Timeline for two tetrahedra in $\mathbb R^4$
Current License: CC BY-SA 4.0
25 events
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| Jul 2, 2021 at 18:21 | history | edited | Michael Hardy | CC BY-SA 4.0 |
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| May 29, 2020 at 21:58 | vote | accept | filipm | ||
| Feb 20, 2014 at 11:58 | answer | added | polyanom | timeline score: 13 | |
| Sep 5, 2013 at 21:48 | review | Suggested edits | |||
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| Aug 7, 2013 at 6:05 | history | edited | Ricardo Andrade |
removed deprecated tag 'geometry' (since this question was bumped to the front page)
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| Aug 7, 2013 at 1:18 | history | rollback | Benjamin Steinberg |
Rollback to Revision 5
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| S Aug 6, 2013 at 19:57 | history | suggested | Mahdi Khosravi | CC BY-SA 3.0 |
title changed
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| Aug 6, 2013 at 19:44 | review | Suggested edits | |||
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| Feb 16, 2011 at 10:45 | history | edited | filipm | CC BY-SA 2.5 |
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| Feb 16, 2011 at 2:00 | comment | added | Tracy Hall | I did check your nice explicit example, and it works, so I was wrong: you have given an example with strictly fewer than $m+n-d$ vertices in common when $m=3$, $n=2$, and $d=4$. I wonder what the correct bound is. | |
| Feb 15, 2011 at 16:42 | comment | added | filipm | Please check the edited version of the question, I tried to add more details. I'm not 100% sure about the construction, so you can check computations if you feel like that. | |
| Feb 15, 2011 at 16:36 | history | edited | filipm | CC BY-SA 2.5 |
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| Feb 12, 2011 at 18:41 | comment | added | Gil Kalai | "It is relatively easy to show..." Filipm, can you give the argument? also can you explain your example of tetrahedron and triangle in R^4 responding to Tracy Hall's question? | |
| Jan 31, 2011 at 21:13 | comment | added | Tracy Hall | Suppose we scale things to edge length $\sqrt{2}$, and place a tetrahedron with vertices at the four standard basis vectors of $\mathbb R^4$. A point whose four components sort to $a \le b \le c \le d$ lies within the allowed region if and only if $(a-1)^2 + b^2 + c^2 + d^2 \le 2$. Where are you proposing to put three such points, distinct from the basis vectors and at pairwise distance $\sqrt2$ from each other? | |
| Jan 24, 2011 at 19:34 | comment | added | filipm | @Tracy Hall: I indeed believe a stronger statement (than the one I proposed above) is true, namely that two $(d-1)$-simplices in $R^d$ must share at least $d-2$ vertices. On the other hand, I don't think your general statement is correct, I think we can embed a triangle and a tetrahedron in $R^4$ without sharing a vertex. Although, some similar generalization is certainly plausible. | |
| Jan 23, 2011 at 22:06 | comment | added | Tracy Hall | The correct generalization appears to be the following: a unit regular $m$-simplex and a unit regular $n$-simplex embedded into $\mathbb R^d$ with union diameter 1 must have at least $m+n-d$ vertices in common. | |
| Dec 16, 2010 at 6:13 | comment | added | Elizabeth S. Q. Goodman | I'm not sure why this should generalize to higher dimensions, since it is not true for line segments in $\mathbb R^2$. In fact, using the bases of two bisecting line segments is the nicest way I can see to fit two equilateral triangles in $\mathbb R^3$ that share only one vertex. (E.g. five vertices $(\pm 1/2, \pm 1/2, 0)$ and $(0,0,\sqrt 3/2)$.) A similar construction for tetrahedra in $\mathbb R^4$ would make them share an edge. A possible attack: consider just the vertices of the two simplices, and study their convex hull. In my construction it's an irregular pryamid (half-octahedron). | |
| Nov 30, 2010 at 19:00 | history | edited | Romeo |
edited tags
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| Nov 30, 2010 at 2:42 | comment | added | Dr Shello | Yikes; what on earth is wrong with the phrase "prove or disprove"? | |
| Nov 29, 2010 at 15:01 | history | edited | filipm | CC BY-SA 2.5 |
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| Nov 29, 2010 at 14:26 | comment | added | sleepless in beantown | @filipm, what work have you done on the problem thus far? Have you played around with it restricted to $\mathbb{R}^3$ and seen what you might find? What kind of envelope would be required such that the restriction of a maximum diameter of $1$ exists? Think about that question in $\mathbb{R}^3$ and $\mathbb{R}^4$ and where that leads you. What have you gotten so far in your thoughts and efforts in this problem? The phrase "To the best of my knowledge it is not known" also encompasses the possibility that you have not thought about it or worked on it. Have you? | |
| Nov 29, 2010 at 14:23 | comment | added | S. Carnahan♦ | filipm: You have the option of changing the question so that it sounds less like homework. Just click the "edit" link below the question, and make the appropriate changes. | |
| Nov 29, 2010 at 14:17 | comment | added | filipm | To the best of my knowledge it is not known whether the statement is true or false. My guess is that it's true. I should have put maybe: "you're kindly asked to prove or disprove the following statement", so that it doesn't sound as a command. I read the faqs and it seems to me that the problem I asked fits well. | |
| Nov 29, 2010 at 12:59 | comment | added | sleepless in beantown | @filipm, Have you done some work on this problem? Do you know whether the statement is provably true or provably false? The command tone in your problem statement "Prove or disprove" makes this look likes it's a homework assignment or problem-set with the "command" being given to the student to solve the problem. If so, then this is not the correct site for homework or unmotivated problems: see FAQs. If you could show what work you done, or some motivation behind this to show that it's a research problem & not homework you're trying to skip doing yourself, that would be better. | |
| Nov 29, 2010 at 12:44 | history | asked | filipm | CC BY-SA 2.5 |