Let $G_1 \subset G$ be the rational points of $p$-adic reductive groups sharing the same derived group. There are some well known results relating representations of $G_1$ to representations of $G$, for example in Marko Tadić's paper Notes on representations of non-Archimedean $\operatorname{SL}(n)$. Consider the following theorem:

Theorem: Let $\pi_1$ be an irreducible admissible representation of $G_1$. There exists an irreducible admissible representation $\pi$ of $G$ such that $\pi_1$ is isomorphic to a subrepresentation of $\pi\rvert_{G_1}$. If $\pi_1$ is generic and supercuspidal, then $\pi$ can be chosen to be generic and supercuspidal.

The proof involves finding a finite rank free group $S_1$ inside $Z(G)$ such that $S_1 \cap G_1 = 1$, and $G/S_1G_1$ is compact. One extends $\pi$ to $S_1G_1$ by making it trivial on $S_1$, and then induces to $G$.

I was wondering how much control we may have for the central character. If $\pi_1$, $\pi_1'$ are irreducible admissible representations of $G_1$ having the same central character, is it always possible to choose the $\pi$, $\pi'$ in the theorem to also have the same central character? If anyone knows a reference that deals with this sort of question, it would be greatly appreciated.

  • 1
    $\begingroup$ Note: Tadić's last name is spelled as so, not with a 'k'. I have edited accordingly, since this is on the front page anyway. $\endgroup$
    – LSpice
    May 3, 2022 at 13:37

1 Answer 1


I think I have something that works, modulo two mild assumptions:

1 . $Z(G_1) \subseteq Z(G)$.

2 . Given a smooth character $\omega_1$ of $Z(G_1)$, there exists an extension $\omega$ of $\omega_1$ to a smooth character of $Z(G)$.

(Sketch): Given a representation $\pi_1$ of $G_1$ with central character $\omega_1$, extend $\omega_1$ to a character $\omega$ of $Z(G)$. Define a representation $\pi^{\vee}$ of $Z(G)G_1$ by $\pi^{\vee}(zg_1) = \omega(z)\pi(g_1)$ for $z \in Z(G)$ and $g_1 \in G_1$. This is well defined. Now, $G/Z(G)G_1$ is a finite abelian group. Induce to get $\operatorname{Ind}_{Z(G)G_1}^G \pi^{\vee}$.

The map $f \mapsto f(1)$ should given an isomorphism of $Z(G)G_1$-representations for an irreducible $G$-subrepresentation $\pi$ of $\operatorname{Ind}_{Z(G)G_1}^G \pi^{\vee}$. Let $\omega_{\pi}$ be the central character of $\pi$. For $f$ in the space of $\pi$ with $f(1) \neq 0$, we have

$$\omega_{\pi}(z)f(1) = f(z) = \omega(z)f(1)$$

and so the central character of $\pi$ is just the character $\omega$. This if we start with $\pi_1, \pi_1'$ with the same central character $\omega_1 = \omega_{\pi_1} = \omega_{\pi_2}$, the central characters of the corresponding $\pi_1$ and $\pi_2$ are the same, equal to $\omega$.

  • $\begingroup$ Smooth characters $\omega_1$ always extend. Choose an open subgroup $U$ of $Z(G)$ whose intersection with $Z(G_1)$ is contained in the kernel of $\omega_1$. Now use the fact that $\mathbb C^\times$ is injective as an Abelian group to define a homomorphism $\omega : Z(G)/U \to \mathbb C^\times$ extending $\omega_1 : Z(G_1)/(U \cap Z(G_1)) \to \mathbb C^\times$. In general, even if $Z(G_1)$ is not contained in $Z(G)$, why not just extend to $Z(G)$ the restriction to $Z(G) \cap Z(G_1)$ of $\omega_1$, then proceed as you do here? $\endgroup$
    – LSpice
    May 3, 2022 at 13:40
  • $\begingroup$ But I am confused; you use $\pi^\vee$ and $\pi$, which I would normally take to mean contragredient representations, but they aren't even defined on the same group. $\endgroup$
    – LSpice
    May 3, 2022 at 13:42

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