# Conjugate points on cut locus

Let $M$ be a Riemannian with nonempty boundary $\partial M$. Define multiplicity of $x\in M$ as the number of minimizing geodesics from $x$ to $\partial M$.

The following fact seems to be standard:

The set of points with multiplicity $\ge 2$ is dense in the cut locus of $M$ with respect to $\partial M$.

Is it proved somewhere?

• I know how to prove it, but if you have a one-line proof I am very interested.

• Any point of multiplicity 1 in the cut locus is conjugate, but not the other way around. Say for a hemisphere, the cut locus contains a single point — it is a conjugate point of multiplicity $\infty$.

• Are cut locus points of multiplicity $1$ identical to conjugate points? (Trying to connect the title to the body of the question.) – Joseph O'Rourke Jan 19 '17 at 19:51
• Let $T_1 \subset T$ the set of points of multiplicity 1. In particular the closure $\bar{T_1}$ is all made of conjugate points. By Sard, $\bar{T_1}$ has zero measure in $M$, hence $\bar{T_1}$ has empty interior. – Raziel Jan 20 '17 at 7:40
• @Raziel it does not work --- the cut locus may have codimension >1. – Anton Petrunin Jan 20 '17 at 17:25
• Anton, did you check: Stephanie B. Alexander, I. David Berg, and Richard L. Bishop, Cut loci, minimizers, and wavefronts in Riemannian manifolds with boundary, Michigan Math Journal, Volume 40, Issue 2 (1993), 229-237. – Misha Jan 31 '17 at 1:20
• @Misha Thank you. Indeed, at the very end, they say "the following proposition is easily verivied" and formulate the needed statement :) – Anton Petrunin Feb 9 '17 at 20:59

Intuitively, this seems plausible if $M$ is complete. Let $d:\partial M \to [0,\infty]$ be the distance function to the cut locus. Then $d$ is continuous (by completeness), and $graph(d) \subset \partial M\times [0,\infty]$ is a copy of $\partial M$. There is an open subset $V\subset \partial M$ on which $d$ is finite.

Then $M$ is obtained by taking $U=\{ (x,t)\subset \partial M\times [0,\infty) | t\leq d(x)\}$ and taking the quotient of $graph(d_{|V})$ by the identification map to the cut locus.

If the multiplicity $\geq 2$ points are not dense in the cut locus, then there is an open subset in the cut locus of multiplicity one. This should pull back to an open subset of $graph(d_{|V})$, which maps homeomorophically under the identification map. But an open ball subset of $graph(d_{|V})$ mapping to the cut locus with multiplicity one will have a collar neighborhood homeomorphic to a half-space, which maps homeomorphically to $M$. However, the boundary of the half-space maps to the cut locus, not to $\partial M$, which is a contradiction to the fact that $M$ is a manifold near the cut locus.

• Thank you very much. [You will not believe me, but we (Yuri Burago and me) found exactly the same proof.] The question was asked by a colleague --- strange we could not find a reference. – Anton Petrunin Jan 30 '17 at 20:37
• Okay, I believe you found this proof, but presumably this wasn't the original one you hinted at in the question? – Ian Agol Jan 31 '17 at 2:59
• Well, I wanted a one-line proof --- now I think it is unlikely one can do a better than this one. – Anton Petrunin Jan 31 '17 at 4:21

I've got a letter from Stephanie Alexander with a complete answer. Let me summarize it here.

The first proof is given in 4.8 of "Schnittort und konvexe Mengen..." by Hermann Karcher (1968). (The formulation is slightly weaker, but from the proof proves our statement follows; the idea is the same as in the answer of Ian Agol.)

An other proof was given in "Decomposition of cut loci" by Richard Bishop (1977).

Latter it appears as Lemma 2 in "Distance function and cut loci..." by Franz-Erich Wolter (1979). The proof is basically the same as Karcher's.

Let me sketch another proof. It use analysis instead of topology. Namely we use existence of flow for continuous vector fields (uniqueness requires more, but we will not use it).

Denote by $$K$$ the subset in the interior of $$M$$ with multiplicity 1 and let $$U\subset K$$ be its interior. We need to show that $$U$$ does not intersect the cut locus.

Consider the function $$f= \mathrm{dist}_{∂M}$$. Note that $$K$$ is the set of points where $$f$$ is differentiable, $$|\nabla_x\,f|=1$$ for any $$x\in K$$ and $$\nabla f$$ is a continuous vector field in $$U$$. In particular $$f$$-gradient flow exist in $$U$$ (in fact, it is unique, but we do not need it).

Fix $$x \in U$$. Moving $$x$$ along the $$f$$-gradient flow, we get a curve $$\gamma$$ such that $$(f\circ\gamma)'\equiv 1$$. In other words, $$\gamma$$ is an extension of the minimizing geodesic from $$∂M$$ to $$x$$. Hence $$x$$ does not lie on the cut locus.

Since $$x\in U$$ is arbitrary, the statement follows.