Finiteness of Obstruction to a Local-Global Principle - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T12:42:57Zhttp://mathoverflow.net/feeds/question/3222http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/3222/finiteness-of-obstruction-to-a-local-global-principleFiniteness of Obstruction to a Local-Global PrincipleJonah Sinick2009-10-29T06:54:51Z2009-10-29T07:26:17Z
<p>Say that a projective variety V over Q satisfies the local-global principle up to finite obstruction (#) if there are only finitely many isomorphism classes of projective varieties over Q that are not isomorphic to V over Q despite being isomorphic to V over every completion of Q..</p>
<p>In section 7 of Barry Mazur's 1993 article titled <em>On the Passage From Local to Global in Number Theory</em>, Mazur describes his attempt to prove that (#) for abelian varieties over Q implies (#) for <em>all</em> projective varieties over Q, and a partial result that he, Yevsey Nisnevich and Ofer Gabber achieved in this direction. Has there been further progress in this direction since 1993?</p>
<p>My understanding is that an effective version of (#) for genus 1 curves (an effective bound on certain Tate-Shafarevich groups) gives a finite algorithm (of <em>a priori</em> bounded running time) for determining whether a genus 1 curve has a rational point, and also that such an effective bound on Tate-Shafarevich groups is expected.</p>
<p>Is an effective version of (#) for general projective varieties over Q expected? If so, how does this relate to Hilbert's 10th problem over Q (which Bjorn Poonen has conjectured to be undecidable)?</p>
http://mathoverflow.net/questions/3222/finiteness-of-obstruction-to-a-local-global-principle/3225#3225Answer by Pete L. Clark for Finiteness of Obstruction to a Local-Global PrinciplePete L. Clark2009-10-29T07:26:17Z2009-10-29T07:26:17Z<p>"Has there been further progress in this area since 1993?"</p>
<p>So far as I know, there has been no direct progress. I feel semi-confident that I would know if there had been a big breakthrough: Mazur was my adviser, this is one of my favorite papers of his, and I still work in this field. Also, I just checked MathReviews and none of the citations to this paper makes a big advance on the problem, although two are somewhat relevant:</p>
<p>MR1905389 Thăńg, Nguyêñ Quoc On isomorphism classes of Zariski dense subgroups of semisimple algebraic groups with isomorphic $p$-adic closures. Proc. Japan Acad. Ser. A Math. Sci. 78 (2002), no. 5, 60--62.</p>
<p>MR2376817 (2009f:14040) Borovoi, M.; Colliot-Thélène, J.-L.; Skorobogatov, A. N. The elementary obstruction and homogeneous spaces. Duke Math. J. 141 (2008), no. 2, 321--364. </p>
<p>I'm not sure what you mean by an effective bound on Shafarevich-Tate groups (henceforth "Sha"). It is certainly expected that the Sha of any abelian variety over a global field is finite. If this is true, then in any given case one can, "in principle", give an explicit upper bound on Sha by the method of n-descents for increasingly large n. (In practice, even for elliptic curves reasonable algorithms have been implemented only for small values of n.) I really can't imagine any algorithm having to do with Sha that has "a priori bounded running time". What do you have in mind here?</p>
<p>As to the final question, let me start by saying that it seems reasonable at least that the set of "companion varieties" (i.e., Q-isomorphism classes of varieties everywhere locally isomorphic to the given variety) of a projective variety V/Q is finite: as above, we believe this for abelian varieties, and Barry Mazur proved in this paper a lot of results in the direction that the conjecture for abelian varieties implies it for arbitrary varieties. (For instance, quoting from memory, I believe he proved the implication for all varieties of general type.)</p>
<p>Here is a key point: suppose you are given a variety V/Q and you are wondering whether it has rational points. If V is itself a torsor under an abelian variety (e.g. a genus one curve), then if you can compute Sha of the Albanese abelian variety of V, you can use this to determine whether or not V has a Q-rational point. In general, the connection between computation of sets of companion varieties of V and deciding whether V has a Q-rational point is less straightforward. If V is a curve, then there are theorem in the direction of the fact that finiteness of Sha(Jac(V)) implies that the Brauer-Manin obstruction is the only one to the existence of rational points on V. In particular, people who believe this (including Bjorn Poonen, I think), believe that there is an algorithm for deciding the existence of rational points on curves. But nowadays we know examples of varieties where the Brauer-Manin obstruction is not sufficient to explain failure of rational points.</p>
<p>So, in summary, it is a perfectly tenable position to believe that companion sets are always finite, even effectively computable, but still there is no algorithm to decide the existence of Q-points on an arbitrary variety.</p>