2
$\begingroup$

For my own convenience I'll work in $\infty$-categories, feel free to answer in whatever framework best suits you. My question is essentially how to show, given an $E_\infty$-ring object $R$ in an $\infty$-category $C$ and another object $M$ with a map $R\otimes M\to M$, that the pair $(R,M)$ is in fact an algebra, as Lurie calls it in Higher Algebra, for the $\infty$-operad $LM^\otimes$. In other words, how to show that $M$ is a module over $R$ with respect to all of $R$'s structure (not just the structure of being "homotopy associative and commutative").

One idea I have is the following: I'd like to produce a map of $\infty$-operads $F:LM^\otimes\to C^\otimes$, where $C^\otimes\to Fin_\ast$ is the coCartesian fibration determining the symmetric monoidal structure on $C$. To produce $F$ I need to have that it takes inert morphisms of $LM^\otimes$ to inert morphisms of $C^\otimes$. Now, recall that the inert morphisms of $LM^\otimes$ are precisely morphisms which map to inert morphisms in $Fin_\ast$ along the forgetful map $LM^\otimes\to Fin_\ast$, which are morphisms $\langle n\rangle\to\langle m\rangle$ that represent $\langle m\rangle$ as a "quotient" of $\langle n\rangle$, where everything we quotient out by gets sent to $\ast\in\langle m\rangle=\{\ast,1,\ldots,m\}$.

However, to describe an $R$-action on $M$, we certainly never send $M$ to $\ast$. As such, the inert morphisms in $LM^\otimes$ shouldn't have anything to do with $M$ (it's just coming along for the ride, so-to-speak). In other words, we need that the morphisms describing the monoidal structure on $R$ (inside of $LM^\otimes$) go to the right place, but we don't have to check anything involving inert morphisms with respect to $M$. Hence a functor $LM^\otimes\to C^\otimes$ which picks out an associative algebra object $R$ of $C$ also picks out an object of $LMod(R)$. If $R$ is in fact $E_\infty$ then we also know that $LMod(R)\simeq Mod^{Comm}(R)$, giving us what we need to know.

Does this seem valid? I have very little experience working with $\infty$-operads, and can't find this sort of thing anywhere anyway, so I might be making some very silly mistakes.

Assuming my argument above doesn't make any sense, does anyone have any other ideas?

Thanks!

$\endgroup$
  • 2
    $\begingroup$ It's not entirely clear to me what you're asking, but if you have a map $a : R \otimes M \to M$ and want to turn it into an $R$-module then you still have an infinite amount of data left to specify: first you want a homotopy between the two maps $R \otimes R \otimes M \to M$ given by "apply $a$ twice" and "multiply and then apply $a$", and then similar things with more $R$s. (In an ordinary category these are just properties of the map $a$, but for $\infty$-categories they are extra data.) I expect you're well aware of this and are actually asking about something else, though! $\endgroup$ – Rune Haugseng Feb 3 '15 at 17:47
  • $\begingroup$ @RuneHaugseng Yeah, definitely. We fleshed this out a little bit more in the chat room (we being Saul and I). I guess the idea is that you've got all the edges and you know the diagrams commute "up to homotopy" but you don't have all the higher structure. $\endgroup$ – Jonathan Beardsley Feb 3 '15 at 19:48
  • $\begingroup$ It seems that 4.4.2.10 of Higher Algebra gives some sort of handle on this kind of thing. $\endgroup$ – Jonathan Beardsley Feb 3 '15 at 19:49
1
$\begingroup$

I don't know that much of the $\infty$ story, so let me phrase my answer in terms of model categories.

Berger-Moerdijk's Resolution of coloured operads and rectification of homotopy algebras is a great reference for colored operads. In 1.5.1 of the arxiv version they point out that for any uncolored operad $P$ there is a 2-colored operad $Mod_P$ whose algebras are pairs $(R,M)$ where $M$ is an $R$-module and $R$ is a $P$-algebra. So for you, $P$ is an $E_\infty$ operad. This colored operad has been studied recently by Batanin and Berger (Tame Polynomial Monads paper) and by Gutierrez-Rondigs-Spitzweck-Ostvaer (a draft on colocalizations).

Notice that the construction of $Mod_P$ is a special case of the Grothendieck construction in category theory. For more, see The Grothendieck construction for model categories by Harpaz and Prasma. In recent work of mine and Batanin's, we generalize a portion of Batanin-Berger and prove that if $P$ is given by a tame polynomial monad then so is $Mod_P$. Rephrased in more classical language this says that if $\mathcal{M}$ is a model category and $P$ is $\Sigma$-cofibrant then so is $Mod_P$ (this is the case for $P=E_\infty$). This might have been known before, but we then use it to get several other punchlines. In particular, a theorem from Spitzweck's thesis (which has been generalized and appeared as Theorem 12.2.A in Fresse's book on operads and modules) proves that $Mod_P$-algebras inherit a semi-model structure from $\mathcal{M}$. The main result in Batanin-Berger bumps this up to a full model structure under certain hypotheses on $\mathcal{M}$. So this category of pairs $(R,M)$ has an $\infty$-category structure too.

Your question seems to be whether or not this $\infty$-category can be realized as the category of algebras over some $\infty$-operad. I am not 100% sure, but I think the cofibrancy on the operad $Mod_P$ implies the answer is yes. Rune Haugseng has a paper where he proves something like this for the operad $Ass$, and I imagine a similar proof will work for any $\Sigma$-cofibrant operad. Setting this up properly seems to be the problem. If I were you I'd talk to Gijs Heuts, since his work with Moerdijk seems the most natural (to a non-expert like me) for proving a statement of this form.

$\endgroup$
  • $\begingroup$ So in fact this category definitely is the $\infty$-category of algebras over an $\infty$-operad (the operad is Lurie's $LM^\otimes$). The issue is proving that a given pair $(R,M)$ is such an algebra. $\endgroup$ – Jonathan Beardsley Feb 3 '15 at 16:56
  • $\begingroup$ (Assuming that $R$ is an algebra for $Ass^\otimes$) $\endgroup$ – Jonathan Beardsley Feb 3 '15 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.