CTC, determinism and valuedness of Riemann tensor

Question

What I understood is that By the math, the Riemann tensor is obtained by parallel-transporting a vector along a closed curve in the considered space, then apply Stoke's theorem.
Now if physics is considered then the space becomes spacetime, and hence it is based on a CTC.
But how could determinism, supported by Einstein, be compatible with the following:
Either there is a well defined value of all vector field at every point at everytime like determinism would like. But then the start and end vector parallel transported along a CTC would be the same, hence no curvature and so no gravity.
Or the start and end vector would be different, meaning that at point $x$ and time $t$ the vector could have many values.
How is this quantization-like aspect reflected in the remaining part of the derivation of the EFE?
I thought it should induce a criterion like only changes in the parallel transported vector that iterated give at some point the identity are allowed, in order to recover determinism.
But since the changes are infinitesimal this can give no quantization rule ?

Anders Sandberg · Answer

There is no conflict, since you are mixing up two things: the geometry of the spacetime manifold, and the physics happening on it.
When you derive the Riemann tensor parallel transport around closed loops is used to motivate the formula (there are presumably other ways to do it too). But these loops are not paths by physical particles, but mathematical curves used to probe the curvature.
In contrast, physics on the manifold may have restrictions such as causality (perhaps) preventing CTCs. But that physics is following a separate set of mathematical rules.
One can compare to how the formalism of special relativity allows superluminal trajectories with bisarre consequences if interpreted as actual particle trajectories: physics may not allow them, but Minkowsky spacetime has no built-in rule against them.

CTC, determinism and valuedness of Riemann tensor

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