1.1 four classes of surfaces - university of oxford...examples 1.1 four classes of surfaces our goal...

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B3.2 GEOMETRY OF SURFACES Mathematical Institute, Oxford. Prof. Alexander F. Ritter. Comments and corrections are welcome: [email protected] Contents 1 Examples 2 2 Definition of surface 12 3 When are two surfaces different? 18 4 The Euler characteristic 20 5 Classification of surfaces 25 6 Orientability 26 7 Local analysis: the inverse and implicit function theorems 30 8 Local analysis: Embedded surfaces are locally graphs 33 9 The tangent space 36 10 Surfaces in R 3 : the first fundamental form 40 11 Surfaces in R 3 : the second fundamental form 49 12 Curvature 54 13 Tangential derivatives and Gauss’ Theorema Egregium 60 14 Geodesic curvature and the Gauss-Bonnet theorem 65 15 Morse functions, Poincar ´ e-Hopf and Hairy Ball Theorem 71 16 Geodesics 76 17 Geodesic normal coordinates 81 18 Surfaces of constant curvature 83 19 Riemann surfaces: holomorphic maps and Riemann-Hurwitz 84 20 Riemann surfaces: Meromorphic functions 89 21 Hyperbolic Geometry: an introduction 94 22 Appendix: Classification of Riemann surfaces 101 B3.2 Course policy: It is essential that you read your notes after each lecture, otherwise you may feel lost. In the third and fourth year courses, the majority of courses will not cover all the material in lectures. You are expected to read the lecture notes to complement the lectures. Geometry of surfaces is a dicult and vast course, and I will do my best to make it digestible. But this will not happen by itself: it requires eort on your part, thinking on your own about the notes, the examples, the exercises. B3.2 Homework policy: Homeworks are typically much harder than the exams: the aim of the homeworks is to make you better mathematicians, to stretch you and to inspire you. The homeworks are not designed to assess your basic understanding of the course (unlike exams). So homework marks are not aiming to predict your exam marks. Date : This version of the notes was created on February 7, 2018. 1 2 B3.2 GEOMETRY OF SURFACES, PROF ALEXANDER F. RITTER 1. Examples 1.1 Four classes of surfaces Our goal is to study and relate three classes of surfaces: (1) Topological surfaces (topological 2-manifolds ), (2) Smooth surfaces (smooth real 2-manifolds ), (a) embedded in R 3 , (b) abstractly (i.e. possibly without a choice of embedding into R N ), (3) Riemann surfaces (complex 1-manifolds ). We will postpone the precise denition to later. For now, the rough idea is that a surface locally looks like a 2-dimensional disc. Whether it looks like the disc continuously, smoothly or holomorphically distinguishes the cases (1), (2), and (3) respectively. The reason for study- ing (2a) before (2b) is that you already know what it means for functions on R 3 to be smooth (innitely dierentiable), whereas in (2b) the denition is a little more dicult because you need to rst dene local smooth coordinates on the surface. Some surfaces are part of all four classes (such as a torus), others only of some, but all our surfaces belong to class (1). Relation to future Part B and Part C courses. You can study Riemann surfaces (and more generally algebraic curves) using tools from alge- bra, in B3.3 Algebraic Curves. The word manifold is the generalization of surface to higher dimensions, so n-manifold means your space locally looks like R n (or rather, like a ball in R n ). C3.3 Dierentiable Manifolds, C3.5 Lie Groups (manifolds which are also groups) and C7.5/7.6 General Relativity all study manifolds from various perspectives. Complex manifolds locally look like C n , and you can study them (in greater generality) using tools from algebra in C3.4 Algebraic Geometry and in C3.7 Elliptic Curves, both of which build upon B3.3. The best tools to study topological manifolds (and more general topologi- cal spaces) come from topology and algebra, and this is done in C3.1: Algebraic Topology. Further references for this course: The notes from 2013 by Prof Nigel Hitchin (online). The notes from 1986 by Prof Graeme Segal (google for “graeme segal geometry of surfaces”). Manfredo P. do Carmo, Dierential Geometry of Curves and Surfaces. Pelham M. H. Wilson, Curved Spaces. Several good references are suggested in the syllabus. Analysis and topology dictionary: On the online course page you will also nd the handout: Analysis and Topology Dictionary – Handout which summarises various useful terminology (e.g. topological space, Hausdor, connected, compact, continuous, homeomorphism, smooth, dieomorphism, holomorphic, etc.)

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B3.2

GEOM

ETRY

OF

SURFACES

Math

ematicalIn

stitute,Oxford

.

Pro

f.AlexanderF.Ritter.

Comments

and

corrections

are

welcome:

ritter@

math

s.ox.ac.uk

Contents

1Examples

2

2Definitionofsu

rfa

ce

12

3W

henaretwo

surfa

cesdifferent?

18

4TheEulercharacteristic

20

5Classificationofsu

rfa

ces

25

6Orientability

26

7Localanaly

sis:

theinverse

and

implicit

functiontheorems

30

8Localanaly

sis:

Embedded

surfa

cesarelocally

graphs

33

9Thetangentspace

36

10Surfa

cesin

R3:thefirst

fundamentalform

40

11Surfa

cesin

R3:these

cond

fundamentalform

49

12Curvature

54

13Tangentialderivativesand

Gauss’TheoremaEgregium

60

14Geodesiccurvatureand

theGauss-B

onnettheorem

65

15Morse

functions,

Poincare-H

opfand

HairyBallTheorem

71

16Geodesics

76

17Geodesicnormalcoordinates

81

18Surfa

cesofconstantcurvature

83

19Riemannsu

rfa

ces:

holomorphic

mapsand

Riemann-H

urwitz

84

20Riemannsu

rfa

ces:

Meromorphic

functions

89

21Hyperbolic

Geometry:anintroduction

94

22Appendix:ClassificationofRiemannsu

rfa

ces

101

B3.2

Coursepolicy:It

isessentialthatyoureadyournotesafter

each

lecture,otherwise

youmayfeel

lost.In

thethirdandfourthyearcourses,themajority

ofcourses

willnotcover

all

thematerialin

lectures.

Youare

expected

toreadthelecture

notesto

complemen

tthe

lectures.

Geometry

ofsurfacesis

adiffi

cultandvast

course,

andIwilldomybest

tomake

itdigestible.

Butthis

willnothappenby

itself:itrequires

effort

onyourpart,thinkingonyour

ownaboutthenotes,

theexamples,

theexercises.

B3.2

Homework

policy:Homew

orksare

typicallymuch

harder

thantheexams:

theaim

of

thehomew

orksis

tomake

youbetter

mathem

aticians,

tostretchyouandto

inspireyou.The

homew

orksare

notdesigned

toassessyourbasicunderstandingofthecourse(unlike

exams).

Sohomew

ork

marksare

notaim

ingto

predictyourexam

marks.

Date:This

versionofth

enoteswascrea

tedonFeb

ruary

7,2018.

1

2B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

1.

Examples

1.1

Fourclassesofsu

rfaces

Ourgoal

isto

studyandrelate

threeclasses

ofsurfaces:

(1)Top

ologicalsurfaces(topological2-m

anifolds),

(2)Smoothsurfaces(smooth

real2-m

anifolds),

(a)em

bedded

inR

3,

(b)ab

stractly

(i.e.p

ossibly

withoutachoiceofem

beddinginto

RN),

(3)Rieman

nsurfaces(complex1-m

anifolds).

Wewillpostpon

etheprecise

definitionto

later.

Fornow

,theroughidea

isthatasurface

locallylook

slikea2-dim

ensionaldisc.

Whether

itlookslikethedisccontinuously,

smoothly

orholom

orphically

distinguishes

thecases(1),(2),and(3)respectively.

Thereasonforstudy-

ing(2a)

before(2b)isthatyoualreadyknow

whatitmeansforfunctionsonR

3to

besm

ooth

(infinitelydifferentiable),

whereasin

(2b)thedefinitionis

alittle

more

diffi

cult

because

you

needto

firstdefinelocalsm

ooth

coordinatesonthesurface.Somesurfacesare

part

ofallfour

classes(such

asatorus),othersonly

ofsome,

butalloursurfacesbelongto

class

(1).

Relation

tofutu

rePart

Band

Part

Ccourses.

You

canstudyRiemannsurfaces(andmore

generallyalgebraic

curves)usingtoolsfrom

alge-

bra,in

B3.3

Algebra

icCurv

es.

Theword

manifold

isthegeneralizationofsurface

tohigher

dim

ension

s,so

n-m

anifold

meansyourspace

locallylookslikeR

n(orrather,likeaballin

Rn).

C3.3

Differe

ntiable

Manifolds,

C3.5

Lie

Gro

ups(m

anifoldswhichare

alsogroups)

andC7.5/7.6

Genera

lRelativityallstudymanifoldsfrom

variousperspectives.Complex

man

ifoldslocallylooklikeC

n,andyoucanstudythem

(ingreatergenerality)usingtools

from

algebra

inC3.4

Algebra

icGeometryandin

C3.7

EllipticCurv

es,

both

ofwhich

buildupon

B3.3.

Thebesttools

tostudytopologicalmanifolds(andmore

generaltopologi-

calspaces)comefrom

topologyandalgebra,andthisisdonein

C3.1:Algebra

icTopology.

Furtherreferencesforth

iscourse:

Thenotes

from

2013byProfNigel

Hitchin

(online).

Thenotes

from

1986byProfGraem

eSegal(google

for“graem

esegalgeometry

ofsurfaces”).

Man

fredoP.doCarm

o,Differen

tialGeometry

ofCurves

andSurfaces.

Pelham

M.H.W

ilson,Curved

Spaces.

Several

goodreferencesare

suggestedin

thesyllabus.

Analysisand

topologydictionary

:Ontheon

linecoursepageyouwillalsofindthehandout:

AnalysisandTopology

Dictionary

–Handout

whichsummarises

varioususefulterm

inology(e.g.topologicalspace,Hausdorff,connected,

compact,

continuou

s,homeomorphism,sm

ooth,diffeomorphism,holomorphic,etc.)

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

3

1.2

Examplesin

each

ofth

eth

reeclasses

(1)TOPOLOGIC

AL

SURFACES:

•Gluingedgesof

asquareyieldsvariou

ssurfaces:

•A

cubeis

atopological

surface,

andso

aretheother

regu

larpolyhed

ra(tetrahedro

n,

octa

hedro

n,icosa

hedro

n,dodecahedro

n).

Butin

fact,they

aretopologicallythesame

(i.e.hom

eomorphic)to

thesphere.

Indeed,thinkingof

thepolyhedra

assittinginsideR

3,

pickahuge

spherewhichcontainsthepolyhed

ronan

dthen

simply

“continuou

slypush”each

pointof

thepolyhedronradiallyou

twardsuntilthepointreaches

thesphere.

This

defines

ahom

eomorphism

(con

vince

yourselfof

this!)1

1It

isusefulto

get

anacq

uaintance

forsp

ottingwhether

somethingis

ahomeo

morp

hism

ornot,

without

thepainstaking

effort

ofwriting

down

an

explicitform

ula

(ifoneis

likely

tomakea

mistakein

spotting

homeo

morp

hisms,

then

oneis

probably

alsolikelyto

write

downaninco

rrectform

ula!).Howev

er,in

this

case

4B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Rem

ark.Theim

ageoftheedges/verticesofthepolyhedrondividethesphereinto

curved

polygons(e.g.on

therightwegotatriangulationofthesphere).

•Theto

rus(upto

homeomorphism)is

alsoapolyhed

ron(non-regular,

non-convex):

(2a)SM

OOTH

SURFACES

INR3:

•TheR

2planeinsideR

3,so

R2=

{(x,y,z)∈R

3:z=

0}.

•Moregenerally

aplanethroughq∈R

3withunit

norm

aln∈R

3:

{p∈R

3:(p

−q)

·n=

0}=

{(x,y,z)∈R

3:n1x+n2y+n3z=

n1q 1

+n2q 2

+n3q 3}.

•Theunit

sphere

inR

3:S2=

{p∈R

3:�p

�=

1}=

{(x,y,z)∈R

3:x2+y2+z2=

1}.

itis

quiteea

sy:if0lies

onth

einsideofth

epolyhed

ron,th

enth

emapx�→

x �x�

isanex

plicitmapfrom

the

polyhed

ronto

theunit

sphere,

oneth

enea

sily

checksth

atit

isaco

ntinuousbijection,andfinallyoneusesth

e

gen

eralth

eorem

thataco

ntinuousbijectionfrom

aco

mpact

space

toaHausd

orff

space

isahomeo

morp

hism.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

5

Noticethat

neareach

pointpyo

uhavetw

oindep

endentlocalsm

oothcoordinates:at

leasttw

oof

thecoordinates

x,y,z

willwork.For

exam

ple,neartheNorth

Polep=

(0,0,1)yo

ucanuse

localcoordinates

x,y

touniquelydescribenearbypoints

since

(x,y):(neigh

bou

rhoodof

p)→

(neigh

bou

rhoodof

0)⊂

R2isasm

oothhom

eomorphism.How

ever

nearpyoucannot

use

x,z

astherearepoints

ofthesphere(x,y,z),

(x,−

y,z),

neartheNorth

Pole(soz≈

1),which

wecannot

tellap

artusingjust

x,z

(sothey

arenogo

odas

coordinates).

•Generalizingtheab

ovequad

raticequation

yieldsellipso

idsan

dhyperb

oloids:

Ellipsoid:

x2

a2+

y2 b2+

z2 c2=

1

Hyperboloidwithon

esheet:

x2

a2+

y2 b2−

z2 c2=

1

Hyperboloidwithtw

osheets:

x2

a2−

y2 b2−

z2 c2=

1,

wherea,b,c

arefixed

constan

ts.Man

yother

surfaces

defined

byquadraticpolynom

ials

will

reduce

toon

eof

theseexam

plesafter

chan

gingcoordinates

(bydiago

nalising).

•A

toru

sin

R3:fixconstan

tsa>

b>

0,then

wecandescribeatorusby

T2

={((a

+bcosψ)cosθ,

(a+

bcosψ)sinθ,

bsinψ):allθ,ψ∈[0,2π]}

•Man

ymoreexam

plesariseassu

rfacesofre

volution,in

whichacurvein

the(x,z)-plane

gets

rotatedab

outthez-axis.For

exam

ple,take:

�Curve=

vertical

line,

then

rotatingaroundthez-axis

gives

acylinder.

�Curve=

straightlinewhichis

neither

vertical

nor

horizontal,then

rotatingarou

nd

thezax

isgives

twoop

positeconestouchingat

avertex.Atthevertex,thesurface

isnot

smooth,so

weneedto

removethevertex.

�Curve=

anellipse

x2

a2+

z2

b2=

1,aparab

olaz=

x2,orahyperbolaxz=

1,then

rotatinggives

anellipsoid,paraboloid,hyperboloid.

Forexam

ple

thepara

boloid

would

be{(x,y,z):z=

x2+

y2}.

•Ruled

surfaces:

thesearesurfacessw

eptou

tmyamov

ingstraightline,

{p(t)+

sn(t):s,t∈R}

sothestraightlineat

timetis

p(t)+

Rn(t)(a

straightlinethrough

thepointp(t)∈

R3

which

isparallelto

theunit

vector

n(t)∈

R3).

How

ever,somecare

isneeded:

not

all

6B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

choicesof

p(t),n(t)willgiveasm

ooth

surface.Forexample,forp(t)=

(cost,sint,0)and

n(t)=

(sin

t,−cost,1)/√2,weget

theone-sheetedhyperboloid

x2+y2−z2=

1:

•Surfacescutoutbyoneequation:

{(x,y,z)∈R

3:f(x,y,z)=

0}.

Man

y,butnot

all,choices

ofsm

ooth

f:R

3→

Rensure

this

isasm

ooth

surface.

(2b)ABSTRACT

SURFACES:

Thequestion

“when

isatopologicalsurface

smooth?”

does

notquitemake

sense.

For

exam

ple,thestan

dard

sphereS2⊂

R3is

asm

ooth

surface,butthecube⊂

R3is

notsm

ooth

dueto

thecorners,

yetboth

are

homeomorphic

topologicalsurfaces.

Thecorrectquestionis

“when

canwedefineasm

ooth

structure

onagiven

topologicalsurface?”.1

Wenow

explain

how

wecangive

acubein

R3a“sm

ooth

structure”.Nearavertex,thecoordinatesx,y,z

donot

vary

smoothly.2

How

ever,wecan

definesm

ooth

localcoordinatesby

composing

thehom

eomorphism

cube→

S2⊂

R3with

smooth

localcoordinatesforS2.

Such

local

coordinates

nearthevertex

ofthecubewillnotbesm

ooth

functionsoftheoriginalx,y,z

coordinates.Noticethatwhatwehavereallydoneis

defineasm

ooth

structure

onthecube

byrequiringthehom

eomorphism

cube→

S2⊂

R3to

besm

ooth!

Thereason

theabovemay

atfirstseem

perplexing,is

thatR

3is

causingunnecessary

confusion

:thereis

noreasonforconsideringsurfacesasalreadysittingsm

oothly

insideR

3.

Indeedsomesm

oothsurfacescannotbeembeddedinsideR

3(hereem

bedded

roughly

means3

asm

oothinjectivemap).

Forexample,theKlein

bottleissm

ooth

(locallyitlookslikeasquare,

sowehavesm

oothlocalx,y

coordinatesfrom

thesquare).

How

ever,it

cannotbeem

bedded

insideR

3(w

ithou

tself-intersections).

•Thetoruscanbeviewed

asthequotientofR

2bythegroupofintegraltranslationsparallel

tothetw

oax

es:

T2=

R2/Z

2=

{[x,y]:(x,y)∈R

2,[x,y]=

[x+n,y

+m]foralln,m

∈Z}

.

1Non-exa

minable:

Theansw

eris,weca

nalw

aysen

dow

atopologicalsu

rface

with

thestru

cture

ofan

abstract

smooth

surface

bych

oosingclev

erloca

lco

ord

inates.

Howev

er,in

higher

dim

ensions(formanifolds)

itca

nhappen

thatatopologicalmanifold

does

notadmit

asm

ooth

stru

cture.This

isaverydiffi

cult

problem

(see

“Relationsh

ipwithtopologicalmanifolds”

athttp://en

.wikiped

ia.org/wiki/Differen

tiable

manifold

).2co

nvince

yourselfofth

is,usingth

atea

chface

isgiven

bysettingoneofth

eco

ord

inatesto

aco

nstant.

3Theprecise

defi

nitionofem

bed

dingis:ahomeo

morp

hism

onto

theim

age.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

7

Wedefinelocalcoordinates

nearapoint[x

0,y

0]∈

T2bysimply

usingthex,y

coordinates

ofR

2nearsomepre-imagepoint(x

0,y

0)∈

R2of

[x0,y

0].

Thereis

aZ2

-worth

ofchoicesof

pre-imagepoints,an

dan

ytw

osuch

choicesof

localcoordinates

willdiffer

byasm

oothmap

(anintegral

tran

slation).

Noticethistorusisnot

sittinginsideR

3,an

dthisconstructionof

thetorusismuch

simpler

toworkwiththan

theab

oveform

ula

forthetorusinsideR

3.For

exam

ple,wehaveanatural

notionof

smoothfunctionf:T

2→

R,nam

elyit

just

1meansasm

oothfunction

� f:R

2→

Rwhichis

tran

slation-invarian

tunder

theab

ovegrou

p,so

� f(x+n,y

+m)=

� f(x,y)forn,m

∈Z.

•Thinkingof

surfaces

asem

bedded

insideR

3also

mak

esitharder

tonoticewhichsurfaces

are

actually

the“sam

e”.For

exam

ple,bycuttingthetorus,

deforming,

andregluing,

weob

tain

thefollow

ingknotted

toru

s:

Thisknottedtorusissm

oothly

hom

eomorphic

totheoriginal

torus(indeed,thinkab

outhow

youwou

ldsetupasm

oothbijection

betweenthem

).Sotheab

stract

surfaces

arethe“sam

e”,

whereasthe“k

nottedness”

isextran

eousinform

ation

havingto

dowith

how

wechoseto

embed

thesurfaceinsideR

3.It

turnsou

tforexam

ple,that

you

cannot

continuou

slydeform

(insideR

3)thetorusinto

theknottedtoruswithou

tcreatingself-intersection

s.2

•Theeff

ortof

definingab

stract

surfaces

isworth

thetrou

ble,an

ditisthemodernviewpoint

ingeom

etry.Thesinfulsecret

ofgeom

etry

isthat

anysm

oothn-m

anifoldcan3besm

oothly

embedded

insideR

Nforlargeenou

ghN

(indeedN

=2n

works).Soon

ecould

inprinciple

only

studysurfaces

embedded

inR

4.For

exam

ple,fortheKlein

bottleyoucanremovethe

self-intersection

that

youseein

R3by“lifting”

onepatch

ofthetw

ointersectingsheets

into

1Compare

this

withth

e1-dim

ensionalca

se:whatdoes

itmea

nto

haveasm

ooth

functiononth

ecircle,

f:S1→

R?It

just

mea

nsasm

ooth

function

� f:R

→R

whichis

1-periodic:� f(x+

1)=

� f(x)(or,

2π-periodic

� f(x+2π)=

� f(x)ifyouth

inkofth

ecircle

asparametrizedbyei

tfort∈

[0,2

π],instea

dofe2

πit

witht∈

[0,1

]).

2This

isatrickyex

ercise

foryou.If

youneedahint,

searchfor“trefoilknot”.

3This

isth

eW

hitney

embeddin

gth

eorem

,andis

wellbey

ondth

escopeofth

isco

urse.

8B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

thefourthdim

ension

.1

•Realpro

jectivesp

aceRP

2(w

hichwealreadymention

edin

(1)ab

ove).Asaset,

RP

2

canbedefined

asthecollection

ofallstraightlines

inR

3through

theorigin.

Such

astraightlineis

determined

byanon-zeropoint(x,y,z)∈R

3\{

0},butrescalingsuch

apointbyan

yλ�=

0∈R

willyield

thesameline.

ThusRP

2arises

asaquotientof

R3\{

0},

whoseequivalence

classesrepresentstraightlines:

RP

2=

{[x,y,z]:(x,y,z)∈R

3\{

0},[x,y,z]=

[λx,λ

y,λ

z]foranyλ�=

0∈R}.

Thesecoordinates

[x,y,z]arecalled

thehomogeneouscoord

inatesforRP

2(they

areonly

defined

upto

rescalingallof

them

byanon-zerorealnumber).

Moreexplicitly,

such

astraightlineisdetermined

bythean

tipodalintersection

points

{x,−

x}

inS2.Sowecanlocallyparametrize

thisspace

byadisc,

just

likeforS2,since

nearbystraight

lines

areparam

etrizedbypoints

inS2close

tox∈S2(thenearbylines

form

adou

ble

cone

withvertex

attheorigin).

Thereforewecan

view

RP

2asthequotientofS2bythegroup

{Id,A

}generated

bythe

antipodal

map

A:S2→

S2,A(x)=

−x:

RP

2=

{x∈S2}/

(x∼

−x).

Noticethis

also

recovers

thedefinitionofRP

2in

(1)above,

since

weon

lyneedtheupper

hem

isphereto

findarepresentativeofeach

point.

(3)RIE

MANN

SURFACES:

Thesearesurfaces

which“h

olomorphically”locallylooklike

D=

{z∈C

:|z|<

1},

solocally

thereis

acomplex

coordinatez.

This

isroughly

thesameas

hav

ingtw

oreal

coordinates

x,y

withanotionof“rotation

by90

◦ ”(m

ultiplication

byi),so

that

z=

x+

iy.

•Complexpro

jectivesp

aceCP

1isthesphereS2asatopolog

ical

surface,wenow

define

localholom

orphic

coordinates.

Onelocalcomplexcoordinatez,defined

everywhereexceptat

theNorth

Pole,

isobtained

byusingthestereographic

projectionfrom

theNorth

Pole:

1Asananalogy:forafigure

8loop(twoloopsjoined

atapoint)

inth

e(x

,y)-planez=

0in

R3,atth

e

crossingyouhavetw

olines

intersecting:youca

nremoveth

eself-intersectionbyslightlyliftingvertica

llyone

ofth

etw

olines.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

9

Explicitly,

S2\(

North

Pole)

isidentified

withC

via

S2\(

North

Pole)

�(X

,Y,Z

)�→

X

1−

Z+

iY

1−

Z=

z∈C,

whereX

2+

Y2+

Z2=

1.Wecanalsodefinealocalcomplexcoordinatew

bytakingthe

conjuga

teof

thestereograp

hicprojectionprojectingfrom

theSouth

Pole.

1SoS2\(Sou

thPole)

isidentified

withC.

Noticewehaveidentified

S2\(North

∪South)in

twodifferentwayswithC\{

0},correspon

ding

tothetw

ocoordinatesz,w

.

Exercise.Show

that

thetw

ocoordinates

arerelatedby

w=

1/z.

Noticethatthis

changeofcoord

inates,

C\0

→C\0

,z�→

1 zis

aholomorphic

map

(i.e.

complexdifferentiab

le).

Infact,wewillseethat

partof

thedefinitionof

Rieman

nsurfacesis

that

coordinatechan

gesmust

beholomorphic

(for

smoothsurfaces

they

must

besm

ooth).

•A

moregeneral

way

tothinkof

CP

1,in

analog

ywithRP

2above,

isas

thesetof

complex

lines

2through

0in

C2.Thus:

CP

1=

{[z 0

:z 1]:(z

0,z

1)∈C

2\{

0},[z 0

:z 1]=

[λz 0

:λz 1]foran

yλ�=

0∈C}.

Then

theregion

z 0�=

0correspon

dsto

S2\(

North

Pole)

above,

andyo

ucanrescale

sothat

[z0:z 1]=

[1:z],so

youob

tain

theab

ovelocalcoordinate

z=

z 1/z

0

withz=

0beingtheSou

thPole.

Whereasz 1

�=0correspon

dsto

S2\(

Sou

thPole),[z

0:z 1]=

[w:1],so

youob

tain

thelocalcoordinatew

=z 0/z

1,an

dw

=0is

theNorth

Pole.

Onthe

overlap(z

0�=

0,z 1

�=0),[1

:z]=

[w:1]

recoverstheab

ovechan

geof

coordinates:w

=1/

z.

Since

zparam

etrizesacopyof

C,yo

ucanthinkof

CP

1asthecompactification

C∪{∞

}wherewead

dtheextrapoint∞

=[0

:1]

=(N

orth

Pole).

•Ellipticcurv

es(overC)arethetori

yougetbyquotientingC

byalattice.

Abovewe

usedthelatticeZ2

⊂R

2≡

C,butwecould

moregenerally

use

anyR-linearlyindep

endent

vectorsω1,ω

2∈R

2an

ddefinethelatticeΛ=

Zω1+Zω

2⊂

R2.Byrescaling,

andrelabelling

ifnecessary,wemay

aswellassumethat

ω1=

1∈C

andthat

ω2=

τ∈C

lies

intheupper

half-planeH

={z

∈C

:Im

(z)>

0}:

1Exercise:w

=X

1+Z

−i

Y1+Z.

2A

complexlineth

rough0in

C2is

aco

mplexvectorsu

bsp

ace

V⊂

C2ofdim

CV

=1.SoV

=C·(z 0

,z1)⊂

C2forsome(z

0,z

1)�=

0∈

C2,andnotice

rescalingdoes

notaffectV

=C·(λz 0

,λz 1

)foranyλ�=

0∈

C.

10

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

•Surfacescutoutbyoneequation:

{(z,w

)∈C

2:f(z,w

)=

0}.

Man

y,butnot

all,choicesof

aholom

orphic

functionf:C

2→

Censure

this

isaRieman

nsurface.

Exam

ple:f=

complexpolynom

ialin

thetw

ovariab

lesz,w

.Wewillshow

that

{(z,w

)∈C

2:w

2=

4(z−

a)(z−

b)(z

−c)},

fordistinct

constants

a,b,c

∈C,is

atoruswithapointremoved

(thereis

anaturalway

toad

dthepoint(z,w

)=

(∞,∞

)to

getthewholetorus).

•Historically,

Rieman

nsurfaces

firstap

pearedin

complexan

alysiswhen

tryingto

dealwith

theproblem

ofmulti-valued

functions.

For

exam

ple,thecomplexloga

rithm

Log

(z)=

log|z|+

iarg(z)

forz∈C\0

has

theproblem

that

theargu

mentison

lydefined

upto

addingmultiplesof

2πi,

since

e2πin

=1forn∈Z.

Therearetw

owaysof

solvingthis

problem:thead

-hocap

proach

isto

mak

eacutin

thecomplexplane,

sowerestrict

toLog

:C\(

−∞

,0]→

Can

dartificially

declare

that

−π<

arg(z)<

π.This

iscalled

abra

nch

oftheLog

“function”.

Apartfrom

thenuisan

ceof

mak

ingartificial

choices,thishas

theproblem

that

foracontinuou

scurvesuch

asthecircle

γ(t)=

e2πit,thefunctionLog

(γ(t))

isnot

continuou

s(itjumpsfrom

πto

−πat,or

rather

isnot

defined

at,t=

1/2).Silly!

Thenaturalremed

yis

really

toconsider

allthese“cut-dom

ains”

(for

thevariou

svalues

ofarg)as

beingglued

1together

accordingto

arg(z)-values

toform

asurface:

1In

each

cut-domain

C\(−

∞,0

],were-insert

two

copiesof(−

∞,0

)alongth

etw

osides

ofth

ecu

t.W

e

glueth

ecu

t-domainsbyiden

tifyingth

eco

piesof(−

∞,0

)in

pairs(soea

chcu

t-domain

isglued

onto

twooth

er

cut-domainsalongth

etw

oco

piesof(−

∞,0

)).Thepoint“0”does

notget

re-inserted

,weremoveit.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

11

Locallythesurfacelook

slikeC,an

din

thepicture

thevertical

axis

keepstrackof

thevalue

arg(z).

Noticehow

thesurfaceis

mad

eou

tof

sheets

(thecut-dom

ains)

and

ifyou

move

towardsthecutfrom

aboveyou

goupthestaircaseto

thenextlevel

inthesheets

(indeed

arg(z)increases),whereasifyouap

proachthecutfrom

underneath

youmovedow

nin

the

staircase(arg(z)decreases).

Thevariou

sbranches

ofLog

(z)thusdetermineawell-defined

complexlogarithm

function

defined

ontheab

oveRieman

nsurface(w

ith

thevertical

axis

coordinatetellingyou

whicharg(z)valueto

use).

•TheRieman

nsurfaceob

tained

from

amulti-valued

holom

orphicfunctionbygluingtogether

the“cut-dom

ains”

may

not

alwaysbeem

beddab

leinsideR

3(w

ithou

tself-intersection

s),the

case

ofLog

zab

ovewas

rather

special.

Con

sider

thesquareroot

z1/2=

e1 2Logz(you

gettw

odistinct

solution

s±w

with(±

w)2

=z,foreach

z�=

0).1

Analogou

slyweob

tain

thesurface:

Theself-intersection

isan

illusion

causedbywan

tingto

view

itin

R3.Wecanthinkof

this

surfaceS

asbeingob

tained

bytw

ocut-domainsC\R

<0,picturedbelow

,whichareglued

alon

gthecut.2Thebottom-cutof

onecut-dom

ainis

identified

withthetop-cutof

theother

dom

ain.Thetw

oshad

edhalf-discs

gluetogether

toform

adiscin

S.Walkingin

thedirection

ofthearrow

intheleft

cut-dom

ainwillmak

euspop

outwherethearrow

points

inrigh

tcut-dom

ain(andthis

shortwalkdoes

not

intersecttheshad

eddiscin

theactual

surface).

00

C\R

<0

C\R

<0

Con

sider

theholom

orphic

map

ϕ:S→

C,w

�→w

2=

z.Thepreim

ageof

asm

alldiscin

Ccentred

atz�=

0consistsof

twodisjointsm

alldiscs

inS

centred

at±√z.Theexception

iswhen

z=

0,in

that

case

thepreim

ageof

asm

alldiscis

just

onediscin

S:trydrawingit

ininsidetheR

3-picture

above.

What

does

this

correspon

dto

inthetw

ocut-dom

ains?

Anequivalentway

todescribeS

isas

the“graph”S

={(z,w

)∈

C2:z−

w2=

0}of

thesquareroot

function.In

this

case,S

→C,(z,w

)�→

wis

thesquareroot

function,an

d(z,w

)�→

zisϕ.Wecanuse

either

zor

was

alocalholom

orphic

coordinateforSnear(z,w

)

1More

explicitly:

the

two

branch

esofth

esquare

rootare

rei

θ�→

√rei

θ/2

and

rei

θ=

re2

πi+

iθ�→

re(

2πi+

iθ)/

2=

√rei

π+iθ

/2=

−√rei

θ/2.

Thesu

rface

islikean

Escher

stairca

se:if

you

goup

twoflights

ofstairs(i.e.θincrea

sesby4π)th

enwewillbeback

towherewestarted

.2Toclarify:ea

chcu

tR<0=

(−∞

,0)getsreplacedbytw

oco

piesofR<0th

atwere-insert

onto

C\R

<0.The

new

boundary

ofth

ecu

t-domain

consistsoftw

ohalf-lines

thatintersectat0,ca

llth

embottom-cut,

top-cut.

Wemodifyth

etopologynea

rth

ecu

ts:th

esh

aded

discabove,

obtained

bygluingtw

ohalf-discs,is

atypical

open

neighbourh

oodofapointofoneofth

etw

oco

piesofR<0.W

hatdoes

aneighbourh

oodof0looklike?

12

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

when

z�=

0,butnear(0,0)wemust

use

wasourlocalcoordinate,notz.Canyouseewhy?

•In

general,given

aholomorphic

functiondefined

onsomeregionofC,theaim

isto

build

aRieman

nsurfacebyanalytic

continuation.

Thatis,you

patch

together

localTaylor

series

fortheholom

orphic

function,andyoutryto

buildthelargestpossible

surface

(which

locallylook

slikeC)on

which

thefunction

can

be(uniquely)extended

to.

Forexample,

theRiemann

hypoth

esisis

aconjecture

aboutthezerosofafunction,theRiemannzeta

function,whichis

constructed

byanalyticcontinuation.

1.3

Non-examples:

spaceswhich

are

notsu

rfaces

•ThediscD

={z

∈C

:|z|<

1}together

withalinesegment:

D∪[1,2)is

notasurface:

atpoints

ofthelinesegment[1,2)thesurface

isnotlocallyhomeomorphic

toadisc.

Such

non

-exam

plesareeasy

tospotsince

surfaceshaveto

be“locally2-dim

ensional”.

•Thedouble

cone(twocones

sharingthevertex)withangle

θ,

{(x,y,z)∈R

3:z2tan2θ=

x2+y2},

isnot

asurface(noteven

topological):atthevertexitisnotlocallyhomeomorphic

toadisc.

1

•Theclosed

disc

D=

{z∈C:|z|≤

1}

isnot

asurface(noteven

topological):atanyboundary

point,itislocallyhomeomorphic

toahalfdisc

D+=

{(x,y)∈R

2:x2+y2<

1,y

≥0}.

Onecould

definesu

rfaceswith

boundary

,byrequiringthatthesurface

islocallyhomeo-

morphic

toD

+at

points

oftheboundary.Then

thecloseddiscD

would

beasurface

(indeed

Rieman

nsurface)withboundary.In

this

course,

wewillnotstudysurfaceswithboundary.

•Theplanewith

twooriginsis

obtained

asthequotientoftw

ocopiesofR

2:

(R2×{1})

�(R

2×{2})/((x,y,1)∼

(x,y,2):forall(x,y)except(0,0)).

This

spaceis

not

Hausdorff:anyopen

setaround(0,0,1)willintersectanyopen

setaround

(0,0,2),

soyou

cannotseparate

thetw

oorigins(0,0,1)�=

(0,0,2).

Locally

thespace

isneverthelesshom

eomorphic

toadisc(forexample

near(0,0,1)it

lookslikeD

×{1}).

Byconvention

,weprohibitsurfacesfrom

beingnon-H

ausdorff.Onereasoniswewantlimitsto

beunique:

(1 n,0,1)∼

(1 n,0,2)converges

totw

odistinct

points:(0,0,1)�=

(0,0,2).

Physically,

itwou

ldbeunrealisticto

haveacommonpath

(t−1,0,1)∼

(t−1,0,2)ofaparticle

attime

t∈[0,1]whichat

timet=

1canbein

twodifferentplaces.

•Quotientspaces

canoften

benon-H

ausdorff.Recallthataquotientspace

isHausdorff

ifan

don

lyiftheequivalence

relation{(x,x

� )∈X

×X

:x∼

x� }

⊂X

×X

isaclosedset.

2.

Definitionofsu

rfa

ce

2.1

Topologicalsu

rfaces

Definition

2.1.A

topologicalsu

rface

isaHausdorff

topologicalspace

Ssuch

thateach

pointp∈S

hasaneighbourhoodhomeomorphic

toanopensubset

ofR

2.

Remark

2.2

(Locallyyouare

adisc).If

theabove

homeomorphism

is

f:(neighbourhoodU

⊂S

ofp)→

V=

f(U

)⊂

R2

1Exercise.Checkth

is.Hint.

whathappensto

connectednesspropertiesif

youremove

thevertex?

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

13

then

wecanshrinkU

byreplacingitwith1� U=

f−1(D

r(f(p)).Then

compose

fwiththemap

R2→

R2,x

�→1 r(x

−f(p))

whichrescalesandtranslatesD

r(f(p))

toourfavourite

unit

disc

D=

D1(0).

Thenew

map

� f:� U→

R2showsthatS

islocallyhomeomorphic

toD

nearp.

Example.Theto

rusasato

pologicalsu

rface.Recallthequotientof

thesquare:

T2=

[0,1]×

[0,1]/((0,y)∼

(1,y)an

d(x,0)∼

(x,1)forallx,y

∈[0,1])

For

aninterior

point(x,y)∈(0,1)×(0,1)of

thesquare,

wecansimply

pickasm

alldiscarou

nd

italso

lyingin

theinterior,then

thehom

eomorphism

fisjust

theidentity.Slightlyharder,fora

point(0,y)on

theleft

edge

ofthesquare,

withy�=

0,consider

thehalf-disc

Dy=

{(X,Y

)∈[0,1]2

:X

2+

(Y−

y)2

<ε}

withcentre(0,y),

radiusε<

min{y

,1−

y}.

(0,y)

Dy

(1,y)

D� y

NoticeD

� y=

{(1−X,Y

)∈[0,1]2

:(X

,Y)∈D

y}isahalf-discwithcentre(1,y),radiusε.

The

twohalf-discs

gluetogether

inthequotientalon

gthecommon

bou

ndaryedge

(0,Y

)∼

(1,Y

)to

form

adiscD

y∪D

� y⊂

T2.Explicitly,wegetahom

eomorphism

f:� U

=D

y∪D

� y⊂

T2� →

� V=

{(X,Y

)∈R

2:X

2+

(Y−

y)2

<ε}

⊂R

2�

withf(X

,Y)=

(X,Y

)on

Dy,andf(X

,Y)=

(X−1,Y)on

D� y.Exercise:

runasimilarargu

ment

forthevertex

(0,0)of

thesquare(gluefourquarter-discs).

Ihop

ethisexam

pleconvincesyouthat

(often)writingtediousform

ulasdoes

not

makean

argu

mentmorerigorousthan

drawingpictures.

Remark

2.3

(Top

olog

icalman

ifolds).An

n-m

anifold

isaHausdorff

topologicalspace

Ssuch

thateach

pointphasaneighbourhoodhomeomorphic

toanopensubset

ofR

n.Asabove,

onecanalways

findalocalhomeomorphism

onto

theballB

1(0)=

{z∈R

n:�z

�<

1}.

Remark

2.4

(Whynot

metricspaces?).Most

topologicalsurfacesarise

asmetricspaces.

2

Soit’s

easy

todescribethetopology:each

opensetis

aunionofopenballs.

Sowhynotstart

offwithametricspace?Themetricis

extradata

whichwedonotcare

about:

wethinkoftwo

topologicalsurfacesasbeingthesameif

thereis

ahomeomorphism

betweenthem

,butthese

rarely

ever

preservedistances.

Onecould

studytopologicalsurfacestogether

withachoiceof

metric,

andrequirehomeomorphismsto

beisometries(sodistance-preserving).

2.2

Localcoord

inatesand

framesofreference

Thehom

eomorphism

fab

ove,

defined

nearp,determines

continuouslocalcoord

inates

onthesurface,

nearp:

q∈S

nearphas

localcoordinates(x,y)=

f(q)∈R

2

1ch

oose

asm

allradiusr>

0forth

eopen

discD

r(f

(p))

={q

∈R2:�q

−f(p)�

<r}so

thatD

r(f

(p))

⊂V.

2Non-exa

minable:oneusu

allyrequires

surfacesandmanifoldsto

bese

cond-counta

ble,andth

isim

plies

(byUrysh

on’s

metriza

tionth

eorem)th

atth

etopologyofasu

rface

oramanifold

isalw

aysinducedbysome

metric.

Second-countablemea

nsth

atth

etopologyhasaco

untable

base

(i.e.th

ereis

aco

untable

familyof

open

sets

Ui,su

chth

atev

eryopen

setis

aunionofsomesu

bfamilyofsu

chUi).

14

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Theab

ovefis

called

ach

art,andtheinverse

F=

f−1:(V

⊂R

2)→

S

iscalled

alocalpara

metrization.Physicistsliketo

callthis

aframeofreference.

For

exam

ple,when

observingaparticle

mov

ingonasurface

Syoudescribeit

interm

sof

somecoordinates

(x(t),y(t))

dep

endingontimet.

Itis

importantthattw

oobservers,

using

differentfram

esof

reference,agreeonwhether

ornottheparticle

ismovingcontinuously.

Let’s

comparetw

oframes

ofreference

F1,F

2ontheoverlapoftheirim

ages.Theparticle’s

localcoordinates

are(x(t),y(t))

and(�x(t),�y(t))forthetw

oobservers.

They

observethesame

particle,

so

F1(x(t),y(t))

=F2(�x(t),�y(t))=

(positionofparticle

inS

attimet).

Therefore,

thech

angeofcoord

inatesfrom

observer

1to

observer

2is:

(�x(t),�y(t))=

(F−1

2◦F

1)(x(t),y(t)).

Since

F1,F

2arehom

eomorphisms,

thetransition

map

F−1

2◦F

1is

continuouswherever

itis

defined.Thus:

Coro

llary

2.5.Foratopologicalsurface,thetransitionmaps(changesofcoordinatesbetween

twolocalparametrizations)

are

always

continuous.

2.3

Smooth

surfacesin

R3

Definition2.6

(Smooth

maps).Given

twoopensubsetsU

⊂R

n,V

⊂R

m,amapf:U

→V

issm

ooth

ifitis

infinitelydifferen

tiable.

1

Given

twoarbitrary

subsetsX

⊂R

n,Y

⊂R

m,amapf:X

→Y

issm

ooth

iflocally2

itis

therestrictionofasm

ooth

functionR

n→

Rm.

Definition

2.7

(Diffeomorphism).

Given

twoarbitrary

subsetsX

⊂R

n,Y

⊂R

m,amap

f:X

→Y

iscalled

diff

eomorp

hism

iffis

ahomeomorphism

andf,f

−1are

both

smooth.

Definition

2.8.A

smooth

surface

inR

3is

asubset

S⊂

R3such

thateach

pointp∈

Shasaneighbourhooddiffeomorphic

toanopensubset

ofR

2.

Remark

s.

�Asbefore,

wecanalwaysarrangeto

havediffeomorphismsf:U

→D

tothedisc.

�Smooth

surfacesin

R3are

alsotopologicalsurfaces,

because

diffeomorphismsare

hom

eomorphisms,

andsubspacesofaHausdorff

space

such

asR

3are

Hausdorff.

1Mea

ning:allpartialderivatives

offofallord

ersex

ist(itfollowsth

atallpartialderivatives

are

continuous,

soth

isdefi

nitionis

thesameasrequiringth

atfhasderivativemapsofallord

ers).

2Explicitly:forea

chp∈

Xth

ereis

anopen

neighbourh

oodU

aroundpandasm

ooth

mapF

:U

→Rm,

such

thatF

=f

on

U∩

X.

Weneed

toex

tend

fto

an

open

set,

because

inord

erto

taketh

elimit

∂xif(p)=

lim

t→0

1 t(f

(p+

tei)−

f(p))

weneedfto

bedefi

ned

alongth

erayp+

teiforsm

allt,

andforallwe

know

this

raymaynotbelongto

thegiven

setX.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

15

�A

localdiffeomorphism

f:S→

R2defined

nearpdetermines

smooth

localcoor-

dinates:

apointqnearphas

coordinates

f(q)=

(x,y)∈R

2.

�TheinverseF

=f−1,whichmap

ssomeop

ensetof

R2to

someop

enneigh

bou

rhood

ofp∈S,is

called

alocalpara

metrization

nearp.

�Todefinesm

ooth

n-m

anifoldsin

Rm

yousimply

replace

R2byR

nab

ove.

Coro

llary

2.9.Forasm

ooth

surface

inR

3,thetransition

maps(changesofcoordinates

betweentwolocalparametrizations)

are

always

smooth.

Proof.

Asin

Section

2.2,

thetran

sition

map

F−1

2◦F

1isdefined

betweentw

oop

ensets

ofR

2.

Since

F1,F

−1

2arediffeomorphisms,

F−1

2◦F

1is

adiffeomorphism

(bythechainrule).

Itis

easy

tocheckwhether

amap

R2→

Sis

smooth:youview

itas

amap

R2→

R3

andcheckthat

itis

infinitelydifferentiab

le.Butcheckingwhether

amapS→

R2is

smooth

isanuisan

ce,becau

sebytheab

ovedefinitionyouwould

needto

firstextendthemap

toa

neigh

bou

rhoodof

S,at

leastlocally.

Wewilllaterprove

(bytheim

plicitfunctiontheorem)

that

this

nuisan

ceis

easily

avoided

usinglinearalgebra:

Theorem

2.10.A

smooth

injectivemapF

:V

→S,defi

ned

onanopensubset

V⊂

R2,is

asm

ooth

localparametrization⇐⇒

∂xF,∂

yF

are

linearlyindepen

den

tateach

pointofV.

Wewilldoon

eexam

ple

explicitly,

butIhop

eyouagreethat

wedonotwan

tto

carryou

tsuch

calculation

severytimeweencounterasm

oothsurface.

Yourtimeis

betterspentat

develop

ingtheab

ilityto

spot

instinctivelywhether

ornotasurfacemay

failto

besm

ooth.

Example.Theto

rusasasm

ooth

surfacein

R3.Recall:

T2=

{((a

+bcosψ)cosθ,

(a+bcosψ)sinθ,

bsinψ)∈R

3:allθ,ψ∈[0,2π]}.

Let’scheckthisisasm

oothsurfacenearthepointp=

(a+b,0,0)

(takingθ=

ψ=

0).Perhaps

unsurprisingly,wewill

trythelocalparam

etrization

F:R

2⊃

(−π,π

)×(−

π,π

)→

T2,(x,y)�→

((a+bcosy)cosx,(a

+bcosy)sinx,bsiny).

Thisismanifestlysm

ooth(since

cos,sinaresm

ooth).

Itisnot

sohardto

checkthat

itisinjective

(checkthis,usingthat

a>

b>

0).

So,

bytheTheorem,wereduce

tolinearalgebra:

weneed

∂xF,∂yF

tobelinearlyindependentin

R3.

Soweneedto

findanon

-zero2×2subdeterminantof

thematrix

�∂xF

∂yF

� =

−(a

+bcosy)sinx

bsinycosx

(a+bcosy)cosx

−bsinysinx

0bcosy

.

Thebottom

tworowsgive

subdeterminant(a

+bcosy)bcosxcosy.Since

a>

b,thefirstbracket

isnon

-zero,

sothissubdeterminantisnon

-zeroexceptwhen

x,y

∈{±

π/2}.

Butin

that

case,the

toptworowsgive

subdeterminant(non-zero)·sin

2xwithsin2x=

1,againnon

-zero.

What

does

itmeanto

haveasm

ooth

mapf:S1→

S2betweensm

oothsurfaces

inR

3?

ByDefinition2.6itmeansfisacontinuou

smap

such

that

locallyneareach

pointof

p∈S1,

fcanbeextended

toasm

ooth

functionR

3→

R3defined

inaneighbou

rhoodof

p∈S1⊂

R3.

16

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

2.4

Abstra

ctsm

ooth

surfaces

Definition

2.11(A

bstract

smoothsurfaces).

Asm

ooth

surface

isaHausdorff

topological

space

S,together

withafamilyofhomeomorphisms,

called

loca

lpara

metrizations,

Fi:(opensubset

Vi⊂

R2)→

(opensubset

Ui⊂

S),

such

thattheUicoverS,so

S=

∪Ui,andonoverlapsthetransition

mapsare

smooth:

F−1

j◦F

i:R

2→

R2

issm

ooth

wherever

defi

ned.1

�Noticethat

Sis

automatically

atopolog

ical

surface.

�Asusual,each

Fidetermines

smooth

localcoord

inates:

q∈S

nearphaslocalcoordi-

nates

(x,y)=

F−1

i(q)∈R

2in

theparam

etrization

Fi.

Remark

2.12(Smoothman

ifolds).Todefi

nesm

ooth

n-m

anifolds,

replace

R2by

Rnabove.

Example.

The

toru

sasan

abstra

ctsm

ooth

surface.

ThequotientT

2=

R2/Z2

isaHausdorff

topolog

ical

space(since

Z2⊂

R2is

aclosed

set).

For

anypointp∈

T2,picka

representative

point�p=

(x,y)∈R

2,meaningp=

[x,y].

Con

sider

D�p=

{(X,Y

)∈R

2:(X

−x)2

+(Y

−y)2

<ε},

thediscwithcentre(x,y)andradiusε=

1/10

0(overkill:ε≤

1/2wou

ldwork).Noticethat

no

twopoints

inD

�pdiffer

byZ2

,thereforethequotientmap

F�p:� V

�p=

D�p⊂

R2� →

� U�p=

{[X,Y

]∈T

2:(X

,Y)∈D

�p}⊂

T2� ,

F�p(X,Y

)=

[X,Y

]

isahom

eomorphism.Since

p∈U

�p,weob

viou

slygetT

2=

∪U�p(takingtheunionover

allchoices

of�p∈

R2forallp∈

T2).

Now

consider

transition

maps.

Supposew

∈U

�p∩U

�q⊂

T2isin

anoverlap.Say

whas

localcoordinates

(X,Y

)∈

V�pand(X

+n,Y

+m)∈

V�qrespectively.By

definitionof

T2then,m

areintegers.Thetransition

map

issm

oothsince

itisthetranslation

F−1

�q◦F

�p:R

2→

R2,(X,Y

)�→

(X+

n,Y

+m).

What

does

itmeanto

haveasm

ooth

mapf:S→

S�betweensm

oothsurfaces?

Wewant:

(1)fis

continuou

sas

amap

oftopolog

ical

spaces,

(2)fis

smoothin

localcoordinates.

Theidea

in(2)is

youpicklocalcoordinates

(x,y)nearp,an

d(�x,�y)nearf(p).

Then

f(x,y)=

(�x(x,y),�y(x,y))

andyo

uwan

t�x(x,y),�y(x,y)to

besm

oothfunctionsofx,y.

1Explicitly,

F−1

j◦F

iis

defi

ned

onF

−1

i(U

i∩Uj)⊂

Vi⊂

R2.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

17

Moreab

stractly:foreach

p∈S,werequirethat

forsomecoordinatepatches

Vi

Fi

−→Ui⊂

SV

� j

F� j

−→U

� j⊂

S�

withp∈Ui,f(p)∈U

� j,themap

fwritten

inlocalcoordinates:

(F� j)−

1◦f

◦Fi:R

2⊃

Vi

Fi

−→Ui

f −→U

� j

(F� j)−

1

−→V

� j⊂

R2

issm

oothwherever

itisdefined

.1A

tediousexercise

isto

show

thatifitholdsforsomeVi,V

� j

asab

ove,

then

itmust

holdforallVi,V

� jas

above(usingthattran

sition

map

saresm

ooth).

Remark

.Anyabstract

surface

that“livesinside”

R3is

asm

ooth

surface

inR

3in

thesense

of

Section2.3.Wefirstclarify

what“livesinside”

means:

youhave

anabstract

smooth

surface

Sandasm

ooth

embeddin

gf:S→

R3(m

eaningahomeomorphism

onto

theim

age,such

thatthemapanditsinverseare

smooth).

Then

f(S

)⊂

R3is

asm

ooth

surface

inR

3:the

localparametrizationsforf(S

)are

givenby

f◦F

iusingtheFiofDefi

nition2.11.

2.5

Riemann

surfaces

InDefinition2.11

,replacingR

2byC,an

d“sm

ooth”by“h

olom

orphic”weob

tain:

Definition

2.13.A

Riemann

surface

isaHausdorff

topologicalspace

S,together

witha

familyofhomeomorphisms,

called

loca

lpara

metrizations,

Fi:(opensubset

Vi⊂

C)→

(opensubset

Ui⊂

S),

such

thattheUicoverS,so

S=

∪Ui,andonoverlapsthetransitionmapsare

holomorphic:

F−1

j◦F

i:C

→C

isholomorphic

wherever

defi

ned.2

�Noticethat

Sis

automatically

atopolog

icalsurface.

�Noticethat

Sis

automatically

asm

oothsurface.

�EachFidetermines

oneholomorp

hic

localcoord

inate:q∈

Snearpcorrespon

dsto

z=

F−1

i(q)∈Cin

theparam

etrization

Fi.

Asthereisjust

one,wecould

callitf i

=F

−1

i∈C.

Remark

2.14(Smoothman

ifolds).Todefi

neco

mplexn-m

anifolds,

replace

Cby

Cnabove.

Example.Theto

rusasaRiemann

surface.Con

sider

T2=

C/Λ

foralattice

Λ=

Zω1+

Zω2⊂

C,

whereω1,ω

2∈C

areR-linearlyindependent(i.e.not

real

multiplesof

each

other).

Thequotient

C/Λ

isaHausdorfftopolog

ical

spacesince

Λ⊂

Cisaclosed

set.

For

anypointp∈T

2,picka

representative

point�p∈C,meaningp=

[�p]in

thequotient.

Con

sider

D�p=

{z∈C

:|z

−�p|

<ε},

thediscwithcentre�p,

radiusε=

min{|ω1|,|ω2|}/1

00.Notwopoints

inD

�pdiffer

byΛ,so

F�p:(V

�p=

D�p⊂

C)→

� U�p=

{[z]∈T

2:z∈D

�p}⊂

T2� ,

F�p(z)=

[z]

isahom

eomorphism.Since

p∈

U�p,

wegetT

2=

∪U�p(theunionover

allchoicesof

�p∈

Cfor

allp∈T

2).

Finally,suppose[w

]∈U

�p∩U

�q⊂

T2isin

anoverlap.Say

[w]has

localcoordinates

1Explicitly,

itis

defi

ned

onF

−1

i(f

−1(U

� j)).

2Explicitly,

F−1

j◦F

iis

defi

ned

onF

−1

i(U

i∩Uj)⊂

Vi⊂

C.

18

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

w∈V�pandw+

nω1+

mω2∈V�qrespectively,wheren,m

∈Z

areintegers.Then

thetransition

map

isholom

orphic

since

itisatranslation:

F−1

�q◦F

�p:C

→C,z�→

z+

nω1+

mω2.

What

isaholomorp

hic

mapf:S→

S�betweenRieman

nsurfaces?Wewan

t:

(1)fis

continuou

sas

amap

oftopolog

ical

spaces,

(2)fis

holom

orphic

inlocalcoordinates.

Themeaningof

(2)isjust

asin

Section

2.4,

replacingR

2byC,“smooth”by“h

olom

orphic”.

3.

Whenaretwo

surfa

cesdifferent?

3.1

Homeomorp

hisms,

diffeomorp

hisms,

biholomorp

hisms

Theclassof

surfaces

affects

when

wewan

tto

view

twosurfaces

asbeingthesameor

different.

You

haveseen

thisin

mathem

aticsbefore:

wethinkof

twosets

as“b

eingthesame”

(iso

morp

hic)ifthereisabijection

:S1→

S2betweenthem

,whereasforvector

spaces

(sets

withad

ditional

structurescalled

additionan

drescaling)

wewan

tthebijection

topreservethe

additional

structures,

sowewan

tabijection

fwithf,f

−1bothlinear.

Wedefine:

(1)Twotopolog

ical

surfaces

areisom

orphic

ifthey

arehomeomorp

hic.

Explicitly:f:S1→

S2is

abijection

,an

df,f

−1arecontinuou

s.(2)Twosm

oothsurfaces

inR

3areisom

orphic

ifthey

arediffeomorp

hic.

Explicitly:f:S1→

S2is

abijection

,an

df,f

−1aresm

ooth.1

(3)Twoab

stract

smoothsurfaces

areisom

orphic

ifthey

arediffeomorp

hic.

(4)TwoRieman

nsurfaces

areisom

orphic

ifthey

arebiholomorp

hic,

Explicitly:f:S1→

S2is

abijection

,an

df,f

−1areholom

orphic.

Noticethat

adiffeomorphism/b

iholom

orphism

isin

particularalso

ahom

eomorphism,so

theunderlyingtopolog

ical

surfaces

arethesame.

How

ever,forallweknow

,theremay

be

several

waysto

turn

atopolog

ical

surfaceinto

asm

oothsurfaceor

aRieman

nsurface,

and

thesedifferentway

smay

not

berelatedbyadiffeomorphism/b

iholom

orphism

even

thou

ghthesurfaces

arehom

eomorphic.Wenow

checkwhat

hap

pensfortori.

3.2

Classification

ofto

ri

Definition3.1

(Torus).A

toru

sisanytopologicalspace

Xwhichishomeomorphic

toS1×S1

(usingtheproduct

topology).

Thereareseveral(hom

eomorphic)waysto

describeS1as

atopolog

ical

space:

(1)S1=

{z∈C

:|z|=

1}⊂

Cwiththesubspacetopolog

y,(2)S1=

R/Z

withthequotienttopolog

y,whereweidentify

x∼

x+

nan

yn∈Z,

(3)S1=

[0,1]/(0

∼1)

withthequotienttopolog

y.

For

exam

ple,ahom

eomorphism

from

(2)to

(1)is

x�→

e2πix.

Thetorusviewed

asthesquarebyidentifyingparallelsides

arises

from

description

(3)of

S1;thetorusas

aquotientR

2/Z

2bythetran

slationgrou

pZ2

arises

from

description(2);

andthetorusS1×

S1⊂

C=

R4arises

naturallyfrom

description

(1).

Asatopolog

ical

surface,

alltori

arethesame:

hom

eomorphic

toS1×

S1.Buthow

man

ydifferentsm

oothsurfaces

arehom

eomorphic

toS1×

S1,an

dhow

man

ydifferentRieman

n

1Non-exa

minable:

Since

surfacesin

R3are

an

abstract

surface

Stogether

with

theadditionalstru

cture

ofanem

bed

dingS

→R3,amore

appropriate

notionofisomorp

hic

isactuallyisoto

py:asm

ooth

familyof

embed

dings.

Explicitly:asm

ooth

mapH

:S×

[0,1

]→

R3su

chth

atH(·,

0):S

→R3,H(·,

1):S

→R3are

thetw

oem

bed

ded

surfaces,

andwewantH(·,

t):S→

R3to

beanem

bed

dingforea

cht∈

[0,1

].

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

19

surfaces

arehom

eomorphic

toS1×

S1?

Wewillnot

provethefollow

inghardtheorem

(aconsequence

oftheclassification

ofcompactsurfaces):

Theorem

3.2.Anysm

ooth

surface

whichis

topologicallyatorusis

diffeomorphic

toS1×S1.

This

turnsou

tto

befalseforRieman

nsurfaces:thereareman

ynon

-biholom

orphic

Rie-

man

nsurfaces

whicharetori.Again,wewillnot

prove

thefollow

inghardtheorem:

Theorem

3.3

(Ellipticcurves

over

C).

AnyRiemannsurface

whichis

topologicallyatorus

isbiholomorphic

toC/Λ

forsomelatticeΛ

whichwemayrescale

sothatΛ=

Z·1

+Z·τ

with

τ∈H

={z

∈C:Im

(z)>

0}.Theseare

called

theellipticcurv

es.

Wewillnow

show

explicitlywhytw

osuch

toriC/Λ

1,C

/Λ2are

diffeomorphic,an

dwewill

show

that

they

arenot

alwaysbiholom

orphic.

Toshow

that

they

arediffeomorphic,wemightas

wellshow

that

allquotients

C/Λ

aredif-

feom

orphicto

R2/Z

2(w

hichisthecase

τ=

i),then

C/Λ

1∼ =

R2/Z

2∼ =

C/Λ

2arediffeomorphic,

asrequired

.IdentifyingC=

R2,z≡

x+iy,write

τ=

a+

ib.Matrixmultiplication

� 1a

0b

� :R

2→

R2

isof

courseasm

oothmap

(itislinear!)an

ditis

bijective(thedeterminan

tb=

Im(τ)>

0is

non

-zero),an

ditmap

sZ2

→Λbijectively(thecolumnsaretheim

ages

ofthestan

dardbasis).

Thusitdefines

adiffeomorphism

R2/Z

2→

C/Λ

.Themapis

how

ever

not

holom

orphic.1

Moregenerally,supposewearegiven

abiholom

orphism

f:C/Λ

→C/Λ

�whereΛ=

Zω1+Zω

2,

Λ�=

Zω� 1+Zω

� 2.

This

meansthat

2itarises

from

quotientingaholom

orphic

map

� f:C→

Cwith

� f(Λ)⊂

Λ�

whichisinjectiveon

anyZω

1+Zω

2-translateof

theunitsquare

(0,1)×i(0,1)

⊂C.It

follow

s

that

� fmap

bijectivelyto

Λ� .

Iteasily

follow

sthat

� fgrow

slinearlyin

z∈

C.

Sothe

Taylorseries

of� fdoes

not

contain

order

z2or

higher

term

s.So

� f(z)=

Az+

Bforsome

A∈C\{

0},B

∈C.Tak

ingz=

0show

sthat

B∈Λ� ,so

bycomposing� fwiththetran

slation

z�→

z−

Bwemay

aswellassumethat

B=

0.So

� f(z)=

Azis

linear.

Thustheproblem

reducesto

classifyinglattices

Λ⊂

Cupto

C-linearbijection

s!Since

Aω1,A

ω2isrequired

tobeaZ-linearbasisfor� ,thetw

obases

Aω1,A

ω2an

dω� 1,ω

� 2

of�differ

byaZ-linearbijection

(thinkof

row-reductionbutworkingover

Z).So

ω� 1=

aAω1+bA

ω2

ω� 2=

cAω1+dAω2.

forsomeinvertible

integer-valued

matrix� a

bcd

� ∈GL(2,Z

).Thus,

usingtheconvention

that

τ=

±ω

1

ω2ad

justingthesign

sothat

τ∈H

={z

∈C

:Im

(z)>

0},

τ�=

±ω� 1

ω� 2

=±aAω1+bA

ω2

cAω1+dAω2=

±±aτ+b

±cτ

+d.

Sothematrixacts

ontheτparam

eter

likeaMob

iusmap

.Byproperties

ofMob

iusmap

s3

since

τ,τ�∈H,wededuce

that

τ�=

M·τ

whereM

=±� ±

ab

±cd

� ∈PSL(2,Z

).

1f(x

+iy)=

(x+ay)+ibyhas:

∂xf=

1,∂yf=

a+ib.TheCauch

y-R

iemanneq

uations∂xf=

−i∂yffail.

2Strictlysp

eaking,weonly

know

this

loca

lly,

asweusedth

equotien

tC

→C/Latticeto

defi

neth

eholo-

morp

hic

loca

lparametriza

tions.

Howev

er,byth

eId

entity

theorem

from

complexanalysisyouknow

that

youca

npatchtogether

theloca

lTaylorseries

uniquelyto

obtain

aglobalholomorp

hic

mapdefi

ned

onC.

3RecallM

obiu

sm

apsare

thebiholomorp

hismsC∪{∞

}=

CP

1→

CP

1=

C∪{∞

},z�→

az+b

cz+d,where

a,b,c,d

∈C

with

ad−

bc�=

0.

Thesemapsdon’t

changeif

you

rescale

all

a,b,c,d

by

thesamenon-zero

complexnumber,so

youmayarrangeth

atad−

bc=

1(w

hichleaves

thefreedom

ofrescalingallby±1).

So

such

mapsare

parametrizedby� a

bc

d

�∈

SL(2,C

)/(±

I)=

PSL(2,C

),andindeedth

emapsco

mpose

accord

ing

20

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Coro

llary

3.4.C/(Z1

+Zτ

)∼ =

C/(Z1

+Zτ

� )are

biholomorphic

ifandonly

ifτ,τ�∈

H=

{z∈C

:Im

(z)>

0}liein

thesameorbitofthePSL(2,Z

)-actiononH

byMobiusmaps.

Coro

llary

3.5.Riemannsurfaceswhichare

topologicallyatorusare

classified

upto

biholo-

morphism

by[τ]∈H/P

SL(2,Z

).

Cultura

lre

mark

:this

moduli

space,H/P

SL(2,Z

),of

modularparam

eters[τ]is

infact

itselfaRieman

nsurfacebiholom

orphic

toC.

4.

TheEulercharacteristic

4.1

Eulerch

ara

cteristic

ofregularpolyhedra

Noticethefollow

ingpattern

inthenumber

ofvertices,edges,

facesof

thePlatonic

solids:

Regularpolyhedron

Facetype

VE

Fχ=

V−

E+

FTetrahedron

Trian

gle

46

42

Cube

Square

812

62

Octah

edron

Trian

gle

612

82

Dodecah

edron

Pentago

n20

3012

2Icosah

edron

Trian

gle

1230

202

Thealternatingdifference

χ=

V−

E+

Fis

called

theEulerch

ara

cteristic.W

hyis

italways2forPlatonicsolids?

Itturnsou

tχisato

pologicalinvariantof

topolog

ical

surfaces,

meaningit

isaquan

tity

whichis

thesameforan

ytw

osurfaces

whicharehom

eomorphic.

Platonic

solidsarehom

eomorphic

tothesphere,

soχ

=χ(S

2)=

2.Thehom

eomorphism

betweenthePlatonicsolidan

dthespheredefines

acellulardecompositionof

thesphere:

asubdivisionof

thesphereinto

region

shom

eomorphic

todiscs

(inthiscase,curved

polygo

ns).

Example.Section

1.2.(1)show

satriangulationinducedby

atetrahedron(get

curved

triangles).

4.2

Cellulardecomposition

Definition4.1

(Cellulardecom

position).

Ace

llulardecompositionofatopologicalsurface

Sis

acollectionofcontinuousmaps,

called

cells,

v i:D

0→

Se j

:D

1→

Sf k

:D

2→

S

respectively

called

0-cells,

1-cells,

2-cells,

where1

Dn

={p

∈R

n:�p

�≤

1}is

the

n-

dim

ensionalunitdisc,

andwerequirethat:

(1)each

maprestricted

totheinteriorofthediscis

ahomeomorphism

onto

theim

age,2

(2)theboundary

ofthediscis

mappedinto

theim

age

ofthelower-dim

ensionalcells,

3

(3)S

ispartitioned

bytheinteriors

ofthecells.

4

Remark

s.�T

hemap

srestricted

tothebou

ndary∂D

narecalled

attach

ingmaps(can

benon

-injective).

�Noticeyo

uarebuildingthespaceinductivelybydim

ension

:yo

ustartwithabunch

ofpoints

X0=

�v i(D

0),then

youattach

linesegm

ents

X1=

X0∪�

e j(D

1)wheretheattachingmap

se j| {0

,1}landinsideX

0,an

dfinally

youattach

2-discs

X2=

X1∪�

f k(D

2)=

Swherethe

tomatrix

multiplica

tion.Soth

egroupofMobiusmapsis

isomorp

hic

toPSL(2,C

).Thesu

bgroupsending

H→

His

PSL(2,R

)(E

xercise:firstforceR

→R,th

enyoujust

needto

ensu

reth

emapsdon’t

flip

Hto

−H).

1SoD0=

{point},D1=

[0,1

]⊂

R,D2=

D=

{(x,y

)∈

R2:x2+

y2≤

1}⊂

R2.Interiors:Int(D0)=

D0,

Int[0,1

]=

(0,1

),Int(D2)=

D=

{z∈

R2:�z

�<

1}.

Boundaries:∂D0=

∅,∂D1=

{0}∪

{1},

∂D2=

S1.

2e j

:(0,1

)→

e j((0,1

))⊂

S,fk:D

→fk(D

)⊂

Sare

homeo

morp

hisms(n

oco

nditiononviasIntD0=

∅).

3e j(0),e j(1)∈

�vi(D

0)andfk(S

1)⊂

�vi(D

0)∪�

e j(D

1).

4S=

�vi(D

0)��

e j(IntD1)��

fk(IntD2)is

adisjointunionofsu

bsets.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

21

attachingmap

sf k| S

1landin

X1.ThesubspaceX

i⊂

Sis

called

thei-sk

eleto

n.

�Con

dition(3)ensurestherearenoredundan

cies

orsillyoverlaps:

thevertices

aredistinct,

novertices

touch

theinterior

ofan

edge

orface,noedge

touches

theinterior

ofaface.

�Theab

ovedefinitionworksmoregenerally

foran

yn-m

anifoldM

,in

whichcase

youcan

havecellsc i

:D

di→

Mof

anydim

ension

di∈{0,1,2,...,n

}.�A

triangulationisacellulardecom

position,wherethefacesareidentified

(via

hom

eomor-

phisms)

withtriangles

andtheattachingmap

sareallinjective,

sothat

thebou

ndaryedgesof

thetriangles

areprecisely

the1-cells,an

dthebou

ndaryverticesof

theedgesareprecisely

the

0-cells.

Theseconditionsarequiteharsh,so

itisusually

very

messy

1to

triangu

late

asurface.

Examples.

Herearethreeexam

plesof

cellulardecom

positionsof

S2:

Heref 1

:D

2∼ =(square)→

(squarewithidentification

s),andthismap

isahom

eomorphism

onthe

interior,buton

thebou

ndaryitisnot

injective.

Noticethat

(just

asforthePlatonicsolids,which

also

yieldcellulardecom

positionsof

S2)thealternatingsum

ofthenumbersof

cells

isalways2:

1−0+1=

21−1+2=

23−2+1=

2.

Averygeneral

(hard)fact

from

algebraic

topologyis:

Theorem

4.2.Anycompact

topologicalmanifold

M(e.g.a

compact

topologicalsurface)“ad-

mits”

2acellulardecompositionandthealternatingsum

ofthenumbers

ofcells

χ(M

)=

(#0-cells)

−(#

1-cells)

+(#

2-cells)

−(#

3-cells)

+···

isthesameforanycellulardecomposition.It

iscalled

theEulerchara

cteristic

ofM

.

Coro

llary

4.3.If

M,N

are

homeomorphic

topologicalmanifoldsthen

χ(M

)=

χ(N

).Sothe

Eulercharacteristicis

atopologicalinvariant.

Proof.

Iff:M

→N

isahom

eomorphism,then

acellulardecom

positionc i

:D

di→

Mof

Mdetermines

acellulardecom

positionf◦c

i:D

di→

Nof

N.Soχ(M

)=

�(−

1)di=

χ(N

).�

Example.Weob

tained

thetorusfrom

asquareby

identifyingtheparalleledges.

Thewhole

squareisa2-cellf 2

:D

2→

T2,thetwonon

-paralleledgesaretwo1-cells

e 1,e

2:D

1→

T2,and

thefourvertices

ofthesquareareidentified

withon

e0-cellv 1

:D

0→

T2.Thus

χ(T

2)=

1−2+1=

0.

1tryto

triangulate

thetoru

s,viewed

asasquare

withparallel

sides

iden

tified

.2Non-examinable:th

estatemen

taswritten

isknownin

alldim

ensionsex

cept4(andin

dim

ension2one

canev

enobtain

atriangulation).

Inreality,oneonly

caresaboutacellulardecomposition“upto

homotopy

equivalence”:loosely,atypeofco

ntinuousdeform

ationth

atis

more

drastic

thanjust

homeo

morp

hisms,

and

ithappen

sto

preserveχ.Asanex

ample,youca

nsquash

acy

linder

totu

rnit

into

acircle,both

haveχ=

0.

That,

upto

homotopy,

manifoldshavecellulardecompositionswasproved

byJohnMilnorin

his

1959paper,

Onspaceshavingthehomotopytype

ofaCW

-Complex.

22

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

For

RP

2thefourvertices

insteaddefinetwo0-cells,so

χ(R

P2)=

2−

2+1=

1.

4.3

Connected

sum

Ingeneral,giventw

osurfaces

S1an

dS2,wecanform

twonew

surfaces:

(1)thedisjointunion:S1�S2,

(2)theconnectedsum:S1#S2.

Easy

exercise.Show

that

anyconnectedcompon

entof

asurfaceis

asurface.

Deduce

that

anysurfaceequalsadisjointunionof

connectedsurfaces.

Theconnected

sum

S1#S2is

obtained

byremov

inga“d

isc”

from

each

ofthetw

osur-

facesan

didentifyingthecircularbou

ndaries.

This

identification

isthesame(upto

hom

eo-

morphism)as

attachingacylinder

bygluingthetw

obou

ndariesof

thecylinder

onto

thetw

obou

ndariesof

theremoved

discs.

Exercise.Checkthat

S1#S2isindeedatopolog

ical

surface.

Con

vince

you

rselfthat

ifS1,S

2

areconnectedthen,upto

hom

eomorphism,it

does

not

matterwhich“d

iscs”youpick.

Exercise.Con

nectedsum

withaspheredoes

nothing.

Con

nectedsum

withatorusis

thesameas

attachingahan

dle

(Section

4.5).

Example.ByExercisesheet1,

RP

2isob

tained

bygluingadiscD

onto

aMob

iusbandM

alon

gthecircularbou

ndary.

SoRP

2\D

=M

.Soconnectedsum

withRP

2isthesameas

attaching

aMob

iusband(Section

4.6).

4.4

Additivityofth

eEulerch

ara

cteristic

Lemma4.4.

(1)χ(S

1�S2)=

χ(S

1)+

χ(S

2),

(2)χ(S

1#S2)=

χ(S

1)+

χ(S

2)−

2.

Proof.

1Pickacellulardecom

positionof

S1,S

2.Then

thisdefines

acellulardecom

positionof

S1�S2so

(1)follow

sim

med

iately.Theidea

in(2)isthat

weremovetw

ofaces,whichmak

esχdropby2,

andweidentify

thecircularbou

ndariesso

welose

onecopyof

S1,whichdoes

not

matterforχsince

χ(S

1)=

0(since

S1is

apointwithan

interval

attached

tothepoint,

soχ=

1−

1=

0).This

idea

iscorrectifyoutriangu

late

S1,S

2an

dremovetw

otriangu

lar

faces.

How

ever,ifweinsteadwan

tto

workwithcellulardecom

positions(w

hicharisemore

naturallythan

triangu

lation

s)then

therigo

rousproof

isalittle

moreinvo

lved,as

follow

s.ThesurfaceS1#S2upto

hom

eomorphism

only

dep

endson

thethechoice

ofconnected

compon

ents

inS1an

din

S2whereyo

upickthe“d

iscs”.Sowemightas

wellpickeach

“disc”

intheinterior

ofa2-cellin

thecellulardecom

position.Todothis

withou

tdestroy

ingthe

cellulardecom

position2wesubdivideeach

original

2-cellf 0

:D

2→

Sias

inthepicture.

1Non-exam

inable

exercise.Forn-m

anifoldsM

,N,explain

how

oneconstru

ctsaconnectedsu

mM

#N

andshow

thatχ(M

#N)=

χ(M

)+

χ(N

)−

χ(S

n),

whereSn

isthen-sphere.

2makingahole

insideth

e“disc”

gives

anannulus(u

pto

homeo

morp

hism),

soit

isnolonger

a2-cell.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

23

Thenew

edgese 1,e

2map

injectivelyinto

Sisince

theoriginal

f 0is

injectiveon

Int(D

2),

andsimilarly

thenew

facesf 1,f

2areinjectiveon

Int(D

2).

How

ever,acommentis

required

abou

tv 1.Iff 0(S

1)alread

ycontainsa0-cell,then

wecanuse

that

forv 1,an

dχwillnot

have

chan

ged.1

Iff 0(S

1)does

not

2contain

a0-cellthen

,bythepartition

ingcondition,f 0(S

1)lies

insidetheim

ageof

anedge,so

creatinganew

v 1meanssubdividingan

edgeinto

two.

Sowe

arealso

creatinganew

edge.Sothisnew

vertex/edge

pairdoes

not

affectχ:1−1=

0.This

invarian

ceof

χis

aspecialinstan

ceof

thevery

general

invarian

ceTheorem

4.2.

Next,weremovethesm

allfaces(f

2in

thepicture)from

S1,S

2,whichmakes

χdropby2.

Finally

weidentify

thetw

obou

ndariesof

thosefaceswhichmeansweidentify

thecopiesin

S1,S

2of

v 2,e

2in

theab

ovepicture,so

χdoes

not

chan

ge(+

1−1=

0).So(2)follow

s.�

Remark

.Ifatopologicalspace

S=

A∪B

(such

asasurface)is

aunionoftwoclosedsubsets,

andsuppose

3S

admitsacellulardecompositionsuch

thatit

inducescellulardecompositions

forA∩B,A

,B,then

bycountingyoudeduce

χ(S

)=

χ(A

)+

χ(B

)−

χ(A

∩B).

Canyousee

how

touse

this

toprove

theabove

form

ula

forχ(S

1#S2)?

4.5

Attach

inghandlesto

asp

here

Observethat

thereis

anaturalway

toorientthebou

ndaryof

a“d

isc”

4in

thesphere:

we

askthat

itob

eystheright-hand

rule

5,withthethumbpointingin

thenormal

outw

ard

direction

(soforvery

smalldiscs,thebou

ndaryis

orientedan

ti-clockwiseifyouarelook

ing

atthesphereS2⊂

R3from

faraw

ay).

Acylinder

isaspacehom

eomorphic

to[0,1]×

S1.Thebou

ndariesareorientedas

follow

s:{1}×

S1is

orientedclockwise,

{0}×

S1is

orientedan

ticlockwise.

Attachingahandle

toS2meansyouremovetw

odisjoint“d

iscs”from

S2,an

dyouglue

thetw

obou

ndarycirclesof

thecylinder

[0,1]×

S1on

tothetw

obou

ndariesof

thediscs

you

removed

inaway

whichpreserves

theab

oveorientation

s(inpractice:

draw

arrowson

the

circularbou

ndaries,

andgluein

away

that

respects

thearrow

direction

s).Theorientation

choicesensure

that

wecanthinkof

thehan

dle

asattached

onto

S2⊂

R3from

the“outside”:

1before

subdivision,f0co

ntributes+1,after

subdivisionv2,e

1,e

2,f

1,f

2co

ntribute

1−

2+

2=

1.

2th

eboundary

ofth

ediscmust

landinsideth

e1-skeleton,butit

maylandin

theinteriorofsome1-cell.

3Non-exa

minable:More

gen

erally,

this

form

ula

forχ

holdswhen

ever

Sis

theunionofth

einteriors

ofth

e

closedsets

A,B

,asaco

nsequen

ceofth

eso-called

Mayer-V

ieto

ris

sequence(see

C3.1

Algeb

raic

Topology).

4“Disc”

willmea

naco

ntinuousmapD2→

Swhichis

ahomeo

morp

hism

onto

itsim

age.

5thumbpointingin

thenorm

aloutw

ard

direction,index

finger

pointingin

theorien

tedcirculardirection,

andmiddle

finger

pointingtoward

sth

ecentreofth

ecircle.

24

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Thus,

startingfrom

asphere,

weob

tain

asequen

ceof

surfaces:

Thenumber

gof

handlesattached

toS2iscalled

thegenusofth

esu

rface,an

dcorrespon

ds

tothenumber

of“dou

ghnutholes”:

Exercise.Supposeyo

uinsteadremovetw

odisjointdiscs

from

thesurface,

andidentify

the

twobou

ndarycircles(inaway

that

preserves

theirorientation

s).Show

that

theresulting

surfaceis

hom

eomorphic

totheab

ovehan

dle-attachment.

Hint.Nearoneofthecirclesyou

canpickaclosedneighbourhoodthatlooks

like

S1×

[−1,0],now

construct

therequired

map.

Lemma4.5.Attachingahandle

decreasesχby

2.

Proof.

Toob

tain

acellulardecom

positionof

acylinder

S1×[0,1],wedeclare

that

e 1=

[0,1]∼ =

{1}×

[0,1]⊂

S1×

[0,1]is

a1-cell.View

each

ofthecirclesS1×

{0}an

dS1×

{1}as

1-cells

e 2,e

3which

havebeen

attached

byidentifyingboth

endpoints

tothesamepoint,

nam

ely

theendpoints

v 1=

(1,0),

v 2=

(1,1)of

e 1.

Thecylinder

itself

defines

a2-cell

bou

nded

bye 1,e

2,e

3.

Thusχ(cylinder)=

2−

3+

1=

0.W

hen

weattach

thecylinder,weruna

constructionsimilar

tothepicture

intheproof

ofLem

ma4.4.

Nam

ely,

weremovetw

odiscs

from

theoriginal

surface(soa2-cell),

whichmak

esχdropby2,

whilst

theidentification

ofthebou

ndarycirclesdoes

not

chan

geχ

(viewingthebou

ndarycircle

asa1-cellwithboth

endpoints

attached

tothesame0-cell,welose

a1-cellan

da0-cell,leav

ingχunaff

ected).

4.6

Attach

ingM

obiusbandsto

asp

here

InExercise

sheet1yo

ustudyMob

iusban

ds.

TheMob

iusban

dM

isthequotientof

the

square[0,1]×

[0,1]byidentifyingthevertical

edgesin

oppositedirection

s,(0,y)∼

(1,1

−y).

Thebou

ndaries[0,1]×

{0}an

d[0,1]×

{1}glueto

give

acircle.

AttachingaM

obiusband

toS2meansyo

uremovea“d

isc”

from

S2,an

dyo

ugluethe

bou

ndarycircle

ofM

onto

thebou

ndaryof

thediscyo

uremoved.Onecannot

draw

this

inR

3withou

tself-intersection

s,so

schem

atically

wewilldraw

Mas

awigglycap:

Theab

ovearethefirsttw

oof

asequen

ceof

surfaces

oneob

tainsbyattachingMob

iusban

ds

toS2(see

Exercise

Sheet1,

Ex.2).

Lemma4.6.AttachingaMobiusbanddecreasesχby

1.

Proof.

Thisissimilar

toLem

ma4.5.

Mhas

a2-cell(thesquare),three1-cells(twoof

thefour

edgesof

thesquareareidentified),tw

o0-cells(verticesareidentified

inpairs).

Soχ(M

)=

0.W

hen

weattach

M,χdropsby1as

weremoveadisc(a

2-cell)from

theoriginal

surface.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

25

5.

Classificationofsu

rfa

ces

5.1

Classification

ofcompactto

pologicalsu

rfaces

Attheendof

Part

ATopologyyo

uplayedwithgluingedges

ofpolygo

ns,

and“p

roved”:

Theore

m5.1.Anycompact

connectedtopologicalsurface

ishomeomorphic

to:

(1)asphereS2withg≥

0handlesattached,or

(2)asphereS2withh≥

1Mobiusbandsattached.

Actually

youprovedtheab

oveunder

thead

ditional

assumption

that

thesurfacecanbe

triangulated(see

theRem

arksin

4.2forthedefinitionof

atriangu

lation).

Ahardtheorem

1of

topolog

yis

that

everytopolog

ical

surfacead

mitsatriangu

lation

.Once

atriangu

lation

has

beenchosen

forthecompactconnectedsurfaceS,bycompact-

nessthereareon

lyfinitelymanytriangles.You

then

inductivelybuilda“p

olygo

n”(upto

hom

eomorphism,so

itmay

havecurved

edges)

intheplaneR

2.Start

withatrianglefrom

S,identify

itwitha“trian

gle”T1(a

hom

eomorphic

copy,

soit

canbecurved)in

theplane.

Then

consider

anad

jacenttrianglein

S,identify

itwitha“trian

gle”T2ad

jacentto

T1,an

dso

on.Since

Sis

connected,once

youhaveexhau

sted

alltrianglesyo

uendupwitha(typi-

callynon

-con

vex)“p

olygo

n”in

R2:theunionof

thetriangles

T1,T

2,....How

ever,theou

ter

bou

ndaryedgeswillbeidentified

inpairs

since

inSeach

edge

belon

gsto

twotriangles.Thus,

upto

hom

eomorphism,theproblem

reducesto

classifyingregu

larpolygo

nswithpairw

iseedge

identifications.

You

solved

this

combinatorialexercise

inPart

ATop

ology.

BySection

4.3,

Theorem

5.1canalso

bestated

asfollow

s:

Coro

llary

5.2.Anycompact

connectedtopologicalsurface

ishomeomorphic

tooneof:

(1)S2#T

2#T

2#

···#

T2forsomenumberg≥

0ofcopiesofT

2,

(2)S2#RP

2#RP

2#

···#

RP

2forsomenumberh≥

1ofcopiesofRP

2.

Coro

llary

5.3.Forthesurfacesin

Theorem

5.1,

(1)χ(S

2withg≥

0handlesattached)=

2−

2g,

(2)χ(S

2withh≥

1Mobiusbandsattached)=

2−

h.

Proof.

FollowsbyCorollary

5.2an

dLem

ma4.4,

usingthat

χ(T

2)=

0andχ(R

P2)=

1.�

TheEulercharacteristic

does

not

distingu

ishthetorusT

2from

theKlein

bottleK:both

haveχ=

0.Onereason

whyT

2an

dK

are

not

hom

eomorphic

isthat

KcontainsaMob

ius

ban

dbutT

2does

not

(ifthey

werehom

eomorphic

then

T2wou

ldalso

contain

one).Indeed,

onecandefineorientabilitybysayingthat

atopolog

ical

surfaceis

orienta

ble

ifan

don

lyif

itdoes

not

contain

aMob

iusban

d.In

Section

6wewilldiscu

ssorientability,

andshow

:

Coro

llary

5.4.TheEulercharacteristicandorien

tability

uniquelydeterminethetopological

surfacesin

Theorem

5.1

upto

homeomorphism

((1)are

orien

table,

(2)are

non-orien

table).

5.2

Classification

ofcompactsm

ooth

surfaces

Thean

alogu

eof

Theorem

5.1is

thefollow

inghardtheorem:

Theore

m5.5.Anycompact

connectedsm

ooth

surface

isdiffeomorphic

to:

(1)asphereS2withg≥

0handlesattached,or

1See

http://math

overflow.net/questions/17578/triangulating-surfaces.

Itis

quite

easy

tofind

cellular

decompositions,

butmuch

hard

erto

triangulate.

Infact,in

higher

dim

ensions,

itis

nottrueth

attopo-

logicalmanifoldsare

alw

ays“triangulable”

(using

the

higher

dim

ensionalanalogues

oftetrahed

ra).

See

http://en

.wikiped

ia.org/wiki/Triangulation

(topology).

26

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

(2)asphereS2withh≥

1Mobiusbandsattached.

Exercise.Con

vince

yourselfthatyoucanmakeattachments

smoothly.

5.3

Classification

ofRiemann

surfaces

This

Section

isnon-examinable.

For

culturalreason

s,Imentionthefollow

ingtheorem

(wecomeback

tothisin

theAppendix).

Theorem

5.6

(Uniform

izationtheorem).

Every

simply-connected1Riemannsurface

isbiholomorphic

tooneof:

(1)CP

1(sometim

escalled

theRiemannsphere),

(2)C

(thecomplexplane)

(3)D

={z

∈C:|z|<

1}(theopendisc).

Rem

ark.Theupper

half-planeH

={z

∈C:Im

(z)>

0}isbiholomorphic

2to

Dvia

z�→

z−i

z+i.

Every

connectedRiemannsurface

isbiholomorphic

toaquotien

tofoneofthose

threesimply-

connectedmodelsby

adiscrete3

groupwhichactsholomorphically,

freely

4andproperly.

5

Later

we’llseehow

this

isrelatedto

geometry:thethreemodelshaverespective

curvatures

+1(Spherical

geom

etry),

0(E

uclideangeometry),

−1(H

yperbolicgeometry).

Example:genus1Riemannsurfacesare,upto

biholomorphism,ellipticcurves

bySection

3.2,

soquotients

ofC

byadiscretegroupoftranslationsdescribed

byalattice.

Coro

llary

5.7.Bystudyingthepossible

groupactionsin

theabove

threecases,

itturnsout

everyconnectedRiemannsurface

isbiholomorphic

tooneof:

(1)CP

1,

(2)C,C

∗=

C\{

0},orC/Λ

foralatticeΛ

⊂C

(thesethreeoptionscomefrom

groups

isomorphic

to{1},Z,

Z2respectively),

(3)H/G

foradiscretesubgroupG

⊂PSL(2,R

)=

{� ab

cd

� :a,b,c,d

∈R,ad−bc

=1}/

±I

actingfreely

byMobiusmapsonH.

6.

Orientability

6.1

Orienta

ble

versusnon-orienta

ble

surfaces

Therearetw

owaysto

orientatriangle,i.e.choosingarrow

salongtheedges

indicatinghow

wewishto

travelon

cearoundthetriangle.Onecould

equivalentlysayasurface

isorienta

ble

ifin

atriangu

lationofthesurface

itis

possible

topickanorientationforeach

triangle,so

that

foran

ytw

otriangleswhichshare

anedgewehavepickedopposite

orientationsforthat

edge

forthetw

otriangles.

How

ever,thenatural6

definitionis

Definition6.1.

1Sim

ply-connected

mea

ns:

connected,and

everyco

ntinuousloop

can

beco

ntinuouslysh

runkto

apoint

(everyco

ntinuousmapS1→

Sca

nbeex

tended

toaco

ntinuousmapD

→S

onth

eclosedunit

disc).

2Via

thatmap,0,i,∞

∈C∪{∞

}mapto

−1,0

,1∈

C∪{∞

},andsince

thedistance

ofz∈

Hfrom

iis

less

thanth

edistance

from

−iwehave|z

−i

z+i|<

1,so

themaplandsinsideD.Now

use

properties

ofMobiusmaps

(such

as,

circles/lines

mapto

circles/lines,andanglesare

preserved

)to

ded

uce

thatit

isabiholomorp

hism.

3Discretemea

nsth

at“points

are

open

”:i.e.

theone-elem

entsu

bsets

{g}are

open

sets,forg∈

G.Example:

Z2⊂

R2withaddition.

4Freelymea

nsallstabilizersare

trivial:

StabG(p)=

{g∈

G:g•p

=p}=

{1}.

5Properly,foradiscretegroupG

actingonasp

ace

S,mea

ns:

anyp,q

∈S

haveneighbourh

oodsUp,U

q

withUp∩(g

•Uq)�=

∅foronly

finitelymanyg∈

G.This

ensu

res,

inparticular,

thatS/G

isHausd

orff

.6Non-exam

inable

Rem

ark.It

isth

eonly

defi

nition

which

gen

eralizesto

higher

dim

ensions.

Atopo-

logicaln-m

anifoldsM

isorienta

ble

if,after

moving

adiscDn

→M

continuously

inM

along

any

path

untilitsim

ageco

incides

with

theoriginalim

age,

thestarting

and

ending

positionsyield

two

embed

dings

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

27

Wewilllooselyuse

theexpressiondiscin

Sto

meanacontinuou

sinjective1

map

D→

S,

andcontinuousfamily

ofdiscsin

Sto

meanacontinuou

smap

F:D×

[0,1]→

S,such

that

each

Ft:D→

S,Ft(z)=

F(z,t)is

adiscin

S.

Definition

6.1

(Orientable

surface).A

topologicalsurface

Sis

orientable

ifforanycon-

tinuousfamilyofdiscs

Ft(D

)in

S,startinganden

dingatthesamediscF0(D

)=

F1(D

),the

circularboundaries

F0(S

1),F1(S

1)are

orien

tedin

thesamedirection.2

This

failsfortheMob

iusban

dM

,so

anysurfacecontainingacopyof

Mis

non

-orientable:

Thusan

ysurfacein

family(2)in

Theorem

5.1is

non

-orientable.

Thesurfaces

infamily(1)in

Theorem

5.1areallorientable,since

they

canbeem

bedded

inR

3withawell-defined

outw

ardnormal

direction

,an

dthen

afamilyof

discs

willhavecircular

bou

ndariesoriented

either

alwaysagreeingor

alwaysdisagreeingwith

therigh

t-han

d-rule

orientation

.3

Remark

.You

may

won

der

whyattachingahan

dle

toasurfaceS

infamily(2)in

Theorem

5.1does

not

givean

ythingnew

.Upto

hom

eomorphism,youcanmoveon

ebou

ndarycircle

ofthecylinder

once

arou

ndaMob

iusban

din

S,whichwillsw

itch

theorientation

ofthe

bou

ndarycircle.If

youthinkof

SinsideR

3with(fictitious)

“self-intersection

s”,then

the

cylinder

isnolongerattached

ontheou

tsideof

thesphere:

oneof

theendsis

attached

from

theinside.

Thecylinder

attached

inthis

way

correspon

dsto

connectedsum

withaKlein

bottleK

=RP

2#RP

2(inthepicture

wedrew

thecircle

alon

gwhichwetaketheconnected

sum

withK).

Soattachingacylinder

toasurfacewhichcontainsacopyof

Misthesame,

up

tohom

eomorphism,as

theconnectedsum

#RP

2#RP

2withtw

ocopiesof

RP

2.Thisiswhat

weexpectedfrom

theclassification

theorem,since

χdropsby2when

attachingacylinder.

Sn−1=

∂Dn

→M

thatdiffer

byahomeo

morp

hism

Sn−1→

Sn−1whichca

nbeco

ntinuouslydeform

edto

theiden

tity

map(see

theanalysishandoutforth

edefi

nitionofdeform

ation).

Foranon-orien

table

n-m

anifold,

therewillex

istsomepath

yieldingahomeo

morp

hism

Sn−1→

Sn−1whichca

nbeco

ntinuouslydeform

edto

thereflection(x

1,...,x

n)�→

(−x1,x

2,...,x

n)(viewingSn−1⊂

Rn).

1W

eco

uld

strength

enth

isto

“em

bed

ding”,mea

ningahomeo

morp

hism

onto

theim

age.

2Mea

ning,F

−1

1◦F0:S1→

S1sendsth

eanticlock

wisepath

eitto

apath

eis(t)forastrictly

increa

sing

functions(t)

(soei

s(t)is

alsoananticlock

wisepath

).3thumbpointingin

thenorm

aldirection,index

finger

pointingin

theorien

tedcirculardirection,andmiddle

finger

pointingtoward

sth

ecentreofth

ecircle.

28

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

6.2

Non-orienta

ble

compactsu

rfacescannotbeembedded

inR3

Definition

6.2

(Embedding).Foratopologicalsurface

S,amapf:S→

R3is

anembed-

din

gifS→

f(S

)⊂

R3is

ahomeomorphism

(inparticularfis

injectiveandcontinuous).

Remark

.Forsm

ooth/RiemannsurfacesS,onerequires

inadditionthatthederivative

map1

Dfis

injectiveateverypoint.

Thinkof

embeddings

inR

3as

givingyouan

identicalcopyof

thesurfacein

R3.

Theore

m6.3.A

non-orien

tablecompact

surface

Scannotbe

embedded

inR

3.

Sketchproof.

Chooseapointp∈R

3nearinfinity,

faraw

ayfrom

S.For

anypointq∈R

3\S

,call

qeven

ifthereis

acontinuou

scurvec:[0,1]→

R3startingat

c(0)

=q,

endingat

c(1)

=p,intersectingS

inafiniteeven

number

ofpoints.Defineqodd

ifthenumber

isodd.For

exam

ple,thestraightlinesegm

entcfrom

qto

poftenworks.

This

definitionis

not

quitecorrect,2butsomealgebraic

topolog

ymachinerybeyon

dthis

courseensuresthat

this

definitioncanbemad

erigo

rousan

dthat

even/o

ddnessisindep

endentof

thechoice

ofc.

But

now

,ifyou

consider

anan

twalkingalon

gtheequator

ofaMob

iusban

d,then

thepositions

q sta

rt,q

endof

theheadof

thean

tbeforean

daftergo

ingarou

ndtheequator

has

chan

gedparity

(tovisualise,

consider

thestraightlinesegm

ents

top).

Butjoiningthecurvefrom

q sta

rtto

q endtraced

outbytheheadof

thean

t(w

hichhas

not

intersectedS)withacurvefrom

q end

topshow

sthat

q sta

rtan

dq e

ndhavethesameparity.

Con

trad

iction

.�

Remark

.Theclosed

Mob

iusban

dem

bedsinto

R3,butis

not

asurface(w

edonot

allow

bou

ndariesin

this

course).Theop

enMob

iusban

dem

bedsinto

R3butis

non

-com

pact.

Remark

.Onecanim

prove

this

proof

toshow

that

thecomplementof

anycompactsurface

Sem

bedded

inR

3has

twoconnected

compon

ents,called

the“inside”

and

the“o

utside”

(the3-dim

ension

alan

alog

ueof

theJord

an

curv

eth

eore

m,whichsaysthat

acontinuou

snon

-self-intersectingclosed

curvedivides

theplaneinto

tworegion

s).

6.3

Orienta

bilityofsm

ooth

surfacesin

term

softh

etransition

maps

Ifwewan

tto

show

that

asm

oothsurfaceS

isorientable

then

wewou

ldhaveto

prove

that

itcontainsnocopyof

theMob

iusdisc–thisishardin

practice.

Itturnsou

tthereisan

equivalentdefinitionof

orientability,

whichis

morepractical.

Definition

6.4

(Orientable

smoothsurface).A

smooth

(abstract)surface

Sis

orien

table,

ifthederivativesofalltransitionmapsF

−1

j◦F

ihave

positive

determinantontheoverlaps:

det(D

F−1

j◦D

Fi)>

0.

(Warning:

thefailure

ofthis

conditiondoesnotim

ply

non-orien

tability)3

Wenow

explain

this.In

R2thepossible

ordered

bases

v 1,v

2comein

twotypes:

(1)right-handed

bases:

thesediffer

from

thestan

dardbasis

e 1=

(1,0),

e 2=

(0,1)bya

linearmap

withpositivedeterminan

t,4

(2)left-handed

bases:

thosewhichdiffer

byalinearmap

withnegativedeterminan

t.

1in

loca

lco

ord

inates,

thematrix

ofpartialderivatives.

2Even

ifS

issm

ooth

someca

reis

needed

,e.g.ifctouch

esS

tangen

tiallyth

enth

eintersectionsh

ould

be

countedmultiple

times.Compare

withth

ephen

omen

onth

atth

epolynomialx2reallyhastw

oroots,notone.

3Youca

nalw

aysco

mpose

Fiwithth

ereflectionR2→

R2,(x

,y)�→

(x,−

y),

toget

anew

loca

lparametriza

-

tion,andth

etransitionmapswillstillallbesm

ooth

.Soneg

ativitydoes

notim

ply

non-orien

tability.

4v1=

Ae 1

,v2=

Ae 2

anddet

A>

0.ExplicitlyA

hasco

lumnsv1,v

2.Soth

eco

nditionis

det(v

1|v

2)>

0.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

29

Thefirsttype,

arecalled

positively

orien

tedbases,an

dcorrespondto

therigh

t-han

d-rule:the

thumbpoints

inthedirection

v 1,theindex

fingerpoints

inthedirectionv 2.Noticetheunique

angleless

than

180◦

from

v 1to

v 2determines

anorientation

ofacircle

centred

at0.

Thus,

theintuition

fordet(D

F−1

j◦DFi)

>0is

that

itensuresthattw

oob

servers(so

Fi,Fj)ag

reeon

whichbasesare

righ

t-han

ded

andwhicharenot:e 1

=(1,0),e 2

=(0,1)is

righ

t-han

ded

forthefirstob

server

intheirlocalcoordinates,an

dthesecorrespon

dto

thebasis

Te 1,T

e 2forthesecondob

server

whereT

=DF

−1

j◦D

Fi(thederivativeofthetran

sition

map

F−1

j◦F

i).Sodet

T>

0ensuresTe 1,T

e 2is

righ

t-han

ded

also

forthesecondob

server.

Example.Thereflection

R2→

R2,(x,y)→

(x,−

y)(correspon

dingto

complexconjugation

)has

det

=−1<

0.Therigh

t-handed

basise 1,e

2mapsto

theleft-handed

basise 1,−

e 2.Notice

thereflection

flipstheorientation

ofthebou

ndaryof

theunitdiscfrom

anticlockwiseto

clockwise,

since

eit�→

e−it.

Exercise.A

smooth

familyofem

bedded

discs

Gt:D

→S

startinganden

dingatthesame

discG

0(D

)=

G1(D

)willflip

theorien

tationoftheboundary

⇔det(D

G−1

1◦D

G0)<

0.

ProofthatDefi

nition

6.4

reallyim

plies

orien

tability.

Let

G:D

→S

beacontinuou

sly

embedded

disc,

andFi:Vi→

Salocalparam

etrisation

withFi(p)=

G(0).

Let

γbea

smallan

ti-clockwiseEuclideancircle

inVi⊂

R2centred

atp.Then

c(t)

=G

−1(F

i(γ(t)))is

acurvein

Davoiding0.Theintegral

ofthean

glevariab

leof

c(t)

iseither

1+2π

or−2π

,or

equivalently:c(t)

iseither

orientedan

ti-clockwiseor

clockwise(ifG

issm

ooth,thetw

ocasescorrespon

drespectivelyto

whether

det(F

−1

i◦G)is

positiveor

negative).

Wecould

havealso

usedapointqdifferentthan0∈D

(bytran

slatingz�→

z−

qan

dthen

calculating

thean

glevariab

le).

Chan

gingG,γor

qcontinuou

slywillchan

gethevalueof

that

integral

continuou

sly,butas

itcanonly

take

values

±2π

itmust

stay

constan

t.Bythesamereason

ing,

wecanalso

allow

moregeneral

continuousloop

sγ⊂

Viavoidingp,notjust

circles,

aslong

asthean

glewith

respectto

thecentrepintegrates

to+2π

(i.e.

“anti-clockwise”

loop

s).

Since

det(D

F−1

j◦D

Fi)

>0fortw

ooverlappingparametrisation

s,theFi,Fjag

reeon

what

orientation

such

γcurves

have.

Soon

thewholesurface

wehaveat

ourdisposal

such

embedded

orientedarbitrarily

small“testcurves”γwhichdeterminewhether

anem

bedded

discG

is“p

ositively-oriented”or

“negatively-oriented”.

Soan

ycontinuou

sfamily

Gt:D

→S

ofem

bedded

discs

areeither

allpositivelyor

allnegativelyoriented,in

particularG

0(D

),G

1(D

)areorientedthesameway.

Lemma6.5.If

asm

ooth

surface

isanorien

tabletopologicalsurface,then

wecanen

sure

(by

composingwithreflections)

thattheparametrizationsFisatisfydet(D

F−1

j◦D

Fi)>

0.

Sketchproof.

Once

you

pickalocalparam

etrization

Fi:Vi→

S,this

willdetermine(onthe

connectedcompon

entof

SwhereFilands)

whether

ornot

youneedto

composeeach

other

Fjwiththereflection

r:R

2→

R2,(x,y)→

(x,−

y).

You

dothisbyhoppingfrom

Fito

other

localparam

etrization

s,each

timecomparingsign

son

theoverlaps(ifyou

getanegativesign

,then

replace

FjbyFj◦r

toensure

thederivativeofthetran

sitionhas

positive

determinan

t).

Theon

lyproblem

isiftherearetw

opathsob

tained

byhop

pingfrom

Fito

Fj,an

don

epathrequires

Fjto

becomposed

withrto

obtain

positivityan

dtheother

pathdoes

not

requireit(see

thepicture).

Butin

that

case,composingthefirstpathwiththereverse

ofthe

secondyieldsaclosed

pathalon

gwhichwecanmoveasm

allsm

oothem

bedded

disc.

Becau

seof

thedisag

reem

entin

sign

s,thebou

ndary

ofthisdiscat

thestartan

dendof

theclosed

path

willhaveflipped

bou

ndaryorientation

.Butthis

contrad

icts

that

Sis

orientable.

1Thatwindingaround0in

theplanemore

than±1times

isim

possible

withoutth

ecu

rveself-intersecting

itselfis

arelativelyea

syco

nsequen

ceofth

eJord

anCurveTheo

rem

(anon-self-intersectingco

ntinuouscu

rve

inth

eplanedivides

theplaneinto

twoco

nnectedco

mponen

ts).

30

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

6.4

Riemann

surfacesare

alw

aysorienta

ble

Theore

m6.6.AnyRiemannsurface

Sis

anorien

tablesurface.

Proof.

Thetran

sition

map

sF

−1

j◦Fiareholom

orphic

so,viewed

asamap

R2→

R2,the

derivativematrixisacompositionof

scalingan

drotation

,so

thedeterminan

tispositive.

7.

Localanaly

sis:

theinverse

andim

plicit

functiontheorems

7.1

Theinversefunction

theorem

Theore

m7.1

(Inversefunctiontheorem).

Foranysm

ooth

mapf:R

n→

Rn,ifthematrix

ofpartialderivativesatp∈R

nis

invertible,then

fis

aloca

ldiff

eomorp

hism

1nearp.

Explicitly:

thetheorem

handsusauniquesm

ooth

mapg:R

n→

Rndefi

ned

nearf(p)such

thatf(g(y))

=yandg(f(x))

=xforallx,y

close

enough

top,f

(p)respectively.

Example.Let

fbethechange

ofvariablesfrom

polar

coordinates

(r,θ)to

(x,y).

Sof(r,θ)=

(rcosθ,rsinθ),so

Df=

� cosθ−rsinθ

sinθ

rcosθ

� ,so

det

Df=

r�=

0forr�=

0.Sonearany(r,θ)∈R

2

withr�=

0,thereis

auniquelocalinverseof

f.Asidefrom

ther=

0issue,

thereis

noglob

alinverseas

fis2π

-periodic

inθ.

Remark

s.

�Arguab

lythemostim

portanttheorem

inan

alysis.

Itsayssimple

linearalgebra

(the

non

-van

ishingof

adeterminan

t)ensuresthesm

oothinvertibilityof

themap

,locally.

�Invertibilityof

Dfisanecessary

condition:2

ifg(f(x))

=xforallxcloseto

pin

Rn,

then

bythechainrule

Dg◦D

f=

D(Id)=

Id(theidentity

map

),so

Dg=

(Df)−

1.

�Since

surfaces

arelocallyparam

etrizedbyR

2,theab

ovetheorem

also

holdsforsm

ooth

map

sbetweensurfaces.

�It

holdsalso

forsm

oothmap

sbetweenman

ifolds,

since

thesearelocallyR

n.

�Invertibilityof

Dfcanbeequivalentlyphrasedas

thelinearindep

enden

ceof

thevec-

tors

∂x1f,...,∂

xnf(w

hichform

thecolumnsof

thematrixDfof

partial

derivatives).

Coro

llary

7.2

(Inversefunctiontheorem

incomplexan

alysis).

Foranyholomorphic

mapf:C

→C,iff� (z 0)�=

0then

fis

alocalbiholomorphism

nearz 0.

1Mea

ning:th

ereare

open

neighbourh

oodsU

⊂Rn

ofpand

V⊂

Rn

off(p)su

chth

atth

erestriction

f| U

:U

→V

isadiffeo

morp

hism,so

thereis

auniquesm

ooth

inverse

f−1:V

→U.

2Asanex

ample,forfunctionsf:R

→R,forinvertibilityyouneedth

atth

eliney=

constantintersects

thegraph

off

inex

actly

onepoint.

By

theinterm

ediate

valueth

eorem,you

ded

uce

thatf

hasto

eith

eralw

aysincrea

seoralw

aysdecrease.Sof�≥

0orf�≤

0.Thereare

bijectivesm

ooth

functionsR

→R

withf�

sometim

eszero,su

chasx�→

x3,butth

eyare

notdiffeo

morp

hisms:

x�→

x1/3is

notsm

ooth

at0,because

the

derivativeblowsupth

ere(thehorizo

ntaltangen

tsto

fbecomevertica

ltangen

tsto

f−1).

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

31

Non-examinableproof.

Theprevioustheorem

implies

thereisasm

oothinversef−1:R

2→

R2

defined

nearf(z

0).

Weneedto

checkf−1isholom

orphic.Soweneedto

checkf−1satisfies

the

Cau

chy-R

ieman

nequations.

Butthisisequivalentto

show

ingthematrix

ofpartial

derivatives

isascalingtimes

arotation(thinkf� (z 0)=

reiθ).

Bythechainrule,Df·D

f−1=

D(Id)=

Id,

soDf−1is

theinverseof

Df,so

itis

ascalingtimes

arotation

since

Dfis.

�Remark

s.

�Asbefore,

theinvertibilityof

Df

(that

is,non

-van

ishingof

f� (z 0))

isanecessary

condition.Explicitly:ifg(f(z))

=zforallzcloseto

z 0in

C,then

bythechainrule

g� (f(z

0))

·f� (z 0)=

1so

g� (f(z

0))

=1/

f� (z 0).

�Since

Rieman

nsurfaces

arelocallyparametrizedbyC,theab

ovetheorem

also

holds

forholom

orphic

map

sbetweenRiemannsurfaces.

�Con

sider

aholom

orphic

map

f:C

n→

Cn(i.e.each

entryf 1,...,f

nis

holom

orphic

ineach

ofthecoordinates

z 1,...,z

n).

Then

linearindep

enden

ceof

thecomplex

derivatives

∂z1f,...,∂

znfat

p∈C

nim

plies

that

fis

alocalbiholomorphism

nearp.

�Theresult

holdsalsoforholomorphic

map

sbetweencomplexman

ifolds,

since

these

arelocallyC

n.

7.2

Theim

plicit

function

theorem

Motivation.When

thedim

ensionsn,m

are

differen

t,thereis

ofcoursenochance

offinding

a(local)

inverseofasm

ooth

mapf:R

n→

Rm.When

n=

mandfis

invertible,f(x)=

chasauniquesolutionf−1(c)=

(g1(c),...,g n

(c)),givingrise

foreach

c∈

Rm

someunique

numbers

g 1,...,g

nforwhichf(g

1,...,g

n)=

c.Now

assumen>

m,then

thenextbest

thing

tofindinganinverse(w

hichcannotexist)

isfindingsomefunctionsg i

whichcanbe

usedto

getridofsomeofthevariables

inR

nandwhichonly

depen

don

theother

variables.

For

example:f:R

2→

R,f(x,y)=

x2−

2y,then

f(x,g(x))

=cif

wetake

g(x)=

1 2(x

2−

c).

Soforthepurpose

ofsolvingf=

cthevariableyis

redundantsince

wecanreplace

itwitha

functiong(x)ofx,butthevariablexis

essential.

Notice

theredundantvariableyis

theone

forwhich∂yf=

−2is

never

zero,whereas∂xf=

2xcanvanish,atx=

0.

Let

f:R

n→

Rm

besm

ooth,an

dn≥

m.Wewan

tto

describethesolution

sof

f(x)=

c

nearagiven

solution

f(p)=

c,wherex,p

∈R

nan

dc∈R

m.

Theore

m7.3

(Implicitfunctiontheorem).

Ifm

columnsofD

pf

are

linearlyindepen

den

t,then

thevariables

xi 1,...,x

i mcorrespondingto

those

columnsare

redundant.

Namely,

they

canbe

replacedby

uniquesm

ooth

functions

g i1,...,g

i m:R

n−m

→R,

depen

dingonly

on

theremainingvariables

xj(soj�=

i 1,...,i

m),

defi

ned

nearx

=pand

satisfyingg i

1(p)=

pi 1,...,g

i m(p)=

pi m,so

that1

f(x)| (

xi1=gi1,...,xim

=gim

)=

c

describes

allsolutionsxnearp.

Examples.

Below

,weseek

solution

sof

f=

0nearp=

(0,...,0).

(1)f(x,y)=

y:∂yf=

1�=

0,so

f(x,g(x))

=0(indeedg(x)=

0).

(2)f(x,y)=

x2−

y:∂yf=

−1�=

0,so

f(x,g(x))

=0(indeedg(x)=

x2).

1More

ped

antica

lly:f−1(c)=

{(x1,x

2,...,x

i 1−1,g

i 1(x

),xi 1

+1,...):x=

(x1,x

2,...,x

i 1−1,x

i 1+1,...)∈

Rn−m},

whereweomit

thevariablesxi 1,x

i 2,...,x

i mandwereplace

them

byth

evalues

ofgi 1,g

i 2,....

32

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

(3)f(x,y)=

(x+

1)2−

1+

y2:∂xf| x=

0,y=0=

2�=

0,so

f(g(y),y)=

0(indeedg(y)=

−1+�

1−

y2,whichisdefined

neary=

0,andnoticeg(0)=

0).

(4)TheunitcircleS1isthesolution

setf=

0forf(x,y)=

x2+y2−1.

For

points(a,b)∈S1

withb�=

0,∂yf=

2y�=

0forycloseenou

ghto

b.Soxis

alocalcoordinate:

S1is

described

by(x,g(x))

near(a,b)(secretlywekn

owg(x)=

√1−

x2,whichis

smooth

away

from

x=

±1,

y=

0).For

a�=

0,∂xf=

2x�=

0forxcloseto

a,so

yisalocal

coordinate:

S1is(g(y),y)near(a,b)(secretlywekn

owg(y)=

�1−

y2).

Sowehave

localcoordinates

everyw

here(a,b

cannot

bothbezero:f(0,0)=

−1�=

0).

(5)In

thepreviousexam

ple,noticethat

wearelocally

param

etrizingS1as

thegraphof

afunction,so

wegeteither

(x,g(x))

or(g(y),y).

(6)For

theunitsphereinR

3,defined

byf(x,y,z)=

x2+y2+z2−1=

0,nearanypointeither

(x,y)or

(x,z)or

(y,z)arelocalsm

oothcoordinates.Indeed,fortheim

plicitfunction

theorem

tofailforf

inthosethreecases,

itwou

ldmeanrespectively

that

∂zf

=0,

∂yf=

0,and∂xf=

0.Butthen

x=

y=

z=

0,whichis

not

apointof

thesphere.

Noticethat

wearelocally

param

erizingS2as

thegraphof

afunction,e.g.

forthe(x,y)

case

(when

∂zf�=

0)wededuce

that

S2islocally

(x,y,g(x,y))

there.

Non-examinable

pro

ofofTheore

m7.3.Byrelabelingcoordinates,wemay

assumethe

last

mcolumnsof

Dpfarelinearlyindep

endent.

Abbreviate

k=

n−

m.Con

sider

F:R

n→

Rn,F(x

1,...,x

n)=

(x1,...,x

k,f

(x1,...,x

n)).

Then

DpF

isinvertible

(try

writingthematrix).

Bytheinversefunctiontheorem,

F−1(x

1,...,x

k,c

1,...,c

m)=

(x1,...,x

k,g

k+1,...,g

n)

foruniquefunctionsg k

+1,...,g

nof

x1,...,x

k,c.

Coro

llary

7.4

(Smoothdep

enden

ceon

cin

Theorem

7.3).Notice

above

g i1,...,g

i mdepen

dsm

oothly

onc.

Sothereare

uniquesm

ooth

functions

Gi 1,...,G

i m:R

n−m×

Rm

→R

defi

ned

nearx

=p,y

=canddepen

dingonly

on

thenon-redundantxjvariables

(soj�=

i 1,...,i

m)andony∈R

m,so

that

f(x)| (

xi1=G

i1,...,xim

=G

im

)=

y

describes

allsolutionsoff(x)=

yforxnearp,andynearc.

Remark

7.5.Thesetofsolutionsoff(x)=

yis

therefore

locallycutoutby

thevanishingof

mfunctions:

xi 1−

Gi 1,...,x

i m−

Gi m.

Con

sider

thechan

geof

coordinates

onR

n(soalocaldiffeomorphism)defined

by

first

permutingthecoordinates

ofR

nso

that

wemay

assumethei 1,...,i

mab

ovearen−

m+

1,...,m,an

dthen

chan

gingthecoordinates

byxj�→

�x j(x)with

�x j(x)=

xjforj≤

man

d�x j(x)=

xj−

g jforj=

n−

m+

1,...,m.

Inthesecoordinates

thesolution

setof

f(x)=

yisparam

etrizedbythefirstn−m

coordinates

�x jan

dis

cutou

tbythem

equationsgiven

bythevanishingof

thelast

mcoordinates:

�x n−m

+1=

0,...,

�x n=

0.

Solocally,yo

ucanthinkof

theinclusion

ofthesolution

set(f(x)=

y)⊂

Rnas

beingsm

oothly

“thesame”

(diffeomorphic)to

thestan

dardinclusion

Rn−m

⊂R

n.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

33

8.

Localanaly

sis:

Embedded

surfa

cesarelocally

graphs

8.1

Criterion

foralocalpara

metrization

ofasm

ooth

surfacein

R3

Ournextgo

al,is

toprove

Theorem

2.10

,an

dto

show

that

asm

oothsurfaceS

⊂R

3is

locallydefined

bythevanishingof

asm

oothfunction(just

likeS2is

locally,

infact

glob

ally,

thezero

setof

thefunctionx2+

y2+

z2−

1:R

3→

R).

Let

S⊂

R3beasm

oothsurface,

F:V

→S

asm

oothmap

,V

⊂R

2an

open

set,

andF(v

0)=

p.

Theore

m8.1.F

isa

smooth

localparametrization

nearp

when

restricted

toa

possibly

smaller

openneighbourhoodV

�⊂

Vofv 0

⇐⇒

∂xF,∂

yF

are

linearlyindepen

den

tatv 0.

Proof.

Theeasy

direction

is⇒

:thereis

asm

oothinverseF

−1,so

F−1◦F

=Id

:V

�→

V� ,

sobythechainrule

DF

−1◦D

F=

Id,so

DF

isinjective,

sothetw

ocolumns∂xF,∂

yF

ofthe

matrixDF

must

belinearlyindep

endentat

each

v∈V

� ,in

particularat

v=

v 0.

Now

theharddirection

⇐.ThematrixDF

isa3×

2matrix,andsince

itscolumnsare

linearlyindep

endentat

v 0,theremust

bea2×2submatrixwithnon

-zerodeterminan

t(basic

linearalgebra).

WLOG

assumeit’s

thefirsttw

orows,

so

�∂xF1

∂yF1

∂xF2

∂yF2

isinvertible

atv=

v 0,whereweclarify:weuse

x,y

coordinates

onV

⊂R

2,an

dweuse

X,Y

,Zcoordinates

onR

3,so

explicitlyF(x,y)=

(F1(x,y),F2(x,y),F3(x,y)).

Com

poseF

withprojectionto

thefirsttw

ocoordinates(X

,Y)ofR

3,

� F=

(X,Y

)◦F

:R

2⊃

V→

R3→

R2,� F(x,y)=

(F1(x,y),F2(x,y)).

Then

D� Fistheab

ove2×2matrix.Bytheinverse

functiontheorem,� Fhas

auniqueinverse

� F−1:R

2→

V⊂

R2defined

near� F(v

0).

Thus� F(x,y)=

(X,Y

)⇔

� F−1(X

,Y)=

(x,y),

so

F(� F−1(X

,Y))

=F(x,y)=

(X,Y

,g(X

,Y))

only

dep

endssm

oothly

on(X

,Y)an

dthusit

determines

auniquesm

oothfunctiong(X

,Y)

(thatis:theZ

coordinate

ofpoints

inS

isdetermined

byX,Y

,nearF(v

0)).

Wenow

definethemap

that

wehop

eis

thechartinverseto

theparametrization

F:

f:R

3→

R2,(X,Y

,Z)�→

� F−1(X

,Y).

Noticethat

fisasm

oothmap

R3→

R2,defined

nearF(v

0),an

dfrestricted

toSnearF(v

0)

becom

esf(X

,Y,g(X

,Y))

=� F−1(X

,Y)=

(x,y)so

itis

theinverse

ofF.This

concludes

the

proof

that

Fisalocaldiffeomorphism

V→

Snearv 0,an

dhence

alocalparam

etrization

.�

(Non-examinable)Exercise.Canyouuse

theidea

intheabove

proofto

state

andprove

ageneralim

plicitfunctiontheorem

forsm

ooth

mapsf:R

n→

Rm

when

n<

m?

Coro

llary

8.2

(Theorem

2.10

).If

F,above,is

also

injectivethen

Fis

asm

ooth

local

parametrizationonallofV

⇐⇒

∂xF,∂

yF

are

linearlyindepen

den

tateach

pointofV.

Proof.

Since

F:V

→F(V

)is

injective,

andbyconstructionsurjective,

itis

bijective,

soit

remainsto

checkthat

F−1:F(V

)→

Vis

smooth.ButF

:V

→F(V

)⊂

Sis

alocal

diffeomorphism

neareach

v 0∈V

bythepreviousTheorem,so

F−1is

smooth.

34

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

8.2

Smooth

surfacesin

R3are

locallygra

phs

Theore

m8.3.Foranysm

ooth

surface

Sin

R3,neareach

pointp∈S,

(1)S

iseither

1asm

ooth

graph(X

,Y,g(X

,Y))

overthecoordinates(X

,Y),

oragraph

overthe(X

,Z)coordinates,

oragraphoverthe(Y

,Z)coordinates,

(2)either

(X,Y

),or(X

,Z),

or(Y

,Z)are

smooth

localcoordinatesforS,

(3)S

islocallycutoutasthezero

setofafunctionR

3→

R.

Proof.

Theproof

ofTheorem

8.1show

edthat

Sis

locally(X

,Y,g(X

,Y)),forsm

ooth

g,if

thefirsttw

orowsof

DF

arelinearlyindep

endent.

Thecases(X

,g(X

,Z),Z),

(g(Y

,Z),Y,Z

)occurifrows1,3,

respectively

rows2,3of

DF

arelinearlyindep

endent.

Thisproves

(1).

Also

(2)follow

ssince,sayin

thefirstcase,(X

,Y)�→

(X,Y

,g(X

,Y))

isalocalparametrization.2

Finally

(3)follow

s,sayin

thefirstcase,byconsideringthefunctionR

3→

R,(X

,Y,Z

)�→

Z−

g(X

,Y)since

Z−

g(X

,Y)=

0cuts

outtheset(X

,Y,g(X

,Y))

asrequired.

8.3

Riemann

surfacesin

C2are

locallygra

phs

Con

sider

asubsetof

C2cutou

tbyaholom

orphic

equation

S=

{(z,w

)∈C

2:f(z,w

)=

0}wheref:C

2→

Cis

holom

orphic

(e.g.acomplexpolynom

ialin

z,w

).Analogouslyto

Theo-

rems8.1an

d8.3(usingSection

7.1to

getholom

orphicity),

wededuce:

Theore

m8.4.S

isaRiemannsurface

near(z

0,w

0)∈S

ifandonly

if

(1)either

∂f

∂z�=

0at(z

0,w

0),

then

Sis

locally(g(w

),w),

(2)or

∂f

∂w�=

0at(z

0,w

0),

then

Sis

locally(z,g(z)),

soS

islocallythegraphofaholomorphic

functiong:C

→C,andS

islocallycutoutby

aholomorphic

function(respectivelyz−

g(w

)=

0,orw−

g(z)=

0).

Example.Con

sider

S=

{(z,w

)∈

C2:f(z,w

)=

w2−

(z−

1)(z

−2)

=0}(recallExercise

sheet1).Then

∂wf=

2wisnon

-zeroexceptat

w=

0.When

w=

0,either

z=

1or

2.Butthen

∂zf=

−(2z−

3)�=

0.SoS

isaRiemannsurface.

Recallwecompactify

Sat

±∞

byrewriting

theequationin

thenew

variables

X=

1/z,

Y=

w/z

,

sothedefiningequationforSbecom

es� f(X,Y

)=

Y2−(1−X)(1−2X

)=

0,and±∞

correspon

dto

thetwonew

points

(X,Y

)=

(0,±

1).Finally

wecheckS∪{±

∞}is

aRiemannsurfaceat

thosenew

points:∂Y� f=

2Y�=

0at

Y=

±1.

Thefollow

ingdefinition

andremarkarenon

-exam

inab

le,butIhop

eit

willinspireyour

interest

inB3.3

Algebra

icCurv

es.

Definition

8.5

(Com

plexalgebraic

curve).A

complex

algebra

iccurv

eS

isthezero

set

f(z,w

)=

0ofacomplexpolynomialfin

twovariables

z,w

.Thesingularpoin

tsare

the

(z0,w

0)∈

Swhereboth

conditionsabove

fail:∂zf=

0,∂wf=

0.Sonon-singularcomplex

algebraic

curves

are

Riemannsurfaces.

Remark

8.6

(Projectivealgebraic

curve).Because

ofthemaximum

modulusprinciple,S⊂

C2cannever

becompact

(otherwise|z|,|

w|w

ould

attain

maxima,so

z,w

would

beconstant

functionson

S).

Asin

theexample

above,S

ismissingsomepoints

atinfinityandone

1W

ealw

aysmea

nnon-exclusive

“eith

er...or...”,so

severaloptionsmayoccur.

2Theco

mpositionofth

ediffeo

morp

hismsF

◦� F−1in

theproofofTheo

rem

8.1,hen

ceadiffeo

morp

hism.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

35

system

aticwayofcompactifyingis

topro

jectivizetheeq

uation.Thatmeans,

youview

Sasasubset

of1

CP

2.SoS⊂

C2are

thepoints

oftheform

[1:z:w]ofthecompactification

S⊂

CP

2.Toprojectivizeapolynomial,youmake

ithomogen

eousby

simply

replacingz=

z 1/z

0,w

=z 2/z

0andthen

rescalingby

theleast

power

ofz 0

togetridofden

ominators.

Example.w

2−

(z−

1)(z

−2)

=0becom

esz2 2−

(z1−

z 0)(z 1

−2z 0)=

0.Wealreadykn

owsolution

swhen

z 0�=

0(w

earethen

allowed

torescalez=

z 1/z 0,w

=z 2/z

0).

Supposeinstead

z 1�=

0,then

wemay

use

localcoordinates

X=

z 0/z

1,Y

=z 2/z

1so

[z0:z 1

:z 2]=

[X:1:Y]

(can

youseewhythesearethesameas

theX,Y

intheab

oveexam

ple?).Theequationbecom

es:

Y2−

(1−

X)(1−

2X)=

0.Sowegettwonew

points

(X,Y

)=

(0,±

1),whichcorrespon

dto

[0:1:±1]

∈CP

2.Finally,supposez 2

�=0(bydefinitionof

CP

2,thez 0,z

1,z

2cannot

all

vanish).

Wecould

use

X=

z 0/z

2,Y

=z 1/z

2,so

[z0:z 1

:z 2]=

[X:Y

:1],andrewrite

the

equation,butbecause

wearenot

intheprevioustwocases,wemay

assumethat

z 0=

z 1=

0,so

iton

lyremainsto

checkwhether

[0:0:1]

isasolution

,anditisn’t.

(Non-examinable)Exercise.

Show

thatif

youprojectivizew

2=

(z−

1)(z

−2)(z

−3),

youneedto

addthepoint∞

=[0

:0:1](andyougetatorus).However,show

thatif

you

projectivizew

2=

(z−

1)(z

−2)···(z−

5)yougetaprojectivecurvewhichis

singularat

infinity.

Abetter

compactificationthanthis,is

described

inthenextexample.

Example.Con

sider

thepolynom

ial

f(w

,z)=

w2−

(z−

a1)(z−

a2)···(z−

an)=

0.

Notice∂wf=

2w�=

0unless

w=

0,andforw

=0wegetz=

ajso

thecondition∂zf�=

0is

equivalentto

requiringthat

non

eof

therootsaj∈C

arerepeated.SowegetaRiemannsurface

when

theajarepairw

isedistinct.

Bythemethodsof

Exercisesheet1,

question

3,on

ecandefineacompactification

atinfinitywhich

yieldsaRiemannsurface.

Atinfinity,weuse

thecoordinates

X=

1 zY

=w zm

wherem

=n/2

ifn

iseven,andm

=(n

+1)/2

ifn

isodd.

Theequationbecom

esY

2=

(1−

a1X)···(1−

anX)forn

even,andY

2=

X(1

−a1X)···(1−

anX)forn

odd.Sowe

compactify

byaddingnew

points

(X,Y

)=

(0,±

1)called±∞

forneven,and(X

,Y)=

(0,0)

1ByanalogywithCP

1,wedefi

ne

CP

2=

{[z 0

:z 1

:z 2

]:(z

0,z

1,z

2)∈

C3\{

0},

[z0:z 1

:z 2

]=

[λz 0

:λz 1

:λz 2

]foranyλ∈

C\{

0}}

(geo

metrica

lly,

thinkofth

epoint[z

0:z 1

:z 2

]asth

eco

mplexlineC·(z 0

,z1,z

2)⊂

C3).

36

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

called∞

fornodd.In

bothcases,wedeclare

that

Yisalocalholom

orphiccoordinateat

infinity.1

Recallfrom

Exercisesheet1,

question

3,that

youcanvisualizetheRiemannsurfaceby

gluingtwo

copiesof

Cwithcutsandthen

compactifyingat

infinity.

Intheab

ovepicture,wefirstcompactified

each

Cto

aCP

1,then

drew

thecuts,2

andin

thelower

picturesweglued

thecuts

toob

tain

the

Riemannsurfaceanddetermined

itsgenusg.

Imagineincreasingnby

2:this

meanswerequiretwoextracuts

intheplanes

CP

1that

we

glue,

givingrise

toan

extrahandle,so

thegenusof

thecompactified

Riemannsurfaceincreases

by1.

For

n=

2wegetasphere(genus0)

andforn=

3wegetatorus(genus1),so

inductively

ingeneral:

genus=

n−

2

2forneven,an

dn−

1

2fornodd

(correspon

dingrespectively

toχ

=4−

nandχ

=3−

n).

TheseRiemannsurfaces

arecalled

hyp

erellip

ticcu

rves.3

9.

Thetangentspace

9.1

Tangentsp

aceforsm

ooth

surfacesin

R3

Let

Fbealocalparam

etrization

nearp∈S,forasm

oothsurfaceSin

R3.RecallbyTheorem

2.10

)that

∂xF,∂yF

arelinearlyindep

endentat

p(w

ewillsuppress

from

thenotationthat

weareevaluatingat

p).

Thusweob

tain

a2-dim

ension

alvector

subspaceof

R3,called

the

tangentsp

ace,as

follow

s

TpS=

span

(∂xF,∂

yF)=

span

(DF

·e1,D

F·e

2)=

DF

·R2⊂

R3

1W

eneed

toch

eckth

atin

asm

all

neighbourh

ood

ofapoint(z,w

)forlargez�=

∞,th

etransition

map

from

theloca

lco

ord

inate

zto

theloca

lco

ord

inate

Yis

biholomorp

hic.

Butnea

r(X

,Y)with

X�=

0we

can

use

eith

erX

orY

asloca

lco

ord

inate

since

weca

nex

press

Yin

term

sofaholomorp

hic

branch

ofth

e

square

rootofapolynomialin

X(this

isth

esameargumen

tth

atproves

thatforth

eRiemannsu

rface

w2=

z

youmaydeclare

thatw

isaholomorp

hic

coord

inate

nea

r(w

,z)=

(0,0

),andzis

aholomorp

hic

coord

inate

elsewhere).Thusit

sufficesto

find

aholomorp

hic

transition

from

zto

X,awayfrom

X=

0.Butth

atwe

know:z�→

X=

1/zis

therequired

loca

lbiholomorp

hism

inz.

2Tounderstandth

ecu

t:forw

2=

(z−

1)(z−

2),

whydowecu

tth

esegmen

t(1,2

)?Theloca

lmodel

nea

r

z=

1andnea

rz=

2is

thatofth

esquare

root,

andforth

esquare

rootwetypicallych

oose

thecu

talongth

e

neg

ativerealaxis.In

ourca

se,wemakecu

ts(−

∞,1

)and

(−∞

,2)and

wepickbranch

esof√z−

1and

of

√z−

2.Foroneco

pyofC,forz=

1+aei

θ=

2+be

iψlet’sdeclare

√z−

1=

a1/2ei

θ/2and√z−

2=

b1/2ei

ψ/2

forθ,ψ

∈(−

π,π

).W

ewould

thinkth

atth

isis

only

accep

table

ifth

ereis

acu

talong(−

∞,2

)⊂

R⊂

C,but

infact

forrealz<

1th

etw

odisco

ntinuitiesca

ncelout:

eiπ/2ei

π/2=

eiπ=

e−iπ

=ei

(−π/2)ei

(−π/2).

3If

you

are

curious

about

why

these

clev

erco

ord

inates

work

at

infinity,

unlike

the

pro-

jectivization

which

typically

gives

rise

toa

singularity

at

infinity,

then

have

alook

at

http://en

.wikiped

ia.org/wiki/Hyperellipticcu

rve

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

37

wheree 1,e

2is

thestan

dardbasis

onthedom

ainR

2ofF.Thinkof

theplaneTpS

asthe

plane1

inR

3whichbestap

proxim

ates

Snearp.

Thepicture

also

show

stheunitvectorn(p)normalto

TpS,obtained

from

thecrossproduct

of∂xF,∂

yF

inR

3an

dthen

normalizing.

Wewilldiscu

ssthis

inSection

9.5.

Theab

ovevector

subspaceTpS

ofR

3is

indep

endentofthechoice

ofparam

etrization

F,

since

foran

other

param

etrization

� F(soan

other

“ob

server”),wehave

D� F·R

2=

D� F·(D

� F−1◦D

F)·R

2=

DF

·R2,

usingthat

thederivativeof

thetran

sition

,Dτ=

D� F−1◦D

F,isalinearisom

orphism

R2→

R2.

9.2

Tangentsp

aceforabstra

ctsm

ooth

surfaces

For

asurfacein

R3,noticethat

DF

identifies

thelocalta

ngentsp

ace

Tp0R

2≡

R2=

span

(e1,e

2)

withthetangentspace

TpS=

span

(∂xF,∂

yF)⊂

R3(w

hereF(p

0)=

p).

Wesaw

abovethat

twoob

serversag

reewhichplaneTpS

⊂R

3is,becau

seDF

·Tp0R

2=

D� F·T

�p 0R

2.Wecan

rewrite

this

as

Dτ·T

p0R

2=

T�p 0R

2

whereτ=

� F−1◦F

isthetran

sition

map

betweenthetw

oob

servers.

For

abstract

smooth

surfaces,wesimply

turn

this

equationinto

thedefinition.For

abstract

surfaces,thereis

no

common

ambientspace(such

asR

3ab

ove)

wherelocalob

serverscancomparetangentspaces,

soon

esimply

workswiththelocaltangentspaces

andoneremem

bersthat

Dτis

themap

whichtran

sformsvectorsfrom

oneob

server’s

coordinatesystem

totheother.

An

equivalentdescription

ofthetangentspace,

isasequivalence

classesofcurv

es.

Nam

ely,

inlocalcoordinates,given

alocaltangentvectorv∈

R2≡

TpU,thereis

asm

ooth

curvec:(−

ε,ε)

→U

passingthrough

c(0)

=pwithinitialvelocity

c�(0)=

v.For

exam

ple,

thestraightlinec(t)

=p+

tv.Weon

lycare

that

thecurveis

defined

forsm

alltimes,so

ε>

0canbesm

all.

Thereareman

ychoices

ofsuch

curves

c(t),since

weon

lyprescribethe

firsttw

oterm

sp+tv

ofaTay

lorseries,e.g.

c(t)

=p+tv

+t2w

isan

other

acceptable

choice

ofcurveforan

yw.Thusvcorrespon

dsto

anequivalence

classof

curves:tw

ocurves

c 1,c

2

areequivalentifc 1(0)=

c 2(0),

c� 1(0)=

c� 2(0).

WecandefinethetangentspaceTpS

asthe

1Tobeprecise,th

eplanewhichbestapproxim

atesS

nea

rpis

thetranslate

p+

TpS

⊂R3,butit

ismore

conven

ientto

work

withavectorsu

bsp

ace

sooneusesTpS.

38

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

collection

ofequivalence

classesof

smooth

curves

cin

Sdefined

forsm

alltimes

withc(0)=

p(theequivalence

relation

gets

checked

inanylocalparametrisationbycomparingvelocities).

Assumingforsimplicity

that

F(0,0)=

p,thereare

twoobviouscurves

inalocalparametri-

sation

t�→

(t,0)an

dt�→

(0,t)correspondingto

thestandard

basisvectors

e 1,e

2∈R

2=

T0U.

InTpS

thesecorrespon

dto

thecurves

t�→

F(t,0)andt�→

F(0,t),

whose

tangentvelocity

vectorsat

pare∂xF

=DF

·e1an

d∂yF

=DF

·e2.In

general,forF(x

0,y

0)=

p,wecan

representthegeneral

tangentvector

v=

a∂xF

+b∂

yF

bythecurveF(x

0+at,y 0

+bt).

Exercise.If

F,� Faretw

olocalparam

etrisationsdefined

nearp,letτ=

� F−1◦F

denote

the

tran

sition

map

,show

that

thelocalcurves

correspondingto

agiven

curvein

Sget

naturally

identified

byτ:U

→� U,an

dthat

thevelocities

transform

byDτ:R

2=

TpU

→Tp� U=

R2,

i.e.

byleft

multiplication

bythematrixofpartialderivatives

ofτ.

For

smoothmap

sϕ:S1→

S2of

surfaces,onecanalsodefinethederivativemappurely

interm

sof

equivalence

classesof

curves:

Dpϕ·[curvec(t)]=

[curveϕ◦c

(t)].

Exercise.Checkthat

inlocalcoordinatesthiscorrespondsto

thematrix

ofpartialderivatives

ofϕat

p,actingbyleft-m

ultiplication

onc�(0)=

v∈R

2,thusD

pϕ:TpS1→

Tϕ(p

)S2isalinear

map

betweenthetangentspaces

(inlocalcoordinates,

e 1,e

2∈R

2mapto

∂xϕ,∂

yϕ∈R

2).

Tan

gentvectors

canalso

bedefined

asdiffere

ntialopera

tors

actingonsm

ooth

functions.

For

exam

ple,iff:S→

Rissm

ooth,then

locallye 1

·fmeansthepartialderivative∂xf∈R.

Atangentvector

vacts

onasm

oothfunctionf:S→

Rbytakingthedirectionalderivative

v·f

=∂t| t=

0(f

◦c(t)),

usingan

yrepresentativecurvec(t)

forv(exercise:show

thatthechoiceofrepresentative

does

not

matter).It

tellsyouhow

much

fvaries

inthev-direction.Explicitlyifv=

a∂xF+b∂

yF

then

v·f

=a∂xf l

oc+

b∂yf l

ocwheref l

oc(x,y)=

(f◦F)(x,y).

Forthis

reason,oneoften

abbreviatesthenotationbysimply

writingv=

a∂x+b∂

y,in

particular∂xF

=∂x,∂yF

=∂y.

9.3

Usingth

eta

ngentplaneto

constru

ctlocalpara

metrizations

Theorem

9.1.Let

Sbe

asm

ooth

surface

inR

3.

Nearanypointp∈

S,wecan

locally

parametrize

Sby

usingtheorthogonalprojectionto

thetangentplaneTpS.In

this

case,S

islocallythegraphofafunctionh:TpS→

Roverthetangentplane(forfixedp).

Proof.

First

apply

arotation

toR

3to

make

thetangentplaneTpS

horizontal:

sovectors

inTpS=

R2⊂

R3havezero

inthethirdentry.

Then

theZ

coordinate

cannotbeusedtogether

withX

orY

togivetw

olocalcoordinates(e.g.ifS

wereagraphF(X

,Z)=

(X,g(X

,Z),Z)

then

∂ZF

wou

ldbeatangentvector

withanon-zerothirdentry).

Therefore

X,Y

must

be

localcoordinates

andSmust

beagrap

h(X

,Y,h

(X,Y

))forasm

ooth

functionh(usingresults

from

Section

8.2).

9.4

Vecto

rfields

Definition

9.2

(Vectorfield).

Atangentvec

torfield

vonS

isasm

ooth

map

v:S→

R3such

thatv(p)∈TpS⊂

R3,

thatis

achoiceoftangentvectorv(p)ateach

pointpofS

varyingsm

oothly

withp∈S.

Locally,

wecanwrite: v(x,y)=

a(x,y)X

1+b(x,y)X

2(v(x,y)∈TF(x

,y)S⊂

R3),

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

39

forsm

oothfunctionsa,b

ofthelocalvariab

lesx,y,whereX

1,X

2isthebasisof

TF(x

,y)Sgiven

bythelocalvector

fields:

X1(x,y)=

∂xF

X2(x,y)=

∂yF.

Rem

ark.Vectorfieldscanalsobe

defi

ned

forabstract

surfaces:

v(x,y)∈

R2=

T(x

,y)V

isa

smooth

functionv:V

→R

2,andthis

must

transform

correctlyifwechange

observer:

�v(τ(x,y))

=(D

(x,y)τ)v(x,y).

Example.For

thecylinder

X2+Y

2=

r2,consider

thevector

fields

E1=

(−sinθ,cosθ,0)

E2=

(0,0,1)

wherep=

F(θ,Z

)=

(rcosθ,rsinθ,Z).

NoticeE

1points

equatorially

inthecircle

direction

ofthecylinder,andE

2points

intheaxisdirection

.Since

X1=

(−rsinθ,rcosθ,0),X

2=

(0,0,1),

E1=

1 rX

1E

2=

X2.

Soageneral

vector

fieldon

thecylinder

has

theform

a(θ,Z

)E

1+b(θ,Z)E

2

foranysm

oothfunctionsa,b

whichare2π

-periodic

inθ.

9.5

Smooth

surfacesin

R3:norm

als

and

theGauss

map

ByTheorem

6.3,

acompactsm

oothsurfaceS

⊂R

3must

beorientable.ByLem

ma6.5,

wecanpickacoverof

S=

∪Fi(Vi)

bylocalparam

etrization

sFi:Vi→

Sso

that

onoverlaps:

det(D

F−1

j◦D

Fi)>

0.

Given

anypointp∈Fi(Vi)

wecandefineaunit

normal

vectorn(p)∈R

3to

Sbyrequiring

that

thethreevectors

DFi·e

1=

∂xFi,

DFi·e

2=

∂yFi,

n(p)

obey

theright-hand

rule:if∂xFi,∂yFiaretheindex

andmiddle

fingerrespectively,then

n(p)points

inthethumbdirection

.Explicitly,

wetakethecrossproduct

andnormalize:

n(p)=

∂xFi×∂yFi

�∂xFi×∂yFi�

∈R

3(w

here∂xFi,∂yFiareevaluated

atp)

Recallin

Section

9.2wedefined

thetangentspace.

Nam

ely(again,evaluatingat

p):

TpS=

span

(∂xFi,∂yFi)=

span

(DFi·e

1,D

Fi·e

2)=

DFi·R

2⊂

R3

Wesaw

that

this

vector

subspaceis

indep

endentof

thechoice

ofparametrization

,since

DFj·R

2=

DFj·(D

F−1

j◦D

Fi)·R

2=

DFi·R

2,usingthat

DF

−1

j◦D

Fiisalinearisom

orphism

R2→

R2.Byconstruction,thevector

n(p)is

aunit

normal

tothe2-dim

ension

alsubspace

TpS.Since

aplanein

R3has

exactlytw

ounit

normals,

thequestion

iswhether

n(p)ever

flipsifwechan

geparam

etrization

s.Butthis

willnever

hap

pen

becau

seS

isorientable:the

tran

sition

shavedet(D

F−1

j◦D

Fi)

>0on

overlaps,

sothetw

ovectors

∂xFi,∂yFihavethe

sameorientation

insideTpS

asthetw

ovectors∂xFj,∂

yFj(indeed,thetw

opairs

differ

by

multiplication

bythetran

sition

map

DF

−1

j◦D

Fi),thereforethecross-product

ofeach

pair

isorientedin

thesamedirection

.

More

intuitivelysaid:twoobservers

willagree

whichplaneTpS

⊂R

3is,they

willagree

aboutwhichordered

basisofTpS

isright-handed,so

they

willagree

aboutwhichofthetwo

norm

als

toTpS

givesrise

toaright-handed

basisforR

3,so

they

compute

thesamen(p).

40

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

TheGauss

mapis

thenorm

alvectorfield

n:S→

R3,p�→

n(p).

Rem

ark.Thefact

thatyouhave

awell-defi

ned

norm

alvectorfieldis

relatedto

thefact

that

thefieldeither

always

points

outwards,

oralways

points

inwardsto

thesurface.

Example.Let

SbethesphereX

2+Y

2+Z

2=

r2.For

acurveγ(t)=

(X(t),Y(t),Z(t))

∈S,

differentiatethedefiningequationto

get:

0=

2XX

� +2YY

� +2ZZ

�=

(X,Y

,Z)·2(X

� ,Y

� ,Z

� ).

For

anytangentvector

v∈TS,thereis1acurveγvin

Swiththat

tangentvector

γ� (0)=

v.So

theab

oveshow

sthat

(X,Y

,Z)isnormal

toTpS(indeedtheradialvector

isou

twardandnormal

tothesphere).Normalizing,

wegetaGauss

map:

n(p)=

(X,Y

,Z)/r.

Lemma9.3.A

choiceoforien

tationonasm

ooth

surface

S⊂

R3is

thesameasachoiceof

asm

ooth

map

n:S

�→S2=

{(X,Y

,Z)∈R

3:X

2+Y

2+Z

2=

1}⊂

R3,

p�→

n(p)

such

thatn(p)is

orthogonalto

TpS

⊂R

3forallp∈

S.More

explicitly,

aparametrization

F:R

2⊃

V→

F(V

)=

U⊂

R3respects

theorien

tationprecisely

if:

n(p)·(∂xF

×∂yF)≡

det

�∂xF

∂yF

n(p)� >

0.

Proof.

Declare

that

abasisv,w

∈TpS⊂

R3is

right-handed⇔

w=

λn(p)forpositive

λ>

0(soin

fact,λ=

�v×w�),so

⇔λ≡

n(p)·(v×w)>

0.

Ourgoal

willbeto

use

theGauss

mapto

definevariousnotionsofcurv

atu

re(G

aussian

curvature,principalcurvatures,

meancurvature).

10.

Surfa

cesin

R3:thefirst

fundamentalform

10.1

Thefirstfundamenta

lform

Let

Sbeasm

ooth

surface

inR

3.Usingthedotproduct

2·onR

3wecandefineaninner

product

onTpS

called

thefirstfundamenta

lform

:

I:TpS×TpS→

R,I(v,w

)=

v·w

=vTw

Theproperties

ofan

inner

product

(bilinearity,symmetry,positive-defi

niten

ess)

all

follow

from

thean

alogou

sproperties

ofthedotproduct.

Inalocalparam

etrizationF

:R

2⊃

V→

F(V

)=

U⊂

R3withF(p

0)=

p,thevectors

v,w

∈TpS=

DpF(R

2)canbewritten

locallyasv F

,wF,where

v=

Dp0F(v

F)

w=

Dp0F(w

F).

Solocallythefundamentalform

is,evaluatingatp0butomittingthatfrom

thenotation,

I F(v,w

)=

DF(v

F)·D

F(w

F)=

vT F(D

FTDF)w

F

1ForF

aparametriza

tion,D

pF

:R2→

TpS

=sp

an(∂

xF,∂

yF)is

surjective.

Soif

Dp0F(v

F)=

v,th

en

c v(t)=

p0+

tvF

∈R2hasc� v

(0)=

vF.Then

γv=

F◦c

vworksbyth

ech

ain

rule:γ� v(0)=

Dp0F(c

� v(0))

=v.

2Recall,in

matrix

notationusingT

fortransp

ose,th

atv·w

=vTw.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

41

Example.For

thestandardbasise 1,e

2of

R2,I(e

1,e

2)=

DF(e

1)·D

F(e

2)=

∂xF·∂

yF.

Locally,

identifyingTp0V

=R

2,theinner

product

becomes,evaluatingat

p0,

R2×R

2→

R,(v,w

)�→

vTAw,whereA

=

�e

ff

g

�=

�∂xF·∂

xF

∂xF·∂

yF

∂yF·∂

xF

∂yF·∂

yF

indeed,thesymmetricmatrixA

=DF

TDF

has

entriesA

ij=

eT iAe j

=I F

(ei,e j),an

dDF(e

i)is

respectively

∂xF

and∂yF

fori=

1,2.

Example.

For

theplaneS

=R

2⊂

R3givenby

Z=

0,andtheparam

etrization

F(r,θ)=

rcosθ

rsinθ

0

weget∂rF

=

cosθ

sinθ

0

and∂θF

=

−rsinθ

rcosθ

0

,thus

A=

�cos2

θ+sin2θ

−rcosθsinθ+rsinθcosθ

−rcosθsinθ+rsinθcosθ

r2sin2θ+r2

cos2

θ

�=

�1

00

r2

soI F

(v,w

)=

v 1w

1+r2v 2w

2(w

ritten

incompon

ents,so

v=

(v1,v

2),etc.)

10.2

Changein

localfirstfundamenta

lform

undercoord

inate

changes

How

does

thelocalexpressionA

ofthefirstfundam

entalform

dep

endon

theob

server?

Con

sider

thepicture

inSection

9.2.

Let

F,� Fbetw

olocalparametrization

s,givingfunda-

mentalform

sA,� A.

Let

τ=

� F−1◦F

:V

→� V

bethetran

sition

map

(defi

ned

ontheoverlapF

−1(� U)⊂

V).

Thelocaltangentspaces

getidentified

via

thelinearisomorphism

Dτ=

D� F−1◦D

F:Tp0V

→T�p 0� V.

Explicitly,

inlocalcoordinates:τ(x,y)=

��x(x,y)

�y(x,y)

�an

dDτ=

�∂x�x

∂y�x

∂x�y

∂y�y

� .

Soweexpectthat

vTAw

=I(v,w

)=

(Dτ(v))

T� A(Dτ(w

))=

vT(D

τ)T

� A(Dτ)w

.Indeed:

A=

DF

TDF

=DF

T(D

� FT)−

1(D

� FTD

� F)D

� F−1DF

=(D

τ)T

� A(D

τ)

Often

,in

practice,

youwillwan

tto

rewrite

this

as:

� A=

(Dτ−1)T

A(D

τ−1).

Example.For

theplaneS

=R

2⊂

R3,wecould

use

theob

viou

sF(x,y)=

(x,y,0)or

polar

coordinates

� F(r,θ)=

(rcosθ,rsinθ,0).Since

τ−1(r,θ)=

(x,y)=

(rcosθ,rsinθ),

(Dτ)−

1=

D(τ

−1)=

� cos

θ−rsinθ

sinθ

rcosθ

� .

Now

I F(v,w

)=

v·w

andA

=I,so

weconfirm

theresultof

thepreviousexam

ple:

� A=

(Dτ−1)T

A(D

τ−1)=

�cosθ

sinθ

−rsinθ

rcosθ

��cosθ

−rsinθ

sinθ

rcosθ

�=

�1

00

r2

�.

Thusthetwolocalexpression

sof

thefirstfundam

entalform

are:

I� F(�v,�w)

=�v 1

�w 1+r2�v 2

�w 2=

v 1w

1+v 2w

2=

I F(v,w

).

42

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

10.3

Application

1:th

elength

ofacurv

eand

isometric

surfaces

Asm

ooth

curv

ein

Sis

asm

oothmap

γ:[a,b]→

S.This

isthesameassayingthat

γ:[a,b]→

R3is

smoothan

dγ(t)∈S

forallt∈[a,b].

Thelength

ofth

ecurv

e(induced

bythenorm

onR

3)is

L(γ)=

�b

a

�γ� (t)�d

t

Lemma10.1.Thelengthofacurveis

indepen

den

tofthechoiceoftime-parametrization.

Proof.

Let

µ(t)=

γ(s(t))

beatime-reparam

etrization

ofγ,so

s:[A

,B]→

[a,b]is

asm

ooth

strictly

increasingfunction.Integratingbysubstitution

(chan

geof

variab

les)

usings�

>0:

L(µ)=

� B A�µ

� (t)�d

t=

� B A�γ

� (s(t))�

s�(t)dt=

� b a�γ

� (s)�d

s=

L(γ).

Tak

inga=

0,wesayacurveγ:[0,b]→

Sispara

metrizedbyarc-length

ifL(γ| [0

,t])=

tforallt.

Bydifferentiatingin

t,this

isequivalentto

hav

ingunit

speed:�γ

� (t)�=

1forallt.

Lemma10.2.Every

smooth

curvewithγ� (t)

�=0foralltcanbe

parametrizedby

arc-len

gth.

Proof.

Let

s(t)

=� t 0

�γ� (t)�d

t.Since

s�>

0,sis

invertible

and(s

−1)�(t)=

1/s�(s

−1(t))

=

1/�γ

� (s−

1(t))�.

Hence

µ(t)=

γ(s

−1(t))

works:

�µ� (t)�=

�γ� (s−

1(t))�(s−

1)�(t)=

1.

Noticethat

foran

ysm

oothcurveγ,thevelocity

vectoralwayslies

inthetangentspace

γ� (t)

∈Tγ(t)S

Indeed:write

γlocallyusingtheparam

etrization

F,sayγloc(t)∈

V⊂

R2,then

γ(t)=

F◦γ

loc(t),

thusγ� (t)

=D

γlo

c(t)F

·γ� loc(t)∈D

γlo

c(t)F(R

2)=

Tγ(t)S.

Theore

m10.3.Len

gthsofcurves

inS

are

determined

bythefirstfundamen

talform

.

Proof.

L(γ)=

� b a�γ

� (t)�d

t=

� b a

�γ� (t)·γ

� (t)dt=

� b a

�I(γ

� (t),γ

� (t))dt.

Assumingthat

γlies

entirely

intheparam

etrization

patch

U(w

ecanad

dlocalcontributions

bycoveringγ[a,b]byseveral

param

etrization

patches),

wecanbeeven

more

explicitusing

thematrixA

=�

ef

fg

�defined

previouslyan

dwritingγloc(t)=

(x(t),y(t))

∈V

⊂R

2:

L(γ)=

� b a

�e� d

x dt

� 2+

2f(dx dt)(

dy

dt)+

g(dy

dt)2

dt

wheree=

e(x(t),y(t)),etc.

dep

endon

thelocalcoordinates,so

dep

endont.

Example.Lettingγloc(t)=

p0+

(t,0)fort∈[0,ε],withε≥

0avariable,by

thefundam

ental

theorem

ofcalculus:

d dεL(γ

loc)=

d dε

�ε

0

�e(p0+

(t,0))dt=

�e(p0+

(ε,0)),

from

whichwerecovere(p0)by

takingε=

0.Soifwekn

owthevalues

L(γ)of

lengthsof

all

curves

inV,werecoverthevalues

ofeon

V.

Exercise.For

anycontinuou

sreal-valued

functionfprove

that

lim

1 ε

� ε 0f(t)dt=

f(0),

as

ε→

0.Deduce

that

lim

1 εL(γ

loc)=

�e(p0)forthecurvein

theexam

ple.

Theore

m10.4.Len

gthsofcurves

determineIlocally.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

43

Proof.

Theexam

ple

aboverecovered

e.Sim

ilarly,werecover

gby

consideringγloc(t)=

p0+

(0,t)an

dwerecover

fbyusingγloc(t)=

p0+

(t,t)(inthelatter

case

d dε

� � ε=0L(γ

loc)=

�e(p0)+

2f(p

0)+

g(p

0),

butweknow

e,gso

werecover

f).

Definition

10.5

(Isometricsurfaces).

TwosurfacesS1,S

2in

R3are

isometric

ifthereis

adiffeomorphism

ϕ:S1→

S2preservinglengthsofcurves:L(ϕ

◦γ)=

L(γ)forγ⊂

S1.

Theore

m10.6.TwosurfacesS1,S

2in

R3are

locallyisometricnearp1,p

2if

andonly

ifthereare

localparametrizationsF1:V

→U1,F2:V

→U2nearp1,p

2yieldingthesamelocal

firstfundamen

talform

I F1=

I F2onV.

Proof.

(⇐):

isim

mediate

from

thelocalexpressionofI,andthelocalcalculation

ofL(γ).

(⇒):

takeF2=

ϕ◦F

1whereϕis

thelocaliso,

andapply

Theorem

10.4.

Example.

Supposewehaveaconemadeou

tof

pap

erandwecutou

tastraightrayfrom

thevertex.

When

weunfold

this

piece

ofpap

erwegetapie-sliced

piece

ofpap

er.

Sothese

twosurfaces

areob

viou

slyisom

etric.

Let’s

proveit.Con

sider

theconeS

={(X,Y

,Z)∈

R3:

X2+

Y2=

a2Z

2,Z

>0}

withangletan−1(a)to

theaxis.Calculate:

F(x,y)=

axcosy

axsiny

x

∂xF

=

acosy

asiny

1

∂yF

=

−axsiny

axcosy

0

A

=

�1+

a2

00

a2x2

Rem

ovetheline(−

aX,0,X

) X∈R

from

S:F

param

etrizesS\(

line)

for(x,y)∈V

=(0,∞

(−π,π

).Weclaim

S\(

line)

isisom

etricto

apie-shap

ein

R2bou

nded

bytworays.Param

etrize

apie-shap

eby

(x,y)�→

(xbcos(cy),xbsin(cy),0)

∈R

3for(x,y)∈

V.TogetthesameA

let

b=

(1+

a2)1

/2,c=

a/b

(note:

c≤

1 2).

10.4

Quadra

ticform

forI,differe

ntials,afast

changeofcoord

inates

Itis

convenientto

abbreviate

thequad

raticform

correspon

dingto

I,written

locally,

by:

I=

edx2+

2fdxdy+

gdy2=

(∂xF·∂

xF)dx2+

2(∂xF·∂

yF)dxdy+

(∂yF·∂

yF)dy2

wheree,f,g

arefunctionsof

thecoordinates

(x,y)∈V

⊂R

2.Thesymbolsdx,dyarecalled

differe

ntials

(ordiffere

ntial1-form

s,or

covecto

rs).

They

areelem

ents

ofthedual

vector

space

(TpS)∗

={linearfunctionsTpS→

R}

called

cota

ngentsp

ace.Locally,

dx,dyarethedual

basisof

thestandardbasis

e 1,e

2∈

R2=

Tp0V.Explicitly,

ifv=

(v1,v

2)∈R

2=

Tp0V:

dx:Tp0V

→R,dx(v)=

v 1,

dy:Tp0V

→R,dy(v)=

v 2.

Example.Asbefore,

writingγloc(t)=

(x(t),y(t)),

dx(γ

� loc(t))

=dx dt,

dy(γ

� loc(t))

=dy dt.

Oneoftendenotes

thestan

dardbasis

e 1,e

2∈R

2=

Tp0V

bythesymbolse 1

=∂ ∂x,e

2=

∂ ∂y,

andso

theconditionof

beingadual

basis

arethemem

orable

look

ingform

ulas

dx(

∂ ∂x)=

1,dx(

∂ ∂y)=

0,dy(

∂ ∂x)=

0,dy(

∂ ∂y)=

1.

Other

form

ulasalso

becom

emoremem

orable:DF(

∂ ∂x)=

∂xF,DF(

∂ ∂y)=

∂yF.

44

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Thechan

geof

Iunder

chan

gesof

coordinates

also

becom

eseasier

inthisnotation.Writing

�γ loc(t)=

(�x(t),�y(t))in

theparam

etrization

� F,bythechainrule

d�x dt=

∂x�xdx dt+

∂y�xdy dt,

d�y dt=

∂x�ydx dt+

∂y�ydy dt

therefore

d�x=

∂x�xdx+

∂y�xdy,

d�y=

∂x�ydx+

∂y�ydy.

Example.For

theplaneS=

R2⊂

R3,F(r,θ)=

(rcosθ,rsinθ,0)

=(x,y,0)=

� F(x,y),

dx=

cosθdr−

rsinθdθ,

dy=

sinθdr+

rcosθdθ.

Recallthat

forτthetran

sition

,Dτ=

� ∂x�x

∂y�x

∂x�y∂y�y

� ,so

�d�x d�y�

=Dτ·�

dx dy

Let’scheckthisisconsistentwithhow

thelocalquad

raticform

Ichan

ges,in

matrix

notation:

(d�xd�y)

��e� f � f�g�

�d�x d�y�

=

�dx dy

� T(D

τ)T

��e� f � f�g�

�dx dy

�=

(dxdy)�ef

fg

��dx dy

�.

Soifyo

uhap

pen

toknow

I=

edx2+

2fdxdy+

gdy2an

dyo

uwan

tto

compute

Iin

the

other

coordinates,I=

�ed�x2

+2� fd

�xd�y+

�gd�y2,then

simply

write

dx=

···,

dy=

···i

nterm

sof

d�x,

d�y,

then

simply

substitute

andform

ally

square/multiply.

Example.For

theplaneS=

R2⊂

R3,F(x,y)=

(x,y,0)and

� F(r,θ)=

(rcosθ,rsinθ,0):

I=

dx2+

dy2=

(cos

θdr−

rsinθdθ)

2+

(sin

θdr+

rcosθdθ)

2=

dr2

+r2

dθ2.

10.5

ExamplesofcalculationsofI

Sphere

ofra

diusa:

I=

a2sin2ydx2+

a2dy2

using:

F(x,y)=

acosxsiny

asinxsiny

acosy

∂xF

=

−asinxsiny

acosxsiny

0

∂yF

=

acosxcosy

asinxcosy

−asiny

Cylinderofra

diusa:

I=

dx2+

a2dy2

using:

F(x,y)=

acosy

asiny

x

∂xF

=

0 0 1

∂yF

=

−asiny

acosy

0

Conewith

angle

tan−1(a)to

theaxis:

I=

(1+

a2)dx2+

a2x2dy2

using:

F(x,y)=

axcosy

axsiny

x

∂xF

=

acosy

asiny

1

∂yF

=

−axsiny

axcosy

0

Surfaceofre

volution,ro

tate

Y=

f(Z

)aboutZ-axis:

I=

(1+

f� (x)2)dx2+

f(x)2dy2

using:

F(x,y)=

f(x)cosy

f(x)siny

x

∂xF

=

f� (x)cosy

f� (x)siny

1

∂yF

=

−f(x)siny

f(x)cosy

0

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

45

10.6

Asu

bstantialexample:ru

led

surfacesin

R3

Rem

ark.Youdonotneedto

mem

orize

forexamstheterm

inology

orform

ulasthatappear

inthis

example,itis

only

supposedto

beaninterestingexample

tosee.

RecallaruledsurfaceS

issw

eptou

tbylines

p(t)+

Rn(t)alon

gacurvet�→

p(t),

where

n(t)is

theunit

directionofthelineat

timet.

Sowecanparametrize

Sby:

F(x,y)=

p(x)+

yn(x)

Aswritten,S

may

self-intersect,

andeven

worse

Smay

notbesm

oothlocally.

Example.Let

Sbethedou

blecone{(X,Y

,Z)∈R

2:X

2+Y

2=

a2Z

2}.

Thisarises

asaruled

surfacep(x)+

yn(x)taking:

p(x)=

acosx

asinx

1

n(x)=

p(x)

�p(x)�

=p(x)

√1+

a2

p(x)+

yn(x)=

(1+

y√1+a2)

acosx

asinx

1

Sdoes

not

self-intersect,butitfails

tobelocally

smoothat

thevertex

(0,0,0).

Ingeneral

Sis

asm

oothsurfacenearF(x,y)⇔

∂xF,∂

yF

are

linearlyindep

endent.

∂xF

=p� (x)+

yn� (x)

and

∂yF

=n(x).

Sosm

oothnessat

F(x,y)is

equivalentto

linearindep

enden

ceof

p� (x)+

yn� (x)an

dn(x),

whichis

equivalentto

thenon

-van

ishingof

thecross-product:1

(p� (x)+

yn� (x))

×n(x)�=

0.

Weoftenuse,withoutmention

ing,

thebasic

trick:

Trick

:If

n(t)∈R

3has

unit

norm

�n(t)�

=1,

then

thevelocity

n� (t)

isperpendicularto

thecurven(t).

Proof:

0=

d dt(1)=

d dt(n(t)·n

(t))

=2n� (t)·n

(t).

�Since

n(x)·n

(x)=

1,wegetn� (x)·n

(x)=

0,thus:

I=

(�p� �

2+

y2�n

� �2+

2yp� ·n� )dx2+

2(p� ·n)dxdy+

dy2

Example.For

thedou

ble

coneS

from

thepreviousexam

ple,

p� (x)=

−asinx

acosx

0

n� (x)=

p� (x)

√1+

a2

so:

�p� �

2=

a2,�n

� �2=

�p� �

2

1+a2

=a2

1+a2,p� ·n�=

�p� �

2

√1+a2

=a2

√1+a2,p� ·n

=p�·p

√1+a2

=0.

Substitutingin

theform

ula

aboveweob

tain:

I=

(a+

ay

√1+a2)2

dx2+

dy2.

Thegeneralform

ula

forIbecom

esI=

(�q��2

+y2�n

� �2)dx2+

2(q

� ·n)dxdy+

dy2if

onereplacesp(x)byaclever

choice

ofcurveq(x)called

lineofstrictionwhichsatisfies

q�(x)·n

� (x)=

0.

1Recallth

ecross-product

inth

estandard

basisi,j,k

ofR3is:

b=

det

a1

b 1i

a2

b 2j

a3

b 3k

which

points

inth

eright-hand

thumb

direction

ifais

theindex

and

bis

themiddle

finger,and

which

has

length

�a×

b�=

�a��

b�|sin

θ|ifθis

theangle

betweena,b.

46

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Let’s

findq(x).

Wewan

tacurve q(

x)=

p(x)+

y(x)n(x)∈S

such

that

q�(x)·n

� (x)=

0.Noticewecanthen

replace

pbyqbecau

sethesurfaceismad

eup

ofthesamestraightlines

p(x)+

Rn(x)=

q(x)+

Rn(x).

Com

pute:

q�·n

�=

(p� +

y� n

+yn� )·n

�=

p� ·n� +

y�n

� �2

sotakingy(x)=

−p� (x)·n

� (x)

�n� (x)�

2works.

Thus:

q(x)=

p(x)−

p� (x)·n

� (x)

�n� (x)�

2n(x)

S=

{q(x)+

yn(x)∈R

3:x,y

∈R}

This

iswell-defined

provided

weassumethat

n� (x)�=

0forallx.Theruledsurfaceis

called

non-cylindricalifitsatisfies

n� (x)�=

0forallx.Onecanbreak

uparuledsurfaceinto

pieces

wherethis

conditionholds,

andthen

separatelystudythecylindricalpieceswheren�=

0for

aninterval

ofvalues

ofx(thesepiecesarevery

simple:n(x)isconstan

tin

x,so

wejust

draw

parallellines

through

thepoints

p(x)in

theconstantdirection

n(x)).

Example.For

thedou

ble

coneS

from

thepreviousexam

ple,

q(x)=

acosx

asinx

1

a2

√1+a2

a2

1+a2

1√1+

a2

acosx

asinx

1

=

0.

Sothelineof

strictionistheconstantcurveat

thevertex

(0,0,0).

Theconditionq�·n

�=

0,mak

esitisalso

easy

tocheckwhereSisalocallysm

oothsurface.

Werequirethenon

-van

ishingof

thecross-product

(q� (x)+

yn� (x))

×n(x)�=

0.Since

n�is

perpendicularto

bothq�,n

,wededuce

1that

q�×

n=

λn�forsomeλ=

λ(x)∈R.Thus:

�(q�+

yn� )×

n�2

=�λ

n� +

yn� ×

n�2

=λ2�n

� �2+

y2�n

� ×n�2

=(λ

2+

y2)�n

� �2,

whereweusedthat

n� ,n� ×

nareperpendicular,

andsomecross-product

tricks.2

Theore

m10.7.Fornon-cylindricalruledsurfacesp(x)+

yn(x)(m

eaningn� (x)�=

0forall

x),

onecanparametrize

Sby

q(x)+yn(x)withq�(x)·n

� (x)=

0,in

whichcase

q(x)is

called

thelineofstriction.MoreoverS

islocallysm

ooth

everyw

hereexceptatthose

points

onthe

lineofstrictionwhereq�,n

becomelinearlydepen

den

t.Thefirstfundamen

talform

is

I=

��q

� �2+

y2�n

� �2

(q� ·n)

(q� ·n)

1

�.

Proof.

Thisfollow

sbytheab

ovecalculation

,since

λ2+y2=

0ifan

don

lyify=

0an

dλ=

0,an

dthelatter

implies

q�×

n=

0,equivalently:q�,n

arelinearlydep

endent.

Example.For

thedou

ble

coneS,thelineof

strictionisq(x)=

(0,0,0),

andq�(x)×

n(x)=

0since

q�=

0.So(0,0,0)istheon

lysingu

larpoint(asexpected).

1This

followsforλ�=

0when

q� ,nare

linea

rlyindep

enden

t,andwhen

they

are

dep

enden

tjust

takeλ=

0.

2Usingth

ecy

clic

symmetry

c·(a×

b)=

det

a1

b 1c 1

a2

b 2c 2

a3

b 3c 3

=

a·(b×

c)

weget

(n� ×

n)·(n� ×

n)=

n� ·(n

×(n

� ×n))

=n� ·n�(thelast

equality

followsbecause

thedirectionsagree,

andth

elength

sagreeusing�a

×b�

=�a

��b�

|sin

θ|,oralternativelyuse:a×

(b×

c)=

(a·c)b

−(a

·b)c).

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

47

10.7

Application

2:th

eangle

betw

een

curv

esin

asu

rface

Thean

gleθbetweentw

ovectors

v,w

∈TS⊂

R3satisfies

v·w

=�v

��w�cosθ,

therefore

cosθ=

v·w

�v��

w�=

I(v,w

)�

I(v,v)�

I(w

,w)

whichon

lydep

endson

I(andthevectorsv,w

).Noticethis

canbeusedto

measure

angles

betweenintersectingcurves:ifγ1,γ

2aresm

oothcurves

inS

whichintersectat

p=

γ1(t)=

γ2(s),then

wecanmeasure

thean

glebetweentheirtangentvectors

v=

γ� 1(t)an

dw

=γ� 2(s).

10.8

Application

3:th

eareaofaregion

inasu

rface

Motivation.Con

sider

theregion

nearpwherethesurfaceSisparametrizedbyF.There-

gion

isap

proxim

ated

byinfinitesim

alparallelog

ramswithedgesthevectors

(∂xF)dx,(∂yF)dy

wherewethinkof

dx,dyas

infinitesim

alincrem

ents

ofx,y.Thearea

ofthisparallelogram

is

(�∂xF�d

x)(�∂yF�d

y)|sin

θ|=

�∂xF

×∂yF�d

xdy,

whereθis

thean

glebetweenthetw

oedges.

Usingthefollow

ingrule

aboutcross-products:1

(a×

b)·(a×

b)=

(a·a)(b·b)−

(a·b)2,

weob

tain,in

term

sof

thefirstfundamentalform

I=

edx2+

2fdxdy+

gdy2=

�e

ff

g

� ,

�∂xF

×∂yF�d

xdy=

�eg

−f2dxdy=

�det(I)dxdy.

Asmathem

aticians,

wejust

turn

this

into

adefinition:

Definition

10.8

(Areaof

aregion

inasurface).TheareaofR

=F(V

)⊂

Sis

defi

ned

as

Area(R)=

� V

�∂xF

×∂yF�d

xdy=

� V

�eg

−f2dxdy=

� V

�det(I)dxdy.

More

generally,

foranyopenregionR

⊂S,wecoverR

by(closuresof)

such

sets

andadd

theareas.

Theore

m10.9.Theareais

well-defi

ned

indepen

den

tlyofchoices

ofparametrization.

Proof.

Suppose(restrictingto

anoverlap)F

:V

→S,� F:V

→S

aretw

oparam

etrization

s,

yieldinglocalfirstfundam

entalform

sA,� A.

Usingthetran

sitionτ,

�det(A

)=

�det((Dτ)T

� A(Dτ))

=|d

etDτ|�

det(� A).

Recall(see

theAnalysishan

dou

t)thechan

geof

variab

lesform

ula

forintegralswillreplace

dxdyby|det

Dτ−1|d

�xd�y,

thus|d

etDτ||det

Dτ−1|=

1disap

pears

when

weintegrate.

Example.For

thesurfaceof

revolution

F(x,y)=

(f(x)cosy,f

(x)siny,x

)(w

heref(x)>

0)wego

tI=

(1+

f� (x)2)dx2+

f(x)2dy2,so

thearea

for(x,y)∈[a,b]×

[0,2π]isthefamiliar

�2π

0

�b

a

f(x)�

1+

f� (x)2

dxdy=

�b

a

2πf(x)�

1+

f� (x)2

dx.

1co

mingfrom

themore

gen

eralru

le:(a

×b)

·(c×

d)=

(a·c)(b·d

)−

(a·d

)(b·c).

48

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

10.9

Riemannian

metric:firstfundamenta

lform

sforabstra

ctsu

rfaces

For

anysurface(orsubmanifold)in

Rnonecandefineafirstfundamentalform

byusing

thedot

product

onR

n.How

ever,foranabstract

smooth

surface

S,thereisnopreferred

way

toem

bed

itin

Rn,so

thereis

nopreferred

inner

product

onTpS.

Definition10.10(R

iemannianmetric).A

Riemannianmetricforasurface

(ormanifold)S

isawell-defi

ned

inner

product

oneach

tangentspace

TpS,whichin

localcoordinatesdepen

ds

smoothly

onp∈S.

Let’s

unpackthis

definition.

Bytangentspace

TpS

wemean

locallyTp0V

=R

2fora

param

etrization

F:V

→S

subject

totherule

thatifwechangeparametrizationto

� Fthen

weidentify

thelocaltangentspacesusingthederivativeofthetransitionτ=

� F−1◦F

:

Tp0V

=R

2Dτ

−→R

2=

T�p 0� V.

TheRieman

nianmetricis

locallygiven

byaninner

product

Tp0V

×Tp0V

=R

2×R

2→

R,(v,w

)�→

vTAw

where

A=

�e

ff

g

�,

andwerequirethatA

=A(p

0)dep

endssm

oothly

onp0,thatis:thefunctionse,f,g

are

smoothfunctionsofthelocalcoordinatesp0=

(x0,y

0).

Inorder

forthis

tobean

inner

product,weneed:

bilinearity

(automatic),symmetry

(automatic),

andpositivedefinitenesswhichbylinearalgebra

isensuredbytheconditions

e>

0and

eg−f2>

0(itfollow

sthatalsog>

0).

Askingthat

theinner

product

iswell-defined

meansthatwewantobserversto

agreeonwhat

inner

product

isbeingusedhavingidentified

theirlocaltangentspacesbyDτasmentioned

above.

SobySection10.2

werequire:

A=

(Dτ)T

� A(D

τ)

Itfollow

sthat,on

ceaRiemannianmetricis

chosenonS(forexample,thefirstfundamental

form

obtained

from

aparticularem

beddingS

→R

n),

wecandefineasbefore:lengthsof

smoothcurves,an

glesbetweenintersectingcurves,andareasofopen

sets

inS.

Examples.

(1)OnthetorusR

2/(Z

ω1+Zω

2)wecanuse

theRiemannianmetric

1

I=

dx2+dy2=

dzdz=

|dz|2

buttherearelots

ofother

choices:

e.g.

rescaletheab

oveby

any(smooth)strictlypositive

dou

bly

periodic

function,meaningf(x

+ω,y

+ω)=

f(x,y)forω∈Zω

1+Zω

2.

(2)Ontheupper

half-planeH

={z

∈C

:Im

(z)>

0},

thehyp

erbolic

metricis

the

Riemannianmetric

I=

dx2+dy2

y2

=dzdz

Im(z)2

=|dz|2

Im(z)2

(3)OntheunitdiscD

={z

∈C:|z|<

1},

thehyp

erbolic

metricistheRiemannianmetric

I=

4(dx2+dy2)

(1−x2−y2)2

=4dzdz

(1−|z|2 )

2=

4|dz|2

(1−|z|2 )

2

1wherez=

x+

iy,dz=

dx+

idy,z=

x−

iy,dz=

dx−

idy.Heredz:TpS

→C

isagain

viewed

asa

linea

rfunctionalonth

etangen

tsp

ace

ofth

egiven

surface

S.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

49

Remark

.(Non

-exam

inab

le)Aconnected1surfacewithaRieman

nianmetricisametricspace,

bydefiningthedista

ncefunction

d:S×

S→

Rbylettingd(p,q)betheinfimum

ofthe

lengthsof

allsm

oothcurves

from

pto

q(you

canalso

allow

piecewisesm

ooth

curves

withou

taff

ectingd).

Inparticular,

theop

enballs

forthis

metricareabasis

forthetopolog

yof

S.

Given

twosm

oothsurfaces

S1,S

2withchoicesof

Rieman

nianmetric,

wesayS1,S

2are

isometric

ifthereis

adiffeomorphism

preservinglengthsofcurves.

Example.Recallweremarkedin

Section

5.3that

thereisabiholom

orphism

2

D→

H,z�→

τ(z)=

iz+

i

−z+

1H

→D,z�→

τ−1(z)=

z−

i

z+

i.

Let’s

checkthisisan

isom

etry

ifweuse

thehyp

erbolic

metrics

from

thepreviousexam

ple.The

abovespecifies

achange

ofcoordinates

z�→

�z=

τ(z).

Rather

than

switchingto

real

coordinates,

recallthat

differentialschange

bymultiplicationby

Dτ,andsince

wemay

identify

Dτwithτ� (z)

when

identifyingR

2≡

C,wededuce:3

d�z=

τ� (z)dz.

Now

calculate:

τ� (z)=

i(−z+1)

−(iz+

i)(−

1)

(−z+

1)2

=2i

(z−

1)2

Im(�z)=

Im(τ(z))

=Im

(iz+

i)(−

z+

1)

|z−

1|2

=Im

i(−|z|2+

2iIm

z+

1)

|z−

1|2

=1−

|z|2

|z−

1|2

I H=

|d�z|

2

Im(�z)2

=4

|z−

1|4

|z−

1|4

(1−

|z|2 )

2|dz|2

=4|dz|2

(1−

|z|2 )

2=

I D.

11.

Surfa

cesin

R3:these

cond

fundamentalform

11.1

Abasictoymodel:

whatis

thecurv

atu

reofacurv

e?

Let

γbeasm

oothcurvein

R3,γ:[0,b]→

R3param

etrizedbyarc-length(i.e.unitspeed:

�γ� (t)�=

1,seeSec.10.3).

Then

theunit

tangentvecto

ris

thevelocity

γ� (t),an

dthe

curv

atu

reis

thenorm

oftheacceleration� (t):

κ(t)=

�� (t)�

Rem

ark.As�γ

� �=

1,thevelocity

γ�is

perpen

dicularto

theacceleration�by

theTrick:

Trick

.�γ

� �=

1⇒

γ� ·

γ�=

1⇒

∂ ∂t(γ

� ·γ� )=

0⇒

2γ��·γ

�=

0⇒

γ��⊥

γ� .

Example.Con

sider

thecircle

X2+

(Y−

r)2=

r2insidetheplaneZ

=0of

R3,withcentre

(0,r,0)andradiusr.

Near0∈R

3,itequalsthecurve

γr(t)=

(rsin

t r,r

−rcos

t r,0)

1Forsu

rfaces(and

manifolds)

connectednessim

plies

path

-connectedness,

soth

ereis

alw

aysa

continu-

ouspath

[0,1

]→

Sbetween

twogiven

points

p,q.

Usingloca

lparametrisationsoneca

napproxim

ate

any

continuouspath

byapiecewisesm

ooth

path

,andth

enoneca

nroundoffco

rnersto

get

asm

ooth

path

.2Recallth

atto

findth

einverse

of

z−i

z+iwejust

compute

theinverse

ofth

ematrix

� 1−i

1i

�whichis

1 2i

�i

i−1

1

andwereadoffth

ematrix

entries:

iz+i

−z+1.

3I’m

sayingweca

niden

tify

�dx

dy

�≡

dzand

hen

ced�z=

�d�x

d�y�

=Dτ·�

dx

dy

�≡

τ� (z)dz.

Although

we

won’t

needit,youmaybecu

rioushow

dzch

anges:th

eru

leis

asex

pected:d�z=

τ� (z)dz.Indeedasalinea

r

function,dz=

dx−

idymaps∂x�→

1,∂

y�→

−i,

wherea

sdz=

dx+

idymaps∂x�→

1,∂

y�→

i.Sodzis

the

conjugate

ofth

elinea

rfunctiondz.Sod�z=

d�z=

τ� (z)dz=

τ� (z)dz

50

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

fortcloseto

0.Weuse

t rinsteadof

tso

that

γrhas

unitspeed:

�γ� r(t)�

=�(cos

t r,sin

t r,0)�

=1.

As� r(t)=

(−1 rsin

t r,1 rcos

t r,0),

thecurvature

is

κr(t)=

�1 r2

=1 r.

Ingeneral,arotation

andtran

slationin

R3willnot

chan

gethecurvature

ofacurveγnor

theproperty

that

γisparam

etrizedbyarc-length(rotationsan

dtran

slationspreservelengths).

Byrotatingan

dtran

slating,

wemay

assumeγ(0)=

0=

γr(0),

γ� (0)

=(1,0,0)=

γ� r(0),

and

� (0)

=(0,κ

(0),0).Pickr=

1/κ(0).

Then

also

� r(0)=

� (0),so

theTaylorseries

forγ,γr

agreeupto

thesecondorder.Sothat

circle

ofradiusr=

1/κ(0)is

thebestquad

raticcurve

whichap

proxim

ates

γat

0(w

hen

κ(0)=

0,wecanthinkof

thecircle

γ∞

ofinfiniteradiusas

thestraightlineequal

tothex-axis,indeedγis

“flat”upto

second-order).

Thecircle

γris

called

theosculatingcircle

forγat

γ(0)=

0,an

dris

called

thera

diusofcurv

atu

re.

Amoregeom

etricway

ofinterpretingthecurvature

isas

follow

s.

Recall�is

orthog

onal

toγ� .

For

this

reason

,

n=

�

�� �

iscalled

thenormal

vectorto

γ(equivalently:γ�� (t)

=κ(t)n(t)).Wenow

ask:byhow

much

does

γ(t)sw

erveaw

ayfrom

thelineγ(0)+

Rγ� (0)

tangentto

γat

t=

0?Thedistance

ofγ(t)from

thestraightlineγ(0)+

Rγ� (t)

is,upto

order

t3errors,1

n(0)·(γ(t)−

γ(0))

=�(0

)�γ

��(0

)�·(γ� (0)t+

1 2� (0)t2

+···)

=1 2�γ

�� (0)�t

2+

···

=1 2κ(0)t

2+

···,

whereweusedthat

γ�� ,γ� a

reorthog

onal.Sothecurvature

κ(0)measureshow

much

thecurve

γ(t)deviatesfrom

thetangentline.

11.2

Thelocalsecond

fundamenta

lform

Let

S⊂

R3beasm

oothsurfacewithaGau

ssmap

n:S

→R

3,so

n(p)·T

pS

=0an

dn(p)·n

(p)=

1.How

much

does

Ssw

erveaw

ayfrom

theplanep+

TpS

tangentto

Sat

p?

1Here

n(t)·(

γ(t)−

γ(0))

would

be

the

correct

distance

ifth

ecu

rve

lies

inside

the

plane

γ(0)+

span(γ

� (0),�(0)).If

wetrunca

teth

eTaylorseries

ofγafter

t3term

s,th

enth

ecu

rvedoes

liein

this

plane.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

51

Let

Fbeaparam

etrization

nearp.Bytran

slatingV

⊂R

2,wecanassumeforsimplicity

that

F(0,0)=

p.

Abbreviate

0=

(0,0).

RecalltheTaylorseries

ofasm

ooth

function

G(x,y)∈R

intw

ovariablesis

G(x,y)

=G(0)

+D

0G

·(x y)

+1 2(x y)T

Hess 0

G(x y)

+···

=G(0)

+x∂xG

+y∂yG

+1 2(x

2∂xxG

+2x

y∂xyG

+y2∂yyG)

+···

wherethepartial

derivatives

ofG

areallevaluated

at(0,0),theHessianHess 0

Gisjust

the

matrixof

second-order

partial

derivatives,andweab

breviate

∂xyG

=∂x(∂

yG),

etc.

Wecan

compute

thisTay

lorseries

takingG

=on

eof

thethreecomponents

ofF

=(F

1,F

2,F

3)∈R

3.

Example.Con

sider

F(x,y)=

(x,y,ax2+

by2)near(x,y)=

0.Theab

ovebecom

es:

F(x,y)=

� 0 0 0

�+

x� 1 0 0

�+

y� 0 1 0

�+

1 2

� x2�

0 0 2a

�+

2xy� 0 0 0

�+

y2�

0 0 2b

��

Therefore,

usingthat

∂xF,∂

yF

∈TpSareorthogo

nal

ton(p),thesign

eddistance

ofanearby

pointF(x,y)from

theplanep+

TpS

is(aga

inevaluatingpartialderivatives

atp):

n(p)·(F(x,y)−

F(0,0))

=1 2(x

2n·∂

xxF+

2xyn·∂

xyF+

y2n·∂

yyF)+

···

Thus,

thean

alog

ueforsurfaces

ofthecurvature

ofacurveis

theform

IIF:R

R2→

R,(v,w

)�→

vT

�n·∂

xxF

n·∂

xyF

n·∂

yxF

n·∂

yyF

�w

whichis

clearlybilinearan

dsymmetric(∂

xyF

=∂yxF

bysm

oothnessof

F).

Example.II

Fneednot

bepositivedefinite,

indeeditcanvanish:fortheplaneR

2⊂

R3taking

F(x,y)=

(x,y,0),

thesecondpartial

derivatives

ofF

arezero.

Example.Con

tinuingwiththeexam

ple

F(x,y)=

(x,y,ax2+

by2),

wepick:

n(0)=

∂xF

×∂yF

�∂xF

×∂yF�� � � � (x

,y)=

(0,0)

=� 1 0 0

�� 0 1 0

� /norm

=� 0 0 1

So2n

(0)·(F(x,y)−

F(0,0))

=2a

x2+

2by2,thereforeat

p=

F(0,0)=

(0,0,0):

IIF(v,w

)=

vT

�2a

00

2b

�w

=2a

v 1w

1+

2bv 2w

2.

Example.

Let

Sbethesphereof

radiusr.

Thenormal

isn(p)=

p.

AttheNorth

pole

p=

(0,0,r),

Sis

locally

thegraphF(X

,Y)=

(X,Y

,√r2

−X

2−

Y2),

andn(p)=

(0,0,1).

Dottingwithn(p)meanswetake

thethirdentry,so

weneedtheHessian

ofh=

√r2

−X

2−

Y2

52

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

at(X

,Y)=

(0,0).

Com

pute:∂Xh=

−X

√r2−X

2−Y

2,so

∂X

Xh| (0

,0)=

−1 rand∂YXh| (0

,0)=

0,

andby

symmetry

also

∂YYh| (0

,0)=

−1 r.Thus

IIF=

vT� −

1 r0

0−

1 r

� w=

−1 rv 1w

1−

1 rv 2w

2.

Wegetthissameform

ateach

pointof

thesphereby

rotation

alsymmetry.1

Toavoidconfusion

later,

wemak

eacleardistinctionbetweenn:S→

R3an

dtheGau

ssmap

inlocalcoordinates

(nF):R

2⊃

V→

R3:

nF(x,y)=

n(F

(x,y)).

Lemma11.1.Locallythematrix

forthesecondfundamen

talform

is,evaluatingatp,

B=

�L

MM

N

�=

�(nF)·∂

xxF

(nF)·∂

xyF

(nF)·∂

yxF

(nF)·∂

yyF

�=

−�

∂x(nF)·∂

xF

∂x(nF)·∂

yF

∂y(nF)·∂

xF

∂y(nF)·∂

yF

Note:∂x(nF)·∂

yF

=∂y(nF)·∂

xF

astheright-handsideis

symmetricusing∂xyF

=∂yxF.

Proof.

Since

nis

orthog

onal

toTS=

span

(∂xF,∂

yF),

(nF)·∂

xF

=0

(nF)·∂

yF

=0.

Differentiatingin

xor

inygives

theequalitybetweenthematricesin

theclaim.

Usingmatrixnotation

,withcolumnsthefirstpartial

derivatives:

DF

=�∂xF

∂yF

�an

dD(nF)=

�∂x(nF)

∂y(nF)� ,

then

bytheLem

ma,

IIF(v,w

)=

vTBw

where:

B=

−D(nF)T

DF

11.3

Thelocalsecond

fundamenta

lform

underach

angeofcoord

inates

Supposewechan

gecoordinates

usingatran

sition

τ=

� F−1◦F

.Since

F(x,y)=

� F(τ(x,y)),

wehavenF(x,y)=

n� F(τ(x,y)).Differentiatingusingthechainrule:

DF

=D

� FDτ

and

D(nF)=

D(n

� F)Dτ.

Thus,

thechan

gein

thesecondfundam

entalform

is:

B=

−D(nF)T

DF

=−(D

(n� F)Dτ)T

(D� FDτ)=

−DτTD(n

� F)T

D� FDτ=

DτT

� BDτ,

thusas

expected:

B=

DτT

� BDτ

11.4

Thesecond

fundamenta

lform

Since

thelocalform

schan

gecorrectlyunder

thetran

sition

(Dτ:TV

→T� V

isthecorrect

identification

betweenthelocaltangentspaces,so

B=

DτT

� BDτis

thecorrectchan

geof

coordinates

forbilinearform

s),thelocalform

s

IIF:R

R2=

TpV

×TpV

→R

determineawell-defined

glob

albilinearform

II:TpS×

TpS→

R

1Apply

arotationR

aboutth

eorigin.Then

forth

eparametriza

tionR◦F

nea

rR(p),

usingn(R

(p))

=R(p),

weget

thesameII

F:n(R

(p))

T∂ij(R

◦F)=

pTR

TR∂ijF

=pT∂ijF

=n(p)T

∂ijF,usingth

atrotationsare

orthogonalmaps(R

TR

=id).

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

53

indep

endentof

theob

server.Moreexplicitly,

ifv=

DF(v

F),w

=DF(w

F)aretheactual

vectorsin

TS⊂

R3,rather

than

thelocalversionsv F

,wF∈TV

=R

2,then

bythechainrule:

IIF(v

F,w

F)=

−vT FD(nF)T

DFw

F=

−(D

nDFv F

)·(DFw

F)=

−Dn(v)·w

=II(v,w

)

isindep

endentof

thechoice

ofF.Therefore:

Theore

m11.2.Thelocalform

sII

Fdetermineawell-defi

ned

symmetricbilinearform

,called

theseco

nd

fundamentalform

,

II:TpS×

TpS→

R,II(v,w

)=

−(D

pn)(v)·w

Weshou

ldnow

clarifybetterwhat

exactly

−Dnmeans.

Wewantto

avoid

usingextension

s1

ofnas

indefinition2.6,

whichismessy

andrequires

usto

mak

echoices(althou

ghthechoice

ofextension

does

not

matterin

theend).

Thederivativemap

Dn:TpS→

Tn(p

)R

3=

R3

ofn:S→

R3isawell-defined

map

indep

endentof

param

etrization

s:in

Section

9.2thismap

was

defined

interm

sofcurves

inS

withou

tusingparam

etrization

s.A

vector

v∈

TS

isan

equivalence

class

ofcurves

[cv]in

S,satisfyingp=

c v(0),

v=

c� v(0),

and

Dn[c

v]=

[n◦c

v].

Noticethat

this

mak

essense

bythechainrule:weidentify

[cv]≡

∂t| t=

0c v

=c� v(0)=

v,so

[n◦c

v]≡

∂t| t=

0(n

◦cv)=

(Dcv(0

)n)(c

� v(0))

=D

pn(v).

Noticetheab

ovedescription

gives

avery

usefulform

ula,whereγ� (t)

=(X

� (t),Y

� (t),Z

� (t))

isageneral

vector

inTS

interm

sofacurveγ(t)=

(X(t),Y(t),Z(t))

∈S:

Dn(γ

� (t))=

Dn

�X

� (t)

Y� (t)

Z� (t)

�=

∂ ∂t� � � � t=

0

n(γ(t))

Example.For

theplaneS=

R2⊂

R3,n=

(0,0,1)isconstant,so

Dn=

0,so

II=

0.

Example.Thecylinder

Sof

radiusr,

X2+

Y2=

r2,has

Gauss

map

n(X

,Y,Z

)=

(X,Y

,0)/r

(thisistheou

twardnormalto

thecylinder).

For

γ(t)=

(X(t),Y(t),Z(t))

inS,wehaven(γ(t))

=(X

(t),Y(t),0)/r

so

Dn

�X

� (t)

Y� (t)

Z� (t)

�=

∂t| t=

0n(γ(t))

=

�X

� (t)/r

Y� (t)/r

0

�.

AbasisforTpS

consistsof

avector

E1tangentto

theequatorialcircle

ofthecylinder

(sotaking

Z(t)=

constant),andavector

E2parallelto

theaxis

ofthecylinder

(sotakingX(t),Y(t)

constant).It

followsby

theab

ovecalculation

that

Dn(E

1)=

E1/r,Dn(E

2)=

0.ThusDn:

TS×

TS→

TS,Dn=

� 1/r0

00

�in

thebasisE

1,E

2.Therefore

II(v,w

)=

−vT

�1 r

00

0

�w

=−1 rv 1w

1.

For

exam

ple,an

orthon

ormal

choice

ofE

1,E

2at

p=

(rcosθ,rsinθ,Z)is

E1=

(−sinθ,cosθ,0)

and

E2=

(0,0,1).

Rem

ark.Eveniftwosurfacesare

isometric,

thesecondfundamen

talform

canbe

substantially

differen

t.E.g.thecylinder

islocallyisometricto

theflatplane,

butit

hasanon-trivialII.

1n

isamapS

→R3,so

firstch

oose

anex

tensionofn

(atleast

loca

lly)to

aneighbourh

oodofS,th

enn

becomes

amapn:R3→

R3defi

ned

nea

rS,hen

ceweknow

whatDn

mea

ns(m

atrix

ofpartialderivatives).

54

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Sothesecondfundamen

talform

depen

dsonextrinsicinform

ation:thechoiceofem

bedding

into

R3.Whereaswethinkofthefirstfundamen

talform

asintrinsic(dueto

Theorem

10.6).

11.5

Summary

offirstand

second

fundamenta

lform

s

Fundam

entalform

TpS×

TpS→

R

v,w

∈TpS⊂

R3

Localfundam

entalform

TpV

×TpV

→R

v F,w

F∈TpV

=R

2

(SoDF(v

F)=

v,etc.)

Localmatrix

Coordinatechan

ge

τ=

� F−1◦F

I(v,w

)=

v·w

I F(v

F,w

F)=

vT FAw

FA

=DF

TDF

A=

DτT

� ADτ

II(v,w

)=

−Dn(v)·w

IIF(v

F,w

F)=

vT FBw

FB

=−D(nF)T

DF

B=

DτT

� BDτ

11.6

Thesecond

fundamenta

lform

isth

evariation

ofth

efirst

Forthepurposesofthis

course,

thefollowingis

notacentralresult,itis

just

acuriosity.

Theore

m11.3.Thesecondfundamen

talform

isthevariationofthefirstfundamen

talform

,

∂ ∂t

� � t=0I t

=−2II,

when

wedeform

thesurface

inthenorm

aldirection,meaningwevary

thelocalparametrization

by(x,y)�→

F(x,y)+

tn(F

(x,y))

interm

softimet∈[0,small].

Proof.

WritingFt=

F+

tnF,weget∂xFt=

∂xF

+t∂

x(nF),

∂yFt=

∂yF

+t∂

y(nF),

so

I t=

DF

T tDFt=

I 0+t

�2∂

x(nF)·∂

xF

∂x(nF)·∂

yF

+∂y(nF)·∂

xF

∂x(nF)·∂

yF

+∂y(nF)·∂

xF

2∂y(nF)·∂

yF

� +order

t2.

Thustheclaim

follow

sbyusingtherelation

sfrom

theproof

ofLem

ma11

.1(inparticu

larthe

symmetry

∂x(nF)·∂

yF

=∂y(nF)·∂

xF).

12.

Curvature

12.1

Thesh

apeopera

torS=

−Dn:TpS→

TpS

Lemma12.1.TpS=

Tn(p

)S2⊂

R3whereS2is

theunitspherein

R3.

Proof.

TpS

isthevector

subspaceof

R3orthog

onal

tothenormal

n(p)at

p∈

S.Sim

ilarly,

Tn(p

)S2is

thevector

subspaceof

R3orthog

onal

tothenormal

tothesphereS2at

n(p).

But

recallthat

thenormal

toS2at

n(p)is

just

n(p).

Sothosevector

subspaces

equal.

Coro

llary

12.2.Dn

:TpS

→Tn(p

)S2can

beview

edasalinearen

domorphism

ofthe2-

dim

ensionalvectorspace

TpS,so

Dn:TpS→

TpS.

Proof.

Weneedto

show

Dpnlandsin

Tn(p

)S2.Let

γ⊂

Sbean

ycurvethrough

γ(0)=

p.

Then

Dn(γ

� )=

∂t(n

◦γ)is

thevelocity

ofacurven◦γ

⊂S2so

itis

tangentto

S2(at

n(γ(0))

=n(p)).Another

proof:differentiaten(γ)·n

(γ)=

1(since

nis

unitlength)in

time:

2n(γ)·D

n(γ

� )=

0so

Dn(γ

� )is

perpendicularto

n,so

itlies

inTn(p

)S2.

Definition

12.3

(Shap

eop

erator).

S=

−Dn:TpS→

TpS

isthesh

apeopera

tor.

Theore

m12.4.Thefundamen

talform

sare

relatedby:

II(v,w

)=

S(v)·w

=I(S(v),w)

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

55

Proof.

II(v,w

)=

−Dn(v)·w

and

Dn(v)∈

Tn(p

)S2=

TpS,so

thatdot

product

can

be

computedusingI:TpS×

TpS→

Ras

both

−Dn(v),w

liein

TpS.

�Coro

llary

12.5.Sis

self-adjoint1

withrespectto

theinner

product

I

I(Sv,w

)=

I(v,Sw).

Proof.

Ian

dII

aresymmetric,

soI(Sv,w

)=

II(v,w

)=

II(w

,v)=

I(Sw,v)=

I(v,S

w).

�Lemma12.6.Let

nF

=n◦F

bethelocalexpressionfornin

theparametrizationF,then

∂x(nF),∂y(nF)∈TS.

(12.1)

Let

[Sij]be

thematrix

2forSin

thebasis∂xF,∂yF

ofTS,then

−∂x(nF)=

S 11∂xF

+S 2

1∂yF

and

−∂y(nF)=

S 12∂xF

+S 2

2∂yF

−D(nF)=

−�∂x(nF)

∂y(nF)� =

�∂xF

∂yF

��S 1

1S 1

2

S 21

S 22

�=

DF

S.(12.2)

Proof.

Bydifferentiatingn·n

=1wededuce

that

∂x(nF)·n

F=

0,∂y(nF)·n

F=

0.Hence

(12.1)

follow

s,since

TS

istheplaneorthog

onal

ton.

Bythechainrule,−Dn(∂

xF)=

−∂x(nF)and−Dn(∂

yF)=

−∂y(nF).

Thus,

abbreviatingx,y

byindices

1,2,

usingthebasis

Xi=

∂iF

wehave:

3−Dn(X

i)=

�2 j=1S j

iXj.SothematrixS i

jrepresents

−Dnin

thebasisX

i=

∂iF

.�

Usingthenotationof

thepreviousproof,Theorem

12.4

canberewritten

locallyas

II(X

i,X

k)=

−Dn(X

i)·X

k=

2 � j=1

S jiX

j·X

k=

2 � j=1

S jiI(X

j,X

k).

Noticethis

implies

B=

STA

(wecomebackto

this

proof

inTheorem

12.8).

IntuitivelySis

“just

thesameas”II:weturned

the2-form

IIinto

alinearmap

usingtheinner

product

I.4

12.2

Principalcurv

atu

res,

mean

curv

atu

re,Gaussian

curv

atu

re

Since

theshap

eop

erator

Sis

self-adjoint,

itcanbediago

nalizedusingan

orthon

ormal

5

basis

E1,E

2of

eigenvectors

forTS

(whereorthonormalityis

computedusingI).

S=

�κ1

00

κ2

1Self-adjointn

essmea

nsth

atth

ematrix

forS

ina

basiswhich

isorthonorm

alwith

resp

ectto

Iwill

be

symmetric.

Howev

er,th

ematrix

S ij

discu

ssed

below

isnotsymmetric

ingen

eral.

The

symmetry

∂iX

k=

∂i∂kF

=∂k∂iF

=∂kX

iim

plies

symmetry

ini,kin

−Dn(X

i)·X

k=

−∂i(n

F)·X

k=

(nF)·∂

iX

k(d

if-

ferentiatingth

eorthogonality

relation(n

F)·X

k=

0).

This

howev

erdoes

notim

ply

thesymmetry

S ij=

S ji,

itonly

implies

thatII(X

i,X

k)=

−Dn(X

i)·X

kis

symmetricin

i,k.

2Soth

eS i

jare

smooth

functionsin

theloca

lvariablesx,y

.3Thei-th

columnofth

ematrix

Sis

theim

ageofth

ei-th

basisvectorX

i,written

inth

ebasisX

j.

4Compare

theraising/loweringofindices

inSect.13.3.W

eare

doingII

ik=

Sj igjk,wheregjk=

I(X

j,X

k).

5W

arnin

g.If

youco

mpute

loca

lly,

youmust

ensu

reorthonorm

ality

I(E

i,E

j)=

δ ij.Forex

ample,ifyou

use

F(λ

x,λ

y)instea

dofF(x

,y),

then

theloca

lmatrix

Bwould

becomeλ2B,so

curvatu

resco

mputed

for

thesematrices

would

loca

llych

angebyλ2orλ4:butwewantth

emto

beindep

enden

tofparametriza

tions!

More

gen

erally,

ifwepickaclev

ernon-orthogonallinea

rtransitionτ=

S,so

Dτ=

S,th

enweca

ntu

rnB

into

amatrix

� B=

STBS

ofth

eform

� 10

01

� ,� 1

00

−1

� ,� ±

10

00

� ,or� 0

00

0

�(a

bilinea

rform

isdetermined

upto

congru

ency

byitssignatu

re).

Soloca

lly,

ifyouignore

orthonorm

ality,youco

uld

arb

itrarily

rescale

bypositive

numbersth

eeigen

values

ofth

eloca

lmatrix

Busingch

anges

ofco

ord

inates.

Soonly

thefollowingsignsare

preserved

:signK

=signdet

B(signsofκ1,κ

2)=

(signsofeigen

values

ofB).

Forex

ample

det(� B)=

det(D

τTBDτ)=

det(D

τ)2

det(B

).Forth

esamereasons,

youhavenoco

ntrolover

the

signofth

emea

ncu

rvatu

re.

56

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Principaldirections

E1,E

2o.n.eigenvectorsof

S

Principalcurv

atu

res

κ1,κ

2eigenvalues

ofS

Mean

curv

atu

reH

=1 2(κ

1+

κ2)

H=

1 2trace(S)

Gaussian

curv

atu

reK

=κ1κ2

K=

det(S)

Wewillseelater,

that

theprincipal

direction

sarethedirection

sthat

acurvein

thesurface

must

travelalon

gto

havemax

imum

andminim

um

curvature

(=thetw

oprincipal

curvatures).

Rem

ark.Explicitly,

weare

diagonalizingthesymmetricbilinearform

II:TS×

TS→

R,so

II(E

i,E

j)=

κiδ

ijand

I(E

i,E

j)=

δ ij

whereδ i

j=

1ifi=

jandδ i

j=

0fori�=

j.If

youthinkofE

i,E

jasvectors

inR

3,then

of

courseorthonorm

ality

just

meansE

i·E

j=

δ ijfortheusualdotproduct.

Example.For

thecylinder

X2+Y

2=

r2at

theendof

Sec.11.4,

atp=

(rcosθ,rsinθ,Z)the

eigenvectorsE

1=

(−sinθ,cosθ,0),E

2=

(0,0,1)areorthon

ormal

w.r.t.dot

product

inR

3,and

wefoundII

=−

1 rdx2,so

κ1=

−1 randκ2=

0(asexpectedsince

intheE

1direction

thesurface

lookslikeacircle

ofradiusr,

andin

theE

2direction

thesurfacelookslikeastraightline).

Example.In

theexam

ple

F(x,y)=

(x,y,ax2+

by2),

wecomputedat

(x,y)=

0:

IIF=

(2a

00

2b)

inthebasisv F

=(1,0),

wF=

(0,1).

Thefirstfundam

entalform

is

I F=

(10

01),

sothebasisv F

,wFisorthon

ormal.Soκ1=

2a,κ2=

2b.

Lemma12.7.If

∂xF,∂yF

∈R

3are

orthonorm

alatp,then

κ1,κ

2=

(theeigenvalues

ofthelocalmatrixB

forII

F)

K=

det

B.

Proof.

Thechan

geof

basis

from

∂xF,∂

yF

toE

1,E

2willbean

orthog

onal

matrixQ,an

dso

IIF=

QT� κ

10

0κ2

� Q=

Q−1� κ

10

0κ2

� Q,usingorthog

onality:Q

T=

Q−1.Butconjuga

tion

does

not

chan

gethecharacteristicpolynom

ial,so

theeigenvalues

arestillκ1,κ

2.

Theore

m12.8.WritingSforthematrix

oftheshape

operatorin

thebasis∂xF,∂

yF

yields

arelationbetweenthelocalmatrices

A,B

forI F

,II F

:1

Weingarten

equations

S=

A−1B

Gaussian

curv

atu

reK

=det

S=

detB

detA

Thesecanbe

written

outexplicitlyin

term

softhecoeffi

cien

tfunctionsofI F

,II F

:

II=

S=

I−1

FII

F=

�e

ff

g

� −1�

LM

MN

�=

1

eg−

f2

�g

−f

−f

e

��

LM

MN

K=

det(S)=

det(II F

)

det(I

F)=

LN

−M

2

eg−

f2

.

1It

isea

syto

checkth

atK

does

notch

angeif

wech

angeloca

lparametriza

tion.More

strikingly,Gauss’

Theo

rem

Egregium

(Section

13.2)saysth

atK

does

notch

angeunder

isometries,so

itis

intrin

sicto

the

surface

withitsRiemannianmetric,

itis

notex

trinsic(=

dep

enden

tonth

ech

oiceofem

bed

ding).

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

57

Proof.

Itis

enou

ghto

show

that

STA

=B,since

then

S=

(BA

−1)T

=(A

T)−

1B

T=

A−1B,

usingthat

A,B

aresymmetric.

Below

aretw

owaysto

checkthat

STA

=B.

Inlocalcoordinates,forv,w

∈R

2,wehaveI(v,w

)=

vTAw,II(v,w

)=

vTBw.

So

I(Sv,w

)=

II(v,w

)becomes

vTST

Aw

=vTBw.Asthisholdsforallv,w

,wededuce

STA

=B.

Alternatively,

combinethematrixequation

inLem

ma12

.6withSection

11.5,

B=

−D(nF)T

DF

=(D

FS)

TDF

=ST

DF

TDF

=ST

A.

12.3

Norm

alcurv

atu

re

Thenorm

alcurv

atu

reof

acurveγ

inS

through

p=

γ(0),

with

γparam

etrized

by

arc-lengthso

speed�γ

� �=

1,isthecompon

entof

theacceleration

�in

thenormal

direction

1

� (0)

·n(p)=

−Dn(γ

� (0))·γ

� (0)

=II(γ

� (0),γ

� (0))

Intuitively,

ifyo

uareracingwithacaron

asurface,

thenormalcurvature

tellsyouhow

much

pullaw

ayfrom

thesurfaceyo

ufeel

when

you

acceleratethecar.

Since

�γ� (0)�=

1,thenormal

curvaturesare

measuredbythequad

raticform

II(v,v)on

unit

vectors

v=

cosθE

1+

sinθE

2∈TpS.Explicitly,

wegettheEulerform

ula:

II(v,v)=

κ1cos2

θ+

κ2sin2θ

soκ1,κ

2aretheextrem

evalues

(min

andmax

)2of

thepossible

normal

curvaturesat

p.

Coro

llary

12.9.Theprincipalcurvaturesκ1,κ

2are

themin

andmaxoftheratioofthetwo

quadraticform

s:

Lx2+

2Mxy+

Ny2

ex2+

2fxy+

gy2

Proof.

Wecanrescale(x,y)withou

taff

ectingtheab

overatioso

that

ex2+2fxy+gy2=

1.But

thisisprecisely

theconditionthat

I F(v

F,v

F)=

1forthelocalvector

v F=

(x,y)∈TV

=R

2.

Then

IIF(v

F,v

F)is

equal

tothenumerator.Weproved

abovethat

forunit

vectors

v∈

R3

themin

andmax

ofthequadraticform

II(v,v)are

theprincipal

curvaturesκ1,κ

2.

12.4

Qualita

tiveinterp

reta

tion

ofth

ecurv

atu

res

Looselyspeaking:

(1)Theprincipal

directionsE

1,E

2tellyou

thedirection

sofsteepestincrease/d

ecrease,

(2)If

κ1>

0,acurveis

SwithtangentE

1is

acceleratingin

thenormal

direction

,since

γ��·n

=II(γ

� ,γ� )

=II(E

1,E

1)=

κ1>

0(usingSec.12.3).Sonearp,thecurvelies

onthesamesideof

TS

asn(p)does.

(3)If

κ1<

0,acurvein

direction

E1acceleratesaw

ayfrom

n(p)so

itlies

ontheother

sideof

TS.

(4)K

>0meansκ1,κ

2havethesamesign

,so

either

allcurves

accelerate

towardsn(p)

oraw

ayfrom

n(p),so

Sislocallyallon

thesamesideof

thetangentplanean

dScan

belocallyap

proxim

ated

byan

ellipsoid.

Wecallapointp∈S

ellipticifκ1,κ

2havethesamesign(⇔

K>

0)(5)Wecallapointp∈

Shyperb

olicif

κ1,κ

2haveop

positesign

(⇔K

<0).

This

meansS

locallylook

slikeasaddle

andS

lies

onbothsides

ofthetangentplane.

1whereweused

theusu

altricks:

ν(t)=

n(γ

(t))

isorthogonalto

TS,so

ν(t)·γ

� (t)

=0,differen

tiating:

ν� ·γ� +

ν·γ

��=

0,andfinallyν� (0)=

∂t| t=

0n(γ

(t))

=Dn(γ

� (0)).Theform

ula

then

follows.

2Proof:

ifκ1≤

κ2,th

enκ1=

κ1(cos2

θ+

sin2θ)≤

κ1co

s2θ+

κ2sin2θ≤

κ2(cos2

θ+

sin2θ)=

κ2.

58

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Example.In

theab

oveexam

ple

F(x,y)=

(x,y,ax2+

by2)near(x,y)=

0:For

a,b

>0,

theorigin

isan

ellipticpoint:

thesurfaceisZ

=aX

2+bY

2,itliesallab

ovethe

tangentplaneZ

=0,andwegetan

ellipse

when

wesliceS

withaplaneZ

=(positiveconstant)

whichisparallelto

thetangentplane:

For

a<

0,b>

0,thesurfacelieson

bothsides

ofthetangentplaneZ

=0,

andwegeta

hyp

erbolawhen

wesliceS:

12.5

Curv

atu

resin

term

softh

eHessian

Theore

m12.10.ParametrizingS⊂

R3nearpasthegraphofafunctionh:TpS→

Rover

thetangentplane(T

heorem

9.1),

IIF

atpbecomes

theHessianofhatp:

IIF=

�L

MM

N

�=

±�

hxx

hxy

hyx

hyy

�=

±Hess p

h,

wherethesign

depen

dsonthechoiceofnorm

al.

InparticulartheGaussiancurvature

atpis

K(p)=

hxxhyy−

h2 xy=

det

Hess p

h.

Proof.

Intheproof

ofTheorem

9.1,

F(x,y)=

(x,y,h

(x,y)),so

∂xF

=(1,0,∂

xh)an

d∂yF

=(0,1,∂

yh)is

abasis

forTpS=

(xy-plane).This

forces

∂xh=

∂yh=

0at

(0,0),

so(0,0)is

acritical

pointof

h.Thusthenormal

is±n(p)=

(1,0,0)×(0,1,0)=

(0,0,1),an

ddot

product

withnjust

meanstaking±

thethirdcompon

entof

avector(thesign

±dep

endson

thechoice

ofGau

ssmap

).Thebasis

∂xF,∂

yF

isorthon

ormal,an

din

this

basis

IIis

thematrixin

the

claim.Now

use

Lem

ma12

.7.

Noticethat,rotatingas

intheproof

withSlocally(X

,Y,h

(X,Y

)),thesign

sof

theHessian

ofhareprecisely

what

youstudiedin

applied

courses

todiscu

ssminim

a,max

imaan

dsaddles

ofafunctionh(X

,Y)of

twovariab

les.

So

κ1,κ

2>

0⇒

Z=

h(X

,Y)has

aminim

um

at0

⇒S

lies

aboveZ

=0

κ1,κ

2<

0⇒

Z=

h(X

,Y)has

amax

imum

at0

⇒S

lies

below

Z=

0K

=κ1κ2<

0⇒

Z=

h(X

,Y)has

asaddle

at0

⇒S

lies

onbothsides

ofZ

=0

Example.Con

sider

thesphereSof

radiusr,

X2+Y

2+Z

2=

r2,neartheNorth

polep=

(0,0,1).

Locally

Sis

thegraphF(X

,Y)=

(X,Y

,√r2

−X

2−

Y2)withnormal

n(p)=

(0,0,1).

The

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

59

Hessian

of√r2

−X

2−Y

2at

0gives

II=

�−

1 r0

0−

1 r

�K

=det

II=

1 r2

This

was

expected:thegreatarcs

areosculatingcirclesof

radiusrwhichaccelerate

away

from

theou

twardnormal

direction

(0,0,1).

Noticethecurvature

becom

essm

allas

r→

∞since

the

surfacelooksmoreandmore“fl

at”forlarger.

12.6

Locallyflatsu

rfaces

Theorem

12.11.If

II=

0nearp,then

Slies

inaplanenearp.

Proof.

Let’suse

thesimplernotationn=

n(x,y)insteadof

n(F

(x,y))

forthelocalexpression

ofn

inthis

proof.Weneedto

show

that

locallytheGau

ssmap

nis

constan

tan

dthat

Ssatisfies

theequationforaplaneorthogon

alto

n:

n·F

=constan

t.(equivalentlyn·(F(x,y)−F(0,0))

=0)

From

II=

0weob

tain

∂xn·∂

xF

=0,

∂xn·∂

yF

=0,

etc.

andhence

bylinearity

∂xn,∂

yn

areorthogon

alto

allof

TS=

span

(∂xF,∂yF).

Differentiatingn·n

=1show

sthey

arealso

orthogon

alto

n:∂xn·n

=0an

d∂yn·n

=0.

Sothey

areorthogon

alto

allof

R3an

dthus

must

vanish.Sonislocallyconstan

tin

x,y.Son·F

isalso

constan

t:∂x(n

·F)=

n·∂

xF

=0

(since

nis

orthogon

alto

∂xF

∈TS),an

dsimilarly

∂y(n

·F)=

0.�

12.7

TheGaussian

curv

atu

reasara

tioofareas

Since

nis

aunit

vector,it

map

sn:S→

S2⊂

R3

into

theunit

sphereS2of

R3.Sowe

cancomparetheareasof

tworegion

sn(U

)⊂

S2an

dU

⊂S.Thefollow

ingsaystheGau

ssian

curvature

Kprecisely

measurestheinfinitesim

alratioof

thosetw

oareas.

Theorem

12.12.TheGaussiancurvature

atp∈S

equals

K(p)=

lim

U→

p±Area(n(U

)⊂

S2)

Area(U

⊂S)

,

wherethelimitis

overshrinkingneighbourhoodsU

ofp,andwhere±

=sign

(K(p)).

Proof.

Worklocallywithaparam

etrization

F=

F(x,y)such

that

F(0,0)=

p.Recallby

Section

10.8,thearea

ofU

=F(V

)⊂

Sis

defined

as

Area(U)=

� V

�∂xF

×∂yF�d

xdy.

Thearea

ofn(U

)is

analogou

sly:1

Area(n(U

))=

� V

�∂x(nF)×∂y(nF)�

dxdy.

1Non-exa

minabletech

nicalremark.If

∂x(n

F),∂y(n

F)are

linea

rlyindep

enden

tat(0,0

)th

ennea

rn(p)∈

S2wehavealoca

lparametriza

tionn◦F

:V

→S2,andth

eform

ula

forArea(n

(U))

isvalid.If∂x(n

F),∂y(n

F)

are

linea

rlydep

enden

tat(0,0

)th

enweca

nnotsayth

at.

Howev

er,oneca

nstillargueth

atth

eratioofth

eareasin

theclaim

converges

tozero

(notice

K(p)=

0here,

asth

eco

lumnsofII

are

linea

rly

dep

enden

t).

Indeed,oneca

nch

eckth

atArea(n

(U))

≤εArea(U

)forsm

allen

oughV

whereε→

0aswesh

rinkV.W

ewill

notca

rryoutth

esedetails.

60

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Wemay

assumethat

Fis

correctlyoriented,so

that

n=

(∂xF

×∂yF)/�∂

xF

×∂yF�.

Using

theshap

eop

erator

Swritten

inthebasis

∂xF,∂

yF,wecompute:

∂x(nF)×

∂y(nF)

=(∂

xFS 1

1+

∂yFS 2

1)×

(∂xFS 1

2+

∂yFS 2

2)

=(S

11S 2

2−

S 21S 1

2)∂xF

×∂yF

=det(S)∂xF

×∂yF

=det(S)�∂

xF

×∂yF�n

.

Since

K(x,y)=

det

Sis

theGau

ssiancurvature

atF(x,y),

and�n

�=

1,

Area(n(U

))=

� V

|K(x,y)|�∂

xF

×∂yF�d

xdy.

Writing|K

(x,y)|=

|K(p)|+

(|K(x,y)|−

|K(p)|)

,weob

tain

Area(n(U

))=

|K(p)|

� V�∂

xF

×∂yF�d

xdy+� V

(|K(x,y)|−|K

(p)|)

�∂xF

×∂yF�d

xdy

=|K

(p)|Area(U)+� V

(|K(x,y)|−|K

(p)|)

�∂xF

×∂yF�d

xdy.

Dividethat

byArea(U),movethefirstterm

ontherigh

tto

theleft,an

dtakeabsolute

values:

� � �Area(n

(U))

Area(U

)−

|K(p)|� � �

≤1

Area(U

max

(x,y)∈

V||K

(x,y)|−

|K(p)||

·� V�∂

xF

×∂yF�d

xdy

=max

(x,y)∈

V||K

(x,y)|−

|K(p)||

andthefinal

expressionconverges

to0as

weshrinkV

to(0,0)since

K(x,y)→

K(0,0)=

K(p)

bycontinuityof

K(sox,y

→0).

Lemma12.13.Werecord

twoform

ulae,

onefrom

Sec.10.8

andonefrom

theproofabove:

∂xF

×∂yF

=±�

det

I Fn

and

∂xn×

∂yn=

±K�

det

I Fn

wherethesign

is+

precisely

if∂xF,∂

yF

isaright-handed

basis((∂xF

×∂yF)·n

>0).

13.

Tangentialderivativesand

Gauss’TheoremaEgregium

13.1

Tangentialderivative(L

evi-Civitaconnection)

Let

Sbeasm

oothsurfacein

R3.Recallin

Section

9.4wedefined

vectorfieldsonS,and

weab

breviatedX

1=

∂xF,X

2=

∂yF

foralocalparam

etrization

Fnearp∈

S.Thus,

at

each

pointp∈S,wehaveabasis

ofR

3given

by:

X1(p),X

2(p),n(p)

since

R3=

TpS⊕

Rn(p),

asnis

orthog

onal

toTpS.Therefore,

thederivatives

ofasm

ooth

localtangentvectorfieldv(x,y)∈

TF(x

,y)S

defined

nearp∈

Scanbewritten

interm

sof

this

basis.In

particular,

theorthog

onal

projectionto

TpS

ofsuch

derivatives

iscalled

the

tangentialderivative,whichyo

ustudyin

Exercise

Sheet2:

∇xv

=orthog

onal

projectionof

∂xvon

toTS

=∂xv−

(n·∂

xv)n

=∂xv+

(∂xn·v)n

wheren=

n(x,y)isthelocalexpressionn(F

(x,y))

fortheGau

ssmap

(alson·∂

xv=

−∂xn·v

bydifferentiatingtheorthog

onalityrelation

n·v

=0,

compareExercise

4ofExercise

Sheet

2).Abbreviate

x,y

byindices

1,2.

Then

∂jv=

∇jv+

(Bv) jn

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

61

since,writingv=

�vi X

ibyabbreviatingthecoeffi

cientfunctionsv1=

a(x,y),v2=

b(x,y),

andrecallingthedefinitionof

thesecondfundam

entalform

Bij=

n·∂

iXj=

−∂in

·Xj,

∂jn·v

=∂jn·�

vi X

i=

−�

Bjiv

i=

−(B

v) j.

Thesymbol

∇is

called

nabla,andtheoperator

∇is

called

aconnectionforthesurface

S.Notice

thataconnectionis

away

todifferentiate

vector

fieldswithoutever

leav

ingthe

tangentspaceTS

(itturnsou

tconnectionsexistalso

forabstract

surfacesan

dman

ifolds).

Noticethat

infact

youcandifferentiatean

yvector

fieldbyan

yother

vector

field.Given

avectorfieldX

=�

ajX

j,wedefine∇

Xlinearlyin

thedifferentiatingvariab

leX:

∇Xv=

�aj∇

jv.

Note

∇Xvis

ofcoursenot

linear(w

ithrespectto

smoothfunctions)

inthevvariab

lesince

∇j(fv)=

(∂jf)v

+f∇

jv

wheref=

f(x,y)isafunction(hereweusedthat

∂j(f)v

isalread

yin

TS,so

doesn’tchan

geunder

orthog

onal

projection).

That

equationis

called,ofcourse,

Leibniz

rule.

Lemma13.1.Thetangentialderivative

only

depen

dsontheRiemannianmetricI(thefirst

fundamen

talform

),so

itis

aninvariantofthesurface

upto

isometries.1

Proof.

This

isacalculation

:

∇iv

=∇

i

�vjX

j=

�∂i(vj)X

j+

vj∇

iXj

sowejust

needto

checkthat

∇iX

jdep

endson

lyon

I F.You

willdothis

rather

explicitlyin

Exercise

Sheet2byexpressing

∇iX

j=

�Γk ijX

k

interm

softhebasisX

k,wherethefunctionsΓk ijare

called

Christoffelsy

mbols

andshow

ing

that

thereis

aform

ula

forthesein

term

sof

thefirstfundam

entalform

anditsderivatives.

Herewewilljust

illustrate

anexample:since

dottingwithX

jkills

thenormal

term

(asnis

orthog

onal

toTS=

span

(X1,X

2)),weget

X1·∇

1X

2=

X1·∂

1X

2=

∂xF·∂

x∂yF

=∂xF·∂

y∂xF

=1 2∂y(∂

xF·∂

xF)=

1 2∂yA

11

whereA

11is

thefirstentryofthematrixA

forI F

.Sim

ilarly,allX

k·∇

iXjaredetermined

byderivatives

ofA

ij,hence

this

determines

∇iX

j(bylinearalgebra).

Cultura

lRemark

.In

ExerciseSheet2youprove

theform

ula:

Γk ij=

1 2

� �

gk�(∂

igj�+

∂jg i

�−

∂�g i

j)

which

givesΓk ij

explicitlyin

term

softheRiemannian

metricg i

j=

I(X

i,X

j)=

Xi·X

j

(thefirstfundamen

talform

).Conversely,

given

anyabstract

surface

ormanifold,witha

Riemannian

metric,

you

simply

defi

neΓk ij

bythatform

ula,thusyou

obtain

aconnection

∇defi

ned

by∇

iXj=

�Γk ijX

k,called

Levi-Civita

connec

tion,whichdefi

nes

tangential

derivatives!

More

ofthis

inC3.3:Differe

ntiable

Manifolds

1i.e.

itdoes

notch

angeev

enifyoupickadifferen

tem

bed

dinginto

R3,provided

thetw

oem

bed

ded

surfaces

are

isometric.

62

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

13.2

Gauss’TheoremaEgregium

Theore

m13.2.TheGaussiancurvature

only

depen

dsonthefirstfundamen

talform

.

Proof2.Let

v(x,y)∈TS

bean

ynon

-zerolocalvector

field(for

exam

ple

v=

∂xF).

Recall∇

iv=

∂iv

−((∂iv)·n

)n.Con

sider

thefollow

ingexpression:

(∇x∇

y−

∇y∇

x)v

=∂x∂yv−

(∂yv·n

)∂xn−

∂y∂xv+

(∂xv·n

)∂yn

=−(∂

yv·n

)∂xn+(∂

xv·n

)∂yn

wherewedropped

alltheterm

sthat

weremultiplesof

nsince

weknow

theresultmust

bein

TS,weusedthat

vissm

oothso

partial

derivatives

commute,an

dweusedthat

∂in

=∇

in∈

TS

(recallthis

isbecau

sedifferentiatingn·n

=1show

sthat

∂in

isorthog

onal

ton

and

hence

isin

TS).

Recallthat

v·∂

in=

−∂iv

·n(bydifferentiatingtheorthog

onalityrelation

v·n

=0),so

theab

ovebecom

es:

(∇x∇

y−

∇y∇

x)v

=(v

·∂yn)∂

xn−

(v·∂

xn)∂

yn

=−(∂

xn×

∂yn)×

vCross-product

tricks1

=∓K√det

I Fn×

vLem

ma12

.13

wherethesign

is−

precisely

if∂xF,∂

yF

isrigh

t-han

ded

.Now

noticethat:byLem

ma13

.1thefunction(∇

x∇

y−∇

y∇

x)v

only

dep

endson

thefirst

fundam

entalform

.Ofcourse√det

I Fon

lydep

endson

thefirstfundam

entalform

,butso

does

∓n×vbecau

sethat

isjust

arotation

by∓90

degrees

ofthevector

vinsideTSan

dwe

know

by10

.7that

thefirstfundam

entalform

canbeusedto

measure

angles.It

follow

sby

theab

oveform

ula

that

also

Kon

lydep

endson

I(usingthat

∓√det

I Fn×v�=

0forv�=

0).

TechnicalRem

ark.Thesign

ambiguityabove

isnotanissue:

thechoiceofnorm

al±naffects

thechoiceoforien

tationforthesurface,andthusthenotionofclockw

ise/anti-clockwiserota-

tionby

90degrees,butthevector∓n×vis

indepen

den

tofthis

choice(thesign

scancel).

Remark

.TheTheoremaEgregium

implies

thattheGaussiancurvature

isanintrinsicin-

variant,i.e.

itis

anisometry

invariant,because

itdepen

dsonly

onthechoiceofRiemannian

metriconthesurface

(whereasother

curvature

invariants,such

asprincipalcurvatures,

are

extrinsic:they

depen

dheavily

onthechoiceofem

beddingofthesurface

into

R3).

Lemma13.3.Werecord

forlater,

theusefulform

ula

from

theabove

proof:

(∇x∇

y−

∇y∇

x)v

=∓K�det

I Fn×

v

with−

sign

precisely

if∂xF,∂yF

isright-handed

(det(∂

xF|∂

yF|n)>

0).

13.3

Riemann

curv

atu

retenso

r

This

Section

isnon-examinable.

Motivation.

Anaturalwayto

thinkofcurvature

isto

ask:how

much

dothetangential

derivatives∇

x,∇

yfailto

commute?

Forasm

alllocalvectorfieldv,if

youthinkintuitively

of∇

xv,∇

yvasbeingsm

allarrowsin

aninfinitesim

alparallelogram,thefailure

ofthis

paral-

lelogram

toclose

upis

measuredby

∇x∇

yv−

∇y∇

xv.This

encodes

how

curved

thespace

is.

TheRiemann

curv

atu

retenso

rR

m ijkis

defined

by:

R(X

i,X

j)X

k=

∇i∇

jX

k−

∇j∇

iXk

=�

Rm ijkX

m

1(a

×b)

×c=

(a·c)b

−(b

·c)a.Forex

ample,(e

e 2)×

e 1=

e 3×

e 1=

e 2=

(e1·e

1)e

2−

(e2·e

1)e

1.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

63

Cultura

lRemark

.Wehave

only

defi

ned

Ron

thebasisX

i=

∂iF

,buttheRiemann

curvature

tensorR(X

,Y)Z

canbe

defi

ned

forgeneralvectorfieldsX,Y

,Z.Themeaningof

tenso

ris

thatitmust

belinearwithrespectto

smooth

functions(notjust

linearwithrespect

toconstants):

R(fX,gY)hZ

=fghR(X

,Y)Z

foranysm

ooth

functionsf,g,h

.Forsake

of

comparison:∇

XY

istensorialonly

inX,whereasin

Yit

satisfies

theLeibn

izrule.

You

cannow

checkby

calculation,usingtheLeibn

izrule

for∇,thatthetensorialconditiononR

implies

thegeneralform

ula

forR

hasanadditionalterm

:

R(X

,Y)Z

=∇

X∇

YZ−∇

Y∇

XZ−∇

[X,Y

]Z

wheretheLie

bra

cket[X

,Y]isdefi

ned

by:[�

vi X

i,�

wjX

j]=

�vi ∂

i(w

j)X

j−�

wj∂j(v

i )X

i.TheLie

bracket

measureshow

much

theflow

oftwovectorfieldsfailsto

commute:youwill

encounterthis

again

inC3.5

Lie

Gro

upsandsecretly

inC2.1

Lie

Algebra

s.In

ourcase:

[Xi,X

j]=

∂i(1)X

j−

∂j(1)X

i=

0since

1is

aconstantcoeffi

cien

t.

InExercise

Sheet3,

youwillshow

that

Rm ijkis

completely

determined

bytheChristoff

el

symbolsΓk ijan

dtheirderivatives,an

dbyExercise

Sheet2theChristoffelsymbolson

lydep

end

ontheRieman

nianmetricI(firstfundam

entalform

).Hence:

Theore

m13.4.TheRiemanncurvature

tensorR

only

depen

dsontheRiemannianmetric

(firstfundamen

talform

),so

itis

aninvariantofthesurface

upto

isometries.1

It’s

oftenusefulto

dot

theab

ovewithan

other

basisvectorX

�,whichdefines

Rijk�

=R(X

i,X

j)X

k·X

=I(R

(Xi,X

j)X

k,X

�)

Thetw

oRieman

ncurvature

tensors

are

relatedbythelowering/raisingofindices

usingthe

Riemannianmetricg i

j=

I ij=

Xi·X

j,explicitly

Rijk�=

�R

m ijkg m

�R

m ijk=

�R

ijk�g�m

wheregij

istheinversematrixof

g ij,thus:

�gijg j

k=

δi k.

Atfirst,

itseem

sthat

Rijk�is

alotofinform

ation(2

4=

16choices

ofvalues

forthefour

indices),

butin

fact

bydefinitionitis

antisymmetricin

i,j,

so

Rijk�=

−R

jik

sowemightas

wellchoosei=

1,j=

2.Then,byExercise

Sheet2you

know

that

tangential

derivatives

arecompatible

withtheRieman

nianmetric:

∂iI(v,w

)=

I(∇

iv,w

)+

I(v,∇

iw)

sousingthesymmetry

∂j∂iI(v,w

)=

∂i∂

jI(v,w

)(since

I(v,w

)isasm

oothfunction),yo

uwill

deduce

inExercise

Sheet3that:

Rijk�=

−R

ij�k

1i.e.

itdoes

notch

angeev

enifyoupickadifferen

tem

bed

dinginto

R3,provided

thetw

oem

bed

ded

surfaces

are

isometric.

64

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

soR

ijk�is

anti-symmetricalsoin

k,�

sowemightaswelltakek=

1,�=

2.Thus,

only

one

valueis

interestingupto

symmetries,1

andyouwillshow

inExercise

Sheet3that:

2

R1212=

−K

det

I F.

Theorem

13.5

(TheoremaEgregium).

TheGaussian

curvature

only

depen

dson

thefirst

fundamen

talform

,notonthesecond(soitdepen

dsontheRiemannianmetricbutnotonthe

particularchoiceofem

beddinginto

R3).

Indeed:

K=

−I(R

(v,w

)v,w

)

|v×w|2

=−

I(R

(v,w

)v,w

)

I(v,v)I(w

,w)−I(v,w

)2

foranytwolinearlyindepen

den

tvectors

v,w

.

Proof.

Thefirstpartfollow

sfrom

R1212=

−K

detI F

since

Rijk�only

dep

endsonI F

and

derivatives

ofI F

.In

thesecondpart,thesecondequality

isjust

theexpansionofthecross-

product

(v×

w)·(v×

w)=

(v·v)(w·w

)−

(v·w

)2.Weonly

needto

checkthesecondpart

forspecificv=

X1,w

=X

2(bylinearalgebra

itthen

holdsforanybasisv,w

:indeedjust

chan

geF

bythechangeofbasiswhichsendsX

1,X

2to

v,w

).Bytheabove:

I(R

(X1,X

2)X

1,X

2)

I(X

1,X

1)I(X

2,X

2)−I(X

1,X

2)2

=R

1212

eg−f2=

−K

det

I Fdet

I F=

−K.

Remark

13.6

(Whyisthatproofconceptuallydifferent?).

Thecalculationin

theabove

proof

isthesameasin

Section13.2

whichcalculatedtheRiemanncurvature

R(∂

xF,∂

yF)v

=∓K�det

I Fn×v.

However,theproofin

Section13.2

madeexplicitreference

tothenorm

alnandthefact

thatS

isem

bedded

inR

3(eventhough

attheen

dweconcludethatK

only

depen

dsontheem

bedding

upto

isometry),

soit

appears

toonly

apply

tosurfacesS

whose

Riemannianmetricarises

from

thefirstfundamen

talform

ofanem

beddingofS

into

R3.However,notallRiemannian

metrics

arise

inthis

way,

forexample:theflattorusT

2=

R2/Z2

,thatis

withRiemannian

metriclocallyinducedby

thestandard

dotproduct

inR

2,is

locallyisometricto

R2so

II=

0(inparticularK

=0),

soit

cannotisometricallyem

bedinto

R3because

byTheorem

12.11

itwould

have

toliein

aplane(m

ore

intuitively:

obviouslyatorusem

bedded

inR

3is

going

tobe

curved!).Theapproach

ofSection13.3

ismore

general,since

itworksforanyabstract

smooth

surface

withanychoiceofRiemannianmetricg i

j.Indeed,in

ExerciseSheet2you

foundaform

ula

forΓk ij

interm

sofg i

j,this

inturn

defi

nes

∇X

iX

j,whichin

turn

defi

nes

R(X

i,X

j)X

kandthusK.Forexample,this

applies

tothehyperbolicplaneH

withoutever

worryingaboutwhether

ornotH

embedsisometricallyinto

R3.

1Culturalrem

ark.In

additionto

theabovesymmetries,th

ereis

onelast

symmetry

thatholdsin

gen

eral

formanifolds.

From

theJacobiid

entity

forLie

brackets,

[X,[Y,Z

]]+

[Y,[Z,X

]]+

[Z,[X,Y

]]=

0

oneobtainsth

efirst

Bianchiid

entity

:

R(X

,Y)Z

+R(Y

,Z)X

+R(Z

,X)Y

=0.

From

this

oneded

ucesth

ecy

clic

symmetry

Rij

k�+

Rjki�

+R

kij

�=

0.Byaddingth

efoureq

uationsyouget

from

this,if

you

reord

erijk�cy

clically,

and

usinganti-symmetries,you

can

ded

uce

thatR

j�ik

=R

ikj�,or

relabelling:R

ijk�=

Rk�ij

soyouca

ninterchangeth

efirstandlast

pair

ofindices.

2From

this

itfollows,

usingR

m ijk=

�R

ijk�g�m,th

at

R2 121=

−Ke,

R2 122=

−Kf,

R1 121=

Kf,

R1 122=

Kg

whereK

isth

eGaussiancu

rvatu

re,e,f,f

,gare

theen

triesofth

efirstfundamen

talform

I F.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

65

Thearetw

omorefamou

scurvatures,

called

Riccicurv

atu

retenso

rRic

ijan

dscalar

curv

atu

reR,they

areob

tained

from

theRieman

ncuravature

bytakingtraces.

TheRiccicurv

atu

reis

thetrace

Ric(X

,Z)=

trace(Y

�→R(X

,Y)Z

)

So,

definingR

ik=

Ric(X

i,X

k),

wehaveto

putX

=X

i,Y

=X

j,Z

=X

kab

ove,

then

take

thej-th

entryof

theresult,an

dsum

over

j(andalso

over

�on

therigh

t):

Rik

=�

Rj ijk=

�R

ijk�g�j

Theloweringof

twoindices

usingtheinverse

metricgij

iscalled

ametric

trace.Sim

ilarly,

thescalarcurv

atu

reis

defined

asthemetrictraceof

Ric:

R=

�gikR

ik

summingover

bothi,k.For

surfaces,

Rij=

−Kg i

jR

=−2K

.

ThecourseC3.3:Differentiable

Manifoldsdevelop

stheseideasfurther.

14.

Geodesiccurvatureand

theGauss-B

onnettheorem

14.1

Geodesiccurv

atu

reand

norm

alcurv

atu

re

Recallthat

foracurveγ:[0,b]→

R3param

etrizedbyarc-length,wedefined

thecurvature

ofγas

thenorm

oftheacceleration:

κ(t)=

�� (t)�.

How

ever,wesaw

that

onasurfacethecorrectnotionof

differentiation(i.e.theon

ewhich

only

dep

endson

theRieman

nianmetric,

andnot

ontheparticularchoiceof

embeddinginto

R3),is

thetangential

derivative: ∇

tγ�

=∂t(γ

� )−(∂

tγ� ·n)n

=γ�� (t)−II(γ

� ,γ� )n

whererecallin

Sec.12.3wealread

ymet

thenormal

curvature

γ��·n

=−Dn(γ

� )·γ

�=

II(γ

� ,γ� ).

Sotheacceleration

break

supinto

twoparts,thetangential

andthenormal

part:

�=

∇tγ� +

II(γ

� ,γ� )n.

Correspon

dinglythecurvature

break

supinto

twoparts

(usingthatnisnormal

to∇

tγ�∈TS),

κ2

=�γ

�� �2

=�∇

tγ� �

2+II(γ

� ,γ� )2

=κ2 geodesic+κ2 norm

al.

Definition14.1

(Geodesiccurvature).

Thegeodesic

curvature

ofacurveγin

Sparametrized

byarc-len

gthis:

κgeodesic=

�∇tγ� �

Acurveγin

Sis

ageo

desicifκgeodesic=

0.

Example.For

theplaneR

2,II

=0so

κgeodesic

=κ.Thereforeκgeodesic

=0im

plies�=

0andhence,integrating,

γ(t)=

p+

tvis

astraightline.

Sogeodesicsin

theplanearestraight

lines.

Example.For

asphereS

ofradiusrin

R3,wesaw

that

thenormal

curvaturesare−1/rin

all

direction

s(asII

=−

1 rid).

Agreatcircle

(acircle

inS

ofmaxim

alradius,

r)has

curvature

1/r

66

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

andnormal

curvature

−1/

r,so

κgeodesic=

0.Sogreatcirclesaregeodesics:

they

look

“straigh

t”from

theview

pointof

thesurface.

Acircle

ofsm

allerradiuss<

rin

Shas

curvature

1/s,

but

normal

curvature

−1/

r,so

κgeodesic=

√s−

2−

r−2�=

0Sosm

allercirclesarenot

geodesics.

Lemma14.2.Forγparametrizedby

arc-len

gth,

±κgeodesic

=γ��·(n×

γ� )

=det(n|γ

� |� )

=∇

tγ� ·(n

×γ� )

Proof.

γis

param

etrizedbyarc-length,so

γ� ·γ�=

1,so

differentiating:

1

∇tγ� ·γ�=

0

(thetangential

acceleration

isperpendicularto

thevelocity).

So∇

tγ�isorthog

onal

toγ�an

dit

isalso

orthog

onal

tonas

itlies

inTS.Son×γ�is

parallelto

∇tγ� .

Moreover,n×γ�has

unitlengthsince

n,γ

�haveunitlengthan

dthey

areperpendicular(since

γ�∈TS).

Therefore

∇tγ�=

±κgeodesicn×

γ� .

Since

nis

orthog

onal

ton×

γ� ,wealso

know

that

γ��·(n×

γ� )=

∇tγ� ·(n

×γ� ).

Coro

llary

14.3.Foracurveγ

inS

whichis

parametrizedby

arc-len

gth,γ

isageodesic

⇔γ��is

norm

alto

S.Foracurveγnotparametrizedby

arc

length,withγ� (t)

�=0,

After

arc-len

gthreparametrizationγbecomes

ageodesic

⇔γ� ,γ�� ,nare

linearlydepen

den

t

Proof.

Ifγis

param

etrizedbyarc-lengththis

follow

sbythelemmasince

det(n|γ

� |� )

=0

precisely

ifn,γ

� ,�arelinearlydep

endent.

If�γ(t)

=γ(s(t))

isareparam

etrization

ofγso

that

�γis

param

etrizedbyarc-length,then

�γ�=

s�γ� (s)

and�γ�

�=

s�� γ

� (s)

+(s

� )2� (s).Since

s�>

0,lineardep

endence

ofn,�γ

� ,�γ�

�is

equivalentto

lineardep

enden

ceof

n,γ

� ,� .

Example.Con

sider

thetorusT

2⊂

R3param

etrizedby

F(θ,ψ

)=

((a+

bcosψ)cosθ,

(a+

bcosψ)sinθ,

bsinψ).

Con

sider

thequotientmap:

R2→

S1×

S1∼ =

T2,(θ,ψ

)�→

(eiθ,e

iψ)�→

F(θ,ψ

)

Con

sider

thestraightline(t,0)in

R2,whichgivesrise

tothefirstcircleS1factor

γ(t)=

F(t,0)=

((a+

b)cost,(a

+b)sint,0)

inT

2.Alongγ,puttingψ=

0,

n=

cost

sint

0

γ�=

∂θF

=(a

+b)

−sint

cost

0

�=

∂θθF

=(a

+b)

−cost

−sint

0

so�isnormal

soγisageodesic.Sim

ilarlyfortheother

circle

factor

γ(t)=

F(0,t)weget�

isamultiple

ofn,so

γisageodesic.How

ever,itisnot

truethat

anystraightlinein

R2will

give

rise

toageodesic

inthistorus:

draw

apicture

forthecircle

γ(t)=

F(π 2,t),

checkthat

n,γ

� ,�

areclearlylinearlyindependent.

Thisisnot

surprisingbecause

theRiemannianmetricis

I=

(a+

bcosψ)2dθ2

+b2dψ2,

sorotation

inθisan

isom

etry

(since

Iwillbeinvariant),butrotation

inψisnot:indeeditcannot

bebecause

youshow

inExerciseSheet3that

theGaussiancurvature

Kvaries

intheψdirection

.

1Forsm

ooth

vectorfieldsv(t),w(t)∈

Tγ(t)S,th

eanalogueofth

eco

mpatibility

equation

∂iI(v,w

)=

I(∇

iv,w

)+I(v,∇

iw)from

Section13.3

holds:

d dt(v

·w)=

∇tv·w

+v·∇

tw.Indeed

d dt(v

·w)=

v� ·w+v·w

� ,andv�=

∇tv+

(n·v

� )nhasv� ·

w=

∇tv·w

since

w∈

TS

isperpen

dicularto

n(andsimilarlyforv·w

� ).

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

67

Thistorusisthereforenot

“flat”,

inthesense

that

thequotientmap

R2→

T2isnot

locally

anisom

etry.

Example.Anexam

ple

ofaflat

torusT

2is

F(θ,ψ

)=

(eiθ,e

iψ)∈S1×S1⊂

C×C=

R4

usingthedot

product

from

R4to

definetheRiemannianmetric,

then

R2→

T2will

bealocal

isom

etry,andgeodesicsin

T2correspon

dto

quotients

ofstraightlines

inR

2.

14.2

ThelocalGauss-B

onnetth

eore

m

Definition

14.4

(Signed

geodesic

curvature).

FollowingLem

ma14.2,wedefi

nethesigned

geo

desiccurv

atu

reκg=

±κgeodesic

ofasm

ooth

curveγin

Sparametrizedby

arc-len

gthby

κg=

∇tγ� ·(n

×γ� )=

γ��·(n×γ� )=

det(n|γ

� |� )

Remark

.If

γis

not

param

etrisedbyarc-length,wedefinethegeodesic

curvature

asthat

obtained

fortheunit-speedreparam

etrisedcurve�γ.

This

yields:

κg=

det(n|γ

� |� )

�γ� �

3.

Proof:

inthenotationof

theproof

ofCorollary

14.3,det(n|�γ

� |�� )

=(s

� )3det(n|γ

� |� )

(using

that

thedeterminan

tis

linearin

each

column,an

dthefact

that

det(n|γ

� |γ� )

=0as

the

columnsarelinearlydep

endent).Finally

thespeed��γ

� �=

1,so

s�=

1/�γ

� �.

Theorem

14.5

(LocalGau

ss-B

onnet

Theorem).

Let

γbe

asm

ooth

simple

1closedcurve,

whichboundsaregion

Rthatlies

entirely

insideaparametrisation

patch.

Assumethatγ

travels

anti-clockwisearoundR.Then

� R

KdA

=2π

−� γ

κgds

wheredsis

theelem

ent2

ofarc-len

gthofγ,anddA

istheelem

ent3

ofareaofS.

Ifγtravels

clockw

isearoundR,then

the−

sign

inthelast

term

is+.If

γis

piecewise

smooth,so

γ� (t)

isdiscontinuousatfinitelymanytimes

t=

t j,then

2πbecomes

2π−

�αj

whereαjis

theanti-clockwiseangle(m

easuredusingI)thatγ�jumpsby

att=

t j.

We’llpro

veth

isafterth

eexamples.

Examples.

(1)For

StheplaneR

2,noticethat

γ� (s)

∈S1⊂

R2since

itis

aunit

lengthvector.So

γ� (s)

=(cos

θ(s),sin

θ(s))foracertainsm

oothangleθ(s).Choosingn=

(0,0,1),

with

R2as

theplaneZ

=0in

R3,then

γ�=

(−sinθ(s),cosθ(s))is

γ�rotatedby

90degrees.Finally

� (s)

=θ�(s)(−sinθ(s),cos

θ(s)),

soκg=

γ��·(n×

γ� )=

θ�(s).

This

confirm

sthetheorem:

� γ

κgds=

�θ�(s)ds=

θ(end)−θ(start)

=2π

=2π−�

0dA.

1sim

ple

mea

nsit

does

notintersectitself,i.e.

γ:[a,b]→

Sis

injective.

2ds

=�

I(γ

� (t),γ

� (t)dt,

reca

llweused

this

todefi

nelength

sL(γ

)=

� γds.

Ifγ

isparametrized

by

arc-len

gth

then

I(γ

� ,γ� )

=1,so

wejust

get

ds=

dt.

3dA

=√det

I Fdxdy,reca

llweusedth

isto

defi

neareas:

Area(R

)=

� RdA.

68

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

(2)For

Sthesphereof

radiusr,

recallK

=1 r2.Takeγ=

theequator:thisisagreatcircle,

soageodesic,so

κg=

0.Thus:

� upperhemisphere

KdA

=1 r2(A

reaofthehem

isphere)

=2π−�

0ds=

2π.

Hence

thearea

ofthehem

isphereis2πr2,so

thearea

ofthesphereis4πr2.

(3)Takeanycurveγin

theunitsphereS2.Thisdivides

thesphereinto

tworegion

sR

1,R

2

withareasA

1,A

2.

Thecurvetravelsanti-clockwisearou

ndon

eregion

,sayR

1,and

clockwisearou

ndR

2.Thus:

Area(S2)=

� R1

KdA+

� R2

KdA

=(2π−� γ

κgds)

+(2π+

� γ

κgds)

=4π.

(4)Thedeepreason

whythe2π

appears,isbecause

thevector

fieldγ�does

not

extendto

anon

-vanishingvector

fieldv=

v(x,y)on

allof

theregion

R.Indeed,in

localcoordinates

γ� loc∈

R2=

TV

swings

arou

ndby

anangle2π,andthenumber

oftimes

thevector

fieldsw

ings

arou

nd(w

hichisan

integer)

dependscontinuou

slyon

thecurve,

hence

this

integerisconstant.

Shrinkingthecurveγlocto

apoint,then

alon

gon

eof

theshrinking

curves

thevector

fieldvshou

ldalso

swingarou

ndon

ce.

Buton

ceyou’veshrunkthe

curveto

apoint,visconstantalon

gtheconstantcurve,

soitsw

ings

arou

ndzero

times.

Con

tradiction

.

Non-examinable

ProofofTheLocalGauss-B

onnet

Theorem.

Step

0.Wemay

assumeγis

parametrizedbyarc-len

gth,so

ds=

dt,

andthatforthe

param

etrization

F(x,y),thebasis

∂xF,∂

yF

isright-handed

(otherwisesw

itch

thesignofy).

Step

1.Wefirstbuildan

orthon

ormalbasisofvectorfieldsforTS

over

theregionR.

Takev=

∂xF/n

orm,wherenorm

=�I(∂

xF,∂

xF).

Then

take

w=

n×v=

(vrotatedby90degrees

insideTS).

Sov,w

isan

orthon

ormal

basis

forTS

over

Rwithv×

w=

n.Differentiatingtherelation

w·w

=1,

wegetthat

∇xw,∇

yw

∈TS

are

orthogonalto

w,so

they

are

proportionalto

v:

∇xw

=Pv

∇yw

=Qv

forsomesm

oothfunctionsP

=P(x,y),Q

=Q(x,y)ofthelocalcoordinates(x,y).

Step

2.Com

pute

theRieman

ncurvature

ofw

intw

oways.

First

byLem

ma13.3,

(∇x∇

y−∇

y∇

x)w

=−K�

det

I Fn×w

=K�

det

I Fv.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

69

Then

explicitlyin

term

sof

P,Q

,

(∇x∇

y−∇

y∇

x)w

=∇

x(Q

v)−∇

y(P

v)

=(∂

xQ

−∂yP)v

+Q∇

xv−P∇

yv

=(∂

xQ

−∂yP)v

wherein

thesecondequalityweknew

thelast

twoterm

shadto

cancelbecau

setheresult

shou

ldbeamultiple

ofv,nam

elyK√det

I Fv,whereas∇

xv,∇

yvareorthogon

alto

v(by

differentiatingtherelation

v·v

=1).

Step

3.Now

wecancompute

theintegral

oftheGau

ssiancurvature:

� RK

dA

=� R

K√det

I Fdxdy

=� R

(∇x∇

y−∇

y∇

x)w

·vdxdy

Usingv·v

=1

=� R

(∂xQ

−∂yP)dxdy

=� R

d(P

dx+Qdy)

=� γ

Pdx+Qdy

Green

’stheorem

inR

2

=� (x

� P+y� Q

)dt

Meaningof

� γ,locallyγ(t)=

(x(t),y(t))

=� (x

� ∇xw+y� ∇

yw)·v

dt

Usingv·v

=1

=�∇

tw·v

dt

Chainrule

∇t=

x� ∇

x+y� ∇

y

Step

4.Now

wecompute

κgin

term

sof

v,w

.Since

γ� (t)

isaunit

vector

inTS,wecanwrite

itas

follow

sin

thebasis

v,w

γ� (t)

=cosθ(t)v+sinθ(t)w

wherev,w

areevaluated

atγ(t)of

course.

Then

∇tγ�=

θ�(t)(−sinθ(t)v+cosθ(t)w)+cosθ(t)∇

tv+sinθ(t)∇

tw.

Thesign

edgeodesic

curvature

becom

es:

κg

=∇

tγ� ·(n

×γ� )

=∇

tγ� ·(cos

θn×

v+sinθn×w)

=∇

tγ� ·(cos

θw−sinθv)

=θ�cos2

θ+

θ�sin2θ+cos2

θ∇

tv·w

−sin2θ∇

tw·v

usingorthon

ormalityof

v,w

,an

dusingthat

∇tvis

orthogon

alto

v(from

differentiating

v·v

=1),similarly

∇tw·w

=0.

Differentiatingv·w

=0weget∇

tv·w

=−v·∇

tw.Thus:

κg=

θ�−∇

tw·v

Step

5.Com

bineStep3an

dStep4:

� R

KdA

=

�θ�(t)dt−

�κg(t)dt

Now

θ(t)

isthean

glebetween

van

dγ� .

Passingto

localcoordinates,v=

∂xF/n

orm

=DF(e

1)/norm

wheree 1

isthestan

dard

basis

vector

(1,0)(sothex-direction

)an

dγ�=

DF(γ

� loc),

solocallyv,γ

�arerepresentedbye 1,γ

� loc.Soθ(t)

isthean

gle(m

easuredusing

I,not

theusual

Euclideanan

gle)

that

thelocaltangentvector

γ� loc(t)∈

R2makes

withthe

positivex-direction

.Since

θ(t)

mak

eson

efullcircle

inthean

ti-clockwisedirection

when

γtravelsan

ti-clockwisearou

ndthesimple

curve,

wemust

haveθ(end)−

θ(start)

=2π

(even

thou

ghθis

not

thean

glewemay

expectwithEuclideaneyes).

Iftherearediscontinuities,

then

theintegral

�θ�(t)dtdoes

not

noticethejumpsbyαjso

itequals

2π−

�αj.W

hen

γtravelsclockwisearou

ndR,simply

consider

thereversedpath�γ(t)

=γ(−

t)an

dnoticethat

κgsw

itches

sign

since

�γ�(t)=

−γ� (−t)

and�γ�

� (t)

=+γ�� (−t).

70

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Cultura

lRemark

.Notice

thekeystep

intheproofis

awayto

pass

from

an

integral

aroundtheboundary

∂R

ofR

toanintegralovertheregionR.This

wasGreen’s

theorem:

� R

dω=

� ∂R

ω

whereω=

Pdx+Qdyisadifferen

tialform

.Observe

thatGreen’stheorem

isthe2-dim

ensional

analogueoftheFundamen

talTheorem

ofCalculus:

�b

a

f� (x)dx=

� [a,b]

df=

� ∂[a,b]

f=

f(b)−

f(a).

Theabove

Green’s

form

ula

holdsin

greatgenerality:R

can

beanysm

ooth

n-dim

ensional

manifold

withboundary,andωcanbe

anydifferen

tialn−1form

(meaningωis

asum

where

each

term

looks

like

fdx∧dy∧···∧

dz

wherefis

asm

ooth

function,andx,y,...,z

are

anyn−1ofthelocalcoordinates,

thesymbol

∧remindsusthat1-form

santi-commute:dx∧dy=

−dy∧dx,just

like

forcross-products).

This

generalizationofGreen’s

form

ula

iscalled

Stokes’stheo

rem,it

isarguablythemost

importantresultin

geometry.More

ofthis

inC3.3

Differe

ntiable

Manifolds.

14.3

Thesu

mofth

eanglesin

ageodesictriangle

Ageodesictriangle

consistsof

aregion

Ran

dapiecewisesm

oothgeodesicγmov

ingan

ti-

clockwisearou

ndthebou

ndaryof

Rhav

ingexactlythreediscontinuitiesat

points

p,q,r

∈S,

called

theverticesof

thetriangle.

Sothebou

ndaryof

Rconsistsof

threegeodesic

arcs

joiningthepoints

p,q,r.

Callαp,α

q,α

rtheinternal

angles

betweenthetw

ogeodesic

arcs

whichmeetat

thepoints

p,q,r

(internal

angle

meanstheon

esw

eptou

tinsideR).

Coro

llary

14.6.Thesum

oftheangles

ofageodesic

trianglethatlies

entirelyin

aparametriza-

tionpatchis

αp+

αq+

αr=

π+

� R

KdA.

Forexample

ifK

>0in

Rthen

thesum

isstrictly

larger

thanπ,ifK

=0in

Rthen

thesum

isπjust

like

inEuclideangeometry,andifK

<0in

Rthen

thesum

isstrictly

less

thanπ.

Proof.

Thean

gles

bywhichγ�jumpsaretheexternal

angles:α1=

π−

αp,α2=

π−

αq,

α3=

π−

αr,so

byTheorem

14.5

withou

tusingthegeodesic

assumption

:

αp+

αq+

αr

=π+

[2π−

(α1+

α2+

α3)]

=π+� R

KdA+

� γκgds.

When

thetriangleis

geodesic

then

thelast

integral

vanishes

since

κg=

0.�

Examples.

Con

sider

ageodesic

trianglewithangles

α,β

,γandvertices

A,B

,C:

(1)Sphericalgeometry.Ontheunitsphere,K

=1,

sothesum

oftheangles

inageodesic

triangleis:

α+

β+

γ=

π+

Area(ABC)≥

π.

(2)Euclidean

geometry.In

theplane,

K=

0,so

thesum

oftheangles

is

α+

β+

γ=

π.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

71

(3)Hyperb

olic

geometry.

Inthehyp

erbolic

plane,

soH

={z

∈C

:Im

z>

0}with

I=

dx2+dy2

y2

=1 y2I E

uclidean,wewill

seelaterthat

K=

−1,

so

α+

β+

γ=

π−

Area(ABC)≤

π.

14.4

TheGauss-B

onnetth

eore

m

Theore

m14.7.Foranysm

ooth

compact

orien

tablesurface

SwithaRiemannianmetric,

χ(S

)=

1 2π

� S

KdA.

Example.For

theunitsphere,

K=

1so

� SK

dA

=Area(S2)=

4π,so

1 2π

� S

KdA

=2=

χ(S

2).

Proof.

Apply

theLocalGau

ss-B

onnet

theorem

toeach

triangleofatriangu

lation

ofS(w

here

wesubdividethetriangu

lation

ifnecessary

sothat

each

triangleis

smallenou

ghto

liein

aparam

etrization

patch,so

that

Theorem

14.5

applies).

Theintegralsof

κgallcancelbecau

seweintegrateκgtw

icealon

geach

edge

butin

oppositedirection

s(see

thecomments

inTheorem

14.5

abou

tan

ti-clockwise/clockwisecurves).

Thus,

asin

theproof

ofCorollary14

.6,

� R

KdA

=�

triangles(�

(internal

angles)−

π).

Let

V,E

,Fdenotethetotalnumber

ofvertices,edges,

triangles.Then

theab

oveequals:

2πV

−πF.

Since

each

edge

belon

gsto

exactlytw

ofaces,an

deach

face

has

exactlythreeedges,3F

=2E

.So2π

V−

πF

=2π

V−

3πF

+2π

F=

2π(V

−E

+F)=

2πχ(S

).�

15.

Morse

functions,

Poincare-H

opfandHairyBallTheorem

72

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

15.1

Criticalpoints

offunctions,

Morsefunctions,

gra

dientvecto

rfield

Afunctionf(x,y)in

twovariab

leshas

acritical

pointat

pifitsfirstderivatives

thereare

zero,equivalentlythedifferential

vanishes:

df=

(∂xf)dx+

(∂yf)dy=

0(evaluated

atp).

Atcritical

points,yo

uwan

tto

know

whether

fhas

aminim

um,max

imum

orasaddle.Recall,

from

calculus,

that

when

theHessian

(evaluated

atp)

Hessf=

� ∂xxf

∂yxf

∂yxf

∂yyf

isnon

-singu

lar(determinan

tis

non

-zero)

then

youknow

thean

swer

bylook

ingat

thesign

s1

oftheeigenvalues

λ1,λ

2(see

Section

12.5).

Acritical

pointpis

non-d

egenera

teiftheHessian

atpis

non

-singu

lar.

Definition

15.1.A

functionis

Morseifallcriticalpoints

are

non-degen

erate.

Itturnsou

tthat

genericallyafunctionis

Morse,in

thesense

that

given

anyfunction(even

thezero

function!)

ifyouwiggleitrandom

lythen

itwillbecom

eMorse.Theword“g

eneric”

has

avery

precise

andrigo

rousmeaningin

mathem

atics2

Itturnsou

tthat

afterachan

geof

coordinates(nam

elyadiago

nalizationargu

mentforthe

Hessian

),aMorse

functionnearacritical

pointpalwayshas

theform

:

f(x,y)=

f(p)+

λ1x2+

λ2y2,

andin

fact

byrescalingx,y

youcanassumetheλjare±1(butyo

ucannot

getridof

the

sign

sbycoordinatechan

ges,

thoseareinvarian

ts).

Sotherearenoother

critical

points

near

p(locallypcorrespon

dsto

(x,y)=

0).Thereforethecritical

points

ofaMorse

functionfare

isolated,so

onacompactsurfacethereareon

lyfinitelyman

ycritical

points.

Definition

15.2

(Gradientvectorfield).

Forasm

ooth

function

f:S

→R,thegradient

vectorfield∇f∈TS

isdefi

ned

3by

theequation

I(·,

∇f)=

df,

whereIis

theRiemannianmetric(firstfundamen

talform

).4

Lemma15.3.

(1)∇fis

orthogonalto

thelevelse

tsf=

constant(thecontourlines),

(2)thepoints

where∇f=

0are

precisely

thecriticalpoints

off,

(3)fincreases5

inthedirectionof∇f.

1Youca

nbypass

findingth

eeigen

values

byfirstfindingdet

Hess=

λ1λ2:ifitis

neg

ativeyouhaveasaddle,

oth

erwiseit

isamax/min.In

themax/min

case:if∂xxf>

0it

must

beamin,if∂xxf<

0it

must

beamax.

2Non-exa

minable:

ABaire

setis

asetwhichco

ntainsaco

untable

intersectionofden

seopen

sets.The

Bairecate

gory

theorem

saysth

atin

aco

mplete

metricsp

ace

everyBairesetis

den

se.Intu

itivelyyouca

nth

inkofBairesetasroughly

mea

ning“ev

eryth

ingoutsideofamea

sure

zero

set”,so

youhave“probability1”

thatapointlies

inth

eBaireset.

Ittu

rnsoutth

atoneca

nputatopologyonallsm

ooth

functionsso

thatth

e

Morsefunctionsform

aBairesu

bset.

3Rem

ark.Exp

licitly,locallyA∇f=

�∂xf

∂yf

�=

∇Euclf

wheref=

f(x

,y)in

localcoordinates,

A=

the

localmatrix

forI,and∇

Euclf

=theEuclidean

gradientyou

are

usedto

(thefirstpa

rtialderivatives).So

(∇f) local=

A−1∇

Euclf

∈R2

=TV.Mappingby

DF

wegettheresu

ltingvectorfield

inTS,so

∇f

=

DF(∇

f) local=

DFA

−1∇

Euclf

∈TS.

4Forex

ample:evaluatingonth

ebasisvectorX

1=

∂xF,weget

I(X

1,∇

f)=

df(X

1)=

∂xfloca

lly.

5Indeed,∇f/�∇

f�is

thedirection

ofmaxim

alincrea

seforf

since

|df(v)|

=|I(v,∇

f)|

≤�∇

f�forunit

vectors

v,byCauch

y-Sch

warz,anddf(∇

f/�∇

f�)

=�∇

f�ach

ieves

equality.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

73

Proof.

(1):

foravector

vtangentto

thelevel

setthefunctionfdoes

not

vary,so

df(v)=

0.Then

bydefinitionI(v,∇

f)=

df(v)=

0,so

vis

orthog

onalto

∇f.(2)holdsbydefinition.

For

(3):

df(∇

f)=

I(∇

f,∇

f)=

�∇f�2

≥0so

fincreasesin

thedirection

of∇f.

For

asurfaceS

⊂R

3,it

turnsou

tthat

alinearfunctional

f:R

3→

R,sayf(p)=

p·v

(for

somefixed

v∈

R3)is

aMorse

functionwhen

restricted

toS

foralmostallchoicesof

v(butnot

allchoices:

v=

0is

bad

).Noticethat

afterrotatingR

3to

mak

ev=

(0,0,1),

youcanthinkof

thesefunctionsas

just

measuringtheheightfunctionZ

forthesurface.

So

generically

(that

is,afterasm

allgeneric

rotation

ofthesurface

ifnecessary),

thefunction

f(X

,Y,Z

)=

Zbecom

esaMorse

function.

Example.For

ourusual

torusT

2in

R3,thefunctionf(X

,Y,Z

)=

Zhas

acircle

ofmaxim

aandacircleof

minim

a.Since

thesecritical

points

arenot

isolated,f:T

2→

Risnot

Morse.But

slightlyrotatingT

2makes

itMorse:

Noticeab

ovethat

#(m

inim

a)−#(sad

dles)+#(m

axim

a)=

0=

χ(T

2).

Isthisacoincidence?

Let’s

checkforthesphere:

nomatterhow

much

you

deform

thesphere,

ageneric

heigh

tfunctionseem

sto

alwayshave#(m

inim

a)−

#(sad

dles)

+#(m

axim

a)=

2=

χ(S

2):

15.2

Criticalpoints

ofaM

orsefunction

recoverth

eEulerch

ara

cteristic

Theore

m15.4.ForanyMorsefunctiononacompact

orien

tedsurface

S,

χ(S

)=

#(m

inim

a)−

#(saddles)

+#(m

axima).

Non-examinableProof1:Sketch.Thehigh-techproof

ofthis,is

toshow

that

aMorse

func-

tion

gives

rise

toacellulardecom

positionof

S(upto

hom

otop

y1),with0-cells,1-cells,2-cells

1Youca

nignore

this

tech

nicalissu

e.Oth

erwiseseeth

efootn

ote

toTheo

rem

4.2.

74

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

correspon

dingbijectivelyto

minim

a,saddlesandmaxim

a.Indeed,each

criticalpointcorre-

spon

dsto

theattachmentof

anew

cell:

fr

attach

ahandle

1-cell

1-handle

Mr+ε

Mr−ε

whereM

r=

{p∈S:f(p)≤

r}is

thesu

blevelset.

Proof2.ByTheorem

14.7,weneedto

show

� SK

dA

=2π(#

(min

andmax)−#(saddles)).

Since

fis

Morse,thecritical

points

are

isolated.Let’s

callm

i,s j,M

k∈

Stheminim

a,

saddles,

andmax

ima.

Picksm

alldisjoint“discs”(inaparametrizationpatch)aroundeach

critical

point,callthesediscs

Di,D

j,D

kandcalltheanti-clockwiseboundary

curves

γi,γj,γ

k.

Sobytheproof

ofTheorem

14.5,letting�runover

allindices

i,j,k,

� �

� D�

KdA

=� �

�θ� �(t)dt−� γ

κgds

whereθ �

isthean

glebetweenv �

andγ� �,andv �

iscorrespondsto

thenorm

alizedfirststandard

basis

vectorin

thelocalcoordinates.

OnthecomplementC

=S\(

thosediscs)wecannotuse

Theorem

14.5

directlyasit

may

not

liewithin

aparam

etrization

patch.How

ever,weknow

how

tobuildaunit

vectorfield:

v c=

∇f/�∇

f�

(note∇f�=

0on

Cas

thecritical

points

lieoutsideC).

Then,asusual,definew

c=

n×v c

tomakev c,w

can

orthon

ormal

basis.Thecalculationin

theproofofTheorem

14.5

stillholds:1

� C

KdA

=−� �

�θ� c

,�(t)dt+

� γ�

κgds

usingthat

theγ�areclockwisebou

ndary

curves

forC,andwhereθ c

,�istheangle

betweenv c

andγ� �.Summingup:

� S

KdA

=� �

� D�

KdA+

� C

KdA

=� �

�(θ

�−θ c

,�)�(t)dt.

Observethat

thean

gledifference

θ �−

θ c,�

equals

theangle

betweenv �

andv c.

Thetotal

chan

gein

this

angle,

asyoutravel

arou

ndγ�,is

aninteger

multiple

of2π.How

ever,because

wedon

’tknow

theRieman

nianmetric(firstfundamentalform

)wedon’t

know

what∇fis

(hen

cewedon

’tknow

v c),so

it’s

hardto

calculate

this

angle.Weuse

adeform

ationtrick:

Ifwedeform

γ�,then

this

integermust

vary

continuously,

butbeinganinteger

itmust

beconstan

t.Bythesameargu

ment,

wemay

alsodeform

themetric:

again

this

integer

must

beconstan

t.Noticein

general

thatif

welinearlyinterpolate

twoinner

products,

sot�·

,·�1+

(1−

t)�·,

·� 2for0

≤t≤

1,then

itstillsatisfies

bilinearity,symmetry,positive

definiteness,

soit

isstillan

inner

product.Sowemay

deform

themetricto

thestandard

1Both

signsonth

erighthandsideare

theopposite

signsofth

ose

foundin

Step5ofth

eproofofTheo

rem

14.5.

This

isbecause

γ�is

aclock

wisecu

rveboundingC,butGreen

’sth

eorem

requires

an

anti-clock

wise

curve.

Soin

Step3ofth

atproof,weget

� CK

dA

=−

�∇

tw

·vdtwithaminussign.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

75

Euclidean

metricin

localcoordinates.

Then

(∇f) localjust

becom

estheusual

Euclidean

∇Euclf.Since

wecanchooselocalcoordinates

sothat

f(x,y)=

f(p)+

λ1x2+

λ2y2,

weget

(∇f) local=

(2λ1x,2λ2y)an

dweknow

v �=

(1,0)is

thefirststan

dardbasis

vector

locally.

Sowejust

needto

understan

dhow

man

ytotalrotationsv c

undergo

eslocally(using

Euclideaneyes)as

wemovealon

gγ�.This

integer

iscalled

theindex

ofth

evecto

rfield

∇faroundthezero

of∇f(thecritical

pointof

f).

Wemay

assumethat

thecurveγ�is

locally(x,y)=

(cos

t,sint)

andso

alon

gγ�:

v c=

1�λ2 1+

λ2 2

�λ1cost

λ2sint

Ifλ1,λ

2arebothpositiveor

bothnegative,

then

thevector

v cwillsw

ingarou

ndon

cean

ti-

clockwise(noticethat

aglob

alminussign

infron

tof

thevector

v cwou

ldbethesameas

chan

gingtto

t+

π,so

itstillrotatesan

ti-clockwise).

Ifλ1,λ

2haveoppositesign

s,then

chan

gingtto

−twou

ldsw

itch

thesign

ofsintab

oveandbringusbackinto

thesituation

whereλ1,λ

2haveequal

signs,so

inthiscase

v csw

ings

clockwisearou

ndonce.Sominim

aan

dmax

imaeach

contribute

+1times

2π,whereassaddlescontribute

−1times

2πto

� SK

dA.

15.3

Indicesofvecto

rfields,

Poincare

-Hopfand

hairyball

theorems

This

Section

isnon-examinable.

Within

thepreviousproof,wedefined

theindex

ofthevectorfield∇f,butthedefinition

worksforan

yvector

fieldvon

S.Given

anisolated

zero

pof

v,pick(right-han

ded

)local

coordinates

nearp,pickasm

allcircularan

ti-clockwisepathγarou

ndpin

thelocalcoordi-

nates,andcounthow

man

ytimes

thevector

fieldvsw

ings

anti-clockwisearou

ndas

you

travel

arou

ndγ(itcounts

as−1times

ifit

swings

clockwise).

Rem

ark.Let’scheckthattheindex

doesnotdepen

dontheobserver.

Inanother

parametriza-

tion,thevectorfieldbecomes

Dτ(v)(w

ithdet

Dτ>

0since

weuse

right-handed

parametriza-

tions).Countingthesw

ings

inthis

parametrizationis

thesameasmeasuringtheanglein

theoriginalparametrization

between

vand

Dτ−1(e

1)(rather

than

e 1=

(1,0)).

Butthe

76

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

(non-vanishing)

vectorfieldDτ−1(e

1)is

defi

ned

onthewhole

parametrization,andwecan1

continuouslydeform

itinto

thevectorfielde 1.Since

theindex

isanintegerwhichdepen

ds

continuouslyonthecurve,

theindex

willnotchange

under

deform

ations.

Theore

m15.5

(Poincare-H

opftheorem).

Given

anycompact

orien

tedsurface

S,andany

vectorfieldvwithisolatedzeros,

χ(S

)=

p∈S

with

v(p

)=0Index

v(p).

Proof.

Thesameproof

asforTheorem

15.4

applies,usingthevector

fieldvonC.

Rem

ark.This

theorem

holdsforanycompact

orien

tedmanifold,iteven

holdsifthemanifold

hasaboundary

provided

thevectorfieldpoints

orthogonallyoutwardsalongtheboundary.

Example.For

thetorusT

2∼ =

S1×S1,thereisanon

-vanishingvector

fieldwhichpoints

alon

gthedirection

ofon

eof

thetwocircle

factors(e.g.pointingin

thelatitudinal

direction

).Soithas

nozeros,so

χ(T

2)=

0.

Coro

llary

15.6

(hairy

balltheorem).

Thereis

nononvanishingcontinuousvectorfieldon

thesphereS2.More

inform

ally:

ifyouattem

ptto

combahairyballto

make

thehair

flat

(tangentto

thesurface),

therewillalways

beatleast

onetuftofhair

somew

here.

Proof.

Ifthevector

fieldhad

nozero,then

byPoincare-H

opf:

χ(S

2)=

0whichis

false.

16.

Geodesics

16.1

Differe

ntiatingvecto

rfieldsdefined

alongacurv

e

Let

w=

w(t)beavector

fieldon

Sdefined

only

alon

gacurveγ(t)∈S,so

w(t)∈Tγ(t)S.

Recallthat

wedefined

∇tw

tobetheorthog

onal

projectionon

toTS:

∇tw

=∂tw−

(n·∂

tw)n.

Wewou

ldexpectthat

theop

erator

∇tsatisfies

thechainrule

∇t=

x� ∇

x+

y� ∇

y

whereγ�=

x� ∂

xF

+y� ∂

yF,usingalocalparam

etrization

F(x,y)(locallyγ� loc=

(x� ,y� )

and

γ�=

DF(γ

� loc)).

Then

wecan

thinkof

∇tas

thetangential

direction

alderivativein

the

direction

γ� ,in

fact

oneoftenalso

writes2

∇γ�to

mean∇

t.How

ever,strictly

speaking∇

xw

isnot

defined

,since

wisavector

fieldon

lydefined

alon

gthecurveγ,so

wecannot“see”

how

wvaries

inthex-direction

,weon

ly“see”how

wvaries

intheγ�direction

.ThenextLem

ma

clarifies

this

isnot

aproblem

(andputs

iton

arigo

rousfootingeven

forabstract

surfaces).

1Fix

areference

point,

andfrom

theremoveoutw

ard

sradiallyandslowly

undoth

etw

istin

Dτ−1e 1

,th

iswilldeform

Dτ−1e 1

into

aco

nstantvectorfield.

2Notice

thatγ�is

notaloca

lvectorfieldsince

itis

only

defi

ned

alongth

ecu

rveγ(t),

sowehaven

’tactually

defi

ned

∇γ�.

Howev

er,since

∇XY

istensorialin

X,th

isis

notan

issu

e.Indeed,letZ

beanyvectorfield

defi

ned

nea

rth

ecu

rvewhichrestrictsto

Z| γ

(t)=

γ� (t).Then

weca

ndefi

ne∇

γ�Y

=(∇

ZY)| γ

(t)evaluating

atγ(t),

and

then

wech

eck

thatth

iseq

uals

�cj(t)(∇

XjY)| γ

(t)indep

enden

tly

ofth

ech

oiceofZ.

Indeed

(∇ZY)| γ

(t)=

∇�

Zj(γ

(t))X

jY

=�

Zj(γ

(t))

(∇X

jY)| γ

(t)=

�cj(t)(∇

XjY)| γ

(t).

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

77

Lemma16.1

(Chainrule).

Let

vbe

anyvectorfieldonS

defi

ned

inaneighbourhoodofthe

curveγ⊂

S,whichextends1

w(t),

i.e.

satisfies

v(γ(t))

=w(t).

Then

1

∇tw

=∇

γ� v

=�

cj∇

jv

whereγ� (t)

=�

cjX

jin

thebasisX

j=

∂jF.

Proof.

γ� (t)

=�

cj(t)X

j| γ(

t)in

thebasis

Xj=

∂jF

evaluated

atγ(t).

Bythechainrule,

∂tw

=∂t(v(γ(t))

=Dv(γ

� (t))=

�∂1v

∂2v

��c1 c2

�=

�cj∂j(v).

So∇

tw

=∂tw−(∂

tw·n

)n=

�cj∂jv−�

cj(∂

jv·n

)n=

�cj∇

Xjv=

∇�

cjX

jv=

∇γ� v.

Remark

16.2.This

Lem

maallowsusto

defi

ne∇

γ�alsoforabstract

smooth

surfacesby

turningtheLem

mainto

aDefi

nition.So∇

tw

or∇

γ� w

means∇

γ� v

foranextensionvas

above,andonechecks

thatthechoiceofextension

vdoesnotmatter.3

Indeed,if

w(t)=

�hk(t)X

k| γ(

t)then

theLeibn

izrule

holds:

∇tw

=�

∂t(h

k)X

k| γ(

t)+

�hk∇

tX

k| γ(

t)where

∇tX

k| γ(

t)=

�cj∇

jX

k| γ(

t)(andthelatter

inturn

iskn

ownsince

∇jX

k=

�Γi jkX

i).

16.2

Equivalentdefinitionsofageodesic

Definition

16.3

(Geodesic).

Asm

ooth

curveγin

Sis

ageo

desicif

∇γ� γ

�=

0

RecallbySection

14.1,ageodesic

isasm

oothcurveγin

Swhichlook

s“straigh

t”from

thepointof

view

ofS,an

dthat

anyof

thefollow

ingconditionsareequivalent:

(1)κg=

0:thegeodesic

curvature

vanishes

(recallκg=

±�∇

tγ� �),

(2)∇

tγ�=

0:thetangential

acceleration

vanishes,

(3)∇

γ� γ

�=

0(usingLem

ma16

.1),

(4)γ��·T

S=

0:theacceleration

�is

normal

toS

(for

γparametrizedbyarc-length),

(5)�=

II(γ

� ,γ� )nwhen

γis

param

etrizedbyarc-length,

(6)γ=

reparam

etrization

ofacurve�γwith�γ�,�γ

�� ,nlinearlydep

endent(C

orollary

14.3).

Rem

ark.Defi

nitions(2)and(3)donotrequirethesurface

Sto

beem

bedded

inR

3:itsuffices

thataRiemannianmetricg i

jis

defi

ned

onS.Indeed,in

ExerciseSheet2youfoundaform

ula

forΓk ij

interm

sofg i

j,this

inturn

defi

nes

∇XY,whichin

turn

defi

nes

∇γ�and∇

t.

Lemma16.4.Geodesicsalways

have

constantspeed:�γ

� (t)�is

constant.

Proof.

∂t�γ

� (t)�2

=∂tI(γ

� ,γ� )=

I(∇

tγ� ,γ� )+

I(γ

� ,∇

tγ� )=

2I(∇

tγ� ,γ� )=

0.�

16.3

Thegeodesicequation

inlocalcoord

inates

Asin

Lem

ma16

.1,

γ� (t)

=�

cj(t)X

j| γ(

t)

inthebasis

Xj=

∂jF

ofTS

evaluatedat

γ(t).

BytheLeibniz

rule,

∇t(�

cjX

j)=

�∂t(c

j)X

j+�

cj∇

tX

j.

1Note,byth

isLem

ma,th

etangen

tialderivative∇

γ�v

does

notdep

endonth

ech

oiceofex

tensionvforw.

3UsingChristoffel

symbols,reca

ll∇

iX

j=

�Γk ijX

k,whereX

kis

abasisforth

etangen

tsp

ace.

Write

v=

�fkX

kin

thatbasis,

wherefkare

functions.

Byco

nstru

ction,w(t)=

�(f

k◦γ(t))

Xk| γ

(t).Byth

e

Leibniz

rule,∇

j(f

kX

k)=

fk∇

jX

k+

∂j(f

k)X

k.Byth

eusu

alch

ain

rule,�

cj∂j(f

k)=

∂t(f

k◦γ).

Thus

�cj∇

jv=

�∂t(f

k◦γ

)Xk+

��

fkcjΓi jkX

iwhichdoes

notdep

endonv.

78

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Bythechainrule

(Lem

ma16.1),summingover

repeatedindices,1

∇tX

j=

�ci∇

iXj=

�ciΓk ijX

k.

Hence

thegeodesic

equation∇

tγ�=

0becomes

∂tck

+�

Γk ijcicj

=0.

Abbreviatingcj

=xj,usingon

e“d

ot”foreach

timederivative,

weobtain:

Coro

llary

16.5.In

localcoordinatesγ(t)=

(x1(t),x2(t))

∈V

⊂R

2thegeodesic

equationis:

xk+�

Γk ijxi x

j=

0(16.1)

ExerciseIn

localcoordinates,letA

=�

ef

fg

�bethematrix

forthefirstfundamentalform

andγloc=

(x(t),y(t))

acurveparam

etrizedbyarc-len

gth,so

theequation(x�

y�)T

A(x�

y�)=

1

holds,

soexplicitly

ex�2+2f

x� y

� +gy�2=

1.Thegeodesic

equationis:

d dt

� A� x

y�

��=

1 2

� x�

y�

� T(∂

xA)� x

y�

� x�

y�

� T(∂

yA)� x

y�

ormoreexplicitly:

d dt(ex� +

fy� )

=1 2(x

�2∂xe+2x� y

� ∂xf+

y�2∂xg)

d dt(fx� +

gy� )

=1 2(x

�2∂ye+2x� y

� ∂yf+y�2∂yg)

Proof:2γisageodesic

precisely

if�isnorm

al,i.e.

perpendicularto

X1,X

2.Thefirstofthe

twoorthogon

alityequations(theother

issimilar)

is

0=

γ��·X

1=

d dt(γ

� ·X

1)−γ� ·

d dt(X

1).

Since

γ�=

x� X

1+y� X

2,thefirstterm

isd dt(γ

� ·X

1)=

d dt(ex� +

fy� )(theleft

handsideofthe

firstequationin

theab

ovebox).

Thesecondterm

,bythechain

rule,is

γ� ·

d dt(X

1)=

(x� X

1+y� X

2)·(x� ∂

xX

1+y� ∂

yX

1)=

(x� ∂

xF

+y� ∂

yF)·(x� ∂

xxF

+y� ∂

yxF).

Now

∂xF

·∂xxF

=1 2∂xe,

∂xF

·∂yxF

+∂yF

·∂xxF

=∂xf,∂yF

·∂yxF

=1 2∂xg.

Example

16.6.When

I=

(G

00

1),

theform

ulasgive:

d dt(G

x� )=

Gx��+

∂xGx�2+

∂yGx� y

�=

1 2x�2∂xG

and

d dt(y

� )=

y��=

1 2x�2∂yG.Soforx(t)=

constantwecanfindasolution:y(t)=

p+ct

forconstants

p,c.

Exercise.(H

arder)Usingtheab

oveform

ulas,

findthegeodesicsin

theupper-halfplaneH.

1reca

llth

atsince

∇iX

j∈

TS,weca

nwrite

∇iX

jin

thebasisX

jandweca

llth

eco

efficien

tfunctionsth

e

Christoffel

symbols:∇

iX

j=

�Γk ijX

k.

2Exercise:Fillin

thedetailsin

thefollowingeq

uivalentproof.

Forγ

parametrized

byarc-len

gth

isa

geo

desic

ifandonly

if�is

norm

al,i.e.

�is

orthogonalto

TS=

span(∂

xF,∂

yF).

This

iseq

uivalentto:

DF

T�=

0.

Since

γ�=

DFγ� locandA

=DF

TDF,sh

ow

thatth

iseq

uationca

nberewritten

as:

d dt(A

γ� loc)=

(d dtDF

T)D

Fγ� loc.

Now

just

dotwith(1,0

)and(0,1

)to

obtain

theaboveeq

uations,

usingth

ech

ain

rule

∂t=

x� ∂

x+

y� ∂

yand

trickslike∂xyF·∂

xF

=∂yxF·∂

xF

=1 2∂y(∂

xF·∂

xF)=

1 2∂ye.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

79

16.4

Existenceand

uniqueness

ofgeodesics

Theore

m16.7.Given

apointp∈

Sandatangentdirection

v∈

TpS,thereis

aunique

geodesic

γp,v:(−

ε,ε)

→S,defi

ned

forsomeε>

0,through

pwithtangentvectorvatp:

γp,v(0)=

pan

dγ� p,v(0)=

v.

Moreover,

γp,v

depen

dssm

oothly

ontheinitialconditionsp,v.

Proof.

xk+

�Γk ijxi x

j=

0areasystem

ofODE’s,therefore(see

AnalysisHan

dou

t):

asolution

exists,it

isunique,

andit

dep

endssm

oothly

ontheinitialconditions.

�Example.

For

S=

R2⊂

R3,usingF(x,y)=

(x,y,0),

wehaveX

1=

(1,0,0)andX

2=

(0,1,0),

so∇

iXj=

0,so

allΓk ij=

0,so

thegeodesic

equationis

xj=

0forj=

1,2.

So(x

1,x

2)=

(p1+

v1t,p2+

v2t)

=p+

tvisastraightlinethrough

palon

gv.

Example.Recallgreatcirclesin

theunitspherearegeodesics.

Con

versely,givenapointp∈S2

and

adirection

v∈

TpS2,thereis

agreatcircle

through

pin

thedirection

v(nam

ely,

the

intersection

oftheplanespan

(p,v)withS2,or

moreexplicitly:rotate

R3so

that

p,v

becom

ep=

(1,0,0)andv=

(0,1,0),then

thegeodesic

istheequator).

So,

byuniqueness,allgeodesics

aregreatcircles.

Remark

.Onecan

show

that

on

acompactsurface,

thegeodesic

isdefined

foralltime:

γp,v:R

→S.In

thenon

-compactcase,this

may

fail.1

Theore

m16.8

(See

C3.3:Differe

ntiable

Manifolds).

Geodesicslocallyminim

izelengthsofcurves

Conversely,

givenapointp∈S,foranypointqsufficien

tlyclose

topthereis

auniquecurve

γfrom

pto

qwhichachievestheminim

um

infL(γ)(takingtheinfimum

oversm

ooth

curves

γfrom

pto

q),andthis

minim

izer

isageodesic.

Example.Theshortest

pathbetweentwopoints

onasphereisthe(short)

arcof

thegreatcircle

through

thosetwopoints.Geodesicsmay

not

beglob

allengthminim

izers:

thefullgreatcircle,

from

theNorth

Poleto

theSou

thPoleandthen

further

totheNorth

Pole,

isageodesic

butof

coursetheconstantgeodesic

attheNorth

Poleistheshortest

pathfrom

theNorth

Poleto

the

North

Pole!

Cultura

lre

mark

.In

physics,

youdeclare

thatlightrays

move

alongshortestpaths(locally).

Solightrays

move

alonggeodesics.

Thus,

itis

nosurprise

thatageodesic

isdetermined

byinitialpositionandvelocity:thinkofpointingaflashlightfrom

apositionpin

adirectionv.

16.5

Examplesofgeodesicsvia

symmetriesand

isometries

Lemma16.9.Anyplaneofsymmetry

locallyintersects

asurface

S⊂

R3in

ageodesic.

Proof.

Thesurface

issymmetricab

outtheplaneP,so

also

thenormal

vector

must

be,

son∈P.Let

γbealocalcurvewhereP

intersects

S,an

dparametrize

γbyarc-length.2

Then

1Forex

ample,forth

enon-compact

surface

R2\{

0}th

estraightlineγ(−

1,0),(1

,0)(t)=

(−1+t,0)is

defi

ned

fort∈

(−∞

,1)butnotatt=

1since

(0,0

)does

notofficiallybelongto

R2\{

0}.

2TechnicalRem

ark.

You

may

worry

thatγ

isnotaregularcu

rve,

i.e.

thatth

eremay

bepoints

with

γ�=

0,in

whichca

seweca

nnotreparametrize

byarc-len

gth

.Suppose

γ� (t 0)=

0.Let

T=

Tγ(t

0)S

beth

e

tangen

tplaneatth

atstationary

point.

AsS

issymmetricaboutP,alsoT

must

be.

TheintersectionP

∩T

ofth

ose

twoplanes

isastraightline.

ByTheo

rem

9.1,weca

nuse

Tto

buildaloca

lparametriza

tionF

for

80

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

γ�∈

Pan

d�is

orthog

onal

toγ�(differentiateγ� ·γ�=

1).

Asγ

isinvariantunder

the

symmetry,so

areγ� ,γ�� ,

soγ��∈

P.

Son,γ

��∈

Parebothorthog

onalto

γ� ,

sothey

are

proportion

al.

Example.For

asurfaceof

revolution

S,anyplaneP

through

theaxisof

revolution

isaplaneof

symmetry.Soiftheaxisof

revolution

istheZ-axis,thecurves

running“vertically”aregeodesics.

For

thespherethecurves

heading“vertically”towardstheNorth

Polearegeodesics:

indeedthey

aremeridiangreatcircles.

Theore

m16.10.If

two

surfacesare

locally

isometric,

then

they

have

locally

the

same

geodesics.

Inparticular,

isometries

ϕ:S1→

S2mapgeodesicsto

geodesics:

1

ϕ◦γ

p,v=

γϕ(p

),Dϕ(v

).

Proof.

Thegeodesic

equationon

lydep

endson

thetangential

derivative,

whichonly

dep

ends

onthefirstfundam

entalform

.�

Example.For

thecylinder

X2+Y

2=

a2,F(x,y)=

(acosy,a

siny,x

)givesfirstfundam

ental

form

I=

dx2+

a2dy2whichis

thesameas

thefirstfundam

entalform

fortheplaneR

2with

F(x,y)=

(x,ay).

Sothecylinder

islocally

isom

etricto

theplane.

Thegeodesicsfortheplane

arestraightlines

(x(t),y(t))

=p+

tv,so

each

non

-con

stantgeodesic

onthecylinder

isahelix:

F(p

+tv)=

(acos(p2+

tv2),asin(p

2+

tv2),p1+

tv1).

Averyusefulgeneralizationof

Lem

ma16

.9,butessentially

based

onthesameidea,is:

Theore

m16.11(Symmetry

implies

geodesic).

Ifϕis

alocalisometry

ofanabstract

smooth

surface,such

thatlocallythefixedpoints

2ofϕ

form

acurveγ(t)withγ��=

0then

γbecomes

ageodesic

after

arc-len

gthreparametrization.

Proof.

γis

fixed

byϕ,so

span

(γ� )⊂

TS

isfixed

3byDϕ.Since

locallyϕ

isnottheidentity

map

,thefixed

locusof

Dγ(t)ϕisprecisely

span

(γ� (t)).

Since

theisom

etry

ϕfixes

γitfollow

s4

that

Dϕ(∇

tγ� )

=∇

tγ� .

Hence

∇tγ�∈

span

(γ� )

⊂TS.Butafterreparametrizingγbyarc-

length,wehave∇

tγ� ·γ�=

0(from

differentiatingγ� ·γ�=

1).So∇

tγ�is

both

proportional

andperpendicularto

γ�in

thetw

o-dim

ension

alspaceTS,so

∇tγ�=

0.Soγisageodesic.

Example.For

theEuclideanplaneR

2,thereflection

inastraightlinepreservesthedot

product,

soitpreservestheRiemannianmetric.

Sostraightlines

aregeodesics.

Example.For

thehyp

erbolicplaneH,weuse

thehyp

erbolicmetric

dx2+dy2

y2

from

Section

10.9.

Snea

rγ(t

0).

Bypickingaregularparametriza

tionofth

estraightlineP

∩T

⊂T,th

eim

agevia

Fwillbea

regularcu

rvein

Sandwejust

use

this

curveinstea

dofγin

theproof.

(Toclarify:th

iscu

rveandγhaveth

esamegeo

metricim

agein

Sbutth

eyare

parametrizeddifferen

tly,

aswegotridofth

estationary

points).

1usingth

enotationfrom

Theo

rem

16.7.

2so

ϕ◦γ(t)=

γ(t),

andϕ(p)�=

pforpclose

toth

eloca

lcu

rveγ

unless

p=

γ(t)forsomet.

Ofco

urse,

wedon’t

wantto

allow

theiden

tity

mapϕ=

idwhichwould

tellusnoth

inginteresting.Notice

thatyouca

nth

inkofϕ

asth

ereflectionin

γ,loca

lly.

3γ�=

∂t(γ

)=

∂t(ϕ

◦γ)=

Dϕ◦γ

� .4Note

thatvia

anisometry,tw

osu

rfaceswithRiemannianmetrics

are

toallintents

andpurp

osesiden

tica

l.More

ped

antica

lly:if

twopatches

ofsu

rfacesare

isometricvia

ϕ:S1→

S2,th

enif

F:V

→S1is

aloca

lparametriza

tion,th

enϕ◦F

isaloca

lparametriza

tionforS2,su

chth

atϕin

loca

lco

ord

inatesjust

becomes

the

iden

tity

map(indeedweuse

thesameloca

lco

ord

inatesx,y

)andth

eRiemannianmetricgij

written

loca

llyis

thesame.

Soth

ecu

rves

γin

S1andϕ◦γ

inS2are

thesamecu

rveγlo

c(t)=

(x(t),y(t))

inloca

lco

ord

inates.

Hen

ce∇

tis

defi

ned

inth

esamewayforboth

surfaces,

andso

∇tγ�is

thesame,

usingLem

ma13.1

that∇

t

only

dep

endsonth

eRiemannianmetric(InExercise

Sheet2youfoundanex

plicitform

ula

forΓk ij

interm

s

ofth

emetricgij,whichdetermines

∇X

iX

j,whichdetermines

∇tbySection16.1).

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

81

Noticethat

thestraightvertical

linex=

0(w

hichisorthogon

alto

thereal

axis)isfixedby

the

reflection

isom

etry

ϕ:H

→H,ϕ(x,y)=

(−x,y)(indeed,put�x=

−x,�y=

y,then

d�x2

=dx2,

etc.).

Sim

ilarly,

forastraightlinex=

c,usingthereflection

ϕ(x,y)=

(2c−

x,y).

Sovertical

straightlines

inH

arehyp

erbolicgeodesics.

For

straightlines

whicharenot

vertical,thisdoes

not

work,

since

thecorrespon

dingreflection

swill

change

yandthemetricwill

noticesuch

changes.

17.

Geodesicnormalcoordinates

17.1

Geodesicnorm

alcoord

inates

Theorem

17.1.Given

p∈S,wecanbuildaright-handed

localparametrizationF(x,y)near

p,such

thatc(x)=

F(x,0)is

ageodesic

andy�→

F(x,y)is

ageodesic

orthogonalto

it.

Proof.

Pickan

ynon

-zerovectorv 0

∈TpS

atp.Let

c(t)

=γp,v

0(t)as

inTheorem

16.7.Let

w(x)=

n×c�(x)=

(c� (x)rotatedby90

degrees).

WedefineF(x,y)to

bethegeodesic

through

c(x)withinitialvelocity

w(x):

F(x,y)=

γc(x

),w(x

)(y)

Bydefinition,thevelocity

vectors

oftherespectivegeodesicsare:

∂xF| y=

0=

c�(x)

∂yF| y=

0=

w(x).

So∂xF,∂

yF

arelinearlyindep

endentat

y=

0(indeedorthogon

al)so

they

must

belinearly

indep

endentalso

forsm

allycloseto

0.Hence

Fisalocalparametrization

forsm

allx,y.

�Coro

llary

17.2.If

wepickv 0

=c�(0)to

have

unitlength,then

fortheabove

F(x,y),

I=

G(x,y)dx2+dy2

forthesm

ooth

functionG(x,y)=

�∂xF�2

>0.

Proof.

Ifwepickv 0

tobeaunit

vector,then

byLem

ma16.4,

�c� (x)�

=�v

0�=

1.

Byconstructionw(x)=

n×c�(x)has

unitlength(asn,c

�areorthogonal),so

byLem

ma16.4,

�∂yF(x,y)�

=�∂

yF(x,0)�

=�w

(x)�

=1.

82

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Wenow

show

∂xF,∂yF

areorthog

onal,i.e.

that

I(∂

xF,∂

yF)=

0.First

wecompute

∂yI(∂

xF,∂

yF)=

I(∇

y∂xF,∂

yF)+

I(∂

xF,∇

y∂yF).

Now

∇y∂yF

=0since

Fis

ageodesic

iny.For

theother

term

:

∇y∂xF

=∂y∂xF

−(n

·∂y∂xF)n

=∂x∂yF

−(n

·∂x∂yF)n

=∇

x∂yF.

SoI(∇

y∂xF,∂

yF)=

I(∇

x∂yF,∂

yF)=

1 2∂xI(∂

yF,∂

yF)=

1 2∂x�∂

yF�2

=1 2∂x(1)=

0.Com-

biningtheab

ove,

∂yI(∂

xF,∂

yF)=

0,thusI(∂

xF,∂

yF)is

constan

tin

yandhence

equalto

zero

(since

∂xF,∂

yF

areorthog

onal

aty=

0,byou

rchoice

ofc�,w

).�

Warn

ing.F(x,y)is

typicallynotageodesic

inthex-coordinate

(fory�=

0),

otherwisewe

would

have

I=

dx2+

dy2,so

Swould

beisometricto

aplane,

soK

=0.

Intheab

oveproof,weshow

edthat

∇y∂xF

=∇

x∂yF

This

isageneral

symmetry

property

that

youprovedin

Exercise

Sheet2:Γk ij=

Γk ji.

Definition

17.3

(Geodesic

normal

coordinates).

Wecalltheabove

coordinatesx,y

geodesic

norm

alcoordinates(takingc�

ofunitlength,so

I=

Gdx2+

dy2).

17.2

TheGaussian

curv

atu

rere

visited

Theore

m17.4.If

I=

Gdx2+

dy2locally(forexample

ingeodesic

norm

alcoordinates),

K=

−∂yy

√G

√G

.

Proof.

Claim

1.Thecurves

(x=

constan

t)aregeodesics,

soF(x,y)is

ageodesic

iny.

Proof.

See

Exam

ple

16.6.Wemay

assumeF(x,y)is

arigh

t-han

ded

parametrization.

Claim

2.Wecancalculate

theRieman

ncurvature

intw

oway

s.Proof.

First,byLem

ma13

.3,

(∇x∇

y−

∇y∇

x)∂

yF

=−K�det

I Fn×

∂yF

=−K√G

∂yF.

Since

∂xF,∂

yF,n

areorthog

onal,n×

∂yF

isparallelto

∂xF,an

dtakinginto

accountorien

-tation

san

dlengths(�∂xF�2

=G,�∂

yF�2

=1):

∂yF

=−∂xF/�

∂xF�=

−1 √G∂xF.

Secon

dly,∇

y∂yF

=0since

Fis

ageodesic

iny,an

dusing∇

x∂yF

=∇

y∂xF:

(∇x∇

y−

∇y∇

x)∂

yF

=−∇

y∇

y∂xF.

Write

∇y∂xF

=a∂xF+b∂

yF

inthebasis∂xF,∂

yF

forTS.Usingthat

I=

Gdx2+dy2,and

that

the∂jF

areorthog

onal

ton:

Ga=

∂xF·∇

y∂xF

=∂xF·∂

y∂xF

=1 2∂y(∂

xF·∂

xF)=

1 2∂yG,

b=

∂yF·∇

y∂xF

=∂yF·∂

y∂xF

=∂yF·∂

x∂yF

=1 2∂x(∂

yF·∂

yF)=

1 2∂x(1)=

0.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

83

Thus∇

y∂xF

=∂yG

2G∂xF

=∂y

√G

√G

∂xF.Combiningan

dusingtheLeibniz

rule:

K∂xF

=−K√G(−

1 √G∂xF)

=−∇

y∇

y∂xF

=−∇

y((

∂y

√G

√G

)∂xF)

=−∂y(∂y

√G

√G

)∂xF

−(∂y

√G

√G

)∇y∂xF

=[−

∂y(∂y

√G

√G

)−

(∂y

√G

√G

)2]∂xF

=[−

∂y∂y

√G

√G

+(∂

y

√G)2

G−

(∂y

√G

√G

)2]∂xF

=−

∂y∂y

√G

√G

∂xF.

Example.For

theunitsphere,

withF(x,y)=

(cos

xsiny,sin

xsiny,cos

y)wesaw

inSection

10.5

that

I=

sin2ydx2+

dy2.Then,as

expected:

K=

−∂yy

�sin2y

�sin2y

=−

∂yysiny

siny

=siny

siny=

1.

17.3

TheGaussian

curv

atu

reofth

ehyperb

olicplane

Recallfrom

Section

10.9thatforthehyperbolic

planeH

={z

∈C

:Im

z>

0},param

etriz-

ingbyF(x,y)=

x+

iy,weuse

theRieman

nianmetric

I=

1 y2dx2+

1 y2dy2=

y−2dx2+

d(log

y)2,

usingthetrickdlogy=

(∂ylogy)dy=

1 ydy.

Coro

llary

17.5.Forthehyperbolicplane,

K=

−1.

Proof.

Inthenew

coordinates�x=

x,�y=

logythemetricbecom

esGd�x2

+d�y2

for

G(�x,�y)=

y−2=

e−2�y .

ByTheorem

17.4,K

=−

∂�y�y√e−

2�y

√e−

2�y

=−

∂�y�ye−

�y

e−

�y=

−e−

�y

e−

�y=

−1.

18.

Surfa

cesofconstantcurvature

Lemma18.1.If

x,y

are

geodesic

norm

alcoordinates,

soI=

Gdx2+

dy2,then

G(x,0)=

1and

∂yG(x,0)=

0.

Proof.

ThecurveF(x,0)isageodesic

ofunitspeed,so

G(x,0)=

�∂xF�2

=1fory=

0.Also

1 2∂yG

=1 2∂y(∂

xF·∂

xF)=

∂xF·∂

y∂xF

=∂xF·∂

x∂yF

=∂xF·∇

x∂yF

=−∇

x∂xF·∂

yF

wherethelast

equalityfollow

sbydifferentiatingtheorthog

onalitycondition∂xF·∂

yF

=0.

Evaluatingat

y=

0,∇

x∂xF

=0since

F(x,0)is

ageodesic

inx.So(∂

yG)| y

=0=

0.�

Theore

m18.2.

(1)If

K=

0,then

thesurface

islocallyisometricto

theplane,

(2)If

K=

1,then

thesurface

islocallyisometricto

theunitsphere,

(3)If

K=

−1,

then

thesurface

islocallyisometricto

thehyperbolicplane.

84

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Proof.

Usinggeodesic

norm

alcoordinates,

I=

Gdx2+dy2,byTheorem

17.4,

∂yy

√G

=−K√G

whereK

=0,1,−1in

thethreerespectivecases.

For

K=

0,integrating:√G(x,y)=

a(x)y

+b(x)forsm

ooth

functionsa,b

oftheother

variab

le,so

G=

(a(x)y+b(x))

2.ByLem

ma18.1,G(x,0)=

1so

b(x)=

±1,and∂yG(x,0)=

0so

2(a(x)0

±1)a(x)=

0so

a(x)=

0.ThusG

=1,so

weget

theplanemetric:

I=

dx2+dy2.

For

K=

±1weneedto

solvethegeneralequationz��±

z=

0.Weknow

1thesolutions:

z=

acost+bsintin

the+

case,andz=

acosh

t+bsinhtin

the−

case.2

SoforK

=+1,G(x,y)=

(a(x)cosy+

b(x)siny)2.

ByLem

ma18.1,G(x,0)=

1so

a(x)=

±1,

and∂yG(x,0)=

0so

2(a(x)1

+b(x)0)(−a(x)0

+b(x)1)=

±2b(x)=

0,so

b(x)=

0.

ThusG

=cos2

y,so

weget

themetricofasphere:

I=

cos2

ydx2+

dy2(bytaking�x=

x,

�y=

π 2−ywegetsin2�yd

�x2+d�y2

asin

Section10.5).

AndforK

=−1,G(x,y)=

(a(x)cosh

y+

b(x)sinhy)2.ByLem

ma18.1,G(x,0)=

1so

a(x)=

±1,

and∂yG(x,0)=

0so

2(a(x)1

+b(x)0)(a(x)0

+b(x)1)=

±2b(x)=

0so

b(x)=

0.

ThusG

=cosh

2y,so

weget

themetric

I=

cosh

2ydx2+dy2.

Weneedaclever

changeofvariablesto

turn

this

into

thehyperbolicplanemetric

� I=

1 �y2(d�x2

+d�y2).

Wetry:3

�x=

extanhy

�y=

exsech

y

Then:

d�x=

extanhydx+exsech

2ydy

d�y=

exsech

ydx−ex

tanhysech

ydy

Squaringan

dad

dingwillmake

thedouble

productscancel,leavingsquares:

d�x2

+d�y2

=e2

x(tanh2y+sech

2y)dx2+e2

x(sech

4y+tanh2ysech

2y)dy2

=e2

xdx2+e2

xsech

2ydy2

=e2

xsech

2y(cosh

2ydx2+dy2)

wherethecoeffi

ciente2

xsech

2y=

�y2asrequired

.�

19.

Riemannsu

rfa

ces:

holomorphic

mapsandRiemann-H

urwitz

1Since

those

are

solutions,

andforappropriate

a,b

they

satisfyanyinitialco

nditionsz(0),z� (0),

then

by

uniquen

essofODE

solutionsth

ereare

nooth

ersolutions.

2Refresh

eronhyperbolicfunctions:

cosh

(t)=

1 2(e

t+e−

t),

sinh(t)=

1 2(e

t−e−

t).

They

satisfyth

eeq

uality

cosh

2(t)−

sinh2(t)=

1andhavederivatives

cosh

�=

sinhandsinh�=

cosh

.3Secondrefresher

onhyperbolicfunctions:

tanht=

sinht

cosh

tandsech

t=

1cosh

thavederivatives

tanh� (t)

=

sech

2(t)andsech

� (t)

=−

sinht

cosh

2t=

−tanhtsech

t.W

ewilluse

theusefuliden

tity

tanh2y+

sech

2y=

1

(which

followsfrom

cosh

2y−

sinh2y

=1).

E.g.

this

showsweca

ninvertth

eabovech

angeofvariables:

�x2+

�y2=

e2xso

werecover

x,th

enwerecover

y.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

85

19.1

Thelocalform

ofaholomorp

hic

map

betw

een

Riemann

surfaces

Theorem

19.1

(Localform

ofaholom

orphic

map

).Foranyholomorphic

mapf:S

→R

betweenRiemannsurfaces,

withf(s)=

r,wecanchoose

localcomplexcoordinatesaround

s∈S,r∈R,so

thatin

localcoordinatesfis

themap1

f:D

→D,f(z)=

zn.

Proof.

Wecan

assume(by

tran

slating)

that

thelocalcoordinates

arechosen

sothat

s,r

correspon

dto

0∈C,so

inlocalcoordinates

f(0)=

0.TheTaylorseries

forfis

f(z)=

anzn+an+1zn+1+···=

anzn(1

+higher

order

term

s)

wherean�=

0is

thefirstnon

-zerocoeffi

cient.

This

n≥

1is

called

theorder

ofthevanishing

f(0)=

0.A

holom

orphic

n-throot

off

isthen

defined

2near0,

withf(z)1

/n=

a1/n

nz+

higher

term

s.Thederivativeof

this

n-throot

atz=

0is

a1/n

n�=

0.Bytheinversefunction

theorem,thereis

alocalholom

orphic

inverseG

:C→

Cdefined

near0,

soG(0)=

0an

d

f(G

(z))

1/n=

z.

Now

wecanchan

gecoordinates

onthedom

ainusingthelocalbiholom

orphism

G,explicitly:

ifF

:C→

S(defined

near0∈C)was

theoriginal

localparam

etrization

nears∈S,then

the

new

oneisF◦G

:C→

C→

S(defi

ned

near0∈C).

Thenew

localexpressionforfbecom

es

z�→

f(G

(z))

=zn.

InExercise

Sheet4,

you

willdeduce

thefollow

ingfrom

Theorem

19.1:

Coro

llary

19.2

(Open

map

pingtheorem).

Anon-constantholomorphic

mapf

:R

→S

betweenRiemannsurfaces,

withR

connected,is

anopenmap:f(anyopenset)

isopen.

InExercise

4,you’lldeduce

thefollow

ing,

forf:R

→Sholom

orphic,R,S

Rieman

nsurfaces:

(1)Iffisnon

-con

stan

t,R

compactconnected,then

f(R

)⊂

Sisaconnectedcompon

ent.

(2)Iffis

non

-con

stan

t,R,S

bothcompactconnected,then

fis

surjective:

f(R

)=

S.

(3)IfR

iscompactconnected,S

non

-com

pactconnected,then

fis

constan

t.(4)A

holom

orphic

map

S→

Con

acompactconnectedRieman

nsurfaceis

constan

t.(5)Fundam

entaltheorem

ofalgebra:non

-con

stan

tcomplexpolynom

ials

havearoot.

Soyou

shou

ldview

theab

oveas

apow

erfulgeneralizationof

thefundam

entaltheorem

ofalgebra.Thefundam

entaltheorem

ofalgebra

infact

saysthat

thenumber

ofroots(cou

nting

multiplicity)equalsthedegreeof

thepolynom

ial.Wenow

generalizethisto

Rieman

nsurfaces.

1Explicitly

and

ped

antica

lly:th

ereare

loca

lparametriza

tionsF

:V

→S,0∈

V⊂

R2,G

:W

→R,

0∈

W⊂

R2,F(0)=

s,G(0)=

randflocal (z)=

G−1◦f

◦F(z)=

zn.W

eabusivelyjust

say“loca

llyf=

...”.

2Fora∈

C,w

∈C

with|w

|<1,(1

+w)a

hasaseries

expansion(N

ewton’s

gen

eralisedbinomialseries)

(1+

w)a

=1+

aw

+a(a

−1)

2!

w2+

a(a

−1)(a−

2)

3!

w3+

···

foranya∈

C,andth

eseries

converges

absolutely.

86

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

19.2

Bra

nch

points

and

ramification

points

Recallthelocalform

Theorem

19.1:f:R

→S

has

thelocalform

z�→

znnearp∈R,where

pcorrespon

dsto

z=

0locally.

Wecallv f

(p)=

nthevalency

offat

p.Geometrically,

ittellsyou

how

man

ysolution

sthereareto

theequation

f(z)=

w

forsm

allw

�=0.

Thisshow

sndoes

not

dep

endon

thechoice

oflocalcoordinates

nearp,f

(p).

Thepointw

=0isbad

becau

setherearemissingsolution

s:on

lyz=

0isasolution

forsm

all

z.Wecallz=

0aramification

pointan

dw

=0abranch

point.

Intuitively,

aramification

pointin

Ris

wherethelocalnumber

ofsolution

sof

f(z)=

whas

sudden

lydropped.The

valueof

w,when

solution

saremissing,

isabranch

point.

Definition

19.3

(Ram

ification

pointan

dbranch

point).Forapointpwherev f

(p)�=

1:

(1)p∈R

iscalled

ramifi

cation

poin

t(2)theim

age

f(p)∈S

iscalled

bra

nch

poin

t(3)v f

(p)is

called

ramifi

cation

index

Equivalently:

�r∈R

isara

mifi

cation

poin

t⇐⇒

thederivative

f� (r)

=0in

localcoordinates,

�s∈S

isabra

nch

poin

t⇐⇒

thepreim

age

f−1(s)⊂

Rcontainsaramificationpoint,

�thera

mifi

cationin

dex=

1+numberofderivativesoffvanishingatrin

localcoordinates.

Lemma19.4.Forcompact

R,v f

(p)=

1forallexceptfinitelymanyp∈R.

Proof.

Locallyf(z)=

zn,so

f� (z)=

nzn−1�=

0forz�=

0nearz=

0.Sothesubsetin

Rof

points

wherev f

(p)>

1is

discrete.

Soit

isfinitewhen

Ris

compact.

19.3

Thedegreeofaholomorp

hic

map

ofcompactRiemann

surfaces

Definition

19.5

(Degree).Thedeg

ree

ofa

non-constantholomorphic

map

f:R

→S

betweencompact

connectedRiemannsurfacesis

deg(f)=

r∈f

−1(s)

v f(r)

wherewefixapoints∈S.

Theore

m19.6.deg(f)doesnotdepen

donthechoiceofpoints∈S.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

87

Proof.

Since

Ris

compact,

f−1(s)⊂

Ris

finite(f

isnot

1constan

tlyr).Picksm

alldisjoint

discs

Dp⊂

R,on

earou

ndeach

pointp∈

f−1(s),

sothaton

each

discfhas

thelocalform

z�→

zvf(p

).Byshrinkingtheradiiof

thediscs

Dp,wecanassumeallD

pmap

surjectively

onto

thesameopen

neigh

bou

rhoodV

ofs.

Byconstruction,f−1(V

)containsallthediscs

Dp,butmay

contain

also

other

points

ofR.How

ever,byfurther

shrinkingtheradiiof

the

Dpwecanensure

thatf−1(V

)=

∪Dpcontainsnothingelse.2

Bythelocalmodel,it

follow

sthat

d=

�p∈f

−1(s)v f

(p)is

thenumber

ofsolution

sto

the

equationf(q)=

wforw

�=s∈V

(ineach

discD

pwefindv f

(p)solution

s).Thus

|f−1(w

)|=

q∈f

−1(w

)

v f(q)=

d,

indep

endentlyof

thechoice

ofw

�=0∈V.Since

Sisconnected,thelocallyconstan

tfunction

w�→

�q∈f

−1(w

)v f

(q)on

Smust

beconstant.

Coro

llary

19.7.Forall

points

s∈

Sexceptbranch

points,thereare

precisely

deg(f)=

|f−1(s)|

points

inR

mappingto

s.

Example.

For

acomplexpolynom

ialf(z)of

degreed,wehaved=

deg(f).

Sothereare

preciselydsolution

sto

f(z)=

0unless

0isabranch

pointof

f.When

0isabranch

point,there

arerepeatedroots,butifwecounttherootswiththecorrectmultiplicityv f

(p)then

therearestill

dsolution

s.

19.4

TheRiemann-H

urw

itzform

ula

Thetotalnumber

of“m

issingsolution

s”that

weexpectedforf(r)=

s,over

alls∈S,is:

b(f)=

� s∈S

(deg(f)−

|f−1(s)|)

=� s∈S

r∈f

−1(s)(v

f(r)−

1)

whichwecallthebra

nch

ing

index

b(f),

whererecallf

:R

→S

isaholom

orphic

map

betweencompactconnectedRiemannsurfacesR,S

.

Theore

m19.8

(Rieman

n-H

urw

itzform

ula).

Foranynon-constantholomorphic

mapf

:R

→S

betweencompact

connectedRiemannsurfaces,

theEulercharacteristics

ofR,S

are

related:

χ(R

)=

deg(f)χ

(S)−

b(f),

whichdetermines

thegenusg(R

)since

χ(R

)=

2−

2g(R

).

z�→

z3

R ST

z�→

z3

R ST

1If

f−1(s)hadinfinitelymanypoints,th

enit

would

havealimit

pointr.Atth

islimit

pointryouco

uld

nothavealoca

lform

oftypez�→

zN,asshasinfinitelymanypreim

ages

nea

rr(inparticularmore

thanN).

2oth

erwise,

byco

ntradiction,th

erewould

beasequen

ceofpoints

inR

bounded

awayfrom

∪{p}=

f−1(s)

forwhichth

ef-values

convergeto

s.Byco

mpactnessofR

asu

bsequen

cewould

convergeto

apointp�with

f(p

� )=

swhichwehadnotincluded

in∪{

p}=

f−1(s),

contradiction.

88

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Proof.

Pickatriangulation

forS

sothatthebranch

points

belongto

thevertices

ofthe

triangu

lation

.Wewantthepreim

ageto

yield

atriangulation

ofR.

Sowesubdividethe

triangles

into

smallertrianglesifnecessary,so

thateach

triangle

T⊂

Slies

insideanopen

setV

⊂S

smallenoughso

thatf−1(V

)→

Vcanbewritten

intheusuallocalform

oneach

connectedcompon

entU

⊂R

off−1(V

)(just

asin

theproofofTheorem

19.6).

Ifthelocal

form

off

:U

→V

isz�→

z,so

v f=

1,then

thepreim

ageofT

isanexact

copyofthat

triangle,

butifthelocalform

isz�→

zn,so

v f=

n,then

f−1(T

)consistsofntriangles(they

shareavertex

ifavertex

ofT

isabranch

point).Thesameholdsforedges.How

ever,for

theverticeswhichare

branch

points

wehavefewer

preim

agevertices

thanexpected(w

elose

v f−1verticesat

each

branch

point).So

V(R

)=

deg(f)V

(S)−

b(f)

E(R

)=

deg(f)E

(S)

F(R

)=

deg(f)F

(S).

Remark

.Theform

ula

alsoholdswhen

Ris

disconnected(apply

theform

ula

oneach

con-

nectedcompon

entofR).

OfcourseS

needsto

beconnected(canyouseewhy?)

Example.Recallfrom

Exercisesheet1that

R=

{(z,w

)∈C

2:w

2=

(z−1)(z−2)(z−3)}

∪{∞

}issecretly

atorus(w

herewecompactify

withapointat

infinity,(X

,Y)=

(0,0)usingX

=1/z,

Y=

w/z

2at

infinity,andwedeclare

that

Yisalocalholom

orphic

coordinate-seeSection

8.3).

Claim

:themap

f:R

→CP

1,f(z,w

)=

zisholom

orphic

(inequivalentnotation:themap

isf(z,w

)=

[1:z]andf(∞

)=

[0:1]).

Proof:zis

alocalcoordinateon

Rwhen

∂wf

�=0,so

when

w�=

0,so

when

z�=

1,2,3.In

that

case

fislocally

themap

f loc(z)=

f(z,w

)=

z,so

holom

orphic!When

z=

1,2,3,wehave

∂zf�=

0,so

wisalocalcoordinate,

soR

islocally

(g(w

),w)foraholom

orphic

functiong,and

sof l

oc(w

)=

f(g(w

),w)=

g(w

)is

holom

orphic.Finally,at

infinity,

fis

locally

thefunction1

f loc(Y

)=

1/f(X

,Y)=

Xso

holom

orphic

because

againR

islocally

agraphX

=g(Y

)forg

holom

orphic,since

∂Xf�=

0at

(X,Y

)=

(0,0).

�For

almostanychoice

ofz,thenumber

ofpreimages

f−1(z)is2since

therewill

betwochoices

ofsquareroot

w.Sodeg(f)=

2.Theon

lytimes

when

thenumber

ofpreimages

drops,

isif

(z−

1)(z

−2)(z

−3)=

0,so

z=

1,2,3,andpossibly

atinfinity.

Since

weon

lyaddon

epoint

atinfinity,

f−1(∞

)on

lycontainson

epointinsteadof

two.

2Thusz=

1,2,3,∞

arethebranch

points,(1,0),(2,0),(3,0),∞

aretheramification

points,theramification

indices

areallv f

=2,

andthebranchingindex

isb(f)=

4·(2−1)=

4.ByRiemann-H

urw

itz,

χ(R

)=

deg(f)·χ

(CP

1)−b(f)=

2·2

−4=

0,

sothegenusisg(R

)=

1so

Risatorus.

Cultura

lRemark

:W

hy

isth

eRiemann-H

urw

itzth

eorem

specta

cular?

Alotof

mathem

atics

youhave

seen

sofar,

atleast

ingeometry,hasinvolved

checkingthings

thatare

obviouslytrue,

oratleast

intuitivelyobvious(e.g.theJordancurvetheorem).

ButRiemann-

Hurw

itzis

differen

t:youare

notverifyingsomethingthatyoualready“kn

ow”is

true.

You

are

provingaglobalresult,namelycomputingthegenusofthesurface,by

simply

doingsome

basiclocalcalculations(order

ofvanishingoflocalderivatives).This

isgeometry

atitsbest:

when

youprove

somethinggeometricalthatyoucannot“see”.

1More

precisely:f=

[1:f(z,w

)]=

[1:z]∈

CP

1so

nea

rz=

∞wemust

write

f=

[1

f(z

,w):1]∈

CP

1so

flo

c(Y

)is

thefirsten

try,

since

thatis

theloca

lco

ord

inate

thatweuse

nea

rth

eNorthPole

[0:1]∈

CP

1.

2Youco

uld

alsoch

eckth

eloca

lmodel

explicitly:Atinfinity,

Ris

loca

llyY

2=

X(1

−X)(1−2X)(1−3X),

soforX

=0th

ereis

alsoadrop.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

89

20.

Riemannsu

rfa

ces:

Meromorphic

functions

20.1

Definition

and

examples

WecanstudyRieman

nsurfaces

Sbyinvestigatingholom

orphic

map

sfrom

Sinto

some

test

Rieman

nsurface.

UsingC

astest

spaceis

essentially

useless:

Theorem

20.1

(See

Section

19.1).

IfS

isacompact

Riemann

surface,then

holomorphic

mapsS→

Care

constantoneach

connectedcomponen

t.

You

canalso

prove

this

via

themax

imum

modulusprinciple,1

andnoticeit

generalizes

2

Liouville’s

theorem

that

abou

nded

holom

orphic

functionC→

Cis

constan

t.Sothesimplest

interestingexam

ple

ofmap

sfrom

aRieman

nsurfaceS

into

atest

space,

isto

consider

thetest

spaceCP

1:

Definition20.2.A

mero

morp

hic

functionis

aholomorphic

mapS→

CP

1,whichis

not

iden

ticallyequalto

infinityonanyconnectedcomponen

tofS.3

Let’s

unpackthedefinition,in

localcoordinates.Recallyoucanview

CP

1=

C∪{∞

}whereyou

use

thelocalcoordinateZ

forC

(correspon

dingto

[Z0:Z1]=

[1:Z]∈CP

1),an

dyouuse

thelocalcoordinateW

=1/Z

near∞

(correspon

dingto

[Z0:Z1]=

[W:1]

∈CP

1).

Con

sider

apointp∈S,an

dpickalocalholom

orphic

coordinateznearp.Then

(1)Iff(p)�=

∞,then

bycontinuityf�=

∞forpoints

ofS

closeto

p,so

locally

Z=

f(z)is

aholom

orphic

functionin

z.

(2)Iff(p)=

∞,then

nearf(p)=

∞∈CP

1wemust

use

W=

1/Z,so

W=

1

f(z)

isaholom

orphic

functionin

z.

Example.Merom

orphic

functionson

Carefunctionsf:C→

C∪{∞

}such

that

(1)f(z)isholom

orphic

inznearpoints

wheref(z)�=

∞,

(2)1/f(z)isholom

orphic

inznearpoints

wheref(z)=

∞,

andfisnot

constantly∞.Thefollowingaremerom

orphic:

f(z)=

z2−3z

+1

z3+z2−z−1

f(z)=

az+b

cz+d

f(z)=

ezf(z)=

1

sinz

f(z)=

holofunctionof

z

holofunctionof

z

Ontheother

handf(z)=

e1/z,withf(0)=

∞,is

not

merom

orphic4nearz

=0because

1/f(z)=

e−1/z=

1−

1 z+

1 2!(

1 z)2

+(higher

order

in1 z)is

not

holom

orphic

inz.Noticethat

z=

0isan

essentialsingularity.

Intuitively,

youcanthinkof

amerom

orphic

functionas

beinglocallyalwaysaquotientof

twoholom

orphic

functions(see

Exercise

sheet4).

Theorem

20.3.Meromorphic

functionsonaRiemannsurface

Sare

precisely

holomorphic

functionsS\P

→C,forsomediscretesetP

⊂S,such

thatfhaspolesatthepoints

inP.

1Theco

ntinuousfunction|f|:

S→

Rattainsamaxim

um

atsomepointp∈

S,butth

enin

loca

lco

ord

inates

|f|w

ould

haveamaxim

um

atpso

itwould

needto

beco

nstantnea

rp.

2Iff:C

→Cisbounded

andholomorp

hic,th

en∞

isaremovable

singularity,so

fex

tendsto

f:CP

1→

C.

3W

eex

cludeth

eco

nstantfunction∞

because

wewantmeromorp

hic

functionsonaco

nnectedRiemann

surface

toform

afield,so

thatfunction

would

beproblematic.

When

Sis

disco

nnected,werequireth

at

meromorp

hic

functionsare

notiden

tica

llyeq

ualto

∞onanyco

nnectedco

mponen

t.4in

fact,it

isnotev

enco

ntinuous:

consider

z=

irforreals

r→

0.

90

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Proof.

Supposefismerom

orphic.In

localcoordinates,supposef(z)=

∞at

z=

0.Expan

dtheholom

orphic

function1/

f(z)nearz=

0in

aTay

lorseries:

1

f(z)=

anzn+

an+1zn+1+

···=

anzn·(1+

higher

order

zterm

s)

Tak

ingthereciprocal,1

weob

tain

aLau

rentseries:

f(z)=

a−1

n

1 zn·(1−

higher

order

zterm

s).

Sof(z)has

apole

atz=

0of

thesameorder

asthevanishingof

1/f(z).

Theclaim

follow

ssince

poles

areisolated

(this

follow

sfrom

theab

ovelocalexpression:f(z)�=

∞exceptat

z=

0).Con

versely,given

f:S\P

→C

asin

theclaim,extendbyf:S→

CP

1,f(p)=

∞forallp∈P,then

fis

merom

orphic

since

1/fis

holom

orphic

atpoints

ofP.

�Coro

llary

20.4.Thesetofmeromorphic

functionsonaconnected2Riemannsurface

Sis

afield,under

pointwiseadditionandmultiplicationoffunctions,

called

thefu

nction

field

K(S

).

Proof.

Thisfollow

sbyexpan

dingf+λg,f·g,1/

fas

Lau

rentseries

inlocalcoordinates

near

anygiven

point,

andusingTheorem

20.3.

�Cultura

lre

mark

.(For

thoseof

youwholikealgebra

andGaloistheory)Studyingcompact

connectedRieman

nsurfaces

isin

fact

equivalentto

studyingfunctionfieldsK(S

)whichare

algebraic

extension

sof

Cof

tran

scenden

cedegree1(a

purely

algebraic

problem).

ThisK(S

)arises

asthefieldof

functionsof

thesm

oothprojectivecurvecorrespon

dingto

S(see

B3.3).

Coro

llary

20.5.Foranymeromorphic

functiononacompact

Riemannsurface,thenumber

ofzerosequals

thenumberofpoles(countedwithmultiplicity).

Proof.

deg(f)=

#poles

=#zeros,

countedwithmultiplicities

v f.

�Example.Merom

orphicfunctionson

CP

1arefunctionsf:C

→C

satisfying(1)and(2)of

the

previousexam

ple,and

(3)iff(∞

)�=

∞,f(1 w)isholom

orphic

inw

nearw

=0,

(4)iff(∞

)=

∞,1/

f(1 w)isholom

orphic

inw

nearw

=0.

wherew

=1/

zisthelocalcoordinatenear∞

∈CP

1.Thefollowingaremerom

orphic:

f(z)=

z2−

3z+

1

z3+

z2−

z−

1f(z)=

az+

b

cz+

df(z)=

(rational

functionin

z)=

polynom

ialin

z

polynom

ialin

z

How

ever

f(z)=

1sinzandf(z)=

ezarenot

merom

orphicat

infinity.

Moregenerallyiff(z)=

a0+

a1z+a2z2+···for

largez,then

inthelocalcoordinatew

=1/

z:f(1 w)=

a0+a11 w+a2(1 w)2+···

sow

=0isapole(orremovable)⇐⇒

only

finitelymanyajarenon

-zero(otherwisew

=0isan

essential

singu

larity).

Theore

m20.6.Allmeromorphic

functionsonCP

1are

rationalfu

nctions.

Proof.

Recallthat

foraholom

orphic

function,zerosareisolated

unless

thefunctionisidenti-

callyzero

(thisistheId

entity

theore

m–seetheAnalysishan

dou

t).Since

CP

1iscompact,

itfollow

sthat

amerom

orphic

functionf:CP

1→

CP

1canhaveon

lyfinitelyman

yzeros

z 1,...,z

n∈C

(unless

fisidenticallyzero)an

don

lyfinitelyman

ypoles

p1,...,p

m∈C

(since

1Recall

11+w

=1−

w+

w2−

w3+

···is

holomorp

hic

inw

∈C,for|w

|<1.

2In

thedisco

nnectedca

se,weca

nnotget

afieldasth

ereare

zero

divisors:ifS

=S1�S2,takefj=

1on

Sjand0oth

erwise,

then

f1·f

2=

0.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

91

thoseare

isolated

zerosof

1/f).

Let

a1,...,a

nandb 1,...,b

mbetheordersof

thezerosan

dof

thepoles.

Let

g(z)=

�(z

−z j)a

j/�(z

−pk)b

k.

Then

f/g

ismerom

orphic,an

dit

nolongerhas

zerosor

poles

inC

(they

areremovab

lesingu

larities).

Ifat

infinityit

also

does

not

haveapole,

then

itis

aholom

orphic

map

f/g

:CP

1→

Can

dhence

isconstan

tbyTheorem

20.1.If

ithas

apoleatinfinity,

then

consider

thereciprocalg/f

:CP

1→

C,whichag

ainmust

beconstant(hereweusedthat

f/g

has

no

zerosin

Cso

g/f

has

nopoles

inC),so

actually

therewas

nopole.Thusf=

constan

t·g,so

fis

aquotientof

polynom

ials.

Noticeab

oveit

iseasy

toverify

Corollary

20.5,since

g(z)anditsreciprocalhavenopole

atinfinity,

so�

aj≤

�b k

and�

aj≥

�b k,hence

equality.

Coro

llary

20.7.ThebiholomorphismsCP

1→

CP

1are

precisely

theMobiusmapsz�→

az+b

cz+d

fora,b,c,d

∈C,ad−

bc�=

0.SothegroupofautomorphismsofCP

1is

PSL(2,C

)=

{� ab

cd

� :a,b,c,d

∈C,ad−bc

=1}

I

Proof.

BythepreviousTheorem,ithasto

bearation

alfunction.Hav

ingcancelled

common

factors,

thepolynom

ialat

thenumerator

andthat

atthedenom

inator

are

each

allowed

tohaveat

moston

eroot

(abijectivemapCP

1→

CP

1hasprecisely

onezero

andon

epole).

20.2

Ellipticfunctions:

mero

morp

hic

functionson

theto

riC/lattice

Recallwemention

edthatsm

oothfunctionsf:S1×

S1=

R2/Z2

→R

onthetorusare

precisely

smoothfunctionsR

2→

Rwhichare1-periodic

ineach

entry:f(x

+n,y

+m)=

f(x,y).

Analogo

usly,

given

amerom

orphic

functionon

anellipticcurveC/(Zω

1+

Zω2),

we

patch

together

thelocalexpressionsforfto

obtain

afunction

f:C

→CP

1

respectingtheequivalence

relation

,so

f(z

+latticepoint)

=f(z).

Thus

f(z

+ω)=

f(z)forallω=

nω1+

mω2wheren,m

∈Z.

Definition

20.8

(Ellipticfunctions).Anellipticfunctionis

ameromorphic

functiononC

whichis

doublyperiodic,thatis

periodic

intwoR-linearlyindepen

den

tdirectionsω1,ω

2∈C.

For

this

reason,Λ=

Zω1+

Zω2is

oftencalled

thelatticeofperiodsoff.

Example.Theseries

f(z)=

� ω∈Λ

1

(z−

ω)3

isan

absolutely

convergentseries

1at

z/∈Λ.Bysomebasic

complexanalysis

whichwerecall

below

,theseries

converges

absolutelyanduniformlyto

aholom

orphicfunctionon

anygivencom-

pactsetavoidingthepoles

ω∈Λ,theseries

canbedifferentiated

term

byterm

,andtheorder

ofsummationdoes

not

matter.

Infact,givenanycompactsetK

⊂C,ifweom

itthefinitelymany

term

sof

theab

oveseries

that

involvepoles

ωthat

liein

K,wededuce

that

therest

oftheseries

converges

holom

orphically,thereforethewholeseries

has

preciselythepoles

determined

bythose

finitelymanyterm

s.Toproveperiodicity,

wejust

needto

reorder

theseries

viaω

�→ω−

ω�to

deduce

that

f(z)=

f(z

+ω� )

foranyω�∈Λ.Atpoints

z=

α∈Λ

thesameargu

mentholdsif

youom

it1

(z−α)3

from

thesum,so

fhas

apoleof

order

3at

α.Sofisan

ellipticfunction.

1Try

checkingth

isbyhand,co

mparingwithth

eco

nvergen

tseries

�∞ n=1

1 n2

(not

1 n3).

Alternatively,

you

candoth

isbyanintegraltest,co

mparingwith� (x

2+

y2)−

3/2dxdy(p

ass

topolarco

ord

inates).

92

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Some

Complex

AnalysisBack

gro

und.Suppose

f n:U

→C

are

continuousfunctions

defined

onan

open

setU

⊂C.W

eierstrass’s

M-test

saysthatif

|f n|≤

MnonU

with

�M

n<

∞,then

�f n

converges

absolutely

anduniform

ly.Sothelimit

iscontinuous(uni-

form

limitsof

continuou

sfunctionsarecontinuous).Suppose

thatf n

isalsoholomorphic,so

thepartial

sumsSn(z)=

�n j=1f j(z)are

holomorphic.Weclaim

thatf

=lim

Snis

holo-

morphic.M

orera

’sth

eorem

saysthatafunctionf:U

→C

isholomorphic

ifandonly

if� ∂

Tf(z)dz=

0forallboundaries

∂T

ofsolidtriangles

T⊂

U.Applyingthis

tooursetup:

� ∂T

f(z)dz=

� ∂T

lim

Sn(z)dz=

lim

� ∂T

Sn(z)dz=

0

whereweuse

abasic

fact

abou

tintegration:forasequen

ceofuniform

lyconvergentfunctions,

thelimit

commuteswiththeintegral.1

Finallyweclaim

thatwecandifferentiate

theseries

term

byterm

.Cauch

y’s

integra

lform

ula

saysif

fis

holomorphic,andw

lies

inthe

interiorofacloseddisc2

entirely

contained

inU,then

f(w

)=

12πi

� γf(z

)z−wdzwhereγis

the

anti-clockwisecurveboundingthedisc.

Itis

relativelystraightforw

ard

toshow

thatonecan

keep

differentiatingin

wto

obtain

form

ulasforthederivatives

aswell,in

particular:

f� (w)=

1 2πi

� γ

f(z)

(z−w)2

dz.

Applied

toou

rsetup,S� n(w

)=

12πi

�Sn(z

)(z

−w)2dz,andusingagain

theabovetheorem

about

commutinglimitsan

dintegrals,

lim

S� n(w

)=

12πi

� γlim

Sn(z

)(z

−w)2

dz=

12πi

� γf(z

)(z

−w)2dz=

f� (w).

Asan

exercise,3

show

that

thederivatives

ofthepartialsumsconvergeuniform

lyto

f� .

20.3

TheW

eierstrass

℘-function

TheW

eierstrass

P-function(orW

eierstrass’s

ellipticfunction)is

℘(z)=

1 z2+

0�=ω∈Λ

�1

(z−ω)2

−1 ω2

Motivation:youcannot

findameromorphic

functiononC/Λ

withonly

onesimple

pole

since

otherwiseC/Λ

wou

ldbebiholom

orphic

toCP

1byExercise

Sheet4,butweknow

that

thetorusan

dthespherearenot

hom

eomorphic.Soyouneedatleast

apoleoforder

2.Soyou

look

at1/z2.Tomakethat

dou

bly

periodic,youwould

consider

�ω∈Λ

1/(z

−ω)2,butthatis

unfortunatelydivergent.4Intuitively,

�0�=ω∈Λ

1/ω2oughtto

havealotofcancellations(for

thelatticeZ·1

+Z·i

weexpectzero

dueto

cancellationsin

pairsvia

thesymmetry

ω�→

iω),

butsadly

that

series

divergesformostchoices

oforderingofthesum.Thefunction℘is

the

difference

ofthesetw

odivergentseries:miracu

louslyit

solves

allconvergence

issues!This

difference

isalso

anaturalchoice

because

then

nearz=

0wehave℘(z)=

1 z2+

g(z)withg

holom

orphic

andg(0)=

0.

Lemma20.9.℘(z)converges

toanellipticfunction,in

thesense

thatitabsolutely

converges

onanycompact

setK

⊂C

once

weomitthefinitelymanyterm

swithpolesin

K.

1provided

weintegrate

over

asetoffinitevolume,

inourca

seth

atis

thelength

of∂T

whichis

finite.

2It

doesn’t

haveto

beadisc,

itca

nbeanysimply

connecteddomain

contained

inU

whose

boundary

isa

piecewisesm

ooth

curve.

3Hint.

Consider

|f� (w)−

S� n(w

)|byestimatingth

eintegral,usingth

etrick|�

g|≤

�|g|.

4Nea

racircle

ofradiusr,centre0,youhaveroughly

2πrterm

s,ea

chofsize

roughly

1 r2,so

thesu

mgrows

roughly

like2π�

1 r,whichgrowslikealogarith

mandthusdiverges.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

93

Proof.

Let’s

firstprove

theinterestingbit,whichis

periodicity,

assumingconvergence.

Rem

ark.Thebrute

forceapproach

toproving℘(z

+ω� )

=℘(z)requires

aclever

reordering

argumen

t,because

youcannotbreakuptheseries

into

twodivergentseries.

Thebetterap

proachis

asfollow

s.Byab

solute

convergence,yo

umay

differentiatetheseries

term

byterm

.Thus:

℘� (z)=

−2� ω∈Λ

1

(z−

ω)3

whichisellipticbythepreviousexam

ple,withorder

3polesat

latticepoints.Now

thetrick:

d dz(℘

(z+

ω� )−

℘(z))

=℘� (z+

ω� )−

℘� (z)=

0

byperiodicity.

So℘(z

+ω� )−℘(z)isconstan

t.Todeterminethat

theconstantiszero,plugin

z=

−ω� /2an

duse

that

℘isan

even

function:℘(−

z)=

℘(z)(reorder

thesum

via

ω�→

−ω).

Wenow

prove

convergence.Compute

1

(z−

ω)2

−1 ω2=

ω2−

(z−

ω)2

ω2(z

−ω)2

=−z2+

2zω

ω2(z

−ω)2

=z(2ω−

z)

ω2(z

−ω)2

For

|ω|>

2|z|,theab

solute

valueof

theab

oveis

bou

nded

by

|z|5 2

|ω|

|ω|2

1 4|ω

|2=

10

|ω|3|z|.

Now

zis

fixed

,an

dthecondition|ω|>

2|z|just

omitsfinitelyman

yterm

sfrom

thesum.Moreover

�0�=ω∈Λ

1 |ω|3

converges

(asin

thepreviousexample,bycomparingwith�

∞ n=1

1 n2).

So,

apart

from

finitelyman

yterm

sin

thesum,wededuce

absolute

convergence.

Thefinitelyman

yterm

sweom

ittedareholom

orphic

exceptforfinitelyman

ypoles,so

theclaim

follow

s.�

Lemma20.10.℘(z)hasthefollowingproperties

(1)℘(z)is

aneven

function:℘(−

z)=

℘(z),

(2)within

thefundamen

talparallelogram

{sω1+

tω2:s,t∈

[0,1]}

⊂C

modulo

edge-

iden

tifications,

℘hasonepole

at0oforder

2(itis

harder

todescribethetwozeros),

(3)deg(℘

)=

2,view

ing℘asamapC/Λ

→CP

1,

(4)℘� (z)=

0athalf-latticepoints

ω 2,where1

ω∈Λ.

(5)℘hasramificationpoints

athalf-latticepoints

andatlatticepoints,so

℘hasprecisely

fourdistinct

ramificationpoints

within

thefundamen

talparallelogram

(modulo

Λ):

0,1 2ω1,

1 2ω2,

1 2(ω

1+

ω2).

(6)Thevalencies

atramificationpoints

are

v ℘=

2,thebranchingindex

b(℘)=

4.

Proof.

(1)follow

sbyreplacingz�→

−zan

dreorderingthesum

via

ω�→

−ω.(2)follow

sby

theproof

ofLem

ma20

.9:forzcloseto

0,thesum

in℘(z)isholom

orphic

once

1 z2isom

itted.

Notethat

theother

poles

ω1,ω

2,ω

1+

ω2in

theparallelog

ram

areallidentified

with0under

thetran

slationgrou

pΛ.(3)follow

sby(2)since

℘−1(∞

)=

0∈C/Λ

withvalency

v ℘(0)=

2.For

(4),

differentiatetheequation℘(−

z)=

℘(z)from

(1):

℘� (z)=

−℘� (−z)

(so℘�is

anoddfunction).

But℘�is

also

dou

bly

periodic,so

ifzsatisfies

z=

−z+

ωin

the

quotientC/Λ

then

℘�must

vanishat

z.Solvingthat

equationwegetz=

1 2ω.Sohalf-lattice

points

satisfy℘� (z)=

0an

darethereforeramification

points,andsince

thepoles

haveorder

2they

arealso

ramification

points

(1/℘

has

azero

oforder

2).Since

valencies

atramification

1Stricly

spea

king℘� (ω)is

notdefi

ned

,asth

ereis

apole,butusingth

eloca

lco

ord

inate

1/(z

−w)one

would

alsofind℘�=

0(indeedth

ose

polesare

ramifica

tionpoints).

94

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

points

satisfy1<

v ℘≤

deg(℘

)=

2,they

must

allbe2.

How

dowekn

ow

thatwehave

foundallramificationpoints?Weuse

Riemann-H

urw

itz:

0=

χ(torus)

=χ(C

/Λ)=

2χ(C

P1)−b(℘)=

4−b(℘),

sothebranchingindex

b(℘)=

4,so

allramificationpoints

are

alreadyaccountedfor.

Thebranch

points

of℘are

denoted:

e 1=

℘(1 2ω1),

e 2=

℘(1 2ω2),

e 3=

℘(1 2(ω

1+ω2)),

∞=

℘(0).

InExercise

sheet4youwillprove:

Theorem

20.11.Thefollowingis

abiholomorphism:

C/Λ

→{(Z,W

)∈C

2:W

2=

4(Z

−e 1)(Z−e 2)(Z−e 3)}

∪{∞

}z

�→(℘

(z),℘� (z))

whereontherightwecompactifyasshownin

Section8.3

(compare

ExerciseSheets

1&

2).

Cultura

lRemark

.Thefunction

fieldofall

meromorphic

functionson

an

ellipticcurve

turnsoutto

beC(℘

,℘� )

=(rationalfunctions1

inthevariables

℘,℘� ).Thekeytrickin

the

proofis

thefact

thatifyoukillallthepolesofameromorpic

function(e.g.by

rescalingwith

polysin

℘,℘� ),then

theresulthasdegree0so

itis

aconstantfunction.TheB3.3

Algebra

icCurv

escoursestudiesmore

generallythefunctionfieldofanyalgebraic

curve.

21.

Hyperbolic

Geometry:anintroduction

21.1

Refresh

eraboutM

obiusmaps

Mob

iusmap

sforhyperbolicgeometry

are

asim

portantasrotationsandtranslationsare

inEuclideangeom

etry.Indeed,forthehyperbolicplaneH

={z

∈C

:Im

z>

0}andthe

hyperbolic

discD

={z

∈C:|z|<

1},

asubgroupoftheMobiusmapswillturn

outto

bethe

grou

pof

allisom

etries.Recallthatwefoundisometries

inSection10.9,betweenD

andH:

D→

H,z�→

τ(z)=

iz+i

−z+1

H→

D,z�→

τ−1(z)=

z−i

z+i.

RecallCorollary

20.7:

Coro

llary

21.1.Thebiholomorphismsϕ:CP

1→

CP

1are

precisely

theMobiusmaps

ϕ(z)=

az+b

cz+d

fora,b,c,d

∈C

withad−

bc�=

0(w

eoften

rescale

numeratorandden

ominatorso

that

ad−bc

=1).In

particular,

ϕ(∞

)=

a/c,

ϕ(−

d/c)

=∞.Theseform

agroupisomorphic

to

PSL(2,C

)=

{� ab

cd

� :a,b,c,d

∈C,ad−bc

=1}/

±Identity.

Werecallsomeusefulproperties

aboutMobiusmaps:

Lemma21.2.

(1)Mobiusmapspreserveangles

(2)Mobiusmapsare

generatedby

translationsϕ(z)=

z+

b,dilationsϕ(z)=

az(for

a�=

0),andinversionsϕ(z)=

1/z(w

hichcorrespondsto

inversionin

theunit

circle

followed

byreflectionin

therealaxis,

since

ϕ(rei

θ)=

1 re−

iθ).

(3)Mobiusmapssendcirclesto

circles(w

hereweallowstraightlines,thoughtofascircles

ofinfiniteradius).

1i.e.

ratiosofpolynomials.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

95

(4)Given

anythreedistinct

points

z 0,z

1,z

2∈

CP

1,thereis

aMobiusmapϕ

such

that

ϕ(z

0)=

0,ϕ(z

1)=

1,ϕ(z

2)=

∞.

(5)A

Mobiusmapis

uniquelydetermined

bywhereit

sendsthreepoints,forexample

itis

determined

bythevalues

ϕ(0),

ϕ(1),

ϕ(∞

).(6)TheMobiusmapswithϕ(H

)=

H,withad−

bc=

1,are

those

witha,b,c,d

allreal:

Mob(H

)=

�az+

b

cz+

d:a,b,c,d

∈R,ad−bc

=1�

∼ ={� a

bcd

� :a,b,c,d

∈R,d

et=

1}/±

Id=

PSL(2,R

)

(7)Given

z∈H

thereis

aMobiusmapϕ

withϕ(H

)=

Handϕ(i)=

z.

Proof.

For(1):

Mobiusmapsϕareholomorphic,so

thederivativeD

zϕisacompositionof

ascalingan

darotation

(see

Analysishan

dout),so

itpreserves

angles.

For(2),

when

c�=

0,

az+

b

cz+

d=

a c−

ad−

bc

c(cz

+d)

soit

isacompositionof

translations,

dilations,

inversions.

Thecase

c=

0is

even

easier.

Con

versely,tran

slations,

dilationsandinversion

sare

Mob

iusmap

s.For(3):

itisenou

ghby(2)to

checkthat

tran

slations,dilation

sandinversionssendcircles

tocircles,

andan

easy

checkshow

sthatthey

do.

For(4),

wecaneven

write

anexplicitform

ula:

ϕ(z)=

(z−

z 0)(z 1

−z 2)

(z−

z 2)(z 1

−z 0)

For(5):

supposeaMob

iusmap

ψsendsdistinct

points

w0,w

1,w

2to

z 0,z

1,z

2.Let

ϕbea

map

asin

(4).

Then

ϕ◦ψ

isaMob

iusmapwhichsendsw

0,w

1,w

2to

0,1,∞

.Let

αbethe

inverseofamapasin

(4)forthepoints

w0,w

1,w

2,so

αsends0,1,∞

tow

0,w

1,w

2.Then

ϕ◦ψ

◦αisaMobiusmapwhichfixes

0,1,∞

.Bylookingattheequation

sthisim

plies

forthe

constan

tsa,b,c,d

whichdefinethemap,yo

ueasily

deduce

that:b=

0(fixes

0),c=

0(fixes

∞)andso

a/d=

1(fixes

1),so

this

map

istheidentity.Soψ=

ϕ−1α−1is

determined

.For(6):

since

Mobiusmapsare

biholomorphisms,ifϕ(H

)=

Hthen

therestrictionϕ:H

→H

isabiholom

orphism

aswell.

Bycontinuity,

thebou

ndaryR

ofH

has

tobemap

ped

into

itself.Asϕ(0)=

r 0,ϕ(1)=

r 1,ϕ(∞

)=

r 2are

realnumbers,

bytheaboveform

ula

wehave

ϕ−1(z)=

(z−r0)(r1−r2)

(z−r2)(r1−r0).Thisisin

PSL(2,R

)so

itsinverseϕisalsoin

PSL(2,R

).Con

versely,

ifa,b,c,d

are

real,then

ϕ(R

)=

R,hence

ϕpermutesthetw

oconnectedcompon

ents

ofC\R

(since

ϕis

abiholomorphism).

Soϕ(H

)=

Hprecisely

ifϕ(i)∈

H.Sowecheckthesign

:sign

Imϕ(i)=

sign

(ad−

bc).

Soϕ(H

)=

Hprecisely

ifdet

>0.

For(7):

z=

b+ia

interm

sof

realan

dim

aginary

partsb,a∈R,then

ϕ(z)=

az+bworks

(sotakec=

0an

dd=

1).

Exercise.

Recall

thehyperbolicdiscD

={z

∈C

:|z|<

1}is

isometricto

H(usingthe

hyperbolicmetric

4|dz|2

(1−|z|2)2

onD).

Show

thattheMobiusmapsforwhichϕ(D

)=

Dare:1

ϕ(z)=

az+

b

bz+

a

1You

could

repea

tth

eproofforH

forD,so

askingyourselfwhich

Mobiusmapssend

∂D

to∂D.

The

shortcu

tis

toobserveth

atisometries

D→

Darise

from

D→

H→

H→

DwhereH

→H

are

theisometries

wefoundabove,

andth

emapsD

→H

(andback

)are

theisometries

τ,τ

−1men

tioned

atth

estart

ofSection

21.1.Youca

nca

lculate

compositionsofMobiusmapsbymultiplyingth

eco

rrespondingmatrices.

96

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

witha,b

∈C

and|a|2−

|b|2=

1.So:1

Mob(D

)=

�az+

b

bz+a:a,b

∈C,|a

|2−

|b|2=

1�∼ =

{� ab

ba

� :a,b

∈C,d

et=

1}/{

eiθId}=

PSU(1,1)

Notice

thata�=

0,so

youcanrescale

numeratorandden

ominatorso

that|a|=

1,so

youcan

replace

a=

eiθ/2.Then

ϕbecomes:

ϕ(z)=

eiθz+

b

bz+1.

withb∈C

and|b|

<1.

Inparticular,

then

ϕ(0)=

eiθbso

pickingb>

0∈R

showsthatthere

isaMobiusmapwithϕ(D

)=

Dandϕ(0)=

somechosenpointin

D.

21.2

Isometriesofth

ehyperb

olicdiscD

and

thehyperb

olicplaneH

Theore

m21.3.Thegroupoforien

tation-preservingisometries

ofH

containsMob(H).

The

groupofallisometries

ofH

containsMob(H)andthereflectionz�→

−z(soit

containsthe

orien

tation-reversingisometries

ψ(z)=

−az+b

−cz+d).

Rem

ark.Later

weshow

that

therearenoother

isom

etries.

Proof.

Westartbycheckingthat

Mob

(H)areisom

etries.Werunthesamecalculation

asat

theendof

Section

10.9:weneedto

show |dz|2

(Im

z)2

=|d(ϕ

(z))|2

(Im

ϕ(z))

2.

First,dϕ(z)=

ϕ� (z)dzwhere,

hav

ingnormalized:ad−

bc=

1,

ϕ� (z)=

a(cz+

d)−

(az+

b)c

(cz+

d)2

=1

(cz+

d)2.

Secon

dly,

Imϕ(z)=

Im(az+

b)(cz+

d)

|cz+

d|2

=Im

(ax+

iay+

b)(cx−

icy+

d)

|cz+

d|2

=(ad−

bc)y

|cz+

d|2

=Im

z

|cz+

d|2.

Therequired

equalityab

ovethen

follow

s.For

thelast

part,

noticethat

|d(−

z)|2

=(−

dz)(−dz)=

|dz|2 ,

andIm

(−z)=

Im(z).

Exercise.ShowthatMob(D)are

orien

tation-preservingisometries

ofD,andthatthereflec-

tionz�→

zis

anorien

tation-reversingisometry

ofD.Notice

inparticularthattherotations

z�→

eiθzare

isometries

ofD,andthatthereflectionin

thelinewithangleθto

therealaxis

isz�→

e2iθz=

eiθe−

iθzso

alsoan(orien

tation-reversing)

isometry.

Now

theissueis:how

canweshow

that

theab

oveareallisom

etries?How

canwebesure

wehavenot

omittedan

y?Theeasiestroute

toprove

this,is

touse

geodesics,

asfollow

s.

21.3

GeodesicsofH

Theore

m21.4.ThegeodesicsofthehyperbolicdiscD

are

circles(includingstraightlines)

whichare

orthogonalto

theboundary

S1=

∂D.

ThegeodesicsofH

are

circles(including

straightlines)whichare

orthogonalto

theboundary

R=

∂H.

1ForSU(2)th

e(2,1

)en

tryofth

ematrix

would

needto

be−b,

soth

atdet

=|a|2

+|b|

2.ForSU(1,1

)th

e

signatu

reofth

equadraticform

hasone+

andone−

sign:det

=+|a|2

−|b|

2.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

97

Proof.

ByTheorem

16.11,

thestraightlinet�→

tvthrough

0withdirection

v∈R

2=

T0D

isageodesicsin

Dbecau

sethe(E

uclidean)reflection

inthat

straightlineis

anisom

etry

ofD.

Sotheseareallthegeodesicsγ0,v

forv∈

T0D

byTheorem

16.7.ByLem

ma21.2

(andthe

fact

that

H∼ =

Dareisom

etric),thereis

aMob

iusisom

etry

ϕ:D

→D

whichsends0to

any

chosen

pointp∈D.ButDϕ:T0D

→TpD

isbijective.

1Soforanyw

∈TpD,γp,w

=ϕ◦γ

0,v

takingv=

Dϕ−1w.Sowehaveob

tained

allgeodesicsin

D.ByLem

ma21.2,ϕ◦γ

0,v

isa

circle

since

γ0,v

isacircle

(line),an

dϕ◦γ

0,v

isorthogon

alto

∂D

becau

seγ0,v

isorthogon

alto

∂D

(usingthat

Mob

iusmap

spreservean

gles,an

dthatϕ(∂D)=

∂D).

Since

H∼ =

Dare

isom

etricvia

aMob

iusmap

,theclaim

forH

follow

sbyTheorem

16.10(again

usingthat

circlesmap

tocirclesvia

Mob

iusmap

s).

Exercise.Use

theexplicitgeodesic

equation,attheen

dofSection16.3,to

obtain

explicit

solutionsyieldingthegeodesicsclaim

edabove.

Hints.For

H,on

eequationbecom

esd dt(x

� /y2)=

0,an

dtheconditionof

beingparam

etrized

byarc-lengthbecom

es(x

�2+

y�2)/y2=

1.You

wan

tan

equationinvolvingon

lyx,y

(not

t),

socalculate

dy/d

x=

y� /x�=

···.

Coro

llary

21.5.Thereare

noother

isometries

forD,H

beyondthose

foundin

Section21.2.

Proof.

SupposeT

isan

isom

etry.Pickaϕ∈Mob

(D)withϕ(0)=

T(0).

Then

ϕ−1◦T

isan

isom

etry

fixing0.

Isom

etries

map

geodesicsto

geodesics(T

heorem

16.10),

andthegeodesics

inD

through

0arestraightlines,so

Tpermutesthestraightlines

through

0.Say

itmap

sthe

geodesic

linesegm

ent[0,1)to

eiθ[0,1).

Then,since

isom

etries

fixlengths,

e−iθϕ−1◦T

fixes

[0,1).

Finally,isom

etries

also

preservean

gles

upto

sign

,2so

e−iθϕ−1◦T

must

either

fixall

straightlines,or

itmust

reflectz�→

z.Thetw

ocasesare

distingu

ished

bywhether

ornot

Tis

orientation

-preserving.

This

proves

thestatem

entforD.ThestatementforH

follow

s3by

usingtheMob

iusisom

etry

D→

H.

21.4

Hyperb

oliclength

sand

hyperb

olicangles

Theorem

21.6.Hyperbolicangles

are

equalto

Euclideanangles.

Proof.

RecallI=

1 y2(dx2+

dy2)in

theusual

param

etrization

F(x,y)=

x+

iyforH,so

thematrixentriesof

Isatisfytheconditionse=

gan

df=

0required

byExercise

Sheet2

togu

aran

teethat

theparam

etrization

Fis

conform

al(i.e.an

gle-preserving).Sohyperbolic

angles

equal

Euclideanan

gles.

Theorem

21.7.In

thehyperbolicdiscD,thedistance

dist D

(0,z)=

2tanh−1|z|.

Indeed

dist D

(p,q)=

2tanh−1

� � � �q−p

1−

pq

� � � �dist H(p,q)=

2tanh−1

� � � �q−p

q−p

� � � �.

Proof.

RecallI=

4(d

x2+dy2)

(1−x2−y2)2

forD

andF(x,y)=

x+iy.Thecurveγ(t)=

(t,0)for0≤

t≤

x

has

γ� (t)

=(1,0),so

I(γ

� (t),γ

� (t))=

4(1

−t2)2.Thus4

L(γ)=

�x

0

�I(γ

� ,γ� )dt=

�x

0

2

1−t2

dt=

2tanh−1x.

1Byth

ech

ain

rule

thederivativeofadiffeo

morp

hism

isbijective:

(Dϕ)−

1=

D(ϕ

−1).

2th

eypreserveth

eRiemannianmetric,

soth

eypreserveco

s(angles),butco

sα=

cos(−α).

3Forth

eorien

tation-preservingones,D

→H

→H

→D

isaco

mpositionofMobiusmaps,

butwealrea

dy

found

all

Mobiusisometries

ofD;forth

eorien

tation-rev

ersingones

just

compose

with

H→

H,z�→

−zto

reduce

toth

eorien

tation-preservingca

se.

4d dxtanh−1(x

)=

1/(tanh� (tanh−1x))

=1/(sech2(tanh−1x))

=1/(1

−tanh2(tanh−1(x

))=

1/(1

−x2).

98

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Since

rotation

sareisom

etries,theform

ula

fordist D

(0,z)isthesamewithx=

|z|inplace

ofx.Given

general

points

p,q

∈D

wesimply

apply

anisom

etry

ϕto

movepto

0an

dthen

use

theab

oveform

ula

withz=

ϕ(q).

Now

ϕ(z)=

z−p

−pz+1works(see

Section

21.1).

Theform

ula

forH

follow

sbyusingtheisom

etry

H→

D,z�→

τ−1(z)=

z−i

z+i:

dist H(p,q)=

2tanh−1

� � � � �

q−i

q+i−

p−i

p+i

1−

p+i

p−iq−i

q+i

� � � � �=2tanh−1

� � � �p+

i

p+

i

p−

q

q−p

� � � �=2tanh−1

� � � �q−

p

q−

p

� � � �.�

Exercise.

Show

bydirectcalculation

thatthelength

inH

ofthesegm

ent(1,y)i

on

the

imaginary

axisis

logy.In

principle

youcould

now

findaMobiusisometry

sendinggeneral

points

p,q

∈H

toi,iy

toobtain

dist H(p,q),

although

this

isnotaseasy

asitwasforD.

21.5

Areasoftriangles,

and

limits

Wecallhyperb

olictriangle

ageodesic

trianglein

thehyperbolic

metric.

Wewillcall

A,B

,Cthevertices,α,β

,γtheinternal

angles,a,b,c

thehyperbolic

lengh

tsof

thesides

oppositeto

thevertices

A,B

,C,so

a=

dist(B,C

)b=

dist(A,C

)c=

dist(A,B

).

Recallthat

geodesicsin

Dpassingthrough

0arestraightlinesegm

ents,so

geodesicspassing

through

points

closeto

0arealmostEuclidean-straigh

t,so

smallgeodesic

triangles

near0are

almostEuclidean.Recallthat

byGau

ss-B

onnet

(usingthat

K=

−1),

α+

β+

γ=

π−

Area(ABC)

soif

thetriangleis

small,

then

thearea

issm

all,

sothesum

ofthean

gles

isalmostπ

asexpectedin

Euclideangeom

etry.Allform

ulasyo

uwrite

dow

nin

thehyperbolic

world

shou

ld,

inthesm

alllimit,resemble

form

ulasforEuclideangeom

etry.

Inthelargelimit,ifweletthevertices

A,B

,Climitto

thebou

ndary∂D,then

thean

gles

α,β

,γconvergeto

zero

becau

sethegeodesicsareperpendicularto

∂D.Geodesic

triangles

withA,B

,C∈∂D

arecalled

idealtriangles,

andbyGau

ss-B

onnet

they

havearea

π.

21.6

Hyperb

olicgeometryand

Euclid’s

axioms

Non-examinable.Forthepurposesofthis

course,

thefollowingis

just

ahistoricalcuriosity.

Recalltheax

ioms(postulates)

ofEuclid

are:

(1)Thereis

auniquestraightlinethrough

anytw

odistinct

points,

(2)Anystraightlinesegm

entcanbeextended

toastraightline,

(3)Given

astraightlinesegm

entAB,thereis

auniquecircle

with

centreA

passing

through

B,

(4)Allrigh

tan

gles

arecongruent,

(5)Given

astraightline�an

dapointpou

tsideof

�,thereis

auniquestraightline

through

pwhichdoes

not

intersect�.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

99

Thepara

llelaxiom

(5)as

aboveiscalled

Playfair’s

axiom.Euclid’soriginal

axiom

(5)

is:Given

astraightline�,

andtw

ostraightlines

L1,L

2form

inginternal

angles

α1,α

2with

α1+

α2<

π=

(tworigh

tan

gles),

then

L1an

dL2intersect.

Thesetw

oform

ulation

sof

(5)

areequivalent.

Thereareother

interestingequivalentform

ulation

sof

axiom

(5):

•Thesum

ofthean

glesin

everytriangleis

π(Saccherian

dLegendre).

•Thereexists

atrianglewhosean

gles

addupto

π(Saccherian

dLegendre).

•Thesum

ofthean

glesis

thesameforeverytriangle.

•Thereexisttw

otriangles

whichare

similar

butnot

congruent(Saccheri).

•Given

atriangle,

onecanconstruct

asimilar

triangleof

anysize

(Wallis).

•Thereexists

atriangleof

arbitrarily

largearea

(Gau

ss).

Example.In

sphericalgeom

etry,that

istheunitsphereS2withthechordalmetricas

inExercise

Sheet2,

axioms(2),(3)and(4)hold.Axiom

(1)fails:thereareinfinitelymanygeodesicsjoining

theNorth

Poleto

theSou

thPole,

andaxiom

(5)fails:anytwogeodesicswill

intersect.

Wecan

makeaxiom

(1)work,

byusingRP

2insteadof

S2:that

is,weidentify

antipodal

points.Then

only

axiom

(5)fails.Thisiscalledellip

ticgeo

metry,andisanon

-Euclideangeom

etry.Itshou

ldbenoted

that

theSaccheri-Legendre

theorem

(thesum

oftheangles

inatriangleisat

mostπ)

holdsin

hyp

erbolic

geom

etry,butnot

inellipticgeom

etry,because

parallellines

donot

existin

ellipticgeom

etry.Theconstructionof

parallellines

ingeom

etry

usesthefact

that

astraightline

divides

theplaneinto

twoconnectedcompon

ents,butthegeodesic

from

theNorth

totheSou

thPolein

RP

2does

not

disconnectRP

2.Theproof

oftheexistence

ofaparallellinein

Euclidean

geom

etry

tacitlyassumes

theaxiom

oforder:that

giventhreepoints

A,B

,Con

astraightline,

oneandon

lyon

epointis

“inbetween”theother

twopoints.This

fails

forRP

2:canyousee

why?

Theore

m21.8.ThehyperbolicdiscD

satisfies

allofEuclid’s

axioms(includingtheaxiom

oforder)exceptforaxiom

(5):

Theon

lythingwestillneedto

prove

isax

iom

(3),

whichfollow

sfrom

thenextLem

ma.

Lemma21.9.Given

anytwopoints

A,B

∈D,thehyperboliccircle

1withcentreA

passing

through

B,is

aEuclideancircle

(whose

Euclidean-cen

treis

typicallynotA).

Proof.

Mob

iusmap

ssendEuclideancirclesto

Euclideancircles(allow

ingstraightlines),an

dhyperbolic

isom

etries

sendhyperbolic

circlesto

hyperbolic

circles.

SobyapplyingaMob

ius

isom

etry

wemay

assumeA

=0∈D.Since

rotation

sabou

t0arehyperbolic

isom

etries,the

Euclideancircle

withcentre0passingthrough

Bcoincides

withthehyperbolic

circle

with

centre0passingthrough

B.

Rem

ark.TheMobiusisometry

willtypicallynotmapEuclideancentres

toEuclideancentres,

indeedthehyperboliccentrewillliecloserto

∂D

than

theEuclidean

centre,

because

short

Euclideandistancesnear∂D

are

actuallyvery

longhyperbolicdistances.

1th

atis

thesetofpoints

{q∈

D:dist D

(A,q)=

r}wherer=

dist D

(A,B

).

100

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

21.7

Thecosineru

leand

thesineru

le

Non-examinable.Forthepurposesofthis

course,

thefollowingis

just

ahistoricalcuriosity.

Recallthat

inEuclideangeom

etry,thecosinerule

states:

c2=

a2+

b2−

2abcosγ(w

hich

isthegeneralizationof

Pythagoras’s

theorem,whichis

thecase

γ=

π/2).

Lemma21.10(C

osineRule).

Forahyperbolictrianglein

D,

cosh

c=

cosh

acosh

b−sinhasinhbcosγ.

Remark

.Forsm

all

x,cosh

x=

ex+e−

x

2∼

1+x+

1 2x2+1−x+

1 2x2

2=

1+

1 2x2and

sinhx

=ex−e−

x

2∼

1+x−1+x

2=

x,so

forsm

all

triangles

theabove

cosinerule

becomes

1+

c2 2

∼(1

+a2 2)(1+

b2 2)−abcosγ,whichis

theEuclideancosinerule

when

droppingorder

4term

s.

Proof.

Usingisom

etries,wemay

assumeC

=0andthen

rotatingwemay

assumeA

>0∈R.

Then

b=

dist(A,C

)=

2tanh−1(A

)byTheorem

21.7.SoA

=tanh

b 2.Rotatingbye−

iγwould

makeB

real

positive,

sothesameform

ula

would

apply,so

B=

eiγtanh

a 2.ByTheorem

21.7,

tanhc 2=

� � � �B−A

1−

AB

� � � �.

Therefore:

1

cosh

c=

1+tanh2

c 2

1−tanh2

c 2

=|1−AB|2+|B

−A|2

|1−AB|2−|B

−A|2

=(1

+|A

|2 )(1

+|B

|2 )−2(A

B+AB)

(1−|A

|2 )(1

−|B

|2 )

Sim

ilarly,cosh

b=

1+ta

nh2

b 2

1−ta

nh2

b 2

=1+|A

|21−|A

|2andcosh

a=

1+|B

|21−|B

|2.Theclaim

now

follow

sfrom

the

final

calculation

:

2(AB

+AB)

(1−|A

|2 )(1

−|B

|2 )=

2tanh

b 2tanh

a 2(e

iγ+e−

iγ)

sech

2b 2sech

2a 2

=2sinh

a 2cosh

a 2sinh

b 2cosh

a 22cosγ

=sinhasinhbcosγ.

Exercise.Show

thatalsoanother

cosinerule

holdsforD:

cosγ=

−cosαcosβ+sinαsinβcosh

c.

1Refresh

er:sinh(ix)=

isin(x

)and

cosh

(ix)=

cos(x)so

anyform

ula

involvingsines

and

cosines

gives

aco

rrespondingform

ula

forhyperbolicfunctionsprovided

youreplace

sinbyisinh.Soyoureplace

cos2,sin2,

tan2byco

sh2,−

sinh2,−

tanh2.Soth

eform

ula

cos2

x=

cos2

x−

sin2x=

cos2

x−sin2x

cos2

x+sin2x=

1−ta

n2x

1+ta

n2x

becomes

cosh

x=

1+

tanh2x

1−

tanh2x.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

101

Exercise.Show

thatin

sphericalgeometry,so

fortheunit

spherewiththechordalmetric

(ExerciseSheet2),

thecosinerulesbecome

cosc

=cosacosb+

sinasinbcosγ

cosγ

=−cosαcosβ+

sinαsinβcosc.

Lemma21.11(SineRule).

Forahyperbolictrianglein

D,

sinα

sinha=

sinβ

sinhb=

sinγ

sinhc

Proof.

BytheCosinerule,

sinh2asinh2bcos2

γ=

(coshc−

cosh

acosh

b)2

=cosh

2c+

cosh

2acosh

2b−

2cosh

acosh

bcosh

c.

Expan

dingcos2

γ=

1−

sin2γan

dsinh2=

cosh

2−1,

then

rearrangingterm

s:

sinh2asinh2bsin2γ

=(cosh2a−

1)(cosh2b−

1)−

cosh

2c−

cosh

2acosh

2b+

2cosh

acosh

bcosh

c

=1−

cosh

2a−

cosh

2b−

cosh

2c−

2cosh

acosh

bcosh

c.

Since

therigh

than

dsideissymmetricin

a,b,c,wededuce

symmetries

fortheleft

han

dside:

sinh2asinh2bsin2γ=

sinh2csinh2asin2β=

sinh2bsinh2csin2α.

22.

Appendix:ClassificationofRiemannsu

rfa

ces

This

Appendix

isnon-examinable.

22.1

Conform

alstru

ctu

reofaRiemann

surface

Recallthat

foraRieman

nsurfaceR

thetran

sition

map

areholom

orphic,so

their

derivatives

Dτ≡

τ� (z)areacompositionof

scalingan

drotation

,so

Dτpreserves

angles.So

Euclideanan

gles

defined

inlocalparam

etrization

sareindep

endentof

theobserver.This

iscalled

theconform

alstru

ctu

reof

R.

Itthereforemak

essense

torestrict

one’sattention

toon

lythoseRieman

nianmetrics

onR

forwhichthean

gles

measuredusingthemetric1

agreewiththeab

ovewell-defined

angles.

Since

inalocalholom

orphic

coordinate

z=

x+

iy

thestan

dardbasis

vectorse 1

=∂xan

de 2

=∂yform

arigh

tangle,such

metrics

haveno

diago

nal

term

sdxdy,so:

I=

f(x,y)(dx2+

dy2)=

f(z)|dz|2 ,

whereof

coursef(x,y)>

0is

apositivesm

oothfunction(positivedefinitenessof

I).

Example.For

H,f(x,y)=

1 y2givesthestandardhyp

erbolic

metric.

Theab

oveisjust

alocalexpression,so

forsuch

Ito

yield

awell-defined

Rieman

nianmetric,

youneedthelocalfunctionsfto

becompatible

withchan

gesof

coordinates.2

Theequivalence

classof

allsuch

metrics

iscalled

theconform

alclass

ofR.(T

wometrics

areequivalentifthey

only

differ

byapositivescalingfunction).

Recallfrom

Section

5.3:

1Recallco

sθ=

I(v

,w)

√I(v

,v)√

I(w

,w)forvectors

v,w

�=0∈

TpS

defi

nes

anangle

±θbetweenv,w

.

2So

� f(�z)

|d�z|

2=

f(z)|dz|2

foraholomorp

hic

transition�z=

τ(z),

so� f(τ(z))

|τ� (z)|2

=f(z).

102

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

Theore

m22.1

(Rieman

nmap

pingtheorem).

Every

simply-connected1Riemannsurface

isbiholomorphic

toeither

CP

1,C

orH.

Aspecialfeature

ofcomplexan

alysis,isthat

aholom

orphic

hom

eomorphism

isau

tomatically

abiholom

orphism

(unlikein

real

analysis,whereR

→R,x�→

x3isasm

oothhom

eomorphism,

butitsinversex�→

x1/3isnot

smooth).

Indeed,since

itisahom

eomorphism,thelocalform

ofthemap

isz�→

z,so

theinversefunctiontheorem

guaran

tees

aholom

orphic

inverse.

Theore

m22.2

(Uniformizationtheorem

dueto

Poincare

andKoeb

e).

Every

compact

Riemann

surface

RhasametricofconstantGaussian

curvature

within

its

conform

alclass.In

particular,

byGauss-B

onnet,itfollowsthatifK

>0then

χ(R

)=

2so

Ris

topologicallyasphere,

ifK

=0then

χ(R

)=

0so

Ris

topologicallyatorus,

andifK

<0

then

χ(R

)<

0so

Ris

topologicallyasurface

ofgenusg≥

2.

Wewillsketch

theproof

ofTheorem

22.2.For

this,weneedatopolog

ical

preliminary.

22.2

Universalcoveringsp

ace

Asshow

nin

thecourseB3.5:Topology

and

Gro

ups,

forareason

able

2connectedan

dpath-con

nectedtopolog

ical

spaceR

(for

exam

ple,foran

yconnectedtopolog

ical

surfaceor

man

ifold)on

ecanform

auniversalcoveringsp

ace

� R.Explicitly,

� Rcanbeconstructed

asthespaceof

equivalence

classesof

continuou

spathsγ:[0,1]→

Rin

Rstartingfrom

afixed

base-pointγ(0)=

p0.Twopathsγ1,γ

2aredefined

tobeequivalentiftheendpoints

agree,

γ1(1)=

γ2(1),

andifγ1canbecontinuou

slydeformed

into

γ2whilst

keepingfixed

boththe

initialpointp0an

dtheend-point.

Theproperties

required

for� Rto

beauniversalcoverare:

(1)

� Ris

asimply-con

nectedtopolog

ical

space(inparticularconnected)

(2)thereis

acontinuou

ssurjectivemap

π:� R→

R,called

pro

jection

map,whichis

alocalhom

eomorphism,

(3)locallyπlook

slikea“stack

ofpan

cakes”:

nam

ely,

arou

ndan

ypointp∈

Rwecan

findasm

allop

enneigh

bou

rhoodV

such

that

π−1(V

)=

�Ujis

adisjointunionof

open

sets

Uj,called

sheets,each

hom

eomorphic

toV

via

π:Uj→

V

Intheexplicitconstructionof

� Rmention

edab

ove,themap

πisπ[γ]=

γ(1):

just

map

thepath

toitsendpoint.

Onecandefineagrou

pG,called

deck

gro

up,consistingof

hom

eomorphisms

ϕ:� R→

� Rcompatible

withtheprojection:π◦ϕ

=π.Solocallyϕpermutesthesheets.One

caneasily

checkthat

ifϕhas

afixed

point(ϕ

(�r)=

�r)then

ϕistheidentity

map

.It

turnsou

t

that

Rcanbeidentified

withthequotientR

=� R/

Gthat

is,thepoints

ofR

canbeviewed

astheorbitsof

theG-actionon

� R.Example.Theuniversalcoverof

thecircle

S1isthereal

lineR,withprojection

π(r)=

e2πir.

Thedeckgrou

pG

isallintegertranslationsr�→

r+

n,n∈Z,

soG

asagrou

pisisom

orphic

toZ

withaddition.Noticethat

indeedS1=

R/Z

.

Theuniversalcoversatisfies

thefollow

inguniversality

property:

given

any

twouniversal

covers

� R 1,� R 2

ofR,thereis

ahom

eomorphism

ψ:� R 1

→� R 2

compatible

withtheprojection

map

s:π2◦ψ

=π1.It

follow

sthat

thedeckgrou

psareisom

orphic:G

1∼ =

G2.

Example.Another

universalcoverforS1isR,withprojection

π2(r)=

eir.Thedeckgrou

pG

2

1Sim

ply-connected

mea

ns:

connected,and

everyco

ntinuousloop

can

beco

ntinuouslysh

runkto

apoint

(everyco

ntinuousmapS1→

Sca

nbeex

tended

toaco

ntinuousmapD

→S

onth

eclosedunit

disc).

2Oneneedsa

tech

nicalco

ndition:

semi-loca

lly

simply-connected.

This

mea

nsev

ery

pointp

hassome

neighbourh

oodV

such

thatloopsin

Vca

nbeco

ntracted

within

Vto

apoint.

B3.2

GEOM

ETRY

OF

SURFACES,

PROF

ALEXANDER

F.RIT

TER

103

isalltranslationsr�→

r+

2πn,n∈Z,

soagainG

2∼ =

(Z,+

).Thehom

eomorphic

identification

withthepreviousexam

ple

isR

→R,r�→

2πr.

22.3

Sketch

pro

ofofth

euniform

ization

theorem

SketchproofofTheorem

22.2.Given

acompactRieman

nsurfaceR,consider

itsuniversal

cover� R.

As� Rislocallyhom

eomorphicto

R,wecanuse

thesamelocalholom

orphiccoordinate

on� Rason

Rvia

this

localidentification.This

mak

es� Rinto

asimply-con

nectedRieman

n

surface,

and

theprojection

map

π:� R

→R

isau

tomaticallyholom

orphic.

Since

πis

alocalhom

eomorphism

(thelocalmodel

isz�→

z),

πhasnobranch

points.Thedeckgrou

p

automatically

consistsof

holomorphichom

eomorphisms� R→

� R,so

they

arebiholom

orphisms.

BytheRieman

n-m

appingtheorem

22.1,� Ris

biholom

orphic

toCP

1,C

orH.

If� R∼ =

CP

1,then

� R,R

arebothcompact

sothereareon

lyfinitelymanysheets

nam

ely

deg

πsheets.Since

πhas

nobranch

points,byRieman

n-H

urw

itz

χ(� R)

=deg(π)χ

(R).

Butχ(� R)

=χ(C

P1)=

2,an

dχ(R

)=

2−

2g∈

{2,0,−

2,−4,...}(since

Rieman

nsurfaces

areorientable),

whichforces

χ(R

)=

2an

dso

deg

π=

1,so

πis

abiholom

orphism.SoR

isbiholom

orphic

toCP

1.SowecangiveR

theusual

metricon

CP

1∼ =

S2⊂

R3withK

=1.

If� R∼ =

C,then

thedeckgroupG

consistsof

asubgrou

pofthebiholomorphismsf:C

→C.

Butsuch

biholomorphismshave1

theform

f(z)=

az+

bfora�=

0,b

∈C.Since

thenon

-identity

deckgrou

ptran

sformationshavenofixed

points,az+

b=

zmust

not

havean

ysolution

,whichforces

a−

1=

0andb�=

0.Sof(z)=

z+

bare

thetran

slations.

These

preservetheflat

metricdx2+

dy2=

|dz|2

on� R,

sothequotientR

=� R/

Gcanbegiven

the

flat

metric,

soK

=0.

If� R∼ =

H,weclaim

that

thedeckgrou

ppreserves

thehyperbolic

metric,

sowecangive

R=

� R/G

thehyperbolic

metric,

soK

=−1.

This

claim

follow

sbyLem

ma22

.4.

�Lemma22.3

(Schwartz’sLem

ma).Anyholomorphic

mapf:D

→D

withf(0)=

0satisfies

|f(z)|≤

|z|forallz∈D.

Proof.

f(0)=

0,so

g(z)=

f(z)/zis

holom

orphic

witharemovab

lesingu

larity

at0,

which

isremoved

bydefiningg(0)=

f� (0).Themax

imum

modulusprinciple

applied

tothedisc

ofradiusrim

plies

that

|g(z)|

≤1 rfor|z|≤

r(usingthat

|f|≤

1,andthat

|z|=

ron

the

bou

ndary

ofthat

disc).Let

r→

1from

below

,then

|g(z)|≤

1forallz∈D,so

|f(z)|≤

|z|.

�Lemma22.4.Anybiholomorphism

f:D

→D

ofthehyperbolicdiscpreserves

thehyperbolic

metric.

Proof.

Com

posingwithaMobiusisom

etry,wemay

assumef(0)=

0.BySchwartz’sLem

ma,

|f(z)|

≤|z|.

Since

fis

invertible,also

|f−1(z)|

≤|z|s

o(rep

lacingzbyf(z))

weget|z|≤

|f(z)|.

Hence

equalityholds:

|f(z)|

=|z|.

Recallbythepreviousproof

that

g(z)=

f(z)/z

isholomorphic.

Since

|g(z)|

=1,

themax

imum

modulusis

attained

atan

interior

point,

thereforethemax

imum

modulusprinciple

implies

that

gis

constant,

sayg(z)=

eiθ.Thus

f(z)=

eiθz.Weknow

this

isan

isom

etry

ofD,so

theclaim

follow

s.�

1Indeed,th

eholomorp

hic

functiong(z)=

f(1/z)defi

ned

onth

epuncturedunitdiscD\{

0}h

asasingularity

at0whichis

eith

erremovable,apole

oranessentialsingularity.If

itwasanessentialsingularity

then

byth

e

Casorati-W

eierstrass

theo

rem

2,gonD

must

attain

values

arb

itrarily

close

tof(0)∈

C.Thisco

ntradicts

thatf

isinjective,

since

g(w

)=

f(1/w)for|1/w|>

1would

attain

avalueth

atf(z)attainsfor|z|<

1(close

tof(0)).

Sog(w

)hasapole

at0,saywithprincipalpart

b nw

−n+

···+

b 1w

−1.Then

f−

b nzn−

···−

b 1z:CP

1→

Cis

aholomorp

hic

function

(wegotrid

ofth

epole

atinfinity)so

itis

constant.

Sof

isapolynomial.

By

injectivityoff,weded

uce

n=

1,so

f(z)=

az+

basrequired

.