Timeline for Supercuspidal, spherical and discrete series representation
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 13, 2022 at 7:19 | comment | added | Amitay | Sorry for the inaccuracy, the word hyperspecial is not the right one. The situation where the spherical transform is more problematic is described in Macdonald, "Spherical Functions on a group of p-adic type", Chapter 5 (at least for simple simply connected groups). The root system is then of type $BC_l$, and the group has one standard maximal compact and one "exceptional". There is a unique discrete spherical function if the conditions of Lemma 5.2.12 are met. It can certainly happen in rank 1 when the building is an irregular tree, but I don't know of an example in higher rank. | |
| May 12, 2022 at 18:02 | comment | added | LSpice | No idea! As soon as I step outside of the realm of supercuspidals, and even of tame supercuspidals, my familiarity deserts me. I don't know how modern usage has changed—there's all sorts of flavours of ‘special’ these days—but Tits - Reductive groups over local fields, §2.4, requires a group with a hyperspecial vertex to be $k^\text{un}$-split. | |
| May 12, 2022 at 17:11 | history | edited | paul garrett | CC BY-SA 4.0 |
added 95 characters in body
|
| May 12, 2022 at 17:09 | comment | added | paul garrett | @LSpice, hm, maybe... My own thinking is probably contaminated to a degree by emphasizing the happier cases... :) And, then, e.g., for a "bad" $U(2,1)$, can it be that all maximal compacts admit some square integrable spherical repns? :) | |
| May 12, 2022 at 17:06 | history | edited | LSpice | CC BY-SA 4.0 |
Link to @Amitay's comment
|
| May 12, 2022 at 17:05 | comment | added | LSpice | To say that a reductive group has a hyperspecial maximal compact, don't you need the group to split over an unramified extension? | |
| May 12, 2022 at 16:25 | history | edited | paul garrett | CC BY-SA 4.0 |
added 802 characters in body
|
| May 12, 2022 at 7:12 | comment | added | Amitay | Two short remarks: 1. There are representations that are square integrable (i.e., discrete series) with Iwahori fixed vector. One is the so-called "Steinberg representation". Those that come from characters of the Iwahori-Hecke algebra are classified in Borel's work (link.springer.com/article/10.1007/BF01390139). 2. Sometimes there are spherical representations that are square integrable. This happens if the maximal compact is not "hyperspecial". See e.g. mathoverflow.net/questions/407630/… | |
| May 11, 2022 at 21:58 | comment | added | paul garrett | @Kimball, I don't really know of any instances, and it would not be harmonious to me, but I might be able to imagine that some people are interested in non-cuspidal discrete series, and at least locally/temporarily want to make "discrete series" exclude supercuspidal. Stranger things have happened. :) | |
| May 11, 2022 at 21:55 | comment | added | Kimball | Many people would count supercuspidals as discrete series - Does anyone not? | |
| May 11, 2022 at 21:04 | vote | accept | Aersk | ||
| May 11, 2022 at 21:02 | vote | accept | Aersk | ||
| May 11, 2022 at 21:04 | |||||
| May 11, 2022 at 20:46 | comment | added | paul garrett | @LSpice, ah, better, I think. :) | |
| May 11, 2022 at 20:40 | comment | added | LSpice | More simply (?) than computing, if $(\sigma, M)$ is smooth with central character $\omega$, $K'$ (any compact open subgroup) admits an Iwahori decomposition with respect to opposite parabolic subgroups $P^\pm$ with common Levi $M$, and $\sigma$ contains a ($K' \cap M$)-fixed vector $v$, then, for every $v^* \ne 0$, the extension by $0$ to $G$ of $k'\cdot m\cdot n^+ \mapsto \langle v^*, \sigma(m)v\rangle$ is a matrix coefficient of $\operatorname{Ind}_P^G \sigma$ that transforms by $\omega$ under the central torus of $M$, hence is not square integrable modulo $Z(G)(F)$ unless $M$ equals $G$. | |
| May 11, 2022 at 20:35 | history | answered | paul garrett | CC BY-SA 4.0 |