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Lorentzian geometry is the geometry of Minkowski spacetime, hence essentially of a Euclidean space, but equipped not with the standard Euclidean Riemannian metric of signature $(+,+,+,…,+)$ (which yields Euclidean geometry) but with the pseudo-Riemannian metric of signature $(−,+,+,…,+).$
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$(M,g)$ is complete iff $(\tilde{M},\tilde{g})$ is complete (non-Riemannian version)
I'm not sure if this question is too low level for Math Overflow (so feel free to move this to SE if you think it is).
Inspired by this and this question I'm wondering if the following statement is a …