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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$\begingroup$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.
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