Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am looking at the Shimura variety for $\mathrm{GSp}_4(\mathbb Q)$, with hyperspecial level structure at $p$. Let $X$ denote the special fiber over $\mathbb F_p$. For simplicity, let us pretend that $X$ is already compact (I prefer to worry about the compactification later; or I could use an inner form of $\mathrm{GSp}_4$ instead.) For non-number theorists, this is probably known as a quotient of the three dimensional Siegel space by a discrete subgroup. Anyways, we have a universal abelian surface $f: A \to X$ and let $\omega$ denote the (rank two) sheaf of invariant differential forms $f_*\Omega_{A/X}^1$.

Consider the projective bundle $Y = \mathbb P(\omega)$; it is a natural $\mathbb P^1$-bundle over $X$. Let $g: Y \to X$ denote the structure morphism. Let $a, b \in \mathbb Z$ be two numbers (possibly negative). I would like to know when is the following line bundle ample over $Y$: $$ \mathcal L_{a,b} = \mathcal O(a) \otimes g^*(\wedge^2\omega)^b. $$ Here $\mathcal O(1)$ is characterized by the property that $g_*\mathcal O(1) = \omega$; it is the universal rank one quotient of $g^*\omega$.

I asked my friend a little bit on this types of questions; and the answer I got is that this has something to do with the stability of the vector bundle $\omega$. Based on this, my guess is that $\mathcal L_{a,b}$ is ample if (and only if) $a>0$ and $a+2b>0$.

As a number theorist, I have very little idea how one should approach such a question. I would appreciate your great help. Thank you! (PS, I do have a good application in mind for this; so it is not a random question.)

share|cite|improve this question
I don't know if we can give a hard bound, but, since $\wedge^2\omega$ is ample over the Siegel threefold, $\mathcal{L}_{a,b}$ will be ample for any $a>0$ and $b>>0$ by Lemma 25.38.7 of the Stacks Project here: –  Keerthi Madapusi Pera Aug 3 '13 at 13:37
Just two remarks: (1) $L_{a,b}$ is not ample if $b=0$, for otherwise $\omega$ would be ample, which it is not (see Moret-Bailly's construction of abelian schemes of relative dimension $2$ over the projective line). (2) if ${\rm det}(\omega)^b\otimes\omega$ is generated by its global sections, then $L_{1,b}$ is very ample and thus ample (but you probably already knew that). –  Damian Rössler Aug 3 '13 at 13:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.