I am a beginner in the field of representation theory. I was reading the book 'The Admissible Dual of $GL(N)$ Via Compact Open Subgroups' by Bushnell and Kutzko.

Let me first describe the book a little.

The book uses hereditary orders $\mathfrak{A}$ arising from lattice chains and then defines simple strata using the special integer $k_0(\beta,\mathfrak{A})$ (cf. page $43$). Then in page $95$ the authors construct the subgroups $H^m(\beta,\mathfrak{A})$ and $J^m(\beta,\mathfrak{A})$ and uses them (specifically $J^0$ and $J^1$ ) in the later chapters to extend representations of $J^1$ to $J^0$ and then they define the simple types $(J,\lambda)$ (cf. page $184$). In page $162$ they define the maximal and minimal orders $\mathfrak{B}_M $ and $\mathfrak{B}_m$ and uses them throughout the theory. Chapter $7$ proceeds with the Iwahori decomposition where they define (cf. page $213$) the parabolic subgroups $P$ with Levi decomposition $P=MU$, where $U$ is the unipotent radical of the group $G$ in consideration. The book finishes (chapter $8$) with the classification of smooth representations: the supercuspidals containing a simple type and the 'atypical' representations containing the split types.

Being a beginner in this field it is necessary for me to have a concrete example to keep in mind while reading the above book because it is too general (construction using lattices and so on). Therefore I would like to have a concrete example of the following:

$(a)$ the hereditary order $\mathfrak{A}$,

$(b)$ the groups $U^n(\mathfrak{A})$ in page $21$,

$(c)$ the integer $k_0(\beta,\mathfrak{A})$,

$(d)$ the groups $H^m(\beta,\mathfrak{A})$ and $J^m(\beta,\mathfrak{A})$, specifically when $m=0,1$,

$(e)$ the utility of the maximal and the minimal hereditary orders $\mathfrak{B}_M$ and $\mathfrak{B}_m$ and there concrete examples,

$(f)$ concrete examples of the groups $P,M,U,U^-$ defined in $(7.1.13)$.

$(g)$ the importance of computing the interwining of representations (cf ($5.1.8$), ($6.1.3)$, e.t.c).

Regarding the Interwining of characters, it is not clear to me why the Interwining is impotant in representation theory. What role does it play? I agree that in chapter $8$ there is a theorem which says that the supercuspidal representations twisted by unramified characters have the same simple types. But I do not understand the impotantance of computing the Interwining of representations.

I know that answering these questions above will consume long typing in latex. I thank you in advance for your efforts.


Your question (f) makes me suspect that you don't really know any of the representation theory of $p$-adic groups at all. You definitely should not try to read the Bushnell--Kutzko book before learning at least the basics of this. Either (the first few sections of) Bushnell--Henniart The local Langlands conjecture for $GL_2$ or one of the various sets of unpublished notes (although a lot of these are themselves very technical) should be read first. The point of the Bushnell--Kutzko book is to prove a very technical explicit description of a large part of the representation theory of $GL_n$, with the real application being a description of a large part of the category of smooth representations of $GL_n$ in terms of affine Hecke algebras. If you don't know much about smooth representations, then I don't know why you'd want to know this (or at least, the details of this). Along the way they describe explicitly the supercuspidal representations. Again, to understand why you'd want to do that, you'll need to know about smooth representations. To answer your question (f), look up parabolic subgroups and Levi factors.

You should understand $GL_2$ first. This is described explicitly (although without giving details backing up the claims) in Henniart's appendix to the paper Multiplicites modulaires des representations... by Breuil--Mézard; filling in the details is a good exercise. Once you understand $GL_2$, you essentially already understand $GL_n$ for any prime $n$. Once you understand $GL_n$ for $n$ prime, you essentially understand a large number of cases of the construction (those arising from "minimal simple strata"), which lets you understand the theory apart from the constant $k_0$, which is probably the hardest part of the construction to understand. In particular, if you don't understand the minimal case then there isn't really much point worrying about it yet.

So rather than explicitly answering each of your questions, here is some motivation of the construction. You identify a class of irreducible representations which you'd like to have a construction for: the supercuspidals. It's conjectured that these look like $Ind_J^G \lambda$ for some open, compact-mod-centre subgroup $J$ of $G$ and some irreducible smooth representation $\lambda$ of $J$. This is where intertwining is important: given $J$ open, compact-mod-centre and $\lambda$ an irrep of $J$, if the $G$-intertwining of $\lambda$ is $J$ then $Ind_J^G \lambda$ is irreducible supercuspidal. This is a crucial observation underlying the whole construction. Central characters are easy, so forgetting about those it's reasonable to study a supercuspidal $\pi$ by how it acts through some good choice of compact open subgroup of $G$.

So you begin to approximate $\pi$. First of all, $\pi$ has an invariant called its (normalized) level: this is a rational number with denominator dividing $n$ which essentially tells you how much of $G$ the action of $\pi$ is interesting on (since it's smooth, restricting to a sufficiently small compact open subgroup isn't going to be very interesting). It turns out that the correct way to define the normalized level is to identify a class of characters $\psi_\beta$ of groups $U^n(\mathfrak{A})$, and identify these in the restriction of $\pi$. These characters are what would probably be called minimal $K$-types these days.

They aren't types in the Bushnell--Kutzko sense, but they're getting there: if $\pi$ contains a given minimal $K$-type, then this tells you something about the level of $\pi$ (although it isn't quite so simple as telling you precisely what the level is). Given $\pi$, you can identify a "correct" conjugacy class of minimal $K$-types in $\pi$, which do tell you the level of $\pi$. This is the first step in the classification, and these $K$-types correspond to technical things which Bushnell and Kutzko call strata. Among strata, there is a subclass of strata called simple strata. To simple strata, you can associate simple characters. A simple character is a character extending the minimal $K$-type of $\pi$ to a larger compact open subgroup, which satisfies nice intertwining properties. This is where the magic happens in the construction, and is the hardest part to understand.

But OK, take the construction of simple characters as given. Take a simple character $\theta$ in $\pi$ (which will be unique up to conjugacy), and consider the representation $Ind_{H^1}^G \theta$. This isn't irreducible, but it's not too bad. Recalling the conditions for induced reps to be irreducible supercuspidal, a pretty good guess is to calculate the intertwining of $\theta$ -- $J$, say -- and then isolate a subrep $\lambda$ of $Ind_{H^1}^J \theta$. Hopefully then $Ind_J^G \lambda$ would be irreducible supercuspidal. This is the case, as long as you take a bit of care with your choices of $\lambda$, and the resulting pairs $(J,\lambda)$ are the maximal simple types. The point is then that $(J,\lambda)$ improves the approximation to $\pi$ enough that it now tells you exactly what $\pi$ is.

(I should note that I've been a bit slack in that my $(J,\lambda)$ is really an extended maximal simple type, the difference doesn't really matter though).

As for the groups? The groups $U^n(\mathfrak{A})$ are easy to understand: $U(\mathfrak{A})$ is a "parahoric subgroup", i.e. the inverse image of a parabolic subgroup in $GL_n$ over the residue field, and is naturally the group of units of a ring (a hereditary order). The powers of the Jacobson radical of this ring then define the filtration $U^n(\mathfrak{A})$ of $U(\mathfrak{A})$. As for the other groups? They're just what work; there isn't some simple insight that makes the definition of $H^1$ look obvious -- this is a vast generalization of the work of Howe, Moy, Kutzko, etc over a number of years. However, if you do follow my advice and look at what happens when $n$ is prime, then you'll see that (as long as you avoid the nasty cases where $k_0$ comes into play), then these groups $H^1,J^1,J$ really aren't too bad at all: they look something like $U^k(E)U^n(\mathfrak{A})$ for some hereditary order $\mathfrak{A}$ and some extension $E/F$ of degree dividing $n$ (and ramification degree the lattice period of $\mathfrak{A}$), where $k=0$ in the case of $J$ and $k=1$ for $H^1,J^1$. In general, depending on your constant $k_0$, you'll have to go through a finite inductive process which makes these groups look similar to this form, but with more terms.

Edit: I just remembered about these notes, which summarize the whole construction without the more technical details pretty quickly. So maybe the thing to do is understand $GL_2$ from Henniart, convince yourself that the same thing works for $GL_n$, $n$ prime, and then compare that process with the one outlined in the notes.

| cite | improve this answer | |
  • $\begingroup$ @P.L Can you give me a reference for the fact that if the interwining of $\lambda$ is $J$ then the induced representation is supercuspidal? $\endgroup$ – MathStudent May 5 '16 at 18:25
  • 2
    $\begingroup$ The one I know is Carayol, Représentations cuspidales du groupe linéaire, Proposition 1.5. He cites Howe's paper on tamely ramified supercuspidals, which I've never read, so I'm not sure where this originates. $\endgroup$ – PL. May 5 '16 at 18:50
  • 1
    $\begingroup$ Oh, something that it occurs to me I should have said: when you look at $GL_n$ for $n$ prime, you can forget about simple strata and $k_0$. Instead, you just consider fundamental strata of the form $[\mathfrak{A},n,n-1,\beta]$, which are much easier to get used to (in particular, the definition of "fundamental" is easy). I think that this is done for $n=2$ in Bushnell--Henniart, but I can't quite remember how much they prove. Maybe one good, although not quite precise, way of understanding $k_0$ is that it measure how far a stratum is from being fundamental. $\endgroup$ – PL. May 6 '16 at 18:06
  • $\begingroup$ @dulalnarugopal, the relevant point is that the representation is supercuspidal if it is irreducible (since then the extension of a matrix coefficient of the inducing subgroup is a compactly supported matrix coefficient); but Mackey theory allows you to test for irreducibility by looking at intertwining. $\endgroup$ – LSpice Nov 26 '16 at 22:34

1) On the groups $U({\mathfrak A}))$ and $U^n ({\mathfrak A})$. Their structure theory are very well explained in:

Bushnell, C. J.; Fröhlich, A. Nonabelian congruence Gauss sums and $p$-adic simple algebras. Proc. London Math. Soc. (3) 50 (1985), no. 2, 207–264.

But as PL wrote, you should start with the case $GL_p$, $p$ prime. The structure of these groups in that case is explained (maybe with a different notation) in Carayol's paper.

2) Before tackling Bushnell and Kutzko's theory, except looking at the $GL_p$ case, you should first read introductory papers :

Henniart, Guy Représentations des groupes réductifs $p$-adiques. (French) [Representations of $p$-adic reductive groups] Séminaire Bourbaki, Vol. 1990/91. Astérisque No. 201-203 (1991), Exp. No. 736, 193–219 (1992).

Bushnell, Colin J.; Kutzko, Philip C. The admissible dual of ${\rm GL}_N$ via restriction to compact open subgroups. Harmonic analysis on reductive groups (Brunswick, ME, 1989), 89–99, Progr. Math., 101,

Kutzko, Philip C. Smooth representations of reductive $p$-adic groups: an introduction to the theory of types. Geometry and representation theory of real and $p$-adic groups (Córdoba, 1995), 175–196, Progr. Math., 158, Birkhäuser Boston, Boston, MA, 1998.

| cite | improve this answer | |
  • $\begingroup$ Except that we should probably call it $\mathrm{GL}_\ell$, $\ell$ prime, rather than $\mathrm{GL}_p$, since (I think) the case where $\ell = p$ is harder. $\endgroup$ – LSpice Nov 26 '16 at 22:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.