How about recasting as - + + + + - + + + + - + + + + - for "obvious" symmetry? To go from your reverse-quaternion-sign matrix to mine, negate the first column and the last three rows; then swap rows 2 & 4. Rich ________________________________________ From: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] on behalf of Dan Asimov [dasimov@earthlink.net] Sent: Friday, July 22, 2011 5:16 PM To: Dan Asimov; math-fun Subject: Re: [math-fun] Geometry question Actually, while I'm fostering topic drift, maybe it's also worth mentioning the even more symmetrical (with respect to the standard basis) orthogonal basis for 4-space: (1 1 1 1) (1 -1 -1 1) (1 1 -1 -1) (1 -1 1 -1). Each axis in the direction of one of these vectors makes the angle of arccos(1/4) with each standard axis. Which is how I should have couched the claim below: in terms of new and old axes. (The angles I said are 45 degrees are actually either 45 or 135 degrees.) As someone recently said: I should ensure brain is engaged before mouth is opened. --Dan I wrote: << << It's an amazing fact about 4-space that there exist orthogonal bases like the one Jakob describes, tilted so that each vector is perpendicular to 2 of the standard basis vectors, and at 45 degrees to the other 2. Jakob wrote: << What about orienting a tesseract with edge length sqrt(2) in the following manner. Let one vertex sit in the origin, and let (1,0,1,0), (-1,0,1,0), (0,1,0,1), (0,-1,0,1) be the coordinates of the four vertices that are (edge-)adjacent to the origin. Since the corresponding vectors are orthogonal and have the same length sqrt(2), this indeed gives rise to a regular tesseract. Now, consider the projection (x,y,z,w) -> (x,y). The vertex set of the tesseract is mapped to the set {-1,0,1} x {-1,0,1}, and the projection of each edge has direction (1,0) or (0,1). Thus the projection of the 1D-skeleton of the tesseract is a 2x2 grid.
Sometimes the brain has a mind of its own. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun