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The 2x3 and 3x4 chessboard complexes (form a square grid of vertices and make a simplex for any set of vertices no two of which are in the same row or column) are a 6-cycle and a triangulated torus with 24 triangles, respectively. The 4x5 chessboard complex is only a pseudomanifold — each vertex has the 3x4 torus as its link, rather than a spherical link that a proper manifold would have — but if you delete its 20 vertices you get a bona-fide cusped hyperbolic 3-manifold, triangulated by 120 regular ideal tetrahedra. It sort of looks like the kind of manifold you might get as the complement of a 20-component link in Euclidean space. Is it a link complement? And if so, which link is it the complement of?

Edit: here are a couple of general references on chessboard complexes.

Ziegler, G. M. (1994). Shellability of chessboard complexes. Israel J. Math. 87: 97–110.

Björner, A.; Lovász, L.; Vrecica, S. T.; Zivaljevic, R. T. (1994). Chessboard complexes and matching complexes. J. London Math. Soc. 49: 25–39.

They also have important applications to the proof of colored Tverberg theorems in discrete geometry: see, e.g.

Pavle V. M. Blagojević, Benjamin Matschke, Günter M. Ziegler (2009). Optimal bounds for the colored Tverberg problem. arXiv:0910.4987.

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@David, can you provide some reference(s) for chessboard complexes? – Joseph O'Rourke Aug 26 '10 at 20:05
Can you provide the triangulation? If you have the triangulation handy you can readily write a script (I have one here, if you like) for the python interface to SnapPea, to look for your manifold in the census of link complements. I could e-mail you the script or you could send me the triangulation, whichever you like. – Ryan Budney Aug 27 '10 at 6:46
I wouldn't expect this to be a link complement - the cusps are rather large, but the systoles of the cusps aren't large enough to rule out outright. You probably won't find this in any census, since the volume is so big. The first thing to check would be the homology. Unfortunately, although there is an algorithm to check if a cusped manifold is a link complement, it's not very practical to implement. There are finitely curves to check on each cusp, so that any link complement must have one of these curves as a meridian. Then you Dehn fill each possible meridian, regeometrize, and induct. – Ian Agol Aug 27 '10 at 7:26
Ah, 120 regular ideal tetrahedra. Do you know the symmetry group of the triangulation -- could this be something as simple as a 120-sheet cover of the Gieseking manifold? – Ryan Budney Aug 27 '10 at 8:08
If anyone discovers which links this is the complement of, I'd love to find out. – Ryan Budney Sep 2 '10 at 7:06
up vote 18 down vote accepted

I met this manifold before. It is a normal cover of the orbifold $\mathbb{H}^3/\mathrm{PSL}(2,\mathbb{Z}[\zeta])$ where $\zeta=e^{\pi i/3}$.

I suspect that it actually is $\mathbb{H}^3/\ker\left(\mathrm{PSL}(2,\mathbb{Z}[\zeta])\rightarrow \mathrm{PSL}(2,\mathbb{Z}[\zeta]/I)\right)$ where $I$ is the ideal $\langle 2+2\zeta\rangle$.

For $I=\langle 2+\zeta\rangle$, the manifold is a link complement, see Ian Agol's paper on the Thurston Congruence Link for a drawing.

For $I=\langle 3\rangle$, I could show that it is a link complement, but failed at constructing the link.

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Here is the file for the manifold for SnapPea/SnapPy:… According to SnapPea/SnapPy, the homology is Z^20, so it might be a link complement. – Matthias Aug 27 '10 at 8:39
Ah. And the symmetry group is apparently octahedral x $\Sigma_5$. – Ryan Budney Aug 27 '10 at 9:18
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. – Ian Agol Aug 27 '10 at 17:07
This is the manifold from my question, right? Not the one with I = <3>. – David Eppstein Aug 27 '10 at 18:18
@ 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. – Ian Agol Aug 27 '10 at 20:39

Figure 1.27 of shows the link.

alt text

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Nice image! And thanks for the cite back to this question. – David Eppstein Feb 10 '12 at 6:32

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