Fourier transform on a simple smooth 1-manifold 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 $\mathbb{R}$.
 A: 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 $\mathbb{Z}$.
