1
$\begingroup$

Let $A$ and $B$ be Abelian Varieties of dimension $d$ over a local field $K$. Let $\phi : A \rightarrow B$ be an isogeny and $\phi^{\prime}$ its dual.

Recall that one has a canonical isomorphism $can_A : H^0(A, {\Omega_A}^d) \cong H^0(A^{\prime}, {\Omega_{A^{\prime}}}^d)$, where $A^{\prime}$ is the dual Abelian variety. Namely, $H^0(A, {\Omega_A}^d) \cong (H^d(A, \mathcal{O}))^* \cong \Lambda^d H^1(A, \mathcal{O})^* \cong H^0(A^{\prime}, {\Omega_{A^{\prime}}}^d)$ (the last isomorphism comes from $H^1(A, \mathcal{O}) \cong Lie(A^{\prime})$).

Is it true that in the following diagram

$\require{AMScd}$ \begin{CD} H^0(A, {\Omega_A}^d) @>can_A>> H^0(A^{\prime}, {\Omega_{A^{\prime}}}^d)\\ @AA \phi^* A @VV {\phi^{\prime}}^* V\\ H^0(B, {\Omega_B}^d) @>>can_B> H^0(B^{\prime}, {\Omega_{B^{\prime}}}^d) \end{CD}

$ {\phi^{\prime}}^* \circ can_A \circ \phi^* = deg(\phi) can_B$?

For example, it is easy to see that the assertion holds when $A$ and $B$ are elliptic curves.

$\endgroup$
1
  • 1
    $\begingroup$ Perhaps you can check what happens for the power of an elliptic curve $E^k$ by using the assertion for elliptic curves. The general case would be similar "by deformation". $\endgroup$
    – Kapil
    May 7, 2021 at 14:49

0

Your Answer

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