Timeline for Symmetries and faces of the associahedron
Current License: CC BY-SA 3.0
11 events
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| Mar 5, 2021 at 15:24 | comment | added | M. Winter | A note on your edit: it is not true that the 1-skeleton if a simple polytope can have no extra symmetries (see e.g. the construction in this answer). Kalai's result states that the combinatorics of the polytopes is uniquely determined yb the 1-skeleton, but it says nothing about the realization of its symmetries. | |
| Oct 5, 2012 at 23:24 | comment | added | Patricia Hersh | 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. | |
| Oct 5, 2012 at 23:05 | vote | accept | Matt Brin | ||
| Oct 5, 2012 at 23:05 | comment | added | Matt Brin | 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. | |
| Oct 5, 2012 at 21:22 | comment | added | Patricia Hersh | 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.) | |
| Oct 5, 2012 at 20:41 | comment | added | Patricia Hersh | 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. | |
| Oct 5, 2012 at 19:46 | comment | added | Matt Brin | 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? | |
| Oct 4, 2012 at 3:09 | history | edited | Patricia Hersh | CC BY-SA 3.0 |
added 2 characters in body
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| Oct 4, 2012 at 2:58 | history | edited | Patricia Hersh | CC BY-SA 3.0 |
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| Oct 4, 2012 at 2:17 | history | edited | Patricia Hersh | CC BY-SA 3.0 |
changed $n$-gon to $(n+1)$-gon
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| Oct 4, 2012 at 1:54 | history | answered | Patricia Hersh | CC BY-SA 3.0 |