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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?

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  • $\begingroup$ The toroidal diameters on a manifold can be of any sizes. $\endgroup$
    – Slereah
    Oct 5 '18 at 20:13
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    $\begingroup$ "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). $\endgroup$ Oct 5 '18 at 20:14
  • $\begingroup$ @AccidentalFourierTransform What do you mean? The circumference of a donut is longer than the "handle" or whatever the term for it is. $\endgroup$
    – John
    Oct 5 '18 at 20:15
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    $\begingroup$ @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. $\endgroup$ Oct 5 '18 at 20:27
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No, the dimensions of the torus can have any size. Sometimes, we visualize the curvature of two-dimensional spaces by thinking of them as surfaces in a flat three-dimensional space. However, physical space does not have to be embedded in a larger space to make sense.

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.

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  • $\begingroup$ Hmm, yeah the mathematical definition is basically just arbitrarily saying when you hit one end go back to the other, right? That feels kind of unnatural... That's why the torus shape makes sense to me, there is no edge that wraps around. I actually read a little bit about that Nash theorem before I asked this question. Can't say I understood it though. Are you saying that you just need to keep adding extra dimensions of space until you can make the torus do what you want? $\endgroup$
    – John
    Oct 5 '18 at 20:38
  • $\begingroup$ @John That's right, you can get anything with enough dimensions. Since it's incredibly hard to picture the results, we prefer to just go with what you call the "arbitrary" definition, which is much more convenient in practice. $\endgroup$
    – knzhou
    Oct 5 '18 at 20:40
  • $\begingroup$ @John I also wrote this answer about how mathematical definitions are chosen. Bottom line: they're chosen so we can actually do math with them, not so they're as intuitive as possible. $\endgroup$
    – knzhou
    Oct 5 '18 at 20:41

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