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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 $rk(\mathcal F)$?

evgeny
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    I believe this is a special case of this question, using some heavy machinery like Hirzebruch-Riemann-Roch. I'd be interested to see if one can deduce the same result in the vector bundle case without it. – msteve Nov 16 '15 at 18:58

1 Answers1

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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).

Remy
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