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$\DeclareMathOperator{\End}{End}$ Let $J=\mathop{Jac}(C)$ be a Jacobian, $\Theta$ the theta divisor. Since it is principal, it induces a bijection between the Néron-Severri group $NS(J)$ and the symmetric (for the Rosatti-involution) endomorphisms $\End^s(J)$. Here we assume that $C$ has a rational point for simplicity.

When $D$ is a divisor on $J$, we can compute the intersection number $D \cdot C$, and via the isomorphism above this induces a form $q$ on $\End^s(J)$. Can we describe this form $q$ explicitly? This is probably well known but I could not find references.

If I am not mistaken, we have $\Theta \cdot C=g$ by the adjunction formula, so $q(m)=m g$. What about the other symmetric endomorphisms? My guess is that in the real multiplication by $K$ case, $q(\alpha)=\mathop{Tr}_{K/\mathbb{Q}}(\alpha)$.

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    $\begingroup$ For $u\in\operatorname{End}^s(J) $, a standard computation shows that the class of $u^*\Theta $ in $\operatorname{NS}(J) $ corresponds to $u^2$. Therefore $q(u^2)=(u^*\Theta \cdot C)$, and by linearity $q(u)=\frac{1}{2} \bigl((1+u)^*\Theta \cdot C\bigr) -(u^*\Theta \cdot C)-g\,$, admittedly not a very nice formula. $\endgroup$
    – abx
    Commented Feb 1, 2021 at 14:56
  • $\begingroup$ I agree with your formula, but I am not sure how this helps to compute $q(u)$ in practice? Do you have a formula for $(u^\ast \Theta \cdot C)$? $\endgroup$ Commented Feb 2, 2021 at 10:00

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