[Math] Hodge laplacian of distance function

hodge-theorylaplacianriemannian-geometry

Let $p$ be a given point on a Riemaniann manifold $\mathcal{M}$. The distance function to point $p$ is denoted $f_p$ :
$$ f_p(q) = \operatorname{dist}(p,q)$$
The exterior derivative is denoted $\mathrm{d}$ and the codifferential is denoted $\delta$. Then the Hodge laplacian is $\Delta = \mathrm{d}\delta + \delta \mathrm{d} = (\mathrm{d} + \delta)^2$ which reduces to $\delta \mathrm{d}$ in the case of functions.

My questions are the following :

  • I think it can be shown that $f_p^2$ (the squared distance function to point $p$) is $C^\infty$ in a neighborhood of $p$. Can anyone confirm ?
  • What is the value of the Hodge laplacian of $f_p^2$ ? By analogy with the case $\mathcal{M} = \mathbb{R}^n$, is my guess that it is a constant function right ? Do we have $\Delta f_p^2 = 2n$ where $n$ is the dimension of $\mathcal{M}$ ?

Best Answer

1) By first removing the cut locus of $p$, the remaining subset of $M$ is an open subset diffeomorphic to some open subset of $T_p M$. Furthermore, if you place yourself in spherical normal coordinates $(r, \sigma)$ (where $\sigma$ encodes all the angular variables), the function $d_p ^2$ becomes $r^2$, which is obviously smoooth.

For a rigorous statement (but following the same idea), see theorem 3.6 at page 166, in chapter IV of "Foundations of Differential Geometry", volume 1, by Kobayashi and Nomizu.

2) In the same spherical normal coordinates, the Laplacian looks like

$$\Delta f \ (q) = \frac {\partial ^2 f} {\partial r^2} (q) + H(p,q) \frac {\partial f} {\partial r} (q) + \frac 1 {r^2} \Delta_S f \ (q) ,$$

where $r = d(p,q)$, $S$ is the geodesic sphere of radius $r$ centered at $p$, $\Delta_S$ is the Laplacian restricted to $S$ and $H(p,q)$ is the mean curvature at $q$ of $S$. It follows that

$$\Delta d_p ^2 \ (q) = \Delta r^2 = 2 + H(p,q) 2r = 2 + 2 H(p,q) d_p (q) .$$

In $\Bbb R^n$, the mean curvature of the sphere of radius $r$ is $H = \frac {n-1} r$, so the above becomes $2n$. Otherwise, as your intuition hinted at, the curvature does play a role and makes the result non-constant, in general.

For an alternative formulation of the Laplacian see also page 48 of "Spectral Theory and Geometry" edited by E.B. Davies and Y. Safarov.

Related Question