# A differentiable isometry is smooth?

I posted this question in MSE but got no response (even after giving a bounty), so I am trying here.

Let $M,N$ be smooth $d$-dimensional Riemannian manifolds.

Suppose $f:M \to N$ is a differentiable isometry ($df_p$ is an isometry at every $p \in M$). I do not assume $f$ is $C^1$.

Is it true that $f$ must be smooth?

This is false when assuming $f$ is only differentiable almost everywhere.

One counter-example is $f(x)=c(x)+x$ where $c$ is the Cantor function. In this case $M=[0,1],N=[0,2]$, $f'=1$ a.e; $M,N$ are not isometric, but both are flat.

Gromov showed similar examples exist for "many" non-flat $M$ and $N=\mathbb{R}^d$.

Some partial results:

By the inverse function theorem for everywhere differentiable maps, $f$ is a local homeomorphism.

We would like to prove it's a local isometry w.r.t the intrinsic distances, then use the Myers-Steenrod theorem (or use the fact geodesics locally minimize length to deduce $f$ maps geodesics to geodesics, hence it factors through its differential via the exponential maps. this works if $f \in C^1$, see here for details).

The devil is in the details: We need to choose suitable length structures on $M,N$ such that $f$ will become an arcwise isometry. We cannot use the class of $C^1$ paths since $f$ does not necessarily map $C^1$ maps to $C^1$ maps.

Also, nothing promises us that for a differentiable path $\gamma$ such that $\|\dot \gamma(t)\|$ is integrable, $\|\dot {f \circ \gamma}(t)\|$ will be integrable.

• Yes, I misunderstood the question, and what I wrote wasn't really an answer. – Igor Belegradek Feb 14 '17 at 19:58

By the nontrivial fact that $f$ is a local homeomorphism, we can assume without loss of generality that $f$ is a bijective homeomorphism. The usual textbook proof of the formula for the differential of the inverse map works here, so that $f^{-1}$ is again a differentiable isometry in the sense of the OP.

So it suffices to show that $f$ is 1-Lipschitz (as a map from $M$ to $N$, which are thought as metric spaces), since then $f^{-1}$ will be 1-Lipschitz as well, i.e. $f:M\to N$ will be a metric space isometry and we will be done thanks to the Myers-Steenrod theorem.

Fix $p,q\in M$ and take any $C^1$ curve $\gamma:[0,1]\to M$ with $\gamma(0)=p$ and $\gamma(1)=q$. Call $L$ the length of $\gamma$. It suffices to prove that $d(f(p),f(q))\le L$ (here $d(\cdot,\cdot)$ denotes the distance), since then, taking the infimum of $L$ among all curves connecting $p$ to $q$, we get $d(f(p),f(q))\le d(p,q)$.

To this aim, fix any $\epsilon>0$.

Claim: For any $t\in [0,1]$ we can find some $\delta(t)>0$ such that it holds $$d(f\circ\gamma(s),f\circ\gamma(u))\le(1+\epsilon)d(\gamma(s),\gamma(t))+(1+\epsilon)d(\gamma(t),\gamma(u))$$ whenever $t\in[s,u]$ and $u-s\le \delta$.

Indeed, using normal coordinates about $\gamma(t)$ and $f\circ\gamma(t)$, we see that for some $\eta(t)>0$ we have $$d(x,\gamma(t))\le\eta(t)\Rightarrow d(f(x),f\circ\gamma(t))\le(1+\epsilon)d(x,\gamma(t)),$$ so we are done by taking $\delta(t)$ such that $\gamma([0,1]\cap \overline B_{\delta(t)}(t))\subseteq \overline B_{\eta(t)}(\gamma(t))$.

Now, by Cousin's theorem, we can find a partition $0=t_0<\cdots<t_N=1$ and suitable $t_i'\in[t_{i-1},t_i]$ (for $i=1,\dots,N$) such that $t_i-t_{i-1}\le\delta(t_i')$. Thus, $$d(f(p),f(q))\le\sum_{i=1}^N d(f\circ\gamma(t_{i-1}),f\circ\gamma(t_i))$$ $$\le(1+\epsilon)\sum_{i=1}^N (d(\gamma(t_{i-1}),\gamma(t_i'))+d(\gamma(t_i'),\gamma(t_i))).$$ The last sum is bounded by $L$. Letting $\epsilon\to 0$ we obtain the thesis.

• Thanks! Your argument is very cool. However, I am having a little trouble completing the details in the "normal coordinates" part of the argument. Can you please elaborate on why there is some $\eta(t)$ such that $d(x,\gamma(t))\le\eta(t)\Rightarrow d(f(x),f\circ\gamma(t))\le(1+\epsilon)d(x,\gamma(t))$? (Of course the dependene on the variable $"t"$ is immaterial here (This is a "pointwise argument"). – Asaf Shachar Feb 24 '17 at 16:41
• Call $z:=\gamma(t)$, $X:=T_zM$ and $Y:=T_{f(z)}N$ ($X$ and $Y$ are endowed with the scalar products given by the Riemannian metrics at $z$ and $f(z)$, respectively). You can find $r,s>0$ small such that $\exp_{z}:B_s(0)\to M$ and $\exp_{f(z)}:B_s(0)\to N$ are diffeomorphisms onto their images (the balls are taken wrt the scalar products on $X$ and $Y$, respectively). As $f$ is continuous, we can assume that $f(\exp_z(B_r(0)))\subseteq\exp_{f(z)}(B_s(0))$. So $\exp_{f(z)}^{-1}\circ f\circ\exp_z$ is defined on $B_r(0)$. Its differential at $0$ is an isometry. – Mizar Feb 24 '17 at 17:26
• (To see this, just recall that, identifying $T_0X\cong X$ and $T_0Y\cong Y$, the differential of $\exp_z$ at $0$ is the identity; the same holds for $\exp_{f(z)}$.) Now, by definition of differentiability, $|\exp_{f(z)}^{-1}\circ f\circ\exp_z(v)|\le(1+\epsilon)|v|$ for any $v\in X$ small enough. This translates to what we wanted. – Mizar Feb 24 '17 at 17:39