A flag complex is contractible iff the underlying graph is....? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T11:49:31Z http://mathoverflow.net/feeds/question/86818 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86818/a-flag-complex-is-contractible-iff-the-underlying-graph-is A flag complex is contractible iff the underlying graph is....? Alireza Abdollahi 2012-01-27T11:57:21Z 2012-01-27T19:28:59Z <p>Let \$G\$ be a finite simple graph and let \$C(G)\$ be the flag complex associated to \$G\$ (the set of vertices of \$C(G)\$ is the vertex set of \$G\$ and the set of all cliques of \$G\$ are its simplexes). </p> <p>Are there characterizations of contractibility of \$C(G)\$ ONLY in terms of the graph theoretical properties of \$G\$?</p> http://mathoverflow.net/questions/86818/a-flag-complex-is-contractible-iff-the-underlying-graph-is/86823#86823 Answer by Benjamin Steinberg for A flag complex is contractible iff the underlying graph is....? Benjamin Steinberg 2012-01-27T12:58:17Z 2012-01-27T12:58:17Z <p>It is known that every induced subcomplex of the flag complex of a graph is contractible iff the graph is chordal (no induced cycles of length 4 or more). I doubt a necessary and sufficient condition that is purely graph theoretic for contractibility of just the flag complex is possible because the barycentric subdivision of any simplicial complex is a flag complex. </p> http://mathoverflow.net/questions/86818/a-flag-complex-is-contractible-iff-the-underlying-graph-is/86854#86854 Answer by Russ Woodroofe for A flag complex is contractible iff the underlying graph is....? Russ Woodroofe 2012-01-27T19:28:26Z 2012-01-27T19:28:26Z <p>A nice graph theory lemma for showing homotopy equivalence that builds on the "clique starring" already discussed is stated as Lemma 3.2 of Alexander Engström's paper <a href="http://arxiv.org/abs/math/0508148" rel="nofollow">arXiv:math/0508148</a>. I'll rephrase in terms of the language the question was asked in (clique complexes rather than independence complexes). </p> <p><b>Lemma:</b> If \$v\$ and \$w\$ are vertices of a graph \$G\$ with \$N[v] \subseteq N[w]\$, then \$C(G)\$ is homotopy equivalent to \$C(G \setminus v)\$.<br> (Here \$N[v]\$ is the closed neighborhood of \$v\$, i.e., \$v\$ and all its neighbors.)</p> <p>The proof technique is that of elementary collapses. See <a href="http://en.wikipedia.org/wiki/Collapse_(topology)" rel="nofollow">http://en.wikipedia.org/wiki/Collapse_(topology)</a>.<br> Even if the lemma of Engström doesn't give you what you need, the broader technique of collapsing can be quite useful. Collapses are the "engine" of discrete Morse theory, for example.<br> It's not too hard to write a computer program (or some are available) that does automatic collapsing, and you could generate a large number of examples this way. If you can collapse a complex to a point, then the complex is contractible. The converse is not true, but the program might at least give you a smaller complex to examine by hand.</p> <p>On the other hand, as Benjamin Steinberg points out, the barycentric subdivision of any simplicial complex is flag, so classifying contractible flag complexes should be as hard as classifying contractible simplicial (and more generally CW) complexes.</p>