Re: [math-fun] Embedding simplicial complexes in Euclidean space
Thanks to Fred, Bill, and Andy for pointing out my blunder with the allegedly 4-triangle Moebius band. --------------------------------------------------------- A natural extension of the question of what dimension a simplicial complex embeds in, affinely on simplices (AOS), is this: QUESTION: Given a simplicial complex K and its abstract symmetry group, what is the smallest N such that K embeds AOS in R^N, such that each original symmetry of the embedded complex is realized by a rotation of R^N ??? 1. E.g., the projective plane can be constructed by identifying the faces of the icosahedron by the antipodal map, giving a simplicial complex K of 10 triangles, having a symmetry group of order 60 (= the smallest non-abelian simple group?). So . . . what is the lowest-dimensional Euclidean space into which this embeds AOS, such that all 60 symmetries are realized by rotations? 2. Ditto for the lovely torus someone mentioned -- given by a simplicial complex of 14 triangles (6 per vertex). Its symmetry group has order 84. (This complex is the dual of the amazing 7-hexagon tiling of the torus.) 3. Likewise for the beautiful rendition of the 3-holed torus as 56 triangles (7 per vertex). Its symmetry group has order 336. (This complex is the dual of the gorgeous 24-heptagon tiling of the surface of genus 3.) (Or, for 2. and 3. one could just ask to realize as rotations the *orientation- preserving* symmetry groups -- of orders 42, and 168 (= the second-smallest non-abelian simple group), respectively.) --Dan
On 11/25/06, Daniel Asimov <dasimov@earthlink.net> wrote:
... 1. E.g., the projective plane can be constructed by identifying the faces of the icosahedron by the antipodal map, giving a simplicial complex K of 10 triangles, having a symmetry group of order 60 (= the smallest non-abelian simple group?).
So . . . what is the lowest-dimensional Euclidean space into which this embeds AOS, such that all 60 symmetries are realized by rotations?
This is presumably the triangulation referred to by Cervone in the quote posted earlier about Brehm's theorem: I cannot at the moment visualise how this construction fails to immerse in 3-space (for which apparently at least 9 vertices are required)! If the embedding in 4-space can be equilateral --- Brehm's paper must discuss this --- then the symmetry would follow. But it somehow doesn't look entirely plausible --- for example, the (eutactic star driven) projection of a hypercubic lattice into a Penrose tiling is down from 5 dimensions, rather than 4. So for what it's worth, I'd put my money on 5-space here. WFL
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Daniel Asimov -
Fred lunnon