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George Keeling

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- Is it true that in three Riemannian dimensions that the curvature scalar determines whether the volume is finite or infinite? Is a flat three-torus a counter example? How do you describe a flat three-torus?

Spatial slices of the Robertson-Walker metrics are maximally symmetric so they must have a constant curvature. Is it true that in three Riemannian dimensions that a constant curvature scalar determines whether the volume is finite or infinite? Carroll seems to have given a counter-example for the flat case when the metric is Euclidean. That could be R³ or, he says, it could be a "more complicated manifold, such as a three-torus ##S^1\times S^1\times S^1##." I know that a two-torus does not have constant curvature so it is not maximally symmetric. I doubt if a three-torus does better. Do we think Carroll meant a "flat three-torus", but surely that would not be ##S^1\times S^1\times S^1##. What would it be?

Here is my simple definition of a flat two-torus:

It is like a rectangle size ##A\times B## with coordinates ##x,y## but if you go off an edge, you come back to the opposite edge, so adding multiples of ##A## to the ##x## coordinate of ##B## to the ##y## coordinate keep you in the same place. It's a bit like a torus ...

Here is my simple definition of a flat two-torus:

It is like a rectangle size ##A\times B## with coordinates ##x,y## but if you go off an edge, you come back to the opposite edge, so adding multiples of ##A## to the ##x## coordinate of ##B## to the ##y## coordinate keep you in the same place. It's a bit like a torus ...