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  1. The dihedral group of order $2n+2$ acts on $K_n$, the $n-2$-dimensional associahedron. Are there any other symmetries? References?

  2. Does the answer to 1 change if we restrict to just the 1-skeleton of $K_n$? References?

  3. 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?

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up vote 13 down vote accepted

The answer to question 1 is no. A reference for this is:

Carl Lee, The associahedron and triangulations of the $n$-gon, European Journal of Combinatorics, 10 (1989), no. 6, 551--560.

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 $v$.

${\bf Edit:}$ 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.

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Patricia: Your comments on 2 and 3 are clear. I don't see how Lee answers 1. I have not digested the paper, but it seems at least partly devoted to geometrically realizing symmetries that are combinatorially obvious. I don't mean that the geometric symmetries are obvious, just the combinatorial ones. I am interested in the combinatorial symmetries. I don't see where more combinatorial symmetries are ruled out. Am I missing something? – Matt Brin Oct 5 '12 at 19:46
Sorry about my confusion. Here is a way to prove 1. The maximal faces of the associahedron may be labeled by single arcs $(i,j)$ connecting vertices $v_i$ and $v_j$ of the $n$-gon. A symmetry must send facet $(i,j)$ to one of the form $(i+r,j+r)$, where vertices of the $n$-gon are indexed mod n. This is because the facet given by $(i,j)$ is a product of two smaller (possibly degenerate) associahedra, one given by triangulations of a $(j-i+1)$-gon and the other given by triangulations of an $(n-j+i+1)$-gon. Now use that flags of faces $F_1\subseteq F_2 \cdots $ must be sent to like flags. – Patricia Hersh Oct 5 '12 at 20:41
I tried to squeeze a whole proof into a single comment above, which wasn't easy -- does it make sense? (I'll check back here in a couple days.) – Patricia Hersh Oct 5 '12 at 21:22
This is fine. The product structures on the faces is the key. Of course, one must prove some sort of uniqueness of factorization, but I assume I can do this with a little work. – Matt Brin Oct 5 '12 at 23:05
Ok, great. I was thinking the fact that there are Catalan many vertices and related face counts should already help with that. You might like looking at the PCMI volume on Geometric Combinatorics and in particular the chapter by Sergey Fomin and Nathan Reading on Associahedra and Cluster Algebras. This whole book is really well written, with many interesting survey articles. – Patricia Hersh Oct 5 '12 at 23:24

I have just seen this question, while looking for something else. The answer to 1 is indeed "no", and an explicit proof appears in (no surprisingly, it follows the same ideas as Patricia's)

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