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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.

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1 Answer 1

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This follows from Schur orthogonality (see Bushnell-Henniart 10a.2).

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 $Z$.

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