Is assigning the central character continuous and surjective? Let $G$ be a topological group and let $H$ be a central subgroup, i.e. $H\subset Z(G)$. Consider the inclusion $\iota:H\hookrightarrow G$.
Denote by $\widehat{G}$ the unitary dual of $G$, i.e. the set of isomorphism classes of irreducible unitary representations of $G$ equipped with the Fell topology, i.e. a neighborhood of $\pi:G\to\text{U}(\mathcal{H})$ is generated by sets of the form $U_{K,\epsilon,x_1,...,x_n}$ consisting of irreducible unitary representations $\rho:G\to\text{U}(\mathcal{V})$ such that there exist $y_1,...,y_n\in\mathcal{V}$ satisfying $|\langle\pi(g)(x_i),x_j\rangle-\langle\rho(g)(y_i),y_j\rangle|<\epsilon$ for all $g\in K$, $i,j=1,...,n$, where $K\subset G$ is compact, $\epsilon>0$, and $x_1,...,x_n\in\mathcal{H}$.
We can get a map $\widehat{\iota}:\widehat{G}\to\widehat{H}$ by setting $\widehat{\iota}(\pi)(h)=\lambda_h^\pi1_\mathbb{C}$, where $\pi(h)=\lambda_h^\pi1_\mathcal{H}$ for some $\lambda_h^\pi\in\mathbb{C}$ by Schur's lemma, $\pi\in\widehat{G}$, $h\in H$.
1. Is $\widehat{\iota}$ continuous?
2. Is $\widehat{\iota}$ surjective?
My attempts:
Ad 1: The Fell topology on $\widehat{H}$ is just the compact-open topology if we identify $\widehat{H}$ with the Pontryagin dual of $H$. So we have to prove that for every $K\subset H$ compact, $U\subset S^1$ open, the set $\widehat{\iota}^{-1}(\{\chi\in\widehat{H}\ |\ \chi(K)\subset U\})$ is open in $\widehat{G}$. This set is just $V:=\{\pi\in\widehat{G}\ |\ \lambda_h^\pi\in U\ \forall h\in K\}$, so let $\pi_0\in V$. We have to find $K',\epsilon,x_1,...,x_n$ as above such that $\pi_0\in U_{K',\epsilon,x_1,...,x_n}\subset V$. My idea: Choose $K'=K$, $n=1$, $x_1\in\mathcal{H}$ such that $\|x_1\|=1$. Since $U$ is open, for all $h\in K$, there exist $\epsilon_h>0$ such that $B_{\epsilon_h}(\lambda_h^{\pi_0})\subset U$. Setting $y_1=x_1$, $\rho\in U_{K,\epsilon_h,x_1}$ just means $|\lambda_h^{\pi_0}-\lambda_h^\rho|<\epsilon_h$, i.e. $\lambda_h^\rho\in B_{\epsilon_h}(\lambda_h^{\pi_0})\subset U$. Now I think the problem is that this is dependent on $h$ and that I can't just take $\inf_{h\in K}\epsilon_h$ or something. Is my attempt correct so far? Is it possible to make this independent of $h$?
Edit: Since a continuous image of a compact set is compact and compact subsets of $\mathbb{R}$ admit an infimum, it would suffice to show that one can choose the $\epsilon_h$, $h\in K$, such that $K\ni h\mapsto\epsilon_h\in\mathbb{R}$ is continuous.
Ad 2: One can consider the case $H=Z(G)$ and the problem is to extend a unirrep of $Z(G)$ to $G$. By Schur's lemma, such a unirrep is just a character of $Z(G)$ but I'm not sure if that's helpful here. I'm aware of the induced representation functor, which constructs from a unirep of $Z(G)$ a unirep of $G$, but this representation is usually not irreducible, so I don't know how to extend in an irreducible way. Any ideas? Or counterexamples?
 A: Addressing the surjectivity part of your question, the answer is yes, and the method of proof is pretty much what you
indicated:  induced representations!
Given a character $\varphi $ on  $H$, we want to produce an irreducible representation $\pi $ of $G$, such that
$$
  \pi (h) = \varphi (h)I, \quad \forall h\in  H.  \tag 1
  $$
Viewing $\varphi $ as a one-dimensional representation of $H$, let us begin by describing the
representation of $G$ induced by $\varphi $, sometimes denoted $\text{Ind}_H^G(\varphi )$, but which I'd rather denote simply by $\pi $.
Consider the space $K(G)$ formed by the continuous, compactly
supported functions of $G$, and let  $\pi _0$ be the (algebraic) representaion of $G$ on $K(G)$ defined by
$$
  \pi _0(r) f |_s = f(r^{-1}s), \quad \forall r, s\in  G, \ \forall f\in  K(G).
  $$
This is sometimes called the left regular representation of $G$, but the inner product to be considered will soon  change this.
We then define an inner product on $K(G)$  by
$$
  \langle f,g\rangle = \int_H \int_G \Delta (t^{-1}) \overline{g(t^{-1})} f(t^{-1}s) \varphi (s) \, dt\,  ds, \quad \forall f,g\in  K(G). \tag 2
  $$
It might be interesting to notice that this may be obtained as the composition of the following processes:

*

*Starting with $f$ and $g$, we take the convolution product  $g^**f$, where $g^*(t) = \Delta (t^{-1}) \overline{g(t^{-1})}$, and
$\Delta $ is the modular function of $G$,


*Next we restrict $g^**f$ to $H$, obtaining a function  $E(g^**f)\in  K(H)$,


*Finally we compute the integrated form $\tilde \varphi $ on $E(g^**f)$, leading up to
$$
  \int_H E(g^**f)(s) \varphi (s) \,ds,
  $$
which is precisely the right-hand-side of (2).
The Hilbert space of $\pi $ is then the completion of $K(G)$ relative to this inner product and the representation $\pi $ of $G$
there is obtained by extending every $\pi _0(t)$ by continuity.
The crucial point is that $\pi $ satisfies (1) because, for every $f,g\in  K(G)$, and every $r\in  H$,  one has that
$$
  \langle  \pi _0(r)f,g\rangle  =
  \int_H \int_G \Delta (t^{-1}) \overline{g(t^{-1})} f(r^{-1}t^{-1}s) \varphi (s) \, dt\,  ds = $$$$ =
  \int_H \int_G \Delta (t^{-1}) \overline{g(t^{-1})} f(t^{-1}s') \varphi (rs') \, dt\,  ds' = $$$$ =
  \varphi (r)\int_H \int_G \Delta (t^{-1}) \overline{g(t^{-1})} f(t^{-1}s') \varphi (s') \, dt\,  ds' = $$$$ =
  \varphi (r)\langle f,g\rangle ,
  $$
where we performed the change of variables  $s=rs'$ in the second step above.
Now, there is no guarantee that $\pi $ is irreducible, but any irreducible representation of $G$ weakly contained
in $\pi $ will also satisfy (1).
