Realization of the $p$-adic Steinberg representation as a subrepresentation

Question

Let $G = \mathrm{GL}_n(F)$ where $F$ = non-archimedean local field. The Langlands Classification tells one that all irreducible admissible reps of $\mathrm{GL}_n(F)$ can be realized as (the unique irreducible) quotient of a parabolicly induced representation.

In particular, if one takes a (multiplicative) character of the diagonal torus $$ \lambda = (\lambda_1, \ldots, \lambda_n) \,\colon \,\, T_n(F) \to \mathbb{C}, \, t = (t_i)^{n}_{i=1} \mapsto \lambda_1(t_1) \cdot \ldots \cdot \lambda_n(t_n), $$ one can inflate $\lambda$ to the standard Borel $B = B_n(F)$ of upper-triangular matrices, and consider the (smoothly normalized) induced representation $$ \mathrm{Ind}^{G}_B(\lambda) = \{f \colon G \stackrel{\text{loc.cst.}}{\to} \mathbb{C} \, | \forall (b,g) \in B \times G \colon f(bg) = (\delta^{1/2} \otimes \lambda)(b) \cdot f(g)\}, $$ where $\delta$ is the modulus character of $B$ and locally constant (which in this case should be the same as smooth) translates as: For every $f$ there exists an open set $U_f$, s.t. $f(gu) = f(g)$ for all $u \in U_f$ (and $g \in G$).

Now the theory tells us that in this case,

1. $\mathrm{Ind}^{G}_B(\lambda)$ is irreducible $\Leftrightarrow \forall i,j \colon \frac{\lambda_i}{\lambda_j} \neq | \cdot |$, where $| \cdot |$ is the absolute value on $F^{\times}$.
2. (Part of the Langlands Classification) There exists exactly one irreducible quotient denoted by $Q(\lambda)$.

Hence, if we take $\lambda := (| \cdot |^{-\frac{n-1}{2}}, | \cdot |^{-\frac{n-3}{2}}, \ldots, | \cdot |^{\frac{n-1}{2}})$, the 'highly reducible' representation $\mathrm{Ind}^{G}_B(\lambda)$ has exactly one quotient $\mathrm{St}_n := Q(\lambda)$, called the \textbf{Steinberg representation}.

Question:

How can one realize $\mathrm{St}_n$ as the subrepresentation of some (parabolically) induced representation? (It would be awesome if somebody could give me an example of what happens f.e. for $n=3$ and how to handle it).

Remark: This should be possible due to the Theorem 6.5. of Prasad-Raghuram's notes (http://www.math.tifr.res.in/~dprasad/ictp2.pdf).

Answer 1

The following works for arbitrary split reductive groups. Let $\mathrm{ind}$ denote the normalized induction, and let $\mathrm{Ind}$ denote the naive induction.

Concretely, the Steinberg representation is the unique quotient of $\mathrm{Ind}_B^G(1)=C_c^\infty(B\backslash G)$, i.e., smooth functions on the flag variety $B\backslash G$ with the usual $G$-action. We quotient out, for each parabolic $P\supset B$, the functions constant on the $P$-cosets, i.e., the sub-representation $C_c^\infty(P\backslash G)\hookrightarrow C_c^\infty(B\backslash G)$.

Example: When $G=\mathrm{GL}_2(F)$, the induced representation $\mathrm{Ind}_B^G(1)$ consists of smooth functions on $\mathbb P^1$ and we quotient out by the constant functions on $\mathbb P^1$ to obtain the Steinberg $\mathrm{St}_G$.

It is known, in general, that $\mathrm{St}_G^\vee\cong\mathrm{St}_G$. Thus the surjection $\mathrm{Ind}_B^G(1)\to\mathrm{St}_G$ can be turned into an injection $\mathrm{St}_G\cong\mathrm{St}_G^\vee\hookrightarrow\mathrm{Ind}_B^G(1)^\vee$. Now, the duality theorem (3.5 of Bushnell-Henniart) tells us $\mathrm{Ind}_B^G(1)^\vee\cong \mathrm{Ind}_B^G(\delta_B^{-1})$. Here $\delta_B(t_1,\dots,t_n)=\|t_1\|^{1-n}\|t_2\|^{3-n}\cdots\|t_n\|^{n-1}$.

In the language of normalized induction, we have $\mathrm{Ind}_B^G(1)=\mathrm{ind}_B^G(\delta_B^{1/2})$, where $\delta_B^{1/2}=\lambda$ in OP's notation. Thus by taking the dual of the surjection $\mathrm{ind}_B^G(\delta_B^{1/2})\to\mathrm{St}_G$ we obtain an injection $\mathrm{St}_G\hookrightarrow\mathrm{ind}_B^G(\delta_B^{-1/2})$. [Recall that contragradients interact much more nicely with normalized induction]