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Let $A$ be a Poisson super algebra ($A$ is a super algebra and $A$ satisfies super Jacobi identity, super commutativity, super Leibniz rule).

Super version of the product of two tensor products is \begin{align} (x \otimes y)(x' \otimes y') = (-1)^{|x'||y|} x x' \otimes y y', \ x, y,x',y' \in A. \end{align}

Are there some references about \begin{align} \{x \otimes y, x' \otimes y'\} = ? \end{align} Here $x, y,x',y' \in A$. Thank you very much.

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    $\begingroup$ $(-1)^{|x'||y|}(\{x,x'\}\otimes yy' + xx'\otimes \{y,y'\})$. $\endgroup$ – Gabriel C. Drummond-Cole Feb 1 '17 at 21:58
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To eliminate this from unanswered questions, let me elaborate on Gabriel's answer.

First of all, you should always remember that in the case of superalgebras all signs appear from the Koszul sign rule. Second, the tensor product of Poisson algebras is a Poisson algebra because the Poisson operad is a Hopf operad. Thinking in terms of operads is advantageous because all identities of the Poisson algebras appear without signs, and signs only arise when evaluating on homogeneous elements.

The Poisson operad $\mathcal{P}$ is generated by operations $\mu$ and $\beta$ (product and bracket) with $\mu(-,-)=\mu(-,-).(12)$ and $\beta(-,-)=-\beta(-,-).(12)$, subject to relations $$ \begin{cases} \mu(\mu(-,-),-)=\mu(-,\mu(-,-)),\\ \beta(\mu(-,-),-)=\mu(\beta(-,-),-).(23)+\mu(-,\beta(-,-)),\\ \beta(\beta(-,-),-)+\beta(\beta(-,-),-).(123)+\beta(\beta(-,-),-).(132)=0. \end{cases} $$ Its Hopf structure is a morphism of operads $\Delta\colon\mathcal{P}\to \mathcal{P}\otimes\mathcal{P}$ defined on generators as follows: $$ \begin{cases} \Delta(\mu)=\mu\otimes\mu,\\ \Delta(\beta)=\mu\otimes\beta+\beta\otimes\mu. \end{cases} $$ What happens when you evaluate all these on elements? Because of the Koszul sign rule, $\mu(-,-)=\mu(-,-).(12)$ and $\beta(-,-)=-\beta(-,-).(12)$ become $\mu(a_1,a_2)=(-1)^{|a_1||a_2|}\mu(a_2,a_1)$ and $\beta(a_1,a_2)=-(-1)^{|a_1||a_2|}\beta(a_2,a_1)$, since the symmetric monoidal structure on super-vector spaces is given by $(12) (a_1\otimes a_2)=(-1)^{|a_1||a_2|}(a_2\otimes a_1)$. Next, the identities are evaluated on $a_1\otimes a_2\otimes a_3$: $$ \begin{cases} \mu(\mu(a_1,a_2),a_3)=\mu(a_1,\mu(a_2,a_3)),\\ \beta(\mu(a_1,a_2),a_3)=(-1)^{|a_2||a_3|}\mu(\beta(a_1,a_3),a_2)+\mu(a_1,\beta(a_2,a_3)),\\ \beta(\beta(a_1,a_2),a_3)+(-1)^{|a_1|(|a_2|+|a_3|)}\beta(\beta(a_2,a_3),a_1)+(-1)^{|a_3|(|a_1|+|a_2|)}\beta(\beta(a_3,a_1),a_2)=0. \end{cases} $$ Finally, the coproduct definition is evaluated on elements of the form $a_1\otimes b_1\otimes a_2\otimes b_2$ of the tensor square of two Poisson algebras $A$ and $B$, which is identified with the product of the tensor square of $A$ and the tensor square of $B$ by means of the transposition of two middle factors, $a_1\otimes b_1\otimes a_2\otimes b_2\mapsto (-1)^{|a_2||b_1|}a_1\otimes a_2\otimes b_1\otimes b_2$: $$ \begin{cases} \Delta(\mu)(a_1\otimes b_1\otimes a_2\otimes b_2)=(-1)^{|a_2||b_1|}\mu(a_1,a_2)\otimes\mu(b_1,b_2),\\ \Delta(\beta)((a_1\otimes b_1\otimes a_2\otimes b_2))=(-1)^{|a_2||b_1|}(\mu(a_1,a_2)\otimes\beta(b_1,b_2)+\beta(a_1,a_2)\otimes\mu(b_1,b_2)). \end{cases} $$

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    $\begingroup$ thank you very much. Could your method be used to solve this problem? $\endgroup$ – Jianrong Li Feb 6 '17 at 14:27
  • $\begingroup$ @JianrongLi This is not a method as such, just unwrapping the definition. I will think about it, and will let you know. $\endgroup$ – Vladimir Dotsenko Feb 7 '17 at 1:16

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