }_{v,{\mu }_0}^2{\phi }_1=0,\hfill \\ \mathrm{order}2:\hfill & {\phi }_1{•}_v{\phi }_1+{\delta }_{v,{\mu }_0}^2{\phi }_2=0.\hfill \end{array}$$

(2)

Let $v_1$ be a vector in $\mathbb{K}\left[{\Sigma }_3\right]$. Then, ${\phi }_1{•}_v{\phi }_1\circ {\Phi }_{v_1}=0$, that is, $(A,{\phi }_1)$ is a $v_{1}v$-algebra, if and only if ${\delta }_{v,{\mu }_0}^2{\phi }_2\circ {\Phi }_{v_1}=0.$ But ${\delta }_{v,{\mu }_0}^2{\phi }_2\circ {\Phi }_{v_1}={\delta }_{H,{\mu }_0}^2{\phi }_2\circ {\Phi }_{v_{1}v}.$ So, we will look when ${\delta }_{H,{\mu }_0}^2{\phi }_2\circ {\Phi }_{v_{1}v}=0$ is satisfied but asking moreover the commutativity of the multiplication ${\mu }_0$. This new hypothesis will be justified in the study of deformation quantization in Section 3.2.

3.2. Case of a Commutative v-Algebra $(A,{\mu }_0)$

Lemma 1. 

Let $(A,{\mu }_0)$ be a commutative algebra with ${\mu }_0\ne 0$ and ${\delta }_{H,{\mu }_0}^2$ the Hochschild coboundary operator:

$${\delta }_{H,{\mu }_0}^2\phi (X,Y,Z)=-X\phi (Y,Z)+\phi (XY,Z)-\phi (X,YZ)+\phi (X,Y)Z$$

where $X,Y,Z\in A$, the map φ is bilinear on A and $XY$ denotes the product ${\mu }_0(X,Y)$. Then, ${\delta }_{H,{\mu }_0}^2\phi \circ {\Phi }_{v_{Lad}}=0.$

Proof. 

It is easy to see that

$$\begin{array}{cc}\hfill {\delta }_{H,{\mu }_0}^2\phi \circ {\Phi }_{v_{Lad}}(X_1,X_2,X_3)=& \sum _{\sigma \in \mathbb{K}\left[{\Sigma }_3\right]}\epsilon \left(\sigma \right){\delta }_{H,{\mu }_0}^2\phi (X_{\sigma \left(1\right)},X_{\sigma \left(2\right)},X_{\sigma \left(3\right)})\hfill \\ \hfill =& \sum _{\sigma \in \mathbb{K}\left[{\Sigma }_3\right]}\epsilon \left(\sigma \right)(\phi (X_{\sigma \left(1\right)},X_{\sigma \left(2\right)})X_{\sigma \left(3\right)}-X_{\sigma \left(1\right)}\phi (X_{\sigma \left(2\right)},X_{\sigma \left(3\right)}))\hfill \\ & +\sum _{\sigma \in \mathbb{K}\left[{\Sigma }_3\right]}\epsilon \left(\sigma \right)\phi (X_{\sigma \left(1\right)}{X}_{\sigma \left(2\right)},X_{\sigma \left(3\right)})\hfill \\ & -\sum _{\sigma \in \mathbb{K}\left[{\Sigma }_3\right]}\epsilon \left(\sigma \right)\phi (X_{\sigma \left(1\right)},X_{\sigma \left(2\right)}{X}_{\sigma \left(3\right)})\hfill \end{array}$$

The commutativity of ${\mu }_0$ implies the cancellation of each term. □

Let us apply Lemma 1 to study, for a commutative v-associative algebra $(A,{\mu }_0)$, the equation

$${\phi }_1{•}_v{\phi }_1+{\delta }_{v,{\mu }_0}^2{\phi }_2=0$$

For any $v_1\in \mathbb{K}\left[{\Sigma }_3\right]$, we have

$${\phi }_1{•}_v{\phi }_1\circ {\Phi }_{v_1}+{\delta }_{v,{\mu }_0}^2{\phi }_2{\Phi }_{v_1}={\phi }_1{•}_{v_{1}v}{\phi }_1+{\delta }_{v_{1}v,{\mu }_0}^2{\phi }_2=0.$$

There is an obvious solution to this equation corresponding to the case $v_{1}v=0$ but which doest not lead to any properties on ${\phi }_1.$ If $M_v$ is the matrix associated with v, then the equation $v_{1}v=0$ corresponds to the linear system $M_v{V}_1=0$ where $V_1$ is the column matrix of the vector $v_1$ and $v_{1}v=0$ if and only if $V_1\in ker{M}_v$. The rank of $M_v$ is the dimension of $Span\left(\mathcal{O}\left(v\right)\right)=F_v$ and it is maximal if and only if $Id\in F_v$. In this case, ${\mu }_0$ is associative. In all the other cases, $rank\left(M_v\right)<6$ and $dimker{M}_v\ge 1$. For example, if $v=v_{Lad}$, then $dim{F}_v=1$ and $v_1{v}_{Lad}\in F_{v_{Lad}}$ for any $v_1\in \mathbb{K}\left[{\Sigma }_3\right]$. More precisely, if $v_1=(a_1,a_2,a_3,a_4,a_5,a_6)\in \mathbb{K}\left[{\Sigma }_3\right]$, then $v_1{v}_{Lad}=\lambda v_{Lad}$ with $\lambda =a_1-a_2-a_3-a_4+a_5+a_6$ and $\{v_1\in \mathbb{K}\left[{\Sigma }_3\right];v_1{v}_{Lad}=0\}$ is a 5-dimensional subspace of $\mathbb{K}\left[{\Sigma }_3\right]$ defined by the linear equation $a_1-a_2-a_3-a_4+a_5+a_6=0$. More generally, let v be in $\mathbb{K}\left[{\Sigma }_3\right]$. We consider the equation $v_{1}v=v_{Lad}$, which corresponds to the linear system $M_v{V}_1=V_{Lad}$. If $v_1\notin ker{M}_v$, the equation $v_{1}v=v_{Lad}$ is equivalent to $v_{Lad}\in \mathrm{Im}{M}_v$.

Proposition 2. 

Let $(A,{\mu }_0)$ be a commutative v-algebra and ${\mu }_t={\mu }_0+t{\phi }_1+t^2{\phi }_2+\cdots$ be a v-formal deformation of ${\mu }_0$. Then, ${\phi }_1$ is a Lie-admissible multiplication on A if $v_{Lad}\in \mathrm{Im}{M}_v$, that is, $v_{Lad}\in F_v$.

Proof. 

If $v_{Lad}\in F_v$, then there exists $v_1$ such that $v_{1}v=v_{Lad}$ and

$${\phi }_1{•}_{v_{1}v}{\phi }_1+{\delta }_{v_{1}v,{\mu }_0}^2{\phi }_2={\phi }_1{•}_{v_{Lad}}{\phi }_1+{\delta }_{v_{Lad},{\mu }_0}^2{\phi }_2=0.$$

From Lemma 1, we deduce

$${\phi }_1{•}_{v_{Lad}}{\phi }_1=0.$$

□

Theorem 1. 

Let $w=a_{1}Id+a_2{\tau }_{12}+a_3{\tau }_{13}+a_4{\tau }_{23}+a_{5}c+a_6{c}^2$ be a vector of $\mathbb{K}\left[{\Sigma }_3\right]$.

1. 

$a_1-a_2-a_3-a_4+a_5+a_6\ne 0$ if and only if $v_{Lad}\in F_w$.

2. 

$a_1+a_2+a_3+a_4+a_5+a_6\ne 0$ if and only if $v_{3Pa}\in F_w$.

Proof. 

In fact, if $\lambda =a_1-a_2-a_3-a_4+a_5+a_6\ne 0$, then $\lambda$ is a non null eigenvalue of $M_w$ and $v_{Lad}$ is an eigenvector corresponding to $\lambda$. In this case, $v_{Lad}\in \mathrm{Im}{M}_w$, that is, $v_{Lad}\in F_w$. Suppose now that $v_{Lad}\in F_w$. Then, it implies that there exists $v={\lambda }_{1}Id+{\lambda }_2{\tau }_{12}+{\lambda }_3{\tau }_{13}+{\lambda }_4{\tau }_{23}+{\lambda }_{5}c+{\lambda }_6{c}^2$ such that $vw=v_{Lad}$, that is,

$$\begin{array}{c}\hfill M_{w}V={}^t(1,-1,-1,-1,1,1)\iff \begin{array}{c}{L}_1\hfill \\ L_2\hfill \\ L_3\hfill \\ L_4\hfill \\ L_5\hfill \\ L_6\hfill \end{array}\left\{\begin{array}{c}{a}_1{\lambda }_1+a_2{\lambda }_2+a_3{\lambda }_3+a_4{\lambda }_4+a_6{\lambda }_5+a_5{\lambda }_6=1\hfill \\ a_2{\lambda }_1+a_1{\lambda }_2+a_5{\lambda }_3+a_6{\lambda }_4+a_4{\lambda }_5+a_3{\lambda }_6=-1\hfill \\ a_3{\lambda }_1+a_6{\lambda }_2+a_1{\lambda }_3+a_5{\lambda }_4+a_2{\lambda }_5+a_4{\lambda }_6=-1\hfill \\ a_4{\lambda }_1+a_5{\lambda }_2+a_6{\lambda }_3+a_1{\lambda }_4+a_3{\lambda }_5+a_2{\lambda }_6=-1\hfill \\ a_5{\lambda }_1+a_4{\lambda }_2+a_2{\lambda }_3+a_3{\lambda }_4+a_1{\lambda }_5+a_6{\lambda }_6=1\hfill \\ a_6{\lambda }_1+a_3{\lambda }_2+a_4{\lambda }_3+a_2{\lambda }_4+a_5{\lambda }_5+a_1{\lambda }_6=1\hfill \end{array}\right..\end{array}$$

Consider $L_1-L_2-L_3-L_4+L_5+L_6$; the system implies that $(a_1-a_2-a_3-a_4+a_5+a_6)({\lambda }_1-{\lambda }_2-{\lambda }_3-{\lambda }_4+{\lambda }_5+{\lambda }_6)=6$, that is, $a_1-a_2-a_3-a_4+a_5+a_6\ne 0$.

We apply the same technique for the 3-power associative case to prove that $a_1+a_2+a_3+a_4+a_5+a_6\ne 0$ if and only if $v_{3Pa}\in F_w.$ □

• Note that this result can be interpreted in terms of representations of $\mathbb{K}\left[{\Sigma }_3\right]$. The projections to the isotypic components of the regular representation of $\mathbb{K}\left[{\Sigma }_3\right]$ are given by right multiplication by the Young symmetrizers. In the case of the trivial and sign component, the Young symmetrizers are proportional to $v_{3Pa}$ and $v_{Lad}$ so that the subrepresentation $F_w$ generated by a vector w contains a copy of the sign representation $F_{v_{Lad}}$ if and only if $w{v}_{Lad}\ne 0.$

Remark 2. 

Consider a commutative product ${\mu }_0$ and $w=a_{1}Id+a_2{\tau }_{12}+a_3{\tau }_{13}+a_4{\tau }_{23}+a_{5}c+a_6{c}^2$ a vector of $\mathbb{K}\left[{\Sigma }_3\right]$. The commutativity of ${\mu }_0$ implies

$$\begin{array}{c}{\delta }_{H,{\mu }_0}^2\phi \circ {\Phi }_w(X,Y,Z)=\hfill \\ (a_5-a_1)X\phi (Y,Z)+(a_3-a_4)X\phi (Z,Y)+(a_4-a_2)Y\phi (X,Z)+(a_6-a_5)Y\phi (Z,X)\hfill \\ +(a_1-a_6)Z\phi (X,Y)+(a_2-a_3)Z\phi (Y,X)+(a_1+a_2)\phi (XY,Z)+(a_3+a_5)\phi (YZ,X)\hfill \\ +(a_4+a_6)\phi (XZ,Y)-(a_1+a_4)\phi (X,YZ)-(a_2+a_5)\phi (Y,XZ)-(a_3+a_6)\phi (Z,XY).\hfill \end{array}$$

(3)

Remark that we reobtain Lemma 1 considering $a_1=-a_2=-a_3=-a_4=a_5=a_6$.

If $dimA=1$, that is, $A=\mathbb{K}$, since ${\mu }_0\ne 0$, any bilinear form writes $\phi =a{\mu }_0$ with $a\in \mathbb{K}$ and satisfies ${\delta }_{H,{\mu }_0}^2\phi \circ {\Phi }_w=0$ with no assumption on $w.$

Suppose that $dimA\ge 2$. Equation (3) with $X=Y=Z$ reduces to

$$(a_1+a_2+a_3+a_4+a_5+a_6)(\phi (XX,X)-\phi (X,XX))=0.$$

Since $A^2\ne 0$, we can choose X such that $XX\ne 0$. If $\phi$ is such that $\phi (XX,X)-\phi (X,XX)\ne 0$, we obtain

$$a_1+a_2+a_3+a_4+a_5+a_6=0$$

and $w=a_{1}Id+a_2{\tau }_{12}+a_3{\tau }_{13}+a_4{\tau }_{23}+a_{5}c-(a_1+a_2+a_3+a_4+a_5)c^2$. Then, from Theorem 1, $v_{3Pa}\notin F_w$. For example, if $\phi$ is skew-symmetric with $\phi (XX,X)\ne 0$ for some $X\in A$, the fact that ${\delta }_{H,{\mu }_0}^2\phi \circ {\Phi }_w=0$ implies that $v_{3Pa}\notin F_w$.

4. Deformation Quantization of the $\mathit{v}$-Algebras with ${\mathit{v}}_{\mathit{Lad}}\in {\mathit{F}}_{\mathit{v}}$

Recall that the rank of a vector $v\in \mathbb{K}\left[{\Sigma }_3\right]$ is the dimension of the vector space $F_v=\mathrm{Span}\left(\mathcal{O}\left(v\right)\right)$. If $v\ne 0$, then $1\le \mathrm{rk}\left(v\right)\le 6.$ If $\mathrm{rk}\left(v\right)=6$, we have $F_v=\mathbb{K}\left[{\Sigma }_3\right]$ and $Id\in F_v$. In this case, any v-associative algebra is associative and we can assume that $v=Id$. Similarly, if $\mathrm{rk}\left(v\right)=1$, then $dim{F}_v=1$ and it is a one-dimensional invariant subspace of $\mathbb{K}\left[{\Sigma }_3\right]$. We have seen that, in this case, $v=v_{Lad}=Id-{\tau }_{12}-{\tau }_{13}-{\tau }_{23}+c+c^2$ or $v=v_{3Pa}=Id+{\tau }_{12}+{\tau }_{13}+{\tau }_{23}+c+c^2$. In this section, we will focus on v-algebras such as $v_{Lad}\in F_v$ because of Proposition 2. In [5], we have the following result:

Theorem 2. 

Every Lie-admissible v-algebra $(A,{\mu }_0)$ corresponds to one of the following types:

1. 

Type (I): $dim{F}_v=1$ and $F_v=F_{v_{Lad}}$.

2. 

Type (II): $dim{F}_v=2$ and $v=Id+c+c^2$. The corresponding v-algebras are also 3-power-associative algebras. These algebras correspond to the $G_5$-associative algebras.

3. 

Type (III): $dim{F}_v=3$. The corresponding v-algebras satisfy:

$$v=\alpha Id-\alpha {\tau }_{12}+(\alpha +\beta -3){\tau }_{13}-\beta {\tau }_{23}+\beta c+(3-\alpha -\beta )c^2$$

with $(\alpha ,\beta )\ne (1,1)$.

4. 

Type (IV): $dim{F}_v=4.$ The v-algebras are of the following type:

(a) 

(IV.1): $v=2Id+(1+t){\tau }_{12}+{\tau }_{13}+c+(1-t)c^2$ with $t\ne 1$,

(b) 

(IV.2): $v=2Id+{\tau }_{12}+{\tau }_{23}+c+c^2.$

5. 

Type (V): $dim{F}_v=5$ and $v=2Id-{\tau }_{12}-{\tau }_{13}-{\tau }_{23}+c.$

6. 

Type (VI): $dim{F}_v=6.$ This corresponds to the class of associative algebras, that is, $v=Id$.

The v-algebras associated with the vector $v=Id+c+c^2$ have $\mathcal{O}\left(v\right)=\{Id+c+c^2,{\tau }_{12}+{\tau }_{13}+{\tau }_{23}\}$ and $dim{F}_v=2.$ For example, $(A,\mu )$ with $\mu$ skew-symmetric is a v-algebra if and only if it is a Lie algebra.

In [5]. we have studied particular classes of v-algebras called G-associative algebras whose defining quadratic relation is associated with subgroups of ${\Sigma }_3$. Consider $G_1=\left\{Id\right\},G_2=\{Id,{\tau }_{12}\},G_3=\{Id,{\tau }_{13}\},G_4=\{Id,{\tau }_{23}\},G_5=A_3=\{Id,c,c^2\}$ and $G_6={\Sigma }_3$ the subgroups of ${\Sigma }_3$. A $G_i$-associative algebra is defined by the relation given by $v_i$-associative algebra with

$$v_i=\sum _{\sigma \in G_i}\epsilon \left(\sigma \right)\sigma .$$

In particular, $G_1$-associative algebras correspond to the associative algebras and $G_6$-associative algebras to the Lie-admissible algebras. These algebras, as well as the case $G_5=A_3$, have been studied previously. The remaining cases are associated with a vector v of rank 3: the $G_2$-associative algebras also called Vinberg algebras and associated with the vector $v=Id-{\tau }_{12}$ correspond to $\alpha =3,\beta =0$; the $G_3$-associative algebras also called Pre-Lie algebras and associated with the vector $v=Id-{\tau }_{13}$ correspond to $\alpha =0,\beta =0$ and finally, the $G_4$-associative algebras, associated with the vector $v=Id-{\tau }_{23}$, correspond to $\alpha =0,\beta =3$. We begin this study by the more classical case corresponding to an associative and commutative multiplication ${\mu }_0$.

4.1. Rank$\left(v\right)=6$: $v=Id$: The Associative Case

The study of deformations of associative algebras was initiated by Gerstenhaber [14] and deformation quantization by Bayen, Flato, Fronsdal, Lichnerowicz and Sternheimer in [1]. In a first step, we summarize this study as part of the v-associative algebras.

When $v=Id$, a v-algebra is an associative algebra. Let $\mu ={\mu }_0+t{\phi }_1+t^2{\phi }_2+\cdots$ be an associative formal deformation of a commutative associative multiplication ${\mu }_0$. In this case, ${\delta }_{Id,{\mu }_0}^2={\delta }_{H,{\mu }_0}^2$ and Equation (2) writes

$$\left\{\begin{array}{cc}\mathrm{order}1\hfill & {\delta }_{H,{\mu }_0}^2{\phi }_1=0,\hfill \\ \mathrm{order}2\hfill & {\phi }_1•{\phi }_1+{\delta }_{H,{\mu }_0}^2{\phi }_2=0.\hfill \end{array}\right.$$

From Proposition 2, as $v_{Lad}\in F_{Id}$, ${\phi }_1$ is a Lie-admissible multiplication.

The bilinear map ${\phi }_1$ also satisfies ${\delta }_{H,{\mu }_0}^2{\phi }_1=0$ and so ${\delta }_{H,{\mu }_0}^2{\phi }_1\circ {\Phi }_v=0$ for any $v\in \mathbb{K}\left[{\Sigma }_3\right]$. Let us determine a vector such that this relation involves a relation on the skew-bilinear map ${\psi }_1$ attached to ${\phi }_1$, that is, ${\psi }_1(x,y)={\phi }_1(x,y)-{\phi }_1(y,x)$. If $v=a_{1}Id+a_2{\tau }_{12}+a_3{\tau }_{13}+a_4{\tau }_{23}+a_{5}c+a_6{c}^2$ with

$$a_5=a_1-a_3+a_4,a_6=a_1+a_2-a_3,$$

then, writing $xy$ for ${\mu }_0(x,y)$, we have $xy=yx$ and ${\delta }_H^2{\phi }_1\circ {\Phi }_v=0$ is equivalent to

$$\begin{array}{c}{a}_1({\psi }_1(xy,z)+{\psi }_1(xz,y)+{\psi }_1(zy,x))+a_2(y{\psi }_1(x,z)+z{\psi }_1(x,y)-{\psi }_1(xz,y)\hfill \\ -{\psi }_1(xy,z))+a_3({\psi }_1(xz,y)+x{\psi }_1(y,z)-z{\psi }_1(x,y))+a_4(-x{\psi }_1(y,z)-y{\psi }_1(x,z)-{\psi }_1(xz,y)\hfill \\ -{\psi }_1(zy,x))=0\hfill \end{array}$$

for any $a_1,a_2,a_3,a_4\in \mathbb{K}$. This is equivalent to

$$\left\{\begin{array}{c}{\psi }_1(xy,z)+{\psi }_1(xz,y)+{\psi }_1(zy,x)=0,\hfill \\ y{\psi }_1(x,z)+z{\psi }_1(x,y)-{\psi }_1(xz,y)-{\psi }_1(xy,z)=0,\hfill \\ {\psi }_1(xz,y)+x{\psi }_1(y,z)-z{\psi }_1(x,y)=0,\hfill \\ -x{\psi }_1(y,z)-y{\psi }_1(x,z)-{\psi }_1(xz,y)-{\psi }_1(zy,x)=0.\hfill \end{array}\right.$$

The third identity is the Leibniz identity between the Lie bracket ${\psi }_1$ and the commutative associative multiplication ${\mu }_0$. Since the other identities are consequences of the Leibniz identity, we find the classical result.

Proposition 3. 

If ${\mu }_t={\mu }_0+t{\phi }_1+t^2{\phi }_2+\cdots$ is an associative formal deformation of the commutative associative multiplication ${\mu }_0$ on A, then $(A,{\mu }_0,{\psi }_1)$ is a Poisson algebra and the formal deformation $(A\left[\left[t\right]\right],{\mu }_t)$ is a deformation quantization of this Poisson algebra.

In this proposition, we see that any associative deformation of the commutative associative algebra $(A,{\mu }_0)$ gives a quantization. But, are there v-formal deformations of ${\mu }_0$ with $v\in \mathbb{K}\left[{\Sigma }_3\right]$ but $v\notin {\Sigma }_3$ which define a deformation quantization of a Poisson algebra $(A,{\mu }_0,\psi )$ for some Lie bracket $\psi$? In [2], we show that there exists a class of nonassociative algebras, called weakly associative algebras, corresponding to the vector $v=Id-{\tau }_{12}+c$ that answers the previous question: any v-formal deformation of a commutative associative algebra defines a deformation quantization of a Poisson algebra. Since the vector $v=Id-{\tau }_{12}+c$, associated with the weakly associative algebra, is of rank 4, we will briefly recall this study in the paragraph dedicated to rank 4.

4.2. Rank$\left(v\right)=1$: $F_v=F_{v_{Lad}}$, That Is, the Lie-Admissible Algebras

Let $(A,{\mu }_0)$ be a commutative $v_{Lad}$-algebra. Let ${\mu }_t={\mu }_0+\sum t^i{\phi }_i$ be a $v_{Lad}$-deformation of ${\mu }_0$. Remark that a commutative product is always Lie-admissible. From Lemma 1, since ${\mu }_0$ is commutative, for any bilinear map $\phi$, we have ${\delta }_{H,{\mu }_0}^2\phi \circ {\Phi }_{v_{Lad}}=0$. This implies that for any $i\ge 1$, ${\delta }_{v_{Lad},{\mu }_0}^2{\phi }_i=0.$ In particular, ${\phi }_1$ is Lie-admissible.

Proposition 4. 

Let $(A,{\mu }_0)$ be a commutative Lie-admissible algebra. For any bilinear map φ, we have

$${\delta }_{v_{Lad},{\mu }_0}^2\phi =0.$$

If ${\mu }_t={\mu }_0+t{\phi }_1+t^2{\phi }_2+\cdots$ is a Lie-admissible formal deformation of ${\mu }_0$, then the algebra $(A,{\phi }_1)$ is Lie-admissible.

Let us note that if ${\mu }_0$ is a skew-symmetric Lie-admissible multiplication (non-necessarily commutative), that is, ${\mu }_0$ is a Lie bracket, then

$${\delta }_{v_{Lad},{\mu }_0}^2\phi =2{\delta }_{CE,{\mu }_0}^2\psi$$

where $\psi$ is the skew-symmetric bilinear map associated to $\phi$ and ${\delta }_{CE,{\mu }_0}^2$ the coboundary operator of the Chevalley Eilenberg cohomology of the Lie algebra $(A,{\mu }_0)$.

4.3. Rank$\left(v\right)=2:F_v=F_{Id+c+c^2}$, That Is, $G_5$-Algebra or $A_3$-Associative Algebra

Recall that for any $w\in \mathbb{K}\left[{\Sigma }_3\right]$, $F_w$ is a ${\Sigma }_3$-invariant vector space so a direct sum of irreducible vector spaces. The irreducible vector spaces are one-dimensional, that is, $F_{v_{Lad}}$ and $F_{v_{3Pa}}$ or two-dimensional, that is, $F_{\alpha (Id-{\tau }_{12})+\beta (c-{\tau }_{23})+(\alpha +\beta )({\tau }_{13}-c^2)}$ with $\alpha ,\beta \in \mathbb{R}.$ As we considered that $v_{Lad}\in F_v$, we have that $F_v=F_{v_{Lad}}\oplus F_{v_{3Pa}}$, and we can assume that $v=Id+c+c^2$.

Any commutative multiplication ${\mu }_0$ satisfies ${\mu }_0•{\mu }_0\circ {\Phi }_v=0$, implying that $(A,{\mu }_0)$ is a v-algebra. We have, for any bilinear map $\phi$ on A:

$${\delta }_{v,{\mu }_0}^2\phi (x,y,z)={\delta }_{H,{\mu }_0}^2\phi \circ {\Phi }_v(x,y,z)=\psi (xy,z)+\psi (yz,x)+\psi (zx,y)$$

for any $x,y,z\in A$, where $\psi$ is the skew-symmetric map associated to $\phi$. Let ${\mu }_t={\mu }_0+\sum t^i{\phi }_i$ be a v-formal deformation of ${\mu }_0$. Since $v_{Lad}\in F_v$ (more precisely $v_{Lad}v=3v_{Lad}$), ${\phi }_1$ is a Lie-admissible multiplication. Moreover, ${\delta }_{v,{\mu }_0}^2{\phi }_1=0$ and the Lie bracket ${\psi }_1$ satisfies

$${\psi }_1(xy,z)+{\psi }_1(yz,x)+{\psi }_1(zx,y)=0$$

for any $x,y,z\in A.$

Proposition 5. 

Consider the vector $v=Id+c+c^2$ of $\mathbb{K}\left[{\Sigma }_3\right]$ and $(A,{\mu }_0)$ a commutative algebra. Then, $(A,{\mu }_0)$ is a v-algebra and for any v-formal deformation $\mu ={\mu }_0+t{\phi }_1+t^2{\phi }_2+\cdots$ of ${\mu }_0$, $(A,{\psi }_1)$ is a Lie algebra such that

$${\psi }_1(xy,z)+{\psi }_1(yz,x)+{\psi }_1(zx,y)=0$$

(4)

for any $x,y,z\in A,$ where ${\psi }_1$ is the skew-symmetric bilinear map attached to ${\phi }_1$.

As the Leibniz rule implies Equation (4), we just have the conditions of a nonassociative Poisson algebra but replacing the Leibniz identity by a weak Leibniz identity (4) and we can define a notion of v-Poisson and nonassociative v-Poisson algebras:

Definition 2. 

A nonassociative v-Poisson algebra is $\mathbb{K}$-vector space A with a Lie bracket ψ and a commutative multiplication μ tied up by the v-Leibniz identity:

$$\mathcal{L}(\mu ,\psi )\circ {\Phi }_v=0.$$

A v-Poisson algebra $(A,\mu ,\psi )$ is a nonassociative v-Poisson such that μ is moreover associative.

We trivially have that a Poisson algebra is a v-Poisson algebra and also a nonassociative v-Poisson algebra. We then obtain

Proposition 6. 

Let $\mu ={\mu }_0+t{\phi }_1+t^2{\phi }_2+\cdots$ be a $(Id+c+c^2)$-formal deformation of a commutative multiplication ${\mu }_0$. Then, $(A\left[\left[t\right]\right],{\mu }_t)$ is a deformation quantization of the nonassociative $(Id+c+c^2)$-Poisson algebra $(A,{\mu }_0,{\psi }_1)$.

If ${\mu }_0$ is commutative associative, $(A,{\mu }_0,{\psi }_1)$ is a $(Id+c+c^2)$-Poisson algebra. Then

Corollary 1. 

Let $\mu ={\mu }_0+t{\phi }_1+t^2{\phi }_2+\cdots$ be a $(Id+c+c^2)$-formal deformation of a commutative associative multiplication ${\mu }_0$. Then, $(A\left[\left[t\right]\right],{\mu }_t)$ is a deformation quantization of the $(Id+c+c^2)$-Poisson algebra $(A,{\mu }_0,{\psi }_1)$.

An example of algebra which is nonassociative $(Id+c+c^2)$-Poisson algebra but not nonassociative Poisson is obtained by considering the following two-dimensional case: let $\{e_1,e_2\}$ be a basis of A and

$${\psi }_1(e_1,e_2)=e_2,{\mu }_0(e_1,e_1)=2\beta e_1,{\mu }_0(e_1,e_2)={\mu }_0(e_2,e_1)=\alpha e_1+\beta e_2,{\mu }_0(e_2,e_2)=2\alpha e_2.$$

The algebra $(A,{\mu }_0,{\psi }_1)$ is a nonassociative $(Id+c+c^2)$-Poisson and it is a Poisson algebra when $\alpha =0$ and $\beta =1.$ Recall (see [7]) that a Poisson algebra $(A,\mu ,\{,\left\}\right)$ is also represented by only one multiplication · which satisfies a nonassociative identity

$$x·(y·z)-(x·y)·z+{\displaystyle \frac{1}{3}}\left[(x·z)·y+(y·z)·x-(y·x)·z-(z·x)·y\right]=0$$

and the two multiplications $\mu$ and $\{,\}$ appearing in the definition of Poisson algebras are reobtained by the depolarization process. This nonassociative multiplication is called Poisson admissible. If we apply this idea to nonassociative $(Id+c+c^2)$-Poisson algebras, we find that the class of nonassociative $(Id+c+c^2)$-Poisson admissible algebras corresponds to the 3-power associative algebras $(A,·)$, that is, the multiplication · satisfies

$${\mathcal{A}}_{·}\circ {\Phi }_{v_{3Pa}}=0.$$

There is a one-to-one correspondence between nonassociative $(Id+c+c^2)$-Poisson algebras and 3-power associative algebras (see Section 8).

4.4. Rank$\left(v\right)=3:v=\alpha Id-\alpha {\tau }_{12}+(\alpha +\beta -3){\tau }_{13}-\beta {\tau }_{23}+\beta c+(3-\alpha -\beta )c^2$ with $(\alpha ,\beta )\ne (1,1)$

We will focus in this section on Vinberg, Pre-Lie and $G_3$-associative algebras after studying the general case which shows that the cases where $\beta =1$ and $\alpha \ne 1$ have additional properties on ${\psi }_1$, the skew-symmetric multiplication associated to ${\phi }_1$ and so are particular in this family.

4.4.1. General Case $v=\alpha Id-\alpha {\tau }_{12}+(\alpha +\beta -3){\tau }_{13}-\beta {\tau }_{23}+\beta c+(3-\alpha -\beta )c^2$ with $(\alpha ,\beta )\ne (1,1)$

As in previous cases, if ${\mu }_t$ is a v-formal deformation of a commutative v-associative multiplication ${\mu }_0$, then ${\phi }_1$ is Lie-admissible and its commutator ${\psi }_1$ is a Lie bracket. The equation ${\delta }_{v,{\mu }_0}^2{\phi }_1=0$ gives additional properties on ${\psi }_1$ if and only if $\beta =1$. In fact

$$\begin{array}{cc}\hfill {\delta }_{v,{\mu }_0}^2{\phi }_1(x,y,z)=& x((\alpha -\beta ){\phi }_1(y,z)-(\alpha +2\beta -3){\phi }_1(z,y))+z(2\alpha +\beta -3){\psi }_1(y,x)\hfill \\ & +y((\alpha +2\beta -3){\phi }_1(z,x)-(\alpha -\beta ){\phi }_1(x,z))+(\alpha -\beta ){\phi }_1(x,yz)\hfill \\ & -(\alpha +2\beta -3){\phi }_1(yz,x)+(\beta -\alpha ){\phi }_1(y,xz)-(3-\alpha -2\beta ){\phi }_1(zx,y)\hfill \\ \hfill =& (\alpha +2\beta -3)(x{\psi }_1(y,z)-z{\psi }_1(x,y)+y{\psi }_1(z,x)-{\psi }_1(yz,x)+{\psi }_1(zx,y))\hfill \\ & -3(\beta -1)(x{\phi }_1(y,z)-y{\phi }_1(x,z)+{\phi }_1(x,yz)-{\phi }_1(y,xz))\hfill \\ & -(\alpha -\beta )z{\psi }_1(x,y).\hfill \end{array}$$

Then, ${\delta }_{v,{\mu }_0}^2{\phi }_1(x,y,z)=0$ gives a relation concerning only ${\psi }_1$ as soon as $\beta =1$. If $\beta \ne 1$, we have to consider an additional condition $x{\phi }_1(y,z)-y{\phi }_1(x,z)+{\phi }_1(x,yz)-{\phi }_1(y,xz))=0$ which does not concern all cocycles ${\phi }_1$. Then, we assume $\beta =1$. Since we assumed that $(\alpha ,\beta )\ne (1,1)$, then $\alpha \ne 1$. Because of this hypothesis, the $G_i$-algebras for $i=2,3,4$ are excluded. In fact, $G_2$-algebras correspond to $\alpha =3,\beta =0$, $G_3$-algebras to $\alpha =0,\beta =3$ and $G_4$-algebras to $\alpha =0,\beta =0$; we will see later on some relations on ${\phi }_1$ or on ${\rho }_1$. For v-associative algebras with $v=\alpha (Id-{\tau }_{12})-{\tau }_{23}+c+(2-\alpha )(c^2-{\tau }_{13})$ and $\alpha \ne 1$, the equation ${\delta }_{v,{\mu }_0}^2{\phi }_1=0$ reduces to

$${\psi }_1(x,yz)-{\psi }_1(y,xz)+x{\psi }_1(y,z)-y{\psi }_1(x,z)-2z{\psi }_1(y,x)=0.$$

Considering the Leibniz operator

$$\mathcal{L}({\mu }_0,{\psi }_1)(x,y,z)={\psi }_1(xy,z)-x{\psi }_1(y,z)-{\psi }_1(x,z)y,$$

the equation ${\delta }_{v,{\mu }_0}^2{\phi }_1=0$ is then equivalent to

$$\mathcal{L}({\mu }_0,{\psi }_1)\circ {\Phi }_{Id-{\tau }_{12}}=0.$$

We then have

Proposition 7. 

If ${\mu }_t$ is a v-deformation of a commutative v-associative algebra $(A,{\mu }_0)$ with $v=\alpha Id-\alpha {\tau }_{12}+(\alpha -2){\tau }_{13}-{\tau }_{23}+c+(2-\alpha )c^2$ and $\alpha \ne 1$, then $(A,{\mu }_0,{\psi }_1$) is a nonassociative $(Id-{\tau }_{12})$-Poisson algebra and $(A\left[\left[t\right]\right],{\mu }_t)$ is a deformation quantization of this nonassociative $(Id-{\tau }_{12})$-Poisson algebra.

4.4.2. $G_2$-Algebras or Vinberg Algebras

For $v=Id-{\tau }_{12},$ a v-algebra is also called a Vinberg algebra. Let ${\mu }_0$ be a commutative Vinberg algebra. For any bilinear map $\phi$, we have

$${\delta }_{v,{\mu }_0}^2\phi (x,y,z)=-x\phi (y,z)+y\phi (x,z)-\phi (x,yz)+\phi (y,xz)+z\psi (x,y)$$

for any $x,y,z\in A$ with $\psi$ the skew-symmetric bilinear map attached to $\phi$. Let ${\mu }_t={\mu }_0+\sum t^i{\phi }_i$ be a $(Id-{\tau }_{12})$-formal deformation of ${\mu }_0$. Using the same notations as above, we have

$$\left\{\begin{array}{c}{\delta }_{v,{\mu }_0}^2{\phi }_1=0,\hfill \\ {\phi }_1{•}_v{\phi }_1+{\delta }_{v,{\mu }_0}^2{\phi }_2=0.\hfill \end{array}\right.$$

As $v_{Lad}\in F_v$, the multiplication ${\phi }_1$ is Lie-admissible and the algebra $(A,{\psi }_1)$ is a Lie algebra.

The equation ${\delta }_{v,{\mu }_0}^2{\phi }_1=0$ writes:

$${\delta }_{v,{\mu }_0}^2{\phi }_1=-{\mathcal{L}}_R({\mu }_0,{\phi }_1)\circ {\Phi }_{Id-{\tau }_{12}}=0,$$

using the right-Leibniz operator:

$${\mathcal{L}}_R({\mu }_0,{\phi }_1)(x,y,z)={\phi }_1(x,{\mu }_0(y,z))-{\mu }_0(y,{\phi }_1(x,z))-{\mu }_0({\phi }_1(x,y),z).$$

Proposition 8. 

Let $(A,{\mu }_0)$ be a commutative Vinberg algebra. Then, any $(Id-{\tau }_{12})$-formal deformation $\mu ={\mu }_0+t{\phi }_1+\cdots$ determines a Lie-admissible algebra $(A,{\phi }_1)$ satisfying

$${\mathcal{L}}_R({\mu }_0,{\phi }_1)\circ {\Phi }_{Id-{\tau }_{12}}=0.$$

4.4.3. $G_4$-Algebra also Called Pre-Lie Algebras: $v=Id-{\tau }_{23}$

This case is similar to the $G_2$-algebra case:

Proposition 9. 

Let $(A,{\mu }_0)$ be a commutative Pre-Lie algebra. Then, any formal $(Id-{\tau }_{23})$-formal deformation $\mu ={\mu }_0+t{\phi }_1+\cdots$ determines a Lie-admissible algebra $(A,{\phi }_1)$ satisfying

$$\mathcal{L}({\mu }_0,{\phi }_1)\circ {\Phi }_{Id-{\tau }_{23}}=0.$$

4.4.4. $G_3$-Algebra: $v=Id-{\tau }_{13}$

A commutative v-algebra is also associative. As $v_{Lad}\in F_v$, the linear term ${\phi }_1$ of a v-formal deformation of ${\mu }_0$ is Lie-admissible. The map ${\phi }_1$ satisfies also

$${\delta }_{v,{\mu }_0}^2{\phi }_1(x,y,z)=-x{\rho }_1(y,z)+z{\rho }_1(x,y)+{\rho }_1(xy,z)-{\rho }_1(x,yz)=0$$

where ${\rho }_1$ is the symmetric map attached to ${\phi }_1$, which can also be written

$${\delta }_{v,{\mu }_0}^2{\phi }_1(x,y,z)=-{\rho }_1(x,yz)+z{\rho }_1(x,y)+y{\rho }_1(x,z)+{\rho }_1(xy,z)-x{\rho }_1(y,z)-y{\rho }_1(x,z)=0,$$

that is,

$$\mathcal{L}({\mu }_0,{\rho }_1)(x,y,z)-\mathcal{L}({\mu }_0,{\rho }_1)(z,y,x)=0.$$

because

$$\mathcal{L}({\mu }_0,{\rho }_1)(x,y,z)={\mathcal{L}}_R({\mu }_0,{\rho }_1)(z,y,x)$$

as ${\rho }_1$ is a commutative multiplication. Remark that if ${\mu }_0$ is commutative and ${\psi }_1$ is skew-symmetric, we have that

$$\mathcal{L}({\mu }_0,{\psi }_1)(x,y,z)=-{\mathcal{L}}_R({\mu }_0,{\psi }_1)(z,y,x).$$

Proposition 10. 

Let $(A,{\mu }_0)$ be a commutative $(Id-{\tau }_{13})$-algebra. Then, if $\mu ={\mu }_0+t{\phi }_1+t^2{\phi }_2+\cdots$ is a $(Id-{\tau }_{13})$-formal deformation of ${\mu }_0$, then if ${\psi }_1$ and ${\rho }_1$ are respectively the skew-symmetric and symmetric bilinear maps associated to ${\phi }_1$

1. 

$(A,{\psi }_1)$ is a Lie algebra,

2. 

the symmetric map ${\rho }_1$ satisfies

$$\mathcal{L}({\mu }_0,{\rho }_1)\circ {\Phi }_{Id-{\tau }_{13}}=0.$$

4.5. Rank$\left(v\right)=4$: $v=2Id+(1+\alpha ){\tau }_{12}+{\tau }_{13}+c+(1-\alpha )c^2$ with $\alpha \ne 1$

Let ${\mu }_t={\mu }_0+t{\phi }_1+\cdots$ be a v-formal deformation of a commutative v-associative algebra. Then, $(A,{\phi }_1)$ is a Lie-admissible algebra and the equation ${\delta }_{{\mu }_0,v}^2{\phi }_1=0$ is equivalent to a quadratic relation on ${\psi }_1$ if and only if $\alpha =-{\displaystyle \frac{1}{2}}$ or ${\delta }_{H,{\mu }_0}^2{\phi }_1=0.$ In fact,

$$\begin{array}{cc}{\delta }_{v,{\mu }_0}^2{\phi }_1(x,y,z)=\hfill & x{\psi }_1(y,z)-\alpha y{\psi }_1(z,x)+z{\psi }_1(x,y)+(2-\alpha ){\psi }_1(z,xy)-2{\psi }_1(yz,x)\hfill \\ & +(-1+\alpha ){\psi }_1(zx,y)+(1+2\alpha ){\delta }_{H,{\mu }_0}^2{\phi }_1(y,x,z).\hfill \end{array}$$

Then, ${\delta }_{v,{\mu }_0}^2{\phi }_1=0$ implies a quadratic relation on ${\psi }_1$ as soon as $(1+2\alpha ){\delta }_{H,{\mu }_0}^2{\phi }_1=0$. Since ${\delta }_{H,{\mu }_0}^2{\phi }_1=0$ is a particular case of ${\delta }_{v,{\mu }_0}^2{\phi }_1=0$, and since we want a generic identity, then $\alpha =-{\displaystyle \frac{1}{2}}$. This corresponds to weakly associative algebra, also called Lie-admissible flexible algebras [2] with vector $v=Id-{\tau }_{12}+c$. In fact, if we denote by w the vector $2Id+(1/2){\tau }_{12}+{\tau }_{13}+c+(3/2)c^2$, then

$$v=Id-{\tau }_{12}+c=w\circ ((1/3)Id-{\tau }_{12}+(7/12){\tau }_{13}+(1/4)c^2)$$

and

$$w=v\circ ((-1/2)Id+(5/2){\tau }_{23}+(3/2)c+(5/2)c^2).$$

We recall results obtained in [2]. Let ${\mu }_0$ be a commutative associative algebra. Then, it is also v-associative and we can consider a v-formal deformation of ${\mu }_0$:

$$\mu ={\mu }_0+t{\phi }_1+t^2{\phi }_2+\cdots$$

We deduce

$$\left\{\begin{array}{cc}\mathrm{order}0\hfill & {\mathcal{A}}_{{\mu }_0}\circ {\Phi }_v=0,\hfill \\ \mathrm{order}1\hfill & {\delta }_{v,{\mu }_0}^2{\phi }_1={\delta }_{H,{\mu }_0}^2{\phi }_1\circ {\Phi }_v=0,\hfill \\ \mathrm{order}2\hfill & {\phi }_1{•}_v{\phi }_1+{\delta }_{H,{\mu }_0}^2{\phi }_2\circ {\Phi }_v=0.\hfill \end{array}\right.$$

(5)

From Proposition 2, ${\phi }_1$ is Lie-admissible and ${\psi }_1$ is a Lie bracket. Let us now investigate the consequences of the equation ${\delta }_{H,{\mu }_0}^2{\phi }_1\circ {\Phi }_v=0$ by considering a vector w canceling the ${\rho }_1$, that is, $w=a_{1}Id+a_2{\tau }_{12}+a_3{\tau }_{13}+a_4{\tau }_{23}+(a_1-a_3+a_4)c+(a_1+a_2-a_3)c^2.$ We have using Leibniz operator associated to ${\mu }_0$ and ${\psi }_1$ that ${\delta }_H^2{\phi }_1\circ {\Phi }_w=0$ is equivalent to

$$\begin{array}{c}\hfill (a_1+a_2)\mathcal{L}({\mu }_0,{\psi }_1)(x,y,z)+(a_1+a_4)\mathcal{L}({\mu }_0,{\psi }_1)(y,z,x)\\ \hfill +(a_1+a_2-a_3+a_4)\mathcal{L}({\mu }_0,{\psi }_1)(z,x,y)=0.\end{array}$$

If the components of w satisfy one of the following conditions

• $a_1+a_2\ne 0$ and $a_1+a_4=a_2-a_3=0$,
• or $a_1+a_4\ne 0$ and $a_1+a_2=a_3-a_4=0$,
• or $a_3-a_4\ne 0$ and $a_1+a_2=a_1+a_4=0,$

• then ${\delta }_H^2{\phi }_1\circ {\Phi }_w=0$ implies $\mathcal{L}({\mu }_0,{\psi }_1)=0.$ For each case, the w vector belongs to $F_{Id-{\tau }_{12}+c}.$ In fact, let us consider the first case. The equation $({\alpha }_{1}Id+{\alpha }_2{\tau }_{12}+{\alpha }_3{\tau }_{13}+{\alpha }_4{\tau }_{23}+{\alpha }_{5}c+{\alpha }_6{c}^2)\circ (Id-{\tau }_{12}+c)=w$ is equivalent to the linear system

  $$\left\{\begin{array}{c}{\alpha }_1-{\alpha }_2+{\alpha }_6=a_1\hfill \\ -{\alpha }_1+{\alpha }_2+{\alpha }_3=a_2\hfill \\ {\alpha }_3+{\alpha }_4-{\alpha }_5=a_2\hfill \\ {\alpha }_2+{\alpha }_4-{\alpha }_6=-a_1\hfill \\ {\alpha }_1-{\alpha }_3+{\alpha }_5=-a_2\hfill \\ -{\alpha }_4+{\alpha }_5+{\alpha }_6=a_1\hfill \end{array}\right.\iff \left\{\begin{array}{c}{\alpha }_4=-{\alpha }_1\hfill \\ {\alpha }_5=-{\alpha }_2\hfill \\ {\alpha }_6=a_1+{\alpha }_2-{\alpha }_1\hfill \\ {\alpha }_3=a_2-{\alpha }_2+{\alpha }_1\hfill \end{array}\right.$$

  which has a nontrivial solution. It is similarly for the other cases. Then, we can find a vector $v_1$ such that $v_{1}v=w$ and ${\delta }_H^2{\phi }_1\circ {\Phi }_w=0$ implies $\mathcal{L}({\mu }_0,{\psi }_1)=0.$ This implies:

Proposition 11. 

Let $(A,{\mu }_0)$ be a commutative algebra and consider the vector $v=Id-{\tau }_{12}+c$. Then, $(A,{\mu }_0)$ is v-associative and for any v-formal deformation ${\mu }_t={\mu }_0+t{\phi }_1+t^2{\phi }_2+\cdots$ of ${\mu }_0$, the algebra $(A,{\mu }_0,{\psi }_1)$ is a nonassociative Poisson algebra where ${\psi }_1$ is the skew-symmetric application associated with ${\phi }_1$.

Corollary 2. 

Let $(A,{\mu }_0)$ be an associative commutative algebra and consider the vector $v=Id-{\tau }_{12}+c$. Then, $(A,{\mu }_0)$ is v-associative and for any v-formal deformation ${\mu }_t={\mu }_0+t{\phi }_1+t^2{\phi }_2+\cdots$ of ${\mu }_0$, the algebra $(A,{\mu }_0,{\psi }_1)$ is a Poisson algebra.

Consequence 1. 

In the usual deformation quantization process, a Poisson algebra $(A,{\mu }_0,{\psi }_1)$ is obtained from a formal deformation ${\mu }_t={\mu }_0+t{\phi }_1+\cdots$ of a commutative associative algebra and the algebra $(A\left[\left[t\right]\right],{\mu }_t)$ is a deformation quantization of the Poisson algebra $(A,{\mu }_0,{\psi }_1)$. Corollary 2 shows that a Poisson algebra $(A,{\mu }_0,{\psi }_1)$ is also obtained from a $(Id-{\tau }_{12}+c)$-formal deformation (that is, a weakly associative formal deformation) ${\mu }_t={\mu }_0+t{\phi }_1+\cdots$ of a commutative associative algebra. Thus, we also consider the algebra $(A\left[\left[t\right]\right],{\mu }_t)$ as a quantization of a Poisson algebra $(A,{\mu }_0,{\psi }_1)$ in this more general case. The v-algebras with $v=Id-{\tau }_{12}+c$ called weakly associative algebras have been introduced in [2], where an algebraic study is presented.

Remark 3. 

In [2], we show that any commutative algebra, any Lie algebra, and any associative algebra is weakly associative. In fact, the class of weakly associative algebras (associated to $v=Id+c-{\tau }_{12}$) is the biggest class containing the Lie algebras and the associative algebras such that the v-deformation of a commutative associative algebra gives a Poisson algebra so quantizations of a Poisson algebra. In [15], we show also that the symmetric Leibniz algebras are also weakly associative.

4.6. Rank$\left(v\right)=5$: $F_v=F_{2Id-{\tau }_{12}-{\tau }_{13}-{\tau }_{23}+c}$

Let ${\mu }_t={\mu }_0+\sum t^i{\phi }_i$ be a v-formal deformation of the commutative v-algebra $(A,{\mu }_0)$. The product ${\phi }_1$ is Lie-admissible from Theorem 1 and ${\psi }_1$ is a Lie bracket. Since ${\mu }_0$ is commutative, we have

$$\begin{array}{cc}{\delta }_{v,{\mu }_0}^2{\phi }_1(x,y,z)=\hfill & -x{\phi }_1(y,z)-y{\phi }_1(z,x)+2z{\phi }_1(x,y)\hfill \\ & +{\phi }_1(xy,z)-{\phi }_1(xz,y)-{\phi }_1(x,yz)+{\phi }_1(z,xy).\hfill \end{array}$$

Choosing $v_1=(a_1,a_2,a_3,a_4,a_1+a_2-a_3,a_1+a_2-a_4)$, we see that ${\delta }_{v,{\mu }_0}^2{\phi }_1\circ {\Phi }_{v_1}$ contains only elements in ${\psi }_1$ so that ${\delta }_{v,{\mu }_0}^2{\phi }_1\circ {\Phi }_{v_1}=0$ reads

$$\begin{array}{c}(a_2-2a_3+a_4)x{\psi }_1(z,y)+(-a_2-a_3+2a_4)y{\psi }_1(z,x)+(2a_2-a_3-a_4)z{\psi }_1(x,y)\hfill \\ +(a_4-a_3){\psi }_1(xy,z)+(a_3-a_2){\psi }_1(xz,y)+(a_2-a_4){\psi }_1(yz,x)=0,\hfill \end{array}$$

that is,

$$\begin{array}{c}{a}_2(-x{\psi }_1(y,z)-y{\psi }_1(z,x)+2z{\psi }_1(x,y)-{\psi }_1(xz,y)+{\psi }_1(yz,x))\hfill \\ +a_3(-2x{\psi }_1(y,z)-y{\psi }_1(z,x)-z{\psi }_1(x,y)-{\psi }_1(xy,z)+{\psi }_1(xz,y)\hfill \\ +a_4(x{\psi }_1(y,z)+2y{\psi }_1(z,x)-z{\psi }_1(x,y)+{\psi }_1(xy,z)-{\psi }_1(yz,x)=0.\hfill \end{array}$$

This is equivalent to the identity

$$-x{\psi }_1(y,z)-y{\psi }_1(z,x)+2z{\psi }_1(x,y)-{\psi }_1(xz,y)+{\psi }_1(yz,x)=0,$$

that is,

$$\mathcal{L}({\mu }_0,{\psi }_1)(y,z,x)-\mathcal{L}({\mu }_0,{\psi }_1)(x,z,y)=0.$$

Proposition 12. 

Let $(A,{\mu }_0)$ be a commutative v-algebra with $v=2Id-{\tau }_{12}-{\tau }_{13}-{\tau }_{23}+c.$ Any v-formal deformation ${\mu }_t={\mu }_0+\sum t^i{\phi }_i$ of $(A,{\mu }_0)$ is a deformation quantization of a nonassociative $(Id-{\tau }_{13})$-Poisson algebra $(A,{\mu }_0,{\psi }_1)$, that is,

1. 

${\mu }_0$ is a commutative multiplication on A,

2. 

${\psi }_1$ is a Lie bracket on A,

3. 

$\mathcal{L}({\mu }_0,{\psi }_1)\circ {\Phi }_{Id-{\tau }_{13}}=0.$

5. A Generalization: ${\left(\mathbb{K}\left[{\Sigma }_{\mathbf{3}}\right]\right)}^{\mathbf{2}}$-Associative Algebras

This generalization has been introduced in [5]. If $(A,\mu )$ is a $\mathbb{K}$-algebra, we denote by ${\mathcal{A}}_{\mu }$ the associator of $\mu$. Let us write this associator in the following form:

$${\mathcal{A}}_{\mu }={\mathcal{A}}_{\mu }^L-{\mathcal{A}}_{\mu }^R$$

where ${\mathcal{A}}_{\mu }^L=\mu \circ (\mu \otimes Id)$ and ${\mathcal{A}}_{\mu }^R=\mu \circ (Id\otimes \mu ).$

Now, instead of considering action of ${\Sigma }_3$-permutation on the associator, we can consider it independently on ${\mathcal{A}}_{\mu }^L$ and ${\mathcal{A}}_{\mu }^R$, which will induce different symmetries.

5.1. Definition

Definition 3. 

Let v and w be two vectors of $\mathbb{K}\left[{\Sigma }_3\right]$. We say that the algebra $(A,\mu )$ is a $(v,w)$-algebra if we have

$$(★)\left\{\begin{array}{c}{\mathcal{A}}_{\mu }^L\circ {\Phi }_v-{\mathcal{A}}_{\mu }^R\circ {\Phi }_w=0\hfill \end{array}\right.$$

If there exists a nontrivial $(v,w)\in \mathbb{K}{\left[{\Sigma }_3\right]}^2$ such that $(A,\mu )$ is a $(v,w)$-algebra, the algebra $(A,\mu )$ is ${\left(\mathbb{K}\left[{\Sigma }_3\right]\right)}^2$-associative.

An interesting case of $(v,w)$-algebras corresponds to the algebras given by the system of equations

$$(★★)\left\{\begin{array}{c}{\mathcal{A}}_{\mu }^L\circ {\Phi }_v=0,\hfill \\ {\mathcal{A}}_{\mu }^R\circ {\Phi }_w=0.\hfill \end{array}\right.$$

In the study of $(v,w)$-algebras, the first study is to know if a $(v,w)$-algebra can be defined as a $v_1$-algebra. The easiest example is when $v=w$. In [5], we study the $(v,w)$-algebras, which are Lie-admissible algebras or Pre-Lie algebras.

Example 1. 

A Leibniz algebra satisfies the quadratic relation

$$\mu \left(\mu \right(x,y),z)=\mu (x,\mu (y,z\left)\right)+\mu \left(\mu \right(x,z),y)$$

or

$${\mathcal{A}}_{\mu }^L\circ {\Phi }_{Id-{\tau }_{23}}-{\mathcal{A}}_{\mu }^R\circ {\Phi }_{Id}=0.$$

Then, Leibniz algebras are $(Id-{\tau }_{23},Id)$-algebras. Symmetric Leibniz algebras are defined by a pair of quadratic relations. They correspond to

$${\mathcal{A}}_{\mu }^L\circ {\Phi }_{Id-{\tau }_{23}}-{\mathcal{A}}_{\mu }^R\circ {\Phi }_{Id},{\mathcal{A}}_{\mu }^L\circ {\Phi }_{Id}-{\mathcal{A}}_{\mu }^R\circ {\Phi }_{Id-{\tau }_{12}}.$$

The notion of $(v,w)$-formal deformation of a $(v,w)$-algebra is similar to this notion for v-algebras. Let $(A,{\mu }_0)$ be a $(v,w)$-algebra defined by a relation of type ($★,★★$). Consider ${\mu }_t={\mu }_0+{\sum }_{i\ge 1}{t}^i{\phi }_i$. We say that ${\mu }_t$ is a $(v,w)$-formal deformation of ${\mu }_0$ if $(A\left[\left[t\right]\right],{\mu }_t)$ is a $(v,w)$-algebra. To describe the relations between the ${\phi }_i$, we need to introduce some notations:

$${\phi }_i{•}_v^L{\phi }_j={\phi }_i\circ ({\phi }_j\otimes Id)\circ {\Phi }_v,{\phi }_i{•}_w^R{\phi }_j={\phi }_i\circ (Id\otimes {\phi }_j)\circ {\Phi }_w,$$

$${\delta }_{v,{\mu }_0}^{2,L}\phi =\phi {•}_v^L{\mu }_0+{\mu }_0{•}_v^L\phi ,{\delta }_{w,{\mu }_0}^{2,R}\phi =\phi {•}_w^R{\mu }_0+{\mu }_0{•}_w^R\phi .$$

Thus, to say that ${\mu }_t$ is a $(v,w)$-formal deformation of ${\mu }_0$ implies in particular:

• order 0: ${\mu }_0$ is a $(v,w)$-algebra,
• order 1: ${\delta }_{v,{\mu }_0}^{2,L}{\phi }_1-{\delta }_{w,{\mu }_0}^{2,R}{\phi }_1=0$,
• order 2: ${\phi }_1{•}_v^L{\phi }_1-{\phi }_1{•}_w^R{\phi }_1+{\delta }_{v,{\mu }_0}^{2,L}{\phi }_2-{\delta }_{w,{\mu }_0}^{2,R}{\phi }_2=0$.

5.2. A Fundamental Example: The Anti-Associative Algebras

Definition 4. 

A $\mathbb{K}$-algebra $(A,\mu )$ is called anti-associative if the multiplication μ satisfies the following identity

$$\mu \left(\mu \right(x,y),z)+\mu (x,\mu (y,z\left)\right)=0$$

for any $x,y,z\in A$.

We will denote by ${\mathcal{AA}}_{\mu }$ the trilinear map

$${\mathcal{AA}}_{\mu }(x,y,z)=\mu (\mu (x,y),z)+\mu (x,\mu (y,z))$$

and $(A,\mu )$ is anti-associative if and only if ${\mathcal{AA}}_{\mu }=0.$ In terms of $(v,w)$-algebra, an anti-associative algebra is an $(Id,-Id)$-algebra. In an anti-associative algebra, all the 4-products are zero. In fact

$$\begin{array}{cc}\hfill x\left(y\right(zt\left)\right)& =-x\left(\right(yz\left)t\right)=\left(x\right(yz\left)\right)t=-\left(\right(xy\left)z\right)t\hfill \\ \hfill \mathrm{and}x\left(y\right(zt\left)\right)& =-\left(xy\right)\left(zt\right)=\left(\right(xy\left)z\right)t\hfill \end{array}$$

so all 4-products are trivial. Anti-associative algebras are therefore always 3-step nilpotent.

Example 2. 

There are ‘natural’ examples of the anti-associativity. For instance, the standard basis elements $\{1,e_1,e_2,e_3,e_4,e_5,e_6,e_7\}$ of the octonions (also called the Cayley algebra) satisfy

$$\left(e_i{e}_j\right)e_k=-e_i\left(e_j{e}_k\right),$$

whenever $e_i{e}_j\ne \pm e_k$ and $1\le i,j,k\le 7$ are distinct. In [8], anti-associative algebras in small dimension are described. For example, in dimension 3, we have obtained the following nonisomorphic nontrivial anti-associative algebras $(A,·)$:

1. 

$e_i·e_i=0,e_1·e_2=-e_2·e_1=e_3$

2. 

$e_1·e_1=e_2,e_1·e_2=-e_2·e_1=e_3$, which happens to be the free anti-associative algebra on one generator,

3. 

$e_1·e_1=e_2,e_1·e_3=a{e}_2,e_3·e_1=b{e}_2,e_3·e_3=e_2,$

4. 

$e_1·e_1=e_2,e_1·e_3=a{e}_2,e_3·e_1=b{e}_2.$

• with $a,b\in \mathbb{K}$ and where $\{e_1,e_2,e_3\}$ is a basis of A.

If $\mathcal{AA}ss$ denotes the quadratic operad corresponding to the anti-associative algebra, then

$$dim\mathcal{AA}ss\left(1\right)=1,dim\mathcal{AA}ss\left(2\right)=2,dim\mathcal{AA}ss\left(3\right)=6\mathrm{and}dim\mathcal{AA}ss\left(n\right)=0\mathrm{for}n\ge 4.$$

Let us note also that this operad is self-dual. In [8], it is proved that the operad $\mathcal{AA}ss$ is not Koszul computing the inverse series of the generating function $g_{\mathcal{AA}ss}\left(t\right)=t+t^2+t^3$.

Concerning the problem of deformation of anti-associative algebras, the ‘standard’ cohomology of an anti-associative algebra A with coefficients in itself is described in [8] and compared to the relevant part of the deformation cohomology based on the minimal model of the anti-associative operad $\mathcal{AA}ss$. Since $\mathcal{AA}ss$ is not Koszul, these two cohomologies differ. The standard cohomology $H_{st}^{*}(A,A)$ is the cohomology of the complex

$$C^1(A,A)\stackrel{{\delta }_{AA}^1}{⟶}{C}^2(A,A)\stackrel{{\delta }_{AA}^2}{⟶}{C}^3(A,A)\stackrel{{\delta }_{AA}^3}{⟶}0\stackrel{0}{⟶}0\stackrel{0}{⟶}\cdots$$

in which $C^p(A,A):=Hom(A^{\otimes p},A)$ for $p=1,2,3$, and all higher $C^p$s are trivial. The two nontrivial pieces of the differential are basically the Hochschild differentials with “wrong” signs of some terms:

$$\begin{array}{cc}\hfill {\delta }_{AA}^1\left(f\right)(x,y)& :=xf\left(y\right)-f\left(xy\right)+f\left(x\right)y, \mathrm{and}\hfill \\ \hfill {\delta }_{AA}^2\left(\phi \right)(x,y,z)& :=x\phi (y,z)+\phi (xy,z)+\phi (x,yz)+\phi (x,y)z,\hfill \end{array}$$

for $f\in Hom(V,V)$, $\phi \in Hom(A^{\otimes 2},V)$ and $x,y,z\in V$. One sees, in particular, that

$$H_{st}^{*}{(A,A)}^p=0$$

for $p\ge 4$.

The deformation cohomology of anti-associative algebras, based of the study of a minimal model, is also studied in [8]. We summarize the results: we consider the complex

$$C_{AAss}^1(A,A)\stackrel{{\delta }^1}{⟶}{C}_{AAss}^2(A,A)\stackrel{{\delta }^2}{⟶}{C}_{AAss}^3(A,A)\stackrel{{\delta }^3}{⟶}{C}_{AAss}^4(A,A)\stackrel{{\delta }^4}{⟶}\cdots$$

-

$C_{AAss}^1(A,A)=Hom(A,A)$

-

$C_{AAss}^2(A,A)=Hom(A^{\otimes 2},A)$

-

$C_{AAss}^3(A,A)=Hom(A^{\otimes 3},A)$, and

-

$C_{AAss}^4(A,A)=Hom(A^{\otimes 5},A)\oplus Hom(A^{\otimes 5},A)\oplus Hom(A^{\otimes 5},A)\oplus Hom(A^{\otimes 5},A)$.

• Observe that $C_{AAss}^p(A,A)=C^p(A,A)$ for $p=1,2,3$, while $C_{AAss}^4(A,A)$ consists of 5-linear maps. The differential ${\delta }^p$ agrees with ${\delta }_{AA}^p$ for $p=1,2$ while, for $g\in C_{AAss}^3(A,A)$, one has

  $${\delta }^3\left(g\right)=({\delta }_1^3\left(g\right),{\delta }_2^3\left(g\right),{\delta }_3^3\left(g\right),{\delta }_4^3\left(g\right)),$$

  where

  $$\begin{array}{cc}\hfill {\delta }_1^3\left(g\right)(x,y,z,t,u)& :=xg(y,z,tu)-g(x,y,z(tu\left)\right)+\left(xy\right)g(z,t,u)-g(xy,zt,u)\hfill \\ \hfill & +g(xy,z,t)u-g\left(\right(xy)z,t,u)+g(x,y,z)\left(tu\right)-g(x,yz,tu),\hfill \\ \hfill {\delta }_2^3\left(g\right)(x,y,z,t,u)& :=g\left(\right(xy)z,t,u)-g(xy,z,t)u+g(x,y,zt)u-g(x,y(zt),u)\hfill \\ \hfill & +xg(y,zt,u)-g(x,y,(zt\left)u\right)+\left(xy\right)g(z,t,u)-g(xy,z,tu),\hfill \\ \hfill {\delta }_3^3\left(g\right)(x,y,z,t,u)& :=g(x,yz,tu)-xg(yz,t,u)+g(x,(yz)t,u)-x\left(g\right(y,z,t\left)u\right)\hfill \\ \hfill & +g(x,y,zt)u-g(xy,z,t)u+\left(g\right(x,y,z\left)t\right)u-g\left(x\right(yz),t,u), \mathrm{and}\hfill \\ \hfill {\delta }_4^3\left(g\right)(x,y,z,t,u)& :=g(xy,zt,u)-g(x,y,(zt\left)u\right)+xg(y,zt,u)-g(x,y(zt),u)\hfill \\ \hfill & +\left(xg\right(y,z,t\left)\right)u-g(x,yz,t)u+\left(g\right(x,y,z\left)t\right)u-g(xy,z,t)u,\hfill \end{array}$$

  for $x,y,z,t,u\in V$.

We consider now formal deformation of anti-associative algebras. If ${\mu }_t={\mu }_0+\sum t^i{\phi }_i$ is an anti-associative formal deformation of ${\mu }_0$, then we have, denoting by $xy$ the product $\mu (x,y)$:

• in degree 0: ${\mu }_0$ is anti-associative,
• In degree 1: ${\phi }_1(x,yz)+x{\phi }_1(y,z)+{\phi }_1(x,y)z+{\phi }_1(xy,z)=0$
• In degree 2:

  $${\phi }_2(x,yz)+x{\phi }_2(y,z)+{\phi }_2(x,y)z+{\phi }_2(xy,z)+{\phi }_1(x,{\phi }_1(y,z))+{\phi }_1({\phi }_1(x,y),z)=0,$$

that is,

$${\delta }_{AA,{\mu }_0}^2\left({\phi }_1\right)=0$$

and

$${\delta }_{AA,{\mu }_0}^2\left({\phi }_2\right)+{\mathcal{AA}}_{{\phi }_1}=0.$$

Assume moreover that ${\mu }_0$ is commutative. It is not difficult to see that ${\delta }_{AA,{\mu }_0}^2\left({\phi }_2\right)\circ {\Phi }_v=0$ implies $v=0$ and we do not have a good deformation quantization framework similar to the associative or v-associative cases.

We are therefore naturally led to consider formal deformation of an anti-associative product ${\mu }_0$, which is also skew-symmetric. In this case, we obtain an anti-commutative version of Lemma 1, which was for a commutative product.

Lemma 2. 

Let $(A,{\mu }_0)$ be an anti-commutative algebra with ${\mu }_0\ne 0$ and ${\delta }_{H,{\mu }_0}^2$ the coboundary operator:

$${\delta }_{AA,{\mu }_0}^2\phi (X,Y,Z)=X\phi (Y,Z)+\phi (XY,Z)+\phi (X,YZ)+\phi (X,Y)Z$$

where $X,Y,Z\in A$, the map φ is bilinear on A and $XY$ denotes the product ${\mu }_0(X,Y)$. Then, ${\delta }_{AA,{\mu }_0}^2\phi \circ {\Phi }_{v_{3Pa}}=0.$

Proof. 

A direct computation proves, like for Lemma 1 for the commutative case, that ${\delta }_{AA,{\mu }_0}^2\phi \circ {\Phi }_{v_{3Pa}}=0.$ □

• As a consequence,

  $${\mathcal{AA}}_{{\phi }_1}\circ {\Phi }_{v_{3Pa}}=0.$$

  We can say that ${\phi }_1$ verify the anti-associative version of the 3-power associative property. We deduce

Proposition 13. 

Let $(A,{\mu }_0)$ be a skew-symmetric anti-associative algebra and $\mu ={\mu }_0+\sum t^k{\phi }_k$ an anti-associative formal deformation of ${\mu }_0$. Then, if ${\rho }_1$ denotes the symmetric part of ${\phi }_1$, the algebra $(A,{\rho }_1)$ is a Jacobi–Jordan algebra, that is,

1. 

$(A,{\rho }_1)$ is a commutative algebra,

2. 

${\rho }_1$ satisfies the “Jacobi” identity:

$${\rho }_1(x,{\rho }_1(y,z))+{\rho }_1(y,{\rho }_1(z,x))+{\rho }_1(z,{\rho }_1(x,y))=0$$

for all $x,y,z\in A$.

Proof. 

The Jacobi identity for ${\rho }_1$ follows from the fact that ${\mathcal{AA}}_{{\phi }_1}\circ {\Phi }_{v_{3Pa}}=0.$ □

Let us examine the first condition ${\delta }_{AA,{\mu }_0}^2\left({\phi }_1\right)=0.$ With a similar proof to the associative case, we show that this identity implies

$${\rho }_1(xy,z)+x{\rho }_1(y,z)+{\rho }_1(x,z)y=0$$

where ${\mu }_0(x,y)=xy.$

Theorem 3. 

Let $(A,{\mu }_0)$ be a skew-symmetric anti-associative algebra and ${\mu }_t={\mu }_0+\sum t^k{\phi }_k$ an anti-associative formal deformation of ${\mu }_0$. Then, $(A,{\mu }_0,{\rho }_1)$ is an anti-Poisson algebra, that is,

1. 

$(A,{\rho }_1)$ is a Jacobi–Jordan algebra,

2. 

The products ${\mu }_0$ and ${\rho }_1$ are tied up by the graded Leibniz identity:

$${\mathcal{L}}_g({\mu }_0,{\rho }_1)(x,y,z)={\rho }_1(xy,z)+x{\rho }_1(y,z)+{\rho }_1(x,z)y=0.$$

We will say that ${\mu }_t$ is a deformation quantization of the anti-Poisson algebra $(A,{\mu }_0,{\rho }_1)$.

Recall that an antiderivation of an algebra $(A,\mu )$ is a linear map f such that

$$f\left(\mu \right(x,y\left)\right)+\mu (x,f(y\left)\right)+\mu \left(f\right(x),y)=0$$

for any $x,y$ in A. The graded Leibniz identity can be interpreted saying that for any $z\in A$, the linear maps $x\to \rho (x,z)$ are an antiderivation of the algebra $(A,{\mu }_0).$

Remark 4. 

Considering an anti-commutative product ${\mu }_0$ and $w=a_{1}Id+a_2{\tau }_{12}+a_3{\tau }_{13}+a_4{\tau }_{23}+a_{5}c+a_6{c}^2$ be a vector of $\mathbb{K}\left[{\Sigma }_3\right]$. The anti-commutativity of ${\mu }_0$ implies

$$\begin{array}{c}{\delta }_{AA,{\mu }_0}^2\phi \circ {\Phi }_w(X,Y,Z)=\hfill \\ (a_1-a_5)X\phi (Y,Z)+(a_4-a_3)X\phi (Z,Y)+(a_2-a_4)Y\phi (X,Z)+(a_5-a_6)Y\phi (Z,X)\hfill \\ +(a_6-a_1)Z\phi (X,Y)+(a_3-a_2)Z\phi (Y,X)+(a_1-a_2)\phi (XY,Z)+(a_5-a_3)\phi (YZ,X)\hfill \\ +(a_6-a_4)\phi (ZX,Y)+(a_1-a_4)\phi (X,YZ)+(a_5-a_2)\phi (Y,ZX)+(a_6-a_3)\phi (Z,XY).\hfill \end{array}$$

(6)

Remark that we reobtain Lemma 2 considering $a_1=a_2=a_3=a_4=a_5=a_6$.

From the $\mathbb{K}\left[{\Sigma }_3\right]$-module structure of $F_w$, we can deduce [5] that $v_{3Pa}\in F_w$ if and only if $F_w$ is odd-dimensional. In terms of the coefficients $a_i$,

$$\begin{array}{cc}\hfill v_{3Pa}\in F_w& \iff \exists v={\lambda }_{1}Id+{\lambda }_2{\tau }_{12}+{\lambda }_3{\tau }_{13}+{\lambda }_4{\tau }_{23}+{\lambda }_{5}c+{\lambda }_6{c}^2,vw=v_{3Pa}\hfill \\ & \iff M_{w}V{=}^t(1,1,1,1,1,1)\hfill \\ \hfill \iff & \left\{\begin{array}{c}{a}_1({\lambda }_1-{\lambda }_5)+a_2({\lambda }_2+{\lambda }_5)+a_3({\lambda }_3+{\lambda }_5)+a_4({\lambda }_4+{\lambda }_5)+a_5({\lambda }_6-{\lambda }_5)=1\hfill \\ a_1({\lambda }_2-{\lambda }_4)+a_2({\lambda }_1+{\lambda }_4)+a_3({\lambda }_6+{\lambda }_4)+a_4({\lambda }_5+{\lambda }_4)+a_5({\lambda }_3-{\lambda }_4)=1\hfill \\ a_1({\lambda }_3-{\lambda }_2)+a_2({\lambda }_5+{\lambda }_2)+a_3({\lambda }_1+{\lambda }_2)+a_4({\lambda }_6+{\lambda }_2)+a_5({\lambda }_4-{\lambda }_2)=1\hfill \\ a_1({\lambda }_4-{\lambda }_3)+a_2({\lambda }_6+{\lambda }_3)+a_3({\lambda }_5+{\lambda }_3)+a_4({\lambda }_1+{\lambda }_3)+a_5({\lambda }_2-{\lambda }_3)=1\hfill \\ a_1({\lambda }_5-{\lambda }_6)+a_2({\lambda }_3+{\lambda }_6)+a_3({\lambda }_4+{\lambda }_6)+a_4({\lambda }_2+{\lambda }_6)+a_5({\lambda }_1-{\lambda }_6)=1\hfill \\ a_1({\lambda }_6-{\lambda }_1)+a_2({\lambda }_4+{\lambda }_1)+a_3({\lambda }_2+{\lambda }_1)+a_4({\lambda }_3+{\lambda }_1)+a_5({\lambda }_5-{\lambda }_1)=1\hfill \end{array}\right.\hfill \end{array}$$

It implies $(a_2+a_3+a_4)({\lambda }_1+{\lambda }_2+{\lambda }_3+{\lambda }_4+{\lambda }_5+{\lambda }_6)=3$, so $a_2+a_3+a_4\ne 0$, and there is a nontrivial solution, for example, ${\lambda }_5={\lambda }_6={\lambda }_1={\lambda }_2={\lambda }_4={\lambda }_3={\displaystyle \frac{1}{2(a_2+a_3+a_4)}}$, that is, $v={\displaystyle \frac{1}{2(a_2+a_3+a_4)}}(Id+{\tau }_{12}+{\tau }_{13}+{\tau }_{23}+c+c^2)$ satisfies $vw=v_{3Pa}$ and $v_{3Pa}\in F_w$.

Theorem 4. 

Let $(A,{\mu }_0)$ be an anti-commutative algebra with ${\mu }_0\ne 0$ and $dim\left(A\right)\ge 2.$ Consider $w=a_{1}Id+a_2{\tau }_{12}+a_3{\tau }_{13}+a_4{\tau }_{23}+a_{5}c+a_6{c}^2$ a vector of $\mathbb{K}\left[{\Sigma }_3\right]$. We have that ${\delta }_{AA,{\mu }_0}^2\phi \circ {\Phi }_w=0$ for every bilinear map φ if $a_1-a_2-a_3-a_4+a_5+a_6=0$ and $a_2+a_3+a_4\ne 0$.

Remark 5 

(Deformation quantization and polarization). In the following section, we recall the notion of polarization/depolarization of a product of an algebra. We will see that when we apply this process to an anti-associative algebra $(A,\mu )$, the associated skew-symmetric map ψ defined by $\psi (x,y)=\mu (x,y)-\mu (y,x)$ and the symmetric map ρ defined by $\rho (x,y)=\mu (x,y)+\mu (y,x)$, which provides $(A,\rho )$ with a Jacobi–Jordan algebra structure, are tied up with the graded Leibniz identity. We develop this point of view in the last section. An algebraic and detailed study of general Jacobi–Jordan algebras is given in [9]. These algebras are also called mock-Lie algebras [10].

5.3. Left-Leibniz Algebras

Recall that $(A,\mu )$ is a left-Leibniz algebra if $\mu$ satisfies the quadratic relation

$$\mu (x,\mu (y,z\left)\right)=\mu \left(\mu \right(x,y),z)+\mu (y,\mu (x,z\left)\right)$$

for any $x,y,z\in A$ what is also written

$${\mathcal{A}}_{\mu }(x,y,z)+\mu (y,\mu (x,z))=0.$$

Let $(A,{\mu }_0)$ be a commutative left-Leibniz algebra, that is, a left-Leibniz algebra with commutative identity. Writing ${\mu }_0(x,y)=xy$, we have

$$\left(xy\right)z-x\left(yz\right)+y\left(xz\right)=0$$

and $v=Id,w=Id-{\tau }_{12}.$ For such multiplication, we have

$${\delta }_{v,w,{\mu }_0}^2\phi (x,y,z)=-x\phi (y,z)+y\phi (x,z)+z\phi (x,y)-\phi (x,yz)+\phi (y,xz)+\phi (xy,z)$$

and if ${\delta }_{v,w,{\mu }_0}^2\phi \circ {\Phi }_{v_1}=0$ for any bilinear map $\phi$, then $v_1=0.$ In fact, if $v_1=(a_1,\cdots ,a_6)$, then ${\delta }_{v,w,{\mu }_0}^2\phi \circ {\Phi }_{v_1}=0$ is considered as a linear equation on the formal variables

$$x\phi (y,z),y\phi (x,z),z\phi (x,y),\phi (x,yz),\phi (y,xz),\phi (xy,z),$$

each one of the coefficients of these variables being 0 implies $a_i=0$ for $i=1,\cdots ,6$. So, if ${\mu }_t$ is a left-Leibniz-formal deformation (that is, a $(Id,Id-{\tau }_{12})$-formal deformation) of ${\mu }_0$, the relation

$${\phi }_1{•}_{Id}^L{\phi }_1-{\phi }_1{•}_{Id-{\tau }_{12}}^R{\phi }_1+{\delta }_{Id,{\mu }_0}^{2,L}{\phi }_2-{\delta }_{Id-{\tau }_{12},{\mu }_0}^{2,R}{\phi }_2=0$$

cannot be reduced.

Remark 6. 

Maybe there exists some bilinear maps ${\phi }_2$ such that ${\delta }_{v,w,{\mu }_0}^2{\phi }_2\circ {\Phi }_{v_1}=0$ for some $v_1\ne 0$. But ${\phi }_2$ have to satisfy the system associated with the deformation equation ${\mu }_t•{\mu }_t=0$, which is very complicated to solve. For these reasons, we consider that the equation ${\delta }_{v,w,{\mu }_0}^2{\phi }_2\circ {\Phi }_{v_1}=0$ is solved for any bilinear map ${\phi }_2$.

Let us consider now the relation

$${\delta }_{v,{\mu }_0}^{2,L}{\phi }_1-{\delta }_{w,{\mu }_0}^{2,R}{\phi }_1=0$$

which is the order 1 consequence of the fact that ${\mu }_t$ is a $(Id,Id-{\tau }_{12})$ deformation of ${\mu }_0$. For any vector $v_1$, we have

$$({\delta }_{v,{\mu }_0}^{2,L}{\phi }_1-{\delta }_{w,{\mu }_0}^{2,R}{\phi }_1)\circ {\Phi }_{v_1}=0.$$

If $v_1=(a_1,a_2,a_3,-a_1,-a_2,-a_3)$, this equation gives:

$$\begin{array}{cc}\hfill a_{1}x{\psi }_1(y,z)+a_{2}y{\psi }_1(x,z)-a_{3}z{\psi }_1(x,y)+(a_3-a_2){\psi }_1(x,yz)& \\ \hfill -(a_1+a_3){\psi }_1(y,xz)+(a_1+a_2){\psi }_1(z,xy)& \hfill =0\end{array}$$

what is also written

$$\begin{array}{c}{a}_1(x{\psi }_1(y,z)-{\psi }_1(y,xz)+{\psi }_1(z,xy))\hfill \\ +a_2(y{\psi }_1(x,z)-{\psi }_1(x,yz)+{\psi }_1(z,xy))\hfill \\ +a_3(-z{\psi }_1(x,y)+{\psi }_1(x,yz)-{\psi }_1(y,xz))=0.\hfill \end{array}$$

and this is equivalent to the relation

$$x{\psi }_1(y,z)-{\psi }_1(y,xz)+{\psi }_1(z,xy)=0.$$

Definition 5. 

A pseudo-Poisson algebra $(A,{\mu }_0,{\psi }_1)$ is a $\mathbb{K}$-vector space A and two bilinear maps satisfying

1. 

${\mu }_0$ is a commutative left-Leibniz multiplication,

2. 

${\psi }_1$ is a skew-symmetric multiplication

3. 

we have the pseudo-Leibniz relation

$$x{\psi }_1(y,z)-{\psi }_1(y,xz)-{\psi }_1(xy,z)=0$$

for any $x,y,z\in A$.

Proposition 14. 

Let $(A,{\mu }_0)$ be a commutative left-Leibniz algebra. It is a $(Id,Id-{\tau }_{12})$-algebra, and any $(Id,Id-{\tau }_{12})$-formal deformation of ${\mu }_0$ is a deformation quantization of a pseudo-Poisson algebra $(A,{\mu }_0,{\psi }_1).$

5.4. Right-Leibniz Algebras

A right-Leibniz algebra $(A,\mu )$, sometimes just called Leibniz algebra, corresponds to the quadratic relation

$$\left(xy\right)z-x\left(yz\right)-\left(xz\right)y=0$$

with $xy$ for $\mu (x,y)$. It is a $(v,w)$ algebra with $v=Id-{\tau }_{23}$ and $w=Id$. Let us note that if $(u,v)\to uv$ is a left-Leibniz product, then $(u,v)\to vu$ is a right-Leibniz product.

Let us consider a commutative right-Leibniz algebra. In this case, the corresponding operator ${\delta }_{v,w,{\mu }_0}$ is given by

$${\delta }_{v,w,{\mu }_0}^2\phi (x,y,z)=z\phi (x,y)+\phi (xy,z)-x\phi (y,z)-\phi (x,yz)-y\phi (x,z)-\phi (xz,y).$$

As before, using the vector $v_1=(a_1,-a_1,a_3,a_4,-a_3,-a_4)$, the equation ${\delta }_{v,w,{\mu }_0}^2\phi (x,y,z)\circ {\Phi }_{v_1}=0$ is reduced to

$$z{\psi }_1(x,y)-{\psi }_1(x,yz)+{\psi }_1(y,xz)=0.$$

Remark that the commutative right-Leibniz multiplication ${\mu }_0$ is also a commutative left-Leibniz multiplication.

Proposition 15. 

Let $(A,{\mu }_0)$ be a commutative Leibniz algebra. It is a $(Id-{\tau }_{23},Id)$-algebra, and any $(Id-{\tau }_{23},Id)$-formal deformation of ${\mu }_0$ is a deformation quantization of a pseudo-Poisson algebra $(A,{\mu }_0,{\psi }_1).$

5.5. Symmetric Leibniz Algebras

A symmetric Leibniz algebra is an algebra $(A,\mu )$ such that for any $x,y,z\in A$, we have

$$\begin{array}{c}\left(xy\right)z-x\left(yz\right)+y\left(xz\right)=0,\hfill \\ \left(xy\right)z-x\left(yz\right)-\left(xz\right)y=0\hfill \end{array}$$

with $\mu (x,y)=xy$. Then, a symmetric Leibniz algebra is an algebra, that is, both left Leibniz and right Leibniz. We deduce immediately that if ${\mu }_t$ is a symmetric Leibniz deformation of a commutative symmetric Leibniz algebra, then

$$x{\psi }_1(y,z)-{\psi }_1(y,xz)-{\psi }_1(xy,z)=0,{\psi }_1(x,yz)-{\psi }_1(y,xz)-z{\psi }_1(x,y)=0$$

where ${\psi }_1$ is the skew-symmetric map associated with ${\phi }_1$ and ${\mu }_0(x,y)=xy$. In particular, we have

$$x{\psi }_1(y,z)+y{\psi }_1(z,x)+z{\psi }_1(x,y)=0.$$

In [2], we have introduced the notion of weakly associative algebras, that is, nonassociative algebras whose multiplication $\mu$ satisfies the identity:

$${\mathcal{A}}_{\mu }(x,y,z)+{\mathcal{A}}_{\mu }(y,z,x)-{\mathcal{A}}_{\mu }(y,x,z)=0.$$

If $(A,{\mu }_0)$ is a symmetric Leibniz algebra and if we denote by ${\mathcal{A}}_{{\mu }_0}$ the associator of the multiplication ${\mu }_0$, the first identity corresponds to

$${\mathcal{A}}_{{\mu }_0}(x,y,z)=-y\left(xz\right)$$

and the second to

$${\mathcal{A}}_{{\mu }_0}(x,y,z)=\left(xz\right)y.$$

We deduce that $(A,{\mu }_0)$ is a symmetric Leibniz algebra if and only if

$$\left\{\begin{array}{c}{\mathcal{A}}_{{\mu }_0}(x,y,z)=-y\left(xz\right),\hfill \\ {\mathcal{A}}_{{\mu }_0}(x,y,z)=\left(xz\right)y.\hfill \end{array}\right.$$

In particular, we deduce

$${\mathcal{A}}_{{\mu }_0}(x,y,z)+{\mathcal{A}}_{{\mu }_0}(y,z,x)=\left(yx\right)z-y\left(xz\right)={\mathcal{A}}_{{\mu }_0}(y,x,z)$$

and $(A,{\mu }_0)$ is also a weakly associative algebra.

Proposition 16 

([15]). Any symmetric Leibniz algebra is weakly associative.

As a consequence, we can consider weakly associative formal deformation of a symmetric Leibniz algebra and, in this case, we find the result of Section 4.5.

As the symmetric Leibniz are weakly associative, they are Lie-admissible, so if $(A,\mu )$ is a symmetric Leibniz, the algebra $(A,{\psi }_{\mu })$ is a Lie algebra, where ${\psi }_{\mu }$ is the skew-symmetric map associated with $\mu$.

Consider the vectors $v=Id,w=Id-{\tau }_{12},v^{\prime }=Id-{\tau }_{23},w^{\prime }=Id$ and a $(v,w,v^{\prime },w^{\prime })$-deformation of ${\mu }_0$, that is, a $(v,w)$-deformation of the $(v,w)$-algebra associated with the equation $\left(xy\right)z-x\left(yz\right)+y\left(xz\right)=0$ and a $(v^{\prime },w^{\prime })$-deformation of the $(v^{\prime },w^{\prime })$-algebra associated with the equation $\left(xy\right)z-x\left(yz\right)-\left(xz\right)y=0$. Since ${\mu }_0$ is a symmetric Leibniz multiplication, the equations coming from the order 2 of a $(v,w,v^{\prime },w^{\prime })$-deformation of ${\mu }_0$ are

$${\delta }_{v,w,{\mu }_0,L}^2{\phi }_2+{\phi }_1{•}_{v,w}{\phi }_1=0$$

and

$${\delta }_{v^{\prime },w^{\prime },{\mu }_0,R}^2{\phi }_2+{\phi }_1{•}_{v^{\prime },w^{\prime }}{\phi }_1=0$$

with

$${\delta }_{v,w,{\mu }_0,L}^2\phi (x,y,z)=x\phi (y,z)-y\phi (x,z)-z\phi (x,y)+\phi (x,yz)-\phi (y,xz)-\phi (xy,z)$$

and

$${\delta }_{v^{\prime },w^{\prime },{\mu }_0,R}^2\phi (x,y,z)=z\phi (x,y)+\phi (xy,z)-x\phi (y,z)-\phi (x,yz)-y\phi (x,z)-\phi (xz,y),$$

The equation

$$[{\delta }_{v,w,{\mu }_0,L}^2{\phi }_2+{\phi }_1{•}_{v,w}{\phi }_1]\circ {\Phi }_{Id-{\tau }_{23}}+[{\delta }_{v^{\prime },w^{\prime },{\mu }_0,R}^2{\phi }_2+{\phi }_1{•}_{v^{\prime },w^{\prime }}{\phi }_1]\circ {\Phi }_{{\tau }_{13}-c}=0$$

implies ${\phi }_1{•}_{v_{Lad}}{\phi }_1={\phi }_1•{\phi }_1\circ {\Phi }_{v_{Lad}}=0.$ We deduce that if ${\mu }_0$ is a Leibniz multiplication, then ${\phi }_1$ is Lie-admissible. As a consequence, if ${\mu }_0$ is a symmetric Leibniz multiplication, then ${\psi }_1$ is a Lie bracket.

Proposition 17. 

Let $(A,{\mu }_0)$ be a commutative symmetric Leibniz algebra. Any symmetric Leibniz-formal deformation of ${\mu }_0$ is a deformation quantization of a pseudo-Poisson algebra $(A,{\mu }_0,{\psi }_1)$.

6. Polarization and Depolarization of $\mathit{v}$-Associative Algebras

Any multiplication $\mu :A\otimes A\to A$ defined by a bilinear application can be decomposed into the sum of a commutative multiplication $\rho$ and a skew-symmetric one $\psi$ via the polarization defined by

$$\rho (x,y)={\displaystyle \frac{1}{2}}(xy+yx)\mathrm{and}\psi (x,y)={\displaystyle \frac{1}{2}}(xy-yx),\mathrm{for}x,y\in A.$$

(7)

where $\mu (x,y)$ is denoted by $xy$. The inverse process of depolarization assembles a commutative multiplication $\rho$ with a skew-symmetric multiplication $\psi$ into the multiplication $\mu$ defined for any $x,y\in V$ by

$$\mu (x,y)=\rho (x,y)+\psi (x,y).$$

(8)

In the following section, we first give the well-known associative example to illustrate the (de)polarization trick before investigating some other classes of algebras.

6.1. Associative Case [7]

Assume that $(A,\mu )$ is an associative algebra. If we polarize the multiplication $\mu$, it writes $\mu (x,y)=\rho (x,y)+\psi (x,y)$, and the associativity condition becomes equivalent to the following two axioms:

$$\begin{array}{ccc}\hfill \psi (x,\rho (y,z\left)\right)& =& \rho \left(\psi \right(x,y),z)+\rho (y,\psi (x,z\left)\right),\hfill \end{array}$$

(9)

$$\begin{array}{ccc}\hfill \psi (y,\psi (x,z\left)\right)& =& -{\mathcal{A}}_{\rho }(x,y,z).\hfill \end{array}$$

(10)

To verify this, observe that associativity is equivalent to

$$R(\rho ,\psi )=(\psi +\rho )\circ (Id\otimes \psi -\psi \otimes Id)+(\psi +\rho )(Id\otimes \rho -\rho \otimes Id)=0.$$

Moreover, it implies that $R(\rho ,\psi )\circ {\Phi }_v=0$ for any $v\in \mathbb{K}\left[{\Sigma }_3\right]$. In particular,

$$R(\rho ,\psi )\circ {\Phi }_{v_{Lad}}=-4\psi \circ (\psi \otimes Id)\circ {\Phi }_{Id+c+c^2}=0$$

and $\psi$ is a Lie bracket. Because of Relation (9), the algebra $(A,\rho ,\psi )$ is a nonassociative Poisson algebra in the general case. It is a Poisson algebra if and only if the Lie bracket $\psi$ is 2-step nilpotent.

Although the associative case is well known, we want to find a systematic method to solve this case which extends to the other identities that interest us. Let $a_{1}Id+a_2{\tau }_{12}+a_3{\tau }_{13}+a_4{\tau }_{23}+a_{5}c+a_6{c}^2$ be $\mathbb{K}\left[{\Sigma }_3\right]$ and let us consider the identity $R(\psi ,\rho )\circ {\Phi }_v=0.$ By grouping the terms $\psi (\mathrm{id}\otimes \psi )$, $\rho (\mathrm{id}\otimes \rho )$, $\rho (\mathrm{id}\otimes \psi )$ and finally $\psi (\mathrm{id}\otimes \rho )$, the coefficients of each of these terms are given by the matricial product

$$N_v\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\\ a_4\\ a_5\\ a_6\end{array}\right)$$

where $N_v$ is the transpose of the matrix

$$\left(\begin{array}{cccccccccccc}1& 0& 1& 1& 0& -1& 1& 0& -1& 1& 0& 1\\ 0& -1& -1& 0& 1& -1& 0& -1& 1& 0& 1& 1\\ -1& 0& -1& -1& 0& 1& 1& 0& -1& 1& 0& 1\\ -1& -1& 0& 1& -1& 0& -1& 1& 0& 1& 1& 0\\ 1& 1& 0& -1& 1& 0& -1& 1& 0& 1& 1& 0\\ 0& 1& 1& 0& -1& 1& 0& -1& 1& 0& 1& 1\end{array}\right)$$

The rank of $N_v$ is 6. Let us search the vectors of this space associated with minimal relations, that is to say with a maximum of 0 among these components. We obtain the independent vectors

$$\left\{\begin{array}{c}(0,0,0,0,0,0,0,1,-1,1,0,0)\hfill \\ (0,0,0,0,0,0,-1,0,1,0,1,0)\hfill \\ (0,0,0,0,0,0,1,-1,0,0,0,1)\hfill \end{array}\right.$$

which correspond to the vector $v=(1,-1,1,1,1,-1)$ and $cv,c^{2}v$ and the relation

$$\psi (x_1,\rho (x_2,x_3))-\rho (x_2,\psi (x_1,x_3))-\rho (x_3,\psi (x_1,x_2))=0.$$

This relation can be written $\mathcal{L}(\psi ,\rho )=0$, where $\psi$ is a Lie bracket and $\rho$ a commutative (nonassociative) multiplication. Similarly, we have the three independent vectors

$$\left\{\begin{array}{c}(0,1,0,-1,0,1,0,0,0,0,0,0)\hfill \\ (0,0,1,1,-1,0,0,0,0,0,0,0)\hfill \\ (1,0,0,0,1,-1,0,0,0,0,0,0)\hfill \end{array}\right.$$

which correspond to $v=(-1,-1,1,-1,1,1)$ and $cv,c^{2}v$ and to the relation

$$\psi (x_1,\psi (x_2,x_3))-\rho (x_3,\rho (x_1,x_2))+\rho (\rho (x_3,x_1),x_2)=0,$$

that is,

$${\mathcal{A}}_{\rho }+{\mathcal{A}}_{\psi }=0.$$

In fact, ${\mathcal{A}}_{\psi }(x_3,x_1,x_2)=-\psi (x_3,\psi (x_1,x_2))+\psi (\psi (x_3,x_1),x_2)=\psi (x_1,\psi (x_2,x_3)).$ This equation implies that $\psi$ is a Lie bracket.

Proposition 18. 

Any associative algebra is associated with polarization/depolarization principle to an algebra $(A,\psi ,\rho )$, where ψ is a Lie bracket, ρ a commutative multiplication satisfying

1. 

$\mathcal{L}(\psi ,\rho )=0$

2. 

${\mathcal{A}}_{\psi }+{\mathcal{A}}_{\rho }=0.$

In particular, if $\rho$ is associative, then $(A,\psi ,\rho )$ is a Poisson algebra with 2-step nilpotent Poisson bracket.

6.2. Lie-Admissible Case

Recall that a nonassociative algebra $(A,\mu )$ is Lie-admissible if the skew-symmetric bilinear map $\psi$ is a Lie bracket. In this case, the polarization principle gives no additional relation.

6.3. Vinberg Algebras

A nonassociative algebra $(A,\mu )$ is a Vinberg algebra if its associator satisfies

$${\mathcal{A}}_{\mu }(x,y,z)-{\mathcal{A}}_{\mu }(y,x,z)=0.$$

Since $v_{Lad}\in Span\left(\mathcal{O}(Id-{\tau }_{12})\right)$, such algebras are Lie-admissible. If we polarize the multiplication $\mu$, we obtain

$$\begin{array}{c}\psi (x_3,\psi (x_1,x_2))+\psi (x_1,\rho (x_2,x_3))-\psi \left(x_{,}\rho (x_1,x_3)\right)+\rho (x_1,\psi (x_2,x_3))+\rho (x_2,\psi (x_3,x_1))-\hfill \\ -2\rho (x_3,\psi (x_1,x_2))+\rho (x_1,\rho (x_2,x_3))-\rho (x_2,\rho (x_1,x_3))=0.\hfill \end{array}$$

This relation is also written as the sum

$$\begin{array}{c}\left\{(x_1•x_3)•x_2-x_1•(x_3•x_2)-[x_3,[x_1,x_2]]\right\}+\left\{[x_1,x_2]•x_3+x_2•[x_1,x_3]-[x_1,x_2•x_3]\right\}\hfill \\ +\left\{[x_2,x_1•x_3]-x_1•[x_2,x_3]-[x_2,x_1]•x_3\right\}=0\hfill \end{array}$$

of three terms which vanish separately if the multiplication is associative, where $x•y=\rho (x,y)$ and $[x,y]=\psi (x,y)$.

6.4. ${\tau }_{13}$-Algebras

These are the algebras $(A,\mu )$ defined by the quadratic relation

$${\mathcal{A}}_{\mu }(x_1,x_2,x_3)-{\mathcal{A}}_{\mu }(x_3,x_2,x_1)=0.$$

This relation is equivalent to

$$\psi (x_1,\psi (x_2,x_3))-{\mathcal{A}}_{\rho }(x_3,x_1,x_2)=0$$

and $\psi$ is a Lie bracket. This relation is minimal. It can also be written

$${\mathcal{A}}_{\psi }+{\mathcal{A}}_{\rho }=0.$$

6.5. $(Id+c+c^2)$-Algebras

We have

$${\mathcal{A}}_{\mu }(x_1,x_2,x_3)+{\mathcal{A}}_{\mu }(x_2,x_3,x_1)+{\mathcal{A}}_{\mu }(x_3,x_1,x_2)=0.$$

In this case, $\psi$ is a Lie bracket and the polarization principe gives after reduction

$$\psi (x_1,\rho (x_2,x_3))+\psi (x_2,\rho (x_3,x_1))+\psi (x_3,\rho (x_1,x_2))=0.$$

It is an identity similar to Equation (4) obtained by deformation, and $(A,\rho ,\psi )$ is a nonassociative $(Id+c+c^2)$-Poisson algebra. Let us note also that a $(Id+c+c^2)$-algebra is Lie-admissible and 3-power-associative.

6.6. Weakly Associative Algebras

This class of nonassociative algebras has been studied in [2] to extend the notion of deformation quantification for associative commutative algebras. Recall that a nonassociative algebra $(A,\mu )$ is weakly associative if we have

$${\mathcal{A}}_{\mu }(x_1,x_2,x_3)+{\mathcal{A}}_{\mu }(x_2,x_3,x_1)-{\mathcal{A}}_{\mu }(x_2,x_1,x_3)=0.$$

From [2], this identity is equivalent to:

• $\psi$ is a Lie bracket,
• $\rho$ is a commutative multiplication satisfying

$$\psi (x_1,\rho (x_2,x_3))-\rho (x_2,\psi (x_1,x_3))-\rho (x_3,\psi (x_1,x_2)).$$

In other words, $(\rho ,\psi )$ satisfy the Leibniz identity: $\mathcal{L}(\psi ,\rho )=0.$

Proposition 19. 

Let $(A,\mu )$ be a weakly associative algebra and $(A,\rho ,\psi )$ its polarized version. Then

1. 

$(A,\psi )$ is a Lie algebra,

2. 

ρ is a commutative multiplication,

3. 

the multiplications ρ and ψ are tied up by the Leibniz identity

$$\mathcal{L}(\rho ,\psi )=0,$$

that is, $(A,\rho ,\psi )$ is a nonassociative Poisson algebra.

Remark 7. 

If we refer to [2,5], weak associativity corresponds to a point of the family of nonassociative algebras corresponding to the identity

$${\mathcal{C}}_{\mu }\left(\alpha \right)(x,y,z)=2{\mathcal{A}}_{\mu }(x,y,z)+(1+\alpha ){\mathcal{A}}_{\mu }(y,x,z)+{\mathcal{A}}_{\mu }(z,y,x)+{\mathcal{A}}_{\mu }(y,z,x)+(1-\alpha ){\mathcal{A}}_{\mu }(z,x,y)=0$$

with $\alpha =-1/2.$ In fact, considering the vectors $v=Id-{\tau }_{12}+c$ and $v^{\prime }=2Id+{\displaystyle \frac{1}{2}}{\tau }_{12}+{\tau }_{13}+c+{\displaystyle \frac{3}{2}}{c}^2$, we have from [5]$dim{F}_{v^{\prime }}=dim{F}_v=4$. Since $(Id+{\displaystyle \frac{3}{2}}{\tau }_{13}+{\tau }_{23}+{\displaystyle \frac{3}{2}}c+c^2)\circ v=v^{\prime }$, we deduce $F_v=F_{v^{\prime }}$.

If we consider the vector $v_1={\displaystyle \frac{1}{3}}Id-{\tau }_{12}+{\displaystyle \frac{7}{12}}{\tau }_{13}+{\displaystyle \frac{1}{4}}{c}^2$, then the polarization of ${\mathcal{C}}_{\mu }\left(\alpha \right)\circ {\Phi }_{v_1}=0$ gives the relation

$$\mathcal{L}(\rho ,\psi )(x_1,x_2,x_3)-\gamma \psi (x_1,\psi (x_2,x_3))-2\psi (x_3,\psi (x_1,x_2))=0$$

with $\gamma ={\displaystyle \frac{2}{3}}(2\alpha -1)$. Since $v_1$ is inversible in the algebra $\mathbb{K}\left[{\Sigma }_3\right]$, this relation is equivalent to ${\mathcal{C}}_{\mu }\left(\alpha \right)=0.$

Proposition 20. 

Let $v=2Id+(1+\alpha ){\tau }_{12}+{\tau }_{13}+c+(1-\alpha )c^2$ with $\alpha \ne 1$. Then, any v-algebra is Lie-admissible and 3-power-associative. The relation ${\mathcal{A}}_{\mu }\circ {\Phi }_v=0$ is equivalent to

$$\mathcal{L}(\rho ,\psi )(x_1,x_2,x_3)-\gamma (\psi (x_1,\psi (x_2,x_3))-2\psi (x_3,\psi (x_1,x_2))$$

with $\gamma ={\displaystyle \frac{2}{3}}(2\alpha -1)$. In particular, if $\alpha ={\displaystyle \frac{1}{2}}$, then $(A,\mu )$ is weakly associative and we have in this case

$$\mathcal{L}(\rho ,\psi )=0,$$

that is, $(A,\psi ,\rho )$ is a nonassociative Poisson algebra.

Remark 8. 

Polarization of a 3-power associative algebra $(A,\mu ).$ The equation corresponding to 3-power associativity

$${\mathcal{A}}_{\mu }(x_1,x_2,x_3)+{\mathcal{A}}_{\mu }(x_2,x_1,x_3)+{\mathcal{A}}_{\mu }(x_1,x_3,x_2)+{\mathcal{A}}_{\mu }(x_3,x_2,x_1)+{\mathcal{A}}_{\mu }(x_2,x_3,x_1)+{\mathcal{A}}_{\mu }(x_3,x_1,x_2)=0$$

is equivalent to

$$\psi (x_1,\rho (x_2,x_3))+\psi (x_2,\rho (x_3,x_1))+\psi (x_3,\rho (x_1,x_2))=0.$$

Thus, the polarized version of $(A,\mu )$ is the algebra $(A,\rho ,\psi )$, where the commutative multiplication ρ and skew-symmetric multiplication ψ (which is not a Lie bracket) are linked by a $(Id+c+c^2)$-Leibniz rule.

6.7. Poisson Algebras

The depolarization of Poisson algebras has been conducted in [7,15]. It reinterprets Poisson algebras as structures with one nonassociative product which satisfies

$$3\left(x\right(yz)-(xy\left)z\right)=-\left(xz\right)y-\left(yz\right)x+\left(yx\right)z+\left(zx\right)y.$$

7. Polarization of ${\left(\mathbb{K}\left[{\Sigma }_{\mathbf{3}}\right]\right)}^{\mathbf{2}}$-Algebras

7.1. Polarization of Anti-Associative Algebras

Let $(A,\mu )$ be an anti-associative algebra:

$${\mathcal{AA}}_{\mu }(x,y,z)=\mu (x,\mu (y,z))+\mu (\mu (x,y),z)=0.$$

(11)

Let $\psi$ and $\rho$ be the skew-symmetric and symmetric bilinear maps associated to $\mu$ by the polarization principle. To simplify the notations, we put $\mu (x,u)=xy,\psi (x,y)=[x,y],\rho (x,y)=x•y$. The identity (11) is equivalent to

$$x•(y•z)+z•(x•y)+x•[y,z]+z•[x,y]+[x,y•z]-[z,x•y]+[x,[y,z\left]\right]-[z,[x,y\left]\right]=0.$$

We denote by $\mathcal{AA}(•,[,\left]\right)(x,y,z)=0$ this identity. If $v=Id+{\tau }_{13}$, then $\mathcal{AA}(•,[,])\circ {\Phi }_v(x,y,z)=0$ is equivalent to

$$x•(y•z)+z•(x•y)+[x,[y,z\left]\right]-[z,[x,y\left]\right]=0$$

which can also be written

$${\mathcal{AA}}_{•}+{\mathcal{AA}}_{[,]}=0.$$

It implies that

$$({\mathcal{AA}}_{•}+{\mathcal{AA}}_{[,]})\circ {\Phi }_{Id+c+c^2}=0,$$

that is,

$$x•(y•z)+y•(z•x)+z•(x•y)=0.$$

We obtain the following result:

Proposition 21. 

Let $(A,\mu )$ be an anti-associative algebra. If ρ is the symmetric map attached to μ, the algebra $(A,\rho )$ is a Jacobi–Jordan algebra.

Recall that these algebras have been studied in [9], where it is proven that these algebras are commutative nilalgebra of index at most three and conversely. The authors give also classifications for the dimension less than or equal to 5.

Let us note that the Jacobi–Jordan algebras, which arise from anti-associative algebras, satisfy also

$$x•(y•(z•t\left)\right)=(x•y)•(z•t)=0,$$

that is, all the product of order 4 are null (see Section 5.2).

Let us now consider a vector v with

$$a_6=a_1-a_2+a_3,a_5=a_1+a_3-a_4.$$

In this case, $\mathcal{AA}(•,[,])\circ {\Phi }_v(x,y,z)=0$ is equivalent to

$$\begin{array}{c}{\lambda }_1(x•[y,z]+y•[z,x]-[x,y•z]+[y,z•x])+{\lambda }_2(z•[x,y]+[x,y•z]-[y,z•x])\hfill \\ +{\lambda }_3(y•[z,x]-[x,y•z]+[z,x•y])=0,\hfill \end{array}$$

with ${\lambda }_1=a_1-a_4,{\lambda }_2=a_1-a_2,{\lambda }_3=a_3-a_2.$ This identity is equivalent to

$$\left\{\begin{array}{c}x•[y,z]+y•[z,x]-[x,y•z]+[y,z•x]=0,\hfill \\ z•[x,y]+[x,y•z]-[y,z•x]=0,\hfill \\ y•[z,x]-[x,y•z]+[z,x•y]=0,\hfill \end{array}\right.$$

or to the axiom

$$x•[y,z]+[x•y,z]+[y,z•x]=0.$$

In fact, if $G(x,y,z)=x•[y,z]+[x•y,z]+[y,z•x]=0,$ then the previous system writes

$$\left\{\begin{array}{c}G(x,y,z)+G(y,z,x)=0,\hfill \\ G(z,x,y)=0,\hfill \\ G(y,z,x)=0.\hfill \end{array}\right.$$

Then, $G(x,y,z)+G(y,z,x)+G(z,x,y)=0$, which gives

$$x•[y,z]+y•[z,x]+z•[x,y]=0.$$

Proposition 22. 

Let $(A,\mu )$ be an anti-associative algebra. If $\psi ,\rho$ are the skew-symmetric and symmetric maps attached to μ, then

1. 

The algebra $(A,\rho )$ is a Jacobi–Jordan algebra.

2. 

This algebra $(A,\rho )$ acts as antiderivation on the skew-symmetric algebra $(A,\psi )$.

In fact, if $f_x\left(y\right)=x•y$, then $x•[y,z]=-[x•y,z]-[y,z•x]$ can be written

$$f_x[y,z]=-[f_x\left(y\right),z]-[y,f_x\left(z\right)].$$

Remark on the graded Leibniz identity. 

The previous identity can be written with the Leibniz identity considering a degree on the operation:

Definition 6. 

If $\varrho ,\eta$ are two multiplications that are symmetric or skew-symmetric, we consider $\left|\varrho \right|$ and $\left|\eta \right|$ their degree, which is 0 if the operation is skew-symmetric and 1 if the operation is skew-symmetric. We call graded Leibniz identity on η and ϱ

$${\mathcal{L}}_g(\eta ,\varrho )(x,y,z)=\varrho (x,\eta (y,z))+{(-1)}^{\left|\varrho \right|}\eta (y,\varrho (x,z))+{(-1)}^{\left|\varrho \right|}\eta (\varrho (x,y),z).$$

(12)

So, we obtain with $\psi$ which is skew-symmetric and $\rho$ which is symmetric,

• $$$$
  $${\mathcal{L}}_g(\rho ,\psi )(x,y,z)=\psi (x,\rho (y,z))-\rho (y,\psi (x,z))-\rho (\psi (x,y),z)=0,$$
  that is, the classical Leibniz identity
  $$[x,y•z]-y•[x,z]-[x,y]•z$$
  and the skew-symmetric $(A,\psi )$ acts as a derivation on the symmetric algebra $(A,\rho )$
• $$$$
  $${\mathcal{L}}_g(\psi ,\rho )(x,y,z)=\rho (x,\psi (y,z))+\psi (y,\rho (x,z))+\psi (\rho (x,y),z)=0,$$
  that is,
  $$x•[y,z]+[y,x•z]+[x•y,z]$$
  and the symmetric $(A,\rho )$ acts as an antiderivation on the skew-symmetric algebra $(A,\psi )$
• $$$$
  $${\mathcal{L}}_g(\psi ,\psi )(x,y,z)=\psi (x,\psi (y,z))-\psi (y,\psi (x,z))-\psi (\psi (x,y),z),$$
  that is,
  $$[x,[y,z\left]\right]+[y,[x,z\left]\right]+[z,[x,y\left]\right]$$
  and the classical Jacobi equation $[x,[y,z\left]\right]+[y,[x,z\left]\right]+[z,[x,y\left]\right]=0$ writes ${\mathcal{L}}_g(\psi ,\psi )=0$
• $$$$
  $${\mathcal{L}}_g(\rho ,\rho )(x,y,z)=\rho (x,\rho (y,z))+\rho (y,\rho (x,z))+\rho (\rho (x,y),z),$$
  that is,
  $$x•(y•z)+y•(z•x)+z•(x•y)$$

• Then, in the graded case, some relations obtained previously become natural.

7.2. A Remark on the Jacobi–Jordan Algebras

In the previous section, we have seen that the operad associated with the anti-associative algebra was not Koszul, implying that the cohomology of deformations of these algebras was the cohomology of the minimal model. It was maybe interesting to look at this problem for the Jacobi–Jordan algebra (this problem is analogous to compare Hochschild and Harrison cohomologies for associative algebras). We denote by $\mathcal{JJ}ss$ the operad associated to the Jacobi–Jordan algebras. It is clear that

$$dim\mathcal{JJ}ss\left(2\right)=1,dim\mathcal{JJ}ss\left(3\right)=2.$$

The vector space $\mathcal{JJ}ss\left(4\right)$ is generated by the element $a\left(b\right(cd\left)\right)$ and their images by $Id\otimes c$ where c is a cycle in ${\Sigma }_3$ and $\left(ab\right)\left(cd\right)$ and their images by $\tau \otimes \tau$ where $\tau$ is the generator of ${\Sigma }_2$. Then, we obtain 15 generators. The commutativity and the Jacobi–Jordan condition imply that we have 10 independent relations. We deduce

$$dim\mathcal{JJ}ss\left(4\right)=5.$$

Recall also that if a quadratic operad $\mathcal{P}$ is Koszul, then its Poincaré series $g_P\left(t\right)=-t+{\sum }_{k\ge 2}{(-1)}^k{\displaystyle \frac{dimP\left(k\right)}{k!}}{t}^k$ and the Poincaré series of its dual ${\mathcal{P}}^{!}$ are tied by the functional equation $g_P(-g_{P!}(-t))=t.$ Since $g_{\mathcal{JJ}ss}\left(t\right)=-t+{\displaystyle \frac{1}{2}}{t}^2-{\displaystyle \frac{1}{3}}{t}^3+{\displaystyle \frac{5}{24}}{t}^4+\cdots$, the inverse series is

$$a\left(t\right)=-t+{\displaystyle \frac{1}{2}}{t}^2-{\displaystyle \frac{1}{6}}{t}^3+\cdots$$

On the other hand, any algebra on the dual operad $\mathcal{JJ}s{s}^{!}$ is anti-associative and skew-symmetric. This can be viewed by computing the ideal of relations of this operad. In fact, if $<,>$ denotes the inner product which defines the dual operad of a quadratic operad, we have, for every $\sigma \in {\Sigma }_3$, $<x_{\sigma \left(1\right)}\left(x_{\sigma \left(2\right)}{x}_{\sigma \left(3\right)}\right),x_{\sigma \left(1\right)}\left(x_{\sigma \left(2\right)}{x}_{\sigma \left(3\right)}\right)>=sgn\left(\sigma \right),<\left(x_{\sigma \left(1\right)}{x}_{\sigma \left(2\right)}\right)x_{\sigma \left(3\right)},\left(x_{\sigma \left(1\right)}{x}_{\sigma \left(2\right)}\right)x_{\sigma \left(3\right)}>=-sgn\left(\sigma \right)$, implying that $<x_1\left(x_2{x}_3\right)+x_2\left(x_3{x}_1\right)+x_3\left(x_1{x}_2\right),x_1\left(x_2{x}_3\right)-x_3\left(x_1{x}_2\right)>=0$ but $\mathcal{JJ}ss$-algebras are commutative so $\mathcal{JJ}s{s}^{!}$-algebras are skew-symmetric and $x_1\left(x_2{x}_3\right)+\left(x_1{x}_2\right)x_3$ is in the ideal of relations and a $\mathcal{JJ}s{s}^{!}$-algebra is anti-associative. We deduce, from the anti-associativity, that $\mathcal{JJ}s{s}^{!}\left(4\right)=0$. Since we have also

$$dim\mathcal{JJ}s{s}^{!}\left(2\right)=1,dim\mathcal{JJ}s{s}^{!}\left(3\right)=3,$$

we deduce that the generating series of $\mathcal{JJ}s{s}^{!}$ is

$$g_{\mathcal{JJ}s{s}^{!}}\left(t\right)=-t+{\displaystyle \frac{1}{2}}{t}^2-{\displaystyle \frac{1}{2}}{t}^3.$$

and cannot be a Poincaré series of a quadratic operad. Then,

Proposition 23. 

The operad $\mathcal{JJ}ss$ of the commutative Jacobi–Jordan algebras is not Koszul. In particular, the cohomology of deformations of a Jacobi–Jordan algebra is the cohomology of the minimal model.

The determination of the minimal model is similar to those proposed in [8].

Let us note that the Koszulness of a quadratic operad can be read on associated free algebras. More precisely, a quadratic operad is Koszul if the corresponding free algebras are Koszul algebras. Let us determine the free Jacobi–Jordan algebra. We denote by $JJ\left(X\right)$ the (nonunitary) free algebra with one generator. Since in a Jacobi–Jordan algebra we have $x^3=0$, then

$$JJ\left(X\right)=\mathbb{K}\{X,X^2\}.$$

Let us denote by $JJ(X,Y)$ the (nonunitary) free algebra with two generators. It is a graded algebra $JJ(X,Y)={\oplus }_{k\ge 1}J{J}_d(X,Y)$ where $J{J}_d(X,Y)$ is the subspace of vectors of degree k. We have

$$J{J}_1(X,Y)=\mathbb{K}\{X,Y\},J{J}_2(X,Y)=\mathbb{K}\{X^2,Y^2,XY\}.$$

To compute $J{J}_3(X,Y)$, we consider the terms

$$X\left(XX\right)=\left(XX\right)X=X^3,Y^3,X\left(XY\right),Y\left(XX\right),Y\left(YX\right),X\left(YY\right).$$

We know that $X^3=Y^3=0.$ We have also

$$2X\left(XY\right)+X^{2}Y=0,2Y\left(XY\right)+X{Y}^2=0.$$

Then

$$J{J}_3(X,Y)=\mathbb{K}\{X^{2}Y,X{Y}^2\}.$$

Let us now consider the terms of degree 4. Recall that for the Jacobi–Jordan algebras arising from anti-associative algebras, any term of degree at least equal to 4 is 0. Let us look now at the general case. It is clear that

$$X^2{X}^2=X{X}^3=Y^2{Y}^2=Y{Y}^3=X{Y}^3=X^{3}Y=0.$$

For the other terms, that is,

$$\left(XY\right)\left(XY\right),X\left(X\left(XY\right)\right),X\left(X^{2}Y\right),X\left(X{Y}^2\right),X\left(Y\left(XY\right)\right),Y\left(Y\left(XY\right)\right),Y\left(X{Y}^2\right),Y\left(X^{2}Y\right)$$

we have

• $X\left(X\right(XY\left)\right)+X\left(X\right(YX\left)\right)+X\left(Y\right(XX\left)\right)=0,$ that is, $2X\left(X\left(XY\right)\right)=-X\left(X^{2}Y\right)$,
• $X\left(X^{2}Y\right)+X^2\left(XY\right)+Y\left(X^{2}X\right)=0,$ that is, $X\left(X^{2}Y\right)=-X^2\left(XY\right)$,
• $X\left(X\right(XY\left)\right)+X\left(\right(XY\left)X\right)+\left(XY\right)\left(XX\right)=0,$ that is, $2X\left(X\left(XY\right)\right)=-X^2\left(XY\right).$

• We deduce

  $$2X\left(X\left(XY\right)\right)=-X\left(X^{2}Y\right)=X^2\left(XY\right)=-X^2\left(XY\right),$$

  that is,

  $$X\left(X\left(XY\right)\right)=X\left(X^{2}Y\right)=X^2\left(XY\right)=0.$$

  Likewise

  $$Y\left(Y\left(XY\right)\right)=Y\left(Y^{2}X\right)=Y^2\left(XY\right)=0.$$

  In other words, if a term contains a variable of degree 3, this term vanishes. As for the other terms, we have

• $X\left(X{Y}^2\right)+X\left(Y\left(XY\right)\right)+X\left(X{Y}^2\right)=0,$ that is, $2X\left(X{Y}^2\right)=-X\left(\left(XY\right)Y\right)$,
• $X\left(X{Y}^2\right)+X\left(Y^{2}X\right)+Y^2\left(XX\right)=0$, that is, $2X\left(X{Y}^2\right)=-X^2{Y}^2$,
• $\left(X\right(XY\left)\right)Y+\left(\right(XY\left)Y\right)X+\left(YX\right)\left(XY\right)=0,$ that is, (XY)(XY) = − (X(XY))Y − X((XY)Y)

• and we deduce

  $$X^2{Y}^2=-2X\left(X{Y}^2\right)=-2Y\left(X^{2}Y\right)=4X\left(\left(XY\right)Y\right)=4(Y\left(\left(XY\right)X\right)=-2{\left(XY\right)}^2.$$

  Then

  $$J{J}_4(X,Y)=\mathbb{K}\left\{X^2{Y}^2\right\}.$$

  Let us consider now the terms of degree 5. Commutativity allows to consider only the products schematized by

  $$★(★(★(★★)),(★★\left)\right(★(★★)),(★★\left)\right(★(★★)).$$

  In the first case, the relations $X\left(X\left(XY\right)\right)=X\left(X^{2}Y\right)=X^2\left(XY\right)=0,Y\left(Y\left(XY\right)\right)=Y\left(Y^{2}X\right)=Y^2\left(XY\right)=0,X\left(\left(XY\right)Y\right)=aX\left(X{Y}^2\right)=b{X}^2{Y}^2$ show that these products of degree 5 are reduced to products of type $X\left(X^2{Y}^2\right)$, which are also null. In the second case, we have to compute products of type $X\left(\right(XY\left)\right(XY\left)\right)$ or $X\left(X^2\left(XY\right)\right)$ or $X\left(Y^2\left(XY\right)\right)$ or $X\left(X^2{Y}^2\right)$. In all these cases, these products are 0. In the third case, the Jacobi–Jordan relation and the computation

  $$\left(X\left(X^{2}Y\right)\right)X=\left(X^2{Y}^2\right)X=0,\left(Y\left(X^{2}Y\right)\right)Y=\left(X^2{Y}^2\right)Y=0$$

  shows also that these products are zero.

Then, we have

$$J{J}_5(X,Y)=0.$$

Proposition 24. 

The free Jacobi–Jordan algebra with two generators $JJ(X,Y)$ if of finite dimension and

$$JJ(X,Y)=\mathbb{K}\{X,Y\}\oplus \mathbb{K}\{X^2,Y^2,XY\}\oplus \mathbb{K}\{X^{2}Y,X{Y}^2\}\oplus \mathbb{K}\left\{X^2{Y}^2\right\}.$$

Let us now look at the dual algebra, that is, anti-associative skew-symmetric algebra. We denote by $AAS(X_1,\cdots ,X_n)$ the free anti-associative skew-symmetric algebra with n-generators. We know that all the product of degree 4 are zero. Then, $AAS(X_1,\cdots ,X_n)$ is a subalgebra of ${\sum }_{1\le k\le 3}AA{S}^k(X_1,\cdots ,X_n)$. Let us determine these algebras.

• $AAS\left(X\right)=\mathbb{K}\left\{X\right\}$. In fact, by antisymmetry $X^2=0$.
• $AAS(X,Y)=\mathbb{K}\{X,Y,XY\}$. In fact, $X^2=Y^2=0$, $X^3=Y^3=0$ and by anti-associativity, $X\left(XY\right)=0.$ In this case, all the products of degree 3 are zero.
• $AAS(X,Y,Z)=\mathbb{K}\{X,Y,Z,XY,XZ,YZ,X(YZ\left)\right\}$.
• $AAS(X_1,\cdots ,X_n)=\mathbb{K}\{X_1,\cdots ,X_n\}\oplus \mathbb{K}\{X_i{X}_j,1\le i<j\le n\}\oplus \mathbb{K}\{X_i\left(X_l{X}_k\right),1\le i<j<k\le n\}$.

• The Hilbert serie of a graded algebra $A=\oplus A_k$ is ${\sum }_{k\ge 0}dim\left(A_k\right)t^k$. Then, these series are

• for $AAS\left(X\right)$: $1+t$
• for $AAS(X,Y)$: $1+2t+t^2={(1+t)}^2$
• for $AAS(X,Y,Z)$: $(1+3t+3t^2+t^3={(1+t)}^3$
• for $AAS(X_1,\cdots ,X_n),n\ge 4$: $1+nt+\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)t^2+\left({\displaystyle \genfrac{}{}{0pt}{}{n}{3}}\right)t^3.$

• From [16], $1+t$, ${(1+t)}^2$, ${(1+t)}^3$ are the Hilbert series of Koszul algebras. For the other cases, this problem will be solved later.

7.3. Leibniz Algebras

Recall that a Leibniz algebra is a quadratic algebra whose multiplication $\mu (x,y)=xy$ satisfies the identity

$$x\left(yz\right)-\left(xy\right)z-y\left(xz\right)=0.$$

(13)

Let $(\rho ,\psi )$ be the pair of bilinear maps given by the polarization of $\mu$. As in the previous case, we write $\mu (x,y)=xy,\psi (x,y)=[x,y],\rho (x,y)=x•y$. Then, (13) is equivalent to

$$\begin{array}{c}R(\rho ,\psi )(x,y,z)=x•(y•z)-(x•y)•z-y•(x•z)+x•[y,z]-[x,y]•z-y•[x,z]\hfill \\ +[x,y•z]-[x•y,z]-[y,x•z]+[x,[y,z\left]\right]-\left[\right[x,y],z]-[y,[x,z\left]\right]=0.\hfill \end{array}$$

Let $v=(a_1,a_2,a_3,a_4,a_5,a_6)$ be in $\mathbb{K}\left[{\Sigma }_3\right]$ where the $a_i$ are the components in the canonical basis. In a first time, we consider the vector $v=(1,1,0,0,0,0).$ Then, $R(\rho ,\psi )\circ {\Phi }_v$ gives in this case the following identity:

$$x•(y•z)=[x,y•z].$$

This implies

$$x•[y,z]-[x,y]•z-y•[x,z]+2[x,y•z]-2[y,x•z]+[x,[y,z\left]\right]-\left[\right[x,y],z]-[y,[x,z\left]\right]=0.$$

Composing this last identity by $\Phi \left(v\right)$ with $v=(a_1,a_2,a_3,a_4,a_1-a_2+a_3,a_1-a_2+a_4)$, we obtain

$$x•[y,z]+y•[z,x]+z•[x,y]+3\left(\right[x,[y,z]]-[[x,y],z]-[y,[x,z]\left]\right)=0.$$

Then, we obtain

$$[x,y•z]+[x•z,y]-z•[x,y]=J_{[,]}(x,y,z)$$

where $J_{[,]}(x,y,z)=[x,[y,z]]-[[x,y],z]-[y,[x,z]]$ is the Jacobi condition for the skew-symmetric map $\psi$.

Proposition 25. 

Let $(A,\mu )$ a Leibniz algebra. It is associated, from the polarization / depolarization principle with a triple $(A,\rho ,\psi )$ where $\rho (x,y)=x•y$ is a commutative multiplication, on A, $\psi (x,y)=[x,y]$ a skew-symmetric multiplication on A satisfying

1. 

$x•[y,z]-[x,y]•z=[x,y•z]$,

2. 

$[x,y•z]+[x•z,y]-z•[x,y]=J\left([,]\right)(x,y,z)$

• where $J\left(\psi \right)$ is the Jacobiator ( $\psi \circ (Id\otimes \psi )\circ {\Phi }_{Id+c+c^2})$, for any $x,y,z\in A$.

Corollary 3. 

With the hypothesis of the previous proposition, assume now that $\psi (x,y)=[x,y]$ is a Lie bracket. In this case, for any $z\in A$, the map $f_z\left(y\right)=x•y$ is a derivation of the Lie algebra $(A,\psi )$.

In fact, $[x,y•z]+[x•z,y]-z•[x,y]=J\left([,]\right)(x,y,z)$ is reduced to

$$[x,f_z\left(y\right)]+[f_z\left(x\right),y]-f_z\left([x,y]\right)=0.$$

Remark 9 

(Case of symmetric Leibniz algebras). Recall that such algebras correspond to the two identities:

$$\left\{\begin{array}{c}x\left(yz\right)-\left(xy\right)z-y\left(xz\right)=0,\hfill \\ \left(xy\right)z-x\left(yz\right)-\left(xz\right)y=0.\hfill \end{array}\right.$$

This pair of relations is equivalent to

$${\mathcal{A}}_{\mu }(x,y,z)-y\left(xz\right)=0,{\mathcal{A}}_{\mu }(y,z,x)+\left(yx\right)z=0$$

that implies

$${\mathcal{A}}_{\mu }(x,y,z)+{\mathcal{A}}_{\mu }(y,z,x)-{\mathcal{A}}_{\mu }(y,x,z)=0.$$

Proposition 26. 

Any symmetric Leibniz algebra is weakly associative. In particular $(A,\rho ,\psi )$ is a nonassociative Poisson algebra.

Let us note that the first half of this proposition is the content of Proposition 16.

8. Deformation Quantization and Polarization

Previous studies show that in many cases, if not almost all, there is a close link between the algebras obtained by the formal deformation process and that of polarization. We will summarize this link.

Type of algebras	Type of algebras appearing	Type of algebras appearing
	in formal deformations	by polarization process
Associative	associative deformation:
	Poisson algebra	Nonassociative Poisson algebra
	weakly associative deformation:
	Nonassociative Poisson algebra
Lie-admissible	Lie-admissible algebra	Lie algebra
$(Id+c+c^2)$-associative	Nonassociative $(Id+c+c^2)$-Poisson algebra	Nonassociative $(Id+c+c^2)$-Poisson algebra
i.e., $G_5$-associative
Vinberg algebras	Lie-admissible algebra with
	$(Id-{\tau }_{12})$-Leibniz condition
Weakly associative	Nonassociative Poisson	Nonassociative Poisson
Anti-associative	Anti-Poisson algebras =	Anti-Poisson algebra =
	Jacobi–Jordan algebra	Jacobi-Jordan algebra
Leibniz	Pseudo-Poisson	Pseudo-Poisson
Symmetric Leibniz	Pseudo-Poisson	Nonassociative Poisson

9. On the Existence of Quantization

In this section, we will call algebra of Poisson type an algebra with two multiplications ${\mu }_0(a,b)=ab,[a,b]$, the first ${\mu }_0$ checks a quadratic relation ${\mu }_0•{\mu }_0=0$, the second is a Lie bracket, these multiplication being connected by a distributive law of Poisson type, for example, v-Leibniz, anti-Leibniz, as seen in the previous sections, or opposite Leibniz, that is,

$$[ab,c]+a[b,c]+[a,c]b=0$$

called also anti-Leibniz in [17]. The problem is whether there exists a formal deformation of ${\mu }_0$ which is a quantization of this Poisson-type algebra. It is sufficient to find a bilinear map ${\phi }_1$ such that

$${\phi }_1•{\mu }_0+{\mu }_0•{\phi }_1=0$$

and this identity has to give the Leibniz-type identity, a second bilinear map satisfying

$${\phi }_1\circ {\phi }_1+{\phi }_2•{\mu }_0+{\mu }_0•{\phi }_2=0$$

and for which we can find a vector $v\in \mathbb{K}\left[{\Sigma }_3\right]$ such that $({\phi }_2•{\mu }_0+{\mu }_0•{\phi }_2)\circ v=0$ and ${\phi }_1$ is Lie-admissible. Any quadratic relation on ${\mu }_0$ is written

$$\begin{array}{cc}{a}_1\left(x_1{x}_2\right)x_3+a_2\left(x_2{x}_1\right)x_3+a_3\left(x_3{x}_2\right)x_1+a_4\left(x_1{x}_3\right)x_2+a_5\left(x_2{x}_3\right)x_1+a_6\left(x_3{x}_1\right)x_2\hfill & \\ +b_1{x}_1\left(x_2{x}_3\right)+b_2{x}_2\left(x_1{x}_3\right)+b_3{x}_3\left(x_2{x}_1\right)+b_4{x}_1\left(x_3{x}_2\right)+b_5{x}_2\left(x_3{x}_1\right)+b_6{x}_3\left(x_1{x}_2\right)\hfill & =0.\hfill \end{array}$$

(14)

Then, ${\phi }_1•{\mu }_0+{\mu }_0•{\phi }_1$ is written, taking into account commutativity of ${\mu }_0$:

$$\begin{array}{c}(a_1+b_6){\phi }_1(x_1,x_2)x_3+(a_1+a_2){\phi }_1(x_1{x}_2,x_3))+(a_2+b_3){\phi }_1(x_2,x_1)x_3+(a_3+b_4){\phi }_1(x_3,x_2)x_1\hfill \\ +(a_3+a_5){\phi }_1(x_3{x}_2,x_1))+(a_4+b_2){\phi }_1(x_1,x_3)x_2+(a_4+a_6){\phi }_1(x_1{x}_3,x_2)+(a_5+b_1){\phi }_1(x_2,x_3)x_1\hfill \\ +(a_6+b_5){\phi }_1(x_3,x_1)x_2+(b_1+b_4){\phi }_1(x_1,x_2{x}_3)+(b_2+b_5){\phi }_1(x_2,x_1{x}_3)+(b_3+b_6){\phi }_1(x_3,x_2{x}_1).\hfill \end{array}$$

The relation ${\phi }_1\circ {\phi }_1+{\phi }_2•{\mu }_0+{\mu }_0•{\phi }_2=0$ gives a relation on ${\psi }_1$ as soon as there is $v\in \mathbb{K}\left[{\Sigma }_3\right]$ such as $({\phi }_2•{\mu }_0+{\mu }_0•{\phi }_2)\circ {\Phi }_v=0.$ Let us consider the matrices

$$A=\left(\begin{array}{cccccc}{a}_1+b_6& a_2+b_3& a_3+b_4& a_4+b_2& a_6+b_5& a_5+b_1\\ a_2+b_3& a_1+b_6& a_5+b_1& a_6+b_5& a_4+b_2& a_3+b_4\\ a_3+b_4& a_6+b_5& a_1+b_6& a_5+b_1& a_2+b_3& a_4+b_2\\ a_4+b_2& a_5+b_1& a_6+b_5& a_1+b_6& a_3+b_4& a_2+b_3\\ a_5+b_1& a_4+b_2& a_2+b_3& a_3+b_4& a_1+b_6& a_6+b_5\\ a_6+b_5& a_3+b_4& a_4+b_2& a_2+b_3& a_5+b_1& a_1+b_6\end{array}\right)$$

and

$$B=\left(\begin{array}{cccccc}{a}_1+a_2& a_1+a_2& a_3+a_5& a_4+a_6& a_4+a_6& a_3+a_5\\ a_3+a_5& a_4+a_6& a_1+a_2& a_3+a_5& a_1+a_2& a_4+a_6\\ a_4+a_6& a_3+a_5& a_4+a_6& a_1+a_2& a_3+a_5& a_1+a_2\\ b_1+b_4& b_2+b_5& b_3+b_6& b_1+b_1& b_3+b_6& b_2+b_5\\ b_2+b_5& b_1+b_4& b_2+b_5& b_3+b_6& b_1+b_4& b_3+b_6\\ b_3+b_6& b_3+b_6& b_1+b_4& b_2+b_5& b_2+b_5& b_1+b_4\end{array}\right).$$

Then, the previous condition is equivalent to say that $kerA\cap kerB\ne 0.$ For example, in the case that ${\mu }_0$ is associative, $a_1=-b_1=1$ and all other constants being zero, then $kerA\cap kerB\ne 0$ is generated by ^t$(1,-1,-1,-1,1,1)$. For any vector $v\in kerA\cap kerB\ne 0$, ${\phi }_1•{\phi }_1\circ {\Phi }_v=0.$ This relation has to give the Lie-admissibility of ${\phi }_1$. This is equivalent to ^t$(1,-1,-1,-1,1,1)\in kerA\cap kerB\ne 0$.

Now, let us examine the consequences of the first relation ${\phi }_1•{\mu }_0+{\mu }_0•{\phi }_1=0.$ This relation gives a relation on ${\psi }_1$ if and only if there is a vector $U={}^t(u_1,u_2,u_3,u_4,u_5,u_6)$ such that $AU={}^t(A_1,-A_1,-A_5,-A_6,A_5,A_6)$. If ^t$(B_1,B_2,B_3,B_4,B_5,B_6)=BU$ then this relation is

$$\begin{array}{c}{A}_1{\psi }_1(x_1,x_2)x_3+A_5{\psi }_1(x_2,x_3)x_1+A_6{\psi }_1(x_3,x_1)x_2+B_1{\psi }_1(x_1{x}_2,x_3)+B_2{\psi }_1(x_2{x}_3,x_1)\hfill \\ +B_3{\psi }_1(x_3{x}_1,x_2)=0\hfill \end{array}$$

Example 3. 

1. Poisson algebras. In this case, we have $A_1=0,A_5=-1,A_6=1$ and $B_1=1,B_2=B_3=0.$ Then, the matrices A and B have to satisfy

$$AU={}^t(0,0,1,-1,-1,1),BU={}^t(1,0,0,0,0,-1),kerA\cap kerB\ne 0.$$

For example, if ${\mu }_0$ is associative, then $U={}^t(1,0,0,-1,0,1)$ and ^t$(1,-1,-1,-1,1,1)$ is in $kerA\cap kerB.$

As another example related to the determination of the matrices A and B, we can assume that the vector ^t$(0,0,1,-1,-1,1)$, which is in the image of A, is a column of this matrix, for example, the first one and that ^t$(1,-1,-1,-1,1,1)$ is in $kerA\cap kerB.$ In this case, $U={}^t(1,0,0,0,0,0)$, implying $B_1=1,B_2=B_3=0$ and

$$\left\{\begin{array}{c}{a}_1+b_6=0\hfill \\ a_2+b_3=0\hfill \\ a_3+b_4=1\hfill \end{array}\right.,\left\{\begin{array}{c}{a}_4+b_2=-1\hfill \\ a_5+b_1=-1\hfill \\ a_6+b_5=1\hfill \end{array}\right.,\left\{\begin{array}{c}{a}_1+a_2=1\hfill \\ a_3+a_5=0\hfill \\ a_4+a_6=0\hfill \end{array}\right..$$

Then

$$\left\{\begin{array}{c}{a}_2=1-a_1\hfill \\ a_5=-a_3\hfill \\ a_6=-a_4\hfill \end{array}\right.,\left\{\begin{array}{c}{b}_2=-1-a_4\hfill \\ b_1=-1+a_3\hfill \\ b_5=1+a_4\hfill \end{array}\right.,\left\{\begin{array}{c}{b}_6=-a_1\hfill \\ b_3=-1+a_1\hfill \\ b_4=1-a_3\hfill \end{array}\right..$$

and the multiplication ${\mu }_0$ satisfies the relation

$$\begin{array}{c}{a}_1\left(x_1{x}_2\right)x_3+(1-a_1)\left(x_2{x}_1\right)x_3+a_3\left(x_3{x}_2\right)x_1+a_4\left(x_1{x}_3\right)x_2-a_3\left(x_2{x}_3\right)x_1-a_4\left(x_3{x}_1\right)x_2\hfill \\ +(-1+a_3)x_1\left(x_2{x}_3\right)+(-1-a_4)x_2\left(x_1{x}_3\right)+(-1+a_1)x_3\left(x_2{x}_1\right)+(1-a_3)x_1\left(x_3{x}_2\right)+\hfill \\ (1+a_4)x_2\left(x_3{x}_1\right)-a_1{x}_3\left(x_1{x}_2\right)=0.\hfill \end{array}$$

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2. Anti-Poisson algebras ([17]). These algebras are related with the “anti-Leibniz” idenity

$${\psi }_1(x_1{x}_2,x_3)+x_1{\psi }_1(x_2,x_3)+{\psi }_1(x_1,x_3)x_2=0.$$

In this case, we have $A_5=1,A_6=-1,A_1=0,B_1=1,B_2=B_3=0.$ Then, the matrices A and B have to satisfy

$$AU={}^t(0,0,-1,1,1,-1),BU={}^t(1,0,0,0,0,-1),kerA\cap kerB\ne 0.$$

Suppose, as in the previous example, that the vector ^t$(0,0,-1,1,1,-1)$, which is in the image of A, is the first column of this matrix. In this case, we obtain

$$\left\{\begin{array}{c}{a}_2=1-a_1\hfill \\ a_5=-a_3\hfill \\ a_6=-a_4\hfill \end{array}\right.,\left\{\begin{array}{c}{b}_2=1-a_4\hfill \\ b_1=1+a_3\hfill \\ b_5=-1+a_4\hfill \end{array}\right.,\left\{\begin{array}{c}{b}_6=-a_1\hfill \\ b_3=-1+a_1\hfill \\ b_4=-1-a_3\hfill \end{array}\right..$$

and ${\mu }_0$ satisfies the identity

$$\begin{array}{c}{a}_1\left(x_1{x}_2\right)x_3+(1-a_1)\left(x_2{x}_1\right)x_3+a_3\left(x_3{x}_2\right)x_1+a_4\left(x_1{x}_3\right)x_2-a_3\left(x_2{x}_3\right)x_1-a_4\left(x_3{x}_1\right)x_2\hfill \\ +(1+a_3)x_1\left(x_2{x}_3\right)+(1-a_4)x_2\left(x_1{x}_3\right)+(-1+a_1)x_3\left(x_2{x}_1\right)+(-1-a_3)x_1\left(x_3{x}_2\right)+\hfill \\ (-1+a_4)x_2\left(x_3{x}_1\right)-a_1{x}_3\left(x_1{x}_2\right)=0.\hfill \end{array}$$

(16)