Let $G$ be a locally compact abelian topological group (hereafter abbreviated ‘LCAG’). Its **Pontryagin dual**^{[α]} $\hat{G}$ is defined as the space of continuous group homomorphisms $\alpha :G\to O$, where $O$ represents the circle group, with a group operation defined pointwise. We will write $\u27e8\alpha ,g\u27e9$ for $\alpha (g)\in O$, and write group operations multiplicatively except for $R$ and its subgroups.

- Any real number $\omega $ defines an element of $\hat{R}$: if we regard $O$ as a multiplicative subgroup of $C$, then this morphism is $$\begin{array}{c}\hfill \alpha :t\mapsto {e}^{i\omega t}\mathrm{\text{.}}\end{array}$$ It is evident from the properties of the complex exponential that this morphism is unique, and so the reals are self-dual. Through a similar argument it is possible to show that dense subgroups of the reals, including the rationals, are also self-dual.

- Next, the integers, also under addition. Like the reals, a group homomorphism $\alpha $ from the integers can be defined for each real number via the complex exponential; however, the homomorphism so defined is not unique. Indeed, ${e}^{i\omega n}={e}^{i(\omega +2\pi )n}$ for all $n\in Z$, so that the value of $\u27e8\alpha ,1\u27e9$ uniquely determines $\alpha $. From this, we deduce that $\hat{Z}\cong O$.

- The circle group itself is also worth considering. An endomorphism of $O$ maps the circle onto itself some integer number of times, with negative numbers corresponding to direction-reversing maps and zero to the constant map. As such, $\hat{O}\cong Z$. This is a special case of the natural isomorphism
^{[β]}between $G$ and $\hat{\hat{G}}$ for any locally compact abelian group.

- The argument for $Z/nZ$ goes similarly to $Z$ itself. Each $\alpha $ is still uniquely determined by $\u27e8\alpha ,1\u27e9$, but the quotient gives an additional constraint. Since $$\begin{array}{c}\hfill {\u27e8\alpha ,1\u27e9}^{n}=\u27e8\alpha ,n\u27e9=\u27e8\alpha ,0\u27e9=1\mathrm{\text{,}}\end{array}$$ $\u27e8\alpha ,1\u27e9$ must be an $n$th root of unity. The $n$th roots of unity form a multiplicative group which is isomorphic to $Z/nZ$ itself, meaning that it too is self-dual.

- It’s not hard to see that $\hat{G\times H}\cong \hat{G}\times \hat{H}$. Indeed, for $\alpha \in \hat{G}$, $\beta \in \hat{H}$, let $\alpha \u2022\beta \in \hat{G\times H}$ such that $\u27e8\alpha \u2022\beta ,(g,h)\u27e9=\u27e8\alpha ,g\u27e9\u27e8\beta ,h\u27e9$. By inspection, this construction is a group monomorphism. Moreover, for any $\gamma \in \hat{G\times H}$, let $\u27e8{p}_{G}(\gamma ),g\u27e9=\u27e8\gamma ,(g,1)\u27e9$ and $\u27e8{p}_{H}(\gamma ),h\u27e9=\u27e8\gamma ,(1,h)\u27e9$. Then ${p}_{G}(\gamma )\u2022{p}_{H}(\gamma )=\gamma $, establishing the isomorphism—in fact a natural isomorphism of bifunctors.

- As a corollary of the above, ${R}^{n}$ is self-dual, as is every finite abelian group. The Pontryagin dual of ${Z}^{n}$ is a torus group and vice versa.

- Consider a group homomorphism $\phi :G\to H$. We can see that it induces a map ${\phi}^{\star}:\hat{H}\to \hat{G}$ by $\u27e8{\phi}^{\star}(\beta ),g\u27e9=\u27e8\beta ,\phi (g)\u27e9$, and that this map is also a group homomorphism. This correspondence makes Pontryagin duality a contravariant functor $P:{LCAG}^{op}\to LCAG$, where $LCAG$ is the category of LCAGs and continuous group homomorphisms.

- Let $\phi $ now be a monomorphism and $\alpha \in \hat{G}$. Define $\beta \in \hat{H}$ by $\u27e8\beta ,h\u27e9=\u27e8\alpha ,{\phi}^{-1}(h)\u27e9$ and we see that $\alpha ={\phi}^{\star}(\beta )$, making ${\phi}^{\star}$ an epimorphism. Conversely, let $\phi $ be an epimorphism, and let $\beta ,\gamma \in \hat{H}$ such that ${\phi}^{\star}(\beta )={\phi}^{\star}(\gamma )$. For any $h\in H$, there exists $g\in G$ with $\phi (g)=h$. Then $$\begin{array}{cc}\hfill \u27e8\beta ,h\u27e9& =\u27e8\beta ,\phi (g)\u27e9\hfill \\ \hfill & =\u27e8{\phi}^{\star}(\beta ),g\u27e9\hfill \\ \hfill & =\u27e8{\phi}^{\star}(\gamma ),g\u27e9\hfill \\ \hfill & =\u27e8\gamma ,\phi (g)\u27e9\hfill \\ \hfill & =\u27e8\gamma ,h\u27e9\mathrm{\text{.}}\hfill \end{array}$$ Since $h$ was arbitrary, $\beta =\gamma $ and ${\phi}^{\star}$ is a monomorphism.

- In particular, if $\iota :H\hookrightarrow G$ is the inclusion map of a subgroup $H\subseteq G$ then ${\iota}^{\star}$ is the corresponding restriction map. If $\pi :G\twoheadrightarrow G/H$ is the projection map onto the quotient group, ${\pi}^{\star}$ is the inclusion map of the annihilator $A=\{\alpha \in \hat{G}\mid \u27e8\alpha ,h\u27e9=1,\forall h\in H\}$. Thus $\hat{H}=\hat{G}/A$ and $\hat{G/H}=A$.

Recall that any locally compact group has a right-invariant real-valued measure, the **Haar measure**, and that it is unique up to a scalar factor. In what follows, $G$ is a LCAG and $\hat{G}$ its dual, and $\mu $ and $\hat{\mu}$ respectively represent a fixed choice of Haar measure for each of the two groups. Let $f:G\to C$ be an absolutely Lebesgue-integrable^{[γ]} map (*not* necessarily a group morphism), and consider the integral $$\begin{array}{c}\hfill \hat{f}(\alpha )={\int}_{G}f(g)\overline{\u27e8\alpha ,g\u27e9}d\mu (g)\mathrm{\text{,}}\end{array}$$ where $O$ is once more considered as a complex subgroup. Since $\left|\u27e8\alpha ,g\u27e9\right|=1$, $$\begin{array}{cc}\hfill |\hat{f}(\alpha )|& \le {\int}_{G}|f(g)\overline{\u27e8\alpha ,g\u27e9}|d\mu (g)\hfill \\ \hfill & ={\int}_{G}|f(g)|d\mu (g)<\infty \mathrm{\text{,}}\hfill \end{array}$$ so the integral does indeed exist. This is the **generalized Fourier transform**: bearing in mind that the Haar measure on discrete groups is simply the counting measure, the cases we have already looked at correspond to common forms of the transform as follows.

$G=\hat{G}=R$ | ordinary (continuous) Fourier transform |

$G=Z$, $\hat{G}=O$ | discrete-time Fourier transform |

$G=O$, $\hat{G}=Z$ | Fourier series |

$G=\hat{G}=Z/nZ$ | discrete Fourier transform |

Provided $\hat{f}$ is also absolutely integrable, the inverse transform can be constructed similarly: $$\begin{array}{c}\hfill f(g)={\int}_{\hat{G}}\hat{f}(\alpha )\u27e8\alpha ,g\u27e9d\hat{\mu}(\alpha )\mathrm{\text{.}}\end{array}$$

[Proof of inversion theorem goes here.]

We can prove a version of the Fourier convolution theorem in this setting. Let $p,q::G\to C$. Their **convolution** $p\star q$ is defined $$\begin{array}{c}\hfill p\star q(g)={\int}_{G}p(h)q(g{h}^{-1})d\mu (h)\mathrm{\text{.}}\end{array}$$ Now consider the integral $$\begin{array}{cc}\hfill I& ={\iint}_{{G}^{2}}|p(h)q(g{h}^{-1})|d\mu (g)d\mu (h)\hfill \\ \hfill & ={\int}_{G}|p(h)|{\int}_{G}|q(g{h}^{-1})|d\mu (g)d\mu (h)\mathrm{\text{.}}\hfill \end{array}$$ Under the change of variable $k=g{h}^{-1}$, bearing in mind the invariance of the Haar measure, $$\begin{array}{cc}\hfill I& ={\int}_{G}|p(h)|{\int}_{G}|q(k)|d\mu (k)d\mu (h)\hfill \\ \hfill & ={\int}_{G}|p(h)|\parallel q{\parallel}_{1}d\mu (h)\hfill \\ \hfill & =\parallel q{\parallel}_{1}{\int}_{G}|p(h)|d\mu (h)\hfill \\ \hfill & =\parallel p{\parallel}_{1}\parallel q{\parallel}_{1}\mathrm{\text{.}}\hfill \end{array}$$ Since the integral is absolute, Fubini’s theorem^{[δ]} tells us that we can reverse the order of integration: $$\begin{array}{cc}\hfill \parallel p\star q{\parallel}_{1}& ={\int}_{G}|{\int}_{G}p(h)q(g{h}^{-1})d\mu (h)|d\mu (g)\hfill \\ \hfill & \le {\iint}_{{G}^{2}}|p(h)q(g{h}^{-1})|d\mu (h)d\mu (g)\hfill \\ \hfill & =I=\parallel p{\parallel}_{1}\parallel q{\parallel}_{1}<\infty \mathrm{\text{,}}\hfill \end{array}$$ meaning that $\hat{p\star q}$ is defined.

To compute it, observe that since $\left|\u27e8\alpha ,g\u27e9\right|=1$ we can apply Fubini’s theorem again: $$\begin{array}{cc}\hfill \hat{p\star q}(\alpha )& ={\int}_{G}{\int}_{G}p(h)q(g{h}^{-1})d\mu (h)\overline{\u27e8\alpha ,g\u27e9}d\mu (g)\hfill \\ \hfill & ={\iint}_{{G}^{2}}p(h)q(g{h}^{-1})d\mu (g)d\mu (h)\hfill \\ \hfill & ={\int}_{G}p(h){\int}_{G}q(g{h}^{-1})\overline{\u27e8\alpha ,g\u27e9}d\mu (g)d\mu (h)\mathrm{\text{.}}\hfill \end{array}$$ Making the same substitution $k=g{h}^{-1}$, $$\begin{array}{cc}\hfill \hat{p\star q}(\alpha )& ={\int}_{G}p(h){\int}_{G}q(k)\overline{\u27e8\alpha ,hk\u27e9}d\mu (k)d\mu (h)\hfill \\ \hfill & ={\int}_{G}p(h)\overline{\u27e8\alpha ,h\u27e9}d\mu (h){\int}_{G}q(k)\overline{\u27e8\alpha ,k\u27e9}d\mu (k)\hfill \\ \hfill & =\hat{p}(\alpha )\hat{q}(\alpha )\mathrm{\text{,}}\hfill \end{array}$$ completing the proof. □

By a similar argument, an equivalent statement holds for the inverse transform as well: $\hat{pq}=\hat{p}\star \hat{q}$.

^{[α]} Named after Soviet mathematician Лев Семёнович Понтрягин *Lev Semyonovich Pontryagin* (1908–1988).^{[β]} Wikipedia assures me that this exists, but when I tried to prove it I didn’t get very far.^{[γ]} Recall that $f$ is absolutely integrable if its ${L}_{1}$ norm $\parallel f{\parallel}_{1}={\int}_{G}|f(g)|d\mu (g)$ exists and is finite.^{[δ]} A generalization to the Lebesgue integral of the fact that double sums can have the order of summation reversed whenever their convergence is absolute.