5
$\begingroup$

In a very captivating introduction to discrete Morse theory, Robin Forman makes the following remark:

...However, that does not explain why so many simplicial complexes that arise in combinatorics are homotopy equivalent to a wedge of spheres. I have often wondered if perhaps there is some deeper explanation for this.

The latter enigma seems to have been addressed elsewhere on MO. I have a more neophyte question: what are practical examples of studying such simplicial complexes, and do any deep combinatorial results stem from this equivalence?

$\endgroup$
5
  • $\begingroup$ You get a lot of alternating-sum identities out of it, at least :) $\endgroup$ Nov 27, 2017 at 23:11
  • 1
    $\begingroup$ I am not sure this qualifies as a "practical example," but shellable simplicial complexes are wedges of spheres of equal dimension. Björner & Wachs generalized to non-pure wedges of spheres (not all equal dimensions), partly to address questions in arrangements of subspaces. "Shellable nonpure complexes and posets. I." Transactions AMS 348.4 (1996): 1299-1327. $\endgroup$ Nov 27, 2017 at 23:41
  • 3
    $\begingroup$ Here's a compelling list. I particularly like Bjorner's paper called "a cell complex in number theory". google.com/… $\endgroup$ Nov 28, 2017 at 2:56
  • $\begingroup$ Lovasz initiated a whole field of determining chromatic properties of graphs from the topological properties of its neighorhood complex or various hom complexes. For the original example of the Kneser graph $K_{n,k}$, Lovasz needed to compute the connectivity of its neghborhood complex, but this ends up being homotopy equivalent to a wedge of spheres of dimensions $n-2k$, so the connectivity is exactly $n-2k-1$ and in turn the chromatic number is exactly $n-2k+2$. $\endgroup$ Nov 29, 2017 at 16:57
  • $\begingroup$ (This way of phrasing the proof was not in Lovasz's original paper, but you can read it in chapter 2 of "A course in Topological combinatorics" by Mark de Longueville) $\endgroup$ Nov 29, 2017 at 16:58

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.