Yup. That’s what I was thinking. You don’t need to warn better, you just need to get readers who pay closer attention. :-)

]]>I think I labelled them correctly according to the conventions I’m describing in this post. For example, here:

the triangle labelled “VE” is obtained from the triangle labelled “V” by reflecting it across the great circle containing the edge labelled “E”. Then the triangle labelled “VEV” is obtained from the triangle labelled “VE” by reflecting it across the great circle containing the edge labelled “V”. And so on.

You may be confused because these conventions are *different* than the conventions I used in Part 2. There, instead of repeatedly reflecting a triangle across fixed great circles, I kept flipping it over one of its edges, each time getting a new triangle that shared one edge with the previous triangle:

Both these approaches are useful; they’re different, but related in an interesting way. I remember Greg Egan getting confused when I used the first approach when he was expecting the second one… and now I’m thinking maybe you’re confused because I’m using the second approach when you were expecting the first one. Is that right?

Clearly I should have done a better job warning everyone that I was talking about things a new way this time! And someday I should explain how the two ways are related (though Greg figured it out).

]]>http://en.wikipedia.org/wiki/Uniform_tiling

http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane

Plus 3D “Honeycombs”, filling space by polyhedra, created from 4-mirror kaleidoscopes:

http://en.wikipedia.org/wiki/Convex_uniform_honeycomb

http://en.wikipedia.org/wiki/Uniform_honeycombs_in_hyperbolic_space

http://en.wikipedia.org/wiki/Paracompact_uniform_honeycombs

And on the hyperbolic front, Italian wikipedian Claudio Rocchini has helped with lots of graphics, including the hyperbolic regular hyperbolic honeycomb {6,3,3}, 3 hexagonal tilings wrapped around an edge, never rendered before as best I know!

http://en.wikipedia.org/wiki/User:Rocchini

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Wow, this is great! My ultimate goal is to go through all the polytopes coming from 4-dot Dynkin diagrams, and say a lot of fun things about them. I hadn’t noticed these relations between the B_{4} and F_{4} series! But now I plan to talk about them when I reach this point.

The 24-cell is one of the wonders of nature, so it’s especially nice to see it partaking in these relations.

It’s also great to see how the 3d relations you noticed earlier are a kind of spinoff of these 4d ones.

Fun, fun, fun…

]]>Thank you for pointing out the counting argument. Makes perfect sense why it’s possible in 3D but not in 4D.

I don’t want to distract folks from the path you’re on with 4D rotations and such. But I think I figured out why this works (in 3D). And found a similar pattern in 4D (though not the one I was expecting).

It seems that this works because an octahedron happens to be the same thing as a rectified tetrahedron. Once you have established that a parent X = a rectified Y, then a truncated X = a cantitruncated Y, and a rectified X = a cantellated Y. Which is (in words) the above pattern.

This is more-or-less obvious from:

http://en.wikipedia.org/wiki/Uniform_polytope

And in particular if you stare long enough at:

The (unexpected) pattern in 4D is between the B_{4} and F_{4} families (the latter being the 24 cell family, unique to 4D):

**o—•—o==o** → **•—o==o—o**

rectified 16 cell → 24 cell

**•—•—•==o** → **•—•==o—o**

cantitruncated 16 cell → truncated 24 cell

**•—o—•==o** → **o—•==o—o**

cantellated 16 cell → rectified 24 cell

(In the interests of sloth, I’m abusing notation somewhat, borrowing from undirected Dynkin diagrams: a double line to denote -4-, and a single line to denote -3-.)

Just as with an octahedron and a rectified tetrahedron, this depends on a 24 cell being the same thing as a rectified 16 cell. Which I guess is just a coincidence.

Though it’s interesting that it carries over so directly. A 16 cell is, after all, a 4D octahedron. And if you just drop the final —o from the above, and you get that same 3D pattern.

]]>Thanks for prodding me to think harder about the relation between the A_{3} family and B_{3} family. It will take me a while to fully work it out, but it will be fun.

Mathematically, the relation starts because the symmetry group of the cube contains the symmetry group of the tetrahedron. The symmetry group of the tetrahedron has 4! = 24 elements since you can permute its vertices any way you like. The symmetry group of the cube has 2^{3} × 3! = 48 elements since you can reflect it along any 3 of the coordinate axes and also permute these axes any way you like. 48 is a multiple of 24, so there’s an *opportunity* for the symmetry group of the cube to contain the symmetry group of the tetrahedron. Nature exploits this opportunity. You can see it here:

where we fit two tetrahedra in a cube, getting a shape Kepler called the stella octangula:

This magic doesn’t happen in higher dimensions! In n dimensions the symmetry group of the n-simplex (the analogue of a tetrahedron) has (n+1)! elements. The symmetry group of the n-cube has n! × 2^{n} elements. And I think the former is never a subgroup of the latter. I guess we can see this by a simple divisibility argument.

For example, in 4 dimensions we get 120 elements in the first group, 24 × 16 = 384 in the second. Since 384 isn’t a multiple of 120, the symmetry group of the 4-cube doesn’t contain the symmetry group of the 4-simplex.

The first group is called the A_{n} Coxeter group; the second is called the B_{n} Coxeter group.

So, I think we’re lucky in 3 dimensions.

]]>If you start with the symmetrical Coxeter diagrams for the A_{3} family, and “shift right”, you get the Coxeter diagrams for the B_{3} family with a “leading zero” (white dot in the V position) representing the same polyhedron:

**•—3—o—3—•** → **o—4—•—3—o** cuboctahedron

** •—3—•—3—•** → **o—4—•—3—•** truncated octahedron

**o—3—•—3—o** → ** o—4—o—3—•** octahedron

The other (asymmetrical) A_{3} diagrams (for tetrahedron and truncated tetrahedron) pair within the A_{3} family.

So all members of the B_{3} family with a white dot in the V position pair with a member of the A_{3} family. And all members of the A_{3} family pair with something. And these seem to cover all the duplications of polyhedra in these families.

I wonder if something like this pattern extends to higher dimensions…

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