Viewing the Fourier transform on $\mathbb{Z}_n$ as a mapping from functions $G\to\mathbb{C}$ to functions $\operatorname{Irr}(G)\to\mathbb{C}$

Question

The Fourier transform on $G:=\mathbb{Z}_n$ is an isomorphism $\mathcal{F}:\mathbb{C}[G]\to\bigoplus_{V\in\operatorname{Irr}(G)}\operatorname{End}(V)$. I'm having trouble understanding how it can be viewed as a mapping from functions $G\to\mathbb{C}$ to functions $\hat{G}:=\operatorname{Irr}(G)\to\mathbb{C}$. I understand that since $G$ is Abelian, all its representations are one dimensional, and therefore, all $\operatorname{End}(V)$, where $V\in\operatorname{Irr}(G)$, are also one dimensional. But how does this induce a mapping from functions $G\to\mathbb{C}$ to functions $\hat{G}\to\mathbb{C}$? I would appreciate being very explicit, as I've been stuck on this for a while.

Answer 1

Let $G$ be a finite abelian group (written multiplicatively, for $x, y \in G$ the group operation is $xy$). Every irreducible complex representation is one-dimensional, so we define the dual group as the set of group homomorphisms $G \to \mathbb{C}^\times$ with the group operation of pointwise multiplication: $$ G^\vee = \{ \chi: G \to \mathbb{C}^\times \mid \chi(xy) = \chi(x) \chi(y) \text{ for all } x, y \in G\}.$$ Note that multiplication in $G^\vee$ is pointwise product of characters, so for $\chi, \eta \in G^\vee$ we have $$ (\chi\eta)(x) := \chi(x) \eta(x).$$ For an arbitrary function $f: G \to \mathbb{C}$, define its Fourier transform $\mathcal{F} f: G^\vee \to \mathbb{C}$ as the function $$ (\mathcal{F}f)(\chi) := \sum_{x \in G} f(x) \chi(x).$$ Writing $F(G)$ and $F(G^\vee)$ for the vector spaces of functions on $G$ and $G^\vee$ respectively, the Fourier transform is a map $\mathcal{F}: F(G) \to F(G^\vee)$. By taking another Fourier transform to get a function on $G^{\vee \vee}$, and using the canonical isomorphism $G \to G^{\vee \vee}$, you can show that $\mathcal{F}$ is in fact invertible.

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But enough abstraction, what does this look like concretely? Let's take $G = \mathbb{Z}_3 = \{0, 1, 2\}$ written additively, so $1 + 2 = 0$ for instance. Let $\omega = \exp(2 \pi i / 3)$ be the principal primitive 3rd root of unity, then the elements of the dual group are $$ G^\vee = \{\chi_0, \chi_1, \chi_2\}, \quad \text{where } \chi_i(n) = \omega^{ni}.$$ Concretely, the elements of the dual group are the maps $$ \begin{aligned} \chi_0: G \to \mathbb{C}^\times & & 0 \mapsto 1 && 1 \mapsto 1 && 2 \mapsto 1, \\ \chi_1: G \to \mathbb{C}^\times & & 0 \mapsto 1 && 1 \mapsto \omega && 2 \mapsto \omega^2, \\ \chi_2: G \to \mathbb{C}^\times & & 0 \mapsto 1 && 1 \mapsto \omega^2 && 2 \mapsto \omega^4=\omega. \end{aligned} $$ A function $f: G \to \mathbb{C}$ can be identified with the vector $(f(0), f(1), f(2)) \in \mathbb{C}^3$, and we can see what happens to it under the Fourier transform. $$ \begin{aligned} (\mathcal{F}f)(\chi_0) &= \sum_{x \in G} f(x) \chi_1(x) = f(0) + f(1) + f(2), \\ (\mathcal{F}f)(\chi_1) &= \sum_{x \in G} f(x) \chi_1(x) = f(0) + \omega f(1) + \omega^2 f(2), \\ (\mathcal{F}f)(\chi_1) &= \sum_{x \in G} f(x) \chi_1(x) = f(0) + \omega^2 f(1) + \omega^4 f(2). \end{aligned} $$ Identifying the function $g: G^\vee \to \mathbb{C}$ with the vector $(g(\chi_0), g(\chi_1), g(\chi_2))$, we see that the matrix of the Fourier transform is $$ [\mathcal{F}] = \begin{pmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega^4 \end{pmatrix},$$ which should be familiar.