Timeline for Non-commuting elements of finite orders in a reductive group over a p-adic field
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 26, 2024 at 20:01 | comment | added | LSpice | Comically, when browsing my old answers looking for a different argument, I found that I had already explicitly described something that can easily be promoted to a counterexample to my implicit belief that the maximal unramified extension of $k$ inside $D$ sits there uniquely. | |
| Oct 15, 2024 at 19:17 | comment | added | Mikhail Borovoi | @WillSawin: Thank you, this is very helpful! | |
| Oct 15, 2024 at 19:16 | vote | accept | Mikhail Borovoi | ||
| Oct 15, 2024 at 17:38 | comment | added | LSpice | Re, yes: I was arguing to myself that $x = \sum_{i = v} \varepsilon_i\varpi^i$ having finite order obviously implies $v = 0$, and then it having prime-to-$p$ order $N$ implies that $1 = x^N \equiv \varepsilon_0^N + N\varepsilon_{i_1}\varpi^{i_1} \pmod{P_D^{i_1 + 1}}$, where $i_1$ is the first positive index with $\varepsilon_{i_1} \ne 0$; but of course that would only work if $\varepsilon_{i_1}$ and $\varpi^{i_1}$ commuted, which they need not. | |
| Oct 15, 2024 at 17:35 | comment | added | Will Sawin | @LSpice Ah, good point about the Zariski density being key. Isn't the point that such a series can be conjugate to the Teichmuller lift $\varepsilon_0$ without being equal to $\varepsilon_0$? | |
| Oct 15, 2024 at 17:32 | comment | added | LSpice | This requires just the additional note that a subgroup that's normalised by $G(k)$ is normal in $G$, since $G(k)$ is Zariski dense in $G$. I otherwise agree with your argument, and yet struggle to reconcile this with a fact that I thought I knew, that every element of a central division algebra $D/k$ can be written as a sum $\sum_{i = v}^\infty \varepsilon_i\varpi^i$, where each $\varepsilon_i$ comes from $\mu_{p'}(L)$, $L/k$ a maximal unramified extension inside $D$, and $\varpi$ belongs to $\operatorname N_D(L) \setminus L$. | |
| Oct 15, 2024 at 16:59 | history | answered | Will Sawin | CC BY-SA 4.0 |