# Does a projective variety have only finitely many associated Hilbert polynomials?

Let $$X$$ be a projective variety over $$\mathbb{C}$$. If $$L$$ is an ample line bundle, then $$h_L$$ denotes the Hilbert polynomial.

Is it true that, if $$L$$ and $$L'$$ are ample line bundles which are equal in the Neron-Severi group, then $$h_L = h_{L'}$$?

Does this imply, together with finite generation of Neron-Severi groups, that the set of polynomials $$\{h_L \ | \ L$$ ample line bundle on $$X \}$$ is finite?

Can anyone recommend a text (book or article) in which these things are explained (to some extent)?

• As Sasha rightly points out, the answer to the second question is negative because you can take linear combinations of line bundles. My answer to this question contains a slightly modified statement you can deduce from finite generation of the Néron–Severi group. Oct 13, 2018 at 15:29
• @R.vanDobbendeBruyn Do I understand correctly that your answer says that, given a choice of generators, one obtains a "universal" polynomial $p$ (in the sense that any other Hilbert polynomial is a very specific type of specialization of this one)? Oct 13, 2018 at 15:54
• Yes, that is the conclusion. (It's universal in a weak sense, because it depends on some choices.) I also worked out one non-trivial example, to show that it's not so easy to get a complete parametrisation of all Hilbert polynomials obtained this way. Oct 13, 2018 at 15:55

No for the second --- even in the simplest case of a projective line, the polynomial $$td + 1$$ is the Hilbert polynomial (with respect to $$L = O(d)$$).
• You don't really need Riemann-Roch, only that $\chi$ is constant in an algebraic family.
• Should that be $td +1$? I thought the degree of the Hilbert polynomial always equals the dimension of the variety. Oct 13, 2018 at 15:53