# Is it true that a map that maps geodesics to geodesics must be an differeomorphism?

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)

• By "unit speed geodesic to unit speed geodesic", do you mean that ||f_* v|| = ||v||$, where$||v||$is the unit vector of a geodesic? – user99914 Mar 26 '15 at 20:10 • @John: I mean like geodesic parametrized by its arc-length. – Bombyx mori Mar 26 '15 at 20:39 • It's certainly not necessarily going to be a diffeomorphism. Consider a covering map with the obvious metrics. – Ted Shifrin Mar 26 '15 at 22:05 • @TedShifrin: But a covering map is not bijective. – Bombyx mori Mar 26 '15 at 22:23 • Oh, oops, I missed the bijective. My apologies. Then it ought to be true. – Ted Shifrin Mar 26 '15 at 22:26 ## 1 Answer Your condition forces$||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)$. • I do not think$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}$is everywhere nonsingular. So I think this finishes the smooth case. – Bombyx mori Mar 28 '15 at 3:09 • I was thinking that Peano curves might cause an issue for the continuous case, but it turns out not to be the case. Indeed we can map$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 3:17
• 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 6:46