TheoremStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedjudge pass (gpt-6.1-sol)

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Compact groups have discrete duals and discrete groups have compact duals

Statement

(1) If $G$ is a compact abelian topological group, then $\hat{G}$ is discrete. (2) If $G$ is a discrete abelian group, then, assuming the Axiom of Choice (The Axiom of Choice) used only through Tychonoff's theorem, $\hat{G}$ is compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

Facts & Assumptions

[F1]

$\hat{G}$ is a Hausdorff topological abelian group, so translations are homeomorphisms; its compact-open subbasis is $S (K, V) = {γ : γ [K] \subseteq V}$ for compact $K \subseteq G$ and open $V \subseteq T$ . (The compact-open character group is a Hausdorff topological abelian group, The Pontryagin dual with the compact-open topology, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace)

[F2]

The arc $D = {z \in T : ∣ z - 1 ∣ < 1}$ contains no nontrivial subgroup of $T$ ; the image of a homomorphism is a subgroup. (The unit-circle arc ${z : ∣ z - 1 ∣ < 1}$ contains no nontrivial subgroup)

[F3]

On a discrete domain the compact-open topology agrees with the topology of pointwise convergence, and on every subset of $C (G, T)$ the compact-open subspace topology is the topology inherited from the product $T^{G}$ . (On a discrete domain the compact-open topology is the topology of pointwise convergence, The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ )

[F4]

Every function from a discrete space is continuous, so for discrete $G$ the dual is the set $Hom (G, T)$ of all homomorphisms, and this set is closed in $T^{G}$ for the product topology. (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Continuity of a map of topological spaces at a point and globally, Pointwise limits of homomorphisms and of equicontinuous characters)

[F5]

Assume the Axiom of Choice: an arbitrary product of compact spaces is compact, and a closed subspace of a compact space is compact. (Tychonoff's theorem: an arbitrary product of compact spaces is compact in the product topology, assuming the Axiom of Choice, A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact, The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, The Axiom of Choice)

[F6]

$T$ is compact and Hausdorff and $1 \in D$ . (The multiplicative unit circle is a compact metrizable topological abelian group)

Proof

Given: An abelian topological group $G$ , compact in case (1) and discrete in case (2).

1.1F1F2F6

Under the hypothesis of (1), $S (G, D) = {γ \in \hat{G} : γ [G] \subseteq D}$ is open in $\hat{G}$ by [F1], since $G$ is compact and $D$ is open and contains $1$ by [F6]; it contains the identity character because $1 [G] = {1} \subseteq D$ . If $γ \in S (G, D)$ , then $γ [G]$ is a subgroup of $T$ contained in $D$ , hence trivial by [F2], so $γ = 1$ ; therefore $S (G, D) = {1}$ and ${1}$ is open in $\hat{G}$ .

1.2F3F4

Under the hypothesis of (2), for every $f : G \to T$ and every open target set $V$ , the preimage $f^{- 1} [V]$ is open because every subset of the discrete source $G$ is open. At any $x$ with $f (x) \in V$ this preimage is the required source neighbourhood, so every such function is continuous by [F4], so the dual is $Hom (G, T)$ with the compact-open topology, which by [F3] is the subspace topology inherited from the product $T^{G}$ ; by [F4] the set $Hom (G, T)$ is closed in $T^{G}$ .

2.1step 1.1F1

Hence $\hat{G}$ is discrete: for any $γ_{0} \in \hat{G}$ the translation $γ \mapsto γ_{0} γ$ is a homeomorphism of $\hat{G}$ by [F1] carrying $1$ to $γ_{0}$ , so ${γ_{0}}$ is the image of the open set ${1}$ and is open; every singleton is open, which is discreteness.

2.2step 1.2F5

The product $T^{G}$ is compact by Tychonoff's theorem under the Axiom of Choice by [F5], and the closed subspace $\hat{G} = Hom (G, T)$ of a compact space is compact by [F5]. This completes (2).

3.1step 2.1step 2.2∎

Clause (1) is step 2.1 and clause (2) is step 2.2, so the theorem is proved.

Depends on

Used by

The circle: the Peter-Weyl basis is the integer characters Example
The Pontryagin dual of a finite cyclic group Example
The Pontryagin dual of the circle is ℤ Example
The Pontryagin dual of ℤ is the circle Example
Duals of finite products and of discrete direct sums Lemma

Dependency tree · two levels

78 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Dikran D. Dikranjan, Introduction to Topological Groups (author lecture notes, Universita di Udine / Universidad Complutense de Madrid, 2007) (standard reference, not scraped)
Manfred Einsiedler and Thomas Ward, Ergodic Theory with a View Towards Number Theory, Appendix C (course-hosted full text) (standard reference, not scraped)
Lynn H. Loomis, Introduction to Abstract Harmonic Analysis, D. Van Nostrand 1953, Chapter VII, Sections 34-35 (printed pp. 134-140) (standard reference, not scraped)