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Let $\mathcal{A}_{g,D}$ be the moduli space of abelian varieties of dimension $g$ and polarization $D$ of type $(d_1, \ldots, d_g)$.

Let $\mathcal{M}$ be the moduli space parametrizing pairs $(A, \mathcal{L})$, where $A \in \mathcal{A}_{g,D}$ and $\mathcal{L}$ is a non-trivial $2$-torsion line bundle on $A$, i.e. a non-zero element of $\textrm{Pic}^0(A)$ such that $\mathcal{L}^{\otimes 2}=\mathcal{O}_A$.

Then there is a covering

$\pi \colon \mathcal{M} \longrightarrow \mathcal{A}_{g,D}$

of degree $2^{2g}-1$, given by $[A, \mathcal{L}] \to A$.

Question Is the monodromy group of $\pi$ transitive? Or, equivalently, is $\mathcal{M}$ connected?

The answer is yes when $g=1$, i.e. for elliptic curves. In fact in this case $\mathcal{M}$ is a particular case of a more general construction called the moduli space of spin curves, which was studied by several authors (Cornalba, Verra, Farkas, etc).

What about the case $g \geq 2$? Is there any reference? I'm particularly interested to the case where $g=2$ and $D=(1,2)$.

EDIT Let me explain better the case I'm interested in, hoping that this can be helpful. Let $(A, D)$ be an abelian surface with polarization of type $(1,2)$, which I assume to be not of product type. The linear system $|D|$ is a pencil, that is $h^0(D)=2$, its general element is irreducible and up to a translation we can take $\mathcal{O}_A(D)$ symmetric, i.e. $(-1)_A^* \mathcal{O}_A(D)= \mathcal{O}_A(D)$. Therefore the base locus of $|D|$ is given by the zero element $o$ of $A$ and by three $2$-division points $e_1$, $e_2$, $e_3$ such that $e_1+e_2=e_3$.

There are exactly three $2$-torsion line bundles $\mathcal{L}_1$, $\mathcal{L}_2$, $\mathcal{L}_3$ on $A$ such that there exists an element in the "translated pencil" $|D \otimes \mathcal{L}_i|$ having a double point at $o$ (which is easily proven to be a node). The set

$\{\mathcal{O}_A, \mathcal{L}_1, \mathcal{L}_2, \mathcal{L}_3\}$

form a subgroup of $\textrm{Pic}^0(A)$ isomorphic to $\mathbb{Z}/(2) \times \mathbb{Z}/(2)$, which is exactly the image of the map

$\phi \colon A[2] \longrightarrow \textrm{Pic}^0(A)[2]$,

see Laurent Moret-Bailly's answer.

edited Dec 21 2010 at 14:51

asked Dec 20 2010 at 16:05

Francesco Polizzi
19k●2●36●56

2

When the polarization has even degree, as in your case, won't some of the 2-torsion points in Pic^0(A) be in the image of A[2] under the polarization, and some not? – JSE Dec 20 2010 at 17:26

I'm not sure whether I understand you correctly, however you are right: the polarization allows one to "select" some of the 2-torsion line bundles. I do not know if this can help, anyway I've edited the answer... – Francesco Polizzi Dec 20 2010 at 17:47

What I meant is what L M-B explained much better below! – JSE Dec 21 2010 at 6:16

Ok, now I understand. Thank you for your comment. – Francesco Polizzi Dec 21 2010 at 14:48

Laurent Moret-Bailly · Accepted Answer · 2010-12-20 18:28:30Z UTC

6

Let's stick to $D=(1,2)$. Associated to any point of $\mathcal{M}$ (say, a complex point) we have an abelian surface $A$ and an isogeny $\phi:A\to A'$ where $A'=\mathrm{Pic}^0(A)$ is the dual. The kernel of $\phi$, in our case, is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$. So, the image of $A[2]$ in $A'$ is a canonically defined subgroup $H$ of $A'[2]$, isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$. On the other hand, the line bundle $\mathcal{L}$ is a nontrivial point in $A'[2]$. Clearly, those $(A,\phi, \mathcal{L})$ such that $\mathcal{L}\in H$ form a nontrivial open and closed subset of $\mathcal{M}$, so $\mathcal{M}$ is not connected.

answered Dec 20 2010 at 18:28

Laurent Moret-Bailly
7,296●19●29

	Thank you for your nice answer. I've checked that the subgroup $H$ actually coincides with the group $\{\mathcal{O}, \mathcal{L}_1, \mathcal{L}_2, \mathcal{L}_3 \}$ in my question. So your argument implies that $\mathcal{M}$ splits as a disjoint union $\mathcal{M}_1 \cup \mathcal{M}_2$, where $\pi_1 \colon \mathcal{M}_1 \to \mathcal{A}$ is a triple cover corresponding to the non-zero elements of $H$ and $\pi_2 \colon \mathcal{M}_2 \to \mathcal{A}$ is a cover of degree $12$. Now it seems to me that both $\mathcal{M}_1$ and $\mathcal{M}_2$ should be connected... – Francesco Polizzi Dec 21 2010 at 14:57
	...since I see no canonical defined subset of $A'[2]$, apart from $H$. May I ask your opinion? Thanks again. – Francesco Polizzi Dec 21 2010 at 15:02
	You should be able to express the moduli space as the quotient of the Siegel upper half plane by some discrete group Gamma commensurable with Sp_{2g}(Q). (I always find Mumford's Tata Lectures on Theta, vol I, a congenial place to look up how this works.) Gamma will come with an embedding into Sp_{2g}(Z_2), and the action of Gamma on the Z_2 / 2Z_2 is the one whose orbits you want to study. – JSE Dec 21 2010 at 16:51
	@JSE Right, Mumford book is already on my desk :-) I'll try to carry out the computation you are suggesting. – Francesco Polizzi Dec 21 2010 at 17:00
	I a m not an expert but I think the general idea is that the disconnectedness of such moduli spaces always has an "obvious" explanation. The easiest way is probably to describe $\mathcal{A}_{g,D}$ as the (stack-theoretic) quotient of Siegel's upper half space $\mathcal{S}$ by some appropriate symplectic group $\Gamma$, making $\Gamma$ the fundamental group of $\mathcal{A}_{g,D}$ as an analytic stack. Then it remains to describe the covering $\mathcal{M}$ as a $\Gamma$-set and count the orbits. I would be surprised if all this were not already in the literature (Igusa?). – Laurent Moret-Bailly Dec 21 2010 at 17:01

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$2$-torsion line bundles on abelian varieties

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