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Let $M$ be a connected complete Riemannian manifold, $N$ a connected Riemannian manifold and $f:M \to N$ a differentiable mapping that is locally an isometry. Assume that any two points of $N$ can be joined by a unique geodesic (parameterized by arc-length of velocity $1$) of $N.$ Prove that $f$ is injective and surjective (and therefore, $f$ is a global isometry.)

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You need to suppose something extra on $M$... connectedness, for example (otherwise, let $M$ be two copies of $N$) –  Mariano Suárez-Alvarez Mar 4 '12 at 2:39
    
Yes, you're right! edited. –  Ehsan M. Kermani Mar 4 '12 at 2:43
    
It is easy to see that $f$ is surjective since $f(N)$ is closed and open on N. –  checkmath Mar 4 '12 at 2:51
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@Jack It is not true that there is a unique geodesic between any two points on the circle. –  yasmar Mar 4 '12 at 21:35
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If $f(x) = f(y)$ and $\gamma: I \to M$ is a geodesic connecting $x$ and $y$, then $f \circ \gamma$ is a geodesic loop on $N$ ... –  yasmar Mar 4 '12 at 21:46

2 Answers 2

up vote 3 down vote accepted

Thanks for all comments.

Injectivity: Let $p\neq q \in M,$ and let $\gamma:I \to M$ be a geodesic $\gamma(0)=p$ and $\gamma(1)=q.$ Cover $\gamma(I),$ by open sets $U_{\alpha}$ where $f|_{U_{\alpha}}:U_{\alpha} \to f(U_{\alpha})$ is an isometry, by compactness, there is a finite covering of $U_i$'s or equivalently, a partition of $I$ where $0=t_0<t_1<...<t_n=1.$ Thus, $\frac{D}{dt}(\frac{d(f \circ \gamma)}{dt}|_{[t_i,t_{i+1}]})=df (\frac{D}{dt}(\frac{d\gamma}{dt}|_{[t_i,t_{i+1}]}))=df(0)=0,$ therefore, $\frac{D}{dt}(\frac{d(f \circ \gamma)}{dt})$ on $I$ implying that $f \circ \gamma$ is a geodesic in $N$ joining $f(p)$ to $f(q).$ If $f(p)=f(q),$ then $f \circ \gamma$ would be a closed geodesic, contradicting the uniqueness assumption.

Surjectivity: Uniqueness of geodesics joining any two points of $N$ implies that $N$ is complete, then $exp_q:T_qN \to N$ is surjective for any $q \in N.$

Let $p \in M$ be fixed. There is an open neighborhood $U \in M$ containing $p$ s.t. $f|_U : U \to f(U)$ is an isometry. We have $f(exp_p(v))=exp_{f(p)}(df_p(v))$ for all $v \in T_pM.$ Let $q \in N$ be arbitrary. There is a $w \in T_{f(p)}N$ s.t. $exp_{f(p)}(w)=q.$ Since, $df_p$ is an isomorphism of vetor spaces, there is a $u \in T_pM$ s.t. $df_p(u)=w.$ Hence, $ f(exp_p(v))=exp_{f(p)}(df_p(v))=q$ and we’re done.

P.S. Sometimes triviality doesn't show itself, when exhaustion has captured our feelings:)

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Good proof! Sorry to nitpick, but just one thing: surjective exponential map does not imply completeness. Consider for instance the open disk $D$ of radius $1$ in $\mathbb R^2$. $D$ is incomplete because geodesics do not exist (in $D$) for all time, yet the exponential map based at any point in the disk is surjective onto the disk. (Your argument still works without completeness) –  treble Mar 5 '12 at 15:49
    
Dear treble, thanks. –  Ehsan M. Kermani Mar 6 '12 at 2:07

For injectivity, pick two distinct points $p,q \in M$ and consider the unique geodesic $\gamma$ from $F(p)$ to $F(q)$. If $p$ and $q$ are distinct, then $\gamma$ is not the constant map, so $F(p)$ and $F(q)$ are distinct. For surjectivity, for any $p \in N$, find some other point $q \in N$ so that $q$ is in the image of $F$ and then consider the unique geodesic $\gamma(t)$ from $q$ to $p$. Then show that the set $t$ for which $\gamma(t)$ is in the image of $F$ is both open and closed.

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Your answer is more like a comment. –  Ehsan M. Kermani Mar 5 '12 at 1:54
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Ehsanmo, I assumed this was a homework exercise, and therefore I didn't want to write down every last detail in my answer (I would feel funny doing someone's homework)- however I would be happy to put in more details if you can describe to me where you are getting stuck when you tried to write the details yourself. –  treble Mar 5 '12 at 2:23
    
"...I would feel funny doing someone's homework..." me too. –  Ehsan M. Kermani Mar 5 '12 at 8:56

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