TITLE: Steinberg reps of reductive groups over local fields vs finite fields
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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.