In GR, we define our manifolds to be locally Euclidean. However, we also demand that our metric tensor have a Lorentzian signature. Since the metric tensor is a measure of curvature, doesn't the first property imply that our metric should be locally Eulcidean (Riemannian)? How does imposing a Lorentzian signature on the metric preserve the locally Euclidean property of the manifold?


1 Answer 1


I believe it is just a matter of notation inconsistency, common in physics.

Manifolds are by definition locally Euclidean (i.e. locally look like $\mathbb{R}^n$, the $n$ dimensional Euclidean space). There is no metric given a priori.

In Special Relativity, we also define a metric of Lorentzian signature, and call this structure Minkowski spacetime. This structure has no curvature.

In General Relativity, what is meant, is that locally, spacetime always resembles that of Special Relativity - in a small enough neighborhood it is like Minkowski.

When one says that in GR manifolds are locally Euclidean, either they over-repeat the definition of a manifold, or are using the description "locally Euclidean" to actually say "locally Minkowski".

  • 2
    $\begingroup$ I think 'locally flat' gets mixed up with 'locally Euclidean' in physics uses, the former of course meaning locally Minkowski in GR contexts. But I agree with your answer. $\endgroup$
    – Eletie
    Apr 14, 2021 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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