Orienting the dual of the associahedron

Question

Let $A_n$ be the dual simplicial complex to the associahedron on $n$ letters. The complex $A_n$ is thus a simplicial triangulation of an $(n-3)$-dimensional sphere. The vertices of $A_n$ correspond to the ways of inserting one nontrivial parentheses into the expression $a_1 a_2 \cdots a_n$, and the $(n-3)$-dimensional simplices correspond to ways of inserting the maximal number ($n-2$) of parentheses into this expression. For instance, the $1$-dimensional simplices of $A_4$ are precisely \begin{align*} &((a_1 a_2) a_3) a_4\\ &(a_1 (a_2 a_3)) a_4\\ &a_1 ((a_2 a_3) a_4)\\ &a_1 (a_2 (a_3 a_4))\\ &(a_1 a_2) (a_3 a_4) \end{align*} Since $A_n$ is a triangulation of an $(n-3)$-dimensional sphere, it can be oriented (in two ways, we pick one).

Question: This orientation should give an orientation on each $(n-3)$-dimensional simplex $\sigma$. Such an orientation corresponds to an ordering (defined up to the action of the alternating group) of the vertices of $\sigma$, i.e., to the pairs of parentheses in $\sigma$. How can we explicitly write down this orientation?

Answer 1

There are many combinatorial models for the associahedron -- parenthesizations of $n$ variables, triangulations of the $(n+1)$-gon, planar binary trees with $n$-leaves. I'll follow the OP's lead and use parenthesizations of $n$-variables.

First, let's remember how we label the vertices of the dual associahedron. We'll label the vertices as $v_{ij}$ for $1 \leq i < j \leq n$, $(i,j) \neq (1,n)$. The vertex $v_{ij}$ occurs in the facet corresponding to a multiplication if the product $a_i a_{i+1} \cdots a_j$ is computed as a partial step in that multiplication. For example, say we have the parenthesization $$((a_1 a_2) a_3) (a_4 a_5).$$ When we compute this, we compute the partial products: $$a_1 a_2,\ a_1 a_2 a_3,\ a_4 a_5$$ and hence this parenthesization will correspond to vertices $v_{12}$, $v_{13}$ and $v_{45}$.

You might notice that I didn't include a vertex $v_{1n}$, which would correspond to the entire product $a_1 a_2 \cdots a_n$. Let's make a new simpicial complex $A_n$ which does use this vertex: So we have $\binom{n}{2}$ vertices $v_{ij}$ for $1 \leq i < j \leq n$ and, for each parenthesization, we have a facet corresponding to the $n-1$ partial products occuring when we compute that parenthesization. So $A_n$ is a closed $n-2$ dimensional ball, with the vertex $v_{1n}$ at the center. And the ordinary dual associahedron is the boundary $\partial A_n$.

With that preamble, here is how to orient $A_n$, and its boundary $\partial A_n$. Take each multiplication and insert the multiplication symbols, so our previous example turns into $((a_1 \ast a_2) \ast a_3) \ast (a_4 \ast a_5)$. Go across the formula from left to right, and write down the partial products in the order that they are output by that multiplication symbol. In our example, we get $a_1 a_2$, $a_1 a_2 a_3$, $a_1 a_2 a_3 a_4 a_5$, $a_4 a_5$.

Theorem Ordering the vertices of each facet of $A_n$ as listed here gives an orientation of the ball $A_n$.

To orient the boundary $\partial A_n$, we have to work a little harder: Take a parenthesization of $n$ variables. This corresponds to a facet $\sigma$ of $A_n$ whose vertices are ordered as above. One of those vertices is $v_{1n}$; let's say that $v_{1n}$ is in the $k$-th position. Delete $v_{1n}$ from the list of vertices of $\sigma$, giving a list of vertices of $\partial \sigma$. Assign that list of vertices, in that order, the sign $(-1)^{k-1}$. In the example above, we had the list of vertices $(v_{12}, v_{13}, v_{15}, v_{45})$. The vertex $v_{15}$ is in the $3$-rd position, so we assign the sign $(-1)^{3-1}= +1$ to the ordering $(v_{12}, v_{13}, v_{45})$ in $\partial A_5$.

Corollary Ordering the vertices of each facet of $\partial A_n$ as listed here gives an orientation of the sphere $\partial A_n$.

Why do I call this a corollary? Because this is the standard recipe for how to orient the boundary of an oriented simplicial complex. So we just need to prove the Theorem, and the Corollary will follow.

Proof of Theorem We just need to take two adjacent facets and check that their orientation is compatible. Two adjacent facets differ by a single association -- say we change $${\big(} (a_{p+1} a_{p+2} \cdots a_q) \ast (a_{q+1} a_{q+2} \cdots a_r) {\big)} \ast (a_{r+1} a_{r+2} \cdots a_s) \qquad (1)$$ to $$(a_{p+1} a_{p+2} \cdots a_q) \ast {\big(} (a_{q+1} a_{q+2} \cdots a_r) \ast (a_{r+1} a_{r+2} \cdots a_s) {\big)} \qquad (2).$$ All the multiplication symbols which I haven't drawn correspond to the same vertices in both $(1)$ and $(2)$. Of the two symbols which I have drawn, the left symbol gives $v_{(p+1) r}$ in $(1)$ and $v_{(p+1) s}$ in $(2)$; the right symbol gives $v_{(p+1) s}$ in $(1)$ and $v_{(q+1) r}$ in $(2)$. So our ordering has changed from $$(\cdots, v_{(p+1) r}, \cdots, v_{(p+1) s},\ \cdots)$$ to $$(\cdots, v_{(p+1) s}, \cdots, v_{(q+1) s},\ \cdots).$$ If we switch the second ordering to $(\cdots, v_{(q+1) s}, \cdots, v_{(p+1) s},\ \cdots)$, we act by a single transposition, introducing a minus sign. Thus, if we rearrange the orders to coincide on the set of vertices we have in common, then the two orders get opposite signs. This is the condition to be an orientation. $\square$

I actually found this in a very different way, but this is answer is already too long, so I'll write more in comments.