Curvature Singularities in Geodesically Complete Manifolds Do there exist manifolds which are geodesically complete, and yet have a curvature singularity? While I don't believe this is the case, I have yet to find a proper proof of the same.
 A: Well, it's no wonder the theorem is difficult to prove false: it's actually true. There do exist manifolds that are geodesically complete and have curvature singularities - a necessary condition for which is that the existence of a divergence in a curvature invariant does not imply geodesic incompleteness.
Heuristically, this is because curvature singularities can occur due to the divergence of curvature invariants which are higher-order in the metric than the first-order condition imposed on the tangent vectors of a geodesic (more specifically, it only depends on the continuously differentiable structure of the metric), so the geodesic cannot "see" the curvature singularity. This result is in some sense converse to the implications of the strong cosmic censorship hypothesis [1]. You may find a few explicit examples in [2], although it's debatable as to whether such solutions are realisable once we impose physically reasonable conditions.
If you are willing to go beyond the framework of GR, then [3] explicitly identifies classes of wormhole solution spacetimes with complete causal geodesics, finite energy density, and curvature singularities in quadratic Palatini $f(R)$ gravity.
References:

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*[1]: Choquet-Bruhat, General Relativity and the Einstein Equations (2009)

*[2]: Geroch, What is a Singularity in General Relativity?

*[3]: Bejarano,
What is a singular black hole beyond General Relativity?
A: Yes there are surfaces that are complete in the means of a geodesic yet they still have curvature singularities. One such great example comes in use with the Poincaré conjecture.
