# If the universe is a $3$-torus, must it be longer in some direction?

So in a finite universe, I've read one possible topology for a flat universe is a 3 torus. On a 2 torus, it's obviously longer in one direction than the other. Would the same hold true for the universe if it was a 3 torus? Can you not have a "cubic" space-time for the universe if it's a 3 torus, would it need to be a rectangular prism instead?

• The toroidal diameters on a manifold can be of any sizes. Oct 5 '18 at 20:13
• "On a 2 torus, it's obviously longer in one direction than the other." no, not necessarily. It may or it may not -- it depends on the specific metric you use. Same for higher-dimensional tori (or any other manifold for that matter). Oct 5 '18 at 20:14
• @AccidentalFourierTransform What do you mean? The circumference of a donut is longer than the "handle" or whatever the term for it is.
– John
Oct 5 '18 at 20:15
• @John You're assuming the torus is embedded in some larger 3D space. That's not assumed in GR; there's no sense in which spacetime is embedded in a larger dimensional space. Oct 5 '18 at 20:27

For example, one way of defining the $$2$$-torus is to consider the plane, with the identifications $$(x, y) \sim (x + L_1, y), \quad (x, y) \sim (x, y + L_2)$$ where $$\sim$$ means that the two points are regarded as the same point. This yields a torus with side lengths $$L_1$$ and $$L_2$$. Similarly, the $$3$$-torus constructed from $$\mathbb{R}^3$$ can have any dimensions $$L_i$$, including having all the $$L_i$$ equal.
By the way, even if you insisted that physical space be embedded in a flat larger space, that still doesn't give you any constraints. The Nash embedding theorem states that any curved space can be embedded in a flat $$\mathbb{R}^n$$ for sufficiently large $$n$$. Unfortunately, $$n$$ is quite high, e.g. $$n = 17$$ for the $$2$$-torus, so the construction is not easy to picture.