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Coxeter Decompositions of Hyperbolic Tetrahedra

a r X i v :m a t h /0212010v 1 [m a t h .M G ] 1 D e c 2002

Coxeter Decompositions of Hyperbolic

Tetrahedra.

A.Felikson

Abstract .In this paper,we classify Coxeter decompositions of hyperbolic tetra-hedra,i.e.simplices in the hyperbolic space I H 3.The paper together with [2]completes the classi?cation of Coxeter decompositions of hyperbolic simplices.

1Introduction

De?nition 1.A convex polyhedron in a space of constant curvature is called a Coxeter polyhedron if all dihedral angles of this polyhedron are the in-teger parts of π.

De?nition 2.A Coxeter decomposition of a convex polyhedron P is a decomposition of P into ?nitely many tiles such that each tile is a Coxeter polyhedron and any two tiles having a common facet are symmetric with respect to this facet.

Coxeter decompositions of hyperbolic triangles were studied in [4],[5],[6],[7]and [1].Coxeter decompositions of hyperbolic simplices of the dimension greater than three were classi?ed in [2].In this paper,we classify Coxeter de-compositions of hyperbolic tetrahedra,i.e.simplices in the hyperbolic space I H 3.The paper completes the classi?cation of the Coxeter decompositions of hyperbolic simplices.

The work was partially written during the stay at the University of Biele-feld.The author is grateful to the hospitality of this University.The author would like to thank O.V.Schwarzman and E.B.Vinberg for their interest to the work.

1.1De?nitions

The tiles in De?nition 2are called fundamental polyhedra .Clearly,any two fundamental polyhedra are congruent to each other.A hyperplane αcon-taining a facet of a fundamental polyhedron is called a mirror if αcontains no facet of P .

De?nition 3.Given a Coxeter decomposition of a polyhedron P ,a dihedral angle of P formed up by facets αand βis called fundamental if no mirror contains α∩β.A vertex A is called fundamental if no mirror contains A .

Notation.

P is a tetrahedron equipped with a Coxeter decomposition;

F is a fundamental polyhedron considered up to an isometry of I H3;

Σ(T)is a Coxeter diagram of a Coxeter tetrahedron T;

N is the number of the fundamental polyhedra inside P.

A Coxeter tetrahedron T can be represented by its Coxeter diagramΣ(T): the nodes v i ofΣ(T)correspond to the facets f i of T,two nodes v i and v j are connected by a k-fold edge if the dihedral angle formed up by f i and f j equalsπ

Lemma1.Let P be a simplex in I H n admitting a Coxeter decomposition of the second type with fundamental polyhedron F.Then the following properties are hold:

1.F is a simplex.

2.If all the vertices of P are fundamental,then N≥2n.

3.Volume property.

V ol(P)

Proof.Suppose that any vertex of P belongs to a unique fundamental tetra-hedron.Since each fundamental tetrahedron has an ideal vertex,N ≤4.This contradicts to the second part of Lemma 1.

To prove the following lemma we need a list of Coxeter decompositions of Euclidean and spherical triangles p such that all the angles of p are fun-damental and N ≤24.See Fig.1for the list.

Figure 1:Decompositions of spherical and Euclidean triangles without non-fundamental angles (N ≤24in Euclidean case)

915

The decompositions 1a–1e are Euclidean,the decomposition 1f is spherical.The numbers of fundamental triangles are written under the decompositions.

Lemma 4.Suppose that any dihedral angle of P is fundamental.Then the decomposition is one of two decompositions shown in Fig.4.

Proof.By Lemma 2the tetrahedron P is unbounded.By Lemma 3P has a non-fundamental vertex.Let A be a non-fundamental vertex of P .

Suppose that A is not an ideal vertex.Consider a small sphere s cen-tered in A .The section of the decomposition by the sphere s is a Coxeter

decomposition of a spherical triangle p =s

P .Clearly,all angles of p are fundamental.The only such a decomposition consists of 15triangles (see Fig.1f).Hence,the number of tiles in P cannot be less than 15.The funda-mental triangle in the decomposition shown in Fig.1f has an angle π

.But there is no pair of unbounded hyperbolic tetrahedra F and P such that V ol (P )

.Therefore,any non-fundamental vertex should be ideal.

Let A be an ideal non-fundamental vertex of P .Consider a small horo-sphere centered in A .The section of the decomposition by this horosphere is a Coxeter decomposition of a Euclidean triangle p without non-fundamental angles.Since the maximal ratio of volumes of unbounded hyperbolic tetra-hedra is 24,the number of triangular tiles f in p cannot be greater than 24.

Therefore,the decomposition of the triangle p is one of the decompositions shown in Fig.1a–1e.Consider these decompositions of p .

1)Suppose that p is decomposed as shown in Fig.1a–1c.Then all the angles of the triangles f and p are equal to π

,π3).All hyperbolic Coxeter tetrahedra

satisfying this condition are shown in Fig.2.The tetrahedra H 24and H 32cannot be fundamental for a Coxeter decomposition of the second type.Indeed,for any Coxeter tetrahedron T we have V ol (T )

V ol (H 32)

<2.The maximal ratio of volumes of the tetrahedra shown

in Fig.2a–2c is 12.Therefore,p cannot be decomposed as the triangle shown in Fig.1c.Thus,F is one of the tetrahedra shown in Fig.2a,P is one of the tetrahedra shown in Fig.2a–2c and p is decomposed as shown in Fig.1a or 1b.

H 12,H 17,H 22,H 26H 24

H 32

k =3,4,5,6?

Figure 2:Diagrams of tetrahedra with subdiagram corresponding to the triangle (π3,π

,since Σ(P )=Σ(F ).Con-sider the set of fundamental tetrahedra having a common ideal vertex whose neighborhood is decomposed as shown in Fig.1a–1c.Each of these tetrahedra has its own edge with dihedral angle π

,where l =4,5,6.Thus,H 12can tessellate only the tetrahedra H 24and H 32.

These tessellations are the decompositions shown in Fig.4.

2)Suppose that p is decomposed as shown in Fig.1d.Then the number of

fundamental tetrahedra in the decomposition cannot be smaller than nine.Further,each of the Coxeter diagramsΣ(F)andΣ(P)has a subdiagram corresponding to a triangle with angles(π

,π

V ol(F)=12

and9+5=14>12.

Figure3:The triangles shaded by distinct ways belong to distinct planes.

These planes intersect each other and form up dihedral angle2π

3,π

)andΣ(P)has a subdiagram corresponding

to a triangle with angles(π

,π

3,the neighborhood of any vertex of P

is decomposed as shown in Fig.1e.Then N≥18·4>24(F has a unique ideal vertex).Since V ol(P)

smallest tetrahedron with the decomposition of the third type,that is for any tetrahedron P′inside P the restriction ofΘ(P)onto P′is a decomposition either of the?rst or of the second type.

LetΘ1andΘ2be decompositions of the?rst type andΘ(P)be a su-perposition ofΘ1andΘ2.It follows from the construction of the inductive algorithm thatΘ(P)is a decomposition of the?rst type.The contradic-tion shows that at least one ofΘ1andΘ2is a decomposition of the second type,i.e.at least one of two decompositions shown in Fig.4.A superposi-tion of any of these two decompositions with any other decomposition is a decomposition of the?rst type.

Tables

The tetrahedra with dihedral angles k iπ

i=1,...,6and subdivide the edges corresponding to the q i

dihedral angles k iπ

Table 1:Hyperbolic Coxeter tetrahedra.The volumes are reprinted from [3].

Notation H 9

H 8

H 7H 6

H 5H 4

H 3H 25

545

Volume

diagram ???

?0.5021308905

0.3586534401

0.2222287320

0.2052887885

0.0933*******.0857*******.0717*******.03905028560.0358850633?

????

???H 1Coxeter

Notation Volume

diagram

Coxeter 5

4444

0.3430033226

0.3053218647H 22

?H 21

0.25373540160.22899139850.21144616800.21144616800.17150166130.16915693440.152********.10572308400.08457846720.0763*******.0422892336H 206

?H 19????6?

??H 186

H 17

H 16H 156

???

H 14

H 136

??H 12

H 11??

???H 10

Notation Volume diagram

Coxeter

1.014916064

0.9159655942

0.8457846720

0.6729858045

0.5562821156

0.5258402692

0.5074708032

0.4579827971

0.4228923360

0.3641071004

H32

H31

H30

H29

H28

H27

H26

H25

H24

H23

Table 2:Decompositions of bounded tetrahedra of the ?rst type.

s s s s

s s s s 1

(2,3;0,0,0,0)

s s s s 2(2,3;0,0,3,3)

s s s s 3

(4,4;1,1,2,2)

¨ §

s s s s

(3,4;1,0,0,1)

|? §

¨ |? §

¨ ￥

s s s s

(6,5;2,2,1,1)

￥

s s s s 1

(2,3;0,0,1,1)

s s s s

(2,3;0,0,2,2)

¤ ?

s s s s

(4,4;2,2,0,0)

s s s s

(2,3;0,0,0,0)

¨#

s s s s

4(4,4;1,1,2,2)

¨#

s s s s

9(8,6;5,5,2,2)

¨§ ¨§

¨ ¤ ?

s s s s 14(12,6;8,8,1,1)

s s s s 15

(16,7;9,9,2,2)

¨ ¤#

? ￥¨# s s s s 17

(20,7;12,11,0,0)

¨ ¤

# ?|s s s s 18

(24,7;14,14,1,1)

¨§# ?|s s s s

19(24,8;16,16,1,1)

¨#

?|2

s s s s

? s s s s

4(4,4;1,1,2,2)

¨ ?

s s s s

(6,5;3,3,0,0)

¨ ¨

s s s s

(12,6;6,6,0,0)

¨ ?

s s s s 0

s s s s

(2,3;0,0,3,3)

¨ ¤

s s s s

(2,3;0,0,0,0)

¨ ?

s s s s

(2,3;0,0,2,2)

¤s s s s 2(3,4;1,0,0,1)

?| ¨§ ? ￥ ¨ ¤#

￥

s s s s

s s s s 1

(2,3;0,0,0,0)

s s s s 2(2,3;0,0,3,3)

s s s s

3(4,4;1,1,2,2)

¨#

s s s s

(2,3;0,0,0,0)

¨ ?

s s

s 0

s s s s

1(2,3;0,0,1,1)

¨# ¨ ¤

s s s s 2

(4,4;1,1,0,0)

¨ ?

s s s s

(2,3;0,0,0,0)

s s

¨s s s s

(2,3;0,0,0,0)

¨ ?

s s s s

(2,3;0,0,0,0)

¨# ?

Figure 4:Hyperbolic tetrahedra of the second type.

(Ideal vertices are marked by small circles.)

????

N =5P =F =?

????

N =12P =F =?b)

References

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