Message #2289

From: schuma <mananself@gmail.com>
Subject: Hyperbolic Honeycomb {7,3,3}
Date: Sun, 24 Jun 2012 08:14:31 -0000

Hi everyone,
I’m continuing talking about my honeycomb/polytope viewer applet. I
added a new honeycomb, and I think it deserves a new topic. This is
{7,3,3}. Each cell is a hyperbolic tiling {7,3}. Please check it here:
http://people.bu.edu/nanma/InsideH3/H3.html
I first heard of this thing together with {3,3,7} in emails with Roice
Nelson. He had been exchanging emails with Andrey Astrelin about them.
We have NOT seen any publication talking about these honeycombs. Even
when Coxeter enumerate the hyperbolic honeycombs, he stopped at
honeycombs like {6,3,3}, where each cell is at most an Euclidean
tessellation like {6,3}. He said, "we shall restrict consideration to
cases where the fundamental region of the symmetry group has a finite
content" (content = volume?), and hence didn’t consider {7,3,3}, where
each cell is a hyperbolic tessellation {7,3}.
We think {3,3,7} and {7,3,3} and other similar objects are
constructable. I derived the edge length of {n,3,3} for general n, and
then computed the coordinates of several vertices of {7,3,3}, then I
plotted them. There’s really nothing so weird about this honeycomb. It
looks just like, or, as weird as, {6,3,3}. The volume of the fundamental
region of {7,3,3} may be infinite, but as long as we talk about the edge
length, face area, everything is finite and looks normal.
I could go and make {8,3,3}, {9,3,3} etc. I also believe {7,3,4} and
{7,3,5} are also pretty well behaved, and looks just like {6,3,4} and
{6,3,5} respectively. Or even {7,4,3}. As long as the vertex figure is
finite (not like {3,4,4}), the image shouldn’t be crazy. Since we are
facing an infinite number of honeycombs here, I feel I should stop at
some point. After all we don’t understand {7,3,3} well, which is the
smallest representative of them. I’d like to spend more energy making
sense of {7,3,3} rather than go further.
It’s not clear for me whether we can identify some heptagons in {7,3} to
make it Klein Quartic, in {7,3,3}. For example, in the hypercube
{4,3,3}, we can replace each cubic cell by hemi-cube by identification.
The result is that all the vertices end up identified as only one
vertex. I don’t know what’ll happen if I replace {7,3} by Klein Quartic
({7,3}_8). It will be awesome if we can fit three KQ around each edge to
make a polytope based on {7,3,3}. If "three" doesn’t work, maybe the one
based on {7,3,4} or {7,3,5} works. I actually also don’t know what’ll
happen if I replace the dodecahedral cells of 120-cell by
hemi-dodecahedra. Does anyone know?
I still suspect people have discussed it somewhere in literature. But I
haven’t found anything really related. Roice found the following
statement and references. I don’t haven’t check them yet.__________
I checked ‘Abstract Regular Polytopes’, and was not able to find
anything on the {7,3,3}. H3 honeycombs make several appearances at
various places in the book, but the language seems to be similar to
Coxeter, and their charts also limited to the same ones. On page 77,
they distinguish between "compact" and "non-compact" hyperbolic types,
and say:
Coxeter groups of hyperbolic type exist only in ranks 3 to 10, and there
are only finitely many such groups in ranks 4 to 10. Groups of compact
hyperbolic type exist only in ranks 3, 4, and 5.
But as best I can tell, "non-compact" still only refers to the same
infinite honeycombs Coxeter enumerated. They reference the following
book:
J. E. Humphreys, Reflection Groups and Coxeter Groups, Cambridge
University Press (Cambridge, 1990).
When researching just now on wikipedia, the page on uniform hyperbolic
honeycombs has a short section on noncompact hyperbolic honeycombs, and
also references the same book by Humphreys. So maybe this book could be
a good reference to dig up, even though I suspect it will still not
mention the {7,3,3}.
Also: Abstract Regular Polytopes, p78:
For the general theory of hyperbolic reflexion groups, the reader is
referred to Vinberg [431-433]. We remark that there are examples of
discrete groups generated by hyperplane reflexions in a hyperbolic space
which are Coxeter groups, but do not have a simplex as a fundamental
region.
These honeycombs fall into that category.
Here are those references:
[431] E. B. Vinberg, Discrete groups in Lobachevskii spaces generated by
reflections, Mat. Sb. 72 (1967), 471-488 (= Math. USSR-Sb. 1 (1967),
429-444). [432] E. B. Vinberg, Discrete linear groups generated by
reflections, Izv. Akad. Nauk. SSSR Ser. Mat. 35 (1971) 1072-1112 (=
Math. USSR-Izv. 5 (1971), 1083-1119).[433] E. B. Vinberg, Hyperbolic
reflection groups. Uspekhi Mat. Nauk 40 (1985), 29-66 (= Russian Math.
Surveys 40 (1985), 31-75).______________
Now I can only say "to the best of our knowledge, I haven’t seen any
discussion about it".
Some more thoughts by Roice:__________We know that for {n,3,3), as n ->
6 from higher values of n, the {n,3} tiling approaches a horosphere,
reaching it at n = 6.
For {7,3,n), as n -> infinity, does the {7,3} tiling approach a
horosphere as well? The curvature definitely flattens out as n
increases. If cells are a horosphere in the limit, a {7,3,infinity}
tiling would have finite cells. It would have an infinite edge-figure,
in addition to an infinite vertex-figure, but as Coxeter did an
enumeration allowing the latter, why not allow the former? I’d like to
understand where in Coxeter’s analysis a {7,3,infinity} tiling does not
fit in. One guess is that even if the {7,3} approaches a horosphere,
it’s volume also goes to 0, so is trivial. The heptagons get smaller
for larger n, so I suppose they must approach 0 size as well.
It would also be interesting to consider how curvature changes for
{n,3,3} as n-> infinity, especially since we already know what the
{infinity,3} tiling looks like._______________
Currently I can’t imagine what {7,3,n} like when n>=6. So I really
cannot comment on {7,3,infinity}. But {infinity, 3, 3} seems to be a
good thing to study.
My formula for the edge length of {n,3,3} is as follows. Following
Coxeter’s notation, if 2*phi is the length of an edge of {n,3,3} (n>=6),
then
cosh(2*phi) = 3*cos^2(pi/n) - 1
Sanity check: when n = 6, this formula gives cosh(2*phi)=5/4, which is
consistent with the number in Coxeter’s table: cosh^2(phi)=9/8.
By sending n to infinity, the edge length of {infinity, 3, 3} is
arccosh(2). I should be able to plot it soon.
By the way, in the applet there’s a "Clifford Torus". It looks much more
beautiful than the polytopes, because the colors of the edges work
pretty well here. Imagine you can fly around a donut, or go into the
donut. The amazing thing is if the space is 3-sphere, the view inside
the donut is exactly as same as the outside.
Nan