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, according to the Myers–Steenrod theorem:
If is distance-preserving and surjective, then it is an isometry (in particular, it is smooth).
For notational simplicity, assume .
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.
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 (2) 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.
It can be checked that
We will postpone this for now.
This is what we 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
To show that is linear, we appeal to a general fact:
Let be normed linear spaces and a map such that for . Then is linear if .
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.
We never proved a fact about the exponential map—the equality
We will briefly sketch the idea here. is the length of the linear path from to in , so it will be sufficient to show:
If is a path in , then as with the derivative staying bounded,
We are abusing notation quite a bit here, but it should not cause confusion.
The reason is that
where in the second equation we are 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.