Sheaves on stacks and interesting functors - MathOverflow most recent 30 from http://mathoverflow.net2013-05-18T09:22:11Zhttp://mathoverflow.net/feeds/question/51598http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/51598/sheaves-on-stacks-and-interesting-functorsSheaves on stacks and interesting functorsAnonymous2011-01-09T23:00:22Z2011-01-11T17:10:20Z
<p>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 $k$.</p>
<p>I hear that there is an isomorphism of stacks $[X/G] \cong [pt/H]$.</p>
<p>I have the following question:</p>
<p>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.</p>
<p>What is this functor taking $G$-equivariant modules over the ring $k[G/H]=k[G]^H$ to vector spaces with $H$ actions?</p>
<p>For example, what happens to the $G$-equivariant $k[G/H]$-module $M=k[G]$?</p>
<p><em>*</em> Edit to more general situation</p>
<p>The answers are getting stuck in a very basic situation, I want to think of a more general situation.</p>
<p>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 $X/(G/H)$.</p>
<p>Question 1: Do we still have $[X/G] \cong [(X/(G/H))/H]$?</p>
<p>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,</p>
<p>$$\mathcal{M}(X \times_{[X/G]} X/(G/H)) \rightarrow \mathcal{M}(X \times_{[X/G]} X/(G/H) \times_{X/(G/H)} X \times_{[X/G]} X/(G/H))$$</p>
<p>What is this functor?</p>
<p>Example: When $H = e$, this equalizer takes the difference between the action and projection pull-backs yielding the functor of invariants.</p>
http://mathoverflow.net/questions/51598/sheaves-on-stacks-and-interesting-functors/51599#51599Answer by David Roberts for Sheaves on stacks and interesting functorsDavid Roberts2011-01-09T23:14:10Z2011-01-09T23:14:10Z<p>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 $H$.</p>
http://mathoverflow.net/questions/51598/sheaves-on-stacks-and-interesting-functors/51622#51622Answer by Sam Gunningham for Sheaves on stacks and interesting functorsSam Gunningham2011-01-10T04:56:55Z2011-01-10T04:56:55Z<p>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.</p>
<p>A $G$-equivariant sheaf on $G/H$ is given by a vector space $V_x$ for each $x\in G/H$, together with isomorphisms $g : V_x \to V_{gx}$. </p>
<p>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 $V_{gH}$.</p>
<p>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 $G/H$). </p>
<p>In your example $k[G]$, $V_{gH} = gk[H]$ and so the $H$-module is $k[H]$. </p>
<p>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... </p>