1.1 Four classes of surfaces - University of Oxford...Examples 1.1 Four classes of surfaces Our goal...
Transcript of 1.1 Four classes of surfaces - University of Oxford...Examples 1.1 Four classes of surfaces Our goal...
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×
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
sΛ
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⇔
v×
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�
Dτ
�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:
a×
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
2×
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
2×
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
1×
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
n×
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
,γ
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
n×
∂yF.
Since
∂xF,∂
yF,n
areorthog
onal,n×
∂yF
isparallelto
∂xF,an
dtakinginto
accountorien
-tation
san
dlengths(�∂xF�2
=G,�∂
yF�2
=1):
n×
∂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
sτ
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
.