Zulip Chat Archive

Stream: general

Topic: finite fourier transform


Kenny Lau (Mar 17 2019 at 18:29):

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

Kenny Lau (Mar 17 2019 at 18:29):

Let GG be a finite abelian group, X(G):=Hom(G,C×)=Hom(G,S1)X(G) := \mathrm{Hom}(G,\Bbb C^\times) = \mathrm{Hom}(G,\Bbb S^1).

Kenny Lau (Mar 17 2019 at 18:30):

1. X(G) is finite.

Kenny Lau (Mar 17 2019 at 18:31):

Now for f:GCf:G \to \Bbb C, define f^:X(G)C\hat{f} : X(G) \to \Bbb C with f^(χ):=1GgGf(g)χ(g)=1GgGf(g)χ(g1)\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 (Mar 17 2019 at 18:32):

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

Kenny Lau (Mar 17 2019 at 18:32):

this is "fourier transform for finite abelian group"

Kenny Lau (Mar 17 2019 at 18:32):

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

Kevin Buzzard (Mar 17 2019 at 18:49):

This is all pretty trivial isn't it?

Kenny Lau (Mar 17 2019 at 18:49):

is it?

Kevin Buzzard (Mar 17 2019 at 18:49):

I doubt you need the structure theorem for finite abelian groups

Kenny Lau (Mar 17 2019 at 18:49):

math-trivial != Lean-trivial

Kenny Lau (Mar 17 2019 at 18:50):

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

Kevin Buzzard (Mar 17 2019 at 18:50):

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

Kenny Lau (Mar 17 2019 at 18:51):

maybe we should prove the structure theorem after all

Kenny Lau (Mar 17 2019 at 18:52):

I mean you still need the fact that 1GχX(G)χ(g)=δ1g\dfrac1{|G|} \sum_{\chi \in X(G)} \chi(g) = \delta_{1g}

Kevin Buzzard (Mar 18 2019 at 00:10):

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 (Mar 18 2019 at 00:10):

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

Kevin Buzzard (Mar 18 2019 at 00:12):

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

Kevin Buzzard (Mar 18 2019 at 00:13):

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


Last updated: Dec 20 2023 at 11:08 UTC