Timeline for Decomposing adelic points using torsors
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 24, 2014 at 10:30 | comment | added | Michael Stoll | To come back to the original question, there are many examples where $X({\mathbb A}_k) \neq \emptyset$, but $\bigcup_\tau f^\tau(Z^\tau({\mathbb A}_k)) = \emptyset$. Then one has a "descent obstruction" (w.r.t. $f$) to the existence of $k$-rational points on $X$. This is a very powerful method of proving that $X(k) = \emptyset$. For example, every instance of the Brauer-Manin obstruction can be realized in this way. See Skorobogatov's book for details. | |
| Nov 23, 2014 at 20:23 | comment | added | Michael Stoll | @user74230: Of course. I wanted to point out where the problem lies, without being too precise. In any case, thanks for your very detailed answer ! | |
| Nov 23, 2014 at 19:37 | comment | added | user74230 | I'm not sure what "essentially equivalent" means, but the question for adelic points is rather weaker than the surjectivity of the global-to-local map as written in the above comment because an "almost everywhere integral" condition is missing on the right side. However, even with that condition imposed on the target, the global-to-local map is generally not surjective for finite $G$, and even for "most" connected reductive $G$. | |
| Nov 23, 2014 at 11:39 | comment | added | Michael Stoll | In fact, when $X$ is a smooth, projective and geometrically irreducible curve, the (or at least my) expectation is that if you consider torsors under finite abelian $G$, then the union of the images of adelic points should converge to the set of $k$-rational points as the degree of the covering tends to infinity (see mathe2.uni-bayreuth.de/stoll/schrift.html#AG16). | |
| Nov 23, 2014 at 11:12 | comment | added | Michael Stoll | This is true for $k$-points. The question for adelic points is essentially equivalent to asking whether $$H^1(k, G) \to \prod_v H^1(k_v, G_{k_v})$$ is surjective, which is certainly wrong for nontrivial finite $G$ (the left hand side is countable, but the right hand side is not). | |
| Nov 23, 2014 at 11:09 | history | edited | Michael Stoll | CC BY-SA 3.0 |
Corrected the counterexample
|
| Nov 23, 2014 at 11:04 | comment | added | user62153 | I was under the impression that you can always lift adelic points along a torsor (i.e. there is a twist of that torsor under which the point lifts). This impression is clearly wrong, but is there any nice setting in which this happens? | |
| Nov 23, 2014 at 11:00 | vote | accept | user62153 | ||
| Nov 23, 2014 at 10:57 | history | answered | Michael Stoll | CC BY-SA 3.0 |