Until recently, I did not believe that it would be possible to have a flat (in the Riemannian sense) torus or doughnut. But then I figured it is possible:

$$(x^1, x^2, x^3, x^4)=\Big(\cos\phi, \sin\phi, \cos\psi, \sin\psi\Big)$$ which is a 2D manifold embedded in the 4D Euclidean space ($\mathbb{E}^4$) with the induced, flat metric $$ds^2=d\phi^2+d\psi^2$$ It also is a torus, clearly! So my question is: Is it possible to have a flat sphere? A manifold with vanishing Riemann tensor everywhere (Not almost everywhere, like a cube) which has a spherical topology? If yes, are there any topologies (high genus, etc.), for which this is not possible? Thanks!


1 Answer 1


Great question!

It is not possible to have a flat sphere. In fact, it follows from the Gauss-Bonnet theorem that any (edit: connected compact orientable) surface which is flat is topologically a torus.

See https://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem

  • $\begingroup$ Wow, beautiful answer; I will remember that! Thank you. $\endgroup$
    – K. Sadri
    Mar 28, 2021 at 15:55
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    $\begingroup$ This is not quite correct. Any connected, compact, orientable flat surface is topologically a torus. But if you don't require orientability, you can also have flat Klein bottles. And if you don't require compactness, you can have planes, cylinders, punctured tori, or any surface that admits an embedding into $\mathbb R^2$, among other things. And of course if you don't require connectedness, you could have disjoint unions of any of the above. $\endgroup$
    – Jack Lee
    Mar 28, 2021 at 17:31
  • $\begingroup$ @JackLee yes sloppy of me, thanks! editing answer $\endgroup$
    – hunter
    Mar 28, 2021 at 17:43

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