Sections of the 600-cell

Thanks to triality we can consistently depart from each vertex of a 24-cell along 4 of the 8 edges giving 4 new vertices for each vertex of the original 24-cell. If the moving points divide the edge in the Golden Ratio we can join the 96 points by new edges of equal length. By adding the 24 vertices of the dual 24-cell and joining these 24 vertices by 12 lines to neighbouring vertices gives a regular polytope with 600 cells. Blue vertices lie on the blue hypercube, green vertices on the green and red vertices on the red hypercube.

The spherical 24-cell with 96 vertices of the 600-cell on its edges.

A closer view omitting the 4 blue vertices of section 15₃ and 4 green vertices of section 14₃ of the 600-cell.

We have 32 red vertices, 32 blue vertices and 32 green vertices. The dual 24-cell with 24 orange vertices completes the number of 120 vertices of the 600-cell.

Vertices of the 600-cell on 2, mutually dual, spherical 24-cells.

A closer view of the vertices of the sections  1₃, 2₃, 3₃, 4₃ , 5₃ , 6₃  , 7₃ and  8₃ .

Around each vertex of the 24-cell originates a tetrahedron with 4 vertices giving 96 new vertices. The 6 orange points are 6 vertices of a cell of the dual 24-cell . Below is a (magnification of the) central (yellow) cell of the 600-cell that is surrounded by 4 brown cells in positions with tetrahedral symmetry and another 4 violet cells in tetrahedral symmetry. The yellow one originated around the central vertex of a 24-cell that is stereographically projected with the north pole at a vertex.

The 24 orange vertices are midpoints of 24 icosahedra. Of one icosahedron we draw 21 edges in color black in the next picture:

In the notation of the book Regular Polytopes by H.S.M.Coxeter we have the sections 1₃, 2₃, 3₃, 4₃, 5₃ in one picture. The blue vertices of the yellow tetrahedron belong to section 1₃, the 4 inner green vertices of the violet tetrahedra belong to section 2₃, the 12 red vertices belong to section 4₃ and the 4x3=12 outer green vertices of the violet tetrahedra belong to section 5₃. The 6 orange vertices (that are not connected to each other but are connected (though not drawn) by 12 edges to other vertices) belong to section 3₃.

In the next picture we first scale down the picture so we can add new layers around the central ones. Leaving out the 4 brown tetrahedra we complete this icosahedron in the color blue by adding section 6₃. Another 5 icosahedra can be added, resulting in 6 icosahedra lying in octahedral symmetry. There can be drawn 12 edges from each orange centre to the 12 vertices of each icosahedron but there are already enough lines so we omitted these edges. The brown tetrahedra share a triangle with an icosahedron and the violet tetrahedra share an edge with an icosahedron. The violet ones are thus of the same type as the central yellow one.

The central tetrahedron and 4 tetrahedra that share a triangular face with it. The blue edges are part of a icosahedron. Here section 1₃ has also blue vertices, but section 2₃ has red vertices. We use projective space now, where the great circles of the 3-sphere S3 become straight lines.

The sections 1₃, 2₃, 3₃ and 4₃ . Here section 3₃ has 2 red, 2 green and 2 blue vertices. Section 4₃ has 12 green vertices. You can see already 6 complete tetrahedral cells in the blue icosahedron and of course 5x6=30 in the other 5 icosahedra. With the 4 tetrahedra with 3 green vertices in section 4₃ we now have 1+4+36+4=45 tetrahedral cells.

The sections 1₃, 2₃, 3₃, 4₃ and  5₃ . Section 5₃ has 12 red vertices. We see 4 new tetrahedra with vertices in one color red. Also 4x3=12 tetrahedra sharing a face with such a cell (3 red and 1 green vertex). Also 4x3x2=24 tetrahedra (inside the icosahedra) sharing an edge with them. This adds to 40 new cells. A subtotal of 85 cells.

The sections 1₃, 2₃, 3₃, 4₃ , 5₃ and  6₃ . Section 6₃ has 12 blue vertices. The blue icosahedron is complete. And so goes for the other 5. This makes 6x10=60 new completed cells. A subtotal of 145 cells.

The sections 1₃, 2₃, 3₃, 4₃ , 5₃ , 6₃ and  7₃ . Now we changed to the spherical 600-cell in stereographic projection. The 4 green vertices of section  7₃ make the 4 tetrahedra with green vertices complete. Mark that the colors red and green are interchanged in the older pictures above and below. We get 4x3=12 cells that share a face with these 4 new cells. A subtotal of 161 cells.

The 4 violet tetrahedra can be completed to 8 that lie around the vertices of a red cube of the original 24-cell. The green vertices belong to section 2₃ and 5₃. The red vertices belong to section 4₃ and 7₃. The color of the vertices indicates that they lie on red and green edges of the 24-cell. There are another 4x4=16 red vertices of 4 tetrahedra mirrored in the sphere with x4=0. This makes a total number of 32 red vertices of 8 tetrahedra. In the same way there are 32 vertices of 8 tetrahedra with green vertices.

Vertices  of the sections 1₃, 2₃, 4₃ , 5₃ , 6₃ and  7₃ on the edges of the 24-cell. 

In the next picture we only keep 2 outer triangles of each of the 6 icosahedra and 4 triangles of 4 tetrahedra. We have skipped the sections 1₃,2₃ 3₃ and 4₃. The 4 red vertices belong to section 7₃, the 12 blue to section 6₃ and the green ones to section 5₃.

We recognize 4 triangles (violet), 12 triangles in blue, green and yellow, and 12 pentagons (colorless). The colorless pentagons are the sections of 12 icosahedra.

Now we add the vertices of section 8₃ (the 12 orange ones making the vertices of a large cuboctahedron), section 9₃ (4 red vertices making a even larger tetrahedron) and section 10₃ (12 blue vertices).

The sections 1₃, 2₃, 3₃, 4₃ , 5₃ , 6₃ , 7₃ and  8₃ . The 12 vertices (4 red, 4 blue and 4 green)  of the cuboctahedron are of section   8₃ . We see 5 cells at each vertex of the cuboctahedron, so we have 12x5=60 new cells. A subtotal of 221 cells.

In the next picture we see sections 6₃,7₃, 8₃, 9₃. The 12 orange vertices are the midpoints of the 12 icosahedra whose colorless pentagonal sections are also visible.

In the next picture we see the sections 6₃,7₃, 8₃, 9₃ and 10₃.

The same sections again, but the connections to the orange vertices omitted.

The 6 red tetrahedra lie around the 6 vertices of the original 24-cell that have 4th coordinate x4 = 0. Together with the yellow tetrahedron of the beginning and an opposite to the yellow one (having opposite x4)(to be seen later) these 8 tetrahedra lie around the 8 vertices of a 16-cell. Mark that these 8 tetrahedra all have blue vertices. These 8x4=32 blue vertices complete the 32 red and 32 green vertices to a total of 96 vertices on the edges of the original 24-cell.

The sections 1₃, 2₃, 3₃, 4₃ , 5₃ , 6₃ , 7₃  , 8₃  , 9₃ and  10₃ . I have counted 6x(1+3+3+5+5)=6x17=102  cells in the 6 coordinate directions and 4x(1+6)=28 cells and 4x(1+6)=28 in the tetrahedral directions. This makes 158 new cells. The 221 cells we had thus far count twice (positive and negative 4th coordinate x4), so we have 2x221+158=600 cells. In the picture we can see the first completed great circles on the projected 3-sphere S3.

That is an awful lot of cells (600) and circles (72) below. There is one blue vertex missing  and some circles are too large to fit in this window:

In the next picture sections 8₃, 9₃ and 10₃.

In the next picture sections 9₃, 10₃ and the 12 green vertices of 11₃.

Compare these sections to the picture of the sections 5₃, 6₃ and 7₃, as these sections are the mirror image when reflected in the sphere x4=0. The 12 pentagons are cross sections of the same 12 icosahedra whose 12 colorless pentagonal sections and orange midpoints we discussed earlier.

Here are sections 10₃, 11₃ and the 12 red vertices of 12₃.

Compare this to the sections 4₃, 5₃ and 6₃.

We can recognize parts of 6 new icosahedra (8 vertices of them in a plane), 4 violet tetrahedra (3 vertices) and 4 brown tetrahedra (3 vertices).

Skipping 10₃ and adding section 13₃ we get 6 orange vertices at the centres of the 6 icosahedra.

In the next picture we add section 14₃ (4 green vertices) and thus complete the 4 violet tetrahedra. Three neighbouring tetrahedra are colored yellow. Again the 4 violet tetrahedra lie in tetrahedral symmetry. Compare this with the picture of the sections 1₃ to 5₃ .

The 4 brown triangles are faces of a set of 4 tetrahedra with red vertices that make a set of 8 violet tetrahedra around the 8 corners of a red cube.

Finally we add 4 blue vertices of section 15₃ that is the mirror image of section 1₃.

The 6 icosahedra are completely flattened in the 6 squares of a cube. (If you want to count the 20 triangular planes, then don't miss the 4 that are stretched out on the 4 edges of the cube)

We recognize the 4 violet and 4 brown tetrahedra as mirror images of the first section picture of this page. How about the mirror tetrahedron of the yellow one ? You're in it yourself!

Section 13₃, 14₃ and 15₃ in the last picture with some adjacent triangles.

The 24-cell can be rotated to get:

To see how around each vertex of the 24-cell 4 vertices of a tetrahedron arise in this situation we show various combinations of sections of the 600-cell, beginning with a vertex.

We can start looking at the sections 1₀, 2₀ and 3₀:

With the coloring of 6 tetrahedra we recognize 6x4 vertices that originate from 6 vetrices of the central cell of the 24-cell. The picture is magnified.

We count 1 + 12 + 8 + 12 = 33 vertices.

Or 1 + 8 + 6x4 = 33.

How the 6 vertices of one cell of the spherical 24-cell multiply into 6x4=24 vertices of 4 tetrahedral cells of the 600-cell. This figure is rotated by 90° relative to the figure above it.

Here are added 9 vertices of the dual 24-cell, where the central vertex is evidently the centre of a icosahedron whose 12 vertices  are on the 12 edges of the octahedron.

The 6 tetrahedra with vertices of one color and the icosahedron they enclose. The centre of the icosahedron is a vertex of the dual 24-cell. We see 70 complete tetrahedra, 2x20 sharing a face with the icosahedron and 30 sharing an edge with the icosahedron. Adding the vertices of the dual 24-cell results in 52 tetrahedra in addition to the 18 we already had. 

Then we look at sections 4₀, 5₀ and 6₀ (also a little bit magnified):

We count 12 + 30 + 12 = 54 vertices.

Or 6 + 12x4 = 54.

The 12 tetrahedra have their origin in the 12 vertices of the cuboctahedron of which the vertices all have 4th coordinate x4 = 0.

How the 12 vertices of the cuboctahedron multiply to 4x12=48 vertices of  12 tetrahedral cells of the 600-cell.

We have 12 tetrahedra with vertices of one color and 2x12=24 tetrahedra sharing a face with such a tetrahedron and a vertex with another. Adding the 6 (blue) vertices of the dual 24-cell gives 24 more tetrahedra. There are 60 tetrahedra in this image. They lie in 12 sets of 5 around a yellow rotation axis where the light green edges of the 5 complete a regular pentagon.

Then the sections 7₀ and 8₀ and 9₀ :

Section 9₀ is at infinity.

We count 20 + 12 +1 = 33 vertices.

Or 8 + 6x4 +1 = 33.

Another 24 vertices of the 600-cell that originated from 6 vertices of one cell of the 24-cell. This figure is rotated by 90° relative to the figure above it.

Again 70 tetrahedral cells, of which 20 have a vertex at infinity. 

Adding the 12 vertices of section  6₀ gives in addition  5x12=60 cells around the yellow rotation axes. We find 60+60=120 more cells, sharing the 12x5=60 triangles in the faces of the red dodecahedron and the 12x5=60 radii of these red pentagons. If you are counting the total of cells we found (remembering that these last 180 cells are also found on the other side of x4=0)  you must have come to a number of 560 cells. The missing 2x20=40 cells can be found on both sides of the light green triangles of the icosidodecahedron at section 5₀ .

Finally, we end with the complete 600-cell in the vertex first projection:

spherical 600-cell in vertex first projection.

To see each section separately you can go to the page about the icosahedral group .

To see the 600-cel in a different projection:

Click Here for an interactive picture of 600-cell 

(made with Cabri Geomètre II and CabriWeb)