New York Journal of Mathematics

New York J Math

Divisible Tilings in the Hyp erb olic Plane

S Allen Broughton Dawn M Haney Lori T McKeough

and Brandy Smith Mayeld

Abstract We consider pairs in the hyp erb olic plane

whichkaleidoscopically tile the plane simultaneously In this case the tiling

by is called a divisible tiling All p ossible such divisible tilings

are classied There are a nite numb er of and parameter families as

well as a nite numb er of exceptional cases

Contents

Intro duction

Tilings and Tiling Groups

Tiling Groups

Divisible Tilings

Overview of Quadrilateral Search

Free Vertices

Constrained Vertices

The Two Search Metho ds

Direct Construction Metho d K

Construction and EliminationNo Interior Hubs

Computer Algorithm and Extension to Interior Hubs

Boundary Construction Metho d K

Geometric Quadrilateral Test

Boundary Word Test

Example Failure of to Tile

Example Successful Tiling of by

Catalogue of Divisible Tilings

App endix A with area

References

Received Octob er

Mathematics Subject Classication B H H F C M

Key words and phrases tiling Fuchsian groups reection groups crystallographic groups

hyp erb olic plane

The last three authors were supp orted byNSFgrant DMS

ISSN

S A Broughton D Haney L McKeough and B Mayeld

Intro duction

Let b e a p olygon in one of the three twodimensional the sphere



S the Euclidean plane E or the hyp erb olic plane H Supp ose also that each

interior angle of the p olygon at vertex P has measure where s is an integer

i i

s

i

The p olygon generates a tiling of the plane by rep eated reections in the sides

of the p olygon Examples are the icosahedral tiling of the sphere by

triangles in Figure and the partially shown tilings of the Euclidean plane

by triangles in Figure and the hyp erb olic plane by

triangles in Figure These tilings are called geodesic kaleidoscopic tilings since

the tiling may be generated by reections in a single tile We explain the term

o dimensional could view the tiling by geo desic so on A denizen of the tw

constructing a p olygon of mirrors meeting at the appropriate angles assuming that

lighttravels in straight lines in the geometry In the Euclidean case the mirrored

p olygons can actually b e physically constructed and the tiling viewed for the three

Euclidean kaleidoscopic triangles and the

one parameter family of kaleidoscopic

The plane maybekaleidoscopically tiled in several dierentways as Figure

shows One way is by triangles A second way is by consisting of four

triangles meeting at the center of the Yet a third way is the tiling by

squares formed from two triangles meeting along a h yp otenuse The tilings by

squares are b oth rened or sub divided bythe tiling of triangles ie each square

is a union of either four nonoverlapping triangles or two nonoverlapping triangles

in the two cases Wesay that the tiling by squares is divisible or that the tiling by

squares is subdivided by the tiling by triangles If are such a triangle and

square resp ectively we call a divisible tiling pair

There are no divisible quadrilateral tilings of the sphere There are innitely

many dierent divisible quadrilateral tilings of the Euclidean plane but they are all

found in the tiling in Figure In this pap er we turn our attention to

the muchricher case of divisible quadrilateral tilings of the hyp erb olic plane The

reader is invited to nd the tiling by quadrilaterals hidden in Figure without

cheating by lo oking at the answers in the tables in Section Each quadrilateral

has triangles

The main result of the pap er Theorem is a complete catalogue of all divis

ible quadrilateral tilings of the plane whichmay b e sub divided by a triangle tiling

For completeness we also include the less complex cases of triangle tilings which

sub divide triangle tilings Theorem and quadrilateral tilings which sub divide

quadrilateral tilings Theorem The classication of tilings of quadrilaterals by

triangles is broken up into two categories constrained and free The main dier

ence b etween the two is that the free tilings o ccur in innite families with simple

parametrizations byintegers but there are only a nite numb er of constrained di

visible tilings The complete lists of the four typ es of divisible tiling pairs describ ed

ab ove are given in Tables in Section More imp ortantly pictures of all

the various tiling pairs are given in Tables of the same section

To put our results in a broader context wesketch an application of the results

to a problem in the classication of Fuchsian groups which in turn is relevantto

the singularity structure of mo duli spaces of Riemann surfaces for certain genera

Let and be tw oFuchsian groups such that has signature l m nand

  

Divisible Tilings in the Hyperbolic Plane

has signature s t u v and let H b e the disc mo del of the hyp erb olic plane



Thus the pro jections H H and H H are branched covers of the sphere

 

branched over p oints of orders l m n and p oints of orders s t u v resp ectively

One may ask under what conditions do es The tiling p olygons and

 

generate groups and such that the conformal subgroups and



  

of index are of the typ e sp ecied in the classication problem The two





groups constructed are real ie the tiling p olygons may b e translated such that

and are real ie b oth groups are invariant under conjugation In turn this

 

implies that the p oints and corresp ond to singular p oints on the mo duli space

 

of Riemann surfaces with real dening equations for certain genera The details

of this will b e discussed in a subsequent pap er which examines divisible tilings

on surfaces The problem of determining pairs with an equal number

 

of branch points was solved in The corresp onding tiling problem is tilings

of triangles by triangles and quadrilaterals by quadrilaterals The solution of this

problem follows easily from though the classication has not b een published

to our knowledge We include b oth of these results for completeness

The remainder of this pap er is structured as follows In Section weintro duce

the necessary background on planar tilings divisible tilings and the tiling groups

In Sections and weintro duce and discuss the two computer algorithms for

determining the divisible tilings and illustrate them with some sample calculations

Finally all results including gures are listed in tables in Section Throughout

the discussion the reader is encouraged to lo ok at these gures to gain aclearer

idea of the denitions and the discussion

We will use the disc mo del for the hyp erb olic plane H in which the p oints are

in the interior of the unit disc the lines are the unit disc portions of circles and

lines p erp endicular to the b oundary of the unit disc and reections are inversions

in the circles dening the lines We will denote the hyp erb olic distance between

twopoints z and z by z z All prop erties weuseabouthyp erb olic geometry

   

in particular the area formula for p olygons may b e found in the text by Beardon

Acknow ledgments The initial part of this researchwork was conducted during the

NSFREU program at RoseHulman Institute of Technology in the summer of

Haney and McKeough and continued in Smith under the direc

tion of Allen Broughton The pro ject worked out the classication under a

restrictivehyp othesis called the corner condition and yielded of the constrained

cases A sub divided quadrilateral satises the corner condition if each corner of

the quadrilateral contains a single triangle For example cases C and C in Ta

ble satisfy this prop erty though cases C and C do not Many of the symmetry

prop erties of the tilings were examined and therefore the symmetry groups of the

tiling pairs are included in the tables in Section The group also under

to ok some of the preliminary work in determining divisible tilings on surfaces the

results of which will app ear in a forthcoming publication The pro ject

to ok a dierent approach that yielded many of the free cases with a small number

of triangles The present work com bines and extends b oth approaches to yield a

complete classication of planar hyp erb olic divisible tilings by quadrilaterals

We thank the numerous participants of the and programs for the useful

conversations and encouragement All numerical calculations were p erformed using

S A Broughton D Haney L McKeough and B Mayeld

Maple The gures were pro duced with Maple and Matlab All Maple and

Matlab scripts used as well as images are available at the tilings website



Figure Icosahedral tiling of S

Figure tiling of H

Figure tiling of E

Tilings and Tiling Groups

A tiling of the spherical Euclidean or hyp erb olic plane is a collection T of

p olygons called tiles that completely cover the plane without overlaps or gaps

The sides or edges of the tiles are called the edges of the tiling and the vertices of

the tiles are called the vertices of the tiling Let E and V denote the collection of

edges and vertices of the tiling

Divisible Tilings in the Hyperbolic Plane

Denition A tiling T of a plane is said to be a kaleidoscopic tiling if the

following condition is met

For eachedgee E of the tiling the reection r in the edge e is an isometry

e

of the plane that maps tiles to tiles In particular it interchanges the two tiles

whose common edge is e

AtilingT is called a geodesic kaleidoscopic tiling if in addition wehave

the following condition

The xed line or mirror fx S r xxg of each reection r is the union

e e

of edges of the tiling Such a line is called a line of the tiling

The tiling of the Euclidean plane by or the do decahedral tiling of the

sphere bypentagons are examples of kaleidoscopic tilings which are not geo desic

For the remainder of the pap er unless sp ecied otherwise all tilings we discuss will

satisfy Denition The following prop osition allows us to easily identify which

give rise to the desired tilings It is easily proven using the Poincare

Polygon Theorem p

Prop osition Let P P P be a ngon Then generates a kalei

  n

doscopic tiling of the plane by repeated reection in its sides only if the interior



angles at the vertices of the polygon have measure where n is an integer If in

i

n

i



addition each n is even say n m so then generates a geodesic

i i i

n m

i i

kaleidoscopic tiling

We shall call a p olygon kaleidoscopic if it generates a kaleidoscopic tiling Through

out the pap er we shall only consider kaleidoscopic tilings that generate geo desic

tilings ie the angles have the form

m

i

Notation A p olygon P P P such that the interior angle at P has radian

  n i

is called an m m m p olygon Note that m tiles meet at the measure

  n i

m

i

vertex P Hence we easily identify the tiles of the icosahedral tiling in Figure

i

as triangles

Tiling Groups The reections in the edges of a tiling generate a group of

isometries of the tiling called the tiling group We describ e this group in some

detail now for the case of a triangle The generalization to the group of a general

p olygon easily follows from the triangle discussion It is easy to show that every

tile in the plane is the image by some element of the tiling group of a single tile

called the master tile pictured in Figure The sides of the master tile are



lab eled p q andr and we denote the vertices opp osite these sides by P Q and R

resp ectively We also denote by p q andr the reection in corresp onding side W e

assume that is an l m ntriangle so that the angles at R P and Q have size



radians radians and radians resp ectivelywherel m and n are integers

l m n

see Figure At each of the vertices of the triangle the pro duct of the

two reections in the sides of the triangle meeting at the vertex is a rotation xing

the vertex The angle of rotation is twice the angle at this vertex For example

the pro duct p q a reection rst through q then through p is a counterclo ckwise



rotation through radians We will refer to this rotation as a pq and use it to

l

lab el the vertex in Figure Rotations around each of the other corners can b e

dened in the same waysothat b qr and c rp are counterclo ckwise rotations

 

through radians and radians resp ectively

m n

S A Broughton D Haney L McKeough and B Mayeld

Q c π /n

p r Δ 0 π /l π /m a b

R q P

Figure The master tile and generators of T and T

From the geometry of the master tile we can derive relations among these group

elements It is clear that since p q and r are reections the order of eachofthese

elements is

  

p q r

From the observations ab out rotations ab ove it is also clear that the orders are

given by

oal obm ocn

and

abc pq q r r p

The reections generate a group T hp q r i and the rotations generate a sub

group T ha b ci which includes only the orientationpreserving isometries in T

The subgroup T is of index in T and hence is also normal in T Here are some

wellknown basic facts ab out T and T

Prop osition Let T and T be derived from a tiling T as above Then the

fol lowing hold

The groups T and T have the fol lowing presentations

   l m n

T p q r p q r pq qr rp

Divisible Tilings in the Hyperbolic Plane

and

l m n

T a b c a b c abc

The ful l tiling group T acts simply transitively on the tiles of T

Pro of These facts are wellknown though wegive a brief pro of sketch for com

pleteness According to the PoincarePolygon Theorem p T is a Fuchsian

group ie a discrete group of isometries and a tile is a fundamental region for T

Any group element xing a tile must therefore x all the interior points of the

tile and equals the identity Thus simple transitivityisproven Now for the rst

part First construct the dual graph by joining incentres of adjacent triangles bya

segment meeting the common edge of the triangles at a right angle It is easy to

see that every word in T corresp onds to an edge path in the dual graph starting

at the incentre of the master tile Words which equal the identity corresp ond to

closed paths Now supp ose we have a word corresp onding to a closed path We

must showthatwe can reduce it to the identityby the relations ab ove Since H is

simply connected a closed edge path in the dual graph is homotopic to the identity

Using the homotopy a closed edgepath can b e deformed to the identity in a series of

moves of the following typ e aintro duce or eliminate a path that crosses a tile edge

and then go es back b replace a path whichmakes a partial clo ckwise turn around

avertex with the complimentary counterclo ckwise turn around the same vertex

  

These twotyp es of homotopies corresp ond to the relations p q r and

l m n

pq qr rp resp ectively This proves that T has no other relations

A similar pro of works for T

Because of the simple transitivity given a master tile there is a unique



isometry g T suchthatg This then allows us to identify a vertex x as



beingavertex of typ e P Q or R dep ending on which vertex it is equivalentto

in the master tile The same applies to edges Of course in scalene triangles the

vertex typ e and edge typ e are easily identied by angle measure and side length

However in the case of an it is necessary to use the tiling group

action to dene typ es Similar remarks apply to the quadrilateral with resp ect to

the corresp onding tiling groups of an s t u v quadrilateral whichwecallQ and

Q

   

w x y z w x y z

Q

s t u v

wx xy yz zw

and

s t u v

Q d e f g d e f g def g

Divisible Tilings

Denition A kaleidoscopic tiling is said to b e divisible if it can be kaleido

scopically divided into a ner tiling Thus wehavetwo tiles b oth of which

generate a kaleidoscopic tiling of the plane Each tile of the tiling is a union of

p olygons from the tiling Wesay that the tiling subdivides the tiling W e

call a divisible tiling pair

We have seen an example of a divisible hyp erb olic quadrilateral tiling in Fig

ure and all examples of divisible tilings of the hyp erb olic plane are given in the

S A Broughton D Haney L McKeough and B Mayeld

gures in Section In each of the gures we only give one quadrilateral though it

may b e easily extended to the plane through reections in the sides of the quadri

lateral In fact the following is easily proven by using reections in the sides of

Lemma Suppose that is a kaleidoscopic tile and that is a larger

polygon such that is a union of the triangles of the tiling dened by and

that the angles of have measure for various integers m Then is a

i

m

i

kaleidoscopic tiling pair

For the remainder of the pap er we shall concentrate on the case where b oth

and are triangles or quadrilaterals and the tilings are geo desic In almost

every case will b e a triangle and will b e a quadrilateral

Remark There are examples of kaleidoscopic tiling pairs where the tilings

may not b e geo desic For example the Euclidean honeycomb tiling by hexagons

may b e sub divided into a tiling by equilateral triangles

Remark For an l m ntriangle to tile an s t u v quadrilateral each

of s t u and v must b e a divisor of one of l m or n For some multiple of an angle

of must t into each corner of the quadrilateral

The number of triangles Let the l m ntriangle and the s t u v quadrilateral

with form a divisible tiling pair The areas of the triangle A and

t

quadrilateral A are given see and by

q

A l m n

t

l m n

A s t u v

q

s t u v

for some p ositive rationals l m nand s t u v Note that wemust therefore

have

and

l m n s t u v



A table of all values of p ossible l m n with l m n is given in App endix A

These are the only values we shall need for our study Now is a union of triangles

congruentto let K denote the numb er of triangles Wehave A KA or

q t

K

s t u v l m n

or alternatively

   

s t u v

s t u v

K

  

l m n

l m n

It turns out that K has an upp er b ound of for all triangles in the hyp erb olic plane

Divisible Tilings in the Hyperbolic Plane

Prop osition Suppose the l m ntriangle tiles the s t u v quadrilateral

Then

   

s t u v

s t u v

K

  

l m n

l m n

Pro of Fix l m and n To maximize K we need to maximize the area of a quadri

lateral Thus each integer s t u and v should be made as large as p ossible

Since eachof s t uand v must divide one of l morn then the largest p ossible

quadrilateral is a b b b bquadrilateral suchthatb is the largest integer selected

from l mandn Now the smallest p ossible value of l m nonthehyp erb olic



plane is for a triangle according to the table in App endix A Picking a



 

quadrilateral as suggested ab ove we get K For any other

 

 

with l m n realized for a triangle triangle wehave l m n

 

But now

   

s t u v

K

  

l m n

Hubs Let b e an arbitrary quadrilateral and let v be anyvertex of the l m n

tiling contained in the interior or on the b oundary of Assume for the moment

that v is of typ e R The collection of l m ntriangles with common vertex at v

and contained in will be called an R hub If the hub occurs at a corner in

then the number of triangles divides l If the hub o ccurs on an edge but not at

a corner then there are exactly l triangles in the hub and a hub o ccurring in the

interior of the quadrilateral has l triangles To distinguish the three typ es of hubs

we call them corner hubs edge hubs and interior hubs resp ectively Since weare

mainly concerned ab out edge hubs we just call them hubs if it will not cause any

confusion An interior hub may b e considered to b e a union of two edge hubs then

it is called a double hub The P hubs and a Qhubs are dened in the same fashion

When l m and n are all distinct we refer to the R hubs P hubs and Qhubs as

l hubsmhubs and nhubs resp ectively The various typ es of hubs are illustrated

in the gures in Table in the next section

The number of R hubs edge hubs and interior hubs with interior hubs counted

as twohubs in a sub divided quadrilateral is the number h given by

R

K c

R

h

R

l

where c is the numb er of triangles o ccurring in the corner R hubs We can clearly

R

see that h mustbeaninteger since the remaining edge hubs have l triangles the

R

interior hubs havel triangles and every triangle b elongs to a unique R hub The

number c is usually easily determined from the numbers s t u v Similar formulas

R

hold for P hubs and Qhubs

K c K c

P Q

h h

P Q

m n

S A Broughton D Haney L McKeough and B Mayeld

Overview of Quadrilateral Search

We shall employtwo dierenttyp es of search algorithms dep ending on whether

K is large or small For lowvalues of K we directly construct tilings of quadrilat

erals without worrying what the angles are For large values of K we will develop

an algorithm that starts with a sp ecic l m ntriangle and determines all

s t u v quadrilaterals that the triangles can tile To understand the rationale

of splitting the searchinto two approaches we need to dene constrained and free

vertices We will also obtain constraints and b ounds on l m n and K that are

helpful in restricting the search These bounds are imp ortant for otherwise the

computer implementation of the search is impractical

Denition Let generate a divisible tiling Then the P typ e vertices

of the tiling are called constrained if at least one b elongs to either an edge hub

or an interior hub ie h Otherwise the P typ e vertices are called free

P

Similar denitions hold for Qtyp e and R typ e vertices

Remark As the same triangle tiling can rene two dierent quadrilateral

tilings freeness is relativeto though we rarely mention

Free Vertices The freeness of vertices is exemplied in the gures in Ta

ble There weshow the rst four quadrilaterals of the innite family of d

tilings of d d d d quadrilaterals where d All the quadrilaterals have the

same divided structure The free vertices may b e freely dragged to the b oundary

of the hyp erb olic plane through an innite discrete set of p ositions In contrast

a constrained vertex cannot since it has a xed measure The angles at the free

vertices b ecome smaller and smaller as we approach the b oundary until we reach

a tiling of an quadrilateral The free vertices are on the

b oundary and have measure As weprove b elow every divisible tiling with

free vertices gives rise to such an innite family of tilings

Remark Free vertices or rather the lackof them playaspecial role in the

interpretation and calculation of the mono dromy group T core Q of the

T

tiling pair This group and its relation to tiling groups of surfaces with

divisible tilings are discussed in greater detail in and in the forthcoming pap er

The search algorithm for quadrilaterals with free vertices has to b e handled dif

ferently from those with constrained vertices only since there are innitely many

p ossibilities Also it turns out that the quadrilaterals with small numb ers of tri

angles have free vertices and those with a large number of triangles have only

constrained vertices and for the midrange of values of K we have both typ es

We dene special K to b e the number such that for any tiling pair with

K special K the divisible tiling has only constrained vertices For values b elow

or equal to special K we will use one algorithm and for those values ab ove special

K we will use another algorithm The following prop osition sp ecies special K

Prop osition Let be an arbitrary kaleidoscopic tiling pair consisting of

an l m ntriangle and an s t u v quadrilateral Let

   

s t u v

K

  

l m n

Divisible Tilings in the Hyperbolic Plane

Table A family of divisible quadrilaterals with free vertices

tiling of tiling of

tiling of tiling of

be the number of triangles covering Then if K then there are no free

vertices in the tiling ie special K equals Furthermore if therearetwotypes

of free vertices then K and if thereare types of free vertices then K

Pro of The examples in Table show that sp ecial K is or greater Now

supp ose that is a tiling pair suc h that Q is a free vertex We shall rst

construct an innite family of divisible quadrilaterals all with the same structure

so Assume that our tile has sides p and q meeting at the origin at R in an angle

l

that q is a p ortion of the p ositive xaxis and p is a diameter in the rst quadrant

The third side r of our triangle is a portion of a euclidean circle in the rst

quadrant meeting q at P in an angle meeting p at Q in an angle and meeting

m n

the b oundary at right angles see the gures in Table Noweach triangle in the

quadrilateral is of the form g where g g T For the quadrilaterals in the

i  K

examples the dierent g s are qqrqrpqrprqrprqpqppr prp prprprprq

Move the side r so that its point of intersection moves along the xaxis but the

angle remains The moving line denoted r will intersect the diameter p in an

N m

S A Broughton D Haney L McKeough and B Mayeld

 

that varies continuously from whenthepointofintersection angle

N l m

is at the origin down to when the side r meets p on the b oundary Let

N N

denote the triangle b ounded by p q r and let p q r also denote the reections

N N

in the sides of Now let g N be the isometry constructed from g by the

N i i

replacements p p q q and r r Because Q is a free vertex then

N

K

g N

i N

i

is a quadrilateral whose sub divided combinatorial structure is indep endent of N

i

where are the numbers of The angles at Qvertices have measure

 k

N

triangles meeting at Qvertices In our example k and

  

Let b e the least common multiple of the s There are now an innite

i

N

numberofinteger values of N d d d d Z suchthat d is an

 

i i

integer for all Q vertices and suchthatthe resulting l m N triple corresp onds

toahyp erb olic triangle The resulting quadrilaterals of typ e s t u v for

N N N N

appropriately selected integers will then form our innite family of kaleidoscopic

quadrilaterals If the other typ es of vertices are free then a multiparameter family

may b e created

Finally let us get a b ound on K Weknow that

   

s t u v

N N N N

K

     

l m N l m N

Nowwemaychose l m N of the form l m dsothat

K lim

    

d

l m d l m

The largest p ossible side for the right hand side is when l m The

next largest size is K for l m If there are two or three typ es of free

vertices then a two or three parameter family limit calculation shows that K

and K resp ectively

Constrained Vertices We have established a limit on K when there are

free vertices wemay establish a limit on l m n when there are only constrained

vertices We shall also show that l m and n must be chosen from the table in

App endix A

Prop osition Let be an arbitrary kaleidoscopic tiling pair consisting of

an l m ntriangle and an s t u v quadrilateral Let

   

s t u v

K

  

l m n

be the number of triangles covering Then if al l the vertices of the tiling are

constrainedthen

l m n K

Pro of If the R vertices are constrained then there is at least one R hub and hence

at least l triangles But then K l The pro ofs of the other inequalities are

identical

Divisible Tilings in the Hyperbolic Plane

The ab ove prop osition implies that the universal b ound of is also a b ound for

l m n when we have constrained vertices However we can do much b etter than

this Assume that l m n Then s t u v n and

s t u v n n n n

s t u v n n n n

It follows that

n n n n

l m n K

l m n

Thus for instance for a n triangle wehave

n

n

n

 

n

 n

Solving this inequality for integer solutions we obtain n Since the triangle

is hyp erb olic then we further restrict n Analogously for n triangles

we get n and for n triangles wehave n

Now let us obtain a lower b ound for K when there are constrained vertices

Prop osition Let be a kaleidoscopic tiling pair with only constrained

vertices Then the number of triangles K is at least six

Pro of Supp ose that has at least one interior hub Then the prop osition is

satised unless the interior hub has four triangles If there are only triangles then

the two other vertices are free Supp ose there are ve triangles Then adjoining

a single triangle to an interior hub of will force us to have at least one edge hub

with three triangles In turn this will force an equivalentvertex to b e a corner hub

with exactly two angles of measure which is not allowed In fact the hub of



four must b e completed to a quadrilateral tiled with six triangles as in case F in

Table

Thus wemay assume that there are no interior hubs and that there is an edge

hub for each dierent typ e of vertex say a P hub H centered at V a Qhub

P P

H centered at V and an R hub H centered at V The number of trian

Q Q R R

gles in at least one of the hubs satises jH H H jK W e shall estimate

P Q R

jH H H j by inclusionexclusion and arrive at a contradiction Wehave

P Q R

jH H H j jH j jH j jH j

P Q R P Q R

jH H jjH H jjH H j

P Q P R Q R

j H H H j

P Q R

l m n

jH H jjH H jjH H j

P Q P R Q R

j H H H j

P Q R

Now jH H j unless V and V are vertices of the same triangle Further

P Q P Q

more jH H j if the edge joining V and V is part of an edge of and

P Q P Q

jH H j otherwise Also jH H H j if V V and V are the

P Q P Q R P Q R

j vertices of the same triangle and it is zero otherwise If jH H j jH H

R P Q P

jH H j then jH H H j and jH H H j l m n

Q R P Q R P Q R

S A Broughton D Haney L McKeough and B Mayeld

On the other hand if jH H H j then one of jH H j jH H j and

P Q R P Q P R

jH H j is zero so j H H H jl m n Thus in all cases wehave

Q R P Q R

l m n jH H H jK or

P Q R

l m n

Nowaswe go through hyp erb olic l m ntriples in App endix A we see that there

is only one triple that satises this namely The quadrilateral must contain

one hub of order but it is imp ossible to add a single triangle to make it into

quadrilateral Wehave eliminated all cases

The inequalitycombined with K shows that we get an area restriction

 

Wenowprove a stronger restriction which allows us to select all our



data from the table in App endix A

Prop osition If an l m ntriangle subdivides an s t u v quadrilateral with



only constrained vertices then l m n



Pro of Supp ose that l m n Let s t u v weknowthat

and so thus

K

Since K must b e an integer and K by our last prop osition then K

Also by Prop osition l m n K so wemay assume without loss of generality

l m n We consider four cases

Case n K There exists exactly one hub and no other triangles

However it is not p ossible to select the other angles to form a quadrilateral

Case n K There is exactly one hub At most one other triangle

can b e added and thus either or p ossibly triangles meet at the vertices Since

there is at least one hub of eachtyp e then the only p ossible hyp erb olic triangle is



However this satises

Case n K Again there is exactly one hub and one or two triangles

must b e added Thus except at the hub or triangles meet at eachvertex

The only hyp erb olic p ossibilities are and triangles

nnnn



The rst satisfy two satisfy and for the last two the upp er b ound

lmn

for K is smaller than

Case n K The only p ossible triangles are and

triangles all of which satisfy the inequality

Since there are no hyp erb olic triangles with n the pro of is complete

Remark After constructing the tables wemay actually verify that l m n





The Two Search Metho ds The existence of free vertices and families of

divisible tilings forces us to split our searchinto two dierent approaches the direct

construction metho d K and the b oundary construction metho d K

We describ e the two approaches very briey here and then devote one section eachto

the full description implementation and computational examples of b oth metho ds

Divisible Tilings in the Hyperbolic Plane

Direct construction method K Assume for the moment that there are no

interior hubs Then each edge of a triangle must either be a part of the side of

the quadrilateral or reach from one side of the quadrilateral to another Thus as

a combinatorial ob ject the quadrilateral may be viewed as a circle with a set

of nonintersecting chords or a b etter yet as a p olygon with K vertices and

with a collection of K diameters see Figure b elow The diameters of the

vertices can then b e lab eled as P Q R vertices via the tiling structure In order to

transform the p olygon into a quadrilateral triangle K corners of the p olygon

must b e attened to straight angles leaving four corners to form a quadrilateral

This imp oses restrictions on l m and n and allows us to compute s t u and v

or conclude that no tiling pair exists The algorithm can b e extended to the case

where there are interior hubs The algorithm dep ends only on the combinatorial

structure of the p olygon and not the actual angles so it can handle the case of free

vertices Unfortunately the computational complexity of the computer search rises

very quickly with K and is not useful for large numb ers of triangles

Boundary construction method K If K then there is only a nite

number of p ossibilities for l m n and hence a nite number of p ossibilities for

s t u v according to Remark For each compatible pair of an l m nand

s t u v we try to construct an s t u v quadrilateral in the l m ntiling by

constructing the p ossible b oundaries of a quadrilateral The tile edges must o ccur

in certain sequences in the tiling thus the b oundary can be constructed combi

natorially By making a geometric check we can tell whether the hyp othesized

b oundary closes up and hence forms a quadrilateral Again there is only a nite

numb er of cases to check

Direct Construction Metho d K

We are rst going to show that all divisible quadrilaterals may b e constructed

from a collection of Euclidean p olygons sub divided into K triangles and their

combinatorial analogs By using the dual graph wewill show the existence of an

us to inductively construct all such p olygons by attaching algorithm that allows

triangles interior hubs and conglomerations of interior hubs to a p olygon with a

fewer numb er of triangles Eachsuch p olygon maythenbetestedtoseeifityields

a quadrilateral

An associated dividedpolygon is constructed as follows See Figures and

below for examples without interior hubs In Figure an asso ciated p olygon

with interior hubs has b een drawn along with its dual graph discussed b elow Let

P P b e the vertices of the triangular tiling of as wemove clo ckwise around

 n

Construct a convex Euclidean ngon whose vertices are lab eled P P Add

 n

in all diagonals P P that corresp ond to edges of the tiling contained in Next

i j

we add points into the interior of the polygon corresp onding to vertices of the

tiling in the interior of We denote these vertices by Q Q see Figure

 s

below We further add in all the segments Q P and Q Q corresp onding to

i j i j

edges of tiling interior to A combinatorial representation of the asso ciated di

vided p olygon or combinatorial divided polygon may then dened as the quintuple

Q gV E V E E where each comp onent fP g fP P g fQ g fQ P g fQ

j i i i i i j i i j i

is taken over the appropriate index set Note that the set E fP P g the set of all

i j

edges connecting b oundary p oints contains all the segments P P i n

i i 

S A Broughton D Haney L McKeough and B Mayeld

Figure Polygon without interior hubs

and P P The combinatorial representation is what we use for computer represen

n 

tation and calculation the Euclidean p olygon realization serves for visualization

Note that the same set of asso ciated p olygons also arises from a p entagon

etc tiled by a triangle The two critical features we need for the asso ciated p olygon

are

it is a convex ngon that is sub divided into triangles and

an even numb er of triangles meet at eachinterior vertex

In order to prove that wemay inductively construct all the asso ciated p olygons

wework with a mo dication of the dual graph of an asso ciated p olygon This is

constructed as follows Place a no de in the interior of each of the triangles Connect

the no des of neighb ouring triangles by an arc crossing the common b oundary The

constructed graph has cycles if and only if there are interior hubs See Figure

for a combined picture of an asso ciated p olygon and its dual graph The graph has

the following prop erties though we dont use them all

It is planar

Each no de is connected to at most three other no des

The arcs of the graph may b e coloured according to whichtyp e of edge they

cross

The minimal cycles haveaneven numb er of no des

tains no other p oint of the graph The region enclosed by a minimal cycle con

We construct the modieddualgraph as follows For eachinterior hub ll in the

p ortion of the dual graph b ounded by the corresp onding cycle We could achieve

this byblackening in the two quadrilaterals in the dual graph in Figure The

resulting ob ject consists of no des arcs isolated hubs and conglomerated hubs An

Divisible Tilings in the Hyperbolic Plane

Figure Polygon and dual graph

isolated hub is one which is connected to another no de or hubs by an arc only as

in Figure A conglomerated hub is one or more hubs joined together as in

cases F F and F in Table There are some restrictions on the hubs

and conglomerated hubs The mo died dual graph has a treelike structure and

therefore can b e constructed by adding one comp onent at a time This leads to the

following prop osition which is the basis of our combinatorial search

Prop osition Let be the associated subdivided polygon constructed from a

tiled quadrilateral Then there is a sequenceof associated subdivided polygons



such that is empty and each is obtainedfrom by adjoin

 s  i  i

ing a triangle hub or conglomerated hub along a single edge

Pro of If we replace eachhub and conglomerated hub by a no de then we obtain a

tree This follows from the fact that the mo died dual graph is simply connected

b ecause it a deformation retract of a quadrilateral Trees may b e constructed by

starting at any single no de and then connecting additional no des one at atime

by exactly one arc each This metho d of tree construction directly translates into

the statement ab out adjoining the comp onents of the asso ciated p olygons

Polygon Construction and EliminationNo Interior Hubs Let us

rst concentrate on those asso ciated p olygons with no interior hubs as the de

velopment is simpler The case for p olygons with interior points requires a few

mo dications which are describ ed b elow Let P b e the set of p olygons sub divided

K

into K triangles without anyinterior vertices The dual graph of any one of these

p olygons is a tree with K no des and K arcs NowK sides are contributed by

the triangles K of which are absorb ed as interior edges corresp onding to the

K arcs Thus there are K triangle sides on the b oundary and that many

S A Broughton D Haney L McKeough and B Mayeld

Figure Asso ciated p olygons for K

vertices as well In this case we are going to geometrically represent the asso ciated

polygon by a regular p olygon for the following reasons The rst is that the number

of such p olygons is well known to b e a Catalan number see p

K

K

jP j cK

K

K

and hence we will call the asso ciated divided p olygons Catalan p olygons This

will help us in making sure that enumeration is complete The second is that we

may use the dihedral symmetries of regular p olygons to reduce the computational

complexity Clearly dihedrally equivalent asso ciated p olygons will lead to the same





divisible quadrilateral if any Since jP j some reduction in com





plexity is desirable The following prop osition demonstrates that the complexityis

reduced for lowvalues of K

Prop osition Let P denote the set of subdivisions of regular K polygons

K

into K triangles Let OP denote the set of dihedral equivalence classes of the

K

subdivided regular polygons Let cK jP j as above ocK jO P j Then

K K

ocK is given by

K 

cK K c



ocK K mod

K

 K 

K  K 

c K c cK

  

ocK K mod

K

P

K

K 

K K

csc cK K c

s

  

K mod ocK

K

and

K 

P

 K  c

K

K 

K K



cK K c csc

s

   

ocK

K

for K mod The growth is stil l exponential but at least is manageable for

K since oc and ocK v cK K The formulas are

given in and further references arelistedinsequence M of the Sloan integer

sequencereference

The adjoining pro cess leads to a combinatorial divided p olygon fP g fP P g

i i j

lead to divisible hyp erb olic The next step is to determine whichl m n

quadrilaterals To do this we need to determine the typ e of the vertices on the

Divisible Tilings in the Hyperbolic Plane

b oundary Supp ose that P P P P and P P are the three edges of an arbitrarily

i j j k k i

chosen triangle P P P contained in Declare the typ es of P P and P

i j k i j k

to b e P Qand R resp ectively The reection of in at least one of its sides must

lie in Supp ose for the sake of argumentthatthissideisP P Then there must

i j

  

be a P such that P P and P P b elong to the edge set of our combinatorial

k j k k i



P is the reected image of P and so they must have the same polygon Now

k k



typ e namely R Since the P is found by examining only the combinatorial data

k



fP g fP P gwesay that P is the combinatorial reection of P Combinatorial

i i j k k

reection maybecontinued until every vertex is assigned a unique typ e

Once the vertices have b een assigned a typ e then the vertices of typ e R P and

Let m be the integer so chosen for P Q maybe assigned an angle of

i i

l m n

Next we assign to eachvertex P the number s of triangles that meet at the vertex

i i

This can b e determined combinatorially by noting that there must b e s edges

i

s

i

of the form P P in the edge set E The angle measure at vertex P is Finally

i j i

m

i

wemust select four vertices in V to serveasthe corners of the quadrilateral or

actually K corners to atten into a straight angle Atthisstagemanyofthe

congurations are eliminated

Let us illustrate the pro cess by discussing the case K Though jP j

there are only dihedrally inequivalent asso ciated p olygons See Figure The

list of vertices in counterclo ckwise order starting at the oclo ckvertex the

vertex typ es and the angle measures divided by are as given in the following

table

Vertex Case Case Case

  

P P P P



m m m

 

P Q Q Q



n n n

  

P P R R



m l l

  

P R Q P

l n m

  

P P P Q

m m n

  

P R R R



l l l

Now from the six vertices wemust select two to atten out to a straight angle

the remaining vertices are corners of the quadrilateral To b e a straight angle we

s

i

must have or s m This automatically forces corners with a single

i i

m

i

triangle to b e a quadrilateral corner as is geometrically obvious In Case we can

either cho ose fP P g or fP P g to atten For P and P cannot b e chosen and

   

if wecho ose one of P or P we are forced to also cho ose the other since the angle



 

If wecho ose fP P g then l and b oth and are measure for b oth is



l n m

recipro cals of integers It follows that l m nmust have the form d eand

s t u v must have the form d e d d The values of d and e are d e

except that d e is disallowed since would then be Euclidean If we

cho ose fP P g then n m and l m n must have the form d



and s t u v must hav e the form dd with d The complete analysis of

S A Broughton D Haney L McKeough and B Mayeld

K without interior hubs is given in the table b elow

Case attened corners l m n s t u v restrictions

P P d e d e d d d e d e



P P d dd d



P P d e d e e d d e

 

P P d dd d



P P d d d d



For larger values of K the wheat to cha ratio decreases drastically so we

need to identify some metho ds of quickly rejecting combinatorial p olygons Let

R R be the vertices of typ e R and let b e the angle multiplicities

 L  L

at these points Dene P P and Q Q be

 M  M  N  N

similarly dened These quantities satisfy the following relations

L M N K

K

 L

K

 M

K

 N

l

Next let l lcm and write There only twoways that we can

 L i

s

i

assign an angle to R vertices so that is an integer submultiple of Either

i

l l

we set the angle to b e in which case the angle of the hub at R is or we

i i

l l s

i

d and then the angle at hub at R is In the rst case set it to

i i

l d l d s d

i

we get an edge hub for each s otherwise wegetacornerhub In the second

i

case every hub is a corner hub and we get a free vertex Similar considerations

apply to the vertices of typ e P and Q

An example will help illustrate Consider the asso ciated p olygon for K in

Figure Pro ceeding counterclo ckwise around the p olygon from the three oclo ck

p osition vertices may b e lab eled R P Q P R Q P R Q R P Q with multiplicities

The sequences of multiplicities are

f g f g f g f g f g f g

i i i

Thus wecancho ose either or corner hubs of typ e R or corner hubsoftyp e

P and or corner hubs of typ e Q Of the eight p ossible choices only one yields

a quadrilateral namely l m and n yielding and corner hubs

of typ es R P and Q resp ectively Thus we get a tiling pair

yielding case C in Table Note for instance that this metho d also allows us

to construct a d d d d hexagon tiled bya d triangle

Computer Algorithm and Extension to Interior Hubs Two Maple

worksheets catp olysmws and hubp olysmws were used to implement the search

for divisible quadrilaterals with K Maple was used so that the graphical

capabilities could be used to draw various p olygons such as the gures in this

section to check the logic of the program during development The algorithm

steps were the following

Create a sequence of les F containing the representatives of each dihedral

K

equivalence class of Catalan p olygons with a given numb er of sides

a Start o with F consisting of a triangle 

Divisible Tilings in the Hyperbolic Plane

b From F create the le G consisting of all p olygons that can be

K K 

created from p olygons in F by adding a triangle to each side of a p olygon

K

from F The p olygons are created as an ordered list

K

c Create an empty le F Place the rst element of G in F

K  K  K 

Create the dihedral orbit of rst elementofG and then removethis

K 

list of p olygons from G Rep eat the pro cedure until G is empty

K  K 

For each p olygon in F do the following

K

a Compute the multiplicity of all vertices

b Lab el P and P R and Q resp ectively Find the unique point P P

  k

which completes the triangle Now rep eatedly use combinatorial reection

to lab el vertices of the p olygon with P Q or R This part can b e written

so that it is guaranteed to terminate

c Construct the f g f g f g and determine all p ossible assignments of

i j k

l m and n that lead to vertices which are corners This part can be

easily mo died to nd tilings of p entagons hexagons etc

d Collect the valid assignments into a le Q

K

The algorithm for the case of interior hubs requires just a few mo dications

Construct by hand a list consisting of a triangle and all hubs and conglomer

ated hubs with twelve or less triangles

Start o with F containing the hub of Next create G b y adding triangles

to the elementin F at all p ossible lo cations

Find representatives of the dihedral orbits and place them in F

Construct further G by attaching a triangle hub or conglomerated hub to

K

p olygons in sets with smaller numb ers of triangles so that the total number

of triangles is K The rst nontriangle addition will b e a hub of to another

hub of Create F to b e a set of representatives of dihedral orbits

K

Create a lab eling scheme for all vertices including interior vertices

Create all valid assignments of angles to b oundary vertices

Check for compatibilitywithinterior vertices since the angles here are pre

determined

Boundary Construction Metho d K

The b oundary construction metho d was used for all divisible tilings for which

K since in this case we are guaranteed that the tiling has no free vertices and

hence there are only a nitely manyl m ntriangles to consider After describing

the metho ds in the next two subsections w e will do some sample calculations to

illustrate the ideas

According to the discussion in the previous sections we shall do the following



Enumerate all of the l m ntriangles for which l m n l m n

l m n and n l m n The table of values in App endix A

n

can b e used here

For each triangle in Step enumerate all candidate quadruples s t u v

formed from l m n The numbers s t u v must b e selected from the divisors

of l m n which are greater than

For each candidate pair of an l m n and an s t u v select those which

pass the K test and the hub tests ie the quotients calculated in

S A Broughton D Haney L McKeough and B Mayeld

and must be integers In addition we eliminate those cases in which

stuv

since these have already b een determined by the direct K

lmn

construction metho d

For each candidate pair resulting from enumerate all p ossible assignments

of vertex typ es of the triangle to the quadrilateral This allows us to take

multiple corners into account

For each candidate pair resulting from p erform the quadrilateral search

algorithm as we describ e next

Quadrilateral Search Having found an l m nandans t u v such that all the

restrictions in hold quadrilaterals tiled by the triangle are sought out using

a Maple program tilequadmws see Two somewhat dierent programs were

develop ed though they dier only in how the quadrilateral test is implemented

The quadrilateral search program is given an l m ntriangle and a quadruple

s t u v and tries to nd to a corresp onding quadrilateral along the lines of the

triangle tiling

Lo ok at Figures or the gures in the tables in Section to help understand

how the algorithm works In addition to the information l m n and s t u v

we need an assignmentofvertex typ es and hub multiplicities to the corners of the

quadrilateral Wemust consider all compatible typ e assignments For example if

l m n and s t u v then the p ossible vertex assignments

are R P R P P P R P R P P P and P P P P with hub multiplicities

and resp ectively Sp ecial attention needs

to b e paid to isosceles triangles For instance in attempting to tile a with

a typ e assignments should b e considered

Supp ose that the typ e assignmentisS S S S where S fP Q R g The

   i

algorithm starts bypicking the vertex A of typ e S on the master tile Pick an edge



of the angle at A and move out along the ray determined bythisedge Wemove

along the ray and observe the triangles we meet on the same side of the rayasour

original triangle We stop at some there are manychoices triangle whose second

vertex encountered is of typ e S Call this vertex B our rst side will be A B



see Figures and To get the second side we turn the corner through

t

radians toward the side of the original triangle and pro ceed along the next rayin

the tiling and stop at a vertex of typ e S say C to pro duce the second side B C



Because of the compatibility conditions when we turn the corner wewillmoveout

along a line of the tiling Turn again through radians to pro duce a third

u

side C D and turn again to pro duce a fourth side D E When wehave nished if

A E and angle D A B has measure then wehave a quadrilateral of the desired

s

typ e Note that there are only a nite numb er of p ossibilities for each side of the

quadrilateral since the number of hubs on the edges is b ounded bythehub numb ers

as illustrated in the attempted tiling of a quadrilateral later in

this section Now the program keeps track of the four combinatorial sides of the

quadrilateral by writing down the sequence of vertex typ es that we pass through as

wemove along an edge of the prop osed b oundary of the quadrilateral This follows

from the following observation

Remark Let b e a line of the tiling and let S S S b e the bi

  

innite sequence of typ es of vertices lying on in the order in whichtheyoccur

Then the entire sequence is determined bythetyp es of two adjacentvertices For

Divisible Tilings in the Hyperbolic Plane

example in a triangle only one sequence o ccurs f RPQRQPg

In a triangle RP Q two dierent sequences occur f RPg and

f PQRQg For a triangle three sequences o ccur f RP g

f R Q g and f PQg In each of the sequences ab ove the entire

sequence is the biinnite cyclic rep etition of the basic pattern shown The pattern

typ es only dep end on parityofthenumb ers l m and n

Nowthatwehave the four edges how can we determine if it is a quadrilateral

with the correct angles There are two metho ds that were used which are explained

o subsections in the next tw

Remark Though one really only needs one typ e of test twowere develop ed

for the following reasons The rst test is faster to calculate and less susceptible to

round o error The second test pro duces b oundary words which can b e used to

determine a surface of minimum genus which supp orts b oth tilings See

Geometric Quadrilateral Test The rst metho d uses the construction of

a quadrilateral given by the following prop osition which is proved in

Prop osition Let be four angles satisfying and



Then for every value of h satisfying

cos cos

h

sin sin

there is a quadrilateral AB C D unique up to congruence such that

mDAB mAB C

mBC D mCDA

and

AB h

The algorithm so far has given us a set of p oints A B C D E such that all

the lengths are known from the lists of p oints on the edges For the lengths of the

sides of the triangles are easily computed and the typ es of all edges comprising a side

of a quadrilateral are known Let h b e the length of the side A B and construct

quadrilateral AB C D as in the prop osition ab ove with

s t u

The algorithm constructs the p oints A B C D in the unit disc and so the

v

lengths BC CD and DA maybenumerically calculated If any of the dierences

BC B C CD C D DA D E exceed an appropriately chosen then

the candidate quadrilateral do es not close up Because Maple is used we can sp ecify

that the calculations b e carried out to a large numb er of decimal places to guarantee

go o d accuracy in the calculation of the lengths and therefore that a small value

of may be chosen If all the dierences are less than the tolerance then the

candidate tiling is directly constructed using the Matlab drawing scripts

Boundary Word Test Our second metho d uses group theory For each

of the p ossible quadrilateral b oundaries the program creates four edge words in

p q and r by making reections through the various hubs along an edge of the

quadrilateral The four edge words are concatenated to form a boundary word

The four edges close up to form a quadrilateral if and only if the b oundary word

S A Broughton D Haney L McKeough and B Mayeld

reduces to the identity The pattern of hubs along the b oundary allowustoquickly

write down the b oundary word The details of determining the boundary word

and showing it is the identity is describ ed in detail in some sample calculations in

Subsection

By attempting to draw a quadrilateral in the triangle tiling as ab ove wehavea

word whichwewant to proveistheidentity There are two p ossible approaches The

rst approach is see if the word can b e reduced to the identity using the relations

that we have in the group T Though in any sp ecic example it always seems

p ossible to nd a reduction to the identity when it exists the authors decided

against attempting to develop a computer algorithm to decide whether a word was

reducible to the identity b ecause of apparent complexity of doing so suggested by

p If the maximum length L of the b oundary words to b e reduced is known

b eforehand the KnuthBendix algorithm p pro duces for eachl m n a

nite list of substitution rules which if applied recursively to a given word of length

at most L will reduce to the identity if and only if the word is the identity The

length of eachedgeword is very crudely b ounded bythenumb er of triangles in

the quadrilateral and so L Thus it is p ossible and interesting in

principle to devise a reduction algorithm however more work is required than by

using the second approachthatwe describ e next

The second approachconverts a pro duct of an even numb er of reections into a

matrix by means of a homomorphism q T PSL C whichwe describ e shortly



The matrix images are computed numerically using Maple whic hintro duces some

error into our calculations due to rounding o the entries in the matrices This

can be remedied that noting that if the computations are carried out with great

accuracysay to decimal places as done in the study see Remarks and

then a pro duct which is not computed to be within a prescrib ed distance of the

identity will not equal the identity

Wenow describ e the map q and state a theorem that guarantees that our conclu

sions based on approximate calculations are justied The reections in the sides

of an l m ntriangle are inversions in the circle dening the side or reection in

a diameter Every such reection has the form T where

A

az b

a b

T z A a bb a

A

b a

bz a

z z

The matrix is determined only up to a scalar multiple of If the inversion is in

a circle centered at z and p erp endicular to the unit disc then A has the form



i

z



p

A

z

z z



 

Otherwise the line is a diameter meeting the p ositive xaxis at the angle and A

has the form

i

e

A

i

e

If two of the edges of the l m ntriangle are diameters through the origin then

the third side is easily determined using analytic geometry see Also note

that the origin is a vertex of the tiling Let P Q R b e the matrices suchthatthe

Divisible Tilings in the Hyperbolic Plane

reections p q r are asso ciated to

p T q T r T

P Q R

Then the elements a pq b qr and c rp are asso ciated to

a T b T and c T

P Q QR RP

since

and T T

A

A

T T T

AB A B

where A is obtained by conjugating the entries of A

Nowaword w in p q r in T representing the identitymust haveaneven number

of factors since an o dd word is orientation reversing Thus w is a word in a pq

  

b qrc rp a qp b rq and c pr and the matrix corresp onding to

w is obtained by making substitutions based on

a P Q

R b Q

c R P

The following theorem now guarantees when an elementwhichisapproximately

equal to the identity is actually equal to the identity We rst dene the L norm

on matrices

kAk maxjAk lj

kl

Prop osition Let T be the group of orientationpreserving isometries gener

atedbyanl m ntriangle in the hyperbolic plane Assume that a vertex of

is at the origin Let A PSL C bea matrix representing an element of g T



according to the substitution in Then there is an depending on and

not on A such that if minkA I k kA I k then g is the identity in T

The prop osition gets used in the following way Supp ose that the matrix A

in the prop osition has b een computed approximately as A Supp ose further that

 

A I by calculation and that A A by controlling the accuracy

 

of the computation Then kA I k A A A I and so that g is

the identity

a b

a b

Pro of Let A and let g ba Observe that is one of

b  a

b a

the vertices of the tiling There is a small hyp erb olic ball of radius h suchthat is

the only vertex contained in that ball and hence h unless Wemay

take h to be any numb er smaller than all the sidelengths of the master tile Let

tanhh b e the Euclidean radius of this ball Thus j j unless

But since minkA I k kA I k then jbj Also as aa bb then

i

e

jbj

This forces and so A jaj and hence j j

i

jaj

e

p

i

cos The smallest value But then min kA I k kA I k e

S A Broughton D Haney L McKeough and B Mayeld

p

cos where b is the largest of l m and n Thus we of this expression is

b

need only to adjust to b e smaller than this

Remark For a given l m n the co ecients of P Q QR and R P all b elong

to nite extension of Q Therefore it is p ossible to exactly calculate words in the

matrices bysymb olic means However the cure of exact symb olic computation is

worse than the disease of approximation

Remark The decimal places used in are probably overkill However

the scripts can be run automatically computing with a large number of decimal

places without an onerous time p enalty It is not to hard to show that an error

L

b ound for q image of a word of length L is M where M is the maximum of

the entries of the matrices in Remark and is the maximum error of individual

matrix entries Thus the required number decimal places is linear in the word

length

Example Failure of to Tile The largest possible

value of K o ccurs for the attempt to tile a quadrilateral by a

triangle The candidate pair passes the K test and the hub tests since K

h h and h Eachofthe four edges of any quadrilateral must

  

havetwo hubs For an edge with only one hub and terminating angles must



closeuptoa triangle as in Figure b elow Similarly there cannot b e zero

hubs as Figure also shows Thus every edge must have exactly twohubs

and hence there is a unique way to construct the quadrilateral But Figure

shows how the attempt to construct such a quadrilateral clearly fails

Fig tiling of Fig A failed b oundary search

Example Successful Tiling of by Now let us ex

amine a quadrilateral which can b e sub divided by triangles Let us sub divide the

quadrilateral with the triangle See Case C in Table for a

picture of the sub divided tiling It is found that K h h and



h By lo oking at the tiling in Section it is clear the any prop osed edge of a

quadrilateral without a hub closes up to a triangle Since h then

the b ound there must be exactly one hub on each side This greatly restricts

ary searches Let us consider the lower b oundary tiles on the lower edge of the

quadrilateral as pictured in Figure

Divisible Tilings in the Hyperbolic Plane

The far left triangle as oriented is a triangle Though the order of l m and

n is unimp ortant for rst three steps of the search the ordering is very imp ortant

in the quadrilateral construction phase Thus for this particular triangle wehave

pq a

qr b

 

rp c

Now let us construct the edge word corresp onding to this part of the b oundary

We lab el the triangles as wemove from left to right It is obvious

  

ver the r edge The reections in sides of that r since we reect o

  

are rpr rq r and r rrr In fact if g for some g T the reections



  

in the sides of are gpg gqg and grg From the picture we see that



is the rq r image of and hence rq rr rq Continuing one more

   

step we see that the reections in the sides of are r q pq r rq r rq q q r and



rq rq r Now is obtained by reecting over the common rq rq r edge Thus

 

rq rq r rq rq rrq rq r The pattern is now evident whichwenow

   

state as a prop osition We omit the easy induction pro of

Figure Boundary tiles of lower edge

Prop osition Let beasequence of tiles such that and

  s i i

meet in edge e Let r the unique edge among fp q r g such that e matches r by

i i i i

the unique transformation in T carrying to Let r also denote the reection

 i i

in r Then

i

r r r for i s

i   i 

Applying the prop osition weget

w rq pqprq r

   

Continuing on with the next edges we get

w w

   

w w w

    

w w w w

    

S A Broughton D Haney L McKeough and B Mayeld

where w w w w r q pq pr q r The fact that all the edge words w w w

     

and w are equal is a consequence of the fourfold rotational symmetry of the

quadrilateral Observe that the edgewords are completely determined by the initial

orientation of and the number and typ e of hubs we pass through along an edge



The pro cess of nding the word can b e sp ed up by observing that passing through

ahub along the edge concatenates a welldened subword to the edge word Thus

one multiplies together a sequence of hub words

The b oundary word wehaveisr q pq pr q r The candidate b oundary closes up

if and only if and only if ie if and only if rq pqprq r Since

   

T acts simply transitively on the triangles then the b oundary closes up if and only

if

rq pqprq r

But wehave

rq pqprq r r q pq pr q rrqpqprqrrqpqprqrrqpqprqr

r q pq pr qqpq pr qqpq pr qqpq pr q r

rq pq pr pq pr pq pr pqprqr

Now pr pr so pr p r and likewise qrqrqrq r Thus we further obtain

r q pq pr q r rq pqrqrqrqpr q r

rq pr pr qr

rq q r

Wehave exactly computed the exp ected reduction Alternativelywe could trans

late the b oundary word into a word in a b c and compute the matrix image in

PSL C One simply takes the letters in b oundary word two at a time with the



    

replacements p q r pq a qr b pr c qp a rq b and



pr c Our word and the replacement matrices are to decimal places

 

rq pqprq r b ac b

i

a A

i

i i

b B

i i

i i

c C

i i

The test matrix is seen to b e approximately a scalar matrix

 

B AC B

The matrix calculations are done to decimal places here to ensure clarity By

computing to a greater numb er of digits enough accuracy can b e achieved to con

clude that the b oundary word is the identity The matrix metho d is preferable

since the computer ie programmer need not get creativeabouthow to reduce

words

Divisible Tilings in the Hyperbolic Plane

Catalogue of Divisible Tilings

This section contains a complete catalogue of all s t u v quadrilaterals which

can b e sub divided byl m ntriangles For completeness wehave also included the

small numb er of tilings of triangles by triangles and quadrilaterals by quadrilaterals

For each category there are two tables a table of data and a table of gures

Theorem Let and denote a hyperbolic kaleidoscopic l m n triangle

  

and l m n triangle respectively Then thereisoneparameter family and ve

  

parameter families of divisible tiling pairs in which therearefree vertices

Thereare two exceptional tiling pairs with constrained vertices only These

families and exceptional cases arelistedinTable and picturedinTable

Theorem Let and denote a hyperbolic kaleidoscopic l m ntriangle

and s t u v quadrilateral respectively Then thereisone parameter family ve

parameter families and parameter families of divisible tiling pairs

in which therearefree vertices These families are listedinTable and pictured

in Table In addition thereare divisible tiling pairs with constrained

vertices only These arelistedTable and picturedinTable

Theorem Let and denote a hyperbolic kaleidoscopic s t u v quad

   

rilateral and s t u v quadrilateral respectively Then thereisoneparameter

   

family and one parameter family of divisible tiling pairs in which there

are free vertices There are no examples with constrained vertices only The two

families are listedinTable and picturedinTable

Some notes on the tables

Although it is usual to arrange for l m n the geometry of a triangle

b eing included in a larger p olygon may force a dierent ordering For b oth

the triangle and the quadrilateral the ordering is obtained bymoving ab out

the gure in a counterclo ckwise sense A dierent orientation is obtained

from the reected quadrilateral These permutations of the ordering are in

the tables of pictures only where comparison of the picture and the ordering

makes sense

The groups T and Q denote the groups generated by reections in the sides

of and resp ectively Then Aut which is equal to the group taking

to itself and Stab T Aut are also given in the tables The

T

groups are computed by visual insp ection of the pictures These groups are

useful in determining the structure of tiling groups of surfaces of minimal

genus supp orting b oth tilings See and the forthcoming pap er for more details

S A Broughton D Haney L McKeough and B Mayeld

Table Triangles sub divided by triangles

Case restrictions K Stab Aut

T

 

TF de d d e Z Z

 

d e

TF d dd d trivial trivial

TF d dd d trivial trivial

TF d d d d d Z Z

 

TF d d d d d D D

 

TF d d d d d Z Z

 

TC Z Z

 

TC Z Z

 

Divisible Tilings in the Hyperbolic Plane

Table Quadrilaterals sub divided by triangles with free vertices

Case restrictions K Stab Aut

T

  

F d e f d e d f Z Z

 

d e f

F d e dd e d e trivial trivial

 

F d e d e d e Z Z if d e

 

d e 

F d e d e d e d e Z Z D if d e

 

F d e d d e d d e Z Z

 

  

F d e d d e e Z if d e Z if d e

 

d e 

F d dd d Z Z

 

F d dd d Z Z

 

F d d d d trivial Z



F d d d d trivial trivial

F d ddd d trivial trivial

F d dd d Z Z

 

F d d d d Z Z

 

F d dd d Z Z

 

F d d d d d trivial trivial

F d d d d d trivial trivial

F d d d d d d Z Z

 

F d d d d trivial trivial

F d d d d trivial trivial

F d dd d Z Z

 

F d d d d Z Z

 

F d d d d d d D D

F d d d d d d Z Z

 

F d d d d d d Z Z

 

F d d d d d trivial trivial

F d d d d d trivial trivial

F d dd d d trivial trivial

F d dd d d trivial trivial

F d d d d d trivial trivial

F d d d d d trivial trivial

F d d d d d d Z Z Z Z

   

F d d d d d d Z Z

 

F d d d d d d trivial trivial

F d d d d d d Z Z

 

S A Broughton D Haney L McKeough and B Mayeld

Table Quadrilaterals sub divided by triangles without free vertices

Case K Stab Aut

T

C Z Z Z

  

C trivial Z



C trivial trivial

C Z Z

 

C Z Z

 

C Z Z

 

C trivial Z



C trivial trivial

C Z Z

 

C Z Z

 

C Z Z Z Z

   

C Z Z

 

C Z Z

C trivial trivial

C Z Z

 

C trivial trivial

C trivial trivial

C trivial trivial

C Z Z

 

C Z Z

 

C Z Z Z Z

   

C Z Z

 

C Z Z

 

C D D

C Z Z

C Z Z

 

C Z Z

 

Table Quadrilaterals sub divided by quadrilaterals

Case restrictions K Stab Aut

T

 

QF de d d e e Z Z

 

d e

QF d d d d d d Z Z D

 

Divisible Tilings in the Hyperbolic Plane

Table Triangles sub divided by triangles

Case TF K Case TF K

d e d d e d d d

Case TF K Case TF K

d d d d d d d

Case TF K Case TF K

d d d d d d d d

S A Broughton D Haney L McKeough and B Mayeld

Table part

Case TC K Case TC K

Divisible Tilings in the Hyperbolic Plane

Table Divisible quadrilaterals with free vertices

Case F K Case F K

d e f d e d f d e dd e

Case F K Case F K

d e d e d e d e d e d e

Case F K Case F K

d e d d d e d e d e e d

S A Broughton D Haney L McKeough and B Mayeld

Table part

Case F K Case F K

d dd d dd

Case F K Case F K

d d d d d d

Case F K Case F K

d ddd d dd

Divisible Tilings in the Hyperbolic Plane

Table part

Case F K Case F K

d d d d dd

Case F K Case F K

d d d d d d d d

Case F K Case F K

d d d d d d d d

S A Broughton D Haney L McKeough and B Mayeld

Table part

Case F K Case F K

d d d d dd

Case F K Case F K

d d d d d d d d

Case F K Case F K

d d d d d d d d d d

Divisible Tilings in the Hyperbolic Plane

Table part

Case F K Case F K

d d d d d d d d

Case F K Case F K

d dd d d d d d

Case F K Case F K

d d d d d d d d

S A Broughton D Haney L McKeough and B Mayeld

Table part

Case F K Case F K

d d d d d d d d d d

Case F K Case F K

d d d d d d d d d d

Divisible Tilings in the Hyperbolic Plane

Table Divisible quadrilaterals with constrained vertices

Case C K Case C K

Case C K Case C K

Case C K Case C K

S A Broughton D Haney L McKeough and B Mayeld

Table part

Case C K Case C K

Case C K Case C K

Case C K Case C K

Divisible Tilings in the Hyperbolic Plane

Table part

Case C K Case C K

Case C K Case C K

Case C K Case C K

S A Broughton D Haney L McKeough and B Mayeld

Table part

Case C K Case C K

Case C K Case C K

Case C K Case C K

Divisible Tilings in the Hyperbolic Plane

Table part

Case C K Case C K

Case C K

Table Quadrilaterals sub divided by quadrilaterals

Case QF K Case QF K

d e d e e d d d d d d

S A Broughton D Haney L McKeough and B Mayeld

App endix A Triangles with area



Angle Description Angle Description



 

 

 

 d

d d

 d

 

 

d 

dd

d 

 

 

d d

dd d d

d d

 

 



 

 

 

 d

d d

 d

d 

d d

d 

 d

d d

 d

d 

d d

d 

 d

d d

 d

d 

d d

d 



 

d d

d d d d

d d

 

 

d d

d d d d

d d

 

 

d d

d d d d

d d

 



d d

dd d d

d d

 



 d

d d

 d

 

 

 d

d d

 d

d 

d d

d 

 d

dd

 d





d 

d d

d

 

d

d d

d

Divisible Tilings in the Hyperbolic Plane

References

A F Beardon The Geometry of Discrete Groups Graduate Texts in Mathematics no

SpringerVerlag New York MR d Zbl

S A Broughton Constructing kaleidoscopic tiling polygons in the hyperbolic plane Amer

Math Monthlytoappear

S A Broughton Kaleidoscopic tiling of surfaces background notes for the RoseHulman

REU Tilings Pro ject

httpwwwrosehulmaneduClassmaHTMLREUTilingspubshtmlKaleido

S A Broughton D Haney L McKeough B Smith Divisible Tilings on Hyperbolic Surfaces

in preparation

J Do chkowa H Harb oth I Mengerson Cut set Catalan numbers Pro ceedings of the

Twentyninth Southeastern International Conference on Combinatorics Graph Theory and

Computing Bo ca Raton FL Congr Numer  CMP

F Levy M S Paterson and D B A Epstein J W Cannon D F Holt S V

W PThurston WordProcessing in Groups Jones and Bartlett Boston MR i

Zbl

R P Grimaldi Discrete and Combinatorial Mathematics fourth ed Addison Wesley Read

ing Zbl

D Haney and L McKeough Quadrilaterals SubdividedbyTriangles in the Hyperbolic Plane

RoseHulman Math Sci Tech Rep no

A Seress AnIntroduction to computational group theory Notices of the AMS  no

JuneJuly MR e Zbl

D Singerman Finitely maximal Fucshian groups J London Math So c

MR Zbl

N J A Sloane S Ploue The Encyclopedia of Integer Sequences Acad Press San Diego

MR a Zbl

B M Smith Triangle tilings of quadrilaterals in the hyperbolic plane in preparation

MAPLEVWaterlo o Maple Inc Waterlo o Canada

MATLAB The Mathworks Natick Massachusetts

RoseHulman NSFREU Tilings w eb site

httpwwwrosehulmaneduClassmaHTMLREUTilingstilingshtml

RoseHulman Institute of Technology Terre Haute IN

allenbroughtonrosehulmanedu httpwwwrosehulmanedubrought

University of Georgia Athens GA

haneydawarchesugaedu

St Pauls School Concord NH

lmckeougspsedu

Cheyenne Court Fairfield Twp OH

brandymayeldhotmailcom

This pap er is available via httpnyjmalbanyedujhtml