††journal:

Symmetric Zinbiel superalgebras111 The work is supported by FCT
UIDB/00212/2020 and UIDP/00212/2020; RFBR 20-01-00030. The authors thank Pasha
Zusmanovich for some useful discussions.

Abstract. The notion of symmetric Zinbiel superalgebras is introduced. We
prove that the nilpotency index of a symmetric Zinbiel superalgebra is not
greater than 4 and describe two-generated symmetric Zinbiel algebras and odd
generated superalgebras. We discuss identities of mono and binary symmetric
Zinbiel and Leibniz algebras. It is proven that each quadratic Zinbiel algebra
is $2$-step nilpotent. Also, we study double extensions of symmetric Zinbiel
algebras.

Keywords: Zinbiel algebra, dual Leibniz algebra, double extension.

MSC2020: 17A30, 17A32.

## Introduction

Loday introduced a class of symmetric operads generated by one bilinear
operation subject to one relation making each left-normed product of three
elements equal to a linear combination of right-normed products:

$(a_{1}a_{2})a_{3}=\sum\limits_{\sigma\in\mathbb{S}_{3}}x_{\sigma}a_{\sigma(1)}(a_{\sigma(2)}a_{\sigma(3)});$

such an operad is called a parametrized one-relation operad. For a particular
choice of parameters $\\{x_{\sigma}\\}$, this operad is said to be regular if
each of its components is the regular representation of the symmetric group;
equivalently, the corresponding free algebra on a vector space $V$ is, as a
graded vector space, isomorphic to the tensor algebra of $V$. Bremner and
Dotsenko classified, over an algebraically closed field of characteristic
zero, all regular parametrized one-relation operads. In fact, they proved that
each such operad is isomorphic to one of the following five operads: the left-
nilpotent operad, the associative operad, the Leibniz operad, the Zinbiel
operad, and the Poisson operad [6]. An algebra $\mathcal{A}$ is called a
(left) Zinbiel algebra if it satisfies the identity

$(xy)z=x(yz+zy).$

Zinbiel algebras were introduced by Loday in [29]. Under the Koszul duality,
the operad of Zinbiel algebras is dual to the operad of Leibniz algebras.
Zinbiel algebras are also known as pre-commutative algebras [28] and
chronological algebras [27]. A Zinbiel algebra is equivalent to a commutative
dendriform algebra [2]. It plays an important role in the definition of pre-
Gerstenhaber algebras. The variety of Zinbiel algebras is a proper subvariety
in the variety of right commutative algebras. Each Zinbiel algebra with the
commutator multiplication gives a Tortkara algebra [16], which has sprung up
in unexpected areas of mathematics [13, 14]. Recently, the notion of matching
Zinbiel algebras was introduced in [21]. Zinbiel algebras also appeared in a
study of rack cohomology [12], number theory [10] and in a construction of a
Cartesian differential category [24]. In recent years, there has been a strong
interest in the study of Zinbiel algebras in the algebraic and the operad
context [4, 8, 15, 23, 16, 17, 18, 21, 26, 30, 32, 33, 34].

Free Zinbiel algebras were shown to be precisely the shuffle product algebra
[30], which is under a certain interest until now [11]. Naurazbekova proved
that, over a field of characteristic zero, free Zinbiel algebras are the free
associative-commutative algebras (without unity) with respect to the
symmetrization multiplication and their free generators are found; also she
constructed examples of subalgebras of the two-generated free Zinbiel algebra
that are free Zinbiel algebras of countable rank [33]. Nilpotent algebras play
an important role in the class of Zinbiel algebras. So, Dzhumadildaev and
Tulenbaev proved that each complex finite-dimensional Zinbiel algebra is
nilpotent [17] and recently Towers proved that each finite-dimensional Zinbiel
algebra is nilpotent [35]; Naurazbekova and Umirbaev proved that in
characteristic zero any proper subvariety of the variety of Zinbiel algebras
is nilpotent [34]. Finite-dimensional Zinbiel algebras with a “big” nilpotency
index are classified in [1, 8]. Central extensions of three-dimensional
Zinbiel algebras were calculated in [4] and of filiform Zinbiel algebras in
[9]. The full system of degenerations of complex four-dimensional Zinbiel
algebras is given in [26].

The present paper is about symmetric (left and right) Zinbiel (super)algebras.
This variety of algebras is dual to symmetric Leibniz algebras (about
symmetric Leibniz algebras see in [5] and references therein) and it is very
thin. As we have proven each symmetric Zinbiel algebra is a $2$-step nilpotent
or $3$-step nilpotent algebra. We also provide a classification of two-
generated symmetric Zinbiel algebras (Theorem 15) and symmetric Zinbiel
superalgebras with two odd generators (Proposition 16). The next part of the
paper is devoted to a study of mono and binary symmetric Zinbiel algebras.
Next, we give the description of quadratic Zinbiel algebras. It has been
proven that each quadratic Zinbiel algebra is $2$-step nilpotent (Proposition
28). The notion of double extensions of quadratic Zinbiel algebras is
introduced and described in the last section.

##  1 Definitions and preliminary results

Let $V=V_{\bar{0}}\oplus V_{\bar{1}}$ be a $\mathbb{Z}_{2}$-graded vector
space over the field $\mathbb{K}$. An element $x$ in $V$ is called homogeneous
if $x\in V_{\bar{0}}$ or $x\in V_{\bar{1}}$. Throughout this paper, all
elements are supposed to be homogeneous unless otherwise stated. For a
homogeneous element $x$ we shall use the standard notation
$|x|\in\mathbb{Z}_{2}=\\{\bar{0},\bar{1}\\}$ to indicate its degree, i.e.
whether it is contained in the even part ($|x|=\bar{0}$) or in the odd part
($|x|=\bar{1}$). All superalgebras considered in this paper are finite-
dimensional.

Let $V$ be a $\mathbb{K}-$vector space and $\gamma:={\wedge V}$ a Grassmann
(or exterior) algebra of $V$. We know that
$\Gamma:=\oplus_{i\in\mathbb{Z}}\wedge^{i}V=\Gamma_{\bar{0}}\oplus\Gamma_{\bar{1}}$
is a $\mathbb{Z}_{2}-$graded associative algebra, where
$\Gamma_{\bar{0}}:=\oplus_{i\in\mathbb{Z}}\wedge^{2i}V$ and
$\Gamma_{\bar{1}}:=\oplus_{i\in\mathbb{Z}}\wedge^{2i+1}V$, such that
$X_{\alpha}X_{\beta}=(-1)^{\alpha\beta}X_{\beta}X_{\alpha},\,\,\forall\,(X_{\alpha},X_{\beta})\in\Gamma_{\alpha}\times\Gamma_{\beta}.$

Let $(\mathcal{A}=\mathcal{A}_{\bar{0}}\oplus\mathcal{A}_{\bar{1}},\cdot)$ be
a superalgebra and $\Gamma(\mathcal{A})$ its Grassmann enveloping algebra
which is a subalgebra of $\mathcal{A}\otimes\Gamma$ given by
$\Gamma(\mathcal{A})=\mathcal{A}_{\bar{0}}\otimes\Gamma_{\bar{0}}\oplus\mathcal{A}_{\bar{1}}\otimes\Gamma_{\bar{1}}$.
Let us assume now that $\mathfrak{B}$ is a homogeneous variety of algebras.
Then, $\mathcal{A}$ is said to be a $\mathfrak{B}$-superalgebra if
$\Gamma(\mathcal{A})$ belongs to $\mathfrak{B}$. So,
$\mathcal{A}=\mathcal{A}_{\bar{0}}\oplus\mathcal{A}_{\bar{1}}$ is a left
(resp. right) Zinbiel superalgebra if $\Gamma(\mathcal{A})$ is a left (resp.
right) Zinbiel algebra. Consequently, we get the following definition.

######  Definition 1.

Let $\mathcal{A}=\mathcal{A}_{\bar{0}}\oplus\mathcal{A}_{\bar{1}}$ be a
$\mathbb{Z}_{2}$-graded vector space and let
$``\cdot":\mathcal{A}\times\mathcal{A}\rightarrow\mathcal{A}$ be a bilinear
map on $\mathcal{A}$ such that
$\mathcal{A}_{i}\cdot\mathcal{A}_{j}\subset\mathcal{A}_{i+j},~{}\forall
i,j\in\mathbb{Z}_{2}$.

  1. 1.

$\mathcal{A}$ is called a left (resp. right) Zinbiel superalgebra if, for all
$x,y,z\in\mathcal{A}_{\bar{0}}\cup\mathcal{A}_{\bar{1}},$

$(xy)z=x(yz)+(-1)^{|y||z|}x(zy),$ (1.1) $({\rm
resp.}~{}~{}x(yz)=(xy)z+(-1)^{|x||y|}(yx)z),$ (1.2)

or, equivalently, $(x,y,z)=(-1)^{|y||z|}x(zy)$ ( resp.
$(x,y,z)=-(-1)^{|x||y|}(yx)z\ )$, where $(x,y,z)=(xy)z-x(yz)$ is the
associator associated to $``\cdot"$.

  2. 2.

$\mathcal{A}$ is called a symmetric Zinbiel superalgebra if it is both a left
and a right Zinbiel superalgebra.

As usual, for $x\in\mathcal{A}_{\bar{0}}\cup\mathcal{A}_{\bar{1}},$ we define
the corresponding endomorphism of $\mathcal{A}$ by $L_{x}(y)=xy$ (resp.
$R_{x}(y)=(-1)^{|x||y|}yx$), $\forall
y\in\mathcal{A}_{\bar{0}}\cup\mathcal{A}_{\bar{1}}$, which is called the left
(resp. right) multiplication by $x$.

The following proposition can be verified by direct calculation.

######  Proposition 2.

Let $(\mathcal{A},\cdot)$ be a symmetric Zinbiel superalgebra. Then,

  1. 1.

$(\mathcal{A},\cdot)$ is a LR-superalgebra (LR-superalgebras are also known as
bicommutative superalgebras), i.e. it satisfies

$(xy)z=(-1)^{|y||z|}(xz)y\mbox{ and }x(yz)=(-1)^{|x||y|}y(xz),\mbox{ for all
}x,y,z\in\mathcal{A}_{\bar{0}}\cup\mathcal{A}_{\bar{1}},$

  2. 2.

$(\mathcal{A},\cdot)$ is an anti-flexible superalgebra, i.e. it satisfies

$(x,y,z)=(-1)^{|x||y|+|z|(|x|+|y|)}(z,y,x),\mbox{ for all
}x,y,z\in\mathcal{A}_{\bar{0}}\cup\mathcal{A}_{\bar{1}},$

  3. 3.

$(\mathcal{A},\cdot)$ satisfies

$(xy)z=-(-1)^{|x|(|y|+|z|)}y(zx),\mbox{ for all
}x,y,z\in\mathcal{A}_{\bar{0}}\cup\mathcal{A}_{\bar{1}},$

  4. 4.

$(\mathcal{A},\cdot)$ satisfies

$(xy)z=-(-1)^{|x||y|+|y||z|+|z||x|}z(yx),\mbox{ for all
}x,y,z\in\mathcal{A}_{\bar{0}}\cup\mathcal{A}_{\bar{1}}.$

In the following, we recall some results about representations of left and
right Zinbiel superalgebras.

######  Definition 3.

Let $(\mathcal{A},\cdot)$ be a non-associative superalgebra, $\mathcal{V}$ be
a $\mathbb{Z}_{2}$-graded vector space and $r,l:\mathcal{A}\rightarrow
End(\mathcal{V})$ be two even linear maps. If $\mathcal{A}$ is a left (resp.
right) Zinbiel superalgebra, then we say that $(r,l)$ is a left (resp. right)
representation of $\mathcal{A}$ in $\mathcal{V}$ if for all
$x,y\in\mathcal{A}_{|x|}\times\mathcal{A}_{|y|}$:

$l(xy)=l(x)l(y)+l(x)r(y);~{}~{}r(x)r(y)=r(xy)+(-1)^{|x||y|}r(yx);~{}~{}l(x)l(y)=(-1)^{|x||y|}[r(y),l(x)],$
$\big{(}{\rm
resp.}~{}~{}l(x)l(y)=l(xy)+(-1)^{|x||y|}l(yx);~{}~{}~{}~{}l(x)r(y)=r(xy)=(-1)^{|x||y|}(r(y)r(x)+r(y)l(x))\big{)}.$

######  Example 4.

Let $(\mathcal{A},\cdot)$ be a left (resp. right) Zinbiel superalgebra.
Consider the even maps $L:\mathcal{A}\rightarrow End(\mathcal{A})$ and
$R:\mathcal{A}\rightarrow End(\mathcal{A})$ defined by:
$L(x):=L_{x},~{}R(x):=R_{x},\forall x\in\mathcal{A}$. Then, $(R,L)$ is a left
(resp. right) representation of $\mathcal{A}$ in $\mathcal{A}$ called the left
(resp. right) adjoint representation of $\mathcal{A}$.

######  Proposition 5.

Let $(\mathcal{A},\cdot)$ be a left (resp. right) Zinbiel superalgebra and
$r,l:\mathcal{A}\rightarrow End(\mathcal{V})$ be two even linear maps. Then,
the $\mathbb{Z}_{2}$-graded vector space
$\overline{\mathcal{A}}=\mathcal{A}\oplus\mathcal{V}$ endowed with the product
defined by:
$(x+u)(y+v)=xy+l(x)(v)+(-1)^{|x||y|}r(y)(u),~{}\forall(x+u,y+v)\in\overline{\mathcal{A}}_{|x|}\times\overline{\mathcal{A}}_{|y|}$,
is a left (resp. right) Zinbiel superalgebra if and only if $(r,l)$ is a left
(resp. right ) representation of $\mathcal{A}$ in $\mathcal{V}$.

######  Proposition 6.

Let $(\mathcal{A},\cdot)$ be a left (resp. right) Zinbiel superalgebra and
$(R,L)$ be the left (resp. right) adjoint representation of $\mathcal{A}$. Let
us consider the even linear maps $L^{*},R^{*}:\mathcal{A}\rightarrow
End(\mathcal{A}^{*})$ defined by:

$L^{*}(x)(f)=(-1)^{|f||x|}f\circ
R(x),~{}~{}~{}~{}R^{*}(x)(f)=(-1)^{|f||x|}f\circ L(x),~{}~{}\forall
x\in\mathcal{A}_{|x|},f\in\mathcal{A}^{*}_{|f|}~{}.$

$(R^{*},L^{*})$ is a left (resp. right) representation of $\mathcal{A}$ in
$\mathcal{A}^{*}$ if and only if $\mathcal{A}$ is $2$-step nilpotent.

Proof. Suppose that $(\mathcal{A},\cdot)$ is a left Zinbiel superalgebra.
Then, for all
$x\in\mathcal{A}_{|x|},y\in\mathcal{A}_{|y|},z\in\mathcal{A}_{|z|}$ and
$f\in\mathcal{A}^{*}_{|f|}$, a simple calculation proves that
$L^{*}(xy)(f)(z)=\big{(}L^{*}(x)L^{*}(y)+L^{*}(x)R^{*}(y)\big{)}(f)(z)$ if and
only if

$y(zx)=-(-1)^{|y||z|}z(yx).$ (1.3)

Moreover,
$R^{*}(x)R^{*}(y)(f)(z)=\big{(}R^{*}(xy)+(-1)^{|x||y|}R^{*}(yx)\big{)}(f)(z)$
if and only if

$y(zx)=-(-1)^{|x|(|y|+|z|)}(xy)z.$ (1.4)

Furthermore, according to (1.3),
$L^{*}(x)L^{*}(y)(f)(z)=\big{(}(-1)^{|x||y|}R^{*}(y)L^{*}(x)-L^{*}(x)R^{*}(y)\big{)}(f)(z)$
if and only if

$z(xy)=(-1)^{|y|(|x|+|z|)}(yz)x.$ (1.5)

By (1.4) and (1.5), we conclude that $(R^{*},L^{*})$ is a left representation
of $\mathcal{A}$ in $\mathcal{A}^{*}$ if and only if $\mathcal{A}$ is $2$-step
nilpotent. By the same way, we prove the result for right Zinbiel
superalgebras. $\square$

In the following, we introduce the representation of symmetric Zinbiel
superalgebras and give some related results.

######  Definition 7.

Let $(\mathcal{A},\cdot)$ be a symmetric Zinbiel superalgebra, $\mathcal{V}$
be a $\mathbb{Z}_{2}$-graded vector space and $r,l:\mathcal{A}\rightarrow
End(\mathcal{V})$ be two even linear maps. Then, we say that $(r,l)$ is a
representation of $\mathcal{A}$ in $\mathcal{V}$ if $(r,l)$ is a left and a
right representation of $\mathcal{A}$ in $\mathcal{V}$. We denote by
$Rep(\mathcal{A},\mathcal{V})$ the set of all representations of $\mathcal{A}$
in $\mathcal{V}$.

######  Example 8.

If $(\mathcal{A},\cdot)$ is a symmetric Zinbiel superalgebra, then:

  1. 1.

$(R,L)$ (see Example 4) is a representation of $\mathcal{A}$ in $\mathcal{A}$
called the adjoint representation of $\mathcal{A}$.

  2. 2.

$(R^{*},L^{*})$ (see Proposition 6) is a representation of $\mathcal{A}$ in
$\mathcal{A}^{*}$ called the co-adjoint representation of $\mathcal{A}$.

######  Remark 9.

Let $(\mathcal{A},\cdot)$ be a symmetric Zinbiel superalgebra, $\mathcal{V}$
be a $\mathbb{Z}_{2}$-graded vector space and $r,l:\mathcal{A}\rightarrow
End(\mathcal{V})$ be two even linear maps. Then, the $\mathbb{Z}_{2}$-graded
vector space $\overline{\mathcal{A}}=\mathcal{A}\oplus\mathcal{V}$ endowed
with the product defined by:

$(x+u)(y+v)=xy+l(x)(v)+(-1)^{|x||y|}r(y)(u),~{}\forall(x+u,y+v)\in\overline{\mathcal{A}}_{|x|}\times\overline{\mathcal{A}}_{|y|}$,

is a symmetric Zinbiel superalgebra if and only if $(r,l)\in
Rep(\mathcal{A},\mathcal{V})$.

##  2 Characterizations of symmetric Zinbiel superalgebras

This section will be devoted to the characterizations of symmetric Zinbiel
superalgebras. In this section, we will consider algebras over the complex
field.

######  Definition 10.

Let $(\mathcal{A},\cdot)$ be a superalgebra, then

  * 1.

$(\mathcal{A},\cdot)$ is a $2$-step nilpotent superalgebra if for all elements
$x,y,z\in\mathcal{A}_{\bar{0}}\cup\mathcal{A}_{\bar{1}}$ we have
$(xy)z=x(yz)=0,$ but the product of the algebra is nonzero.

  * 2.

$(\mathcal{A},\cdot)$ is a $3$-step nilpotent superalgebra if for all elements
$x,y,z,t\in\mathcal{A}_{\bar{0}}\cup\mathcal{A}_{\bar{1}}$ we have

$(xy)(zt)=x((yz)t)=x(y(zt))=((xy)z)t=(x(yz))t=0,$

but the algebra is non-$2$-step nilpotent.

######  Example 11.

Each $2$-step nilpotent superalgebra is a symmetric Zinbiel superalgebra.

######  Theorem 12.

If $(\mathcal{A},\cdot)$ is a nonzero non-$2$-step nilpotent symmetric Zinbiel
superalgebra, then

  1. 1.

$(\mathcal{A},\cdot)$ is a $3$-step nilpotent superalgebra, i.e. it is a
central extension of a $2$-step nilpotent superalgebra;

  2. 2.

$x^{3}=0$ for each $x\in\mathcal{A}_{|x|}.$

Proof. For the first part of the theorem we consider the following relations:

$(xy)(zt)$ | $=$ | $(-1)^{|z|(|x|+|y|)}z((xy)t)=-(-1)^{(|z|+|t|)(|x|+|y|)+|x||y|}z(t(yx)),$
---|---|---
$(xy)(zt)$ | $=$ | $-(-1)^{(|x|+|y|)(|z|+|t|)+|z||t|}(tz)(xy)=-(-1)^{|t|(|x|+|y|+|z|)}(t(xy))z=(-1)^{(|z|+|t|)(|x|+|y|)}z(t(xy)).$

Hence,

$0=z(t(xy+(-1)^{|x||y|}yx))=z((tx)y),$

which gives that all products of $4$ elements are equal to zero and
$(\mathcal{A},\cdot)$ is a $3$-step nilpotent superalgebra.

Now, for the second part of the theorem, we are using (1.1) and (1.2) and
substituting $y=z=x\in\mathcal{A}_{|x|}$. We get:

  1. 1.

If $x\in\mathcal{A}_{\bar{0}}$, then $x^{2}x=2xx^{2}$ and $xx^{2}=2x^{2}x$,
which implies that $x^{2}x=0=xx^{2}$. So, $x^{3}=x^{2}x=xx^{2}=0$.

  2. 2.

If $x\in\mathcal{A}_{\bar{1}}$, then $x^{2}x=0=xx^{2}$. So, $x^{3}=0$.

$\square$

######  Corollary 13.

If $(\mathcal{A},\cdot)$ is a $d$-generated symmetric Zinbiel superalgebra,
then $dim(\mathcal{A},\cdot)\leq-d+d^{2}+2d^{3}+d^{4}.$

Proof. It is easy to see that the maximal dimension of a central extension
$\hat{\mathcal{A}}$ of the $d$-dimensional zero product algebra (with a basis
$\\{e_{1},\ldots,e_{d}\\}$) is $d+d^{2}.$ Let say that the last algebra has a
basis $\\{e_{1},\ldots,e_{d},e_{1,1},\ldots,e_{d,d}\\},$ such that
$e_{i,j}=e_{i}e_{j}.$ ${\mathcal{A}}$ is a central extension of
$\hat{\mathcal{A}}.$ Note now, that $x^{3}=0$ for each $x\in\mathcal{A},$
hence $e_{i}e_{i,i}=e_{i,i}e_{i}=0$ and ${\rm dim}\
\hat{\mathcal{A}}^{2}=d^{2}.$ It is easy to see that the dimension of the
cohomology space of $\hat{\mathcal{A}}$ is not greater than
$(d+d^{2})^{2}-d^{2}-2d.$ Hence, the dimension of a central extension of a
central extension of the $d$-generated zero product algebra is not greater
than $d+d^{2}+(d+d^{2})^{2}-d^{2}-2d=-d+d^{2}+2d^{3}+d^{4}.$ $\square$

######  Corollary 14.

If $(\mathcal{A},\cdot)$ is a one-generated symmetric Zinbiel superalgebra,
then there are only two following opportunities

  1. 1.

$dim(\mathcal{A},\cdot)=(2,0)$ and the multiplication table is given by
$e_{1}^{2}=e_{2}.$

  2. 2.

$dim(\mathcal{A},\cdot)=(1,1),$ $\mathcal{A}_{\bar{0}}=\langle e_{1}\rangle,$
$\mathcal{A}_{\bar{1}}=\langle e_{2}\rangle,$ and the multiplication table is
given by $e_{2}^{2}=e_{1}.$

######  Theorem 15.

If $(\mathcal{A},\cdot)$ is a two-generated symmetric Zinbiel algebra, then
there are only following opportunities:

  1. 1.

if $(\mathcal{A},\cdot)$ is a $2$-step nilpotent algebra, then $3\leq
dim(\mathcal{A},\cdot)\leq 6$ and

    1. (a)

if $dim(\mathcal{A},\cdot)=3,$ then $(\mathcal{A},\cdot)$ is isomorphic to one
of the following algebras

$\begin{array}[]{ll
llll}\mathfrak{N}^{3}_{1}&:&e_{1}e_{1}=e_{2}&e_{1}e_{2}=0&e_{2}e_{1}=0&e_{2}e_{2}=0\\\
\mathfrak{N}^{3}_{2}&:&e_{1}e_{1}=e_{3}&e_{1}e_{2}=0&e_{2}e_{1}=0&e_{2}e_{2}=e_{3}\\\
\mathfrak{N}^{3}_{3}&:&e_{1}e_{1}=0&e_{1}e_{2}=e_{3}&e_{2}e_{1}=-e_{3}&e_{2}e_{2}=0\\\
\mathfrak{N}^{3}_{4,\lambda}&:&e_{1}e_{1}=\lambda
e_{3}&e_{1}e_{2}=0&e_{2}e_{1}=e_{3}&e_{2}e_{2}=e_{3}\\\ \end{array}$

    2. (b)

if $dim(\mathcal{A},\cdot)=4,$ then $(\mathcal{A},\cdot)$ is isomorphic to one
of the following algebras

$\begin{array}[]{ll
llll}\mathfrak{N}^{4}_{1}&:&e_{1}e_{1}=0&e_{1}e_{2}=e_{3}&e_{2}e_{1}=e_{4}&e_{2}e_{2}=0\\\
\mathfrak{N}^{4}_{2}&:&e_{1}e_{1}=e_{3}&e_{1}e_{2}=0&e_{2}e_{1}=e_{4}&e_{2}e_{2}=0\\\
\mathfrak{N}^{4}_{3}&:&e_{1}e_{1}=e_{3}&e_{1}e_{2}=0&e_{2}e_{1}=0&e_{2}e_{2}=e_{4}\\\
\mathfrak{N}^{4}_{4}&:&e_{1}e_{1}=e_{3}&e_{1}e_{2}=e_{3}&e_{2}e_{1}=e_{4}&e_{2}e_{2}=e_{3}\\\
\mathfrak{N}^{4}_{5,\lambda}&:&e_{1}e_{1}=e_{3}&e_{1}e_{2}=0&e_{2}e_{1}=e_{4}&e_{2}e_{2}=e_{3}+\lambda
e_{4}\\\
\mathfrak{N}^{4}_{6,\lambda}&:&e_{1}e_{1}=e_{3}&e_{1}e_{2}=e_{4}&e_{2}e_{1}=\lambda
e_{4}&e_{2}e_{2}=0\\\ \end{array}$

    3. (c)

if $dim(\mathcal{A},\cdot)=5,$ then $(\mathcal{A},\cdot)$ is isomorphic to one
of the following algebras

$\begin{array}[]{ll
llll}\mathfrak{N}^{5}_{1}&:&e_{1}e_{1}=0&e_{1}e_{2}=e_{3}&e_{2}e_{1}=e_{4}&e_{2}e_{2}=e_{5}\\\
\mathfrak{N}^{5}_{2}&:&e_{1}e_{1}=e_{5}&e_{1}e_{2}=e_{3}&e_{2}e_{1}=e_{4}&e_{2}e_{2}=e_{5}\\\
\mathfrak{N}^{5}_{3}&:&e_{1}e_{1}=e_{3}&e_{1}e_{2}=e_{4}&e_{2}e_{1}=e_{4}&e_{2}e_{2}=e_{5}\\\
\mathfrak{N}^{5}_{4,\lambda}&:&e_{1}e_{1}=e_{3}+\lambda
e_{5}&e_{1}e_{2}=e_{3}&e_{2}e_{1}=e_{4}&e_{2}e_{2}=e_{5}\\\ \end{array}$

    4. (d)

if $dim(\mathcal{A},\cdot)=6,$ then $(\mathcal{A},\cdot)$ is isomorphic to the
following algebra

$\begin{array}[]{ll
llll}\mathfrak{N}^{6}_{1}&:&e_{1}e_{1}=e_{3}&e_{1}e_{2}=e_{4}&e_{2}e_{1}=e_{5}&e_{2}e_{2}=e_{6}\end{array}$

  2. 2.

if $(\mathcal{A},\cdot)$ is a $3$-step nilpotent algebra, then $6\leq
dim(\mathcal{A},\cdot)\leq 8$ and

    1. (a)

if $dim(\mathcal{A},\cdot)=6,$ then $(\mathcal{A},\cdot)$ is isomorphic to one
of the following algebras:

$\mathfrak{Z}^{6}_{1}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=e_{5}$ | $e_{1}e_{5}=e_{6}$ | $e_{5}e_{1}=-e_{6}$
---|---|---|---|---|---|---
|  | $e_{2}e_{4}=-2e_{6}$ | $e_{4}e_{2}=-e_{6}$ | $e_{2}e_{3}=e_{6}$ | $e_{3}e_{2}=2e_{6}$ |
$\mathfrak{Z}^{6}_{2}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=e_{5}+e_{6}$ | $e_{1}e_{5}=e_{6}$ | $e_{5}e_{1}=-e_{6}$
|  | $e_{2}e_{4}=-2e_{6}$ | $e_{4}e_{2}=-e_{6}$ | $e_{2}e_{3}=e_{6}$ | $e_{3}e_{2}=2e_{6}$ |
    2. (b)

if $dim(\mathcal{A},\cdot)=7,$ then $(\mathcal{A},\cdot)$ is isomorphic to the
following algebra

$\mathfrak{Z}^{7}_{1}$ | $:$ | $e_{1}e_{1}=e_{7}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=e_{5}$ | $e_{1}e_{5}=e_{6}$
---|---|---|---|---|---|---
|  | $e_{5}e_{1}=-e_{6}$ | $e_{2}e_{4}=-2e_{6}$ | $e_{4}e_{2}=-e_{6}$ | $e_{2}e_{3}=e_{6}$ | $e_{3}e_{2}=2e_{6}$
    3. (c)

if $dim(\mathcal{A},\cdot)=8,$ then $(\mathcal{A},\cdot)$ is isomorphic to the
following algebra

$\mathfrak{Z}^{8}_{1}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{4}=2e_{7}$ | $e_{1}e_{5}=-e_{7}$ | $e_{1}e_{6}=e_{8}$ | $e_{2}e_{1}=e_{5}$
---|---|---|---|---|---|---|---
|  | $e_{2}e_{2}=e_{6}$ | $e_{2}e_{3}=-e_{7}$ | $e_{2}e_{4}=e_{8}$ | $e_{2}e_{5}=-2e_{8}$ | $e_{3}e_{2}=e_{7}$ | $e_{4}e_{1}=e_{7}$
|  | $e_{4}e_{2}=2e_{8}$ | $e_{5}e_{1}=-2e_{7}$ | $e_{5}e_{2}=-e_{8}$ | $e_{6}e_{1}=-e_{8}$ |  |

Proof. The first part of the theorem is following from [7]. The proof of the
second part of the theorem will be done the usual way by calculations of
symmetric Zinbiel central extensions (it is known as the Skjelbred-Sund method
(about the description of the method for an arbitrary variety see [25] and for
Zinbiel case see [4]).

Thanks to [4], there are no non-trivial symmetric Zinbiel central extensions
of algebras $\mathfrak{N}^{3}_{i}$ and $\mathfrak{N}^{4}_{i}.$ Hence, we will
consider only algebras $\mathfrak{N}^{5}_{i}$ and $\mathfrak{N}^{6}_{1}.$

  * 1.

The algebra $\mathfrak{N}^{5}_{1}$ has only two $1$-dimensional and one
$2$-dimensional central extensions:

$\mathfrak{Z}^{6}_{1}$ | : | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=e_{5}$ | $e_{1}e_{5}=e_{6}$ | $e_{5}e_{1}=-e_{6}$ |
---|---|---|---|---|---|---|---
|  | $e_{2}e_{4}=-2e_{6}$ | $e_{4}e_{2}=-e_{6}$ | $e_{2}e_{3}=e_{6}$ | $e_{3}e_{2}=2e_{6}$ |  |
$\mathfrak{Z}^{6}_{2}$ | : | $e_{1}e_{1}=e_{6}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=e_{5}$ | $e_{1}e_{5}=e_{6}$ |
|  | $e_{5}e_{1}=-e_{6}$ | $e_{2}e_{4}=-2e_{6}$ | $e_{4}e_{2}=-e_{6}$ | $e_{2}e_{3}=e_{6}$ | $e_{3}e_{2}=2e_{6}$ |
$\mathfrak{Z}^{7}_{1}$ | : | $e_{1}e_{1}=e_{7}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=e_{5}$ | $e_{1}e_{5}=e_{6}$ |
|  | $e_{5}e_{1}=-e_{6}$ | $e_{2}e_{4}=-2e_{6}$ | $e_{4}e_{2}=-e_{6}$ | $e_{2}e_{3}=e_{6}$ | $e_{3}e_{2}=2e_{6}$ |
  * 2.

The algebras $\mathfrak{N}^{5}_{2},\mathfrak{N}^{5}_{3}$ and
$\mathfrak{N}^{5}_{4,\lambda}$ have no central extensions.

  * 3.

The algebra $\mathfrak{N}^{6}_{1}$ has only one $2$-dimensional central
extension:

$\mathfrak{Z}^{8}_{1}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{4}=2e_{7}$ | $e_{1}e_{5}=-e_{7}$ | $e_{1}e_{6}=e_{8}$ | $e_{2}e_{1}=e_{5}$
---|---|---|---|---|---|---|---
|  | $e_{2}e_{2}=e_{6}$ | $e_{2}e_{3}=-e_{7}$ | $e_{2}e_{4}=e_{8}$ | $e_{2}e_{5}=-2e_{8}$ | $e_{3}e_{2}=e_{7}$ | $e_{4}e_{1}=e_{7}$
|  | $e_{4}e_{2}=2e_{8}$ | $e_{5}e_{1}=-2e_{7}$ | $e_{5}e_{2}=-e_{8}$ | $e_{6}e_{1}=-e_{8}$ |  |

$\square$

######  Proposition 16.

If $(\mathcal{A},\cdot)$ is a symmetric Zinbiel superalgebra with two odd
generators, then $(\mathcal{A},\cdot)$ is a $2$-step nilpotent superalgebra
given in the first part of the theorem 15 with the following grading
$\mathcal{A}_{1}=\langle e_{1},e_{2}\rangle$ and $\mathcal{A}_{0}=\langle
e_{3},\ldots,e_{2+i}\rangle,$ where $i=dim(\mathcal{A}_{0}).$

Proof. The description of $2$-step nilpotent symmetric Zinbiel superalgebras
with two odd generators is following from the theorem 15.

If $(\mathcal{A},\cdot)$ is a $3$-step nilpotent symmetric Zinbiel
superalgebra, then the product of three elements $x,y,z$ of
$(\mathcal{A},\cdot)$ is equal to zero if an elements from its is not a linear
combination of generators. Hence, superalgebra $(\mathcal{A},\cdot)$ without
grading is equivalent to an algebra with the following identities

$\Omega=\\{(xy)z=x(yz-zy)\mbox{ and }x(yz)=(xy-yx)z\\}.$

By the standard Skjelbred-Sund method, it is possible to verification, that
there are no non-trivial extensions of algebras from the first part of the
theorem 15 in the variety defined by identities from $\Omega.$ It is following
that there are no $3$-step nilpotent symmetric Zinbiel superalgebras with two
odd generators. $\square$

##  3 Mono and binary symmetric Zinbiel and Leibniz algebras

In this section, we will consider algebras over the complex field.

######  Definition 17.

Let $\Omega$ be a variety of algebras defined by a family of polynomial
identities, then we say that an algebra $\mathcal{A}\in\Omega_{i}$ if and only
if each $i$-generated subalgebra of $\mathcal{A}$ gives an algebra from
$\Omega.$ In particular, if $\mathcal{A}\in\Omega_{1},$ then $\mathcal{A}$ is
a mono-$\Omega$ algebra, if $\mathcal{A}\in\Omega_{2},$ then $\mathcal{A}$ is
a binary-$\Omega$ algebra.

For example, let ${\rm Ass}$ be the class of associative algebras. Then by
Artin’s theorem, the class ${\rm Ass}_{2}$ coincides with the class of
alternative algebras. Albert’s theorem follows that the class ${\rm Ass}_{1}$
coincides with the class of power-associative algebras. It is easy to see that
${\rm Lie}_{1}$ coincides with anticommutative algebras, i.e. they satisfy the
identity $x^{2}=0.$ The identities of ${\rm Lie}_{2}$ are described by Gainov
[19]. Below we will consider mono and binary symmetric Leibniz and Zinbiel
algebras.

Thanks to [20, 22], we have the description of mono and binary left Leibniz
algebras.

######  Theorem 18.

An algebra $\mathcal{A}$ is mono left Leibniz if and only if it satisfies the
following identities:

$x^{2}x=0,$ $x^{2}x^{2}=0.$

An algebra $\mathcal{A}$ is binary left Leibniz if and only if it satisfies
the following identities:

$x^{2}y=0,$ $x(yx)=(xy)x+yx^{2},$ $x(y(xy))=(xy)(xy)+y(x(xy)).$

Thanks to [23], we have the description of mono and binary left Zinbiel
algebras.

######  Theorem 19.

An algebra $\mathcal{A}$ is mono left Zinbiel if and only if it satisfies the
following identities:

$xx^{2}=2x^{2}x,$ $x^{2}x^{2}=3(x^{2}x)x.$

An algebra $\mathcal{A}$ is binary left Zinbiel if and only if it satisfies
the following identities:

$x(yx)=(xy+yx)x,$ $x(xy)=2x^{2}y.$

As some trivial corollaries from the previous theorems, we have the following
propositions.

######  Proposition 20.

The variety of mono symmetric Zinbiel algebras coincides with the variety of
mono symmetric Leibniz algebras and it is defined by the following identities

$x^{2}x=0,$ $xx^{2}=0,$ $x^{2}x^{2}=0.$

######  Proposition 21.

The variety of binary symmetric Zinbiel algebras is defined by the following
identities

$x(yx)=(xy+yx)x,$ $x(xy)=2x^{2}y,$

$(xy)x=x(yx+xy),$ $(yx)x=2yx^{2}.$

######  Proposition 22.

The variety of binary symmetric Zinbiel algebras is defined by the following
identities

$x^{2}y=0,$ $x(yx)=(xy)x,$ $x(y(xy))=(xy)(xy)+y(x(xy))$

$yx^{2}=0,$ $(xy)(xy)=((xy)x)y+x((xy)y).$

######  Proposition 23.

The intersection of varieties of binary symmetric Zinbiel and Leibniz algebras
it is defined by the following identities

$(x_{1}x_{2})x_{3}=(-1)^{\sigma}(x_{\sigma(1)}x_{\sigma(2)})x_{\sigma(3)}$ and
$x_{1}(x_{2}x_{3})=(-1)^{\sigma}x_{\sigma(1)}(x_{\sigma(2)}x_{\sigma(3)}),$
$\sigma\in\mathbb{S}_{3}.$

Proof. Thanks to two previous propositions, we have that each algebra from the
intersection of symmetric Zinbiel and Leibniz algebras satisfies the following
identities

$x^{2}y=0,$ $x(xy)=0,$ $yx^{2}=0,$ $(yx)x=0.$

Hence, by linearization, it satisfies

$(x_{1}x_{2})x_{3}=-(x_{2}x_{1})x_{3},$
$x_{1}(x_{2}x_{3})=-x_{2}(x_{1}x_{3}),$
$x_{1}(x_{2}x_{3})=-x_{1}(x_{3}x_{2}),$
$(x_{1}x_{2})x_{3}=-(x_{1}x_{3})x_{2}.$

It is easy to see that each algebra that satisfies the previous identities
also satisfies all identities from two previous propositions. Hence, we
complete the proof of the statement. $\square$

######  Theorem 24.

Let ${\rm SZ}$ and ${\rm SL}$ be the varieties of symmetric Zinbiel and
symmetric Leibniz algebras. Then, the following inclusions are strict.

$\begin{array}[]{ccccccc}&&{\rm SZ}&\subset&{\rm SZ}_{2}\\\
&\rotatebox{45.0}{$\subset$}&&&\rotatebox{90.0}{$\subset$}&\rotatebox{-45.0}{$\subset$}\\\
{\rm SL}\cap{\rm SZ}={\rm SL}\cap{\rm SZ}_{2}={\rm SZ}\cap{\rm
SL}_{2}&\subset&{\rm Ass}\cap{\rm Lie}_{1}&\subset&{\rm SL}_{2}\cap{\rm
SZ}_{2}&\subset&{\rm SZ}_{1}={\rm SL}_{1}\\\
&\rotatebox{-45.0}{$\subset$}&&&\rotatebox{-90.0}{$\subset$}&\rotatebox{45.0}{$\subset$}\\\
&&{\rm SL}&\subset&{\rm SL}_{2}\end{array}$

Proof. Let us prove that ${\rm SL}\cap{\rm SZ}_{2}$ is defined by $(xy)z=0$
and $x(yz)=0.$ Obviously, that each algebra from ${\rm SL}\cap{\rm SZ}_{2}$
satisfies identities from the previous proposition and two Leibniz identities.
Hence,

$2(xy)z=(xy)z+(xz)y+x(yz)=x(yz)$ and $2x(yz)=x(yz)+(xy)z+y(xz)=(xy)z,$

which gives $(xy)z=0$ and $x(yz)=0.$

${\rm SZ}\cap{\rm SL}_{2}$ is also defined by $(xy)z=0$ and $x(yz)=0,$ because
each algebra from ${\rm SZ}\cap{\rm SL}_{2}$ satisfies identities from the
previous proposition. Hence, it satisfies

$(xy)z=x(yz+zy)=0$ and $x(yz)=(xy+yx)z=0.$

Obviously, we have the equalities ${\rm SL}\cap{\rm SZ}={\rm SL}\cap{\rm
SZ}_{2}={\rm SZ}\cap{\rm SL}_{2}.$

All indicated inclusions are following from previous propositions and other
known results. To prove the strict inclusion $\omega\subset\Omega$ of
varieties of algebras, we will indicate an algebra
$\mathcal{A}\in\Omega\setminus\omega.$

${\rm SL}\cap{\rm SZ}$ | $\subset$ | ${\rm SZ}$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{5}=e_{6}$ | $e_{2}e_{1}=e_{4}$ |
---|---|---|---|---|---|---
${\rm SL}_{2}\cap{\rm SZ}_{2}$ | $\subset$ | ${\rm SZ}_{2}$ | $e_{2}e_{2}=e_{5}$ | $e_{2}e_{3}=e_{6}$ | $e_{2}e_{4}=-2e_{6}$ |
${\rm SL}_{2}$ | $\subset$ | ${\rm SL}_{1}$ | $e_{3}e_{2}=2e_{6}$ | $e_{4}e_{2}=-e_{6}$ | $e_{5}e_{1}=-e_{6}$ |
${\rm SL}\cap{\rm SZ}$ | $\subset$ | ${\rm SL}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{2}e_{3}=e_{4}$ | $e_{3}e_{2}=-e_{4}$
${\rm SL}\cap{\rm SZ}$ | $\subset$ | ${\rm Ass}\cap{\rm Lie}_{1}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{1}e_{6}=e_{7}$ | $e_{2}e_{1}=-e_{4}$
|  |  | $e_{2}e_{3}=e_{6}$ | $e_{2}e_{5}=-e_{7}$ | $e_{3}e_{1}=-e_{5}$ | $e_{3}e_{2}=-e_{6}$
|  |  | $e_{3}e_{4}=e_{7}$ | $e_{4}e_{3}=e_{7}$ | $e_{5}e_{2}=-e_{7}$ | $e_{6}e_{1}=e_{7}$
${\rm Ass}\cap{\rm Lie}_{1}$ | $\subset$ | ${\rm SL}_{2}\cap{\rm SZ}_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{1}e_{6}=e_{7}$ |
${\rm SL}$ | $\subset$ | ${\rm SL}_{2}$ | $e_{2}e_{1}=-e_{4}$ | $e_{2}e_{3}=e_{6}$ | $e_{2}e_{5}=-e_{7}$ |
${\rm SZ}$ | $\subset$ | ${\rm SZ}_{2}$ | $e_{3}e_{1}=-e_{5}$ | $e_{3}e_{2}=-e_{6}$ | $e_{3}e_{4}=e_{7}$ |
${\rm SZ}_{2}$ | $\subset$ | ${\rm SZ}_{1}$ | $e_{1}e_{2}=e_{2}$ |  |  |
${\rm SL}_{2}\cap{\rm SZ}_{2}$ | $\subset$ | ${\rm SL}_{2}$ | $e_{2}e_{1}=-e_{2}$ |  |  |

$\square$

From the proof of the present theorem it is easy to see that the variety ${\rm
SL}_{2}\cap{\rm SZ}_{2}$ does not contain ${\rm SL}$ and ${\rm SZ}$ as
subvarieties. It gives the following interesting open question.

######  Question 1.

Is ${\rm SL}_{1}$ the minimal variety that contains the varieties ${\rm SL}$
and ${\rm SZ}$ (resp. ${\rm SL}_{2}$ and ${\rm SZ}_{2}$) as subvarieties?

##  4 Quadratic Zinbiel superalgebras

######  Definition 25.

A triple $(\mathcal{A},\cdot,\mathfrak{B})$ is said to be a quadratic Zinbiel
superalgebra if $(\mathcal{A},\cdot)$ is a left (or a right) Zinbiel
superalgebra endowed with a non-degenerate bilinear form $\mathfrak{B}$
satisfying the following properties:

  1. (i)

$\mathfrak{B}(x,y)=(-1)^{|x||y|}\mathfrak{B}(y,x),~{}\forall
x\in\mathcal{A}_{|x|},~{}y\in\mathcal{A}_{|y|}$, i.e. $\mathfrak{B}$ is
supersymmetric;

  2. (ii)

$\mathfrak{B}(xy,z)=\mathfrak{B}(x,yz),~{}\forall x,y,z\in\mathcal{A}$, i.e.
$\mathfrak{B}$ is invariant;

  3. (iii)

$\mathfrak{B}(\mathcal{A}_{\bar{0}},\mathcal{A}_{\bar{1}})=\mathfrak{B}(\mathcal{A}_{\bar{1}},\mathcal{A}_{\bar{0}})=\\{0\\}$,
i.e. $\mathfrak{B}$ is even.

In this case, $\mathfrak{B}$ is called an invariant scalar product on
$\mathcal{A}$.

######  Proposition 26.

Let $(\mathcal{A},\cdot)$ be a left (resp. right) Zinbiel superalgebra. If
$\mathcal{A}$ is quadratic, then $\mathcal{A}$ is symmetric.

Proof. Suppose that $(\mathcal{A},\cdot)$ is a left Zinbiel superalgebra.
Then, for all $x,y,z,t\in\mathcal{A}_{\bar{0}}\cup\mathcal{A}_{\bar{1}},$ we
have

$0=\mathfrak{B}((xy)z-x(yz)-(-1)^{|y||z|}x(zy),t)=\mathfrak{B}(x,y(zt)-(yz)t-(-1)^{|y||z|}(zy)t).$

The fact that $\mathfrak{B}$ is non-degenerate implies that
$y(zt)-(yz)t-(-1)^{|y||z|}(zy)t=0$. Consequently, $\mathcal{A}$ is a right
Zinbiel superalgebra. By the same way, we prove the result for right Zinbiel
superalgebra. $\square$

######  Proposition 27.

Let $\mathcal{A}=\mathcal{A}_{\bar{0}}\oplus\mathcal{A}_{\bar{1}}$ be a
symmetric Zinbiel superalgebra. Then, $\mathcal{A}$ is quadratic if and only
if the adjoint and the co-adjoint representations of $\mathcal{A}$ are
equivalent and $dim(\mathcal{A}_{\bar{1}})$ is even.

Proof. The proof is similar to that in [3]. $\square$

The following result comes from Proposition 6.

######  Proposition 28.

Let $(\mathcal{A},\cdot)$ be a left (resp. right) Zinbiel superalgebra. If
$\mathcal{A}$ is quadratic, then $\mathcal{A}$ is $2$-step nilpotent.

##  5 Extensions of quadratic Zinbiel superalgebras

In this section, we introduce the notion of a double extension of quadratic
Zinbiel superalgebras by the one-dimensional $\mathbb{Z}_{2}$-graded vector
space, which is performed in two distinct steps, a central extension followed
by a semi-direct product. In the first step, we are going to introduce the
notion of central extension by the one-dimensional $\mathbb{Z}_{2}$-graded
vector space.

######  Proposition 29.

Let $(\mathcal{A},\cdot,\mathfrak{B})$ be a quadratic Zinbiel superalgebra and
$\Omega:\mathcal{A}\times\mathcal{A}\rightarrow\mathbb{K}$ be a bilinear map
of degree $\alpha$. Then, there exists a homogeneous endomorphism $\delta$ of
$\mathcal{A}$ of degree $\alpha$ such that
$\Omega(x,y)=\mathfrak{B}(\delta(x),y),~{}\forall x,y\in\mathcal{A}$. The map
$\Omega$ is a homogeneous scalar Zinbiel 2-cocycle of $\mathcal{A}$ of degree
$\alpha$ if and only if, for all
$x,y\in\mathcal{A}_{|x|}\times\mathcal{A}_{|y|}$, the following assertions
hold:

${\bf(i)}~{}\delta(xy)=\delta(x)y+(-1)^{|y|(|x|+\alpha)}y\delta(x)\quad
and\quad{\bf(ii)}~{}\delta(xy)=-(-1)^{\alpha|x|}x\delta(y).$

Proof. ${\bf(i)}$ It is easy to see the following

$\Omega(x,yz)+(-1)^{|y||z|}\Omega(x,zy)-\Omega(xy,z)=\mathfrak{B}(\delta(x)y+(-1)^{|y|(|x|+\alpha)}y\delta(x)-\delta(xy),z).$

${\bf(ii)}$ Note that $~{}\forall
x,y,z\in\mathcal{A}_{|x|}\times\mathcal{A}_{|y|}\times\mathcal{A}_{|z|}$ we
have

$\Omega(x,yz)+(-1)^{|z|(|x|+|y|)}\Omega(zx,y)=(-1)^{|x|(|y|+|z|)+\alpha|y|}\mathfrak{B}(y,(-1)^{\alpha|z|}z\delta(x)+\delta(zx)).$

$\square$

######  Proposition 30.

Let $(\mathcal{A},\cdot,\mathfrak{B})$ be a quadratic Zinbiel superalgebra,
$\mathcal{V}=\mathbb{K}d$ be the one-dimensional $\mathbb{Z}_{2}$-graded
vector space and $\delta$ be a homogeneous endomorphism of $\mathcal{A}$ of
degree $\alpha$. Define the bilinear map
$\Omega:\mathcal{A}\times\mathcal{A}\rightarrow\mathcal{V^{*}}$ by
$\Omega(x,y)=\mathfrak{B}(\delta(x),y)d^{*}$. Then, the
$\mathbb{Z}_{2}$-graded vector space
$\widetilde{\mathcal{A}}=\mathcal{A}\oplus\mathcal{V^{*}}$ endowed with the
product defined by

$(x+\lambda
d^{*})\star_{\Omega}(y+\lambda^{\prime}d^{*})=xy+\Omega(x,y),~{}~{}\mbox{for~{}all}~{}(x+\lambda
d^{*}),(y+\lambda^{\prime}d^{*})\in\widetilde{\mathcal{A}},$

is a symmetric Zinbiel superalgebra if and only if $\Omega$ is a homogeneous
Zinbiel 2-cocycle of $\mathcal{A}$ on the trivial $\mathcal{A}-$module
$\mathcal{V^{*}}$ of degree $\alpha$. It is termed the central extension of
$\mathcal{A}$ by $\mathcal{V^{*}}$ by means of $\Omega$.

Proof. We only need calculation, where we use Proposition 29. $\square$

We will define, in the following, another type of extension of symmetric
Zinbiel superalgebras. Let us consider $(\mathcal{A},\cdot)$ a symmetric
Zinbiel superalgebra and $\mathcal{V}=\mathbb{K}d$ a one-dimensional
$\mathbb{Z}_{2}$-graded vector space. Let $\delta,D$ be two homogeneous
endomorphisms of $\mathcal{A}$ of degree $|d|$ and $\mathfrak{a}_{0}\in{\rm
Ann}(\mathcal{A})\cap\mathcal{A}_{\overline{0}}$. Consider the
$\mathbb{Z}_{2}$-graded vector space
$\overline{\mathcal{A}}=\mathcal{A}\oplus\mathcal{V}$ on which we define the
following product:

$d\mathrel{\scalebox{0.7}{$\odot$}}d=\mathfrak{a}_{0};\quad
x\mathrel{\scalebox{0.7}{$\odot$}}y=xy;\quad
d\mathrel{\scalebox{0.7}{$\odot$}}x=\delta(x);\quad
x\mathrel{\scalebox{0.7}{$\odot$}}d=D(x),~{}~{}\forall x,y\in\mathcal{A}.$
(5.1)

######  Proposition 31.

The $\mathbb{Z}_{2}$-graded vector space $\overline{\mathcal{A}}$ endowed with
the product (5.1) is a symmetric Zinbiel superalgebra if and only if, for all
$x,y\in\mathcal{A}_{|x|}\times\mathcal{A}_{|y|}$, the following conditions
hold:

$\displaystyle\delta(xy)=(-1)^{|d||x|}x\delta(y)=\delta(x)y+(-1)^{|d||x|}D(x)y=-(-1)^{|y|(|x|+|d|)}D(y)x;$
$\displaystyle\delta(x)y=(-1)^{|x||y|}\delta(y)x=-(-1)^{|d|(|x|+|y|)}xD(y);$
$\displaystyle D(xy)=(-1)^{|d||y|}D(x)y;$
$\displaystyle\delta^{2}(x)=(-1)^{|d|}\delta^{2}(x)=-D^{2}(x)=(1+(-1)^{|d|})\mathfrak{a}_{0}x;$
$\displaystyle\delta\circ
D(x)=-(-1)^{|d|}D\circ\delta(x)=(-1)^{|d||x|}x\mathfrak{a}_{0}=-(-1)^{|d|(|x|+|d|)}\mathfrak{a}_{0}x;~{}~{}\delta(\mathfrak{a}_{0})=D(\mathfrak{a}_{0})=0.$

In this case, $(\delta,D,\mathfrak{a}_{0})$ is called ”an admissible triple”
and the symmetric Zinbiel superalgebra $\overline{\mathcal{A}}$ is termed the
semi-direct product of $\mathcal{A}$ by $\mathcal{V}$ by means of
$(\delta,D,\mathfrak{a}_{0})$.

Proof. The proof is a straightforward computation. $\square$

Now, we are in a position to introduce the double extensions of quadratic
Zinbiel superalgebras. We start by presenting the concept of even double
extension of these superalgebras by the one-dimensional Lie algebra.

######  Theorem 32.

Let $(\mathcal{A},\cdot,\mathfrak{B})$ be a quadratic Zinbiel superalgebra,
$\mathcal{V}=\mathbb{K}d$ be the one-dimensional vector space and
$(\delta,\delta^{*},\mathfrak{a}_{0})$ be an admissible triple of
$\mathcal{A}$ such that $\delta(\mathcal{A})\subseteq{\rm
Ann}(\mathcal{A}),~{}\delta(\mathcal{A}^{2})=\\{0\\},~{}\delta^{2}=\delta\circ\delta^{*}=\delta^{*}\circ\delta=\\{0\\},~{}\mathfrak{a}_{0}\in{\rm
Ann}(\mathcal{A})\cap\mathcal{A}_{\bar{0}}$ and
$\mathfrak{B}(\mathfrak{a}_{0},\mathfrak{a}_{0})=0$, where $\delta^{*}$ is the
adjoint of $\delta$ with respect to $\mathfrak{B}$. Then, the
$\mathbb{Z}_{2}$-graded vector space
$\overline{\mathcal{A}}=\mathcal{V}^{*}\oplus\mathcal{A}\oplus\mathcal{V}$
endowed with the following product:

$d\mathrel{\scalebox{0.7}{$\odot$}}d=\mathfrak{a}_{0}+\alpha d^{*};$ | $d\mathrel{\scalebox{0.7}{$\odot$}}x=\delta(x)+\mathfrak{B}(x,\mathfrak{a}_{0})d^{*};$ |
---|---|---
$x\mathrel{\scalebox{0.7}{$\odot$}}y=xy+\mathfrak{B}(\delta(x),y)d^{*};$ | $x\mathrel{\scalebox{0.7}{$\odot$}}d=\delta^{*}(x)+\mathfrak{B}(x,\mathfrak{a}_{0})d^{*},$ | $\forall x,y\in\mathcal{A},~{}\alpha\in\mathbb{K},$

is a symmetric Zinbiel superalgebra. Moreover, the even bilinear form
$\overline{\mathfrak{B}}:\overline{\mathcal{A}}\times\overline{\mathcal{A}}\rightarrow\mathbb{K}$,
defined by
$\overline{\mathfrak{B}}|_{\mathcal{A}\times\mathcal{A}}=\mathfrak{B};~{}~{}\overline{\mathfrak{B}}(d,d^{*})=\overline{\mathfrak{B}}(d^{*},d)=1,$
is an associative scalar product on $\overline{\mathcal{A}}$.

The Zinbiel superalgebra $(\overline{\mathcal{A}},\overline{\mathfrak{B}})$ is
called the even double extension of $\mathcal{A}$ by $\mathcal{V}$ by means of
the admissible triple $(\delta,\delta^{*},\mathfrak{a}_{0})$.

Proof. Since $\delta(\mathcal{A}^{2})=\\{0\\}$ and
$\delta(\mathcal{A})\subseteq{\rm Ann}(\mathcal{A})$, then Proposition 29
implies that the even map
$\Omega:\mathcal{A}\times\mathcal{A}\rightarrow\mathcal{V}^{*},~{}(x,y)\longmapsto\Omega(x,y)=\mathfrak{B}(\delta(x),y)d^{*}$,
is an even Zinbiel 2-cocycle of $\mathcal{A}$ on the trivial
$\mathcal{A}-$module $\mathcal{V}^{*}$. So,
$\widetilde{\mathcal{A}}=\mathcal{A}\oplus\mathcal{V}^{*}$ endowed with the
product defined by:

$(x+f)\star_{\Omega}(y+g)=xy+\mathfrak{B}(\delta(x),y)d^{*},~{}\forall
x,y\in\mathcal{A},~{}f,g\in\mathcal{V}^{*},$

is a symmetric Zinbiel superalgebra, central extension of $\mathcal{A}$ by
means of $\Omega$. Now, we define two even endomorphisms $\delta_{1}$ and
$\delta_{2}$ of $\widetilde{\mathcal{A}}$, respectively, by

$\delta_{1}(x+\lambda d^{*})=\delta(x)+\mathfrak{B}(x,\mathfrak{a}_{0})d^{*}$
and $\delta_{2}(x+\lambda
d^{*})=\delta^{*}(x)+\mathfrak{B}(x,\mathfrak{a}_{0})d^{*},~{}\forall
x\in\mathcal{A},~{}\lambda\in\mathbb{K}.$

Let us consider the element $\mathfrak{a}_{1}=\mathfrak{a}_{0}+\alpha
d^{*}\in\widetilde{\mathcal{A}}$, where $\alpha$ is a fixed scalar in
$\mathbb{K}$. So, $\mathfrak{a}_{1}\in{\rm
Ann}(\widetilde{\mathcal{A}})\cap\widetilde{\mathcal{A}}_{\bar{0}}$. Since

$\delta(\mathcal{A})\subseteq{\rm
Ann}(\mathcal{A}),~{}\delta(\mathcal{A}^{2})=\\{0\\},~{}\delta^{2}=\delta\circ\delta^{*}=\delta^{*}\circ\delta=\\{0\\},~{}\mathfrak{a}_{0}\in{\rm
Ann}(\mathcal{A})\cap\mathcal{A}_{\bar{0}},~{}\mathfrak{B}(\mathfrak{a}_{0},\mathfrak{a}_{0})=0$,

then a simple computation proves that the triple
$(\delta_{1},\delta_{2},\mathfrak{a}_{1})$ is an admissible triple of
$\widetilde{\mathcal{A}}$. Consequently, we can consider the symmetric Zinbiel
superalgebra
$\overline{\mathcal{A}}=\widetilde{\mathcal{A}}\oplus\mathcal{V}=\mathcal{V}^{*}\oplus\mathcal{A}\oplus\mathcal{V}$,
semi-direct product of $\widetilde{\mathcal{A}}$ by $\mathcal{V}$. In
addition, it is easy to show that the bilinear form
$\overline{\mathfrak{B}}:\overline{\mathcal{A}}\times\overline{\mathcal{A}}\rightarrow\mathbb{K}$,
defined by
$\overline{\mathfrak{B}}|_{\mathcal{A}\times\mathcal{A}}=\mathfrak{B};~{}\overline{\mathfrak{B}}(d,d^{*})=1$,
is an associative scalar product on $\overline{\mathcal{A}}$. $\square$

Our next goal is to establish the converse of Theorem 32.

######  Theorem 33.

Let $(\mathcal{A},\mathrel{\scalebox{0.7}{$\odot$}},\mathfrak{B})$ be a
quadratic Zinbiel superalgebra of dimension $(n,m)$ such that $n\geq 2$. If
${\rm Ann}(\mathcal{A})\cap\mathcal{A}_{\bar{0}}\neq\\{0\\}$, then
$\mathcal{A}$ is isomorphic to an even double extension of a quadratic Zinbiel
superalgebra $(\mathcal{H},\cdot,\mathfrak{B}_{\mathcal{H}})$ of dimension
$(n-2,m)$ by the one-dimensional vector space.

Proof. Suppose that ${\rm
Ann}(\mathcal{A})\cap\mathcal{A}_{\bar{0}}\neq\\{0\\}$, then there exists a
non-zero element $e$ of ${\rm Ann}(\mathcal{A})\cap\mathcal{A}_{\bar{0}}$.
Denote by $\mathcal{J}=\mathbb{K}e$ and by $\mathcal{J}^{\bot}$ the orthogonal
of $\mathcal{J}$ with respect to $\mathfrak{B}$. Then,
$\mathcal{J}\subseteq\mathcal{J}^{\bot}$. Since $\mathfrak{B}$ is even and
non-degenerate, then there exists $d\in\mathcal{A}_{\bar{0}}\backslash\\{0\\}$
such that $\mathfrak{B}(e,d)=\mathfrak{B}(d,e)=1$. Let
$\mathcal{V}=\mathbb{K}d$ and
$\mathcal{H}=(\mathcal{J}\oplus\mathcal{V})^{\bot}$, then
$\mathcal{A}=\mathcal{J}\oplus\mathcal{H}\oplus\mathcal{V}$ and
$\mathcal{J}^{\bot}=\mathcal{J}\oplus\mathcal{H}$ is an ideal of
$\mathcal{A}$. Let $x,y\in\mathcal{H}_{|x|}\times\mathcal{H}_{|y|}$, then
$x\mathrel{\scalebox{0.7}{$\odot$}}y=\Phi(x,y)e+xy$, where
$\Phi:\mathcal{H}\times\mathcal{H}\rightarrow\mathbb{K}$ is an even bilinear
form and $``\cdot":\mathcal{H}\times\mathcal{H}\rightarrow\mathcal{H}$ is an
even bilinear map. It is easy to show that $(\mathcal{H},\cdot)$ is a
symmetric Zinbiel superalgebra and that the even bilinear form
$\mathfrak{B}_{\mathcal{H}}=\mathfrak{B}|_{\mathcal{H}\times\mathcal{H}}$ is
an associative scalar product on $(\mathcal{H},\cdot)$. Therefore, the even
bilinear form $\Phi$ is an element of $(Z^{2}_{\rm
SZinb}(\mathcal{H},\mathbb{K}))_{\bar{0}}$. The fact that
$\mathcal{J}^{\bot}=\mathcal{J}\oplus\mathcal{H}$ is an ideal of $\mathcal{A}$
implies that

$\displaystyle d\mathrel{\scalebox{0.7}{$\odot$}}d=\alpha
e+\mathfrak{a}_{0}+\lambda
d,~{}\hbox{where}~{}\alpha,\lambda\in\mathbb{K},~{}\mathfrak{a}_{0}\in\mathcal{H}_{\bar{0}},$
$\displaystyle
x\mathrel{\scalebox{0.7}{$\odot$}}d=\Psi(x)e+D(x),~{}\hbox{where}~{}D\in(End(\mathcal{H}))_{\bar{0}},~{}\Psi\in(\mathcal{H}^{*})_{\bar{0}},$
$\displaystyle
d\mathrel{\scalebox{0.7}{$\odot$}}x=\Theta(x)e+\delta(x),~{}\hbox{where}~{}\delta\in(End(\mathcal{H}))_{\bar{0}},~{}\Theta\in(\mathcal{H}^{*})_{\bar{0}}.$

Let $x,y\in\mathcal{H}_{|x|}\times\mathcal{H}_{|y|}$. Since $\mathfrak{B}$ is
supersymmetric and associative, then
$\mathfrak{B}(x\mathrel{\scalebox{0.7}{$\odot$}}y,d)=\mathfrak{B}(x,y\mathrel{\scalebox{0.7}{$\odot$}}d)=\mathfrak{B}(d\mathrel{\scalebox{0.7}{$\odot$}}x,y)$.
It follows that $\Phi(x,y)=\mathfrak{B}(x,D(y))=\mathfrak{B}(\delta(x),y)$.
So, $D=\delta^{*}$. Moreover,
$\Psi(x)=\mathfrak{B}(x\mathrel{\scalebox{0.7}{$\odot$}}d,d)=\mathfrak{B}(d,d\mathrel{\scalebox{0.7}{$\odot$}}x)=\Theta(x)=\mathfrak{B}(x,\mathfrak{a}_{0})$
and
$\lambda=\mathfrak{B}(d\mathrel{\scalebox{0.7}{$\odot$}}d,e)=\mathfrak{B}(d,d\mathrel{\scalebox{0.7}{$\odot$}}e)=0$.
Then, $d\mathrel{\scalebox{0.7}{$\odot$}}d=\alpha e+\mathfrak{a}_{0}$. Since
$\mathcal{A}$ is a left Zinbiel superalgebra satisfying:

$\mathscr{X}\mathrel{\scalebox{0.7}{$\odot$}}(\mathscr{Y}\mathrel{\scalebox{0.7}{$\odot$}}\mathscr{Z})+(-1)^{|z|(|x|+|y|)}(\mathscr{Z}\mathrel{\scalebox{0.7}{$\odot$}}\mathscr{X})\mathrel{\scalebox{0.7}{$\odot$}}\mathscr{Y}=0,~{}~{}\mbox{for~{}all}~{}\mathscr{X}\in\mathcal{A}_{{|x|}},\mathscr{Y}\in\mathcal{A}_{{|y|}},\mathscr{Z}\in\mathcal{A}_{{|z|}},$

then a simple computation proves that $(\delta,D,\mathfrak{a}_{0})$ is an
admissible triple of $\mathcal{H}$,

$\delta(\mathcal{H})\subseteq{\rm
Ann}(\mathcal{H}),~{}\delta(\mathcal{H}^{2})=\\{0\\},~{}\delta^{2}=\delta\circ
D=D\circ\delta=\\{0\\},~{}\mathfrak{a}_{0}\in{\rm
Ann}(\mathcal{H})\cap\mathcal{H}_{\bar{0}}$ and
$\mathfrak{B}(\mathfrak{a}_{0},\mathfrak{a}_{0})=0$.

Therefore, we can consider the quadratic Zinbiel superalgebra
$\overline{\mathcal{A}}=\mathcal{V}^{*}\oplus\mathcal{H}\oplus\mathcal{V}$,
even double extension of $(\mathcal{H},\cdot,\mathfrak{B}_{\mathcal{H}})$ by
$\mathcal{V}$ by means of $(\delta,D,\mathfrak{a}_{0})$.

It is clear that the map

$\Gamma:\mathcal{A}\rightarrow\overline{\mathcal{A}},~{}\lambda
e+x+\lambda^{\prime}d\longmapsto\lambda d^{*}+x+\lambda^{\prime}d$,

is an isomorphism of Zinbiel superalgebras. Then, $\mathcal{A}$ is isomorphic
to the even double extension of $\mathcal{H}$ by $\mathcal{V}$. $\square$

In the second part of this section, we will present the notion of an odd
double extension of quadratic Zinbiel superalgebras by the one-dimensional
$\mathbb{Z}_{2}$-graded vector space with even part zero.

######  Theorem 34.

Let $(\mathcal{A},\cdot,\mathfrak{B})$ be a quadratic Zinbiel superalgebra,
$\mathcal{N}=\mathcal{N}_{\bar{1}}=\mathbb{K}d$ be the one-dimensional
$\mathbb{Z}_{2}$-graded vector space with even part zero and
$(\delta,D,\mathfrak{a}_{0})$ be an admissible triple of $\mathcal{A}$ such
that

$\delta(\mathcal{A})\subseteq{\rm
Ann}(\mathcal{A}),~{}\delta(\mathcal{A}^{2})=\\{0\\},~{}\delta\circ
D=D\circ\delta=\\{0\\},~{}\mathfrak{a}_{0}\in{\rm
Ann}(\mathcal{A})\cap\mathcal{A}_{\bar{0}}$ and
$\mathfrak{B}(\mathfrak{a}_{0},\mathfrak{a}_{0})=0$,

where $D\in(End(\mathcal{A}))_{\bar{1}}$ verifying
$\mathfrak{B}(\delta(x),y)=(-1)^{|x|+|y|}\mathfrak{B}(x,D(y))$ for all
$x,y\in\mathcal{A}_{|x|}\times\mathcal{A}_{|y|}$. Then, the
$\mathbb{Z}_{2}$-graded vector space
$\overline{\mathcal{A}}=\mathcal{N}^{*}\oplus\mathcal{A}\oplus\mathcal{N}$
endowed with the following product:

$d\mathrel{\scalebox{0.7}{$\odot$}}d=\mathfrak{a}_{0};$ | $d\mathrel{\scalebox{0.7}{$\odot$}}x=\delta(x)-\mathfrak{B}(x,\mathfrak{a}_{0})d^{*};$ |
---|---|---
$x\mathrel{\scalebox{0.7}{$\odot$}}y=xy-\mathfrak{B}(\delta(x),y)d^{*};$ | $x\mathrel{\scalebox{0.7}{$\odot$}}d=D(x)+\mathfrak{B}(x,\mathfrak{a}_{0})d^{*},$ | $\forall x,y\in\mathcal{A},$

is a symmetric Zinbiel superalgebra. Moreover, the even bilinear form
$\overline{\mathfrak{B}}:\overline{\mathcal{A}}\times\overline{\mathcal{A}}\rightarrow\mathbb{K}$,
defined by
$\overline{\mathfrak{B}}|_{\mathcal{A}\times\mathcal{A}}=\mathfrak{B};~{}~{}\overline{\mathfrak{B}}(d^{*},d)=-\overline{\mathfrak{B}}(d,d^{*})=1$,
is an associative scalar product on $\overline{\mathcal{A}}$.

The Zinbiel superalgebra $(\overline{\mathcal{A}},\overline{\mathfrak{B}})$ is
termed the odd double extension of $\mathcal{A}$ by $\mathcal{N}$ by means of
the admissible triple $(\delta,D,\mathfrak{a}_{0})$.

Proof. The proof is similar to that of Theorem 32. $\square$

The following theorem is the converse of Theorem 34.

######  Theorem 35.

Let $(\mathcal{A},\mathrel{\scalebox{0.7}{$\odot$}},\mathfrak{B})$ be a
quadratic Zinbiel superalgebra of dimension $(n,m)$ such that $m\geq 2$. If
${\rm Ann}(\mathcal{A})\cap\mathcal{A}_{\bar{1}}\neq\\{0\\}$, then
$\mathcal{A}$ is isomorphic to an odd double extension of a quadratic Zinbiel
superalgebra $(\mathcal{H},\cdot,\mathfrak{B}_{\mathcal{H}})$ of dimension
$(n,m-2)$ by the one-dimensional Lie superalgebra with even part zero.

Proof. Suppose that ${\rm
Ann}(\mathcal{A})\cap\mathcal{A}_{\bar{1}}\neq\\{0\\}$, then there exists a
non-zero element $e$ of ${\rm Ann}(\mathcal{A})\cap\mathcal{A}_{\bar{1}}$.
Denote by $\mathcal{J}=\mathbb{K}e$ and by $\mathcal{J}^{\bot}$ the orthogonal
of $\mathcal{J}$ with respect to $\mathfrak{B}$. Then,
$\mathcal{J}\subseteq\mathcal{J}^{\bot}$. Since $\mathfrak{B}$ is even and
non-degenerate, then there exists $d\in\mathcal{A}_{\bar{1}}\backslash\\{0\\}$
such that $\mathfrak{B}(e,d)=-\mathfrak{B}(d,e)=1$. Let
$\mathcal{N}=\mathbb{K}d$ and
$\mathcal{H}=(\mathcal{J}\oplus\mathcal{N})^{\bot}$, then
$\mathcal{A}=\mathcal{J}\oplus\mathcal{H}\oplus\mathcal{N}$ and
$\mathcal{J}^{\bot}=\mathcal{J}\oplus\mathcal{H}$ is an ideal of
$\mathcal{A}$. Let $x,y\in\mathcal{H}_{|x|}\times\mathcal{H}_{|y|}$, then
$x\mathrel{\scalebox{0.7}{$\odot$}}y=\Phi(x,y)e+xy$, where
$\Phi:\mathcal{H}\times\mathcal{H}\rightarrow\mathbb{K}$ is an odd bilinear
form and $``\cdot":\mathcal{H}\times\mathcal{H}\rightarrow\mathcal{H}$ is an
even bilinear map. It is easy to show that $(\mathcal{H},\cdot)$ is a
symmetric Zinbiel superalgebra and that the even bilinear form
$\mathfrak{B}_{\mathcal{H}}=\mathfrak{B}|_{\mathcal{H}\times\mathcal{H}}$ is
an associative scalar product on $(\mathcal{H},\cdot)$. Therefore, the odd
bilinear form $\Phi$ is an element of $(Z^{2}_{\rm
SZinb}(\mathcal{H},\mathbb{K}))_{\bar{1}}$. Since
$\mathcal{J}^{\bot}=\mathcal{J}\oplus\mathcal{H}$ is an ideal of
$\mathcal{A}$, then

$\displaystyle
d\mathrel{\scalebox{0.7}{$\odot$}}d=\mathfrak{a}_{0},~{}\hbox{where}~{}\mathfrak{a}_{0}\in\mathcal{H}_{\bar{0}},$
$\displaystyle
x\mathrel{\scalebox{0.7}{$\odot$}}d=\Psi(x)e+D(x),~{}\hbox{where}~{}D\in(End(\mathcal{H}))_{\bar{1}},~{}\Psi\in(\mathcal{H}^{*})_{\bar{0}},$
$\displaystyle
d\mathrel{\scalebox{0.7}{$\odot$}}x=\Theta(x)e+\delta(x),~{}\hbox{where}~{}\delta\in(End(\mathcal{H}))_{\bar{1}},~{}\Theta\in(\mathcal{H}^{*})_{\bar{0}}.$

Let $x,y\in\mathcal{H}_{|x|}\times\mathcal{H}_{|y|}$. Since $\mathfrak{B}$ is
supersymmetric and associative, then
$\mathfrak{B}(d,x\mathrel{\scalebox{0.7}{$\odot$}}y)=\mathfrak{B}(d\mathrel{\scalebox{0.7}{$\odot$}}x,y)=(-1)^{|x|+|y|}\mathfrak{B}(x,y\mathrel{\scalebox{0.7}{$\odot$}}d)$,
which implies that
$\Phi(x,y)=-\mathfrak{B}(\delta(x),y)=-(-1)^{|x|+|y|}\mathfrak{B}(x,D(y))$.
Furthermore,
$\Psi(x)=\mathfrak{B}(x\mathrel{\scalebox{0.7}{$\odot$}}d,d)=\mathfrak{B}(d,d\mathrel{\scalebox{0.7}{$\odot$}}x)=-\Theta(x)=\mathfrak{B}(x,\mathfrak{a}_{0})$.
Since $\mathcal{A}$ is a left Zinbiel superalgebra satisfying:

$\mathscr{X}\mathrel{\scalebox{0.7}{$\odot$}}(\mathscr{Y}\mathrel{\scalebox{0.7}{$\odot$}}\mathscr{Z})+(-1)^{|z|(|x|+|y|)}(\mathscr{Z}\mathrel{\scalebox{0.7}{$\odot$}}\mathscr{X})\mathrel{\scalebox{0.7}{$\odot$}}\mathscr{Y}=0,~{}~{}\mbox{for~{}all}~{}\mathscr{X}\in\mathcal{A}_{{|x|}},\mathscr{Y}\in\mathcal{A}_{{|y|}},\mathscr{Z}\in\mathcal{A}_{{|z|}},$

then an easy calculation proves that $(\delta,D,\mathfrak{a}_{0})$ is an
admissible triple of $\mathcal{H},$

$\delta(\mathcal{H})\subseteq{\rm
Ann}(\mathcal{H}),~{}\delta(\mathcal{H}^{2})=\\{0\\},~{}\delta\circ
D=D\circ\delta=\\{0\\},~{}\mathfrak{a}_{0}\in{\rm
Ann}(\mathcal{H})\cap\mathcal{H}_{\bar{0}}$ and
$\mathfrak{B}(\mathfrak{a}_{0},\mathfrak{a}_{0})=0$.

Consequently, we can consider the quadratic Zinbiel superalgebra
$\overline{\mathcal{A}}=\mathcal{N}^{*}\oplus\mathcal{H}\oplus\mathcal{N}$,
odd double extension of $(\mathcal{H},\cdot,\mathfrak{B}_{\mathcal{H}})$ by
$\mathcal{N}$ by means of $(\delta,D,\mathfrak{a}_{0})$.

Clearly, the map

$\Gamma:\mathcal{A}\rightarrow\overline{\mathcal{A}},~{}\lambda
e+x+\lambda^{\prime}d\longmapsto\lambda d^{*}+x+\lambda^{\prime}d$,

is an isomorphism of Zinbiel superalgebras. Then, $\mathcal{A}$ is isomorphic
to the odd double extension of $\mathcal{H}$ by $\mathcal{N}$. $\square$

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