Timeline for What is the Shafarevich-Tate group of GL(2)?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 28, 2010 at 3:33 | comment | added | BCnrd | Just to explain the origin of Marty's comments, I had made some comments addressing an earlier version of the question (as Marty's comments do...as long as they remain posted above), and after the revision my comments were no longer relevant, so I deleted them. | |
| Oct 27, 2010 at 20:27 | history | edited | Dror Speiser | CC BY-SA 2.5 |
corrected wording; deleted 5 characters in body
|
| Oct 27, 2010 at 16:10 | history | edited | Dror Speiser | CC BY-SA 2.5 |
made distinction between Sha and integral Sha
|
| Oct 27, 2010 at 6:53 | history | edited | Dror Speiser | CC BY-SA 2.5 |
made question more explicit
|
| Oct 27, 2010 at 4:42 | comment | added | Marty | I agree with Brian about the problems in the question as stated. I think of $Sha(G)$, for conn. red. groups $G$ over global fields $K$, as the kernel of the "Hasse" map $H^1(G_K, G(\bar K)) \rightarrow \prod_v H^1(G_v, G(\bar K_v))$, where the product is over all places $v$. Kottwitz (Stab. Trace. Form., Duke) has proven (finished off by someone else for groups with $E_8$ factors) that this $Sha(G)$ is canonically isomorphic to the Pontrjagin dual of $H^1(K, Z(\hat G))$, where $Z(\hat G)$ is the dual group (group with dual root system). Of course, when $G = GL_n$, $Z(\hat G) = GL(1)$. | |
| Oct 26, 2010 at 18:54 | comment | added | Kevin Buzzard | I'm not entirely sure of myself in this area, but I suspect that this might be an easy question if you think about it in the right way. I am wondering whether "pure thought" tells you that for $GL(n)$ this Sha set is parametrising isomorphism classes of locally free modules over the integers $R$ of $K$, of rank~$n$, and by some standard calculation each such thing is isomorphic to $R^{n-1}+I$ for $I$ a locally free rank 1 module whose isomorphism class is uniquely determined. If you can fill in the details of this and it turns out to be OK, it leads to a proof that you get the class gp again. | |
| Oct 26, 2010 at 17:33 | history | asked | Dror Speiser | CC BY-SA 2.5 |