I Riemann Curvature: Understanding Parallel Transport on 1D Rings

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Everyone who is currently studying GR must be familiar with this picture. We find Riemann curvature by paraller transport a "test vector" around and see whether the vector changes its direction.

My question. How does it work with one dimensional Ring? A geomteric ring is intuitively curved but the only parallel transport possible for a vector to the point where it previously started, just give the sampe direction.
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You are mixing up intrinsic and extrinsic curvature. A 1d space has no intrinsic curvature (as you appear to have deduced), but you can embed it in a higher dimensional space where its tangent vector field (also embedded in that space) need not always point in the same direction. This latter is what you are calling "intuitively" curved.

GR cares about intrinsic curvature. Spacetime isn't embedded in a higher dimensional space that we are aware of, so extrinsic curvature isn't a useful concept.
 
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Narasoma said:
How does it work with one dimensional Ring?
It doesn't. A one-dimensional manifold cannot have any intrinsic curvature.
 
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