### Where does the Pi go?

A very short, sort of random thought.

It is fairly well known that experts in partial differential equations and experts in harmonic analysis prefer to define the Fourier transform differently. For harmonic analysts, the Fourier transform and its inverse are naturally

$\displaystyle \mathcal{F}[f](\xi) = \int f(x) \exp(-2\pi i \xi\cdot x) dx,~ \mathcal{F}^{-1}[g](x) = \int g(\xi) \exp(2\pi i \xi \cdot x) d\xi$

This definition has the advantange that the formulae, up to the single minus sign, is virtually identical for the forwards and backwards transforms. Furthermore, the transform is an isometry on the (Hilbert) space of square integrable functions. For people who do PDEs, the Fourier transform is often written

$\displaystyle \mathcal{F}[f](\xi) = \int_{\mathbb{R}^n} f(x) \exp(- i \xi \cdot x) dx, ~\mathcal{F}^{-1}[g](\xi) = \frac{1}{(2\pi)^{n}}\int_{\mathbb{R}^n} g(\xi) \exp(i \xi \cdot x) d\xi$

so that the Fourier transform of the derivative $\nabla f$ is given by $i\xi \mathcal{F}[f]$. This convention is more convenient for pseudo-differential calculus where you don’t want to carry around the factors of $2\pi$ everywhere. Sometimes to keep the transform as an isometry, the factor of $1 / (2\pi)^n$ gets distributed half on the forward transform and half on the backwards transform.

I’ve long taken as granted that these are two different point-of-views, and that depending on whether one uses more often the transforms themselves or the differentiation-multiplication-duality properties, one choose a convention that minimizes the spurious factors of $2\pi$ that floats around in the formula. More importantly, I’ve always thought that some factors of $2\pi$ are a necessary evil.

And yesterday, while trying to finally learn the Atiyah-Singer index theorem from the monograph by Peter Gilkey, I came across a wonderful definition that just blew my mind. On page 3, he just defines the notation $dx, dy, d\xi$, etc to be $1 / (2\pi)^{n/2}$ times the standard Lebesgue measure. And problem solved.