0
$\begingroup$

The following exercise is exercise 2.2.3 from these lecture notes by Daniel Bump. If $G$ is a finite abelian group, then denote by $G^{\ast}$ the set of its characters, further if $x \in G$ let $\check x$ be the function on $G^{\ast}$ given by $\check x(\chi) = \chi(x)$.

Let $\mathcal F : L^2(G) \to L^2(G^{\ast})$ be the Fourier transform, defined by $\mathcal{F}f = \hat f$, where $\hat{f} : L^2(G^{\ast}) \to \mathcal C$ $$ \hat f(\chi) = \frac{1}{\sqrt{|G|}} \sum_{x\in G} \chi(x) f(x). $$ Prove that $$ f(x) = \frac 1{\sqrt{|G^\ast|}} \sum_{\chi \in G^\ast} \overline{\check x(\chi)}\hat f(\chi). $$ (I recently asked a question because the formula is messed up in the original document, see here).

Now I tried to prove it, using that the characters form an orthonormal set we can write $$ f = \sum_i c_i \chi_i $$ where $G^{\ast} = \{ \chi_i : i = 1,\ldots, n \}$ are the characters in some order. And pluggin in the given formula for $\hat f$ with $n = |G| = |G^{\ast}|$: \begin{align*} \frac{1}{\sqrt n} \sum_{\chi \in G^{\ast}} \overline{\check x(x)} \hat f(x) & = \frac{1}{\sqrt n} \sum_{\chi \in G^{\ast}} \overline{\check x(x)} \left( \frac{1}{\sqrt n} \sum_{g\in G} \chi(g) f(g) \right) \\ & = \frac{1}{n} \sum_{\chi \in G^{\ast}} \sum_{g\in G} \overline{\check x(x)} \chi(g) f(g) \\ & = \sum_{\chi \in G} \overline{\check x(x)} \frac{1}{n}\left( \sum_{g\in G} \chi(g) f(g) \right) \\ & = \sum_{\chi \in G} \overline{\check x(x)} \langle \chi, f \rangle. \end{align*} Now with the above $\langle \chi, f \rangle = \sum_i c_i \langle \chi, \chi_i \rangle = c_k$ with $\chi = \chi_k$. So all I got is $$ \frac{1}{\sqrt n} \sum_{\chi \in G^{\ast}} \overline{\check x(x)} \hat f(x) = \sum_{\chi \in G} \overline{\check x(x)} c_k = \sum_{\chi \in G} \overline{\chi(x)} c_k $$ with $\check x(x) = \chi(x)$, but I have to establish that this equals $f(x)$, so how to proceed?

$\endgroup$
1
$\begingroup$

I think it's not so good to use natural number indices for the representation of $f$ as a linear combination of characters. I believe it's better to write

$$f = \sum_{\psi \in G^\ast} c_{\psi}\cdot \psi.\tag{1}$$

And then compute $\hat{f}(\chi)$ using $(1)$:

\begin{align} \hat{f}(\chi) &= \frac{1}{\sqrt{\lvert G\rvert}}\sum_{g\in G} \chi(g)f(g)\\ &= \frac{1}{\sqrt{\lvert G\rvert}} \sum_{g\in G} \chi(g)\Biggl(\sum_{\psi\in G^\ast} c_\psi \psi(g)\Biggr)\\ &= \frac{1}{\sqrt{\lvert G\rvert}} \sum_{\psi \in G^\ast} c_\psi \Biggl(\sum_{g\in G} \chi(g)\psi(g)\Biggr), \end{align}

noting

$$\sum_{g\in G} \chi(g)\psi(g) = \begin{cases} \lvert G\rvert &, \psi = \overline{\chi} \\ 0 &, \psi \neq \overline{\chi}\end{cases}$$

to conclude

$$\hat{f}(\chi) = \sqrt{\lvert G\rvert}\cdot c_{\overline{\chi}}.\tag{2}$$

We note in passing that it would have been more convenient to define the Fourier coefficients with $\overline{\chi(g)}$ rather than $\chi(g)$.

Now we compute

\begin{align} \frac{1}{\sqrt{\lvert G^\ast\rvert}} \sum_{\chi \in G^\ast} \overline{\check{x}(\chi)}\hat{f}(\chi) &= \frac{1}{\sqrt{\lvert G^\ast\rvert}} \sum_{\chi \in G^\ast} \overline{\chi(x)} \cdot \sqrt{\lvert G\rvert} c_{\overline{\chi}}\\ &= \frac{\sqrt{\lvert G\rvert}}{\sqrt{\lvert G^\ast\rvert}} \sum_{\chi \in G^\ast} c_{\overline{\chi}}\cdot \overline{\chi}(x)\\ &= \frac{\sqrt{\lvert G\rvert}}{\sqrt{\lvert G^\ast\rvert}} \sum_{\psi \in G^\ast} c_\psi\cdot \psi(x)\\ &= \frac{\sqrt{\lvert G\rvert}}{\sqrt{\lvert G^\ast\rvert}} \cdot f(x). \end{align}

And now we need the fact that for finite groups we have $\lvert G\rvert = \lvert G^\ast\rvert$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.