Let $G$ be a reductive group over a non-archimedean field $F$ with reisdue field $f$.

Edit: The statements only make sense modulo tensoring by one-dimensional representations.

Are the unitary, square-integrable representation (modulo tensoring by one-dimensional reps) of $G(F)$, which are not supercuspidal, in one-to-one correspondence with a certain subclass of representations of $G(f)$ (modulo tensoring by one-dimensional reps)?

I am mostly interested in the case of $G=GL(n)$. The question has an easy answer if $n=2$.

For the case GL(2): The sq.int, non-supercuspidal reps are isomorphic to the Steinberg tensored by a one-dimensional representation $G(F)$. There is precisely one irreducible rep of $G(f)= G(o/p)$ contained in the Steinberg rep of $G(F)$, i.e., it is the Steinberg of $G(f)$.

More concretely, do they also in general admit a $\Gamma(p) = \{ \gamma \in G(o) : \gamma = 1 \bmod p \}$-invariant vector modulo tensoring by one-d. reps? Is their restriction to the Iwahori(= pullback of $B(f)$ to $G(o)$ for a fixed Borel subgroup) or its "Levi component" a one-dimensional representation?

  • 2
    $\begingroup$ Here is something worth looking at: square integrable representations in the unramified principal series of the p-adic group correspond to certain representations if affine Hecke algebras (classified by Kazhdan and Lusztig in their 1997 Inventiones paper. On the other hand representations of Weyl groups were classified by Springer in his 1978 paper in the same journal. $\endgroup$ Apr 29, 2013 at 11:22
  • $\begingroup$ @Marc: It would help a non-specialist like me to have some indication of what is true for $n=2$ about the subclass of finite group representations involved, since the latter are well understood in that special case. (For larger ranks much is known but much is also unknown about the finite group representations.) $\endgroup$ Apr 29, 2013 at 13:38

2 Answers 2


For a general $G$, it is false that a general square integrable representation has a fixed non zero vector under the first congruence subgroup (even after a suitable twisting by a character) (there is a counter-example for e.g. ${\rm GL}(4)$). So in general you have to restrict to "level $0$" square integrable representations. For certain groups it is known that level zero square integrable irreducible representation are somehow parametrized by certain representation of $G(f)$. For exemple this is done by Silberger and Zink for the group ${\rm GL}(m,D)$, $D$ a division algebra:

Silberger, Allan J.(1-CVLS); Zink, Ernst-Wilhelm(D-HUMB-IM) An explicit matching theorem for level zero discrete series of unit groups of $p$-adic simple algebras. (English summary) J. Reine Angew. Math. 585 (2005),

P.S. For ${\rm GL}(p)$, $p$ prime, the non supercuspidal square integrable irreducible representations are up to twisting level $0$ representations.


As a follow-up to Paul Broussous's answer, not every representation is depth-0, up to a central twist, but those that are are described by representations of finite groups associated to, but not necessarily of the same type as, $G$. Specifically, they are quotients of parahoric subgroups, and the representations are cuspidal. This is described in Proposition 6.8 of the second Moy–Prasad paper.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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