6
$\begingroup$

At the moment I'm reading through Edward B. Curtis, 'Simplicial Homotopy Theory' (Advances in Mathematics 6, 107-209 (1971)) in order to learn about simplicial sets and I run into a problem where the cone is defined.

He defines the cone of a simplicial set $K$ as follows: $(CK)_n = \{(x,q)\mid x\in K_{n-q}, 0\leq k\leq n\}$ with face maps $$d_i(x,q)=\left\{\begin{array}{ll} (x,q-1) & \text{for }0\leq i<q\\ (d_{i-q}x,q) & \text{for }q\leq i\leq n,\end{array}\right. $$ and the degeneracy maps in a similar fashion as $$s_i(x,q)=\left\{\begin{array}{ll} (x,q+1) & \text{for }0\leq i<q\\ (s_{i-q}x,q) & \text{for }q\leq i\leq n.\end{array}\right. $$

Apart from the $k$ in the definition of $(CK)_n$ that looks like it should be a $q$ (and that indeed is $q$ in 'Introductory Lectures on Braids, Configurations and their Applications' by A. Jon Berrick e.a.) , these maps are where confusion starts.

The degeneracy maps $s_i$ are supposed to go from $(CK)_n$ to $(CK)_{n+1}$, but the first part of the definition doesn't give me the impression it does that. We find that $(CK)_{n+1} = \{(x,q)\mid x\in K_{(n+1)-q}, 0\leq k\leq n+1\}$, which we can rewrite as $\{(x,q)\mid x\in K_{n-(q-1)}, 0\leq k\leq n+1\}$. However, this would suggest that for $i<q$, the first part of the definition of $s_i$ actually should map $(x,q)$ to $(x,q-1)$. As it is now, I'd expect it to map to $(CK)_{n-1}$ instead of $(CK)_{n+1}$.

Did Curtis really swap the plus and minus signs in the degeneracy and face maps (which then got copied wrongly by other authors) or is there something relevant I'm totally missing?

$\endgroup$
0
1
$\begingroup$

The definition of the facemaps and degeneracies as given is actually correct (except for the wrong $k$, which you already noticed). Note that the nondegenerate simplices in $CK$ are given by $$(CK)_n^\text{ND} = \{(x,q)\mid x\in K_{n-q}^\text{ND}, 0\leq q\leq 1\}$$ Where the original $n$-simplices are represented by tuples $(x,0)$, and the tuples $(x,1)$ correspond to $n-1$-simplices joined by one vertex (the tip of the cone). Now given any degenerate simplex $(x,q)$, the $q$ kind of descibes how degenerate the simplex is with respect to the new vertex. E.g. for some $x\in K_n$, we have $$(x,2) = s_0(x,1) \\ (x,3) = s_0(x,2) = s_1(x,2) = s_0(s_0(x,1)) = s_1(s_0(x,1)) $$

$\endgroup$
1
  • $\begingroup$ Thanks. I already realised the question was somewhat silly when I couldn't sleep last night. Basically, the answer can be summarised as $x\in K_{n-q} = K_{(n+1)-(q+1)}$. $\endgroup$ – HSN Jul 31 '13 at 9:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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