nLab
Myers-Steenrod theorem

Idea

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

It follows that if $d,d\prime$ are the metrics on $M,N$ induced by the Riemannian metrics $g,h$, then $d\prime (\varphi (x),\varphi (y))=d(x,y)$ for $x,y\in M$—that is, $\varphi$ is distance-preserving. Interestingly, a version of the converse is true, according to the Myers–Steenrod theorem:

Theorem (Myers–Steenrod)

If $\varphi :M\to N$ is distance-preserving and surjective, then it is an isometry (in particular, it is smooth).

For notational simplicity, assume $N=M$.

$\varphi$ 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 $\varphi \circ \gamma$ is a geodesic in $N$.

The geodesic $\gamma$ can be assumed to be parametrized by unit length. We have for all $t$,

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

Thus

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

In a Riemannian manifold $M$, if $q,r$ lie in a small neighborhood of $p$ and

(3)$$d(q,r)+d(r,p)=d(q,p),$$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 (2) $\varphi$ 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 $\varphi$ as a map $\tilde{\varphi }:U\to V$ in these exponential coordinates.
It is obtained as follows: if $v\in U$, take a geodesic ${\gamma }_v$ with ${\gamma }_v\prime (0)=v$ and consider $(\varphi \circ {\gamma }_v)\prime (0)$. In fact this induces more generally a map

(4)$$\tilde{\varphi }:T_p(M)\to T_q(M).$$\tilde{\phi}: T_p(M) \to T_q(M).

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

$\tilde{\varphi }$ is an isometry

It can be checked that

(5)$$\boxed{\underset{A,B\to 0\in T_p(M)}{\lim }\frac{d({\exp }_p(A),{\exp }_p(B))}{\mid A-B\mid }}=1.$$\boxed{\lim_{A,B \to 0 \in T_p(M)} \frac{d(\exp_p(A),\exp_p(B)) }{|A-B|} } = 1.

We will postpone this for now.

This is what we will use to show that $\tilde{\varphi }:T_p(M)\to T_q(M)$ 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 ${\gamma }_{A\prime },{\gamma }_{B\prime }$ where $A\prime =\tilde{\varphi }(A),B\prime =\tilde{\varphi }(B)$. Then $d({\gamma }_A(t),{\gamma }_B(t))=d({\gamma }_{A\prime }(t),{\gamma }_{B\prime }(t))$. So

(6)$$1=\underset{t\to 0}{\lim }\frac{d({\exp }_p(tA),{\exp }_p(tB))}{t\mid A-B\mid }=\underset{t\to 0}{\lim }\frac{d({\exp }_q(tA\prime ),{\exp }_q(tB\prime ))}{t\mid A-B\mid }=\frac{\mid A\prime -B\prime \mid }{\mid A-B\mid }.$$1= \lim_{t \to 0} \frac{d(\exp_p(t A),\exp_p(t B)) }{t|A-B|} = \lim_{t \to 0} \frac{d(\exp_q(t A'),\exp_q(t B')) }{t|A-B|} = \frac{|A'-B'|}{|A-B|}.

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

Let $X,Y$ be normed linear spaces and $T:X\to Y$ a map such that $\mid \mathrm{Tx}-\mathrm{Tx}\prime \mid =\mid x-x\prime \mid$ for $x,x\prime \in X$. Then $T$ is linear if $T(0)=0$.

Conclusion of the proof

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

A lemma

We never proved a fact about the exponential map—the equality

(7)$$\underset{A,B\to 0\in T_p(M)}{\lim }\frac{d({\exp }_p(A),{\exp }_p(B))}{\mid A-B\mid }=1.$$\lim_{A,B \to 0 \in T_p(M)} \frac{d(\exp_p(A),\exp_p(B)) }{|A-B|} = 1.

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

Lemma

If $c:(0,1)\to T_p(M)$ is a path in $T_p(M)$, then as $c((0,1))\to 0$ with the derivative $\mid c\prime \mid$ staying bounded,

(8)$$\frac{l({\exp }_p(c))}{l(c)}=1.$$\frac{ l(\exp_p(c))}{l(c)} = 1.

We are abusing notation quite a bit here, but it should not cause confusion.

The reason is that

(9)$$l(c)=\int \mid c\prime \mid$$l(c) = \int |c'|

while

(10)$$l({\exp }_p(c))=\int \mid ({\exp }_p{)}_{*c(t)}c\prime (t){\mid }_{{\exp }_p(c(t))}$$l(\exp_p(c)) = \int | (\exp_p)_{*c(t)} c'(t) |_{\exp_p(c(t))}

where in the second equation we are 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 $1\pm ϵ$) to that of $T_p(M)$ if we work in local coordinates and agree to identify the tangent spaces.