Pontryagin Duality for Locally Compact Abelian Groups

Choose an $H$-open neighbourhood $V$ of $e$ with $C:={\mathrm{c}\mathrm{l}}_H(V)$ compact in $H$, and write $V=H\cap O$ with $O$ open in $G$. Then $C$ is compact in $G$ and closed in $G$, and $V\subseteq C\subseteq H$ with ${\mathrm{c}\mathrm{l}}_G(V)\subseteq C$ because $C$ is closed in $G$ and contains $V$.

Let $x\in {\mathrm{c}\mathrm{l}}_G(H)$. The set $x-O=\{x-o:o\in O\}$ is an open neighbourhood of $x$ by [F2], so by the closure characterisation it meets $H$: there are $h\in H$ and $o\in O$ with $h=x-o$, that is $x=h+o$.

With $h,o$ as in step 2.1, the point $o=(-h)+x$ lies in ${\mathrm{c}\mathrm{l}}_G(H)$, because $-h+H=H$ and translations preserve closures; and $o\in O$, so $o\in {\mathrm{c}\mathrm{l}}_G(H)\cap O\subseteq {\mathrm{c}\mathrm{l}}_G(H\cap O)={\mathrm{c}\mathrm{l}}_G(V)\subseteq C\subseteq H$, using $V=H\cap O$ and the inclusion of [F4].

Since $o\in H$ and $x=h+o$ with $h\in H$, the subgroup $H$ contains $x$. Hence every $x\in {\mathrm{c}\mathrm{l}}_G(H)$ lies in $H$, that is ${\mathrm{c}\mathrm{l}}_G(H)=H$ and $H$ is closed in $G$. Conversely, suppose $G$ is locally compact and $H$ is closed. For $h\in H$, a compact neighbourhood $N$ of $h$ in $G$ gives a compact neighbourhood $N\cap H$ in $H$: it is closed in the compact space $N$ and contains the trace on $H$ of an open neighbourhood of $h$. The Hausdorff property and continuous group operations restrict to $H$, as does abelianness. Thus a closed subgroup of an LCA group is LCA.

The quotient map $q$ is open: for open $U\subseteq G$ one has $q^{-1}(q(U))=U+H={\bigcup }_{h\in H}(U+h)$, a union of open translates, hence open in $G$; by the definition of the quotient topology $q(U)$ is open in $G\mathrm{/}H$.

$G\mathrm{/}H$ is Hausdorff: let $q(x)\mathrm{\ne }q(y)$, so $x-y\mathrm{\notin }H$. Since $H$ is closed, choose an open neighbourhood $W$ of $x-y$ with $W\cap H=\mathrm{\varnothing }$; by continuity of subtraction there are open neighbourhoods $U,V$ of $0$ with $(x-y)+U-V\subseteq W$. Then $q(x+U)$ and $q(y+V)$ are open by step 1.1 and are disjoint: if $x+u+H=y+v+H$ with $u\in U$, $v\in V$, then $x-y+u-v\in H\cap W=\mathrm{\varnothing }$, a contradiction.

$G\mathrm{/}H$ is locally compact: given $x\in G$, choose a compact neighbourhood $N$ of $x$ and an open $U$ with $x\in U\subseteq N$. Then $q(N)$ is compact as a continuous image of $N$, and it is closed because $G\mathrm{/}H$ is Hausdorff by step 2.1; $q(U)$ is an open neighbourhood of $q(x)$ contained in $q(N)$. Hence $q(N)$ is a compact neighbourhood of $q(x)$.

The quotient operations are continuous: the product $q\times q:G\times G\to G\mathrm{/}H\times G\mathrm{/}H$ is a continuous open surjection (images of basic open rectangles are open rectangles), hence a quotient map, and $m_{G\mathrm{/}H}\circ (q\times q)=q\circ m_G$ where $m_G$ and $m_{G\mathrm{/}H}$ are the respective addition maps. The quotient universal property therefore makes addition on $G\mathrm{/}H$ continuous, and inversion descends in the same way from inversion in $G$. Thus $G\mathrm{/}H$ is an abelian topological group which is Hausdorff and locally compact, as claimed.

In a compact Hausdorff totally disconnected space $X$, let $U$ be an open neighbourhood of $x$. Let $\mathcal{C}$ be the family of all clopen sets containing $x$. Total separatedness implies the open sets $X\setminus C$, $C\in \mathcal{C}$, cover the compact set $X\setminus U$: each $y\mathrm{\notin }U$ is excluded by at least one such $C$. A finite subcover gives $C_1,\dots ,C_n\in \mathcal{C}$ with $X\setminus U\subseteq {\bigcup }_j(X\setminus C_j)$. Thus $V={\bigcap }_j{C}_j$ is clopen and $x\in V\subseteq U$. If the complement is empty take $V=X$. The family includes all separators, so no point-indexed choice is made.

Let $E\subseteq G$ be compact and open; if $E=\mathrm{\varnothing }$ take $W:=G$, which is symmetric and satisfies $E+W=\mathrm{\varnothing }=E$. Assume $E\mathrm{\ne }\mathrm{\varnothing }$. The set $D:=\{(x,y)\in E\times G:x+y\mathrm{\notin }E\}$ is closed in $E\times G$: it is the trace on $E\times G$ of the preimage of the closed set $G\setminus E$ under the continuous addition map $G\times G\to G$. It misses $E\times \{0\}$ because $x+0=x\in E$ for $x\in E$. Consider the family $\mathcal{R}$ of all pairs $(U,V)$ with $U$ open in $E$, $V$ open in $G$ containing $0$, and $U\times V$ disjoint from $D$. The product topology and continuity ensure their first coordinates cover $E$, without choosing one rectangle for each point.

Compactness gives finitely many pairs $(U_j,V_j)\in \mathcal{R}$ whose first coordinates cover $E$. Put $W_0={\bigcap }_j{V}_j$ and $W=W_0\cap (-W_0)$. Then $W$ is a symmetric open neighbourhood of $0$, and each $x\in E$, $w\in W$ belongs to some admissible rectangle, giving $x+w\in E$. Since $0\in W$, $E+W=E$, proving (i). Only a finite subfamily of the specified family was selected.

Let $U$ be a neighbourhood of $0$ in $G$. Choose an open neighbourhood $U_0\subseteq U$ of $0$ and a compact neighbourhood $N$ of $0$ with $N\subseteq U_0$. Then $N$ is a compact Hausdorff space, and it is totally disconnected: for $x\in N$ the component of $x$ in $N$ is a connected subset of $G$ containing $x$, hence is contained in the component $C(x)=\{x\}$ of $G$. Applying step 1.1 in $N$ to the relatively open set $N\cap {\mathrm{i}\mathrm{n}\mathrm{t}}_G(N)\cap U_0$, which contains $0$, we obtain a set $V$ that is clopen in $N$ with $0\in V\subseteq {\mathrm{i}\mathrm{n}\mathrm{t}}_G(N)\cap U_0$.

Now let $E$ be compact and open with $0\in E$, and let $W$ be as in step 2.1, so that $W=-W$, $0\in W$ and $E+W=E$. Put $F:=\{x\in G:x+E=E\}$. Then $F$ is a subgroup: if $x+E=E$ and $y+E=E$ then $(x+y)+E=x+(y+E)=x+E=E$ by associativity, and from $x+E=E$ we get $-x+E=-x+(x+E)=E$; certainly $0\in F$. Also $F\subseteq E$, because for $x\in F$ one has $x=x+0\in x+E=E$. Moreover $F\supseteq W$: for $w\in W$, both $E+w\subseteq E$ and $E-w\subseteq E$ hold, so $E+w=E$, so $F$ contains the neighbourhood $W$ of $0$ and is therefore open; and $F$ is closed because $\{x:x+E\subseteq E\}={\bigcap }_{e\in E}(E-e)$ and $\{x:E\subseteq x+E\}={\bigcap }_{e\in E}(e-E)$ are intersections of translates of the closed set $E$, hence closed, and $F$ is their intersection. Being a closed subset of the compact set $E$, the subgroup $F$ is compact. Thus $F$ is a compact open subgroup of $G$ contained in $E$, proving (ii).

The set $V$ of step 2.2 is compact in $G$: it is closed in $N$ and $N$ is compact. It is open in $G$: being open in $N$ it has the form $V=N\cap O$ with $O$ open in $G$, and $V\subseteq {\mathrm{i}\mathrm{n}\mathrm{t}}_G(N)$, so the criterion of [F8] applies with $U_1:={\mathrm{i}\mathrm{n}\mathrm{t}}_G(N)$ and $V=U_1\cap O$. Thus $V$ is a compact open subset of $G$ with $0\in V\subseteq U$.

Since $E$ is a union of cosets of the subgroup $F$ (for $f\in F$ one has $f+E=E$) and $F$ is open, each coset is open and the cosets partition $E$; compactness of $E$ makes this open cover of $E$ finite, so $E$ is a finite union of open cosets of $F$. This proves (iii).

Applying step 3.1 to the compact open set $E:=V$ of step 3.2, which contains $0$ and is contained in $U$, produces a compact open subgroup $F_0\subseteq V\subseteq U$. As $U$ was an arbitrary neighbourhood of $0$, every neighbourhood of $0$ contains a compact open subgroup of $G$; together with parts (i), (ii) and (iii) proved in steps 2.1, 3.1 and 4.1 this is the statement.

Choose a compact symmetric neighbourhood $C$ of the identity $e$ of $G$ and put $H_0:=⟨C⟩={\bigcup }_{k\ge 1}{C}^k$. Then $H_0$ is an open subgroup of $G$ and is generated by the compact set $C$, so $H_0$ is compactly generated.

By [F1] there are $m,n\ge 0$, a compact group $K$ and a topological isomorphism $\phi :H_0\to {\mathbb{R}}^m\times {\mathbb{Z}}^n\times K$. Let $H:={\phi }^{-1}({\mathbb{R}}^m\times \{0\}\times K)$. Since $\{0\}$ is open in the discrete group ${\mathbb{Z}}^n$, the set ${\mathbb{R}}^m\times \{0\}\times K$ is open in ${\mathbb{R}}^m\times {\mathbb{Z}}^n\times K$, so $H$ is an open subgroup of $H_0$ and hence open in $G$; it is closed as well, because its complement in $G$ is the union of the remaining cosets of $H$, each of which is open by [F3].

$H$ is compactly generated. Indeed $H$ is isomorphic under $\phi$ to ${\mathbb{R}}^m\times \{0\}\times K$, which is homeomorphic to ${\mathbb{R}}^m\times K$; a closed ball of large radius in ${\mathbb{R}}^m$ is compact and generates ${\mathbb{R}}^m$ as a group, because every $v\in {\mathbb{R}}^m$ is an integer multiple of a vector of sufficiently small norm, and $K$ generates itself, so the compact product of that ball with $K$ generates ${\mathbb{R}}^m\times K$.

Let $L\le H$ be an open subgroup of $H$. Then $H\mathrm{/}L$ is discrete, because every coset of the open subgroup $L$ is open. The composite ${\mathbb{R}}^m\to H\to H\mathrm{/}L$ of the inclusion of the connected factor (under the identification of step 2.1) with the quotient map is continuous, so its image is connected in the discrete space $H\mathrm{/}L$, hence a single point; therefore ${\mathbb{R}}^m\subseteq L$. It follows that every coset of $L$ meets $\{0\}\times K$, so the quotient map restricts to a continuous surjection $K\to H\mathrm{/}L$, and $H\mathrm{/}L$ is compact as a continuous image of the compact group $K$; being compact and discrete it is finite. Hence every open subgroup $L$ of $H$ has finite index in $H$.

The subgroup $H$ of steps 1.1-4.1 is open in $G$, compactly generated, and contains no open subgroup of infinite index. If $G$ is discrete, then $H=\{0\}$ is open, compactly generated as the subgroup generated by the compact set $\{0\}$, and its only subgroup is itself, of index $1$.

Let $\phi :H\to {\mathbb{R}}^m\times {\mathbb{Z}}^n\times K$ be a topological isomorphism as in [F1], and suppose $n\ge 1$. Then $L:={\phi }^{-1}({\mathbb{R}}^m\times \{0\}\times K)$ is an open subgroup of $H$ by [F3], and taking images under $\phi$ identifies the quotient $H\mathrm{/}L$ with $({\mathbb{R}}^m\times {\mathbb{Z}}^n\times K)\mathrm{/}({\mathbb{R}}^m\times \{0\}\times K)$, which is isomorphic to ${\mathbb{Z}}^n$ by [F2]; as ${\mathbb{Z}}^n$ is infinite for $n\ge 1$, the index $[H:L]$ is infinite by [F4]. This contradicts the hypothesis that $H$ has no open subgroup of infinite index. Hence $n=0$.

With $n=0$, [F1] gives a topological isomorphism $H\cong {\mathbb{R}}^m\times \{0\}\times K\cong {\mathbb{R}}^m\times K$, and hence $H\cong K\oplus {\mathbb{R}}^m$ with the compact group $W:=K$ and the exponent $m\ge 0$. This is the statement.

The map $\widehat{q}(\chi )=\chi \circ q$ is a group homomorphism: for $x\in G$ one has $\widehat{q}({\chi }_1{\chi }_2)(x)={\chi }_1(q(x)){\chi }_2(q(x))=(\widehat{q}{\chi }_1)(x)(\widehat{q}{\chi }_2)(x)$, since evaluation is pointwise. It is continuous by the functoriality clause of [F3] applied to the continuous homomorphism $q$ of [F1]. Its image lies in $H^{\perp }$, because for $h\in H$ one has $\widehat{q}(\chi )(h)=\chi (q(h))=\chi (0)=1$, and it is injective because $q$ is surjective.

By the closed-subgroup clause of [F3] the map $\widehat{q}$ is a topological group isomorphism from $\widehat{G\mathrm{/}H}$ onto $H^{\perp }$, where the annihilator is the subgroup displayed in [F2] and is closed in $\widehat{G}$.

Combining steps 1.1 and 1.2, $\widehat{q}:\widehat{G\mathrm{/}H}\to \widehat{G}$ is an isomorphism of topological groups onto $H^{\perp }$, and $H^{\perp }$ is a closed subgroup of $\widehat{G}$ topologically isomorphic to $\widehat{G\mathrm{/}H}$; this is the statement.

Apply [F1] to $G$: there is an open subgroup $H\le G$ which is compactly generated and contains no open subgroup of infinite index.

Apply [F2] to $H$: there are $n\ge 0$ and a compact group $W$ with $H\cong W\oplus {\mathbb{R}}^n$ as topological groups. Since $H$ is open in $G$ it is closed by [F3], and the isomorphism is the required one. Thus $G$ contains an open subgroup isomorphic to ${\mathbb{R}}^n\times W$, which is the statement.

Because $x\mathrm{\ne }0$ and $G$ is Hausdorff, addition is continuous at $(0,0)$ and $G\setminus \{x\}$ is open, so there are open neighbourhoods $W_1,W_2$ of $0$ with $W_1+W_2\subseteq G\setminus \{x\}$; replacing them by their intersections with their negatives and with each other, we obtain a symmetric open neighbourhood $W$ of $0$ with $(W+W)\cap \{x\}=\mathrm{\varnothing }$.

Choose a symmetric compact neighbourhood $S$ of $0$ with $S\subseteq W$: a neighbourhood basis of open sets with compact closure at $0$ supplies an open $V$ with $0\in V\subseteq {\mathrm{c}\mathrm{l}}_G(V)\subseteq W$, and $S:={\mathrm{c}\mathrm{l}}_G(V)\cap (-{\mathrm{c}\mathrm{l}}_G(V))$ is compact and symmetric with $0\in S\subseteq W$. Then $S-S\subseteq W+W$, so $x\mathrm{\notin }S-S$; hence $S\cap (x+S)=\mathrm{\varnothing }$, since $s=x+s^{\mathrm{\prime }}$ would give $x=s-s^{\mathrm{\prime }}\in S-S$.

Apply the cutoff of [F2] with $K=\{0\}$ and $U={\mathrm{i}\mathrm{n}\mathrm{t}}_G(S)$ to obtain $g\in C_c(G)$ with $g\ge 0$, $g(0)=1$ and $\mathrm{supp}g\subseteq S$. Then $h:=g\ast \tilde{g}$ is continuous with compact support, positive definite, and $h(0)={\int }_G\mathrm{\mid }g{\mathrm{\mid }}^{2}d{m}_G>0$ because $g\equiv ̸0$ and $g\ge 0$.

For this $h$ one has $h(x)=0$: by the convolution formula and $\tilde{g}(x-y)=\overline{g(y-x)}$ from [F5], $h(x)={\int }_{G}g(y)\overline{g(y-x)}d{m}_G(y)$, and the integrand vanishes identically because $g(y)\mathrm{\ne }0$ forces $y\in S$ while $g(y-x)\mathrm{\ne }0$ forces $y\in x+S$, and $S\cap (x+S)=\mathrm{\varnothing }$.