I already asked this on Math.SE, but didn't receive an answer yet.

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 :)

  • $\begingroup$ 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. $\endgroup$ Commented Jun 26, 2023 at 19:41
  • $\begingroup$ @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 $L$. $\endgroup$ Commented Jun 26, 2023 at 20:13
  • $\begingroup$ What you write about Picard groups is why I'm cautious with just intersecting $D$ with the $E_i$. $\endgroup$ Commented Jun 26, 2023 at 20:15

1 Answer 1


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 $(2,2)$.↯


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