AREA OF POLYGONS IN HYPERBOLIC GEOMETRY

By

BENJAMIN AARON BAILEY, B.S.

A THESIS

IN

MATHEMATICS

Submitted to the Graduate Faculty of Texas Tech University in

Partial Fulfillment of the Requirements for

the Degree of

MASTER OF SCIENCE

Approved

Qiairperson of thefcommittee

Accepted

Dean ofthe Graduate School

August, 2004

ACKNOWLEDGMENTS

My sincerest thanks and appreciation go to my thesis committee members, Dr.

Roger Barnard, Dr. Petros Hadjicostas (chair), Dr. Magdalena Toda, and Dr. Brock

Williams, for the support and knowledge they have offered over the last semester.

11

CONTENTS

ACKNOWLEDGMENTS ii

ABSTRACT iv

LIST OF FIGURES v

I. INTRODUCTION 1

II. ANALYTIC DERIVATION OF MAIN RESULTS 2

III. GEOMETRIC DERIVATION OF MAIN RESULTS 9

IV. IDENTITIES AND APPLICATIONS 13

V. CONCLUDING REMARKS 17

BIBLIOGRAPHY 18

111

ABSTRACT

Consider the Poincare model for hyperbolic geometry on the unit disc and an

arbitrary n-gon in this geometry. Chapter I gives a brief introduction to the confor-

mal metric, which generates hyperbohc geometry. In Chapter H, there is an analytic

derivation of a convenient computational formula for the hyperbolic area of a hy­

perbolic n-gon in terms of the coordinates of its vertices, as well as an insightful

geometric interpretation of this formula in terms of naturally occuring angles of the

n-gon. Chapter III extends the formulas in Chapter II to the closed unit disc with

an alternative, more geometrically motivated proof. Chapter IV uses the results of

Chapter I to establish identities between hyperbolic area and perimeter of an n-gon.

A proof of the existence of a solution for the following isoperimetric problem also is

given: Maximize (if such a maximum exists) the area of an n-gon with fixed perimeter.

IV

LIST OF FIGURES

2.1 Illustration of inward and outward sides 7

2.2 Angle 6k traversed by the tangent vector 8

3.1 Example of (pkitk+i) ~ (t>k{tk) for outward (left) and inward (right) sides. 11

3.2 Illustration for equation (3.12) 12

CHAPTER I

INTRODUCTION

Consider the Poincare model of hyperbolic geometry on the unit disc {z G C :

\z\ < 1}. According to [1], p.97, the hyperbolic arclength P of a curve 7 of bounded

variation is defined by

7

Also, by [1], p. 138, the hyperbolic area of a region 7?, is given by

^ = / / ( T ^ ) ^ d - d v ^ / / ( ^ 3 ^ ) ^ < l - * - (1-2)

Equation (1.1) in differential form is

This gives a function, p(^), which relates the hyperbolic separation dP of two nearby

points to their Euclidean separation |dz|. In a certain sense, this function describes

the "density" of space at a given point. This function serves to define distance and

the notion of a line (more properly called a geodesic) in this geometry. Take z and

w in the interior of the unit disc. By [1], p.89, the distance between these two points

is the infimum of the hyperbolic arclengths of all curves joining these two points. A

geodesic between z and u; is a curve whose length attains this infimum. Prom [4],

p.316, there is a unique geodesic between two points, and this geodesic is an arc of a

circle, which intersects the unit circle at right angles. If one of the points lies on the

boundary, the geodesic can still be defined in this geometric sense even though the

distance to any other point is infinite.

We can say that p{z) = _? p generates hyperbolic geometry as it completely

determines geodesies between all points in the unit disc. This background prepares

us for the following formulas and results.

CHAPTER II

ANALYTIC DERIVATION OF MAIN RESULTS

Prom [1], p. 137, the formula for area of a hyperbolic n-gon (henceforth referred to

as an n-gon) is given by n

A^n{n-2)-J20k, (2-1) k=l

where 6k represents the interior angle of the k^^ vertex. This formula, though geo­

metrically elegant, is not computational in nature unless the interior angles are given.

The following theorem provides a simple computational formula for the hyperbolic

area (henceforth referred to as area), where the coordinates of the vertices are given

that also yields an equally elegant geometric interpretation. The following proof and

interpretation assume that the vertices are in the interior of the unit disc, though in

Chapter III a proof will be given that removes this restriction.

Theorem 2.0.1. Consider the Poincare disc model for hyperbolic geometry. Let

(x i ,y i ) , . . . , (xn^Vn) defined in the interior of the unit disc be the positively oriented

vertices of an n-gon. Define (x„+i,y„_|_i) to be (xi,yi). The area is given by the

formula

A = y 2 Tan-^ ( ^^^^^^ " ^^^^^^ ] . (2.2) ^ \ l - XkXk^x - ykyk+ij

Proof. We seek to evaluate equation (1.2) where TZ is the region bounded by 7, the

boundary of the n-gon. The proof is divided as follows:

1. Conversion of the area double integral to a line integral via Green's theorem.

2. Determining a parameterization 7 for the boundary of an n-gon.

3. Evaluation of the area as a line integral over 7.

(1) Green's theorem in the plane states that if B = M(x, y)i + N{x, j/)j is a contin­

uously differentiable vector field on TZ and 7 = dTZ, the boundary of TZ, is a Jordan

curve of bounded variation, then

II i^-^) ""'' = / 1- + -^v- P-3) n OR

Applying Green's formula for x^ -|- y^ < 1, let

2y . 2x — iH

1 — x^ — y^ 1 — x'^ — y' B = M{x, y)\ -h N{x, y)j = , J ,i + ^ ^J- (2-4)

Substituting this quantity into equation (2.3) we obtain

^ = / / ( r 3 i r 7 ) ' d x d ! , = / j - | - ^ ( . d y - v d x ) . (2.5)

n 7

Consider 7 = dTZ as a subset of C; then x^ + 2/ = 1 : and zdz = {x — iy){dx + idy)

implies xdy — ydx = Im.{zdz) so that the right hand side of the area equation (2.5)

becomes

A = Im<p ^--dz. (2.6) 2z

7

The above expression for area is sufficient for the proof of equation (2.2), but a

stronger statement is more fruitful for the later development of identities.

Claim: I = § iqljjd^ is purely imaginary. 7

Proof of claim: Clearly T ~ § iltvidz. Adding and converting to real form, we have 7

f + '=/(r3^<'. + -J^d,). (2.7) 7

Applying Green's formula for x^ + y^ < 1, let

B = M(x, y)i + N{x, y)j = ^^ , i + ^ - ^ j . (2.8) I — x-^ — y^ 1 — x^ — y^

Substituting this quantity into equation (2.3), we obtain

° = / ( r r ^ d - + I 3 ^ d y ) = ^ + / . (2.9)

We now have Re(/) = (/ + / ) /2 = 0, which implies / = i Im(/). That is, / is purely

imaginary. This concludes the proof of the claim.

The immediate consequence of this result is

^ - / r ^ d z . (2.10)

(2) By [6], p.89, the isometries in this model of hyperbolic geometry are the Mobius

transformations of the following form Ma{z) — e*^^f , where a is a point in the open

unit disc and ^ is a real number. These mappings are disc automorphisms and form

a group with respect to composition. These properties will be used to parameterize

the geodesic from Zk = (xk.yk) to Zk-^i = (x^+i,2/^+1)- In this chapter it is assumed

that consecutive points are distinct.

Let 7fc : [0,1] —> C be the geodesic from Zk to Zk+i in the open unit disc.

The hyperbolic arclength of 7^, equation (1.1), is invariant under the transformation

^f^i^) = f^tz' Note that Mk{zk) = 0 and Mk{zk+i) = ^ ^ f g ; ^ 0. Define the

quantity

a. = jrr—^ - ^ ^ ^ ^ ^ - (2.11)

Now, l fc+il < 1 implies |Mfc(2:fc+i)| < 1, which forces \ak\ > 1. Since Mk is an isometry,

the image of 7^ under Mk must be the geodesic from 0 to Mk{zk+i)^ Geodesies

firom the origin to a point are Euclidean lines, so an appropriate equation for this

ray is Mk{^k{t)) = t/ok where t G [0,1]. We now have 7fc(t) = Mj^^{t/ak)j where

Ml^^{z) = ^'~+x' ^^ *^^^ ^^ hsYe the following parameterization

lk{t) = l ^ ^ where t G [ 0 , l ] . (2.12) tzk + ak

(3) Given equation (2.12) for each side of the n-gon, the right hand side of equa­

tion (2.10) can be expressed as

A i

k=l

1^ r 2z ^ , , lE/l^gi"^. (213)

7fc

which, after substitution, becomes

^ k=l - 0 1

2 (t±2kSi.\ A (t+akZk\ \tZk+B.kJ dt \tzk+ak)

di. (2.14)

tzk+ak

Algebraic simplification of the integrand yields

1 "" rl 2ak{t -H akZk)

(i fc + afc)(|afe|2-t2)

After partial fraction decomposition, we have

di. (2.15)

1 " r

= 1^1 _2zk/ak__

tzk/ak + 1 lofel +t ' \ak\ -t_

1 1 -I- di. (2.16)

£fc* ak

< 1 For the first integrand, recall that \ak\ > 1, so that for all t G [0,1] we have

This shows that the denominator has a positive real part, so that we may integrate

using the principal branch of the logarithm. Combining this with the last two inte­

grands, we have

(2.17) fc=l fe=l

Gathering real and imaginary parts.

Zk A^T.^^.,[l.fJ., E'» fc=l

1 + Zk_

dk + ^ l n W

k=l lafcP - 1

(2.18)

This implies

A = X : 2 a r g f l + ^ ) = 5 ; : 2 a r g ( - ^ . Zk\ V - . / 1 - \Zk? \

k=l ZkZk+l

n / 1 I |2 \

= 5 Z 2 arg ( u--^^ 12 ( ~ ZkZk+i) ) = ^ 2 arg (1 - ZfcZfc+i)

= V 2 T a n - ^ f ^ ^ ^ ^ i ^ ^ ^ ' ) (since Re(l - . . z ^ i ) > 0); ^ \\3^{1 - ZkZk+i) J

= E^Tan-' . A ; = l ^

^kVk-^i - Uk^k+i

- XkXk-\.i - ykVk+i

n

We now have a convenient computational formula for area. The use of arg and

Tan" , however, suggests the geometric interpretation should be based on naturally

occuring angles in the n-gon. This is the case, but before the formula is given, some

definitions need to be introduced.

Definition 1.) A side of an n-gon is said to be an inward side if it is concave with

respect to the interior of the n-gon in the Euclidean sense.

Definition 2.) A side of an n-gon is said to be an outward side if it is convex with

respect to the interior of the n-gon in the Euclidean sense. A side that corresponds

to a Euclidean line segment will be considered an outward side.

For example, in Figure 2.1, the geodesic from zi to Z2 is outward, and the geodesic

from Z2 to Zz is inward.

Theorem 2.0.2. Consider an n-gon with I inward sides and m outward sides. Then

the area of the n-gon is given by the formula

I m

A = Y.^k-J^Pk (2.19) k=l k=l

where ak and /3k denote the angle measure of the arcs of the inward and outward

sides, respectively.

The proof of this theorem calls for an analytic condition for a side to be inward or

outward. Refer to Figure 2.1. A side is inward if the smaller angle (magnitude-wise,

measured in polar coordinates and from the origin) swept from Zk to Zk-\-i is positive

but not equal to TT. Similarly, a side is outward if the smaller angle (magnitude-wise,

measured in polar coordinates and from the origin) swept from Zk to Zk-\-i is negative,

0, or ±7r. Let k be the positive normal to the complex plane. If Zk and Zk-^i are

visualized as vectors in the plane, then the sign of the cross product {zk x Zfc+i) • k =

^kyk+i—yk^k+i in the k direction will determine the side's inwardness (sign is positive)

Figure 2.1: Illustration of inward and outward sides.

or outwardness (sign is not positive), so that

^kVk+i — yk^k+i > 0 imphes 7jt is inward, and (2.20)

^^kUk+i - ykXk+i < 0 implies jk is outward. (2.21)

Note that this condition is valid even if the vertices lie on the boundary. This imphes

that a term in the sum of equation (2.2) is positive if jk is inward, and non-positive

if 7fe is outward. The sum may be written as

^kUk+i - yjb^fc+i ^=E 7fc-in

2 Tan - 1

1 - XkXk+i - ykVk+i -E 2 Tan-i ( _ M ^ ± L Z ^ ^ ^ ^ ± 2 _ ] \ 1 - XfcXfe+i - ykyk+ij

(2.22)

or equivalently as

^=E 7fe-lTl

2arg 1 + - E Ik-out

2 arg 1-f Ofc

(2.23)

2 a r g ( l + f t ) The only step that remains is to geometrically interpret the angles

Since the geodesies are actually segments of circles, the angular measure of the arc

from Zk to Zk-^i is simply the angle swept out by the tangent vector from i = 0 to t = 1,

that is the magnitude of the angle between the vectors 7][.(0) and 7^(1) (Figure 2.2).

If we represent these vectors by (ai,6i) and (a2,62)) respectively, then the cosine of

the angle 6k between them is cos(6k) = ^ ° +y^^ ^. If the vectors are represented

7

Figure 2.2: Angle 6k traversed by the tangent vector,

by Ol -f ibi and 02 + ib2, then

cos (4 ) = Re(7U0)7Ul))

W(o)IK(i)l Substituting the values for 7^(0) and 7j[.(l), we obtain

COs(^fc) Re

= Re

= Re

Re Qfckfc + QfcP

ak{zk + ak)'

«^«l(l + Ht)' = Re (i + ft)

1 + ^

I H - ^

= cos arg 1 -I-

1 + ^ Ofe.

- \ 2'

(i + t ) (i + ft)

(i + ft) Re 1 + ?

ak

COS 2 arg 1 + dk

Now, both jb and 2 a r g ( l + l t )

Ok =

(2.24)

(2.25)

are between 0 and TF, SO we conclude that

Zk^ 2 arg 1-t-

Ofc (2.26)

By virtue of equation (2.23), the theorem is proven, and we have an expression for

area with a convenient analytic and geometric form. •

8

CHAPTER III

GEOMETRIC DERIVATION OF MAIN RESULTS

The derivation in Chapter II was based on an analytic approach to hyperbolic

geometry. The alternative derivation of equations (2.19) and (2.2) in this chapter is

more geometrically motivated and results in an extension of these equations to the

closed unit disc. The respective strengths and weaknesses of these methods of proof

will be discussed in the concluding remarks. The following theorem (modified for our

purposes) is found in [3], p.250.

Theorem 3.0.3. Let z{t) = a{t)-\-ib{t), t G [to^ti], with a,b e C^[to,ti\ and \z{t)\

identically 1. There exists (f){t) G C^to^^i] '^ith z{t) = e^^^^\

Note that the difference between any two functions (f){t) is a constant multiple of

27r. In the context of our problem, z{t) will be the tangent vector Tk{t) = rHtTI-

We have only defined a parameterization when all vertices are in the interior of the

unit disc, but a parameterization certainly exists when this restriction is removed.

Essentially, 0(t) is the differentiable function yielding the angle with respect to the x-

axis of Tfc(f) as it travels over the boundary of the n-gon. The boundary is piecewise

C^, so that except at vertices, 0(i) exists. Adhering to convention, clockwise angles

will be negative, and counterclockwise angles will be positive. Define exterior angles

at the vertices in the following way:

6ext = TT — dint, (3.1)

where 6int is the positive interior angle at each vertex. These quantities are also

defined for vertices which lie on the boundary of the unit disc. Note that 6ext can be

either positive or negative, depending on whether the interior angle is less than or

greater than TF.

The following statements and formula are found in [3], pp.265-268.

Let 7 : [0, /] -> E^ be a positively oriented parameterization of a Jordan curve where

(a) 7 is one-to-one.

(b) Let 0 = tQ,ti,... ,tn = I he di partition of [0, /] where 7 is C^ on (to, ti)(J ...U

(tn-iytn) and 7'(t) ^ 0 for all t in this range. Also let 7 ; have non-zero left and right

derivatives at i i , . . . , in-i with a non-zero right derivative at to and a non-zero left

derivative at t„. Let 0 • • • ^n-i denote the external angles of 7, where 0i '• [U^ti^i] -^ R

is a function (described in Theorem 3.0.3) which describes the rotation ofthe tangent

vector on [fi,ti_|_i].

If the above is true, the following formula holds:

n—1 n—1

J2 (Mtk+i) - Mtk)) + J2^k = 27r. (3.2) fc=0 k=Q

This formula, though ultimately topological, states the intuitive notion that the total

angle swept by the tangent vector, including jumps is equal to 27r.

Let 7 be the boundary of an n-gon. Refer to Figure 3.1. In this instance, the

geometric interpretation of 0fc( fe+i) - <l>k{tk) is |0fc(*fc+i) - <t>kitk)\ = otk, where a^,

0 < QJfc < TF, is the angle measure of the arc of the geodesic from ^{tk) to 7(tfc_|_i). If

the geodesic is inward, then the angular difference is negative, whereas ifthe geodesic

is outward, the angular difference is positive or 0. This yields the following:

(t>k{tk+i) - <f>k{tk) = -o^k if the geodesic is inward; (3.3)

(f>k{tk-\-i) — <f>k{tk) — Oik if the geodesic is outward. (3.4)

If, in equation (3.2), we break the first sum into the sum over I inward sides,

ao,...,Q;/_i and m outward sides, /3o,...,^m-i ^^<^ note equation (3.1), then we

have l—l m—l n—1

-Y^'^k + Y^Pk + Y^iT^- {eint)k) = 27r. (3.5) k=0 k=0 k=0

After rearrangement and reindexing, beginning from A; = 1, we have

n I 771

7r(n - 2) - ^{9int)k = J ] ^ ^ " ! E ^ ^ ' (3-6) A ; = l fc=l k=l

but the left hand side is simply equation (2.1) for the area, so that we have proven,

in an alternative fashion, equation (2.19). Additionally, this formula is now valid for

10

Figure 3.1: Example of (pk{tk-\-i) — <f>k{tk) for outward (left) and inward (right) sides.

the closed unit disc.

We will now derive equation (2.2) geometrically so that it will be valid on the closed

disc. In light of criteria (2.20) and (2.21), all that is necessary to show is that the angle

2 Tan" •G ^kyk-\-\-yk^k+\ -xkXk-\-i~ykyk+i

This measure a^ of the geodesic between Zk and z^+i is

is clearly the case if Zk and z^+i lie on some ray throught the origin i.e., when Xkyk-\-i ~

VkXk+i = 0, which is equivalent to Im.{zkZk-]-i) = 0. So, assume that Im{zkZk+i) 7 0.

Let Zk^Zk^i G 5i(0)2. Refer to Figure 3.2. The Euclidean line segment (zk^Zk^i) has

midpoint ^^^'^^^. The geodesic from Zk to Zk+i must be a segment of a circle C which

intersects the unit disc at a right angle. Let c, a complex number, denote the center

of C. If r is the radius of C, then clearly |cp = r^ -h 1. A parameterization of C is

z = c-\- \ / | cP — 1 e* , where z is any point on C and ^ is a real number. (3.7)

Basic algebra shows that

zc + zc= 1 + \z\ (3.8)

In particular, this is satisfied for z = Zk and z — Zk+v

ZkC + ZkC = H-I^fcl^,

Zk+lC + Zk+lC = l-h|^fc+lP

(3.9)

(3.10)

11

Figure 3.2: Illustration for equation (3.12)

Solving for c we obtain

^ ^ {l + \Zk\'^)Zk+l-{l-\-\Zk-^l\'')Zk

ZkZk+l — ZkZk+l

Geometrically, we have ak given by the following formula,

(3.11)

ak = 2 Tan"^

After substitution of c,

ak = 2 Tan"^

2

c — Zk+Zk+l

= 2 Tan - 1 Zk+l — Zk

2c — Zk- Zk+i (3.12)

( fc+l - Zk){ZkZk+l — ZkZk+l)

2(1 -I- \zk\^)zk+i - 2(1 + \zk+i\'^)zk - {zk -h Zk+i){zkZk+i - ZkZk+i) (3.13)

Simplification yields

ak = 2 Tan"^

We now have the following,

lm{zkZk+i)

(Zjt+i - Zk){ZkZk+l - ZkZk+l)

{zk+i - Zk){2 — Zk-\.iZk — Zk-\-iZk) (3.14)

ttfc = 2 Tan"^ R e ( l - ZkZk+i)

2 Tan - 1 / XkVk+i - VkXk+i

11 - XkXk+i - ykVk+i (3.15)

This shows that equation (2.2) is indeed valid for the closed disc.

12

CHAPTER IV

IDENTITIES AND APPLICATIONS

In this chapter, several identities will be developed which hinge on the analytic

method of Chapter II, in particular equations (2.10) and (2.18). Among these identi­

ties is a relatively simple expression which couples area and perimeter, as well as two

alternative formulas for area, depending upon whether 0 is in the interior or exterior

of the n-gon.

If 7 is the parameterization for the n-gon, then, after calculation of equation (1.1)

the perimeter is

P = T\n (Pl^) =J22 Tanh- JL. (4.1) tt v K i - i ; ^ ki

We can rewrite this as

{Wk\ + iy\ A , / /Ki + i \ V kP ' ' = EM^E^^HEi"

so that we have

p-lfSftrj-'+EKiSr) When substituted into equation (2.17), this yields

. 1 V^^i f..Zk\,P, 1 v ^ , f W

which simplifies to

^ = i y : 2 1 o g 1 + ^ - F ^ - F i V l n ^ l ^ , (4.4)

^ ^ V|afcl' + |afci; '

Surprisingly, we have a nice expression with the real part equal to area and the

imaginary part equal to perimeter.

Equation (2.2) rewritten as

A = I m y " 2 1 o g f l - F — ^ (4.6)

fe V ak) 13

also yields two more area formulas:

A = -An + y ^ 2 arg ( iH ) if 0 is in the interior of the n-gon (4,7) t l V <^kZkJ

A = y 2 arg ( iH ) if 0 is in the exterior of the n-gon. (4.8) t i V <^kZkJ

To prove this result we need the following claim.

Claim: Let a G C \ [-1,0] then f^S^ = log (l-f- ^) , where log denotes the principal

branch of the logarithm.

Proof of claim: If a G C \ [—1,0] and a is real, then clearly the claim is true. If

a G C \ [—1,0] and a is complex, then t + a is non-negative for t G [0,1], so that

we can integrate with the principal branch of the logarithm. We have that /Q 4 ^ =

log(l-t-a)—log(a). The claim is proven if we can show that arg(H-a)—arg(a) G (—TT, TT).

Now, Im (1 -h a) = Im (a) implies

arg(l + a) e [0,7r) 4=> arg(a) G [0,7r), and

arg(l + a) G (-TT, 0) <=^ arg(a) G (-TT, 0).

In either case, it is clear that arg(l -\-a) — arg(a) G (-TF, TT). This concludes the proof

of the claim.

To prove equations (4.7) and (4.8), consider / = f \^z, then I = Y11=IIQ ^tH^*' 7

where ^k{t) = It^t-Z ^^^ * [ ' ^]- Applying algebra,

i-f:[\ "^^^'^''''^ ,di (4.9)

friJo {t + cikZk){tZk + ak)

and partial fractions,

i = y -^^dt - y / ^^^-dt. (4.10) ^^JQ ^ + 0-kZk frt Jo t^k + Ofc

14

By the claim and previous work, this becomes

^ = E>o<i + i ) - E i o g ( i + | ) . (4.n)

Taking the imaginary part of each side of the equation and rearranging terms we have

A = -2 Im(/) + V 2 arg f 1 + - ^ " j . (4.12)

Now I = 27rm(7), where n(7) is the wrapping number of 7 about 0, so in the first

case, / = 27ri, and in the second case, 7 = 0. This proves the result.

These identities are interesting because they perhaps suggest that sufficient alge­

braic relations exist between area and perimeter to yield a purely analytic solution to

the isoperimetric problem: Given an n-gon with fixed perimeter, maximize (if such

a maximum exists) the area of the n-gon. This has been proven by [2], but a proof

of existence of a solution is not given. By a personal communication from Dr. Roger

Barnard via Dr. Petros Hadjicostas (Texas Tech University), we seek to show that

A<.{n-2)- 2n Sin- f ? ^ ) , (4.13) - ^ ^ Vcosh(P/2n)y ^ ^

which with minimal effort can be shown to be equivalent to

Re cos — +i-z-]> cos(7r/n). (4.14) \ 2n 2n /

This is similar in spirit to the above identities.

Another use of equation (2.2) is the proof of existence of a solution to the isoperi­

metric problem. Denote r G 5i(0)" by r = (^i, . . . jZn)- Denote the area equa­

tion (2.2) by A{r) and equation (1.1) by P{r). The area and perimeter functions are

certainly defined for any such configuration r G 5i(0)", though the points Zx,...,Zn

may correspond to self intersecting n-gons or n-gons with negative orientation. Work­

ing on this expanded set and given a fixed perimeter, the maximum value of the area

function exists. This is expressed by the following theorem.

15

Theorem 4.0.4. If P > 0 is fixed, then A{r) attains its supremum over the set

S = {reBi{0)^:P = P{r)},

Proof Define Si = {r e Bi(0)^ : P = P(r),zi = 0}. Denote the range of A over S

and ^i as A{S) and ^ (^ i ) , respectively. Also, Si C S implies A{Si) C A{S). We will

now show the opposite inclusion. If a G ^4(5), then there exist ra = {zi,.. ^ ,Zn) E: S

such that a = A{ra). Let M{z) be an isometry of the disc such that M{zi) = 0.

Define M(ra) = M{zi,...,Zn) tobe {M{zi),... ,M{zn)). Then M(ra) can be written

as M{ra) = (0,4? • • *) -s n) = ^1 ^ ^i- M is an isometry, meaning that, in this case,

a = A{ra) = A{M{ra)) = A{r^J G A{Si). This shows that A{S) C A{Si). That is,

A{S) = A{Si). Note that A{r) is continuous on ^i .

If we show that ^i is closed, then Si is compact, and ^(51) (hence A{S)) attains

its supremum, and we have proven the theorem. In order to prove that Si is closed,

we first need to show that Si C J5i(0)^. Let r = {zi = 0,2:2, • - • ,Zn) G Si. Denote

the hyperbolic distance between two points Zj^Zk by d{zj,Zk)' Explicitly, d{zj,Zk) =

2 Tanh"^ Zfc-Z. If j is an integer from 1 to n, then by the triangle inequality and l - Z j Z f c

the fact that P = Yl^=i d{zk, Zk+i),

d{0,Zj) < d{0,Z2)-\- ...-\-d{zj.uZj)

< P. (4.15)

This yields 2 Tanh~^|zj| < P, implying \zj\ < tanhP/2, so that Zj e BtanhP/2(0).

Now r e 5tanhP/2(0)" lesults with Si C 5tanhP/2(0)", hence 5^ C 5tanhP/2(0)". This

means 5T C Bi(0)" since for all real P, t anhP/2 < 1.

We will now show that ^i is closed. If / = ( z i , . . . , ^n) is a limit point of Si, then

there exists {rm}^=i C ^i such that lim^^oo rm = I- For each r„i, the first coordinate

is 0, so that Zi = 0, the first condition for I E Si. Now, P{r) is continuous at I so

lim^^oo P{rm) = limm- oo P = P = P{1), the second condition for I e Si. Therefore

^i is closed, and the theorem is proven. •

16

CHAPTER V

CONCLUDING REMARKS

The two proofs of equation (2.2) and its geometric interpretation equation (2.19)

have respective strengths and weaknesses. The analytic derivation (Chapter II) has

the advantage of yielding some unexpected relationships between area and perimeter,

and even (with relatively little work) alternative formulas for the area provided that

the origin does not lie on the boundary of the n-gon. This proof relies on the fact

that hyperbolic geometry is generated by a conformal metric. The drawback to this

method is that the proof is only valid if all of the vertices are in the interior of

the unit disc. Otherwise the conformal metric has a singularity which destroys the

validity of Green's theorem. This proof, in particular equation (2.10), allows for some

unexpected identities involving the perimeter and the area, and it suggests a direct

link to the solution of the isoperimetric problem. One such link is the existence

of a maximum configuration if the set of Jordan n-gons is expanded to include self

intersections. Potentially, this method could also be applied to geometries generated

by other conformal metrics, for instance, higher dimensional hyperbolic and spherical

geometries by [5].

The geometric proof (Chapter III) relies only on the equations (2.1), (3.2) and a

geometric knowledge of the geodesies. The benefit of this proof is its brevity, and it

avoids the singularity of the conformal metric. This proof establishes equations (2.2)

and (2.19) on the entire unit disc, but does not suggest any ofthe identities developed

from the analytic method. In conclusion, we see that apart from their inherent appeal,

equations (2.2) and (2.19) have ranging applicability to hyperbohc geometry as a

whole.

17

BIBLIOGRAPHY

[1] Anderson, J. W. Hyperbolic Geometry. London, Springer. 1999.

[2] Bezdek, K. Ein elementarer Beweis fiir die isoperimetrische Ungleichung in der Eucklidischen und hyperbolischen Ebene. [An elementary proof for the isoperi­metric inequality in the Euclidean and Hyperbolic planes.] Ann. Univ. Sci. Budapest Eotvos Sect. Math. 27 (1984) 107-112 (1985).

[3] Do Carmo, M. P. Differential Geometry of Curves and Surfaces. Upper Saddle River, New Jersey, Prentice-HaU Inc. 1976.

[4] Needham, T. Visual Complex Analysis. New York, New York, Oxford Univer­sity Press. 1997.

[5] Ratcliffe, J. G. Foundations of Hyperbolic Manifolds. New York, Springer-Verlag. 1994.

[6] Stillwell, J. Geometry of Surfaces. New York, Springer-Verlag. 1992.

18

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