# Fourier transform in curved spacetimes

When in a flat spacetime, one can use the identity

$$\int^\infty_{-\infty} d^3k~ e^{i \bf{ k \cdot r}} f(k)=\int^\infty_{-\infty} dk ~ k f(k)\sin(kr)$$

Does this generalise to curved spacetimes, for example de Sitter?

• Care to check: physics.stackexchange.com/q/56551 – PML Nov 2 '17 at 23:03
• On this paper (rspa.royalsocietypublishing.org/content/319/1539/509) the author discusses one way to define one sort of Fourier transform in a curved spacetime for the purposes of defining multipole moments for the distribution of matter and charge. I don't know if it can be of use for you, but it might be good to know about. – user1620696 Nov 2 '17 at 23:47
• Just realised, the Harish-Chandra Schwartz space allows one to define a Schwartz space on a semisimple Lie group. Since Fourier transforms are automorphisms of a Schwartz space, this may allow one to generalise the Fourier transform to some manifolds. – JamalS Nov 3 '17 at 23:08

I am unaware of a generalisation of the Fourier transform to arbitrary Riemannian manifolds. But it can be generalised in the context of Lie groups, which of course, are manifolds by definition.

For certain groups, if $f$ is some function in $L^1(G)$, we can define a Fourier transform $\hat f$ on the Pontryagin dual $\hat G$ by,

$$\hat f(\chi) = \int_G f(x)\overline{\chi(x)} \, \mathrm d\mu$$

using the Haar measure, where $\chi$ is the character. By considering $U(1)$, one can recover standard Fourier analysis. This procedure also descends to finite groups:

If $\rho : G \to \mathrm{GL}(V_\rho)$ is a representation, and $\varphi$ is a function on $G$, we can define the Fourier transform $\varphi(\rho)$ in $\mathrm{End}(V_\rho)$ as,

$$\hat\varphi(\rho) = \sum_{g\in G} \varphi(g)\rho(g).$$

Hope this can be of use.

• What you defined as the Fourier transform of a function only makes sense for $f$ being a class function. It's fine if you're talking about Abelian groups. – childofsaturn Nov 3 '17 at 1:22
• @childofsaturn Yes, as in a function on the conjugacy classes? – JamalS Nov 3 '17 at 7:33
• Yeah, a function invariant under conjugation. $L^1(G)$ on the other hand denotes functions on $G$ with no such restriction if my understanding of the notation is right. – childofsaturn Nov 3 '17 at 18:19
• @childofsaturn From Fulton and Harris' representation theory text though, the characters are stated be functions on the group which can equivalently be interpreted as functions on the conjugacy classes. Did you downvote? I don't think it merits that as nothing is wrong, and I've clarified this point. – JamalS Nov 3 '17 at 20:03