# Characters and the dual group

**Lesson HA-LCA-02.** A character records a frequency without choosing coordinates on a group. This lesson constructs the topology on the set of characters, computes the classical examples, and proves local compactness directly. Every group below is Hausdorff, locally compact and abelian unless another hypothesis is stated.

The freely accessible sources are D. H. Fremlin, [*Measure Theory*, §445A–B, version of 20 March 2008](https://www1.essex.ac.uk/maths/people/fremlin/mt4.2013/mt445.tex), and Terence Tao, [245C, Notes 2, Definition 8 and Exercises 9–11](https://terrytao.wordpress.com/2009/04/06/the-fourier-transform/), dated 6 April 2009. Source hints and elementary auxiliary claims are proved below. The earlier programme proofs we use are [compactness and elementary calculus](../prerequisites/src/banach-spectrum.md), [compact cutoffs and representation](../prerequisites/src/finite-radon-representation.md), [uniform approximation](../prerequisites/src/uniform-approximation.md), [integration](../prerequisites/src/integration-and-l1.md), and [Haar measure](../prerequisites/src/haar-measure.md).

Adaptation and new exposition: GPT-6 Astra (OpenAI), Ultra, October 2026; checked by its author. This combined lesson is under the [Design Science License](../assets/fremlin/DESIGN-SCIENCE-LICENSE.txt). Fremlin's original copyright 1998 and source notices are retained in the original source package.

## 1. The topology of frequencies

Write \(\mathbb T=\{z\in\mathbb C:|z|=1\}\), with multiplication. A **character** of \(G\) is a continuous homomorphism \(\gamma:G\to\mathbb T\). Their set is \(\widehat G\), with pointwise multiplication and inverse \(\gamma^{-1}=\overline\gamma\). The identity character is denoted \(1\). For compact \(K\subseteq G\), put
\[
 d_K(\gamma,\eta)=\sup_{x\in K}|\gamma(x)-\eta(x)|,
 \qquad d_\varnothing=0.
\]
The topology of **compact convergence** is generated by these pseudometrics. Thus a neighbourhood basis at \(1\) consists of
\[
 N(K,\varepsilon)=\{\gamma:d_K(\gamma,1)<\varepsilon\}.
\]
Finite intersections are refined by using the union of the compact sets and the smallest tolerance.

<a id="ha-lca-02-proposition-1-1"></a>
### Proposition 1.1. The dual is a Hausdorff topological group

With compact convergence, \(\widehat G\) is a Hausdorff abelian topological group.

**Proof.** Pointwise products and conjugates of continuous homomorphisms are continuous homomorphisms; the group identities hold at each point. For complex numbers of modulus one, the triangle inequality gives
\[
 d_K(\gamma\eta,\gamma_0\eta_0)
 \le d_K(\gamma,\gamma_0)+d_K(\eta,\eta_0),\qquad
 d_K(\gamma^{-1},\gamma_0^{-1})=d_K(\gamma,\gamma_0).
\]
These prove continuity of multiplication and inversion at every point. If \(\gamma\ne\eta\), choose \(x\) where their values differ. The pseudometric for the compact singleton \(\{x\}\) separates them by disjoint sufficiently small balls. \(\square\)

<a id="ha-lca-02-lemma-1-2"></a>
### Lemma 1.2. Small powers on the circle

If \(z\in\mathbb T\) satisfies \(|z^j-1|\le1\) for every positive integer \(j\), then \(z=1\). In particular a subgroup contained in \(\{|z-1|<1\}\) is trivial.

If \(n\ge1\) and \(|z^j-1|\le1/2\) for \(1\le j\le n\), then
\[
 |z-1|\le \frac1n.                                    \tag{1.1}
\]

**Proof.** In the first assertion \(|z^j-1|^2=2-2\Re(z^j)\), so \(\Re(z^j)\ge1/2\). If \(z\ne1\), the geometric sum gives
\[
 \left|\frac1n\sum_{j=1}^n z^j\right|
 \le\frac{2}{n|1-z|}\longrightarrow0,
\]
contradicting its real part being at least \(1/2\). This applies to every element of the stated subgroup and all its positive powers.

For the second assertion, \(\Re(z^j)\ge1-|z^j-1|\ge1/2\) for \(1\le j<n\); the term \(j=0\) has real part one. Therefore \(|\sum_{j=0}^{n-1}z^j|\ge n/2\). Multiplying this sum by \(z-1\) gives \(z^n-1\), whose modulus is at most \(1/2\). Division by the lower bound proves (1.1). \(\square\)

<a id="ha-lca-02-lemma-1-3"></a>
### Lemma 1.3. Circle parametrization and the quotient \(\mathbb R/\mathbb Z\)

With \(2\pi=\tau\) and \(e(t)=\exp(it)\) as constructed in the earlier Banach reading, Lemma 1.3, the map
\[
 E(t)=e(2\pi t)
\]
is a surjective continuous homomorphism \(\mathbb R\to\mathbb T\) with kernel \(\mathbb Z\). It induces a topological group isomorphism \(\mathbb R/\mathbb Z\cong\mathbb T\). Also
\[
 |e(t)-e(s)|\le|t-s|.                                 \tag{1.2}
\]

**Proof.** In the earlier construction, \(\alpha=\tau/4\) is the first positive zero of \(c(t)=\Re e(t)\); \(c>0\) on \([0,\alpha)\), and \(e(\alpha)=i\). The function \(s(t)=\Im e(t)\) has derivative \(c(t)\), so is strictly increasing from zero to one on \([0,\alpha]\): the difference between two values is the integral of \(c\), positive on the interior of that interval. The intermediate-value theorem from the earlier Lemma 1.1 shows that every point of the closed first quadrant of the circle occurs exactly once on this arc. Its real part is the nonnegative square root of \(1-s(t)^2\). The identity \(e(t+\alpha)=ie(t)\) gives the other three quadrants. With half-open arcs, every circle point occurs exactly once for \(0\le t<\tau\).

For any real \(t\), choose the integer \(k\) with \(k\tau\le t<(k+1)\tau\). Periodicity and the preceding uniqueness show \(e(t)=1\) exactly when \(t=k\tau\). This proves the assertions about \(E\). The quotient map \(q:\mathbb R\to\mathbb R/\mathbb Z\) is open, since \(q^{-1}(q(U))=\bigcup_{k\in\mathbb Z}(U+k)\) is open when \(U\) is open. Its induced bijection to \(\mathbb T\) is continuous by the definition of the quotient topology. The quotient is compact because it is the image of \([0,1]\), and \(\mathbb T\) is Hausdorff. A continuous bijection from a compact space to a Hausdorff space has continuous inverse: a closed subset of the domain is compact, and its image is compact and therefore closed. Thus the bijection is a homeomorphism.

Finally \(e'=ie\), with \(|e|=1\). The fundamental theorem and norm bound for the continuous Riemann integral, proved in the earlier Lemma 1.2, give (1.2). \(\square\)

<a id="ha-lca-02-proposition-1-4"></a>
### Proposition 1.4. Evaluation and pullback

The pairing
\[
 G\times\widehat G\longrightarrow\mathbb T,\qquad
 (x,\gamma)\longmapsto\gamma(x)
\]
is jointly continuous. If \(T:G\to H\) is a continuous homomorphism, its pullback \(T^\wedge:\widehat H\to\widehat G\), defined by \(T^\wedge(\eta)=\eta\circ T\), is a continuous homomorphism.

**Proof.** Fix \((x_0,\gamma_0)\) and a compact neighbourhood \(K\) of \(x_0\). For \(x\in K\),
\[
 |\gamma(x)-\gamma_0(x_0)|
 \le d_K(\gamma,\gamma_0)+|\gamma_0(x)-\gamma_0(x_0)|.
\]
Choose \(\gamma\) close on \(K\), and choose \(x\) in the interior of \(K\) close enough to \(x_0\) for the second term to be small. This proves joint continuity. For pullback, \(T(K)\) is compact, and
\(d_K(\eta\circ T,\eta_0\circ T)=d_{T(K)}(\eta,\eta_0)\).
The homomorphism identity is pointwise. \(\square\)

## 2. Compactness and discreteness

<a id="ha-lca-02-theorem-2-1"></a>
### Theorem 2.1. Compact and discrete groups exchange roles

If \(G\) is compact, then \(\widehat G\) is discrete. If \(G\) is discrete, then \(\widehat G\) is compact.

**Proof.** In the compact case, \(N(G,1)=\{1\}\) by Lemma 1.2, since the image of each character is a subgroup. Thus the identity is open, and translations in the dual make every singleton open.

In a discrete space, compact sets are finite: the cover by all singletons has a finite subcover only for a finite set. Conversely finite sets are compact. Compact convergence is therefore pointwise convergence, the subspace topology from \(\mathbb T^G\). Every homomorphism on a discrete group is continuous. The conditions
\[
 \gamma(0)=1,\qquad \gamma(x+y)=\gamma(x)\gamma(y)
 \quad(x,y\in G)
\]
are closed conditions in the product, so their simultaneous solution set is closed. The circle is compact by the earlier finite-dimensional compactness proof, and its arbitrary product is compact by the earlier Banach reading, Lemma 4.1. A closed subset of that product is compact. \(\square\)

## 3. The line, the integers and the circle

<a id="ha-lca-02-theorem-3-1"></a>
### Theorem 3.1. All continuous characters of the real line

Every continuous homomorphism \(\chi:\mathbb R\to\mathbb T\) is
\[
 \chi_\xi(x)=e^{2\pi i\xi x}
\]
for a unique \(\xi\in\mathbb R\). The map \(\xi\mapsto\chi_\xi\) is a topological group isomorphism \(\mathbb R\to\widehat{\mathbb R}\).

**Proof.** Continuity at zero gives a strictly positive \(\delta\) such that \(\Re\chi(t)>0\) whenever \(|t|\le\delta\). The quadrant parametrization in Lemma 1.3 gives a unique \(\theta\in(-\pi/2,\pi/2)\) with \(\chi(\delta)=e^{i\theta}\). We claim
\[
 \chi(2^{-n}\delta)=e^{i2^{-n}\theta}\quad(n\ge0).       \tag{3.0}
\]
The case \(n=0\) is the choice of \(\theta\). If it holds for \(n\), then the homomorphism law makes \(\chi(2^{-n-1}\delta)\) a square root of \(e^{i2^{-n}\theta}\). The two possible roots are \(e^{i2^{-n-1}\theta}\) and its negative: dividing any root by the first gives a number whose square is one, hence \(1\) or \(-1\). The first root has positive real part, and the second negative real part. Our choice of \(\delta\) selects the first. This proves (3.0) by induction.

Integer multiples and inverses in the homomorphism law now give
\(\chi(k2^{-n}\delta)=e^{ik2^{-n}\theta}\) for every integer \(k\). These points are dense in \(\mathbb R\): for each \(x\), the integer \(k_n=\lfloor 2^n x/\delta\rfloor\) satisfies
\(0\le x-k_n2^{-n}\delta<2^{-n}\delta\).
Continuity on both sides therefore gives
\(\chi(x)=e^{i(\theta/\delta)x}=E(\xi x)\), where \(\xi=\theta/(2\pi\delta)\).

Conversely every displayed function is a continuous homomorphism by Lemma 1.3. If \(\chi_\xi=\chi_\eta\) and \(\xi\ne\eta\), evaluating at \(x=1/(2(\xi-\eta))\) would give \(-1=1\); this proves uniqueness.

For \(|x|\le R\), (1.2) gives
\[
 |\chi_\xi(x)-\chi_\eta(x)|\le2\pi R|\xi-\eta|,
\]
so the parametrization is continuous for compact convergence. For its inverse, fix \(\varepsilon>0\). If \(|\xi|\ge\varepsilon\), then \(x=1/(2\xi)\) lies in the compact interval \([-1/(2\varepsilon),1/(2\varepsilon)]\) and \(\chi_\xi(x)=-1\). Thus the neighbourhood defined by being within distance one of the identity on that interval forces \(|\xi|<\varepsilon\). This proves inverse continuity at zero, and the group law gives it everywhere. \(\square\)

<a id="ha-lca-02-corollary-3-2"></a>
### Corollary 3.2. The integer, circle and finite cyclic duals

There are topological group isomorphisms
\[
 \widehat{\mathbb Z}\cong\mathbb T,\quad
 \widehat{\mathbb T}\cong\mathbb Z,\quad
 \widehat{\mathbb Z/N\mathbb Z}\cong\mathbb Z/N\mathbb Z
 \quad(N\ge1).
\]
Their pairings are respectively \(z^n\), \(z^n\), and \(e^{2\pi i kn/N}\).

**Proof.** A homomorphism on \(\mathbb Z\) is determined by \(z=\chi(1)\), and is \(\chi(n)=z^n\). Every such map is continuous because \(\mathbb Z\) is discrete. Evaluation at \(1\) is continuous in the dual topology, and the inverse map \(z\mapsto(n\mapsto z^n)\) is continuous on every finite set of integers. This proves the first topological identification.

For a circle character \(\chi\), the function \(t\mapsto\chi(E(t))\) is a real-line character, hence \(E(\xi t)\). Its period one implies \(E(\xi)=1\), so \(\xi=n\in\mathbb Z\) by Lemma 1.3. Surjectivity of \(E\) now gives \(\chi(z)=z^n\). These characters are distinct, and \(\widehat{\mathbb T}\) is discrete by Theorem 2.1; hence the algebraic identification with discrete \(\mathbb Z\) is a homeomorphism.

On \(\mathbb Z/N\mathbb Z\), the value at \(1\) must satisfy \(z^N=1\). Lemma 1.3 writes \(z=E(t)\) and shows \(Nt\in\mathbb Z\); exactly the \(N\) values \(E(k/N)\), \(0\le k<N\), occur. They all define characters with the stated formula. The dual is discrete by Theorem 2.1 because the finite group is compact, so this is also a topological identification. \(\square\)

<a id="ha-lca-02-lemma-3-3"></a>
### Lemma 3.3. Lebesgue measure and its normalization

There is a unique Borel measure \(\lambda\) on \(\mathbb R\) with
\(\lambda((a,b])=b-a\) for \(a<b\). Its ordinary completion is Lebesgue measure. It is a Haar measure and satisfies, for \(s\ne0\),
\[
 \lambda(sA+b)=|s|\lambda(A),\qquad
 \int_{\mathbb R}h(sx+b)\,d\lambda(x)
   =|s|^{-1}\int_{\mathbb R}h(t)\,d\lambda(t)              \tag{3.1}
\]
for measurable \(A\) and nonnegative measurable or integrable \(h\).

**Proof.** The earlier [Haar reading, Theorems 4.1 and 5.1](../prerequisites/src/haar-measure.md#ha-lca-pre-haar-theorem-4-1), supplies a nonzero translation-invariant measure, finite on compact sets and positive on nonempty open sets. Since \(\mathbb R\) is a countable union of compact intervals, its completed domain is the ordinary completion of its Borel domain.

Every singleton has the same mass \(c\). Arbitrarily many distinct points fit in \([-1,1]\), so \(nc\le\mu([-1,1])<\infty\) for every \(n\); hence \(c=0\). The number \(\mu((0,1])\) is finite and positive. Divide the measure by that number. Partitioning \((0,1]\) into \(q\) translates of \((0,1/q]\) gives mass \(1/q\) for the latter interval, and then mass \(p/q\) for \((0,p/q]\). Monotonicity and rational upper and lower approximations to a positive real \(t\) give \(\mu((0,t])=t\). Translation gives all the asserted interval values.

These values determine the Borel measure. On each finite window \((-n,n]\), the intersections with half-open intervals, together with the empty set and the window, form a family closed under intersections that generates its Borel sets. Indeed open intervals, and then all open sets, are countable unions of intervals with rational endpoints of this kind, after intersection with the window. The measures restricted to the window are finite and have equal total mass. The proved \(\pi\)-\(\lambda\) argument in the earlier [integration reading, Lemma 4.1](../prerequisites/src/integration-and-l1.md#ha-lca-pre-integral-lemma-4-1), therefore makes the restrictions equal. Increasing the windows proves uniqueness on \(\mathbb R\).

For \(s\ne0\), the Borel measure \(A\mapsto |s|^{-1}\lambda(sA+b)\) has the same interval values; when \(s<0\), the possible change of included endpoint has zero mass. Uniqueness proves the first formula on Borel sets. Affine maps take Borel null sets to Borel null sets, so the formula extends to the completion. Applying it to indicators, then finite simple functions, and then increasing simple approximations proves the integral formula for nonnegative functions. Decomposing an integrable function into real and imaginary positive and negative parts gives the remaining case. \(\square\)

<a id="ha-lca-02-example-3-4"></a>
### Example 3.4. Two coordinates for the same frequency

Using \(e^{isx}\) in place of \(e^{2\pi i\xi x}\) amounts to the change of coordinate \(s=2\pi\xi\) on \(\widehat{\mathbb R}\). These are the same characters. Lebesgue measure \(d\xi\) in the first frequency coordinate becomes \(ds/(2\pi)\) in the second.

**Verification.** Theorem 3.1 identifies the character in either notation, and \(2\pi\xi=s\). Formula (3.1), applied to \(h(s)=F(s/(2\pi))\), gives
\[
 \int_{\mathbb R}F(\xi)\,d\xi
   =\frac1{2\pi}\int_{\mathbb R}F(s/(2\pi))\,ds .
\]
Thus any later integral formula using this frequency measure changes by precisely that factor. No Fourier inversion theorem is assumed here. \(\square\)

## 4. Products and sums

<a id="ha-lca-02-proposition-4-1"></a>
### Proposition 4.1. Finite products

For finitely many LCA groups \(G_1,\ldots,G_r\), restriction to coordinate axes gives a topological group isomorphism
\[
 \widehat{G_1\times\cdots\times G_r}
       \cong \widehat G_1\times\cdots\times\widehat G_r .
\]
Its inverse sends \((\gamma_1,\ldots,\gamma_r)\) to
\((x_1,\ldots,x_r)\mapsto\prod_{j=1}^r\gamma_j(x_j)\).

**Proof.** A finite product is again LCA: distinct points are separated in a differing coordinate, the group operations are continuous coordinatewise, and products of compact neighbourhoods are compact neighbourhoods by the earlier Radon reading, Lemma 1.6. Each coordinate-axis inclusion is continuous, so restriction gives a continuous homomorphism by Proposition 1.4. Every element of the product is the sum of its coordinate-axis elements. Applying a character to this sum proves the inverse formula, and the displayed product is a continuous character.

For completeness, continuity of the inverse follows directly on compact sets. If \(K\) is compact in the product and \(K_j\) is its compact projection to \(G_j\), telescoping products of unit complex numbers gives
\[
 \sup_{x\in K}\left|\prod_j\gamma_j(x_j)-\prod_j\eta_j(x_j)\right|
 \le\sum_j d_{K_j}(\gamma_j,\eta_j).
\]
Hence finitely many compact-convergence conditions on the coordinates control any such condition on the product. \(\square\)

<a id="ha-lca-02-corollary-4-2"></a>
### Corollary 4.2. Vector groups and tori

For every nonnegative integer \(r\),
\[
 \widehat{\mathbb R^r}\cong\mathbb R^r,\qquad
 \widehat{\mathbb Z^r}\cong\mathbb T^r,\qquad
 \widehat{\mathbb T^r}\cong\mathbb Z^r .
\]
The first pairing is \(e^{2\pi i\xi\cdot x}\); the other two are
\(\prod_{j=1}^r z_j^{n_j}\).

**Proof.** Apply Proposition 4.1 to the identifications in Theorem 3.1 and Corollary 3.2. Multiplying the scalar exponentials adds their exponents, giving the dot product. The empty product when \(r=0\) is the trivial group, whose sole character is the identity. \(\square\)

<a id="ha-lca-02-theorem-4-3"></a>
### Theorem 4.3. Arbitrary compact products and discrete direct sums

Let \((K_i)_{i\in I}\) be compact Hausdorff abelian groups. With the product topology on \(\prod_iK_i\) and the discrete topology on the direct sum,
\[
 \widehat{\prod_{i\in I}K_i}\cong
       \bigoplus_{i\in I}\widehat K_i .                 \tag{4.1}
\]
If \((D_i)_{i\in I}\) are discrete abelian groups and \(\bigoplus_iD_i\) is given the discrete topology, then
\[
 \widehat{\bigoplus_{i\in I}D_i}\cong
       \prod_{i\in I}\widehat D_i                      \tag{4.2}
\]
with the product topology on the right.

**Proof.** The compact product is compact by the previously proved arbitrary product theorem. It is Hausdorff because distinct points differ in some coordinate; coordinatewise continuity of the group operations makes it a compact abelian topological group. If \(\gamma\) is a continuous character on it, continuity at zero supplies a basic open cylinder, restricting only a finite set \(F\subseteq I\) of coordinates, on which \(|\gamma-1|<1\). The subgroup whose \(F\)-coordinates are all zero lies inside that cylinder. Its image is a subgroup contained in the small arc, so Lemma 1.2 makes this image trivial. Consequently \(\gamma\) depends only on its \(F\)-coordinates. Proposition 4.1 identifies it with a finite product of coordinate characters. This representation is unique by restriction to each coordinate axis. Conversely every such finite product is continuous. Both sides of (4.1) are discrete, the left by Theorem 2.1; the algebraic isomorphism is therefore a homeomorphism.

For (4.2), every element \(x=(x_i)\) of the direct sum has finite support. Any family \((\gamma_i)\) determines
\(\gamma(x)=\prod_{i:x_i\ne0}\gamma_i(x_i)\).
This is a homomorphism, automatically continuous on the discrete direct sum. Restriction to the summands recovers the family, proving a bijective homomorphism. The right side is compact: each factor is compact by Theorem 2.1, and arbitrary products preserve compactness. The displayed map to the left side is continuous, because evaluation at any \(x\) involves only finitely many coordinates, and compact convergence on a discrete group is pointwise convergence. The left side is Hausdorff by Proposition 1.1. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, as proved in Lemma 1.3. \(\square\)

<a id="ha-lca-02-example-4-4"></a>
### Example 4.4. The binary product

The continuous characters of the compact group
\(K=\prod_{j\ge1}\mathbb Z/2\mathbb Z\) are exactly
\[
 \gamma_F(x)=(-1)^{\sum_{j\in F}x_j},
 \qquad F\subseteq\mathbb N\text{ finite}.
\]
Its dual is the discrete group \(\bigoplus_{j\ge1}\mathbb Z/2\mathbb Z\).

**Verification.** Corollary 3.2 gives precisely two characters of each binary factor. Theorem 4.3 says a character of the product uses nonidentity characters in only finitely many factors, giving the formula. Distinct finite sets give distinct characters by evaluation on a coordinate where they differ. Multiplication corresponds to symmetric difference of the sets, which is addition in the direct sum. \(\square\)

<a id="ha-lca-02-example-4-5"></a>
### Example 4.5. A circle inside the two-dimensional torus

Let
\[
 H=\{(E(t),E(2t)):t\in\mathbb R\}\subseteq\mathbb T^2 .
\]
Then \(H\) is a closed subgroup isomorphic to \(\mathbb T\). Its annihilator, the set of characters of \(\mathbb T^2\) that equal one on \(H\), is
\[
 H^\perp=\{(-2n,n):n\in\mathbb Z\}\subseteq\mathbb Z^2 .
\]

**Verification.** The map \(z\mapsto(z,z^2)\) is a continuous injective homomorphism from the compact circle to \(\mathbb T^2\); its image is compact and closed, and the inverse on its image is the first-coordinate projection. By Corollary 4.2, the character indexed by \((m,n)\) restricts to
\(t\mapsto E((m+2n)t)\).
Theorem 3.1 says this is identically one exactly when \(m+2n=0\), giving the asserted integer subgroup.

More generally, for the image \(t\mapsto(E(pt),E(qt))\) with integers \(p,q\), the same computation gives the annihilator
\(\{(m,n)\in\mathbb Z^2:mp+nq=0\}\).
The image is closed because it is the image of the compact circle under \(z\mapsto(z^p,z^q)\). These conclusions do not require a classification theorem for closed subgroups. \(\square\)

## 5. Why the dual is locally compact

<a id="ha-lca-02-proposition-5-1"></a>
### Proposition 5.1. One compact neighbourhood controls equicontinuity

Fix a compact neighbourhood \(V\) of zero in \(G\). Let \(C_V\) be the family of all homomorphisms \(\gamma:G\to\mathbb T\), initially without any continuity assumption, such that
\[
 |\gamma(x)-1|\le\tfrac12\quad(x\in V).
\]
For every \(\varepsilon>0\), there is an open neighbourhood \(W\) of zero with
\[
 |\gamma(w)-1|<\varepsilon
 \quad(w\in W,\ \gamma\in C_V).                         \tag{5.1}
\]
In particular every member of \(C_V\) is continuous, and this family is equicontinuous at every point.

**Proof.** Choose an integer \(n\) with \(1/n<\varepsilon\), and an open neighbourhood \(U\) of zero contained in \(V\). For \(1\le j\le n\), the map \(x\mapsto jx\) is continuous, being a repeated sum. Thus
\(W=\bigcap_{j=1}^n\{x:jx\in U\}\)
is an open neighbourhood of zero. For \(w\in W\), all powers
\(\gamma(w)^j=\gamma(jw)\), \(1\le j\le n\), are within \(1/2\) of one. Lemma 1.2 gives
\(|\gamma(w)-1|\le1/n<\varepsilon\), uniformly in \(\gamma\).
This is continuity at zero for every member. The identity
\[
 |\gamma(x+w)-\gamma(x)|=|\gamma(w)-1|
\]
both extends continuity to every point and proves equicontinuity there. \(\square\)

<a id="ha-lca-02-theorem-5-2"></a>
### Theorem 5.2. A compact neighbourhood in the dual

The set \(C_V\) of Proposition 5.1 is a compact neighbourhood of the identity in \(\widehat G\). Hence the dual of every LCA group is again an LCA group.

**Proof.** Regard \(C_V\) first as a subset of the compact product \(\mathbb T^G\). The homomorphism equations are closed, as in Theorem 2.1, and the inequalities at all \(x\in V\) are closed coordinate conditions. Their common solution set is therefore compact in the pointwise topology. Proposition 5.1 shows that all its members are continuous characters.

We show that the pointwise and compact-convergence topologies coincide on this set. Fix compact \(K\subseteq G\) and \(\varepsilon>0\). By Proposition 5.1 choose an open neighbourhood \(W\) such that
\(|\gamma(w)-1|<\varepsilon/3\) for all \(w\in W\) and all \(\gamma\in C_V\).
Finitely many translates \(x_1+W,\ldots,x_r+W\) cover \(K\). If \(x=x_j+w\) belongs to such a translate, then for \(\gamma,\eta\in C_V\),
\[
 |\gamma(x)-\eta(x)|
 \le|\gamma(x)-\gamma(x_j)|
       +|\gamma(x_j)-\eta(x_j)|
       +|\eta(x_j)-\eta(x)|
 <2\varepsilon/3+|\gamma(x_j)-\eta(x_j)|.
\]
Thus closeness within \(\varepsilon/3\) at finitely many points ensures \(d_K(\gamma,\eta)<\varepsilon\). Compact-convergence neighbourhoods on \(C_V\) are consequently pointwise neighbourhoods. The reverse implication follows by using singleton compact sets. Compactness of \(C_V\) in the dual topology now follows from its product compactness.

Finally \(N(V,1/2)\subseteq C_V\) is a neighbourhood of the identity. Proposition 1.1 already proved that the dual is Hausdorff and is an abelian topological group. This compact neighbourhood proves local compactness. \(\square\)

<a id="ha-lca-02-lemma-5-3"></a>
### Lemma 5.3. Irrational rotations have dense positive orbits

If \(\alpha\) is irrational, then \(\{E(k\alpha):k\ge0\}\), and each of its tails, is dense in \(\mathbb T\).

**Proof.** For an integer \(Q\ge2\), put the \(Q+1\) fractional parts of \(k\alpha\), \(0\le k\le Q\), in the \(Q\) half-open intervals of length \(1/Q\) partitioning \([0,1)\). Two lie in the same interval. Subtracting their indices yields an integer \(q\), \(1\le q\le Q\), for which \(q\alpha\) differs from an integer by a number \(\delta'\) with
\(0<|\delta'|<1/Q\). The difference is nonzero because \(\alpha\) is irrational.

Write \(\delta=|\delta'|\). The points
\(0,\delta,2\delta,\ldots,\lfloor1/\delta\rfloor\delta\)
have successive gaps at most \(\delta\), including the gap back to one. Their circle images therefore lie within \(2\pi\delta\) of every circle point, by (1.2). If \(\delta'<0\), their reflected images have the same property. In either case these points occur among the positive multiples of \(q\alpha\), modulo integers. Since \(Q\) can be arbitrarily large, the positive orbit is dense. A tail beginning at index \(M\) is the rotation by \(E(M\alpha)\) of the entire positive orbit; this rotation is a homeomorphism, so the tail is dense too. \(\square\)

<a id="ha-lca-02-example-5-4"></a>
### Example 5.4. Pointwise convergence does not determine the dual topology

There is a net of real-line characters converging pointwise to \(1\) whose frequencies tend to \(+\infty\). It does not converge to \(1\) uniformly on compact sets. There is also a net of such escaping characters with a discontinuous homomorphism as a pointwise cluster point in \(\mathbb T^{\mathbb R}\).

**Verification.** First let \(F=\{x_1,\ldots,x_m\}\) be finite, let \(\varepsilon>0\), and let \(R\ge1\) be an integer. Choose an integer
\(Q>2\pi(R+1)/\varepsilon\).
The \(Q^m+1\) vectors
\[
 (\{kx_1\},\ldots,\{kx_m\}),\qquad 0\le k\le Q^m,
\]
fall into \(Q^m\) half-open cubes of side \(1/Q\). Subtracting two in the same cube gives an integer \(q\ge1\) such that each \(qx_j\) differs from an integer by less than \(1/Q\). Set
\(a=\lfloor R/q\rfloor+1\) and \(t=aq\).
Then \(t>R\), \(a\le R+1\), and (1.2) gives
\[
 |E(tx_j)-1|<2\pi(R+1)/Q<\varepsilon .
\]
For empty \(F\), any \(t>R\) works. Direct triples \((F,\varepsilon,R)\) by enlargement of \(F\), decrease of \(\varepsilon\), and increase of \(R\), and choose such a \(t\) for each triple. The resulting \(\chi_t\) converge pointwise to \(1\), while \(t\to+\infty\). Theorem 3.1 rules out compact convergence to \(1\), since that would require \(t\to0\). This proves that the two topologies on the character set are different.

For the cluster-point assertion, take \(\alpha=\sqrt2\). This number is irrational: a lowest-terms expression \(p/q\) would imply \(p^2=2q^2\), forcing \(p\) even and then \(q\) even, a contradiction. For each triple of positive integers \((n,m,R)\), Lemma 5.3 applied to \(n!\alpha\) allows a positive integer \(k\) with
\[
 t=n!k>R,\qquad |E(t\alpha)+1|<1/m .
\]
Order the triples coordinatewise. For every rational \(r=p/q\), the values \(E(tr)\) are eventually one, since \(q\) divides \(n!\) when \(n\ge q\). The values at \(\alpha\) tend to \(-1\).

In the compact space \(\mathbb T^{\mathbb R}\), the closures of the tails of this net have the finite-intersection property: a tail beyond the coordinatewise maximum lies in any specified finite collection of tails. Compactness therefore supplies a point \(\psi\) in their intersection. This is a pointwise cluster point. The closed homomorphism equations hold at every net term and hence at \(\psi\); similarly \(\psi(r)=1\) on \(\mathbb Q\) and \(\psi(\alpha)=-1\). If \(\psi\) were continuous, Theorem 3.1 would give \(\psi(x)=E(\xi x)\). The equalities at all \(1/N\) force \(\xi/N\) to be an integer for every positive integer \(N\); choosing \(N>|\xi|\) forces \(\xi=0\). That contradicts the value at \(\alpha\). Thus the cluster point is discontinuous. The assertion concerns a deliberately chosen net; it does not claim that every escaping net has only discontinuous pointwise cluster points. \(\square\)

## 6. Exercises with complete solutions

<a id="ha-lca-02-exercise-6-1"></a>
### Exercise 6.1. Circle endomorphisms

Classify the continuous homomorphisms \(\mathbb T\to\mathbb T\), and determine their kernels and images.

**Solution.** Corollary 3.2 proves that every such map is \(z\mapsto z^n\) for a unique integer \(n\). If \(n=0\), the kernel is all of \(\mathbb T\) and the image is \(\{1\}\). If \(n\ne0\), write \(z=E(t)\) by Lemma 1.3. Then \(z^n=1\) exactly when \(nt\in\mathbb Z\), so the kernel consists of the \(|n|\) distinct roots \(E(k/|n|)\), \(0\le k<|n|\). For any target \(E(s)\), the point \(E(s/n)\) maps to it. The image is therefore all of \(\mathbb T\). \(\square\)

<a id="ha-lca-02-exercise-6-2"></a>
### Exercise 6.2. Agreement with finite-group characters

Let \(F\) be a finite abelian group with its Hausdorff topology. Show that its topological characters are exactly its homomorphisms to \(\mathbb C^\times\), and that its dual is finite and discrete.

**Solution.** A finite Hausdorff space is discrete: each singleton is closed, so its complement, a finite union of closed singletons, is closed as well. Thus all maps from \(F\) are continuous. For every \(x\in F\), two of \(0,x,\ldots,|F|x\) coincide, so \(dx=0\) for some integer \(d\), \(1\le d\le |F|\). A homomorphism \(\rho:F\to\mathbb C^\times\) satisfies \(\rho(x)^d=1\), hence \(|\rho(x)|=1\). It is therefore a character in the present sense; the converse is immediate from \(\mathbb T\subset\mathbb C^\times\).

Let \(N=|F|!\). Since each such \(d\) divides \(N\), every character value is an \(N\)-th root of unity. Corollary 3.2 proved that there are precisely \(N\) such roots. Thus the dual is a subset of the finite set of all maps from \(F\) to these roots, and is finite. Finally \(F\) is compact, so Theorem 2.1 gives the discrete dual topology. This establishes agreement of the definitions without presupposing a decomposition of \(F\) into cyclic factors. \(\square\)

<a id="ha-lca-02-exercise-6-3"></a>
### Exercise 6.3. The compactness argument behind equicontinuity

Give the compactness proof for \(C_V\) directly, explaining why equicontinuity upgrades pointwise control to uniform control on every compact set.

**Solution.** Proposition 5.1 proves equicontinuity without assuming continuity of the initial homomorphisms. In the product \(\mathbb T^G\), impose the equations \(\gamma(0)=1\) and \(\gamma(x+y)=\gamma(x)\gamma(y)\), and impose \(|\gamma(v)-1|\le1/2\) for every \(v\in V\). Each equation or inequality specifies a closed set because it involves continuous coordinate maps. Their intersection \(C_V\) is closed in a compact product, hence compact pointwise. Every member is continuous by Proposition 5.1.

For compact \(K\) and \(\delta>0\), choose an open neighbourhood \(W\) on which every member differs from one by less than \(\delta/4\). Cover \(K\) by finitely many translates \(x_j+W\). If \(\gamma,\eta\in C_V\) and
\(|\gamma(x_j)-\eta(x_j)|<\delta/2\) at each centre, then
\[
 |\gamma(x)-\eta(x)|
 \le |\gamma(x_j+w)-\gamma(x_j)|
     +|\gamma(x_j)-\eta(x_j)|
     +|\eta(x_j)-\eta(x_j+w)|
 <\delta
\]
for \(x=x_j+w\). The finite maximum of the centre errors is strictly less than \(\delta/2\), so the same estimate bounds the supremum on \(K\) strictly below \(\delta\). This proves continuity of the inclusion from \(C_V\) with its pointwise topology to \(C_V\) with compact convergence. Since singleton evaluations are continuous in the latter topology, the two topologies agree. Hence \(C_V\) is compact in the dual. It contains \(N(V,1/2)\), completing the local-compactness argument. No sequential compactness assumption, metrizability assumption or unproved compactness criterion is used. \(\square\)

<a id="ha-lca-02-exercise-6-4"></a>
### Exercise 6.4. Measurable characters of the line

Prove that a Lebesgue-measurable homomorphism \(\chi:\mathbb R\to\mathbb T\) is continuous and therefore is \(x\mapsto E(\xi x)\) for a unique real \(\xi\).

**Solution.** Set \(g=\overline\chi\,\mathbf1_{[0,1]}\). By Lemma 3.3 this is an \(L^1\) function and
\(\int g\chi\,d\lambda=1\).
The earlier [integration reading, Theorem 3.1](../prerequisites/src/integration-and-l1.md#ha-lca-pre-integral-theorem-3-1), gives a continuous compactly supported \(f\) with \(\|f-g\|_1<1/2\). Thus
\(c=\int f\chi\,d\lambda\)
satisfies \(|c-1|<1/2\), so \(c\ne0\).

Translation invariance and the homomorphism identity give, for every \(x\),
\[
 \int_{\mathbb R} f(y-x)\chi(y)\,dy
   =\int_{\mathbb R}f(t)\chi(t+x)\,dt
   =c\chi(x).                                         \tag{6.1}
\]
All integrals exist because \(|\chi|=1\). The earlier [Haar reading, Lemma 6.1](../prerequisites/src/haar-measure.md#ha-lca-pre-haar-lemma-6-1), proves continuity of \(x\mapsto L_xf\) in \(L^1\), where \(L_xf(y)=f(y-x)\). Consequently the left side of (6.1) is continuous: its change between \(x\) and \(x_0\) is bounded by \(\|L_xf-L_{x_0}f\|_1\). Division by \(c\) proves continuity of \(\chi\), and Theorem 3.1 supplies its form and uniqueness. \(\square\)

<a id="ha-lca-02-exercise-6-5"></a>
### Exercise 6.5. A compact metrizable group has countable dual

Prove countability of \(\widehat K\) when \(K\) is a compact metrizable abelian group. Supply the separability argument needed for orthogonal characters.

**Solution.** Normalize Haar measure \(\mu\) so that \(\mu(K)=1\), using the earlier Haar existence and uniqueness theorem. Fix a metric \(d\) inducing the compact topology.

First \(C(K)\) is separable in the uniform norm. For each positive integer \(n\), compactness gives a finite \(1/n\)-net in \(K\); their union \(D\) is countable and dense. The real continuous functions \(x\mapsto d(x,a)\), \(a\in D\), separate points: for \(x\ne y\), choose \(a\in D\) with \(d(x,a)<d(x,y)/3\), and the triangle inequality gives \(d(y,a)>2d(x,y)/3\). The real algebra generated by these functions and constants is uniformly dense in \(C(K,\mathbb R)\), by the proved [uniform approximation reading, Theorem 2.1](../prerequisites/src/uniform-approximation.md#ha-lca-pre-approx-theorem-2-1). Its polynomials with rational coefficients in finitely many generators form a countable dense subset of this algebra: for any one polynomial, all its finitely many monomials are bounded on \(K\), so sufficiently close rational coefficients give any prescribed uniform accuracy. The resulting countable family is dense in all of \(C(K,\mathbb R)\). Taking its pairs as real and imaginary parts proves separability of \(C(K)\).

We next prove that \(C(K)\) is dense in \(L^2(\mu)\), where functions agreeing almost everywhere are identified and
\(\|f\|_2=(\int|f|^2\,d\mu)^{1/2}\).
The inequalities used here require no completeness theorem for \(L^2\). For \(f,g\in L^2\), the product \(f\overline g\) is integrable because \(2|fg|\le |f|^2+|g|^2\). If \(\|g\|_2>0\), substituting
\(a=(\int f\overline g\,d\mu)/\|g\|_2^2\)
in \(0\le\int|f-ag|^2\,d\mu\) gives
\[
 \left|\int f\overline g\,d\mu\right|\le\|f\|_2\|g\|_2 .
\]
If \(\|g\|_2=0\), then \(g=0\) almost everywhere, and the same inequality follows. Expanding \(\|f+g\|_2^2\) and using this inequality proves
\(\|f+g\|_2\le\|f\|_2+\|g\|_2\).
The other norm properties follow from the integral; zero norm means equality almost everywhere by the earlier integration reading, Corollary 1.3.

For \(f\in L^2\), the bounded truncations \(f\mathbf1_{\{|f|\le N\}}\) approach \(f\) in \(L^2\) by dominated convergence applied to the squared errors, dominated by \(|f|^2\). Each bounded measurable complex function can be uniformly approximated by finite measurable simple functions, by dividing its bounded real and imaginary ranges into intervals of small length. Because \(\mu(K)=1\), uniform error bounds also bound \(L^2\) error.

For each completed measurable \(A\subseteq K\) and \(\varepsilon>0\), replace \(A\) by a Borel representative modulo a null set. Compact inner regularity and outer regularity supply a compact \(F\) and open \(U\) with
\(F\subseteq A_{\mathrm B}\subseteq U\) and \(\mu(U\setminus F)<\varepsilon\).
These regularity properties on compact \(K\) are proved in the earlier Haar reading, Theorem 4.1; the Borel representative is supplied by the integration reading, Lemma 1.1. The earlier [Radon reading, Lemma 1.3](../prerequisites/src/finite-radon-representation.md#ha-lca-pre-radon-lemma-1-3), gives \(u\in C(K)\) with \(0\le u\le1\), \(u=1\) on \(F\) and \(u=0\) outside \(U\). Thus
\[
 \|\mathbf1_A-u\|_2^2\le \mu(U\setminus F)<\varepsilon .
\]
Finite linear combinations and the proved triangle inequality approximate every simple function by continuous functions in \(L^2\). The preceding truncation and simple approximations therefore prove density. A countable uniformly dense subset of \(C(K)\) is also \(L^2\)-dense there, since \(\|h\|_2\le\|h\|_\infty\); it is consequently dense in \(L^2\). This proves the needed separability.

Finally characters have \(L^2\) norm one. If \(\gamma\ne\eta\), choose \(a\in K\) with \(\gamma(a)\overline{\eta(a)}\ne1\). Translation invariance gives
\[
 \int_K\gamma(x)\overline{\eta(x)}\,d\mu(x)
 =\gamma(a)\overline{\eta(a)}
       \int_K\gamma(x)\overline{\eta(x)}\,d\mu(x).
\]
The integral is zero. Expanding the squared norm gives
\(\|\gamma-\eta\|_2=\sqrt2\).
Balls of radius \(1/3\) about a countable dense subset cover \(L^2\), and no such ball contains two characters, since its diameter is at most \(2/3<\sqrt2\). Assign each character the first ball containing it; this is an injection into a countable set. Thus \(\widehat K\) is countable. \(\square\)

## 7. What comes next

The arguments above already show that the dual of the discrete group \(\mathbb Q\) is compact. Computing that compact group explicitly, and describing the duals of \(p\)-adic integers and Prüfer groups, are later tasks in the course. The present lesson does not use those identifications.

The next step is to connect a frequency \(\gamma\) with the functional
\(f\mapsto\int_G f(x)\gamma(x)\,d\mu(x)\)
on the convolution algebra \(L^1(G)\). The dual topology just proved will allow that connection to be made for arbitrary nets and arbitrary LCA groups, without a countability assumption.