# Super version of Poisson brackets of tensor products

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.

• $(-1)^{|x'||y|}(\{x,x'\}\otimes yy' + xx'\otimes \{y,y'\})$. – Gabriel C. Drummond-Cole Feb 1 '17 at 21:58

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, $a_1\otimes b_1\otimes a_2\otimes b_2\mapsto (-1)^{|a_2||b_1|}a_1\otimes a_2\otimes b_1\otimes b_2$: $$\begin{cases} \Delta(\mu)(a_1\otimes b_1\otimes a_2\otimes b_2)=(-1)^{|a_2||b_1|}\mu(a_1,a_2)\otimes\mu(b_1,b_2),\\ \Delta(\beta)((a_1\otimes b_1\otimes a_2\otimes b_2))=(-1)^{|a_2||b_1|}(\mu(a_1,a_2)\otimes\beta(b_1,b_2)+\beta(a_1,a_2)\otimes\mu(b_1,b_2)). \end{cases}$$