Volume of the hypersphere in 4D
===============================
Denote the volume of a ball of radius R in n-dimensions by w_n*R^n.
Compute the volume w_{n+2}*R^(n+2) of a ball of radius R in dimension n+2
by a double integral:
w_{n+2}*R^(n+2)
= integral_0^R [integral_0^(2pi) [w_n*(R^2-r^2)^(n/2)]] r*dphi*dr
= pi*w_n*integral_0^R [-(R^2-r^2)^(n/2)]*(-2r)*dr
= pi*w_n*(R^2)^(n/2+1)/(n/2+1)
= 2pi*w_n/(n+2)*R^(n+2).
Thus, w_{n+2} = 2pi*w_n/(n+2), which can be rewritten as
m*w_m*R^(m-1) = 2pi*R*w_n*R^n, where m = n+2,
which relates the volume of the ball in n-dimesnions, w_n*R^n, to the volume of
the sphere in (n+2)-dimensions, m*w_m*R^(m-1), where m = n+2. In particular,
w_4 = 2pi*w_2/4 = 2pi*pi/4 = pi^2/2. Thus, in 4D, the interior of a hyperball
of radius R has a 4D-measure pi^2/2*R^4, while its boundary, the hypersphere
S^3, has 3D-measure 2*pi^2*R^3.
dimension n boundary n*w_n*R^(n-1) interior w_n*R^n
1 2 2R
2 2pi*R pi*R^2
3 4pi*R^2 4/3*pi*R^3
4 2pi^2*R^3 1/2*pi^2*R^4
5 8/3pi^3*R^4 8/15*pi^2*R^5
6 pi^3*R^5 1/6*pi^3*R^6
The hypercube in 4D
===================
The cube in 4D, called a tesseract, has 16 vertices, 32 edges, 24 2-faces
and 8 3-faces. Note that 16-32+24-8 = 0. The tesseract has 8 diagonals.
The group rotational symmetries of the tesseract is of order 192. Its
double group is represented by the group of unit quaternions (a,b), where
a an b are in the double group of the 3D-cube, of order 48, and are due to
certain constraints.
The parallel projection of the tesseract in 3D, with vertex closest to
the viewer, is a rhombic dodecahedron. An edge closest to the viewer,
gives a hexagonal prism. For web-pages of the tesseract, see
http://pweb.netcom.com/~hjsmith/WireFrame4/tesseract.html
http://www.uni-bonn.de/~uzs083/versch.html
http://dogfeathers.com/java/hyprcube.html
http://www.uccs.edu/~eswab/hcubsect.htm
http://www.cs.reading.ac.uk/people/jpb/publications/ndcubes.html
http://www.geom.umn.edu/docs/forum/polytope/
Regular polyhedra in 3D
=======================
The cube in 3D has 8 vertices, 12 edges and 6 faces. Note that 8-12+6 = 2.
The cube has 4 diagonals, whose permutations give the group of rotational
symmetries of the cube, S_4.
Denote the number of vertices by N0, edges by N1 and faces by N2.
For all convex polyhedra in 3D, N0-N1+N2 = 2. For all regular polyhedra,
with regular p-gons as faces and q edges meeting at a vertex, we have
q*N0 = 2*N1 = p*N2.
For the 3D-cube: Each vertex is a meeting point of three edges each with two
endpoints, so 3*N0 = 2*N1. Each edge is a boundary of two faces, each with
four sides, so 2*N1 = 4*N2. Thus, 3*N0 = 2*N1 = 4*N2.
The following table gives numbers Nk, k = 0,1,2:
N0 N1 N2 Schläfli symbol
Tetrahedron 4 6 4 {3,3}
Octahedron 6 12 8 {3,4}
Cube 8 12 6 {4,3}
Icosahedron 12 30 20 {3,5}
Dodecahedron 20 30 12 {5,3}
Note that N0-N1+N2 = 2.
Regular polytopes in 4D
=======================
Schläfli symbol N0 N1 N2 N3 Face Vertex
{3,3,3} 5 10 10 5 Tetrahedron Tetrahedron
{3,3,4} 8 24 32 16 Tetrahedron Octahedron
{4,3,3} 16 32 24 8 Cube Tetrahedron
{3,4,3} 24 96 96 24 Octahedron Cube
{3,3,5} 120 720 1200 600 Tetrahedron Icosahedron
{5,3,3} 600 1200 720 120 Dodecahedron Tetrahedron
N0-N1+N2-N3 = 0.
The group of rotational symmetries of the cube in 3D
====================================================
The group of rotational symmetries of the cube is of order 24 and the group
of orthogonal symmetries of the cube is of order 48.
There is another group of order 48 related to the cube: the group of unit
quaternions whose factor group is the group of rotational symmetries of the
cube. This double group of the group of rotational symmetries is interesting:
the unit quaternions of it do not form a regular polytope in 4D, in contrast
to double groups of groups of rotational symmetries of the regular tetrahedron
and dodecahedron (and of the icosahedron). It can be composed, though, of two
dually drawn 24-cells, of Schläfli symbol {3,4,3}, inscribed in the unit
sphere S^3 of R^4.
Exercise: While the faces of the 24-cell are regular
octahedra, what are the faces of the 48-cell, whose
vertices are unit quaternions in the double group of
the group of rotational symmetries of the cube?
The faces are all isometric to the tetrahedron with vertices
1, (1+i)/sqrt(2), (1+j)/sqrt(2), (1+i+j+k)/2
which is not regular, while its edges have lengths 1 and sqrt(2-sqrt(2)). The
edge 1, (1+i)/sqrt(2) (of length sqrt(2-sqrt(2)) is surrounded by 8 tetrahedra:
1, (1+i)/sqrt(2), (1+j)/sqrt(2), (1+i+j+k)/2
1, (1+i)/sqrt(2), (1+j)/sqrt(2), (1+i+j-k)/2
1, (1+i)/sqrt(2), (1-j)/sqrt(2), (1+i-j+k)/2
1, (1+i)/sqrt(2), (1-j)/sqrt(2), (1+i-j-k)/2
1, (1+i)/sqrt(2), (1+k)/sqrt(2), (1+i+j+k)/2
1, (1+i)/sqrt(2), (1+k)/sqrt(2), (1+i-j+k)/2
1, (1+i)/sqrt(2), (1-k)/sqrt(2), (1+i+j-k)/2
1, (1+i)/sqrt(2), (1-k)/sqrt(2), (1+i-j-k)/2.
The edge 1, (1+i+j+k)/2 (of length 1) is surrounded by 3 tetrahedra:
1, (1+i+j+k)/2, (1+i)/sqrt(2), (1+j)/sqrt(2)
1, (1+i+j+k)/2, (1+i)/sqrt(2), (1+k)/sqrt(2)
1, (1+i+j+k)/2, (1+j)/sqrt(2), (1+k)/sqrt(2).
The polytope which is the subject of your exercise has the following counts:
N0 = 48
N1 = 336
N2 = 576
N3 = 288.
Schläfli's generalization of Euler's formula:
N0-N1+N2-N3 = 48-336+576-288 = 0.
Each of the N3 tetrahedra has 4 faces, and each face is common to two
tetrahedra, so:
N2 = (4*N3)/2 = (4*288)/2 = 576.
Each of the four short edges of a cell is surrounded by 8 tetrahedra. And
each of the two long edges is surrounded by 3 tetrahedra. So, the total
number of edges in the polytope is:
N1 = 288*4/8 + 288*2/3 = 336.
And N0 is twice the number of vertices in a 24-cell:
N0 = 2*24 = 48.
The faces of the 48-vertex are the above irregular tetrahedra, of edge-lengths
1 and sqrt(2-sqrt(2)). (Thus, n2(face) = 4.)
The vertex figure of the 48-vertex is composed of reciprocally posited
octahedron and cube, whose diagonal is (sqrt(2)-1)sqrt(3) times the diagonal
of the octohedron (so we do not have a rhombidodecahedron but). The vertex
figure is a polyhedron with 14 vertices, 36 edges and 24 faces
(thus n0(vertex) = 14).
N0, N1, N2, N3 satisfy, for a convex polytope, N0-N1+N2-N3 = 0,
and for a regular polytope, n0(vertex)*N0 = 2N1 and 2N2 = n2(face)*N3.