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Let $X$ be a scheme and let $G$ be an abstract group acting on $X$ by scheme automorphisms. I'm happy to assume finiteness conditions on $X$ (such as locally Noetherian) and on $G$ (such as $G$ is finite) as necessary.

I understand that it is possible to enlarge the category of schemes in such a way that there is a good "quotient space" $X/G$ associated to this data.

A. What is the correct framework in which to study $X/G$? Is this an example of a quotient stack or perhaps an algebraic space?

There should be a good category of sheaves on $X/G$. Let's agree that the "right" category of sheaves on the scheme $X$ is the category of quasi-coherent sheaves of $\mathcal{O}_X$-modules.

B. What is the "right" category of sheaves on $X/G$?

This should be an abelian category with enough injectives. Assuming the answer to this question is "yes", let's call this category $Qcoh(X/G)$.

C. Is it possible to realise $Qcoh(X/G)$ as a subcategory of the category of abelian sheaves $Sh(\mathcal{C})$ on some Grothendieck site $\mathcal{C}$?

One candidate for such a $\mathcal{C}$ is the set of $G$-stable open subsets of $X$, but presumably this is too naive to work in general.

D. Is there always a morphism of abelian categories $Qcoh(X) \to Qcoh(X/G)$?

On the other hand, it is possible to form the category of $G$-equivariant quasi-coherent sheaves $G-Qcoh(X)$ on $X$.

E. What, if any, is the connection between $G-Qcoh(X)$ and $Qcoh(X/G)$?

I suspect that the answer to this last question is "$Qcoh(X/G)$ is equal to $G-Qcoh(X)$ by definition", in which case the question becomes "Can you shed some light on this definition"?

Finally, references to the literature where similar questions are studied would be greatly appreciated.

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Hello anon, I expect that you will find the following notes by Justin Hilburn very interesting pages.uoregon.edu/njp/hilburn.pdf These notes accompanied a very nice talk which discussed Qcoh for derived stacks, but don't let the word derived scare you off. Just as a brief response to your first two specific questions: A) Why not both? B) Justin's notes(and others) will take the Qcoh's to just be modules on "affine pieces" and glue them together. –  B. Bischof Aug 25 '12 at 17:56
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You could also have a look at Angelo's Notes on Grothendieck topologies, fibered categories and descent theory homepage.sns.it/vistoli/descent.pdf –  Niels Aug 25 '12 at 21:00
    
@Niels, what did you have in mind? As far as I remember he doesn't really go into these questions since these questions are asking about Qcoh on stacks, where as Vistoli talks about Qcoh on schemes forming stacks. Although he does talk a bit about torsors, so maybe this is what you were referring to. –  B. Bischof Aug 25 '12 at 23:04
    
@B. Bischof, well, these notes contain a very complete treatment of $G$-sheaves. As you understood, the section on $G$-torsors will apply mutatis mutandis for $X\to [X/G]$, thus establishing the correspondence between quasi-coherent $G$-sheaves and quasi-coherent sheaves on the quotient stack. –  Niels Aug 26 '12 at 7:28
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Maybe mathoverflow.net/questions/54496 can be helpful. –  Dan Petersen Aug 26 '12 at 7:36

1 Answer 1

up vote 3 down vote accepted

I think this question is a good one, but don't expect an encyclopedic answer — MO is not an encyclopedia. Here are some answers, with the disclaimer that I'm a category theorist but not an algebraic geometer.

To question A, by and large the 21st perspective will probably say that it is definitely a stack, in some notion of the word. Certainly there are applications where you do want to consider the "space" quotient $X/G$, in which points in the same orbit are honestly identified. This is like taking a form of "$\pi_0$" of the stack. (Not etale $\pi_0$, certainly, but a form of $\pi_0$ that's valued in spaces rather than sets.)

To questions B, D, and E, the answer is that, as you guessed, the best definition of $\operatorname{QCoh}(X/G)$ is the category of $G$-modules in $\operatorname{QCoh}(X)$, at least when $G$ is a finite group. The geometric intuition is that a quasicoherent sheaf on $X$ is something like a vector bundle over $X$. In the quotient $X/G$, we add an isomorphism between any two points for each way that they are related by an element of $G$. So a vector bundle over $X/G$ should have a fiber over each point of $X$, and an isomorphism between these fibers for each pair of $G$-related points.

There is a quotient morphism $X \to X/G$. The $\operatorname{QCoh}$ functor is best understood as contravariant, just like $\mathcal{O}$ is contravariant. Namely, a geometric morphisms $f: X \to Y$ correspond (modulo details) to symmetric monoidal "linear" functors $f^\ast : \operatorname{QCoh}(Y) \to \operatorname{QCoh}(X)$, which pull back a "vector bundle" along the map. This is certainly true for $\operatorname{QCoh}(X/G) = \operatorname{QCoh}(X)^G$, with the quotient morphism corresponding to the functor "forget the $G$-action". That said, each such functor $f^\ast$ also has a right adjoint $f_\ast : \operatorname{QCoh}(X) \to \operatorname{QCoh}(Y)$, which is not usually symmetric monoidal — it is that takes a "vector bundle" over $X$ and makes it into the "vector bundle" over $Y$ whose stock over $y\in Y$ is the space of sections over $f^{-1}(y)$ of the corresponding bundle on $X$. In the case of the quotient map $X \to X/G$, its right adjoint is the "free" functor, assigning to a quasicoherent module $M$ the corresponding free $G$-module $G \otimes M$. You ask for $f_\ast$ to be "a morphism of abelian categories", which is vague to me. The best definition I know of "morphism of abelian categories" is a right-exact functor (if I have left and right correct), in which case in general pushforward maps are not morphisms — they are instead left-exact. I think that if $G$ is finite, then in fact the pushforward along the quotient map is exact; maybe I need to include that the characteristic of the ground field does not divide the order of $G$.

As for C, as just a subcategory, I'm sure the answer is yes. If you ask for more conditions, the answer is probably still yes, at least in the finite-group case: you should be able to take the category of cocommutative coalgebras in $\operatorname{QCoh}(X/G)$ and find this as sheaves-of-sets on something. But I'd have to think more about details.

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Thanks! Re D I guess I wanted a good way of pulling back objects from $Qcoh(X/G$ to those in $Qcoh(X)$ but also pushing forward from $Qcoh(X)$ to $Qcoh(X/G)$. I would have called it "morphism of topoi" except that $Qcoh(X)$ isn't a topos since we're taking only quasi-coherent sheaves and not all sheaves. –  anonymous Aug 26 '12 at 9:53
    
@anonymous: Thinking in terms of topoi can provide very good intuition. In arxiv.org/abs/1105.3104, we develop the idea (not really due to us) that the 2-category of {presentable categories, left adjoints, and natural transformations} is a higher-categorical analog of {abelian groups, homomorphisms}, and that {symmetric closed monoidal presentable categories, symmetric monoidal left adjoints, monoidal natural transformations} is an analog of {commutative rings, homomorphisms}. In this sense, QCoh is directly analogous to the functor of global functions. It then makes sense to ask ... –  Theo Johnson-Freyd Aug 26 '12 at 12:38
    
... to what extent are algebraic stacks "2-affine". This question is a deep one. On the one hand, in arxiv.org/abs/math/0412266 Lurie shows that at the 1-morphism level QCoh is faithful with essential image those functors satisfying a hard-to-check and slightly inelegant "tameness" condition (at the 2-morphism level Lurie only considers isomorphisms, which we believe I believe is a mistake). On the other hand, Brandenburg and Chirvasitu show (without using Lurie's results) that for quasicompact and quasiseparated schemes, QCoh is an equivalence onto its image. ... –  Theo Johnson-Freyd Aug 26 '12 at 12:45
    
... It seems to be unknown whether there are any examples of symmetric monoidal left adjoints between QCoh(stacks) that do not satisfy Lurie's tameness condition. Anyway, especially in my paper with Alex Chirvasitu (op. cit), a few of our definitions are very motivated by the situation in Grothendieck topoi. –  Theo Johnson-Freyd Aug 26 '12 at 13:00
    
@Theo: Meanwhile Daniel Schäppi has also shown it for X = BG, where G is a finite flat group scheme. –  Martin Brandenburg Sep 8 '12 at 14:01

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