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Let $ G $ be a locally compact abelian (LCA) group and $ \widehat{G} $ the Pontryagin dual of $ G $, i.e., the set of all continuous homomorphisms $ G \to \mathbb{R} / \mathbb{Z} $. Clearly, $ \widehat{G} $ is an abelian group. The topology on $ \widehat{G} $ is generated by the sub-basic sets $$ U_{K,\xi_{0},\epsilon} \stackrel{\text{def}}{=} \left\{ \xi \in \widehat{G} ~ \Big| ~ |\xi(x) - {\xi_{0}}(x)| < \epsilon ~ \text{for all} ~ x \in K \right\} $$ for compact subsets $ K \subseteq G $, $ \xi_{0} \in \widehat{G} $ and $ \epsilon > 0 $. Show that $ \widehat{G} $ equipped with this topology is an LCA group.

Anyone is welcome to write a proof or to give a link to a proof. Thanks!

Berrick Caleb Fillmore
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user112564
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1 Answers1

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Added in edit: Keith Conrad points out that the proof can be found in Hewitt and Ross's Abstract Harmonic Analysis, volume I, Theorem 23.15 (p. 361). The proof is essentially as I describe below, except Hewitt and Ross do not bother with the C$^*$-completion of $L^1(G)$, they just define a topology directly on $\widehat{G}$ agreeing with the weak-* topology on the spectrum of $L^1(G)$ as a Banach algebra (Theorem 23.13).

My original post follows:

I don't know if this is the standard proof or not. I wasn't able to find a proof in the standard reference, Hewitt and Ross, and someone worried me a bit elsewhere by claiming that $\widehat{G}$ wasn't necessarily locally compact if $G$ was. Nonetheless, when I had looked into this previously I came up with this argument.

If $G$ is locally compact abelian, the convolution algebra $L^1(G)$ is commutative and can be completed to a commutative (non-unital unless G is discrete) C$^*$-algebra $C^*(G)$. The spectrum of $C^*(G)$ is locally compact (Gelfand duality for nonunital commutative C$^*$-algebras), and it is isomorphic to the spectrum of $L^1(G)$. There is an isomorphism (as sets) between the spectrum of $L^1(G)$ and the group $\widehat{G}$. However, $\mathrm{Spec}(C^*(G))$ has the weak-* topology, which is not prima facie the same as the usual Pontryagin topology on $\widehat{G}$.

If we examine Dixmier's C$^*$-algebras, Theorem 13.5.2, we have that the weak-* topology for states coincides with uniform convergence on compact subsets (of $G$) for the corresponding positive-definite functions, since the set of states is bounded. (Multiplicative) Characters of a group live a double life as representations and positive definite functions (just as characters of a C$^*$-algebra live a double life as representations and pure states), so uniform convergence on compact subsets of $G$ is the Pontryagin topology, as was required.

Robert Furber
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    There is a proof in "the standard reference, Hewitt and Ross" Volume I. See Theorem 23.15 (p. 361). It uses the maximal ideal space of the Banach algebra $L^1(G)$ and involves showing two differently define topologies on $\widehat{G}$ agree, so it is essentially the argument you describe. – KCd Mar 22 '19 at 16:33
  • @KCd Thank you, Keith. I must have just failed to read to the end of those 5 lines the first time I skimmed through. – Robert Furber Mar 23 '19 at 20:46
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    A proof of local compactness that was inspired by Corollary 23.16 in Hewitt & Ross vol. 1 is at https://kconrad.math.uconn.edu/blurbs/gradnumthy/loccptascoli.pdf. It uses Ascoli's theorem to get local compactness and does not directly use Banach algebra methods. – KCd Mar 26 '19 at 11:52