Strange polytope in H3

Higher-dimensional geometry (previously "Polyshapes").

Strange polytope in H3

Postby wendy » Mon Feb 12, 2018 9:02 am

I've been playing around with hexagons of late.

The first is, i suppose, hooh4oHHo&#tq. H here is q*h = sqrt(6). It's a bevelled square prism, such that the top and bottom become squares, the vertices being on the mid-points of the original squares, and the rectangles are converted into hexagons.

As a vertex figure, this leads to a tiling, with four {4,6} and two {6,4} at the vertex, and eight hexagon-prisms, which serve to connect {6,4}, in pairs.

I have not walked enough of this figure as yet, but the edge is quite large: it's nearly four times the edge of {5,4}, for instance.
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Re: Strange polytope in H3

Postby ubersketch » Tue Feb 13, 2018 12:13 am

Seems to be uniform from what I can understand.
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Re: Strange polytope in H3

Postby wendy » Tue Feb 13, 2018 10:00 am

It's uniform, to be sure. But it has cells like {6,4} and {4,6} which are infinite (specifically, bollic).

Of course, if we understand this symmetry, which *might* be a subset of {6,4,3}, then it might self-intersect with itself in the way that {8,3,4} and {8,4,A} do in their uniform tilings. It has no horotopic cells, which means that a full-cover is possible. Horotopic cells generally prevent this.
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Re: Strange polytope in H3

Postby ubersketch » Wed Feb 14, 2018 10:42 pm

wendy wrote:It's uniform, to be sure. But it has cells like {6,4} and {4,6} which are infinite (specifically, bollic).

Of course, if we understand this symmetry, which *might* be a subset of {6,4,3}, then it might self-intersect with itself in the way that {8,3,4} and {8,4,A} do in their uniform tilings. It has no horotopic cells, which means that a full-cover is possible. Horotopic cells generally prevent this.

Yeah, my mind twists and turns at euclidean/hyperbolic tilings as cells.
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