Timeline for Over a finite field, does a torsor under the component group of G lift to a torsor under G?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 29, 2014 at 23:29 | vote | accept | Question Mark | ||
| Jun 29, 2014 at 21:27 | comment | added | user27920 | As an explicit example for the end of the above comment, if $K = \mathbf{R}$ then there's the $K$-anisotropic form corresponding to the compact Lie group ${\rm{SU}}(n)$ (so the fiber of ${\rm{H}}^1(\mathbf{R}, {\rm{Aut}}_{{\rm{SL}}_n/\mathbf{R}}) \rightarrow {\rm{H}}^1(\mathbf{R},\mathbf{Z}/(2))$ over the nontrivial class in the target has more than one point even though the kernel is trivial). | |
| Jun 29, 2014 at 21:22 | comment | added | user27920 | Showing fibers have $\le 1$ point reduces (by cocycle-twisting) to ${\rm{H}}^1(k,\mathscr{G})=1$ for every $k$-form $\mathscr{G}$ of $G^0$; just for $G^0$ is insufficient. Classifying $K$-forms of ${\rm{SL}}_n$ for a general $K$ uses $G={\rm{Aut}}_{{\rm{SL}}_n/K}$, a semi-direct product of $\mathbf{Z}/(2)$ against $G^0 = {\rm{PGL}}_n$. Here, ${\rm{H}}^1(K,G^0)=1$ but the fiber of ${\rm{H}}^1(K,G)\rightarrow {\rm{H}}^1(K,\mathbf{Z}/(2))$ over separable quadratic $K'/K$ is ${\rm{SU}}(h)$'s for $(-1)^{n+1}$-hermitian $h:{K'}^n\times{K'}^n\rightarrow K'$; one is quasi-split but more can arise! | |
| Jun 29, 2014 at 21:18 | answer | added | Daniel Loughran | timeline score: 5 | |
| Jun 29, 2014 at 19:44 | history | asked | Question Mark | CC BY-SA 3.0 |