Intuition behind Clifford Fourier transform

Question

I have a question regarding the Clifford Fourier transform. For the classical Fourier transform we have a clear spectral interpretation and the power of Fourier transform is obvious. However, I am curious to know the intuitive meaning behind the Clifford Fourier transform and its applications. What about its spectral interpretation? I will be thankful if you provide explanations for these things behind Clifford algebra.

Answer 1

In your pdf the finite dimensional real algebra $A=G_3$ is a $\Bbb{C}$-algebra, where the complex number $i$ is what they call $i_3$ (ie. $i_3$ commutes with every element of $A$ and it satisfies $i_3^2=-1$).

Let $f: \Bbb{R}^n\to A$. Take a complex basis $A=\sum_{j=0}^J e_j \Bbb{C}$ so that $$f=\sum_{j=0}^J e_j f_j=\sum_{j=0}^J f_j e_j $$ with $f_j:\Bbb{R^n\to C}$. Then the Fourier transform of $f$ is just $$F(f)(\omega)= \sum_{j=0}^J e_j F(f_j)(\omega)= \sum_{j=0}^J F(f_j) e_j=\int_{\Bbb{R}^n} f(x)e^{i (\omega^\top x)}dx $$ where $F(f_j)$ is the standard Fourier transform of complex valued functions and $i=i_3\in A$ and $\omega^\top x$ is a dot product vectors of $\Bbb{R}^n$ and the complex number $e^{i (\omega^\top x)}$ is in $A$.

Obviously the inverse Fourier transform is $$F^{-1}(f)=\sum_{j=0}^J e_j F^{-1}(f_j)$$ What is it useful for ? Let $g:\Bbb{R}^n\to A,g=\sum_{j=0}^J e_j g_j$. We have the convolution $$f\star g(x)=\int_{\Bbb{R}^n} f(y)g(x-y)dy=\sum_{j=0}^J \sum_{l=0}^L e_j e_l f_j\ast g_l(x)$$ where $\ast$ is the standard convolution of complex valued functions and I used that $\Bbb{C}\subset A$ commutes with $A$ to move the $e_l$ on the left.

Then the Fourier transform "diagonalizes" the convolution, ie. $F(f\star g)=F(f)F(g)$ where the latter is a non-commutative pointwise product.