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

Suppose $f:M \to N$ is a $C^1$ isometry (i.e $f$ is continuously differentiable, and $df_p$ is an isometry for every $p \in M$).

I think that $f$ must be smooth (I describe a way of showing this below). Is there a simpler way?

It suffices to show $f$ maps geodesic to geodesic, that is: $\alpha$ is a geodesic $\Rightarrow$ $f \circ \alpha$ is a geodesic.

Indeed, if this is the case and $\alpha(0)=p,\alpha'(0)=v$, then $f \circ \alpha(0)=f(p),(f \circ \alpha)'(0)=df_p(v)$, so by uniqueness of geodeics, we conclude that $f \circ exp_p^M(tv)=f \circ \alpha(t)=exp_{f(p)}^N(t \cdot df_p(v))$. In other words, $$f \circ exp_p^M=exp_{f(p)}^N\circ df_p,$$

so $f$ is a local diffeomorphism, hence smooth.

The problem is: How do we know $f$ maps geodesic to geodesic?

If we take the differntial definition, $\nabla \dot\alpha=0$, then we hit a problem since a-priori $f \circ \alpha$ is differentiable only once (the geodesic equation is of second order).

However, we could use the characterization of geodesics as length minimizers, but this requires that we know that geodesics "beat" any $C^1$ paths, not just (piecewise) smooth ones. This is essentially saying that the Riemannian distance does not change when using only $C^1$ paths.

Another way to proceed is to use the Myers-Steenrod theorem:

We consider the lenght structure $(\mathcal{A}_M,L_M)$ where $\mathcal{A}$ is the set of all $C^1$ paths in $M$, and $L_M:\mathcal{A}_M \to \mathbb{R}$ is the standard Riemannian length: $L(\gamma)=\int \|\dot \gamma(t)\|$. (and similarly for $N$).

By our assumption, $f$ maps paths in $\mathcal{A}_M$ to paths in $\mathcal{A}_N$, and it's also length-preserving, i.e an arcwise isometry.

By the inverse function theorem, $f$ is a local homeomorphism, hence a local isometry w.r.t the intrinsic distances defined using the class of $C^1$ paths. (By intrinsic I mean we use only paths that stay inside the neighbourhood and take the associated distance, not the restricted distance from the manifolds).

Since the Riemannian distance does not change when using only $C^1$ paths, it follows that $f$ is a local isometry w.r.t the intrinsic distances (defined as usual using the class of piecsewise smooth paths paths).

Now the smoothness of $f$ follows from the Myers-Steenrod theorem.

  • $\begingroup$ Myers-Steenrod theorem seems a bit too heavy for this; a more elementary proof comes from the fact that (for any $p\in M$) the square composed of $f$, $d_pf$, and of the exponential maps at $p\in M$ and $f(p)\in N$, is a commutative diagram (and since every map in it, except possibly $f$, is a local diffeomorphism, $f$ must also be a local diffeo). $\endgroup$
    – user8268
    Feb 5, 2017 at 22:20
  • $\begingroup$ I agree that what you suggest is the natural way to do this. The problem lies with the details however: How do you know the diagram commutes? i.e How do you show $f$ maps geodesic to geodesic? (Using the differential formulation doesn't work since it's a second order equation, and our $f$ is only $C^1$). I think perhaps the fastest way is to use the fact that geodesics locally minimize length, even when compared to $C^1$ paths. (i.e the Riemannian distance does not change if we use $C^1$ paths). I have added some details in the question. $\endgroup$ Feb 7, 2017 at 15:00
  • $\begingroup$ @user8268 Do you have an easier way of showing $f$ maps geodesics to geodescis? $\endgroup$ Feb 7, 2017 at 15:00
  • $\begingroup$ I don't see an easier way than using local length minimization, but that is a reasonably elementary thing (the standard proof using the exponential map works perfectly well for $C^1$ paths) $\endgroup$
    – user8268
    Feb 7, 2017 at 15:29

2 Answers 2


$f$ maps a geodesic to a geodesic :

Since $d_pf$ has full rank so there is open ball $B:=B_\epsilon (p)$ s.t. $f$ is bijective on it

For $q\in B$, there is unique geodesic $c$ from $p$ to $q$ Then $$ L:=d_M(p,q)={\rm length}\ c={\rm length}\ f\circ c \geq d_N(f(p),f(q))$$

If $c_1$ is a unique geodesic from $f(p)$ to $f(q)$ then we have a curve $ c_2$ s.t. $f\circ c_2=c_1$

Since $f^{-1}$ is $1$-Lipschitz so $$L\geq d_N(f(p),f(q))={\rm length}\ c_1={\rm length}\ c_2 \geq d_M(p,q) =L$$

So we complete the proof

  • $\begingroup$ Thanks. I think the reasoning might be simpler(?) (as discussed with user8268 in the comments). All we need to see is that the image of a geodesic locally beats any $C^1$ path, which is easy. (You can see my added answer now for details). $\endgroup$ Feb 8, 2017 at 14:13

As mentioned in the quesiton, it's enough to show that if $\alpha$ is a geodesic, then $f \circ \alpha$ is a geodesic.

Let $p \in M$, and let $\alpha$ be a geodesic emanating from $p$.

Then for sufficiently small $t$, it holds that $L(\alpha|_{[0,t]}) \le L(\beta|_{[0,t]})$ for any $C^1$ path $\beta$ connecting $\alpha(0),\alpha(t)$. (This follows from the standard proof that geodesics are locally length minimizing, which works for $C^1$ paths).

By the inverse funciton theorem there exist open neighbourhoods $p \in U,f(p) \in V$ s.t $\, \, f:U \to V$ is invertible.

We claim that $f \circ \alpha$ is also length minimizing for sufficiently small $t$. Indeed, if there was a shorter $C^1$ path $c$ between $f(p),f(\alpha(t))$, we could assume W.L.O.G that it lies completely inside $V$, hence $f^{-1} \circ c$ was a $C^1$ path shorter than $\alpha$, a contradiction.

Thus, $f \circ \alpha$ is a geodesic, as required. (Since it's length minimizing and parametrized by arc-length).


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