# DJing From My Phone

Last Friday was the second time I DJed at a lindy hop party from my phone. The first time was in January, at a bar in Copenhagen during a local weekly swing night, after a wonderful weekend of teaching taster classes with Lee at the fantastic festival LHTA. That time I had used my special iOS pyhton program (written in Pythonista) and played from my small-screened iPhone 4S. This time, in Ljubljana,  I used my huge-screened Note 4, and a program I whipped up in the two days leading to the party. Did I test the program thoroughly beforehand? No. Was my set ok? Umm, yeah, sure.

Here are the abilities I need for a (good) set:

1. External sound card
2. Preview songs
3. Queue songs
4. Lock buttons on playing device
5. Search in media library by title, artist, bpm and comments.

This project of playing from my Note 4 that runs Android started when I realised that I can connect an external usb sound card to it using something called an OTG cable, as in On-The-Go. Brilliant! This solves point no. 1.

A few months ago I spent quite some time in trying to write a native iOS app that will play two songs in the two channels of headphones. There already exists such an app in the app store, but it’s expensive, doesn’t have a lock buttons feature, doesn’t search in comments, and it has turn tables. Enough with the turn tables already! I’m not that kind of DJ… Anyway, I didn’t figure it out before I gave up, and I was sad. But if I play music form my phone – I can just preview on my iPad, using my old python program. Great. This solves point no. 2.

I made the program with two tabs: one to show the media library and one the current playlist. Clicking a song in the media library tab queues a song in the playlist tab, and clicking a song in the playlist tab starts to play it. Unless the lock is on. Which is crucial. Personally, I am always worried about accidental clicks. It doesn’t matter if I’m using my phone, my iPad, or even my MacBook. Having a program that just ignores all clicks, when a lock is on, reduces my anxiety by a big factor. And I know a few other DJs who have confessed to the same stress. But when the lock is on I can still queue songs and change their position in the playlist. So points 3 and 4 are good too.

Of course I’ll need to search for songs. If I’ve already previewed a song on another device and I just want to add it now then I can search by title. But I really like seeing my comments… So I made them show up on the screen. And if they’re there, I should also be able to search the comments tag as well. What if I’m riding shotgun and have to provide the music? I’m not going to preview songs. Being able to search for a low energy groovy song would be nice.

Here’s a screenshot from the end of my set:

Though everything was solid, the set was stressful for me. I wasn’t sure the app would hold up with no crashes, and if the battery would survive supplying power to a usb card and having the screen on almost constantly. I’m happy to say that the app didn’t crash, and the battery only went down by 20%! That’s it! Granted, I turned off everything else and went into airplane mode (can you imagine getting a call in the middle of a set?). The only snafu was during the first song: I forgot that I made a lock-all-buttons button, and my headphones cable accidentally touched the phone’s screen – that’s it, it just kinda grazed the screen for a split second. And the song changed. Somehow, it was kind of on beat, and with no pause. So it was ok, and then I clicked the lock button.

One more thing about my set: following this post, I decided to use a normal distribution for my bpms. I had python opened up on my MacBook (which I brought fearing that my phone would explode and I would need another device) with numpy imported, and I got my bpms from its random module. For the first half hour, which was between band sets, I used a mean of 170 and sigma 16.66. Then, after the band went off for the last time around 1:20 AM, I reduced the mean to 150. Finally, after an hour and until the end of my set I set the mean to 140. Only two times during the set did I decide to overrule the random procedure. Since I had say in the energy and subgenre of songs, no bpm was supposed to be problem. Except for those two times, that two songs above 190 were just played, and I wanted a slower song to put on next. Overall, I think the room was pretty much swinging and bouncing until the end, at which point a jam of three musicians started on stage. The highest bpm was 202 and the lowest was 124.

Being able to go have a set without carrying a heavy computer is awesome. Once I add some features, like tagging and saving playlists and maybe even previewing, I will upload the app to the Google play store. But who knows when that will be…

# Quadratic Twists for Representations of Finite Symplectic Groups

I started mentioning descent in this post. During the course of writing my masters results in article form, I found a cute descent-like result that can be proven with similar ideas.

Theorem. Let $\sigma$ be a representation of $Sp_{2m}(\mathbb{F}_q)$ in general position. Then ${\textup{Ind}_{P_{1,2m,1}}^{Sp_{2m+2}(\mathbb{F}_q)}(1_{\mathbb{F}_q^\times}\times\sigma)}$ is reducible and has a unique semisimple component, call it $\pi(\sigma)$. Then

$\tilde{\sigma}:=(\textup{Res}_{Sp_{2m}\ltimes H_{2m}}(\pi(\sigma))\otimes\omega_\psi)^{H_{2m}}$

is the “quadratic twist” of $\sigma$, i.e. the characters in its Deligne-Lusztig parameter are the same, but multiplied by the quadratic character. A representation is said to be in general position if its character is plus or minus a Deligne-Lusztig character in general position.

Here $\psi$ is an additive character of $\mathbb{F}_q$, $\omega_\psi$ is the Weil representation of $Sp_{2m}\ltimes H_{2m}$, and $P_{1,2m,1}$ is a rational parabolic subgroup with Levi subgroup isomorphic to $\mathbb{F}_q^\times\times Sp_{2m}(\mathbb{F}_q)$.

The strategy of proof is as follows:

1. Express $\pi(\sigma)$ as a combination of Deligne-Lusztig characters.
2. Compute the degree of $\pi(\sigma)$.
3. Deduce a bound on the degree of $\tilde{\sigma}$
4. Compute the character value of $latex \tilde{\sigma}$ at all non-unipotent elements.
5. Deduce the dimension of $\tilde\sigma$.
6. Deduce character values at unipotent elements.

The first four points are pretty straightforward computations with Deligne-Lusztig characters. The fifth point is an application of the following crucial lemma:

Lemma 1. Let $\textup{\textbf{G}}$ be a connected reductive algebraic group over $\mathbb{F}_q$ with Frobenius morphism $F$. Then, as $n\rightarrow\infty$, for any pair of representations $\rho_1,\rho_2$ of $\textup{\textbf{G}}^{F^n}$, such that ${\textbf{dim}\ \rho_1\ne \emph{dim}\ \rho_2}$ and ${\chi_{\rho_1}-\chi_{\rho_2}}$ is zero on all non-unipotent elements, we have:

$\emph{dim}\ \rho_1\oplus\rho_2 \gg \sqrt{|\textup{\textbf{G}}^{F^n}|(q^n)^{r(\textup{\textbf{G}})}}$

where the implied constant depends only on $\textup{\textbf{G}}$.

In our case, $\rho_1$ is $\tilde\sigma$, and $\rho_2$ is the quadratic twist of $\sigma$. Assume that $q$ is large, and that ${\emph{dim}\ \rho_1\ne \emph{dim}\ \rho_2}$. Then the hypotheses of the lemma are satisfied. But then the conclusion of the lemma contradicts the bound on the dimension of $\tilde\sigma$. So, for all large enough $q$, we must have ${\emph{dim}\ \rho_1= \emph{dim}\ \rho_2}$. But the dimension of $\tilde\sigma$, as well as the dimension of the quadratic twist of $\sigma$, are polynomials in $q$ ($\sigma$ going over a family of Deligne-Lusztig characters in general position for the same torus). Hence, for all $q$ we have equality of dimensions.

We get the desired result by applying the following lemma:

Lemma 2. Let $\rho_1,\rho_2$ be two representations of a finite group $G$ with equal dimensions, such that their characters agree on a subset $H\subset G$, and that $\rho_2$ is irreducible. Assume there exists a $z\in Z(G)\cap H$, such that $z\cdot(G\backslash H)\subset H$. Then $\chi_{\rho_1}=\chi_{\rho_2}$ on all of $G$.

Our result says that we can twist representations of $Sp_{2m}(\mathbb{F}_q)$ with an explicit representation theoretic construction. I am not aware of such a result over local and global fields. All proofs that I have seen for twisting general automorphic forms use converse theorems. I will note that Shimura gave a geometric meaning to twisting modular forms by a general Dirichlet character, but being geometric in nature precludes it from working for Maass forms.

——————

Proof of Lemma 2: Since $\rho_2$ is irreducible, and $z\in Z(G)$, $z$ acts as a scalar on $\rho_2$, say $\lambda_z$. So

$\chi_{\rho_1}(z) = \chi_{\rho_2}(z) = \lambda_z\chi_{\rho_2}(1)=\lambda_z\chi_{\rho_1}(1)$

which implies that $z$ also acts as $\lambda_z$ on $\rho_1$.

Let $g\in G\backslash H$. Then we have

$\chi_{\rho_1}(g)=\chi_{\rho_1}(z^{-1}zg)=\lambda_z^{-1}\chi_{\rho_1}(zg)=\lambda_z^{-1}\chi_{\rho_2}(zg)=\chi_{\rho_2}(g).$
$\blacksquare$

Proof of Lemma 1: Let $n$ be a positive integer, and $\rho_1$,$\rho_2$ be as in the statement of the lemma. The assumption ${\textup{dim} \rho_1\ne \textup{dim} \rho_2}$ implies that ${(\chi_{\rho_1}-\chi_{\rho_2},\textup{reg}_{\textup{\textbf{G}}})\ne 0}$. Using Corollary 7.7 in [DL], there exists a pair $({\textup{\textbf{T}}},\theta)$, $\theta\in\widehat{{\textup{\textbf{T}}}^{F^n}}$, such that

$(\chi_{\rho_1}-\chi_{\rho_2}, R_{{\textup{\textbf{T}}}}^{\textup{\textbf{G}}}(\theta)) \ne 0.$

Since ${\chi_{\rho_1}-\chi_{\rho_2}}$ is supported only at unipotent elements, this character pairing depends only on the values at unipotent elements. But this in turn is independent of $\theta$ by the character formula, Theorem 4.2 in [DL], so

$(\chi_{\rho_1}-\chi_{\rho_2}, \varepsilon_{\textup{\textbf{G}}}\varepsilon_{\textup{\textbf{T}}} R_{\textup{\textbf{T}}}^{\textup{\textbf{G}}}(\theta')) \ne 0.$

for all $\theta'\in\widehat{{\textup{\textbf{T}}}^{F^n}}$.

For any $\theta'\in\widehat{{\textup{\textbf{T}}}^{F^n}}$ in general position ${\varepsilon_{\textup{\textbf{G}}}\varepsilon_{\textup{\textbf{T}}} R_{\textup{\textbf{T}}}^{\textup{\textbf{G}}}(\theta')}$ is the character of an irreducible representation by Theorem 6.8 and Theorem 7.1 in [DL], thus the above shows that it is a constituent of at least one of $\rho_1,\rho_2$. The number of such $\theta'$ is

$\gg (q^n)^{r({\textup{\textbf{G}}})}$

and each $\varepsilon_{\textup{\textbf{G}}}\varepsilon_{\textup{\textbf{T}}} R_{{\textup{\textbf{T}}}}^{\textup{\textbf{G}}}(\theta')$ has dimension

$\varepsilon_{\textup{\textbf{G}}}\varepsilon_{\textup{\textbf{T}}} R_{\textup{\textbf{T}}}^{\textup{\textbf{G}}}(\theta')(1) = |{\textup{\textbf{G}}}^{F^n}/{\textup{\textbf{T}}}^{F^n}|_{p'}\gg \sqrt{|{\textup{\textbf{G}}}^{F^n}|(q^n)^{-r({\textup{\textbf{G}}})}}$

by Theorem 7.1 in [DL]. This proves the lemma.
$\blacksquare$

After finishing with my masters results, I’ll write the first four points as well. Stay tuned!

——————

[DL] P. Deligne, G. Lusztig, Representations of reductive groups over finite fields, Ann. of Math. (2) 103, no. 1, 103?161. (1976).

# Beats Distribution

When another post on Jive Junction’s Facebook group asked about BPMs for lindy hop parties, I decided to check the distribution from recent saved DJed playlists, 14 in total. These include two from big events (Lindy Shock, Vintage Swing Festival), three from local (to Slovenia) events, and the rest are weekly local dance nights in Tel Aviv and Ljubljana. Here’s the plot:

$\mu=146, \sigma=17.1$

Except for the peak in the middle, the distribution is pretty normal. When I DJ, I try to match the speed to what I think will get the most dancers on the floor, in the current song and the next. Over time, I have consciously come to think that 146 is the most ideal BPM for this task – whether after a fast song or a slow song, or even of the same speed, it will bring out the most dancers for this and the next song.

But I haven’t really documented how many people are dancing to each song, and maybe I have it off. So I’ve now started, and I will also randomise my playlists a bit more with the help of a computer, which hopefully is unbiased. I definitely want to try Shana Worel’s suggestion of $\mu=170,\sigma=16.6$ for a few sessions (“with a few outliers sprinkled in for good measure”).

It would be pretty cool to get a result similar to that well known experiment in the west coast world. 122 was it?

————————–

The above plot was made using a python script that I wrote, that goes over playlists in the iTunes Media Library file. This is an xml file, which on my mac was found in ‘~/Music/iTunes/’. You can run the script as well to see your own statistics. If there are playlists that are not relevant, you can name them and the script will disregard them. I used numpy and matplotlib to generate the plot, and those few lines are not included.

from xml.dom.minidom import parse, parseString
from math import sqrt

# Change this to wherever your 'iTunes Music Library.xml' is.
_itunes = &quot;~/Music/iTunes/iTunes Music Library.xml&quot;

# Irrelevant playlists. For example, the bpms from my
# wedding playlist should not be counted.
irrelevant = ['April','From Videos','Wedding!','WeirdThing']

###########################################################
###########################################################

itunes = parse(_itunes)

_tracks = itunes.childNodes[1].childNodes[1].childNodes[24]
_playlists = itunes.childNodes[1].childNodes[1].childNodes[28]

def getKeyIndex(idict, key):
cn = idict.childNodes
for i in range(len(cn)):
if cn[i].hasChildNodes() and cn[i].tagName == u'key' and cn[i].childNodes[0].data == key:
return i
return -1

def getValue(idict, key):
i = getKeyIndex(idict, key)
if i != -1:
cn = idict.childNodes
i = i+1
while not cn[i].hasChildNodes():
i = i+1
if i &gt;= len(cn):
return -1
return cn[i].childNodes[0].data
return -1

# Parse tracks
tracks = {}
for track in _tracks.childNodes:
if track.hasChildNodes() and track.tagName == u'dict':
id = int(track.childNodes[2].childNodes[0].data)
bpm = int(getValue(track, u'BPM'))
tracks[id] = (bpm,)

# Parse playlists
playlists = {}
for playlist in _playlists.childNodes:
badtags = ['Distinguished Kind','Smart Info', 'Folder', 'Master']
name = getValue(playlist, 'Name')
itemsIndex = getKeyIndex(playlist, 'Playlist Items')

if set([getKeyIndex(playlist, tag) for tag in badtags]) == set([-1]) and name != -1 and itemsIndex != -1:
while not playlist.childNodes[itemsIndex].hasChildNodes() or playlist.childNodes[itemsIndex].tagName != u'array':
itemsIndex += 1
if itemsIndex &gt;= len(playlist.childNodes):
break

if itemsIndex &gt;= len(playlist.childNodes):
continue

ptracks = [int(i.childNodes[0].data) for i in playlist.childNodes[itemsIndex].getElementsByTagName('integer')]

playlists[name] = ptracks

for irr in irrelevant:
playlists.pop(irr)

allbpms = []
for p in playlists:
for id in playlists[p]:
if tracks[id][0] &gt; 0:
allbpms.append(tracks[id][0])

mu = (sum(allbpms)+0.0)/len(allbpms)
sigma = sqrt(sum([(x-mu)**2 for x in allbpms])/len(allbpms))

print 'mu:', mu
print 'sigma:', sigma


# Counting Symplectic Matrices Satisfying Something

While working on my thesis, I wanted to bound the dimension of a Jacquet module of a uniform character (i.e. a combination of Deligne-Lusztig characters) using an idea of Gelfand, appearing in ‘Representations of the full linear group over finite fields’ (1970). In modern terms, Gelfand shows that the cuspidal characters of $G=GL_n(\mathbb{F}_q)$ are generic/regular/have a Whittaker model.

For a cuspidal representation $\tau$, and $\psi\in \hat{\mathbb{F}_q}$, a non-trivial additive character, Gelfand shows that (1): $(\textup{res}_U^G\ \tau,\psi)_U$ is a rational function of $q$ of degree $0$, and (2): it equals $1$ for all large enough $q$. Hence it is always $1$.

The first property is proved by a rather explicit computation. Gelfand shows that the number of upper triangular unipotent matrices in a given conjugacy class is a polynomial in $q$, and computes its degree. This is enough, since the character values of a cuspidal character at unipotent elements was already known, due to Green (I think). The second property essentially follows from the first and a multiplicity one result.

I wanted to do the same with a representation of $Sp_{2n}(\mathbb{F}_q)$: show that the dimension of the Jacquet module with respect to some unipotent subgroup was a polynomial and that I can compute its degree. Specifically, I needed to compute the number of matrices such that

$u=\begin{pmatrix} u'&M&Z \\ &I_{2\ell}&M' \\ & &u'^* \end{pmatrix}\in Sp_{2m}(\mathbb{F}_q),$
$\textup{rank}(u-1)=k,$
$u'_{1,2}+u'_{2,3}+...+u'_{m-\ell,m-\ell+1}=c$

for any positive integers $m, k, \ell$, and $c\in\mathbb{F}_q$. (The symplectic form is defined in the linked file below.)

So I did. I knew that I didn’t have a formula for the values of my character at unipotent elements, but I knew that the values are polynomials, and I thought I knew their degrees. This would be enough to prove what I needed. But I was wrong. I did not know the degrees of the polynomials. I couldn’t find a way to compute them without doing some heavy lifting, like in Lusztig’s ‘On the Green Polynomials of Classical Groups’ (1976). Essentially, I was missing the following result: degrees of values at unipotent elements of “small” representations are smaller than degress of values from cuspidal representations. Small here means a semisimple representation that is not also regular. I still don’t know if this is true or not.

But then I found a different way to bound the dimension of my Jacquet module: compute the character’s wave front set and use Lusztig’s results from ‘A unipotent support for irreducible representations’ (1992).

Back to counting matrices. I use the same ideas from Gelfand’s paper, only extending the amount of variables I need to keep track of (not just rank). If you’re interested, here it is:

http://www.math.tau.ac.il/~drorspei/Unipotent Symplectic Matrices Satisfying Something.pdf

# A Primate’s Post

About a year ago I started watching videos of a course on stress called Stress and Your Body, produced by The Great Courses, and given by Robert Sapolsky. I would say it is mid-level, in that the lectures are far from technical, and yet quite a bit of the neurobiology is explained. For some reason  I didn’t take the viewing seriously until a couple of weeks ago. But then I decided I really wanted to know about stress, and watched a lecture every day, until I have now finally finished the course.

And it was brilliant. Saplosky is fun. For example, he tells us that he really hates Wagner’s music, “can’t stand him!”, in order to give an example of psychological stress for him specifically. And then he goes on to talk about the sympathetic and parasympathetic nervous systems and the specific biological effects of his stress – even more fun.

The one thing that I keep finding more evidence for is that the brain is essentially a prediction-reaction box. It just predicts all the time, and reacts in a way that it predicts will make for better predictions later. As in, there isn’t necessarily a “good” state to want – other than the ability to predict itself. But maybe I keep “finding” this since I’m already predicting it to be true…

Figuring out that I really liked this professor, I decided to also listen to his (audio*)book ‘A Primate’s Memoir: A Neuroscientist’s Unconventional Life Among the Baboons‘. In this instance there was no long neglect, and I finished listening in two weeks (mostly during walks and car drives). I highly recommend the book, but only as a fun memoir of a neuroscientist who worked in Africa. It is not a book on Sapolsky’s work, and by far most of the chapters are not about his baboon troop. It does not reflect an objective view of Africa or anything like that. I say this since some bad reviews on Amazon expected the opposite.

Most of all, I like that the book has a large number of characters. Second to that, is the contrast between the cultures (the many cultures – I’m not implying that there are only two). On the one hand I feel I know more about Kenya now (maybe Kenya of twenty years ago). On the other, I realise that if anything, I only know of Sapolsky’s Kenya. But I’m satisfied with that for now. Finally, listening to long descriptions of behaviour of baboons, with clear personalities, friendships, rivalries, is quite the eye opener.

* I’ll note that the audiobook, narrated by Mike Chamberlane, has a good pace and fun accentuation. I don’t know if this is really how it was, but I enjoyed his soft and slow rendering of the locals speech.

# A Master’s Problem

I’d like to tell you about my masters thesis problem, given to me by my advisor David Soudry. It is essentially about translating the Descent construction of Ginzburg-Rallis-Soudry to the case of finite fields. Their work, over local and global fields, has had a major impact on the field of Langlands Functoriality. Here’s one of their main results, in its simplest form possible, as they have proved in greater generality:

Theorem (Ginzburg-Rallis-Soudry ’99). Let $\pi$ be an irreducible, supercuspidal, self-dual representation of $GL_{2n}(\mathbb{Q}_p)$ such that $L^s(\pi,\Lambda^2,s)$has a pole at $0$. Then there exists an explicit construction of a representation $\sigma_\pi$ of $\widetilde{Sp}_{2n}(\mathbb{Q}_p)$ with the right (according to the Langlands conjectures) gamma factor.

One of the main reasons why descent is so powerful is that it can give us lots of explicit information on endoscopic representations, without using proxies such trace formulas or converse theorems.

It is not hard to figure out what the theorem should be over finite fields. In this post I will explain, and prove a bit of, the smallest case.

Fix an odd prime power $q$. For convenience (mine mostly) we will present everything with matrices. Let

$J = \begin{pmatrix} &&&1\\ &&1&\\ &1&&\\ 1&&& \end{pmatrix}$

and let $\hat{G}:=SO_4(\mathbb{F}_q)=\{g\in GL_5(\mathbb{F}_q)\ |\ {}^t gJg=J\}$. We have the usual parabolic subgroup with Levi subgroup isomorphic to $G:=GL_2(\mathbb{F}_q)$:

$P = \{\begin{pmatrix} g & M \\ & J {}^tg^{-1} J \end{pmatrix}\ |\ g\in GL_2(\mathbb{F}_q),\ M\in M_{2}(\mathbb{F}_q),\ {}^t(Jg^{-1}M)=-Jg^{-1}M \}$.

We will also need a specific vector, and my advisor’s favourite is $v= {}^t(0,1,1,0)$. Then it is easy to see that the subgroup of $SO_4(\mathbb{F}_q)$ fixing $v$ (* I like to multiply from the left) is isomorphic to $H:=SO_3(\mathbb{F}_q)$. Denote by $\textup{Irr}_{sd}(G)$ the irreducible self dual representations (well, isomorphism classes) of $G$.

Definition For $\tau\in \textup{Irr}_{sd}(G)$, extend to $P$, induce to $\hat{G}$, and take the component of least dimension, which we’ll call $\pi(\tau)$. The descent of $\tau$ is the restriction $\textup{res}_H^{\hat{G}}\ \pi(\tau)\in \textup{Reps}(H)$.

The part where we “take the component of least dimension” is analogous to taking the residue of Eisenstein series in the original Ginzburg-Rallis-Soudry construction. With this definition, we have the following result:

Theorem The map $\tau\mapsto \textup{res}_H^{\hat{G}}\ \pi(\tau)$ is a bijection between the $\textup{Irr}_{sd}(G)$ and $\textup{Irr}(H)$, that preserves dimension. Moreover, for each such $\tau$ there is a $s\in Sp_2(\mathbb{F}_q)\le G$, such that $\tau$ is in the Lusztig family of $s$, and then $\textup{res}_H^{\hat{G}}\ \pi(\tau)$ is in the Lusztig family of $s$, considered as an element of the dual of $H$, which is isomorphic to $Sp_2(\mathbb{F}_q)$.

It is easy to compute the descent of the trivial representation. Indeed, the induction to $\hat{G}$ will contain the trivial representation, and so $\pi(\tau)=1$, giving that the descent is the trivial representation of $H$. Great!

Let’s compute one general case, that of “principal series” in general position. These are the representation of the form $\tau=\textup{Ind}_B^G(\theta\times\theta^{-1})$, where $\theta^2\ne1$. The $\theta^{-1}$ comes from the self-dual condition. Note that the existence of such $\theta$ implies that $q>3$. So we’ll assume that. ($B$ is a Borel subgroup.)

At this point I want to start using Deligne-Lusztig theory. So $\tau=R_{T_{1^2}}^G(\theta\times\theta^{-1})$. The induction, by transitivity of Deligne-Lusztig inductions, is $\rho:=R_{T_{1^2}}^{\hat{G}}(\theta\times\theta^{-1})$. Here $T_{1^2}$ is the obvious torus rationally isomorphic to $\mathbb{G}_m^2$, in $G$ or $\hat{G}$ depending on the context.

The element $J$, which can thought of as in the Weyl group of $\hat{G}$, fixes $T_{1^2}$, and also $\theta\times\theta^{-1}$. It is to go over the four Weyl elements and see that $1,J$ are the only such elements. So $(\rho,\rho)=2$, by a basic result in Deligne-Lusztig theory, Theorem 6.8 in [DL].

We also have the virtual character $\rho':=-R_{T_{2}}^{\hat{G}}(\theta\circ\textup{det})$. By the same argument as above, $(\rho',\rho')=2$. Going to a quadratic extension, the two tori become rationally conjugate, and it is a simple computation to show that $(\theta\times\theta^{-1})\circ\textup{Norm}$ and $\theta\circ\textup{det}\circ\textup{Norm}$ are conjugate. By standard results we get that $\rho$ and $\rho$ must share at least one component. Since also $(\rho,\rho')=0$, we must have that

$\rho=W+\pi,\ \rho'=W-\pi$

for two irreducible proper characters $W, \pi$. But we can also calculate the dimension of $\rho$ and $\rho'$, using Theorem 7.1 of [DL]:

$\rho(1)=\frac{|\hat{G}|_{p'}}{|T_{1^2}^F|}=\frac{(q^2-1)^2}{(q-1)^2} = (q+1)^2$
$\rho'(1)=\frac{|\hat{G}|_{p'}}{|T_{2}^F|}=\frac{(q^2-1)^2}{q^2-1} = q^2-1$

Since the dimension of $\rho'$ is positive, $\pi$ must have dimension smaller than $W$, and we get

$\pi=\frac{\rho-\rho'}{2}$
$\pi(1)=\frac{\rho(1)-\rho'(1)}{2}=q+1$

Alright! Let’s finish with a computation of the character at another semi-simple element. This means that there is an $a\in \mathbb{F}_q^\times$ such that $a^2\ne 1$. Choose such, and let

$g=\begin{pmatrix} a&&& \\ &1&& \\ &&1& \\ &&&a^{-1} \end{pmatrix}\in H$

It is easy to see that $g$ is not conjugate to any element of $T_{2}$, so by the character formula for Deligne-Lusztig characters, Theorem 4.2 in [DL], we get

$\pi(g)=\frac{1}{2}R_{T_{1^2}}^{\hat{G}}(\theta\times\theta^{-1})(g)=\theta(a)+\theta^{-1}(a)$

Looking at the character table of $H\cong PGL_2(\mathbb{F}_q)$, for example here, we can see that there is exactly one representation of dimension $q+1$ that can have the above character value. Working with Deligne-Lusztig characters, this representation’s character must be

$\textup{res}_H^{\hat{G}}\ \pi(R_{T^2}^G(\theta\times\theta^{-1}))=R_{T_{1^2}}^H(\theta\times\theta^{-1}).$

Which is exactly what one would expect.

My thesis work is to prove the analogue of the above result to the general case, but only for those $\tau$ that descend to cuspidal representations that are associated to the Coxeter torus. The strategy of the proof works for all irreducible self-dual representations that are not unipotent, and I am preparing an article on this.

The computation of descent for unipotent characters is a bit weird, and I’m not sure what the correct result even is. I will sometime soon write a post on computations with $GL_4$, in which I will also define descent in general.

—————————

[DL] P. Deligne, G. Lusztig, Representations of reductive groups over finite fields, Ann. of Math. (2) 103, no. 1, 103?161. (1976).