Any finite *simplicial* complex with n vertices is a subcomplex of the (n-1)-simplex, since each simplex in the complex is determied by its set of vertices. Therefore it embeds affinely in R^(n-1). It seems that by a general position argument you can embed a k- dimensional complex affinely in (2k+1)-space. You just need to make sure that a bunch of linear subspaces don't have On Nov 23, 2006, at 4:53 AM, Daniel Asimov wrote:
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.
This is not a simplicial complex, because any middle vertex along an edge is connected twice to one of the identified vertices --- once horizontally, and once kitty-cornered (diagonally). Similar mistake to the version below. This illustrates why simplicial complexes are not the most graphically intuitive concept. Bill On Nov 23, 2006, at 12:42 PM, Daniel Asimov wrote:
Andy wrote:
<< More simply, take two equilateral triangles, identify all three corners of one with all three corners of the other. This does not have an affine embedding in R^n for all n. This (along with identifying some of the edges of the triangles) is exactly what you've done with the two center triangles of your strip, in your more complex example.
Even simpler, in one dimension less, take two line segments, and identify their endpoints. This does not embed in R^n affinely for any n.
. . .
The problem is, in a simplicial complex any two simplices that intersect must do so in one common face (of any dimension). So two 2-simplices that intersect in all three 0-faces doesn't qualify.
Ditto, of course, for two 1-simplices that intersect in both their endpoints.
--Dan
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