Arbitrarily choose an element $e\in E(N)$, and let $G$ be a graph representing
$N\backslash e$. Then $|V(G)|=r$. By Lemma 7.5, $N$ does not have a line
containing more than three lonely elements, so neither does $N\backslash e$.
Thus $G$ has no more than three edges linking each pair of its vertices. By
Lemma 2.2 every parallel class of $N$ has at most two elements, no vertex of
$G$ has more than two incident loops. By Lemma 7.6, $N$ has no more than three
pairs of elements in parallel, so $N\backslash e$ has no more than three pairs
of elements in parallel. Thus $G$ has at most three vertices that have two
incident loops. Thus $|E(G)|\leq 3{r\choose 2}+r+3$, and so $|E(N)|\leq
3{r\choose 2}+r+4$. ∎

Theorem 7.1 now follows easily.

###### Proof of Theorem 7.1.

All non-empty rank-0 and all rank-1 matroids are bicircular. A rank-2 matroid
is bicircular if and only if it has at most two rank-1 flats of size greater
than one. It is easy to see that a rank-2 matroid has more than two rank-1
flats containing at least two elements if and only if it contains $M(2C_{3})$
as a minor. So $M(2C_{3})$ is the only excluded minor of rank 2. For
$r\in\\{3,4,\ldots,7\\}$, apply Lemma 7.7: an excluded minor of rank $r$ for
the class of bicircular matroids has at most $3{r\choose 2}+r+4$ elements. ∎

## References

  * [1] Nathan Bowler, Daryl Funk, and Daniel Slilaty, _Describing quasi-graphic matroids_ , European J. Combin. 85 (2020), 103062, 26. MR 4037634
  * [2] Collette R. Coullard, John G. del Greco, and Donald K. Wagner, _Representations of bicircular matroids_ , Discrete Applied Mathematics 32 (1991), no. 3, 223 – 240.
  * [3] Jim Geelen, Bert Gerards, and Geoff Whittle, _Quasi-graphic matroids*_ , Journal of Graph Theory 87 (2018), no. 2, 253–264.
  * [4] Dillon Mayhew and Gordon Royle, _Personal communication_ , 2009.
  * [5] James Oxley, _Matroid theory_ , second ed., Oxford Graduate Texts in Mathematics, vol. 21, Oxford University Press, Oxford, 2011. MR 2849819 (2012k:05002)
  * [6] Donald K. Wagner, _Connectivity in bicircular matroids_ , J. Combin. Theory Ser. B 39 (1985), no. 3, 308–324. MR 815399
  * [7] Thomas Zaslavsky, _Frame matroids and biased graphs_ , European J. Combin. 15 (1994), no. 3, 303–307. MR 1273951 (95a:05021)


Poisson

# Anti-associative algebras

Elisabeth Remm  Université de Haute-Alsace, IRIMAS UR 7499, F-68100 Mulhouse,
France<EMAIL_ADDRESS>
###### Abstract.

An anti-associative algebra is a nonassociative algebra whose multiplication
satisfies the identity $a(bc)+(ab)c=0.$ Such algebras are nilpotent. We
describe the free anti-associative algebras with a finite number of
generators. Other types of nonassociative algebras, obtained either by the
process of polarization, such as Jacobi-Jordan algebras, or obtained by
deformation quantization, are associated with this class of algebras.
Following Markl-Remms work [6], we describe the operads associated with these
algebra classes and in particular the cohomology complexes in relation to
deformations.

###### Contents

  1. 1 Definition and basic properties
    1. 1.1 Definition
    2. 1.2 Nilpotency of anti-associative algebras
    3. 1.3 The associative and Lie multiplication algebras
    4. 1.4 Derivations and anti-derivations
  2. 2 Free anti-associative algebras
    1. 2.1 The free algebra $\mathcal{F_{AA}}(1)$
    2. 2.2 The free algebra $\mathcal{F_{AA}}(2)$
    3. 2.3 The free algebra $\mathcal{F_{AA}}(k)$
  3. 3 Homology of anti-associative algebras
    1. 3.1 The standard complex of homology
    2. 3.2 The homology of the free algebra of one generator
  4. 4 Commutative and anti-commutative anti-associative algebras
    1. 4.1 Case commutative
    2. 4.2 Case Anti-commutative
  5. 5 Polarization and depolarization of anti-associative algebras
    1. 5.1 Polarization of an anti-associative algebra: the commutative part
    2. 5.2 Polarization of an anti-associative algebra: the anti-commutative part
    3. 5.3 Deformation quantization of anti-Poisson algebras
  6. 6 The operad $\mathcal{AA}ss$
    1. 6.1 The operad $\mathcal{AA}ss$
    2. 6.2 The minimal model

##  1\. Definition and basic properties

Let $\mathbb{K}$ be a field of characteristic $0$. Recall that an algebra
$(A,\mu)$ over $\mathbb{K}$ (often simply called an algebra) is a
$\mathbb{K}$-vector space $A$ equipped with a bilinear product $\mu$.

###  1.1. Definition

######  Definition 1.

We say that the algebra $(A,\mu)$ is anti-associative if the product $\mu$
satisfies the quadratic identity

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

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

Such algebra cannot be unitatry

Examples

  1. (1)

In dimension $2$, the only non trivial anti-associative algebra is isomorphic
to the algebra given by

$e_{1}^{2}=e_{2},\ e_{1}e_{2}=e_{2}e_{1}=e_{2}^{2}=0.$

  2. (2)

In dimension $3$ we have the following non isomorphic nontrivial anti-
associative algebras $(A,\cdot)$:

    1. (a)

$e_{i}\cdot e_{i}=0,\ e_{1}\cdot e_{2}=-e_{2}\cdot e_{1}=e_{3}$

    2. (b)

$e_{1}\cdot e_{1}=e_{2},\ e_{1}\cdot e_{2}=-e_{2}\cdot e_{1}=e_{3}$

    3. (c)

$e_{1}\cdot e_{1}=e_{2},\ e_{1}\cdot e_{3}=ae_{2},\ e_{3}\cdot e_{1}=be_{2},\
e_{3}\cdot e_{3}=e_{2},$

    4. (d)

$e_{1}\cdot e_{1}=e_{2},\ e_{1}\cdot e_{3}=ae_{2},\ e_{3}\cdot e_{1}=be_{2}.$

with $a,b\in\mathbb{K}$ and where $\\{e_{1},e_{2},e_{3}\\}$ is a basis of $A$.

  3. (3)

Let us denote by $(\mathcal{F}_{AA}(X),\cdot)$ the free anti-associative
algebra with $1$-generator. It is a graded algebra

$\mathcal{F}_{AA}(X)=\oplus_{k\geq 0}\mathcal{F}_{AA}(X)(k)$

where $\mathcal{F}_{AA}(X)(k)$ consists of the vector space of elements of
degree $k$. The anti-associativity implies that the component of degree $0$ is
null. Let us determine each of these components.

    1. (a)

Degree $2$: We put $X\cdot X=X^{2}$. Thus
$\mathcal{F}_{AA}(X)(2)=\mathbb{K}\\{X^{2}\\}.$

    2. (b)

Degree $3$: We have $X\cdot(X^{2})=-X^{2}\cdot X$. We put $X^{3}=X\cdot
X^{2}$. Then $X^{2}\cdot X=-X^{3}$ and
$\mathcal{F}_{AA}(X)(3)=\mathbb{K}\\{X3\\}$.

    3. (c)

Degree $4$:

$\left\\{\begin{array}[]{l}X\cdot X^{3}=-X^{2}\cdot X^{2}\\\ X\cdot
X^{3}=-X\cdot(X^{2}\cdot X)=(X\cdot X^{2})\cdot X=-(X^{2}\cdot X)\cdot
X=X^{2}\cdot X^{2}.\end{array}\right.$

Then $X^{2}\cdot X^{2}=0$ and any product of degree $4$ is zero. Then
$\mathcal{F}_{AA}(X)(4)=\\{0\\}$

We deduce that $\dim\mathcal{F}_{AA}(X)=3.$ It corresponds to the previous
example (b).

###  1.2. Nilpotency of anti-associative algebras

Recall that an algebra $A$ is called nilpotent if there exists an integer $k$
such that any product of $k$ elements of $A$ is $0$. The smaller integer $k$
such that the subspace $A^{k}$ generated by all the products of degree $k$ is
zero is called the nilindex of $A$.

######  Proposition 2.

Let $A$ be an anti-associative algebra. Then $A$ is nilpotent of nilindex $4$.

Proof. To simplify the notations, we write $xy$ in place of $x\cdot y$. Let
$a,b,c,d$ be elements of $A$. We denote We have

$a(b(cd))=-(ab)(cd)=-a((bc)d)=(a(bc))d=-((ab)c)d=(ab)(cd).$

Then $(ab)(cd)=0$ and all the other products also are $0$.

###  1.3. The associative and Lie multiplication algebras

Let $a$ be an element of the anti-associative algebra $A$. We denote by
$L_{a}$ and $R_{a}$ the left and right translation in $A$:

$L_{a}(x)=ax,\ \ R_{a}(x)=xa.$

Since $A$ is a nilpotent algebra of nilindex $4$, these linear maps are
nilpotent of degree $3$:

$L_{a}^{3}=R_{a}^{3}=0.$

Let $\mathcal{M}(A)$ the subalgebra of $End(A)$ generated by the linear maps
$L_{a}$ and $R_{a}$ for any $a\in A$. It is an associative algebra called the
multiplication algebra of $A$. Since

$\left\\{\begin{array}[]{l}L_{a}L_{b}=-L_{ab},\\\ R_{a}R_{B}=-R_{ba},\\\
L_{a}R_{b}=-R_{b}L_{a},\\\ R_{a}L_{b}=-L_{b}R_{a},\\\ \end{array}\right.$

these identities resulting from the anti-associative , any element of
$\mathcal{M}(A)$ can be written of the form $L_{x}R_{y}$,with $x,y\in A$. Then
$\mathcal{M}(A)$ is nilpotent associative algebra of nilindex $3$.

Example. Let us consider the free anti-associative algebra with one generator.
It is of dimension $3$ and its multiplication satisfies

$e_{1}e_{1}=e_{2},\ e_{1}e_{2}=-e_{2}e_{1}=e_{3}$

(with $e_{1}=X,e_{2}=X^{2},e_{3}=XX^{2}$). The associative algebra
$\mathcal{M}(A)$ is $3$-dimensional generated by
$L_{e_{1}},R_{e_{1}},L_{e_{2}}.$

We can consider also the Lie subalgebra $\mathcal{L}(A)$ of $gl(A)$ generated
by the left and right multiplication of $A$. It corresponds to the Lie algebra
associated with the associative algebra $\mathcal{M}(A)$. Since it is
generated by the Lie brackets between these nilpotent linear maps, we have

######  Proposition 3.

The Lie multiplication algebra associated with the anti-associative algebra is
a $2$-step nilpotent Lie algebra.

Example. In the previous example, the Lie algebra $\mathcal{L}(A)$ is
isomorphic to the $3$-dimensional Heisenberg algebra.

###  1.4. Derivations and anti-derivations

Recall that a derivation of the algebra $A$ is a linear operator which
satisfies

$f(xy)=xf(y)+f(x)y.$

The set of derivations of $A$, $D(A)$, is a Lie algebra. A derivation $f$ is
called inner if $f\in D(A)\cap\mathcal{L}(A).$ For example, we consider for
any $x\in A$ the linear map $L_{x}-R_{x}$ which belongs to $\mathcal{L}(A)$.
In [7], it is proved that this linear map is a derivation of $A$ if and only
if $A$ is weakly associative. This is equivalent to say that the anti-
associative algebra $A$ satisfies also the identity

$x(yz)+y(zx)+z(yx)=0$

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

######  Definition 4.

Let $A$ be an algebra (not necessry anti-associative). A linear map $f\in
End(A)$ is an anti-derivation if this map satisfies

$f(xy)+xf(y)+f(x)y=0$

for any $x,y\in A.$

We denote by $\widetilde{D}(A)$ the set of anti-derivation of $A$.

######  Proposition 5.

For any algebra $A$ (anti-associative or not),
$[\widetilde{D}(A),\widetilde{D}(A)]\subset D(A).$

Proof. Let $f,g$ be two anti-derivations of $A$.

$f\circ g(xy)=f(g(xy))=-f(xg(y)+g(x)y)=f(x)g(y)+xf(g(y))+f(g(x))y+g(x)f(y).$

Then

$\begin{array}[]{ll}[f,g](xy)&=f(x)g(y)+xf(g(y))+f(g(x))y+g(x)f(y)-g(x)f(y)-xg(f(y))\\\
&\ \ \ \ -g(f(x))y-f(x)g(y)\\\ &=x(f(g(y))-g(f(y))+(f(g(x))-g(f(x))y\\\
&=x[f,g](y)+[f,g](x)y\end{array}$

and $[f,g]$ is a derivation of $A$.

######  Proposition 6.

Let $A$ be an anti-associative algebra. Then for any $x\in A$, the linear map
$L_{x}+R_{x}$ is an anti-derivation of $A$.

Proof.
$(L_{x}+R_{x})(yz)+(L_{x}+R_{x})(y)z+y(L_{x}+R_{x})(z)=x(yz)+(yz)x+(xy)z+y(zx)=0$
since $A$ is anti-associative .

Remark. Let $A$ be an algebra (not necessarly anti-associative ). Then
$L_{x}+R_{x}$ is an anti-derivation if and only if the multiplication of $A$
satisfies:

$\widetilde{\mathcal{A}}(x,y,z)+\widetilde{\mathcal{A}}(y,x,z)+\widetilde{\mathcal{A}}(y,z,x)=0$

where $\widetilde{\mathcal{A}}$ is the trilinear map

$\widetilde{\mathcal{A}}(x,y,z)=x(yz)+(xy)z.$

An algebra whose the operator $\widetilde{\mathcal{A}}$ satisfies this
identity will be called weakly anti-associative .

An anti-derivation $f$ of an anti-associative algebra $A$ of type
$f=L_{x}+R_{x}$ for some $x\in A$ will be called inner. We will denote $f_{x}$
such inner anti-derivation.

For any anti-derivation $f$ of $A$ we have

$\begin{array}[]{ll}[f_{x},f](y)&=f_{x}(f(y))-f(xy+yx)\\\
&=xf(y)+f(y)x-f(x)y-xf(y)-yf(x)-f(y)x\\\ &=-f(x)y-yf(x)\\\
&=-f_{f(x)}(y).\end{array}$

Then $[f_{x},f]=f_{f(x)}$.

######  Proposition 7.

$[\widetilde{\mathcal{I}}(A),D(A)]\subset\widetilde{\mathcal{I}}(A).$

Assume that $L_{x}+R_{x}\in D(A)$. This is equivalent to write

$x(yz)+(yz)x-y(xz+zx)-(xy+yx)z=0\ \ \forall y,z\in A$

that is

$x(yz)+(yz)x=0\ \forall y,z\in A$

Then $L_{x}+R_{x}\in D(A)$ if and only if $(L_{x}+R_{x})(u)=0\ \ \forall u\in
A^{2}.$ We deduce:

######  Proposition 8.

$\widetilde{\mathcal{I}}(A)\cap D(A)=\\{L_{x}+R_{x}\ {\rm such\ that}\
(L_{x}+R_{x})(u)=0\ \ \forall u\in A^{2}\\}.$

In particular for any $x\in A^{2}$, the inner anti-derivation $L_{x}+R_{x}$ is
also a derivation, which we had already noted. More generally, let $f$ be in
$\widetilde{D}(A)\cap D(A)$. Then

$f(xy)=xf(y)+f(x)y=-xf(y)-f(x)y$

or equivantly

$f(x^{2})=0,\ \ xf(y)+f(x)y=0$

for any $x,y\in A$.

To simplify notations, we will write $\widetilde{ad}(x)$ for the inner anti-
derivation $L_{x}+R_{x}$. For two endomorphisms $f$ and $g$ of $A$, we dene
the symmetric product

$f\bullet g=f\circ g+g\circ f.$

Then

$\begin{array}[]{ll}\widetilde{ad}(x)\bullet\widetilde{ad}y&=L_{x}L_{y}+L_{x}R_{y}+R_{x}L_{y}+R_{x}R_{y}+L_{y}L_{x}+L_{y}R_{x}+R_{y}L_{x}+R_{y}R_{x}\\\
&=-L_{xy+yx}-R_{xy+yx}\\\ &=-\widetilde{ad}(xy+yx)\end{array}$

because
$(L_{x}R_{y}+R_{x}L_{y}+L_{y}R_{x}+R_{y}L_{x})(z)=x(zy)+(yz)x+y(zx)+(xz)y=0$
for any $z\in A$, the algebra $A$ being anti-associative . Likewise, if
$f\in\widetilde{D}(A)$,

$f\bullet\widetilde{ad}(x)=-\widetilde{ad}(f(x)).$

######  Proposition 9.

Let $A$ be an anti-associative algebra, $f\in\widetilde{D}(A)$. Then

  1. (1)

$\widetilde{ad}(x)\bullet\widetilde{ad}y=-\widetilde{ad}(xy+yx)$ for any
$x,y\in A$,

  2. (2)

$f\bullet\widetilde{ad}(x)=-\widetilde{ad}(f(x))$ for any $x\in A$.

It should be noted that in general, such a product of two anti-associative
derivations is not an anti-associative derivation. In fact, if
$f,g\in\widetilde{D}(A)$,

$\begin{array}[]{lll}(f\bullet
g)(xy)&=&f(g(xy))+g(f(xy))=-f(g(x)y+xg(y))-g(f(x)y+xf(y))\\\
&=&f(g(x))y+g(x)f(y)+f(x)g(y)+xf(g(y))+g(f(x))y+f(x)g(y)\\\
&&+g(x)f(y)+xg(f(y))\\\
&=&2(f(x)g(y)+g(x)f(y))+(f(g(x))+g(f(x))y+x(f(g(y))+g(f(y)).\\\ \end{array}$

and

$x(f\bullet g)(y)+(f\bullet g)(x)y=x(f(g(y))+g(f(x))+(f(g(x))+g(f(x))y.$

Thus $f\bullet g$ is an anti-associative derivation if and only if these
linear maps satisfy

$f(x)g(y)+g(x)f(y)=-(f(g(x))+g(f(x))y-x(f(g(y))+g(f(y)).$

Remark: on the inner derivations of nonassociative algebras Usually we cannot
define directly the notion of inner derivation on a nonassociative algebra. If
$U$ is such algebra, we consider as above the Lie multiplication algebra
$\mathcal{L}(U)$ generated by the left and right multiplication of $U$. We say
that a derivation $f$ of $U$ is inner if $f\in\mathcal{L}(U)$. The set
$\mathcal{L}(U)\cap D(U)$ is an ideal of $D(U)$. In case of $U$ is a Lie
algebra, then $\mathcal{L}(U)\cap D(U)$ corresponds to the set of adjoint
operators, that is $L_{x}-R_{x}$. For associative algebra, an element of
$\mathcal{L}(U)\cap D(U)$ is of type $L_{x}-R_{x}$ as soon as $U$ is an
unitary associative algebra. It is not always the case when $U$ is not
unitary. For example, if we consider the $2$-dimensional associative algebra
given in a basis $\\{e_{1},e_{1}\\}$ by $e_{1}e_{1}=0$ (other non defined
products are $0$), then any derivation of $L_{x}-R_{x}$ are zero, $D(U)$ is
the set of endomorphism $f$ satisfying $f(e_{1})=ae_{1}+be_{2},\ f(e_{2})=0$,
$\mathcal{L}(U)$ is the set of endomorphism $f$ satisfying $f(e_{1})=be_{2},\
f(e_{2})=0$ and $\mathcal{L}(U)\cap D(U)$ is the set of endomorphism $f$
satisfying $f(e_{1})=be_{2},\ f(e_{2})=0$. Recall also that $L_{x}-R_{x}$ is a
derivation of $U$ if and only if $U$ is a weakly associative algebra.

In case of the nonassociative algebra $A$ is anti-associative the brackets
$[f_{x},f_{y}]$, $[L_{x}-R_{x},f_{y}]$, $[L_{x}-R_{x},L_{y}-R_{y}]$ are always
derivations of $A$ and also anti-associative derivation because any product of
$4$ elements of $A$ is zero. For the same raison, $L_{x}R_{y}$ and
$R_{x}L_{y}$ are derivations and anti-derivations of $A$. This implies that
$L_{x}+R_{y}$ is an anti-derivation as soon as $y=x+v$ with $z(vt)=0$ for any
$z,t\in A.$ In particular $L_{x}+R_{x+v}$ is an anti-derivation for any $v\in
A^{2}$ and $\mathcal{L}(A)=\widetilde{\mathcal{I}}(A)$ as soon as $R_{v}=0$
for any $v\in A^{2}$.

##  2\. Free anti-associative algebras

We denote by $\mathcal{F_{AA}}(k)$ the free anti-associative algebra with $k$
generators. This algebra is graded by the degree of its elements:

$\mathcal{F_{AA}}(k)=\oplus_{1\leq n\leq 3}\mathcal{F_{AA}}(k)^{(n)}$

because all products of degree $4$ are null. Recall that the anti-
associativity prevents having elements of degree $0$.

###  2.1. The free algebra $\mathcal{F_{AA}}(1)$

We have already described this algebra in the first paragraph. Let us recall
this construction briefly. Let $X$ be the generator. Then
$X^{2}\in\mathcal{F_{AA}}(1)$. For the elements of degree $3$, since we have
$X.X^{2}=-X^{2}.X$, we deduce that
$\mathcal{F_{AA}}(1)^{(3)}=\mathbb{K}\\{X.X^{2}\\}$. Then
$\dim\mathcal{F_{AA}}(1)$ and a basis is given by

$X,\ X^{2},\ XX^{2}$

and the multiplication

$X.X=X^{2},X.X^{2}=XX^{2},X^{2}.X=-XX^{2}.$

To use conventional conventions we denote
$e_{1}^{(1)}=X,e_{1}^{(2)}=X^{2},e_{1}^{(3)}=X.X^{2}.$ Then we have

$e_{1}^{(1)}\cdot e_{1}^{(1)}=e_{1}^{(2)},e_{1}^{(1)}\cdot
e_{1}^{(2)}=-e_{1}^{(2)}\cdot e_{1}^{(1)}=e_{1}^{(3)}.$

The Lie algebra $D(\mathcal{F_{AA}}(1))$ is $3$-dimensional constituted of
matrices $\begin{pmatrix}\alpha&0&0\\\ \beta&-\alpha&0\\\
\gamma&0&\alpha\end{pmatrix}$ while $\widetilde{D}(\mathcal{F_{AA}}(1))$ is
the space of matrices $\begin{pmatrix}\alpha&0&0\\\ \beta&-2\alpha&0\\\
\gamma&0&\alpha\end{pmatrix}$ and an anti-associative derivation is a
derivation when it is trivial on
$(\mathcal{F_{AA}}(1))^{2}=\mathbb{K}\\{e_{1}^{(2)},e_{1}^{(3)}\\}.$ Any inner
anti-associative derivation is collinear to $L_{e_{1}^{(1)}}+R_{e_{1}^{(1)}}.$

###  2.2. The free algebra $\mathcal{F_{AA}}(2)$

Let $\mathcal{F_{AA}}(2)=\oplus_{1\leq n\leq 3}\mathcal{F_{AA}}(2)^{(n)}$ the
graded decomposition of $\mathcal{F_{AA}}(2)$. If we denote by $X,Y$ the
generators of this free algebra, we shall have

  1. (1)

$\mathcal{F_{AA}}(2)^{(1)}=\mathbb{K}\\{X,Y\\}$

  2. (2)

$\mathcal{F_{AA}}(2)^{(2)}=\mathbb{K}\\{X^{2},Y^{2},XY,YX\\}$

  3. (3)

$\mathcal{F_{AA}}(2)^{(3)}=\mathbb{K}\\{X(X^{2}),X(XY),X(YX),Y(XX),X(Y^{2}),Y(XY),Y^{2}X,YY^{2}\\}$

and $\dim\mathcal{F_{AA}}(2)=14.$ Moreover, we have

$\left\\{\begin{array}[]{l}\mathcal{F_{AA}}(2)^{(1)}\cdot\mathcal{F_{AA}}(2)^{(1)}=\mathcal{F_{AA}}(2)^{(2)}\\\
\mathcal{F_{AA}}(2)^{(1)}\cdot\mathcal{F_{AA}}(2)^{(2)}=\mathcal{F_{AA}}(2)^{(2)}\cdot\mathcal{F_{AA}}(2)^{(1)}=\mathcal{F_{AA}}(2)^{(3)}\\\
\end{array}\right.$

all other non defined products being zero. Now let’s look at the anti-
derivations. Any anti-derivation of $A=\mathcal{F_{AA}}(2)$ is entirely
determined by its values $f(X)$ and $f(Y)$. In fact for any generators $uv$ of
$\mathcal{F_{AA}}(2)^{(2)}$ with $u,v\in\\{X,Y\\}$ we have
$f(uv)=-uf(v)-f(u)v.$ Any generators of $\mathcal{F_{AA}}(2)^{(3}$ is written
$u(vw)$ with $u,v,w\in\\{X,Y\\}$. Since $(uv)w=-u(vw)$ we have

$f(u(vw))=u(f(v)w),\ \ u(vf(w))-f(u)(vw)=0.$

If we put $f(X)=a_{1}X+a_{2}Y+Z_{1},\ f(Y)=b_{1}X+b_{2}Y+Z_{2}$ with
$Z_{i}\in\mathcal{F_{AA}}(2)^{(2)}\oplus\mathcal{F_{AA}}(2)^{(3)}$, then the
previous equations give $a_{2}=b_{1}=0,\ a_{1}=b_{2}.$ As consequence

$f(u)=a_{1}u,\ \forall u\in\mathcal{F_{AA}}(2)^{(3)}$

and for any $v\in\mathcal{F_{AA}}(2)^{(2)}$,

$f(v)=-2a_{1}v+Z,\ Z\in\mathcal{F_{AA}}(2)^{(3)}.$

In particular

$\dim\widetilde{D}(\mathcal{F_{AA}}(2))=13=\dim\mathcal{F_{AA}}(2)^{(2)}\oplus\mathcal{F_{AA}}(2)^{(3)}+1.$

Now, concerning the inner anti-derivation, we can easily see that the linear
maps $L_{x}+R_{x}$ for any $x=X,Y,X^{2},Y^{2},XY,YX$ are linearly independent.
But these anti-derivations also satisfy

$L_{X}+R_{X}(\mathcal{F_{AA}}(2)^{(1)})\subset\mathcal{F_{AA}}(2)^{(2)},\
L_{Y}+R_{Y}(\mathcal{F_{AA}}(2)^{(1)})\subset\mathcal{F_{AA}}(2)^{(2)}.$

We deduce that

$\dim\widetilde{I}(\mathcal{F_{AA}}(2))=6.$

###  2.3. The free algebra $\mathcal{F_{AA}}(k)$

The free algebra with $k$ generators $X_{1},\cdots,X_{k}$ admits the following
grading associated with the degree of the products of generators:

$\mathcal{F_{AA}}(k)=\oplus_{1\leq n\leq 3}\mathcal{F_{AA}}(k)^{(n)}$

and we have

  1. (1)

$\mathcal{F_{AA}}(k)^{(1)}=\mathbb{K}\\{X_{1},\cdots,X_{k}\\},$

  2. (2)

$\mathcal{F_{AA}}(k)^{(2)}=\mathbb{K}\\{X_{i}X_{j},1\leq i,j\leq k\\},$

  3. (3)

$\mathcal{F_{AA}}(k)^{(3)}=\mathbb{K}\\{X_{i}(X_{j}X_{l}),\ 1\leq i,j,l\leq
k\\}.$

Then

  1. (1)

$\dim\mathcal{F_{AA}}(k)^{(1)}=k,$

  2. (2)

$\dim\mathcal{F_{AA}}(k)^{(2)}=k^{2},$

  3. (3)

$\dim\mathcal{F_{AA}}(k)^{(3)}=k^{3}.$

and

$\dim\mathcal{F_{AA}}(k)=k+k^{2}+k^{3}.$

By induction, we prove that any anti-derivation $f$ is completely defined by
the vectors $f(X_{i}),i=1,\cdots,k$ and these vector satisfy

$X_{i}(X_{j}f(X_{k}))-f(X_{i})(X_{j}X_{k})=0$

which implies that the restriction of $f$ to $\mathcal{F_{AA}}(k)^{(1)}$ is
equal to $a\cdot Id.$ In particular

$\dim\widetilde{D}(\mathcal{F_{AA}}(k))=1+\dim\mathcal{F_{AA}}(k)^{(2)}\oplus\mathcal{F_{AA}}(k)^{(3)}=1+k^{2}+k^{3}.$

Concerning the inner anti-derivations, we can easily see that the linear maps
$L_{x}+R_{x}$ for any generator
$x\in\mathcal{F_{AA}}(k)^{(1)}\oplus\mathcal{F_{AA}}(k)^{(2)}$ are linearly
independent. We deduce that

$\dim\widetilde{I}(\mathcal{F_{AA}}(2))=k+k^{2}.$

The free algebra $\mathcal{F_{AA}}(k)$ admits a natural grading associated
with the nilpotency property:

$\mathcal{F_{AA}}(k)=\mathcal{F_{AA}}(k)^{(1)}\oplus\mathcal{F_{AA}}(k)^{(2)}\oplus\mathcal{F_{AA}}(k)^{(3)}$

with

$\mathcal{F_{AA}}(k)^{(1)}\mathcal{F_{AA}}(k)^{(1)}=\mathcal{F_{AA}}(k)^{(2)},\mathcal{F_{AA}}(k)^{(1)}\mathcal{F_{AA}}(k)^{(2)}=\mathcal{F_{AA}}(k)^{(3)},\mathcal{F_{AA}}(k)^{(i)}\mathcal{F_{AA}}(k)^{(3)}=0.$

If $\\{e_{1},\cdots,e_{k}\\}$ is a basis of $\mathcal{F_{AA}}(k)^{(1)}$, that
is a basis of generators of the free algebra, $\\{f_{i,j}=e_{i}e_{j},1\leq
i,j\leq k\\}$ a basis of $\mathcal{F_{AA}}(k)^{(2)}$,
$\\{g_{i,j,l}=e_{i}f_{j,l}\\}$ a basis of $\mathcal{F_{AA}}(k)^{(3)}$, then we
have

$\left\\{\begin{array}[]{l}e_{i}e_{j}=f_{i,j}\\\ e_{i}f_{j,l}=g_{i,j,l}\\\
e_{i}g_{j,l,s}=g_{j,l,s}e_{i}=f_{i,j}f_{l,s}=f_{i,j}g_{l,s,t}=g_{l,s,t}f_{i,j}=0\\\
f_{i,j}e_{l}=-e_{i}f_{j,l}\end{array}\right.$

Application: Construction of anti-associative algebras. From this free anti-
associative algebra, we deduce the anti-associative algebras with $k$
generators:

$A=A^{(1)}\oplus A^{(2)}\oplus A^{(3)}$

where $A^{(1)}=AA,\ A^{(2)}=A.A^{(1)}\oplus A^{(1)}A,\
A^{(3)}=A^{(1)}.A^{(2)}\oplus A^{(2)}.A^{(1)}$ and
$A^{(1)}=\mathcal{F_{AA}}(k)^{(1)}.$ This allows to construct the anti-
associative algebras from the data of the generators. For example, if $\dim
A^{(1)}=2$, then $\dim A^{(2)}\leq 4.$ If this dimension is equal to $4$, we
have $\mathcal{F_{AA}}(2)$. If this dimension is equal to $1$, then $\dim
A^{(3)}\leq 2$. We complete the table of multiplication putting
$f_{i,j}e_{k}=-e_{i}f_{j,k}$.

Example $\dim A^{(1)}=2$. In this case $\dim A\leq 14$. If $\dim A=14$, then
$A$ is isomorphic to $\mathcal{F_{AA}}(2)$. We can describe the other anti-
associative algebras considering the dimensions of $A^{(2)}$ and $A^{(3)}$. We
will not develop here a complete classification but show how to obtain it. We
put $A^{(1)}=\mathbb{K}\\{e_{1},e_{2}\\}$.

  1. (1)

If $\dim A^{(2)}=1$, then $A^{(3)}=0.$ In fact

1\. If $A^{(2)}=\mathbb{K}\\{f_{1,1}\\}$, then

$e_{1}f_{1,1}=f_{1,1}e_{1}=0,e_{2}f_{1,1}=\alpha
f_{1,1}e_{1}=0,f_{1,1}e_{2}=-\beta e_{1}f_{1,1}=0$

and $A^{(3)}=0.$

2\. If $A^{(2)}=\mathbb{K}\\{f_{1,2}\\}$, then we can assume that
$e_{1}e_{1}=e_{2}e_{2}=0$ if not we are in the previous case. Then

$e_{1}f_{1,2}=0,\ f_{1,2}e_{1}=0,\ f_{1,2}e_{2}=0,\
e_{2}f_{1,2}=-bf_{1,2}e_{2},=0$

This implies that $A^{(3)}=0.$ We obtain the following $3$-dimensional anti-
associative algebras:

    1. (a)

$e_{i}e_{j}=\alpha_{ij}f_{1,1},\alpha_{1,1}=1,e_{i}f_{1,1}=f_{1,1}e_{i}=0$.

    2. (b)

$e_{i}e_{i}=0,e_{1}e_{2}=-e_{2}e_{1}=f_{1,2},e_{i}f_{1,2}=f_{1,2}e_{i}=0$.

  2. (2)

If $A^{(2)}=\mathbb{K}\\{f_{1,1},f_{1,2}\\}$ then $\dim A^{(3)}\leq 1$ and if
this dimension is not $0$, then $A^{(3)}=\mathbb{K}\\{g_{1,1,2}\\}$. We obtain
the following algebras

    1. (a)

$e_{1}e_{1}=f_{1,1},\ e_{1}e_{2}=f_{1,2},\ e_{2}e_{1}=af_{1,1}+bf_{1,2},\
e_{2}e_{2}=cf_{1,1}+df_{1,2}.$

    2. (b)

$\left\\{\begin{array}[]{l}e_{1}e_{1}=f_{1,1},e_{1}e_{2}=f_{1,2},e_{2}e_{1}=af_{1,1}+bf_{1,2},e_{2}e_{2}=cf_{1,1}+df_{1,2}\\\
e_{1}f_{1,1}=0,e_{1}f_{1,2}=g_{1,1,2},e_{2}f_{1,1}=b^{2}g_{1,1,2},e_{2}f_{1,2}=(a+bd)g_{1,1,2}\\\
f_{1,1}e_{1}=0,f_{1,2}e_{1}=-bg_{1,1,2},f_{1,1}e_{2}=-g_{1,1,2},f_{1,2}e_{2}=-dg_{1,1,2}\\\
\end{array}\right.$

This algebra is of dimension $5$ and can be viewed as the quotient of the free
algebra $\mathcal{F_{AA}}(2)$ modulo the relations $YX-aX^{2}-bXY,\
Y^{2}-cX^{2}-dXY.$

    3. (c)

Let us note that if we assume always that $\dim A^{(2)}=2$, and if $A$ is not
isomorphic with the previous one, then such algebra is isomorphic to the
quotient of $\mathcal{F_{AA}}(2)$ modulo the relations $XY=aX^{2},YX=cX^{2}$.
In this case $A^{(3)}=0$ and we have the algebra

$\left\\{\begin{array}[]{l}e_{1}e_{1}=f_{1,1},e_{1}e_{2}=af_{1,1},e_{2}e_{1}=cf_{1,1},e_{2}e_{2}=f_{2,2}\\\
e_{i}f_{j,k}=0,f_{j,k}e_{i}=0.\end{array}\right.$

This algebra is of dimension $4$.

All the other cases can be traited in a similar way. It will be a next work.

##  3\. Homology of anti-associative algebras

###  3.1. The standard complex of homology

Let $A$ be a finite dimensional anti-associative algebra and
$\widetilde{\Lambda^{k}(A)}$ the vector space spanned by elements of the form
$a_{1}\widetilde{\wedge}a_{2}\widetilde{\wedge}\cdots\widetilde{\wedge}a_{k}$
for any $k\geq 1$. If $\\{e_{1},\cdots,e_{n}\\}$, then
$\widetilde{\Lambda^{k}(A)}$ is the set of linear combinations of vectors
$e_{i_{1}}\widetilde{\wedge}e_{i_{2}}\widetilde{\wedge}\cdots\widetilde{\wedge}e_{i_{k}}$,
$1\leq i_{1},\cdots,i_{k}\leq n.$

Instead of defining the ordinary exterior product associated with the relation
$a\wedge b=-b\wedge a$, we consider the relations

  1. (1)

$a_{\sigma(1}\widetilde{\wedge}a_{\sigma(2)}\widetilde{\wedge}\cdots\widetilde{\wedge}a_{\sigma(k)}=(-1)^{k}\epsilon(\sigma)a_{1}\widetilde{\wedge}a_{2}\wedge\cdots\widetilde{\wedge}a_{k},$

where $\sigma$ is a permutation of $\\{1,\cdots,k\\}$ and $\epsilon(\sigma)$
its signum. In particular we have

$a_{2}\widetilde{\wedge}a_{1}=-a_{2}\widetilde{\wedge}a_{1},\
a_{2}\widetilde{\wedge}a_{3}\widetilde{\wedge}a_{1}=a_{1}\widetilde{\wedge}a_{2}\widetilde{\wedge}a_{3}$

  2. (2)

If $b_{1}=a_{1}a_{2},b_{i-1}=a_{i}a_{i+1}$, and for $l=2,\cdots
i-1,b_{l}=a_{l-1}$ and for $l\geq i+1,b_{l}=a_{i+2}$,

$\begin{array}[]{ll}a_{1}a_{2}\widetilde{\wedge}a_{3}\widetilde{\wedge}\cdots\widetilde{\wedge}a_{i}a_{i+1}\widetilde{\wedge}\cdots
a_{k}=b_{1}\widetilde{\wedge}\cdots\widetilde{\wedge}b_{i-1}\widetilde{\wedge}b_{i}\widetilde{\wedge}\cdots\widetilde{\wedge}b_{k-2}\\\
=(-1)^{k}\epsilon(\sigma_{1})\epsilon({\sigma_{2}})b_{\sigma_{1}(1)}\widetilde{\wedge}\cdots\widetilde{\wedge}b_{\sigma_{1}(i)}\widetilde{\wedge}\cdots\widetilde{\wedge}b_{\sigma_{1}(k-2)}\end{array}$

for any permutation $\sigma_{1}$ of $k_{2}$ elements and $\sigma_{2}$ is the
permutation of $k$ elements defined from $\sigma_{1}$

We denote by $H_{*}(A,A)$ the homology groups of the chain complex

$(C_{q}(A,A)=\widetilde{\Lambda^{q+1}(A)},b_{q})$

where $b_{q}$ is the differential

$b_{q}:C_{q}(A,A)\rightarrow C_{q-1}(A,A)$

given by

$b_{q}(a_{1}\widetilde{\wedge}a_{2}\widetilde{\wedge}\cdots\widetilde{\wedge}a_{q}\widetilde{\wedge}a_{q+1})=\sum_{i=1}^{i=q-1}a_{i}a_{i+1}\widetilde{\wedge}a_{1}\widetilde{\wedge}\cdots\widetilde{\wedge}a_{i-1}\widetilde{\wedge}a_{i+2}\widetilde{\wedge}\cdots\widetilde{\wedge}a_{q+1}+a_{q+1}a_{1}\widetilde{\wedge}a_{2}\widetilde{\wedge}\cdots\widetilde{\wedge}a_{q-1}.$

Low dimensional computations:

  1. (1)

The map

$b_{1}:C_{1}(A,A)=A\widetilde{\wedge}A\rightarrow C_{0}(A,A)=A$

is given by

$b_{1}(a_{1}\widetilde{\wedge}a_{2})=a_{1}a_{2}+a_{2}a_{1}.$

  2. (2)

The map

$b_{2}:C_{2}(A,A)=A^{\widetilde{\wedge}^{3}}\rightarrow C_{1}(A,A)$

is given by

$b_{2}(a_{1}\widetilde{\wedge}a_{2}\widetilde{\wedge}a_{3})=a_{1}a_{2}\widetilde{\wedge}a_{3}+a_{2}a_{3}\widetilde{\wedge}a_{1}+a_{3}a_{1}\widetilde{\wedge}a_{2}.$

We have

$\begin{array}[]{lll}b_{1}(b_{2}((a_{1}\widetilde{\wedge}a_{2}\widetilde{\wedge}a_{3})&=&b_{1}(a_{1}a_{2}\widetilde{\wedge}a_{3})+b_{1}(a_{2}a_{3}\widetilde{\wedge}a_{1})+b_{1}(a_{3}a_{1}\widetilde{\wedge}a_{2})\\\
&=&(a_{1}a_{2})a_{3}+a_{3}(a_{1}a_{2})+(a_{2}a_{3})a_{1}+a_{1}(a_{2}a_{3})+(a_{3}a_{1})a_{2}+a_{2}(a_{1}a_{3})\\\
&=&0.\end{array}$

  3. (3)

The map

$b_{3}:C_{3}(A,A)=A^{\widetilde{\wedge}^{3}}\rightarrow C_{2}(A,A)$

is given by

$b_{3}(a_{1}\widetilde{\wedge}a_{2}\widetilde{\wedge}a_{3}\widetilde{\wedge}a_{4})=a_{1}a_{2}\widetilde{\wedge}a_{3}\widetilde{\wedge}a_{4}+a_{2}a_{3}\widetilde{\wedge}a_{1}\widetilde{\wedge}a_{3}+a_{3}a_{4}\widetilde{\wedge}a_{1}\widetilde{\wedge}a_{2}+a_{4}a_{1}\widetilde{\wedge}a_{2}\widetilde{\wedge}a_{3}.$

We have

$\begin{array}[]{lll}b_{2}(b_{3}(a_{1}\widetilde{\wedge}a_{2}\widetilde{\wedge}a_{3}\widetilde{\wedge}a_{4})&=&b_{2}(a_{1}a_{2}\widetilde{\wedge}a_{3}\widetilde{\wedge}a_{4})+b_{2}(a_{2}a_{3}\widetilde{\wedge}a_{4}\widetilde{\wedge}a_{1})+b_{2}(a_{3}a_{4}\widetilde{\wedge}a_{1}\widetilde{\wedge}a_{2})\\\
&&+b_{2}(a_{4}a_{1}\widetilde{\wedge}a_{2}\widetilde{\wedge}a_{3})\\\
&=&(a_{1}a_{2})a_{3}\widetilde{\wedge}a_{4}+a_{3}a_{4}\widetilde{\wedge}a_{1}a_{2}+a_{4}(a_{1}a_{2})\widetilde{\wedge}a_{3}+(a_{2}a_{3})a_{4}\widetilde{\wedge}a_{1}\\\
&&+a_{4}a_{1}\widetilde{\wedge}a_{2}a_{3}+a_{1}(a_{2}a_{3})\widetilde{\wedge}a_{4}+(a_{3}a_{4})a_{1}\widetilde{\wedge}a_{2}+a_{1}a_{2}\widetilde{\wedge}a_{3}a_{4}\\\
&&+a_{2}(a_{3}a_{4})\widetilde{\wedge}a_{1}+(a_{4}a_{1})a_{2}\widetilde{\wedge}a_{3}+a_{2}a_{3}\widetilde{\wedge}a_{4}a_{1}+a_{3}(a_{4}a_{1})\widetilde{\wedge}a_{2}\\\
&=&0.\end{array}$

In fact

$a_{3}a_{4}\widetilde{\wedge}a_{1}\widetilde{\wedge}a_{2}=\alpha_{1}\widetilde{\wedge}\alpha_{2}=(-1)^{2}\epsilon(\sigma_{1})\epsilon(\sigma_{2})(\alpha_{2}\widetilde{\wedge}\alpha_{1})=-a_{1}a_{2}\widetilde{\wedge}a_{3}a_{4}$

where $\sigma_{1}$ is the transposition $\tau_{12}$ and $\sigma_{2}$ the
cyclic permutation $(3,4,1,2)$.

  4. (4)

Before the study of the general case, we will develop the particular case
$k=5$ because we will differ the case $k$ odd and even. . When we develop
$b_{3}(b_{4}(a_{1}\widetilde{\wedge}a_{2}\widetilde{\wedge}a_{3}\widetilde{\wedge}a_{4}\widetilde{\wedge}a_{5})$,
we have terms of type
$(a_{i}(a_{j}a_{k})\widetilde{\wedge}a_{l}\widetilde{\wedge}a_{s}$ and
$(a_{i}a_{j})a_{k}\widetilde{\wedge}a_{l}\widetilde{\wedge}a_{s}$, and these
terms are eliminated by anti-associative identity. All the other terms are of
type $a_{i}a_{j}\widetilde{\wedge}a_{k}\widetilde{\wedge}a_{l}a_{s}$ which can
be regrouped by pairs. For example, we have in the development the two terms
$a_{3}a_{4}\widetilde{\wedge}a_{5}\widetilde{\wedge}a_{1}a_{2}$ and
$a_{1}a_{2}\widetilde{\wedge}a_{3}\widetilde{\wedge}a_{4}a_{5}$. In this case
$\sigma_{1}$ is the permutation $\begin{pmatrix}1&2&3\\\ 3&1&2\end{pmatrix}$
which is an even cycle, and $\sigma_{2}=\begin{pmatrix}3&4&5&1&2\\\
1&2&3&4&5\end{pmatrix}$ which is also an even cycle. But
$(-1)^{k}\epsilon(\sigma_{1})\epsilon(\sigma_{2})=-1$ and this two termes
eliminate themselves. We deduce $b_{3}b_{4}=0$.

######  Lemma 10.

For any $i\geq 1$, $b_{i}\circ b_{i+1}=0.$

Proof. As in the previous example, when we develop this identity we obtain
terms of kind
$(a_{1}a_{2})a_{3}\widetilde{\wedge}b_{4}\widetilde{\wedge}\cdots\widetilde{\wedge}a_{i+1}$
and
$a_{1}(a_{2}a_{3})\widetilde{\wedge}a_{4}\widetilde{\wedge}\cdots\widetilde{\wedge}a_{i+1}$
that cancel each other out using the identity related to the anti-
associativity Concerning the other terms we have to distinguish the cases
$i+1$ odd and even. In fact, in this development we have exactly two terms
which are equal up a permutation. If $i+1$ is even, then $\sigma_{1}$ is odd
and $\sigma_{2}$ is even and the
$(-1)^{i+1}\epsilon(\sigma_{1})\epsilon(\sigma_{2})=-1$ and the sum of these
two terms is $0$, if $i+1$ is odd, the permutation $\sigma_{1}$ and
$\sigma_{2}$ have the same signum and
$(-1)^{i+1}\epsilon(\sigma_{1})\epsilon(\sigma_{2})=-1$ and in this case also
the sum of these two terms is zero. Then $b_{i}\circ b_{i+1}=0.$

We deduce that ${\rm im}\ b_{i+1}$ is a subspace of $\ker b_{i}$ and the
homology corresponds to the quotient ${\rm im}\ b_{i+1}/\ker b_{i}$.

###  3.2. The homology of the free algebra of one generator

Let $\mathcal{F_{AA}}(1)$ be the free anti-associative algebra with one
generator. It admits a basis $\\{e_{1},e_{2},e_{3}\\}$ with

$e_{1}e_{1}=e_{2},\ e_{1}e_{2}=-e_{2}e_{1}=e_{3}$

all the other non defined products are zero. We have

  1. (1)

$\widetilde{\Lambda^{1}(\mathcal{F_{AA}}(1))}=\mathcal{F_{AA}}(1)$

  2. (2)

$\widetilde{\Lambda^{2}(\mathcal{F_{AA}}(1))}=\mathbb{K}\\{e_{1}\widetilde{\wedge}e_{1},e_{1}\widetilde{\wedge}e_{2},e_{1}\widetilde{\wedge}e_{3},e_{2}\widetilde{\wedge}e_{2},e_{2}\widetilde{\wedge}e_{3},e_{3}\widetilde{\wedge}e_{3}\\}$

  3. (3)

$\widetilde{\Lambda^{3}(\mathcal{F_{AA}}(1))}$ is generated by the vectors
$e_{1}\widetilde{\wedge}e_{1}\widetilde{\wedge}e_{1},e_{1}\widetilde{\wedge}e_{1}\widetilde{\wedge}e_{2},e_{1}\widetilde{\wedge}e_{1}\widetilde{\wedge}e_{3},e_{1}\widetilde{\wedge}e_{2}\widetilde{\wedge}e_{2}$,

$e_{1}\widetilde{\wedge}e_{2}\widetilde{\wedge}e_{3},e_{1}\widetilde{\wedge}e_{3}\widetilde{\wedge}e_{3},e_{2}\widetilde{\wedge}e_{2}\widetilde{\wedge}e_{2},e_{2}\widetilde{\wedge}e_{2}\widetilde{\wedge}e_{3},e_{2}\widetilde{\wedge}e_{3}\widetilde{\wedge}e_{3},e_{3}\widetilde{\wedge}e_{3}\widetilde{\wedge}e_{3}$.

Then

${\rm im}\
b_{2}=\mathbb{K}\\{e_{1}\widetilde{\wedge}e_{2},e_{2}\widetilde{\wedge}e_{2},e_{2}\widetilde{\wedge}e_{3},e_{3}\widetilde{\wedge}e_{3}\\},\
\ \ker
b_{1}=\mathbb{K}\\{e_{1}\widetilde{\wedge}e_{2},e_{2}\widetilde{\wedge}e_{2},e_{2}\widetilde{\wedge}e_{3},e_{3}\widetilde{\wedge}e_{3}\\}$

and we have

$H_{1}(\mathcal{F_{AA}}(1),\mathcal{F_{AA}}(1))=0.$

##  4\. Commutative and anti-commutative anti-associative algebras

###  4.1. Case commutative

Let $A$ be an anti-associative algebra. Assume that the product is
commutative. Then

$a(bc)=-(ab)c=-c(ab)=-c(ba)=(ca)b=b(ca)=-(bc)a=-a(bc)=-(cb)a=c(ba)$

Then

$a(bc)=0.$

We deduce that any commutative anti-associative algebra is $3$-nilpotent. It
is also a commutative associative algebra. The classification of nilpotent
associative algebras is established up the dimension $5$ [5]. We deduce from
these lists the following examples:

  1. (1)

Dimension $2$:

$e_{1}e_{1}=e_{2}.$

  2. (2)

Dimension $3$:

$e_{1}e_{2}=e_{2}e_{1}=e_{3}$

  3. (3)

Dimension $4$

    1. (a)

$e_{1}e_{1}=e_{2},\ e_{1}e_{3}=e_{3}e_{1}=e_{4}$

    2. (b)

$e_{1}e_{1}=e_{2},\ e_{3}e_{3}=e_{4}$

    3. (c)

$e_{1}e_{1}=e_{4},\ e_{2}e_{3}=e_{3}e_{2}=e_{4}.$

The free algebra with $p$ generators $X_{1},\cdots X_{p}$ admits a basis
$\\{X_{i},1\geq i\leq p,X_{ij},1\leq i\leq j\leq p\\}$ with the multiplication

$X_{i}X_{j}=X_{j}X_{i}=X_{ij},X_{i}X_{jk}=0,\ X_{jk}X_{rs}=0.$

Remark. If for any $x\in A$ we have $x^{2}=0$, then $xy=0$ for any $x,y\in A$
and $A$ is abelian.

###  4.2. Case Anti-commutative

An anti-associative algebra is said anti-commutative if for any $a,b\in A$:

$ab=-ba.$

In particular $a^{2}=0$ for any $a\in A$. The first non abelian example is in
dimension $3$ and given by

$e_{1}e_{2}=-e_{2}e_{1}=e_{3}.$

In this case, this algebra is also a Lie algebra isomorphic to the Heisenberg
algebra.

######  Proposition 11.

An anti-commutative anti-associative algebra is a Lie algebra if and only if
this algebra is $2$-step nilpotent.

In fact, let $A$ be an anti-commutative anti-associative algebra. For any
$a,b,c\in A$ we have
$a(bc)=-a(cb)=-(bc)a=(cb)a=b(ca)=-b(ac)=-c(ba)=c(ab)=-(ab)c=-(ca)b=(ac)b=-(ba)c.$

We deduce

$a(bc)+b(ca)+c(ab)=3a(bc).$

Then $a(bc)+b(ca)+c(ab)=0$ if and only if $a(bc)=0$ for any $a,b,c\in A$.

We deduce that any $2$-step nilpotent Lie algebra is an ant-commutative anti-
associative algebra. This gives a lot of examples of anti-commutative anti-
associative algebras.

Concerning a general classification of the finite dimensional anti-commutative
anti-associative algebras, the previous result show that this problem is not
easy to solve because still to day we don’t know the classification of the
$2$-step nilpotent Lie algebras. For the very small dimension, the computation
are not very difficult (the problem arrives as soon as the dimension is
greater that $7$). We obtain $A$ is abelian or isomorphic to

  1. (1)

Dimension $3$: $e_{1}e_{2}=e_{3}.$

  2. (2)

Dimension $4$: $e_{1}e_{2}=e_{3},e_{i}e_{4}=0$

  3. (3)

Dimension $5$: $e_{1}e_{2}=e_{3},e_{1}e_{4}=e_{5},$

We can also describe the free anti-commutative anti-associative algebras
$\mathcal{F}_{\overline{AA}}(p)$ with $p$ generators.

1\. The free anti-commutative algebra with one generator $X$ is of dimension
$1$ ($x^{2}=0$) and corresponds to the one dimensional Lie algebra.

2\. The free algebra with two generators $X,Y$ is the vector space
$\mathbb{K}\\{X,Y,XY\\}$ (any product of degree $3$ are zero). It is
isomorphic to the $3$-dimensional Heisenberg algebra.

3\. The free algebra with $p$ generators $X_{1},\cdots,X_{p}$ admits the basis
$X_{i},1\leq i\leq p,X_{i}X_{j},1\leq i<j\leq p,X_{i}(X_{j}X_{k}),1\leq
i<j<k\leq p.$ If we denote by
$e_{i}=X_{i},e_{ij}=X_{i}X_{j},e_{i,j,k}=X_{I}(X_{j}X_{k})$, the anti-
associative multiplication is given by

$\left\\{\begin{array}[]{l}e_{i}e_{j}=e_{ij},\ 1\leq i<j\leq p\\\
e_{i}e_{jk}=e_{ijk},1\leq i<j<k\leq p\\\
e_{i}e_{jkl}=e_{ij}e_{kl}=e_{ij}e_{kls}=e_{ijk}e_{lsr}=0.\end{array}\right.$

If $p\geq 3$, this algebra is not a Lie algebra. For example, if $p=3$, the
dimension is $7$, a basis is given by the family
$\\{e_{1},e_{2},e_{3},e_{12},e_{13},e_{23},e_{123}$ and the multiplication:

$\left\\{\begin{array}[]{l}e_{1}e_{2}=e_{12},e_{1}e_{3}=e_{13},e_{2}e_{3}=e_{23}\\\
e_{1}e_{23}=-e_{2}e_{13}=e_{3}e_{12}=e_{123}.\end{array}\right.$

It is not $2$-step nilpotent then it is not a Lie algebra.

##  5\. Polarization and depolarization of anti-associative algebras

Let $(A,\mu)$ a $\mathbb{K}$-algebra whose multiplication is $\mu$. The
polarization process determines two other multiplications $\rho_{\mu}$ and
$\psi_{\mu}$ on $A$ such that $\rho_{\mu}$ is commutative and the second
$\psi_{\mu}$ is anti-commutative, these multiplications being given by

$\rho(x,y)=\mu(x,y)+\mu(y,x),\ \ \ \psi_{\mu}(x,y)=\mu(x,y)-\mu(y,x)$

for any $x,y\in A$.

The depolarization process permits to find again $\mu$ starting of a
commutative multiplication $\rho$ and of an anti-commutative multiplication by

$\mu(x,y)=\frac{1}{2}(\rho(x,y)+\psi(x,y).$

For example, if $\mu$ is associative, then $\rho_{\mu}$ is a Jordan
multiplication and $\psi_{\mu}$ a Lie bracket. An example of depolarization
process concerns the Poisson algebras which can be viewed by this process as
nonassociative algebras.

###  5.1. Polarization of an anti-associative algebra: the commutative part

Let $A$ be an anti-associative algebra whose multiplication is denoted by
$xy$. Let $\rho$ and $\psi$ the associated multiplication by the polarization
process:

$\rho(x,y)=xy+yx,\ \ \psi(x,y)=xy-yx.$

To simplify we will write

$x\bullet y=\rho(x,y)=xy+yx,\ \ [x,y]=\psi(x,y)=xy-yx.$

We have, for any $x_{1},x_{2},x_{3}\in A$:

$x_{1}\bullet(x_{2}\bullet x_{3})+(x_{1}\bullet x_{2})\bullet
x_{3}=x_{1}(x_{3}x_{2})-x_{2}(x_{3}x_{1})-x_{2}(x_{1}x_{3})+x_{3}(x_{1}x_{2}).$

If we denote by $\mathcal{AA}_{\bullet}$ the trilinear map

$\mathcal{AA}_{\bullet}(x_{1},x_{2},x_{3})=x_{1}\bullet(x_{2}\bullet
x_{3})+(x_{1}\bullet x_{2})\bullet x_{3}$

then, for any $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\in\mathbb{K}$ we
have

$\begin{array}[]{l}\alpha_{1}(\mathcal{AA}_{\bullet}(x_{1},x_{2},x_{3})+\mathcal{AA}_{\bullet}(x_{2},x_{3},x_{1})+\mathcal{AA}_{\bullet}(x_{3},x_{1},x_{2}))\\\
+\alpha_{2}(\mathcal{AA}_{\bullet}(x_{2},x_{1},x_{3})-\mathcal{AA}_{\bullet}(x_{3},x_{1},x_{2})\\\
\alpha_{3}(\mathcal{AA}_{\bullet}(x_{3},x_{2},x_{1})+\mathcal{AA}_{\bullet}(x_{2},x_{3},x_{1})+\mathcal{AA}_{\bullet}(x_{3},x_{1},x_{2})\\\
+\alpha_{4}(\mathcal{AA}_{\bullet}(x_{1},x_{3},x_{1})-\mathcal{AA}_{\bullet}(x_{2},x_{3},x_{1}))=0\end{array}$

that is equivalent to the system

$\left\\{\begin{array}[]{l}\mathcal{AA}_{\bullet}(x_{1},x_{2},x_{3})+\mathcal{AA}_{\bullet}(x_{2},x_{3},x_{1})+\mathcal{AA}_{\bullet}(x_{3},x_{1},x_{2}))=0,\\\
\mathcal{AA}_{\bullet}(x_{1},x_{2},x_{3})-\mathcal{AA}_{\bullet}(x_{3},x_{2},x_{1}).\end{array}\right.$

Since $\bullet$ is commutative, this system is equivalent to

(2) $x_{1}\bullet(x_{2}\bullet x_{3})+x_{2}\bullet(x_{3}\bullet
x_{1})+x_{3}\bullet(x_{1}\bullet x_{2})=0$

######  Proposition 12.

Let $(A;\mu)$ be an anti-associative algebra. Then the commutative
multiplication $\bullet$ associated with $\mu$ satisfies the Jacobi identity

$x_{1}\bullet(x_{2}\bullet x_{3})+x_{2}\bullet(x_{3}\bullet
x_{1})+x_{3}\bullet(x_{1}\bullet x_{2})=0$

Such algebras are called Jacobi-Jordan algebras.

Usually, a Jacobi-Jordan algebra over $\mathbb{K}$ ( [1, 2]) is a
nonassociative algebra which satisfies the axiom

  1. (1)

$xy=yx$

  2. (2)

$x(yz)+y(zx)+z(xy)=0$

Then, given an anti-associative algebra $A$, the new product $x\bullet
y=xy+yx$ is a Jacobi-Jordan product on $A$. We will call these Jacobi-Jordan
algebras, as well as any subalgebras of these Jacobi-Jordan algebras, special
Jacobi-Jordan algebras.

Examples on Jacobi-Jordan algebras [1]:

  1. (1)

Dimension $2$: $e_{1}e_{1}=e_{2}$

  2. (2)

Dimension $3$: $e_{1}e_{1}=e_{2},\ e_{3}e_{3}=e_{2}$

  3. (3)

Dimension $4$:

    1. (a)

$e_{1}e_{1}=e_{2},\ e_{1}e_{3}=e_{4}$

    2. (b)

$e_{1}e_{1}=e_{2},\ e_{3}e_{4}=e_{2}$

An algebraic study of these algebras can be found in [1, 2]. A general Jacobi-
Jordan algebra is a nilalgebra of nilindex $3$. In fact for any $x\in A$ we
have $x^{3}=0$. More precisely, in [1], one proves that the class of Jacobi-
Jordan algebras and the class of commutative nilalgebras of nilindex at most
three coincides. This implies, from an Albert result, that

######  Proposition 13.

Any finite dimensional Jordan-Jacobi algebra is nilpotent.

######  Definition 14.

We will said that an algebra $(A,\mu)$ is Jacobi-Jordan admissible is the
commutative algebra $(A,\bullet)$ define by $x\bullet y=\mu(x,y)+\mu(y,x)$ is
a Jordan-Jacobi algebra.

A necessary and sufficient condition for a multiplication $\mu$ to be Jacobi-
Jordan algebra is

$\sum_{\sigma\in\Sigma_{3}}\mathcal{AA}_{\mu}(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)})=0$

where $\Sigma_{3}$ is the symmetric group of degree $3$.

In particular, anti-associative algebras are Jacobi-Jordan admissible. A
Jacobi-Jordan algebra what comes from an anti-associative algebras will be
called a special Jacobi-Jordan algebra. In the following, we will give some
examples of non special Jacobi-Jordan algebras.

######  Proposition 15.

If a Jacobi-Jordan algebra is special, then its nilindex is smaller or equal
to $4$.

Example: the free Jacobi-Jordan algebra $\mathcal{F}_{JJ}(p)$. Let us consider
the free Jacobi-Jordan algebra $\mathcal{F}_{JJ}(p)$ with $p$ generators. For
$p=1$, we have

$\mathcal{F}_{JJ}(1)=\mathbb{K}\\{X\\}\oplus\mathbb{K}\\{X^{2}\\}$

and the multiplication corresponds to

$e_{1}e_{1}=e_{2}.$

This algebra is special because it is a commutative anti-associative algebra
and any commutative anti-associative algebra is a special Jordan-Jacobi
algebra.

For $p=2$, we have

$\mathcal{F}_{JJ}(2)=\oplus_{k\geq 1}\mathcal{F}_{JJ}(2)^{(k)}$

where $\mathcal{F}_{JJ}(2)^{(k)}$ is the vector space of terms of degree $k$
and

$\mathcal{F}_{JJ}(2)^{(k)}\cdot\mathcal{F}_{JJ}(2)^{(k^{\prime})}=\mathcal{F}_{JJ}(2)^{(k+k^{\prime}).}$

We have

  * •

$\mathcal{F}_{JJ}(2)^{(1)}=\mathbb{K}\\{X,Y\\},$

  * •

$\mathcal{F}_{JJ}(2)^{(2)}=\mathbb{K}\\{X^{2},XY,Y^{2}\\},$

  * •

$\mathcal{F}_{JJ}(2)^{(3)}=K\\{X(X(XY)),X(Y(XY)),Y(X(XY)),Y(Y(XY))\\}.$ In
fact, we haVE:

$\left\\{\begin{array}[]{l}X(Y(XX))=-2X(X(XY))\\\ X(X(YY))=-2X(Y(XY))\\\
Y(X(YY))=-2Y(Y(XY))\end{array}\right.$

  * •

For the terms in degree $4$ we have to compute the terms of the form
$a(b(cd))$ and $(ab)(cd)$ with $a,b,c,d=X$ or $Y$. We have

$\left\\{\begin{array}[]{l}(ab)(cd)+c(d(ab))+d((ab)c)=0,\\\
(cd)(ab)+a(b(cd))+b((cd)a)=0.\end{array}\right.$

implying

$c(d(ab))+d((ab)c)=a(b(cd))+b((cd)a).$

This identity implies when $c=d=X,a=b=Y$:

$X(XY^{2}))=Y(YX^{2})$

and when $c=X,a=b=d=Y$:

$Y(Y^{2}X)=2Y(Y(XY)).$

But $Y(XY^{2})+Y(Y(XY))+Y(XY^{2})=2Y(XY^{2})+Y(Y(XY))=0.$ Then

$2Y(XY^{2})=-Y(Y(XY)).$

We deduce

$Y(XY^{2})=Y(Y(XY))=0$

and also

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

In summary:

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

For the other terms we have

$(XY)(XY)=-X((XY)Y)-Y((XY)X)=-Y(XY))X-((XY)X)Y$

implying $X(Y(XY))=Y(X(XY))$. In the same way we have

$X^{2}Y^{2}=-2X(XY^{2})=-2Y(X^{2}Y).$

In conclusion we have

$\left\\{\begin{array}[]{l}Y(X^{2}Y)=(XY)^{2}=X(XY^{2}),\\\
X(Y(XY))=Y(X(XY))=-\frac{1}{2}X(XY^{2}),\\\
X^{2}Y^{2}=-2X(XY^{2})\end{array}\right.$

and all the terms of degree $4$ are zero or equal to $\alpha X(XY^{2}).$ We
remark also that all the terms of partial degree $3$ (that is the maximum of
the degree of $X$ and the degree of $Y$) are zero.

We deduce

$\dim\mathcal{F}_{JJ}(2)^{(4)}=1.$

######  Proposition 16.

The free Jordan-Jacobi algebra with two generators is not special.

Using the previous remark which precise that any term of partiel degree
greater or equal to $3$ is zero, we deduce that
$\mathcal{F}_{JJ}(2)^{(5)}=\\{0\\}$ and more generaly
$\mathcal{F}_{JJ}(p)^{(k)}=\\{0\\}$ as soon as $k\geq 2p+1$.

###  5.2. Polarization of an anti-associative algebra: the anti-commutative
part

If we denote by $\bullet$ and $[,]$ the commutative and anti-commutative
multiplication associated with the multiplication of the anti-associative
algebra $A$, that is

$x\bullet y=xy+yx,\ \ [x,y]=xy-yx$

then the axiom of antiassociativity is equivalent to

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

This equation is solved as soon as $(A,\bullet)$ is nilpotent with an nilindex
equal to $3$ and $[x,y]=0$ for any $x,y\in A$. Then

######  Proposition 17.

Any Jacobi-Jordan algebra with a nilindex equal to $3$ is special.

We can also analize some consequences of the equation (3). Using the natural
action of the symmetric group $\Sigma_{3}$ on the triple $(x,y,z)$, we obtain

######  Proposition 18.

The commutative and anti-commutative multiplication associated with the anti-
associative multiplication satisfy the identities

  1. (1)

$[x,y\bullet z]+[y,z\bullet x]+[z,x\bullet y]=0,$

  2. (2)

$[x,[y,z]]-2[y,[z,x]]+[z,[x,y]]=x\bullet(y\bullet z)-z\bullet(x\bullet y),$

  3. (3)

$x\bullet[y,z]+[y,z\bullet x]+[x\bullet y,z]=0$

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

Remarks.

  1. (1)

The identity $x\bullet[y,z]=-[y,z\bullet x]-[x\bullet y,z]$ implies

$x\bullet[y,z]+y\bullet[z,x]+z\bullet[x,y]=0.$

Since $[x,y\bullet z]+[y,z\bullet x]+[z,x\bullet y]=0$ and
$x(yz)=x\bullet(y\bullet z)+[x,y\bullet z]+x\bullet[y,z]+[x,[y,z]]$, we have

$x(yz)+y(zx)+z(xy)=[x,[y,z]]+[y,[z,x]]+[z,[x,y]]$

and the algebra $(A,[,])$ is a Lie algebra if and only if the anti-associative
algebra $A$ is a non-commutative Jacobi-Jordan algebra.

  2. (2)

If we denote by $f_{x}$ the linear map $f_{x}(y)=x\bullet y$, then the second
relation is written

$f_{x}[y,z]+[y,f_{x}(z)]+[f_{x}(y),z]=0$

that is $f_{x}$ is an anti-derivation of the algebra $(A,[,])$. If
$\widetilde{\mathcal{D}er}(A,[,])$ denotes the subvector space of $End(A)$
whose elements are the anti-derivation of $(A,[,]$, then

$[\widetilde{\mathcal{D}er}(A,[,]),\widetilde{\mathcal{D}er}(A,[,])]\subset\mathcal{D}er(A,[,]).$

If we call inner anti-derivation the anti-derivations of type $f_{x}$ for
$x\in A$, then

$f_{x\bullet y}=-f_{x}\circ f_{y}-f_{y}\circ f_{x}$

and the subspace of inner derivation is a subalgebra of the Jordan algebra
$End(A)$.

######  Proposition 19.

Let $A$ be a Jacobi-Jordan algebra and $f$ an anti-derivation of $A$. We can
provide on the one dimensional extension $A^{\prime}=A\oplus\mathbb{K}\\{X\\}$
of $A$ a Jordan-Jacobi structure $\mu$ given by

$\left\\{\begin{array}[]{l}\mu(x,y)=xy,\ \forall x,y\in A\\\ \mu(x,X)=f(x)\
\forall x\in A,\\\ \mu(X,X)=0.\end{array}\right.$

###  5.3. Deformation quantization of anti-Poisson algebras

######  Definition 20.

An anti-Poisson algebra is a triple $(A,\psi,\rho)$ where $A$ is a
$\mathbb{K}$-vector space, $\psi$ an anti-commutative (or skew-symmetric)
multiplication on $A$, $\rho$ a Jacobi-Jordan multiplication on $A$ such that
these two bilinear maps satisfy the anti-Leibniz identity:

$\mathcal{L}_{g}(\psi,\rho)(x,y,z)=\rho(\psi(x,y),z)+\psi(x,\rho(y,z))+\psi(\rho(x,z),y)=0.$

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

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

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

for any $x,y$ in $A$. The anti-Leibniz identity can be interpreted saying that
for any $z\in A$, the linear maps $x\rightarrow\rho(x,z)$ is an antiderivation
of the algebra $(A,\psi).$

A formal deformation of an anti-associative algebra $(A,\mu_{0})$ is given by
a family

$\varphi_{i}:A\otimes A\rightarrow A,\ \ i\in\mathbb{N}$

satisfying $\varphi_{0}=\mu_{0}$ and

$(D_{k}):\ \ \ \ \ \sum_{i+j=k,i,j\geq
0}\varphi_{i}(\varphi_{j}(a,b),c)+\sum_{i+j=k,i,j\geq
0}\varphi_{i}(a,\varphi_{j}(b,c))=0$

for any $a,b,c\in A$ and for each $k\geq 1$. Now expand $\mu(a,b)$, for
$a,b\in A$, into the power series

$\mu(a,b)=\mu_{0}(a,b)+t\varphi_{1}(a,b)+t^{2}\varphi_{2}(a,b)+\cdots$

for some $k$-bilinear functions $\varphi_{i}:A\otimes A\rightarrow A$. Then
$\mu$ is anti-associative if and only if $(D_{k})$ are satisfied for each
$k\geq 1$.

In we denote as before $\mu_{0}(a,b)=ab$ the multiplication of the anti-
associative algebra $A$, then

$(D_{1}):\ \ \ \ \
\varphi_{1}(a,bc)+a(\varphi_{1}(b,c))+\varphi_{1}(ab,c)+(\varphi_{1}(a,b))c=0,$
$(D_{2}):\ \ \ \ \
\varphi_{1}(a,\varphi_{1}(b,c))+\varphi_{1}(\varphi_{1}(a,b),c))+\varphi_{2}(a,bc)+a(\varphi_{2}(b,c))+\varphi_{2}(ab,c)+(\varphi_{2}(a,b))c=0.$

To make a link with the classical cohomological approach of deformation
theories, we put

$Z^{2}_{\mathcal{AA}}(A,A)=\\{\varphi:A\otimes A\rightarrow
A,\delta^{2}(\varphi)(a,b,c)=\varphi(a,bc)+a(\varphi(b,c))+\varphi(ab,c)+(\varphi(a,b))c=0.\\}$

Then $(D_{1})$ is satisfied as soon as $\varphi_{1}\in
Z^{2}_{\mathcal{AA}}(A,A)$. For example, if $f$ is a linear endomorphism of
$A$, the bilinear map $\varphi_{f}(a,b)=f(ab)-f(a)b-af(b)$ is in
$Z^{2}_{\mathcal{AA}}(A,A)$.

Assume now that $\mu_{0}$ is anti-commutative: $ba=-ab$ and let us consider
the condition $D_{2}$ what do we write

$\varphi_{1}(x_{1},\varphi_{1}(x_{2},x_{3}))+\varphi_{1}(\varphi_{1}(x_{1},x_{2}),x_{3}))+\delta^{2}(\varphi_{2})(x_{1},x_{2},x_{3})=0.$

Let us determine the constants $a_{i}$ such that

$\sum_{\sigma\in\Sigma_{3}}a_{\sigma}\delta^{2}(\varphi_{2})(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)})=0.$

If we develop this equation we obtain the non trivial solution $a_{\sigma}=1$
for every $\sigma\in\Sigma_{3}$. We deduce from $D_{2}$ that

$\sum_{\sigma\in\Sigma_{3}}\varphi_{1}(x_{\sigma(1)},\varphi_{1}(x_{\sigma(2)},x_{\sigma(3)}))+\varphi_{1}(\varphi_{1}(x_{\sigma(1)},x_{\sigma(2)}),x_{\sigma(3)})=0$

and $\varphi_{1}$ is an anti-associative admissible multiplication. We deduce
that the commutative multiplication $\rho_{1}$ associated with $\varphi_{1}$
is an anti-associative multiplication. Now, from $(D_{1})$ we have

$\sum_{\sigma\in\Sigma_{3}}a_{\sigma}\delta^{2}(\varphi_{1})(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)})=0$

for any $a_{\sigma}$. In particular, if

$a_{Id}=a_{\tau(12)}=a_{c},\ a_{\tau(13)}=a_{\tau(23)}=a_{c^{2}}$

where $\tau(ij)$ is the transposition between $i$ and $j$ and $c$ the cycle
$(123)$, then we have

$\varphi_{1}(x_{1},x_{2}x_{3})+x_{2}\varphi_{1}(x_{3},x_{1})-x_{3}\varphi_{1}(x_{1},x_{2})+x_{2}\varphi_{1}(x_{1},x_{3})-x_{3}\varphi_{1}(x_{2},x_{1})+\varphi_{1}(x_{2}x_{3},x_{1})=0$

that is

$\rho_{1}(x_{1},x_{2}x_{3})+x_{2}\rho_{1}(x_{1},x_{3})-x_{3}\rho_{1}(x_{1},x_{2})=0$

or since $\mu_{0}$ is anti-commutative,

$\rho_{1}(x_{1},x_{2}x_{3})+x_{2}\rho_{1}(x_{1},x_{3})+\rho_{1}(x_{1},x_{2})x_{3}=0.$

We shall say that the multiplication $\rho_{1}$ and $\mu_{0}$ are connected by
the anti-Leibniz relation.

######  Theorem 21.

Let $(A,\mu_{0})$ be an anti-symmetric anti-associative algebra and
$\mu_{t}=\mu_{0}+\sum t^{k}\varphi_{k}$ an anti-associative formal deformation
of $\mu_{0}$, that is $\mu_{t}$ is an anti-associative multiplication. Then
$(A,\mu_{0},\rho_{1})$ is an anti-Poisson algebra, that is

  1. (1)

$(A,\rho_{1})$ is a Jordan-Jacobi algebra,

  2. (2)

The products $\mu_{0}$ and $\rho_{1}$ are connected by the anti-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 $(A,\mu_{0},\rho_{1})$.

##  6\. The operad $\mathcal{AA}ss$

###  6.1. The operad $\mathcal{AA}ss$

Recall that deformations of an associative algebra are controlled by the
Hochschild cohomology $H^{*}(A,A)$. This means that $H^{1}(A,A)$ classifies
infinitesimal deformations and $H^{2}(A,A)$ contains obstructions for their
extensions. In this case this cohomology coincides with the operadic
cohomology because the corresponding quadratic operad $\mathcal{A}ss$ is a
Koszul operad. In this section we recall some results given in [6] and we
refer to this paper for the all the proofs. Let $\mathcal{AA}ss=\oplus_{n\geq
1}\mathcal{AA}ss(n)$ the quadratic operad associated with anti-associative
algebras. We have

  1. (1)

$\mathcal{AA}ss(1)=\mathbb{K}$

  2. (2)

$\mathcal{AA}ss(2)=\mathbb{K}\\{x_{1}x_{2},x_{2}x_{1}\\}$

  3. (3)

$\mathcal{AA}ss(3)=\mathbb{K}\\{x_{1}(x_{2}x_{3}),x_{2}(x_{1}x_{3}),x_{3}(x_{2}x_{1}),x_{1}(x_{3}x_{2}),x_{2}(x_{3}x_{1}),x_{3}(x_{1}x_{2})\\}$

  4. (4)

$\mathcal{AA}ss(n)=\\{0\\}$ for $n\geq 4$.

The Poincaré or generating series of $\mathcal{AA}ss$ is the series

$g_{\mathcal{AA}ss(t)}=\sum\frac{(-1)^{n}}{n!}\dim(\mathcal{AA}ss(n))t^{n}=-t+t^{2}-t^{3}.$

Since $\mathcal{AA}ss(1)=\mathbb{K}$, $\mathcal{AA}ss$ admits a minimal model,
unique up to isomorphism. The generating series $g_{M}$ associated with this
minimal model is the formal inverse of $g_{\mathcal{AA}ss(t)}$ taken with the
opposite sign

$g_{\mathcal{AA}ss}(-g_{M}(t))=t.$

We deduce

$g_{M}(t)=-t+t^{2}-t^{3}+4t^{5}-14t^{6}+30t^{7}-33t^{8}-55t^{9}+\cdots$

and this series cannot be the Poincare Series of the dual operad. In fact, if
$\mathcal{R}$ is the ideal of relations which defines $\mathcal{AA}ss$ and
which is generated as a $\Sigma_{3}$-module by the relation
$(x_{1}(x_{2}x_{3})+(x_{1}x_{2})x_{3})$, the ideal $\mathcal{R}^{\bot}$ which
generates the dual operad $\mathcal{AA}ss^{!}$ is the orthogonal of
$\mathcal{R}$ for the classical inner product

(13) $\displaystyle\left\\{\begin{array}[]{l}<(x_{i}\cdot x_{j})\cdot
x_{k},(x_{i^{\prime}}\cdot x_{j^{\prime}})\cdot x_{k^{\prime}}>=0,\ {\rm if}\
(i,j,k)\neq(i^{\prime},j^{\prime},k^{\prime}),\\\ <(x_{i}\cdot x_{j})\cdot
x_{k},(x_{i}\cdot x_{j})\cdot x_{k}>={\varepsilon(\sigma)},\qquad{\rm with}\
\sigma=\left(\begin{array}[]{lll}1&2&3\\\ i&j&k\end{array}\right)\\\
<x_{i}\cdot(x_{j}\cdot x_{k}),x_{i^{\prime}}\cdot(x_{j^{\prime}}\cdot
x_{k^{\prime}})>=0,\ {\rm if}\
(i,j,k)\neq(i^{\prime},j^{\prime},k^{\prime}),\\\ <x_{i}\cdot(x_{j}\cdot
x_{k}),x_{i}\cdot(x_{j}\cdot x_{k})>=-{\varepsilon(\sigma)}\qquad{\rm with}\
\sigma=\left(\begin{array}[]{lll}1&2&3\\\ i&j&k\end{array}\right),\\\
<(x_{i}\cdot x_{j})\cdot x_{k},x_{i^{\prime}}\cdot(x_{j^{\prime}}\cdot
x_{k^{\prime}})>=0,\end{array}\right.$

where $\varepsilon(\sigma)$ is the signature of $\sigma$. Then
i$\mathcal{R}^{\bot}$ is also generated as a $\Sigma_{3}$-module by the
relation $(x_{1}(x_{2}x_{3})+(x_{1}x_{2})x_{3})$ and $\mathcal{AA}ss$ is
selfdual, it was also the case for the operad $\mathcal{A}ss$ of associative
algebra. We deduce that the generating series of $\mathcal{AA}ss^{!}$ is
$-t+t^{2}-t^{3}$. We deduce

######  Theorem 22.

[6] . The quadratic operad $\mathcal{AA}ss$ is not a Koszul operad.

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 [6] 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

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

in which $C^{p}(A,A):=Hom(A^{\otimes^{n}},A)$ 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:

$\displaystyle\delta^{1}_{AA}(f)(x,y)$
$\displaystyle:=xf(y)-f(xy)+f(x)y,\mbox{ and}$
$\displaystyle\delta^{2}_{AA}(\varphi)(x,y,z)$
$\displaystyle:=x\varphi(y,z)+\varphi(xy,z)+\varphi(x,yf)+\varphi(x,y)z,$

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

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

for $p\geq 4$.

###  6.2. The minimal model

In case of anti-associative algebras, the deformation cohomology is described
by the cohomology defined by the minimal model.This minimal model is an
homology isomorphism $(\Gamma(E),\partial)\rightarrow\mathcal{AA}ss$ from the
free operad $\Gamma(E)$ equipped to a differential and $\mathcal{AA}ss$. The
Poincaré series, studied above as the inverse of the Poincaré series of
$\mathcal{AA}ss$, that is

$g_{M}(t)=-t+t^{2}-t^{3}+4t^{5}-14t^{6}+30t^{7}-33t^{8}-55t^{9}+\cdots$

corresponds to the series associated with the minimal model. The deformation
cohomology of anti-associative algebras, based of the study of a minimal model
is also studied in [6]. We summarize the results: we consider the complex

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

– $C^{1}_{AAss}(A,A)=Hom(A,A)$

– $C^{2}_{AAss}(A,A)=Hom(A^{\otimes^{2}},A)$

– $C^{3}_{AAss}(A,A)=Hom(A^{\otimes^{3}},A)$, and

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

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

$\delta^{3}(g)=(\delta_{1}^{3}(g),\delta_{2}^{3}(g),\delta_{3}^{3}(g),\delta_{4}^{3}(g)),$

where

$\displaystyle\delta_{1}^{3}(g)(x,y,z,t,u)$
$\displaystyle:=xg(y,z,tu)-g(x,y,z(tu))+(xy)g(z,t,u)-g(xy,zt,u)$
$\displaystyle\ +g(xy,z,t)u-g((xy)z,t,u)+g(x,y,z)(tu)-g(x,yz,tu),$
$\displaystyle\delta_{2}^{3}(g)(x,y,z,t,u)$
$\displaystyle:=g((xy)z,t,u)-g(xy,z,t)u+g(x,y,zt)u-g(x,y(zt),u)$
$\displaystyle\ +xg(y,zt,u)-g(x,y,(zt)u)+(xy)g(z,t,u)-g(xy,z,tu),$
$\displaystyle\delta_{3}^{3}(g)(x,y,z,t,u)$
$\displaystyle:=g(x,yz,tu)-xg(yz,t,u)+g(x,(yz)t,u)-x(g(y,z,t)u)$
$\displaystyle\ +g(x,y,zt)u-g(xy,z,t)u+(g(x,y,z)t)u-g(x(yz),t,u),\mbox{ and}$
$\displaystyle\delta_{4}^{3}(g)(x,y,z,t,u)$
$\displaystyle:=g(xy,zt,u)-g(x,y,(zt)u)+xg(y,zt,u)-g(x,y(zt),u)$
$\displaystyle\ +(xg(y,z,t))u-g(x,yz,t)u+(g(x,y,z)t)u-g(xy,z,t)u,$

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

Case of Jacobi-Jordan algebra. Let $\mathcal{J}a\mathcal{J}o=\oplus_{n\geq
2}\mathcal{J}a\mathcal{J}o(n)$ be the quadratic operad associated with the
Jacobi-Jordan multiplication. It is clear that

  1. (1)

$\dim\mathcal{J}a\mathcal{J}o(2)=1$ from the commutativity of the
multiplication,

  2. (2)

$\dim\mathcal{J}a\mathcal{J}o(3)=2$ and more precisely
$\mathcal{J}a\mathcal{J}o(3)=\mathbb{K}\\{x_{1}(x_{2}x_{3}),x_{2}(x_{3}x_{1})\\}$

The vector space $\mathcal{J}a\mathcal{J}o(4)$ is generated by the products
$(x_{i}(x_{j}(x_{h}x_{k}))$ and $(x_{i}x_{j})(x_{k}x_{l})$. From the Jacobi-
Jordan axiom, we have

$\begin{array}[]{ll}(x_{i}x_{j})(x_{k}x_{l})&=-x_{i}(x_{j}(x_{j}x_{k}))-x_{j}(x_{i}(x_{k}x_{l}))\\\
&=-x_{k}(x_{l}(x_{i}x_{j}))-x_{l}((x_{i}x_{j})x_{k})\end{array}$

then $(x_{i}(x_{j}(x_{h}x_{k}))$ generates $\mathcal{J}a\mathcal{J}o(4)$.
Moreover, the previous identity implies

$\begin{array}[]{l}x_{1}(x_{2}(x_{3}x_{4}))+x_{2}(x_{1}(x_{3}x_{4})-x_{3}(x_{4}(x_{1}x_{2}))-x_{4}(x_{3}(x_{1}x_{2}))=0,\\\
x_{3}(x_{2}(x_{1}x_{4}))+x_{2}(x_{3}(x_{1}x_{4})-x_{1}(x_{4}(x_{3}x_{2}))-x_{4}(x_{1}(x_{3}x_{2}))=0,\\\
x_{4}(x_{2}(x_{3}x_{1}))+x_{2}(x_{4}(x_{3}x_{1})-x_{3}(x_{1}(x_{4}x_{2}))-x_{1}(x_{3}(x_{4}x_{2}))=0.\end{array}$

This shows that $(x_{i}(x_{j}(x_{h}x_{k}))$ generates
$\mathcal{J}a\mathcal{J}o(4)$ for $i=1,2,3$. Since

$(x_{i}(x_{k}(x_{i}x_{j}))=-(x_{i}(x_{j}(x_{h}x_{k}))-(x_{i}(x_{h}(x_{k}x_{i}))$

we deduce, by adding the $3$ identities, that

$\left\\{\begin{array}[]{l}x_{1}(x_{4}(x_{2}x_{3}))+x_{2}(x_{4}(x_{1}x_{3}))+x_{3}(x_{4}(x_{1}x_{2}))=0,\\\
x_{1}(x_{3}(x_{2}x_{4}))+x_{2}(x_{3}(x_{1}x_{4}))+x_{4}(x_{3}(x_{1}x_{2}))=0,\\\
x_{1}(x_{2}(x_{3}x_{4}))+x_{3}(x_{2}(x_{1}x_{4}))+x_{4}(x_{2}(x_{1}x_{3}))=0,\\\
x_{2}(x_{1}(x_{3}x_{4}))+x_{3}(x_{1}(x_{2}x_{4}))+x_{4}(x_{1}(x_{3}x_{4}))=0.\\\
\end{array}\right.$

Since the rank of this system is equal to $3$, then
$\mathcal{J}a\mathcal{J}o(4)$ is the space

$\mathbb{K}\\{(x_{1}(x_{2}(x_{3}x_{4})),(x_{1}(x_{3}(x_{4}x_{2})),(x_{2}(x_{1}(x_{3}x_{4})),(x_{2}(x_{3}(x_{4}x_{1})),(x_{3}(x_{1}(x_{2}x_{4}))\\}$

and

$\dim\mathcal{J}a\mathcal{J}o(4)=5.$

Since the product is commutative, $\mathcal{J}a\mathcal{J}o(5)$ is generated
by the products represented by

$\bullet(\bullet(\bullet(\bullet\bullet))),\ \
\bullet((\bullet\bullet)(\bullet\bullet)),\
(\bullet\bullet)((\bullet\bullet)\bullet).$

The Jordan-Jacobi condition gives (we use now the letters $a,b,c\cdots$ in
place of $x1,x_{2},x_{3},\cdots$ to short the formulae):

$\begin{array}[]{l}(ab)((cd)e)+a(b((cd)e)+b(a((cd)e))=0\\\
a((bc)(de))+(bc)((de)a)+(de)(a(bc))=0\end{array}$

this shows that $\mathcal{J}a\mathcal{J}o(5)$ is generated by the products
$\bullet(\bullet(\bullet(\bullet\bullet)))$. Since
$\dim\mathcal{J}a\mathcal{J}o(4)=5$, we have $25$ generators of this type. We
have

$(ab)((cd)e)+a(b((cd)e)+b(a((cd)e))=0$

and also

$\begin{array}[]{ ll}(ab)((cd)e)&=-(cd)(e(ab))-e((ab)(cd))\\\
&=c(d(e(ab)))+d(c(e(ab)))+e(a(b(cd)))+e(b(a(cd)))\end{array}$

that implies

$a(b((cd)e)+b(a((cd)e))+c(d(e(ab)))+d(c(e(ab)))+e(a(b(cd)))+e(b(a(cd)))=0$

and by symmetry

$a(c((bd)e)+c(a((bd)e))+b(d(e(ac)))+d(b(e(ac)))+e(a(c(bd)))+e(c(a(bd)))=0.$

Since the other relations are consequences of the previous relations, we
deduce that

$\dim\mathcal{J}a\mathcal{J}o(5)=23.$

At this step, the generating series is

$g(t)=-t+\frac{t^{2}}{2}-\frac{t^{3}}{3}+\frac{t^{4}}{6}+\frac{18}{5!}t^{5}+o(t^{6})$

and this inverse

$\widetilde{g}(t)=-t+\frac{t^{2}}{2}-\frac{t^{3}}{6}+\frac{13}{5!}t^{5}+o(t^{6})$

If this series is the Poincaré series of an operad
$\mathcal{E}=\oplus\mathcal{E}(n)$, then

$\dim\mathcal{E}(2),\dim\mathcal{E}(3)=1,\dim\mathcal{E}(4)=0.$

Computing the ideal of relations of the dual operad
$\mathcal{J}a\mathcal{J}o^{!}$, we see that the associated multiplication is
anti-symmetric and satisfies

$x_{1}(x_{2}x_{3})-x_{3}(x_{1}x_{2})=0$

From the skew-symmetry, this relation is equivalent to

$x_{1}(x_{2}x_{3})=x_{2}(x_{3}x_{1})=x_{3}(x_{1}x_{2})$

and any $\mathcal{J}a\mathcal{J}o$-algebra is a anti-symmetric anti-
commutative algebra. Then

$\mathcal{J}a\mathcal{J}o^{!}(4)=\\{0\\}.$

Its generating series is

$g_{\mathcal{J}a\mathcal{J}o^{!}}=-t+\frac{t^{2}}{2}-\frac{t^{3}}{6}$

and the inverse series

$-t+\frac{t^{2}}{2}-\frac{t^{3}}{3}+\frac{5t^{4}}{24}-\frac{t^{5}}{12}-\frac{7t^{6}}{144}+\frac{13t^{7}}{72}+O(t^{8}).$

Such series cannot be the generating series of a quadratic operad because the
sign of $t^{6}$ is negative. Consequently, $\mathcal{J}a\mathcal{J}o^{!}$ is
not a Koszul operad and

######  Theorem 23.

The quadratic operad $\mathcal{J}a\mathcal{J}o$ corresponding to the Jacobi-
Jordan algebra is not a Koszul operad.

We deduce,as in the previous case, that the deformation cohomology is
described by the cohomology defined by the minimal model. Since the Poincaré
series of $\mathcal{J}a\mathcal{J}o$ is
$g(t)=-t+\frac{t^{2}}{2}-\frac{t^{3}}{3}+\frac{5t^{4}}{4!}+o(t^{5})$, the
inverse is $\widetilde{g}(t)=-t+\frac{t^{2}}{2}-\frac{t^{3}}{6}+o(t^{5})$. We
consider the complex

$C^{1}_{\mathcal{J}a\mathcal{J}o}(A,A)\stackrel{{\scriptstyle\delta^{1}}}{{\longrightarrow}}C^{2}_{\mathcal{J}a\mathcal{J}o}(A,A)\stackrel{{\scriptstyle\delta^{2}}}{{\longrightarrow}}C^{3}_{\mathcal{J}a\mathcal{J}o}(A,A)\stackrel{{\scriptstyle\delta^{3}}}{{\longrightarrow}}C^{4}_{\mathcal{J}a\mathcal{J}o}(A,A)\stackrel{{\scriptstyle\delta^{4}}}{{\longrightarrow}}\cdots$

with

– $C^{1}_{\mathcal{J}a\mathcal{J}o}(A,A)=Hom(A,A)$

– $C^{2}_{\mathcal{J}a\mathcal{J}o}(A,A)=Sym(A^{\otimes^{2}},A)$

– $C^{3}_{\mathcal{J}a\mathcal{J}o}(A,A)=Sym(A^{\otimes^{3}},A)$, and

– $C^{4}_{\mathcal{J}a\mathcal{J}o}(A,A)=Sym(A^{\otimes^{5}},A)\oplus
Sym(A^{\otimes^{5}},A)\oplus Sym(A^{\otimes^{5}},A)\oplus
Sym(A^{\otimes^{5}},A)$

where $Sym(A^{\otimes^{n}},A)$ denotes the space of symmetric $n$-linear maps,
that is invariant by the action of the symmetric group $\Sigma_{n}$. The
differentials $\delta^{i}$ of this cohomological complex coincide with the
standard cohomology in degree $1$ and $2$:

$\delta^{1}(f)(x_{1},x_{2})=f(x_{1}x_{2})-x_{1}f(x_{2})-f(x_{1})x_{2}$

and

$\delta^{2}\varphi(x_{1},x_{2},x_{3})=x_{1}\varphi(x_{2},x_{3})+\varphi(x_{1},x_{2}x_{3})+x_{2}\varphi(x_{3},x_{1})+\varphi(x_{2},x_{3}x_{1})+x_{3}\varphi(x_{1},x_{2})+\varphi(x_{3},x_{1}x_{2})$

where $\varphi$ is a symmetric bilinear form, that is invariant by
$\Sigma_{2}$. We have

$\delta^{2}\varphi(x_{1},x_{2},x_{3})=\sum_{\sigma\in\Sigma_{3}}\delta^{2}_{AA}\varphi(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)})$

where $\delta^{2}_{AA}$ is the differential associated with the cohomological
complex of the anti-associative algebras. Then, in degree $4$, we will have

$\delta^{3}(\psi)=(\sum_{\sigma\in\Sigma_{4}}\delta_{1,AA}^{3}(\psi),\delta_{2,AA}^{3}(\psi),\delta_{3,AA}^{3}(\psi),\delta_{4,AA}^{3}(\psi)),$

where $\psi$ is invariant by $\Sigma_{4}$ and $\delta_{i,AA}^{3}$
corresponding to the anti-associative case.

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