4
$\begingroup$

I'm studying Differential Manifolds using Manfredo do Carmo's Book (Riemannian Geometry) and although I see no mention of this in Do Carmo's book, it's really easy to see a Riemannian Manifold as a metric space (if $M$ is the manifold, using $d_M(x,y) = \inf\{L(\gamma) \colon \gamma$ is a curve on $M$ and $a,b \in \gamma([0,1])\}$, where $L(\gamma)$ is the length of the curve $\gamma$.

Is the topology induced in $M$ (viewed as a set) by the metric $d_M$ the same of $M$ as topological space?

My problem to solve this self-proposed question was analyzing the open balls of $(M,d_M)$. References and Hints are appreciated.

$\endgroup$
9
$\begingroup$

First, you need to stipulate that $M$ is connected for the distance function to turn $M$ into a metric space.

When $M$ is connected, the answer is yes, the metric topology is the same as the given manifold topology. The proof boils down to showing that the Riemannian metric is uniformly comparable to the Euclidean metric in small coordinate balls. You can read a proof in my Introduction to Smooth Manifolds (2nd ed., Theorem 13.29).

$\endgroup$
-1
$\begingroup$

If I remember correctly, what you are asking follows from Hopf-Rinow Theorm. Here is the link from wiki. But I recommend you to read another book by Do Carmo called Riemmanian Geometry.

$\endgroup$
  • 2
    $\begingroup$ The Hopf-Rinow theorem doesn't imply this. In fact, the proof of Hopf-Rinow typically uses the fact that the metric topology is equal to the manifold topology. $\endgroup$ – Jack Lee Oct 19 '14 at 19:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.