Let me take $G$ to be a simple (connected) split reductive group over a local field $K$. One way I might go about constructing a (smooth, admissible) complex representation $\sigma$ of $G$ is as follows:

  • pick a parahoric subgroup $P$ of $G$, with pro-unipotent radical $U$,
  • form the quotient $P/U$, which is a (connected) reductive group over a finite field,
  • write down a representation $\overline{\rho} \colon P/U \to \mathrm{GL}_n(\mathbb{C})$,
  • inflate this to a representation $\rho \colon P \to \mathrm{GL}_n(\mathbb{C})$,
  • induce $\rho$, giving $\sigma = \mathrm{ind}_P^G(\rho)$ (compact induction).

How does this procedure work out? Excepting that I might have to take possibly smaller $U_0 \subset U$, is this expected to produce all the (smooth admissible irreducible complex) representations of $G$? When do I get something irreducible, or a supercuspidal? What is the proper formulation when $G$ is not assumed to be split over $K$?

  • $\begingroup$ Bushnell&Henniart "Local Langlands for GL(2)$ us a good place to start. I am not sure about the terminology you use, but I guess you ask something like mathoverflow.net/questions/101067/…? $\endgroup$
    – Marc Palm
    Commented Oct 15, 2012 at 9:59
  • $\begingroup$ As Paul Broussous mentions below (mathoverflow.net/a/109797/2383), you will get all representations this way (by shrinking $U_0$), but in a very impractical way: your induced representations will be enormous (in particular, not irreducible), and there will probably be no effective description of the decomposition. As Paul mentions, for the complete spectrum, you want the theory of types; but, for supercuspidals, you probably want to see the work of Yu ams.org/mathscinet-getitem?mr=1824988 and its precursors (Adler, Corwin–Howe, Howe–Moy). $\endgroup$
    – LSpice
    Commented Jan 6, 2015 at 17:29

1 Answer 1


This procedure allows to construct all "level $0$" irreducible representations of G. They appear as subquotients of your compactly induced representations. Here "level $0$ means that the representation has a non-zero fixed vector under the pro-unipotent subgroup of some parahoric. The answers to your questions are is the following paper:

Morris, Lawrence Level zero $\bf G$-types. Compositio Math. 118 (1999), no. 2, 135–157

If you use smaller groups $U_o \subset U$, then you can indeed get any irreducible representation as subquotient of a compactly induced representation. However when the compactly induced representation is irreducible it is automatically supercuspidal (see e.g. Bushnell-Henniart for a proof of that). All explicitely known supercuspidal representations are indeed obtained by compact induction. But it is still conjectural that they all are.

In general a compactly induced representation from an irreducible representation of a compact open subgroup splits in two part. An admissible part which is a finite sum of supercuspidal representations and a non admissible part which contains non-supercuspidal as irreducible subquotients.

To describe the non supercuspidal representations by compact open data, a good point of view is that of "types". You may read Bushnell and Kutzko's papers on that subject.

  • $\begingroup$ Thanks, I'm going to be reading that minutiously. What about taking smaller $U_0$ inside $U$, does this give higher level representations, and are those expected to be all of them? $\endgroup$
    – Will
    Commented Oct 16, 2012 at 10:20
  • 1
    $\begingroup$ At least in the tame (i.e., large residual characteristic) case, it is now known that all supercuspidals do arise by compact induction (indeed, by Yu's construction), by Kim's work on exhaustion ams.org/mathscinet-getitem?mr=2276772. $\endgroup$
    – LSpice
    Commented Jan 6, 2015 at 17:31

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