I already asked this on [Math.SE](https://math.stackexchange.com/questions/4722773/how-to-determine-the-type-of-a-divisor-d-on-a-product-of-elliptic-curves), but didn't receive an answer yet.




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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 $\iota\_i(E\_i)$?




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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 :)



How to determine the type of a divisor on a product of elliptic curves?
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 $(2,2)$.↯


