# Coefficients of irreducible supercuspidal representation

I am going through Getz and Hahn's Intro to Automorphic Reps, and I got stuck on a question about the existence of a certain kind of coefficient (Ex 8.11 of https://sites.duke.edu/heekyounghahn/files/2022/04/GTM.pdf). The question reads as follows:

Suppose $$(\pi, V)$$ is an irreducible supercuspidal representation of $$G(F)$$, where $$G$$ is a reductive group over a non-archimedean local field $$F$$. Show that there exists a function $$f_\pi \in C_c^\infty(G(F))$$ satisfying (1) $$\operatorname{tr} \pi(f_\pi) = 1$$ and (2) if $$\pi'$$ is another irreducible admissible representation $$\pi'$$ of $$G(F)$$ such that $$\operatorname{tr} \pi'(f_\pi)\ne 0$$, then $$\pi \cong \pi'\otimes \chi$$ for some quasi-character $$\chi:G(F)\to \mathbb C^\times$$.

My rough idea was as follows:

As $$\pi$$ is irreducible, there exists $$v\in V$$ such that $$V =G(F) \cdot v$$. Consider any parabolic $$P$$ with unipotent radical $$N$$ and Levi $$M$$. As $$\pi$$ is supercuspidal, the Jacquet module $$V_N$$ is zero, and so in particular there exists a compact subgroup $$K = K(P)$$ such that $$\int_K \pi(n) v dn = 0$$. Take $$f_P := 1_{K}$$. By running through 'enough' parabolics, we define $$f_\pi$$ to be the sum of all of these $$f_P$$, renormalized so that the trace of $$f_\pi$$ is 1.

If $$(\pi', W)$$ is some other irreducible supercuspidal representation with $$\operatorname{tr} \pi'(f_\pi)\ne 0$$, then from the non-archimidean Langlands classification we know that $$\pi'$$ is the unique irreducible quotient of a parabolically induced representation $$I(\sigma, \lambda)$$ for some $$\sigma$$ a unitary tempered representation of a parabolic subgroup $$P$$ and some $$\lambda\in X^*(G) \otimes\mathbb C$$. Computing traces, we see that $$\operatorname{tr} \pi'(f_\pi) = \operatorname{tr} (\sigma \otimes |\lambda| )(f_\pi)$$. Taking $$K = K(P)$$-fixed points and using admissibility, we know that $$V^K \cong W^K \otimes \chi$$ for some quasi-character $$\chi$$. Since this is true for 'many' parabolics $$P$$, and using the irreducibility as well as supercuspidality of $$\pi'$$, we can conclude that $$V\cong W\otimes \chi$$.

This is clearly a very hand-wavy 'proof', and I'm not even sure this works. Is the above idea salvageable? If not, I would appreciate any hints on how to proceed.

More generally, I'm still not entirely sure how to think about either supercuspidality or about coefficients (smooth compactly supported functions $$f_\pi$$ such that $$\operatorname{tr}\pi(f_\pi)\ne 0$$ but $$\operatorname{tr}\pi'(f_\pi)=0$$ for $$\pi'\not\cong \pi$$) of an irreducible representation, and I would appreciate any insight into either of these beyond the definitions.

Let $$K\subset G$$ be a compact open subgroup such that $$V^K\ne0$$. Pick a basis $$\{v_1,\dots,v_N\}$$ of $$V^K$$ and a dual basis $$\{v_1^\vee,\dots,v_N^\vee\}$$ of $$(V^K)^\vee$$.
Now $$f_\pi(g):=\langle \pi^\vee(g)v_1^\vee,v_1\rangle$$ is a matrix coefficient of $$\pi^\vee$$, which is $$K$$-invariant.
Now, by definition $$\mathrm{tr}\ \pi(f_\pi)$$ is the trace of the endomorphism of $$V^K$$ given by $$v\mapsto \int_{G/Z}\langle \pi^\vee(g)v_1^\vee,v_1\rangle \pi(g)vdg.$$ Thus, \begin{align*}\mathrm{tr}\ \pi(f_\pi)&=\sum_{i=1}^n\int_{G/Z}\langle\pi^\vee(g)v_1^\vee,v_1\rangle\langle v_i^\vee,\pi(g)v_i\rangle dg\\ &=d(\pi)^{-1}\sum_{i=1}^n\langle v_1^\vee,v_i\rangle\langle v_i^\vee,v_1\rangle\\ &=d(\pi)^{-1}. \end{align*} Thus normalizing appropriately, $$f_\pi$$ satisfies condition (1) of the problem.
Moreover, if $$\pi'$$ is another irreducible admissible representation, then we can twist $$\pi'$$ by a suitable character so that $$\pi$$ and $$\pi'$$ has the same central character. Now, if $$\pi\ncong\pi'$$, then by Schur orthogonality \begin{align*}\mathrm{tr}\ \pi'(f_\pi)&=\sum_{i=1}^n\int_{G/Z}\langle\pi^\vee(g)v_1^\vee,v_1\rangle\langle v_i^\vee,\pi'(g)v_i\rangle dg\\ &=0. \end{align*}
Remark: Supercuspidality is used to say the integral converges, since it says the matrix coefficient $$f_\pi(g)$$ is compactly supported modulo $$Z$$.