TITLE: Tensor product of two Poisson modules
QUESTION [1 upvotes]: Let $H$ be a Poisson algebra. A Poisson $H$-module is a vector space $V$ with two bilinear maps
\begin{align}
H \otimes V \to V \\
(h,v) \mapsto h.v,
\end{align}
\begin{align}
H \otimes V \to V \\
(h,v) \mapsto \{h,v\},
\end{align}
such that
\begin{align}
& (xy).v = x.(y.v),  \quad (1)　\\
& \{\{x,y\},v\} = \{x,\{y,v\}\} - \{y,\{x,v\}\}　\quad (2)\\
& \{xy,v\} = x.\{y,v\} + y.\{x,v\} \quad (3)\\
& \{x,y\}.v = \{x,y.v\} - y.\{x,v\},　\quad (4)
\end{align}
see for example, Section 2 on pages 2,3 in the paper.
Let $V, W$ be two Poisson modules. Is $V \otimes W$ a Poisson module? I think that we can define $\{x, v \otimes w\} = \{x, v\} \otimes w + v \otimes \{x, w\}$. Then $v \otimes w$ satisfies (2). How to define $x.(v \otimes w)$ such that $v \otimes w$ satisfies (1), (3), (4)?
If $v \otimes w$ satisfies (1), then we have to assume that $H$ is a bialgebra and define $x.(v \otimes w) = x_{(1)}.v \otimes x_{(2)}.w$. But it seems that (3) is not satisfied?
Thank you very much.

REPLY [1 votes]: In article you presented there is an assumption that our algebra is commutative, so tensor product of left $H$-modules is an left $H$-module with
$$x.(v\otimes w)=(x.v)\otimes w=v\otimes (x.w)$$ and simple calculations give you that required conditions are satisfied. 
As an example I give you a proof of (3):
$$\mathrm{LHS}=\{xy,v\otimes w\}=\{xy,v\}\otimes w+v\otimes \{xy,w\}\stackrel{(3)}{=}x.\{y,v\}\otimes w+y.\{x,v\}\otimes w +v\otimes x.\{y,w\}+v\otimes y.\{x,w\}=x.\left(\{y,v\}\otimes w + v\otimes \{y,w\}\right)+ \\+y.\left(\{x,v\}\otimes w+v\otimes\{x,w\}\right)=x.\{y,v\otimes w\}+y.\{x,v\otimes w\}=\mathrm{RHS}$$ 
Similarly for another identities.
In case $H$ is noncommutative you cannot define a left module in this way. You need to know that $V$ is a right module and $W$ is a left module, or you need to have a bialgebra structure on $H$ (see e.g. here) but the coalgebra maps should be compatible with Poisson structure, i.e. $\Delta(\{a,b\})=\{\Delta(a),\Delta(b)\}_{H\otimes H}$ and similarly for counit $\varepsilon$.
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