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 oneexplicitly 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!