I'm trying to use the language of strata to organize $K$-types of irreducible smooth representations of $GL(2)$ (and then hopefully prove things). Unfortunately, I'm still new to it, so I might be making some mistakes.

My main reference is the Bushnell-Henniart book. For anyone who doesn't have access to it and wants to "play along at home", this essay contains a distillation of the important ideas.

Let $k$ be a $p$-adic field with ring of integers $\mathfrak o$, $G=GL_2(k)$, $K=GL_2(\mathfrak o)$. The principal congruence subgroup of level $\varpi^N$ (or level $N$, by abuse of notation) is $K_N=1+\varpi^N M_2(\mathfrak o)$ (and $K_0=K$). Let $\psi$ be an additive character on $k$ with conductor $\mathfrak o$ and extend it to $M_2(k)$ by composing with the trace: $\psi_M(x):=\psi({\rm tr}\ x)$. To further abuse notation, we'll suppress the subscript $M$ in $\psi_M$.

We'll say that the level of an irreducible (smooth) representation $\sigma$ of $K$ is the largest $N$ such that $\sigma$ is nontrivial on $K_N$ and trivial on $K_{N+1}$. Assume $\sigma$ has level $N\ge 1$. Since $K_N/K_{N+1}$ is a finite abelian group, $\sigma|_{K_N}$ decomposes into a direct sum of characters. Using the isomorphism $K_N/K_{N+1}\simeq M_2(\mathfrak o/\varpi)$ (given by $x\rightarrow x-1$), these characters can be written in the form $\psi_a(x):=\psi\big(a(x-1)\big)$ for some $a\in M_2(\mathfrak o/\varpi)$. For our purposes, a stratum is the level $N$ and the character $\psi_a$ on $K_N$.

What strata appear in the $K$-types of irreducible representations of $G$?

My possibly-incorrect understanding is that for a representation that can be compactly-induced from $ZK$, the fundamental stratum of the representation will appear in a $K$-type of lowest level. What happens for higher levels? And what happens for representations associated to the other chain order (than $M_2(\mathfrak o)$)?

For facts about $K$-types of irreducible representations, see Casselman's Restriction paper (this version should be more easily available), or Henniart's Appendix to this paper. For the supercuspidal case, I am aware Hansen's paper (though I don't understand it). The subtext of this paragraph is that I know (in principle) the $K$-types I'm interested in, yet I am still unable to transform this into knowledge of the corresponding strata. Whether this is due to me overlooking something simple or to a more serious issue, I do not know.

As an example, let $\chi_1$ and $\chi_2$ be characters of $k^\times$, with $\chi_1$ unramified and $\chi_2$ ramified with conductor $N_0\ge 1$ (so $\chi_2$ is nontrivial on $1+\varpi^{N_0}\mathfrak o$ but trivial on $1+\varpi^{N_0+1}\mathfrak o$). Set $\chi=\chi_1\otimes\chi_2$, and let ${\rm Ind}_P^G\chi$ be the corresponding ramified principal series. Then ${\rm Ind}_P^G\chi$ contains $K$-types $\sigma_N(\chi)$ of level $N\ge N_0$, where $\sigma_N(\chi)$ is the subrepresentation of ${\rm Ind}_{P\cap K}^K(\chi)$ of level $N$ (we aren't distinguishing between $\chi$ and its restriction to $K\cap P$).

Take $g=\bigg(\matrix{a & b\cr c & d}\bigg)\in K_N$, so that $c=u\varpi^N$, with $u\in\mathfrak o^\times$, then (for example) $$g=\bigg(\matrix{adu^{-1}-b\varpi^N&b\cr &d}\bigg)\bigg(\matrix{1&\cr \varpi^N&1}\bigg)\bigg(\matrix{ud^{-1}&\cr &1}\bigg)$$ Thus, for $v\in \sigma_N(\chi)$ and $g=\bigg(\matrix{a & b\cr c & d}\bigg)\in K_N$ $$\sigma_{N,\chi}(g)\cdot v=\chi_2(d)\sigma_{N,\chi}\Bigg(\bigg(\matrix{1&\cr \varpi^N&1}\bigg)\bigg(\matrix{ud^{-1}&\cr &1}\bigg)\Bigg)\cdot v$$ Since $\chi_2$ has conductor $N_0$, there exists $a_2$ such that $\chi_2(d)=\psi\big(a_2(d-1)\big)$. When $d\in 1+\varpi^N$ for $N>N_0$, the character will be trivial. I'm happy with this, but I don't understand how to get the rest of the calculation to work out, though I feel it should be a straight-forward exercise.

  • $\begingroup$ I have written something up refering heavily to Silberger, Casselman and Bushnell-Henniart, especially with a paragraph relating K-types and strata for GL(2). If you are interested, send me an email marc.palm@uni-hamburg.de. The philosophy is that you should not try to understand all $K$-types, but only the interesting ones (possibly varying $K$ though). This is in particular useful for GL(n), because we do not have a classification of irrep of GL(n, o). $\endgroup$ – Marc Palm Oct 22 '12 at 8:09
  • $\begingroup$ Btw, the relation between $K$-types and strata is best done via Clifford theory (I found Stasinski pretty useful here arxiv.org/abs/0807.4684) $\endgroup$ – Marc Palm Oct 22 '12 at 8:16
  • $\begingroup$ At the lowest level I think you will see all $K$-conjugates of the fundamental stratum $\psi_a$, that is to say all strata $\psi_{a'}$ with $a' \in M_2(\mathfrak{o}/\varpi)$ which are $GL_2(\mathfrak{o}/\varpi)$ conjugate to $a$. I don't know what happens at higher levels but it seems Hansen's paper indeed gives the answer. $\endgroup$ – François Brunault Oct 22 '12 at 9:16
  • $\begingroup$ Yes, the strata of lowest type correspond to the one you induce to get supercuspidal representation and is the fundamental one. (I didn't address this below). $\endgroup$ – Marc Palm Oct 22 '12 at 12:38
  • $\begingroup$ @MarcPalm, if your 'something' is of the form of general notes, rather than of an answer specifically to this question, then it sounds very interesting! Would you consider putting it on the arXiv? $\endgroup$ – LSpice Jul 11 '18 at 11:32

I will try to answer the question as far as I have understood it. Please comment.

Clifford's theorem

Let $G$ be a finite group. Let $H$ be a normal subgroup of $G$. The groups $G$ resp. $G/H$ act on the irreducible representations of $H$ via conjugation on $H$.

  1. Let $\rho$ be an irreducible representation of $G$, then the restriction ( \Res_{H} \rho ) contains precisely one $G$-orbit of irreducible representations of $H$.
  2. Let $\psi$ be an irreducible representation of $H$, and let $G_\psi$ be its stabilizer in $G$. We have a one-to-one correspondence between irreducible representations $\rho_0$ of $G_\psi$, contained in $Ind_{H}^{G_\psi} \psi$, and irreducible representations $\rho$ of $G$, contained in $Ind_{G_\psi}^{G} \psi$. The correspondence is given by $$\rho_0 \mapsto \rho = Ind_{G_\psi}^{G} \rho_0.$$

We want to apply Clifford theory to the irreducible representations of $GL_2(o)$ of level $n \geq 1$, we follow Stasinski. The quotient $\Gamma(p^{n}) / \Gamma(p^{n+1})$ is abelian and isomorphic to the endomorphism ring $M_2(F_q)$ of $F_q \oplus F_q$ via $$ \iota_n \colon \Gamma(p^{n}) / \Gamma(p^{n+1}) \xrightarrow\cong M_2(F_q), \qquad x \mapsto (x-1)/\pi^{n} \bmod p.$$ According to Clifford theory, the restriction $$ Res_{\Gamma(p^{n})} \rho $$ decomposes into a $GL_2(o)$-orbit of one-dimensional representations $$\psi: \Gamma(p^n) \rightarrow \mathbb{C}, \qquad \psi|_{\Gamma(p^{n+1})} = 1.$$ The Pontryagin dual $\widehat{M_2(F_q)}$ is canonically isomorphic to $M_2(F_q) = \Gamma(p^{n}) / \Gamma(p^{n+1})$ via $$ M_2(F_q) \xrightarrow{\cong} \widehat{M_2(F_q)}, \qquad x \mapsto \psi_x, \qquad \psi_x(y) := \psi_{p} \circ tr \left( x \cdotp \right).$$ The orbit space of the conjugation action of $GL_2(o)$ on $\widehat{ \Gamma(p^{n-1}) / \Gamma(p^n)} $ is isomorphic to the orbit space of the action of $GL_2(F_q)$ on the ring $M_2(F_q)$ by conjugation with inverse elements.

Via induction and restriction, Frobenius reciprocity, you can figure out: Every strata with $\mathfrak{e}=1$ gives you a $GL_2(o)$-type. Every $GL_2(o)$-type occurs in the restriction of some unitarizabile $GL_2(F)$-representation.

This is a very standard fact. Perhaps use that $GL_2(F)$ unitary representation are all admissible, and that the compact induction is unitarizabile together with Frobenius reciprocity, and general decomposition theorems for unitarizabile representations into irreducibles.

For the normalizer of the Iwahori subgroup, the relation betwenn $K$-type and strata is somewhat more difficult, but follows the same principle (Clifford theory).

Silberger has considered the story for both in odd residue characteristic (the oddness assumption seems irrelevant). He is also a little more detailed about the $K$-types regarding the last section of your question than Casselman.

Usually everything you want to show can be done via Frobenius reciprocity, the Mackey Induction-Restriction formula, Clifford theory, and coset decomposition (Bruhat over the residue field/Iwahori decomposition), of course exploited in the right fashion.

So regarding the last section, you only want to compute $Res_{\Gamma(p^{N-1})} Ind_{\Gamma_0(p^N)}^K \chi$ via Mackey IndRes formula, and you will get the conjugacy class of strata (via Clifford theory). The necessary coset space is computed via Bruhat and Iwahori decompositions.

  • 1
    $\begingroup$ Thanks for this nice answer. There is something I don't get yet : is it possible to compute also $\operatorname{Res}_K \operatorname{cInd}_{KZ}^G \chi$ when $G=\mathrm{GL}_2(k)$ and $\chi$ is a representation of $KZ$? You've taken the case $G$ is finite, but presumably Clifford theory extends to the case of compact induction? $\endgroup$ – François Brunault Oct 22 '12 at 13:00
  • $\begingroup$ (I mean $\operatorname{Res}_K (\operatorname{cInd}_{KZ}^G)$ where $\operatorname{cInd}$ denotes compact induction.) $\endgroup$ – François Brunault Oct 22 '12 at 13:02
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    $\begingroup$ Hi Marc, I sent you an email. I'll write some more comments here later (and I finally got around to commenting on your previous answer to a question of mine), but I have to go run some errands. Thanks for the answer! $\endgroup$ – B R Oct 22 '12 at 17:23
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    $\begingroup$ @Marc : Thanks for your comments. I'm not familiar with technical differences between $\operatorname{cInd}$ and $\operatorname{Ind}$ but I guess you're right! $\endgroup$ – François Brunault Oct 23 '12 at 9:02
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    $\begingroup$ @François Brunault: I see now that I didn't actually address your representation! Clifford theory extends to profinite (mod center) groups. Computing $Res_K Ind_{KZ}^G \chi$ is possible with Mackey's Induction Restriction Theorem, but the theory of types is precisely the gadget to avoid such long and messy computations, at least that is how I try to use it! $\endgroup$ – Marc Palm Nov 30 '12 at 14:04

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