Is there a "derived" Free $P$-algebra functor for an operad $P$?

2011-08-03T21:09:26Z

Recall that an operad (in vector spaces, say) $P$ consists of a collection of vector spaces $P(n)$ for $n\geq 0$, such that $P(n)$ is equipped with an action by the symmetric group $S_n$, with maps $P(n) \otimes P(k_1) \otimes \dots \otimes P(k_n) \to P(k_1+\dots k_n)$ for any $n,k_1,\dots,k_n$, subject to natural associativity constraints. Recall also that a $P$-algebra for an operad $P$ consists of a vector space $V$ and for each $n$ a map $P(n) \to \hom(V^{\otimes n},V)$ subject to natural associativity constraints. Recall finally that for any operad $P$, the functor from vector spaces to $P$-algebras that is adjoint to the Forgetful functor is: $$ \operatorname{Free}: V \mapsto \bigoplus_n \bigl(P(n) \otimes V^n \bigr)_{S_n} $$ where by $W_{S_n}$ I mean the coinvariants of the $S_n$-module $W$, i.e. $W_{S_n} = W \otimes_{\mathbb{K} S_n} 1$.

Let me switch to working with chain complexes rather than vector spaces. Then I can do a more refined operation than taking coinvariants. Namely, I can use the derived functor of coinvariants: you replace $W$ by a projective resolution, tensor, and consider the result up to quasiisomorphism. Which is to say "the complex that computes $\operatorname{Tor}$".

Question: What is the meaning of the ($\infty$-)functor that takes derived coinvariants rather than coinvariants in the above formula? What relationship does it play to the operad, etc.?

The type of answer I'm looking might be the following: "The $(\infty,1)$-functor so described is adjoint to the functor that forgets from the $(\infty,1)$-category of strongly homotopy $P$-algebras to the $(\infty,1)$-category of chain complexes."

Another direction that you could take the following question might be:

Question 2: What type of "Koszul duality" statements are there using the "derived coinvariants" version of $\operatorname{Free}$?

Answer by David Ben-Zvi for Is there a "derived" Free $P$-algebra functor for an operad $P$?

2011-08-04T14:57:56Z

Yes this is an expression of the free algebra functor, left adjoint to the forgetful functor, see Section 3.1.3 of Lurie's Higher Algebra.
Koszul duality in the (oo,1)-setting is treated very nicely in the work of John Francis, see in particular math/1104.0181, where eg the relation with deformation theory is explained (the tangent complex to an augmented O-algebra carrying an action of the Koszul dual operad, etc)

Is there a "derived" Free $P$-algebra functor for an operad $P$? - MathOverflow

Is there a "derived" Free $P$-algebra functor for an operad $P$?

Answer by David Ben-Zvi for Is there a "derived" Free $P$-algebra functor for an operad $P$?