Let $M$ be a compact, connected, oriented smooth Riemannian manifold with non-empty boudary.

Let $f:M \to M$ be a smooth orientation preserving isometric immersion.

Is it true that $f$ is locally distance preserving?

(in the sense that around every $p \in M$ there exist a neighbourhood $U$ s.t $d(f(x),f(y))=d(x,y)$ for all $x,y \in U$)

This question is really about the boundary points of $M$.

In the interior this holds: Since $f(M^o) \subseteq M^o$ we can consider it as a map $M^o \to M^o$, and as such it is an arcwise isometry which is also a local homeomorphism, hence a local isometry w.r.t the intrinsic distance on $M^o$, hence also w.r.t the Riemannian distance on $M$.

A positive answer to this question, will settle these two questions.

Edit 1-Generalization:

Perhaps to better understand the question, let's see what happens when we broaden it a little.

Suppose $F:M \to N$ is an isometric immersion between two equidimensional Riemannian manifolds. Is it necessarily locally distance preserving?

(I do not assume here that $M,N$ are compact,connected,or oriented. In this more general case $F$ does not need to be a local diffeomorphism since it doesn't have to be open; Take e.g $M=N=[0,\infty),F(x)=x+1$).

Edit 2 - A proposed solution:

Let $p \in \partial M$. The idea is to use the inverse function theorem for points in $M^o$ which are close to $p$: Around each point of $M^o$, we can find a neighbourhood where $f$ is invertible, hence an isometry (since $f$ and its inverse are arcwise isometries, hence $1$-Lipschitz).

If we could obtain some lower bound on the size of these neighbourhoods*, then perhaps we can "push them" to the boundary, and deduce local distance preservation around it (via a continuity argument).


*By looking at the standard proof of the inverse function theorem, it turns out that in the Euclidean case, when $M=\mathbb{R}^d$, if

$$ B_r(p) \subseteq \{x \in \mathbb{R}^d \, | \, \|df_x-df_p\| \le \frac{1}{2}\|(df_p)^{-1} \|\} \tag{1}$$ then $$f \text{ is invertible at } B_r(p).$$

Moreover, in this case, $df_x \in O(n)$ for all $x$ implies $f$ is affine, i.e $df$ is a constant orthogonal matrix (A corollary of Liouville's theorem). Hence the premise in $(1)$ holds vacuously for every $r$.

Even when $M$ is flat (i.e only localy Euclidean), one can use this approach as follows:

Given $p \in \partial M$ we can take a sequence $p_n \to p, p_n \in M^o$. Then, around each $p_n$ there is a ball where $M$ is isometric to Euclidean space. Then, Liouville's theorem imply $df$ is locally constant around each $p_n$, hence (by connectivity) it is in fact a constant matrix. (Here there is a subtlety, I am identifying all the tangent spaces near $p$, with $\mathbb{R}^d$, since I am implicitly assuming that there is a neighbourhood of $p$ isometric to an open subset of $\mathbb{R}^d$ together with smooth parts of its boundary. Unfortunately, It is not clear I can asume this...).

Now for the general case, we can use normal coordinates around the point $p \in \partial M$, and hopefully obtain some estimate on the neighbourhoods size. ("Every metric is locally approximately Euclidean").


1 Answer 1


I believe the answer is, yes, $f$ is locally distance preserving when $M$ is compact and, no, when $M$ is not compact.

Let me start with the noncompact case. Consider the example $$ M = \{ (x,y) \in \mathbb R^2 \,:\, y \leq x^2 \} \,$$ and define $$ f(x,y) = (x,y-1)\,. $$ Then $f$ is an isometric immersion, but it is not locally distance preserving for points along the boundary, because the boundary is concave. For $p_1, p_2 \in \partial M$ close enough together we have $$ d(f(p_1),f(p_2)) = |p_1 - p_2| < d(p_1,p_2)\,.$$ The problem is that $f$ maps points from the boundary into the interiour.

Now assume $M$ is compact. The claim is that if $f : M \to M$ is an isometric immersion, then $f(\partial M) \subseteq \partial M$. I do not have a fully worked out proof for this, but the idea is that compactness implies that $M$ has finite volume and an isometric immersion cannot "lose" volume. However, if $f(p) \in M^\circ$ for some $p \in \partial M$, then there would be half of the space, that is locally not reached.

If you know that $f$ maps the boundary to the boundary then it should be possible to show that $f$ is a local diffeomorphism and hence locally distance preserving.

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    $\begingroup$ Thanks! Your counter-example for the non-compact case is great! However, the compact case is more subtle. First, a manifold can be isometrically immersed in a manifold with less volume (wrap a circle with radius $2$ onto a circle with radius $1$). So, the "volume loss" argument is a little problematic. Secondly, you are right that the whole point is whether or not $f$ maps boundary to boundary. Indeed, in that case it would be a local diffeomorphism (see here math.stackexchange.com/questions/1984346/…) $\endgroup$ Commented Mar 14, 2017 at 17:08
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    $\begingroup$ ..Also, if $f(\partial M) \subseteq \partial M$, then it must be a Riemannian isometry! This is explained, in a short argument here: mathoverflow.net/questions/253930/… , below the "Remarks" section. Finally, I am not convined by your argument "that we would indeed lose half of the space"... Again, I think you are in the right direction, and again all these properties (isometry, boundary preserrvation, local diffoemorphism) are related\equivalent, but the problem is giving a rigorous arguemt for the validity of any of them... $\endgroup$ Commented Mar 14, 2017 at 17:12
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    $\begingroup$ You are right, this is more subtle than it appears at first sight. My latest guess is that $f:M^\circ \to f(M^\circ)$ is a covering map and by connectedness every point has the same number of preimages. Can one use topology to show that $f|_{M^\circ}$ is surjective? $\endgroup$ Commented Mar 15, 2017 at 7:47

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