Timeline for when the derived group of the group of $k$-rational points has nonempty interior in the strong topology
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| May 26, 2014 at 4:00 | comment | added | Rupert | Many thanks for your help. Peter McNamara, can I ask you to elaborate further about the anisotropic case. I was told that the only examples of anisotropic absolutely quasi-simple algebraic groups over local fields of positive characteristic were of the form $\mathrm{SL}_{1}(\Delta)$ for central simple division algebras $\Delta$, and that the derived group $[G(k),G(k)]$ was equal to the group of norm-1 units in the unique maximal order of $\Delta$, which I thought was open in the strong topology? Does this include your anisotropic form of $\mathrm{PGL_p}$? | |
| May 26, 2014 at 3:57 | vote | accept | Rupert | ||
| May 20, 2014 at 13:24 | comment | added | Jim Humphreys | @Rupert: The Steinberg lecture notes are online at math.ucla.edu/~rst (though of course the presentation comes from earlier papers of Steinberg). | |
| May 20, 2014 at 11:50 | answer | added | Peter McNamara | timeline score: 1 | |
| May 20, 2014 at 11:27 | comment | added | Peter McNamara | Steinberg: Lectures on Chevalley Groups contains the presentation | |
| May 19, 2014 at 11:25 | comment | added | Rupert | Many thanks for your help. Yes, strong topology does mean the topology coming from the topology on $k$. I think I heard that phrase used somewhere in an algebraic geometry textbook but I'm not sure. Do you know of anywhere where I can read more about the Steinberg presentation? | |
| May 17, 2014 at 8:53 | comment | added | Peter McNamara | and I don't understand your comment about the anisotropic case since if you take an anisotropic form of PGL_p, the commutator subgroup is not open either. Does strong topology mean what I think it does (usual topology from topology of k)? I've never heard the words strong topology in this context before. | |
| May 17, 2014 at 0:06 | comment | added | Peter McNamara | If G=PGL_p and k has char p, the commutator subgroup only contains elements whose determinant is a p-th power and so doesn't contain any open neighbourhood of 1. If G is split, semisimple and simply connected then G(k) is perfect for all fields k with |k|>3 by the Steinberg presentation so you're fine there. The most likely general criterion seems to be if the kernel of the canonical map from the simply connected cover is a group scheme of order prime to p=char(k). | |
| May 15, 2014 at 19:49 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| May 15, 2014 at 19:47 | comment | added | Jim Humphreys | Probably it's enough to deal with Chevalley groups (split case): when $G$ is $k$-isotropic, $\S7$ of the Borel-Tits paper Groupes reductifs constructs a split subgroup having as maximal torus a maximal $k$-split torus of $G$. (When $G$ is simply connected this subgroup is also of simply connected type, but that doesn't seem to be relevant here.) | |
| May 15, 2014 at 11:45 | history | asked | Rupert | CC BY-SA 3.0 |