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Suppose $U$ is a complex algebraic unipotent group. Let $X$ be a projective variety with a $U$-action. For simplicity, we may assume that there are only finite many $U$ orbits on $X$. The primary example that I have in mind is when $X=G/B$, the flag variety associated to a reductive group $G$, with $U\subset B$ the unipotent radical of a Borel subgroup acting on the left. In this case, the $U$-orbits are indexed by the Weyl group elements.

My question is:

Is there a concrete description of $D^b(Coh^U(X))$, the derived category of $U$-equivariant coherent sheaves on $X$? By concrete, I mean is there a way to construct a collection of generators and explicitly calculate the morphisms between them? I'm also particularly interested in the case for $X=G/B$ as above.

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    $\begingroup$ The structure sheaves of the closures of strata have a natural $U$-equivariant structure, and they should generate, but I don't know how to calculate the homomorphisms. The line bundles on $G/B$ have a $G$-equivariant structure, hence $G/B$-equivariant, and you can probably calculate the $U$-equivariant homomorphisms from Borel-Weil-Bott, but it's not obvious to me how to choose a subset that generates. $\endgroup$
    – Will Sawin
    Commented Apr 1, 2021 at 4:09
  • $\begingroup$ @WillSawin: Dear Will, thanks for your comments! Maybe I will first try to calculate the U-equivariant morphisms between the line bundles as you suggested. Is it true that the $U$-invariant space in $Ext^i(\mathcal{L}(\lambda), \mathcal{L}(\mu))$ (using Borel-Weil-Bott) calculates the $U$-equivariant morphisms? Or the latter is more complicated than that? $\endgroup$
    – Amanda Taylor
    Commented Apr 1, 2021 at 17:17
  • $\begingroup$ I thought it should just be the $U$-equivariant space, but then I realized that the cohomology should probably also involve the $U$-cohomology of the Borel-Weil-Bott Ext-space (which is itself the Ext of the constant representation by that representation in the category of $U$-representations) and thus might be a bit more complicated. $\endgroup$
    – Will Sawin
    Commented Apr 1, 2021 at 17:26
  • $\begingroup$ @WillSawin: I see. Thanks! $\endgroup$
    – Amanda Taylor
    Commented Apr 1, 2021 at 17:52

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