# Existence of a Riemannian metric inducing a given distance.

Let $M$ be a smooth, finite-dimensional manifold. Suppose $M$ is also a metric space, with a given distance function $d: M \times M \rightarrow \mathbb{R}_{+}$, which is compatible with the original (manifold) topology on $M$.

Question: is there a Riemannian metric $g$ on $M$ such that the distance $$d_g(p, q) = \inf_{\gamma \in \Omega(p, q)} L_g(\gamma)$$ coincides with $d$?

I believe that this setting is very standard, but for the sake of completeness: $L_g$ denotes the Riemannian length of the curve $\gamma$, and $\Omega(p, q)$ the set of all piecewise smooth curves $\gamma : [a, b] \rightarrow M$ s.t. $\gamma(a) = p$ and $\gamma(b) = q$. Thanks in advance.

I believe the answer is no in general.

Consider the taxicab metric on $$\mathbb{R}^2$$. That is $$d((x_1,y_1),(x_2,y_2)) = |x_1 - x_2| + |y_1-y_2|$$. This metric induces the same topology on $$\mathbb{R}^2$$ as the standard metric.

The 2 points $$(0,0)$$ and $$(a,a)$$ with $$a>0$$ are a distance $$2a$$ from each other in this metric. The key point is that there are infinitely many shortest "geodesics" between these 2 points - any monotonic staircase picture is an example of one.

On the other hand, there is a well known consequence of the Gauss Lemma that, given a Riemannian metric, for a small enough neighborhood around any point there are unique shortest geodesics between any 2 points in the neighborhood.

But any neighborhood of $$(0,0)$$ contains at least one point of the form $$(a,a)$$, so the taxicab metric cannot be induced from a Riemannian metric.

This is an old post and Jason already gave a satisfactory (negative) answer to it, but some stubborn people do not take no for an answer! One of such people was A.D.Alexandov who asked (in 1940s or 1950s, I think) about synthetic geometric conditions on metric spaces $$(M,d)$$ which ensure that the distance function $$d$$ comes from a Riemannian metric (no a priori assumption that $$M$$ is homeomorphic to a manifold!). The first obvious necessary condition is that $$M$$ is locally compact (every topological manifold, of course, satisfies this property) and $$d$$ is a path-metric, i.e, $$d(x,y)=\inf_{p} L(p)$$ where the infimum is taken over the length of all paths $$p$$ connecting $$x$$ to $$y$$. (The $$l_1$$-metric in Jason's answer does pass this test.) Every Riemannian metric has Riemannian metric tensor (which makes no sense for general path-metric spaces, of course) as well as the (sectional) curvature. The latter still has no meaning in the setting of arbitrary path-metric spaces. However, Alexandrov realized that for arbitrary path-metric spaces one can define the notions of upper and lower curvature bounds. Every Riemannian manifold, of course, does have curvature locally bounded above and below (the metric in Jason's answer fails this test). One needs to impose one more geometric condition on $$(M,d)$$: Every finite geodesic in $$(M,d)$$ has to extend in both directions (at least a little bit) to a longer geodesic. This eliminates, for instance, manifolds with boundary.

Alexandrov then asked if the curvature conditions plus extendibility of geodesics, plus some obvious topological restrictions (see below) is sufficient for the path-metric to be Riemannian. (In the 1930s A. Wald found a metric characterization of two-dimensional Riemannian manifolds.)

Remarkably, the answer to Alexandrov's question turned out to be positive:

[1] Suppose that $$(M,d)$$ is a locally compact finite dimensional path-metric space with curvature locally bounded above and below and extendible geodesics. Then $$M$$ is homeomorphic to a smooth manifold $$M'$$ and, under this homeomorphism, the distance function $$d$$ is isometric to the distance function coming from a Riemannian metric $$g$$ on $$M'$$, the regularity of the metric tensor $$g$$ is $$C^{1,\alpha}$$.

[2] Under further synthetic geometric "curvature-type" restrictions on $$d$$, the metric $$g$$ is $$C^\infty$$-smooth.

See:

[1] I. Nikolaev, Smoothness of the metric of spaces with bilaterally bounded curvature in the sense of A. D. Aleksandrov. Sibirsk. Mat. Zh. 24 (1983), no. 2, 114–132.

[2] I. Nikolaev, A metric characterization of Riemannian spaces. Siberian Adv. Math. 9 (1999), no. 4, 1–58.

• Thank for you for this answer - while I'm somewhat familiar with Alexandrov spaces, this implies they are a lot closer to being manifolds than how I usually perceive them. Incidentally, if $M$ is a topological manifold with no smooth structure (so, not homeomorphic to any smooth manifold), then this says surprising (to me) things about what kind of metrics you can have on it. Feb 12, 2014 at 16:18
• @JasonDeVito If I remember correctly from Burago et al's book on metric geometry math.psu.edu/petrunin/papers/alexandrov/bbi.pdf , every Alexandrov Space is a topologically stratified space en.wikipedia.org/wiki/Topologically_stratified_space , hence quite close to being a manifold. Jul 23, 2017 at 18:08
• For statement [1], we have that, "$M$ is homeomorphic to a smooth manifold $M'$." Is it possible to show a stronger relation than homeomorphism? Intuitively, topological homeomorphism permits all kinds of crazy continuous deformation, but it feels as though we have a much more constrained relation here. More like an isometry. (Although maybe that is already covered by the statement that "distance function $d$ is isometric to the distance function coming from Riemannian metric $g$.") Sorry if this question is naive - I do not have a rigorous understanding of these ideas. Jul 18, 2020 at 20:56
• @user12262: (1) Nothing is known in the semi-Riemannian setting, it's not even clear what the right setup is. (2) Yes, in principle, but it's hard even for surfaces. Jan 18, 2022 at 22:30
• Moishe Kohan: "(1) Nothing is known in the semi-Riemannian setting, it's not even clear what the right setup is." -- Thanks. (Still, it seems that some stubborn people don't take "nothing" for an answer ...) "(2) Yes, in principle, but it's hard even for surfaces." -- Part of the difficulty is the search for an appropriate "smooth structure", I suppose ... Jan 18, 2022 at 23:53

Every compact metric space of covering dimension $n$ can be embedded isometrically in $\mathbb{R}^{2n+1}$, as far as I can tell. This is discussed in this MO post. In particular, a compact, metric manifold can be embedded in this way and then inherits a Riemannian metric from the ambient Euclidean space.

• The link you reference doesn't saying anything about embedding isometrically in $\mathbb{R}^{2n+1}$. In fact, there are known metrics on, say, $S^1\times S^1$ which are not inherited from any embedding into $\mathbb{R}^3$ (say, the flat metric). On the other hand, Nash proved that every Riemannian manifold does inherit its metric from some embedding into $\mathbb{R}^N$ for $N$ sufficiently large (If I recall, $N$ grows roughly as $n^2$ or half that). Feb 7, 2012 at 3:03
• 1. Round circle does not admit an isometric embedding into any Euclidean space. 2. One can weaken the notion of an isometric embedding and require it to be a "length-preserving" topological embedding, in line with the notion of a Riemannian isometric embedding. Even in this sense, some Finsler manifolds do not embed ("Lipschitz and path isometric embeddings of metric spaces", by E.Le Donne). Dec 20, 2014 at 22:31

The chord metric on $S^n$ is a counter example

Please see the following RG converßation

https://www.researchgate.net/post/A_particular_Riemannian_metric_on_Rn