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

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2390 views

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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Thanks all for the links which gives the answer from Kostya: In non-trivial topology, the light rays will "wrap around" our universe multiple times and you'll be able to see the same (similar) copies of galaxies. @LubošMotl great point on the small enough torus. Also, I know that the three torus doesn't need to be a sub manifold and we can make sense of it without a 4D space. But wonder if the observable multiple images can also suggest the existence of an extra spatial dimension. I mean keeping open the possibility of the embedding.

This post imported from StackExchange Physics at 2014-08-14 08:28 (UCT), posted by SE-user Victor Vahidi Motti
And Lubos's comments aside, if you are absolutely committed to the immersion of the manifold in a bigger space, there is a theorem that says that every manifold an be embedded in a larger flat space.

This post imported from StackExchange Physics at 2014-08-14 08:28 (UCT), posted by SE-user Jerry Schirmer
@JerrySchirmer thanks. in Carroll's you see that immersion is introduced as to be different from being embedded. That is yet another question of mine. Carroll's explanation is rather vague and unclear in the errata page. Couldn't make sense of the difference.

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

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