\chi_{1}(x)+\cdots+\chi_{n}(x)-\tfrac{n}{2}
\text{ is even }
\end{equation*}

Hence, the vertices are all the point $x\in\Aff$ such that $\chi_{j}(x)\in\tfrac{1}{2}\Z$ for all $j$ and the number of non-integer coordinates is neither $1$ nor $n-1$.	In particular, we have

\begin{equation}\label{eq:verticesInZn}
\mcal{V}\subsetneq
o + \tfrac{1}{2}\Z^n\subsetneq
o + \tfrac{1}{2}\mcal{P}^{\vee}.
\end{equation}

[notation]omega i prime@$\omega_{i}'$
[notation]X 00@$\mcal{X}^{00}$
[notation]X 01@$\mcal{X}^{10}$
[notation]X 10@$\mcal{X}^{01}$
[notation]X 11@$\mcal{X}^{11}$
[notation]X 0@$\mcal{X}^{(0)}$
[notation]X 1@$\mcal{X}^{(1)}$
Let $\omega_{i}'=h_{i}^{-1}\omega_{i}$. Consider the following sets:

\begin{equation}\label{eq:SetsX:Dn}
\begin{aligned}
\mcal{X}^{00}	&:=
o+\Z\omega_{1}'+\cdots+\Z\omega_{n}', &
\mcal{X}^{01}	&:=
\mcal{X}^{00}-\tfrac{1}{2}(\omega_{n-1}+\omega_{n}),\\
\mcal{X}^{10}	&:=
\mcal{X}^{00}-\tfrac{1}{2}\omega_{1}, &
\mcal{X}^{11}	&:=
\mcal{X}^{00}-\tfrac{1}{2}(\omega_{1}+\omega_{n-1}+\omega_{n}),\\
\mcal{X}^{(0)}	&:=
\mcal{X}^{00}\cup\mcal{X}^{10}, &
\mcal{X}^{(1)}	&:=
\mcal{X}^{01}\cup\mcal{X}^{11}.
\end{aligned}
\end{equation}

We have $\mcal{X}^{(0)}\cup\mcal{X}^{(1)} = o+\tfrac{1}{2}\Z^n$.

It is clear that $\mcal{X}^{(0)}\cup\mcal{X}^{(1)} \subset o + \tfrac{1}{2}\Z^n$.
Conversely, we have

\begin{align*}
\tfrac{1}{2}\vect{e}_{1} &=
\tfrac{1}{2}\omega_{1},\\
\tfrac{1}{2}\vect{e}_{2} &=
\omega_{2}'-\tfrac{1}{2}\omega_{1},\\
\tfrac{1}{2}\vect{e}_{i} &=
\omega_{i}'-\omega_{i-1}',\\
\tfrac{1}{2}\vect{e}_{n-1} &=
\tfrac{1}{2}(\omega_{n-1}+\omega_{n})-\omega_{n-2}',\\
\tfrac{1}{2}\vect{e}_{n} &=
\omega_{n}-\tfrac{1}{2}(\omega_{n-1}+\omega_{n}).
\end{align*}

Then the statement follows.

To better describe the vertices, we need the notion introduced in <ref>.
Then the complement of $\mcal{V}$ in $\mcal{X}^{(0)}\cup\mcal{X}^{(1)}$ can be described as follows.

A point $x\in \mcal{X}^{(0)}$ is not a vertex if and only if either $J_{x}=\Set*{j_{1},j_{2}}$ and $j_{2}-j_{1}=1$, or $J_{x}=\Set*{1}$.

[notation]Xi 0@$\Xi^{(0)}$
We will use $\Xi^{(0)}$ to denote the set of points $x\in\Aff$ having the property in the lemma.

First, points in $\Xi^{(0)}$ cannot be vertices.
If $J_{x}=\Set*{1}$, then $x$ is not a vertex since either all $\chi_{j}(x)$ are integers except $j=1$ (when $a_{n-1}(x)+a_{n}(x)$ is even), or all $\chi_{j}(x)$ are non-integers except $j=1$ (when $a_{n-1}(x)+a_{n}(x)$ is odd).
Next, suppose $J_{x}=\Set*{j_{1},j_{2}}$ and $j_{2}-j_{1}=1$.
If $j_{2}<n-1$, then $x$ is not a vertex since either all $\chi_{j}(x)$ are integers except $j=j_{2}$ (when $a_{n-1}(x)+a_{n}(x)$ is even), or all $\chi_{j}(x)$ are non-integers except $j=j_{2}$ (when $a_{n-1}(x)+a_{n}(x)$ is odd).
If $j_{2}=n-1$, then $x$ is not a vertex since $\chi_{n}(x)\notin\tfrac{1}{2}\Z$.
If $j_{2}=n$, then $J_{x}=\Set*{n-1,n}$ and we leave this situation in <ref>.

Conversely, suppose $x\in \mcal{X}^{(0)}$ and $x\notin\Xi^{(0)}$.
Then there are four cases:

* $\abs*{J_{x}}\ge 3$. Then among the coordinates of $x$, at least two of them are integers and two of them are non-integers.

* $J_{x}=\Set*{j_{1},j_{2}}$ and $1<j_{2}-j_{1}<n-2$.
Then $\chi_{j_{1}+1}(x),\cdots,\chi_{j_{2}}(x)$ are either all the integer coordinates of $x$, or all the non-integer coordinates of $x$.

* $J_{x}=\Set*{j_{1}}$ and $1<j_{1}<n-1$.
Then $\chi_{1}(x),\cdots,\chi_{j_{1}}(x)$ are either all the integer coordinates of $x$, or all the non-integer coordinates of $x$.

* $x$ has no jumps. Then it is a special vertex.

In any of above cases, $x$ is a vertex by our characterization.

A point $x\in \mcal{X}^{(1)}$ is not a vertex if and only if $J_{x}\subset\Set*{n-2,n-1,n}$.

Note that $x\in \mcal{X}^{(1)}$ implies that $\Set*{n-1,n}\subset J_{x}$.
[notation]Xi 1@$\Xi^{(1)}$
We will use $\Xi^{(1)}$ to denote the set of points $x\in\Aff$ having the property that $\Set*{n-1,n}\subset J_{x}\subset\Set*{n-2,n-1,n}$.

First, points in $\Xi^{(1)}$ cannot be vertices.
Indeed, if $x\in \Xi^{(1)}$, then $a_{j}(x)\in\Z$ for all $j<n-2$.
Hence, $\chi_{1}(x),\cdots,\chi_{n-2}(x)$ are either all integers or all non-integers.
Hence, by <ref>, for $x$ to be a vertex, we must have that $2\chi_{n-1}(x)$ and $2\chi_{n}(x)$ are integers in the same parity.
But $\Set*{n-1,n}\subset J_{x}$ implies that they are not.

Conversely, suppose $x\in \mcal{X}^{(1)}$ and $x\notin\Xi^{(1)}$.
Then $\Set*{n-1,n}\subset J_{x}$ implies that exactly one of $\chi_{n-1}(x)$ and $\chi_{n}(x)$ is an integer.
If $j<n-2$ is an index in $J_{x}$, then exactly one of $\chi_{j}(x)$ and $\chi_{j+1}(x)$ is an integer.
Hence, among the coordinates of $x$, at least two of them are integers and two of them are non-integers.

We illustrate the structure of $\mcal{V}$ by the following diagrams:

\begin{equation*}
\begin{tikzcd}[column sep=tiny]
&& {	\mcal{X}^{(0)}\cup\mcal{X}^{(1)}	}
\ar[dl,no head]\ar[dr,no head]
&	{	\Xi^{(0)}\cup\Xi^{(1)}	}
\ar[d,no head] \\
&	{	\mcal{V}	}
\ar[dl,no head]\ar[dr,no head]
&&	{	(\mcal{X}^{(0)}\cap\Xi^{(0)})\cup(\mcal{X}^{(1)}\cap\Xi^{(1)})	} \\
{	\mcal{V}_{\dagger}	}
&&	{	\bigcup\limits_{i=2,\cdots,n-2}\mcal{V}_{i}	}	&
\end{tikzcd}
\end{equation*}

where $\dagger$ denotes “being special”.

Let $x\in\mscr{D}[\vC]$ be a vertex.
We divide into two cases: $x\in\mcal{X}^{(0)}$ or $x\in\mcal{X}^{(1)}$.

First, let us assume $x\in\mcal{X}^{(0)}$ and suppose $x$ has jumps $j_{1},\cdots,j_{s}$, ordered from smallest to largest.
To construct a path between $x$ and a special vertex $x_{0}$ in $\mscr{D}[\vC]$ verifying <ref>, we need the following lemmas.

Suppose either $j_{s}-j_{s-1}>1$ or $s>3$. Then $y=x-\tfrac{1}{2}\omega_{j_{1}}$ is a vertex in $\mscr{D}[\vC]\cap\mcal{X}^{(0)}$ adjacent to $x$.

It is clear that $y\in\mcal{X}^{(0)}$.
Then the proof is similar to <ref> except that the root $a_{[j_{s}]}$ is defined as follows:

\begin{equation*}
a_{[j_{s}]} =
\begin{dcases*}
a_{j_{s}-1}+2a_{j_{s}}+\cdots+2a_{n-2}+a_{n-1}+a_{n} & if $j_{s}-j_{s-1}>1$,\\
a_{j_{s-2}}+\cdots+a_{j_{s}-1}+2a_{j_{s}}+\cdots+2a_{n-2}+a_{n-1}+a_{n} & otherwise,\\
\end{dcases*}
\end{equation*}

Hence, we omit the proof here.

Note that $a_{0}(y)\in\Z$ and $\ceil{a_{0}(x)}-a_{0}(x) = 1$.
Hence, by repeating using <ref>, we can reduce our problem to the case where $s=3$, or further $s=1$ if we start with $j_{s}-j_{s-1}>1$.

Now, we may assume either $s\le 3$ with $j_{s}-j_{s-1}=1$ or $s=1$.

Suppose $s=3$ and $j_{1}>1$.
Then $x_{0}=x-\tfrac{1}{2}(\omega_{j_{1}}-\omega_{j_{2}}+\omega_{j_{3}})$ is a special vertex in $\mscr{D}[\vC]$ adjacent to $x$ and verifying <ref>.

The proof is the same as <ref>.

Suppose $s=3$, $j_{3}-j_{2}=1$ and $j_{1}=1$, then $y=x-\tfrac{1}{2}(-\omega_{j_{1}}+\omega_{j_{2}})$ is a vertex in $\mscr{D}[\vC]\cap\mcal{X}^{(0)}$ adjacent to $x$.

It is clear that $y\in\mcal{X}^{(0)}$.
Then the proof is similar to <ref> except that the root $a_{[j_{s}]}$ is defined as follows:

\begin{equation*}
a_{[j_{s}]} =
\end{equation*}

Hence, we omit the proof here.

Note that the only jump of $y$ is $j_{3}>1$.
Hence, $a_{0}(y)\in\Z$ and $\ceil{a_{0}(x)}-a_{0}(y)=1$.
Therefore, <ref> reduces our problem to the case where $s=1$.

Note that, by <ref>, $s=2$ and $j_{s}-j_{s-1}=1$ contradict to each other.
Therefore, we may assume $s=1$ now.
By <ref> again, we must have $j_{1}>1$.
Let $x_{0}=x-\tfrac{1}{2}\omega_{j_{1}}$.
Then it is in a special vertex in $\mscr{D}[\vC]$ since $a_{1}(x_{0}),\cdots,a_{n}(x_{0})$ are non-negative integers.
Applying <ref> to the simple roots $a_{1},\cdots,a_{n}$ and the vertices $x$ and $x_{0}$, we see that they are adjacent. Moreover, we have $a_{0}(x)-a_{0}(x_{0})=1$ verifying <ref>.
This finishes the proof of <ref> when $x\in\mcal{X}^{(0)}$.

Next, let us assume $x\in\mcal{X}^{(1)}$. Then we must have $\Set*{n-1,n}\subset J_{x}$.
Suppose $x$ has jumps $j_{1},\cdots,j_{s},n-1,n$, ordered from smallest to largest.
To construct a path between $x$ and a special vertex $x_{0}$ in $\mscr{D}[\vC]$ verifying <ref>, we need the following lemmas.

Suppose either $j_{s}-j_{s-1}>1$ or $s>2$. Then $y=x-\tfrac{1}{2}\omega_{j_{1}}$ is a vertex in $\mscr{D}[\vC]\cap\mcal{X}^{(1)}$ adjacent to $x$.

It is clear that $y\in\mcal{X}^{(1)}$.
When $j_{s}-j_{s-1}>1$ or $s>3$, the proof is similar to <ref> except that we have to define $a_{[n-1]}$ and $a_{[n]}$ as follows:

\begin{align*}
a_{[n-1]} &=
a_{[n]} &=
\end{align*}

When $s=3$ and $j_{3}-j_{2}=1$, the proof still works if we define $a_{[j_{s}]}$ as follows:

\begin{equation*}
a_{[j_{3}]} = a_{j_{2}}+\cdots+a_{n}. \qedhere
\end{equation*}

Note that $a_{0}(y)\in\Z$ and $\ceil{a_{0}(x)}-a_{0}(x) = 1$.
Hence, by repeating using <ref>, we can reduce our problem to the case where $s=2$, or further $s=1$ if we start with $j_{s}-j_{s-1}>1$.

Now, we may assume either $s\le 2$ with $j_{s}-j_{s-1}=1$ or $s=1$.

Suppose $s=2$ and $j_{1}>1$.
Then $x_{0}=x-\tfrac{1}{2}(\omega_{j_{1}}-\omega_{j_{2}}+\omega_{n-1}+\omega_{n})$ is a special vertex in $\mscr{D}[\vC]$ adjacent to $x$ and verifying <ref>.

First note that $a_{0}(x)\in\Z$ and

\begin{equation*}
a_{0}(x)-a_{0}(x_{0}) = \tfrac{1}{2}a_{0}(\omega_{j_{1}}-\omega_{j_{2}}+\omega_{n-1}+\omega_{n}) = 1.
\end{equation*}

Then the proof is similar to <ref> except that there is no $j_3$ and that we need to define $a_{[n-1]}$ and $a_{[n]}$ as follows:

\begin{align*}
a_{[n-1]} &=
a_{[n]} &=
\end{align*}

Then the statement follows.

Suppose $s=2$ and $j_{1}=1$, then $x_{0}=x-\tfrac{1}{2}(-\omega_{j_{1}}+\omega_{j_{2}}-\omega_{n-1}+\omega_{n})$ is a special vertex in $\mscr{D}[\vC]$ adjacent to $x$ and verifying <ref>.

First note that $a_{0}(x)\in\tfrac{1}{2}+\Z$ and

\begin{equation*}
a_{0}(x)-a_{0}(x_{0}) = \tfrac{1}{2}a_{0}(-\omega_{j_{1}}+\omega_{j_{2}}-\omega_{n-1}+\omega_{n}) = \tfrac{1}{2}.
\end{equation*}

Then the proof is similar to <ref> except that $a_{[n]}$ is defined as follows:

\begin{equation*}
a_{[n]} =
\end{equation*}

Then the statement follows.

Suppose $s=1$.
Then $x_{0}=x-\tfrac{1}{2}(\omega_{j_{1}}-\omega_{n-1}+\omega_{n})$ is a special vertex in $\mscr{D}[\vC]$ adjacent to $x$ and verifying <ref>.

First note that

\begin{equation*}
a_{0}(x)-a_{0}(x_{0}) = \tfrac{1}{2}a_{0}(\omega_{j_{1}}-\omega_{n-1}+\omega_{n}) = \tfrac{1}{2}a_{0}(\omega_{j_{1}}).
\end{equation*}

Hence, $\ceil{a_{0}(x)}-a_{0}(x_{0})=1$.
Then the proof is similar to <ref> or <ref> except that there is no $j_2$ and that the root $a_{[n]}$ is defined as follows:

\begin{equation*}
a_{[n]} =
\end{equation*}

Here we need $j_1<n-2$, which is guaranteed by <ref>.

This finishes the proof of <ref>.

Let $I$ be a type and follow <ref>.
Using the notations introduced in <ref>, we have $\mcal{V}[I,r]=\partial(r,\vC,I)$.
We also have introduced the sets $\mcal{X}^{00}$, $\mcal{X}^{10}$, $\mcal{X}^{01}$, $\mcal{X}^{11}$, $\mcal{X}^{(0)}$, and $\mcal{X}^{(1)}$ in <ref>.
Inspired <ref> and <ref>, we consider the following sets for each $J\subset\Set*{1,\cdots,n}$:
[notation]Xi J@$\Xi_{J}$
[notation]X J@$\mcal{X}_{J}$

\begin{align*}
\Xi_{J} &:=
\Set*{	x\in\Aff	\given	J_{x}=J	}
\mcal{X}_{J} &:=
\Xi_{J}\cap o+\tfrac{1}{2}\Z^n.
\end{align*}

Note that $o+\tfrac{1}{2}\Z^n = \mcal{X}^{(0)}\cup\mcal{X}^{(1)}$.
Then we have

\begin{align*}
\MoveEqLeft
o+\tfrac{1}{2}\Z^n \setminus \mcal{V} \\
\mcal{X}_{\Set*{1}}\cup\mcal{X}_{\Set*{1,2}}\cup\cdots\cup\mcal{X}_{\Set*{n-3,n-2}}\cup\mcal{X}_{\Set*{n-1,n}}\cup\mcal{X}_{\Set*{n-2,n-1,n}}.
\end{align*}

Note that, for any $J$, we have

\begin{equation}\label{eq:XJfromX0Dn}
\mcal{X}_{J} = \mcal{X}_{\emptyset}-\sum_{j\in J}\tfrac{1}{2}\omega_{j}.
\end{equation}

Moreover, it is clear that $\mcal{X}_{\emptyset}$ is precisely $o+\mcal{P}^{\vee}$, the set of special vertices.

Next, we consider $\mcal{X}[I]$ for above sets.
First, if $\Set*{n-1,n}\cap I\neq\emptyset$ then $\mcal{V}[I]\subset\mcal{X}^{(0)}[I]$.
Otherwise, $\mcal{V}[I]\cap\mcal{X}^{(1)}[I]\neq\emptyset$.
In each case, we have $\mcal{V}[I]\cap(\mcal{X}^{10}[I]\cup\mcal{X}^{11}[I])=\emptyset$ if and only if $\ell_{1}>1$.
For any $J$, it is clear that $\mcal{X}_{J}[I]\neq\emptyset$ if and only if $I\cap J=\emptyset$. If this is the case, we have the following refinement of <ref>:

\begin{equation}\label{eq:XIJfromXI0Dn}
\mcal{X}_{J}[I] = \mcal{X}_{\emptyset}[I]-\sum_{j\in J}\tfrac{1}{2}\omega_{j}.
\end{equation}

Finally, we consider $\mcal{X}[I,r]$. First, it is clear that

\begin{equation}
\label{eq:IndexBsrciDn:pu00}
\mcal{X}^{00}[I,r] =
\Set*{
x = o+c_{1}\omega_{\ell_{1}}'+\cdots+c_{t}\omega_{\ell_{t}}'
\given
\begin{array}{c}
c_{1}+\cdots+c_{t} = r
\end{array}
\end{equation}

Then the followings follow from <ref>:

\begin{align}
\label{eq:IndexBsrciDn:pu01}
\mcal{X}^{01}[I,r] &=
\mcal{X}^{00}[I,r+1]-\tfrac{1}{2}(\omega_{n-1}+\omega_{n}),\\
\label{eq:IndexBsrciDn:pu10}
\mcal{X}^{10}[I,r] &=
\mcal{X}^{00}[I,r]-\tfrac{1}{2}\omega_{1},\\
\label{eq:IndexBsrciDn:pu11}
\mcal{X}^{11}[I,r] &=
\mcal{X}^{00}[I,r+1]-\tfrac{1}{2}(\omega_{1}+\omega_{n-1}+\omega_{n}).
\end{align}

Then we need to work out $\mcal{X}_{J}[I,r]$. For $\mcal{X}_{\emptyset}[I,r]$, an explicit description is given in <ref>.
For general $J$, we have the following refinement of <ref>:

[notation]deltal (J)@$\delta(J)$
Suppose $I\cap J=\emptyset$.
Then we have

\begin{equation*}
\mcal{X}_{J}[I,r] =
\mcal{X}_{\emptyset}[I,r+\abs*{J}-\delta(J)]-\sum_{j\in J}\tfrac{1}{2}\omega_{j},
\end{equation*}

where $\delta(J)$ is defined as follows:

\begin{equation*}
\delta(J) =
\begin{dcases*}
2 & if $\Set*{1,n-1,n}\subset J$,\\
0 & if $\Set*{1,n-1,n}\cap J=\emptyset$,\\
1 & otherwise.
\end{dcases*}
\end{equation*}

By <ref>, it suffices to show that for any $x\in\mcal{X}_{\emptyset}[I,r]$,

\begin{equation*}\label{eq:tag:lem:XIJfromXI0Dn}\tag{$\ast$}
\ceil*{a_{0}\left(x-\sum_{j\in J}\tfrac{1}{2}\omega_{j}\right)} =
\end{equation*}

Note that $a_{0}(x)\in\Z$ and that

\begin{equation*}
a_{0}\left(\sum_{j\in J}\tfrac{1}{2}\omega_{j}\right) =
\begin{dcases*}
\abs*{J}-\tfrac{3}{2} & if $\Set*{1,n-1,n}\subset J$,\\
\abs*{J}-1 & if $1\notin J$ but $\Set*{n-1,n}\subset J$,\\
\abs*{J}-\tfrac{1}{2} & if $1\in J$ but $\Set*{n-1,n}\cap J=\emptyset$,\\
\abs*{J} & if $\Set*{1,n-1,n}\cap J=\emptyset$.
\end{dcases*}
\end{equation*}

Then <ref> follows.

We illustrate above discussions by the following diagrams:

\begin{equation*}
\begin{tikzcd}[column sep=0pt]
& {	\mcal{X}^{00}[I,r]	}
\ar[dl,no head]\ar[dr,no head]
&	{	\bigcup\limits_{J=\Set*{2,3},\cdots,\Set*{n-3,n-2}}\Xi_{J}	}
\ar[d,no head] \\
\mathclap{	\mcal{V}[I,r]	}\ar[d,no head]
&&	{	\bigcup\limits_{J=\Set*{2,3},\cdots,\Set*{n-3,n-2}}\mcal{X}_{J}[I,r]	} \\
\mathclap{	\mcal{V}_{\dagger}[I,r]=\mcal{X}_{\emptyset}[I,r]	}
\end{tikzcd}
\end{equation*}

Vertices of type $I$ in $\Apt[D_{n}]$ ($1\in I$ and $\Set*{n-1,n}\cap I\neq\emptyset$)

\begin{equation*}
\begin{tikzcd}[column sep=0pt]
& {	\mcal{X}^{00}[I,r]\cup\mcal{X}^{01}[I,r]	}
\ar[dl,no head]\ar[dr,no head]
&	{	\bigcup\limits_{
\crampedsubstack{
\Set*{n-1,n},\Set*{n-2,n-1,n}
}}\Xi_{J}	}
\ar[d,no head] \\
\mathclap{	\mcal{V}[I,r]	}\ar[d,no head]
&&	{	\bigcup\limits_{
\crampedsubstack{
\Set*{n-1,n},\Set*{n-2,n-1,n}
}}\mcal{X}_{J}[I,r]	} \\
\mathclap{	\mcal{V}_{\dagger}[I,r]=\mcal{X}_{\emptyset}[I,r]	}
\end{tikzcd}
\end{equation*}

Vertices of type $I$ in $\Apt[D_{n}]$ ($1\in I$ and $\Set*{n-1,n}\cap I=\emptyset$)

\begin{equation*}
\begin{tikzcd}[column sep=0pt]
& {	\mcal{X}^{00}[I,r]\cup\mcal{X}^{10}[I,r]	}
\ar[dl,no head]\ar[dr,no head]
&	{	\bigcup\limits_{J=\Set*{1},\Set*{1,2},\cdots,\Set*{n-3,n-2}}\Xi_{J}	}
\ar[d,no head] \\
\mathclap{	\mcal{V}[I,r]	}\ar[d,no head]
&&	{	\bigcup\limits_{J=\Set*{1},\Set*{1,2},\cdots,\Set*{n-3,n-2}}\mcal{X}_{J}[I,r]	} \\
\mathclap{	\mcal{V}_{\dagger}[I,r]=\mcal{X}_{\emptyset}[I,r]	}
\end{tikzcd}
\end{equation*}

Vertices of type $I$ in $\Apt[B_{n}]$ ($1\notin I$ and $\Set*{n-1,n}\cap I\neq\emptyset$)

\begin{equation*}
\begin{tikzcd}[column sep=0pt]
& {	\mcal{X}^{(0)}[I,r]\cup\mcal{X}^{(1)}[I,r]	}
\ar[dl,no head]\ar[dr,no head]
&	{	\bigcup\limits_{
\crampedsubstack{
\Set*{n-1,n},\Set*{n-2,n-1,n}
}}\Xi_{J}	}
\ar[d,no head] \\
\mathclap{	\mcal{V}[I,r]	}\ar[d,no head]
&&	{	\bigcup\limits_{
\crampedsubstack{
\Set*{n-1,n},\Set*{n-2,n-1,n}
}}\mcal{X}_{J}[I,r]	} \\
\mathclap{	\mcal{V}_{\dagger}[I,r]=\mcal{X}_{\emptyset}[I,r]	}
\end{tikzcd}
\end{equation*}

Vertices of type $I$ in $\Apt[B_{n}]$ ($1\notin I$ and $\Set*{n-1,n}\cap I=\emptyset$)

§ ASYMPTOTIC ANALYSIS

This section aims to provide tools to analyze the asymptotic behavior of the simplicial volume and the simplicial surface area.

We have already seen that $\fun{SV}[r]\asymp\fun{SSA}[r]$ in <ref>, the introduction.
Hence, in order to prove <ref>, it suffices to prove the simplicial surface area part.
Likewise, the simplicial volume part of <ref> can be deduced from the simplicial surface area part, either by <ref> or direct computation.
Therefore, it suffices to consider the asymptotic analysis of the simplicial surface area $\fun{SSA}[\:\cdot\:]$ only.

In the formulas <ref>, there are only finitely many $I\subset\Delta$ and each $\mscr{P}_{\Phi;I}[q]$ is an integral polynomial.
Hence, the asymptotic study of $\fun{SSA}[\:\cdot\:]$ can be reduced to summations of the following form:

\begin{equation}\label{eq:SX}
\fun{S}_{\mcal{X}[I]}[r] :=
\sum_{x\in\mcal{X}[I,r]}\prod_{a\in\Phi^{+}}q^{\ceil{a(x)}},
\end{equation}

where $\mcal{X}$ is a set of points, and the notations $\mcal{X}[I]$ and $\mcal{X}[I,r]$ follow <ref>.

The growth of $\fun{S}_{\mcal{X}[I]}[r]$ varies for different types $I$.
For the purpose of asymptotic analysis, only the dominant ones are relevant.
To better analyze their growth, we introduce the following auxiliary functions:

\begin{equation}\label{eq:SXasymp}
\fun{S}^{\asymp}_{\mcal{X}[I]}[r] :=
\sum_{x\in\mcal{X}[I,r]}q^{2\rho(x)},
\end{equation}

where $2\rho$ is the sum of positive roots.
Note that

\begin{equation*}
2\rho(x)=\sum_{a\in\Phi^{+}}a(x)	\le
\sum_{a\in\Phi^{+}}\ceil{a(x)}	\le
\sum_{a\in\Phi^{+}}(a(x)+1)=
\end{equation*}

Hence, we have $\fun{S}_{\mcal{X}[I]}[r]\asymp\fun{S}^{\asymp}_{\mcal{X}[I]}[r]$.
But the later one is easier to study.

This section is structured as follows.
In <ref>, we will introduce $q$-numbers and $q$-functions and discuss the discrete calculus on them.
We will then only focus on the $q$-functions defined by (super) $q$-exponential polynomials.
To study them algebraically, we will review gradings and filtrations in <ref>.
Then in <ref>, we will introduce (super) $q$-exponential polynomials  and study the asymptotic properties of the $q$-functions defined by them.
Finally, with those notions in hand, we will study the asymptotic growth of multi-summations in <ref>.

§.§ Discrete calculus of q-functions

It is often more convenient to treat $q$ as a formal variable when we apply algebraic operations to $\fun{S}_{\mcal{X}[I]}[r]$ and $\fun{S}^{\asymp}_{\mcal{X}[I]}[r]$.
But to carry out the asymptotic analysis, we need to view $q$ as a real number.
Inspired by this, we have the following definition.

q-number@$q$-number!level of
q-function@$q$-function!level of
[notation]Q (q,-)@$\mbb{Q}[q;-]$
Let $q$ be a formal variable and $h$ a positive integer.
Then a $q$-number (of level $h$) is a rational function of $q^{1/h}$ over $\Q$ having no poles on the half real line $\R_{>1}:=\Set*{	r\in\R	\given r>1	}$.
Let $\mbb{Q}[q;h]$ denote the ring of $q$-numbers of level $h$.
Then a $q$-function (of level $h$) is a function defined for sufficiently large integers and valued in $\mbb{Q}[q;h]$.

Let $h$ be a positive integer larger than $1$.
Then $(q^{1/h}-1)^{-1}$ is a $q$-number of level $h$, while $(q-h)^{-1}$ is not a $q$-number.

[notation]R (>1)@$\R_{>1}$
A rational function of $q^{1/h}$ is in particular an algebraic function of $q$ and hence we can talk about its poles.
On the half real line $\R_{>1}$, the function $q^{1/h}$ has a unique real-valued branch.
This allows us to treat $q$-numbers as real-valued continuous functions on $\R_{>1}$.

q-number@$q$-number!pointwise topology of
q-function@$q$-function!limit of
[notation]f (q)@$f_{q}$
Each $\mbb{Q}[q;h]$ is a principal ideal domain.
When the level $h$ varies, they form an inductive system.
Let $\mbb{Q}[q;-]$ denote the inductive limit.
We will view it as the ring of all $q$-numbers.
On this principal ideal domain, we will consider the pointwise topology inheriting from the algebra $\mscr{C}[\R_{>1}]$ of real-valued continuous functions on $\R_{>1}$.
In particular, if $f$ is a $q$-function, then the limit of $f(z)$ as $z\to\infty$ is defined pointwise:

\begin{equation*}
\lim_{z\to\infty}f(z) =
\left(\lim_{z\to\infty}f(z)(q)\right)_{q>1}.
\end{equation*}

Then we can view each $q$-function $f$ as a family of discrete functions $(f_{q})_{q>1}$ indexed by the half real line $\R_{>1}$, where $f_{q}(z):=f(z)(q)$.

q-function@$q$-function!asymptotic equality of
asymptotically equal
Let $f$ and $g$ be two $q$-functions.
We say that they are asymptotically equal and that $f$ has asymptotic growth $g$, denoted by $f(z)\sim g(z)$, if

\begin{equation*}
\lim_{z\to\infty}\frac{f(z)}{g(z)} = 1.
\end{equation*}

We also need asymptotic dominant relations of $q$-functions.
Like the topology, these notions are defined pointwise.

q-function@$q$-function!eventually positive
q-function@$q$-function!eventually non-negative
A $q$-number $C$ is said to be positive (resp. non-negative) if for all $q>1$, $C(q)$ is a positive (resp. non-negative) real number.
A $q$-function $f$ is said to be eventually positive (resp. eventually non-negative) if  for all $q>1$, $f_{q}$ is an eventually positive (resp. eventually non-negative) function, namely: $f_{q}(z)>0$ (resp. $f_{q}(z)\ge 0$) for sufficiently large $z$.

asymptotic dominant relation
Let $f$ and $g$ be two $q$-functions.
We say that $f$ dominates $g$, denoted by $f(z)\gg g(z)$, if there exists a positive $q$-number $C$ such that $\abs*{f} - C\cdot\abs*{g}$ is an eventually non-negative $q$-function.
We will denote $f(z)\asymp g(z)$ if both $f(z)\gg g(z)$ and $g(z)\gg f(z)$.

Then we can consider the discrete calculus on $q$-numbers.

q-function@$q$-function!difference of
[notation]dif (f)@$\adif{f}$
Let $f$ be a $q$-function.
Its difference $\adif{f}$ is the following $q$-function:

\begin{equation*}
\adif{f}(z) := f(z+1) - f(z).
\end{equation*}

The difference operator $\adif$ is $\mbb{Q}[q;-]$-linear and satisfies the Leibniz rule:

\begin{equation}
\label{eq:LeibnizRule}
\adif[fg] = f\cdot\adif{g} + g\cdot\adif{f} + \adif{f} \cdot\adif{g}.
\end{equation}

q-function@$q$-function!eventually strictly increasing
A $q$-function $f$ is said to be eventually strictly increasing if for all $q>1$, $f_{q}$ is an eventually strictly increasing function.

Clearly, $f$ is eventually strictly increasing if and only if $\adif{f}$ is eventually positive.

A $q$-function $f$ is said to be unbounded if for all $q$-number $C$, the $q$-function $\abs*{f}-C$ is eventually positive.

Let $f$ and $g$ be two eventually strictly increasing unbounded $q$-functions.
Then we have $f(z)\sim g(z)$ if and only if $\adif{f}(z) \sim \adif{g}(z)$.

Apply Stolz-Cesàro theorem (see e.g. <cit.>*theorem 2.7.2) to $f_{q}$ and $g_{q}$ for all $q>1$.
Then the statement follows.

We also need the discrete version of integrals.

q-function@$q$-function!anti-difference of
[notation]sum (f)@$\asum{f}$
[notation]sum (f) anchor (a)@$\asum_{a}{f}$
Let $f$ be a $q$-function.
Then an anti-difference of $f$ is a $q$-function $g$ such that $\adif{g}=f$.
Since $\fun{Ker}[\adif]$ consists of constant $q$-functions, we see that the anti-difference is not unique but unique up to a constant $q$-function.
By an abuse of notation, we will use $\asum{f}$ to denote an anti-difference of $f$.

Let $a$ be an integer in the domain of $f$.
Then the anti-difference of $f$ with anchor $a$, denoted by $\asum_{a}{f}$, is defined as follows:

\begin{equation*}
\asum_{a}{f}(z) := (\asum{f})(z) - (\asum{f})(a).
\end{equation*}

Note that $\asum_{a}{f}$ is well-defined although $\asum{f}$ is not.

Note that, if $a,b$ are two integers, then we have the following summation formula:

\begin{equation}\label{eq:FTC}
\sum_{z=a}^{b-1}f(z) =
(\asum{f})(b) - (\asum{f})(a) =
\end{equation}

We will consider the following notions of $q$-numbers and $q$-functions.

A $q$-number is said to be primary if it is of level one.
Then a $q$-function is said to be primary if its values are primary $q$-numbers.

§.§ Weakly graded algebras

Before moving on, let's review gradings and then the filtrations induced by them.
In the study of (super) $q$-exponential polynomials, it is the filtration induced by a grading, rather than the grading itself, will play an essential role.

Throughout this subsection, $R$ is a commutative ring and $\Gamma$ is an additive monoid.
In the applications later, $\Gamma$ will be $\N$, $\Q$, $\F_2$, or products of them.

graded module
grading!homogeneous component
grading!homogeneous element
[notation]M (g)@$M_{g}$
A $\Gamma$-grading on an $R$-module $M$ is a decomposition into a direct sum

\begin{equation*}
M = \bigoplus_{g\in\Gamma}M_{g},
\end{equation*}

where each $M_{g}$ is an $R$-submodule, called the homogeneous component of grade $g$.
Elements of $M_{g}$ are said to be homogeneous of grade $g$.
A general element $m$ of $M$ is decomposed into homogeneous elements $m_{g}$ ($g\in\Gamma$), each $m_{g}$ is called its homogeneous component of grade $g$.
An $R$-module equipped with a $\Gamma$-grading is called a $\Gamma$-graded module over $R$.

operator on graded module!respect the grading
operator on graded module!shift the grading!homogeneously
Let $M$ be a $\Gamma$-graded module over $R$ and $h\in\Gamma$.
Then an operator $T$ on $M$ is said to respect the grading if $T(M_{g})\subset M_{g}$ for all $g\in\Gamma$ and shift the grading homogeneously by $h$ if $T(M_{g})\subset M_{g+h}$ for all $g\in\Gamma$.

Suppose $\Gamma$ is a group and $h\in\Gamma$.
Let $M$ be a $\Gamma$-graded projective module over $R$ and $T$ a surjective $R$-linear operator $T$ on $M$ shifting the grading homogeneously by $h$.
Then there is a section of $T$ shifts the grading homogeneously by $-h$.

The conditions on $T$ imply that its restriction to each $M_{g}$ is surjective onto $M_{g+h}$.
Since $M_{g+h}$ is projective, $T|_{M_{g}}$ admits a section $S_{g}\colon M_{g+h}\to M_{g}$.
Then the desired section of $T$ is the direct sum of the sections $S_{g}$.

[notation]Grothendieck (Gamma)@$\mcal{G}[\Gamma]$
The Grothendieck group $\mcal{G}[\Gamma]$ of $\Gamma$ is the universal Abelian group under $\Gamma$.
If $M$ is a $\Gamma$-graded module, then we will treat it as a $\mcal{G}[\Gamma]$-graded module by defining $M_{g}=\Set*{0}$ if $g\in\mcal{G}[\Gamma]\setminus\Gamma$.
With this convention, <ref> holds even without assuming that $\Gamma$ is a group.

Let $M$ be a $\Gamma$-graded projective module over $R$ and $T$ a surjective $R$-linear operator $T$ on $M$ respecting the grading.
Then there is a section of $T$ respects the grading.

graded algebra
A $\Gamma$-graded algebra over $R$ is an $R$-algebra $A$ equipped with a $\Gamma$-grading such that

\begin{equation*}
A_{g}A_{h} \subset A_{g+h},
\qquad\text{for all}\qquad
\end{equation*}

Note that $A_{0}$ is a subalgebra and each $A_{g}$ ($g\in\Gamma$) is an $A_{0}$-bimodule.
The subalgebra $A_{0}$ is called the subalgebra of grade $0$.

graded algebra!graded module of
If $A$ is a $\Gamma$-graded algebra over $R$, then a $\Gamma$-graded (left) $A$-module is a $\Gamma$-graded module $M$ over $R$ equipped with a (left) $A$-action such that

\begin{equation*}
A_{g}M_{h} \subset M_{g+h},
\qquad\text{for all}\qquad
\end{equation*}

graded algebra!homogeneous basis of
graded algebra!coefficient attached to $e_{g}$
Let $A$ be an $R$-algebra $A$ equipped with a $\Gamma$-grading.
Suppose that $A_{0}$ is a subalgebra of $A$ and that each homogeneous component $A_{g}$ ($g\in\Gamma$) is a free $A_{0}$-module of rank one.
Then the grading can be written as follows:

\begin{equation*}
A = \bigoplus_{g\in\Gamma}A_{0}e_{g},
\end{equation*}

where each $e_{g}\in A_{g}$ is a generator of the $A_{0}$-module $A_{g}$ and $e_{0}=1\in A_{0}$.
The family $(e_{g})_{g\in\Gamma}$ is called a homogeneous basis of $A$ over $A_{0}$.
Note that a homogeneous basis of $A$ determines the  $\Gamma$-grading on it.
Let $a\in A$.
Then its homogeneous component of grade $g$ is of the form $a_{g}=c_{g}e_{g}$ with $c_{g}\in A_{0}$.
The element $c_{g}$ is called the coefficient of $a_{g}$ and is said to be a coefficient of $a$ and attached to $e_{g}$ in $a$.

The assumption in <ref> does not require $A$ to be a $\Gamma$-graded algebra.

In the rest of this subsection, $\Gamma$ is a totally ordered additive monoid (for instance, $\Gamma$ is a submonoid of $(\R,+)$).

[notation]M (le g)@$M_{\le g}$

[notation]M (approx g)@$M_{\approx g}$
A $\Gamma$-filtration on an $R$-module $M$ is a family of $R$-submodules $(M_{\le g})_{g\in\Gamma}$ of $M$ such that $M_{\le g} \subset M_{\le h}$ whenever $g\le h$ and that $\bigcup_{g\in\Gamma}M_{\le g}=M$.

We will use $M_{\approx g}$ to denote the set of elements in $M_{\le g}$ but not in any $M_{\le h}$ with $h<g$.

grading!filtration induced by
A $\Gamma$-grading on an $R$-module $M$ induces a $\Gamma$-filtration as follows:

M_≤g    := ⊕_h≤gM_h.     (g∈Γ)

Then we say an element $m\in M$ is of grade $g$ if $m\in M_{\approx g}$.
If an element is of grade $g$, then its homogeneous component of grade $g$ will be called its leading term.

Note that a nonzero homogeneous element of grade $g$ is of grade $g$.
For a general nonzero element $m\in M$, its grade is the largest $g$ such that the homogeneous component of grade $g$ of $m$ is nonzero.

If $M$ is $\N$-graded, we will say that $0\in M$ has grade $-1$.
If $M$ is $\Q$-graded, we will say that $0\in M$ has grade $-\infty$.

operator on graded module!respect the filtration strictly
operator on graded module!shift the grading
Let $M$ be a $\Gamma$-graded module over $R$ and $h\in\Gamma$.
Then an operator $T$ on $M$ is said to respect the filtration strictly if $T(M_{\approx g})\subset M_{\approx g}$ for all $g\in\Gamma$ and shift the grading by $h$ if $T(M_{\approx g})\subset M_{\approx g+h}$ for all $g\in\Gamma$.

Suppose $\Gamma$ is a totally ordered group and $h\in\Gamma$.
Let $M$ be a $\Gamma$-graded module over $R$ and $T$ a surjective $R$-linear operator $T$ on $M$ shifting the grading by $h$.
Then any section of $T$ shifts the grading by $-h$.

Since $M_{\approx g}\cap M_{\approx g'}=\emptyset$ whenever $g\neq g'$, the conditions on $T$ imply that the preimage of $M_{\approx g+h}$ under $T$ is precisely $M_{\approx g}$.
Hence, if $S$ is a section of $T$, we have $S(M_{\approx g+h})\subset M_{\approx g}$ as expected.

The total order on $\Gamma$ can be canonically extended to its Grothendieck group $\mcal{G}[\Gamma]$.
Then <ref> holds without assuming that $\Gamma$ is a group.

Let $M$ be a $\Gamma$-graded module over $R$ and $T$ a surjective $R$-linear operator $T$ on $M$ respecting the filtration strictly.
Then any section of $T$ respects the filtration strictly.

weakly graded algebra
weakly graded module
A weakly $\Gamma$-graded algebra over $R$ is an $R$-algebra $A$ equipped with a $\Gamma$-grading such that

\begin{equation*}
A_{\approx g}A_{\approx h} \subset A_{\approx g+h}
\qquad\text{for all}\qquad
\end{equation*}

If $A$ is a weakly $\Gamma$-graded algebra over $R$, then

a weakly $\Gamma$-graded (left) $A$-module is a $\Gamma$-graded module $M$ over $R$ equipped with a (left) $A$-action such that

\begin{equation*}
A_{\approx g}M_{\approx h} \subset M_{\approx g+h}
\qquad\text{for all}\qquad
\end{equation*}

If $A$ is a weakly $\Gamma$-graded algebra over $R$.
Then a free (left) module over $A$ is naturally a weakly $\Gamma$-graded (left) $A$-module.

degree of a polynomial
The ring of polynomials $R[z]$ over $R$ is naturally a $\N$-graded algebra, where the monomials $(z^{n})_{n\in\N}$ forms a homogeneous basis of it.
However, if the characteristic of $R$ is $0$, then the induced $\N$-filtration on $R[z]$, namely the degree filtration, can also be induced from the following alternative grading:

\begin{equation*}
R[z] = \bigoplus_{n\in\N}R\binom{z}{n},
\end{equation*}

[notation]z choose n@$\binom{z}{n}$

\begin{equation*}
\binom{z}{n}	:=	\frac{1}{n!}z(z-1)\cdots(z-n+1).
\end{equation*}

In this paper, this grading will be called the degree.
Note that this convention is different from the usual one.
In particular, this grading does not make $R[z]$ into a $\N$-graded algebra over $R$, only a weakly $\N$-graded algebra.

§.§ q-exponential polynomials

In this and the next section, we will introduce (super) $q$-exponential polynomials and study their interaction with anti-difference operators.

[notation]Qq [z]@$\mbb{Q}[q;-][z]$
A $q$-polynomial is a polynomial with $q$-number coefficients.
Following the usual notation, we will denote the ring of $q$-polynomials by $\mbb{Q}[q;-][z]$.

q-exponential polynomial@$q$-exponential polynomial
[notation]Qq [z]qQz@$\mbb{Q}[q;-][z]q^{\Q z}$

A $q$-exponential polynomial is a finite formal sum

\begin{equation}
\label{eq:ExponentialPolynomial}
f(z) = \sum_{\nu}f_{\nu}(z)q^{\nu z},
\end{equation}

where $\nu\in\Q$ and each $f_{\nu}(z)$ is a $q$-polynomial.
The ring of $q$-exponential polynomials will be denoted by $\mbb{Q}[q;-][z]q^{\Q z}$.

q-polynomial@$q$-polynomial!degree of
q-polynomial@$q$-polynomial!leading coefficient of
Following <ref>, we will consider the following grading on $\mbb{Q}[q;-][z]$ and call it the degree:

\begin{equation*}
\mbb{Q}[q;-][z] = \bigoplus_{n\in\N}\mbb{Q}[q;-]\binom{z}{n}.
\end{equation*}

That is to say, an element of grade $n$ in the sense of <ref> will be said to be of degree $n$.
However, note that this grading only makes $\mbb{Q}[q;-][z]$ a weakly $\N$-graded algebra over $\mbb{Q}[q;-]$.
Let $f$ be a $q$-polynomial.
We will use $\fun{deg}[f]$ to denote its degree.
The leading coefficient $\fun{lead}[f]$ of $f$ is defined to be the coefficient of its leading term, namely the $q$-number attached to $\binom{z}{\fun{deg}[f]}$ in $f$.

q-exponential polynomial@$q$-exponential polynomial!order of
q-exponential polynomial@$q$-exponential polynomial!degree of
q-exponential polynomial@$q$-exponential polynomial!leading coefficient of
[notation]ord (f)@$\fun{ord}[f]$
[notation]deg (f)@$\fun{deg}[f]$
[notation]lead (f)@$\fun{lead}[f]$
The following grading on $\mbb{Q}[q;-][z]q^{\Q z}$ will be called the order:

\begin{equation*}
\mbb{Q}[q;-][z]q^{\Q z} =
\bigoplus_{\nu\in\Q}\mbb{Q}[q;-][z]q^{\nu z}.
\end{equation*}

That is to say, an element of grade $\nu$ in the sense of <ref> will be said to be of order $\nu$.
Note that this grading makes $\mbb{Q}[q;-][z]q^{\Q z}$ a $\Q$-graded algebra over $\mbb{Q}[q;-][z]$.
We will use $\fun{ord}[f]$ to denote the order of a $q$-exponential polynomial $f$.

Each homogeneous component $\mbb{Q}[q;-][z]q^{\nu z}$ is
a free module of rank one over the weakly $\N$-graded algebra $\mbb{Q}[q;-][z]$ and thus naturally a weakly $\N$-graded module.
The degree $\fun{deg}[f]$ and the leading coefficient $\fun{lead}[f]$ of a $q$-exponential polynomial $f$ are defined to be the degree and the leading coefficient of its leading term in the order grading.

Let $f$ be a $q$-exponential polynomial as in <ref>.
Then its order is the largest $\nu\in\Q$ such that $f_{\nu}\neq 0$, its degree and leading coefficient is the degree and the leading coefficient of the $q$-polynomial $f_{\fun{ord}[f]}$.

Every $q$-polynomial will be viewed as a $q$-exponential polynomial which is homogeneous of order $0$.

It is clear that a $q$-exponential polynomial $f$ defines a $q$-function.
We will use the same notion to denote this $q$-function.
Then it has the following asymptotic growth:

\begin{equation}
\label{eq:AsymptoticOfEP}
\end{equation}

We thus introduce the following convention.

Let $\fun{S}$ be a $q$-function and $f$ a $q$-exponential polynomial.
If $\fun{S}$ can be defined by a $q$-exponential polynomial whose leading term is the same as $f(z)$, then we will say that $\fun{S}[z]$ has asymptotic growth $f(z)$ and write

\begin{equation*}
\fun{S}[z] \sim f(z)
\end{equation*}

by an abuse of language.
Note that this implies that $\fun{S}$ is asymptotically equal to the $q$-function defined by $f$.

Now, we turn to the discrete calculus.

q-polynomial@$q$-polynomial!difference of
q-exponential polynomial@$q$-exponential polynomial!difference of
[notation]dif (f)@$\adif{f}$
The difference operator $\adif$ on $q$-polynomials is the $\mbb{Q}[q;-]$-linear operator vanishing on constant $q$-polynomials and satisfying the following:

zn	   =

This operator extends to $q$-exponential polynomials as follows:

[znq^νz]    :=
(q^ν-1)zn + q^νzn-1

It is straightforward to verify that the difference operator $\adif$ satisfies the Leibniz rule <ref>.
For a $q$-exponential polynomial $f$, the $q$-function defined by $\adif{f}$ is precisely the difference of the $q$-function defined by $f$.

The following lemma follows from the definition.

The linear operator $\adif$ respects the order grading on $\mbb{Q}[q;-][z]q^{\Q z}$.
On each homogeneous component $\mbb{Q}[q;-][z]q^{\nu z}$ ($\nu\neq 0$), the operator $\adif$ respects the degree filtration strictly.
On the subalgebra $\mbb{Q}[q;-][z]$, the operator $\adif$ shifts the degree homogeneously by $-1$.

Next, we will introduce the anti-difference operators.

[notation]sum (f)@$\asum{f}$
The linear operators $\adif$ admits a section $\asum$ such that

*  it respects the order;

*  on each homogeneous component of order $\nu\neq 0$, it respects the degree filtration strictly;

*  on the subalgebra of order $0$, it shifts the degree homogeneously by $1$.

Moreover, if $f$ is a $q$-exponential polynomial, then we have the following formula:

\begin{equation}\label{eq:asumEP}
\fun{lead}[\asum{f}] =
\begin{dcases*}
\left(q^{\fun{ord}[f]}-1\right)^{-1}\fun{lead}[f] &
if $\fun{ord}[f]\neq 0$,\\
\fun{lead}[f] &
if $\fun{ord}[f] = 0$.
\end{dcases*}
\end{equation}

First note that the restriction of the linear operator $\adif$ to each homogeneous component $\mbb{Q}[q;-][z]q^{\nu z}$ ($\nu\neq 0$) is bijective, while its restriction to the subalgebra $\mbb{Q}[q;-][z]$ is surjective.
In particular, the linear operator $\adif$ itself is surjective.
Since $\mbb{Q}[q;-][z]q^{\Q z}$ is a free module over $\mbb{Q}[q;-]$,
the statements on orders and degrees follows from <ref> by applying <ref> to the operator $\adif$.
As for the leading coefficients, $\fun{ord}[f] = 0$ case follows from <ref>.
If $\fun{ord}[f] > 0$, by <ref>, we have

\begin{equation*}
\fun{lead}[\adif{f}] =
\left(q^{\fun{ord}[f]}-1\right)\fun{lead}[f].
\end{equation*}

Replacing $f$ by $\asum{f}$, <ref> follows.

q-polynomial@$q$-polynomial!anti-difference of
q-exponential polynomial@$q$-exponential polynomial!anti-difference of
[notation]sum (f) anchor (a)@$\asum_{a}{f}$
The linear operator $\asum$ in <ref> is called the free anti-difference operator.
Let $a$ be an integer.
The anti-difference operator with anchor $a$, denoted by $\asum_{a}$, is the linear operator $\asum-\fun{ev}_{a}\circ\asum$, where $\fun{ev}_{a}$ evaluates a $q$-exponential polynomial $f(z)$ at $z=a$.

For a $q$-exponential polynomial $f$, the $q$-function defined by  $\asum_{a}{f}$ is the anti-difference with anchor $a$ of the $q$-function defined by $f$.

We end this subsection with discussions of primarity.

Let $f$ be a $q$-exponential polynomial as in <ref>.
Then $f$ is said to be primary if its coefficients are primary $q$-numbers and $f_{\nu} = 0$ for all $\nu\notin\Z$.

Clearly, primary $q$-exponential polynomials define primary $q$-functions.

Let $f$ be a $q$-exponential polynomial.
If $\fun{ord}[f] \ge 0$, then the $q$-functions defined by $\asum{f}$ for all $a\in\Z$ are asymptotically equal to the $q$-function defined by $\asum{f}$.
If $f$ is a primary, then so are $\adif{f}$, $\asum{f}$, and $\asum_{a}{f}$ ($a\in\Z$).

We have $\asum_{a}{f}-\asum{f}\in\mbb{Q}[q;-]$.
Note that elements of $\mbb{Q}[q;-]$ have order $0$ and degree $0$, while $\asum{f}$ has nonzero order or degree by <ref>.
Hence, we have $\asum_{a}{f}\sim\asum{f}$.

The last statement is evident.

§.§ Super q-exponential polynomials

parity function
parity q-function@parity $q$-function
A parity function is a function which is defined on integers and factors through the projection $\Z\to\mathbb{F}_{2}$.
A parity function valued in $q$-numbers is called a parity $q$-function.

super q-polynomial@super $q$-polynomial
super q-exponential@super $q$-exponential
super q-exponential polynomial@super $q$-exponential polynomial
A super $q$-polynomial is a polynomial with coefficients in parity $q$-functions.
A super $q$-exponential polynomial is a finite formal sum

\begin{equation}
\label{eq:superExponentialPolynomial}
f(z) = \sum_{\nu}f_{\nu}(z)q^{\nu z},
\end{equation}

where $\nu\in\Q$ and each $f_{\nu}(z)$ is a super $q$-polynomial.

To better understand the structure of the ring of super $q$-exponential polynomials, we recall the following notions.

superalgebra!even part of
superalgebra!odd part of
A superalgebra over a commutative ring $R$ is a $\F_{2}$-graded algebra $A$ over $R$.
The $\F_2$-grading $A = A_{0}\oplus A_{1}$ is called the parity.
In particular, the subalgebra $A_{0}$ is called the even part and the $A_{0}$-module $A_{1}$ is called the odd part.
For any element $a\in A$, its homogeneous component of parity $0$ (resp. $1$) is called its even part (resp. odd part).
If $A$ is a superalgebra, then an $A$-supermodule is a $\F_{2}$-graded $A$-module.

We will use $(-1)^{z}$ to denote the parity function mapping even numbers to $1$ and odd numbers to $-1$.

Then the following lemma is easy to verify.

Let $e(z)$ be a parity function.
Then we have

\begin{equation*}
e(z) =	\frac{1}{2}(e(0)+e(1)) + \frac{1}{2}(e(0)-e(1))(-1)^{z}.
\end{equation*}

By this lemma, we have the following.

[notation]Qq [-1]@$\mbb{Q}[q;-][(-1)^{z}]$
The ring of parity $q$-functions with formal variable $z$ is precisely the $\mbb{Q}[q;-]$-algebra generated by $(-1)^{z}$ and is a superalgebra decomposed into even and odd parts as follows:

\begin{equation*}
\mbb{Q}[q;-][(-1)^{z}] =
\mbb{Q}[q;-] \oplus \mbb{Q}[q;-](-1)^{z}.
\end{equation*}

[notation]Qq [-1,z]@$\mbb{Q}[q;-][(-1)^{z},z]$
Then the ring of super $q$-polynomials can be denoted by $\mbb{Q}[q;-][(-1)^{z},z]$.

The ring of super $q$-polynomials is a superalgebra decomposed into even and odd parts as follows:

\begin{equation*}
\mbb{Q}[q;-][(-1)^{z},z] =
\mbb{Q}[q;-][z] \oplus \mbb{Q}[q;-][z](-1)^{z}.
\end{equation*}

super q-polynomial@super $q$-polynomial!degree of
super q-polynomial@super $q$-polynomial!even degree of
super q-polynomial@super $q$-polynomial!odd degree of
super q-polynomial@super $q$-polynomial!leading coefficient of
super q-polynomial@super $q$-polynomial!even leading coefficient of
super q-polynomial@super $q$-polynomial!odd leading coefficient of
The degree grading on $\mbb{Q}[q;-][(-1)^{z},z]$ is defined similarly to that on $\mbb{Q}[q;-][z]$ in <ref> and makes it a weakly $\N$-graded algebra over $\mbb{Q}[q;-][(-1)^{z}]$.
Since this weakly graded algebra is also a superalgebra, the degree grading induces a grading on its even part and one on its odd parts.
They are called the even degree and the odd degree.
Note that the even degree is precisely the degree grading on $\mbb{Q}[q;-][z]$ defined in <ref>, and the odd degree is precisely the natural grading on a free module of rank one over $\mbb{Q}[q;-][z]$.

[notation]deg (f)@$\fun{deg}[f]$
[notation]deg (f) 0@$\fun{deg}_{0}[f]$
[notation]deg (f) 1@$\fun{deg}_{1}[f]$
[notation]lead (f)@$\fun{lead}[f]$
[notation]lead (f) 0@$\fun{lead}_{0}[f]$
[notation]lead (f) 1@$\fun{lead}_{1}[f]$
Let $f$ be a super $q$-polynomial.
We will use $\fun{deg}[f]$ to denote its degree.
Its even degree $\fun{deg}_{0}[f]$ (resp. odd degree $\fun{deg}_{1}[f]$) is the even degree (resp. odd degree) of its even part (resp. odd part).
The leading coefficient $\fun{lead}[f]$ of $f$ is defined to be the coefficient of its leading term, namely the parity $q$-function attached to $\binom{z}{\fun{deg}[f]}$ in $f$.
Its even leading coefficient $\fun{lead}_{0}[f]$ (resp. odd leading coefficient $\fun{lead}_{1}[f]$) is the leading coefficient of its even part (resp. odd part).

Let $f$ be a super $q$-polynomial as follows:

\begin{equation*}
f(z) = f_{0}(z) + f_{1}(z)(-1)^{z},
\end{equation*}

where $f_{0}$ and $f_{1}$ are $q$-polynomials.
Then the even degree of $f$ is $\fun{deg}[f_{0}]$, the odd degree of $f$ is $f$ is $\fun{deg}[f_{1}]$, and the degree of $f$ is the larger one of them.
If $\fun{deg}[f_{0}]>\fun{deg}[f_{1}]$, then the leading coefficient of $f$ is precisely its even leading coefficient, which is $\fun{lead}[f_{0}]$.
If $\fun{deg}[f_{0}]<\fun{deg}[f_{1}]$, then the leading coefficient of $f$ is precisely its odd leading coefficient multiplied by $(-1)^{z}$, which is $\fun{lead}[f_{1}](-1)^{z}$.
If $\fun{deg}[f_{0}]=\fun{deg}[f_{1}]$, then the leading coefficient of $f$ is the parity $q$-function $\fun{lead}[f_{0}]+\fun{lead}[f_{1}](-1)^{z}$.

[notation]Qq [-1,z]qQz@$\mbb{Q}[q;-][(-1)^{z},z]q^{\Q z}$
Similarly to <ref>, the ring of super $q$-exponential polynomials will be denoted by  $\mbb{Q}[q;-][(-1)^{z},z]q^{\Q z}$.

super q-exponential polynomial@super $q$-exponential polynomial!order of
super q-exponential polynomial@super $q$-exponential polynomial!degree of
super q-exponential polynomial@super $q$-exponential polynomial!even degree of
super q-exponential polynomial@super $q$-exponential polynomial!odd degree of
super q-exponential polynomial@super $q$-exponential polynomial!leading coefficient of
super q-exponential polynomial@super $q$-exponential polynomial!even leading coefficient of
super q-exponential polynomial@super $q$-exponential polynomial!odd leading coefficient of
[notation]ord (f)@$\fun{ord}[f]$
[notation]deg (f)@$\fun{deg}[f]$
[notation]deg (f) 0@$\fun{deg}_{0}[f]$
[notation]deg (f) 1@$\fun{deg}_{1}[f]$
[notation]lead (f)@$\fun{lead}[f]$
[notation]lead (f) 0@$\fun{lead}_{0}[f]$
[notation]lead (f) 1@$\fun{lead}_{1}[f]$
The order grading on $\mbb{Q}[q;-][(-1)^{z},z]q^{\Q z}$ is defined similarly to that on $\mbb{Q}[q;-][z]q^{\Q z}$ in <ref> and makes it a $\Q$-graded algebra over the superalgebra $\mbb{Q}[q;-][(-1)^{z},z]$.
We will use $\fun{ord}[f]$ to denote the order of a super $q$-exponential polynomial $f$.

Each homogeneous component $\mbb{Q}[q;-][(-1)^{z},z]q^{\nu z}$ is a free supermodule of rank one over the weakly $\N$-graded superalgebra $\mbb{Q}[q;-][(-1)^{z},z]$ and thus naturally a weakly $\N$-graded supermodule.
The degree $\fun{deg}[f]$, the even degree $\fun{deg}_{0}[f]$, the odd degree $\fun{deg}_{1}[f]$, the leading coefficient $\fun{lead}[f]$, the even leading coefficient $\fun{lead}_{0}[f]$, and the odd leading coefficient $\fun{lead}_{1}[f]$ of a super $q$-exponential polynomial $f$ are defined to be the degree, the even degree, the odd degree, the leading coefficient, the even leading coefficient, and the odd leading coefficient of its leading term in the order grading.

Let $f$ be a $q$-exponential polynomial as in <ref>.
Then the order of $f$ is the largest $\nu\in\Q$ such that $f_{\nu}\neq 0$ and its leading term is the product of the super $q$-polynomial $f_{\fun{ord}[f]}$ and $q^{\fun{ord}[f] z}$.

Every super $q$-polynomial will be viewed as a super $q$-exponential polynomial which is homogeneous of order $0$.

It is clear that a super $q$-exponential polynomial $f$ defines a $q$-function.
We will use the same notion to denote this $q$-function.
Then we have the following asymptotic equalities:

\begin{equation}\label{eq:AsymptoticOfSEP}
\begin{aligned}
f(z) &\sim
\left(
\fun{lead}_{0}[f] \binom{z}{\fun{deg}_{0}[f]} +
\fun{lead}_{1}[f] \binom{z}{\fun{deg}_{1}[f]} (-1)^{z}
\right) q^{\fun{ord}[f]z} \\
\fun{lead}[f] \binom{z}{\fun{deg}[f]} q^{\fun{ord}[f]z}.
\end{aligned}
\end{equation}

Note that the leading coefficient $\fun{lead}[f]$ is a parity $q$-function rather than a $q$-number.
In particular, the asymptotic behaviors of $f(z)$ along even integers and odd integers are different if $\fun{deg}_{0}[f] = \fun{deg}_{1}[f]$.

Let $\fun{S}$ be a $q$-function and $f$ a super $q$-exponential polynomial.
If $\fun{S}$ can be defined by a super $q$-exponential polynomial whose leading term is the same as $f(z)$, then we will say that $\fun{S}$ has asymptotic growth $f$ and write

\begin{equation*}
\fun{S}[z] \sim f(z)
\end{equation*}

by an abuse of language.
Note that this implies that $\fun{S}$ is asymptotically equal to the $q$-function defined by $f$.

Now, we turn to the discrete calculus.

super q-polynomial@super $q$-polynomial!difference of
super q-exponential polynomial@super $q$-exponential polynomial!difference of
[notation]dif (f)@$\adif{f}$
The difference operator $\adif$ on super $q$-exponential polynomials is the extension of the difference operator defined in <ref> satisfying the Leibniz rule <ref> and acts on parity $q$-functions as in <ref>.

By <ref>, the action of $\adif$ on parity $q$-functions is determined by its action on $(-1)^{z}$.
Note that $\adif{(-1)^{z}}=-2(-1)^{z}$.
Hence, the action of $\adif$ on the superalgebra of super $q$-exponential polynomials respects the parity.

For a super $q$-exponential polynomial $f$, the $q$-function defined by $\adif{f}$ is precisely the difference of the $q$-function defined by $f$.

The following lemma follows from the definition.

The linear operator $\adif$ respects the parity and the order grading on $\mbb{Q}[q;-][(-1)^{z},z]q^{\Q z}$.
On each homogeneous component $\mbb{Q}[q;-][(-1)^{z},z]q^{\nu z}$ ($\nu\neq 0$), the operator $\adif$ respects the degree filtration, the even degree filtration on its even part, and the odd degree filtration on its odd parts strictly.
On the subalgebra $\mbb{Q}[q;-][(-1)^{z},z]$, the operator $\adif$ shifts the even degree homogeneously by $-1$ and respects the odd degree filtration strictly.

Next, we will introduce the anti-difference operators.

[notation]sum (f)@$\asum{f}$
The linear operators $\adif$ admits a section $\asum$ such that

*  it respects the parity and the order;

*  on each homogeneous component of order $\nu\neq 0$, it respects the degree filtration, the even degree filtration on its even part, and odd degree filtration on its odd part strictly;

*  on the subalgebra of order $0$, it shifts the even degree homogeneously by $1$ and respects the odd degree filtration strictly.

Moreover, if $f$ is a $q$-exponential polynomial, then we have  the following formulas:

\begin{equation}\label{eq:asumSEP}
\begin{aligned}
\fun{lead}_{0}[\asum{f}] &=
\begin{dcases*}
\left(q^{\fun{ord}[f]}-1\right)^{-1}\fun{lead}_{0}[f]
& if $\fun{ord}[f]\neq 0$,\\
\fun{lead}_{0}[f]
& if $\fun{ord}[f] = 0$,
\end{dcases*}
\\
\fun{lead}_{1}[\asum{f}] &=
\end{aligned}
\end{equation}

First note that the restriction of the linear operator $\adif$ to each homogeneous component $\mbb{Q}[q;-][(-1)^{z},z]q^{\nu z}$ ($\nu\neq 0$) is bijective, while its restriction to the subalgebra $\mbb{Q}[q;-][(-1)^{z},z]$ is surjective.
In particular, the linear operator $\adif$ itself is surjective. Since $\mbb{Q}[q;-][(-1)^{z},z]q^{\Q z}$ is a free module over $\mbb{Q}[q;-]$, the statements on orders and degrees follows from <ref> by applying <ref> to the operator $\adif$.
The statements on even leading coefficients follows from <ref>.
As for the odd leading coefficients,
first note that for each $\nu\in\Q$, we have

\begin{equation*}
\adif[(-1)^{z}\binom{z}{n}q^{\nu z}] =
- \left(q^{\nu}+1\right)(-1)^{z}\binom{z}{n}q^{\nu z}
- q^{\nu}(-1)^{z}\binom{z}{n-1}q^{\nu z}.
\end{equation*}

Therefore, we have

\begin{equation*}
\fun{lead}_{1}[\adif{f}] =
\end{equation*}

Replacing $f$ by $\asum{f}$, <ref> follows.

super q-polynomial@super $q$-polynomial!anti-difference of
super q-exponential polynomial@super $q$-exponential polynomial!anti-difference of
[notation]sum (f) anchor (a)@$\asum_{a}{f}$
The linear operator $\asum$ in <ref> is called the free anti-difference operator.
Let $a$ be an integer.
The anti-difference operator with anchor $a$, denoted by $\asum_{a}$, is the linear operator $\asum-\fun{ev}_{a}\circ\asum$, where $\fun{ev}_{a}$ evaluates a super $q$-exponential polynomial $f(z)$ at $z=a$.

For a super $q$-exponential polynomial $f$, the $q$-function defined by $\asum_{a}{f}$ is the anti-difference with anchor $a$ of the $q$-function defined by $f$.

We end this subsection with discussions of primarity.

Let $f$ be a super $q$-exponential polynomial as in <ref>.
Then $f$ is said to be primary if its coefficients are primary $q$-numbers and $f_{\nu} = 0$ for all $\nu\notin\Z$.

Clearly, primary super $q$-exponential polynomials define primary $q$-functions.

Let $f$ be a $q$-exponential polynomial.
If $\fun{ord}[f] > 0$ or $\fun{ord}[f] = 0$ with either $\fun{deg}_{0}[f] \ge 0$ or $\fun{deg}_{1}[f] > 0$, then the $q$-functions defined by $\asum{f}$ for all $a\in\Z$ are asymptotically equal to the $q$-function defined by $\asum{f}$.
If $f$ is primary, then so are $\adif{f}$, $\asum{f}$, and $\asum_{a}{f}$ ($a\in\Z$).

We have $\asum_{a}{f}-\asum{f}\in\mbb{Q}[q;-]$.
Elements of $\mbb{Q}[q;-]$ have order $0$ and even degree $0$, while the assumption on $f$ implies that $\asum{f}$ has nonzero order or degree by <ref>.
Hence, we have $\asum_{a}{f}\sim\asum{f}$.

The last statement is evident.

§.§ Asymptotic growth of multi-summations

To analyze the growth of the $q$-functions $\fun{S}_{\mcal{X}[I]}[r]$ and $\fun{S}^{\asymp}_{\mcal{X}[I]}[r]$, we need to write them as $q$-exponential polynomials.
This can be done by considering multi-summations of homogeneous (super) $q$-exponential polynomials.
In this subsection, we give some general results.

The strategy is: we will inductively construct a sequence of (super) $q$-exponential polynomials where the final one defines the desired $q$-function, and then we will compare the orders, the degrees, and the leading coefficients of them.

To better describe the results, let's introduce some conventions.

[notation]X index@$X^{\mfrak{i}}$
[notation]mu dot c@$\bm{\mu}\cdot\vect{c}$
Let $\mfrak{i}$ be an index set.
The set of functions from $\mfrak{i}$ to another set $X$ will be denoted by $X^{\mfrak{i}}$.
Such a function $\vect{c}$ will be identified with a sequence $(c_i)_{i\in\mfrak{i}}$ indexed by $\mfrak{i}$, where $c_i=\vect{c}(i)$.
The constant sequence mapping all $i\in\mfrak{i}$ to $1$ will be denoted by $\vect{1}$.
If $\bm{\mu}$ and $\vect{c}$ are two sequences of real numbers indexed by $\mfrak{i}$, then $\bm{\mu}\cdot\vect{c}$ denotes their dot product, namely $\sum_{i\in\mfrak{i}}\mu_ic_i$.

Let $\fun{S}$ be the $q$-function defined by the following multi-summation

\begin{equation*}
\fun{S}[z] =
\sum_{
\vect{c}\in\Z_{>0}^{\mfrak{i}}\colon
\vect{1}\cdot\vect{c} = z
\end{equation*}

where $\bm{\mu}$ is a sequence of non-negative rational numbers.
Define the following notations:

[notation]mu max@$\mu_{\max}$
[notation]index max@$\mfrak{i}_{\max}$

* $\mu_{\max}$ is the maximum of $\bm{\mu}$;

* $\mfrak{i}_{\max}$ is the set of indices $i\in\mfrak{i}$ such that $\mu_i = \mu_{\max}$.

Then $\fun{S}$ can be defined by a $q$-exponential polynomial so that

\begin{equation*}
\fun{S}[z] \sim
\prod_{	i\notin \mfrak{i}_{\max}	}
\left(	q^{\mu_{\max}-\mu_{i}}-1	\right)^{-1}
\cdot\binom{z}{\abs{\mfrak{i}_{\max}}-1} q^{\mu_{\max} z}.
\end{equation*}

Moreover, if $\bm{\mu}$ takes integral values, then the $q$-exponential polynomial is primary.

Note that the $q$-function $\fun{S}$ is eventually positive since the leading coefficient of the $q$-exponential polynomial defining it is positive.

First note that the condition on the sequence $\vect{c}$ of variables is stable under reindexing.
Hence, we may assume $\mfrak{i}=\Set*{1,\cdots,t}$ and $\mfrak{i}_{\max} = \Set*{1,\cdots,i_0}$ by reindexing the sequence $\bm{\mu}$ if necessary.
We change the variables from $\vect{c}$ to $\vect{b}$ as follows:

b_i    := c_1 + ⋯+ c_i.     (1≤i≤t)

Then we can write $\fun{S}[z]$ as follows:

\begin{equation*}
\fun{S}[z] =
q^{\nu_t z}\sum_{b_{t-1}=t-1}^{z-1}q^{\nu_{t-1} b_{t-1}}\cdots\sum_{b_1=1}^{b_2-1}q^{\nu_1 b_1},
\end{equation*}

where $\nu_i = \mu_i - \mu_{i+1}$ for $1 \le i \le t-1$ and $\nu_t = \mu_t$.

To analyze the growth of $\fun{S}[z]$, we define $f_1,\cdots,f_t$ inductively as follows:

f_1(z)    = q^ν_1 z,

f_i(z)    = q^ν_i z_i-1f_i-1(z).

(1 < i ≤t)

Then each $f_i$ is a $q$-exponential polynomial, and we can analyze them by induction.
In particular, $f_t$ defines the $q$-function $\fun{S}$ by <ref>.
Moreover, if $\mu_1,\cdots,\mu_t$ are integers, then every $f_i$ is primary by <ref>.

For $1< i \le i_0$, repeatedly applying <ref>, we have

\begin{equation*}
\fun{ord}[f_{i}] =
\nu_{i} + \fun{ord}[f_{i-1}] =
\nu_{i} =
\begin{dcases*}
0 & if $i<i_0$,\\
\mu_{i_0} - \mu_{i_0+1} & if $i=i_0$.
\end{dcases*}
\end{equation*}

By <ref>, we have the following recurrence relations:

\begin{align*}
\fun{deg}[f_{i}]& =
\fun{deg}[f_{i-1}] + 1,
\fun{lead}[f_{i}] &=
\fun{lead}[f_{i-1}].
\end{align*}

In particular, we have $\fun{ord}[f_{i_0}] = \mu_{i_0} - \mu_{i_0+1}$, $\fun{deg}[f_{i_0}] = i_0-1$, and $\fun{lead}[f_{i_0}] = 1$.

For $i_0 < i \le t$, repeatedly applying <ref>, we have

\begin{equation*}
\fun{ord}[f_{i}] =
\nu_{i} + \fun{ord}[f_{i-1}] =
\nu_{i} + \mu_{i_0} -\mu_{i} =
\begin{dcases*}
\mu_{i_0} - \mu_{i+1} & if $i<t$,\\
\mu_{i_0} & if $i=t$.
\end{dcases*}
\end{equation*}

In particular, they are positive.
Then by <ref>, we have the following recurrence relations:

\begin{align*}
\fun{deg}[f_{i}] &=
\fun{deg}[f_{i-1}],
\fun{lead}[f_{i}] &=
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}\fun{lead}[f_{i-1}].
\end{align*}

In particular, we have $\fun{ord}[f_{t}]=\mu_{i_0}$, $\fun{deg}[f_{t}]=i_0-1$, and $\fun{lead}[f_{t}]$ equals the product of $\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}$ for $i_0+1\le i\le t$.

Therefore, we have the following:

\[
\fun{S}[z] \sim
\prod_{i=i_0+1}^{t}
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot\binom{z}{i_0-1} q^{\mu_{i_0} z}.
\]

This proves the lemma.

In the rest of this subsection, we will consider multi-summations involving parity functions.
We first extend <ref> to the following definition.

multivariable parity function
A multivariable parity function (indexed by $\mfrak{i}$) is a function defined on $\Z^{\mfrak{i}}$ factoring through the projection $\Z^{\mfrak{i}}\to\F_2^{\mfrak{i}}$.

[notation]z bar@$\overline{z}$
By an abuse of notation, we will use the same notation to denote a multivariable parity function indexed by $\mfrak{i}$ and a function on defined on $\F_2^{\mfrak{i}}$.
In other words, we will treat any sequence in $\F_2$ as a sequence in $\Z$ by viewing $0\in\F_2$ and $1\in\F_2$ as their standard representatives $0\in\Z$ and $1\in\Z$.

Let $\fun{S}$ be the $q$-function defined by the following multi-summation

\begin{equation*}
\fun{S}[z] =
\sum_{
\vect{c}\in\Z_{>0}^{\mfrak{i}}\colon
\vect{1}\cdot\vect{c} = z
\end{equation*}

where $\bm{\mu}$ is a sequence of non-negative rational numbers and $e$ is a multivariable parity function.
Define the following notations:

[notation]mu max@$\mu_{\max}$
[notation]index max@$\mfrak{i}_{\max}$

* $\mu_{\max}$ is the maximum of $\bm{\mu}$;

* $\mfrak{i}_{\max}$ is the set of indices $i\in\mfrak{i}$ such that $\mu_i = \mu_{\max}$.

Then $\fun{S}$ can be defined by a super $q$-exponential polynomial so that

\begin{equation*}
\fun{S}[z] \sim
\left(	C_{\bm{\mu},e,0}+C_{\bm{\mu},e,1}(-1)^{z}	\right)
\cdot\binom{z}{\abs{\mfrak{i}_{\max}}-1} q^{\mu_{\max} z},
\end{equation*}

where the constants $C_{\bm{\mu},e,0}$ and $C_{\bm{\mu},e,1}$ are defined as follows:

\begin{align*}
C_{\bm{\mu},e,0} &:=
\sum_{\vect{s}\in\F_2^{\mfrak{i}}}
C_{\bm{\mu},e,1} &:=
\sum_{\vect{s}\in\F_2^{\mfrak{i}}}
\end{align*}

[notation]mu max minus mu@$\mu_{\max}-\bm{\mu}$
where $\mu_{\max}-\bm{\mu}$ denotes the sequence $(\mu_{\max}-\mu_i)_{i\in\mfrak{i}}$ and $C_{\bm{\mu}}$ the following the constant:

\begin{equation*}
C_{\bm{\mu}} :=
\frac{1}{2^{\abs{\mfrak{i}_{\max}}}}
\prod_{i\notin \mfrak{i}_{\max}}
\left(	q^{2(\mu_{\max}-\mu_{i})}-1	\right)^{-1}.
\end{equation*}

Moreover, if $\bm{\mu}$ and $e$ take integral values, then the super $q$-exponential polynomial is primary.

Note that the even leading coefficient $C_{\bm{\mu},e,0}$ is positive and the odd leading coefficient $C_{\bm{\mu},e,1}$ satisfies $\abs*{C_{\bm{\mu},e,1}}<C_{\bm{\mu},e,0}$.
Hence, the $q$-function $\fun{S}$ is eventually positive.
Note that $C_{\bm{\mu},e,1}$ could be $0$, in which case the asymptotic growth of $\fun{S}[z]$ along even integers and odd integers coincide.

To prove <ref>, we begin with some special cases.

Let $\fun{S}$ be the $q$-function defined by the following multi-summation

\begin{equation*}
\fun{S}[z] =
\sum_{
\vect{c}\in\Z_{>0}^{\mfrak{i}}\colon
\vect{1}\cdot\vect{c} = z
\end{equation*}

where $\vect{s}$ is a sequence of integers.
Define the following notations:

[notation]index 0up@$\mfrak{i}^{0}$
[notation]index 1up@$\mfrak{i}^{1}$

* $\mfrak{i}^{0}$ (resp. $\mfrak{i}^{1}$) is the set of indices $i\in\mfrak{i}$ such that $s_i$ is even (resp. odd).

Then $\fun{S}$ can be defined by a primary super $q$-polynomial so that

\begin{equation*}
\fun{S}[z] \sim
\left(-\tfrac{1}{2}\right)^{\abs*{\mfrak{i}^{1}}}\binom{z}{\abs*{\mfrak{i}^{0}}-1} +
\left(-\tfrac{1}{2}\right)^{\abs*{\mfrak{i}^{0}}}\binom{z}{\abs*{\mfrak{i}^{1}}-1}(-1)^{z}.
\end{equation*}

First, if either $\vect{s}$ contains no even numbers or no odd numbers, then the statement follows from <ref>.
We may assume that the sequence $\vect{s}$ contains at least one even number and one odd number.
Since the condition on the sequence $\vect{c}$ of variables is stable under reindexing, we may assume $\mfrak{i} = \Set*{1,\cdots,t}$ and $\mfrak{i}^{\circ} = \Set*{2,\cdots,i_0+1}$ by reindexing the sequence $\vect{s}$ if necessary.
We change the variables from $\vect{c}$ to $\vect{b}$ as follows:

b_i    := c_1+⋯+c_i.     (1≤i≤t)

Then we can write $\fun{S}[z]$ as follows:

\begin{equation*}
\fun{S}[z] =
(-1)^{r_t z}
\sum_{b_{t-1}=t-1}^{z-1}
(-1)^{r_{t-1} b_{t-1}}\cdots
\sum_{b_1=1}^{b_2-1}
(-1)^{r_1 b_1},
\end{equation*}

where $r_i = s_i - s_{i+1}$ for $1\le i\le t-1$ and $r_t = s_t$.
Then our assumption implies that $r_i$ is even whenever $i\notin\Set*{1,i_0+1,t}$.

To analyze the growth of $\fun{S}[z]$, we can define $f_1,\cdots,f_t$ inductively as follows:

f_1(z)    = (-1)^r_1 z,

f_i(z)    = (-1)^r_i z_i-1f_i-1(z).

(1 < i ≤t)

Note that $f_1$ fails the condition of <ref>.
But we can compute $\asum_{1}f_1$ directly:

\begin{equation*}
\asum_{1}f_1(z) = -\tfrac{1}{2}-\tfrac{1}{2}(-1)^{z}.
\end{equation*}

Then each $f_i$ is a primary super $q$-polynomial by <ref>.
Moreover, $f_t$ defines the $q$-function $\fun{S}$ by <ref>.

For $i\notin\Set*{1,i_0+1,t}$, since $r_i$ is even, by <ref>, we have the following recurrence relations:

\begin{align*}
\fun{deg}_{0}[f_i] &=
\fun{deg}_{0}[f_{i-1}] + 1, &
\fun{deg}_{1}[f_i] &=
\fun{deg}_{1}[f_{i-1}], \\
\fun{lead}_{0}[f_i] &=
\fun{lead}_{0}[f_{i-1}], &
\fun{lead}_{1}[f_i] &=
\end{align*}

On the other hand, when $r_{i}$ is odd, we have

\begin{align*}
\fun{deg}_{0}[f_i] &=
\fun{deg}_{1}[f_{i-1}], &
\fun{deg}_{1}[f_i] &=
\fun{deg}_{0}[f_{i-1}] + 1, \\
\fun{lead}_{0}[f_i] &=
-\tfrac{1}{2}\fun{lead}_{1}[f_{i-1}], &
\fun{lead}_{1}[f_i] &=
\fun{lead}_{0}[f_{i-1}].
\end{align*}

If $i_0 = t-1$, then $r_t=r_{i_0+1}=s_t$ is also even, and we have

\begin{align*}
\fun{deg}_{0}[f_{t}] &=
\fun{deg}_{0}[f_{t-1}] + 1 = \cdots
= \fun{deg}_{0}[f_{2}] + t - 2 = t - 2, \\
\fun{deg}_{1}[f_{t}] &=
\fun{deg}_{1}[f_{t-1}] = \cdots
= \fun{deg}_{1}[f_{2}] = 0, \\
\fun{lead}_{0}[f_{t}] &=
\fun{lead}_{0}[f_{t-1}] = \cdots
= \fun{lead}_{0}[f_{2}] = -\tfrac{1}{2}, \\
\fun{lead}_{1}[f_{t}] &=
-\tfrac{1}{2}\fun{lead}_{1}[f_{t-1}] = \cdots
= \left(-\tfrac{1}{2}\right)^{t-2}\fun{lead}_{1}[f_{2}] = \left(-\tfrac{1}{2}\right)^{t-1}.
\end{align*}

Otherwise, both $r_t$ and $r_{i_0+1}$ are odd, and we have

\begin{align*}
\fun{deg}_{0}[f_{t}] &=
\fun{deg}_{1}[f_{t-1}] = \cdots =
\fun{deg}_{1}[f_{i_0+1}] \\ &=
\fun{deg}_{0}[f_{i_0}] + 1 = \cdots =
\fun{deg}_{0}[f_{2}] + i_0 - 2 + 1 = i_0 - 1, \\
\fun{deg}_{1}[f_{t}] &=
\fun{deg}_{0}[f_{t-1}] + 1 = \cdots =
\fun{deg}_{0}[f_{i_0+1}] + t - i_0 - 2 + 1 \\ &=
\fun{deg}_{1}[f_{i_0}] + t - i_0 - 1 = \cdots =
\fun{deg}_{1}[f_{2}] + t - i_0 - 1 = t - i_0 - 1, \\
\fun{lead}_{0}[f_{t}] &=
-\tfrac{1}{2}\fun{lead}_{1}[f_{t-1}] = \cdots =
-\tfrac{1}{2}\left(-\tfrac{1}{2}\right)^{t - i_0 - 2}\fun{lead}_{1}[f_{i_0+1}] \\ &=
\left(-\tfrac{1}{2}\right)^{t - i_0 - 1}\fun{lead}_{0}[f_{i_0}] = \cdots =
\left(-\tfrac{1}{2}\right)^{t - i_0 - 1}\fun{lead}_{0}[f_{2}] = \left(-\tfrac{1}{2}\right)^{t - i_0}, \\
\fun{lead}_{1}[f_{t}] &=
\fun{lead}_{0}[f_{t-1}] = \cdots =
\fun{lead}_{0}[f_{i_0+1}] \\ &=
-\tfrac{1}{2}\fun{lead}_{1}[f_{i_0}] = \cdots =
-\tfrac{1}{2}\left(-\tfrac{1}{2}\right)^{i_0-2}\fun{lead}_{1}[f_{2}] = \left(-\tfrac{1}{2}\right)^{i_0}.
\end{align*}

Then the lemma follows.

Next, we consider the following situation.

Let $\fun{S}$ be the $q$-function defined by the following multi-summation

\begin{equation*}
\fun{S}[z] =
\sum_{
\vect{c}\in\Z_{>0}^{\mfrak{i}}\colon
\vect{1}\cdot\vect{c} = z
\end{equation*}

where $\vect{s}$ is a sequence of integers and $\bm{\mu}$ is a sequence of non-negative rational numbers.
Define the following notations:

[notation]mu max@$\mu_{\max}$
[notation]index max@$\mfrak{i}_{\max}$
[notation]index max 0up@$\mfrak{i}_{\max}^{0}$
[notation]index max 1up@$\mfrak{i}_{\max}^{1}$
[notation]index 0up@$\mfrak{i}^{0}$
[notation]index 1up@$\mfrak{i}^{1}$

* $\mu_{\max}$ is the maximum of $\bm{\mu}$;

* $\mfrak{i}_{\max}$ is the set of indices $i\in\mfrak{i}$ such that $\mu_i = \mu_{\max}$;

* $\mfrak{i}_{\max}^{0}$ (resp. $\mfrak{i}_{\max}^{1}$) is the set of indices $i\in\mfrak{i}_{\max}$ such that $s_i$ is even (resp. odd);

* $\mfrak{i}^{0}$ (resp. $\mfrak{i}^{1}$) is the set of indices $i\in\mfrak{i}\setminus\mfrak{i}_{\max}$ such that $s_i$ is even (resp. odd).

Then $\fun{S}$ can be defined by a super $q$-exponential polynomial so that

\begin{equation*}
\fun{S}[z] \sim
\left(
f_{\vect{s},\bm{\mu},0}(z) +
\right)q^{\mu_{\max} z},
\end{equation*}

where $f_{\vect{s},\bm{\mu},\square}(z)$ ($\square=0,1$) is the following $q$-polynomial:

\begin{equation*}
f_{\vect{s},\bm{\mu},\square}(z) :=
\prod_{i\notin\mfrak{i}_{\max}}
\left(	(-1)^{\square+s_i}q^{\mu_{\max}-\mu_{i}}-1	\right)^{-1}
\cdot
\left(-\tfrac{1}{2}\right)^{\abs*{\mfrak{i}_{\max}^{1-\square}}}
\binom{z}{\abs*{\mfrak{i}_{\max}^{\square}}-1}.
\end{equation*}

Moreover, if $\bm{\mu}$ takes integral values, then the super $q$-exponential polynomial is primary.

Note that the leading coefficient may be negative.
However, if $\vect{s}$ contains no even numbers (resp. odd numbers), then the even (resp. odd) leading coefficient is zero and the odd (resp. even) leading coefficient is positive.

First note that the condition on the sequence $\vect{c}$ of variables is stable under reindexing.
Hence, we may assume $\mfrak{i} = \Set*{1,\cdots,t}$, $\mfrak{i}_{\max} = \Set*{1,\cdots,i_0}$, and $\mfrak{i}^{0} = \Set*{i_0+1,\cdots,i_1}$ by reindexing the sequences $\bm{\mu}$ and $\vect{s}$ if necessary.
We change the variables from $\vect{c}$ to $\vect{b}$ as follows:

b_i    := c_1+⋯+c_i.     (i_0≤i≤t)

Then we can write $\fun{S}[z]$ as follows:

\begin{equation*}
\fun{S}[z] =
(-1)^{r_t z}q^{\nu_t z}
\sum_{b_{t-1}=t-1}^{z-1}
(-1)^{r_{t-1} b_{t-1}}q^{\nu_{t-1} b_{t-1}}\cdots
\sum_{b_{i_0}=i_0}^{b_{i_0+1}-1}
(-1)^{-s_{i_0+1} b_{i_0}}q^{\nu_{i_0} b_{i_0}}
\fun{S}_{\vect{s}|_{\mfrak{i}_{\max}}}[b_{i_0}],
\end{equation*}

* $r_i = s_i - s_{i+1}$ for $i_0+1 \le i\le t-1$ and $r_t = s_t$;

* $\nu_i = \mu_i - \mu_{i+1}$ for $i_0\le i\le t-1$ and $\nu_t = \mu_t$;

* $\vect{s}|_{\mfrak{i}_{\max}}$ is the subsequence of $\vect{s}$ indexed by $\mfrak{i}_{\max}$ and the $q$-function $\fun{S}_{\vect{s}|_{\mfrak{i}_{\max}}}$ is defined as follows:

\begin{equation*}
\fun{S}_{\vect{s}|_{\mfrak{i}_{\max}}}[z] :=
\sum_{
\vect{c}\in\Z_{>0}^{\mfrak{i}_{\max}}\colon
\vect{1}\cdot\vect{c} = z
\end{equation*}

By <ref>, the $q$-function $\fun{S}_{\vect{s}|_{\mfrak{i}_{\max}}}$ can be defined by a primary super $q$-polynomial $f$, for which we have

\begin{align*}
\fun{deg}_{0}[f] &=
\abs*{\mfrak{i}_{\max}^{0}}-1, &
\fun{deg}_{1}[f] &=
\abs*{\mfrak{i}_{\max}^{1}}-1, \\
\fun{lead}_{0}[f] &=
\left(-\tfrac{1}{2}\right)^{\abs*{\mfrak{i}_{\max}^{1}}}, &
\fun{lead}_{1}[f] &=
\left(-\tfrac{1}{2}\right)^{\abs*{\mfrak{i}_{\max}^{0}}}.
\end{align*}

To analyze the growth of $\fun{S}[z]$, we can define $f_{i_0},\cdots,f_t$ inductively as follows:

f_i_0(z)    =
(-1)^-s_i_0+1 zq^ν_i_0 zf(z),

f_i(z)    =
(-1)^r_i zq^ν_i z_i-1f_i-1(z).

(i_0 < i ≤t)

Then each $f_i$ is a super $q$-exponential polynomial, and we can analyze them by induction.
In particular, $f_t$ defines the $q$-function $\fun{S}$ by <ref>.
Moreover, if $\mu_1,\cdots,\mu_t$ are integers, then every $f_i$ is primary by <ref>.

For each $i_0 < i \le t$, repeatedly applying <ref>, we have

\begin{equation*}
\fun{ord}[f_{i}] =
\nu_{i} + \fun{ord}[f_{i-1}] =
\nu_{i} + \mu_{i_0} -\mu_{i} =
\begin{dcases*}
\mu_{i_0} - \mu_{i+1} & if $i<t$,\\
\mu_{i_0} & if $i=t$.
\end{dcases*}
\end{equation*}

When $i\notin\Set*{i_1,t}$, since $r_i$ is even, by <ref>, we have the following recurrence relations:

\begin{align*}
\fun{deg}_{0}[f_i] &=
\fun{deg}_{0}[f_{i-1}], &
\fun{deg}_{1}[f_i] &=
\fun{deg}_{1}[f_{i-1}], \\
\fun{lead}_{0}[f_i] &=
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}\fun{lead}_{0}[f_{i-1}], &
\fun{lead}_{1}[f_i] &=
\left(-q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}\fun{lead}_{1}[f_{i-1}].
\end{align*}

On the other hand, when $r_{i}$ is odd, we have

\begin{align*}
\fun{deg}_{0}[f_i] &=
\fun{deg}_{1}[f_{i-1}], &
\fun{deg}_{1}[f_i] &=
\fun{deg}_{0}[f_{i-1}], \\
\fun{lead}_{0}[f_i] &=
\left(-q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}\fun{lead}_{1}[f_{i-1}], &
\fun{lead}_{1}[f_i] &=
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}\fun{lead}_{0}[f_{i-1}].
\end{align*}

If $i_1=t$, namely $\mfrak{i}^{1}=\emptyset$, then both $s_{i_0+1}$ and $r_t=s_t$ are even, and we have

\begin{align*}
\fun{deg}_{0}[f_{t}] &=
\fun{deg}_{0}[f_{t-1}] = \cdots =
\fun{deg}_{0}[f_{i_0}] =
\abs*{\mfrak{i}_{\max}^{0}}-1 = t-1, \\
\fun{deg}_{1}[f_{t}] &=
\fun{deg}_{1}[f_{t-1}] = \cdots =
\fun{deg}_{1}[f_{i_0}] =
\abs*{\mfrak{i}_{\max}^{1}}-1 = -1, \\
\fun{lead}_{0}[f_{t}] &=
\left(q^{\mu_{i_0}-\mu_{t}}-1\right)^{-1}
\cdot\fun{lead}_{0}[f_{t-1}] = \cdots \\ &=
\prod_{i=i_0+1}^{t}
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot\fun{lead}_{0}[f_{i_0}] =
\prod_{i=i_0+1}^{t}
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot
\left(-\tfrac{1}{2}\right)^{\abs*{\mfrak{i}_{\max}^{1}}}.
\end{align*}

If $i_1=i_0$, namely $\mfrak{i}^{0}=\emptyset$, then both $s_{i_0+1}$ and $r_t=s_t$ are odd, and we have

\begin{align*}
\fun{deg}_{0}[f_{t}] &=
\fun{deg}_{1}[f_{t-1}] = \cdots =
\fun{deg}_{1}[f_{i_0}] =
\abs*{\mfrak{i}_{\max}^{0}}-1 = -1, \\
\fun{deg}_{1}[f_{t}] &=
\fun{deg}_{0}[f_{t-1}] = \cdots =
\fun{deg}_{0}[f_{i_0}] =
\abs*{\mfrak{i}_{\max}^{1}}-1 = t-1, \\
\fun{lead}_{1}[f_{t}] &=
\left(q^{\mu_{i_0}-\mu_{t}}-1\right)^{-1}
\cdot\fun{lead}_{0}[f_{t-1}] = \cdots \\ &=
\prod_{i=i_0+1}^{t}
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot\fun{lead}_{0}[f_{i_0}] =
\prod_{i=i_0+1}^{t}
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot
\left(-\tfrac{1}{2}\right)^{\abs*{\mfrak{i}_{\max}^{0}}}.
\end{align*}

If $i_0 < i_1 < t$, then $s_{i_0+1}$ is even while $r_{i_1}$ and $r_t=s_t$ are odd. We thus have

\begin{align*}
\fun{deg}_{0}[f_{t}] &=
\fun{deg}_{1}[f_{t-1}] = \cdots =
\fun{deg}_{1}[f_{i_1}] \\ &=
\fun{deg}_{0}[f_{i_1-1}] = \cdots =
\fun{deg}_{0}[f_{i_0}] =
\abs*{\mfrak{i}_{\max}^{0}}-1 \ge 0, \\
\fun{deg}_{1}[f_{t}] &=
\fun{deg}_{0}[f_{t-1}] = \cdots =
\fun{deg}_{0}[f_{i_1}] \\ &=
\fun{deg}_{0}[f_{i_1-1}] = \cdots =
\fun{deg}_{0}[f_{i_0}] =
\abs*{\mfrak{i}_{\max}^{1}}-1 \ge 0,\\
\fun{lead}_{0}[f_{t}] &=
\left(-q^{\mu_{i_0}-\mu_{t}}-1\right)^{-1}
\cdot\fun{lead}_{1}[f_{t-1}] = \cdots =
\prod_{i=i_1+1}^{t}
\left(-q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot\fun{lead}_{1}[f_{i_1}] \\	&=
\prod_{i=i_1+1}^{t}
\left(-q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot
\left(q^{\mu_{i_0}-\mu_{i_1}}-1\right)^{-1}
\cdot\fun{lead}_{0}[f_{i_1-1}] = \cdots \\ &=
\prod_{i=i_1+1}^{t}
\left(-q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot
\prod_{i=i_0+1}^{i_1}
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot\fun{lead}_{0}[f_{i_0}] \\ &=
\prod_{i=i_1+1}^{t}
\left(-q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot
\prod_{i=i_0+1}^{i_1}
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot
\left(-\tfrac{1}{2}\right)^{\abs*{\mfrak{i}_{\max}^{1}}}, \\
\fun{lead}_{1}[f_{t}] &=
\left(q^{\mu_{i_0}-\mu_{t}}-1\right)^{-1}
\cdot\fun{lead}_{0}[f_{t-1}] = \cdots =
\prod_{i=i_1+1}^{t}
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot\fun{lead}_{0}[f_{i_1}] \\ &=
\prod_{i=i_1+1}^{t}
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot
\left(-q^{\mu_{i_0}-\mu_{i_1}}-1\right)^{-1}
\cdot\fun{lead}_{1}[f_{i_1-1}] = \cdots \\ &=
\prod_{i=i_1+1}^{t}
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot
\prod_{i=i_0+1}^{i_1}
\left(-q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot\fun{lead}_{1}[f_{i_0}] \\ &=
\prod_{i=i_1+1}^{t}
\left(q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot
\prod_{i=i_0+1}^{i_1}
\left(-q^{\mu_{i_0}-\mu_{i}}-1\right)^{-1}
\cdot
\left(-\tfrac{1}{2}\right)^{\abs*{\mfrak{i}_{\max}^{0}}}.
\end{align*}

Then the lemma follows.

To deduce <ref> from <ref>, we need the following notions.

multivariable parity function!Fourier transform of
[notation]e hat@$\widehat{e}$
Let $e$ be a multivariable parity function indexed by $\mfrak{i}$.
Then its Fourier transform $\widehat{e}$ is the following:

\begin{equation*}
\widehat{e}(-)	:=
\sum_{\vect{s}\in\F_2^{\mfrak{i}}}
\end{equation*}

[notation]index 1@$\mfrak{i}_{1}$
[notation]index 2@$\mfrak{i}_{2}$
[notation]cequences, union of them@$\vect{c}_{1}\sqcup\vect{c}_{2}$
It is often the case that the index set $\mfrak{i}$ admits a partition $\mfrak{i} = \mfrak{i}_{1} \sqcup \mfrak{i}_{2}$.
For any sequence $\vect{c}$ indexed by $\mfrak{i}$, we will use $\vect{c}_{1}$ and $\vect{c}_{2}$ to denote the subsequences of $\vect{c}$ indexed by $\mfrak{i}_{1}$ and $\mfrak{i}_{2}$ respectively.
Conversely, if $\vect{c}_{1}$ and $\vect{c}_{2}$ are two sequences indexed by $\mfrak{i}_{1}$ and $\mfrak{i}_{2}$ respectively, then we will use $\vect{c}_{1}\sqcup\vect{c}_{2}$ to denote the sequence indexed by $\mfrak{i}$ obtained from them.

We have the following multivariable version of <ref>.

Let $e$ be a multivariable parity function indexed by $\mfrak{i}$.
Then we have

\begin{equation*}
e(\vect{c})	=
\frac{1}{2^{\abs*{\mfrak{i}}}}
\sum_{\vect{s}\in\F_2^{\mfrak{i}}}
\widehat{e}(\vect{s})(-1)^{\vect{s}\cdot\vect{c}}.
\end{equation*}

Moreover, if $\mfrak{i}$ admits a partition $\mfrak{i} = \mfrak{i}_{1} \sqcup \mfrak{i}_{2}$, then we have

\begin{equation*}
\sum_{\vect{s}_{1}\in\F_2^{\mfrak{i}_{1}}}
\frac{1}{2^{\abs*{\mfrak{i}_{2}}}}
\sum_{\vect{s}_{2}\in\F_2^{\mfrak{i}_{2}}}
\widehat{e}(\vect{c}_{1}\sqcup\vect{s}_{2})(-1)^{\vect{s}_{2}\cdot\vect{c}_{2}}.
\end{equation*}

This follows from the general theory of Fourier transforms on finite Abelian groups (see e.g. <cit.>*chap. XVIII, 5 and 6).
To verify the lemma directly, note that for any index set $\mfrak{i}$ and any sequence $\vect{c}$ indexed by $\mfrak{i}$, we have

\begin{equation*}
\sum_{\vect{s}\in\F_2^{\mfrak{i}}}
(-1)^{\vect{s}\cdot\vect{c}} =
\begin{dcases*}
2^{\abs*{\mfrak{i}}} & if $\vect{\overline{c}}=0$,\\
0 & otherwise.
\end{dcases*}
\end{equation*}

Then the statement follows by straightforward computations.

We are now able to prove <ref>.

By <ref>, we can write $\fun{S}$ as follows:

\begin{equation*}
\fun{S}[z] =
\frac{1}{2^{\abs*{\mfrak{i}}}}
\sum_{\vect{s}\in\F_2^{\mfrak{i}}}
\widehat{q^{e(-)}}(\vect{s})\fun{S}_{\vect{s}}[z],
\end{equation*}

where each $\fun{S}_{\vect{s}}$ is defined as follows:

\begin{equation*}
\fun{S}_{\vect{s}}[z] =
\sum_{
\vect{c}\in\Z_{>0}^{\mfrak{i}}\colon
\vect{1}\cdot\vect{c} = z
\end{equation*}

By <ref>, each $\fun{S}_{\vect{s}}$ can be defined by a super $q$-exponential polynomial of order $\mu_{\max}$, even degree $\abs*{\mfrak{i}_{\max}^{0}}-1$, and odd degree $\abs*{\mfrak{i}_{\max}^{1}}-1$ (see there for the notations). Moreover, if $\bm{\mu}$ takes integral values, then these super $q$-exponential polynomials are primary.

Note that the degree of $\fun{S}_{\vect{s}}$ achieves its maximum $\abs*{\mfrak{i}_{\max}}-1$ if and only if $\vect{s}$ contains no odd numbers or no even numbers.
Furthermore, in that case, both the even and odd leading coefficients of $\fun{S}_{\vect{s}}$ are non-negative (indeed, one is zero and another is positive).

Therefore, $\fun{S}$ can be defined by a super $q$-exponential polynomial of order $\mu_{\max}$ and degree $\abs*{\mfrak{i}_{\max}}-1$, and if $\bm{\mu}$ and $e$ take integral values, then this super $q$-exponential polynomial is primary.
Moreover, we have

\begin{equation*}\label{eq:lem:MultiSumSEP:ast}\tag{$\ast$}
\fun{S}[z] \sim
\frac{1}{2^{\abs*{\mfrak{i}}}}
\fun{E}[z]\binom{z}{\abs*{\mfrak{i}_{\max}}-1}q^{\mu_{\max} z},
\end{equation*}

where the parity $q$-function $\fun{E}$ is given as follows:

\begin{align*}
\fun{E}[z] &:=
\sum_{\vect{s}\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\widehat{q^{e(-)}}(\vect{0}\sqcup\vect{s})
\prod_{i\notin\mfrak{i}_{\max}}
\left(
\right)^{-1} \\
&\qquad +
\sum_{\vect{s}\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\widehat{q^{e(-)}}(\vect{1}\sqcup\vect{s})
\prod_{i\notin\mfrak{i}_{\max}}
\left(
\right)^{-1}
\cdot (-1)^{z}.
\end{align*}

To deduce the formula in <ref> from above one, note that

\begin{align*}
\MoveEqLeft
\prod_{i\notin\mfrak{i}_{\max}}
\left(
\right)
\prod_{i\notin\mfrak{i}_{\max}}
\left(
\right)^{-1} \\
\prod_{i\notin\mfrak{i}_{\max}}
\left(
\right)
% &=
% \sum_{\vect{s}'\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
% 	\prod_{i\notin\mfrak{i}_{\max}}
% 		\left((-1)^{s_i}q^{\mu_{\max}-\mu_{i}}\right)^{s'_{i}} \\
\sum_{\vect{s}'\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\end{align*}

and similarly

\begin{equation*}
% \MoveEqLeft
\prod_{i\notin\mfrak{i}_{\max}}
\left(
\right)
\prod_{i\notin\mfrak{i}_{\max}}
\left(
\right)^{-1} =
\sum_{\vect{s}'\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\end{equation*}

Therefore, we have

\begin{align*}
\MoveEqLeft
\prod_{i\notin\mfrak{i}_{\max}}
\left(
\right)
\cdot
\fun{E}[z]\\
\sum_{\vect{s}\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\widehat{q^{e(-)}}(\vect{0}\sqcup\vect{s})
\sum_{\vect{s}'\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
q^{(\mu_{\max}-\bm{\mu})\cdot\vect{s}'} \\
&\qquad +
\sum_{\vect{s}\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\widehat{q^{e(-)}}(\vect{1}\sqcup\vect{s})
\sum_{\vect{s}'\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\cdot (-1)^{z} \\
\sum_{\vect{s}'\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\left(
\sum_{\vect{s}\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\widehat{q^{e(-)}}(\vect{0}\sqcup\vect{s})
\right)
q^{(\mu_{\max}-\bm{\mu})\cdot\vect{s}'} \\
&\qquad +
\sum_{\vect{s}'\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\left(
\sum_{\vect{s}\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\widehat{q^{e(-)}}(\vect{1}\sqcup\vect{s})
\right)
\cdot (-1)^{z+\vect{1}\cdot\vect{s}'}.
\end{align*}

By <ref>, we have

\begin{align*}
\MoveEqLeft
\frac{1}{2^{\abs*{\mfrak{i}\setminus\mfrak{i}_{\max}}}}
\cdot
\prod_{i\notin\mfrak{i}_{\max}}
\left(
\right)
\cdot
\fun{E}[z]\\
\sum_{\vect{s}'\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\left(
\sum_{\vect{s}\in\F_2^{\mfrak{i}_{\max}}}
\right)
q^{(\mu_{\max}-\bm{\mu})\cdot\vect{s}'} \\
&\qquad +
\sum_{\vect{s}'\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\left(
\sum_{\vect{s}\in\F_2^{\mfrak{i}\setminus\mfrak{i}_{\max}}}
\right)
\cdot (-1)^{z+\vect{1}\cdot\vect{s}'} \\
\sum_{\vect{s}\in\F_2^{\mfrak{i}}}
\left(
1 +
\right).
\end{align*}

Apply this to <ref>, then <ref> follows.

§.§ Asymptotic growth of non-balanced multi-summations

This subsection aims to apply the results in <ref> to get asymptotic growth of multi-summations which are non-balanced in the sense that the summation condition of variables is no longer $\vect{1}\cdot\vect{c}=z$.
In our applications, the coefficients in the summation condition can only be either $1$ or $2$, see <ref>.
Hence, we will assume that the index set $\mfrak{i}$ admits a partition $\mfrak{i} = \mfrak{i}_{1} \sqcup \mfrak{i}_{2}$ and then follow <ref>.

Let $\fun{S}$ be the $q$-function defined by the following multi-summation

\begin{equation*}
\fun{S}[z] =
\sum_{
\vect{c}\in\Z_{>0}^{\mfrak{i}}\colon
\vect{1}\cdot\vect{c}_{1} +
2(\vect{1}\cdot\vect{c}_{2}) = z
\end{equation*}

where $\bm{\mu}$ is a sequence of non-negative rational numbers.
For $\square=1,2$,
define the following notations:

[notation]mu square max@$\mu_{\square\max}$
[notation]index square max@$\mfrak{i}_{\square\max}$

* $\mu_{\square\max}$ is the maximum of $\bm{\mu}_{\square}$;

* $\mfrak{i}_{\square\max}$ is the set of indices $i\in\mfrak{i}_{\square}$ such that $\mu_i = \mu_{\square\max}$.

Then $\fun{S}$ can be defined by a super $q$-exponential polynomial.

If $2\mu_{1\max} > \mu_{2\max}$, then we have

\begin{equation*}
\fun{S}[z] \sim
\sum_{\vect{s}\in\F_2^{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}}
\cdot
\binom{z}{\abs{\mfrak{i}_{1\max}}-1} q^{\mu_{1\max} z},
\end{equation*}

where $\mu_{1\max}-\bm{\mu}|_{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}$ denotes the sequence $(\mu_{1\max}-\mu_i)_{i\in\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}$ and the constant $C_{\bm{\mu}}$ is defined as follows:

\begin{equation*}
C_{\bm{\mu}} :=
\prod_{	i\in \mfrak{i}_{1}\setminus\mfrak{i}_{1\max}	}
\left(	q^{2\mu_{1\max}-2\mu_{i}}-1	\right)^{-1}
\prod_{	i\in \mfrak{i}_{2}	}
\left(	q^{2\mu_{1\max}-\mu_{i}}-1	\right)^{-1}.
\end{equation*}

If $2\mu_{1\max} < \mu_{2\max}$, then we have

\begin{equation*}
\fun{S}[z] \sim
\left(	C_{\bm{\mu},0}+C_{\bm{\mu},1}(-1)^{z}	\right)
\cdot
\binom{z}{\abs{\mfrak{i}_{2\max}}-1} q^{\tfrac{1}{2}\mu_{2\max} z},
\end{equation*}

where the constants $C_{\bm{\mu},0}$ and $C_{\bm{\mu},1}$ are defined as follows:

\begin{align*}
C_{\bm{\mu},0} &=
\sum_{\vect{s}\in\F_2^{\mfrak{i}_{1}}}
C_{\bm{\mu},1} &=
\sum_{\vect{s}\in\F_2^{\mfrak{i}_{1}}}
\end{align*}

where $\tfrac{1}{2}\mu_{2\max}-\bm{\mu}_{1}$ denotes the sequence $(\tfrac{1}{2}\mu_{2\max}-\mu_i)_{i\in\mfrak{i}_{1}}$ and the constant $C_{\bm{\mu}}$ is defined as follows:

\begin{equation*}
C_{\bm{\mu}} :=
\frac{1}{2^{\abs{\mfrak{i}_{2\max}}}}
\prod_{	i\in \mfrak{i}_{1}	}
\left(	q^{\mu_{2\max}-2\mu_{i}}-1	\right)^{-1}
\prod_{	i\in \mfrak{i}_{2}\setminus\mfrak{i}_{2\max}	}
\left(	q^{\mu_{2\max}-\mu_{i}}-1	\right)^{-1}.
\end{equation*}

If $2\mu_{1\max} = \mu_{2\max}$, then we have

\begin{equation*}
\fun{S}[z] \sim
\sum_{\vect{s}\in\F_2^{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}}
\cdot
\binom{z}{\abs{\mfrak{i}_{1\max}}+\abs{\mfrak{i}_{2\max}}-1} q^{\mu_{1\max} z},
\end{equation*}

where $\mu_{1\max}-\bm{\mu}|_{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}$ denotes the sequence $(\mu_{1\max}-\mu_i)_{i\in\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}$ and the constant $C_{\bm{\mu}}$ is defined as follows:

\begin{equation*}
C_{\bm{\mu}} :=
\frac{1}{2^{\abs{\mfrak{i}_{2\max}}}}
\prod_{	i\in \mfrak{i}_{1}\setminus\mfrak{i}_{1\max}	}
\left(	q^{2\mu_{1\max}-2\mu_{i}}-1	\right)^{-1}
\prod_{	i\in \mfrak{i}_{2}\setminus\mfrak{i}_{2\max}	}
\left(	q^{2\mu_{1\max}-\mu_{i}}-1	\right)^{-1}.
\end{equation*}

Moreover, if both $\bm{\mu}_{1}$ and $\tfrac{1}{2}\bm{\mu}_{2}$ take integral values, then the super $q$-exponential polynomial is primary.

Note that the even leading coefficient $C_{\bm{\mu},0}$ is positive and the odd leading coefficient $C_{\bm{\mu},1}$ satisfies $\abs*{C_{\bm{\mu},1}}<C_{\bm{\mu},0}$.
Hence, the $q$-function $\fun{S}$ is eventually positive.
Note that $C_{\bm{\mu},1}$ could be $0$, in which case the asymptotic growth of $\fun{S}[z]$ along even integers and odd integers coincide.

In the proof of above lemma and its many applications, a fundamental trick is to extend the domain of a $q$-function to include non-integers.
If this $q$-function is defined by a (super) $q$-exponential polynomial, then it is clear how to do this: simply evaluate this (super) $q$-exponential polynomial.
On the other hand, when the $q$-function is given by a (multi-)summation, it is natural to define its value at non-integer points being zero.
However, keep these two conventions may cause confusions especially when an asymptotic equality connecting a (multi-)summation and a (super) $q$-exponential polynomial is provided.
Hence, we will abandon the second convention and use the following one instead.

[notation]S one half z@$\fun{S}[\tfrac{1}{2}z]$
Suppose $\fun{S}$ is a $q$-function such that it can be defined by a (super) $q$-exponential polynomial $f$.
When we write $\fun{S}[\tfrac{1}{2}n]$, where $n\in\Z$, we actually mean the evaluation of $f(z)$ at $z=\tfrac{1}{2}n$.
Note that this may cause $\fun{S}$ having nonzero value at a half-integer even though $\fun{S}$ may be given by a (multi-)summation.

Then by <ref>, the $q$-function that gives $\fun{S}[\tfrac{1}{2}z]$ when $z$ is even and $0$ when $z$ is odd is the following one:

\begin{equation*}
\tfrac{1}{2}(1+(-1)^{z})\fun{S}[\tfrac{1}{2}z].
\end{equation*}

By introducing new variables $\vect{s}\in\F_2^{\mfrak{i}_{}}$ and replacing $\vect{c}_{1}$ by $2\vect{c}_{1}-\vect{s}$, we can write the $q$-function $\fun{S}$ as follows:

\begin{equation*}
\fun{S}[z] =
\sum_{
\crampedsubstack{
\vect{c}\in\Z_{>0}^{\mfrak{i}},
\vect{s}\in\F_2^{\mfrak{i}_{1}}\\
2(\vect{1}\cdot\vect{c}) = z + \vect{1}\cdot\vect{s}
}q^{	-\bm{\mu}_{1}\cdot\vect{s}
+	(2\bm{\mu}_{1}\sqcup\bm{\mu}_{2})\cdot\vect{c}	}.
\end{equation*}

Consider the following $q$-function:

\begin{equation*}
\fun{S}^{\prime}[z] :=
\sum_{
\vect{c}\in\Z_{>0}^{\mfrak{i}}\colon
\vect{1}\cdot\vect{c} = z
}q^{	\bm{\mu'}\cdot\vect{c}	},
\end{equation*}

where $\bm{\mu'}$ is the sequence $2\bm{\mu}_{1}\sqcup\bm{\mu}_{2}$.
Then we have

\begin{equation*}
\fun{S}[z] =
\sum_{
\vect{s}\in\F_2^{\mfrak{i}_{1}}\colon
z+\vect{1}\cdot\vect{s}\in 2\Z
}q^{	-\bm{\mu}_{1}\cdot\vect{s}	}
\fun{S}^{\prime}[\tfrac{1}{2}(z+\vect{1}\cdot\vect{s})].
\end{equation*}

Note that the summation only takes over those sequence $\vect{s}\in\F_2^{\mfrak{i}_{1}}$ satisfying $z+\vect{1}\cdot\vect{s}\in 2\Z$.
Hence, following <ref>, we have

\begin{equation*}\label{eq:SintoBanlace}\tag{$\ast$}
\fun{S}[z] =
\sum_{
\vect{s}\in\F_2^{\mfrak{i}_{1}}
\tfrac{1}{2}\left(1+(-1)^{z+\vect{1}\cdot\vect{s}}\right)
q^{	-\bm{\mu}_{1}\cdot\vect{s}	}
\fun{S}^{\prime}[\tfrac{1}{2}(z+\vect{1}\cdot\vect{s})].
\end{equation*}

By <ref>, the $q$-function $\fun{S}^{\prime}$ can be defined by a $q$-exponential polynomial $f$ which is primary when $\bm{\mu'}$ only contains integers.
Moreover, we have

\begin{equation*}
\fun{S}^{\prime}[z] \sim
\prod_{	i\notin \mfrak{i}^{\prime}_{\max}	}
\left(	q^{\mu'_{\max}-\mu'_{i}}-1	\right)^{-1}
\cdot\binom{z}{\abs{\mfrak{i}^{\prime}_{\max}}-1} q^{\mu'_{\max} z},
\end{equation*}

where $\mu'_{\max}$ is the maximum of the sequence $\bm{\mu'}$ and $\mfrak{i}^{\prime}_{\max}$ is the set of indices $i\in\mfrak{i}$ such that $\mu'_{i}$ achieves this maximum.
Applying this to <ref> and noticing that

\begin{equation*}
\binom{\tfrac{1}{2}(z+\vect{1}\cdot\vect{s})}{\abs{\mfrak{i}^{\prime}_{\max}}-1}
q^{	\mu'_{\max} \tfrac{1}{2}(z+\vect{1}\cdot\vect{s})	}
\sim
\left(\tfrac{1}{2}\right)^{\abs{\mfrak{i}^{\prime}_{\max}}-1}
q^{	\tfrac{1}{2}\mu'_{\max}\vect{1}\cdot\vect{s} }
\binom{z}{\abs{\mfrak{i}^{\prime}_{\max}}-1}q^{	\tfrac{1}{2}\mu'_{\max} z	},
\end{equation*}

we obtain the following asymptotic equality:

\begin{align*}
\fun{S}[z] &\sim
\tfrac{1}{2^{\abs{\mfrak{i}^{\prime}_{\max}}}}
\prod_{	i\notin \mfrak{i}^{\prime}_{\max}	}
\left(	q^{\mu'_{\max}-\mu'_{i}}-1	\right)^{-1}\\
\cdot
\sum_{	\vect{s}\in\F_2^{\mfrak{i}_{1}}	}
\left(1+(-1)^{z+\vect{1}\cdot\vect{s}}\right)
q^{	(\tfrac{1}{2}\mu'_{\max} - \bm{\mu}_{1})\cdot\vect{s}	}
\cdot\binom{z}{\abs{\mfrak{i}^{\prime}_{\max}}-1}
q^{	\tfrac{1}{2}\mu'_{\max} z	}.
\end{align*}

In the sequence $\bm{\mu'}$, the maximum $\mu'_{\max}$ is $\max\Set*{	2\mu_{1\max},	\mu_{2\max}	}$, and we have

\begin{equation*}
\mfrak{i}^{\prime}_{\max} =
\begin{dcases*}
\mfrak{i}_{1\max}
& if $2\mu_{1\max} > \mu_{2\max}$,\\
\mfrak{i}_{2\max}
& if $2\mu_{1\max} < \mu_{2\max}$,\\
\mfrak{i}_{1\max}\sqcup\mfrak{i}_{2\max}
& if $2\mu_{1\max} = \mu_{2\max}$.
\end{dcases*}
\end{equation*}

In the first and the third case, the sequence $\tfrac{1}{2}\mu'_{\max} - \bm{\mu}_{1}$ contains a zero.
Hence, we have

\begin{equation*}
\sum_{	\vect{s}\in\F_2^{\mfrak{i}_{1}}	}
q^{	(\tfrac{1}{2}\mu'_{\max} - \bm{\mu}_{1})\cdot\vect{s}	}
= 0.
\end{equation*}

Then the asymptotic relations in the lemma follows.
Note that the proof of <ref> also shows that $f(\tfrac{1}{2}z)$ is a primary $q$-exponential polynomial if $\bm{\mu'}$ only contains even integers.
Then the last statement follows.

The following lemma will not be used in this paper.
It can be deduced from <ref> similarly to <ref>

Let $\fun{S}$ be the $q$-function defined by the following multi-summation

\begin{equation*}
\fun{S}[z] =
\sum_{
\vect{c}\in\Z_{>0}^{\mfrak{i}}\colon
\vect{1}\cdot\vect{c}_{1} +
2(\vect{1}\cdot\vect{c}_{2}) = z
\end{equation*}

where $\bm{\mu}$ is a sequence of non-negative rational numbers and $e$ is a multivariable parity function.
For $\square=1,2$,
define the following notations:

[notation]mu square max@$\mu_{\square\max}$
[notation]index square max@$\mfrak{i}_{\square\max}$

* $\mu_{\square\max}$ is the maximum of $\bm{\mu}_{\square}$;

* $\mfrak{i}_{\square\max}$ is the set of indices $i\in\mfrak{i}_{\square}$ such that $\mu_i = \mu_{\square\max}$.

Then $\fun{S}$ can be defined by a super $q$-exponential polynomial.

If $2\mu_{1\max} > \mu_{2\max}$, then we have

\begin{equation*}
\fun{S}[z] \sim
\left(	C_{\bm{\mu},e,0}+C_{\bm{\mu},e,1}(-1)^{z}	\right)
\cdot
\binom{z}{\abs{\mfrak{i}_{1\max}}-1} q^{\mu_{1\max} z},
\end{equation*}

where the constants $C_{\bm{\mu},e,0}$ and $C_{\bm{\mu},e,1}$ are defined as follows:

\begin{align*}
C_{\bm{\mu},e,0} &=
\sum_{
\vect{s}_{1}\in\F_2^{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}
q^{	2(\mu_{1\max}-\bm{\mu}|_{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}})\cdot\vect{s}_{1}	} \\
\sum_{
\vect{s}_{0}\in\F_2^{\mfrak{i}_{1}},
\vect{s}_{2}\in\F_2^{\mfrak{i}_{2}}
q^{	e(\vect{s}_{0}\sqcup\vect{s}_{2})	+	(\mu_{1\max}-\bm{\mu}_{1})\cdot\vect{s}_{0} + (2\mu_{1\max}-\bm{\mu}_{2})\cdot\vect{s}_{2}	},
\\
C_{\bm{\mu},e,1} &=
\sum_{
\vect{s}_{1}\in\F_2^{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}
q^{	2(\mu_{1\max}-\bm{\mu}|_{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}})\cdot\vect{s}_{1}	} \\
\sum_{
\vect{s}_{0}\in\F_2^{\mfrak{i}_{1}},
\vect{s}_{2}\in\F_2^{\mfrak{i}_{2}}
q^{	e(\vect{s}_{0}\sqcup\vect{s}_{2})	+	(\mu_{1\max}-\bm{\mu}_{1})\cdot\vect{s}_{0} + (2\mu_{1\max}-\bm{\mu}_{2})\cdot\vect{s}_{2}	},
\end{align*}

where $\mu_{1\max}-\bm{\mu}|_{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}$ denotes the sequence $(\mu_{1\max}-\mu_i)_{i\in\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}$, $\mu_{1\max}-\bm{\mu}_{1}$ the sequence $(\mu_{1\max}-\mu_i)_{i\in\mfrak{i}_{1}}$, $2\mu_{1\max}-\bm{\mu}_{2}$ the sequence $(2\mu_{1\max}-\mu_i)_{i\in\mfrak{i}_{2}}$, and $C_{\bm{\mu}}$ the following constant:

\begin{equation*}
C_{\bm{\mu}} :=
\frac{1}{2^{\abs{\mfrak{i}_{1\max}}}}
\prod_{	i\in \mfrak{i}_{1}\setminus\mfrak{i}_{1\max}	}
\left(	q^{4\mu_{1\max}-4\mu_{i}}-1	\right)^{-1}
\prod_{	i\in \mfrak{i}_{2}	}
\left(	q^{4\mu_{1\max}-2\mu_{i}}-1	\right)^{-1}.
\end{equation*}

If $2\mu_{1\max} < \mu_{2\max}$, then we have

\begin{equation*}
\fun{S}[z] \sim
\left(	C_{\bm{\mu},e,0}+C_{\bm{\mu},e,1}(-1)^{z}	\right)
\cdot
\binom{z}{\abs{\mfrak{i}_{2\max}}-1} q^{\tfrac{1}{2}\mu_{2\max} z},
\end{equation*}

where the constants $C_{\bm{\mu},e,0}$ and $C_{\bm{\mu},e,1}$ are defined as follows:

\begin{align*}
C_{\bm{\mu},e,0} &=
\sum_{
\vect{s}_{0},\vect{s}_{1}\in\F_2^{\mfrak{i}_{1}},
\vect{s}_{2}\in\F_2^{\mfrak{i}_{2}}
\left(
1 +
\vect{1}\cdot\vect{s}_{0}+\vect{1}\cdot\vect{s}_{1}+\vect{1}\cdot\vect{s}_{2}
\right)\Big(	\\
q^{	e(\vect{s}_{0}\sqcup\vect{s}_{2})	+	(\tfrac{1}{2}\mu_{2\max}-\bm{\mu}_{1})\cdot\vect{s}_{0} + 2(\tfrac{1}{2}\mu_{2\max}-\bm{\mu}_{1})\cdot\vect{s}_{1} + (\mu_{2\max}-\bm{\mu}_{2})\cdot\vect{s}_{2}	}\Big),
\\
C_{\bm{\mu},e,1} &=
\sum_{
\vect{s}_{0},\vect{s}_{1}\in\F_2^{\mfrak{i}_{1}},
\vect{s}_{2}\in\F_2^{\mfrak{i}_{2}}
\left(
\vect{1}\cdot\vect{s}_{0}
} +
\vect{1}\cdot\vect{s}_{1}+\vect{1}\cdot\vect{s}_{2}
\right)\Big(	\\
q^{	e(\vect{s}_{0}\sqcup\vect{s}_{2})	+	(\tfrac{1}{2}\mu_{2\max}-\bm{\mu}_{1})\cdot\vect{s}_{0} + 2(\tfrac{1}{2}\mu_{2\max}-\bm{\mu}_{1})\cdot\vect{s}_{1} + (\mu_{2\max}-\bm{\mu}_{2})\cdot\vect{s}_{2}	}\Big),
\end{align*}

where $\tfrac{1}{2}\mu_{2\max}-\bm{\mu}_{1}$ denotes the sequence $(\tfrac{1}{2}\mu_{2\max}-\mu_i)_{i\in\mfrak{i}_{1}}$, $\mu_{2\max}-\bm{\mu}_{2}$ the sequence $(\mu_{2\max}-\mu_i)_{i\in\mfrak{i}_{2}}$, and $C_{\bm{\mu}}$ the following the constant:

\begin{equation*}
C_{\bm{\mu}} :=
\frac{1}{2^{2\abs{\mfrak{i}_{2\max}}}}
\prod_{	i\in \mfrak{i}_{1}	}
\left(	q^{2\mu_{2\max}-4\mu_{i}}-1	\right)^{-1}
\prod_{	i\in \mfrak{i}_{2}\setminus\mfrak{i}_{2\max}	}
\left(	q^{2\mu_{2\max}-2\mu_{i}}-1	\right)^{-1}.
\end{equation*}

If $2\mu_{1\max} = \mu_{2\max}$, then we have

\begin{equation*}
\fun{S}[z] \sim
\left(	C_{\bm{\mu},e,0}+C_{\bm{\mu},e,1}(-1)^{z}	\right)
\cdot
\binom{z}{\abs{\mfrak{i}_{1\max}}+\abs{\mfrak{i}_{2\max}}-1} q^{\mu_{1\max} z},
\end{equation*}

where the constants $C_{\bm{\mu},e,0}$ and $C_{\bm{\mu},e,1}$ are defined as follows:

\begin{align*}
C_{\bm{\mu},e,0} &=
\sum_{
\vect{s}_{1}\in\F_2^{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}
q^{	2(\mu_{1\max}-\bm{\mu}|_{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}})\cdot\vect{s}_{1}	} \\
\sum_{
\vect{s}_{0}\in\F_2^{\mfrak{i}_{1}},
\vect{s}_{2}\in\F_2^{\mfrak{i}_{2}}
q^{	e(\vect{s}_{0}\sqcup\vect{s}_{2})	+	(\mu_{1\max}-\bm{\mu}_{1})\cdot\vect{s}_{0} + (\mu_{2\max}-\bm{\mu}_{2})\cdot\vect{s}_{2}	},
\\
C_{\bm{\mu},e,1} &=
\sum_{
\vect{s}_{1}\in\F_2^{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}
q^{	2(\mu_{1\max}-\bm{\mu}|_{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}})\cdot\vect{s}_{1}	} \\
\sum_{
\vect{s}_{0}\in\F_2^{\mfrak{i}_{1}},
\vect{s}_{2}\in\F_2^{\mfrak{i}_{2}}
q^{	e(\vect{s}_{0}\sqcup\vect{s}_{2})	+	(\mu_{1\max}-\bm{\mu}_{1})\cdot\vect{s}_{0} + (2\mu_{1\max}-\bm{\mu}_{2})\cdot\vect{s}_{2}	},
\end{align*}

where $\mu_{1\max}-\bm{\mu}|_{\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}$ denotes the sequence $(\mu_{1\max}-\mu_i)_{i\in\mfrak{i}_{1}\setminus\mfrak{i}_{1\max}}$, $\mu_{1\max}-\bm{\mu}_{1}$ the sequence $(\mu_{1\max}-\mu_i)_{i\in\mfrak{i}_{1}}$, $\mu_{2\max}-\bm{\mu}_{2}$ the sequence $(\mu_{2\max}-\mu_i)_{i\in\mfrak{i}_{2}}$, and $C_{\bm{\mu}}$ the following constant:

\begin{equation*}
C_{\bm{\mu}} :=
\frac{1}{2^{\abs{\mfrak{i}_{1\max}}+2\abs{\mfrak{i}_{2\max}}}}
\prod_{	i\in \mfrak{i}_{1}\setminus\mfrak{i}_{1\max}	}
\left(	q^{4\mu_{1\max}-4\mu_{i}}-1	\right)^{-1}
\prod_{	i\in \mfrak{i}_{2}\setminus\mfrak{i}_{2\max}	}
\left(	q^{2\mu_{2\max}-2\mu_{i}}-1	\right)^{-1}.
\end{equation*}

Moreover, if both $\bm{\mu}_{1}$ and $\tfrac{1}{2}\bm{\mu}_{2}$ take integral values, then the super $q$-exponential polynomial is primary.

Note that the even leading coefficient $C_{\bm{\mu},e,0}$ is positive and the odd leading coefficient $C_{\bm{\mu},e,1}$ satisfies $\abs*{C_{\bm{\mu},e,1}}<C_{\bm{\mu},e,0}$.
Hence, the $q$-function $\fun{S}$ is eventually positive.
Note that $C_{\bm{\mu},e,1}$ could be $0$, in which case the asymptotic growth of $\fun{S}[z]$ along even integers and odd integers coincide.

§ SIMPLICIAL VOLUME IN BUILDINGS OF AN TYPE

In this section, we will prove the $A_{n}$ part of <ref>.
More precisely, we will prove the following stronger theorem.

Let $\mathscr{B}$ be a Bruhat-Tits building of split classical type $A_n$ over a local field $K$ with residue cardinality $q$.
Then the simplicial volume $\fun{SV}[\:\cdot\:]$ and the simplicial surface area $\fun{SSA}[\:\cdot\:]$ in it can be defined by primary $q$-exponential polynomials whose leading terms are of the form:

\begin{align*}
\fun{SV}[r] &\sim
\tilde{C}(n) \cdot r^{\varepsilon(n)}q^{	\pi(n) r	},&
\fun{SSA}[r] &\sim
C(n) \cdot r^{\varepsilon(n)}q^{	\pi(n) r	},
\end{align*}

where $\varepsilon(n)=0$ and $\pi(n)=(\tfrac{n+1}{2})^{2}$ if $n$ is odd, while $\varepsilon(n)=1$ and $\pi(n)=\frac{n}{2}(\frac{n}{2}+1)$ if $n$ is even.

Moreover, we will obtain explicit formulas for the constants $\tilde{C}(n)$ and $C(n)$.

By the discussion at the beginning of <ref>,
this can be done as follows.
First, we will compute the asymptotic growth of $\fun{S}_{\mcal{V}[I]}[r]$ for each type $I\subset\Delta$.
This allows us to find the dominant ones.
On the other hand, by <ref>, we have

\begin{equation}\label{eq:SSA=SumSI:An}
\fun{SSA}[r] =
\sum_{I\subset\Delta}
\frac{
\mscr{P}_{A_n;I}[q]
\end{equation}

Then we can obtain the asymptotic growths of $\fun{SSA}[r]$ and  $\fun{SV}[r]$.

§.§ Asymptotic growth of SVIr

Now, let $I$ be a type and follow <ref>.
We are going to compute the asymptotic growth of $\fun{S}_{\mcal{V}[I]}[r]$.
Since all vertices are special, we have

\begin{equation*}
\fun{S}_{\mcal{V}[I]}[r] =
\fun{S}^{\asymp}_{\mcal{V}[I]}[r] =
\sum_{x\in\mcal{V}[I,r]}q^{2\rho(x)}.
\end{equation*}

Then by <ref>, we have

\begin{equation*}
\fun{S}_{\mcal{V}[I]}[r] =
\sum_{\crampedsubstack{
c_1+\cdots+c_t = r
}}q^{	\sum\limits_{i=1}^{t}\ell_{i}(n+1-\ell_{i})c_i	}.
\end{equation*}

Now, we apply <ref> to above summation, where the index set $\mfrak{i}$ is $\Set*{1,\cdots,t}$ and the sequence $\bm{\mu}$ is

μ_i   =ℓ_i(n+1-ℓ_i).     (1 ≤i ≤t)

Since all members of $\bm{\mu}$ are integers, $\fun{S}_{\mcal{V}[I]}$ can be defined by a primary $q$-exponential polynomial.
Note that, by <ref>, this already implies that $\fun{SV}[\:\cdot\:]$ and  $\fun{SSA}[\:\cdot\:]$ can be defined by primary $q$-exponential polynomials.

The knowledge of quadratic function shows that either $\mfrak{i}_{\max}$ is a singleton $\Set*{i_0}$ or it consists of two consecutive indices $\Set*{i_0,i_0+1}$.

If $\mfrak{i}_{\max}=\Set*{i_0}$.
Then $\mu_{\max} = \ell_{i_0}(n+1-\ell_{i_0})$,
$\mu_{\max} - \mu_i = (\ell_{i_0}-\ell_{i})(n+1-\ell_{i_0}-\ell_{i})$,
and we have

\begin{equation}
\label{eq:AsymptoticOfSI:An:1}
\fun{S}_{\mcal{V}[I]}[r] \sim
\prod_{i\neq i_0}
\left(
\right)^{-1}
\cdot q^{\ell_{i_0}(n+1-\ell_{i_0}) r}.
\end{equation}

In particular, $\fun{S}_{\mcal{V}[I]}$ has order $\ell_{i_0}(n+1-\ell_{i_0})$ and degree $0$.

If $\mfrak{i}_{\max}=\Set*{i_0,i_0+1}$.
Then $\mu_{\max} = \ell_{i_0}(n+1-\ell_{i_0})$,
$\mu_{\max} - \mu_i = (\ell_{i_0}-\ell_{i})(n+1-\ell_{i_0}-\ell_{i})$,
and we have

\begin{equation}
\label{eq:AsymptoticOfSI:An:2}
\fun{S}_{\mcal{V}[I]}[r] \sim
\prod_{i\neq i_0,i_0+1}
\left(
\right)^{-1}
\cdot rq^{\ell_{i_0}(n+1-\ell_{i_0}) r}.
\end{equation}

In particular, $\fun{S}_{\mcal{V}[I]}$ has order $\ell_{i_0}(n+1-\ell_{i_0})$ and degree $1$.

§.§ Dominant types

dominant type
Now, we are able to figure out for which type $I$, the $q$-function $\fun{S}_{\mcal{V}[I]}$ is dominant among its siblings. We will say that such a type is dominant.

When $n$ is odd, we have

\begin{equation*}
\mu_{\max}=\ell_{i_0}(n+1-\ell_{i_0})
\le (\frac{n+1}{2})^2.
\end{equation*}

The equality achieves exactly when $\ell_{i_0}=\frac{n+1}{2}$.
Therefore, $\fun{S}_{\mcal{V}[I]}$ is dominant exactly when $\tfrac{n+1}{2}\notin{I}$.
In this case, we have $\mfrak{i}_{\max}=\Set*{i_0}$ and $\ell_{i_0}=\frac{n+1}{2}$.

When $n$ is even, we have

\begin{equation*}
\mu_{\max}=\ell_{i_0}(n+1-\ell_{i_0})
\le \frac{n}{2}\left(\frac{n}{2}+1\right).
\end{equation*}

The equality achieves exactly when $\ell_{i_0}=\frac{n}{2}$ or $\frac{n}{2}+1$.
Therefore, $\fun{S}_{\mcal{V}[I]}$ is dominant only if $\Set*{\tfrac{n}{2},\tfrac{n}{2}+1} \not\subset {I}$.
There are three cases:
if $\Set*{\tfrac{n}{2},\tfrac{n}{2}+1}\cap{I} = \Set*{\tfrac{n}{2}+1}$, we have $\mfrak{i}_{\max} = \Set*{i_0}$ and $\ell_{i_0}=\frac{n}{2}$;
if $\Set*{\tfrac{n}{2},\tfrac{n}{2}+1}\cap{I} = \Set*{\tfrac{n}{2}}$, we have $\mfrak{i}_{\max} = \Set*{i_0}$ and $\ell_{i_0}=\frac{n}{2}+1$;
if $\Set*{\tfrac{n}{2},\tfrac{n}{2}+1}\cap{I} = \emptyset$, we have $\mfrak{i}_{\max} = \Set*{i_0,i_0+1}$, $\ell_{i_0}=\frac{n}{2}$, and $\ell_{i_0+1}=\frac{n}{2}+1$.
Among them, the last one gives the dominant type since $\fun{S}_{\mcal{V}[I]}$ has degree $1$ in that case while it has degree $0$ in the first two cases.

§.§ Asymptotic growths of SSA and SV

We are now going to obtain the asymptotic growth of $\fun{SSA}[r]$.
By <ref>, we have

\begin{equation*}
\fun{SSA}[r]	\sim
\sum_{I\text{ is dominant}}
\frac{
\mscr{P}_{A_n;I}[q]
\end{equation*}

When $n$ is odd, by <ref> and <ref>, we see that $\fun{SSA}[\:\cdot\:]$ can be defined by a primary $q$-exponential polynomial so that

\begin{equation}\label{eq:Asymp:An:odd:SSA}
\fun{SSA}[r] \sim
C(n) \cdot q^{	(\frac{n+1}{2})^2 r	},
\end{equation}

where the constant $C(n)$ is defined as follows:

\begin{equation}\label{eq:Asymp:An:odd:C}
\index[notation]{C (n)@$C(n)$}%
C(n) :=
\sum_{\crampedsubstack{
I\subset\Delta \\
\tfrac{n+1}{2}\notin{I}
\frac{
\mscr{P}_{A_n;I}[q]
\prod_{\crampedsubstack{
1 \le i \le t_{I} \\
\ell_{i}(I) \neq \tfrac{n+1}{2}
\left(
q^{	\left(\ell_{i}(I)-\tfrac{n+1}{2}\right)^2}-1
\right)^{-1}.
\end{equation}

As a consequence, we see that $\fun{SV}[\:\cdot\:]$ can be defined by a primary $q$-exponential polynomial so that

\begin{equation}\label{eq:Asymp:An:odd:SV}
\fun{SV}[r] =
\sum_{z=0}^{r}\fun{SSA}[z] \sim
\frac{
q^{(\frac{n+1}{2})^2} - 1
}C(n) \cdot q^{	(\frac{n+1}{2})^2 r	},
\end{equation}

When $n$ is even, by <ref> and <ref>, we see that $\fun{SSA}[\:\cdot\:]$ can be defined by a primary $q$-exponential polynomial so that

\begin{equation}\label{eq:Asymp:An:even:SSA}
\fun{SSA}[r] \sim
C(n) \cdot rq^{	\frac{n}{2}(\frac{n}{2}+1) r	},
\end{equation}

where the constant $C(n)$ is defined as follows:

\begin{equation}\label{eq:Asymp:An:even:C}
\index[notation]{C (n)@$C(n)$}%
C(n) :=
\sum_{\crampedsubstack{
I\subset\Delta \\
\tfrac{n}{2},\tfrac{n}{2}+1\notin{I}
\frac{
\mscr{P}_{A_n;I}[q]
\prod_{\crampedsubstack{
1 \le i \le t_{I} \\
\ell_{i}(I) \neq \tfrac{n}{2},\tfrac{n}{2}+1
\left(
q^{	(\ell_{i}(I)-\frac{n}{2})(\ell_{i}(I)-\frac{n}{2}-1)	}-1
\right)^{-1}.
\end{equation}

As a consequence, we see that $\fun{SV}[\:\cdot\:]$ can be defined by a primary $q$-exponential polynomial so that

\begin{equation}\label{eq:Asymp:An:even:SV}
\fun{SV}[r] =
\sum_{z=0}^{r}\fun{SSA}[z] \sim
\frac{
q^{\frac{n}{2}(\frac{n}{2}+1)} - 1
}C(n) \cdot q^{	\frac{n}{2}(\frac{n}{2}+1) r	},
\end{equation}

Since all vertices are special, we have $\fun{SSA}_{\dagger}[r]=\fun{SSA}[r]$ and $\fun{SV}_{\dagger}[r]=\fun{SV}[r]$, where $\dagger$ denotes “being special”.

By <ref>,
we have proved <ref>.
Moreover, by <ref>, we have the following explicit formulas for the first factor of $C(n)$:

\begin{align*}
\mscr{P}_{A_n;I}[q]
\qbinom{n+1}{
\ell_{1}(I),\ell_{2}(I)-\ell_{1}(I),\cdots,\ell_{t}(I)-\ell_{t-1}(I),n+1-\ell_{t}(I)
}(q), \\
\frac{
\end{align*}

See <ref> for the definition of the symbol $\qbinom{\:\cdot\:}{\:\cdot\:,\cdots,\:\cdot\:}$.

§ SIMPLICIAL VOLUME IN BUILDINGS OF CN TYPE

In this section, we will prove the $C_{n}$ part of <ref>.
More precisely, we will prove the following stronger theorem.

Let $\mathscr{B}$ be a Bruhat-Tits building of split classical type $C_n$ over a local field $K$ with residue cardinality $q$.
Then the simplicial volume $\fun{SV}[\:\cdot\:]$ and the simplicial surface area $\fun{SSA}[\:\cdot\:]$ in it can be defined by primary super $q$-exponential polynomials whose leading terms are of the form:

\begin{align*}
\fun{SV}[r] &\sim
\tilde{C}(n) \cdot q^{	\tfrac{n(n+1)}{2} r	},&
\fun{SSA}[r] &\sim
C(n) \cdot q^{	\tfrac{n(n+1)}{2} r	},
\end{align*}

where $\tilde{C}(n)$ and $C(n)$ are primary $q$-numbers, not just parity $q$-functions.

Moreover, we will obtain explicit formulas for the constants $\tilde{C}(n)$ and $C(n)$.

By the discussion at the beginning of <ref>,
this can be done as follows.
First, we estimate the asymptotic growth of $\fun{S}_{\mcal{V}[I]}[r]$ for each type $I\subset\Delta$ using the auxiliary function $\fun{S}^{\asymp}_{\mcal{V}[I]}$ in <ref>.
This allows us to find the dominant ones in <ref>.
Then we can compute the leading coefficient of $\fun{S}_{\mcal{V}[I]}[r]$ for dominant ones in <ref>.
Finally, by <ref>, we have

\begin{equation}\label{eq:SSA=SumSI:Cn}
\fun{SSA}[r] =
\sum_{I\subset\Delta}
\frac{
\mscr{P}_{C_n;I}[q]
\sim
\sum_{I\text{ is dominant}}
\frac{
\mscr{P}_{C_n;I}[q]
\end{equation}

Then we can obtain the asymptotic growths of $\fun{SSA}[r]$ and $\fun{SV}[r]$.

Along the discussion, we will also consider the asymptotic growths of $\fun{SSA}_{\dagger}[r]$ and $\fun{SV}_{\dagger}[r]$, where $\dagger$ denotes “being special”. Namely, we will also prove the following theorem (in <ref>).

Let $\mathscr{B}$ be a Bruhat-Tits building of split classical type $C_n$ over a local field $K$ with residue cardinality $q$.
Then the special simplicial volume $\fun{SV}_{\dagger}[\:\cdot\:]$ and the special simplicial surface area $\fun{SSA}_{\dagger}[\:\cdot\:]$ in it can be defined by primary super $q$-exponential polynomials whose leading terms are of the form:

\begin{align*}
\fun{SV}_{\dagger}[r] &\sim
\tilde{C}_{\dagger}(n) \cdot q^{	\tfrac{n(n+1)}{2} r	},&
\fun{SSA}_{\dagger}[r] &\sim
C_{\dagger}(n) \cdot q^{	\tfrac{n(n+1)}{2} r	},
\end{align*}

where $\tilde{C}_{\dagger}(n)$ and $C_{\dagger}(n)$ are primary $q$-numbers, not just parity $q$-functions.

Moreover, we will obtain explicit formulas for the constants $\tilde{C}_{\dagger}(n)$ and $C_{\dagger}(n)$.

§.§ Asymptotic growth of SasympVIr

Now, let $I$ be a type and follow <ref>.
We are going to estimate the asymptotic growth of $\fun{S}_{\mcal{V}[I]}[r]$ up to the leading coefficient.

By <ref>, we have

\begin{equation*}
\fun{S}^{\asymp}_{\mcal{V}[I]}[r] =
\sum_{\crampedsubstack{
c_1+\cdots+c_t = r
}}q^{	\sum\limits_{i=1}^{t}\tfrac{1}{2}\ell_{i}(2n+1-\ell_{i})c_i	}.
\end{equation*}

Now, we apply <ref> to above summation, where the index set $\mfrak{i}$ is $\Set*{1,\cdots,t}$ and the sequence $\bm{\mu}$ is

μ_i   =12ℓ_i(2n+1-ℓ_i).     (1 ≤i ≤t)

Since all members of $\bm{\mu}$ are integers, $\fun{S}^{\asymp}_{\mcal{V}[I]}$ can be defined by a primary $q$-exponential polynomial.
The knowledge of quadratic function shows that $\mfrak{i}_{\max}=\Set*{t}$ with $\mu_{\max}=\tfrac{1}{2}\ell_{t}(2n+1-\ell_{t})$.
Then we have

\begin{equation*}
\fun{S}^{\asymp}_{\mcal{V}[I]}[r] \sim
\prod_{	i=1	}^{t-1}
\left(	q^{\tfrac{1}{2}(\ell_{t}-\ell_{i})(2n+1-\ell_{t}-\ell_{i})} - 1	\right)^{-1}
\cdot q^{	\tfrac{1}{2}\ell_{t}(2n+1-\ell_{t}) r	}.
\end{equation*}

Since $\fun{S}_{\mcal{V}[I]}[r]\asymp\fun{S}^{\asymp}_{\mcal{V}[I]}[r]$, we see that $\fun{S}_{\mcal{V}[I]}$ has order $\tfrac{1}{2}\ell_{t}(2n+1-\ell_{t})$ and degree $0$.

§.§ Asymptotic growth of SVdaggerIr

Next, we are going to compute the asymptotic growth of $\fun{S}_{\mcal{V}_{\dagger}[I]}[r]$.

If $\ell_{t}<n$, then by <ref>, we have

\begin{equation*}
\fun{S}_{\mcal{V}_{\dagger}[I]}[r] =
\sum_{\crampedsubstack{
2c_1+\cdots+2c_t = r
}}q^{	\sum\limits_{i=1}^{t}\ell_{i}(2n+1-\ell_{i})c_i	}.
\end{equation*}

Now, we apply <ref> to this summation, where the index set $\mfrak{i}$ is $\Set*{1,\cdots,t}$, the partition $\mfrak{i}=\mfrak{i}_{1}\sqcup\mfrak{i}_{2}$ is $\Set*{1,\cdots,t} = \emptyset\sqcup\Set*{1,\cdots,t}$, and the sequence $\bm{\mu}$ is

μ_i   =ℓ_i(2n+1-ℓ_i).     (1 ≤i ≤t)

Since all members of $\bm{\mu}$ are even integers, $\fun{S}_{\mcal{V}_{\dagger}[I]}$ can be defined by a primary super $q$-exponential polynomial.
The knowledge of quadratic function shows that $\mfrak{i}_{\max}=\mfrak{i}_{2\max}=\Set*{t}$ with $\mu_{\max}=\mu_{2\max}=\ell_{t}(2n+1-\ell_{t})$.
Then by <ref>, we have

\begin{equation*}
\fun{S}_{\mcal{V}_{\dagger}[I]}[r] \sim
\prod_{	i=1	}^{t-1}
\left(	q^{(\ell_{t}-\ell_{i})(2n+1-\ell_{t}-\ell_{i})} - 1	\right)^{-1}
\cdot	\tfrac{1}{2}\left(1+(-1)^{r}\right)
\cdot q^{	\tfrac{1}{2}\ell_{t}(2n+1-\ell_{t}) r	}.
\end{equation*}

In particular, it has order $\tfrac{1}{2}\ell_{t}(2n+1-\ell_{t})$ and degree $0$.

If $\ell_{t}=n$, then by <ref>, we have

\begin{equation*}
\fun{S}_{\mcal{V}_{\dagger}[I]}[r] =
\sum_{\crampedsubstack{
2c_1+\cdots+2c_{t-1}+c_t = r
}}q^{	\sum\limits_{i=1}^{t-1}\ell_{i}(2n+1-\ell_{i})c_i + \tfrac{n(n+1)}{2}c_t	}.
\end{equation*}

Now, we apply <ref> to this summation, where the index set $\mfrak{i}$ is $\Set*{1,\cdots,t}$, the partition $\mfrak{i}=\mfrak{i}_{1}\sqcup\mfrak{i}_{2}$ is $\Set*{1,\cdots,t} = \Set*{t}\sqcup\Set*{1,\cdots,t-1}$, and the sequence $\bm{\mu}$ is

μ_i    = ℓ_i(2n+1-ℓ_i),
(1 ≤i < t)

μ_t    = n(n+1)2.

Since $\mu_{t}$ is an integer and all members of $\bm{\mu}_{2}$ are even integers, $\fun{S}_{\mcal{V}_{\dagger}[I]}$ can be defined by a primary super $q$-exponential polynomial.
The knowledge of quadratic function shows that $\mfrak{i}_{2\max}=\Set*{t-1}$, $\mu_{2\max}=\ell_{t-1}(2n+1-\ell_{t-1})$, and $2\mu_{1\max}>\mu_{2\max}$.
Then by <ref>, we have

\begin{equation}\label{eq:AsymptoticOfSdaggerI:Cn}
\fun{S}_{\mcal{V}_{\dagger}[I]}[r] \sim
\prod_{	i=1	}^{t-1}
\left(	q^{(n-\ell_{i})(n+1-\ell_{i})} - 1	\right)^{-1}
\cdot q^{	\tfrac{n(n+1)}{2} r	}.
\end{equation}

In particular, it has order $\tfrac{n(n+1)}{2}$ and degree $0$.

§.§ Dominant types

Now, we are able to figure out which type is dominant.

We begin with $\fun{S}_{\mcal{V}[I]}$, since $\ell_t \le n$, we have

\begin{equation*}
\mu_{\max} =
\tfrac{1}{2}\ell_{t}(2n+1-\ell_{t}) \le \tfrac{n(n+1)}{2}.
\end{equation*}

The equality achieves exactly when $\ell_{t}=n$.
Therefore, $\fun{S}_{\mcal{V}[I]}$ is dominant exactly when $n\notin{I}$. In this case, we have $\mfrak{i}_{\max}=\Set*{t}$ and $\ell_{t}=n$.

Next, we consider $\fun{S}_{\mcal{V}_{\dagger}[I]}$.
Then similar argument shows that it is dominant exactly when $n\notin{I}$.

§.§ Asymptotic growths of SSAdagger and SVdagger

We are now able to obtain the asymptotic growths of $\fun{SSA}_{\dagger}[r]$ and $\fun{SV}_{\dagger}[r]$.

By <ref>, <ref>, and <ref>, we see that $\fun{SSA}_{\dagger}[\:\cdot\:]$ can be defined by a primary super $q$-exponential polynomial so that

\begin{equation}\label{eq:Asymp:Cn:SSAdagger}
\fun{SSA}_{\dagger}[r] \sim
C_{\dagger}(n) \cdot q^{	\tfrac{n(n+1)}{2} r},
\end{equation}

where the constant $C_{\dagger}(n)$ is defined as follows:
[notation]C dagger (n)@$C_{\dagger}(n)$

\begin{equation}\label{eq:Asymp:Cn:Cdagger}
C_{\dagger}(n)	:=
\sum_{n\notin I}
\frac{
\mscr{P}_{C_n;I}[q]
\prod_{i=1}^{t_{I}-1}
\left(	q^{	(n-\ell_{i}(I))(n+1-\ell_{i}(I))	}-1	\right)^{-1}.
\end{equation}

As a consequence, we see that $\fun{SV}_{\dagger}[\:\cdot\:]$ can be defined by a primary super $q$-exponential polynomial so that

\begin{equation}\label{eq:Asymp:Cn:SVdagger}
\fun{SV}_{\dagger}[r] =
\sum_{z=0}^{r}\fun{SSA}_{\dagger}[z] \sim
\frac{
q^{\tfrac{n(n+1)}{2}} - 1
}C_{\dagger}(n) \cdot q^{	\tfrac{n(n+1)}{2} r}.
\end{equation}

§.§ Asymptotic growth of dominant SVIr

Let $I$ be a type and follow <ref>.
We are going to compute the asymptotic growth of $\fun{S}_{\mcal{V}[I]}[r]$ when $I$ is dominant.

First, we need to write $\fun{S}_{\mcal{V}[I]}[r]$ into a multi-summation.
To do this, we pick an arbitrary $x\in\mcal{V}[I]$ and investigate the difference between $2\rho(x)$ and the sum of $\ceil{a(x)}$ for $a(x)>0$.
To better describe this sum, we introduce the following conventions.

[notation]l t+1@$\ell_{t+1}=n$
For any $j\in\Set*{1,\cdots,n}$, we will use $\ell^{-1}(j)$ to denote the index $i\in\Set*{1,\cdots,t+1}$ such that $\ell_{i-1}< j \le \ell_{i}$, where $\ell_{t+1}=n$.

parity function!standard
[notation]z bar@$\vect{\overline{z}}$

We will use $\overline{z}$ to denote standard parity function mapping even numbers to $0$ and odd numbers to $1$.
Note that $\ceil*{\tfrac{1}{2}z} = \tfrac{1}{2}(z+\overline{z})$.

The summation $c_i+\cdots+c_j$ is read to be $0$ when $i>j$.

Now, suppose $x = o+c_1\omega_{\ell_{1}}'+\cdots+c_t\omega_{\ell_{t}}'\in\mcal{V}[I]$. By <ref>, we have

(χ_j-χ_j')(x)    =
(1 ≤j < j' ≤n)

(χ_j+χ_j')(x)    =
) +
(1 ≤j < j' ≤n)

2χ_j(x)    =
(1 ≤j ≤n)

Therefore, we have

\begin{align*}
\sum_{a\in\Phi^{+}}\ceil{a(x)} &=
\sum_{1 \le j < j' \le n}
\left(
\ceil{(\chi_{j}-\chi_{j'})(x)} +
\ceil{(\chi_{j}+\chi_{j'})(x)}
\right) +
\sum_{j=1}^{n}\ceil{2\chi_{j}(x)} \\
% &=
% \sum_{1 \le j < j' \le n}
% \Bigg(
% 	\ceil*{\tfrac{1}{2}
% 	\left(
% 		c_{\ell^{-1}(j)}+\cdots+c_{\ell^{-1}(j')-1}
% 	\right)} \\
% 	&\qquad\qquad +
% 	\ceil*{\tfrac{1}{2}
% 	\left(
% 		c_{\ell^{-1}(j)}+\cdots+c_{\ell^{-1}(j')-1}
% 	\right) +
% 	c_{\ell^{-1}(j')}+\cdots+c_{t}}
% \Bigg) \\
% &\qquad\qquad +
% \sum_{j=1}^{n}\ceil*{c_{\ell^{-1}(j)}+\cdots+c_{t}} \\
&= 2\rho(x) +
\sum_{1 \le j < j' \le n}
\overline{	c_{\ell^{-1}(j)}+\cdots+c_{\ell^{-1}(j')-1}	}.
\end{align*}

From above analysis, we can define the parity function $e_{I}$ as follows:

\begin{equation}
\label{eq:DefineEpsilon:Cn}
e_{I}(c_1,\cdots,c_{t}) :=
\sum_{1 \le i < i' \le t+1}
\overline{c_i+\cdots+c_{i'-1}}.
\end{equation}

Then we have

\begin{equation*}
\sum_{a\in\Phi^{+}}\ceil{a(x)} =
2\rho(x) + e_{I}(c_1,\cdots,c_{t}).
\end{equation*}

Now, we apply <ref> to the following summation.

\begin{equation*}
\fun{S}_{\mcal{V}[I]}[r] =
\sum_{\crampedsubstack{
c_1+\cdots+c_t = r
}}q^{	\sum\limits_{i=1}^{t}\tfrac{1}{2}\ell_{i}(2n+1-\ell_{i})c_i	+ e_{I}(c_1,\cdots,c_{t})}.
\end{equation*}

Note that the index set $\mfrak{i}$ is $\Set*{1,\cdots,t}$ and the sequence $\bm{\mu}$ is

μ_i   =12ℓ_i(2n+1-ℓ_i).     (1 ≤i ≤t)

Since all members of $\bm{\mu}$ are integers and $e_{I}$ is valued in integers, $\fun{S}_{\mcal{V}[I]}$ can be defined by a primary super $q$-exponential polynomial.
Note that, by <ref>, this already implies that $\fun{SV}[\:\cdot\:]$ and  $\fun{SSA}[\:\cdot\:]$ can be defined by primary super $q$-exponential polynomials.

Now, let $I$ be a dominant type, namely $n\notin I$.
Then we have $\mfrak{i}_{\max}=\Set*{t}$, $\ell_{t}=n$, and $\mu_{\max} = \tfrac{n(n+1)}{2}$.

\begin{equation*}
\fun{S}_{\mcal{V}[I]}[r] \sim
\left(	C_{I,0} + C_{I,1}(-1)^{r}	\right)
\cdot q^{\tfrac{n(n+1)}{2} r},
\end{equation*}

where the constants $C_{I,0}$ and $C_{I,1}$ are defined as follows:

\begin{align*}
C_{I,0} &:=
\sum_{\vect{s}\in\F_2^{t}}
\fun{E}_{I}(\vect{s}),
C_{I,1} &:=
\sum_{\vect{s}\in\F_2^{t}}
\fun{E}_{I}(\vect{s}),
\end{align*}

where the constant $C_{I}$ and the function $\fun{E}_{I}\colon\F_2^{t}\to\mbb{Q}[q;-]$ are defined as follows:

\begin{align}
\label{eq:EIandCI:Cn}%
C_{I} &:=
\frac{1}{2}
\prod_{i=1}^{t-1}
\left(	q^{	(n-\ell_{i})(n+1-\ell_{i})	}-1	\right)^{-1},
\fun{E}_{I}(\vect{s}) &:=
q^{	e_{I}(\vect{s}) +
\sum\limits_{i=1}^{t-1}
\tfrac{1}{2}(n-\ell_{i})(n+1-\ell_{i})s_i	}.
\end{align}

From the definition <ref> of $e_{I}$, it is clear that

\begin{equation*}
\fun{E}_{I}(s_1,\cdots,s_{t-1},0) =
\fun{E}_{I}(s_1,\cdots,s_{t-1},1).
\end{equation*}

Therefore, $C_{I,1} = 0$ and we thus have

\begin{equation}
\label{eq:AsymptoticOfSI:Cn}
\fun{S}_{\mcal{V}[I]}[r] \sim
\sum_{\vect{s}\in\F_2^{t-1}}
2\fun{E}_{I}[\vect{s}\sqcup 0]
\cdot q^{\tfrac{n(n+1)}{2} r},
\end{equation}

where $\vect{s}\sqcup 0$ is the sequence $s_1,\cdots,s_{t-1},0$.

§.§ Asymptotic growths of SSA and SV

We are now going to obtain the asymptotic growth of $\fun{SSA}[r]$.

By <ref>, <ref>, and <ref>, we see that $\fun{SSA}[\:\cdot\:]$ can be defined by a primary super $q$-exponential polynomial so that

\begin{equation}\label{eq:Asymp:Cn:SSA}
\fun{SSA}[r] \sim
C(n) \cdot q^{	\tfrac{n(n+1)}{2} r},
\end{equation}

where the constant $C(n)$ is defined as follows:

\begin{equation}\label{eq:Asymp:Cn:C}
\index[notation]{C (n)@$C(n)$}%
C(n)	:=
\sum_{n\notin I}
\left(
\frac{
\mscr{P}_{C_n;I}[q]
\prod_{i=1}^{t_{I}-1}
\left(	q^{	(n-\ell_{i}(I))(n+1-\ell_{i}(I))	}-1	\right)^{-1}
\sum_{\vect{s}\in\F_2^{t_{I}-1}}
\fun{E}_{C_n;I}[\vect{s}]
\right),
\end{equation}

where the function $\fun{E}_{C_n;I}\colon\F_2^{t_{I}-1}\to\mbb{Q}[q;-]$ is defined as follows:

\begin{align}\label{eq:Asymp:Cn:E}
% \index[notation]{E Cn I@$\fun{E}_{C_n;I}$}%
\fun{E}_{C_n;I}[\vect{s}] &:=
q^{	\sum\limits_{1 \le i < i' \le t_{I}}
\overline{s_i+\cdots+s_{i'-1}}	}\\
\nonumber&\quad\cdot
q^{	\sum\limits_{i=1}^{t_{I}-1}
\tfrac{1}{2}(n-\ell_{i}(I))(n+1-\ell_{i}(I))s_i	}.
\end{align}

As a consequence, we see that $\fun{SV}[\:\cdot\:]$ can be defined by a primary super $q$-exponential polynomial so that

\begin{equation}\label{eq:Asymp:Cn:SV}
\fun{SV}[r] =
\sum_{z=0}^{r}\fun{SSA}[z] \sim
\frac{
q^{\tfrac{n(n+1)}{2}} - 1
}C(n) \cdot q^{	\tfrac{n(n+1)}{2} r}.
\end{equation}

By <ref>,
we have proved <ref>.
Moreover, by <ref>, we have the following explicit formulas for the first factor of $C(n)$:

\begin{align*}
\mscr{P}_{C_n;I}[q]
\frac{
\prod\limits_{i=1}^{t}
\frac{
\prod\limits_{i=1}^{t}
\end{align*}

See <ref> for the definitions of the symbols $[\:\cdot\:]!$ and $[2\:\cdot\:]!!$.

§ SIMPLICIAL VOLUME IN BUILDINGS OF BN TYPE

In this section, we will prove the $B_{n}$ part of <ref>.
More precisely, we will prove the following stronger theorem.

Let $\mathscr{B}$ be a Bruhat-Tits building of split classical type $B_n$ over a local field $K$ with residue cardinality $q$.
Then the simplicial volume $\fun{SV}[\:\cdot\:]$ and the simplicial surface area $\fun{SSA}[\:\cdot\:]$ in it can be defined by super $q$-exponential polynomials whose leading terms are of the form:
[The leading terms may give an impression that these $q$-functions can be defined by primary $q$-exponential polynomials when $n=3$. However, we will see this is false in <ref>.]

\begin{align*}
\fun{SV}[r] &\sim
\tilde{C}(n) \cdot q^{	\pi(n) r	},&
\fun{SSA}[r] &\sim
C(n) \cdot q^{	\pi(n) r	},
\end{align*}

where $\pi(n)=\tfrac{n^2}{2}$ when $n \ge 4$ and $\pi(3)=5$.
The leading coefficients $\tilde{C}(3)$ and $C(3)$ are primary $q$-numbers, not just parity $q$-functions.
Moreover, the four $q$-functions $\fun{SV}[2\:\cdot\:]$, $\fun{SV}[2\:\cdot\:+1]$, $\fun{SSA}[2\:\cdot\:]$, and $\fun{SSA}[2\:\cdot\:+1]$ can be defined by primary $q$-exponential polynomials whose leading terms are of the form:

\begin{align*}
\fun{SV}[2r] &\sim
\tilde{C}_{0}(n) \cdot q^{	2\pi(n) r	},&
\fun{SSA}[2r] &\sim
C_{0}(n) \cdot q^{	2\pi(n) r	},\\
\fun{SV}[2r+1] &\sim
\tilde{C}_{1}(n) \cdot q^{	2\pi(n) r	},&
\fun{SSA}[2r+1] &\sim
C_{1}(n) \cdot q^{	2\pi(n) r	}.
\end{align*}

We will obtain explicit formulas for the parity functions $\tilde{C}(n)$ and $C(n)$, and the constants $\tilde{C}_{0}(n)$, $C_{0}(n)$, $\tilde{C}_{1}(n)$, and $C_{1}(n)$.

But before proving <ref>, we will first analyze the asymptotic growths of $\fun{SSA}_{\dagger}[r]$ and $\fun{SV}_{\dagger}[r]$, where $\dagger$ denotes “being special”. We will prove the following.

Let $\mathscr{B}$ be a Bruhat-Tits building of split classical type $B_n$ over a local field $K$ with residue cardinality $q$.
Then the special simplicial volume $\fun{SV}_{\dagger}[\:\cdot\:]$ and the special simplicial surface area $\fun{SSA}_{\dagger}[\:\cdot\:]$ in it can be defined by super $q$-exponential polynomials whose leading terms are of the form:<ref>

\begin{align*}
\fun{SV}_{\dagger}[r] &\sim
\tilde{C}_{_{\dagger}}(n) \cdot q^{	\pi(n) r	},&
\fun{SSA}_{\dagger}[r] &\sim
C_{_{\dagger}}(n) \cdot q^{	\pi(n) r	},
\end{align*}

where $\pi(n)=\tfrac{n^2}{2}$ when $n \ge 4$ and $\pi(3)=5$.
The leading coefficients $\tilde{C}_{\dagger}(3)$ and $C_{\dagger}(3)$ are primary $q$-numbers, not just parity $q$-functions.
Moreover, the four $q$-functions $\fun{SV}_{\dagger}[2\:\cdot\:]$, $\fun{SV}_{\dagger}[2\:\cdot\:+1]$, $\fun{SSA}_{\dagger}[2\:\cdot\:]$, and $\fun{SSA}_{\dagger}[2\:\cdot\:+1]$ can be defined by primary $q$-exponential polynomials whose leading terms are of the form:

\begin{align*}
\fun{SV}_{\dagger}[2r] &\sim
\tilde{C}_{{\dagger}0}(n) \cdot q^{	2\pi(n) r	},&
\fun{SSA}_{\dagger}[2r] &\sim
C_{{\dagger}0}(n) \cdot q^{	2\pi(n) r	},\\
\fun{SV}_{\dagger}[2r+1] &\sim
\tilde{C}_{{\dagger}1}(n) \cdot q^{	2\pi(n) r	},&
\fun{SSA}_{\dagger}[2r+1] &\sim
C_{{\dagger}1}(n) \cdot q^{	2\pi(n) r	}.
\end{align*}

We will also give explicit formulas for the parity functions $\tilde{C}_{\dagger}(n)$ and $C_{\dagger}(n)$, and the constants $\tilde{C}_{{\dagger}0}(n)$, $C_{{\dagger}0}(n)$, $\tilde{C}_{{\dagger}1}(n)$, and $C_{{\dagger}1}(n)$.
The proof of <ref> turns out to play an essential role in the study of $\fun{SSA}[r]$ and $\fun{SV}[r]$.

This section is structured as follows.
In <ref>, we will compute the asymptotic growth of $\fun{S}_{\mcal{V}_{\dagger}[I]}[r]$ for each type $I\subset\Delta$. This allows use to find the dominant ones of $\fun{S}_{\mcal{V}_{\dagger}[I]}[r]$, which will be done in <ref>.
Then in <ref>, we will obtain the asymptotic growths of $\fun{SSA}_{\dagger}[r]$ and $\fun{SV}_{\dagger}[r]$.
In <ref>, we will analyze $\fun{S}_{\mcal{V}_{\dagger}[I]}[2r]$ and $\fun{S}_{\mcal{V}_{\dagger}[I]}[2r+1]$. Combine <ref>, we finish proving <ref>.
After that, in <ref>, we will estimate the asymptotic growths of $\fun{S}_{\mcal{X}^{0}[I]}[r]$ and $\fun{S}_{\mcal{X}^{1}[I]}[r]$ using the auxiliary functions $\fun{S}^{\asymp}_{\mcal{X}^{0}[I]}$ and $\fun{S}^{\asymp}_{\mcal{X}^{1}[I]}$.
Note that $\mcal{V}$ is between $\mcal{V}_{\dagger}$ and $\mcal{X}^{0}\cup\mcal{X}^{1}$.
Therefore, we can combine <ref> and <ref> to estimate the asymptotic growth of each $\fun{S}_{\mcal{V}[I]}[r]$ and find the dominant ones, which will be done in <ref>.
Once we found the dominant types, we can proceed to compute the asymptotic growth of dominant $\fun{S}_{\mcal{V}[I]}[r]$.
This will be done in three steps:
in <ref>, we will compute the asymptotic growths of $\fun{S}_{\mcal{X}^{0}[I]}[r]$ and $\fun{S}_{\mcal{X}^{1}[I]}[r]$; in <ref>, we will deduce the asymptotic growth of $\fun{S}_{\mcal{X}_{J}[I]}[r]$ from that of $\fun{S}_{\mcal{V}_{\dagger}[I]}[r]$; finally in <ref>, the asymptotic growth of $\fun{S}_{\mcal{V}[I]}[r]$ will be deduced from them.
Then in <ref>, we will obtain the asymptotic growths of $\fun{SSA}[r]$ and $\fun{SV}[r]$.
In <ref>, we will analyze $\fun{S}_{\mcal{X}_{J}[I]}[2r]$ and $\fun{S}_{\mcal{X}_{J}[I]}[2r+1]$. Combine <ref>, we finish proving <ref>.

Throughout this section, we will heavily use the various index sets $\mcal{V}$, $\mcal{V}_{\dagger}$, $\mcal{X}^{0}$, $\mcal{X}^{1}$, and $\mcal{X}_{J}$.
We refer to <ref> for the structure of them.

§.§ Asymptotic growth of SVdaggerIr

Now, let $I$ be a type and follow <ref>.
We are going to compute the asymptotic growth of $\fun{S}_{\mcal{V}_{\dagger}[I]}[r]$.
We will separate the discussion into two cases: <ref> $\ell_{1}>1$ and <ref> $\ell_{1}=1$.

If $\ell_{1}>1$, then by <ref>, we have

\begin{equation*}
\fun{S}_{\mcal{V}_{\dagger}[I]}[r] =
\sum_{\crampedsubstack{
2c_1+\cdots+2c_t = r
}}q^{	\sum\limits_{i=1}^{t}\ell_{i}(2n-\ell_{i})c_i	}.
\end{equation*}

Now, we apply <ref> to this summation, where the index set $\mfrak{i}$ is $\Set*{1,\cdots,t}$, the partition $\mfrak{i}=\mfrak{i}_{1}\sqcup\mfrak{i}_{2}$ is $\Set*{1,\cdots,t} = \emptyset\sqcup\Set*{1,\cdots,t}$, and the sequence $\bm{\mu}$ is

μ_i   =ℓ_i(2n-ℓ_i).     (1 ≤i ≤t)

The knowledge of quadratic function shows that $\mfrak{i}_{\max}=\mfrak{i}_{2\max}=\Set*{t}$ with $\mu_{\max}=\mu_{2\max}=\ell_{t}(2n-\ell_{t})$.
Then by <ref>, we have

\begin{equation}\label{eq:Bn:AsymptoticOfSdaggerI}
\fun{S}_{\mcal{V}_{\dagger}[I]}[r] \sim
\prod_{	i=1	}^{t-1}
\left(	q^{(\ell_{t}-\ell_{i})(2n-\ell_{t}-\ell_{i})} - 1	\right)^{-1}
\cdot	\tfrac{1}{2}\left(1+(-1)^{r}\right)
\cdot q^{	\tfrac{1}{2}\ell_{t}(2n-\ell_{t}) r	}.
\end{equation}

In particular, it has order $\tfrac{1}{2}\ell_{t}(2n-\ell_{t})$ and degree $0$.

If $\ell_{1}=1$, then by <ref>, we have

\begin{equation*}
\fun{S}_{\mcal{V}_{\dagger}[I]}[r] =
\sum_{\crampedsubstack{
c_1+2c_2+\cdots+2c_t = r
}}q^{	\sum\limits_{i=1}^{t}\ell_{i}(2n-\ell_{i})c_i	}.
\end{equation*}

Now, we apply <ref> to this summation, where the index set $\mfrak{i}$ is $\Set*{1,\cdots,t}$, the partition $\mfrak{i}=\mfrak{i}_{1}\sqcup\mfrak{i}_{2}$ is $\Set*{1,\cdots,t} = \Set*{1}\sqcup\Set*{2,\cdots,t}$, and the sequence $\bm{\mu}$ is

μ_i   =ℓ_i(2n-ℓ_i).     (1 ≤i ≤t)

The knowledge of quadratic function shows that $\mfrak{i}_{2\max}=\Set*{t}$ and $\mu_{2\max}=\ell_{t}(2n-\ell_{t})$.

Depending on $n$ and $\ell_{t}$, there are two possibilities: $2(2n-1) > \ell_{t}(2n-\ell_{t})$ and $2(2n-1) < \ell_{t}(2n-\ell_{t})$.
Note that $2(2n-1) = \ell_{t}(2n-\ell_{t})$ is impossible since the left-hand side has remainder $2$ modulo $4$ while the right-hand side is either odd or a multiple of $4$.

If $2(2n-1) > \ell_{t}(2n-\ell_{t})$, then by <ref>, we have

\begin{equation}\label{eq:Bn:AsymptoticOfSdaggerI:l}
\fun{S}_{\mcal{V}_{\dagger}[I]}[r] \sim
\prod_{	i=2	}^{t}
\left(	q^{2(2n-1)-\ell_{i}(2n-\ell_{i})} - 1	\right)^{-1}
\cdot q^{	(2n-1) r	}.
\end{equation}

In particular, it has order $2n-1$ and degree $0$.

If $2(2n-1) < \ell_{t}(2n-\ell_{t})$, then by <ref>, we have

\begin{align}\label{eq:Bn:AsymptoticOfSdaggerI:s}
\fun{S}_{\mcal{V}_{\dagger}[I]}[r] &\sim
\left(	q^{\ell_{t}(2n-\ell_{t})-2(2n-1)} - 1	\right)^{-1}
\prod_{	i=2	}^{t-1}
\left(	q^{(\ell_{t}-\ell_{i})(2n-\ell_{t}-\ell_{i})} - 1	\right)^{-1} \\
\cdot\tfrac{1}{2}
\left(
\left(
\right)
\left(
\right)(-1)^{r}
\right) \\
\cdot q^{	\tfrac{1}{2}\ell_{t}(2n-\ell_{t}) r	}.
\end{align}

In particular, it has order $\tfrac{1}{2}\ell_{t}(2n-\ell_{t})$ and degree $0$.

§.§ Dominant types for SVdaggerIr

Now, we are able to figure out for which type $I$, $\fun{S}_{\mcal{V}_{\dagger}[I]}[r]$ is dominant.

When $n=3$, we have $\ell_{t}(I)\le 3$ for all $I$.
Therefore, $2(2n-1) > \ell_{t}(I)(2n-\ell_{t}(I))$.
Hence, $\fun{S}_{\mcal{V}_{\dagger}[I]}[r]$ is dominant exactly when $1\notin I$.
Note that, such a type $I$ must be one of the following: $\Set*{2,3}$, $\Set*{2}$, $\Set*{3}$, and $\emptyset$.
Using <ref>, we can deduce the asymptotic growth of dominant $\fun{S}_{\mcal{V}_{\dagger}[I]}[r]$ as follows.

\begin{align}
\label{eq:B3:AsymptoticOfSdaggerI:23}
\fun{S}_{\mcal{V}_{\dagger}[\Set*{2,3}]}[r] &=
q^{	(2n-1) r	} = q^{	5r	}, \\
\label{eq:B3:AsymptoticOfSdaggerI:2}
\fun{S}_{\mcal{V}_{\dagger}[\Set*{2}]}[r] &\sim
\left(	q^{2(2n-1)-\ell_{2}(2n-\ell_{2})} - 1	\right)^{-1}
\cdot q^{	(2n-1) r	} \\
\nonumber&=
\frac{1}{\left(	q - 1	\right)}
% \left(	q - 1	\right)^{-1}
\cdot q^{	5 r	}, \\
\label{eq:B3:AsymptoticOfSdaggerI:3}
\fun{S}_{\mcal{V}_{\dagger}[\Set*{3}]}[r] &\sim
\left(	q^{2(2n-1)-\ell_{2}(2n-\ell_{2})} - 1	\right)^{-1}
\cdot q^{	(2n-1) r	}\\
\nonumber&=
\frac{1}{\left(	q^2 - 1	\right)}
% \left(	q^2 - 1	\right)^{-1}
\cdot q^{	5 r	}, \\
\label{eq:B3:AsymptoticOfSdaggerI:emptyset}
\fun{S}_{\mcal{V}_{\dagger}[\emptyset]}[r] &\sim
\prod_{	i=2	}^{3}
\left(	q^{2(2n-1)-\ell_{i}(2n-\ell_{i})} - 1	\right)^{-1}
\cdot q^{	(2n-1) r	} \\
\nonumber&=
\frac{1}{\left(	q - 1	\right)\left(	q^2 - 1	\right)}
% \left(	q - 1	\right)^{-1}
% \left(	q^2 - 1	\right)^{-1}
\cdot q^{	5 r	}.
\end{align}

If $n\ge 4$, then we have $2(2n-1) < \ell_{t}(I)(2n-\ell_{t}(I))$ when $\ell_{t}(I)\ge 4$.
On the other hand, since $\ell_{t}(I)\le n$, we have

\begin{equation*}
\mu_{2\max} = \ell_{t}(I)(2n-\ell_{t}(I)) \le n^2.
\end{equation*}

The equality achieves when $\ell_{t}(I)=n$.
Hence, $\fun{S}_{\mcal{V}_{\dagger}[I]}[r]$ is dominant exactly when $n\notin I$.
In that case, its asymptotic growth is given by <ref>.

§.§ Asymptotic growths of SSAdagger and SVdagger