# trapped geodesics

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,

• 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. Feb 27, 2019 at 12:02
• 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.)
– Ali
Feb 27, 2019 at 12:19
• 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). Feb 27, 2019 at 12:25
• 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.
– Ali
Feb 27, 2019 at 13:46
• Would you mind posting the argument as an answer? I'm quite curious to see it. Apr 16, 2019 at 2:21

## 1 Answer

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.