Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
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 '11 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 '11 at 18:12
2  
Armstrong says "in general position". –  Tom Goodwillie Apr 9 '11 at 4:02
1  
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 '11 at 23:54

2 Answers 2

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.