MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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. – Laurent Berger Feb 27 '12 at 11:42
up vote 3 down vote accepted

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 $$ |f_ng_n^{-1}-fg^{-1}| \leq |f_n(g_n^{-1}-g^{-1})|+|(f_n-f)g^{-1}| = |g_n-g| + |f_n-f|. $$

share|cite|improve this answer
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 $G$. – user21706 Feb 27 '12 at 13:33
michael, on a uniform space, like a topological group, the two notions are the same (see – B R Feb 27 '12 at 15:00
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? – Gerald Edgar Feb 27 '12 at 15:01
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. – B R Feb 27 '12 at 15:08
You are right, then actually the question is very easy. – user21706 Feb 27 '12 at 15:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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