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Assume a very simple smooth 1-manifold, with a single chart covering, What I'd like to know is, can we use and Fourier transform for functions on this manifold just as we did for the case of $\mathbb{R}$ with all the properties of the Fourier transform applicable ? If so, Is the proof simple ?

EDIT [in response to comment from Pete L. Clark]

Consider the maps $y : (0,1) \to \mathbb{R}$ and a bijective map $z : (0,1) \to (0,1)$ where $z$ is continuous in $(0,1)$. consider $\tau \in (0,1)$ where $\tau = z(t)$ for some $t \in (0,1)$. Let $x(t) = y(\tau)$, I want to represent $y(\tau)$ using Fourier transform (with all the properties of FT) rather than taking FT of $x(t)$. This would ease me in avoiding some derivations.

EDIT 2 [earlier it should be read as](sorry for the error in earlier edit)

Let $M$ be smooth 1-manifold. Consider the maps $y : M \to \mathbb{R}$ and a bijective map $z : (0,1) \to M$ where $z$ is continuous in $(0,1)$ is a co-ordinate chart which covers $M$. Can i take the Fourier Transform of $y$ as though it ($y$) is acting on $\mathbb{R}$.

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    A smooth $n$-manifold with a single coordinate chart is, up to diffeomorphism, an open subset of $\mathbb{R}^n$. But what do you mean by the Fourier transform in this case -- say for the open interval $(0,1)$? The usual formalism of the Fourier transform requires a group structure which is not present here.2011-07-08
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    @Pete L. Clark : I do not intend to go deep into the formalism as my purpose would serve if the functions defined on the manifold are all reasonably well behaved as piece-wise continuous and are of bounded variation. please clarify if i could avoid the formalism in this case.2011-07-08
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    $@$Rajesh: no, the issue is not with the well-behavedness of the functions: you just (as far as I can see; I am not a real expert in this branch of mathematics) need an algebraic structure that you don't have in order to define the transform. My question above is a serious one: suppose you have a function $f: (0,1) \rightarrow \mathbb{R}$, as nice as you want: e.g. extending smoothly to $[0,1]$ (well, we had better not assume that $f(0) = f(1)$!). I just don't know what "the Fourier transform of $f$" means: what do *you* mean by it?2011-07-08

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If you take a smooth $1$-manifold with a single covering, it is homeomorphic to $\mathbb{R}$, and so by fixing a specific homeomorphism, you can do fourier transforms. This would be a little forced, as Fourier transforms don't transform in any meaningful way under diffeomorphisms of $\mathbb{R}$

The Fourier transform makes essential use of the fact that $\mathbb{R}$ is a (locally compact abelian) group. You have Fourier transforms swap multiplication and convolution, but you can't even define convolution without using the group structure (and you also want that the Lebesgue measure on $\mathbb{R}$ is translation invariant, and thus a Haar measure).

In general, you can define the Fourier transform on a locally compact abelian compact group $G$, and it sends functions on $G$ to functions on the Pontryagin dual. In fact, the Fourier coefficients of periodic functions are just what you get when you do the Fourier transform on $S^1$, as the Pontryagin dual of the circle is just $\mathbb{Z}$.

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    could you please give a quick reference book where i can find all the basics required for a formalism of Fourier transform including the group structure.2011-07-08
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    I don't know of a good reference for Fourier analysis on locally compact groups, and unfortunately an analysis book that does Fourier on $\mathbb{R}^n$ is going to stress the analytical point instead of the group theoretic ones, but Rudin's "Real and Complex Analysis" has a chapter on Fourier transforms, and a cursory look at the basic properties reveal that they use addition/subtraction everywhere. This IS the group structure. What is less obvious at first is that the functions $e^{iax}$ are exactly the characters of $\mathbb{R}$.2011-07-08
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    Another possible reference is the [Fourier analysis book by By Elias M. Stein and Rami Shakarchi](http://books.google.com/books?id=FAOc24bTfGkC&printsec=frontcover&dq=stein+shakarchi+fourier+analysis&hl=en&ei=NqwWTrOKLvGNsALG4Z08&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCkQ6AEwAA#v=onepage&q&f=false). Again, they probably do not discuss group theoretic concerns directly, but it will at least give you a good sense of how Fourier transforms are used2011-07-08
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    I have not looked at it, but Rudin's [Fourier Analysis on Groups](http://books.google.com/books?id=WlzLFUo0VzYC&printsec=frontcover&dq=rudin+fourier+analysis+on+groups&hl=en&src=bmrr&ei=N68WTqW-CtGHsALKyKB3&sa=X&oi=book_result&ct=book-thumbnail&resnum=1&ved=0CCoQ6wEwAA#v=onepage&q&f=false) might have what you want.2011-07-08
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    I do not intend to go deep into the formalism as my purpose would serve if the functions defined on the manifold are all reasonably well behaved as piece-wise continuous and are of bounded variation. please clarify if i could avoid the formalism in this case.2011-07-08
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    @Aaron: +1: your answer is essentially my comment, but better explained. Let me just say though that I have no reason to believe sending the OP to advanced texts on Fourier analysis on locally compact groups is going to help him: at that level, the group $G$ is explicitly part of the setup, but the main problem here is that the group is conspicuously missing. Right?2011-07-08
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    @Pete @Rajesh It really depends on what Rajesh wants to do with the Fourier transform. Of course if he has a single coordinate chart, he can just fix his chart once and for all and pretend he is working on $\mathbb{R}$. In this case the problem becomes whether he has any additional structure on the manifold (a Riemannian metric etc) that needs to be factored in.2011-07-08
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    More precisely, in one-dimension, the "group structure" manifests as the translation invariance of the measure, which can easily be restored on a one-dimensional smooth manifold with some volume form as long as we assume that the manifold doesn't have finite volume. It is much more tricky in higher dimensions.2011-07-08
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    @Pete I agree that sending OP to an advanced book dealing with Fourier analysis on groups probably won't be the most useful thing, but he asked for a reference, and if he doesn't want to settle for "the formula for convolution uses addition," I would rather he have somewhere thorough but inappropriate to look, just so he can skim it and say "Yes, there is the group structure."2011-07-08
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    @Rajesh You don't need to go into details of the formalism at all if you are willing to accept that you do Fourier transforms on $\mathbb{R}^n$. The point of looking into the formalism is just to see what you do need, and to appreciate that you do NOT have it on a manifold, even if the manifold is homeomorphic to $\mathbb{R}$:you need three things, a group structure, a translation invariant measure (so that you have a notion of integration), and an analogue of the functions $x\mapsto e^ax$. These are all missing in your setup, even when you restrict your functions.2011-07-08
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    @Aaron : In my understanding, you say that I cannot pretend that $y(\tau)$ is operating on $\mathbb{R}$ and take a Fourier transform. Please let me know whether I got it right. Comments from Willie seem to suggest that I can. I do like to go through the book by Rudin, but currently I need an answer of the form yes/no.2011-07-09
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    @Aaron : also in case if i did get the set up, does it involve putting some restrictions on $z(t)$ ?2011-07-09
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    @Rajesh I'm not quite sure what you mean by your edit or your question, but if you're asking what I think you are: A function on $(0,1)$ is a function defined on $\mathbb{R}$, and as long as $\int_{\mathbb R} |f(x)|dx <\infty$, you can take the Fourier transform. You don't need any continuity conditions beyond measurability (which is implied by piece-wise continuity). HOWEVER, Fourier transforms don't behave well with respect to composition, and so you are going to lose all the nice properties that Fourier has if you do what you are suggesting.2011-07-09
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    @Rajesh: Phrased differently, convolution, differentiation, and multiplication by $x$ all change in very different ways under change of variables, so depending on what you mean by "the properties of fourier transforms" everything will break down, and the relations you have on each side will look very different. If you are judicious, you can write out the formulas that you do get between the regular transformation and the transformation after change of variables, but the formulas will be messy and won't be in any standard references.2011-07-09
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    @Aaron : sorry for the mistake in my earlier edit which was due to some confusion. please see the latest edit, I think in this case your answer is 'no', but I really appreciate some help in creating a Group structure.2011-07-09
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    @Aaron let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/761/discussion-between-rajesh-d-and-aaron)2011-07-09