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If I have a function $f(\mathbf{x})$ defined over $\mathbb{R}^3$ and wish to make a Fourier transform of this function, I do

\begin{equation} f(\mathbf{x})=\int_{\mathbb{R}^3}\hat{f}(\mathbf{k})e^{2\pi i \mathbf{x} \cdot \mathbf{k}} d^3k, \end{equation} up to some normalization factor, which is equivalent to expanding $f(\mathbf{x})$ in a basis of plane waves.

The question is, if $f$ is defined instead over the three-torus $\mathbb{T}^3$ or perhaps $\mathbb{T}^2 \times \mathbb{R}$ or $S^1 \times \mathbb{R}^2$, then how do I go about doing this Fourier transform?

My first instinct is to say that a periodic boundary condition in the $i^\text{th}$ dimension would make the wave modes in that direction discrete:

\begin{equation} k=\frac{2\pi}{L_i}n_i, \end{equation}

for all integers $n_i$ given the torus length $L_i$. This would then change the integral over wave-vectors in that direction to a summation:

\begin{equation} \int dk \to \frac{1}{L_i}\sum_{k}, \end{equation} and continue with the analysis as before. Is this the correct approach? Thanks for any help!

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Yes, that is the correct approach. – Robert Israel Feb 13 '13 at 20:17
In the general setting of duality of locally compact abelian groups, the dual of a compact group is discrete, and vice versa. – user53153 Feb 13 '13 at 20:30
up vote 1 down vote accepted

For function on the circle i.e $2\pi$-perdiodic functions, you can use the Fourier Series. And the 3-torus is just $\rm S^1 \times \rm S^1 \times \rm S^1$.

For $\rm S^1 \times \mathbb R^2$ use Fourier Series on the first variable and Fourier transform for the last two.

Justification : the characters of $\mathbb R$ are the functions $x \to e^{iax}$ for $a \in \mathbb R$. While the characters of $\mathbb R/2\pi\mathbb Z$ are the functions $x \to e^{inx}$ with $n \in \mathbb N$.

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Thanks Damien. Can you expand on the "character" comment or point me to a reference? I'm not familiar with this concept and Google is pointing me to irrelevant pages. – cosmoguy Feb 13 '13 at 20:34
Take any topological group $\rm G$ with an Haar mesure $d\mu$ on it. Note $\widehat{\rm G}$ its set of characters i.e continuous morphisms $\rm G \to \rm S^1$. Then for any function $f \in \rm L^1(\rm G)$, its Fourier transform is $\widehat{f}(\chi) = \int_{g \in \rm G} f(g) \chi(g)d\mu$. – Damien L Feb 13 '13 at 20:37

Another geometry for 3 dimension is $\mathbb{R}^1 \times S^2$, otherwise known as spherical coordinates. The variables over the sphere are periodic with the spherical polynomials as a basis. Over $\mathbb{R}^1$, the Fourier basis is discrete if the object has finite support. In $n$ dimensions, this can be generalized to $\mathbb{R} \times S^{n-1}$ with the ultra spherical polynomials.

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