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Let $(M,g)$ be a connected Riemannian manifold. Let $d_g$ be the induced distance metric of $g$. Now let $d$ be some other metric on $M$.

Suppose that for each $x \in M$, there is a neighborhood $U$ of $x$ so that $d = d_g$ on $U \times U$.

Question: Does this imply that $d = d_g$ on $M \times M$?

I am not assuming that $d$ is the metric of some Riemannian metric; I know the answer is yes in that case.

Remarks:

  1. It's clear to me that $d = d_g$ on a neighborhood of the diagonal in $M \times M$; this essentially is just a rephrasing of the hypothesis.

  2. Connectedness is necessary to avoid stupid counter examples. For example, we could take two disjoint points $\{x,y\}$. Then $d_g$ between them is infinity, but we can take $d(x,y) = 1$. They both agree in a neighborhood of $x$, namely $x$. (We can build similar examples by taking disjoint unions of non-zero dimensional manifolds.)

  3. Maybe the 'right' question is actually: let X be a (path) connected topological space and consider two metrics $g$ and $h$ on X, compatible with the topology on X. If these metrics are locally equal in the sense of the question (i.e. agree on a neighborhood of the diagonal), are they equal? (I feel like this either has an obvious proof or some terrible counter example. It appears to be true on graphs metrized by assigning edge lengths - oops, this is wrong by the cut off metric example.)

  4. I (think) I can prove a special case. See motivation section below.

Motivation: Let $(N,h)$ be a connected Riemannian manifold, with geodesic distance function $d_N$. Let $G$ be the group of a covering space action by isometries on $N$, with quotient map $\pi$, with $G$ finite. Then $N / G$ inherits a Riemannian metric $g$ by using local trivializations. $N / G$ also inherits a metric, defined by $d([x], [y]) = \inf_{g \in G} d_N(gx, y)$. Locally $d = d_g$ because of local trivializations. I want to know if $d = d_g$ on all of $N / G$.

I think this is true if the geodesic distance between any two points in $N$ and $N / G$ is always realized by some shortest path, by using the covering space path lifting + being a geodesic is a local condition + isometry of covering actions in order to relate shortest paths in $N$ to those in $N / G$. That is, if $g$ minimizes $d(gx, y)$ let $\gamma : [0,1] \to N$ be a shortest path in $N$ between $gx$ and $y$. Then $\pi( \gamma)$ is a (potentially self intersecting) geodesic, of the same length as $\gamma$, connecting $[x]$ and $[y]$. Hence $d_g \leq d$. On the other hand, if we have a geodesic path $\gamma$ from $[x]$ to $[y]$, we can lift it to a geodesic from $x$ to $hy$ of the same length, for some $h \in G$, and hence $d_g \geq d$.

I would like to be able to drop this condition about there always being shortest path witnesses to shortest distances (I mean that the inf in geodesic distance is achieved). I think that maybe in this covering space case one can achieve this by taking a sequence of paths approximating the geodesic distance, but the argument starts to get a lot fuzzier, and I am already outside my comfort zone and procrastinating on my actual work as it is...

I would appreciate a reference for either the main question or the motivation.

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  • $\begingroup$ You might be looking for the notion of “length metric”. See Bridson—Haefliger’s book Metric spaces of non-positive curvature for details. $\endgroup$ – HJRW May 22 at 5:55
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The answer is no. Take any non-convex region in the plane, and let the Riemannian metric be the ordinary Euclidean metric $ds^2=dx^2+dy^2$. Then define the new metric as the infimum of Euclidean diameters of curves connecting $a$ to $b$. This new metric coincides with the Riemannian metric locally, but does not coincide globally. This metric even has a name: Mazurkiewicz metric.

One can also construct a compact example by taking $M$ to be a ramified covering of the sphere, and pullback of the spherical metric. This Riemannian metric in general does not coincide with the Mazurkiewicz metric corresponding to it. (Diameter used in the definition is the diameter of the projection of the curve on the sphere. Projection of the Riemannian-shortest path on $M$ can have small diameter but large length).

In general, any distance $d$ which locally coincides with the Riemannian distance $d_R$ must satisfy $d\leq d_R$. Just break the curve on which the Riemannian distance is (almost) achieved into small pieces, and use the triangle inequality for $d$.

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  • $\begingroup$ Thank you! What about the special case described in the motivation section? (The counter example you suggests that something about the convexity like condition I impose is necessary.) $\endgroup$ – Lorenzo May 22 at 2:23
  • $\begingroup$ There is no convexity in my second example. $\endgroup$ – Alexandre Eremenko May 22 at 2:31
  • $\begingroup$ Thanks, I really like your examples! They definitely answer my question. In the motivation section I'm looking at a particular metric, which is just induced by the quotient action (without the Mazurkiewicz metric idea). Since the more general question has a negative answer, I'd be interested in knowing if the case I'm considering there is correct. (I'm sure I can write out a detailed proof if it's necessary at some point, this question just came up talking to a friend about their research. Mostly I'm concerned about missing some key counter example, like the one you shared.) $\endgroup$ – Lorenzo May 22 at 2:46
  • $\begingroup$ Never mind -- this seems to be answered by Nik Weavers comment below. Thanks a lot for your help! $\endgroup$ – Lorenzo May 22 at 17:07
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Take $d(x,y) = {\rm min}(|x-y|, 1)$ on $\mathbb{R}$,

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  • $\begingroup$ Thanks! I definitely should have thought of this, because I've seen this trick recently in the construction of a metric on path space $C([0,\infty), \mathbb{R})$. As a follow up -- do you know if what I was sketching in the 'motivation section' (the particular case when the Riemannian metric and the metric arise as the quotient of a group action) is correct? $\endgroup$ – Lorenzo May 22 at 4:31
  • $\begingroup$ Yes, more generally the quotient of any length space by a proper isometric action is again a length space, which implies that the two metrics agree globally. See Proposition 4.1 of these nice notes. $\endgroup$ – Nik Weaver May 22 at 15:14
  • $\begingroup$ Great - thank you! $\endgroup$ – Lorenzo May 22 at 17:08
  • $\begingroup$ You are welcome! $\endgroup$ – Nik Weaver May 22 at 19:26

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