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Let $X$ be a smooth projective threefold. Let $I_n(X,\beta)$ be the Hilbert scheme parametrizing subschemes $Z \subset X$ with curve class $\beta \in H_2(X,\mathbb{Z})$ and $\chi(\mathcal{O}_Z)=n$. Can one understand the second condition $\chi(\mathcal{O}_Z)=n$ geometrically?

I am aware that when $\beta=0$, $I_n(X,0)=X^{[n]}$ is the Hilbert schemes of $n$ points on $X$. How should one understand $\chi(\mathcal{O}_Z)=n$ for general $\beta$?

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  • $\begingroup$ Note that if $Z$ is a smooth curve, then $\chi(\mathcal{O_Z}) = 1-g$ where $g$ is the genus of $Z$ (by Serre duality). $\endgroup$ – Piotr Achinger Jun 22 '13 at 10:10
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As Piotr explained if $Z$ is a smooth curve of genus $g$ then $\chi(O_Z) =1 - g$. Now you can increase the Euler characteristic by adding nilpotent elements to the structure sheaf. To be more precise, let $I_Z$ be the ideal of $Z$, choose a point $x \in X$ and consider a surjective map $I_Z \to O_x$. Its kernel is an ideal of a new subscheme $Z' \subset Z$ with the same curve class but with $\chi(O_{Z'}) = \chi(O_Z) + 1$. Continuing in this way you can increase $\chi$ as much as you want.

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