Is the projection from $\mathbb R^n$ locally isometric to the flat torus?

Question

I’m reading a problem asking to introduce a Riemannian metric on $T^n$ in such a way that the projection $\pi(x_1,...,x_n)=(\exp(ix_1),...,\exp(ix_n))$ from $\mathbb R^n$ to $T^n$ is a local isometry. Does the flat metric apply here?

By definition, flat torus is defined as the product of n copies of the Riemannian manifold $S^1$ with its metric induced by $\mathbb R$.

Maybe you ought to clarify what the projection $\pi$ you refer to is? — SBK, Dec 13 '18 at 15:16
isn't the flat metric exactly defined via this property? related to this: https://math.stackexchange.com/questions/450110/metric-of-the-flat-torus?rq=1 — Balou, Dec 13 '18 at 15:19
I’m reading DoCarmo’s book and the definition there is different @Balou — , Dec 13 '18 at 15:23
maybe you should state then how "the flat metric" is defined in DoCarmo — Balou, Dec 13 '18 at 15:29
ok, then it is less obvious then if you defined it via $T^n = R^n / Z^n$ and the inherited metric... — Balou, Dec 14 '18 at 10:21

score 1 · Accepted Answer · answered Dec 13 '18 at 15:15

1

Yes; both of these spaces are "flat" in the sense that the curvature of the Riemannian metric is identically zero.

answered Dec 13 '18 at 15:15

SBK

3,041

Why does DoCarmo following this question want to prove that the torus with the metric introduced is isometric to the flat torus? Is it not redundant? – Dec 13 '18 at 15:17
I imagine it's just building familiarity with the rigorous definitions. Just because you know it's true, it certainly doesn't mean that its a pointless exercise – SBK Dec 13 '18 at 19:08

score -1 · Answer 2 · answered Sep 25 '19 at 20:18

Notice that the natural projection written above maps a point of $R^n$ to a point in $T^n$. This may seem odd at first but notice how the coordinate curves of $R^n$ get mapped to $T^n$. Because $e^{ix}$ is a point in a complex circle, the projection essentially wraps each axis of $R^n$ into a circle of the torus. This maps coordinate curves into coordinate curves. Since the projection is evidently the cartesian product of a number of complex circles (which are the same as $S^1$) you can introduce the product metric the same as if you were working with real numbers exclusively. Since you can take the vectors to be coordinate curves locally, the product metric is a local isometry.

Is the projection from $\mathbb R^n$ locally isometric to the flat torus?

2 Answers2