$\newcommand{\Pois}{\mathit{Pois}}\newcommand{\Comm}{\mathit{Comm}}\newcommand{\Lie}{\mathit{Lie}}$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 $n$.

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    $\begingroup$ 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. $\endgroup$ Sep 14 at 17:07
  • $\begingroup$ I should add that the decomposition of the $\mathrm{Pois}_n$ I describe is $\mathrm{Pois}_n= \mathrm{com} \circ \mathrm{lie}[n]$. $\endgroup$ Sep 14 at 17:21
  • $\begingroup$ 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 $Lie$. $\endgroup$ Sep 18 at 7:03
  • $\begingroup$ @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. $\endgroup$ Sep 18 at 14:44
  • $\begingroup$ @AlessandroNanto that depends on what you mean by $\mathcal{O}_\infty$. To me it always means Cobar over the Koszul dual cooperad. $\endgroup$ Sep 18 at 15:50

2 Answers 2


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 $As\circ Lie$.

In this case, the only distributive law is the trivial one (see http://www.tac.mta.ca/tac/volumes/34/41/34-41.pdf). Therefore you won't get anything interesting.


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