Symmetrical simplex coordinates and more from George

Some time ago it occurred to me to try to figure out an "elegant" set of
four-dimensional Cartesian coordinates for the vertices of a regular
pentachoron. One set, for example, comprises the five points (1,1,1,0),
(1,-1,-1,0), (-1,1,-1,0), (-1,-1,1,0), and (0,0,0,sqrt(5)). But these
coordinates are "unbalanced": The coordinates of the fifth vertex are unlike

the coordinates of the first four vertices, and the center of the
pentachoron
is not at the origin.

I soon came up with a much more elegant set of coordinates, with the
pentachoron centered at the origin. We know that the edges of a regular
pentachoron will, in a certain orientation, project two-dimensionally into a

regular pentagon plus all its diagonals; and that in this orientation it
will
project into the same figure in the absolutely perpendicular plane as well.
The difference between the two projections is that the edges that project
into the pentagon's >edges< in one project into the pentagon's >diagonals<
in
the other.

Since the coordinates of the vertices of a regular pentagon in the xy-plane
are

(cos[2i*pi/5], sin[2i*pi/5])

for integer i=0,...,4,

and the coordinates of the vertices of a regular pentagram are

(cos[4i*pi/5], sin[4i*pi/5])

for integer i=0,...,4,

we can combine the two sets of two-dimensional coordinates into one set of
four-dimensional vertex-coordinates of a skew pentagon in E(4) that projects

into a pentagon in the xy-plane and a pentagram in the zw-plane; that is, it

projects into the edges of a pentagon in the xy-plane and the diagonals of a

pentagon in the zw-plane:

(cos[2i*pi/5], sin[2i*pi/5], cos[4i*pi/5], sin[4i*pi/5])

for integer i=0,...,4.

This skew polygon (and it is no mere zigzag!) is in fact the Petrie polygon
of a regular pentachoron, so we're immediately done: we have a neat,
balanced
set of coordinates for a regular pentachoron whose edge is sqrt(5).

This is such a cute set of coordinates that I figured Coxeter must have
described it somewhere in >Regular Polytopes<. Sure enough, in a single
paragraph on p. 245 he provides >generalized< "cute coordinates" for the
vertices of a regular n-dimensional simplex; the case n=4 gives the
coordinates derived above.

The pattern for an n-dimensional simplex is quite simple. Just string
together the coordinate pairs

cos(2r*i*pi/[n+1]), sin(2r*i*pi/[n+1])

for integer r=1,...,n/2 if n is even. If n is odd, string the same
coordinate
pairs together for integer r=1,...,(n-1)/2 and then append as the nth
coordinate
(-1)^i/sqrt(2). This yields the coordinates of vertex i of a simplex of edge

sqrt(n+1), for integer i=0,...,n. The regular n-dimensional simplex with
these vertex coordinates always projects into a regular (n+1)-gon in the
xy-plane.

For example, for n=3 (the regular tetrahedron), the pattern yields the four
points

(cos[0pi/4],sin[0pi/4],1/sqrt[2])
(cos[2pi/4],sin[2pi/4],-1/sqrt[2])
(cos[4pi/4],sin[4pi/4],1/sqrt[2])
(cos[6pi/4],sin[6pi/4],-1/sqrt[2])

or, since cos(pi/2) = 0, sin(pi/2) = 1, etc.,

(1,0,1/sqrt[2])
(0,1,-1/sqrt[2])
(-1,0,1/sqrt[2])
(0,-1,-1/sqrt[2])

These are the vertices of a regular tetrahedron of edge sqrt(3+1) = 2. It
projects into a square with both diagonals in the xy-plane. (The
two-dimensional case, of course, easily yields an equilateral triangle of
edge sqrt[3] in the xy-plane; and the one-dimensional case trivially yields
the two points [+-1/sqrt{2}], the end points of a line segment of length
sqrt[2].)

For the regular five-dimensional simplex, or regular hexatetron, we have
these six vertices:

(cos[0pi/6],sin[0pi/6],cos[0pi/6],sin[0pi/6],1/sqrt[2])
(cos[2pi/6],sin[2pi/6],cos[4pi/6],sin[4pi/6],-1/sqrt[2])
(cos[4pi/6],sin[4pi/6],cos[8pi/6],sin[8pi/6],1/sqrt[2])
(cos[6pi/6],sin[6pi/6],cos[12pi/6],sin[12pi/6],-1/sqrt[2])
(cos[8pi/6],sin[8pi/6],cos[16pi/6],sin[16pi/6],1/sqrt[2])
(cos[10pi/6],sin[10pi/6],cos[20pi/6],sin[20pi/6],-1/sqrt[2])

We can reduce the fractions to lowest terms, and noting than cos(pi/3)=1/2,
sin(pi/3)=sqrt(3)/2, etc., we obtain the six vertices (assuming I've not
made
a typographical error!)

(1,0,1,0,1/sqrt[2])
(1/2,sqrt[3]/2,-1/2,sqrt[3]/2,-1/sqrt[2])
(-1/2,sqrt[3]/2,-1/2,-sqrt[3]/2,1/sqrt[2])
(-1,0,1,0,-1/sqrt[2])
(-1/2,-sqrt[3]/2,-1/2,sqrt[3]/2,1/sqrt[2])
(1/2,-sqrt[3]/2,-1/2,-sqrt[3]/2,-1/sqrt[2])

This gives a regular hexatetron with an edge length of sqrt(6). It projects
into a regular hexagon with all its diagonals in the xy-plane.

Proving that these coordinates always yield an n-dimensional simplex of edge

sqrt(n+1) for every integer n>0 requires some fairly fancy trigonometry. You

must show that the distance between points i and j is always sqrt(n+1) for
any i~=j. Basically what happens is that all the messy trigonometric sums
and
products in the distance summation expression eventually cancel out, leaving

a string of n+1 1's to add up and take the square root of.

If you want similarly symmetric vertex coordinates for the dispentachoron,
simply find the midpoints of all ten edges of the pentachoron:

(cos[2i*pi/5]+cos[2j*pi/5], sin[2i*pi/5]+sin[2j*pi/5],
cos[4i*pi/5]+cos[4j*pi/5], sin[4i*pi/5]+sin[4j*pi/5])/2

for integers i<j in {0,...,4}. If you prefer radicals, note that

sin[2*pi/5] = (sqrt[10+2sqrt[5]])/4
sin[4*pi/5] = (sqrt[10-2sqrt[5]])/4
cos[2*pi/5] = (sqrt[5]-1)/4
cos[4*pi/5] = -(sqrt[5]+1)/4

and so on.