Mathematics desk
< August 16	<< Jul \| August \| Sep >>	Current desk >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

August 17

4-D Anti(?)-anti Prism question...

Consider the following 4 dimensional objects with 16 vertices. The Hypercube is a cubical prism and there Square (in xy) anti-prism (in z) prism(in w). For the first, all the squares in the xy direction are oriented the same, and in the second, two of the xy squares are oriented 45 degrees off from each other two, but the ones that are oriented the same have the same w coordinate. If it possible to have a 4-D figure where the two xy squares that are identically oriented have both different z and w values, so that in both the z and w direction they are anti-prisms?Naraht (talk) 01:31, 17 August 2018 (UTC)[reply]

The first is a square antiprism prism. Double sharp (talk) 09:51, 17 August 2018 (UTC)[reply]

Are you asking about a figure in which two "ortho" squares have w=z=-h and w=z=+h, and two "oblique" squares sit on the other two corners of the w,z square? Certainly you could generate the convex hull of these various squares; whether it would have any interesting properties, I cannot say. —Tamfang (talk) 02:38, 20 August 2018 (UTC)[reply]

Likely it's analogous to the grand antiprism, the only uniform member of that family. —Tamfang (talk) 06:19, 20 August 2018 (UTC)[reply]

I think the relevant family is instead that of the great duoantiprism, ht_0,1,2,3{5,2,5/3}. Naraht's second proposed polychoron would then be ht_0,1,2,3{4,2,4}, which I suppose might be called the 4,4-duoantiprism following the nomenclature proposed here at the Higher Dimensions Forum (which also explains why most such figures cannot be made uniform). Analogous to 3D antiprisms, it would have as its cells 16 square antiprisms connected by 32 tetrahedra. Double sharp (talk) 15:08, 20 August 2018 (UTC)[reply]

To be made more clear. (using r2 as the square root of 2) (x,y,z,w) = {(1,1,-1,-1), (1,1,-1,1), (1,1,1,-1), (1,1,1,1), (-1,-1,-1,-1),(-1,-1,-1,1),(-1,-1,1,-1),(-1,-1,1,1), (1,-1,0,r2),(1,-1,r2,0),(1,-1,0,-r2),(1,-1,-r2,0), (-1,1,0,r2),(-1,1,r2,0),(-1,1,0,-r2),(-1,1,-r2,0)} Naraht (talk) 17:21, 20 August 2018 (UTC)[reply]