# Poisson-like cohomologies associated with some Lie superalgebras

Kentaro Mikami Akita University (This work is supported by JSPS KAKENHI Grant
Number 22K03306.) Tadayoshi Mizutani Saitama University Hajime Sato Nagoya
University

###### Contents

  1. 1 Introduction
  2. 2 Poisson-like cohomologies of Lie superalgebra
    1. 2.1 Examples of Poisson-like cohomologies of superalgebra by a Lie algebra
  3. 3 Poisson-like cohomology of superalgebra of differential forms
    1. 3.1 Poisson-like cohomology groups by 2-forms of dual Lie algebra
  4. 4 Dual Poisson-like coboundary operator and cohomologies
    1. 4.1 Multi-vector fields and differential forms with polynomial coefficients on $\mathbb{R}^{n}$

##  1 Introduction

First, we recall briefly the Poisson cohomology of a Poisson manifold $M$. A
Poisson tensor $\pi$ is a 2-vector field on $M$ by which
$\\{f,g\\}_{\pi}=\langle\pi,df\wedge dg\rangle$ becomes the Poisson bracket of
functions $f$ and $g$ on $M$. This is equivalent to
$[\pi,\pi]_{\text{\footnotesize S}}=0$, where
$[\cdot,\cdot]_{\text{\footnotesize S}}$ is the Schouten(-Nijenhuis) bracket
on multi-vector fields. A quick introduction of the Schouten bracket is

(1.1) $[X_{1}\wedge\cdots\wedge X_{r},Y_{1}\wedge\cdots\wedge
Y_{s}]_{\text{\footnotesize
S}}=\sum(-1)^{i+j}[X_{i},Y_{j}]\wedge(X_{1}\wedge\cdots\widehat{X_{i}}\cdots\wedge
X_{r})\wedge(Y_{1}\wedge\cdots\widehat{Y_{j}}\cdots\wedge Y_{s})$

where $[X_{i},Y_{j}]$ is the Jacobi-Lie bracket of two vector fields $X_{i}$
and $Y_{j}$. In short, the Schouten bracket is a Lie superalgebra bracket on
$\sum_{j=1}^{\textrm{dim}M}\Lambda^{j}\mathrm{T}(M)$, where the grading of
$\Lambda^{j}\mathrm{T}(M)=j-1$.

For a Poisson tensor $\pi$, we define $d_{\pi}(U)=[\pi,U]_{\text{\footnotesize
S}}$ for $\displaystyle U\in\Lambda^{j}\mathrm{T}(M)$. Then $\displaystyle
d_{\pi}\left(\Lambda^{j}\mathrm{T}(M)\right)\subset\Lambda^{j+1}\mathrm{T}(M)$,
and $\displaystyle d_{\pi}\circ d_{\pi}=0$ holds by super bracket property of
the Schouten bracket. Thus, $d_{\pi}$ is a coboundary operator of degree $1$
and we obtain the so-called Poisson cohomology groups. We easily expect that
an even $p$-multivector field $\Pi\in\Lambda^{p}\mathrm{T}(M)$ with
$[\Pi,\Pi]_{\text{\footnotesize S}}=0$ defines
$d_{\Pi}(U)=[\Pi,U]_{\text{\footnotesize S}}$, satisfies $\displaystyle
d_{\Pi}\left(\Lambda^{j}\mathrm{T}(M)\right)\subset\Lambda^{j+p-1}\mathrm{T}(M)$
and $\displaystyle d_{\Pi}\circ d_{\Pi}=0$. In this note, we translate notion
of Poisson cohomology groups of Poisson manifolds to those of corresponding
concepts of Lie superalgebras. Each vector field $X$ on $M$ is a section of
$\mathrm{T}(M)$ and exactly written as $X\in\Gamma\mathrm{T}(M)$, but here we
denote it simply as $X\in\mathrm{T}(M)$. This abbreviation also applies to all
multi-vector fields and differential forms.

##  2 Poisson-like cohomologies of Lie superalgebra

The abstract definition of a Lie superalgebra is the following:

######  Definition 2.1 (Lie superalgebra).

Suppose a real vector space $\mathfrak{g}$ is graded by integers as
$\displaystyle\mathfrak{g}=\sum_{j\in\mathbb{Z}}\mathfrak{g}_{j}$ and there
exits a $\mathbb{R}$-bilinear operation $[\cdot,\cdot]$ satisfying

(2.1)
$\displaystyle[\mathfrak{g}_{i},\mathfrak{g}_{j}]\subset\mathfrak{g}_{i+j}$
(2.2) $\displaystyle[X,Y]+(-1)^{xy}[Y,X]=0\quad\text{ where
}X\in\mathfrak{g}_{x}\text{ and }Y\in\mathfrak{g}_{y}$ (2.3)
$\displaystyle(-1)^{xz}[[X,Y],Z]+(-1)^{yx}[[Y,Z],X]+(-1)^{zy}[[Z,X],Y]=0\;.$

Then we call $\mathfrak{g}$ a $\mathbb{Z}$-graded (or pre) Lie superalgebra.

######  Remark 2.1.

Super Jacobi identity (2.3) above is equivalent to one of the following two.

(2.4) $\displaystyle[[X,Y],Z]$ $\displaystyle=[X,[Y,Z]]+(-1)^{yz}[[X,Z],Y]$
(2.5) $\displaystyle[X,[Y,Z]]$ $\displaystyle=[[X,Y],Z]+(-1)^{xy}[Y,[X,Z]]$

We translate and generalize the notion of Poisson cohomology groups above on
Lie superalgebras, and we call the generalized notion to Poisson-like
cohomology groups.

######  Proposition 2.1.

Let $\mathfrak{g}=\sum_{i}{\mathfrak{g}}_{i}$ be a Lie superalgebra. For odd
$p$, assume $\pi\in{\mathfrak{g}}_{p}$ satisfies $[\pi,\pi]=0$. Then the map
$d_{\pi}$ defined by

(2.6) $\displaystyle d_{\pi}(U)=[\pi,U]\qquad(U\in\mathfrak{g})$ satisfies
(2.7) $\displaystyle
d_{\pi}({\mathfrak{g}}_{i})\subset{\mathfrak{g}}_{i+p}\quad\text{and}\quad
d_{\pi}\circ d_{\pi}=0\;.$

Proof: For $\pi\in{\mathfrak{g}}_{p}$, Jacobi identity shows
$[\pi,[\pi,U]]=[[\pi,\pi],U]+(-1)^{p^{2}}[\pi,[\pi,U]]$. Since $p$ is odd,
$2[\pi,[\pi,U]]=[[\pi,\pi],U]$ holds, and the assumption $[\pi,\pi]=0$ implies
$2[\pi,[\pi,U]]=0$. $\blacksquare$

######  Remark 2.2.

We may call $d_{\pi}$ defined by (2.6) the coboundary operator associated with
$\pi$, or Poisson-like coboundary operator, simply.

######  Remark 2.3.

Contrary to Proposition 2.1 above, we start from $P\in{\mathfrak{g}}_{p}$
where $p$ is even. Then $[P,P]=0$ holds automatically because
$[P,P]+(-1)^{p^{2}}[P,P]=0$ by symmetric property of super bracket and $p$
being even. Since

$(d_{P}\circ
d_{P})(U)=[P,[P,U]]=[[P,P],U]+(-1)^{p^{2}}[P,[P,U]]=0+[P,[P,U]]=(d_{P}\circ
d_{P})(U)\;,$

there is no information about $d_{P}\circ d_{P}$ in general for each even
grade $P$. We refer to Example 3.1 for a concrete example.

######  Definition 1 (generalized Poisson cohomology).

In a more general context than Proposition 2.1, let $\Phi$ be a linear
operator on a Lie superalgebra $\mathfrak{g}=\sum_{i}{\mathfrak{g}}_{i}$ into
itself satisfying

(2.8)
$\Phi({\mathfrak{g}}_{i})\subset{\mathfrak{g}}_{i+p}\quad\text{and}\quad\Phi\circ\Phi=0\;.$

We call $\Phi$ the coboundary operator of degree $p$. Then $\Phi$ defines

(2.9)
$\mathbb{H}^{i,p}=\ker(\Phi:{\mathfrak{g}}_{i}\to{\mathfrak{g}}_{i+p})/\Phi({\mathfrak{g}}_{i-p})\;.$

We call (2.9) Poisson-like cohomology groups defined by $\Phi$.

If ${\mathfrak{g}}_{i}$ are finite dimensional, the dimension of
$\mathbb{H}^{i,p}$ makes sense and we may call them Betti numbers. By adding
one more assumption, $\textrm{dim}{\mathfrak{g}}_{i}=0$ for $|i|>b$, i.e., the
length of Lie superalgebra is finite, then we get the following result.

######  Theorem 2.1.

Consider a Lie superalgebra $\mathfrak{g}=\sum_{|i|\leqq b}{\mathfrak{g}}_{i}$
and each ${\mathfrak{g}}_{i}$ is finite dimensional. Let $\Phi$ be a linear
map satisfying (2.8) in the definition above, i.e., a coboundary operator of
degree $p$. Then we have an integer

(2.10) $\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}\left(\ker(\Phi:{\mathfrak{g}}_{i}\to{\mathfrak{g}}_{i+p})/\Phi({\mathfrak{g}}_{i-p})\right)\;.$

The above (2.10) is independent of the operator $\Phi$ and agrees with the
value

(2.11) $\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}{\mathfrak{g}}_{i}\;.$

Proof:

The (2.10) above $\displaystyle=\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}(\textrm{dim}\ker(\Phi:{\mathfrak{g}}_{i}\to{\mathfrak{g}}_{i+p})-\textrm{dim}(\Phi({\mathfrak{g}}_{i-p}))$
by the rank theorem of linear map, we see
$\textrm{dim}{\mathfrak{g}}_{i}-\textrm{dim}\ker(\Phi:{\mathfrak{g}}_{i}\to{\mathfrak{g}}_{i+p})=\textrm{dim}\Phi({\mathfrak{g}}_{i})$
and we get $\displaystyle=\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}(\textrm{dim}{\mathfrak{g}}_{i}-\textrm{dim}\Phi({\mathfrak{g}}_{i})-\textrm{dim}\Phi({\mathfrak{g}}_{i-p})$
$\displaystyle=\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}{\mathfrak{g}}_{i}-\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}\Phi({\mathfrak{g}}_{i})-\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}\Phi({\mathfrak{g}}_{i-p})$ we replace $i-p$ by $j$, we
get $\displaystyle=\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}{\mathfrak{g}}_{i}-\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}\Phi({\mathfrak{g}}_{i})-\sum_{k}(-1)^{k}\sum_{k|p|\leqq
p+j<(k+1)|p|}\textrm{dim}\Phi({\mathfrak{g}}_{j})$ we may put
$p=\varepsilon|p|$ where $\varepsilon=\pm 1$, and
$\displaystyle=\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}{\mathfrak{g}}_{i}-\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}\Phi({\mathfrak{g}}_{i})-\sum_{k}(-1)^{k}\sum_{(k-\varepsilon)|p|\leqq
j<(k-\varepsilon+1)|p|}\textrm{dim}\Phi({\mathfrak{g}}_{j})$
$\displaystyle=\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}{\mathfrak{g}}_{i}-\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}\Phi({\mathfrak{g}}_{i})-\sum_{k}(-1)^{k+\varepsilon}\sum_{k|p|\leqq
j<(k+1)|p|}\textrm{dim}\Phi({\mathfrak{g}}_{j})$
$\displaystyle=\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}{\mathfrak{g}}_{i}-\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}\Phi({\mathfrak{g}}_{i})+\sum_{k}(-1)^{k}\sum_{k|p|\leqq
j<(k+1)|p|}\textrm{dim}\Phi({\mathfrak{g}}_{j})$
$\displaystyle=\sum_{k}(-1)^{k}\sum_{k|p|\leqq
i<(k+1)|p|}\textrm{dim}{\mathfrak{g}}_{i}\;.$

######  Remark 2.4.

When $p=1$, (2.11) is just the Euler number, and so (2.11) is a generalization
of the Euler number.

###  2.1 Examples of Poisson-like cohomologies of superalgebra by a Lie
algebra

Although we have a typical superalgebra
$\sum_{p=1}^{\textrm{dim}M}\Lambda^{p}\mathrm{T}(M)$ for a manifold $M$, their
chain spaces are infinite dimensional, and it is hard to manipulate those
(co)homology groups. If $M$ is a finite dimensional Lie group, then its
invariant multivector fields are finite dimensional, and provides a
superalgebra, which is the direct sum of
${\mathfrak{g}}_{i}=\Lambda^{i+1}\mathfrak{m}$ for $i=0,\ldots,n-1$, where $M$
is a $n$-dimensional Lie group whose Lie algebra is $\mathfrak{m}$.

######  Example 2.1.

We focus on the 4-dimensional Lie algebra spanned by $y_{i}$ for $i=1..4$
satisfying (minimum) bracket relations
$\text{Type}[1]=\\{[y_{2},y_{4}]=y_{1},[y_{3},y_{4}]=y_{2}\\}$ which is listed
in [1], and take a 2-vector field

$\pi=c_{1}y_{1}\wedge y_{2}+c_{2}y_{1}\wedge y_{3}+c_{3}y_{1}\wedge
y_{4}+c_{4}y_{2}\wedge y_{3}+c_{5}y_{2}\wedge y_{4}+c_{6}y_{3}\wedge y_{4}$

where $c_{j}$ are real constants. Since

$[\pi,\pi]_{\text{\footnotesize
S}}/2=(-{c}_{2}{c}_{6}+{c}_{4}{c}_{5})\mathit{\&^{\wedge}}(y_{1},y_{2},y_{3})+(-{c}_{3}{c}_{6}+{c}_{5}^{2})\mathit{\&^{\wedge}}(y_{1},y_{2},y_{4})+{c}_{5}{c}_{6}\mathit{\&^{\wedge}}(y_{1},y_{3},y_{4})+{c}_{6}^{2}\mathit{\&^{\wedge}}(y_{2},y_{3},y_{4})\;,$

Poisson condition $[\pi,\pi]_{\text{\footnotesize S}}=0$ implies

$\left\\{-c_{2}c_{6}+c_{4}c_{5}=0,-c_{3}c_{6}+c_{5}^{2}=0,c_{5}c_{6}=0,c_{6}^{2}=0\right\\}.$

Just solving Poisson condition, we get $c_{1},c_{2},c_{3},c_{4}$ are free, and
$c_{5}{=}0,c_{6}{=}0$ . Thus,

$\pi=y_{1}\wedge(c_{1}y_{2}+c_{2}y_{3}+c_{3}y_{4})+c_{4}y_{2}\wedge
y_{3}\;,\text{ and
}\;\pi\wedge\pi=2c_{3}c_{4}{\&}^{\wedge}\\!\left(y_{1},y_{2},y_{3},y_{4}\right)\;.$

Since

$\displaystyle
d_{\pi}[y_{1},\ldots,y_{4}]=[d_{\pi}(y_{1}),\ldots,d_{\pi}(y_{4})]=\left[0,0,-c_{3}y_{1}{\wedge}y_{2},c_{4}y_{1}{\wedge}y_{3}+c_{2}y_{1}{\wedge}y_{2}\right]$
$\displaystyle d_{\pi}[y_{1}\wedge y_{2},\ldots,y_{3}\wedge
y_{4}]=\left[0,0,0,0,c_{4}{\&^{\wedge}}\\!\left(y_{1},y_{2},y_{3}\right),-c_{2}{\&^{\wedge}}\\!\left(y_{1},y_{2},y_{3}\right)-c_{3}{\&^{\wedge}}\\!\left(y_{1},y_{2},y_{4}\right)\right]$
$\displaystyle d_{\pi}[y_{1}\wedge y_{2}\wedge y_{3},\ldots,y_{2}\wedge
y_{3}\wedge y_{4}]=\left[0,0,0,0\right]\;,$

their matrix representations of $d_{\pi}$ are

$A(1,2)=\left[\begin{array}[]{cccccc}0&0&0&0&0&0\\\ 0&0&0&0&0&0\\\
-c_{3}&0&0&0&0&0\\\ c_{2}&c_{4}&0&0&0&0\end{array}\right],\quad
A(2,3)=\left[\begin{array}[]{cccc}0&0&0&0\\\ 0&0&0&0\\\ 0&0&0&0\\\ 0&0&0&0\\\
c_{4}&0&0&0\\\ -c_{2}&-c_{3}&0&0\end{array}\right],\quad
A(3,4)=\left[\begin{array}[]{c}0\\\ 0\\\ 0\\\ 0\end{array}\right],$

so their ranks are
$\operatorname{rank}A(1,2)=\operatorname{rank}A(2,3)=\operatorname{rank}\begin{bmatrix}{c}_{2}&{c}_{3}\\\
{c}_{4}&0\end{bmatrix}$, and $\operatorname{rank}A(3,4)=0$.

If $c_{3}c_{4}\neq 0$, i.e., if $\pi$ is a “symplectic structure”, then
$\operatorname{rank}A(1,2)=\operatorname{rank}A(2,3)=2,\operatorname{rank}A(3,4)=0$
and the Betti numbers are (2,2,2,1).

If $c_{3}=c_{4}=0$ and $c_{2}=0$, then $\pi=c_{1}y_{1}\wedge y_{2}$ and
$\operatorname{rank}A(1,2)=\operatorname{rank}A(2,3)=\operatorname{rank}A(3,4)=0$
and the Betti numbers are (4,6,4,1).

If $c_{3}=c_{4}=0$ and $c_{2}\neq 0$, or if $c_{3}c_{4}=0$ and
$c_{3}^{2}+c_{4}^{2}>0$, then then
$\operatorname{rank}A(1,2)=\operatorname{rank}A(2,3)=1$ and
$\operatorname{rank}A(3,4)=0$ and the Betti numbers are (3,4,3,1).

######  Remark 2.5.

The Euler number is always $1$ because ${\mathfrak{g}}_{0}=\mathbb{R}^{4}$ and
${\mathfrak{g}}_{j}=\Lambda^{j+1}{\mathfrak{g}}_{0}$ for $j=0,\ldots,3$, but
$\Lambda^{0}{\mathfrak{g}}_{0}=\mathbb{R}$ is not counted.

######  Remark 2.6.

[1] gives us 12 different types of 4-dimensional Lie algebra. Type[1] has been
introduced above. For instance, the second one is given by
$\text{Type}[2]=\\{[y_{1},y_{4}]=ay_{1},[y_{2},y_{4}]=y_{2},[y_{3},y_{4}]=y_{2}+y_{3}\\}$.
It has a parameter $a$, the Poisson condition has five cases and so the story
becomes a little complicated, but still the same discussion works well as in
Type[1].

##  3 Poisson-like cohomology of superalgebra of differential forms

Let $M$ be a differential manifold. The direct sum
$\sum_{j=0}^{\textrm{dim}M}\Lambda^{j}\mathrm{T}^{*}(M)$ with the bracket

(3.1) $[\alpha,\beta]=(-1)^{a}d(\alpha\wedge\beta)\quad\text{where
}\quad\alpha\in\Lambda^{a}\mathrm{T}^{*}(M)$

forms a Lie superalgebra with the grading of $a$-form to be $-(a+1)$ (cf.
[3]). However, another grading of $a$-form by $a+1$ makes sense. It is clear
the identity map gives Lie superalgebra isomorphism between the two kinds of
grading.

We follow Proposition 2.1 and apply the same argument as in Example 2.1 to
$\sum_{j=0}^{\textrm{dim}M}\Lambda^{j}\mathrm{T}^{*}(M)$, and get the
following fact.

######  Proposition 3.1.

Assume $\varphi$ is an even $p$-form (i.e., the grade is $p+1$) on a manifold
$M$.

  1. 1.

Let $\varphi$ satisfy $[\varphi,\varphi]=d(\varphi\wedge\varphi)=0$. Then
$d_{\varphi}$ satisfies $d_{\varphi}\circ d_{\varphi}=0$ and defines Poisson-
like cohomology groups. The coboundary operator is
$d_{\varphi}(U)=d(\varphi\wedge U)$, in short, $d_{\varphi}=d\circ
e_{\varphi}=e_{d\varphi}+e_{\varphi}\circ d$ where $e_{\psi}(U)=\psi\wedge U$.

  2. 2.

Let $\varphi$ be a closed even $p$-form. Then $d_{\varphi}=e_{\varphi}\circ d$
and it defines Poisson-like cohomology groups.

In particular, let $\varphi$ be a constant function $1$, which is a 0-form on
$M$. The associated Poisson-like cohomology groups are just the de Rham
cohomology groups of $M$.

Although $\sum_{p=0}^{\textrm{dim}M}\Lambda^{p}\mathrm{T}^{*}(M)$ for a
manifold $M$ is a typical superalgebra, its chain spaces are infinite
dimensional, and it may be hard to manipulate those (co)homology groups. If
$M$ is a finite dimensional Lie group, then its invariant forms are finite
dimensional, and we would like to handle the superalgebra $\mathfrak{h}$
defined by $\mathfrak{h}_{[-1-j]}=\Lambda^{j}\mathfrak{m}^{*}$ for
$j=0,\ldots,\textrm{dim}\mathfrak{m}$ where $\mathfrak{m}$ is the Lie algebra
of a Lie group $M$ and $\mathfrak{m}^{*}$ is the dual space of $\mathfrak{m}$.
The super bracket is given by $[\alpha,\beta]=(-1)^{a}d(\alpha\wedge\beta)$
where $\alpha\in\Lambda^{a}\mathfrak{m}^{*}$.

If $\varphi\in\Lambda^{p}\mathfrak{m}^{*}$ with $p$ odd, then
$[\varphi,\varphi]=0$ holds automatically but it is not clear that
$d_{\varphi}\circ d_{\varphi}=0$ or not, when
$d_{\varphi}(U)=[\varphi,U]=(-1)^{p}d(\varphi\wedge U)$. One concrete example
is the following:

######  Example 3.1.

Start with 4-dimensional Lie algebra of
$\text{Type}[12]=\\{[y_{1},y_{2}]=0,[y_{1},y_{3}]=y_{1},[y_{1},y_{4}]=-y_{2},[y_{2},y_{3}]=y_{2},[y_{2},y_{4}]=y_{1},[y_{3},y_{4}]=0\\}$.
Then $d$ is given by $d(z_{1})=-z_{1}\wedge z_{3}-z_{2}\wedge z_{4}$,
$d(z_{2})=z_{1}\wedge z_{4}-z_{2}\wedge z_{3}$, $d(z_{3})=d(z_{4})=0$ where
$z_{i}$ are the dual of $y_{j}$ . Take a 1-form
$\varphi=\sum_{i=1}^{4}{c}_{i}z_{i}$. The matrix representations of
$d_{\varphi}$ are the followings.

$A(0,2)=\left[\begin{array}[]{cccccc}0&{c}_{1}&-{c}_{2}&{c}_{2}&{c}_{1}&0\end{array}\right],\
A(1,3)=\left[\begin{array}[]{cccc}2{c}_{2}&0&-{c}_{4}&{c}_{3}\\\
-2{c}_{1}&0&-{c}_{3}&-{c}_{4}\\\ 0&0&{c}_{2}&-{c}_{1}\\\
0&0&{c}_{1}&{c}_{2}\end{array}\right],\
A(2,4)=\left[\begin{array}[]{c}-2{c}_{4}\\\ 0\\\ 2{c}_{2}\\\ 0\\\ -2{c}_{1}\\\
0\end{array}\right],$

and A(3,5) = A(4,6) = O. Since
$d_{\varphi}^{2}(\Lambda^{0}\mathfrak{m}^{*})=-2\left({c}_{1}^{2}+{c}_{2}^{2}\right)\&^{\wedge}\left(z_{1},z_{2},z_{3},z_{4}\right)$
and $d_{\varphi}^{2}(\Lambda^{j}\mathfrak{m}^{*})=$ trivial for $j=1\ldots 4$.
Thus, if ${c}_{1}^{2}+{c}_{2}^{2}=0$ then $d_{\varphi}$ defines “cohomology
groups”. When ${c}_{3}={c}_{4}=0$ then the coboundary operator is trivial and
the sequence of Betti number is $[1,4,6,4,1]$, which are $\binom{4}{j}$ for
$j=0\ldots 4$.

If ${c}_{3}\neq 0$ and ${c}_{4}=0$ then non-trivial rank is
$\operatorname{rank}A(1,3)=2$ and the sequence of Betti numbers is
[1,2,6,2,1]. If ${c}_{4}\neq 0$ then non-trivial ranks are
$\operatorname{rank}A(1,3)=2$ and $\operatorname{rank}A(2,4)=1$, so the
sequence of Betti numbers is [1,2,5,2,0].

###  3.1 Poisson-like cohomology groups by 2-forms of dual Lie algebra

We study Poisson-like coboundary operator defined by a 2-form $\varphi$
satisfying $[\varphi,\varphi]=0$ on the Lie superalgebra
$\sum_{j=0}^{\textrm{dim}\mathfrak{m}}\Lambda^{j}\mathfrak{m}^{*}$ generated
by the dual space of finite dimensional Lie algebra $\mathfrak{m}$. Since
$[\varphi,\varphi]\in\Lambda^{2+2+1}\mathfrak{m}^{*}$ in general, if
$\textrm{dim}\mathfrak{m}\leqq 4$, then $[\varphi,\varphi]=0$ holds
automatically. This means every 2-form is a Poisson-like tensor if
$\textrm{dim}\mathfrak{m}\leqq 4$.

######  Example 3.2.

Let $\varphi$ be a 2-form on a dual space of a 4-dimensional Lie algebra.
Since $[\varphi,\varphi]$ is 5-form, $[\varphi,\varphi]=0$ holds in general.
Thus, every 2-form $\varphi$ defines coboundary operator $d_{\varphi}$ and
Poisson-like cohomologies.

######  Example 3.3.

Take a 5-dimensional Lie algebra

$\displaystyle\text{Type}[8]:=\\{$
$\displaystyle[y_{1},y_{2}]=[y_{1},y_{3}]=[y_{1},y_{4}]=[y_{1},y_{5}]=[y_{2},y_{3}]=[y_{2},y_{4}]=0,$
$\displaystyle[y_{2},y_{5}]=y_{1},[y_{3},y_{4}]=0,[y_{3},y_{5}]=y_{3},[y_{4},y_{5}]=uy_{4}\\}\;.$

The dual space has the relations

$\left[d(z_{1}),\ldots,d(z_{5})\right]=\left[-z_{2}{\wedge}z_{5},0,-z_{3}{\wedge}z_{5},-uz_{4}{\wedge}z_{5},0\right]$

where $u$ is a real constant. Let $\varphi$ be a 2-form

$\displaystyle\varphi=$ $\displaystyle{c}_{1}z_{1}\wedge
z_{2}+{c}_{2}z_{1}\wedge z_{3}+{c}_{3}z_{1}\wedge z_{4}+{c}_{4}z_{1}\wedge
z_{5}+{c}_{5}z_{2}\wedge z_{3}$ $\displaystyle+{c}_{6}z_{2}\wedge
z_{4}+{c}_{7}z_{2}\wedge z_{5}+{c}_{8}z_{3}\wedge z_{4}+{c}_{9}z_{3}\wedge
z_{5}+{c}_{10}z_{4}\wedge z_{5}$

where ${c}_{i}$ are constant for $i=1,\ldots,10$.

$[\varphi,\varphi]=0$ is equivalent to

(3.2) $2(1+u)({c}_{1}{c}_{8}-{c}_{2}{c}_{6}+{c}_{3}{c}_{5})=0\;.$

If $1+u=0$, any 2-form $\varphi$ satisfies $[\varphi,\varphi]=0$ and defines
coboundary operators and Poisson-like cohomologies. The matrix representations
of $d_{\varphi}$ are as follows.

$\displaystyle A(0,3)$
$\displaystyle=\left[\begin{array}[]{*{12}{c}}0&0&0&0&{c}_{2}&-{c}_{3}&0&{c}_{2}+{c}_{5}&{c}_{3}-{c}_{6}&0\end{array}\right]$
$\displaystyle A(1,4)$
$\displaystyle=\left[\begin{array}[]{*{5}{c}}0&-{c}_{5}&{c}_{6}&0&-{c}_{8}\\\
0&{c}_{2}&-{c}_{3}&0&0\\\ 0&-{c}_{1}&0&0&{c}_{3}\\\ 0&0&{c}_{1}&0&-{c}_{2}\\\
0&0&0&0&0\\\ \end{array}\right]$ $\displaystyle A(i,i+3)$
$\displaystyle=0\qquad(i=2,\ldots).$

If $1+u\neq 0$, then ${c}_{1}{c}_{8}-{c}_{2}{c}_{6}+{c}_{3}{c}_{5}=0$ holds if
and only if $[\varphi,\varphi]=0$.

If ${c}_{1}={c}_{2}={c}_{3}={c}_{5}={c}_{6}={c}_{8}=0$ then $d_{\varphi}=0$,
i.e., $A(i,i+3)=0$ for $i=0,\ldots$. Otherwise, we may assume ${c}_{8}\neq 0$
and so ${c}_{1}=({c}_{2}{c}_{6}-{c}_{3}{c}_{5})/{c}_{8}$. Then the matrix
representations of $d_{\varphi}$ are as follows.

$\displaystyle A(0,3)$
$\displaystyle=\left[\begin{array}[]{cccccccccc}0&0&0&0&{c}_{2}&u{c}_{3}&0&{c}_{2}+{c}_{5}&{c}_{3}+u{c}_{6}&{c}_{8}\left(u+1\right)\end{array}\right]\;,\quad\operatorname{rank}(A(0,3))=1\;,$
$\displaystyle A(1,4)$
$\displaystyle=\left[\begin{array}[]{ccccc}0&-{c}_{5}&-u{c}_{6}&-(u+1){c}_{8}&-{c}_{8}\\\
0&{c}_{2}&u{c}_{3}&0&-(u+1){c}_{8}\\\
0&\displaystyle\frac{-{c}_{2}{c}_{6}+{c}_{3}{c}_{5}}{{c}_{8}}&0&\left(u+1\right){c}_{3}&{c}_{3}+(u+1){c}_{6}\\\
0&0&\displaystyle-\frac{\left({c}_{2}{c}_{6}-{c}_{3}{c}_{5}\right)u}{{c}_{8}}&-(u+1){c}_{2}&-{c}_{2}-\left(u+1\right){c}_{5}\\\
0&0&0&0&0\end{array}\right]\;,\quad\operatorname{rank}(A(1,4))=2\;,$
$\displaystyle A(2,5)$
$\displaystyle=(u+1)\left[\begin{array}[]{*{10}{c}}{c}_{8}&-{c}_{6}&{c}_{5}&0&{c}_{3}&-{c}_{2}&0&\displaystyle\frac{\left({c}_{2}{c}_{6}-{c}_{3}{c}_{5}\right)}{{c}_{8}}&0&0\end{array}\right]^{T}\;,\qquad\operatorname{rank}(A(2,5))=1\;.$

##  4 Dual Poisson-like coboundary operator and cohomologies

We already handle two kinds of Lie superalgebras, one is the direct sum of
tangential multi-vector fields with the Schouten bracket and the other is the
direct sum of differential forms with a Lie super bracket given by
$\displaystyle[\alpha,\beta]=(-1)^{a}d(\alpha\wedge\beta)$ where $\alpha$ is
an $a-$form. In the context of Poisson-like cohomology theory, we have two
long “exact sequences” as follows.

${\Lambda^{1}\mathfrak{m}}$${\Lambda^{2}\mathfrak{m}}$${\cdots}$${\Lambda^{m-1}\mathfrak{m}}$${\Lambda^{m}\mathfrak{m}}$${{0}}$${\Lambda^{0}\mathfrak{m}^{*}}$${\Lambda^{1}\mathfrak{m}^{*}}$${\Lambda^{2}\mathfrak{m}^{*}}$${\cdots}$${\Lambda^{m-1}\mathfrak{m}^{*}}$${\Lambda^{m}\mathfrak{m}^{*}}$${{0}}$$\scriptstyle{d_{\pi}}$$\scriptstyle{d_{\pi}}$$\scriptstyle{d_{\pi}}$$\scriptstyle{d_{\pi}}$$\scriptstyle{d_{\pi}}$$\scriptstyle{d}$$\scriptstyle{d}$$\scriptstyle{d}$$\scriptstyle{d}$$\scriptstyle{d}$$\scriptstyle{d}$

where $\displaystyle\mathfrak{m}=\mathrm{T}(M)$ for a differentiable
$m$-dimensional manifold $M$. There may be a duality between multi-vector
fields and differential forms of a finite dimensional Lie algebra in some
naive sense, we may consider the dual object of Poisson-like coboundary
operators and Poisson-like cohomology groups in the case.

######  Example 4.1.

As the first experiment, we handle the tangential Lie superalgebra associated
with the 4-dimensional Lie algebra $\mathfrak{m}$ which is called Type[1] in
[1] explained in Example 2.1, and the Poisson-like coboundary operator which
comes from the constant function 1 explained in Proposition 3.1 (2) on the
cotangent Lie superalgebra associated with the 4-dimensional Lie algebra
$\mathfrak{m}$. Namely, we have two sequences as above. In a “naive” sense, we
may define

(4.1) $\langle\Omega,d_{\pi}U\rangle=\langle\delta\Omega,U\rangle\quad\text{
for }\text{ a $p$-form }\Omega\text{ and a $(p-1)$-vector }U$

where $\langle\cdot,\cdot\rangle$ is the natural pairing.

We revisit Example 2.1. The Poisson structures are given by

$\displaystyle\pi$
$\displaystyle=y_{1}\wedge(c_{1}y_{2}+c_{2}y_{3}+c_{3}y_{4})+c_{4}y_{2}\wedge
y_{3}\;,\text{ and
}\;\pi\wedge\pi=2c_{3}c_{4}{\&}^{\wedge}\\!\left(y_{1},y_{2},y_{3},y_{4}\right)\;.$
$\displaystyle
d_{\pi}[y_{1},\ldots,y_{4}]=[d_{\pi}(y_{1}),\ldots,d_{\pi}(y_{4})]=\left[0,0,-c_{3}y_{1}{\wedge}y_{2},c_{4}y_{1}{\wedge}y_{3}+c_{2}y_{1}{\wedge}y_{2}\right]$
$\displaystyle d_{\pi}[y_{1}\wedge y_{2},\ldots,y_{3}\wedge
y_{4}]=\left[0,0,0,0,c_{4}{\&^{\wedge}}\\!\left(y_{1},y_{2},y_{3}\right),-c_{2}{\&^{\wedge}}\\!\left(y_{1},y_{2},y_{3}\right)-c_{3}{\&^{\wedge}}\\!\left(y_{1},y_{2},y_{4}\right)\right]$
$\displaystyle d_{\pi}[y_{1}\wedge y_{2}\wedge y_{3},\ldots,y_{2}\wedge
y_{3}\wedge y_{4}]=\left[0,0,0,0\right]\;.$

The Lie algebra structure defined by $[y_{2},y_{4}]=y_{1}$,
$[y_{3},y_{4}]=y_{2}$ induces the exterior differentiation $d$ given by
$dz_{1}=-z_{2}\wedge z_{4}\;,\ dz_{2}=-z_{3}\wedge z_{4}\;,\ dz_{3}=0\;,\
dz_{4}=0\;$, where $\\{\;z_{i}\mid i=1..4\;\\}$ are the dual basis of
$\\{\;y_{i}\mid i=1..4\;\\}$. From (4.1), $\delta\Omega$ is given by
$\displaystyle\delta\Omega=\sum_{U}\langle\Omega,d_{\pi}U\rangle\text{Dual}\;U$.
Thus,

$\displaystyle\delta[z_{1},\ldots,z_{4}]=[\delta(z_{1}),\ldots,\delta(z_{4})]=[0,0,0,0]$
$\displaystyle\delta[z_{1}\wedge z_{2},\ldots,z_{3}\wedge
z_{4}]=[-{c}_{3}z_{3}+{c}_{4}z_{4},{c}_{4}z_{4},0,0,0,0]$
$\displaystyle\delta[z_{1}\wedge z_{2}\wedge z_{3},\ldots,z_{2}\wedge
z_{3}\wedge z_{4}]=[{c}_{4}z_{2}\wedge z_{4}-{c}_{2}z_{3}\wedge
z_{4},-{c}_{3}z_{3}\wedge z_{4},0,0]$ $\displaystyle\delta[z_{1}\wedge
z_{2}\wedge z_{3}\wedge z_{4}]=[0]$

And we get the following table.

$\begin{array}[]{c|*{5}{c}}&0&1&2&3&4\\\ \hline\cr\text{ Dim}&1&4&6&4&1\\\
\text{ ImageDim}&0&0&r&r&0\\\ \text{ Betti}&1&4-r&2(3-r)&4-r&1\end{array}$

where $r=\operatorname{rank}\text{ of 2x2 matrix
}\begin{bmatrix}{c}_{2}&{c}_{3}\\\ {c}_{4}&0\end{bmatrix}$. By taking
$\displaystyle\delta=\delta_{\pi}=$ the dual of $d_{\pi}$, we get
“2-dimensional cochain complex” with an anchor the cochain complex of de Rham
as below.

${\Lambda^{0}\mathfrak{m}^{*}}$${\Lambda^{1}\mathfrak{m}^{*}}$${\Lambda^{2}\mathfrak{m}^{*}}$${\Lambda^{3}\mathfrak{m}^{*}}$${\Lambda^{4}\mathfrak{m}^{*}}$${{0}}$${\Lambda^{0}\mathfrak{m}^{*}}$${\Lambda^{1}\mathfrak{m}^{*}}$${\Lambda^{2}\mathfrak{m}^{*}}$${\Lambda^{3}\mathfrak{m}^{*}}$${\Lambda^{4}\mathfrak{m}^{*}}$${{0}}$$\scriptstyle{d}$$\scriptstyle{d}$$\scriptstyle{\delta_{\pi}}$$\scriptstyle{d}$$\scriptstyle{\delta_{\pi}}$$\scriptstyle{d}$$\scriptstyle{\delta_{\pi}}$$\scriptstyle{d}$$\scriptstyle{\delta_{\pi}}$$\scriptstyle{d}$$\scriptstyle{d}$$\scriptstyle{d}$$\scriptstyle{d}$$\scriptstyle{d}$

In this example, $\delta\circ d=0$ and $d\circ\delta=0$ on each stage.

######  Example 4.2.

The 4-dimensional Lie algebra $\mathfrak{m}$ Type[2] (
$[y_{1},y_{4}]=ay_{1},[y_{2},y_{4}]=y_{2},[y_{3},y_{4}]=y_{1}+y_{2}$) in [1]
gives non trivial case. The exterior differentiation $d$ is described as
follows.

$\displaystyle d([z_{1},\ldots,z_{4}])$ $\displaystyle=[-az_{1}\wedge
z_{4},-z_{2}\wedge z_{4}-z_{3}\wedge z_{4},-z_{3}\wedge z_{4},0]$
$\displaystyle d([z_{1}\wedge z_{2},\ldots])$
$\displaystyle=[(a+1)\mathit{\&^{\wedge}}(z_{1},z_{2},z_{4})+\mathit{\&^{\wedge}}(z_{1},z_{3},z_{4}),(a+1)\mathit{\&^{\wedge}}(z_{1},z_{3},z_{4}),0,2\mathit{\&^{\wedge}}(z_{2},z_{3},z_{4}),0,0]$
$\displaystyle d([\mathit{\&^{\wedge}}(z_{1},z_{2},z_{3}),\ldots])$
$\displaystyle=[(-a-2)\mathit{\&^{\wedge}}(z_{1},z_{2},z_{3},z_{4}),0,0,0]$
$\displaystyle d([\mathit{\&^{\wedge}}(z_{1},z_{2},z_{3},z_{4})])$
$\displaystyle=[0]$

A candidate of Poisson tensor is $\pi={C}_{1}y_{1}\wedge
y_{2}+{C}_{2}y_{1}\wedge y_{3}+{C}_{3}y_{1}\wedge y_{4}+{C}_{4}y_{2}\wedge
y_{3}+{C}_{5}y_{2}\wedge y_{4}+{C}_{6}y_{3}\wedge y_{4}$ for constants
${{C}_{1},{C}_{2},{C}_{3},{C}_{4},{C}_{5},{C}_{6}}$. Since

$\displaystyle[\pi,\pi]=$
$\displaystyle(-a{C}_{1}{C}_{6}+a{C}_{2}{C}_{5}-{C}_{1}{C}_{6}+{C}_{2}{C}_{5}-{C}_{2}{C}_{6}-2{C}_{3}{C}_{4})\mathit{\&^{\wedge}}(y_{1},y_{2},y_{3})$
$\displaystyle+(a{C}_{3}{C}_{5}-{C}_{3}{C}_{5}-{C}_{3}{C}_{6})\mathit{\&^{\wedge}}(y_{1},y_{2},y_{4})$
$\displaystyle+(a{C}_{3}{C}_{6}-{C}_{3}{C}_{6})\mathit{\&^{\wedge}}(y_{1},y_{3},y_{4})+{C}_{6}^{2}\mathit{\&^{\wedge}}(y_{2},y_{3},y_{4})$

the Poisson conditions are,

$\displaystyle-a{C}_{1}{C}_{6}+a{C}_{2}{C}_{5}-{C}_{1}{C}_{6}+{C}_{2}{C}_{5}-{C}_{2}{C}_{6}-2{C}_{3}{C}_{4}=0,$
$\displaystyle a{C}_{3}{C}_{5}-{C}_{3}{C}_{5}-{C}_{3}{C}_{6}=0,\qquad
a{C}_{3}{C}_{6}-{C}_{3}{C}_{6}=0,\qquad{C}_{6}^{2}=0\;.$

Solving the above, we get 5 cases as follows.

$\displaystyle\\{a=-1,{C}_{1}={C}_{1},{C}_{2}={C}_{2},{C}_{3}=0,{C}_{4}={C}_{4},{C}_{5}={C}_{5},{C}_{6}=0\\}\;,$
$\displaystyle\\{a=1,{C}_{1}={C}_{1},{C}_{2}={C}_{2},{C}_{3}={C}_{3},{C}_{4}={C}_{2}{C}_{5}/{C}_{3},{C}_{5}={C}_{5},{C}_{6}=0\\}\;,$
$\displaystyle\\{a=a,{C}_{1}={C}_{1},{C}_{2}=0,{C}_{3}=0,{C}_{4}={C}_{4},{C}_{5}={C}_{5},{C}_{6}=0\\}\;,$
$\displaystyle\\{a=a,{C}_{1}={C}_{1},{C}_{2}={C}_{2},{C}_{3}=0,{C}_{4}={C}_{4},{C}_{5}=0,{C}_{6}=0\\}\;,$
$\displaystyle\\{a=a,{C}_{1}={C}_{1},{C}_{2}={C}_{2},{C}_{3}={C}_{3},{C}_{4}=0,{C}_{5}=0,{C}_{6}=0\\}\;.$

Case 1: $\pi={C}_{1}y_{1}\wedge y_{2}+{C}_{2}y_{1}\wedge
y_{3}+{C}_{4}y_{2}\wedge y_{3}+{C}_{5}y_{2}\wedge y_{4}$, where
${a=-1,{C}_{3}=0,{C}_{6}=0}$.
$\pi\wedge\pi=-2{C}_{2}{C}_{5}\mathit{\&^{\wedge}}(y_{1},y_{2},y_{3},y_{4})$.
Poisson coboundary operator $d_{\pi}$ is

$\displaystyle d_{\pi}[y_{1},\ldots,y_{4}]$
$\displaystyle=[-{C}_{5}y_{1}\wedge y_{2},0,-{C}_{5}y_{2}\wedge
y_{3},{C}_{2}y_{1}\wedge y_{2}+2{C}_{4}y_{2}\wedge y_{3}+{C}_{5}y_{2}\wedge
y_{4}]$ $\displaystyle d_{\pi}[y_{1}\wedge y_{2},\ldots]$
$\displaystyle=[0,0,-2{C}_{5}\mathit{\&^{\wedge}}(y_{1},y_{2},y_{4})-2{C}_{4}\mathit{\&^{\wedge}}(y_{1},y_{2},y_{3}),0,0,-{C}_{2}\mathit{\&^{\wedge}}(y_{1},y_{2},y_{3})]$
$\displaystyle d_{\pi}[\mathit{\&^{\wedge}}(y_{1},y_{2},y_{3}),\ldots]$
$\displaystyle=[0,0,-{C}_{5}\mathit{\&^{\wedge}}(y_{1},y_{2},y_{3},y_{4}),0]\;.$

The dual operator $\delta$ of $d_{\pi}$ is given by

$\displaystyle\delta[\mathit{\&^{\wedge}}(z_{1},\ldots,z_{4})]$
$\displaystyle=[-{C}_{5}\mathit{\&^{\wedge}}(z_{1},z_{3},z_{4})]$
$\displaystyle\delta[\mathit{\&^{\wedge}}(z_{1},z_{2},z_{3}),\ldots]$
$\displaystyle=[-2{C}_{4}z_{1}\wedge z_{4}-z_{3}\wedge
z_{4}{C}_{2},-2{C}_{5}z_{1}\wedge z_{4},0,0]$ $\displaystyle\delta[z_{1}\wedge
z_{2},\ldots]$
$\displaystyle=[{C}_{2}z_{4}-{C}_{5}z_{1},0,0,2{C}_{4}z_{4}-{C}_{5}z_{3},{C}_{5}z_{4},0]$
$\displaystyle\delta[z_{1},\ldots,z_{4}]$ $\displaystyle=[0,0,0,0]$

We verify $\delta\circ\delta=0$ directly from the above relations. In this
case, $d\circ\delta$ and $\delta\circ d$ are not trivial and satisfy

$\displaystyle(d\circ\delta)\text{ and }(\delta\circ d)\text{ of
}[\mathit{\&^{\wedge}}(z_{1},\ldots,z_{4})]$ $\displaystyle=[0]\text{ and
}[0]$ $\displaystyle(d\circ\delta)\text{ and }(\delta\circ d)\text{ of
}[\mathit{\&^{\wedge}}(z_{1},z_{2},z_{3}),\ldots]$
$\displaystyle=[0,0,0,0]\text{ and
}[{C}_{5}\mathit{\&^{\wedge}}(z_{1},z_{3},z_{4}),0,0,0]$
$\displaystyle(d\circ\delta)\text{ and }(\delta\circ d)\text{ of }[z_{1}\wedge
z_{2},\ldots]$ $\displaystyle=[-{C}_{5}z_{1}\wedge
z_{4},0,0,{C}_{5}z_{3}\wedge z_{4},0,0]\text{ and }[0,0,0,0,0,0]$
$\displaystyle(d\circ\delta)\text{ and }(\delta\circ d)\text{ of
}[z_{1},z_{2},\ldots]$ $\displaystyle=[0,0,0,0]\text{ and
}[0,-{C}_{5}z_{4},0,0]\;.$

We leave the handling of other cases to readers. Case 2, 3 and 5 are non-
trivial, namely $\delta\circ d\neq 0$ and $d\circ\delta\neq 0$ in some stage.
Even $\delta$ in Case 4 is non-trivial but $\delta\circ d=0$ and
$d\circ\delta=0$.

######  Remark 4.1.

We constructed the dual $\delta$ of Poisson cohomology operator $d_{\pi}$
formally. It may be interesting to study the geometric meanings of
$\delta\circ d$ or $d\circ\delta$ for the original Poisson structure $\pi$.

###  4.1 Multi-vector fields and differential forms with polynomial
coefficients on $\mathbb{R}^{n}$

As explained several times, two prototypes of Lie superalgebras are

  1. 1.

$\sum_{i=1}^{\textrm{dim}M}\Lambda^{i}\mathrm{T}(M)$ with the Schouten bracket
$[\cdot,\cdot]_{\text{\footnotesize S}}$ (cf. [2]), and

  2. 2.

$\sum_{i=0}^{\textrm{dim}M}\Lambda^{i}\mathrm{T}^{*}(M)$ with
$[\alpha,\beta]=(-1)^{a}d(\alpha\wedge\beta)$ where
$\alpha\in\Lambda^{a}\mathrm{T}^{*}(M)$ (cf. [3]).

As concrete examples of finite dimensional sub superalgebras, we know the
space of multivector fields with homogeneous polynomials as coefficient on
$n$-dimensional number space $\mathbb{R}^{n}$ with the Cartesian coordinates
$x_{1},\ldots,x_{n}$ by the Schouten bracket (cf. [2]), and the differential
forms with homogeneous polynomials with the modified super bracket by the
exterior differentiation $d$. We denote $d(x_{i})=z_{i}$, then $d(z_{i})=0$
holds and $d(P)=\sum_{i=1}^{n}\frac{\partial P}{\partial x_{i}}z_{i}$ for a
polynomial $P$, and we see the following.

(4.2)
$d({\mathfrak{C}}^{m}_{k})\subset{\mathfrak{C}}^{m+1}_{k-1}\quad\text{where}\quad{\mathfrak{C}}^{m}_{k}=m\text{-forms
with }k\text{-homogeneous polynomials as coefficient}\;.$

In the same way, we set $\displaystyle C_{k}^{m}=m\text{-vector fields with
}k\text{-homogeneous polynomials}$.

######  Proposition 4.1.

(4.3) $\textrm{dim}{\mathfrak{C}}^{m}_{k}\ \text{ and }\
\textrm{dim}C_{k}^{m}\quad\text{are both equal to }\
\tbinom{n}{m}\tbinom{n-1+k}{n-1}.$

Thus, $C_{k}^{m}=\\{0\\}$ and ${\mathfrak{C}}^{m}_{k}=\\{0\\}$ when $m>n$, and
$C_{k}^{m}=\\{0\\}$ and ${\mathfrak{C}}^{m}_{k}=\\{0\\}$ when $k<0$.

######  Remark 4.2.

The relation (4.2) shows the sum $m+k$ of two indices $m$ and $k$ of
${\mathfrak{C}}^{m}_{k}$ is “invariant” under the operation $d$.

######  Proposition 4.2.

Let $\pi\in C_{h}^{2}$ with $[\pi,\pi]=0$, namely $h$-homogeneous Poisson
2-vector field on $\mathbb{R}^{n}$, and let
$d_{\pi}=[\pi,\cdot]_{\text{\footnotesize S}}$. It holds that

(4.4) $d_{\pi}(C_{k}^{m})\subset C_{h+k-1}^{m+1}\;.$

######  Remark 4.3.

Looking for some “invariant” by two indices $m,k$ of $C_{k}^{m}$ in (4.4), the
sum $m+k$ is invariant when $h=0$ and the difference $m-k$ is invariant when
$h=2$.

######  Definition 4.1.

Take a 2-homogeneous Poisson 2-tensor field $\pi$ on $\mathbb{R}^{n}$. Let
$\mathit{Vol}=\mathit{\&^{\wedge}}(z_{1},\ldots,z_{n})$ (the standard volume
form). By the natural pairing $\langle\mathit{Vol},U\rangle$ of $U\in
C_{k}^{m}$ and $\mathit{Vol}$, we get
$\langle\mathit{Vol},U\rangle\in{\mathfrak{C}}^{n-m}_{k}$ and get the dual
$\delta$ of $d_{\pi}$ as below.

${C_{k}^{m}\ni U}$${[\pi,U]_{\text{\footnotesize S}}\in
C_{k+1}^{m+1}}$${{\mathfrak{C}}^{n-m}_{k}\ni\langle\mathit{Vol},U\rangle}$${\langle\mathit{Vol},[\pi,U]_{\text{\footnotesize
S}}\rangle\in{\mathfrak{C}}^{n-(m+1)}_{k+1}}$$\scriptstyle{d_{\pi}}$$\scriptstyle{vol}$$\scriptstyle{vol}$$\scriptstyle{\delta}$

Thus, there is a chain complex:

(4.5)
${\mathfrak{C}}^{n}_{k}\mathop{\longrightarrow}^{\delta}{\mathfrak{C}}^{n-1}_{k+1}\mathop{\longrightarrow}^{\delta}\ldots\mathop{\longrightarrow}^{\delta}{\mathfrak{C}}^{1}_{k+n-1}\mathop{\longrightarrow}^{\delta}{\mathfrak{C}}^{0}_{k+n}\;.$

Since $C_{2}^{2}$ is spanned by the product of $\\{x_{i}x_{j}\mid 1\leqq
i\leqq j\leqq n\\}$ and $\\{y_{i}\wedge y_{j}\mid 1\leqq i<j\leqq n\\}$, a
general 2-homogeneous 2-vector field $\pi$ is given by

(4.6) $\pi=\sum_{i\leqq j,k<\ell}C[i,j:k,\ell]x_{i}x_{j}y_{k}\wedge
y_{\ell}\quad\text{
with}\quad\tbinom{n}{2}\tbinom{n-1+2}{2}\;\text{-parameters}\quad
C[i,j:k,\ell]\ .$

The Poisson conditions are given by

$\displaystyle[\pi,\pi]=$ $\displaystyle\sum\sum
C[i,j:k,\ell]C[i^{\prime},j^{\prime}:k^{\prime},\ell{}^{\prime}][x_{i}x_{j}y_{k}\wedge
y_{\ell},x_{i^{\prime}}x_{j^{\prime}}y_{k^{\prime}}\wedge y_{\ell^{\prime}}]$
$\displaystyle=$ $\displaystyle\sum\sum
C[i,j:k,\ell]C[i^{\prime},j^{\prime}:k^{\prime},\ell{}^{\prime}]\left(x_{i}x_{j}[y_{\ell},x_{i^{\prime}}x_{j^{\prime}}]\mathit{\&^{\wedge}}(y_{k},y_{k^{\prime}},y_{\ell^{\prime}})-x_{i}x_{j}[y_{k},x_{i^{\prime}}x_{j^{\prime}}]\mathit{\&^{\wedge}}(y_{\ell},y_{k^{\prime}},y_{\ell^{\prime}})\right.$
$\displaystyle\left.\hskip 156.49014pt-
x_{i^{\prime}}x_{j^{\prime}}[y_{k^{\prime}},x_{i}x_{j}]\mathit{\&^{\wedge}}(y_{k},y_{\ell},y_{\ell^{\prime}})+x_{i^{\prime}}x_{j^{\prime}}[y_{\ell^{\prime}},x_{i}x_{j}]\mathit{\&^{\wedge}}(y_{k},y_{\ell},y_{k^{\prime}})\right)\;.$

For instance, let $n=4$. Then we need 60 parameters $C[i,j:k,\ell]$. Now we
denote them in 1-dimensional way like $C[1],\ldots,C[60]$. The Poisson
conditions are given by 80 bi-homogeneous equations. Additionally, we care
about the rank of Poisson structure. The rank is 4 if $\pi\wedge\pi\neq 0$ in
this case, and is 2 if $\pi\wedge\pi=0$ and $\pi\neq 0$. Still we have gotten
the equations, it is hard to show the all of them here, nor to solve the given
equations by hand, so we ask help by Symbol calculus. It provides general
solutions of 51 types. One of them is the following.

The independent variables are
$\\{\text{C}_{4},\text{C}_{7},\text{C}_{10},\text{C}_{14},\text{C}_{37},\text{C}_{40}\\}$.
The dependencies are

$\displaystyle\text{C}_{1}=0,\text{C}_{2}=0,\text{C}_{3}=0,\text{C}_{5}=0,\text{C}_{6}=0,\text{C}_{8}=0,\text{C}_{9}=0,\text{C}_{11}=0,\text{C}_{12}=0,\text{C}_{13}=0,\text{C}_{15}=0,\text{C}_{16}=0,$
$\displaystyle\text{C}_{17}=\frac{\text{C}_{7}\left(\text{C}_{14}-\text{C}_{37}\right)}{\text{C}_{4}},\text{C}_{18}=0,\text{C}_{19}=0,\text{C}_{20}=-\frac{\text{C}_{7}\text{C}_{40}-\text{C}_{10}\text{C}_{14}+\text{C}_{10}\text{C}_{37}}{\text{C}_{4}},\text{C}_{21}=0,\text{C}_{22}=0,\text{C}_{23}=0,$
$\displaystyle\text{C}_{24}=0,\text{C}_{25}=0,\text{C}_{26}=0,\text{C}_{27}=0,\text{C}_{28}=0,\text{C}_{29}=0,\text{C}_{30}=0,\text{C}_{31}=0,\text{C}_{32}=0,\text{C}_{33}=0,\text{C}_{34}=0,\text{C}_{35}=0,$
$\displaystyle\text{C}_{36}=0,\text{C}_{38}=0,\text{C}_{39}=0,\text{C}_{41}=0,\text{C}_{42}=0,\text{C}_{43}=0,\text{C}_{44}=0,\text{C}_{45}=0,\text{C}_{46}=0,\text{C}_{47}=0,\text{C}_{48}=0,\text{C}_{49}=0,$
$\displaystyle\text{C}_{50}=0,\text{C}_{51}=0,\text{C}_{52}=0,\text{C}_{53}=0,\text{C}_{54}=0,\text{C}_{55}=0,\text{C}_{56}=0,\text{C}_{57}=0,\text{C}_{58}=0,\text{C}_{59}=0,\text{C}_{60}=-\text{C}_{37}\;.$

Assume
$\text{C}_{4}=\text{C}_{37}=1,\text{C}_{10}=\text{C}_{14}=\text{C}_{40}=0$.
Then the Poisson tensor is

(4.7) $\pi=+(x_{1}+\text{C}_{7}x_{2})x_{4}y_{1}\wedge
y_{2}-\text{C}_{7}x_{2}x_{4}y_{1}\wedge y_{3}+x_{2}x_{4}y_{2}\wedge
y_{3}-x_{4}^{2}y_{3}\wedge y_{4}\;.$

######  Remark 4.4.

When given (4.7), $[\pi,\pi]=0$ and
$\pi\wedge\pi=-2(x_{1}+\text{C}_{7}x_{2})x_{4}^{3}\mathit{\&^{\wedge}}(y_{1},y_{2},y_{3},y_{4})$
are straightforward reconfirmations.

We only show the table of the Poisson cohomology groups derived from the
Poisson structure (4.7) the following left.

$\begin{array}[]{c|*{9}{c}}&C_{0}^{1}&\rightarrow&C_{1}^{2}&\rightarrow&C_{2}^{3}&\rightarrow&C_{3}^{4}\\\
\hline\cr\text{Dim}&4&&24&&40&&20\\\ \hline\cr\text{Rank}&3&&15&&10&&0\\\
\hline\cr\text{Betti}&1&&6&&15&&10\end{array}\qquad\qquad\begin{array}[]{c|*{10}{c}}&{\mathfrak{C}}^{3}_{0}&\rightarrow&{\mathfrak{C}}^{2}_{1}&\rightarrow&{\mathfrak{C}}^{1}_{2}&\rightarrow&{\mathfrak{C}}^{0}_{3}\\\
\hline\cr\text{Dim}&4&&24&&40&&20\\\ \hline\cr\text{Rank}&3&&15&&10&&0\\\
\hline\cr\text{Betti}&1&&6&&15&&10\end{array}$

The table right above shows the behavior of the dual of the Poisson cohomology
with respect the volume form
$\mathit{Vol}=\mathit{\&^{\wedge}}(z_{1},..,z_{4})$. There is the de Rham
cochain complex by $d$ on
${\mathfrak{C}}^{0}_{3}\rightarrow{\mathfrak{C}}^{1}_{2}\rightarrow{\mathfrak{C}}^{2}_{1}\rightarrow{\mathfrak{C}}^{3}_{0}$,
which is always trivial in the sense that the each Betti number is 0. And
there is a “double” complex as the following. It may be interesting to study
how the double complex shows properties of 2-homogeneous Poisson 2-vector
field.

${{\mathfrak{C}}^{0}_{3}}$${{\mathfrak{C}}^{1}_{2}}$${{\mathfrak{C}}^{2}_{1}}$${{\mathfrak{C}}^{3}_{0}}$${{0}}$${{\mathfrak{C}}^{0}_{3}}$${{\mathfrak{C}}^{1}_{2}}$${{\mathfrak{C}}^{2}_{1}}$${{\mathfrak{C}}^{3}_{0}}$${{0}}$$\scriptstyle{d}$$\scriptstyle{d}$$\scriptstyle{\delta_{\pi}}$$\scriptstyle{d}$$\scriptstyle{\delta_{\pi}}$$\scriptstyle{d}$$\scriptstyle{\delta_{\pi}}$$\scriptstyle{d}$$\scriptstyle{d}$$\scriptstyle{d}$$\scriptstyle{d}$

## References

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  * [3] Kentaro Mikami and Tadayoshi Mizutani  “Superalgebra structure on differential forms of manifold” , 2021  arXiv:2105.09738
  * [4] Kentaro Mikami and Tadayoshi Mizutani  “The second Betti number of doubly weighted homology groups of some pre Lie superalgebra”  In _Tohoku Math. J._ 74.2, 2022, pp. 301–311  DOI: 10.2748/tmj.20210208
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