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I have a general question in Riemannian geometry: Let M be a compact manifold and $\partial M \neq \emptyset$. Then shoot a geodesic from any boundary point perpendicularly into the interior of M. How can one prove it will end at boundary? If so, it induces a transformation of $\partial M$, does anyone know any result about this transformation? For example, one can ask rigidity property, i.e. if the transformation is an isometry, can one say anything about M?

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Will it end at the boundary? – Mariano Suárez-Alvarez Nov 19 '09 at 20:44
up vote 8 down vote accepted

It doesn't seem true to me that the geodesic will always return to the boundary. It may end up accumulating around a closed geodesic in the interior.

For example, consider the hyperboloid of one sheet $$x^2+y^2-z^2=1$$ in $\mathbb{R}^3$. This is a surface of revolution and so one can talk of the angular momentum of a path with respect to the axis. For a geodesic this must be constant (Clairault's theorem). If a geodesic has momentum exactly one (in some well-chosen units!) it will accumulate on the "waist" of the surface, where $z=0$.

Of course this is a non-compact example, but if a geodesic accumulates on the waist, it stays inside some compact region. So it is easy to imagine a compact surface with boundary which contains the relevant compact region and for which this accumulating geodesic meets the boundary at a right angle. This gives an example where the geodesic doesn't return to the boundary. (I think it's even possible to find an example of a metric on the disc of this sort.)

(In case you're curious, a geodesic on the hyperboloid starting with $z>0$, pointing "down" and with momentum less than 1 will pass through the waist and end up asymptotic to a meridian at z = -infinity. If the momentum is bigger than 1 it will not reach the waist and swing back upwards, ending up asymptotic to a meridian at z = +infinity.)

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In fact it is not true that the geodesic will shoot again the boundary. You can already construct such a contexample on a disc. First thing that you should do is to understand that if we have a hyperbolic annulus i.e. H^2/Z it has one closed geodesics and there are other (inifinite) geodesics that spriral and finally tend to the closed one. Such a geodesic is contained in a half of the cylinder. Now you can make a cut on the cylinder that will intersect the geondesic by the angle pi/ end through the rest. Finally replace the other infinite half cylinder by a disk.

There are some cases when this map is a piescewise isomtetry -- there is a whole science about this topic, called interval exchange transformations. This happen for surfaces of any genus >1.

It is not clear how much you can say about M if the map is well defined. For example you can take MxI -- a direct product. But in the case when the boundary of M is connetced I don't see for the moment any other example than just a standard disk D^n (say with metric of constant curvature)

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Our examples look almost identitcal. I guess we were even writing them down at the same time! – Joel Fine Nov 19 '09 at 22:17

You can also see this indirectly in some examples:

Take a hyperbolic surface $S$ with a single geodesic boundary component $\gamma$. Choosing inward pointing normals to $\gamma$, you can lift $\gamma$ to the unit tangent bundle. If all paths perpendicular to $\partial S$ ended in $\partial S$, the geodesic flow (suitably interpreted) would produce a nontrivial homotopy from $\partial S$ into $\partial S$. This would produce a nontrivial conjugacy $t\gamma t^{-1} = \gamma^k$ in the fundamental group, which can't happen.

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