Re: [math-fun] Embedding simplicial complexes in Euclidean space
Make a band of four equilateral triangles in the plane, forming a parallelogram. Now identify the left and right edges of the parallelogram, creating a simplicial complex K, containing four 2-simplices, that's topologcially a Moebius band. For all n, there is no topological embedding h: K -> R^n such that h is affine on simplices. (Exercise: prove this.) That settles one question. --Dan
Right, now I've got the problem straight ... On 11/23/06, Daniel Asimov <dasimov@earthlink.net> wrote:
Make a band of four equilateral triangles in the plane, forming a parallelogram.
Minor quibble --- it's irrelevant that they're equilateral, since the embedding is affine [good word, that!]
Now identify the left and right edges of the parallelogram, creating a simplicial complex K, containing four 2-simplices, that's topologcially a Moebius band.
But, as AL implicitly points out, this is not a complex --- denoting vertices along the top line by A,C,B and the bottom by B',D,A', the triangles CDA and CDA' coincide, as well as edges AB' and A'B. WFL
Pondering this more constructively, and following up on Q1, it's plausible that we can restrict attention to triangulations of n-manifolds, since these have the least freedom to "flap around" during the embedding. And for the same reason, to minimal triangulations, such as the Csaszar (dual Szilassi) 7-point triangulation of the torus. Googling "minimal+triangulation+surface" turned up a nice page on this topic by Davide P. Cervone at http://www.math.union.edu/~dpvc/professional/research/traditional.html containing lots of surprising results --- I'm tempted to quote chunks, but the following must give the flavour: "For example, the real projective plane can be triangulated using only six vertices, but Brehm showed that it requires nine vertices to immerse this surface in three-space." [Hard to imagine the (equivalent) Steiner Roman surface requiring fewer than 10?] Fred Lunnon
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Fred lunnon