I am just learning what a topology is and from what I have understood up till now is that a topological space is nothing but a set with a notion of nearness that is given introducing open sets.

Ok, so, in the definition of manifold that I have seen (in Wald's general relativity book) a manifold is constructed mapping subsets of the manifold to be set with open subsets of $\mathbb{R}^n$.

This way we introduce a topology in our manifold to be, since we can use the notion of open balls in $\mathbb{R}^n$ to define open balls of the manifold and hence we get a topology.

Now, we can add more structure to the manifold endowing it with a metric and making it a metric space.

Now my question. I know that a metric induces naturally a topology. But we already had a topology before itroducing the metric structure, so, are this two topologies the same topology?

  • $\begingroup$ Doesn't it matter if the metric is a riemannian one or a lorentzian one? $\endgroup$ – PhoenixPerson Aug 8 '14 at 16:06
  • 2
    $\begingroup$ Of course, the metric must be Riemannian. You don't get a metric space otherwise. $\endgroup$ – Zhen Lin Aug 8 '14 at 16:06
  • $\begingroup$ in place of a riemmanian metric you could (as is done in general relativity) define a pseudo-riemmanian metric right? $\endgroup$ – PhoenixPerson Aug 8 '14 at 16:08
  • $\begingroup$ No, that wouldn't work. In a pseudo-Riemannian metric there can be distinct points that are distance zero apart, which cannot happen in a metric space. $\endgroup$ – Zhen Lin Aug 8 '14 at 16:10
  • 1
    $\begingroup$ In terms of pseudo-Riemannian manifolds, you may be interested in the Alexandrov topology. $\endgroup$ – Willie Wong Aug 21 '14 at 10:44

For the Riemannian case this is surely true, you may find it in "Foundations of Differential Geometry" by Kobayashi and Nomizu, volume 1, page 166, proposition 3.5.

In the pseudo-Riemannian case this is not true as indicated in a comment above by Zhen Lin (the topology induced by the pseudometric will no longer be Hausdorff, for instance).


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.