For $F \subset G$ two algebraic groups, consider a homogeneous space $H$ of the form $G/F$. Now every vector bundle over $H$ is a coherent sheaf, but the converse is not true. What happens in the equivariant setting? Do there exist equivariant sheafs that are not equivariant vector bundles?


I assume that the algebraic group $G$ is smooth and connected, and that you are asking about equivariance for the natural action of $G$ on $H$. There is a quotient morphism $q:G\to H$ that is $G$-equivariant and faithfully flat. Thus, to prove that a $G$-equivariant coherent sheaf on $H$ is locally free, it suffices to check that the pullback $G$-equivariant coherent sheaf on $G$ is locally free. Since $G$ is smooth, for every coherent sheaf $\mathcal{F}$ on $G$, there exists a maximal open subscheme $U$ of $G$ on which $\mathcal{F}$ is locally free (possibly zero), and $U$ is dense in $G$. If $\mathcal{F}$ is $G$-equivariant, then $U$ is $G$-invariant. The only dense $G$-invariant open subset of $G$ is all of $G$. Thus every $G$-equivariant coherent sheaf on $G$ is locally free. Therefore every $G$-equivariant coherent sheaf on $H$ is locally free.

Edit. As Sasha points out, it is unnecessary to pullback to $G$ (it just makes it easier for me to think about).

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    $\begingroup$ Jason, what was the reason of considering the pullback of $\mathcal{F}$ to $G$? You could use the same argument directly on $H$ --- since the action of $G$ on $H$ is transitive the only $G$-invariant open subset of $H$ is all $H$. $\endgroup$ – Sasha Dec 2 '13 at 17:36
  • $\begingroup$ . . . just a thought though - does this mean that the category of equivariant vector bundles is an abelian category? $\endgroup$ – Christian Fischmann Dec 2 '13 at 17:52
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    $\begingroup$ Yes, and moreover it is equivalent to representations of $F$ (the equivalence is given by taking a fiber of a vector bundle at some chosen point of $H$). $\endgroup$ – Sasha Dec 2 '13 at 18:11

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