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This exercise is from Do Carmo, Riemannian Geometry.

Introduce a Riemannian metric on the torus $T^n$ in such a way that the natural projection $\pi:\mathbb{R}^n\to T^n$ given by $$\pi(x_1,...,x_n)=(e^{2\pi ix_1},...,e^{2\pi ix_n})$$ is a local isometry. Show that with this metric $T^n$ is isometric to the flat torus.*

*The flat torus is just $T^n=S^1\times ...\times S^1$ with the product Riemannian metric.

I don't know if this is correct, but in the Lee's book Introduction to Smooth Manifolds, there's something about proper actions that may be related:

The discrete group $\mathbb{Z}^n$ acts smoothly, freely and properly on $\mathbb{R}^n$ by translations. Then there is a unique smooth structure making the quotient map into a smooth covering map. To verify that this smooth structure is the same as the one we defined proviously (thinking of $T^n$ as the product manifold $S^1\times ...\times S^1$) we just check that the covering map $\pi:\mathbb{R}^n\to T^n$ [defined above] is a local diffeomorphism with the product smooth structure on $T^n$. Hence $\mathbb{R}^n/\mathbb{Z}^n$ is diffeomorphic to $T^n$.

So basically me question is if we can use the fact that $\mathbb{Z}^n$ acts on $\mathbb{R}^n$ to give $\mathbb{R}^n/\mathbb{Z}^n$ a Riemannian structure? Is there any bibliography that can be useful?

Thank you.

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  • $\begingroup$ Yes, that works. $\endgroup$ – Qiaochu Yuan May 15 '16 at 2:57

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