# How to determine the type of a divisor on a product of elliptic curves?

Say $$E_1, \dotsc, E_n$$ are elliptic curves (everything over $$\mathbb C$$), and $$D \subset E_1 \times \dotsc \times E_n$$ is an effective divisor. How can I determine the type $$(d_1, \dotsc, d_n)$$ of the line bundle $$\mathcal O(D)$$?

If $$D = \sum_i \operatorname{pr}_i^* D_i$$ for divisors $$D_i \subset E_i$$, then $$d_i = \deg D_i = \deg (D \cdot \iota_i(E_i))$$, where $$\iota_i: E_i \to E_1 \times \dotsc \times E_n$$ embeds $$E_i$$ by $$z \mapsto (0, \dotsc, z, \dotsc, 0)$$. Is it true in general that one gets $$d_i$$ as the degree of the intersection of $$D$$ with $$\iota_i(E_i)$$?

Concretely, I'm interested in the following two cases:

1. Let $$D_0 \subset E \times E$$ be the union of the diagonal and the anti-diagonal,i.e. $$D_0 = \{(z,z) : z \in E\} \cup \{(z,-z):z\in E\}.$$
2. Take the quotient $$A = E \times E / \langle (\frac 1 2,\frac 1 2)\rangle$$, by the $$2$$-torsion point $$(\frac 1 2, \frac 1 2)$$, and divisor $$D_1 = \overline{D_0}$$. Using the isomorphism $$E \times E \to E \times E, (z_1, z_2) \mapsto (z_1, z_1 - z_2)$$ one sees that $$A$$ is isomorphic to $$E / \langle \frac 1 2 \rangle \times E$$, so again a product of ellpitic curves.

Any help would be appreciated :)

• What do you mean by the type of a divisor on $\prod_{i=1}^n E_i$? In other words, what is the definition of $d_i$? It is not true that $\operatorname{Pic}(\prod_i E_i) \cong \prod_i \operatorname{Pic}(E_i)$, and the same statement for $\operatorname{NS}$ also fails. Commented Jun 26, 2023 at 19:41
• @R.vanDobbendeBruyn If $L$ is a line bundle on a complex torus $X = V/\Gamma$, it's first Chern class can be thought of as a Hermitian form on $V$, whose imaginary part $E=\Im H$ is an alternating real form that takes integral values on the lattice $\Gamma$. By the theorem on elementary divisors, there is a symplectic basis of $\Gamma$ such that $E = \left( \begin{matrix} 0 & D \\ -D & 0\end{matrix} \right)$, and $D = \operatorname{diag}(d_1, \dotsc, d_n)$. Then $(d_1,\dots,d_n)$ is called the type of $L$. Commented Jun 26, 2023 at 20:13
• What you write about Picard groups is why I'm cautious with just intersecting $D$ with the $E_i$. Commented Jun 26, 2023 at 20:15

I managed to calculate my examples. In the first one, $$D_0$$ has indeed polarization type $$(2,2)$$. To see this, let $$E = \mathbb C / (\mathbb Z + \tau \mathbb Z)$$ be an elliptic curve, and consider the isogeny $$\varphi: E \times E \to E \times E, (z_1, z_2) \mapsto (z_1 + z_2, z_1 - z_2).$$ Then $$\varphi^2(z_1, z_2) = (2z_1, -2z_2)$$, so $$\deg \varphi^2 = 16$$ and $$\deg \varphi = 4$$.
Note that $$D_0 = \varphi^* \tilde D$$, where $$\tilde D$$ is the divisor $$\tilde D = E \times \{0\} \cup \{0\} \times E.$$ Clearly $$\tilde D$$ has type $$(1,1)$$, and the corresponding hermitian form on $$\mathbb C^2$$ is $$H\left(\left(\begin{smallmatrix}u_1 \\ u_2 \end{smallmatrix}\right),\left(\begin{smallmatrix}v_1 \\ v_2 \end{smallmatrix}\right)\right) = \frac 1 {\operatorname{Im}\tau} (u_1 \bar v_1 + u_2 \bar v_2).$$ If $$\phi: \mathbb C^2 \to \mathbb C^2$$ is the analytic representation of $$\varphi$$, an easy calculation shows $$\phi^* H = 2H$$. Hence $$D$$ has type $$(2,2)$$.
I think my second example actually serves as a counterexample to the claim that the $$d_i$$ can be computed as $$E_i \cdot D$$. The projection $$E \times E \to A$$ has degree $$2$$, and the above isogeny $$\varphi$$ factors over $$A$$, so $$\tilde D$$ has to pull-back to a polarization of type $$(1,2)$$ on $$A$$ (for combinatorial reasons). But under the identification $$A \cong F \times E$$, with $$F = E / \langle \frac 1 2 \rangle$$, $$D_1$$ is given by $$D_1 = (F \times 0) \cup \{(z,2z)\} \subset F \times E.$$ Then $$0 \times E$$ intersects both components in the point $$(0,0)$$; $$F \times 0$$ doesn't intersect itself, but intersects the second component in the points $$(0,0)$$ and $$(\frac \tau 2, 0)$$. All intersections are transversal, so this would give a polarization type $$(2,2)$$.↯