MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer

Manin's ICM talk at Nice (1970) might be a good place to start.

If you read French, an elementary introduction to the local-to-global principle and to the Manin obstruction can be found on David Harari's homepage. This article also appeared in the Gazette des Mathématiciens (Janvier 2006).

A detailed treatment can be found in the book

MR1845760 (2002d:14032) Skorobogatov (Alexei), Torsors and rational points. Cambridge Tracts in Mathematics, 144. Cambridge University Press, Cambridge, 2001. viii+187 pp.

share|cite|improve this answer
For reference: I. Manin, "Le groupe de Brauer-Grothendieck en géométrie diophantienne", Proc. Internat. Congr. Math. (Nice), 1970. – Timo Keller Jan 10 '10 at 12:49
You might be especially interested in Theorème 4.1 in Harari's article. – Chandan Singh Dalawat Jan 10 '10 at 12:57
Strictly speaking, there is no obstruction whatsoever to the existence of rational point on an abelian variety : the origin is a rational point ! What you have in mind is that if C is smooth projective curve of genus >0 with jacobian J, or if C is a torsor under an abelian variety J, everything being defined over some number field K, and if C has K_v-points at every place v of K, then the Manin obstruction to the existence of a K-point is the only one as long as Sha(J) is finite. – Chandan Singh Dalawat Jan 10 '10 at 13:26
@Chandan This has not been proved for curves, only for torsors of abelian varieties. – Felipe Voloch Jan 10 '10 at 15:13
@Felipe : You are right. It is perhaps expected to be true for curves of genus >1. – Chandan Singh Dalawat Jan 11 '10 at 4:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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