An isometry between two Riemannian manifolds ${(M,g), (N,h)}$ is a diffeomorphism ${\phi: M \rightarrow N}$ such that ${\phi^* h = g}$. In other words, ${\phi}$ respects the Riemannian structure as well as the differentiable structure.

It follows that if ${d,d'}$ are the metrics on ${M,N}$ induced by the Riemannian metrics ${g,h}$, then ${d'(\phi(x),\phi(y)) = d(x,y)}$ for ${x,y \in M}$—that is, ${\phi}$ is distance-preserving. Interestingly, a version of the converse is true:

Theorem 1 (Myers-Steenrod) If ${\phi: M \rightarrow M}$ 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 ${\phi: M \rightarrow N}$! 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.

${\phi}$ is evidently a homeomorphism. Now pick ${p \in M}$ and a neighborhood ${D_r(p)}$ such that for any ${q \in D_r(p)}$, there is a unique geodesic in that neighborhood connecting ${p,q}$. Call it ${\gamma}$. I claim that ${\phi \circ \gamma}$ is a geodesic in ${N}$. The geodesic ${\gamma}$ can be assumed to be parametrized by unit length. We have for all ${t}$,

$\displaystyle d(p,q) = d(\gamma(t),p) + d(\gamma(t), q).$

Thus

$\displaystyle d(\phi(p),\phi(q)) = d(\phi(\gamma(t)),\phi(p)) + d(\phi(\gamma(t)), \phi(q)) \ \ (*).$

In other words, we have strict equality in the triangle inequality.

Geodesic preservation

I will state a small but useful fact: in a Riemannian manifold ${M}$, if ${q,r}$ lie in a small neighborhood of ${p}$ and

$\displaystyle d(q,r) + d(r,p) = d(q,p) ,$

then ${r}$ lies on a geodesic betwen ${p,q}$. Indeed, draw a geodesic ${\alpha}$ from ${q}$ to ${r}$ and a geodesic ${\beta}$ from ${r}$ to ${p}$ that travel at unit speed and minimze the distances; we can do this locally even without the assumption of completeness. The catenation ${\alpha \beta}$ minimizes the length from ${q}$ to ${p}$, so it is an unbroken geodesic. Thus ${r}$ lies on the geodesic.

So, returning to the discussion above, we see that by (*) ${\phi}$ sends geodesics to geodesics.

Exponential coordinates

Now fix open ${U \subset T_p(M),V \subset T_q(N)}$ containing the origin such that ${\exp_p}$ takes ${U}$ diffeomorphically to a neighborhood ${D_r(p)}$, and ${\exp_q}$ takes ${V}$ diffeomorphically to ${D_r(q)}$. There is an expression for ${\phi}$ as a map ${\tilde{\phi}: U \rightarrow V}$ in these exponential coordinates. It is obtained as follows: if ${v \in U}$, take a geodesic ${\gamma_v}$ with ${\gamma_v'(0) = v}$ and consider ${(\phi \circ \gamma_v)'(0)}$. In fact this induces more generally a map

$\displaystyle \tilde{\phi}: T_p(M) \rightarrow T_q(M).$

Since ${\gamma_v}$ moves at constant speed ${|v|}$ and ${\phi \circ \gamma_v}$ moves at the same speed, it follows that ${\tilde{\phi}}$ is norm-preserving. Also ${\tilde{\phi}(0) = 0}$. If we can show that ${\tilde{\phi}}$ is linear, then we’ll get smoothness in this exponential coordinate system—hence smoothness in general.

${\tilde{\phi}}$ is an isometry

It can be checked that

$\displaystyle \boxed{\lim_{A,B \rightarrow 0 \in T_p(M)} \frac{d(\exp_p(A),\exp_p(B)) }{|A-B|} } = 1.$

I will postpone this for now.

This is what I will use to show that ${\tilde{\phi}: T_p(M) \rightarrow T_q(M)}$ is an isometry, where each tangent space is given the norm from the Riemannian metric. Pick ${A,B \in T_p(M)}$ and consider the geodesics ${\gamma_{A},\gamma_{B}}$ at ${p}$ and ${\gamma_{A'}, \gamma_{B'}}$ where ${A'=\tilde{\phi}(A), B' = \tilde{\phi}(B)}$. Then ${d(\gamma_{A}(t), \gamma_{B}(t)) = d( \gamma_{A'}(t), \gamma_{B'}(t))}$. So

$\displaystyle 1= \lim_{t \rightarrow 0} \frac{d(\exp_p(tA),\exp_p(tB)) }{t|A-B|} = \lim_{t \rightarrow 0} \frac{d(\exp_q(tA'),\exp_q(tB')) }{t|A-B|} = \frac{|A'-B'|}{|A-B|}.$

Isometries are linear maps

To show that ${\tilde{\phi}}$ is linear, we appeal to a general fact:

Theorem 2 Let ${X,Y}$ be normed linear spaces and ${T: X \rightarrow Y}$ a map such that ${|Tx - Tx'| = |x-x'|}$ for ${x,x' \in X}$. Then ${T}$ is linear if ${T(0) = 0}$.

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 ${\phi}$ is smooth, and that for ${v \in T_p(M)}$, ${|v|_p = |\phi_*(v)|_{\phi(p)}}$, i.e. ${\phi}$ preserves lengths on tangent vectors. By the polarization identity, ${\phi}$ preserves the inner product, and is thus an isometry.

A lemma

I never proved a fact I stated about the exponential map—the equality

$\displaystyle \lim_{A,B \rightarrow 0 \in T_p(M)} \frac{d(\exp_p(A),\exp_p(B)) }{|A-B|} = 1.$

I will briefly sketch the idea here. ${|A-B|}$ is the length of the linear path from ${A }$ to ${B}$ in ${T_p(M)}$, so it will be sufficient to show:

Lemma 3

If ${c:(0,1) \rightarrow T_p(M)}$ is a path in ${T_p(M)}$, then as ${c((0,1)) \rightarrow 0}$ with the derivative ${|c'|}$ staying bounded,$\displaystyle \frac{ l(\exp_p(c))}{l(c)} = 1.$

I’m abusing notation quite a bit here, but I do not think this should cause confusion.

Basically, the reason is that

$\displaystyle l(c) = \int |c'|$

while

$\displaystyle l(\exp_p(c)) = \int | (\exp_p)_{*c(t)} c'(t) |_{\exp_p(c(t))}$

where in the second equation I am referring to the norms induced by the metric on the various tangent spaces of ${M}$. The difference between these two is ${o(l(c))}$, because ${\exp_p}$ has derivative the identity at ${0 \in T_p(M)}$, and the metric on ${T_{\exp_p(c(t))}(M)}$ is very close (say by a factor of ${1 \pm \epsilon}$) to that of ${T_p(M)}$ if we work in local coordinates and agree to identify the tangent spaces.