# Retraction of a Riemannian manifold with boundary to its cut locus

This question is edited following the comment of Joseph. He pointed out that the main object of the first version of this question is the cut locus.

Recall that the cut locus of a set $S$ in a geodesic space $X$ is the closure of the set of all points $p \in X$ that have two or more distinct shortest paths in $X$ from $S$ to $p$. http://en.wikipedia.org/wiki/Cut_locus

A simple lemma shows that, for a disk $D^2$ with a Riemannian metric, the cut locus of $D^2$ with respect to its boundary is a tree. A picture of such tree can be found on page 542, figure 17 of the article of Thurston "Shapes of polyhedra". The tree is white. http://arxiv.org/PS_cache/math/pdf/9801/9801088v2.pdf For an ellipse on the 2-plane, the tree is the segment that joins its focal points.

More generically for a Riemannian manifold $M^n$ with boundary, the cut locus of $\partial M$ should be a deformation retract of $M$. (I guess it is a $CW$ complex of dimension less than $n$.) To prove this lemma, notice that $M^n\setminus\operatorname{cut-locus}(\partial M^n)$ is canonically foliated by geodesic segments that join $X$ with $\partial M$.

I wonder if this lemma has a name or maybe is contained in some textbook on Riemannian geometry?

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Dima, could you please comment why "it is a CW complex". – Petya Jul 14 '10 at 1:22
@Dimitri: Are you defining the cut locus of the boundary of the disk? – Joseph O'Rourke Jul 14 '10 at 1:36
@Dimitri: Pardon me for repeating the point, but I think the cut locus of an ellipse is a segment. It also goes under the name medial axis. See the figure at the Wikipedia entry: en.wikipedia.org/wiki/Medial_axis . – Joseph O'Rourke Jul 14 '10 at 10:13
If it really is a retract of a smooth compact manifold (equivalently, a retract of a finite CW complex), then that's almost as good as being finite CW. By the way, what's an example of such a space (compact ENR) not admitting a CW structure? – Tom Goodwillie Jul 14 '10 at 11:22
Beware that cut loci can be non-triangulable, even on strictly convex revolution surfaces, as has been shown by H. Gluck and D. Singer ams.org/journals/bull/1976-82-04/S0002-9904-1976-14125-0/… – BS. Jul 14 '10 at 12:56