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August 12[edit]

Cutting a tesseract with realms, and so on[edit]

Do the lazy caterer's sequence (cutting a 2-cube into pieces) and cake numbers (cutting a 3-cube into pieces) extend naturally to higher dimensions? In general, is the "hypercake sequence" $H^{n}$ in n dimensions given by $H_{k}^{n}=\sum _{i=0}^{n}{k \choose i}$ ?

And since these sequences are for hypercubes only, what happens for other polytopes? You can topologically transform any convex polytope into any other, and so they should have the same sequences, but what about toroidal polytopes and other non-simply connected polytopes? Double sharp (talk) 09:24, 12 August 2012 (UTC)[reply]

Yes the sequence generalizes and it applies to any convex n dimensional figure or the whole space. If they are not convex you can get any number at all with one cut. Dmcq (talk) 19:28, 12 August 2012 (UTC)[reply]

How? Double sharp (talk) 14:31, 13 August 2012 (UTC)[reply]

Not literally any number, but definitely more than 2. The precise maximum number with one cut depends on the precise form of the boundary - the wigglier, the more.←86.139.64.77 (talk) 18:17, 13 August 2012 (UTC)[reply]

Polychoron vertex figures[edit]

In polychora, the vertex figure is a polyhedron. The faces of these polyhedra are polygons, which are themselves vertex figures of the polyhedra that are the cells of the polychoron. Does the vertex figure contain the information necessary to write the "edge configuration" (analogous to the vertex configuration for polyhedra) of polyhedra surrounding an edge of the polychoron? And does this hold further for all n-polytopes? Double sharp (talk) 10:42, 12 August 2012 (UTC)[reply]

Maybe I'm not understanding the question. Each vertex of the VF represents an edge of the polychoron, surrounded by polygons representing cells. In general, of course, a uniform polychoron won't have just one edge configuration, unless its VF is itself ~~uniform~~ vertex-transitive. —Tamfang (talk) 15:37, 12 August 2012 (UTC)[reply]

So the vertex configuration of a polyhedron is just another of those coincidences that only work in 3D? Double sharp (talk) 14:29, 13 August 2012 (UTC)[reply]

Maybe I'm not understanding the question. The definition of a uniform polytope, of any dimension, implies that there's only one kind of vertex figure. If there's only one kind of edge figure, that's a higher degree of uniformity for which there may not be a name in general; polyhedra with that property are quasiregular; such polychora and tilings (of S3, E3 or H3) include the regular, rectified and alternated {p,q,r}; runcinated and bitruncated {p,q,p}; also the quarter-cubic honeycomb and some analogous hyperbolic forms that I won't list here. —Tamfang (talk) 06:06, 14 August 2012 (UTC)[reply]

I arguably didn't give you a straight answer the first time. Yes, the vertex figure tells you what surrounds each edge. —Tamfang (talk) 06:06, 14 August 2012 (UTC)[reply]

So the vertex figure of a polychoron tells what surrounds each edge, but this may be different between one edge and another? Double sharp (talk) 06:54, 18 August 2012 (UTC)[reply]

Yes. (Didn't I say that in the first place?) For example, consider the truncated cubic honeycomb. The vertex figure is a square pyramid. The pyramid's base represents an octahedron, and the flanks represent truncated cubes. The apex represents an edge shared by the octagons of four truncated cubes. The other four corners represent edges where two truncated cubes meet an octahedron. —Tamfang (talk) 20:07, 18 August 2012 (UTC)[reply]

OK, thanks. (We need similar vertex figure images for the uniform polychora, not just the uniform honeycombs.) Double sharp (talk) 04:19, 19 August 2012 (UTC)[reply]

Since User:Tomruen replaced my "perspective" VFs for honeycombs with his own "topological" VFs, I'm less motivated to try to make the polychoral series, which are less easy anyway. —Tamfang (talk) 04:28, 19 August 2012 (UTC)[reply]

I prefer the perspective ones, myself. (But I think they should all be labelled, like this one.) Double sharp (talk) 08:40, 19 August 2012 (UTC)[reply]