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Let $G$ be a finite group and $H \subset G$ a normal subgroup. Consider $G$, $H$, and $X=G/H$ as affine algebraic groups over some algebraically closed base field $k$.

I hear that there is an isomorphism of stacks $[X/G] \cong [pt/H]$.

I have the following question:

To give a sheaf (of vector spaces) on the stack $[X/G]$ is the same as giving a $G$-equivariant sheaf on $X$. By the isomorphism above, it is the same as giving a vector space with an $H$ action.

What is this functor taking $G$-equivariant modules over the ring $k[G/H]=k[G]^H$ to vector spaces with $H$ actions?

For example, what happens to the $G$-equivariant $k[G/H]$-module $M=k[G]$?

* Edit to more general situation

The answers are getting stuck in a very basic situation, I want to think of a more general situation.

Suppose that $G$ is an affine algebraic group over an algebraically closed subfield, $H$ a normal subgroup, and $X$ an affine $G$-variety with action factoring through $G/H$. Suppose that $G/H$ acts properly and freely on $X$. The stack $[X/(G/H)]$ is representable by a scheme $X/(G/H)$.

Question 1: Do we still have $[X/G] \cong [(X/(G/H))/H]$?

If so, Question 2: For any $G$-equivariant sheaf $\mathcal{M}$ on the space $X$, by descent for Cartesian sheaves $\mathcal{M}(X/(G/H))$ is computed by the kernel of the diagram,

$$\mathcal{M}(X \times_{[X/G]} X/(G/H)) \rightarrow \mathcal{M}(X \times_{[X/G]} X/(G/H) \times_{X/(G/H)} X \times_{[X/G]} X/(G/H))$$

What is this functor?

Example: When $H = e$, this equalizer takes the difference between the action and projection pull-backs yielding the functor of invariants.

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2 Answers 2

I am not sure from your comment whether this is all already clear to you or not, but here is how I thought about this anyway.

A $G$-equivariant sheaf on $G/H$ is given by a vector space $V_x$ for each $x\in G/H$, together with isomorphisms $g : V_x \to V_{gx}$.

In particular, $V_H$ is a $H$-module and in general $V_{gH}$ is a $gHg^{-1}$-module. Each $g \in G$ gives you an isomorphism from the $H$-module $V_H$ to the $gHg^{-1}$-module $V_{gH}$.

These isomorphisms are exactly the descent data from $G/H$ to $pt/H$. So, the $H$-module you want is $V_H$ (i.e. the fibre of the sheaf above the point $H$ in $G/H$).

In your example $k[G]$, $V_{gH} = gk[H]$ and so the $H$-module is $k[H]$.

Reading your comment and David Robert's answer again, I don't think I have added anything, but I'll leave this here in case it helps...

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Thanks for the response, Please see the new edit. –  Anonymous Jan 11 '11 at 17:10
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Well as far as the stacks go, they have obvious presentations by finite groupoids. There is a functor $\mathbf{B}H := (H\rightrightarrows \ast) \to X\rtimes G := (X \times G \rightrightarrows X)$ sending the one object of $\mathbf{B}H$ to the coset $H \in X$. The sheaf of vector spaces over $X\rtimes G$ is then a vector bundle on $X$ with a $G$-action i.e. a family of vector spaces indexed by $X$. The functor you are after is related to the restriction functor sending $\{V_x | x\in X\} \mapsto V_H$. So you are looking for some functor of modules which is restriction to a submodule invariant under $H$.

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Thanks for the comment. I am aware what the answer should be. As in my example, $k[G] \cong \oplus_{\overline{g} \in G/H} g k[H]$. We want to pick out one of any of these copies $k[H]$ (with $H$ action), all of which are isomorphic under the action of $G/H$. There should be descent data describing how to pass to a sheaf on the point. If $H=\{e\}$ then it is just taking $G$-invariants. It should be something like taking $G/H$-invariants -- but this doesn't make any sense on most $G$ modules (e.g. $k[G]$). I am curious how the descent data would encode this interesting example. –  Anonymous Jan 10 '11 at 3:33
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