3
$\begingroup$

Let $X$ be a finite, regular CW complex, and let $X'$ be its barycentric subdivision (i.e. the order complex of the face poset of $X$). Assume $X$ is collapsible.

Is $X'$ collapsible? Is $X'$ non-evasive?

By Theorem 2.10 of Welker the answer to both questions is positive if $X$ is a simplicial complex. But what about more general, regular CW complexes? I believe at least the first question has a positive and well-known answer, but is it stated explicitly anywhere in the literature?

$\endgroup$
2
$\begingroup$

The first question is answered (modulo some details) by Forman in

R Forman, Morse theory for cell complexes. Advances in Mathematics, 134 pp 90 - 145, (1998).

Theorem 12.1 shows that a discrete Morse function $f$ on a polyhedron $X$ induces a discrete Morse function $\hat{f}$ on the subdivision $\hat{X}$ produced by bisecting a single $d$-cell $\sigma$ of dimension $d$ into two $d$-cells $\sigma_1$ and $\sigma_2$ which share a $d-1$-face $\tau$, and that (among other things) this $\hat{f}$ has precisely the same number of critical cells as $f$.

There are three things to check:

  1. $X$ is collapsible if and only if it admits a discrete Morse function with precisely one critical cell (this is true).

  2. Barycentric subdivision of $X$ may be achieved via a sequence of bisections so that each intermediate step produces a regular CW complex (I think this is true), and

  3. The proof of this theorem does not require anything more from our polyhedron $X$ than the fact that it is a finite regular CW complex (this is true).

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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