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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.

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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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