User anonymous - MathOverflow

The etale site of a closed subscheme and its etale Grothendieck subtopology

2011-02-28T01:24:17Z

There is a very basic theorem for the Zariski topology.

Let X = Spec(R) and Y=Spec(R/I) for I some reduced ideal. Y obtains a topology two ways, one is the subspace topology as a subset of X and another as the spectrum of a ring. These topologies are the same by the correspondence between ideals of R containing I and ideals in R/I.

Is there a close statement to this in the etale toplogy? There are two natural ways to understand open sets on Y, those which come from etale neighborhoods of X base changed to Y and those which are etale neighborhoods of Y.

I did a computation today in a very special case and it seems that both of these topologies seem to be 'the same'.

Does anyone know if this statement is true in a general context and where I might locate this resource?

Thanks.

Sheaves on stacks and interesting functors

2011-01-09T23:00:22Z

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.

Comment by Anonymous

2011-02-28T05:08:37Z

Thanks for rephrasing it in this language. I'll browse SGA for statement like this tomorrow.

Comment by Anonymous

2011-02-28T05:06:16Z

Thanks. This proof was very close to my original proof. I felt that my extra assumptions had more to do with personal comfort than actual mathematics.

Comment by Anonymous

2011-01-11T17:10:43Z

Thanks for the response, Please see the new edit.

Comment by Anonymous

2011-01-10T03:33:49Z

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.