Describing the depth zero compact induction of a representation which is not cuspidal Let $G$ be a reductive $p$-adic group. I would like to describe a representation of the type
$$\rho = \mathrm{c-Ind}_K^G\,\sigma$$
where $K$ is a maximal parahoric subgroup of $G$ and $\sigma$ is an irreducible representation of the maximal reductive quotient of $K$, which I denote by $\mathcal K$. Thus, the group $\mathcal K$ is a finite group of Lie type, whose representation theory is described by the work of Deligne and Lusztig.
I see in the literature many tools to describe this induction when $\sigma$ is assumed to be cuspidal, which makes sense as cuspidal representations are the "building blocks" in Harish-Chandra / Deligne-Lusztig theories. However, I fail to find a way to reduce to "the cuspidal case" if my $\sigma$ is more general. Is there a reference explaining what kind of strategy can be used in this case ? Is there anything to say in particular when $\sigma$ is unipotent ?

Some of my thoughts.

*

*By twisting $\sigma$ with a properly chosen character for the center $Z$ of $G$, one can reduce to the description of the $\chi$-part of $\rho$, that is the largest subquotient where $Z$ acts like $\chi$. This turns out to be isomorphic to $\mathrm{c-Ind}_{ZK}^G \sigma'$ where $\sigma'$ is our extension by $\chi$. Thus, we may assume our group $K$ to be open, compact modulo the center, containing the center. In this context, if $\sigma$ is cuspidal, then the induction of $\sigma'$ will be irreducible and supercuspidal if I'm not mistaken.


*If $\sigma$ is not cuspidal, just irreducible, then it has a cuspidal support $(\mathcal L,\tau)$ with $\mathcal L$ a Levi in $\mathcal K$ and $\tau$ a cuspidal representation of $\mathcal L$. We may assume $\mathcal L$ to be standard and choose a standard parabolic $\mathcal P$ in $\mathcal K$ containing $\mathcal L$. The preimage of $\mathcal P$ in $K$ would be a parahoric subgroup $K'$ of $G$, to which a Levi subgroup $L'$ can be associated. It is so that $K'$ is a maximal parahoric of $L'$.
Now, at the level of finite groups, $\sigma$ is a component inside the Harish-Chandra induction of $\tau$, from $\mathcal L$ to $\mathcal K$. Then, $\sigma$ is inflated to $K$ and compactly induced to $G$ to give $\rho$. But is it possible to go the other way round ?
That is, I could consider $\tau$ the cuspidal representation of $\mathcal L$, inflate it to $K'$, compacly induce it to $L'$, then take the Harish-Chandra induction to $G$.
How would $\rho$ relate to the representation one gets with this recipe ?
 A: I think that I found a partial answer to my question in Moy-Prasad and Morris' papers, so I'll write it down here. There may be mistakes, so please let me know in the comments if I make one.
For the beginning, I recall the notations I used above. All representations are assumed to be smooth.
The group $G$ is any reductive connected $p$-adic group. For $K$ a parahoric subgroup, we denote by $K^+$ its pro-unipotent normal subgroup, it is also denoted $K_{0^+}$ in the notations of the Moy-Prasad filtration. The quotient $K/K^+$ is denoted by $\mathcal K$, it is called the maximal reductive quotient of $K$. It is the group of rational points of a reductive group over the (finite) residue field. We fix such a parahoric $K$, and we consider a non-cuspidal irreducible representation $\sigma$ of $\mathcal K$.
The representation $\sigma$ of the finite group of Lie type $\mathcal K$ has a cuspidal support for which we fix a representative $(\mathcal L,\tau)$. It means that $\mathcal L$ is a Levi complement in $\mathcal K$ and $\tau$ is an irreducible cuspidal representation of $\mathcal L$, such that $\sigma$ is an irreducible component inside the Harish-Chandra induction of $\tau$. This is well-defined : up to isomorphism the Harish-Chandra induction does not depend on the choice of a parabolic subgroup containing $\mathcal L$ ; and the conjugation-class of the pair $(\mathcal L,\tau)$ is determined by $\sigma$. Because $\sigma$ is not cuspidal, the Levi $\mathcal L$ is proper in $\mathcal K$.
Fix a parabolic $\mathcal P$ of $\mathcal K$ containing $\mathcal L$ as a Levi complement. The preimage of $\mathcal P$ inside $K$ is a proper parahoric subgroup $K'$. The maximal reductive quotient of $K'$ is no other than the levi complement $\mathcal L$ by construction, that is we have $\mathcal L \simeq K'/(K')^+$.
Now, the Harish-Chandra induction of $\tau$ means that we first extend $\tau$ to a representation of $\mathcal P$ by letting the unipotent radical act trivially, then we use ordinary induction of finite groups to obtain a representation of $\mathcal K$. The representation $\sigma$ is some irreducible component inside. We can do the same at the level of the parahorics, that is modulo $K^+$.
We still denote by $\tau$ its inflation to $K'$. The parabolic $\mathcal P$ is the image of $K'$ inside $\mathcal K \simeq K/K^+$. So we extend $\tau$ to a representation $\tilde{\tau}$ of $K'K^+$ by letting $K^+$ act trivially : it is possible because $K'\cap K^+ \subset (K')^+$ already acts trivially. Then, we induce $\tilde{\tau}$ from $K'K^+$ to $K$ : whether we take compact induction or not doesn't matter as $K$ is already compact. The resulting representation of $K$ is then precisely the inflation of the Harish-Chandra induction we described above, at the level of the finite groups.
Thus, the inflation of $\sigma$ to $K$ is an irreducible component of the representation $\mathrm{Ind}_{K'K^+}^{K} \tilde{\tau}$. This can be identified with the $K^+$-invariant of the representation $\mathrm{Ind}_{K'}^{K} \tau$. By left exacteness and transitivity of compact induction, it follows that $\mathrm{c-Ind}_K^{G} \sigma$ is some subspace of $\mathrm{c-Ind}_{K'}^G \tau$. Because $\tau$ is inflated from a cuspidal representation of the maximal reductive quotient of $K'$, the latter induction can be described to some extent, with the theory of types.
Consider pairs $(M,\rho)$ where $M$ is a Levi complement in $G$ and $\rho$ is an irreducible supercuspidal representation of $M$. Two such pairs $(M,\rho)$ and $(M',\rho')$ are said to be inertially equivalent if they differ up to conjugation in $G$ and multiplication by an unramified quasi-character. The inertial class of a pair is denoted by $[M,\rho]$. Let us denote by $\mathcal B(G)$ the set of inertial classes in $G$. Given an irreducible representation $\pi$, it determines a unique inertial class denoted $\ell(\pi) \in \mathcal B(G)$.
Any pair of a parahoric subgroup and a representation inflated from a cuspidal of the maximal reductive quotient is called a unrefined depth zero type of $G$. For instance, $(K',\tau)$ is such a pair. It is known that any unrefined depth zero type is an $\mathfrak S$-type for some finite set $\mathfrak S\subset \mathcal B(G)$. Let us denote by $\mathfrak S_{\tau}$ the finite set associated to $(K',\tau)$. It means that given an irreducible representation $\pi$ of $G$, it is an irreducible subquotient of $\mathrm{c-Ind}_{K'}^{G} \tau$ if and only if its inertial class $\ell(\pi)$ is in $\mathfrak S_{\tau}$. Thus, it remains to describe this set $\mathfrak S_{\tau}$, which is what follows.
There is a canonical way to associate a Levi complement of $G$ to any parahoric subgroup. So, we consider $L$ the Levi complement associated to $K'$. Because $K'$ is not a maximal parahoric, $L$ is a proper subgroup. By the construction of $L$, it is known that $L \cap K'$ is a maximal parahoric subgroup of $L$ whose maximal reductive quotient is isomorphic to the one of $K'$, that is $\mathcal L$. So, we have an identification
$$\mathcal L \simeq K'/(K')^+ \simeq (L\cap K')/(L\cap K')^+$$
We can now consider $\tau$ as an inflated representation of $L\cap K'$. The pair $(L\cap K',\tau)$ is an unrefined depth zero type of $L$, we denote by $\mathfrak S_{\tau}^L \subset \mathcal B(L)$ its associated finite set so that it is a $\mathfrak S_{\tau}^L$-type. Because $L\cap K'$ is a maximal parahoric of $L$, this set can be described. It consists of all inertial classes $[L,\rho]_L$ in $L$ where $\rho$ is a supercuspidal representation of the form
$$\rho = \mathrm{c-Ind}_{\mathrm{N}_{L}(L\cap K')}^L \hat{\tau}$$
where $\hat{\tau}$ is any representation of the normalizer of $L\cap K'$ in $L$ which contains $\tau$ on restriction to $L\cap K'$. Any compact induction of this form is supercuspidal irreducible. Moreover, the set $\mathfrak S_{\tau}^L$ is not necessarily a singleton ; it is one however if $L\cap K'$ coincide with the maximal compact subgroup of its normalizer $\mathrm{N}_{L}(L\cap K')$.
Eventually, the set $\mathfrak S_{\tau}$ consists of all inertial classes $[L,\rho]$ in $G$ for each element $[L,\rho]_L$ in $\mathfrak S_{\tau}^L$.
So here is my partial answer at last. If no mistake was made, then every irreducible subquotient of $\mathrm{c-Ind}_K^G \sigma$ has inertial support an element of $\mathfrak S_{\tau}$. In particular, no such subquotient is a supercuspidal representation of $G$. I don't know if more can be said (admissibility ? A more precise description of the irreducible subquotients ?).
