TITLE: definition of the Fourier transform of function on the sphere
QUESTION [6 upvotes]: Let $f: S^{n-1}\longrightarrow R^n$ be  even continuous function.
What is the Fourier transform of $f$?

REPLY [2 votes]: One way of defining the Fourier transform over the $n-$dimensional sphere, is to define it over $\mathbb{R}^{n-1}$ and use the Cayley transform to map the operator to $\mathbb{S}^{n-1}$. If we do this,
then we define the Fourier transform $\mathcal{F}$ of an $L^2-$integrable $f$ over $\mathbb{S}^{n-1}$ with respect to the Clifford algebra $A_n$ as
\begin{equation*}
\mathcal{F}_{\mathbb{S}^{n-1}}(f)(u)=\int_{\mathbb{S}^{n-1}}H_{\mathbb{S}^{n-1}}(x,u)f(x)d\sigma(x)
\end{equation*}
where
\begin{equation*}
H_{\mathbb{S}^{n-1}}=c(n)J(\psi,u)e^{-i\langle \psi(x),\psi(u)\rangle}\overline{J}(\psi,x).
\end{equation*}
Here, $d\sigma$ is the respective measure, $c(n)$ is a normalisation constant, and $\psi$ is the inverse of the Cayley transformation.
Note: When we are doing any kind of harmonic analysis over the sphere, we need to introduce spherical harmonics.