LAMBERT CUBE AND LOBELL POLYHEDRON REVISITED

PETER BUSER, ALEXANDER MEDNYKH, AND ANDREI VESNIN

Dedicated to Ernest Borisovich Vinberg on the occasion of his academic anniversary

Abstract. The first example of a closed orientable hyperbolic 3-manifold was constructedby F. Lobell in 1931 from eight copies of the right-angled 14-hedron. We consider the familyof hyperbolic polyhedra which generalize the Lambert cube and the Lobell polyhedron. Forpolyhedra from this family we give trigonometric relations between essential dihedral anglesand lengths and obtain volume formulae in various forms.

1. Introduction

The theory of polyhedra in non-Euclidean spaces is one of many areas of mathematics thatreceived a powerful influence from E.B. Vinberg (see, for example, [1, 25, 36, 37, 38]). Hisresults on existence of hyperbolic Coxeter groups and on volume formulae for a wide classof hyperbolic polyhedra gave new life to this theory and attracted many investigators to it.

In the present paper we obtain volume formulae for a family of almost right-angled poly-hedra in Lobachevsky space H3, i.e. for polyhedra with only a few dihedral angles not equalto π/2. The interest in this class of polyhedra is motivated by two classical examples: theLambert cube and the Lobell polyhedron.

Both polyhedra can be considered as natural generalizations of a Euclidean right-angledcube. In the first case, the combinatorial structure of the cube is preserved while dihedralangles vary. More precisely, let L(α1, α2, α3) be a cube with dihedral angles α1, α2 and α3

as presented in Fig. 1 while all other dihedral angles are equal to π/2. It is well-knownthat if each of the αi’s belongs to (0, π/2) then L(α1, α2, α3) can be realized as a boundedpolyhedron in H3. Each such polyhedron is called a Lambert cube in honor of J.H. Lambert(1729–1777) who investigated hyperbolic quadrilaterals with three right angles arising asfaces of L(α1, α2, α3).

α1

α2

α3

Figure 1. Lambert cube L(α1, α2, α3).

In the second case, right dihedral angles are preserved while the combinatorial structurevaries. More precisely, for each n > 5 one can consider a right-angled polyhedron R(n) withn-gonal top and bottom and 2n pentagons on the lateral surface (see Fig. 3, where R(5),which is a dodecahedron, and R(6) are presented). The right-angled 14-hedron R(6) is calledthe Lobell polyhedron in honor of F.R. Lobell (1893–1964). We recall, that in 1890 Klein,

Date: February 24, 2012.1991 Mathematics Subject Classification. 57M60, 51M25, 51M10, 51F15.Key words and phrases. hyperbolic polyhedron, right-angled polyhedron, volume, orbifold.A.M. and A.V. are supported by RFBR grant 10-01-00642 and by the grant SO RAN – UrO RAN.

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Figure 2. Polyhedra R(5) and R(6).

inspired by a number of examples due to Clifford, formulated the problem of describingall compact, connected, Riemannian manifolds of constant curvature. Then Killing showedthat these manifolds are of the form X/Γ, where X is a complete simply-connected space ofconstant curvature and Γ is a co-compact discrete group of isometries of X, acting withoutfixed points. He called them Clifford – Klein space forms.

Examples of three-dimensional Clifford – Klein space forms of negative curvature wereunknown for a long time [11, p.269-270]: “In bezug auf die hyperbolische Geometrie wollenwir nur hervorheben, daß eine geschlossene dreidimensionale hyperbolische Raumform vonendlichem Volumen bis jetzt nicht gefunden zu sein scheint.”

Answering in the affirmative a question on the existence of Clifford – Klein space formsof constant negative curvature, Lobell [12] in 1931 constructed the first example of a closedorientable hyperbolic 3-manifold by gluing together eight copies of the Lobell polyhedronR(6). Some modification of Lobell’s construction was done in [29].

Lobell’s construction was reformulated algebraically and an infinite series of finite setsL(n), n > 5, of closed hyperbolic 3-manifolds were constructed in [16, 19, 33]. Each manifoldfrom L(n), called a Lobell manifold, is constructed from eight copies of the right-angled(2n+ 2)-hedron R(n). In particular, L(6) contains the first example constructed by Lobell.Moreover, L(n) contains both orientable and non-orientable manifolds. Essentially, theapproach from [16, 19, 33] gives a method for constructing a closed orientable hyperbolic 3-manifold from eight copies of a right-angled polyhedron in H3, based on a coloring of its facesin four colors. A similar method was independently suggested by Takahashi [32]. Isometrygroups of Lobell manifolds of a special kind were described in [15, 19]. The volume formulafor Lobell manifolds was obtained in [34].

Recently Lobell manifolds, as well as right-angled polyhedra, have become objects ofextensive investigations [2, 7, 9, 13, 14, 18, 28, 31]. In particular, arithmetical propertiesof these manifolds were investigated in [2]. Upper and lower bounds for the complexityof Lobell manifolds were obtained in [13, 14]. An arrangement of right-angled hyperbolicpolyhedra by volume was done in [9]. It turns out that the smallest volume is attained bythe regular right-angled dodecahedron R(5) and the second – by the 14-hedron R(6). Four-dimensional generalizations of Lobell’s construction were considered in [7]. Right-angledpolyhedra arising as convex cores of quasi-Fuchsian groups were investigated in [18].

Computations of volumes of polyhedra in H3 is a difficult problem and explicit formulae forvolumes are obtained only for some classes of polyhedra. Initial results in this direction wereobtained by N.I. Lobachevskij. Nice expressions for volumes of some classes of polyhedra wereobtained by Vinberg [1, 36, 38] and Kellerhals [10] in terms of the Lobachevsky function Λ(x).

The aim of the present paper is to obtain explicit formulae for volumes of an infinitefamily of remarkable polyhedra in H3. In particular, we generalize the following previouslyobtained results: formulae for volumes of Lambert cubes which were described in [3, 10],the volume formula for non-compact Lobell type polyhedra which was found in [18], and the

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volume formula for the polyhedra R(n) (and so, the volume formula for Lobell manifolds)in terms of Λ(x) which was obtained in [34] (its particular case, for n divisibly by 6, wasindependently obtained in [31]).

The method we use to obtain volume formulae is the following. First, we relate theessential dihedral angles and lengths of polyhedra by a number of trigonometric relationssimilar to the Sine and Tangent rules for a triangle, using one more parameter (called aprincipal parameter). Then we find an algebraic equation, depending on the combinatorialtype of the polyhedra, to evaluate this principal parameter. Finally, we apply the Schlaflivariation formula to find an explicit expression for the volume. This fruitful approach hasbeen successfully applied in our previous papers [3, 4, 5, 17, 20] to find volume formulae forsimplest knot and link cone-manifolds and polyhedra.

The paper is organized as follows. In Section 2 we recall some basic facts about thevector model of H3, the Rivin theorem on existence of polyhedra in H3, and the Schlafliformula. In Section 3 we investigate the hyperbolic Lambert cube L(α1, α2, α3) obtainingtrigonometric relations between dihedral angles and lengths in Propositions 3.1 and 3.2and volume formulae in Theorem 3.5. In Section 4 we consider a hyperbolic polyhedronL(α1, α2, α3, α4) with two vertices at infinity. Trigonometric relations between its dihedralangles α1, α2, α3, α4 and lengths of corresponding edges are obtained in Propositions 4.1, 4.2,and 4.3. These relations are used in Theorem 4.6 to obtain volume formulae. In Section 5we consider an infinite family of almost right-angled hyperbolic polyhedra generalizing R(n).Using a trigonometric relation between dihedral angles and lengths (see Proposition 5.1) weobtain volume formula for these polyhedra in Theorem 5.2.

The authors are thankful to Professor E.B. Vinberg for bringing to their attention Lobell’spaper [12] in 1977. This paper became a source of new ideas and results during the last thirtyyears.

2. Preliminary facts

2.1. The model of the hyperbolic 3-space. We will use the following vector model ofhyperbolic 3-space H3 [1, 26]. A scalar product in R4 defined for vectors x = (x0, x1, x2, x3)and y = (y0, y1, y2, y3) by the formula

(x, y) = −x0y0 + x1y1 + x2y2 + x3y3,

turns R4 into a pseudo-Euclidean space R3,1. A basis e0, e1, e2, e3 in R3,1 is said to beorthonormal if (e0, e0) = −1 and (ei, ei) = 1 for i = 1, 2, 3, and (ei, ej) = 0 for i 6= j,i, j ∈ 0, 1, 2, 3.

ConsiderH3 = x ∈ R3,1 : (x, x) = −1, x0 > 0

with the Riemannian metric induced by the pseudo-Euclidean metric on R3,1:

ds2 = −dx20 + dx2

1 + dx22 + dx2

3.

Any hyperplane H ⊂ H3 is of the form He = x ∈ H3 : (x, e) = 0, where e ∈ R3,1 and(e, e) = 1. Let H−e = x ∈ H3 : (x, e) ≤ 0 be one of two subspaces determined by He.Two hyperplanes He and Hf are intersecting if |(e, f)| < 1, and dihedral angle αe,f betweenH−e and H−f is given by cosαe,f = −(e, f). Two hyperplanes He and Hf are parallel if|(e, f)| = 1. Two hyperplanes He and Hf are divergent if |(e, f)| > 1, and distance ρe,fbetween them (realized by a geodesic in H−e ∩H−f ) is given by cosh ρe,f = −(e, f).

A convex polyhedron P ⊂ H3 is defined as an intersection of finitely many half-spaces:P =

⋂si=1 H

−i , where each hyperplane Hi is determined by its normal vector ei as above. It

may always be assumed that none of half-spaces H−i contains the intersection of all others.Under this condition half-spaces H−i are uniquely determined by P . The Gram matrix ofthe system of vectors e1, . . . , es is said to be Gram matrix of the polyhedron P .

3

2.2. The Rivin theorem. To demonstrate existence of polyhedra in H3 we will use thefollowing characterization due to Rivin [8], that is done in terms of a Gaussian image of apolyhedron.

Let P be a compact convex polyhedron in H3. The Gauss map G from the boundary ∂Pof P to the unit sphere S2 is a set-valued function which assign to each point p ∈ ∂P the setof outward unit normals to planes supporting ∂P at p. Thus, the whole of a face f of ∂Pis mapped under G to a single point which is outward unit normal to f . An edge e of ∂Pis mapped to a geodesic segment G(e) on S2, whose length is easily seen to be the exteriordihedral angle at the edge e. A vertex v of ∂P is mapped by G to a spherical polygon G(v),whose sides are the images under G of edges incident to v and whose angles are easily seento be the angles supplementary to the planar angles of the faces incident to v; that is, G(e1)and G(e2) meet at angle π − α whenever e1 and e2 meet at angle α. Glue the resultingpolygons together into a closed surface, using the rule that faces G(v1) and G(v2) are gluingisometrically whenever vertices v1 and v2 share an edge.

The resulting metric space G(∂P ) is topologically the sphere S2 and the complex is com-binatorially dual to ∂P . But metrically it is a cone-manifold with the underlying space S2.In general case the sum of angles of a bounded hyperbolic n-gon f is less than 2π(n − 2).Hence the sum of angles around a vertex G(f) is greater than 2π. Thus vertices G(f) ofG(∂P ) are cone points with cone angle greater than 2π.

The following theorem gives a precise characterization of those cone-manifolds that canarise as an image G(∂P ) for a compact convex polyhedron P in H3.

Theorem 2.1. [8] A metric space (M, g) homeomorphic to S2 can arise as the Gaussianimage G(∂P ) of boundary of a compact convex polyhedron P in H3 if and only if the followingconditions hold:

(a) The metric g has constant curvature 1 away from a finite collection of cone points ci.(b) The cone angles at ci are greater than 2π.(c) The lengths of closed geodesics of (M, g) are all strictly greater than 2π.

Furthermore the metric of G(∂P ) determines the hyperbolic polyhedron P uniquely (up to amotion).

2.3. The Schlafli formula. To obtain volume formulae we will use the following Schlaflivariation formula (see [1, p. 119] for proof and more discussions).

Theorem 2.2. Let a convex polyhedron in Hn be deformed in such a way that its combina-torial structure is preserved while its dihedral angles vary in a differentiable manner. Underthese conditions its volume is also varying in a differentiable manner, and its differential isgiven by the Schlafli formula

d volP = − 1

n− 1

∑dimF=n−2

volF dαF ,

where the sum is taken over all (n − 2)-dimensional faces F of the polyhedron P and αFdenotes the dihedral angle at the face.

In the case of polyhedra in H3 the Schlafli formula reduces to

(1) d volP = −1

2

∑i

`idαi,

where the sum is taken over all edges of P , and `i denotes the length of the i-th edge andαi is the dihedral angle at this edge.

4

3. The Lambert cube L(α1, α2, α3)

3.1. The existence. Consider the Lambert cube L(α1, α2, α3) presented by its projectionin Fig. 4 with three outgoing edges meeting in a point. Combinatorially it is a cube. Threeof its dihedral angles are essential and equal to α1, α2, α3, while all other dihedral angles aresuppose to be π/2. Combinatorially, the Gaussian image G(∂L(α1, α2, α3)) is an octahedron.

α1

α3

α2e3m e1m

e2m

f2m

f1m f3

mα2

α2

α1

α1

α3

α3

α2

α3

α1

Figure 3. L(α1, α2, α3) and G(∂L(α1, α2, α3)).

Three its edges are of lengths αi = π−αi, i = 1, 2, 3, while all other edges are of length π/2. Ithas six cone points: for each i there are two cone points with the cone angle 3π/2+(π−αi) =2π+(π/2−αi). Thus, to obtain cone angles greater than 2π we need to assume that αi < π/2for all i. Obviously, these inequalities also guarantee that lengths of any closed geodesic isgreater than 2π. Indeed, any closed geodesic consists of at least four segments with length> π/2 with the strong inequality for at least one of segments. Therefore, by Theorem 2.1 theLambert cube L(α1, α2, α3) is a bounded hyperbolic polyhedron if and only if all its essentialangles are less than π/2. (Recall that L(π/2, π/2, π/2) is the Euclidean cube.)

3.2. Relations between angles and lengths. Let us denote by `1, `2, and `3 the lengthsof the edges of the hyperbolic Lambert cube L(α1, α2, α3) corresponding to the essentialangles α1, α2, and α3 respectively. We call `1, `2, and `3 essential lengths. The followingproposition gives relations between these essential parameters.

Proposition 3.1 (Sine-cosine Rule). The essential angles and lengths of the hyperbolic Lam-bert cube L(α1, α2, α3) satisfy the following relation:

(2)sinαisinh `i

· sinαi+1

sinh `i+1

· cosαi+2

cosh `i+2

= 1.

where i = 1, 2, 3 and all suffices are taken by mod 3.

Proof. Denote by e1, e2, e3, f1, f2, and f3 normal vectors in R3,1 determine hyperplanes He1 ,He2 , He3 , Hf1 , Hf2 , Hf3 supporting faces of L(α1, α2, α3) (see Fig. 4). Assume that vectorsare oriented in such a way that

L(α1, α2, α3) = H−e1 ∩H−e2∩H−e3 ∩H

−f1∩H−f2 ∩H

−f3.

By the definition of L(α1, α2, α3) we have

(ei, ei) = 1, (fi, fi) = 1, (ei, ei+1) = 0, (fi, fi+1) = 0,

(ei, fi) = − cosh li, (ei+1, fi) = − cosαi+2,5

for i = 1, 2, 3, where all suffices are taken by mod 3. As one can see from Fig. 4, facesdetermined by vectors e1, e2, and e3 are mutually orthogonal and are intersecting in onepoint that is a vertex of the cube. Let e0 ∈ R3,1 be the vector presenting this vertex. Sincee0 ∈ He1 ∩He2 ∩He3 we get (ei, e0) = 0 for i = 1, 2, 3. Hence

(ei, ej) = 0, 0 ≤ i < j ≤ 3; (ei, ei) = 1, i = 1, 2, 3; (e0, e0) = −1,

and vectors e0, e1, e2, e3 are forming an orthonormal basis in R3,1. Hence, vectors f1, f2,and f3 can be expressed as linear combinations of these basis vectors.

Suppose that f1 is a linear combination of the form

f1 = d · e0 + a · e1 + b · e2 + c · e3

for some a, b, c, d ∈ R. Then

a = (f1, e1) = − cosh `1, b = (f1, e2) = − cosα3, c = (f1, e3) = 0,

and

1 = (f1, f1) = −d2 + a2 + b2 + c2.

Therefore

d2 = a2 + b2 − 1 = cosh2 `1 + cos2 α3 − 1 = cosh2 `1 − sin2 α3.

Since e0 ∈ H−f1 , we have d = −(f1, e0) > 0, and d =√

cosh2 `1 − sin2 α3.Applying similar considerations for f2 and f3 we obtain:

f1 = d1 · e0 − cosh `1 · e1 − cosα3 · e2 +0 · e3

f2 = d2 · e0 +0 · e1 − cosh `2 · e2 − cosα3 · e3

f3 = d3 · e0 − cosα2 · e1 +0 · e2 − cosh `3 · e3

with di given by

di =

√cosh2 `i − sin2 αi+2.

For i = 1, 2, 3 from (fi, fi+1) = 0 we get

(3) didi+1 = cosαi+2 cosh `i+1.

Therefore

d2i d

2i+1 = cos2 αi+2 cosh2 `i+1,

and using d2i+1 = cosh2 `i+1 − sin2 αi, we get

d2i cosh2 `i+1 − d2

i sin2 αi = cos2 αi+2 cosh2 `i+1.

Hence

(d2i − cos2 αi+2) cosh2 `i+1 = d2

i sin2 αi.

Using d2i − cos2 αi+2 = sinh2 `i, we get

sinh `i cosh `i+1 = di sinαi,

whence

di =sinh `i cosh `i+1

sinαi.

Substituting expressions for di and di+1 into (3) we have

sinh `i cosh `i+1

sinαi· sinh `i+i cosh `i+2

sinhαi+1

= cosαi+2 cosh `i+1,

that implies the formula (2). 6

Proposition 3.2 (Tangent Rule). The essential angles and lengths of the hyperbolic Lambertcube L(α1, α2, α3) satisfy the following relations:

(4)tanα1

tanh `1

=tanα2

tanh `2

=tanα3

tanh `3

= T,

where T is a positive root of the equation

(5) T 2 =3∏i=1

T 2 − tan2 αi1 + tan2 αi

.

Proof. Formula (2) implies

sinαisinh `i

· sinαi+1

sinh `i+1

· sinαi+2

sinh `i+2

=tanαi+2

tanh `i+2

for i = 1, 2, 3 with all suffices taken by mod 3. Thus, we have three equations with the sameexpression on the left. Denote it by T :

T =tanα1

tanh `1

=tanα2

tanh `2

=tanα3

tanh `3

.

Relation (5) follows from an elementary observation:

sin2 αi

sinh2 `i=T 2 − tan2 αi1 + tan2 αi

.

Remark 3.3. To simplify notations, let us denote Ai = tanαi for i = 1, 2, 3. It is easy tosee that (5) is equivalent to the equation (T 2 + 1)P (T 2) = 0, where

P (x) = x2 −

(1 +

3∑i=1

A2i

)x−

3∏i=1

A2i .

Hence, the principal parameter T from Proposition 3.2 is the positive root of the equation

T 2 = K +

(K2 +

3∏i=1

A2i

)1/2

,

where K = (1 +∑3

i=1A2i )/2. We observe that T is the largest real root of the equation

P (x2) = 0 and P (x2) is positive if x2 tends to infinity. Hence P (x2) > 0 for any x > T .Straightforward calculations shows that

P (tan2 αi) = − tan2 αi(1 + tan2 αi−1)(1 + tan2 αi+1) < 0,

that implies T > tanαi for i = 1, 2, 3.

3.3. The volume formulae. Recall [1, 26] that volumes of polyhedra in H3 traditionallyare expressed in terms of the Lobachevsky function

Λ(x) = −∫ x

0

log | 2 sin ζ | dζ.

Let us introduce the function

∆(λ, σ) = Λ(λ+ σ)− Λ(λ− σ)

which admits to simplify volume formulae in many cases due to the following Lemma.7

Lemma 3.4. [18] Consider

I(L, S) =

∫ +∞

S

log

∣∣∣∣t2 − L2

1 + L2

∣∣∣∣ dt

1 + t2,

where L = tanλ, S = tanσ, and 0 < λ, σ < π2. Then

I(L, S) = ∆(λ, σ)−∆(π/2, σ).

Theorem 3.5. The volume of the hyperbolic Lambert cube L(α1, α2, α3), 0 < α1, α2, α3 <π/2, is given by formulae (6) and (7):

(6) volL(α1, α2, α3) =1

4

∫ ∞T

log

[1

t2

3∏i=1

t2 − tan2 αi1 + tan2 αi

]dt

1 + t2,

where T is a positive root of the equation

T 4 −

(1 +

3∑i=1

tan2 αi

)T 2 −

3∏i=1

tan2 αi = 0;

(7) volL(α1, α2, α3) =1

4

[3∑i=1

∆(αi, θ) −∆(0, θ)− 2∆(π/2, θ)

],

where θ, 0 < θ < π/2, is given by tan θ = T .

Proof. Let V = V (α1, α2, α3) = volL(α1, α2, α3) be the volume function. To find V we aregoing to use the Schlafli formula (1) and then to integrate the differential form

ω = dV = −1

2

3∑i=1

`idαi

which is defined for 0 < α1, α2, α3 < π/2. To simplify notations, denote Ai = tanαi. Thendαi = dAi/(1 +A2

i ), and by (4) we have `i = arctanh (Ai/T ). Thus we need to integrate thedifferential form

ω = −1

2

3∑i=1

arctanh (Ai/T )dAi

1 + A2i

with T satisfying (5). In order to do this we will find an extended differential form Ω of fourindependent variables T , A1, A2, and A3:

Ω = −1

2

(3∑i=1

arctanh (Ai/T )dAi

1 + A2i

+ `(A1, A2, A3, T )dT

1 + T 2

)with the following properties:(10) Ω is smooth and exact in the region

U = (A1, A2, A3, T ) ∈ R4 : T > Ai > 0, i = 1, 2, 3;(20) Ω = ω for all (A1, A2, A3, T ) ∈ U satisfying the equation (5).

Let ` = `(A1, A2, A3, T ). Since Ω is supposed to be exact we have

1

1 + T 2

∂`

∂Ai=

1

1 + A2i

∂

∂T(arctanh (Ai/T )) =

Ai(1 + A2

i )(A2i − T 2)

.

Hence

` =

∫(1 + T 2)Ai

(1 + A2i )(A

2i − T 2)

dAi =1

2log

T 2 − A2i

1 + A2i

+ C,

8

where C depends on the variables T,Aj, j 6= i. Since the considerations are symmetric withrespect to A1, A2 and A3 we obtain

` =1

2log

[3∏i=1

T 2 − A2i

1 + A2i

]+ φ(T ),

where the function C = φ(T ) depends of T only. The choice φ(T ) = − log T assures thatcondition (20) will be satisfied. So, we have

` = `(A1, A2, A3, T ) =1

2log

[1

T 2

3∏i=1

T 2 − A2i

1 + A2i

].

Now we are able to prove formula (6). Consider the function

W = W (A1, A2, A3, T ) =1

4

∫ ∞T

log

[1

t2

3∏i=1

t2 − A2i

1 + A2i

]dt

1 + t2,

where T is a positive root of (5). By Remark 3.3 for t ≥ T we have t2 − A2i ≥ T 2 − A2

i > 0for i = 1, 2, 3. Hence, the expression under the log in the above integral is positive.

Straightforward calculations show that

∂W

∂Ai= −arctanh (Ai/T )

2(1 + A2i )

for i = 1, 2, 3 and W (A1, A2, A3, T )→ 0 as A1, A2, A3 →∞. By (1) and Proposition 3.2 thevolume function V = V (α1, α2, α3) = V (A1, A2, A3) satisfies the same conditions:

∂V

∂Ai=∂V

∂αi

dαidAi

= −`i2· 1

1 + A2i

= −arctanh (Ai/T )

2(1 + A2i )

for i = 1, 2, 3 and V (A1, A2, A3)→ 0 as A1, A2, A3 →∞. Hence the functions V (A1, A2, A3)and W (A1, A2, A3, T ) coincide and (6) holds.

The formula (7) follows immediately from (6) by Lemma 3.4.

Remark 3.6. Propositions 3.1 and 3.2 were already observed by Pashkevich in [24], wheretaking 1/T as a principal parameter the analog of (6) was obtained. Formula (7) wasoriginally obtained by Mednykh [17]. This is a refined version of Kellerhals formula [10] forthe volume of the hyperbolic Lambert cube in terms of the Lobachevsky function. In [10]the parameter θ was calculated in a slightly more complicated way by making use of dihedralangles and lengths of polyhedron. For the spherical version of this result see [3].

4. The polyhedron L(α1, α2, α3, α4)

4.1. The existence. Consider a polyhedron L(α1, α2, α3, α4) presented by its projection inFig. 5 with four outgoing edges meeting in a point. This polyhedron has two 4-valent verticesand eight 3-valent vertices, and all its faces are quadrilaterals. If 0 < α1, α2, α3, α4 < π thenby [27] L(α1, α2, α3, α4) can be realized in H3 in such a way that 4-valent vertices are idealpoints; 3-valent vertices are finite points; four essential dihedral angles are equal to α1, α2,α3, and α4, while all other dihedral angles are π/2.

We point out that the polyhedron L(α1, π/2, α3, π/2) arose in [18] as the convex core ofa quasi-Fuchsian punctured torus group, where some relations between dihedral angles andlengths of the polyhedron and the volume of the polyhedron were obtained for this particularcase.

9

1

2

3

4

5

6 7

8

∞mα3

α1

α4

α2α1 α2 α3 α4α4

1 234

5 678

∞

∞

Figure 4. The polyhedron L(α1, α2, α3, α4).

4.2. Relations between angles and lengths. Let us denote by `1, `2, `3, and `4 thelengths of the edges corresponding to the essential dihedral angles α1, α2, α3, and α4 respec-tively. We call these lengths essential. In this subsection we will find relations between theessential angles and lengths of the polyhedron.

Consider two adjacent quadrilaterals of L(α1, α2, α3, α4) pictured in Fig. 6. One has anglesαi, 0, π/2, π/2 and another – angles αi+1, 0, π/2, π/2. Denote the length of the commonedge by xi.

`i

`i+1

xiαi

αi+1

0

0

Figure 5. Adjacent quadrilaterals.

From the equations for hyperbolic quadrilaterals [6] we havecos 0 = − cosαi+1 cosh `i + sinαi+1 sinh `i sinhxi,

cos 0 = − cosαi cosh `i+1 + sinαi sinh `i+1 sinhxi.

Hencesinαi+1 sinh `i

1 + cosαi+1 cosh `i=

sinαi sinh `i+1

1 + cosαi cosh `i+1

.

Using the identity

(sinα sinh `)2 + (1 + cosα cosh `)2 = (cosα + cosh `)2,

we get

(8)cosαi+1 + cosh `i

1 + cosαi+1 cosh `i=

cosαi + cosh `i+1

1 + cosαi cosh `i+1

.

10

Note that cosαi+1 + cosh `i > cosαi+1 + 1 > 0 and

1 + cosαi+1 cosh `i = sinαi+1 sinh `i sinhxi > 0.

Analogously, cosαi + cosh `i+1 > 0 and 1 + cosαi cosh `i+1 > 0. Rearranging (8) we obtain

1− cosh `i cosαicosh `i − cosαi

=1− cosh `i+1 cosαi+1

cosh `i+1 − cosαi+1

that holds for i = 1, . . . , 4 with suffices taken by mod 4. Denoting this ratio by

(9) v =1− cosh `i cosαicosh `i − cosαi

we get

(10) cosh `i =1 + v cosαiv + cosαi

and

cosαi =1− v cosh `icosh `i − v

.

Suppose that the faces of L(α1, α2, α3, α4) are numbered as in Fig. 5. Denote by ρ(i, j)the distance between faces numbered by i and j. Let us introduce notations v = cosh ρ(1, 7),w = cosh ρ(2, 8), t = cosh ρ(3, 5), s = cosh ρ(4, 6). Let G be the Gram matrix of thepolyhedron L(α1, α2, α3, α4), i.e. the 8× 8-matrix:

1 0 −1 0 0 − cosα4 −v − cosh `10 1 0 −1 − cosh `4 0 − cosα3 −w−1 0 1 0 −t − cosh `3 0 − cosα2

0 −1 0 1 − cosα1 −s − cosh `2 00 − cosh `4 −t − cosα1 1 0 −1 0

− cosα4 0 − cosα3 −s 0 1 0 −1−v − cosα3 0 − cosh `2 −1 0 1 0

− cosh `1 −w − cosα2 0 0 −1 0 1

By [1] the rank of G is at most four. Denote by G(i1, i2, i3, i4, i5), the main minor of G,

formed by lines and columns with numbers i1, i2, i3, i4, i5. The vanishing of the determinantsof the 5× 5–minors of G gives us relations between the entries of G. Taking the minors cor-responding to the columns (1, 2, 4, 5, 6), (1, 2, 3, 5, 6), (3, 4, 5, 7, 8), (3, 4, 6, 7, 8), respectively,we obtain following four equations:

s2(cosh2 `4 − 1) = (cosh `4 + cosα1)2(1− cos2 α4),(11)

t2(1− cos2 α4) = (cosh `3 + cosα2)2(cosh2 `4 − 1),(12)

t2(cosh2 `2 − 1) = (cosh `2 + cosα1)2(1− cos2 α2),(13)

s2(1− cos2 α2) = (cosh `3 + cosα2)2(cosh2 `2 − 1).(14)

Compare the product of left sides of (11) and (12) with the product of right sides of thesame equations and, analogously, compare the product of left sides of (13) and (14) with theproduct of right sides of the same equations:

s t = (cosh `3 + cosα4)(cosh `4 + cosα1),

s t = (cosh `2 + cosα1)(cosh `3 + cosα2).

Hence

(15) (cosh `3 + cosα4)(cosh `4 + cosα1) = (cosh `2 + cosα1)(cosh `3 + cosα2)11

Substituting expressions (10) for `i in terms of v and αi, we have(1 + v cosα3

v + cosα3

+ cosα4

)(1 + v cosα4

v + cosα4

+ cosα1

)=

(1 + v cosα2

v + cosα2

+ cosα1

)(1 + v cosα3

v + cosα3

+ cosα2

).

Hence

(16) (1− v2)2 =4∏i=1

(v + cosαi).

Denote α = (α1, α2, α3, α4). Since cosh `i − cosαi > 0, the function v = v(α) definedby (9) is continuous in the region U = α = (α1, α2, α3, α4) : 0 < αi < π, i = 1, . . . , 4.Note that if α ∈ U then v(α) 6= 0 and v(α) 6= 1. Indeed, if v(α) = 0 then from (16) we get1 =

∏4i=1 cosαi and so α 6∈ U . Similar, if v(α) = 1 then from (16) we get 0 =

∏4i=1(1+cosαi)

and again α 6∈ U . For particular value α = (π/2, π/2, π/2, π/2) we get v(α) = 1/ cosh `i < 1and (1 − v(α)2)2 = 1, whence v(α) = 1/

√2 ∈ (0; 1). By continuity, we conclude that

v(α) ∈ (0; 1) for all α ∈ U .Thus we have

Proposition 4.1. The essential angles and lengths of the hyperbolic polyhedron L(α1, α2, α3, α4)satisfy the following relation for all i = 1, . . . , 4:

(17)1− cosh `i cosαicosh `i − cosαi

= v,

where v, 0 < v < 1, is a root of the equation (16).

Taking into account a new principal parameter

T 2 =1 + v

1− v=

tan2(αi/2)

tanh2(`i/2),

we obtain the following two propositions.

Proposition 4.2 (Tangent Rule). The essential angles and lengths of the hyperbolic poly-hedron L(α1, α2, α3, α4) satisfy the following relations:

(18)tan(α1/2)

tanh(`1/2)=

tan(α2/2)

tanh(`2/2)=

tan(α3/2)

tanh(`3/2)=

tan(α4/2)

tanh(`4/2)= T,

where T , T > 1, is a root of the equation

(19) T 4 =4∏i=1

T 2 − tan2(αi/2)

1 + tan2(αi/2).

Proposition 4.3 (Sine Rule). The essential angles and lengths of the hyperbolic polyhedronL(α1, α2, α3, α4) satisfy the following relation:

(20)sinα1

sinh `1

· sinα2

sinh `2

· sinα3

sinh `3

· sinα4

sinh `4

= 1.

Proof. Using (18) we have

T 2 − tan2(αi/2)

T (1 + tan2(αi/2))=

sinαisinh `i

.

Hence (19) can be rewritten in the form (20). 12

Remark 4.4. To simplify notations, let us denote Ai = tan(αi/2) for i = 1, . . . , 4. Theequation (19) is equivalent to the equation (T 2 + 1)Q(T 2) = 0, where Q(x) is a cubicpolynomial

Q(x) = x3 −

(1 +

4∑i=1

A2i

)x2 −

( ∑1≤i<j<k≤4

A2iA

2jA

2k +

4∏i=1

A2i

)x+

4∏i=1

A2i .

Since Q(0) > 0 and Q(1) < 0, there are roots x1, x2, and x3 of Q(x) = 0 such that x1 < 0,0 < x2 < 1, and x3 > 1. Therefore, the equation (19) has only one root satisfying T > 1.Similar to Remark 3.3 we observe that T is the largest real root of the equation Q(x2) = 0and Q(x2) is positive if x2 tends to infinity. Hence Q(x2) > 0 for any x > T . Straightforwardcalculations shows that

Q(

tan2 αi2

)= − tan4 αi

2·(

1 + tan2 αi+1

2

)·(

1 + tan2 αi+2

2

)·(

1 + tan2 αi+3

2

)< 0,

that implies T > tan(αi/2) for i = 1, . . . , 4.

Remark 4.5. The equation (16) is equivalent to Q(x) = 0 under the substituting v =(x − 1)/(x + 1). Hence, there is only one root v of (16) satisfying 0 < v < 1. If T is thelargest real root of Q(T 2) = 0 then v = (T 2 − 1)/(T 2 + 1) and

v + cosαi =T 2 − 1

T 2 + 1+

1− tan2(αi/2)

1 + tan2(αi/2)=

2 (T 2 − tan2(αi/2))

(T 2 + 1)(1 + tan2(αi/2)).

Therefore, v + cosαi > 0 for i = 1, . . . , 4.

4.3. The volume.

Theorem 4.6. The volume of the hyperbolic polyhedron L(α1, α2, α3, α4), with 0 < αi < π,i = 1, . . . , 4, is given by formulae (21), (22) and (23):

(21) volL(α1, α2, α3, α4) =1

2

∫ 1

v

log

[1

(1− ν2)2

4∏i=1

(ν + cosαi)

]dν√

1− ν2,

where v satisfying 0 < v < 1 is a root of equation (16);

(22) volL(α1, α2, α3, α4) =

∫ ∞T

log

[1

t4

4∏i=1

t2 − tan2(αi/2)

1 + tan2(αi/2)

]dt

1 + t2,

where T satisfying T > 1 is a root of equation (19);

(23) volL(α1, α2, α3, α4) =4∑i=1

∆(αi/2, θ) − 2∆(0, θ)− 2∆(π/2, θ),

where θ, 0 < θ < π/2, is such that tan θ = T .

Proof. Formulae (21), (22) and (23) give us different expressions for the volume of the samehyperbolic polyhedra depending on the choice of a principal parameter. Formula (22) corre-sponds to the case where the principal parameter T is taken from the Tangent Rule (Propo-sition 4.2). The method to obtain the volume formula from the Tangent Rule is the sameas in the proof of Theorem 3.5. Obviously, formula (22) arises under the same method withslight modifications. Formula (21) corresponds to the case where the principal parameter vis taken from the relation between essential angles and lengths described in Proposition 4.1.Here we would like to demonstrate how to obtain the volume formula using this kind ofrelations between essential parameters.

13

Let V = V (α1, α2, α3, α4) = volL(α1, α2, α3, α4) be the volume function. To find V wewill use the Schlafli variation formula. We will need to integrate the differential form

ω = dV = −1

2

4∑i=1

`idαi

which is defined for 0 < α1, α2, α3, α4 < π. In order to do this we will find an extendeddifferential form Ω of five independent variables v and αi, i = 1, . . . , 4:

Ω = −1

2

(4∑i=1

`idαi + ` dv

),

where `i = arccosh1 + v cosαiv + cosαi

, i = 1, . . . , 4, with ` = `(v, α1, α2, α3, α4). Here v plays a role

of a principal parameter. We look for Ω satisfying the following properties:(10) Ω is smooth and exact in the region

U = (v, α1, . . . , α4) : 0 < v < 1, 0 < αi < π, v + cosαi > 0, i = 1, . . . , 4;

(20) Ω = ω for all (v, α1, . . . , α4) ∈ U satisfying equation (16).

Since Ω is supposed to be exact, fixing i we have

∂`

∂αi=∂`i∂v

=∂

∂v

(arccosh

1 + v cosαiv + cosαi

)=

− sinαi

(v + cosαi)√

1− v2.

Hence

` =

∫− sinαi dαi

(v + cosαi)√

1− v2=

1√1− v2

log(v + cosαi) + C,

where C is depended only of of v and αj, j 6= i. Therefore,

` =1√

1− v2log

(1

(1− v2)2

4∏i=1

(v + cosαi)

)We note that in region U equation (16) is equivalent to the following condition:

1

(1− v2)2

4∏i=1

(v + cosαi) = 1.

If this condition is satisfied, we have ` = 0 and consequently Ω = ω.By the Schlafli formula the volume function V satisfies the following system of differential

equations∂V

∂αi= −`i

2, i = 1, . . . , 4. Moreover, it was shown in [18] that if α2 = α4 = π/2

and α1 = α3 → π then polyhedron L(α1, α2, α3, α4) is collapsing to a subset of a hyperbolicplane. Hence, V → 0 as α2 = α4 = π/2 and α1 = α3 → π.

Let us consider the function

W =1

2

∫ 1

v

log

[1

(1− ν2)2

4∏i=1

(ν + cosαi)

]dν√

1− ν2.

Taking into account equations (10) and (16) by the Leibniz rule we obtain

∂W

∂αi=

1

2

∫ 1

v

∂

∂αi

(log

[1

(1− ν2)2

4∏i=1

(ν + cosαi)

])dν√

1− ν2

=1

2

∫ 1

v

− sinαiν + cosαi

dν√1− ν2

= −1

2arccosh

1 + v cosαiv + cosαi

= −`i2.

14

We note that if v is a root of (16) satisfying 0 < v < 1 then v → 1 as α2 = α4 = π/2 andα1 = α3 → π. Then the condition W → 0 follows from convergence of the integral. Thus,W satisfies that same conditions as V , whence W = V .

Formula (22) follows from (21) by the substitution ν = (t2− 1)/(t2 + 1), and formula (23)is a consequence of (22) in virtue of Lemma 3.4.

Remark 4.7. Results of numerical calculation of volumes of hyperbolic polyhedraL(α1, α2, α3, α4) by Theorem 4.6 coincide with those obtained by making use of computerprogram Orb created by C. Hodgson and D. Heard [23].

5. The case n ≥ 5

5.1. The existence. Let us consider for n ≥ 5 a polyhedron L(α1, α2, . . . , αn) with n-gonaltop and n-gonal bottom and with two levels of pentagons as a lateral surface (see Fig. 7 forthe projection of L(α1, α2, . . . , α5)). We will assume that essential dihedral angles equal toαi, 0 < αi < π, i = 1, . . . , n, while all other dihedral angles equal to π/2.

It follows from the Rivin theorem that L(α1, α2, . . . , αn) can be realized in H3. Let usdemonstrate arguments of the proof in the case n = 5. Combinatorially, the Gaussian image

α1

α2

α3

α4

α5

α1

α1

α2

α2

α3

α3α4

α4

α5

α5

α1 α2

α3

α4

α5

Figure 6. The polyhedron L(α1, α2, α3, α4, α5) and its Gaussian image.

G(L(α1, α2, . . . , α5)) is an icosahedron with five edges of lengths αi = π − αi, i = 1, . . . , 5,while all other edges of length π/2. It has twelve cone points: two for each cone angle4π/2 + (π − αi) = 2π + (π − αi), i = 1, . . . , 5 and two with angles 5π/2. Therefore, at eachcone point the sum of angles is greater than 2π. Moreover, lengths of closed geodesics aregreater than 2π, because any geodesic consists of at least five segments such that length ofat least four of them is equal to π/2. Therefore, L(α1, α2, . . . , α5) is hyperbolic polyhedronif and only if all of its essential angles are less than π/2.

If all αi → π then the polyhedron collapse to a right-angled prism in H2 × R1 which hasthe right-angled regular n-gon as the bottom.

Below we will consider the symmetrical case: all essential dihedral angles are equal to α.Denote the length of an edge corresponding to α by `α. For such a polyhedron with a cyclicsymmetry of order n we will use notation Ln(α) = L(α, α, . . . , α).

5.2. Relations between angles and lengths. Since Ln(α) has a cyclic symmetry of ordern, it is enough to consider a local fragment of it. Let us denote by Π0 and Σ0 the bottomplane and the top plane of Ln(α) respectively, and by Π1, Π2, Σ1, Σ2 its lateral planes as inFig. 8. Let O and O′ be centers on the bottom and the top, respectively, which are right-angled regular n-gons. Let OB be the line perpendicular to the side of the bottom polygon,and OA be the line passing through the a vertex of the bottom polygon. Obviously, theangle between OA and OB is equal to ν = π/n.

15

α α α α αΠ1

Σ1

Π2

Σ2

Σ0

Π0A B

Figure 7. The polyhedron L5(α)

Let k and m be unit vectors in Π0 such that k is normal to OB, m is normal to OA,and they are oriented “outside” from the triangle AOB: (k,m) = − cos ν. Let p1, p2, s1, s2

be unit vectors normal to Π1, Π2, Σ1, Σ2, respectively. Since dihedral angles (if planes areintersecting) and distances (if plane are diverging) between pairs of planes Π1, Π2, Σ1 andΣ2 are known, we have

(p1, p2) = 0, (s1, p1) = 0, (s2, p1) = − cosh `α, (s1, p2) = − cosα, (s1, s2) = 0.

Since Ln(α) has cyclic symmetry of order n, the plane Σ2 is the image of the plane Σ1

under the rotation ρ about OO′ on the angle 2ν = 2π/n. Obviously, ρ can be presented asa composition of the reflection in the plane orthogonal to k with the reflection in the planeorthogonal to m.

According to a general theory of reflections in vector spaces [1], the reflection rµ in theplane orthogonal to a vector µ can be presented in the form

rµ(x) = x− 2(x, µ) · µ.Therefore,

ρ(x) = rm(rk(x)) = x− 2(x, k) · k − 2(x− 2(x, k) · k,m) ·m.Since Σ2 = ρ(Σ1), we get:

s2 = s1 − 2(s1, k) · k − 2(s1 − 2(s1, k) · k,m) ·m= s1 − 2(s1, k) · k − 2(s1,m) ·m+ 4(s1, k)(k,m) ·m.

Using this equality, we can represent the scalar product of vectors s2 and p1 as the following:

(s2, p1) = (s1, p1)− 2(s1, k)(k, p1)− 2(s1,m)(m, p1) + 4(s1, k)(k,m)(m, p1).

Since (s1, p1) = 0 and (k, p1) = 0, we conclude:

(s2, p1) = −2(s1,m)(m, p1) + 4(s1, k)(k,m)(m, p1).

Recalling that (k,m) = − cos ν, (m, p1) = − cos(π/4) = −1/√

2, and (s2, p1) = − cosh `α,we obtain

− cosh `α =√

2 · (s1,m) + 2√

2 · cos ν · (s1, k).

Representing m = (p2 − p1)/√

2 and using (s1, p2) = − cosα and (s1, p1) = 0, we get

(24) cosh `α = cosα− 2√

2 cos ν · (s1, k).

To find (s1, k) let us take the scalar product of vectors s2 and s1, where s2 is presented asthe linear combinations of s1, k, and m, we have

(s2, s1) = (s1, s1)− 2(s1, k)(k, s1)− 2(s1,m)(m, s1) + 4(s1, k)(k,m)(m, s1).

Since (s2, s1) = 0, (s1, s1) = 1, (k,m) = − cos ν, and (s1,m) = − 1√2

cosα, we obtain

2(s1, k)2 − 2√

2 · cosα · cos ν · (s1, k) + cos2 α− 1 = 0.16

Solving this quadratic equation we have

(s1, k) =1√2

(cosα cos ν ±

√cos2 α cos2 ν + sin2 α

),

and by (24):

cosh `α = cosα− 2 cosα cos2 ν ∓ 2 cos ν√

cos2 α cos2 ν + sin2 α

= − cosα cos(2ν)∓ 2 cos ν√

cos2 α cos2 ν + sin2 α

To guarantee cosh `α > 1 we have to choose “+” for the square root. Thus, we obtained

Proposition 5.1. The essential angles and lengths of the hyperbolic polyhedron Ln(α), n >5, with 0 < α < π, satisfy the following relation:

(25) cosh `α = − cosα cos(2π/n) + 2 cos(π/n)√

cos2 α cos2(π/n) + sin2 α.

In particular, if α = π/2 then cosh `α = 2 cos(π/n).The formula (25) also holds for n = 4 with 0 < α < π and for n = 3 with 0 < α < π/2.

5.3. The volume formula. Substituting the expression for the length `α into the Schlafliformula, we will obtain the expression for the volume of the polyhedron.

Theorem 5.2. The volume volLn(α) of the hyperbolic polyhedron Ln(α), n > 5, where0 < α < π, is given by the following formula:

n

2

∫ π

α

arccosh

[− cosµ cos(2π/n) + 2 cos(π/n)

√cos2 µ cos2(π/n) + sin2 µ

]dµ.

Recall that the polyhedron Ln(π/2) is right-angled and is the same as the polyhedronR(n) discussed in the introduction. Polyhedra L5(π/2) = R(5) and L6(π/2) = R(6) arepresented in Fig. 3. The volume formula for Ln(π/2) was obtained in [34] in the context ofvolumes of Lobell manifolds.

Proposition 5.3. [34] The volume of the polyhedron Ln(π/2), n > 5, is given by

volLn(π/2) =n

2(2Λ(θ) + Λ (θ + π/n) + Λ (θ − π/n)− Λ (2θ − π/2)) ,

where θ = π/2− arccos 12 cos(π/n)

.

In particular, if L is the Lobell manifold obtained in [12] from eight copies of R(6) then

volL = 24 (2Λ(θ) + Λ (θ + π/6) + Λ (θ − π/6)− Λ (2θ − π/2)) ,

where θ = π/2− arccos 1√3. Hence, volL = 48.184368 . . . .

Also, it is easy to see the following asymptotic formula:

volLn(π/2) ∼ 5

4· vmax3 · n n→∞,

where vmax3 = 2Λ(π/6) = 1.014 . . . is the maximal volume of a hyperbolic 3-simplex, that is,the volume of a regular ideal tetrahedron in H3.

References

[1] D.V. Alekseevskij, E.B. Vinberg and A.S. Solodovnikov, Geometry of spaces of constant curvature. In:Geometry, II, 1–138, Encyclopaedia Math. Sci., 29, Springer, Berlin, 1993.

[2] O. Antolin-Camarena, G.R. Maloney and R.K.W. Roeder, Computing arithmetic invariants for hyper-bolic reflection groups. In: Complex Dynamics: Families and Friends (D. Schleicher ed.), 597–631, A.K.Peters, 2008.

[3] D. Derevnin and A. Mednykh, The volume of the Lambert cube in spherical space. Math. Notes 86(1-2)(2009) 176–186.

17

[4] D. Derevnin, A. Mednykh and M. Mulazzani, Volumes for twist link cone-manifolds. Bol. Soc. Mat.Mexicana (3) 10 (2004), 129–146.

[5] D.A. Derevnin, A.D. Mednykh and M.G. Pashkevich, On the volume of symmetric tetrahedron inhyperbolic and spherical spaces. Siberian Math. J. 45 (2004), 840–848.

[6] W. Fenchel, Elementary Geometry in Hyperbolic Space, de Gruyter, 1989.[7] A. Garrison and R. Scott, Small covers of the dodecahedron and the 120-cell. Proc. Amer. Math. Soc.

131(3) (2003), 963–971.[8] C.D. Hodgson, I. Rivin, W.D. Smith, A characterization of convex hyperbolic polyhedra and of convex

polyhedra inscribed in the sphere. Bull. Amer. Math. Soc. 27(2) (1992), 246–251.[9] T. Inoue, Organizing volumes of right-angled hyperbolic polyhedra. Algebraic & Geometric Topology 8

(2008), 1523–1565.[10] R. Kellerhals, On the volume of hyperbolic polyhedra. Math. Ann. 285 (1989), 541–569.[11] F. Klein, Vorlesungen uber Nicht-Euklidische Geometrie, Chelsea Publishing Company, New York.[12] F. Lobell, Beispiele geschlossene dreidimensionaler Clifford–Kleinischer Raume negative Krummung.

Ber. Verh. Sachs. Akad. Lpz., Math.-Phys. Kl. 83 (1931), 168–174.[13] S. Matveev, C. Petronio and A. Vesnin, Two-sided complexity bounds for Lobell manifolds. Doklady

Mathematics 76(2) (2007), 689–691.[14] S. Matveev, C. Petronio and A. Vesnin, Two-sided asymptotic bounds for the complexity of some closed

hyperbolic three-manifolds. J. Australian Math. Soc. 86(2) (2009), 205–219.[15] A. Mednykh, Automorphism groups of three-dimensional hyperbolic manifolds. Sov. Math. Dokl. 32

(1985), 633–636.[16] A.D. Mednykh, Automorphism groups of three-dimensional hyperbolic manifolds. Amer. Math. Soc.

Transl. (2) 151 (1992) 107–119.[17] A. Mednykh, On hyperbolic and spherical volumes for knot and link cone-manifolds. In: Kleinian

Groups and Hyperbolic 3-Manifolds (Y. Komori, V. Markovic, and C. Series, eds.), Lond. Math. Soc.Lec. Notes 299, Cambridge Univ. Press, 2003, 145–165.

[18] A. Mednykh, J. Parker and A. Vesnin, On hyperbolic polyhedra arising as convex cores of quasi-Fuchsianpunctured torus groups. Bol. Soc. Mat. Mexicana (3) 10 (2004), 357–381.

[19] A. Mednykh and A. Vesnin, On three-dimensional hyperbolic manifolds of Lobell type. In: Complexanalysis and applications’85 (Varna, 1985), 440–446, Publ. House Bulgar. Acad. Sci., Sofia, 1986.

[20] A. Mednykh and A. Vesnin, On the volume of hyperbolic Whitehead link cone-manifolds. SCIENTIA,Series A: Mathematical Science, 8 (2002), 1–11.

[21] A. Mednykh and A. Vesnin, Colourings of polyhedra and hyperelliptic 3-manifolds. In: Recent advancesin group theory and low-dimensional topology (Pusan, 2000), 123–131, Res. Exp. Math., 27, Heldermann,Lemgo, 2003.

[22] J. Milnor, The Schlafli differential equality. In: J. Milnor, Collected papers, Vol. 1, Geometry, Publishor Perish, Houston, 1994, 281–295.

[23] Orb, computer program available from www.ms.unimelb.edu.au/˜snap/orb.html.[24] M. Pashkevich, Spontaneous surgery on the Borromean rings. Siberian Math. J. 44(4) (2003), 645–658.[25] L. Potyagailo and E. Vinberg, On right-angled reflection groups in hyperbolic spaces. Comment. Math.

Helv. 80(1) (2005), 63–73.[26] J. Ratcliffe, Foundations of hyperbolic manifolds. Second edition. Graduate Texts in Mathematics, 149.

Springer, New York, 2006. xii+779 pp.[27] I. Rivin, A characterization of ideal polyhedra in hyperbolic 3-space. Ann. of Math. (2) 143(1) (1996),

51–70.[28] R.K.W. Roeder, Constructing hyperbolic polyhedra using Newton’s Method. Experimental Mathematics

16(4) (2007), 463–492.

[29] T. Salenius, Uber dreidimensionale geschlossene Raume konstanter negative Krummung. In: Den 11-teSkandinavske Matematikerkongress, Trondheim, 1949, 107–112.

[30] H. Seifer and C. Weber, Die beiden Dodekaederraume. Math. Z. 37 (1933), 237–253.[31] D. Surchat, Infinite de valeurs propres sous le spectre essentiel du Laplacien d’un graphe, Presetee au

Department de Mathematiques Ecole Polytechnique Federale de Lausanne pour l’obtention du grade dedocteur es sciences, Lausanne, EPFL 1993, 103 pp.

[32] M. Takahashi, On the concrete construction of hyperbolic structures of 3-manifolds. Tsukuba J. Math.textbf9(1) (1985), 41–83.

[33] A. Vesnin, Three-dimensional hyperbolic manifolds of Lobell type. Siberian Math. J. 28(5) (1987),731–734.

[34] A. Vesnin, Volumes of three-dimensional hyperbolic Lobell manifolds. Math. Notes 64(1) (1998), 15–19.

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[35] A. Vesnin and A. Mednykh, Three-dimensional hyperelliptic manifolds and Hamiltonian graphs. SiberianMath. J. 40(4) (1999), 628–643.

[36] E.B. Vinberg, The volume of polyhedra on a sphere and in Lobachevsky space. In: Algebra and analysis(Kemerovo, 1988), 15–27, Amer. Math. Soc. Transl. Ser. 2, 148, Amer. Math. Soc., Providence, RI,1991.

[37] E.B. Vinberg and O.V. Shvartsman, Discrete groups of motions of spaces of constant curvature. In:Geometry, II 139–248, Encyclopaedia Math. Sci., 29, Springer, Berlin, 1993.

[38] E.B. Vinberg, Volumes of non-Euclidean polyhedra. Russian Math. Surveys 48(2) (1993), 15–45.

EPFL SB IGAT GEOM, Station 8 - Batiment MA, CH-1015 Lausanne, SwitzerlandE-mail address: peter.buser@epfl.ch

Sobolev Institute of Mathematics, Novosibirsk 630090, RussiaE-mail address: mednykh@math.nsc.ru

Sobolev Institute of Mathematics, Novosibirsk 630090, Russia, and Omsk State TechnicalUniversity, Omsk 644050, Russia,

E-mail address: vesnin@math.nsc.ru

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