6
$\begingroup$

Hello everyone,

I'm currently working on dg-categories, in particular I'm looking for some convenient characterization of the dg-category $\mathcal{RH}om(\mathcal A,\mathcal B)$, for two given dg-categories $\mathcal A$ and $\mathcal B$, at least in cases when $\mathcal A$ is "easy". For example, if $\mathcal A$ is the category $\Delta^1$ with two objects $0$ and $1$ and freely generated over $k$ (a fixed ground commutative ring) by one nontrivial morphism $0 \to 1$, then $\mathcal{RH}om(\mathcal A,\mathcal B) = \mathcal{M}or(\mathcal B)$ (equality up to quasi-equivalence, I suppose), where $\mathcal{M}or(\mathcal B)$ is the dg-category of morphisms of $\mathcal B$ as defined in Drinfeld's article "dg quotients of dg categories" (Drinfeld assumes $\mathcal B$ pretriangulated, but it is not really necessary).

In Keller's survey "on differential graded categories" it is stated that $\mathcal{RH}om(\mathcal A,\mathcal B)$ is (quasi equivalent to) the category of strictly unital $A_\infty$ functors from $\mathcal A$ to $\mathcal B$, both viewed as $A_\infty$-categories.

Moreover, it seems a well-known fact that a given $A_\infty$-category is equivalent - in the $A_\infty$ sense - to some dg-category. Drinfeld himself (first paragraph of Appendix IV of "Dg quotients of dg categories") sketches a procedure to associate a dg-functor to a $A_\infty$-functor between dg-categories, upon changing the source dg-category, in a simple case: as far as I can understand, the idea is to formally add morphisms and coboundary relations to the source category, in a smart way.

My question is the following: how do you explain the above procedure, in general? Namely, what morphisms and what differentials I really have to add, keeping Drinfeld's example in mind? I am looking for something as elementary as possible, and living solely in the "world of dg categories". Even an explanation in some simple situations would be of great help. A caveat: I know nearly nothing about $A_\infty$-categories!

Thanks in advance; I hope everything above is clear enough.

$\endgroup$
2
  • $\begingroup$ This doesn't stay solely in the dg world, but the proof I know of "every Aoo category is equivalent to a dg category" is more conceptual--it's the Yoneda Lemma. Consider the category of all contravariant Aoo functors from $\mathcal{A}$ to $Ch$, the category of chain complexes. The category of Aoo functors would normally be an Aoo category, but because $Ch$ is a dg category, this functor category is actually a dg category. A version of the usual Yoneda argument shows that $\mathcal{A}$ is in fact equivalent as an Aoo category to its image inside this dg category. $\endgroup$ Dec 6, 2012 at 19:31
  • $\begingroup$ Let me quote: "The category of Aoo functors would normally be an Aoo category, but because Ch is a dg category, this functor category is actually a dg category" This is, in fact, quite the point. I would like to <i>describe in an elementary way</i> such functor category, and functor categories of this kind. $\endgroup$ Dec 7, 2012 at 9:32

1 Answer 1

2
$\begingroup$

I don't quite know what kind of properties do you want the DG-category $RHom(A,B)$ to have, but there is a natural nice DG-category which may work. Namely, the category of right quasi-representable h-projective $(A^{op}\otimes B)$-DG-modules.

$\endgroup$
1
  • $\begingroup$ The dg-category $\mathcal{RH}om(\mathcal A, \mathcal B)$ is the internal hom in the homotopy category of dg-categories. As far as I could see, it is quite difficult to describe in general, but I still hope that some useful characterizations can be obtained when the source dg-category is "simple" enough. What I really seek is a... "hands on" description, something I'm able to do computations with. One example is outlined above: when $\mathcal A = \Delta^1$, then $\mathcal{RH}om(\mathcal A, \mathcal B)$ is the dg-category of morphisms in $\mathcal B$, which is "elementary" enough. $\endgroup$ Dec 7, 2012 at 9:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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