An isometry between two Riemannian manifolds is a diffeomorphism such that . In other words, respects the Riemannian structure as well as the differentiable structure.
It follows that if are the metrics on induced by the Riemannian metrics , then for —that is, is distance-preserving. Interestingly, a version of the converse is true:
Theorem 1 (Myers-Steenrod) If is distance-preserving and surjective, then it is an isometry (in particular, it is smooth).
I should observe that the proof applies to the case of a map of Riemannian manifolds ! I don’t know why Helgason states it with the extra hypotheses. Initially I wrote this post with this question unresolved under the goal of figuring out what I had missed and why this hypothesis was necessary by blogging, but I was unable to do so. With that complete, however, a search reveals that it’s stated in the additional generality in Petersen’s book (the relevant parts of which can be viewed at Google Books). However, it’s easier to do the notations in the case of one manifold.
is evidently a homeomorphism. Now pick and a neighborhood such that for any , there is a unique geodesic in that neighborhood connecting . Call it . I claim that is a geodesic in . The geodesic can be assumed to be parametrized by unit length. We have for all ,
In other words, we have strict equality in the triangle inequality.
I will state a small but useful fact: in a Riemannian manifold , if lie in a small neighborhood of and
then lies on a geodesic betwen . Indeed, draw a geodesic from to and a geodesic from to that travel at unit speed and minimze the distances; we can do this locally even without the assumption of completeness. The catenation minimizes the length from to , so it is an unbroken geodesic. Thus lies on the geodesic.
So, returning to the discussion above, we see that by (*) sends geodesics to geodesics.
Now fix open containing the origin such that takes diffeomorphically to a neighborhood , and takes diffeomorphically to . There is an expression for as a map in these exponential coordinates. It is obtained as follows: if , take a geodesic with and consider . In fact this induces more generally a map
Since moves at constant speed and moves at the same speed, it follows that is norm-preserving. Also . If we can show that is linear, then we’ll get smoothness in this exponential coordinate system—hence smoothness in general.
is an isometry
It can be checked that
I will postpone this for now.
This is what I will use to show that is an isometry, where each tangent space is given the norm from the Riemannian metric. Pick and consider the geodesics at and where . Then . So
Isometries are linear maps
To show that is linear, we appeal to a general fact:
Theorem 2 Let be normed linear spaces and a map such that for . Then is linear if .
This is a very interesting theorem, but it is easier to leave it for a later post.
Conclusion of the proof
Now we have seen that is smooth, and that for , , i.e. preserves lengths on tangent vectors. By the polarization identity, preserves the inner product, and is thus an isometry.
I never proved a fact I stated about the exponential map—the equality
I will briefly sketch the idea here. is the length of the linear path from to in , so it will be sufficient to show:
Lemma 3If is a path in , then as with the derivative staying bounded,
I’m abusing notation quite a bit here, but I do not think this should cause confusion.
Basically, the reason is that
where in the second equation I am referring to the norms induced by the metric on the various tangent spaces of . The difference between these two is , because has derivative the identity at , and the metric on is very close (say by a factor of ) to that of if we work in local coordinates and agree to identify the tangent spaces.