I. Lambert cube $\mathfrak L(\alpha_1,\alpha_2,\alpha_3)$

In this paper (p.8), we find the volume $V$ of the hyperbolic Lambert cube for the special case $\alpha=\alpha_1 = \alpha_2 = \alpha_3$ as

$$V_\alpha = \int_x^\infty \ln\left(\frac1{t^2}\left(\frac{t^2-\tan^2\alpha}{1+\tan^2\alpha}\right)^3\right)\frac{dt}{t^2+1}$$

where $x$ is the positive root of $x^4-(1+3\tan^2\alpha)x^2-\tan^6\alpha =0$. Hence,

$$\begin{array}{|c|c|c|} \hline \alpha &x& V_{\alpha} & 8V_{\alpha}\\ \hline \pi/3 &\left(\frac{1+\sqrt{13}}2\right)^{3/2}& 0.324423 & \color{brown}{2.59538}\\ \pi/4 &\left(\frac{1+\sqrt{5}}2\right)^{3/2}& 0.538275 & \color{blue}{4.30620}\\ \pi/5 &\left(\frac{5-\sqrt{5}}2\right)^{3/2}& 0.658081 & 5.26465\\ \hline \end{array}$$

II. Lobell polyhedron $L(n)$

In this paper (p.33), we find the volume (using a different formula) of $L(5)$ as

$$\begin{array}{|c|c|c|} \hline n & L(n) & V_n\\ \hline 5 & L(5) & \color{blue}{4.30620}\\ 6 & L(6) & 6.02304\\ \hline \end{array}$$

III. Closed hyperbolic 3-manifolds

On a hunch, I checked those volumes in the Hodgson & Weeks census and found,

$$\begin{array}{|c|c|c|} \hline \text{Dehn filling}& \text{Symmetry} & \text{Volume}\\ \hline m160(-3, 2) & D6 & \color{brown}{2.59538}\\ \hline s648(-5, 1) & D4 & \color{blue}{4.30620}\\ s921(-3, 1) & D2 & 4.30620\\ m400( 4, 1) & Z/2 & 4.30620\\ \hline \end{array}$$

IV. Questions

  1. Why are the volumes for $\alpha =\pi/3, \pi/4$ and $L(5)$ found in the census?
  2. Conversely, why for $\alpha =\pi/5$ and $L(6)$ are they NOT present?
  • $\begingroup$ A first check for a possible explanation might be trying to figure out whether the fundamental domains of those 3-manifolds could be (scissors congruent to) said polyhedras. $\endgroup$ – ThiKu Jun 20 '19 at 16:15

Let me make a remark about the first question and then try to answer the second.

For the first question, using snappy's arithmetic computations (available when using snappy as a sage module), it looks like m400(4,1), s648(-5,1), s921(-3,1) are likely 60 fold covers of the quotient of $H^3$ by an arithmetic tetrahedral group $\Gamma(3,3,5,2,2,2)=\langle x,y,z| x^3,y^5,z^3, (yz^{-1})^2,(zx^{-1})^2,(xy^{-1})^2\rangle$ (see

Maclachlan, Colin; Reid, Alan W., The arithmetic of hyperbolic 3-manifolds, Graduate Texts in Mathematics. 219. New York, NY: Springer. xiii, 463 p. (2003). ZBL1025.57001. (Especially pages 144, 415 for more information on tetrahedral groups.)

For an exact representation of this group, you can use Grant Lakeland's notes available from his website: Matrix group computations. It's possible this group (or its index 2 supergroup $\Gamma(3,2,5,2,2,4)$) is the symmetries of the dodecahedron corresponding to $L(5)$, but I don't know if that is written down anywhere.

For the second question, $\alpha=\pi/5$ and $𝐿(6)$ are likely not in the Hodgson-Weeks census because they are too complicated. The Hodgson-Weeks census is a set of 11,031 fillings of ideal triangulations with 7 or fewer tetrahedra such that (experimentally) the injectivity radius is sufficiently large (I think above .3). So if $L(6)$ can only be obtained from filling an ideal triangulation with 8 or more tetrahedra then it won't appear here.

This appears to be the case. Or more precisely, as the geodesic cut-off doesn't seem to be a restriction for $L(6)$, it's seems the answer is that nothing with a fundamental domain with a volume like that of $L(6)$ can be obtained from filling a cusped manifold with 7 or fewer tetrahedra. I would conjecture that the same could be said for $\alpha=\pi/5$, but I don't know what its shortest geodesic is in that case.

  • 1
    $\begingroup$ Your guess about the tetrahedral group is true, simply because this tetrahedron tiles the dodecahedron (the stabiliser of one of the vertices is $A_5$ or $S_5$ and its orbit gives the tiling). It is possible to identify exactly which group this is. I'm not sure all details are in MacLachlan--Reid. $\endgroup$ – Jean Raimbault Jun 21 '19 at 7:17

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