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? 



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).


