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Is there an example of a pair $M,N$ of connected, oriented equidimensional Riemannian manifolds with the following properties:

  1. $M$ is everywhere non-flat, $N$ is flat.

  2. There exist a map $f:M \to N$ which is differentiable almost everywhere (a.e), and $df$ is an orientation-preserving isometry a.e.

(An easier goal: Find a pair of manifolds which are not locally isometric, but which admit a map as in 2. We should probably restrict here to manifolds without boundary, since otherwise $M=[0,1],N=\mathbb{R},f(x)=x$ is an example. )

Context:

The point is to see whether curvature differences obstruct existence of a.e orientation-preserving isometries.

If we omit the requirement on the orientation, then there is a lot of flexibility; Gromov showed that for any metric $g$ on the unit $d$-dimensional disk $\mathbb{D}^d$ there is an a.e isometry $f:(\mathbb{D}^d,g) \to (\mathbb{R}^d,e)$. ($e$ is the Euclidean metric).


Further comments:

  1. Gromov's a.e isometry cannot be orientation-preserving:

It's $1$-Lipschitz, and hence in $W^{1,\infty}(M,N)$, and every map $f \in W^{1,\infty}(M,N)$ satisfying $df \in \text{SO}$ a.e is smooth. (Thus Gromov's map cannot be orientation-preserving or orientation-reversing on any open subset of the domain. It must "switch" orientations in an"infinite rate").

  1. An a.e orientation preserving isometry does not need to be smooth:

For an example take $M=[0,1],N=[0,2],f(x)=c(x)+x$ where $c$ is the Cantor function. Then $f'=1$ a.e.

This example can be used to show that there is an a.e orientation-preserving isometry from a circle of radius $1$ into a circle of radius $2$. (Of course, there is no smooth local isometry from the former into the latter).

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There is a discontinuous map $f\colon\mathbb{S}^2\to\mathbb{R}^2$ such that $d_xf$ is defined and isometric for almost all $x$. (If you want a continuous one then I am sure the answer is "no")

To construct such $f$ do the following:

  1. Start with a sequence of finer and finer subdivision $(K_n)$ of $\mathbb{S}^2$ into polygons; say next subdivision divedes each polygon in 4 nearly equal pieces.

  2. Construct a maximal tree $T_n$ of cuts in in the 1-skeleton of $K_n$, in such a way that $T_n$ is obtained from $T_{n-1}$ by adding minimal length of cuts.

  3. Finally consider the development of these polygons in the plane and pass to the limit.

The construction

There are few estimates in the construction you have to take care of, but I am sure everything should work.

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  • $\begingroup$ Thanks. However, I am afraid I do not understand too many parts in your idea; (1) Do you consider $\mathbb{S}^2$ with the round metric? (2) I am not sure what is the role of the maximal tree of cuts? (you said cuts of $1$-skeleton, so I guess you just mean "edges"? (3) By "development", do you mean to a "gluing diagram" (a set of polygons that is assigned the way in which they should be glued to each other along sides and vertices)? Is such a diagram always unique? (Perhaps this what the tree for...) $\endgroup$ – Asaf Shachar Mar 18 '17 at 8:20
  • $\begingroup$ @AsafShachar Yes, I consider sphere with round metric. After cutting along the tree you can develop sphere into plane with very small distortion. However you may loose all this in the limit. It happens if you cut the sphere along the meridians from the north pole almost to the south pole. In order to keep the property, you need to cut along a fractal tree, so that most of the time one can travel between close points without going too far and passing cuts. $\endgroup$ – Anton Petrunin Mar 18 '17 at 14:43
  • $\begingroup$ I am still not sure what exactly do you mean by the cuts and the tree. Even more basically, by polygons on the sphere, do you mean union of geodesic triangles (or graphs that can be decomposed into a triangulation, that is perhaps with some of the edges missing?). If I understood correctly the spirit of your argument, you are trying to build a "very accurate" map of the sphere. (in the sense of geographic maps). $\endgroup$ – Asaf Shachar Mar 18 '17 at 19:39
  • $\begingroup$ @AsafShachar "cut" means "cut", with scissors --- look at the picture; yes geodesic triangles will work for you; I am not sure what do you mean by "accurate", I would say "fractal". $\endgroup$ – Anton Petrunin Mar 18 '17 at 20:50
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    $\begingroup$ How did you do those nice cartoons? $\endgroup$ – Artur Jackson Apr 14 '17 at 5:32
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I will add an explanation of Anton Petrunin's comment

Question : There is no continuous map $$f :B\rightarrow \mathbb{E}^2$$ where $B$ is a geodesic ball in $S^2(1)$ of radius $\varepsilon$ s.t. $df$ is isometric a.e.

Proof : Assume that $f$ is a continuous map. Note that ${\rm length}\ f(\partial B)={\rm length}\ \partial B$. In further, $f(B)$ is enclosed by $f(\partial B)$. Since $f$ is a volume preserving map, by isoperimetric inequality, it is a contradiction.

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  • $\begingroup$ I think your claim contradicts "Gromov's rumpling theorem", which says there is a length-preserving map $\mathbb{S}^2 \to \mathbb{R}^2$. (See here). Any such map will be $1$-Lipschitz, hence continuous and isometric a.e. If I am wrong, would you mind to elaborate on some points?: (1) I do not see why ${\rm length}\ f(\partial B)={\rm length}\ \partial B$ ($df$ does not need to be isometric on $\partial B$ which is of measure zero). $\endgroup$ – Asaf Shachar Apr 14 '17 at 8:46
  • $\begingroup$ (2) In what sense do you mean "enclosed"? If you refer to the Jordan curve theorem, the separating curve should be an injective image of the circle, while here we do not know $f|_{\partial B}$ is injective. (3) I do not see why $f$ should be volume preserving (there is no regularity assumption except for continuity, why is that enough?). Finally, I don't think you used the assumption $df$ is orientation-preserving a.e... $\endgroup$ – Asaf Shachar Apr 14 '17 at 8:46
  • $\begingroup$ (1) Assume that $df_p$ exists. Then there is $\epsilon$ s.t. $f$ is injective on $B_\epsilon (p)$ Proof : If $x_n\rightarrow p$ and $f(x_n)=f(p)$, then $df_p$ is not isometric. Consider the case where $x_n,\ y_n\rightarrow p$ and $f(x_n)=f(y_n)$ Then $df_p$ is not isometric. As far as I know, in Gromov's rumpling theorem the map is not injective. (2) $f$ is volume preserving by 197p. in A Course in Metric Geometry - BBI $\endgroup$ – Hee Kwon Lee Apr 14 '17 at 9:14
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    $\begingroup$ Thanks. I must say I am not convinced that $f$ must be injective on some small ball. If this argument was valid, then it would also work for Gromov's map: $\tilde f:\mathbb{S}^2 \to \mathbb{R}^2$ is length-preserving, hence $d \tilde f$ is an isometry almost everywhere. Then, you claim it is injective on a small ball. I still think you are trying to prove a claim which is too strong, that contradicts Gromov's result. (Gromov's map does not need to be orientaion-preserving a.e however...). $\endgroup$ – Asaf Shachar Apr 14 '17 at 15:24
  • $\begingroup$ I agree with you. For injectiveness I think that orientation preserving is needed. $\endgroup$ – Hee Kwon Lee Apr 14 '17 at 16:01

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