# Is an ellipsoid surface, with shortest path length used as metric, a metric space?

Is an ellipsoid surface, with shortest-path length used as metric, a metric space?

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## 1 Answer

Yes of course.

In any metric space in which any two points can be connected by a rectifiable curve (i.e curves $c:[0,1] \to X$ of finite length $\ell(c)$), define the shortest length metric by $$p(x,y) = \inf_{c:x \to y} \ell(c(x,y)),$$ where the infimum is taken over all rectifiable curves connecting $x$ to $y$. I claim that it is a metric, I'm not claiming that it induces the right topology, though.

Since we have $\ell(c) \geq d(x,y)$, we have $p(x,y) \gt 0$ whenever $x \neq y$. Symmetry is obvious since the length of a rectifiable curve is invariant under reparameterization $t \mapsto 1-t$.

The triangle inequality follows immediately from the the definition of the infimum: Choose paths connecting $x$ to $y$ and paths connecting $y$ to $z$ whose length realizes the infimum up to $\varepsilon/2$. Their concatenation is a path of length at most $p(x,y) + p(y,z) + \varepsilon$ and connects $x$ to $z$ and since $\varepsilon$ was arbitrary we have $p(x,z) \leq p(x,y) + p(y,z)$.

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In fact, this metric induces the "subspace topology" on the surface: it is a theorem in Riemannian geometry that every metric tensor that is smooth on a given atlas induces a metric (of metric spaces) in the way described in the answer, and this induced metric is compatible with the topology of the manifold. (The proof can be found on Lee's book "Riemannian geometry: an introduction to curvature", in the chapter "geodesics and distance"). –  Ronaldo May 11 '11 at 3:14
@Ronaldo: Thanks for pointing this out. I know, but not at the level of generality I stated it. You need some local compactness. –  t.b. May 11 '11 at 3:19
You're right. I didn't pay so much attention to that when I wrote the comment; I was thinking about the specific case of surfaces. –  Ronaldo May 13 '11 at 23:13
@Ronaldo: To be fair, I just didn't have the patience of going into this argument, so I "cheated" my way around it :) Thanks, it certainly is a good and helpful point you're making! –  t.b. May 13 '11 at 23:25