Timeline for answer to A curious relation between angles and lengths of edges of a tetrahedron by Greg Egan
Current License: CC BY-SA 4.0
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36 events
| when toggle format | what | by | license | comment | |
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| Jul 27, 2019 at 2:07 | comment | added | Ian Agol | Cool (re sphere slope). Considering the dual equation, this gives another relation between the $l_{ij}$ and $\alpha_{ij}$. | |
| Jul 25, 2019 at 12:30 | comment | added | Arseniy Akopyan | Thanks. According to the presentation above we can not catch spherical volume from a Gram matrix. and therefore from $\alpha$ and $\beta$. Miracle did not happen :-( | |
| Jul 25, 2019 at 12:13 | comment | added | Greg Egan | @ArseniyAkopyan In the Euclidean case, it was just a direct computation based on the formula for the tan of half the solid angle that Robin Houston quoted. I turned the dot products between edge vectors in that formula into expressions depending only on the edge lengths, using the cosine law. In the spherical case, I constructed the vertices of a tet with arbitrary edge lengths, and computed all the geometrical features needed to find the four points $(\cot(\Omega_i/2),\cot(P_i/2))$, and the line containing them all. | |
| Jul 25, 2019 at 12:08 | comment | added | Daniil Rudenko | @ArseniyAkopyan It is equivalent to MY formula: write down everything with exponents, apply linearity of logarithm and you will see sum of dilogarithms. | |
| Jul 25, 2019 at 11:47 | comment | added | Arseniy Akopyan | @GregEgan How you get all these formulas? | |
| Jul 25, 2019 at 9:16 | comment | added | Arseniy Akopyan | Look on the formula from Theorem 3. It is hyperbolic, but can be calculated. homepages.warwick.ac.uk/~masgar/Seminar/WGTS/Abrosimov.pdf | |
| Jul 25, 2019 at 5:14 | history | edited | Greg Egan | CC BY-SA 4.0 |
Corrected mistake in beta_S
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| Jul 25, 2019 at 4:56 | history | edited | Greg Egan | CC BY-SA 4.0 |
Removed unnecessary algebra
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| Jul 25, 2019 at 4:41 | history | edited | Greg Egan | CC BY-SA 4.0 |
Tidied up extraneous material
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| Jul 25, 2019 at 4:23 | history | edited | Greg Egan | CC BY-SA 4.0 |
Added expression for beta
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| Jul 25, 2019 at 1:34 | comment | added | Greg Egan | Let us continue this discussion in chat. | |
| Jul 25, 2019 at 1:32 | history | edited | Greg Egan | CC BY-SA 4.0 |
Determinant form for gamma_S
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| Jul 25, 2019 at 1:23 | comment | added | Daniil Rudenko | @GregEgan Yes, this is the paper. I am sorry for the delay: I will put the proof on arXiv by the end of this week. But it is based on algebraic geometry, I don't know any elementary proof. And your result about $\alpha_S$ is very beautiful! | |
| Jul 25, 2019 at 1:13 | comment | added | Greg Egan | @DaniilRudenko The only Murakami and Yano paper I have is “On the volume of a hyperbolic and spherical tetrahedron”. Is that the one you are referring to? If you can see an easy way to derive the slope from their result, I hope you will post it in an answer! | |
| Jul 25, 2019 at 1:03 | comment | added | Greg Egan | @IanAgol I have something reasonable for the slope now in the spherical case: a manifestly invariant expression that agrees with the Euclidean version in the limit of small edge lengths. | |
| Jul 25, 2019 at 0:58 | history | edited | Greg Egan | CC BY-SA 4.0 |
Added spherical case
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| Jul 25, 2019 at 0:07 | comment | added | Daniil Rudenko | And the polynomial should be a Gram matrix determinant. | |
| Jul 24, 2019 at 23:55 | comment | added | Daniil Rudenko | Let me mention that the slope can be easily computed in terms of $y_i$ parameters from MY formula. | |
| Jul 24, 2019 at 22:51 | comment | added | Greg Egan | @IanAgol I don’t know any formula for a spherical tetrahedron’s volume other than that given by Murakami and Yano, which is pretty horrendous. I’ve now managed to get the slope of the line expressed as a ratio of polynomials of trig functions of edge lengths, times a square root of another polynomial, but there are 2269 terms in all, so it's still rather opaque. | |
| Jul 24, 2019 at 18:39 | comment | added | Ian Agol | @GregEgan: I see, sounds complicated. One could also wonder: do the solid angles, the volume and dual volume determine the simplex? This is six data points, so at least enough dimensions. One has Schlafli's formula (and its dual) which give the variation of volume (and dual volume) in terms of the edge lengths and dihedral angles. But I'm not sure if this helps. | |
| Jul 24, 2019 at 16:23 | comment | added | Greg Egan | @IanAgol No, not without knowing a lot more about how to calculate those volumes. At present all I can do is construct a spherical tetrahedron's vertices and normals from its edge lengths, and compute the solid angles at each vertex. I've derived formulas for the cots of half the solid angles in terms of the edge lengths, but they are literally megabytes long. | |
| Jul 24, 2019 at 16:05 | comment | added | Ian Agol | @GregEgan can you plot the relationship keeping volume and dual volume fixed? | |
| Jul 24, 2019 at 6:02 | comment | added | Greg Egan | @IanAgol Sorry, I just realised that the cross-ratios mentioned in the question make it clear that it’s $(\cot(\Omega_i/2),\cot(P_i/2))$ that are colinear — and I do see that numerically for arbitrary sizes. | |
| Jul 24, 2019 at 5:40 | comment | added | Greg Egan | @IanAgol Do you have any suggestion for the form the colinear points ought to take in the spherical version? Numerically, I see $(\cot(\Omega_i/2),\csc(P_i))$ approximately colinear for small tetrahedra, as you’d expect from the Euclidean case, but this doesn’t hold for finite spherical tetrahedra. | |
| Jul 23, 2019 at 18:44 | comment | added | Arseniy Akopyan | In spherical case "$6V\alpha$" and "$\beta$" have algebraic forms (from, say, tangents of halves of "$l_{ij}$"). It will be absolutely awesome, if in spherical case instead of V we have something like f(V), where f - is not algebraic function, but some understandable function from the volume. | |
| Jul 23, 2019 at 18:37 | comment | added | Arseniy Akopyan | @IanAgol I burned my laptop, but did not find anything readable. | |
| Jul 23, 2019 at 16:45 | comment | added | Ian Agol | It would be interesting to see if a similar formula holds in the spherical case (as Akopyan hints at). If so, it might have a more symmetric form (between a spherical tetrahedron and its dual). | |
| Jul 23, 2019 at 10:10 | history | edited | Greg Egan | CC BY-SA 4.0 |
Added Permament form for beta
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| Jul 23, 2019 at 9:41 | history | edited | Greg Egan | CC BY-SA 4.0 |
More manifestly invariant formula for beta
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| Jul 23, 2019 at 8:25 | comment | added | Daniil Rudenko | Also, a relation between cotangent and lengths already appeared once in a very special case: a computation of a Dehn invariant of a Schläfli orthoscheme in the proof of Sydler’s theorem. | |
| Jul 23, 2019 at 8:23 | history | edited | Greg Egan | CC BY-SA 4.0 |
Correction
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| Jul 23, 2019 at 8:22 | comment | added | Daniil Rudenko | I would compare your beautiful result with the relation Schläfli differential relation: it should lead to something interesting. | |
| Jul 23, 2019 at 8:19 | comment | added | Greg Egan | @DaniilRudenko I’ve added some remarks about beta, though it would be nice to have something more intuitive and geometrical. | |
| Jul 23, 2019 at 8:17 | history | edited | Greg Egan | CC BY-SA 4.0 |
Remarks on the construction of beta.
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| Jul 23, 2019 at 8:02 | comment | added | Daniil Rudenko | Brilliant! I wonder, what is the geometric meaning of beta. | |
| Jul 23, 2019 at 6:08 | history | answered | Greg Egan | CC BY-SA 4.0 |