#  $q$-Rious identities for parafermion theories

Peter Bouwknegt  Mathematical Sciences Institute, Australian National
University, Canberra, ACT 2601, Australia<EMAIL_ADDRESS>,  Shane
Chern  Department of Mathematics and Statistics, Dalhousie University,
Halifax, NS, B3H 4R2, Canada<EMAIL_ADDRESS><EMAIL_ADDRESS>and
Bolin Han  Mathematical Sciences Institute, Australian National University,
Canberra, ACT 2601, Australia<EMAIL_ADDRESS>
###### Abstract.

In this paper we will prove a series of $q$-identities suggested by the
realisation of certain conformal field theories by so-called ‘coupled free
fermions’. We will consider $q$-series arising from coupled free fermions
constructed by the parafermion coset construction as well as from scaled root
lattices, and some interesting relations between the two.

##  1\. Introduction

It has been demonstrated in the literature that combining number theory and
mathematical physics techniques can produce useful and interesting, or even
unexpected results for both areas. Especially, $q$-identities have played
important roles in conformal field theories and vice versa. For instance, we
have the famous Jacobi triple product directly related to the Weyl denominator
formula, the Hecke type identities motivated by the string functions of affine
Lie algebra developed in [17] and then proved by the classical theory of
$q$-series in [2], and more recently some Generalised Rogers–Ramanujan
identities derived from a theory of parafermions of the type
$(\widehat{{\mathfrak{sl}}_{n+1}})_{2}/\widehat{\mathfrak{u}(1)}^{n}$ [4].

In this paper, we will reveal some new $q$-identities, including some of
Rogers-Ramanujan type, observed from universal chiral partition functions of
parafermions of the type
$(\widehat{{\mathfrak{sl}}_{n+1}})_{2}/\widehat{\mathfrak{u}(1)}^{n}$ and
characters of parafermions constructed from scaled root lattices of the type
$A_{n}/\sqrt{2}$. They are then rigorously proved using various techniques of
$q$-series. With these identities, we are also able to derive explicit
expressions for some string functions, which can be hard to compute even with
the machinery given in [18], in a relatively straightforward manner. The
success of proving these specialised $q$-identities further motivates us to
investigate more general $q$-identities. This work once again illustrates the
vibrant collaboration between number theory and mathematical physics.

The physical motivation behind the $q$-series arising in this paper is
discussed in [15, 6].

##  2\. String functions for level-2 ${\mathfrak{sl}}_{n}$ parafermions

###  2.1. String functions and parafermionic characters

Let ${\mathfrak{g}}$ be a finite-dimensional Lie algebra. Denote the
(untwisted) affine Kac–Moody algebra at level $k$, associated to
${\mathfrak{g}}$, by $\widehat{\mathfrak{g}}_{k}$. Let $\Lambda$ be an
integrable dominant weight of $\widehat{\mathfrak{g}}_{k}$, and $\lambda$ be a
maximal weight in the representation with highest weight $\Lambda$. Then the
so-called string functions $c^{\Lambda}_{\lambda}$ are defined in [17] by Kac
and Peterson:

$c^{\Lambda}_{\lambda}(\tau)=q^{s_{\Lambda}(\lambda)}\sum_{n\geq
0}\text{mult}_{\Lambda}(\lambda-n\delta)e^{2\pi i\tau},$

where

$s_{\Lambda}(\lambda)=\dfrac{|\Lambda+\rho|^{2}}{2(k+h^{\vee})}-\dfrac{|\rho|^{2}}{2h^{\vee}}-\dfrac{|\lambda|^{2}}{2k},$

$\rho$ is half of the sum of all positive roots of $\hat{{\mathfrak{g}}}$, and
$h^{\vee}$ is the dual Coxeter number of $\hat{{\mathfrak{g}}}$. The modular
transformation law of string functions are provided in [17] as well.

It is also known that given $\widehat{\mathfrak{g}}_{k}$, one can construct a
parafermionic conformal field theory as described in [12]. Algebraically, this
theory can be thought of the GKO coset construction of the (generalized) WZW
model (see, for example, [10]) based on $\widehat{\mathfrak{g}}_{k}$
constrained by $\widehat{\mathfrak{u}(1)}^{n}$, where
$n=\text{rank}(\mathfrak{g})$ [13]. We denote such a construction as
$\text{PF}_{k}({\mathfrak{g}})$. Explicitly, we define the parafermions
$\psi_{\alpha}$ for every $\alpha\in Q$, where $Q$ is the root lattice of
$\mathfrak{g}$. Identify $\psi_{\alpha}=\psi_{\beta}$ if
$\alpha\equiv\beta\mod kQ$. The OPEs of these parafermions are

$\displaystyle\psi_{\alpha}(z)\psi_{-\alpha}(w)\sim\dfrac{1}{(z-w)^{2-|\alpha|^{2}/k}},$
(2.1)
$\displaystyle\psi_{\alpha}(z)\psi_{\beta}(w)\sim\dfrac{c_{\alpha,\beta}\psi_{\alpha+\beta}(w)}{(z-w)^{1+(\alpha,\beta)/k}},$
(2.2)

where $c_{\alpha,\beta}$ are some constants to be fixed.

We notice that when $k=2$ and ${\mathfrak{g}}$ is simply-laced, this
construction is of particular interest because then the OPEs (2.1) and (2.2)
will reduce to

$\displaystyle\psi_{\alpha}(z)\psi_{\alpha}(w)\sim\dfrac{1}{(z-w)},$ (2.3)
$\displaystyle\psi_{\alpha}(z)\psi_{\beta}(w)\sim\dfrac{c_{\alpha,\beta}\psi_{\alpha+\beta}(w)}{(z-w)^{1/2}},$
(2.4)

if $\alpha+\beta\mod 2Q$ is a root of ${\mathfrak{g}}$, so we have real free
fermions $\psi_{\alpha}$ that couple to each other in a certain way. We may
call such a system as a “coupled free fermion system.”

The modules of $\text{PF}_{k}({\mathfrak{g}})$ are again characterised by
$\Lambda$ and $\lambda$, denoted by $L^{\Lambda}_{\lambda}$. The modules
$L^{\Lambda}_{\lambda}$ and $L^{\Lambda^{\prime}}_{\lambda^{\prime}}$ are
equivalent if

$\lambda^{\prime}\equiv\lambda\mod kQ$

or

$\Lambda^{\prime}=\sigma\Lambda\;\;\text{and}\;\;\lambda^{\prime}=\sigma\lambda,$

where $\sigma$ is an automorphism of the Dynkin diagram of
$\hat{{\mathfrak{g}}}$. The characters (or partition functions) of the modules
are given by

$b^{\Lambda}_{\lambda}(\tau):=\text{Tr}\;e^{\left(L_{0}-c/24\right)2\pi
i\tau}$

restricting to the module $L^{\Lambda}_{\lambda}$, where $L_{0}$ is the
operator that diagonalises the fock space of $\text{PF}_{k}({\mathfrak{g}})$.
It is proven in [12] that the parafermionic characters $b^{\Lambda}_{\lambda}$
of $\text{PF}_{k}({\mathfrak{g}})$ are simply related to the string functions
$c^{\Lambda}_{\lambda}$ by

$b^{\Lambda}_{\lambda}(\tau)=\eta(\tau)^{n}c^{\Lambda}_{\lambda}(\tau),$

where $\eta(\tau)$ is the Dedekind eta function.

Given such a close relation between string functions of
$\widehat{{\mathfrak{g}}}_{k}$ and characters of
$\text{PF}_{k}({\mathfrak{g}})$, we can study them all at once and take
advantages from either side as needed.

In the remaining of this section, we list a few examples of string functions
and parafermionic characters for later reference in this paper. In the
examples, we write $q=e^{2\pi i\tau}$ by convention and use Dynkin lables to
represent the weights.

###  2.2. ${\mathfrak{sl}}_{2}$

The string functions of $(\widehat{\mathfrak{sl}_{2}})_{2}$ have been
explicitly computed in [18] (Example 3 in Section 4.6). Hence we have explicit
expressions for all characters $b^{\Lambda}_{\lambda}$ of
$\text{PF}_{2}(\mathfrak{sl}_{2})$ as follows:

$\displaystyle b^{20}_{20}(\tau)$
$\displaystyle=\frac{1}{2}\big{(}\eta(2\tau)^{-1}\eta(\tau/2)^{-1}\eta(\tau)^{2}+\eta(\tau)^{-1}\eta(\tau/2)\big{)},$
$\displaystyle b^{20}_{02}(\tau)$
$\displaystyle=\frac{1}{2}\big{(}\eta(2\tau)^{-1}\eta(\tau/2)^{-1}\eta(\tau)^{2}-\eta(\tau)^{-1}\eta(\tau/2)\big{)},$
$\displaystyle b^{11}_{11}(\tau)$ $\displaystyle=\eta(\tau)^{-1}\eta(2\tau),$

###  2.3. ${\mathfrak{sl}}_{3}$

The string functions of $(\widehat{\mathfrak{sl}_{3}})_{2}$ have been
explicitly computed in [18] (Example 4 in Section 4.6). Hence we have explicit
expressions for all characters $b^{\Lambda}_{\lambda}$ of
$\text{PF}_{2}(\mathfrak{sl}_{3})$. We claim that they can be rewritten as
follows:

$\displaystyle b_{200}^{200}(\tau)$
$\displaystyle=\eta(\tau)^{-2}q^{\frac{1}{30}}\big{(}(q^{\frac{1}{2}};q^{\frac{1}{2}})_{\infty}(q,q^{\frac{3}{2}},q^{\frac{5}{2}};q^{\frac{5}{2}})_{\infty}+q^{\frac{1}{2}}(q^{2};q^{2})_{\infty}(q^{2},q^{8},q^{10};q^{10})_{\infty}\big{)},$
(2.5a) $\displaystyle b_{011}^{200}(\tau)$
$\displaystyle=\eta(\tau)^{-2}q^{\frac{8}{15}}(q^{2};q^{2})_{\infty}(q^{2},q^{8},q^{10};q^{10})_{\infty},$
(2.5b) $\displaystyle b_{110}^{110}(\tau)$
$\displaystyle=\eta(\tau)^{-2}q^{\frac{2}{15}}(q^{2};q^{2})_{\infty}(q^{4},q^{6},q^{10};q^{10})_{\infty},$
(2.5c) $\displaystyle b_{002}^{110}(\tau)$
$\displaystyle=\eta(\tau)^{-2}q^{\frac{2}{15}}\big{(}(q^{2};q^{2})_{\infty}(q^{4},q^{6},q^{10};q^{10})_{\infty}-(q^{\frac{1}{2}};q^{\frac{1}{2}})_{\infty}(q^{\frac{1}{2}},q^{2},q^{\frac{5}{2}};q^{\frac{5}{2}})_{\infty}\big{)},$
(2.5d)

where the $q$-Pochhammer symbols are defined for
$n\in\mathbb{N}\cup\\{\infty\\}$:

$\displaystyle(a;q)_{n}$ $\displaystyle:=\prod_{j=0}^{n-1}(1-aq^{j}),$
$\displaystyle(a_{1},a_{2},\ldots,a_{r};q)_{n}$
$\displaystyle:=(a_{1};q)_{n}(a_{2};q)_{n}\cdots(a_{r};q)_{n}.$

Note that for notational convenience, we often write hereinafter, with $E$
standing for Euler, that

$\displaystyle E(q):=(q;q)_{\infty}.$

It is worth pointing out that the above four expressions will be examined as
part of the proof of Theorem 3.

###  2.4. ${\mathfrak{sl}}_{4}$

Neither the explicit expressions of string functions of
$(\widehat{\mathfrak{sl}_{4}})_{2}$ nor those of characters of
$\text{PF}_{2}({\mathfrak{sl}}_{4})$ are known in the literature. We claim
that

######  Theorem 1.

Let

$\begin{array}[]{lp{1.6cm}l}\lx@intercol
b^{2000}_{2000}(\tau)+b^{2000}_{0020}(\tau)=\dfrac{\eta(4\tau)^{4}\eta(6\tau)^{8}}{\eta(\tau)\eta(2\tau)^{4}\eta(3\tau)^{3}\eta(12\tau)^{4}}+\dfrac{\eta(2\tau)^{8}\eta(3\tau)\eta(12\tau)^{4}}{\eta(\tau)^{5}\eta(4\tau)^{4}\eta(6\tau)^{4}},\hfil\lx@intercol\\\
b^{2000}_{2000}(\tau)-b^{2000}_{0020}(\tau)=\dfrac{\eta(\tau)^{2}}{\eta(2\tau)^{2}},&&b^{2000}_{0101}(\tau)=\dfrac{\eta(2\tau)^{2}\eta(6\tau)^{2}}{\eta(\tau)^{3}\eta(3\tau)},\\\
b^{0101}_{2000}(\tau)=b^{0101}_{0002}(\tau)=\dfrac{3\eta(6\tau)^{3}}{\eta(\tau)^{2}\eta(2\tau)},&&b^{0101}_{0101}(\tau)=\dfrac{\eta(2\tau)^{3}\eta(3\tau)^{2}}{\eta(\tau)^{4}\eta(6\tau)},\\\
b^{1100}_{1100}(\tau)=\dfrac{\eta(4\tau)^{5}}{\eta(\tau)^{3}\eta(8\tau)^{2}},&&b^{1100}_{0011}(\tau)=\dfrac{2\eta(2\tau)^{2}\eta(8\tau)^{2}}{\eta(\tau)^{3}\eta(4\tau)}.\end{array}$

Then each function $\eta(\tau)^{-3}b^{\Lambda}_{\lambda}$ with
$(\Lambda,\lambda)$ appearing above is an expression of the corresponding
string function $c^{\Lambda}_{\lambda}$ of
$(\widehat{\mathfrak{sl}_{4}})_{2}$. Hence each $b^{\Lambda}_{\lambda}$ is an
expression for the character of the module $L^{\Lambda}_{\lambda}$ of
$\emph{PF}_{2}(\mathfrak{sl}_{4})$.

###### Proof.

Given that these expressions of $b^{\Lambda}_{\lambda}$ are all in terms of
Dedekind eta functions, one can easily compute their modular transformations
under $\tau\mapsto\tau+1$ and $\tau\mapsto-1/\tau$ and compare the results
computed using Theorem A(1) in [18] as follows:

$\displaystyle
b^{2000}_{2000}\left(-\frac{1}{\tau}\right)-b^{2000}_{0020}\left(-\frac{1}{\tau}\right)$
$\displaystyle=2\left(b^{1100}_{1100}(\tau)+b^{1100}_{0011}(\tau)\right)$
$\displaystyle
b^{2000}_{2000}\left(-\frac{1}{\tau}\right)+b^{2000}_{0020}\left(-\frac{1}{\tau}\right)$
$\displaystyle=\dfrac{1}{2\sqrt{3}}\left(b^{2000}_{2000}(\tau)+b^{2000}_{0020}(\tau)+6b^{2000}_{0101}(\tau)+6b^{0101}_{0101}(\tau)+2b^{0101}_{2000}(\tau)\right)$
$\displaystyle
b^{0101}_{2000}\left(-\frac{1}{\tau}\right)=b^{0101}_{0002}\left(-\frac{1}{\tau}\right)$
$\displaystyle=\dfrac{1}{2\sqrt{3}}\left(b^{2000}_{2000}(\tau)+b^{2000}_{0020}(\tau)+6b^{2000}_{0101}(\tau)-3b^{0101}_{0101}(\tau)-b^{0101}_{2000}(\tau)\right)$
$\displaystyle b^{2000}_{0101}\left(-\frac{1}{\tau}\right)$
$\displaystyle=\dfrac{1}{4\sqrt{3}}\left(b^{2000}_{2000}(\tau)+b^{2000}_{0020}(\tau)-2b^{2000}_{0101}(\tau)-2b^{0101}_{0101}(\tau)+2b^{0101}_{2000}(\tau)\right)$
$\displaystyle b^{0101}_{0101}\left(-\frac{1}{\tau}\right)$
$\displaystyle=\dfrac{1}{2\sqrt{3}}\left(b^{2000}_{2000}(\tau)+b^{2000}_{0020}(\tau)-2b^{2000}_{0101}(\tau)+b^{0101}_{0101}(\tau)-b^{0101}_{2000}(\tau)\right)$
$\displaystyle b^{1100}_{1100}\left(-\frac{1}{\tau}\right)$
$\displaystyle=\dfrac{1}{4}\left(b^{2000}_{2000}(\tau)-b^{2000}_{0020}(\tau)+2b^{1100}_{1100}(\tau)-2b^{1100}_{0011}(\tau)\right)$
$\displaystyle b^{1100}_{0011}\left(-\frac{1}{\tau}\right)$
$\displaystyle=\dfrac{1}{4}\left(b^{2000}_{2000}(\tau)-b^{2000}_{0020}(\tau)-2b^{1100}_{1100}(\tau)+2b^{1100}_{0011}(\tau)\right)$

Also, using Theorem A(4)111With a closer look at the proof of Theorem A(3), we
believe Theorem A(4) holds true if we replace all the groups $\Gamma(Nm)$ and
$\Gamma(N(m+g))$ by $\Gamma_{0}(Nm)$ and $\Gamma_{0}(N(m+g))$, respectively.
and Sturm’s bound for modular forms (see, for example, Theorem 6.4.7 in [22]),
we know that comparing coefficients of $\eta(\tau)^{-3}b^{\Lambda}_{\lambda}$
with the numerical result of $c^{\Lambda}_{\lambda}$, computed from its
definition, up to the order of $q^{100}$ will be sufficient to conclude that
$\eta(\tau)^{-3}b^{\Lambda}_{\lambda}=c^{\Lambda}_{\lambda}$. All of above
have been verified with the help of SageMath and Mathematica.∎

In fact, these expressions of $b^{\Lambda}_{\lambda}$ arise in the way of
proving Theorem 4 and 5. The results are now summarised as (3.14), (3.15) and
(3.16).

##  3\. Universal Chiral partition functions for level-2 ${\mathfrak{sl}}_{n}$
parafermions

###  3.1. Universal chiral partition functions

It can be seen from [12] that the representation theory of
$\text{PF}_{2}(\mathfrak{sl}_{n})$ involves intertwiners between various
representations, so it is natural to think of fractional exclusion statistics
where the statistical interactions of intertwiners are encoded into a matrix
$\mathbf{G}$ [14].

The so-called universal chiral partition function (UCPF) is in the form of

$\text{UCPF}\left(\textbf{G};\textbf{a}\right)=q^{\delta}\sum_{N_{i}\geq
0}\dfrac{q^{\frac{1}{2}\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}\cdot\mathbf{N}-\mathbf{a}\cdot\mathbf{N}}}{\prod_{i}(q;q)_{N_{i}}}$

where $q^{\delta}$ is a prefactor,
$\mathbf{N}=(N_{1},N_{2},\dots,N_{m})^{\mathsf{T}}$,
$\mathbf{a}=(a_{1},a_{2},\dots,a_{m})$ and $m$ is the number of quasi-
particles [8].

In the case of $\text{PF}_{2}(\mathfrak{sl}_{n})$, generalised commutation
relations [23, 15, 6] allow us to analyse the exclusion statistics of
intertwiners. This motivates us to conjecture a statistical interaction matrix
$\mathbf{G}$ for $\text{PF}_{2}(\mathfrak{sl}_{n})$. We will examine a few
examples to see the connection between UCPFs and parafermionic characters
$b^{\Lambda}_{\lambda}$.

###  3.2. ${\mathfrak{sl}}_{2}$

In this case the exclusion statistic is trivial, so we define
$\textbf{G}_{2}=\frac{1}{2}(2)$ and we have

######  Theorem 2.

$\displaystyle\emph{UCPF}\left(\mathbb{G}_{2};\mathbb{0}\right)$
$\displaystyle=q^{-\frac{1}{48}}\sum_{N\geq
0}\dfrac{q^{\frac{1}{2}N^{2}}}{(q;q)_{N}}=b^{20}_{20}+b^{20}_{02},$ (3.1a)
$\displaystyle\emph{UCPF}\left(\mathbb{G}_{2};(1/2)\right)$
$\displaystyle=q^{\frac{1}{16}-\frac{1}{48}}\sum_{N\geq
0}\dfrac{q^{\frac{1}{2}(N^{2}-N)}}{(q;q)_{N}}=2b^{11}_{11}.$ (3.1b)

###### Proof.

The two relations are instances of Euler’s second summation formula (A.3b).
For (3.1a), we use (A.3b) with $z=-q^{\frac{1}{2}}$. For (3.1b), we again use
(A.3b) but with $z=-1$. ∎

###  3.3. ${\mathfrak{sl}}_{3}$

According to the explicit analysis of $\text{PF}_{2}(\mathfrak{sl}_{3})$ in
[3, 15, 6], we conjecture that

$\displaystyle\mathbf{G}_{3}:=\dfrac{1}{2}\begin{pmatrix}2&1&1\\\ 1&2&1\\\
1&1&2\end{pmatrix}$ (3.2)

and, noticing that there are (up to equivalence) one untwisted and one twisted
representations of $\text{PF}_{2}(\mathfrak{sl}_{3})$, we consider

$\displaystyle\text{UCPF}\left(\mathbb{G}_{3};\mathbb{0}\right)$
$\displaystyle=q^{-\frac{1}{20}}\sum_{N_{i}\geq
0}\dfrac{q^{\frac{1}{2}\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{3}\cdot\mathbf{N}}}{\prod_{i}(q;q)_{N_{i}}},$
$\displaystyle\text{UCPF}\left(\mathbb{G}_{3};1/2(0,1,1)\right)$
$\displaystyle=q^{\frac{1}{10}-\frac{1}{20}}\sum_{N_{i}\geq
0}\dfrac{q^{\frac{1}{2}\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G_{3}}\cdot\mathbf{N}-\frac{1}{2}(N_{2}+N_{3})}}{\prod_{i}(q;q)_{N_{i}}},$

where

$\displaystyle\mathbf{N}^{\mathsf{T}}=(N_{1},N_{2},N_{3}).$ (3.3)

It is easy to observe relations between these UCPFs and the parafermionic
characters $b^{\Lambda}_{\lambda}$ numerically. In particular, we have the
following result.

######  Theorem 3.

$\displaystyle\emph{UCPF}\left(\mathbb{G}_{3};\mathbb{0}\right)$
$\displaystyle=b_{200}^{200}+3b_{011}^{200},$ (3.4a)
$\displaystyle\emph{UCPF}\left(\mathbb{G}_{3};1/2(0,1,1)\right)$
$\displaystyle=b_{002}^{110}+3b_{110}^{110}.$ (3.4b)

###### Proof.

We begin by noticing that (3.4a) and (3.4b) are, respectively, equivalent to

$\displaystyle\sum_{N_{i}\geq
0}\frac{q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{3}\cdot\mathbf{N}}}{\prod_{i}(q^{2};q^{2})_{N_{i}}}$
$\displaystyle=\frac{(q;q)_{\infty}(q^{2},q^{3},q^{5};q^{5})_{\infty}}{(q^{2};q^{2})^{2}_{\infty}}+\frac{4q(q^{4};q^{4})_{\infty}(q^{4},q^{16},q^{20};q^{20})_{\infty}}{(q^{2};q^{2})^{2}_{\infty}},$
(3.5a) $\displaystyle\sum_{N_{i}\geq
0}\frac{q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{3}\cdot\mathbf{N}-N_{2}-N_{3}}}{\prod_{i}(q^{2};q^{2})_{N_{i}}}$
$\displaystyle=-\frac{(q;q)_{\infty}(q,q^{4},q^{5};q^{5})_{\infty}}{(q^{2};q^{2})^{2}_{\infty}}+\frac{4(q^{4};q^{4})_{\infty}(q^{8},q^{12},q^{20};q^{20})_{\infty}}{(q^{2};q^{2})^{2}_{\infty}}.$
(3.5b)

For (3.5a), we first show that

$\displaystyle\operatorname{LHS}\eqref{eq:triple-
AG-1-2}=\frac{(q^{2};q^{2})_{\infty}(q^{2},q^{8},q^{10};q^{10})_{\infty}(q^{6},q^{10},q^{14};q^{20})_{\infty}}{(q;q)_{\infty}(q^{4};q^{4})_{\infty}(q^{3},q^{5},q^{7};q^{10})_{\infty}}+\frac{2q(q^{4};q^{4})_{\infty}(q^{4},q^{16},q^{20};q^{20})_{\infty}}{(q^{2};q^{2})^{2}_{\infty}},$
(3.6)

where the right-hand side is adopted from [18]. To see this, we take
$(z_{1},z_{2},z_{3})=(1,1,1)$ in (7). Thus,

$\displaystyle\operatorname{LHS}\eqref{eq:triple-AG-1-2}$
$\displaystyle=(-q;q^{2})_{\infty}\sum_{M\geq
0}\frac{q^{3M^{2}}}{(q;q^{2})_{M}(q^{4};q^{4})_{M}}+(-q^{2};q^{2})_{\infty}\sum_{M\geq
0}\frac{2q^{3M^{2}+3M+1}}{(q^{2};q^{2})_{M}(q^{2};q^{4})_{M+1}}.$

Now, using (A.9b) with $q\mapsto-q$ and (A.9c) with $q\mapsto q^{2}$, we
arrive at (3.6). We further rewrite (3.6) with the Rogers–Ramanujan functions
(A.2a) and (A.2b):

$\displaystyle\operatorname{LHS}\eqref{eq:triple-
AG-1-2}=\frac{E(q^{2})^{4}}{E(q)^{2}E(q^{4})^{2}}G(-q)+\frac{2qE(q^{4})^{2}}{E(q^{2})^{2}}H(q^{4}).$

To show (3.5a), it suffices to prove

$\displaystyle\frac{E(q^{2})^{4}}{E(q)^{2}E(q^{4})^{2}}G(-q)G(q^{4})-\frac{E(q)^{2}}{E(q^{2})^{2}}G(q)G(q^{4})=\frac{2qE(q^{4})E(q^{20})}{E(q^{2})^{2}}.$
(3.7)

By (A.8a) and (A.8b), we may obtain an expression of $G(q)G(q^{4})$ and thus
also of $G(-q)G(q^{4})$. Then,

$\displaystyle\operatorname{LHS}\eqref{eq:G-modular-eqn}$
$\displaystyle=\frac{E(q^{2})^{3}E(q^{5})^{2}}{2E(q)^{2}E(q^{4})^{2}E(q^{10})}-\frac{E(q)^{2}E(q^{10})^{5}}{2E(q^{2})^{3}E(q^{5})^{2}E(q^{20})^{2}}$
$\displaystyle=\frac{\phi(q)\phi(-q^{5})-\phi(-q)\phi(q^{5})}{2E(q^{2})^{2}},$

thereby confirming (3.7) via the following modular equation of degree $5$ (See
[5, p. 278]):

$\displaystyle\phi(q)\phi(-q^{5})-\phi(-q)\phi(q^{5})=4qE(q^{4})E(q^{20}).$
(3.8)

Here, Ramanujan’s theta function $\phi(q)$ is defined in (A.1a).

For (3.5b), we first establish that

$\displaystyle\operatorname{LHS}\eqref{eq:triple-
AG-2-2}=\frac{(q^{2};q^{2})_{\infty}(q^{4},q^{6},q^{10};q^{10})_{\infty}(q^{2},q^{10},q^{18};q^{20})_{\infty}}{(q;q)_{\infty}(q^{4};q^{4})_{\infty}(q,q^{5},q^{9};q^{10})_{\infty}}+\frac{2(q^{4};q^{4})_{\infty}(q^{8},q^{12},q^{20};q^{20})_{\infty}}{(q^{2};q^{2})^{2}_{\infty}},$
(3.9)

where the right-hand side is also adopted from [18]. To see this, we take
$(z_{1},z_{2},z_{3})=(1,q^{-1},q^{-1})$ in (7). Thus,

$\displaystyle\operatorname{LHS}\eqref{eq:triple-AG-2-2}$
$\displaystyle=(-q;q^{2})_{\infty}\sum_{M\geq
0}\frac{q^{3M^{2}-2M}}{(q;q^{2})_{M}(q^{4};q^{4})_{M}}+(-q^{2};q^{2})_{\infty}\sum_{M\geq
0}\frac{2q^{3M^{2}+M}}{(q^{2};q^{2})_{M}(q^{2};q^{4})_{M+1}}.$

We then make use of (A.9a) with $q\mapsto-q$ and (A.9d) with $q\mapsto q^{2}$
to derive (3.9). We further rewrite (3.9) in terms of the Rogers–Ramanujan
functions:

$\displaystyle\operatorname{LHS}\eqref{eq:triple-
AG-2-2}=\frac{E(q^{2})^{4}}{E(q)^{2}E(q^{4})^{2}}H(-q)+\frac{2E(q^{4})^{2}}{E(q^{2})^{2}}G(q^{4}).$

So (3.5b) is equivalent to

$\displaystyle\frac{E(q^{2})^{4}}{E(q)^{2}E(q^{4})^{2}}H(-q)H(q^{4})+\frac{E(q)^{2}}{E(q^{2})^{2}}H(q)H(q^{4})=\frac{2E(q^{4})E(q^{20})}{E(q^{2})^{2}}.$
(3.10)

We also deduce from (A.8a) and (A.8b) an expression of $H(q)H(q^{4})$. Then,

$\displaystyle\operatorname{LHS}\eqref{eq:H-modular-
eqn}=\frac{\phi(q)\phi(-q^{5})-\phi(-q)\phi(q^{5})}{2qE(q^{2})^{2}},$

thereby confirming (3.10) with recourse to (3.8). Thus, (3.5b) holds true. ∎

###  3.4. ${\mathfrak{sl}}_{4}$

We can generalise the analysis above to the case of
$\text{PF}_{2}(\mathfrak{sl}_{4})$. In this case, extra work is needed to deal
with the intertwiners related to the module $L^{2000}_{0020}$. Details are
given in [15, 6]. In short, we conjecture that

$\displaystyle\mathbf{G}_{4}:=\frac{1}{2}\begin{pmatrix}2&1&1&0&1&1\\\
1&2&1&1&0&1\\\ 1&1&2&1&1&2\\\ 0&1&1&2&1&1\\\ 1&0&1&1&2&1\\\
1&1&2&1&1&2\end{pmatrix}$ (3.11)

and define, for (up to equivalence) one untwisted and two twisted
representations,

$\displaystyle\text{UCPF}\left(\mathbb{G}_{4};\mathbb{0}\right)$
$\displaystyle:=q^{-\frac{1}{12}}\sum_{N_{i}\geq
0}\dfrac{q^{\frac{1}{2}\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}}}{\prod_{i}(q;q)_{N_{i}}},$
$\displaystyle\text{UCPF}\left(\mathbb{G}_{4};1/2(1,0,1,1,0,1)\right)$
$\displaystyle:=q^{\frac{1}{6}-\frac{1}{12}}\sum_{N_{i}\geq
0}\dfrac{q^{\frac{1}{2}\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-\frac{1}{2}(N_{1}+N_{3}+N_{4}+N_{6})}}{\prod_{i}(q;q)_{N_{i}}},$
$\displaystyle\text{UCPF}\left(\mathbb{G}_{4};1/2(1,1,0,0,0,1)\right)$
$\displaystyle:=q^{\frac{1}{8}-\frac{1}{12}}\sum_{N_{i}\geq
0}\dfrac{q^{\frac{1}{2}\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-\frac{1}{2}(N_{1}+N_{2}+N_{6})}}{\prod_{i}(q;q)_{N_{i}}},$

where

$\displaystyle\mathbf{N}^{\mathsf{T}}=(N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}).$
(3.12)

Then we obtain more identities by connecting UCPFs and parefermionic
characters of $\text{PF}_{2}(\mathfrak{sl}_{4})$.

######  Theorem 4.

$\displaystyle\emph{UCPF}\left(\mathbb{G}_{4};\mathbb{0}\right)$
$\displaystyle=b^{2000}_{2000}+b^{2000}_{0020}+6b^{2000}_{0101},$ (3.13a)
$\displaystyle\emph{UCPF}\left(\mathbb{G}_{4};1/2(1,0,1,1,0,1)\right)$
$\displaystyle=2\left(b^{0101}_{2000}+3b^{0101}_{0101}\right),$ (3.13b)
$\displaystyle\emph{UCPF}\left(\mathbb{G}_{4};1/2(1,1,0,0,0,1)\right)$
$\displaystyle=4\left(b^{1100}_{1100}+b^{1100}_{0011}\right).$ (3.13c)

###### Proof.

We start by noticing that (3.13a) is equivalent to

$\displaystyle\sum_{N_{i}\geq
0}\frac{q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}}}{\prod_{i}(q^{2};q^{2})_{N_{i}}}=\frac{(q^{2};q^{2})^{3}_{\infty}(q^{6};q^{6})^{5}_{\infty}}{(q;q)^{2}_{\infty}(q^{3};q^{3})^{2}_{\infty}(q^{4};q^{4})^{2}_{\infty}(q^{12};q^{12})^{2}_{\infty}}+\frac{4q(q^{4};q^{4})^{2}_{\infty}(q^{12};q^{12})^{2}_{\infty}}{(q^{2};q^{2})^{3}_{\infty}(q^{6};q^{6})_{\infty}}.$
(3.14)

To see this equivalence, we shall also use the fact that

$\displaystyle\frac{E(q^{2})^{3}E(q^{6})^{5}}{E(q)^{2}E(q^{3})^{2}E(q^{4})^{2}E(q^{12})^{2}}+\frac{4qE(q^{4})^{2}E(q^{12})^{2}}{E(q^{2})^{3}E(q^{6})}$
$\displaystyle\quad=\frac{E(q^{8})^{4}E(q^{12})^{8}}{E(q^{2})E(q^{4})^{4}E(q^{6})^{3}E(q^{24})^{4}}+\frac{q^{2}E(q^{4})^{8}E(q^{6})E(q^{24})^{4}}{E(q^{2})^{5}E(q^{8})^{4}E(q^{12})^{4}}+\frac{6qE(q^{4})^{2}E(q^{12})^{2}}{E(q^{2})^{3}E(q^{6})},$

which follows by substituting (A.6c) into the left-hand side of the above.
Now, for (3.14), it suffices to use (5.2) with $\delta_{1}=\delta_{2}=0$ and
$z_{1}=z_{2}=1$, and then apply (20) with $q$ replaced by $q^{2}$.

Next, (3.13b) is equivalent to

$\displaystyle\sum_{N_{i}\geq
0}\frac{q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-(N_{1}+N_{3}+N_{4}+N_{6})}}{\prod_{i}(q^{2};q^{2})_{N_{i}}}=\frac{6(q^{3};q^{3})^{3}_{\infty}}{(q;q)_{\infty}(q^{2};q^{2})^{2}_{\infty}}.$
(3.15)

Further, the equivalence comes from

$\displaystyle\frac{E(q^{3})^{3}}{E(q)E(q^{2})^{2}}$
$\displaystyle=\frac{E(q^{4})^{3}E(q^{6})^{2}}{E(q^{2})^{4}E(q^{12})}+\frac{qE(q^{12})^{3}}{E(q^{2})^{2}E(q^{4})},$

which is true by invoking (A.6g). Now, for (3.15), we use (5.2) with
$\delta_{1}=1$, $\delta_{2}=0$ and $(q,z_{1},z_{2})\mapsto(q^{2},q^{2},1)$.
Then,

$\displaystyle\operatorname{LHS}\eqref{eq:VOA-1-2}$
$\displaystyle=\frac{1}{(q^{2};q^{2})_{\infty}^{2}}\sum_{M_{1},M_{2}=-\infty}^{\infty}q^{M_{1}^{2}+M_{2}^{2}-M_{1}M_{2}-M_{1}}(1+q^{2M_{1}})$
$\displaystyle=\frac{2}{(q^{2};q^{2})_{\infty}^{2}}\sum_{M_{1},M_{2}=-\infty}^{\infty}q^{M_{1}^{2}+M_{2}^{2}-M_{1}M_{2}-M_{1}}$
(by (20))
$\displaystyle=\frac{2}{E(q^{2})^{2}}\left(\frac{E(q^{2})^{7}E(q^{3})E(q^{12})}{E(q)^{3}E(q^{4})^{3}E(q^{6})}+\frac{2E(q^{4})^{3}E(q^{6})^{2}}{E(q^{2})^{2}E(q^{12})}\right)$
(by (A.7a)) $\displaystyle=\frac{6E(q^{3})^{3}}{E(q)E(q^{2})^{2}}.$

Finally, (3.13c) is equivalent to

$\displaystyle\sum_{N_{i}\geq
0}\frac{q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-(N_{1}+N_{2}+N_{6})}}{\prod_{i}(q^{2};q^{2})_{N_{i}}}=\frac{4(q^{2};q^{2})_{\infty}^{2}}{(q;q)_{\infty}^{2}}.$
(3.16)

For the equivalence, we need

$\displaystyle\frac{E(q^{2})^{2}}{E(q)^{2}}$
$\displaystyle=\frac{E(q^{8})^{5}}{E(q^{2})^{3}E(q^{16})^{2}}+\frac{2qE(q^{4})^{2}E(q^{16})^{2}}{E(q^{2})^{3}E(q^{8})},$

which can be shown with recourse to (A.6a). Further, for (3.16), we only need
to take $(z_{1},z_{2})=(1,1)$ in (5.3a). ∎

##  4\. Lattice characters for $\mathbb{Z}_{2}$-orbifolds of scaled
${\mathfrak{sl}}_{n}$-root lattices

###  4.1. Fermionic lattice construction

Denote the root lattice of ${\mathfrak{sl}}_{n+1}$ by $A_{n}$. There is a
standard process to construct a bosonic conformal field theory from $A_{n}$
(see for example, Section 15.6.3 in [10]). We notice that scaling the root
lattice $A_{n}$ by a factor of $1/\sqrt{2}$ results in a conformal field
theory with coupled free fermions [15, 6], denoted by $A_{n}/\sqrt{2}$ . The
modules of $A_{n}/\sqrt{2}$ are in correspondence with
$\left(A_{n}^{*}/\sqrt{2}\right)/\left(A_{n}/\sqrt{2}\right)$, where $L^{*}$
denotes the dual lattice of $L$. The expressions of characters of these
modules are also straightforward. We have

$\chi_{\gamma}(\tau)=\eta(\tau)^{-n}\sum_{v\in
A_{n}/\sqrt{2}+\gamma}q^{\frac{1}{2}|v|^{2}},$ (4.1)

where $\gamma\in\left(A_{n}^{*}/\sqrt{2}\right)/\left(A_{n}/\sqrt{2}\right)$.

For example, in the case of ${\mathfrak{sl}}_{2}$, we have
$\left(A_{1}^{*}/\sqrt{2}\right)/(A_{1}/\sqrt{2})\cong\\{\gamma_{0}=0,\gamma_{1}=\alpha/2\sqrt{2}\\}\cong\\{0,\frac{1}{2}\\}$,
with corresponding characters

$\displaystyle\chi_{0}$
$\displaystyle=\frac{1}{\eta(\tau)}\sum_{n\in\mathbb{Z}}q^{\frac{1}{2}n^{2}}=\eta(\tau)^{4}\eta(\tau/2)^{-2}\eta(2\tau)^{-2}$
$\displaystyle\chi_{1}$
$\displaystyle=\frac{1}{\eta(\tau)}\sum_{n\in\mathbb{Z}}q^{\frac{1}{2}(n+\frac{1}{2})^{2}}=\eta(\tau)^{-2}\eta(2\tau)^{2}.$

In general, it turns out that in order to recover the full set of string
functions, we need to consider a $\mathbb{Z}_{2}$ orbifold of the above
lattice construction, involving both twisted sectors as well as twist
projections. This is described in detail in [15, 6]. In the following we just
list several resulting characters and the relation to string functions and
UCPFs.

###  4.2. ${\mathfrak{sl}}_{3}$

Let $\\{\alpha_{1},\alpha_{2}\\}$ be simple roots of ${\mathfrak{sl}}_{3}$ and
they form a basis of $A_{2}/\sqrt{2}$. Then the dual lattice
$A_{2}^{*}/\sqrt{2}$ has a basis
$\left\\{\frac{1}{3\sqrt{2}}(\alpha_{1}-\alpha_{2}),\frac{1}{\sqrt{2}}\alpha_{1}\right\\}$,
so there are three cosets in
$\left(A_{2}^{*}/\sqrt{2}\right)/\left(A_{2}/\sqrt{2}\right)$, represented by
$\gamma_{0}:=0$, $\gamma_{1}:=\frac{1}{3\sqrt{2}}(\alpha_{1}-\alpha_{2})$ and
$\gamma_{2}:=\frac{2}{3\sqrt{2}}(\alpha_{1}-\alpha_{2})$.

The characters of the orbifold modules without the twist factor are

$\displaystyle\chi_{\gamma_{0}}(\tau)$
$\displaystyle=\eta(\tau)^{-2}\sum_{m,n\in\mathbb{Z}}q^{\frac{1}{2}(m^{2}+n^{2}-mn)},$
$\displaystyle\chi_{\gamma_{1}}(\tau)=\chi_{\gamma_{2}}(\tau)$
$\displaystyle=\eta(\tau)^{-2}q^{\frac{2}{3}}\sum_{m,n\in\mathbb{Z}}q^{\frac{1}{2}(m^{2}+n^{2}-mn+2m)},$
$\displaystyle\chi_{orb}(\tau)$
$\displaystyle=\eta(\tau)^{2}\eta(\tau/2)^{-2}.$

We notice that they are closely related to the UCPFs of
$\text{PF}_{2}(\mathfrak{sl}_{4})$, as we should expect by studying the
conformal field structure of these two theories.

######  Theorem 5.

$\displaystyle\chi_{\gamma_{0}}$
$\displaystyle=\emph{UCPF}\left(\mathbb{G}_{4};\mathbb{0}\right),$ (4.2a)
$\displaystyle 2\chi_{\gamma_{1}}$
$\displaystyle=\emph{UCPF}\left(\mathbb{G}_{4};1/2(1,0,1,1,0,1)\right),$
(4.2b) $\displaystyle 4\chi_{orb}$
$\displaystyle=\emph{UCPF}\left(\mathbb{G}_{4};1/2(1,1,0,0,0,1)\right).$
(4.2c)

###### Proof.

We obtain (4.2a) by taking $\delta_{1}=\delta_{2}=0$ and $z_{1}=z_{2}=1$ in
(5.2). For (4.2b), we know from (5.2) that

$\displaystyle\text{UCPF}\left(\mathbb{G_{4}};1/2(1,0,1,1,0,1)\right)$
$\displaystyle=\frac{2q^{\frac{1}{6}}}{\eta(\tau)^{2}}\sum_{m,n=-\infty}^{\infty}q^{\frac{m^{2}}{2}+\frac{n^{2}}{2}-\frac{mn}{2}-\frac{m}{2}}.$

Making the change of variables $(m,n)\mapsto(n-m,-m-1)$ in the above gives the
desired result. Finally, (4.2c) is a direct consequence of (3.16). ∎

The characters of the orbifold modules, with the twist factor inserted, are

$\displaystyle\chi_{\gamma_{0}}^{T}$
$\displaystyle=\eta^{-2}\sum_{m,n\in\mathbb{Z}}(-1)^{m+n}q^{\frac{1}{2}(m^{2}+n^{2}-mn)},$
(4.3a) $\displaystyle\chi_{\gamma_{1}}^{T}=\chi_{\gamma_{2}}^{T}$
$\displaystyle=\eta^{-2}q^{\frac{2}{3}}\sum_{m,n\in\mathbb{Z}}(-1)^{m+n}q^{\frac{1}{2}(m^{2}+n^{2}-mn+2m)},$
(4.3b) $\displaystyle\chi_{orb}^{T}$
$\displaystyle=\eta(2\tau)^{2}\eta(\tau/2)^{2}\eta(\tau)^{-4}.$ (4.3c)

We again notice some identities between these lattice characters and the
parafermionic characters.

######  Theorem 6.

$\displaystyle\chi_{\gamma_{0}}^{T}$
$\displaystyle=b^{2000}_{2000}+b^{2000}_{0020}-b^{2000}_{0101},$ (4.4a)
$\displaystyle\chi_{\gamma_{1}}^{T}$
$\displaystyle=b^{0101}_{2000}-b^{0101}_{0101},$ (4.4b)
$\displaystyle\chi_{orb}^{T}$ $\displaystyle=b^{1100}_{1100}-b^{1100}_{0011}.$
(4.4c)

###### Proof.

We see that the right-hand side of (4.3a) is an instance of (20):

$\displaystyle\sum_{m,n\in\mathbb{Z}}(-1)^{m+n}q^{\frac{1}{2}(m^{2}+n^{2}-mn)}=\frac{E(q^{\frac{1}{2}})^{2}E(q^{\frac{3}{2}})^{2}}{E(q)E(q^{3})}.$

Therefore, (4.4a) is equivalent to

$\displaystyle\frac{E(q)^{2}E(q^{3})^{2}}{E(q^{2})^{3}E(q^{6})}=\frac{E(q^{8})^{4}E(q^{12})^{8}}{E(q^{2})E(q^{4})^{4}E(q^{6})^{3}E(q^{24})^{4}}+\frac{q^{2}E(q^{4})^{8}E(q^{6})E(q^{24})^{4}}{E(q^{2})^{5}E(q^{8})^{4}E(q^{12})^{4}}-\frac{2qE(q^{4})^{2}E(q^{12})^{2}}{E(q^{2})^{3}E(q^{6})},$

which follows by substituting (A.6d) into the left-hand side of the above.

For the right-hand side of (4.3b), we first change the variables
$(m,n)\mapsto(-n-1,m-n-1)$. Thus,

$\displaystyle\sum_{m,n\in\mathbb{Z}}(-1)^{m+n}q^{\frac{1}{2}(m^{2}+n^{2}-mn+2m)}$
$\displaystyle=q^{-\frac{1}{2}}\sum_{m,n\in\mathbb{Z}}(-1)^{m}q^{\frac{1}{2}(m^{2}+n^{2}-mn-m)}$
(by (20))
$\displaystyle=q^{-\frac{1}{2}}\left(\frac{E(q)^{7}E(q^{\frac{3}{2}})E(q^{6})}{E(q^{\frac{1}{2}})^{3}E(q^{2})^{3}E(q^{3})}-\frac{2E(q^{2})^{3}E(q^{3})^{2}}{E(q)^{2}E(q^{6})}\right)$
(by (A.7b))
$\displaystyle=-\frac{q^{-\frac{1}{2}}E(q^{\frac{1}{2}})^{3}E(q^{3})^{2}}{E(q)^{2}E(q^{\frac{3}{2}})}.$

We then see that (4.4b) is equivalent to

$\displaystyle\frac{E(q)^{3}E(q^{6})^{2}}{E(q^{2})^{4}E(q^{3})}=\frac{E(q^{4})^{3}E(q^{6})^{2}}{E(q^{2})^{4}E(q^{12})}-\frac{3qE(q^{12})^{3}}{E(q^{2})^{2}E(q^{4})},$

which is true by invoking (A.6e).

Finally, (4.4c) is equivalent to

$\displaystyle\frac{E(q)^{2}E(q^{4})^{2}}{E(q^{2})^{4}}=\frac{E(q^{8})^{5}}{E(q^{2})^{3}E(q^{16})^{2}}-\frac{2qE(q^{4})^{2}E(q^{16})^{2}}{E(q^{2})^{3}E(q^{8})},$

which can be shown with recourse to (A.6b). ∎

##  5\. A $q$-rious perspective

As we will see in this section, the previous results can be nicely established
from a $q$-series perspective. They are also curious to the community of
$q$-theorists for their deep connections with identities of Rogers–Ramanujan
type.

We begin with a general transform that rewrites a triple summation related to
$\mathbf{G}_{3}$ in (3.2) into a single summation.

######  Theorem 7.

Let $\mathbf{G}_{3}$ and $\mathbf{N}$ be as in (3.2) and (3.3), respectively.
Then,

$\displaystyle\sum_{N_{1},N_{2},N_{3}\geq
0}\frac{z_{1}^{N_{1}}z_{2}^{N_{2}}z_{3}^{N_{3}}q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{3}\cdot\mathbf{N}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}}$
$\displaystyle\qquad=(-z_{1}q;q^{2})_{\infty}\sum_{M\geq
0}\frac{z_{2}^{M}z_{3}^{M}q^{3M^{2}}(-z_{2}z_{3}^{-1}q;q^{2})_{M}(-z_{2}^{-1}z_{3}q;q^{2})_{M}}{(-z_{1}q;q^{2})_{M}(q^{2};q^{2})_{2M}}$
$\displaystyle\qquad\quad+(-z_{1}q^{2};q^{2})_{\infty}\sum_{M\geq
0}\frac{(z_{2}+z_{3})z_{2}^{M}z_{3}^{M}q^{3M^{2}+3M+1}(-z_{2}z_{3}^{-1}q^{2};q^{2})_{M}(-z_{2}^{-1}z_{3}q^{2};q^{2})_{M}}{(-z_{1}q^{2};q^{2})_{M}(q^{2};q^{2})_{2M+1}}.$
(5.1)

For $\mathbf{G}_{4}$, we are facing sextuple summations, but we unfortunately
have to give up some generality. That is, to bring out valid transforms, it is
necessary to reduce the number of free parameters. However, we still have a
compensation with two sets of entirely different transforms to be established.
In our first result, the sextuple summation is rewritten as a double bilateral
sum.

######  Theorem 8.

Let $\mathbf{G}_{4}$ and $\mathbf{N}$ be as in (3.11) and (3.12),
respectively. Then for $\delta_{1},\delta_{2}\in\\{0,1\\}$,

$\displaystyle\sum_{N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}\geq
0}\frac{z_{1}^{N_{1}+N_{3}-N_{4}-N_{6}}z_{2}^{N_{2}+N_{3}-N_{5}-N_{6}}q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-2\delta_{1}(N_{4}+N_{6})-2\delta_{2}(N_{5}+N_{6})}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}(q^{2};q^{2})_{N_{6}}}$
$\displaystyle\qquad=\frac{1}{(q^{2};q^{2})_{\infty}^{2}}\sum_{M_{1},M_{2}=-\infty}^{\infty}z_{1}^{M_{1}}z_{2}^{M_{2}}q^{M_{1}^{2}+M_{2}^{2}-M_{1}M_{2}}(1+\delta_{1}q^{2M_{1}})(1+\delta_{2}q^{2M_{2}}).$
(5.2)

For the second result, its flavour is similar to Theorem 7.

######  Theorem 9.

Let $\mathbf{G}_{4}$ and $\mathbf{N}$ be as in (3.11) and (3.12),
respectively. Then,

$\displaystyle\sum_{N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}\geq
0}\frac{z_{1}^{N_{1}+N_{3}+N_{4}+N_{6}}z_{2}^{N_{2}+N_{3}+N_{5}+N_{6}}q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-(N_{1}+N_{2}+N_{6})}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}(q^{2};q^{2})_{N_{6}}}$
$\displaystyle\qquad=\frac{(-z_{1}z_{2};q)_{\infty}}{(q;q^{2})_{\infty}}\sum_{M\geq
0}\frac{(-1)^{M}q^{M^{2}}(z_{1}^{2};q^{2})_{M}(z_{2}^{2};q^{2})_{M}}{(q^{2};q^{2})_{M}(-z_{1}z_{2};q)_{2M}}$
$\displaystyle\qquad\quad+\frac{(-z_{1}z_{2};q)_{\infty}}{(q;q^{2})_{\infty}}\sum_{M\geq
0}\frac{(-1)^{M}\big{(}z_{1}+z_{2}\big{)}q^{M^{2}+2M}(z_{1}^{2};q^{2})_{M}(z_{2}^{2};q^{2})_{M}}{(q^{2};q^{2})_{M}(-z_{1}z_{2};q)_{2M+1}},$
(5.3a) $\displaystyle\sum_{N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}\geq
0}\frac{(-1)^{N_{1}+N_{2}+N_{6}}z_{1}^{N_{1}+N_{3}+N_{4}+N_{6}}z_{2}^{N_{2}+N_{3}+N_{5}+N_{6}}q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-(N_{1}+N_{2}+N_{6})}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}(q^{2};q^{2})_{N_{6}}}$
$\displaystyle\qquad=\frac{(z_{1}z_{2};-q)_{\infty}}{(-q;q^{2})_{\infty}}\sum_{M\geq
0}\frac{q^{M^{2}}(z_{1}^{2};q^{2})_{M}(z_{2}^{2};q^{2})_{M}}{(q^{2};q^{2})_{M}(z_{1}z_{2};-q)_{2M}}$
$\displaystyle\qquad\quad-\frac{(-z_{1}z_{2};-q)_{\infty}}{(-q;q^{2})_{\infty}}\sum_{M\geq
0}\frac{\big{(}z_{1}+z_{2}\big{)}q^{M^{2}+2M}(z_{1}^{2};q^{2})_{M}(z_{2}^{2};q^{2})_{M}}{(q^{2};q^{2})_{M}(-z_{1}z_{2};-q)_{2M+1}}.$
(5.3b)

###  5.1. Proof of Theorem 7

Starting with the $N_{1}$-summation, we have

$\displaystyle\operatorname{LHS}\eqref{eq:G3-general}$
$\displaystyle=\sum_{N_{2},N_{3}\geq
0}\frac{z_{2}^{N_{2}}z_{3}^{N_{3}}q^{\binom{N_{2}+N_{3}}{2}+\binom{N_{2}}{2}+\binom{N_{3}}{2}+N_{2}+N_{3}}}{(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}}\sum_{N_{1}\geq
0}\frac{z_{1}^{N_{1}}q^{2\binom{N_{1}}{2}+(N_{2}+N_{3}+1)N_{1}}}{(q^{2};q^{2})_{N_{1}}}$
(by (A.3b)) $\displaystyle=\sum_{N_{2},N_{3}\geq
0}\frac{z_{2}^{N_{2}}z_{3}^{N_{3}}q^{\binom{N_{2}+N_{3}}{2}+\binom{N_{2}}{2}+\binom{N_{3}}{2}+N_{2}+N_{3}}}{(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}}(-z_{1}q^{N_{2}+N_{3}+1};q^{2})_{\infty}$
($M=N_{2}+N_{3}$) $\displaystyle=\sum_{M\geq
0}(-z_{1}q^{M+1};q^{2})_{\infty}q^{\binom{M}{2}+M}\sum_{\begin{subarray}{c}N_{2},N_{3}\geq
0\\\
N_{2}+N_{3}=M\end{subarray}}\frac{z_{2}^{N_{2}}z_{3}^{N_{3}}q^{\binom{N_{2}}{2}+\binom{N_{3}}{2}}}{(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}}$
(by (B.1b)) $\displaystyle=\sum_{M\geq
0}(-z_{1}q^{2M+1};q^{2})_{\infty}q^{\binom{2M}{2}+2M}$
$\displaystyle\quad\quad\times\frac{z_{2}^{M}z_{3}^{M}q^{M^{2}-M}(-z_{2}z_{3}^{-1}q;q^{2})_{M}(-z_{2}^{-1}z_{3}q;q^{2})_{M}}{(q^{2};q^{2})_{2M}}$
$\displaystyle\quad+\sum_{M\geq
0}(-z_{1}q^{2M+2};q^{2})_{\infty}q^{\binom{2M+1}{2}+(2M+1)}$
$\displaystyle\quad\quad\times\frac{(z_{2}+z_{3})z_{2}^{M}z_{3}^{M}q^{M^{2}}(-z_{2}z_{3}^{-1}q^{2};q^{2})_{M}(-z_{2}^{-1}z_{3}q^{2};q^{2})_{M}}{(q^{2};q^{2})_{2M+1}}$
$\displaystyle=(-z_{1}q;q^{2})_{\infty}\sum_{M\geq
0}\frac{z_{2}^{M}z_{3}^{M}q^{3M^{2}}(-z_{2}z_{3}^{-1}q;q^{2})_{M}(-z_{2}^{-1}z_{3}q;q^{2})_{M}}{(-z_{1}q;q^{2})_{M}(q^{2};q^{2})_{2M}}$
$\displaystyle\quad+(-z_{1}q^{2};q^{2})_{\infty}\sum_{M\geq
0}\frac{(z_{2}+z_{3})z_{2}^{M}z_{3}^{M}q^{3M^{2}+3M+1}(-z_{2}z_{3}^{-1}q^{2};q^{2})_{M}(-z_{2}^{-1}z_{3}q^{2};q^{2})_{M}}{(-z_{1}q^{2};q^{2})_{M}(q^{2};q^{2})_{2M+1}},$

thereby establishing the desired result. ∎

###  5.2. Proof of Theorem 8

We begin with the following change of variables

$\displaystyle\left\\{\begin{aligned} M_{1}&=N_{1}+N_{3}-N_{4}-N_{6},\\\
M_{2}&=N_{2}+N_{3}-N_{5}-N_{6},\\\ M_{3}&=N_{4}+N_{6},\\\
M_{4}&=N_{5}+N_{6},\end{aligned}\right.\quad\Longleftrightarrow\quad\left\\{\begin{aligned}
N_{1}&=M_{1}+M_{3}-N_{3},\\\ N_{2}&=M_{2}+M_{4}-N_{3},\\\
N_{4}&=M_{3}-N_{6},\\\ N_{5}&=M_{4}-N_{6}.\end{aligned}\right.$ (5.4)

Then the summation on the left-hand side of (5.2) is rewritten over indices
$(M_{1},M_{2},M_{3},M_{4},N_{3},N_{6})$:

$\displaystyle\operatorname{LHS}\eqref{eq:VOA-z-1}$
$\displaystyle=\sum_{M_{1},M_{2}=-\infty}^{\infty}z_{1}^{M_{1}}z_{2}^{M_{2}}\sum_{M_{3},M_{4}\geq
0}q^{-2\delta_{1}M_{3}-2\delta_{2}M_{4}}$ $\displaystyle\quad\times
q^{M_{1}^{2}+M_{2}^{2}+2M_{3}^{2}+2M_{4}^{2}+M_{1}M_{2}+2M_{1}M_{3}+2M_{1}M_{4}+2M_{2}M_{3}+2M_{2}M_{4}+4M_{3}M_{4}}$
$\displaystyle\quad\times\sum_{N_{3}\geq
0}\frac{q^{2N_{3}^{2}-2N_{3}(M_{1}+M_{2}+M_{3}+M_{4})}}{(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{M_{1}+M_{3}-N_{3}}(q^{2};q^{2})_{M_{2}+M_{4}-N_{3}}}$
$\displaystyle\quad\times\sum_{N_{6}\geq
0}\frac{q^{2N_{6}^{2}-2N_{6}(M_{3}+M_{4})}}{(q^{2};q^{2})_{N_{6}}(q^{2};q^{2})_{M_{3}-N_{6}}(q^{2};q^{2})_{M_{4}-N_{6}}}$
(by (A.11))
$\displaystyle=\sum_{M_{1},M_{2}=-\infty}^{\infty}z_{1}^{M_{1}}z_{2}^{M_{2}}\sum_{M_{3},M_{4}\geq
0}q^{-2\delta_{1}M_{3}-2\delta_{2}M_{4}}$ $\displaystyle\quad\times
q^{M_{1}^{2}+M_{2}^{2}+2M_{3}^{2}+2M_{4}^{2}+M_{1}M_{2}+2M_{1}M_{3}+2M_{1}M_{4}+2M_{2}M_{3}+2M_{2}M_{4}+4M_{3}M_{4}}$
$\displaystyle\quad\times\frac{q^{-2(M_{1}+M_{3})(M_{2}+M_{4})}}{(q^{2};q^{2})_{M_{1}+M_{3}}(q^{2};q^{2})_{M_{2}+M_{4}}}\cdot\frac{q^{-2M_{3}M_{4}}}{(q^{2};q^{2})_{M_{3}}(q^{2};q^{2})_{M_{4}}}$
$\displaystyle=\sum_{M_{1},M_{2}=-\infty}^{\infty}z_{1}^{M_{1}}z_{2}^{M_{2}}q^{M_{1}^{2}+M_{2}^{2}-M_{1}M_{2}}$
$\displaystyle\quad\times\sum_{M_{3}\geq
0}\frac{q^{2M_{3}^{2}+2(M_{1}-\delta_{1})M_{3}}}{(q^{2};q^{2})_{M_{3}}(q^{2};q^{2})_{M_{1}+M_{3}}}\sum_{M_{4}\geq
0}\frac{q^{2M_{4}^{2}+2(M_{2}-\delta_{2})M_{4}}}{(q^{2};q^{2})_{M_{4}}(q^{2};q^{2})_{M_{2}+M_{4}}}$
(by (A.12))
$\displaystyle=\frac{1}{(q^{2};q^{2})_{\infty}^{2}}\sum_{M_{1},M_{2}=-\infty}^{\infty}z_{1}^{M_{1}}z_{2}^{M_{2}}q^{M_{1}^{2}+M_{2}^{2}-M_{1}M_{2}}(1+\delta_{1}q^{2M_{1}})(1+\delta_{2}q^{2M_{2}}),$

confirming the desired result. ∎

###  5.3. Proof of Theorem 9

The proof of Theorem 9 is entirely different from that for Theorem 8. For
(5.3a), starting with the inner summation in $N_{6}$ by using (A.3b), we have,
by also recalling Lemma 22,

$\displaystyle\operatorname{LHS}\eqref{eq:VOA-1-3}$
$\displaystyle=\sum_{N_{1},N_{2},N_{3},N_{4},N_{5}\geq
0}\frac{z_{1}^{N_{1}+N_{3}+N_{4}}z_{2}^{N_{2}+N_{3}+N_{5}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}}$
$\displaystyle\quad\times
q^{N_{1}^{2}+N_{2}^{2}+N_{3}^{2}+N_{4}^{2}+N_{5}^{2}+N_{1}N_{2}+N_{1}N_{3}+N_{1}N_{5}+N_{2}N_{3}+N_{2}N_{4}+N_{3}N_{4}+N_{3}N_{5}+N_{4}N_{5}-N_{1}-N_{2}}$
$\displaystyle\quad\times\sum_{N_{6}\geq
0}\frac{(z_{1}z_{2})^{N_{6}}q^{2\binom{N_{6}}{2}+(N_{1}+N_{2}+2N_{3}+N_{4}+N_{5})N_{6}}}{(q^{2};q^{2})_{N_{6}}}$
$\displaystyle=\sum_{N_{1},N_{2},N_{3},N_{4},N_{5}\geq
0}\frac{z_{1}^{N_{1}+N_{3}+N_{4}}z_{2}^{N_{2}+N_{3}+N_{5}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}}$
$\displaystyle\quad\times
q^{N_{1}^{2}+N_{2}^{2}+N_{3}^{2}+N_{4}^{2}+N_{5}^{2}+N_{1}N_{2}+N_{1}N_{3}+N_{1}N_{5}+N_{2}N_{3}+N_{2}N_{4}+N_{3}N_{4}+N_{3}N_{5}+N_{4}N_{5}-N_{1}-N_{2}}$
$\displaystyle\quad\times(-z_{1}z_{2}q^{N_{1}+N_{2}+2N_{3}+N_{4}+N_{5}};q^{2})_{\infty}$
$\displaystyle=\sum_{M\geq 0}(-z_{1}z_{2}q^{M};q^{2})_{\infty}T_{M}$
$\displaystyle=\sum_{M\geq
0}(-z_{1}z_{2}q^{2M};q^{2})_{\infty}T_{2M}+\sum_{M\geq
0}(-z_{1}z_{2}q^{2M+1};q^{2})_{\infty}T_{2M+1}$
$\displaystyle=(-z_{1}z_{2};q^{2})_{\infty}\sum_{M\geq
0}\frac{z_{1}^{M}z_{2}^{M}q^{M^{2}}(-z_{1}z_{2}^{-1};q^{2})_{M}(-z_{1}^{-1}z_{2};q^{2})_{M}}{(q^{2};q^{2})_{M}(q;q^{2})_{M}(-z_{1}z_{2};q^{2})_{M}}$
$\displaystyle\quad+(-z_{1}z_{2}q;q^{2})_{\infty}\sum_{M\geq
0}\frac{\big{(}z_{1}+z_{2}\big{)}z_{1}^{M}z_{2}^{M}q^{M(M+1)}(-z_{1}z_{2}^{-1}q;q^{2})_{M}(-z_{1}^{-1}z_{2}q;q^{2})_{M}}{(q^{2};q^{2})_{M}(q;q^{2})_{M+1}(-z_{1}z_{2}q;q^{2})_{M}}$
$\displaystyle=(-z_{1}z_{2};q^{2})_{\infty}\cdot\lim_{\tau\to
0}{}_{3}\phi_{2}\left(\begin{matrix}1/\tau,-z_{1}z_{2}^{-1},-z_{1}^{-1}z_{2}\\\
-z_{1}z_{2},q\end{matrix};q^{2},-z_{1}z_{2}q\tau\right)$
$\displaystyle\quad+(-z_{1}z_{2}q;q^{2})_{\infty}\cdot\frac{z_{1}+z_{2}}{1-q}\cdot\lim_{\tau\to
0}{}_{3}\phi_{2}\left(\begin{matrix}1/\tau,-z_{1}z_{2}^{-1}q,-z_{1}^{-1}z_{2}q\\\
-z_{1}z_{2}q,q^{3}\end{matrix};q^{2},-z_{1}z_{2}q^{2}\tau\right)$
$\displaystyle=(-z_{1}z_{2};q^{2})_{\infty}\cdot\frac{(-z_{1}z_{2}q;q^{2})_{\infty}}{(q;q^{2})_{\infty}}\lim_{\tau\to
0}{}_{3}\phi_{2}\left(\begin{matrix}1/\tau,z_{2}^{2},z_{1}^{2}\\\
-z_{1}z_{2},-z_{1}z_{2}q\end{matrix};q^{2},q\tau\right)$
$\displaystyle\quad+(-z_{1}z_{2}q;q^{2})_{\infty}\cdot\frac{z_{1}+z_{2}}{1-q}\cdot\frac{(-z_{1}z_{2}q^{2};q^{2})_{\infty}}{(q^{3};q^{2})_{\infty}}\lim_{\tau\to
0}{}_{3}\phi_{2}\left(\begin{matrix}1/\tau,z_{2}^{2},z_{1}^{2}\\\
-z_{1}z_{2}q,-z_{1}z_{2}q^{2}\end{matrix};q^{2},q^{3}\tau\right)$
$\displaystyle=\frac{(-z_{1}z_{2};q)_{\infty}}{(q;q^{2})_{\infty}}\sum_{M\geq
0}\frac{(-1)^{M}q^{M^{2}}(z_{1}^{2};q^{2})_{M}(z_{2}^{2};q^{2})_{M}}{(q^{2};q^{2})_{M}(-z_{1}z_{2};q)_{2M}}$
$\displaystyle\quad+\frac{(-z_{1}z_{2};q)_{\infty}}{(q;q^{2})_{\infty}}\sum_{M\geq
0}\frac{(-1)^{M}\big{(}z_{1}+z_{2}\big{)}q^{M^{2}+2M}(z_{1}^{2};q^{2})_{M}(z_{2}^{2};q^{2})_{M}}{(q^{2};q^{2})_{M}(-z_{1}z_{2};q)_{2M+1}},$

confirming (5.3a), where we make use of (A.10).

For (5.3b), we also calculate the inner summation over $N_{6}$ by using (A.3b)
and then apply Lemma 22. Thus,

$\displaystyle\operatorname{LHS}\eqref{eq:VOA-1-3-T-1}$
$\displaystyle=\sum_{N_{1},N_{2},N_{3},N_{4},N_{5}\geq
0}\frac{(-1)^{N_{1}+N_{2}}z_{1}^{N_{1}+N_{3}+N_{4}}z_{2}^{N_{2}+N_{3}+N_{5}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}}$
$\displaystyle\quad\times
q^{N_{1}^{2}+N_{2}^{2}+N_{3}^{2}+N_{4}^{2}+N_{5}^{2}+N_{1}N_{2}+N_{1}N_{3}+N_{1}N_{5}+N_{2}N_{3}+N_{2}N_{4}+N_{3}N_{4}+N_{3}N_{5}+N_{4}N_{5}-N_{1}-N_{2}}$
$\displaystyle\quad\times(z_{1}z_{2}q^{N_{1}+N_{2}+2N_{3}+N_{4}+N_{5}};q^{2})_{\infty}$
$\displaystyle=\sum_{M\geq 0}(z_{1}z_{2}q^{M};q^{2})_{\infty}\tilde{T}_{M}$
$\displaystyle=\sum_{M\geq
0}(z_{1}z_{2}q^{2M};q^{2})_{\infty}\tilde{T}_{2M}+\sum_{M\geq
0}(z_{1}z_{2}q^{2M+1};q^{2})_{\infty}\tilde{T}_{2M+1}$
$\displaystyle=(z_{1}z_{2};q^{2})_{\infty}\sum_{M\geq
0}\frac{z_{1}^{M}z_{2}^{M}q^{M^{2}}(z_{1}z_{2}^{-1};q^{2})_{M}(z_{1}^{-1}z_{2};q^{2})_{M}}{(q^{2};q^{2})_{M}(-q;q^{2})_{M}(z_{1}z_{2};q^{2})_{M}}$
$\displaystyle\quad-(z_{1}z_{2}q;q^{2})_{\infty}\sum_{M\geq
0}\frac{\big{(}z_{1}+z_{2}\big{)}z_{1}^{M}z_{2}^{M}q^{M(M+1)}(z_{1}z_{2}^{-1}q;q^{2})_{M}(z_{1}^{-1}z_{2}q;q^{2})_{M}}{(q^{2};q^{2})_{M}(-q;q^{2})_{M+1}(z_{1}z_{2}q;q^{2})_{M}}$
$\displaystyle=(z_{1}z_{2};q^{2})_{\infty}\cdot\lim_{\tau\to
0}{}_{3}\phi_{2}\left(\begin{matrix}1/\tau,z_{1}z_{2}^{-1},z_{1}^{-1}z_{2}\\\
z_{1}z_{2},-q\end{matrix};q^{2},-z_{1}z_{2}q\tau\right)$
$\displaystyle\quad-(z_{1}z_{2}q;q^{2})_{\infty}\cdot\frac{z_{1}+z_{2}}{1+q}\cdot\lim_{\tau\to
0}{}_{3}\phi_{2}\left(\begin{matrix}1/\tau,z_{1}z_{2}^{-1}q,z_{1}^{-1}z_{2}q\\\
z_{1}z_{2}q,-q^{3}\end{matrix};q^{2},-z_{1}z_{2}q^{2}\tau\right)$
$\displaystyle=(z_{1}z_{2};q^{2})_{\infty}\cdot\frac{(-z_{1}z_{2}q;q^{2})_{\infty}}{(-q;q^{2})_{\infty}}\lim_{\tau\to
0}{}_{3}\phi_{2}\left(\begin{matrix}1/\tau,z_{2}^{2},z_{1}^{2}\\\
z_{1}z_{2},-z_{1}z_{2}q\end{matrix};q^{2},-q\tau\right)$
$\displaystyle\quad-(z_{1}z_{2}q;q^{2})_{\infty}\cdot\frac{z_{1}+z_{2}}{1+q}\cdot\frac{(-z_{1}z_{2}q^{2};q^{2})_{\infty}}{(-q^{3};q^{2})_{\infty}}\lim_{\tau\to
0}{}_{3}\phi_{2}\left(\begin{matrix}1/\tau,z_{2}^{2},z_{1}^{2}\\\
z_{1}z_{2}q,-z_{1}z_{2}q^{2}\end{matrix};q^{2},-q^{3}\tau\right)$
$\displaystyle=\frac{(z_{1}z_{2};-q)_{\infty}}{(-q;q^{2})_{\infty}}\sum_{M\geq
0}\frac{q^{M^{2}}(z_{1}^{2};q^{2})_{M}(z_{2}^{2};q^{2})_{M}}{(q^{2};q^{2})_{M}(z_{1}z_{2};-q)_{2M}}$
$\displaystyle\quad-\frac{(-z_{1}z_{2};-q)_{\infty}}{(-q;q^{2})_{\infty}}\sum_{M\geq
0}\frac{\big{(}z_{1}+z_{2}\big{)}q^{M^{2}+2M}(z_{1}^{2};q^{2})_{M}(z_{2}^{2};q^{2})_{M}}{(q^{2};q^{2})_{M}(-z_{1}z_{2};-q)_{2M+1}},$

confirming (5.3b), where we also apply (A.10). ∎

##  6\. Outlook

As we have seen above, the universal chiral partition function can be a bridge
to obtain closed forms of string functions, which is generally a hard
question. We are interested in generalising this method to higher rank affine
Lie algebras, but currently we are facing some obstacles of determining the
statistical interaction matrix G for higher rank cases analogously. We may
need more tools such as tensors to achieve the goal.

It is also intriguing that we have found an equivalence between the coset
construction $\text{PF}_{2}(\mathfrak{sl}_{4})$ and the
$\mathbb{Z}_{2}$-orbifold of the lattice construction $A_{2}/\sqrt{2}$, which
integrates three major methods of constructing conformal field theories. The
structure of these two theories is examined in more details in [15, 6].

From the $q$-series side, this work provides some insights into the modular
properties of $q$-summations. It is in general not easy to construct
$q$-summations with modular properties, but with the considerations of
Kac–Moody algebra, one may get a basic idea of how these summations would look
like, thereby even leading to interesting $q$-hypergeometric transforms.

Yet another dynamic area that has a close bond with the Rogers–Ramanujan type
identities is the theory of integer partitions. It is known that one of the
two original identities due to Rogers [28] and Ramanujan [24],

$\displaystyle\sum_{n\geq
0}\frac{q^{n^{2}}}{(q;q)_{n}}=\frac{1}{(q,q^{4};q^{5})_{\infty}},$

has the following partition-theoretic interpretation (See Corollary 7.6 in
[1]):

> The number of partitions of a nonnegative integer $n$ into parts congruent
> to $\pm 1$ modulo $5$ is the same as the number of partitions of $n$ such
> that the adjacent parts differ by at least $2$.

We also expect that the summation side of our identities of Rogers–Ramanujan
type would have similar meaning as partitions, perhaps under analogous but
more complicated gap conditions for the parts.

Finally, some conjectural Rogers–Ramanujan type identities in terms of integer
partitions were proposed in a recent seminal paper due to Kanade and Russell
[19]. They had pointed out that three of their Mod $9$ Conjectures, which
remain open, are related to the principally specialized characters of level 3
standard modules for the affine Lie algebra $D_{4}^{(3)}$. More recently, the
analytic counterparts of these conjectures were independently constructed by
Kurşungöz [20], and Chern and Li [9]. Again, these analytic identities have
the form of “$q$-summation $=$ $q$-series infinite product.” We hope that a
deeper look at the summations in the analytic counterparts of the
Kanade–Russell Conjectures from the Lie-algebraic side would reveal more
information and therefore open a window for their proofs.

##  Appendix A $q$-Series prerequisites

For the proofs of the transforms in Section 5, as well as the aforementioned
Rogers–Ramanujan type identities, it is necessary to record some results from
the standard theory of $q$-series. Recall that the $q$-hypergeometric function
${}_{r}\phi_{s}$ is defined by

$\displaystyle{}_{r}\phi_{s}\left(\begin{matrix}A_{1},A_{2},\ldots,A_{r}\\\
B_{1},B_{2},\ldots,B_{s}\end{matrix};q,z\right):=\sum_{n\geq
0}\frac{(A_{1},A_{2},\ldots,A_{r};q)_{n}\big{(}(-1)^{n}q^{\binom{n}{2}}\big{)}^{s-r+1}z^{n}}{(q,B_{1},B_{2},\ldots,B_{s};q)_{n}}.$

Meanwhile, we introduce Ramanujan’s theta functions

$\displaystyle\phi(q)$
$\displaystyle:=\sum_{n=-\infty}^{\infty}q^{n^{2}}=\frac{(q^{2};q^{2})^{5}_{\infty}}{(q;q)^{2}_{\infty}(q^{4};q^{4})^{2}_{\infty}},$
(A.1a) $\displaystyle\psi(q)$ $\displaystyle:=\sum_{n\geq
0}q^{\frac{n(n+1)}{2}}=\frac{(q^{2};q^{2})^{2}_{\infty}}{(q;q)_{\infty}},$
(A.1b)

and the Rogers–Ramanujan functions

$\displaystyle G(q)$ $\displaystyle:=\frac{1}{(q,q^{4};q^{5})_{\infty}},$
(A.2a) $\displaystyle H(q)$
$\displaystyle:=\frac{1}{(q^{2},q^{3};q^{5})_{\infty}}.$ (A.2b)

######  Lemma 10 (Euler’s summations).

We have

$\displaystyle\frac{1}{(z;q)_{\infty}}$ $\displaystyle=\sum_{n\geq
0}\frac{z^{n}}{(q;q)_{n}},$ (A.3a) $\displaystyle(z;q)_{\infty}$
$\displaystyle=\sum_{n\geq 0}\frac{(-1)^{n}z^{n}q^{\binom{n}{2}}}{(q;q)_{n}}.$
(A.3b)

###### Proof.

See Andrews [1, (2.2.5) and (2.2.6)]. ∎

######  Lemma 11 (Jacobi’s triple product identity).

We have

$\displaystyle(q,z,q/z;q)_{\infty}=\sum_{n=-\infty}^{\infty}(-1)^{n}z^{n}q^{\binom{n}{2}}.$
(A.4)

###### Proof.

See Andrews [1, (2.2.10)]. ∎

######  Lemma 12 ($q$-Binomial theorem).

We have

$\displaystyle(z;q)_{N}=\sum_{n\geq
0}(-1)^{n}z^{n}q^{\binom{n}{2}}\begin{bmatrix}N\\\ n\end{bmatrix}_{q}.$ (A.5)

###### Proof.

See Andrews [1, (3.3.6)]. ∎

######  Lemma 13.

We have

$\displaystyle\frac{1}{E(q)^{2}}$
$\displaystyle=\frac{E(q^{8})^{5}}{E(q^{2})^{5}E(q^{16})^{2}}+\frac{2qE(q^{4})^{2}E(q^{16})^{2}}{E(q^{2})^{5}E(q^{8})},$
(A.6a) $\displaystyle E(q)^{2}$
$\displaystyle=\frac{E(q^{2})E(q^{8})^{5}}{E(q^{4})^{2}E(q^{16})^{2}}-\frac{2qE(q^{2})E(q^{16})^{2}}{E(q^{8})},$
(A.6b) $\displaystyle\frac{1}{E(q)E(q^{3})}$
$\displaystyle=\frac{E(q^{8})^{2}E(q^{12})^{5}}{E(q^{2})^{2}E(q^{4})E(q^{6})^{4}E(q^{24})^{2}}+\frac{qE(q^{4})^{5}E(q^{24})^{2}}{E(q^{2})^{4}E(q^{6})^{2}E(q^{8})^{2}E(q^{12})},$
(A.6c) $\displaystyle E(q)E(q^{3})$
$\displaystyle=\frac{E(q^{2})E(q^{8})^{2}E(q^{12})^{4}}{E(q^{4})^{2}E(q^{6})E(q^{24})^{2}}-\frac{qE(q^{4})^{4}E(q^{6})E(q^{24})^{2}}{E(q^{2})E(q^{8})^{2}E(q^{12})^{2}},$
(A.6d) $\displaystyle\frac{E(q)^{3}}{E(q^{3})}$
$\displaystyle=\frac{E(q^{4})^{3}}{E(q^{12})}-\frac{3qE(q^{2})^{2}E(q^{12})^{3}}{E(q^{4})E(q^{6})^{2}},$
(A.6e) $\displaystyle\frac{E(q^{3})}{E(q)^{3}}$
$\displaystyle=\frac{E(q^{4})^{6}E(q^{6})^{3}}{E(q^{2})^{9}E(q^{12})^{2}}+\frac{3qE(q^{4})^{2}E(q^{6})E(q^{12})^{2}}{E(q^{2})^{7}},$
(A.6f) $\displaystyle\frac{E(q^{3})^{3}}{E(q)}$
$\displaystyle=\frac{E(q^{4})^{3}E(q^{6})^{2}}{E(q^{2})^{2}E(q^{12})}+\frac{qE(q^{12})^{3}}{E(q^{4})}.$
(A.6g)

###### Proof.

For (A.6a), see [16, (1.9.4)]. Also, (A.6b) follows by taking $q\mapsto-q$ in
(A.6a). For (A.6c) and (A.6d), see [16, (30.12.3) and (30.12.1)]. For (A.6e),
see [16, (22.7.3)]. Also, (A.6f) follows by taking $q\mapsto-q$ in (A.6e). For
(A.6g), see [16, (22.7.5)]. ∎

######  Lemma 14.

We have

$\displaystyle\frac{E(q^{2})^{7}E(q^{3})E(q^{12})}{E(q)^{3}E(q^{4})^{3}E(q^{6})}+\frac{2E(q^{4})^{3}E(q^{6})^{2}}{E(q^{2})^{2}E(q^{12})}$
$\displaystyle=\frac{3E(q^{3})^{3}}{E(q)},$ (A.7a)
$\displaystyle\frac{E(q^{2})^{7}E(q^{3})E(q^{12})}{E(q)^{3}E(q^{4})^{3}E(q^{6})}-\frac{2E(q^{4})^{3}E(q^{6})^{2}}{E(q^{2})^{2}E(q^{12})}$
$\displaystyle=-\frac{E(q)^{3}E(q^{6})^{2}}{E(q^{2})^{2}E(q^{3})}.$ (A.7b)

###### Proof.

For (A.7a), we substitute (A.6f) and (A.6g) into this relation and find that
the two sides are equal. Also, (A.7b) follows by invoking (A.6e) and (A.6f). ∎

######  Lemma 15.

We have

$\displaystyle G(q)G(q^{4})+qH(q)H(q^{4})$
$\displaystyle=\frac{E(q^{2})^{4}}{E(q)^{2}E(q^{4})^{2}},$ (A.8a)
$\displaystyle G(q)G(q^{4})-qH(q)H(q^{4})$
$\displaystyle=\frac{E(q^{10})^{5}}{E(q^{2})E(q^{5})^{2}E(q^{20})^{2}}.$
(A.8b)

###### Proof.

The two identities are among a set of 40 identities compiled by Ramanujan in
his Lost Notebook [25, pp. 236–237]. See also Hirschhorn [16, (17.4.8) and
(17.4.9)]. ∎

######  Lemma 16.

We have

$\displaystyle\sum_{n\geq
0}\frac{(-1)^{n}q^{3n^{2}-2n}}{(-q;q^{2})_{n}(q^{4};q^{4})_{n}}$
$\displaystyle=\frac{(q,q^{4},q^{5};q^{5})_{\infty}}{(q^{2};q^{2})_{\infty}},$
(A.9a) $\displaystyle\sum_{n\geq
0}\frac{(-1)^{n}q^{3n^{2}}}{(-q;q^{2})_{n}(q^{4};q^{4})_{n}}$
$\displaystyle=\frac{(q^{2},q^{3},q^{5};q^{5})_{\infty}}{(q^{2};q^{2})_{\infty}},$
(A.9b) $\displaystyle\sum_{n\geq
0}\frac{q^{(3n^{2}+3n)/2}}{(q;q)_{n}(q;q^{2})_{n+1}}$
$\displaystyle=\frac{(q^{2},q^{8},q^{10};q^{10})_{\infty}}{(q;q)_{\infty}},$
(A.9c) $\displaystyle\sum_{n\geq
0}\frac{q^{(3n^{2}+n)/2}}{(q;q)_{n}(q;q^{2})_{n+1}}$
$\displaystyle=\frac{(q^{4},q^{6},q^{10};q^{10})_{\infty}}{(q;q)_{\infty}}.$
(A.9d)

###### Proof.

These identities are all due to Rogers. See [29, p. 330 (5)] for (A.9a), [27,
p. 339, Ex. 2] for (A.9b), [29, p. 330 (2), line 2] for (A.9c), and [29, p.
330 (2), line 1] for (A.9d). The first three identities also appear as
equations (15), (19) and (44) on Slater’s famous list [30]. See also equations
(2.5.5), (2.5.7), (2.10.4) and (2.10.6) on a more comprehensive and up-to-date
list due to Mc Laughlin, Sills and Zimmer [21]. ∎

######  Lemma 17.

We have

$\displaystyle{}_{3}\phi_{2}\left(\begin{matrix}a,b,c\\\
d,e\end{matrix};q,\frac{de}{abc}\right)=\frac{(e/a,de/(bc);q)_{\infty}}{(e,de/(abc);q)_{\infty}}{}_{3}\phi_{2}\left(\begin{matrix}a,d/b,d/c\\\
d,de/(bc)\end{matrix};q,\frac{e}{a}\right).$ (A.10)

###### Proof.

See Gasper and Rahman [11, (III.9)]. ∎

######  Lemma 18.

For any integers $M,N\geq 0$,

$\displaystyle\sum_{n\geq
0}\frac{q^{n^{2}-(M+N)n}}{(q;q)_{n}(q;q)_{M-n}(q;q)_{N-n}}=\frac{q^{-MN}}{(q;q)_{M}(q;q)_{N}}.$
(A.11)

###### Proof.

This is the $(a,c)\mapsto(q^{-M},0)$ specialization of the second
$q$-Chu–Vandermonde summation [11, (II.6)]:

$\displaystyle{}_{2}\phi_{1}\left(\begin{matrix}a,q^{-N}\\\
c\end{matrix};q,q\right)=\frac{(c/a;q)_{N}a^{N}}{(c;q)_{N}}.$

See also [7, Lemma D.3]. ∎

######  Lemma 19.

Let $\delta\in\\{0,1\\}$. For any integer $N$,

$\displaystyle\sum_{n\geq
0}\frac{q^{n^{2}+n(N-\delta)}}{(q;q)_{n}(q;q)_{n+N}}$
$\displaystyle=\frac{1+\delta q^{N}}{(q;q)_{\infty}}.$ (A.12)

###### Proof.

We first consider the case where $N\geq 0$. For $\delta=0$, we have

$\displaystyle\sum_{n\geq 0}\frac{q^{n^{2}+nN}}{(q;q)_{n}(q;q)_{n+N}}$
$\displaystyle=\frac{1}{(q;q)_{N}}\sum_{n\geq
0}\frac{q^{2\binom{n}{2}+n(N+1)}}{(q;q)_{n}(q^{N+1};q)_{n}}$
$\displaystyle=\frac{1}{(q;q)_{N}}\lim_{\tau\to
0}{}_{2}\phi_{1}\left(\begin{matrix}1/\tau,1/\tau\\\
q^{N+1}\end{matrix};q,q^{N+1}\tau^{2}\right)$
$\displaystyle=\frac{1}{(q;q)_{N}}\frac{1}{(q^{N+1};q)_{\infty}}$
$\displaystyle=\frac{1}{(q;q)_{\infty}}.$

Here we apply the $q$-Gauß summation [11, (II.8)]:

$\displaystyle{}_{2}\phi_{1}\left(\begin{matrix}a,b\\\
c\end{matrix};q,\frac{c}{ab}\right)=\frac{(c/a,c/b;q)_{\infty}}{(c,c/(ab);q)_{\infty}}.$

For $\delta=1$, we have

$\displaystyle\sum_{n\geq 0}\frac{q^{n^{2}+n(N-1)}}{(q;q)_{n}(q;q)_{n+N}}$
$\displaystyle=\frac{1}{(q;q)_{N}}\lim_{\tau\to
0}{}_{2}\phi_{1}\left(\begin{matrix}1/\tau,1/\tau\\\
q^{N+1}\end{matrix};q,q^{N}\tau^{2}\right)$
$\displaystyle=\frac{1}{(q;q)_{\infty}}\lim_{\tau\to
0}{}_{2}\phi_{1}\left(\begin{matrix}q^{-1},1/\tau\\\
q^{N}\tau\end{matrix};q,q^{N+1}\tau\right)$
$\displaystyle=\frac{1}{(q;q)_{\infty}}\sum_{n\geq
0}\frac{(q^{-1};q)_{n}(-1)^{n}q^{\binom{n}{2}+(N+1)n}}{(q;q)_{n}}$
$\displaystyle=\frac{1}{(q;q)_{\infty}}\left(1-\frac{(1-q^{-1})q^{N+1}}{1-q}\right),$

confirming the desired result. Here we make use of Heine’s second
transformation [11, (III.2)]:

$\displaystyle{}_{2}\phi_{1}\left(\begin{matrix}a,b\\\
c\end{matrix};q,z\right)=\frac{(c/b,bz;q)_{\infty}}{(c,z;q)_{\infty}}{}_{2}\phi_{1}\left(\begin{matrix}abz/c,b\\\
bz\end{matrix};q,\frac{c}{b}\right).$

When $N<0$, we make the substitution $n\mapsto n-N$ and carry out a similar
calculation. We shall also notice the fact that $1/(q;q)_{k}=0$ for integers
$k<0$. ∎

######  Lemma 20.

We have

$\displaystyle\sum_{m,n=-\infty}^{\infty}z_{1}^{m}z_{2}^{n}q^{\frac{m^{2}}{2}+\frac{n^{2}}{2}-\frac{mn}{2}}$
$\displaystyle=(-z_{1}^{2}z_{2}q^{\frac{3}{2}},-z_{1}^{-2}z_{2}^{-1}q^{\frac{3}{2}},q^{3};q^{3})_{\infty}(-z_{2}q^{\frac{1}{2}},-z_{2}^{-1}q^{\frac{1}{2}},q;q)_{\infty}$
$\displaystyle\quad+z_{1}q^{\frac{1}{2}}(-z_{1}^{2}z_{2}q^{3},-z_{1}^{-2}z_{2}^{-1},q^{3};q^{3})_{\infty}(-z_{2},-z_{2}^{-1}q,q;q)_{\infty}.$
(A.13)

###### Proof.

Considering the parity of $m$, we have

$\displaystyle\sum_{m,n=-\infty}^{\infty}z_{1}^{m}z_{2}^{n}q^{\frac{m^{2}}{2}+\frac{n^{2}}{2}-\frac{mn}{2}}$
$\displaystyle=\sum_{u,v=-\infty}^{\infty}z_{1}^{2u}z_{2}^{v}q^{\frac{(2u)^{2}}{2}+\frac{v^{2}}{2}-\frac{(2u)v}{2}}$
$\displaystyle\quad+\sum_{u,v=-\infty}^{\infty}z_{1}^{2u+1}z_{2}^{v}q^{\frac{(2u+1)^{2}}{2}+\frac{v^{2}}{2}-\frac{(2u+1)v}{2}}$
$\displaystyle=\sum_{u,v=-\infty}^{\infty}z_{1}^{2u}z_{2}^{u-(u-v)}q^{\binom{u-v}{2}+\frac{u-v}{2}+3\binom{u}{2}+\frac{3u}{2}}$
$\displaystyle\quad+z_{1}q^{\frac{1}{2}}\sum_{u,v=-\infty}^{\infty}z_{1}^{2u}z_{2}^{u-(u-v)}q^{\binom{u-v}{2}+(u-v)+3\binom{u}{2}+3u}$
($w=u-v$)
$\displaystyle=\sum_{u=-\infty}^{\infty}(z_{1}^{2}z_{2})^{u}q^{3\binom{u}{2}+\frac{3u}{2}}\sum_{w=-\infty}^{\infty}z_{2}^{-w}q^{\binom{w}{2}+\frac{w}{2}}$
$\displaystyle\quad+z_{1}q^{\frac{1}{2}}\sum_{u=-\infty}^{\infty}(z_{1}^{2}z_{2})^{u}q^{3\binom{u}{2}+3u}\sum_{w=-\infty}^{\infty}z_{2}^{-w}q^{\binom{w}{2}+w}$
(by (A.4))
$\displaystyle=(-z_{1}^{2}z_{2}q^{\frac{3}{2}},-z_{1}^{-2}z_{2}^{-1}q^{\frac{3}{2}},q^{3};q^{3})_{\infty}(-z_{2}q^{\frac{1}{2}},-z_{2}^{-1}q^{\frac{1}{2}},q;q)_{\infty}$
$\displaystyle\quad+z_{1}q^{\frac{1}{2}}(-z_{1}^{2}z_{2}q^{3},-z_{1}^{-2}z_{2}^{-1},q^{3};q^{3})_{\infty}(-z_{2},-z_{2}^{-1}q,q;q)_{\infty},$

which is our desired result. ∎

##  Appendix B Some auxiliary series

For the purpose of showing Theorems 7 and 9, we also need to establish
explicit expressions of some auxiliary series.

We start with an easier one.

######  Lemma 21.

Define

$\displaystyle S_{M}:=\sum_{\begin{subarray}{c}m,n\geq 0\\\
m+n=M\end{subarray}}\frac{x^{m}y^{n}q^{\binom{m}{2}+\binom{n}{2}}}{(q^{2};q^{2})_{m}(q^{2};q^{2})_{n}}.$
(B.1a)

Then for $M\geq 0$,

$\displaystyle
S_{M}=\frac{y^{M}q^{\binom{M}{2}}(-xy^{-1}q^{1-M};q^{2})_{M}}{(q^{2};q^{2})_{M}}.$
(B.1b)

###### Proof.

Noticing that $n=M-m$, we have

$\displaystyle S_{M}$ $\displaystyle=\sum_{m\geq
0}\frac{x^{m}y^{M-m}q^{\binom{m}{2}+\binom{M-m}{2}}}{(q^{2};q^{2})_{m}(q^{2};q^{2})_{M-m}}$
$\displaystyle=\frac{y^{M}q^{\binom{M}{2}}}{(q^{2};q^{2})_{M}}\sum_{m\geq
0}(xy^{-1})^{m}q^{2\binom{m}{2}+(1-M)m}{M\brack m}_{q^{2}}$ (by (A.5))
$\displaystyle=\frac{y^{M}q^{\binom{M}{2}}}{(q^{2};q^{2})_{M}}(-xy^{-1}q^{1-M};q^{2})_{M},$

which is exactly our desired result. For our purpose of using this relation in
the proof of Theorem 7, it would be better to restate it according to the
parity of $M$:

$\displaystyle S_{2M}$
$\displaystyle=\frac{x^{M}y^{M}q^{M^{2}-M}(-xy^{-1}q;q^{2})_{M}(-x^{-1}yq;q^{2})_{M}}{(q^{2};q^{2})_{2M}},$
(B.2) $\displaystyle S_{2M+1}$
$\displaystyle=\frac{(x+y)x^{M}y^{M}q^{M^{2}}(-xy^{-1}q^{2};q^{2})_{M}(-x^{-1}yq^{2};q^{2})_{M}}{(q^{2};q^{2})_{2M+1}},$
(B.3)

where $M\geq 0$. ∎

The next result is more surprising.

######  Lemma 22.

Define

$\displaystyle T_{M}$
$\displaystyle:=\sum_{\begin{subarray}{c}N_{1},N_{2},N_{3},N_{4},N_{5}\geq
0\\\
N_{1}+N_{2}+2N_{3}+N_{4}+N_{5}=M\end{subarray}}\frac{z_{1}^{N_{1}+N_{3}+N_{4}}z_{2}^{N_{2}+N_{3}+N_{5}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}}$
$\displaystyle\ \times
q^{N_{1}^{2}+N_{2}^{2}+N_{3}^{2}+N_{4}^{2}+N_{5}^{2}+N_{1}N_{2}+N_{1}N_{3}+N_{1}N_{5}+N_{2}N_{3}+N_{2}N_{4}+N_{3}N_{4}+N_{3}N_{5}+N_{4}N_{5}-N_{1}-N_{2}}.$
(B.4)

Then for $M\geq 0$,

$\displaystyle
T_{M}=\frac{\big{(}z_{1}+z_{2}\big{)}z_{2}^{M-1}q^{\binom{M}{2}}(-z_{1}z_{2}^{-1}q^{2-M};q^{2})_{M}}{\big{(}1+z_{1}z_{2}^{-1}q^{M}\big{)}(q;q)_{M}}.$
(B.5)

Also, define

$\displaystyle\tilde{T}_{M}$
$\displaystyle:=\sum_{\begin{subarray}{c}N_{1},N_{2},N_{3},N_{4},N_{5}\geq
0\\\
N_{1}+N_{2}+2N_{3}+N_{4}+N_{5}=M\end{subarray}}\frac{(-1)^{N_{1}+N_{2}}z_{1}^{N_{1}+N_{3}+N_{4}}z_{2}^{N_{2}+N_{3}+N_{5}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}}$
$\displaystyle\ \times
q^{N_{1}^{2}+N_{2}^{2}+N_{3}^{2}+N_{4}^{2}+N_{5}^{2}+N_{1}N_{2}+N_{1}N_{3}+N_{1}N_{5}+N_{2}N_{3}+N_{2}N_{4}+N_{3}N_{4}+N_{3}N_{5}+N_{4}N_{5}-N_{1}-N_{2}}.$
(B.6)

Then for $M\geq 0$,

$\displaystyle\tilde{T}_{M}=-\frac{\big{(}z_{1}-(-1)^{M}z_{2}\big{)}z_{2}^{M-1}(-q)^{\binom{M}{2}}(z_{1}z_{2}^{-1}q^{2-M};q^{2})_{M}}{\big{(}1-z_{1}z_{2}^{-1}q^{M}\big{)}(-q;-q)_{M}}.$
(B.7)

###### Proof.

We may compute the recurrences for $T_{M}$ and $\tilde{T}_{M}$ with recourse
to the Mathematica package qMultiSum implemented by Riese [26] of Research
Institute for Symbolic Computation (RISC) of Johannes Kepler University. This
package can be downloaded at the website of RISC:

https://www3.risc.jku.at/research/combinat/software/ergosum/index.html

However, to reduce the amount of calculations, it is necessary to rewrite
$T_{M}$ as

$\displaystyle T_{M}$ $\displaystyle=\sum_{N_{1},N_{2},N_{3}\geq
0}\frac{z_{1}^{N_{1}+N_{3}}z_{2}^{M-N_{1}-N_{3}}(-z_{1}z_{2}^{-1}q^{1+2N_{2}+2N_{3}-M};q^{2})_{M-N_{1}-N_{2}-2N_{3}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{M-N_{1}-N_{2}-2N_{3}}}$
$\displaystyle\times
q^{M^{2}-MN_{1}-2MN_{2}-3MN_{3}+N_{1}^{2}+2N_{2}^{2}+3N_{3}^{2}+2N_{1}N_{2}+2N_{1}N_{3}+4N_{2}N_{3}-N_{1}-N_{2}}.$
(B.8)

To see this, we recall $N_{5}=M-N_{1}-N_{2}-2N_{3}-N_{4}$ and start with the
inner summation in $N_{4}$ by using (A.5),

$\displaystyle T_{M}$ $\displaystyle=\sum_{N_{1},N_{2},N_{3}\geq
0}\frac{z_{1}^{N_{1}+N_{3}}z_{2}^{M-N_{1}-N_{3}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}}$
$\displaystyle\quad\times
q^{M^{2}-MN_{1}-2MN_{2}-3MN_{3}+N_{1}^{2}+2N_{2}^{2}+3N_{3}^{2}+2N_{1}N_{2}+2N_{1}N_{3}+4N_{2}N_{3}-N_{1}-N_{2}}$
$\displaystyle\quad\times\sum_{N_{4}\geq
0}\frac{z_{1}^{N_{4}}z_{2}^{-N_{4}}q^{2\binom{N_{4}}{2}+(1+2N_{2}+2N_{3}-M)N_{4}}}{(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{M-N_{1}-N_{2}-2N_{3}-N_{4}}}$
$\displaystyle=\sum_{N_{1},N_{2},N_{3}\geq
0}\frac{z_{1}^{N_{1}+N_{3}}z_{2}^{M-N_{1}-N_{3}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{M-N_{1}-N_{2}-2N_{3}}}$
$\displaystyle\quad\times
q^{M^{2}-MN_{1}-2MN_{2}-3MN_{3}+N_{1}^{2}+2N_{2}^{2}+3N_{3}^{2}+2N_{1}N_{2}+2N_{1}N_{3}+4N_{2}N_{3}-N_{1}-N_{2}}$
$\displaystyle\quad\times\sum_{N_{4}\geq
0}z_{1}^{N_{4}}z_{2}^{-N_{4}}q^{2\binom{N_{4}}{2}+(1+2N_{2}+2N_{3}-M)N_{4}}{M-N_{1}-N_{2}-2N_{3}\brack
N_{4}}_{q^{2}}$ $\displaystyle=\sum_{N_{1},N_{2},N_{3}\geq
0}\frac{z_{1}^{N_{1}+N_{3}}z_{2}^{M-N_{1}-N_{3}}(-z_{1}z_{2}^{-1}q^{1+2N_{2}+2N_{3}-M};q^{2})_{M-N_{1}-N_{2}-2N_{3}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{M-N_{1}-N_{2}-2N_{3}}}$
$\displaystyle\quad\times
q^{M^{2}-MN_{1}-2MN_{2}-3MN_{3}+N_{1}^{2}+2N_{2}^{2}+3N_{3}^{2}+2N_{1}N_{2}+2N_{1}N_{3}+4N_{2}N_{3}-N_{1}-N_{2}}.$

Now we compute the recurrence for $T_{M}$ according to the parity of $M$. In
particular, (B.5) is equivalent to

$\displaystyle T_{2M}$
$\displaystyle=\frac{z_{1}^{M}z_{2}^{M}q^{M^{2}}(-z_{1}z_{2}^{-1};q^{2})_{M}(-z_{1}^{-1}z_{2};q^{2})_{M}}{(q^{2};q^{2})_{M}(q;q^{2})_{M}},$
(B.9a) $\displaystyle T_{2M+1}$
$\displaystyle=\frac{\big{(}z_{1}+z_{2}\big{)}z_{1}^{M}z_{2}^{M}q^{M(M+1)}(-z_{1}z_{2}^{-1}q;q^{2})_{M}(-z_{1}^{-1}z_{2}q;q^{2})_{M}}{(q^{2};q^{2})_{M}(q;q^{2})_{M+1}}.$
(B.9b)

We further write $z=z_{1}z_{2}^{-1}$ and define

$\displaystyle T^{\star}_{M}$ $\displaystyle:=\sum_{N_{1},N_{2},N_{3}\geq
0}\frac{z^{N_{1}+N_{3}}(-zq^{1+2N_{2}+2N_{3}-M};q^{2})_{M-N_{1}-N_{2}-2N_{3}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{M-N_{1}-N_{2}-2N_{3}}}$
$\displaystyle\ \times
q^{-MN_{1}-2MN_{2}-3MN_{3}+N_{1}^{2}+2N_{2}^{2}+3N_{3}^{2}+2N_{1}N_{2}+2N_{1}N_{3}+4N_{2}N_{3}-N_{1}-N_{2}}.$

Then $T_{M}=T^{\star}_{M}\cdot z_{2}^{M}q^{M^{2}}$. Therefore, it suffices to
show

$\displaystyle T_{2M}^{\star}$
$\displaystyle=\frac{z^{M}q^{-3M^{2}}(-z;q^{2})_{M}(-z^{-1};q^{2})_{M}}{(q^{2};q^{2})_{M}(q;q^{2})_{M}},$
(B.10a) $\displaystyle T_{2M+1}^{\star}$
$\displaystyle=\frac{\big{(}1+z\big{)}z^{M}q^{-3M(M+1)}(-zq;q^{2})_{M}(-z^{-1}q;q^{2})_{M}}{q(q^{2};q^{2})_{M}(q;q^{2})_{M+1}}.$
(B.10b)

We first treat (B.10a). To avoid ambiguity, we call
$\mathscr{T}_{M}^{\star}:=T_{2M}^{\star}$ for $M\geq 0$. Let us import the
qMultiSum package:

⬇

<<RISC‘qMultiSum‘

Then the recurrence for $\mathscr{T}_{M}^{\star}$ can be computed
automatically by calling the commands:

⬇

ClearAll[M, N1, N2, N3, z];

summand =

z^(N1 + N3) q^(- N1 - M N1 + N1^2 - N2 - 2 M N2 + 2 N1 N2 + 2 N2^2 - 3 M N3 +
2 N1 N3 + 4 N2 N3 + 3 N3^2)

* qPochhammer[-z q^(1 + 2 N2 + 2 N3 - M), q^2, M - N1 - N2 - 2 N3]

/ (qPochhammer[q^2, q^2, N1] qPochhammer[q^2, q^2, N2] qPochhammer[q^2, q^2,
N3] qPochhammer[q^2, q^2, M - N1 - N2 - 2 N3]) /. {M -> 2 M};

stru = qFindStructureSet[summand, {M}, {N1, N2, N3}, {1}, {2, 2, 2}, {2, 2,
2}, qProtocol -> True];

rec = qFindRecurrence[summand, {M}, {N1, N2, N3}, {1}, {2, 2, 2}, {2, 2, 2},
qProtocol -> True, StructSet -> stru[[1]]];

sumrec = qSumRecurrence[rec]

The output gives a fifth order recurrence for $\mathscr{T}_{M}^{\star}$. So it
suffices to confirm (B.10a), or equivalently (B.9a), for $0\leq M\leq 4$ and
to verify that the right-hand side of (B.10a) fulfills the same recurrence.
The latter calculation can be done directly due to the fact that the ratio of
the $(M+1)$-th and $M$-th terms on the right-hand side of (B.10a) is a
rational function of $q^{M}$. Here we will not display the concrete recurrence
for simplicity, but all Mathematica codes can be found at the following url
(notice that in the output recurrence, $\operatorname{SUM}[M]$ stands for
$\mathscr{T}_{M}^{\star}$):

https://github.com/shanechern/VOA

For (B.10b), we only need to replace the summand in the codes with

⬇

summand =

z^(N1 + N3) q^(- N1 - M N1 + N1^2 - N2 - 2 M N2 + 2 N1 N2 + 2 N2^2 - 3 M N3 +
2 N1 N3 + 4 N2 N3 + 3 N3^2)

* qPochhammer[-z q^(1 + 2 N2 + 2 N3 - M), q^2, M - N1 - N2 - 2 N3]

/ (qPochhammer[q^2, q^2, N1] qPochhammer[q^2, q^2, N2] qPochhammer[q^2, q^2,
N3] qPochhammer[q^2, q^2, M - N1 - N2 - 2 N3]) /. {M -> 2 M + 1};

The rest can be proceeded analogously.

In the same way, we rewrite (22) as

$\displaystyle\tilde{T}_{M}$ $\displaystyle=\sum_{N_{1},N_{2},N_{3}\geq
0}\frac{(-1)^{N_{1}+N_{2}}z_{1}^{N_{1}+N_{3}}z_{2}^{M-N_{1}-N_{3}}(-z_{1}z_{2}^{-1}q^{1+2N_{2}+2N_{3}-M};q^{2})_{M-N_{1}-N_{2}-2N_{3}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{M-N_{1}-N_{2}-2N_{3}}}$
$\displaystyle\times
q^{M^{2}-MN_{1}-2MN_{2}-3MN_{3}+N_{1}^{2}+2N_{2}^{2}+3N_{3}^{2}+2N_{1}N_{2}+2N_{1}N_{3}+4N_{2}N_{3}-N_{1}-N_{2}}.$

Also, (B.7) is equivalent to

$\displaystyle\tilde{T}_{2M}$
$\displaystyle=\frac{z_{1}^{M}z_{2}^{M}q^{M^{2}}(z_{1}z_{2}^{-1};q^{2})_{M}(z_{1}^{-1}z_{2};q^{2})_{M}}{(q^{2};q^{2})_{M}(-q;q^{2})_{M}},$
(B.11a) $\displaystyle\tilde{T}_{2M+1}$
$\displaystyle=-\frac{\big{(}z_{1}+z_{2}\big{)}z_{1}^{M}z_{2}^{M}q^{M(M+1)}(z_{1}z_{2}^{-1}q;q^{2})_{M}(z_{1}^{-1}z_{2}q;q^{2})_{M}}{(q^{2};q^{2})_{M}(-q;q^{2})_{M+1}}.$
(B.11b)

The remaining computer-assisted calculations are similar. ∎

##  Appendix C More identities of Rogers–Ramanujan type

In Section 3, we have already witnessed several Rogers–Ramanujan type
identities as specialisations of the transforms in Section 5. Since such
instances are curious to the community of $q$-theorists, we will record more
in this appendix.

First, we reload the two identities from Theorem 7, namely, (3.5a) and (3.5b).

######  Theorem 23.

Let $\mathbf{G}_{3}$ and $\mathbf{N}$ be as in (3.2) and (3.3), respectively.
Then,

$\displaystyle\sum_{N_{1},N_{2},N_{3}\geq
0}\frac{q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{3}\cdot\mathbf{N}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}}$
$\displaystyle\qquad=\frac{(q;q)_{\infty}(q^{2},q^{3},q^{5};q^{5})_{\infty}}{(q^{2};q^{2})^{2}_{\infty}}+\frac{4q(q^{4};q^{4})_{\infty}(q^{4},q^{16},q^{20};q^{20})_{\infty}}{(q^{2};q^{2})^{2}_{\infty}},$
(C.1a) $\displaystyle\sum_{N_{1},N_{2},N_{3}\geq
0}\frac{q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{3}\cdot\mathbf{N}-N_{2}-N_{3}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}}$
$\displaystyle\qquad=-\frac{(q;q)_{\infty}(q,q^{4},q^{5};q^{5})_{\infty}}{(q^{2};q^{2})^{2}_{\infty}}+\frac{4(q^{4};q^{4})_{\infty}(q^{8},q^{12},q^{20};q^{20})_{\infty}}{(q^{2};q^{2})^{2}_{\infty}}.$
(C.1b)

We then take a look at the sextuple summations in Theorems 8. First, we should
keep in mind that the right-hand side of (5.2) is always possible to be
represented as a linear combination of products of theta series by invoking
Lemma 20. However, some cases, including (3.14) and (3.15) are more
interesting.

######  Theorem 24.

Let $\mathbf{G}_{4}$ and $\mathbf{N}$ be as in (3.11) and (3.12),
respectively. Then,

$\displaystyle\sum_{N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}\geq
0}\frac{q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}(q^{2};q^{2})_{N_{6}}}$
$\displaystyle\qquad=\frac{(q^{2};q^{2})^{3}_{\infty}(q^{6};q^{6})^{5}_{\infty}}{(q;q)^{2}_{\infty}(q^{3};q^{3})^{2}_{\infty}(q^{4};q^{4})^{2}_{\infty}(q^{12};q^{12})^{2}_{\infty}}+\frac{4q(q^{4};q^{4})^{2}_{\infty}(q^{12};q^{12})^{2}_{\infty}}{(q^{2};q^{2})^{3}_{\infty}(q^{6};q^{6})_{\infty}},$
(C.2a) $\displaystyle\sum_{N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}\geq
0}\frac{(-1)^{N_{1}+N_{2}+N_{4}+N_{5}}q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}(q^{2};q^{2})_{N_{6}}}$
$\displaystyle\qquad=\frac{(q;q)^{2}_{\infty}(q^{3};q^{3})^{2}_{\infty}}{(q^{2};q^{2})^{3}_{\infty}(q^{6};q^{6})_{\infty}},$
(C.2b) $\displaystyle\sum_{N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}\geq
0}\frac{q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-(N_{1}+N_{3}+N_{4}+N_{6})}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}(q^{2};q^{2})_{N_{6}}}$
$\displaystyle\qquad=\frac{6(q^{3};q^{3})^{3}_{\infty}}{(q;q)_{\infty}(q^{2};q^{2})^{2}_{\infty}},$
(C.2c) $\displaystyle\sum_{N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}\geq
0}\frac{(-1)^{N_{1}+N_{3}+N_{4}+N_{6}}q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-(N_{1}+N_{3}+N_{4}+N_{6})}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}(q^{2};q^{2})_{N_{6}}}$
$\displaystyle\qquad=-\frac{2(q;q)^{3}_{\infty}(q^{6};q^{6})^{2}_{\infty}}{(q^{2};q^{2})^{4}_{\infty}(q^{3};q^{3})_{\infty}},$
(C.2d) $\displaystyle\sum_{N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}\geq
0}\frac{(-1)^{N_{1}+N_{2}+N_{4}+N_{5}}q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-(N_{1}+N_{3}+N_{4}+N_{6})}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}(q^{2};q^{2})_{N_{6}}}$
$\displaystyle\qquad=\frac{2(q;q)^{3}_{\infty}(q^{6};q^{6})^{2}_{\infty}}{(q^{2};q^{2})^{4}_{\infty}(q^{3};q^{3})_{\infty}}.$
(C.2e)

###### Proof.

We only need to establish (C.2b), (C.2d) and (C.2e). For (C.2b), we use (5.2)
with $\delta_{1}=\delta_{2}=0$ and $z_{1}=z_{2}=-1$ and then apply (20) with
$q$ replaced by $q^{2}$. For (C.2d), we use (5.2) with $\delta_{1}=1$,
$\delta_{2}=0$ and $(q,z_{1},z_{2})\mapsto(q^{2},-q^{2},1)$, and then take
advantage of (20) followed by an application of (A.7b). Finally, (C.2e) is the
$\delta_{1}=1$, $\delta_{2}=0$ and $(q,z_{1},z_{2})\mapsto(q^{2},-q^{2},-1)$
case of (5.2). ∎

Finally, we notice that the sextuple summations in Theorem 9 reduce to linear
combinations of infinite products if either of $z_{1}$ and $z_{2}$ is $\pm 1$.
Due to the symmetry, we only record the specialisation of $z_{2}\in\\{\pm
1\\}$.

######  Theorem 25.

Let $\mathbf{G}_{4}$ and $\mathbf{N}$ be as in (3.11) and (3.12),
respectively. Then,

$\displaystyle\sum_{N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}\geq
0}\frac{z^{N_{1}+N_{3}+N_{4}+N_{6}}q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-(N_{1}+N_{2}+N_{6})}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}(q^{2};q^{2})_{N_{6}}}$
$\displaystyle\qquad=\frac{2(-z;q)_{\infty}}{(q;q^{2})_{\infty}},$ (C.3a)
$\displaystyle\sum_{N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}\geq
0}\frac{(-1)^{N_{2}+N_{3}+N_{5}+N_{6}}z^{N_{1}+N_{3}+N_{4}+N_{6}}q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-(N_{1}+N_{2}+N_{6})}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}(q^{2};q^{2})_{N_{6}}}$
$\displaystyle\qquad=0,$ (C.3b)
$\displaystyle\sum_{N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}\geq
0}\frac{(-1)^{N_{1}+N_{2}+N_{6}}z^{N_{1}+N_{3}+N_{4}+N_{6}}q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-(N_{1}+N_{2}+N_{6})}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}(q^{2};q^{2})_{N_{6}}}$
$\displaystyle\qquad=\frac{(z;-q)_{\infty}}{(-q;q^{2})_{\infty}}-\frac{(-z;-q)_{\infty}}{(-q;q^{2})_{\infty}},$
(C.3c) $\displaystyle\sum_{N_{1},N_{2},N_{3},N_{4},N_{5},N_{6}\geq
0}\frac{(-1)^{N_{1}+N_{3}+N_{5}}z^{N_{1}+N_{3}+N_{4}+N_{6}}q^{\mathbf{N}^{\mathsf{T}}\cdot\mathbf{G}_{4}\cdot\mathbf{N}-(N_{1}+N_{2}+N_{6})}}{(q^{2};q^{2})_{N_{1}}(q^{2};q^{2})_{N_{2}}(q^{2};q^{2})_{N_{3}}(q^{2};q^{2})_{N_{4}}(q^{2};q^{2})_{N_{5}}(q^{2};q^{2})_{N_{6}}}$
$\displaystyle\qquad=\frac{(z;-q)_{\infty}}{(-q;q^{2})_{\infty}}+\frac{(-z;-q)_{\infty}}{(-q;q^{2})_{\infty}}.$
(C.3d)

###### Proof.

In (5.3a) and (5.3b), we take $(z_{1},z_{2})\mapsto(z,1)$, then (C.3a) and
(C.3c) follow, respectively. For (C.3b) and (C.3d), the only difference is
that we change $z_{2}$ to $-1$. ∎

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