# Integral Representation of Dirac Delta on a Manifold

In Cartesian coordinates on an $\Bbb{R}^d$ plane the Dirac delta can be represented as a Fourier transform: $$\delta^d(\vec r-\vec r')=\int \frac{d^d \vec k}{(2\pi)^d}\;\;\exp\left({i\vec k\cdot (\vec r-\vec r')}\right)\tag{1}$$

Correct me if I am wrong, but on a general manifold it is typical to define the Dirac delta in two ways*:

• As: $$\int d^d\xi \; \sqrt{g(\vec \xi)} \delta(\xi'-\xi)=1$$

• Or as: $$\int d^d\xi \; \delta(\xi'-\xi)=1$$

Do either of these permit an integral representation like the form of $(1)$?

$^*$I believe this is what is indicated in this answer on PSE.

• Just for completeness you can always use quantum mechanics (physicist here :) ) to write the Dirac delta as a linear combination of e.g. energy eigenstates. In the case of a sphere this would take the form of a Spherical Harmonic expansion. – Quantum spaghettification May 29 '18 at 8:14

I find it to be most natural to define the Dirac delta $\delta_{x'}$ on a manifold $M$ so that for any smooth function $f:M\to\mathbb C$ the formal integral $\int_Mf(x)\delta_{x'}(x)dV(x)$ (or more precisely, the duality pairing $\langle\delta_{x'},f\rangle$) equals $f(x')$. Here $V$ denotes the volume measure of the manifold (which may be Riemannian or Lorentzian, for example). In fact, I find this the most natural way to go even over a Euclidean space. I would call this the definition of the Dirac delta.
I wrote $\delta_{x'}(x)$ instead of $\delta(x-x')$ on purpose. The difference $x-x'$ doesn't mean anything on a general manifold. However, $x-x'$ does mean something in local coordinates, but then both $x$ and $x'$ have to be within the same coordinate chart. In local coordinates $dV(x)=\sqrt{|g(x)|}dx$. Once you are in a local coordinate chart, you get $$\int\sqrt{|g(x)|} \delta_{x'}(x)f(x)dx=f(x'), \tag{2}$$ which in particular implies your first option. (Again, this is just a formal integral.) However, I would not treat this as the definition, but as a local coordinate representation of the Dirac delta.
Once you are on a local coordinate chart, you can indeed formally write $$\delta_{x'}(x) = |g(x')|^{-1/2}\int_{\mathbb R^d}\frac{1}{(2\pi)^d}e^{ik\cdot (x-x')}dk.$$ Here $x$ and $x'$ are vectors in $\mathbb R^d$ and are identified with two points on $M$ via a coordinate chart. (If you plug this representation to (2) with a smooth function $f$ supported in the coordinate chart, you obtain the correct result.) The integral is taken over the entire Euclidean space. I doubt this representation is of much use, since the Fourier transform tends to be less useful on manifolds. (It can be defined on coordinate charts, but it does not have the nice and simple global properties the usual Fourier transform does.)
Euclidean spaces have a lot of special structure. This structure allows us to speak of the difference $x-x'$ and also allows for (simple) Fourier analysis. On general manifolds you lose both, so a Fourier representation and difference of two points are kind of lost at the same time. Locally a manifold can be identified with (a subset of) a Euclidean space, but only locally.