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# A matrix solution to any polygon equation

Zheyan Wan  Beijing Institute of Mathematical Sciences and Applications,
Beijing 101408, China
<EMAIL_ADDRESS>
###### Abstract.

In this article, we construct matrices associated to Pachner
$\frac{n-1}{2}$-$\frac{n-1}{2}$ moves for odd $n$ and matrices associated to
Pachner $(\frac{n}{2}-1)$-$\frac{n}{2}$ moves for even $n$. The entries of
these matrices are rational functions of formal variables in a field. We prove
that these matrices satisfy the $n$-gon equation for any $n$.

##  1\. Introduction

In this article, an $n$-gon means a simplicial complex with $n$ vertices.
Pachner moves [10, 8] are operations on a simplicial complex that changes its
triangulation to another one. For example, a Pachner 2-2 move is a flip of
diagonals in a quadrilateral, a Pachner 2-3 move is an operation that replaces
two tetrahedra 1245, 2345 with three tetrahedra 1234, 1235, 1345, see Fig. 1,
and so on.

Figure 1. Pachner 2-3 move.

The polygon equation is closely related to the Pachner moves. More precisely,
the $n$-gon equation is an equation between two products of Pachner moves. If
$n$ is odd, then the left-hand side of the $n$-gon equation is the product of
$\frac{n-1}{2}$ Pachner $\frac{n-1}{2}$-$\frac{n-1}{2}$ moves, while the
right-hand side of the $n$-gon equation is the product of $\frac{n+1}{2}$
Pachner $\frac{n-1}{2}$-$\frac{n-1}{2}$ moves. If $n$ is even, then the left-
hand side of the $n$-gon equation is the product of $\frac{n}{2}$ Pachner
$(\frac{n}{2}-1)$-$\frac{n}{2}$ moves, while the right-hand side of the
$n$-gon equation is the product of $\frac{n}{2}$ Pachner
$(\frac{n}{2}-1)$-$\frac{n}{2}$ moves. For example, the pentagon equation is
an equation between the product of 2 Pachner 2-2 moves and the product of 3
Pachner 2-2 moves, the hexagon equation is an equation between the product of
3 Pachner 2-3 moves and the product of 3 Pachner 2-3 moves, and so on.

The polygon equation itself can also be regarded as a Pachner move. More
precisely, if $n$ is odd, then the $n$-gon equation can be regarded as a
$\frac{n-1}{2}$-$\frac{n+1}{2}$ move (that replaces the left-hand side with
the right-hand side, both sides consist of a sequence of Pachner
$\frac{n-1}{2}$-$\frac{n-1}{2}$ moves and each Pachner
$\frac{n-1}{2}$-$\frac{n-1}{2}$ move in both sides corresponds to a face of
the $n$-gon (regarded as a simplicial complex)). If $n$ is even, then the
$n$-gon equation can be regarded as a $\frac{n}{2}$-$\frac{n}{2}$ move (that
replaces the left-hand side with the right-hand side, both sides consist of a
sequence of Pachner $(\frac{n}{2}-1)$-$\frac{n}{2}$ moves and each Pachner
$(\frac{n}{2}-1)$-$\frac{n}{2}$ move in both sides corresponds to a face of
the $n$-gon (regarded as a simplicial complex)).

In [1], a matrix solution for any odd polygon (odd-gon) equation was
constructed. Namely, for odd $n$, a $\frac{n-1}{2}\times\frac{n-1}{2}$ matrix
(whose entries are rational functions of formal variables in a field) was
constructed for each Pachner $\frac{n-1}{2}$-$\frac{n-1}{2}$ move, and it was
proved in [1] that these matrices satisfy the odd-gon equation. In the present
article, we do similar things for any even polygon (even-gon) equation.
Namely, for even $n$, we construct a $\frac{n}{2}\times(\frac{n}{2}-1)$ matrix
(whose entries are rational functions of formal variables in a field) for each
Pachner $(\frac{n}{2}-1)$-$\frac{n}{2}$ move, and we prove that these matrices
satisfy the even-gon equation. Our method also applies to any odd-gon equation
and our solution for any odd-gon equation is exactly a partial case of the one
in [1]. A similar but different solution for any odd-gon equation was given in
[4]. A different solution to the hexagon equation was given in [7]. These
solutions to the polygon equations are useful to study polygon relations
appearing from simplicial cocycles and invariants of manifolds, like in [2,
5].

This article is structured as follows: In Sec. 2, we formulate the polygon
equation. In Sec. 3, we formulate our main results (a matrix solution for any
polygon equation). In Sec. 4, we prove our main results. In Sec. 5, we give an
example: solving the pentagon equation. In Sec. 6, we give an example: solving
the hexagon equation. In Sec. 7, we give an example: solving the heptagon
equation.

###  1.1. Acknowledgements

The author is grateful to I.G. Korepanov for the helpful discussion and
comments.

##  2\. The polygon equation

In this section, we formulate the polygon equation.

The $n$-gon equation involves two sequences of Pachner moves. Each sequence is
a path from the initial triangulation to the final triangulation of the
$n$-gon. The number of $(n-3)$-simplices in each step of the two paths does
not change for odd $n$ and increases by 1 for even $n$. Each $(n-3)$-simplex
has $n-2$ vertices, it corresponds to a pair $(i,j)$ where $1\leq i<j\leq n$
and $i,j$ are not the vertices of the $(n-3)$-simplex.

The initial triangulation of the $n$-gon consists of $(n-3)$-simplices which
correspond to the pairs

(1) $\displaystyle(i,j)=(n+1-2k,n+2-2l)\text{ for }1\leq l\leq
k\leq\left\lfloor\frac{n-1}{2}\right\rfloor$

where $\left\lfloor x\right\rfloor$ means the largest integer less than or
equal to $x$. There are
$\dfrac{\left\lfloor\frac{n-1}{2}\right\rfloor(\left\lfloor\frac{n-1}{2}\right\rfloor+1)}{2}$
$(n-3)$-simplices in the initial triangulation of the $n$-gon. For example,
the initial triangulation of the pentagon consists of triangles 123, 134, and
145 which correspond to the pairs (4,5), (2,5), and (2,3). The initial
triangulation of the hexagon consists of tetrahedra 1234, 1245, and 1256 which
correspond to the pairs (5,6), (3,6), and (3,4). See Sec. 5 and Sec. 6.

We can do two Pachner moves for the initial triangulation of the $n$-gon,
leading to two sequences of Pachner moves on the left-hand side and right-hand
side of the $n$-gon equation. Note that there are
$\left\lfloor\frac{n-1}{2}\right\rfloor$ pairs in (1) with the smallest $i$
and $\left\lfloor\frac{n-1}{2}\right\rfloor$ pairs in (1) with the largest
$j$. The two Pachner moves for the initial triangulation of the $n$-gon act on
the $(n-3)$-simplices corresponding to the
$\left\lfloor\frac{n-1}{2}\right\rfloor$ pairs in (1) with the smallest $i$
and $\left\lfloor\frac{n-1}{2}\right\rfloor$ pairs in (1) with the largest $j$
respectively. We denote these two sets of pairs by

(2)
$\displaystyle(i_{\min},j_{1}),(i_{\min},j_{2}),\dots,(i_{\min},j_{\left\lfloor\frac{n-1}{2}\right\rfloor})$

and

(3)
$\displaystyle(i_{1},j_{\max}),(i_{2},j_{\max}),\dots,(i_{\left\lfloor\frac{n-1}{2}\right\rfloor},j_{\max})$

respectively. Each Pachner move in the $n$-gon equation only involves $n-1$
vertices, we denote the remaining vertex by $q$. Let
$\\{1,2,\dots,n\\}\setminus\\{q\\}=\\{a_{1},a_{2},\dots,a_{n-1}\\}$. We divide
the set $\\{a_{1},a_{2},\dots,a_{n-1}\\}$ into two parts:

(4)
$\displaystyle\\{a_{1},a_{2},\dots,a_{n-1}\\}=\\{b_{1},b_{2},\dots,b_{\left\lfloor\frac{n-1}{2}\right\rfloor}\\}\cup\\{c_{1},c_{2},\dots,c_{\left\lceil\frac{n-1}{2}\right\rceil}\\}$

where $\left\lceil x\right\rceil$ means the smallest integer greater than or
equal to $x$.

We denote the Pachner move that replaces the $(n-3)$-simplices corresponding
to the pairs $(b_{1},q)$, $(b_{2},q)$, $\dots$,
$(b_{\left\lfloor\frac{n-1}{2}\right\rfloor},q)$ with the $(n-3)$-simplices
corresponding to the pairs $(c_{1},q)$, $(c_{2},q)$, $\dots$,
$(c_{\left\lceil\frac{n-1}{2}\right\rceil},q)$ by $d_{b_{1}b_{2}\cdots
b_{\left\lfloor\frac{n-1}{2}\right\rfloor},c_{1}c_{2}\cdots
c_{\left\lceil\frac{n-1}{2}\right\rceil}}$ or $d_{b_{1}b_{2}\cdots
b_{\left\lfloor\frac{n-1}{2}\right\rfloor}}^{(q)}$ for short. Then the two
Pachner moves for the initial triangulation of the $n$-gon are
$d_{j_{1}j_{2}\cdots j_{\left\lfloor\frac{n-1}{2}\right\rfloor}}^{(i_{\min})}$
and $d_{i_{1}i_{2}\cdots
i_{\left\lfloor\frac{n-1}{2}\right\rfloor}}^{(j_{\max})}$ respectively.

######  Convention 2.1.

In this article, we list the elements of all sets always in increasing order,
if not explicitly stated otherwise.

For odd $n$, $i_{\min}=2$ and
$\\{j_{1},j_{2},\dots,j_{\frac{n-1}{2}}\\}=\\{3,5,\dots,n\\}$, $j_{\max}=n$
and $\\{i_{1},i_{2},\dots,i_{\frac{n-1}{2}}\\}=\\{2,4,\dots,n-1\\}$. The
$n$-gon equation is

(5) $\displaystyle d_{35\cdots n}^{(2)}d_{257\cdots n}^{(4)}\cdots
d_{24\cdots(n-3)n}^{(n-1)}=d_{24\cdots(n-1)}^{(n)}d_{24\cdots(n-3)n}^{(n-2)}\cdots
d_{35\cdots n}^{(1)}.$

For even $n$, $i_{\min}=3$ and
$\\{j_{1},j_{2},\dots,j_{\frac{n}{2}-1}\\}=\\{4,6,\dots,n\\}$, $j_{\max}=n$
and $\\{i_{1},i_{2},\dots,i_{\frac{n}{2}-1}\\}=\\{3,5,\dots,n-1\\}$. The
$n$-gon equation is

(6) $\displaystyle d_{46\cdots n}^{(3)}d_{368\cdots n}^{(5)}\cdots
d_{35\cdots(n-3)n}^{(n-1)}d_{35\cdots(n-1)}^{(1)}=d_{35\cdots(n-1)}^{(n)}d_{35\cdots(n-3)n}^{(n-2)}\cdots
d_{46\cdots n}^{(2)}.$

For odd $n$, the final triangulation of the $n$-gon consists of
$(n-3)$-simplices which correspond to the pairs

(7) $\displaystyle(i,j)=(n-2k,n+1-2l)\text{ for }1\leq l\leq
k\leq{\frac{n-1}{2}}.$

There are $\dfrac{{(\frac{n-1}{2})}({\frac{n+1}{2}})}{2}$ $(n-3)$-simplices in
the final triangulation of the $n$-gon. For even $n$, the final triangulation
of the $n$-gon consists of $(n-3)$-simplices which correspond to the pairs

(8) $\displaystyle(i,j)=(n-2k,n+1-2l)\text{ for }1\leq l\leq
k\leq{\frac{n}{2}}-1\text{ and }(1,n+2-2l)\text{ for }1\leq l\leq\frac{n}{2}.$

There are $\dfrac{{\frac{n}{2}}({\frac{n}{2}+1})}{2}$ $(n-3)$-simplices in the
initial triangulation of the $n$-gon. For example, the final triangulation of
the pentagon consists of triangles 125, 235, and 345 which correspond to the
pairs (3,4), (1,4), and (1,2). The final triangulation of the hexagon consists
of tetrahedra 1236, 1346, 1456, 2345, 2356, and 3456 which correspond to the
pairs (4,5), (2,5), (2,3), (1,6), (1,4), and (1,2). See Sec. 5 and Sec. 6.

The $n$-gon equation is shown in Fig. 2.

Figure 2. The $n$-gon equation involves two sequences of Pachner moves from
the initial triangulation to the final triangulation of the $n$-gon. Each
arrow with label $q$ denotes the Pachner move which does not involve the
vertex $q$.

##  3\. Main results

In this section, we formulate our main results (a matrix solution for any
polygon equation), and we will give some examples later.

###  3.1. Construction of matrices associated to Pachner moves

For any integer $n\geq 5$, let $\zeta_{i}$ ($i=1,2,\dots,n$) be pairwise
distinct variables in a field $F$. We need to assume that the characteristic
of $F$ is zero to prove our main results. For $1\leq k\leq n$ and $1\leq
i_{1}<i_{2}<\cdots<i_{k}\leq n$, we denote

(9) $\displaystyle V(i_{1},i_{2},\dots,i_{k})=\begin{vmatrix}1&1&\cdots&1\\\
\zeta_{i_{1}}&\zeta_{i_{2}}&\cdots&\zeta_{i_{k}}\\\
\zeta_{i_{1}}^{2}&\zeta_{i_{2}}^{2}&\cdots&\zeta_{i_{k}}^{2}\\\
\vdots&\vdots&\ddots&\vdots\\\
\zeta_{i_{1}}^{k-1}&\zeta_{i_{2}}^{k-1}&\cdots&\zeta_{i_{k}}^{k-1}\\\
\end{vmatrix}$

as the Vandermonde determinant.

If $n$ is odd, we associate a $\frac{n-1}{2}\times\frac{n-1}{2}$ matrix
$P_{a_{2}a_{4}a_{6}\cdots a_{n-1},a_{1}a_{3}a_{5}\cdots a_{n-2}}$ ($1\leq
a_{1}<a_{3}<\cdots<a_{n-2}\leq n$, $1\leq a_{2}<a_{4}<\cdots<a_{n-1}\leq n$,
$\\{a_{1},a_{3},\dots,a_{n-2}\\}\cap\\{a_{2},a_{4},\dots,a_{n-1}\\}=\emptyset$)
to the Pachner $\frac{n-1}{2}$-$\frac{n-1}{2}$ move for the $(n-1)$-gon
$a_{1}a_{2}a_{3}\cdots a_{n-1}$ (regarded as a simplicial complex) that
replaces the $(n-3)$-simplices $a_{1}a_{2}a_{3}\cdots a_{n-2}$,
$a_{1}a_{2}a_{3}\cdots a_{n-4}a_{n-2}a_{n-1}$, $\dots$, $a_{1}a_{3}a_{4}\cdots
a_{n-2}a_{n-1}$ with the $(n-3)$-simplices $a_{1}a_{2}a_{3}\cdots
a_{n-3}a_{n-1}$, $a_{1}a_{2}a_{3}\cdots a_{n-5}a_{n-3}a_{n-2}a_{n-1}$,
$\dots$, $a_{2}a_{3}a_{4}\cdots a_{n-1}$ as follows: the $(i,j)$-entry of the
matrix $P_{a_{2}a_{4}a_{6}\cdots a_{n-1},a_{1}a_{3}a_{5}\cdots a_{n-2}}$ is

(10) $\displaystyle(P_{a_{2}a_{4}a_{6}\cdots a_{n-1},a_{1}a_{3}a_{5}\cdots
a_{n-2}})_{i}^{j}=(-1)^{j+\frac{n-1}{2}}\frac{V(a_{n-2i},a_{2},a_{4},\dots,\hat{a}_{n+1-2j},\dots,a_{n-1})}{V(a_{2},a_{4},a_{6},\dots,a_{n-1})}$

where $\hat{a}_{k}$ means that $a_{k}$ is omitted.

If $n$ is even, we associate a $\frac{n}{2}\times(\frac{n}{2}-1)$ matrix
$P_{a_{1}a_{3}a_{5}\cdots a_{n-3},a_{2}a_{4}a_{6}\cdots a_{n-2}a_{n-1}}$
($1\leq a_{1}<a_{3}<\cdots<a_{n-3}\leq n$, $1\leq
a_{2}<a_{4}<\cdots<a_{n-2}<a_{n-1}\leq n$,
$\\{a_{1},a_{3},\dots,a_{n-3}\\}\cap\\{a_{2},a_{4},\dots,a_{n-2},a_{n-1}\\}=\emptyset$)
to the Pachner $(\frac{n}{2}-1)$-$\frac{n}{2}$ move for the $(n-1)$-gon
$a_{1}a_{2}a_{3}\cdots a_{n-1}$ (regarded as a simplicial complex) that
replaces the $(n-3)$-simplices $a_{1}a_{2}a_{3}\cdots a_{n-4}a_{n-2}a_{n-1}$,
$a_{1}a_{2}a_{3}\cdots a_{n-6}a_{n-4}a_{n-3}a_{n-2}a_{n-1}$, $\dots$,
$a_{2}a_{3}a_{4}\cdots a_{n-2}a_{n-1}$ with the $(n-3)$-simplices
$a_{1}a_{2}a_{3}\cdots a_{n-2}$, $a_{1}a_{2}a_{3}\cdots a_{n-3}a_{n-1}$,
$\dots$, $a_{1}a_{3}a_{4}\cdots a_{n-1}$ as follows: the $(i,j)$-entry of the
matrix $P_{a_{1}a_{3}a_{5}\cdots a_{n-3},a_{2}a_{4}a_{6}\cdots
a_{n-2}a_{n-1}}$ is

(13) $\displaystyle(P_{a_{1}a_{3}a_{5}\cdots a_{n-3},a_{2}a_{4}a_{6}\cdots
a_{n-2}a_{n-1}})_{i}^{j}=\left\\{\begin{array}[]{ll}(-1)^{j+\frac{n}{2}-1}\frac{V(a_{n+2-2i},a_{1},a_{3},\dots,\hat{a}_{n-1-2j},\dots,a_{n-3})}{V(a_{1},a_{3},a_{5},\dots,a_{n-3})}&\text{if
}i>1\\\
(-1)^{j+\frac{n}{2}-1}\frac{V(a_{n-1},a_{1},a_{3},\dots,\hat{a}_{n-1-2j},\dots,a_{n-3})}{V(a_{1},a_{3},a_{5},\dots,a_{n-3})}&\text{if
}i=1\end{array}\right.$

where $\hat{a}_{k}$ means that $a_{k}$ is omitted.

Note that the sum of the entries of each row of the matrix $P$ is 1. This
construction is obtained by induction on $n$ and inspired by [6], see later
sections for examples.

We can extend the matrices we constructed for Pachner moves above by adding a
row and a column so that the entry is 1 at the intersection and 0 elsewhere
corresponding to each fixed $(n-3)$-simplex in the Pachner moves.

######  Theorem 3.1.

The extended matrices satisfy the $n$-gon equation (5) or (6).

###  3.2. Comparison of our solution to the odd-gon equation with the one in
[1]

Note that our solution to the odd-gon equation is exactly a partial case of
the one in [1]. To show this, we introduce Grassmannian and Plücker
coordinates. Let $F$ be a field, $F^{n}$ the row vector space of dimension $n$
over $F$, $\mathsf{e}_{1},\ldots,\mathsf{e}_{n}$ the standard basis in
$F^{n}$.

We assume that $n$ is odd. Let

(14)
$\mathcal{M}=\begin{pmatrix}\alpha_{1,1}&\alpha_{1,2}&\cdots&\alpha_{1,n}\\\
\alpha_{2,1}&\alpha_{2,2}&\cdots&\alpha_{2,n}\\\
\vdots&\vdots&\ddots&\vdots\\\
\alpha_{{\frac{n-1}{2}},1}&\alpha_{{\frac{n-1}{2}},2}&\cdots&\alpha_{{\frac{n-1}{2}},n}\\\
\end{pmatrix}$

and

(15)
$\mathcal{M}^{\perp}=\begin{pmatrix}\beta_{1,1}&\beta_{1,2}&\cdots&\beta_{1,n}\\\
\beta_{2,1}&\beta_{2,2}&\cdots&\beta_{2,n}\\\ \vdots&\vdots&\ddots&\vdots\\\
\beta_{{\frac{n+1}{2}},1}&\beta_{{\frac{n+1}{2}},2}&\cdots&\beta_{{\frac{n+1}{2}},n}\\\
\end{pmatrix}$

be matrices defining elements $L$ of the Grassmannian
$\mathrm{Gr}({\frac{n-1}{2}},n)$ and $L^{\perp}$ of the Grassmannian
$\mathrm{Gr}({\frac{n+1}{2}},n)$ where $L^{\perp}$ is the orthogonal
complement of $L$, that is, matrices whose rows are of the full rank
${\frac{n-1}{2}}$ and ${\frac{n+1}{2}}$ respectively, which can also be
written as

$v_{i}=\beta_{i,1}\mathsf{e}_{1}+\dots+\beta_{i,n}\mathsf{e}_{n},$

span an ${\frac{n+1}{2}}$-dimensional plane $L^{\perp}\subset F^{n}$.

The _multivectors_ are elements of the _exterior algebra_ $\bigwedge F^{n}$
over $F^{n}$. We introduce an ${\frac{n+1}{2}}$-vector—the exterior product of
all $v_{i}$:

$w=v_{1}\wedge v_{2}\wedge\cdots\wedge
v_{{\frac{n+1}{2}}}=\sum_{k_{1}<\cdots<k_{{\frac{n+1}{2}}}}p_{k_{1},\ldots,k_{{\frac{n+1}{2}}}}(L^{\perp})\mathsf{e}_{k_{1}}\wedge\dots\wedge\mathsf{e}_{k_{{\frac{n+1}{2}}}},$

where $p_{k_{1},\ldots,k_{{\frac{n+1}{2}}}}(L^{\perp})$ are determinants made
of $k_{1}$-th, …, $k_{{\frac{n+1}{2}}}$-th columns of $\mathcal{M}^{\perp}$,
called also _Plücker coordinates_ of the Grassmannian
$\mathrm{Gr}({\frac{n+1}{2}},n)$.

######  Assumption 3.2.

Below, we assume that all Plücker coordinates are nonzero.

In Sec. 4, we will consider a partial case: Grassmannian defined by the
Vandermonde matrix, which satisfies Assumption 3.2.

The Plücker coordinates define an embedding, called _Plücker embedding_ , of
the Grassmannian $\mathrm{Gr}({\frac{n+1}{2}},n)$ into the projectivization
$\mathbb{P}(\bigwedge^{{\frac{n+1}{2}}}F^{n})$ of the ${\frac{n+1}{2}}$-th
exterior power of $F^{n}$. Note that the Plücker coordinates are homogeneous
coordinates, we may multiply them by a common factor.

######  Lemma 3.3.

Let
$\\{l_{1},\dots,l_{{\frac{n-1}{2}}}\\}=\\{1,2,\dots,n\\}\setminus\\{k_{1},\ldots,k_{{\frac{n+1}{2}}}\\}$
and $L^{\perp}$ be the orthogonal complement of $L$ in $F^{n}$, then the
Plücker coordinates of $L$ and $L^{\perp}$ are related as follows:

(16) $\displaystyle
p_{k_{1},\ldots,k_{{\frac{n+1}{2}}}}(L^{\perp})=(-1)^{\\#(\\{l_{1},\dots,l_{{\frac{n-1}{2}}}\\}\cap\\{1,2,\dots,{\frac{n+1}{2}}\\})}p_{l_{1},\dots,l_{{\frac{n-1}{2}}}}(L)$

up to a common factor.

###### Proof.

We write $\mathcal{M}=(A,B)$ where $A$ is a
${\frac{n+1}{2}}\times{\frac{n+1}{2}}$ matrix and $B$ is a
${\frac{n+1}{2}}\times{\frac{n-1}{2}}$ matrix. Since $A$ is invertible, we can
multiply $\mathcal{M}$ by $A^{-1}$ from the left, this changes the Plücker
coordinates by a common factor. Without loss of generality, we may assume that
$A=I$ is the identity ${\frac{n+1}{2}}\times{\frac{n+1}{2}}$ matrix and
$\mathcal{M}=(I,B)$. The matrix defining $L^{\perp}$ can be chosen as the
transpose of $\begin{pmatrix}-B\\\ I\end{pmatrix}$ where $I$ is the identity
${\frac{n-1}{2}}\times{\frac{n-1}{2}}$ matrix. Then a direct computation
proves this lemma. ∎

We rewrite the $(i,j)$-entry of the matrix in [1, eq. (44)] in our notation as

(17)
$\displaystyle(-1)^{i}\frac{p_{a_{2j},a_{1},a_{3},\dots,\hat{a}_{2i-1},\dots,a_{n-2},q}(L^{\perp})}{p_{a_{1},a_{3},\dots,a_{n-2},q}(L^{\perp})}$

where $\\{1,2,\dots,n\\}\setminus\\{q\\}=\\{a_{1},a_{2},\dots,a_{n-1}\\}$. We
compare this with (10) and use Lemma 3.3, we find that our solution to the
odd-gon equation is exactly a partial case of the one in [1].

##  4\. Proof of the main results

In this section, we prove Theorem 3.1 by the following steps:

  1. Step 1

We associate a vector $f^{(n)}_{1\cdots\hat{i}\cdots\hat{j}\cdots n}$ to each
$(n-3)$-simplex where $i$ and $j$ are not vertices of the $(n-3)$-simplex.

  2. Step 2

The $n$-gon equation involves two sequences of Pachner moves. Each sequence is
a path from the initial triangulation to the final triangulation of the
$n$-gon. The number of $(n-3)$-simplices in each step of the two paths does
not change for odd $n$ and increases by 1 for even $n$.

We can extend the matrices we constructed for Pachner moves in Sec. 3.1 by
adding a row and a column so that the entry is 1 at the intersection and 0
elsewhere corresponding to each fixed $(n-3)$-simplex in the Pachner moves.

By Proposition 4.5, the extended matrices map the vectors associated to
$(n-3)$-simplices in the triangulation in each step to the vectors associated
to $(n-3)$-simplices in the triangulation in the next step. So the products of
matrices on both sides of the $n$-gon equation act on the vectors associated
to $(n-3)$-simplices in the initial triangulation in the same way.

  3. Step 3

By Proposition 4.6, the vectors $f^{(n)}_{1\cdots\hat{i}\cdots\hat{j}\cdots
n}$ associated to $(n-3)$-simplices in the initial triangulation are linearly
independent, so we conclude that the products of matrices on both sides of the
$n$-gon equation are equal.

###  4.1. Construction of vectors associated to $(n-3)$-simplices

In this subsection, we construct vectors associated to $(n-3)$-simplices. This
construction is inspired by [6].

Let

(18) $\displaystyle\mathcal{M}=\begin{pmatrix}1&1&\cdots&1\\\
\zeta_{1}&\zeta_{2}&\cdots&\zeta_{n}\\\
\zeta_{1}^{2}&\zeta_{2}^{2}&\cdots&\zeta_{n}^{2}\\\
\vdots&\vdots&\ddots&\vdots\\\
\zeta_{1}^{\left\lfloor\frac{n}{2}\right\rfloor-1}&\zeta_{2}^{\left\lfloor\frac{n}{2}\right\rfloor-1}&\cdots&\zeta_{n}^{\left\lfloor\frac{n}{2}\right\rfloor-1}\end{pmatrix},$

then the Plücker coordinates are the Vandermonde determinants
$V(k_{1},\ldots,k_{\left\lfloor\frac{n}{2}\right\rfloor})$ which are nonzero.

Given $n$ pairwise distinct variables $\zeta_{i}$ ($i=1,2,\dots,n$) and a
$(n-3)$-simplex $a_{1}a_{2}\cdots a_{n-2}$ ($1\leq
a_{1}<a_{2}<\cdots<a_{n-2}\leq n$), we can construct a vector
$(f_{a_{1}}^{(n)},f_{a_{2}}^{(n)},\dots,f_{a_{n-2}}^{(n)})$ such that

(19) $\displaystyle\sum_{i=1}^{n-2}f_{a_{i}}^{(n)}\zeta_{a_{i}}^{m}=0\text{
for }m=0,1,\dots,\left\lfloor\frac{n}{2}\right\rfloor-1$

where $\left\lfloor x\right\rfloor$ means the largest integer less than or
equal to $x$. For example, we can choose

(20) $\displaystyle
f_{a_{i}}^{(n)}=f^{(n)}(a_{i},a_{1},a_{2},\dots,\hat{a}_{i},\dots,a_{n-2})$

where $\hat{a}_{i}$ means that $a_{i}$ is omitted and $f^{(n)}$ is a to-be-
defined function. To define $f^{(n)}$, we first define a function $g^{(n)}$ as
follows:

(21) $\displaystyle
g_{a_{i}}^{(n)}=g^{(n)}(a_{i},a_{1},a_{2},\dots,\hat{a}_{i},\dots,a_{n-2})$

and

(22) $\displaystyle
g^{(n)}(a_{1},a_{2},\dots,a_{n-2})=\frac{1}{\prod\limits_{i=2}^{n-2}(\zeta_{a_{1}}-\zeta_{a_{i}})}.$

Note that $g^{(n)}$ is symmetric with respect to the variables $a_{i}$ for
$2\leq i\leq n-2$. Then we define $f^{(n)}$ as follows:

(23) $\displaystyle f^{(n)}(a_{1},a_{2},\dots,a_{n-2})=\sum_{2\leq
i_{1}<i_{2}<\cdots<i_{\left\lfloor\frac{n}{2}\right\rfloor}\leq
n-2}g^{(\left\lfloor\frac{n}{2}\right\rfloor+3)}(a_{1},a_{i_{1}},a_{i_{2}},\dots,a_{i_{\left\lfloor\frac{n}{2}\right\rfloor}})$

where $\left\lfloor x\right\rfloor$ means the largest integer less than or
equal to $x$. For example,

$\displaystyle f^{(5)}(a_{1},a_{2},a_{3})$ $\displaystyle=$ $\displaystyle
g^{(5)}(a_{1},a_{2},a_{3}),{}$ $\displaystyle
f^{(6)}(a_{1},a_{2},a_{3},a_{4})$ $\displaystyle=$ $\displaystyle
g^{(6)}(a_{1},a_{2},a_{3},a_{4}),{}$ $\displaystyle
f^{(7)}(a_{1},a_{2},a_{3},a_{4},a_{5})$ $\displaystyle=$
$\displaystyle\sum_{i=2}^{5}g^{(6)}(a_{1},a_{2},\dots,\hat{a}_{i},\dots,a_{5}),{}$
(24) $\displaystyle f^{(8)}(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6})$
$\displaystyle=$
$\displaystyle\sum_{i=2}^{6}g^{(7)}(a_{1},a_{2},\dots,\hat{a}_{i},\dots,a_{6})$

where $\hat{a}_{i}$ means that $a_{i}$ is omitted.

######  Lemma 4.1.

The function $g^{(n)}$ defined above satisfies

(25) $\displaystyle\sum_{i=1}^{n-2}g_{a_{i}}^{(n)}\zeta_{a_{i}}^{m}=0\text{
for }m=0,1,\dots,n-4.$

###### Proof.

Note that

(28) $\displaystyle
g^{(n)}(a_{i},a_{1},a_{2},\dots,\hat{a}_{i},\dots,a_{n-2})=(-1)^{i+1}\frac{\prod\limits_{\begin{array}[]{cc}1\leq
j<k\leq n-2\\\ j,k\neq
i\end{array}}(\zeta_{a_{j}}-\zeta_{a_{k}})}{\prod\limits_{1\leq j<k\leq
n-2}(\zeta_{a_{j}}-\zeta_{a_{k}})}$

where both the denominator and the numerator are Vandermonde determinants.
Also, note that

(29)
$\displaystyle\begin{vmatrix}\zeta_{a_{1}}^{m}&\zeta_{a_{2}}^{m}&\cdots&\zeta_{a_{n-2}}^{m}\\\
1&1&\cdots&1\\\ \zeta_{a_{1}}&\zeta_{a_{2}}&\cdots&\zeta_{a_{n-2}}\\\
\zeta_{a_{1}}^{2}&\zeta_{a_{2}}^{2}&\cdots&\zeta_{a_{n-2}}^{2}\\\
\vdots&\vdots&\ddots&\vdots\\\
\zeta_{a_{1}}^{n-4}&\zeta_{a_{2}}^{n-4}&\cdots&\zeta_{a_{n-2}}^{n-4}\\\
\end{vmatrix}=0$

for $m=0,1,\dots,n-4$. Then expanding this determinant along the first row
proves this lemma. ∎

######  Proposition 4.2.

The function $f^{(n)}$ defined above satisfies (19).

###### Proof.

By definition,

(32) $\displaystyle
f^{(n)}(a_{i},a_{1},a_{2},\dots,\hat{a}_{i},\dots,a_{n-2}){}$ $\displaystyle=$
$\displaystyle\sum_{\begin{array}[]{cc}1\leq
i_{1}<i_{2}<\cdots<i_{\left\lfloor\frac{n}{2}\right\rfloor}\leq n-2\\\
i_{l}\neq i,\forall
l\end{array}}g^{(\left\lfloor\frac{n}{2}\right\rfloor+3)}(a_{i},a_{i_{1}},a_{i_{2}},\dots,a_{i_{\left\lfloor\frac{n}{2}\right\rfloor}})$

where $\left\lfloor x\right\rfloor$ means the largest integer less than or
equal to $x$. Therefore,

$\displaystyle\sum_{i=1}^{n-2}f^{(n)}(a_{i},a_{1},a_{2},\dots,\hat{a}_{i},\dots,a_{n-2})\zeta_{a_{i}}^{m}{}$
$\displaystyle=$ $\displaystyle\sum_{i=1}^{n-2}\sum_{\begin{array}[]{cc}1\leq
i_{1}<i_{2}<\cdots<i_{\left\lfloor\frac{n}{2}\right\rfloor}\leq n-2\\\
i_{l}\neq i,\forall
l\end{array}}g^{(\left\lfloor\frac{n}{2}\right\rfloor+3)}(a_{i},a_{i_{1}},a_{i_{2}},\dots,a_{i_{\left\lfloor\frac{n}{2}\right\rfloor}})\zeta_{a_{i}}^{m}{}$
$\displaystyle=$ $\displaystyle\sum_{1\leq
j_{1}<j_{2}<\cdots<j_{\left\lfloor\frac{n}{2}\right\rfloor+1}\leq
n-2}\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor+1}g^{(\left\lfloor\frac{n}{2}\right\rfloor+3)}(a_{j_{k}},a_{j_{1}},\dots,\hat{a}_{j_{k}},\dots,a_{j_{\left\lfloor\frac{n}{2}\right\rfloor+1}})\zeta_{a_{j_{k}}}^{m}{}$
(36) $\displaystyle=$ $\displaystyle 0$

where $\hat{a}_{i}$ means that $a_{i}$ is omitted,
$\\{j_{1},j_{2},\dots,j_{\left\lfloor\frac{n}{2}\right\rfloor+1}\\}=\\{i,i_{1},i_{2},\dots,i_{\left\lfloor\frac{n}{2}\right\rfloor}\\}$,
and we have used Lemma 4.1 in the last equality. Hence we have proved that the
function $f^{(n)}$ satisfies (19). ∎

We add zeros in the vector
$(f_{a_{1}}^{(n)},f_{a_{2}}^{(n)},\dots,f_{a_{n-2}}^{(n)})$ at the positions
which are not the vertices of the $(n-3)$-simplex to extend the vector
$(f_{a_{1}}^{(n)},f_{a_{2}}^{(n)},\dots,f_{a_{n-2}}^{(n)})$ to a vector of
length $n$. We denote the extended vector by $f^{(n)}_{a_{1}a_{2}\cdots
a_{n-2}}$.

Proposition 4.2 implies that the vectors $f^{(n)}_{a_{1}a_{2}\cdots a_{n-2}}$
are orthogonal to the row vectors of $\mathcal{M}$ (18), namely the vectors
$f^{(n)}_{a_{1}a_{2}\cdots a_{n-2}}$ are in $L^{\perp}$ where $L$ is the
$\left\lfloor\frac{n}{2}\right\rfloor$-plane in $F^{n}$ defined by
$\mathcal{M}$.

Each Pachner move in the $n$-gon equation only involves $n-1$ vertices, we
denote the remaining vertex by $q$. Let
$\\{1,2,\dots,n\\}\setminus\\{q\\}=\\{a_{1},a_{2},\dots,a_{n-1}\\}$.

######  Proposition 4.3.

For any $\left\lfloor\frac{n-1}{2}\right\rfloor$ different values $j_{1}$,
$\dots$, $j_{\left\lfloor\frac{n-1}{2}\right\rfloor}$ in
$\\{1,2,\dots,n-1\\}$, the vectors $f^{(n)}_{a_{1}\cdots\hat{a}_{j_{m}}\cdots
a_{n-1}}$ $(m=1,\dots,\left\lfloor\frac{n-1}{2}\right\rfloor)$ are linearly
independent.

###### Proof.

By definition, the $k$-th component of the vector $f^{(n)}_{a_{1}a_{2}\cdots
a_{n-2}}$ is

(39) $\displaystyle(f^{(n)}_{a_{1}a_{2}\cdots
a_{n-2}})^{k}=\left\\{\begin{array}[]{ll}f^{(n)}(a_{i},a_{1},\dots,\hat{a}_{i},\dots,a_{n-2})&\text{if
}k=a_{i}\\\ 0&\text{if }k\not\in\\{a_{1},\dots,a_{n-2}\\}\end{array}\right.$

and

(42) $\displaystyle f^{(n)}(a_{i},a_{1},\dots,\hat{a}_{i},\dots,a_{n-2})$
$\displaystyle=$ $\displaystyle\sum_{\begin{array}[]{cc}1\leq
i_{1}<i_{2}<\cdots<i_{\left\lfloor\frac{n}{2}\right\rfloor}\leq n-2\\\
i_{l}\neq i,\forall
l\end{array}}\frac{V(a_{i_{1}},a_{i_{2}},\dots,a_{i_{\left\lfloor\frac{n}{2}\right\rfloor}})}{V(a_{i},a_{i_{1}},a_{i_{2}},\dots,a_{i_{\left\lfloor\frac{n}{2}\right\rfloor}})}{}$
(45) $\displaystyle=$ $\displaystyle\sum_{\begin{array}[]{cc}1\leq
i_{1}<i_{2}<\cdots<i_{\left\lfloor\frac{n}{2}\right\rfloor}\leq n-2\\\
i_{l}\neq i,\forall
l\end{array}}\frac{1}{\prod\limits_{l=1}^{\left\lfloor\frac{n}{2}\right\rfloor}(\zeta_{a_{i}}-\zeta_{a_{i_{l}}})}$

where $V(a_{i_{1}},a_{i_{2}},\dots,a_{i_{l}})$ is the Vandermonde determinant
(9) and $\left\lfloor x\right\rfloor$ means the largest integer less than or
equal to $x$.

We need to prove that if

(46)
$\displaystyle\sum_{m=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\lambda_{m}f^{(n)}_{a_{1}\cdots\hat{a}_{j_{m}}\cdots
a_{n-1}}=0,$

then $\lambda_{m}=0$ for $m=1,\dots,\left\lfloor\frac{n-1}{2}\right\rfloor$.
The $j_{m_{0}}$-th component is

(51) $\displaystyle\sum_{\begin{array}[]{cc}m=1\\\ m\neq
m_{0}\end{array}}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\lambda_{m}\sum_{\begin{array}[]{cc}1\leq
i_{1}<i_{2}<\cdots<i_{\left\lfloor\frac{n}{2}\right\rfloor}\leq n-1\\\
i_{l}\neq j_{m},j_{m_{0}},\forall
l\end{array}}\frac{1}{\prod\limits_{l=1}^{\left\lfloor\frac{n}{2}\right\rfloor}(\zeta_{a_{j_{m_{0}}}}-\zeta_{a_{i_{l}}})}=0.$

The coefficient of the term corresponding to
$\\{i_{1},\dots,i_{\left\lfloor\frac{n}{2}\right\rfloor}\\}=\\{1,2,\dots,n-1\\}\setminus\\{j_{1},\dots,j_{\left\lfloor\frac{n-1}{2}\right\rfloor}\\}$
is

(54) $\displaystyle\sum_{\begin{array}[]{cc}m=1\\\ m\neq
m_{0}\end{array}}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\lambda_{m}=0.$

Hence $\lambda_{m}=0$ for $m=1,\dots,\left\lfloor\frac{n-1}{2}\right\rfloor$.
We have used the assumption that the characteristic of $F$ is zero. ∎

######  Proposition 4.4.

The vectors $f^{(n)}_{a_{1}a_{2}\cdots\hat{a}_{i}\cdots a_{n-1}}$
($i=1,2,\dots,n-1$) span a
$\left\lfloor\frac{n-1}{2}\right\rfloor$-dimensional linear space.

###### Proof.

By Proposition 4.2, the vectors $f^{(n)}_{a_{1}a_{2}\cdots\hat{a}_{i}\cdots
a_{n-1}}$ are in $L^{\perp}$ where $L$ is the
$\left\lfloor\frac{n}{2}\right\rfloor$-plane in $F^{n}$ defined by
$\mathcal{M}$ (18). Also, by definition, the $q$-th component of the vectors
$f^{(n)}_{a_{1}a_{2}\cdots\hat{a}_{i}\cdots a_{n-1}}$ is zero. So the vectors
$f^{(n)}_{a_{1}a_{2}\cdots\hat{a}_{i}\cdots a_{n-1}}$ are in
$(L\oplus\mathsf{e}_{q})^{\perp}$ where $\mathsf{e}_{q}$ is the $q$-th
standard basis of $F^{n}$. Therefore, the dimension of the linear space
spanned by the vectors $f^{(n)}_{a_{1}a_{2}\cdots\hat{a}_{i}\cdots a_{n-1}}$
is less than or equal to
$n-1-\left\lfloor\frac{n}{2}\right\rfloor=\left\lfloor\frac{n-1}{2}\right\rfloor$.
By Proposition 4.3, the dimension of the linear space spanned by the vectors
$f^{(n)}_{a_{1}a_{2}\cdots\hat{a}_{i}\cdots a_{n-1}}$ is exactly
$\left\lfloor\frac{n-1}{2}\right\rfloor$. ∎

By Proposition 4.4, the vectors $f^{(n)}_{a_{1}a_{2}\cdots\hat{a}_{i}\cdots
a_{n-1}}$ associated to the $(n-3)$-simplices in the triangulation after a
Pachner move are linear combinations of the vectors
$f^{(n)}_{a_{1}a_{2}\cdots\hat{a}_{i}\cdots a_{n-1}}$ associated to the
$(n-3)$-simplices in the triangulation before that Pachner move.

######  Proposition 4.5.

For odd $n$,

(55) $\displaystyle P_{a_{2}a_{4}a_{6}\cdots a_{n-1},a_{1}a_{3}a_{5}\cdots
a_{n-2}}\cdot\begin{pmatrix}f^{(n)}_{a_{1}a_{2}a_{3}\cdots a_{n-2}}\\\
f^{(n)}_{a_{1}a_{2}a_{3}\cdots a_{n-4}a_{n-2}a_{n-1}}\\\ \vdots\\\
f^{(n)}_{a_{1}a_{3}a_{4}\cdots
a_{n-2}a_{n-1}}\end{pmatrix}=\begin{pmatrix}f^{(n)}_{a_{1}a_{2}a_{3}\cdots
a_{n-3}a_{n-1}}\\\ f^{(n)}_{a_{1}a_{2}a_{3}\cdots
a_{n-5}a_{n-3}a_{n-2}a_{n-1}}\\\ \vdots\\\ f^{(n)}_{a_{2}a_{3}a_{4}\cdots
a_{n-1}}\end{pmatrix}.$

For even $n$,

(56) $\displaystyle P_{a_{1}a_{3}a_{5}\cdots a_{n-3},a_{2}a_{4}a_{6}\cdots
a_{n-2}a_{n-1}}\cdot\begin{pmatrix}f^{(n)}_{a_{1}a_{2}a_{3}\cdots
a_{n-4}a_{n-2}a_{n-1}}\\\ f^{(n)}_{a_{1}a_{2}a_{3}\cdots
a_{n-6}a_{n-4}a_{n-3}a_{n-2}a_{n-1}}\\\ \vdots\\\
f^{(n)}_{a_{2}a_{3}a_{4}\cdots
a_{n-2}a_{n-1}}\end{pmatrix}=\begin{pmatrix}f^{(n)}_{a_{1}a_{2}a_{3}\cdots
a_{n-2}}\\\ f^{(n)}_{a_{1}a_{2}a_{3}\cdots a_{n-3}a_{n-1}}\\\ \vdots\\\
f^{(n)}_{a_{1}a_{3}a_{4}\cdots a_{n-1}}\end{pmatrix}.$

###### Proof.

For odd $n$, note that

(61) $\displaystyle(P_{a_{2}a_{4}a_{6}\cdots a_{n-1},a_{1}a_{3}a_{5}\cdots
a_{n-2}})_{i}^{j}$ $\displaystyle=$
$\displaystyle(-1)^{j+\frac{n-1}{2}}\frac{V(a_{n-2i},a_{2},a_{4},\dots,\hat{a}_{n+1-2j},\dots,a_{n-1})}{V(a_{2},a_{4},a_{6},\dots,a_{n-1})}{}$
$\displaystyle=$
$\displaystyle\frac{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n-1}{2}}(\zeta_{a_{n-2i}}-\zeta_{a_{n+1-2j^{\prime}}})}{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n-1}{2}}(\zeta_{a_{n+1-2j}}-\zeta_{a_{n+1-2j^{\prime}}})},$

we need to prove that

(66)
$\displaystyle\sum_{j=1}^{\frac{n-1}{2}}\frac{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n-1}{2}}(\zeta_{a_{n-2i}}-\zeta_{a_{n+1-2j^{\prime}}})}{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n-1}{2}}(\zeta_{a_{n+1-2j}}-\zeta_{a_{n+1-2j^{\prime}}})}f^{(n)}_{a_{1}a_{2}\cdots\hat{a}_{n+1-2j}\cdots
a_{n-1}}{}$ (67) $\displaystyle=$ $\displaystyle
f^{(n)}_{a_{1}\cdots\hat{a}_{n-2i}\cdots a_{n-1}}$

where $\hat{a}_{i}$ means that $a_{i}$ is omitted. In particular, the
$a_{n-2i}$-th component is

(74)
$\displaystyle\sum_{j=1}^{\frac{n-1}{2}}\frac{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n-1}{2}}(\zeta_{a_{n-2i}}-\zeta_{a_{n+1-2j^{\prime}}})}{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n-1}{2}}(\zeta_{a_{n+1-2j}}-\zeta_{a_{n+1-2j^{\prime}}})}\sum_{\begin{array}[]{cc}1\leq
i_{1}<i_{2}<\cdots<i_{{\frac{n-1}{2}}}\leq n-1\\\ i_{l}\neq
n+1-2j,n-2i,\forall
l\end{array}}\frac{1}{\prod\limits_{l=1}^{{\frac{n-1}{2}}}(\zeta_{a_{n-2i}}-\zeta_{a_{i_{l}}})}=0.$

The $a_{n-2i^{\prime}}$-th component ($i^{\prime}\neq i$) is

(81)
$\displaystyle\sum_{j=1}^{\frac{n-1}{2}}\frac{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n-1}{2}}(\zeta_{a_{n-2i}}-\zeta_{a_{n+1-2j^{\prime}}})}{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n-1}{2}}(\zeta_{a_{n+1-2j}}-\zeta_{a_{n+1-2j^{\prime}}})}\sum_{\begin{array}[]{cc}1\leq
i_{1}<i_{2}<\cdots<i_{{\frac{n-1}{2}}}\leq n-1\\\ i_{l}\neq
n+1-2j,n-2i^{\prime},\forall
l\end{array}}\frac{1}{\prod\limits_{l=1}^{{\frac{n-1}{2}}}(\zeta_{a_{n-2i^{\prime}}}-\zeta_{a_{i_{l}}})}{}$
(84) $\displaystyle=$ $\displaystyle\sum_{\begin{array}[]{cc}1\leq
i_{1}<i_{2}<\cdots<i_{{\frac{n-1}{2}}}\leq n-1\\\ i_{l}\neq
n-2i,n-2i^{\prime},\forall
l\end{array}}\frac{1}{\prod\limits_{l=1}^{{\frac{n-1}{2}}}(\zeta_{a_{n-2i^{\prime}}}-\zeta_{a_{i_{l}}})}.$

The $a_{n+1-2j_{0}}$-th component is

(93) $\displaystyle\sum_{\begin{array}[]{cc}j=1\\\ j\neq
j_{0}\end{array}}^{\frac{n-1}{2}}\frac{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n-1}{2}}(\zeta_{a_{n-2i}}-\zeta_{a_{n+1-2j^{\prime}}})}{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n-1}{2}}(\zeta_{a_{n+1-2j}}-\zeta_{a_{n+1-2j^{\prime}}})}\sum_{\begin{array}[]{cc}1\leq
i_{1}<i_{2}<\cdots<i_{{\frac{n-1}{2}}}\leq n-1\\\ i_{l}\neq
n+1-2j,n+1-2j_{0},\forall
l\end{array}}\frac{1}{\prod\limits_{l=1}^{{\frac{n-1}{2}}}(\zeta_{a_{n+1-2j_{0}}}-\zeta_{a_{i_{l}}})}{}$
(96) $\displaystyle=$ $\displaystyle\sum_{\begin{array}[]{cc}1\leq
i_{1}<i_{2}<\cdots<i_{{\frac{n-1}{2}}}\leq n-1\\\ i_{l}\neq
n-2i,n+1-2j_{0},\forall
l\end{array}}\frac{1}{\prod\limits_{l=1}^{{\frac{n-1}{2}}}(\zeta_{a_{n+1-2j_{0}}}-\zeta_{a_{i_{l}}})}.$

We can use the fact

(99)
$\displaystyle\sum_{j=1}^{\frac{n-1}{2}}\frac{1}{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n-1}{2}}(\zeta_{a_{n+1-2j}}-\zeta_{a_{n+1-2j^{\prime}}})}=0$

to prove (74). We can use the trick

(100)
$\displaystyle\frac{\zeta_{a_{n-2i}}-\zeta_{a_{n+1-2j^{\prime}}}}{(\zeta_{a_{n-2i^{\prime}}}-\zeta_{a_{n-2i}})(\zeta_{a_{n-2i^{\prime}}}-\zeta_{a_{n+1-2j^{\prime}}})}=\frac{1}{\zeta_{a_{n-2i^{\prime}}}-\zeta_{a_{n-2i}}}-\frac{1}{\zeta_{a_{n-2i^{\prime}}}-\zeta_{a_{n+1-2j^{\prime}}}}$

to prove (81). The proof for (93) is straightforward.

For even $n$, note that

(103) $\displaystyle(P_{a_{1}a_{3}a_{5}\cdots a_{n-3},a_{2}a_{4}a_{6}\cdots
a_{n-2}a_{n-1}})_{i}^{j}$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}(-1)^{j+\frac{n}{2}-1}\frac{V(a_{n+2-2i},a_{1},a_{3},\dots,\hat{a}_{n-1-2j},\dots,a_{n-3})}{V(a_{1},a_{3},a_{5},\dots,a_{n-3})}&\text{if
}i>1\\\
(-1)^{j+\frac{n}{2}-1}\frac{V(a_{n-1},a_{1},a_{3},\dots,\hat{a}_{n-1-2j},\dots,a_{n-3})}{V(a_{1},a_{3},a_{5},\dots,a_{n-3})}&\text{if
}i=1\end{array}\right.{}$ (114) $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\frac{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n}{2}-1}(\zeta_{a_{n+2-2i}}-\zeta_{a_{n-1-2j^{\prime}}})}{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n}{2}-1}(\zeta_{a_{n-1-2j}}-\zeta_{a_{n-1-2j^{\prime}}})}&\text{if
}i>1\\\ \frac{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\ j^{\prime}\neq
j\end{array}}^{\frac{n}{2}-1}(\zeta_{a_{n-1}}-\zeta_{a_{n-1-2j^{\prime}}})}{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n}{2}-1}(\zeta_{a_{n-1-2j}}-\zeta_{a_{n-1-2j^{\prime}}})}&\text{if
}i=1\end{array}\right.,$

we need to prove that for $i>1$,

(119)
$\displaystyle\sum_{j=1}^{\frac{n}{2}-1}\frac{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n}{2}-1}(\zeta_{a_{n+2-2i}}-\zeta_{a_{n-1-2j^{\prime}}})}{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n}{2}-1}(\zeta_{a_{n-1-2j}}-\zeta_{a_{n-1-2j^{\prime}}})}f^{(n)}_{a_{1}\cdots\hat{a}_{n-1-2j}\cdots
a_{n-1}}{}$ (120) $\displaystyle=$ $\displaystyle
f^{(n)}_{a_{1}\cdots\hat{a}_{n+2-2i}\cdots a_{n-1}},$

and

(125)
$\displaystyle\sum_{j=1}^{\frac{n}{2}-1}\frac{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n}{2}-1}(\zeta_{a_{n-1}}-\zeta_{a_{n-1-2j^{\prime}}})}{\prod\limits_{\begin{array}[]{cc}j^{\prime}=1\\\
j^{\prime}\neq
j\end{array}}^{\frac{n}{2}-1}(\zeta_{a_{n-1-2j}}-\zeta_{a_{n-1-2j^{\prime}}})}f^{(n)}_{a_{1}\cdots\hat{a}_{n-1-2j}\cdots
a_{n-1}}{}$ (126) $\displaystyle=$ $\displaystyle f^{(n)}_{a_{1}\cdots
a_{n-2}}$

where $\hat{a}_{i}$ means that $a_{i}$ is omitted. The proof is similar to
that for odd $n$. ∎

The $n$-gon equation involves two sequences of Pachner moves. Each sequence is
a path from the initial triangulation to the final triangulation of the
$n$-gon. The number of $(n-3)$-simplices in each step of the two paths does
not change for odd $n$ and increases by 1 for even $n$.

We can extend the matrices we constructed for Pachner moves in Sec. 3.1 by
adding a row and a column so that the entry is 1 at the intersection and 0
elsewhere corresponding to each fixed $(n-3)$-simplex in the Pachner moves.

By Proposition 4.5, the extended matrices map the vectors associated to
$(n-3)$-simplices in the triangulation in each step to the vectors associated
to $(n-3)$-simplices in the triangulation in the next step. So the products of
matrices on both sides of the $n$-gon equation act on the vectors associated
to $(n-3)$-simplices in the initial triangulation in the same way.

######  Proposition 4.6.

The vectors $f^{(n)}_{1\cdots\hat{i}\cdots\hat{j}\cdots n}$ associated to
$(n-3)$-simplices in the initial triangulation of the $n$-gon (1) are linearly
independent, that is, span an
$\dfrac{\left\lfloor\frac{n-1}{2}\right\rfloor(\left\lfloor\frac{n-1}{2}\right\rfloor+1)}{2}$-dimensional
linear space.

###### Proof.

By definition,

(131) $\displaystyle f^{(n)}_{1\cdots\hat{i}\cdots\hat{j}\cdots
n}=\sum_{\begin{array}[]{cc}r=1\\\ r\neq
i,j\end{array}}^{n}\mathsf{e}_{r}\sum_{\begin{array}[]{cc}1\leq
i_{1}<\cdots<i_{\left\lfloor\frac{n}{2}\right\rfloor}\leq n\\\ i_{m}\neq
i,j,r,\forall
m\end{array}}\frac{1}{\prod\limits_{m=1}^{\left\lfloor\frac{n}{2}\right\rfloor}(\zeta_{r}-\zeta_{{i_{m}}})}$

where $\mathsf{e}_{r}$ ($r=1,2,\dots,n$) are the standard basis of $F^{n}$. We
need to prove that if

(132)
$\displaystyle\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\sum_{l=1}^{k}\lambda_{n+1-2k,n+2-2l}f^{(n)}_{1\cdots\widehat{n+1-2k}\cdots\widehat{n+2-2l}\cdots
n}=0,$

then $\lambda_{n+1-2k,n+2-2l}=0$ for $1\leq l\leq
k\leq\left\lfloor\frac{n-1}{2}\right\rfloor$. For each fixed $l_{0}$, the
coefficient of $\mathsf{e}_{n+2-2l_{0}}$ is

(137)
$\displaystyle\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\sum_{\begin{array}[]{cc}l=1\\\
l\neq
l_{0}\end{array}}^{k}\lambda_{n+1-2k,n+2-2l}\sum_{\begin{array}[]{cc}1\leq
i_{1}<\cdots<i_{\left\lfloor\frac{n}{2}\right\rfloor}\leq n\\\ i_{m}\neq
n+1-2k,n+2-2l,n+2-2l_{0},\forall
m\end{array}}\frac{1}{\prod\limits_{m=1}^{\left\lfloor\frac{n}{2}\right\rfloor}(\zeta_{n+2-2l_{0}}-\zeta_{{i_{m}}})}.$

For each fixed $k_{0}$, the coefficient of the term corresponding to
$\\{i_{1},\dots,i_{\left\lfloor\frac{n}{2}\right\rfloor}\\}=\\{1,2,\dots,n\\}\setminus(\\{n+2-2l\mid
l=1,2,\dots,\left\lfloor\frac{n-1}{2}\right\rfloor\\}\cup\\{n+1-2k_{0}\\})$ in
the coefficient of $\mathsf{e}_{n+2-2l_{0}}$ is

(140) $\displaystyle\sum_{\begin{array}[]{cc}l=1\\\ l\neq
l_{0}\end{array}}^{k_{0}}\lambda_{n+1-2k_{0},n+2-2l}=0.$

Therefore,

(141)
$\displaystyle\lambda_{n+1-2k_{0},n+2-2l_{0}}=\sum_{l=1}^{k_{0}}\lambda_{n+1-2k_{0},n+2-2l}$

is independent of $l_{0}$. Again, by (140), we have
$\lambda_{n+1-2k_{0},n+2-2l_{0}}=0$ for $k_{0}\geq 2$. We have used the
assumption that the characteristic of $F$ is zero.

Similarly, for each fixed $k_{0}$, the coefficient of
$\mathsf{e}_{n+1-2k_{0}}$ is

(146)
$\displaystyle\sum_{l=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\sum_{\begin{array}[]{cc}k=l\\\
k\neq
k_{0}\end{array}}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\lambda_{n+1-2k,n+2-2l}\sum_{\begin{array}[]{cc}1\leq
i_{1}<\cdots<i_{\left\lfloor\frac{n}{2}\right\rfloor}\leq n\\\ i_{m}\neq
n+1-2k,n+2-2l,n+1-2k_{0},\forall
m\end{array}}\frac{1}{\prod\limits_{m=1}^{\left\lfloor\frac{n}{2}\right\rfloor}(\zeta_{n+1-2k_{0}}-\zeta_{{i_{m}}})}.$

For each fixed $l_{0}$, the coefficient of the term corresponding to
$\\{i_{1},\dots,i_{\left\lfloor\frac{n}{2}\right\rfloor}\\}=\\{1,2,\dots,n\\}\setminus(\\{n+1-2k\mid
k=1,2,\dots,\left\lfloor\frac{n-1}{2}\right\rfloor\\}\cup\\{n+2-2l_{0}\\})$ in
the coefficient of $\mathsf{e}_{n+1-2k_{0}}$ is

(149) $\displaystyle\sum_{\begin{array}[]{cc}k=l_{0}\\\ k\neq
k_{0}\end{array}}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\lambda_{n+1-2k,n+2-2l_{0}}=0.$

Therefore,

(150)
$\displaystyle\lambda_{n+1-2k_{0},n+2-2l_{0}}=\sum_{k=l_{0}}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\lambda_{n+1-2k,n+2-2l_{0}}$

is independent of $k_{0}$. Again, by (149), we have
$\lambda_{n+1-2k_{0},n+2-2l_{0}}=0$ for
$l_{0}\leq\left\lfloor\frac{n-1}{2}\right\rfloor-1$. We have used the
assumption that the characteristic of $F$ is zero.

We combine these results and conclude that $\lambda_{n+1-2k,n+2-2l}=0$ for
$1\leq l\leq k\leq\left\lfloor\frac{n-1}{2}\right\rfloor$. ∎

By Proposition 4.6, the vectors $f^{(n)}_{1\cdots\hat{i}\cdots\hat{j}\cdots
n}$ associated to $(n-3)$-simplices in the initial triangulation are linearly
independent, so we conclude that the products of matrices on both sides of the
$n$-gon equation are equal. Hence we have proved Theorem 3.1.

##  5\. Example: solving the pentagon equation

###  5.1. Construction of vectors associated to triangles and matrices
associated to Pachner 2-2 moves

In this subsection, we construct vectors associated to triangles and matrices
associated to Pachner 2-2 moves. This construction is inspired by [6].

Given 3 pairwise distinct variables $\zeta_{i}$ ($i=1,2,3$) and a triangle
123, we can construct a vector $(f_{1},f_{2},f_{3})$ such that

(151) $\displaystyle f_{1}+f_{2}+f_{3}=0\text{ and
}f_{1}\zeta_{1}+f_{2}\zeta_{2}+f_{3}\zeta_{3}=0.$

For example,

(152)
$\displaystyle(f_{1},f_{2},f_{3})=(\frac{1}{(\zeta_{1}-\zeta_{2})(\zeta_{1}-\zeta_{3})},\frac{1}{(\zeta_{2}-\zeta_{1})(\zeta_{2}-\zeta_{3})},\frac{1}{(\zeta_{3}-\zeta_{1})(\zeta_{3}-\zeta_{2})}).$

For the quadrilateral 1234, the Pachner 2-2 move replaces the triangles 123
and 134 with the triangles 124 and 234. We will construct a $2\times 2$ matrix
$P_{24,13}$ for this Pachner move such that

(155) $\displaystyle
P_{24,13}\left(\begin{array}[]{cccc}f_{1}(123)&f_{2}(123)&f_{3}(123)&0\\\
f_{1}(134)&0&f_{2}(134)&f_{3}(134)\end{array}\right)$ (158) $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cccc}f_{1}(124)&f_{2}(124)&0&f_{3}(124)\\\
0&f_{1}(234)&f_{2}(234)&f_{3}(234)\end{array}\right).{}$

We solve this matrix equation and get

(161) $\displaystyle
P_{24,13}=\left(\begin{array}[]{cc}\frac{\zeta_{2}-\zeta_{3}}{\zeta_{2}-\zeta_{4}}&\frac{\zeta_{3}-\zeta_{4}}{\zeta_{2}-\zeta_{4}}\\\
\frac{\zeta_{2}-\zeta_{1}}{\zeta_{2}-\zeta_{4}}&\frac{\zeta_{1}-\zeta_{4}}{\zeta_{2}-\zeta_{4}}\end{array}\right).$

Note that the matrix $P_{24,13}$ is invertible and the sum of the entries of
each row of the matrix $P_{24,13}$ is 1. This gives a solution to the pentagon
equation which is essentially the one in [9] after taking the transpose.

###  5.2. A solution to the pentagon equation

In this subsection, we check that (161) gives a solution to the pentagon
equation.

The pentagon equation is

(162) $\displaystyle d_{35,14}d_{25,13}=d_{24,13}d_{25,14}d_{35,24}$

where $d_{kl,ij}$ ($1\leq i<j\leq 5$, $1\leq k<l\leq 5$,
$\\{i,j\\}\cap\\{k,l\\}=\emptyset$) corresponds to the Pachner 2-2 move in a
quadrilateral $ikjl$ that replaces the triangles $ijk$, $ijl$ with the
triangles $ikl$, $jkl$ (the order of vertices can be rearranged from small to
large).

We will construct matrices $a_{kl,ij}$ for each Pachner 2-3 move satisfying
the pentagon equation (162).

For the quadrilateral 1234, the Pachner 2-2 move replaces the triangles 123
and 134 with the triangles 124 and 234. We can construct a $2\times 2$ matrix
$P_{24,13}$ (161) for this Pachner move as we did in Sec. 5.1.

We can extend this matrix by adding a row and a column so that the entry is 1
at the intersection and 0 elsewhere corresponding to each fixed tetrahedron in
the Pachner 2-2 moves. The extended matrix is denoted as $a_{24,13}$.

Figure 3. Triangulation graph for the pentagon.

The pentagon equation (162) involves 5 Pachner 2-2 moves: (1)$\to$(2)$\to$(5)
and (1)$\to$(3)$\to$(4)$\to$(5) where (1)-(5) are triangulations of a pentagon
as follows (see Fig. 3):

  1. (1)

123, 134, 145,

  2. (2)

123, 135, 345,

  3. (3)

124, 234, 145,

  4. (4)

125, 234, 245,

  5. (5)

125, 235, 345.

We can check that the matrices $a_{kl,ij}$ for these Pachner 2-2 moves satisfy
the pentagon equation (162), namely

(169)
$\displaystyle\left(\begin{array}[]{ccc}\frac{\zeta_{3}-\zeta_{2}}{\zeta_{5}-\zeta_{2}}&\frac{\zeta_{1}-\zeta_{2}}{\zeta_{5}-\zeta_{2}}&\\\
\frac{\zeta_{3}-\zeta_{5}}{\zeta_{2}-\zeta_{5}}&\frac{\zeta_{1}-\zeta_{5}}{\zeta_{2}-\zeta_{5}}&\\\
&&1\end{array}\right)\left(\begin{array}[]{ccc}1&&\\\
&\frac{\zeta_{4}-\zeta_{3}}{\zeta_{5}-\zeta_{3}}&\frac{\zeta_{1}-\zeta_{3}}{\zeta_{5}-\zeta_{3}}\\\
&\frac{\zeta_{4}-\zeta_{5}}{\zeta_{3}-\zeta_{5}}&\frac{\zeta_{1}-\zeta_{5}}{\zeta_{3}-\zeta_{5}}\end{array}\right){}$
(179) $\displaystyle=$ $\displaystyle\left(\begin{array}[]{ccc}1&&\\\
&\frac{\zeta_{3}-\zeta_{4}}{\zeta_{3}-\zeta_{5}}&\frac{\zeta_{4}-\zeta_{5}}{\zeta_{3}-\zeta_{5}}\\\
&\frac{\zeta_{3}-\zeta_{2}}{\zeta_{3}-\zeta_{5}}&\frac{\zeta_{2}-\zeta_{5}}{\zeta_{3}-\zeta_{5}}\end{array}\right)\left(\begin{array}[]{ccc}\frac{\zeta_{2}-\zeta_{4}}{\zeta_{2}-\zeta_{5}}&&\frac{\zeta_{4}-\zeta_{5}}{\zeta_{2}-\zeta_{5}}\\\
&1&\\\
\frac{\zeta_{2}-\zeta_{1}}{\zeta_{2}-\zeta_{5}}&&\frac{\zeta_{1}-\zeta_{5}}{\zeta_{2}-\zeta_{5}}\end{array}\right)\left(\begin{array}[]{ccc}\frac{\zeta_{2}-\zeta_{3}}{\zeta_{2}-\zeta_{4}}&\frac{\zeta_{3}-\zeta_{4}}{\zeta_{2}-\zeta_{4}}&\\\
\frac{\zeta_{2}-\zeta_{1}}{\zeta_{2}-\zeta_{4}}&\frac{\zeta_{1}-\zeta_{4}}{\zeta_{2}-\zeta_{4}}&\\\
&&1\end{array}\right).$

Note that the matrices act from the left, so the order in the products is
reversed in comparison to (162).

##  6\. Example: solving the hexagon equation

###  6.1. Construction of vectors associated to tetrahedra and matrices
associated to Pachner 2-3 moves

In this subsection, we construct vectors associated to tetrahedra and matrices
associated to Pachner 2-3 moves. This construction is inspired by [6].

Given 4 pairwise distinct variables $\zeta_{i}$ ($i=1,2,3,4$) and a
tetrahedron 1234, we can construct a vector $(f_{1},f_{2},f_{3},f_{4})$ such
that

(180) $\displaystyle
f_{1}\zeta_{1}^{m}+f_{2}\zeta_{2}^{m}+f_{3}\zeta_{3}^{m}+f_{4}\zeta_{4}^{m}=0\text{
for }m=0,1,2.$

For example,

$\displaystyle(f_{1},f_{2},f_{3},f_{4})$ $\displaystyle=$
$\displaystyle(\frac{1}{(\zeta_{1}-\zeta_{2})(\zeta_{1}-\zeta_{3})(\zeta_{1}-\zeta_{4})},\frac{1}{(\zeta_{2}-\zeta_{1})(\zeta_{2}-\zeta_{3})(\zeta_{2}-\zeta_{4})},$
$\displaystyle\frac{1}{(\zeta_{3}-\zeta_{1})(\zeta_{3}-\zeta_{2})(\zeta_{3}-\zeta_{4})},\frac{1}{(\zeta_{4}-\zeta_{1})(\zeta_{4}-\zeta_{2})(\zeta_{4}-\zeta_{3})}).{}$

For the polyhedron 12345, the Pachner 2-3 move (see Fig. 1) replaces the
tetrahedra 1245 and 2345 with the tetrahedra 1234, 1235, and 1345. We will
construct a $3\times 2$ matrix $P_{13,245}$ for this Pachner move such that

(184) $\displaystyle
P_{13,245}\left(\begin{array}[]{ccccc}f_{1}(1245)&f_{2}(1245)&0&f_{3}(1245)&f_{4}(1245)\\\
0&f_{1}(2345)&f_{2}(2345)&f_{3}(2345)&f_{4}(2345)\end{array}\right)$ (188)
$\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccc}f_{1}(1234)&f_{2}(1234)&f_{3}(1234)&f_{4}(1234)&0\\\
f_{1}(1235)&f_{2}(1235)&f_{3}(1235)&0&f_{4}(1235)\\\
f_{1}(1345)&0&f_{2}(1345)&f_{3}(1345)&f_{4}(1345)\end{array}\right).{}$

We solve this matrix equation and get

(192) $\displaystyle
P_{13,245}=\left(\begin{array}[]{ccc}\frac{\zeta_{1}-\zeta_{5}}{\zeta_{1}-\zeta_{3}}&\frac{\zeta_{5}-\zeta_{3}}{\zeta_{1}-\zeta_{3}}\\\
\frac{\zeta_{1}-\zeta_{4}}{\zeta_{1}-\zeta_{3}}&\frac{\zeta_{4}-\zeta_{3}}{\zeta_{1}-\zeta_{3}}\\\
\frac{\zeta_{1}-\zeta_{2}}{\zeta_{1}-\zeta_{3}}&\frac{\zeta_{2}-\zeta_{3}}{\zeta_{1}-\zeta_{3}}\end{array}\right).$

Note that the sum of the entries of each row of the matrix $P_{13,245}$ is 1.

###  6.2. A solution to the hexagon equation

In this subsection, we check that (192) gives a solution to the hexagon
equation.

The hexagon equation is

(193) $\displaystyle
d_{(46,125)}d_{(36,124)}d_{(35,246)}=d_{(35,124)}d_{(36,125)}d_{(46,135)}$

where $d_{(ij,klm)}$ ($1\leq i<j\leq 6$, $1\leq k<l<m\leq 6$,
$\\{i,j\\}\cap\\{k,l,m\\}=\emptyset$) corresponds to the Pachner 2-3 move in a
polyhedron $ijklm$ with 6 faces where each face is a triangle, $ij$ is the
vertical axis, and $klm$ is the horizontal triangle. It replaces the
tetrahedra $iklm$, $jklm$ with the tetrahedra $ijkl$, $ijkm$, $ijlm$ (the
order of vertices can be rearranged from small to large).

We will construct matrices $a_{ij,klm}$ for each Pachner 2-3 move satisfying
the hexagon equation (193).

For the polyhedron 12345, the Pachner 2-3 move (see Fig. 1) replaces the
tetrahedra 1245 and 2345 with the tetrahedra 1234, 1235, and 1345. We can
construct a $3\times 2$ matrix $P_{13,245}$ (192) for this Pachner move as we
did in Sec. 6.1.

We can extend this matrix by adding a row and a column so that the entry is 1
at the intersection and 0 elsewhere corresponding to each fixed tetrahedron in
the Pachner 2-3 moves. The extended matrix is denoted as $a_{13,245}$.

Figure 4. Triangulation graph for the shifted octahedron.

The hexagon equation (193) involves 6 Pachner 2-3 moves:
(1)$\to$(2)$\to$(4)$\to$(6) and (1)$\to$(3)$\to$(5)$\to$(6) where (1)-(6) are
triangulations of a shifted octahedron as follows (see Fig. 4):

  1. (1)

1234, 1245, 1256,

  2. (2)

1234, 1246, 1456, 2456,

  3. (3)

1235, 1256, 1345, 2345,

  4. (4)

1236, 1346, 1456, 2346, 2456,

  5. (5)

1236, 1345, 1356, 2345, 2356,

  6. (6)

1236, 1346, 1456, 2345, 2356, 3456.

We can check that the matrices $a_{ij,klm}$ for these Pachner 2-3 moves
satisfy the hexagon equation (193), namely

(209) $\displaystyle\left(\begin{array}[]{cccccc}1&0&0&0&0\\\ 0&1&0&0&0\\\
0&0&1&0&0\\\
0&0&0&\frac{\zeta_{3}-\zeta_{6}}{\zeta_{3}-\zeta_{5}}&\frac{\zeta_{6}-\zeta_{5}}{\zeta_{3}-\zeta_{5}}\\\
0&0&0&\frac{\zeta_{3}-\zeta_{4}}{\zeta_{3}-\zeta_{5}}&\frac{\zeta_{4}-\zeta_{5}}{\zeta_{3}-\zeta_{5}}\\\
0&0&0&\frac{\zeta_{3}-\zeta_{2}}{\zeta_{3}-\zeta_{5}}&\frac{\zeta_{2}-\zeta_{5}}{\zeta_{3}-\zeta_{5}}\end{array}\right)\left(\begin{array}[]{ccccc}\frac{\zeta_{3}-\zeta_{4}}{\zeta_{3}-\zeta_{6}}&\frac{\zeta_{4}-\zeta_{6}}{\zeta_{3}-\zeta_{6}}&0&0\\\
\frac{\zeta_{3}-\zeta_{2}}{\zeta_{3}-\zeta_{6}}&\frac{\zeta_{2}-\zeta_{6}}{\zeta_{3}-\zeta_{6}}&0&0\\\
0&0&1&0\\\
\frac{\zeta_{3}-\zeta_{1}}{\zeta_{3}-\zeta_{6}}&\frac{\zeta_{1}-\zeta_{6}}{\zeta_{3}-\zeta_{6}}&0&0\\\
0&0&0&1\end{array}\right)\left(\begin{array}[]{cccc}1&0&0\\\
0&\frac{\zeta_{4}-\zeta_{5}}{\zeta_{4}-\zeta_{6}}&\frac{\zeta_{5}-\zeta_{6}}{\zeta_{4}-\zeta_{6}}\\\
0&\frac{\zeta_{4}-\zeta_{2}}{\zeta_{4}-\zeta_{6}}&\frac{\zeta_{2}-\zeta_{6}}{\zeta_{4}-\zeta_{6}}\\\
0&\frac{\zeta_{4}-\zeta_{1}}{\zeta_{4}-\zeta_{6}}&\frac{\zeta_{1}-\zeta_{6}}{\zeta_{4}-\zeta_{6}}\end{array}\right)$
(225) $\displaystyle=$ $\displaystyle\left(\begin{array}[]{cccccc}1&0&0&0&0\\\
0&\frac{\zeta_{4}-\zeta_{5}}{\zeta_{4}-\zeta_{6}}&\frac{\zeta_{5}-\zeta_{6}}{\zeta_{4}-\zeta_{6}}&0&0\\\
0&\frac{\zeta_{4}-\zeta_{3}}{\zeta_{4}-\zeta_{6}}&\frac{\zeta_{3}-\zeta_{6}}{\zeta_{4}-\zeta_{6}}&0&0\\\
0&0&0&1&0\\\ 0&0&0&0&1\\\
0&\frac{\zeta_{4}-\zeta_{1}}{\zeta_{4}-\zeta_{6}}&\frac{\zeta_{1}-\zeta_{6}}{\zeta_{4}-\zeta_{6}}&0&0\end{array}\right)\left(\begin{array}[]{ccccc}\frac{\zeta_{3}-\zeta_{5}}{\zeta_{3}-\zeta_{6}}&\frac{\zeta_{5}-\zeta_{6}}{\zeta_{3}-\zeta_{6}}&0&0\\\
0&0&1&0\\\
\frac{\zeta_{3}-\zeta_{2}}{\zeta_{3}-\zeta_{6}}&\frac{\zeta_{2}-\zeta_{6}}{\zeta_{3}-\zeta_{6}}&0&0\\\
0&0&0&1\\\
\frac{\zeta_{3}-\zeta_{1}}{\zeta_{3}-\zeta_{6}}&\frac{\zeta_{1}-\zeta_{6}}{\zeta_{3}-\zeta_{6}}&0&0\end{array}\right)\left(\begin{array}[]{cccc}\frac{\zeta_{3}-\zeta_{4}}{\zeta_{3}-\zeta_{5}}&\frac{\zeta_{4}-\zeta_{5}}{\zeta_{3}-\zeta_{5}}&0\\\
0&0&1\\\
\frac{\zeta_{3}-\zeta_{2}}{\zeta_{3}-\zeta_{5}}&\frac{\zeta_{2}-\zeta_{5}}{\zeta_{3}-\zeta_{5}}&0\\\
\frac{\zeta_{3}-\zeta_{1}}{\zeta_{3}-\zeta_{5}}&\frac{\zeta_{1}-\zeta_{5}}{\zeta_{3}-\zeta_{5}}&0\end{array}\right).{}$

Note that the matrices act from the left, so the order in the products is
reversed in comparison to (193). A different solution to the hexagon equation
was given in [7].

##  7\. Example: solving the heptagon equation

###  7.1. Construction of vectors associated to 4-simplices and matrices
associated to Pachner 3-3 moves

In this subsection, we construct vectors associated to 4-simplices and
matrices associated to Pachner 3-3 moves. This construction is inspired by
[6].

Given 5 pairwise distinct variables $\zeta_{i}$ ($i=1,2,3,4$) and a 4-simplex
12345, we can construct a vector $(f_{1},f_{2},f_{3},f_{4},f_{5})$ such that

(226) $\displaystyle
f_{1}\zeta_{1}^{m}+f_{2}\zeta_{2}^{m}+f_{3}\zeta_{3}^{m}+f_{4}\zeta_{4}^{m}+f_{5}\zeta_{5}^{m}=0\text{
for }m=0,1,2.$

For example,

(227)
$\displaystyle(f_{1},f_{2},f_{3},f_{4},f_{5})=(f(1,2,3,4,5),f(2,1,3,4,5),f(3,1,2,4,5),f(4,1,2,3,5),f(5,1,2,3,4))$

where

(228) $\displaystyle
f(i,j,k,l,m):=\frac{4\zeta_{i}-\zeta_{j}-\zeta_{k}-\zeta_{l}-\zeta_{m}}{(\zeta_{i}-\zeta_{j})(\zeta_{i}-\zeta_{k})(\zeta_{i}-\zeta_{l})(\zeta_{i}-\zeta_{m})}.$

Let $m$ be the matrix

(235)
$\displaystyle\left(\begin{array}[]{cccccc}0&f(2,3,4,5,6)&f(3,2,4,5,6)&f(4,2,3,5,6)&f(5,2,3,4,6)&f(6,2,3,4,5)\\\
f(1,3,4,5,6)&0&f(3,1,4,5,6)&f(4,1,3,5,6)&f(5,1,3,4,6)&f(6,1,3,4,5)\\\
f(1,2,4,5,6)&f(2,1,4,5,6)&0&f(4,1,2,5,6)&f(5,1,2,4,6)&f(6,1,2,4,5)\\\
f(1,2,3,5,6)&f(2,1,3,5,6)&f(3,1,2,5,6)&0&f(5,1,2,3,6)&f(6,1,2,3,5)\\\
f(1,2,3,4,6)&f(2,1,3,4,6)&f(3,1,2,4,6)&f(4,1,2,3,6)&0&f(6,1,2,3,4)\\\
f(1,2,3,4,5)&f(2,1,3,4,5)&f(3,1,2,4,5)&f(4,1,2,3,5)&f(5,1,2,3,4)&0\end{array}\right),$

then the rank of $m$ is 3.

For the polyhedron 123456, the Pachner 3-3 move replaces 3 of 6 4-simplices
with the other 3, i.e. the 4-simplices 12345, 12356, and 13456 with the
4-simplices 12346, 12456, and 23456. We will construct a $3\times 3$ matrix
$P_{246,135}$ for this Pachner move such that

(240) $\displaystyle P_{246,135}\cdot$
$\displaystyle\left(\begin{array}[]{cccccc}f(1,2,3,4,5)&f(2,1,3,4,5)&f(3,1,2,4,5)&f(4,1,2,3,5)&f(5,1,2,3,4)&0\\\
f(1,2,3,5,6)&f(2,1,3,5,6)&f(3,1,2,5,6)&0&f(5,1,2,3,6)&f(6,1,2,3,5)\\\
f(1,3,4,5,6)&0&f(3,1,4,5,6)&f(4,1,3,5,6)&f(5,1,3,4,6)&f(6,1,3,4,5)\end{array}\right){}$
(244) $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cccccc}f(1,2,3,4,6)&f(2,1,3,4,6)&f(3,1,2,4,6)&f(4,1,2,3,6)&0&f(6,1,2,3,4)\\\
f(1,2,4,5,6)&f(2,1,4,5,6)&0&f(4,1,2,5,6)&f(5,1,2,4,6)&f(6,1,2,4,5)\\\
0&f(2,3,4,5,6)&f(3,2,4,5,6)&f(4,2,3,5,6)&f(5,2,3,4,6)&f(6,2,3,4,5)\end{array}\right).{}$

We solve this matrix equation and get

(248) $\displaystyle P_{246,135}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccc}\frac{V(2,4,5)}{V(2,4,6)}&\frac{V(2,5,6)}{V(2,4,6)}&\frac{V(5,4,6)}{V(2,4,6)}\\\
\frac{V(2,4,3)}{V(2,4,6)}&\frac{V(2,3,6)}{V(2,4,6)}&\frac{V(3,4,6)}{V(2,4,6)}\\\
\frac{V(2,4,1)}{V(2,4,6)}&\frac{V(2,1,6)}{V(2,4,6)}&\frac{V(1,4,6)}{V(2,4,6)}\end{array}\right)$

where $V(i,j,k)=\begin{vmatrix}1&1&1\\\ \zeta_{i}&\zeta_{j}&\zeta_{k}\\\
\zeta_{i}^{2}&\zeta_{j}^{2}&\zeta_{k}^{2}\end{vmatrix}=(\zeta_{i}-\zeta_{j})(\zeta_{i}-\zeta_{k})(\zeta_{j}-\zeta_{k})$
is the Vandermonde determinant. Note that the matrix $P_{246,135}$ is
invertible and the sum of the entries of each row of the matrix $P_{246,135}$
is 1 because $\begin{vmatrix}1&1&1&1\\\ 1&1&1&1\\\
\zeta_{1}&\zeta_{2}&\zeta_{3}&\zeta_{4}\\\
\zeta_{1}^{2}&\zeta_{2}^{2}&\zeta_{3}^{2}&\zeta_{4}^{2}\end{vmatrix}=V(2,3,4)-V(1,3,4)+V(1,2,4)-V(1,2,3)=0$.
This gives a solution to the heptagon equation which is exactly a partial case
of the one in [1]. A similar solution to the heptagon equation was given in
[3] and similar solutions to general odd-gon equations were given in [4].

## References

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