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The universal covering group $G$ of $\mathrm{SL}_2({\mathbb R})$ has infinite center. Is there an irreducible unitary representation $\pi$ of $G$, whose central character is injective? Or does every $\pi$ factor through a finite cover of $\mathrm{SL}_2({\mathbb R})$?

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Yes.

(a) This is equivalent to ask about the existence of an extremal normalized positive-definite function $\phi$ on $G$ that is "faithful" on its infinite cyclic center $Z$, that is, $\phi^{-1}(\{1\})\cap Z=\{0\}$. This is just because these functions are precisely the functions of the form $x\mapsto\phi(x)=\langle \xi,\pi(x)\xi\rangle$ for unitary representations $\pi$ and unit vectors $\xi$. (Extremal means it lies in an extremal axis in the cone of positive-definite functions.)

(b) In turn, this is equivalent to ask about the existence of a nonzero unitary representation of $G$ for which $Z$ acts as a faithful character. Indeed if $\phi$ is as in (a) then the corresponding $\pi$ works. Conversely, if we have such $\pi$ and $\phi$ is any normalized positive-definite function associated to $\pi$, then $\phi$ in the convex hull of extremal positive-definite functions. Let $\chi$ be the restriction $\phi|_Z$: this is by assumption a faithful character. Since $\phi$ is extremal in restriction to $Z$, we deduce that $\phi$ is in the convex hull of extremal positive-definite functions that equal $\chi$ on $Z$. Whence the existence of such a nonzero function.

(c) Now we need to construct $\pi$ as in (b). Fix a faithful unit character $\chi$ on $Z$ (i.e., an injective homomorphism into the unit circle in $\mathbf{C}$). Let $H_0$ be the space of locally measurable functions $G\to\mathbf{C}$ (modulo equality on null subsets), and $H_\chi$ its subspace of functions $f$ satisfying (1) $f(zx)=\chi(z)f(x)$ for every $z\in Z$ and almost every $x\in G$; (2) given (1), $|f|$ is $Z$-invariant, and we assume that the resulting function $|f|:G/Z\to\mathbf{C}$ is in $L^2$. This is a Hilbert space under the scalar product computed on a fundamental domain $X$ of the action of $Z$ on $G$, i.e., we identify $H_\chi$ to $L^2(\chi)$ to make it a Hilbert space. The group $G$ acts on $H_\chi$ by shifting: $g\cdot f(x)=f(g^{-1}x)$. This is a unitary representation, for which $Z$ acts as the scalar $\chi$, and $H_\chi\neq\{0\}$.


Note that I just assumed, say, that $G$ is a second-countable locally compact group and $Z$ is a central discrete subgroup and proved that every character of $Z$ can be obtained by restricting some irreducible unitary representation of $G$, and hence the conclusion holds as soon as $Z$ possesses a faithful character.

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  • $\begingroup$ Why is it irreducible? $\endgroup$
    – user473423
    Jan 23, 2023 at 16:07
  • $\begingroup$ @Echo the representation in (c) is not irreducible. Have you read (b) ? $\endgroup$
    – YCor
    Jan 23, 2023 at 16:07
  • $\begingroup$ OK, I got it. Thanks! $\endgroup$
    – user473423
    Jan 23, 2023 at 18:18

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