0
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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).

$\endgroup$
4
  • $\begingroup$ Dear @Alex M., I cannot get your point. I think my relation reduces to the simple WT in $R$. What do you mean by the resulting transform would not have nice properties?! BTW, do you think the above relation can be valid on a simple manifolds e.g. on a manifold which is diffeomorphic to $R^n$?! $\endgroup$
    – AFZQ
    Commented May 15, 2016 at 12:52
  • $\begingroup$ @AFZQ: Read that paragraph on Wikipedia again: the Weierstrass transform $W[f]$ is then given by following the solution of the heat equation for one time unit, starting with the initial "temperature distribution" $f$. This is exactly what I'm doing above. Working as you suggest would make "your" Weierstrass transform not have many nice properties that the one constructed with the heat kernel has. Been there, done that (in a different setting, though) - you won't find anything interesting in that direction. Keep in mind that the naive generalizations are not always fruitful ones. $\endgroup$
    – Alex M.
    Commented May 15, 2016 at 12:58
  • $\begingroup$ @AFZQ: To clarify my comment above: what you are hoping for is like hoping that $\frac 1 {\sqrt {4 \pi t}} \textrm e ^{- \frac {d(x,y)^2} {4t}}$ will solve the heat equation on a manifold just because it happens to do so in $\Bbb R^n$. It is an educated hope, but it is destined to fail. The same type of thinking applies here: since in $\Bbb R^n$ the Weierstrass transform is a byproduct of the heat equation, it turns out that the same applies on a manifold - even if it initially shocks our expectations. $\endgroup$
    – Alex M.
    Commented May 15, 2016 at 17:59
  • $\begingroup$ I think everything is starting to be a bit clearer now! $\endgroup$
    – AFZQ
    Commented May 15, 2016 at 19:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .