Timeline for Regular solids and $\mathbb{Z}_5$
Current License: CC BY-SA 4.0
21 events
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| Nov 29, 2023 at 23:02 | comment | added | David I. McIntosh | @GerryMyerson : lol. This question arose because of my (intentionally) comical attempt to map the 5 regular solids (models of which are sitting on the mantle, made from my kids' magnetic toys) to the 5 possible "topologically" unique train track layouts made with exactly two Y connectors. Interesting problem to prove there are exactly 5, and then classify all infinite train paths around each layout according to various properties. | |
| Nov 29, 2023 at 20:05 | review | Close votes | |||
| Dec 4, 2023 at 3:07 | |||||
| Nov 29, 2023 at 19:46 | comment | added | Ryan Budney | @DavidI.McIntosh I've voted to move the question. Hopefully that helps. | |
| Nov 29, 2023 at 6:45 | comment | added | მამუკა ჯიბლაძე | Fwiw, the smallest modulus $m$ such that numbers of faces of regular 4-polytopes are all distinct mod $m$ is $21$; the smallest prime such modulus is $29$ | |
| Nov 29, 2023 at 5:12 | comment | added | David I. McIntosh | Arrg, seems this should be on math.stackexchange not mathoverflow?? | |
| Nov 29, 2023 at 4:32 | comment | added | Ryan Budney | en.wikipedia.org/wiki/Regular_4-polytope But in short, there's the simplex(pentachoron), the cube (sometimes called the octachoron or 8-cell), the dual of the cube (hexadecachoron or 16-cell) then there's the dual pair of the 600-cell and 120-cell, and lastly there's the self-dual 24-cell. | |
| Nov 29, 2023 at 4:27 | history | edited | David I. McIntosh | CC BY-SA 4.0 |
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| Nov 29, 2023 at 4:18 | comment | added | David I. McIntosh | @RyanBudney : Sorry, I missed Ryan's comment above about $n=4$. Ryan, what are the regular polytopes in $n=4$? And in reply to your comment, see Christophe Leuridan's comment above and my correction above for other cases then. | |
| Nov 29, 2023 at 4:10 | comment | added | David I. McIntosh | @ChristopheLeuridan : You made a slight mistake: you changed $2^n$ to $n$. So for $n\ge5$, if what you say is true and there are only three regular polytopes with the number of faces given by $2^n$, $2n$ and $n+1$, then these are distinct modulo 3 if and only f $n$ is odd and NOT 1 modulo 3 (i.e. $n\ne1|3$). Thus, my observation seems to hold true for all $n=3|6$ or $n=5|6$ (using the notation $|$ for "modulo"). But I can't say about $n=4$ without knowing more. | |
| Nov 25, 2023 at 5:51 | comment | added | David Roberts♦ | Not sure any of that is useful, but you really might just be running into a small-number phenomenon. Or perhaps the regular graph description (instead of face counting) in higher dimensions does something.... I can't check right now. | |
| Nov 25, 2023 at 5:47 | comment | added | David Roberts♦ | I'm tempted to say there might be something about the Euler characteristic that helps here. The number of faces is 2 + E - V, and having the mod 5 count of faces be an injective function is the same as saying the difference E - V is an injective function mod 5. This is then a fact about the underlying edge graph, and regularity means the degree D of the vertices to be constant. Then E = DV/2, so that we are left with considering the function V(D/2 - 1). 2 is invertible mod 5, so we are considering the function V(D - 2). We know 1≤D-2≤3 and V≥4 from elementary considerations. | |
| Nov 24, 2023 at 23:03 | history | edited | YCor |
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| Nov 24, 2023 at 21:55 | comment | added | Gerry Myerson | Kepler tried to match the five Platonic solids to the five spaces between the six planets that were known in his day. It didn't end well. en.wikipedia.org/wiki/Mysterium_Cosmographicum | |
| Nov 24, 2023 at 20:59 | comment | added | Christophe Leuridan | The observation holds only in odd dimensions. If $n \ge 5$, there are only three regular polytopes in dimension $n$, the hypercube with $2n$ faces, the hyperoctahedron with $2^n$ faces, the simplex with $n+1$ faces. The numbers $n$, $2n$ and $n+1$ are different modulo 3 if and only if $n$ is odd. | |
| Nov 24, 2023 at 20:13 | comment | added | Sam Hopkins | I think this question is potentially interesting. The classification of regular polyhedra is related to other classifications like of simple Lie algebras, etc., and so it’s not impossible to imagine a somewhat deep reason for this face-counting bijection to work. | |
| Nov 24, 2023 at 20:10 | comment | added | David I. McIntosh | Yes, there are 5! bijections, that is pretty trivial. But a bijection that selects such a fundamental (integer) property of the polyhedra, and then maps to $Z_5$ using the most basic, obvious of maps, namely mod 5, is too much of a coincidence. | |
| Nov 24, 2023 at 20:09 | comment | added | Ryan Budney | Maybe it would be more interesting if you had a similar phenomenon for the regular solids in $\mathbb R^n$ for all $n$ ? You get a similar correspondence for $n=2$, but not $n=4$ (where there are six). | |
| Nov 24, 2023 at 20:06 | comment | added | Ryan Budney | Probably the most significant reason is the sets have the same cardinality. There has to be some bijection, let it be this one. | |
| Nov 24, 2023 at 20:01 | history | edited | David I. McIntosh | CC BY-SA 4.0 |
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| Nov 24, 2023 at 19:58 | history | edited | David I. McIntosh | CC BY-SA 4.0 |
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| Nov 24, 2023 at 19:52 | history | asked | David I. McIntosh | CC BY-SA 4.0 |