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}} = A^t(K)_{p^{\infty}}$$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 ...