the Fourier transformation of a scalar function with respect to one variable might be defined as

$\mathcal{F}\left[w\right](\omega )\equiv \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}w(t)e^{-\mathrm{i}\omega t}dt$

In physics, this transformation along with its generalization, the Laplace transformation, has a tremendous importance because of its feature to turn linear partial differential equations into algebraic ones.

Now, suppose that we have a pseudo-Riemannian manifold $\mathcal{M}$ where $\det{g_{\mu\nu}} = -1$ holds like in special and general relativity.

I am wondering, what would be the generalization of the Fourier transformation of scalar functions or forms?

The difficulty I have with this question is that a three-dimensional slicing of $\mathcal{M}$ is not unique, so how to take the integral in an invariant form? What will happen to the differntials $dx^{\mu}$, e.g. $dt,dx^i\rightarrow d\omega dx^i$ in some sense?

Any insight would be well appreciated.
Thank you in advance


Robert Filter
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    What you need to do is analyze the spectrum of the [Laplace-Beltrami operator](http://en.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator) on that space, that is what gives you the functions which allow you to define the Fourier transform. The $e^{(\vec{k}\cdot\vec{x}-\omega t)}$ are basically solutions of the Laplace equation in Euclidean space and expressed in ordinary cartesian coordinates. – Raskolnikov Dec 11 '10 at 14:06
  • @Raskolnikov: Thank you for the comment which pointed directly in the right direction. Nevertheless my experience in functional analysis on manifolds is very limited so I could not really grasp the idea. I think Willie Wong pointed out these lines of thought in a very explicit and decent way. – Robert Filter Dec 12 '10 at 08:08
  • I am interested in such a generalisation to apply in this MO post https://mathoverflow.net/questions/356784/a-fourier-elliptic-vector-field-on-a-riemannian-manifold – Ali Taghavi Apr 07 '20 at 20:58

2 Answers2


There are three main ways of interpreting the Fourier transform.

Decomposition relative to eigenfunctions of the Laplacian

On $\mathbb{R}^n$, the plane waves $E_\xi(x) = \exp( i \xi\cdot x)$ can be interpreted as generalized eigenfunctions of the Laplacian. That is, let

$$ \triangle = \sum_{i=1}^n \left(\frac{\partial}{\partial x_i}\right)^2 $$

then we have

$$ \triangle E_\xi(x) = -|\xi|^2 E_\xi(x) $$

Notice that I say "generalized eigenfunctions". This is because usually it is preferred to do linear algebra/functional analysis on a complete inner product space (a Hilbert space), and in the case of functions on Euclidean space, it is usually preferred to work over the Hilbert space of square integrable functions. The functions $E_\xi(x)$, however, do not belong in the Hilbert space.

But roughly speaking, the Fourier representation of a function

$$ f(x) = \int_{\mathbb{R}^n} \hat{f}(\xi) E_\xi(x) d\xi $$

can be thought of as a decomposition of $f(x)$ in terms of a linear combination of eigenfunctions of the Laplacian.

In this form, as Raskolnikov mentioned in the comment, there is a natural generalization to compact Riemannian manifolds. Let $(M,g)$ be a compact Riemannian manifold, and let $\triangle_g$ denote the associated Laplace-Beltrami operator. One has that $\triangle_g$ is a densely defined, unbounded, self-adjoint operator on $L^2(M)$, the space of square integrable functions on the manifold $M$. As it turns out (using a bit of linear partial differential equations theory and a bit of functional analysis), $\triangle_g$ is invertible if we restrict the co-domain to functions of mean 0 on $M$. The operator $\triangle_g^{-1}$ furthermore turns out to be a compact operator, and thus has discrete eigenvalues with finite dimensional eigen-spaces in $L^2$ by the spectral theorem for compact operators. It is clear by definition that the eigenfunctions of $\triangle_g^{-1}$ will be also eigenfunctions for $\triangle_g$ (just change the eigenvalue $\lambda \to \lambda^{-1}$. Using the spectral theorem we can write an arbitrary square integrable function on $M$ as a unique linear combination of the set of eigenfunctions.

This is one interpretation of the Fourier transform.

You can further refine this theory in two directions. For functions (and to a small extent tensors), instead of explicitly decomposing using an eigenbasis of $L^2$, you can study "smooth frequency cut-offs" by considering the heat-flow on the manifold. That is, given a function $f$, consider the solution to the heat equation $u(t,x)$ defined on $\mathbb{R}_+ \times M$

$$ \partial_t u = \triangle_g u, \quad u(0,x) = f(x) $$

and study, the cut-off of $f$ by a bump-function $\zeta$ in frequencies in the following sense: take a non-negative function $\zeta\in C^\infty_0(\mathbb{R}_+)$ and consider the one-parameter family of functions

$$ f_\lambda(x) = \int_0^\infty \zeta(\lambda t) u(t,x) dt $$

roughly speaking $f_\lambda(x)$ here corresponds to, in the Euclidean space case, the restriction

$$ \int_{\frac12 \lambda <|\xi|< 2\lambda} \hat{f}(\xi) E_\xi(x) d\xi $$

of the function to the spatial frequencies that are close, in absolute value, to $\lambda$. This theory is developed more in Eli Stein's book "Topics in Harmonic Analysis related to Littlewood-Paley Theory".

Another generalization of this decomposition is applicable to not just functions, but also differential forms. In this generalization, instead of looking at the Laplace-Beltrami operator, we consider the Hodge-Laplace operator defined for differential forms. As it turns out, essentially the same argument as for the scalar case give rise to a decomposition of square-integrable differential forms into three categories, the harmonic forms, the exact forms, and the co-exact forms. This goes under the name of Hodge theory, and I refer you to Wikipedia, and also S. Agmon's "Lectures on Elliptic Boundary Value Problems", among many other very good books in this subject.

Fourier transform and representation theory

Another way of treating the Fourier transform on Euclidean space is from the point of view of Euclidean space as an abelian Lie group (in particular a locally compact topological group) which is "self-dual".

Very roughly speaking, for an abelian locally compact topological group $G$, you can consider the set of group homomorphisms from $G\to T$, where $T$ is the one-dimensional circle group. In the case $G = \mathbb{R}^n$, we see that each of the functions $E_\xi(x)$ is one such group homomorphism. This set of group homomorphisms can be made into another topological group, known as the dual group of $G$. In the case of Euclidean space, its dual group has the same structure as another copy of the Euclidean space, hence we say it is self-dual.

Now, given an invariant measure $\mu$ on $G$, in the sense that group translations in $G$ leaves the measure the same (I am being very rough here and sweeping a lot of stuff under the rug), it turns out that we can define the analogue of the Fourier transform by the same formula

$$ \int_G f(x) E_{-\xi}(x) d\mu(x) $$

where $\xi$ is a natural parameter of the dual group of $G$. So in particular, this theory, which I have only sketched above in the Abelian case, can be generalized to be applicable to the case when we start with a Riemannian manifold $(M,g)$ that is either a Lie group or a symmetric space. This notion is widely applicable in representation theory and analytic number theory, among other things.

See this Wikipedia article, Walter Rudin's "Fourier analysis on groups", S. Helgason's "Differential geometry, Lie groups, and symmetric spaces", and the article by Atle Selberg called "Harmonic analysis and discontinuous groups in weakly symmetric riemannian spaces, with applications to Dirichlet series" for some standard references.

Symmetries on the manifold and partial Fourier transforms

A much more pedestrian way of getting the Fourier transform for functions on manifolds is to just integrate using the usual formula. But as you noted in your question, depending on how you "slice" the manifold, you can get different representations, and whether those representations are meaningful can be questionable.

Fortunately, in one special case we can have a meaningful formula. Let $(M,g)$ be a (pseudo)Riemannian manifold (note that this is the only case I am familiar with that has a natural extension to the case of pseudo-Riemannian manifolds), and assume that it admits a continuous symmetry. That is, there exists a nowhere vanishing vector field $T$ such that the Lie derivative $\mathcal{L}_Tg = 0$; in other words $T$ is a Killing vector. Assume further that the symmetry being considered is a free and proper $\mathbb{R}$ action, and hence that the integral curves of $T$ are diffeomorphic to $\mathbb{R}$ (so in particular $T$ does not have closed orbits. You can also similarly assume the symmetry comes from a free and proper $\mathbb{S}^1$ action, in which case all integral curves are closed, so $T$ does not have open orbits; in this latter case you will deal with Fourier series instead of Fourier transform, I'll leave the obvious changes necessary as an exercise).

Then the action of $T$ defines an equivalence relation on $M$, and we can consider the quotient manifold $Q = M / T$. By the quotient manifold theorem, this quotient is a smooth manifold and also inherits a Riemannian structure, but that is not too important right now.

The point is the following, we can write a point $x\in M$ as $(t,q)\in \mathbb{R}\times Q$, where $t$ is a natural coordinate on $T$ so that $\partial_t$ corresponds to $\mathcal{L}_T$ in $M$. Using this splitting, you can take the Fourier transform in $t$ the usual way, writing $f(x) = f(t,q)$

$$ \hat{f}(\tau,q) = \int f(t,q) \exp( - it\tau) dt $$

That $T$ is a Killing vector is used here so that if you pick a coordinate system adapted to the coordinate $t$, the actual volume form

$$ \sqrt{|g|} dt dq $$

will be independent of $t$, so when you try to do analysis with this notion of the Fourier transform, the formula above will not look like it is missing a factor coming from the volume form (since anything independent of $t$ can be factored out, and taken outside of the integral, so the worst-case scenario is the entire expression is off be a fixed factor depending on $q$, which can always be inserted back later).

This formula can also be easily generalized to the case where $(M,g)$ admits several mutually commuting Killing vector fields $T_1, T_2, \ldots, T_k$. In the case of $\mathbb{R}^n$, the standard coordinate vector fields form a complete set of $n$ mutually commuting Killing vector fields that span the tangent space at every point, and this allows us to take the Fourier transform "in all directions".

This last formulation is useful especially in studying linear partial differential equations on a space-time that is stationary (for example, a stationary black-hole background). See the papers of Dafermos and Rodnianski for example (paper 1, paper 2 ).

Willie Wong
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    I should add that in the case your manifold can be decomposed as a Cartesian product $\mathbb{R}^{p,q}\times M$ with the product metric $\eta\oplus g$, where $\eta$ is the signature $(p,q)$ flat metric on $\mathbb{R}^{p,q}$, you can simply use the "normal" Fourier transform on the $\mathbb{R}^{p,q}$ part, and use one of the notions above for the $M$ part. So in particular, if you have a preferred time direction $\mathbb{R}$ with some spatial foliation $M$ that is compact Riemannian, the "Fourier basis" will be formed by taking $E_\tau(t) \cdot F_k(q)$, where $E_\tau$ are defined as before ... – Willie Wong Dec 11 '10 at 20:17
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    ... on $\mathbb{R}^{p,q}$, and $F_k$ are eigenfunctions of the Laplacian on $M$. – Willie Wong Dec 11 '10 at 20:18
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    Nice answer, you really put a lot of work into it. I'd give you a +10 if I could. I also learned about the Lie group approach long ago but it was somewhere in a corner of my mind. Thanks for refreshing it. – Raskolnikov Dec 11 '10 at 21:40
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    Thank you very much for your thoughtful and detailed answer. If I got you correctly, the notion only makes sense for the existence of a timelike killing vector, say $\partial_t$. This should imply that a time <> frequency FT only makes sense for stationary solutions in which the quotient manifold $M/\psi_t$ ($\psi_t$ being the local flux to $\partial_t$) exists. Hence, the FT is only feasible for dynamical field theories if the dynamics are independent from the (background) metric. – Robert Filter Dec 11 '10 at 23:35
  • The moral of the story is like you said. Of course, nothing is to stop you from cheating by just fixing a preferred 3+1 splitting or something similar (or just fix a global chart and work there when allowed). So the words "only feasible" should be interpreted to mean something with the condition that "with sufficiently nice properties in regards to the geometrical structure etc." – Willie Wong Dec 12 '10 at 01:05
  • Regarding the Killing vector field and consideration of quotint $Q=M/T$, is not a Kronecker flow a Killing vector field since its flow is isometry, all orbits are open but we have a bad (non Hausdorff) quotint? – Ali Taghavi Apr 07 '20 at 02:30
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    @AliTaghavi: thanks, I implicitly assumed that the group action is proper. I now added that and make it explicit. – Willie Wong Apr 07 '20 at 14:23
  • This is a really nice answer! One thing I might add to it is that you only discuss abelian groups in the second section but, if the particular manifold in question is a non-Abelian Lie group, the Peter-Weyl theorem is a generalization in this direction. – David E Speyer Aug 05 '21 at 14:15
  • @DavidESpeyer: good point. I intentionally did not want to even sketch the argument of the Peter-Weyl theorem since I wanted to keep this answer basic. The various generalizations are however touched upon in the references that I gave in the end of the section. I will add a few more words to emphasize that the general theory covers more than just the abelian case. – Willie Wong Aug 05 '21 at 18:28

Maybe you would be interested to look at this paper too: Fourier transform on 2 step Liegroups

Ali Taghavi
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