A Proof of Seidel's Conjectures on the Volume of Ideal Tetrahedra in Hyperbolic 3-Space / Omar Chavez Cussy; Orientador Carlos Henrique Grossi Ferreira

A proof of Seidel’s conjectures on the volume of ideal tetrahedra in hyperbolic 3-space

Omar Chavez Cussy

SERVIÇO DE PÓS-GRADUAÇÃO DO ICMC-USP

Data de Depósito:

Assinatura: ______

Omar Chavez Cussy

A proof of Seidel’s conjectures on the volume of ideal tetrahedra in hyperbolic 3-space

Master dissertation submitted to the Instituto de Ciências Matemáticas e de Computação – ICMC- USP, in partial fulﬁllment of the requirements for the degree of the Master Program in Mathematics. EXAMINATION BOARD PRESENTATION COPY Concentration Area: Mathematics Advisor: Prof. Dr. Carlos Henrique Grossi Ferreira

USP – São Carlos May 2017 Ficha catalográfica elaborada pela Biblioteca Prof. Achille Bassi e Seção Técnica de Informática, ICMC/USP, com os dados fornecidos pelo(a) autor(a)

Chavez Cussy, Omar C512p A proof of Seidel's conjectures on the volume of ideal tetrahedra in hyperbolic 3-space / Omar Chavez Cussy; orientador Carlos Henrique Grossi Ferreira. -- São Carlos, 2017. 85 p.

Dissertação (Mestrado - Programa de Pós-Graduação em Matemática) -- Instituto de Ciências Matemáticas e de Computação, Universidade de São Paulo, 2017.

1. Seidel's conjectures. 2. Volume of ideal tetrahedra. 3. Real hyperbolic space. I. Grossi Ferreira, Carlos Henrique, orient. II. Título. Omar Chavez Cussy

Uma demonstração de conjecturas de Seidel sobre o volume de tetraedros ideais no 3-espaço hiperbólico

Dissertação apresentada ao Instituto de Ciências Matemáticas e de Computação – ICMC-USP, como parte dos requisitos para obtenção do título de Mestre em Ciências – Matemática. EXEMPLAR DE DEFESA Área de Concentração: Matemática Orientador: Prof. Dr. Carlos Henrique Grossi Ferreira

USP – São Carlos Maio de 2017

Este trabalho e´ dedicado as` crianc¸as adultas que, quando pequenas, sonharam em se tornar cientistas. Em especial, ao pesquisadores do Instituto de Cienciasˆ Matematicas´ e de Computac¸ao˜ (ICMC).

ACKNOWLEDGEMENTS

I would like to express my gratitude to my advisor, Carlos Henrique Grossi Ferreira, who very kindly introduced me to his work area and gave me the guidance and necessary tools for this project. Also, I am very grateful to all the professors of the Mathematics department at the ICMC-USP who developed lectures or seminars which I attended to in the last two years.

“As invenc¸oes˜ sao,˜ sobretudo, o resultado de um trabalho de teimoso.” (Santos Dumont)

RESUMO

OMAR CHAVEZ C.. A proof of Seidel’s conjectures on the volume of ideal tetrahedra in hyperbolic 3-space. 2017. 85 f. Master dissertation (Master student Program in Mathematics) – Instituto de Ciências Matemáticas e de Computação (ICMC/USP), São Carlos – SP.

Provamos duas conjecturas apresentadas por J. J. Seidel em “On the volume of a hyperbolic simplex”, Stud. Sci. Math. Hung. (21, 243–249, 1986). Estas conjecturas referem ao volume de tetraedros ideais no 3-espaço hiperbólico e estão relacionadas com o seguinte quadro geral. Como fórmulas expl´ıcitas para grandezas geométricas no espaço hiperbólico (distância, área, volume, etc.) tipicamente envolvem funções transcendentais sofisticadas, édesejável (e, na prática, bastante útil) expressar tais grandezas geométricas como aplicações monótonas de ma- pas algébricos. A Especulação 1 de Seidel diz que o volume de um tetraedro ideal no 3-espaço hiperbólico depende apenas do determinante e do permanente da matriz de Gram duplamente estocástica G de seus vértices; a Especulação 4 afirma que o referido volume émonótono tanto no determinante quanto no permanente de G. Damos respostas afirmativas às Especulações 1 e 4 ao parametrizar o espaço classificador de tetraedros ideais (marcados) de maneira adequada.

Palavras-chave: Espac¸o hiperb´olico real, Volume de tetraedros ideais, Conjecturas de Seidel .

ABSTRACT

We prove a couple of conjectures raised by J. J. Seidel in ”On the volume of a hyperbolic simplex”, Stud. Sci. Math. Hung. (21, 243–249, 1986). These conjectures concern the volume of ideal hyperbolic tetrahedra in hyperbolic 3-space and are related to the following general framework. Since explicit formulae for geometric quantities in hyperbolic space (distance, area, volume, etc.) typically involve sophisticated transcendental functions, it is desirable (and quite useful in practice) to expresses these geometric quantities as monotonic functions of algebraic maps. Seidel’s Speculation 1 says that the volume of an ideal tetrahedron in hyperbolic 3-space depends only on the determinant and permanent of the doubly stochastic Gram matrix of its vertices; Speculation 4 claims that the mentioned volume is monotone in both the determinant and permanent. We are able to give afﬁrmative answers to Speculations 1 and 4 by parameterizing the classifying space of (labelled) ideal tetrahedra in a suitable way.

Key-words: Real hyperbolic space, Volume of ideal tetrahedra, Seidel’s conjectures.

LIST OF FIGURES

Figure1 – Theopenset ∆...... 23 Figure 2 – Young tableau ...... 24 Figure 3 – The open region U, domain of the parameters a and b...... 37 Figure4 – Theopenset V , domain of the parameters...... 48 Figure 5 – Bijection between W and V ...... 49 Figure6 – Theopenset ∆, domain of the parameters (c,d)...... 51 Figure7 – Theset R...... 83 Figure8 – Theset S...... 85

LIST OF TABLES

Table 1 – Points obtained by reﬂecting (c,d) ∆1...... 52 ∈

CONTENTS

1 INTRODUCTION ...... 21

2 PRELIMINARIES ...... 27 2.1 The projective space ...... 27 n 2.1.1 The real hyperbolic space HR ...... 27 2.1.2 Tangent space ...... 29 2.1.3 Metric ...... 30 2.2 Linear objects and duality ...... 30 2.2.1 Geodesics ...... 30 2.2.2 Totally geodesic planes ...... 31 2.2.3 Duality ...... 31 2.2.4 Dihedral angles ...... 32

3 TETRAHEDRA ...... 35 3.1 Ideal and labelled ideal tetrahedra in H3 ...... 35 3.1.1 A ﬁrst classifying space ...... 35 3.1.2 Determinant and permanent as coordinates ...... 44 3.1.3 The classifying space ∆ ...... 50 3.2 ProofofSpeculation1 ...... 55 3.3 Solving system 3.13. A few derivatives...... 58 3.4 Volume formulae ...... 63 3.5 Proof of a particular case of Speculation 3 ...... 64 3.6 ProofofSpeculation4 ...... 64

BIBLIOGRAPHY ...... 75

APPENDIX A HERMITIAN FORMS ...... 77 A.1 Hermitian forms ...... 77

APPENDIX B AN EXPLICIT DESCRIPTION OF R AND S ..... 81 B.1 An explicit description of R and S ...... 81

CHAPTER 1

INTRODUCTION

It is well known that transcendental methods are typically involved in calculating even simple geometric invariants in hyperbolic geometry (say, distance, area, volume, etc.). This has been observed already by Gauss, who referred to volume-related problems in hyperbolic geometry as a ‘jungle’. One way to deal with this kind of difﬁculty is to express a given geometric invariant as a monotonic function of algebraic expressions. As a toy example, let us consider the projective model of hyperbolic n-space. We take an (n + 1)-dimensional R-linear space V equipped with a bilinear symmetric form of signature +; the hyperbolic n-space is nothing but the open n-ball of positive points −−···− Hn := p P V p, p > 0 ∈ R |h i inside the projective space. Hyperbolicn-space is endowed with the (squared) distance function

p,q 2 d2(p,q) := arcCosh h i p, p q,q h ih i for p,q Hn. Clearly, distance is a monotonic function of the tance ∈ p,q 2 ta(p,q) := h i . p, p q,q h ih i In practical applications, it is usually much simpler to consider the tance instead of the distance. J. J. Seidel’s conjectures concern applying an analogous idea to the case of the volume formula of non-degenerate ideal tetrahedra in Hn. A non-degenerate ideal tetrahedron is noth- n ing but an n-tuple of points (v1,...,vn) in the ideal boundary (the absolute) of H such that the points v1,...,vn do not belong to a same hyperplane. We will actually call (v1,...,vn) a labelled non-degenerate ideal tetrahedron because the vertices are speciﬁed in a particular order.

Choosing representatives vi V, i = 1...,n, we associate to a non-degenerate ideal tetrahe- ∈ dron a Gram matrix G :=(gij), gij := vi,v j . Clearly, a Gram matrix of an ideal tetrahedron h i depends on the choice of the representatives vi V . Among all the Gram matrices of a given ∈ 22 Chapter 1. Introduction non-degenerate ideal tetrahedron, there is a single one, DSG, that is doubly stochastic (a matrix is called doubly stochastic if all its entries are non-negative and the sum of entries in every row and every column equals 1). Seidel claims that:

Speculation 1. The determinant and the permanent1 of DSG completely determine the volume of the associated ideal tetrahedron. Speculation 4. The volume is a monotonic function of the determinant and of the permanent of DSG.

We prove that both conjectures are true in the case n = 3, i.e., in the case of 3-dimensional hyperbolic space.2 In fact, we prove a stronger version of Speculation 1: the determinant and permanent of DSG determine not only the volume of the corresponding tetrahedron, they determine the tetrahedron itself (modulo isometries). Roughly speaking, the proof goes as follows. We begin by describing a classifying (mod- uli) space of all labelled non-degenerate ideal tetrahedra in H3 (see Subsection 3.1.3). This is accomplished by invoking a simple linear algebra fact: Let (v1,...,vn) and (w1,...,wn) be n- tuples in a finite dimensional linear space equipped with a symmetric bilinear form. Assume that the kernel of the form restricted to the subspaces Rv1 + + Rvn and Rw1 + + Rwn ··· ··· is null. Then the n-tuples differ by an element I of the orthogonal group O V := J GL V ∈ | Jv,Jw = v,w , Ivi = wi, if and only if their Gram matrices are the same. This fact plus an ap- h i h i propriate choice of coordinates allow us to describe the space of non-degenerate labelled ideal tetrahedron as the interior of an equilateral triangle ∆ in the Euclidean plane. The altitudes of ∆ divide the triangle into six smaller triangles — each is a copy of the space of (non-labelled) non-degenerate ideal tetrahedra. The fact that we obtain six copies of the space of non-degenerate ideal tetrahedra is simple to explain. There is a natural action of the 4-symmetric group S4 on the space of labelled non-degenerate ideal tetrahedra (permutation of vertices). But this action has a kernel which is isomorphic to Klein’s four group H and we arrive at an effective S3 = S4/H action. This S3- action simply permutes the three dihedral angles of an ideal tetrahedron: an ideal tetrahedron in hyperbolic 3-space has only three possibly distinct dihedral angles because opposite angles in the tetrahedron are necessarily equal. At the level of ∆, the mentioned S3-action corresponds to reflections in the altitudes. Once ∆ is obtained, Speculation 1 can be proved as follows. It is easy to obtain a necessary and sufficient condition on a pair of real numbers α,β such that α and β are respectively the determinant and permanent of the doubly stochastic matrix of at least one non-degenerate ideal tetrahedron (this is an application of the so-called Sylvester’s criterion in linear algebra).

1 The permanent of a matrix G = (gi j) is deﬁned by the expression perG := ∑σ Sn g1σ(1)g2σ(2) ...gnσ(n), where ∈ Sn stands for the symmetric n-group. 2 We also partially prove Speculation 3 which concerns the minimum possible value of the permanent of some doubly stochastic matrices. 23

∆

∆6 ∆1

b bb

∆5 b b ∆2

b b

∆4 ∆3

Figure 1 – The open set ∆.

Source: Elaborated by the author.

Writing down the formulae stating that the determinant and permanent of the doubly stochastic matrix of an arbitrary non-degenerate ideal tetrahedron respectively equal α and β leads to solving (in the generic case) a polynomial of degree six. It turns out that, once a single solution of this polynomial is obtained, the other solutions are the obvious ones: they correspond to the Gram matrices of tetrahedra that differ only as labelled tetrahedra.3 Figuring out the region ∆ was a crucial step in the solution of Speculation 1 since the equations relating the determinant and permanent of a doubly stochastic matrix to coordinates in other parameterizations of the classifying space of tetrahedra tend to be quite involved (say, it does not look feasible to solve this problem in the coordinates a, b of the region U, see Subsection 3.1.2). Solving the above mentioned polynomial of degree six allows us to express the entries of the double stochastic matrix DSG of an ideal tetrahedron in terms of its determinant α and permanent β. In this way, one ﬁnds an explicit formula for the volume in terms of α and β. Indeed, one of the simplest ways to calculate the volume of an ideal hyperbolic tetrahedron was discovered by Milnor, who found an expression in terms of Lobachevsky’s function and the dihedral angles of the tetrahedron:

vol(T ) = l(θ1) + l(θ2) + l(θ3).

It is quite simple4 to express the dihedral angles in terms of the entries of DSG and there- 3 It could be that Seidel did not state Speculation1 in its full generality because he did not make a clear distinction between labelled and non-labelled tetrahedra. His Speculation 1 was possibly made after an observation that there existed tetrahedra with different Gram matrices but with the same volume. However, these tetrahedra are geometrically the same as non-labelled tetrahedra.) 4 Curiously, the entries of DSG constitute lengths of the sides of an Euclidean triangle whose internal angles are 24 Chapter 1. Introduction fore we arrive at a volume formula that depends only on α and β. Differentiation plus a (non- straightforward) manipulation of inequalities leads to a proof of Speculation 4. It should be mentioned that Speculation 1 was given a ‘counter-example’ in [1]. Actu- ally, the authors of [1] do not deal with the doubly stochastic Gram matrix of vertices of an ideal tetrahedron. Instead, they consider a normalized Gram matrix of the points which are polar to the faces of the tetrahedron. Seidel’s speculations combine very well with some aspects of the elementary representation theory of the symmetric group which we recall below (our approach follows [4] almost literally).

Let n N be a natural number. A partition λ =(λ1,...,λk) of n is a choice of non-null ∈ k natural numbers λ1,...,λk satisfying n = ∑ j=1 λ j and λi > λi+1 for every i. The Young diagram of a partition λ is a union of squares organized as follows: the ﬁrst horizontal line consists of

λ1 adjacent squares, the second horizontal line consists of λ2 adjacent squares, etc.; the k-th horizontal line consists of λk adjacent squares. The squares in consecutive lines are drawn in such a way that the first square of the upper line is adjacent to the first square of the lower line. A Young tableau Tλ of a Young diagram is obtained by filling the squares of the diagram with the numbers 1,...,n (without repetition). The results we are interested in do not depend on a particular Young tableau associated to a Young diagram, so we assume that the diagram is always filled in some particular way (say, from 1 to n following lines). Below we illustrate the 5 Young tableau related to the partitions of the number 4.

Figure 2 – Young tableau

Source: Elaborated by the author.

Let Tλ be a Young tableau. We deﬁne two subgroups Lλ ,Cλ 6 Sn of the symmetric group as follows: the subgroup Lλ consists of all permutations that preserve the rows of the Young tableau and the subgroup Cλ consists of all permutations that preserve the columns of the Young tableau.

exactly the dihedral angles of the corresponding ideal tetrahedron, see Subsection 3.1.2. 25

In the group algebra CSn we introduce the Young projectors 1 1 aλ := ∑ g, bλ := ∑ (sgng)g, Lλ g L Cλ g C | | ∈ λ | | ∈ λ where sgng denotes the sign of the permutation g Sn. The Schur symmetrizer is deﬁned to be ∈ the element c := a b CSn. (Note that the Schur symmetrizer is non-null because L C = λ λ λ ∈ λ ∩ λ 1.)

The subspace Vλ :=(CSn)cλ 6 CSn is an irreducible representation of Sn (and every irreducible representation of Sn is isomorphic to some Vλ ). We are now ready to introduce the Schur functors.

λ n Given a partition λ of n, we denote S V := V ⊗ cλ . In particular, when λ =(n) we have Sλ = §n, i.e., Sλ is the n-th symmetric power functor. When λ =(1,1,...,1), we have Sλ = n, λ λ that is, S is n-th exterior power functor. For other partitions, S ‘interpolates’ between theseV extremal cases.

The space SλV can be visualized as follows: their elements are (ﬁnite) linear combina- tions of decomposable terms of the type

v1 ...vn :=(v1 vn)c . ⊗···⊗ λ

For instance, in the exterior power case, v1 ...vn = ∑g Sn (sgng)vg1 vgn. In the symmetric ∈ ⊗···⊗ power case, v1 ...vn = ∑g Sn vg1 vgn. ∈ ⊗···⊗ One can readily see that Sλ is indeed a functor from the category of finite dimensional linear spaces to itself. Indeed, we already know how Sλ behaves at the level of objects. At the level of morphisms, let f : V W be a linear map. It suffices to define Sλ f : SλV SλW →λ → for decomposable elements: (S f )(v1 ...vn) := f v1 ... f vn. We have just arrived at the Schur functors.

It should be noted that, if V is endowed with a symmetric bilinear form , , then h− −i SλV has a naturally induced symmetric bilinear form. This is well known, for example, in the case of the exterior power functor: the induced symmetric bilinear form on n V is given, at the level of decomposible elements, by V

v1 vn,w1 wn := detaij, h ∧···∧ ∧···∧ i where aij := vi,w j . In general, we have: h i

4.1. Deﬁnition — Theorem [5] Let λ be a partition of n, let Vλ be the associated λ representation of the symmetric group Sn, and let S be the corresponding Schur functor. Then

v1 ...vn,w1 ...wn := χga a a h i ∑ 1g(1) · 2g(2) ····· ng(n) g Sn ∈ 26 Chapter 1. Introduction

λ gives a induced symmetric bilinear form on S V, where aij := vi,w j and χ is the character h i of the representation.

The above formula, in the cases of the exterior and symmetric powers, corresponds respectively the determinant and permanent of the matrix with entries aij. Littlewood-Richardson call χga a a ∑ 1g(1) · 2g(2) ····· ng(n) g Sn ∈ the immanant of the matrix with entries aij (see [7]).

There are deep connections between Schur functors and the geometry of Hn. For in- 3 3 stance, let p1, p2, p3 H be pairwise distinct points. Then p1 p2 p3 V corresponds ∈ ∧ ∧ ∈ (under the Hodge star operator ⋆ : 3 V 1 V = V) to the polar point of the plane gener- → V ated by p1, p2, p3 (see Subsection 2.2.4V ). In orderV to calculate the angle θ between the planes generated by, say, p1, p2, p3 and q1,q2,q3, one essentially has to calculate the tance between 3 p := p1 p2 p3 and q := q1 q2 q3 in V ; explicitly, the mentioned angle is given by ∧ ∧ ∧ ∧ V p,q 2 cos2 θ = h i , p, p q,q h ih i where the symmetric bilinear form in the previous formula is the one induced by the exterior power functor in 3 V . Hence, isV it no accident that the determinant and permanent (of a suitably normalized Gram matrix of ideal points) play an important role in volume problems. Since non-degenerate ideal tetrahedra in H3 form a 2-dimensional manifold, determinant and permanent sufﬁce. In the general case of a tetrahedron whose vertices are all inside H3, the classifying space is 6- dimensional. There are, as we saw above, 5 Schur functors related to the representations of the symmetric group S4. Therefore, we make the following 3 Speculation. Let T be a tetrahedron whose vertices (v1,v2,v3,v4) are all in H . Then T is determined, up to isometry, by the trace and all immanants of the doubly stochastic Gram matrix associated to the vertices. The volume of T is a monotonic function of the trace and of all immanants of this matrix. 27

CHAPTER 2

PRELIMINARIES

2.1 The projective space

Let V be an R-linear space of dimension n + 1 endowed with the linear topology, i.e., the topology induced from any norm in V . We denote by PRV the projective space. We have the quotient map π : V P V, where V := V 0 is the linear space punctured at the origin. ∗ → R ∗ \{ } Let b0,b1,...,bn be an ordered basis in V. Let x0,x1,...,xn :V R be linear coordinates n → in regard to this basis, that is, xi(v) = xi where v = ∑i=0 xibi. In this case, we also write v = (x0,x1,...,xn) and π(v)=[x0 : x1 : ... : xn]. Finally, we introduce some notation. We frequently write p instead of π(p), where p ∈ V ∗. In such cases, by this notation we mean that our considerations do not change if we re- choose representatives in V of points in the projective space. One more convention. Given a subset S V we denote by P S := π(S 0 ) P V the image of S V under the quotient ⊂ R \{ } ⊂ R ⊂ map π : V P V . In this way, for every linear subspace K V , we can consider the projective ∗ → R ≤ space PRK inside PRV.

n 2.1.1 The real hyperbolic space HR From now on, we suppose that V is equipped with an hermitian form , :V V R h− −i × → of signature (n,0,1), see Deﬁnition A.1.23. As V is a real linear space we have that , is a h− −i bilinear symmetric form.

Deﬁnition 2.1.1. The signature of p P V is the sign of p, p . It is well deﬁned since, for ∈ R h i another representative rp V, r R , we have rp,rp = r2 p, p . The projective space P V ∈ ∈ ∗ h i h i R is divided into three disjoint parts consisting of positive, negative, and isotropic points:

BV := p P V p, p > 0 , EV := p P V p, p < 0 , SV := p P V p, p = 0 . { ∈ R |h i } { ∈ R |h i } { ∈ R |h i }

The isotropic points constitute the absolute SV of PRV . 28 Chapter 2. Preliminaries

Proposition 2.1.2. Let V be a space of signature (n,0,1). The following are satisﬁed in PRV :

(i) The positive and negative points, BV and EV , are open subsets of PRV . The absolute SV is a closed subset in PRV . (ii) BV = BV SV and EV = EV SV . ∪ ∪ (iii) The absolute SV is the boundary of BV and EV.

Proof. The function V R, p p, p is a composition of continuous functions. Then, the ∗ → 7→ h i set p V p, p > 0 is open in V . As p V p, p > 0 = π 1(BV ) and π is a quotient { ∈ ∗|h i } ∗ { ∈ ∗|h i } − map we have that BV is open in PRV . In a same way, EV is an open set in PRV . Then, SV = P V (BV EV ) is a closed subset in P V . R \ ∪ R It is clear that BV BV SV because BV SV = P V EV is a closed set that con- ⊂ ∪ ∪ R \ tains BV . Let p BV SV = p PRV p, p 0 . Let U be a neighborhood of p in PRV. 1 ∈ ∪ { ∈ |h i ≥ } Then, π− (U) is a neighborhood of p in V ∗. Because of the signature of the space V, we can choose b V such that b,b > 0. Let r R such that r p,b 0 and p + rb π 1(U). Thus, ∈ h i ∈ ∗ h i ≥ ∈ − p + rb, p + rb = p, p + 2r p,b + r2 b,b > 0, i.e., π(p + rb) BV U. This shows that h i h i h i h i ∈ ∩ p BV. We conclude that BV = BV SV . The proof of EV = EV SV is analogous. ∈ ∪ ∪ The boundary of BV is

BV P V BV = BV EV SV = BV EV SV =(BV SV ) (EV SV ) = SV. ∩ R \ ∩ ∪ ∩ ∪ ∪ ∩ ∪ It is similar to show that the boundary of EV is SV .

Proposition 2.1.3. BV is homeomorphic to the closed ball Bn, BV is homeomorphic to the open n n 1 ball B and SV is homeomorphic to the sphere S − .

Proof. Let b0,b1,...,bn be an orthonormal basis in V such that b0 is positive and the other vectors are negative. Let p BV SV and take any representative p = x0b0 +x1b1 +...+xnbn. It 2 2 ∈ ∪2 satisﬁes p, p = x x ... x 0, thus x0 = 0. We can therefore assume that x0 = 1 andsuch h i 0 − 1 − − n ≥ 6 n a representative of p is unique. Then, the map B BV SV , (α1,...,αn) [1: α1 : ... : αn] → ∪ 7→ is a continuous bijection. As the closed ball Bn is compact and BV SV is a Hausdorff space ∪ n n 1 the map is a homeomorphism. By restricting the map to the open ball B and the sphere S − we obtain the homeomorphism between BV and Bn and the homeomorphism between SV and n 1 S − respectively.

Let p P V be nonisotropic. We introduce the following notation for the orthogonal ∈ R decomposition:1

V = Rp p⊥, v = π1(p)v + π2(p)v, ⊕ where v, p v, p π1(p)v := h i p Rp, π2(p)v := v h i p p⊥. p, p ∈ − p, p ∈ h i h i 1 We denote by p the subspace (Rp) = v V v, p = 0 . ⊥ ⊥ { ∈ |h i } 2.1. The projective space 29

The deﬁnitions of the functions π1(p) and π2(p) do not depend on the choice of a representative p V. ∈

2.1.2 Tangent space

∞ Let p PRV . Let f C (U), where U PRV is an open neighborhood of p. Let ∈ ∈ ∞ 1 ⊂ f denote the map f π 1 C (π− (U)). The function f satisﬁes f (rp) = f (p), for all ◦ |π− (U) ∈ r R∗. Since f is smooth it deﬁnes the linear map Dp f : V R given by e∈ → e e e e f (p +tv) f (p) e d Dp f (v) = lim − = f (p +tv), t 0 t dt t=0 → e e for all v V . e e ∈

Proposition 2.1.4. The map t : Lin(Rp,V) TpPRV, ϕ tϕ , where tϕ f := Dp f (ϕ p), for all ∞ → 7→ f C (U), is linear. Also, tϕ = 0 if and only if ϕ p Rp. ∈ ∈ e

Proof. Note that tϕ TpPRV because ]f + g = f + g, rf = r f , and fg = f g, for all f ,g ∞ 1 ∈ ∈ C (π− (U)) and r R. The linearity of the map Dp f guarantees the linearity of the map t. ∈ e e e e f ee Suppose that t = 0. Let b ,b ,...,b be any basis in V . Write p = α b + α b + ... + ϕ 0 1 n e 0 0 1 1 αnbn. For simplicity we assume that α0 = 0. So, p U0 = [x0 : x1 : ... : xn] x0 = 0 . The 6 ∈ { | 6 } xi function fi : U0 R, [x0 : x1 : ... : xn] is smooth for each i = 1,...,n, and fi : x0b0 +x1b1 + → 7→ x0 xi ... + xnbn x . Let ϕ p = γ0b0 + γ1b1 + ... + γnbn. By hypothesis, 7→ 0 e d αi +tγi γiα0 αiγ0 0 = tϕ fi = = −2 . dt t=0 α0 +tγ0 α 0

γ0 γ0 Note that γi = αi for each i = 0,1,...,n, then ϕ p = p Rp. α0 α0 ∈

Corollary 2.1.5. The map j : Lin(Rp,V/Rp) TpP V , ϕ j(ϕ) such that j(ϕ) : f → R 7→ 7→ Dp f ϕ p , where ϕ p V is any representative of the class ϕ p V/Rp, is a linear isomor- ∈ ∈ phism. e c c

Proof. Note that j(ϕ) is well deﬁned. The linearity of the map Dp f guarantees the linearity of the map j. As the linear spaces Lin(Rp,V/Rp) and TpPRV have the same dimension, the e injectivity of j which comes from the last proposition which shows the isomorphism between them.

Finally, for a nonisotropic p P V, we have V/Rp = p and conclude that ∈ R ∼ ⊥

TpP V = Lin(Rp, p⊥) = , p p⊥, R ∼ ∼ h− i where , p p := , p v : x x, p v . h− i ⊥ {h− i 7→ h i } 30 Chapter 2. Preliminaries

2.1.3 Metric

Let p P V be a nonisotropic point. Given v p , we deﬁne ∈ R ∈ ⊥

tp v := , p v TpP V. , h− i ∈ R

Note that tp,v does depend on the choice of a representative p V : if we pick a new representa- 1 ∈ tive rp V, r R , then we must take v V in place of v in order to keep tp v the same. ∈ ∈ ∗ r ∈ , The tangent space TpPRV is equipped with the hermitian form

tp v ,tp v := p, p v1,v2 . (2.1) , 1 , 2 −h ih i

This deﬁnition is correct as the formula is independent of the choice of representatives p,v1,v2 ∈ V providing the same tp,v1,tp,v2 . One can readily see that this hermitian form, called a hermitian metric, depends smoothly on a nonisotropic p. We are going to work on the open set of positive points BV P V. With the metric ⊂ R deﬁned in equation 2.1, the space BV becomes a Riemannian manifold. We will denote this n space by HR and call it the hyperbolic space.

2.2 Linear objects and duality

2.2.1 Geodesics

Deﬁnition 2.2.1. Let W V be a two-dimensional linear subspace such that the hermitian form, ≤ being restricted to W, is nonnull. We call P W P V a geodesic. R ⊂ R

Let b1,b2 be an orthonormal basis in W . As the signature of V is (n,0,1), by Claim A.1.27, the possible signatures of b1,b2 are (2,0,0), (1,0,1), and (1,1,0). Also, we suppose that negative vectors, null vectors and positive vectors are listed in that order in the basis b1,b2. Let p P W and write p = rb1 + sb2 for some r,s R. ∈ R ∈ 2 2 Let b1,b2 be of signature (2,0,0). We have p, p = r s < 0, then p is negative. Thus, the h i − − geodesic PRW is totally contained in EV. 2 2 Let b1,b2 be of signature (1,0,1). The solutions of the equation p, p = r + s = 0 are h i − s = r. We can suppose that r = 1. Then, the isotropic points in the geodesic p P W are the ± ∈ R classes of b1 + b2 and b1 b2. These points are the vertices of the geodesic. Note that if we − write an element w W as w = r(b1 + b2) + s(b1 b2), where r,s R, then the point w BV ∈ − ∈ ∈ when r and s are of the same sign and w EV when r and s are of distinct sign. ∈ 2 Let b1,b2 be of signature (1,1,0). We have p, p = r . The unique isotropic point in the h i − geodesic PRW is b2. All the other points in PRW are in EV . 2.2. Linear objects and duality 31

2.2.2 Totally geodesic planes

Deﬁnition 2.2.2. Let W V be a three-dimensional linear subspace. We call P W P V a ≤ R ⊂ R plane.

Let b1,b2,b3 be an orthonormal basis in W. As the signature of V is (n,0,1), by Claim

A.1.27, the possible signatures of b1,b2,b3 are (3,0,0), (2,0,1), and (2,1,0). Also, we suppose that the negative vectors, the null and the positive vectors are listed in that order in the basis b1,b2,b3. Let p P W and write p = rb1 + sb2 +tb3 for some r,s,t R. ∈ R ∈ 2 2 2 Let b1,b2,b3 be of signature (3,0,0). We have p, p = r s t < 0, then p is negative. h i − − − Thus, the plane PRW is totally contained in EV. 2 2 2 Let b1,b2,b3 be of signature (2,0,1). The solutions for the equation p, p = r s +t = 0 h i − − are t2 = r2 +s2. Thus, we can take the unique representative of p such that t = 1,i.e., r2 +s2 = 1. It follows that P W SV is homeomorphic to the sphere S1. In the same way we can show that R ∩ P W BVis homeomorphic to the open ball B2. R ∩ 2 2 Let b1,b2,b3 be of signature (2,1,0). We have p, p = r s . The unique isotropic point in h i − − the plane PRW is b3. All the other points in PRW are in EV .

2.2.3 Duality

Let W V be an n-dimensional linear subspace. As the signature of V is (n,0,1), the ≤ possible signatures of W are (n,0,0), (n 1,0,1), and (n 1,1,0). In the following we use − − frequently Claim A.1.6.

Let W be of signature (n,0,0), then W = Rp, where p BV. Also, if p BV , we have p = W, ⊥ ∈ ∈ ⊥ where W is of signature (n,0,0).

Let W be of signature (n 1,0,1), then W = Rp, where p EV . Also, if p BV , we have − ⊥ ∈ ∈ p = W, where W is of signature (n 1,0,1). ⊥ − Let W be of signature (n 1,1,0), by Claim A.1.5we have dim W = 1, then W = Rp. By − ⊥ ⊥ Claim A.1.3, (Rp) =(W ) W. Assuming that p is positive or negative, we have that Rp ⊥ ⊥ ⊥ ⊃ in nondegenerate and then (Rp)⊥ has dimension n, thus (Rp)⊥ = W which is a contradiction because W is degenerate. We conclude that p SV . In theother sense, if p SV . By Claim A.1.5 ∈ ∈ we know that dim p 3. As the form in V is nondegenerate we conclude that dim p = 3. ⊥ ≥ ⊥ As Rp p and the possible signatures for p are (n,0,0), (n 1,0,1), and (n 1,1,0) we ⊂ ⊥ ⊥ − − conclude that the signature of p is (n 1,1,0). ⊥ − When n = 2, we have the duality between points and geodesics in PRV.

When n = 3, we have the duality between points and planes in PRV . 32 Chapter 2. Preliminaries

2.2.4 Dihedral angles

Let us apply the duality between points and planes in real hyperbolic space in order to ﬁnd an expression for the dihedral angle between planes that we will apply in the proof of Seidel’s Speculation 4. First, we need to discuss a few basic facts about the Hodge star operator. There is no need to specify the dimension of V neither the signature of the non-degenerate symmetric bilinear form until the very end of the subsection. The results in this subsection are related to Section 2.2 in [9].

Let V be an R-linear N-dimensional space equipped with a non-degenerate symmetric bilinear form , of signature (n,0,m), where N = n + m. Let σ = 1 if n = 0 mod2 and h− −i σ = 1 otherwise. − The k-th exterior power k V ,1 6 k 6 N, is equipped with the symmetric bilinear form deﬁned by V v1 vk,w1 wk := det(gij), h ∧···∧ ∧···∧ i k where gij := vi,w j . Note that this form on V is non-degenerate. Indeed, let (b1,...,bN) h i be an orthonormal basis for V , that is, bi,bi V= σi and bi,b j = 0 for i = j, where σi = 1. h i h i 6 k ± Then (bi bi 1 6 i1 < < ik 6 N), is an orthonormal basis for V since bi 1 ∧···∧ k | ··· h 1 ∧···∧ bi ,bi bi = σi ...σi = 1 and bi bi ,b j b j vanishes for i1 < < i , k 1 ∧···∧ k i 1 k ± h 1 ∧···∧ k 1 ∧···∧ k i V ··· k j1 < < jk and (i1,...,ik) =( j1,..., jk) because it is the determinant of a matrix that has at ··· 6 least one null row.

Fix ω N V such that ω,ω = σ. The Hodge star operator is the R-linear map : k N k ∈ h i k ∗ V − V, b b, deﬁned by requiring that a b = a,b ω for every a V . Clearly, → V 7→ ∗ ∧∗ h i ∈ ω = σ since ω ω = ω,ω ω = σω. V∗ V ∧∗ h i V

Lemma 2.2.3. The Hodge star operator is injective. It satisﬁes the following identities:

a b = b a for every a,b k V • ∧∗ ∧∗ ∈ a,b = σ (a b) = σ (b V a) for every a,b k V •h i ·∗ ∧∗ ·∗ ∧∗ ∈ k(N k) k V N k a, b =( 1) a,b , for every a V and b − V •h ∗ i − − h∗ i ∈ ∈ 2 =( 1)k(N k)σ V V •∗ − −

Proof. If b = 0, then a b = a,b ω = 0 for every a k V which implies b = 0 because the ∗k ∧∗ h i ∈ form on V is nondegenerate. V TheV ﬁrst identity is obvious. The second follows from (a b) = a,b w = σ a,b . ∗ ∧∗ h i·∗ h i The third follows from the second:

k(N k) k(N k) a, b = σ ( b a)=( 1) − σ ( a b)=( 1) − a,b . h ∗ i ·∗ ∗ ∧∗ − ·∗ ∗ ∧∗ − h∗ i 2.2. Linear objects and duality 33

Finally, let us show the last equility. Take an orthonormal basis b1,...,bN in V , that is, bi,bi = h i σi = 1 and bi,b j = 0 for i = j. We can assume that ω = b1 bN. Fix i1 < i2 < < ± h i 6 ∧···∧ ··· ik and let j1 < j2 < < jN k be the indices complementary to i1 < i2 < < ik. Let us ··· − ··· show that (bi1 bik ) = sgn(h) σi1 σik b j1 b jN k , where h is the permutation ∗ ∧···∧ · ··· · ∧···∧ − (i1,i2,...,ik, j1, j2,..., jN k) (1,2,...,N). Indeed, for every l1 < < lk, − 7→ ···

0, if (l1,...,lk) =(i1,...,ik) (bl1 blk ) (sgnh σi1 σik b j1 b jN k ) = 6 ∧···∧ ∧ · ··· · ∧···∧ −  σi ...σi ω if (l1,...,lk)=(i1,...,ik)  1 k ·

= bl bl ,bi ...bi ω. h 1 ∧···∧ k 1 ∧ k i Finally, it remains to observe that

k(N k) (bi1 bik ) = sgn(h) σi1 ...σik (b j1 b jN k )=( 1) − σ bi1 bik , ∗ ∗ ∧···∧ · ·∗ ∧···∧ − − · ∧···∧ k(N k) where ( 1) − = sgn(h) sgn(h′) and h′ is the permutation ( j1,..., jN k,i1,...,ik) − · − 7→ (1,2,...,N).

From now on, assume that V is of signature (3,0,1). Let v1,v2,v3 SV be pairwise ∈ distinct isotropic points that generate the plane P := PRW1 for W1 = Rv1 + Rv2 + Rv3. We claim that u := (v1 v2 v3) is the polar point to P. Indeed, by the previous lemma, ∗ ∧ ∧

vi, (v1 v2 v3) ω = vi (v1 v2 v3) = vi v1 v2 v3 = 0 ∗ ∧ ∧ · ∧∗ ∗ ∧ ∧ ∧ ∧ ∧ for i = 1,2,3. We are now ready to calculate the expression for the angle between (half-)planes that we will need in Chapter 3.

Let v1,v2,w1,w2 SV be pairwise distinct isotropic points. Let u1 := (v1 v2 w1) be ∈ ∗ ∧ ∧ the polar point to the plane P1 generated by v1,v2,w1 and let u2 := (v1 v2 w2) be the polar ∗ ∧ ∧ point to the plane P2 generated by v1,v2,w2. The geodesic G joining v1,v2 is common to P1 and P2; let p G. We take a representative p V such that p, p = 1. Our intention is to measure ∈ ∈ h i the angle θ [0,π] between the half-plane in P1 which contains w1 and is determined by G and ∈ the half-plane in P2 which contains w2 and is determined by G.

By Lemma 4.2.2. in [3], ni := , p ui is a normal vector to Pi at p, i = 1,2. The angle h− i θ [0,π] is exactly the angle between n1 and n2, i.e., ∈ n1,n2 u1,u2 cosθ = h i = −h i n1,n1 n2,n2 u1,u1 u2,u2 h i h i −h i −h i From Lemma 2.2.3 we obtainp p p p

0 v1,v2 v1,w2 h i h i u1,u2 = (v1,v2,w1), (v1,v2,w2) = det v2,v1 0 v2,w2 , −h i − ∗ ∗ h i h i w1,v1 w1,v2 w1,w2 h i h i h i   34 Chapter 2. Preliminaries

0 v1,v2 v1,w1 h i h i u1,u1 = (v1,v2,w1), (v1,v2,w1) = det v2,v1 0 v2,w1 , −h i − ∗ ∗ h i h i w1,v1 w1,v2 0, h i h i  and  

0 v1,v2 v1,w2 h i h i u2,u2 = (v1,v2,w2), (v1,v2,w2) = det v2,v1 0 v2,w2 . −h i − ∗ ∗ h i h i w2,v1 w2,v2 0 h i h i    since 2 = 1. ∗ In Chapter 3, we will calculate the above angle θ in terms of some speciﬁc representatives for v1,v2,w1,w2 V. ∈ 35

CHAPTER 3

TETRAHEDRA

In this chapter, we introduce the space of labelled ideal tetrahedra as well as a couple of parameterizations of this space. The main objective is to arrive at the doubly stochastic parameterization which is used to formulate Seidel’s conjecture.

From now on, we deal only with an R-linear space V of signature (3,0,1). As in the previous chapter, it gives rise to the extended real hyperbolic space divided into the proper hyperbolic space H3, its absolute SV and elsewhere EV.

3.1 Ideal and labelled ideal tetrahedra in H3

An ideal tetrahedron in P V consists of four ideal points v1,v2,v3,v4 , vi SV .A la- R { } ∈ belled ideal tetrahedron is an ideal tetrahedron with ordered vertices, i.e., a 4-tuple (v1,v2,v3,v4) of ideal points. The set of ideal tetrahedra with ordered vertices is denoted by

T := (v1,v2,v3,v4) v1,v2,v3,v4 are points in SV . | 3.1.1 A ﬁrst classifying space

The main result in this subsection is Theorem 3.1.5, where a parameterization of the space of labelled non-degenerate ideal tetrahedra is given. The theorem is then applied in Propo- sition 3.1.8 to ﬁnd, in an explicit form, the unique doubly stochastic matrix associated to a non-degenerate labelled ideal tetrahedron.

Deﬁnition 3.1.1. Given (v1,v2,v3,v4) T , we denote the whole class of Gram matrices of ∈ representatives of v1,v2,v3,v4 by

M (v1,v2,v3,v4) := G(v1,v2,v3,v4) vi is a representative of vii for i = 1,2,3,4 . | 36 Chapter 3. Tetrahedra

Of course, rechoosing representatives, one can describe M (v1,v2,v3,v4) using the Gram matrix G(v1,v2,v3,v4) of any given representatives v1,v2,v3,v4 of the vertices. This simple fact is stated and proved below.

Proposition 3.1.2. Let (v1,v2,v3,v4) T . Fix representatives v1,v2,v3,v4 of the vertices of the ∈ labelled ideal tetrahedron. Then

M (v1,v2,v3,v4) = DG(v1,v2,v3,v4)D D = diag(x1,x2,x3,x4), xi R∗ for i = 1,2,3,4 . | ∈ Proof. Let u1,u2,u3,u4 be arbitrary representatives of v1, v2, v3, v4. This means that ui = xivi where xi = 0, i = 1,2,3,4. The Gram matrix 6

G(u1,u2,u3,u4) = G(x1v1,x2v2,x3v3,x4v4)

x1v1,x1v1 x1v1,x2v2 x1v1,x3v3 x1v1,x4v4 h i h i h i h i  x2v2,x1v1 x2v2,x2v2 x2v2,x3v3 x2v2,x4v4  = h i h i h i h i  x3v3,x1v1 x3v3,x2v2 x3v3,x3v3 x3v3,x4v4  h i h i h i h i  x4v4,x1v1 x4v4,x2v2 x4v4,x3v3 x4v4,x4v4  h i h i h i h i   = DG(v1,v2,v3,v4)D, where D is the diagonal matrix diag(x1,x2,x3,x4), xi = 0. 6

Observation 3.1.3. Suppose that vi and ui are representatives of the vertex vi of the ideal tetrahedron (v1,v2,v3,v4). Then (v1,v2,v3,v4) is linearly independent if and only if (u1,u2,u3,u4) is linearly independent.

A labelled tetrahedron (v1,v2,v3,v4) T is degenerate if its vertices lie in a common plane 3 ∈ in PRV. In terms of representatives v1,v2,v3,v4, this means exactly that (v1,v2,v3,v4) is linearly dependent. Observation 3.1.4. Let u and v be different points in SV . Then u,v = 0. Indeed, assuming h i 6 u,v = 0, the space spanned by u and v has induced null form and is two-dimensional; however, h i in the case of signature (3,0,1), the highest possible dimension of a subspace with null form is 1 (see Claim A.1.26).

Let (v1,v2,v3,v4) T be a non-degenerate ideal tetrahedron. We are interested in dou- ∈ bly stochastic Gram matrices in M (v1,v2,v3,v4). (A matrix is called doubly stochastic if all its entries are non-negative and the sum of coefﬁcients along any column or row equals 1. As we will see later, Seidel’s conjecture is formulated in terms of doubly stochastic matrices.) As a preliminary step, we introduce the following parameterization of the space T .

Theorem 3.1.5. The classifying space (modulo isometries) of non-degenerate labelled ideal tetrahedra can be identiﬁed with the open region

U := (a,b) R2 (a 1)2 < b < (a + 1)2 ∈ | − 3.1. Ideal and labelled ideal tetrahedra in H3 37 in R2. Explicitly, this identiﬁcation is as follows. Given a non-degenerate labelled ideal tetrahedron

(v1,v2,v3,v4) T , there exists a unique Gram matrix G(v1,v2,v3,v4) M (v1,v2,v3,v4) of the ∈ ∈ form 0 1 a b 1 0 1 a G(v1,v2,v3,v4) = a 1 0 1   b a 1 0   with (a,b) U. Conversely, for (a,b) U, there exists a non-degenerate labelled ideal tetrahe- ∈ ∈ dron whose Gram matrix is of the above form. Tetrahedra with a same Gram matrix differ by an isometry.

(1,1) b

Figure 3 – The open region U, domain of the parameters a and b.

Source: Elaborated by the author.

Proof. Let v1,v2,v3,v4 be representatives of v1,v2,v3,v4. Since the ideal tetrahedron (v1,v2,v3,v4) is non-degenerate its vertices are pairwise distinct pointsinSV , Observation 3.1.4 implies that

vi,v j = 0 when i = j. So we can take the new representatives 6 6 v2,v3 v3,v4 1 u1 := h i v1, u2 := h iv2, u3 := v3, u4 := v4. v1,v2 v3,v4 v2,v3 v3,v4 h ih i h i h i These satisfy ui,ui 1 = 1 for i = 1,2,3. h + i 1 4 u2,u4 Now let α := h i = 0 and deﬁne u1,u3 h i 6

1 1 w := αu , w := u , w := αu , w := u . 1 1 2 α 2 3 3 4 α 4 38 Chapter 3. Tetrahedra

Clearly, wi,wi 1 = 1 for i = 1,2,3; note that h + i 2 w1,w3 = α u1,u3 = u1,u3 u2,u4 |h i| |h i| |h ih i| p and 1 w2,w4 = u2,u4 = u1,u3 u2,u4 |h i| α2 |h i| |h ih i| imply w1,w3 = w2,w4 . We have just arrived atp a Gram matrix |h i| |h i| 0 1 a b 1 0 1 c G(w1,w2,w3,w4) = , a 1 0 1   b c 1 0     where a,b,c R and a = c . ∈ ∗ | | | | In what follows we will repeatedly apply Sylvester’s criterion (see Proposition A.1.24) to different ordered bases.

On one hand, the Gram matrix of the basis β =(v1 + v2,v2,v3,v4) is

2 1 a + 1 b + c  1 0 1 c  Gββ = . a + 11 0 1    b + c c 1 0      The determinants of the ﬁrst three principal sub-matrices of Gββ are

detGββ = 2, detGββ = 1, detGββ = 2a. 1 2 − 3 ββ ββ They are non-null. The determinant detG4 = detG is also non-null because V is non- degenerate (see Lemma A.1.21). By Sylvester’s criterion, three and one are respectively the amount of negative and positive numbers in the sequence

detGββ 1 detGββ detGββ detGββ = 2, 2 = , 3 = 2a, 4 . 1 ββ −2 ββ − ββ detG1 detG2 detG3 This implies that a > 0.

On the other hand, the Gram matrix of the basis γ :=(v3 + v4,v4,v2,v1) is

2 11 + c a + b 1 0 c b Gγγ =  . 1 + c c 0 1    a + b b 1 0      The determinants of the principal sub-matrices

detGγγ = 2, detGγγ = 1, detGγγ = 2c, detGγγ = detG, 1 2 − 3 4 3.1. Ideal and labelled ideal tetrahedra in H3 39 are non-null. Since three and one are respectively the amount of negative and positive numbers in the sequence

γγ γγ γγ γγ detG2 1 detG3 detG4 detG1 = 2, γγ = , γγ = 2c, γγ detG1 −2 detG2 − detG3 we obtain c > 0 and detGγγ = detGγγ < 0. We already knew that a = c and a > 0, hence 4 | | | | a = c. It remains to observe that

2 11 + a a + b 1 0 a b 0 > detGγγ = = a4 + b2 + 1 2a2b 2a2 2b = − − − 1 + a a 0 1

a + b b 1 0

= (a + 1)2 b (a 1)2 b − − − implies the desired inequalities (a 1)2 < b < (a+1)2 because (a 1)2 b < (a+1)2 b and − − − − the numbers (a 1)2 b, (a + 1)2 b have distinct signs. − − − In order to prove the uniqueness of the Gram matrix G(v1,v2,v3,v4), suppose that G(u1,u2,u3,u4) M (v1,v2,v3,v4) has the form ∈

0 1 a′ b′  1 0 1 a′ G(u1,u2,u3,u4) = a′ 1 0 1   b′ a′ 1 0     with (a ,b ) U. By Proposition 3.1.2 we have G(u1,u2,u3,u4) = DG(v1,v2,v3,v4)D, where ′ ′ ∈ D = diag(x1,x2,x3,x4), xi = 0 for i = 1,2,3,4. So 6

0 1 a′ b′ 0 x1x2 ax1x3 bx1x4  1 0 1 a′  x2x1 0 x2x3 ax2x4 = , a′ 1 0 1 ax3x1 x3x2 0 x3x4      b′ a′ 1 0 bx4x1 ax4x2 x4x3 0          i.e.,

1 = x1x2 = x2x3 = x3x4  a′ = ax1x3 = ax2x4   b′ = bx1x4

It follows from the ﬁrst line that x1 = x3 and x2 = x4 as well as that x1,x2 are of a same sign 2 2 and x2,x3 are of a same sign. Since a = 0, the second line implies x = x , that is, x1 = x2. 6 1 2 We obtain x1 = x2 = x3 = x4 = 1 and conclude, in both cases, that a = a and b = b, i.e., ± ′ ′ G(u1,u2,u3,u4) = G(v1,v2,v3,v4). 40 Chapter 3. Tetrahedra

Given (a,b) U, let us show that there exists a non-degenerate labelled ideal tetrahe- ∈ dron (v1,v2,v3,v4) T such that ∈ 0 1 a b 1 0 1 a G(v1,v2,v3,v4) = , a 1 0 1   b a 1 0     for some representatives v1,v2,v3,v4. Let b1,b2,b3,b4 be an orthonormal basis in V with b1,b1 = h i 1 and b2,b2 = b3,b3 = b4,b4 = 1. We deﬁne h i h i h i −

v1 :=b1 + b2, 1 1 v2 := b1 b2, 2 − 2 a a v3 := 1 + b1 + 1 b2 + √2ab3, 2 − 2 b b a2 + b 1 v4 := a + b1 + a b2 + − b3+ 2 − 2 √2a a + √b + 1 a + √b 1 a √b + 1 a + √b + 1 − − − v b4. u 2a u t The inequalities (a 1)2 < b < (a + 1)2 imply a + √b 1,a √b + 1, a + √b + 1 > 0; in − − − − other words, v4 is well deﬁned. The Gram matrix of v1,v2,v3,v4 is the required one. Since its determinant is non-null (in fact, the inequalities (a 1)2 < b < (a+1)2 imply that it is negative), − the corresponding labelled ideal tetrahedron (v1,v2,v3,v4) T is non-degenerate. ∈ Finally, the fact that tetrahedra with a same Gram matrix differ by an isometry follows directly from [2], Lemma 4.8.1.

Observation 3.1.6. ?? The region (a,b) R2 a,b > 0 and b =(a 1)2 corresponds to all ∈ | ± degenerate labelled ideal tetrahedra with pairwise distinct vertices. Indeed, note that if v1,v2,v3 are pairwise distinct points in SV , then W := Rv1 + Rv2 + Rv3 is a space of signature (2,0,1). The rest follows applying Sylvester’s criterion as before. Therefore, the region

U (0,1),(1,0) = (a,b) R2 a,b > 0 and (a 1)2 6 b 6 (a + 1)2 \{ } ∈ | − parameterizes all labelled ideal tetrahedra with pairwise distinct vertices. The remaining points

(0,1) and (1,0) correspond respectively to the degenerate labelled tetrahedra (v1,v2,v1,v2) and (v1,v2,v3,v1). In connection with this Observation, see also Subsection 3.1.3.

Deﬁnition 3.1.7. A doubly stochastic matrix is a square matrix A =(aij) of nonnegative real numbers such that ∑i aij = ∑ j aij = 1. 3.1. Ideal and labelled ideal tetrahedra in H3 41

Proposition 3.1.8. Let (v1,v2,v3,v4) T be a labelled ideal tetrahedron with pairwise distinct ∈ vertices. There exists a unique doubly stochastic Gram matrix M M (v1,v2,v3,v4). Explicitly, ∈ 0 1 a √b 1 1 0 √b a M =  , √ √ a + b + 1  a b 0 1    √b a 1 0      where (a,b) U (0,1),(1,0) . Conversely, given (a,b) U (0,1),(1,0) , there exists ∈ \{ } ∈ \{ } a labelled ideal tetrahedron (v1,v2,v3,v4) T with pairwise distinct vertices whose doubly ∈ stochastic matrix is the above M.

Proof. By Observation ?? there exits a Gram matrix G(v1,v2,v3,v4) M (v1,v2,v3,v4) of the ∈ form 0 1 a b 1 0 1 a G(v1,v2,v3,v4) = a 1 0 1   b a 1 0   with (a,b) U (0,1),(1,0) . Using Proposition 3.1.2, we look for a diagonal matrix D = ∈ \{ } diag(w,x,y,z) such that w,x,y,z R and ∈ ∗ 0 wx awy bwz  wx 0 xy axz DG(v1,v2,v3,v4)D = awy xy 0 yz    bwz axz yz 0    is doubly stochastic.  

First, note that w,x,y,z are of a same sign because, otherwise, at least one entry in

DG(v1,v2,v3,v4)D would be negative. Also, DG(v1,v2,v3,v4)D =( D)G(v1,v2,v3,v4)( D) − − and therefore we can take D with positive coefﬁcients. Thus, we only have to solve wx + awy + bwz = 1  wx + xy + axz = 1   awy + xy + yz = 1  bwz + axz + yz = 1  for positive numbers w,x,y,z.   1 From the ﬁrst equation, w = . This implies x + ay + bz 1 x + xy + axz = 1 x + ay + bz   1  ay + xy + yz = 1 . (3.1)  x + ay + bz  1 bz + axz + yz = 1  x + ay + bz    42 Chapter 3. Tetrahedra

1 axz Summing these three equations we obtain xy + yz + axz = 1, that is, y = − . In particular, x + z 1 axz = 0. Replacing y in the first and third equations of 3.1 leads to − 6 x + z 1 axz x + − x + axz = 1 (x + bz)(x + z) + a(1 axz) x + z  − . (3.2)  x + z 1 axz  bz + axz + − z = 1 (x + bz)(x + z) + a(1 axz) x + z −  Multiplying the first equation by bz, the second equation by x, and subtracting the results, we obtain 1 axz 1 axz − bxz + abxz2 ax2z − xz = bz x x + z − − x + z − which implies (1 axz)xz − (b 1) + axz(bz x) = bz x x + z − − − that is (1 axz)xz − (b 1)=(bz x)(1 axz). x + z − − − Since 1 axz = 0, we arrive at xz(b 1)=(bz x)(x + z), i.e., at x2 = bz2. In other words, − 6 − − x = √bz. Replacing x in the first equation of 3.2 gives

√bz + z 1 a√bz2 √bz + − √bz + a√bz2 = 1 √bz + bz √bz + z + a 1 a√bz2 √bz + z − which is nothing but

b + √b z2 + a√bz2 1 b + √b z2 a√bz2 + 1 = 0. (3.3) − − We claim that b + √b z2 a√bz2 + 1 = 0. Indeed, if we assume otherwise, then − 6 1 = √b a 1 √b z2; − − this is a contradiction: the inequality (a 1)2 6 b satisﬁed by (a,b) U (0,1),(1,0) implies − ∈ \{ } a 1 √b < 0. By 3.3, − − b + √b z2 + a√bz2 1 = 0 − and we arrive at 1 z = . √4 b a + √b + 1 Therefore, the unique (positive) solution is p

1 √4 b w = z = , x = y = √4 b a + √b + 1 a + √b + 1 p p 3.1. Ideal and labelled ideal tetrahedra in H3 43 and the corresponding doubly stochastic matrix is

0 1 a √b 1  1 0 √b a  G(u ,u ,u ,u ) = DG(v ,v ,v .v )D = , 1 2 3 4 1 2 3 4 √ √ a + b + 1  a b 0 1    √b a 1 0      where (a,b) U (0,1),(1,0) . ∈ \{ } The converse is immediate.

Observation 3.1.9. Let v,u be different points in SV. We can choose representatives such that 1 u,v = , then the Gram matrix h i 2 1 1 0 0 2 2 1 1 0 0  G(v,v,u,u) = 2 2 1 1 0 0  2 2   1 1   0 0  2 2    is doubly stochastic. It is not difﬁcult to verify that it is the unique doubly stochastic Gram matrix in M (v,v,u,u). Similarly, G(v,u,v,u) and G(v,u,u,v) are respectively unique doubly stochastic Gram matrices in M (v,u,v,u) and M (v,u,u,v). In Proposition 3.1.8, the points (0,1),(1,0) U can be associated respectively to the ideal tetrahedra (v,u,v,u) and (v,u,u,v), ∈ but there is no point in U which can be associated to the ideal tetrahedron (v,v,u,u).

Observation 3.1.10. Let (v1,v2,v3,v4) T be a labelled ideal tetrahedron. If the number of dis- ∈ tinct vertices of (v1,v2,v3,v4) is 1 or 3, there is no a doubly stochasticmatrix in M (v1,v2,v3,v4). Indeed, if (v1,v2,v3,v4) has one distinct vertex, M (v1,v2,v3,v4) consists only on the null matrix. If (v1,v2,v3,v4) has three distinct vertices (say v1 = v2 — the other cases are similar), assume that there exist representatives v1,v1′ ,v3,v4 of v1,v1,v3,v4 with doubly stochastic Gram matrix. Note that v = rv1, where r R . The doubly stochastic Gram matrix is 1′ ∈ ∗

0 0 v1,v3 v1,v4 h i h i  0 0 r v1,v3 r v1,v4  G(v1,rv1,v3,v4) = h i h i .  v1,v3 r v1,v3 0 v3,v4  h i h i h i   v1,v4 r v1,v4 v3,v4 0  h i h i h i    Then v1,v3 + v1,v4 = 1 h i h i  r v1,v3 + r v1,v4 = 1  h i h i .  v1,v3 + r v1,v3 + v3,v4 = 1  h i h i h i v1,v4 + r v1,v4 + v3,v4 = 1  h i h i h i Summing the third and forth equations, we have 

v1,v3 + v1,v4 + r v1,v3 + r v1,v4 + 2 v3,v4 = 2. h i h i h i h i h i 44 Chapter 3. Tetrahedra

Then from the ﬁrst and the second equation we get v3,v4 = 0, i.e., v3 = v4 which is a contra- h i diction with the number of distinct vertices of the ideal tetrahedron.

Let (v1,v2,v3,v4) T be a labelled ideal tetrahedron with 2 or 4 distinct vertices. We ∈ denote by DSG(v1,v2,v3,v4) the unique doubly stochastic Gram matrix in M (v1,v2,v3,v4) and by

DS := DSG(v1,v2,v3,v4) (v1,v2,v3,v4) T with an even number of distinct vertices | ∈ the space of doubly stochastic Gram matrices of labelled ideal tetrahedra with 2 or 4 distinct vertices. Clearly, the space DS is simply another way of parameterizing labelled ideal tetrahedra with 2 or 4 distinct vertices via the injection (and surjective except for one point) ϕ : U DS → deﬁned by 0 1 a √b 1  1 0 √b a  ϕ(a,b) = . √ √ a + b + 1  a b 0 1    √b a 1 0      The space of non-degenerate labelled ideal tetrahedra contains six components that are ‘copies’ of the space of (non-labelled) ideal tetrahedra; these copies correspond to those permutations of the vertices that cannot be accomplished by applying an isometry (see Section ???). They are not easy to describe as subsets of U. In the next section, in order to solve Seidel’s ﬁrst conjecture, we will essentially express a and b in terms of the permanent and the determinant of the doubly stochastic matrix associated to a non-degenerate labelled ideal tetrahedron. Curiously, this will lead to a suitable reparametrization of U where the mentioned six components will be quite easy to spot.

3.1.2 Determinant and permanent as coordinates

In the Introduction, we discussed that rational representations of the symmetric group give rise to Schur functors, which are functors from the category of finite-dimensional linear spaces to itself. Moreover, given a finite-dimensional linear space V equipped with a symmetric bilinear form and a Schur functor F, there is a natural way to endow FV with a symmetric bilinear form. This construction leads to the concept of immanant of a matrix. The particular cases corresponding to the trivial and to the alternating representations are respectively the permanent and the determinant. In Seidel’s conjecture for hyperbolic 3-space, only these two cases are needed and we remind the definition of the first (for the general definition of immanant, see the Introduction).

Deﬁnition 3.1.11. Let A =(aij) be an n n matrix. The permanent of A is given by × perA = a a a , ∑ 1σ(1) 2σ(2) ··· nσ(n) σ Sn ∈ 3.1. Ideal and labelled ideal tetrahedra in H3 45

where Sn is the symmetric group.

The ﬁrst of Seidel’s conjectures, Speculation 1 (see the Introduction), says that the determinant and the permanent of the doubly stochastic matrix associated to a labelled ideal tetrahedron uniquely determines its volume. This conjecture is solved by showing that there exists exactly six labelled non-degenerate ideal tetrahedra (hence, with different doubly stochastic matrices) with the same permanent and determinant; however, these six labelled tetrahedra differ only by the order of their vertices. The fourth conjecture, Speculation 4 (see the Introduction), claims that the volume of an ideal tetrahedron is a monotonic function of the determinant (respectively, the permanent) of the doubly stochastic matrix associated to the tetrahedron. The volume of an ideal tetrahedron

(v1,v2,v3,v4) T can be expressed in terms of the entries of any matrix in M (v1,v2,v3,v4) ∈ corresponding to the tetrahedron. Indeed, it is easy to express the dihedral angles of the tetrahedron in terms of the coefﬁcients of the Gram matrix and it is well-known that the volume of an ideal tetrahedron is a function of its dihedral angles — see Section ???.

Hence, our ﬁrst step towards these Speculations will be trying to express a,b in terms of the determinant and the permanent of the corresponding doubly stochastic matrix (see Proposi- tion 3.1.8).

Denote by R R2 the set of ordered pairs of values of determinant and permanent of ⊂ doubly stochastic Gram matrices of labelled ideal tetrahedra, i.e., R := (detM,perM) M DS . (3.4) | ∈ In terms of (a,b) U and the function ϕ introduced at the end of the previous subsection, R ∈ can be described as R = detϕ(a,b),perϕ(a,b) (a,b) U detDSG(v,v,u,u),perDSG(v,v,u,u) , | ∈ where v,u are distinct points in SV (see Observation [ 3.1.9). We have 0 1 a √b 1  1 0 √b a  detϕ(a,b) = det 4 · √ a + √b + 1  a b 0 1    √b a 1 0    a4 + b2 + 1 2a2b 2a2 2b  = − − 4 − a + √b + 1 a + √b + 1 a + √b + 1 a √b + 1 a + √b 1 = − − − − 4 a + √b + 1 a + √b + 1 a √b + 1 a + √b 1 = − − − . (3.5) − 3 a + √b + 1 46 Chapter 3. Tetrahedra and 0 1 a √b 1  1 0 √b a  perϕ(a,b) = per 4 · √ a + √b + 1  a b 0 1    √b a 1 0    a4 + b2 + 1 + 2a2b +2a2 + 2b  = 4 a + √b + 1 2 a2 + b+ 1 = 4 . (3.6) a + √b + 1 Therefore, if we want to express a and b in terms of α := detϕ(a,b) and β := perϕ(a,b) for (α,β) R, we need to solve ∈ a + √b + 1 a √b + 1 a + √b 1 − − − = α  − 3  a + √b + 1  (3.7)  2 .  a2 + b + 1  4 = β a + √b + 1   It turns out that this is not a straightforward task. To accom plish this task, we will reparameterize  the space DS in a more symmetric fashion.

1 1 Proposition 3.1.12. Let V := , [1,∞).Themap f =( f1, f2) : V U deﬁned by − 2 2 × → 1 1 2 f (u,v) = v u , v + u − − 2 − 2 ! is bijective.

Proof. First, we show that f has the appropriate codomain, i.e., f (u,v) U = (a,b) (a ∈ | − 2 6 2 6 2 1) b (a + 1) . We have 2 2 2 2 1 3 f2(u,v) f1(u,v) 1 = v + u v u = 2(v 1)(2u + 1) > 0 − − − 2 − − − 2 − and 2 2 2 2 1 1 f1(u,v) + 1 f2(u,v) = v u + v + u = 2v( 2u + 1) > 0. − − 2 − − 2 − √b a a + √b + 1 Let (a,b) U. The ordered pair − , V . Indeed, a 1 6 √b 6 a + 1 ∈ 2 2 ! ∈ − 1 √b a 1 a + √b + 1 implies 6 − 6 and 1 a 6 √b implies 1 6 . The inverse of f is given −2 2 2 − 2 1 √b a a + √b + 1 by f − (a,b) = − , . 2 2 ! 3.1. Ideal and labelled ideal tetrahedra in H3 47

Observation 3.1.13. In terms of the coordinates (u,v) V of the above Proposition we have ∈ 1 1 0 1 v u v + u − − 2 − 2   1 1  1 0 v + u v u  1  − 2 − − 2 ϕ f (u,v) =  , ◦ 2v  1 1  v u v + u 0 1   − − 2 − 2       1 1  v + u v u 1 0   − 2 − − 2    2 (2u + 1)(2u 1)(v 1) 4u2 + 4v2 4v + 3 det ϕ f (u,v) = − − , per ϕ f (u,v) = − . 4v3 64v4 ◦ ◦ ◦ ◦ Fix a point (α,β) R. Now, our task became solving ∈ (2u + 1)(2u 1)(v 1) − − = α 4v3  2 .  4u2 + 4v2 4v + 3  − = β 64v4   2 (2u + 1)(2u 1)(v 1) 4u2 + 4v2 4v + 3 Deﬁne F : V R2,(u,v) − − , − . The Jacobian 4v3 64v4 → 7→ ! matrix of F at a point (u,v) V is ∈ 2u(v 1) 4u2 1 (2v 3) − − − v3 4v4  −  JF (u,v) = . u 4u2 + 4v2 4v + 3 4u2 + 4v2 4v + 3 4u2 2v + 3     4− − 5 −   4v − 16v    Its determinant equals

u(2v 3 + 2u)(2v 3 2u) 4u2 +(2v 1)2 + 2 detJ (u,v) = − − − − F 16v8 3 3 and vanishes at the points (u,v) V such that u = 0 or v = u or v = + u. So F has a local ∈ 2 − 2 inverse at all points in V except at those lying on the three concurrent lines in Figure 4. The central vertical line u = 0 is a visible line of symmetry in V ; this means that the points (u,v) and ( u,v) correspond to different doubly stochastic matrices with the same deter- − minant and permanent (in fact, they correspond to distinct ideal tetrahedra which are isometric as non-labelled ideal tetrahedron). We can already see in Figure 4 the above mentioned six ‘copies’ of the space of (non-labelled) nondegenerate ideal tetrahedra, but the corresponding 3 3 symmetries around the lines v = u and v = + u (this is line L in Figure 4) are not as 2 − 2 simple to express as the one around u = 0. Let us show how to ﬁnd a more ‘homogeneous’ classifying space. 48 Chapter 3. Tetrahedra

V ( u,v) (u,v) − b b 1 ,2 L 2

1 1 ,1 ,1 − 2 2

1 1 − 2 2 Figure 4 – The open set V, domain of the parameters.

Source: Elaborated by the author.

Let W R2 be the equilateral triangle (without the vertex on the top) in Figure 5 with ⊂ 1 1 basis the segment joining ,1 and ,1 . We want to ﬁnd a bijective map from W to −2 −2 V . For this purpose, from some point on the vertical axis (in Figure 5, this is the point below √3 the triangle basis) we will project segments joining 0,1 + and (s,1) onto the vertical 2 ! 1 1 line s [1,∞), where s , . The blue dashed line in Figure 5 shows the idea of this { } × ∈ −2 2 projection. We would like the altitudes of the triangle to become the three concurrent lines in 1 √3 1 V ; in particular, the point ,1 + should map to ,2 . Thus, the point from which we 4 4 2 ! 1 √3 must project is the intersection of the vertical axis with the straight line that joins ,1 + 4 4 ! 1 √3 and ,2 , i.e., the point 0, . 2 2 ! This construction is summarized in the following proposition.

Proposition 3.1.14. Let W be the triangle (without the vertex on the top) in Figure 5 and let V √3s √3 be as in Proposition 3.1.12. The function g : W V,(s,t) , , is → 7→ 2 + √3 2t 2 + √3 2t ! a bijection. − − 3.1. Ideal and labelled ideal tetrahedra in H3 49

1 ,2 L 2 √3 0,1 + 2

b 1 √3 4 ,1 + 4

1 ,1 1 ,1 − 2 (s,1) 2 √3 b 0, 2 Figure 5 – Bijection between W and V.

Source: Elaborated by the author.

Finally, we make the translation W ∆ := W 0,1 + √3 so that the orthocenter → ′ − 6 √3 0,1 + 6 of the triangle gets centred at the origin. The region ∆′ is the classifying space of labelled ideal tetrahedra (except for one point ideal tetrahedron) that we are going to use in the proofs of Speculations 1 and 4.

Proposition 3.1.15. Let

√3 √3 √3 ∆ := (c,d) R2 d > , d 6 √3c + , and d 6 √3c + ( ∈ | − 6 − 3 3 ) stand for the triangle in Figure 6. The function φ : ∆ DS deﬁned by → 0 2 2√3d 2 3c + √3d 2 + 3c + √3d − − 1  2 2√3d 0 2 + 3c + √3d 2 3c + √3d φ(c,d) = − − , 6 √ √ √ 2 3c + 3d 2 + 3c + 3d 0 2 2 3d   − −  2 + 3c + √3d 2 3c + √3d 2 2√3d 0   − −    is a bijection. The determinant and permanent of the matrix φ(c,d) are 2 (2√3d + 1)(3c + √3d 1)(3c √3d + 1) 3c2 + 3d2 + 2 detφ(c,d) = − − , perφ(c,d) = . 27 6 √3 Proof. First note that φ 0, = DSG(v,v,u,u). The function φ is a bijection by Proposition 3 ! 3.1.14. The rest follows directly from Observation 3.1.13. 50 Chapter 3. Tetrahedra

The non-null entries of the matrix φ(c,d) are expressions that will appear repeatedly in the future and we name them r,s,t : ∆ R, where → 2 2√3d 2 3c + √3d 2 + 3c + √3d r(c,d) := − , s(c,d) := − , t(c,d) := . (3.8) 6 6 6 We have detφ = (r + s +t)( r + s +t)(r s +t)(r + s t) and per φ =(r2 + s2 +t2)2. − − − − We found it a beautiful fact (see the next proposition) that the above three positive numbers constitute the sides of an Euclidean triangle. Another form of this fact can also be found in Seidel’s work, at the end of Section 2 in [10]. It has the following interpretation. Opposite dihedral angles of a non-degenerate ideal tetrahedron are equal; hence, such a tetrahedron has at most three distinct dihedral angles, θ1, θ2, and θ3 and these dihedral angles are exactly the internal angles of the Euclidean triangle with sides r(c,d), s(c,d), and t(c,d). In particular,

θ1 +θ2 +θ3 = π (a fact discovered by Milnor, see [8], Lemma 2). All these facts will be proved in Section ???.

Proposition 3.1.16. Let (c,d) ∆. The positive numbers r(c,d),s(c,d), and t(c,d) constitute ∈ the sides of a (non-degenerate) Euclidean triangle.

Proof. We denote r := r(c,d), s := s(c,d), and t := t(c,d). The matrix φ(c,d) is the Gram matrix of four linearly independent vectors (a basis for V ). The signature of the hermitian form on V is (3,0,1). By Proposition A.1.19,det φ(c,d) hasthe same signas det diag( 1, 1, 1,1) = − − − 1. Thus − det φ(c,d) < 0, i.e., (r + s +t)( r + s +t)(r s +t)(r + s t) < 0, − − − − that is, ( r + s +t)(r s +t)(r + s t) > 0. − − − There is an even number of negative factors on the left side of the last inequation. If we assume that there are two negative factors (say, the ﬁrst two ones — the other cases are similar) then r + s + t < 0 and r s + t < 0. But this implies t < 0 which is a contradiction. We conclude − − that there are no negative factors in the above expression. So, the triangle inequalities r < s +t, s < r +t, and t < r + s are satisﬁed.

3.1.3 The classifying space ∆

The region ∆ described in Proposition 3.1.15 is the classifying space of labelled non- degenerate ideal tetrahedra that we are going to use in order to prove Speculations 1 and 4. It is divided by its altitudes into six congruent triangles. Each such triangle is, as we show below, the space of non-labelled non-degenerate ideal tetrahedra. The reﬂections in the altitudes of ∆ correspond to the relabellings that cannot be achieved by means of isometries. In this way, 3.1. Ideal and labelled ideal tetrahedra in H3 51 a generic non-labelled tetrahedron (three distinct dihedral angles) has six copies in ∆ while more symmetric tetrahedra (exactly two equal dihedral angles) have only three copies; ﬁnally, the regular tetrahedron has a single copy (since, in this case, there are enough symmetries to permute the vertices in every possible way). The generic case is illustrated in Figure 6.

∆

∆6 ∆1

b bb

∆5 b b ∆2

b b

∆4 ∆3

Figure 6 – The open set ∆, domain of the parameters (c,d).

Source: Elaborated by the author.

We call ∆i, i = 1,...,6, the six congruent triangles into which the equilateral triangle ∆ is divided by its altitudes. We recall that ∆ is an open region and does not contain its boundary

(which correspond to some degenerate tetrahedra). Therefore, each ∆i does not contain one of its legs (the longest). Explicitly,

2 √3 √3 ∆1 := (c,d) R c 0, d c, and d < √3c + . (3.9) ( ∈ ≥ ≥ 3 − 3 )

For future reference, we display in the table below the reﬂections in altitudes of ∆ of a point

(c,d) ∆1: ∈

Proposition 3.1.17. Let (c,d) ∆1. Then ∈ (i) r(c,d) 6 s(c,d) 6 t(c,d)

(ii) r(c,d) = s(c,d) if, and only if, (c,d) is on the smallest leg of ∆1 (iii) s(c,d) = t(c,d) if, and only if, (c,d) is on the hypotenuse of ∆1

Proof. We have

2 3c + √3d 2 2√3d c + √3d s(c,d) r(c,d) = − − = − > 0 − 6 − 6 6 52 Chapter 3. Tetrahedra

Triangle Point

∆1 (c,d)

c+√3d √3c d ∆2 2 , 2− c+√3d √3c d ∆3 − 2 , − 2 − c √3d √3c d ∆4 −2 , − 2 − c √3d √3c d ∆5 − −2 , 2− ∆6 ( c,d) −

Table 1 – Points obtained by reﬂecting (c,d) ∆1. ∈

with the equality holding if, and only if, (c,d) is on the smallest leg of ∆1. Analogously,

2 + 3c + √3d 2 3c + √3d t(c,d) s(c,d) = − = c > 0, − 6 − 6 with the equality holding if, and only if, (c,d) is on the hypotenuse of ∆1.

Let us see which permutations of vertices can always be performed through isometries.

Theorem 3.1.18. Let (v1,v2,v3,v4) T be a non-degenerate labelled ideal tetrahedron. Let ∈ S4 denote the 4-symmetric group and let H := 1,(12)(34),(13)(24),(14)(23) ⊳ S4 be the Klein four group. The tetrahedra v ,v ,v ,v and v ,v ,v ,v σ1(1) σ1(2) σ1(3) σ1(4) σ2(1) σ2(2) σ2(3) σ2(4) 1 are isometric if (and only if, in the generic case) σ1σ − H. Equivalently, at the level of Gram 2 ∈ matrices, the quotient group S3 = S4/H acts on the space DS by permuting the entries r, s, and t. At the level of the classifying space ∆, this action of S3 is nothing but reﬂecting on the altitudes.

Proof. Let (v1,v2,v3,v4) T be a non-degenerate labelled ideal tetrahedron, let σ S4 be an ∈ ∈ element of the 4-symmetric group, and let G(v1,v2,v3,v4) be the doubly stochastic Gram matrix of representatives of the vertices of the tetrahedron, i.e., DSG(v1,v2,v3,v4) = G(v1,v2,v3,v4).

Since the doubly stochastic Gram matrix corresponding to vσ(1) ,vσ(2),vσ(3),vσ(4) is unique, we have DSG vσ(1) ,vσ(2),vσ(3),vσ(4) = G vσ(1),vσ(2),vσ(3),vσ(4) . By Proposition 3.1.15, 0 r s t r 0 t s DSG v1 ,v2,v3,v4 = s t 0 r   t s r 0     3.1. Ideal and labelled ideal tetrahedra in H3 53 and 0 r′ s′ t′ r′ 0 t′ s′ DSG vσ(1) ,vσ(2),vσ(3),vσ(4) = , s′ t′ 0 r′   t′ s′ r′ 0     for some r,s,t,r ,s ,t R>0. We have r,s,t = r ,s ,t , therefore σ simply permutes the ′ ′ ′ ∈ { } { ′ ′ ′} entries r,s,t of DSG(v1,v2,v3,v4).

A direct calculation shows that DSG vσ(1),vσ(2),vσ(3),vσ(4) = DSG v1 ,v2,v3,v4 for σ H = Id,(12)(34),(13)(24),(14)(23) . In the generic case, r,s,t are pairwise distinct and, ∈ therefore, DSG v ,v ,v ,v = DSG v1,v2,v3,v4 only if σ H. σ(1) σ(2) σ(3) σ(4) ∈ It remains to show that S4/H acts on ∆ by reﬂections on the altitudes. Let (c,d) ∆1. ∈ The matrices corresponding to the points in the table above are (the proof is a tedious calculation that can be seen below):

0 r s t r 0 t s φ(c,d) = = DSG v ,v ,v ,v ,σ H σ(1) σ(2) σ(3) σ(4) ∈ s t 0 r   t s r 0   0 s r t  c + √3d √3c d s 0 t r φ , − = = DSG v ,v ,v ,v ,σ (23)H 2 2 σ(1) σ(2) σ(3) σ(4) ∈ ! r t 0 s   t r s 0   0 t r s c + √3d √3c d t 0 s r φ − , − − = = DSG v ,v ,v ,v ,σ (123)H 2 2 σ(1) σ(2) σ(3) σ(4) ∈ ! r s 0 t    s r t 0   0 t s r c √3d √3c d t 0 r s φ − , − − = = DSG v ,v ,v ,v ,σ (13)H 2 2 σ(1) σ(2) σ(3) σ(4) ∈ ! s r 0 t    r s t 0   0 s t r c √3d √3c d s 0 r t  φ − − , − = = DSG v ,v ,v ,v ,σ (234)H, 2 2 σ(1) σ(2) σ(3) σ(4) ∈ ! t r 0 s   r t s 0   0 r t s r 0 s t  φ ( c,d) = = DSG v ,v ,v ,v ,σ (12)H. − σ(1) σ(2) σ(3) σ(4) ∈ t s 0 r   s t r 0     54 Chapter 3. Tetrahedra

The ﬁrst equality follows from 3.8. The entries (multiplied by 6) of the matrix corre- c+√3d √3c d sponding to the point , − ∆2 are the following: 2 2 ∈ c + √3d √3c d √3c d 6r , − = 2 2√3 − = 2 3c + √3d = 6s, 2 2 ! − 2 − c + √3d √3c d c + √3d √3c d 6s , − = 2 3 + √3 − = 2 2√3d = 6r, 2 2 ! − 2 2 − c + √3d √3c d c + √3d √3c d 6t , − = 2 + 3 + √3 − = 2 + 3c + √3d = 6t. 2 2 ! 2 2

c+√3d √3c d For the point − , − − ∆3, 2 2 ∈ c + √3d √3c d √3c d 6r − , − − = 2 2√3− − = 2 + 3c + √3d = 6t, 2 2 ! − 2 c + √3d √3c d c + √3d √3c d 6s − , − − = 2 3− + √3− − = 2 2√3d = 6r, 2 2 ! − 2 2 − c + √3d √3c d c + √3d √3c d 6t − , − − = 2 + 3− + √3− − = 2 3c + √3d = 6s. 2 2 ! 2 2 −

c √3d √3c d For the point − , − − ∆4, 2 2 ∈ c √3d √3c d √3c d 6r − , − − = 2 2√3− − = 2 + 3c + √3d = 6t, 2 2 ! − 2 c √3d √3c d c √3d √3c d 6s − , − − = 2 3 − + √3− − = 2 3c + √3d = 6s, 2 2 ! − 2 2 − c √3d √3c d c √3d √3c d 6t − , − − = 2 + 3 − + √3− − = 2 2√3d = 6r. 2 2 ! 2 2 −

c √3d √3c d For the point − − , − ∆5, 2 2 ∈ c √3d √3c d √3c d 6r − − , − = 2 2√3 − = 2 3c + √3d = 6s, 2 2 ! − 2 − c √3d √3c d c √3d √3c d 6s − − , − = 2 3− − + √3 − = 2 + 3c + √3d = 6t, 2 2 ! − 2 2 c √3d √3c d c √3d √3c d 6t − − , − = 2 + 3− − + √3 − = 2 2√3d = 6r. 2 2 ! 2 2 −

Finally, for the point ( c,d) ∆6, − ∈ 6r ( c,d) = 2 2√3d = 6r, − − 6s( c,d) = 2 3( c) + √3d = 6t, − − − 6t ( c,d) = 2 + 3( c) + √3d = 6s. − − 3.2. Proof of Speculation 1 55

We arrived at the space of non-degenerate ideal tetrahedra modulo isometries:

Corollary 3.1.19. Each ∆i,i = 1,...,6, is a copy of the space of nondegenerate non-labelled ideal tetrahedra.

Corollary 3.1.20. The doubly stochastic Gram matrices corresponding to the labelled ideal tetrahedra that differ by the action of S3 in the previous theorem have the same determinant and permanent.

Proof. The expressions for the determinant and permanent (see the formulae right after Propo- sition 3.1.15 are symmetric in r,s,t.

We will prove in Corollary 3.2.3 that the function G : ∆ R deﬁned by → G(c,d) = detφ(c,d),perφ(c,d) (3.10) is injective when restricted to ∆1. For now, let us show that G is locally invertible outside the altitudes. Indeed, the Jacobian matrix of G at a point (c,d) is

1 2c 2√3d + 1 2 √3c2 √3d2 + d J (c,d) = − G 3 2 2 2 2 c 3c + 3d + 2 d 3c + 3d + 2 ! and 2√3c √3d + c √3d c 3c2 + 3d2 + 2 detJ (c,d) = − . G 9 Thus, detJG = 0 if, and only if, (c,d) ∆ does not belong to an altitude. 6 ∈

3.2 Proof of Speculation 1

In Theorem 3.1.18 we saw that the points in ∆ that differ by reﬂections on the altitudes of ∆ (these points are listed in Table 1) correspond to doubly stochastic matrices in DS with the same determinant and permanent (see Corollary 3.1.20). This means that, for a given (α,β) R ∈ (see the deﬁnition of R in 3.4), the points in Table 1 are solutions of

(2√3d + 1)(3c + √3d 1)(3c √3d + 1) − − = α 27  2 . (3.11)  3c2 + 3d2 + 2  = β 6  Speculation 1 claims that the volume of an ideal tetrahedron is determined by the determinant and permanent of its doubly stochastic Gram matrix. In fact, a stronger result holds: the determinant and permanent in question completely determine the non-degenerate ideal tetrahedron 56 Chapter 3. Tetrahedra up to isometry (see Theorem 3.2.5). This is a consequence of Corollary 3.1.19 and of the fact, proved below, that the points in Table 1 are all the solutions of 3.11. We deﬁne S := α, β (α,β) R . (3.12) ∈ n p o Lemma 3.2.1. The solutions of 3.11 are the same as the solutions of

(2√3d + 1)(3c + √3d 1)(3c √3d + 1) − − = α 27  , (3.13)  3c2 + 3d2 + 2  = ω 6  where ω = β. There are at most six different solutions (c,d) of the above system of equations. p Proof. It is immediate that the system above is equivalent to 3.11. The ﬁrst equation implies

2 2√3d + 1 9c2 √3d 1 = 27α. − − 1 Using the second equation, we may replace c2 by (6ω 3d2 2) in the previous expression. 3 − − We obtain a cubic in d which has at most three different solutions. By the second equation, each such solution gives rise to at most two different solutions of the form (c,d).

Let ∆ denote the closure of ∆ in R2. We consider the function H : ∆ R2 deﬁned by → 2 (2√3d + 1)(3c + √3d 1)(3c √3d + 1) 3c2 + 3d2 + 2 H(c,d) = − − , . 27 6 !

Proposition 3.2.2. The function H : ∆i H ∆i is a homeomorphism. |∆i → Proof. Note that the formulas in Table 1 are also valid for the points in ∆. Let σ be an element of the reﬂection group S3, represented by one of the formulas in Table 1. We claim that H(c,d) =

H σ(c,d) for all (c,d) ∆1. Indeed, we can write (c,d) = limn ∞ (cn,dn), where (cn,dn) ∆1 ∈ → ∈ for each n. As σ is a continuous function σ(c,d) = limn ∞ σ (cn,dn), where σ(c,d) ∆i and → ∈ (cn,dn) ∆i for some i = 1,...,6. Then ∈

H(c,d) = lim H (cn,dn), and H σ(c,d) = lim H σ (cn,dn) . n ∞ n ∞ → → Since H coincides with G in ∆ we have H(cn,dn) = G(cn,dn) and H σ(cn,dn) = G σ(cn,dn) . By Corollary 3.1.20 G(cn,dn) = G σ(cn,dn) , so H(c,d) = lim G(cn,dn) = lim G σ (cn,dn) = H σ(c,d) . n ∞ n ∞ → → 3.2. Proof of Speculation 1 57

Let (c,d) ∆1, the points in Table 1 are solutions of the system 3.11 with (α,β) = ∈ H(c,d), and then they are solutions of the system 3.13 where ω = β

It sufﬁces to show that H : ∆1 H ∆1 is a homeomorphism.p In order to show that |∆1 → H is injective, suppose that H(c,d) = H(c′,d′) where (c,d),(c′,d′) ∆1. In particular, (c,d) |∆1 ∈ and (c′,d′) are solutions of 3.13. If some of the points, say (c,d), is in the interior of the angle formed by the hypotenuse and the shortest leg, we have (c,d)=(c′,d′) because otherwise the system of equations 3.13 would have more than six solutions, contradicting Lemma 3.2.1. We can therefore assume that (c,d) belongs to either the shortest leg of ∆1 or to its hypotenuse. Note that, according to the second equation of 3.13, the solutions of the system lie in a circumference centered at the origin. So, if (c,d)=(0,0), then (c ,d )=(0,0). If (c,d) =(0,0) ′ ′ 6 is, say, on the hypotenuse(we can arrive to this case if it lies on the shortest leg) of ∆1, then c = 0.

Assume that (c,d) =(c ,d ). Since (c ,d ) cannot be in the interior of ∆1, it has to be a point 6 ′ ′ ′ ′ p on the intersection of the shortest leg of ∆1 with the circumference of radius d. Reﬂecting p on the appropriate altitude of ∆ we arrive at the point (0, d); therefore G(0,d) = G(0, d). In − − particular, the determinants of the doubly stochastic matrices in question are equal, that is,

2√3d + 1 √3d 1 √3d + 1 = 2√3( d) + 1 √3( d) 1 √3( d) + 1 , − − − − − − i.e., 6√3d3 + 9d2 1 = 6√3d3 + 9d2 1. − − − leading to d = 0, which is a contradiction.

Corollary 3.2.3. The restriction G : ∆1 R is a bijection, where G and ∆1 are respectively |∆1 → the function and the region deﬁned in 3.10 and 3.9.

Observation 3.2.4. Given (α,β) R, there exists a unique (c,d) ∆1 such that G(c,d)=(α,β). ∈ ∈ The other solutions are obtained by reﬂecting (c,d) on the altitudes of ∆. Thus, if (c,d) is in the interior of ∆1, there are six different solutions in ∆; if (c,d) =(0,0) is on the hypotenuse or on 6 the shortest leg of ∆1, we have exactly three different solutions in ∆; if (c,d)=(0,0), there is a unique solution.

We arrive at

Theorem 3.2.5 (Seidel, [10], Speculation 1). For each (α,β) R there exists, up to isome- ∈ try, a unique non-degenerate ideal tetrahedron T T such that α = detDSG(T ) and β = ∈ perDSG(T). In particular, the volume of a non-degenerate ideal tetrahedron is completely determined by the determinant and permanent of its doubly stochastic matrix.

Proof. By Observation 3.2.4, there exists a unique (c,d) ∆1 such that the determinant and per- ∈ manent of the matrix φ(c,d) are respectively α and β. The conclusion follows from Corollary 3.1.19. 58 Chapter 3. Tetrahedra

In Proposition 3.4.3 we will give an explicit expression for the volume of an ideal tetrahedron in terms of the determinant and permanent of the doubly stochastic Gram matrix of its ideal vertices.

3.3 Solving system 3.13. A few derivatives.

Let us ﬁnd an explicit expression for (c,d) ∆ in terms of given permanent and deter- ∈ minant of a doubly stochastic matrix of a labelled (non-degenerate) ideal tetrahedron. This is the expression that is going to be used in order to express the volume of the tetrahedron as a function of determinant and permanent. We remind that the region S alluded to in this section is the one deﬁned in 3.12 and described in Appendix B.

By Observation 3.2.4, for a given (α,ω) S, there exists a unique (c,d) ∆1 satisfying ∈ ∈ the system of equations 3.13:

(2√3d + 1)(3c + √3d 1)(3c √3d + 1) − − = α 27   3c2 + 3d2 + 2  = ω 6   We will write c and d in terms of α and ω such that (c,d) ∆1. As we have seen, the ∈ other solutions can be obtained from this one.

Observation 3.3.1. Let (α,ω) S. Let (c,d) be the unique solution of 3.13 that belongs to ∆1. 1 1 ∈ Then (α,ω) = , if, and only if, (c,d)=(0,0). −27 3 b2 a3 Lemma 3.3.2. Let a,b R be such that a = 0 and + 6 0. The cubic equation x3 +ax+b = ∈ 6 4 27 0 has the three real roots

a 1 3√3b 2kπ xk = 2 − cos arccos − + , k = 0,1,2. 3 3 2√ a3 3 ! r − 1 1 Proposition 3.3.3. Let (α,ω) be a ﬁxed point in the interior of S , . The solution \ −27 3 of 3.13 lying in ∆1 is given by

2 1 27α + 18ω 7 c = √3ω 1sin arccos − −3 (3.14) r3 − 3 4√2(3ω 1) 2 ! − and

2 1 27α + 18ω 7 d = √3ω 1cos arccos − −3 . (3.15) r3 − 3 4√2(3ω 1) 2 ! − 3.3. Solving system 3.13. A few derivatives. 59

Proof. From the second equation of the system we have

2 c2 = 2ω d2 . − − 3

As we are looking for (c,d) ∆1, we choose the positive root ∈ 2 c = 2ω d2 . (3.16) r − − 3 From the ﬁrst equation follows that

2 27α = 2√3d + 1 9c2 √3d 1 . − − Equation 3.16 implies that

27α = 2√3d + 1 18ω 9d2 6 3d2 + 2√3d 1 − − − − i.e.,

0 = 24√3d3 +(1 3ω)12√3d +(27α 18ω + 7). − − We therefore get the cubic equation

0 = x3 + ax + b where x := 2√3d and

a := 6(1 3ω), (3.17) − b := 27α 18ω + 7. (3.18) − b2 a3 In order to solve the cubic, we ﬁrst have to determine the sign of + . Since 4 27

3c2 + 3d2 + 2 a = 6 1 3 = 9 c2 + d2 < 0 (3.19) − 6 − and 2√3d + 1 3c + √3d 1 3c √3d + 1 3c2 + 3d2 + 2 b = 27 − − 18 + 7 = 27 6 −

= 6√3d 3c2 d2 , (3.20) − we get

2 3 b a 2 + = 27c2 3d2 c2 6 0. (3.21) 4 27 − − 60 Chapter 3. Tetrahedra

Hence, by Lemma 3.3.2, the solutions of the cubic equation are a 1 3√3b 2kπ xk = 2 cos arccos − + , k = 0,1,2. −3 3 2√ a3 3 ! r − Deﬁning d := 1 x , we have k 2√3 k √ a 1 3√3b 2kπ dk = − cos arccos − + , k = 0,1,2. 3 3 2√ a3 3 ! − From Equation 3.16, the c corresponding to each dk is, for each k = 0,1,2,

2 2 ck = 2ω dk r − − 3 a c = − d2 k 9 k r − √ a 1 3√3b 2kπ c = − 1 cos2 arccos − + k 3 v 3 3 3 u − 2√ a ! u − √ at 1 3√3b 2kπ ck = − sin arccos − + . 3 3 2√ a3 3 ! −

Note that requiring (c,d) to be in the interior of ∆1 implies that the point (d,c), written in polar coordinates, has angular coordinate in the interval (0,π/3). As the angular coordinates of the points (d1,c1) and (d2,c2) are not in (0,π/3), the solution we are looking for is the one corresponding to k = 0.

From now on, we will see c and d as functions c,d : S R such that c(α,ω),d(α,ω) → is the solution of 3.13 in ∆1, i.e., 1 1 0 , if (α,ω) = , −27 3 c(α,ω) =   2 1 27α + 18ω 7 1 1  √3ω 1sin arccos − − , if (α,ω) = ,  3 3 3 27 3 r − 4√2(3ω 1) 2 ! 6 − −  (3.22)  and 1 1 0 , if (α,ω) = , −27 3 d(α,ω) =  .  2 1 27α + 18ω 7 1 1  √3ω 1cos arccos − − , if (α,ω) = ,  3 3 3 27 3 r − 4√2(3ω 1) 2 ! 6 − −  (3.23)  Note that both c and d are continuous in S and smooth in intS. We frequently denote c := c(α,ω) and d := d(α,ω). We will now ﬁnd the partial derivatives of c and d with respect to α and ω. The derived expressions will be used in the proof of Speculation 4. 3.3. Solving system 3.13. A few derivatives. 61

Proposition 3.3.4. Let (α,ω) intS. Then ∈ ∂c d (α,ω) = , (3.24) ∂α 6c(3d2 c2) − ∂d 1 (α,ω) = . (3.25) ∂α − 6(3d2 c2) − Proof. The partial derivative of c with respect to α is

∂c √2√3ω 1 1 27α + 18ω 7 (α,ω) = − cos arccos − −3 , ∂α k 3 4√2(3ω 1) 2 ! − where 2 3 2 k := 4√2(3ω 1) 2 ( 27α + 18ω 7) . − − − − r √ 1 27α+18ω 7 √ From Equation 3.15, 2√3ω 1cos 3 arccos − −3 = 3d(α,ω). So, − 4√2(3ω 1) 2 − ∂c √3 (α,ω) = d(α,ω). (3.26) ∂α k Similarly, the partial derivative of d with respect to α is

∂d √2√3ω 1 1 27α + 18ω 7 (α.ω) = − sin arccos − −3 , ∂α − k 3 4√2(3ω 1) 2 ! − and, from Equation 3.14, we obtain ∂d √3 (α,ω) = c(α,ω). (3.27) ∂α − k We can write k in terms of c(α,ω) and d(α,ω) as follows. First, it is possible to write k in terms of a and b because equations 3.54 and 3.18 imply that 3ω 1 = a and 27α + − − 6 − 18ω 7 = b. Therefore, − − k = 32(3ω 1)3 ( 27α + 18ω 7)2, − − − − q a 3 k = 32 ( b)2 −6 − − r b2 a3 k = 2 + . − 4 27 s From Equation 3.21,

k = 2 27c2 (3d2 c2)2. − q 2 2 Since (c,d) ∆1, we have 3d c . So ∈ ≥ k = 6√3c 3d2 c2 . (3.28) − Finally, replacing k in equations 3.26and 3.27, we arrive at the desired formulae. 62 Chapter 3. Tetrahedra

Proposition 3.3.5. Let (α,ω) intS. Then ∈ ∂c 3d2 3c2 √3d (α,ω) = − − (3.29) ∂ω 3c(3d2 c2) − ∂d 6d + √3 (α,ω) = . (3.30) ∂ω 3(3d2 c2) − Proof. The partial derivative of c respect to ω is

∂c √3 1 27α + 18ω 7 (α,ω) = sin arccos − −3 ∂ω √2√3ω 1 3 4√2(3ω 1) 2 !− − − 3√3(9α 2ω + 1) 1 27α + 18ω 7 − cos arccos − −3 . − k√2√3ω 1 3 4√2(3ω 1) 2 ! − − As in the proof of the previous proposition, we write

∂c 3 9(9α 2ω + 1) (α,ω) = c(α,ω) − d(α,ω) = ∂ω 2(3ω 1) − 2(3ω 1)k − − 3 = (kc(α,ω) 3(9α 2ω + 1)d(α,ω)). (3.31) 2(3ω 1)k − − − Since c(α,ω),d(α,ω) is a solution of 3.13, we obtain (2√3d + 1)(3c + √3d 1)(3c √3d + 1) 3c2 + 3d2 + 2 9α 2ω + 1 = 9 − − 2 + 1 − 27 − 6 = 2 3√3c2d + c2 √3d3 + d2 . (3.32) − From 3.32 and 3.28, we get

∂c 3 (α,ω) = 6√3c2 3d2 c2 6d 3√3c2d + c2 √3d3 + d2 ∂ω 2(3ω 1)6√3c(3d2 c2) − − − − − √3 = c2 + d2 √3d2 √3c2 d . 2(3ω 1)c(3d2 c2) − − − − We arrive at ∂c 3d2 3c2 √3d (α,ω) = − − ∂ω 3c(3d2 c2) − because 2(3ω 1) = 3 c2 + d2 . − The partial derivative ofd with respect to ω is

∂d √3 1 27α + 18ω 7 (α,ω) = cos arccos − −3 ∂ω √2√3ω 1 3 4√2(3ω 1) 2 ! − − 3√3(9α 2ω + 1) 1 27α + 18ω 7 + − sin arccos − −3 . k√2√3ω 1 3 4√2(3ω 1) 2 ! − − 3.4. Volume formulae 63

Then, ∂d 3 9(9α 2ω + 1) (α,ω) = d(α,ω) + − c(α,ω) ∂ω 2(3ω 1) 2(3ω 1)k − − 3 = (kd(α,ω) + 3(9α 2ω + 1)c(α,ω)). 2(3ω 1)k − − From 3.32 and 3.28, we obtain ∂d 3 (α,ω) = 6√3cd 3d2 c2 + 6c 3√3c2d + c2 √3d3 + d2 ∂ω 2(3ω 1)6√3c(3d2 c2) − − − − 3 ∂d 6d + √3 = c c2 + d2 2√3d + 1 = (α,ω) = . 2(3ω 1)√3c(3d2 c2) ∂ω 3(3d2 c2) − − −

3.4 Volume formulae

There are several known formulae for the volume of an ideal tetrahedron in hyperbolic 3-space. Perhaps the simplest one was obtained by Milnor. It expresses the volume as a function of the dihedral angles of the tetrahedron. In order to present Milnor’s form, we introduce the Lobachevsky function:

Deﬁnition 3.4.1. The Lobachevsky function l : R R is given by → x l(x) := log 2sint dt. − 0 | | Z Clearly, l is differentiable at each x = kπ, k Z, and 6 ∈ dl (x) = log 2sinx . dx − | | l If 0 < x < π, then d (x) = log(2sinx). dx − Theorem 3.4.2 (Milnor, [8] Lemma 4). The volume of an ideal non-degenerate tetrahedron T with dihedral angles θ1,θ2,θ3 is given by

vol(T ) = l(θ1) + l(θ2) + l(θ3), where l is the Lobachevsky function.

It followsfrom section ??? that the dihedral angles of the labelled tetrahedron (v1,v2,v3,v4) in terms of the entries r,s,t of its doubly stochastic matrix are given by r2 + s2 +t2 r2 s2 +t2 r2 + s2 t2 a = arccos − , a = arccos − , a = arccos − . 1 2st 2 2rt 3 2rs

Indeed, let us calculate the dihedral angle between the faces generated be v1,v2,v3 and v1,v2,v4. 64 Chapter 3. Tetrahedra

Note that the dihedral angles of the polyhedron are, by the law of cosines, exactly the internal angles of the Euclidean triangle with sides of lengths r,s,t. Summarizing, we have the following

Proposition 3.4.3. The volume function vol : S R is given by →

vol(α,ω) = l(a1) + l(a2) + l(a3), (3.33) where

r2 + s2 +t2 r2 s2 +t2 r2 + s2 t2 a = arccos − , a = arccos − , a = arccos − , 1 2st 2 2rt 3 2rs

2 2√3d 2 3c + √3d 2 + 3c + √3d r = − , s = − , t = , 6 6 6 0 , if (α,ω) = 1 , 1 − 27 3 c = 2 1 27α+18ω 7 1 1 ,  3 √3ω 1sin 3 arccos − −3 , if (α,ω) = 27 , 3  − 4√2(3ω 1) 2 6 − − q  0 , if (α,ω) = 1 , 1 − 27 3 d = 2 1 27α+18ω 7 1 1 .  3 √3ω 1cos 3 arccos − −3 , if (α,ω) = 27 , 3  − 4√2(3ω 1) 2 6 − − q  3.5 Proof of a particular case of Speculation 3

Recall the deﬁnition of R from 3.4. Given (α,β) R, we have ∈ 2 α = (r + s +t)( r + s +t)(r s +t)(r + s t) and β = r2 + s2 +t2 , − − − − for some r,s,t 0 such that r + s +t = 1. ≥ Note that β is the size (to the forth grade) of the vector (r,s,t) R3. Then the permanent 1 ∈ has a unique minimum when r = s = t = . 3 From Proposition 3.1.16 we also have that r,s,t are the sides of an euclidean triangle. Let A denote the area of this triangle. By the Heron’s formula α = 16A2. Thus, the determinant 1 − function has a unique minimum when r = s = t = . 3

3.6 Proof of Speculation 4

Let O be the open set O := (r,s,t) R3 r < s +t, s < r +t, t < r + s . Consider the ∈ | functions (c,d) : S ∆1 ∆ and (r,s,t) : ∆ O as they are in Deﬁnition 3.4.3, i.e., as in → ⊂ → equations 3.14, 3.15. Consider the function f : O R deﬁned by →

f (r,s,t) = l(a1) + l(a2) + l(a3), (3.34) 3.6. Proof of Speculation 4 65 where r2 + s2 +t2 r2 s2 +t2 r2 + s2 t2 a = arccos − , a = arccos − , a = arccos − . 1 2st 2 2rt 3 2rs The function equality vol = f (r,s,t) (c,d) is satisﬁed. ◦ ◦ We can calculate the partial derivatives of the volume functions at all the points of intS. The Chain rule ensures that ∂vol ∂ ( f (r,s,t)) ∂c ∂ ( f (r,s,t)) ∂d = ◦ + ◦ ∂α ∂c ∂α ∂d ∂α ∂vol ∂ f ∂r ∂ f ∂s ∂ f ∂t ∂c ∂ f ∂r ∂ f ∂s ∂ f ∂t ∂d = + + + + + . (3.35) ∂α ∂r ∂c ∂s ∂c ∂t ∂c ∂α ∂r ∂d ∂s ∂d ∂t ∂d ∂α In the same way

∂vol ∂ f ∂r ∂ f ∂s ∂ f ∂t ∂c ∂ f ∂r ∂ f ∂s ∂ f ∂t ∂d = + + + + + . (3.36) ∂ω ∂r ∂c ∂s ∂c ∂t ∂c ∂ω ∂r ∂d ∂s ∂d ∂t ∂d ∂ω The partial derivatives of r, s, t respect to c are

∂r ∂s 1 ∂t 1 (c,d) = 0, (c,d) = , (c,d) = . ∂c ∂c −2 ∂c 2 And the partial derivatives of r, s, t respect to d are

∂r √3 ∂s √3 ∂t √3 (c,d) = , (c,d) = , (c,d) = . ∂d − 3 ∂d 6 ∂d 6 Then, replacing these values in equations 3.35 and 3.36, we obtain

∂vol 1 ∂ f ∂ f ∂c √3 ∂ f ∂ f ∂ f ∂d = + + 2 + + , (3.37) ∂α 2 − ∂s ∂t ∂α 6 − ∂r ∂s ∂t ∂α and

∂vol 1 ∂ f ∂ f ∂c √3 ∂ f ∂ f ∂ f ∂d = + + 2 + + . (3.38) ∂ω 2 − ∂s ∂t ∂ω 6 − ∂r ∂s ∂t ∂ω Proposition 3.6.1. The partial derivatives of f respect to the variables r,s,t are

∂ f r logr + cosa3slogs + cosa2t logt (r,s,t) = 2 − , (3.39) ∂r (r + s +t)( r + s +t)(r s +t)(r + s t) − − − ∂ f cosa3r logr slogs + cosa1t logt (r,s,t) = 2p − , (3.40) ∂s (r + s +t)( r + s +t)(r s +t)(r + s t) − − − ∂ f cosa2r logr + cosa1slogs t logt (r,s,t) = 2p − . (3.41) ∂t (r + s +t)( r + s +t)(r s +t)(r + s t) − − − Proof. From Observation ?? andp the Chain rule, we have that d 1 (l(arccosx)) = log 2 1 x2 . dx √1 x2 − − p 66 Chapter 3. Tetrahedra

Denote by h := (r + s +t)( r + s +t)(r s +t)(r + s t). The following identities will be − − − used. p r2 + s2 +t2 2 (r + s +t)( r + s +t)(r s +t)(r + s t) h 1 − = − − − = , s − 2st 2st 2st p r2 s2 +t2 2 (r + s +t)( r + s +t)(r s +t)(r + s t) h 1 − = − − − = , s − 2rt 2rt 2rt p r2 + s2 t2 2 (r + s +t)( r + s +t)(r s +t)(r + s t) h 1 − = − − − = . s − 2rs 2rs 2rs p Let us ﬁnd the partial derivative of the f respect to r in Equation 3.34. We have that

∂ ∂ r2 + s2 +t2 2r h (l(a1)) = l arccos − = log , ∂r ∂r 2st − h st ∂ ∂ r2 s2 +t2 r2 + s2 t2 h (l(a )) = l arccos − = − log , ∂r 2 ∂r 2rt rh rt ∂ ∂ r2 + s2 t2 r2 s2 +t2 h (l(a )) = l arccos − = − log . ∂r 3 ∂r 2rs rh rs Adding these equations we get

∂ f 1 r2 + s2 t2 r2 s2 +t2 (r,s,t) = 2r logr + − logs + − logt ∂r h − r r ∂ f r logr + cosa3slogs + cosa2t logt (r,s,t) = 2 − . ∂r (r + s +t)( r + s +t)(r s +t)(r + s t) − − − Because of the symmetry of pf respect to the variables we get the other two identities.

∂ f ∂ f ∂ f Observation 3.6.2. We need the values of ∂r , ∂s , ∂t only at the points that are image by the function (r,s,t), i.e., (r,s,t)=(r,s,t)(c(α,ω),d(α,ω)), for some (α.ω) intS. In this case, ∈ the expression

0 r s t r 0 t s (r + s +t)( r + s +t)(r s +t)(r + s t) = det  = α. (3.42) − − − − − s t 0 r   t s r 0     Also, for these kind of points (r,s,t), we have that r + s +t = 1. Thus,

r2 + s2 +t2 r2 +(s +t)2 2st r2 +(1 r)2 1 2r cosa1 = − = − − = − − 1 = − 1. 2st 2st 2st − 2st − Similarly, 1 2s cosa2 = − 1, 2rt − 1 2t cosa3 = − 1. 2rs − 3.6. Proof of Speculation 4 67

Then, equations 3.39, 3.40, 3.41 become

∂ f 2 1 2t 1 2s (r,s,t) = r logr + − s logs + − t logt , (3.43) ∂r √ α − 2r − 2r − − ∂ f 2 1 2t 1 2r (r,s,t) = − r logr slogs + − t logt , (3.44) ∂s √ α 2s − − 2s − − ∂ f 2 1 2s 1 2r (r,s,t) = − r logr + − s logs t logt . (3.45) ∂t √ α 2t − 2t − − − Proposition 3.6.3. Let (α,ω) intS. Then, ∈ ∂vol 1 st t (α,ω) = M log + N log , ∂α 1944c(3d2 c2)rst√ α r2 s − − where c, d, r, s, t are as in Deﬁnition 3.4.3 and

M :=27c3d 9√3c3 + 63cd3 9√3cd2 18cd + 2√3c, − − − N := 27√3c4 + 27√3c2d2 + 27c2d + 9√3c2 + 18√3d4 + 27d3 9√3d2 6d. − − −

∂c Proof. By equations 3.44 and 3.45, the factor of ∂α in Equation 3.37 is

1 ∂ f ∂ f 1 1 2t 1 2s 1 2r 1 2r + = − + − logr + − logs − logt 2 − ∂s ∂t √ α − 2s 2t 2t − 2s − r (s(1 2s) t(1 2t))logr + rs(1 2r)logs rt(1 2r)logt = − − − − − − . 2rst√ α − Then, 1 ∂ f ∂ f Alogr + Blogs +C logt + = , (3.46) 2 − ∂s ∂t 2rst√ α − where A := r (s(1 2s) t(1 2t)), B := rs(1 2r), and C := rt(1 2r). − − − − 2 2√3d − 2 3−c+√3d 2+3c+√3d We can write A, B, C in terms of c and d replacing r = − 6 , s = − 6 , and t = 6 . After operating, we get

1 A = 6cd2 + √3cd + c , 9 − 1 B = 18cd2 3√3cd 3c 6√3d3 9d2 + 3√3d + 2 , 54 − − − − 1 C = 18cd2 3√3cd 3c + 6√3d3 + 9d2 3√3d 2 . 54 − − − − From Equation 3.24, the ﬁrst term of the sum in Equation 3.37 is

1 ∂ f ∂ f ∂c Alogr + Blogs +C logt d + = 2 − ∂s ∂t ∂α 2rst√ α 6c(3d2 c2) − − 1 ∂ f ∂ f ∂c dAlogr + dBlogs + dC logt + = . (3.47) 2 − ∂s ∂t ∂α 12c(3d2 c2)rst√ α − − 68 Chapter 3. Tetrahedra

∂d On the other hand, by equations 3.43, 3.44, and 3.45, the factor of ∂α in Equation 3.37 is √3 ∂ f ∂ f ∂ f √3 1 2t 1 2s 1 2t 1 2r 2 + + = − + − logr + − + − logs 6 − ∂r ∂s ∂t 3√ α 2s 2t − r 2t − 1 2s 1 2r + − + − logt − r 2s √3 = (r (t(1 2t) + s(1 2s))logr+ 6rst√ α − − − s(r(1 2r) 2t(1 2t))logs +t (r(1 2r) 2s(1 2s))logt). − − − − − − Then, √3 ∂ f ∂ f ∂ f Dlogr + E logs + F logt 2 + + = , (3.48) 6 − ∂r ∂s ∂t 6rst√ α − where we call D := √3r (t(1 2t) + s(1 2s)), E := √3s(r(1 2r) 2t(1 2t)), and − − − − − F := √3t (r(1 2r) 2s(1 2s)). − − − As before, we can write D, E, F in terms of c and d. After operating, we obtain 1 D = 27c2d 9√3c2 + 9d3 9d + 2√3 , 27 − − 1 E = 27√3c3 27c2d + 9√3c2 + 27√3cd2 + 27cd + 9√3c 9d3 + 9d 2√3 , 54 − − − − 1 F = 27√3c3 27c2d + 9√3c2 27√3cd2 27cd 9√3c 9d3 + 9d 2√3 . 54 − − − − − − From Equation 3.25, the second term of the sum in Equation 3.37 is √3 ∂ f ∂ f ∂ f ∂d Dlogr + E logs + F logt 1 2 + + = 6 − ∂r ∂s ∂t ∂α 6rst√ α −6(3d2 c2) − − √3 ∂ f ∂ f ∂ f ∂d cDlogr cE logs cF logt 2 + + =− − − (3.49) 6 − ∂r ∂s ∂t ∂α 36c(3d2 c2)rst√ α − − Summing equations 3.47 and 3.49, we obtain ∂vol (3dA cD)logr +(3dB cE)logs +(3dC cF)logt = − − − , ∂α 36c(3d2 c2)rst√ α − − where 1 3dA cD = 27c3d + 9√3c3 63cd3 + 9√3cd2 + 18cd 2√3c , − 27 − − − 1 3dB cE = 27√3c4 + 27c3d 9√3c3 27√3c2d2 27c2d 9√3c2 + 63cd3 9√3cd2 − 54 − − − − − 18cd + 2√3c 18√3d4 27d3 + 9√3d2 + 6d , − − − 1 3dC cF = 27√3c4 + 27c3d 9√3c3 + 27√3c2d2 + 27c2d + 9√3c2 + 63cd3 9√3cd2 − 54 − − − 18cd + 2√3c + 18√3d4 + 27d3 9√3d2 6d . − − − At this point, note that M M N M + N 3dA cD = , 3dB cE = − , 3dC cF = . − −27 − 54 − 54 3.6. Proof of Speculation 4 69

Then,

∂vol 1 M M N M + N (α,ω) = logr + − logs + logt ∂α 36c(3d2 c2)rst√ α −27 54 54 − − ∂vol 1 st t (α,ω) = M log + N log . ∂α 1944c(3d2 c2)rst√ α r2 s − −

Proposition 3.6.4. Let (α,ω) intS. Then, ∈ ∂vol 2√3d + 1 st t (α,ω) = Plog + Qlog , ∂ω 324c(3d2 c2)rst√ α r2 s − − where c, d, r, s, t are as in Deﬁnition 3.4.3 and

P := 18√3c3d + 18c3 + 6√3cd3 18cd2 + 6√3cd 2c, − − − Q :=27c4 36c2d2 12√3c2d 3c2 + 9d4 9d2 + 2√3d. − − − −

Proof. The ﬁrst term in the sum of Equation 3.38 is obtained replacing equations 3.29 and 3.46 as follows:

1 ∂ f ∂ f ∂c Alogr + Blogs +C logT 3d2 3c2 √3d + = − − 2 − ∂s ∂t ∂ω 2rst√ α 3c(3d2 c2) − − 1 ∂ f ∂ f ∂c 3 3d2 3c2 √3d (Alogr + Blogs +C logT ) + = − − . (3.50) 2 − ∂s ∂t ∂ω 18c(3d2 c2)rst√ α − − The second term in the sum of Equation 3.38 is obtained equations 3.30 and 3.48 as follows:

√3 ∂ f ∂ f ∂ f ∂d Dlogr + E logs + F logt 6d + √3 2 + + = 6 − ∂r ∂s ∂t ∂ω 6rst√ α 3(3d2 c2) − − √3 ∂ f ∂ f ∂ f ∂d c 6d + √3 (Dlogr + E logs + F logt) 2 + + = . (3.51) 6 − ∂r ∂s ∂t ∂ω 18c(3d2 c2)rst√ α − − Summing equations 3.50 and 3.51, we obtain

∂vol J logr + K logs + Llogt = , ∂ω 18c(3d2 c2)rst√ α − − where

J :=3 3d2 3c2 √3d A + c 6d + √3 D, − − K :=3 3d2 3c2 √3d B + c 6d + √3 E, − − L :=3 3d2 3c2 √3d C + c 6d + √3 F. − − 70 Chapter 3. Tetrahedra

We replace the values of A, B, C, D, E, F in terms of c and d obtained in the proof of Proposition 3.6.3. After operating, we obtain 2√3d + 1 J = 18√3c3d 18c3 6√3cd3 + 18cd2 6√3cd + 2c , 9 − − − 2√3d + 1 K = 27c4 18√3c3d + 18c3 + 36c2d2 + 12√3c2d + 3c2 + 6√3cd3 18cd2 18 − − − +6√3cd 2c 9d4 + 9d2 2√3d , − − − 2√3d + 1 L = 27c4 18√3c3d + 18c3 36c2d2 12√3c2d 3c2 + 6√3cd3 18cd2 18 − − − − − +6√3cd 2c + 9d4 9d2 + 2√3d . − − And note that 2√3d + 1 2√3d + 1 2√3d + 1 J = ( P), K = (P Q), L = (P + Q). 9 − 18 − 18 Finally, we have that ∂vol 2√3d + 1 = ( 2Plogr +(P Q)logs +(P + Q)logt) ∂ω 324c(3d2 c2)rst√ α − − − − ∂vol 2√3d + 1 st t = Plog + Qlog . ∂ω 324c(3d2 c2)rst√ α r2 s − −

Observation 3.6.5. We claim that for a given (α,ω) intS, we have that (c,d) int∆1. In- ∈ ∈ deed, if we suppose that (c,d)∆1 int∆1, Proposition B.1.3 implies that (α,ω) = I(c,d) / intS. \ st t ∈ This fact explains that the factor of M log r2 + N log s in Proposition 3.6.3 and the factor of st t ∂vol ∂vol Plog r2 + Qlog s in Proposition 3.6.4 are both positive. With this, the signs of ∂α and ∂ω st t st t are determined by the signs of M log r2 + N log s and Plog r2 + Qlog s respectively. Proposition 3.6.6. Let (α,ω) intS. Then, ∈ st t M log + N log < 0. r2 s

1 √3d 2 3c+√3d 2+3c+√3d Proof. We will write c and d in terms of r, s, t. Since r = −3 , s = − 6 , t = 6 , 1 3r we have that c = t s and d = − . Note that there are other formulas to express c and d in − √3 terms of r, s, t because r + s +t = 1, but we will use the mentioned formulas. Now,

M =27c3d 9√3c3 + 63cd3 9√3cd2 18cd + 2√3c − − − M =c 3d √3 9c2 + 21d2 + 4√3d 2 − − 13r 1 3r 2 1 3r M =(t s) 3 − √3 9(t s)2 + 21 − + 4√3 − 2 − √3 − − √3 √3 − ! M = 27√3r(t s) 7r2 + s2 +t2 2st 6r + 1 . (3.52) − − − − 3.6. Proof of Speculation 4 71

The same for

N = 27√3c4 + 27√3c2d2 + 27c2d + 9√3c2 + 18√3d4 + 27d3 9√3d2 6d − − − N =3 9√3c4 + 3c2 3√3d2 + 3d + √3 + d 2√3d + 1 √3d + 2 √3d 1 − − 1 3r 2 1 3r N =3 9√3(t s)4 + 3(t s)2 3√3 − + 3 − + √3 − − − √3 √3 ! 1 3r 1 3r 1 3r 1 3r + − 2√3 − + 1 √3 − + 2 √3 − 1 √3 √3 √3 √3 − N = 27√3 (t s)4 (t s)2 3r2 3r + 1 + r(1 3r)(1 r)(1 2r) . (3.53) − − − − − − − − From Proposition3.1.16, the positive numbers r, s, t are the sides of an Euclidean triangle. Also,

(α,ω) intS implies that (c,d) int∆1, then by Proposition 3.1.17, we have that r < s < t. ∈ ∈ We deﬁne a,ε R as ∈ s a := , (3.54) r t s ε := − . (3.55) r Then, a > 1 and0 < ε < 1. Also, we have that t = a + ε. (3.56) r The equation 1 = r + s +t implies that 1 = 1 + 2a + ε. (3.57) r Now, we write Equation 3.52 as

t s s2 t2 2st 6 1 M = 27√3r4 − 7 + + + − r r2 r2 − r2 − r r2 M = 27√3r4ε 7 + a2 +(a + ε)2 2a(a + ε) 6(1 + 2a + ε)+(1 + 2a + ε)2 − − − 4 2 2 3 2 M = 54√3r 2a ε + 2aε 4aε + ε 2ε + ε . − − − Then st M log = 54√3r4 2a2ε + 2aε2 4aε + ε3 2ε2 + ε loga(a + ε). (3.58) r2 − − − In the same way for Equation 3.53.

t s 4 t s 2 3 1 1 1 1 N = 27√3r4 − − 3 + + 1 2 3 − r − r − r r2 r − r − r − ! N = 27√3r4 ε4 + ε2 3 3(1 + 2a + ε)+(1 + 2a + ε)2 − − +(1 + 2a + ε 1)(1 + 2a + ε 2)(1 + 2a + ε 3)) − − − N = 54√3r4 4a3 2a2ε2 + 6a2ε 6a2 2aε3 + 4aε2 6aε + 2a + ε3 2ε2 + ε . − − − − − − 72 Chapter 3. Tetrahedra

Then

t N log = 54√3r4 4a3 2a2ε2 + 6a2ε 6a2 2aε3 + 4aε2 6aε + 2a + ε3 2ε2 + ε s − − − − − − a + ε log . (3.59) a Summing equations 3.58 and 3.59, we have that st t M log + N log = 54√3r4X, r2 s − where

X := 2a2ε + 2aε2 4aε + ε3 2ε2 + ε loga(a + ε)+ − − a + ε 4a3 2a2ε2 + 6a2ε 6a2 2aε3 + 4aε2 6aε + 2a + ε3 2ε2 + ε log − − − − − a X =4a3 2a2ε2 + 8a2ε 6a2 2aε3 + 6aε2 10aε + 2a + 2ε3 4ε2 +2ε log(a + ε)+ − − − − − 3 2 2 2 2 3 2 4a + 2a ε 4a ε + 6a + 2aε 2aε + 2aε 2a loga − − − − 2 2 X =2 (a + ε)(a 1) 2a ε + ε 1 log(a + ε) a(a + ε 1) 2a ε 1 loga . − − − − − − − (3.60) We claim that X > 0. Indeed, we will show in a ﬁrst step that

X ′ :=(a + ε)(a 1)log(a + ε) a(a + ε 1)loga (3.61) − − − is a positive number. For this, we use the following series for the natural logarithm. For each x 1, ≥ ∞ 1 x 1 n logx = − . ∑ n x n=1 Then,

∞ 1 (a + ε 1)n ∞ 1 (a 1)n X ′ =(a + ε)(a 1) ∑ − n a(a + ε 1) ∑ −n − n=1 n (a + ε) − − n=1 n a ∞ n 1 ∞ n 1 1 (a + ε 1) − 1 (a 1) − X ′ =(a 1)(a + ε 1) ∑ − n 1 (a 1)(a + ε 1) ∑ −n 1 − − n=1 n (a + ε) − − − − n=1 n a − ∞ n 1 n 1 1 a + ε 1 − a 1 − X ′ =(a 1)(a + ε 1) − − , − − ∑ n a + ε − a n=1 ! a+ε 1 a 1 ε is a positive number because − − = > 0. a+ε − a a(a+ε) For a second step, X ′ > 0 implies that

(a + ε)(a 1)log(a + ε) > a(a + ε 1)loga. − − As 2a ε2 + ε 1 > 2a ε2 1 > 0, we have that − − − − (a + ε)(a 1) 2a ε2 + ε 1 log(a + ε) > a(a + ε 1) 2a ε2 1 loga. − − − − − − 3.6. Proof of Speculation 4 73

Seeing Equation 3.60, we have that X is a positive number. Finally, st t M log + N log = 54√3r4X < 0. r2 s −

The last Proposition shows that the volume function is decreasing in the determinant.

Proposition 3.6.7. Let (α,ω) intS. Then, ∈ st t Plog + Qlog > 0. r2 s Proof. We implement the same idea used to prove Proposition 3.6.6. So, we consider a and ε as they were deﬁned in equations 3.54 and 3.55 respectively and their consequences. 1 3r Remember that c = t s and d = − . So, − √3 P = 18√3c3d + 18c3 + 6√3cd3 18cd2 + 6√3cd 2c − − − P = 2c 9c2 √3d 1 3√3d3 + 9d2 3√3d + 1 − − − − 1 3r 1 3r 3 1 3r 2 P = 2(t s) 9(t s)2 √3 − 1 3√3 − + 9 − − − − √3 − − √3 √3 1 3r 3√3 − + 1 − √3 P = 54r(t s)(r s +t)(r + s t) − − − − t s P = 54(r s +t)(r + s t)r2 − − − − r P = 54(r s +t)(r + s t)r2ε. − − − Then, st Plog = 54(r s +t)(r + s t)r2ε loga(a + ε). (3.62) r2 − − − The same for Q =27c4 36c2d2 12√3c2d 3c2 + 9d4 9d2 + 2√3d − − − − Q =27c4 3c2 12d2 + 4√3d + 1 + 9d4 9d2 + 2√3d − − 1 3r 2 1 3r 1 3r 4 Q =27(t s)4 3(t s)2 12 − + 4√3 − + 1 + 9 − − − − √3 √3 √3 ! 1 3r 2 1 3r 9 − + 2√3 − − √3 √3 Q =54(r s +t)(r + s t)(2st rs rt) − − − − 2st s t Q =54(r s +t)(r + s t)r2 − − r2 − r − r Q =54(r s +t)(r + s t)r2 (2a(a + ε) a (a + ε)) − − − − Q =54(r s +t)(r + s t)r2 2a2 + 2aε 2a ε . − − − − 74 Chapter 3. Tetrahedra

Then, t a + ε Qlog = 54(r s +t)(r + s t)r2 2a2 + 2aε 2a ε log . (3.63) s − − − − a Summing equations 3.62 and 3.63, we have that st t Plog + Qlog = 54(r s +t)(r + s t)r2Y, r2 s − − where a + ε Y := ε loga(a + ε) + 2a2 + 2aε 2a ε log − − − a Y = 2a2 + 2aε 2a 2ε log(a + ε) + 2a2 2aε + 2a loga − − − − Y =2((a + ε)(a 1)log(a+ ε) a(a + ε 1)loga). − − −

From equation 3.61, we note that Y = 2X ′. Since X ′ > 0, we conclude that st t Plog + Qlog = 54(r s +t)(r + s t)r2Y > 0. r2 s − −

The last Proposition shows that the volume function is increasing in the permanent of the matrix. 75 BIBLIOGRAPHY

1 ABROSIMOV, N. V. Seidel’s problem on the volume of a non-euclidean tetrahedron. Dok- lady Mathematics, v. 82, p. 843–846, 2010. Cited on page 24.

2 ANAN’IN, S.; GROSSI, C. H. Basic coordinate-free non-euclidean geometry. 2011. Cited on page 40.

3 ANAN’IN, S.; GROSSI, C. H.; GUSEVSKII, N. Complex hyperbolic structures on disc bundles over surfaces. Cited on page 33.

4 FULTON, W.; HARRIS, J. Representation Theory. [S.l.]: Springer-Verlag, 1991. Cited on page 24.

5 GROSSI, C. H. Schur functors in classic geometries. Cited on page 25.

6 LEE,J.M. Introduction to Smooth Manifolds. [S.l.]: Springer, 2013. No citation.

7 LITTLEWOOD, D. E.; RICHARDSON, A. R. Group characters and algebra. Philosophical Transactions of the Royal Society A, v. 233, p. 99–141, 1934. Cited on page 26.

8 MILNOR, J. Hyperbolic geometry: the ﬁrst 150 years. Bulletin of the American Mathe- matical Society, v. 6, p. 9–24, 1982. Cited 2 times on pages 50 and 63.

9 MOHANTY, Y. Z. Hyperbolic polyhedra: volume and scissors congruence. 2002. Cited on page 32.

10 SEIDEL, J. J. On the volume of a hyperbolic simplex. Studia Scientiarum Mathemati- carum Hungarica, v. 21, p. 243–249, 1986. Cited 2 times on pages 50 and 57.

APPENDIX A

HERMITIAN FORMS

A.1 Hermitian forms

We deal with ﬁnite-dimensional linear spaces over R or C. To cover both cases, denote the scalars by K. The symbol k stands for the conjugate to the (complex) number k K. ∈ Deﬁnition A.1.1. Let V be a K-linear space. A hermitian form is a map , : V V K, h− −i × → (x,y) x,y linear in x and such that x,y = y,x for all x,y V . In particular, x,y + y = 7→ h i h i h i ∈ h ′i x,y + x,y and x,ky = k x,y . h i h ′i h i h i

If W V is a subspace, we can restrict the form , to W W , getting a linear space W ≤ h− −i × equipped with the induced hermitian form.

Deﬁnition A.1.2. Let V be a linear space equipped with a hermitian form and let W V be a ≤ subspace. We deﬁne1 W := v V v,w = 0 for all w W , the orthogonal to W . We call ⊥ { ∈ |h i ∈ } V ⊥ the kernel of the form on V . If the kernel vanishes, we say that the form is nondegenerate. If the induced form on a subspace W V is nondegenerate, W is said to be nondegenerate. For ≤ U,W V the orthogonal of W relatively to U is given by W U := W U. ≤ ⊥ ⊥ ∩

Claim A.1.3. If W V , then W V and W W ⊥. If W1,W2 V, then (W1 + W2) = ≤ ⊥ ≤ ⊂ ⊥ ≤ ⊥ W W and W +W (W1 W2) . 1⊥ ∩ 2⊥ 1⊥ 2⊥ ⊂ ∩ ⊥ Claim A.1.4. It is possible to deﬁne an induced form on V/V by [x],[y] := x,y for [x],[y] ⊥ h i h i ∈ V/V ⊥. The linear space V/V ⊥ equipped with the induced hermitian form is nondegenerate. Decomposing V = V W , the linear function W V/V , w [w] is a natural isomorphism ⊥ ⊕ → ⊥ 7→ that preserves the forms.

Claim A.1.5. If W V, then dimW + dimW dimV. ≤ ⊥ ≥ 1 We also write v,W = 0 to denote v W . h i ∈ ⊥ 78 APPENDIX A. Hermitian forms

Indeed, using induction on dimW , decompose W = W Kw for some w W . Being W ′ ⊕ ∈ ∗ ′⊥ ∩ (Kw) the kernel of the functional W K given by the rule x x,w . By the Claim A.1.3 ⊥ ′⊥ → 7→ h i we have dimW = dim(W (Kw) ) and the Rank-nullity theorem implies that dim(W ⊥ ′⊥ ∩ ⊥ ′⊥ ∩ (Kw) ) dimW 1, then dimW +1 dimW . The rest follows from dimW 1 = dimW ⊥ ≥ ′⊥ − ⊥ ≥ ′⊥ − ′ and by induction.

Claim A.1.6. If W V is a nondegenerate subspace, then V = W W . ≥ ⊕ ⊥ Claim A.1.7. Suppose that both W and V are nondegenerate, where W V. Then W ⊥ = W. ≥ ⊥ Claim A.1.8. Suppose that both W and V are nondegenerate, where W V. Then W is non- ≥ ⊥ degenerate.

Claim A.1.9. If the form on V is not identically null, then there exists a nonisotropic2 v V , ∈ i.e., v,v = 0. h i 6

Indeed, assuming that v,v = 0 for all v V, we obtain v1 + v2,v1 + v2 = 0 and, hence, h i ∈ h i Re v1,v2 = 0 for all v1,v2 V . If K = C, it remains to apply the last identity to iv1,v2 in h i ∈ order to get Im v1,v2 = 0. h i Claim A.1.10. Suppose that both W and V are nondegenerate, where W V. Then, there exists a nondegenerate subspace W V such that W W and dimW = dimW + 1. ′ ≤ ≤ ′ ′

Definition A.1.11. A flag of subspaces is a chain of subspaces V0 V1 V2 ... Vn such that ≤ ≤ ≤ ≤ Vn = V and dimVi = i for all i. If V is equipped with a hermitian form, a flag is nondegenerate when all Vi’s are nondegenerate.

Claim A.1.12. Every nondegenerate linear space admits a nondegenerate ﬂag of subspaces.

Deﬁnition A.1.13. A linear basis β : b1,b2,...,bn is orthonormal if bi,bi 1,0,1 and h i∈{− } bi,b j = 0 for all i and j such that i = j. Denote by β , β0, β+ the amount of elements in the 6 − basis β such that bi,bi = 1, bi,bi = 0, bi,bi = 1, respectively. The triple (β ,β0,β+) is h i − h i h i − the signature of the basis.

Claim A.1.14. Let β : b1,b2,...,bn be an orthonormal basis in V . Then the set of isotropic vectors in β is a basis for V ⊥. In particular, the dimension of the kernel of the form in V is β0, dimV ⊥ = β0.

Proposition A.1.15 (Gram-Schmidt orthogonalization). Let V0 V1 V2 ... Vn be a non- ≤ ≤ ≤ ≤ degenerate ﬂag of subspaces in V. Then there exists an orthonormal basis b1,b2,...,bn in V such that b1,b2,...,bk is a basis in Vk for all k.

Corollary A.1.16. Every linear space with a hermitian form admits an orthonormal basis.

2 If v,v = 0, v is called isotropic. h i A.1. Hermitian forms 79

Deﬁnition A.1.17. Let v1,v2,...,vk V . The matrix G := G(v1,v2,...,vk) :=[gij], where gij := ∈ vi,v j , is called the Gram matrix of v1,v2,...,vk.

t Obviously, G = G, where Mt denotes the transpose matrix of M and M denotes the matrix M with conjugate entries. In other words, G is hermitian (symmetric when K = R). Thus detG is always a real number. The following Proposition shows that the Gram matrix of some basis in V determines the hermitian form on V.

ββ Proposition A.1.18. Let G := G(b1,b2,...,bn) be theGram matrix of somebasis β : v1,v2,..., vn in V . Then t ββ v,v′ =[v]β G [v′]β for all v,v′ V , where [v]β denotes the column matrix whose entries are the coefﬁcients ci ∈ n appearing in the linear combination v = ∑i=1 cibi.

A basis is orthonormal if and only if its Gram matrix is diagonal with diagonal entries 1,0,1. − β Proposition A.1.19. Let α : a1,a2,...,an and β : b1,b2,...,bn be bases in V and let Mα =[mij] n be the matrix representing a change of basis from α to β, that is, b j = ∑i=1 mijai for all j. Then

ββ β t αα β G =(Mα ) G Mα .

In particular, it follows that the sign of detGββ does not depend on the choice of the basis because detGββ = detMβ 2 detGαα . | α | ββ Corollary A.1.20. Let β : b1,b2,...,bn be any basis in V . Then dimV = n rk G , where ⊥ − rk Gββ is the rank of the matrix Gββ .

Lemma A.1.21. Let Gββ bethe Gram matrixof abasis in a linearspace V. Then V is degenerate if, and only, if detGββ = 0.

Proposition A.1.22 (Sylvester’s law of inertia). The signature does not depend on the choice of an orthonormal basis.

Deﬁnition A.1.23. The signature of a space V is the signature of any orthonormal basis in V.

How do we measure it? By Claim A.1.4 we begin by ﬁnding the signature of the non- degenerate space W V/V where V = V W. By Claim A.1.14 this signature will be of ≃ ⊥ ⊥ ⊕ the type (n ,0,n+) and the signature of V ⊥ is (0,dimV ⊥,0), where dimV ⊥ can be found using − Corollary A.1.20. Noting that the signature of V is the sum of the signatures of W and V ⊥, the signature of V is (n ,dimV ⊥,n+). So the problem can be reduced to the case of a nondegener- − ate V . 80 APPENDIX A. Hermitian forms

Proposition A.1.24 (Sylvester criterion). Let γ : c1,c2,...,cn be a basis in V with a known γγ γγ γγ matrix G . Let G be the matrix of c1,c2,...,ck for every k such that 1 k n. If detG = 0 k ≤ ≤ k 6 for every k, then the signature of the space equals (n ,0,n+), where n is the amount of negative − − numbers in the sequence

γγ γγ γγ γγ detG2 detG3 detGn detG1 , γγ , γγ ,..., γγ detG1 detG2 detGn 1 − and n+ is the amount of positive numbers in the same sequence.

The following results concern the study of the possible signatures of a subspace when the signature of the space is given.

Claim A.1.25. Let V be a space of signature (n ,n0,n+). If V contains a subspace W of sig- − nature (m ,m0,m+), then the space V/V ⊥ (of signature (n ,0,n+)) possesses a subspace of − − signature (m ,m0 m,m+), where m = dimW V ⊥. If the space V /V ⊥ contains a subspace U − − ∩ 1 of signature (m ,m0 m,m+), where m = dimπ− (U) V ⊥ and the map π : V V /V ⊥ takes − − ∩ → v to its class, then V contains a subspace of signature (m ,m0,m+). −

Claim A.1.26. Let V be a space of signature (n ,0,n+). The highest possible dimension of a − subspace W with the null induced form is min(n ,n+). −

Claim A.1.27. Let V be a space of signature (n ,0,n+). Then V contains a subspace W of − signature (m ,m0,m+) if and only if −

m n , m+ n+, m0 n m , m0 n+ m+. − ≤ − ≤ ≤ − − − ≤ −

Claim A.1.28. Let V be a space of signature (n ,n0,n+). Then V contains a subspace of signa- − ture (m ,m0,m+) if and only if −

m n , m+ n+, m + m0 n + n0 , m0 + m+ n0 + n+. − ≤ − ≤ − ≤ − − ≤ 81

APPENDIX B

AN EXPLICIT DESCRIPTION OF R AND S

B.1 An explicit description of R and S

In what follows we use the notation from Section 3.1.3; in particular, we need deﬁnitions 3.9, 3.10, and 3.4. We will give explicit descriptions of the sets R and S introduced in 3.4 and 3.12. The following proposition describes the sets R and intR.

2 2 Proposition B.1.1. Let ∆1 be the closure of ∆1 in R and let H : ∆1 R be deﬁned by → 2 2√3d + 1 3c + √3d 1 3c √3d + 1 3c2 + 3d2 + 2 H(c,d) = − − , . 27 6 ! Then (i) H is a homeomorphism onto its image. 2 (ii) C1 C2 C3 is a simple closed curve in R , where ∪ ∪ √3 1 √3 C1 := H (0,0), 0, , C2 := H (0,0), , , " 3 !! " 4 12 !!

1 √3 √3 C3 := H , , 0, . " 4 12 ! 3 !# 2 (iii) The bounded component of R (C1 C2 C3) is \ ∪ ∪ 1 1 1 2 A := (x,y) R2 < y < , x > 2 6√y 2 + 1 1 6√y 2 ( ∈ 9 4 −27 − − − ) q q \ 2 1 9 1 (x,y) R2 < y , x < 1 2 6√y 2 6√y 2 + 1 ( ∈ 9 ≤ 64 −27 − − − ) q q 9 1 (x,y) R2 y < , x < 0 . ∈ 64 ≤ 4 [

82 APPENDIX B. An explicit description of R and S

(iv) R = A C1 C2. ∪ ∪ (v) intR = A.

Proof. (i) The injectivity of H can be shown as in the proof of Proposition 3.2.3. Since H : ∆1 → H ∆1 is a continuous bijective function and ∆1 is compact, we have that H is a homeomorphism. √3 1 √3 1 √3 √3 (ii) The boundary bd∆1 = (0,0), 0, 3 (0,0), 4 , 12 4 , 12 , 0, 3 , √3 so H bd∆1 = C1 C2 C3. Parameterizingh the hypotenuse Sh h:= (0,0)S, h0, of∆1 byi ∪ ∪ 3 h √3 t 0, 0 t < 1 , ( 3 ! ≤ )

one obtains √3 H(h) = C1 = H 0, t 0 t < 1 = ( 3 ! ≤ )

1 2 1 2 = (2t + 1)(1 t) , t2 + 2 0 t < 1 . −27 − 36 ≤ Reparameterizing, we get

1 2 1 1 C1 = 2 6√y 2 + 1 1 6√y 2 ,y y < . (B.1) ( −27 − − − ! 9 ≤ 4) q q

1 √3 1 √3 Writing the smallest leg l1 := (0,0), , of ∆1 as t , 0 t < 1 , we have 4 12 4 12 ≤ h n o

t √3 H(l1) = C2 = H , t 0 t < 1 = ( 4 12 ! ≤ )

2 1 t 2 1 t2 = (1 t) + 1 , + 2 0 t < 1 = ( −27 − 2 36 4 ! ≤ )

1 2 1 9 = 1 2 6√y 2 6√y 2 + 1 ,y y < . ( −27 − − − ! 9 ≤ 64) q q

1 √3 √3 Parameterizing the largest leg l2 := 4 , 12 , 0, 3 of ∆1 by h i 1 √3 √3 (1 t) , +t 0, 0 t 1 , ( − 4 12 ! 3 ! ≤ ≤ )

we get

2 1 t √3 t2 + 3 C3 = H − , (3t + 1) 0 t 1 = 0, 0 t 1 = ( 4 12 ! ≤ ≤ ) ( 64 ! ≤ ≤ )

9 1 = (0,y) y . 64 ≤ ≤ 4

B.1. An explicit description of R and S 83

1 0, 4

A 9 0, 64

1 , 1 − 27 9 Figure 7 – The set R.

Source: Elaborated by the author.

The boundary of ∆1 is a simple closed curve, so H bd∆1 = C1 C2 C3 is a simple closed ∪ ∪ curve in R2 that joins the points H(0,0) = 1 , 1 , H 1 , √3 = 0, 9 , H 0, √3 = 0, 1 . − 27 9 4 12 64 3 4 2 (iii) By Jordan’s theorem, R (C1 C2 C3) has exactly two connected components \ ∪ ∪ B1,B2 with B1 bounded and B2 unbounded; the curve C1 C2 C3 is the boundary of each ∪ ∪ component (see Figure 7). In this way, B1 is the open region A delimited by the curves C1, C2, and C3.

(iv) We claim that A = H int∆1 . First, note that A H ∆1 . Indeed, if ⊂ 2 A R H ∆1 = ∅, ∩ \ 6 then the simple closed curve C1 C2 C3 = H bd∆1 H ∆1 is not null homotopic in H(∆1). ∪ ∪ ⊂ This is a contradiction because, by (i), H ∆1 is simply connected.

Now, we have A H ∆1 H bd∆1 = H ∆1 bd∆1 = H int∆1 . Conversely, H int∆1 = 2⊂ \ \ H ∆1 H bd∆1 R (C1 C2 C3) = A B and H int∆1 is a connected space that inter- \ ⊂ \ ∪ ∪ ∪ sects A. It follows that H int∆1 A and, therefore, H int∆1 = A. ⊂ Finally, Proposition 3.2.3 implies R = G(∆1). Since Hcoincides with G in ∆1, we have 84 APPENDIX B. An explicit description of R and S

√3 1 √3 R = H (∆1) = H int∆1 (0,0), 0, (0,0), , = A C1 C2. 3 4 12 ∪ ∪ Sh Sh (v) C1 C2 bdA implies intR = int (A C1 C2) = A. ∪ ⊂ ∪ ∪

Observation B.1.2. The bijective function G : ∆1 R of Proposition 3.2.3 is a homeomor- |∆1 → phism because G ∆ = H ∆ . | 1 | 1 The following result describes the sets S and intS. The proof is very similar to that of Proposition B.1.1 and will be omitted.

2 Proposition B.1.3. Let I : ∆1 R be deﬁned by → 2√3d + 1 3c + √3d 1 3c √3d + 1 3c2 + 3d2 + 2 I(c,d) = − − , . 27 6 ! Then (i) I is a homeomorphism onto its image. 2 (ii) D1 D2 D3 is a simple closed curve in R , where ∪ ∪ √3 1 √3 D1 = I (0,0), 0, , D2 = I (0,0), , , " 3 !! " 4 12 !!

1 √3 √3 D3 = I , , 0, . " 4 12 ! 3 !# 2 (iii) The bounded component of R (D1 D2 D3) is \ ∪ ∪ 1 1 1 2 B = (x,y) R2 < y < , x > 2 6y 2 + 1 1 6y 2 ∈ 3 2 −27 − − − \ 1 3 1 p p 2 (x,y) R 2 < y , x < 1 2 6y 2 6y 2 + 1 ∈ 3 ≤ 8 −27 − − − 3 1 p p (x,y) R2 y < , x < 0 . ∈ 8 ≤ 2 [

(iv) S = B D1 D2. ∪ ∪ (v) intS = B.

Figure 8 displays the set S. B.1. An explicit description of R and S 85

1 0, 2

B 3 0, 8

1 , 1 − 27 3 Figure 8 – The set S.

Source: Elaborated by the author.