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The Pontryagin dual of the discrete integers is the circle

Statement

Let $Z$ have the discrete topology, let $T = {z \in C : ∣ z ∣ = 1}$ have the subspace topology from $C$ , and give $\hat{Z}$ the compact-open topology of the Pontryagin dual. Then $Ψ : T ⟶ \hat{Z}, Ψ (z) (n) : = z^{n},$ is a group isomorphism and a homeomorphism. Thus in the angular coordinate $z = e^{i λ}$ , the characters of $Z$ are exactly $n \mapsto e^{i n λ}$ , with $λ$ taken modulo $2 π$ .

Facts & Assumptions

Given: The additive integer group with the discrete topology, the multiplicative unit circle, and the Pontryagin dual with its compact-open topology.

[F1]

The Pontryagin dual with the compact-open topology: $\hat{Z}$ is the set of continuous group homomorphisms $Z \to T$ , with pointwise multiplication and the compact-open topology.

[F2]

The integers as equivalence classes of pairs of naturals, Arithmetic on the integers, The integers form a commutative ring, and The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies: the integers form an additive group, and the discrete topology makes every subset open; every map from a discrete space is continuous.

[F3]

Monoid homomorphism and group homomorphism: a group homomorphism preserves multiplication.

[F4]

Integer powers in the complex field and Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute: for $z \neq 0$ , integer powers satisfy $z^{m + n} = z^{m} z^{n}$ and $z^{- n} = (z^{n})^{- 1}$ .

[F5]

The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane and 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: $T$ carries the metric and topology inherited from $C$ .

[F6]

$C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ and Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive: multiplication on $C$ is commutative, $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ , $z \overline{z} = ∣ z ∣^{2}$ , and conjugation is an involution. Hence $∣ \overline{z} ∣^{2} = \overline{z} z = z \overline{z} = ∣ z ∣^{2}$ , and nonnegative moduli give $∣ \overline{z} ∣ = ∣ z ∣$ . Thus for $∣ z ∣ = 1$ , $z \overline{z} = 1$ and $∣ \overline{z} ∣ = 1$ , so the unit circle is an abelian group under multiplication with inverse $z^{- 1} = \overline{z}$ .

[F7]

On a discrete domain the compact-open topology is the topology of pointwise convergence and The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ : when the domain is discrete, the compact-open topology on a family of functions is the subspace topology from the product $T^{Z}$ .

[F8]

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 and Basis and subbasis for a topology, and the topology generated by a family of sets: product basic open sets restrict only finitely many coordinates, and the projections are continuous.

[F9]

Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological: a homeomorphism is a continuous bijection with continuous inverse.

[F10]

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

[F11]

$\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential and $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ : the complex exponential satisfies $\exp (u + v) = \exp (u) \exp (v)$ and $\exp (i λ) = \cos λ + i \sin λ$ .

[F12]

$t \mapsto (\cos t, \sin t)$ is a bijection from $[0, 2 π)$ onto the real unit circle: under $z = x + i y \leftrightarrow (x, y)$ , every point of $T$ has the form $\exp (i λ)$ for a unique $λ \in [0, 2 π)$ .

Proof

Proof technique: Identify each character by its value at $1$ , then verify continuity in the pointwise topology.

1.1F1F2F3F4

For $z \in T$ , define $χ_{z} (n) : = z^{n}$ . By [F4], $χ_{z} (m + n) = z^{m + n} = z^{m} z^{n} = χ_{z} (m) χ_{z} (n)$ , so $χ_{z}$ is a group homomorphism by [F3]. It is continuous because its domain $Z$ is discrete by [F2]. Hence $χ_{z} \in \hat{Z}$ by [F1], and $Ψ (z) = χ_{z}$ is well defined.

1.2F5F6F8algebra

Multiplication on $T$ is continuous. For $z, w, z_{0}, w_{0} \in T$ , $∣ z w - z_{0} w_{0} ∣ \leq ∣ z ∣ ∣ w - w_{0} ∣ + ∣ w_{0} ∣ ∣ z - z_{0} ∣ = ∣ w - w_{0} ∣ + ∣ z - z_{0} ∣ .$ Thus a product-neighborhood restricting both coordinates within $ε / 2$ of $(z_{0}, w_{0})$ maps inside the $ε$ -ball about $z_{0} w_{0}$ . Inversion is continuous because $z^{- 1} = \overline{z}$ by [F6] and $∣ \overline{z} - \overline{w} ∣ = ∣ z - w ∣$ by conjugation preserving modulus [F6].

2.1F1F2F3F4F6step 1.1

The map $Ψ$ is a group homomorphism: for $z, w \in T$ and $n \in Z$ , $(z w)^{n} = z^{n} w^{n}$ because $T$ is abelian, by [F4] and [F6]. It is injective since $Ψ (z) (1) = z$ . If $χ \in \hat{Z}$ , put $z : = χ (1) \in T$ . First $χ (0) = χ (0)^{2}$ ; since $χ (0) \in T$ is invertible, $χ (0) = 1$ . Repeatedly applying the homomorphism law then gives $χ (n) = z^{n}$ for $n \geq 0$ . Thus for $n > 0$ , $χ (n) χ (- n) = χ (0) = 1$ , so $χ (- n) = z^{- n}$ ; hence $χ = Ψ (z)$ and $Ψ$ is surjective.

2.2F5F7F8F10step 1.2

Each coordinate map $z \mapsto Ψ (z) (n) = z^{n}$ is continuous: for $n = 0$ it is constant; for $n > 0$ it is a finite composite of the continuous multiplication map from step 1.2 and the identity; for $n < 0$ it is a finite composite followed by the continuous inverse map from step 1.2. Composite continuity follows from [F10]. A basic open set in the product $T^{Z}$ restricts only finitely many coordinates by [F8], so its preimage is a finite intersection of open coordinate preimages. Thus $Ψ$ is continuous into the product, and [F7] gives continuity into the subspace $\hat{Z}$ .

3.1F7F8F9step 2.1

The inverse map sends $χ$ to $χ (1)$ . Evaluation at coordinate $1$ is continuous as the restriction of a product projection by [F7]–[F8], and $Ψ (χ (1)) = χ$ by step 2.1. Thus the inverse is continuous; $Ψ$ is a homeomorphism by [F9], and with step 2.1 it is the asserted group isomorphism.

4.1F4F11F12step 1.1∎

By [F12], every $z \in T$ has a unique angle $λ$ modulo $2 π$ with $z = e^{i λ}$ , using [F11]. The exponential addition law [F11] and integer-power laws [F4] give $Ψ (e^{i λ}) (n) = (e^{i λ})^{n} = e^{i n λ}$ for every $n \in Z$ .

Depends on

Used by

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Sources

Dikran D. Dikranjan, Introduction to Topological Groups, Section 7.1, Example 7.1(2) (standard reference, not scraped)
Lynn H. Loomis, Introduction to Abstract Harmonic Analysis, Chapter VII, Section 35E (standard reference, not scraped)