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Let $M,N$ be smooth, connected, oriented $d$-dimensional Riemannian manifolds (perhaps with boundary).

Suppose there exist a map $f:M \to N$ which is differentiable almost everywhere on $M$, and its differential is an orientation-preserving isometry almost everywhere.

Is it true that $M$ can be isometrically immersed in $N$? (i.e does there exist a smooth isometric immersion $M \to N$?)


If we omit the requirement of orientation-preserving, then this is false:

Gromov showed that for any metric $g$ on the unit $d$-dimensional disk $\mathbb{D}^d$ there is an arcwise isometry $f:(\mathbb{D}^d,g) \to (\mathbb{R}^d,e)$ (which has to be an isometry a.e).

If $g$ is non-flat, then no smooth isometric immersion exist, of course.

Note: In that case, Gromov's arcwise isometry $f$ cannot preserve orientation: It is $1$-Lipschitz, and hence in $W^{1,\infty}(M,N)$. My colleagues and I showed here (section 3) that any weakly differentiable map $f \in W^{1,\infty}(M,N)$ satisfying $df \in SO(n)$ a.e is weakly harmonic, hence smooth. (This shows $f$ cannot be orientation-preserving or orientation-reversing on any open subset of the domain. It must "switch" rotations infinitely often).

Edit: It turns out $M$ is not necessarily isometrically immersible in $N$. However, in the example given $M,N$ were both flat.

This still leaves open the question whether examples of such maps exist for pairs of maniolds which have different curvatures.

Specifically, is there an example for a non-flat $M$, and a flat $N$, such that there exist an a.e orientation-preserving isometric immersion $M \to N$?

The point is to see how much flexibility this notion allows. Gromov's maps show that curvature differences between manifolds do not obstruct existence of a.e isometries.

Do such differences obstruct a.e-orientation-preserving isometries?


Note that in general an "almost everywhere isometry" does not need to be smooth.

An example is $M=[0,1],N=[0,2],f(x)=c(x)+x$ where $c$ is the Cantor function. In this case, $f'=1$ a.e, and $M$ can be isometrically immersed in $N$, via the natural inclusion map.

To refute the claim, one could try to use $f^{-1}:[0,2] \to [0,1]$, but it is not differentiable a.e on $[0,2]$;

$f ^{-1}$ is differentiable on $f(O)$, when $O$ is the complement of the Cantor set (since $O$ is the set where $f$ is differentiable). By the definition of $f$, it is not hard to see that $m(f(O))=m(O)=1$, hence is not of full measure in $[0,2]$.

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  • $\begingroup$ Hint: Use your example of function $f$ to construct a homeomorphism from the circle $M$ of length $1$ to the circle $N$ of length $2$, which is an isometric immersion a.e.. But, of course, there is no smooth isometric immersion $M\to N$. $\endgroup$ – Moishe Kohan Mar 7 '17 at 18:49
  • $\begingroup$ Thanks! Your idea is very nice. (In afterthought, it looks very natural to use the example I gave to "wrap" one circle onto the other). I must say I am still interested in finding examples in higher dimensions, though. (I would be happy to see examples when the curvatures of $M,N$ differ, which is impossible in dimension $1$, since everything is flat there. I think your idea, to use this "seed example" as a way to "pathologically double" a manifold, can probably be lifted to two dimensional examples. $\endgroup$ – Asaf Shachar Mar 7 '17 at 20:24
  • $\begingroup$ However, I think it only naturally extends itself to cases when $M$ is a scaled version of $N$ (indeed, a double), so the signs of the curvatures would be identical. Again, I think it is interesting to see how much flexibility this notion allows. Gromov's maps show curvature differences do not obstruct a.e isometries. Do such differences obstruct a.e-orientation-preserving isometries? (e.g can we have $M$ non-flat and $N$ flat or vice versa?) $\endgroup$ – Asaf Shachar Mar 7 '17 at 20:24
  • $\begingroup$ The example generalizes naturally to a map between $n$-dimensional tori or, more generally, manifolds containing flat pieces. However, the case of maps between manifolds of negative curvature looks challenging. $\endgroup$ – Moishe Kohan Mar 8 '17 at 0:43
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This answer is based on a comment by Moishe Cohen.

Let $M,N$be circles with radiuses $1,2$ respectively. We define $g:M \to N$ by

$$g(e^{2\pi it})=2e^{\pi if(t)}$$

where $f(x)=x+c(x)$ is the example function given in the question.

$g$ is differentiable at $\{e^{2\pi it} \, | \,t \in [0,1]\setminus C \}$, which is a set of full measure on the unit circle $M$.

Let $ \alpha(s)=e^{2\pi i(t+s)},\alpha'(s)=2\pi i e^{2\pi i(t+s)},\alpha'(0)=2\pi i e^{2\pi it}$: Then

$$dg_{(e^{2\pi it})}(2\pi i e^{2\pi it})=\frac{d}{ds}|_{s=0}g(\alpha(s))=\frac{d}{ds}|_{s=0} 2e^{\pi if(t+s)}=2e^{\pi if(t)}f'(t)=2e^{\pi if(t)}$$

for every $t \in [0,1]\setminus C $.

So $dg_{(e^{2\pi it})}$ maps a vector with length $2\pi$ to a vector of the same length, and since the tangent spaces are one-dimensional this implies it's an isometry a.e.

It is also easy to see it's differential is orientaion-preserving whenever it is defined.

Thus, $g$ satisfies the requirements, but $M$ is not isometrically immersible in $N$.

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