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Suppose a (linear algebraic) group $G$ acts on a variety $X$ and that $U$ is a $G$-invariant open subvariety. My question is: under what conditions is the restriction functor

$i^*: Vect^G(X) \rightarrow Vect^G(U)$

an equivalence of categories of $G$-equivariant vector bundles? Obviously not in general, but we do have: if $X$ is normal (I think Gorenstein or even less will do actually) and $U$ is dense with complement of codimension at least 2 then $i^*$ is fully faithful. What more is needed to make it surjective? What if the complement of $U$ is actually a fixed point of the $G$ action?

The example I have in my head here is that of $G$ being the product of a finite subgroup of $SL(2,\mathbb{C})$ and a $\mathbb{C}^*$, acting canonically on $X = \mathbb{C}^2$. Then the $G$-equivariant vector bundles on the punctured plane should be equivalent to those on the whole $\mathbb{C}^2$, that is to representations of $G$. So I guess I'm asking: why is this true?

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1 Answer 1

up vote 2 down vote accepted

The statement is true if $X$ is regular of dimension 2 (an in very few other cases, I would guess). Anyway, this certainly applies to your example.

The point is that every locally free sheaf on $U$ has an extension to a reflexive sheaf on $X$, which in this case is locally free. Then you need to show that the extension of a $G$-equivariant sheaf is again $G$-equivariant; the $G$-equivariant structure is given by an isomorphism of locally free sheaves on the product $G \times U$, and this isomorphism extends uniquely to an isomorphism on $G \times X$, because the restriction functor is fully faithful.

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