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The $n$-dimensional torus can be obtained as a quotient: $T^n=\mathbb{R}^n/\mathbb{Z}^n$.

As pointed out here, the standard metric on $\mathbb{R}^n$ is invariant under translation by the elements of $\mathbb{Z}^n$ so it descends to the quotient (i.e there is a unique Riemannian metric on $T^n$ making the canonical projection a Riemannian isometry).

Is there an easy way to see that the curvature (of that metric on $T^n$) is zero? I am aware of O'Neill's formula, but I would like to find a more elementary way to prove this without using it. Is it possible? (at least for $n=2$ )?

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Asaf Shachar
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    That looks like an $n$-dimensional torus :-) – robjohn May 12 '15 at 20:06
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    Curvature is a local invariant, $\Bbb R^n$ is flat, and the projection $\Bbb R^n \to T^n$ is (by construction of the metric you chose on $T^n$) a local isometry. Is that satisfying? –  May 12 '15 at 20:07
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    Isn't it locally isometric to $\mathbb{R}^n$? – robjohn May 12 '15 at 20:07
  • @MikeMiller: great minds think alike :-) – robjohn May 12 '15 at 20:08
  • @MikeMiller: You are right of course. The important point is that $Z^n$ is discrete, so the fact that the projection is a submersion automatically implies its an immersion, and immersions are locally embeddings. This is what I was missing. – Asaf Shachar May 12 '15 at 20:19

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For sake of completeness I am writing the full answer, as suggested by Mike Miller:

$Z^n$ is discrete implies dim($\mathbb{T}^n$)=dim($\mathbb{R}^n$), so the fact that the projection is a submersion automatically implies its an immersion, and immersions are locally embeddings.

It is a local isometry (not just an embedding) by the construction of the metric.

Asaf Shachar
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