# Poisson and homotopy Poisson operads

$$\newcommand{\Pois}{\mathit{Pois}}\newcommand{\Comm}{\mathit{Comm}}\newcommand{\Lie}{\mathit{Lie}}$$For my thesis, I'm trying to understand the Poisson operad (I'll call it $$\Pois$$) and its homotopy counterpart $$\Pois_\infty$$.

In Loday-Valette, it is stated that $$\Pois=\Comm\circ \Lie$$, the operad structure relying on the existence of a distributive law $$\lambda:\Comm\circ \Lie\rightarrow \Lie\circ \Comm$$. Somewhere on the internet, I've read that $$\Pois_\infty$$ can be defined in a similar way, as the composition $$E_\infty \circ \Lie$$ (if I remember correctly).

I was wondering if there's any reference for this (if it is correct). Also, can we define $$\Pois_n:=E_n\circ \Lie$$? What if we use the cofibrant replacement $$\Lie_\infty$$ (the $$L_\infty$$ operad) instead of $$Lie$$, do we still have an operad structure? I assume by homotopy theory of operads that using $$\Lie_\infty$$ is not a problem, but I'm not so sure.

EDIT: As @Connor Malin rightfully points out, I'm using a different notation than usual. Here $$\Pois_n$$ means a Poisson structure up to homotopy, while normally it means a Poisson structure with Lie brackets of degree $$n$$.

• This is an interesting question, I just want to point out that you are intending $\mathrm{Pois}_n$ to mean "a homotopy Poisson algebra with coherences up to degree n", but I think standard practice is that $\mathrm{Pois}_n$ means a (graded) commutative algebra with a compatible Lie bracket of degree $n$ (or perhaps $n-1$), though someone can correct me if I am wrong. Sep 14 at 17:07
• I should add that the decomposition of the $\mathrm{Pois}_n$ I describe is $\mathrm{Pois}_n= \mathrm{com} \circ \mathrm{lie}[n]$. Sep 14 at 17:21
• I doubt it. The Koszul dual cooperad of $Pois=Comm\circ Lie$ should be something like $Pois^¡=Lie^{¡}\circ Comm^{¡}$. In order to get $Pois_\infty$, you have to apply the Cobar construction to this cooperad, so you loose the circle product and you have $Lie_\infty$ inside rather than just $Lie$. Sep 18 at 7:03
• @FernandoMuro, good point about $Pois_\infty$ containing $Lie_\infty$ rather than $Lie$, makes sense to me. I thought that the composition $E_\infty\circ Lie$ was something like an alternative model for $Pois_\infty$ than the one given by the cobar construction. Sep 18 at 14:44
• @AlessandroNanto that depends on what you mean by $\mathcal{O}_\infty$. To me it always means Cobar over the Koszul dual cooperad. Sep 18 at 15:50

I'll use the standard notation that $$\mathrm{Pois}_n$$ is the usual $$n$$-Poisson operad. I'll assume that you mean $$(\mathrm{Pois}_n)_\infty = \Omega(\mathrm{Pois}_n^¡)$$. Then no, $$(\mathrm{Pois}_n)_\infty$$ is not equal to either $$E_\infty \circ \mathrm{Lie}$$, $$E_\infty \circ L_\infty$$, or $$E_n \circ \mathrm{Lie}$$.
• The last one, $$E_n \circ \mathrm{Lie}$$, if it can even be defined meaningfully (you'd need a rewriting rule $$\mathrm{Lie} \circ E_n \to E_n \circ \mathrm{Lie}$$ and I don't know many interesting candidates) would have an altogether different type - it'd be an $$E_n$$-algebra equipped with an extra Lie bracket. Not what we want. The Lie bracket is already in $$E_n$$.
• $$E_\infty \circ \mathrm{Lie}$$ is not right: in the cofibrant resolution $$(\mathrm{Pois}_n)_\infty$$, the Lie bracket is relaxed up to homotopy, not just the commutative product. So that can't be right.
• Finally, in $$E_\infty \circ L_\infty$$, the commutative product and the Lie bracket would be relaxed up to homotopy. But that's not all there is to $$\mathrm{Pois}_n$$. The equality $$[a b, c] = a [b, c] \pm [a, c] b$$ in $$\mathrm{Pois}_n$$, a relation in the operad, does not hold strictly in a homotopy Poisson algebra. It becomes a homotopy. And then, per the usual yoga, there are homotopies between homotopies (to "kill" extraneous homology classes), homotopies between homotopies between homotopies, and so on.
As Fernando explained in the comment, $$(\mathrm{Pois}_n)_\infty$$ is the cobar construction of the Koszul dual of $$\mathrm{Pois}_n$$. That Koszul dual is isomorphic to $$\mathrm{Lie}_n^¡ \circ \mathrm{Com}^¡ = \mathrm{Com}_n^c \circ \mathrm{Lie}^c$$ (maybe with shifts...). In the cobar construction, you do get a copy of $$\Omega(\mathrm{Lie}_n^¡) = (\mathrm{Lie}_n)_\infty$$ and a copy of $$\Omega(\mathrm{Com}^¡) = \mathrm{Com}_\infty$$. But there's much more than that: there are also trees that involves operations from both cooperads.
In order to define an operad structure on $$E_n\circ Lie$$, you need to specify a distributive law. When $$n=1$$, let me use that $$E_n$$ is equivalent to $$As$$ and rather consider $$As\circ Lie$$.