Timeline for Branching laws for smooth representations
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Dec 22, 2020 at 15:21 | history | bounty ended | David Loeffler | ||
| S Dec 22, 2020 at 15:21 | history | notice removed | David Loeffler | ||
| Dec 22, 2020 at 15:21 | vote | accept | David Loeffler | ||
| Dec 22, 2020 at 12:48 | answer | added | user171030 | timeline score: 5 | |
| Dec 22, 2020 at 10:47 | comment | added | Paul Broussous | Did you ask Nadir Matringe ? ([email protected]) | |
| Dec 18, 2020 at 12:52 | comment | added | David Loeffler | Yes, I'm sure it's non-zero, because one can construct a canonical element of this space using the Asai zeta integral. Over a finite field the reps are all semisimple, so the exact seq is obviously exact on the right; however smooth reps over local fields are not semisimple in general, so the sequence continues into a long exact sequence of Ext groups, and the content of my question is whether or not the map $Hom_{GL_2(F)}(\pi, \mathrm{St}) \to Ext^1_{GL_2(F)}(\pi, \mathbb{C})$ is non-zero. | |
| Dec 18, 2020 at 11:07 | comment | added | Dror Speiser | If $E/F$ is a finite field extension (not characteristic 2) and $\pi$ is a generic, principal, non-Steinberg representation, that is not the normalised induction of a pair of characters of $E^\times$ which are distinct, and both trivial on $F^\times$, then I think that $(\pi, 1)_{B(F)}=0$. Are you sure that over local fields the hom-space is at least one dimensional? And if $\pi$ is the normalised induction ... then $(\pi, 1)_{B(F)}=2$, so I would guess that over local fields too $dim\ Hom=2$. | |
| S Dec 18, 2020 at 8:24 | history | bounty started | David Loeffler | ||
| S Dec 18, 2020 at 8:24 | history | notice added | David Loeffler | Draw attention | |
| Dec 16, 2020 at 13:49 | history | edited | David Loeffler | CC BY-SA 4.0 |
added 54 characters in body
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| Dec 15, 2020 at 23:21 | history | asked | David Loeffler | CC BY-SA 4.0 |