Why is the Pontraygin dual a locally compact group?

Question

Let $G$ be a locally compact topological group, and let $\hat{G}$ be the set of continuous characters $G \rightarrow S^1$. We give $\hat{G}$ the topology for which a basis of open sets is $$T(\epsilon, \chi_0, K) = \{ \chi \in \hat{G} : |\chi(x) - \chi_0(x)| < \epsilon, \forall x \in K\}$$ for $\epsilon > 0, \chi_0 \in \hat{G},$ and $K$ compact. I know how to prove that $\hat{G}$ is a Hausdorff topological group, but I'm having trouble proving that it is locally compact. Someone posted an answer here Prove that the Pontryagin dual of a locally compact abelian group is also a locally compact abelian group., but I didn't understand it. The hint in the notes I'm reading (Terrence Tao's online notes) is that if $U$ is any neighborhood of the identity in $G$, then for small $\epsilon > 0$, $$\{ \chi \in \hat{G} : |\chi(x) - 1| \leq \epsilon, \forall x \in U\}$$ is compact. I tried supposing such a set was contained in a union of basis elements $T(\epsilon_i, \chi_i, K_i)$, but I didn't get anywhere.

Answer 1

I'm not sure whether directly trying to show that every open cover of the given set has a finite subcover leads to anything. I remember running into one wall after the other for a long time trying to prove that $\hat{G}$ is locally compact too.

I think it's best to take a small detour. Consider the space $X = (S^1)^G$ of all maps $G \to S^1$. We endow it with the product topology, or equivalently the topology of pointwise convergence. By Tíkhonov's theorem, $X$ is compact. In $X$, consider the subspace $\tilde{G}$ of characters (continuous or not) on $G$. Since

$$\tilde{G} = \bigcap_{x,y \in G} \{ f\in X : f(x)f(y) = f(xy)\},$$

we see that $\tilde{G}$ is a closed subspace of $X$, hence compact. Also, for every $S\subset G$ and $\epsilon > 0$, the set

$$A(S,\epsilon) = \bigl\{ f\in X : \bigl(\forall x\in S\bigr)\bigl(\lvert f(x) - 1\rvert \leqslant \epsilon\bigr)\bigr\}$$

is closed in $X$, hence compact. In particular, for a neighbourhood $U$ of the identity in $G$, and $\epsilon > 0$, the set

$$V(U,\epsilon) = \tilde{G} \cap A(U,\epsilon)$$

is a closed subset of $X$, hence compact.

Now we note that $V(U,\epsilon) \subset \hat{G}$ for small enough $\epsilon$:

For simplicity, we use the arc-length distance on $S^1$, not the Euclidean distance. Assume $\epsilon < \frac{\pi}{2}$. Starting with $U_0 = U$, choose a sequence $(U_n)$ of neighbourhoods of the identity in $G$ such that $U_{n+1}\cdot U_{n+1} \subset U_n$ for all $n \in \mathbb{N}$, and set $\epsilon_n = 2^{-n}\cdot \epsilon$. For $x \in U_{n+1}$ and $\chi \in V(U_{n}, \epsilon_{n})$, we have $\lvert \chi(x) - 1\rvert \leqslant \epsilon_n$ since $x\in U_n$, and $\lvert \chi(x)^2 - 1\rvert = \lvert \chi(x^2) - 1\rvert \leqslant \epsilon_n$, since $x^2 \in U_n$. The latter implies $\lvert \chi(x) - 1\rvert \leqslant \epsilon_n/2 = \epsilon_{n+1}$ since $\epsilon_n < \frac{\pi}{2}$. Hence $V(U_n,\epsilon_n) \subset V(U_{n+1}, \epsilon_{n+1})$.

In particular $V(U,\epsilon) \subset V(U_n,\epsilon_n)$ for all $n$, and since for every neighbourhood $W$ of $1$ in $S^1$ there is an $n\in\mathbb{N}$ such that $\{ z : \lvert z-1\rvert \leqslant \epsilon_n\} \subset W$, we see that every $\chi \in V(U,\epsilon)$ is continuous at the identity of $G$, and as a homomorphism between topological groups, it is continuous everywhere.

We saw above that $V(U,\epsilon)$ is compact in the topology of pointwise convergence, but what we need is that $V(U,\epsilon)$ is compact in the topology of compact convergence (or locally uniform convergence, since $G$ is locally compact).

However, if we look at the proof of continuity above a little closer, we see that we proved more than just $V(U,\epsilon) \subset \hat{G}$. In fact, we proved that $V(U,\epsilon)$ is a uniformly equicontinuous family. For all $\chi \in V(U,\epsilon)$ and $x,y\in G$ with $x^{-1}y \in U_n$, we have

$$\lvert \chi(y) - \chi(x)\rvert = \lvert \chi(x)^{-1}\chi(y) - 1\rvert = \lvert \chi(x^{-1}y) - 1\rvert \leqslant \epsilon_n.$$

Now we can invoke a theorem of Ascoli-Bourbaki,

Let $X$ be a topological space and $Y$ a uniform space. Let $H \subset C(X,Y)$ be equicontinuous. Then on $H$, the uniform structures of compact convergence and of pointwise convergence coincide.

In particular, the topologies of compact convergence and of pointwise convergence coincide on $H$.

And thus we see that $V(U,\epsilon)$ is compact in the topology of compact convergence.

Finally, if $U$ is a compact neighbourhood of the identity in $G$, then $V(U,\epsilon)$ is a neighbourhood of the identity in $\hat{G}$, and we see that $\hat{G}$ with the topology of compact convergence is locally compact.

Answer 2

The other answer posted here uses Ascoli's theorem as the method to prove a set of functions is compact. Another proof using Ascoli's theorem is here. The basic idea is that if $K \subset G$ is compact with $\mu(K) > 0$ and $0 < \varepsilon < 1$ then the characters $\chi \in \widehat{G}$ satisfying $|\chi(x) - 1| < \varepsilon$ for all $x \in K$ also satisfy $$\left|\int_K \chi(x)\,d\mu(x)\right| \geq (1-\varepsilon)\mu(K) > 0,$$ and the set of such $\chi$ is has compact closure in $\widehat{G}$ by Ascoli's theorem. In fact, the set of such $\chi$ is compact since it is closed in $\widehat{G}$ by continuity of the Fourier transform.

The compactness of the set of such $\chi$ can be regarded as a form of the Riemann-Lebesgue lemma on the locally compact abelian group $G$: for each $f \in L^1(G)$, the number $\int_G f(x)\chi(x)\,d\mu(x)$ is arbitrarily small for all $\chi$ outside some compact subset of $\widehat{G}$. The proof of local compactness uses this idea with $f$ being the characteristic function of the compact set $K$.