Hilbert polynomial and Chern classes

Question

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.

Answer 1

This is a good question.

For sheaves on $\mathbf{P}^n_k$ where $k$ is a field, it is true that the Hilbert polynomial of a sheaf determines and is determined by its Chern character. See https://arxiv.org/abs/math/0406588 for a proof. A simple proof can be given by using a free resolution and use induction on the length of the resolution. Note that the Chern classes alone cannot recover the Chern or Hilbert polynomial, since they do not tell us about the rank of the sheaf. To summarize, in terms of information encoded: Hilbert polynomial is equivalent to the Chern polynomial + rank which is equivalent to the Chern character.

The above discussion was for the case $X = \mathbf{P}^n_k$. In the general setting where $X$ is smooth projective, the Chern classes take value in the Chow ring $A^\bullet(X)$. The Hilbert polynomial of a sheaf $\mathscr{F}$ gives partial information about the degrees of the Chern classes $c_i(\mathscr{F}).H^{n-i}$, where $H$ is the hyperplane class of $\mathscr{O}(1)$ and $n = \dim X$. There is little hope in recovering the classes $c_i(\mathscr{F})$ just from these intersection numbers. To elaborate, the Hirzebruch-Riemann-Roch states that $$\chi(\mathscr{F}(l)) = \int_X \operatorname{ch}(\mathscr{F}(l))\operatorname{td}(X).$$ We are hoping to recover $\operatorname{ch}(\mathscr{F}(l))$ from the intersection numbers $\chi(\mathscr{F}(l))$ - in general this is an uphill battle, although I can imagine this to work for some simple toric varieties $X$ (e.g. $\mathbf{P^n_k}$). It would be quite interesting to see examples of $(X, \mathscr{O}(1))$ where the Hilbert polynomial determines the Chern character of all sheaves other than the case of projective spaces.

In the converse direction, the $K$-group modulo torsion $K(X)\otimes \mathbb{Q}$ is isomorphic to the Chow ring modulo torsion $A^\bullet(X)\otimes \mathbb{Q}$ via the map of taking the Chern character. This means that knowing the Chern character of $\mathscr{F}$ gives you the class of $[\mathscr{F}]\in K(X)$ modulo torsion. Since taking the Hilbert polynomial is a group homomorphism $K(X)\to \mathbb{Z}[t]$, the torsion classes have zero Hilbert polynomial. Thus to know the Hilbert polynomial of $\mathscr{F}$, it is sufficient to know the Hilbert polynomials of elements in $K(X)$ that generate a subgroup containing $[\mathscr{F}]$. If (*) $\mathscr{F}$ is a sheaf on $X$ admitting a finite free resolution of the form $\mathscr{L}_\bullet\to \mathscr{F}$, where $\mathscr{L}_i $ is of the form $\bigoplus_{i = 1}^n \mathscr{O}(-a_i)$, then the Hilbert polynomial of $\mathscr{F}$ can be determined (theoretically) from the Chern character of $\mathscr{F}$.

Note: I tend to believe that if $\mathscr{O}(1)$ is very ample and $X$ is smooth, then every sheaf $\mathscr{F}$ satisfies this property (*) but I am not 100% sure.