The Pontryagin dual of a compact abelian group is discrete

Question

Suppose $G$ is a compact abelian group. Show that the Pontryagin dual $\hat{G}$ is discrete.

I came up with this exercise when I read an introduction to Fourier analysis on locally compact abelian groups. A quick search on the site does not return any result. I walked through my thoughts on this exercise, but I ended up with a satisfactory proof. So I put it as an answer below. Any alternative approach/reference will be welcome.

Answer 1

The first step one is supposed to do is unwrapping all the related definitions:

• A map $\chi:G\to S^1$ is called a character of $G$ if it is a group homomorphism: $$ \chi(g_1g_2) = \chi(g_1)\chi(g_2),\quad \ g_1,g_2\in G\;, $$ and it is continuous. Here $S^1$ donotes the circle group.

• One can check that the (pointwise) product of two characters is again a character; the set $\hat{G}$ (together with the product) of all characters of $G$ is a group.

• One introduces a topology on $\hat{G}$ by defining the neighborhoods of a given $\chi_0\in\hat{G}$ as follows. Let $K\subset G$ be a compact set and $\epsilon>0$. Then set the neighborhood $$ V_{K,\epsilon} = \{\chi \in \hat{G}:\sup_{g\in K}|\chi(g)-\chi_0(g)|<\epsilon\}\;. $$ So this is essentially the topology of uniform convergence on compact sets.

$\def\gd{\hat{G}} $ To show that $\gd$ is discrete, it suffices, by definition, to show that given any $\chi_0\in\gd$, $\{\chi_0\}$ is a neighborhood of $\chi_0$. So I should look for a compact subset $K$ of $G$ and $\epsilon$, such that $V_{K,\epsilon}=\{\chi_0\}$. Since by assumption $G$ is compact, we can take $K=G$. Recall that our goal now is getting $$ V_{G,\epsilon}(\chi_0) =\{\chi_0\} \;. $$ If we have the following implication, this would be done:

if $|\chi(g)-\chi_0(g)|<\epsilon$ for all $g\in G$, then $\chi=\chi_0$.

Observe that $$ |\chi(g)-\chi_0(g)| = \bigg|\chi_0(g)\big((\chi_0^{-1}\chi)(g)-1_{\gd}(g)\big)\bigg| =\big|(\chi_0^{-1}\chi)(g)-1\big|\;. $$

So we can further reduce the problem to the following (so that we can take $\epsilon=1$ above):

let $\chi$ be a character of $G$ such that $|\chi(g)-1|\leq 1$ for every $g\in G$. Then $\chi = 1_{\gd}$, i.e., $\chi(g)=1$ for every $g\in G$.

This can be done through proof by contradiction. Suppose $\chi(g)\neq 1$. On the one hand, $\chi(g^n) = (\chi(g))^n$ for every positive integer $n$ since $\chi$ is a homomorphism. On the other hand, for sufficiently large $n$, one must have $|(\chi(g))^n-1|>1$ since $\chi(g)\neq 1$; this can be seen by the geometry of the circle: multiplication is essentially a rotation.

Answer 2

For convenience I will write the circle group as $S=\mathbf{R}/3\mathbf{Z}$, with its geodesic distance $d_S(s,s')=|s-s'|\in [0,3/2]$, where $|s|=\min_t|t|$ where $t\in\mathbf{R}$ ranges over lifts of $s$. (Here $3$ could be replaced by any real $r\ge 3$.)

By definition, we have $\hat{G}=\mathrm{Hom}(G,S)$, endowed with additive group law $(u+v)(g)=u(g)+v(g)$, and the topology induced by the distance $d(u,v)=\|u-v\|$, where $\|u\|=\sup_{g\in G}|u(g)|$ (it induces uniform convergence, which coincide with uniform convergence on compact subsets).

Let $\pi:\mathbf{R}\to\mathbf{R}/3\mathbf{Z}$ be the projection. Let $\pi'$ be its restriction to $]-1,1[$: then $\pi'$ is injective: let $j:\pi(]-1,1[)\to\mathopen]-1,1[$ be its inverse. Then ($\sharp$) $j$ is a partial group homomorphism: $j(x+y)=j(x)+j(y)$ for all $x,y\in\pi(]-1,1[)$ such that $x+y\in \pi(]-1,1[)$.

(While $j$ could have been defined on $\pi([-1,1])$, the above partial additivity would fail: indeed we have in $\mathbf{R}/3\mathbf{Z}$ the equality $\bar{1}+\bar{1}=\overline{-1}$ but $1+1\neq -1$.)

For $u\in \hat{G}$ such that $|u|<1$, $u$ is valued in $\pi(]-1,1[)$, and hence, by ($\sharp$), $j\circ\pi$ is a continuous homomorphism $G\to\mathbf{R}$. Its image is a bounded subgroup of $\mathbf{R}$, hence reduced to $\{0\}$, so $u=0$. This proves that $\hat{G}$ is discrete.

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In general, if $G$ is an arbitrary topological group and we consider the group of continuous homomorphisms $G\to S^1$ with uniform convergence topology, this proves that this topology is discrete (note that in this case the "norm" $\|\cdot\|$ takes infinite values).