The icosahedral group
For understanding the following it is advisable to first read the page on the tetrahedral and octahedral group.
As seen on that page the rotations of platonic polyhedra can be represented by the vertices of 4-dimensional polytopes. The relation between the angle of rotation ß and the distance to the origin in 3-dim space is:
distance = tan(ß/4) = sin (ß/2)/(1+cos(ß/2)).
The value of the fourth dimension x4 is calculated by :
x4 = cos(ß/2).
The icosahedral group of rotations contains all 12 elements of the tetrahedral group. Instead of adding rotations about 4-fold symmetry-axes, as we did in the case of the octahedral group, we now add rotations about 5-fold axes. The orientation of these axes can be found by dividing the 24 edges of the 24-cell, that we previously used for the 4-fold axes of the octahedral group, in the Golden Ratio 1/GR:
1/GR = ½ ( SQR(5) -1) = 0,61803398875
(where GR = ½ ( SQR(5) + 1) = 1,61803398875
.
(made with POV-Ray)
This is the spherical 24-cell rotated by 90° .The distance the spheres on the circles have from the vertices of the 24-cell is 22.01812173°.
The 96 points on the 96 edges of this 24-cell gives us 4 new 24-cells that together with the original 24-cell of the tetrahedral group gives 120 vertices of a so-called 600-cell.
The 24 points in the color light red, representing the tetrahedral group, are the vertices of a 24-cell that is dual to the one whose edges are divided in the Golden Ratio:
We must be a little bit cautious to show the complete 600-cell at once because the 600-cell has 720 edges... So we better build it up in pieces. First we search for the vertices that represent the 5-fold rotations:
We have made 4 sections of constant value of the fourth coordinate x4. Only 48 of the 96 edges of the 24-cell are shown. They are the edges on which the 4x12 vertices of the 4 icosahedra lie.
We can now also omit these 48 edges and see the 4 icosahedra more clearly:
The 2x30=60 blue lines are edges of the 600-cell. The 2x30=60 grey lines are connecting lines whose length is GR( = 1,61803398875 ) times as big as the edge of the 600-cell.
In each of the 12 directions we get 4 vertices that represent rotations about 72°, 144°, 216° and 288°. The distance to the origin is calculated as:
tan ( 72°/4 ) = tan ( 18° ) = 0.3249196962329 corresponding to a value of x4 = ½ GR = cos ( 36° )
tan ( 144°/4 ) = tan ( 36° ) = 0.7265425280054 corresponding to a value of x4 = ½ ( GR - 1 ) = cos ( 72º )
tan ( 216°/4 ) = tan (54° ) = 1.376381920471 corresponding to a value of x4 = - ½ ( GR - 1 ) = cos ( 108º )
tan ( 288°/4 ) = tan ( 72° ) = 3.077683537175 corresponding to a value of x4 = - ½ GR = cos ( 144º ) .
The 3-fold rotations are found as 2x20 = 40 vertices consisting of 16 vertices of a hypercube ( 16 light red ) together with 3x2x4 = 24 points on 3x2 squares:
Omitting the 24 edges of the 24-cell:
In each of the 20 directions we get 2 vertices with a distance to the origin:
tan ( 120°/4 ) = tan ( 30° ) = 0.5773502691896 corresponding to a value of x4 = cos ( 60° ) = 0.5
tan ( 240°/4 ) = tan ( 60° ) = 1.732050807569 corresponding to a value of x4 = cos ( 120° ) = - 0.5.
Comparing with the tetrahedral group we have 10 3-fold rotation axes instead of 4. This corresponds to fitting 5 cubes in a dodecahedron by rotating the initial red cube 4 times about body diagonals:
Remember that in the tetrahedral group the 3-fold axes correspond to the vertices of a hypercube that looks like a set of 2 cubes with their 8 vertices joined by 8 lines. I omitted the smaller cube of each hypercube to avoid getting too many lines.
This compound of 5 cubes is also found at www.dogfeathers.com/java/hyperstar.html where the following picture can be found:
We can choose alternating vertices of each cube in 2 different ways to get 2 sets of 5 tetrahedra that are reciprocal ( one is the mirror image of the other if reflected in a spherical surface) and enantiomorphous ( if one is "left-handed" then the other is "right-handed"):
To complete the enumeration of the symmetry rotations of the icosahedral group we look for the 2-fold rotations.
They are represented by the 30 vertices of an icosidodecahedron.There are 24 vertices lying on the edges of a cuboctahedron and the remaining 6 ( light red ) are the vertices of an octahedron (figure on the left). Omitting the 24 edges of the cuboctahedron (24 edges of the 24 cell ) we get the icosidodecahedron on the right:
The 30 vertices represent the 15 rotations about 2-fold axes with distance to the origin:
tan ( 180°/4 ) = tan ( 45° ) = 1 corresponding to a value of x4 = cos ( 90° ) = 0 .
The 30 vertices also distribute in 5 times the 6 vertices of 5 octahedra: (see at www.dogfeathers.com/java/hyperstar.html )
Finally we have 2 vertices in the origin and at infinity to complete the 60 rotations that leave the icosahedron (or dodecahedron) invariant.
Resuming results, we have:
24 5-fold rotations represented by 4 icosahedra,
20 3-fold rotations represented by 2 dodecahedra,
15 2-fold rotations represented by the icosidodecahedron and
1 identity rotation represented by the vertices at the origin and at infinity.
