# Representations of groups with the same derived group, how much control do we have over the central character?

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.

• 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. May 3 at 13:37

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 $$\omega$$.

• 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? May 3 at 13:40
• 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. May 3 at 13:42