# Proof that the Pontryagin dual of a topological group is a topological group

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

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