2
$\begingroup$

I'm looking for a proof that the Pontryagin dual $G^*$ of a topological group $G$ is a topological group.

It's very easy to prove that $G^*$ is a group, my troubles are in proving that the map $G^* \times G^* \to G^* : (f,g) \mapsto fg^{-1}$ is continuous and so $G^*$ is topological.

I read in "Rudin - Fourier Analysis on Groups" a proof that $G^*$ is a Locally Compact Abelian group when $G$ is LCA, but it's too much for my purposes and the proof involves the Fourier transform and so the Haar measure, I think these tools are not necessary.

Thanks very much for any suggestions.

$\endgroup$
1
  • 1
    $\begingroup$ Take a look at Lefschetz' book "Algebraic topology", the beginning has a lot of detailed background on topological groups and Pontryagin duality. It's a little old fashioned, but I found it very useful. $\endgroup$ Feb 27, 2012 at 11:42

1 Answer 1

4
$\begingroup$

I don't think this is a research level question, but here is an argument.

The topology of $G^*$ is given by uniform convergence on compact subsets of $G$. Let $K\subset G$ be compact, then we need to show that if $f_n\to f$ and $g_n\to g$ uniformly on $K$, then $f_ng_n^{-1}\to fg^{-1}$ uniformly on $K$. This is immediate from the pointwise bound $$ |f_ng_n^{-1}-fg^{-1}| \leq |f_n(g_n^{-1}-g^{-1})|+|(f_n-f)g^{-1}| = |g_n-g| + |f_n-f|. $$

$\endgroup$
5
  • $\begingroup$ According to the definitions that I use the topology of $G^∗$ is given by the subspace topology induced from the compact-open topology of $C(G,\mathbb{C})$ (the space of continuous function $G \to \mathbb{C}$). But in general the compact-open topology of $C(G,\mathbb{C})$ is not equal to the topology of uniform convergence on compact subsets of $G$. So your argument does not work for every $G$. $\endgroup$
    – user21706
    Feb 27, 2012 at 13:33
  • $\begingroup$ michael, on a uniform space, like a topological group, the two notions are the same (see en.wikipedia.org/wiki/Compact-open_topology). $\endgroup$
    – B R
    Feb 27, 2012 at 15:00
  • 1
    $\begingroup$ However, $\mathbb C$ is metrizable, so the compact-open topology for $C(G,\mathbb C)$ is the topology of uniform convergence on compact sets. Isn't that right? $\endgroup$ Feb 27, 2012 at 15:01
  • $\begingroup$ Bah, I garbled my comment, which should have been more like Gerald's: "When the target space is a metric space, the two notions are the same". See wikipedia. $\endgroup$
    – B R
    Feb 27, 2012 at 15:08
  • $\begingroup$ You are right, then actually the question is very easy. $\endgroup$
    – user21706
    Feb 27, 2012 at 15:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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