# Transposed Poisson superalgebra

Viktor Abramov
Institute of Mathematics and Statistics,
University of Tartu, Estonia Olga Liivapuu
Institute of Forestry and Engineering,
Estonian University of Life Sciences, Estonia

###### Abstract

In this paper we propose the notion of a transposed Poisson superalgebra. We
prove that a transposed Poisson superalgebra can be constructed by means of a
commutative associative superalgebra and an even degree derivation of this
algebra. Making use of this we construct two examples of transposed Poisson
superalgebra. One of them is the graded differential algebra of differential
forms on a smooth finite dimensional manifold, where we use the Lie derivative
as an even degree derivation. The second example is the commutative
superalgebra of basic fields of the quantum Yang-Mills theory, where we use
the BRST-supersymmetry as an even degree derivation to define a graded Lie
bracket. We show that a transposed Poisson superalgebra has six identities
that play an important role in the study of the structure of this algebra.

##  1 INTRODUCTION

Poisson algebras play an important role in many branches of differential
geometry, theoretical mechanics, and field theory. Thus this area of research
is actively developing, as evidenced by various generalizations of the concept
of Poisson algebra. One direction of development in the theory of Poisson
algebras is the extension of the concept of Poisson algebra to structures with
an $n$-ary multiplication law. Initially such a generalization was proposed by
Nambu in his paper [7] and the model of quarks in the theory of elementary
particles served as a stimulus for such a generalization. Later, this
generalization called Nambu-Poisson algebra was studied in a number of papers,
and an excellent survey of this direction in the development of the theory of
Poisson algebras is the paper by Takhtajan [9]. A Nambu-Poisson algebra is a
$n$-Lie algebra, where $n$-Lie algebra is based on an extension of binary Lie
bracket to $n$-ary Lie bracket. The concept of a $n$-Lie algebra was proposed
by Filippov [5]. The elements of a $n$-Lie algebra satisfy the Filippov-Jacobi
identity, which is a generalization of Jacobi identity to $n$-ary bracket. The
classification of simple linearly compact $n$-Lie superalgebras is given in
[4]. A class of ternary Lie superalgebras constructed by means of the
supertrace and applications of these ternary Lie superalgebras in BRST-
supersymmetries are proposed in [1, 2].

Recently, the notion of a transposed Poisson algebra has been proposed. A
transposed Poisson algebra can be considered as a structure dual to the
concept of a Poisson algebra. It was shown in [3] that a transposed Poisson
algebra in its properties is very similar to a Poisson algebra. For example,
the class of transposed Poisson algebras is closed under the tensor product of
such algebras. In [3] it is also indicated that there is an important
connection between transposed Poisson algebras and Novikov-Poisson algebras.
More exactly, if we equip a Novikov-Poisson algebra with the Lie commutator
constructed with the help of multiplication in Novikov-Poisson algebra, then
the Novikov-Poisson algebra becomes a transposed Poisson algebra. In [3] it is
shown that a transposed Poisson algebra possesses a number of identities and
this is an important property of a transposed Poisson algebra. In [6] all
structures of complex transposed Poisson algebras on Galilean type Lie
algebras and superalgebras are described.

In this paper we propose the notion of a transposed Poisson superalgebra. In
analogy with transposed Poisson algebra a transposed Poisson superalgebra is a
vector space endowed with two structures, where one of them is a commutative
associative superalgebra and the second is a Lie superalgebra. These two
structures satisfy the graded compatibility condition. We prove that if
$\mathfrak{G}$ is a commutative associative superalgebra and $\delta$ is an
even derivation of $\mathfrak{G}$ then the graded Lie bracket

$[u,v]_{\delta}=u\cdot\delta(v)-(-1)^{|u|\,|v|}\,v\cdot\delta(u),$ (1)

where $u,v\in{\mathfrak{G}}$, $|u|,|v|$ are degrees of elements $u,v$
respectively, defines the structure of a transposed Poisson superalgebra on
$\mathfrak{G}$. We propose the classification of transposed Poisson
superalgebras in low-dimensional case, that is, in the case of
(1,1)-superalgebras. We give two examples of transposed Poisson superalgebra
based on the graded Lie bracket (1). The first example is a transposed Poisson
superalgebra constructed with the help of graded differential algebra.
Particularly, we show that the algebra of differential forms on a smooth
finite-dimensional manifold $M^{n}$ can be equipped with a structure of a
transposed Poisson superalgebra if we use (1), where $u,v$ are differential
forms and $\delta$ is the Lie derivative $\cal L_{\mathtt{X}}$, $\mathtt{X}$
is a vector field on $M^{n}$. We prove that the graded Lie bracket of two
differential forms induces a graded Lie bracket on the cohomology classes of
differential forms. The second example of a transposed Poisson superalgebra is
related to quantum Yang-Mills theory. The BRST-supersymmetry of the quantum
Yang-Mills theory can be considered as an even derivation of the superalgebra
of basic fields of theory. We construct the graded Lie bracket on this
superalgebra with the help of operator of BRST-supersymmetry. Finally we prove
that a transposed Poisson superalgebra possesses a rich class of identities
which can be considered as graded versions of identities proposed in [3].

##  2 Transposed Poisson Superalgebra

In this section we propose a notion of a graded transposed Poisson algebra. We
construct an example of a graded transposed Poisson algebra by means of a
graded differential algebra.

Let $\mathbb{K}$ be a field either of real or complex numbers. A transposed
Poisson algebra $\mathfrak{P}$ [3] is a vector space over $\mathbb{K}$ with
two algebraic operations. One of them will be denoted by
$(x,y)\in{\mathfrak{P}}\times{\mathfrak{P}}\to x\cdot y\in{\mathfrak{P}}$ and
it defines a structure of associative commutative algebra on $\mathfrak{P}$.
The second is a Lie bracket
$(x,y)\in{\mathfrak{P}}\times{\mathfrak{P}}\to[x,y]\in{\mathfrak{P}}$, that
is, bilinear, skew-symmetric mapping which satisfies the Jacoby identity.
These two structures, i.e. an associative commutative algebra and a Lie
algebra, define a structure of transposed Poisson algebra if they are
compatible and the condition of compatibility of these two structures has the
following form

$2z\cdot[x,y]=[z\cdot x,y]+[x,z\cdot y],\quad x,y,z\in{\mathfrak{P}}.$ (2)

An important example of a transposed Poisson algebra [3] is an associative
commutative algebra $\cal A$ endowed with the Lie bracket
$[u,v]_{\delta}=u\cdot\delta(v)-v\cdot\delta(u)$, where $u,v\in{\cal A}$,
$(u,v)\to u\cdot v$ is a multiplication in $\cal A$ and $\delta:{\cal
A}\to{\cal A}$ is a derivation. Particularly if $M^{n}$ is a smooth
$n$-dimensional manifold, ${\cal F}(M^{n})$ is the algebra of smooth functions
on $M^{n}$ and $\tt X$ is a vector field then

$[f,g]_{\tt X}=f\,{\tt X}(g)-g\,{\tt X}(f),\;\;\;f,g\in{\cal F}(M^{n}),$ (3)

defines transposed Poisson algebra of smooth functions on a manifold $M^{n}$.

We extend the notion of transposed Poisson algebra to a super case by the
following definition.

######  Definition 2.1.

Let ${\mathfrak{G}}={\mathfrak{G}}_{0}\oplus{\mathfrak{G}}_{1}$ be a super
vector space over $\mathbb{K}$, where the degree of a homogeneous element
$x\in{\mathfrak{G}}$ will be denoted by $|x|$, that is,
$|x|\in{\mathbb{Z}}_{2}$. A transposed Poisson superalgebra is a triple
$(\mathfrak{G},\,\cdot\,,[\,,\,])$, where $(\mathfrak{G},\,\cdot\,)$ is an
associative commutative superalgebra and $(\mathfrak{G},[\,.\,])$ is a Lie
superalgebra. Hence we have the following properties

$|x\cdot y|=|x|+|y|,\;\;x\cdot y=(-1)^{|x||y|}\,y\cdot x,$

and

$\displaystyle|[x,y]|=|x|+|y|,$ $\displaystyle[x,y]=-(-1)^{|x|\,|y|}[y,x],$
$\displaystyle[x,[y,z]]=[[x,y],z]+(-1)^{|x|\,|y|}[y,[x,z]].$

The compatibility condition for these two structures has the form

$2z\cdot[x,y]=[z\cdot x,y]+(-1)^{|x|\,|z|}[x,z\cdot y].$ (4)

Obviously $({\mathfrak{G}}_{0},\cdot)$ is an associative commutative algebra
and $({\mathfrak{G}}_{0},[\,,\,])$ is a Lie algebra. In the case of even
degree elements, i.e. elements of $\mathfrak{G}_{0}$, the compatibility
condition (4) takes on the form (2). Hence
$({\mathfrak{G}}_{0},\cdot,[\,,\,])$ is a transposed Poisson algebra.

An important example of a transposed Poisson algebra is an algebra constructed
by means of an associative commutative algebra and its derivation. The next
theorem shows that this construction can be extended to the case of
superalgebras.

######  Theorem 1.

Let ${\cal A}={\cal A}_{0}\oplus{\cal A}_{1}$ be an associative commutative
superalgebra and $\delta:{\cal A}\to{\cal A}$ be an even degree derivation of
$\cal A$. Define bracket by

$[u,v]_{\delta}=u\cdot\delta(v)-(-1)^{|u|\,|v|}v\cdot\delta(u).$ (5)

Then (5) defines the structure of Lie superalgebra on $\cal A$ and this
structure satisfies the compatibility condition (4), i.e. the bracket (5)
defines the structure of transposed Poisson superalgebra on $\cal A$.

###### Proof.

It is easy to see that the bracket (5) is graded skew-symmetric, i.e.
$[u,v]_{\delta}=-(-1)^{|u|\,|v|}[v,u]_{\delta}$ and $|[u,v]|=|u|+|v|$. Then
the double brackets can be expressed as follows

$\displaystyle\big{[}w,[u,v]_{\delta}\big{]}_{\delta}$ $\displaystyle=$
$\displaystyle
w\,u\,\delta^{2}(v)-\delta(w)\,u\,\delta(v)-w\,\delta^{2}(u)\,v+\delta(w)\,\delta(u)\,v,$
$\displaystyle\big{[}[w,u]_{\delta},v\big{]}_{\delta}$ $\displaystyle=$
$\displaystyle-w\,\delta^{2}(u)\,v+w\,\delta(u)\,\delta(v)+\delta^{2}(w)\,u\,v-\delta(w)\,u\,\delta(v),$
$\displaystyle(-1)^{|u|\,|w|}\big{[}u,[w,v]_{\delta}\big{]}_{\delta}$
$\displaystyle=$ $\displaystyle
w\,u\,\delta^{2}(v)-w\,\delta(u)\,\delta(v)-\delta^{2}(w)\,u\,v+\delta(w)\delta(u)\,v.$

From this it follows that the graded bracket (5) satisfies the graded Jacobi
identity

$[w,[u,v]_{\delta}]_{\delta}=[[w,u]_{\delta},v]_{\delta}+(-1)^{|u|\,|w|}[u,[w,v]_{\delta}]_{\delta}.$

Hence the graded bracket (5) defines the Lie superalgebra on commutative
superalgebra $\cal A$. We check the compatibility condition (4) by
straightforward calculations. Indeed we find

$\displaystyle 2\,w\cdot[u,v]_{\delta}$ $\displaystyle=$ $\displaystyle
2\,w\cdot u\cdot\delta(v)-2\,w\cdot\delta(u)\cdot v,$ $\displaystyle{[}w\cdot
u,v]_{\delta}$ $\displaystyle=$ $\displaystyle w\cdot
u\cdot\delta(v)-\delta(w)\cdot u\cdot v-w\cdot\delta(u)\cdot v,$
$\displaystyle(-1)^{|u|\,|w|}[u,w\cdot v]_{\delta}$ $\displaystyle=$
$\displaystyle-w\cdot\delta(u)\cdot v+\delta(w)\cdot u\cdot v+w\cdot
u\cdot\delta(v).$

Thus a commutative superalgebra $\cal A$ endowed with the graded Lie bracket
(5) is a transposed Poisson superalgebra. ∎

##  3 Classification of Transposed Poisson Superalgebras in Dimension (1,1)

In this section we consider a low-dimensional case of a transposed Poisson
superalgebra over the field of complex numbers and give a classification of
transposed Poisson superalgebras in this case. We consider transposed Poisson
superalgebras $\mathfrak{G}=\mathfrak{G}_{0}\oplus\mathfrak{G}_{1}$, where
$\mbox{dim}\,\mathfrak{G}_{0}=\mbox{dim}\,\mathfrak{G}_{1}=1$. Let
${e}_{0},{e}_{1}$ be a basis for a super vector space $\mathfrak{G}$ such that
$e_{0}$ is an even degree element which spans the even subspace
$\mathfrak{G}_{0}$ and $e_{1}$ is an odd degree element which spans the odd
subspace $\mathfrak{G}_{1}.$ We assume that there is a commutative associative
multiplication on a super vector space $\mathfrak{G}$, which will be denoted
by $(x,y)\to x\cdot y,x,y\in\mathfrak{G}$. From the structure of superalgebra
and commutativity of a multiplication it follows

$e_{0}\cdot e_{0}=a\,e_{0},\;\;e_{0}\cdot e_{1}=e_{1}\cdot
e_{0}=b\,e_{1},\;\;e_{1}\cdot e_{1}=0,$ (6)

where $a,b$ are complex numbers. In this simple case there is only one
combination of three generators which leads to a non-trivial condition derived
from associativity. This combination is $e_{0},e_{0},e_{1}$. We have

$\displaystyle e_{0}\cdot(e_{0}\cdot e_{1})=(e_{0}\cdot e_{0})\cdot
e_{1}\;\Rightarrow\;\;b\;e_{0}\cdot e_{1}=a\,e_{0}\cdot
e_{1}\;\Rightarrow\;b^{2}\,e_{1}=ab\,e_{1}.$ (7)

Hence a multiplication is associative in three cases $a=b=0$, $a=b\neq 0$ and
$a\neq 0,b=0$.

Now we consider the second structure of a transposed Poisson superalgebra,
that is, a Lie superalgebra. It is known that in the case of Lie superalgebras
of (1,1)-type there are three non-isomorphic Lie superalgebras

  * I)

$[e_{0},e_{0}]=0,\;\;[e_{0},e_{1}]=0,\;\;[e_{1},e_{1}]=0,$

  * II)

$[e_{0},e_{0}]=0,\;\;[e_{0},e_{1}]=0,\;\;[e_{1},e_{1}]=e_{0},$

  * III)

$[e_{0},e_{0}]=0,\;\;[e_{0},e_{1}]=e_{1},\;\;[e_{1},e_{1}]=0.$

In the case of the Abelian Lie superalgebra I we have two transposed Poisson
superalgebras

$\displaystyle e_{0}\cdot e_{0}=e_{0},\;e_{0}\cdot e_{1}=e_{1}\cdot
e_{0}=0,\;e_{1}\cdot e_{1}=0,$
$\displaystyle[e_{0},e_{0}]=0,\;[e_{0},e_{1}]=0,\;[e_{1},e_{1}]=0,$

and

$\displaystyle e_{0}\cdot e_{0}=e_{0},\;e_{0}\cdot e_{1}=e_{1}\cdot
e_{0}=e_{1},\;e_{1}\cdot e_{1}=0,$
$\displaystyle[e_{0},e_{0}]=0,\;[e_{0},e_{1}]=0,\;[e_{1},e_{1}]=0.$

In the case of the Lie superalgebra II we have $a=b=0$. Hence we have only one
transposed Poisson superalgebra

$\displaystyle e_{0}\cdot e_{0}=e_{0},\;e_{0}\cdot e_{1}=0\cdot
e_{0}=0,\;e_{1}\cdot e_{1}=0,$
$\displaystyle[e_{0},e_{0}]=0,\;[e_{0},e_{1}]=0,\;[e_{1},e_{1}]=e_{0}.$

In the case of the Lie superalgebra III it follows from the compatibility
condition $a=b$ and we get one more transposed Poisson superalgebra

$\displaystyle e_{0}\cdot e_{0}=e_{0},\;e_{0}\cdot e_{1}=e_{1}\cdot
e_{0}=e_{1},\;e_{1}\cdot e_{1}=0,$
$\displaystyle[e_{0},e_{0}]=0,\;[e_{0},e_{1}]=e_{1},\;[e_{1},e_{1}]=0.$

##  4 Infinite Dimensional Transposed Poisson Superalgebras

In this section we apply Theorem (1) to construct a transposed Poisson
superalgebra by means of differential forms on a smooth $n$-dimensional
manifold $M^{n}$.

Let us consider a graded commutative differential algebra
$({\mathfrak{A}},d)$, where
${\mathfrak{A}}=\oplus_{p\in{\mathbb{Z}}}{\mathfrak{A}}^{p}$, and
$d:{\mathfrak{A}}^{i}\to{\mathfrak{A}}^{i+1}$ is a differential of
$\mathfrak{A}$, that is, an anti-derivation of degree 1. Let us denote a
multiplication in $\mathfrak{A}$ by $(u,v)\to u\cdot v$, where
$u,v\in{\mathfrak{A}}$ and the degree of a homogeneous element $u$ by $|u|$.
Then for $u\in{\mathfrak{A}}^{p}$ we have $|u|=p$ and

$|u\cdot v|=|u|+|v|,\;\;u\cdot v=(-1)^{|u|\,|v|}v\cdot u,\;\;d(u\cdot
v)=(du)\cdot v+(-1)^{|u|}u\cdot dv.$

Assume that $\delta$ is a degree (-1) anti-derivation of $\mathfrak{A}$, i.e.
$\delta:{\mathfrak{A}}^{p}\to{\mathfrak{A}}^{p-1}$ and $\delta(u\cdot
v)=(\delta u)\cdot v+(-1)^{|u|}u\cdot\delta v$. Then
$D=d\circ\delta+\delta\circ d$ is the 0-degree derivation of $\mathfrak{A}$.
We can consider $\mathfrak{A}$ as a commutative superalgebra if we define the
parity of a homogeneous element $u$ as $|u|(\mbox{mod}\;2)$. Then $d,\delta$
are odd derivations and $D$ is the even derivation of $\mathfrak{A}$. Define
the bracket by

$[u,v]_{D}=u\cdot D(v)-(-1)^{|u|\,|v|}\,v\cdot D(u).$ (8)

Then according to Theorem 1 the triple $({\mathfrak{A}},\cdot,[\,,\,]_{D})$ is
a transposed Poisson superalgebra.

As an example of the structure described above we consider the algebra of
differential forms on a smooth $n$-dimensional manifold $M^{n}$. Let
$\Omega(M^{n})=\oplus_{p}\Omega^{p}(M^{n})$ be the graded algebra of
differential forms on $M^{n}$. If we denote by ${\cal F}(M^{n})$ the
commutative algebra of smooth functions on a manifold $M^{n}$ then
$\Omega^{0}(M^{n})$ is identified with the algebra of functions ${\cal
F}(M^{n})$, i.e. $\Omega^{0}(M^{n})={\cal F}(M^{n})$. The graded algebra of
differential forms $\Omega(M^{n})$ can be considered as a superalgebra if we
define the degree of $p$-form $\omega$ by $|\omega|=p\;(\mbox{mod}\;2)$. Then
it follows from the property of the wedge product of two forms
$\omega\wedge\theta=(-1)^{pq}\theta\wedge\omega$, where
$\omega\in\Omega^{p}(M^{n}),\theta\in\Omega^{q}(M^{n})$, that $\Omega(M^{n})$
is a commutative superalgebra. Let $\tt X$ be a smooth vector field on a
manifold $M^{n}$. Then the operator of contraction of a differential form with
a vector field $i_{\mathtt{X}}:\Omega^{p}\to\Omega^{p-1}$ and the exterior
differential $d:\Omega^{p}\to\Omega^{p+1}$ are odd derivations of the
commutative superalgebra of differential forms. Then the Lie derivative

${\cal L}_{\tt X}=d\circ i_{\mathtt{X}}+i_{\mathtt{X}}\circ d$

is an even degree derivation of the commutative superalgebra of differential
forms, that is, ${\cal L}_{\tt X}:\Omega^{p}(M^{n})\to\Omega^{p}(M^{n})$ and
${\cal L}_{\tt X}(\omega\wedge\theta)={\cal L}_{\tt
X}(\omega)\wedge\theta+\omega\wedge{\cal L}_{\tt X}(\theta)$. Then the bracket
(8) takes on the form

$\displaystyle[\omega,\theta]_{\tt X}=\omega\wedge{\cal L}_{\tt
X}(\theta)-(-1)^{|\omega|\,|\theta|}\theta\wedge{\cal L}_{\tt X}(\omega),$ (9)

and the commutative superalgebra of differential forms $\Omega(M^{n})$ endowed
with the bracket (9) is a transposed Poisson superalgebra. Particularly if we
consider two 0-forms, that is, two functions $f,g\in{\cal F}(M^{n})$ then the
graded Lie bracket (9) takes on the form

$[f,g]_{\tt X}=f\,{\cal L}_{\tt X}(g)-g\,{\cal L}_{\tt X}(f)=f\,{\tt
X}(g)-g\,{\tt X}(f).$

Hence the graded bracket (9) restricted to the subalgebra of smooth functions
${\cal F}(M^{n})$ makes it a transposed Poisson algebra (3). Let
$\Omega^{p}_{\tt c}(M^{n})$ be the vector space of closed $p$-differential
forms and $\Omega^{p}_{\tt e}(M^{n})\subset\Omega^{p}_{\tt c}(M^{n})$ be the
vector space of exact $p$-differential forms.

######  Proposition 1.

If $\omega\in\Omega^{p}_{\tt c}(M^{n}),\theta\in\Omega^{q}_{\tt c}(M^{n})$
then $[\omega,\theta]_{\tt X}\in\Omega^{p+q}_{\tt c}(M^{n})$, i.e. the graded
Lie bracket (9) of two closed forms is a closed form. If
$\omega\in\Omega^{p}_{\tt e}(M^{n})$ and $\theta\in\Omega^{q}_{\tt c}(M^{n})$
then $[\omega,\theta]_{\tt X}\in\Omega^{p+q}_{\tt e}(M^{n})$, i.e. if one of
two closed differential forms is exact then the graded Lie bracket (9) of
these two forms is an exact differential form.

###### Proof.

Let $\omega\in\Omega^{p}_{\tt c}(M^{n}),\theta\in\Omega^{q}_{\tt c}(M^{n})$,
that is, $d\omega=d\theta=0$. We have

$\displaystyle[\omega,\theta]_{\tt X}$ $\displaystyle=$
$\displaystyle\omega\wedge{\cal L}_{\tt
X}(\theta)-(-1)^{|\omega|\,|\theta|}\theta\wedge{\cal L}_{\tt X}(\omega)$
$\displaystyle=$ $\displaystyle\omega\wedge i_{\tt X}\circ
d(\theta)+\omega\wedge d\circ i_{\tt
X}(\theta)-(-1)^{|\omega|\,|\theta|}(\theta\wedge i_{\tt X}\circ
d(\omega)+\theta\wedge d\circ i_{\tt X}(\omega))$ $\displaystyle=$
$\displaystyle\omega\wedge d\circ i_{\tt
X}(\theta)-(-1)^{|\omega|\,|\theta|}\theta\wedge d\circ i_{\tt X}(\omega).$

Now differentiating the graded Lie bracket we obtain

$\displaystyle d([\omega,\theta]_{\tt X})$ $\displaystyle=$ $\displaystyle
d(\omega)\wedge d\circ i_{\tt X}(\theta)+(-1)^{|\omega|}\omega\wedge
d^{2}\circ i_{\tt X}(\theta)$
$\displaystyle-(-1)^{|\omega|\,|\theta|}d(\theta)\wedge d\circ i_{\tt
X}(\omega)-(-1)^{(|\omega|+1)\,|\theta|}\theta\wedge d^{2}\circ i_{\tt
X}(\omega)=0$

If $\omega$ is an exact form, that is, $\omega\in\Omega^{p}_{\tt
e}(M^{n}),\omega=d\zeta,\zeta\in\Omega^{p-1}(M^{n})$ then making use of
$d\circ{\cal L}_{\tt X}={\cal L}_{\tt X}\circ d$ we get

$\displaystyle[\omega,\theta]_{\tt X}=[d\zeta,\theta]_{\tt X}$
$\displaystyle=$ $\displaystyle d\zeta\wedge{\cal L}_{\tt
X}(\theta)-(-1)^{|\omega|\,|\theta|}\theta\wedge{\cal L}_{\tt X}(d\zeta)$
$\displaystyle=$ $\displaystyle d\zeta\wedge{\cal L}_{\tt
X}(\theta)-(-1)^{|\omega|\,|\theta|}\theta\wedge d\circ{\cal L}_{\tt
X}(\zeta)$ $\displaystyle=$ $\displaystyle d\big{(}\zeta\wedge{\cal L}_{\tt
X}(\theta)-(-1)^{|\zeta|\,|\theta|}\theta\wedge{\cal L}_{\tt
X}(\zeta)\big{)}.$

∎

Thus the proved proposition shows that the graded Lie bracket (9) induces the
graded Lie bracket on the graded algebra of de Rham cohomologies, which
defines the structure of transposed Poisson superalgebra.

Our next example of a transposed Poisson superalgebra is related to the
quantum theory of Yang-Mills fields [8]. The gauge group of the Yang-Mills
theory is $\mbox{SU}(2)$. Let ${\mathfrak{su}}(2)$ be the Lie algebra of
$\mbox{SU}(2)$, $t_{a}$ be a basis for ${\mathfrak{su}}(2)$ and
$[t_{a},t_{b}]=K^{c}_{ab}t_{c}$, i.e. $K^{c}_{ab}$ are structure constants of
${\mathfrak{su}}(2)$. The basic fields of the quantum Yang-Mills theory are
${\mathfrak{su}}(2)$-valued functions $A_{\mu},c,\bar{c}$ defined on a space-
time, where $A_{\mu}$ ($\mu=0,1,2,3$) are Yang-Mills fields, $c,\bar{c}$ are
ghost fields. Since the basic fields are ${\mathfrak{su}}(2)$-valued functions
we can express them in the terms of a basis as follows
$A_{\mu}=A^{a}_{\mu}\,t_{a},c=c^{a}\,t_{a},\bar{c}=\bar{c}^{a}\,t_{a}$. We
consider the algebra of fields of the quantum Yang-Mills theory as an algebra
generated by $A^{a}_{\mu},b^{a},c^{a},\bar{c}^{a}$, their all orders partial
derivatives with respect to coordinates of space-time and Grassmann variables
$\xi,\eta,\ldots$ which do not depend on a point of space-time. We assume that
the gauge fields $A^{a}_{\mu}$ and an auxiliary field $b^{a}$ and all their
partial derivatives are commuting generators of algebra and they also commute
with ghost fields, their partial derivatives and Grassmann variables
$\xi,\eta,\ldots$. The ghost fields $c^{a},\bar{c}^{a}$, their partial
derivatives and $\xi,\eta,\ldots$ are generators of Grassmann type, that is,
they anticommute with each other. Now attributing degree zero to the gauge
fields $A^{a}_{\mu}$ and an auxiliary field $b^{a}$ (and their all partial
derivatives), degree one to ghost fields (and their partial derivatives) and
Grassmann variables $\xi,\eta,\ldots$, we turn the algebra of the fields of
quantum Yang-Mills theory into a commutative superalgebra. An important role
in the quantum theory of Yang-Mills fields is played by BRST supersymmetry,
which has the form [8]

$\displaystyle A^{a}_{\mu}\;\to\;A^{a}_{\mu}+\delta
A^{a}_{\mu},\;c^{a}\;\to\;c^{a}+\delta
c^{a},\;\bar{c}^{a}\;\to\;\bar{c}^{a}+\delta\bar{c}^{a},$ (10)

where

$\delta A^{a}_{\mu}=\xi\,D_{\mu}c^{a},\;\;\delta
c^{a}=-\frac{1}{2}\,K^{a}_{bd}\,\xi\,c^{b}\,c^{d}\;,\;\delta\bar{c}^{a}=\xi\,b^{a},$
(11)

and $D_{\mu}c^{a}=\partial_{\mu}c^{a}+K^{a}_{bd}A^{b}_{\mu}\,c^{d}$ is the
covariant derivative. The BRST supersymmetry can be considered as an even
derivation of the commutative superalgebra of the fields of the quantum Yang-
Mills theory. Indeed let us define the mapping $s$ on the basic fields as
follows

$\displaystyle
sA^{a}_{\mu}=(D_{\mu}c)^{a}\;,\;sc^{a}=-\frac{1}{2}\,K^{a}_{bd}\,c^{b}\,c^{d}\;,\;s\bar{c}^{a}=b^{a}\;,\;sb^{a}=0,$
(12)

and extend this mapping to the odd derivation of the commutative superalgebra
of fields by assuming that it acts on products of fields according to the
graded Leibniz rule. We also assume that the derivation $s$ commutes with
partial derivatives and Grassmann variables $\xi,\eta,\ldots$, i.e.
$s\circ\partial_{\mu}=\partial_{\mu}\circ s,s\xi=\xi s$. It can be verified
that the odd derivation $s$ has a very important property $s^{2}=0$. Then
$\delta=\xi\,s$ is the even derivation of the commutative superalgebra of
fields. Indeed for any product of basic fields $\phi,\psi$ we have

$\displaystyle\delta(\phi\,\psi)$ $\displaystyle=$
$\displaystyle\xi\,s(\phi\,\psi)=\xi\big{(}\delta(\phi)\,\psi+(-1)^{|\phi|}\phi\,\delta(\psi)\big{)}$
$\displaystyle=$
$\displaystyle\xi\,\delta(\phi)\,\psi+(-1)^{|\phi|}\xi\,\phi\,\delta(\psi)$
$\displaystyle=$
$\displaystyle\xi\,\delta(\phi)\,\psi+(-1)^{2|\phi|}\phi\,\xi\,\delta(\psi)=\delta(\phi)\,\psi+\phi\,\delta(\psi).$

Let $\Phi,\Psi$ be two elements of the commutative superalgebra of the fields
of quantum Yang-Mills theory, that is, finite polynomials on the basic fields
and their partial derivatives. We define the graded Lie bracket of these two
polynomials by

$[\Phi,\Psi]_{\delta}=\Phi\,\delta(\Psi)-(-1)^{|\Phi|\,|\Psi|}\,\Psi\,\delta(\Phi).$
(13)

Then the algebra of the fields of quantum Yang-Mills theory endowed with the
graded Lie bracket (13) is a transposed Poisson superalgebra.

##  5 Identities in Transposed Poisson Superalgebra

The aim of this section is to show that the identities, which hold in the case
of a transposed Poisson algebra [3], can be extended to a transposed Poisson
superalgebra if we modify the identities proposed in [3] with the help of the
rule of signs.

######  Theorem 2.

Let $({\mathfrak{G}},\cdot,[\;,\;])$ be a transposed Poisson superalgebra.
Then for any $h,x,y,z\in{\mathfrak{G}}$ we have the following identities

$\displaystyle(-1)^{|x|\,|z|}\,x\cdot[y,z]+(-1)^{|x|\,|y|}\,y\cdot[z,x]+(-1)^{|y|\,|z|}\,z\cdot[x,y]\\!\\!\\!$
$\displaystyle=$ $\displaystyle\\!\\!\\!0,$ (14)
$\displaystyle(-1)^{|x|\,|z|}\,[h\cdot[x,y],z]+(-1)^{|x|\,|y|}\,[h\cdot[y,z],x]+(-1)^{|y|\,|z|}\,[h\cdot[z,x],y]\\!\\!\\!$
$\displaystyle=$ $\displaystyle\\!\\!\\!0,$ (15)
$\displaystyle(-1)^{|x|\,|z|}\,[h\cdot x,[y,z]]+(-1)^{|x|\,|y|}\,[h\cdot
y,[z,x]]+(-1)^{|y|\,|z|}\,[h\cdot z,[x,y]]\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\\!0,$ (16)
$\displaystyle(-1)^{|x|\,|z|}\,[h,x]\,[y,z]+(-1)^{|x|\,|y|}\,[h,y]\,[z,x]+(-1)^{|y|\,|z|}\,[h,z]\,[x,y]\\!\\!\\!$
$\displaystyle=$ $\displaystyle 0,$ (17) $\displaystyle 2\,u\cdot
v\cdot[x,y]=(-1)^{|x|\,|v|}[u\cdot x,v\cdot y]+(-1)^{|u|\,(|x|+|v|)}[v\cdot
x,u\cdot y],$ (18) $\displaystyle(-1)^{|u|\,|yv|}\,x\cdot[u,y\cdot
v]+(-1)^{|v|\,|xy|}\,v\cdot[x\cdot y,u]+(-1)^{|x|\,|yv|}y\cdot[v,x]\cdot
u\\!\\!\\!$ $\displaystyle=$ $\displaystyle\\!\\!\\!0.$ (19)

###### Proof.

In order to prove identity (14) we make use of the compatibility condition (4)
as follows

$\displaystyle(-1)^{|x|\,|z|}\;2x\cdot[y,z]$ $\displaystyle=$
$\displaystyle(-1)^{|x|\,|z|}[x\cdot y,z]+(-1)^{|x|\,|z|+|x|\,|y|}[y,x\cdot
z],$ $\displaystyle(-1)^{|x|\,|y|}\;2y\cdot[z,x]$ $\displaystyle=$
$\displaystyle(-1)^{|x|\,|y|}[y\cdot z,x]+(-1)^{|x|\,|y|+|y|\,|z|}[x,y\cdot
x],$ $\displaystyle(-1)^{|y|\,|z|}\;2z\cdot[x,y]$ $\displaystyle=$
$\displaystyle(-1)^{|y|\,|z|}[z\cdot x,y]+(-1)^{|y|\,|z|+|x|\,|z|}[x,z\cdot
y].$

Making use of the commutativity of a superalgebra $\mathfrak{G}$ and the
graded skew-symmetry of bracket we can represent the right-hand sides of the
above identities as follows

$\displaystyle(-1)^{|x|\,|z|}\;2x\cdot[y,z]$ $\displaystyle=$
$\displaystyle(-1)^{|x|\,|z|}[x\cdot y,z]-(-1)^{|y|\,|z|}[z\cdot x,y],$
$\displaystyle(-1)^{|x|\,|y|}\;2y\cdot[z,x]$ $\displaystyle=$
$\displaystyle(-1)^{|x|\,|y|}[y\cdot z,x]-(-1)^{|x|\,|z|}[x\cdot y,z],$
$\displaystyle(-1)^{|y|\,|z|}\;2z\cdot[x,y]$ $\displaystyle=$
$\displaystyle(-1)^{|y|\,|z|}[z\cdot x,y]-(-1)^{|x|\,|y|}[y\cdot z,x].$

Summing up the left-hand sides and right-hand sides of the obtained
identities, we get identity (14). The proof of identity (15) is similar to the
proof of identity (14). First we use the compatibility condition (4) to obtain
the following three identities

$\displaystyle(-1)^{|x|\,|z|}\,2\;h\cdot[[x,y],z]$ $\displaystyle=$
$\displaystyle(-1)^{|x|\,|z|}\,[h\cdot[x,y],z]+(-1)^{|x|\,|z|+(xy,h)}[[x,y],h\cdot
z],$ $\displaystyle(-1)^{|x|\,|y|}\,2\;h\cdot[[y,z],x]$ $\displaystyle=$
$\displaystyle(-1)^{|x|\,|y|}\,[h\cdot[y,z],x]+(-1)^{|x|\,|y|+(yz,h)}[[y,z],h\cdot
x],$ $\displaystyle(-1)^{|y|\,|z|}\,2\;h\cdot[[z,x],y]$ $\displaystyle=$
$\displaystyle(-1)^{|y|\,|z|}\,[h\cdot[z,x],y]+(-1)^{|y|\,|z|+(xz,h)}[[z,x],h\cdot
y],$

where
$(xy,h)=|x|\,|h|+|y|\,|h|$,$(yz,h)=|y|\,|h|+|z|\,|h|$,$(xz,h)=|x|\,|h|+|z|\,|h|$.
Due to the super Jacobi identity the sum of left-hand sides of the above
identities is equal to zero. Hence we get

$\displaystyle(-1)^{|x|\,|z|}\,[h\cdot[x,y],z]+(-1)^{|x|\,|y|}\,[h\cdot[y,z],x]+(-1)^{|y|\,|z|}\,[h\cdot[z,x],y]$
$\displaystyle\quad\quad\quad\qquad+(-1)^{|x|\,|z|+(xy,h)}[[x,y],h\cdot
z]+(-1)^{|x|\,|y|+(yz,h)}[[y,z],h\cdot x]$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+(-1)^{|y|\,|z|+(xz,h)}[[z,x],h\cdot
y]=0.$ (20)

The sum of the first three terms in the above identity is the left-hand side
of identity (15). Thus in order to prove identity (15) we need to show that
the sum of the last three terms in the above identity is zero. For the first
one of them, that is, $[[x,y],h\cdot z]$ (we temporarily omit the coefficient
$(-1)^{|x|\,|z|+(xy,h)}$ , which will be taken into account later) we have the
super Jacobi identity

$(-1)^{|x|\,|h|+|x|\,|z|}[[x,y],h\cdot z]+(-1)^{|x|\,|y|}[[y,h\cdot
z],x]+(-1)^{|y|\,|h|+|y|\,|z|}[[h\cdot z,x],y]=0.$ (21)

Applying to the middle term at the left-hand side of this equality the
compatibility condition

$2\,h\cdot[y,z]=[h\cdot y,z]+(-1)^{|y|\,|h|}[y,h\cdot
z]\;\;\Rightarrow\;\;[y,h\cdot z]=(-1)^{|y|\,|h|}(2\,h\cdot[y,z]-[h\cdot
y,z]),$

and substituting the obtained expression multiplied by
$(-1)^{|x|\,|z|+(xy,h)}$ into super Jacobi identity (21) we get

$\displaystyle(-1)^{|y|\,|h|}[[x,y],h\cdot
z]+(-1)^{(yzh,x)}([2\,h\cdot[y,z],x]\\!\\!\\!\\!$ $\displaystyle-$
$\displaystyle\\!\\!\\!\\![[h\cdot y,z],x])$ (22) $\displaystyle+$
$\displaystyle\\!\\!\\!\\!\\!(-1)^{|y|\,|z|+(zh,x)}[[h\cdot z,x],y]=0,$

where $(yzh,x)=|y|\,|x|+|z|\,|x|+|h|\,|x|$ and $(zh,x)=|z|\,|x|+|h|\,|x|$.
Applying the same calculations to the next two terms in (20), we get two more
identities

$\displaystyle(-1)^{|z|\,|h|}[[y,z],h\cdot
x]+(-1)^{(xzh,y)}([2\,h\cdot[z,x],y]\\!\\!\\!\\!$ $\displaystyle-$
$\displaystyle\\!\\!\\!\\![[h\cdot z,x],y])$ (23) $\displaystyle+$
$\displaystyle\\!\\!\\!\\!\\!(-1)^{|x|\,|z|+(xh,y)}[[h\cdot x,y],z]=0,$
$\displaystyle(-1)^{|x|\,|h|}[[y,z],h\cdot
x]+(-1)^{(xyh,z)}([2\,h\cdot[z,x],y]\\!\\!\\!\\!$ $\displaystyle-$
$\displaystyle\\!\\!\\!\\![[h\cdot z,x],y])$ (24) $\displaystyle+$
$\displaystyle\\!\\!\\!\\!\\!(-1)^{|x|\,|y|+(yh,z)}[[h\cdot x,y],z]=0,$

Summing up the identities (22), (23), (24) multiplied by
$(-1)^{|x|\,(|z|+|h|)},(-1)^{|y|\,(|x|+|h|)}$ and $(-1)^{|z|\,(|y|+|h|)}$
respectively, we obtain

$\displaystyle(-1)^{|x|\,|z|+(xy,h)}[[x,y],h\cdot
z]+(-1)^{|x|\,|y|+(yz,h)}[[y,z],h\cdot x]+(-1)^{|y|\,|z|+(xz,h)}[[z,x],h\cdot
y]$ $\displaystyle
2\,\big{(}(-1)^{|x|\,|z|}\,[h\cdot[x,y],z]+(-1)^{|x|\,|y|}\,[h\cdot[y,z],x]+(-1)^{|y|\,|z|}\,[h\cdot[z,x],y]\big{)}=0.$
(25)

Subtracting the identity (20) from (25) we get the identity (15). By virtue of
the proven identity (15), the sum of the first three terms in (20) will be
equal to zero. The remaining three terms, after a suitable permutation of the
arguments, give the identity (16).

In order to prove the last identity (17) we apply the compatibility condition
to the following products

$\displaystyle 2\,[x,y]\cdot[h,z]$ $\displaystyle=$ $\displaystyle[[x,y]\cdot
h,z]+(-1)^{|h|(|x|+|y|)}[h,[x,y]\cdot z],$ $\displaystyle 2\,[y,z]\cdot[h,x]$
$\displaystyle=$ $\displaystyle[[y,z]\cdot
h,x]+(-1)^{|h|(|y|+|z|)}[h,[y,z]\cdot x],$ $\displaystyle 2\,[z,x]\cdot[h,y]$
$\displaystyle=$ $\displaystyle[[z,x]\cdot
h,y]+(-1)^{|h|(|x|+|z|)}[h,[z,x]\cdot y].$

In the first brackets on the right-hand sides of these relations, we rearrange
$h$ and $[x,y]$, $h$ and $[y,z]$, $h$ and $[z,x]$. Next we multiply the first
relation by $(-1)^{|x|\,|h|+|y|\,|h|+|x|\,|z|}$, the second by
$(-1)^{|y|\,|h|+|z|\,|h|+|x|\,|y|}$ and the third by
$(-1)^{|x|\,|h|+|z|\,|h|+|y|\,|z|}$. Finally we rearrange the brackets at the
left-hand side of every relation. We obtain

$\displaystyle(-1)^{|y|\,|z|}\,2\,[h,z]\cdot[x,y]$ $\displaystyle=$
$\displaystyle(-1)^{|x|\,|z|}[h\cdot[x,y],z]+[h,(-1)^{|x|\,|z|})[x,y]\cdot
z],$ $\displaystyle(-1)^{|x|\,|z|}\,2\,[h,x]\cdot[y,z]$ $\displaystyle=$
$\displaystyle(-1)^{|x|\,|y|}[h\cdot[y,z],x]+[h,(-1)^{|x|\,|y|}[y,z]\cdot x],$
$\displaystyle(-1)^{|x|\,|y|}2\,[h,y]\cdot[z,x]$ $\displaystyle=$
$\displaystyle(-1)^{|y|\,|z|}[h\cdot[z,x],y]+[h,(-1)^{|y|\,|z|}[z,x]\cdot y].$

Summing up the left-hand sides and the right-hand sides of these relations we
get the relation, whose left-hand side (multiplied by 1/2) coincides with the
left-hand side of the identity (17), but the right-hand side is equal to zero.
Indeed the sum of the first terms at the right-hand side of every relation
vanishes because of the identity (15). The sum of the second terms vanishes
because of the identity (14). The identity (18) is proved by means of the
compatibility condition. In order to prove the identity (19) we take the sum
of the three following relations: the compatibility condition for the products
$2\,x\cdot[u,y\cdot v]$, $2\,v\cdot[x\cdot y,u]$ and the identity (18)
multiplied by 2. This ends the proof of Theorem 2. ∎

##  6 CONCLUSIONS

The impetus for writing this article was the paper [3], which introduces the
concept of a transposed Poisson algebra, describes its structure, gives
important examples and relations to other algebraic structures. Our goal in
this article is to show that the notion of a transposed Poisson algebra can be
extended to superalgebras in substance merely by means of Manin’s rule of
signs, and also to show that important structures of a transposed Poisson
algebra (example constructed with the help of derivation and identities) also
hold in a transposed Poisson superalgebra in a graded form. We think that the
example of transposed Poisson superalgebra constructed by means of
differential forms and Lie derivative reveals a geometric meaning of the
notion of transposed Poisson superalgebra. In the paper [3] the authors show
that the identities in a transposed Poisson algebra play an important role in
establishing a relation with Hom-Lie algebras. We hope that the graded
versions of these identities proved in this paper play the same important role
in establishing connections between a transposed Poisson superalgebra and Hom-
Lie superalgebras. This, and a notion of a transposed Poisson 3-Lie
superalgebra, is our next goal in developing this direction.

## ACKNOWLEDGEMENTS

The authors express their deep gratitude to the colleagues and students
participating in the Seminar of Geometry and Topology at the Institute of
Mathematics and Statistics of the University of Tartu for useful discussions
of the structures presented in this article.

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