Timeline for Any finite dimensional admissible(smooth) irreducible representation of GL(2,Q_p) is 1-dim
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 3, 2010 at 15:10 | comment | added | AlgSoul | to BCnrd: I was just too stupid at that time, being unable to figure out why it is enough to prove only for SL2(k)...your illustration is clear. Thanks! | |
| Aug 3, 2010 at 13:59 | comment | added | BCnrd | Here's the proof that ${\rm{SL}}_2$ case (for which Kevin's argument works over any non-arch. local field $k$) implies case of any $k$-split conn'd reductive $G$. Let $G' \rightarrow D(G)$ be the $k$-split simply conn'd central cover of derived gp, so $G'(k) \rightarrow G(k)$ has normal image with abelian quotient. By Schur can replace $G$ with $G'$ to reduce to showing the only finite-dim. irred. adm. rep'n of $G(k)$ is trivial one when $G$ is simply conn'd. The cocharacter lattice of split max. $k$-torus is gen'td by coroots since $G$ is s.c., so $G(k)$ is gen'td by ${\rm{SL}}_2(k)$'s. QED | |
| Aug 3, 2010 at 13:59 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
Questioner changed question slightly (removed assertion that only 1 rep existed) so I changed answer to reflect this.
|
| Aug 3, 2010 at 13:35 | vote | accept | AlgSoul | ||
| Aug 3, 2010 at 13:35 | vote | accept | AlgSoul | ||
| Aug 3, 2010 at 13:35 | |||||
| Aug 3, 2010 at 13:30 | history | answered | Kevin Buzzard | CC BY-SA 2.5 |