finite fourier transform

Stream: general

[Kenny Lau]

How feasible is proving fourier transform for a finite abelian group? (more details to come)

[Kenny Lau]

Let $G$ be a finite abelian group, $X(G) := \mathrm{Hom}(G,\Bbb C^\times) = \mathrm{Hom}(G,\Bbb S^1)$.

[Kenny Lau]

1. X(G) is finite.

[Kenny Lau]

Now for $f:G \to \Bbb C$, define $\hat{f} : X(G) \to \Bbb C$ with $\hat{f}(\chi) := \dfrac1{|G|} \sum_{g \in G} f(g) \overline{\chi(g)} = \dfrac1{|G|} \sum_{g \in G} f(g) \chi(g^{-1})$

[Kenny Lau]

2. theorem: $f(g) = \sum_{\chi \in X(G)} \hat{f}(\chi) \chi(g)$

[Kenny Lau]

this is "fourier transform for finite abelian group"

[Kenny Lau]

(more details to come if people are interested, but that's currently the end of the details)

[Kevin Buzzard]

This is all pretty trivial isn't it?

[Kenny Lau]

is it?

[Kevin Buzzard]

I doubt you need the structure theorem for finite abelian groups

[Kenny Lau]

math-trivial != Lean-trivial

[Kenny Lau]

I wouldn't know how to prove it without the structure theorem

[Kevin Buzzard]

You prove by induction that G-hat has the same size as G

[Kenny Lau]

maybe we should prove the structure theorem after all

[Kenny Lau]

I mean you still need the fact that $\dfrac1{|G|} \sum_{\chi \in X(G)} \chi(g) = \delta_{1g}$

[Kevin Buzzard]

A standard undergraduate course in representation theory proves all this and more -- even an analogue in the non-abelian case. All it needs is the structure of semisimple modules over an alg closed field.

[Kevin Buzzard]

You cannot possibly need the structure theorem for finite abelian groups, because the result generalises to the non-abelian case.

[Kevin Buzzard]

If G acts on itself by left multiplication and you extend to an action on $\mathbb{C}[G]$ then this is just the statement that the identity has $n$ fixed points and everything else has none.

[Kevin Buzzard]

Now you prove that every character shows up in the regular representation and you're done by counting.