A lemma for proving Myers-Steenrod theorem

So here is what I am having problems with. The lemma is that if we suppose $(M,g)$ is a Riemannian manifold, and let $p\in M$. Then if $X,Y\in T_pM$ we have the following

$$\lim_{t\rightarrow 0+}\frac{d(\exp_p(ut),exp_p(vt))}{t|u-v|}=1$$

The hint I was originally given was that in exponential coordinates we had $g_{ij}=g_{\mathbb{R^n}ij}+O(|x|^2)$. So I applied the hint by assuming first that $t$ is sufficiently small so that there is a minimal geodesic $\exp(f_t)$ connecting $\exp(ut)$ and $\exp(vt)$ using convex neighborhoods, where $f_t$ is a path in $T_pM$. Then we notice that

$$\ell(\gamma_t)=\int\sqrt{g_ij(0)\dot{f_t}^i\dot{f_t}^j+O(|f_t|)^2\dot{f_t}^i\dot{f_t}^j}dt$$ So what I want to show is that $\frac{O(|f_t|^2)f_t^if_t^j}{\ell(f_t)^2}\rightarrow 0$ as $t\rightarrow 0$, which would be sufficient to establishing the above result, but I don't know how to bound this term. Any hints would be appreciated. Thanks.

Solution 1:

We shall use the following result: in the normal coordinates about $p\in M$, we have $$g_{ij}(x)=\delta_{ij}+O(\|x\|^2).$$

Let $B_{\epsilon}(p)$ be a normal coordinate ball about $p$, in which $\exp_p$ is the chart. Suppose $u,v$ are unit length vectors in $T_pM\cong\Bbb R^n$, we claim that

$$\lim_{t\to 0+}\frac{d(\exp_p(tu),\exp_p(tv))}{t}\to \|u-v\|.$$

Proof:

Let $\gamma:I_1\to M$ be the unit speed geodesic segment from $\exp_p(tu)$ to $\exp_p(tv)$, and $f:I_2\to T_pM$ be the straight segment from $tu$ to $tv$ parametrised in a way that $\exp_p\circ f$ is unit speed. Then we have \begin{align} d(\exp_p(tu),\exp_p(tv))&=L[\gamma]\\ L[\exp_p^{-1}\circ\gamma]&\ge L[f]=t\|u-v\|\\ L[\exp_p\circ f]&\ge L[\gamma]=d(\exp_p(tu),\exp_p(tv)) \tag{*}\end{align} Moreover, we set $t<\epsilon/3$, so that the whole $\gamma$ lies completely within the normal coordinate ball. In fact, $\gamma$ must lie completely within $\bar B_{3t}(p)$ which is easily confirmed by the triangle inequality for $d$.

We consider the difference between $L[\exp_p^{-1}\circ\gamma]$ and $L[\gamma]$. To simplify notations let $\eta:=\exp_p^{-1}\circ\gamma:I_1\to \Bbb R^n$. Then $$ L[\eta]=\int_{I_1}\|\eta'(s)\|\mathrm ds,\quad L[\gamma]=\int_{I_1}\|(\eta^i)'(s)\partial_i\|_g\mathrm ds=\int_{I_1}1\mathrm ds=|I_1|. $$ Note that $$\|(\eta^i)'(s)\partial_i\|_g=(\eta^i)'(s)(\eta^j)'(s)g_{ij}(\eta(s))=\|\eta'(s)\|^2+(\eta^i)'(s)(\eta^j)'(s)O(t^2)$$ and that $\eta$ can be written as $$\eta(s)=\exp_p^{-1}\circ\gamma(s)=\exp_p^{-1}\circ\exp((tu,s\exp_{tu}^{-1}(tv))),$$ (we can assume $tv$ always lies in at least one normal coordinate ball about $tu$ for each $t$ small enough, by the uniformly normal neighbourhood lemma.) where all the maps are smooth, thus $\|\eta'(s)\|$ can be uniformly bounded as $t$ varies. So we can find $M>0$ so that $$ \|\eta'(s)\|^2-Mt^2\le1=\|(\eta^i)'(s)\partial_i\|^2_g\le \|\eta'(s)\|^2+Mt^2.$$ Or $1-Mt^2\le \|\eta'(s)\|^2\le 1+Mt^2$. As such, $$L[\eta]=\int_{I_1}\|\eta'(s)\|\mathrm ds \begin{cases} \le\int_{I_1}(1+Mt^2)^{1/2}\mathrm ds\le\int_{I_1}(1+Mt^2)\mathrm ds=(1+Mt^2)L[\gamma]\\ \ge\int_{I_1}(1-Mt^2)^{1/2}\mathrm ds\ge\int_{I_1}(1-Mt^2)\mathrm ds=(1-Mt^2)L[\gamma] \end{cases}$$ Hence $L[\eta]=L[\gamma](1+O(t^2))$. Using similar arguments we know $L[f]=L[\exp_p\circ f](1+O(t^2))$. Therefore, using $(*)$ we have that \begin{align*} d(\exp_p(tu),\exp_p(tv)) &=L[\gamma]=L[\eta](1+O(t^2))\\ &\le L[f](1+O(t^2))=t\|u-v\|(1+O(t^2))\\ &\le L[\exp_p\circ f](1+O(t^2))\\ &\le L[\gamma](1+O(t^2))=d(\exp_p(tu),\exp_p(tv))(1+O(t^2)) \end{align*} Dividing them by $t$, we obtain $$\frac{d(\exp_p(tu),\exp_p(tv))}{t}\le \|u-v\|(1+O(t^2))\le \frac{d(\exp_p(tu),\exp_p(tv))}{t}\cdot(1+O(t^2)).$$ Hence by letting $t\to 0+$ we prove our claim.

(Myers-Steenrod theorem) Let $(M,g)$ be a Riemannian manifold, $F$ be a distance isometry on $M$. Then $F$ is a diffeomorphism.

Scheme of Proof:

Step one: prove a distance isometry $F$ preserves geodesics, so that for $p\in M,v\in T_pM$, $F\circ\exp_p(tv)$ traces out a local geodesic passing through $F(p)$ for $t$ small enough. You can prove this property by noting that for $p,q$ close enough, $\exists ! w$ such that $$d(F(p),F(q))=d(F(p),w)+d(w,F(q)),$$ and such a $w$ must lie on the geodesic segment connecting $F(p)$ and $F(q)$. Repeat this argument indefinitely, and use the density of dyadic numbers to conclude $F$ locally preserves geodesics.

Step two: Given any $p\in M$, by step one we can define a map $$D_p^F:T_pM\ni v\mapsto (t\mapsto F\circ\exp_p(tv))'_{t=0}\in T_{F(p)}M.$$ Use the previous result to prove that $D_p^F$ is an isometry. Then use this answer to prove $D_p^F$ is actually a linear isometry. Finally note that locally we have, for $q$ close enough to $p$, that $F(q)=\exp_{F(p)}\circ D_p^F\circ\exp_p^{-1}(q)$. Thus $F$ is a local diffeomorphism. But $F$ is bijective so must be a global diffeomorphism.