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?



Hilbert polynomial of an abelian scheme
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...


