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?

$\begingroup$ 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 $K_0$. $\endgroup$– Chris BravAug 4, 2010 at 7:38

$\begingroup$ 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 symmetricgroupequivariant sheaves on affine 2n space. This might help. $\endgroup$– Alexander WooAug 5, 2010 at 2:16
2 Answers
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, 13391360 (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.

$\begingroup$ Can the higher $Hilb_d(P^2)$'s also be expressed as projective bundles? $\endgroup$ Nov 1, 2018 at 16:16

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