What is the precise relationship between the Selmer group of an abelian variety and that of its dual? For instance, does the vanishing of one not imply the same for the other?

To fix ideas, let $A$ be an abelian variety defined over a number field $K$, with $A^t$ denote the corresponding dual abelian variety. Fix a rational prime $p$. Suppose for instance that we consider the compactified Selmer group \begin{align*} \mathfrak{S}(A/K) &= \ker \left( H^1(G_S(K), A_{p^n}) \longrightarrow \bigoplus_{v \in S} H^1(K_v, A(\overline{K}_v) )(p) \right).\end{align*} Here, $S$ is any finite set of primes of $K$ containing the primes above $p$ and the primes where $A$ has bad reduction; $G_S(K) = \operatorname{Gal}(K^S/K)$, where $K^S$ is the maximal extension of $K$ unramified outside of $S$ and the archimedean primes of $K$, and $A_{p^n} = \ker \left([p^n]:A \longrightarrow A\right)$ denotes the $p^n$ torsion of $A$. Does $\mathfrak{S}(A/K) =0$ if and only if $\mathfrak{S}(A^t/K)=0$? Clearly this will be the case if $A$ is principally polarized (in which case there is an isomorphism $A \cong A^t$). However, the general case seems tricky to prove by inspection. I am aware that if the $p$-primary part of the Tate-Shafarevich group $\operatorname{Sha}(A/K)$ is finite, then its follows from basic properties of the Cassels-Tate pairing that $\mathfrak{S}(A/K) \cong \mathfrak{S}(A^t/K)$ if and only if $A(K)_{p^{\infty}} \cong A^t(K)_{p^{\infty}}$ (where $A(K)_{p^{\infty}} = \bigcup_{n \geq 0} A(K)_{p^n}$). But, when is it the case that this latter condition is known (not) to be true? Is there not a better, perhaps unconditional, deduction? Surely this ought to be classical ...

  • $\begingroup$ Is the last $\cong$ in your question, just any isomorphism of groups or do you want to take the one induced by an isogeny from $A$ to its dual ? If $p$ divides the degree of this isogeny then it might be that only in one direction the isogeny gives an isomorphism (and maybe none, I don't know), but I have no examples at hand. $\endgroup$ – Chris Wuthrich Mar 9 '11 at 14:50
  • $\begingroup$ I think that any isomorphism should do. And I agree that if p divides the isogeny, then it is not at all clear whether or not such an isomorphism should hold! $\endgroup$ – jvo Mar 9 '11 at 17:28
  • 1
    $\begingroup$ Did you really mean to define $\mathfrac{S}(A/K)$ the way you did ? The compact Selmer group, usually denoted with $\mathfrak{S}(A/K)$, is in the Galois cohomology of $T_p A$, not $A_{p^{\infty}}$. If you change this in the definition then your remarks afterwards are correct. If you want to keep this definition, then the answer of Remke Kloosterman tells you what should be true. $\endgroup$ – Chris Wuthrich Mar 10 '11 at 0:47
  • $\begingroup$ Ah yes, thanks for pointing this out! $\endgroup$ – jvo Mar 10 '11 at 7:12

Let $\varphi:A\to A^t$ be a polarization. Then $\varphi$ is an isogeny. In order to study the difference between the Selmer groups of $A$ and of $A^t$ you need to study the torsion subgroups of $A(K)$ and $A^t(K)$, and to study the difference between the Tate-Shafarevich groups of $A$ and $A^t$. The comparison of the torsion subgroups is quite straightforward.

If you believe the Birch and Swinnerton-Dyer conjecture then $$ \frac{ | Sha(A/K)| }{ | Sha(A^t/K)|} = \frac{\Omega_{A^t}}{\Omega_A}\prod_\ell \frac{c_{A^t,\ell}}{c_{A,\ell}} $$ holds, where $\Omega_A$ is the real period and the $c_{A,\ell}$ are the Tamagawa numbers. Actually, this formula is proven under the hypothesis that $Sha(A/K)$ is finite.

Now, the quotient of the product of Tamagawa numbers might be different from one. Actually, I expect that this product can be arbitrary large. (A similar result can be used to construct elliptic curves with large Tate-Shafarevich groups.)

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.