#  Topological invariance of complex Chern-Simons and Teichmüller TQFT
perturbation theory

Stavros Garoufalidis  International Center for Mathematics, Department of
Mathematics
Southern University of Science and Technology
Shenzhen, China
http://people.mpim-bonn.mpg.de/stavros<EMAIL_ADDRESS>,  Matthias
Storzer  Max Planck Institute for Mathematics
Vivatsgasse 7, 53111 Bonn, Germany
http://guests.mpim-bonn.mpg.de/storzer<EMAIL_ADDRESS>and  Campbell
Wheeler  Max Planck Institute for Mathematics
Vivatsgasse 7, 53111 Bonn, Germany
http://guests.mpim-bonn.mpg.de/cjwh<EMAIL_ADDRESS>
(Date: 23 May 2023)

###### Abstract.

We prove the topological invariance of a formal power series defined using an
ideal triangulation of a 3-manifold with torus boundary components (together
with some further choices). This formal power series is conjectured to agree
to all orders in perturbation theory with two important topological invariants
of hyperbolic knots, namely the Kashaev invariant and the Andersen–Kashaev
invariant (also known as the state-integral) of Teichmüller TQFT.

Key words and phrases: 3-manifolds, knots, complex Chern–Simons theory,
Teichmüller TQFT, hyperbolic geometry, asymptotic expansions, perturbation
theory, Faddeev quantum dilogarithm, state-integrals, volume conjecture, ideal
triangulations, Neumann-Zagier data, half-symplectic matrices, 2–3 Pachner
moves, Fourier transform, pentagon.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 The Faddeev quantum dilogarithm
    2. 2.2 Neumann–Zagier data
    3. 2.3 Geometric aspects
  3. 3 Formal Gaussian integration
    1. 3.1 Basics on formal Gaussian integration
    2. 3.2 $q$–holonomic aspects of $\psi_{\hbar}$
    3. 3.3 Fourier transform of $\psi_{\hbar}$
    4. 3.4 Pentagon identity for $\psi_{\hbar}$
    5. 3.5 Inversion formula for $\psi_{\hbar}$
  4. 4 Elementary invariance properties
  5. 5 Invariance under the choice of quad
  6. 6 Invariance under Pachner moves
    1. 6.1 The case of $B$ with full rank
    2. 6.2 The case of $B$ with nullity one
    3. 6.3 The case of $B$ with nullity two
  7. 7 The series of the simplest hyperbolic $4_{1}$ knot
  8. A Complements on the Fourier transform
  9. B Complements on the pentagon identity

##  1\. Introduction

This paper concerns the topological invariance of a formal power series
associated to an ideally triangulated 3-manifold $M$ with a torus boundary
component [10]. The series is defined by formal Gaussian integration of a
finite dimensional integral (a so-called “state-integral”) and it is expected
to coincide to all orders with the asymptotic expansion of three important
topological invariants of 3-manifolds.

The first one is the Kashaev invariant of a hyperbolic knot [24], where
Kashaev’s famous volume conjecture is refined to an asymptotic statement to
all orders in perturbation theory using the above formal power series. This
was studied in detail in [20] based on extensive numerical computations, but
it is only proven for a handful of hyperbolic knots.

A second appearance of the formal power series above is in the asymptotic
expansion of the Andersen–Kashaev state integral [2], as was shown for the
simplest hyperbolic knot, $4_{1}$, in [2, Sec.12], and also observed
numerically for the first three simplest hyperbolic knots in [19]. The state-
integrals of [2] are finite-dimensional integrals whose integrand is a product
of Faddeev quantum dilogarithms times an exponential of a quadratic form,
assembled from an ideal triangulation of a 3-manifold with torus boundary
components. Andersen–Kashaev proved that these are topological invariants that
are the partition function of the Teichmüller TQFT [2, 1], which is a
3-dimensional version of a quantization of Teichmüller space [25, 16].

A third connection with the formal power series mentioned above is with
complex Chern–Simons theory, the latter introduced by Witten [35] and studied
extensively by Gukov [22]. For reasons that are not entirely understood, the
partition function of complex Chern–Simons theory for 3-manifolds with torus
boundary reduces to a finite-dimensional state-integral, as if some unknown
localization principle holds. The corresponding state-integrals were
introduced and studied by Hikami [23], Dimofte [9] and others [11].
Unfortunately, in those works the integration contour was not pinned down, and
this problem was finally dealt with in [2] and, among other things, implied
topological invariance of the state-integrals, which were coined to be the
partition function of Teichmüller TQFT. But ignoring the contour of
integration, and focusing on a critical point of the action, which is a
solution to a system of polynomial equations, allowed [10] to give a
definition of the formal power series that are the main focus of our paper.

The above discussion points out several aspects of these formal power series
and conjectural relations to perturbation theory of complex Chern–Simons and
of Teichüller TQFT. Aside from conjectures, this paper concerns a theorem, the
topological invariance of the above formal power series.

Let us briefly recall the key ingredients that go into the definition of the
series, and discuss those in detail in later sections. The first ingredient is
an ideal triangulation $\mathcal{T}$ with $N$ ideal tetrahedra of a 3-manifold
$M$ with torus boundary. Each tetrahedron has shapes
$z\in\mathbbm{C}\setminus\\{0,1\\}$, $z^{\prime}=1/(1-z)$ and
$z^{\prime\prime}=1-1/z$ attached to its three pairs of opposite edges, and
the shapes satisfy a system of polynomial equations (so-called “gluing
equations” [32] or Neumann-Zagier equations [28]) determined by the
combinatorics of the triangulation, one for each edge and peripheral curve.
After some choices are made (such as an ordering of the tetrahedra and their
edges, a choice of shape for each tetrahedron, a choice of an edge to remove
from the gluing equations, and a choice of peripheral curve to include), one
obtains two matrices $A$ and $B$ with integer entries such that $(A\,|\,B)$ is
the upper-half of a symplectic $2N\times 2N$ matrix, as well as a vector
$\nu\in\mathbbm{Z}^{N}$. In addition, we choose a solution
$z=(z_{1},\dots,z_{N})$ of the gluing equations as well as a flattening
$(f,f^{\prime\prime})$, i.e., an integer solution to the equation
$\nu=Af+Bf^{\prime\prime}$. The power series $\Phi^{\Xi}(\hbar)$ depends on
the tuple $\Xi=(A,B,\nu,z,f,f^{\prime\prime})$, which we collectively call a
NZ-datum.

The next ingredient that goes in the definition of $\Phi^{\Xi}(\hbar)$ is an
auxiliary formal power series

$\psi_{\hbar}(x,z)\;:=\;\exp\bigg{(}\\!-\\!\\!\sum_{\underset{k+\frac{\ell}{2}>1}{k,\ell\in\mathbbm{Z}_{\geq
0}}}\frac{B_{k}\,x^{\ell}\,\hbar^{k+\frac{\ell}{2}-1}}{\ell!\,k!}\operatorname{Li}_{2-k-\ell}(z)\bigg{)}\in
1+\hbar^{\frac{1}{2}}\mathbbm{Q}[z,(1-z)^{-1}][x]\llbracket\hbar^{\frac{1}{2}}\rrbracket\,.$
(1)

This series, which differs from the one given in [10, Eq. 1.9] by a factor of
$\exp\big{(}\frac{1}{12}\hbar-\frac{1}{2}x\hbar^{\frac{1}{2}}\big{)}$, comes
from the asymptotics of the infinite Pochhammer symbol (also known as the
quantum dilogarithm)

$(x;q)_{\infty}=\prod_{k=0}^{\infty}(1-q^{k}x)=\exp\Big{(}-\sum_{n=1}^{\infty}\frac{x^{n}}{n(1-q^{n})}\Big{)}\,.$
(2)

Explicitly, for complex numbers $x$ and $z$ with $|z|<1$ and
$\mathrm{Re}(\hbar)<0$ we have (see [36] and also [21, Lem.2.1])

$(ze^{x\hbar^{1/2}};e^{\hbar})_{\infty}^{-1}\;\sim\;\frac{1}{\sqrt{1-z}}\exp\Big{(}-\frac{\operatorname{Li}_{2}(z)}{\hbar}-\frac{\operatorname{Li}_{1}(z)x}{\hbar^{\frac{1}{2}}}-\frac{\operatorname{Li}_{0}(z)x^{2}}{2}\Big{)}\psi_{\hbar}(x,z)\,,\qquad(\hbar\rightarrow
0)\,.$ (3)

Assuming that the matrix $B$ is invertible, (i.e., that $\det(B)\neq 0$), we
introduce the function

$f^{\Xi}_{\hbar}(x,z)\;=\;\exp\bigg{(}\frac{\hbar^{\frac{1}{2}}}{2}x^{t}1-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}B^{-1}\nu+\frac{\hbar}{8}f^{t}B^{-1}Af\bigg{)}\prod_{i=1}^{N}\psi_{\hbar}(x_{i},z_{i})\in\mathbbm{Q}(z)[x]\llbracket\hbar^{\frac{1}{2}}\rrbracket\,,$
(4)

which is the product of one quantum dilogarithm per tetrahedron (each with its
own integration variable), with some additional terms coming from the NZ-datum
where $x=(x_{1},...,x_{N})^{t}$ and $z=(z_{1},...,z_{N})$.

The last ingredient is the formal Gaussian integration [4]

$\displaystyle\left\langle
f_{\hbar}(x,z)\right\rangle_{x,\Lambda}\;:=\;\exp\left({\frac{1}{2}\sum_{i,j=1}^{N}(\Lambda^{-1})_{i,j}\frac{\partial}{\partial
x_{i}}\frac{\partial}{\partial
x_{j}}}\right)f_{\hbar}(x,z)\Big{|}_{x=0}=\frac{\int
e^{-\frac{1}{2}x^{t}\Lambda\,x}f_{\hbar,z}(x)\,dx}{\int
e^{-\frac{1}{2}x^{t}\Lambda\,x}\,dx}\in\mathbbm{Q}(z)\llbracket\hbar\rrbracket$
(5)

of a power series
$f_{\hbar}(x,z)\in\mathbbm{Q}(z)[x]\llbracket\hbar^{\frac{1}{2}}\rrbracket$,
with respect to an invertible matrix $\Lambda$. Assuming further that the
symmetric matrix

$\Lambda\;=\;-B^{-1}A+\mathrm{diag}(1/(1-z_{j}))$ (6)

is invertible, we define

$\Phi^{\Xi}(\hbar)\;=\;\langle
f^{\Xi}_{\hbar}(x,z)\rangle_{x,\Lambda}\in\mathbbm{Q}(z)\llbracket\hbar\rrbracket\,.$
(7)

Using a solution $z$ of the Neumann–Zagier equations that corresponds to the
discrete faithful representation
$\rho^{\mathrm{geom}}:\pi_{1}(M)\to\mathrm{PSL}_{2}(\mathbbm{C})$ of a cusped
hyperbolic 3-manifold $M$, one obtains a series $\Phi^{\Xi}(\hbar)$ with
coefficients in the invariant trace field of $M$. Our main theorem is the
following.

######  Theorem 1.1.

$\Phi^{\Xi}(\hbar)$ is a topological invariant of a cusped hyperbolic
3-manifold.

A corollary of the above theorem is that every coefficient of the above series
is a topological invariant of a cusped hyperbolic 3-manifold. These
topological invariants have a geometric origin, since they take values in the
invariant trace field of the manifold, and hence they are algebraic, but not
(in general) rational, numbers. This series seems to come from hyperbolic
geometry (though a definition of these invariants in terms of the hyperbolic
geometry of the 3-manifold is not known), and perhaps not from enumerative
quantum field theory (such as the Gromov–Witten or any of the 4-dimensional
known theories). The behavior of these series under finite cyclic coverings of
cusped hyperbolic 3-manifolds is given in [18], and the formulas presented
there (e.g., for the coefficient of $\hbar^{2}$) point to unknown connections
with the spectral theory of hyperbolic 3-manifolds.

The proof of our theorem combines the ideas of the topological invariance of
the Andersen–Kashaev state-integrals [2] with those of the Aarhus integral
[3]. We briefly recall that the building block for the state-integral is the
Faddeev quantum dilogarithm, the state-integral is a finite-dimensional
integral of a product of Faddeev quantum dilogarithm [14], one for each
tetrahedron of an (ordered) ideal triangulation. Aside from elementary
choices, two important parts in the proof of topological invariance of the
state-integral is invariance under (a) the choice of a nondegenerate quad, and
(b) 2–3 ordered Pachner moves, the latter connecting one ideal triangulation
with another. In [2], (a) and (b) are dealt with a Fourier transform and a
pentagon identity for the Faddeev quantum dilogarithm.

The power series $\Phi^{\Xi}(\hbar)$ is given instead by a formal Gaussian
integral (as opposed to an integral over Euclidean space), where the building
block is the formal power series $\psi_{\hbar}$ instead of the Faddeev quantum
dilogarithm. The topological invariance of the $\Phi^{\Xi}(\hbar)$ under the
choice of quad and under the 2–3 Pachner moves follows from a Fourier tranform
identity and a pentagon identity for $\psi_{\hbar}$; see Theorems 3.4 and 3.6
below. Although these identities are, in a sense, limits of the corresponding
identities for the Faddeev quantum dilogarithm (just as $\psi_{\hbar}$ is a
limit of the Faddeev quantum dilogarithm), it would requires additional
analytic work to do so, and instead we give algebraic proofs of theorems 3.4
and 3.6 using properties of the formal Gaussian integration, together with
holonomic properties of the involved formal power series.

Having discussed the similarities between the proof of Theorem 1.1 and the
corresponding theorem for the state-integral of [2], we now point out some
differences. In [2], Andersen–Kashaev use ordered triangulations, and the
state-integral is obtained by the push-forward of a distribution with
variables at the faces and the tetrahedra of the ordered triangulation. Part
of the distribution is a product of delta functions in linear forms of the
face-variables. In our Theorem 1.1, and in the formal Gaussian integration, we
carefully avoided the need to use delta functions, although such a
reformulation of our results are possible, with additional effort.

We end this section by pointing out a wider context for the asymptotic series
$\Phi^{\Xi}(\hbar)$ and Theorem 1.1. It was clear from [10] that a NZ-datum
$\Xi=(A,B,\nu,z,f,f^{\prime\prime})$ depends on two square matrices $A$ and
$B$ such that $AB^{t}$ is symmetric that may or may not come from topology,
and that the series $\Phi^{\Xi}(\hbar)$ is defined under the assumption that
$\det(B)\neq 0$ and $\det(\Lambda)\neq 0$. Doing so, the proof of Theorem 1.1
shows that the series $\Phi^{\Xi}(\hbar)$ is invariant under the moves that
appear in [20, Sec.6]; see also Section 4.

##  2\. Preliminaries

###  2.1. The Faddeev quantum dilogarithm

In this subsection, we recall some basic properties of the Faddeev quantum
dilogarithm, which are motivations for Theorems 3.4 and 3.6 below. At the same
time, we will ignore additional properties of the Faddeev quantum dilogarithm
that play no role in our paper, such as the fact that it is a meromorphic
function with precise zeros, poles and residues.

The Faddeev quantum dilogarithm [14]
$\varphi:=\operatorname{\Phi}_{\mathsf{b}}{}$ satisfies a key integral
pentagon identity [15]

$e^{2\pi
ixy}\widetilde{\varphi}(x)\widetilde{\varphi}(y)=\int_{\mathbbm{R}}e^{\pi
iz^{2}}\widetilde{\varphi}(x-z)\widetilde{\varphi}(z)\widetilde{\varphi}(y-z)\operatorname{d}\\!z$
(8)

where both sides are tempered distributions on $\mathbbm{R}$ and
$\widetilde{\varphi}$ denotes the distributional (inverse) Fourier
transformation

$\widetilde{\varphi}(x):=\int_{\mathbbm{R}}e^{-2\pi
ixy}\varphi(y)\operatorname{d}\\!y\,.$ (9)

It turns out that the inverse Fourier transform of $\varphi^{\pm 1}$ is
expressed in terms of $\varphi$ as given in [2, Sec.13.2]

$\displaystyle\int_{\mathbbm{R}}e^{-2\pi ixy}\varphi(y)\operatorname{d}\\!y$
$\displaystyle=\zeta_{8}(q/\widetilde{q})^{\frac{1}{24}}e^{-\pi
ix^{2}}\varphi(-x+c_{\mathsf{b}})$ (10)
$\displaystyle\int_{\mathbbm{R}}e^{-2\pi
ixy}\varphi(y)^{-1}\operatorname{d}\\!y$
$\displaystyle=\zeta_{8}^{-1}(\widetilde{q}/q)^{\frac{1}{24}}e^{\pi
ix^{2}}\varphi(x-c_{\mathsf{b}})$

where $q=e^{2\pi i\mathsf{b}^{2}}$, $\widetilde{q}=e^{-2\pi
i/\mathsf{b}^{2}}$, $c_{\mathsf{b}}=\frac{i}{2}(\mathsf{b}+\mathsf{b}^{-1})$
and $\zeta_{8}=e^{2\pi i/8}$. The Faddeev quantum dilogarithm satisfies the
inversion formula

$\varphi(x)\varphi(-x)=(\widetilde{q}/q)^{\frac{1}{24}}e^{\pi ix^{2}}$ (11)

see for example [2, App.A]. In a certain domain, the Faddeev quantum
dilogarithm is given as a ratio of two Pochhammer symbols

$\operatorname{\Phi}_{\mathsf{b}}(x)\;=\;\frac{(-q^{\frac{1}{2}}e^{2\pi
i\mathsf{b}x};q)_{\infty}}{(-\widetilde{q}^{\frac{1}{2}}e^{2\pi
i\mathsf{b}^{-1}x};\widetilde{q})_{\infty}}\,,$ (12)

from which it follows that its asymptotic expansion as $\mathsf{b}\to 0$ is
given by replacing the denominator by $1$ and the numerator by the asymptotic
expansion of the Pochhammer symbol.

###  2.2. Neumann–Zagier data

In this section, we discuss in detail Neumann–Zagier data following [10]. We
start with a 3-manifold $M$ with torus boundary component (all manifolds and
their triangulations will be oriented throughout the paper) equipped with a
concrete oriented ideal triangulation, that is with a triangulation such that
each tetrahedron $\Delta$ of $\mathcal{T}$ comes with a bijection of its
vertices with those of the standard 3-simplex. (All triangulations that are
used in the computer programs SnapPy [7] and Regina [6] are concrete).

Every concrete tetrahedron $\Delta$ has shape parameters
$(z,z^{\prime},z^{\prime\prime})$ assigned to pairs of opposite edges as in
Figure 1, where the complex numbers $z^{\prime}=1/(1-z)$ and
$z^{\prime\prime}=1-1/z$ satisfy the equations

$z^{\prime\prime}+z^{-1}=1,\qquad zz^{\prime}z^{\prime\prime}=-1\,.$ (13)

$0$$1$$2$$3$$z$$z$$z^{\prime}$$z^{\prime}$$z^{\prime\prime}$$z^{\prime\prime}$
Figure 1. The shapes of an ideal tetrahedron.

If $\mathcal{T}$ is a triangulation as above, an Euler characteristic argument
shows that the number of tetrahedra equals to the number of edges. Fix an
ordering of the tetrahedra $\Delta_{j}$ for $j=1,\dots,N$ and of the edges
$e_{1},\dots,e_{N}$ of $\mathcal{T}$, and assign a shape $z_{j}$ to the
tetrahedron $\Delta_{j}$ for $j=1,\dots,N$. To describe the complete
hyperbolic structure of $M$ (when it exists) and its deformations, Thurston
[32] introduced the gluing equations for the variables $z=(z_{1},\dots,z_{N})$
around each edge $e_{i}$, for $i=1,\dots,N$. In logarithmic form, these
equations have the form

$\sum_{j=1}^{N}G_{i,j}\log z_{j}+G^{\prime}_{i,j}\log
z^{\prime}_{j}+G^{\prime\prime}_{i,j}\log z^{\prime\prime}_{j}\;=\;2\pi
i,\qquad(i=1,\dots,N)$ (14)

where $G_{i,j}$ (and likewise for $G^{\prime}_{i,j}$ and
$G^{\prime\prime}_{i,j}$) denote the number of times that an edge of
$\Delta_{j}$ labelled $z_{j}$ winds around the edge $e_{i}$. Every peripheral
curve $\gamma$ in the boundary also gives rise to a gluing equation of the
same form as (14), except the right hand side is $0$ instead of $2\pi i$.
Choosing a symplectic basis for $H_{1}(\partial M,\mathbbm{Z})$, we can
enhance the equations (14) by adding the peripheral equations

$\sum_{j=1}^{N}G_{N+c,j}\log z_{j}+G^{\prime}_{N+c,j}\log
z^{\prime}_{j}+G^{\prime\prime}_{N+c,j}\log
z^{\prime\prime}_{j}\;=\;0,\qquad(c=1,2)\,.$ (15)

It turns out that the $(N+2)\times N$ matrices $G$, $G^{\prime}$ and
$G^{\prime\prime}$ have both symmetry and redundancy. Any one of the edge
equations is implies by the others, and we make a choice to remove one of
them, and replace it by one peripheral equation for a fixed peripheral curve,
resulting into $N\times N$ matrices $\mathbf{G}$, $\mathbf{G}^{\prime}$ and
$\mathbf{G}^{\prime\prime}$. Using the second Equation (13) in logarithmic
form $\log z_{j}+\log z^{\prime}_{j}+\log z^{\prime\prime}_{j}=\pi i$, we can
eliminate one of the three shapes of each tetrahedron (this is a choice of
quad). For example, eliminating the shape $z^{\prime}_{j}$ for $j=1,\dots,N$
then results into a system of equations

$\sum_{j=1}^{N}A_{i,j}\log z_{j}+B_{i,j}\log z^{\prime\prime}_{j}\;=\;\pi
i\nu\qquad(i=1,\dots,N)$ (16)

where

$A\;=\;\mathbf{G}-\mathbf{G}^{\prime},\qquad
B\;=\;\mathbf{G}^{\prime\prime}-\mathbf{G}^{\prime}$ (17)

are the Neumann–Zagier matrices [28] and
$\nu=(2,\dots,2,0)^{t}-\mathbf{G}^{\prime}(1,\dots,1,1)^{t}\in\mathbbm{Z}^{N}$.
The Neumann–Zagier matrices $(A\,|\,B)$ have an important symplectic property,
they are the upper part of a symplectic matrix over $\mathbbm{Z}[1/2]$ (and
even a symplectic matrix over $\mathbbm{Z}$ if one divides the peripheral
gluing equation by 2 while keeping the integrality of its coefficients) [28].
It follows that $AB^{t}$ is symmetric, that $(A\,|\,B)$ has full rank, and
that one can always choose a quad such that $B$ is invertible–for this see
[10, Lem.A.3].

A further ingredient of a Neumann–Zagier datum is a flattening, that is a
triple $(f,f^{\prime},f^{\prime\prime})$ of vectors in $\mathbbm{Z}^{N}$ that
satisfy the conditions

$f+f^{\prime}+f^{\prime\prime}=(1,\dots,1)^{t},\qquad\mathbf{G}f+\mathbf{G}^{\prime}f^{\prime}+\mathbf{G}^{\prime\prime}f^{\prime\prime}=(2,\dots,2,0)^{t}\,.$
(18)

Using our choice of quad, we can eliminate $f^{\prime}$ and thus obtain a pair
$(f,f^{\prime\prime})$ of vectors in $\mathbbm{Z}^{N}$ that satisfy the
condition

$Af+Bf^{\prime\prime}=\nu\,.$ (19)

This defines all the terms that appear in a NZ-datum
$\Xi=(A,B,\nu,z,f,f^{\prime\prime})$. The definition of the series
$\Phi^{\Xi}(\hbar)$ requires a nondegenerate NZ datum $\Xi$, that is one that
satisfies the condition $\det(B)\neq 0$, which as we discussed above, can
always be achieved, as well as the condition $\det(\Lambda)\neq 0$. We discuss
this choice next, and connect it to the geometric representation of a
hyperbolic 3-manifold $M$.

We end this section with a comment regarding Neumann–Zagier data of
3-manifolds with several (as opposed to one) torus boundary components. Their
ideal triangulations with equal number $N$ of tetrahedra as edges and the edge
gluing equations have the same shape (14) as above. If $r$ denotes the number
of boundary components, then there are $2r$ peripheral equations (15) and
after a choice of one peripheral curve per boundary component, this leads to
$(N+r)\times N$ matrices $G$, $G^{\prime}$ and $G^{\prime\prime}$. The edge
gluing equations have redundancy, and although it is not true that we can
remove any $r$ of them, it is shown in [17, Sec.4.6] that one can remove $r$
of them and and replace them by $r$ peripheral equations so as to obtain
$N\times N$ matrices $\mathbf{G}$, $\mathbf{G}^{\prime}$ and
$\mathbf{G}^{\prime\prime}$ such that the corresponding matrices $(A\,|\,B)$
defined in (17) have the same symplectic properties as in the case of $r=1$.
Moreover, any two choices of removal of the $r$ edge equations are related to
each other by an invertible matrix in $\mathrm{GL}_{r}(\mathbbm{Z})$. Finally,
flattenings satisfy Equations (18) where the right hand side of the second
equation is the vector $(2,\dots,2,0,\dots,0)^{t}\in\mathbbm{Z}^{N+r}$ with
the first $N$ coordinates equal to $2$ and the remaining $r$ coordinates equal
to zero.

For simplicity in the presentation (related to the choice of peripheral curves
and the flattenings), we will give the proof of Theorem 1.1 assuming that the
3-manifold $M$ has one cusp. The proof applies verbatim to the general case of
arbitrary number of cusps.

###  2.3. Geometric aspects

In this section, we discuss some geometric aspects of Theorem 1.1 related to
the choice of the shapes $z$ in a Neumann–Zagier datum. Let us fix an ideal
triangulation $\mathcal{T}$. A solution
$z\in(\mathbbm{C}\setminus\\{0,1\\})^{N}$ to the Neumann–Zagier equations
gives rise, via a developing map, to a representation
$\rho_{z}:\pi_{1}(M)\to\mathrm{PSL}_{2}(\mathbbm{C})$. For a detailed
discussion, see the appendix of [5]. However, not every representation
$\rho:\pi_{1}(M)\to\mathrm{PSL}_{2}(\mathbbm{C})$ is “seen” by $\mathcal{T}$,
that is, is in the image of the map $z\to\rho_{z}$. What’s more, if $\rho$ is
in the image of the above map and we do a 2–3 Pachner move on the
triangulation, it may no longer be in the image of the map corresponding to
the new triangulation. The reason is that the shapes of the two triangulations
are related by a rational map, which may send a complex number different from
$0$ and $1$ to $0$, $1$ or $\infty$. To phrase the problem differently, every
two ideal triangulations (each with at least two tetrahedra, as we will always
assume) of a 3-manifold with non-empty boundary are related by a sequence of
2–3 Pachner moves [27, 31]. However, it is not known that the set of ideal
triangulations that see the discrete faithful representation
$\rho^{\mathrm{geom}}:\pi_{1}(M)\to\mathrm{PSL}_{2}(\mathbbm{C})$ is connected
under 2–3 Pachner moves, nor is it known whether the set of nondegenerate NZ
data is connected under 2–3 Pachner moves.

A solution to these issues was found in [10], and this was used to prove the
topological invariance of the 1-loop function, and was also used in [17] to
prove the topological invariance of the 3D-index. Let us recall the geometric
details here. Every cusped hyperbolic 3-manifold $M$ (complete, finite volume)
has a canonical ideal cell decomposition whose cells are 3-dimensional convex
ideal polytopes given by Epstein–Penner [12]. It is easy to see that every
convex ideal polyhedron can be triangulated into ideal tetrahedra by
connecting an ideal vertex to all other ideal vertices (thus reducing the
problem to the ideal cone of an ideal polygon), and then triangulating further
the ideal polygon into ideal triangles. Doing so, the triangulation of the
common faces of the 3-dimensional convex ideal polytopes may not match, in
which case one can pass from one triangulation of a polygonal face to another
by adding flat tetrahedra.

The question is whether every two such triangulations are related by a
sequence of 2–3 moves. This is a combinatorial problem of convex geometry,
which we summarize below. For a detailed discussion, the reader may consult
the book [8] and references therein.

Fix a convex polytope $P$ in $\mathbbm{R}^{d}$. One can consider the set of
triangulations of $P$. When $d=2$, $P$ is a polygon and it is known that every
two triangulations are related by a sequence of flips. For general $d$, flips
are replaced by geometric bistellar moves. When $d\geq 5$, it is known that
the graph of triangulations (with edges given by geometric bistellar flips) is
not connected, and has isolated vertices. For $d=3$, it is not known whether
the graph is connected.

The situation is much better when one considers regular triangulations of $P$.
In that case, the corresponding graph of regular triangulations is connected
and, in fact, it is the edge set of the secondary polytope of $P$. When $d=3$
and $P$ is convex and in general position, then the only geometric bistellar
move is the 2–3 move where the added edge that appears in the move is an edge
that connects two vertices of $P$. When $d=3$ and $P$ is not in general
position, the same conclusion holds as long as one allows for tetrahedra that
are flat, i.e., lie on a 2-dimensional plane.

Returning to the Epstein-Penner ideal cell decomposition, let
$\mathcal{T}^{\mathrm{EP}}$ denote a regular ideal triangulations of the
canonical ideal cell decomposition of $M$. (For a detailed discussion of this
set, see also [17, Sec.6]). The set $\\{\mathcal{T}^{\mathrm{EP}}\\}$ of
regular ideal triangulations is connected by 2–3 Pachner moves. Moreover, such
triangulations see $\rho^{\mathrm{geom}}$ since by the geometric construction,
the shapes are always nondegenerate, i.e., not equal to $0$ or $1$ and in fact
always have nonnegative (though sometimes zero) imaginary part.

Finally, we need to show that $\det(\Lambda)$ is nonzero. This follows from
the fact that
$\det(B)\det(\Lambda)=\det(-A+B\mathrm{diag}(1/(1-z_{j}))$ equals (up to
multiplication by a monomial in $z$ and $z^{\prime\prime}$) to the 1-loop
invariant of [10, Sec.1.3]. The nonvanishing of the latter follows from
Thurston’s hyperbolic Dehn surgery theorem [32], which implies that
$\rho^{\mathrm{geom}}\in X_{M}^{\mathrm{geom}}\cup P_{M}$ is an isolated
smooth point of the geometric component $X^{\mathrm{geom}}_{M}$ of the
$\mathrm{PSL}_{2}(\mathbbm{C})$-character variety of $M$, intersected with the
locus $P_{M}$ of boundary-parabolic
$\mathrm{PSL}_{2}(\mathbbm{C})$-representations, i.e., representations $\rho$
satisfying $\mathrm{tr}(\rho(\gamma))^{2}=4$ for all peripheral elements
$\gamma\in\pi_{1}(M)$. Since the 1-loop invariant is the determinant of the
Hessian of the defining NZ equations of $\rho^{\mathrm{geom}}$, it follows
that the 1-loop invariant is nonzero.

##  3\. Formal Gaussian integration

###  3.1. Basics on formal Gaussian integration

In this section, we review the basic properties of formal Gaussian
integration, which is a combinatorial analogue of integration of analytic
functions. This theory was introduced and studied in detail in [3], where it
was used to define a universal perturbative finite type invariant of rational
homology 3-spheres, and to identify it with the trivial connection
contribution of Chern–Simons perturbation theory.

As a warm-up, the formal Gaussian integral of a monomial $x^{n}$ in one
variable with respect to the quadratic form $\lambda\neq 0$ is defined by

$\displaystyle\left\langle
x^{n}\right\rangle_{x,\lambda}\;=\;\begin{cases}\lambda^{-n/2}(n-1)!!&n\,\,\text{even}\\\
0&n\,\,\text{odd}\,.\end{cases}$ (20)

When $\lambda>0$, then the above bracket equals to a normalized Gaussian
integral

$\left\langle
x^{n}\right\rangle_{x,\lambda}\;=\;\frac{\int_{\mathbbm{R}}e^{-\frac{1}{2}\lambda
x^{2}}x^{n}\,dx}{\int_{\mathbbm{R}}e^{-\frac{1}{2}\lambda x^{2}}\,dx}\,,$ (21)

explaining the naming of formal Gaussian integration. The formal Gaussian
integration can be extended by linearity to $\langle f(x)\rangle_{x,\lambda}$
for any polynomial $f(x)$, or even further to a power series $f(x)=\sum_{n\geq
0}a_{n}x^{n}$ whose coefficients tend to zero in a local ring (such as the
ring $\mathbbm{Q}\llbracket\hbar^{\frac{1}{2}}\rrbracket$).

The formal Gaussian integral (20) has a multivariable extension given in
Equation (5) where $x$ is a vector of variables and $\Lambda$ is an invertible
matrix over a field matching the size of $x$. Throughout this paper, the
entries of $\Lambda$ are elements of the field $\mathbbm{Q}(z)$ where $z$ is a
vector of variables, the integrable functions $f_{\hbar}(x,z)$ that we apply
the formal Gaussian integration with respect to $x$ with be elements of
$\mathbbm{Q}(z)[x]\llbracket\hbar^{\frac{1}{2}}\rrbracket$, and the result of
formal Gaussian integration will be an element of the ring
$\mathbbm{Q}(z)\llbracket\hbar\rrbracket$ or
$\mathbbm{Q}(z)[x]\llbracket\hbar^{\frac{1}{2}}\rrbracket$. When we specialize
to the case of a vector of algebraic numbers $z$, then $\mathbbm{Q}(z)$
defines the corresponding number field.

Just like integration of sufficiently smooth functions satisfies certain
invariance properties (such as change of variables, iterated integration, and
even integration by parts [3]), so does formal Gaussian integration. The
corresponding identities of formal Gaussian integration are combinatorial
statements about polynomials or rational functions and often follow from the
corresponding statements of integration of functions.

We now give some elementary properties of formal Gaussian integration that we
will use in our paper.

######  Lemma 3.1.

(a) For all invertible matrices $\Lambda$ and $P$ over $\mathbbm{Q}(z)$, we
have

$\displaystyle\left\langle f_{\hbar}(Px,z)\right\rangle_{x,P^{t}\Lambda
P}\;=\;\left\langle f_{\hbar}(x,z)\right\rangle_{x,\Lambda}\,.$ (22)

(b) For all invertible matrices $\Lambda$ and vectors $G$ over
$\mathbbm{Q}(z)$, we have

$\langle\exp(-G^{t}\Lambda
x\hbar^{\frac{1}{2}})f_{\hbar}(x+G\hbar^{\frac{1}{2}},z)\rangle_{x,\Lambda}\;=\;\exp\left(\frac{G^{t}\Lambda
G}{2}\hbar\right)\langle f_{\hbar}(x,z)\rangle_{x,\Lambda}$ (23)

(c) If $\Lambda=\left(\begin{smallmatrix}A&B\\\
B^{t}&C\end{smallmatrix}\right)$, and $\Lambda$ and $A$ invertible, then for
any $F$, we have

$\displaystyle\left\langle\exp(Fx^{\prime}\hbar^{\frac{1}{2}})f_{\hbar}(x^{\prime\prime},z)\right\rangle_{x,\Lambda}\;=\;\exp\left(\frac{FA^{-1}F^{t}}{2}\hbar\right)\left\langle\exp(-FA^{-1}Bx^{\prime\prime}\hbar^{\frac{1}{2}})f_{\hbar}(x^{\prime\prime},z)\right\rangle_{x^{\prime\prime},C-B^{t}A^{-1}B}\\!\\!.$
(24)

###### Proof.

Part (a) follows from the fact the integration is unchanged under a linear
change of variables.

Part (b) follows from the fact the integration is unchanged under an affine
change of variables.

Part (c) follows from Fubini’s theorem [3, Prop.2.13], combined with Equation
(23). ∎

The next lemma, which will be important in the application of $q$-holonomic
methods in Section 3.2 and in the proofs of Theorems 3.4 and 3.6 below,
concerns the behavior of formal Gaussian integration when
$z=(z_{1},\dots,z_{r})$ is shifted to
$e^{\varepsilon\hbar}z:=(e^{\varepsilon_{1}\hbar}z_{1},\dots,e^{\varepsilon_{r}\hbar}z_{r})$
for integers $\varepsilon_{j}$.

######  Lemma 3.2.

For all invertible matrices $\Lambda(z)$ over $\mathbbm{Q}(z)$ and all
integrable functions $f_{\hbar}$, we have

$\langle f_{\hbar}(x,z)\rangle_{x,\Lambda(z)}|_{z\mapsto
e^{\varepsilon\hbar}z}\;=\;\sqrt{\frac{\det\Lambda(e^{\varepsilon\hbar}z)}{\det\Lambda(z)}}\Big{\langle}\exp\bigg{(}\frac{x^{t}\big{(}\Lambda(z)-\Lambda(e^{\varepsilon\hbar}z)\big{)}x}{2}\bigg{)}f_{\hbar}(x,e^{\varepsilon\hbar}z)\Big{\rangle}_{x,\Lambda(z)}\,.$
(25)

###### Proof.

The lemma follows from recentering the Gaussian after multiplying $z$ by
$e^{\varepsilon\hbar}$. ∎

###  3.2. $q$–holonomic aspects of $\psi_{\hbar}$

It is well-known that identities of holonomic functions can be proven
algorithmically [34, 30]. We will use these ideas, adapted to our needs, to
prove fundamental identities between certain Gaussian integrals involving the
building block $\exp(\psi_{\hbar})$ of our series $\Phi^{\Xi}(\hbar)$. Since
$\psi_{\hbar}$ is related to the infinite Pochhammer symbol given in Equation
(2), its functional equations will be of fundamental importance. From its
definition, it is clear that the infinite Pochhammer symbol satisfies a simple
first order linear $q$–difference equation

$(x;q)_{\infty}\;=\;(1-x)(qx;q)_{\infty}\,.$ (26)

To convert this into an identity of formal $\hbar$-series where $q=e^{\hbar}$,
we use the fact that there is an action of the quantum plane (also known as
$q$–Weyl [13, Ex. 1.7] algebra) on the space
$\mathbbm{Q}(z)[x]\llbracket\hbar^{\frac{1}{2}}\rrbracket$ of integrable
functions by an action on the $z$-variable given by

$(Lf_{\hbar})(x,z)\;=\;f_{\hbar}(x,e^{\hbar}z),\qquad(Mf_{\hbar})(x,z)\;=\;e^{\hbar}f_{\hbar}(x,e^{\hbar}z)$
(27)

where $LM=qML$. The next lemma asserts that the completion
$\widehat{\psi}_{\hbar}(x,z)$

$\displaystyle\widehat{\psi}_{\hbar}(x,z)\;:$
$\displaystyle=\;\exp\Big{(}\\!-\frac{\operatorname{Li}_{2}(z)}{\hbar}-\frac{\operatorname{Li}_{1}(z)x}{\hbar^{\frac{1}{2}}}+\frac{1}{2}\operatorname{Li}_{1}(z)-\frac{\operatorname{Li}_{0}(z)x^{2}}{2}\Big{)}\psi_{\hbar}(x,z)$
(28) $\displaystyle=\;\exp\Big{(}\\!-\sum_{k,\ell\in\mathbbm{Z}_{\geq
0}}\frac{B_{k}\,x^{\ell}\,\hbar^{k+\frac{\ell}{2}-1}}{\ell!\,k!}\operatorname{Li}_{2-k-\ell}(z)\Big{)}$
$\displaystyle\quad\in\exp\Big{(}\\!-\frac{\operatorname{Li}_{2}(z)}{\hbar}-\frac{\operatorname{Li}_{1}(z)x}{\hbar^{\frac{1}{2}}}+\frac{1}{2}\operatorname{Li}_{1}(z)-\frac{\operatorname{Li}_{0}(z)x^{2}}{2}\Big{)}(1+\hbar^{\frac{1}{2}}\mathbbm{Q}(z)[x]\llbracket\hbar^{\frac{1}{2}}\rrbracket)$

of $\psi_{\hbar}(x,z)$, as well as $\psi_{\hbar}(x,z)$ itself, satisfy a
corresponding first order linear $q$–difference equation, albeit with
complicated coefficients.

######  Lemma 3.3.

(a) We have:

$\widehat{\psi}_{\hbar}(x,e^{\hbar}z)\;=\;(1-ze^{x\hbar^{1/2}})\,\widehat{\psi}_{\hbar}(x,z)\,.$
(29)

(b) We have:

$\displaystyle\psi_{\hbar}(x,e^{\hbar}z)\;=\;(1-ze^{x\hbar^{1/2}})\,\psi_{\hbar}(x,z)\sqrt{\frac{1-e^{\hbar}z}{1-z}}$
(30)
$\displaystyle\times\exp\Big{(}\frac{\operatorname{Li}_{2}(e^{\hbar}z)}{\hbar}-\frac{\operatorname{Li}_{2}(z)}{\hbar}+\frac{\operatorname{Li}_{1}(e^{\hbar}z)x}{\hbar^{\frac{1}{2}}}-\frac{\operatorname{Li}_{1}(z)x}{\hbar^{\frac{1}{2}}}+\frac{1}{2}\operatorname{Li}_{0}(e^{\hbar}z)x^{2}-\frac{1}{2}\operatorname{Li}_{0}(z)x^{2}\Big{)}\,.$

Note that the identity (29) takes place in a larger ring, which includes the
symbols $\exp(\frac{\operatorname{Li}_{2}(z)}{\hbar})$,
$\exp(\frac{\operatorname{Li}_{1}(z)x}{\hbar^{\frac{1}{2}}})$,
$\exp(\frac{\operatorname{Li}_{0}(z)x^{2}}{2})$ and
$\exp(\frac{\operatorname{Li}_{0}(z)x^{2}}{2})$ which one can adjoin in the
differential field $\mathbbm{Q}(z)((\hbar))$ as is standard in differential
Galois theory of linear differential equations [33]. The symbols
$\operatorname{Li}_{k}(z)$ for $k=0,1,2$ can be interpreted as normalized
solutions to the linear differential equations
$z\frac{d}{dz}\operatorname{Li}_{k}(z)=\operatorname{Li}_{k-1}(z)$ with
$\operatorname{Li}_{0}(z)=z/(1-z)$ and satisfy the usual properties
$\operatorname{Li}_{k}(e^{\hbar}z)=\sum_{r=0}^{\infty}\frac{1}{r!}\operatorname{Li}_{k-r}(z)\hbar^{r}$.
On the other hand, the coefficients in identity (30) are elements of the ring
$\mathbbm{Q}(z)[x]\llbracket\hbar^{\frac{1}{2}}\rrbracket$.

###### Proof.

Part (a) can be proved directly using the facts that for the Bernoulli
polynomials $B_{r}(x)$ we have $B_{r}(1)=B_{r}+\delta_{r,1}$ and
$B_{r}(x)=\sum_{k=0}^{r}\binom{r}{k}B_{n-k}x^{k}$. Applying these identities
we find that

$\displaystyle\exp\Big{(}-\sum_{k,\ell\in\mathbbm{Z}_{\geq
0}}\frac{B_{k}x^{\ell}\hbar^{k+\frac{\ell}{2}-1}}{\ell!k!}\operatorname{Li}_{2-k-\ell}(ze^{\hbar})\Big{)}$
(31) $\displaystyle\;=\;\exp\Big{(}-\sum_{r,\ell\in\mathbbm{Z}_{\geq
0}}\sum_{k=0}^{r}B_{k}\binom{r}{k}\frac{x^{\ell}\hbar^{r+\frac{\ell}{2}-1}}{\ell!r!}\operatorname{Li}_{2-r-\ell}(z)\Big{)}$
$\displaystyle\;=\;\exp\Big{(}-\sum_{r,\ell\in\mathbbm{Z}_{\geq
0}}\frac{B_{r}(1)x^{\ell}\hbar^{r+\frac{\ell}{2}-1}}{\ell!r!}\operatorname{Li}_{2-r-\ell}(z)\Big{)}$
$\displaystyle\;=\;\exp\Big{(}-\operatorname{Li}_{1}(ze^{x\hbar^{1/2}})-\sum_{r,\ell\in\mathbbm{Z}_{\geq
0}}\frac{B_{r}x^{\ell}\hbar^{r+\frac{\ell}{2}-1}}{\ell!r!}\operatorname{Li}_{2-r-\ell}(z)\Big{)}$
$\displaystyle\;=\;(1-ze^{x\hbar^{1/2}})\exp\Big{(}-\sum_{r,\ell\in\mathbbm{Z}_{\geq
0}}\frac{B_{r}x^{\ell}\hbar^{r+\frac{\ell}{2}-1}}{\ell!r!}\operatorname{Li}_{2-r-\ell}(z)\Big{)}\,.$

Part (b) follows from (a), using Equation (28) and expanding in $\hbar$ to
find that

$\displaystyle\sqrt{\frac{1-e^{\hbar}z}{1-z}}\exp\Big{(}\frac{\operatorname{Li}_{2}(e^{\hbar}z)}{\hbar}-\frac{\operatorname{Li}_{2}(z)}{\hbar}+\frac{\operatorname{Li}_{1}(e^{\hbar}z)x}{\hbar^{\frac{1}{2}}}-\frac{\operatorname{Li}_{1}(z)x}{\hbar^{\frac{1}{2}}}+\frac{1}{2}\operatorname{Li}_{0}(e^{\hbar}z)x^{2}-\frac{1}{2}\operatorname{Li}_{0}(z)x^{2}\Big{)}\in\mathbbm{Q}(z)[x]\llbracket\hbar^{\frac{1}{2}}\rrbracket\,.$
(32)

∎

We end this section by discussing a useful relation between
$\psi_{\hbar}(x,z)$ and its specialization at $x=0$. It is easy to see that
the completion $\widehat{\psi}_{\hbar}(x,z)$ of $\psi_{\hbar}(x,z)$ is regular
at $x=0$ and satisfies

$\widehat{\psi}_{\hbar}(x,z)\;=\;\widehat{\psi}_{\hbar}(0,ze^{x\hbar^{1/2}})\,.$
(33)

This implies a corresponding statement

$\displaystyle\psi_{\hbar}(x,z)\;=\;\psi_{\hbar}\big{(}0,ze^{x\hbar^{1/2}}\big{)}C_{\hbar}(x,z),$
(34)

for $\psi_{\hbar}$ where

$\displaystyle C_{\hbar}(x,z)$
$\displaystyle\;=\;\exp\Big{(}\\!-\sum_{\ell\geq
3}\frac{\hbar^{\frac{\ell}{2}-1}}{\ell!}\operatorname{Li}_{2-\ell}(z)x^{\ell}+\frac{1}{2}\sum_{\ell\geq
1}\frac{h^{\frac{\ell}{2}}}{\ell!}\operatorname{Li}_{1-\ell}(z)x^{\ell}\Big{)}$
(35)
$\displaystyle\;=\;\exp\Big{(}\\!-\frac{\operatorname{Li}_{2}(ze^{x\hbar^{\frac{1}{2}}})}{\hbar}+\frac{\operatorname{Li}_{2}(z)}{\hbar}-\frac{\log(1-z)}{\hbar^{\frac{1}{2}}}x+\frac{z}{1-z}\frac{x^{2}}{2}$
$\displaystyle\qquad\qquad-\frac{1}{2}(\log(1-ze^{z\hbar^{\frac{1}{2}}})-\log(1-z))\Big{)}\,.$

###  3.3. Fourier transform of $\psi_{\hbar}$

In this section, we prove two functional identities for
$\psi_{\hbar}(\bullet,z)$, $\psi_{\hbar}(\bullet,z^{\prime})$ and
$\psi_{\hbar}(\bullet,z^{\prime\prime})$ where $z^{\prime}:=1/(1-z)$ and
$z^{\prime\prime}:=1-1/z$ are related to the
$\mathbbm{Z}/3\mathbbm{Z}$-symmetry of the shapes of a tetrahedron, and
reflect the $\mathbbm{Z}/3\mathbbm{Z}$-symmetry of the dilogarithm function.

The next theorem is a formal Gaussian integration version of the Fourier
transform of the Faddeev quantum dilogarithm given in Equation (9). The proof,
however, does not follow from the distributional identity (9), and instead
uses $q$–holonomic ideas and the basics of formal Gaussian integration.

This theorem and its Corollary 3.5 are used in Section 5 to show that the
power series $\Phi^{\Xi}(\hbar)$ are independent of the choice of a
nondegenerate quad.

######  Theorem 3.4.

We have:

$\displaystyle\psi_{\hbar}(x,z)\;=\;e^{-\frac{\hbar}{24}}\Big{\langle}\exp\Big{(}\Big{(}y+\frac{xz}{1-z}\Big{)}\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}\,\psi_{\hbar}\Big{(}y+\frac{xz}{1-z},\frac{1}{1-z}\Big{)}\Big{\rangle}_{y,1-z^{-1}}$
(36)

in the ring $\mathbbm{Q}(z)[x]\llbracket\hbar^{\frac{1}{2}}\rrbracket$.

In fact, both sides of of Equation (36) lie in the subring
$\mathbbm{Q}[z^{\pm},(1-z)^{-1}][x]\llbracket\hbar^{\frac{1}{2}}\rrbracket$.

###### Proof.

It it clear from the definition that both sides of Equation (36) are elements
of the ring $\mathbbm{Q}(x)[z]\llbracket\hbar\rrbracket$. First, we will prove
the specialization of Equation (36) when $x=0$, i.e., we will show that

$\displaystyle\psi_{\hbar}(0,z)\;=\;e^{-\frac{\hbar}{24}}\Big{\langle}\exp\Big{(}\frac{y}{2}\hbar^{\frac{1}{2}}\Big{)}\,\psi_{\hbar}\Big{(}y,\frac{1}{1-z}\Big{)}\Big{\rangle}_{y,1-z^{-1}}\in\mathbbm{Q}(z)[[\hbar]]\,.$
(37)

To prove this, we will combine $q$-holonomic ideas with formal Gaussian
integration. From Equation (29), we see that the left hand side of Equation
(37) multiplied by
$\frac{\exp(-\operatorname{Li}_{2}(z)\hbar^{-1})}{\sqrt{1-z}}$ satisfies a
simple first order $q$–difference equation. We want to show the same for the
right hand side. To do this, consider the function

$\displaystyle
I_{m,\hbar}(w,z)\;=\;w^{m}e^{-\frac{\hbar}{24}+\frac{\pi^{2}}{6\hbar}}\exp\Big{(}\log(w)\Big{(}\\!-\frac{\log(w)}{2\hbar}+\frac{\pi
i}{\hbar}-\frac{\log(z)}{\hbar}+\frac{1}{2}\Big{)}\Big{)}\,\widehat{\psi}_{\hbar}(0,w)$
(38)

which is an element of a larger ring discussed in relation to
$\widehat{\psi}_{\hbar}$ of Equation (28). Using Equation (34) and the
symmetry

$\operatorname{Li}_{2}(z)\;=\;\operatorname{Li}_{2}\Big{(}\frac{1}{1-z}\Big{)}-\frac{1}{2}\log({1-z})^{2}+\pi
i\log({1-z})-\log(z)\log({1-z})-\frac{\pi^{2}}{6}$ (39)

of the dilogarithm [36], it is easy to see that the right hand side of
Equation (37) is given by

$\Big{\langle}\sqrt{-z}\exp\Big{(}\frac{\operatorname{Li}_{2}(z)}{\hbar}+(1-z^{-1})\frac{w^{2}}{2}\Big{)}I_{0,\hbar}\Big{(}\frac{1}{1-z}e^{w\hbar^{1/2}},z\Big{)}\Big{\rangle}_{w,1-z^{-1}}\,.$
(40)

The function $I_{m,\hbar}$ satisfies the linear $q$–difference equations

$\displaystyle I_{m,\hbar}(e^{\hbar}w,z)$
$\displaystyle\;=\;-e^{m\hbar}w^{-1}z^{-1}(1-w)I_{m,\hbar}(w,z)\,,$ (41)
$\displaystyle I_{m,\hbar}(w,e^{\hbar}z)$
$\displaystyle\;=\;w^{-1}I_{m,\hbar}(w,z)\,,$ $\displaystyle I_{m,\hbar}(w,z)$
$\displaystyle\;=\;wI_{m,\hbar}(w,z)\,,$

which imply that

$(1-e^{-m\hbar}z)I_{m,\hbar}\Big{(}\frac{1}{1-z}e^{w\hbar^{1/2}+\hbar},z\Big{)}\;=\;I_{m}\Big{(}\frac{1}{1-z}e^{w\hbar^{1/2}},e^{\hbar}z\Big{)}.$
(42)

If we multiply both sides of (42) with the factors from Equation (40) and take
the bracket of both sides and apply Lemma 3.1 and Lemma 3.2 to change
coordinates and the Gaussian, we find that when $m=0$ both sides of Equation
(37) satisfy the same $q$–difference equation (29). Moreover, it is easy to
see that both sides of Equation (40) are power series in $\hbar$ with
coefficients rational functions of $z$ of nonpositive degree, thus we can work
with the ring $\mathbbm{Q}\llbracket
z^{-1}\rrbracket\llbracket\hbar\rrbracket$ instead. In this case, the Newton
polygon of this first order linear $q$-difference equation implies that it has
a unique solution in $\mathbbm{Q}\llbracket
z^{-1}\rrbracket\llbracket\hbar\rrbracket$ determined by its value at
$z=\infty$. Therefore, the identity in Equation (37) follows from its
specialization at $z=\infty$. Since

$\displaystyle\psi_{\hbar}(0,\infty)$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{12}\Big{)}\,,$ (43) $\displaystyle
e^{\frac{-\hbar}{24}}\left\langle\exp\Big{(}\frac{w}{2}\hbar^{\frac{1}{2}}\Big{)}\psi_{\hbar}\left(w,0\right)\right\rangle_{w,1}$
$\displaystyle\;=\;e^{\frac{-\hbar}{24}}\left\langle\exp\left(\frac{w}{2}\hbar^{\frac{1}{2}}\right)\right\rangle_{w,1}\;=\;\exp\Big{(}\frac{\hbar}{12}\Big{)}\,,$

this completes the proof of Equation (37).

Going back to the general case of $z$, Equation (36) follows from Equation
(34) together with (37) using Lemma 3.2 to shift the Gaussian and Lemma 3.1 to
change integration variable via the transformation

$w\mapsto
w-\frac{1}{\hbar^{\frac{1}{2}}}\log\left(\frac{1-z}{1-ze^{x\hbar^{1/2}}}\right)-\frac{x}{1-z^{-1}}\,.$
(44)

The detailed calculation is given in Appendix A. ∎

Theorem 3.4 implies the following identity relating $\psi_{\hbar}(\bullet,z)$
and $\psi_{\hbar}(\bullet,z^{\prime\prime})$.

######  Corollary 3.5.

We have:

$\displaystyle\psi_{\hbar}(x,z)\;=\;e^{\frac{\hbar}{24}}\Big{\langle}\exp\Big{(}\\!-\frac{x}{2}\hbar^{\frac{1}{2}}\Big{)}\,\psi_{\hbar}\Big{(}y-\frac{x}{1-z},1-z^{-1}\Big{)}\Big{\rangle}_{y,z-1}$
(45)

in the ring $\mathbbm{Q}(z)[x]\llbracket\hbar^{\frac{1}{2}}\rrbracket$.

###### Proof.

Apply Equation (36) to the $\psi_{\hbar}$ that appears on the right hand side
of the same Equation (36). Then apply a change of variables and Fubini’s
theorem of Lemma 3.1 to calculate the bracket for the variable that doesn’t
appear in the argument of the remaining $\psi_{\hbar}$. ∎

###  3.4. Pentagon identity for $\psi_{\hbar}$

In this section, we prove a pentagon identity for the functions
$\psi_{\hbar}(\bullet,z)$ where $z$ takes the five values

$[z_{1}]+[z_{2}]\mapsto[z_{1}z_{0}^{-1}]+[z_{0}]+[z_{2}z_{0}^{-1}],\qquad
z_{0}\;=\;z_{1}+z_{2}-z_{1}z_{2}$ (46)

what appear in the 5-term relation for the dilogarithm. The next theorem is a
formal Gaussian integration version of the pentagon identity (8) of the
Faddeev quantum dilogarithm and will be used in Section 6 to prove that the
series $\Phi^{\Xi}(\hbar)$ is independent of 2–3 Pachner moves. Let us denote

$\delta\;=\;2+\operatorname{Li}_{0}(z_{1}z_{0}^{-1})+\operatorname{Li}_{0}(z_{0})+\operatorname{Li}_{0}(z_{2}z_{0}^{-1})\;=\;\frac{(z_{1}+z_{2}-z_{1}z_{2})^{2}}{z_{1}z_{2}(1-z_{1})(1-z_{2})}\,.$
(47)

######  Theorem 3.6.

We have:

$\displaystyle\psi_{\hbar}(x,z_{1})\,\psi_{\hbar}(y,z_{2})\;=\;e^{-\frac{\hbar}{24}}\Big{\langle}\psi_{\hbar}\Big{(}-w-y+\frac{xz_{2}+yz_{1}}{z_{0}},z_{1}z_{0}^{-1}\Big{)}$
(48)
$\displaystyle\psi_{\hbar}\Big{(}w+x+y-\frac{xz_{2}+yz_{1}}{z_{0}},z_{0}\Big{)}\psi_{\hbar}\Big{(}-w-x+\frac{xz_{2}+yz_{1}}{z_{0}},z_{2}z_{0}^{-1}\Big{)}\Big{\rangle}_{w,\delta}$

in the ring
$\mathbbm{Q}(z_{1},z_{2})[x,y]\llbracket\hbar^{\frac{1}{2}}\rrbracket$.

###### Proof.

It is clear from the definition that both sides of Equation (48) are elements
of the ring
$\mathbbm{Q}(z_{1},z_{2})[x,y]\llbracket\hbar^{\frac{1}{2}}\rrbracket$. To
prove this, we follow the same approach we used to prove Theorem 3.4. First,
we prove the identity (48) when $u=v=0$, i.e., we will show that

$\psi_{\hbar}(0,z_{1})\,\psi_{\hbar}(0,z_{2})\;=\;e^{-\frac{\hbar}{24}}\big{\langle}\psi_{\hbar}(-w,z_{1}z_{0}^{-1})\,\psi_{\hbar}(w,z_{0})\,\psi_{\hbar}(-w,z_{2}z_{0}^{-1})\big{\rangle}_{w,\delta}$
(49)

in the ring $\mathbbm{Q}(z_{1},z_{2})\llbracket\hbar\rrbracket$ and to prove
this we will again use $q$–holonomic methods. To do so, consider the function

$\displaystyle
I_{m,\hbar}(z_{1},z_{2},z)\;=\;e^{\frac{\pi^{2}}{6\hbar}-\frac{\hbar}{24}}z^{m}\,\widehat{\psi}_{\hbar}(0,z_{1}z^{-1})\,\widehat{\psi}_{\hbar}(0,z)\,\widehat{\psi}_{\hbar}(0,z_{2}z^{-1})$
(50)
$\displaystyle\times\exp\Big{(}-\frac{\log(z_{1})\log(z_{2})}{\hbar}+\frac{\log(z)\log(z_{1})}{\hbar}+\frac{\log(z)\log(z_{2})}{\hbar}-\frac{\log(z)^{2}}{\hbar}\Big{)}\,,$

which is again an element of a larger ring discussed in relation to
$\widehat{\psi}_{\hbar}$ of Equation (28). Using Equation (34) and the five
term relation for the dilogarithm [36], it is easy to see that the right hand
side of Equation (49) is given by

$\Big{\langle}\frac{\sqrt{z_{1}z_{2}}(1-z_{1})(1-z_{2})}{z_{1}+z_{2}-z_{1}z_{2}}\exp\Big{(}\frac{\operatorname{Li}_{2}(z_{1})}{\hbar}+\frac{\operatorname{Li}_{2}(z_{2})}{\hbar}+\frac{\delta}{2}z^{2}\Big{)}I_{0,\hbar}(z_{1},z_{2},z_{0}e^{w\hbar^{1/2}})\Big{\rangle}_{w,\delta}\,.$
(51)

The function $I_{m,\hbar}$ satisfies the system of linear $q$–difference
equations

$\displaystyle I_{m,\hbar}(e^{\hbar}z_{1},z_{2},z)$
$\displaystyle\;=\;z\,z_{2}^{-1}(1-z_{1}z^{-1})\,I_{m,\hbar}(z_{1},z_{2},z)\,,$
(52) $\displaystyle I_{m,\hbar}(z_{1},e^{\hbar}z_{2},z)$
$\displaystyle\;=\;z\,z_{1}^{-1}(1-z_{2}z^{-1})\,I_{m,\hbar}(z_{1},z_{2},z)\,,$
$\displaystyle I_{m,\hbar}(z_{1},z_{2},e^{\hbar}z)$
$\displaystyle\;=\;(1-z)\,z_{1}z_{2}z^{-2}e^{(m-1)\hbar}I_{m,\hbar}(z_{1},z_{2},z)\,,$
$\displaystyle I_{m+1,\hbar}(z_{1},z_{2},z)$
$\displaystyle\;=\;z\,I_{m,\hbar}(z_{1},z_{2},z)\,,$

which can be derived from Equation (29). In fact, the function $I_{m,\hbar}$
is holonomic of rank $1$, a fact that we will not use explicitly [30].
Therefore, we see that

$\displaystyle
I_{m,\hbar}(e^{\hbar}z_{1},z_{2},z)\;=\;z_{2}^{-1}I_{m+1,\hbar}(z_{1},z_{2},z)-z_{1}z_{2}^{-1}I_{m,\hbar}(z_{1},z_{2},z)\,,$
(53) $\displaystyle
I_{m,\hbar}(e^{\hbar}z_{1},z_{2},(e^{\hbar}z_{1}+z_{2}-e^{\hbar}z_{1}z_{2})e^{w\hbar^{1/2}})$
$\displaystyle\;=\;z_{2}^{-1}I_{m+1,\hbar}(z_{1},z_{2},(e^{\hbar}z_{1}+z_{2}-e^{\hbar}z_{1}z_{2})e^{w\hbar^{1/2}})-z_{1}z_{2}^{-1}I_{m,\hbar}(z_{1},z_{2},(e^{\hbar}z_{1}+z_{2}-e^{\hbar}z_{1}z_{2})e^{w\hbar^{1/2}})\,,$
$\displaystyle
I_{m,\hbar}(z_{1},z_{2},(z_{1}+z_{2}-z_{1}z_{2})e^{z\hbar^{1/2}+\hbar})$
$\displaystyle\;=\;z_{1}z_{2}e^{(m-1)\hbar}I_{m-2,\hbar}(z_{1},z_{2},(z_{1}+z_{2}-z_{1}z_{2})e^{z\hbar^{1/2}})-z_{1}z_{2}e^{(m-1)\hbar}I_{m-1,\hbar}(z_{1},z_{2},(z_{1}+z_{2}-z_{1}z_{2})e^{z\hbar^{1/2}})\,.$

Let us define $\widehat{J}_{m,\hbar}$ and $J_{m,\hbar}$ by the equation

$\displaystyle\widehat{J}_{m,\hbar}(z_{1},z_{2})$
$\displaystyle\;=\;\frac{1}{\sqrt{(1-z_{1})(1-z_{2})}}\exp\Big{(}-\frac{\operatorname{Li}_{2}(z_{1})}{\hbar}-\frac{\operatorname{Li}_{2}(z_{2})}{\hbar}\Big{)}J_{m,\hbar}(z_{1},z_{2})$
(54)
$\displaystyle\;=\;\Big{\langle}\frac{\sqrt{z_{1}z_{2}(1-z_{1})(1-z_{2})}}{z_{1}+z_{2}-z_{1}z_{2}}\exp\Big{(}\frac{\delta}{2}z^{2}\Big{)}I_{m,\hbar}(z_{1},z_{2},z_{0}e^{w\hbar^{1/2}})\Big{\rangle}_{w,\delta}\,.$

If we multiply both sides of the Equations (53) with the factors from Equation
(51), take the bracket of both sides, and apply Lemma 3.2 to change
coordinates and the Gaussian, we find that
$\widehat{J}_{\hbar,m}(z_{1},z_{2})$ satisfies the $q$–difference equations

$\displaystyle\widehat{J}_{m,\hbar}(e^{\hbar}z_{1},z_{2})$
$\displaystyle\;=\;z_{2}^{-1}\widehat{J}_{m+1,\hbar}(z_{1},z_{2})-z_{1}z_{2}^{-1}\widehat{J}_{m,\hbar}(z_{1},z_{2})\,,$
(55) $\displaystyle\widehat{J}_{m,\hbar}(z_{1},e^{\hbar}z_{2})$
$\displaystyle\;=\;z_{1}^{-1}\widehat{J}_{m+1,\hbar}(z_{1},z_{2})-z_{1}^{-1}z_{2}\widehat{J}_{m,\hbar}(z_{1},z_{2})\,,$
$\displaystyle\widehat{J}_{m,\hbar}(z_{1},z_{2})$
$\displaystyle\;=\;z_{1}z_{2}e^{(m-1)\hbar}\widehat{J}_{m-2,\hbar}(z_{1},z_{2})-z_{1}z_{2}e^{(m-1)\hbar}\widehat{J}_{m-1,\hbar}(z_{1},z_{2})\,.$

From these relations we can derive the equations

$\displaystyle 0$
$\displaystyle\;=\;(z_{1}-z_{1}e^{(m+1)\hbar}+z_{1}^{2}e^{(m+1)\hbar}-z_{1})\widehat{J}_{m,\hbar}(z_{1},z_{2})$
(56)
$\displaystyle\qquad+(z_{1}z_{2}e^{(m+1)\hbar}-z_{2}+qz_{1})\widehat{J}_{m,\hbar}(e^{\hbar}z_{1},z_{2})+z_{2}\widehat{J}_{m,\hbar}(e^{2\hbar}z_{1},z_{2})\,,$
$\displaystyle 0$
$\displaystyle\;=\;(z_{2}-z_{2}e^{(m+1)\hbar}+z_{2}^{2}e^{(m+1)\hbar}-z_{2})\widehat{J}_{m,\hbar}(z_{1},z_{2})$
$\displaystyle\qquad+(z_{2}z_{1}e^{(m+1)\hbar}-z_{1}+qz_{2})\widehat{J}_{m,\hbar}(z_{1},e^{\hbar}z_{2})+z_{1}\widehat{J}_{m,\hbar}(z_{1},e^{2\hbar}z_{2})\,.$

These $q$–difference equations give rise to equations for $J_{m,\hbar}$ given
by

$\displaystyle 0$
$\displaystyle\;=\;(z_{1}-z_{1}e^{\hbar}+z_{1}^{2}e^{\hbar}-z_{1})J_{m,\hbar}(z_{1},z_{2})$
(57)
$\displaystyle\qquad+(z_{1}z_{2}e^{\hbar}-z_{2}+qz_{1})\sqrt{\frac{1-z_{1}}{1-e^{\hbar}z_{1}}}\exp\Big{(}\frac{\operatorname{Li}_{2}(z_{1})-\operatorname{Li}_{2}(e^{\hbar}z_{1})}{\hbar}\Big{)}J_{m,\hbar}(e^{\hbar}z_{1},z_{2})$
$\displaystyle\qquad+z_{2}\sqrt{\frac{1-z_{1}}{1-e^{2\hbar}z_{1}}}\exp\Big{(}\frac{\operatorname{Li}_{2}(z_{1})-\operatorname{Li}_{2}(e^{2\hbar}z_{1})}{\hbar}\Big{)}J_{m,\hbar}(e^{2\hbar}z_{1},z_{2})\,,$
$\displaystyle 0$
$\displaystyle\;=\;(z_{2}-z_{2}e^{\hbar}+z_{2}^{2}e^{\hbar}-z_{2})J_{m,\hbar}(z_{1},z_{2})$
$\displaystyle\qquad+(z_{2}z_{1}e^{\hbar}-z_{1}+qz_{2})\sqrt{\frac{1-z_{2}}{1-e^{\hbar}z_{2}}}\exp\Big{(}\frac{\operatorname{Li}_{2}(z_{2})-\operatorname{Li}_{2}(e^{\hbar}z_{2})}{\hbar}\Big{)}J_{m,\hbar}(z_{1},e^{\hbar}z_{2})$
$\displaystyle\qquad+z_{1}\sqrt{\frac{1-z_{2}}{1-e^{2\hbar}z_{2}}}\exp\Big{(}\frac{\operatorname{Li}_{2}(z_{2})-\operatorname{Li}_{2}(e^{2\hbar}z_{2})}{\hbar}\Big{)}J_{m,\hbar}(z_{1},e^{2\hbar}z_{2})\,.$

These equations are also satisfied by the left hand side of Equation (49),
which follows from repeated application of Equation (30).

It is easy to see that both sides of Equation (49) are formal power series in
$\hbar$ with coefficients rational functions of $(z_{1},z_{2})$ of nonpositive
degree with respect to both $z_{1}$ and $z_{2}$, thus their coefficients embed
in $\mathbbm{Q}\llbracket z_{1}^{-1},z_{2}^{-1}\rrbracket$, and it suffices to
prove (49) in the ring $\mathbbm{Q}\llbracket
z_{1}^{-1},z_{2}^{-1}\rrbracket\llbracket\hbar\rrbracket$. In this case, we
consider the linear $q$-difference equations (57) over the ring
$\mathbbm{Q}\llbracket
z_{1}^{-1},z_{2}^{-1}\rrbracket\llbracket\hbar\rrbracket$. Looking at the
corresponding Newton polytopes, the leading monomials that appear as
coefficients of this $q$–difference equation (57) at $\infty$ are
$z_{1}^{2}(1+\mathrm{O}(z_{1}^{-1})\mathrm{O}(z_{2}^{0}))$ and
$z_{1}^{2}z_{2}(1+\mathrm{O}(z_{1}^{-1})\mathrm{O}(z_{2}^{0}))$ for the first
equation, and $z_{2}^{2}(1+\mathrm{O}(z_{1}^{0})\mathrm{O}(z_{2}^{-1}))$ and
$z_{1}z_{2}^{2}(1+\mathrm{O}(z_{1}^{0})\mathrm{O}(z_{2}^{-1}))$ for the second
equation, respectively. The structure of these monomials and their Newton
polytopes shows that there is a unique solution to these equations in
$\mathbbm{Q}\llbracket
z_{1}^{-1},z_{2}^{-1}\rrbracket\llbracket\hbar\rrbracket$ determined by the
value at $z_{1}=z_{2}=\infty$. Therefore, the identity in Equation (49)
reduces to its specialization at $z_{1}=z_{2}=\infty$. Since

$\displaystyle\psi_{\hbar}(0,\infty)\,\psi_{\hbar}(0,\infty)$
$\displaystyle\;=\;e^{\frac{\hbar}{6}}\quad\text{and}$ (58) $\displaystyle
e^{-\frac{\hbar}{24}}\big{\langle}\psi_{\hbar}(-w,0)^{2}\,\psi_{\hbar}(w,\infty)\big{\rangle}_{w,\delta}$
$\displaystyle\;=\;e^{-\frac{\hbar}{24}}\Big{\langle}\exp\Big{(}\frac{\hbar}{12}-\frac{z\hbar^{\frac{1}{2}}}{2}\Big{)}\Big{\rangle}_{w,\delta}\;=\;e^{\frac{\hbar}{6}}\,,$

this completes the proof of Equation (49).

Going back to the general case of $x,y$, Equation (48) follows from Equation
(34) together with (49) using Lemma 3.2 to shift the Gaussian and with a
change of integration variable

$w\mapsto
w+\frac{1}{\hbar^{\frac{1}{2}}}\log\left(\frac{z_{1}+z_{2}-z_{1}z_{2}}{z_{1}e^{x_{1}\hbar^{1/2}}+z_{2}e^{x_{2}\hbar^{1/2}}-z_{1}z_{2}e^{(x_{1}+x_{2})\hbar^{1/2}}}\right)+x+y-\frac{xz_{1}+yz_{2}}{z_{0}}\,.$
(59)

The detailed calculation is given in Appendix B. ∎

###  3.5. Inversion formula for $\psi_{\hbar}$

In this section, we give an inversion formula for $\psi_{\hbar}$ analogous to
the inversion formula (11) for the Faddeev quantum dilogarithm. Although we
will not use this formula explicitly in our paper, we include it for
completeness.

######  Lemma 3.7.

We have:

$\psi_{\hbar}(x,z)\frac{1-z^{-1}e^{-x\hbar^{1/2}}}{1-z^{-1}}\psi_{\hbar}(-x,1/z)\;=\;\exp\Big{(}-\frac{x\hbar^{\frac{1}{2}}}{2}+\frac{\hbar}{12}\Big{)},$
(60)
$\widehat{\psi}_{\hbar}(0,z)\widehat{\psi}_{\hbar}(0,1/z)\;=\;\frac{\sqrt{-z}}{1-z}\exp\Big{(}\frac{\pi^{2}}{6\hbar}+\frac{1}{2\hbar}\log(-z)^{2}+\frac{\hbar}{12}\Big{)}\,.$
(61)

###### Proof.

These formulas all follow from the well-known inversion formulas for the
polylogarithms (see for example [29]):

$\displaystyle\operatorname{Li}_{2}(z)+\operatorname{Li}_{2}(1/z)$
$\displaystyle\;=\;-\frac{\pi^{2}}{6}-\frac{1}{2}\log(-z)^{2}\,,$ (62)
$\displaystyle\operatorname{Li}_{1}(z)-\operatorname{Li}_{1}(1/z)$
$\displaystyle\;=\;-\log(-z)\,,$
$\displaystyle\operatorname{Li}_{0}(z)+\operatorname{Li}_{0}(1/z)$
$\displaystyle\;=\;-1\,,$
$\displaystyle\operatorname{Li}_{-n}(z)+(-1)^{n}\operatorname{Li}_{-n}(1/z)$
$\displaystyle\;=\;0\qquad(n>0)\,.$

Using these relations, we have

$\displaystyle\widehat{\psi}_{\hbar}(0,z)\widehat{\psi}_{\hbar}(0,1/z)$
$\displaystyle\;=\;\exp\Big{(}\\!-\sum_{k\in\mathbbm{Z}_{\geq
0}}\frac{B_{k}\,\hbar^{k-1}}{k!}(\operatorname{Li}_{2-k}(z)+\operatorname{Li}_{2-k}(1/z))\Big{)}$
(63)
$\displaystyle\;=\;\frac{\sqrt{-z}}{1-z}\exp\Big{(}\frac{\pi^{2}}{6\hbar}+\frac{1}{2\hbar}\log(-z)^{2}+\frac{\hbar}{12}\Big{)}\,.$

Similarly, we find that

$\displaystyle\psi_{\hbar}(x,z)\frac{1-z^{-1}e^{-x\hbar^{1/2}}}{1-z^{-1}}\psi_{\hbar}(-x,1/z)$
(64)
$\displaystyle\;=\;\exp\bigg{(}\\!-\\!\\!\sum_{\underset{k+\frac{\ell}{2}>1}{k,\ell\in\mathbbm{Z}_{\geq
0}}}\frac{B_{k}\,x^{\ell}\,\hbar^{k+\frac{\ell}{2}-1}}{\ell!\,k!}\big{(}\operatorname{Li}_{2-k-\ell}(z)+(-1)^{k+\ell}\operatorname{Li}_{2-k-\ell}(1/z)\big{)}\bigg{)}$
$\displaystyle\;=\;\exp\Big{(}-\frac{x\hbar^{\frac{1}{2}}}{2}+\frac{\hbar}{12}\Big{)}\,.$

Notice that this is exactly the factor that appeared in the remark after
Equation (1), which gave the relation between the $\psi_{\hbar}$ used in the
current paper and the ones used in [10, Eq. 1.9]. Indeed,
$\frac{1-z^{-1}}{1-z^{-1}e^{-x\hbar^{1/2}}}\psi_{\hbar}(-x,1/z)^{-1}$ is
exactly the series used in [10] where the factor
$\frac{1-z^{-1}}{1-z^{-1}e^{-x\hbar^{1/2}}}$ amounts to swapping the sign of
$B_{1}$ as done there. ∎

##  4\. Elementary invariance properties

In this section, we review some basic choices needed to define the the
Neumann–Zagier data for a triangulation with $N$ tetrahedra, namely:

  * (a)

an ordering of the $N$ tetrahedra,

  * (b)

an ordering of the $N$ edges,

  * (c)

an edge equation to remove,

  * (d)

a path to represent the meridian curve,

  * (e)

a flattening.

A change of these choices changes the corresponding Neumann–Zagier data in a
simple way, which we now describe. Fix a triangulation and the choices needed
to define Neumann–Zagier data

$\Xi\;=\;(A,B,\nu,z,f,f^{\prime\prime})\,.$ (65)

Regarding choice (a), suppose that $\sigma\in S_{N}$ is a permutation (and
also the associated matrix) of our labelling of the tetrahedra. Then the
Neumann–Zagier matrices transform as follows

$\Xi\;=\;(A,B,\nu,z,f,f^{\prime\prime})\mapsto(A\sigma,B\sigma,\nu,\sigma^{-1}z,\sigma^{-1}f,\sigma^{-1}f^{\prime\prime})\;=\;\Xi\cdot\sigma\,.$
(66)

This implies that the integrand $f^{\Xi}_{\hbar}(x,z)$ of Equation (4) and the
quadratic form $\Lambda$ of Equation (6) satisfy

$f_{\Xi\cdot\sigma,\hbar}(\sigma^{-1}x,\sigma^{-1}z)\;=\;f^{\Xi}_{\hbar}(x,z)\qquad\Lambda^{\Xi\cdot\sigma}\;=\;\sigma^{t}\Lambda^{\Xi}\sigma\,,$
(67)

which combined with the fact that integration is invariant under a linear
change of variables (see part (a) of Lemma 3.1), implies that
$\Phi^{\Xi\cdot\sigma}(\hbar)=\Phi^{\Xi}(\hbar)$.

Choices (b), (c) and (d) are a special case of the following transformation of
$P\in\mathrm{GL}_{N}(\mathbbm{Z})$ acting on Neumann–Zagier data via

$\Xi\;=\;(A,B,\nu,z,f,f^{\prime\prime})\mapsto(PA,PB,P\nu,z,f,f^{\prime\prime})\;=\;P\cdot\Xi\,.$
(68)

It follows that the integrand and the quadratic form satisfy

$f^{P\cdot\Xi}_{\hbar}(x,z)\;=\;f^{\Xi}_{\hbar}(x,z),\qquad\Lambda^{P\cdot\Xi}\;=\;\Lambda^{\Xi}\,,$
(69)

which implies again that $\Phi^{\Xi\cdot\sigma}(\hbar)=\Phi^{\Xi}(\hbar)$.

Finally, if $\Xi$ and $\widetilde{\Xi}$ differ by a choice of flattening, then
it is easy to see that

$f^{\widetilde{\Xi}}_{\hbar}(x,z)\;=\;e^{c\hbar}f^{\Xi}_{\hbar}(x,z),\qquad\Lambda^{\widetilde{\Xi}}\;=\;\Lambda^{\Xi}$
(70)

for some $c\in\frac{1}{8}\mathbbm{Z}$, which implies that
$\Phi^{\widetilde{\Xi}}(\hbar)=e^{c\hbar}\,\Phi^{\Xi}(\hbar)$.

##  5\. Invariance under the choice of quad

The definition of the series $\Phi^{\Xi}(\hbar)$ requires some choices, some
of which were described and dealt with in the previous Section 4. What remains
to complete the proof of Theorem 1.1 is the independence of
$\Phi^{\Xi}(\hbar)$ under the choice of a nondegenerate quad, and the
independence under the 2–3 Pachner moves that connect two ideal
triangulations.

In this section, we will prove that $\Phi^{\Xi}(\hbar)$ is independent of the
choice of a nondegenerate quad. Recall that to each pair of opposite edges of
an ideal tetrahedron there is an associated variable called a shape variable.
When defining Neumann–Zagier data (see Section 2.2), we must choose an edge
for each tetrahedron, which associates one of these shapes. There are three
possible choices, which leads to the action on $\mathbbm{Z}/3\mathbbm{Z}$ on
the each column of the Neumann–Zagier data. All in all, for a triangulation
with $N$ tetrahedra, this leads to $3^{N}$ choices on of Neumann–Zagier data.
On the other hand, the definition of the series $\Phi^{\Xi}(\hbar)$ requires a
choice of a nondegenerate quad, i.e., one for which $\det(B)\neq 0$ (such
quads always exist [10, Lem.A.3]). In this section, we will show that any of
the $3^{N}$ choices of quad with $\det(B)\neq 0$ lead to the same
$\Phi^{\Xi}(\hbar)$ series.

######  Theorem 5.1.

The series $\Phi^{\Xi}(\hbar)$ is independent of the choice of a nondegenerate
quad.

###### Proof.

Fix two non–degenerate NZ data $\Xi=(A,B,\nu,z,f,f^{\prime\prime})$ and
$\widetilde{\Xi}=(\widetilde{A},\widetilde{B},\widetilde{\nu},\widetilde{z},\widetilde{f},\widetilde{f}^{\prime\prime})$
related by a quad change of the same ideal triangulation. The nondegeneracy
assumption implies that $\det(B)\neq 0$ and $\det(\widetilde{B})\neq 0$.

After reordering the tetrahedra (which does not change the $\Phi^{\Xi}(\hbar)$
series, as follows from Section 4), we can assume that $\widetilde{\Xi}$ is
obtained by applying a change in quad that fixes the first $N_{0}$ shapes
$z^{(0)}$, replaces the next $N_{1}$ shapes $z^{(1)}$ by $z^{\prime(1)}$ and
replaces the next $N_{2}$ shapes $z^{(2)}$ by $z^{\prime(2)}$. (Recall that
$z^{\prime}=1/(1-z)$ and $z^{\prime\prime}=1-1/z$).

This partitions the shapes $z=(z^{(0)},z^{(1)},z^{(2)})$ into three sets of
size $N_{0},N_{1},N_{2}$ and the matrices $A$ and $B$ into three block
matrices $A_{i}$ and $B_{i}$ of size $N\times N_{i}$ for $i=0,1,2$

$\displaystyle A$ $\displaystyle\;=\;(A_{0}\,|\,A_{1}\,|\,A_{2})\,,\qquad B$
$\displaystyle\;=\;(B_{0}\,|\,B_{1}\,|\,B_{2})\,,$ (71)

and similarly for the flattening $f=(f^{(0)},f^{(1)},f^{(2)})$ and
$f^{\prime\prime}=(f^{\prime\prime(0)},f^{\prime\prime(1)},f^{\prime\prime(2)})$.
After the quad moves, the corresponding matrices and vectors are given by

$\displaystyle\tilde{A}$
$\displaystyle\;=\;(A_{0}\,|-B_{1}\,|-A_{2}+B_{2})\,,\quad$
$\displaystyle\tilde{B}$
$\displaystyle\\!\\!\\!=\;(B_{0}\,|\,A_{1}-B_{1}\,|-A_{2})\,,$ (72)
$\displaystyle\tilde{\nu}$ $\displaystyle\;=\;\nu-B_{1}1-A_{2}1\,,\quad$
$\displaystyle\tilde{z}$
$\displaystyle\\!\\!\\!=\;(z^{(0)},z^{\prime(1)},z^{\prime\prime(2)})\,,$
$\displaystyle\tilde{f}$
$\displaystyle\;=\;(f^{(0)},1-f^{(1)}-f^{\prime\prime(1)},f^{\prime\prime(2)})\,,\quad$
$\displaystyle\tilde{f}^{\prime\prime}$
$\displaystyle\\!\\!\\!=\;(f^{\prime\prime(0)},f^{(2)},1-f^{(2)}-f^{\prime\prime(2)})\,.$

We also partition the vector of formal Gaussian integration variables
$x=(x^{(0)},x^{(1)},x^{(2)})$, as well as the symmetric matrix $Q:=B^{-1}A$

$Q\;=\;\begin{pmatrix}Q_{00}&Q_{01}&Q_{02}\\\ Q_{01}^{t}&Q_{11}&Q_{12}\\\
Q_{02}^{t}&Q_{12}^{t}&Q_{22}\\\ \end{pmatrix}\,.$ (73)

With the above notation, we have

$\Phi^{\Xi}(\hbar)\;=\;\langle I_{0}\rangle_{x,\Lambda_{0}}$ (74)

where

$I_{0}\;=\;\exp\Big{(}\frac{\hbar}{8}f^{t}B^{-1}Af-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}(B^{-1}\nu-1)\Big{)}\prod_{j=1}^{N}\psi_{\hbar}(x_{j},z_{j})$

and

$\Lambda_{0}\;=\;\mathrm{diag}(z^{\prime})-Q\,.$

Applying the first quad move as in Theorem 3.4 to the $\psi_{\hbar}$ with
arguments of $z^{(1)}$ and the second quad move as in its Corollary 3.5 to the
$\psi_{\hbar}$ with arguments $z^{(2)}$, we obtain that

$\Phi^{\Xi}(\hbar)\;=\;\langle I_{1}\rangle_{(x,y,w),\Lambda_{1}}$ (75)

where

$\displaystyle I_{1}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{8}f^{t}B^{-1}Af-\frac{\hbar^{\frac{1}{2}}}{2}{x^{(0)}}^{t}(B^{-1}\nu-1)^{(0)}-\frac{N_{1}\hbar}{24}-\frac{\hbar^{\frac{1}{2}}}{2}{x^{(1)}}^{t}(B^{-1}\nu-1)^{(1)}$
$\displaystyle\qquad\qquad\quad+\sum_{j=1}^{N_{1}}\Big{(}y_{j}+\frac{x_{j}^{(1)}z_{j}^{(1)}}{1-z_{j}^{(1)}}\Big{)}\frac{\hbar^{\frac{1}{2}}}{2}+\frac{N_{2}\hbar}{24}-\frac{\hbar^{\frac{1}{2}}}{2}{x^{(2)}}^{t}(B^{-1}\nu-1)^{(2)}-\sum_{j=1}^{N_{2}}x_{j}^{(2)}\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}$
$\displaystyle\times\prod_{j=1}^{N_{0}}\psi_{\hbar}(x_{j}^{(0)},z_{j}^{(0)})\prod_{j=1}^{N_{1}}\psi_{\hbar}\Big{(}y_{j}+\frac{x_{j}^{(1)}z_{j}^{(1)}}{1-z_{j}^{(1)}},\frac{1}{1-z_{j}^{(1)}}\Big{)}\prod_{j=1}^{N_{2}}\psi_{\hbar}\Big{(}w_{j}-\frac{x_{j}^{(2)}}{1-z_{j}^{(2)}},1-{z_{j}^{(2)}}^{-1}\Big{)},$

$y$ and $w$ are vectors of size $N_{1}$ and $N_{2}$, respectively, and

$\Lambda_{1}\;=\;\begin{pmatrix}\Lambda&0&0\\\
0&\mathrm{diag}(z^{\prime\prime(1)})&0\\\
0&0&\mathrm{diag}(z^{(2)})-I\end{pmatrix}\,.$

Making the change of variables $y_{j}\mapsto
y_{j}-\frac{x_{j}^{(1)}z_{j}^{(1)}}{1-z_{j}^{(1)}}$ and $w_{j}\mapsto
w_{j}+\frac{x_{j}^{(2)}}{1-z_{j}^{(2)}}$ and using Lemma 3.1 we obtain that

$\Phi^{\Xi}(\hbar)\;=\;\langle I_{2}\rangle_{(x,y,w),\Lambda_{2}}$ (76)

where

$\displaystyle I_{2}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{8}f^{t}B^{-1}Af-\frac{\hbar^{\frac{1}{2}}}{2}{x^{(0)}}^{t}(B^{-1}\nu-1)^{(0)}-\frac{N_{1}\hbar}{24}-\frac{\hbar^{\frac{1}{2}}}{2}{x^{(1)}}^{t}(B^{-1}\nu-1)^{(1)}$
$\displaystyle\qquad\qquad\quad+y^{t}1\frac{\hbar^{\frac{1}{2}}}{2}+\frac{N_{2}\hbar}{24}-\frac{\hbar^{\frac{1}{2}}}{2}{x^{(2)}}^{t}(B^{-1}\nu)^{(2)}\Big{)}$
$\displaystyle\qquad\qquad\times\prod_{j=1}^{N_{0}}\psi_{\hbar}(x_{j}^{(0)},z_{j}^{(0)})\prod_{j=1}^{N_{1}}\psi_{\hbar}\Big{(}y_{j},\frac{1}{1-z_{j}^{(1)}}\Big{)}\prod_{j=1}^{N_{2}}\psi_{\hbar}\big{(}w_{j},1-{z_{j}^{(2)}}^{-1}\big{)}$

and

$\Lambda_{2}\;=\;\left(\\!\\!\begin{array}[]{ccc|cc}\mathrm{diag}(z^{\prime(0)})-Q_{00}&-Q_{01}&-Q_{02}&0&0\\\
-Q_{01}^{t}&I-Q_{11}&-Q_{12}&I&0\\\ -Q_{02}^{t}&-Q_{12}^{t}&-Q_{22}&0&-I\\\
\hline\cr 0&I&0&\mathrm{diag}(z^{\prime\prime(1)})&0\\\
0&0&-I&0&\mathrm{diag}(z^{(2)})-I\end{array}\\!\\!\right)\,.$

Note that $x^{(1)}$ and $x^{(2)}$ do not appear in the arguments of
$\psi_{\hbar}$. Moreover, since $BQ=A$, we see that

$B\begin{pmatrix}I&Q_{01}&Q_{02}\\\ 0&Q_{11}-I&Q_{12}\\\
0&Q_{12}^{t}&Q_{22}\end{pmatrix}\begin{pmatrix}I&0&0\\\ 0&I&0\\\
0&0&-I\end{pmatrix}\;=\;(B_{0}\,|\,A_{1}-B_{1}\,|-A_{2})\;=\;\widetilde{B}\,,$
(77)

which implies that

$\det\begin{pmatrix}Q_{11}-I&Q_{12}\\\ Q_{12}^{t}&Q_{22}\end{pmatrix}\neq
0\,,\quad\text{and}\quad\begin{pmatrix}Q_{11}-I&Q_{12}\\\
Q_{12}^{t}&Q_{22}\end{pmatrix}^{-1}\;=\;\begin{pmatrix}0&I&0\\\ 0&0&-I\\\
\end{pmatrix}\widetilde{B}^{-1}B\begin{pmatrix}0&0\\\ I&0\\\
0&I\end{pmatrix}\,.$ (78)

Therefore, we can apply Fubini’s Theorem (Lemma 3.1) with the integration
variables $x^{(1)},x^{(2)}$, and use Lemma 5.2 and the equality
$Qf+f^{\prime\prime}=B^{-1}\nu$ to obtain that

$\Phi^{\Xi}(\hbar)\;=\;e^{c\hbar}\langle
I_{3}\rangle_{\widetilde{x},\Lambda_{3}}\,.$ (79)

where $\widetilde{x}=(x^{(0)},y,w)$, $c\in\frac{1}{24}\mathbbm{Z}$

$I_{3}\;=\;\exp\Big{(}\frac{\hbar}{8}\widetilde{f}^{t}\widetilde{B}^{-1}\widetilde{A}\widetilde{f}-\frac{\hbar^{\frac{1}{2}}}{2}\widetilde{x}^{t}(\widetilde{B}^{-1}\widetilde{\nu}-1)\Big{)}\prod_{j=1}^{N}\psi_{\hbar}(\widetilde{x}_{j},\widetilde{z}_{j})$

and

$\Lambda_{3}\;=\;\mathrm{diag}(\widetilde{z}^{\prime})-\widetilde{B}^{-1}\widetilde{A}\,.$

The right hand side of Equation (79) is exactly equal to
$e^{c\hbar}\Phi^{\widetilde{\Xi}}(\hbar)$, completing the proof of Theorem
5.1. ∎

######  Lemma 5.2.

With the notation as in the proof of Theorem 5.1, we have the following
identities

$\displaystyle\widetilde{B}^{-1}\widetilde{A}$
$\displaystyle\;=\;\begin{pmatrix}Q_{00}&0&0\\\ 0&0&0\\\
0&0&-I\end{pmatrix}+\begin{pmatrix}Q_{01}&Q_{02}\\\ -I&0\\\
0&I\end{pmatrix}\begin{pmatrix}Q_{11}-I&Q_{12}\\\
Q_{12}^{t}&Q_{22}\end{pmatrix}^{-1}\begin{pmatrix}Q_{01}^{t}&-I&0\\\
Q_{02}^{t}&0&I\end{pmatrix},$ (80)
$\displaystyle\widetilde{B}^{-1}\widetilde{\nu}-1$
$\displaystyle\;=\;\begin{pmatrix}I&0&0\\\ 0&0&0\\\
0&0&0\end{pmatrix}(B^{-1}\nu-1)-\begin{pmatrix}0\\\ 1\\\ 0\end{pmatrix}$
$\displaystyle\qquad+\begin{pmatrix}-Q_{01}&-Q_{02}\\\ I&0\\\
0&-I\end{pmatrix}\begin{pmatrix}Q_{11}-I&Q_{12}\\\
Q_{12}^{t}&Q_{22}\end{pmatrix}^{-1}\left(\begin{pmatrix}0&I&0\\\
0&0&I\end{pmatrix}B^{-1}\nu-\begin{pmatrix}1\\\ 0\end{pmatrix}\right)$ (81)

and for some $d\in\mathbbm{Z}$

$\displaystyle\widetilde{f}\widetilde{B}^{-1}\widetilde{\nu}\;=\;f^{t}B^{-1}\nu+d$
(82) $\displaystyle-\Big{(}(B^{-1}\nu)^{t}\begin{pmatrix}0&0\\\ I&0\\\
0&I\end{pmatrix}-\begin{pmatrix}1&0\end{pmatrix}\Big{)}\begin{pmatrix}Q_{11}-I&Q_{12}\\\
Q_{12}^{t}&Q_{22}\end{pmatrix}^{-1}\left(\begin{pmatrix}0&I&0\\\
0&0&I\end{pmatrix}B^{-1}\nu-\begin{pmatrix}1\\\ 0\end{pmatrix}\right).$

###### Proof.

We will show that boths sides of Equation (80) and Equation (81) are equal
after multiplying by the invertible matrix $\widetilde{B}$. Denote

$\begin{pmatrix}Q_{11}-I&Q_{12}\\\
Q_{12}^{t}&Q_{22}\end{pmatrix}^{-1}\;=\;\begin{pmatrix}\Gamma_{11}&\Gamma_{12}\\\
\Gamma_{12}^{t}&\Gamma_{22}\end{pmatrix}\,,$ (83)

and note that

$\displaystyle\widetilde{B}\begin{pmatrix}-Q_{01}&-Q_{02}\\\ I&0\\\
0&-I\end{pmatrix}\begin{pmatrix}Q_{11}-I&Q_{12}\\\
Q_{12}^{t}&Q_{22}\end{pmatrix}^{-1}\;=\;\widetilde{B}\begin{pmatrix}-Q_{01}\Gamma_{11}-Q_{02}\Gamma_{12}^{t}&-Q_{01}\Gamma_{12}-Q_{02}\Gamma_{22}\\\
\Gamma_{11}&\Gamma_{12}\\\ -\Gamma_{12}^{t}&-\Gamma_{22}\end{pmatrix}$ (84)
$\displaystyle\;=\;(-B_{0}Q_{01}\Gamma_{11}-B_{0}Q_{02}\Gamma_{12}^{t}+(A_{1}-B_{1})\Gamma_{11}+A_{2}\Gamma_{12}^{t}\;|\;-B_{0}Q_{01}\Gamma_{12}-B_{0}Q_{02}\Gamma_{22}+(A_{1}-B_{1})\Gamma_{12}+A_{2}\Gamma_{22})$
$\displaystyle\;=\;((B_{1}(Q_{11}-I)+B_{2}Q_{12}^{t})\Gamma_{11}+(B_{1}Q_{12}+B_{2}Q_{22})\Gamma_{12}^{t}\;|\;(B_{1}(Q_{11}-I)+B_{2}Q_{12}^{t})\Gamma_{12}+(B_{1}Q_{12}+B_{2}Q_{22})\Gamma_{22})$
$\displaystyle\;=\;(B_{1}\;|\;B_{2})\,.$

Therefore, since

$\widetilde{B}\begin{pmatrix}Q_{00}&0&0\\\ 0&0&0\\\
0&0&-I\end{pmatrix}\;=\;(B_{0}Q_{00}\;|\;0\;|-A_{2})\,,$ (85)

we see that multiplying the right hand side of Equations (80) on the left by
$\widetilde{B}$ we have

$(B_{0}Q_{00}+B_{1}Q_{01}^{t}+B_{2}Q_{02}^{t}\;|-B_{1}\;|-A_{2}+B_{2})\;=\;\widetilde{A}\,,$
(86)

which completes the proof of Equation (80). Similarly, since

$\widetilde{B}\begin{pmatrix}I&0&0\\\ 0&0&0\\\
0&0&0\end{pmatrix}(B^{-1}\nu-1)-\widetilde{B}\begin{pmatrix}0\\\ 1\\\
0\end{pmatrix}\;=\;B_{0}(B^{-1}\nu)^{(0)}-B_{0}1-A_{1}1+B_{1}1$ (87)

we see that multiplying the right hand side of Equations (81) on the left by
$\widetilde{B}$ we have

$B_{0}(B^{-1}\nu)^{(0)}-B_{0}1-A_{1}1+B_{1}1+B_{1}(B^{-1}\nu)^{(1)}-B_{1}1+B_{2}(B^{-1}\nu)^{(2)}\;=\;\widetilde{\nu}-\widetilde{B}1\,,$
(88)

which completes the proof of Equation (81). For Equation (82), we will compute
the three terms that appear there. Firstly, use Equation (81) to obtain that

$\begin{pmatrix}0&I&0\\\
0&0&-I\end{pmatrix}(\widetilde{B}^{-1}\widetilde{\nu}-1)\;=\;-\begin{pmatrix}1\\\
0\end{pmatrix}+\begin{pmatrix}Q_{11}-I&Q_{12}\\\
Q_{12}^{t}&Q_{22}\end{pmatrix}^{-1}\left(\begin{pmatrix}0&I&0\\\
0&0&I\end{pmatrix}B^{-1}\nu-\begin{pmatrix}1\\\ 0\end{pmatrix}\right),$ (89)

to obtain that

$\displaystyle\left((B^{-1}\nu)^{t}\begin{pmatrix}0&0\\\ I&0\\\
0&I\end{pmatrix}-\begin{pmatrix}1&0\end{pmatrix}\right)\left(\begin{pmatrix}0&I&0\\\
0&0&-I\end{pmatrix}(\widetilde{B}^{-1}\widetilde{\nu}-1)+\begin{pmatrix}1\\\
0\end{pmatrix}\right)$ (90)
$\displaystyle\;=\;({(B^{-1}\nu)^{(1)}}^{t}-1)(\widetilde{B}^{-1}\widetilde{\nu})^{(1)}-{(B^{-1}\nu)^{(2)}}^{t}(\widetilde{B}^{-1}\widetilde{\nu})^{(2)}+{(B^{-1}\nu)^{(2)}}^{t}1\,.$

Secondly, expanding out

$\displaystyle\widetilde{f}^{t}\widetilde{B}^{-1}\widetilde{\nu}$
$\displaystyle\;=\;f^{(0)}(\widetilde{B}^{-1}\widetilde{\nu})^{(0)}+(1-f^{(1)}-f^{\prime\prime(1)})(\widetilde{B}^{-1}\widetilde{\nu})^{(1)}+f^{\prime\prime(2)}(\widetilde{B}^{-1}\widetilde{\nu})^{(2)}$
(91) $\displaystyle\;=\;f^{(0)}(\widetilde{B}^{-1}\widetilde{\nu})^{(0)}$
$\displaystyle\qquad+(1-f^{(1)}-{(B^{-1}\nu)^{(1)}}^{t}+f^{(0)}Q_{01}+f^{(1)}Q_{11}+f^{(2)}Q_{12}^{t})(\widetilde{B}^{-1}\widetilde{\nu})^{(1)}$
$\displaystyle\qquad+({(B^{-1}\nu)^{(2)}}^{t}-f^{(0)}Q_{02}-f^{(1)}Q_{12}-f^{(2)}Q_{22})(\widetilde{B}^{-1}\widetilde{\nu})^{(2)}\,.$

Finally, using Equation (77) we obtain that

$\displaystyle f^{t}B^{-1}\nu-f^{(1)}1+f^{\prime\prime(2)}1$
$\displaystyle\;=\;f^{t}B^{-1}\nu-f^{(1)}1-f^{t}Q(0,0,1)^{t}+{(B^{-1}\nu)^{(2)}}^{t}1$
(92) $\displaystyle\;=\;f^{t}B^{-1}\widetilde{\nu}+{(B^{-1}\nu)^{(2)}}^{t}1$
$\displaystyle\;=\;f^{(0)}(\widetilde{B}^{-1}\widetilde{\nu})^{(0)}$
$\displaystyle\qquad+(f^{(0)}Q_{01}+f^{(1)}(Q_{11}-I)+f^{(2)}Q_{12}^{t})(\widetilde{B}^{-1}\widetilde{\nu})^{(1)}$
$\displaystyle\qquad-(f^{(0)}Q_{02}+f^{(1)}Q_{12}+f^{(2)}Q_{22})(\widetilde{B}^{-1}\widetilde{\nu})^{(2)}+{(B^{-1}\nu)^{(2)}}^{t}1\,.$

∎

##  6\. Invariance under Pachner moves

In this section, we will prove that $\Phi^{\Xi}$ is invariant under 2–3
Pachner moves. There are several versions of the 2–3 move (and of the
corresponding pentagon identity in Teichmüller TQFT [26]) and the one we
choose in the next theorem is slightly different from the one in [10, Sec.3.6]
and can be related by composing with quad moves.

The 2–3 move involves two triangulations $\mathcal{T}$ and
$\widetilde{\mathcal{T}}$ with $N+2$ and $N+3$ tetrahedra, respectively, shown
in Figure 2.

$z_{1}$$z_{1}^{\prime}$$z_{1}^{\prime\prime}$$z_{2}$$z_{2}^{\prime}$$z_{2}^{\prime\prime}$$\widetilde{z}_{0}$$\widetilde{z}_{0}^{\prime}$$\widetilde{z}_{0}^{\prime\prime}$$\widetilde{z}_{1}$$\widetilde{z}_{1}^{\prime}$$\widetilde{z}_{1}^{\prime\prime}$$\widetilde{z}_{2}$$\widetilde{z}_{2}^{\prime\prime}$$\widetilde{z}_{2}^{\prime}$
Figure 2. The 2–3 Pachner move.

Using $z_{0}=z_{1}+z_{2}-z_{1}z_{2}$ from Equation (46) used in Theorem 3.6,
it follows that the shapes $z$ of $\mathcal{T}$ and $\widetilde{z}$ of
$\widetilde{\mathcal{T}}$ are related by

$\displaystyle z\;=\;(z_{1},z_{2},z^{*})\mapsto\widetilde{z}\;=$
$\displaystyle\big{(}\widetilde{z}_{0},\;\widetilde{z}_{1},\;\widetilde{z}_{2},\;\widetilde{z}^{*}\big{)}$
(93) $\displaystyle\;=$
$\displaystyle\big{(}z_{0},\;z_{1}z_{0}^{-1},\;z_{2}z_{0}^{-1},\;z^{*}\big{)}\,.$

Similarly to [10, Eq. (3.20) and (3.21)], these shapes satisfy the relations

$\displaystyle\widetilde{z}_{0}^{\prime}$
$\displaystyle\;=\;z_{1}^{\prime}z_{2}^{\prime}\,,\quad$
$\displaystyle\widetilde{z}_{1}^{\prime\prime}$
$\displaystyle\;=\;z_{1}^{\prime\prime}z_{2}\,,\quad$
$\displaystyle\widetilde{z}_{2}^{\prime\prime}$
$\displaystyle\;=\;z_{1}z_{2}^{\prime\prime}\,,$ (94) $\displaystyle z_{1}$
$\displaystyle\;=\;\widetilde{z}_{0}\widetilde{z}_{1}\,,\quad$ $\displaystyle
z_{1}^{\prime}$
$\displaystyle\;=\;\widetilde{z}_{1}^{\prime}\widetilde{z}_{2}\,,\quad$
$\displaystyle z_{1}^{\prime\prime}$
$\displaystyle\;=\;\widetilde{z}_{0}^{\prime\prime}\widetilde{z}_{2}^{\prime}\,,$
$\displaystyle z_{2}$
$\displaystyle\;=\;\widetilde{z}_{0}\widetilde{z}_{2}\,,\quad$ $\displaystyle
z_{2}^{\prime}$
$\displaystyle\;=\;\widetilde{z}_{1}\widetilde{z}_{2}^{\prime}\,,\quad$
$\displaystyle z_{2}^{\prime\prime}$
$\displaystyle\;=\;\widetilde{z}_{0}^{\prime\prime}\widetilde{z}_{1}^{\prime}\,.$

If we write the Neumann–Zagier matrices of $\mathcal{T}$ in the form

$A\;=\;(a_{1}\,|\,a_{2}\,|\,a_{*})\,,\qquad
B=\;(b_{1}\,|\,b_{2}\,|\,b_{*})\,,$ (95)

where $a_{1}$, $a_{2}$ are the first two columns of $A$ and $a_{*}$ is the
block $(N+2)\times N$ matrix of the remaining $N$ columns of $A$, and likewise
for $B$, then using Equations (94) the corresponding Neumann–Zagier matrices
of $\widetilde{\mathcal{T}}$ are given by

$\widetilde{A}\;=\;\begin{pmatrix}-1&0&0&0\\\
-b_{1}-b_{2}+a_{1}+a_{2}&a_{1}-b_{2}&a_{2}-b_{1}&a_{*}\end{pmatrix},\qquad\widetilde{B}\;=\;\begin{pmatrix}-1&1&1&0\\\
0&b_{1}&b_{2}&b_{*}\end{pmatrix}$ (96)

and the corresponding vector $\widetilde{\nu}=(1,\nu)$. Analogously to the
shapes, we will fix flattenings
$(\widetilde{f}_{0},\widetilde{f}_{1},\widetilde{f}_{2},\widetilde{f}^{*})$
and
$(\widetilde{f}_{0}^{\prime\prime},\widetilde{f}_{1}^{\prime\prime},\widetilde{f}_{2}^{\prime\prime},\widetilde{f}^{\prime\prime*})$
for $(\widetilde{A}\,|\,\widetilde{B})$ and
$(f_{1},f_{2},f^{*})=(\widetilde{f}_{0}+\widetilde{f}_{1},\widetilde{f}_{0}+\widetilde{f}_{2},f^{*})$
and
$(f^{\prime\prime}_{1},f^{\prime\prime}_{2},f^{\prime\prime*})=(\widetilde{f}_{0}^{\prime\prime}+\widetilde{f}_{2}^{\prime},\widetilde{f}_{0}^{\prime\prime}+\widetilde{f}_{1}^{\prime},f^{\prime\prime*})$
for $(A\,|\,B)$. The data of the flattenings then satisfy the additive
versions of Equations (94).

######  Theorem 6.1.

The series $\Phi^{\Xi}(\hbar)$ is invariant under 2–3 Pachner moves.

The proof involves an application of the pentagon identity for $\psi_{\hbar}$
of Theorem 3.6.

For two triangulations $\mathcal{T}$ and $\widetilde{\mathcal{T}}$ with $N+2$
and $N+3$ tetrahedra and NZ matrices $(\mathcal{A}\,|\,\mathcal{B})$,
respectively, related by a Pachner 2–3 move. To define the corresponding
series $\Phi^{\Xi}(\hbar)$ and $\Phi^{\widetilde{\Xi}}(\hbar)$, we need to
possibly change quads on $\mathcal{T}$ and $\widetilde{\mathcal{T}}$ so that
both $\mathcal{B}$ and $\widetilde{\mathcal{B}}$ are invertible. By a quad
move $q$, we can replace $(\mathcal{A}\,|\,\mathcal{B})$ by
$(\mathbf{A}\,|\,\mathbf{B})$ where $\det(\mathbf{B})\neq 0$. Recalling that
quad moves act on tetrahedra and actions on different tetrahedra commute, we
can write the move $q=q_{2}\times q_{N}$ as a product of quad moves $q_{2}$ on
the first 2 tetrahedra times moves $q_{N}$ on the remaining $N$ tetrahedra of
$\mathcal{T}$. Let $(A\,|\,B)$ denote the result of applying the move $1\times
q_{N}$ on the $N+2$ tetrahedra of $\mathcal{T}$.

Since the $(N+2)\times(N+2)$ matrix $\mathbf{B}$ has full rank and $B$ and
$\mathbf{B}$ have the same last $N$ columns, it follows that $B$ has nullity
$0$, $1$ or $2$.

Now on $\widetilde{\mathcal{T}}$, we can apply the identity move on the first
three tetrahedra and the $q_{N}$ moves on the remaining $N$ tetrahedra, which
transforms the NZ matrices
$(\widetilde{\mathcal{A}}\,|\,\widetilde{\mathcal{B}})$ to
$(\widetilde{A}\,|\,\widetilde{B})$, and further apply the $q_{2}$ moves on
the second and third tetrahedra to obtain the NZ matrices
$(\widetilde{\mathbf{A}}\,|\,\widetilde{\mathbf{B}})$. By looking at how the
matrix $B$ transforms under a 2–3 move (see Equation (96)), it follows that
$\widetilde{\mathbf{B}}$ has the same rank as $\mathbf{B}$, and hence
$\det(\widetilde{\mathbf{B}})\neq 0$.

The above discussion can be summarized in the following commutative diagram.

${(\mathcal{A}\;|\;\mathcal{B})}$${(\widetilde{\mathcal{A}}\;|\;\widetilde{\mathcal{B}})}$${(A\;|\;B)}$${(\widetilde{A}\;|\;\widetilde{B})}$${(\mathbf{A}\;|\;\mathbf{B})}$${(\widetilde{\mathbf{A}}\;|\;\widetilde{\mathbf{B}})}$$\scriptstyle{2\to
3}$$\scriptstyle{1\times q_{N}}$$\scriptstyle{q_{2}\times
q_{N}}$$\scriptstyle{\\!\\!\\!\\!1\times 1\times q_{N}}$$\scriptstyle{1\times
q_{2}\times q_{N}}$$\scriptstyle{2\to 3}$$\scriptstyle{q_{2}\times
1}$$\scriptstyle{1\times q_{2}\times 1}$ (97)

where $\det(\mathbf{B})\neq 0$ and $\det(\widetilde{\mathbf{B}})\neq 0$. Now
$\Phi^{\Xi}(\hbar)$ and $\Phi^{\widetilde{\Xi}}(\hbar)$ can be defined using
the nondegenerate NZ data with matrices $(\mathbf{A}\,|\,\mathbf{B})$ and
$(\widetilde{\mathbf{A}}\,|\,\widetilde{\mathbf{B}})$, respectively. We will
show that the two $\hbar$-series are equal using a diagram

${\Phi^{\Xi}(\hbar)}$${\bullet}$${\bullet}$${\bullet}$${\Phi^{\widetilde{\Xi}}(\hbar)}$$\scriptstyle{\otimes^{i}\text{Fourier}}$Pentagon$\scriptstyle{\otimes^{i}\text{Fourier}^{-1}}$$\scriptstyle{2i-\text{Fubini}}$
(98)

where $i=0,1,2$ denotes the nullity of $B$. We will treat each case in a
separate section. With this discussion the Theorem 6.1 follows from
Propositions 6.2, 6.4, and 6.6. Before proceeding with the proof, we will set
some notation. Similarly to the proof of the quad invariance, denote
$\mathbf{Q}=\mathbf{B}^{-1}\mathbf{A}$

$\displaystyle\mathbf{Q}\;=\;\left(\\!\\!\begin{array}[]{cc|c}\mathbf{Q}_{11}&\mathbf{Q}_{12}&\mathbf{Q}_{1}^{*}\\\
\mathbf{Q}_{12}&\mathbf{Q}_{22}&\mathbf{Q}_{2}^{*}\\\
\hline\cr\mathbf{Q_{1}^{*}}^{t}&\mathbf{Q_{2}^{*}}^{t}&\mathbf{Q}^{*}\\\
\end{array}\\!\\!\right),$ (99)

where $\mathbf{Q}_{ij}$ are matrices of size $1\times 1$, $\mathbf{Q}_{j}^{*}$
are matrices of size $1\times N$ and $\mathbf{Q}^{*}$ is a matrix of size
$N\times N$.

###  6.1. The case of $B$ with full rank

In this section, we prove Theorem 3.6 under the assumption that the matrix $B$
has full rank. In this case, Equation (98) simplifies to the following one

${\Phi^{\Xi}(\hbar)}$${\Phi^{\widetilde{\Xi}}(\hbar)}$Pentagon (100)

since we do not need to apply Fourier transform, nor Fubini’s theorem.

######  Proposition 6.2.

If $B$ has full rank, then $\Phi^{\Xi}(\hbar)$ is invariant under the 2–3
Pachner move given in Equation (96).

###### Proof.

To begin with, noting that in this case $A=\mathbf{A}$ and $B=\mathbf{B}$, we
have:

$\Phi^{\Xi}(\hbar)\;=\;\langle I_{0}\rangle_{x,\Lambda}$ (101)

where

$I_{0}\;=\;\exp\Big{(}\frac{\hbar}{8}f^{t}B^{-1}Af-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}(B^{-1}\nu-1)\Big{)}\prod_{j=1}^{N+2}\psi_{\hbar}(x_{j},z_{j})$

and

$\Lambda_{0}\;=\;\mathrm{diag}(z^{\prime})-\mathbf{Q}$

and $x=(x_{1},x_{2},x^{*})$ are the integration variables. Applying the
pentagon identity of Theorem 3.6, to the $\psi_{\hbar}$ with arguments
$x_{1},x_{2}$ and introducing a new integration variable $x_{0}$, we obtain
that

$\Phi^{\Xi}(\hbar)\;=\;\langle I_{1}\rangle_{(x_{0},x),\Lambda_{1}}$ (102)

where

$\displaystyle I_{1}$
$\displaystyle\;=\;\exp\Big{(}\\!-\frac{\hbar}{24}+\frac{\hbar}{8}f^{t}B^{-1}Af-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}(B^{-1}\nu-1)\Big{)}\,\psi_{\hbar}\Big{(}-x_{0}-x_{2}+\frac{x_{1}z_{2}+x_{2}z_{1}}{z_{0}},z_{1}z_{0}^{-1}\Big{)}$
$\displaystyle\\!\\!\times\psi_{\hbar}\Big{(}x_{0}+x_{1}+x_{2}-\frac{x_{1}z_{2}+x_{2}z_{1}}{z_{0}},z_{0}\Big{)}\,\psi_{\hbar}\Big{(}-x_{0}-x_{1}+\frac{x_{1}z_{2}+x_{2}z_{1}}{z_{0}},z_{2}z_{0}^{-1}\Big{)}\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\Lambda_{1}\;=\;\begin{pmatrix}\frac{({z_{1}}+{z_{2}}-z_{1}z_{2})^{2}}{({z_{1}}-1){z_{1}}({z_{2}}-1){z_{2}}}&0\\\
0&\Lambda\end{pmatrix}\,.$

Making a change of variables $x_{0}\mapsto
x_{0}+x_{1}(-1+z_{2}z_{0}^{-1})+x_{2}(-1+z_{1}z_{0}^{-1})$ using Lemma 3.1, we
obtain that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{2}\rangle_{(x_{0},x),\Lambda_{2}},$ (103)

where

$\displaystyle I_{2}$
$\displaystyle\;=\;\exp\Big{(}-\frac{\hbar}{24}+\frac{\hbar}{8}f^{t}B^{-1}Af-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}(B^{-1}\nu-1)\Big{)}\,\psi_{\hbar}(-x_{0}+x_{1},z_{1}z_{0}^{-1})$
$\displaystyle\qquad\times\psi_{\hbar}(x_{0},z_{0})\,\psi_{\hbar}(-x_{0}+x_{2},z_{2}z_{0}^{-1})\,\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\Lambda_{2}\;=\;\left(\\!\\!\begin{array}[]{ccc|c}\frac{({z_{1}}+{z_{2}}-z_{1}z_{2})^{2}}{({z_{1}}-1){z_{1}}({z_{2}}-1){z_{2}}}&\frac{{z_{1}+z_{2}-z_{1}z_{2}}}{z_{2}({z_{1}}-1)}&\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{{z_{1}}({z_{2}}-1)}&0\\\
\frac{{z_{1}+z_{2}-z_{1}z_{2}}}{z_{2}({z_{1}}-1)}&-{\mathbf{Q}_{11}}{-}\frac{{z_{1}+z_{2}-z_{1}z_{2}}}{z_{2}({z_{1}}-1)}&1-{\mathbf{Q}_{12}}&-{\mathbf{Q}_{1}^{*}}\\\
\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{{z_{1}}({z_{2}}-1)}&1-{\mathbf{Q}_{12}}&-\mathbf{Q}_{22}{-}\frac{{z_{1}+z_{2}-z_{1}z_{2}}}{z_{1}({z_{2}}-1)}&-{\mathbf{Q}_{2}^{*}}\\\
\hline\cr
0&-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{\mathrm{diag}({z^{*}}^{\prime})}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right).$

Making a change of variables $x_{1}\mapsto x_{1}+x_{0}$ and $x_{2}\mapsto
x_{2}+x_{0}$ using Lemma 3.1, and denoting $\widetilde{x}=(x_{0},x)$,
$\widetilde{z}=(z_{0},z_{1}z_{0}^{-1},z_{2}z_{0}^{-1},z^{*})$ we obtain that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;e^{c\hbar}\langle
I_{3}\rangle_{\widetilde{x},\Lambda_{3}},$ (104)

where

$I_{3}\;=\;\exp\Big{(}\frac{\hbar}{8}\widetilde{f}^{t}\widetilde{B}^{-1}\widetilde{A}\widetilde{f}-\frac{\hbar^{\frac{1}{2}}}{2}\widetilde{x}^{t}(\widetilde{B}^{-1}\widetilde{\nu}-1)\Big{)}\prod_{j=1}^{N+3}\psi_{\hbar}(\widetilde{x}_{j},\widetilde{z}_{j})$

and

$\Lambda_{3}\;=\;\mathrm{diag}(\tilde{z}^{\prime})-\tilde{B}^{-1}\tilde{A}\,.$

∎

The next lemma identifies vectors and matrices of the two triangulations
$\mathcal{T}$ and $\widetilde{\mathcal{T}}$.

######  Lemma 6.3.

With the notation used in the proof of the previous Proposition 6.2,

$\displaystyle\widetilde{B}^{-1}\widetilde{A}$
$\displaystyle\;=\;\left(\\!\\!\begin{array}[]{ccc|c}{\mathbf{Q}_{11}}+2{\mathbf{Q}_{12}}+{\mathbf{Q}_{22}}-1&{\mathbf{Q}_{11}}+{\mathbf{Q}_{12}}-1&{\mathbf{Q}_{12}}+{\mathbf{Q}_{22}}-1&{\mathbf{Q}_{1}^{*}}+{\mathbf{Q}_{2}^{*}}\\\
{\mathbf{Q}_{11}}+{\mathbf{Q}_{12}}-1&{\mathbf{Q}_{11}}&{\mathbf{Q}_{12}}-1&{\mathbf{Q}_{1}^{*}}\\\
{\mathbf{Q}_{12}}+{\mathbf{Q}_{22}}-1&{\mathbf{Q}_{12}}-1&{\mathbf{Q}_{22}}&{\mathbf{Q}_{2}^{*}}\\\
\hline\cr{\mathbf{Q}_{1}^{*}}^{t}+{\mathbf{Q}_{2}^{*}}^{t}&{\mathbf{Q}_{1}^{*}}^{t}&{\mathbf{Q}_{2}^{*}}^{t}&{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right)$ (109)
$\displaystyle\widetilde{B}^{-1}\widetilde{\nu}$
$\displaystyle\;=\;\begin{pmatrix}(B^{-1}\nu)_{1}+(B^{-1}\nu)_{2}-1\\\
B^{-1}\nu\end{pmatrix}$ (110)
$\displaystyle\widetilde{f}^{t}\widetilde{B}^{-1}\widetilde{\nu}$
$\displaystyle\;=\;f^{t}B^{-1}\nu-\widetilde{f}_{0}$ (111)

###### Proof.

We will show that both sides of Equations (109) and Equations (110) are equal
after multiplying by the invertible matrix $\widetilde{B}$. For Equation
(109), we have

$\displaystyle\left(\\!\\!\begin{array}[]{ccc|c}-1&1&1&0\\\
0&b_{1}&b_{2}&b_{*}\end{array}\\!\\!\right)\left(\\!\\!\begin{array}[]{ccc|c}{\mathbf{Q}_{11}}+2{\mathbf{Q}_{12}}+{\mathbf{Q}_{22}}-1&{\mathbf{Q}_{11}}+{\mathbf{Q}_{12}}-1&{\mathbf{Q}_{12}}+{\mathbf{Q}_{22}}-1&{\mathbf{Q}_{1}^{*}}+{\mathbf{Q}_{2}^{*}}\\\
{\mathbf{Q}_{11}}+{\mathbf{Q}_{12}}-1&{\mathbf{Q}_{11}}&{\mathbf{Q}_{12}}-1&{\mathbf{Q}_{1}^{*}}\\\
{\mathbf{Q}_{12}}+{\mathbf{Q}_{22}}-1&{\mathbf{Q}_{12}}-1&{\mathbf{Q}_{22}}&{\mathbf{Q}_{2}^{*}}\\\
\hline\cr{\mathbf{Q}_{1}^{*}}^{t}+{\mathbf{Q}_{2}^{*}}^{t}&{\mathbf{Q}_{1}^{*}}^{t}&{\mathbf{Q}_{2}^{*}}^{t}&{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right)$ (112)
$\displaystyle\;=\;\left(\\!\\!\begin{array}[]{ccc|c}-1&0&0&{0}\\\
\begin{subarray}{c}b_{1}(-1+\mathbf{Q}_{11}+\mathbf{Q}_{12})+b_{*}{\mathbf{Q}_{1}^{*}}^{t}\\\
+b_{2}(-1+\mathbf{Q}_{12}+\mathbf{Q}_{22})+b_{*}{\mathbf{Q}_{2}^{*}}^{t}\end{subarray}&\begin{subarray}{c}b_{1}\mathbf{Q}_{11}+b_{*}{\mathbf{Q}_{1}^{*}}^{t}\\\
+b_{2}(-1+\mathbf{Q}_{12})\end{subarray}&\begin{subarray}{c}b_{2}\mathbf{Q}_{22}+b_{*}{\mathbf{Q}_{2}^{*}}^{t}\\\
+b_{1}(-1+\mathbf{Q}_{12})\end{subarray}&b_{1}\mathbf{Q}_{1}^{*}+b_{2}\mathbf{Q}_{2}^{*}+b_{*}\mathbf{Q}_{*}\end{array}\\!\\!\right)$
$\displaystyle\;=\;\widetilde{A}.$

For Equation (110), we have

$\begin{pmatrix}-1&1&1&0\\\
0&b_{1}&b_{2}&b_{*}\end{pmatrix}\begin{pmatrix}(B^{-1}\nu)_{1}+(B^{-1}\nu)_{2}-1\\\
B^{-1}\nu\end{pmatrix}\;=\;\begin{pmatrix}1\\\ \nu\end{pmatrix}.$ (113)

Finally, for Equation (111),

$\displaystyle\widetilde{f}^{t}\widetilde{B}^{-1}\nu$
$\displaystyle\;=\;\widetilde{f}^{t}\begin{pmatrix}(B^{-1}\nu)_{1}+(B^{-1}\nu)_{2}-1\\\
B^{-1}\nu\end{pmatrix}$ (114)
$\displaystyle\;=\;(\widetilde{f}_{0}+\widetilde{f}_{1})(B^{-1}\nu)_{1}+(\widetilde{f}_{0}+\widetilde{f}_{2})(B^{-1}\nu)_{2}+f^{*}(B^{-1}\nu)_{1}-\widetilde{f}_{0}\,,$

and we recall we choose $f_{i}=\widetilde{f}_{0}+\widetilde{f}_{i}$ for
$i=1,2$. ∎

###  6.2. The case of $B$ with nullity one

In this section, we prove Theorem 3.6 under the assumption that the matrix $B$
has nullity 1. In this case, starting from the series $\Phi^{\Xi}(\hbar)$,
there are three intermediate formulas (shown as bullets) in Equation (98) that
eventually identify the result with $\Phi^{\widetilde{\Xi}}(\hbar)$. The
intermediate formulas involve a Fourier transform (adding one integration
variable), a pentagon (adding a second variable), an inverse Fourier transform
(adding a third), and an application of Fubini’s theorem that removes two
integration variables. The detailed computation is given in the next
proposition.

######  Proposition 6.4.

If $B$ has rank $N+1$ then $\Phi^{\Xi}(\hbar)$ is invariant under the 2–3
Pachner move given in Equation (98).

###### Proof.

Following the discussion above we can assume that
$\operatorname{rank}(b_{1}\,|\,b_{2}\,|\,b_{*})=\operatorname{rank}(b_{2}\,|\,b_{*})=N+1$.
Then by [10, Lem.A.3], the matrix $(a_{1}\,|\,b_{2}\,|\,b_{*})$ has full rank
and we can apply a quad move to the first columns of $(A\,|\,B)$ to obtain

$\displaystyle\mathbf{A}\;=\;(-b_{1}\,|\,a_{2}\,|\,a_{*})\,,\qquad\mathbf{B}\;=\;(a_{1}-b_{1}\,|\,b_{2}\,|\,b_{*})\,,\qquad\boldsymbol{\nu}\;=\;\nu-
b_{1}\,,$ (115)

where $\mathbf{B}$ has full rank. The proof will use the following sequence of
intermediate matrices

$(\mathbf{A}\,|\,\mathbf{B}\,|\,\boldsymbol{\nu})\,=\left(\\!\\!\begin{array}[]{c|c|c|c|c|c|c}-b_{1}&a_{2}&a_{*}&a_{1}-b_{1}&b_{2}&b_{*}&\nu-
b_{1}\end{array}\\!\\!\right)$$q_{1}^{-1}\times
1$$(A\,|\,B\,|\,\nu)\,=\left(\\!\\!\begin{array}[]{c|c|c|c|c|c|c}a_{1}&a_{2}&a_{*}&b_{1}&b_{2}&b_{*}&\nu\end{array}\\!\\!\right)$$2\to
3$$(\widetilde{A}\,|\,\widetilde{B}\,|\,\widetilde{\nu})\,=\left(\\!\\!\begin{array}[]{c|c|c|c|c|c|c|c|c}-1&0&0&0&-1&1&1&0&1\\\
a_{1}+a_{2}-b_{1}-b_{2}&a_{1}-b_{2}&a_{2}-b_{1}&a_{*}&0&b_{1}&b_{2}&b_{*}&\nu\end{array}\\!\\!\right)$$1\times
q_{1}\times
1$$(\widetilde{\mathbf{A}}\,|\,\widetilde{\mathbf{B}}\,|\,\widetilde{\boldsymbol{\nu}})\,=\left(\\!\\!\begin{array}[]{c|c|c|c|c|c|c|c|c}-1&-1&0&0&-1&-1&1&0&0\\\
a_{1}+a_{2}-b_{1}-b_{2}&-b_{1}&a_{2}-b_{1}&a_{*}&0&a_{1}-b_{1}-b_{2}&b_{2}&b_{*}&\nu-
b_{1}\end{array}\\!\\!\right)$ (116)

where $\mathbf{B}$ and $\widetilde{\mathbf{B}}$ are invertible. With
$x=(x_{1},x_{2},x^{*})$ and $z=(z_{1},z_{2},z^{*})$ vectors of size $N+2$,
$\Phi^{\Xi}(\hbar)$ is defined by

$\Phi^{\Xi}(\hbar)\;=\;\langle
I_{0}\rangle_{(x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{0}}$ (117)

where with $\mathbf{f}=(f_{1}^{\prime},f_{2},f^{*})$

$\displaystyle I_{0}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x(\mathbf{B}^{-1}\boldsymbol{\nu}-1)\Big{)}\,\psi_{\hbar}\Big{(}x_{1},\frac{1}{1-z_{1}}\Big{)}\,\psi_{\hbar}(x_{2},z_{2})\,\prod_{j=1}^{N}\psi_{\hbar}(x^{*}_{j},z^{*}_{j})$

and

$\mathbf{\Lambda}_{0}\;=\;\mathrm{diag}(z_{1}^{\prime\prime},z_{2}^{\prime},{z^{*}}^{\prime})-\mathbf{Q}\,.$

We apply Corollary 3.5 to the $\psi_{\hbar}$ with argument $x_{1}$ and
introduce a new variable $w_{1}$ to obtain

$\Phi^{\Xi}(\hbar)\;=\;\langle
I_{1}\rangle_{(w_{1},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{1}}$ (118)

where

$\displaystyle I_{1}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{24}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-x_{1}\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}\Big{(}w_{1}-x_{1}(1-z_{1}^{-1}),z_{1}\Big{)}\,\psi_{\hbar}(x_{2},z_{2})\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{1}\;=\;\begin{pmatrix}\frac{z_{1}}{1-z_{1}}&0\\\
0&\mathbf{\Lambda}\end{pmatrix}.$

We substitute $w_{1}\mapsto w_{1}+x_{1}(1-z_{1}^{-1})$ and obtain, from Lemma
3.1, that

$\Phi^{\Xi}(\hbar)\;=\;\langle
I_{2}\rangle_{(w_{1},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{2}}$ (119)

where

$\displaystyle I_{2}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{24}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-x_{1}\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}\,\psi_{\hbar}(w_{1},z_{1})\,\psi_{\hbar}(x_{2},z_{2})\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and with $\mathbf{Q}_{11}=0$ from Lemma 6.5

$\mathbf{\Lambda}_{2}\;=\;\left(\\!\\!\begin{array}[]{ccc|c}\frac{z_{1}}{1-{z_{1}}}&-1&0&0\\\
-1&0&-{\mathbf{Q}_{12}}&-{\mathbf{Q}_{1}^{*}}\\\
0&-{\mathbf{Q}_{12}}&\frac{1}{1-{z_{2}}}-{\mathbf{Q}_{22}}&-{\mathbf{Q}_{2}^{*}}\\\
\hline\cr
0&-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right).$

We apply the pentagon identity of Theorem 3.6 to the $\psi_{h}$ with arguments
$w_{1}$ and $x_{1}$ and obtain, with the new integration variable $x_{0}$,
that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{3}\rangle_{(x_{0},w_{1},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{3}}$ (120)

where

$\displaystyle I_{3}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-x_{1}\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}\,\psi_{\hbar}\Big{(}-x_{0}-x_{2}+\frac{w_{1}z_{2}+x_{2}z_{1}}{z_{0}},z_{1}z_{0}^{-1}\Big{)}$
$\displaystyle\psi_{\hbar}\Big{(}x_{0}+w_{1}+x_{2}-\frac{w_{1}z_{2}+x_{2}z_{1}}{z_{0}},z_{0}\Big{)}\,\psi_{\hbar}\Big{(}-x_{0}-w_{1}+\frac{w_{1}z_{2}+x_{2}z_{1}}{z_{0}},z_{2}z_{0}^{-1}\Big{)}\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{3}\;=\;\left(\\!\\!\begin{array}[]{cccc|c}\frac{({z_{1}}+{z_{2}}-z_{1}z_{2})^{2}}{({z_{1}}-1){z_{1}}({z_{2}}-1){z_{2}}}&0&0&0&0\\\
0&\frac{z_{1}}{1-{z_{1}}}&-1&0&0\\\
0&-1&0&-{\mathbf{Q}_{12}}&-{\mathbf{Q}_{1}^{*}}\\\
0&0&-{\mathbf{Q}_{12}}&\frac{1}{1-{z_{2}}}-{\mathbf{Q}_{22}}&-{\mathbf{Q}_{2}^{*}}\\\
\hline\cr
0&0&-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right).$

We use Lemma 3.1 to change the variables $x_{0}\mapsto
x_{0}-w_{1}-x_{2}+\frac{w_{1}z_{2}+x_{2}z_{1}}{z_{0}}$ and obtain that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{4}\rangle_{(x_{0},w_{1},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{4}}$ (121)

where

$\displaystyle I_{4}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-x_{1}\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}(-x_{0}+w_{1},z_{1}z_{0}^{-1})\,\psi_{\hbar}(x_{0},z_{0})\,\psi_{\hbar}(-x_{0}+x_{2},z_{2}z_{0}^{-1})\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{4}\;=\;\left(\\!\\!\begin{array}[]{cccc|c}\frac{({z_{1}}+{z_{2}}-z_{1}z_{2})^{2}}{({z_{1}}-1){z_{1}}({z_{2}}-1){z_{2}}}&\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{({z_{1}}-1){z_{2}}}&0&\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{{z_{1}}({z_{2}}-1)}&0\\\
\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{({z_{1}}-1){z_{2}}}&\frac{{z_{1}}}{{z_{2}}(1-{z_{1}})}&-1&1&0\\\
0&-1&0&-{\mathbf{Q}_{12}}&-{\mathbf{Q}_{1}^{*}}\\\
\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{{z_{1}}({z_{2}}-1)}&1&-{\mathbf{Q}_{12}}&-\mathbf{Q}_{22}-\frac{z_{1}+z_{2}-z_{1}z_{2}}{z_{1}(z_{2}-1)}&-{\mathbf{Q}_{2}^{*}}\\\
\hline\cr
0&0&-{Q_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right).$

We substitute $w_{1}\mapsto w_{1}+x_{0}$ and $x_{2}\mapsto x_{2}+x_{0}$ to
obtain, with Lemma 3.1, that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{5}\rangle_{(x_{0},w_{1},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{5}}$ (122)

where

$\displaystyle I_{5}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-\frac{\hbar^{\frac{1}{2}}}{2}x_{0}(\mathbf{B}^{-1}\boldsymbol{\nu}_{2}-1)-x_{1}\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}(w_{1},z_{1}z_{0}^{-1})\,\psi_{\hbar}(x_{0},z_{0})\,\psi_{\hbar}(x_{2},z_{2}z_{0}^{-1})\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{5}\;=\;\left(\\!\\!\begin{array}[]{cccc|c}-{\mathbf{Q}_{22}}+\frac{1}{({z_{1}}-1)({z_{2}}-1)}&0&-1-{\mathbf{Q}_{12}}&1-{\mathbf{Q}_{22}}&-{\mathbf{Q}_{2}^{*}}\\\
0&\frac{{z_{1}}}{{z_{2}}(1-{z_{1}})}&-1&1&0\\\
-1-{\mathbf{Q}_{12}}&-1&0&-{\mathbf{Q}_{12}}&-{\mathbf{Q}_{1}^{*}}\\\
1-{\mathbf{Q}_{22}}&1&-{\mathbf{Q}_{12}}&-\mathbf{Q}_{22}+\frac{z_{1}+z_{2}-z_{1}z_{2}}{z_{1}(z_{2}-1)}&-{\mathbf{Q}_{2}^{*}}\\\
\hline\cr-{\mathbf{Q}_{2}^{*}}^{t}&0&-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right).$

By applying the first quad move from Theorem 3.4 to
$\psi_{\hbar}(w_{1},z_{1}z_{0}^{-1})$ we obtain, with a new integration
variable $y_{1}$, that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;e^{-\frac{\hbar}{24}}\langle
I_{6}\rangle_{(y_{1},x_{0},w_{1},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{6}}$
(123)

where

$\displaystyle I_{6}$
$\displaystyle\;=\;\exp\Big{(}-\frac{\hbar}{24}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x(\mathbf{B}^{-1}\boldsymbol{\nu}-1)$
$\displaystyle\quad-\frac{\hbar^{\frac{1}{2}}}{2}x_{0}((\mathbf{B}^{-1}\boldsymbol{\nu})_{2}-1)-x_{1}\frac{\hbar^{\frac{1}{2}}}{2}+\Big{(}y_{1}+\frac{w_{1}z_{1}z_{0}^{-1}}{1-z_{1}z_{0}^{-1}}\Big{)}\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}\Big{(}y_{1}+\frac{w_{1}z_{1}z_{0}^{-1}}{1-z_{1}z_{0}^{-1}},\frac{1}{1-z_{1}z_{0}^{-1}}\Big{)}\,\psi_{\hbar}(x_{0},z_{0})\,\psi_{\hbar}(x_{2},z_{2}z_{0}^{-1})\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{6}\;=\;\left(\\!\\!\begin{array}[]{ccccc|c}\frac{(z_{1}-1)z_{2}}{z_{1}}&0&0&0&0&0\\\
0&-{\mathbf{Q}_{22}}+\frac{1}{({z_{1}}-1)({z_{2}}-1)}&0&-1-{\mathbf{Q}_{12}}&1-{\mathbf{Q}_{22}}&-{\mathbf{Q}_{2}^{*}}\\\
0&0&\frac{{z_{1}}}{{z_{2}}(1-{z_{1}})}&-1&1&0\\\
0&-1-{\mathbf{Q}_{12}}&-1&0&-{\mathbf{Q}_{12}}&-{\mathbf{Q}_{1}^{*}}\\\
0&1-{\mathbf{Q}_{22}}&1&-{\mathbf{Q}_{12}}&-\mathbf{Q}_{22}+\frac{z_{1}+z_{2}-z_{1}z_{2}}{z_{1}(z_{2}-1)}&-{\mathbf{Q}_{2}^{*}}\\\
\hline\cr
0&-{\mathbf{Q}_{2}^{*}}^{t}&0&-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\end{array}\\!\\!\right).$

We change variables $y_{1}\mapsto
y_{1}-w_{1}\frac{z_{1}z_{0}^{-1}}{1-z_{1}z_{0}^{-1}}$ using Lemma 3.1 and
obtain that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{7}\rangle_{(y_{1},x_{0},w_{1},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{7}}$
(124)

where

$\displaystyle I_{7}$
$\displaystyle\;=\;\exp\Big{(}-\frac{\hbar}{24}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-\frac{\hbar^{\frac{1}{2}}}{2}x_{0}((\mathbf{B}^{-1}\boldsymbol{\nu})_{2}-1)-x_{1}\frac{\hbar^{\frac{1}{2}}}{2}+y_{1}\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}\Big{(}y_{1},\frac{1}{1-z_{1}z_{0}^{-1}}\Big{)}\,\psi_{\hbar}(x_{0},z_{0})\,\psi_{\hbar}(x_{2},z_{2}z_{0}^{-1})\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{7}\;=\;\left(\\!\\!\begin{array}[]{ccccc|c}\frac{(z_{1}-1)z_{2}}{z_{1}}&0&1&0&0&0\\\
0&-{\mathbf{Q}_{22}}+\frac{1}{({z_{1}}-1)({z_{2}}-1)}&0&-1-{\mathbf{Q}_{12}}&1-{\mathbf{Q}_{22}}&-{\mathbf{Q}_{2}^{*}}\\\
1&0&0&-1&1&0\\\
0&-1-{\mathbf{Q}_{12}}&-1&0&-{\mathbf{Q}_{12}}&-{\mathbf{Q}_{1}^{*}}\\\
0&1-{\mathbf{Q}_{22}}&1&-{\mathbf{Q}_{12}}&-\mathbf{Q}_{22}+\frac{z_{1}+z_{2}-z_{1}z_{2}}{z_{1}(z_{2}-1)}&-{\mathbf{Q}_{2}^{*}}\\\
\hline\cr
0&-{\mathbf{Q}_{2}^{*}}^{t}&0&-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right).$

Therefore, we can apply Fubini’s Theorem (Lemma 3.1) with the integration
variables $w_{1},x_{1}$, to obtain that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{8}\rangle_{(y_{1},x_{0},x_{2},x^{*}),\mathbf{\Lambda}_{8}}$ (125)

where

$\displaystyle I_{8}$
$\displaystyle\;=\;\exp\Big{(}-\frac{\hbar}{24}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}(x_{0}((\mathbf{B}^{-1}\boldsymbol{\nu})_{2}-1)+y_{1}((\mathbf{B}^{-1}\boldsymbol{\nu})_{1}-1)$
$\displaystyle\qquad\qquad+x_{2}((\mathbf{B}^{-1}\boldsymbol{\nu})_{2}+(\mathbf{B}^{-1}\boldsymbol{\nu})_{1}-1)+x^{*}((\mathbf{B}^{-1}\boldsymbol{\nu})^{*}-1))\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}\Big{(}y_{1},\frac{1}{1-z_{1}z_{0}^{-1}}\Big{)}\,\psi_{\hbar}(x_{0},z_{0})\,\psi_{\hbar}(x_{2},z_{2}z_{0}^{-1})\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{8}\;=\;\left(\\!\\!\begin{array}[]{ccc|c}\frac{({z_{1}}-1){z_{2}}}{{z_{1}}}&-{\mathbf{Q}_{12}}-1&-{\mathbf{Q}_{12}}&-{Q_{1}^{*}}\\\
-{\mathbf{Q}_{12}}-1&\frac{1}{({z_{1}}-1)({z_{2}}-1)}-{\mathbf{Q}_{22}}&-{\mathbf{Q}_{12}}-{\mathbf{Q}_{22}}&-{\mathbf{Q}_{2}^{*}}\\\
-{\mathbf{Q}_{12}}&-{\mathbf{Q}_{12}}-{\mathbf{Q}_{22}}&-2{\mathbf{Q}_{12}}-\mathbf{Q}_{22}+\frac{z_{1}+z_{2}-z_{1}z_{2}}{z_{1}(1-z_{2})}&-{\mathbf{Q}_{1}^{*}}-{\mathbf{Q}_{2}^{*}}\\\
\hline\cr-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&-{\mathbf{Q}_{1}^{*}}^{t}-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right)\,.$

Using Lemma 6.5 we obtain, with respect to the integration variables
$\tilde{x}=(x_{0},y_{1},x_{2},x^{*})$ and some $c\in\frac{1}{24}\mathbbm{Z}$,
that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;e^{c\hbar}\langle
I_{9}\rangle_{\tilde{x},\mathbf{\Lambda}_{9}}$ (126)

where

$I_{9}\;=\;\exp\Big{(}\frac{\hbar}{8}\widetilde{\mathbf{f}}^{t}\widetilde{\mathbf{B}}^{-1}\widetilde{\mathbf{A}}\widetilde{\mathbf{f}}-\frac{\hbar^{\frac{1}{2}}}{2}\tilde{x}^{t}(\widetilde{\mathbf{B}}^{-1}\widetilde{\boldsymbol{\nu}}-1)\Big{)}\,\prod_{j=0}^{N+2}\psi_{\hbar}(\widetilde{x}_{j}^{*},\widetilde{z}_{j}^{*})$

and with $\widetilde{\mathbf{A}}$, $\widetilde{\mathbf{B}}$ as in (116)

$\mathbf{\Lambda}_{9}\;=\;\mathrm{diag}(\widetilde{z}_{0}^{\prime},\widetilde{z}_{1}^{\prime\prime},\widetilde{z}_{2}^{\prime},\widetilde{z}^{*}{}^{\prime})-{\widetilde{\mathbf{B}}}^{-1}\widetilde{\mathbf{A}}.$

∎

######  Lemma 6.5.

With the notation used in the previous proof of Proposition 6.4 we have
$\mathbf{Q}_{11}=0$ and the following equalities:

$\displaystyle\widetilde{\mathbf{B}}^{-1}\widetilde{\mathbf{A}}$
$\displaystyle\;=\;\left(\\!\\!\begin{array}[]{ccc|c}\mathbf{Q}_{22}&\mathbf{Q}_{12}+1&\mathbf{Q}_{12}+\mathbf{Q}_{22}&\mathbf{Q}_{2}^{*}\\\
\mathbf{Q}_{12}+1&0&\mathbf{Q}_{12}&\mathbf{Q}_{1}^{*}\\\
\mathbf{Q}_{12}+\mathbf{Q}_{22}&\mathbf{Q}_{12}&2\mathbf{Q}_{12}+\mathbf{Q}_{22}&\mathbf{Q}_{1}^{*}+\mathbf{Q}_{2}^{*}\\\
\hline\cr{\mathbf{Q}_{2}^{*}}^{t}&{\mathbf{Q}_{1}^{*}}^{t}&{\mathbf{Q}_{1}^{*}}^{t}+{\mathbf{Q}_{2}^{*}}^{t}&\mathbf{Q}^{*}\end{array}\\!\\!\right),$
(131) $\displaystyle\widetilde{\mathbf{B}}^{-1}\widetilde{\boldsymbol{\nu}}$
$\displaystyle\;=\;\left(\\!\\!\begin{array}[]{c}(\mathbf{B}^{-1}\boldsymbol{\nu})_{2}\\\
(\mathbf{B}^{-1}\boldsymbol{\nu})_{1}\\\
(\mathbf{B}^{-1}\boldsymbol{\nu})_{1}+(\mathbf{B}^{-1}\boldsymbol{\nu})_{2}\\\
(\mathbf{B}^{-1}\boldsymbol{\nu})^{*}\end{array}\\!\\!\right),$ (136)
$\displaystyle\widetilde{\mathbf{f}}^{t}\widetilde{\mathbf{B}}^{-1}\widetilde{\boldsymbol{\nu}}$
$\displaystyle\;=\;\mathbf{f}^{t}\mathbf{B}^{-1}\boldsymbol{\nu}\,.$ (137)

###### Proof.

Similarly to the proof of Lemma 6.3 we will proof Equations (131) and (136) by
showing that both sides are equal after multiplying by the invertible matrix
$\widetilde{\mathbf{B}}$. Using $\mathbf{Q}=\mathbf{B}^{-1}\mathbf{A}$ and the
fact that $(a_{1}\,|\,b_{2}\,|\,b_{*})$ are linearly independent we conclude
$\mathbf{Q}_{11}=0$. For (131) we compute

$\displaystyle\begin{pmatrix}-1&-1&1&0\\\
0&a_{1}-b_{1}-b_{2}&b_{2}&b_{*}\end{pmatrix}\left(\\!\\!\begin{array}[]{ccc|c}\mathbf{Q}_{22}&\mathbf{Q}_{12}+1&\mathbf{Q}_{12}+\mathbf{Q}_{22}&\mathbf{Q}_{2}^{*}\\\
\mathbf{Q}_{12}+1&0&\mathbf{Q}_{12}&\mathbf{Q}_{1}^{*}\\\
\mathbf{Q}_{12}+\mathbf{Q}_{22}&\mathbf{Q}_{12}&2\mathbf{Q}_{12}+\mathbf{Q}_{22}&\mathbf{Q}_{1}^{*}+\mathbf{Q}_{2}^{*}\\\
\hline\cr{\mathbf{Q}_{2}^{*}}^{t}&{\mathbf{Q}_{1}^{*}}^{t}&{\mathbf{Q}_{1}^{*}}^{t}+{\mathbf{Q}_{2}^{*}}^{t}&\mathbf{Q}^{*}\end{array}\\!\\!\right)$
(138) $\displaystyle\;=$ $\displaystyle\begin{pmatrix}-1&-1&0&0\\\
\begin{subarray}{c}(a_{1}-b_{1})\mathbf{Q}_{12}\\\
+b_{2}\mathbf{Q}_{22}+b_{*}{\mathbf{Q}_{2}^{*}}^{t}\end{subarray}&\begin{subarray}{c}b_{2}\mathbf{Q}_{12}+b_{*}{\mathbf{Q}_{1}^{*}}^{t}\end{subarray}&\begin{subarray}{c}b_{2}\mathbf{Q}_{12}+b_{*}{\mathbf{Q}_{1}^{*}}^{t}+\\\
(a_{1}-b_{1})\mathbf{Q}_{12}+b_{2}\mathbf{Q}_{22}+b_{*}{\mathbf{Q}_{1}^{*}}^{t}\end{subarray}&\begin{subarray}{c}(a_{1}-b_{1}-b_{2})\mathbf{Q}_{1}^{*}\\\
+b_{2}(\mathbf{Q}_{1}^{*}+\mathbf{Q}_{2}^{*})+b_{*}\mathbf{Q}^{*}\end{subarray}\end{pmatrix}$
$\displaystyle\;=$ $\displaystyle\widetilde{\mathbf{A}}$

For (136) we have

$\displaystyle\begin{pmatrix}-1&-1&1&0\\\
0&a_{1}-b_{1}-b_{2}&b_{2}&b_{*}\end{pmatrix}\left(\\!\\!\begin{array}[]{c}(\mathbf{B}^{-1}\boldsymbol{\nu})_{2}\\\
(\mathbf{B}^{-1}\boldsymbol{\nu})_{1}\\\
(\mathbf{B}^{-1}\boldsymbol{\nu})_{1}+(\mathbf{B}^{-1}\boldsymbol{\nu})_{2}\\\
(\mathbf{B}^{-1}\boldsymbol{\nu})^{*}\end{array}\\!\\!\right)$ (139)
$\displaystyle\;=\;\left(\\!\\!\begin{array}[]{c}0\\\
\boldsymbol{\nu}\end{array}\\!\\!\right)$

For Equation (137), we have

$\displaystyle\widetilde{\mathbf{f}}^{t}\left(\\!\\!\begin{array}[]{c}(\mathbf{B}^{-1}\boldsymbol{\nu})_{2}\\\
(\mathbf{B}^{-1}\boldsymbol{\nu})_{1}\\\
(\mathbf{B}^{-1}\boldsymbol{\nu})_{1}+(\mathbf{B}^{-1}\boldsymbol{\nu})_{2}\\\
(\mathbf{B}^{-1}\boldsymbol{\nu})^{*}\end{array}\\!\\!\right)$ (140)
$\displaystyle\;=\;(\widetilde{f}_{1}^{\prime}+\widetilde{f}_{2})(\mathbf{B}^{-1}\boldsymbol{\nu})_{1}+(\widetilde{f}_{0}-\widetilde{f}_{2})(\mathbf{B}^{-1}\boldsymbol{\nu})_{2}+\widetilde{f}^{*}(\mathbf{B}^{-1}\boldsymbol{\nu})^{*}\,.$

Then, using the analogous relations between the flattening as given in
Equation (94) for the shapes, we have
$\widetilde{f}_{1}^{\prime}+\widetilde{f}_{2}=f_{1}^{\prime}=\mathbf{f}_{1}$
and $\widetilde{f}_{0}-\widetilde{f}_{2}=f_{2}=\mathbf{f}_{2}$. ∎

###  6.3. The case of $B$ with nullity two

In this section, we prove Theorem 3.6 under the assumption that the matrix $B$
has nullity 2. Similarly to the proof of Proposition 6.4, we use three
intermediate formulas in Equation (98). They involve two Fourier transforms
(adding two variables), a pentagon (adding a third variable), two inverse
Fourier transforms (adding two variables) and an applications of Fubini’s
Theorem (removing four integration variables). The details are given in the
following proposition.

######  Proposition 6.6.

If $B$ has rank $N$ then $\Phi^{\Xi}(\hbar)$ is invariant under the 2–3
Pachner move given in Equation (98).

###### Proof.

Following the discussion above, we can assume that
$\operatorname{rank}(b_{1}\,|\,b_{2}\,|\,b_{*})=\operatorname{rank}(\,b_{*})=N$.
Then by [10, Lem.A.3], the matrix $(a_{1}-b_{1}\,|\,a_{2}-b_{2}\,|\,b_{*})$
has full rank and we can apply a quad move to the first columns of $(A\,|\,B)$
to obtain

$\displaystyle\mathbf{A}\;=\;(-b_{1}\,|\,-b_{2}\,|\,a_{*}),\qquad\mathbf{B}\;=\;(a_{1}-b_{1}\,|\,a_{2}-b_{2}\,|\,b_{*})$
(141)

where $\mathbf{B}$ has full rank. The proof will use the following sequence of
intermediate matrices

$(\mathbf{A}\,|\,\mathbf{B}\,|\,\boldsymbol{\nu})\,=\left(\\!\\!\begin{array}[]{c|c|c|c|c|c|c}-b_{1}&-b_{2}&a_{*}&a_{1}-b_{1}&a_{2}-b_{2}&b_{*}&\nu-
b_{1}-b_{2}\end{array}\\!\\!\right)$$q_{2}^{-1}\times
1$$(A\,|\,B\,|\,\nu)\,=\left(\\!\\!\begin{array}[]{c|c|c|c|c|c|c}a_{1}&a_{2}&a_{*}&b_{1}&b_{2}&b_{*}&\nu\end{array}\\!\\!\right)$$2\to
3$$(\widetilde{A}\,|\,\widetilde{B}\,|\,\widetilde{\nu})\,=\left(\\!\\!\begin{array}[]{c|c|c|c|c|c|c|c|c}-1&0&0&0&-1&1&1&0&1\\\
a_{1}+a_{2}-b_{1}-b_{2}&a_{1}-b_{2}&a_{2}-b_{1}&a_{*}&0&b_{1}&b_{2}&b_{*}&\nu\end{array}\\!\\!\right)$$1\times
q_{2}\times
1$$(\widetilde{\mathbf{A}}\,|\,\widetilde{\mathbf{B}}\,|\,\widetilde{\boldsymbol{\nu}})=$$\left(\\!\\!\begin{array}[]{c|c|c|c|c|c|c|c|c}-1&-1&-1&0&-1&-1&-1&0&-1\\\
a_{1}+a_{2}-b_{1}-b_{2}&-b_{1}&-b_{2}&a_{*}&0&a_{1}-b_{1}-b_{2}&a_{2}-b_{1}-b_{2}&b_{*}&\nu-
b_{1}-b_{2}\end{array}\\!\\!\right)$ (142)

where $\mathbf{B}$ and $\widetilde{\mathbf{B}}$ are invertible. With
$x=(x_{1},x_{2},x^{*})$ and $z=(z_{1},z_{2},z^{*})$ vectors of size $N+2$,
$\Phi^{\Xi}(\hbar)$ is defined by

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{0}\rangle_{(x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{0}}$ (143)

where with $\mathbf{f}=(f_{1}^{\prime},f_{2}^{\prime},f^{*})$

$\displaystyle I_{0}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}(\mathbf{B}^{-1}\boldsymbol{\nu}-1)\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}\Big{(}x_{1},\frac{1}{1-z_{1}}\Big{)}\,\psi_{\hbar}\Big{(}x_{2},\frac{1}{1-z_{2}}\Big{)}\,\prod_{j=1}^{N}\psi_{\hbar}(x^{*}_{j},z^{*}_{j})$

and

$\mathbf{\Lambda}_{0}\;=\;\mathrm{diag}(1-z_{1}^{-1},1-z_{2}^{-1},{z^{*}}^{\prime})-\mathbf{Q}\,.$

By applying Corollary 3.5 to both $\psi_{\hbar}(x_{1},\frac{1}{1-z_{1}})$ and
$\psi_{\hbar}(x_{1},\frac{1}{1-z_{1}})$ we obtain with new integration
variables $w_{1}$ and $w_{2}$ that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{1}\rangle_{(w_{1},w_{2},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{1}}$ (144)

where

$\displaystyle I_{1}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{12}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-(x_{1}+x_{2})\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}\Big{(}w_{1}-x_{1}(1-z_{1}^{-1}),z_{1}\Big{)}\,\psi_{\hbar}\Big{(}w_{2}-x_{2}(1-z_{2}^{-1}),z_{1}\Big{)}\,\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{1}\;=\;\begin{pmatrix}\frac{z_{2}}{1-z_{2}}&0&0\\\
0&\frac{z_{1}}{1-z_{1}}&0\\\ 0&0&\mathbf{\Lambda}\end{pmatrix}$

The substitutions $w_{1}\mapsto w_{1}+x_{1}(1-z_{1}^{-1})$ and $w_{2}\mapsto
w_{2}+x_{2}(1-z_{2}^{-1})$ imply according to Lemma 3.1 that

$\Phi^{\Xi}(\hbar)\;=\;\langle
I_{2}\rangle_{(w_{1},w_{2},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{2}}$ (145)

where

$\displaystyle I_{2}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{12}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-(x_{1}+x_{2})\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}(w_{1},z_{1})\,\psi_{\hbar}(w_{2},z_{2})\,\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and with $\mathbf{Q}_{11}=\mathbf{Q}_{12}=\mathbf{Q}_{22}=0$ from Lemma 6.7

$\mathbf{\Lambda}_{2}\;=\;\left(\\!\\!\begin{array}[]{cccc|c}\frac{z_{1}}{1-{z_{1}}}&0&-1&0&0\\\
0&\frac{z_{2}}{1-{z_{2}}}&0&-1&0\\\ -1&0&0&0&-{\mathbf{Q}_{1}^{*}}\\\
0&-1&0&0&-{\mathbf{Q}_{2}^{*}}\\\ \hline\cr
0&0&-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right).$

We apply the pentagon identity of Theorem 3.6 to the $\psi_{h}$ with arguments
$w_{1}$ and $w_{1}$ and obtain with the new integration variable $x_{0}$

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{3}\rangle_{(x_{0},w_{1},w_{2},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{3}}$
(146)

where

$\displaystyle I_{3}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{24}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-(x_{1}+x_{2})\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}\Big{(}-x_{0}-w_{2}+\frac{w_{1}z_{2}+w_{2}z_{1}}{z_{0}},z_{1}z_{0}^{-1}\Big{)}\,\psi_{\hbar}\Big{(}x_{0}+w_{1}+w_{2}-\frac{w_{1}z_{2}+w_{2}z_{1}}{z_{0}},z_{0}\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}\Big{(}-x_{0}-w_{1}+\frac{w_{1}z_{2}+w_{2}z_{1}}{z_{0}},z_{2}z_{0}^{-1}\Big{)}\,\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{3}\;=\;\left(\\!\\!\begin{array}[]{ccccc|c}\frac{({z_{1}}+{z_{2}}-z_{1}z_{2})^{2}}{({z_{1}}-1){z_{1}}({z_{2}}-1){z_{2}}}&0&0&0&0&0\\\
0&\frac{z_{1}}{1-{z_{1}}}&0&-1&0&0\\\ 0&0&\frac{z_{2}}{1-{z_{2}}}&0&-1&0\\\
0&-1&0&0&0&-{\mathbf{Q}_{1}^{*}}\\\ 0&0&-1&0&0&-{\mathbf{Q}_{2}^{*}}\\\
\hline\cr
0&0&0&-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right).$

With change of variables $x_{0}\mapsto
x_{0}-w_{1}-w_{2}+\frac{w_{1}z_{2}+w_{2}z_{1}}{z_{0}}$ Lemma 3.1 gives that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{4}\rangle_{(x_{0},w_{1},w_{2},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{4}}$
(147)

where

$\displaystyle I_{4}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{24}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-(x_{1}+x_{2})\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}(-x_{0}+w_{1},z_{1}z_{0}^{-1})\,\psi_{\hbar}(x_{0},z_{0})\,\psi_{\hbar}(-x_{0}+w_{2},z_{2}z_{0}^{-1})\,\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{4}\;=\;\left(\\!\\!\begin{array}[]{ccccc|c}\frac{({z_{1}}+{z_{2}}-z_{1}z_{2})^{2}}{({z_{1}}-1){z_{1}}({z_{2}}-1){z_{2}}}&\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{({z_{1}}-1){z_{2}}}&\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{({z_{2}}-1){z_{1}}}&0&0&0\\\
\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{({z_{1}}-1){z_{2}}}&\frac{z_{1}}{1-{z_{1}}}&1&-1&0&0\\\
\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{({z_{2}}-1){z_{1}}}&1&\frac{z_{2}}{1-{z_{2}}}&0&-1&0\\\
0&-1&0&0&0&-{\mathbf{Q}_{1}^{*}}\\\ 0&0&-1&0&0&-{\mathbf{Q}_{2}^{*}}\\\
\hline\cr
0&0&0&-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right).$

We substitute $w_{1}\mapsto w_{1}+x_{0}$ and $w_{2}\mapsto w_{2}+x_{0}$ to
obtain with Lemma 3.1 that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{5}\rangle_{(x_{0},w_{1},w_{2},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{5}}$
(148)

where

$\displaystyle I_{5}$
$\displaystyle\;=\;\exp\Big{(}\frac{\hbar}{24}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-(x_{1}+x_{2})\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}(w_{1},z_{1}z_{0}^{-1})\,\psi_{\hbar}(x_{0},z_{0})\,\psi_{\hbar}(w_{2},z_{2}z_{0}^{-1})\,\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{5}\;=\;\left(\\!\\!\begin{array}[]{ccccc|c}\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{({z_{1}}-1)({z_{2}}-1)}&0&0&-1&-1&0\\\
0&\frac{z_{1}}{(1-{z_{1}})z_{2}}&1&-1&0&0\\\
0&1&\frac{z_{2}}{(1-{z_{2}})z_{1}}&0&-1&0\\\
-1&-1&0&0&0&-{\mathbf{Q}_{1}^{*}}\\\ -1&0&-1&0&0&-{\mathbf{Q}_{2}^{*}}\\\
\hline\cr
0&0&0&-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right).$

By applying the first quad move from Theorem 3.4 to both
$\psi_{\hbar}(w_{1},z_{1}z_{0}^{-1})$ and
$\psi_{\hbar}(w_{2},z_{2}z_{0}^{-1})$ we obtain with new integration variables
$y_{1}$ and $y_{2}$ that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{6}\rangle_{(y_{1},y_{2},x_{0},w_{1},w_{2},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{6}}$
(149)

where

$\displaystyle I_{6}$
$\displaystyle\;=\;\exp\Big{(}-\frac{\hbar}{24}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-(x_{1}+x_{2})\frac{\hbar^{\frac{1}{2}}}{2}$
$\displaystyle\qquad\qquad+\Big{(}y_{1}+\frac{w_{1}z_{1}z_{0}^{-1}}{1-z_{1}z_{0}^{-1}}\Big{)}\frac{\hbar^{\frac{1}{2}}}{2}+\Big{(}y_{2}+\frac{w_{2}z_{2}z_{0}^{-1}}{1-z_{2}z_{0}^{-1}}\Big{)}\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$
$\displaystyle\qquad\times\psi_{\hbar}\Big{(}y_{1}+\frac{w_{1}z_{1}z_{0}^{-1}}{1-z_{1}z_{0}^{-1}},\frac{1}{1-z_{1}z_{0}^{-1}}\Big{)}\,\psi_{\hbar}\Big{(}y_{2}+\frac{w_{2}z_{2}z_{0}^{-1}}{1-z_{2}z_{0}^{-1}},\frac{1}{1-z_{2}z_{0}^{-1}}\Big{)}\,\psi_{\hbar}(x_{0},z_{0})$

and

$\mathbf{\Lambda}_{6}\;=\;\left(\\!\\!\begin{array}[]{ccccccc|c}\frac{(z_{1}-1)z_{2}}{z_{1}}&0&0&0&0&0&0&0\\\
0&\frac{(z_{2}-1)z_{1}}{z_{2}}&0&0&0&0&0&0\\\
0&0&\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{({z_{1}}-1)({z_{2}}-1)}&0&0&-1&-1&0\\\
0&0&0&\frac{z_{1}}{(1-{z_{1}})z_{2}}&1&-1&0&0\\\
0&0&0&1&\frac{z_{2}}{(1-{z_{2}})z_{1}}&0&-1&0\\\
0&0&-1&-1&0&0&0&-{\mathbf{Q}_{1}^{*}}\\\
0&0&-1&0&-1&0&0&-{\mathbf{Q}_{2}^{*}}\\\ \hline\cr
0&0&0&0&0&-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right).$

We change variables $y_{1}\mapsto
y_{1}-w_{1}\frac{z_{1}z_{0}^{-1}}{1-z_{1}z_{0}^{-1}}$ and $y_{2}\mapsto
y_{2}-w_{2}\frac{z_{2}z_{0}^{-1}}{1-z_{2}z_{0}^{-1}}$ using Lemma 3.1 to
obtain that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{7}\rangle_{(y_{1},y_{2},x_{0},w_{1},w_{2},x_{1},x_{2},x^{*}),\mathbf{\Lambda}_{7}}$
(150)

where

$\displaystyle I_{7}$
$\displaystyle\;=\;\exp\Big{(}-\frac{\hbar}{24}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}-\frac{\hbar^{\frac{1}{2}}}{2}x^{t}(\mathbf{B}^{-1}\boldsymbol{\nu}-1)-(x_{1}+x_{2})\frac{\hbar^{\frac{1}{2}}}{2}+(y_{1}+y_{2})\frac{\hbar^{\frac{1}{2}}}{2}\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}\Big{(}y_{1},\frac{1}{1-z_{1}z_{0}^{-1}}\Big{)}\,\psi_{\hbar}\Big{(}y_{2},\frac{1}{1-z_{2}z_{0}^{-1}}\Big{)}\,\psi_{\hbar}(x_{0},z_{0})\,\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{7}\;=\;\left(\\!\\!\begin{array}[]{ccccccc|c}\frac{(z_{1}-1)z_{2}}{z_{1}}&0&0&1&0&0&0&0\\\
0&\frac{(z_{2}-1)z_{1}}{z_{2}}&0&0&1&0&0&0\\\
0&0&\frac{{z_{1}}+{z_{2}}-z_{1}z_{2}}{({z_{1}}-1)({z_{2}}-1)}&0&0&-1&-1&0\\\
1&0&0&0&1&-1&0&0\\\ 0&1&0&1&0&0&-1&0\\\
0&0&-1&-1&0&0&0&-{\mathbf{Q}_{1}^{*}}\\\
0&0&-1&0&-1&0&0&-{\mathbf{Q}_{2}^{*}}\\\ \hline\cr
0&0&0&0&0&-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{z^{*}}^{\prime}-{\mathbf{Q}^{*}}\\\
\end{array}\\!\\!\right).$

Therefore, we can apply Fubini’s Theorem (Lemma 3.1) with the integration
variables $w_{1},w_{2},x_{1},x_{2}$, to obtain that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;\langle
I_{8}\rangle_{(y_{1},y_{2},x_{0},x^{*}),\mathbf{\Lambda}_{8}}$ (151)

where

$\displaystyle I_{8}$
$\displaystyle\;=\;\exp\Big{(}-\frac{\hbar}{4}(\mathbf{B}^{-1}\boldsymbol{\nu})_{1}(\mathbf{B}^{-1}\boldsymbol{\nu})_{2}-\frac{\hbar}{24}+\frac{\hbar}{8}\mathbf{f}^{t}\mathbf{B}^{-1}\mathbf{A}\mathbf{f}+\frac{\hbar^{\frac{1}{2}}}{2}\big{(}x_{0}\left((\mathbf{B}^{-1}\boldsymbol{\nu})_{1}+(\mathbf{B}^{-1}\boldsymbol{\nu})_{2}\right)\big{)}$
$\displaystyle\qquad\qquad\quad-\frac{\hbar^{\frac{1}{2}}}{2}y_{1}\left((\mathbf{B}^{-1}\boldsymbol{\nu})_{1}-1\right)-\frac{\hbar^{\frac{1}{2}}}{2}y_{2}\left((\mathbf{B}^{-1}\boldsymbol{\nu})_{2}-1\right)$
$\displaystyle\qquad\qquad\quad-\frac{\hbar^{\frac{1}{2}}}{2}{x^{*}}^{t}\left((\mathbf{B}^{-1}\boldsymbol{\nu})^{*}-(\mathbf{B}^{-1}\boldsymbol{\nu})_{2}\mathbf{Q}_{1}^{*}-(\mathbf{B}^{-1}\boldsymbol{\nu})_{1}\mathbf{Q}_{2}^{*}-1\right)\Big{)}$
$\displaystyle\qquad\times\psi_{\hbar}\Big{(}y_{1},\frac{1}{1-z_{1}z_{0}^{-1}}\Big{)}\,\psi_{\hbar}\Big{(}y_{2},\frac{1}{1-z_{2}z_{0}^{-1}}\Big{)}\,\psi_{\hbar}(x_{0},z_{0})\,\prod_{j=1}^{N}\psi_{\hbar}(x_{j}^{*},z_{j}^{*})$

and

$\mathbf{\Lambda}_{8}\;=\;\left(\\!\\!\begin{array}[]{ccc|c}\frac{({z_{1}}-1){z_{2}}}{{z_{1}}}&0&-1&-{\mathbf{Q}_{1}^{*}}\\\
0&\frac{({z_{2}}-1){z_{1}}}{{z_{2}}}&-1&-{\mathbf{Q}_{2}^{*}}\\\
-1&-1&\frac{{z_{1}}+{z_{2}}-z_{2}z_{2}}{({z_{1}}-1)({z_{2}}-1)}+2&{\mathbf{Q}_{1}^{*}}+{\mathbf{Q}_{2}^{*}}\\\
\hline\cr-{\mathbf{Q}_{1}^{*}}^{t}&-{\mathbf{Q}_{2}^{*}}^{t}&{\mathbf{Q}_{1}^{*}}^{t}+{\mathbf{Q}_{2}^{*}}^{t}&-{\mathbf{Q}^{*}}+{\mathbf{Q}_{1}^{*}}{\mathbf{Q}_{2}^{*}}+{\mathbf{Q}_{2}^{*}}{\mathbf{Q}_{1}^{*}}\end{array}\\!\\!\right).$

Using Lemma 6.7, we obtain with respect to the integration variables
$\tilde{x}=(x_{0},y_{1},y_{2},x^{*})$ and some $c\in\frac{1}{24}\mathbb{Z}$
that

$\displaystyle\Phi^{\Xi}(\hbar)\;=\;e^{c\hbar}\langle
I_{9}\rangle_{\tilde{x},\mathbf{\Lambda}_{9}}$ (152)

where with $\widetilde{\mathbf{A}},\widetilde{\mathbf{B}}$ and
$\widetilde{\boldsymbol{\nu}}$ as in Equation (142) and

$\displaystyle
I_{9}\;=\;\exp\Big{(}\frac{\hbar}{8}\widetilde{\mathbf{f}}^{t}\widetilde{\mathbf{B}}^{-1}\widetilde{\mathbf{A}}\widetilde{\mathbf{f}}-\frac{\hbar^{\frac{1}{2}}}{2}\tilde{x}^{t}(\widetilde{\mathbf{B}}^{-1}\widetilde{\boldsymbol{\nu}}-1)\Big{)}\,\prod_{j=0}^{N+2}\psi_{\hbar}(\widetilde{x}_{j}^{*},\widetilde{z}_{j}^{*})$
(153)

and

$\displaystyle\mathbf{\Lambda}_{9}$
$\displaystyle\;=\;\mathrm{diag}(\widetilde{z}_{0}^{\prime},\,\widetilde{z}_{1}^{\prime\prime},\,\widetilde{z}_{2}^{\prime\prime},\,\widetilde{z}^{*}{}^{\prime})-{\widetilde{\mathbf{B}}}^{-1}\widetilde{\mathbf{A}}.$
(154)

∎

######  Lemma 6.7.

With the notation used in the proof of the previous Proposition 6.6 we have
$\mathbf{Q}_{11}=\mathbf{Q}_{12}=\mathbf{Q}_{22}=0$ and the following
equalities:

$\displaystyle\widetilde{\mathbf{B}}^{-1}\widetilde{\mathbf{A}}$
$\displaystyle\;=\;\left(\\!\\!\begin{array}[]{ccc|c}-1&1&1&-\mathbf{Q}_{1}^{*}-\mathbf{Q}_{2}^{*}\\\
1&0&0&\mathbf{Q}_{1}^{*}\\\ 1&0&0&\mathbf{Q}_{2}^{*}\\\
\hline\cr-{\mathbf{Q}_{1}^{*}}^{t}-{\mathbf{Q}_{2}^{*}}^{t}&{\mathbf{Q}_{1}^{*}}^{t}&{\mathbf{Q}_{2}^{*}}^{t}&\mathbf{Q}^{*}-{\mathbf{Q}_{1}^{*}}^{t}{\mathbf{Q}_{2}^{*}}-{\mathbf{Q}_{2}^{*}}^{t}{\mathbf{Q}_{1}^{*}}\end{array}\\!\\!\right),$
(159) $\displaystyle\widetilde{\mathbf{B}}^{-1}\widetilde{\boldsymbol{\nu}}$
$\displaystyle\;=\;\left(\\!\\!\begin{array}[]{c}-({\mathbf{B}}^{-1}\boldsymbol{\nu})_{1}-({\mathbf{B}}^{-1}\boldsymbol{\nu})_{2}+1\\\
({\mathbf{B}}^{-1}\boldsymbol{\nu})_{1}\\\
{(\mathbf{B}}^{-1}\boldsymbol{\nu})_{2}\\\
({\mathbf{B}}^{-1}\boldsymbol{\nu})^{*}-({\mathbf{B}}^{-1}\boldsymbol{\nu})_{1}\mathbf{Q}_{2}^{*}-({\mathbf{B}}^{-1}\boldsymbol{\nu})_{2}\mathbf{Q}_{1}^{*}\end{array}\\!\\!\right),$
(164)
$\displaystyle\widetilde{\mathbf{f}}^{t}\widetilde{\mathbf{B}}^{-1}\widetilde{\boldsymbol{\nu}}$
$\displaystyle\;=\;\mathbf{f}^{t}\mathbf{B}^{-1}\boldsymbol{\nu}+\widetilde{f}_{0}-2({\mathbf{B}}^{-1}\boldsymbol{\nu})_{1}({\mathbf{B}}^{-1}\boldsymbol{\nu})_{2}\,.$
(165)

###### Proof.

The relation $\mathbf{B}^{-1}\mathbf{A}=\mathbf{Q}$ and the fact that the
columns of $(a_{1}\,|\,a_{2}\,|\,b_{*})$ are linearly independent imply that
$\mathbf{Q}_{11}=\mathbf{Q}_{12}=\mathbf{Q}_{22}=0$. We will prove Equation
(159) by showing the identity after multiplying by the invertible matrix
$\widetilde{\mathbf{B}}$. For (159) we compute

$\displaystyle\left(\\!\\!\begin{array}[]{ccc|c}-1&-1&-1&0\\\
0&a_{1}-b_{1}-b_{2}&a_{2}-b_{1}-b_{2}&b_{*}\end{array}\\!\\!\right)\left(\\!\\!\begin{array}[]{ccc|c}-1&1&1&-\mathbf{Q}_{1}^{*}-\mathbf{Q}_{2}^{*}\\\
1&0&0&\mathbf{Q}_{1}^{*}\\\ 1&0&0&\mathbf{Q}_{2}^{*}\\\
\hline\cr-{\mathbf{Q}_{1}^{*}}^{t}-{\mathbf{Q}_{2}^{*}}^{t}&{\mathbf{Q}_{1}^{*}}^{t}&{\mathbf{Q}_{2}^{*}}^{t}&\mathbf{Q}^{*}-{\mathbf{Q}_{1}^{*}}^{t}{\mathbf{Q}_{2}^{*}}-{\mathbf{Q}_{2}^{*}}^{t}{\mathbf{Q}_{1}^{*}}\end{array}\\!\\!\right)$
(166) $\displaystyle\;=\;\left(\\!\\!\begin{array}[]{ccc|c}-1&-1&-1&{0}\\\
\hline\cr\begin{subarray}{c}a_{1}+a_{2}-2b_{1}-2b_{2}\\\
-b_{*}{\mathbf{Q}_{1}^{*}}^{t}-b_{*}{\mathbf{Q}_{2}^{*}}^{t}\end{subarray}&{b_{*}{\mathbf{Q}_{1}^{*}}^{t}}&{b_{*}{\mathbf{Q}_{2}^{*}}^{t}}&\begin{subarray}{c}(a_{1}-b_{1}){\mathbf{Q}_{1}^{*}}^{t}+(a_{2}-b_{2}){\mathbf{Q}_{2}^{*}}^{t}\\\
-b_{2}{\mathbf{Q}_{1}^{*}}^{t}-b_{1}{\mathbf{Q}_{2}^{*}}^{t}+b_{*}{\mathbf{Q}^{*}}-b_{*}{\mathbf{Q}_{1}^{*}}^{t}{\mathbf{Q}_{2}^{*}}-b_{*}{\mathbf{Q}_{2}^{*}}^{t}{\mathbf{Q}_{1}^{*}}\end{subarray}\end{array}\\!\\!\right)$
$\displaystyle\;=\;\widetilde{\mathbf{A}}.$

For Equation (164), we compute

$\displaystyle\left(\\!\\!\begin{array}[]{ccc|c}-1&-1&-1&0\\\
0&a_{1}-b_{1}-b_{2}&a_{2}-b_{1}-b_{2}&b_{*}\end{array}\\!\\!\right)\left(\\!\\!\begin{array}[]{c}-({\mathbf{B}}^{-1}\boldsymbol{\nu})_{1}-({\mathbf{B}}^{-1}\boldsymbol{\nu})_{2}+1\\\
({\mathbf{B}}^{-1}\boldsymbol{\nu})_{1}\\\
({\mathbf{B}}^{-1}\boldsymbol{\nu})_{2}\\\
({\mathbf{B}}^{-1}\boldsymbol{\nu})^{*}-({\mathbf{B}}^{-1}\boldsymbol{\nu})_{1}\mathbf{Q}_{2}^{*}-({\mathbf{B}}^{-1}\boldsymbol{\nu})_{2}\mathbf{Q}_{1}^{*}\end{array}\\!\\!\right)$
(167) $\displaystyle\;=\;\left(\\!\\!\begin{array}[]{c}-1\\\
\boldsymbol{\nu}\end{array}\\!\\!\right)\,.$

For Equation (165), we note that for $i=1,2$ we have

$\widetilde{\mathbf{f}}^{*}{}^{t}\mathbf{Q}_{i}^{*}\;=\;(\widetilde{\mathbf{B}}^{-1}\widetilde{\boldsymbol{\nu}})_{i}-\widetilde{\mathbf{f}}_{0}-\widetilde{\mathbf{f}}_{i}^{\prime\prime}$
(168)

and so

$\displaystyle\widetilde{\mathbf{f}}^{t}\widetilde{\mathbf{B}}^{-1}\widetilde{\boldsymbol{\nu}}\;=\;\widetilde{f}_{0}+(\widetilde{f}_{1}^{\prime}+\widetilde{f}_{2})({\mathbf{B}}^{-1}\boldsymbol{\nu})_{1}+(\widetilde{f}_{1}+\widetilde{f}_{2}^{\prime})({\mathbf{B}}^{-1}\boldsymbol{\nu})_{2}+\widetilde{f}^{*}{}^{t}({\mathbf{B}}^{-1}\boldsymbol{\nu})^{*}-2({\mathbf{B}}^{-1}\boldsymbol{\nu})_{1}({\mathbf{B}}^{-1}\boldsymbol{\nu})_{2}$
(169)

Then, using the analogous relations between the flattening as given in
Equation (94) for the shapes, we have
$\widetilde{f}_{1}^{\prime}+\widetilde{f}_{2}=f_{1}^{\prime}=\mathbf{f}_{1}$
and
$\widetilde{f}_{1}+\widetilde{f}_{2}^{\prime}=f_{2}^{\prime}=\mathbf{f}_{2}$.
∎

##  7\. The series of the simplest hyperbolic $4_{1}$ knot

In this section, we discuss an effective computation of the power series
$\Phi^{\Xi}(\hbar)$ for the simplest hyperbolic knot, namely the $4_{1}$ knot.
This example was studied extensively in [10]. From [10, Ex. 2.6], we obtain
that the NZ datum $\Xi_{4_{1}}$ is given by

$A\;=\;\begin{pmatrix}2&2\\\ 1&1\end{pmatrix},\quad
B\;=\;\begin{pmatrix}1&1\\\ 1&0\end{pmatrix},\quad\nu\;=\;\begin{pmatrix}2\\\
1\end{pmatrix},\quad f\;=\;\begin{pmatrix}0\\\ 1\end{pmatrix},\quad
f^{\prime\prime}\;=\;\begin{pmatrix}0\\\ 0\end{pmatrix},\quad$ (170)

and $z_{1}=z_{2}=\zeta_{6}=e^{2\pi i/6}$. Therefore, we have

$\Phi^{\Xi_{4_{1}}}(\hbar)\;=\;e^{\frac{\hbar}{8}}\langle\psi_{\hbar}(x_{1},\zeta_{6})\psi_{\hbar}(x_{2},\zeta_{6})\rangle_{(x_{1},x_{2}),\Lambda_{0}}\,,$
(171)

where

$\Lambda_{0}\;=\;\begin{pmatrix}\zeta_{6}-1&-1\\\
-1&\zeta_{6}-1\end{pmatrix}\,.$ (172)

This is a two dimensional formal Gaussian integral which can be simplified.
Using $\zeta_{6}=1/(1-\zeta_{6})=1-\zeta_{6}^{-1}$ and applying a change of
coordinates $x_{1}\mapsto x_{1}-\zeta_{6}x_{2}$, we find that

$\Phi^{\Xi_{4_{1}}}(\hbar)\;=\;e^{\frac{\hbar}{8}}\langle\psi_{\hbar}(x_{1}-\zeta_{6}x_{2},\zeta_{6})\psi_{\hbar}(x_{2},\zeta_{6})\rangle_{(x_{1},x_{2}),\Lambda_{1}}\,,$
(173)

where

$\Lambda_{1}\;=\;\begin{pmatrix}\zeta_{6}-1&0\\\
0&2\zeta_{6}-1\end{pmatrix}\,.$ (174)

We can apply Fubini’s theorem [3, Prop.2.13] and Corollary 3.5 to perform the
integral over $x_{1}$. After renaming the variable $x_{2}$ by $x$, we express
$\Phi^{\Xi_{4_{1}}}(\hbar)$ by a one-dimensional formal Gaussian integral

$\Phi^{\Xi_{4_{1}}}(\hbar)\;=\;e^{\frac{\hbar}{6}}\Big{\langle}\exp\Big{(}\frac{x}{2}\hbar^{\frac{1}{2}}\Big{)}\psi_{\hbar}(x,\zeta_{6})^{2}\Big{\rangle}_{x,2\zeta_{6}-1}\,.$
(175)

Using the definition of $\psi_{\hbar}$ from Equation (1) and expanding to
$O(\hbar^{5/2})$, we obtain that

$\displaystyle\exp\Big{(}\frac{x}{2}\hbar^{\frac{1}{2}}\Big{)}\psi_{\hbar}(x,\zeta_{6})^{2}$
(176)
$\displaystyle\;=\;1+\bigg{(}\frac{1}{3}x^{3}+\Big{(}\zeta_{6}-\frac{1}{2}\Big{)}x\bigg{)}\hbar^{1/2}+\bigg{(}\frac{1}{18}x^{6}+\Big{(}\frac{1}{2}\zeta_{6}-\frac{1}{4}\Big{)}x^{4}-\frac{7}{8}x^{2}+\Big{(}-\frac{1}{6}\zeta_{6}+\frac{1}{6}\Big{)}\bigg{)}\hbar$
$\displaystyle\qquad+\bigg{(}\frac{1}{162}x^{9}+\Big{(}\frac{1}{9}\zeta_{6}-\frac{1}{18}\Big{)}x^{7}-\frac{1}{2}x^{5}+\Big{(}-\frac{73}{72}\zeta_{6}+\frac{77}{144}\Big{)}x^{3}+\Big{(}\frac{1}{12}\zeta_{6}+\frac{1}{4}\Big{)}x\bigg{)}\hbar^{3/2}$
$\displaystyle\qquad+\bigg{(}\frac{1}{1944}x^{12}+\Big{(}\frac{5}{324}\zeta_{6}-\frac{5}{648}\Big{)}x^{10}-\frac{37}{288}x^{8}+\Big{(}-\frac{1337}{2160}\zeta_{6}+\frac{1357}{4320}\Big{)}x^{6}$