TITLE: Is rank of vector bundle encoded in its Hilbert polynomial
QUESTION [6 upvotes]: Let $\mathcal F$ be a vector bundle over a projective variety $(X, \mathcal O_X(1))$, and $P_\mathcal F(m)=\chi(X, \mathcal F(m))$ be its Hilbert polynomial. Then can I define from $P_\mathcal F$ the value of the rank $rk(\mathcal F)$?

REPLY [4 votes]: Let $(X,\mathcal O_X(1))$ be projective of dimension $n$. Let $\eta \in X$ be the generic point. For a sheaf $
\mathcal F$ write
$$P_{\mathcal F}(m) = \sum_{i=0}^n a_i(\mathcal F) \frac{m^i}{i!}.$$
Let $d = \deg X = a_n(\mathcal O_X)$ be the degree of $(X, \mathcal O_X(1))$.

Lemma. Let $\mathcal E$ be a vector bundle of rank $r$ over $X$. Then $a_n(\mathcal E) = rd$.

Proof. For $\mathcal E = \mathcal O_X$, this is the definition of $d$. By additivity of the Hilbert polynomial, this proves the result when $\mathcal E$ is the trivial bundle of rank $r$.
Now let $\mathcal E$ be any vector bundle. Let $s_1, \ldots , s_r \in \mathcal E_\eta$ be a basis for the generic fibre of $\mathcal E$. Then $s_i$ is a rational section defined away from some divisor $D_i$. Setting $D = \sum D_i$, we see that the map
\begin{align*}
\mathcal O_X^n &\to \mathcal E(D)\tag{1}\label{1}\\
e_i &\mapsto s_i
\end{align*}
is defined. This gives a short exact sequence
$$0 \to \mathcal O_X^n \to \mathcal E(D) \to \mathcal F \to 0,$$
where $\mathcal F$ is supported on some divisor $D'$ (since (\ref{1}) is generically an isomorphism). By additivity of the Hilbert polynomial, we get
$$P_{\mathcal E(D)} = P_{\mathcal O_X^n} + P_\mathcal F.$$
But $\mathcal F$ is supported in lower dimension, so it does not contribute to the leading coefficient. A similar argument allows us to replace $\mathcal E(D)$ by $\mathcal E$. $\square$
Remark. If you don't like rational sections, you can also use Serre's theorem to conclude that $\mathcal E(m)$ is globally generated for $m \gg 0$, and then choose $s_i$ that generate $\mathcal E(m)_\eta$.
Remark. It seems that we haven't really used that $\mathcal E$ is a vector bundle. The same argument should work for any coherent sheaf (where the rank is, by definition, the dimension of the generic fibre).