# Constructing an explicit extension of a continuous character on a closed subgroup of a certain locally compact abelian group

Let $G$ be a locally compact abelian group and $\omega: G \times G \to \mathbb{T}$ a continuous multiplier on $G$, i.e., \begin{align} \forall r,s,t \in G: \qquad \omega(s,t) ~ \omega(r,s + t) & = \omega(r,s) ~ \omega(r + s,t), \\ \omega(0_{G},r) & = 1 = \omega(r,0_{G}). \end{align} Let $G^{\omega}$ be the group with $\mathbb{T} \times G$ as its underlying set and the group operations defined by \begin{align} \forall (z,r),(z',s) \in \mathbb{T} \times G: \qquad (z,r) (z',s) & \stackrel{\text{df}}{=} (z z' \omega(r,s),r + s), \\ (z,r)^{-1} & \stackrel{\text{df}}{=} \left( \overline{z ~ \omega(r,-r)},-r \right). \end{align} Note: $G^{\omega}$ is abelian if and only if $\omega$ is symmetric, i.e., $\omega(r,s) = \omega(s,r)$ for all $r,s \in G$.

Next, equip $G^{\omega}$ with the obvious product topology so that $G^{\omega}$ becomes a locally compact group.

Now, $\mathbb{T}$ is homeomorphic to the closed subgroup $H \stackrel{\text{df}}{=} \mathbb{T} \times \{ 0_{G} \}$ of $G^{\omega}$, so there is a one-to-one correspondence between continuous characters on $\mathbb{T}$ and continuous characters on $H$. Therefore, for every continuous character $\varphi$ on $H$, there exists an $n \in \mathbb{Z}$ such that $$\forall z \in \mathbb{T}: \qquad \varphi(z,0_{G}) = z^{n}.$$

My question is this:

Question. Let $\omega$ be symmetric so that $G^{\omega}$ is now a locally compact abelian group. Then given a continuous character $\varphi$ on $H$, can one explicitly construct, in terms of $\omega$, an extension of $\varphi$ to a continuous character on all of $G^{\omega}$?

Using Pontryagin Duality, it is not hard to prove the existence of such an extension, but I would like to know if one can do so by furnishing an explicit formula in terms of $\omega$.

Thank you very much!

It is a known fact (e.g. Baggett & Kleppner 1973, p. 308) that a (continuous) multiplier $\omega$ on a locally compact abelian group is symmetric iff it is trivial, i.e. $\omega(r,s) = \xi(r)\xi(s)\xi(r+s)^{-1}$ for some (continuous) $\xi:G\to\mathbf T$. Then one checks without trouble that $(z,r)\mapsto (z\xi(r))^n$ is a character extending your $\varphi$ to $G^\omega$.
Edit: Here is the Baggett-Kleppner argument, fleshed out to hopefully address your objections in the comments. Trivial $\Rightarrow$ symmetric is clear. Conversely, assume $\omega$ symmetric. Then as you noted, $G^\omega$ with the product topology is a locally compact abelian group. (Weil's result is not needed here.) By e.g. Hewitt-Ross 1963 (24.4) it admits a continuous character $\chi$ extending $(z,0)\mapsto z$. Putting $\xi(r)=\chi(1,r)$ we then obtain, as desired, $$\xi(r)\xi(s)=\chi((1,r)(1,s))=\chi((\omega(r,s),0)(1,r+s))=\omega(r,s)\xi(r+s).$$ Remark: Admittedly this is a lot like what you asked to avoid. Indeed (24.4) specializes a corollary (24.12) of Pontryagin duality (24.8), and one might as well use it directly to extend your $\varphi$ (by $\chi^n$). This special case (compact subgroup) is much easier but still not an "explicit construction" (which I doubt exists, in this generality).
• @Transcendental : Have you looked up the two proofs of this fact given in Kleppner 1965 and (the above-quoted) Baggett-Kleppner 1973 (both free access)? I find the latter particularly transparent. But perhaps it does not as "explicitly construct" ($\xi$ from $\omega$) as you'd like? – Francois Ziegler Oct 26 '16 at 2:01
• I’ve seen both proofs. Kleppner’s 1965 proof relies upon the construction of the so-called Weil topology on $G^{\omega}$, where $\omega: G \times G \to \mathbb{T}$ is merely assumed to be a Borel-measurable multiplier. I’m not entirely convinced by the proof because Paul Halmos, in his book ‘Measure Theory’, appears to say that in order for the Weil topology to exist on a given measurable group, certain measure-theoretical conditions ($\sigma$-finiteness being one of them) must first hold, which were somehow passed over in silence by Kleppner. – Transcendental Oct 29 '16 at 3:56
• The 1973 Baggett-Kleppner proof appears succinct but relies upon the existence of an irreducible $\omega$-representation of $G$, for which I haven’t seen an argument yet. One might reason that because there is a one-to-one correspondence between (i) $\omega$-representations of $G$ and (ii) unitary representations of $G^{\omega}$ that satisfy a certain normalization condition, one can work with (ii) instead. However, this correspondence is based upon the existence of a locally compact Hausdorff group topology on $G^{\omega}$, which requires a legitimate application of Weil’s result! – Transcendental Oct 29 '16 at 4:21