I want to show that isometries map geodesics to geodesic using that an isometry of two riemannian manifolds also implies an isometry of metric spaces.
I have found proofs using the length minimizing properties of geodesics but I don't really see how this works yet:
Let $(M,g), (N,h)$ be Riemannian manifolds and $F: M \rightarrow N$ an isometry.
The length minimizing property is local, correct? So for some point I have a normal neighborhood $U$ for some point $p \in M$, then for each point $q\in U$ there exists a unique geodesic $\gamma$ connecting $p$ and $q$. I also know that than it holds
$d(p,q)=L(\gamma)=\int\limits_a^b \sqrt{g(\gamma'(s),\gamma'(s))} \ ds$
Now if I have a geodesic from $p$ to $q$, then
$L(F(\gamma))=L(\gamma)=d(p,q)=d(F(p),F(q))$
can I then conclude, that $F(\gamma)$ is a geodesic, correct?
How can I deduce the statement for general geodesics (not only those in a normal neighborhood)?
Thanks in advance for any help!
Best Answer
For an intervall $I$ a smooth curve $\gamma: I\to M$ is a geodesic iff the following two conditions hold:
Now if $\gamma: I\to M$ is a geodesic and $F:M\to N$ an isometry then $$|(F\circ\gamma)'|=|dF(\gamma')|=|\gamma'|$$ so $|(F\circ\gamma)'|$ is constant. Also for $t\in I$ and $J,r,s$ as in the definition above $$d(F(\gamma(r)),F(\gamma(s)))=d(\gamma(r),\gamma(s))=L(\gamma_{|[r,s]})=L((F\circ\gamma)_{|[r,s]}) $$ so $F\circ \gamma$ is locally shortest. Hence $F\circ\gamma$ is a geodesic.