Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I would like to proof the following theorem:

Let $\pi:X\rightarrow X/G$ be a principal $G$-bundle (say of varieties, Zariski locally trivial), then $\pi^*$ induces an equivalence between modules on the quotient and equivariant modules on the total space.

Now I would like to know if this can be proven along the following lines:

  1. Define a model structure on some category $V$ containing the category of simplicial varieties. Maybe the category simplicial presheaves?

  2. Define the category of quasi coherent modules over an object of $V$. For an ordinary variety $X$ this should be the category of quasicoherent $\mathcal O_X$-modules. For the "action simplex" $$... G\times G\times X \Rrightarrow G\times X \Rightarrow X$$ this should be the category of $G$ - equivariant qc $\mathcal O_X$-modules, where $G$ is an algebraic group and $X$ is an ordinary variety.

  3. Proof that in the situation of the theorem the map from the action simplex to $X/G$ is a weak equivalence.

  4. Proof that weak equivalences between objects of $V$ induce equivalences between the corresponding module categories.

I'm not yet experienced enough with model categories to judge if this program has a chance to go through. I am quite optimistic about the first three steps. However about the last one my only vague intuition is that homotopy equivalent topological spaces have "same" vector bundles. Maybe one can define a model structure such that weak equivalences are precisely the morphisms which induce equivalences between module categories?

So my questions are:

Has this a chance to work?

If yes: What is the model structure and does one proof 3+4? Is it done somewhere?
If no: Why not? Why is it a bad idea?

P.S. I know how to proof the theorem in a much easier, hands on way. In this question I am mainly interested in a proof along the above lines.

share|improve this question
    
Do you call the proof in Vistoli's lectures on descent theory (see arxiv) hands on ? It is natural generality/statement and general nonsense proof. In your proof sketch you go into wider category in which you have some relation of equivalence which reduces to "having the same categories of quasicoherent sheaves" on representables. There you nowhere use to have axioms of model category, just that it is a relation of equivalence, hence your requirement to be actually a weak equivalence of a model structure is an overkill. –  Zoran Skoda Aug 8 '11 at 13:12
1  
Model structures are for homotopy theory, and there's (essentially) no homotopy theory in this problem! This would be a more useful approach if you were concerned about derived categories. The point is that for abelian categories, all you need is the groupoid underlying your simplicial scheme, which is significantly less data. By the way, if you were trying to give a proof along these lines, then statement 4 of your outline is just what's called descent, and (classical) fppf descent is exactly the non-trivial step of the "easier" proof you cite. –  Moosbrugger Aug 8 '11 at 18:26
    
The hands on proof I had in mind goes roughly as follows: We want to construct a quasi inverse functor to $\pi^*$. Our candidate is the invariant section functor: $M\mapsto \pi_*(M)^G$. After constructing candidates for the unit and counit it is enough to show that they are isomorphisms locally on the base. Hence we can assume that the bundle is trivial and everything is affine. In this situation $M\mapsto \pi_*(M)^G$ is isomorphic to pulling back along a section of the bundle $i^*$. With a bit of commutative algebra one shows that $i^*$ and $\pi^*$ are mutually quasi inverses. –  Jan Weidner Aug 9 '11 at 7:43
    
Do your comments imply that a proof along the above lines is impossible? The reason why I wanted to do it this way is the following: Suppose we have an algebraic group acting $G$ on a variety. However the action is not very nice and we only know for a normal subgroup $H$ that $X\rightarrow X/H$ exists as a principal bundle. What I would like to proof is that there is an equivalence of categories: $\mathcal O_X-mod^G=\mathcal O_{X/H}-mod^{G/H}$. And I thought maybe this could be done by showing $X//G=(X//H)//(G//H)$ where the $//$ mean "homotopy quotients" in the category $V$. –  Jan Weidner Aug 9 '11 at 8:00
1  
Ah. Perhaps a less hands-on proof is to show that the datum of equivariance is equivalent to the datum of descent datum (hint: $X\underset{X/G}{\times}X\simeq G\times X$). Then just apply faithfully flat descent. There's a generalization of this claim for "bad" actions of $G$ on $X$: it says that $G$-equivariant quasi-coherent sheaves on $X$ are the same as quasi-coherent sheaves on the stack $X/G$. This is basically formal once you have the definitions and the case of principal bundles. It also immediately implies the statement you specified about $G$ and $H$. –  Moosbrugger Aug 9 '11 at 15:12
show 2 more comments

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.