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Let $S^{[2]}$ be the Hilbert scheme of two points on a smooth projective surface (actually, right now I am particularly interested in del Pezzo surfaces). Let $B$ be the exceptional divisor of the Hilbert-Chow morphism $S^{[2]} \rightarrow \operatorname{Sym}^2 S$. Let $L$ be a divisor on $S$ and $\tilde L$ the corresponding "symmeterized" divisor on $S^{[2]}$, i.e. the set of $D \in S^{[2]}$ such that $\operatorname{Supp} D \cap \operatorname{Supp} L \neq 0$.

I would like to have a formula for $\chi(\tilde L+\frac{n}{2}B)$ in terms of $n$ and the invariants of $S$ and $L$. (We have a conjecture for del Pezzo surfaces by doing a different computation that we expect to match.)

In, e.g. [Li, Qin, and Wang], they say that there is an algorithm to compute the cup product of any two cohomology classes for an arbitrary $S$. Furthermore, Boissiere has a paper about finding "universal formulas" for chern classes of tangent bundles of $S^{[n]}$.

If I understand the chern classes of the the tangent bundle, maybe I could understand the todd class, and assuming I could convert $\tilde L + \frac{n}{2}B$ into generators suitable for the algorithm mentioned by L,Q,W, maybe I could use GRR to compute the desired Euler characteristic.

I've been looking at these and the surrounding papers, but I'm having a hard time getting a feel for what they can do. I don't mind putting in some time in to learn some new stuff, but I'd' like to be sure I'm going in the right direction first.

Question Is the plan described above realistic? If so, do you have any recommendations for where to start reading? Or is there a different strategy that looks better?

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    $\begingroup$ Using that stuff for this computation would be utterly absurd. The Hilbert scheme of two points is just the blowup of the diagonal of the symmetric product, which should give you an explicit handle on all the classes you need for RR. $\endgroup$ – Vivek Shende Oct 24 '14 at 7:20
  • $\begingroup$ So, for example, you can write down a formula for todd S[2] that is explicit enough to do the intersections with? $\endgroup$ – Drew Oct 24 '14 at 15:33
  • $\begingroup$ Blow-up $S\times S$ along the diagonal; then there is a $finite$ cover to $S^{[2]}$. Now use Leray! $\endgroup$ – Walter Neff Oct 25 '14 at 9:14

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