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.)
How to go from local to global isometry
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2You need to suppose something extra on $M$... connectedness, for example (otherwise, let $M$ be two copies of $N$) – 2012-03-04
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0Yes, you're right! edited. – 2012-03-04
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0It is easy to see that $f$ is surjective since $f(N)$ is closed and open on N. – 2012-03-04
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2@Jack It is not true that there is a unique geodesic between any two points on the circle. – 2012-03-04
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1If $f(x) = f(y)$ and $\gamma: I \to M$ is a geodesic connecting $x$ and $y$, then $f \circ \gamma$ is a geodesic loop on $N$ ... – 2012-03-04
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0@Jack Lee: You example does work here. Look at the assumptions again. Take north and south pole of $S^1,$ there are two different geodesics joining them. – 2012-03-04
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0there are two distinct geodesics joining *any* two points on the circle, or a sphere for that matter. Antipodal points are only needed if you want an example of two minimizing geodesics joining a pair of points. – 2012-03-05
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0Of course, by north and south pole, I meant two antipodal points. – 2012-03-05
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0@yasmar: You're right, I didn't look closely enough at the hypothesis. Sorry, I take it back. – 2012-03-05
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0Can $M,N$ have some boundary components? – 2015-12-14
2 Answers
Thanks for all the 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
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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1Good 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) – 2012-03-05
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0Dear treble, thanks. – 2012-03-06
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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0Your answer is more like a comment. – 2012-03-05
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1Ehsanmo, 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. – 2012-03-05
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0"...I would feel funny doing someone's homework..." me too. – 2012-03-05