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This is coming out of Mumford's GIT, section 7.2, page 131.

$A/S$ an abelian scheme of dimension $g$ with polarization $\bar{\omega}$ of degree $d^2$. Then $\pi_*(L^\Delta(\bar{\omega})^3)$ is locally free on $S$ of rank $6^gd$ which defines the closed immersion $\varphi_3 : A \rightarrow \mathbb{P}(\pi_{*}(L^\Delta(\bar{\omega})^3))$. Equip this with a linear rigidification $\phi : \mathbb{P}(\pi_{*}(L^\Delta(\bar{\omega})^3)) \rightarrow \mathbb{P_m} \times S$ so that we get an embedding $I : A \rightarrow \mathbb{P_m} \times S$.

Mumford then states the Hilbert Polynomial of $I(A)$ is easily computed to be $P(X) = 6^gdX^g$.

Exactly how does one go about finding this Hilbert polynomial?

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[edited only to insert "r" into the title's "Hilbert"] –  Noam D. Elkies Feb 25 '12 at 20:05
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up vote 2 down vote accepted

Let us look at a single geometric fiber $X$. Let $L^\Delta(\bar\omega)|_X = \mathcal{O}_X(D)$. The Riemann-Roch theorem for abelian varieties (Mumford "Abelian Varieties", Chap. 3 Section 16) states that $$ \chi(\mathcal{O}_X(D)) = D^g/g!$$ and moreover that $\chi(\mathcal{O}_X(D))^2 = \deg \phi$, where $\phi$ is the polarization map defined by $\mathcal{O}_X(D)$. So the Hilbert polynomial $\chi(\mathcal{O}_X(3nD)) = 3^gn^gD^g/g! = 3^g n^g \chi(\mathcal{O}_X(D)) = 3^g n^g d$. I got almost the right answer (where did $2^g$ go?), so maybe I misunderstood the question, but I hope this is still helpful.

EDIT. Is the superscript $\Delta$ the symmetrization of the line bundle in question? Then it would explain why the $2^g$ above is missing...

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$L^\Delta(\bar{\omega}) = \Delta^*((1_A \times \bar{\omega})^*(P))$, $P$ the Poincare bundle, $\Delta$ the diagonal. –  rghthndsd Feb 25 '12 at 18:47
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