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Do all metric tensors have signature (-,+,+,+) or (+,-,-,-) in the Pseudo-riemannian manifold describing spacetime in the Theory of General Relativity?

If yes:

In this answer by John Rennie, it is stated that:

Lorentzian manifolds are a special case of pseudo-Riemannian manifolds where the signature of the metric is (3,1) (or (1,3) depending on your sign convention).

Since Lorentzian manifolds ≡ signature (1,3), if the answer to my question is yes, it means that the General relativity spacetime is a 4D Lorentzian manifold.

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    $\begingroup$ Do you have any particular reason to think that the answer is "no"? $\endgroup$
    – Javier
    Feb 14, 2022 at 12:58
  • $\begingroup$ Yes i do, because in this comment by Valter Moretti he suggests to look for Pseudo-riemannian manifolds instead of Lorentzian manifolds when in General relativity; and concerning the question "Is the Schwarzschild metric a metric of Lorentzian manifold", he replies with "It is a metric of Pseudo-riemannian manifolds" $\endgroup$
    – TrentKent6
    Feb 14, 2022 at 13:09

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Yes. In GR, spacetime is a 4D Lorentzian manifold

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