Let $X$ be a smooth projective variety over $\mathbb{C}$ satisfying $H^1(\mathcal{O}_X)=0$. Fix $i:X \to \mathbb{P}^n$ a closed immersion and let $\mathcal{O}_X(1)$ be the corresponding very ample line bundle. This means that the Picard group is isomorphic to the Neron-severi group of $X$. By Severi's theorem of base, we know that the rank of the Neron-Severi group is finite. Does this mean that for a fixed Hilbert polynomial $P$, there are only finitely many invertible sheaves (upto isomorphism) on $X$ with Hilbert polynomial $P$ i.e., is $$\#\{\mathcal{L} \in \mbox{Pic}(X)|\chi(\mathcal{L}(m))=P(m) \mbox{ for } m \gg 0\}< \infty?$$

  • 1
    $\begingroup$ What is $\mathscr{L}(m)$? Do you fix some ample class? Francesco's answer addresses a different question, taking as Hilbert polynomial $\chi (\mathscr{L}^{\otimes m})$. $\endgroup$ – abx Apr 9 '15 at 10:06
  • $\begingroup$ @abx: Yes, I do fix an ample line bundle. And as you guessed, $\mathcal{L}(m) \cong \mathcal{L} \otimes \mathcal{O}_X(m)$. $\endgroup$ – Ron Apr 9 '15 at 11:56
  • $\begingroup$ Oh, sorry. I misinterpreted your notation $\mathcal{L}(m)$ as $\mathcal{L}^m$ in additive notation, since you did not say that you were choosing an ample class. So my answer does not really answer your original question. $\endgroup$ – Francesco Polizzi Apr 9 '15 at 14:16
  • $\begingroup$ @Polizzi: May be I am wrong, it seems to me the Hilbert polynomial of all these $(-1)$ curves are still the same (the formula is different from the one you mentioned). So, it seems to be still a counterexample. Am I wrong? $\endgroup$ – Ron Apr 9 '15 at 15:04
  • $\begingroup$ If you fix an ample divisor $H$, you cannot conclude in principle that $HE$ is independent from the $(-1)$-curve $E$ that you are choosing. $\endgroup$ – Francesco Polizzi Apr 9 '15 at 15:07


For instance, there are rational surfaces containing infinitely many $(-1)$-curves. An example in given by the blow-up $X$ of $\mathbb{P}^2$ at nine points that are the base locus of a general pencil of cubics: indeed, $\textrm{Aut}(X)$ contains a copy of $\mathbb{Z}^8$ generated by translations by differences of the nine sections of the elliptic fibration coming from the pencil, and the orbit of a $(-1)$-curve by this subgroup gives infinitely many of them.

Two such $(-1)$-curves are not linearly equivalent, since any $(-1)$-curve $E$ is isolated into its linear equivalence class, however by the Riemann-Roch theorem they have the same Hilbert polynomial, namely $$P(m)=\frac{mE(mE-K_X)}{2}+ \chi(\mathcal{O}_X) = -\frac{1}{2}(m^2-m)+1.$$


ADDED. I was confused by the OP's notation in the first version of the question and I took as "Hilbert polynomial" the quantity $\chi(\mathcal{L}^m)$. It appears from the comments (and the edit) above that the OP was actually thinking to a more standard Hilbert polynomial of type $\chi(\mathcal{L} \otimes H^m)$, were $H$ is a (fixed) ample class. Then my answer does not provide a counterexample in this case.

  • $\begingroup$ @Polizzi: A stupid question, to contradict finiteness of the above mentioned set, is it sufficient to show that any two such $(-1)$-curves are not linearly equivalent? $\endgroup$ – Ron Apr 9 '15 at 8:36
  • 1
    $\begingroup$ Well, if the curves are pairwise linearly inequivalent, the corresponding line bundles are pairwise non-isomorphic, hence you have infinitely many non-isomorphic invertible sheaves on $X$ with the same Hilbert polynomial, right? $\endgroup$ – Francesco Polizzi Apr 9 '15 at 8:39
  • $\begingroup$ @Polizzi: My confusion was, this is only showing pairwise non-isomorphic i.e., the cardinality of the set is greater than $2$. Now how do I conclude that the cardinality is not finite? $\endgroup$ – Ron Apr 9 '15 at 8:44
  • 1
    $\begingroup$ Let $\mathcal{S}$ be the set of $(-1)$-curves and let $$f \colon \mathcal{S} \to \textrm{Pic}(X)$$ be the map sending each curve into its linear equivalence class. Since any two $E_1$ and $E_2$ in $\mathcal{S}$ are not linear equivalent, $f$ is an injective map. So its image is infinite, because $\mathcal{S}$ is infinite by construction. $\endgroup$ – Francesco Polizzi Apr 9 '15 at 8:48
  • $\begingroup$ @Polizzi: Thank you very much. Very helpful. Do you know of any criterion, under which the above mentioned set in my question will be finite? $\endgroup$ – Ron Apr 9 '15 at 8:51

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.