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!



Constructing an explicit extension of a continuous character on a closed subgroup of a certain locally compact abelian group
It is a known fact (e.g. [Baggett & Kleppner 1973](http://www.ams.org/mathscinet-getitem?mr=364537), 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](http://www.ams.org/mathscinet-getitem?mr=156915) (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).


