Timeline for Is the 4x5 chessboard complex a link complement?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Aug 28, 2010 at 11:10 | comment | added | Bruno Martelli | Yesterday I tried the same fillings as Ian: when I saw that the fundamental group had no generators and no relators, I interpreted it as "SnapPea cannot find a presentation because it's too complicate". :-D | |
| Aug 27, 2010 at 22:04 | vote | accept | David Eppstein | ||
| Aug 27, 2010 at 21:00 | comment | added | Matthias | 3. By construction, the symmetry group of your manifold acts transitively on the tetrahedra, even the oriented flags. The triangulation of a cusp cross section has the same property. Furthermore there are 6 triangles meetings at a vertex in a cusp cross section. Regular maps of the torus of type {3,6} are classified by ideals in $\mathbb{Z}[\zeta]$. Because you have 24 triangles, the norm needs to be 12. The only ideal (up to conjugacy) is $\langle 2+2\zeta\rangle$. | |
| Aug 27, 2010 at 21:00 | comment | added | Matthias | Sorry, I was over the character limit. 2. The order of $\mathrm{PSL}(2,\mathbb{Z}[\zeta]/\langle 2+2\zeta\rangle)$ is twice that of $\mathrm{SL}(2,\mathbb{Z}[\zeta]/\langle 2+2\zeta\rangle)$ which is the product of $\mathrm{SL}$ for $2$ and $1+\zeta$. We get $60*24/2=720$. So my manifold really is the one for $I=\langle 2+2\zeta$. Remark: For large enough ideals $I$, the largest normal cover with fixed cusp structure seems to be larger than the manifold with symmetry group $\mathrm{PSL}(2,\mathbb{Z}[\zeta]/I)$. | |
| Aug 27, 2010 at 20:59 | comment | added | Matthias | @ David: Yes. I double checked this morning: 1. My manifold has 120 regular tetrahedra, 20 cusps, and each cusp has 24 triangles such that they form a regular map {3,6} of the torus. The cusp looks like the lattice $\mathbb{Z}[\zeta]$ divided by $I=\langle 2+2\zeta\rangle$. It is a normal cover of the Bianchi orbifold for $\mathbb{Z}[\zeta]$. It covers it 720 times. By construction (through my Python script), it is the largest normal cover of this Bianchi orbifold with this fixed cusp structure. | |
| Aug 27, 2010 at 20:39 | comment | added | Ian Agol | @ David: Yes, your manifold appears to be a link complement. However, it might be quite a pain to find an explicit link diagram. It also seems to have many distinct embeddings into S^3. | |
| Aug 27, 2010 at 18:18 | comment | added | David Eppstein | This is the manifold from my question, right? Not the one with I = <3>. | |
| Aug 27, 2010 at 17:07 | comment | added | Ian Agol | I just plugged this into SnapPea, and did 1/0 surgery on each component, and it gave the trivial group. So Matthias' manifold is a link complement. | |
| Aug 27, 2010 at 9:18 | comment | added | Ryan Budney | Ah. And the symmetry group is apparently octahedral x $\Sigma_5$. | |
| Aug 27, 2010 at 8:39 | comment | added | Matthias | Here is the file for the manifold for SnapPea/SnapPy: math.berkeley.edu/~matthias/… According to SnapPea/SnapPy, the homology is Z^20, so it might be a link complement. | |
| Aug 27, 2010 at 8:26 | history | answered | Matthias | CC BY-SA 2.5 |