Suppose $(M,g)$ is a compact Riemannian manifold with smooth boundary. We call a point $p \in M$ regular if there exists a geodesic of finite arc length passing through $p$ with end points on $\partial M$ (By end points I mean two points on the geodesic which also lie on $\partial M$ and such that the geodesic hits the boundary non-tangentially at these points).

(a) Is it true that the set of points $p \in M$ which are not regular has measure zero?

If the answer to part (a) is affirmative I would also like to pose a second part to question as follows:

Given any regular point $p \in M$, we consider the set of all geodesics of the above type passing through $p$. We say $p$ is optimal if there is no conjugate points to $p$ on at least one of these geodesics.

(b) Is it true that the set of non-optimal points in $M$ have measure zero?

Thanks, Thanks,

  • 1
    $\begingroup$ I haven't completely worked it through, but I think I have a counterexample. Start with a $2$-sphere; remove 2 small circles - one at the equator, one at a pole. Attach two more spheres (with one circle removed each). Smooth this a little bit, so the new manifold is Riemannian. Remove a small circle from the third sphere. Then many of the points in the first sphere should not be regular; the geodesics joining points on the first sphere to the third sphere are heavily restricted in what they can do on the second sphere. $\endgroup$
    – user44191
    Feb 27, 2019 at 12:02
  • $\begingroup$ If I understood this correctly then this produces the counter example due to $M$ not being connected? (I actually implicitly was assuming that $M$ is connected although I forgot to state this clearly.) $\endgroup$
    – Ali
    Feb 27, 2019 at 12:19
  • $\begingroup$ This example is connected; it's the connected sum (en.wikipedia.org/wiki/Connected_sum) of three spheres (minus a small disk), and is homeomorphic to a sphere (minus a small disk). $\endgroup$
    – user44191
    Feb 27, 2019 at 12:25
  • $\begingroup$ Yes, I see the issue. Maybe if we additionally impose that $p$ is close to $\partial M$ (in terms of some uniform constant depending on $(M,g)$) then at least (a) should hold. $\endgroup$
    – Ali
    Feb 27, 2019 at 13:46
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    $\begingroup$ Would you mind posting the argument as an answer? I'm quite curious to see it. $\endgroup$
    – user44191
    Apr 16, 2019 at 2:21

1 Answer 1


Let us shoot a geodesic from a random point in a random direction. Note that with probability 0 it is one-sides finite + other-side infinite.

However, if you shoot a geodesic from a nonregular point, then with positive probability it is one-sides finite + other-side infinite. It implies that probability to choose a nonregular point has to be zero. In other words, we get "yes" for (a).

For (b), a sphere without a small disc seems to be a counterexample.


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