# Reference request: Recovering a Riemannian metric from the distance function

Let $M = (M, g)$ be a Riemannian manifold, and let $p \in M$.

Writing $d$ for the geodesic distance in $M$, there is a function $$d(-, p)^2 : M \to \mathbb{R}.$$ This function is smooth near $p$. Hence for each point $x \in M$ sufficiently close to $p$, we have the Hessian $$\text{Hess}_x(d(-, p)^2)$$ (defined using the Levi-Civita connection), which is a bilinear form on $T_x M$. In particular, we can take $x$ to be equal to $p$ itself, giving a bilinear form $$\text{Hess}_p(d(-, p)^2)$$ on $T_p M$. But of course, we already have another bilinear form on $T_p M$, namely, the Riemannian metric $g_p$ itself. And the fact is that up to a constant factor, these two forms are equal: $$g_p = \frac{1}{2} \text{Hess}_p(d(-, p)^2).$$ I'm looking for a reference for this fact. For the purposes of what I'm writing, it would ideally be a reference that states this fact in the same simple direct terms as above, without involving any other differential-geometric concepts (e.g. normal coordinates).

I understand that this is a basic fact of Riemannian geometry, so I've already looked for it in various introductions to the subject, including those by do Carmo, Jost, Lee, and Petersen. But I haven't found it stated in any of those sources (which isn't to say it's not there). I have found more sophisticated stuff about $\text{Hess}_x(d(-, p)^2)$ for points $x$ different from $p$, but not the simple fact I'm looking for.

Requests for references often result in people giving their favourite proofs rather than a reference. While that doesn't do any harm (and can be quite interesting), I emphasize that it's a reference I'm looking for, not a proof.

• I asked a similar question a while ago on MSE, see math.stackexchange.com/questions/1161589/…. My answer there seems to be related to what you are looking for. In addition, there is another answer that I haven't been able to make sense of. – S.Surace Aug 9 '18 at 20:47
• @S.Surace: thanks, I hadn't seen that MSE question. Nothing there answers my question (i.e. provides a reference to the stated equation), but it seems that you're interested in this stuff for similar reasons to me. In particular, I'd seen the some of that literature on contrast functions that you mention in your MSE answer, which seems to take as its starting point the result that I want a reference for. – Tom Leinster Aug 9 '18 at 21:28
• @S.Surace: It seems that you have read only the title of this post, but not its content. Indeed, the title suggests a completely diffferent question - the one that you have asked. – Alex M. Aug 9 '18 at 21:58
• @AlexM. Am I sure it's true? I believe it's true because someone whose expertise I trust tells me that it is. For a proof, they pointed me to p.4-5 of the paper "Hessian of the Riemannian squared distance" by Pennec: www-sop.inria.fr/members/Xavier.Pennec/… . But the fact I'm interested in isn't stated directly there; you have to do a bit of work to dig it out. I'm looking for a reference where it's stated directly. – Tom Leinster Aug 9 '18 at 22:09
• It seems to me that the answer is stated explicitly in equation (5) in the Pennec paper, if you take into account the displayed equation after (2). I'm not sure you'll get anything more explicit than that, My preferred approach to this is the equation of Villani stated in the last sentence before section 2.2. – Deane Yang Aug 10 '18 at 0:48

While it does not answer your question, the following direct argument may clarify certain things:

Since the Hessian is a symmetric bilinear form, it suffices to show $\frac{1}{2}Hess_p(d^2(\cdot,p))(v,v)=|v|^2$.

If $f$ is a smooth function near $p$ in $\mathbb{R}^n$, then $Hess_p(f)(v,v)=\frac{d^2}{dt^2}\vert_{t=0}f(\gamma(t))$, where $\gamma$ is any smooth path with $\gamma(0)=p$ and $\gamma'(0)=v$. This formula continues to hold, if $p$ is a critical point of $f$ on a manifold (in this case the definition of the Hessian does not rely on the choice of a Riemannian metric).

If $\gamma$ is the geodesic through $p$ with $\gamma'(0)=v$, then, since $\gamma$ is locally distance minimizing, $d(\gamma(t),p)=|tv|$ for $t$ near $0$. Combined with the above this gives the result.

(If we use a Riemannian metric $g$ and its associated Levi-Civita connection $\nabla$ to define $Hess_p^g(f)$ at a noncritical point $p$ of $f$, then the formula $\frac{d^2}{dt^2}\vert_{t=0}f(\gamma(t)) =Hess^g_p(f)(v,v)$ still holds, if $\nabla_t\gamma'(0)=0$. This is however not used above).

This is described in painstaking detail in the paper of Xavier Pennec (2017). (Hessian of he Riemannian Squared Distance).

• Thanks. I've seen that paper (and mentioned it in my conversation with Alex M in the comments above), but I don't see the fact I'm concerned with stated explicitly. You can get it by combining the first equation on p.4 with equation (5) on p.5, at least if you're fluent in normal coordinates. But do you see anywhere in this paper where Pennec states it explicitly? – Tom Leinster Aug 9 '18 at 23:38

For the requested reference: I believe it should follow from inequalities (5.6.6) in Jost (2011, p. 235) (plus user_1789’s polarization argument) because $r(x)\to0$ as $x\to p$.

• Funny: I was looking at that exact theorem when your post popped up. What values of $\lambda$ and $\mu$ are you taking in Jost's theorem? – Tom Leinster Aug 9 '18 at 23:59
• I think it doesn’t matter, they simplify in a (l’Hospital) limit, no? – Francois Ziegler Aug 10 '18 at 0:13
• I see what you mean! Good, so that's one reference - thanks. Though it does remind me slightly of the famous mathoverflow.net/a/42519.... I mean, I'd love to have a reference where the statement is made directly rather than derived from something much more sophisticated. – Tom Leinster Aug 10 '18 at 0:55

I believe that the reason why you cannot find the result that you are asking about printed anywhere is that it is, after all, a mere exercise in Riemannian computation. First, it is easy to show that if $f$ is smooth around $p$, then $(Hf)_{ij} = \partial^2_{ij}f - \Gamma_{ij} ^k \partial_k f$ in any system of coordinates around $p$. Now, since your $f = d_p^2$ has radial symmetry, it is natural to continue the work in spherical normal coordinates, i.e. you go in $T_pM$ through $\exp_p ^{-1}$ and there you introduce spherical coordinates $r, \sigma_1, \dots, \sigma_n$, with $n = \dim M$. Since $\Gamma_{ij}^k (p) = 0$ as a consequence of your coordinates being normal, you will have $(Hf)_{ij} (p) = (\partial^2_{ij}f) (p) = (\partial ^2 _{rr} r^2) (p) = 2$ (all the other second-order partial derivatives vanish at $p$ because $f=r^2$ does not contain the variables $\sigma_1, \dots, \sigma_n$).

On the other hand, it is known that in normal spherical coordinates the expression of the metric tensor is $g_{ij} = \delta_{ij} + o(r)$, so that $g_{ij} (p) = \delta_{ij} (p)$ (the Kronecker symbol), whence it follows that $(Hf)(p) = 2g(p)$ (the metric evaluated at $p$). See p.114 of I. Chavel, "Riemannian Geometry - A Modern Introduction", 2006, or the more general theorem 2.53 of Cartan on p.83 of S. Rosenberg, "The Laplacian on a Riemannian Manifold", 1997, or Petersen's book cited here.

• Thanks for the references. I'll look up Chavel and Rosenberg (but as I said in the question, I've already tried Petersen). – Tom Leinster Aug 10 '18 at 12:42
• My guess is that you won't find it anywhere, simply because it's an easy exercise. It doesn't "deserve" to be a theorem. If you want to use it in an article, do like it's often done in differential geometry: prepend the statement by the words "it's well known that". – Alex M. Aug 10 '18 at 13:45
• Hmm. I'm not convinced by the principle of your argument. There are many easy exercises that are found in introductions to the subject concerned - e.g. it's an easy exercise that inverses in a group are unique, and you'll find that stated & proved in every intro to group theory. Personally, I like to give good references when I can. I think the culture of saying "it's well known" without giving a reference is tremendously off-putting to non-expert readers, and contributes to the harmful atomization of mathematics. All that said, you may be right that I won't find what I want. – Tom Leinster Aug 10 '18 at 13:57

If I am not mistaken, in order to define the Hessian you need to fix a connection. I suspect the Riemannian metric you get will depend on this connection, as Finsler metrics also have distance functions.

• Right; I meant the Hessian with respect to the Levi-Civita connection. – Tom Leinster Aug 9 '18 at 20:16
• Actually, since $p$ is a critical point of $d(-,p)^2$, one can define the Hessian in $p$ without any choice of connection (just locally as the Hessian of that function in a chart). – Panagiotis Konstantis Aug 9 '18 at 20:25
• I believe the way to avoid using the Levi-Civita connection (which already determines the metric) is to use the distance function to define geodesics as length minimizing curves and define the Hessian using the second derivative of $d^2$ in the direction of each geodesic. However, there are lots of details to work out. There's a slight chance that this is worked out in the book by Gromov et all, Metric Structure for Riemannian and non-Riemannian Spaces. – Deane Yang Aug 9 '18 at 20:38
• It might appear in a paper that is trying to show that a length space with some additional properties is in fact a Riemannan manifold. Alas, I don’t know of any offhand. – Deane Yang Aug 9 '18 at 22:53
• I suggest looking at the papers by Karcher with his collaborators that are cited in Smith, P. D.; Yang, Deane, "Removing point singularities of Riemannian manifolds", especially those that discuss what they call "almost linear coordinates", which are coordinates constructed using only the distance function. What you want should be at the very least a corollary of something proved in one of these papers. – Deane Yang Aug 9 '18 at 23:03