# Is it possible to define linear $A_\infty$-categories as special $\infty$-categories?

A functor $$N\colon\mathrm{Cat}_{A_\infty}\longrightarrow\mathrm{Cat}_\infty$$ is constructed in a paper [1] by Faonte. This gives a way to get an $$\infty$$-category by starting with an $$A_\infty$$-category.

Going the other way, is it possible to define linear $$A_\infty$$-categories as special $$\infty$$-categories?

## References

[1] Simplicial nerve of an A-infinity category (Giovanni Faonte, arXiv:1312.2127), suggested by DamienC in an answer to MO152370.

• If Faonte's functor is fully faithful, then yes. – David Roberts Jan 13 at 20:03
• I'm pretty sure that an $A_∞$-category is just going to be simply a $k$-linear $∞$-category, but I don't know if anyone has written a proof yet. I'm pretty sure also that it is well known that an $A_∞$-algebra (whatever definition you're using) is just an $E_1$-algebra in the derived category of your base ring, although I'll let someone more familiar than me with algebraic operads hunt down the references. – Denis Nardin Jan 13 at 20:55
• @DenisNardin Thanks for the pointer! Do you know good sources for learning about $k$-linear $\infty$-categories? In particular, do they appear also on Lurie's Higher Algebra or only on SAG? (The relevant nLab page points only to Section 6 of DAG-VII for linear $\infty$-categories.) – Théo de Oliveira Santos Jan 14 at 1:43
• @Untitled I don't know a particular reference, I'd define them as $D(k)$-enriched $∞$-categories, but I'm not aware of whether someone has actually tried to use them for something. – Denis Nardin Jan 14 at 11:31
• @DenisNardin, I think the term $k$-linear $\infty$-categories usually refers to stable $\infty$-categories which are modules over $Perf(k)$. These are automatically $D(k)$-enriched, but not the other way around. In either case the embedding in $\infty$-categories is not fully-faithful, and in the case of $D(k)$-enriched categories (which is what $A_\infty$-categories are probably a model of) it is not even conservative. In particular, the answer to the OP's question is no. – Yonatan Harpaz Jan 14 at 21:13

Conjectural Answer


Conjecture. $$A_\infty$$-categories are $$D(k)$$-enriched $$\infty$$-categories, where $$D(k)$$ is the derived category of our base ring $$k$$.

Corollary. The functor $$N\colon\mathrm{Cat}_{A_\infty}\rightarrow\mathrm{Cat}_\infty$$ (in whichever way it might be constructed) is not fully faithful.

Proof. This follows from the fact that $$D(k)$$-enriched categories are not equivalent to $$\infty$$-categories.

This is meant as a synthesis of the question comments. For this reason it is marked as community wiki. (Many thanks to Yonatan Harpaz and Denis Nardin!) Please tell me if this is inappropriate! (In which case I apologize)

• I just want to say that, while it is very likely that $A_∞$-categories are a model for $D(k)$-enriched $∞$-categories, I'm not aware of anyone actually having proved it. Indeed, if someone knew a reference, I'd love to see it. – Denis Nardin Jan 15 at 15:40
• @DenisNardin Thanks for clearing the misunderstanding; I updated the answer to reflect that (I think this might lead to a (non-conjectural) answer quicker). I'm interested in such a reference too, specially since they seem to be very few. – Théo de Oliveira Santos Jan 15 at 15:58
• I believe this comparison follows, indirectly, from results in the literature: $D(k)$-enriched $\infty$-categories are equivalent to dg-categories (arxiv.org/abs/1312.3881), and there are functors between the categories of dg-categories and $A_\infty$-categories (with $A_\infty$-morphisms) that give an equivalence after inverting the weak equivalences. – Rune Haugseng Feb 24 at 14:02
• A little less vaguely: dg-categories can be regarded as a subcategory of $A_\infty$-categories; call the inclusion $I$. If $C$ is an $A_\infty$-category one can define a (large) dg-category $\mathrm{Mod}(C)$ of $C$-modules in chain complexes, with a Yoneda embedding $C \to I\mathrm{Mod}(C)$; the essential image $M(C)$ is then a dg-category with an $A_\infty$-morphism $C \to IM(C)$, which is a weak equivalence. Now you just need to find a natural weak equivalence between $MI(D)$ and $D$ for $D$ a dg-category, and probably the Yoneda map $I(D) \to I\mathrm{Mod}(I(D))$ is just the usual one... – Rune Haugseng Feb 24 at 14:09
• @Rune If you wanted to flesh out your comments into an answer, I'm sure the OP would appreciate it (I'd sure do!) – Denis Nardin Feb 24 at 21:23