Timeline for A flag complex is contractible iff the underlying graph is....?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 7, 2013 at 3:27 | vote | accept | Alireza Abdollahi | ||
| Jan 27, 2012 at 19:28 | history | edited | j.c. | CC BY-SA 3.0 |
typos
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| Jan 27, 2012 at 19:28 | answer | added | Russ Woodroofe | timeline score: 5 | |
| Jan 27, 2012 at 12:58 | answer | added | Benjamin Steinberg | timeline score: 6 | |
| Jan 27, 2012 at 12:28 | comment | added | David E Speyer | Actually, James comment undersells the situation. Let $G$ be any graph, whether or not $C(G)$ is contractible, and let $H$ be the cone on $G$. Then $C(H)$ is the cone on $C(G)$, and hence contractible. | |
| Jan 27, 2012 at 12:26 | comment | added | David E Speyer | Also, if $G$ is any tree, then $C(G) = G$ and is contractible. I'm having trouble thinking of a good class of graphs that contains trees and is closed under coning. | |
| Jan 27, 2012 at 12:21 | comment | added | James Griffin | Suppose you have a graph G and we know that the flag complex C(G) is contractible. Pick a clique K inside G and define H to be the graph given by 'taking the cone' over K, so adding an extra vertex v and adding edges (v,k) for each k in K. Then C(H) is still contractible. My guess is that any G with C(G) contractible can always be built by repeatedly taking cones over subcliques. I think this counts as a graph theoretic condition. Can anyone verify my guess? | |
| Jan 27, 2012 at 11:57 | history | asked | Alireza Abdollahi | CC BY-SA 3.0 |