Let $M,N$ be smooth Riemannian manifolds with boundary (In particular, we assume the boundaries are smooth).

Suppose we have a map $\phi:M \to N$ which satisfies the following properties:

$$(1) \, \, \phi:M \to N \, \, \text{is a bijection}$$

$$ (2) \, \, \phi(\operatorname{int}M)=\operatorname{int}N,\phi(\partial M)=\partial N $$

$$ (3) \, \, \phi:M \to N \, \,\text{is a metric isometry}$$

By the Myers-steenrod theorem, applied to $\phi|_{\operatorname{int}M} :\operatorname{int} M \to \operatorname{int}N $, $\phi$ is a diffeomorphism between $\operatorname{int} M , \operatorname{int}N$.

Question: Is $\phi$ necessarily smooth as a map $M \to N$?

When looking at the proof of Myers-steenrod theorem here, the problem seems to be that initial conditions do not determine a unique geodesic, if the starting point is on the boundary.

The basic idea of the proof is to express the map $\phi$ in exponential coordinates, then show this representation is linear, hence smooth. However, constructing this representation relies on the uniqueness of geodesics.

I suspect there might be a counter example where singularity occurs at the boundary, but I could not fine one.


1 Answer 1


Yes, $\phi$ is smooth. Indeed, fix any $x\in\partial M$ and take any nieghbourhood $x\in U\subseteq\partial M$ with compact closure. Calling $\nu:\partial M\to TM$ the (unit) inward-pointing normal vector, we can find some $\epsilon>0$ so small that

  • the map $\alpha:U\times [0,\epsilon]\to M$, $\alpha(y,t):=\exp(t\nu(y))$ is well-defined and is a diffeomorphism onto its image;
  • $\text{dist}(\alpha(y,t),\partial M)=t$ for all $(y,t)\in U\times [0,\epsilon]$.

For any $y\in U$, the curve $t\mapsto\phi\circ\alpha(y,t)$ is a unit-speed geodesic: this is true on the interval $(0,\epsilon]$ by interior smoothness of $\phi$, so it is true on $[0,\epsilon]$ by continuity. Since $\text{dist}(\phi\circ\alpha(y,t),\partial N)=t$, we deduce that it is a minimizing geodesic from $\partial N$ to $\phi\circ\alpha(y,\epsilon)$, implying that $\frac{d}{dt}(\phi\circ\alpha)|_{t=0}\perp\partial M$, i.e. $$ \phi\circ\alpha(y,t)=\exp(t\nu(\phi(y)))\qquad (*)$$ (now $\nu$ denotes the inward-pointing normal in $N$).

Up to shrinking $\epsilon$, we can assume that $(z,t)\mapsto\exp(t\nu(z))$ gives a diffeomorphism from $\phi(U)\times [0,\epsilon]$ onto its image, as well. We call $(\beta,\tau)$ its smooth inverse.

Finally, $\phi|_U$ is smooth (as $\phi(y)=\beta\circ\phi\circ\alpha(y,\epsilon)$ and as $\phi$ is smooth on $\text{int}(M)$), so also $\phi|_{\alpha(U\times[0,\epsilon])}$ (by $(*)$).

  • $\begingroup$ Thanks. I am trying to fill in some details which I am not sure about: 1) You implicitly assume there are unique geodesics emanating from the boundary and orthogonal to it. I guess the only situation of non-uniqeuness that can happen (i.e two different geodesics with identical initial velocity) is if the velocity is tangent to the bounday? Do you know a reference for this? $\endgroup$ Nov 5, 2016 at 17:54
  • $\begingroup$ 2) Do you have a refernce for the fact there is a small neighbourhood $U$, and $\epsilon$ such that $\alpha$ (that is the normal exponential map) is a diffeomorphism? I know this is true when there is no boundary, and what you have described sounds to me like a "half" of a tubular neighbourhood. Is there an analogue of theat theorem for here? $\endgroup$ Nov 5, 2016 at 17:55
  • 1
    $\begingroup$ (1) By "geodesic" I mean a smooth curve $\gamma$ such that $\nabla_{\dot\gamma}\dot\gamma\equiv 0$. So a geodesic is uniquely determined by starting point and initial velocity (if you write the equation explicitly in a chart, you are solving a Cauchy problem). $\endgroup$
    – Mizar
    Nov 5, 2016 at 18:07
  • $\begingroup$ (2) I have no reference, but the idea is that it suffices to prove $\alpha$ injective (by the inverse function theorem). You can cover $\overline{U}$ with finitely many open sets $V_i$ such that $\alpha$ is injective on $V_i\times [0,\epsilon]$ for a unique, very small $\epsilon>0$. Now, if $\alpha(y,s)=\alpha(y',t)$ and $\epsilon$ is very small, then $y$ is very close to $y'$, so they both lie in some $V_i$, i.e. $y=y'$ and $s=t$. $\endgroup$
    – Mizar
    Nov 5, 2016 at 18:09
  • $\begingroup$ To prove uniqueness of the geodesic normal to the boundary, it suffices to extend the metric smoothly to a neighborhood of $\partial M$ and observe that there exists a unique geodesic through any point in $\partial M$ and normal to $\partial M$ in that open manifold. Since the geodesic is transversal to $\partial M$, there exists a sufficiently small interval starting from but not including that point, which lies in the interior of $M$. $\endgroup$
    – Deane Yang
    Nov 5, 2016 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.