Construction algorithm of 120 and 600 cell?

[Secret]

Construction algorithm of:
n simplex: Take a n-1 dimensional 'base' , locate it's center, and pull the center into a new othrogonal direction
n hypercube: Take a n-1 dimensional 'base' , translate the 'base' in a new othrogonal direction
n cross polytope: Take a n-1 dimensional 'base', locate its center, and pull the center into a pair of opposite new othrogonal directions
n hypersphere: take a n-1 dimensional object, locate its center, produce a n-2 dimensional axis which passes through the center and coplanar with the n-1 dimensional object, revolve the n-1 object around the n-2 axis

Question:
What about,
24 cell?
dodecaplexes?
isosacherdron and it's higher dimension anologues?

[wendy]

The 24ch has a construction of o3x3o4o, 'CO', or o3m3o4o 'double-tesseract'

The first involves truncating the 16ch until the vertices fall at the midpoints of the edges thereof.

The second involves creating pyramids on the faces of the tesseract, until the sloping faces of the pyramids fall by pairs.

The 600ch, has gosset's construction, by means of

a. Take a 24ch, its vertices belong to three 16ch. One can colour these eg red,blue,green. Each octahedral face then has three diagonals, red,blue,green. One then can set arrows on the edges in the direction red->blue, blue->green, green->red. This is 'coherent indexing', and can be done for any xPoQoRo.., where Q is even.

b. Divide the edges into the ratio of 0.618, 0.382 (the same for all). This gives for the octahedron, a s3s4o, or icosahedron. New faces form, specifically an always regular tetrahedron, (at the original vertices), and four extra triangular pyramids. When the icosahedron becomes regular, so do the triangular pyramids, giving a figure with 1.00 tetrahedra, and 24 icosahedra. This is s3s4o3o, or snub 24choron.

c. By raising pyramids on the icosahedra, one replaces the 24 icosa with 24*20 tetrahedra, or 4.00 further pyramids, or together 500 tetrahedral faces.

The twelftychoron or x5o3o3o, comes by replacing the tetrahedra in c by points at the centre, giving a 500 vertices, 1000 edges, 600 pentagons, and 100 dodecahedra.

[quickfur]

One simple, "intuitive" way of constructing a 24-cell is to take two tesseracts, cut one into 8 cubical pyramids (4D pyramids with cubical bases), and stick the pyramids onto the facets of the other tesseract. The square pyramid cells of the cubical pyramids will merge with square pyramids from adjacent cubical pyramids into octahedra. Since each cubical pyramid has 6 square pyramid cells, and there are 8 cubical pyramids, and since the square pyramids form pairs, you get 6*8/2 = 24 octahedra.

This construction is analogous to the way you can take two cubes in 3D, cut one into square pyramids and attach them to the faces of the other, and form a rhombic dodecahedron.

[quickfur]

The 120-cell and 600-cell are a bit more difficult to construct in an "intuitive" sorta way.

Perhaps the easiest way to visualize the 120-cell is by "layers". Start with one dodecahedron (first layer), stick 12 dodecahedra around it and fold it into 4D until they touch each other (second layer). This forms a kind of 4D bowl shape with depressions into which you can fit another 20 dodecahedra (corresponding to the corners of the original dodecahedron), the 3rd layer. This 3rd layer has 12 depressions which can fit another 12 dodecahedra (the 3.5th layer, 'cos the 12 dodecahedra in this layer are not attached to each other, or you can simply consider them as part of the 3rd layer).

Now take two copies of the 3 layers that you've made so far, which form two hemi-hyperspherical bowl shapes, and join them together at the far ends of the 3.5th layer's 12 dodecahedra. You now have almost the entire 120-cell, except that the joining left some gaps (the 3rd layers of each bowl don't touch each other). These gaps are filled in by another 30 dodecahedra, corresponding to the edges of the original dodecahedron.

This gives you (1 + 12 + 20 + 12)*2 + 30 = 120 dodecahedra.

It's really hard to describe this without pictures, so I'll have to shamelessly point you to my website, which explains this structure in detail:

http://eusebeia.dyndns.org/4d/120-cell.html

Here's a picture that shows you the first layer (blue), the second layer (green), and the 3rd and 3.5th layers (red). The 30 "equatorial" cells are not shown here because this is a perspective projection, and they are hidden from the 4D viewpoint.

[quickfur]

As for the 600-cell, you can use the same layers method to visualize it. Start with a single point, call it the North Pole, and then take 20 tetrahedra and attach them to this point in the geometry of an icosahedron. In 3D, this leaves some tiny gaps between them; this gives you a bit of room to fold them slightly into 4D so that the gaps close up, and you get a sorta 4D "floret" with an icosahedral boundary. This is the 1st layer.

Then you stick on another 20 tetrahedra to this icosahedral floret, which gives you a spiky bowl which you can fit 30 tetrahedra into (corresponding to the edges of the icosahedron). This gives you a sorta 4D bowl with 12 pentagonal gaps; each gap can take 5 tetrahedra. You may consider this as the "2nd layer", in a sense. The "rim" of this bowl is now a pentakis dodecahedron (a dodecahedron with 12 pentagonal pyramids stuck to its faces).

From here, it gets really hard to describe the next layer in words... you take 30 pairs of tetrahedra and stick them on so that they straddle the edges where the pentagonal pyramids touch. You then get grooves radiating from the tips of the pentagonal pyramids, going between each of these pairs. Since there are 5 such grooves per pentagonal pyramid, and there are 12 pyramids, you stick on another 5*12 = 60 tetrahedra to fill in these grooves. Now you end up with a 4D bowl where the "rim" has 20 conspicuous tetrahedron-shaped gaps, so you fill in the 20 tetrahedra to complete the 3rd layer.

Now take two copies of these 3 layers, and join them together where you filled in the last 20 tetrahedra. However, just as with the 120-cell, the other 3rd layer cells from each bowl don't join up with each other; they leave 12 gaps shaped like flat pentagonal bipyramids: each of these gaps are filled up by 5 tetrahedra -- so that's another 60 tetrahedra to fill in the "equator".

If you add everything up, you'll find that there are 600 cells.

And yes, most of this probably went right over your head because this is really hard to describe in words, so I'll just have to shamelessly point out my website again, where you get to see the pictures that hopefully are worth 1000 words:

http://eusebeia.dyndns.org/4d/600-cell.html

Now, mind you, there are other, less complex ways to derive the 600-cell, but they involve geometric operations that aren't easy to visualize, so I personally didn't find it very helpful until after the fact. One way is to take a 24-cell, divide its edges in the Golden Ratio in such a way that cutting the 24-cell down to those edges gives you regular icosahedra. (See, I told you it wasn't easy to visualize.) This process introduces a bunch of new tetrahedral cells, so you end up with something called the "snub 24-cell", which has 24 icosahedral cells and a whole bunch of tetrahedra joining them (120 of them, to be precise). Now take 480 tetrahedra, and form them into 24 icosahedral pyramids (just like the 1st layer I described above), and stick these pyramids onto the icosahedral cells of the snub 24-cell. Bingo! You get the 600-cell.

Yeah, that's not really that much easier to understand.

The best way to get the 600-cell is just to take the dual of the 120-cell: take a 120-cell, find the center of all 120 dodecahedra, discard the dodecahedra but keep the centers, now wrap some 4D shrink-wrap around all 120 points and shrink-wrap them. Your 4D shrinkwrap is now in the shape of a 600-cell, brand new, shrinkwrapped.

[wendy]

For the twelftychora, simply take six bi-decagonal prisms, and arrange them such that the squares fall by fives in the 120 faces.

[Hodge8]

Gosset's construction for 600 cell.
Wendy, you said:

New faces form, specifically an always regular tetrahedron, (at the original vertices), and four extra triangular pyramids.

Now, I recognise Gosset's construction for the 600 cell from Coxeter's "Regular Polytopes". P153 my edition, he says:

Let us place such a pyramid on each icosahedron of s{3,4,3}. The effect is to replace each icosahedron by a cluster of twenty tetrahedra.

I think I have a reasonable ability to visuaize polytopes, but I am having great difficulty to vizualize this particular part of the construction. I can see that an icosahedron could be replaced by 20 tetrahedra in 4D - merely connect the vertices of the icosahedron to the centre, and I assume that in 4D the twenty so formed tetrahedra are regular (as opposed to 3D where they are irregular). But why do you need the pyramid ? Is it only to fill in the gaps ?
I would appreciate a little more explanation.

[quickfur]

These are 4D pyramids with an icosahedron-shaped base, and 20 tetrahedra connecting the base to the apex. It's a rather shallow pyramid because regular tetrahedra almost tile the interior of the icosahedron. Once these pyramids are fitted onto the snub 24-cell, the icosahedral base lies in the interior of the polytope, leaving only the 20 tetrahedra as part of the surface.

This is, of course, equivalent to just adding 20 tetrahedra to each icosahedral cell of the snub 24-cell, because if the tetrahedra are regular, they will not lie flat on the icosahedral facet but will jut upwards slightly, thus the icosahedral cell becomes the base of a 4D pyramid in which the sides are the 20 tetrahedra. The base, of course, lies in the interior of the resulting 600-cell.

It's important to understand that when dealing with 4D objects, one should think in terms of volumes (3D hypersurfaces) rather than 2D faces. Adding 20 tetrahedra to an icosahedral cell is NOT putting them "inside" the icosahedron, because regular tetrahedra do not fit inside an icosahedron. There is 4D depth involved here; the 20 tetrahedra do not lie inside the icosahedron at all, but rather rise above it at a slight incline in the 4th direction. That's why we speak of pyramids: the situation is analogous to erecting 5 equilateral triangles over a regular pentagon: the triangles do not lie inside the pentagon, but rise above it in the 3rd direction, so that one may consider them to be a pentagonal pyramid (since their base is in the shape of a pyramid).

It is inaccurate to think that the 20 tetrahedra lie on the "inside" of the icosahedral cell; in fact, such would be the case if the tetrahedra were not regular, but slightly flattened. In that case, they would actually lie flat on the icosahedral hypersurface of the snub 24-cell, but you would not get a 600-cell out of that, merely the same snub 24-cell except with each icosahedral cell tiled by 20 (non-regular) tetrahedra lying flat on it.

[Hodge8]

quickfur, thank you for your detailed comments, which have helped me to think about the construction from a different point of view.

[Klitzing]

Thought that the following question might be placed best within this thread.

Infact it kind of asks for a specific subsymmetry of .3.3.5., esp. of "ex":

• Is it possible to globally combine the 600 tets (tetrahedra) of ex (hexacosachoron) into rosettes of 5 each, in such a way, that any tet would be used and OTOH the rosettes would not have overlaps?
• Could that be done in a symmetrical way (as well / only)?
• What would be the according (sub)symmetry of .3.3.5.?
  I.e. how are those to be omitted central edges of those rosettes to be selected?
• Or, otherwise, is there a proof that such a selection cannot be done symmetrically?

If that could be done systematically, this then would provide a quite nice non-convex regular faced polychoron with 240 peppy cells (pentagonal pyramids) only! The dihedral angle between these at the triangles should be the same as that of 2 neighbouring ikes (icosahedra) in an according ex diminishing. Therefore that one would be 120 degrees. But as the here introduced pentagons are burried underneath the to be omitted central edges of these rosettes, the according dihedral angle there should be non-convex.

Any help onto that idea?

--- rk

[wendy]

It is of course, possible to divide the x3o3o5o in this manner.

You note that for example, that the tetrahedron has six edges, and 6*120 = 720.

You construct a set of twelve clifford-parallel decagons. These would produce in the GAP, the two removed decagons (which make the pentagonal antiprisms), and two sets of five that swirl around the triangles in the paps. When you put the extra two back, you have twelve decagons, each of which with five pentagons at each edge, give the 120 vertices and 600 tetrahedra of the {3,3,5}.

The 120 pentagons that form orthogonal to the edges here are the 120 edges of a 5(3)5.

So it has an overall symmetry of 600, which is 1/24 of the {3,3,5}, of which the tetrahedra by fives or the pentagon 120 make the poincare-dodecahedron. The pentagonal-tegums are alternate cells to the same symmetry that leads to the cells of {5,3,3} making the poincare-dodecahedra.

I showed this to JHC by random discussion.

[Klitzing]

Silly me, that non-convex object with its 240 pentagonal pyramids within the strut of the hexacosachoron, which nonetheless has its full surface exposed, already had been discovered in "the early mid-2000's", as is stated on Jonathan's own website, cf. the page on swirlprismatic scaliforms. He there also displays the according vertex figure, which indeed goes completely conform with your description, Wendy.

Within the "polychoron list-archive" I did not find a corresponding post of Dinogeorge, but at least in Jonathan's listing of scaliforms from end of december 2000 it was already contained. So I assume that it just had been a private note from George to Jonathan, which he refers.

Therefore, having digged out both these references, I should have known before; I simply had completely forgotten about it - and now had "rediscovered" it 16 years thereafter.

Nota bene, that figure not only is scaliform, it even happens to be a noble polychoron, as it uses not only a single cell type but a single vertex type too.

--- rk

[quickfur]

This makes me wonder, is it possible to derive a convex CRF from this polytope via Stott expansion? Can it be done while preserving the pentagonal pyramid cells?

[Klitzing]

Stott expansion always preserves the orientation of the cells or at least their faces, at most it just pulls them somehow apart (kind of "explode") and put new cells inbetween. Therefore a convex dihedral angle phi either remains the same (nothing inserted inbetween) or becomes phi/2+90 deg. (prism inserted). Note that this value then equates to (phi+180 deg.)/2, i.e. it becomes interpretable as the medium of itself and the flat angle (180 deg.). The same holds true for concave angles. Either phi remains as is, or it becomes the medium with the flat angle. Therefore convex angles always give rise to convex ones, while concave ones give rise to concave ones only.

--- rk

[quickfur]

Ah, I see what you mean. You're right, concave angles will never become convex by mere Stott expansion.

[Klitzing]

You construct a set of twelve clifford-parallel decagons. These would produce in the GAP, the two removed decagons (which make the pentagonal antiprisms), and two sets of five that swirl around the triangles in the paps. When you put the extra two back, you have twelve decagons, each of which with five pentagons at each edge, give the 120 vertices and 600 tetrahedra of the {3,3,5}.

Even so I know how to construct gap, e.g. cf. its description here, and thus ought understand those 12 Clifford parallel decagons (fibres of the Hopf fibration), I still struggle with those in a way.

We have 12 decagons. And 12 x 10 = 120. Thus those would make a full dissection of the vertex set of the 600-cell into according subsets. The number 12 relates to the vertices of an ike, I suppose. But then, the mere cross-product of an ike and a decagon would be 5 dimensional. So that is it not.

If we chop off one vertex of the 600-cell, we get there an ike. When assigning the decagons to that ike, we would get some assymmetry, for a single such decagon would have to run through that chopped off vertex, while the others should not. And even if it is not that unit ike, but the f-scaled ike, occuring 2 vertex layers deeper, the very same argument applies.

So, could anyone bring some light in that mess, please?

--- rk

[wendy]

The vertex figure of {3,3,5} is of course an icosahedron. When you take to account that there is a great circle running through here, two of the pentagons are a pap or pentagonal antiprism. The remaining five appear as f-chords, which go from the top vertex of a pap, around 3/10 or 108 degrees of the circle to the base of the pap.

If you consider the lacing-edges of the pap in xo5ox&#x, it forms a zigzag decagon. If you take alternate edges of this, you will see there is an 'f' diagonal perpendicular to these, this makes the five pentagons that are incident on the vertex.

[Klitzing]

This relation to pap was already known so far. It is already outlined in the given link, and also can be obtained from the vertex figure of spysp. But that only relates one decagon to its 5 neighbouring ones. - My question rather was onto the global distribution of all 12 decagons of the fibration, esp. how that one could be obtained in the most local way.
--- rk

[wendy]

If you start with a column of 10 paps, this forms the vertices directly connected to a set in the centre of each pap.

So a 'string' is not just only the ten girthing edges, but rather includes ten pentagons, in alternation x5o then o5x. The outside belongs to two sets of spirals of length 10. You can generate these locally with the right-hand rule. As you move along the edge of the decagon by 36 degrees, the pentagon turns by 36 degrees from x5o to o5x (by clock-face, to add 1 from 02468 to 13579. So as you progress up the tower there is a set of spirals that are eg 0123456789, 2345678901, &c.

Each of the edges eg from 0 to 1, 'owns' two triangles on each side, this is part of a pentagonal tegum. We see that these do not intersect, and lie completely between pairs of pentagons in the tower. In essence, this shows that these spirals are also straight in E3, ie girthing decagons. Looking at this, we see the central column is a right-hand spiral around any of the outer rings.

We see that colour-wise, each ring of 10, in right turning, represents a vertex surrounded by five. This means that the twelve sets of ten have the same colour-relation as the twelve vertices of an icosahedron. The actual connection to an icosahedron is a little subtle, but it's not a direct product.

CE2

To fully grasp the idea, we turn to CE2, which is the euclidean two-space, with complex lines. The lines are actually argand diagrams. The plot of a line through a point 0,0 is y=ax.

We look at a number w, which is exp ivt, where i = sqrt(-1), v is a velocity in angle/time, and t is time. If we multiply both sides by w, then we get yw = a.wx, we notice that this sweeps the whole argand diagram in x and y, but the slope of the line does not change. This is a swirlybob.

Since all of these are complex numbers, then a directly corresponds to the 'slope' of the line, is also complex. We take the argand plane as x,y,0, and put a sphere whose diameter is 0,0,0 to 0,0,1. We draw a ray from some x,y,0 to 0,0,1, and note where it intersects the sphere. The points directly antipedal on this sphere represents planes that are perfectly perpendicular.

The points on this sphere, directly correspond to a point on the argand diagram, which in turn, directly correspond to individual great circles on a clifford parallel.

What happens is that the twelve sets of great circles in a clifford set, directly corresponds to the twelve vertices of the icosahedron on this sphere.