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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 $\mathcal{M}$ ?
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2 Answers 2

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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.

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  • $\begingroup$ Awesome dude ! Thanks a lot, especially since I had given up hope for this question. $\endgroup$ Sep 2, 2015 at 15:18
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The answer to your first question is yes. You can view the distance function you defined $$f_p(q) = d(p, q)$$ as the radial component of the inverse of the exponential map at $p$. The exponential map is known to be a diffeomorphism in a neighborhood $p$. You can define the full neighborhood in which it is smooth in terms of its conjugate points according to this Mathoverflow Thread:

The distance function is differentiable at $(p,q) \in M \times M$ if and only if there is a unique length-minimizing geodesic from p to q. Furthermore, the distance function is $C^\infty$ in a neighborhood of $(p,q)$ if and only if $p$ and $q$ are not conjugate points along this minimizing geodesic.

Regarding the answer to your second question, I can't find a specific counter-example but I see no reason why it would be true. I did do various searches of literature and found references to bounding the Laplacian of distance functions. e.g. In On the differential structure of metric measure spaces and applications it is stated on page $2$ that $$\Delta f_p(q) \leq \frac{n - 1}{f_p(q)}$$ when $M$ is a manifold with non-negative Ricci curvature.

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  • $\begingroup$ Thanks to the fist part of your answer, it occured to me that you could write the square distance function as $f_p(q)^2 = ||\ln_p(q)||_p^2$, which might help to work out the computation (though I did not manage to complete it). The second part left me more puzzled as I do not know which Laplacian your reference is using. $\endgroup$ Jul 16, 2015 at 11:19
  • $\begingroup$ Another similar reference to your second one can be found here : www.math.zju.edu.cn/swm/rg_section_6.pdf $\endgroup$ Jul 16, 2015 at 11:21
  • $\begingroup$ After a quick research, it seems to me that your reference is dealing with the Laplace-Beltrami operator and not the Hodge laplacian (also cf www.maths.bath.ac.uk/~feb/papers/icms/paper.pdf). BUT since I specifically need the case of 0-forms (functions), they do coincide (cf mathoverflow.net/questions/29226/… ) ! $\endgroup$ Jul 16, 2015 at 11:32
  • $\begingroup$ @G.Fougeron Happy to help. If you found this useful, please vote it up one. Cheers. $\endgroup$
    – muaddib
    Jul 16, 2015 at 11:57
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    $\begingroup$ Since the formula $\Delta f_p^2 = 2n$ holds in $\mathbb{R}^n$ (with canonical metric) , what causes its failure in a general Riemann manifold ? What is the relationship between $\Delta f_p^2$ and the Ricci curvature ? is there a way to simply express $\Delta \frac{1}{2} f_p^2 - n$ in terms of the Ricci curvature tensor ? $\endgroup$ Jul 16, 2015 at 15:53

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