Timeline for Hasse principle for $H^2$ of a maximal torus of a split simply connected group?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 3, 2016 at 13:31 | answer | added | Mikhail Borovoi | timeline score: 2 | |
| Aug 2, 2016 at 19:57 | comment | added | Mikhail Borovoi | @JasonStarr: Let $Z$ denote the centralizer of $T$ in $M_{2n}$, then $Z$ is a commutative $k$-algebra of dimension $2n$, and the $2n$-dimensional representation gives $2n$ eigenspaces, which we divide into $n$ pairs using our nondegenerate skew-symmetric form. The Galois group acts on this set of pairs, and to each orbit of the Galois group we associate a one-dimensional torus over the field corresponding to the stabilizer of a pair in the orbit. Thus $T$ is indeed a product of Weil restrictions of one-dimensional tori. | |
| Aug 2, 2016 at 19:19 | comment | added | Mikhail Borovoi | @JasonStarr: Yes, it seems that any maximal torus $T$ of $\mathrm{Sp}_{2n}$ is a product $\prod T_i$, where each $T_i$ is a Weil restriction of a one-dimensional torus. The argument above shows that $Ш^2$ is trivial for a one-dimensional torus, and hence, for $T$. | |
| Aug 2, 2016 at 19:19 | comment | added | Jason Starr | Conclude that the squaring map on $\Sha^2(k,T)$ is trivial. I do not know what comes next ... I guess that argument (at best) shows that $\Sha^2(k,T)$ is a $2$-torsion group. | |
| Aug 2, 2016 at 19:14 | comment | added | Jason Starr | Here is my line of thinking, probably wrong. Let $(\bullet)^t:\text{GL}_{2n}\to \text{GL}_{2n}$ be the "transpose" involution associated to the perfect nondegenerate pairing, so that $\text{Sp}_{2n}$ is the fixed point locus of the involution $\iota(B) = (B^t)^{-1}$. Let $T$ be a maximal torus in $\text{Sp}_{2n}$. Denote by $C_{GL}(T)$ the centralizer of $T$ in $\text{GL}_{2n}$. Consider the morphism $f:C_{GL}(T)\to C_{GL}(t)$ by $f(B) = \iota(B)B$. Argue (geometrically) that $f$ is a group homomorphism onto $T$ that restricts to the squaring homomorphism on $T$ . . . | |
| Aug 2, 2016 at 18:34 | comment | added | Jason Starr | "Could you please elaborate?" No, probably I cannot. I just wrote something that came into my head. I will think it through more carefully and write more if I have an actual argument. | |
| Aug 2, 2016 at 18:26 | comment | added | Mikhail Borovoi | @JasonStarr: Could you please elaborate? | |
| Aug 2, 2016 at 18:22 | comment | added | Jason Starr | Presumably the same argument holds for $G=\text{Sp}_{2n}$. | |
| Aug 2, 2016 at 17:42 | history | asked | Mikhail Borovoi | CC BY-SA 3.0 |