[Math] Brauer-Manin obstruction and Tate-Shafarevich group of an Abelian variety

Question

I read that the Brauer-Manin obstruction $A(\mathbb{A}_K)^{\mathbf{Br}}$ of an Abelian variety $A$ over a number field $K$ equals (naturally?) its Tate-Shafarevich group $\mathrm{III}(A)$.

Is this true? And if so, where can I find a proof?

Answer 1

The quotient of what you called the Brauer-Manin obstruction by the closure of $A(K)$ within it is related to the divisible part of Sha. In particular, if Sha has no divisible part (e.g. if it is finite) then the Brauer-Manin obstruction is the closure of $A(K)$. See L. Wang, Brauer-Manin obstruction to weak approximation on abelian varieties, Israel J. Math. 94 (1996), 189–200.

Note that these two groups in your question are very different, so they can't be equal. For instance, Sha is torsion, but the Brauer-Manin obstruction usually isn't.