Volumes of hyperbolic truncated tetrahedra

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Content VOLUMES OF HYPERBOLIC TRUNCATED TETRAHEDRA
by
Jihoon Sohn
A Dissertation Presented to the
FACULTY OF THE USC GRADUATE SCHOOL
UNIVERSITY OF SOUTHERN CALIFORNIA
In Partial Fulfillment of the
Requirements for the Degree
DOCTOR OF PHILOSOPHY
(MATHEMATICS)
August 2022
Copyright 2022 Jihoon Sohn
This thesis is dedicated to
my advisor Francis
who made me a better mathematician
and my wife Hye Min
who made me a better person.
ii
Acknowledgements
First and foremost, I would like to thank my advisor Francis Bonahon for introducing me to the
beautiful world of hyperbolic tetrahedra. Since the day one at USC, he has taught me how to think,
write and present mathematically while being friendly, patient, and supportive. Our minimum one-
and-a-half-hour-long weekly meetings always inspired me and made me grow mathematically. In
the meantime, I cannot forget to mention his delicious handmade foods including tamales and fig
jam. Without his academic and personal support, this thesis would not have been possible. Huge
thanks go out to my committee members Ken Alexander, Aravind Asok, Muhao Chen, Edmond
Jonckheere, and Aaron Lauda. I also would like to thank Muhao Chen and Mingyu Derek Ma for
our idea-provoking discussions on hyperbolic knowledge graph representations.
Special thanks go out to my academic brother Daniel Douglas. He has always been an affable
colleague. I also thank Tommasso Cremaschi for introducing me to Teichm¨ uller spaces and his in-
credible barbeque and cakes. I would like to thank my USC math friends Gene Kim and Hyunjung
Kim who helped me settle down at Kaprielian Hall and in Los Angeles. My other USC friends,
Tae Hyung Kim, Kyeongsuk Lee, and Seungjong Lee and UCLA friends, Jiwoo Han, Younghak
Kwon, and Jaehoon Lee deserve my great thanks for our enjoyable moments in LA.
Last but not least, I would like to thank my family and family-in-law for the endless support and
the unconditional love. Finally, I would like to thank my wonderful wife Hye Min Kim, who has
always encouraged me. During our Ph.D. studies, we celebrated achievements together, overcame
challenges together, and most importantly, survived our long journey together. I could not have
finished my study without you. Thank you and love you.
iii
Table of Contents
Dedication ii
Acknowledgements iii
List of Tables v
List of Figures vi
Abstract vii
Chapter 1: Introduction 1
1.1 Motivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2.1 Lorentzian spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2.2 Hyperbolic truncated tetrahedra . . . . . . . . . . . . . . . . . . . . . . . 8
1.2.3 V olume functions of truncated tetrahedra . . . . . . . . . . . . . . . . . . 15
Chapter 2: Volumes of proper truncated tetrahedra 18
2.1 Proper truncated tetrahedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.2 Combinatorial structures of truncated tetrahedra . . . . . . . . . . . . . . . . . . . 20
2.3 Classification of truncated tetrahedra . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.4 Main theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.4.1 Basic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.4.2 Quadratic equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.4.3 V olumes of hyperbolic truncated tetrahedra . . . . . . . . . . . . . . . . . 35
2.5 Properties of Gram matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
2.6 Schl¨ afli’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
2.7 Derivatives of Bloch-Wigner dilogarithm and the volume function . . . . . . . . . 43
2.8 Proof of main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
Chapter 3: Conclusion and future direction 67
References 68
iv
List of Tables
2.1 Table of classification of combinatorial properties of projective tetrahedra. Row
and column represent the number of hyperideal vertices and hyperideal edges of
projective tetrahedra, respectively. The red vertices/edges/faces represent finite
vertices/edges/faces respectively, and the black represents hyperideal. . . . . . . . 28
2.2 For some triples(v,e, f) (the number of hyperideal vertices, edges, and faces, re-
spectively), they are matched to more than one classes of projective tetrahedra.
The table lists all such triples and denotes each of them using its combinatorial
characteristics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.3 The classification of combinatorial structures of truncated tetrahedra. (* is the class
which is paired with itself.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2.4 The pairs of z
i j
and R
i j
are determined by the combinatorial properties of∆. . . . . 55
v
List of Figures
1.1 The red tetrahedron ∆, the blue tetrahedron ∆
′
, and the green tetrahedron ∆
′′
are
inRP
3
and they all have the same vertices v
1
,··· ,v
4
. The two tetrahedra∆ and
∆
′
share one face and there are three more tetrahedra that share one face with∆.
On the other hand, ∆ and ∆
′′
share two opposite edges and we have total three
tetrahedra that share two opposite edges with∆. In summary, given four vertices v
i
for i= 1,··· ,4, there are eight different tetrahedra inRP
3
: one∆-like tetrahedron,
four∆
′
-like tetrahedra, and three∆
′′
-like tetrahedra. . . . . . . . . . . . . . . . . . 9
1.2 The figure on the left is the truncated tetrahedron ∆
trunc
obtained from a tetrahedron
∆ by truncation at v
i
. The figure on the right is the intersection of ∆ and ∆
∗ .
Observe that those two agree. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.3 Observe that ∆
trunc
in the left has hyperideal vertices but finite (red) edges. On
the other hand,∆
trunc
in the right has two finite vertices v
1
, v
4
, and two hyperideal
vertices v
2
, v
3
, and the hyperideal edge e
i j
. We thus has a new edge that came from
the intersection of two truncation faces v
⊥
2
and v
⊥
3
. The new edge is also contained
in the edge e
∗ 14
connecting two vertices n
1
and n
4
of∆
∗ . . . . . . . . . . . . . . . . 16
2.1 Two examples of nonempty, nonideal, and proper truncated tetrahedra. . . . . . . . 22
2.2 The tetrahedron∆ in the left has three hyperideal vertices v
1
, v
2
, v
3
, three hyper-
ideal edges e
12
, e
13
, e
23
, and one hyperideal face f
4
, followed by∆∈ T
3,3,1
. Note
that each hyperideal vertex and its truncation face have the same color. The trun-
cated tetrahedron∆
trunc
derived from∆ in T
3,3,1
was unaddressed in earlier works.
The projective tetrahedron∆ in the right has only hyperideal vertices, hyperideal
edges, and hyperideal faces. Thus, we have ∆
trunc
∈ T
4,6,4
. Since the dual tetra-
hedron∆
∗ is finite, the corresponding truncated tetrahedron ∆
trunc
is finite as well.
Hence, in the view of∆
∗ , the truncated tetrahedron is derived from a tetrahedron
in T
0,0,0
. Therefore, the classes T
4,6,4
and T
0,0,0
of tetrahedra are giving the same
combinatorial structure of corresponding truncated tetrahedron. . . . . . . . . . . . 29
2.3 The Euler dilogarithm Li
2
(z) is defined as an integral along the path connecting O
and z. Since the domain is simply connected, it does not depend on the path. . . . . 32
vi
Abstract
A hyperbolic truncated tetrahedron is a 3-dimensional polyhedron in the projective modelH
3
⊂ RP
3
of the hyperbolic space, which is obtained by polar truncation at the vertices that lie outside of
H
3
. In this paper, we present two different computations of the volume of a hyperbolic truncated
tetrahedron, one in terms of its edge lengths and another in terms of its dihedral angles. Among
various cases of hyperbolic truncated tetrahedron, our method provides a unified approach to cases
covered by earlier results of J. Murakami, M. Yano, A. Ushijima, and A. Kolpakov, and extends to
cases that had been left unaddressed. Our key technical idea is to rely on the (real-valued) Bloch-
Wigner dilogarithm function instead of the (complex-valued) Euler dilogarithm, which eliminates
subtle phase issues and has better analytic properties.
vii
Chapter 1
Introduction
1.1 Motivations
A hyperbolic tetrahedron is the fundamental and essential object in three dimensional hyperbolic
manifold and is closely related to the V olume Conjecture. In [1], Kashaev conjectured that there is a
relation between quantum invariants of hyperbolic knots and the volumes of the knot complements
of them, and checked it numerically for some examples. H. Murakami observed in [2] that this
relation holds for some closed hyperbolic 3-manifolds. These two papers suggest that there is a
certain relation between the hyperbolic volume of a three-manifold and its Turaev-Viro invariant
(see [3]), which is an invariant given by the sum of quantum 6 j-symbols associated to the tetrahedra
of simplicial decomposition of the manifold. Later, according to [4] and [5], it was proved that the
asymptotics of quantum 6 j-symbols are strongly related to the V olume Conjecture. Particularly,
in [6], the hyperbolic volume of an ideal tetrahedron is given by computing the asymptotics of the
quantum 6 j-symbols.
The first closed formula that computes the volume of hyperbolic tetrahedron was given in [7]
and another formula was given in [8]. Both formulas used geometric approaches with decompo-
sitions into different sets of ideal tetrahedra. More symmetric version was given in [9]. All these
formulas are given in terms of dihedral angles of the tetrahedron. [10] generalized the concept of
tetrahedra by moving the vertices to the boundary of hyperbolic space, and even beyond the bound-
ary. In other words, considering the projective modelH
3
⊂ RP
3
of hyperbolic space, vertices were
1
moved fromH
3
toRP
3
\H
3
. For each vertex located outside of the hyperbolic space, we truncate
the tetrahedron with the polar plane of the vertex, which are orthogonal to all faces containing the
vertex. Given a projective tetrahedron, the polyhedron that is obtained by polar truncations at all
vertices that lie outside ofH
3
is called the truncated tetrahedron. Note that any truncated tetrahe-
dron has a finite volume. Ushijima showed in [10] that the formula from [9] can be applied to the
certain types of truncated tetrahedra: when all six edges of a projective tetrahedron meetsH
3
, then
the volume of the corresponding truncated tetrahedron is given by the same formula as in [9].
On the other hand, J. Murakami and Ushijima proposed in [11] another volume formula of a
hyperbolic tetrahedron in terms of edge lengths and this formula is related to the V olume Conjec-
ture for a hyperbolic 3-manifold. The formula in terms of dihedral angles in [9] comes from the
6 j-symbols as well, and the two formulas share the key function, which will be discussed later.
Subsequently, [12] introduced a formula for a particular case of truncated tetrahedron which is
sharing the same key function but having both dihedral angles and edge lengths as its variables.
Inspired by the series of formulas that give hyperbolic volumes of a few cases of truncated
tetrahedra, we propose a formula that computes the volume of a proper truncated tetrahedron,
which was introduced in [13]. The main contribution of the thesis is the following:
• Our formula simplifies the earlier approaches. The earlier formulas were based on the
complex-valued Euler dilogarithm function. Our methods, instead, rely on the real-valued
Bloch-Wigner dilogarithm function and this modification deletes the phase issue of the Euler
dilogarithm functions.
• Our formula unifies the earlier approaches. The earlier formulas vary and depend on the com-
binatorial structures of truncated tetrahedra. In particular, some of earlier formulas computes
the volume in terms of dihedral angles and the other computes in terms of edge lengths. In
our formula, replacing the tetraheron to its dual tetrahedron does not affect the correspond-
ing truncated tetrahedron and its volume but exchanges the dihedral angles and the edge
lengths. Given that the truncated tetrahedra covered in earlier results are all proper, our
formula agrees with them.
2
• Our formula further covers unaddressed proper truncated tetrahedra.
• Future research may explore that our formula is applicable to a nonproper truncated tetra-
hedron. A nonproper truncated tetrahedron is composed of a number of polyhedra, where
adjacent polyhedra are connected via their common edge. The formula computes the alter-
nating sum of volumes of polyhedra.
1.2 Preliminaries
1.2.1 Lorentzian spaces
LetR
3,1
be a Lorentzian space with the Lorentzian inner product
⟨x,y⟩=− x
0
y
0
+ x
1
y
1
+ x
2
y
2
+ x
3
y
3
for x=(x
0
,x
1
,x
2
,x
3
),y=(y
0
,y
1
,y
2
,y
3
)∈R
4
. LetR
3,1
T
be the set of time-like vectors,R
3,1
L
be the
set of light-like vectors, andR
3,1
S
be the set of space-like vectors, which are defined as follows:
R
3,1
T
=

x∈R
3,1
|⟨x,x⟩< 0

, R
3,1
L
=

x∈R
3,1
|⟨x,x⟩= 0

,
R
3,1
S
=

x∈R
3,1
|⟨x,x⟩> 0

.
Then we haveR
3,1
=R
3,1
T
F
R
3,1
L
F
R
3,1
S
, orR
3,1
can be divided into these three disjoint subsets.
Note thatR
3,1
S
is connected, but not simply connected, andR
3,1
T
is composed of two disjoint simply
connected components:
R
3,1
T,+
:=

x∈R
3,1
|⟨x,x⟩< 0,x
0
> 0

,
R
3,1
T,− :=

x∈R
3,1
|⟨x,x⟩< 0,x
0
< 0

.
3
Proposition 1. For two time-like vectors x, y∈R
3,1
T
=R
3,1
T,+
F
R
3,1
T,− , the sign of their Lorentzian
inner product depends on the components x and y are contained in. In particular,







⟨x,y⟩< 0, if x, y∈R
3,1
T,+
or x, y∈R
3,1
T,− ,
⟨x,y⟩> 0, if x∈R
3,1
T,+
, y∈R
3,1
T,− or x∈R
3,1
T,− , y∈R
3,1
T,+
.
Definition 2. For a vector x∈R
3,1
, its Lorentzian norm∥x∥ is defined as
∥x∥=
p
⟨x,x⟩=















p
⟨x,x⟩ if x∈R
3,1
S
,
0 if x∈R
3,1
L
,
i
p
− ⟨x,x⟩ if x∈R
3,1
T
.
Let H =

x∈R
3,1
|⟨x,x⟩=− 1

be the hyperboloid of two sheets. The restriction of the
quadratic form induced by the Lorentzian inner product to the tangent spaces of H is positive
definite and gives a Riemannian metric on H. The space obtained from the upper sheet H
+
=

x∈R
3,1
|⟨x,x⟩=− 1,x
0
> 0

of H equipped with the above metric is called the hyperboloid
model of 3-dimensional hyperbolic space.
Consider a projection π fromR
3,1
\{O} ontoRP
3
along the line passing through the origin
O ofR
3,1
. Then, H
+
andR
3,1
T
are projected to the subset

[x]∈RP
3
|⟨x,x⟩< 0

ofRP
3
by π.
In the standard affine chart in R
3
⊂ R
3,1
given by (x
1
,x
2
,x
3
)→[1,x
1
,x
2
,x
3
] the set π(H
+
) is a
unit ball B={(x
1
,x
2
,x
3
)|x
2
1
+ x
2
2
+ x
2
3
< 1}⊂ R
3
⊂ RP
3
. Along with the metric onRP
3
pushed
forward by π from the hyperbolic metric on H
+
, we have another hyperbolic space B, which is
called the Beltrami-Klein model of 3-dimensional hyperbolic space. This model is the main model
of hyperbolic space in this paper, therefore we simply call itH
3
instead of B.
On the other hand, let H
′
={x∈R
3,1
|⟨x,x⟩ = 1} be the hyperboloid of one sheet. Then
dS= H
′
/{±} andR
3,1
S
are projected to the subset dS
3
={[x]∈RP
3
|⟨x,x⟩> 0} ofRP
3
byπ. The
space dS
3
is called de Sitter space endowed with the induced metric. Then we have dS
3
=RP
3
\H
3
as sets.
4
Definition 3. For a point p∈RP
3
, we say that p is finite if p∈H
3
, ideal if p∈∂H
3
, and hyperideal
if p∈ dS
3
.
Note. Time-like vectors, light-like vectors, and space-like vectors in R
3,1
correspond to finite
points, ideal points, and hyperideal points inRP
3
viaπ.
One of the most useful properties of the Beltrami-Klein model is the duality inH
3
which comes
from the duality ofRP
3
.
Definition 4. Let P be a k-dimensional plane (for 0≤ k≤ 2) inRP
3
and Q=π
− 1
(P)∪{O} be a
union of the inverse imageπ
− 1
(P) and the origin O. Then Q is a(k+ 1)-dimensional subspace in
R
3,1
. Let Q
⊥
be a(3− k)-dimensional orthogonal subspace of Q with respect to⟨· ,·⟩:
Q
⊥
=

x∈R
3,1
|⟨x,y⟩= 0 ∀y∈ Q

.
The dual space P
⊥
is defined as P
⊥
=π(Q
⊥
\{O})⊂ RP
3
.
Note. The dual P
⊥
of k-dimensional plane P (for 0≤ k≤ 2) is a(2− k)-dimensional plane.
For x∈RP
3
, lete x∈R
3,1
be an arbitrary point inπ
− 1
(x). Note thatπ
− 1
(x)⊂ R
3,1
\{O} is a
line passing through the origin O and is composed of two disjoint rays that have the same starting
point O with opposite directions. The hyperplanee x
⊥
⊂ R
3,1
is uniquely determined by x regardless
of a choice ofe x.
Proposition 5. For x∈RP
3
, the hyperplanee x
⊥
meetsR
3,1
T
if and only if x is hyperideal.
Definition 6. For x∈RP
3
, pick an arbitrary pointe x∈π
− 1
(x). The pointe x inR
3,1
is called a lift of
x. Consider an oriented hyperplanee x
⊥
ofR
3,1
where the orientation ofe x
⊥
is given by its normal
vectore x. Define the polar plane P
e x
ofe x asπ

e x
⊥
∩R
3,1
T,+

.e x is called a pole of its polar plane P
e x
.
Note. By Proposition 5, a polar plane P
e x
is nonempty if and only if x is hyperideal. For a hyperideal
point x and its lifte x, the polar plane P
e x
is an oriented hyperplane ofH
3
, where its orientation is
derived from the orientation ofe x
⊥
.
5
Note. Given a hyperideal point x, we have two options for the orientation of the hyperplanee x
⊥
.
For two differente x
1
,e x
2
∈ π
− 1
(x), there exists a nonzero scalar c∈R such thate x
1
= ce x
2
. If
c> 0, the corresponding oriented hyperplanese x
⊥
1
ande x
⊥
2
are identical. If c< 0,e x
⊥
1
ande x
⊥
2
share
the hyperplane but have the opposite orientations. The same holds for the corresponding polar
planes P
e x
1
and P
e x
2
. Hence the orientation of the polar plane P
e x
of x depends on which connected
component ofπ
− 1
(x) the lifte x of x belongs to.
We have the following propositions on relationships between the Lorentzian inner product and
geometric objects. See, for example, [14] for the details.
Proposition 7. Let x and y be two finite points in H
3
. Then, for any liftse x ande y that are in the
same component of R
3,1
T
, the hyperbolic distance d between these two points x and y in H
3
is
calculated by
⟨e x,e y⟩=∥e x∥∥e y∥coshd.
Note. For x,y∈H
3
, their liftse x,e y are time-like vectors on the same component ofR
3,1
T
and thus
⟨e x,e y⟩< 0 by Proposition 1. Furthermore, their Lorentzian norms are in iR
+
by Definition 2. Hence
the multiplication of their Lorentzian norms is always negative and we thus get coshd≥ 0 in the
formula. Furthermore, by Cauchy’s inequality, we actually get coshd≥ 1. Therefore, for any lifts,
the hyperbolic distance d between two finite points is well-defined.
Note. On the other hand, if the liftse x ande y are in the different components, then the hyperbolic
distance d between x and y can be calculated by
⟨e x,e y⟩=− ∥e x∥∥e y∥coshd.
In this case, we have⟨e x,e y⟩> 0 by Proposition 1, and their Lorentzian norms are positive imaginary,
followed by∥e x∥∥e y∥< 0. Hence, we again have coshd≥ 0 and the distance is well defined.
Proposition 8. Let x and y be two hyperideal points in dS
3
. Then, for any liftse x ande y, one of the
following holds:
6
(a) |⟨e x,e y⟩| <∥e x∥∥e y∥ if and only if two polar planes P
e x
and P
e y
intersect. Furthermore, the
hyperbolic (exterior) angleα between these two polar planes is calculated by
⟨e x,e y⟩=∥e x∥∥e y∥cosα.
(b) |⟨e x,e y⟩|>∥e x∥∥e y∥ if and only if two polar planes P
e x
and P
e y
are disjoint inH
3
. Furthermore,
the hyperbolic distance d along the common orthogonal line between these two polar planes
is calculated by
|⟨e x,e y⟩|=∥e x∥∥e y∥coshd.
(c) |⟨e x,e y⟩|=∥e x∥∥e y∥ if and only if two polar planes P
e x
and P
e y
intersect on ∂H
3
. The distance
between these two polar planes is 0. The hyperbolic exterior angle between the two planes
is 0 if the orientations of two polar planes agree, andπ otherwise.
Note. In case (b),⟨e x,e y⟩< 0 if and only if the two half-spaces ofH
3
defined by the oriented polar
planes P
e x
and P
e y
, respectively, are disjoint.
Note. When x and y are both hyperideal, their lifts are both space-like vectors and thus their norms
are positive real. Hence we get∥e x∥∥e y∥> 0. In case (c), the exterior angle is 0 when⟨e x,e y⟩< 0 or
⟨e x,e y⟩=− ∥e x∥∥e y∥, and isπ when⟨e x,e y⟩> 0 or⟨e x,e y⟩=∥e x∥∥e y∥. Therefore, according to case (c),
the equation in case (a) can be extended from α∈(0,π) to α∈[0,π]. The equation in case (b)
also can be extended from d > 0 to d≥ 0 by case (c).
Proposition 9. Let x be a finite point and y be a hyperideal point. Then, for any lifte x ande y, the
signed hyperbolic distance d between x and P
e y
is calculated by
⟨e x,e y⟩= i∥e x∥∥e y∥sinhd,
where⟨e x,e y⟩< 0 if and only ife x ande y lie in opposite sides of the hyperplanee y
⊥
.
Note. When⟨e x,e y⟩> 0, we have sinhd < 0, or the distance d between x and P
e y
is negative.
7
1.2.2 Hyperbolic truncated tetrahedra
Now we explicitly define our main object, a truncated tetrahedron. Consider a tetrahedron in R
3,1
whose verticese v
i
, i= 1, . . . , 4 are all distinct and their projections π(e v
i
)∈RP
3
, i= 1, . . . , 4 are
distinct as well. We further assume that the projections are not coplanar. In summary, suppose that
four verticese v
i
are linearly independent inR
3,1
. Let
e
∆ be a 4-dimensional cone inR
3,1
where its
vertex is the origin O and its base is the tetrahedron with verticese v
i
. Then,
e
∆ can be expressed as
follows:
e
∆={a
1
e v
1
+ a
2
e v
2
+ a
3
e v
3
+ a
4
e v
4
̸= O| a
i
≥ 0∀i}. (1.1)
Denote the projections of
e
∆ ande v
i
by∆ :=π(
e
∆) and v
i
:=π(e v
i
), respectively. Then∆ is a tetrahe-
dron inRP
3
with vertices v
i
. The cone
e
∆ is called a lift of the tetrahedron∆, ande v
i
is called a lift
of the vertices v
i
.
Definition 10. For a projective tetrahedron∆ and its vertices v
i
, we say it is well-defined when the
liftse v
i
, i= 1, . . . , 4 are linearly independent.
For the rest part of the paper, assume that every projective tetrahedron is well-defined.
Proposition 11. Given a tetrahedron∆⊂ RP
3
with vertices v
i
, there are exactly two 4-dimensional
cones that are lifts of∆. In other words, to define a tetrahedron ∆ inRP
3
, we need to clarify the
lifts of vertices.
Here, we give a brief note with examples regarding the Proposition 11. Given a projective
tetrahedron∆ and its vertices v
i
, let
e
∆ ande v
i
be the lifts of∆ and v
i
, respectively. Consider another
set of lifts c
i
e v
i
of v
i
, where c
i
are nonzero real numbers. Regardless of a choice of c
i
, its projection
π(c
i
e v
i
) agrees with v
i
.
If c
i
> 0 for all i, then the 4-dimensional cone corresponding to c
i
e v
i
is the same cone as
e
∆, hence
the tetrahedron derived from c
i
e v
i
agrees with∆ inRP
3
. When c
i
< 0 for all i, the 4-dimensional
cone corresponding to these c
i
e v
i
are located on the opposite side of
e
∆ and thus its projection is
identical with∆. Therefore, given a tetrahedron∆, there are two lifts
e
∆ of∆: one derived from c
i
v
i
with c
i
> 0 for all i, and the other derived from c
i
v
i
with c
i
< 0 for all i.
8
∆
∆
′
∆
′′
v
1
v
2
v
3
v
4
RP
3
Figure 1.1: The red tetrahedron ∆, the blue tetrahedron ∆
′
, and the green tetrahedron ∆
′′
are in
RP
3
and they all have the same vertices v
1
,··· ,v
4
. The two tetrahedra ∆ and ∆
′
share one face
and there are three more tetrahedra that share one face with∆. On the other hand,∆ and∆
′′
share
two opposite edges and we have total three tetrahedra that share two opposite edges with ∆. In
summary, given four vertices v
i
for i= 1,··· ,4, there are eight different tetrahedra inRP
3
: one
∆-like tetrahedron, four∆
′
-like tetrahedra, and three∆
′′
-like tetrahedra.
However, when c
i
are neither all positive nor all negative, the cone obtained from c
i
e v
i
is dif-
ferent from
e
∆ and hence the corresponding tetrahedron is different from∆ as well. For example,
in Figure 1.1, observe that∆,∆
′
, and∆
′′
are different tetrahedra. In fact, if we consider a cone
e
∆
′
with verticese v
′
i
= c
′
i
e v
i
where c
′
1
< 0 and c
′
2
, c
′
3
, c
′
4
> 0, then the projection∆
′
=π(
e
∆
′
) is another
tetrahedron inRP
3
. Note that∆
′
and∆ share one face which does not contain v
1
. The vertices c
′
i
e v
i
where c
′
1
> 0 and c
′
2
, c
′
3
, c
′
4
< 0 define the cone located on the opposite side of
e
∆
′
and thus define
the same tetrahedron∆
′
. On the other hand, the cone
e
∆
′′
with verticese v
′′
i
= c
′′
i
v
i
where c
′′
i
satisfy
either c
′′
1
, c
′′
2
< 0, c
′′
3
, c
′′
4
> 0, or c
′′
1
, c
′′
2
> 0, c
′′
3
, c
′′
4
< 0 projects to another tetrahedron∆
′′
⊂ RP
3
where∆
′′
shares two opposite edges v
1
v
2
and v
3
v
4
with∆.
Note that ∆, ∆
′
, and ∆
′′
still sharing the same four vertices v
i
∈RP
3
. Therefore we have
multiple tetrahedra which have the same 4 vertices inRP
3
. Since there is a choice of sign for the
lift of each vertex, we have 16 different cones inR
3,1
. Among them, the pair of cones located on
the opposite sides project to the same tetrahedron inRP
3
, and we thus have total eight tetrahedra
that share the same vertices.
9
Now we define the dual tetrahedron. Let ∆ be a tetrahedron inRP
3
with its vertices v
i
and
e
∆,
e v
i
be the lifts of∆, v
i
, respectively. Then define a polar set
e
∆
∗ of∆ as
e
∆
∗ =
n
y∈R
3,1
\{O}|⟨y,x⟩≤ 0∀x∈
e
∆
o
.
Since x∈
e
∆ can be represented as a
1
e v
1
+a
2
e v
2
+a
3
e v
3
+a
4
e v
4
with a
i
≥ 0, i= 1, . . . , 4, the polar set
e
∆
∗ can be rewritten as
e
∆
∗ =

y∈R
3,1
\{O}|⟨y,a
1
e v
1
+ a
2
e v
2
+ a
3
e v
3
+ a
4
e v
4
⟩≤ 0∀a
i
≥ 0, i= 1,...,4

=

y∈R
3,1
\{O}|⟨y,e v
i
⟩≤ 0∀i

,
which makes
e
∆
∗ a 4-dimensional cone. Consider a projection ∆
∗ =π(
e
∆
∗ ) of
e
∆
∗ toRP
3
and let
e n
i
∈R
3,1
for i= 1, . . . , 4 be points satisfying

e n
i
,e v
j

= 0 ∀ j̸= i, ⟨e n
i
,e v
i
⟩=− 1.
Lemma 12. ∆
∗ is a projective tetrahedron with vertices n
i
:=π(e n
i
) and the liftse n
i
define ∆
∗ .
Proof. First, we show that∆
∗ is a well-defined projective tetrahedron, or all e n
i
are linearly inde-
pendent. In fact, by definition of e n
i
, we have
N
⊤
JV =− I
4
,
where I
4
is the identity matrix of dimension 4 and
N=






| |
e n
1
··· e n
4
| |






, J=









− 1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1









, V =






| |
e v
1
··· e v
4
| |






.
10
Hence, we have N =− V
− 1
J. Since alle v
i
are linearly independent, suche n
i
always exist and are
linearly independent as well. Therefore,∆
∗ is well-defined.
For a point y in
e
∆
∗ ⊂ R
3,1
, using the basis{e n
i
∀i}, it can be represented as
y= b
1
e n
1
+ b
2
e n
2
+ b
3
e n
3
+ b
4
e n
4
, b
i
∈R, i= 1,...,4.
Then we have the following equivalences from the definition of e n
i
:
⟨y,e v
i
⟩≤ 0 ⇔ ⟨b
1
e n
1
+ b
2
e n
2
+ b
3
e n
3
+ b
4
e n
4
,e v
i
⟩≤ 0 ⇔ b
i
⟨e n
i
,e v
i
⟩≤ 0 ⇔ b
i
≥ 0.
Hence we get
e
∆
∗ ={b
1
e n
1
+ b
2
e n
2
+ b
3
e n
3
+ b
4
e n
4
̸={O}| b
i
≥ 0∀i}.
In other words,
e
∆
∗ is the 4-dimensional cone inR
3,1
where its vertex is the origin and its base is
the tetrahedron with verticese n
i
. Therefore, the projection ∆
∗ is a projective tetrahedron inRP
3
with verticesπ(e n
i
)= n
i
and∆
∗ is defined by the lifts e n
i
.
Definition 13. We call∆
∗ a dual tetrahedron of∆.
Note. The three verticese n
j
, j̸= i are contained in the orthogonal subspace ofe v
⊥
i
by the definition
ofe n
i
. Thus, n
j
, j̸= i are contained in the dual space v
⊥
i
. In other words, the dual space v
⊥
i
contains
the opposite face of n
i
in∆
∗ . Likewise, the dual space n
⊥
i
contains the opposite face of v
i
in∆.
Lemma 14. Let∆ be a projective tetrahedron and∆
∗ be its dual tetrahedron. The dual tetrahedron
of∆
∗ is∆.
Proof. For the verticese n
i
of the base of the cone
e
∆
∗ , we need to find vertices e w
i
that defines ∆
∗∗ .
The verticese w
i
satisfy the following relations withe n
i
:

e n
i
,e w
j

= 0 ∀ j̸= i, ⟨e n
i
,e w
i
⟩=− 1.
11
From Lemma 12, we know that the verticese v
i
satisfy theses relations and they are unique up to
positive scalar multiplication. Therefore, we have
e
∆
∗∗ =
e
∆.
Now we define a truncation at a vertex of a projective tetrahedron. Let ∆ be a tetrahedron in
RP
3
with vertices v
i
. Lete v
i
be the lifts of v
i
that define ∆ and let
e
∆ be the lift of∆. The tetrahedron
truncated at v
i
is defined as
π

e
∆∩

y∈R
3,1
\{O}|⟨y,e v
i
⟩≤ 0

,
andπ

e
∆∩

y∈R
3,1
\{O}|⟨y,e v
i
⟩= 0

is called a truncation face at v
i
. Note that the truncation
face is contained in v
⊥
i
. It is called a truncation because nonhyperbolic parts of∆ is truncated by
taking the intersection. In fact, if a vertex v
i
of∆ is hyperideal, the truncation face meetsH
3
by
Proposition 5. Taking intersection with

y∈R
3,1
\{O}|⟨y,e v
i
⟩≤ 0

truncates the space-like part
of
e
∆ neare v
i
by the orthogonal planee v
⊥
i
. On the other hand, if a vertex v
i
is finite, the truncation
face does not meet the truncation face at v
i
again by Proposition 5. Therefore, in the view ofRP
3
,
the truncation occurs only at hyperideal vertices and the nonhyperbolic part near the hyperideal
vertex v
i
is truncated by the dual space v
⊥
i
.
The tetrahedron truncated at its all vertices v
i
can be simplified as follow: (see Figure 1.2)
π

e
∆∩
4
\
i=1

y∈R
3,1
\{O}|⟨y,e v
i
⟩≤ 0

!
=π

e
∆∩

y∈R
3,1
\{O}|⟨y,e v
i
⟩≤ 0∀i

=π

e
∆∩
e
∆
∗
.
Definition 15. For a projective tetrahedron∆, the truncated tetrahedron∆
trunc
associated with∆
is defined as ∆
trunc
=π

e
∆∩
e
∆
∗
, or a tetrahedron truncated at its all vertices v
i
.
Note. ∆
trunc
is different from ∆∩∆
∗ . Specifically, for a projective tetrahedron ∆, there are two
distinct lifts, where one is in the opposite position of the other with respect to the origin. Say
e
∆
and− e
∆ are two lifts. For these lifts, there exist polar sets
e
∆
∗ and− e
∆
∗ , respectively. These two
polar sets are two distinct lifts of∆
∗ . Hence, ∆∩∆
∗ consists of two parts: one is π

e
∆∩
e
∆
∗
=
12
H
3
v
1
v
2
v
3
v
4
∆
trunc
v
1
v
2
v
3
v
4
n
1
n
2
n
3
n
4
∆
∆
∗ ∆
trunc
Figure 1.2: The figure on the left is the truncated tetrahedron ∆
trunc
obtained from a tetrahedron∆
by truncation at v
i
. The figure on the right is the intersection of ∆ and∆
∗ . Observe that those two
agree.
π

− e
∆∩− e
∆
∗
and the other is π

e
∆∩− e
∆
∗
=π

− e
∆∩
e
∆
∗
. Therefore, ∆∩∆
∗ includes ∆
trunc
and the inclusion is strict if
e
∆∩− e
∆
∗ is nonempty.
Let g
i
be the face of the dual tetrahedron∆
∗ which is opposite to n
i
. As we mentioned in the
note after Definition 13, g
i
is contained in the hyperplane v
⊥
i
and so is the truncation face at v
i
.
Hence a truncation at v
i
can be considered as a truncation by the face g
i
of∆
∗ . The notation will
be clear in the next chapter.
Lemma 16. The truncated tetrahedron∆
trunc
is contained in hyperbolic space.
Proof. Pick a point x∈∆
trunc
and its lifte x∈R
3,1
. By definition of ∆
trunc
, we havee x∈
e
∆∩
e
∆
∗ . It
means thate x satisfies ⟨e x,e y⟩≤ 0 for alle y∈
e
∆, in particular, fore x∈
e
∆. Therefore we get⟨e x,e x⟩≤ 0,
or x∈H
3
, and finally ∆
trunc
⊂ H
3
.
Note. A truncated tetrahedron∆
trunc
maybe empty. For instance, if∆ does not intersect withH
3
,
then we have∆
trunc
⊂ π(
e
∆)⊂ H
3
c
. Since∆
trunc
∈H
3
by the Lemma 16,∆
trunc
is empty. We are
only interested in a nonempty∆
trunc
and want to compute its volume.
13
Lemma 17. For a tetrahedron ∆ in RP
3
, the truncated tetrahedron ∆
trunc
associated with ∆ is
identical with the truncated tetrahedron(∆
∗ )
trunc
associated with∆
∗ .
Proof. By definition, we have ∆
trunc
=π

e
∆∩
e
∆
∗
and (∆
∗ )
trunc
=π

e
∆
∗ ∩
e
∆
∗∗
. From Lemma
14, we have
e
∆
∗∗ =
e
∆ and thus∆
trunc
and(∆
∗ )
trunc
are identical.
Definition 18. Let∆ be a tetrahedron inRP
3
with vertices v
i
, edges e
i j
connecting two vertices
v
i
and v
j
, and faces f
i
opposite to v
i
. Let l
i j
be the projective line that contains e
i j
. An edge e
i j
is called finite if l
i j
intersectsH
3
, ideal if l
i j
is tangent to∂H
3
, and hyperideal if l
i j
does not meet
H
3
. Similarly, let P
i
be a projective plane that contains f
i
. A face f
i
is called finite if P
i
intersects
H
3
, ideal if P
i
is tangent to∂H
3
, and hyperideal if P
i
does not meetH
3
. Finally, when∆∈H
3
, the
projective tetrahedron∆ is called finite .
Lemma 19. Let∆ be a tetrahedron inRP
3
with vertices v
i
and edges e
i j
connecting v
i
and v
j
, and
∆
∗ be its dual with vertices n
i
and edges e
∗ i j
connecting n
i
and n
j
. Then, e
i j
is finite/ideal/hyperideal
if and only if e
∗ kl
is hyperideal/ideal/finite respectively, where indices i, j, k, l are all distinct.
Proof. For edge e
i j
, let m be the projective line that contains e
i j
. Then the lifte m is
e m=

a
i
e v
i
+ a
j
e v
j
∈R
3,1
| a
i
,a
j
∈R

.
Then, its orthogonal space is
e m
⊥
=

b
k
e n
k
+ b
l
e n
l
∈R
3,1
| b
k
,b
l
∈R

,
where the indices i, j, k, and l are all distinct. Note that e m
⊥
is a lift of the projective line m
⊥
that
contains e
kl
. Then m
⊥
is the dual space of m. Therefore, m
⊥
is finite/ideal/hyperideal if and only
if m is hyperideal/ideal/fintie, respectively. Therefore, the same relation holds for e
i j
and e
∗ kl
.
14
1.2.3 Volume functions of truncated tetrahedra
In this subsection, we introduce the known formulas that are applicable to some truncated tetrahera.
To be specific, [9] gave the formula for a finite tetrahedron and [10] showed that this formula also
can be applied to a truncated tetrahedron with six edges meetingH
3
. (See the left in Figure 1.3.)
Define a complex-valued function P
Li
= P
Li
(ζ,ξ
12
,ξ
13
,ξ
14
,ξ
23
,ξ
24
,ξ
34
) as
P
Li
:= Li
2
(ζ)+ Li
2
(ξ
12
ξ
13
ξ
24
ξ
34
ζ)+ Li
2
(ξ
12
ξ
14
ξ
23
ξ
34
ζ)+ Li
2
(ξ
13
ξ
14
ξ
23
ξ
24
ζ)
− Li
2
(− ξ
12
ξ
13
ξ
14
ζ)− Li
2
(− ξ
12
ξ
23
ξ
24
ζ)− Li
2
(− ξ
13
ξ
23
ξ
34
ζ)− Li
2
(− ξ
14
ξ
24
ξ
34
ζ)
(1.2)
whereξ
i j
, i̸= j, andζ are complex values. Here, Li
2
(z) is the dilogarithm function defined by the
analytic continuation of the following integral:
Li
2
(x)=− Z
x
0
log(1− t)
t
dt for a real number x< 1.
Let x
i j
, i̸= j be complex values and z
± be the two solutions
z
− =
− q− p
q
2
− 4pr
2p
, z
+
=
− q+
p
q
2
− 4pr
2p
of the quadratic equation pz
2
+ qz+ r= 0, where
p= x
12
x
13
x
14
x
23
x
24
x
34
+ x
12
x
34
+ x
13
x
24
+ x
14
x
23
+ x
12
x
13
x
23
+ x
12
x
14
x
24
+ x
13
x
14
x
34
+ x
23
x
24
x
34
,
q=−
x
12
− 1
x
12

x
34
− 1
x
34

+

x
13
− 1
x
13

x
24
− 1
x
24

+

x
14
− 1
x
14

x
23
− 1
x
23

,
r=
1
x
12
x
13
x
14
x
23
x
24
x
34
+
1
x
12
x
34
+
1
x
13
x
24
+
1
x
14
x
23
+
1
x
12
x
13
x
23
+
1
x
12
x
14
x
24
+
1
x
13
x
14
x
34
+
1
x
23
x
24
x
34
.
15
v
1
v
2
v
3
v
4
∆
trunc
v
1
v
2
v
3
v
4
∆
trunc
Figure 1.3: Observe that ∆
trunc
in the left has hyperideal vertices but finite (red) edges. On the
other hand,∆
trunc
in the right has two finite vertices v
1
, v
4
, and two hyperideal vertices v
2
, v
3
, and
the hyperideal edge e
i j
. We thus has a new edge that came from the intersection of two truncation
faces v
⊥
2
and v
⊥
3
. The new edge is also contained in the edge e
∗ 14
connecting two vertices n
1
and
n
4
of∆
∗ .
Theorem 20 (Murakami-Yano, Ushijima). Let ∆
trunc
be a truncated tetrahedron with dihedral
anglesα
i j
of edge e
i j
connecting two vertices v
i
of∆, whose edges are all finite. When x
i j
= e
iα
i j
,
the hyperbolic volume of∆
trunc
is obtained as
V ol(∆
trunc
)= Im
P
Li
(z
− ,x
12
,x
13
,x
14
,x
23
,x
24
,x
34
)− P
Li
(z
+
,x
12
,x
13
,x
3
,x
4
,x
5
,x
6
)
4
.
Murakami and Ushijima found out another volume formula for a finite tetrahedron in terms of
edge lengths instead of dihedral angles. Interestingly, they used the same function P
Li
and z
± as
the formula in terms of dihedral angles.
Define a complex-valued function Q
Li
(ξ
12
,ξ
13
,ξ
14
,ξ
23
,ξ
24
,ξ
34
) as follows:
Q
Li
(ξ
12
,ξ
13
,ξ
14
,ξ
23
,ξ
24
,ξ
34
)
:=
i
4
(
P
Li
(z
− ,ξ
12
,ξ
13
,ξ
14
,ξ
23
,ξ
24
,ξ
34
)− z
− ∂P
Li
∂ζ

ζ=z
− · logz
− !
−
P
Li
(z
+
,ξ
12
,ξ
13
,ξ
14
,ξ
23
,ξ
24
,ξ
34
)− z
+
∂P
Li
∂ζ

ζ=z
+
· logz
+
!)
(1.3)
16
Theorem 21 (Murakami-Ushijima). Let∆ be a finite tetrahedron with edge lengths l
i j
of edge e
i j
connecting two vertices v
i
of∆. When x
i j
= e
l
i j
, the hyperbolic volume of∆ is obtained as
V ol(∆)= V
l
− ∑
{i, j}⊂{ 1,2,3,4}
l
i j
· ∂V
l
∂l
i j
,
where V
l
:= Q
Li
(− x
34
,− x
24
,− x
23
,− x
14
,− x
13
,− x
12
).
Finally, Kolpakov and Murakami introduced at [12] and modified at [15] the formula for a
truncated tetrahedron where two vertices are inH
3
and the other two are outsideH
3
, and the edge
connecting these two hyperideal vertices are also hyperideal. In this case, two truncation faces meet
inside of the given tetrahedron and their intersection is a new edge of the truncated tetrahedron.
(See the right in Figure 1.3) The volume formula of this type of tetrahedron also uses the same
function P
Li
and Q
Li
.
Theorem 22 (Kolpakov-Murakami). Let∆
trunc
be a truncated tetrahedron as given in above de-
scription. In other words, let v
1
, v
2
be two hyperideal vertices, and e be the intersection of two
truncation faces v
⊥
1
and v
⊥
2
. Letα
i j
be dihedral angles of edges e
i j
except for e
34
, l andα be the
length of e and the corresponding dihedral angle, and V ol(∆
trunc
) be its hyperbolic volume. Then,
when x
34
= e
l
and x
i j
= e
iα
i j
for the others,
V ol(∆
trunc
)= Re

− V+ x
34
· ∂Q
Li
∂ξ
34

ξ
i j
=x
i j
· logx
34
!
,
where V = Q
Li
(x
12
,x
13
,x
14
,x
23
,x
24
,x
34
).
17
Chapter 2
Volumes of proper truncated tetrahedra
2.1 Proper truncated tetrahedra
In this section, we define the main object of the paper, a nonideal proper truncated tetrahedron.
The properness of a truncated tetrahedron was first introduced in [13].
Definition 23. Let∆ be a tetrahedron inRP
3
with its vertices v
i
, i= 1, . . . , 4, and∆
∗ be its dual
tetrahedron with its vertices n
i
, i= 1, . . . , 4. Let f
i
be a face of∆ opposite to v
i
and g
i
be a face of
∆
∗ opposite to n
i
. We say a truncation at v
i
is proper if f
i
and g
i
do not meet inH
3
. A truncated
tetrahedron∆
trunc
is proper if all truncations are proper.
Note. The properness is a property of the truncation. In fact, for a hyperideal vertex v
i
, when the
truncation at v
i
is proper, the truncation face is contained in g
i
of∆
∗ .
Note. A properness at v
i
is defined in H
3
⊂ RP
3
. In the view ofR
3,1
, it can be defined similarly as
well. For example, define two sets
e
f
4
ande g
4
inR
3,1
as follows:
e
f
4
=

a
1
e v
1
+ a
2
e v
2
+ a
3
e v
3
∈R
3,1
|a
1
,a
2
,a
3
≥ 0

, − e
f
4
=
n
− x∈R
3,1
|x∈
e
f
4
o
e g
4
=

b
1
e n
1
+ b
2
e n
2
+ b
3
e n
3
∈R
3,1
|b
1
,b
2
,b
3
≥ 0

, − e g
4
=

− x∈R
3,1
|x∈e g
4

Then we haveπ(
e
f
4
)=π(− e
f
4
)= f
4
and the same holds fore g
4
and− e g
4
. When f
4
and g
4
meet in
H
3
, we have four possibilities in terms of
e
f
4
ande g
4
:
18
(a)
e
f
4
ande g
4
meet inR
3,1
T
,
(b) − e
f
4
and− e g
4
meet inR
3,1
T
,
(c)
e
f
4
and− e g
4
meet inR
3,1
T
, and
(d) − e
f
4
ande g
4
meet inR
3,1
T
.
Note that (a) holds if and only if (b) holds, and (c) holds if and only if (d).
We claim that the case (c) and (d) cannot happen. We only need to show that (c) cannot happen.
Let x be a point in f
4
∩g
4
⊂ H
3
and lete x be a lift of x contained in
e
f
4
∩− e g
4
. Thene x can be written
ase x=∑
3
i=1
a
i
e v
i
=∑
3
j=1
b
j
e n
j
, where a
i
> 0 and b
j
< 0 for i, j= 1, . . . , 3. Hence we have
⟨e x,e x⟩=
*
3
∑
i=1
a
i
e v
i
,
3
∑
j=1
b
j
e n
j
+
=
3
∑
i=1
a
i
b
i

e v
i
,e n
j

+
∑
i̸= j
a
i
b
j

e v
i
,e n
j

=− 3
∑
i=1
a
i
b
i
> 0.
Thuse x always lies inR
3,1
S
, which contradicts to x∈H
3
. Therefore, if f
4
and g
4
meet inH
3
, in the
view ofR
3,1
, the lifts
e
f
4
ande g
4
meet inR
3,1
T
.
In conclusion, in the view ofR
3,1
, the truncation at v
i
is proper if
e
f
i
ande g
i
do not meet inR
3,1
T
.
Definition 24. Let∆ be a tetrahedron inRP
3
with vertices v
i
, edges e
i j
connecting two vertices v
i
and v
j
, and faces f
i
opposite to v
i
. We say a tetrahedron∆ is nonideal when all vertices, edges,
faces are not ideal. A truncated tetrahedron∆
trunc
associated with a nonideal tetrahedron∆ is also
called nonideal.
Note. By Definition 18, neither the projective lines containing edges e
i j
nor the projective planes
containing f
i
are tangent to∂H
3
.
Lemma 25. If a tetrahedron∆ is nonideal, then its dual tetrahedron∆
∗ is also nonideal.
Proof. For the vertex n
i
of the dual tetrahedron∆
∗ , it is ideal if and only if n
⊥
i
is tangent to the
hyperbolic space H
3
. Since n
⊥
i
contains f
i
and f
i
is nonideal, we get n
⊥
i
is not tangent to H
3
.
Therefore, n
i
is not ideal. Similarly, the face g
i
of ∆
∗ is nonideal because the vertex v
i
of ∆ is
nonideal.
19
For the edges, by Lemma 19, since all the edges of∆ are nonideal, the edges of∆
∗ are nonideal
as well. In sum, all vertices, edges, and faces of the dual tetrahedron∆
∗ is nonideal.
In the remainder of the paper, we assume that every truncated tetrahedron is nonempty, proper,
and nonideal unless it is specified.
2.2 Combinatorial structures of truncated tetrahedra
The combinatorial structure of a (nonempty, nonideal, and proper) truncated tetrahedron∆
trunc
is
the information of∆
trunc
as a polyhedron including numbers of vertices, edges, and faces. On the
other hand, the combinatorial property of a projective tetrahedron∆ is the information of numbers
of finite/hyperideal vertices, edges, and faces. In this section, we show that the combinatorial
structure of∆
trunc
is decided by the combinatorial properties of∆.
First of all, we clarify the notations. Let∆ be a projective tetrahedron with vertices v
i
, edges
e
i j
, and faces f
i
. From Lemma 12, the dual tetrahedron∆
∗ has n
i
as its vertices. Let e
∗ i j
be an edge
of∆
∗ connecting two vertices n
i
and n
j
, where i, j are two distinct indices, and g
i
be a face of∆
∗ opposite to n
i
.
Among components of the combinatorial structure of a truncated tetrahedron∆
trunc
, we start
from the faces. Since∆
trunc
is an intersection of a projective tetrahedron∆ and its dual tetrahedron
∆
∗ , there are two types of faces of∆
trunc
: one from∆ and the other from∆
∗ . The number of faces
of∆
trunc
is between 4 and 8 because both∆ and∆
∗ have four faces.
Definition 26. In a truncated tetrahedron ∆
trunc
, we call a face of ∆
trunc
a original face if it is
contained in a face of∆, and a truncation face if it is contained in a face of∆
∗ . See Figure 2.1a.
Secondly, we inspect how the edges of∆
trunc
look like. Since a truncated tetrahedron∆
trunc
is
a polyhedron, every edge of∆
trunc
is always an intersection of two faces. Since there are two types
of faces, either an original face or a truncation face, there are three types of edges in truncated
tetrahedron:
20
(a) an intersection of two original faces (an edge derived from original tetrahedron∆),
(b) an intersection of two truncation faces (an edge derived from dual tetrahedron∆
∗ ), and
(c) an intersection of an original face and a truncation face.
Definition 27. In a truncated tetrahedron ∆
trunc
, we say an edge is an original edge if it is an
intersection of two original faces, a dual edge if it is an intersection of two truncation faces, and a
truncation edge if it is an intersection of one original face and one truncation face. We say an edge
is essential if it is either original edge or dual edge. See Figure 2.1a.
Note. We are interested in essential edges rather than truncation edges. The reason will be covered
in the later sections.
Lemma 28. For an original edge e of a truncated tetrahedron∆
trunc
, where e is derived from an
edge e
i j
of∆, then exactly one of the following holds (see Figure 2.1b):
(1) e= e
i j
and the endpoints of e are v
i
and v
j
.
(2) one endpoint of e is one vertex (v
i
or v
j
) and the other endpoint is the intersection of e
i j
and
the orthogonal plane (v
⊥
j
or v
⊥
i
, respectively) of the other vertex.
(3) the endpoints of e are intersections of e
i j
and the orthogonal planes v
⊥
i
and v
⊥
j
.
The same argument holds for a dual edge.
Proof. Let e be an original edge of a truncated tetrahedron∆
trunc
. Then, there exists an edge e
i j
of
∆ which contains e. If e
i j
of∆ does not meet the faces of∆
∗ , then neither does e (⊂ e
i j
). Hence e
i j
is not truncated by the dual tetrahedron and we have e= e
i j
. In this case, the endpoints of e are v
i
and v
j
, which corresponds to the case (1).
Suppose e
i j
meets some faces of∆
∗ . Then among four faces of∆
∗ , only g
i
⊂ v
⊥
i
and g
j
⊂ v
⊥
j
can meet e
i j
. In fact, e
i j
is an intersection of two faces f
k
, f
l
⊂ ∆, where i, j, k, l are all distinct
indices. Since the truncated tetrahedron is proper, e
i j
cannot meet the faces g
k
and g
l
of∆
∗ .
21
v
1
v
2
v
3
v
4
∆
trunc
(a) In this example, red faces are original
faces and blue faces are truncation faces.
In terms of edges, the red edges are original
edges, the black edges are truncation edges,
and the blue edge is a dual edge.
v
1
v
2
v
3
v
4
v
⊥
1
v
⊥
4
∆
trunc
(b) In this example, black edges are trunca-
tion edges and all the other colored edges
are original. The green edge is of type (1)
in Lemma 28, the orange edges are of type
(2), and the yellow edge is of type (3).
Figure 2.1: Two examples of nonempty, nonideal, and proper truncated tetrahedra.
First, when e
i j
meets one face of∆
∗ , say g
i
, then e
i j
is divided into two parts by the intersection
of e
i j
and g
i
, and e is one of them. Let x be the intersection of e
i j
and g
i
. For the case of x= v
i
or
x= v
j
, it can be considered similarly as the previous case. Hence, we may assume that x is different
from v
i
and v
j
. Then, there exists a lifte x of x that can be represented ase x= te v
i
+(1− t)e v
j
for
some real number t∈(0,1). Note that t satisfies the following equation:

te v
i
+(1− t)e v
j
,e v
i

= t⟨e v
i
,e v
i
⟩+(1− t)

e v
i
,e v
j

= 0.
If e is the part between v
i
and x, then v
i
is contained in∆
trunc
⊂ ∆
∗ . From the definition of ∆
∗ , we
get⟨e v
i
,e v
i
⟩≤ 0 and

e v
i
,e v
j

≤ 0. Combined with above equation, we conclude that⟨e v
i
,e v
i
⟩= 0 and

e v
i
,e v
j

= 0. Then, the edge e
i j
is contained in v
⊥
i
and v
⊥
i
is tangent toH
3
. Therefore, the edge e
i j
is not in the hyperbolic space, and therefore it cannot be a part of the truncated tetrahedron∆
trunc
.
Therefore, as in the case (2), e should be the edge between v
j
and x.
Lastly, when e
i j
meets both g
i
and g
j
, with intersection x
i
and x
j
, x
i
is located between v
i
and
x
j
. This is because, otherwise, the whole e
i j
is not contained in∆
trunc
. Then, e cannot be the edge
22
between v
i
and x
i
or the edge between v
j
and v
j
for the same reason as above case. Hence, e is the
edge between x
i
and x
j
as mentioned in the case (3).
In summary, for an original edge e⊂ e
i j
, the endpoints of e is either the endpoint of e
i j
or the
intersection of e
i j
and the orthogonal planes v
⊥
i
or v
⊥
j
. Similar argument holds for a dual edge.
Lemma 29. Let∆ be a nonideal projective tetrahedron with vertices v
i
, ande v
i
be the lifts of v
i
that define ∆. Let∆
trunc
be the proper truncated tetrahedron and e be an original edge of∆
trunc
that is contained in e
i j
, which is connecting v
i
and v
j
. Then,

e v
i
,e v
j

< 0. Similarly, for the dual
edge e of∆
trunc
that is contained in e
∗ i j
, we have

e n
i
,e n
j

< 0.
Proof. If e
i j
does not meet the faces of ∆
∗ , then the endpoints of e are v
i
and v
j
as we saw in
previous Lemma 28. In this case, we have v
i
, v
j
∈H
3
and especially, their lifts are located in the
same components ofR
3,1
T
. Therefore, by Proposition 1,

e v
i
,e v
j

< 0.
Suppose e
i j
meets some faces of∆
∗ . Again, as we saw in above Lemma 28, either v
⊥
i
or v
⊥
j
meets the edge e
i j
. Without loss of generality, assume that v
⊥
i
meets e
i j
. Then, since v
⊥
i
meets the
hyperbolic space, v
i
is hyperideal, or⟨e v
i
,e v
i
⟩> 0. In the view ofR
3,1
, the edge connectinge v
i
and
e v
j
is divided by eithere v
⊥
i
, ande v
i
∈R
3,1
S
. Sincee v
j
is located on the other side ofe v
i
with respect to
e v
⊥
i
, the Lorentzian product

e v
i
,e v
j

is negative. The similar argument holds for the dual edge.
Lemma 30. Let e be a truncation edge of a truncated tetrahedron∆
trunc
. Then the dihedral angle
associated with e is the right angle.
Proof. Let e be a truncation edge and say it is the intersection of a face f
i
from∆ and a face g
j
from ∆
∗ . By properness of ∆
trunc
, two indices i and j are distinct. Also, from Lemma 16, both
f
i
and g
j
are finite, or both n
i
and v
j
are hyperideal. Then, since we have f
i
⊂ n
⊥
i
, g
j
⊂ v
⊥
j
, and

n
i
,v
j

= 0, the dihedral angle associated with e isπ/2 by Proposition 8.
Finally, we classify the vertices of ∆
trunc
. Note that all vertices are an intersection of three
or more faces of ∆
trunc
. Suppose there is a vertex that is an intersection of four or more faces.
Each face came from either∆ or∆
∗ . Since four faces from the same tetrahedron do not meet at
23
one vertex, the maximum number of faces from one tetrahedron is three. There are three possible
cases:
(a) one face of∆ and three faces of∆
∗ meet.
(b) three faces of∆ and one face of∆
∗ meet.
(c) two faces of∆ and two faces of∆
∗ meet.
Lemma 31. Let ∆ be a nonideal projective tetrahedron, ∆
∗ be its dual, and ∆
trunc
be a proper
truncated tetrahedron. Then,
(a) a face of∆ cannot contain a vertex of∆
∗ .
(b) a face of∆
∗ cannot contain a vertex of∆.
(c) an edge of∆ cannot meet the edge of∆
∗ .
Proof. Starting with (a), suppose a face f
i
of∆ contain a vertex n
i
′ of∆
∗ . Since n
i
′ is the inter-
section of three faces other than g
i
′, these three faces of∆
∗ meet f
i
. By properness of∆
trunc
, the
one face of∆
∗ that does not meet f
i
should be g
i
. We thus have i
′
= i and v
i
∈ g
i
⊂ v
⊥
i
. Hence
⟨v
i
,v
i
⟩= 0, or v
i
is ideal, which contradicts to the assumption that∆ is nonideal. The same argu-
ment holds for (b).
Now we move on to (c). Suppose e
i j
and e
∗ i
′
j
′
meet. Since e
i j
is the edge that the faces f
k
, f
l
meet where i, j, k, l are distinct indices, g
k
and g
l
cannot meet e
i j
because of the properness. Then,
the edge e
∗ i
′
j
′
should be the intersection of g
i
and g
j
, which is the edge e
∗ kl
. From Lemma 19, when
one of e
i j
and e
∗ kl
is finite, then the other is hyperideal. Thus they can meet in H
3
only if both are
ideal. It contradicts again to the assumption that∆ is nonideal.
From above Lemma, all vertices of∆
trunc
is an intersection of three faces. Then there are four
types of edges because a vertex of∆
trunc
is an intersection of three faces of either∆ or∆
∗ . The
four types are as follows:
(a) an intersection of three faces of∆. All three adjacent edges of the vertex are original edges.
24
(b) an intersection of three faces of∆
∗ . All three adjacent edges are dual edges.
(c) an intersection of two faces of∆ and one face of∆
∗ . Among three adjacent edges, one is an
original edge and the other two are truncation edges.
(d) an intersection of one face of∆ and two faces of∆
∗ . Among three adjacent edges, one is a
dual edge and the other two are truncation edges.
Note that the vertices of types (c) and (d) came from the truncation.
Lemma 32. For any truncated tetrahedron∆
trunc
, the number of essential edges is always six.
Proof. We use the classification of the vertices, edges, and faces. Regarding vertices, let v
1
, v
2
, v
3
,
and v
4
be the number of vertices of types (a), (b), (c), and (d), respectively. For edges, let e
o
, e
d
,
and e
t
be the number of original edges, dual edges, and truncation edges, respectively. Finally, let
f
o
, f
t
be the number of original faces, truncation faces, respectively.
Consider the vertices v
i
of ∆. If v
i
is finite, then it is counted as a vertex of type (a). If v
i
is hyperideal, then it is truncated by the dual tetrahedron and counted as a truncation face at the
hyperideal vertex. Therefore, v
1
+ f
t
always agree with the number of vertices of∆, which is 4.
Similarly, we have v
2
+ f
o
= 4 from the view of∆
∗ .
On the other hand, in the viewpoint of four types of vertices of∆
trunc
and their three adjacent
edges, the truncation edges appear only in a vertices of types (c) and (d), and exactly two adjacent
edges of these vertices are truncation edges. Since each truncation edge is counted twice from its
two endpoints, we have
2e
t
= 2v
3
+ 2v
4
, or e
t
= v
3
+ v
4
.
The Euler characteristic of three dimensional polyhedron is 2. Thus we have
2=(v
1
+ v
2
+ v
3
+ v
4
)− (e
o
+ e
d
+ e
t
)+( f
o
+ f
t
)
=(v
1
+ f
t
)+(v
2
+ f
o
)+(v
3
+ v
4
− e
t
)− (e
o
+ e
d
)= 8− (e
o
+ e
d
),
25
or e
o
+ e
d
= 6. Therefore, the sum of numbers of original edges and dual edges, or the number of
essential edges, is always 6.
Now we show that the combinatorial property of∆ decides the combinatorial structure of∆
trunc
.
Lemma 33. Let∆
trunc
be a (nonempty, nonideal, and proper) truncated tetrahedron. Then, there
exists exactly one essential edge denoted as e
ess
i j
associated with two distinct indices i and j. The
essential edge e
ess
i j
is original if and only if the edge e
i j
is finite. In other words, for any two distinct
indices i and j, exactly one of the following holds:
• there exists an original edge in∆
trunc
which is contained in e
i j
.
• there exists a dual edge in∆
trunc
which is contained in e
∗ kl
, where i, j, k, l are all distinct.
Proof. Given two distinct indices i and j, only one of e
i j
and e
∗ kl
is finite by Lemma 19, where i, j,
k, l are all distinct. Therefore, at most one of the above two option holds for all distinct sets of two
indices, and the number of essential edges is at most six. However, from Lemma 32, the number
of essential edges are exactly six. Therefore, the equality holds for all sets of two distinct indices.
In other words, for any distinct pair of indices, one of the above two options holds.
26
2.3 Classification of truncated tetrahedra
In this section, we list up all the possible combinatorial structures of truncated tetrahedra. Tech-
nically, our formula brings one algebraic computation of the hyperbolic volume of a truncated
tetrahedron. However, our formula has different geometric interpretations depending on the com-
binatorial structure of a truncated tetrahedron. Hence, we need to classify truncated tetrahedra
based on their combinatorial structures. According to Lemma 33 in the previous section, the
classification is based on the combinatorial properties of ∆, including the number of hyperideal
vertices, hyperideal edges, and hyperideal faces.
First, we classify based on the number of hyperideal vertices of a projective tetrahedron∆.
(a) If there is no hyperideal vertex, then∆ is a finite tetrahedron that lies in H
3
.
(b) When ∆ has only one hyperideal vertex, there is also one simple case with one truncation
face.
(c) When ∆ has two hyperideal vertices v
1
and v
2
, we have two cases based on the combina-
torial property of the edge e
12
connecting v
1
and v
2
: one with finite e
12
and the other with
hyperideal e
12
. Since all the other edges are containing at least one finite vertex, those edges
are all finite.
(d) When∆ has three hyperideal vertices v
1
, v
2
, and v
3
, we have total five cases based on the
combinatorial properties of three edges e
12
, e
13
, and e
23
, and the face f
4
opposite to v
4
. Note
that if at least one of the three edges are finite, then so is f
4
.
i. All three edges e
12
, e
13
, and e
23
are finite.
ii. Two of them are finite and the other one is hyperideal.
iii. One of them is finite and the other two are hyperideal.
iv. All three edges are hyperideal, but the face f
4
is finite.
v. All three edges are hyperideal and f
4
is also hyperideal.
27
0 1 2 3 4
0
1
2
3
4
5
6
Table 2.1: Table of classification of combinatorial properties of projective tetrahedra. Row and
column represent the number of hyperideal vertices and hyperideal edges of projective tetrahedra,
respectively. The red vertices/edges/faces represent finite vertices/edges/faces respectively, and the
black represents hyperideal.
(e) When all four vertices of∆ are hyperideal, we have total 19 cases. See Table 2.1.
In Table 2.1, the column represents the number of hyperideal vertices and the row represents
the number of hyperideal edges. Red vertices, red edges and red faces of tetrahedra in the table
mean finite vertices, finite edges, and finite faces, respectively. On the other hand, black vertices,
black edges and black faces represent hyperideal vertices, hyperideal edges, and hyperideal faces,
respectively. In summary, we have total 28 possible combinatorial properties of tetrahedra.
Now we identify above classifications of tetrahedra as T
v,e, f
, where v, e, and f represents for
the number of hyperideal vertices, edges, and faces, respectively. For instance, the left in Figure
2.2 is one example of truncated tetrahedron∆
trunc
derived from a tetrahedron∆ in the class T
3,3,1
,
which was unaddressed in earlier works. In most cases, a triple (v,e, f) is identified as exactly
one class of tetrahedra. However, for some triples, they are matched with more than one classes.
28
v
1
v
2
v
3
v
4
v
1
v
2
v
3
v
4
Figure 2.2: The tetrahedron∆ in the left has three hyperideal vertices v
1
, v
2
, v
3
, three hyperideal
edges e
12
, e
13
, e
23
, and one hyperideal face f
4
, followed by∆∈ T
3,3,1
. Note that each hyperideal
vertex and its truncation face have the same color. The truncated tetrahedron∆
trunc
derived from
∆ in T
3,3,1
was unaddressed in earlier works. The projective tetrahedron ∆ in the right has only
hyperideal vertices, hyperideal edges, and hyperideal faces. Thus, we have∆
trunc
∈ T
4,6,4
. Since the
dual tetrahedron∆
∗ is finite, the corresponding truncated tetrahedron ∆
trunc
is finite as well. Hence,
in the view of∆
∗ , the truncated tetrahedron is derived from a tetrahedron in T
0,0,0
. Therefore, the
classes T
4,6,4
and T
0,0,0
of tetrahedra are giving the same combinatorial structure of corresponding
truncated tetrahedron.
For example, when the tetrahedron identified as the triple (v,e, f)=(4,2,0) has two black (hy-
perideal) edges. There are two subclasses: the one with adjacent black edges and the one with
diagonal (non-adjacent) black edges. Denote the first one by T
4,2a,0
(adjacent) and the second one
by T
4,2d,0
(diagonal). The tetrahedron with (v,e, f)=(4,4,0) has four black edges and two red
edges. Focusing on red edges, we have two subclasses for this triple as well: the one with adjacent
red edges and the one with diagonal red edges. Similarly, denote the first one by T
4,4a,0
(adjacent)
and the second one by T
4,4d,0
(diagonal). Finally, for the tetrahedron with(v,e, f)=(4,3,0), there
are three subclasses of tetrahedra with three black edges: the one with three edges sharing one
vertex, the one with three edges in z-shape, and the one with three edges surrounding one face.
Denote the first one by T
0,3v,4
, the second one by T
0,3z,4
, and the third one by T
0,3 f,4
. These classes
are summarized in Table 2.2.
As we saw in Lemma 17, the truncated tetrahedron ∆
trunc
can be obtained from either ∆ or
∆
∗ . In other words, one truncated tetrahedron may be obtained from two different classes in Table
29
(v,e, f)=(4,2,0) (v,e, f)=(4,3,0) (v,e, f)=(4,4,0)
T
4,2a,0
T
4,2d,0
T
4,3v,0
T
4,3z,0
T
4,3 f,0
T
4,4d,0
T
4,4a,0
Table 2.2: For some triples (v,e, f) (the number of hyperideal vertices, edges, and faces, respec-
tively), they are matched to more than one classes of projective tetrahedra. The table lists all such
triples and denotes each of them using its combinatorial characteristics.
2.1: one class that∆ is contained in and another class that∆
∗ is contained in. For example, the
right truncated tetrahedron in Figure 2.2 is obtained from an original tetrahedron∆ in T
4,6,4
whose
vertices, edges, and faces are all hyperideal. Then, the dual tetrahedron∆
∗ is a finite tetrahedron in
T
0,0,0
. Hence this truncated tetrahedron∆
trunc
can be obtained from a tetrahedron in either T
4,6,4
or
T
0,0,0
. We pair these two classes T
4,6,4
and T
0,0,0
of tetrahedra because the combinatorial structures
of two corresponding truncated tetrahedra are the same.
Now we find all pairs of classes of tetrahedra that have the same combinatorial structures of
corresponding truncated tetrahedra. By Proposition 5, the dual of a hyperideal vertex of∆ is a finite
face of∆
∗ and the dual of a hyperideal face of∆ is a finite vertex of ∆
∗ . Furthermore, by Lemma
19, the dual of a hyperideal edge of∆ is a finite edge of ∆
∗ . In sum, if∆ has v hyperideal vertices, e
hyperideal edges, and f hyperideal faces, then∆
∗ has(4− f) hyperideal vertices,(6− e) hyperideal
edges, and(4− v) hyperideal faces. Hence, in general, T
v,e, f
are paired with T
4− f,6− e,4− v
. Denote
classes of truncated tetrahedra derived from T
v,e, f
by T
trunc
v,e, f
. Then we have T
trunc
v,e, f
= T
trunc
4− f,6− e,4− v
.
There are a number of special cases.
(a) According to the general rule, the class T
4,2,0
is paired with T
4,4,0
and both have two sub-
classes. By Lemma 19, the subclass T
4,2a,0
is paired with T
4,4a,0
, and T
4,2d,0
is paired with
T
4,4d,0
.
(b) Similarly, the class T
4,3,0
is paired with itself. Because of Lemma 19, the subclass T
4,3v,0
is
paired with T
4,3v,0
. On the other hand, the subclass T
4,3z,0
is paired with itself.
(c) There is one more class that is paired with itself: T
3,3,1
.
30
T
trunc
0,0,0
= T
trunc
4,6,4
T
trunc
2,1,0
= T
trunc
4,5,2
T
trunc
3,2,0
= T
trunc
4,4,1
T
trunc
4,0,0
= T
trunc
4,6,0
T
trunc
4,2d,0
= T
trunc
4,4d,0
T
trunc
1,0,0
= T
trunc
4,6,3
T
trunc
3,0,0
= T
trunc
4,6,1
T
trunc
3,3,0
= T
trunc
4,3,1
T
trunc
4,1,0
= T
trunc
4,3,0
T
trunc
4,3v,0
= T
trunc
4,3 f,0
T
trunc
2,0,0
= T
trunc
4,6,2
T
trunc
3,1,0
= T
trunc
4,5,1
T
trunc
3,3,1
= T
trunc
3,3,1
* T
trunc
4,2a,0
= T
trunc
4,4a,0
T
trunc
4,3z,0
= T
trunc
4,3z,0
*
Table 2.3: The classification of combinatorial structures of truncated tetrahedra. (* is the class
which is paired with itself.)
Table 2.3 lists all pairs of classes of tetrahedra that derive the same combinatorial structures of
corresponding truncated tetrahedra. We conclude that there are 15 classes of truncated tetrahedra.
Note that the earlier works dealt with only a number of cases of these 15 classes. [9] and [11]
gave volume formulas for compact tetrahedra that correspond to T
trunc
0,0,0
in terms of dihedral angles
and edge lengths, respectively. The formula in [10], which are in terms of dihedral angles, can be
applied to multiple classes of truncated teterahedra whose six edges are all finite, including T
trunc
0,0,0
,
T
trunc
1,0,0
, T
trunc
2,0,0
, T
trunc
3,0,0
, and T
trunc
4,0,0
. Finally, [12] suggested a formula that gives the volume of any
truncated tetrahedra in T
trunc
2,1,0
.
Truncated tetrahedra that are in the other nine classes were unaddressed so far. We provide
two different geometric formula for such classes of truncated tetrahedra, one in terms of dihedral
angles and the other in terms of edge lengths. For the class T
trunc
0,0,0
of compact tetrahedra, these two
formulas agree with the formulas given in [9] and [11]. For the other covered classes T
trunc
1,0,0
, T
trunc
2,0,0
,
T
trunc
3,0,0
, T
trunc
4,0,0
, and T
trunc
2,1,0
, the formula in terms of dihedral angles agree with existing formulas. We
are giving one more computation of volumes for the tetrahedra in these classes in terms of edge
lengths.
31
2.4 Main theorems
2.4.1 Basic functions
The Euler dilogarithm Li
2
(z) is defined as the following (See Figure 2.3):
Li
2
(z)=
Z
z
0
− log(1− u)
u
du for z∈C\[1,∞).
Note that the Euler dilogarithm is complex-valued and is discontinuous along the cut. To be pre-
cise, the Euler dilogarithm Li
2
(z) jumps by 2πilog|z| as z crosses the cut.
z
O
C
1
path of integral
cut
Figure 2.3: The Euler dilogarithm Li
2
(z) is defined as an integral along the path connecting O and
z. Since the domain is simply connected, it does not depend on the path.
The Bloch-Wigner dilogarithm D(z) :C→R is defined as
D(z)= Im[Li
2
(z)]+ arg(1− z)log|z|,
where ‘Im’ denotes the imaginary part and ‘arg’ denotes the branch of the argument lying in
(− π,π]. The Bloch-Wigner dilogarithm D(z) is not only continuous onC but also real analytic on
C except at the two points 0 and 1. For more details, see [16].
32
Now we define a function that plays an important role in computation of the hyperbolic volume
V ol(∆
trunc
) of a truncated tetrahedron∆
trunc
. For complex variables z
i j
∈C and z∈C, let P be a
linear combination of the Bloch-Wigner dilogarithm functions D defined as
P(z,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
)
= D(z)+ D(z
12
z
13
z
24
z
34
z)+ D(z
12
z
14
z
23
z
34
z)+ D(z
13
z
14
z
23
z
24
z)
− D(− z
12
z
13
z
23
z)− D(− z
12
z
14
z
24
z)− D(− z
13
z
14
z
34
z)− D(− z
23
z
24
z
34
z).
The function P is inpsired by [9], where the Euler dilogarithm functions are replaced by the Bloch-
Wigner dilogarithm functions. We call P a modified Murakami-Yano function .
2.4.2 Quadratic equation
Given six parameters z
12
, z
13
, . . . , z
34
, consider a quadratic equation
pz
2
+ qz+ r= 0 (2.1)
where the coefficients p, q, and r are given as
p= z
12
z
13
z
14
z
23
z
24
z
34
+ z
12
z
34
+ z
13
z
24
+ z
14
z
23
+ z
12
z
13
z
14
+ z
12
z
23
z
24
+ z
13
z
23
z
34
+ z
14
z
24
z
34
,
q=−
z
12
− z
− 1
12

z
34
− z
− 1
34

+

z
13
− z
− 1
13

z
24
− z
− 1
24

+

z
14
− z
− 1
14

z
23
− z
− 1
23

,
r=(z
12
z
13
z
14
z
23
z
24
z
34
)
− 1
+(z
12
z
34
)
− 1
+(z
13
z
24
)
− 1
+(z
14
z
23
)
− 1
+(z
12
z
13
z
14
)
− 1
+(z
12
z
23
z
24
)
− 1
+(z
13
z
23
z
34
)
− 1
+(z
14
z
24
z
34
)
− 1
.
(2.2)
This quadratic equation is derived from
(1− z)(1− z
12
z
13
z
24
z
34
z)(1− z
12
z
14
z
23
z
34
z)(1− z
13
z
14
z
23
z
24
z)
=(1+ z
12
z
13
z
23
z)(1+ z
12
z
14
z
24
z)(1+ z
13
z
14
z
34
z)(1+ z
23
z
24
z
34
z).
(2.3)
33
Lemma 34. Lete v
i
, i= 1, . . . , 4 be four points inR
3,1
that are linearly independent. Furthermore,
supposee v
i
have nonzero norms. Define z
i j
, i̸= j be the complex value determined as a solution of
z
i j
+ z
− 1
i j
=− 2·

e v
i
,e v
j

∥e v
i
∥∥e v
j
∥
. (2.4)
Assuming pr̸= 0 for p, r in (2.2) defined by such z
i j
, the equation (2.1) defined by such z
i j
has
degree 2 and has two distinct nonzero solutions.
Remark. The complex value z
i j
is either a real number with absolute value greater than 1 or a
complex number with absolute value 1.
Proof. We should prove that the quadratic equation (2.1) has two distinct solutions. In other words,
we show that the discriminant of (2.1) is nonzero. In fact, we have
q
2
− 4pr= 16· det









1 − z
12
+z
− 1
12
2
− z
13
+z
− 1
13
2
− z
14
+z
− 1
14
2
− z
12
+z
− 1
12
2
1 − z
23
+z
− 1
23
2
− z
24
+z
− 1
24
2
− z
13
+z
− 1
13
2
− z
23
+z
− 1
23
2
1 − z
34
+z
− 1
34
2
− z
14
+z
− 1
14
2
− z
24
+z
− 1
24
2
− z
34
+z
− 1
34
2
1









. (2.5)
Therefore, when z
i j
satisfy (2.4), we get
q
2
− 4pr= 16· det

e v
i
,e v
j

∥e v
i
∥∥e v
j
∥
!
i, j=1,...,4
=
16
∥e v
1
∥
2
∥e v
2
∥
2
∥e v
3
∥
2
∥e v
4
∥
2
detG,
where G is the Gram matrix

e v
i
,e v
j

i, j=1,...,4
. We have detG < 0 because of the signature of
the inner product inR
3,1
. Combining with∥e v
i
∥
2
̸= 0 for all indices, the discriminant q
2
− 4pr is
nonzero.
In summary, since p̸= 0 and the discriminant is nonzero, the quadratic equation has two distinct
solutions. Since r̸= 0, the solutions are both nonzero.
When z
i j
are given as Lemma 34, let z
+
and z
− be two distinct solutions of the quadratic
equation (2.1). Now we determine which of two solutions is called z
+
. To do so, we should
34
precisely define
p
q
2
− 4pr. First, we fix
√
detG as
√
detG= i
√
− detG where i is given as i=
√
− 1. Then, from
p
q
2
− 4pr=
4· √
detG
∥e v
1
∥∥e v
2
∥∥e v
3
∥∥e v
4
∥
,
we have
p
q
2
− 4pr∈























iR
+
all four vertexe v
i
are inR
3,1
T
orR
3,1
S
,
R
+
one vertex is inR
3,1
T
and the other three are inR
3,1
S
,
iR
− two vertices are inR
3,1
T
and the other two are inR
3,1
S
,
R
− three vertices are inR
3,1
T
and the other is inR
3,1
S
.
(2.6)
Then, z
− and z
+
are defined as follows:
z
+
=
− q+
p
q
2
− 4pr
2p
, z
− =
− q− p
q
2
− 4pr
2p
. (2.7)
2.4.3 Volumes of hyperbolic truncated tetrahedra
Now we show that the algebraic formula in the following theorem computes the volume of nonempty,
nonideal and proper truncated tetrahedron followed by geometric corollaries.
Theorem 35. Let ∆⊂ RP
3
be a projective tetrahedron and v
i
∈RP
3
, i= 1, . . . , 4 be the four
noncoplanar and nonideal vertices of ∆. Suppose that ∆
trunc
is a (nonideal) proper truncated
tetrahedron associated with∆. Let z
i j
be a complex solution of (2.4) and z
− and z
+
be given as
(2.7). Then the hyperbolic volume of∆
trunc
is computed as
V ol(∆
trunc
)=
1
4
P(z
− ,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
)− 1
4
P(z
+
,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
). (2.8)
35
Theorem 35 can be interpreted geometrically by using edge lengths. Define d
gen
i j
as follows:
d
gen
i j
=























































(hyperbolic length of the original edge e
i j
),
when v
i
,v
j
∈H
3
,
(hyperbolic length of the original edge contained in e
i j
)+ i
π
2
,
when v
i
∈H
3
,v
j
∈ dS
3
,
(hyperbolic length of the original edge contained in e
i j
)+ iπ,
when v
i
,v
j
∈ dS
3
and the line connecting v
i
,v
j
meetsH
3
,
i(hyperbolic (exterior) dihedral angle of the dual edge e
∗ kl
),
when the line connecting v
i
,v
j
does not meetH
3
,
(2.9)
Definition 36. For two distinct indices i and j, the quantity d
gen
i j
∈C defined as above is called
an generalized edge length of the essential edge e
ess
i j
of the nonideal proper truncated tetrahedron
∆
trunc
.
In particular, from Proposition 7, 8, and 9, d
gen
i j
satisfies
1
2

e
d
gen
i j
+ e
− d
gen
i j

=

e v
i
,e v
j

∥e v
i
∥∥e v
j
∥
.
The name came from the first case of (2.9): when ∆⊂ H
3
, the generalized edge length d
gen
i j
is the
hyperbolic length of the edge e
i j
of∆. Then, we have the following corollary where the volume is
given as formula in terms of generalized edge lengths.
Corollary 37. Let d
gen
i j
be the generalized edge length of the essential edge e
ess
i j
. Then hyperbolic
volume of a nonideal proper truncated tetrahedron∆
trunc
is given by
V ol(∆
trunc
)= Q

− e
d
gen
12
,− e
d
gen
13
,− e
d
gen
14
,− e
d
gen
23
,− e
d
gen
24
,− e
d
gen
34

,
36
where Q(z
12
,z
13
,z
14
,z
23
,z
24
,z
34
) is the function given in the right hand side of (2.8).
On the other hand, Theorem 35 is applicable for the dual tetrahedron∆
∗ , and in this version,
we use dihedral angles instead of edge lengths. Define α
gen
i j
as follows:
α
gen
i j
=























































π - (hyperbolic (exterior) dihedral angle of the original edge e
i j
),
when the line connecting n
i
, n
j
does not meetH
3
.
i(hyperbolic length of the dual edge e
∗ kl
),
when n
k
,n
l
∈ dS
3
and the line connecting n
k
, n
l
meetsH
3
,
i(hyperbolic length of the dual edge e
∗ kl
)+
π
2
,
when n
k
∈H
3
,n
l
∈ dS
3
,
i(hyperbolic length of the dual edge e
∗ kl
)+π,
when n
k
,n
l
∈ dS
3
.
(2.10)
Definition 38. For two distinct indices i and j, the quantityα
gen
i j
∈C defined as above is called an
generalized dihedral angle of the essential edge e
ess
i j
of the nonideal proper truncated tetrahedron
∆
trunc
.
Similarly as above,α
gen
i j
suffices
− 1
2

e
iα
gen
i j
+ e
− iα
gen
i j

=
⟨e n
k
,e n
l
⟩
∥e n
k
∥∥e n
l
∥
.
The name again came from the first case of (2.10): when ∆⊂ H
3
, the generalized dihedral angle
α
i j
is the hyperbolic interior angle between two faces n
⊥
k
and n
⊥
l
, or dihedral angle at e
i j
of ∆.
Then, we have another corollary where the volume is given as formula in terms of dihedral angles.
37
Corollary 39. Let α
gen
i j
be the generalized dihedral angle of the essential edge e
ess
i j
. Then hyper-
bolic volume of a nonideal proper truncated tetrahedron∆
trunc
is given by
V ol(∆
trunc
)= Q

e
iα
gen
34
,e
iα
gen
24
,e
iα
gen
14
,e
iα
gen
23
,e
iα
gen
13
,e
iα
gen
12

,
where Q(z
12
,z
13
,z
14
,z
23
,z
24
,z
34
) is the function given in the right hand side of (2.8).
2.5 Properties of Gram matrices
In this section, we introduce the properties of Gram matrix in the Lorentzian spaceR
3,1
. Consider
a tetrahedron inR
3,1
with distinct pointse v
i
, i= 1, . . . , 4 that are linearly independent. Suppose that
alle v
i
are not on light-coneR
3,1
L
, or∥e v
i
∦= 0. We further assume that neither 2-dimensional planes
containing two vertices and the origin O nor hyperplanes containing three vertices and the origin
O are tangent to the light-coneR
3,1
L
. Let
e
∆ be the 4-dimensional cone inR
3,1
where its vertex is
the origin O and its base is the tetrahedron with verticese v
i
. Note that
e
∆ is a lift of the nonideal
projective tetrahedron∆.
Definition 40. The Gram matrix G of
e
∆ is the symmetric matrix of Lorentzian products, whose
entries are given by G
i j
=

e v
i
,e v
j

∈R, where i, j= 1, . . . , 4.
Lemma 41. The determinant of the Gram matrix G of
e
∆ is always negative.
Proof. The Gram matrix G of
e
∆ can be written as
G=






− e v
⊤
1
− .
.
.
− e v
⊤
4
− 














− 1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1















| |
e v
1
··· e v
4
| |






,
38
or G= V
⊤
JV where
V =






| |
e v
1
··· e v
4
| |






, J=









− 1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1









.
Here, V is invertible sincee v
i
are linearly independent. Therefore, we get detG=− (detV)
2
<
0.
Definition 42. The normalized Gram matrix
b
G of
e
∆ is a symmetric matrix of Lorentzian products
divided by the Lorentzian norms of vertices, whose entries are given by
b
G
i j
=

e v
i
,e v
j

∥e v
i
∥∥e v
j
∥
, i, j= 1,...,4.
Remark. The normalized Gram matrix is well-defined when ∥e v
i
∦= 0 for all i. By definition, the
normalized Gram matrix reduces the effect of norms∥e v
i
∥. Since the norm of a vertex is real or pure
imaginary, the normalized Gram matrix is a complex-valued matrix where the entries are either real
or pure imaginary. Note that the normalized Gram matrix is not Hermitian but symmetric.
Lemma 43. The determinant of the normalized Gram matrix
b
G of
e
∆ is real and nonzero.
Proof. The normalized Gram matrix
b
G of
e
∆ can be written as
b
G= D
− 1
v
GD
− 1
v
, where
D
v
=









∥e v
1
∥ 0 0 0
0 ∥e v
2
∥ 0 0
0 0 ∥e v
3
∥ 0
0 0 0 ∥e v
4
∥









.
Then we get
det
b
G=
detG
∏
4
i=1
∥e v
i
∥
2
.
39
Since ∥e v
i
∦= 0, the squared norm ∥e v
i
∥
2
is a nonzero real number. Because of Lemma 41, we
conclude that det
b
G is nonzero.
Note. det
b
G is either positive or negative, depending on the number of space-like vertices among
e v
i
: the determinant is negative when the number of space-like vertices are even, and is positive
when the number is odd.
Lemma 44. Any non-diagonal entries of normalized Gram matrix
b
G of∆ are different from± 1.
Proof. Because of the assumption in the beginning of the section, we have that the 2-dimensional
plane containing two verticese v
i
,e v
j
and the origin O is not tangent to the light-coneR
3,1
L
. In other
words, the set{se v
i
+ te v
j
|s,t∈R} is not tangent to the light-cone. Then the discriminant of the
quadratic equation

se v
i
+te v
j
,se v
i
+te v
j

is nonzero, i.e.,

e v
i
,e v
j

2
−∥ e v
i
∥
2
∥e v
j
∥
2
̸= 0.
Therefore, the non-diagonal entry
b
G
i j
is different from± 1.
Now we consider the polar set
e
∆
∗ of
e
∆ which is defined as
e
∆
∗ =
n
y∈R
3,1
\{O}|⟨y,x⟩≤ 0∀x∈
e
∆
o
.
Lete n
i
be another set of four vertices such that

e n
i
,e v
j

= 0 ∀ j̸= i, ⟨e n
i
,e v
i
⟩=− 1,
or N
⊤
JV =− I
4
, where N is defined as
N=






| |
e n
1
··· e n
4
| |






.
40
Since V is invertible, N can be written as N =− J(V
⊤
)
− 1
. Then N is invertible as well, or the
verticese n
i
, i= 1, . . . , 4 are linearly independent as well ase v
i
. Note thate n
i
is the normal vector of
the hyperplane that contains O and three verticese v
j
, j̸= i. By the assumption in the beginning of
the section, the hyperplane is not tangent toR
3,1
L
, ore n
i
/ ∈R
3,1
L
. Thus, we get∥e n
i
∦= 0. Eache n
i
is
uniquely determined by four verticese v
i
. By Lemma 12,
e
∆
∗ is another 4-dimensional cone inR
3,1
where its vertex is the origin O and its base is the tetrahedron with verticese n
i
.
Lemma 45. An entry of the normalized Gram matrix
b
G
∗ of
e
∆
∗ satisfies

b
G
∗ i j

2
=

e n
i
,e n
j

2
∥e n
i
∥
2
∥e n
j
∥
2
=
b
C
2
i j
b
C
ii
b
C
j j
,
where
b
C is the cofactor matrix of
b
G.
Proof. The Gram matrix G
∗ of
e
∆
∗ can be expressed as
G
∗ = N
⊤
JN=

− V
− 1
J

· J·
− J(V
⊤
)
− 1

= V
− 1
J(V
⊤
)
− 1
=

V
⊤
JV

− 1
= G
− 1
= D
− 1
v
b
G
− 1
D
− 1
v
,
and then the normalized Gram matrix
b
G
∗ of
e
∆
∗ is given as
b
G
∗ = D
− 1
n
G
∗ D
− 1
n
= D
− 1
n
D
− 1
v
· b
G
− 1
· D
− 1
v
D
− 1
n
.
Then we have

e n
i
,e n
j

∥e n
i
∥∥e n
j
∥
=
b
G
∗ i j
=
1
∥e v
i
∥∥e n
i
∥
· b
C
ji
det
b
G
· 1
∥e v
j
∥∥e n
j
∥
=
1
∥e v
i
∥∥e v
j
∥∥e n
i
∥∥e n
j
∥
· b
C
i j
det
b
G
,
where
b
C
i j
is the(i, j)-cofactor of
b
G. Therefore, we get
b
C
ii
= det
b
G·∥e v
i
∥
2
∥e n
i
∥
2
,
b
C
i j
= det
b
G·

e n
i
,e n
j

·∥e v
i
∥∥e v
j
∥ for i̸= j,
(2.11)
41
followed by
b
C
2
i j
b
C
ii
b
C
j j
=

e n
i
,e n
j

2
∥e n
i
∥
2
∥e n
j
∥
2
.
Lemma 46. The(i,i)-cofactor matrix
b
C
ii
is a nonzero real number.
Proof. Since ∆ is a nonideal tetrahedron, both ∥e v
i
∥
2
and ∥e n
i
∥
2
are nonzero. Combining with
Lemma 43 and (2.11), we get
b
C
ii
∈R and
b
C
ii
̸= 0.
Lemma 47. For the cofactor matrix
b
C of
b
G, we have
b
C
2
kl
− b
C
kk
b
C
ll
=

e v
i
,e v
j

2
∥e v
i
∥
2
∥e v
j
∥
2
− 1
!
· det
b
G,
where i, j, k, l are all distinct.
Proof. From taking the determinant of the following equality of matrices,









det
b
G 0 0 0
0 det
b
G 0 0
b
C
31
b
C
32
b
C
33
b
C
34
b
C
41
b
C
42
b
C
43
b
C
44









=









1
⟨e v
1
,e v
2
⟩
∥e v
1
∥∥e v
2
∥
⟨e v
1
,e v
3
⟩
∥e v
1
∥∥e v
3
∥
⟨e v
1
,e v
4
⟩
∥e v
1
∥∥e v
4
∥
⟨e v
2
,e v
1
⟩
∥e v
2
∥∥e v
1
∥
1
⟨e v
2
,e v
3
⟩
∥e v
2
∥∥e v
3
∥
⟨e v
2
,e v
4
⟩
∥e v
2
∥∥e v
4
∥
0 0 1 0
0 0 0 1









· 








b
C
11
b
C
12
b
C
13
b
C
14
b
C
21
b
C
22
b
C
23
b
C
24
b
C
31
b
C
32
b
C
33
b
C
34
b
C
41
b
C
42
b
C
43
b
C
44









,
we have
det
b
G
2
·
b
C
33
b
C
44
− b
C
2
34

=

1− ⟨e v
1
,e v
2
⟩
2
∥e v
1
∥
2
∥e v
2
∥
2
!
· det
b
G
3
,
or generally,
b
C
2
kl
− b
C
kk
b
C
ll
=

e v
i
,e v
j

2
∥e v
i
∥
2
∥e v
j
∥
2
− 1
!
· det
b
G.
42
2.6 Schl¨ afli’s formula
Recall Schl¨ afli’s differential formula (see, for example, [17]), which is going to be used in the
proof of the main theorem:
Theorem 48 (Schl¨ afli) . Let P
t
be a differentiable 1-parameter family of polyhedra inH
3
for t near
0. Then the derivative of the volume V
t
of P
t
at t= 0 is
d
dt
V
t
=
1
2
∑
e∈E
0
l
0
(e)
d
dt
α
t
(e),
where E
t
is a set of edges of P
t
, l
0
(e) is the length of the edge e andα
t
(e) is the (external) dihedral
angle of P
t
along the edge e.
Note. Here,α
t
is the external dihedral angle (π minus internal dihedral angle) of the corresponding
edge.
2.7 Derivatives of Bloch-Wigner dilogarithm and the volume
function
In this section, we compute the derivatives of Bloch-Wigner dilogarithm functions and especially
the volume function, the right hand side of (2.8). These derivatives will be used in Schl¨ afli’s
formula to verify that (2.8) has the same variation as the hyperbolic volume of a nonideal proper
truncated tetrahedron.
First, we deal with the derivative of Bloch-Wigner dilogarithm function
D(z)= Im[Li
2
(z)]+ arg(1− z)log|z|,
where ‘arg’ denotes the branch (− π,π] of the argument of the complex number. Since D is a
real-valued function unlike its complex domain, we should take derivatives with respect to real
43
variables. In that sense, we consider D(c(t)), where c(t) is a complex-valued real analytic function
of the real variable t. Then D(c(t)) is real analytic with respect to t except at{t∈R| c(t)= 0,1},
where its first derivative is given as follows:
Lemma 49. The first derivative of D (c(t)) is given as
d
dt
D(c(t))= log|c(t)|· d
dt
arg(1− c(t))− log|1− c(t)|· d
dt
arg(c(t)),
except at the set of non-differentiable points{t∈R| c(t)= 0,1}.
Note.
d
dt
argc(t) and
d
dt
arg(1− c(t)) are well-defined regardless of the range of principal values
because both 1− c(t) and c(t) are non-zero complex numbers.
Proof. Let the complex logarithm ‘log’ have its imaginary value in(− π,π]. From argz= Im[logz]
for z∈C, we have
d
dt
D(c(t))=
d
dt
Im[Li
2
(c(t))]+ arg(1− c(t))· d
dt
log|c(t)|+ log|c(t)|· d
dt
arg(1− c(t))
= Im

d
dt
Li
2
(c(t))+ log(1− c(t))· d
dt
log|c(t)|

+ log|c(t)|· d
dt
arg(1− c(t)). (2.12)
Let u(t),v(t) be two real-valued and real analytic functions that satisfies
c(t)= e
u(t)+iv(t)
.
Especially, by adjusting the branch of argument, we can always find the real analytic function v(t).
Then, we have c
′
(t)= c(t)· (u
′
(t)+ iv
′
(t)). Applying these to the inside of Im[− ] of (2.12), we get
d
dt
Li
2
(c(t))+ log(1− c(t))· d
dt
log|c(t)|=− log(1− c(t))
c(t)
· c
′
(t)+ log(1− c(t))· u
′
(t)
=− log(1− c(t))·
u
′
(t)+ iv
′
(t)

+ log(1− c(t))· u
′
(t)
=− ilog(1− c(t))· v
′
(t)
44
Note that v
′
(t)=
d
dt
argc(t)∈R is well-defined regardless of the branch. Therefore, we get
d
dt
D(c(t))= Im[− ilog(1− c(t))]· d
dt
argc(t)+ log|c(t)|· d
dt
arg(1− c(t)).
From Im[− ilogz]=− Re[logz]=− log|z|, we get
d
dt
D(c(t))=− log|1− c(t)|· d
dt
arg(c(t))+ log|c(t)|· d
dt
arg(1− c(t)).
Now we move on to the derivative of the volume function. Let ∆ be a nonideal projective
tetrahedron ande v
i
∈R
3,1
, i= 1, . . . , 4 be the lifts that defines ∆. Assume that the corresponding
truncated tetrahedron∆
trunc
is proper. Let z
i j
, i̸= j be non-zero complex variables that satisfies
z
i j
+ z
− 1
i j
=− 2·

e v
i
,e v
j

∥e v
i
∥

e v
j

.
Recall the quadratic equation pz
2
+ qz+ r= 0 where the coefficients are given as
p= z
12
z
13
z
14
z
23
z
24
z
34
+ z
12
z
34
+ z
13
z
24
+ z
14
z
23
+ z
12
z
13
z
14
+ z
12
z
23
z
24
+ z
13
z
23
z
34
+ z
14
z
24
z
34
,
q=−
z
12
− z
− 1
12

z
34
− z
− 1
34

+

z
13
− z
− 1
13

z
24
− z
− 1
24

+

z
14
− z
− 1
14

z
23
− z
− 1
23

,
r=(z
12
z
13
z
14
z
23
z
24
z
34
)
− 1
+(z
12
z
34
)
− 1
+(z
13
z
24
)
− 1
+(z
14
z
23
)
− 1
+(z
12
z
13
z
14
)
− 1
+(z
12
z
23
z
24
)
− 1
+(z
13
z
23
z
34
)
− 1
+(z
14
z
24
z
34
)
− 1
.
By Lemma 34, it has two distinct nonzero solutions z
− and z
+
. Recall the modified Murakami-
Yano function P:
P(z,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
)
= D(z)+ D(z
12
z
13
z
24
z
34
z)+ D(z
12
z
14
z
23
z
34
z)+ D(z
13
z
14
z
23
z
24
z)
− D(− z
12
z
13
z
23
z)− D(− z
12
z
14
z
24
z)− D(− z
13
z
14
z
34
z)− D(− z
23
z
24
z
34
z).
45
Let Q be the function given in the main theorem as follows:
Q(z
12
,z
13
,z
14
,z
23
,z
24
,z
34
)
=
1
4
P(z
− ,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
)− 1
4
P(z
+
,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
).
(2.13)
Since P is a linear sum of Bloch-Wigner dilogarithm functions, so is Q.
Lemma 50. For each Bloch-Wigner dilogarithm function in Q, its input is neither 0 nor 1, i.e.,
(a) both z
− and z
+
are neither 0 nor 1,
(b) for distinct indices i, j, k, l, we have z
i j
z
ik
z
jl
z
kl
z
± ̸= 0,1, and
(c) for distinct indices i, j, k, we have− z
i j
z
ik
z
jk
z
± ̸= 0,1.
Proof. First of all, by assumption that all z
i j
and the solutions z
± are nonzero, we get that the input
of each Bloch-Wigner dilogarithm function in Q cannot be zero.
Now we show that these inputs cannot be 1 as well. Starting with (a), suppose the quadratic
equation (2.1) has 1 as a solution. Then, we have p+ q+ r= 0, or
(1+ z
12
z
13
z
23
)(1+ z
12
z
14
z
24
)(1+ z
13
z
14
z
34
)(1+ z
23
z
24
z
34
)= 0.
Then one of the above four terms should be zero. Without loss of generality, assume that the first
term is zero, or z
12
z
13
z
23
=− 1. The(4,4)-cofactor matrix
b
C
44
of the normalized Gram matrix
b
G
of
e
∆ is given as
b
C
44
= det






1 − z
12
+z
− 1
12
2
− z
13
+z
− 1
13
2
− z
12
+z
− 1
12
2
1 − z
23
+z
− 1
23
2
− z
13
+z
− 1
13
2
− z
23
+z
− 1
23
2
1






=(z
12
z
13
z
23
+ 1)

1+
z
23
z
12
z
13

1+
z
13
z
12
z
23

1+
z
12
z
13
z
23

.
(2.14)
46
Therefore, applying z
12
z
13
z
23
=− 1 to above, we get
b
C
44
= 0, which contradicts to Lemma 46.
Hence, the solution z
± of Equation (2.1) cannot be 1.
Moving on to (b), suppose
z
12
z
13
z
24
z
34
z
− = 1. (2.15)
Since z
− satisfies Equation (2.3), we get
(1+ z
12
z
13
z
23
z
− )(1+ z
12
z
14
z
24
z
− )(1+ z
13
z
14
z
34
z
− )(1+ z
23
z
24
z
34
z
− )= 0. (2.16)
Then, one of the above four terms should be zero. Assume that the first term is zero, or
z
23
z
24
z
34
z
− =− 1. (2.17)
By Equations (2.15) and (2.17), we have z
12
z
13
=− z
23
. Applying this again to (2.14) we get
b
C
44
= 0, which contradicts again to Lemma 46. When another term of (2.16) is zero, we get the
similar results. The same holds for z
+
. Hence, for any distinct indices i, j, k, and l, we have
z
i j
z
ik
z
jl
z
kl
z
± ̸= 1.
Lastly, we investigate if (c) holds or not. Suppose z
12
z
13
z
23
z
− =− 1. Similarly as above, from
Equation (2.3), we get
(1− z
− )(1− z
12
z
13
z
24
z
34
z
− )(1− z
12
z
14
z
23
z
34
z
− )(1− z
13
z
14
z
23
z
24
z
− )= 0.
From (a), we know that z
− ̸= 1. From (b), we know that any of the latter three terms cannot be
zero. Therefore, we conclude that z
12
z
13
z
23
z
− =− 1 cannot hold. The same holds for other indices
and z
+
.
47
From Lemma 50, the inputs of Bloch-Wigner dilogarithm functions in Q are neither 0 nor 1.
Therefore, Q is real analytic on the space of z
i j
defined as Equation (2.4) in Lemma 34. Before we
compute the partial derivatives of Q, define r
i j
∈R andθ
i j
∈(− π,π] as
r
i j
= log|z
i j
|, θ
i j
= argz
i j
.
Then, we have z
i j
= e
r
i j
+iθ
i j
. Since the solution z
± are functions of complex variables z
i j
, it can be
considered as functions of real variables r
i j
andθ
i j
.
Lemma 51. Consider z
− and z
+
as functions of r
i j
andθ
i j
. Then, we compute the followings:
∂
∂r
i
0
j
0
Q

e
r
12
+iθ
12
,e
r
13
+iθ
13
,e
r
14
+iθ
14
,e
r
23
+iθ
23
,e
r
24
+iθ
24
,e
r
34
+iθ
34

=
1
4
∑
1≤ i< j≤ 4
r
i j
· ∂
∂r
i
0
j
0
arg
ψ
i j
(z
− )ϕ
i j
(z
+
)
ϕ
i j
(z
− )ψ
i j
(z
+
)
,
∂
∂θ
i
0
j
0
Q

e
r
12
+iθ
12
,e
r
13
+iθ
13
,e
r
14
+iθ
14
,e
r
23
+iθ
23
,e
r
24
+iθ
24
,e
r
34
+iθ
34

=− 1
4
log

ψ
i
0
j
0
(z
− )ϕ
i
0
j
0
(z
+
)
ϕ
i
0
j
0
(z
− )ψ
i
0
j
0
(z
+
)

+
1
4
∑
1≤ i< j≤ 4
r
i j
· ∂
∂θ
i
0
j
0
arg
ψ
i j
(z
− )ϕ
i j
(z
+
)
ϕ
i j
(z
− )ψ
i j
(z
+
)
,
where i
0
̸= j
0
,ϕ
i j
(z) andψ
i j
(z) are defined as
ψ
i j
(z)=(1− z
i j
z
ik
z
jl
z
kl
z)(1− z
i j
z
jk
z
il
z
kl
z),
ϕ
i j
(z)=(1+ z
i j
z
ik
z
jk
z)(1+ z
i j
z
il
z
jl
z),
(2.18)
and i, j, k, l are distinct indices.
Proof. We compute the second equation here because it is more complicated. Since the indices are
symmetric, it is enough to see what
∂
∂θ
12
Q is. We will use Lemma 49 to compute. For
∂
∂θ
12
D(z
− ),
∂
∂θ
12
D(z
− )= log|z
− |· ∂
∂θ
12
arg(1− z
− )− log|1− z
− |· ∂
∂θ
12
arg(z
− )
= r
− · ∂
∂θ
12
arg(1− z
− )− log|1− z
− |· ∂θ
− ∂θ
12
,
48
where r
− , θ
− are the logarithm of absolute value and the argument of z
− = e
r
− +iθ
− , respectively.
Secondly, we compute
∂
∂θ
12
D(z
12
z
13
z
24
z
34
z
− ).
∂
∂θ
12
D(z
12
z
13
z
24
z
34
z
− )= log|z
12
z
13
z
24
z
34
z
− |· ∂
∂θ
12
arg(1− z
12
z
13
z
24
z
34
z
− )
− log|1− z
12
z
13
z
24
z
34
z
− |· ∂
∂θ
12
arg(z
12
z
13
z
24
z
34
z
− )
=(r
12
+ r
13
+ r
24
+ r
34
+ r
− )· ∂
∂θ
12
arg(1− z
12
z
13
z
24
z
34
z
− )
− log|1− z
12
z
13
z
24
z
34
z
− |· ∂
∂θ
12
(θ
12
+θ
13
+θ
24
+θ
34
+θ
− )
=(r
12
+ r
13
+ r
24
+ r
34
+ r
− )· ∂
∂θ
12
arg(1− z
12
z
13
z
24
z
34
z
− )
− log|1− z
12
z
13
z
24
z
34
z
− |·
1+
∂θ
− ∂θ
12

.
Similarly,
∂
∂θ
12
D(− z
12
z
13
z
23
z
− )=(r
12
+ r
13
+ r
23
+ r
− )· ∂
∂θ
12
arg(1+ z
12
z
13
z
23
z
− )
− log|1+ z
12
z
13
z
23
z
− |·
1+
∂θ
− ∂θ
12

.
Combining with
∂
∂θ
12
arg(xy)=
∂
∂θ
12
arg(x)+
∂
∂θ
12
arg(y) for any x,y∈C, we get
∂
∂θ
12
P

z
− ,e
r
12
+iθ
12
,e
r
13
+iθ
13
,e
r
14
+iθ
14
,e
r
23
+iθ
23
,e
r
24
+iθ
24
,e
r
34
+iθ
34

= r
− · ∂
∂θ
12
argS(z
− ,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
)
+
∑
1≤ i< j≤ 4
r
i j
· ∂
∂θ
12
arg
(1− z
i j
z
ik
z
jl
z
kl
z
− )(1− z
i j
z
jk
z
il
z
kl
z
− )
(1+ z
i j
z
ik
z
jk
z
− )(1+ z
i j
z
il
z
jl
z
− )
− log

(1− z
12
z
13
z
24
z
34
z
− )(1− z
12
z
14
z
23
z
34
z
− )
(1+ z
12
z
13
z
23
z
− )(1+ z
12
z
14
z
24
z
− )

− log|S(z
− ,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
)|· ∂θ
− ∂θ
12
,
(2.19)
49
where k,l are the remaining two indices in{1,2,3,4} except for i and j, and
S(z,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
)
=
(1− z)(1− z
12
z
13
z
24
z
34
z)(1− z
12
z
14
z
23
z
34
z)(1− z
13
z
14
z
23
z
24
z)
(1+ z
12
z
13
z
23
z)(1+ z
12
z
14
z
24
z)(1+ z
13
z
14
z
34
z)(1+ z
23
z
24
z
34
z)
.
Since z
− is the solution of S(z,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
)= 1 by Equation (2.3), we have
log|S|= 0, and argS= 0.
Then, (2.19) can be simplified as
∂
∂θ
12
P

z
− ,e
r
12
+iθ
12
,e
r
13
+iθ
13
,e
r
14
+iθ
14
,e
r
23
+iθ
23
,e
r
24
+iθ
24
,e
r
34
+iθ
34

=
∑
1≤ i< j≤ 4
r
i j
· ∂
∂θ
12
arg
ψ
i j
(z
− )
ϕ
i j
(z
− )
− log

ψ
12
(z
− )
ϕ
12
(z
− )

.
The same holds for z
+
, so we can get the second equation of the Lemma. Similarly, we can show
that the first equation holds.
Before we move on to the proof of the main theorem, we compute the ratio
ψ
i j
(z
− )ϕ
i j
(z
+
)
ϕ
i j
(z
− )ψ
i j
(z
+
)
which
came up in the previous lemma.
Lemma 52. The ratio R
i j
=
ψ
i j
(z
− )ϕ
i j
(z
+
)
ϕ
i j
(z
− )ψ
i j
(z
+
)
satisfies
R
i j
+ 2+
1
R
i j
= 4· ⟨e n
k
,e n
l
⟩
2
∥e n
k
∥
2
∥e n
l
∥
2
,
where the indices i, j, k, l are all distinct ande n
i
∈R
3,1
are the four lifts that define the dual
tetrahedron∆
∗ ⊂ RP
3
of∆.
50
Proof. By algebraic computation, we have
ψ
i j
(z
− )ϕ
i j
(z
+
)
=
4
p
2
z
2
i j
z
kl
(z
ik
z
kl
+ z
il
)(z
il
z
kl
+ z
ik
)(z
jk
z
kl
+ z
jl
)(z
jl
z
kl
+ z
jk
)

b
C
kl
− z
i j
− z
− 1
i j
2
p
det
b
G
!
,
ϕ
i j
(z
− )ψ
i j
(z
+
)
=
4
p
2
z
2
i j
z
kl
(z
ik
z
kl
+ z
il
)(z
il
z
kl
+ z
ik
)(z
jk
z
kl
+ z
jl
)(z
jl
z
kl
+ z
jk
)

b
C
kl
+
z
i j
− z
− 1
i j
2
p
det
b
G
!
,
where p is defined by (2.2), the indices i, j, k, l are all distinct, and
b
C
kl
is the (k,l)-entry of the
cofactor matrix
b
C of the normalized Gram matrix
b
G of ∆. Similar to (2.6), the term
p
det
b
G is
defined as follows:
p
det
b
G=
√
detG
∥e v
1
∥∥e v
2
∥∥e v
3
∥∥e v
4
∥
(2.20)
∈























iR
+
all four vertexe v
i
are inR
3,1
T
orR
3,1
S
,
R
+
one vertex is inR
3,1
T
and the other three are inR
3,1
S
,
iR
− two vertices are inR
3,1
T
and the other two are inR
3,1
S
,
R
− three vertices are inR
3,1
T
and the other is inR
3,1
S
.
Therefore, the ratio R
i j
becomes
R
i j
=
ψ
i j
(z
− )ϕ
i j
(z
+
)
ϕ
i j
(z
− )ψ
i j
(z
+
)
=
b
C
kl
− z
i j
− z
− 1
i j
2
p
det
b
G
b
C
kl
+
z
i j
− z
− 1
i j
2
p
det
b
G
.
On the other hand, from Lemma 47 and the definition of z
i j
given in (2.4), we get
b
C
2
kl
− b
C
kk
b
C
ll
=

z
i j
− z
− 1
i j
2
!
2
· det
b
G. (2.21)
51
Then, we get
R
i j
+ 2+
1
R
i j
=
b
C
kl
− z
i j
− z
− 1
i j
2
p
det
b
G
b
C
kl
+
z
i j
− z
− 1
i j
2
p
det
b
G
+ 2+
b
C
kl
+
z
i j
− z
− 1
i j
2
p
det
b
G
b
C
kl
− z
i j
− z
− 1
i j
2
p
det
b
G
=
2
b
C
2
kl
+ 2(
b
C
2
kl
− b
C
kk
b
C
ll
)
b
C
kk
b
C
ll
+ 2= 4· b
C
2
kl
b
C
kk
b
C
ll
From Lemma 45, we get
R
i j
+ 2+
1
R
i j
= 4· ⟨e n
k
,e n
l
⟩
2
∥e n
k
∥
2
∥e n
l
∥
2
.
Lemma 53. If we replace the solution z
i j
of Equation (2.4) by the other solution z
− 1
i j
, then the ratio
R
i j
gets replaced by R
− 1
i j
.
Proof. We observe how R
i j
changes when z
i j
is replaced with z
− 1
i j
. Since the normalized Gram
matrix
b
G remains the same, the cofactor matrix
b
C and the determinant det
b
G of
b
G also do not
change. Thus, the new ratio becomes
b
C
kl
− z
− 1
i j
− z
i j
2
p
det
b
G
b
C
kl
+
z
− 1
i j
− z
i j
2
p
det
b
G
=
b
C
kl
+
z
i j
− z
− 1
i j
2
p
det
b
G
b
C
kl
− z
i j
− z
− 1
i j
2
p
det
b
G
=
1
R
i j
.
Note. Lemma 52 and 53 are highly non-obvious or geometrically unclear.
We can find all pairs of z
i j
and R
i j
according to combinatorial property of∆ and see the ge-
ometric interpretations of these algebraically defined objects. To do so, we define =w geometric
notations. The complex variable z
i j
is defined using e v
i
ande v
j
, i.e. each z
i j
is paired with the
edge e
i j
of∆. When e
i j
is finite, the corresponding essential edge e
ess
i j
of∆
trunc
is contained in e
i j
.
Denote the (exterior) dihedral angle of e
i j
by α
i j
and the edge length of e
ess
i j
by d
i j
. When e
i j
is
hyperideal, the corresponding essential edge e
ess
i j
is contained in the finite edge e
∗ kl
of ∆
∗ . Here,
i, j, k, and l are all distinct indices. Denote the (exterior) dihedral angle of e
∗ kl
by α
∗ kl
and the
edge length of e
ess
i j
by d
∗ kl
. Note that the edge length of an essential edge is defined as the distance
between endpoints designated in Lemma 28.
52
Lemma 54. Let∆ be a projective tetrahedron with vertices v
i
ande v
i
be the lifts of vertices that
define ∆. Let e
i j
be the edge of∆ connecting v
i
and v
j
. Then,

e v
i
,e v
j

∥e v
i
∥

e v
j

=























coshd
i j
when e
i j
is finite and v
i
,v
j
∈H
3
,
isinhd
i j
when e
i j
is finite and v
i
∈H
3
,v
j
∈ dS
3
/ v
j
∈H
3
,v
i
∈ dS
3
,
− coshd
i j
when e
i j
is finite and v
i
,v
j
∈ dS
3
,
cosα
∗ kl
when e
i j
is hyperideal.
Proof. The first case is equivalent to the case (1) of Lemma 28. Since the edge e
i j
lie inH
3
, the
liftse v
i
ande v
j
are in the same component ofR
3,1
T
. Then, by Proposition 7, we get

e v
i
,e v
j

∥e v
i
∥

e v
j

= coshd
i j
.
The second case is equivalent to the case (2) of Lemma 28. We have

e v
i
,e v
j

< 0 from Lemma
29. Thus, by Proposition 9, we get

e v
i
,e v
j

∥e v
i
∥

e v
j

= isinhd
i j
.
As expected, the third case is equivalent to the case (3) of Lemma 28. Similarly, we have

e v
i
,e v
j

< 0 from Lemma 29. Then, by Proposition 8 (b), we get

e v
i
,e v
j

∥e v
i
∥

e v
j

=− coshd
i j
.
Finally, in the fourth case, since e
i j
is hyperideal, the orthogonal planes v
⊥
i
and v
⊥
j
meet inH
3
and their intersection is the edge e
∗ kl
of the dual tetrahedron∆
∗ . By Proposition 8 (a), we get

e v
i
,e v
j

∥e v
i
∥

e v
j

= cosα
∗ kl
.
53
Once the normalized Lorentzian product is decided, there are two possibilities for z
i j
:
z
i j
=



















− e
d
i j
or − e
− d
i j
when e
i j
is finite and v
i
,v
j
∈H
3
,
− ie
d
i j
or ie
− d
i j
when e
i j
is finite and v
i
∈H
3
,v
j
∈ dS
3
/ v
j
∈H
3
,v
i
∈ dS
3
,
e
d
i j
or e
− d
i j
when e
i j
is finite and v
i
,v
j
∈ dS
3
,
− e
iα
∗ kl
or − e
− iα
∗ kl
when e
i j
is hyperideal.
The ratio R
i j
has two choices as well once the normalized Lorentzian product ofe n
k
ande n
l
is
decided. Applying Lemma 54 fore n
k
ande n
l
, then
R
i j
=



















e
2d
∗ kl
or e
− 2d
∗ kl
when e
∗ kl
is finite and n
k
,n
l
∈H
3
,
− e
2d
∗ kl
or − e
− 2d
∗ kl
when e
∗ kl
is finite and n
k
∈H
3
,n
l
∈ dS
3
/ n
l
∈H
3
,n
k
∈ dS
3
,
e
2d
∗ kl
or e
− 2d
∗ kl
when e
∗ kl
is finite and n
k
,n
l
∈ dS
3
,
e
2iα
i j
or e
− 2iα
i j
when e
∗ kl
is hyperideal.
In summary, all possible pairs of z
i j
and R
i j
depends on the combinatorial properties of∆ as
given in the following.
Lemma 55. According to the location of vertices v
i
, v
j
, n
k
, and n
l
for four distinct indices i, j, k,
l, the corresponding z
i j
and R
i j
are paired as in the Table 2.4.
Proof. All the variables z
i j
and the ratio R
i j
have two candidates, respectively, which makes total
four possible pairs of z
i j
and R
i j
, when the following two combinatorial properties are given: (1)
whether the edge e
ess
i j
is original or dual, and (2) where the vertices v
i
, v
j
, n
k
, and n
l
are located.
By Lemma 53, R
i j
is determined by the choice of z
i j
. Therefore, only two among the four pairs are
valid. We need to find valid pairs z
i j
and R
i j
.
First of all, using (2.11) and (2.20), the numerator of the ratio R
i j
is
b
C
kl
− z
i j
− z
− 1
i j
2
p
det
b
G=
1
∥e v
k
∥∥e v
l
∥
· "
detG
∥e v
i
∥
2
∥e v
j
∥
2
· ⟨e n
k
,e n
l
⟩−
z
i j
− z
− 1
i j
2
!
· √
detG
∥e v
i
∥∥e v
j
∥
#
.
54
Table 2.4: The pairs of z
i j
and R
i j
are determined by the combinatorial properties of∆.
Edge Vertices
⟨
e v
i
,e v
j⟩
∥e v
i
∥∥e v
j
∥
⟨e n
k
,e n
l
⟩
∥e n
k
∥∥e n
l
∥
z
i j
R
i j
e
ess
i j
: original
i.e.
e
i j
: finite,
e
∗ kl
: hyperideal
v
i
, v
j
∈H
3
,
n
k
, n
l
∈ dS
3
coshd
i j
cosα
i j
− e
d
i j
e
2iα
i j
− e
− d
i j
e
− 2iα
i j
v
i
∈H
3
, v
j
∈ dS
3
,
n
k
, n
l
∈ dS
3
isinhd
i j
cosα
i j
− ie
d
i j
e
2iα
i j
ie
− d
i j
e
− 2iα
i j
v
i
, v
j
∈ dS
3
,
n
k
, n
l
∈ dS
3
− coshd
i j
cosα
i j
e
d
i j
e
2iα
i j
e
− d
i j
e
− 2iα
i j
e
ess
i j
: dual
i.e.
e
i j
: hyperideal,
e
∗ kl
: finite
v
i
, v
j
∈ dS
3
,
n
k
, n
l
∈ dS
3
cosα
∗ kl
− coshd
∗ kl
− e
iα
∗ kl
e
− 2d
∗ kl
− e
− iα
∗ kl
e
2d
∗ kl
v
i
, v
j
∈ dS
3
,
n
k
∈H
3
, n
l
∈ dS
3
cosα
∗ kl
isinhd
∗ kl
− e
iα
∗ kl
− e
− 2d
∗ kl
− e
− iα
∗ kl
− e
2d
∗ kl
v
i
, v
j
∈ dS
3
,
n
k
, n
l
∈H
3
cosα
∗ kl
coshd
∗ kl
− e
iα
∗ kl
e
− 2d
∗ kl
− e
− iα
∗ kl
e
2d
∗ kl
Similarly applying to the denominator of R
i j
, the ratio becomes
R
i j
=
detG
∥e v
i
∥
2
∥e v
j
∥
2
· ⟨e n
k
,e n
l
⟩− z
i j
− z
− 1
i j
2
· √
detG
∥e v
i
∥∥e v
j
∥
detG
∥e v
i
∥
2
∥e v
j
∥
2
· ⟨e n
k
,e n
l
⟩+
z
i j
− z
− 1
i j
2
· √
detG
∥e v
i
∥∥e v
j
∥
,
which depends only on four liftse v
i
,e v
j
,e n
k
, ande n
l
. Especially, it does not depend one v
k
ande v
l
, so
we may assume that v
k
, v
l
∈ dS
3
without loss of generality.
Starting with the case of e
ess
i j
is original. Then, e
i j
is finite and e
∗ kl
is hyperideal, and thus n
k
and n
l
always lie in dS
3
. There are three separate cases depending on where v
i
and v
j
are located
in, and the normalized Lorentzian product and z
i j
are determined by Lemma 54 as follows:
(a) if v
i
, v
j
∈H
3
, then we have
⟨
e v
i
,e v
j⟩
∥e v
i
∥·∥e v
j
∥
= coshd
i j
, and z
i j
is either− e
d
i j
or− e
− d
i j
.
(b) if v
i
∈H
3
, v
j
∈ dS
3
, then we have
⟨
e v
i
,e v
j⟩
∥e v
i
∥·∥e v
j
∥
= isinhd
i j
, and z
i j
is either− ie
d
i j
or ie
− d
i j
.
(c) if v
i
, v
j
∈ dS
3
, then we have
⟨
e v
i
,e v
j⟩
∥e v
i
∥·∥e v
j
∥
=− coshd
i j
, and z
i j
is either e
d
i j
or e
− d
i j
.
55
We first see the case (a), and assume that we pick z
i j
=− e
d
i j
. Because v
i
, v
j
∈H
3
and v
k
,
v
l
∈ dS
3
, we get det
b
G< 0 and
p
det
b
G∈ iR
− . We further have
z
i j
− z
− 1
i j
2
=− v
u
u
t

z
i j
− z
− 1
i j
2
!
2
∈R
− .
Then, we get from (2.21) that
R
i j
=
b
C
kl
− z
i j
− z
− 1
i j
2
p
det
b
G
b
C
kl
+
z
i j
− z
− 1
i j
2
p
det
b
G
=
b
C
kl
− i
s
−
z
i j
− z
− 1
i j
2

2
det
b
G
b
C
kl
+ i
s
−
z
i j
− z
− 1
i j
2

2
det
b
G
=
b
C
kl
− i
q
b
C
kk
b
C
ll
− b
C
2
kl
b
C
kl
+ i
q
b
C
kk
b
C
ll
− b
C
2
kl
.
Applying (2.11),∥e v
k
∥> 0,∥e v
l
∥> 0, and det
b
G< 0, we get
R
i j
=
det
b
G·∥e v
k
∥∥e v
l
∥· ⟨e n
k
,e n
l
⟩− i
r
det
b
G
2
·∥e v
k
∥
2
∥e v
l
∥
2
·
∥e n
k
∥
2
∥e n
l
∥
2
− ⟨e n
k
,e n
l
⟩
2

det
b
G·∥e v
k
∥∥e v
l
∥· ⟨e n
k
,e n
l
⟩+ i
r
det
b
G
2
·∥e v
k
∥
2
∥e v
l
∥
2
·
∥e n
k
∥
2
∥e n
l
∥
2
− ⟨e n
k
,e n
l
⟩
2

=
⟨e n
k
,e n
l
⟩+ i
q
∥e n
k
∥
2
∥e n
l
∥
2
− ⟨e n
k
,e n
l
⟩
2
⟨e n
k
,e n
l
⟩− i
q
∥e n
k
∥
2
∥e n
l
∥
2
− ⟨e n
k
,e n
l
⟩
2
From∥e n
k
∥> 0,∥e n
l
∥> 0, and
⟨e n
k
,e n
l
⟩
∥e n
k
∥·∥e n
l
∥
= cosα
i j
, above is simplified as
R
i j
=
cosα
i j
+ i
p
1− cos
2
α
i j
cosα
i j
− i
p
1− cos
2
α
i j
= e
2iα
i j
.
If we pick z
i j
=− e
− d
i j
, we get R
i j
= e
− 2iα
i j
from Lemma 53.
Through the similar approach, we can find R
i j
for the case (b) and (c), and also for the case of
e
ess
i j
is dual. Results are summarized in Table 2.4.
56
2.8 Proof of main theorem
In this section, we prove Theorem 35, the main theorem of the paper. Before that, we need a
number of lemmas to apply Schl¨ afli’s formula in the proof.
Let∆⊂ RP
3
be a nonideal projective tetrahedron whose four vertices v
i
∈RP
3
are not coplanar.
Let∆
∗ be a dual of projective tetrahedron of∆. Suppose the truncated tetrahedron∆
trunc
associated
with∆ is proper. Let e
i j
be the edge of∆ connecting v
i
and v
j
, and e
∗ kl
be the edge of∆
∗ connecting
n
k
and n
l
, where i, j, k, l are all distinct. By Lemma 19, only one of e
i j
and e
∗ kl
is finite and the
other is hyperideal. By Lemma 33, for two distinct indices i and j, there exists exactly one essential
edge e
ess
i j
in∆
trunc
which is contained in either e
i j
or e
∗ kl
.
Definition 56. An essential dihedral angle α
ess
i j
is the (exterior) dihedral angle of the essential
edge e
ess
i j
. An essential edge length d
ess
i j
is the edge length of the essential edge e
ess
i j
, which is a
distance between two endpoints clarified in Lemma 28.
Note. When the essential edge e
ess
i j
agrees exactly with an original edge and contained in e
i j
, then
α
ess
i j
and α
i j
are the same variables and d
ess
i j
and d
i j
are the same variables. We use these bad
notations to emphasize that e
i j
is a finite edge. Similarly, when the essential edge e
ess
i j
is a dual
edge and contained in e
∗ kl
, where the indices i, j, k and l are distinct, α
ess
i j
and α
∗ kl
are the same
variables and d
ess
i j
and d
∗ kl
are the same variables. In summary,







α
ess
i j
=α
i j
, d
ess
i j
= d
i j
when e
ess
i j
⊂ e
i j
,
α
ess
i j
=α
∗ kl
, d
ess
i j
= d
∗ kl
when e
ess
i j
⊂ e
∗ kl
.
(2.22)
Lemma 57. Let α
ess
be a list of dihedral angles α
ess
i j
of six essential edges of ∆
trunc
. Given
the combinatorial structure of∆
trunc
including whether each essential edge is original or dual, the
function from∆
trunc
toα
ess
is injective up to isometry. In other words, if we know the combinatorial
structure of∆
trunc
and its dihedral angles at six essential edges, then we can uniquely determine a
truncated tetrahedron∆
trunc
(α
ess
) up to isometry. Furthermore, the inverse function fromα
ess
to
∆
trunc
(α
ess
) is differentiable with respect to each dihedral angleα
ess
i j
.
57
Proof. Fix one combinatorial structure of truncated tetrahedra and let T T be a space of truncated
tetrahedra of given combinatorial structure up to isometry. Thus it is determined whether each
e
i j
is finite or hyperideal. Let f : T T →R
6
+
be a function defined as f(T)=α
ess
. Note that we
know all the dihedral angles of∆
trunc
: it is the right angles for the truncation edges andα
ess
i j
for the
essential edges e
ess
i j
. Then, by the Cauchy Lemma (for the proof of this lemma, see [18] or [19]),
the truncated tetrahedron∆
trunc
is determined up to isometry. Hence f is injective. Furthermore,
we have the injectivity of differential map d f from the infinitesimal version of the Cauchy Lemma.
Let’s move on to the differentiability of f
− 1
. In fact, both the space of dihedral angles and
the space of truncated tetrahedra have dimension 6. Since the differential d f between these two
tangent spaces is injective, we get that d f is isomorphism, hence f is a diffeomorphism. Therefore,
the inverse map f
− 1
is differentiable with respect toα
ess
i j
.
Lemma 58. Given a combinatorial structure, the space of truncated tetrahedra up to isometry is
connected.
Proof. We first consider the set F of finite tetrahedra up to isometry. We show that F is connected.
For one element∆=∆
trunc
of F, with vertices v
1
, v
2
, v
3
, v
4
, let f
4
be the face that does not contain
v
4
. Assume that the vertex v
4
is located positive over f
4
of∆. In other words, if we considerH
3
as a sphere inR
3
, the Klein model, and relocate f
4
on xy-plane by some isometry ofH
3
, then the
vertex v
4
is located in upper half-ball{(x,y,z)| x
2
+ y
2
+ z
2
< 1,z> 0}. Again, by some isometry
ofH
3
, we can relocate one edge e
13
of f
4
on x-axis and the remaining vertex v
2
of f
4
on positive
y-axis. This isometry preserves the xy-plane, hence v
4
is still in the upper half-ball after it is sent
by isometry. Following this process, every element in F can be identified as six real numbers: two
real numbers in (− 1,1) as coordinates of two vertices v
1
,v
3
on x-axis, one positive real number
in(0,1) as coordinate of one vertex v
2
on y-axis, and three real numbers as coordinates of the last
one vertex v
4
in upper half-ball. Hence we conclude that F is path-connected, and therefore, is
connected.
For other combinatorial properties of projective tetrahedra with finite edges, there are four
possible structures according to the number of hyperideal vertices. Similarly as the previous finite
58
tetrahedron case, we can conclude that every single projective tetrahedra can be identified as six
real numbers, which possibly have absolute values greater than 1. Hence, the space of projective
tetrahedra inH
3
for such combinatorial property is path-connected, and therefore, is connected.
Since a truncated tetrahedron are determined by its original tetrahedron, the space of truncated
tetrahedra is also connected.
Let’s move on to the combinatorial properties with finite faces and at least one hyperideal edge.
Whenever e
13
is finite, we relocate e
13
on x-axis, and whenever e
13
is hyperideal, we make the y-
coordinate of e
13
be− 2 so that the dual of e
13
passes through(0,− 0.5,0). By similar arguments
for the all hyperideal edges of an original tetrahedron ∆ in RP
3
, we conclude that every single
truncated tetrahedron with faces meeting H
3
also can be identified as six real numbers. Hence
the space of tetrahedra of such combinatorial properties with faces meetingH
3
is also connected,
followed by the space of truncated tetrahedra of such combinatorial structure is connected.
Finally, the combinatorial properties with at least one hyperideal face can be covered similarly
as before. To sum up, for all combinatorial properties of∆ (whether each vertex, edge, and face is
finite or hyperideal), the space of corresponding truncated tetrahedron is connected.
Note. To define the formula in the main theorem, we needed one more nontrivial assumption that
pr̸= 0 in Lemma 34. From the previous Lemma, the set of truncated tetrahedra given a combina-
torial structure is of real dimension 6. Since pr is a complex number, the set of truncated tetrahedra
with pr= 0 is a closed subset of codimension 2. Therefore, the set of truncated tetrahedra of pr̸= 0
is still connected.
Now we are ready to prove Theorem 35. Recall a function Q of six complex variables z
i j
with
1≤ i< j≤ 4 from (2.13):
Q(z
12
,z
13
,z
14
,z
23
,z
24
,z
34
)
=
1
4
P(z
− ,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
)− 1
4
P(z
+
,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
).
59
Q can be considered as a function of twelve real variables r
i j
and θ
i j
where r
i j
and θ
i j
are a
logarithm of absolute value and an argument of z
i j
= e
r
i j
+iθ
i j
. For the notational convenience,
denote it by Q(R,Θ) where R=(r
12
,r
13
,r
14
,r
23
,r
24
,r
34
) andΘ=(θ
12
,θ
13
,θ
14
,θ
23
,θ
24
,θ
34
).
Fix one combinatorial structure and let ∆
trunc
(α
ess
) be a proper truncated tetrahedron where
α
ess
=

α
ess
i j

1≤ i< j≤ 4
is a collection of dihedral angles at essential edges. The truncated tetrahe-
dron∆
trunc
is uniquely defined by α
ess
from Lemma 57. Lete v
i
be lifts of v
i
that define ∆. Recall
that we define z
i j
by (2.4) and denote it by z
i j
(∆
trunc
(α
ess
)). Furthermore, let r
i j
(∆
trunc
(α
ess
))
and θ
i j
(∆
trunc
(α
ess
)) be the logarithm of absolute value and the argument of z
i j
(∆
trunc
(α
ess
)).
Then, by Lemma 55, we have two choices for each z
i j
. Then R(α
ess
) and Θ(α
ess
), the sets of
r
i j
(∆
trunc
(α
ess
)) andθ
i j
(∆
trunc
(α
ess
)), respectively, are given as follows:
If e
i j
is finite,







r
i j
(∆
trunc
(α
ess
))= d
i j
, θ
i j
(∆
trunc
(α
ess
))=− π
2
× n, or
r
i j
(∆
trunc
(α
ess
))=− d
i j
, θ
i j
(∆
trunc
(α
ess
))=
π
2
× n.
If e
i j
is hyperideal,







r
i j
(∆
trunc
(α
ess
))= 0, θ
i j
(∆
trunc
(α
ess
))=α
∗ kl
, or
r
i j
(∆
trunc
(α
ess
))= 0, θ
i j
(∆
trunc
(α
ess
))=− α
∗ kl
.
(2.23)
where n is the number of vertices inH
3
among v
i
and v
j
, i.e.
n=















0 when v
i
,v
j
∈H
3
,
1 when v
i
∈H
3
,v
j
∈ dS
3
or v
i
∈ dS
3
,v
j
∈H
3
,
2 when v
i
,v
j
∈ dS
3
.
Note that the complex variable z
i j
is in fact a 1-dimensional real variable because, in any case, only
one of r
i j
andθ
i j
varies while the other is constant.
Now we show that Q(α
ess
)= Q(R(α
ess
),Θ(α
ess
)) calculates the volume of the truncated tetra-
hedron∆
trunc
(α
ess
). Since we have the connectedness of the space of one specific combinatorial
structure of truncated tetrahedra from Lemma 58, we can apply Schl¨ afli’s differential formula.
60
To show that the function Q(α
ess
) is a volume function, we need to see if Schl¨ afli’s differential
formula (Theorem 48) holds for all edges of∆
trunc
(α
ess
). Among essential edges and truncation
edges in ∆
trunc
(α
ess
), we compute the derivatives of Q(α
ess
) only for six essential edges e
ess
i j
of
∆
trunc
(α
ess
) because the truncation edges have fixed dihedral angles
π
2
by Lemma 30. Therefore,
we need to show the following propositions:
(a) When the essential edge e
ess
i j
is original, we have
∂
∂α
ess
i j
Q(α
ess
)=
1
2
d
i j
.
(b) When the essential edge e
ess
i j
is dual, we have
∂
∂α
ess
i j
Q(α
ess
)=
1
2
d
∗ kl
.
(c) At some value∆
trunc
of a truncated tetrahedron, Q(α
ess
) computes its actual volume.
Proposition 59. The derivative of Q(α
ess
) with respect toα
ess
i j
is computed as
∂Q(α
ess
)
∂α
ess
i j
=
1
2
d
ess
i j
.
Remark. This proposition proves the proposition (a) and (b) at once.
Proof. From Chain rule, we get
∂Q(α
ess
)
∂α
ess
i j
=
∑
1≤ i
1
< j
1
≤ 4
∂r
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· ∂Q(α
ess
)
∂r
i
1
j
1

R(α
ess
),Θ(α
ess
)
+
∑
1≤ i
1
< j
1
≤ 4
∂θ
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· ∂Q(α
ess
)
∂θ
i
1
j
1

R(α
ess
),Θ(α
ess
)
Given a combinatorial property of∆, we saw from (2.23) that r
i
1
j
1
(∆
trunc
(α
ess
)) is constant when
e
i
1
j
1
is hyperideal andθ
i
1
j
1
(∆
trunc
(α
ess
)) is constant when e
i
1
j
1
is finite. Then the above becomes
∂Q(α
ess
)
∂α
ess
i j
=
∑
e
i
1
j
1
:finite
∂r
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· ∂Q(α
ess
)
∂r
i
1
j
1

R(α
ess
),Θ(α
ess
)
+
∑
e
i
1
j
1
:hyperideal
∂θ
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· ∂Q(α
ess
)
∂θ
i
1
j
1

R(α
ess
),Θ(α
ess
)
(2.24)
We compute the first term of RHS in step 1, compute the second term in step 2, and combine them
in step 3.
61
Step 1 Let’s see the first term of the right hand side of (2.24). From Lemma 51, we get
∂Q(α
ess
)
∂r
i
1
j
1

R(α
ess
),Θ(α
ess
)
=
1
4
∑
1≤ i
2
< j
2
≤ 4
r
i
2
j
2
(∆
trunc
(α
ess
))· ∂
∂r
i
1
j
1
argR
i
2
j
2
. (2.25)
We divide into two cases: (1-a) e
i
2
j
2
is finite, or (1-b) e
i
2
j
2
is hyperideal.
(1-a) When e
i
2
j
2
is finite, we have two possibilities from (2.23) and obtain R
i
2
j
2
accordingly
from Lemma 55:







r
i
2
j
2
(∆
trunc
(α
ess
))= d
i
2
j
2
, R
i
2
j
2
= e
2iα
i
2
j
2
, or
r
i
2
j
2
(∆
trunc
(α
ess
))=− d
i
2
j
2
, R
i
2
j
2
= e
− 2iα
i
2
j
2
.
In both cases, we get
r
i
2
j
2
(∆
trunc
(α
ess
))· ∂
∂r
i
1
j
1
argR
i
2
j
2
=







d
i
2
j
2
· ∂
∂r
i
1
j
1
2α
i
2
j
2
− d
i
2
j
2
· ∂
∂r
i
1
j
1
(− 2α
i
2
j
2
)
= 2d
i
2
j
2
· ∂α
i
2
j
2
∂r
i
1
j
1
.
(1-b) On the other hand, when e
i
2
j
2
is hyperideal, we have r
i
2
j
2
(∆
trunc
(α
ess
))= 0.
Therefore, (2.25) becomes
∂Q(α
ess
)
∂r
i
1
j
1

R(α
ess
),Θ(α
ess
)
=
1
2
∑
e
i
2
j
2
:finite
d
i
2
j
2
· ∂α
i
2
j
2
∂r
i
1
j
1
.
Step 2 In the second term of RHS of (2.24), e
i
1
j
1
is now hyperideal. From Lemma 51, we get
∂Q(α
ess
)
∂θ
i
1
j
1

R(α
ess
),Θ(α
ess
)
=− 1
4
log

R
i
1
j
1

+
1
4
∑
1≤ i
2
< j
2
≤ 4
r
i
2
j
2
(∆
trunc
(α
ess
))· ∂
∂θ
i
1
j
1
argR
i
2
j
2
. (2.26)
For the latter part of RHS of (2.26), we can divide into two cases, similarly as before, based on
whether e
i
2
j
2
is finite or hyperideal. The results are similar as well. When e
i
2
j
2
is finite, regardless
of a choice of z
i
2
j
2
, we have
r
i
2
j
2
(∆
trunc
(α
ess
))· ∂
∂θ
i
1
j
1
argR
i
2
j
2
= 2d
i
2
j
2
· ∂α
i
2
j
2
∂θ
i
1
j
1
.
62
When e
i
2
j
2
is hyperideal, the summand disappears because of r
i
2
j
2
(∆
trunc
(α
ess
))= 0.
Then the second term of RHS of (2.24) becomes
∑
e
i
1
j
1
:hyperideal
∂θ
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· ∂Q(α
ess
)
∂θ
i
1
j
1

R(α
ess
),Θ(α
ess
)
=− 1
4
∑
e
i
1
j
1
:hyperideal
∂θ
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· log|R
i
1
j
1
|
+
1
2
∑
e
i
1
j
1
:hyperideal
∂θ
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· 

∑
e
i
2
j
2
:finite
d
i
2
j
2
· ∂α
i
2
j
2
∂θ
i
1
j
1


.
(2.27)
Both θ
i
1
j
1
(∆
trunc
(α
ess
)) and |R
i
1
j
1
| depend on a choice of z
i
1
j
1
from (2.23) and Lemma 55,
respectively:







θ
i
1
j
1
(∆
trunc
(α
ess
))=α
∗ k
1
l
1
, |R
i
1
j
1
|= e
− 2d
∗ k
1
l
1
when z
i
1
j
1
= e
iα
∗ k
1
l
1
,
θ
i
1
j
1
(∆
trunc
(α
ess
))=− α
∗ k
1
l
1
, |R
i
1
j
1
|= e
2d
∗ k
1
l
1
when z
i
1
j
1
= e
− iα
∗ k
1
l
1
,
where the indices i
1
, j
1
, k
1
, l
1
are all distinct. Then, the first term of the RHS of (2.27) becomes
− 1
4
∑
e
i
1
j
1
:hyperideal
∂θ
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· log|R
i
1
j
1
|=
1
2
∑
e
i
1
j
1
:hyperideal
d
∗ k
1
l
1
· ∂α
∗ k
1
l
1
∂α
ess
i j
,
regardless of a choice of z
i
1
j
1
.
Therefore, the second term of RHS of (2.24) becomes
∑
e
i
1
j
1
:hyperideal
∂θ
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· ∂Q(α
ess
)
∂θ
i
1
j
1

R(α
ess
),Θ(α
ess
)
=
1
2
∑
e
i
1
j
1
:hyperideal
d
∗ k
1
l
1
· ∂α
∗ k
1
l
1
∂α
ess
i j
+
1
2
∑
e
i
1
j
1
:hyperideal
∂θ
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· 

∑
e
i
2
j
2
:finite
d
i
2
j
2
· ∂α
i
2
j
2
∂θ
i
1
j
1


.
63
Step 3 Taken together, (2.24) becomes
∂Q(α
ess
)
∂α
ess
i j
=
1
2
∑
e
i
1
j
1
:finite
∂r
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· 

∑
e
i
2
j
2
:finite
d
i
2
j
2
· ∂α
i
2
j
2
∂r
i
1
j
1


+
1
2
∑
e
i
1
j
1
:hyperideal
d
∗ k
1
l
1
· ∂α
∗ k
1
l
1
∂α
ess
i j
+
1
2
∑
e
i
1
j
1
:hyperideal
∂θ
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· 

∑
e
i
2
j
2
:finite
d
i
2
j
2
· ∂α
i
2
j
2
∂θ
i
1
j
1


=
1
2
∑
e
i
1
j
1
:hyperideal
d
∗ k
1
l
1
· ∂α
∗ k
1
l
1
∂α
ess
i j
+
1
2
∑
e
i
2
j
2
:finite
d
i
2
j
2
· 

∑
e
i
1
j
1
:finite
∂r
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· ∂α
i
2
j
2
∂r
i
1
j
1
+
∑
e
i
1
j
1
:hyperideal
∂θ
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· ∂α
i
2
j
2
∂θ
i
1
j
1


(2.28)
On the other hand, consider α
i
2
j
2
as a function f(R(α
ess
),Θ(α
ess
))=α
i
2
j
2
for a finite edge
e
i
2
j
2
. Then f is indeed differentiable. Hence,
∂α
i
2
j
2
∂α
ess
i j
=
∂
∂α
ess
i j
f(R(α
ess
),Θ(α
ess
))
=
∑
1≤ i
1
< j
1
≤ 4
∂r
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· ∂α
i
2
j
2
∂r
i
1
j
1
+
∑
1≤ i
1
< j
1
≤ 4
∂θ
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· ∂α
i
2
j
2
∂θ
i
1
j
1
=
∑
e
i
1
j
1
:finite
∂r
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· ∂α
i
2
j
2
∂r
i
1
j
1
+
∑
e
i
1
j
1
:hyperideal
∂θ
i
1
j
1
(∆
trunc
(α
ess
))
∂α
ess
i j
· ∂α
i
2
j
2
∂θ
i
1
j
1
.
The second equality comes from the Chain rule and the third equality comes from (2.23) that
r
i
1
j
1
(∆
trunc
(α
ess
)) is constant when e
i
1
j
1
is hyperideal andθ
i
1
j
1
(∆
trunc
(α
ess
)) is constant when e
i
1
j
1
is finite. Thus, (2.28) becomes
∂Q(α
ess
)
∂α
ess
i j
=
1
2
∑
e
i
1
j
1
:hyperideal
d
∗ k
1
l
1
· ∂α
∗ k
1
l
1
∂α
ess
i j
+
1
2
∑
e
i
2
j
2
:finite
d
i
2
j
2
· ∂α
i
2
j
2
∂α
ess
i j
.
64
Each edge e
i j
is either finite or hyperideal. Combining with (2.22), we get
∂Q(α
ess
)
∂α
ess
i j
=
1
2
∑
1≤ i
1
< j
1
≤ 4
d
ess
i
1
j
1
· ∂α
ess
i
1
j
1
∂α
ess
i j
.
Sinceα
ess
i
1
j
1
are independent variables ofα
ess
, we have
∂α
ess
i
1
j
1
∂α
ess
i j
=







1 when{i
1
, j
1
}={i, j},
0 when{i
1
, j
1
}̸={i, j}.
Therefore, we get
∂Q(α
ess
)
∂α
ess
i j
=
1
2
d
ess
i j
.
Finally, we show the proposition (c). Let O be a regular ideal octahedron. Note that O is the
truncated tetrahedron by truncating all vertices of the regular tetrahedron inRP
3
where all edges
are tangent toH
3
. Hence the regular ideal octahedron is on the boundary of the space of proper
truncated tetrahedra with hyperideal vertices and faces meetingH
3
.
Proposition 60. Let O be the regular ideal octahedron. Then, V ol(O), defined in Theorem 35 and
evaluated at O, computes the volume of O.
Proof. O can be considered as a truncated tetrahedron by truncating all vertices of the regular
tetrahedron inRP
3
circumscribingH
3
. Its dihedral angles are all π/2. For any four vertices of
O except for two vertices in the opposite position, there exists a square that includes these four
vertices. Note that there are three squares of these types. O can be sliced by any two squares of
these types, and divided into four identical ideal tetrahedra with its dihedral angles
π
2
,
π
4
, and
π
4
.
Therefore, from Milnor’s volume formula for an ideal tetrahedron (see, for example, [20]), the
actual volume of O is
4·
Λ

π
2

+Λ

π
4

+Λ

π
4

= 8Λ

π
4

,
65
where the Lobachevsky functionΛ(θ) is defined as
Λ(θ)=− Z
θ
0
log|2sinu|du.
The Lobachevsky function has the following relation with dilogarithm function:
2Λ

θ
2

= Im Li
2

e
iθ

= D

e
iθ

.
On the other hand, O can be regarded as the truncated tetrahedron, where its all six essential
edges converge to the vertices of O, i.e. the essential edges with lengths 0. Then, we get z
i j
= e
0
= 1
for all i̸= j. Thus, the quadratic equation (2.1) becomes
8z
2
+ 8= 0,
followed by z
− =− i, z
+
= i. Furthermore, in this case, the modified Murakami-Yano function P
becomes
P(z,z
12
,z
13
,z
14
,z
23
,z
24
,z
34
)= 4D(z)− 4D(− z).
Therefore,
Q(O)=
1
4
[(4D(i)− 4D(− i))− (4D(− i)− 4D(i))]
= 2D(i)− 2D(− i)= 4
h
Λ

π
4

− Λ

− π
4
i
= 8Λ

π
4

,
which shows that Q(O) is actually the volume of truncated tetrahedron O.
To sum up, we proved the propositions (1), (2) and (3), and thus Theorem 35 is proved.
66
Chapter 3
Conclusion and future direction
In this paper, we gave the formula that computes the hyperbolic volume of a nonideal and proper
truncated tetrahedron. Our formula simplified and unified the earlier approaches. Replacing Eu-
ler dilogarithm functions to Bloch-Wigner dilogarithm functions eliminated the phase issue of
complex-valued Euler dilogarithm functions. Exchange of a original tetrahedron to its dual tetra-
hedron preserves the correponding truncated tetrahedron but switches two different types of vari-
ables: dihedral angles and edge lengths. Among fifteen classes of proper truncated tetrahedron,
our method provides another computation of volumes in terms of edge lengths to classes covered
by earlier results, and extends to nine unaddressed classes of truncated tetrahedra.
Our formula may extend to proper truncated tetrahedra with a number of ideal vertices. Be-
cause of the analyticity of our formula, it is applicable to the boundary cases: truncated tetrahedra
derived from projective tetrahedra that includes ideal vertices, ideal edges, and ideal faces.
Future research may explore the volume formula of nonproper truncated tetrahedra. For the
nonproper cases, we define the truncated tetrahedron in different way: instead of starting from
one projective tetrahedron, we deal with the eight tetrahedra at once that are sharing the same
vertices. The nonproper truncated tetrahedron is the union of the intersection of each tetrahedron
and its corresponding dual. Then a nonproper truncated tetrahedron is composed of a number of
polyhedra, where adjacent polyhedra are connected via their common edge. Our formula computes
the alternating sum of volumes of these polyhedra.
67
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Creator Sohn, Jihoon (author)

Core Title Volumes of hyperbolic truncated tetrahedra

Contributor Electronically uploaded by the author (provenance)

School College of Letters, Arts and Sciences

Degree Doctor of Philosophy

Degree Program Mathematics

Degree Conferral Date 2022-08

Publication Date 07/26/2022

Defense Date 05/12/2022

Publisher University of Southern California (original), University of Southern California. Libraries (digital)

Tag Bloch-Wigner dilogarithm function,hyperbolic geometry,OAI-PMH Harvest,quantum topology,truncated tetrahedra,volumes

Format application/pdf (imt)

Language English

Advisor Bonahon, Francis (committee chair), Chen, Muhao (committee member), Lauda, Aaron (committee member)

Creator Email jhsohn11@gmail.com,jihoonso@usc.edu

Permanent Link (DOI) https://doi.org/10.25549/usctheses-oUC111375410

Unique identifier UC111375410

Legacy Identifier etd-SohnJihoon-10994

Document Type Dissertation

Format application/pdf (imt)

Rights Sohn, Jihoon

Type texts

Source 20220728-usctheses-batch-962 (batch), University of Southern California (contributing entity), University of Southern California Dissertations and Theses (collection)

Access Conditions The author retains rights to his/her dissertation, thesis or other graduate work according to U.S. copyright law. Electronic access is being provided by the USC Libraries in agreement with the author, as the original true and official version of the work, but does not grant the reader permission to use the work if the desired use is covered by copyright. It is the author, as rights holder, who must provide use permission if such use is covered by copyright. The original signature page accompanying the original submission of the work to the USC Libraries is retained by the USC Libraries and a copy of it may be obtained by authorized requesters contacting the repository e-mail address given.

Repository Name University of Southern California Digital Library

Repository Location USC Digital Library, University of Southern California, University Park Campus MC 2810, 3434 South Grand Avenue, 2nd Floor, Los Angeles, California 90089-2810, USA

Repository Email uscdl@usc.edu

Abstract (if available)

Abstract A hyperbolic truncated tetrahedron is a 3-dimensional polyhedron in the projective model of the hyperbolic space, which is obtained by polar truncation at the vertices that lie outside of the hyperbolic space. In this paper, we present two different computations of the volume of a hyperbolic truncated tetrahedron, one in terms of its edge lengths and another in terms of its dihedral angles. Among various cases of hyperbolic truncated tetrahedron, our method provides a unified approach to cases covered by earlier results of J. Murakami, M. Yano, A. Ushijima, and A. Kolpakov, and extends to cases that had been left unaddressed. Our key technical idea is to rely on the (real-valued) Bloch-Wigner dilogarithm function instead of the (complex-valued) Euler dilogarithm, which eliminates subtle phase issues and has better analytic properties.