Yes, this is a wonderful (if degenerate) way of looking at the hypercube. Degenerate because the nearest and furthest vertices appear to coincide at (0,0,0), when in fact they are far apart in 4-space. This is exactly analogous to the isometric projection of the 3-cube into 2-space (or any isometric projection of an N-cube into (N-1)-space). The standard optical illusions that use 3d to 2d isometric projection work just fine up a dimension using the rhombic icosahedral forms, namely: the reversible staircase (concave convex), the impossible triangle (which becomes the impossible skew quadrilateral), and the endlessly rising staircase (which works poorly in 3d to 2d — the loop is a kite when really it wants to be a digon — but works beautifully in 4d to 3d — the loop is a triangle). As you probably know, the fact that you were able to use such nice integer coordinates for the vertices of the tetrahedron in 3-space is a consequence of the dimension of our space being one less than a power of 2. This does not happen in general…for instance the isometric projection of the cube into a plane results in a regular hexagon, which does NOT align nicely with a square grid. The 2-d analog of your idea is to map the four basis vectors u, v, w, and x onto four segments radiating from a point at 45° angles, which leads to one of the most standard (and pretty) pictures of the hypercube. When I first started exploring Hypercubes as a kid, I realized you could of course generalize this for any dimension, and draw yet higher dimensional cubes. The six cube drawing is particularly pretty; I drew up to 9 by hand. Pushing your idea upward, there is no nice way to project 5 basis vectors into 3-space, but there is a completely symmetrical way to project 6 basis vectors into 3 space, namely vectors radiating out from the origin to half the vertices of an icosahedron (the other six vectors point in the opposite directions). And this leads to the rhombic triacontahedron, which I'm sure Jeanine and Fred are familiar with, with 30 golden mean rhombic faces. This projection is the one that leads to the structure of quasicrystals. I haven't tried pushing this higher…I wonder what zonohedron one would get starting from 15 vectors that point out to half the vertices of a dodecahedron? That seems like the next thing to try. Does anyone know? — Scott On Sat, May 12, 2018 at 8:07 PM, James Propp <jamespropp@gmail.com> wrote:
I can answer my own question:
Let the vectors u, v, w, and x be (1,1,1), (-1,-1,1), (-1,1,-1), and (1,-1,-1). Then u+v = (0,0,2), u+w = (0,2,0), u+x = (2,0,0), v+w = (-2,0,0), v+x = (0,-2,0), w+x = (0,0,-2), u+v+w = (-1,1,1), u+v+x = (1,-1,1), u+w+x = (1,1,-1), and v+w+x = (-1,-1,-1), which exactly correspond to the coordinates given in https://en.wikipedia.org/wiki/Rhombic_dodecahedron#Cartesian_coordinates .
Jim
On Sat, May 12, 2018 at 10:40 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Coxeter's "Regular Polytopes" has a lot about this and similar zonohedral projections of orthotopes. WFL
On 5/13/18, James Propp <jamespropp@gmail.com> wrote:
But the rhombic dodecahedron has 14 vertices, and I thought mine had
Did I miscount?
... Oh wait, 0 is interior to the tetrahedron with vertices u, v, w, and x, so it can't be a vertex!
Can someone explain to me why Jeannine is right (now that I believe her)?
Jim
On Sat, May 12, 2018 at 7:22 PM, Jeannine Mosely <j9mosely@gmail.com> wrote:
I believe you are talking about the rhombic dodecahedron.
--Jeannine
On Sat, May 12, 2018, 7:09 PM James Propp <jamespropp@gmail.com> wrote:
Let u, v, w, and x be the vectors from the center of a tetrahedron to the four vertices. Does the zonohedron formed by the fifteen vectors u, v, w, x, u+v, u+w, u+x, v+w, v+x, w+x, u+v+w, u+v+x, u+w+x, v+w+x, and u+v+w+x=0 have a name? Better, are there images of it on the web? Better still, are there gifs or videos that show it rotating in a way that brings its three-dimensionality to the fore?
I ask because this seems like a nice way of looking at a hypercube (leaving aside the defect that two of the vertices of the hypercube coincide).
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