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Let $p\in M$ be a point in a closed riemannian manifold $M$. Recall that the cut locus of $p$ is the subset of $M$ consisting of all points that are connected to $p$ by at least 2 distance-minimizing geodesics.

I will start with a general question:

Is it true that for generic metrics on $M$ and generic points $p$, the cut locus of $p$ is a $(n-1)$-dimensional polyhedron with generic singularities?

By "generic singularities" I mean that $M$ is a simple polyhedron. See for instance this paper of Alexander and Bishop.

This property is certainly not satisfied for some important specific metrics: for instance, if $M$ is a round sphere the cut locus is a point, no matter where $p$ is. If $M$ is a flat torus, we get a generic polyhedron for generic flat metrics. What about hyperbolic manifolds? So, this is my question:

Let $M^n$ be a hyperbolic $n$-manifold. Is the cut locus of a generic point a $(n-1)$-polyhedron with generic singularities?

Of course I am mostly interested in the case $n=3$. In dimension $n=2$ one may also pick a generic hyperbolic metric.

Edit: In dimension 1, a simple polyhedron is a graph with vertices of valence 2 or 3. In dimension 2, it is a polyhedron such that the link of a point is either a circle, a circle with a diameter, or a circle with three radii.

In general, a $n$-dimensional compact polyhedron is simple if every point has a neighborhood which is the cone over the $(k-1)$-skeleton of the $(k+1)$-simplex, times a $(n-k)$-disc.

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@Bruno: IF $C(p)$ is the cut locus from $p$, $M \setminus C(p)$ is homeomorphic to an open ball. So on a 2-manifold, $C(p)$ is a tree. Would you call a tree a 1-dimensional polyhedron? Just trying to undertand your terminology... –  Joseph O'Rourke Oct 26 '10 at 18:20
    
A 1-dimensional polyhedron is simple if it is a 3-valent graph (or a circle). Vertices have valence 2 or 3. I was trying unsuccessfully to embed a picture in the text... I will try again later (probably tomorrow morning) –  Bruno Martelli Oct 26 '10 at 19:12
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Bruno:Have you looked at the papers of Michael Buchner.He showed the cut locus of any compact real analytic manifold with real analytic metric is subanalytic. He also classified the cut loci in the generic situation upto dimension six. –  Mohan Ramachandran Oct 26 '10 at 19:18
    
Following Mohan's lead, the paper by Buchner entitled "Simplicial structure of the real analytic cut locus" [Proc. Amer. Math. Soc. 64 (1977), no. 1, 118–121.] "establishes that the cut locus of a compact real analytic Riemannian manifold of dimension $n$ is homeomorphic to a finite $(n-1)$-dimensional simplicial complex." –  Joseph O'Rourke Oct 26 '10 at 20:21
    
@Joseph: This doesn't address Bruno's main question, since nothing says the simplicial complex must be simple. –  Dylan Thurston Oct 26 '10 at 22:45
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3 Answers 3

up vote 6 down vote accepted

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.)

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Diaz and Ushijima are right: Proofs in the paper of Jorgensen and Marden are hopelessly wrong (they confused algebraic and semialgebraic sets). Counterexamples to one of their lemmas are constructed in front.math.ucdavis.edu/1201.3129 where a weaker genericity result is proven in dimension 3. –  Misha Jan 14 '13 at 7:38
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As indicated by Mohan, Buchner has written some papers on the subject in the 70s, for instance here. In his study, he fixes the manifold $M$ and the point $p$, and he varies the metrics $g$. He defines a notion of cut-stable metric. Cut-stable metrics form a dense open subset of the set of all riemannian metrics.

The cut locus $C(g)$ determined by a cut-stable metric $g$ is locally stable (it is the same PL object as one varies $g$ locally). Moreover, the local structure of $C(g)$ is indeed generic when $M$ has low dimensions: however the "generic polyhedra" one can obtain are a strictly larger class than the "simple polyhedra" I defined above.

For instance, if $M^2$ is a surface the cut-locus $C(g)$ of a cut-stable metric $g$ is a graph with vertices having valence 1, 2, or 3. Vertices with valence 1 are inavoidable for instance on a 2-sphere, for obvious reasons (the cut locus is a tree). However a simple polyhedron contains only vertices of valence 2 and 3.

Analogously, if $M^3$ is a 3-manifold the cut-locus $C(g)$ of a cut-stable metric $g$ is a 2-dimensional polyhedron whose local structure belongs to finitely many types. There are five links one can get: the three arising in the definition of a simple polyhedron (circle, circle with diameter, circle with three radii), plus two more (segment and a circle with a radius). Therefore the answer to my first question in low dimensions is:

Let $M^n$ has dimension 2 or 3. For generic metrics on $M$ and generic points $p$, the cut locus of $p$ is a (n−1)-dimensional polyhedron with generic singularities. However, the polyhedron may not be simple.

On the other hand, if the cut-stable metric $g$ has non-positive curvature, there are no conjugate points and the non-simple singularities cannot arise: thus we really get a simple polyhedron in this case (at least in dimension $n=2$ and $n=3$).

In all this discussion the metric $g$ is generic, so it gives no information for hyperbolic 3-manifolds (i.e. the second question).

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If curvature $\le -1$ then cut locus is is glued from the boundary of fundamental domain of $\pi_1$-action on the universal cover. If curvature $=-1$ then the fundamental domain is a convex polyhedral. The gluing maps are piecewise isometries, so the result is $(n-1)$-polyhedron, But one construct an action which gives a complicated link. Take $\mathbb Z^2$ action on hyperbolic $3$-space such that it has one fixed point on the absolute and action on its horosphere is standard action of $\mathbb Z^2$ on the Euclidean plane. (I do not see how to make a compact example.)

For the first question, in some sense the answer is "YES" if $\dim \ge 3$. I.e. there is a $G_\delta$-set in $C^\infty$-topology which satisfies your condition and dense in Gromov--Hausdorff metric.

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I don't understand the answer to the first part; can you explain it some more? How do you know that you won't have simple domains for a good choice of $p$? –  Dylan Thurston Oct 27 '10 at 3:29
    
I made a correction. –  Anton Petrunin Oct 27 '10 at 14:51
    
In the second question, the manifold $M^n$ is closed (while your counterexample is homemorphic to $S^1\times S^1\times \mathbb R$, if I have understood). Concerning your answer to the first question, could you say something more or give some reference? Thank you, –  Bruno Martelli Oct 27 '10 at 21:37
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