Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been studying the Fourier Transform on Riemannian Manifolds. I got the general idea (I think) and I started by trying to define a Fourier Transform on the circle, $S^1$. I have found that this result is already known: $$F(n)=\frac{1}{2\pi}\int_0^{2\pi}f(\theta)e^{-in\theta}d\theta,$$ the same result that I got. But then I got confused... Would the inverse Fourier transform be: $f(\theta)=\sum_{n\in\mathbb Z}^\infty F(n) e^{in\theta}$?

If so, if the circle doesn't have unitary radius but some value $R\in \mathbb R$ I imagine that in the definition of the Fourier transform I would have to change $d\theta \to Rd\theta$...is this correct? And what about the inverse Fourier transform?

share|improve this question
1  
What is the Fourier Transform on Riemannian Manifolds? –  timur Mar 16 '13 at 0:50
    
@timur You can check this other question that I asked on Physics.SE. A more complete answer would be the one found in here. –  PML Mar 16 '13 at 1:02
    
Thanks. There is no "the" Fourier transform on a general Riemannian manifold. What you are asking about is Fourier series. –  timur Mar 16 '13 at 2:49
    
@timur I'm confused... Isn't the Fourier transform a generalization of Fourier series to non-periodic functions? Are you telling me that I can no longer define a Fourier Transform on a Riemannian Manifold? I realize that in a general Riemannian Manifold one can't define a Fourier Transform (series?) but in some specific cases when the Riemannian manifold is either a Lie group or a symmetric space one can, right? –  PML Mar 16 '13 at 11:05
1  
As for the Fourier series, what you say is true in a certain sense. Your question is entirely about the Fourier series, and I don't see why you need to mention Riemannian manifolds etc. It might be a good idea to try to understand Fourier series fairly well before talking about generalizations of Fourier transform to Riemannian manifolds. –  timur Mar 17 '13 at 20:55

2 Answers 2

up vote 3 down vote accepted

One way to think about the FT on the circle is that it is really the FT of a periodic function. Let's think about the FT of a function $f$ that has period $p$. We may write $f$ as

$$f(x) = \sum_{n=-\infty}^{\infty} f_0(x+n p)$$

where

$$f_0(x) = \begin{cases} \\ f(x) & |x|<p/2 \\ 0 & |x| > p/2\end{cases}$$

We may rewrite this expression for $f$ as a convolution:

$$f(x) = f_0(x) \otimes \sum_{n=-\infty}^{\infty} \delta(x+n p) $$

where $\delta$ is the Dirac delta function and $\otimes$ denotes convolution. The series of delta functions on the right is known as a comb function. We may then find the FT of $f$, $\hat{f}$:

$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: f(x) \,e^{i k x}$$

by the convolution theorem. Note that the FT of the comb function is another comb function; the result is

$$\hat{f}(k) = 2 \pi \hat{f_0}(k) \sum_{n=-\infty}^{\infty} \delta(p k - n) = \frac{2 \pi}{p} \sum_{n=-\infty}^{\infty} \hat{f_0}\left(\frac{n}{p}\right) \delta\left(k-\frac{n}{p}\right) $$

That last step results from the sampling property of delta functions. Taking the inverse FT gives us a new representation for $f$:

$$f(x) = \sum_{n=-\infty}^{\infty} \frac{1}{p} \hat{f_0}\left(\frac{n}{p}\right) \exp{\left(i \frac{n x}{p}\right)}$$

and, from above,

$$\frac{1}{p} \hat{f_0}\left(\frac{n}{p}\right) = \frac{1}{p} \int_{-p/2}^{p/2} dx \: f_0(x) \exp{\left(i \frac{n x}{p}\right)}$$

Now, finally, to answer your question: the period $p$ is indicative of a map from a circle of circumference $p$ to the interval $x \in [-p/2,p/2)$. From the above, the coefficients of a Fourier series are simply the FT of a single period of the periodic function. Thus, the inverse transforms for the circle and the real line are one and the same.

share|improve this answer

There is a general theorem of Pontryagin for locally compact abelian groups.

The dual of $\mathbb R$, meaning the group of all continuous maps $\mathbb R \to \mathbb U$ is $\widehat{\mathbb R} \simeq \mathbb R$.

For the circle $\widehat{\mathbb U} = \mathbb Z$.

And here goes the general definition of the Fourier transform. Let $f \in \mathrm L^2(\mathrm G)$ with $\mathrm G$ a locally compact abelian group, for $\chi \in \widehat{\mathrm G}$, $$ \widehat{f}(\chi) = \int_{g \in \mathrm G} f(g) \overline{\chi(g)} d\mu$$ where $d\mu$ is a Haar measure on $\mathrm G$.

With this, you get back all the theorems known for the Fourier transform on $\mathbb R$. Especially the Fourier Inverse is $$ \hat{h}(g) = \int_{\chi \in \widehat{\mathrm G}} h(g) \chi(g) d\mu$$.

This explains you why in one case you have an integral on $\mathbb R$ (for the Fourier Transform on $\mathbb R$) and on the other case you have a series (for the Fourier Transform on the circle).

share|improve this answer
    
Although both answers appear to me to be right. Ron's answer is more complete. I'm having a bit of a dilemma, though, (bounty award wise) because your answer is clearer to me - since I've been studying the compact groups and Characters point of view of the Fourier Series/transform - and it allows for generalization for other Lie groups... If you could edit your question to make it more complete, not just for me but to some one else who stumbles on this kind of problems I would consider giving you the award. =) –  PML Mar 22 '13 at 15:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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