Timeline for Do plane projections determine a convex polytope?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 13, 2013 at 17:15 | answer | added | Anton Petrunin | timeline score: 13 | |
| Feb 12, 2013 at 11:44 | comment | added | Mark Meckes | @Igor: I haven't read it yet, but this preprint apparently gives a new short proof: arxiv.org/abs/1302.2354 | |
| Feb 21, 2010 at 2:49 | vote | accept | Igor Pak | ||
| Feb 19, 2010 at 14:34 | answer | added | Mark Meckes | timeline score: 22 | |
| Feb 19, 2010 at 2:55 | comment | added | Igor Pak | Thanks, David, I haven't seen this result. Will take a look. | |
| Feb 19, 2010 at 1:17 | comment | added | David E Speyer | There is a similar, but different, lemma in ams.org/mathscinet-getitem?mr=733052 : If S is a subset of R^n such that the projection of S to every R^{d+1} is a union of finitely many d-dimensional polyhedra, then S is a union of finitely many d-dimensional polyhedra. | |
| Feb 18, 2010 at 4:46 | comment | added | Steve Huntsman | I was originally thinking there might be some Bolzano-Weierstrassy thing that might reduce the problem, figured I'd comment since there's nothing else up here. The problem's considerably subtler than it appears at first. | |
| Feb 18, 2010 at 4:19 | comment | added | Igor Pak | Let me make an easy general comment: any proof must use substantially the fact that this is a $3$-dim problem (it obviously fails in $\Bbb R^2$). Also, use the fact that $P$ is bounded, since all projections of a circular cone are polyhedral cones. | |
| Feb 18, 2010 at 4:18 | comment | added | Igor Pak | I am not sure I follow. You assume that $P$ is not a polytope and you conclude with a property implying that $P$ is really not a polytope. Now what? | |
| Feb 18, 2010 at 0:08 | comment | added | Steve Huntsman | Let Q' be obtained by augmenting with supporting half-spaces such that the projections of Q and P coincide. By hypothesis, Q' still strictly contains P. By induction, we may therefore assume there is an infinite sequence of distinct points in Q that are not contained in P. Similarly, Q cannot be chosen so that all its orthogonal projections are the same as those of P. | |
| Feb 18, 2010 at 0:08 | comment | added | Steve Huntsman | Suppose not. By hypothesis, any finite intersection Q of supporting half-spaces of P strictly contains P. Because Q is a polytope, its orthogonal projections are polytopes. Let x be a point in Q\P. There is an orthogonal projection such that the image of x and that of P are disjoint. The corresponding polytope obtained by projecting Q strictly contains the projection of P. | |
| Feb 17, 2010 at 19:18 | history | asked | Igor Pak | CC BY-SA 2.5 |