# Unitary Central Character by Schur's Lemma

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.

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.

• I don't think this answers OP's question – D_S Jan 31 at 1:53

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.

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