# Hilbert polynomial and Chern classes

Motivation: moduli spaces of semistable sheaves.

Let $(X,\mathcal O_X(1))$ be a smooth projective variety over a field $k=\overline k$. When one defines the moduli functor $\mathcal M_P:\textrm{Sch}_k\to \textrm{Sets}$ of semistable sheaves, one fixes the Hilbert polynomial $P$; more precisely, I recall how the moduli functor is defined: an $S$-family $[\mathscr F]\in\mathcal M_P(S)$ is an equivalence class of a sheaf $\mathscr F\in\textrm{Coh}(X\times S)$ such that $\mathscr F_s$ is semistable on $X_s$, and has Hilbert polynomial $P$ for every closed point $s\in S$.

Question: is fixing $P$ equivalent to fixing the Chern classes, and/or to fixing the Chern character?

The Hilbert polynomial of a sheaf $E\in\textrm{Coh}(X)$ is $P(E,m)=\chi(X,E(m))=\int_X\textrm{ch}(E(m))\cdot\textrm{td}(X)$, and the Chern character is a polynomial in the Chern classes. Thus, given the Hilbert polynomial, one recovers the Chern classes. But do the Chern classes determine the Hilbert polynomial? In particular, I would like to know if the moduli problem of semistable sheaves stays unchanged if we replace "Hilbert polynomial" by "Chern character", or "Chern classes".

Thank you.

• Dear @Brenin, I read you interesting question. Coul you explain me why $\chi(X,E(m))= \int_X ch(E(m)) \dot td(X)$? Thank you very much in advance! – ArthurStuart Nov 27 '13 at 13:15
• @ArthurStuart: Thanks for asking! that equality is one way of stating Hirzebruch-Riemann-Roch theorem, which holds for any vector bundle (so in that equality you can replace $E(m)$ by any vector bundle $F$). – Brenin Nov 27 '13 at 13:59
• Dear @Brenin can you give me a reference in which this equality is proved? Thank you! – ArthurStuart Nov 28 '13 at 10:12