Let $\mathbb{F}_q$ be a finite field and let $B$ be the standard Borel subgroup of $GL(2, \mathbb{F}_q)$ consisting of upper triangular matrices. Is there an explicit description of the restriction to $B$ of the cuspidal representations of $GL(2, \mathbb{F}_q)$? All the constructions of the cuspidal representations I know are fairly complicated, but for what I’m doing I only need to know their restrictions to $B$, so I was hoping there was something easier.
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1$\begingroup$ You can find a construction in chapter 9 of my notes dpmms.cam.ac.uk/~sjw47/2023Lectures.pdf. More precisely you want the things I call $\mu_\theta$ at the end of section 9.3. $\endgroup$– Simon WadsleyCommented Dec 1, 2023 at 18:22
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$\begingroup$ @SimonWadsley: You calculate the character table (which can also be found in many other places), but on page 67 you say re the cuspidal representations “we won’t be able to explicitly construct the representations”. $\endgroup$– LindaCommented Dec 1, 2023 at 18:29
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$\begingroup$ If you look closely you'll see that I do construct the cuspidal representations after restriction to $B$ though not on the whole of $GL_2(\mathbb{F}_q)$. They are constructed exactly as Paul Broussous indicates in his answer but more details are provided. You can see from the characters that the $\mu_\theta$ which are constructed as representations coincide with the restrictions of the cuspidal representations. $\endgroup$– Simon WadsleyCommented Dec 1, 2023 at 18:31
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$\begingroup$ @SimonWadsley: Ah, I understand what you were getting at now! I am working on my phone, so I didn’t see the edit you made to your comment giving a more precise reference until after I replied. $\endgroup$– LindaCommented Dec 1, 2023 at 18:35
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$\begingroup$ @Linda. There is no simple construction of cuspidal representations. You have Deligne-Lusztig theory which involves very sophisticated tools from algebraic geometry. You also have the construction via the Weil representation (in characteristic not $2$). This latter is simpler. $\endgroup$– Paul BroussousCommented Dec 1, 2023 at 19:06
1 Answer
Let $Z$ be the center of $G={\rm GL}(2,{\mathbb F}_q )$ and $$ M=\left\{ \left( \begin{array}{cc} a & b \\ 0 & 1\end{array}\right)\in G\right\} $$ be the mirabolic subgroup of $G$, so that $B=ZM$. Fix a non trivial character $\psi$ of $U$. By the theory of Kirillov models, the representation $\mu ={\rm Ind}_U^M \, \psi$ is irreducible (of dimension $q-1$), and for any irreducible representation $\pi$ of $G$, of dimension $>1$, we have $$ {\rm dim}\, {\rm Hom}_M (\pi ,\mu ) =1 $$ If $\pi$ is cuspidal, then ${\rm dim}\, \pi =q-1$, so that, $\pi_{\mid M}\simeq \mu$. So the restriction of $\pi$ to $M$ does not depends on the irreducible cuspidal representation $\pi$. The restriction of those $\pi$ to $B$ only depends on the central character $\omega_\pi$ of $\pi$.
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$\begingroup$ This is really nice! Two questions: 1. Is there a friendly place to read about Kirillov models (I’m not a representation theorist)? and 2. I assume in your last sentence you mean the restriction of $\pi$ to $B$, right? $\endgroup$– LindaCommented Dec 1, 2023 at 17:48
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$\begingroup$ @Linda. For point 2, you're right! I corrected the mistake. For point 1, I don't know. $\endgroup$ Commented Dec 1, 2023 at 19:03