4
$\begingroup$

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."

  1. How the set of point can be a tree? What are the edges?
  2. Is $p$ an element of $\operatorname{cut-locus}(p)$?

I can not find any paper of Myers in this case. Thanks!

$\endgroup$
  • $\begingroup$ Dear azita, I changed the tags and the formatting of your question, I hope you don't mind. Perhaps you could add a reference for the quote? Also, re 2: The answer is surely no, by definition of the cut-locus? $\endgroup$ – Mark Grant Dec 29 '13 at 8:59
  • $\begingroup$ Dear @Mark Grant, it is always a good idea to add a top level tag (these are tags with a two letter code, such as 'at.algebraic-topology') whenever possible. See the meta thread meta.mathoverflow.net/questions/1075/… for some more information. $\endgroup$ – Ricardo Andrade Dec 29 '13 at 18:55
  • $\begingroup$ Dear Rikardo .thank you. when I was studying paper about algebraic-topology meet this case. $\endgroup$ – azita lekpour Dec 30 '13 at 7:08
8
$\begingroup$

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:

  1. The complement to the cut locus is a 2-disk.

  2. 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.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

If you will excuse me substituting a polyhedron for the Riemannian manifold (imagine rounding the vertices), this figure shows how the cut locus (red) from source $x$ is a tree. (The green arcs are equidistant from $x$.)
     2x1x1Box
     (Figure from Discrete and Computational Geometry)

Concerning, "What are the edges?": They are geodesics, such that there are two distinct shortest paths from $x$ to every interior point of an edge of the cut locus. Points of the cut locus of degree $k$ have $k$ distinct shortest paths from $x$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ And see this earlier MO question for an image of a cut locus on a Zoll surface. $\endgroup$ – Joseph O'Rourke Dec 29 '13 at 13:31
  • $\begingroup$ Dear josef.I see your answer. this figure must be useful .i need thinking about it.thank you so much. $\endgroup$ – azita lekpour Dec 30 '13 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.