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It is well-know that the category of coherent D-modules over a smooth algebraic $k$-scheme is a Tannakian category. So it is equivalent to the category of finite representations of some affine group scheme G/k. My question is do we have the similar statement for quasi-coherent D-modules? I hope that the category of quasi-coherent D-modules is equivalent to the representation (not necessarily finite) category of some affine group schemes. Is that true, is there any reference for that?

I hope that any quasi-coherent D-module is the union of its coherent sub D-modules. If the answer to the above question is true then this is true. If the above is wrong. I still believe this is true. At least I think it is true for char$k$=0, where a quasi-coherent D-module is a quasi-coherent sheaf with a flat connection. If this is true, could you give me any reference?

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Do you mean that the category of D-modules that are coherent as O-modules, i.e. vector bundles with connection is Tannakian? Because I don't know what the tensor structure would be on coherent D-modules –  YBL May 8 '11 at 10:31
    
Yes, coherent O_X-module with the action of Der(X/k). In chark=0 it is just a coherent sheaf with a flat connection. To see the tensor structure in char 0, you can look at "Nipotent Connections and the monodromy theorem application of a result of Turrittin" by N.Katz –  Lei May 8 '11 at 11:08
    
It is certainly not true that a quasi-coherent $\mathcal D$-module is the union of its $\mathcal D$-modules that are coherent as $\mathcal O$-modules; indeed, the latter set often contains no non-zero elements. On the other hand, coherent $\mathcal D$-module normally has another meaning, i.e. coherent as a $\mathcal D$-module. –  Emerton May 8 '11 at 12:22
    
I just saw the the above comment is explained more carefully in Lars's answer! –  Emerton May 8 '11 at 12:23

3 Answers 3

up vote 3 down vote accepted

Hi Lei, I think your second statement is false, even in char. $0$: Let $X=\mathbb{A}_k^1$ for some field $k$, and $U:=X\setminus \{0\}$. Denote the open immersion by $i$. Then $i_*\mathcal{O}_U$ is $\mathcal{O}_X$-quasi-coherent and not coherent. Consider the canonical connection on $i_*\mathcal{O}_U$: If $x$ is a coordinate on $\mathbb{A}^1_k$, then $\nabla(x):=dx$. This is a connection on $i_*\mathcal{O}_U$: if we plug in $\frac{1}{x}$ we get $-\frac{1}{x^2}dx\in i_*\mathcal{O}_U\otimes \Omega^1_{X/k}$. But the section $\frac{1}{x}$ is not contained in a $\mathcal{O}_X$-coherent sub $D_X$-module: The smallest sub $D_X$-module containing $\frac{1}{x}$ contains $\frac{1}{x^n}$ for all $n\in \mathbb{Z}$. This is not finitely generated over $\mathcal{O}_X$.

What is true however, is that on a smooth variety any $\mathcal{O}_X$-quasi-coherent $D_X$-module is the union of its $D_X$-coherent submodules, but these do not need to be $\mathcal{O}_X$-coherent. A reference for this is this:

D-Modules, Perverse Sheaves, and Representation Theory: Cor. 1.4.17

http://books.google.com/books?id=8ewkW5SC7DcC&lpg=PP1&dq=hotta%20takeuchi&pg=PA29#v=onepage&q&f=false)

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Thanks Lars and Emerton –  Lei May 8 '11 at 12:45

The proper Tannakian theory that has a chance to encompass categories of coherent or quasicoherent D-modules deals with tensor categories WITHOUT a faithful fiber functor - for example the categories of (quasi)coherent sheaves on quasicompact varieties, or more generally geoemtric stacks (see Lurie's article or this MO question). (Once you have D-modules that are not local systems, measuring them at a single point can't be faithful any more..) Of course to talk about quasicoherent sheaves we need to drop rigidity since they are not dualizable objects, but that's not a serious problem, since quasicoherent sheaves (eg D-modules) are unions of coherent ones (eg coherent D-modules, which does NOT mean coherent as O-modules).

In any case in this context one can hope to recover not an affine group scheme but rather the underlying variety or stack again (the usual Tannakian story is the case of the classifying stack pt/G of an affine group scheme G). Here one builds a stack of all fiber functors (not necessarily faithful), and you can hope to reconstruct the original variety/stack by this procedure (the content of Lurie's theorem).

So how might this apply to D-modules? D-modules on X (coherent or quasicoherent) are the same as O-modules (coherent or quasicoherent) on the de Rham space (see e.g. this MO question). So it's reasonable to ask whether Tannakian reconstruction rebuilds the de Rham space of X from the tensor category of D-modules on X. I believe the answer is yes - certainly there's a natural map from the de Rham space of X to the Tannakian functor of D-modules (for every point in X you get a fiber functor and infinitesimally nearby points give isomorphic fiber functors). I would imagine this is an equivalence but haven't thought it through.

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"or this MO question"? –  Allen Knutson May 8 '11 at 20:20
    
Thanks, link fixed.. –  David Ben-Zvi May 9 '11 at 0:41

Consider a coherent $D_X$-module $M$. Then being coherent over $D_X$ is not asking much, $M$ can have arbitrary singularities (its characteristic variety $Ch(M)$ can be pretty much any involutive closed conic subset of $T^*X$).

On the other hand, asking $M$ to be coherent as an $O_X$-module is a very strong restriction. It implies that $M$ is actually locally free of finite rank over $O_X$. This is also equivalent to having a trivial characteristic variety $Ch(M) = T^*_XX$ (the zero section of the cotangent bundle), i.e. $M$ has no singularity at all. It's just a finite rank vector bundle with a connexion.

In this last case you have the tensor structure by tensoring over $O_X$ and you get a tannakian category $(Conn(X),\otimes)$. But here $\otimes$ is an exact functor only because our modules are locally free over it. It would not be the case for general coherent D-modules even less quasi-coherent ones.

What you can do is to consider the category of ind-objects $Ind(Conn(X),\otimes)$. Its objects are (possibly infinite dimensional) vector bundles with a connexion. The tensor product extends as an exact functor. But you don't get a tannakian category because you don't have duals anymore.

Just take $X = Spec(k)$ a point. Then $(Conn(X),\otimes) = (Vect_f(k),\otimes)$ the category of finite dimensionsional vector spaces and $Ind(Vect_f(k),\otimes) = (Vect(k),\otimes)$. But it is well know that duality breaks down for infinite dimensional vector spaces: for example $X \to (X^\vee)^\vee$ is no longer an isomorphism.

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