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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 also true for Pseudo-Riemannian manifolds: A Pseudo-Riemannian manifold $(M,g)$ is complete if and only if its universal covering $(\tilde{M},\tilde{g})$ is complete.

The reason I ask is that in the above links the theorem of Hopf Rinow plays an important role in the proof of this statement and since this theorem is not available for Pseudo-Riemannian manifolds, I'm wondering if the statement also holds for Pseudo-Riemannian manifolds.

If this statement is not true, are there special instances where it holds nevertheless?

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Completeness of a pseudo-Riemannian manifold (or a manifold with affine connection, more generally) means precisely the completeness of its geodesic flow on its tangent bundle. But the tangent bundle of any covering space is a covering space of the tangent bundle. The geodesic flow is the flow of the geodesic vector field on the tangent bundle. A vector field on a manifold is complete if and only if its lift to some (hence any) covering space is complete, which is just because the flow lines both lift (as a covering space) and project.

In the Riemannian setting, Hopf-Rinow says that completeness of the flow of the geodesic vector field is precisely completeness as a metric space. But completeness as a metric space is not defined on pseudo-Riemannian manifolds; the automorphism group of the pseudo-Riemannian geometry on Minkowski space doesn't preserve any metric, as the stabilizer of point is not compact.

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  • $\begingroup$ There are some things I don't understand (yet): I know how one can lift a diffeomorphism of $M$ to a diffeomorphism of the frame bundle of $M$, but how do you lift the geodesic flow of $TM$ to the frame bundle? And by geodesic vector field you mean the vector field of this geodesic flow of the frame bundle, right? Is there maybe a reference in which the geodesic flow on the frame bundle of a pseudo-Riemannian manifold is defined? $\endgroup$
    – user450093
    Feb 13, 2019 at 15:20
  • $\begingroup$ You could view the geodesic flow as a flow on $TM$ instead of the frame bundle; that will work just as well, with the same proof as above, just replacing the expression frame bundle by the expression $TM$. The geodesic vector field on $TM$ is the vector field whose flow is the geodesic flow. $\endgroup$
    – Ben McKay
    Feb 13, 2019 at 15:58

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