4
$\begingroup$

Let $O$ be an operad, and let $D$ be a symmetric monoidal category. Then there is a forgetful functor $\operatorname{Alg}_O(D) \to D$. This functor is an equivalence in either of the following cases:

  1. If $D$ is cocartesian monoidal and $O = E_\infty$;

  2. If $D$ is cocartesian monoidal and $O = E_1$.

Question: For which operads $O$ does the statement

For all cocartesian monoidal $D$, the forgetful functor $\operatorname{Alg}_O(D) \to D$ is an equivalence.

hold?

Notes:

  • If $D$ is cocartesian monoidal, we have $D = \operatorname{Alg}_{E_\infty}(D)$, so that $\operatorname{Alg}_O(D) = \operatorname{Alg}_O(\operatorname{Alg}_{E_\infty}(D)) = \operatorname{Alg}_{O \otimes E_\infty}(D)$. So a sufficient condition is that the Boardman-Vogt tensor product have $O \otimes E_\infty = E_\infty$. By a theorem of Schlank and Yanovski, this holds whenever $O$ is reduced, i.e. $O$ is single-colored and $O(0) = O(1) = \ast$.

This covers a great many examples, but maybe it's true even more generally? And I'd be interested to understand the case of enriched operads as well.

Improve this question
$\endgroup$
6
  • $\begingroup$ I'm slightly confused by your use of the $E_\infty$ operad and not simply the "commutative" operad and the reference you cite. Is $D$ intended to be an $\infty$-category ? Otherwise I don't quite understand what you mean. At least one example not covered by the result you cite would be any operads with $O(1)=*$ and $O(i) = \emptyset$ for all $i >1$ , but with $O(0)$ arbitrary. $\endgroup$
    – Simon Henry
    Commented Nov 20, 2022 at 5:24
  • $\begingroup$ Oh sure -- I suppose I'm using an "implicit $\infty$" convention. $\endgroup$
    – Tim Campion ♦
    Commented Nov 20, 2022 at 15:57
  • $\begingroup$ I think the condition $E_\infty \otimes O \simeq E_\infty$ you are proposing is also necessary. In fact, it seems that two operads have the same co-algebras in all cartesian categories if and only if $E_\infty \otimes O \simeq E_\infty \otimes O'$. But I don't know if there is a way to rephrase it in a more practical way. $\endgroup$
    – Simon Henry
    Commented Nov 20, 2022 at 19:37
  • $\begingroup$ The idea being that, at least for 1-categories, the category of $E_\infty$-coalgebra is cartesian monoidal and model of $E_\infty \otimes O$ are models of $O$ in this cartesian monoidal category... I'm not sure if this sort of thing has been developed for $\infty$-categories though. $\endgroup$
    – Simon Henry
    Commented Nov 20, 2022 at 19:52
  • $\begingroup$ (By "model" I meant coalgebras) $\endgroup$
    – Simon Henry
    Commented Nov 20, 2022 at 19:59

0

Reset to default