Timeline for Characters of simply connected semsimple algebraic groups over local fields
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 20, 2016 at 9:29 | vote | accept | Daniel Loughran | ||
| May 19, 2016 at 19:44 | answer | added | Mikhail Borovoi | timeline score: 6 | |
| May 19, 2016 at 13:02 | comment | added | YCor | You're right. The absolute simplicity is not important, but the isotropic assumption is essential in the argument, which uses generation by unipotents. (As Mikhail, in the anisotropic case I'm not aware of an example with nontrivial finite abelian quotient.) | |
| May 19, 2016 at 13:01 | comment | added | Mikhail Borovoi | For $G=SL(1,D)$ the group $G(k)$ is compact, hence profinite, hence it has a lot of finite quotients. | |
| May 19, 2016 at 12:44 | comment | added | Daniel Loughran | @YCor: Do you know a precise reference? I looked at Platonov-Rapinchuk's book, but I only found the weaker statement mentioned by Mikhail. | |
| May 19, 2016 at 12:24 | comment | added | Daniel Loughran | @Mikhail: Thanks. The result you state has many assumptions. Are you implying that the answer to my question is "no" in general? E.g. what happens for the group $\mathrm{SL}_1(D)$? | |
| May 19, 2016 at 11:26 | history | edited | Paul Broussous |
edited tags
|
|
| May 19, 2016 at 11:16 | comment | added | Mikhail Borovoi | A version of @YCor's comment: Let $G$ be a simply connected absolutely simple $k$-group, where $k$ is a nonarchimedean local field (a $p$-adic field or the field of rational functions in one variable over a finite field). Assume that $G$ is isotropic (i.e., not isomorphic to $\mathrm{SL}(1,D)$ of a central division algebra $D$ over $k$). Then any nontrivial normal subgroup of $G(k)$ is central, hence finite. For a proof see the book by Platonov and Rapinchuk. Therefore, $G(k)$ admits no nontrivial homomorphisms into abelian groups. | |
| May 19, 2016 at 10:46 | comment | added | Daniel Loughran | @YCor: Don't you mean that the answer to my question is in fact "yes"? Which book of Margulis are you referring to? | |
| May 19, 2016 at 10:36 | history | edited | Daniel Loughran | CC BY-SA 3.0 |
added 120 characters in body
|
| May 19, 2016 at 10:35 | comment | added | Daniel Loughran | @Jason and Peter: Yes sorry, I forgot to compose det with a character; I have changed the statement. | |
| May 19, 2016 at 10:15 | comment | added | Peter McNamara | For PGL_n, det takes values in H^1(k,mu_n)=k*/(k*)^n. But det is surjective and its target admits nontrivial homomorphisms to the circle when k is the p-adics so you've still got your topological character. | |
| May 19, 2016 at 9:58 | comment | added | Jason Starr | Are you taking $n=p-1$, so that the reduction of the determinant on $\textbf{GL}_{p-1}$ is a well-defined homomorphism $\textbf{PGL}_{p-1}(\mathbb{Z}_p) \to \mathbb{F}_p^\times$? If so, how do you extend this to $\textbf{PGL}_n(\mathbb{Q}_p)$? | |
| May 19, 2016 at 9:55 | comment | added | Jason Starr | What do you mean by $g\mapsto \text{det}(g)$? For every linear representation that I can think of, the image of $\textbf{PGL}_n$ is contained in $\textbf{SL}_M$. | |
| May 19, 2016 at 9:37 | history | asked | Daniel Loughran | CC BY-SA 3.0 |