Let $X$ be a smooth, projective variety over an algebraically closed field of characteristic zero. Fix a polarisation on $X$. Let $\mathcal{L}$ be an invertible sheaf on $X$ with Hilbert polynomial, say $P(t)$. Denote by $P'(t)$ the Hilbert polynomial of the dual $\mathcal{L}^\vee$. Is there any relation between the roots of $P(t)$ and $P'(t)$? In particular, I am looking for examples when $P(t)=P'(t+b)$ for some integer $b$.
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1$\begingroup$ Are you measuring the Hilbert polynomial of $\mathcal{L}$ with respect to a specified ample invertible sheaf $\mathcal{A}$ on $X$, i.e., $P(d) = \chi(X,\mathcal{A}^{\otimes d}\otimes \mathcal{L})$ for every integer $d\geq 0$? Or is $P(t)$ the Hilbert polynomial of $\mathcal{L}$ "with respect to itself", i.e., $P(d)=\chi(X,\mathcal{L}^{\otimes d})$ for every integer $d\geq 0$? $\endgroup$– Jason StarrCommented May 23, 2018 at 18:22
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$\begingroup$ @JasonStarr with respect to a specified ample invertible sheaf. $\endgroup$– JanaCommented May 23, 2018 at 19:30
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$\begingroup$ In general there is no direct relationship between $P(t)$ and $P'(t)$. However, by Snapper's Lemma, the function $p(a,b)=\chi(X,\mathscr L^{\otimes a} \otimes \mathcal O(b))$ is a polynomial in $\mathbb Q[a,b]$; you are interested in comparing $p(1,b)$ with $p(-1,b)$. Of course it is easy to make examples where these are very different (already on $\mathbb P^n \times \mathbb P^m$, I believe). $\endgroup$– R. van Dobben de BruynCommented May 23, 2018 at 20:34
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1 Answer
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For the sake of simplicity, let us assume that $\mathcal{L}$ is a polarization, and that we are computing the Hilbert polynomial of $\mathcal{L}$ with respect to itself, i.e. $P(t)=\chi(X, \, \mathcal{L}^{\otimes t})$.
If $\mathcal{L}$ is such that $K_X=(\mathcal{L^{\vee}})^{\otimes b}$, then by Serre duality we obtain $$P'(t+b)=\chi(X, \, (\mathcal{L}^{\vee})^{\otimes t+b}) = \chi(X, K_X \otimes \mathcal{L}^{\otimes t+b})=\chi(X, \, \mathcal{L}^{\otimes t})= P(t).$$
For instance, if $X=\mathbb{P}^{n-1}$ and $\mathcal{L}=\mathcal{O}_{\mathbb{P}^{n-1}}(1)$, we have $K_X = (\mathcal{L}^{\vee})^{\otimes n}$ and so $$P'(t+n)=P(t).$$
answered May 23, 2018 at 18:31