cool uniform polychora

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

cool uniform polychora

Postby quickfur » Mon Jul 11, 2005 2:22 am

What do you get if you glue 120 buckyballs together in 4D and fill in the gaps with truncated tetrahedra?


You get a uniform polytope with 120 buckyballs and 600 truncated tetrahedra, called the Truncated-icosahedral hexacosihecatonicosachoron. This cool polychoron has 1200 triangular ridges, 720 pentagons, and a whopping 2400 hexagons. It contains 7200 edges. :-)

OK, OK, so all this is already known on George Olshevsky's uniform polytope page, but it's still cool 'cos I independently discovered this just earlier today. I was playing around with truncating the regular polychora, and found some interesting uniform polychora. Here's a list of the ones I found:

* The rectified 5-cell consists of 5 tetrahedra and 5 octahedra.

* The bitruncate (or mesotruncate) of the 5-cell is an interesting polychoron with 10 truncated tetrahedral cells. This is interesting 'cos all the cells are identical, and the hexagonal faces of the truncated tetrahedra are joined in complementary orientation, so that every edge is only touching one triangle.

* The rectified 8-cell consists of 16 tetrahedra and 8 cuboctahedra.

* The bitruncate (or mesotruncate) of the 8-cell is the same as the bitruncate of the 16-cell, and consists of 16 truncated tetrahedra and 8 truncated octahedra. Like the bitruncated 5-cell, the hexagonal faces of the cells are joined in complementary orientation.

* The rectified 16-cell is, of course, the 24-cell; and the rectified 24-cell consists of 24 cubes and 24 cuboctahedra.

* What do you get if you take 48 cubes, cut off their corners, and fold them up into 4D? You get the bitruncate (mesotruncate) of the 24-cell, which is made of 48 truncated cubes. Like the other mesotruncates, the octagonal faces of the 48 cells are joined in complementary orientation so that only one triangular face is attached to each edge. Again, I found this particularly interesting 'cos all 48 cells are identical.

* The coolest part is with the 120-cell and the 600-cell. Rectifying the 120-cell yields a polychoron with 600 tetrahedra and 120 icosidodecahedra; and rectifying the 600-cell gives a polychoron with 600 octahedra and 120 icosahedra. If you make a deeper cut, you get the bitruncate, made of 120 buckyballs and 600 truncated tetrahedra.
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Postby wendy » Wed Jul 20, 2005 12:28 am

You should look, say at the figure x3x3o5o. This has 120 icosahedra, and 720 truncated tetrahedra. The faces are closely the same size, and the resulting mess is most like the x3x5o, in that they have lots of faces of similar size.

The example o3x3x5o, or bitruncated {3,3,5} more resembles in appearence something like a truncated dodecahedron, with faces relatively large and small.
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Postby quickfur » Wed Aug 03, 2005 3:40 am

wendy wrote:You should look, say at the figure x3x3o5o. This has 120 icosahedra, and 720 truncated tetrahedra. The faces are closely the same size, and the resulting mess is most like the x3x5o, in that they have lots of faces of similar size.

Hmm. It'll take me a while to visualize that one. I've basically taken the approach of manipulating 4D objects by understanding their 3D projections. Maybe one of these days I'll discuss my method of manipulating uniform (Archimedean) polychora, using cell-branching cycles and edge-cycles.

The example o3x3x5o, or bitruncated {3,3,5} more resembles in appearence something like a truncated dodecahedron, with faces relatively large and small.

Yeah I realize that. I just thought it was cool 'cos it'd look like a bunch of pearls embedded in a spherical shell or something. :-)

OTOH, I find the 48-cell (the mesotruncate of the 24-cell) eminently interesting. I've realized that its projection into 2D is a fascinating figure with a high degree of octagonal symmetry. I just need to sit down and work out a way of coaxing my calculator program to draw this figure. The 3D projection also involves lots of octagons, and is similar in symmetry to the rhombicuboctahedron.
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Postby wendy » Wed Aug 03, 2005 11:16 pm

It's one of the three shapes that support poincare cells. These shapes correspond to the symmetries of three dimensions, in rather interesting ways:

rectangular (order 8) ~~ tesseract
prismatic (order 4p) ~~ 2p-2p prism
tetrahedral (order 24) ~~ 24-choron
octahedral (order 48) ~~ octagonny = bitruncated 24choron
icosahedral (order 120) ~~ twelftychoron

Each of these can be cells of some kind of swirl-prism symmetry, the whole cell being one repeated unit. So for example, the icosahedral group has rotations of order 2, 3, 5. Here we have rotations of order 4+, 6+, 10+ (some kind of alternation applies: clifford-rotation does it nicely).

So if one makes right-clifford rotation the only translation, one can get periods of 4, 6, 10, that make the vertices of a ID, dodeca and icosa in some space, and the object moved around is the face of a 120-choron.

This is the same group as i see in the quartrions, and so the vertices of the duals make units in these systems. The span of these give rise to the quaterion integer systems.

The octagonny also tiles in a peicewise and discrete (read infinitely-dense), manner. Its a very interesting group, it can be made entirely of reflections, but it has no schläfli symbol. It's the one i call BB. It's pretty interesting lattice.

I done some rather interesting lace-cities on it. The symmetry is not so much octagonal, because these octagons are actually the group 8+, formed by sets of alternating squares presented on the same axis [rather like the top-down view of a square antiprism].

It's kind of like looking at a chessboard, and seeing eight rows with four black squares each.

By doing a bit of tweaking with the Schläfli symbol, i found out what fraction of solid space is occupied by the vertex: (vertex angle). It is quite large: 30s 50f = 73/288 of all-space.
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Postby quickfur » Thu Aug 04, 2005 5:42 am

wendy wrote:It's one of the three shapes that support poincare cells. These shapes correspond to the symmetries of three dimensions, in rather interesting ways:

rectangular (order 8 ) ~~ tesseract
prismatic (order 4p) ~~ 2p-2p prism
tetrahedral (order 24) ~~ 24-choron
octahedral (order 48 ) ~~ octagonny = bitruncated 24choron
icosahedral (order 120) ~~ twelftychoron

Cool stuff. I think I need to go back and revisit symmetry and group theory in math. I grasp the basic concepts, but have little in-depth knowledge of how it all fits together.

BTW, I like the name "octagonny". Neatly captures its appearance in 2D projections. :-)

[...]The octagonny also tiles in a peicewise and discrete (read infinitely-dense), manner. Its a very interesting group, it can be made entirely of reflections, but it has no schläfli symbol. It's the one i call BB. It's pretty interesting lattice.

I done some rather interesting lace-cities on it. The symmetry is not so much octagonal, because these octagons are actually the group 8+, formed by sets of alternating squares presented on the same axis [rather like the top-down view of a square antiprism].

Yeah, there are some cool things going on in it, given that it is cell-uniform, edge-uniform, and also vertex-uniform. The way 14 cells surround each cell in a 6+8 arrangement is fascinating. The projection of this arrangement into 3D is an interesting exercise in spatial decomposition in its own right.

The cell-first projection into 2D causes 6 layers of cells to collapse onto the same octagonal pattern, which is what prompted me to say that there's a lot of octagonal things going on. Of course, what really is happening is better captured by the 3D projection, where you see that there are 14 cycles that pass through each cell, 6 being the cells joined by opposite octagonal faces and 8 being the cells joined by opposite triangular faces. The 6 cycles each consists of 8 cells in alternating orientation, and the other 8 cycles each consists of 6 cells. This seems reminiscient of the relationship between the gonality and the number of cells in a duoprism (n cells as m-gonal prisms, and m cells as n-gonal prisms).

This structure is rather close to the 4D analogue of the rhombicuboctahedron: the runcinated tetracube (made of 32 cubes, 16 tetrahedra, and 32 triangular prisms). This polychoron also has similar cycles: an 8-membered cube-tetrahedron-triangular prism cycle along the same axes as the octagonny's 6-membered cycles, and the 8-membered cycles of cubes along the same axes as the octagonny's 8-membered cycles. The octagonny essentially replaces the tetrahedral configuration of prisms around each tetrahedral "corner" of the runcinated tetracube with a single cell each, the alternating orientations of all the cells just working out in such a way that everything fits together perfectly.

As you can see, 8-membered cycles pop up very frequently with these two polychora, and they also give rise to very similar cell-first projections. In fact, I independently deduced the structure of the runcinated tetracube using a purely visual method, after I learned the structure of the 48-cell. I later confirmed with George Olshevsky's page that my results matched his.

It's kind of like looking at a chessboard, and seeing eight rows with four black squares each.

By doing a bit of tweaking with the Schläfli symbol, i found out what fraction of solid space is occupied by the vertex: (vertex angle). It is quite large: 30s 50f = 73/288 of all-space.

Interesting. I wish I had a stronger foundation in rigorous math... my method of handling 4D objects is essentially visual, which does help greatly in intuition, but lacking when it comes to these types of derivations.
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Postby quickfur » Thu Aug 04, 2005 5:59 am

Another cool thing about the octagonny is that its cells corresponds to the vertices of the 24-cell and its dual, which means that they correspond with the root system of type F<sub>4</sub>. One might even think of the octagonny as the representation of the F<sub>4</sub> root system. (Although, I only have a vague idea what this means... :-P I basically learned this on wikipedia.)

On another note, I suppose it is a general fact that the mesotruncate of any self-dual polytope will have the property that it is facet-uniform. In 3D, the mesotruncate of the tetrahedron is the octahedron, which happens to be regular; in 4D, the mesotruncate of the pentatope consists of 10 truncated tetrahedra. The 24-cell happens to be self-dual and non-simplicial, which gives its mesotruncate, the octagonny, such special properties. Are there any other non-simplicial self-dual polytopes in higher dimensions? I wonder what interesting properties their mesotruncates would have, if indeed they exist. (Is it even possible for a non-regular polytope to be self-dual??)
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Postby wendy » Thu Aug 04, 2005 6:45 am

Don't worry, i don't get too heavy into the maths. I am just clever enough to avoid it!

It's worth noting there is an 'octagrammy' as well. This is the isomorph, on the octagonal scale, of the octagonny, where the octagons get replaced by octagrams, and vice versa.

octagrammy is o3x4/3x3o, or the quasi-bitruncated 24choron.

When i was playing with the octagon/octagram, i spent a fair bit of time with these two, because they lead to a stable set of quartrions.

octagonny is also a known tiler in the hyperbolic geometry H4, this was discovered by me. It tiles with 64 at a vertex, the dual being a tiling of bi-octagonal prisms, 288 at a vertex. What is interesting about this, is that it is a mirror-edge/mirror-margin dual, where neither of the tilings are regular.

This group has the wythoff-frieze {8,3,4,3} as subgroups.

W

By the way, i rely more in intuition than mathematics. Were it not for the circle-drawing, i should never have ventured into hyperbolic geometry.

Here's another cool thing for you.

{3,3} d2 = 1.5 * {3} d2 = 1.33333 product = 2 {3,3,4}
{3,4} d2 = 2 * {4} d2 = 2.00000 product = 4 {3,4,3}
{4,3} d2 = 3 * {3} d2 = 1.33333 product = 4 {3,4,3}
{3,5} d2 = 3.618033 {5} d2 = 2.894427191 product = 10.472136
{5,3} d2 = 7.854102 {3} d2 = 1.333333333 product = 10.472136
10.472136 is d2 of {3,3,5}.

What i have done here is multiplied the diameter square of the regular
polyhedra (in 3d), by the diameters of their vertex-figures. The product
is the same throughout, and correspond to the diameters of figures
with 8, 24, and 120 vertices.

This is no particular accident, and from what i guage, is never seen in higher dimensions again.
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Postby quickfur » Fri Aug 05, 2005 4:45 pm

Alright, here's a diagram of the projection of the octagonny into 2D:
Image
It is interesting to note that the area of each of the smaller 8 octagons is exactly half the area of the central octagon.

As you can see, there are lots of octagonal things going on here... the entire projection has an octagonal envelope, there is a central octagon, and there are 8 surrounding octagons. There are also 2 sets of 8 triangles arranged in octagonal formation.
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