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Does the Metric tensor define a metric on the topology of the manifold/space it's on, and/or does the metric on the topology of the space define a metric tensor, if you go through all the trouble of defining a vector space/tangent space for it to exist on?

LlwlL
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  • in complete Riemannian manifolds, yes, a metric tensor induces a distance in the manifold through minimum length of paths betwwen two points. But not in general. –  May 23 '22 at 06:28
  • As written, your question is unclear. It appears as something created by google translate or a similar program. I am voting to close due to lack of clarity. – Moishe Kohan May 23 '22 at 10:04
  • Okay but does a metric in the topological sense define a metric tensor. – LlwlL May 24 '22 at 04:07
  • What is "a metric in the topological sense"? Do you mean: If $g$ is a Riemannian metric and $d_g$ the corresponding distance function on the manifold, does $d_g$ determine $g$? – Moishe Kohan May 24 '22 at 04:50
  • All the topology books I’ve read started the first chapter discussing “Metric spaces” which are sets with a distance function. So by “a metric in a topological sense” I do mean a set or manifold with a distance function. I asked the question because some distance functions don’t seem like they would correspond to tensors, like the “taxi cab metric.” – LlwlL May 25 '22 at 06:21
  • Your question is very unclear, I truly do not understand what are you asking about. Edit it to explain what do you really mean. In particular, do not try to fit a question in a single sentence. Also, check if answers to this question address the issue you are interested in. – Moishe Kohan May 25 '22 at 09:38
  • Yeah that’s it. Thank you so much! – LlwlL May 25 '22 at 18:11
  • Note that it doesn’t even make sense to talk about a metric tensor unless you’re on a differentiable (at least $C^1$) manifold. – Ted Shifrin May 26 '22 at 03:59

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