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How to test that a flat metric represents a global three-torus geometry

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When introducing Robertson-Walker metrics, Carroll's suggests that we

consider our spacetime to be $R \times \Sigma$, where $R$ represents the time direction and $\Sigma$ is a maximally symmetric three-manifold.

He then goes on to discuss the curvature on $\Sigma$ which yields the metric on this three-surface

$d\sigma^2=\frac{d \bar{r}^2}{1-k\bar{r}^2}+\bar{r}^2d\Omega^2$

Case $k=0$ corresponds to no curvature and is called flat. So the metric, after introducing a new radial coordinate $\chi$ defined by $d\chi=\frac{d\bar{r}}{\sqrt{1-k\bar{r}^2}}$, the flat metric on $\Sigma$ becomes

$d\sigma^2=d\chi^2+\chi^2d\Omega^2$

$d\sigma^2=dx^2+dy^2+dz^2$

which is simply flat Euclidean space.

Carroll then points out that

Globally, it could describe $R^3$ or a more complicated manifold, such as the three-torus $S^1 \times S^1 \times S^1$.

I see that the metric on $S^1 \times S^1 \times S^1$ is also given by $d\theta^2+d\phi^2+d\psi^2$ and therefore there could be a fourth spatial dimension in which $\Sigma$ is a submanifold.

However, I am unsure how can we test by experiments or cosmological observations to know for sure whether the flat metric is indeed Euclidean or to conclude a more complicated global three-torus geometry.

This post imported from StackExchange Physics at 2014-08-14 08:28 (UCT), posted by SE-user Victor Vahidi Motti
asked Aug 12, 2014 in Theoretical Physics by Victor Vahidi Motti (20 points) [ no revision ]
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Related, and possibly a duplicate: Is topology of universe observable?

This post imported from StackExchange Physics at 2014-08-14 08:28 (UCT), posted by SE-user John Rennie
In general relativity, we never imagine that the space or spacetime is a submanifold in a higher-dimensional one (your "4D space") - the embedding is just a tool to visualize curved spaces in popular texts for children. ... Otherwise, the toroidal shape - with its finite volume and periodicity - clearly has lots of potential consequences. One may sail around like Magellan, see multiple images of the same celestial objects in many directions, etc. For any of these to be observable, the torus must be small enough - not too much bigger than the visible Universe.

This post imported from StackExchange Physics at 2014-08-14 08:28 (UCT), posted by SE-user Luboš Motl
possible duplicate of What is known about the topological structure of spacetime?

This post imported from StackExchange Physics at 2014-08-14 08:28 (UCT), posted by SE-user Danu
Related: physics.stackexchange.com/q/111670

This post imported from StackExchange Physics at 2014-08-14 08:28 (UCT), posted by SE-user Danu
See this paper : Cosmic microwave background anisotropies in multi-connected flat spaces

This post imported from StackExchange Physics at 2014-08-14 08:28 (UCT), posted by SE-user Trimok
Most recent comments show all comments
@VictorVahidiMotti: and I will say that almost no one uses these embeddings, except for very specialized purposes.

This post imported from StackExchange Physics at 2014-08-14 08:28 (UCT), posted by SE-user Jerry Schirmer
@JerrySchirmer is this one of those very specialized purposes: If you induce, i.e. pulling back, a flat 4D metric on the three sphere you get the spatial metric and the topology of de Sitter space: $R \times S^3$

This post imported from StackExchange Physics at 2014-08-14 08:28 (UCT), posted by SE-user Victor Vahidi Motti

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