Does there exist an exceptional collection of coherent sheaves on the Hilbert scheme of points on projective plane? If so, Could it be a strong full collection?

  • Maybe. I would look in Nakajima's book or his papers on quiver varieties, where he constructs a resolution of the diagonal for at least some quiver varieties. Probably the answer to your question can be dug out of some statement about a basis for $K_0$. – Chris Brav Aug 4 '10 at 7:38
  • Could you give the definitions for 'exceptional collection' and 'strong full collection' or a pointer please? For the Hilbert scheme of points on the affine (not projective) plane, there is a derived equivalence of categories to symmetric-group-equivariant sheaves on affine 2n space. This might help. – Alexander Woo Aug 5 '10 at 2:16

It's more than 7 years after the question, but better late than never. The result of @Sasha holds for any number of points, and for surfaces more general than $\mathbb{P}^2$: once $\mathbf{D}^{\mathrm{b}}(S)$ has a full exceptional collection, then so does $\mathbf{D}^{\mathrm{b}}(\mathop{\mathrm{Hilb}}^nS)$. This is proposition 1.3 of

Krug, Andreas; Sosna, Pawel, On the derived category of the Hilbert scheme of points on an Enriques surface, Sel. Math., New Ser. 21, No. 4, 1339-1360 (2015). ZBL1331.18015.

One can easily check the case of $Hilb_2(P^2)$ --- there is a map $Hilb_2(P^2) \to (P^2)^\vee$ (the dual plane) which takes a subscheme to the unique line containing it. The fibers are $Hilb_2(P^1) = P^2$, so $Hilb^2(P^2)$ is a $P^2$ bundle over $(P^2)^\vee$, hence has an exceptional collection.

  • Can the higher $Hilb_d(P^2)$'s also be expressed as projective bundles? – S. S. Nov 1 at 16:16
  • 1
    @SaS: I don't think so. – Sasha Nov 2 at 7:47

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