## polytopes with catalan hulls

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

### polytopes with catalan hulls

I'm making a video about the catalan solids, and I wanted to mention some of the higher dimensional shapes they relate to as convex hulls.

For example, how a tesseract looks like a rhombic dodecahedron when viewed vertex first.

I think a 5-cube or a 6-cube looks like a rhombic triacontahedron, but I'm not sure which one..

Also, I found this:

"A 24-cell viewed under a vertex-first perspective projection has a surface topology
of a tetrakis hexahedron and the geometric proportions of the rhombic dodecahedron,
with the rhombic faces divided into two triangles."

does this mean that a 24-cell looks like a tetrakis hexahedron, or it is just related to it topologically?
(I can't quite tell from looking at one.)

Are there any other interesting connections?

Dionian

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Joined: Sat Jan 19, 2008 12:11 pm

### Re: polytopes with catalan hulls

The general uniform polytope is bounded by faces that are an intersection of primary catalans (ie the ones with one mirror-node). In effect, the figure is an intersection of several primary catalans. One finds the face-nodes, by removing the nodes in the dynkin-symbol, one at a time, and see if the remaining set is still connected to marked nodes. (eg x--x--o removing 1, 3 still leaves bits connected to x-nodes, but 2 leaves x o, one of which has no connection).

The figures like rhombo-CO, and rhombo-truncated CO have faces formed by rhombic faces.

For example, the primary catalans for the octahedral group are cube, rhombo-dodecahedron, octahedron. Taking the cube and octahedron, one can start from a miniture octahedron, and make it become large as the common cube becomes smaller. One looks at the intersection: it starts as a octahedron, which first is truncated to tO, then to cuboctahedron, then to truncated cube, then to cube (as the octahedron goes large and the cube goes small). This sequence can be read in four dimensions as the cross sections of the cube (or octahedral) anti-tegum.

The dual of the anti-tegum series can be visualised by supposing that there is a skin covering the two members, but held tight. We start with a cube. As the octahedron starts to touch the face, the faces of the cube are pushed outwards into low pyramids (tetrikis cube = fourfold cube). Eventually, the pyramids become high enough that they join by pairs, to form a figure with rhombic faces [diagonals = edges of cube and octahedron]. After this point, the faces split along the edges of the octahedron, giving trikis octahedron (thrice octahedron), and these pyramids eventually fall to zero height.

In higher dimensions, the primary catalans are the surtegmic (surface + tegum) figures.

The reason for the equation of the topological form of 24ch to the tetrakis cube (rather than the geometric shape = rhombo-dodecahedron), is that some edges appear as the short diagonals of the rhombus. The rhombi are in fact the edge-first presentation of the octahedron.

The face-first presentation of the 120ch gives a tricontahedron, where the 30 faces are replaced by the edge-first presentation of a dodecahedron, the figure is intersected by a dodecahedron of unit edge (giving the pleat margin of two dodecahedra, whose margin appears full on.

eg PG margin, bevel
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

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### Re: polytopes with catalan hulls

Thanks, Wendy. After re-reading this a couple times it now makes a lot of sense.

The skin analogy helps a lot!

So if a rhombic dodecahedron can be the outer hull of a hypercube, does this mean that we can have a rhombic dodec in 4-d by itself that is made out of square faces? Would this be a skewed?

Dionian

Posts: 27
Joined: Sat Jan 19, 2008 12:11 pm

### Re: polytopes with catalan hulls

One should remember that in 3d, the corresponding projection of the cube and rhombic dodecahedron is a hexagon. You can have a skew hexagon on both of these figures, but i can't see what it makes.

When you look at the rhombic dodecahedron projection of the tesseract in 3d, ye see that the 12 bounding rhombohedra are indeed the twelve faces of the four cubes, that do not contain the common vertex of these four curves (eg the twelve faces not containing the point (1,1,1,1).

It works in higher dimensions: but the rhombic dodecahedron is both m3o3m (the dual of the runcinated simplex), and o3m4o (double-cube = dual of rectified cross). In four dimensions, the m3o3o3m (20chora) and o3m3o4o (double-tesseract = surtegmated measure-polytope), are different: this remains so for higher dimesnions.
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
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Location: Brisbane, Australia

### Re: polytopes with catalan hulls

Oh man!! That is so awesome!

Does the operation that gives you the polytope half-way between duals change as you go into higher and higher dimensions?

For example, from what I just read, a bitruncated tesseract gives you a shape half way between the tesseract and its dual (the 16-cell). (and so a bitruncated 16-cell is the same shape)..
If you bitruncate a polyteron, is it the same type of thing?

(I'm excited now!)

Dionian

Posts: 27
Joined: Sat Jan 19, 2008 12:11 pm

### Re: polytopes with catalan hulls

You should understand that the dynkin symbol is not so much just a handy way of writing some polytopes, but an actual coordinate system, no less than the usual x,y,z system. A figure like x3o3x corresponds to 1,0,1 (all changes of sign). The big difference here is that instead of 8 changes of sign, ye can have 24 or 120 or whatever changes of sign.

It then becomes possible to think of a shape that is 20% cube and 80% cuboctahedron or whatever. This is what makes "progressions" tick. You can even have something that's 30% C + 50% CO + 20% O. It's a kind of truncated-cuboctahedron, but the three kinds of edges are not equal.

The standard coordinates for a cube is (1,1,1) ACHS. This gives a edge of 2 (eg 1,1,1 to 1,1,-1). For a dynkin symbol, the standard is to make the edges 2, which makes the plane v=0 to v=1 differ by unit-thickness. Since 0,0,0 to 1,0,0 is a sloping line, this distance gives no direct relation to the thickness.

The relevant axies for the octahedral group are: [q=sqrt(2)] O = (q,0,0), CO = (q,q,0) and C = (1,1,1). The effect of symmetry is all sign, all permutations, applied after the sum. One can see that the tC is o3x3x = 0,1,1 = CO + C = (q+1,1,1). Of course, you can go entirely gradually, so that one goes from C to CO smoothly. The trick here is that it's a progression of x* C + y*z*CO, where x+y = 1, and z is some free scale-factor. Going from the CO to the O, requires some fiddling, but the same thing occurs.

You can of course progress any figure to any other, since at x% of the change you have x% (TOP) + y% (BOTTOM), where x+y=100.

The class of figures formed by this form of progression are 'lace prisms', and their duals are 'lace tegums'. In 3D, you can have T, B as any of (point, figure, dual, truncate). The relevant lace-prisms are point-figure = pyramid, figure-figure = prism, figure-dual = antiprism, figure-truncate = cupola. Of course, if T and B can be different sizes, you can have figure-big figure = rostrum.

For example, the progression of a dodecahedron, pentagon first is:

x5o [f] f5o [1] o5f [f] o5x. f = 1.61803398875&c.

This is a pentagonal (rostrum + antiprism + rostrum), which can be seen by drawing in the three layers. An icosahedron goes pentagonal (pyramid + antiprism + pyramid). Once you get the different heights right (here shown [f], [1]), it's quite easy to draw something like a pentagon of size 1.309017 for a distance of 0.309017 of the total height of 4.236067.

[we note that 0.309017 is half of f, so the figure is 50% (x5o) + 50% (f5o), or (0.5+0.5f, 0) on the pentagonal coordinates. This is a pentagon of that size (ie 2.618 / 2 = 1.309017).]

One can calculate the kinds of faces of any lace prism, because all the sloping faces are lace-prisms. You apply the same rule as for ordinary dynkin-symbol, but match out the corresponding nodes top and bottom, eg

x3o5x || x3x5o
x3o5x || ----- top = rhombo-ID
------- "|| x3x5o bottom = trunc icosa
* o5x || * x5o anti-prisms (12)
x * x || x * o triangular prism (30), literally rectangle on-top line
x3o * || x3x * triangular cupola (triangle-hexagon of cuboctahedron)

This is the usual "cover a node to reveal a face".
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
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Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: polytopes with catalan hulls

Wonderful.

I'm just now learning the dynkin symbols so I haven't processed all of this just yet, but it sounds promising.