From:eppstein@ics.uci.edu (David Eppstein)Newsgroups:sci.mathSubject:Do buckyballs fill hyperbolic space?Date:3 Apr 1996 17:34:53 -0800Organization:UC Irvine Dept. Information and Computer Science

Suppose you make a buckyball (truncated icosohedron, soccer ball) in hyperbolic space, of the right volume so that the dihedral angles are all 120 degrees. It looks like it should fill hyperbolic space (with two hexagons and one pentagon meeting at each edge). Is this right? (I've just been looking at Sascha Rogmann's page http://www-public.rz.uni-duesseldorf.de/~rogmann/hyperbolic.html which includes some stuff on hyperbolic buckyballs but it looks from his pictures like he's using 90 degree instead of 120 degree dihedrals.) -- David Eppstein UC Irvine Dept. of Information & Computer Science eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

From:fc3a501@rzaixsrv1.uni-hamburg.de (Hauke Reddmann)Newsgroups:sci.mathSubject:Re: Do buckyballs fill hyperbolic space?Date:7 Apr 1996 17:38:24 GMTOrganization:University of Hamburg -- Germany

David Eppstein (eppstein@ics.uci.edu) wrote: : Suppose you make a buckyball (truncated icosohedron, soccer ball) : in hyperbolic space, of the right volume so that the dihedral angles : are all 120 degrees. It looks like it should fill hyperbolic space : (with two hexagons and one pentagon meeting at each edge). : Is this right? Hm, wild guess: ANY archimedean polyhedra should do. -- Hauke Reddmann <:-EX8 fc3a501@math.uni-hamburg.de UP AGAIN! HOORAY! fc3a501@rzaixsrv1.rrz.uni-hamburg.de IF NOT reddmann@chemie.uni-hamburg.de SCIENCE ONLY

To:fc3a501@math.uni-hamburg.deSubject:Do buckyballs fill hyperbolic space?Date:Mon, 08 Apr 1996 17:06:39 -0700From:David Eppstein <eppstein@ICS.UCI.EDU>

Hm, wild guess: ANY archimedean polyhedra should do. Regular tetrahedra do not, unless you allow cusps (vertices at infinity). The tetrahedron has too small a solid angle to fill the hyperbolic space around a point (with a symmetric vertex neighborhood). I haven't thought about the others in detail, but I'm also not convinced about snub cubes and snub dodecahedra. Maybe the rest are ok. You'd have to be careful how you glue them together (e.g. for the buckyball you have to give the hexagon faces a 1/6 twist before gluing so that you don't get two pentagons on the same edge). By the way, there was a minor detail wrong in my post -- buckyballs don't have one dihedral, they have two, one (x) between pairs of hexagons and one (y) between hexagons and pentagons. The volume one wants is the one for which x+2y=360. -- David Eppstein UC Irvine Dept. of Information & Computer Science eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

From:Hauke Reddmann <fc3a501@geomat.math.uni-hamburg.de>Subject:Re: Do buckyballs fill hyperbolic space?To:David Eppstein <eppstein@ICS.UCI.EDU>Date:Tue, 9 Apr 1996 08:37:32 +0200 (DFT)

> > Hm, wild guess: ANY archimedean polyhedra should do. > > Regular tetrahedra do not, unless you allow cusps (vertices at infinity). > The tetrahedron has too small a solid angle to fill the hyperbolic space > around a point (with a symmetric vertex neighborhood). > Oh yes, of course. I was only thinking of topological restraints (whether the glueing should fit). Maybe a coordinate approach would clear things up? HR

Newsgroups:sci.mathFrom:Olaf Delgado <delgado@mathematik.uni-bielefeld.de>Subject:Re: Do buckyballs fill hyperbolic space?Date:Tue, 9 Apr 1996 12:29:13 GMTOrganization:Universitaet Bielefeld, Rechenzentrum

David Eppstein wrote: > Suppose you make a buckyball (truncated icosohedron, soccer ball) > in hyperbolic space, of the right volume so that the dihedral angles > are all 120 degrees. It looks like it should fill hyperbolic space > (with two hexagons and one pentagon meeting at each edge). > Is this right? I suppose you are talking about a proper archimedean buckyball, which means that all the faces are regular pentagons or hexagons, respectively? I don't think it's possible in hyperbolic space that all the dihedral angles are 120 degrees, but what you really need is just that the sum of twice the pentagon-hexagon angle plus the hexagon-hexagon angle is 360 degrees, which should be perfectly possible. Then, indeed, such a buckyball would fill hyperbolic space.

From:zare@cco.caltech.edu (Douglas J. Zare)Date:9 Apr 1996 21:03:52 GMTNewsgroups:sci.mathSubject:Re: Do buckyballs fill hyperbolic space?

Hauke Reddmann <fc3a501@rzaixsrv1.uni-hamburg.de> wrote: >David Eppstein (eppstein@ics.uci.edu) wrote: >: Suppose you make a buckyball (truncated icosohedron, soccer ball) >: in hyperbolic space, of the right volume so that the dihedral angles ^^^^^^^^^^^^^^^^^^^ >: are all 120 degrees. It looks like it should fill hyperbolic space >: (with two hexagons and one pentagon meeting at each edge). >: Is this right? > >Hm, wild guess: ANY archimedean polyhedra should do. Can you do this? The symmetry group does not act transitively on the edges; for the snub cube (a major benefactor of Caltech--see http://www.caltech.edu/map/images/beck_inst.gif) and snub dodecahedron there are 3 types of edges. I think you would actually want to get a linear combination of the two to be 360 degrees, and within that locus fix the edge lengths. Douglas Zare http://www.cco.caltech.edu/~zare

From:mathwft@math.canterbury.ac.nz (Bill Taylor)Newsgroups:sci.mathSubject:Hyperspace tiling again.Date:15 Apr 1996 05:21:34 GMTOrganization:Department of Mathematics and Statistics, University of Canterbury, Christchurch, NZ.

There was a recent question about tiling hyperbolic 3-space with soccer-ball solids, (truncated icosahedra). And whether all Archimedeans could do the same. Here's a rather simpler question. Is it possible to tile the hyperbolic plane with squares, all meeting 5 at a vertex? (Or similarly, any regular congruent polygons with constant excessive valence?) It would seem so, if my memory of Escher-like patterns is anything to rely on. There's certainly no combinatorial objection; so it's just a matter of whether hyperbolic geometry allows it; i.e. allows them all to be squares. So my question really is: is it "easy" to prove it possible, using synthetic hyperbolic geometry? I guess yes, but I'm no expert at handling such proofs. Amyone? ------------------------------------------------------------------------------- Bill Taylor wft@math.canterbury.ac.nz ------------------------------------------------------------------------------- The Balkans have always produced more history than they can consume locally. -------------------------------------------------------------------------------

From:Doug Dunham <ddunham@d.umn.edu>Subject:Filling hyperbolic space with buckyballsTo:David Eppstein <eppstein@ics.uci.edu>Date:Thu, 25 Jul 1996 18:06:26 -0500 (CDT)

I just discovered your web page and spent a bit of time thinking about the hyperbolic buckyball tessellation. I apologize if the following is know to you (and the others) -- in which case, please ignore it. I think the tessellation that you describe exists -- with 2 hexagons and one pentagon meeting along each edge. (In the following, I assume that the truncated icosahedron is "regular" -- i.e. the faces are regular pentagons and hexagons -- and use that term instead of buckyball, which may be less regular.) In general, there seem to be a couple of considerations for the existence of a tiling by a polyhedron: 1) topological or combinatorial -- can the tiling be extended in a consistent manner, and 2) "metric" -- does there exist a polyhedron with the right dihedral angles, edge lengths, etc. In a highly symmetric tiling (with large symmetry group), which I think is what you are proposing, the combinatorial constraints can be checked locally. In particular, 4 truncated icosahedra can be placed (combinatorially) around a vertex in a consistent way with 2 hexagons and a pentagon around each edge -- the vertex figure is a non-regular tetrahedron with 2 "skew" (non-meeting) edges corresponding to pentagons and 4 longer "hexagon" edges. So there seems to be no combinatorial problem. The other consideration is: is there an edge length for which the dihedral angles add up to 360 degrees (the edge length determines all the other geometry of a "regular" hyperbolic truncated icosahedron)? I would make the usual continuity argument for the existence of the proper edge length: for very small edge lengths, things are almost Euclidean and the 3 dihedral angles add up to more than 360 degrees; the dihedral angles decrease continuously with increasing edge length (which can increase to infinity for an asymptotic truncated icosahedron). However, the dihedral angles don't decrease to 0. So the question is: does the dihedral angle sum decrease to less than 360 degrees? If so, the Intermediate Value Theorem says that there is an edge length for which the dihedral angle sum is exactly 360 degrees. I think the answer to the question is "yes". The hexagon- hexagon dihedral angle is larger than the hexagon-pentagon dihedral angle in the Euclidean limit (of small edge length) and I think that this always remains true as the edge length is increased (this is the shaky part of the argument). But there is a finite edge length at which the hexagon-hexagon dihedral angle is 120 degrees -- simply truncate an icosahedron of the regular tessellation of hyperbolic space by icosahedra meeting 12 at a vertex. For this edge length, the dihedral angle sum is less than 360 degrees.[

There may be other symmetric tessellations with more truncated icosahedra around the edges (and more complex vertex figures). And there may also be non-period space-filling arrangements -- like the Penrose tilings. Other Archimedean solids may also ("regularly") tile hyperbolic space (or Euclidean 3-space or the 3-sphere). I looked at Sascha Rogmann's buckyball tessellation. It appears to me to be as follows (but I may be misunderstanding it): there are planes of pentagons and planes of hexagons. At each vertex there is one pentagon plane and two hexagon planes -- all mutually perpendicular. So all dihedral angles are 90 degrees, in each polygon plane 4 polygons meet at a vertex, and the vertex figure is an octahedron (but non-regular). The tiling of the pentagon planes is regular, but the tiling of the hexagon planes is not regular -- the hexagon edges in pentagons planes are shorter than the hexagon edges perpendicular to the pentagon planes. The hexagon tiling could not be by regular hexagons, since two different regular tilings with the same vertex valence have different edge lengths. Thus the buckyballs are not "regular" truncated icosahedra. PS: Here is the answer to Bill Taylor's question: for any positive integers p and q with (p - 2)*(q - 2) > 4, there is a tiling of the hyperbolic plane by regular p-gons meeting q at a vertex (i.e. q is the vertex valence). The edge length is 2*b where cosh(b) = cos(PI/q) / sin(PI/p) So for fixed q, edge length increases with p (the fact used in the previous paragraph -- where q was 4 and p was 5 or 6). -- Doug Dunham phone: (218) 726-7510 ddunham@d.umn.edu FAX: (218) 726-8240 Dept. of Computer Science, UMD 320HH Duluth, MN 55812-2496

To:ddunham@d.umn.eduSubject:Filling hyperbolic space with buckyballsDate:Fri, 26 Jul 1996 10:51:11 -0700From:David Eppstein <eppstein@ICS.UCI.EDU>

Thanks for the interesting message. I agree with you re the combinatorial and metric/dihedral considerations -- which is why I suggested the tiling in the first place. Your argument about the truncation of the icosohedron tiling giving a lower bound on dihedrals seems to come close to completing the picture. One thing I'm not sure about: The hexagon- hexagon dihedral angle is larger than the hexagon-pentagon dihedral angle in the Euclidean limit (of small edge length) and I think that this always remains true as the edge length is increased (this is the shaky part of the argument). Is this true for truncated ideal icosohedra? I think the hex-hex dihedral is 2pi/5. I'm having a hard time visualizing the hex-pent dihedrals but I'm wondering if they may be greater than pi/2, since they would equal pi/2 in the limit of no truncation and seem to decrease towards that limit.[

I think the same argument about only needing to test local combinatorial and metric considerations means one doesn't have to worry about the solid angles at the vertices. But...I'd be more comfortable if I could e.g. get SnapPea to find a hyperbolic manifold with the right fundamental domain. -- David Eppstein UC Irvine Dept. of Information & Computer Science eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

To:ddunham@d.umn.eduSubject:followupDate:Fri, 26 Jul 1996 13:18:40 -0700From:David Eppstein <eppstein@ICS.UCI.EDU>

I wrote: I think the hex-hex dihedral [for truncated ideal icosohedra] is 2pi/5 Of course this should be 3pi/5, 72 degrees. I got the interior and exterior dihedrals switched... Anyway, the fact we were trying to prove, that the dihedrals add to less than 2 pi for some hyperbolic truncated icosohedron, seems easiest to see for the ideal truncated icosohedron rather than the truncated ideal icosohedron. For the ITE, I don't know what the dihedrals are, but the three dihedrals at any vertex sum to pi. -- David Eppstein UC Irvine Dept. of Information & Computer Science eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

From:Doug Dunham <ddunham@d.umn.edu>Subject:Re: followupTo:David Eppstein <eppstein@ICS.UCI.EDU>Date:Fri, 26 Jul 1996 17:45:49 -0500 (CDT)

In re-thinking the argument I made, I have decided that I got it backward -- namely, the pent-hex dihedrals are larger than the hex-hex dihedrals -- so my previous argument won't work. However, I think that your observation below closes the gap. > Anyway, the fact we were trying to prove, that the dihedrals > add to less than 2 pi for some hyperbolic truncated icosohedron, > seems easiest to see for the ideal truncated icosohedron > rather than the truncated ideal icosohedron. > For the ITE, I don't know what the dihedrals are, > but the three dihedrals at any vertex sum to pi. This seems to give us what we need -- a sum of 2 pent-hex and one hex-hex dihedral angles, which is just what we are trying to calculate. This sum starts out to be more than 2 pi (at small edge length) and then decreases to pi as the edge length goes to infinity (so for some intermediate edge length, the sum is exactly 2 pi). Incidentally, out of curiosity, I borrowed a paper model of a truncated icosahedron (somewhat the worse for wear) and made very crude measurements with a protractor of the dihedral angles. What I found was that the pent-hex angle was about 147 degrees and the hex-hex angle was about 135 degrees (so the dihedral angle sum starts out in the Euclidean limit to be about 430 degrees). I think the same argument can be used to show that the truncated dodecahedron can also tile hyperbolic space. The dihedral angles of the truncated tetrahedron and truncated cube are too small to tile hyperbolic (or Euclidean) space in this way (with a tetrahedral vertex figure -- but there may be other ways with more complex vertex figures); however I would guess that they do tile the 3-sphere in this way. I also think the truncated octahedron tiles Euclidean 3-space in this way (in fact I can even visualize it by looking at a model) -- I think that this is well known. -- Doug Dunham phone: (218) 726-7510 ddunham@d.umn.edu FAX: (218) 726-8240 Dept. of Computer Science, UMD 320HH Duluth, MN 55812-2496

To:ddunham@d.umn.eduSubject:followupDate:Fri, 26 Jul 1996 15:58:44 -0700From:David Eppstein <eppstein@ICS.UCI.EDU>

I also think the truncated octahedron tiles Euclidean 3-space in this way (in fact I can even visualize it by looking at a model) -- I think that this is well known. Right. More generally, this is an example of a permutahedron [= convex hull of permutations of the vector (0,1,2,...,d); note that this satisfies the equation sum(v_i)=const so it is d-dimensional rather than (d+1)-dimensional] and these are apparently known to tesselate Euclidean space in any dimension (I think the crystallographers call it the hyperbolic lattice, not to be confused with hyperbolic space -- see http://alife.santafe.edu/alife/topics/cas/archives/92/0071.html for some information or misinformation (ignore the bit calling it a truncated cuboctahedron). -- David Eppstein UC Irvine Dept. of Information & Computer Science eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

From:Warren.Krieger@defence.gov.auTo:eppstein@ICS.UCI.EDUDate:Fri, 14 Jan 2000 11:31:09 +1000Subject:SEC: UNCLASSIFIED:-Do Buckyballs tile Hyperbolic Space.

I notice on the Geometric Junkyard page, a discussion on whether Buckyballs (truncated icosahedra) and other Archimedean polyhedra tile hyperbolic space. Here's my answer. One can truncate the regular honeycombs to yield a figure consisting entirely of truncated polytopes, a truncated {p, q} is a cell of a certian truncation of {q, p, q}. Thus we find a figure with truncated tetrahedra and truncated cubes in normal 4D, in {3,3,3} and {3,4,3}, truncated octahedra tesselate 3-space, derived from {4,3,4}, and truncated dodecahedra and icosahedra from {3,5,3} and {5,3,5}. In the case of the rhombotruncated icosahedron and rt cuboctahedron, there are symmetry groups whose fundemental regions recriprocate into tilings of these figures. One can also get the truncated polytopes to tile using laminahedra, which are polytopes bounded by unbounded planes. For example, in 2D, a laminahedra is a strip. One can fill the strip with anything, eg triangles or squares, without changing the surface of the strip (division into equal segments). If one wants to make a uniform lattice, one can stack these strips up in any way that preserves uniform verticies: either all squares, all triangles, or alternating square and triangle strips. In three dimensions, the same thing can be done with plane slabs. The sides of these slabs are inscribed with the verticies of any of the two-dimensional uniform lattices. We can join different slabs together, as long as they have the same markings on the side: the most common case is the triangles. Here, there are triangular prisms, and slabs of oct-tet trusses. These can be arrayed in different ways, eg alternating slabs of prisms and oct-tet trusses. Since the oct-tet truss exists in every dimension as a slab-lattice, one can create alternating lattices in any dimension. In hyperbolic space, one can not place two flat surfaces to run equidistant from each other, but one can still construct figures bounded by unbounded planes. In hyperbolic space, a surface that is equidistant from a flat plane is a curve. If one puts poles up on the plane, the tops of the poles are further than the base of them. Let us suppose that a plane is inscribed with an {8, 3}, and we erect poles on its. If the poles are high enough, the octagon that span the tops of this is the same as that from an {8, 4}. The resultant figure is a right-angled {8, 3}. If we keep the flat surface on the back, we get a slab-like figure, with a rough surface on one side and a flat one on the other, the two never crossing or meeting. We can now take our right-angled {8, 3}, and place eight of these at each corner, to yield an {8, 3, 4}. This joins all of the rough sides of the slab figures, and we're left with a figure bounded by unbounded flat {8,3}'s, and detail completey inside, representing the verticies, edges and planes of a {8,3,4}. If now new faces were to start expandinging from the verticies of the {8, 3, 4}, they would eventually fill up the space within the laminae, and assume the shape of a truncated cube. The octagonal faces of this are internal to the laminahdra, and the triangular faces appear on the flat surfaces, to make the faces inscribed with the dual {3, 8}. Using the flat surfaces as mirrors, we fill all space with truncated cubes. Of course, we can use any hyperbolic lattice {p,q} and any elliptic polyhedron {q,r} in the same way, this means that from, for example, a {120, 5, 3}, we find a figure bounded by laminahedra bounded by flat {5, 120}, the interior being buckyballs that share their hexagonal face with another internal buckyball, and expose the pentagonal face to the pentagonal surface. In the case of laminahedra bounded by {p, 3}, the {p,3} are the same size regardless of how it comes (this is because angle-angle-angle is a congurance in hyperbolic geometry), and so it is possible to tile space with alternate laminahedra inscribed with, say {8, 3, 3} and {8, 3, 5}, yielding a honeycomb made of truncated tetrahedra and truncated icosahedra, eight of each to a vertix. Of the cuboctahedron, the icosadodecahedron and rhombi-icosadodecahedron, the vertix figure of these are reptangles where the opposite sides are of the same nature, different to the other pair. If the vertix figure closes, it would be made of made entirely of quadralaterals (eg cube, rhombic dodecahedron). Consider a line that joins the opposite sides of these quadralaterals. Since they have no means to end or to cross themselves, the tiling must be done in such a way that these lines are unbounded (either circles or open lines) and non-crossing. One possible tiling is a solution topologically equal to a {4, 2p}. With p=1, we get a solution where the polyhedra are stuck back to back. With the case of p=2, the figure honeycomb is hyperbolic, but the verticies reside at infinity. When p>2, the lines run perpendicular to a plane, and thence we can pass infinite planes through this. This cuts off the corners of the figure, and replaces these by a quadralateral. This rhombitruncates the icosadodecahedra and cuboctahedra, but he rbombi-icosahedra does not become anything uniform. Of the rhombicuboctahedron, the vertix figure is a reptangle, with three sides becominging squares, and the fourth a triangle. We can 'mark' three edges of the cube with a '3' so that each face has just one marked side. Rhombic Dodecahedra and tricontahedra might also be so done. Whether these yield figures is open. The snub cube and dodecahedron have pentagonal corners, and so a dodecahedron is the the likely vertix figure. One marks six of the sides with a 4 or 5, representing the six sides that become squares or pentagons. There is more than one way to do this, for example, the edges containing the inscribed octahedron, or the six edges that are parallel to the equator when the poles are at the verticies. In the case of the snub tetrahedron, this yields the {3,5,3}. It may be possible, but the thing would be relatively large. In conclusion. We find the truncated and rhombitruncated polyhedra (2 p | q and (2 p q |) in many different forms, the semiregular ones (2 | 3 4), (2| 3 5) and (3 5|2) have none, and the snub figurus (|2 3 q) and the (3 4|2) may have one or so. Warren Krieger