Timeline for When is the cut locus a finite tree?
Current License: CC BY-SA 4.0
16 events
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| Jul 15 at 17:31 | comment | added | Joseph O'Rourke | @MikhailKatz: See the three references in the Wikipedia article on Blum. | |
| Nov 18, 2023 at 18:45 | comment | added | Mikhail Katz | Do you have some references for this work by Blum? | |
| Jan 31, 2022 at 15:31 | history | edited | JHM | CC BY-SA 4.0 |
deleted 3446 characters in body
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| Jan 31, 2022 at 15:05 | history | edited | JHM | CC BY-SA 4.0 |
added 733 characters in body
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| Feb 2, 2021 at 1:07 | comment | added | JHM | By the phrase "the infinite branches do not have a common intersection" i mean the "graph-edges" containing the corners do not share a common "vertex". | |
| Feb 2, 2021 at 0:39 | comment | added | JHM | @GabeK You're correct that any infinite-sided polygon inscribed in unit disk has cut loci and medial axes with infinitely many branches. Indeed $M(A)$ intersects the boundary at points with divergent curvature (the corners). I'm claiming that the infinite branches do not have a common intersection except when they are located on a spherical arc and focus to a common point. You're correct this deserves to be clarified. | |
| Feb 1, 2021 at 17:17 | comment | added | Gabe K | But compactness won't rule out something like a curve which is piece-wise linear and touches the unit disk at angles $\theta/ n$. Now that might not be enough on it's own to make the cut locus at the origin have infinite degree, but it seems like you could modify the construction to make that happen. | |
| Feb 1, 2021 at 16:40 | comment | added | JHM | @Gabe K. Maybe the correct argument is simply this: $C$ is compact when $A$ is open and bounded. QED. There are no accumulation points. | |
| Feb 1, 2021 at 14:29 | comment | added | Gabe K | I still don't understand why the degree of the vertices needs to be finite. One could imagine a region which is tangent to a disk at infinitely many points, but which has no circular arcs. The intersections will have an accumulation point, but it seems that the case of regular polygons is very special. | |
| Feb 1, 2021 at 13:05 | history | edited | JHM | CC BY-SA 4.0 |
Added defn of contravariant functor
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| Feb 1, 2021 at 0:48 | history | edited | JHM | CC BY-SA 4.0 |
Elaborated. Tried to answer the OPs question more directly.
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| Jan 31, 2021 at 16:22 | history | edited | JHM | CC BY-SA 4.0 |
Added comment on upper semicontinuity of $C$, $M(A)$.
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| Jan 31, 2021 at 16:05 | history | edited | JHM | CC BY-SA 4.0 |
Addressed the OP's question more directly.
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| Jan 31, 2021 at 15:18 | comment | added | JHM | @Leo Moos. Yes i will elaborate my above answer. But briefly, if the cut locus had a point ("vertex") which had infinite "degree", then the maximal disk centred at that point would intersect the boundary at infinitely many points, and then we'd conclude the boundary would consist of circular arcs (quarter circles, etc.). But this leads to contradiction because the cut locus of circular arcs are points, not edges. Maybe thats still not clear.... | |
| Jan 30, 2021 at 14:53 | comment | added | Leo Moos | Would you mind adding some information? I am still unsure about some details. Why is the cut locus locally a finite graph? What sort of deformations will in- or decrease the number of edges? What do you mean by 'branches' of $C$? I am confused about your last paragraph. When you say that the 'intersection with a disc is finite', presumably you mean that it has finitely many connected components; wouldn't that always be the case for a regular curve? In short, I still don't see what deformations will guarantee that the cut locus is a finite graph, and for which 'pathological' curves this fails. | |
| Jan 30, 2021 at 12:06 | history | answered | JHM | CC BY-SA 4.0 |