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Let $K$ be a number field, $\bar{K}$ a fixed algebraic closure, $G_K$ the absolute galois group, $O_\bar{K}$ the ring of integers of the algebraic closure, $\mathfrak{p}$ runs over prime ideals and subscripts mean p-adic completions.

Define the "Integral Shafarevich-Tate group" of $GL_2(O_\bar{K})$, a $G_K$-module, to be (normal galois cohomology follows): $$\mathrm{III}^{int}(GL_2) = Ker\\ \lgroup\\ H^1(G_K, GL_2(O_\bar{K}) \rightarrow \prod_{\mathfrak{p}} H^1(G_{K_{\mathfrak{p}}},GL_2(O_\bar{K_\mathfrak{p}})) \rgroup$$

For $GL_1(O_\bar{K})$ it is known* to be isomorphic to the class group of $K$. My question is: $$\text{What is }\mathrm{III}^{int}(GL_2)?$$

It is not clear that the above defines a group since $GL_2$ is not abelian. The article "Abelianization of the First Galois Cohomology of Reductive Groups", M. Borovoi, shows that reductive groups over a number field have abelian Sha (over the field - not "integral" Sha). It might not be related, but I suspect something similar would work for the above.

The same question for $GL_n$, quaternion units, etc. is also very interesting.

(*) The case of $GL_1$ is proved in "Visibility of Ideal Classes" by Schoof and Washington (arxiv:0809.5209).

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    $\begingroup$ 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. $\endgroup$
    – Kevin Buzzard
    Commented Oct 26, 2010 at 18:54
  • $\begingroup$ 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 $G = GL_n$, $Z(\hat G) = GL(1)$. $\endgroup$
    – Marty
    Commented Oct 27, 2010 at 4:42
  • $\begingroup$ 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. $\endgroup$
    – BCnrd
    Commented Oct 28, 2010 at 3:33

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