When is the cut locus a finite tree?

Question

Let $\Omega \subset \mathbf{R}^2$ be a bounded, simply connected domain, with a regular boundary, say of class $C^2$ at least. Let the cut locus $C$ of $\Omega$ be the set of points $x \in \Omega$ for which there exist two or more points $z_1,z_2 \in \partial \Omega$ for which \begin{equation} \lvert x - z_i \rvert = \operatorname{dist}(x,\partial \Omega). \end{equation}

It is claimed in a paper of Panov and Petrunin that an arbitrarily small perturbation of the boundary guarantees that the cut locus $C$ is a finite graph, embedded inside $\Omega$. (In fact, once the graph structure of $C$ is established one can show that $C$ is a tree.) They attribute this fact to Ionin and Pestov, but unfortunately this is available only in Russian.

Question. How does this perturbation argument go? (And for which 'pathological' domains is it necessary in the first place?)

Answer 1

Edited Jan31, 2022: the cut locus of the curve $\gamma$ is a finite tree if the boundary is smooth and if the curve bounds a contractible domain, i.e. $\Omega \approx \{pt\}$. This follows from Blum's MAT and the basic ideas of the theory. The observation about $C$ being a locally finite tree requires the boundary be smooth or $C^1$. In this case the ``branches" of the tree cannot accumulate, and the principle curvatures of the boundary hypersurface are bounded. Otherwise if the boundary is only $C^0$, then the cut locus $C$ might be a tree with infinite degree vertices.

N.B. It does not really make sense to speak of trees in this category, so terms like "branches" and "tree" are equally informal.

Let $A\subset \mathbb{R}^n$ be a bounded open subset. The cut locus $C$ of $A$ is defined as the domain of nondifferentiability of $dist_{\partial A}: A\to \mathbb{R}$.

Blum's Medial Axis Transform (MAT) is the set $$M(A):=\{x\in A ~|~ card|argmin_{y\in \partial A} dist(x,y)|\geq 2\}.$$

We find $M(A) \subset C$, although $M(A)$ is not always closed, with $\overline{M(A)}=C$.

(1) To answer the question posed in the OP's title: Both the cut locus $C$ and $M(A)$ have the homotopy type of the domain $A$, so if $A$ (or $\Omega$ in the notation of the OP) is contractible, then $C$ and $M(A)$ is contractible.

(2) To answer the OPs specific question about pathologies in $C$ and $M(A)$: the cut locus $C$ branches wherever the principal curvatures of the boundary $\partial A$ are positive (convex) with respect to the interior $A$. Saul Rodrigues Martin gives interesting example in Is the max-centre map continuous for open bounded domains?

My original answer to the OP was not satisfactory. But essentially I would argue that the cut locus, if interpreted as a type of tree graph, has finitely many graph-edges if the boundary $\gamma$ is basically $C^1$. But once you have infinitely many edges, it's possible to get those edges to accumulate or to share a common intersection, i.e. a vertex of infinite degree.