Isogonal swirl polychora

[Mercurial, the Spectre]

[...]
How about the alternated truncated octahedral prism (aka the pyritohedral icosahedral antiprism)? It's not uniform. Coordinates are:
(0, ±1/2, ±0.80901699437494742410, 0.35355339059327376220) (all even permutations for the first three coordinates, for a total of 12 vertices)
(±1/2, 0, ±0.80901699437494742410, -0.35355339059327376220) (same as above for the first three coordinates)
And also its dual.

Here's the same projection in a different orientation:



I'm not sure what to make of this; it looks like the prism of an irregular hexagonal prism. Is there a specific 4D viewpoint you have in mind?

Anyway, here's the dual:



Not sure what to make of this either. It does look like a pretty little diamond shape, though.

Again, if you have a specific 4D viewpoint in mind, do let me know.

Wait, that's not the shape I had in mind. This shape is actually a distorted 4-6 duoprism and its dual, a 4-6 duopyramid. Sorry.

The coordinates of the alternated truncated octahedral prism are:
(0, ±1/2, ±0.80901699437494742410, 0.35355339059327376220) (4)
(±0.80901699437494742410, 0, ±1/2, 0.35355339059327376220) (4)
(±1/2, ±0.80901699437494742410, 0, 0.35355339059327376220) (4)
They define a regular icosahedron (alternated truncated octahedron) of length 1.

(±1/2, 0, ±0.80901699437494742410, -0.35355339059327376220) (4)
(±0.80901699437494742410, ±1/2, 0, -0.35355339059327376220) (4)
(0, ±0.80901699437494742410, ±1/2, -0.35355339059327376220) (4)
They define another regular icosahedron (alternated truncated octahedron) of length 1, this time antialigned like how a compound of two icosahedra would result in a nonuniform truncated octahedron.

[quickfur]

Aaaahhh, I'm such an idiot!!! I mistyped the permutation operator (apecs instead of epacs: so it was giving me all permutations of coordinate and even changes of sign, instead of the intended even permutations of coordinate and all changes of sign).

Here's the rendering of the real polytope originally requested:



Now, this viewpoint probably isn't what you intended , so here's an icosahedron-first projection:



To prevent obscuring the view, I changed the texture on the nearest cell to just an edge-outline instead.

And now here's the dual:



This dual has the most fascinating cell shape: a kind of long shard-like antibipyramid shape with a long apex and a short one. Like a gem. These cells are clustered around the two apices in dodecahedral arrangement. Very pretty!

Now, one thing that might not have been very obvious, is that this dual is actually very long, as can be seen from this side view:



The overall shape is also very gem-like. A truly fascinating polychoron!

[Mercurial, the Spectre]

To quickfur:
How about this figure and its dual: https://polytope.miraheze.org/wiki/6-cubic_swirlprism (the coordinates are in the page)

[quickfur]

Hmm, the coordinates are a bit ambiguous. When it says "reflections through the x=y and z=w hyperplanes", what order are w, x, y, z? For example, does <0, 0, √2/2, √2/2> generate <0, ±√2/2, 0, ±√2/2>, <±√2/2, 0, ±√2/2, 0>, etc., or are we swapping the 1st and 2nd coordinates and the 3rd and 4th coordinates?

[quickfur]

It's confusing because the first 4 coordinates in the second section seem to generate all permutations of coordinates, so I'm not sure why they are specifically listed only to undergo the x=y, w=z reflections.

[Mercurial, the Spectre]

Hmm, the coordinates are a bit ambiguous. When it says "reflections through the x=y and z=w hyperplanes", what order are w, x, y, z? For example, does <0, 0, √2/2, √2/2> generate <0, ±√2/2, 0, ±√2/2>, <±√2/2, 0, ±√2/2, 0>, etc., or are we swapping the 1st and 2nd coordinates and the 3rd and 4th coordinates?

Yes, for both of them. Say you have (x,y,z,w), then the symmetry implies (y,x,w,z) is equal. Also, you have to do "all" sign changes, not permutations.

[Klitzing]

in other words: Mercurial's "x=y and z=w reflections" thus equates into "permutations within first pair of coords as well as permutations within second pair of coords"

--- rk

[quickfur]

in other words: Mercurial's "x=y and z=w reflections" thus equates into "permutations within first pair of coords as well as permutations within second pair of coords"

--- rk

Does that mean <x,y,z,w> generates <y,x,z,w>, <y,x,w,z>, and <x,y,w,z>? Or just <y,x,w,z>?

[quickfur]

Something is not right with the coordinates as presented on the wiki page. I'm doing permutations of the first and second pairs of coordinates, and it's generating 208 vertices instead of the 144 stated on the page.

Also, why are <0, 0, √2/2, √2/2> listed in essentially all permutations? If we're swapping the first and second pairs of coordinates freely, then the second and third rows are redundant, and the first 4 rows taken together generate all permutations. Clearly, I'm not understanding something here, or the wiki page is inaccurate.

[Mercurial, the Spectre]

Something is not right with the coordinates as presented on the wiki page. I'm doing permutations of the first and second pairs of coordinates, and it's generating 208 vertices instead of the 144 stated on the page.

Also, why are <0, 0, √2/2, √2/2> listed in essentially all permutations? If we're swapping the first and second pairs of coordinates freely, then the second and third rows are redundant, and the first 4 rows taken together generate all permutations. Clearly, I'm not understanding something here, or the wiki page is inaccurate.

If you understand, then the coordinates <0, 0, √2/2, √2/2>, when transposed through x=y and z=w rotations, yield the same value, because both x-y and z-w are the same. But if you have <0, √2/2, 0, √2/2>, you have the x=y and z=w rotation yielding <√2/2, 0, √2/2, 0>.

So the coordinates of the swirlprism are:
(0,0,0,1) (all permutations and sign changes = 8)
(1/2,1/2,1/2,1/2) (all permutations and sign changes = 16)

The two coordinates are the same under x=y and z=w rotation symmetry.
(0, 0, sqrt(2)/2, sqrt(2)/2) (all sign changes = 4)
(sqrt(2)/2, sqrt(2)/2, 0, 0) (all sign changes = 4)

The next six pairs of coordinates are related through x=y and z=w rotation symmetry.
(0, sqrt(2)/2, 0, sqrt(2)/2) (all sign changes = 4)
(sqrt(2)/2, 0, sqrt(2)/2, 0) (all sign changes = 4)

(0, sqrt(2)/2, sqrt(2)/2, 0) (all sign changes = 4)
(sqrt(2)/2, 0, 0, sqrt(2)/2) (all sign changes = 4)

(0, 0, sqrt(2-sqrt(3))/2, sqrt(2+sqrt(3))/2) (all sign changes = 4)
(0, 0, sqrt(2+sqrt(3))/2, sqrt(2-sqrt(3))/2) (all sign changes = 4)

(sqrt(2-sqrt(3))/2, sqrt(2+sqrt(3))/2, 0, 0) (all sign changes = 4)
(sqrt(2+sqrt(3))/2, sqrt(2-sqrt(3))/2, 0, 0) (all sign changes = 4)

(0, 0, 1/2, sqrt(3)/2) (all sign changes = 4)
(0, 0, sqrt(3)/2, 1/2) (all sign changes = 4)

(1/2, sqrt(3)/2, 0, 0) (all sign changes = 4)
(sqrt(3)/2, 1/2, 0, 0) (all sign changes = 4)

The next four pairs of coordinates are again related through x=y and z=w rotation symmetry.
((sqrt(3)-1)/4, (sqrt(3)+1)/4, (sqrt(3)-1)/4, (sqrt(3)+1)/4) (all even sign changes = 8)
((sqrt(3)+1)/4, (sqrt(3)-1)/4, (sqrt(3)+1)/4, (sqrt(3)-1)/4) (all even sign changes = 8)

(sqrt(2)/4, sqrt(6)/4, sqrt(2)/4, sqrt(6)/4) (all even sign changes = 8)
(sqrt(6)/4, sqrt(2)/4, sqrt(6)/4, sqrt(2)/4) (all even sign changes = 8)

((sqrt(3)-1)/4, (sqrt(3)+1)/4, (sqrt(3)+1)/4, (sqrt(3)-1)/4) (all odd sign changes = 8)
((sqrt(3)+1)/4, (sqrt(3)-1)/4, (sqrt(3)-1)/4, (sqrt(3)+1)/4) (all odd sign changes = 8)

(sqrt(2)/4, sqrt(6)/4, sqrt(6)/4, sqrt(2)/4) (all odd sign changes = 8)
(sqrt(6)/4, sqrt(2)/4, sqrt(2)/4, sqrt(6)/4) (all odd sign changes = 8)

Anyway, here is the off file of the polychoron mentioned above:
https://cdn.discordapp.com/attachments/ ... rl_144.off

alongside the other two cubic swirlprisms:
https://cdn.discordapp.com/attachments/ ... irl_96.off
https://cdn.discordapp.com/attachments/ ... rl_120.off

You might want to check them and their duals.

[quickfur]

[...]
If you understand, then the coordinates <0, 0, √2/2, √2/2>, when transposed through x=y and z=w rotations, yield the same value, because both x-y and z-w are the same. But if you have <0, √2/2, 0, √2/2>, you have the x=y and z=w rotation yielding <√2/2, 0, √2/2, 0>.

Aha, see this was the part that was not clear. What you're saying is that you apply both x=y and z=w reflections together as a unit, not as all combinations (i.e., reflect x=y independently of reflect x=w, and close over all combinations). The way it's worded is not clear that this is what is meant.

Thanks, that makes it a lot clearer.

[...]
The two coordinates are the same under x=y and z=w rotation symmetry.
(0, 0, sqrt(2)/2, sqrt(2)/2) (all sign changes = 4)
(sqrt(2)/2, sqrt(2)/2, 0, 0) (all sign changes = 4)

The next six pairs of coordinates are related through x=y and z=w rotation symmetry.
(0, sqrt(2)/2, 0, sqrt(2)/2) (all sign changes = 4)
(sqrt(2)/2, 0, sqrt(2)/2, 0) (all sign changes = 4)

(0, sqrt(2)/2, sqrt(2)/2, 0) (all sign changes = 4)
(sqrt(2)/2, 0, 0, sqrt(2)/2) (all sign changes = 4)
[...]

The above is needlessly confusing, because it is exactly the same as all permutations and sign changes of (0, 0, sqrt(2)/2, sqrt(2)/2). It could have been written as one line, instead of 4 lines that looks like it's supposed to be somehow different from all permutations and sign changes (in contrast to the first two rows on the wiki page).

[quickfur]

Anyway, regardless of that, here's an initial render:



The colored cells above show two of the 8 rings of trigonal antiprisms in this pretty polychoron. The rings are antipodal to each other. I colored the cells with alternating colors by hand, in order to show the structure better. Very interesting indeed!

I left the tetrahedral cells unhighlighted in order to minimize clutter.

Note that the magenta/cyan ring is actually not visible from this 4D viewpoint (clipping is on, and this is using perspective projection), but the ridges that lie on the projection envelope is visible here. So you're not seeing the actual cells, just a subset of their ridges that happen to lie on the projection envelope. You can also see where the ring has a crossover (twist) on the left side of the image (some of the ridges exchange places with other ridges across the twist).

[Mercurial, the Spectre]

Anyway, regardless of that, here's an initial render:



The colored cells above show two of the 8 rings of trigonal antiprisms in this pretty polychoron. The rings are antipodal to each other. I colored the cells with alternating colors by hand, in order to show the structure better. Very interesting indeed!

I left the tetrahedral cells unhighlighted in order to minimize clutter.

Note that the magenta/cyan ring is actually not visible from this 4D viewpoint (clipping is on, and this is using perspective projection), but the ridges that lie on the projection envelope is visible here. So you're not seeing the actual cells, just a subset of their ridges that happen to lie on the projection envelope. You can also see where the ring has a crossover (twist) on the left side of the image (some of the ridges exchange places with other ridges across the twist).

So, you can apparently put vertices onto each of the 192 octahedra and take the resulting convex hull. The result should have 192 vertices forming eight 24-gons. Can you render it?

[quickfur]

And here's a rendering of the dual:



Again, these cells come in rings of 24 members each. Polyview didn't have a function for coloring alternating cells like this, so I had to copy-n-paste a little selection snippet 24 times to extract the list of cells in a ring, then have polyview output a graph of their ridge-connectivity and use GraphViz to visualize the ring and pick out every other member by hand.

Anyway, I didn't check, but looks like this is cell-transitive.

[...]
So, you can apparently put vertices onto each of the 192 octahedra and take the resulting convex hull. The result should have 192 vertices forming eight 24-gons. Can you render it?

OK, this was not trivial because polyview doesn't have an automated function for doing this. But thanks to shell-scripting shenanigans involving running polyview twice to extract the list of cell centroids, I managed to do it:

I type this:

Code: Select all

$ echo 'list cells with 6 v' | polyview -q swirlprism_6cubic.def | sed -e's/^cell \([0-9][0-9]*\).*/centroid(cell \1)/' | polyview -q -Pmax swirlprism_6cubic.def | makepoly centroids - centroids.def

And get this:

Code: Select all

Number type: double
Computing dual representation...done.
Polytope has 192 vertices, 432 facets
Creating definition of polytope 'centroids' in 'centroids.def'...
Writing out vertices... Done.
Writing out face lattice (may take a while)...
Done.
Finished.


That generated the desired polytope definition. And now the render:


Very similar to the dual, except that it's not cell-transitive because there are tetrahedral cells between the rings of antiprisms. Also, these are square antiprisms, whereas in the dual the cells in a ring are joined by octagons rather than squares, and have trapezoidal faces as well.

[Mercurial, the Spectre]

And here's a rendering of the dual:



Again, these cells come in rings of 24 members each. Polyview didn't have a function for coloring alternating cells like this, so I had to copy-n-paste a little selection snippet 24 times to extract the list of cells in a ring, then have polyview output a graph of their ridge-connectivity and use GraphViz to visualize the ring and pick out every other member by hand.

Anyway, I didn't check, but looks like this is cell-transitive.

[...]
So, you can apparently put vertices onto each of the 192 octahedra and take the resulting convex hull. The result should have 192 vertices forming eight 24-gons. Can you render it?

OK, this was not trivial because polyview doesn't have an automated function for doing this. But thanks to shell-scripting shenanigans involving running polyview twice to extract the list of cell centroids, I managed to do it:

I type this:

Code: Select all

$ echo 'list cells with 6 v' | polyview -q swirlprism_6cubic.def | sed -e's/^cell \([0-9][0-9]*\).*/centroid(cell \1)/' | polyview -q -Pmax swirlprism_6cubic.def | makepoly centroids - centroids.def

And get this:

Code: Select all

Number type: double
Computing dual representation...done.
Polytope has 192 vertices, 432 facets
Creating definition of polytope 'centroids' in 'centroids.def'...
Writing out vertices... Done.
Writing out face lattice (may take a while)...
Done.
Finished.


That generated the desired polytope definition. And now the render:


Very similar to the dual, except that it's not cell-transitive because there are tetrahedral cells between the rings of antiprisms. Also, these are square antiprisms, whereas in the dual the cells in a ring are joined by octagons rather than squares, and have trapezoidal faces as well.

What would be the dual of the last polychoron with 192 vertices? And what would be its cell type?

[quickfur]

Here's a render of the dual of the 192-vertex polychoron:



It's kinda hard to see, but the ring/twistor here has roughly triangular shape, even though the ridges between ring members are actually distorted hexagons. So it's a kind of distorted triangular prism whose bases are somewhat twisted and split into hexagons, but not quite to the extent of an antiprism, and it also has some rhomboid faces.

I didn't check, but I think it should be cell-transitive.

[Mercurial, the Spectre]

Another thing is, if you look at the rectified 24-cell, you can find an inscribed 24-cell of length sqrt(3). If you remove one of them, I think you should get an isogonal 72-vertex polychoron with tetrahedral swirl symmetry. Can you take a look at its elements?

Coordinates on this page: https://polytope.miraheze.org/wiki/Swir ... etrachoron (remember, the ± sign before the parentheses means the coordinate and its central inversion, such as ±(±1, 0.2, 0.3, 0.4) = (±1, 0.2, 0.3, 0.4) and (±1, -0.2, -0.3, -0.4)).