# Asymptotics of quantum $6j$-symbols and generalized hyperbolic tetrahedra

Giulio Belletti and Tian Yang

###### Abstract

We establish the geometry behind the quantum $6j$-symbols under only the
admissibility conditions as in the definition of the Turaev-Viro invariants of
$3$-manifolds. As a classification, we show that the $6$-tuples in the quantum
$6j$-symbols give in a precise way to the dihedral angles of (1) a spherical
tetrahedron, (2) a generalized Euclidean tetrahedron, (3) a generalized
hyperbolic tetrahedron or (4) in the degenerate case the angles between four
oriented straight lines in the Euclidean plane. We also show that for a large
proportion of the cases, the $6$-tuples always give the dihedral angles of a
generalized hyperbolic tetrahedron and the exponential growth rate of the
corresponding quantum $6j$-symbols equals the suitably defined volume of this
generalized hyperbolic tetrahedron. It is worth mentioning that the volume of
a generalized hyperbolic tetrahedron can be negative, hence the corresponding
sequence of the quantum $6j$-symbols could decay exponentially. This is a
phenomenon that has never been aware of before.

##  1 Introduction

Quantum $6j$-symbols are the main building blocks of the Turaev-Viro
invariants of $3$-manifolds [23]; and the asymptotic behavior of the former
plays a central role in understanding that of the latter [7, 2, 3]. It is
proved in [8] (see also [6, 3]) that if the $6$-tuples of a sequence of
quantum $6j$-symbols correspond in a precise way to the dihedral angles of a
hyperbolic hyperideal tetrahedron, then the quantum $6j$-symbols grow
exponentially, and the growth rate is given by the volume of this hyperbolic
hyperideal tetrahedron. It is believed that this should hold without the
condition on the type of the vertices of the tetrahedron. On the other hand,
the case where the sequence of the $6$-tuples does not correspond to a
hyperbolic tetrahedron has not been studied; although it is equally crucial to
the asymptotics of the Turaev-Viro invariants and seems to hold abundant
geometric significance, as numerical computation shows that the sequence of
quantum $6j$-symbols could either grow or decay exponentially in this case!
The goal of this paper is to answer the following fundamental question.

######  Question 1.1.

What is the asymptotic behavior of a sequence of quantum $6j$-symbols
$\Bigg{|}\begin{matrix}a_{1}^{(r)}&a_{2}^{(r)}&a_{3}^{(r)}\\\
a_{4}^{(r)}&a_{5}^{(r)}&a_{6}^{(r)}\end{matrix}\Bigg{|}$ evaluated at
$q=e^{\frac{2\pi\sqrt{-1}}{r}}$ under only the $r$-admissibility conditions of
the $6$-tuples $\\{(a_{1}^{(r)},\dots,a_{6}^{(r)})\\}?$

Under the condition of Question 1.1, for each $k\in\\{1,\dots,6\\},$ let

$\alpha_{k}=\lim_{r\to\infty}\frac{2\pi a_{k}^{(r)}}{r};$

and let

$\theta_{k}=|\pi-\alpha_{k}|,$

or equivalently,

$\alpha_{k}=\pi\pm\theta_{k}.$

Then $(\alpha_{1},\dots,\alpha_{6})$ satisfies the _admissibility conditions_.
(See Definition 4.1.) To answer Question 1.1, we as one of the main results of
this paper in Theorem 1.2 classify all the admissible $6$-tuples
$(\alpha_{1},\dots,\alpha_{6})$ into the dihedral angles of a spherical, a
generalized Euclidean or a generalized hyperbolic tetrahedron, together with
one degenerate extra case. We defined these terms respectively in Sections 3.1
and 4.1, and give criteria of them in Theorem 3.7 and Theorem 4.6 in terms of
the Gram matrix of the $6$-tuples. For the generalized hyperbolic tetrahedra,
there is a natural way to define the _edge lengths_ (see Definition 3.13) and
the _volume_ (see Definition 3.17), which extend the edge lengths and volume
of a hyperbolic tetrahedron, and still satisfy the Schläfli formula (see
Proposition 3.21) and additivity (see (7.4)). We then obtain an explicit
formula for the volume of a generalized hyperbolic tetrahedron in terms of the
admissible $6$-tuple $(\alpha_{1},\dots,\alpha_{6}).$ (See Theorem 5.1 and
Theorem 6.5.) Another main result of this paper is Theorem 1.4 that under one
extra condition of the $6$-tuple $(\alpha_{1},\dots,\alpha_{6})$ other than
the necessary admissibility conditions, $(\theta_{1},\dots,\theta_{6})$ is the
set of dihedral angles of a generalized hyperbolic tetrahedron, and the
exponential growth rate of the corresponding sequence of quantum $6j$-symbols
equals the volume of this generalized hyperbolic tetrahedron. It worths
mentioning that the volume of a generalized hyperbolic tetrahedron can be
negative, which implies that the sequence of quantum $6j$-symbols
exponentially decays. This phenomenon has never been aware of before. We also
would like to mention that in the only exceptional case that Theorem 1.4 does
not cover (Case (1) of Proposition 3.15), there is something more interesting
happening, namely, we found $6$-tuples giving the dihedral angles of a
generalized hyperbolic tetrahedron with positive volume, yet the exponential
growth rate of the corresponding sequence of quantum $6j$-symbols is negative.
(See Example 8.2). This deserves a further study.

######  Theorem 1.2.

Let $(\alpha_{1},\dots,\alpha_{6})$ be an admissible $6$-tuple real numbers,
and for $k\in\\{1,\dots,6\\}$ let

$\theta_{k}=|\pi-\alpha_{k}|.$

Then $(\theta_{1},\dots,\theta_{6})$ is of one of the following four
possibilities:

  1. (1)

the set of dihedral angles of a spherical tetrahedron,

  2. (2)

the set of dihedral angles of a generalized Euclidean tetrahedron,

  3. (3)

the set of dihedral angles of a generalized hyperbolic tetrahedron, and

  4. (4)

the set of angles between four oriented straight lines in the Euclidean plane.

######  Remark 1.3.

A hyperbolic tetrahedron was called a generalized hyperbolic tetrahedron in
Ushijima [24] to indicate that the vertices can be not only regular, but also
ideal or hyperideal. In this article, we will reserve the terminology
_generalized hyperbolic tetrahedra_ for a more general set of geometric
objects. See Definition 3.1. The definition for generalized Euclidean
tetrahedron will be given in Section 4.1.

######  Theorem 1.4.

Let $\\{(a_{1}^{(r)},\dots,a_{6}^{(r)})\\}$ be a sequence of $r$-admissible
$6$-tuples. For each $k\in\\{1,\dots,6\\},$ let

$\alpha_{k}=\lim_{r\to\infty}\frac{2\pi a_{k}^{(r)}}{r},$

and let

$\theta_{k}=|\pi-\alpha_{k}|.$

Let $G$ be the Gram matrix of $(\theta_{1},\dots,\theta_{6}),$ and for
$i\in\\{1,2,3,4\\}$ let $G_{ii}$ be the $ii$-th cofactor of $G.$ If

$G_{ii}<0$

for at least one $i\in\\{1,2,3,4\\},$ then

  1. (1)

$(\theta_{1},\dots,\theta_{6})$ is the set of dihedral angles of a generalized
hyperbolic tetrahedra $\Delta,$ and

  2. (2)

as $r$ varies over all positive odd integers,

$\lim_{r\to\infty}\frac{2\pi}{r}\ln\Bigg{|}\begin{matrix}a_{1}^{(r)}&a_{2}^{(r)}&a_{3}^{(r)}\\\
a_{4}^{(r)}&a_{5}^{(r)}&a_{6}^{(r)}\end{matrix}\Bigg{|}_{q=e^{\frac{2\pi\sqrt{-1}}{r}}}=\mathrm{{Vol}}(\Delta).$

In a sequel paper, we will address the asymptotic expansion of sequences of
quantum $6j$-symbols, which is a refinement of Theorem 1.4. See Theorem 1.5
below. To be precise, let $\\{(a_{1}^{(r)},\dots,a_{6}^{(r)})\\}$ be a
sequence of $r$-admissible $6$-tuples such that for any $k\in\\{1,\dots,6\\},$
either $a_{k}^{(r)}>\frac{r}{2}$ for all $r$ or $a_{k}^{(r)}<\frac{r}{2}$ for
all $r.$ In the former case we let $\mu_{k}=1$ and in the latter case we let
$\mu_{k}=-1,$ and we let

$\theta^{(r)}_{k}=\mu_{k}\bigg{(}\frac{2\pi a_{k}^{(r)}}{r}-\pi\bigg{)}$

for each $k\in\\{1,\dots,6\\}.$ For each $r,$ let $G^{(r)}$ be the Gram matrix
of the $6$-tuple $(\theta^{(r)}_{1},\dots,\theta^{(r)}_{6}),$ and for
$i\in\\{1,2,3,4\\},$ let $G_{ii}^{(r)}$ be the $ii$-th cofactor of $G^{(r)}.$
If $G_{ii}^{(r)}<0$ for for at least one $i\in\\{1,2,3,4\\},$ then by Theorem
1.4 (1), $(\theta^{(r)}_{1},\dots,\theta^{(r)}_{6})$ is the set of dihedral
angles of a generalized hyperbolic tetrahedron $\Delta^{(r)}.$ We let
$\mathrm{Vol}(\Delta^{(r)})$ and $(l^{(r)}_{1},\dots,l^{(r)}_{6})$
respectively be the volume and the set of edge lengths of $\Delta^{(r)},$ and
also denote by $G(\Delta^{(r)})$ the Gram matrix of $\Delta^{(r)}.$ Finally we
assume that $\\{(\theta^{(r)}_{1},\dots,\theta^{(r)}_{6})\\}$ converges as $r$
tends to infinity, to the set of dihedral angles of a generalized hyperbolic
tetrahedron without ideal vertices.

######  Theorem 1.5.

Under the above assumptions, as $r$ varies over all positive odd integers,

$\Bigg{|}\begin{matrix}a_{1}^{(r)}&a_{2}^{(r)}&a_{3}^{(r)}\\\
a_{4}^{(r)}&a_{5}^{(r)}&a_{6}^{(r)}\end{matrix}\Bigg{|}_{q=e^{\frac{2\pi\sqrt{-1}}{r}}}=\frac{\sqrt{2}\pi}{r^{\frac{3}{2}}}\frac{e^{-\frac{1}{2}\sum_{k=1}^{6}\mu_{k}l_{k}^{(r)}}}{\sqrt[4]{-\det
G(\Delta^{(r)})}}e^{\frac{r}{2\pi}\mathrm{Vol}(\Delta^{(r)})}\bigg{(}1+O\Big{(}\frac{1}{r}\Big{)}\bigg{)}.$

Acknowledgments. The authors would like to thank Francis Bonahon, Jihoon Sohn
and Ka Ho Wong for helpful discussions. The second author is supported by NSF
Grants DMS-1812008 and DMS-2203334.

##  2 Preliminaries

###  2.1 Hyperbolic geometry in dimension $3$

Hyperboloid model. The Lorentzian space $\mathbb{E}^{3,1}$ is the vector space
$\mathbb{R}^{4}$ with the inner product $\langle,\rangle$ defined for
$\mathbf{x}=(x_{1},x_{2},x_{3},x_{4})$ and
$\mathbf{y}=(y_{1},y_{2},y_{3},y_{4})$ by

$\langle\mathbf{x},\mathbf{y}\rangle=x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}-x_{4}y_{4}.$

The de Sitter space is

$\mathbb{S}(1)=\\{\mathbf{v}\in\mathbb{E}^{3,1}\ |\
\langle\mathbf{v},\mathbf{v}\rangle=1\\}.$

The light cone is

$\mathbb{S}(0)=\\{\mathbf{v}\in\mathbb{E}^{3,1}\ |\
\langle\mathbf{v},\mathbf{v}\rangle=0\\},$

and the upper- and lower-light cones are respectively

$\mathbb{L}^{3}_{+}=\\{\mathbf{v}\in\mathbb{S}(0)\ |\
v_{4}>0\\}\quad\text{and}\quad\mathbb{L}^{3}_{-}=\\{\mathbf{v}\in\mathbb{S}(0)\
|\ v_{4}<0\\}$

so that
$\mathbb{S}(0)=\mathbb{L}^{3}_{+}\cup\mathbb{L}^{3}_{-}\cup\\{\mathbf{0}\\}.$
The hyperboloid is

$\mathbb{S}(-1)=\\{\mathbf{v}\in\mathbb{E}^{3,1}\ |\
\langle\mathbf{v},\mathbf{v}\rangle=-1\\},$

and the upper- and lower-hyperboloids are respectively

$\mathbb{H}^{3}_{+}=\\{\mathbf{v}\in\mathbb{S}(-1)\ |\
v_{4}>0\\}\quad\text{and}\quad\mathbb{H}^{3}_{-}=\\{\mathbf{v}\in\mathbb{S}(-1)\
|\ v_{4}<0\\}$

so that $\mathbb{S}(-1)=\mathbb{H}^{3}_{+}\cup\mathbb{H}^{3}_{-}.$ The inner
product $\langle,\rangle$ on $\mathbb{E}^{3,1}$ restricts to a Riemannian
metric on $\mathbb{S}(-1)$ with constant sectional curvature $-1,$ call the
_hyperbolic metric_. We also consider the spaces

$\mathbb{B}^{3,1}=\\{\mathbf{v}\in\mathbb{E}^{3,1}\ |\
\langle\mathbf{v},\mathbf{v}\rangle<0\\},$

and

$\overline{\mathbb{B}^{3,1}}=\mathbb{B}^{3,1}\cup\mathbb{S}(0)=\\{\mathbf{v}\in\mathbb{E}^{3,1}\
|\ \langle\mathbf{v},\mathbf{v}\rangle\leqslant 0\\},$

For each $\mathbf{v}\in\mathbb{S}(1),$ let

$\mathbf{\Pi}_{\mathbf{v}}=\\{\mathbf{w}\in\mathbb{E}^{3,1}\ |\
\langle\mathbf{v},\mathbf{w}\rangle=0\\}$

be the hyperplane planing containing all the vectors perpendicular to
$\mathbf{v}.$ Then $\mathbf{\Pi}_{\mathbf{v}}$ intersects $\mathbb{B}^{3,1}.$
This is gives a one-to-one correspondence between hyperplanes $\mathbf{\Pi}$
intersecting $\mathbb{B}^{3,1}$ and pairs of opposite vectors
$\\{\pm\mathbf{v}\\}$ in $\mathbb{S}(1)$ perpendicular to $\Pi$. We defined
the _orientation_ of a hyperplane intersecting $\mathbb{B}^{3,1}$ to be a
specification of one of the two vectors $\\{\pm\mathbf{v}\\}.$ Suppose
$\mathbf{v}$ defines the orientation of $\mathbf{\Pi}_{\mathbf{v}},$ then we
call $\mathbf{v}$ the _outward normal vector_ of $\mathbf{\Pi}_{\mathbf{v}},$
and call the intersection

$\Pi_{\mathbf{v}}=\mathbf{\Pi}_{\mathbf{v}}\cap\mathbb{S}(-1)$

the _plane of truncation_ at $\mathbf{v}.$

Projective model. Let the affine hyperplane
$\mathbb{P}^{3}_{1}=\\{\mathbf{x}\in\mathbb{E}^{3,1}\ |x_{4}=1\\},$ and let
$\mathrm{p}:\mathbb{E}^{3,1}{\smallsetminus}\\{\mathbf{x}\in\mathbb{E}^{3,1}\
|x_{4}=0\\}\to\mathbb{P}^{3}_{1}$ be the radial projection along the ray from
the origin $\mathbf{0}.$ Then $\mathrm{p}$ continuously extends to

$\mathrm{p}:\mathbb{E}^{3,1}{\smallsetminus}\\{\mathbf{0}\\}\to\mathbb{P}^{3}_{1}\cup\mathbb{P}^{3}_{\infty},$

where $\mathbb{P}^{3}_{\infty}$ is the set of lines in the linear subspace
$\\{\mathbf{x}\in\mathbb{E}^{3,1}\ |x_{4}=0\\}$ passing through the origin
$\mathbf{0}.$ The projective model is then

$\mathbb{P}^{3}=\mathbb{P}^{3}_{1}\cup\mathbb{P}^{3}_{\infty}.$

The radial projection $\mathrm{p}$ restricts to a two-to-one map from
$\mathbb{S}(-1)$ to the open unit ball in $\mathbb{P}^{3}_{1},$ which is a
homeomorphism on each of $\mathbb{H}^{3}_{+}$ and $\mathbb{H}^{3}_{-},$ hence
the unit ball inherits the hyperbolic metric from them via $\mathrm{p}.$ The
Klein model of the hyperbolic space, denoted by $\mathbb{H}^{3},$ is the unit
ball of $\mathbb{P}^{3}_{1}$ with this induced metric. The image of the light
cone $\mathbb{S}(0){\smallsetminus}\\{\mathbf{0}\\}$ is the unit circle
$\mathbb{S}^{2}_{\infty}$ in $\mathbb{P}^{3}_{1},$ and the closure of the
hyperbolic space in the projective model is

$\overline{\mathbb{H}^{3}}=\mathbb{H}^{3}\cup\mathbb{S}^{2}_{\infty}.$

We notice that the spaces $\mathbb{B}^{3,1}$ and $\overline{\mathbb{B}^{3,1}}$
are respectively the pre-images of $\mathbb{H}^{3}$ and
$\overline{\mathbb{H}^{3}}$ under the radial projection $\mathrm{p}.$

Hyperbolic tetrahedra. A _hyperbolic tetrahedron_ is the convex hull of the
vertices of a quadruple of linearly independent vectors $\mathbf{v}_{1},$
$\mathbf{v}_{2},$ $\mathbf{v}_{3}$ and $\mathbf{v}_{4}$ in
$\mathbb{H}^{3}_{+}\cup\mathbb{L}^{3}_{+}\cup\mathbb{S}(1)$ or in
$\mathbb{H}^{3}_{-}\cup\mathbb{L}^{3}_{-}\cup\mathbb{S}(1)$ such that: (1) for
each pair $\\{i,j\\}\subset\\{1,2,3,4\\},$ the straight line segment
$L^{+}_{ij}$ passing through the vertices of $\mathbf{v}_{i}$ and
$\mathbf{v}_{j}$ intersects $\overline{\mathbb{B}^{3,1}},$ and (2) for any
$\mathbf{v}_{i}\in\mathbb{S}(1),$ all the other vertices
$\mathbf{v}_{j},\mathbf{v}_{k},\mathbf{v}_{l}$ are on the other side of the
hyperplane $\mathbf{\Pi}_{i}$ perpendicular to $\mathbf{v}_{i}.$ We call the
vertex $\mathbf{v}_{i}$ of a hyperbolic tetrahedron a _regular vertex_ if
$\mathbf{v}_{i}\in\mathbb{S}(-1),$ an _ideal vertex_ if
$\mathbf{v}_{i}\in\mathbb{S}(0)$ and a _hyperideal vertex_ if
$\mathbf{v}_{i}\in\mathbb{S}(1).$ We call a hyperbolic tetrahedron _regular_ ,
_ideal_ or _hyperideal_ if all the vertices of it are respectively so.

In the projective model, by considering the radial projection $\mathrm{p},$ a
hyperbolic tetrahedron is the convex hull of four points
$v_{1},v_{2},v_{3},v_{4}$ in the general positions in $\mathbb{P}^{3}$ such
that, (1) for any $v_{i}$ and $v_{j}$ that are not in
$\overline{\mathbb{H}^{3}},$ the straight line segment (inherited from the
affine structure of $\mathbb{P}^{3}_{1}$) connecting them intersects
$\overline{\mathbb{H}^{3}},$ and (2) for each $v_{i}$ that is not in
$\overline{\mathbb{H}^{3}},$ all the other vertices $v_{j},v_{k},v_{l}$ are on
the other side of the radial projection of the plane of truncation $\Pi_{i}.$
In the projective model, a vertex $v_{i}$ is _regular_ if
$v_{i}\in\mathbb{H}^{3},$ _ideal_ if $v_{i}\in\mathbb{S}^{2}_{\infty}$ and
_hyperideal_ if $v_{i}\notin\overline{\mathbb{H}^{3}};$ and a hyperbolic
tetrahedron is _regular_ , _ideal_ or _hyperideal_ if all the vertices of it
are respectively so.

Back to the hyperboloid model, the _face_ $F_{i}$ opposite to the vertex
$\mathbf{v}_{i}$ is the plane containing the vertices of
$\mathbf{v}_{j},\mathbf{v}_{k},\mathbf{v}_{l},$
$\\{j,k,l\\}=\\{1,2,3,4\\}{\smallsetminus}\\{i\\},$ and the dihedral angle
$\theta_{ij}$ is the angle between the faces $F_{i}$ and $F_{j}.$ The _Gram
matrix_ of a $6$-tuple $(\theta_{12},\dots,\theta_{34})$ of real numbers is
the following $4\times 4$ matrix

$G=\left[\begin{matrix}1&-\cos\theta_{12}&-\cos\theta_{13}&-\cos\theta_{14}\\\
-\cos\theta_{12}&1&-\cos\theta_{23}&-\cos\theta_{24}\\\
-\cos\theta_{13}&-\cos\theta_{23}&1&-\cos\theta_{34}\\\
-\cos\theta_{14}&-\cos\theta_{24}&-\cos\theta_{34}&1\\\ \end{matrix}\right].$

If $(\theta_{12},\dots,\theta_{34})$ is the set of dihedral angles of a
hyperbolic tetrahedron $\Delta,$ then $G$ is called the _Gram matrix_ of
$\Delta.$

######  Theorem 2.1 (Luo [17], Ushijima [24]).

Suppose $(\theta_{12},\dots,\theta_{34})$ is a $6$-tuple of numbers in
$[0,\pi].$ Then the following statements are equivalent.

  1. (1)

$(\theta_{12},\dots,\theta_{34})$ is the set of dihedral angles of a
hyperbolic tetrahedron.

  2. (2)

The Gram matrix $G$ of $(\theta_{12},\dots,\theta_{34})$ satisfies the
following two conditions:

    1. (a)

$G$ has signature $(3,1),$

    2. (b)

the $ij$-th cofactor $G_{ij}>0$ for any $\\{i,j\\}\subset\\{1,2,3,4\\}.$

Moreover, when the above conditions are satisfied, the vertex $\mathbf{v}_{i}$
is regular (resp. ideal, hyperideal) if $G_{ii}>0$ (resp. $G_{ii}=0,$
$G_{ii}<0$).

######  Theorem 2.2 (Bonahon-Bao [1]).

Suppose $(\theta_{12},\dots,\theta_{34})$ is a $6$-tuple of numbers in
$[0,\pi].$ Then the following statements are equivalent.

  1. (1)

$(\theta_{12},\dots,\theta_{34})$ is the set of dihedral angles of a
hyperideal hyperbolic tetrahedron.

  2. (2)

For each $i\in\\{1,2,3,4\\},$ $\theta_{jk}+\theta_{jl}+\theta_{kl}<\pi,$ where
$\\{j,k,l\\}=\\{1,2,3,4\\}{\smallsetminus}\\{i\\}.$

In particular, the spaces of hyperideal hyperbolic tetrahedra parametrized by
the dihedral angles is a convex open polytope in $[0,\pi]^{6}.$

###  2.2 Quantum $6j$-symbols

Let $r$ be an odd integer and $q$ be an $r$-th root of unity. For the context
of this paper we are only interested in the case
$q=e^{\frac{2\pi\sqrt{-1}}{r}},$ but the definitions in this section work with
any choice of $q.$

As is customary we define $[n]=\frac{q^{n}-q^{-n}}{q-q^{-1}},$ and the quantum
factorial

$[n]!=\prod_{k=1}^{n}[k].$

A triple $(a_{1},a_{2},a_{3})$ of integers in $\\{0,\dots,r-2\\}$ is _$r$
-admissible_ if

  1. (1)

$a_{i}+a_{j}-a_{k}\geqslant 0$ for $\\{i,j,k\\}=\\{1,2,3\\}.$

  2. (2)

$a_{1}+a_{2}+a_{3}\leqslant 2(r-2),$

  3. (3)

$a_{1}+a_{2}+a_{3}$ is even.

For an $r$-admissible triple $(a_{1},a_{2},a_{3}),$ define

$\Delta(a_{1},a_{2},a_{3})=\sqrt{\frac{[\frac{a_{1}+a_{2}-a_{3}}{2}]![\frac{a_{2}+a_{3}-a_{1}}{2}]![\frac{a_{3}+a_{1}-a_{2}}{2}]!}{[\frac{a_{1}+a_{2}+a_{3}}{2}+1]!}}$

with the convention that $\sqrt{x}=\sqrt{-1}\sqrt{|x|}$ when the real number
$x$ is negative.

A 6-tuple $(a_{1},\dots,a_{6})$ is _$r$ -admissible_ if the triples
$(a_{1},a_{2},a_{3}),$ $(a_{1},a_{5},a_{6}),$ $(a_{2},a_{4},a_{6})$ and
$(a_{3},a_{4},a_{5})$ are $r$-admissible.

######  Definition 2.3.

The _quantum $6j$-symbol_ of an $r$-admissible 6-tuple $(a_{1},\dots,a_{6})$
is

$\bigg{|}\begin{matrix}a_{1}&a_{2}&a_{3}\\\
a_{4}&a_{5}&a_{6}\end{matrix}\bigg{|}=\sqrt{-1}^{-\sum_{i=1}^{6}a_{i}}\Delta(a_{1},a_{2},a_{3})\Delta(a_{1},a_{5},a_{6})\Delta(a_{2},a_{4},a_{6})\Delta(a_{3},a_{4},a_{5})\\\
\sum_{z=\max\\{T_{1},T_{2},T_{3},T_{4}\\}}^{\min\\{Q_{1},Q_{2},Q_{3}\\}}\frac{(-1)^{z}[z+1]!}{[z-T_{1}]![z-T_{2}]![z-T_{3}]![z-T_{4}]![Q_{1}-z]![Q_{2}-z]![Q_{3}-z]!},$

where

$T_{1}=\frac{a_{1}+a_{2}+a_{3}}{2},\quad
T_{2}=\frac{a_{1}+a_{5}+a_{6}}{2},\quad
T_{3}=\frac{a_{2}+a_{4}+a_{6}}{2},\quad T_{4}=\frac{a_{3}+a_{4}+a_{5}}{2},$
$Q_{1}=\frac{a_{1}+a_{2}+a_{4}+a_{5}}{2},\quad
Q_{2}=\frac{a_{1}+a_{3}+a_{4}+a_{6}}{2},\quad
Q_{3}=\frac{a_{2}+a_{3}+a_{5}+a_{6}}{2}.$

Let $\big{\\{}(a^{(r)}_{1},\dots,a^{(r)}_{6})\big{\\}}$ be a sequence of
quantum $6j$-symbols. For each $i\in\\{1,\dots,6\\},$ let

$\alpha_{i}=\lim_{r\to\infty}\frac{2\pi a_{i}^{(r)}}{r};$

and let

$\theta_{i}=|\pi-\alpha_{i}|,$

or equivalently,

$\alpha_{i}=\pi\pm\theta_{i}.$

######  Theorem 2.4 (Costantino[8], Chen-Murakami [6]).

If

  1. (1)

$(\theta_{1},\dots,\theta_{6})$ is the set of dihedral angles of a hyperbolic
tetrahedron $\Delta,$

  2. (2)

$\alpha_{i}=\pi-\theta_{i}$ for all $i\in\\{1,\dots,6\\},$ and

  3. (3)

at least one vertex of $\Delta$ is ideal or hyperideal,

then as $r$ varies over all positive odd integers,

$\lim_{r\to\infty}\frac{2\pi}{r}\ln\Bigg{|}\begin{matrix}a_{1}^{(r)}&a_{2}^{(r)}&a_{3}^{(r)}\\\
a_{4}^{(r)}&a_{5}^{(r)}&a_{6}^{(r)}\end{matrix}\Bigg{|}_{q=e^{\frac{2\pi\sqrt{-1}}{r}}}=\mathrm{Vol}(\Delta).$

######  Theorem 2.5 (Costantino [8], Belletti-Detcherry-Kalfagianni-Yang [3]).

If for each triple $(i,j,k)$ around a vertex:

  1. (1)

$0\leqslant\alpha_{i}+\alpha_{j}-\alpha_{k}\leqslant 2\pi,$ and

  2. (2)

$2\pi\leqslant\alpha_{i}+\alpha_{j}+\alpha_{k}\leqslant 4\pi,$

then

  1. (1)

$(\theta_{1},\dots,\theta_{6})$ is the set of dihedral angles of a hyperbolic
tetrahedron $\Delta$ with all the vertices ideal or hyperideal, and

  2. (2)

as $r$ varies over all positive odd integers,

$\lim_{r\to\infty}\frac{2\pi}{r}\ln\Bigg{|}\begin{matrix}a_{1}^{(r)}&a_{2}^{(r)}&a_{3}^{(r)}\\\
a_{4}^{(r)}&a_{5}^{(r)}&a_{6}^{(r)}\end{matrix}\Bigg{|}_{q=e^{\frac{2\pi\sqrt{-1}}{r}}}=\mathrm{Vol}(\Delta).$

######  Remark 2.6.

We notice that both Theorem 2.4 and Theorem 2.5 are special cases of Theorem
1.4.

###  2.3 A remark on the notations

Keep in mind that as ordered $6$-tuples, $(\theta_{12},\dots,\theta_{34})$ and
$(\theta_{1},\dots,\theta_{6})$ are different, and are related by their Gram
matrices

$\begin{split}\left[\begin{matrix}1&-\cos\theta_{12}&-\cos\theta_{13}&-\cos\theta_{14}\\\
-\cos\theta_{12}&1&-\cos\theta_{23}&-\cos\theta_{24}\\\
-\cos\theta_{13}&-\cos\theta_{23}&1&-\cos\theta_{34}\\\
-\cos\theta_{14}&-\cos\theta_{24}&-\cos\theta_{34}&1\\\
\end{matrix}\right]=\left[\begin{matrix}1&-\cos\theta_{1}&-\cos\theta_{2}&-\cos\theta_{6}\\\
-\cos\theta_{1}&1&-\cos\theta_{3}&-\cos\theta_{5}\\\
-\cos\theta_{2}&-\cos\theta_{3}&1&-\cos\theta_{4}\\\
-\cos\theta_{6}&-\cos\theta_{5}&-\cos\theta_{4}&1\\\
\end{matrix}\right],\end{split}$

i.e., $\theta_{1}=\theta_{12},$ $\theta_{2}=\theta_{13},$
$\theta_{3}=\theta_{23},$ $\theta_{4}=\theta_{34},$ $\theta_{5}=\theta_{24}$
and $\theta_{6}=\theta_{14}.$ See Figure 1. For example, in the former, the
first three angles $(\theta_{12},\theta_{13},\theta_{14})$ are at the edges
around the face $F_{1}$ opposite to the vertex $\mathbf{v}_{1},$ whereas in
the latter, the first three angles $(\theta_{1},\theta_{2},\theta_{3})$ are at
the edges around the vertex $\mathbf{v}_{4}.$ In this paper, we will use both
notations for difference purpose; and when we switch from one to the other, we
change the ordered $6$-tuples.

Figure 1:

##  3 Generalized hyperbolic tetrahedra

######  Definition 3.1 (Generalized hyperbolic tetrahedron).

A _generalized hyperbolic tetrahedron_ is a quadruple of linearly independent
vectors $\mathbf{v}_{1},$ $\mathbf{v}_{2},$ $\mathbf{v}_{3}$ and
$\mathbf{v}_{4}$ in
$\mathbb{S}(-1)\cup\mathbb{S}(0)\cup\mathbb{S}(1)\subset\mathbb{E}^{3,1}$ such
that for each pair $\\{i,j\\}\subset\\{1,2,3,4\\},$ the straight line $L_{ij}$
passing through the vertices of $\mathbf{v}_{i}$ and $\mathbf{v}_{j}$
intersects $\overline{\mathbb{B}^{3,1}}.$

######  Remark 3.2.

The last condition is equivalent to that the radial projection of $L_{ij}$
into the projective model $\mathbb{P}^{3}$ intersects
$\overline{\mathbb{H}^{3}}.$

######  Remark 3.3.

The main difference between a generalized hyperbolic tetrahedron and a
hyperbolic tetrahedron is that: (1) we allow the vertices to be in
$\mathbb{H}^{3}_{+}\cup\mathbb{L}^{3}_{+}$ and
$\mathbb{H}^{3}_{-}\cup\mathbb{L}^{3}_{-}$ at the same time, not only in one
of them, (2) we allow the whole straight line $L_{ij}$ passing through
$\mathbf{v}_{i}$ and $\mathbf{v}_{j}$ to intersect
$\overline{\mathbb{B}^{3,1}},$ not only the line segment $L^{+}_{ij}$ between
them, and (3) we allow $\mathbf{v}_{i}$ and other vertices $\mathbf{v}_{j},$
$\mathbf{v}_{k}$ or $\mathbf{v}_{l}$ to be on the same side of the hyperplane
$\mathbf{\Pi}_{i}$ perpendicular to $\mathbf{v}_{i}.$

######  Remark 3.4.

Such objects were also studied by Bonahon and Sohn [5], where they also
considered the case with deep truncations, that is to say, the case where some
$L_{ij}$s do not intersect $\mathbb{B}^{3,1}.$ For the purpose of studying the
asymptotics of quantum $6j$-symbols, we will not consider those cases in this
article. The quantum content of deeply truncated tetrahedra can be found in
Belletti-Yang [4].

Two generalized hyperbolic tetrahedra are _isometric_ if the two quadruples
are related by an element of $O(3,1).$

###  3.1 Dihedral angles, Gram matrix and a criterion

The vectors $\mathbf{v}_{1},\dots,\mathbf{v}_{4}$ are the _vertices_ of the
generalized hyperbolic tetrahedron. We call a vertex _regular_ if it is in
$\mathbb{S}(-1),$ _ideal_ if it is in $\mathbb{S}(0),$ and _hyperideal_ if is
in $\mathbb{S}(1).$ A regular or ideal vertex is _positive_ if it is in
$\mathbb{H}^{3}_{+}\cup\mathbb{L}^{3}_{+},$ and is _negative_ if it is in
$\mathbb{H}^{3}_{-}\cup\mathbb{L}^{3}_{-}.$

For each $i\in\\{1,\dots,4\\},$ the _face_ opposite to $\mathbf{v}_{i}$ is the
plane $F_{i}$ containing the vertices of $\mathbf{v}_{j},$ $\mathbf{v}_{k}$
and $\mathbf{v}_{l},$ $\\{j,k,l\\}=\\{1,2,3,4\\}{\smallsetminus}\\{i\\},$
i.e.,

$F_{i}=\\{c_{j}\mathbf{v}_{j}+c_{k}\mathbf{v}_{k}+c_{l}\mathbf{v}_{l}\ |\
c_{j},c_{k},c_{l}\in\mathbb{R},c_{j}+c_{k}+c_{l}=1\\}.$

The _outward normal vector_ of $F_{i}$ is the vector $\mathbf{u}_{i}$ such
that

  1. (1)

$\mathbf{u}_{i}\in\mathbb{S}(1),$

  2. (2)

$\langle\mathbf{u}_{i},\mathbf{v}\rangle=0$ for any vector $\mathbf{v}\in
F_{i},$ and

  3. (3)

$\langle\mathbf{u}_{i},\mathbf{v}_{i}\rangle<0.$

Since $L_{jk},$ $L_{jl}$ and $L_{kl}$ intersect $\mathbb{B}^{3,1},$ so does
the plane $F_{i}$ containing them. Therefore, the outward normal vector
$\mathbf{u}_{i}$ of $F_{i}$ exists uniquely.

######  Definition 3.5 (Dihedral angles).

For $\\{i,j\\}\subset\\{1,2,3,4\\}.$ the _dihedral angle_ $\theta_{ij}$
between $F_{i}$ and $F_{j}$ is defined as $\pi$ minus the angle between the
outward normal vectors $\mathbf{u}_{i}$ and $\mathbf{u}_{j}.$ I.e.,

$\theta_{ij}=\pi-\cos^{-1}\langle\mathbf{u}_{i},\mathbf{u}_{j}\rangle,$

or equivalently,

$\langle\mathbf{u}_{i},\mathbf{u}_{j}\rangle=-\cos\theta_{ij}.$

We see from the definition that for $i\in\\{1,2,3,4\\},$
$\\{\theta_{jk},\theta_{jl},\theta_{kl}\\}$ is the set of dihedral angles at
the edges around the vertex $\mathbf{v}_{i},$ where
$\\{j,k,l\\}=\\{1,2,3,4\\}{\smallsetminus}\\{i\\}.$

######  Definition 3.6 (Gram matrix).

The _Gram matrix_ of a $6$-tuple $(\theta_{12},\dots,\theta_{34})$ of real
numbers is the following $4\times 4$ matrix

$G=\left[\begin{matrix}1&-\cos\theta_{12}&-\cos\theta_{13}&-\cos\theta_{14}\\\
-\cos\theta_{12}&1&-\cos\theta_{23}&-\cos\theta_{24}\\\
-\cos\theta_{13}&-\cos\theta_{23}&1&-\cos\theta_{34}\\\
-\cos\theta_{14}&-\cos\theta_{24}&-\cos\theta_{34}&1\\\ \end{matrix}\right].$

If $(\theta_{12},\dots,\theta_{34})$ is the set of dihedral angles of a
generalized hyperbolic tetrahedron $\Delta,$ then $G$ is called the _Gram
matrix_ of $\Delta.$

######  Theorem 3.7.

Suppose $(\theta_{12},\dots,\theta_{34})$ is a $6$-tuple of numbers in
$[0,\pi].$ Then the following statements are equivalent.

  1. (1)

$(\theta_{12},\dots,\theta_{34})$ is the set of dihedral angles of a
generalized hyperbolic tetrahedron.

  2. (2)

The Gram matrix $G$ of $(\theta_{12},\dots,\theta_{34})$ has signature
$(3,1).$

###### Proof.

The proof follows the same idea of that of Luo [17, Theorem] and Ushijima [24,
Theorem 3.2].

Suppose (1) holds. Then $G$ is the Gram matrix of a generalized hyperbolic
tetrahedron $\Delta.$ Let $\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ in
$\mathbb{S}(1)$ be the outward normal vectors of the faces of $\Delta$ and let

$U=[\mathbf{u}_{1},\mathbf{u}_{2},\mathbf{u}_{3},\mathbf{u}_{4}]$

be the $4\times 4$ matrix with $\mathbf{u}_{i}$’s as the columns. Then

$G=U^{T}\cdot I_{3,1}\cdot U,$

where $U^{T}$ is the transpose of $U,$ and $I_{3,1}$ is the matrix

$I_{3,1}=\left[\begin{matrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&-1\\\
\end{matrix}\right].$

Hence $G$ has signature $(3,1),$ and (b) holds.

Suppose (2) holds, so that $G$ has signature $(3,1).$ Then

$\det G<0,$

and by Sylvester’s Law of Inertia,

$G=U^{T}\cdot I_{3,1}\cdot U$

for an invertible matrix $U,$ uniquely determined up to an action of elements
of $O(3,1).$ Let $\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ be the columns of $U$
considered as vectors in $\mathbb{E}^{3,1}$ and let

$\mathbf{w}_{i}=\sum_{j=1}^{4}G_{ij}\mathbf{u}_{j},$

where $G_{ij}$ is the $ij$-th cofactor of $G.$ Then we have

$\langle\mathbf{w}_{i},\mathbf{u}_{j}\rangle=\delta_{ij}\det G,$

where $\delta_{ij}$ is the Kronecker symbol, and as a consequence,

$\langle\mathbf{w}_{i},\mathbf{w}_{j}\rangle=G_{ij}\det G.$

For $i\in\\{1,2,3,4\\},$ we define the vector $\mathbf{v}_{i}$ as follows. If
$G_{ii}=0,$ then let

$\mathbf{v}_{i}=\mathbf{w}_{i};$

and if $G_{ii}\neq 0,$ then let

$\mathbf{v}_{i}=\frac{\mathbf{w}_{i}}{\sqrt{|G_{ii}\det G|}}.$

Next we show that

  1. (a)

$\mathbf{v}_{1},\dots,\mathbf{v}_{4}$ define a generalized hyperbolic
tetrahedron $\Delta.$

  2. (b)

$\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ are the outward normal vectors of
$\Delta$, so that $G$ is the Gram matrix of $\Delta.$

From this it follows that $(\theta_{12},\dots,\theta_{34})$ is the set of
dihedral angles of $\Delta,$ and (a) holds.

To prove (a), we need following three steps.

  1. Step 1.

We show that $\mathbf{v}_{1},\dots,\mathbf{v}_{4}$ are linearly independent.
Indeed, let

$W=[\mathbf{w}_{1},\mathbf{w}_{2},\mathbf{w}_{3},\mathbf{w}_{4}]$

be the $4\times 4$ matrix containing $\mathbf{v}_{i}$’s as the columns, then

$W=U\cdot\mathrm{Ad}(G),$

where $\mathrm{Ad}(G)$ is the adjoint matrix of $G.$ By Cramer’s rule,

$\mathrm{Ad}(G)=\det G\cdot G^{-1}.$

Therefore, $\det\mathrm{Ad}(G)\neq 0,$ and as a consequence,

$\det W=\det U\cdot\det\mathrm{Ad}(G)\neq 0,$

and the columns $\mathbf{w}_{1},\dots,\mathbf{w}_{4}$ of $W$ are linearly
independent. Since $\mathbf{v}_{1},\dots,\mathbf{v}_{4}$ are non-zero scalar
multiples of $\mathbf{w}_{1},\dots,\mathbf{w}_{4},$ they are linearly
independent.

  2. Step 2.

We show that for each $i,$
$\mathbf{v}_{i}\in\mathbb{S}(-1)\cup\mathbb{S}(0)\cup\mathbb{S}(1).$ Indeed,
if $G_{ii}>0,$ then

$\langle\mathbf{v}_{i},\mathbf{v}_{i}\rangle=\frac{\langle\mathbf{w}_{i},\mathbf{w}_{i}\rangle}{-G_{ii}\det
G}=-1,$

and

$\mathbf{v}_{i}\in\mathbb{S}(-1);$

if $G_{ii}=0,$ then

$\langle\mathbf{v}_{i},\mathbf{v}_{i}\rangle=\langle\mathbf{w}_{i},\mathbf{w}_{i}\rangle=G_{ii}\det
G=0,$

and

$\mathbf{v}_{i}\in\mathbb{S}(0);$

and if $G_{ii}<0,$ then

$\langle\mathbf{v}_{i},\mathbf{v}_{i}\rangle=\frac{\langle\mathbf{w}_{i},\mathbf{w}_{i}\rangle}{G_{ii}\det
G}=1,$

and

$\mathbf{v}_{i}\in\mathbb{S}(1).$

  3. Step 3.

We show that for any $\mathbf{v}_{i}$ and $\mathbf{v}_{j}$, $L_{ij}$
intersects $\overline{\mathbb{B}^{3,1}}.$ If one of $\mathbf{v}_{i}$ or
$\mathbf{v}_{j}$ is already in $\mathbb{S}(-1)\cup\mathbb{S}(0),$ then
$L_{ij}$ intersects $\overline{\mathbb{B}^{3,1}}.$ Hence we only need to
consider the case that $\mathbf{v}_{i},\mathbf{v}_{j}\in\mathbb{S}(1).$ By
Step (2) above, this is equivalent to that $G_{ii}<0$ and $G_{jj}<0.$ Now by
Jacobi’s Theorem (see [20, 2.5.1. Theorem]) and that $\det G<0,$ we have

$G_{ij}^{2}-G_{ii}G_{jj}=(\cos^{2}\theta_{ij}-1)\det G\geqslant 0.$

Then either $G_{ij}\geqslant\sqrt{G_{ii}G_{jj}}$ or
$G_{ij}\leqslant-\sqrt{G_{ii}G_{jj}}.$ In either of the cases, by Lemma 3.8
below, $L_{ij}$ intersects $\overline{\mathbb{B}^{3,1}}.$

This completes the proof of (a).

For (b), we verify the conditions of an outward normal vector in the following
steps.

  1. Step 1.

Since $\langle\mathbf{u}_{i},\mathbf{u}_{i}\rangle$ equals the $i$-th diagonal
entry of $G$ which equals $1,$

$\mathbf{u}_{i}\in\mathbb{S}(1).$

  2. Step 2.

For $j\neq i,$ we have

$\langle\mathbf{u}_{i},\mathbf{v}_{j}\rangle=c_{j}\cdot\langle\mathbf{u}_{i},\mathbf{w}_{j}\rangle=\delta_{ij}\det
G=0,$

where $c_{j}=1$ if $G_{jj}=0,$ and $c_{j}=\frac{1}{\sqrt{|G_{jj}\det G|}}$ if
$G_{jj}\neq 0.$ Therefore, for any vector $\mathbf{v}$ in the plane $F_{i}$
containing $\mathbf{v}_{j},$ $\mathbf{v}_{k}$ and $\mathbf{v}_{l},$
$\\{j,k,l\\}=\\{1,2,3,4\\}{\smallsetminus}\\{i\\},$

$\langle\mathbf{u}_{i},\mathbf{v}\rangle=0.$

  3. Step 3.

For each $i\in\\{1,2,3,4\\},$ since $c_{i}>0$ and $\det G<0,$ we have

$\langle\mathbf{u}_{i},\mathbf{v}_{i}\rangle=c_{i}\cdot\langle\mathbf{u}_{i},\mathbf{w}_{i}\rangle=c_{i}\det
G<0.$

This completes the proof of (b). ∎

######  Lemma 3.8.

Suppose $G_{ii}<0$ and $G_{jj}<0.$ Let $L_{ij}^{+}$ be the line segment
connecting $\mathbf{v}_{i}$ and $\mathbf{v}_{j},$ and let
$L_{ij}^{-}=L_{ij}{\smallsetminus}L_{ij}^{+}.$ Then

  1. (1)

$L_{ij}^{+}$ intersects $\overline{\mathbb{B}^{3,1}}$ if and only if

$G_{ij}\geqslant\sqrt{G_{ii}G_{jj}};$ (3.1)

and

  2. (2)

$L_{ij}^{-}$ intersects $\overline{\mathbb{B}^{3,1}}$ if and only if

$G_{ij}\leqslant-\sqrt{G_{ii}G_{jj}}.$ (3.2)

###### Proof.

First, notice that a point on $L^{+}_{ij}$ has the form
$t\mathbf{v}_{i}+(1-t)\mathbf{v}_{j}$ for $t$ in the interval $(0,1),$ and a
point on $L^{-}_{ij}$ has the same form for $t$ in
$(-\infty,0)\cup(1,\infty).$

By the computation

$\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle=-\frac{G_{ij}}{\sqrt{G_{ii}G_{jj}}},$

we have that (3.1) is equivalent to
$\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle\leqslant-1$ and that (3.2) is
equivalent to $\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle\geqslant 1.$ Since
$\langle\mathbf{v}_{i},\mathbf{v}_{i}\rangle=\langle\mathbf{v}_{j},\mathbf{v}_{j}\rangle=1,$
we have that

$\langle
t\mathbf{v}_{i}+(1-t)\mathbf{v}_{j},t\mathbf{v}_{i}+(1-t)\mathbf{v}_{j}\rangle=2(1-\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle)t^{2}+2(\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle-1)t+1.$

Then (1) follows from the fact that the quadratic inequality
$2(1-\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle)t^{2}+2(\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle-1)t+1\leqslant
0$ has a solution in $(0,1)$ if and only if
$\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle\leqslant-1;$ and (2) follows from
the fact that the same quadratic inequality has a solution in
$[-\infty,0)\cup(1,\infty]$ if and only if
$\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle\geqslant 1.$ ∎

Similar to Theorem 3.7, we have the following criterion for Gram matrices with
signature $(2,1),$ which will be need in the classification of admissible
angles in Section 4.

######  Theorem 3.9.

Let $(\theta_{12},\dots,\theta_{34})$ be a $6$-tuple of numbers in $[0,\pi],$
and let $G$ be its Gram matrix. If the signature of $G$ is $(2,1),$ then
$(\theta_{12},\dots,\theta_{34})$ is the set of angles between four oriented
geodesics in the hyperbolic plane $\mathbb{H}^{2}$ that mutually intersect in
$\overline{\mathbb{H}^{2}}.$

######  Remark 3.10.

Here the orientation of a geodesic is defined to the a specification of its
normal vector, and the angle between two geodesics is $\pi$ minus the angle
between the two normal vectors that define the orientation of the geodesics.
See Figure 2.

Figure 2: In the figure, the grey disc is the closure
$\overline{\mathbb{H}^{2}}$ of the hyperbolic plane in the projective model.

###### Proof.

By Sylvester’s Law of Inertia,

$G=W^{T}\cdot I_{2,1}\cdot W$

for a $4\times 4$ matrix $W,$ where $I_{2,1}$ is the matrix

$I_{2,1}=\left[\begin{matrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&-1&0\\\ 0&0&0&0\\\
\end{matrix}\right].$

Let $\mathbf{w}_{1},\dots,\mathbf{w}_{4}$ be the columns of $W,$ and for each
$i\in\\{1,\dots,4\\},$ let $\mathbf{u}_{i}$ be the vector in
$\mathbb{E}^{2,1}$ obtained from $\mathbf{w}_{i}$ by erasing the last
component. If $\langle,\rangle$ denotes the inner product of signature $(2,1)$
on $\mathbb{E}^{2,1},$ then we have

$\langle\mathbf{u}_{i},\mathbf{u}_{i}\rangle=\mathbf{w}_{i}^{T}\cdot
I_{2,1}\cdot\mathbf{w}_{i}=1$

for $i\in\\{1,2,3,4\\},$ and

$\langle\mathbf{u}_{i},\mathbf{u}_{j}\rangle=\mathbf{w}_{i}^{T}\cdot
I_{2,1}\cdot\mathbf{w}_{j}=-\cos\theta_{ij}$

for $\\{i,j\\}\subset\\{1,2,3,4\\}.$ In particular, for each
$i\in\\{1,2,3,4\\},$

$\mathbf{u}_{i}\in\mathbb{S}(1),$

the de Sitter space in $\mathbb{E}^{2,1}$ consisting of vectors of norm $1.$
For each $i\in\\{1,2,3,4\\},$ let $\Pi_{i}$ be the plane in $\mathbb{E}^{2,1}$
perpendicular to $\mathbf{u}_{i}$ and let $L_{i}$ be the intersection of
$\Pi_{i}$ with $\mathbb{H}^{2}_{+},$ the upper sheet of the hyperboloid
consisting of vectors of norm $-1$ and the last component positive. We orient
$L_{i}$ as the direction of $\mathbf{u}_{i}.$ Then the angle between $L_{i}$
and $L_{j}$ is
$\pi-\cos^{-1}\langle\mathbf{u}_{i},\mathbf{u}_{j}\rangle=\theta_{ij},$ and
$L_{1},\dots,L_{4}$ with this orientation are the desired oriented geodesics.
∎

To see properties of the dihedral angles and to define the edge lengths and
the volume of generalized hyperbolic tetrahedra, we need to use the projective
model $\mathbb{P}^{3}.$

First, we look at the dihedral angles of the generalized hyperbolic tetrahedra
respectively with vertices
$\\{-\mathbf{v}_{i},\mathbf{v}_{j},\mathbf{v}_{k},\mathbf{v}_{l}\\}$ and
$\\{-\mathbf{v}_{i},-\mathbf{v}_{j},\mathbf{v}_{k},\mathbf{v}_{l}\\},$
$\\{i,j,k,l\\}=\\{1,2,3,4\\}.$ We denote by $\Delta_{0}$ the radial projection
of the convex hull of the vertices of
$\\{\mathbf{v}_{1},\dots,\mathbf{v}_{4}\\}.$ For $i\in\\{1,\dots,4\\},$ let
$\Delta_{i}$ be the radial projection of the convex hull of the vertices of
the quadruple
$\\{-\mathbf{v}_{i},\mathbf{v}_{j},\mathbf{v}_{k},\mathbf{v}_{l}\\};$ and for
$\\{i,j\\}\subset\\{1,2,3,4\\},$ let $\Delta_{ij}$ be the radial projection of
the convex hull of the vertices of the quadruple
$\\{-\mathbf{v}_{i},-\mathbf{v}_{j},\mathbf{v}_{k},\mathbf{v}_{l}\\}.$ Then
$\Delta_{0},\Delta_{1},\Delta_{2},\Delta_{3},\Delta_{4},\Delta_{12},\Delta_{13},\Delta_{23}$
provide a decomposition of $\mathbb{P}^{3}$ as they are the connected
components of the complement of the radial projections of the faces
$F_{1},\dots,F_{4}$ of $\Delta.$ As an immediate consequence, we have the
following

######  Proposition 3.11.

Let $\theta_{12},\dots,\theta_{34}$ be the dihedral angles of $\Delta.$

  1. (1)

Suppose $\theta^{*}_{12},\dots,\theta^{*}_{34}$ is the set of dihedral angles
of $\Delta_{i},$ $i\in\\{1,2,3,4\\}.$ Then for $j\neq i,$

$\theta^{*}_{ij}=\pi-\theta_{ij};$

and for $j,k\neq i,$

$\theta^{*}_{jk}=\theta_{jk}.$

  2. (2)

Suppose $\theta^{**}_{12},\dots,\theta^{**}_{34}$ is the set of dihedral
angles of $\Delta_{ij},$ $\\{i,j\\}\in\\{1,2,3,4\\}.$ Then for
$\\{s,t\\}=\\{i,j\\}$ or $\\{1,2,3,4\\}{\smallsetminus}\\{i,j\\},$

$\theta^{**}_{st}=\theta_{st};$

and for $\\{s,t\\}\neq\\{i,j\\}$ nor $\\{1,2,3,4\\}{\smallsetminus}\\{i,j\\},$

$\theta^{**}_{st}=\pi-\theta_{st}.$

We call the operation sending a $6$-tuple
$\\{\theta_{st}\\}_{\\{s,t\\}\subset\\{1,2,3,4\\}}$ to
$\\{\theta^{*}_{st}\\}_{\\{s,t\\}\subset\\{1,2,3,4\\}}$ the _change of angles
operation_ opposite to the vertex $\mathbf{v}_{i}.$ In the case that
$\theta_{st}$’s are the dihedral angles of generalized hyperbolic tetrahedron
with vertices $\mathbf{v}_{1},\dots,\mathbf{v}_{4},$ this operation
corresponds to changing the dihedral angles this generalized hyperbolic
tetrahedron to that of the generalized hyperbolic tetrahedron with vertices
$\\{-\mathbf{v}_{i},\mathbf{v}_{j},\mathbf{v}_{k},\mathbf{v}_{l}\\}.$ We also
notice that sending $\\{\theta_{st}\\}_{\\{s,t\\}\subset\\{1,2,3,4\\}}$ to
$\\{\theta^{**}_{st}\\}_{\\{s,t\\}\subset\\{1,2,3,4\\}}$ corresponds to doing
the change of angles operations twice, respectively opposite to
$\mathbf{v}_{i}$ and $\mathbf{v}_{j}.$ Later we will need the following lemma
which says that a change of angles operation does not change the signature of
the Gram matrix.

######  Lemma 3.12.

Let $G$ be the Gram matrices of the $6$-tuple
$\\{\theta_{ij},\theta_{ik},\theta_{il},\theta_{jk},\theta_{jl},\theta_{kl}\\},$
and let $G^{*}$ be the Gram matrix of the $6$-tuple
$\\{\pi-\theta_{ij},\pi-\theta_{ik},\pi-\theta_{il},\theta_{jk},\theta_{jl},\theta_{kl}\\},$
$\\{i,j,k,l\\}=\\{1,2,3,4\\}.$ Then $G$ and $G^{*}$ have the same signature.

###### Proof.

Suppose $G$ has signature $(p,q).$ Then by Sylvester’ Law of Inertia,

$G=U^{T}\cdot I_{p,q}\cdot U$

for some $4\times 4$ matrix $U$ with columns
$\mathbf{u}_{1},\mathbf{u}_{2},\mathbf{u}_{3},\mathbf{u}_{4},$ where $I_{p,q}$
is the $4\times 4$ diagonal matrix with the first $p$ diagonal entries $1,$
the next $q$ diagonal entries $-1$ and other diagonal entries $0.$ Let $U^{*}$
be the matrix obtained by changing the $i$-th column $\mathbf{u}_{i}$ of $U$
to $-\mathbf{u}_{i}.$ Then we have

$G^{*}=U^{*T}\cdot I_{p,q}\cdot U^{*},$

and hence $G^{*}$ has the same signature $(p,q)$ as $G$. ∎

###  3.2 Edge lengths

Next, we define the distance between the vertices, and define the edge lengths
of a generalized hyperbolic tetrahedron.

The _distance_ $d_{ij}$ between vertices $\mathbf{v}_{i}$ and $\mathbf{v}_{j}$
is defined as follows. Let $G$ be the Gram matrix of a generalized hyperbolic
tetrahedra $\Delta$ with vertices $\mathbf{v}_{1},$ $\mathbf{v}_{2},$
$\mathbf{v}_{3}$ and $\mathbf{v}_{4},$ and let $G_{ij}$ be its $ij$-th
cofactor.

  1. (1)

If $G_{ii}>0$ and $G_{jj}>0,$ then
$\mathbf{v}_{i},\mathbf{v}_{j}\in\mathbb{S}(-1),$ and $d_{ij}$ is defined to
be the hyperbolic distance between their radial projections $v_{i}$ and
$v_{j}$ in $\mathbb{H}^{3},$ i.e.,

$d_{ij}=\cosh^{-1}\frac{|G_{ij}|}{\sqrt{G_{ii}G_{jj}}}.$ (3.3)

  2. (2)

If $G_{ii}>0$ and $G_{jj}<0,$ then $\mathbf{v}_{i}\in\mathbb{S}(-1)$ and
$\mathbf{v}_{j}\in\mathbb{S}(1),$ and $d_{ij}$ is the defined to be the
hyperbolic distance between the radial projections $v_{i}$ and $\Pi_{j}$ in
$\mathbb{H}^{3},$ i.e.,

$d_{ij}=\sinh^{-1}\frac{|G_{ij}|}{\sqrt{-G_{ii}G_{jj}}}.$ (3.4)

  3. (3)

If $G_{ii}<0$ and $G_{jj}<0,$ then
$\mathbf{v}_{i},\mathbf{v}_{j}\in\mathbb{S}(1),$ and $d_{ij}$ is defined to be
the hyperbolic distance between the radial projections $\Pi_{i}$ and $\Pi_{j}$
in $\mathbb{H}^{3},$ i.e.,

$d_{ij}=\cosh^{-1}\frac{|G_{ij}|}{\sqrt{G_{ii}G_{jj}}}.$ (3.5)

We define the _geometric edges_ of $\Delta$ as follows. If
$\mathbf{v}_{i},\mathbf{v}_{j}\in\mathbb{S}(-1)\cup\mathbb{S}(0),$ then
$v_{i},v_{j}\in\overline{\mathbb{H}^{3}}$ and the geometric edge connecting
$\mathbf{v}_{i}$ and $\mathbf{v}_{i}$ is the geodesic in $\mathbb{H}^{3}$
connecting $v_{i}$ and $v_{j};$ if $\mathbf{v}_{i}\in S(-1)\cup\mathbb{S}(0)$
and $\mathbf{v}_{j}\in\mathbb{S}(1),$ then $v_{i}\in\overline{\mathbb{H}^{3}}$
and $\Pi_{j}\subset\mathbb{H}^{3},$ and the geometric edge connecting
$\mathbf{v}_{i}$ and $\mathbf{v}_{i}$ is the shortest geodesic in
$\mathbb{H}^{3}$ between $v_{i}$ and $\Pi_{j};$ and if
$\mathbf{v}_{i},\mathbf{v}_{j}\in\mathbb{S}(1),$ then
$\Pi_{i},\Pi_{j}\subset\mathbb{H}^{3},$ and the geometric edge connecting
$\mathbf{v}_{i}$ and $\mathbf{v}_{i}$ is the shortest geodesic in
$\mathbb{H}^{3}$ between $\Pi_{i}$ and $\Pi_{j}.$ Then the distance $d_{ij}$
between $\mathbf{v}_{i}$ and $\mathbf{v}_{j}$ is the hyperbolic length of the
geometric edge connecting them. Suggested by Proposition 3.14 below, we make
the following

######  Definition 3.13 (Edge lengths).

The _length_ $l_{ij}$ of the geometric edge between the faces $F_{i}$ and
$F_{j}$ (which is the geometric edge connecting the vertices $\mathbf{v}_{k}$
and $\mathbf{v}_{l}$) is defined by

$l_{ij}=\left\\{\begin{array}[]{rcl}d_{kl}&\text{if }&G_{kl}>0,\\\
-d_{kl}&\text{if }&G_{kl}\leqslant 0,\end{array}\right.$

where $\\{k,l\\}=\\{1,2,3,4\\}{\smallsetminus}\\{i,j\\}.$

We call a geometric edge _positive_ , _non-positive_ or _negative_ if its
length is respectively so.

######  Proposition 3.14.

Let $G$ be the Gram matrix of a generalized hyperbolic tetrahedra $\Delta$
with vertices $\mathbf{v}_{1},$ $\mathbf{v}_{2},$ $\mathbf{v}_{3}$ and
$\mathbf{v}_{4},$ and let $G_{ij}$ be the $ij$-th cofactor of $G.$

  1. (1)

If $G_{ii}>0$ and $G_{jj}>0,$ or equivalently
$\\{\mathbf{v}_{i},\mathbf{v}_{j}\\}\subset\mathbb{S}(-1)=\mathbb{H}^{3}_{+}\cup\mathbb{H}^{3}_{-},$
then $G_{ij}\neq 0.$

    1. (a)

If $G_{ij}>0,$ then
$\\{\mathbf{v}_{i},\mathbf{v}_{j}\\}\subset\mathbb{H}^{3}_{+}$ or
$\mathbb{H}^{3}_{-}.$ (See Figure 3 (a) for the radial projection in
$\overline{\mathbb{H}^{3}}.$)

    2. (b)

If $G_{ij}<0,$ then
$\\{\mathbf{v}_{i},-\mathbf{v}_{j}\\}\subset\mathbb{H}^{3}_{+}$ or
$\mathbb{H}^{3}_{-}.$ (See Figure 3 (b) for the radial projection in
$\overline{\mathbb{H}^{3}}.$)

Figure 3: The positive geometric edge in (a) is colored in blue, and the
negative geometric edge in (b) is colored in red.

  2. (2)

Suppose $G_{ii}>0$ and $G_{jj}<0,$ or equivalently
$\mathbf{v}_{i}\in\mathbb{S}(-1)=\mathbb{H}^{3}_{+}\cup\mathbb{H}^{3}_{1}$ and
$\mathbf{v}_{j}\in\mathbb{S}(1).$

    1. (a)

If $G_{ij}>0,$ then $v_{i}$ and $v_{j}$ are on different sides of $\Pi_{j}.$
(See Figure 4 (a).)

    2. (b)

If $G_{ij}\leqslant 0,$ then $v_{i}$ and $v_{j}$ are on the same side of
$\Pi_{j}.$ (See Figure 4 (b).)

Figure 4: The positive geometric edge in (a) is colored in blue, and the non-
positive geometric edge in (b) is colored in red.

  3. (3)

If $G_{ii}<0$ and $G_{jj}<0,$ or equivalently
$\\{\mathbf{v}_{i},\mathbf{v}_{j}\\}\subset\mathbb{S}(1),$ then $G_{ij}\neq
0.$

    1. (a)

If $G_{ij}>0,$ then $L_{ij}^{+}$ intersects $\Pi_{i}$ and $\Pi_{j}.$ (See
Figure 5 (a).)

    2. (b)

If $G_{ij}<0,$ then $L_{ij}^{-}$ intersects $\Pi_{i}$ and $\Pi_{j}.$ (See
Figure 5 (b).)

Figure 5: The positive geometric edge in (a) is colored in blue, and the
negative geometric edge in (b) is colored in red.

###### Proof.

For (1), let $d_{ij}$ be the distance between $\mathbf{v}_{i}$ and
$\mathbf{v}_{j},$ i.e., the hyperbolic distance between $v_{i}$ and $v_{j}$ in
$\mathbb{H}^{3}.$ If $\mathbf{v}_{i},\mathbf{v}_{j}\in\mathbb{H}^{3}_{+}$ or
$\mathbb{H}^{3}_{-},$ then

$\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle=-\cosh d_{ij}<0;$

and if $\mathbf{v}_{i}\in\mathbb{H}^{3}_{+}$ and
$\mathbf{v}_{j}\in\mathbb{H}^{3}_{-},$ then

$\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle=\cosh d_{ij}>0.$

Then the result follows from the computation that

$\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle=-\frac{G_{ij}}{\sqrt{G_{ii}G_{jj}}}.$

For (2), we use the computation

$\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle=-\frac{G_{ij}}{\sqrt{-G_{ii}G_{jj}}}.$

If $G_{ij}>0,$ then $\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle<0,$ and
$\mathbf{v}_{i}$ and $\mathbf{v}_{j}$ are on different sides of the hyperplane
$\mathbf{\Pi}_{j}$ in $\mathbb{E}^{3,1},$ hence $L^{+}_{ij}$ intersects
$\mathbf{\Pi}_{j}.$ As a consequence, the radial projection $L^{+}_{ij}$
intersects the radial projection $\Pi_{j}.$ If $G_{ij}\leqslant 0,$ then
$\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle\geqslant 0,$ and $\mathbf{v}_{i}$
and $\mathbf{v}_{j}$ are on the same side of $\mathbf{\Pi}_{j},$ and hence
$L^{-}_{ij}$ intersects it. As a consequence, the radial projection
$L^{-}_{ij}$ intersects the radial projection $\Pi_{j}.$

For (3), let $d_{ij}$ be the distance between $\mathbf{v}_{i}$ and
$\mathbf{v}_{j},$ i.e., the hyperbolic distance between the radial projections
$\Pi_{i}$ and $\Pi_{j}$ in $\mathbb{H}^{3}.$ Then comparing the formulas

$|\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle|=\cosh d_{ij}$

and

$\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle=-\frac{G_{ij}}{\sqrt{G_{ii}G_{jj}}},$

we have $G_{ij}\neq 0.$ Now if $G_{ij}>0,$ then
$\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle<0,$ and $\mathbf{v}_{i}$ and
$\mathbf{v}_{j}$ are on different sides of the hyperplanes $\mathbf{\Pi}_{i}$
and $\mathbf{\Pi}_{j}$ in $\mathbb{E}^{3,1},$ hence $L^{+}_{ij}$ intersects
$\mathbf{\Pi}_{i}$ and $\mathbf{\Pi}_{j}.$ As a consequence, the radial
projection $L^{+}_{ij}$ intersects the radial projections $\Pi_{i}$ and
$\Pi_{j}.$ If $G_{ij}<0,$ then
$\langle\mathbf{v}_{i},\mathbf{v}_{j}\rangle>0,$ and $\mathbf{v}_{i}$ and
$\mathbf{v}_{j}$ are on the same side of $\mathbf{\Pi}_{i}$ and
$\mathbf{\Pi}_{j},$ and hence $L^{-}_{ij}$ intersects them. As a consequence,
the radial projection $L^{-}_{ij}$ intersects the radial projections $\Pi_{i}$
and $\Pi_{j}.$ ∎

###  3.3 A classification

We give a classification of generalized hyperbolic tetrahedra $\Delta$ in
terms of their Gram matrix. Let $\mathbf{v}_{1},\dots,\mathbf{v}_{4}$ be the
vertices of $\Delta.$ Recall that the change of angles operation opposite to
the vertex $\mathbf{v}_{i}$ sends the set of dihedral angles
$\\{\theta_{ij},\theta_{ik},\theta_{il},\theta_{jk},\theta_{jl},\theta_{kl}\\}$
to
$\\{\pi-\theta_{ij},\pi-\theta_{ik},\pi-\theta_{il},\theta_{jk},\theta_{jl},\theta_{kl}\\},$
$\\{i,j,k,l\\}=\\{1,2,3,4\\},$ which are the dihedral angles of the
generalized hyperbolic tetrahedron with set of vertices
$\\{-\mathbf{v}_{i},\mathbf{v}_{j},\mathbf{v}_{k},\mathbf{v}_{l}\\}.$

######  Proposition 3.15.

Let $G$ be the Gram matrix of a generalized hyperbolic tetrahedron.

  1. (1)

If $G_{ii},G_{jj},G_{kk},G_{ll}\geqslant 0,$ then there are the following
cases. See Figure 6.

    1. (a)

$G_{ij},G_{ik},G_{il},G_{jk},G_{jl},G_{kl}>0.$

    2. (b)

$G_{il},G_{jl},G_{kl}<0$ and $G_{ij},G_{ik},G_{jk}>0.$

    3. (c)

$G_{ik},G_{il},G_{jk},G_{jl}<0$ and $G_{ij},G_{kl}>0.$

    4. (d)

Otherwise, by doing a sequence of change of angles operations, we are in case
(a), (b) or (c).

Figure 6: In (a), all the vertices
$\mathbf{v}_{i},\mathbf{v}_{j},\mathbf{v}_{k},\mathbf{v}_{l}$ are positive. In
(b), the vertices $\mathbf{v}_{i},$ $\mathbf{v}_{j}$ and $\mathbf{v}_{k}$ are
positive and $\mathbf{v}_{l}$ is negative. In (c), the vertices
$\mathbf{v}_{i}$ and $\mathbf{v}_{j}$ are positive and $\mathbf{v}_{k}$ and
$\mathbf{v}_{l}$ are negative. The positive geometric edges are colored in
blue and the negative geometric edges are colored in red.

  2. (2)

If $G_{ii},G_{jj},G_{kk}\geqslant 0$ and $G_{ll}<0,$ then there are the
following cases. See Figure 7.

    1. (a)

$G_{ij},G_{ik},G_{il},G_{jk},G_{jl},G_{kl}>0.$

    2. (b)

$G_{kl}\leqslant 0$ and $G_{ij},G_{ik},G_{il},G_{jk},G_{jl}>0.$

    3. (c)

$G_{jl},G_{kl}\leqslant 0$ and $G_{ij},G_{ik},G_{il},G_{jk}>0.$

    4. (d)

$G_{il},G_{jl},G_{kl}\leqslant 0$ and $G_{ij},G_{ik},G_{jk}>0.$

    5. (e)

Otherwise, by doing a sequence of change of angles operations, we are in case
(a) or (b).

Figure 7: All the regular or ideal vertices $\mathbf{v}_{i},$ $\mathbf{v}_{j}$
and $\mathbf{v}_{k}$ are positive. The triangles of truncation are colored in
grey, the positive geometric edges are colored in blue and the non-positive
geometric edges are colored in red.

  3. (3)

If $G_{ii},G_{jj}\geqslant 0$ and $G_{kk},G_{ll}<0,$ then there are the
following cases. See Figure 8.

    1. (a)

$G_{ij},G_{ik},G_{il},G_{jk},G_{jl},G_{kl}>0.$

    2. (b)

$G_{jk}\leqslant 0$ and $G_{ij},G_{ik},G_{il},G_{jl},G_{kl}>0.$

    3. (c)

$G_{ik},G_{jk}\leqslant 0$ and $G_{ij},G_{il},G_{jl},G_{kl}>0.$

    4. (d)

$G_{il},G_{jk}\leqslant 0$ and $G_{ij},G_{ik},G_{jl},G_{kl}>0.$

    5. (e)

Otherwise, by doing a sequence of change of angles operations, we are in case
(a), (b), (c) or (d).

Figure 8: All the regular or ideal vertices $\mathbf{v}_{i}$ and
$\mathbf{v}_{j}$ are positive. The triangles of truncation are colored in
grey, the positive geometric edges are colored in blue and the negative
geometric edges are colored in red.

  4. (4)

If $G_{ii}\geqslant 0$ and $G_{jj},G_{kk},G_{ll}<0,$ then there are the
following cases. See Figure 9.

    1. (a)

$G_{ij},G_{ik},G_{il},G_{jk},G_{jl},G_{kl}>0.$

    2. (b)

$G_{ij}\leqslant 0$ and $G_{ik},G_{il},G_{jk},G_{jl},G_{kl}>0.$

    3. (c)

$G_{kl}<0$ and $G_{ij},G_{ik},G_{il},G_{jk},G_{jl}>0.$

    4. (d)

$G_{ij}\leqslant 0,$ $G_{kl}<0$ and $G_{ik},G_{il},G_{jk},G_{jl}>0.$

    5. (e)

Otherwise, by doing a sequence of change of angles operations, we are in case
(a), (b), (c) or (d).

Figure 9: The regular or ideal vertice $\mathbf{v}_{i}$ is positive. The
triangles of truncation are colored in grey, the positive geometric edges are
colored in blue and the non-positive geometric edges are colored in red.

  5. (5)

If $G_{ii},G_{jj},G_{kk},G_{ll}<0,$ then there are the following cases. See
Figure 10.

    1. (a)

$G_{ij},G_{ik},G_{il},G_{jk},G_{jl},G_{kl}>0.$

    2. (b)

$G_{kl}<0$ and $G_{ij},G_{ik},G_{il},G_{jk},G_{jl}>0.$

    3. (c)

$G_{ij},G_{kl}<0$ and $G_{ik},G_{il},G_{jk},G_{jl}>0.$

    4. (d)

Otherwise, by doing a sequence of change of angles operations, we are in case
(a), (b) or (c).

Figure 10: The triangles of truncation are colored in grey, the positive
geometric edges are colored in blue and the negative geometric edges are
colored in red.

######  Remark 3.16.

In (2), (c) actually can be obtained from (b) be doing a change of angles
operation along the face $F_{l},$ and (d) can be obtained from (a) be doing a
change of angles operation along the face $F_{l}.$ We still list these two
redundant cases here for the purpose of proving Proposition 4.10 in Section 4,
which is a crucial step in the proof of Theorem 1.4 (1).

###### Proof of Proposition 3.15.

In (1), all the vertices are regular or ideal. Then up to a sequence of change
of angles operations, either all the vertices are in
$\mathbb{H}^{3}_{+}\cup\mathbb{L}^{3}_{+}$ as in (a); or exactly one vertex is
in $\mathbb{H}^{3}_{-}\cup\mathbb{L}^{3}_{-}$ as in (b); or exactly two
vertices are in $\mathbb{H}^{3}_{-}\cup\mathbb{L}^{3}_{-}$ as in (c).

In (2), (3), (4), (5), up to a sequence of change of angles operations
opposite to the negative regular vertices, we may assume that all the regular
vertices are positive as in (a), (b), (c), (d) of each of the cases. Then up
to the change of angles operations around the faces opposite to the hyperideal
vertices, it is to look at in the projective model the relative positions
between the $v_{i}$’s for regular or ideal vertices and $\Pi_{j}$’s for
hyperideal vertices in $\overline{\mathbb{H}^{3}}.$

In (2), $\mathbf{v}_{i},\mathbf{v}_{j},\mathbf{v}_{k}$ are regular or ideal,
and $\mathbf{v}_{l}$ is hyperideal. Then in the projective model, either all
$v_{i},v_{j},v_{k}$ are on one side of $\Pi_{l}$ as in (a); or one of them,
say, $v_{k}$ and the other two $v_{i},v_{j}$ are one different sides of
$\Pi_{l}$ as in (b).

In (3), $\mathbf{v}_{i},\mathbf{v}_{j}$ are regular or ideal, and
$\mathbf{v}_{k},\mathbf{v}_{l}$ are hyperideal. Then in the projective model,
$\Pi_{k}$ and $\Pi_{l}$ divide $\overline{\mathbb{H}^{3}}$ into three piece
$D_{k}$ that is only adjacent to $\Pi_{k},$ $D_{l}$ that is only adjacent to
$\Pi_{l},$ and $D_{kl}$ that is adjacent to both $\Pi_{k}$ and $\Pi_{l}.$ Then
either $v_{i},v_{j}$ are in $D_{kl}$ as in (a); or $v_{i}$ is in $D_{kl}$ and
$v_{j}$ is in, say, $D_{k}$ as in (b); or $v_{i},v_{j}$ are in, say, $D_{k}$
as in (c); or $v_{i}$ is in, say, $D_{k}$ and $v_{j}$ is in $D_{l}$ as in (d).

In (4), $\mathbf{v}_{i}$ is regular or ideal, and
$\mathbf{v}_{j},\mathbf{v}_{k},\mathbf{v}_{l}$ are hyperideal. Then in the
projective model, $\Pi_{j},$ $\Pi_{k}$ and $\Pi_{l}$ divide
$\overline{\mathbb{H}^{3}}$ into four piece in two different ways. In one way
as in (a) and (b), there is a piece $D_{jkl}$ that is adjacent to all of
$\Pi_{j},\Pi_{k},\Pi_{l},$ and there are three pieces $D_{j},$ $D_{k},$
$D_{l}$ that are respective only adjacent to $\Pi_{j},$ $\Pi_{k},$ $\Pi_{l}.$
Then either $v_{i}$ is in $D_{jkl}$ as in (a); or $v_{i}$ is in, say, $D_{j}$
as in (b). in the other way as in (c) and (d), there are two pieces $D_{jk},$
$D_{kl}$ that are respectively adjacent to $\Pi_{j},\Pi_{k}$ and
$\Pi_{k},\Pi_{l},$ and to pieces $D_{j},$ $D_{l}$ that are respectively only
adjacent to $\Pi_{j}$ and $\Pi_{l}.$ Then either $v_{i}$ is in, say, $D_{jk}$
as in (c); or $v_{i}$ is in, say, $D_{j}$ as in (d).

In (5), all the vertices are hyperideal, and $\Pi_{1},\dots,\Pi_{4}$ divede
$\overline{\mathbb{H}^{3}}$ in three different ways as in (a), (b) and (c). ∎

###  3.4 Volume and the Schläfli formula

Finally, let us define the volume of a generalized hyperbolic tetrahedron. A
_geometric face_ of $\Delta$ is a region in the radial projection of the faces
$F_{i}$’s bounded by the geometric edges of $\Delta$ and the intersections
$\\{F_{i}\cap\Pi_{j}\\}_{i,j\in\\{1,2,3,4\\}},$ and a _geometric piece_ of
$\Delta$ is a region in $\mathbb{H}^{3}$ bounded by the geometric faces of
$\Delta$ and the planes $\\{\Pi_{i}\\}_{i\in\\{1,2,3,4\\}}.$ Then each
geometric piece of $\Delta$ is a polyhedron in $\mathbb{H}^{3}.$ We also
observe that the intersection of each $\Pi_{i}$ with the union of the
geometric pieces is a hyperbolic triangle which we call a _triangle of
truncation_ , and that each geometric piece of $\Delta$ is contained in
exactly one of $\Delta_{0},$ $\Delta_{i}$’s and $\Delta_{ij}$’s. See Figures
6, 7, 8, 9, 10.

######  Definition 3.17 (Volume).

For a geometric piece $P$ of a generalized hyperbolic tetrahedron $\Delta,$
let $\mathrm{Vol}(P)$ be the hyperbolic volume of $P$ considered as a
hyperbolic polyhedron. If $P$ is contained in
$\Delta_{0}\cup\Delta_{12}\cup\Delta_{13}\cup\Delta_{14},$ then let

$\mathrm{Vol}^{P}(\Delta)=\mathrm{Vol}(P);$

and if $P$ is contained in
$\Delta_{1}\cup\Delta_{2}\cup\Delta_{3}\cup\Delta_{4},$ then let

$\mathrm{Vol}^{P}(\Delta)=-\mathrm{Vol}(P).$

The _volume_ $\mathrm{Vol}(\Delta)$ of $\Delta$ is defined as

$\mathrm{Vol}(\Delta)=\sum_{P}\mathrm{Vol}^{P}(\Delta),$

where the sum if over all the geometric pieces of $\Delta.$

######  Example 3.18.

In Figure 6, let $P$ be the only geometric piece. Then for (a) and (c),

$\mathrm{Vol}(\Delta)=\mathrm{Vol}(P),$

and for (b),

$\mathrm{Vol}(\Delta)=-\mathrm{Vol}(P).$

######  Example 3.19.

In Figure 8 (a), let $P$ be the only geometric piece. Then

$\mathrm{Vol}(\Delta)=\mathrm{Vol}(P).$

In Figure 8 (b), let $P_{ikl}$ be the geometric piece bounded by $v_{i},$
$\Pi_{k}$ and $\Pi_{l}$ and let $P_{jk}$ be the geometric piece bounded by
$v_{j}$ and $\Pi_{k},$ then

$\mathrm{Vol}(\Delta)=\mathrm{Vol}(P_{ikl})-\mathrm{Vol}(P_{jk}).$

In Figure 8 (c), let $P_{kl}$ be the geometric piece bounded by $\Pi_{k}$ and
$\Pi_{l}$ and let $P_{ijk}$ be the geometric piece bounded by $v_{i},$ $v_{j}$
and $\Pi_{k},$ then

$\mathrm{Vol}(\Delta)=\mathrm{Vol}(P_{kl})-\mathrm{Vol}(P_{ijk}).$

In Figure 8 (d), let $P_{il}$ be the geometric piece bounded by $v_{i}$ and
$\Pi_{l},$ let $P_{kl}$ be the geometric piece bounded by $\Pi_{k}$ and
$\Pi_{l}$ and let $P_{jk}$ be the geometric piece bounded by $v_{j}$ and
$\Pi_{k},$ then

$\mathrm{Vol}(\Delta)=-\mathrm{Vol}(P_{il})+\mathrm{Vol}(P_{kl})-\mathrm{Vol}(P_{jk}).$

######  Example 3.20.

In each generalized hyperbolic tetrahedron in Figure 11, let $P_{ij}$ be the
geometric piece bounded by $\Pi_{i}$ and $\Pi_{j}.$ Then for (a) and (c),

$\mathrm{Vol}(\Delta)=-\mathrm{Vol}(P_{12})+\mathrm{Vol}(P_{23})-\mathrm{Vol}(P_{34}),$

and for (b),

$\mathrm{Vol}(\Delta)=\mathrm{Vol}(P_{12})-\mathrm{Vol}(P_{23})+\mathrm{Vol}(P_{34}).$

Figure 11: The triangles of truncation are colored in grey, the positive
geometric edges are colored in blue and the non-positive geometric edges are
colored in red.

######  Proposition 3.21 (Schläfli Formula).

For a generalized hyperbolic tetrahedron $\Delta$ with either regular or
hyperideal vertices, let $\mathrm{Vol}(\Delta)$ be the volume of $\Delta$ as
defined in Definition 3.17, and for $\\{i,j\\}\subset\\{1,\dots,4\\},$ let
$\theta_{ij}$ and $l_{ij}$ be the dihedral angle at and the length of the edge
$e_{ij}$ between the faces $F_{i}$ and $F_{j}$ respectively defined in
Definition 3.5 and Definition 3.13. Then

$\frac{\partial\mathrm{Vol}(\Delta)}{\partial\theta_{ij}}=-\frac{l_{ij}}{2}.$
(3.6)

To prove Proposition 3.21, we need the following Lemma 3.22 whose proof
follows immediately from Definition 3.13 and Proposition 3.14.

######  Lemma 3.22.

Let $\Delta$ be a generalized hyperbolic tetrahedron with only positive
regular vertices, and let $e_{st}$ be a geometric edge of $\Delta$
intersecting a geometric piece $P,$ $\\{s,t\\}\subset\\{1,2,3,4\\}.$

  1. (1)

If $P$ is contained in $\Delta_{0},$ then $e_{st}$ is positive.

  2. (2)

If $P$ is contained in $\Delta_{i},$ $i=1,2,3,4,$ then $e_{st}$ is positive
for $\\{s,t\\}=\\{i,j\\}$ with $j\in\\{1,2,3,4\\}{\smallsetminus}\\{i\\};$ and
$e_{st}$ is non-positive for $\\{s,t\\}=\\{j,k\\}$ with
$\\{j,k\\}\subset\\{1,2,3,4\\}{\smallsetminus}\\{i\\}.$

  3. (3)

If $P$ is contained in $\Delta_{ij},$ $\\{i,j\\}=\\{1,2\\},$ $\\{1,3\\}$ or
$\\{1,4\\},$ then $e_{st}$ is positive for $\\{s,t\\}=\\{i,j\\}$ or
$\\{1,2,3,4\\}{\smallsetminus}\\{i,j\\};$ and $e_{st}$ is non-positive for
$\\{s,t\\}\neq\\{i,j\\}$ nor $\\{1,2,3,4\\}{\smallsetminus}\\{i,j\\}.$

###### Proof of Proposition 3.21.

We first consider the special case that all the regular vertices of $\Delta$
are in $\mathbb{H}^{3}_{+}\cup\mathbb{L}^{3}_{+}.$ Then

$\mathrm{Vol}(\Delta)=\sum_{P}\mathrm{Vol}^{P}(\Delta),$ (3.7)

where the sum is over all the geometric pieces of $\Delta,$ and
$\mathrm{Vol}^{P}(\Delta)$ is as defined in Definition 3.17. For each $P,$ let
$e^{P}_{ij}$ be the intersection of $P$ with the geometric edge $e_{ij}$ of
$\Delta,$ which is between the faces $F_{i}$ and $F_{j}.$ Let
$\theta^{P}_{ij}$ be the dihedral angle of $P$ at $e^{P}_{ij}$ considered as a
hyperbolic polyhedron, and defined the length $l^{P}_{ij}$ of $e^{P}_{ij}$ as
follows. If $d^{P}_{ij}$ be the hyperbolic length of $e^{P}_{ij},$ then
$l^{P}_{ij}=d^{P}_{ij}$ if $e_{ij}$ is a positive geometric edge, and
$l^{P}_{ij}=-d^{P}_{ij}$ if $e_{ij}$ is a non-positive geometric edge.

Then for each $\\{i,j\\}\subset\\{1,2,3,4\\},$

$l_{ij}=\sum_{P}l^{P}_{ij},$ (3.8)

where the sum is over all the geometric pieces of $\Delta.$

Next we prove that for each geometric piece $P$ of $\Delta,$

$\frac{\partial\mathrm{Vol}^{P}(\Delta)}{\partial\theta_{ij}}=-\frac{l^{P}_{ij}}{2}.$
(3.9)

Then the result follows from (3.7), (3.8) and (3.9).

To prove (3.9) we have the following three cases:

  1. (1)

If $P$ is contained in $\Delta_{0},$ then by Lemma 3.22, we have

$(\mathrm{Vol}^{P}(\Delta),\theta^{P}_{ij},l^{P}_{ij})=(\mathrm{Vol}(P),\theta_{ij},d^{P}_{ij}),$

and the Schläfli formula for $P$ considered as a hyperbolic polyhedron implies

$\frac{\partial\mathrm{Vol}^{P}(\Delta)}{\partial\theta_{ij}}=\frac{\partial\mathrm{Vol}(P)}{\partial\theta^{P}_{ij}}=-\frac{d^{P}_{ij}}{2}=-\frac{l^{P}_{ij}}{2}.$

  2. (2)

If $P$ is contained in $\Delta_{i},$ $i=1,2,3,4,$ then for
$j\in\\{1,2,3,4\\}{\smallsetminus}\\{i\\},$ by Proposition 3.11 and Lemma 3.22
we have

$(\mathrm{Vol}^{P}(\Delta),\theta^{P}_{ij},l^{P}_{ij})=(-\mathrm{Vol}(P),\pi-\theta_{ij},d^{P}_{ij}),$

and the Schläfli formula for $P$ considered as a hyperbolic polyhedron implies

$\frac{\partial\mathrm{Vol}^{P}(\Delta)}{\partial\theta_{ij}}=\frac{\partial(-\mathrm{Vol}(P))}{\partial(\pi-\theta^{P}_{ij})}=\frac{\partial\mathrm{Vol}(P)}{\partial\theta^{P}_{ij}}=-\frac{d^{P}_{ij}}{2}=-\frac{l^{P}_{ij}}{2};$

and for $\\{j,k\\}\subset\\{1,2,3,4\\}{\smallsetminus}\\{i\\},$ by Proposition
3.11 and Lemma 3.22 we have

$(\mathrm{Vol}^{P}(\Delta),\theta^{P}_{jk},l^{P}_{jk})=(-\mathrm{Vol}(P),\theta_{jk},-d^{P}_{jk}),$

and the Schläfli formula for $P$ considered as a hyperbolic polyhedron implies

$\frac{\partial\mathrm{Vol}^{P}(\Delta)}{\partial\theta_{jk}}=\frac{\partial(-\mathrm{Vol}(P))}{\partial\theta^{P}_{jk}}=\frac{d^{P}_{ij}}{2}=-\frac{l^{P}_{ij}}{2}.$

  3. (3)

If $P$ is contained in $\Delta_{ij},$ $\\{i,j\\}=\\{1,2\\},$ $\\{1,3\\}$ or
$\\{1,4\\},$ then for $\\{s,t\\}=\\{i,j\\}$ or
$\\{1,2,3,4\\}{\smallsetminus}\\{i,j\\},$ by Proposition 3.11 and Lemma 3.22
we have

$(\mathrm{Vol}^{P}(\Delta),\theta^{P}_{st},l^{P}_{st})=(\mathrm{Vol}(P),\theta_{ij},d^{P}_{st}),$

and the Schläfli formula for $P$ considered as a hyperbolic polyhedron implies

$\frac{\partial\mathrm{Vol}^{P}(\Delta)}{\partial\theta_{st}}=\frac{\partial\mathrm{Vol}(P)}{\partial\theta^{P}_{st}}=-\frac{d^{P}_{st}}{2}=-\frac{l^{P}_{st}}{2};$

and $\\{s,t\\}\neq\\{i,j\\}$ nor $\\{1,2,3,4\\}{\smallsetminus}\\{i,j\\},$ by
Proposition 3.11 and Lemma 3.22 we have

$(\mathrm{Vol}^{P}(\Delta),\theta^{P}_{st},l^{P}_{st})=(\mathrm{Vol}(P),\pi-\theta_{st},-d^{P}_{st}),$

and the Schläfli formula for $P$ considered as a hyperbolic polyhedron implies

$\frac{\partial\mathrm{Vol}^{P}(\Delta)}{\partial\theta_{st}}=\frac{\partial\mathrm{Vol}(P)}{\partial(\pi-\theta^{P}_{st})}=\frac{d^{P}_{ij}}{2}=-\frac{l^{P}_{ij}}{2}.$

This completes the proof under the assumption that all the regular vertices of
$\Delta$ are in $\mathbb{H}^{3}_{+}\cup\mathbb{L}^{3}_{+}.$

For the general case, we observe that replacing a regular vertex
$\mathbf{v}_{i}$ by its negative $-\mathbf{v}_{i}$ changes $\Delta_{0}$ to
$\Delta_{i},$ hence switches the roles of
$\Delta_{0}\cup\Delta_{12}\cup\Delta_{13}\cup\Delta_{14}$ and
$\Delta_{1}\cup\Delta_{2}\cup\Delta_{3}\cup\Delta_{4}.$ Then from Definition
3.17, the volume changes sign; and by Proposition 3.11 (1) and Lemma 3.22 (2),
for each edge $e_{st},$ exactly one of $\partial\theta_{st}$ and $l_{st}$
changes sign and the other do not change sign. Then the result follows from
the previous special case and an induction on the number of vertices in in
$\mathbb{H}^{3}_{-}\cup\mathbb{L}^{3}_{-}.$ ∎

##  4 Classification of admissible $6$-tuples

The goal of this section is to understand the geometry of $6$-tuples
$(\alpha_{12},\dots,\alpha_{34})$ satisfying the following _admissibility
conditions_ , which come from sequences of the quantum $6j$-symbols. The main
result of this section is the following Theorem 4.3, which is a refinement of
Theorem 1.2, where the term _generalized Euclidean tetrahedron_ will be
explained in Section 4.1, after which we will prove Theorem 4.3 in Section
4.2. As an application of Theorem 4.3, we prove Theorem 1.4 (1) at the end of
Section 4.2.

######  Definition 4.1 (Admissibility conditions).

A triple of real numbers $(\alpha_{1},\alpha_{2},\alpha_{3})$ is _admissible_
if

  1. (1)

$\alpha_{i}\in[0,2\pi]$ for $i\in\\{1,2,3\\},$

  2. (2)

$\alpha_{i}+\alpha_{j}-\alpha_{k}\geqslant 0$ for $\\{i,j,k\\}=\\{1,2,3\\},$
and

  3. (3)

$\alpha_{1}+\alpha_{2}+\alpha_{3}\leqslant 4\pi.$

A $6$-tuple $(\alpha_{12},\dots,\alpha_{34})$ of real numbers is _admissible_
if for each $i\in\\{1,2,3,4\\},$ the triple
$(\alpha_{jk},\alpha_{jl},\alpha_{kl})$ is admissible, where
$\\{j,k,l\\}=\\{1,2,3,4\\}{\smallsetminus}\\{i\\}.$

We notice that the admissibility conditions are a system of linear
inequalities. To have a clearer picture, we also need to consider the
following strict admissibility conditions where all the inequalities in the
admissibility conditions are required to be strict.

######  Definition 4.2 (Strict admissibility conditions).

A triple of real numbers $(\alpha_{1},\alpha_{2},\alpha_{3})$ is _strictly
admissible_ if

  1. (1)

$\alpha_{i}\in(0,\pi)\cup(\pi,2\pi)$ for $i\in\\{1,2,3\\},$

  2. (2)

$\alpha_{i}+\alpha_{j}-\alpha_{k}>0$ for $\\{i,j,k\\}=\\{1,2,3\\},$ and

  3. (3)

$\alpha_{1}+\alpha_{2}+\alpha_{3}<4\pi.$

A $6$-tuple $(\alpha_{12},\dots,\alpha_{34})$ of real numbers is _strictly
admissible_ if for each $i\in\\{1,2,3,4\\},$ the triple
$(\alpha_{jk},\alpha_{jl},\alpha_{kl})$ is strictly admissible, where
$\\{j,k,l\\}=\\{1,2,3,4\\}{\smallsetminus}\\{i\\}.$

######  Theorem 4.3.

Let $(\alpha_{12},\dots,\alpha_{34})$ be a $6$-tuple real numbers, and for
$\\{i,j\\}\subset\\{1,\dots,4\\},$ let

$\theta_{ij}=|\pi-\alpha_{ij}|.$

  1. (1)

If $(\alpha_{12},\dots,\alpha_{34})$ is strictly admissible, then there are
the following three possibilities:

    1. (a)

$(\theta_{12},\dots,\theta_{34})$ is the set of dihedral angles of a spherical
tetrahedron, in which case its Gram matrix has signature $(4,0),$ i.e., is
positive definite.

    2. (b)

$(\theta_{12},\dots,\theta_{34})$ is the set of dihedral angles of a
generalized Euclidean tetrahedron, in which case its Gram matrix has signature
$(3,0).$

    3. (c)

$(\theta_{12},\dots,\theta_{34})$ is the set of dihedral angles of a
generalized hyperbolic tetrahedron, in which case its Gram matrix has
signature $(3,1).$

  2. (2)

If $(\alpha_{12},\dots,\alpha_{34})$ is admissible, then there is an extra
possibility: $(\theta_{12},\dots,\theta_{34})$ is the set of angles between
four oriented straight lines in the Euclidean plane $\mathbb{E}^{2},$ in which
case its Gram matrix has signature $(2,0)$ or $(1,0).$

The proof of Theorem 4.3 makes an intensive use of the Cauchy Interlace
Theorem, which we recall below. The proof can be found in e.g. [13, p.411],
[14, p.185] or [20, p.186].

######  Theorem 4.4 (Cauchy Interlace Theorem).

Let $A$ be an $n\times n$ Hermitian matrix and let $B$ be an
$(n-1)\times(n-1)$ principal submatrix of $A.$ If
$\lambda_{1}\leqslant\cdots\leqslant\lambda_{n}$ list the eigenvalues of $A$
and $\mu_{1}\leqslant\cdots\leqslant\mu_{n-1}$ list the eigenvalues of $B,$
then

$\lambda_{1}\leqslant\mu_{1}\leqslant\lambda_{2}\leqslant\mu_{2}\leqslant\cdots\leqslant\lambda_{n-1}\leqslant\mu_{n-1}\leqslant\lambda_{n}.$

###  4.1 Generalized Euclidean tetrahedra

######  Definition 4.5 (Generalized Euclidean tetrahedron).

A _generalized Euclidean tetrahedron_ is a quadruple of vectors
$\mathbf{u}_{1},\mathbf{u}_{2},\mathbf{u}_{3},\mathbf{u}_{4}$ in the unit
sphere $\mathbb{S}^{2}$ of the Euclidean space $\mathbb{E}^{3}$ that span
$\mathbb{E}^{3}.$

The _face_ orthogonal to $\mathbf{u}_{i}$ is the plane $F_{i}$ in
$\mathbb{E}^{3}$ tangent to $\mathbb{S}^{2}$ at $\mathbf{u}_{i},$ and
$\mathbf{u}_{i}$ is the _outward normal vector_ of $F_{i}.$ See Figure 12 (a),
(b), (c) for the generic cases and (d), (e), (f) for some non-generic cases.
In particular, in (a), the vectors $\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ are
not contained in any half-space of $\mathbb{E}^{3},$ then the faces
$F_{1},\dots,F_{4}$ bound a Euclidean tetrahedron in the usual sense (the
convex hull of four points in $\mathbb{E}^{3}$ in a general position) with
$\mathbf{u}_{1},\dots,\mathbf{u}_{4}.$

Figure 12: In each of the figures, the disc represents the unit sphere
$\mathbb{S}^{2}$ in $\mathbb{E}^{3}.$ In (a), all the vectors
$\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ point outwards of the Euclidean
tetrahedron, in (b), the vector $\mathbf{u}_{1}$ points inwards the
“tetrahedron” on the top and the vectors
$\mathbf{u}_{2},\mathbf{u}_{3},\mathbf{u}_{4}$ point outwards, and in (c), the
vector $\mathbf{u}_{1},\mathbf{u}_{2}$ point inwards the “tetrahedron” on the
top and the vectors $\mathbf{u}_{3},\mathbf{u}_{4}$ point outwards. In (d) and
(e), the vectors $\mathbf{u}_{1},\mathbf{u}_{2},\mathbf{u}_{3}$ lie in the
same plan of $\mathbb{E}^{3},$ and in (f), the vectors $\mathbf{u}_{1}$ and
$\mathbf{u}_{2}$ coincide. In all these non-generic cases,
$\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ still span $\mathbb{E}^{3}.$

The _dihedral angle_ $\theta_{ij}$ between the faces $F_{i}$ and $F_{j}$ is
defined to be $\pi$ minus the angle between $\mathbf{u}_{i}$ and
$\mathbf{u}_{j},$ i.e.,

$\theta_{ij}=\pi-\cos^{-1}\langle\mathbf{u}_{i},\mathbf{u}_{j}\rangle,$

where $\langle,\rangle$ here is the standard inner product on
$\mathbb{E}^{3}.$

######  Theorem 4.6.

Suppose $(\theta_{12},\dots,\theta_{34})$ is a $6$-tuple of numbers in
$[0,\pi].$ Then the following statements are equivalent.

  1. (1)

$(\theta_{12},\dots,\theta_{34})$ is the set of dihedral angles of a
generalized Euclidean tetrahedron.

  2. (2)

The Gram matrix $G$ of $(\theta_{12},\dots,\theta_{34})$ has signature
$(3,0).$

###### Proof.

The proof follows the idea of that of Luo [17, Theorem].

Suppose (1) holds. Let $\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ be the vectors in
$\mathbb{S}^{2}$ defining the generalized Euclidean tetrahedron. Since they
span $\mathbb{E}^{3},$ there are three of them, say,
$\mathbf{u}_{1},\mathbf{u}_{2},\mathbf{u}_{3}$ forming a basis of
$\mathbb{E}^{3}.$ Let

$U=[\mathbf{u}_{1},\mathbf{u}_{2},\mathbf{u}_{3}].$

Then

$G_{1}=U^{T}\cdot U,$

where $G_{1}$ is the first $3\times 3$ principal submatrix $G.$ Since the tips
of $\mathbf{u}_{1},\mathbf{u}_{2},\mathbf{u}_{3}$ are the vertices of a
spherical triangle, $G_{1}$ is positive definite. As a consequence of the
Cauchy Interlace Theorem, $G$ has at least three positive eigenvalues.

Next we prove that $\det G=0.$ Indeed, since
$\mathbf{u}_{1},\mathbf{u}_{2},\mathbf{u}_{3}$ form a basis of
$\mathbb{E}^{3},$

$\mathbf{u}_{4}=a_{1}\mathbf{u}_{1}+a_{2}\mathbf{u}_{2}+a_{3}\mathbf{u}_{3}$

for some real numbers $a_{1},a_{2},a_{3}.$ If we let $\mathbf{g}_{i}$ be the
$i$-th column of $G,$ then

$\mathbf{g}_{4}=a_{1}\mathbf{g}_{1}+a_{2}\mathbf{g}_{2}+a_{3}\mathbf{g}_{3},$

and as a consequence, $\det G=0.$ Now $G$ has at least three positive
eigenvalues and $\det G=0,$ hence the signature equals $(3,0),$ and (2) holds.

Suppose (2) holds, so that $G$ has signature $(3,0).$ Then by Sylvester’s Law
of Inertia,

$G=W^{T}\cdot I_{3,0}\cdot W$

for some $4\times 4$ matrix $W,$ where $I_{3,0}$ is the matrix

$I_{3,0}=\left[\begin{matrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&0\\\
\end{matrix}\right].$

Let $\mathbf{w}_{1},\dots,\mathbf{w}_{4}$ be the columns of $W,$ and for each
$i\in\\{1,\dots,4\\},$ let $\mathbf{u}_{i}$ be the vector in $\mathbb{E}^{3}$
obtained from $\mathbf{w}_{i}$ by erasing the last component. If
$\langle,\rangle$ denotes the standard inner product on $\mathbb{E}^{3},$ then
we have

$\langle\mathbf{u}_{i},\mathbf{u}_{i}\rangle=\mathbf{w}_{i}^{T}\cdot
I_{3,0}\cdot\mathbf{w}_{i}=1$

for $i\in\\{1,2,3,4\\},$ and

$\langle\mathbf{u}_{i},\mathbf{u}_{j}\rangle=\mathbf{w}_{i}^{T}\cdot
I_{3,0}\cdot\mathbf{w}_{j}=-\cos\theta_{ij}$ (4.1)

for $\\{i,j\\}\subset\\{1,2,3,4\\}.$ In particular,

$\mathbf{u}_{i}\in\mathbb{S}^{2}$

for each $i\in\\{1,2,3,4\\}.$

Next we will show that $\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ span
$\mathbb{E}^{3}$; we proceed by contradiction. Suppose they do not span; then
$\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ lie in a subspace of $\mathbb{E}^{3}$
spanned by two vectors, say, $\mathbf{u}_{1}$ and $\mathbf{u}_{2}.$ Then

$\mathbf{u}_{3}=a_{1}\mathbf{u}_{1}+a_{2}\mathbf{u}_{2}\quad\text{and}\quad\mathbf{u}_{4}=b_{1}\mathbf{u}_{1}+b_{2}\mathbf{u}_{2}$

for some real numbers $a_{1},a_{2},b_{1},b_{2}.$ As a consequence, the columns
$\mathbf{g}_{1},\dots,\mathbf{g}_{4}$ of $G$ satisfy

$\mathbf{g}_{3}=a_{1}\mathbf{g}_{1}+a_{2}\mathbf{g}_{2}\quad\text{and}\quad\mathbf{g}_{4}=b_{1}\mathbf{g}_{1}+b_{2}\mathbf{g}_{2}.$

This implies that the rank of $G$ is at most $2,$ contradicting the assumption
that $G$ has signature $(3,0).$ Therefore,
$\mathbf{u}_{1},\dots,\mathbf{u}_{4}$ are four vectors in $\mathbb{S}^{2}$
spanning $\mathbb{E}^{3},$ hence define a generalized Euclidean tetrahedron
$\Delta,$ and (4.1) shows that the dihedral angles of $\Delta$ are
$\theta_{12},\dots,\theta_{34}.$ ∎

######  Theorem 4.7.

Let $(\theta_{12},\dots,\theta_{34})$ be a $6$-tuple of numbers in $[0,\pi],$
and let $G$ be its Gram matrix. If the signature of $G$ is $(2,0)$ or $(1,0),$
then $(\theta_{12},\dots,\theta_{34})$ is the set of angles between four
oriented straight lines in the Euclidean plane $\mathbb{E}^{2}.$ Moreover, the
signature of $G$ is $(1,0),$ then the four straight lines are parallel.

######  Remark 4.8.

Here the orientation of a straight line is defined by a specification of its
normal vector, and the angle between two straight lines is $\pi$ minus the
angle between the two normal vectors that define the orientation of the
straight lines; See Figure 13.

Figure 13: In the figure, the circle represents the unit circle
$\mathbb{S}^{1}$ in $\mathbb{E}^{2}.$

###### Proof.

Suppose $G$ has signature $(2,0),$ then by Sylvester’s Law of Inertia,

$G=W^{T}\cdot I_{2,0}\cdot W$

for some $4\times 4$ matrix $W,$ where $I_{2,0}$ is the matrix

$I_{2,0}=\left[\begin{matrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&0&0\\\ 0&0&0&0\\\
\end{matrix}\right].$

Let $\mathbf{w}_{1},\dots,\mathbf{w}_{4}$ be the columns of $W,$ and for each
$i\in\\{1,\dots,4\\},$ let $\mathbf{u}_{i}$ be the vector in $\mathbb{E}^{2}$
obtained from $\mathbf{w}_{i}$ by erasing the last two components. If
$\langle,\rangle$ denotes the standard inner product on $\mathbb{E}^{2},$ then
we have

$\langle\mathbf{u}_{i},\mathbf{u}_{j}\rangle=\mathbf{w}_{i}^{T}\cdot
I_{3,0}\cdot\mathbf{w}_{j}=-\cos\theta_{ij}$

for $\\{i,j\\}\subset\\{1,2,3,4\\}.$ Let $L_{i}$ be the straight line in
$\mathbb{E}^{2}$ tangent to the unit circle $\mathbb{S}^{1}$ at
$\mathbf{u}_{i},$ oriented as the direction of $\mathbf{u}_{i}.$ Then the
angle between $L_{i}$ and $L_{j}$ is
$\pi-\cos^{-1}\langle\mathbf{u}_{i},\mathbf{u}_{j}\rangle=\theta_{ij},$ and
$L_{1}\dots,L_{4}$ with this orientation are the desired oriented straight
lines.

Suppose $G$ has signature $(1,0),$ then by Sylvester’s Law of Inertia,

$G=W^{T}\cdot I_{2,0}\cdot W$

for some $4\times 4$ matrix $W,$ where $I_{1,0}$ is the matrix

$I_{1,0}=\left[\begin{matrix}1&0&0&0\\\ 0&0&0&0\\\ 0&0&0&0\\\ 0&0&0&0\\\
\end{matrix}\right].$

Let $\mathbf{w}_{1},\dots,\mathbf{w}_{4}$ be the columns of $W,$ and for each
$i\in\\{1,\dots,4\\},$ let $a_{i}$ be the first entry of $\mathbf{w}_{i}$ and
let

$\mathbf{u}_{i}=\left[\begin{matrix}a_{i}\\\ 0\\\ \end{matrix}\right].$

If $\langle,\rangle$ denotes the standard inner product on $\mathbb{E}^{2},$
then for each $i\in\\{1,2,3,4\\},$ we have

$a_{i}^{2}=\langle\mathbf{u}_{i},\mathbf{u}_{i}\rangle=\mathbf{w}_{i}^{T}\cdot
I_{3,0}\cdot\mathbf{w}_{i}=1,$

and for each $\\{i,j\\}\subset\\{1,2,3,4\\},$ we have

$a_{i}a_{j}=\langle\mathbf{u}_{i},\mathbf{u}_{j}\rangle=\mathbf{w}_{i}^{T}\cdot
I_{3,0}\cdot\mathbf{w}_{j}=-\cos\theta_{ij}.$

As a consequence, $a_{i}=\pm 1$ and
$\mathbf{u}_{i}=\pm\left[\begin{matrix}1\\\ 0\\\ \end{matrix}\right]$ for each
$i.$ Then the desired parallel oriented straight lines are the vertical lines
passing through the points $\pm(1,0).$ ∎

###  4.2 A proof of Theorem 4.3 and Theorem 1.4 (1)

The key ingredient in the proof of Theorem 4.3 is the following Lemma.

######  Lemma 4.9.

Let $(\alpha_{1},\alpha_{2},\alpha_{3})$ be triple of real numbers. For
$i\in\\{1,2,3\\},$ let

$\theta_{i}=|\pi-\alpha_{i}|,$

and let

$G=\left[\begin{matrix}1&-\cos\theta_{1}&-\cos\theta_{2}\\\
-\cos\theta_{1}&1&-\cos\theta_{3}\\\ -\cos\theta_{2}&-\cos\theta_{3}&1\\\
\end{matrix}\right].$

  1. (1)

If $(\alpha_{1},\alpha_{2},\alpha_{3})$ is strictly admissible and $\det
G\leqslant 0,$ then

$\theta_{1}+\theta_{2}+\theta_{3}\leqslant\pi.$

  2. (2)

If $(\alpha_{1},\alpha_{2},\alpha_{3})$ is admissible and $\det G<0,$ then

$\theta_{1}+\theta_{2}+\theta_{3}<\pi.$

###### Proof.

For (1), we first observe that the principal minors are strictly positive due
to the strict admissibility condition (1), thus $\det G$ is positive if and
only if $G$ is positive definite. The latter is equivalent is equivalent to
$(\theta_{1},\theta_{2},\theta_{3})$ being the set of dihedral angles of a
spherical triangle, which holds if and only if
$(\theta_{1},\theta_{2},\theta_{3})$ satisfies the system of the following
four equations

$\left\\{\begin{array}[]{c}\theta_{1}+\theta_{2}+\theta_{3}>\pi\\\
(\pi-\theta_{1})+(\pi-\theta_{2})+\theta_{3}>\pi\\\
(\pi-\theta_{1})+\theta_{2}+(\pi-\theta_{3})>\pi\\\
\theta_{1}+(\pi-\theta_{2})+(\pi-\theta_{3})>\pi\\\ \end{array}.\right.$ (4.2)

Therefore, if $\det G\leqslant 0,$ then one of the following four inequalities
is satisfied.

$\theta_{1}+\theta_{2}+\theta_{3}\leqslant\pi,$ (4.3)
$(\pi-\theta_{1})+(\pi-\theta_{2})+\theta_{3}\leqslant\pi,$ (4.4)
$(\pi-\theta_{1})+\theta_{2}+(\pi-\theta_{3})\leqslant\pi,$ (4.5)
$\theta_{1}+(\pi-\theta_{2})+(\pi-\theta_{3})\leqslant\pi.$ (4.6)

In the rest of the proof, we will show that (4.4) is impossible under the
admissibility conditions. Then by similar reasons, we can rule out (4.5) and
(4.6), and leaves (4.3) as the only possibility. We consider the following
cases.

  1. (a)

$\alpha_{1}<\pi,$ $\alpha_{2}<\pi$ and $\alpha_{3}<\pi.$ Then
$\theta_{1}=\pi-\alpha_{1},$ $\theta_{2}=\pi-\alpha_{2}$ and
$\theta_{3}=\pi-\alpha_{3},$ and (4.4) is equivalent to

$\alpha_{1}+\alpha_{2}-\alpha_{3}\leqslant 0,$

which contradicts the strict admissibility condition
$\alpha_{1}+\alpha_{2}-\alpha_{3}>0.$

  2. (b)

$\alpha_{1}>\pi,$ $\alpha_{2}<\pi$ and $\alpha_{3}<\pi.$ (The case
$\alpha_{2}>\pi,$ $\alpha_{1}<\pi$ and $\alpha_{3}<\pi$ is similar by
symmetry.) Then $\theta_{1}=\alpha_{1}-\pi,$ $\theta_{2}=\pi-\alpha_{2}$ and
$\theta_{3}=\pi-\alpha_{3},$ and (4.4) is equivalent to

$\alpha_{1}-\alpha_{2}+\alpha_{3}\geqslant 2\pi.$

Then we have

$\alpha_{1}-\alpha_{2}-\alpha_{3}=(\alpha_{1}-\alpha_{2}+\alpha_{3})-2\alpha_{3}\geqslant
2\pi-2\pi=0,$

which contradicts the strict admissibility condition
$\alpha_{2}+\alpha_{3}-\alpha_{1}>0.$

  3. (c)

$\alpha_{1}<\pi,$ $\alpha_{2}<\pi$ and $\alpha_{3}>\pi.$ Then
$\theta_{1}=\pi-\alpha_{1},$ $\theta_{2}=\pi-\alpha_{2}$ and
$\theta_{3}=\alpha_{3}-\pi,$ and (4.4) is equivalent to

$\alpha_{1}+\alpha_{2}+\alpha_{3}\leqslant 2\pi.$

Then we have

$\alpha_{1}+\alpha_{2}-\alpha_{3}=(\alpha_{1}+\alpha_{2}+\alpha_{3})-2\alpha_{3}\leqslant
2\pi-2\pi=0,$

which contradicts the strict admissibility condition
$\alpha_{1}+\alpha_{2}-\alpha_{3}>0.$

  4. (d)

$\alpha_{1}>\pi,$ $\alpha_{2}>\pi$ and $\alpha_{3}<\pi.$ Then
$\theta_{1}=\alpha_{1}-\pi,$ $\theta_{2}=\alpha_{2}-\pi$ and
$\theta_{3}=\pi-\alpha_{3},$ and (4.4) is equivalent to

$\alpha_{1}+\alpha_{2}+\alpha_{3}\geqslant 4\pi,$

which contradicts the strict admissibility condition
$\alpha_{1}+\alpha_{2}+\alpha_{3}<4\pi.$

  5. (e)

$\alpha_{1}>\pi,$ $\alpha_{2}<\pi$ and $\alpha_{3}>\pi.$ (The case
$\alpha_{2}>\pi,$ $\alpha_{1}<\pi$ and $\alpha_{3}>\pi$ is similar by
symmetry.) Then $\theta_{1}=\alpha_{1}-\pi,$ $\theta_{2}=\pi-\alpha_{2}$ and
$\theta_{3}=\alpha_{3}-\pi,$ and (4.4) is equivalent to

$\alpha_{2}+\alpha_{3}-\alpha_{1}\leqslant 0,$

which contradicts the strict admissibility condition
$\alpha_{2}+\alpha_{3}-\alpha_{1}>0.$

  6. (f)

$\alpha_{1}>\pi,$ $\alpha_{2}>\pi$ and $\alpha_{3}>\pi.$ Then
$\theta_{1}=\alpha_{1}-\pi,$ $\theta_{2}=\alpha_{2}-\pi$ and
$\theta_{3}=\alpha_{3}-\pi,$ and (4.4) is equivalent to

$\alpha_{1}+\alpha_{2}-\alpha_{3}\geqslant 2\pi.$

Then we have

$\alpha_{1}+\alpha_{2}+\alpha_{3}=(\alpha_{1}+\alpha_{2}-\alpha_{3})+2\alpha_{3}\geqslant
2\pi+2\pi=4\pi,$

which contradicts the strict admissibility condition
$\alpha_{1}+\alpha_{2}+\alpha_{3}<4\pi.$

For (2), by the discussion in the beginning of the proof of Part (1), we see
that if $\theta_{1},$ $\theta_{2},$ $\theta_{3}$ are numbers in $(0,\pi),$
then $\det G>0$ if and only if the system of equations (4.7) hold. Therefore,
for $\theta_{1},$ $\theta_{2},$ $\theta_{3}$ in $[0,\pi],$ if the following
system of equalities

$\left\\{\begin{array}[]{c}\theta_{1}+\theta_{2}+\theta_{3}\geqslant\pi\\\
(\pi-\theta_{1})+(\pi-\theta_{2})+\theta_{3}\geqslant\pi\\\
(\pi-\theta_{1})+\theta_{2}+(\pi-\theta_{3})\geqslant\pi\\\
\theta_{1}+(\pi-\theta_{2})+(\pi-\theta_{3})\geqslant\pi\\\
\end{array}\right.$ (4.7)

hold, then $\det G\geqslant 0.$ As a consequence, if $\det G<0,$ then one of
the following four mutually exclusive strict inequalities is satisfied.

$\theta_{1}+\theta_{2}+\theta_{3}<\pi,$ (4.8)
$(\pi-\theta_{1})+(\pi-\theta_{2})+\theta_{3}<\pi,$ (4.9)
$(\pi-\theta_{1})+\theta_{2}+(\pi-\theta_{3})<\pi,$ (4.10)
$\theta_{1}+(\pi-\theta_{2})+(\pi-\theta_{3})<\pi.$ (4.11)

Similar to the proof of Part (1), we will show that (4.9) is impossible under
the admissibility conditions. Then by similar reasons, we can also rule out
(4.10) and (4.11), and leaves (4.8) as the only possibility. The argument is
very similar to that of Part (1), and we include the details for the readers’
convenience. We consider the following cases.

  1. (a)

$\alpha_{1}\leqslant\pi,$ $\alpha_{2}\leqslant\pi$ and
$\alpha_{3}\leqslant\pi.$ Then $\theta_{1}=\pi-\alpha_{1},$
$\theta_{2}=\pi-\alpha_{2}$ and $\theta_{3}=\pi-\alpha_{3},$ and (4.9) is
equivalent to

$\alpha_{1}+\alpha_{2}-\alpha_{3}<0,$

which contradicts the admissibility condition
$\alpha_{1}+\alpha_{2}-\alpha_{3}\geqslant 0.$

  2. (b)

$\alpha_{1}>\pi,$ $\alpha_{2}\leqslant\pi$ and $\alpha_{3}\leqslant\pi.$ (The
case $\alpha_{1}\leqslant\pi,$ $\alpha_{2}>\pi$ and $\alpha_{3}\leqslant\pi$
is similar by symmetry.) Then $\theta_{1}=\alpha_{1}-\pi,$
$\theta_{2}=\pi-\alpha_{2}$ and $\theta_{3}=\pi-\alpha_{3},$ and (4.9) is
equivalent to

$\alpha_{1}-\alpha_{2}+\alpha_{3}>2\pi.$

Then we have

$\alpha_{1}-\alpha_{2}-\alpha_{3}=(\alpha_{1}-\alpha_{2}+\alpha_{3})-2\alpha_{3}>2\pi-2\pi=0,$

which contradicts the admissibility condition
$\alpha_{2}+\alpha_{3}-\alpha_{1}\geqslant 0.$

  3. (c)

$\alpha_{1}\leqslant\pi,$ $\alpha_{2}\leqslant\pi$ and $\alpha_{3}>\pi.$ Then
$\theta_{1}=\pi-\alpha_{1},$ $\theta_{2}=\pi-\alpha_{2}$ and
$\theta_{3}=\alpha_{3}-\pi,$ and (4.9) is equivalent to

$\alpha_{1}+\alpha_{2}+\alpha_{3}<2\pi.$

Then we have

$\alpha_{1}+\alpha_{2}-\alpha_{3}=(\alpha_{1}+\alpha_{2}+\alpha_{3})-2\alpha_{3}<2\pi-2\pi=0,$

which contradicts the admissibility condition
$\alpha_{1}+\alpha_{2}-\alpha_{3}\geqslant 0.$

  4. (d)

$\alpha_{1}>\pi,$ $\alpha_{2}>\pi$ and $\alpha_{3}\leqslant\pi.$ Then
$\theta_{1}=\alpha_{1}-\pi,$ $\theta_{2}=\alpha_{2}-\pi$ and
$\theta_{3}=\pi-\alpha_{3},$ and (4.9) is equivalent to

$\alpha_{1}+\alpha_{2}+\alpha_{3}>4\pi,$

which contradicts the admissibility condition
$\alpha_{1}+\alpha_{2}+\alpha_{3}\leqslant 4\pi.$

  5. (e)

$\alpha_{1}>\pi,$ $\alpha_{2}\leqslant\pi$ and $\alpha_{3}>\pi.$ (The case
$\alpha_{1}\leqslant\pi,$ $\alpha_{2}>\pi$ and $\alpha_{3}>\pi$ is similar by
symmetry.) Then $\theta_{1}=\alpha_{1}-\pi,$ $\theta_{2}=\pi-\alpha_{2}$ and
$\theta_{3}=\alpha_{3}-\pi,$ and (4.9) is equivalent to

$\alpha_{2}+\alpha_{3}-\alpha_{1}<0,$

which contradicts the admissibility condition
$\alpha_{2}+\alpha_{3}-\alpha_{1}\geqslant 0.$

  6. (f)

$\alpha_{1}>\pi,$ $\alpha_{2}>\pi$ and $\alpha_{3}>\pi.$ Then
$\theta_{1}=\alpha_{1}-\pi,$ $\theta_{2}=\alpha_{2}-\pi$ and
$\theta_{3}=\alpha_{3}-\pi,$ and (4.9) is equivalent to

$\alpha_{1}+\alpha_{2}-\alpha_{3}>2\pi.$

Then we have

$\alpha_{1}+\alpha_{2}+\alpha_{3}=(\alpha_{1}+\alpha_{2}-\alpha_{3})+2\alpha_{3}>2\pi+2\pi=4\pi,$

which contradicts the admissibility condition
$\alpha_{1}+\alpha_{2}+\alpha_{3}\leqslant 4\pi.$

∎

###### Proof of Theorem 4.3.

For (1), due to the strict admissibility condition (1),
$\cos\theta_{ij}\neq\pm 1$ and hence all the $2\times 2$ principal submatrices
of $G$ are positive definite. Then by the Cauchy Interlace Theorem, $G$ has at
least two positive eigenvalues. We consider the following cases:

  1. (a)

If for at least one $i\in\\{1,2,3,4\\},$ $G_{ii}>0,$ then due to the strict
admissibility condition that $\cos\theta_{jk}\neq\pm 1,$ the $i$-th $3\times
3$ principal submatrix is positive definite, and hence all its eigenvalues are
positive. As a consequence of the Cauchy Interlace Theorem, $G$ has at least
three positive eigenvalues hence the signature is $(4,0),$ $(3,0)$ or $(3,1).$

  2. (b)

If for all $i\in\\{1,2,3,4\\},$ $G_{ii}<0,$ then by Lemma 4.9 (2) and Bonahon-
Bao [1], $(\theta_{12},\dots,\theta_{34})$ is the set of dihedral angles of a
hyperideal tetrahedron, and $G$ has signature $(3,1).$

  3. (c)

If some $G_{ii}=0,$ then $(\theta_{12},\dots,\theta_{34})$ lies in the closure
of the region given by cases of (a) and (b), hence besides $(4,0),$ $(3,0)$ or
$(3,1),$ the signature of $G$ can have extra possibilities $(2,0)$ and
$(2,1).$

Next we rule out the extra possibilities as follows. By Theorem 4.7 and
Theorem 3.9, $(\theta_{12},\dots,\theta_{34})$ is the set of angles between
four straight lines in $\mathbb{E}^{2}$ or four intersecting geodesics in
$\overline{\mathbb{H}^{2}}.$ This implies that $G_{ii}\leqslant 0$ for all
$i\in\\{1,2,3,4\\},$ and from Figure 13 and Figure 2 we see that there is
always a triple of angles (e.g. $(\theta_{12},\theta_{13},\theta_{23})$ )
satisfying one of the inequalities (4.4), (4.5) and (4.6), which by Lemma 4.9
(1) is impossible.

For (2), $(\theta_{12},\dots,\theta_{34})$ lies in the closure of the region
given by cases of (a) and (b), hence besides $(4,0),$ $(3,0)$ or $(3,1),$ the
signature of $G$ can have extra possibilities $(2,0),$ $(1,0),$ $(2,1)$ and
$(1,1).$

By Theorem 4.7, if the signature of $G$ is $(2,0)$ or $(1,0),$ then
$(\theta_{12},\dots,\theta_{34})$ is the set of angles between four straight
lines in $\mathbb{E}^{2},$ which is what Part (2) claims.

Next we rule out the other two possibilities that $G$ has signature $(2,1)$
and $(1,1).$

By Theorem 3.9 and Lemma 4.9 (2), if $G$ has signature $(2,1),$ then
$(\theta_{12},\dots,\theta_{34})$ is the set of angles between four
intersecting geodesics in $\overline{\mathbb{H}^{2}}.$ This implies that
$G_{ii}\leqslant 0$ for all $i\in\\{1,2,3,4\\},$ and from Figure 2 we see
there is always a triple of angles satisfying one of the inequalities (4.9),
(4.10) and (4.11), which by Lemma 4.9 (2) is impossible.

If $G$ has signature $(1,1),$ then $\theta_{ij}=0$ or $\pi$ for all
$\\{i,j\\}\subset\\{1,2,3,4\\},$ because otherwise some $2\times 2$ submatrix
of $G$ would be positive definite and by the Cauchy Interlace Theorem $G$
would have at least two positive eigenvalues, whose signature cannot be
$(1,1).$ Then a case by case computation shows that the signature of $G$ can
only be $(3,1),$ $(2,1)$ or $(1,0).$ Indeed, by Lemma 3.12, by a sequence of
change of angles operations we only need to compute the following three cases:

  1. (1)

$(0,0,0,0,0,0)$ whose Gram matrix has signature $(3,1).$ This is the set of
dihedral angles of the hyperbolic tetrahedron whose geometric piece is the
regular ideal octahedron.

  2. (2)

$(\pi,0,0,0,0,0)$ whose Gram matrix has signature $(2,1).$ This is the set of
angles between four geodesics $L_{1},$ $L_{2}$ $L_{3}$ and $L_{4}$ in
$\overline{\mathbb{H}^{2}}$ such that $L_{1}$ and $L_{2}$ coincide and
$L_{2},$ $L_{3}$ and $L_{4}$ are the edges of an ideal hyperbolic triangle.

  3. (3)

$(\pi,0,0,0,0,\pi)$ whose Gram matrix has signature $(1,0).$ This is the set
of angles between four parallel straight lines $L_{1},$ $L_{2}$ $L_{3}$ and
$L_{4}$ in $\mathbb{E}^{2},$ where $L_{1}$ and $L_{2}$ have the same
orientation which is opposite to that of $L_{3}$ and $L_{4}.$

Therefore, $G$ cannot have signature $(1,1).$ ∎

###### Proof of Theorem 1.4 (1).

By Theorems 4.3, 3.7, 4.6 and 4.7, the signature of $G$ can only be $(4,0),$
$(3,0),$ $(2,0),$ $(1,0)$ or $(3,1).$ By Cauchy Interlace Theorem, if
$G_{ii}<0$ for some $i\in\\{1,2,3,4\\},$ then there must be at least one
negative eigenvalue. As a consequence, the only possibility is $(3,1),$ and by
Theorem 3.7, $(\theta_{1},\dots,\theta_{6})$ is the set of dihedral angles of
a generalized hyperbolic tetrahedron. ∎

######  Proposition 4.10.

The space of $(\alpha_{1},\dots,\alpha_{6})$ satisfying the conditions of
Theorem 1.4 is connected.

###### Proof.

By Theorem 1.4, $(\theta_{1},\dots,\theta_{6})$ is the set of dihedral angles
of a generalized hyperbolic tetrahedron $\Delta$ with at least one hyperideal
vertex. Then by Lemma 4.9 (2), the only possibilities are those in Proposition
3.15 (2a), (2b), (2c), (2d), (3a), (3b), (3c), (3d), (4a), (4b) and (5a). We
observe that in any of these cases, the negative edges always connect a
regular or ideal vertex with a hyperideal vertex. Then along each negative
edge, by pushing the regular or ideal vertex away from the hyperideal one, one
deforms $\Delta$ into cases Proposition 3.15 (2a), (3a) (4a) and (5a). By
further pushing all the regular and ideal vertices out of
$\overline{\mathbb{H}^{3}},$ we deform $\Delta$ into case (5a), a hyperbolic
tetrahedron with four hyperideal vertices. By Bonahon-Bao [1], the space of
such hyperbolic tetrahedra is connected. This completes the proof. ∎

##  5 A volume formula

The main results of this section are Theorem 5.1, where we obtain a formula
for the volume of a generalized hyperbolic tetrahedron in terms of the
dihedral angles. The formula is in the same spirit of Murakami-Yano [18] and
Ushijima [24], and essentially coincides with [18, Theorem 2] in the case of
hyperbolic tetrahedra with only regular vertices. In Section 6, by studying
the asymptotics of quantum $6j$-symbols, we obtain a simpler presentation of
the volume formula when at least one vertex of the generalized hyperbolic
tetrahedron is hyperideal. See Theorem 6.5. It worth mentioning that a
different volume formula was also obtained in Sohn [21] and Bonahon-Sohn [5]
for generalized hyperbolic tetrahedra, which also works for the cases with
deep truncations.

Our formula is closely related to the critical values of a function defined
using the dilogarithm function. Let
$\log:\mathbb{C}{\smallsetminus}(-\infty,0]\to\mathbb{C}$ be the standard
logarithm function defined by

$\log z=\log|z|+\sqrt{-1}\cdot\arg z$

with $-\pi<\arg z<\pi.$ The dilogarithm function
$\mathrm{Li}_{2}:\mathbb{C}{\smallsetminus}(1,\infty)\to\mathbb{C}$ is defined
by

$\mathrm{Li}_{2}(z)=-\int_{0}^{z}\frac{\log(1-u)}{u}du$

where the integral is along any path in $\mathbb{C}{\smallsetminus}(1,\infty)$
connecting $0$ and $z,$ which is holomorphic in
$\mathbb{C}{\smallsetminus}[1,\infty)$ and continuous in
$\mathbb{C}{\smallsetminus}(1,\infty).$ The dilogarithm function satisfies the
following property (see eg. Zagier [25]). On the unit circle
$\big{\\{}z=e^{2\sqrt{-1}\theta}\,\big{|}\,0\leqslant\theta\leqslant\pi\big{\\}},$

$\mathrm{Li}_{2}(e^{2\sqrt{-1}\theta})=\frac{\pi^{2}}{6}+\theta(\theta-\pi)+2\sqrt{-1}\cdot\Lambda(\theta).$
(5.1)

Here $\Lambda:\mathbb{R}\to\mathbb{R}$ is the Lobachevsky function defined by

$\Lambda(\theta)=-\int_{0}^{\theta}\log|2\sin t|dt,$

which is an odd function of period $\pi$ (see eg. Thurston’s notes [22,
Chapter 7]).

Suppose $(\theta_{1},\dots,\theta_{6})$ is the set of dihedral angles of a
generalized truncated hyperbolic tetrahedron $\Delta,$
$\alpha_{i}=\pi\pm\theta_{i}$ for $i\in\\{1,\dots,6\\}$ and
$\bm{\alpha}=(\alpha_{1},\dots,\alpha_{6}).$ Let

$U(\bm{\alpha},\xi)=-\frac{1}{2}\sum_{i=1}^{4}\sum_{j=1}^{3}\mathrm{L}(\eta_{j}-\tau_{i})+\frac{1}{2}\sum_{i=1}^{4}\mathrm{L}(\tau_{i}-\pi)-\mathrm{L}(\xi-\pi)+\sum_{i=1}^{4}\mathrm{L}(\xi-\tau_{i})+\sum_{j=1}^{3}\mathrm{L}(\eta_{j}-\xi),$
(5.2)

where $\mathrm{L}:\mathbb{R}\to\mathbb{C}$ is the function defined by

$\mathrm{L}(x)=\mathit{Li}\big{(}e^{2\sqrt{-1}x}\big{)}+x^{2},$

and

$\tau_{1}=\frac{\alpha_{1}+\alpha_{2}+\alpha_{3}}{2},\quad\tau_{2}=\frac{\alpha_{1}+\alpha_{5}+\alpha_{6}}{2},\quad\tau_{3}=\frac{\alpha_{2}+\alpha_{4}+\alpha_{6}}{2},\quad\tau_{4}=\frac{\alpha_{3}+\alpha_{4}+\alpha_{5}}{2},$
$\eta_{1}=\frac{\alpha_{1}+\alpha_{2}+\alpha_{4}+\alpha_{5}}{2},\quad\eta_{2}=\frac{\alpha_{1}+\alpha_{3}+\alpha_{4}+\alpha_{6}}{2},\quad\eta_{3}=\frac{\alpha_{2}+\alpha_{3}+\alpha_{5}+\alpha_{6}}{2}.$

Then a direct computation shows that

$\frac{\partial
U(\bm{\alpha},\xi)}{\partial\xi}=2\sqrt{-1}\cdot\log\frac{(1-z)(1-zu_{1}u_{2}u_{4}u_{5})(1-zu_{1}u_{3}u_{4}u_{6})(1-zu_{2}u_{3}u_{5}u_{6})}{(1-zu_{1}u_{2}u_{3})(1-zu_{1}u_{5}u_{6})(1-zu_{2}u_{4}u_{6})(1-zu_{3}u_{4}u_{5})}\quad\quad(\mathrm{mod}\
4\pi)$

where $u_{i}=e^{\sqrt{-1}\alpha_{i}}$ for $i\in\\{1,\dots,6\\}$ and
$z=e^{-2\sqrt{-1}\xi};$ and the equation

$\frac{\partial U(\bm{\alpha},\xi)}{\partial\xi}=0\quad\quad(\mathrm{mod}\
4\pi)$

is equivalent to

$\frac{(1-z)(1-zu_{1}u_{2}u_{4}u_{5})(1-zu_{1}u_{3}u_{4}u_{6})(1-zu_{2}u_{3}u_{5}u_{6})}{(1-zu_{1}u_{2}u_{3})(1-zu_{1}u_{5}u_{6})(1-zu_{2}u_{4}u_{6})(1-zu_{3}u_{4}u_{5})}=1,$

which simplifies to the following quadratic equation

$Az^{2}+Bz+C=0,$ (5.3)

where

$\begin{split}A=&u_{1}u_{4}+u_{2}u_{5}+u_{3}u_{6}-u_{1}u_{2}u_{6}-u_{1}u_{3}u_{5}-u_{2}u_{3}u_{4}-u_{4}u_{5}u_{6}+u_{1}u_{2}u_{3}u_{4}u_{5}u_{6},\\\
B=&-\Big{(}u_{1}-\frac{1}{u_{1}}\Big{)}\Big{(}u_{4}-\frac{1}{u_{4}}\Big{)}-\Big{(}u_{2}-\frac{1}{u_{2}}\Big{)}\Big{(}u_{5}-\frac{1}{u_{5}}\Big{)}-\Big{(}u_{3}-\frac{1}{u_{3}}\Big{)}\Big{(}u_{6}-\frac{1}{u_{6}}\Big{)},\\\
C=&\frac{1}{u_{1}u_{4}}+\frac{1}{u_{2}u_{5}}+\frac{1}{u_{3}u_{6}}-\frac{1}{u_{1}u_{2}u_{6}}-\frac{1}{u_{1}u_{3}u_{5}}-\frac{1}{u_{2}u_{3}u_{4}}-\frac{1}{u_{4}u_{5}u_{6}}+\frac{1}{u_{1}u_{2}u_{3}u_{4}u_{5}u_{6}}.\end{split}$

Let

$z=\frac{-B+\sqrt{B^{2}-4AC}}{2A}\quad\text{and}\quad
z^{*}=\frac{-B-\sqrt{B^{2}-4AC}}{2A}$ (5.4)

be the two solutions of (5.3). Here as a convention, we let
$\sqrt{x}=\sqrt{-1}\sqrt{|x|}$ if $x$ is a negative real number. Then by a
direct computation (see also [18, 24]), we have

$B^{2}-4AC=16\det G,$ (5.5)

where $G$ is the Gram matrix of $\Delta.$ By Theorem 3.7, $G$ has signature
$(3,1)$ and hence $\det G<0.$ As a consequence, $B^{2}-4AC<0.$ Due to the fact
that $B$ is real and $A$ and $C$ are complex conjugate, we have

$|z|=|z^{*}|=1;$

and as a consequence the equations

$e^{-2\sqrt{-1}\xi}=z\quad\text{and}\quad e^{-2\sqrt{-1}\xi}=z^{*}$

respectively have a unique real solution

$\xi=\xi(\bm{\alpha})\quad\text{and}\quad\xi=\xi^{*}(\bm{\alpha})$

in the interval $[\pi,2\pi).$ Then we have

$\frac{\partial
U}{\partial\xi}\Big{|}_{\xi=\xi(\bm{\alpha})}=4k\pi\quad\text{and}\quad\frac{\partial
U}{\partial\xi}\Big{|}_{\xi=\xi^{*}(\bm{\alpha})}=4k^{*}\pi$ (5.6)

for some integers $k$ and $k^{*}.$

For a fixed $\bm{\alpha}=(\alpha_{1},\dots,\alpha_{6}),$ let

$V(\xi)=\frac{1}{2}\mathrm{Im}U(\bm{\alpha},\xi).$

Then by the relationship between the dilogarithm function and the Lobachevsky
function,

$\begin{split}V(\xi)=&\delta(\alpha_{1},\alpha_{2},\alpha_{3})+\delta(\alpha_{1},\alpha_{5},\alpha_{6})+\delta(\alpha_{2},\alpha_{4},\alpha_{6})+\delta(\alpha_{3},\alpha_{4},\alpha_{5})\\\
&+\Lambda(2\pi-\xi)+\sum_{i=1}^{4}\Lambda(\xi-\tau_{i})+\sum_{j=1}^{3}\Lambda(\eta_{j}-\xi),\end{split}$
(5.7)

where

$\delta(\alpha,\beta,\gamma)=-\frac{1}{2}\Lambda\Big{(}\frac{-\alpha+\beta+\gamma}{2}\Big{)}-\frac{1}{2}\Lambda\Big{(}\frac{\alpha-\beta+\gamma}{2}\Big{)}-\frac{1}{2}\Lambda\Big{(}\frac{\alpha+\beta-\gamma}{2}\Big{)}+\frac{1}{2}\Lambda\Big{(}\frac{\alpha+\beta+\gamma}{2}\Big{)}.$

######  Theorem 5.1.

Suppose $\Delta$ is a generalized hyperbolic tetrahedron with dihedral angles
$(\theta_{1},\dots,\theta_{6})$ and
$\bm{\alpha}=(\pi\pm\theta_{1},\pi\pm\theta_{2},\pi\pm\theta_{3},\pi\pm\theta_{4},\pi\pm\theta_{5},\pi\pm\theta_{6}).$
Then

$\mathrm{Vol}(\Delta)=V(\xi(\bm{\alpha})).$

###### Proof.

The proof follows the same idea of Ushijima [24, Theorem 1.1], which is to
verify the Schläfli formula (3.6). To this end, let

$W(\bm{\alpha})=U(\bm{\alpha},\xi(\bm{\alpha}))\quad\text{and}\quad
W^{*}(\bm{\alpha})=U(\bm{\alpha},\xi^{*}(\bm{\alpha})),$

and let

$F(\bm{\alpha})=\frac{1}{2}(W(\bm{\alpha})-W^{*}(\bm{\alpha}))$

and

$F^{*}(\bm{\alpha})=\frac{1}{2}(W(\bm{\alpha})+W^{*}(\bm{\alpha})).$

First, we will prove that when $G_{ii}\neq 0$ for all $i\in\\{1,2,3,4\\},$

$\frac{\partial\mathrm{Im}F}{\partial\theta_{k}}=-l_{k}$ (5.8)

and

$\frac{\partial\mathrm{Im}F^{*}}{\partial\theta_{k}}=0$ (5.9)

for each $k\in\\{1,\dots,6\\}.$ If these identities hold, then

$\frac{\partial\mathrm{Im}W}{\partial\theta_{k}}=\frac{\partial\mathrm{Im}F}{\partial\theta_{k}}+\frac{\partial\mathrm{Im}F^{*}}{\partial\theta_{k}}=-l_{k};$

and since

$V(\xi(\bm{\alpha}))=\frac{1}{2}W(\bm{\alpha}),$

we have

$\frac{\partial V(\xi(\bm{\alpha}))}{\partial\theta_{k}}=-\frac{l_{k}}{2},$

which satisfies the Schläfli formula in Proposition 3.21. Hence

$V(\xi(\bm{\alpha}))=\mathrm{Vol}(\Delta)+C$ (5.10)

for some constant $C.$ Then by continuity, this identity extends to
$\bm{\alpha}$ with some $G_{ii}=0,$ and extends to the closure of the space of
all $\bm{\alpha}$s coming from the dihedral angles of generalized hyperbolic
tetrahedra. At the end of the proof, we will show that $C=0$ by doing a direct
computation for a certain choice of degenerate $\bm{\alpha}$ that liesagaion
the boundary of the space.

Now to prove (5.8) and (5.9), we have

$\frac{\partial W}{\partial\alpha_{k}}=\frac{\partial
U}{\partial\alpha_{k}}\Big{|}_{\xi=\xi(\bm{\alpha})}+\frac{\partial
U}{\partial\xi}\Big{|}_{\xi=\xi(\bm{\alpha})}\cdot\frac{\partial\xi(\bm{\alpha})}{\partial\alpha_{k}}=\frac{\partial
U}{\partial\alpha_{k}}\Big{|}_{\xi=\xi(\bm{\alpha})}+4k\pi\cdot\frac{\partial\xi(\bm{\alpha})}{\partial\alpha_{k}},$

and

$\frac{\partial W^{*}}{\partial\alpha_{k}}=\frac{\partial
U}{\partial\alpha_{k}}\Big{|}_{\xi=\xi^{*}(\bm{\alpha})}+\frac{\partial
U}{\partial\xi}\Big{|}_{\xi=\xi^{*}(\bm{\alpha})}\cdot\frac{\partial\xi(\bm{\alpha})}{\partial\alpha_{k}}=\frac{\partial
U}{\partial\alpha_{k}}\Big{|}_{\xi=\xi^{*}(\bm{\alpha})}+4k^{*}\pi\cdot\frac{\partial\xi(\bm{\alpha})}{\partial\alpha_{k}}.$

Then

$\frac{\partial F}{\partial\alpha_{k}}=\frac{1}{2}\Big{(}\frac{\partial
U}{\partial\alpha_{k}}\Big{|}_{\xi=\xi(\bm{\alpha})}-\frac{\partial
U}{\partial\alpha_{k}}\Big{|}_{\xi=\xi^{*}(\bm{\alpha})}\Big{)}+4k\pi\cdot\frac{\partial\xi(\bm{\alpha})}{\partial\alpha_{k}}-4k^{*}\pi\cdot\frac{\partial\xi(\bm{\alpha})}{\partial\alpha_{k}},$

and

$\frac{\partial F^{*}}{\partial\alpha_{k}}=\frac{1}{2}\Big{(}\frac{\partial
U}{\partial\alpha_{k}}\Big{|}_{\xi=\xi(\bm{\alpha})}+\frac{\partial
U}{\partial\alpha_{k}}\Big{|}_{\xi=\xi^{*}(\bm{\alpha})}\Big{)}+4k\pi\cdot\frac{\partial\xi(\bm{\alpha})}{\partial\alpha_{k}}+4k^{*}\pi\cdot\frac{\partial\xi(\bm{\alpha})}{\partial\alpha_{k}};$

and hence

$\frac{\partial\mathrm{Im}F}{\partial\alpha_{k}}=\frac{1}{2}\mathrm{Im}\Big{(}\frac{\partial
U}{\partial\alpha_{k}}\Big{|}_{\xi=\xi(\bm{\alpha})}-\frac{\partial
U}{\partial\alpha_{k}}\Big{|}_{\xi=\xi^{*}(\bm{\alpha})}\Big{)}$ (5.11)

and

$\frac{\partial\mathrm{Im}F^{*}}{\partial\alpha_{k}}=\frac{1}{2}\mathrm{Im}\Big{(}\frac{\partial
U}{\partial\alpha_{k}}\Big{|}_{\xi=\xi(\bm{\alpha})}+\frac{\partial
U}{\partial\alpha_{k}}\Big{|}_{\xi=\xi^{*}(\bm{\alpha})}\Big{)}.$ (5.12)

In the rest of the proof, we look at $\alpha_{1},$ and the argument for the
other $\alpha_{k}$’s follows verbatim. By a direct computation, we have

$\begin{split}\frac{\partial
U}{\partial\alpha_{1}}=&\frac{\sqrt{-1}}{2}\cdot\log\frac{(1-u_{1}u_{2}u_{3}^{-1})(1-u_{1}u_{2}^{-1}u_{3})(1-u_{1}u_{5}u_{6}^{-1})(1-u_{1}u_{5}^{-1}u_{6})}{u_{1}^{4}(1-u_{1}^{-1}u_{2}u_{3})(1-u_{1}^{-1}u_{2}^{-1}u_{3}^{-1})(1-u_{1}^{-1}u_{5}u_{6})(1-u_{1}^{-1}u_{5}^{-1}u_{6}^{-1})}\\\
&+\sqrt{-1}\cdot\log\frac{u_{4}(1-zu_{1}u_{2}u_{3})(1-zu_{1}u_{5}u_{6})}{(1-zu_{1}u_{2}u_{4}u_{5})(1-zu_{1}u_{3}u_{4}u_{6})}\quad\quad(\mathrm{mod}\
\pi),\end{split}$ (5.13)

and as a consequence of (5.11) and (5.13), we have

$\frac{\partial\mathrm{Im}F}{\partial\alpha_{1}}=\mathrm{Im}\bigg{(}\frac{\sqrt{-1}}{2}\cdot\log\frac{(1-zu_{1}u_{2}u_{3})(1-zu_{1}u_{5}u_{6})(1-z^{*}u_{1}u_{2}u_{4}u_{5})(1-z^{*}u_{1}u_{3}u_{4}u_{6})}{(1-z^{*}u_{1}u_{2}u_{3})(1-z^{*}u_{1}u_{5}u_{6})(1-zu_{1}u_{2}u_{4}u_{5})(1-zu_{1}u_{3}u_{4}u_{6})}\bigg{)}.$

Let $\mathrm{R}$ and $\mathrm{S}$ respectively be the terms in
$(1-zu_{1}u_{2}u_{3})(1-zu_{1}u_{5}u_{6})(1-z^{*}u_{1}u_{2}u_{4}u_{5})(1-z^{*}u_{1}u_{3}u_{4}u_{6})$
not containing and containing $\sqrt{B^{2}-4AC}.$ Then by a direct computation
(see also Murakami-Yano [18] and Ushijima [24]),

$\mathrm{R}=8\mathrm{Q}G_{34},$

where

$\mathrm{Q}=\frac{1}{4}A^{-2}u_{1}^{2}u_{4}^{-1}(u_{4}u_{5}-u_{3})(u_{3}u_{4}-u_{5})(u_{2}u_{4}-u_{6})(u_{4}u_{6}-u_{2})$

and $G_{ij}$ is the $ij$-th cofactor of the Gram matrix $G;$ and

$\mathrm{S}=\mathrm{Q}\big{(}u_{1}^{-1}-u_{1}\big{)}\sqrt{B^{2}-4AC}=4\mathrm{Q}\big{(}u_{1}^{-1}-u_{1}\big{)}\sqrt{\det
G},$

where the last equality comes from (5.5.) Here recall that the placement of
the entries of $G$ follows the rule that if $-\cos\theta_{k}$ is in row $i$
and column $j,$ then $\theta_{k}$ is the dihedral angle between the faces
$F_{i}$ and $F_{j}.$ For example, $\theta_{1}$ is at the edge between $F_{1}$
and $F_{2},$ which is also the edge connecting the vertices $\mathbf{v}_{3}$
and $\mathbf{v}_{4}.$

By Jacobi’s Theorem (see [20, 2.5.1. Theorem]),

$G_{34}^{2}-G_{33}G_{44}=(\cos^{2}\theta_{1}-1)\det G=-\sin^{2}\theta_{1}\det
G.$

  1. (1)

If $\alpha_{1}=\pi+\theta_{1},$ then

$\mathrm{S}=4\mathrm{Q}\cdot 2\sqrt{-1}\sin\theta_{1}\cdot\sqrt{\det
G}=-8\mathrm{Q}\sqrt{G_{34}^{2}-G_{33}G_{44}}.$

Therefore, we have

$\begin{split}\frac{\partial\mathrm{Im}F}{\partial\alpha_{1}}=&\mathrm{Im}\bigg{(}\frac{\sqrt{-1}}{2}\cdot\log\frac{G_{34}-\sqrt{G_{34}^{2}-G_{33}c_{44}}}{G_{34}+\sqrt{G_{34}^{2}-G_{33}G_{44}}}\bigg{)}.\end{split}$

Let $d_{34}$ be the distance between the vertices $\mathbf{v}_{3}$ and
$\mathbf{v}_{4}.$ Then we consider the following cases.

    1. (a)

If $G_{33}G_{44}>0$ and $G_{34}>0,$ then $l_{1}=d_{34}$ and

$0<\frac{G_{34}-\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}+\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}<1.$

Then by (3.3) and (3.5)

$\frac{G_{34}-\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}+\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}=e^{-2d_{34}}=e^{-2l_{1}},$

and

$\frac{\partial\mathrm{Im}F}{\partial\theta_{1}}=\frac{\partial\mathrm{Im}F}{\partial\alpha_{1}}=\mathrm{Im}\bigg{(}\frac{\sqrt{-1}}{2}\cdot\log
e^{-2l_{1}}\bigg{)}=-l_{1}.$

    2. (b)

If $G_{33}G_{44}>0$ and $G_{34}<0,$ then $l_{1}=-d_{34}$ and

$\frac{G_{34}-\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}+\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}>1.$

Then by (3.3) and (3.5)

$\frac{G_{34}-\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}+\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}=e^{2d_{34}}=e^{-2l_{1}},$

and

$\frac{\partial\mathrm{Im}F}{\partial\theta_{1}}=\frac{\partial\mathrm{Im}F}{\partial\alpha_{1}}=\mathrm{Im}\bigg{(}\frac{\sqrt{-1}}{2}\cdot\log
e^{-2l_{1}}\bigg{)}=-l_{1}.$

    3. (c)

If $G_{33}G_{44}<0$ and $G_{34}>0,$ then $l_{1}=d_{34}$ and

$-1<\frac{G_{34}-\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}+\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}<0.$

Then by (3.4)

$\frac{G_{34}-\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}+\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}=-e^{-2d_{34}}=-e^{-2l_{1}},$

and

$\frac{\partial\mathrm{Im}F}{\partial\theta_{1}}=\frac{\partial\mathrm{Im}F}{\partial\alpha_{1}}=\mathrm{Im}\bigg{(}\frac{\sqrt{-1}}{2}\cdot\log(-e^{-2l_{1}})\bigg{)}=-l_{1}.$

    4. (d)

If $G_{33}G_{44}<0$ and $G_{34}\leqslant 0,$ then $l_{1}=-d_{34}$ and

$\frac{G_{34}-\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}+\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}\leqslant-1.$

Then by (3.4)

$\frac{G_{34}-\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}+\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}=-e^{2d_{34}}=-e^{-2l_{1}},$

and

$\frac{\partial\mathrm{Im}F}{\partial\theta_{1}}=\frac{\partial\mathrm{Im}F}{\partial\alpha_{1}}=\mathrm{Im}\bigg{(}\frac{\sqrt{-1}}{2}\cdot\log(-e^{-2l_{1}})\bigg{)}=-l_{1}.$

  2. (2)

If $\alpha_{1}=\pi-\theta_{1},$ then

$\mathrm{S}=-4\mathrm{Q}\cdot 2\sqrt{-1}\sin\theta_{1}\cdot\sqrt{\det
G}=8\mathrm{Q}\sqrt{G_{34}^{2}-G_{33}G_{44}}.$

Therefore, we have

$\begin{split}\frac{\partial\mathrm{Im}F}{\partial\alpha_{1}}=&\mathrm{Im}\bigg{(}\frac{\sqrt{-1}}{2}\cdot\log\frac{G_{34}+\sqrt{G_{34}^{2}-G_{33}c_{44}}}{G_{34}-\sqrt{G_{34}^{2}-G_{33}G_{44}}}\bigg{)}.\end{split}$

Let $d_{34}$ be the distance between the vertices $\mathbf{v}_{3}$ and
$\mathbf{v}_{4}.$ Then we consider the following cases.

    1. (a)

If $G_{33}G_{44}>0$ and $G_{34}>0,$ then $l_{1}=d_{34}$ and

$\frac{G_{34}+\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}-\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}>1.$

Then by (3.3) and (3.5),

$\frac{G_{34}+\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}-\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}=e^{2d_{34}}=e^{2l_{1}},$

and

$\frac{\partial\mathrm{Im}F}{\partial\theta_{1}}=-\frac{\partial\mathrm{Im}F}{\partial\alpha_{1}}=-\mathrm{Im}\bigg{(}\frac{\sqrt{-1}}{2}\cdot\log
e^{2l_{1}}\bigg{)}=-l_{1}.$

    2. (b)

If $G_{33}G_{44}>0$ and $G_{34}<0,$ then $l_{1}=-d_{34}$ and

$0<\frac{G_{34}+\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}-\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}<1.$

Then by (3.3) and (3.5),

$\frac{G_{34}+\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}-\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}=e^{-2d_{34}}=e^{2l_{1}},$

and

$\frac{\partial\mathrm{Im}F}{\partial\theta_{1}}=-\frac{\partial\mathrm{Im}F}{\partial\alpha_{1}}=-\mathrm{Im}\bigg{(}\frac{\sqrt{-1}}{2}\cdot\log
e^{2l_{1}}\bigg{)}=-l_{1}.$

    3. (c)

If $G_{33}G_{44}<0$ and $G_{34}>0,$ then $l_{1}=d_{34}$ and

$\frac{G_{34}+\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}-\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}<-1.$

Then by (3.4),

$\frac{G_{34}+\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}-\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}=-e^{2d_{34}}=-e^{2l_{1}},$

and

$\frac{\partial\mathrm{Im}F}{\partial\theta_{1}}=-\frac{\partial\mathrm{Im}F}{\partial\alpha_{1}}=-\mathrm{Im}\bigg{(}\frac{\sqrt{-1}}{2}\cdot\log(-e^{2l_{1}})\bigg{)}=-l_{1}.$

    4. (d)

If $G_{33}G_{44}<0$ and $G_{34}\leqslant 0,$ then $l_{1}=-d_{34}$ and

$-1\leqslant\frac{G_{34}+\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}-\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}<0.$

Then by (3.4),

$\frac{G_{34}+\sqrt{{G^{\prime}}_{34}^{2}-G_{33}G_{44}}}{G_{34}-\sqrt{{G}_{34}^{2}-G_{33}G_{44}}}=-e^{-2d_{34}}=-e^{2l_{1}},$

and

$\frac{\partial\mathrm{Im}F}{\partial\theta_{1}}=-\frac{\partial\mathrm{Im}F}{\partial\alpha_{1}}=-\mathrm{Im}\bigg{(}\frac{\sqrt{-1}}{2}\cdot\log(-e^{2l_{1}})\bigg{)}=-l_{1}.$

This proves (5.8).

Next we prove (5.9). By (5.13) and (5.9), we have

$\begin{split}\frac{\partial
U}{\partial\alpha_{1}}=\mathrm{Im}\bigg{(}&\sqrt{-1}\cdot\log\frac{(1-u_{1}u_{2}u_{3}^{-1})(1-u_{1}u_{2}^{-1}u_{3})(1-u_{1}u_{5}u_{6}^{-1})(1-u_{1}u_{5}^{-1}u_{6})}{u_{1}^{4}(1-u_{1}^{-1}u_{2}u_{3})(1-u_{1}^{-1}u_{2}^{-1}u_{3}^{-1})(1-u_{1}^{-1}u_{5}u_{6})(1-u_{1}^{-1}u_{5}^{-1}u_{6}^{-1})}\\\
&+\sqrt{-1}\cdot\log\frac{u_{4}^{2}(1-zu_{1}u_{2}u_{3})(1-z^{*}u_{1}u_{2}u_{3})(1-zu_{1}u_{5}u_{6})(1-z^{*}u_{1}u_{5}u_{6})}{(1-zu_{1}u_{2}u_{4}u_{5})(1-z^{*}u_{1}u_{2}u_{4}u_{5})(1-zu_{1}u_{3}u_{4}u_{6})(1-z^{*}u_{1}u_{3}u_{4}u_{6})}\bigg{)}.\end{split}$

Then (5.9) follows from the following direct computations

$(1-zu_{1}u_{2}u_{3})(1-z^{*}u_{1}u_{2}u_{3})=\frac{1}{A}\frac{(u_{1}u_{2}u_{3})^{2}}{u_{4}u_{5}u_{6}}\Big{(}1-\frac{u_{4}u_{5}}{u_{3}}\Big{)}\Big{(}1-\frac{u_{4}u_{6}}{u_{2}}\Big{)}\Big{(}1-\frac{u_{5}u_{6}}{u_{1}}\Big{)}\Big{(}1-\frac{1}{u_{1}u_{2}u_{3}}\Big{)},$
$(1-zu_{1}u_{5}u_{6})(1-z^{*}u_{1}u_{5}u_{6})=\frac{1}{A}\frac{(u_{1}u_{5}u_{6})^{2}}{u_{2}u_{3}u_{4}}\Big{(}1-\frac{u_{2}u_{4}}{u_{6}}\Big{)}\Big{(}1-\frac{u_{3}u_{4}}{u_{5}}\Big{)}\Big{(}1-\frac{u_{2}u_{3}}{u_{1}}\Big{)}\Big{(}1-\frac{1}{u_{1}u_{5}u_{6}}\Big{)},$
$(1-zu_{1}u_{2}u_{4}u_{5})(1-z^{*}u_{1}u_{2}u_{4}u_{5})=\frac{1}{A}\frac{(u_{1}u_{2}u_{4}u_{5})^{2}}{u_{3}u_{6}}\Big{(}1-\frac{u_{3}}{u_{4}u_{5}}\Big{)}\Big{(}1-\frac{u_{6}}{u_{2}u_{4}}\Big{)}\Big{(}1-\frac{u_{6}}{u_{1}u_{5}}\Big{)}\Big{(}1-\frac{u_{3}}{u_{1}u_{2}}\Big{)},$
$(1-zu_{1}u_{3}u_{4}u_{6})(1-z^{*}u_{1}u_{3}u_{4}u_{6})=\frac{1}{A}\frac{(u_{1}u_{3}u_{4}u_{6})^{2}}{u_{2}u_{5}}\Big{(}1-\frac{u_{2}}{u_{4}u_{6}}\Big{)}\Big{(}1-\frac{u_{5}}{u_{3}u_{4}}\Big{)}\Big{(}1-\frac{u_{2}}{u_{1}u_{3}}\Big{)}\Big{(}1-\frac{u_{5}}{u_{1}u_{6}}\Big{)}.$

We are left to show that the constant $C$ in (5.10) equals $0.$ First by a
direct computation we have for
$(\theta_{1},\dots,\theta_{6})=(\pi,0,0,\pi,0,0)$ or $(\pi,\pi,\pi,0,0,0)$
that $\xi(\bm{\alpha})=\frac{3\pi}{2}$ and $V(\xi(\bm{\alpha}))=0,$ which
equals the volume of a “flat tetrahedron”. Hence $C=0$ in these degenerate
cases.

Then we claim that each generalized hyperbolic tetrahedron $\Delta$ without
ideal vertices can be continuously deformed into one of the previous two cases
without changing the type of the vertices along the way. We consider the
following three cases.

  1. (1)

If $\Delta$ has at least one regular vertex, say, $\mathbf{v}_{1},$ then
project $\mathbf{v}_{1}$ into $F_{1}$ along the shortest geodesic connecting
the two provides the desired deformation (with a small perturbation in the
non-generic case that the projection of $\mathbf{v}_{1}$ lines on an edge of
$\Delta$).

  2. (2)

If all the vertices of $\Delta$ are hyperideal and all the edges are positive,
then by Bonahon-Bao [1] that the space of such hyperbolic tetrahedra is
connected, $\Delta$ can be continuously deformed into a flat one.

  3. (3)

If all the vertices of $\Delta$ are hyperideal and at least one edge, say,
$e_{34}$ (that connects $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$) is negative,
then moving $\mathbf{v}_{1}$ sufficiently close to $\mathbf{v}_{2}$ along the
line segment $L^{+}_{ij}$ connecting them and then projecting $\mathbf{v}_{1}$
to the plane to $F_{1}$ provides the desired deformation.

Finally, by the continuity of both the volume function and
$V(\xi(\bm{\alpha})),$ $C=0$ for all generalized hyperbolic tetrahedron, and

$\mathrm{Vol}(\Delta)=V(\xi(\bm{\alpha})).$

∎

##  6 Growth of quantum $6j$-symbols

The main results of this section is Theorem 1.4, which consider the growth
rate of quantum $6j$-symbols. For each $k\in\\{1,\dots,6\\},$ let

$\alpha_{k}=\lim_{r\to\infty}\frac{2\pi a_{k}^{(r)}}{r};$

and let

$\theta_{k}=|\pi-\alpha_{k}|,$

or equivalently,

$\alpha_{k}=\pi\pm\theta_{k}.$

We observe that if the $6$-tuple $(a_{1}^{(r)},\dots,a_{6}^{(r)})$ is
$r$-admissible for each $r\geqslant 3$, then the $6$-tuple
$(\alpha_{1},\dots,\alpha_{6})$ is admissible. Let

$T^{(r)}_{1}=\frac{a^{(r)}_{1}+a^{(r)}_{2}+a^{(r)}_{3}}{2},\
T^{(r)}_{2}=\frac{a^{(r)}_{1}+a^{(r)}_{5}+a^{(r)}_{6}}{2},\
T^{(r)}_{3}=\frac{a^{(r)}_{2}+a^{(r)}_{4}+a^{(r)}_{6}}{2},\
T^{(r)}_{4}=\frac{a^{(r)}_{3}+a^{(r)}_{4}+a^{(r)}_{5}}{2},$
$Q^{(r)}_{1}=\frac{a^{(r)}_{1}+a^{(r)}_{2}+a^{(r)}_{4}+a^{(r)}_{5}}{2},\ \
Q^{(r)}_{2}=\frac{a^{(r)}_{1}+a^{(r)}_{3}+a^{(r)}_{4}+a^{(r)}_{6}}{2},\ \
Q^{(r)}_{3}=\frac{a^{(r)}_{2}+a^{(r)}_{3}+a^{(r)}_{5}+a^{(r)}_{6}}{2}.$

For $i\in\\{1,2,3,4\\},$ let

$\tau_{i}=\lim_{r\to\infty}\frac{2\pi T^{(r)}_{i}}{r}$

and for $j\in\\{1,2,3\\},$ let

$\eta_{j}=\lim_{r\to\infty}\frac{2\pi Q^{(r)}_{j}}{r}.$

Then

$\tau_{1}=\frac{\alpha_{1}+\alpha_{2}+\alpha_{3}}{2},\quad\tau_{2}=\frac{\alpha_{1}+\alpha_{5}+\alpha_{6}}{2},\quad\tau_{3}=\frac{\alpha_{2}+\alpha_{4}+\alpha_{6}}{2},\quad\tau_{4}=\frac{\alpha_{3}+\alpha_{4}+\alpha_{5}}{2},$
$\eta_{1}=\frac{\alpha_{1}+\alpha_{2}+\alpha_{4}+\alpha_{5}}{2},\quad\eta_{2}=\frac{\alpha_{1}+\alpha_{3}+\alpha_{4}+\alpha_{6}}{2},\quad\eta_{3}=\frac{\alpha_{2}+\alpha_{3}+\alpha_{5}+\alpha_{6}}{2}.$

One of the main ingredients used to estimate the quantum $6j$-symbol is the
following lemma, first appeared in Garoufalidis-Le [12, Proposition 8.2] for
$q=e^{\frac{\pi\sqrt{-1}}{r}},$ and then in the other roots of unity in
Detcherry-Kalfagianni [10, Proposition 4.1].

######  Lemma 6.1.

For any integer $0<n<r,$ let $\\{n\\}=q^{n}-q^{-n}$ and
$\\{n\\}!=\prod_{k=1}^{n}\\{k\\}.$ Then at $q=e^{\frac{2\pi\sqrt{-1}}{r}},$

$\log\left|\\{n\\}!\right|=-\frac{r}{2\pi}\Lambda\bigg{(}\frac{2n\pi}{r}\bigg{)}+O(\log
r),$

where the term $O(\log r)$ is such that there exist constants $C,r_{0}$
independent of $n$ and $r$ such that $O(\log r)\leqslant C\log r$ whenever
$r>r_{0}.$

The other main ingredient is the following symmetry of quantum $6j$-symbols
established in Detcherry-Kalfagianni-Yang [11, Lemma A.3].

######  Lemma 6.2.

For $a\in\\{0,...,r-2\\},$ let $a^{\prime}=r-2-a.$ Then at
$q=e^{\frac{2\pi\sqrt{-1}}{r}},$

$\bigg{|}\begin{matrix}a_{1}&a_{2}&a_{3}\\\
a_{4}&a_{5}&a_{6}\end{matrix}\bigg{|}=\bigg{|}\begin{matrix}a_{1}&a_{2}&a_{3}\\\
a_{4}^{\prime}&a_{5}^{\prime}&a_{6}^{\prime}\end{matrix}\bigg{|}=\bigg{|}\begin{matrix}a_{1}&a_{2}^{\prime}&a_{3}^{\prime}\\\
a_{4}&a_{5}^{\prime}&a_{6}^{\prime}\end{matrix}\bigg{|}.$

We call the operation that changes the three $a_{i}$’s at the edges around a
face to $a_{i}^{\prime}$s a _change of colors operation around a face_ , and
the operation that changes the four $a_{i}$’s at the edges around a
quadrilateral to $a_{i}^{\prime}$s a _change of colors operation around a
quadrilateral_. Then Lemma 6.2 says that a quantum $6j$-symbol remains
unchanged under a change of colors operation around a face or a quadrilateral.

######  Lemma 6.3.

Every quantum $6j$-symbol is equal to one that is in one of the following
three cases.

  1. (1)

$a_{i}<\frac{r-2}{2}$ for all $i\in\\{1,\dots,6\\}.$

  2. (2)

$a_{i}>\frac{r-2}{2}$ for exactly one $i\in\\{1,\dots,6\\}.$

  3. (3)

$a_{i}>\frac{r-2}{2}$ and $a_{j}>\frac{r-2}{2}$ for exactly one opposite pair
$\\{i,j\\}\in\big{\\{}\\{1,4\\},\\{2,5\\},\\{3,6\\}\big{\\}}.$

###### Proof.

In the proof, we will call an $a_{i}$ _big_ if it is greater than
$\frac{r-2}{2}.$

If the quantum $6j$-symbol contains exactly zero, one, or two opposite big
$a_{i}$’s, then the result holds automatically.

If the quantum $6j$-symbol contains exactly two adjacent big $a_{i}$’s, then
they must belong to a common face. Doing a change of colors operation around
that face will reduce the number of big $a_{i}$’s to one.

If the quantum $6j$-symbol contains exactly three big $a_{i}$’s, then either
they are around a face in which case a change of colors operation around that
face will reduce the number of big $a_{i}$’s to zero, or they are contained in
a quadrilateral in which case a change of colors operation around that
quadrilateral will reduce the number of big $a_{i}$’s to one, or they are
around a vertex in which case two of them are in a common face and a change of
colors operation around that face will change the big $a_{i}$’s into an
opposite pair.

If the quantum $6j$-symbol contains exactly four big $a_{i}$’s, then either
they belong to a quadrilateral in which case a change of colors operation
around that quadrilateral will reduce the number of big $a_{i}$’s to zero, or
they contain a face in which case a change of colors operation around that
face will reduce the number of big $a_{i}$’s to one,

If the quantum $6j$-symbol contains exactly five or six big $a_{i}$’s, then
they must contain a quadrilateral in which case a change of colors operation
around that quadrilateral will reduce the number of big $a_{i}$’s to one in
the former case, and change the big $a_{i}$’s into an opposite pair in the
latter case. ∎

###### Proof of Theorem 1.4 (2).

The proof follows the same idea of Costantino [8, Theorem 1.2]. For the
simplicity of the notations, we will write $T_{i}$ for $T^{(r)}_{i}$ for
$i\in\\{1,2,3,4\\},$ and write $Q_{j}$ for $Q^{(r)}_{j}$ for
$j\in\\{1,2,3\\}.$

By Lemma 6.1, we have

$\lim_{r\to\infty}\frac{2\pi}{r}\log\Delta(a^{(r)}_{i},a^{(r)}_{j},a^{(r)}_{k})|=\delta(\alpha_{i},\alpha_{j},\alpha_{k}).$
(6.1)

Next, we study the asymptotics of

$S=\sum_{z=\max\\{T_{1},T_{2},T_{3},T_{4}\\}}^{\min_{j}\\{Q_{1},Q_{2},Q_{3}\\}}\frac{(-1)^{z}[z+1]!}{\prod_{i=1}^{4}[z-T_{i}]!\prod_{j=1}^{3}[Q_{j}-z]!}.$

Let

$S_{z}=\frac{(-1)^{z}[z+1]!}{\prod_{i=1}^{4}[z-T_{i}]!\prod_{j=1}^{3}[Q_{j}-z]!}.$

The proof contains the following three steps.

  1. Step 1.

Since $[z+1]!=0$ when $z>r-2,$ $S_{z}=0$ for those $z.$ Hence we only need to
consider $S_{z}$ for $z$ in between $\max\\{T_{1},T_{2},T_{3},T_{4}\\}$ and
$\min\\{Q_{1},Q_{2},Q_{3},r-2\\}.$ We will show that for those $z,$ all
$S_{z}$ have the same sign so the growth rate of the sum is determined by that
of the largest term.

  2. Step 2.

If $\lim_{r\to\infty}\frac{2\pi z^{(r)}}{r}=\xi,$ then by Lemma 6.1 we have

$\lim_{r\to\infty}\frac{2\pi}{r}\log|S_{z^{(r)}}|=\sum_{i=1}^{4}\Lambda(\xi-\tau_{i})+\sum_{j=1}^{3}\Lambda(\eta_{j}-\xi)-\Lambda(\xi).$

We will show that the function $s$ defined by

$s(\xi)=\sum_{i=1}^{4}\Lambda(\xi-\tau_{i})+\sum_{j=1}^{3}\Lambda(\eta_{j}-\xi)-\Lambda(\xi)$

has a unique maximum point $\xi_{0}$ on the interval
$I=[\max\\{\tau_{1},\tau_{2},\tau_{3},\tau_{4}\\},\min\\{\eta_{1},\eta_{2},\eta_{3},2\pi\\}]$
so that the growth rate of $S$ equals $s(\xi_{0}),$ and hence the growth rate
of the quantum $6j$-symbol equals $V(\xi_{0}),$ where $V$ is the function
defined in (5.7).

  3. Step 3.

We will show that $\xi_{0}=\xi(\bm{\alpha})$ so that
$V(\xi_{0})=V(\xi(\bm{\alpha})),$ which by Theorem 5.1 equals
$\mathrm{Vol}(\Delta).$

We accomplish Step 1 by showing that the ratio between to consecutive summands
$S_{z}$ and $S_{z-1}$ is positive. By a direct computation, we have

$\frac{S_{z}}{S_{z-1}}=-\frac{\sin{\frac{2\pi(z+1)}{r}}\sin{\frac{2\pi(Q_{1}-z+1)}{r}}\sin{\frac{2\pi(Q_{2}-z+1)}{r}}\sin{\frac{2\pi(Q_{3}-z+1)}{r}}}{\sin{\frac{2\pi(z-T_{1})}{r}}\sin{\frac{2\pi(z-T_{2})}{r}}\sin{\frac{2\pi(z-T_{3})}{r}}\sin{\frac{2\pi(z-T_{4})}{r}}}$

for $z$ satisfying $\max\\{T_{1},T_{2},T_{3},T_{4}\\}+1\leqslant
z\leqslant\min\\{Q_{1},Q_{2},Q_{3},r-2\\}.$ The heuristic idea here is to look
at the sign of the following function

$h(\xi)=-\frac{\sin(\xi)\sin(\eta_{1}-\xi)\sin(\eta_{2}-\xi)\sin(\eta_{3}-\xi)}{\sin(\xi-\tau_{1})\sin(\xi-\tau_{2})\sin(\xi-\tau_{3})\sin(\xi-\tau_{4})}.$

If $h(\xi)>0$ for every $\xi,$ than for $r$ sufficiently large,
$\frac{S_{z}}{S_{z-1}}>0$ for every $z.$ To this end, by Lemma 6.3, we only
need to consider the three possibilities listed there.

  1. (1)

In this case, we have $\alpha_{1},\dots,\alpha_{6}\leqslant\pi.$ We mention
here that this case is the only place where we need condition (2) that
$G_{ii}<0$ for some $i\in\\{1,2,3,4\\}.$

By reindexing $\tau_{i}$’s and $\eta_{j}$’s if necessary, we assume that the
vertex $\mathbf{v}_{4}$ is a hyperideal vertex. Then by Lemma 4.9 (2),

$\theta_{3}+\theta_{4}+\theta_{5}<\pi.$ (6.2)

Since $\alpha_{i}\leqslant\pi$ for each $i\in\\{1,\dots,6\\},$ we have
$\theta_{i}=\pi-\alpha_{i}$ and (6.2) is equivalent to

$\tau_{4}>\pi.$ (6.3)

As a consequence of (6.3), we have $I\subset[\pi,2\pi],$ and in particular,

$\xi\in[\pi,2\pi].$ (6.4)

By the condition that all $\alpha_{i}\leqslant\pi$, we have that for all
$i\in\\{1,2,3,4\\}$ and $j\in\\{1,2,3\\},$

$\eta_{j}-\tau_{i}=\frac{\alpha_{k}+\alpha_{l}-\alpha_{m}}{2}\leqslant\frac{\alpha_{k}+\alpha_{l}}{2}\leqslant\pi$
(6.5)

for some triple $(k,l,m)$ around a vertex. As a consequence, we have

$0\leqslant\xi-\tau_{i}\leqslant\eta_{1}-\tau_{i}\leqslant\pi$ (6.6)

for all $i\in\\{1,2,3,4\\},$ and

$0\leqslant\eta_{j}-\xi\leqslant\eta_{j}-\tau_{1}\leqslant\pi$ (6.7)

for all $j\in\\{1,2,3\\}.$

  2. (2)

In this case, we assume, say, $\alpha_{1}\geqslant\pi$ and
$\alpha_{2},\dots,\alpha_{6}\leqslant\pi.$ Then we have

$\tau_{1}=\frac{\alpha_{1}+\alpha_{2}+\alpha_{3}}{2}=\alpha_{1}+\frac{\alpha_{2}+\alpha_{3}-\alpha_{1}}{2}\geqslant\alpha_{1}\geqslant\pi,$
(6.8)

where the penultimate inequality comes from the admissibility conditions. As a
consequence of (6.8), we have $I\subset[\pi,2\pi],$ and in particular,

$\xi\in[\pi,2\pi].$ (6.9)

By bounding $\xi$ from above by $\eta_{3},$ we have

$\xi-\tau_{i}\in[0,\pi]$ (6.10)

for all $i\in\\{1,2,3,4\\}.$ Indeed,

$0\leqslant\xi-\tau_{1}\leqslant\eta_{3}-\tau_{1}=\frac{\alpha_{5}+\alpha_{6}-\alpha_{1}}{2}\leqslant\frac{\alpha_{5}+\alpha_{6}}{2}\leqslant\pi.$
$0\leqslant\xi-\tau_{2}\leqslant\eta_{3}-\tau_{2}=\frac{\alpha_{2}+\alpha_{3}-\alpha_{1}}{2}\leqslant\frac{\alpha_{2}+\alpha_{3}}{2}\leqslant\pi.$
$0\leqslant\xi-\tau_{3}\leqslant\eta_{3}-\tau_{3}=\frac{\alpha_{3}+\alpha_{5}-\alpha_{4}}{2}\leqslant\frac{\alpha_{3}+\alpha_{5}}{2}\leqslant\pi.$
$0\leqslant\xi-\tau_{4}\leqslant\eta_{3}-\tau_{4}=\frac{\alpha_{2}+\alpha_{6}-\alpha_{4}}{2}\leqslant\frac{\alpha_{2}+\alpha_{6}}{2}\leqslant\pi.$

Also, by bounding $\xi$ from below by $\tau_{1},$ we have

$\eta_{j}-\xi\in[0,\pi]$ (6.11)

for all $j\in\\{1,2,3\\}.$ Indeed,

$0\leqslant\eta_{1}-\xi\leqslant\eta_{1}-\tau_{1}=\frac{\alpha_{4}+\alpha_{5}-\alpha_{3}}{2}\leqslant\frac{\alpha_{4}+\alpha_{5}}{2}\leqslant\pi.$
$0\leqslant\eta_{2}-\xi\leqslant\eta_{2}-\tau_{1}=\frac{\alpha_{4}+\alpha_{6}-\alpha_{2}}{2}\leqslant\frac{\alpha_{4}+\alpha_{6}}{2}\leqslant\pi.$
$0\leqslant\eta_{3}-\xi\leqslant\eta_{2}-\tau_{1}=\frac{\alpha_{2}+\alpha_{3}-\alpha_{1}}{2}\leqslant\frac{\alpha_{2}+\alpha_{3}}{2}\leqslant\pi.$

  3. (3)

In this case, we have, say, $\alpha_{1},\alpha_{4}\geqslant\pi$ and
$\alpha_{2},\alpha_{3},\alpha_{5},\alpha_{6}\leqslant\pi.$ We claim that

$\pi\leqslant\tau_{i}\leqslant 2\pi$ (6.12)

for each $i\in\\{1,2,3,4\\},$ and

$0\leqslant\eta_{j}-\tau_{i}\leqslant\pi$ (6.13)

for each $i\in\\{1,2,3,4\\}$ and $j\in\\{1,2,3\\}.$ Indeed, the second half of
(6.12) and the first half of (6.13) come from the admissibility conditions of
$(\alpha_{1},\dots,\alpha_{6}).$ For first half of (6.12), we have

$\tau_{i}=\frac{\alpha_{j}+\alpha_{k}+\alpha_{l}}{2}$

for some triple $(j,k,l)$ around a vertex. Since $\alpha_{1}$ and $\alpha_{4}$
are angles of opposite edges, without loss of generality, we may assume that
$(j,k,l)=(1,2,3).$ Then

$\frac{\alpha_{1}+\alpha_{2}+\alpha_{3}}{2}=\alpha_{1}+\frac{\alpha_{2}+\alpha_{3}-\alpha_{1}}{2}\geqslant\pi,$
(6.14)

where the last inequality comes from the admissibility conditions. For the
second half of (6.13), we have

$\eta_{j}-\tau_{i}=\frac{\alpha_{k}+\alpha_{l}-\alpha_{m}}{2}$

for some triple $(k,l,m)$ around a vertex. Again, without of generality,
assume that $\\{j,k,l\\}=\\{1,2,3\\}.$ Then we have

$\frac{\alpha_{2}+\alpha_{3}-\alpha_{1}}{2}\leqslant\frac{\alpha_{2}+\alpha_{3}}{2}\leqslant\pi,$
$\frac{\alpha_{1}+\alpha_{2}-\alpha_{3}}{2}\leqslant\frac{(\alpha_{2}+\alpha_{3})+(\alpha_{2}-\alpha_{3})}{2}=\alpha_{2}\leqslant\pi,$

and

$\frac{\alpha_{1}+\alpha_{3}-\alpha_{2}}{2}\leqslant\frac{(\alpha_{2}+\alpha_{3})+(\alpha_{3}-\alpha_{3})}{2}=\alpha_{3}\leqslant\pi,$

where the first inequalities in the last two cases come from the admissibility
conditions. As a consequence of (6.12), we have $I\subset[\pi,2\pi],$ and in
particular,

$\xi\in[\pi,2\pi].$ (6.15)

As a consequence of (6.13), we have

$0\leqslant\xi-\tau_{i}\leqslant\eta_{1}-\tau_{i}\leqslant\pi$ (6.16)

for all $i\in\\{1,2,3,4\\},$ and

$0\leqslant\eta_{j}-\xi\leqslant\eta_{j}-\tau_{1}\leqslant\pi$ (6.17)

for all $j\in\\{1,2,3\\}.$

From (6.4), (6.6) and (6.7) in Case (1), (6.9), (6.10) and (6.11) in Case (2),
and (6.15), (6.16) and (6.17) in Case (3), we have for sufficiently large $r$
that

$\frac{r-2}{2}<z<r-2,$ $0<z-T_{i}<\frac{r-2}{2}$

for all $i\in\\{1,2,3,4\\},$ and

$0<Q_{j}-z<\frac{r-2}{2}$

for all $j\in\\{1,2,3\\}.$ As a consequence, we have $\frac{S_{z}}{S_{z-1}}>0$
for all $z$ in the range, and all the $S_{z}$ have the same sign. This
completes Step 1.

We accomplish Step 2 as follows. If
$\max\\{\tau_{1},\tau_{2},\tau_{3},\tau_{4}\\}=\min\\{\eta_{1},\eta_{2},\eta_{3},2\pi\\},$
then the interval $I$ is a single point and the result holds automatically. If
$\max\\{\tau_{1},\tau_{2},\tau_{3},\tau_{4}\\}<\min\\{\eta_{1},\eta_{2},\eta_{3},2\pi\\},$
then we show that the function $s(\xi)$ is strictly concave down on the
interval $I$ and that the derivative $s^{\prime}(\xi)$ has different signs at
the two end points of $I.$ In this case, we first have

$2\pi-\tau_{i}>0$ (6.18)

for each $i\in\\{1,2,3,4\\},$ and

$\eta_{j}-\tau_{i}>0$ (6.19)

for each $i\in\\{1,2,3,4\\}$ and $j\in\\{1,2,3\\}.$ Then we compute

$s^{\prime}(\xi)=\log\bigg{(}\frac{\sin(2\pi-\xi)\sin(\eta_{1}-\xi)\sin(\eta_{2}-\xi)\sin(\eta_{3}-\xi)}{\sin(\xi-\tau_{1})\sin(\xi-\tau_{2})\sin(\xi-\tau_{3})\sin(\xi-\tau_{4})}\bigg{)},$
(6.20)

and

$s^{\prime\prime}(\xi)=-\sum_{i=1}^{4}\cot(\xi-\tau_{i})-\sum_{j=1}^{3}\cot(\eta_{j}-\xi)-\cot(2\pi-\xi).$
(6.21)

As a consequence, we have

$\lim_{\xi\to\max\\{\tau_{1},\tau_{2},\tau_{3},\tau_{4}\\}^{+}}s^{\prime}(\xi)=+\infty\quad\text{and}\quad\lim_{\xi\to\min\\{\eta_{1},\eta_{2},\eta_{3},2\pi\\}^{-}}s^{\prime}(\xi)=-\infty.$
(6.22)

We still consider the three cases of Lemma 6.3.

  1. (1)

Recall in this case, we have $\alpha_{1},\dots,\alpha_{6}\leqslant\pi.$ Then
by (6.3) and (6.18), we have

$0<(\xi-\tau_{4})+(2\pi-\xi)=2\pi-\tau_{4}<\pi,$ (6.23)

and by (6.5) and (6.19), we have for each $i\in\\{1,2,3\\}$ that

$0<(\xi-\tau_{i})+(\eta_{i}-\xi)=\eta_{i}-\tau_{i}\leqslant\pi.$ (6.24)

In particular, both inequalities in (6.23) are strict. Therefore, by (6.21),