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Let $\pi:P\to B$ is a $G$-principal bundle, which means $G$ acts on $P$ freely and $\pi$ is a locally trivial fibration. Here is a well-known theorem:

THeorem: The inverse image functor $\pi^{*}$ gives an equivalence from $\mathbf{Sh}_G(P)$ to $\mathbf{Sh}(B)$, and the inverse functor is given by $\pi_* ^G$.

Let me explain some notations: an object in $\mathbf{Sh}_G(P)$ is a pair $(\mathcal{F},\alpha)$, where $\alpha:p^{*}\mathcal{F}\simeq a^{*}\mathcal{F}$, and satisfies the cocycle condition. Here $p:G\times P\to P$ is the projection, and $a:G\times P\to P$ is the action. The functor $\pi_{*}^G$ is given by assigning open subset $U$ to $\mathcal{F}(\pi^{-1}(U))^G$.
We need to show that:

  1. $\pi^{*}$ is fully-faithful, i.e., for any two sheaves $\mathcal{F}$ and $\mathcal{G}$, $Hom(\mathcal{F},\mathcal{G})\simeq Hom_G(\pi^{*}\mathcal{F},\pi^{*}\mathcal{G})$.
  2. $\pi^{*}$ is essentially surjective, i.e., for any $G$-equivariant sheaf $\mathcal{H}$, there exists an isomorphism $\mathcal{H}\simeq\pi^{*}\pi_{*}^G\mathcal{H}$.

(1) follows from $\pi_{*}^G\circ\pi^{*}\simeq Id$, which is relatively easy.

For (2), I have checked when $B$ is a point. For general case, I can reduce it to the following isomorphism: $\pi_{*}(\mathcal{H})_b^G\simeq\Gamma\left(\pi^{-1}(b), \mathcal{H}|_{\pi^{-1}(b)}\right)^G$, where $b\in B$.

I know for nice group $G$, for example when $G$ is compact, this is really an isomorphism, because we have base change. For general group, I have no idea at all.

Who can help to finish this proof? Or give some other idea?

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2 Answers 2

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A general setup and general nonsense proof for this kind of theorems for the "descent along torsors" is in

  • Angelo Vistoli, Grothendieck topologies, fibered categories and descent theory, FGA explained, 1–104

For a clean elementary categorical approach I would also recommend recent article

  • Tomasz Brzezinski, On synthetic interpretation of quantum principal bundles, arxiv/0912.0213

and some $n$lab remarks at codomain fibration.

In the algebraic setup, torsors are usually considered in the flat topology. Related issues are J. Milne's book on etale cohomology. Basic theorems on the descent of sheaves along torsors are from Grothendieck and there are also analogues in various cohomologies, and for equivariant K-theory (look also at Thomason). A noncommutative version is possible with Hopf algebras instead of algebraic groups, what in affine case reduces to celebrated Schneider's theorem. In this regard cf. Hopf-Galois extension in $n$lab and for nonaffine generalizations general discussion in my article

  • Z. Škoda, Some equivariant constructions in noncommutative algebraic geometry, Georgian Mathematical Journal 16 (2009), No. 1, 183--202, arXiv:0811.4770.
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  • $\begingroup$ Thanks. THese are much more advanced. I will go into these details after I have understood the classical situation completely. $\endgroup$
    – JJH
    Apr 6, 2010 at 22:38
  • $\begingroup$ Yeah, I've gotta says, this answer seems to be completely out of line with the question. It's great to reach beyond the question when you give an answer, but I would say this isn't really useful for completing a proof in the language the problem is stated in. $\endgroup$
    – Ben Webster
    Apr 6, 2010 at 23:42
  • $\begingroup$ Brzezinski's paper quoted is more or less in the same terms as your question, but analysing them in all detail. I mean just the first third of the paper or so. The rest is about generalizations including noncommutative. $\endgroup$ Apr 7, 2010 at 0:38
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One thing you didn't specify is what topology your sheaves are in; I don't know if this result is even true in the Zariski topology.

In the either the etale topology (for working over arbitrary base fields) or the complex topology (if you're working over the complex numbers), it's easy to finish this proof; any principal bundle is locally trivial (by the inverse function theorem), so your local statement only needs to be proven for the trivial bundle, where it's obvious.

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  • $\begingroup$ My question is for sheaves on usual topology. I also don't think it is true in Zariski topology, since the topology on the product is not usual product topology. I think I have figured out the proof, Just like you said to prove the trivial bundle first, which is almost the same proof as the case when $B$ is a point. What made me stuck is that I wanted to reduce the problem to the point case, and then one have to use base change, which is not true for any group $G$. Essentially it is enough to construct a canonical isomorphism $\mathcal{H}|{e}\times B\simeq \pi_\* ^G \mathcal{H}$. $\endgroup$
    – JJH
    Apr 7, 2010 at 12:00
  • $\begingroup$ I will think of etale topology situation later, but the problem is that I'm not sure I will be able to work with this language. $\endgroup$
    – JJH
    Apr 7, 2010 at 12:01

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