1
$\begingroup$

Consider the following situation: $X/k$ is a smooth projective geometrically connected curve with Albanese $A$ and Abel-Jacobi map $AJ: X \to A$, and $B/k$ an abelian variety.

Let $\alpha \in \mathrm{Hom}_k(A,B)$ and $\beta \in \mathrm{Hom}_k(B,A)$ be homomorphisms with graphs $\Gamma_\alpha \subseteq A \times_k B$ and $\Gamma_\beta \subseteq B \times_k A$.

Set $\gamma(\alpha): X \to A \to B$ and $\gamma'(\beta): X \to A \to A^\vee \to B^\vee$ where $A \to A^\vee$ is the canonical principal polarisation ($X$ is a curve!) and $\beta^\vee: A^\vee \to B^\vee$.

Now I want to relate $\mathrm{deg}_{X \times_k B}([\Gamma_{\gamma(\alpha)}] . {}^t[\beta])$ (where $[\beta]$ is the line bundle given by $B^\vee(X) = \mathrm{Pic}^0(B \times_k X)$ and ${}^t$ switches the factors) and $\mathrm{deg}_{A \times_k B}([\Gamma_\alpha] . [{}^t\Gamma_\beta])$. Are they equal?

EDIT: Here is what I made of Harry's response:

Let $AJ_B = AJ \times_k \mathrm{id}_B: X \times_k B \hookrightarrow A \times_k B$ be the closed immersion [$g>0$]. When one takes the base change of $\Gamma_\alpha \subseteq A \times_k B$ by $AJ_B$, one gets $\Gamma_{\gamma(\alpha)} = \Gamma_{\alpha \circ AJ} \subseteq X \times_k B$. $[\beta]$ is the line bundle associated to $\gamma'(\beta) \in \mathrm{Pic}^0(B \times_k X)$ and ${}^t$ indicates the switch of the two factors.

$\deg_{X \times_k B}([\Gamma_{\gamma(\alpha)}] . {}^t[\beta])$

$= \deg_{X \times_k B}(AJ_B^*[\Gamma_{\alpha}] . {}^t[\beta])$

$= \deg_{A \times_k B}(AJ_{B,*}(AJ_B^*[\Gamma_{\alpha}] . {}^t[\beta])) $ (as $AJ_B$ is a closed immersion)

$= \deg_{A \times_k B}([\Gamma_{\alpha}] . AJ_{B,*}({}^t[\beta]))$ by the projection formula ($AJ_B$ is proper)

$= \deg_{A \times_k B}([\Gamma_{\alpha}] . [{}^t\Gamma_{\beta}])$ (???)

$\endgroup$
2
  • $\begingroup$ Would you mind elaborating on the definition of the line bundle $[\beta]$? $\endgroup$
    – Harry
    Mar 20, 2012 at 15:18
  • 2
    $\begingroup$ Did you try applying the projection formula from $X\times_k B \to A\times_k B$? $\endgroup$
    – Harry
    Mar 20, 2012 at 15:21

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.