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.*



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> 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)?
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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 $G(f)$.



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> 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?
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Steinberg reps of reductive groups over local fields vs finite fields
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. 


