This question is very underdeveloped, but I was wondering if there was a map from the sphere to the torus which preserves length of closed curves? I was just thinking about taking a walk on a sphere and coming back to my starting point. Could I possibly parametrize a similar walk on the torus having the same length. I tried to get a better understanding of this question by making deforming the sphere until it was almost identical to the torus, topologically of course, but since the torus in genus-1 there is only but so much I can do.

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It is quite amazing but such maps do exist!

Call a map between two metric spaces (say, Riemannian manifolds) $f: M\to N$ a path-isometry if it preserves lengths of all paths. Recall that if $p: [a,b]\to M$ is a path (say, Lipschitz-continuous), then its length is defined as $$ \int_a^b |p'(t)|dt. $$

Example. $f: R\to R$, $f(x)=|x|$.

This map is not a homeomorphism even locally, but it preserves lengths of paths (since the derivative of $f$ has absolute value $1$ a.e.). Of course, $f$ is not $C^1$ but merely Lipschitz, but that's the natural regularity condition since all path-isometries are 1-Lipschitz.

M.Gromov in his book "Partial Differential Relations" proved the following amazing theorem (see 2.4.11 in the book):

Theorem. For every Riemannian surface $M$ there exists a path-isometry $f: M\to R^2$ (where $R^2$ is equipped with the standard flat metric).

Gromov proved much more, but the above theorem is enough for us.

Therefore, take Gromov's path-isometry $g: S^2\to R^2$; its image is contained in a disk $D$. Now, the torus admits a flat metric and $D$ admits an isometric map $i$ to such flat torus $T^2$. Lastly, the composition $$ f=i\circ g: S^2\to T^2 $$ is the required path-isometry.

Edit. Suppose you want to impose an extra regularity assumption on your maps, namely, $f\in C^1$ (continuously differentiable). Then one can verify that all $C^1$-smooth path-isometries are actually Riemannian isometries in the sense that the derivative $df: TM\to TN$ preserves lengths of tangent vectors. In particular, such $f$ is an immersion: $df_m: T_mM\to T_{f(m)}N$ is injective for all $m\in M$.

Then one observes that an immersion $f: S^2\to T^2$ do not exist. One can either prove it by verifying that such $f$ would have to be a covering map (and the sphere does not cover the torus) or simply take a nowhere vanishing vector field $X$ on $T^2$ and take its pull-back $Y=f^*(X)$ under $f$:

$Y(m)\in T_mS^2$ is the unique vector such that $df(Y(m))=X(f(m))$.

Then $Y$ would be a nowhere vanishing vector field on $S^2$, which contradicts the Hairy Ball theorem.

  • $\begingroup$ That's amazing indeed-I'm glad you saw this. $\endgroup$ – Kevin Carlson Jul 7 '14 at 10:00

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