Let $pi:Pmapsto B$ 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 $Sh_G(P)$ to $Sh(B)$, and the inverse functor is given by $pi_* ^G$.
Let me explain some notations. the object in $Sh_G(P)$ is a pair $(mathcal{F},alpha)$, where $alpha:p^*mathcal{F}simeq a^* mathcal{F}$, and satisfy cocycle condition. Here $p:Gtimes Pmapsto P$ is the projection, and $a:Gtimes Pmapsto P$ is the action.
The functor $pi_* ^G$ is given by assigning open subset V 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 $pi_*^Gcircpi^*simeq Id$, 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(pi^{-1}(b), mathcal{H}|_{pi^{-1}(b)})^G$, where $bin B$.
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?
No comments:
Post a Comment