Let $\Delta$ be a flag complex on $n$ vertices. Let $r$ be the smallest size of the facets of $\Delta$. Suppose that $2r>n$. Must $\Delta$ be acyclic?


It looks to me like you can prove the stronger property of contractibility by induction, as follows. Let $\Delta$ be the independence complex of graph $G$, as guaranteed by the flag property.

If $\Delta$ is a cone, then $\Delta$ is contractible. This will be the base case, along with dimension 0 (where $\Delta$ has a single point) and dimension 1 (where $\Delta$ is a 1-simplex).

Otherwise, every vertex $v$ has degree at least 1 in $G$. Consider a fixed pair of vertices $v,w$ that are adjacent in $G$. We'll use that the link of $v$ in $\Delta$ is the independence complex of $G \setminus N[v]$ (and similarly for $w$). Now $\operatorname{link}_\Delta v$, $\operatorname{link}_\Delta w$ are contained in the independence complex of $G \setminus \{v,w\}$.

Since we remove at most one vertex from each maximal face in each case, $r$ goes down by at most one in each considered subcomplex, while the number of vertices goes down by at least 2. So by induction, $\operatorname{link}_\Delta v$, $\operatorname{link}_\Delta w$, and the independence complex of $G \setminus {v,w}$ are all contractible.

Now $\Delta$ is the union of the faces that contain $v$, those that contain $w$, and those that contain neither. So $\Delta$ is the union of the subcomplexes $\Delta_1 = v*\operatorname{link} v$, $\Delta_2 = w*\operatorname{link} w$, and $\Delta_0$ the independence complex of $G \setminus \{v,w\}$. It now follows by e.g. Lemma 10.4(ii) of Björner's Topological methods that $\Delta$ is contractible.

I didn't immediately see the answer to the question of whether there's a sensible generalization of some sort to non-flag complexes. The flag property is used above in where $\operatorname{link}_\Delta v$ is the independence complex of $G \setminus N[v]$; also in finding a pair of vertices that are in no common face.

(Updated over initial version to fix a problem with the facet sizes in the induction, in response to a comment of and off-MO discussion with @Hailong Dao.)

  • $\begingroup$ Great, thanks Russ. This remind me of my paper with Jay, perhaps I should dig it out. $\endgroup$ – Hailong Dao Nov 19 '18 at 16:18
  • $\begingroup$ Perhaps the non-evasiveness has to be addressed further given that the proof changed. 1) the ones described in the problem. 2) the ones obtained from removing a vertex, which give an additional local condition to relax the larger one, i.e there is a vertex with a pleasant link! It seems like this imposes many conditions on the graph. Do you have a systematic way to construct interesting examples? $\endgroup$ – José Alejandro Samper Dec 9 '18 at 20:55
  • $\begingroup$ @JoséAlejandroSamper, as far as examples, look at the graph theory literature, under the name independent domination. A nontrivial example can be constructed by attaching a large number of pendant edges to each vertex of a complete graph. See e.g. Independent domination in graphs: a survey and recent results by Goddard and Henning. $\endgroup$ – Russ Woodroofe Dec 10 '18 at 12:14
  • $\begingroup$ @JoséAlejandroSamper, I thought when I updated the post that I saw how to easily recover nonevasiveness, but now I no longer see it. I'll delete my earlier comment claiming that it holds. $\endgroup$ – Russ Woodroofe Dec 10 '18 at 12:37
  • $\begingroup$ I still feel it should hold! Anyways, i like this property quite a bit. Do you know how close to tight is this? $\endgroup$ – José Alejandro Samper Dec 10 '18 at 17:27

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