MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a smooth variety, $\mathcal D$ the sheaf of algebraic differentail operators, $\Omega$ the algebraic deRham complex and $\mathcal M$ a quasi coherent $\mathcal O_X$-module.

Now there is a bijection between $\mathcal D$-module structures on $\mathcal M$ and $\Omega$ dg-module structures on $\Omega \otimes_{\mathcal O_X} \mathcal M$.

For example given a $\mathcal D$-module structure on $\mathcal M$ we can define the corresponding differential by imposing the rule $d(m)(X)=X.m$ on $\Omega\otimes_{\mathcal O_X} \mathcal M$. (X denotes a vectorfield and $m$ a local section of $\mathcal M$.

A similar statement holds if one takes instead $\mathcal D=U(\mathfrak g)$ the universal envelope of a Lie-algebra and $\Omega=\bigwedge \mathfrak g^*$ the standard complex.

Now in both of there cases $\mathcal D$ and $\Omega$ are Koszul-dual, meaning that there are equivalences between carefully defined versions of their "derived categories". Yet formulations of Koszul duality I am aware of, seem not to extend the above correspondence. Roughly speaking they are $\mathcal M \mapsto Hom(\Omega, \mathcal M)$ instead of $\mathcal M \mapsto \Omega \otimes \mathcal M$.

So my questions are:

What does the above observation have to do with Kosul duality?

Is there a formulation of Koszul duality extending the above correspondence?

share|cite|improve this question
Is it necessary to call the bijection "easy to see"? – S. Carnahan Nov 25 '11 at 14:50
Maybe not, sorry. I automatically think of a $D$ modules as an integrable connections. And I internalized that a connection is integrable iff the map $M\rightarrow \omega^1 \otimes M$ can be extended to a differential. So with these two things in mind it is "easy to see", but I agree that it is not so obvious from other viewpoints. – Jan Weidner Nov 25 '11 at 16:30

There are, basically, two answers to your question that I am aware of.

The first one amounts to saying that there is no difference between tensoring with $\Omega$ and taking $Hom$ from $\Omega$. More precisely, the difference between these things corresponds to the difference between the left and right $\mathcal D$-modules: the former are supposed to be tensored with $\Omega$, while for the latter one takes $Hom$ from $\Omega$. Indeed, the $\mathcal O$-algebra $\Omega$ being Frobenius, tensoring with it only differs from taking the $Hom$ from it by the twist with the line bundle of top forms and the homological shift. This twist just transforms left $\mathcal D$-modules into right $\mathcal D$-modules. The three functors form a commutative triangle (up to the shift).

Similarly, if your Lie algebra $\mathfrak g$ is finite-dimensional, tensoring $\mathfrak g$-modules with the cohomological or the homological standard complex of $\mathfrak g$ is almost the same functor, the twist with the one-dimensional $\mathfrak g$-module of top exterior forms on $\mathfrak g$ and the homological shift being the only difference between the two.

The second answer purports to construct the kind of duality that you are looking for in the more general situations when $\Omega$ is no longer Frobenius. The first question that you can ask yourself in this case is, what would be the functor in the opposite direction, i.e., the adjoint functor to tensoring with $\Omega$ over $\mathcal O$? The answer is, it is the functor $Hom_{\mathcal O}(\mathcal D,{-})$.

The latter functor looks somewhat problematic when the variety $X$ is not affine, as the internal $Hom$ from a noncoherent quasi-coherent sheaf may be not a well-behaved operation. Perhaps this problem can be dealt with, but at the moment I do not know how to do it.

When $X$ is affine, however, the theory of derived $D$-$\Omega$ duality for the functors $\Omega\otimes_{\mathcal O}{-}$ and $Hom_{\mathcal O}(\mathcal D,{-})$ is developed in Appendix B to my AMS Memoir "Two kinds of derived categories, Koszul duality, and comodule-contramodule correspondence", In the case of a (possibly infinite-dimensional) Lie algebra $\mathfrak g$, this is done in Section 6.6 of the same paper.

Basically, you want to look on the module-contramodule side of the commutative triangle of "Koszul triality", as presented in my paper. In particular, the appropriate category of $\Omega$-modules that you want to consider (when $\Omega$ is infinite-dimensional, e.g., $\Omega=\bigwedge\mathfrak g^\ast$ and $\mathfrak g$ is infinite-dimensional) is that of contramodules (modules with the infinite summation operations). And the appropriate version of the derived category of $\Omega$-modules is the contraderived category.

share|cite|improve this answer

As far as I remember this is explained (maybe in differet terms) in Kapranov's paper

"On DG-modules over the de rham complex and the vanishing cycles functor", Lecture Notes in Mathematics, 1991, Volume 1479.

You could also have a look at the discussion below this entry of the everything seminar.

I think the kind of statement you are expecting ("equivalences between carefully defined versions of their "derived categories") are of a type you can find in this paper.

In any way, the equivalence is given by a $\otimes$-Hom adjunction. I might have misunderstood something but I don't really understand where is the problem with having $Hom(\Omega,-)$ and $\Omega\otimes -$.

EDIT: more precisely, I have the feeling that the statement is the following.

If $A$ is Koszul and if $B$ is its Koszul dual then $Hom(A,-)$ and $A\otimes-$ defines an equivalence of DG categories between

  • $B$-modules

  • the category $\mathcal P_A$ as it is defined e.g. in Block's paper.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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