Nakajima gives a definition of the Hilbert polynomial on p. 5 here that I can't explain or find explanation for anywhere else. It goes, $X$ is a projective scheme over algebraically closed field with ample line bundle $\mathcal{O}_X(1)$. We look at closed subschemes $Z$ of the product $X\times U$ such that the projection $\pi: Z\to U$ is flat. Let $Z_u=\pi^{-1}(u)$. Now the Hilbert polynomial in $u$ is $$p_u(m)=\chi(\mathcal{O}_{Z_u}\otimes\mathcal{O}_X(m))$$ where $\chi$ is the Euler characteristic (which just becomes the rank of the global sections for large $m$ by Serre vanishing theorem).

I'm ok with the usual $p(m)=\chi(\mathcal{O}_X(m))$, or $p_\mathscr{F}(m)=\chi(\mathscr{F}(m))$ for quasi-coherent sheaves $\mathscr{F}$ but I dont know what this sheaf $\mathcal{O}_{Z_u}$ is exactly. I see from Hartshorne that it could be the structure sheaf of the fiber product $Z_u=Z\times_U \operatorname{Spec} k(u)$ where $k(u)$ is the residue field of $u$. But I guess I don't unerstand enough about products to know why this can also be a quasi-coherent sheaf on $X$ (?) tensorable (over $\mathcal{O}_X$ ?) with the twisted sheaf $\mathcal{O}_X(m)$.

Thinking again of $Z_u$ as homeomorphic to the subset $\pi^{-1}(u)\subset X$, maybe were throwing the structure sheaf onto this subset and doing something skyscraper-y to get just the $m^{th}$ graded global sections in their restriction? Or is $Z_u$ actually a closed subscheme because we only take closed points $u$ making $\mathcal{O}_{Z_u}$ an honest quotient of the structure sheaf of $X$ by an ideal sheaf?

Something like this would seem to solve my problems, because then $\mathcal{O}_{Z_u}\otimes \mathcal{O}_X(m)=\mathcal{O}_{Z_u}(m)$ and we have the usual Hilbert polynomial, but I don't know why the fibers should be closed.

Please halp.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.