The ground state metric, after an extra dimension of space is compactified (to a circle) in Einsteinian gravity, is the metric which corresponds to the R_4 × S_1 geometry of the separated dimensions. How does one arrive at this metric, from the cross product of 4 dimensional gravity and the extra dimension?

For reference:

  1. JM Overduin's cover of Kaluza-Klein Gravity , on page 21 mentions the ground state metric, as the "vacuum expectation value of the full metric", which "determines the topology of the compact space". It says that it is, topologically, R_4 × S_1.

  2. In answer to a question of mine, Luboš Motl explained that the ground state metric, in practice, is the one for which the geometry is the Cartesian product R_4 × S_1.

So, how could one derive the ground state metric, knowing the topologies R_4 and S_1, and taking their cross product?

  • $\begingroup$ I'm not sure what the question is - the usual treatments of Kaluza-Klein rather explicitly do the dimensional reduction and end up with a 4D metric. Can you make more explicit what you're asking about (by giving references, examples, and formulae)? $\endgroup$ – ACuriousMind Jun 28 '16 at 13:39
  • $\begingroup$ Supposing you knew the topologies of R_4, and S_1, and took their cross product. How would you find the resultant metric? $\endgroup$ – J.P. Escarcega Jun 28 '16 at 17:37
  • $\begingroup$ @J.P.Escarcega Given Riemannian manifolds $(M,g)$, $(N,h)$ there is no unique way to give $M\times N$ a metric. For instance, $g+h$ works, but so does $g+fh$ where $f$ is a function that is never zero. $\endgroup$ – Ryan Unger Jun 29 '16 at 16:09

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