# Oriented posets, Rank Matrices and q-deformed Markov Numbers

EZGİ KANTARCI OĞUZ  Dept. of Mathematics, Galatasaray University, İstanbul
<EMAIL_ADDRESS>
###### Abstract.

We define _oriented posets_ with corresponding _rank matrices_ , where linking
two posets by an edge corresponds to matrix multiplication. In particular,
linking chains via this method gives us fence posets, and taking traces gives
us circular fence posets. As an application, we give a combinatorial model for
$q$-deformed Markov numbers. We also resolve a conjecture of Leclere and
Morier-Genoud and give several identities between circular rank polynomials.

The author was supported by Tübitak BİDEB grant 121C285.

##  1\. Introduction

In this paper, we introduce a new tool motivated by the recent work on
$q$-deformations of rationals ([1], [2]), fence posets and their rank
polynomials ([3], [4]). Oriented posets are posets with specialized end points
that can be connected together to build larger posets. We can keep track of
this building operation by matrix multiplication, which allows for fast and
easy calculations. In this paper, we develop their combinatorics and consider
applications on rank polynomial identities and Markov numbers. Oriented posets
so far seem to be versatile objects that are applicable to other branches of
combinatorics. See [5] and [6] for applications in calculating cluster algebra
coefficients and studying polytopes respectively.

Let us start with our motivating example of fence posets. Fence posets are
combinatorial objects that come up in a variety of settings, including cluster
algebras, quiver representations, snake graphs and continued fractions (See
[7], [8]). Consider composition
$\alpha=(\alpha_{1},\alpha_{2},\ldots,\alpha_{s})$ of $n$, where the number
$s$ of parts is called the length of $\alpha$, denoted $\ell(\alpha)$ and $n$
is called the size of $\alpha$, denoted $|\alpha|$. We define an $n+1$ element
poset $F(\alpha)$, called the fence poset of $\alpha$ by taking the transitive
closure of the following cover relations:

$x_{1}\preceq x_{2}\preceq\cdots\preceq x_{\alpha_{1}+1}\succeq
x_{\alpha_{1}+2}\succeq\cdots\preceq x_{\alpha_{1}+\alpha_{2}+1}\preceq
x_{\alpha_{1}+\alpha_{2}+2}\preceq\cdots\preceq
x_{\alpha_{1}+\alpha_{2}+\alpha_{3}+1}\succeq\cdots$

The maximal chain corresponding to each $\alpha_{i}$ is called a segment of
the poset. If the length of $\alpha$ is even, we also define the circular
fence poset $\overline{F}(\alpha)$ corresonding to $\alpha$ by setting
$x_{n+1}=x_{1}$ in $F(\alpha)$. Note that unlike the regular fence poset, the
number of nodes in $\overline{F}(\alpha)$ matches the size of $\alpha$.

######  Example 1.1.

For $\alpha=(2,1,1,3)$ the regular and circular fence posets are depicted in
Figure 1 below.

$x_{1}$$x_{2}$$x_{3}$$x_{4}$$x_{5}$$x_{6}$$x_{7}$$x_{8}$

$x_{1}$$x_{2}$$x_{3}$$x_{4}$$x_{5}$$x_{6}$$x_{7}$

Figure 1. Posets $F(2,1,1,3)$ (left) and $\overline{F}(2,1,1,3)$ (right).

Down-closed subsets (subsets $I$ which satisfy $x\in I$, $y\preceq
x\Rightarrow y\in I$) of a fence are called its ideals. Ideals of a fence
poset $F(\alpha)$ (resp. circular fence poset $\overline{F}(\alpha)$) ordered
by inclusion have the structure of a distributive lattice ranked by the number
of elements in each ideal. We denote this lattice by $J(\alpha)$ (resp.
$\overline{J}(\alpha)$). The generating polynomials of $J(\alpha)$ and
$\overline{J}(\alpha)$ are called the rank polynomial of $\alpha$ and the
circular rank polynomial of $\alpha$ respectively:

$\displaystyle\mathfrak{R}(\alpha;q)=$ $\displaystyle\sum_{I\in
J(\alpha)}q^{|I|},\qquad\qquad\overline{\mathfrak{R}}(\alpha;q)=\sum_{I\in\overline{J}(\alpha)}q^{|I|}.$

######  Example 1.2.

For the regular and circular fence posets for $\alpha=(2,1,1,3)$ illustrated
in Figure 1, we get the following rank polynomials:

$\displaystyle\mathfrak{R}((2,1,1,3);q)=$ $\displaystyle
1+3q+5q^{2}+6q^{3}+6q^{4}+5q^{5}+3q^{6}+2q^{7}+q^{8},$
$\displaystyle\overline{\mathfrak{R}}((2,1,1,3);q)=$ $\displaystyle
1+2q+3q^{2}+4q^{3}+4q^{4}+3q^{5}+2q^{6}+q^{7}=[5]_{q}[4]_{q}.$

Rank polynomials recently came up in Ovsienko and Morier-Genoud’s work [1]
defining a $q$-deformation of rational numbers using their continued fraction
expansions. For a given rational number $r/t$, the $q$-deformation $[r/t]_{q}$
equals $R(q)/T(q)$ where $R(q),T(q)\in\mathbb{Z}[q]$ are polynomials that, at
$q=1$, evaluate to $r$ and $t$ respectively. When $\frac{r}{t}\geq 1$, one can
find a corresponding composition $\alpha_{r/t}$ such that $R(q)$ equals the
rank polynomial of $\mathfrak{R}(\alpha_{r/t};q)$ and $T(q)$ equals the rank
polynomial of the same fence poset with leftmost segment deleted. The exact
nature of these constructions is described in Section 2 below. Ovsienko and
Morier-Genoud’s observations on this front led them to conjecture that the
rank polynomials of fence posets are unimodal. This conjecture was extended to
describe the interlacing properties in [3], and the extended version was
proved in [4] using circular fence posets. It was further shown that:

######  Theorem 1.3 ([4]).

Circular rank polynomials
$\overline{\mathfrak{R}}((\alpha_{1},\alpha_{2},\ldots,\alpha_{2s});q)$ are
symmetric with non-negative integer coefficients that remain constant under
cyclic shifts of parts of $\alpha$.

(1)
$\displaystyle\overline{\mathfrak{R}}((\alpha_{1},\alpha_{2},\ldots,\alpha_{2s});q)=\overline{\mathfrak{R}}((\alpha_{2s},\alpha_{1},\alpha_{2},\ldots,\alpha_{2s-1});q).$

The polynomials are also unimodal if one of the following is satisfied:

  * •

$\alpha_{1}+\alpha_{2}+\cdots+\alpha_{2s}$ is odd,

  * •

$\alpha$ has two consecutive parts that are larger than $1$,

  * •

$\alpha$ has three consecutive parts $k,1,l$ with $|k-l|>1$.

It was in fact conjectured in [4] that the circular rank polynomial is
unimodal if the number of segments is different than $4$.

######  Conjecture 1.4 ([4]).

$\overline{\mathfrak{R}}(\alpha;q)$ is unimodal except for the cases
$\alpha=(1,k,1,k)$ or $(k,1,k,1)$ for some positive integer $k$.

In [2], Leclere and Morier-Genoud work with a $q$-deformation of
$\text{PSL}(2,\mathbb{Z})$. For a $r/t$ with corresponding regular and
negative continued fraction expressions $[a_{1},a_{2},\ldots,a_{2m}]$ and
$\llbracket c_{1},c_{2},\ldots,c_{k}\rrbracket$ respectively, they work with
matrices $M_{q}(c_{1},c_{2},\ldots,c_{k})$ and
$M_{q}^{+}(a_{1},a_{2},\ldots,a_{2m})$ in $\text{PSL}_{q}(2,\mathbb{Z})$,
previously defined in [1]. The top and bottom left entries of
$M_{q}(c_{1},c_{2},\ldots,c_{k})$ are given by $R(q)$ and $T(q)$ whereas the
top and bottom left entries of $M_{q}^{+}(a_{1},a_{2},\ldots,a_{2m})$ are
given by $qR(q)$ and $qT(q)$ for some $z\in\mathbb{Z}$.

The traces of $M_{q}(c_{1},c_{2},\ldots,c_{k})$, up to a multiple $\pm q^{N}$,
are shown to be symmetric polynomials on $q$ with non-negative integer
entries. Also, in a similar flavor to the circular rank polynomials, they are
invariant under cyclic shifts of the sequence $\llbracket
c_{1},c_{2},\ldots,c_{k}\rrbracket$. Leclere and Morier-Genoud further
conjecture the following:

######  Conjecture 1.5 ([2], Conjecture 3.12).

For any sequence $c_{1},c_{2},\ldots,c_{k}$ of integers, the trace
$\mathrm{tr}(M_{q}(c_{1},c_{2},\ldots,c_{k}))$ is a polynomial in
$\mathbb{Z}[q,q^{-1}]$ with unimodal coefficients.

The matrices $M_{q}$ and $M^{+}_{q}$ as rank matrices of oriented posets,
which leads us to conclude the $\mathrm{tr}(M(c_{1},c_{2},\ldots,c_{k}))$ is
indeed a circular rank polynomial. In fact, one can show that (see Proposition
5.5) for $r/t\geq 1$ with regular and negative continued fraction expressions
given by $[a_{1},a_{2},\cdots,a_{2m}]$ and $\llbracket
c_{1},c_{2},\ldots,c_{k}\rrbracket$ respectively, we have:

$\displaystyle\mathrm{tr}(M_{q}(c_{1},c_{2},\ldots,c_{k})$
$\displaystyle=\overline{\mathfrak{R}}((a_{1}-1,a_{2},\ldots,a_{2m});q),$
$\displaystyle\mathrm{tr}(M_{q}^{+}(a_{1},a_{2},\ldots,a_{2m}))$
$\displaystyle=\overline{\mathfrak{R}}((a_{1},a_{2},\ldots,a_{2m});q).$

This implies in particular that Conjecture 1.5 does not hold, but one can
apply the results from Theorem 1.3 to characterize large families where we
have unimodality.

In Section 3, we define _oriented posets_
$\overrightarrow{P}:=(P,x_{L},x_{R})$ with specialized start and end points.
Instead of rank polynomials, these posets come with $2\times 2$ _rank
matrices_ ${\triangledown}\mathfrak{R}_{q}(\overrightarrow{P})$, where
multiplication corresponds to combining two posets, and we can connect the two
endpoints of a poset by simply taking the trace. In particular, any fence
poset can be built from copies of the poset $\overrightarrow{\bullet}$
consisting of a single node and made circular by taking the trace (See Section
5). This allows us prove Proposition 5.5 as well as construct new identities
between infinite families of rank polynomials in Section 6.

###### Theorem 6.3.

Let $X$ be a palindromic composition with an even number of parts. For $k\geq
1$, $r\geq 1$ we have:

(Id 1) $\displaystyle\overline{\mathfrak{R}}((1,k,r+1,X,r);q)$
$\displaystyle=[k+1]_{q}\cdot\overline{\mathfrak{R}}((r+2,X,r);q),$ (Id 2)
$\displaystyle\overline{\mathfrak{R}}((k,1,k+r,X,r);q)$
$\displaystyle=[k+1]_{q}\cdot\overline{\mathfrak{R}}((k+r+1,X,r);q).$

As an application, we give a combinatorial model for the $q$-deformed Markov
numbers in terms of fence posets. Solutions of the Markov equation
$x^{2}+y^{2}+z^{2}=3xyz$ are called Markov triples. Defined by A. Markoff in
his thesis in 1880 [9], they are motivated by the study of rational
approximations and Diophantine analysis. All Markov triples can be calculated
from the initial solution $(1,1,1)$ via operations of the form
$(x_{0},y_{0},z_{0})\rightarrow(x_{0},y_{0},(x_{0}^{2}+y_{0}^{2})/z_{0})$. The
solutions form an infinite binary tree called the _Markov tree_. Some
equivalent constructions are Christoffel words under the operation
$(u,uv,v)\rightarrow(u,uuv,uv)$, Cohn matrices under matrix multiplication,
simple fractions in the interval $[0,1]$ under the Farey sum, and mutations of
the quiver representation for a torus with one marked point ([10], [11],
[12]). Frobenius conjectured in $1912$ that each Markov number appears as the
maximum of a unique Markov triple, a conjecture still open to this day (The
interested reader is refered to the 2012 book [13] on the subject). Markov
numbers were recently $q$-deformed using Cohn matrices ( [2], [14]). In
Section 7, we use oriented posets to give a combinatorial model to directly
calculate $q$-deformed Markov numbers.

###### Theorem 7.3.

The $q$-deformed Markov number corresponding to the Christoffel $ab$-word
$w(a,b)=a\hat{w}(a,b)b$ is the rank polynomial of the circular fence poset
$\overline{F}(3,1,\alpha(\hat{w}))$, where $\alpha(\hat{w})$ is formed by
replacing each $a$ by the pair $1,1$ and each $b$ by the pair $2,2$ in
$\hat{w}$.

We finish with listing some future research directions and references to some
recent work using oriented posets including a resolution of Conjecture 1.4
([6]).

##  2\. Preliminaries and Notation

For visual clarity, we at times use the notation $\cdot$ for both regular and
matrix multiplication.

###  2.1. Fence Posets and Rank Polynomials

A composition $\alpha=(\alpha_{1},\alpha_{2},\ldots,\alpha_{s})$ of $n$ is a
list of non-negative integers that add up to $n$. Here $n$ is called the size
of $\alpha$, denoted $|\alpha|$, and $s$ is called the length of $\alpha$.
During this work, there are multiple instances when we add extra parts to a
composition from the left. In that case, we slightly abuse notation and use
$(i,\alpha)$ to denote the composition
$(i,\alpha_{1},\alpha_{2},\ldots,\alpha_{s})$. We also use the notation
$a^{b}$ to denote $b$ consecutive parts of size $a$.

We follow the convention of [3] to start our fence posets with an up step. The
rank polynomials for the posets starting with down steps can easily be
inferred from these for the following reason: The complement of any ideal is
an up-closed subset of the poset (satisfies $x\in I$, $y\succeq x\Rightarrow
y\in I$). If we flip a fence poset $F(\alpha)$ upside down, the resulting
poset has the rank polynomial given by reversing the coefficients of $\alpha$:

$\displaystyle\mathfrak{R}^{\updownarrow}(\alpha;q)$
$\displaystyle=q^{|\alpha+1|}\cdot\mathfrak{R}(\alpha;q^{-1}).$

When the fence we are considering is circular, turning the poset upside down
is the same thing as cyclicly shifting segments by one, so the symmetry of the
rank polynomial is actually equivalent to the invariance under cyclic shifts.

$\displaystyle\overline{\mathfrak{R}}^{\updownarrow}(\alpha;q)$
$\displaystyle=\overline{\mathfrak{R}}((\alpha_{2s},\alpha_{1},\alpha_{2},\ldots,\alpha_{2s-1});q)=q^{|\alpha|}\cdot\overline{\mathfrak{R}}(\alpha,q^{-1})=\overline{\mathfrak{R}}(\alpha;q).$

$x_{1}$$x_{2}$$x_{3}$$x_{4}$$x_{5}$$x_{6}$$x_{7}$$x_{8}$

$x_{1}$$x_{2}$$x_{3}$$x_{4}$$x_{5}$$x_{6}$$x_{7}$

Figure 2. Posets $F^{\updownarrow}(2,1,1,3)$ (left) and
$\overline{F}^{\updownarrow}(2,1,1,3)=\overline{F}(1,1,3,2)$ (right).

######  Example 2.1.

The upside down reflection of the regular and circular fence posets for
$\alpha=(2,1,1,3)$ are illustrated in Figure 2. The corresponding rank
polynomials are:

$\displaystyle\mathfrak{R}^{\updownarrow}((2,1,1,3);q)$
$\displaystyle=q^{8}\cdot\mathfrak{R}((2,1,1,3);q^{-1})=1+2q+3q^{2}+5q^{3}+6q^{4}+6q^{5}+5q^{6}+3q^{7}+q^{8},$
$\displaystyle\overline{\mathfrak{R}}^{\updownarrow}((2,1,1,3);q)$
$\displaystyle=\overline{\mathfrak{R}}((2,1,1,3);q)=1+2q+3q^{2}+4q^{3}+4q^{4}+3q^{5}+2q^{6}+q^{7}=[5]_{q}[4]_{q}.$

###  2.2. $q$-Rationals

For a given rational number $r/t$, two natural ways to express $r/t$ as a
continued fraction are:

(2)
$\displaystyle\displaystyle\frac{r}{t}=a_{1}+\cfrac{1}{a_{2}+\cfrac{1}{a_{3}+\cfrac{1}{\ddots+\cfrac{1}{a_{2m}}}}}=c_{1}-\cfrac{1}{c_{2}-\cfrac{1}{c_{3}-\cfrac{1}{\ddots-\cfrac{1}{c_{k}}}}}$

The expressions $[a_{1},a_{2},\ldots,a_{2m}]$ and $\llbracket
c_{1},c_{2},\ldots,c_{k}\rrbracket$ are called the regular and negative
continued fractions for $r/t$ respectively. They are unique under the
conditions $a_{i}\in\mathbb{Z}$, $a_{i}\geq 1$ for $i\geq 2$ for the regular
case and $c_{i}\in\mathbb{Z}$, $c_{i}\geq 2$ for $i\geq 2$ for the negative
case. Furthermore, one can easily convert one expression into the other via:

(3) $\displaystyle\llbracket c_{1},c_{2},\ldots,c_{k}\rrbracket=\llbracket
a_{1}+1,2^{a_{2}-1},a_{3}+2,2^{a_{4}-1},a_{5}+2,\ldots,a_{2m-1}+2,2^{a_{2m}-1}\rrbracket$

where $2^{t}$ stands for $t$ parts of size $2$. The $q$-analogue for
$\frac{r}{t}$ is then obtained by replacing the terms with powers of $q$ and
$q$-integers in the following way:

(4)
$\displaystyle\displaystyle\left[\frac{r}{t}\right]_{q}:=[a_{1}]_{q}+\cfrac{q^{a_{1}}}{[a_{2}]_{q^{-1}}+\cfrac{q^{-a_{2}}}{\ddots+\cfrac{q^{a_{2m-1}}}{[a_{2m}]_{q^{-1}}}}}=[c_{1}]_{q}-\cfrac{q^{c_{1}-1}}{[c_{2}]_{q}-\cfrac{q^{c_{2}-1}}{\ddots-\cfrac{q^{c_{k-1}-1}}{[c_{k}]_{q}}}}$

Here, for a negative integer $-n$, we take
$[-n]_{q}=-q^{-1}-q^{-2}-\cdots-q^{-n}$.

The non-trivial fact that the two $q$-expressions given in Equation 2 give the
same $q$-analogue is shown in [1]. It is further shown that the said
$q$-analogue equals a rational function $\frac{R(q)}{T(q)}$ with polynomials
$R(q),T(q)$ evaluating to $r$ and $s$ respectively at $q=1$.

######  Example 2.2.

The fraction $32/7$ has the regular and negative continued fraction
expressions $[3,1,1,4]$ and $[4,3,2,2,2]$.

$\displaystyle\frac{32}{9}=3+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{4}}}=4-\cfrac{1}{3-\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2}}}}.$

We can calculate its $q$-analogue via Equation 4.

$\displaystyle\displaystyle\left[\frac{32}{9}\right]_{q}$
$\displaystyle=[3]_{q}+\cfrac{q^{3}}{[1]_{q^{-1}}+\cfrac{q^{-1}}{[1]_{q}+\cfrac{q}{[4]_{q^{-1}}}}}=[4]_{q}-\cfrac{q^{4}}{[3]_{q}-\cfrac{q^{3}}{[2]_{q}-\cfrac{q^{2}}{[2]_{q}-\cfrac{q^{2}}{[2]_{q}}}}}$
$\displaystyle=\frac{1+3q+5q^{2}+6q^{3}+6q^{4}+5q^{5}+3q^{6}+2q^{7}+q^{8}}{1+2q+2q^{2}+2q^{3}+q^{4}+q^{5}}=\frac{\mathfrak{R}((2,1,1,3);q)}{\mathfrak{R}((1,3);q)}.$

In the case $\frac{r}{t}\geq 1$, the coefficient sequences satisfy $a_{1}\geq
1$ and $c_{1}\geq 2$, and it is possible to describe $R(q)$ and $T(q)$ in
terms of rank polynomials as in Example 2.2 above (The other cases can also be
reduced to this case using the identity $[x+1]_{q}=q[x]_{q}+1$ from [1]).

For the rest of this work, we assume $\frac{r}{t}\geq 1$ with continued
fraction expressions $[a_{1},a_{2},\ldots,a_{2m}]$ and $\llbracket
c_{1},c_{2},\ldots,c_{k}\rrbracket$. In [1], it is shown that one has:

$\displaystyle R(q)$
$\displaystyle=\mathfrak{R}((a_{1}-1,a_{2},a_{3},\ldots,a_{2m}-1);q),$
$\displaystyle T(q)$
$\displaystyle=\mathfrak{R}((0,a_{2}-1,a_{3},\ldots,a_{2m}-1);q).$

where if the first non-zero part of the composition is the second part, we
take the fence poset that starts with a down step instead of an up step.

This means, if we set $\alpha_{r/t}:=(a_{1}-1,a_{2},a_{3},\ldots,a_{2m}-1)$,
$R(q)$ is given by the rank polynomial of $F(\alpha_{r/t})$ and $T(q)$ is the
rank polynomial of the same poset with the leftmost segment deleted.
Conversely, for a composition
$\alpha=(u_{1},d_{1},u_{2},d_{2},\ldots,u_{s},d_{s})$ (here we allow $d_{s}=0$
for full generality), $\alpha=\alpha_{r/t}$ for the rational number $r/t$ with
the following continued fraction expressions:

(5) $\displaystyle[u_{1}+1,d_{1},u_{2},d_{2},\ldots,u_{s},d_{s}+1]=\llbracket
u_{1}+2,2^{d_{1}-1},u_{2}+2,2^{d_{2}-1},\ldots,u_{s}+1,2^{d_{s}}\rrbracket.$

Note that if we further allow $u_{1}=0$, corresponding to the case of the
fence posets starting with a down step, we can recover all possible $r/t\geq
1$ in this manner.

###  2.3.  $\text{PSL}_{q}(2,\mathbb{Z})$

In [2], Leclere and Morier-Genoud work with the $q$-deformation of
$\text{PSL}(2,\mathbb{Z})$ defined by the following $q$-versions of the
standard generators considered modulo $\pm I$:

$\displaystyle R_{q}=\begin{bmatrix}q&1\\\ 0&1\end{bmatrix},\qquad
S_{q}=\begin{bmatrix}0&-q^{-1}\\\ 1&0\end{bmatrix}.$

They set

(6) $\displaystyle M_{q}(c_{1},c_{2},\ldots,c_{k})$
$\displaystyle:=R_{q}^{c_{1}}S_{q}R_{q}^{c_{2}}S_{q}\cdots
R_{q}^{c_{k}}S_{q},$ (7) $\displaystyle M^{+}_{q}(a_{1},a_{2},\ldots,a_{2m})$
$\displaystyle:=R_{q}^{a_{1}}(R_{q}S_{q}R_{q})^{a_{2}}R_{q}^{a_{3}}(R_{q}S_{q}R_{q})^{a_{4}}\cdots
R_{q}^{a_{2m-1}}(R_{q}S_{q}R_{q})^{a_{2m}}.$

Here, the top left and bottom left entries of
$M_{q}(c_{1},c_{2},\ldots,c_{k})$ are given by $R(q)$ and $T(q)$ respectively;
whereas top left and bottom left entries of
$M_{q}^{+}(a_{1},a_{2},\ldots,a_{2m})$ are given by $qR(q)$ and $qT(q)$ ([1],
[2]).

######  Lemma 2.3 ([2]).

Every matrix $M\in\mathrm{PSL}_{q}(2,\mathbb{Z})$ can be written as
$M_{q}(c_{1},c_{2},\ldots,c_{k})$ for some integers $c_{i}\in\mathbb{Z}$.
Furthermore, the trace of $M$ equals, up to a multiplicative factor of $q^{\pm
N}$ for some $N\in\mathbb{Z}$ the trace of some
$M_{q}(c_{1},c_{2},\ldots,c_{k})$ with all $c_{i}\geq 2$.

This lemma implies that, especially when considering Conjecture 1.5, it makes
sense to limit our attention to the case $r/t\geq 1$, which is exactly when
all $c_{i}$ are $\geq 2$.

The traces of matrices in $\mathrm{PSL}_{q}(2,\mathbb{Z})$ also have the
following properties in common with circular rank polynomials ([2], Lemmas
3.8, 3.9, 3.10):

  * •

$\operatorname{tr}(M_{q}(c_{1},c_{2},\ldots,c_{k}))$ is a symmetric
polynomial.

  * •

It is invariant under cyclic shifts of $\llbracket
c_{1},c_{2},\ldots,c_{k}\rrbracket$ so that:

$\operatorname{tr}(M_{q}(c_{1},c_{2},\ldots,c_{k}))=\operatorname{tr}(M_{q}(c_{2},c_{3},\ldots,c_{k},c_{1})).$

  * •

When all $c_{i}$ are at least $2$, the polynomial
$\operatorname{tr}(M_{q}(c_{1},c_{2},\ldots,c_{k}))$ has positive integer
coefficients.

######  Example 2.4.

We have seen in Example 2.2 that $32/9$ has the continued fraction expressions
$[3,1,1,4]$ and $\llbracket 4,3,2,2,2\rrbracket$. The matrices corresponding
to these expressions are:

$\displaystyle M_{q}(4,3,2,2,2)$
$\displaystyle=\begin{bmatrix}1\\!+\\!3q\\!+\\!5q^{2}\\!+\\!6q^{3}\\!+\\!6q^{4}\\!+\\!5q^{5}\\!+\\!3q^{6}\\!+\\!2q^{7}\\!+\\!q^{8}&-q(1\\!+\\!3q\\!+\\!5q^{2}\\!+\\!5q^{3}\\!+\\!5q^{4}\\!+\\!3q^{5}\\!+\\!q^{6})\\\
1\\!+\\!2q\\!+\\!2q^{2}\\!+\\!2q^{3}\\!+\\!q^{4}\\!+\\!q^{5}&-q(1\\!+\\!2q\\!+\\!2q^{2}\\!+\\!q^{3}\\!+\\!q^{4})\end{bmatrix}$
$\displaystyle=\begin{bmatrix}\mathfrak{R}((2,1,1,3);q)&-q\mathfrak{R}((2,1,1,2);q)\\\
\mathfrak{R}((1,3);q)&-q\mathfrak{R}((1,2);q).\\\ \end{bmatrix},$
$\displaystyle M^{+}_{q}(3,1,1,4)$
$\displaystyle=\begin{bmatrix}q(1\\!+\\!3q\\!+\\!5q^{2}\\!+\\!6q^{3}\\!+\\!6q^{4}\\!+\\!5q^{5}\\!+\\!3q^{6}\\!+\\!2q^{7}\\!+\\!q^{8})&1\\!+\\!2q\\!+\\!2q^{2}\\!+\\!q^{3}\\!+\\!q^{4}\\\
q(1\\!+\\!2q\\!+\\!2q^{2}\\!+\\!2q^{3}\\!+\\!q^{4}\\!+\\!q^{5})&1\\!+\\!q\end{bmatrix}$
$\displaystyle=\begin{bmatrix}q\mathfrak{R}((2,1,1,3);q)&\mathfrak{R}((2,1);q)\\\
q\mathfrak{R}((1,3);q)&\mathfrak{R}((1);q)\\\ \end{bmatrix}.$

The traces of the two matrices are given by circular rank polynomials:

$\displaystyle\operatorname{tr}(M_{q}(4,3,2,2,2))$
$\displaystyle=\mathfrak{R}((2,1,1,3);q)-q\mathfrak{R}((1,2);q)=\overline{\mathfrak{R}}((2,1,1,4);q)),$
$\displaystyle\operatorname{tr}(M^{+}_{q}(3,1,1,4))$
$\displaystyle=q\mathfrak{R}((2,1,1,3);q)-1-q=\overline{\mathfrak{R}}((3,1,1,4);q)).$

.

In Section 5, we will show that the traces of the matrices $M_{q}$ and
$M^{+}_{q}$ are always given by circular rank polynomials.

##  3\. Oriented Posets

We define an oriented poset to be a triple
$\overrightarrow{P}:=(P,x_{L},x_{R})$ where $P$ is a poset with two
specialized nodes $x_{L}$, $x_{R}$. One can think of them as the left end node
and the right end node respectively. We will connect oriented posets via
connecting the left end of one to the right end of the other by adding a
relation. More precisely speaking, we add oriented posets via the operation:

$\displaystyle(P,x_{L},x_{R})\searrow(Q,y_{L},y_{R}):=(S,x_{L},y_{R})$

where $S$ is the unique poset given by connecting the posets $P$ and $Q$ by
setting $x_{R}\succeq y_{L}$. We also use the notation
$\circlearrowright\\!(\overrightarrow{P})$ to denote the structure obtained by
adding the relation $x_{R}\succeq x_{L}$. This is generally a poset, though in
the case $x_{R}\preceq x_{L}$ in the original poset we get some degeneracy.

There are two natural ways to see the length $n$ chain poset as an oriented
poset: The increasing chain $\overrightarrow{U_{n}}$ with $x_{L}$ taken to be
the minimal element and $x_{R}$ the maximum, and the decreasing chain
$\overrightarrow{D_{n}}$ with the opposite orientation. See Figure 3 below for
an example.

$x_{L}$$x_{R}$$x_{L}$$y_{R}$$y_{L}$$y_{R}$$\overrightarrow{U_{3}}$$\overrightarrow{D_{3}}$$\overrightarrow{U_{3}}\searrow\overrightarrow{D_{3}}$
Figure 3.  $\overrightarrow{U_{3}}\searrow\overrightarrow{D_{3}}$ gives us the
fence poset for $(3,4)$.

The motivation behind oriented posets is to build up larger families of posets
modularly from smaller pieces, using the addition operation connecting the
right end of one poset to the left end of another. In particular, we will use
them to build regular and circular fence posets and calculate their rank
polynomials in Section 5. Though this building method is not sufficient to
capture the complicated nature of a general poset, it works well when the
underlying graph naturally comes with edge-cuts that makes the structure
disconnected, as in the cases of the Dynkin diagrams for Lie groups. With this
paradigm, we can use $2\times 2$ matrices to calculate the rank polynomials of
the larger posets we obtain, where addition and connecting the ends operations
correspond to matrix multiplication and taking the trace respectively.

Consider the subsets of the distributive lattice $J(P)$ determined by the
inclusion or exclusion of vertices $x_{L}$ and $x_{R}$. Our notation is that
the subscript on the left describes the exclusion ($0$) or inclusion ($1$) of
$x_{L}$ and the subscript on the right describes the inclusion or exclusion of
$x_{R}$:

$\displaystyle\mathfrak{R}(\overrightarrow{P};q):=$ $\displaystyle\sum_{I\in
J(P)}q^{|I|}\qquad\qquad$
${}_{0}\mathfrak{R}(\overrightarrow{P};q):=\sum_{\begin{subarray}{c}I\in
J(P)\\\ x_{L}\notin I\end{subarray}}q^{|I|}$
$\displaystyle\mathfrak{R}_{0}(\overrightarrow{P};q):=$
$\displaystyle\sum_{\begin{subarray}{c}I\in J(P)\\\ x_{R}\notin
I\end{subarray}}q^{|I|}\qquad$
${}_{0}\mathfrak{R}_{0}(\overrightarrow{P};q):=\sum_{\begin{subarray}{c}I\in
J(P)\\\ x_{L},x_{R}\notin I\end{subarray}}q^{|I|}$

The superscripts can also be taken to be $1$, to specify $x_{L}$ or $x_{R}$ is
going to be included.

######  Definition 3.1.

We define the rank matrix of an oriented poset $(P,x_{L},x_{R})$ as follows:

$\displaystyle{\triangledown}\mathfrak{R}_{q}(\overrightarrow{P}):=\begin{bmatrix}\mathfrak{R}(\overrightarrow{P};q)&-\mathfrak{R}_{1}(\overrightarrow{P};q)\\\
{}_{0}\mathfrak{R}(\overrightarrow{P};q)&-\,_{0}\mathfrak{R}_{1}(\overrightarrow{P};q)\end{bmatrix}$

The increasing and decreasing length $n$ chains have the following rank
matrices:

$\displaystyle{\triangledown}\mathfrak{R}_{q}(\overrightarrow{U_{n}})$
$\displaystyle=\begin{bmatrix}[n+2]_{q}&-q^{n+1}\\\
1&0\end{bmatrix}=R_{q}^{n+2}S_{q}.$
$\displaystyle{\triangledown}\mathfrak{R}_{q}(\overrightarrow{D_{n}})$
$\displaystyle=\begin{bmatrix}[n+2]_{q}&-q[n+1]_{q}\\\
[n+1]_{q}&-q[n]_{q}\end{bmatrix}=(R_{q}^{2}S_{q})^{n+1}.$

Note that in both cases, the resulting matrix lies in
$\mathrm{PSL_{q}}(2,\mathbb{Z})$. For simplicity, we will use the notation
$D_{0}$ for the matrix corresponding to the case
$\overrightarrow{\bullet}=\overrightarrow{U_{0}}=\overrightarrow{U_{0}}$ of
having only one vertex. This can be seen as the generator of a down-step.

$\displaystyle D_{0}:=$
$\displaystyle{\triangledown}\mathfrak{R}_{q}(\overrightarrow{\bullet})=\begin{bmatrix}[2]_{q}&-q\\\
1&0\end{bmatrix}=R_{q}^{2}S_{q}.$

As mentioned above, the motivation behind rank matrices is to be able to
calculate the rank polynomials corresponding to
$\circlearrowright\\!(\overrightarrow{P})$ and
$\overrightarrow{P}\searrow\overrightarrow{Q}$ using simple matrix operations.

######  Proposition 3.2.

Let $\overrightarrow{P}=(P,x_{L},x_{R})$ and
$\overrightarrow{Q}=(Q,y_{L},y_{R})$ be two oriented posets. Then we have:

(8)
$\displaystyle{\triangledown}\mathfrak{R}_{q}(\overrightarrow{P}\searrow\overrightarrow{Q})={\triangledown}\mathfrak{R}_{q}(\overrightarrow{P})\cdot{\triangledown}\mathfrak{R}_{q}(\overrightarrow{!})$
(9)
$\displaystyle\mathfrak{R}(\circlearrowright\\!(\overrightarrow{P}))=\mathrm{tr}({\triangledown}\mathfrak{R}_{q}(\overrightarrow{P}))$

###### Proof.

The way we form
$\overrightarrow{P}=(P,x_{L},x_{R})\searrow\overrightarrow{Q}=(Q,y_{L},y_{R})$
is to add a new relation $x_{R}\succeq y_{L}$. So any ideal of
$\overrightarrow{P}\searrow\overrightarrow{Q}$ can be seen as a pair
$(I_{1},I_{2})$ where $I_{1}$ is an ideal of $P$, $I_{2}$ is an ideal of $Q$
and we have $x_{R}\in I_{1}\Rightarrow y_{L}\in I_{2}$. On the rank polynomial
side, this means:

$\displaystyle\mathfrak{R}(\overrightarrow{P}\searrow\overrightarrow{Q};q)=\mathfrak{R}(\overrightarrow{P};q)\,\mathfrak{R}(\overrightarrow{Q};q)-\mathfrak{R}_{1}(\overrightarrow{P};q)\,_{0}\mathfrak{R}(\overrightarrow{Q};q).$

Note that this is exactly the upper left entry of
${\triangledown}\mathfrak{R}_{q}(\overrightarrow{P})\cdot{\triangledown}\mathfrak{R}_{q}(\overrightarrow{q})$.
Other entries are calculated by adding the extra conditions $x_{L}$ is in and
$y_{R}$ is out to the above equation, and similarly match the matrix
multiplication.

For the second equation, note that the trace of
${\triangledown}\mathfrak{R}_{q}(\overrightarrow{P})$ is given by
$\mathfrak{R}(\overrightarrow{P};q)-\,_{0}\mathfrak{R}_{1}(\overrightarrow{P};q)$,
which counts the ideals with $x_{L}\in I\Rightarrow x_{R}\in I$. This is the
same thing as adding the relation $x_{L}\succeq x_{R}$. ∎

######  Remark 3.3.

The down chain $\overrightarrow{D_{n}}$ can be built by adding up $n+1$ copies
of $\overrightarrow{\bullet}$. Accordingly we have:

(10) $\displaystyle{\triangledown}\mathfrak{R}_{q}(\overrightarrow{D_{n}})=$
$\displaystyle(D_{0})^{n+1}$

######  Example 3.4.

For the poset $(3,4)$ given in Figure 3, the rank polynomial is given by the
top left entry of

$\displaystyle\displaystyle{\triangledown}\mathfrak{R}_{q}(\overrightarrow{U_{3}})\,D_{0}^{4}=$
$\displaystyle\begin{bmatrix}([5]_{q})^{2}-q^{4}[4]_{q}&q^{5}[3]_{q}-q[5]_{q}[4]_{q}\\\
[5]_{q}&-q[4]_{q}\,\end{bmatrix}$ $\displaystyle\mathfrak{R}((3,4);q)=$
$\displaystyle[5]_{q}^{2}-q^{4}[4]_{q}=1+2q+3q^{2}+4q^{3}+4q^{4}+3q^{5}+2q^{6}+q^{7}+q^{8}.$

##  4\. Dual Addition

In this section, we look at a dual definition of addition where we add the
relation $x_{R}\preceq y_{L}$ instead of $x_{R}\succeq y_{L}$ to connect two
posets $(P,x_{L},x_{R})$ and $(Q,y_{L},y_{R})$. We denote the directed poset
$(S,x_{L},y_{R})$ we obtain this way by
$(P,x_{L},x_{R})\nearrow(Q,y_{L},y_{R})$.We also use the notation
$\circlearrowleft\\!(\overrightarrow{P})$ to denote the structure obtained by
adding the relation $x_{R}\preceq x_{L}$ to $(P,x_{L},x_{R})$.

$x_{L}$$x_{R}$

Figure 4. The posets $\overrightarrow{U_{3}}\nearrow\overrightarrow{D_{3}}$
(left) and
$\circlearrowleft\\!(\overrightarrow{U_{3}}\nearrow\overrightarrow{D_{3}})$
(right).

######  Definition 4.1.

The dual rank matrix of a oriented poset $(P,x_{L},x_{R})$ is defined as
follows:

(11)
$\displaystyle\displaystyle{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P}):=\begin{bmatrix}\mathfrak{R}_{1}(\overrightarrow{P};q)&\mathfrak{R}_{0}(\overrightarrow{P};q)\\\
{}_{0}\mathfrak{R}_{1}(\overrightarrow{P};q)&{}_{0}\mathfrak{R}_{0}(\overrightarrow{P};q)\end{bmatrix}$

######  Proposition 4.2.

Let $\overrightarrow{P}=(P,x_{L},x_{R})$ and
$\overrightarrow{Q}=(Q,y_{L},y_{R})$ be two oriented posets. Then we have:

(12)
$\displaystyle{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P}\nearrow\overrightarrow{Q})={\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P})\cdot{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{Q}),$
(13) $\displaystyle
R(\circlearrowleft\\!(\overrightarrow{P}))=\mathrm{trace}({\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P})).$

The proof is similar to the regular case and will be omitted.

For the chains $\overrightarrow{U_{n}}$ and $\overrightarrow{D_{n}}$, we get
the following expressions for dual rank matrices:

(14)
$\displaystyle{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{U_{n}})$
$\displaystyle=\begin{bmatrix}q^{n+1}&[n+1]_{q}\\\
0&1\end{bmatrix}=R_{q}^{n+1}\in\mathrm{PSL_{q}}(2,\mathbb{Z}),$ (15)
$\displaystyle{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{D_{n}})$
$\displaystyle=\begin{bmatrix}q[n+1]_{q}&1\\\
q[n]_{q}&1\end{bmatrix}=R_{q}(R_{q}S_{q}R_{q})^{n}\in\mathrm{PSL_{q}}(2,\mathbb{Z}).$

We use the notation $U_{0}$ for the dual rank matrix of
$\overrightarrow{\bullet}=\overrightarrow{U_{0}}=\overrightarrow{D_{0}}$:

$\displaystyle U_{0}$ $\displaystyle=\begin{bmatrix}q&1\\\
0&1\end{bmatrix}=R_{q}.$

######  Remark 4.3.

The increasing chain $\overrightarrow{U_{n}}$ can be built by adding up $n+1$
copies of $\overrightarrow{\bullet}$ in this dual way. Accordingly we have:

(16)
$\displaystyle{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{U_{n}})=$
$\displaystyle(U_{0})^{n+1}.$

######  Example 4.4.

As seen in Figure 4 above,
$\overrightarrow{U_{3}}\nearrow\overrightarrow{D_{3}}$ gives us the fence
poset $F(4,3)$ and
$\circlearrowleft\\!(\overrightarrow{U_{3}}\nearrow\overrightarrow{D_{3}})$
gives us the circular fence poset $\overline{F}(5,3)$. The corresponding rank
polynomials can be calculated as follows:

$\displaystyle{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{U_{3}}\nearrow\overrightarrow{D_{3}})$
$\displaystyle=\begin{bmatrix}q^{4}&[4]_{q}\\\
0&1\end{bmatrix}\cdot\begin{bmatrix}q[4]_{q}&1\\\
q[3]_{q}&1\end{bmatrix}=\begin{bmatrix}(q+q^{2}+q^{3}+q^{5})[4]_{q}&[5]_{q}\\\
q[3]_{q}&1\end{bmatrix}$ $\displaystyle\mathfrak{R}((4,3);q)$
$\displaystyle={\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{U_{3}}\nearrow\overrightarrow{D_{3}})[1,1]+{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{U_{3}}\nearrow\overrightarrow{D_{3}})[1,2]$
$\displaystyle=(q+q^{2}+q^{3}+q^{5})[4]_{q}+[5]_{q}=[5]_{q}[4]_{q}+q^{8}.$
$\displaystyle\overline{\mathfrak{R}}((5,3);q)$
$\displaystyle=\mathrm{tr}(\overrightarrow{U_{3}}\nearrow\overrightarrow{D_{3}}))=1+q+2q^{2}+3q^{3}+3q^{4}+3q^{5}+2q^{6}+q^{7}+q^{8}.$

We will use the expression _rank matrices_ of $\overrightarrow{P}$ when we
need to refer to the rank matrix
${\triangledown}\mathfrak{R}_{q}(\overrightarrow{P})$ and the dual rank matrix
${\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P})$
together.

######  Lemma 4.5.

The rank matrices of an oriented poset $\overrightarrow{P}$ are related by:

$\displaystyle{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P})$
$\displaystyle={\triangledown}\mathfrak{R}_{q}(\overrightarrow{P})\begin{bmatrix}0&1\\\
-1&1\end{bmatrix}\qquad\qquad{\triangledown}\mathfrak{R}_{q}(\overrightarrow{P})={\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P})\begin{bmatrix}1&-1\\\
1&0\end{bmatrix}.$

As a consequence, for any two oriented posets
$\overrightarrow{P}=(P,x_{L},x_{R})$ and $\overrightarrow{Q}=(Q,y_{L},y_{R})$
we have:

(17)
$\displaystyle{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P})\cdot{\triangledown}\mathfrak{R}_{q}(\overrightarrow{Q})$
$\displaystyle={\triangledown}\mathfrak{R}_{q}(\overrightarrow{P}\nearrow\overrightarrow{Q}),$
(18)
$\displaystyle{\triangledown}\mathfrak{R}_{q}(\overrightarrow{P})\cdot{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{Q})$
$\displaystyle={\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P}\searrow\overrightarrow{Q}).$

###### Proof.

The first part follows from the fact that
$\mathfrak{R}(q)=\mathfrak{R}_{0}(q)+\mathfrak{R}_{1}(q)$ and
${}_{0}\mathfrak{R}(q)=\,_{0}\mathfrak{R}_{0}(q)+\,_{0}\mathfrak{R}_{1}(q)$
and the second part follows from the associativity of matrix multiplication. ∎

##  5\. The Case of Fence Posets

As we are able to build up any fence poset via adding up up and down steps, we
can use the corresponding matrices to calculate rank polynomials. For a
composition $\alpha$ of $n$, we can see the fence poset as an oriented poset
by taking $x_{0}$ to be the left end, and $x_{n}$ to be the right end. One can
build the fence poset of $\alpha$ from up and down steps. Looking at the
corresponding multiplication of the matrices $U_{0}$ and $D_{0}$ gives us the
rank matrices for $\overrightarrow{\alpha}$. Which matrix we obtain is
determined by whether we pick $U_{0}$ or $D_{0}$ for the rightmost node.

######  Proposition 5.1.

Consider the oriented poset $\overrightarrow{\alpha}$ corresponding to
$\alpha=(u_{1},d_{1},u_{2},d_{2},\ldots,u_{s},d_{s})$ where we allow $d_{s}=0$
for full generality. Then $\overrightarrow{\alpha}$ has rank matrices:

(19) $\displaystyle{\triangledown}\mathfrak{R}(\overrightarrow{\alpha})$
$\displaystyle:=(U_{0})^{u_{1}}(D_{0})^{d_{1}}(U_{0})^{u_{2}}(D_{0})^{d_{2}}\cdots(U_{0})^{u_{s-1}}(D_{0})^{d_{s-1}}(U_{0})^{u_{s}}(D_{0})^{d_{s}+1},$
(20)
$\displaystyle{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}(\overrightarrow{\alpha})$
$\displaystyle:=(U_{0})^{u_{1}}(D_{0})^{d_{1}}(U_{0})^{u_{2}}(D_{0})^{d_{2}}\cdots(U_{0})^{u_{s-1}}(D_{0})^{d_{s-1}}(U_{0})^{u_{s}}(D_{0})^{d_{s}}(U_{0}),.$

The rank polynomial $\mathfrak{R}(\alpha,q)$ is given by the top left entry of
${\triangledown}\mathfrak{R}(\overrightarrow{\alpha})$. Alternatively, one can
take the sum of the entries on the top row of
${\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}(\overrightarrow{\alpha})$.

######  Example 5.2.

We can construct the fence poset for $\alpha=(2,1,1,3)$ from Figure 1 by
connecting
$\overrightarrow{U_{2}}\searrow\overrightarrow{U_{1}}\searrow\overrightarrow{D_{2}}$
or equivalently
$\overrightarrow{\bullet}\nearrow\overrightarrow{\bullet}\nearrow\overrightarrow{\bullet}\searrow\overrightarrow{\bullet}\nearrow\overrightarrow{\bullet}\searrow\overrightarrow{\bullet}\searrow\overrightarrow{\bullet}\searrow\overrightarrow{\bullet}$.

$x_{L}$$x_{R}$$\overrightarrow{U_{2}}\searrow\overrightarrow{U_{1}}\searrow\overrightarrow{D_{2}}$

$\quad=\quad$
$x_{L}$$x_{R}$$\overrightarrow{\bullet}\nearrow\overrightarrow{\bullet}\nearrow\overrightarrow{\bullet}\searrow\overrightarrow{\bullet}\nearrow\overrightarrow{\bullet}\searrow\overrightarrow{\bullet}\searrow\overrightarrow{\bullet}\searrow\overrightarrow{\bullet}$

The corresponding rank matrix is given by:

$\displaystyle{\triangledown}\mathfrak{R}(\overrightarrow{(2,1,1,3)})=\begin{bmatrix}[4]_{q}&-q^{3}\\\
1&0\end{bmatrix}\begin{bmatrix}[3]_{q}&-q^{2}\\\
1&0\end{bmatrix}\begin{bmatrix}[4]_{q}&-q[3]_{q}\\\
[3]_{q}&-q[2]_{q}\end{bmatrix}=U_{0}^{2}D_{0}U_{0}D_{0}^{4}$
$\displaystyle=\begin{bmatrix}1\\!+\\!3q\\!+\\!5q^{2}\\!+\\!6q^{3}\\!+\\!6q^{4}\\!+\\!5q^{5}\\!+\\!3q^{6}\\!+\\!2q^{7}\\!+\\!q^{8}&-q(1\\!+\\!3q\\!+\\!5q^{2}\\!+\\!5q^{3}\\!+\\!5q^{4}\\!+\\!3q^{5}\\!+\\!q^{6})\\\
1\\!+\\!2q\\!+\\!2q^{2}\\!+\\!2q^{3}\\!+\\!q^{4}\\!+\\!q^{5}&-q(1\\!+\\!2q\\!+\\!2q^{2}\\!+\\!q^{3}\\!+\\!q^{4})\end{bmatrix}$

Note that the top left hand side evaluates to $\mathfrak{R}((2,1,1,3);q)$ that
we calculated in Example 1.1. Furthermore, the whole matrix matches
$M_{q}(4,3,2,2,2)$ from Example 2.4.

After building up the fences this way, we can use the $\circlearrowright\\!\,$
operation to make our fences circular. One should note that the operation
$\circlearrowright\\!$ adds a new relation $x_{R}\succeq x_{L}$ to our poset.
If $\alpha$ has an even number of parts, this corresponds to increasing the
last part by $1$, whereas if it has an odd number of parts (larger than $1$)
an extra part of size $1$ is added. In any case we end up with an even number
of parts in our circular fence. That is why the matrix for the circular fence
that we define in Equation 21 below differs from the one in Equation 19 in the
last entry.

######  Proposition 5.3.

Let $P$ be the circular fence poset corresponding
$\alpha=(u_{1},d_{1},u_{2},d_{2},\ldots,u_{s},d_{s})$. The circular rank
polynomial $\overline{\mathfrak{R}}(\alpha,q)$ is given by the trace of the
following matrix:

(21) $\displaystyle\overline{\mathfrak{M}}_{q}(\alpha)$
$\displaystyle:=(U_{0})^{u_{1}}(D_{0})^{d_{1}}(U_{0})^{u_{2}}(D_{0})^{d_{2}}\cdots(U_{0})^{u_{s}}(D_{0})^{d_{s}}$
$\displaystyle={\triangledown}\mathfrak{R}_{q}(\overrightarrow{\alpha})\cdot
D_{0}^{-1}={\triangledown}\mathfrak{R}_{q}(\overrightarrow{\alpha})\cdot
U_{0}^{-1}.$

######  Example 5.4.

For the circular fence poset corresponding to $\alpha=(2,1,1,3)$ from Figure 1
we connect the two ends of
$\overrightarrow{U_{2}}\searrow\overrightarrow{U_{1}}\searrow\overrightarrow{D_{1}}$.
We can equivalently look at the trace of the matrix
$\overline{\mathfrak{M}}_{q}((2,1,1,3))=U_{0}^{2}\cdot D_{0}\cdot U_{0}\cdot
D_{0}^{2}$. Note that taking the trace gives us the final down-step connecting
the two ends.

$\circlearrowright\\!(\overrightarrow{U_{2}}\searrow\overrightarrow{U_{1}}\searrow\overrightarrow{D_{1}})$

$\quad=\quad$
$\circlearrowright\\!(\overrightarrow{\bullet}\nearrow\overrightarrow{\bullet}\nearrow\overrightarrow{\bullet}\searrow\overrightarrow{\bullet}\nearrow\overrightarrow{\bullet}\searrow\overrightarrow{\bullet}\searrow\overrightarrow{\bullet})$

$\displaystyle\overline{\mathfrak{M}}_{q}((2,1,1,3))=\begin{bmatrix}[4]_{q}&-q^{3}\\\
1&0\end{bmatrix}\begin{bmatrix}[3]_{q}&-q^{2}\\\
1&0\end{bmatrix}\begin{bmatrix}[3]_{q}&-q[2]_{q}\\\
[2]_{q}&-q[1]_{q}\end{bmatrix}$
$\displaystyle=\begin{bmatrix}1+3q+5q^{2}+6q^{3}+6q^{4}+5q^{5}+3q^{6}+2q^{7}+q^{8}&-q[3]_{q}(1+q)(1+q+q^{3})\\\
1+2q+2q^{2}+q^{3}+q^{4}&-q(1+2q+q^{2}+q^{3}).\end{bmatrix}$

The trace of the matrix is equal to
$[5]_{q}[4]_{q}=\overline{\mathfrak{R}}((2,1,1,3);q)$ from Example 1.1.

######  Proposition 5.5.

Let $\llbracket c_{1},c_{2},\ldots,c_{k}\rrbracket$ and
$[a_{1},a_{2},\cdots,a_{2m}]$ be the continued fraction expressions for
$r/t\geq 1$. Then we have:

(22) $\displaystyle M_{q}(c_{1},c_{2},\ldots,c_{k})$
$\displaystyle=\overline{\mathfrak{M}}_{q}(\alpha_{r/q})\cdot
D_{0}={\triangledown}\mathfrak{R}_{q}(\overrightarrow{\alpha_{r/q}}),$ (23)
$\displaystyle M^{+}_{q}(a_{1},a_{2},\cdots,a_{2m})$
$\displaystyle=\overline{\mathfrak{M}}_{q}(\alpha_{r/q})\cdot D_{0}\cdot
U_{0}={\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{(a_{1}-1,a_{2},\ldots,a_{2m},1)})).$

Furthermore,

(24) $\displaystyle\mathrm{tr}(M_{q}(c_{1},c_{2},\ldots,c_{k}))$
$\displaystyle=\overline{\mathfrak{R}}((a_{1}-1,a_{2},\ldots,a_{2m};q),$ (25)
$\displaystyle\mathrm{tr}(M_{q}^{+}(a_{1},a_{2},\ldots,a_{2m}))$
$\displaystyle=\overline{\mathfrak{R}}((a_{1},a_{2},\ldots,a_{2m});q).$

###### Proof.

Remember that by Equation 3, we have $\llbracket
c_{1},c_{2},\ldots,c_{k}\rrbracket=\llbracket
a_{1}+1,2^{a_{2}-1},a_{3}+2,2^{a_{4}-1},a_{5}+2,\ldots,a_{2m-1}+2,2^{a_{2m}-1}\rrbracket$.
We plug in $D_{0}=R_{q}^{2}S_{q}$ and $U_{0}=R_{q}$ in Equations 6 7 to get a
rank matrix of a fence poset.

$\displaystyle M_{q}(c_{1},c_{2},\ldots,c_{k})$
$\displaystyle=R_{q}^{a_{1}+1}S_{q}(R_{q}^{2}S_{q})^{a_{2}-1}R_{q}^{a_{3}+2}S_{q}(R_{q}^{2}S_{q})^{a_{4}-1}\cdots
R_{q}^{a_{2m-1}+2}S_{q}(R_{q}^{2}S_{q})^{a_{2m}-1}$
$\displaystyle=(U_{0})^{a_{1}-1}(D_{0})^{a_{2}}(U_{0})^{a_{3}}(D_{0})^{a_{4}}\cdots(U_{0})^{a_{2m-1}}(D_{0})^{a_{2m}}$
$\displaystyle={\triangledown}\mathfrak{R}_{q}(\overrightarrow{\alpha_{r/t}})),$
$\displaystyle M_{q}^{+}(a_{1},a_{2},\ldots,a_{2m})$
$\displaystyle=R_{q}^{a_{1}}(R_{q}S_{q}R_{q})^{a_{2}}R_{q}^{a_{3}}(R_{q}S_{q}R_{q})^{a_{4}}\cdots
R_{q}^{a_{2m-1}}(R_{q}S_{q}R_{q})^{a_{2m}}$
$\displaystyle=R_{q}^{a_{1}-1}(R_{q}^{2}S_{q})^{a_{2}}R_{q}^{a_{3}}(R_{q}^{2}S_{q})^{a_{4}}\cdots(R_{q})^{a_{2m-1}}(R_{q}^{2}S_{q})^{a_{2m}}R_{q}$
$\displaystyle=(U_{0})^{a_{1}-1}(D_{0})^{a_{2}}(U_{0})^{a_{3}}(D_{0})^{a_{4}}\cdots(U_{0})^{a_{2m-1}}(D_{0})^{a_{2m}}(U_{0})$
$\displaystyle={\triangledown}\mathfrak{R}_{q}(\overrightarrow{\alpha_{r/t}}))\cdot(U_{0}).$

The trace formula for $M_{q}(c_{1},c_{2},\ldots,c_{k})$ follows directly. For
the case of $M_{q}^{+}(a_{1},a_{2},\ldots,a_{2m})$, we take advantage of the
fact that trace remains unchanged if we move $U_{0}$ from one side to the
other.

$\displaystyle\mathrm{tr}(M_{q}^{+}(a_{1},a_{2},\ldots,a_{2m}))$
$\displaystyle=\mathrm{tr}(\overline{\mathfrak{M}}_{q}(\alpha_{r/q})\cdot
D_{0}\cdot
U_{0})=\mathrm{tr}(U_{0}\cdot\overline{\mathfrak{M}}_{q}(\alpha_{r/q})\cdot
D_{0})$
$\displaystyle=\mathrm{tr}(\overline{\mathfrak{M}}_{q}((a_{1},a_{2},\ldots,a_{2m})))=\overline{\mathfrak{R}}((a_{1},a_{2},\ldots,a_{2m});q).$

Alternatively, one can consider the continuant expressions for the matrices
$M_{q}$ and $M_{q}^{+}$ (see Propositions 5.2 and 5.3 from [1]) to verify
directly that the entries match those of the rank matrices. ∎

The above proposition allows us to apply the discussion of unimodality of rank
polynomials of circular fences to the traces of matrices in
$\mathrm{PSL}_{q}(2,\mathbb{Z})$. We have one degenerate case of $\llbracket
c_{1},c_{2},\ldots,c_{k}\rrbracket$ with $c_{i}=2$ for all $i$. In that case
the corresponding fence is a chain of length $k$, and taking the trace adds
the relation $x_{1}\succeq x_{k}$ where $x_{2}$ is the minimal node and
$x_{k}$ is the maximal. The resulting structure has all nodes equal to each
other, so the only two ideals consist of the empty set and the whole chain.
Another violation of unimodality comes from the cases $(1,k,1,k)$ and
$(k,1,k,1)$ whose circular rank polynomials are not unimodal.

######  Corollary 5.6.

The polynomials $\mathrm{tr}(M_{q}(c_{1},c_{2},\ldots,c_{k}))$ are not always
unimodal. For $c_{i}\geq 2$ we get the following counterexamples:

(CE 1) $\displaystyle\operatorname{tr}(M_{q}(2^{k}))=$ $\displaystyle 1+q^{k}$
$\displaystyle\text{ where }k\geq 2,$ (CE 2)
$\displaystyle\operatorname{tr}(M_{q}(n+2,n+2)=$ $\displaystyle
1+2q+3q^{2}+\cdots+(n+1)q^{n}$ $\displaystyle+n\,q^{n+1}+(n+1)q^{n+2}+\ldots$
$\displaystyle\ldots+3q^{2n-2}+2q^{2n-1}+q^{2n}$ $\displaystyle\text{ where
}n\geq 1,$ (CE 2’)
$\displaystyle\operatorname{tr}(M_{q}(2^{n-1},3,2^{n-1},3))=$
$\displaystyle\operatorname{tr}(M_{q}(n+2,n+2))$ $\displaystyle\text{ where
}n\geq 1.$

######  Conjecture 5.7.

Up to cyclic shifts of $\llbracket c_{1},c_{2},\ldots,c_{k}\rrbracket$ , the
examples listed in Corollary 5.6 are the only counterexamples to unimodality
for $\mathrm{tr}(M_{q}(c_{1},c_{2},\ldots,c_{k}))$, $c_{i}\geq 2$.

As mentioned before in Section 2, this actually is a statement concerning the
traces of all matrices in $\mathrm{PSL}_{q}(2,\mathbb{Z})$, as all such
matrices are given by some $M_{q}(c_{1},c_{2},\ldots,c_{k})$ with $c_{i}\geq
2$ up to a factor of $q^{\pm N}$.

We can also apply the partial unimodality results from Theorem 1.3 using the
rank polynomial descriptions of traces of $M_{q}$ and $M_{q}^{+}$. For the
continued fraction expression $[a_{1},a_{2},\ldots,a_{s}]$ one only needs to
check the conditions for $\alpha=(a_{1},a_{2},\ldots,a_{s})$. For the
corresponding expression $\llbracket c_{1},c_{2},\ldots,c_{k}\rrbracket$, one
checks $\alpha-1$ formed by subtracting $1$ from the leftmost part of $\alpha$
instead.

##  6\. Rank Polynomial Identities

In this section we will be concerned with some multiplicative identities
concerning rank polynomials.

######  Example 6.1.

Looking at the rank polynomials for $\overline{\mathfrak{R}}((k,2,1,1,1,1);q)$
one can spot a common factor.

$\displaystyle\overline{\mathfrak{R}}((1,2,1,1,1,1;q)$
$\displaystyle=(q^{6}+2q^{5}+2q^{4}+3q^{3}+2q^{2}+2q+1)(q+1)$
$\displaystyle\overline{\mathfrak{R}}((2,2,1,1,1,1;q)$
$\displaystyle=(q^{6}+2q^{5}+2q^{4}+3q^{3}+2q^{2}+2q+1)(q^{2}+q+1)$
$\displaystyle\overline{\mathfrak{R}}((3,2,1,1,1;q)$
$\displaystyle=(q^{6}+2q^{5}+2q^{4}+3q^{3}+2q^{2}+2q+1)(q^{3}+q^{2}+q+1)$

That polynomial matches
$\overline{\mathfrak{R}}((3,1,1,1;q)=(q^{6}+2q^{5}+2q^{4}+3q^{3}+2q^{2}+2q+1)$.
The pattern persists when we have a larger number of pieces of size $1$.

(Id 0.1) $\displaystyle\overline{\mathfrak{R}}((k,2,1^{2s+2});q)$
$\displaystyle=[k+1]_{q}\cdot\overline{\mathfrak{R}}((3,1^{2s+1});q).$

We also have a companion identity when the $3$ on the right hand side varies.

(Id 0.2) $\displaystyle\overline{\mathfrak{R}}((k+1,1,k,1^{2s+1});q)$
$\displaystyle=[k+1]_{q}\cdot\overline{\mathfrak{R}}((k+2,1^{2s+1});q).$

$\qquad\longleftrightarrow$

Figure 5. Illustration of (Id 0.1) at $s=4$, $k=3$

We will see next that the identities described are actually part of a larger
family of rank polynomial identities concerning palindromic compositions. For
this purpose, we extend the notation
$\overline{\mathfrak{M}}_{q}(\alpha):=(U_{0})^{u_{1}}(D_{0})^{d_{1}}(U_{0})^{u_{2}}(D_{0})^{d_{2}}\cdots(U_{0})^{u_{s}}(D_{0})^{d_{s}}$
from Equation 21 to include the cases where $u_{1}$, $d_{s}$ or both may be
zero. In the language of fences, this means that we include the cases of
starting with a down step as well. See Section 2 for a discussion on the
corresponding rank polynomials.

######  Lemma 6.2.

Let $\rho=(\rho_{1},\rho_{2},\ldots,\rho_{s})$ be a palindromic composition (A
composition satisfying $\rho_{1}=\rho_{s}$, $\rho_{2}=\rho_{s-1}$, $\ldots$
etc.). Consider the following maps on $2\times 2$ matrices:

$\displaystyle\Psi^{+}(X)$
$\displaystyle:=\operatorname{tr}\left(\begin{bmatrix}1&-1-q\\\
1&-1\end{bmatrix}\cdot\begin{bmatrix}x_{11}&x_{12}\\\
x_{21}&x_{22}\end{bmatrix}\right)=x_{11}+x_{12}-(1+q)x_{21}-x_{22}.$
$\displaystyle\Psi^{-}(X)$
$\displaystyle:=\operatorname{tr}\left(\begin{bmatrix}1&-1-q^{2}\\\
1-q&-1\end{bmatrix}\cdot\begin{bmatrix}x_{11}&x_{12}\\\
x_{21}&x_{22}\end{bmatrix}\right)=x_{11}+(1-q)x_{12}-(1+q^{2})x_{21}-x_{22}.$

For $\rho$ with an even number of parts we have:

(26)
$\displaystyle\Psi^{+}(\overline{\mathfrak{M}}_{q}(\rho))=\Psi^{+}(\overline{\mathfrak{M}}_{q}(0,\rho,0))=0,$

For $\rho$ with an odd number of parts we have:

(27)
$\displaystyle\Psi^{-}(\overline{\mathfrak{M}}_{q}(0,\rho))=\Psi^{+}(\overline{\mathfrak{M}}_{q}(\rho,0))=0.$

###### Proof.

The equalities may be proved by induction. Equation 26 is quite
straightforward as the matrix $\overline{\mathfrak{M}}_{q}(\rho)$ for any
palindromic composition $\rho$ can be formed by starting with identity and
multiplying by $U_{0}$ and $D_{0}$ from the sides. It is quite straightforward
to verify via direct calculation that $\Psi^{+}(I)=0$ and for all $2\times 2$
matrices $X$ one has

$\displaystyle\Psi^{+}(U_{0}\cdot X\cdot D_{0})=\Psi^{+}(D_{0}\cdot X\cdot
U_{0})=q\Psi^{+}(X).$

The situation of Equation 27 is only slightly more complicated as the middle
part of an odd palindromic composition can be odd or even. The result follows
as $\Psi^{-}(I)=\Psi^{-}(U_{0})=\Psi^{-}(D_{0})=0$ and for all $X$,

$\displaystyle\Psi^{-}(U_{0}\cdot X\cdot U_{0})=\Psi^{-}(D_{0}\cdot X\cdot
D_{0})=q\Psi^{-}(X).$

∎

######  Theorem 6.3.

Let $\rho$ be a palindromic composition with an even number of parts. For
$k\geq 1$, $r\geq 1$ we have:

(Id 1) $\displaystyle\overline{\mathfrak{R}}((1,k,r+1,\rho,r);q)$
$\displaystyle=[k+1]_{q}\cdot\overline{\mathfrak{R}}((r+2,\rho,r);q),$ (Id 2)
$\displaystyle\overline{\mathfrak{R}}((k,1,k+r,\rho,r);q)$
$\displaystyle=[k+1]_{q}\cdot\overline{\mathfrak{R}}((k+r+1,\rho,r);q).$

###### Proof.

We will prove the statement by reducing both identities to
$\Psi^{+}(\overline{\mathfrak{M}}_{q}(\rho))=0$ from Lemma 6.2. Note that
$(r,\rho,r)$ is palindromic if and only if $\rho$ is palindromic. So, if we
denote the composition formed by adding $n$ to the _leftmost_ part of $\rho$
by $\rho+n$, it suffices to show the following hold for any palindromic
composition $\rho$ with an even number of parts:

(Id 1′) $\displaystyle\overline{\mathfrak{R}}((1,k,\rho+1);q)$
$\displaystyle=[k+1]_{q}\cdot\overline{\mathfrak{R}}((\rho+2);q),$ (Id 2′)
$\displaystyle\overline{\mathfrak{R}}((k,1,\rho+k);q)$
$\displaystyle=[k+1]_{q}\cdot\overline{\mathfrak{R}}((\rho+k+1);q).$

By Equation 21, (Id $1^{\prime}$) can be expressed in terms of
$\operatorname{tr}$ as:

$\displaystyle\operatorname{tr}(U_{0}\cdot(D_{0})^{k}\cdot
U_{0}\cdot\overline{\mathfrak{M}}_{q}(\rho))=[k+1]_{q}\operatorname{tr}(U_{0}^{2}\cdot\overline{\mathfrak{M}}_{q}(\rho)),$
$\displaystyle\Leftrightarrow$
$\displaystyle\operatorname{tr}(U_{0}\cdot((D_{0})^{k}-[k+1]_{q})\cdot
U_{0}\cdot\overline{\mathfrak{M}}_{q}(\rho))=0.$

This expression is actually independent of $k$ as we have:

$\displaystyle(D_{0})^{k}-[k+1]_{q}=\begin{bmatrix}0&-q[k]_{q}\\\
[k]_{q}&-q[k-1]_{q}-[k+1]_{q}\end{bmatrix}=\begin{bmatrix}0&-q[k]_{q}\\\
[k]_{q}&-(1+q)[k]_{q}\end{bmatrix}=[k]_{q}\begin{bmatrix}0&-q\\\
1&-1-q\end{bmatrix}.$

This means $\overline{\mathfrak{M}}_{q}(\rho)$ satisfies (Id $1^{\prime}$) if
and only if:

(28) $\displaystyle\operatorname{tr}\left(\begin{bmatrix}q&1\\\
0&1\end{bmatrix}\begin{bmatrix}0&-q\\\
1&-1-q\end{bmatrix}\begin{bmatrix}q&1\\\
0&1\end{bmatrix}\overline{\mathfrak{M}}_{q}(\rho)\right)=q\Psi^{+}(\overline{\mathfrak{M}}_{q}(\rho))=0.$

Similarly, we can rewrite (Id 2′) in terms of trace of rank matrices:

$\displaystyle\operatorname{tr}(U_{0}^{k}\cdot(D_{0})\cdot(U_{0})^{k}\cdot\overline{\mathfrak{M}}_{q}(\rho))=[k+1]_{q}\operatorname{tr}(U_{0}^{k+1}\cdot\overline{\mathfrak{M}}_{q}(\rho))$
$\displaystyle\Leftrightarrow$
$\displaystyle\operatorname{tr}(U_{0}\cdot(U_{0}^{k-1}\cdot
D_{0}-[k+1]_{q})\cdot(U_{0})^{(k-1)}\cdot
U_{0}\cdot\overline{\mathfrak{M}}_{q}(\rho))=0.$

Again, this expression ends up not being dependent on the value of $k$.

$\displaystyle(U_{0}^{k-1}\cdot
D_{0}-[k+1]_{q})\cdot(U_{0})^{k-1})=q^{k-1}\begin{bmatrix}0&-q\\\
1&-1-q\end{bmatrix},$

which means by Equation 28, the condition to satisfy (Id 2′) is also that
$\Psi^{+}(\overline{\mathfrak{M}}_{q}(\rho))=0$. ∎

######  Question 6.4.

Equation 26 for palindromic compositions of even length gives rise to the rank
polynomial identities listed above. Are there similar consequences to Equation
27? Can we find rank polynomial identities for palindromic compositions of odd
length?

######  Theorem 6.5.

For $k\geq 2$, $s\geq 0$ we have:

(Id 3) $\displaystyle\overline{\mathfrak{R}}((2k,k^{2s+1});q)$
$\displaystyle=(1+q^{k})\cdot\overline{\mathfrak{R}}((k+1,k-1,k^{2s});q).$

###### Proof.

Set $A:=U_{0}^{k}D_{0}^{k}$. By Equation 21, we can rewrite (Id 3) in terms of
traces:

$\displaystyle\operatorname{tr}(U_{0}^{2k}D_{0}^{k}A^{s})=(1+q^{k})\operatorname{tr}(U_{0}^{k+1}D_{0}^{k-1}A^{s})$
$\displaystyle\Leftrightarrow$
$\displaystyle\operatorname{tr}((U_{0}^{2k}D_{0}^{k}-(1+q^{k})U_{0}^{k+1}D_{0}^{k-1})\cdot
A^{s})=0.$

It is in fact enough to prove the identity for $A^{0}$ and $A^{1}$ as the
characteristic polynomial of $A$ has degree $2$. At $s=0$ we have:

$\displaystyle
U_{0}^{2k}D_{0}^{k}-(1+q^{k})U_{0}^{k+1}D_{0}^{k-1}=\displaystyle\frac{q^{k-1}}{q-1}\begin{bmatrix}-1+q+q^{2}-q^{k}&1-q^{2}-q^{3}+q^{k}+q^{k+2}-q^{k+3}\\\
1+2q-q^{k}&1-q-q^{2}+q^{k}.\end{bmatrix}$

The trace of the above matrix is $0$. One can do a similar calculation to show
the trace is zero for the $s=1$ case as well. ∎

######  Question 6.6.

It might be interesting to see if (Id 3) above generalizes to an identity on
palindromic compositions.

##  7\. A Combinatorial Model for $q$-deformed Markov Numbers

In the introduction, we have mentioned that all positive integer triples
solving the Markov Diophantine equation $x^{2}+y^{2}+z^{2}=3xyz$ can be
reached from the inital solution $(1,1,1)$ by operations of the type
$(a,b,c)\rightarrow(a,b,(a^{2}+b^{2})/c)$. Equivalently, one can make use of
Christoffel words. Starting with the triple $(a,ab,b)$, all the triples we can
obtain by the operations:

$\displaystyle(u,uv,v)\longrightarrow(u,uuv,uv)$
$\displaystyle(u,uv,v)\longrightarrow(uv,uvv,v)$

are called Christoffel triples, and the words $w(a,b)$ that occur as
coordinates in these triples are called Christoffel words. Let $W_{C}(a,b)$
denote the set of Christoffel words. Given a Christoffel word $w(a,b)$, the
corresponding Markov number can be obtained by calculating the matrix
$w(A,B)$, and then dividing the trace by three, where $A$ and $B$ are the Cohn
matrices defined as:

$\displaystyle A:=\begin{bmatrix}2&1\\\ 1&1\end{bmatrix}\qquad\qquad
B:=\begin{bmatrix}5&2\\\ 2&1\\\ \end{bmatrix}$

In [2], the following $q$-transformation is given for the Cohn matrices:

$\displaystyle[A]_{q}:=\begin{bmatrix}q+q^{2}&1\\\
q&1\end{bmatrix}\qquad\qquad[B]_{q}:=\begin{bmatrix}q+2q^{2}+q^{3}+q^{4}&1+q\\\
q+q^{2}&1\\\ \end{bmatrix}$

A similar deformation involving negative powers of $q$ is given in [14] where
$[A]_{q}$ is replaced with $q^{-1}[A]_{q}$ and $[B]_{q}$ is replaced with
$q^{-2}[B]_{q}$. The two deformations are equivalent up to a multiplication
with a power of $q$, we are following the [2] version that takes $q^{0}$ as
the smallest term for ease of notation.

For a given $w\in W_{C}(a,b)$, we will denote the length of $w$ by
$\ell(w)=\ell_{a}(w)+\ell_{b}(w)$ where $\ell_{a}(w)$ (resp. $\ell_{b}(w)$)
stands for the number of times $a$ (resp. $b$) occurs in $w$. If $w(a,b)\in
W_{C}$ is not equal to the trivial words $a$ and $b$, then we also define
$\hat{w}(a,b)$ to be the palindromic word satisfying
$w(a,b)=a\,\hat{w}(a,b)\,b$. Finally, we set
$w_{q}(A,B):=w({[A]_{q}},{[B]_{q}})$ for ease of notation.

######  Theorem 7.1 ([14], Theorem 2.6).

For a Christoffel triple $(w,w\,w^{\prime},w^{\prime})$,

$(x,y,z):=([3]_{q})^{-1}\cdot(\mathrm{tr}(w_{q}(A,B),\mathrm{tr}(w_{q}(A,B)w^{\prime}_{q}(A,B)),\mathrm{tr}(w^{\prime}_{q}(A,B)))$

satisfies the following $q$-deformed Markov equation :

(29)
$\left(\frac{x}{q^{c_{x}}}\right)^{2}+\left(\frac{y}{q^{c_{y}}}\right)^{2}+\left(\frac{z}{q^{c_{z}}}\right)^{2}+\frac{(q-1)^{3}}{q^{3}}=[3]_{q}\frac{xyz}{q^{c_{x}+c_{y}+c_{z}}}.$

where $c_{x}:=\ell(w)+\ell_{b}(w)$,
$c_{z}:=\ell(w^{\prime})+\ell_{b}(w^{\prime})$ and $c_{y}:=c_{x}+c_{z}$.

We will now give a combinatorial description for entries of the Cohn matrices
using oriented posets. Consider the following posets:

$x_{L}$$x_{R}$$y_{L}$$y_{R}$$\overrightarrow{P_{A}}$$\overrightarrow{P_{B}}$

These deformations will help us view any Cohn matrix as a rank matrix of a
fence poset, whose trace is given by a circular rank polynomial. For a given
word $w(A,B)$, form a corresponding composition $\alpha(w)$ by replacing each
$B$ with a pair of parts of size $2$ and each $A$ with a pair of parts of size
$1$. For example, if $w(A,B)=AAAB$ we get $\alpha(w)=(1,1,1,1,1,1,2,2).$

######  Proposition 7.2.

The $q$-deformations $[A]_{q}$ and $[B]_{q}$ correspond to the dual rank
matrices for the oriented posets $\overrightarrow{P_{A}}$ and
$\overrightarrow{P_{B}}$ respectively. As a consequence, for any word $w(A,B)$
we have

(30) $\mathrm{tr}(w_{q}(A,B))=\overline{\mathfrak{R}}(\alpha(w))$

###### Proof.

Calculating the corresponding rank polynomials shows us that
${\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P_{A}})=[A]_{q}$
and
${\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P_{B}})=[B]_{q}$.
The matrix product $[A]_{q}^{k}\,[B]_{q}$ therefore corresponds to the poset
formed by connecting copies of $A$ and $B$. With the connected edges included,
any copy of $B$ corresponds to $2,2$ and any copy of $A$ corresponds to $1,1$
. As seen in the example
$\overrightarrow{P_{A}}\nearrow\overrightarrow{P_{A}}\nearrow\overrightarrow{P_{A}}\nearrow\overrightarrow{P_{B}}$
below, we end up with the fence poset for $\alpha(w)$ with the leftmost edge
missing. Taking the trace is equivalent to forming the missing edge by
connecting the two ends, giving us $\overline{\mathfrak{R}}(\alpha(w))$.

$x_{L}$$x_{R}$$\overrightarrow{P_{A}}\nearrow\overrightarrow{P_{A}}\nearrow\overrightarrow{P_{A}}\nearrow\overrightarrow{P_{B}}$

$\circlearrowleft\\!(\overrightarrow{P_{A}}\nearrow\overrightarrow{P_{A}}\nearrow\overrightarrow{P_{A}}\nearrow\overrightarrow{P_{B}})$

∎

######  Theorem 7.3.

For a Christoffel $ab$-word $w(a,b)=a\hat{w}(a,b)b$ we have:

(31)
$\mathrm{tr}(w_{q}(A,B))=\overline{\mathfrak{R}}((1,1,\alpha(\hat{w}),2,2);q)=[3]_{q}\cdot\overline{\mathfrak{R}}((3,1,\alpha(\hat{w});q).$

In particular,

$\mathrm{tr}(w_{q}(A,B))\in[3]_{q}\mathbb{N}[q].$

Consequentally, the $q$-deformed Markov number corresponding to
$w(a,b)=a\hat{w}(a,b)b$ is given by the rank polynomial of the circular fence
poset $\overline{F}(3,1,\alpha(\hat{w}))$. It is symmetric and unimodal.

###### Proof.

As the word $\hat{w}$ is palindromic and consists of pairs of parts, the
corresponding permutation $\alpha(\hat{w})$ is palindromic with an even number
of parts. Applying (Id 1) with $k=2$ and $r=1$ gives us:

$\displaystyle\overline{\mathfrak{R}}((1,2,2,\alpha(\hat{w}),1);q)=[3]_{q}\overline{\mathfrak{R}}((3,\alpha(\hat{w}),1);q).$

This is equivalent to Equation 31 as moving matrices cyclically does not
change trace and we can reverse the order by the symmetry of circular rank
polynomials. The symmetry and unimodality are shown in [4] and [6]
respectively. ∎

Figure 6. The circular fence poset $\overline{F}(3,1,1,1,2,2,1,1)$

For example, for the Christoffel word $a^{2}bab=a(aba)b$, we get the circular
fence poset for $(3,1,1,1,2,2,1,1)$ shown in Figure 6. The corresponding
$q$-deformed Markov number is given by:

$\overline{\mathfrak{R}}((3,1,1,1,2,2,1,1);q)=1+4q+9q^{2}+16q^{3}+23q^{4}+29q^{5}+30q^{6}+29q^{7}+23q^{8}+16q^{9}+9q^{10}+4q^{11}+q^{12}.$

See Table 1 for other examples of $q$-deformed numbers with corresponding
Christoffel words and circular fence posets.

$ab$ | $\overline{F}(3,1)$ | $1+q+q^{2}+q^{3}+q^{4}$
---|---|---
$a^{2}b$ | $\overline{F}(3,1,1,1)$ | $1+2q+2q^{2}+3q^{3}+2q^{4}+2q^{5}+q^{6}$
$ab^{2}$ | $\overline{F}(3,1,2,2)$ | $1+2q+4q^{2}+5q^{3}+5q^{4}+5q^{5}+4q^{6}+2q^{7}+q^{8}$
$a^{3}b$ | $\overline{F}(3,1,1,1,1,1)$ | $1+3q+4q^{2}+6q^{3}+6q^{4}+6q^{5}+4q^{6}+3q^{7}+q^{8}$
$abab^{2}$ | $\overline{F}(3,1,2,2,1,1,2,2)$ | $1+4q+11q^{2}+22q^{3}+36q^{4}+50q^{5}+60q^{6}+65q^{7}+60q^{8}+\cdots+q^{14}$
$ab^{2}$ | $\overline{F}(3,1,2,2,2,2)$ | $1+3q+8q^{2}+14q^{3}+20q^{4}+25q^{5}+27q^{6}+25q^{7}+20q^{8}+\cdots+q^{12}$
$a^{4}b$ | $\overline{F}(3,1,1,1,1,1,1,1)$ | $1+4q+7q^{2}+11q^{3}+14q^{4}+15q^{5}+14q^{6}+11q^{7}+7q^{8}+4q^{9}+q^{10}$
$ab^{4}$ | $\overline{F}(3,1,2,2,2,2,2,2)$ | $1+4q+13q^{2}+29q^{3}+53q^{4}+82q^{5}+110q^{6}+131q^{7}+139q^{8}+\cdots+q^{16}$
Table 1. Christoffel Words with corresponding fences and Markov polynomials.

##  8\. Generalized Oriented Posets

In this section, we will consider a generalization of oriented posets where
instead of one specialized vertex for each end, we pick a list of left end
vertices $L=(L_{1},L_{2},\ldots,L_{t})$ and a list of right end vertices
$(R_{1},R_{2},\ldots,R_{s})$ on the poset $P$. We call the structure
$\overrightarrow{P}=(P,L,R)$ we obtain a  generalized (t,s)-oriented poset.

As in Section 3, we consider subsets of the distributive lattice $J(P)$. We
use a subscript on the right to describe which nodes on $R$ should be
included, and a subscript on the left to describe which nodes on $L$ should be
_excluded_. Let $A$ be a subset of $[t]$ and $B$ be a subset of $[s]$. We set:

$\displaystyle\mathfrak{R}(\overrightarrow{P};q):=$ $\displaystyle\sum_{I\in
J(P)}q^{|I|}\qquad\qquad$
${}_{{A}}\mathfrak{R}(\overrightarrow{P};q):=\sum_{\begin{subarray}{c}I\in
J(P)\\\ i\in A\Rightarrow L_{i}\notin I\end{subarray}}q^{|I|}$ $\displaystyle
R_{B}(\overrightarrow{P};q):=$ $\displaystyle\sum_{\begin{subarray}{c}I\in
J(P)\\\ i\in B\Rightarrow\mathfrak{R}_{i}\in I\end{subarray}}q^{|I|}\qquad$
${}_{A}R_{B}(\overrightarrow{P};q):=\sum_{\begin{subarray}{c}I\in J(P)\\\ i\in
A\Rightarrow L_{i}\notin I\\\ i\in B\Rightarrow R_{i}\in
I\end{subarray}}q^{|I|}$
$\displaystyle\mathfrak{R}_{{B}^{*}}(\overrightarrow{P};q):=$
$\displaystyle\sum_{\begin{subarray}{c}I\in J(P)\\\
i\in{{B}}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{\Leftrightarrow}}R_{i}\in
I\end{subarray}}q^{|I|}\qquad$
${}_{A}\mathfrak{R}_{{B}^{*}}(\overrightarrow{P};q):=\sum_{\begin{subarray}{c}I\in
J(P)\\\ i\in A\Rightarrow L_{i}\notin I\\\
i\in{{B}}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{\Leftrightarrow}}\mathfrak{R}_{i}\in
I\end{subarray}}q^{|I|}$

Generalized $(t,s)$-oriented posets have rank matrices of size $2^{t}\times
2^{s}$, where rows are indexed by subsets of $[t]$ and columns are indexed by
subsets of $s$. When $t=s=4$ we get the following matrices:

$\displaystyle{\triangledown}\mathfrak{R}_{q}(\overrightarrow{P})$
$\displaystyle=\begin{bmatrix}R&-R_{\\{1\\}}&-R_{\\{2\\}}&R_{\\{1,2\\}}\\\
{}_{\\{1\\}}R&-_{\\{1\\}}R_{\\{1\\}}&-_{\\{1\\}}R_{\\{2\\}}&{}_{1}R_{\\{1,2\\}}\\\
{}_{\\{2\\}}R&-_{\\{2\\}}R_{\\{1\\}}&-_{\\{2\\}}R_{\\{2\\}}&{}_{\\{2\\}}R_{\\{1,2\\}}\\\
{}_{\\{1,2\\}}R&-_{\\{1,2\\}}R_{\\{1\\}}&-_{\\{1,2\\}}R_{\\{2\\}}&{}_{\\{1,2\\}}R_{\\{1,2\\}}\end{bmatrix},$
$\displaystyle{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P})$
$\displaystyle=\begin{bmatrix}R_{{\\{1,2\\}}^{*}}&R_{{\\{2\\}}^{*}}&R_{{\\{1\\}}^{*}}&R_{{\varnothing}^{*}}\\\
{}_{\\{1\\}}R_{{\\{1,2\\}}^{*}}&{}_{\\{1\\}}R_{{\\{2\\}}^{*}}&{}_{\\{1\\}}R_{{\\{1\\}}^{*}}&{}_{1}R_{{\varnothing}^{*}}\\\
{}_{\\{2\\}}R_{{\\{1,2\\}}^{*}}&{}_{\\{2\\}}R_{{\\{2\\}}^{*}}&{}_{\\{2\\}}R_{{\\{1\\}}^{*}}&{{}_{\\{2\\}}R_{{\varnothing}^{*}}}\\\
{}_{\\{1,2\\}}R_{{\\{1,2\\}}^{*}}&{}_{\\{1,2\\}}R_{{\\{2\\}}^{*}}&{{}_{\\{1,2\\}}R_{{\\{1\\}}^{*}}}&{}_{\\{1,2\\}}R_{{\varnothing}^{*}}.\end{bmatrix}$

More generally, the rank matrix and the dual rank matrix of a generalized
(t,s)-oriented poset $\overrightarrow{P}$ are matrices of size $2^{|L|}\times
2^{|R|}$ defined as follows:

$\displaystyle{\triangledown}\mathfrak{R}_{q}(\overrightarrow{P}):=((-1)^{|B|}\,_{A}\mathfrak{R}_{B})_{A,B},\qquad{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P}):=\,(_{A}\mathfrak{R}_{{B^{c}}^{*}})_{A,B}.$

Here $A$ and $B$ range over subsets of $[t]$ and $[s]$ respectively:
$\varnothing\subseteq A\subsetneq[t]$, $\varnothing\subsetneq B\subseteq[s]$.
Note that when $t=s=1$, we recover the rank matrices we previously defined.

For a generalized $(t,s)$-oriented poset $\overrightarrow{P}=(P,L_{P},R_{P})$
and a generalized $(s,r)$-oriented poset $\overrightarrow{Q}=(Q,L_{Q},R_{Q})$,
we set $\overrightarrow{P}\searrow\overrightarrow{Q}=(S,L_{P},R_{Q})$ where
$S$ is the poset given by connecting $P$ and $Q$ by adding relations
${R_{P}}_{i}\succeq{L_{P}}_{i}$ for all $i\in S$. Also, when $t=s$ we denote
the structure obtained by adding the relations ${R}_{i}\succeq{L}_{i}$ for all
$i\in S$ by $\circlearrowright\\!(\overrightarrow{P})$.

######  Proposition 8.1.

The rank matrices for generalized oriented posets satisfy the following:

$\displaystyle{\triangledown}\mathfrak{R}_{q}(\overrightarrow{P}\searrow\overrightarrow{Q})$
$\displaystyle={\triangledown}\mathfrak{R}_{q}(\overrightarrow{P})\cdot{\triangledown}\mathfrak{R}_{q}(\overrightarrow{Q}),\qquad\qquad\mathfrak{R}(\circlearrowright\\!(\overrightarrow{P});q)=\mathrm{tr}({\triangledown}\mathfrak{R}_{q}(\overrightarrow{P})),$
$\displaystyle{\triangledown}\mathfrak{R}_{q}(\overrightarrow{P}\nearrow\overrightarrow{Q})$
$\displaystyle={\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P})\cdot{\triangledown}\mathfrak{R}_{q}(\overrightarrow{Q}),$
$\displaystyle{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P}\nearrow\overrightarrow{Q})$
$\displaystyle={\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P})\cdot{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{Q}),\qquad\qquad\mathfrak{R}(\circlearrowleft\\!(\overrightarrow{P});q)=\mathrm{tr}({\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P})),$
$\displaystyle{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{P}\searrow\overrightarrow{Q})$
$\displaystyle={\triangledown}\mathfrak{R}_{q}(\overrightarrow{P})\cdot{\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}(\overrightarrow{Q}).$

The proofs are similar to the case with $t=1$ and $s=1$. The multiplication of
matrices corresponds to adding the connecting relations, by inclusion
exclusion for ${\triangledown}\mathfrak{R}_{q}$ case, and adding up disjoint
possiblities for the
${\rotatebox[origin={c}]{180.0}{$\triangledown$}}\mathfrak{R}_{q}$ case. The
conditions on $L_{P}$ and $R_{Q}$ remain unchanged, so those sets are carried
over as left and right connecting node sets of the product. The trace works
similarly. One can go from one matrix to the other by right multiplication
with an invertible matrix as in the $t=s=1$ case, so that multiplying regular
and dual rank matrices behaves nicely.

The choice of inclusion-exclusion for the regular rank matrix and taking
disjoint possiblities for the dual addition is not dictated by what we want to
achieve, but a particular choice made in order to match the existing notation.

As the size of the matrix increases exponentially with the size of $t$ and
$s$, the setting of generalized oriented matrices might not be very practical
for studying posets. There are cases where it comes in very handy however. One
case is to describe end-points of the poset where no more addition is planned.
In that case one can take $s$ or $t$ to be $0$, and simplify calculations. If
this is done at both left and right ends, one ends up with only the rank
polynomial as the output. There is also the option of simplifying the matrices
when the selected left and right end-points have relations among them, making
some of the rows redundant. The simplest case is when one adds the condition
that the nodes on $L$ and $R$ are chains. In that case, one can use
$(s+1)\times(t+1)$ matrices to convey the information needed for adding up
posets.

##  9\. Comments and Further Directions

  * •

Rank Polynomials of Self-Dual Partitions: The Ferrers diagram of a partition
$\lambda$ has a natural poset structure given by replacing the boxes with
nodes of the poset, as seen in Figure 7 below. The ideals of the poset
correspond to nodes below $\lambda$ in the Hasse diagram of the Young Lattice,
so the rank polynomial can be seen as the generating function of the size
statistic on an interval in the Young lattice. It was conjectured in 1990 (see
[15]) that if $\lambda$ is a self-dual partition (it is equal to its
transpose) then the corresponding rank polynomial is unimodal.

Figure 7. The Young diagram for $(7,6,4,3,3,1)$ (left) with corresponding
poset (middle) and decomposition (right).

One can use the generalized rank matrices defined in Section 8 to decompose
the poset for $\lambda$ into one central rectangular lattice corresponding to
the Durfee square of $\lambda$ multiplied with chains of decreasing sizes from
both sides. If $\lambda$ is self-dual, as seen in Figure 7, the added chains
will be paired up, as will the corresponding matrices. It might be interesting
to examine whether the symmetry of the structure can be used to make an
argument towards unimodality.

  * •

Resolution of Conjecture 1.4: After this paper was shared on arXiv, the
author, together with Ravichandran and Özel, was able to resolve Conjecture
1.4 about the unimodality of circular rank polynomials using characteristic
polynomials of rank matrices. This is one of the results in the upcoming paper
[6]. This also proves Conjecture 5.7, fully characterizing the cases where
$\mathrm{tr}(M_{q}(c_{1},c_{2},\ldots,c_{k}))$ for $c_{i}\geq 2$ is not
unimodal.

  * •

Generating functions and Cluster Algebras: The rank polynomial keeps track of
the number of elements of each ideal of a poset, but the machinery described
here would work equally as well with any other statistic. In particular, for a
poset on nodes $x_{1},x_{2},\cdots,x_{k}$, one can replace the rank polynomial
by the generating polynomial $\sum_{I}\prod_{i\in I}x_{i}$ at each step to get
the generating function of the ideals of the matrix.

This particular case has an application in cluster algebras, namely, it allows
us to calculate the expansion formula of a curve in terms of a chosen basis
using matrices. In the recent work [5], the author and Emine Yıldırım were
able to use oriented posets to obtain a matrix formulation for expansion
formulas for curves in cluster algebras. This method is easily programmable
and is able to give the results fairly quickly, especially in larger cases.

Connection to recent work:  After an initial version of this work was shared
on the arXiv, there were some developments that might lead to future
connections. The recent paper on higher dimer covers has a larger matrix
expression related to cluster algebras [16]. We believe that an expansion of
the weight matrices from [5] to the generalized case that takes into account
the particular nature of the underlying posets might provide a $q$-deformation
for said matrix, as well as resolve some questions posed by the authors.
Differential operators corresponding to $T_{q}$ and $S_{q}$ are used in [17]
to give a $q$-deformation of the Witt algebra. Leclere and Morier-Genoud also
developed a new combinatorial model to study the numerators and denominators
of $q$-deformed rationals using triangulated polygons and annuli. The
connections between this model and the related cluster expansions might be
interesting to explore.

## Acknowledgements

The author was supported by Tübitak BİDEB 2218 grant 121C285. The author would
like to thank Valentin Ovsienko for pointing out the similarity between rank
polynomials of circular fence posets and matrices from 1.5. The author also
would like to thank Mohan Ravichandran, Emine Yıldırım and Can Ozan Oğuz for
helpful discussions on finalizing the notation for oriented posets and the
anonymous referees for their thoughtful suggestions and pointing out related
literature.

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