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?

`$n=2$`

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.) $\endgroup$ – Jim Humphreys Apr 29 '13 at 13:38