nLab Myers-Steenrod theorem




In their 1939 paper, Myers and Steenrod proved two theorems on Riemannian manifolds.

The first theorem shows that distance-preserving maps between Riemannian manifolds are differentiable isometries.

In 1956 Palais simplified the proof and extended the result to show that a Riemannian manifold can be reconstructed from its metric space. That is to say, the functor that sends Riemannian manifolds and isometries to metric spaces and isometries is a fully faithful functor.

The second theorem proves that the group of isometries of a Riemannian manifold is a Lie group.


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

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

Theorem (Myers–Steenrod)

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

For notational simplicity, assume N=MN=M.

ϕ\phi is evidently a homeomorphism. Now pick pMp \in M and a neighborhood D r(p)D_r(p) such that for any qD r(p)q \in D_r(p), there is a unique geodesic in that neighborhood connecting p,qp,q. Call it γ\gamma. I claim that ϕγ\phi \circ \gamma is a geodesic in NN.

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

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


(1)d(ϕ(p),ϕ(q))=d(ϕ(γ(t)),ϕ(p))+d(ϕ(γ(t)),ϕ(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 MM, if q,rq,r lie in a small neighborhood of pp and

d(q,r)+d(r,p)=d(q,p), d(q,r) + d(r,p) = d(q,p) ,

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

So, returning to the discussion above, we see that by (1) ϕ\phi sends geodesics to geodesics.

Exponential coordinates

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

ϕ˜:T p(M)T q(M). \tilde{\phi}: T_p(M) \to T_q(M).

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

ϕ˜\tilde{\phi} is an isometry

It can be checked that

lim A,B0T p(M)d(exp p(A),exp p(B))|AB|=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 ϕ˜:T p(M)T q(M)\tilde{\phi}: T_p(M) \to T_q(M) is an isometry, where each tangent space is given the norm from the Riemannian metric. Pick A,BT p(M)A,B \in T_p(M) and consider the geodesics γ A,γ B\gamma_{A},\gamma_{B} at pp and γ A,γ B\gamma_{A'}, \gamma_{B'} where A=ϕ˜(A),B=ϕ˜(B)A'=\tilde{\phi}(A), B' = \tilde{\phi}(B). Then d(γ A(t),γ B(t))=d(γ A(t),γ B(t))d(\gamma_{A}(t), \gamma_{B}(t)) = d( \gamma_{A'}(t), \gamma_{B'}(t)). So

1=lim t0d(exp p(tA),exp p(tB))t|AB|=lim t0d(exp q(tA),exp q(tB))t|AB|=|AB||AB|. 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 ϕ˜\tilde{\phi} is linear, we appeal to a general fact:

Let X,YX,Y be normed linear spaces and T:XYT: X \to Y a map such that |TxTx|=|xx||Tx - Tx'| = |x-x'| for x,xXx,x' \in X. Then TT is linear if T(0)=0T(0) = 0.

Conclusion of the proof

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

A lemma

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

lim A,B0T p(M)d(exp p(A),exp p(B))|AB|=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. |AB||A-B| is the length of the linear path from AA to BB in T p(M)T_p(M), so it will be sufficient to show:


If c:(0,1)T p(M)c:(0,1) \to T_p(M) is a path in T p(M)T_p(M), then as c((0,1))0c((0,1)) \to 0 with the derivative |c||c'| staying bounded,

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

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


l(exp p(c))=|(exp p) *c(t)c(t)| 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 MM. The difference between these two is o(l(c))o(l(c)), because exp p\exp_p has derivative the identity at 0T p(M)0 \in T_p(M), and the metric on T exp p(c(t))(M)T_{\exp_p(c(t))}(M) is very close (say by a factor of 1±ϵ1 \pm \epsilon) to that of T p(M)T_p(M) if we work in local coordinates and agree to identify the tangent spaces.


A simplified proof that isometries are differentiable is given by

  • Richard S. Palais, On the differentiability of isometries, Proceedings of the American Mathematical Society 8:4 (1956), 805–807. doi.

Last revised on December 2, 2022 at 19:46:24. See the history of this page for a list of all contributions to it.