# Are metric isometries smooth at the boundary?

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.

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 $(*)$).

• 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? – Asaf Shachar Nov 5 '16 at 17:54
• 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? – Asaf Shachar Nov 5 '16 at 17:55
• (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). – Mizar Nov 5 '16 at 18:07
• (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 $s=t$. – Mizar Nov 5 '16 at 18:09
• 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 $M$. – Deane Yang Nov 5 '16 at 18:22