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Character Groups and Elementary LCA Duals

1 · Prerequisites

2 · Summary

This page builds the character group of an abelian topological group with the compact-open topology and proves the elementary properties of the Pontryagin dual that do not require inversion theory: the dual is a Hausdorff topological abelian group, evaluation is jointly continuous for locally compact Hausdorff domains, pointwise limits preserve homomorphisms and preserve continuous characters along equicontinuous families, duals of finite products and of discrete direct sums are computed, and the two one-way implications between compactness and discreteness are proved. Characters take values in the multiplicative unit circle $T = {z \in C : ∣ z ∣ = 1}$ , which is identified with the published circle $R / Z$ by an explicit topological group isomorphism; the dual is written multiplicatively and groups are written additively.

The local prerequisites are proved on this page rather than cited from outside it: the unit circle is a compact metrizable topological abelian group; on a discrete domain the compact-open topology is the topology of pointwise convergence; the arc ${∣ z - 1 ∣ < 1}$ contains no nontrivial subgroup; continuous characters of the real line are exactly the exponentials $t \mapsto \exp (2 π i ξ t)$ ; and pointwise limits along equicontinuous families of characters are characters. The compact-open neighbourhood $N = {γ : γ (K) \subseteq {∣ z - 1 ∣ \leq 1 / 2}}$ of the identity is equicontinuous and compact, which yields local compactness of the dual of a locally compact abelian group by Ascoli's sufficiency theorem; the Axiom of Choice is used exactly there, in Tychonoff's theorem, and in the compact-lift theorem behind the annihilator computation, and is declared on the items that use it.

The dual homomorphism lemma states continuity of pullback along an arbitrary continuous homomorphism and, for a closed subgroup of a locally compact abelian group, identifies the dual of the quotient with the annihilator. The two compact/discrete implications are deliberately one-way, and the direct-sum statement is made only for the discrete topology on an algebraic direct sum: it explicitly disclaims the subspace topology of a product of non-discrete factors. The companion page collects the four elementary dual computations (finite cyclic groups, the circle, the integers, and Euclidean space).

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The multiplicative unit circle is a compact metrizable topological abelian group

Statement

Let $T : = {z \in C : ∣ z ∣ = 1}$ carry the subspace topology of $C$ (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, The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane) and the multiplication of $C$ , and let $ε : R / Z \to T$ , $ε ([t]) : = \exp (2 π i t)$ , for the published one-dimensional torus $R / Z$ (The one-dimensional torus and its normalized Haar integral). Then $T$ is a compact metrizable topological abelian group (Topological group: multiplication and inversion are continuous), $ε$ is an isomorphism of topological groups, and $∣ z w - z ∣ = ∣ w - 1 ∣, ∣ z^{- 1} - w^{- 1} ∣ = ∣ z - w ∣$ for all $z, w \in T$ .

Facts & Assumptions

[F1]

For all complex $z, w$ , $\exp (z + w) = \exp z \exp w$ , and for real $x, y$ , $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ with $∣ \exp (x + i y) ∣ = e^{x}$ . The real exponential satisfies $e^{0} = 1$ (its defining series has constant term $1$ and all other terms $0$ ). ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , The real exponential function and the number $e$ by a power series)

[F2]

$\sin$ and $\cos$ are differentiable on $R$ , hence continuous, and $\sin 0 = 0$ , $\cos 0 = 1$ . (The derivatives of sine and cosine are cosine and minus sine, A function differentiable at $c$ is continuous at $c$ )

[F3]

$\sin$ and $\cos$ have period $2 π$ : $\sin (x + 2 π) = \sin x$ and $\cos (x + 2 π) = \cos x$ for every real $x$ . (The zero sets of sine and cosine and the least positive common period 2 pi)

[F4]

$t \mapsto (\cos t, \sin t)$ is a bijection from $[0, 2 π)$ onto the Euclidean unit circle $S^{1} = {(a, b) : a^{2} + b^{2} = 1}$ . ( $t \mapsto (\cos t, \sin t)$ is a bijection from $[0, 2 π)$ onto the real unit circle)

[F5]

$Φ : C \to R^{2}$ , $Φ (a + b i) = (a, b)$ , is a bijection compatible with addition and multiplication; $d_{C} (z, w) = ∣ z - w ∣ = ∥ Φ (z) - Φ (w) ∥_{2}$ . For all $z, w \in C$ , $z \overline{z} = ∣ z ∣^{2}$ , $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ . Continuity of maps between subsets of $C$ is continuity for the metric $d_{C}$ . ( $C$ is the real coordinate plane, with coordinate arithmetic, The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane, Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive)

[F6]

The canonical projection $q : R \to R / Z$ is continuous and open, $[s] = [t]$ exactly when $s - t \in Z$ , every class has exactly one representative in $[0, 1)$ , and $R / Z$ is the quotient group of the additive group $R$ by its subgroup $Z$ , with $[s] + [t] = [s + t]$ . Moreover $R / Z$ is compact. (The one-dimensional torus and its normalized Haar integral, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , $R / Z$ is compact and path-connected)

[F7]

Quotient universal property: a continuous map $g : R \to W$ constant on the fibres of $q$ factors uniquely as $g = \overset{ˉ}{g} \circ q$ with $\overset{ˉ}{g}$ continuous. (For a quotient map $q : X \to Y$ , a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$ ; and a composite of quotient maps is a quotient map)

[F8]

The product topology on $R^{2}$ is the Euclidean metric topology. A map into a product is continuous exactly when all its components are; a composite of continuous maps is continuous; the identity and scalar multiples $α i d$ of real functions are continuous. (A subset of $R^{n}$ with the product topology is compact exactly when it is closed and bounded, the product topology being the Euclidean metric topology, A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function)

[F9]

Continuous images of compact spaces are compact; a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism. A metric space is Hausdorff, and the metric topology of a metric subspace is its subspace topology. (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, Distinct points of a metric space have disjoint balls around them, Isometry, isometric embedding, and the subspace metric on a subset)

[F10]

A topological group is a group whose multiplication and inversion are continuous for the product topology. (Topological group: multiplication and inversion are continuous)

Proof

Given: The multiplicative unit circle $T \subseteq C$ with the subspace topology, and $ε ([t]) = \exp (2 π i t)$ on the published torus $R / Z$ .

1.1F1F2F3

For every integer $k$ , $\exp (2 π i k) = 1$ : by [F1] and [F3] with [F2], $\exp (2 π i k) = e^{0} (\cos (2 π k) + i \sin (2 π k)) = \cos 0 + i \sin 0 = 1$ , because $2 π k$ is an integer multiple of the period $2 π$ of sine and cosine.

1.2F1

The image of $ε$ lies in $T$ : for real $t$ , $∣ \exp (2 π i t) ∣ = e^{0} = 1$ by [F1].

1.3F1F4F5

$ε$ is surjective onto $T$ : if $z = a + b i \in T$ then $a^{2} + b^{2} = ∣ z ∣^{2} = 1$ by [F5], so $(a, b) \in S^{1}$ and [F4] gives $θ \in [0, 2 π)$ with $(a, b) = (\cos θ, \sin θ)$ ; putting $t : = θ / (2 π) \in [0, 1)$ and using [F1] and [F5] gives $ε ([t]) = \exp (2 π i t) = \cos θ + i \sin θ = a + b i = z$ .

1.4F5

$T$ is closed under multiplication and inversion, and the two displayed identities hold: for $z, w \in T$ , $∣ z w ∣ = ∣ z ∣ ∣ w ∣ = 1$ and $∣ z^{- 1} ∣ = ∣ z ∣^{- 1} = 1$ by [F5], so $z w, z^{- 1} \in T$ ; also $∣ z w - z ∣ = ∣ z ∣ ∣ w - 1 ∣ = ∣ w - 1 ∣$ and $∣ z^{- 1} - w^{- 1} ∣ = ∣ w - z ∣ / (∣ z ∣ ∣ w ∣) = ∣ z - w ∣$ by [F5].

2.1step 1.1F1F6

$ε$ is well defined on classes and is a group homomorphism: if $[s] = [t]$ then $s - t = k \in Z$ by [F6], so $\exp (2 π i s) = \exp (2 π i t) \exp (2 π i k) = \exp (2 π i t)$ by [F1] and step 1.1; and $ε ([s] + [t]) = \exp (2 π i (s + t)) = \exp (2 π i s) \exp (2 π i t) = ε ([s]) ε ([t])$ by [F1] and [F6].

3.1step 2.1F1F4F5F6

$ε$ is injective: if $ε ([s]) = ε ([t])$ , replace the classes by their unique representatives $s, t \in [0, 1)$ by [F6]; then $\cos (2 π s) = \cos (2 π t)$ and $\sin (2 π s) = \sin (2 π t)$ by [F1] and [F5], so the bijectivity in [F4] applied to $2 π s, 2 π t \in [0, 2 π)$ gives $2 π s = 2 π t$ , hence $s = t$ , hence $[s] = [t]$ .

3.2step 2.1step 1.2F2F5F7F8

$ε$ is continuous as a map $R / Z \to C$ : the map $g (t) : = \exp (2 π i t)$ is continuous on $R$ because $t \mapsto 2 π t$ , $\sin$ and $\cos$ are continuous by [F2] and [F8], hence $t \mapsto (\cos 2 π t, \sin 2 π t)$ is continuous into $R^{2}$ by [F8], and $g = Φ^{- 1} (\cos 2 π \cdot, \sin 2 π \cdot)$ is continuous by [F5], [F8] and the distance identity $d_{C} (z, w) = ∥ Φ (z) - Φ (w) ∥_{2}$ read as $ε$ - $δ$ continuity of $Φ^{- 1}$ ; by step 2.1 $g$ is constant on the fibres of $q$ , so the quotient universal property [F7] makes $ε$ continuous into $C$ , and its corestriction to the subspace $T$ is continuous by the subspace topology.

4.1step 1.3step 3.2F6F9

$T$ is compact: it is the image $ε (R / Z)$ by step 1.3 of the compact space $R / Z$ under the continuous map of step 3.2, and continuous images of compact spaces are compact by [F9].

4.2step 1.2step 3.1step 1.3step 3.2F5F6F9

$ε$ is a homeomorphism onto $T$ : it is a continuous bijection by steps 2.1, 3.1, 1.3 and 3.2 whose domain is compact by [F6] and whose image lies in $T$ by step 1.2, and $T$ is Hausdorff as a subspace of the metric space $C$ by [F5] and [F9]; the compact-to-Hausdorff clause of [F9] applies to the corestriction.

5.1step 4.2step 1.4F5F9F10∎

Multiplication and inversion on $T$ are continuous, so $T$ is a topological abelian group: for $z, z_{0}, w, w_{0} \in T$ , $∣ z w - z_{0} w_{0} ∣ \leq ∣ z - z_{0} ∣ ∣ w ∣ + ∣ z_{0} ∣ ∣ w - w_{0} ∣ = ∣ z - z_{0} ∣ + ∣ w - w_{0} ∣$ by [F5], so the open rectangle $(B (z_{0}, δ) \cap T) \times (B (w_{0}, δ) \cap T)$ with $δ = ε / 2$ is mapped into $B (z_{0} w_{0}, ε) \cap T$ , which is continuity of multiplication at $(z_{0}, w_{0})$ ; and $∣ z^{- 1} - w^{- 1} ∣ = ∣ z - w ∣$ by step 1.4 makes inversion distance preserving, hence continuous. Group axioms and commutativity are inherited from $C$ by steps 2.1, 3.1 and 1.3, and metrizability of $T$ is [F9] applied to the metric subspace $T \subseteq C$ .

LemmaStatement: Literature-sourcedProof: Literature-sourcedOpen item page →

On a discrete domain the compact-open topology is the topology of pointwise convergence

Statement

Let $X$ be a discrete topological space (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) and $Y$ a topological space. Every compact subset of $X$ is finite (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), and on $C (X, Y)$ the compact-open topology (The compact-open topology on $C (X, Y)$ for arbitrary topological spaces) coincides with the topology of pointwise convergence (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ ); hence on every $F \subseteq C (X, Y)$ the compact-open subspace topology is the subspace topology inherited from the product $Y^{X}$ . In particular finite intersections of sets ${f : f (x) \in V}$ , with $x \in X$ and $V \subseteq Y$ open, form a basis; these are open-coordinate constraints, not requirements that a coordinate equal a prescribed value.

Facts & Assumptions

[F1]

In the discrete topology on $X$ every subset is open, and a subset $K \subseteq X$ is compact exactly when every open cover of $K$ by open sets of $X$ (equivalently of the subspace $K$ ) has a finite subcover; the empty space and every finite space are compact. (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it)

[F2]

The compact-open topology on $C (X, Y)$ is generated by the subbasis of sets $S (K, V) = {f \in C (X, Y) : f [K] \subseteq V}$ with $K \subseteq X$ compact and $V \subseteq Y$ open. (The compact-open topology on $C (X, Y)$ for arbitrary topological spaces)

[F3]

The topology of pointwise convergence on $C (X, Y)$ is the subspace topology inherited from the product $Y^{X}$ ; its subbasis consists of the traces of the sets $π_{x}^{- 1} [V] = {f \in Y^{X} : f (x) \in V}$ , $x \in X$ , $V \subseteq Y$ open, and its basic open sets impose open-set constraints at finitely many points of $X$ . (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ , 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)

[F4]

In the product topology on $Y^{X}$ the basic open sets are the boxes that restrict only finitely many coordinates. (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)

Proof

Given: A discrete space $X$ , a topological space $Y$ , and the two topologies on $C (X, Y)$ .

1.1F1

Every compact subset $K \subseteq X$ is finite: the family of singleton subsets ${x}$ , $x \in K$ , is an open cover of the subspace $K$ by [F1]. Compactness of $K$ gives a finite subcover, exhibiting $K$ as the union of finitely many singletons; for $K = \emptyset$ , the empty family suffices.

1.2F2F3

For every finite $K \subseteq X$ and every open $V \subseteq Y$ the subbasic compact-open set is the finite intersection $S (K, V) = ⋂_{x \in K} {f \in C (X, Y) : f (x) \in V}$ .

2.1step 1.1step 1.2F1F2F3

Each compact-open subbasic set is a finite intersection of pointwise subbasic sets by steps 1.1 and 1.2, and each pointwise subbasic set ${f \in C (X, Y) : f (x) \in V}$ equals $S ({x}, V)$ , which is compact-open subbasic because the singleton ${x}$ is compact by [F1]. A topology containing a family contains the topology generated by it, so the two topologies on $C (X, Y)$ each contain the other, hence are equal; by [F3] this common topology is the subspace topology inherited from $Y^{X}$ .

3.1step 2.1F3F4∎

Restricting an equality of topologies to a subset preserves it: for $F \subseteq C (X, Y)$ the traces on $F$ of the two topologies coincide. The pointwise topology is the subspace topology from $Y^{X}$ by [F3], and its basic open sets impose open-set constraints on finitely many coordinates by [F3, F4]; hence on $C (X, Y)$ , and on every $F \subseteq C (X, Y)$ , those sets form a basis.

LemmaStatement: Literature-sourcedProof: Literature-sourcedjudge pass (gpt-6.1-sol)Open item page →

The unit-circle arc ${z : ∣ z - 1 ∣ < 1}$ contains no nontrivial subgroup

Statement

Let $T$ be the multiplicative unit circle (The multiplicative unit circle is a compact metrizable topological abelian group) and $D : = {z \in T : ∣ z - 1 ∣ < 1}$ . Every subgroup $H \leq T$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups) with $H \subseteq D$ is trivial; equivalently, for every $z \in T$ with $z \neq 1$ there is a positive integer $n$ with $z^{n} \notin D$ .

Facts & Assumptions

[F1]

$ε : R / Z \to T$ , $ε ([t]) = \exp (2 π i t)$ , is an isomorphism of topological groups; in particular it is injective and surjective, and $∣ z^{- 1} - w^{- 1} ∣ = ∣ z - w ∣$ for all $z, w \in T$ . (The multiplicative unit circle is a compact metrizable topological abelian group)

[F2]

$\exp (x + i y) = e^{x} (\cos y + i \sin y)$ for real $x, y$ , and $\exp (0) = 1$ from the defining series of the complex exponential. Also $\cos (2 u) = 1 - 2 \sin^{2} u$ for every real $u$ . ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , The complex exponential by its power series, Double-angle and quadratic power-reduction identities)

[F3]

$\sin$ and $\cos$ are differentiable on $R$ , with $\sin 0 = 0$ and $\cos 0 = 1$ . (The derivatives of sine and cosine are cosine and minus sine)

[F4]

Sine is strictly increasing on $[- π / 2, π / 2]$ . (Signs, monotonicity intervals, and ranges of sine and cosine)

[F5]

For every real $x$ , $\sin (- x) = - \sin x$ , $\cos (- x) = \cos x$ , and $\cos (x + π / 2) = - \sin x$ . (Pythagorean and parity identities for all six trigonometric functions on their natural domains, Quarter-turn values and shifts by pi/2 and pi)

[F6]

For every real $x$ there is a unique integer $⌊ x ⌋$ with $⌊ x ⌋ \leq x < ⌊ x ⌋ + 1$ . (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ )

[F7]

Every class in $R / Z$ has exactly one representative in $[0, 1)$ , and $R / Z$ is the quotient group of the additive group $R$ by its subgroup $Z$ , so that $[s] + [t] = [s + t]$ and in particular $[1 - t_{0}] = [- t_{0}]$ . (The one-dimensional torus and its normalized Haar integral, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ )

Proof

Given: The multiplicative unit circle $T$ , the arc $D = {z \in T : ∣ z - 1 ∣ < 1}$ , and a subgroup $H \leq T$ .

1.1F2F3F4F5

For real $u$ , $\exp (2 π i u) - 1 = (\cos 2 π u - 1) + i \sin 2 π u$ by [F2], so $∣ \exp (2 π i u) - 1 ∣^{2} = (\cos 2 π u - 1)^{2} + \sin^{2} 2 π u = 2 - 2 \cos 2 π u = 4 \sin^{2} (π u)$ ; hence $∣ \exp (2 π i u) - 1 ∣ = 2 ∣ \sin (π u) ∣$ . In particular $\sin (π / 6) = 1 / 2$ : writing $s : = \sin (π / 6)$ , we have $s > 0$ because $0 < π / 6 < π / 2$ and sine is strictly increasing on $[- π / 2, π / 2]$ with $\sin 0 = 0$ by [F4] and [F3]; the double-angle identity of [F2] gives $\cos (π / 3) = 1 - 2 s^{2}$ , while $\cos (π / 3) = \cos (π / 2 - π / 6) = - \sin (- π / 6) = \sin (π / 6) = s$ by [F5]; thus $2 s^{2} + s - 1 = 0$ , that is $(2 s - 1) (s + 1) = 0$ , and $s > 0$ forces $s = 1 / 2$ .

1.2F1F2F7

Let $z \in T$ , $z \neq 1$ , with $∣ z - 1 ∣ < 1$ . By [F1] and [F7] write $z = ε ([t_{0}]) = \exp (2 π i t_{0})$ with $t_{0} \in [0, 1)$ , and put $u : = z$ and $t : = t_{0}$ if $t_{0} \leq 1 / 2$ , while if $t_{0} > 1 / 2$ put $u : = z^{- 1}$ and $t : = 1 - t_{0} \in (0, 1 / 2)$ ; in the second case $u = ε ([- t_{0}]) = ε ([1 - t_{0}]) = \exp (2 π i t)$ because $[1 - t_{0}] = [- t_{0}]$ in $R / Z$ by [F1] and [F7]. Then $0 < t \leq 1 / 2$ , $u \neq 1$ (as $t \neq 0$ , $ε$ being injective with $ε ([0]) = \exp (0) = 1$ by [F1] and [F2]), and $∣ u - 1 ∣ = ∣ z - 1 ∣ < 1$ by [F1]; moreover $∣ u^{n} - 1 ∣ = ∣ z^{n} - 1 ∣$ for every $n \geq 1$ , again by [F1].

2.1step 1.1step 1.2F4

In the situation of step 1.2 we have $∣ u - 1 ∣ = 2 ∣ \sin (π t) ∣$ by step 1.1, so $∣ \sin (π t) ∣ < 1 / 2 = \sin (π / 6)$ by step 1.1 and the hypothesis; since $0 < π t \leq π / 2$ and sine is strictly increasing on $[0, π / 2]$ by [F4], this gives $π t < π / 6$ , that is $0 < t < 1 / 6$ .

3.1step 1.1step 1.2step 2.1F4F6

Put $n : = ⌊ 1 / (6 t) ⌋ + 1 \geq 1$ for the $t$ of step 2.1 by [F6]. Then $n > 1 / (6 t)$ , so $n t > 1 / 6$ , and $n \leq 1 / (6 t) + 1$ , so $n t \leq 1 / 6 + t < 1 / 3 < 1 / 2$ ; hence $π / 6 < π n t < π / 2$ and strict monotonicity of sine on $[0, π / 2]$ by [F4] gives $\sin (π n t) > \sin (π / 6) = 1 / 2$ . Therefore $∣ u^{n} - 1 ∣ = 2 \sin (π n t) > 1$ by step 1.1, so $u^{n} \notin D$ ; by step 1.2 also $z^{n} \notin D$ when $u = z^{- 1}$ , and plainly $z^{n} \notin D$ when $u = z$ .

4.1step 3.1F1∎

Every $z \in T$ with $z \neq 1$ therefore has a positive power outside $D$ : if $∣ z - 1 ∣ < 1$ this is step 3.1, and if $∣ z - 1 ∣ \geq 1$ then $z \notin D$ already. Conversely let $H \leq T$ with $H \subseteq D$ and suppose $h \in H$ , $h \neq 1$ ; then $h^{n} \notin D$ for some $n \geq 1$ , while $h^{n} \in H \subseteq D$ , a contradiction, so $H = {1}$ and every subgroup contained in $D$ is trivial.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Continuous characters of the real line are exponentials

Statement

Every continuous group homomorphism $φ : R \to T$ from the additive line to the multiplicative unit circle (The multiplicative unit circle is a compact metrizable topological abelian group) is $φ (t) = \exp (2 π i ξ t)$ for a unique $ξ \in R$ ; conversely every such map $φ_{ξ} (t) = \exp (2 π i ξ t)$ is a continuous character of the additive line.

Facts & Assumptions

[F1]

$ε : R / Z \to T$ , $ε ([t]) = \exp (2 π i t)$ , is an isomorphism of topological groups; in particular $ε$ and its inverse are continuous, $ε ([0]) = 1$ , and $ε ([t])$ has modulus $1$ for every real $t$ . (The multiplicative unit circle is a compact metrizable topological abelian group)

[F2]

$p : R \to R / Z$ , $p (t) = [t]$ , is a covering map, it is the quotient homomorphism of the additive group $R$ modulo $Z$ , so $p (u + v) = p (u) + p (v)$ , and $p (u) = [0]$ exactly when $u \in Z$ . ( $p : R \to R / Z$ is a covering map with translated interval sheets, The one-dimensional torus and its normalized Haar integral)

[F3]

$R$ is path-connected and locally path-connected, and it is simply connected: its fundamental group at $0$ is trivial. ( $R^{n}$ is polygonally connected, connected, locally path-connected and locally connected, Every nonempty convex subset of $R^{n}$ is simply connected, Simply connected topological spaces, Paths, path-connected spaces and path components)

[F4]

Lifting criterion: for a path-connected and locally path-connected $Y$ , a based map $f : (Y, y_{0}) \to (B, b_{0})$ and a covering $p : (E, e_{0}) \to (B, b_{0})$ , a based lift of $f$ exists if and only if $f_{*} π_{1} (Y, y_{0}) \subseteq p_{*} π_{1} (E, e_{0})$ , and it is unique. (Lifting criterion for maps from path-connected locally path-connected spaces, Lifts of maps, paths, and homotopies through a covering map)

[F5]

The continuous image of a connected space is connected, and a connected subset of $R$ is order-convex. (A continuous image of a connected space is connected, and connectedness is a topological property, The connected subspaces of $R$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $R$ ", Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets)

[F6]

An additive function $f : R \to R$ that is continuous at a single point satisfies $f (x) = f (1) x$ for every real $x$ . (Six regularity conditions each force an additive $f : R \to R$ to be $x \mapsto f (1) x$ : continuity at a single point, monotonicity on a nondegenerate interval, boundedness above on one, boundedness below on one, constancy of sign on one, and a graph that is not dense in $R^{2}$ )

[F7]

$\exp (z + w) = \exp z \exp w$ for complex $z, w$ ; $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , so $∣ \exp (i y) ∣ = 1$ and $\exp (0) = 1$ ; and $e^{i π} + 1 = 0$ . ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , The complex exponential by its power series)

[F8]

Composites of continuous maps are continuous, and the identity together with its real scalar multiples $t \mapsto ξ t$ are continuous. (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function)

Proof

Given: A continuous group homomorphism $φ : R \to T$ of the additive line.

1.1F1F3F8

The map $ψ : = ε^{- 1} \circ φ : R \to R / Z$ is a continuous group homomorphism with $ψ (0) = [0]$ : $ε^{- 1}$ is continuous by [F1], $R$ is path-connected by [F3], and $ψ (0) = ε^{- 1} (φ (0)) = ε^{- 1} (1) = [0]$ because a group homomorphism sends the identity to the identity (Monoid homomorphism and group homomorphism) and $ε ([0]) = 1$ by [F1].

2.1step 1.1F2F3F4

There is a continuous lift $θ : R \to R$ with $p \circ θ = ψ$ and $θ (0) = 0$ : the domain $R$ is path-connected and locally path-connected and its fundamental group at $0$ is trivial by [F3], so $ψ_{*} π_{1} (R, 0) \subseteq p_{*} π_{1} (R, 0)$ holds vacuously, and the lifting criterion [F4] applied to $ψ$ and the covering $p$ of [F2] supplies the based lift.

3.1step 1.1step 2.1F2F8

For fixed real $t$ the map $h (s) : = θ (s + t) - θ (s) - θ (t)$ is continuous and takes values in $Z$ : continuity is by [F8], and $p (θ (s + t)) = ψ (s + t) = ψ (s) + ψ (t) = p (θ (s)) + p (θ (t)) = p (θ (s) + θ (t))$ by [F2] and step 1.1, so $θ (s + t) - θ (s) - θ (t) \in \ker p = Z$ by [F2].

4.1step 2.1step 3.1F5

The image $h [R]$ is connected by [F5], being the continuous image of the connected space $R$ ; being a connected subset of $R$ it is order-convex by [F5], and an order-convex subset of $Z$ with two distinct elements $a < b$ would contain $a + 1 / 2 \notin Z$ , so $h [R]$ is a singleton. Since $h (0) = θ (t) - θ (0) - θ (t) = 0$ by step 2.1, that singleton is ${0}$ , so $θ (s + t) = θ (s) + θ (t)$ for all real $s, t$ : the lift $θ$ is additive.

5.1step 2.1step 4.1F6

By steps 2.1 and 4.1 the map $θ$ is additive and continuous, hence $θ (t) = ξ t$ for every real $t$ , where $ξ : = θ (1)$ , by [F6].

6.1step 2.1step 5.1F1F2

Consequently $φ (t) = ε (ψ (t)) = ε (p (θ (t))) = ε ([θ (t)]) = \exp (2 π i θ (t)) = \exp (2 π i ξ t)$ for every real $t$ , by [F1], [F2], step 2.1 and step 5.1.

7.1step 6.1F7

The parameter $ξ$ is unique: if $\exp (2 π i ξ t) = \exp (2 π i ξ^{'} t)$ for all real $t$ , then $\exp (2 π i (ξ - ξ^{'}) t) = 1$ for all $t$ by [F7]; were $ξ \neq ξ^{'}$ , the choice $t = 1 / (2 ∣ ξ - ξ^{'} ∣) > 0$ would give $\exp (\pm π i) = - 1$ by [F7], contradicting $\exp (\pm π i) = 1$ ; hence $ξ = ξ^{'}$ .

8.1step 6.1F1F2F7F8∎

Conversely, for every real $ξ$ the map $φ_{ξ} (t) : = \exp (2 π i ξ t)$ is a continuous group homomorphism $R \to T$ : it is a homomorphism by the addition formula [F7], it takes values in $T$ because $∣ \exp (2 π i ξ t) ∣ = 1$ by [F7], and it is the composite $ε \circ p \circ (t \mapsto ξ t)$ of the continuous maps $t \mapsto ξ t$ [F8], $p$ [F2] and $ε$ [F1], hence continuous by [F8].

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The Pontryagin dual with the compact-open topology

Definition

Let $G$ be an abelian topological group (Topological group: multiplication and inversion are continuous) written additively, and let $T = {z \in C : ∣ z ∣ = 1}$ be the multiplicative unit circle (The multiplicative unit circle is a compact metrizable topological abelian group).

A character of $G$ is a continuous group homomorphism $γ : G \to T$ (Monoid homomorphism and group homomorphism, Continuity of a map of topological spaces at a point and globally). The Pontryagin dual of $G$ is $\hat{G} : = {Hom}_{c t s} (G, T) : = {γ : G \to T : γ is a continuous group homomorphism},$ the set of characters, equipped with:

Pointwise multiplication. For $γ_{1}, γ_{2} \in \hat{G}$ the product is $(γ_{1} γ_{2}) (x) : = γ_{1} (x) γ_{2} (x)$ for every $x \in G$ , with the constant character $x \mapsto 1$ as its identity and $x \mapsto γ (x)^{- 1}$ as the inverse of $γ$ . With these operations the set of characters is a group, and it is abelian; this and the continuity of the two operations are proved in the next item, so no separate well-definedness obligation is left open here.
Compact-open topology. The topology is the compact-open topology inherited from $C (G, T)$ (The compact-open topology on $C (X, Y)$ for arbitrary topological spaces), that is the subspace 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) for the subbasis $S (K, V) : = {γ \in \hat{G} : γ [K] \subseteq V}, K \subseteq G compact, V \subseteq T open .$

The evaluation pairing is written $⟨ x, γ ⟩ : = γ (x) \in T, x \in G, γ \in \hat{G} .$

The dual is written multiplicatively, so products of characters are written $γ_{1} γ_{2}$ and the identity is written $1$ . Through the topological group isomorphism $ε : R / Z \to T$ of The multiplicative unit circle is a compact metrizable topological abelian group, characters may equivalently be viewed as continuous homomorphisms into the published circle $R / Z$ ; all statements below use the multiplicative circle $T$ and the compact-open subbasis displayed above.

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The compact-open character group is a Hausdorff topological abelian group

Statement

Let $G$ be an abelian topological group and let $\hat{G}$ be its Pontryagin dual (The Pontryagin dual with the compact-open topology). With pointwise multiplication and inversion, $\hat{G}$ is a Hausdorff topological abelian group (Topological group: multiplication and inversion are continuous, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not). Explicitly, for $f, g \in \hat{G}$ , compact $K \subseteq G$ and $ε > 0$ , writing $U_{K} (f, r) : = {h \in \hat{G} : ∣ h (x) - f (x) ∣ < r for every x \in K}$ for $r > 0$ , these sets form a neighbourhood basis at $f$ , and $U_{K} (f, ε / 2) \cdot U_{K} (g, ε / 2) \subseteq U_{K} (f g, ε), U_{K} (f, ε)^{- 1} = U_{K} (f^{- 1}, ε) .$

Facts & Assumptions

[F1]

$T$ is a compact metrizable topological abelian group; in particular multiplication and inversion are continuous and every element has modulus $1$ . For all $z, w \in T$ one has $∣ z^{- 1} - w^{- 1} ∣ = ∣ z - w ∣$ and $∣ z w - z ∣ = ∣ w - 1 ∣$ . (The multiplicative unit circle is a compact metrizable topological abelian group)

[F2]

$\hat{G} = {Hom}_{c t s} (G, T)$ is the set of continuous homomorphisms $γ : G \to T$ , with pointwise multiplication and the compact-open topology with subbasis $S (K, V) = {γ : γ [K] \subseteq V}$ for compact $K \subseteq G$ and open $V \subseteq T$ . (The Pontryagin dual with the compact-open topology)

[F3]

For all complex $z, w$ : $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ , $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ , and $z \overline{z} = ∣ z ∣^{2}$ ; also $∣ \exp (i y) ∣ = 1$ for real $y$ . (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ )

[F4]

A map into a product space is continuous exactly when all its components are, and composites of continuous maps are continuous. (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous)

[F5]

Subbasic open sets on a subspace are the traces of subbasic open sets of the ambient space, and singletons are compact. (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, The compact-open topology on $C (X, Y)$ for arbitrary topological spaces, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right)

[F6]

$T$ is Hausdorff: distinct points of a metric space are separated by disjoint open balls. (Distinct points of a metric space have disjoint balls around them, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not)

[F7]

Continuous images of compact sets are compact; continuous real-valued functions on nonempty compact spaces attain their maxima; closed subsets of compact spaces and finite unions of compact subsets are compact. Compact subsets admit finite subcovers from ambient open covers. (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact, A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it)

Proof

Given: An abelian topological group $G$ and its dual $\hat{G}$ with the compact-open topology.

1.1F1F2F4

Pointwise multiplication and inversion are well defined on $\hat{G}$ and give it the structure of an abelian group: for $f, g \in \hat{G}$ the maps $f g$ and $f^{- 1}$ are group homomorphisms because $(f g) (x + y) = f (x + y) g (x + y) = f (x) f (y) g (x) g (y) = f (x) g (x) f (y) g (y) = (f g) (x) (f g) (y)$ and $f^{- 1} (x + y) = f (x + y)^{- 1} = f (x)^{- 1} f (y)^{- 1}$ by [F2] and commutativity of $T$ ; they are continuous because $x \mapsto (f (x), g (x))$ is continuous into the product by [F4], multiplication on $T$ is continuous by [F1], and $f g$ is the composite of these two maps, while $f^{- 1}$ is the composite of $f$ with the continuous inversion of $T$ by [F1] and [F4]. The group axioms for $\hat{G}$ hold pointwise because $T$ is an abelian group by [F1], with pointwise constant $1$ as identity.

1.2F2F3F7

Each $U_{K} (f, r)$ is compact-open open. First, for any character $h$ and $a > 0$ , cover the compact image $h [K]$ by finitely many balls $B (z_{j}, a / 3)$ with centres in $T$ , and put $L_{j} : = {x \in K : ∣ h (x) - z_{j} ∣ \leq a / 3}$ . These closed subsets of $K$ are compact and cover $K$ by [F7]. The open set $W_{h} : = ⋂_{j} S (L_{j}, B (z_{j}, 2 a / 3))$ contains $h$ and lies in $U_{K} (h, a)$ by the triangle inequality. Now if $h \in U_{K} (f, r)$ and $K \neq \emptyset$ , the continuous function $x \mapsto ∣ h (x) - f (x) ∣$ attains a maximum $m < r$ by [F7]; its continuity follows from $∣ ∣ u - v ∣ - ∣ u^{'} - v^{'} ∣ ∣ \leq ∣ u - u^{'} ∣ + ∣ v - v^{'} ∣$ . Choose $0 < a < r - m$ . The preceding $W_{h}$ is contained in $U_{K} (f, r)$ , so every member of $U_{K} (f, r)$ has an open neighbourhood inside it. If $K = \emptyset$ , $U_{K} (f, r) = \hat{G}$ .

1.3F1F3

The displayed estimates hold: for $f^{'} \in U_{K} (f, ε / 2)$ , $g^{'} \in U_{K} (g, ε / 2)$ and $x \in K$ , $∣ f^{'} (x) g^{'} (x) - f (x) g (x) ∣ \leq ∣ f^{'} (x) - f (x) ∣ + ∣ g^{'} (x) - g (x) ∣ < ε$ , since all values have modulus $1$ . Also $∣ f^{'} (x)^{- 1} - f (x)^{- 1} ∣ = ∣ f^{'} (x) - f (x) ∣$ ; inversion is involutive, so the second displayed equality follows.

1.4F2F5F6

$\hat{G}$ is Hausdorff: if $f \neq g$ in $\hat{G}$ , there is $x \in G$ with $f (x) \neq g (x)$ ; by [F6] choose disjoint open $V, W \subseteq T$ with $f (x) \in V$ , $g (x) \in W$ ; then $S ({x}, V)$ and $S ({x}, W)$ are open in $\hat{G}$ by [F5], they contain $f$ and $g$ respectively, and they are disjoint because no function can take the same value in both $V$ and $W$ .

2.1step 1.2F2F3F7

These sets form a neighbourhood basis at $f$ . If $f \in S (K, V)$ , cover $f [K]$ by finitely many balls $B (z_{j}, a_{j})$ with $a_{j} > 0$ and $B (z_{j}, 2 a_{j}) \subseteq V$ , using compactness of $f [K]$ and openness of $V$ . For $r : = \min_{j} a_{j} > 0$ , the triangle inequality gives $U_{K} (f, r) \subseteq S (K, V)$ . For empty $K$ take any $r > 0$ . Any finite intersection of such subbasic neighbourhoods contains $U_{⋃_{j} K_{j}} (f, \min_{j} r_{j})$ , and the union is compact by [F7]; an empty intersection is the whole dual. Together with step 1.2 this proves the basis assertion.

3.1step 1.1step 2.1step 1.3F4

Multiplication and inversion on $\hat{G}$ are continuous. By step 2.1 it suffices to test $U_{K} (f g, ε)$ and $U_{K} (f^{- 1}, ε)$ at arbitrary $f, g$ . Step 1.3 maps the open rectangle $U_{K} (f, ε / 2) \times U_{K} (g, ε / 2)$ into the first set and maps the open neighbourhood $U_{K} (f, ε)$ into the second.

4.1step 1.1step 2.1step 1.3step 1.4step 3.1∎

The pointwise group of step 1.1 is Hausdorff by step 1.4 and has continuous operations by step 3.1, so it is a Hausdorff topological abelian group. The asserted neighbourhood basis and estimates are steps 2.1 and 1.3.

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Evaluation of characters is jointly continuous

Statement

Let $G$ be a locally compact Hausdorff abelian group (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space). The evaluation pairing $\hat{G} \times G \to T, (γ, x) \mapsto γ (x),$ is continuous for the compact-open topology on $\hat{G}$ (The Pontryagin dual with the compact-open topology) and the given topology on $G$ (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, Continuity of a map of topological spaces at a point and globally).

Facts & Assumptions

[F1]

$G$ is a topological group whose translations and inversion are homeomorphisms, and $G$ is locally compact: every point has a compact neighbourhood. (Left and right translations and inversion in a topological group are homeomorphisms, Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space)

[F2]

In a locally compact Hausdorff space, every open neighbourhood $U$ of a point $x$ contains an open set $V$ with $x \in V \subseteq \overline{V} \subseteq U$ and $\overline{V}$ compact. (In a locally compact Hausdorff space every open set containing a point contains an open set containing it whose closure is compact and still inside; such a space is regular, Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open)

[F3]

The dual consists of the continuous homomorphisms $γ : G \to T$ , with the compact-open subbasis $S (K, V) = {γ : γ [K] \subseteq V}$ for compact $K \subseteq G$ and open $V \subseteq T$ ; every such set is open in the subspace topology and contains every character mapping $K$ into $V$ . (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)

[F4]

Continuous images of compact sets are compact, and a translate of a compact set is compact; a translate of an open set is open. (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, Left and right translations and inversion in a topological group are homeomorphisms, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right)

[F5]

For all $z, w \in T$ : $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ , $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ , $z^{- 1}$ has modulus $1$ , and $∣ z^{- 1} - w^{- 1} ∣ = ∣ z - w ∣$ ; in particular every $γ (x)$ and every $γ_{0} (x_{0})$ has modulus $1$ . (The multiplicative unit circle is a compact metrizable topological abelian group, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive)

Proof

Given: A locally compact Hausdorff abelian group $G$ , a character $γ_{0} \in \hat{G}$ , a point $x_{0} \in G$ , and $ε > 0$ .

1.1F1F2F3

Choose an open neighbourhood $U$ of $x_{0}$ with $γ_{0} [U] \subseteq B (γ_{0} (x_{0}), ε / 2)$ , possible because $γ_{0}$ is continuous at $x_{0}$ by [F3]; by [F2] choose an open $V$ with $x_{0} \in V \subseteq \overline{V} \subseteq U$ and $K : = \overline{V}$ compact. Then $K$ is a compact neighbourhood of $x_{0}$ with $γ_{0} [K] \subseteq B (γ_{0} (x_{0}), ε / 2)$ .

2.1step 1.1F1F3F4F5

The translate $K - x_{0} = {x - x_{0} : x \in K}$ is compact and is a neighbourhood of $0$ : it is the image of $K$ under the homeomorphism $x \mapsto x - x_{0}$ of [F1, F4], and it contains the open translate $V - x_{0}$ of $V$ , which contains $0$ . Moreover $γ_{0} [K - x_{0}] \subseteq B (1, ε / 2)$ , because for $x \in K \subseteq U$ one has $γ_{0} (x - x_{0}) = γ_{0} (x) γ_{0} (x_{0})^{- 1}$ and $∣ γ_{0} (x) γ_{0} (x_{0})^{- 1} - 1 ∣ = ∣ γ_{0} (x) - γ_{0} (x_{0}) ∣ < ε / 2$ by [F5].

3.1step 2.1F3F5

Let $γ \in S ({x_{0}}, B (γ_{0} (x_{0}), ε / 2)) \cap S (K - x_{0}, B (1, ε / 2))$ and $x \in K$ . Since $γ$ is a homomorphism, $γ (x) = γ (x_{0}) γ (x - x_{0})$ , so $γ (x) - γ_{0} (x_{0}) = (γ (x_{0}) - γ_{0} (x_{0})) γ (x - x_{0}) + γ_{0} (x_{0}) (γ (x - x_{0}) - 1)$ and hence $∣ γ (x) - γ_{0} (x_{0}) ∣ \leq ∣ γ (x_{0}) - γ_{0} (x_{0}) ∣ + ∣ γ (x - x_{0}) - 1 ∣ < ε / 2 + ε / 2 = ε$ by [F5] and the choice of $γ$ .

4.1step 1.1step 2.1step 3.1F2F3∎

The set $W : = (S ({x_{0}}, B (γ_{0} (x_{0}), ε / 2)) \cap S (K - x_{0}, B (1, ε / 2))) \times V$ is a neighbourhood of $(γ_{0}, x_{0})$ in $\hat{G} \times G$ : it is a product of an open set containing $γ_{0}$ and an open set containing $x_{0}$ , the first because $γ_{0} (x_{0}) \in B (γ_{0} (x_{0}), ε / 2)$ and $γ_{0} [K - x_{0}] \subseteq B (1, ε / 2)$ by step 2.1, the second because $V$ is open and $x_{0} \in V \subseteq K$ . By step 3.1 the pairing maps $W$ into $B (γ_{0} (x_{0}), ε)$ ; since open balls form a neighbourhood base at $γ_{0} (x_{0})$ , the pairing is continuous at the arbitrary point $(γ_{0}, x_{0})$ , hence continuous.

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Pointwise limits of homomorphisms and of equicontinuous characters

Statement

(1) Let $G$ be a group, written additively. A pointwise limit of group homomorphisms $G \to T$ is a homomorphism; precisely, $Hom (G, T)$ is closed in $T^{G}$ for the topology of pointwise convergence (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ , 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). (2) For an abelian topological group $G$ , if a pointwise limit of continuous homomorphisms is taken along an equicontinuous family, then the limit is continuous, hence a character. (3) If the abelian topological group $G$ is discrete (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), then $\hat{G} = Hom (G, T)$ is closed in $T^{G}$ for the product topology.

Facts & Assumptions

[F1]

Multiplication in $T$ is continuous. $T^{G}$ is the product of the constant family with one factor $T$ per $x \in G$ ; a map into a product is continuous exactly when all its components are, and the projection $f \mapsto f (x)$ is continuous for every $x$ . (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, A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, The multiplicative unit circle is a compact metrizable topological abelian group, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous)

[F2]

$T$ is Hausdorff, and the diagonal $Δ = {(z, z) : z \in T}$ is closed in $T \times T$ : if $z \neq w$ , disjoint open neighbourhoods of $z$ and $w$ give an open rectangle around $(z, w)$ missing $Δ$ . (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Distinct points of a metric space have disjoint balls around them, 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)

[F3]

A point lies in the closure of a set exactly when some net in the set converges to it. (A point lies in the closure of a set if and only if a net in the set converges to it, Directed preorders and nets)

[F4]

The closure in $Y^{X}$ with the topology of pointwise convergence of an equicontinuous family $F \subseteq C (X, Y)$ into a metric space $Y$ is equicontinuous, and every member of that closure is continuous; the topology of pointwise convergence on $C (X, Y)$ is the subspace topology inherited from $Y^{X}$ . (The pointwise closure of an equicontinuous family is equicontinuous and consists of continuous maps, Equicontinuity on a topological domain and pointwise relative compactness, The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ )

[F5]

A discrete topology makes every subset open, and continuity means that for each point and open neighbourhood of its image there is an open source neighbourhood mapped into it. (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Continuity of a map of topological spaces at a point and globally)

[F6]

The dual consists of the continuous homomorphisms $G \to T$ with the compact-open topology, and the compact-open topology on a discrete domain agrees with the topology of pointwise convergence, i.e. with the subspace topology from $T^{G}$ . (The Pontryagin dual with the compact-open topology, On a discrete domain the compact-open topology is the topology of pointwise convergence)

Proof

Given: A group $G$ , the product space $T^{G}$ with the topology of pointwise convergence, and the set $Hom (G, T)$ of all group homomorphisms $G \to T$ .

1.1F1F2

$Hom (G, T)$ is closed in $T^{G}$ : it is the intersection over all $x, y \in G$ of the sets $E_{x, y} : = {f : f (x + y) = f (x) f (y)}$ , and each $E_{x, y}$ is the preimage of the diagonal $Δ \subseteq T \times T$ under the map $φ_{x, y} (f) : = (f (x + y), f (x) f (y))$ , which is continuous because both components are continuous by [F1]; preimages of the closed set $Δ$ under continuous maps are closed by [F2], and arbitrary intersections of closed sets are closed.

2.1step 1.1F3

Consequently a pointwise limit of group homomorphisms is a homomorphism: if a net $(γ_{j})$ in $Hom (G, T)$ converges pointwise to $γ$ , then $γ$ lies in the closure of $Hom (G, T)$ by [F3], and that closure equals the closed set $Hom (G, T)$ by step 1.1, so $γ$ is a homomorphism.

2.2step 1.1F5F6

If $G$ is discrete, then for any map $f : G \to T$ , point $x \in G$ and open set $V$ containing $f (x)$ , the preimage $f^{- 1} [V]$ is a subset of $G$ , hence open and contains $x$ ; it maps into $V$ , so the continuity definition [F5] makes $f$ continuous at every $x$ . Thus every map $G \to T$ is continuous, so the continuous characters are exactly the homomorphisms, $\hat{G} = Hom (G, T)$ , and this set is closed in $T^{G}$ by step 1.1; by [F6] the compact-open topology on $\hat{G}$ is its subspace topology from $T^{G}$ , so $\hat{G}$ is a closed subset of $T^{G}$ as asserted.

3.1step 2.1F4

If the homomorphisms $γ_{j}$ above are continuous and the family ${γ_{j}}$ is equicontinuous, then $γ$ is continuous: the family lies in $C (G, T)$ , its pointwise closure is equicontinuous and consists of continuous functions by [F4], and $γ$ belongs to that closure.

4.1step 2.1step 3.1step 2.2∎

Clauses (1), (2) and (3) of the statement are steps 2.1, 3.1 and 2.2 respectively.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Dual homomorphisms: continuity, and the annihilator of a closed subgroup

Statement

Assume the Axiom of Choice (The Axiom of Choice), used only in part (b), through the compact-lift theorem for closed subgroup quotients.

(a) If $φ : G \to H$ is a continuous homomorphism of abelian topological groups, the pullback $\hat{φ} : \hat{H} \to \hat{G}$ , $\hat{φ} (γ) : = γ \circ φ$ , is a continuous group homomorphism.

(b) If $H$ is a closed subgroup of a locally compact Hausdorff abelian group $G$ and $q : G \to G / H$ is the quotient homomorphism, then $\hat{q} : \hat{G / H} \to \hat{G}$ is a topological group isomorphism onto the annihilator $H^{⊥} = {γ \in \hat{G} : γ (h) = 1 for all h \in H}$ , a closed subgroup of $\hat{G}$ . No stronger claim is made for pullbacks along non-proper maps.

Facts & Assumptions

[F1]

Characters are the continuous homomorphisms into $T$ ; $\hat{G}$ carries pointwise multiplication and the compact-open topology with subbasis $S (K, V) = {γ : γ [K] \subseteq V}$ for compact $K \subseteq G$ and open $V \subseteq T$ . (The Pontryagin dual with the compact-open topology)

[F2]

$\hat{G}$ is a Hausdorff topological abelian group; translations $γ \mapsto γ_{0} γ$ are homeomorphisms. (The compact-open character group is a Hausdorff topological abelian group, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological)

[F3]

Composites of continuous maps are continuous, and continuous images of compact sets are compact. (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right)

[F4]

Quotient universal property: a continuous map on $G$ constant on the fibres of the quotient map $q$ factors uniquely through $G / H$ , and continuity of a map out of $G / H$ is equivalent to continuity after composition with $q$ . (For a quotient map $q : X \to Y$ , a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$ ; and a composite of quotient maps is a quotient map, The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection)

[F5]

If $H$ is a subgroup of an abelian group $G$ , its cosets form the abelian quotient group $G / H$ with $(x + H) + (y + H) = x + y + H$ . The quotient topology makes $q$ a continuous quotient surjection. (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , Every quotient group of an abelian group is abelian, The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection)

[F6]

An open continuous surjection is a quotient map. (A continuous open surjection, a continuous closed surjection, and a continuous surjection admitting a continuous section are all quotient maps)

[F7]

Multiplication on $T$ is continuous and $1$ is closed in the metric space $T$ . (The multiplicative unit circle is a compact metrizable topological abelian group)

[F8]

A locally compact space gives each point a compact neighbourhood containing an open neighbourhood. Compact subsets of a Hausdorff space are closed; finite unions of compact subsets are compact (combine the finitely many finite subcovers). In a topological group, translations and inversion are homeomorphisms and group operations are continuous. Products have the basis of finite open-coordinate constraints and maps into products are continuous coordinatewise. (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space, Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Topological group: multiplication and inversion are continuous, Left and right translations and inversion in a topological group are homeomorphisms, 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, A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice)

[F9]

A compact subset has a finite subcover from every family of ambient open sets covering it, also in indexed form; choosing from finitely many listed nonempty sets requires no Axiom of Choice. (A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, Every natural-number-indexed list of nonempty sets has a choice function on its family of values)

[A1]

The Axiom of Choice supplies a choice from each member of an arbitrary family of nonempty sets. (The Axiom of Choice)

Proof

Given: Continuous homomorphisms of abelian topological groups as in (a) and (b), and the Axiom of Choice for the compact-lift theorem.

1.1F1F2F3

Part (a): for $γ \in \hat{H}$ the composite $γ \circ φ$ is continuous by [F3] and is a homomorphism, so $\hat{φ} (γ) \in \hat{G}$ ; $\hat{φ}$ is a group homomorphism because $\hat{φ} (γ_{1} γ_{2}) (x) = γ_{1} (φ (x)) γ_{2} (φ (x)) = (\hat{φ} γ_{1}) (x) (\hat{φ} γ_{2}) (x)$ . For compact $K \subseteq G$ and open $V \subseteq T$ the preimage satisfies ${\hat{φ}}^{- 1} (S_{G} (K, V)) = {γ \in \hat{H} : γ [φ [K]] \subseteq V} = S_{H} (φ [K], V)$ , and $φ [K]$ is compact by [F3]; this set is subbasic open in $\hat{H}$ by [F1], so $\hat{φ}$ is continuous. Pullback preserves the identity map and reverses composition: for $ψ : H \to J$ , associativity gives $γ \circ (ψ \circ φ) = (γ \circ ψ) \circ φ$ , hence $\hat{ψ \circ φ} = \hat{φ} \circ \hat{ψ}$ .

1.2F3F4F5F6F8

Part (b), quotient topology. For every open $O \subseteq G$ , $q^{- 1} (q (O)) = O + H = ⋃_{h \in H} (O + h)$ is open by translations, hence $q (O)$ is open by the quotient topology. Thus $q$ is open. Distinct cosets $q (x), q (y)$ have $x - y \notin H$ . Closedness of $H$ gives an open neighbourhood $W$ of $x - y$ disjoint from $H$ . By continuity of subtraction choose identity neighbourhoods $U, V$ with $(x - y) + U - V \subseteq W$ . The open sets $q (x + U)$ and $q (y + V)$ are disjoint: an intersection would give $x + u - y - v \in H \cap W$ . Therefore $G / H$ is Hausdorff. For any $x \in G$ , choose a compact neighbourhood $N$ of $x$ and open $U$ with $x \in U \subseteq N$ . Then $q (N)$ is compact by continuity, closed because the quotient is Hausdorff, and contains the open neighbourhood $q (U)$ of $q (x)$ . Thus the quotient is locally compact. The product $q \times q$ is a continuous open surjection: images of basic open rectangles are open rectangles, and arbitrary opens are unions of those rectangles. It is therefore quotient by [F6]. The quotient multiplication is continuous since its composite with $q \times q$ is the continuous map $q \circ m_{G}$ , and the quotient universal property [F4] applies; similarly inversion descends through $q$ . Hence $G / H$ is an abelian topological group.

1.3F1F7

$H^{⊥}$ is closed in $\hat{G}$ : it is the intersection over $h \in H$ of the sets ${γ : γ (h) = 1}$ , each of which is the preimage of the closed set ${1} \subseteq T$ under the evaluation map $γ \mapsto γ (h)$ , and that evaluation is continuous because ${γ : γ (h) \in V} = S ({h}, V)$ is subbasic open for every open $V \subseteq T$ .

2.1A1F5F8F9step 1.2

Compact lifts, proved locally. Let $L \subseteq G / H$ be compact. If $L = \emptyset$ , take $K = \emptyset$ . Otherwise, for every $l \in L$ the set of triples $(x, N, U)$ with $q (x) = l$ , $N$ a compact neighbourhood of $x$ , and $U$ open with $x \in U \subseteq N$ is nonempty by surjectivity and [F8]. Use [A1] to choose such a triple $(x_{l}, N_{l}, U_{l})$ for each $l$ ; this is the only invocation of Choice in this argument. By step 1.2 the sets $q (U_{l})$ form an open cover of $L$ . The indexed ambient-cover criterion [F9] gives finitely many indices whose sets cover $L$ . Take their associated triples, and let $K$ be the union of the corresponding $N_{l}$ . Finite unions of compact subsets are compact by [F8], and $L \subseteq ⋃_{l} q (U_{l}) \subseteq q (K)$ . This proves the compact-lift theorem needed below from earlier topology alone.

2.2step 1.1step 1.2F1F5

The pullback $\hat{q} : \hat{G / H} \to \hat{G}$ is a continuous group homomorphism by step 1.1 applied to the continuous homomorphism $q$ ; its image lies in $H^{⊥}$ because $\hat{q} (γ) (h) = γ (q (h)) = γ (0) = 1$ for $h \in H$ , and $\hat{q}$ is injective because $γ \circ q = 1$ forces $γ = 1$ , the quotient map $q$ being surjective.

3.1step 2.2F1F4F5

The image of $\hat{q}$ equals $H^{⊥}$ : if $γ \in \hat{G}$ satisfies $γ [H] = {1}$ , then $γ$ is constant on the fibres of $q$ , for $q (x) = q (x^{'})$ means $x - x^{'} \in H$ and then $γ (x) = γ (x^{'}) γ (x - x^{'}) = γ (x^{'})$ ; the quotient universal property [F4] factors $γ = γ^{'} \circ q$ with $γ^{'}$ continuous, and $γ^{'}$ is a homomorphism because for cosets $a = q (x)$ , $b = q (y)$ one has $γ^{'} (a + b) = γ^{'} (q (x + y)) = γ (x + y) = γ (x) γ (y) = γ^{'} (a) γ^{'} (b)$ . Hence $γ = \hat{q} (γ^{'}) \in \hat{q} [\hat{G / H}]$ .

4.1step 2.1step 2.2step 3.1F2F3F5F7F9

The inverse ${\hat{q}}^{- 1} : H^{⊥} \to \hat{G / H}$ is continuous: fix $γ_{0} \in H^{⊥}$ and put $δ_{0} : = {\hat{q}}^{- 1} (γ_{0})$ , and let $S (L, V) = {δ : δ [L] \subseteq V}$ be a subbasic open set containing $δ_{0}$ , with $L \subseteq G / H$ compact and $V \subseteq T$ open. Since $L$ is compact and $δ_{0}$ is continuous, $δ_{0} [L]$ is compact by [F3]; consider all pairs $(A, B)$ of open subsets of $T$ with $1 \in B$ and $A B \subseteq V$ . The first coordinates $A$ of these pairs cover $δ_{0} [L]$ : for every $t \in δ_{0} [L] \subseteq V$ , continuity of multiplication at $(t, 1)$ provides such a pair with $t \in A$ . The ambient-cover criterion [F9] selects finitely many first coordinates $A_{1}, \dots, A_{n}$ covering $δ_{0} [L]$ , and finite choice [F9] selects their associated $B_{i}$ . Set $W = ⋂_{i = 1}^{n} B_{i}$ , an open neighbourhood of $1$ with $δ_{0} (l) W \subseteq V$ for every $l \in L$ ; for empty $L$ , use $W = T$ . This construction uses no arbitrary-index choice. By the compact-lift theorem in step 2.1 choose compact $K \subseteq G$ with $L \subseteq q [K]$ ; then $N : = (γ_{0} \cdot S_{G} (K, W)) \cap H^{⊥}$ is a neighbourhood of $γ_{0}$ in $H^{⊥}$ , because $S_{G} (K, W)$ is a subbasic open neighbourhood of $1$ in $\hat{G}$ and translation by $γ_{0}$ is a homeomorphism by [F2]. For $γ = γ_{0} η \in N$ with $η \in S_{G} (K, W)$ and $l = q (k) \in L$ with $k \in K$ one has ${\hat{q}}^{- 1} (γ) (l) = γ (k) = γ_{0} (k) η (k) = δ_{0} (l) η (k) \in δ_{0} (l) W \subseteq V$ ; hence ${\hat{q}}^{- 1} [N] \subseteq S (L, V)$ and ${\hat{q}}^{- 1}$ is continuous at the arbitrary point $γ_{0}$ .

5.1step 1.1step 2.2step 3.1step 1.3step 4.1∎

Conclusion of (b): by steps 2.2, 3.1 and 4.1 the map $\hat{q}$ is a group isomorphism of $\hat{G / H}$ onto $H^{⊥}$ that is continuous and has continuous inverse, hence a topological group isomorphism onto $H^{⊥}$ ; $H^{⊥}$ is closed in $\hat{G}$ by step 1.3, and it is a subgroup because it is the kernel of the homomorphism $γ \mapsto (γ (h))_{h \in H}$ restricted to the abelian group $\hat{G}$ . Together with part (a), proved in step 1.1, this is the statement.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

A compact identity neighbourhood in the dual

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $G$ be a locally compact Hausdorff abelian group, $K \subseteq G$ a symmetric compact neighbourhood of $0$ , and $D : = {z \in T : ∣ z - 1 ∣ \leq 1 / 2}$ . Then $N : = {γ \in \hat{G} : γ [K] \subseteq D}$ is equicontinuous (Equicontinuity on a topological domain and pointwise relative compactness), is compact in the compact-open topology of $C (G, T)$ (The compact-open topology on $C (X, Y)$ for arbitrary topological spaces), and is a neighbourhood of the identity character in $\hat{G}$ .

Facts & Assumptions

[F1]

$T$ is a compact metrizable topological abelian group with continuous multiplication; the map $z \mapsto ∣ z - 1 ∣$ is continuous, so $D = {∣ z - 1 ∣ \leq 1 / 2}$ is closed in $T$ , and $1 \in D$ . (The multiplicative unit circle is a compact metrizable topological abelian group, The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane)

[F2]

Every subgroup $H \leq T$ with $H \subseteq D$ is trivial; equivalently, for every $z \in T$ with $z \neq 1$ there is a positive integer $k$ with $z^{k} \notin D$ . (The unit-circle arc ${z : ∣ z - 1 ∣ < 1}$ contains no nontrivial subgroup)

[F3]

In a compact space every family of closed sets with the finite intersection property has nonempty intersection. (A space is compact exactly when every family of closed subsets with the finite intersection property has nonempty intersection, Finite intersection property, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right)

[F4]

The dual consists of the continuous homomorphisms $γ : G \to T$ with the compact-open topology and pointwise multiplication; $S (K, V) = {γ : γ [K] \subseteq V}$ is subbasic open; the compact-open topology is finer than the topology of pointwise convergence, whose subbasic sets on $C (G, T)$ are the $S ({x}, V)$ . (The Pontryagin dual with the compact-open topology, The compact-open topology on $C (X, Y)$ for arbitrary topological spaces, The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ )

[F5]

Pointwise limits of characters are characters: the pointwise limit of continuous homomorphisms taken along an equicontinuous family is a character, and the pointwise closure of an equicontinuous family of continuous maps consists of continuous maps. (Pointwise limits of homomorphisms and of equicontinuous characters, The pointwise closure of an equicontinuous family is equicontinuous and consists of continuous maps)

[F6]

[F7]

Assume the Axiom of Choice. For any topological space $X$ and compact metric space $Y$ , the compact-open closure of an equicontinuous family $F \subseteq C (X, Y)$ is compact. (Under Choice, an equicontinuous family into a compact metric target has compact compact-open closure, Under Choice, equicontinuity and pointwise relative compactness give compact compact-open closure, The Axiom of Choice)

[F8]

Translations and the maps $u \mapsto k u$ $(k \geq 1)$ on a topological group are continuous, and a finite intersection of open neighbourhoods of $0$ is an open neighbourhood of $0$ ; composites of continuous maps are continuous. (Left and right translations and inversion in a topological group are homeomorphisms, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Continuity of a map of topological spaces at a point and globally)

Proof

Given: A locally compact Hausdorff abelian group $G$ , a symmetric compact neighbourhood $K$ of $0$ , the set $D = {∣ z - 1 ∣ \leq 1 / 2}$ , and $N = {γ \in \hat{G} : γ [K] \subseteq D}$ .

1.1F1F4

$N$ is a neighbourhood of the identity character: the constant character $1$ satisfies $1 [K] = {1} \subseteq D$ by [F1], so $1 \in N$ ; and the open arc $D^{\circ} = {z \in T : ∣ z - 1 ∣ < 1 / 2}$ contains $1$ . Hence $S (K, D^{\circ})$ is a subbasic compact-open neighbourhood of the identity by [F4], and $S (K, D^{\circ}) \subseteq N$ because $D^{\circ} \subseteq D$ . Thus $N$ is a neighbourhood, although it need not be open.

1.2F1F2F3F8

$N$ is equicontinuous: fix $ε > 0$ and put $O : = B (1, ε) \cap T$ , an open neighbourhood of $1$ in $T$ ; for $k \geq 1$ put $F_{k} : = {z \in T : z^{j} \in D for j = 1, \dots, k}$ . Each $F_{k}$ is closed, being a finite intersection of preimages of the closed set $D$ under the power maps $z \mapsto z^{j}$ , which are continuous by induction on $j$ from the continuity of multiplication on $T$ by [F1]; the sequence $F_{1} \supseteq F_{2} \supseteq \dots$ has intersection ${1}$ by [F2], because any $z \neq 1$ satisfies $z^{j} \notin D$ for some $j$ . Hence $F_{k} \subseteq O$ for some $k$ : otherwise the sets $F_{k} ∖ O$ would be nonempty closed sets with the finite intersection property, being decreasing, and [F3] would produce $z \in ⋂_{k} (F_{k} ∖ O) \subseteq {1} ∖ O = \emptyset$ .

2.1step 1.2F1F8

With $k$ as in step 1.2 put $U : = {u \in G : j u \in K for j = 0, 1, \dots, k}$ , a neighbourhood of $0$ : for $j \geq 1$ the set ${u : j u \in K}$ contains the open set ${u : j u \in K^{\circ}}$ , which is the preimage of the open $K^{\circ}$ under the continuous map $u \mapsto j u$ and contains $0$ , while $0 \in K$ for $j = 0$ ; a finite intersection of neighbourhoods of $0$ is a neighbourhood of $0$ by [F8]. For $u \in U$ and $γ \in N$ one has $γ (u)^{j} = γ (j u) \in D$ for $j = 1, \dots, k$ , so $γ (u) \in F_{k} \subseteq O$ , that is $∣ γ (u) - 1 ∣ < ε$ . Thus $N$ is equicontinuous at $0$ .

3.1step 2.1F1

$N$ is equicontinuous at every point $x_{0} \in G$ : for $x - x_{0} \in U$ and $γ \in N$ , $γ (x) = γ (x_{0}) γ (x - x_{0})$ and hence $∣ γ (x) - γ (x_{0}) ∣ = ∣ γ (x_{0}) ∣ ∣ γ (x - x_{0}) - 1 ∣ = ∣ γ (x - x_{0}) - 1 ∣ < ε$ by step 2.1 and [F1].

4.1step 1.1step 3.1F1F4F5F6

$N$ is closed in $C (G, T)$ for the compact-open topology: let $γ$ lie in the compact-open closure of $N$ . The compact-open topology is finer than the pointwise topology by [F4], so $γ$ lies in the pointwise closure of $N$ ; by [F6] some net $(γ_{j})$ in $N$ converges to $γ$ pointwise. All $γ_{j}$ are characters and the family $N$ is equicontinuous by step 3.1, so $γ$ is a character by [F5]; and $γ [K] \subseteq D$ because each $γ_{j} [K] \subseteq D$ and $D$ is closed by [F1]. Hence $γ \in N$ and $N$ is compact-open closed.

5.1step 3.1step 4.1F1F7

$N$ is compact in the compact-open topology: $N \subseteq C (G, T)$ is equicontinuous by step 3.1 and $T$ is a compact metric space by [F1], so the compact-open closure of $N$ is compact by [F7]; by step 4.1 that closure is $N$ itself.

6.1step 1.1step 3.1step 5.1∎

By steps 1.1, 3.1 and 5.1 the set $N$ is equicontinuous, compact in the compact-open topology, and a neighbourhood of the identity character in $\hat{G}$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The dual of a locally compact abelian group is locally compact abelian

Statement

Assume the Axiom of Choice (The Axiom of Choice). If $G$ is a locally compact Hausdorff abelian group (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space), then $\hat{G}$ is a locally compact Hausdorff abelian topological group, and the sets ${γ : γ [K] \subseteq V}$ with $K \subseteq G$ compact and $V \subseteq T$ open with $1 \in V$ form a basis of neighbourhoods of the identity character (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

Facts & Assumptions

[F1]

$\hat{G}$ is a Hausdorff topological abelian group, translations are homeomorphisms, and its topology is generated by the subbasis $S (K, V) = {γ : γ [K] \subseteq V}$ with $K \subseteq G$ compact and $V \subseteq T$ open. (The compact-open character group is a Hausdorff topological abelian group, The Pontryagin dual with the compact-open topology, The compact-open topology on $C (X, Y)$ for arbitrary topological spaces, Left and right translations and inversion in a topological group are homeomorphisms)

[F2]

Assume the Axiom of Choice. For every symmetric compact neighbourhood $K$ of $0$ in $G$ , the set $N_{K} : = {γ \in \hat{G} : γ [K] \subseteq D}$ with $D = {∣ z - 1 ∣ \leq 1 / 2}$ is equicontinuous, is compact in the compact-open topology of $C (G, T)$ , and is a neighbourhood of the identity character in $\hat{G}$ . (A compact identity neighbourhood in the dual, The Axiom of Choice)

[F3]

In a locally compact Hausdorff space every point has a compact neighbourhood, and we may take it symmetric in a topological group: if $N$ is a compact neighbourhood of $0$ then so is $N \cap (- N)$ . (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space, In a locally compact Hausdorff space every open set containing a point contains an open set containing it whose closure is compact and still inside; such a space is regular, Left and right translations and inversion in a topological group are homeomorphisms, In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism)

[F4]

A composite of continuous maps is continuous. (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous)

Proof

Given: A locally compact Hausdorff abelian group $G$ , and $\hat{G}$ with the compact-open topology.

1.1F1

$\hat{G}$ is a Hausdorff abelian topological group by [F1].

1.2F1

The sets $S (K, V)$ with $K$ compact and $V$ open containing $1$ form a basis of neighbourhoods of $1$ in $\hat{G}$ : any open neighbourhood of $1$ contains a finite intersection $⋂_{i \leq n} S (K_{i}, V_{i})$ of subbasic sets each containing $1$ by [F1], and that intersection contains the subbasic set $S (⋃_{i \leq n} K_{i}, ⋂_{i \leq n} V_{i})$ , where the union of finitely many compact sets is compact, the finite intersection of open sets is open and contains $1$ , and $1 (K) = {1}$ lies in every $V$ with $1 \in V$ .

2.1step 1.1F1F2F3F4

$\hat{G}$ is locally compact: by [F3] choose a symmetric compact neighbourhood $K$ of $0$ in $G$ (a compact neighbourhood intersected with its inverse image under the continuous inversion is compact and symmetric). By [F2] the set $N_{K} = {γ : γ [K] \subseteq D}$ is a compact neighbourhood of $1$ in $\hat{G}$ ; translations are homeomorphisms of $\hat{G}$ by [F1], so every character has a compact neighbourhood, and $\hat{G}$ is locally compact by [F1].

3.1step 1.1step 1.2step 2.1∎

By steps 1.1, 1.2 and 2.1 the dual $\hat{G}$ is a locally compact Hausdorff abelian topological group and the compact-open sets $S (K, V)$ form a basis of neighbourhoods of the identity character, as asserted.

TheoremStatement: Literature-sourcedProof: Literature-sourcedjudge pass (gpt-6.1-sol)Open item page →

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.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Duals of finite products and of discrete direct sums

Statement

(1) For locally compact Hausdorff abelian groups $G_{1}, \dots, G_{n}$ the map $Φ : \hat{G_{1}} \times \dots \times \hat{G_{n}} \to \hat{G_{1} \times \dots \times G_{n}}, Φ (γ_{1}, \dots, γ_{n}) ((x_{1}, \dots, x_{n})) : = \prod_{j} γ_{j} (x_{j}),$ is an isomorphism of topological groups for the product topologies (choice-free, finite $n$ ). (2) If $(G_{i})_{i \in I}$ is a family of discrete abelian groups and $G = ⨁_{i} G_{i}$ is their algebraic direct sum equipped with the discrete topology (The direct sum of an indexed family of modules), then, assuming the Axiom of Choice (The Axiom of Choice), $\hat{G}$ is topologically isomorphic to the product $\prod_{i} \hat{G_{i}}$ with the product topology. No claim is made here about a direct sum carrying the subspace topology of the product of non-discrete factors.

Facts & Assumptions

[F1]

A finite product $H_{1} \times \dots \times H_{n}$ of topological groups with the product topology is a topological group: multiplication and inversion are continuous because each component is a composite of projections, which are continuous, with the continuous operations of the factors, and a map into a product is continuous exactly when its components are. (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, A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, Topological group: multiplication and inversion are continuous, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous)

[F2]

The dual of an abelian topological group is a Hausdorff topological abelian group; the dual of a discrete abelian group is compact. (The compact-open character group is a Hausdorff topological abelian group, Compact groups have discrete duals and discrete groups have compact duals)

[F3]

Pullback along a continuous homomorphism is a continuous homomorphism of duals; on a discrete domain the compact-open topology is the topology of pointwise convergence, i.e. the subspace topology from the product. (Dual homomorphisms: continuity, and the annihilator of a closed subgroup, On a discrete domain the compact-open topology is the topology of pointwise convergence, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Continuity of a map of topological spaces at a point and globally)

[F4]

Tychonoff's theorem (Choice) makes arbitrary products of compact spaces compact, finite products of compact spaces are compact, and a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism. (Tychonoff's theorem: an arbitrary product of compact spaces is compact in the product topology, assuming the Axiom of Choice, A product of finitely many compact spaces is compact in the product topology, A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological, The Axiom of Choice)

[F5]

In $T$ multiplication is continuous and $1$ has an open neighbourhood basis; the finite product of open sets containing $1$ contains an open neighbourhood of $(1, \dots, 1)$ and the product of $n$ factors all lying in an open neighbourhood $W$ of $1$ lies in $W$ whenever they lie in a suitable smaller open neighbourhood. (The multiplicative unit circle is a compact metrizable topological abelian group, Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open)

Proof

Given: Locally compact Hausdorff abelian groups $G_{1}, \dots, G_{n}$ , and a family $(G_{i})_{i \in I}$ of discrete abelian groups.

1.1F1algebra

Part (1), the map $Φ$ is a bijective group homomorphism: it is a homomorphism because both sides multiply pointwise, $Φ (γ_{1} γ_{1}^{'}, \dots, γ_{n} γ_{n}^{'}) (x) = \prod_{j} γ_{j} (x_{j}) γ_{j}^{'} (x_{j}) = (Φ (γ) Φ (γ^{'})) (x)$ ; it is injective because $γ_{j}$ is recovered from $χ = Φ (γ)$ by restricting $χ$ to the $j$ -th coordinate axis, and it is surjective because every character $χ$ of the product gives characters $γ_{j} (x_{j}) : = χ (0, \dots, 0, x_{j}, 0, \dots, 0)$ with $χ (x) = \prod_{j} γ_{j} (x_{j})$ for every $x$ , by multiplicativity of $χ$ and the decomposition of $x$ into its coordinate vectors; $γ_{j}$ is a continuous homomorphism because the coordinate inclusion $x_{j} \mapsto (0, \dots, x_{j}, \dots, 0)$ is continuous.

1.2F3

Part (2), the restriction map: let $G = ⨁_{i} G_{i}$ carry the discrete topology and let $ι_{i} : G_{i} \to G$ be the coordinate inclusion, a homomorphism and continuous because its source $G_{i}$ is discrete: every open target set has an open preimage, as every subset of $G_{i}$ is open. The map $ρ : \hat{G} \to \prod_{i} \hat{G_{i}}$ , $ρ (χ) : = (χ \circ ι_{i})_{i}$ , is a group homomorphism, and it is bijective: injective because a homomorphism on the direct sum is determined by its values on the summands, and surjective because for any family $(γ_{i})$ the formula $χ (x) : = \prod_{i} γ_{i} (x_{i})$ is a finite product over the support of $x$ , is a homomorphism, is continuous because every open target set has an open preimage in the discrete source $G$ , and satisfies $χ \circ ι_{i} = γ_{i}$ .

2.1step 1.1F1F2F5

Part (1), $Φ$ is continuous at the identity: let $K \subseteq G_{1} \times \dots \times G_{n}$ be compact and $W \subseteq T$ open with $1 \in W$ ; the projections $K_{j} : = π_{j} [K]$ are compact, and by [F5] choose an open neighbourhood $V$ of $1$ with $V^{n} \subseteq W$ , so that whenever $γ_{j} [K_{j}] \subseteq V$ for all $j$ and $x \in K$ one has $\prod_{j} γ_{j} (x_{j}) \in V^{n} \subseteq W$ . Hence $Φ (S (K_{1}, V) \times \dots \times S (K_{n}, V)) \subseteq S (K, W)$ , and the product is an open neighbourhood of the identity of $\hat{G_{1}} \times \dots \times \hat{G_{n}}$ by [F1] and [F2]; since $Φ$ is a homomorphism of topological groups and translations are homeomorphisms, continuity at the identity gives continuity everywhere.

2.2step 1.2F3

$ρ$ is continuous: each component $χ \mapsto χ \circ ι_{i}$ is the pullback along the continuous homomorphism $ι_{i}$ , hence continuous by [F3]; a map into the product $\prod_{i} \hat{G_{i}}$ is continuous exactly when all its components are.

3.1step 1.1step 2.1F1F2F5

Part (1), $Φ^{- 1}$ is continuous at the identity: let $K_{j} \subseteq G_{j}$ be compact and $U_{j} \subseteq T$ open with $1 \in U_{j}$ , and put $K : = (K_{1} \cup {0}) \times \dots \times (K_{n} \cup {0})$ , a compact subset of the product; if $χ \in S (K, ⋂_{j} U_{j})$ and $γ_{j} (x_{j}) : = χ (0, \dots, x_{j}, \dots, 0)$ for $x_{j} \in K_{j}$ , then $γ_{j} (x_{j}) \in ⋂_{j} U_{j} \subseteq U_{j}$ , so $Φ^{- 1} (χ) \in S (K_{1}, U_{1}) \times \dots \times S (K_{n}, U_{n})$ ; hence $Φ^{- 1}$ maps a subbasic identity neighbourhood into a basic identity neighbourhood and is continuous at the identity, hence everywhere. Since $Φ$ is a continuous bijective homomorphism with continuous inverse, it is an isomorphism of topological groups, completing (1).

3.2step 1.2step 2.2F2F4

Both sides of $ρ$ are compact Hausdorff: $\hat{G}$ is compact by [F2] because $G$ is discrete, each $\hat{G_{i}}$ is compact by [F2], and the product $\prod_{i} \hat{G_{i}}$ is compact by Tychonoff's theorem by [F4]; both are Hausdorff being duals of topological groups by [F2] and products of Hausdorff spaces. Therefore the continuous bijection $ρ$ from the compact space $\hat{G}$ onto the Hausdorff space $\prod_{i} \hat{G_{i}}$ is a homeomorphism by [F4], so $\hat{G}$ is topologically isomorphic to the product of the duals; this completes (2).

4.1step 3.1step 3.2∎

Parts (1) and (2) are steps 3.1 and 3.2 respectively, so the lemma is proved.

5 · Examples, counterexamples and false statements

None yet.

Character Groups and Elementary LCA Duals

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The multiplicative unit circle is a compact metrizable topological abelian group

Statement

Facts & Assumptions

Proof

On a discrete domain the compact-open topology is the topology of pointwise convergence

Statement

Facts & Assumptions

Proof

The unit-circle arc {z:∣z−1∣<1}{z:∣z−1∣<1} contains no nontrivial subgroup

Statement

Facts & Assumptions

Proof

Continuous characters of the real line are exponentials

Statement

Facts & Assumptions

Proof

The Pontryagin dual with the compact-open topology

Definition

The compact-open character group is a Hausdorff topological abelian group

Statement

Facts & Assumptions

Proof

Evaluation of characters is jointly continuous

Statement

Facts & Assumptions

Proof

Pointwise limits of homomorphisms and of equicontinuous characters

Statement

Facts & Assumptions

Proof

Dual homomorphisms: continuity, and the annihilator of a closed subgroup

Statement

Facts & Assumptions

Proof

A compact identity neighbourhood in the dual

Statement

Facts & Assumptions

Proof

The dual of a locally compact abelian group is locally compact abelian

Statement

Facts & Assumptions

Proof

Compact groups have discrete duals and discrete groups have compact duals

Statement

Facts & Assumptions

Proof

Duals of finite products and of discrete direct sums

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources

The unit-circle arc ${z : ∣ z - 1 ∣ < 1}$ contains no nontrivial subgroup