2
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Hi François. Thank you for your response. Actually, I wanted to use an answer to my question to prove the fact that a continuous multiplier on a locally compact Hausdorff abelian group is symmetric if and only if it is trivial, i.e., equivalent to the trivial multiplier. $\endgroup$ – Transcendental Oct 25 '16 at 22:53
  • $\begingroup$ @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? $\endgroup$ – Francois Ziegler Oct 26 '16 at 2:01
  • $\begingroup$ 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. $\endgroup$ – Transcendental Oct 29 '16 at 3:56
  • $\begingroup$ 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! $\endgroup$ – Transcendental Oct 29 '16 at 4:21
  • $\begingroup$ I’ve ordered a copy of Kleppner’s 1960 PhD thesis (through ProQuest) to see if he has done things in more detail there. In the meantime, I’ll accept your answer. Thank you so much for your time! $\endgroup$ – Transcendental Oct 29 '16 at 4:28

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.