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On page 120 of his Basic Topology, Armstrong defines the $k$-simplex in $\mathbb{E}^n$ with verices $v_0,\ldots,v_k$ to be complex hull of said vertices. (A similar definition is given on Wikipedia).

Here's what bothers me: Given a $k$-simplex $\sigma$ ($k>0$) so defined, and $m$ other points (all distinct from each other and from the vertices of $\sigma$) in $\sigma$, the above definition says that the convex hull $\sigma'$ of the $k+1$ vertices plus the $m$ additional points is a $k+m$-simplex. But $\sigma=\sigma'$. So, $\sigma$ is both a $k$- and $k+m$-simplex.

Am I missing something?

flag	asked Apr 8 2011 at 17:32 Avi Steiner 377●8

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Typically one requires the vertices $v_0,\ldots,v_k$ to be in general position. Are you sure there is no such requirement in Armstrong's text? I guess if not, then one can speak of a "degenerate k-simplex", in the same sense that a line (1-simplex) is a degenerate triangle (2-simplex). – Faisal Apr 8 2011 at 17:42

The standard definition is that $v_0,\dots,v_k$ are affinely independent. (The definition on Wikipedia uses the equivalent condition that it has dimension $k$.) – Emil Jeřábek Apr 8 2011 at 18:12

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Armstrong says "in general position". – Tom Goodwillie Apr 9 2011 at 4:02

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Perhaps you're making a spelling mistake. Surely Armstrong wrote convex hull not the complex hull. I've never seen the term complex hull before. – Ryan Budney Jun 11 2011 at 23:54

Igor Rivin · Answer 1 · 2011-04-09 02:38:09Z UTC

I can't speak for Armstrong's book but the wikipedia article http://en.wikipedia.org/wiki/Simplex defines an $n$-simplex as an $n$-dimensional polytope which is the hull of of its $n+1$ vertices. This is correct, and adding $m$ points will break the above definition.

Erik P. · Answer 2 · 2011-04-08 18:03:48Z UTC

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I would agree and would require that no $v_i$ be in the complex hull of the others.

answered Apr 8 2011 at 18:03

Erik P.
36●2

Collinear vertices and definition of k-simplex

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