## The canonical divisor of the Hilbert scheme $Hilb^n P^2$?

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Hey everyone,

I was wondering if anyone knows what the canonical divisor of the Hilbert scheme $Hilb^n P^2$ is --$Hilb^n P^2$ is the Hilbert scheme of degree-n zero dimensional subschemes of the projective plane $P^2$. Any references?

Many thanks in advance.

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## 2 Answers

There is an easy formula for the canonical divisor on the Hilbert scheme of $n$ poin ts on any any smooth projective surface $X$. Let's first fix some notations. Denote by $X^{n}$ the $n-$fold product with projections $pr_i X^{n}\to X$. We can consider line bundles of the form $$L^{[n]}=pr_1^* L \otimes\cdots \otimes pr_n^* L$$It is not hard to show that this decends to a line bundle on the symmetric product ${X}^{[n]}$ and this defines a homomorphism $Pic(X)\to Pic X^{[n]}$. In this notation, the canonical line bundle is given by $$\omega_{X^{[n]}}=\omega_{X}^{[n]}.$$I think this is in Göttsche's book on Hilbert schemes of points.

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Thank you. It was very helpful. – Turkelli Feb 18 2011 at 18:37

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$n=1$ already tells you that the anticanonical divisor is going to be nicer than the canonical, in that it's effective. There, the divisor given by the three coordinate lines is anticanonical.

Next step, look at the Chow variety of $n$ points in $P^2$, with an anticanonical given by "some point is on some coordinate line".

Then use the fact that the morphism from Hilb to Chow is crepant, to say that we can pull the anticanonical back. So: the divisor given by "some point is on some coordinate line" is again anticanonical up on the Hilbert scheme.

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Oh, I see, thank you. In fact, I care about the anti-canonical divisor (rather than the canonical) as I am interested in Batyrev-Manin conjectures for Hilbert schemes. I wonder if the effective cone of the Hilbert schemes is known?  – Turkelli Feb 19 2011 at 21:32
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