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Let $M,N$ be connected Riemannian manifolds. Let $f:M\rightarrow N$ be a bijective smooth map that maps any unit speed geodesic in $M$ to unit speed geodesic in $N$.

Question: Is this suffice to show that $f$ induces a differeomorphism $M\cong N$? Can $f$ even be an (local) isometry? If either is not true, what additional assumption is needed (like $M,N$ are simply-connected)?

(originally asked by a classmate and I was stuck)

Bombyx mori
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  • By "unit speed geodesic to unit speed geodesic", do you mean that ||f_* v|| = ||v||$, where $||v||$ is the unit vector of a geodesic? –  Mar 26 '15 at 20:10
  • @John: I mean like geodesic parametrized by its arc-length. – Bombyx mori Mar 26 '15 at 20:39
  • It's certainly not necessarily going to be a diffeomorphism. Consider a covering map with the obvious metrics. – Ted Shifrin Mar 26 '15 at 22:05
  • @TedShifrin: But a covering map is not bijective. – Bombyx mori Mar 26 '15 at 22:23
  • Oh, oops, I missed the bijective. My apologies. Then it ought to be true. – Ted Shifrin Mar 26 '15 at 22:26
  • @TedShifrin: Yes, if it is not bijective then $\mathbb{R}^{1}\rightarrow \mathbb{R}^{2}$ is a trivial counter-example. My confusion is I do not see how it is true. The exponential map only works locally and I need something global to patch it up together. – Bombyx mori Mar 26 '15 at 22:28

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Your condition forces $||f_*v|| = ||v||$ for any vector $v$, as pointed out by John in the comments. (Quick proof: given $v$, let $\lambda = 1/||v||$ and consider the geodesic $\exp(t\lambda v)$; $f$ maps this to a unit speed geodesic, so $||f_*(\lambda v)|| = 1$.)

Thus the derivative $f_* : T_pM \to T_{f(p)} N$ is nonsingular at every point $p$, and so the inverse function theorem implies that $f$ is a local diffeomorphism. Since you've assumed $f$ is bijective, $f$ must be a diffeomorphism. $f$ is then evidently also an isometry since inner products are determined by their associated norm.

Supplemental Interesting Stuff About Smoothness:

I believe this is true even if $f$ isn't assumed smooth, that is, a bijective map $f: M \to N$ of connected Riemannian manifolds that takes unit speed geodesics to unit speed geodesics is necessarily a smooth isometry. The rest of this answer will be an outline of a proof of this fact, so if you're not interested you can stop reading now.

Your condition that geodesics go to geodesics certainly forces $d(f(x), f(y)) \leq d(x,y)$, so $f$ is continuous. I claim that $f$ is a local metric isometry in the sense that given $x \in M$, there exists a neighborhood $U$ of $x$ such that for $y, z \in U$ we have $d(f(y), f(z)) = d(y,z)$.

To see this fix $r$ so that the geodesic ball $B_r(f(x))$ about $f(x)$ of radius $r$ in $N$ is strongly convex in the sense that there is a unique $N$-geodesic between any two of its points living entirely within the ball, and this geodesic is minimizing in $N$. By making $r$ smaller if necessary we may also assume $B_r(x)$ is strongly convex in $M$. Note that $f$ maps $B_r(x)$ into $B_r(f(x))$.

Fix $y,z \in B_r(x)$; let $\gamma$ be the minimizing geodesic from $y$ to $z$. Then $f \circ \gamma$ is a geodesic from $f(y)$ to $f(z)$, which is contained within $B_r(f(x))$, hence is the unique such geodesic, hence is the unique minimizing geodesic between these points. Thus $d(f(y), f(z)) = d(y,z)$ as claimed.

But then fix $s < r$, for $r$ small enough so that the above holds, and consider the restriction of $f$ to the sphere $S_s(x)$ of radius $s$ about $x$. This is homeomorphic to an $(n-1)$-sphere and its image lies in the $(n-1)$-sphere $S_s(f(x))$ about $f(x)$ in $N$. Since $f$ is injective and continuous and $S_s(x)$ is compact, the restriction of $f$ to $S_s(x)$ is a homeomorphism onto its image. But the $(n-1)$-sphere does not embed as a proper subset of itself, so the image of $f$ must contain all of $S_s(f(x))$. In particular, $f$ is a metric-preserving homeomorphism $B_r(x) \to B_r(f(x))$.

Now apply the Meyers-Steenrod theorem to the map $f : B_r(x) \to B_r(f(x))$ to see that $f$ is smooth on $B_r(x)$.

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mollyerin
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    I do not think $df:T_{p}(M)\rightarrow T_{p}(N)$ is nonsingular implies it must be a local differeomorphism, since we have trivial examples like $id: \mathbb{R}^{1}\rightarrow \mathbb{R}^{2}$. One need invariance of domain to show the dimension is the same and $f$ is a local homeomorphism, then you can use inverse function theorem to show it is a local differeomorphism. I agree that if a map is bijective and a local differeomorphism, then it must be a differeomorphism. I still need to take a look at the second part. As you suspected the original question just assumed $f$ to be continuous. – Bombyx mori Mar 27 '15 at 15:26
  • You're right; I had in my head the assumption that $M$ and $N$ have the same dimension, which is not an assumption you're making. The constant rank theorem does apply to show that $f$ is locally an embedding, though, and then it seems to me that if the dimension of $N$ is too large the image of $M$ has measure zero, hence can't be all of $N$. (I don't see how invariance of domain helps us here, since the problem is showing that $N$ isn't large dimension.) – mollyerin Mar 27 '15 at 20:49
  • Locally we can compose a map $\mathbb{R}^{n}\rightarrow M\rightarrow N\rightarrow \mathbb{R}^{m}$, where $n,m$ are the dimension of $M,N$ respectively and the first/last maps are the coordinate maps. You are right if $m>n$ invariance of domain does not apply, and constant rank theorem would help as $df_{p}$ is everywhere nonsingular. So I think this finishes the smooth case. – Bombyx mori Mar 28 '15 at 03:09
  • I was thinking that Peano curves might cause an issue for the continuous case, but it turns out not to be the case. Indeed we can map $D^{n}\rightarrow M\rightarrow N\rightarrow D^{m}$ under appropriate choice of neighborhood and scaling. Then since any continuous bijection from a compact space to a Hausdauff space is an homeomorphism, we can conclude that $m=n$. So we get $f$ to be a local homeomorphism and hence must be a global homeomorphism as it is bijective. I did not know Myers–Steenrod theorem and it is wonderful to know. Thanks! – Bombyx mori Mar 28 '15 at 03:17
  • I read through what you did and found we basically adopted the same strategy, but your proof is much more detailed. Thanks! – Bombyx mori Mar 28 '15 at 06:46