Timeline for Is there a simpler proof of the key lemma in the paper by Hiroshi Iriyeh and Masataka Shibata on the 3D Mahler conjecture?
Current License: CC BY-SA 3.0
11 events
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| S Dec 23, 2017 at 19:20 | history | suggested | dvitek |
added convex-geometry tag
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| Dec 23, 2017 at 17:49 | review | Suggested edits | |||
| S Dec 23, 2017 at 19:20 | |||||
| Jun 12, 2017 at 18:15 | comment | added | Paata Ivanishvili | it might be of interest: in the first part of the proof of Theorem 1 keithmball.files.wordpress.com/2014/11/… Ball constructs a similar map to simplex as Makeev does but he uses Browers Fixed Point Theorem (BFPT) unlike Makeev. So it quite might be that one needs to apply BFPT in a proper way which I don't see yet how. | |
| Jun 12, 2017 at 16:19 | comment | added | Fedor Petrov | @fedja yes, the claim is indeed bit different, but the topological argument is still the same | |
| Jun 12, 2017 at 15:30 | comment | added | fedja | @FedorPetrov Except in our case the measures to partition depend on the planes. However the idea that if the solution is unique in some generic position, then there is an odd number of solutions in every generic position is amazing and certainly very promising. It should be rather standard, of course, but the good side of ignorance (mine) is the possibility to get surprised with the facts everybody else considers routine :-). | |
| Jun 12, 2017 at 14:14 | comment | added | Fedor Petrov | (continued) Makeev partitions two measures which are absolutely continuous w.r.t. Lebesgue measure, but actually this absolute continuity is not essential. | |
| Jun 12, 2017 at 13:42 | comment | added | Fedor Petrov | Roman Karasev (private communication) suggests that this essentially follows from the results of [V.V. Makeev. Equipartition of a continuous mass distribution. Journal of Mathematical Sciences, 140:4 (2007), 551--557, mathnet.ru/links/a1722f058ed63ed36cb13d9e67fd255f/znsl299.pdf, theorems 5 and 6] | |
| Jun 12, 2017 at 13:33 | history | edited | fedja | CC BY-SA 3.0 |
edited title
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| Jun 12, 2017 at 2:32 | comment | added | dhy | Now one can hope that geometric considerations will show that (in this "generic" situation) the intersection class in $H^6((\mathbb{RP}^2)^3,\mathbb{Z}/2\mathbb{Z})$ will be nontrivial, and that furthermore that there is no contribution from some degenerate case e.g. two planes coinciding. Then it probably still remains necessary to pass to some perturbation of $K$, and to run this argument there instead. So even if this sketch can be made to work, I doubt that a fully rigorous proof would fit in a half-page... | |
| Jun 12, 2017 at 2:28 | comment | added | dhy | Some naive thoughts: We can try and derive this from an intersection theory statement on $(\mathbb{RP}^2)^3$. Points correspond to triples of planes, and the condition that three planes split $K$ into $8$ parts of equal volume will give a codimension $3$ cycle for "generic" $K$. (We get codimension $3$ instead of codimension $7$ because diametrically opposite regions have the same area, by reflection symmetry of $K$). Similarly each of the conditions that two planes split a cross-section into four parts of equal area should lead to a codimension $1$ cycle. | |
| Jun 12, 2017 at 1:05 | history | asked | fedja | CC BY-SA 3.0 |