600-cell in projective geometry
In a
similar way as we treated the desmic
tetrahedra, where the cube and octahedron also appeared, one can treat the
other 2 platonic polyhedra: the dodecahedron and the icosahedron. They have 15
pairs of parallel edges giving 15 points of intersection with the plane at
infinity. In projective space it is a configuration of a pentagram and a
straight line.
The
straight line in dark green contains 5 points that are intersected by a grey and
a light green line. These pairs of lines become parallel lines in the affine
plane that you get by making the dark green line the line at infinity.
This leads
to an elliptic configuration known as an icosidodecahedron with 15 elliptic
points as 15 pairs (=30) of vertices. Instead of great circles I’ve drawn chords
from point to point. The picture on the right also shows the grey lines (now as
great “circles” in black) that connect 3 elliptic points (=6 vertices):
With a
different choice for the line at infinity we see in 2 variations (one with
colored pentagons) 5 halves of black great circles:
You can draw 6 great circles at the surface that contain 10 vertices. Through each vertex intersect 2 circles. They contain 10 vertices. There are 6 lines through the origin of 3-space, normal to the planes that contain these circles. They each contain 10 vertices. They pierce through the midst of the pentagonal faces of the icosidodecahedron.
Through the pairs of vertices of the pair of pentagons that lie in planes parallel to the plane of a decagon one can draw 5 great circles lying on the hypersphere. Through the remaining 5 pairs of opposite vertices one can do the same. This results in 10+10+5x10+5x10=120 vertices that make the vertices of a 600-cell.
On these 12 great circles lie 120 edges. There are 6 such sets of 12 circles through the same 120 vertices, making 6x120=720 edges of the 600-cell.
There are 6 great circles passing through each vertex
corresponding to 12 edges leaving from one vertex. So, through the vertices of
the green icosidodecahedron are passing 4 great circles besides the 2 green ones
of the icosidodecahedron itself. Two of them, the grey ones, are passing through
a pair of blue edges of the blue dodecahedron and two of them, the yellow ones,
are passing through a pair of red edges of the red icosahedron. This can all be
checked in the next image in which for clarity only the vertices with fourth
coordinate x4<=0 :
The 30
great circles that make the 30 edges of the dodecahedron have 2 elliptic points
of the elliptic dodecahedron and an elliptic point of the icosidodecahedron. The
other 2 elliptic points lie on a (Euclidean) icosahedron with an edge length
that has a ratio to the edge length of the (Euclidean) 600-cell exactly equal to
the Golden Ratio (= 1,618034.. ). Resuming, on these circles we have 4 vertices
that belong to the dodecahedron, 2 to the icosidodecahedron and 4 to two large
icosahedra.
The 30
great circles that make the 30 edges of the icosahedron have 2 elliptic points
of the elliptic icosahedron and an elliptic point of the icosidodecahedron. The
other 2 elliptic points lie on the aforementioned dodecahedron. Resuming, we
have 4 vertices belong the icosahedron, 2 to the icosidodecahedron and 4 to the
dodecahedron. Recall that you see only 2 of the 4 vertices of the 2 icosahedra
and the 2 dodecahedra.
Concerning
the vertices of the large icosahedron we can remark that 5 great circles come
from a vertex of a dodecahedron and 1 from the small red icosahedron. This
sufficiently amounts to 12 edges of the 600-cell because the large edges of the
large icosahedron do not belong to this enumeration. It is more appropriate to
see the 12 grey vertices as belonging to an elliptic small stellated
dodecahedron:
Its core is
the blue dodecahedron.
On the
other hand the blue dodecahedron is the case of an elliptic great stellated
dodecahedron, of which the red icosahedron is the core :
It is easy
to count the 600 tetrahedral cells:
20 inside
the red icosahedron with 1 vertex at the centre;
20 on the
20 faces of the red icosahedron;
30 on the
edges of the red icosahedron with the opposite edge belonging to the blue
dodecahedron;
60 (12x5)
at the vertices of the red icosahedron sharing the straight yellow edge to the
grey vertex of the elliptic small
stellated dodecahedron;
60 in pairs
on the edges of the blue dodecahedron sharing a triangular face with 2 blue
vertices and 1 green vertex;
20 on the
vertices of the blue dodecahedron and the 20 green triangles of the
icosidodecahedron;
60 with 1
grey edge of a triangle on the elliptic small stellated dodecahedron and an
opposite green edge on the icosidodecahedron;
This
amounts to 20+20+30+60+60+20+60 = 270 tetrahedra. There are another 270
tetrahedra mirrored in the 2-sphere x4=0. We sum up to 540 tetrahedra.
The remaining 60 tetrahedra lie in 12 groups of 5 in the 12 green pentagons of the icosidodecahedron sharing a yellow straight edge from the grey vertex of the elliptic small stellated dodecahedron to its mirrored image with opposite fourth coordinate x4.
To see the the connection with the projective geometry I repeat the elliptic picture of the complete 600-cell, where I use only 3 colors: green for the icosidodecahedron, red for the small stellated dodecahedron and blue for the great stellated dodecahedron.
In POV-Ray (rotated by 90°):
In POV-Ray (rotated by 90°):
Making a slightly different choice for the absolute plane we get:
Only one triangle with green edges and violet edges is in the plane at infinity in this affine choice. The 3 pairs of parallel green lines meet at these 3 points.
There are 2 vertices of the pentagram at infinity and 1 of the 5 points of incidence with the line that completes the projective pentagram. In another plane the projective complete pentagram with its green line as absolute line looks like this:
