-----Original Message----- From: math-fun-bounces+andy.latto=pobox.com@mailman.xmission.com [mailto:math-fun-bounces+andy.latto=pobox.com@mailman.xmission .com] On Behalf Of Daniel Asimov Sent: Thursday, November 23, 2006 4:54 AM To: math-fun Subject: 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.

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. If a complex is to be embedded affinely in R^N, it must not contain two distinct simplices with the same vertex set. Is this necessary condition also sufficient? Andy Latto andy.latto@pobox.com