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? 



 When is restriction an equivalence of categories of equivariant vector bundles?
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.


