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

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.

• Related, and possibly a duplicate: Is topology of universe observable? – John Rennie Aug 12 '14 at 8:29
• 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. – Luboš Motl Aug 12 '14 at 8:40
• possible duplicate of What is known about the topological structure of spacetime? – Danu Aug 12 '14 at 8:49
• – Danu Aug 12 '14 at 8:51
• 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. – Jerry Schirmer Aug 12 '14 at 17:10