Is there a "derived" Free $P$-algebra functor for an operad $P$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T15:31:56Z http://mathoverflow.net/feeds/question/72035 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72035/is-there-a-derived-free-p-algebra-functor-for-an-operad-p Is there a "derived" Free $P$-algebra functor for an operad $P$? Theo Johnson-Freyd 2011-08-03T21:09:26Z 2011-08-04T14:57:56Z <p>Recall that an <em>operad</em> (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 <em>$P$-algebra</em> 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. $W_{S_n} = W \otimes_{\mathbb{K} S_n} 1$.</p> <p>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 <em>derived</em> 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 $\operatorname{Tor}$".</p> <blockquote> <p><strong>Question:</strong> 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.?</p> </blockquote> <p>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."</p> <p>Another direction that you could take the following question might be:</p> <blockquote> <p><strong>Question 2:</strong> What type of "Koszul duality" statements are there using the "derived coinvariants" version of $\operatorname{Free}$?</p> </blockquote> http://mathoverflow.net/questions/72035/is-there-a-derived-free-p-algebra-functor-for-an-operad-p/72090#72090 Answer by David Ben-Zvi for Is there a "derived" Free $P$-algebra functor for an operad $P$? David Ben-Zvi 2011-08-04T14:57:56Z 2011-08-04T14:57:56Z <ol> <li><p>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. </p></li> <li><p>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) </p></li> </ol>