Pontryagin Duality and the Fourier Transform

1 Introduction

We follow Rudin [3] and Terras [4], and refer to sp4comm [2] and Folland [1].

2 $\mathbb{Z}/n\mathbb{Z}$

Let $n$ be a positive integer.

$$\mathbb{Z}/n\mathbb{Z}=\{k+n\mathbb{Z}:k\in \mathbb{Z}\}=\{k+n\mathbb{Z}:0\le k\le n-1\}$$

$Z_n=\mathbb{Z}/n\mathbb{Z}$ is a ring. $|Z_n|=n$. We focus on the additive group $(Z_n,+)$.

3 Haar measure

Let

$${ℂ}^{Z_n}$$

be the set of functions $Z_n\to ℂ$.

We use counting measure on the group $Z_n$ as Haar measure (which is discrete and compact) with total volume $n$, and Since $Z_n$ has finite Haar measure (being a compact group) and every function $Z_n\to ℂ$ is continuous (being a discrete group), we have

$${ℂ}^{Z_n}=L^1(Z_n)=L^2(Z_n).$$

We specify function space for emphasis.

For $f,g\in L^2(Z_n)$, define

$${(f,g)}_{L^2(Z_n)}=\sum _{x\in Z_n}f(x)\overline{g(x)}$$

Define

$${|f|}_{L^2(Z_n)}=\sqrt{{(f,f)}_{L^2(Z_n)}}=\sqrt{\sum _{x\in Z_n}f(x)\overline{f(x)}}.$$

Define

$${|f|}_{L^1(Z_n)}=\sum _{x\in Z_n}|f(x)|.$$

4 ${\delta }_a$

For $a\in Z_n$, define ${\delta }_a:Z_n\to ℂ$ by

For $a,b\in Z_n$,

${({\delta }_a,{\delta }_b)}_{L^2(Z_n)}$ $={\displaystyle \sum _{x\in Z_n}}{\delta }_a(x)\overline{{\delta }_b(x)}$

$={\displaystyle \sum _{x\in Z_n}}{\delta }_a(x){\delta }_b(x)$

For $f\in L^2(Z_n)$,

$$f(x)=\sum _{a\in Z_n}f(a){\delta }_a(x)=\sum _{a=0}^{n-1}f(a){\delta }_a(x),x\in Z_n.$$

Indeed, $\{{\delta }_0,\mathrm{\dots },{\delta }_{n-1}\}$ is an orthonormal basis of $L^2(Z_n)$.

5 Convolution

For $f,g\in L^1(Z_n)$, define the convolution $f*g\in L^1(Z_n)$ by

$$(f*g)(x)=\sum _{y\in Z_n}f(y)g(x-y),x\in Z_n.$$

$$f*g=g*f,f,g\in L^1(Z_n).$$

$$f*(g*h)=(f*g)*h,f,g,h\in L^1(Z_n).$$

For $f\in L^1(Z_n)$ and for $a\in Z_n$,

$$(f*{\delta }_a)(x)=f(x-a),x\in Z_n.$$

For $a,b\in Z_n$, and for $x\in Z_n$,

$$({\delta }_a*{\delta }_b)(x)={\delta }_a(x-b)={\delta }_{a+b}(x).$$

5.1 Example

Take $n=15$. Define $f={\delta }_0+{\delta }_1+{\delta }_2$. For $x\in Z_{15}$,

$(f*f)(x)$ $=({\delta }_0+{\delta }_1+{\delta }_2)*({\delta }_0+{\delta }_1+{\delta }_2)$

$={\delta }_0*{\delta }_0+{\delta }_1*{\delta }_1+{\delta }_2*{\delta }_2$

$+2{\delta }_0*{\delta }_1+2{\delta }_0*{\delta }_2+2{\delta }_1*{\delta }_2$

$={\delta }_0+{\delta }_2+{\delta }_4$

$+2{\delta }_1+2{\delta }_2+2{\delta }_3$

$={\delta }_0+2{\delta }_1+3{\delta }_2+2{\delta }_3+{\delta }_4$

6 Dual group

Let $S^1=\{z\in ℂ:|z|=1\}$, which is a multiplicative group.

Let $\widehat{Z_n}$ be the set of group homomorphisms $Z_n\to S^1$.

We use normalized counting measure on the group $\widehat{Z_n}$ as Haar measure (which is discrete and compact) with total volume 1.

For $a\in Z_n$, define $e_a:Z_n\to S^1$ by

$$e_a(x)=\exp \left(2\pi i\frac{ax}{n}\right),x\in Z_n.$$

$e_a$ is an element of $\widehat{Z_n}$.

Define $\psi :Z_n\to \widehat{Z_n}$ by $\psi (a)=e_a$.

$\widehat{Z_n}$ $=\{e_a:a\in Z_n\}$

$=\{\psi (a):a\in Z_n\}$

$=\psi (Z_n)$

$\psi :Z_n\to \widehat{Z_n}$ is an isomorphism of groups.

7 Fourier transform

Define the Fourier transform ${ℱ}_n:L^2(Z_n)\to L^2(\widehat{Z_n)}$ by

$$({ℱ}_{n}f)(e_a)=\sum _{x\in Z_n}f(x)e_a(-x),e_a\in \widehat{Z_n}.$$

8 Pullback

We introduce the operator $F_n:L^2(Z_n)\to L^2(Z_n)$, which is defined via composition with the Fourier transform ${ℱ}_n$ and the function $\psi$ as

$$F_{n}f=({ℱ}_{n}f)\circ \psi .$$

We pullback ${ℱ}_{n}f:\widehat{Z_n}\to ℂ$ to a function $Z_n\to ℂ$.

We remind ourselves (a) that for $a\in Z_n$, the function $e_a:Z_n\to S^1$ is defined by

$$e_a(x)=\exp \left(2\pi i\frac{ax}{n}\right),x\in Z_n,$$

(b) that $\psi :Z_n\to \widehat{Z_n}$ is defined by $\psi (a)=e_a$,

and (c) that $\psi :Z_n\to \widehat{Z_n}$ is an isomorphism of groups, by

$\widehat{Z_n}$ $=\{e_a:a\in Z_n\}$

$=\{\psi (a):a\in Z_n\}$

$=\psi (Z_n)$

$({ℱ}_{n}f)(\psi (a))$ $={\displaystyle \sum _{x\in Z_n}}f(x)e_a(-x)$

$={\displaystyle \sum _{x\in Z_n}}f(x)\overline{e_a(x)}$

$={(f,e_a)}_{L^2(Z_n)}$

Thus

$$(F_{n}f)(x)={(f,e_x)}_{L^2(Z_n)},x\in Z_n.$$

In the sequel, we use the term Fourier transform to refer both to ${ℱ}_n$ and to $F_n$, but preserve the distinction for calculations.

8.1 Example: $n\mathrm{=}\mathrm{7}$ and $f\mathrm{=}{\delta }_{\mathrm{3}}$

By

$$(F_{n}f)(x)={(f,e_x)}_{L^2(Z_n)},x\in Z_n$$

we have

$$(F_7{\delta }_3)(x)={({\delta }_3,e_x)}_{L^2(Z_7)},x\in Z_7.$$

The inner product ${({\delta }_3,e_x)}_{L^2(Z_7)}$ is given by

${({\delta }_3,e_x)}_{L^2(Z_7)}$ $={\displaystyle \sum _{y\in Z_7}}{\delta }_3(y)\overline{e_x(y)}$

$={\displaystyle \sum _{y=0}^6}{\delta }_3(y)\overline{\exp \left(2\pi i{\displaystyle \frac{xy}{7}}\right)}.$

Since ${\delta }_3(y)=1$ only when $y=3$ and $0$ otherwise, the sum collapses to a single term:

${({\delta }_3,e_x)}_{L^2(Z_7)}$ $=\overline{\exp \left(2\pi i{\displaystyle \frac{3x}{7}}\right)}$

$=\exp \left(-2\pi i{\displaystyle \frac{3x}{7}}\right)$

$=e_{-3}(x)$

Thus,

$$(F_7{\delta }_3)(x)=e_{-3}(x),x\in Z_7,$$

namely,

$$F_7{\delta }_3=e_{-3}.$$

9 $\widehat{Z_n}$

$\widehat{Z_n}$ is the set of group homomorphisms $Z_n\to S^1$.

$\widehat{Z_n}$ is a group using pointwise multiplication of functions $Z_n\to S^1$, the Pontryagin dual group of $Z_n$.

For $a\in Z_n$, define $e_a\in \widehat{Z_n}$ by

$$e_a(x)=\exp \left(2\pi i\frac{ax}{n}\right),x\in Z_n,$$

Define $\psi :Z_n\to \widehat{Z_n}$ by

$$\psi (a)=e_a,a\in Z_n.$$

We have

$\widehat{Z_n}$ $=\{e_a:a\in Z_n\}$

$=\{\psi (a):a\in Z_n\}$

$=\psi (Z_n)$

Thus, $\psi :Z_n\to \widehat{Z_n}$ is an isomorphism of groups.

10 Haar measure

Let $G$ be a locally compact abelian group.

$\widehat{G}$ is the set of continuous group homomorphisms $G\to S^1$. It is a group with operation $({\varphi }_1{\varphi }_2)(x)={\varphi }_1(x){\varphi }_2(x)$, ${\varphi }_1,{\varphi }_2\in \widehat{G}$, $x\in G$ (namely, pointwise multiplication).

We assign $\widehat{G}$ the coarsest topology such that for each $x\in G$, the map $\varphi \mapsto \varphi (x)$ is continuous $\widehat{G}\to S^1$ (namely, the final topology on $\widehat{G}$).

One proves that $\widehat{G}$ is a locally compact abelian group.

If $G$ is a discrete LCA group, then $\widehat{G}$ is a compact LCA group.

10.1 Finite LCA groups

Let $G$ be a finite locally compact abelian group. $G$ must have the discrete topology. Hence the Borel $\sigma$-algebra of $G$ is equal to the power set of $G$, denoted $𝒫(G)$.

Because $G$ has the discrete topology, $\widehat{G}$ is equal to the set of group homomorphisms $G\to S^1$.

Assign $G$ the Haar measure $m_G$ defined by $m_G(A)=|A|$ for $A\in 𝒫(G)$. One checks that $m_G$ indeed is a Haar measure. (Counting measure.)

Assign $\widehat{G}$ the Haar measure $m_{\widehat{G}}$ defined by $m_{\widehat{G}}(A)=\frac{1}{|\widehat{G}|}\cdot |A|$ for $A\in 𝒫(\widehat{G})$. (Normalized counting measure.)

$L^2(G)$ is equal to the set of functions $G\to ℂ$ and $L^2(\widehat{G})$ is equal to the set of functions $\widehat{G}\to ℂ$.

11 $L^2(Z_n)$

For $f,g\in L^2(Z_n)$, define

$${(f,g)}_{L^2(Z_n)}=\sum _{x\in Z_n}f(x)\overline{g(x)}$$

Define

$${|f|}_{L^2(Z_n)}=\sqrt{{(f,f)}_{L^2(Z_n)}}=\sqrt{\sum _{x\in Z_n}f(x)\overline{f(x)}}.$$

For $a\in Z_n$, define ${\delta }_a:Z_n\to ℂ$ by

For $a,b\in Z_n$,

${({\delta }_a,{\delta }_b)}_{L^2(Z_n)}$ $={\displaystyle \sum _{x\in Z_n}}{\delta }_a(x)\overline{{\delta }_b(x)}$

$={\displaystyle \sum _{x\in Z_n}}{\delta }_a(x){\delta }_b(x)$

For $f\in L^2(Z_n)$,

$$f(x)=\sum _{a\in Z_n}f(a){\delta }_a(x)=\sum _{a=0}^{n-1}f(a){\delta }_a(x),x\in Z_n.$$

12 Fourier transform

Define the Fourier transform ${ℱ}_n:L^2(Z_n)\to L^2(\widehat{Z_n)}$ by

$$({ℱ}_{n}f)(e_a)=\sum _{x\in Z_n}f(x)\overline{e_a(x)}=\sum _{x\in Z_n}f(x)e_a(-x),e_a\in \widehat{Z_n}.$$

13 Pullback

We introduce the operator $F_n:L^2(Z_n)\to L^2(Z_n)$, which is defined via composition with the Fourier transform ${ℱ}_n$ and the function $\psi$ as

$$F_{n}f=({ℱ}_{n}f)\circ \psi .$$

That is, for $a\in Z_n$,

$$(F_{n}f)(a)=({ℱ}_{n}f)(e_a).$$

We pullback ${ℱ}_{n}f:\widehat{Z_n}\to ℂ$ to a function $F_{n}f:Z_n\to ℂ$.

We have

$({ℱ}_{n}f)(\psi (a))$ $={\displaystyle \sum _{x\in Z_n}}f(x)e_a(-x)$

$={\displaystyle \sum _{x\in Z_n}}f(x)\overline{e_a(x)}$

$={(f,e_a)}_{L^2(Z_n)}$

Thus

$$(F_{n}f)(x)={(f,e_x)}_{L^2(Z_n)},x\in Z_n.$$

14 Inverse Fourier transform

Define the Haar measure $m_{\widehat{Z_n}}$ on $\widehat{Z_n}$ by $m_{\widehat{Z_n}}(A)=\frac{1}{n}\cdot |A|$ for $A\in 𝒫(\widehat{Z_n})$.

Let $f\in L^2(Z_n)$ and let $x\in Z_n$.

${\displaystyle {\int }_{\widehat{Z_n}}}({ℱ}_{n}f)(\gamma )\gamma (x)𝑑m_{\widehat{Z_n}}(\gamma )$ $={\displaystyle \frac{1}{n}}{\displaystyle \sum _{\gamma \in \widehat{Z_n}}}({ℱ}_{n}f)(\gamma )\gamma (x)$

$={\displaystyle \frac{1}{n}}{\displaystyle \sum _{a\in Z_n}}({ℱ}_{n}f)(e_a)e_a(x)$

$={\displaystyle \frac{1}{n}}{\displaystyle \sum _{a\in Z_n}}\left({\displaystyle \sum _{y\in Z_n}}f(y)\overline{e_a(y)}\right)e_a(x)$

$={\displaystyle \frac{1}{n}}{\displaystyle \sum _{a\in Z_n}}{\displaystyle \sum _{y\in Z_n}}f(y)\overline{e_a(y)}{e}_a(x)$

$={\displaystyle \frac{1}{n}}{\displaystyle \sum _{y\in Z_n}}f(y)\left({\displaystyle \sum _{a\in Z_n}}{e}_a(x)\overline{e_a(y)}\right)$

We use the orthogonality relations for characters of finite abelian groups. For $a,b\in Z_n$ we have $e_a,e_b\in \widehat{Z_n}$, and

$$\sum _{x\in Z_n}{e}_a(x)\overline{e_b(x)}=n{\delta }_{a,b}.$$

Then, as $e_a(x)=e_x(a)$ and $e_a(y)=e_y(a)$,

${\displaystyle \frac{1}{n}}{\displaystyle \sum _{y\in Z_n}}f(y)\left({\displaystyle \sum _{a\in Z_n}}{e}_a(x)\overline{e_a(y)}\right)$ $={\displaystyle \frac{1}{n}}{\displaystyle \sum _{y\in Z_n}}f(y)\left({\displaystyle \sum _{a\in Z_n}}{e}_x(a)\overline{e_y(a)}\right)$

$={\displaystyle \frac{1}{n}}{\displaystyle \sum _{y\in Z_n}}f(y)\cdot n{\delta }_{x,y}$

$={\displaystyle \sum _{y\in Z_n}}f(y){\delta }_{x,y}$

$={\displaystyle \sum _{y\in Z_n}}f(y){\delta }_x(y)$

$=f(x).$

We have established that for $f\in L^2(Z_n)$ and for $x\in Z_n$,

$${\int }_{\widehat{Z_n}}({ℱ}_{n}f)(\gamma )\gamma (x)𝑑m_{\widehat{Z_n}}(\gamma )=\frac{1}{n}\sum _{a\in Z_n}({ℱ}_{n}f)(e_a)e_a(x)=f(x).$$

Also,

$$\frac{1}{n}\sum _{a\in Z_n}(F_{n}f)(a)e_a(x)=\frac{1}{n}\sum _{a\in Z_n}({ℱ}_{n}f)(e_a)e_a(x)=f(x).$$

We have established the Fourier inversion formula for $f\in L^2(Z_n)$:

$$f(x)=\frac{1}{n}\sum _{a\in Z_n}(F_{n}f)(a)e_a(x),x\in Z_n.$$

15 ${\mathrm{\ell }}^1(\mathbb{Z})$

Let ${ℂ}^{\mathbb{Z}}$ be the set of functions $\mathbb{Z}\to ℂ$.

Let $x\in {\mathrm{\ell }}^1(\mathbb{Z})$ be the set of those $x\in {ℂ}^{\mathbb{Z}}$ such that

and define

$${|x|}_{{\mathrm{\ell }}^1(\mathbb{Z})}=\sum _{n\in \mathbb{Z}}|x[n]|.$$

Let ${\mathrm{\ell }}^2(\mathbb{Z})$ be the set of those $x\in {ℂ}^{\mathbb{Z}}$ such that

16 ${\mathrm{\ell }}^2(\mathbb{Z})$

For $x\in {\mathrm{\ell }}^2(\mathbb{Z})$, define

$${|x|}_{{\mathrm{\ell }}^2(\mathbb{Z})}=\sqrt{\sum _{n\in \mathbb{Z}}{|x[n]|}^2},$$

and for $x,y\in {\mathrm{\ell }}^2(\mathbb{Z})$, define

$${(x,y)}_{{\mathrm{\ell }}^2(\mathbb{Z})}=\sum _{n\in \mathbb{Z}}x[n]\overline{y[n]}.$$

For $k\in \mathbb{Z}$, define $T_k:{ℂ}^{\mathbb{Z}}\to {ℂ}^{\mathbb{Z}}$ by

$$(T_{k}x)[n]=x[n-k],n\in \mathbb{Z}.$$