Unitary Central Character by Schur's Lemma

Question

Consider an irreducible smooth representation $\pi$ of the group $G=GL_n(\mathbb{Q}_p)$ with center $Z$. Does there exist a unitary central character for $\pi$?

More precisely, is there a (quasi-)character $\omega: G \to \mathbb{C}^{\times}$ such that $\pi \otimes \omega$ when restricted to the center $Z$ is a unitary character for $Z$? I find this result casually stated in many references, where they say it follows from Schur's lemma. But I am unable to see it directly from Schur's lemma.

Answer 1

Recall the following from Garrett's Automorphic Representations and L-functions:

Definition: A central character is simply a continuous group homomorphism $\omega : Z_A \to $ $\mathbb C^\times$ where $Z$ is the center of a reductive linear group $G$ over a field $k$ and $Z_A$ denotes the adele points of $Z$.

Triviality of $\omega \in Z_k$ is often assumed, resulting in the mapping $\omega : Z_A \to Z_k \diagup Z_A \to $ $\mathbb C^\times$. Furthermore, $\omega$ is said to be unitary when, $\forall z \in Z_A$, $|\omega(z)| = 1$

The key theorems are as follows:

Theorem 4.1 (Schur's Lemma over $\mathbb C$). If $V$ is an irreducible complex $G$-representation, then every linear operator $\phi : V \to V$ that commutes with $G$ is a scalar.

Proof. Let $\lambda$ be an eigenvalue of $\phi$ and assume that the eigenspace $E_\lambda$ is $G$-invariant. Then $v \in E_\lambda \implies \phi(v) = \lambda v,$ whence $\phi(gv) = g\phi(v) = g(\lambda v) = \lambda * gv$, so $gv \in E_\lambda$ (g was arbitrary). However, by irreducibility, this implies that $E_\lambda = V$, thus $\phi = \lambda \text{Id}$.

So, consider an irreducible smooth $G$-representation $\pi$.

Theorem 3.8 (Schur's Lemma). For a td-group $G$, any irreducible smooth $G$-representation $\pi$ satisfies End$_G$($\pi$) = $\mathbb C$.

Since a unitary central character is, by definition, a continuous group homomorphism $\omega : Z_A \to \mathbb C^\times$ such that $|\omega(z)| = 1 \space \forall z \in Z_A$, and, by Theorem 3.8, any irreducible smooth $G$-representation $\pi$ satisfies End$_G$($\pi$) = $\mathbb C$, we can associate a central character $\omega_\pi(z) : Z \to \mathbb C^\times$ such that, for all matrix coefficients $f$ of $\pi$, $f(zh) = \omega_\pi(z)f(h).$ (commuting with G as a scalar, cf. Thm 4.1)

A matrix coefficient defined in Definition 5.1 of 2.

Answer 2

There is a central character $\omega:Z \rightarrow \mathbb{C}^{\times}$ by Schur, but it need not be unitary. E.g., consider $∣\text{det}∣$.

Answer 3

Let $Z$ be the center of $G$, and $\varpi: Z \rightarrow \mathbb C^{\ast}$ the central quasicharacter of $\pi$. We can identify $Z$ with $\mathbb Q_p^{\ast}$. You are asking whether there exists a quasicharacter $\chi$ of $G$ such that the quasicharacter

$$z \mapsto \varpi(z)\chi(z)$$

of $Z$ is unitary. The answer is yes. It is a fact that every quasicharacter of $\mathbb Q_p^{\ast}$ is of the form $x \mapsto c(x)|x|^s$, where $c$ is a quasicharacter of $\mathbb Z_p^{\ast}$ (automatically unitary since $\mathbb Z_p^{\ast}$ is compact), extended to all of $\mathbb Q_p^{\ast}$ via a choice of uniformizer, and $s$ a complex number.

Write $\varpi(x) = c(x)|x|^s$ as above, and let $\chi$ be the character of $G$ given by

$$g \mapsto |\operatorname{det}(g)|^{-\frac{s}{n}}$$

Upon restriction to $Z = \mathbb Q_p^{\ast}$, $\chi$ becomes the character $x \mapsto |x|^{-s}$, so we have $\varpi(x)\chi(x) = c(x)$, which is unitary.