Re: [math-fun] Geometry question
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. --Dan 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.
David wrote: << Can a regular tessertact can be oriented in R^4 so that the orthogonal projection of its 1D edges onto a 2D plane have the shape of a 2 x 2 grid?
Sometimes the brain has a mind of its own.
I withdraw my earlier ill-considered riposte. [Memo to self: start brain before engaging mouth / pen / keyboard.] WFL On 7/22/11, Dan Asimov <dasimov@earthlink.net> 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.
--Dan
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.
David wrote: << Can a regular tessertact can be oriented in R^4 so that the orthogonal projection of its 1D edges onto a 2D plane have the shape of a 2 x 2 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
On 7/22/2011 6:42 PM, Fred lunnon wrote:
I withdraw my earlier ill-considered riposte. [Memo to self: start brain before engaging mouth / pen / keyboard.]
WFL Fred:
As far as I am concerned, the considerable mileage I have gotten from your numerous insightful posts grants you standing absolution for the rare faux pas. I was pretty sure the tesseract mapped to the grid, but I wasn't smart enough to confirm it. So rather than assume it was true and reveal what an idiot I am, I tossed the question out for expert analysis, and you ended up taking the bullet for me. My apologies. So, I suppose a motivation is in order. I had the following conjecture, probably nothing more than a corollary to projective geometers: Given an n-dimensional hypercube H, it is possible to orthogonally project H to some suitably coordinated complex plane P so that some vertex V of H maps to 0 and the vertices H-edge-adjacent to V map onto the n-th roots of unity. I leave it to the geometers to state this correctly, and if true, I imagine it is a fairly elementary theorem of projective geometry. Now I could see the conjecture was true in dimensions 0 through 3, but I wanted a warm fuzzy about dimension 4 before I was willing to accept it was true for higher dimensions. The dimension 4 conjecture required the tesseract vertices and edges to map to the aforementioned 2x2 grid, which I guess is confirmed. So suppose the conjecture is indeed true. Then the vertices of H map to the sums of subsets of the nth roots of unity, which implies that the number of vertices of H that map to 0 is Sloane's A103314(n).
On 7/23/11, David Wilson <davidwwilson@comcast.net> wrote:
Fred:
As far as I am concerned, the considerable mileage I have gotten from your numerous insightful posts grants you standing absolution for the rare faux pas.
Hey, somebody actually reads my posts! The dodgy ones, anyway ...
... I had the following conjecture, probably nothing more than a corollary to projective geometers: Given an n-dimensional hypercube H, it is possible to orthogonally project H to some suitably coordinated complex plane P so that some vertex V of H maps to 0 and the vertices H-edge-adjacent to V map onto the n-th roots of unity. I leave it to the geometers to state this correctly, and if true, I imagine it is a fairly elementary theorem of projective geometry.
Sticking my neck out again (some people never learn), this looks to be what are called "eutactic stars" in Coxeter's book "Regular Polytopes". I strongly recommend you to take a look at this before spending more time on this topic. There is a eutactic star projecting the 5-cube into a regular decagon, and projecting the 6-cube into an icosahedron. Neither is straightforward to construct, being precursors to Penrose tilings in 2 and 3 dimensions. Fred Lunnon
participants (3)
-
Dan Asimov -
David Wilson -
Fred lunnon