
Re: Tesseract Projected along a Main Diagonal
Posted:
Sep 16, 1994 1:44 PM


In article <35ar9f$cfn@newsbf01.news.aol.com> hopd@aol.com (HopD) writes: >In article <348a6r$eij@forum.swarthmore.edu>, asimov@wk223.nas.nasa.gov >(Daniel A. Asimov) writes: > >Re: Tesseract Projected along a main diagonal. >Take the points of the tesseracts vertices to be 16 ordered quadruples >whose elements were either one or zero (0,0,0,0), (1,0,0,0), (0,1,0,0), >etc. Describe these quadruples in terms of 4 unit vectors, the first >pointing in the direction of the diagonal and the other 3 perpendicular to >the first and each other. Take the vertices described with this basis and >eliminate the components pointing in the direction of the diagonal. I >believe you are left with ordered triples in a space perpendicular to the >diagonal. Sketching it these triples seem to form a rhombic dodecahedron. > From Hop David Ajo, Arizona a055@amug.org 
In case anyone forgot my original question, I'll paraphrase it as follows:
QUESTION: If an ordinary cube is projected orthogonally onto the plane perpendicular to its main diagonal, the result is a regular hexagon (as is wellknown).
What is the result of projecting a 4dimensional cube (also called a tesseract) onto a hyperplane (which is a copy of 3space) that is perpendicular to one of its main diagonals?
* * *
The answer given above by Hop David is correct. This was the only response offered by anyone, so this was apparently a fairly difficult problem.
My solution (partial):
Assume that the hypercube's 16 vertices lie at the points (x,y,z,w) where each of x,y,z,w take the values 0 or 1 in all combinations.
I.e., the vertices are (0,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1), and the 11 other vectors obtained from these by permuting their coordinates.
A vector in the direction of a main diagonal is (1,1,1,1). We can project any point onto the 3plane P perpendicular to this vector if we can find an orthonormal basis for P.
Fortuitously, this is easier than it may sound. In fact, the four vectors
( 1, 1, 1, 1), ( 1, 1, 1, 1), ( 1, 1, 1, 1), ( 1, 1, 1, 1)
are mutually orthogonal, since the dot product of any two of them is 0.
These four vectors each have length = 2, so { a, b, c } form an orthonormal basis for the hyperplane P, where
a = ( 1/2, 1/2, 1/2, 1/2), b = ( 1/2, 1/2, 1/2, 1/2), c = ( 1/2, 1/2, 1/2, 1/2).
We can choose these vectors a, b, c as the directions for a set of three orthogonal axes for the plane P.
So, using this coordinate system, the projection of any vertex v of the hypercube onto the hyperplane P is ( v*a, v*b, v*c ).
Now we can calculate the projections of all 16 vertices of the hypercube in this way. (For example, v = (1,0,1,1) is projected onto (v*a, v*b, v*c) = (1/2, 1/2, 1/2).)
As it turns out, (0,0,0,0) and (1,1,1,1) project to (0,0,0); the other fourteen vertices of the hypercube project to the vertices of a rhombic dodecahedron in 3space. A little more thought will convince you that this (solid) rhombic dodecahedron is in fact the projection of the hypercube onto P.
Dan Asimov Mail Stop T27A1 NASA Ames Research Center Moffett Field, CA 940351000
asimov@nas.nasa.gov (415) 6044799 w (415) 6043957 fax

