### M Theory Lesson 138

The expression $F_{n}$ for the codimension 1 faces of the associahedron is

$F_{n}(1) = \frac{1}{2} n (n + 3) = \frac{1}{2} n (n + 1) + n = (\sum_{i=1}^{n} i) + n$

This corresponds to rewriting the sequence in the form

2

2 + 2 + 1

2 + 2 + 1 + 3 + 1

2 + 2 + 1 + 3 + 1 + 4 + 1

After eliminating the bold $n$ terms this becomes

2

1 + 2

1 + 2 + (1 + 2)

1 + 2 + 3 + (1 + 1 + 2)

and so on. The previous post also separated the set of $F_{n}(1)$ Young diagrams into a set of $n$ yellow tiled diagrams and another set of $1 + 2 + \cdots + n$ diagrams with at most two purple tiles in a row.

Now consider the $k = 2$ sequence $F_{n}(2)$, which counts the number of divisions of an $(n+3)$-gon into three pieces by two diagonals. The terms of this sequence are given by the formula

$F_{n}(2) = \frac{1}{3} B(n + 4, 2) B(n,2) = \frac{1}{12} n(n-1)(n+3)(n+4) = \frac{1}{6} (n-1)(n+4) F_{n}(1)$

What set of diagrams counts this sequence? The $F_{n}(1)$ factor says that before cutting an $n$-gon into three pieces we must cut it into two. Note also that $F_{2}(2) = F_{2}(1) = 5$ since having cut a pentagon into two pieces, there is only one way to cut it into three. It is therefore more natural to write

$F_{n+1}(2) = \frac{1}{6} n(n+5) F_{n+1}(1) = [\frac{1}{3} (\sum_{i=1}^{n} i) + \frac{2}{3} n] F_{n+1}(1) $

$= [ \frac{1}{3} F_{n}(1) + \frac{1}{3} n] F_{n+1}(1) = \frac{1}{3} F_{n+1}(1)(n + F_{n}(1))$

This says that once we have chopped the $(n+4)$-gon into two pieces, either we have a triangle and an $(n+3)$-gon to chop up, or we have at least an $\frac{1}{2}(n+6)$-gon (ignoring odd $n$ for now) for which we can choose $n$ diagonals meeting the existing diagonal. For example, the $21$ edges of the 3d Stasheff polytope arise from the relation

$21 = \frac{1}{3} 9 (2 + 5)$

which says that once a hexagon is cut into two, in one of $9$ ways, either there is a pentagon to chop up, in one of $5$ ways, or the hexagon is split into two squares, one of which may be cut up in two ways. The full set of $9$ diagrams for $n=3$ appears, along with the pentagon subset and the $n$ yellow pieces. Note that either $F_{n+1}(1)$ is divisible by $3$, or $F_{n}(1) + n$ is divisible by $3$. For example, $14 + 4 = 18$ and $35 + 7 = 42$. The overcount factor of $3$ comes from the familiar cyclic symmetry of a central triangle in the hexagon, the three bisections of which mark the three possible choices for a square face on the 3d polytope. Similarly, the factor of $2$ in the $F_{n}(1)$ sequence came from the two diagonals of a square, which obey an $S_2$ symmetry under rotation.

Young diagrams are usually used to label irreducible representations of the symmetric group $S_n$. If certain collections of Young diagrams are used to label associahedra (which may be obtained from the permutohedra with vertices the elements of $S_n$) then there is a close connection between the collection of groups $\{ S_n \}$ and its representations. However, the whole heirarchy in all dimensions needs to be considered if we want to understand this correspondence between a theory and its models.

$F_{n}(1) = \frac{1}{2} n (n + 3) = \frac{1}{2} n (n + 1) + n = (\sum_{i=1}^{n} i) + n$

This corresponds to rewriting the sequence in the form

2

2 + 2 + 1

2 + 2 + 1 + 3 + 1

2 + 2 + 1 + 3 + 1 + 4 + 1

After eliminating the bold $n$ terms this becomes

2

1 + 2

1 + 2 + (1 + 2)

1 + 2 + 3 + (1 + 1 + 2)

and so on. The previous post also separated the set of $F_{n}(1)$ Young diagrams into a set of $n$ yellow tiled diagrams and another set of $1 + 2 + \cdots + n$ diagrams with at most two purple tiles in a row.

Now consider the $k = 2$ sequence $F_{n}(2)$, which counts the number of divisions of an $(n+3)$-gon into three pieces by two diagonals. The terms of this sequence are given by the formula

$F_{n}(2) = \frac{1}{3} B(n + 4, 2) B(n,2) = \frac{1}{12} n(n-1)(n+3)(n+4) = \frac{1}{6} (n-1)(n+4) F_{n}(1)$

What set of diagrams counts this sequence? The $F_{n}(1)$ factor says that before cutting an $n$-gon into three pieces we must cut it into two. Note also that $F_{2}(2) = F_{2}(1) = 5$ since having cut a pentagon into two pieces, there is only one way to cut it into three. It is therefore more natural to write

$F_{n+1}(2) = \frac{1}{6} n(n+5) F_{n+1}(1) = [\frac{1}{3} (\sum_{i=1}^{n} i) + \frac{2}{3} n] F_{n+1}(1) $

$= [ \frac{1}{3} F_{n}(1) + \frac{1}{3} n] F_{n+1}(1) = \frac{1}{3} F_{n+1}(1)(n + F_{n}(1))$

This says that once we have chopped the $(n+4)$-gon into two pieces, either we have a triangle and an $(n+3)$-gon to chop up, or we have at least an $\frac{1}{2}(n+6)$-gon (ignoring odd $n$ for now) for which we can choose $n$ diagonals meeting the existing diagonal. For example, the $21$ edges of the 3d Stasheff polytope arise from the relation

$21 = \frac{1}{3} 9 (2 + 5)$

which says that once a hexagon is cut into two, in one of $9$ ways, either there is a pentagon to chop up, in one of $5$ ways, or the hexagon is split into two squares, one of which may be cut up in two ways. The full set of $9$ diagrams for $n=3$ appears, along with the pentagon subset and the $n$ yellow pieces. Note that either $F_{n+1}(1)$ is divisible by $3$, or $F_{n}(1) + n$ is divisible by $3$. For example, $14 + 4 = 18$ and $35 + 7 = 42$. The overcount factor of $3$ comes from the familiar cyclic symmetry of a central triangle in the hexagon, the three bisections of which mark the three possible choices for a square face on the 3d polytope. Similarly, the factor of $2$ in the $F_{n}(1)$ sequence came from the two diagonals of a square, which obey an $S_2$ symmetry under rotation.

Young diagrams are usually used to label irreducible representations of the symmetric group $S_n$. If certain collections of Young diagrams are used to label associahedra (which may be obtained from the permutohedra with vertices the elements of $S_n$) then there is a close connection between the collection of groups $\{ S_n \}$ and its representations. However, the whole heirarchy in all dimensions needs to be considered if we want to understand this correspondence between a theory and its models.

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