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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.

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

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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 en.wikipedia.org/wiki/Compact-open_topology). –  B R Feb 27 '12 at 15:00
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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
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