The n-cube |
The
n-cube Cn is the n-dimensional version of the 2-cube (square)
and 3-cube (cube). The most convenient way to think of it is as the set
of all real n-vectors (a1, a2,...,an)
with -1 < = ai <= 1 for all i.
Cn has 2n vertices (two choices +1 or -1 for each coordinate), n2n-1 edges (n at each vertex, counting each edge twice), and 2n boundary hyperplanes (n-1-dimensional boundary elements), also called facets. Note that each facet is a Cn-1 obtained by keeping one coordinate constant. [Check how this works for n = 2 and 3] Cn has n2 reflection hyperplanes, which are the solution sets of the homogeneous linear equations:
|
. |
Symmetries of the n-cube |
Now
we show by induction that Cn has at most 2n n! symmetries.
We have seen this is true for n = 2 and 3. Assume true for Cn-1.
Choose one of the 2n facets to go to the bottom (i.e. the facet which meets
the xn-axis at the point (0,0,...,-1)). Use a symmetry of Cn-1
as a symmetry of the bottom facet. This will permute the other facets of
Cn. Now every symmetry of Cn is obtained in this
way because any symmetry maps facets to facets. So there are at most 2n
x 2n-1 x (n-1)! = 2nn! symmetries of Cn.
Now we show that there at least 2nn! symmetries of Cn. Choose a vertex v. There are n facets containing v. Each of the n! permutations of these facets gives a symmetry of Cn fixing v, the even permutations being direct (rotations about an n-2-dimensional hyperplane) and the odd ones indirect (reflections in a hyperplane of symmetry). Then choose a vertex v' to map v to. This can be done in 2n ways. Every symmetry that maps v to v' can be obtained by conjugating a permutation of facets by one of these 2n maps. Conclusion: Cn has exactly 2n n! symmetries, and we know how they act on vertices and adjacent facets. |
. |
Go to Table of Contents.
Last update: 12 January, 2000
Author: Phill Schultz, schultz@maths.uwa.edu.au