Weierstrass transform on the Riemannian manifold

Question

I've read on this Wikipedia article that Weierstrass transform (WT) can be defined on any Riemannian manifold $(M,g)$, but it seems a bit complicated to me. I'm not sure but I guess one can write the Weierstrass transform as follow: $$ (4\pi\epsilon)^{\frac{n}{2}}\exp(\epsilon\Delta)f(x)\mid_{x=x_0}=\int d\Omega_{x}\,\exp(-\frac{d(x,x_0)^2}{4\epsilon})\,f(x) $$ where $\Delta$ is the Laplace–Beltrami operator, $d\Omega_{x}$ is the invariant valume element, $d\Omega_{x}=\sqrt{det~g}d^nx$, $d(x,x_0)$ is the geodesic distance between arbitrary point of $x$ and base point of $x_0$, $f$ is scalar function and $\epsilon$ is a constant. Could anyone help me how to write the WT and prove the relation?! I think I should use Normal coordinates.

Answer 1

No, not quite. The trick is that the quantity $\frac 1 {\sqrt {4 \pi}} \textrm e ^{- \frac {(x-y)^2} {4}}$ from $\Bbb R$ does not get replaced with what you suggest (the resulting transform would not have nice properties), but rather with $k(1,x,y)$ where $(t,x,y) \mapsto k(t,x,y)$ is the kernel of the heat equation (it does not have an explicit expression and it is not always unique - which means that on "wild" manifolds you get several Weierstrass transforms, each one corresponding to a kernel). Therefore, $F(x) = \int k(1,x,y) f(y) \ \textrm {vol}(y)$ (with $\textrm {vol}$ being the volume form).