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So the Hopf-Rinow theorem tells us that if we have a connected Riemannian manifold then it is equivalent saying that $M$ is geodesically complete and $(M,d)$ is a complete metric space, where $d(p,q)$ is the infimum of the lenght of all curves connecting the 2 points $p$ and $q$. My question is suppose we have $M=\mathbb{R}^n$ or for the matter any complete metric space with respect to some distance $\rho$, is there any way that we can relate the distance $d$ that is induced by the metric to the distance $\rho$ that makes the space into a complete metric space, so we might be able to conclude that it is geodesically complete? Also these 2 distances $d$ and $\rho$ can give us very different topologies on $M$ right ?

Alp Uzman
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Someone
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The issue is that a Riemannian manifold $(M,g)$ gives rise to a metric space $(M,d_g)$. Now, if you have an arbitrary distance function $d$ on a manifold $M$, compatible with its topology, it is not necessarily the case that $d = d_g$ for some Riemannian metric $g$ (e.g., one necessary condition is that $d\colon M\times M \to \Bbb R$ is smooth in a neighborhood of the diagonal $\Delta \subseteq M\times M$).

So in general, if you start with an arbitrary distance function, there is no need for this to be related to any Riemannian metric at all. Also, there are examples of two distance functions inducing the same topology on a given set with one of them complete but the other not, and we can have two complete distance functions inducing different topologies.

Ivo Terek
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  • Alright , so let me try to think of this in practice terms , suppose we have the torus with the metric induced by the euclidean metric and we want to know if it is geodesically complete, my first approach would be to say yes since it is a complete metric space because it is a compact set, but this is relative to the subspace topology of the euclidean metric of $\mathbb{R}^3$. Is this argument correct or do we need to go about it in another way ? – Someone Jan 19 '20 at 09:23
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    That's what I would say, but there's one subtle thing: the manifold has a topology before having a Riemannian metric. So a priori you have two topologies: the manifold topology and the metric space topology. But they're equal, and this is noteworthy. – Ivo Terek Jan 19 '20 at 14:56
  • Alright yes before we introduce the metric the manifold has a topology , how do we now that they're equal? – Someone Jan 19 '20 at 14:59
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    Because every open set contains a metric ball, and every metric ball is open in the manifold topology. This is an argument with the exponential map and convex neighborhoods of points. – Ivo Terek Jan 19 '20 at 15:02