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Let $k$ be a local field and $n \in \mathbb{N}$.

Question: I would like to know precisely, which irreducible (admissible) representations of $G := \textrm{GL}_n(k)$ are generic, i.e. which admit Whittaker-models.

Preliminaries: Recall that if $\psi \colon k \to \mathbb{C}^{\times}$ is a (unitary?) additive character, $\psi$ lifts to a character of the (standard) maximal unipotent subgroup $U:=U_n(k) \subset G$ of the standard Borel $B := B_n(k) \subset G$ (upper-triangular matrices), and the space of Whittaker functions is the smooth induction $$ \mathfrak{W}(\psi) := \textrm{Ind}^{G}_{U}(\psi). $$ An irreducible admissible representation $(\pi,V)$ of $G$ is called generic, iff there exists an embedding $$ \mathfrak{W} \colon (\pi,V) \hookrightarrow \mathfrak{W}(\psi). $$ By Gelfand-Kajdan's Multiplicity One-Theorem, the multiplicity of $(\pi,V)$ inside $\mathfrak{W}(\psi)$ is $\leq 1$, so if $(\pi,V)$ exists, we call its (unique) image $\mathfrak{W}(\pi, \psi)$ to be the Whittaker-model of $(\pi,V)$.

Reference: My result resumee is Prasad-Raghuram's notes on "Representation Theory of $GL(n)$ over Non-Arch.fields" (http://www.math.tifr.res.in/~dprasad/ictp2.pdf). In Thm.9.3 (reference is Theorem 9.7. in Zelevinsky's 'Induced Representations of reductive $\mathfrak{p}$-adic groups II'), the authors state that

An irreducible admissible representation $\pi$ (realized as the unique irreducible quotient of some parabolically induced representation) is generic, if and only if $\pi$ itself is parabolically induced (from essentially square-integrable) and (thus, the parabolic induction itself is irreducible).

"Thus, in particular, discrete series representations are generic". I don't get this, as Steinberg representation is a discrete series, so it should be generic, but on the other hand, it is not of the form $\textrm{Ind}^{G}_{B}(\chi_1,\ldots,\chi_n)$, but the unique irreducible quotient of the reducible $$ \textrm{Ind}^{G}_{B}(| \cdot |^{-\frac{n-1}{2}}, | \cdot |^{-\frac{n-3}{2}}, \ldots, | \cdot |^{\frac{n-1}{2}}), $$ so it directly contradicts Theorem 9.3 of Prasad-Raghuram. What did I misunderstand?

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    $\begingroup$ The Steinberg is parabolically induced from an essentially square-integrable representation, because $G$ itself is a parabolic in $G$, and the Steinberg is the induction of itself. (You can also write it as a subquotient of a reducible principal series, but that doesn't contradict the statement in my previous sentence.) $\endgroup$
    – David Loeffler
    Commented Jun 12, 2023 at 12:11
  • $\begingroup$ @DavidLoeffler thank you! So you say, I take the segment $\Delta := (| \cdot |^{-\frac{n-1}{2}}, | \cdot |^{-\frac{n-3}{2}}, \ldots, | \cdot |^{\frac{n-1}{2}})$, and the irreducible quotient $Q(\Delta)$ of $\mathrm{Ind}^{G}_{B}(\Delta)$ is ess.square-integrable, and therefore $Q(\Delta) = \mathrm{Ind}^{G}_{G}(Q(\Delta))$ is generic, right? $\endgroup$
    – Maty Mangoo
    Commented Jun 12, 2023 at 12:28
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    $\begingroup$ Yes, it is discrete-series so it is generic. $\endgroup$
    – David Loeffler
    Commented Jun 12, 2023 at 12:30
  • $\begingroup$ @DavidLoeffler I interpreted the Steinberg as follows: I have the segments $\Delta_1 := | \cdot |^{-\frac{n-1}{2}}, \ldots, \Delta_n := | \cdot |^{\frac{n-1}{2}}$. Then each segment itself is irreducible, thus $Q(\Delta_i) = \textrm{Ind}^{k^{\times}}_{k^{\times}}(\Delta_i)$. But I think this was my misinterpretation; probably Prasad-Raghuram already want the does not precede condition for the segments, i.e. $\Delta_1 \otimes \ldots \otimes \Delta_n$ shall not be a segment itself. $\endgroup$
    – Maty Mangoo
    Commented Jun 12, 2023 at 12:52
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    $\begingroup$ Yes, if you write the characters that way around, the unique quotient is the trivial representation (which is not generic). $\endgroup$
    – David Loeffler
    Commented Jun 12, 2023 at 13:09

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