Timeline for is a linear algebraic group over an extension of $\mathbb{Q}_p$ a locally pro finite group?
Current License: CC BY-SA 4.0
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| Sep 9 at 3:43 | comment | added | Antonius | A slightly sharper formulation of van Dantzig's theorem is that every totally disconnected locally compact group has a neighbourhood base of compact open subgroups. As a compact totally disconnected Hausdorff group is the same thing as a profinite group, you are done. | |
| Sep 8 at 10:44 | comment | added | Mario | @YCor , Thank you very much for your answer. It is not clear to me how the Van Danzig theorem proves that the $F$-points of $G$ form a locally profinite group. | |
| Sep 8 at 9:51 | comment | added | YCor | You mean, the group of $F$-points. The answer is yes, by an old theorem of Van Dantzig: every totally disconnected locally compact group has a compact (hence profinite) open subgroup. Van Dantzig's theorem can be found at many places and can also be proved as an exercise (once it is granted that there is a basis of clopen compact neighborhoods of the identity). | |
| Sep 8 at 8:19 | history | asked | Mario | CC BY-SA 4.0 |