most recent 30 from http://mathoverflow.net 2013-06-19T22:26:46Z http://mathoverflow.net/feeds/question/108764 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108764/symmetries-and-faces-of-the-associahedra Matt Brin 2012-10-03T23:19:25Z 2012-10-04T03:09:08Z

<ol> <li><p>The dihedral group of order $2n+2$ acts on $K_n$, the $n-2$-dimensional associahedron. Are there any other symmetries? References?</p></li> <li><p>Does the answer to 1 change if we restrict to just the 1-skeleton of $K_n$? References?</p></li> <li><p>It is "obvious" that any simple circuit (simple closed walk, simple closed path, whatever terminology you prefer) of length 4 or 5 is a 2-dimensional face of $K_n$. Is this true? Proof? Reference?</p></li> </ol>

http://mathoverflow.net/questions/108764/symmetries-and-faces-of-the-associahedra/108772#108772 Patricia Hersh 2012-10-04T01:54:56Z 2012-10-04T03:09:08Z

<p>The answer to question 1 is no. A reference for this is:</p> <p>Carl Lee, The associahedron and triangulations of the $n$-gon, European Journal of Combinatorics, 10 (1989), no. 6, 551--560.</p> <p>The answer to question 3 is yes. I think this is clear from the viewpoint where you think of vertices of the associahedron as triangulations of an $(n+1)$-gon and you obtain higher dimensional faces containing such a vertex by deleting edges from the triangulation. This is the viewpoint e.g. discussed by Carl Lee. A 4-cycle involving a vertex $v$ of the associahedron implies that the two edges $e_1,e_2$ in the 4-cycle containing $v$ correspond to the deletion of a pair of edges $E_1,E_2$ from the triangulation corresponding to $v$ such that the concurrent deletion of $E_1, E_2$ yields two quadrilateral regions in the resulting subdivision; a 5-cycle involving a vertex $v$ of the associahedron likewise results from two edges $E_1, E_2$ of the corresponding triangulation whose concurrent deletion yields a single pentagonal region. In either case, the 4-cycle or 5-cycle then clearly bounds a face of the associahedron, namely the one given by the subdivision in which $E_1$ and $E_2$ are deleted from the triangulation corresponding to $v$. </p> <p>${\bf Edit:}$ I just realized we can deduce that the answer to 2 is also no, by virtue of a result of Gil Kalai. Kalai proved that any $d$-dimensional simple polytope is determined by its 1-skeleton. So we can use that the associahedron is a simple polytope to see that its 1-skeleton can't have any extra symmetries not present in the associahedron itself.</p>