dg.differential-geometryriemannian-geometrysurfaces
Let $M$ be a 2-dimensional closed Riemmanian manifold diffeomorphic to $S^2$.
S.B.Myers says "the cut-locus of every point $x\in M$ is a finite tree."
- How the set of point can be a tree? What are the edges?
- Is $p$ an element of $\operatorname{cut-locus}(p)$?
I can not find any paper of Myers in this case.
Thanks!
Let me continue the comments above here so I can include a figure. Here are examples of the medial axis of two different convex polygons (from my own work):
Franz-Erich Wolter wrote his Ph.D. dissertation on "Cut loci in bordered and unbordered Riemannian manifolds" (Technische Universität Berlin, 1985). That might contain some useful information.
The question you pose is stated as an open question (in the 3-dimensional hyperbolic case) in the following paper:
Díaz, Raquel; Ushijima, Akira On the properness of some algebraic equations appearing in Fuchsian groups. Topology Proc. 33 (2009), 81–106.
Quoting from the review on mathscinet:
[the paper] takes its motivation from the fact that, apparently, the statement about the genericity of Dirichlet fundamental polyhedra is open for $\mathbb{H}^3$. (According to the authors, the paper of T. Jørgensen and A. Marden [in Holomorphic functions and moduli, Vol. II (Berkeley, CA, 1986), 69--85, Springer, New York, 1988] has a gap which the present authors have so far been unable to fix.)
Best Answer
I think Myers only considered analytic metrics, see his papers "Connections between differential geometry and topology I and II", Duke Math. J. 1 (1935), 376-391, and 2 (1936), 95-102.
For arbitrary metrics on $S^2$ the cut locus is indeed a tree. This can be deduced from e.g. in [Shiohama and Tanaka, Cut loci and distance spheres on Alexandrov surfaces] who work in the (more general) settings of Alexandrov spaces homeomorphic to surfaces and prove that the cut locus is a local tree. (This paper is available online, search by title). Specializing to the case when the surface is a Riemannian sphere we observe:
The complement to the cut locus is a 2-disk.
If the cut locus contains an embedded circle, then by Jordan curve theorem the circle separates $S^2$ into two disks, and by part 1 one of the disks must lie in the cut locus, so the cut locus cannot be local tree.
One should be careful with what is meant by a local tree. By a result of Gluck and Singer [Scattering of Geodesic Fields II, Annals of Mathematics, Second Series, Vol. 110, No. 2 (Sep., 1979), pp. 205-225] there is a positively curved Riemannian metric on $S^2$, in fact a convex surface of revolution, for which the cut locus cannot be triangulated, so it is definitely not a finite tree.
For recent works in this area see papers of Itoh, e.g. http://arxiv.org/pdf/1103.1758 and references therein.