In how many ways can one define a Riemannian metric on $T^n$ such that the projection map $\pi:\mathbb R^n\rightarrow T^n$ defined by $\pi(x_1,...,x_n)=(e^{ix_1},...,e^{ix_n})$ is a local isometry?


You keep asking the same question, effectively, over and over.

If you have a covering projection $\pi\colon M\to N$ and $M$ is a Riemannian manifold whose Riemannian metric $g$ is invariant under the group of deck (covering) transformations, then the only metric on $N$ which makes $\pi$ a local isometry is the obvious one, namely $(\pi^{-1})^*g$. This is computed locally (since $\pi$ has smooth local inverses), but gives a well-defined metric on $N$ because of the assumption on the invariance of $g$. To emphasize: There is a unique metric on $N$ so that $\pi$ is a local isometry.

  • $\begingroup$ Thanks. I am just a bad beginner in the realm of differential geometry having started Riemannian Geometry without any sufficient background on the geometry of manifolds nor the classical geometry of curves and surfaces so I may ask questions trivial to experts $\endgroup$ – user555729 Dec 13 '18 at 18:26
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    $\begingroup$ I understand. I personally would recommend that you study some classic curves and surfaces material before doing the graduate Riemannian geometry, as you'll get intuition and basic knowledge that will help. My undergraduate text is freely available at a link in my profile. $\endgroup$ – Ted Shifrin Dec 13 '18 at 18:41
  • $\begingroup$ I downloaded it and it contains all necessary information well-written concisely, so due to the limitation of my time this is the perfect classical differential geometry book for me. Thanks $\endgroup$ – user555729 Dec 14 '18 at 8:29

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