19
$\begingroup$

Does anyone know of a simplicial complex which is not collapsible but whose barycentric subdivision is?

Every collapsible complex is necessarily contractible, and subdivision preserves the topological structure, so we are certainly looking for a complex which is contractible, but not collapsible. The only complexes I know of which are contractible but not collapsible are the dunce cap and Bing's house with two rooms. Neither of these have any free faces, and so no iterated subdivision will result in a collapsible complex.

Improve this question
$\endgroup$
1
  • 6
    $\begingroup$ Main Theorem 5 of arxiv.org/pdf/1107.5789.pdf states that if a $d$-complex admits a geometric realization as a convex subset of $\mathbb{R}^d$, then its $(d-2)$-nd barycentric subdivision is collapsible. $\endgroup$
    – Richard Stanley
    Feb 5, 2012 at 3:08

2 Answers 2

Reset to default
21
$\begingroup$

Lickorish and Martin constructed, for each $r$, a triangulation of the $3$-ball whose $r$th barycentric subdivision collapses, but $(r-1)$th doesn't. The basic idea, going back to Furch and Bing, is to triangulate a cube with a knotted hole, where the missing knot has bridge index $2^r+1$, and then fill back a small part of the hole - so that topologically, no hole remains, but the $3$-ball now contains a knot triangulated by a single edge.

Added later: Kearton and Lickorish also constructed triangulations of the $n$-ball, $n\ge 3$, whose $r$th barycentric subdivision is not collapsible. On the other hand, every triangulation of a ball becomes collapsible after some number of barycentric subdivisions, according to a recent preprint by Adiprassito and Benedetti (see their Corollary 3.5).

Improve this answer
$\endgroup$
1
  • $\begingroup$ ... and it was known from the very beginning that every triangulation of a ball has a stellar subdivision that is collapsible (Theorem 7 in Simplicial spaces, nuclei, and $m$-groups). Whitehead's papers are always an inspiring reading. maths.ed.ac.uk/~aar/papers/jhcw9.pdf $\endgroup$
    – Sergey Melikhov
    Jun 21, 2012 at 0:26
11
$\begingroup$

@Andy: there is now a small explicit example available, with 15 vertices and 55 tetrahedra, in case you're still interested: it's the 3-ball called $B_{15,66}$ in here. (Sorry for the huge delay and the shameless self-promotion.)

Improve this answer
$\endgroup$

Your Answer

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

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