User:J D Bowen/LA Term

Abstract
In this article, we will discuss the finite Fourier transform of functions on finite fields. We will examine when such transforms exist, what applications they have, and how we can represent the transform in various ways.

We will need to examine some properties of finite graphs and spaces over them. We intend to investigate the effect of the discrete Fourier transform on the function space of a finite set, including convolutions of finite functions and their similarities and differences to more familiar continuous concepts. We describe the fast Fourier transform, and examine how and why this provides a computational speed-up over the original formulation of the transform.

Definitions and First Results
The Fourier transform of a function from $$\mathbb{C} \to \mathbb{C} $$ is studied in any undergraduate course in differential equations. But this concept can be modified for the finite field $$\mathbb{Z}_p \ $$. Given a function $$f:\mathbb{Z}_p \to \mathbb{C} \ $$, we define the finite Fourier transform of this function as

$$\mathfrak{F}(f)(x)=\hat{f}(x)= \sum_{y\in\mathbb{Z}_p} f(y)e^{\frac{-2\pi i xy}{p}} \ $$.

Note that this constitutes a transformation of the vector space $$L^2(\mathbb{Z}_p) \ $$ over $$\mathbb{C} \ $$. If we define $$\delta_i(j):=\delta_{ij} \ $$, it is clear that this forms a basis for $$L^2(\mathbb{Z}_p) \ $$. Under this basis, a function $$f:\mathbb{Z}_p \to \mathbb{C} \ $$ has vector representation

$$f = (f(0),f(2),\dots,f(p-1))^T \ $$,

because

$$f(0)\delta_{x,0}+f(1)\delta{x,1}+\dots+f(p-1)\delta_{x,p-1} = f(x) \ $$.

Notice that the definition of $$\hat{f} \ $$ implies that we can represent the finite Fourier transform as a matrix:

$$[(e^{\frac{-2\pi i (i-1)(j-1)}{p}})]_{i,j=1,1}^{p,p} (f(0),f(2),\dots,f(p-1))^T = \sum_{y\in\mathbb{Z}_p} f(y)e^{\frac{-2\pi i xy}{p}} = \hat{f}(x) \ $$.

So $$\mathfrak{F} \ $$ is a linear transformation of $$L^2 (\mathbb{Z}_p) \ $$. It follows from the form of the matrix that this is a bijection on $$L^2(\mathbb{Z}_p) \ $$.

It becomes natural to ask what the eigenvalues of this transformation are. To answer this questions, observe that

$$\mathfrak{F} \delta_a(-x)=e^{2\pi i ax/p} \implies \mathfrak{F}^2(\delta_a(x)) = \sum_{y\in\mathbb{Z}_p} e^{2\pi i y(x-a)/p} \ $$.

But $$\sum_{y\in\mathbb{Z}_p} e^{2\pi i xy/p} = p \ \text{if} \ x=0 \ \text{mod} \ p, \ 0 \ \text{otherwise}. \ $$

So $$\mathfrak{F}^2 ( \delta_a(-x)) = p\delta_a(x) \ $$.

This implies $$\mathfrak{F}^4 = p^2I \ $$, and so the eigenvalues of the discrete Fourier transform are the solutions of $$\lambda^4-p^2=0 \ $$, ie,

$$\sqrt{p},-\sqrt{p}, i\sqrt{p}, -i\sqrt{p} \ $$. [1]

Graph Theory
Finite fields, or for that matter, finite groups $$\mathbb{Z}/n\mathbb{Z} \ $$, can be thought of as circulant graphs. Given the success of

For a graph of $$n \ $$ vertices labelled $$0,\dots, n-1 \ $$, we can form the adjacency matrix $$A=(a_{ij})_{nn} \ $$, where $$a_{ij} \ $$ is 1 if $$i, j \ $$ are directly connected in the graph, and $$0 \ $$ if they are not. Obviously, such a matrix will be diagonally symmetric, since $$a_{ij}=a_{ji} \ $$.

Define the Cayley graph as being a graph with with an adjacency matrix that is symmetric about the other diagonal as well, ie, $$a_{p+1-j,p+1-i}=a_{i,j} \ $$.

[1] Terras, Audrey. Fourier analysis on finite groups and application.