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Let $M$ be a complete simply-connected Riemannian manifold with non-positive sectional curvature. Let $\gamma$ be a geodesic in $M$ and let $p$ be a point which does not lie on $\gamma$. Prove that the shortest distance between $\gamma$ and $p$ is realized by a unique geodesic which is perpendicular to $\gamma$

a solution is provided here in problem 3

However, I do not think this solution is right. Because it claims at first that we can always achieve a minimum by quoting the result from problem 2, while problem 2 only applies to closed submanifolds. How do we actually show that there is always a critical point?

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  • $\begingroup$ We can always attain a minimum, that can be proved quite easily and has not much to do with RG. $\endgroup$ – user99914 Mar 15 '17 at 7:13
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    $\begingroup$ Why can you always achieve a minimum? The geodesic is not assumed to be a closed set. $\endgroup$ – koch Mar 15 '17 at 19:10
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The solution is not perfectly worded; though the ingredients needed are all there already:

Let $\gamma$ be written as $\gamma(t) = \exp_x (tv)$ for some $x\in M, v\in T_xM$. Since $M$ is complete simply connected with non-positive sectional curvatures, Hadamard's theorem imply that $\exp_x : T_xM \to M$ is a global diffeomorphism. Since $\{ tv : t\in \mathbb R\}$ is a closed set in $T_xM$, $\gamma$ is a closed set in $M$.

Consider $q\in M $ not in $\gamma$. Then the previous question gives you $q'\in \gamma$ so that the distance $\{ d(q, y) : y\in \gamma\}$ is minimized at $q'$. Then the minimal geodesic joining $p, q'$ will be perpendicular to $\gamma$.

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  • $\begingroup$ That is very helpful, thank you very much! $\endgroup$ – koch Mar 16 '17 at 13:31

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