Consider a smooth tame Deligne-Mumford stack $[Y/G]$, a point $[p]$ on it with stabilizer group $H$. Is it true that every representation of $H$ can be extended to a locally free sheaf on $[Y/G]$?

Alternatively, consider a smooth scheme (or algebraic space) $Y$ with a $G$ action. Let $p$ be a point on $Y$ with a finite stabilizer group $H$. Is the restriction map from the Grothendieck group of $G$-equivariant locally free sheaves on $Y$ to the representation ring of $H$

$$K^0_G(Y)\to Rep(H)$$ $$E\mapsto E|_p$$

surjective? (An affirmative answer with possibly some additional conditions, or a negative answer with an explicit counterexample are both very welcome.)


1 Answer 1


Unfortunately that is not always possible. For instance, let $Y$ be $\mathbb{A}^n$, and let $\rho:G\times \mathbb{A}^n\to \mathbb{A}^n$ be a faithful, linear representation. Then the locally free sheaves on $[Y/G]$ are the same as $G$-representations. If there exists a point $p$ of $Y$ whose stabilizer is $H$, then you are asking whether every linear representation of $H$ is the restriction of a linear representation of $G$.

So let $\rho$ be the standard linear representation of $\mathfrak{S}_n$ on $\mathbb{A}^n$, let $G$ be the alternating subgroup $\mathfrak{A}_n$, and let $p$ be the vector $(0,0,1,1,x_5,\dots,x_n)$, where $0,1,x_5,\dots,x_n$ are all distinct. The stabilizer subgroup of $p$ is the cyclic subgroup generated by the involution $(12)(34)$. There is a nontrivial character of this group. Yet there is only the trivial character of $\mathfrak{A}_n$.


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