Differential Geometry- curves and surfaces -

I.

Introduction

it.

What is

DifferentialGeometry

How can you tell if you"

live"

on the surface of a ball

--

- --

.

.

§called a sphere

-or 2- sphere

or the plane

?one way is to look at"

straight lines' '

in the plane of it two people walk in the

✓"

same direction"

from different

points they stay a fixed

distance app art

( parallel lines don't intersect )

but on the sphere

µ the distances get

..closer together

that is,

the"

geometry" of

"

lines"

on the sphere is different

from the geometry of the plane .

SO we see the"

curvature" of the sphere by looking at

straight lines in the space

True scion : What about the 3 - dimensional space in which we live ?

isit' ' flat Euclidean space" ?

is it a

"

3 - dimensional sphere" ?

something else ?

General Relativity postulates that gravity can be understood as

a

"

curvature "in space lain e)

.

The language to study all these ideas is RiemannianGeometry

or more generally DifferentGeometry

and it all starts with studying curvesandstraightlines

This course is an introduction to Riemann coin Geometry

through curves and surfaces in Euclidean spaceC see list of topics on the web page )

B.

The geometry of Euclidean Space

IR "= { I p, , . . . , p n ) I pi a real number

,i.e

. p ,E IR }

we can think of D= Ip , , . . .

, Pn ) as a pointin IR"

or atech ai IR"

• F

when thinking of B as a vector we will frequently write it as a

column vector

op

given pig EIR"

then their dot product is

F.9- = pig ,t pzgze . . . t paan = ⇐

,

Pi 9 i

The

we sometimes write 48,97 for F.5 and thisgives an inner

product on IR"

,that is l .

,. ) satisfies

D 4 pig ) = II. p )symmetric

2) tap , g) = a sp. 97=48 , af )

I Ft9- if ftp.ry-sg.ry ) linear

'

3) I F, p ) Z O

I F , p > = o ⇒ p } positive definite

geometry is about lengthsandangles ,with a dot product we

can define the length of F to be

lip H - LEFTand the angle between B and of to be

cos a = CBI it

1181111911 €p

note for this to be well-defined,

we need

lemma I ( Cauchy - Schwartz inequality)

for all F. g- E IR"

KpiIHe 1101111511

with equality it and only if I and g- are linearly dependent

Proof nice trick : compute the length of a linear combination

Of Hap tbg IE tap tbg , apt bop )

= a 'll pl ft b 'll g H 't Zab 9.5 )

so it a = Hghand b = I HFH,

then we have

0 I 2118112119712IZHPTIH-gtkp.gl= 21181111911 ( lip 1111911 I 6,5I ) } ④

so if Hp Ht 0411911,

then

± tph > E 11,51111511 ( if either HFH -

- o or 11511=0

then E is obvious )and ltpllllglkmax ftp.T )

, -45,57 )= 145.9731

thus the E is the lemma is true

note : assuming HFIH of 11911 then

trig 7 = Itp1111911⇐

we have equality in ④

⇐

Http49 I

HgYp-H" on -

i

aeaeracy⇒

of I er product

krillg- I Hgtlp = O

-tie

. p and g- are

linear- ly dependent #

The standard distance between points in' IR

"

is .

d I pig ) = Itp - g- It

a metric on a set X is a function

( metrics describegli stanced : X x X → IR

between pointssuch that

i ) dlp ,g) Z O with equality ⇒ p= q

a ) dip . 9) =D (9. P )

triangle inequality3) dcp , g) E d Cp .

rt t dlr, 9 )

forexercise . Show that d (pig ) above is ap #

qmetric on IR "

given two metric spaces I M,

,d , ) and ( Mr ,da ) an isometry

is a surjective function

to : M,

→ Me

such thatdip ha

,fly ) ) = d. I x

, y) for all X. y EM,

Isometries identify points of M, with points of Mz so that

distances are preserved . They are" symmetries " of

spaces with metric 's

We are interested in isometries from DR"

,d ) to itself

Notice any"

geometric quantity" should not change

under isometries leg length of a curve . . . )

An orthogonal transform is a linear map

A : IR"

→ pinsuch that

ftp.Agl-hp.gl for all pig

Theorem 2 :

If f : IR"

→ IR"

is an isometry ,then there is

some I E IR"

and orthogonal transform A

such that tip ) = a- t Ap

Proof let I (B) = fcp ) - f- CE )

it we show Iis ① linear andn ~

② satisfies Lf IF),

f ta ) ) =L pig 7?

then we are done since we can set A = t and I = flotto get

tipi' Atta

note LI - I , I -77=115114117112-245,77

So245

,7--115112+117112 - HI-FIT

thusz C

ftp.FGD-xfiptlftHFGTH-HFirst-

Fight= Http ) - f- CEN HI f Cgt - f IoTH

'

- It f Cpt - f GT ITisometry? ftp.olf-illq-olf-llp -9112

-

= 110112+119112 - lip - g- If = up . I >

so f satisfies ②

now let E,

. . . ,En be an orthonormal basis for R

"

leg. It:o) , E = LI:o) ,

. -. )

- -

exercise : f let), . . . ,

f ten ) isalso an orthonormal basis

for IR" because of ②

so for an

c peg ,,Ice, > = Lptg, = IF .-9745.57- ~ ~ ~

= http ),

f ( Edt Hlf ),

f let ) )~ a

~

= ( f (f) tf Cg- ),

f (Ed ) for all i

~ ~~

and thus f ( ptg ) = ftF)t f CE)

exercise . Prove this it it is not clear to you

Hint : To,

. . . In an orthonormal basis ,then

E = I ⇐ 48,5;> stew ,Ii > for all i←

~

scimitar ly I flop ),

flea ) ) = top ,Ez > = Csp , -97

- - - ~

= ch f ( p ),

f (Ee ) ) = L of IF ),

f IED- -

so t ( c p ) = c f CF )~

and thus f is linear#

So any isometry of IR"

( also called a rigidmotion ) is

a composition of

① an orthogonal transformation

f- C p ) = Af and

② a translationf- I ft F ta

we understand ②.

let 's explore ①

Recall : given a linearmap A : IR "

→ IR"

we can express it as an nxn matrix

e. g .let I ,

. . . En bethe standardbasis for IR"

AE,= a

, f-it . . .

t an ien

I etma .

. laida

:

any vector can be written

I = Viet . . . then =

then AT corresponds to the vector MAL ?;)from now on we will think of A as the matrix above

that rep resets it in this basis

now with Tr and I =

" te

L f,-w , = f. in = Ftw where IT means the transposeof v

re.

switch rows and columns

for any matrix A

( Atv,

I > = LATE) = TEA'T tu = E' AT = to,

AE >

if A is an orthogonal transform,

then

IT,

AT ) = LATE,

I ) - LA Atv,

AT )

SO

( T . A ATF,

AT) = O = L 8

,AT ) for a1

wi. if we let I run through an orthonormal basis I

,. . .

Ten

we seeJ - AAT F =D

fun identitymatrix

soA ATF = I = I dnt

and

A AT = Idn

this implies 1- - detlldn ) = det ( AA' ) - (detA)

'

SOdef A =

I 1

it def A = I,

we call A a special orthogonal transform

A side . Old = I orthogonal transforms of IR "

}

SO In = I special " " }

are examples of Liegroups_ ,the study of these is

a beautiful and deep area of math

Isometries of IR ?

If A = ( Edb) is a special orthogonal transformation

then I = det A -

- ad - be

and too .tl :ballscat %a

so we have a 2+5=1at d 2=1

.( d. c)

act bd -

- Ono

ad - be = I

← unit circleF unique angle 0 set

.

D= cos o

( = sin'

0

now [ ba) . Idc) = O so [ ba ) is a unit vector

orthogonal to [ I ]

f- sure , lose )• .

I cost,

Sino )b a µ

dc

•

(

cost,

- Sino )

b a

finally ad - be = 1 ⇒ a = co so

b =- Sino

so A = ( cost - sin a

Sino cost)

and A corresponds to arotationabout the origin by angle 0

exercise it A not special, but just orthogonal, then

A = ('

o? , ) . ( cos o - since

~sin cos o

) some a

reflect about x - axis

so rigid motions of 1122 are compositions of :

rotationstranslations

,

and

reflectionsaboutx-axis

exercise . Isometries of IR'

are compositions of

rotations about some line,

translations,

reflections about xy - planereflections through the origin

exercise . let E . . . En be anyorthonormal basisfor IR

"based

at a point p E IR"

and I . . . In be another orthonormal basis for IR"

based

at a point g- E IR"

Then there is an isometry to : IR"

→ IR"

suchthat

e-a lol p ) =

gif -HE

,I Dotplez ) -

-f

,%I we total derivative of 4 at I£

Recall . given a function

F : IR"

→ Mm

we can write it

FIX, , . . . Xn ) = ( f,

I x, . . .Xu )

,. . .

, fmlx, , . . . xn ))then

D Ep : IR"

→ IRm

r avectors based actors basedat p at g-

Hint . consider the case

where E, . . .En is the is a linear map

that can be expressedstandard basis and f -

- O as the mxn matrix

then consider 2 f - ,1- I p- ( p

¢ to) = g- t AT

me . nai:÷s÷i÷÷.

. . . . .

÷÷÷÷÷÷¥⇐÷m

I Curves

A. Curves in IR"

A curve C is the coinage

ofa coat in in a ou s function

I : I → IR"

where

I = I a. b) or Ca. b ]

example . I It ) = C t,

t ) t E lo,

I ]

§ I t ) = ( It,

I t ) t E Eo ,

' Iz ]

@

J It ) = I t'

,t

' ) TECO , D c

@

note .

D all 3 functions have the same linage ,so a given

curve can be described by many different functions

2) we say that I I or I or J . . . ) parameterize the

curve C

3) We can think of C as the pathofaparticle or

a piece of wire in IR"

4) we frequently confuse C and I but remember we

are really interested in C,

I is just a convenient

way to describe C mathematically5) Curves do not need to be given by a parameterization

e. g . g I e y'

= I

of course we can also parameterize it : I HI =L cost ,sin t )

t E [ o,

21T ]

Remark There is a subtlety in the definition of curve

really it is not just the linage of I but the"

trajectory "

of the particle traveling along C

<

example .

the coinage of I and f are the same

but the order in which parts of the pathare traversed is different so we will

say cheese are differentcurve

So really we should think of a curve C as

the image of some I :C a. 53 → IR "

together with the order in which the

points on C are encounter e

examples :

it

Straight-line

given points p and g- in IR"

I think of them as vectors )then

I th = (I - t ) f t tf t Eco,

I ]

parameterize the line from p togy z

e. g .

C o

@

x I ,@

ICH -

- l-f) lift 13 ) x

-

- ft

I;] Titled -A t tf?)

"

circles

=L' ¥ I

given r EIR, r > o

F C- IR2

then I Ltt = Ft ( r cost,

r s int ) t t lo .2T ]

parameterize s the

firdeofradiusraboutpe

's .

a-us .

- I ? ) - if =L ? I iii.ITy

:x

3) Helix

given r,

h E IR,

r > 0,

h Z O

set I I t ) = ( r cost,

r s int, ht ) t e IR

2-

↳9 y

x

4) I It ) = ( t ? t' ) cusp

t e IR ✓T

given a curve C, parameterized by a function

I.I → HM

IIt ) = It,

Itt, Htt

,. . .

,data )

recall from calculus that at the point I Lto ) c- C a tangentvector to C is given by

I ' Ito) =L dittos,- . .

,a

'

nIto D

Tito,

•→ It

Remarks : it Actually I'

Hot is a vector based at to,

o,

. . . ,o )

When we say a vector T is based at f,

then we

shift I so Its' ' tail

"

is at p

? off to based atp✓

so we should say I '

Ito ) based at Itto) is tangent to C

2) If T to,

then the line spanned by I is

↳ = { rt I r E IR }

it E is based at F then the line through F in

the direction of T is

{ r Ttp I re IR }

fir• pI

so the tangent line to the curve C at D= It to )

is given byTp C = { It to ) tr I

'

Ito ) I r E IR }

Tpc"

Tpc

Recall : the tangent line to C at p is the"

line that best

approximates C at B"

a parameterized curve I HI is called regular if I' let

existsand isnon-Zero for all t

we need this to talk about tangent lines and

manyother thingsso we usually assume curves are regular

( except maybe at afinite number of points )

note . let I : I → IR"

parameterize a curve

if I ' '

It ) = O for all t,

then I parameter I Ees ( part

of ) a line

to see this note

I' '

Its = I Ii' Itt

,. . . , In

"

I t ) ) = I O,

. . .

,O )

SO

x'j It ) = O di . I → IR

integrate to get

ditty = f ai'

It ) d t t vz = Vi for some constant Vi

SO

wtf Sai Hdt t pi = vet t pi

that is

* , ftp.fef?rnI--p- + it

so the second derivative tells us how far I is

from being a line

Problem : I' '

is not a geometric quantity !

i.e .it depends on the parameterization not just C

e. g .

I It ) = I t,

t ) te Eo ,it

f- Lt ) = It ? E ) te Eo.

, ]

I "

I t ) = I o.o ) t I"

I t ) = ( 2,

2)

also HI " 11=0 117" 11=252

but both give .

⑧

so I" doesn't necessarily give us information

about the curve C it parametrizes

( e.g .can't tell example is a line from I " )

to fix this we need to consider arelengthRecall from calculus that if C is parameterized by

I :[a. b ] → IR "

then the length of C is given by

length (c) = Sba HI ' # Hdt

or more generally the length along C from

the end point I Ca ) to Ics ) is

lemmaI .

lls ) = Sas HIGH It

It I is a regular parameterization of a curve C,

then we can re parameterize C by another function

To:[ o

, e) → IR"

such thatup

'

I s ) It = 1 for all s

Remarks :

, ) if I :[ a. b) → IR " parameterize s a curve C and

B- :c c. d) → IR" is another parameterization of C

then we say I is a reparamecerizats.co f C

note : if I is one - to - one ( re.

I CE,

) = I Cta )

then E,

= ta )

then for each s E Ead ],

there is a unique

ts Ela . b) such that Icts ) = J Cs )

F fist Icts )

[ c ,d ] - ~• C

t f[ a. b ]

so set f : I c. d) → Ca . b ] : s t t,

and we see

f- C s ) = I ( fist ) = I of C s )

exercise : Convince yourself that you can

find f- as long as I is regular( i.e

.don't need one - to - one )

Conversely , given anyfunction h :[ he ] → La. b ]

Joh : ( h.e) → IR

"

is a re parameterization of I

2) We say that I :{ a. b) → IR"

parametrizes C by arc length

it It Its)H=1 for all s e [ o.

l ]

note given such an I we have

list = S !HI'

lxllldx = s

i.e. length of C from Ilo ) to Its ) is s

so lemma says regular curves can alwaysbe parameterized by arc length

Proof given I :[ a , 63 → IR"

parameter it ing C

let f CE ) = Sat II I'M Hdx

the fundamental theorem ofcalculus says

date = HI' kill -5%35%7so f is increasing on [ a. b ]

:. f is one - to - one

it f- fl b) = length of C then f also onto [ o,

l ]

e -

J t 't '

I I t

a b

exercise fat has an inverse

I recall this is a function

g : [ o, e) → La ,

b ]

Sf. fog C s ) = s and got Lt ) = t )

chain rule givesdads Htt ) ) date ) -

- dat ( go f) HI -

- Fett = I

SO dates ) = daft, where s - Ht ) and t =

g Cs)

now set B Csl = Ilg HD

so B : foie ] → IR"

parameterize s C

and

Itp'

It -

- HI '

lgcsb g' Coll = HI '

Ist dat, all = " I ' kN ¥*y= HI ' IHH ¥ HIT

= 1EH

example . Helix

Itt ) = ( r cost,

r suit,

be ) te Co ,x )

I 'It ) = ( - r suit

, r cost ,b)

It I '

HN -

- r'sin2ter7os7t5# = FELT

so t

f- It ) = So ¥5 dx - FEE t

the inverse of f is

f- II s ) = bzS

SOBcs ) = I tf - '

I s )) = ( r costs s,

r sin Is , bf¥ )

is a parameterization by arc length

Notation .

When we use the

variablesis a parameterization of

a curve we will always mean we have parameterized byarelength, where t is used for any parameterization

now given a parameterization To : lo, e ] → IN " of a curve C by

are length we say

IN = pics )

is the unit tangent vector

lemma 2 .

I '

Cs ) is perpendicularto Test

ProofTis ) . Fcs) = It Fcs ) It

'

= I

the product rule gives

O -

- Ist = Is IF = (IsF) It to lads'T)

= 2 F '

19 . Fcs)

so I '. F = O

¥7

We call NTS) = Its ) the unit normal vector to C at pics)

Fl s )

¥7 .

>

s)

F' 1st

Wecall Kis ) -

- It I '

Call the curvature of C at post

note : Kcs ) = HI"

Csl It is the ace Iteration of a particle movingon C with unit speedso you feel the curvature of a road while driving !

exercise :

1) Show the curvature of a curve C is independent of param .

this says 2) Show it I : IR"

→ M " is an isometry then the curvature

K isreally

ggE off' Ifis the same as the curvature of Icc ,

tellingyou

how 4

C sitsin IN

3) It I '.

Iab ] → IR"

is any regular parameterization of C

then show you can calculate the curvature of C at Ict )

BY* = attest

4) Show directly that the formula in. 3) is independent of

parameterizationTh " 3 .

aregular curve C is C part of ) a line

if

curvature of C is 0 at all points

Remark .

So curvature is preciselythe measure of how

far a curve is from being a line !

Proof :# If C is aline from p tog then

Test -

- tf p t IF

for s E lo ,e ] l -

- Hp - IH

is an are length parameterization of C

pts) =- tep t

tfS o

Kcs ) = It I " call -

- toll =

g

⇐ ) let Gls ) be an are length parameterization of C

assume KCSI = O so B "

Cs ) = 8

then we saw earlier that Bcs ) parameter ites part

of a lineat

example : recall an are length parameterization of the Helix is

I Csl = ( r cos ¥+5,

r sin SEE ,

"

is'

is ( -

asin ¥5

,

us Fees,

¥+E )

and

I" 1st =L - ¥5 cos ¥5 , ¥5 sin ¥I ,

o )

thus

KC s ) = Hf " call = rr¥b

Again let§ : lo . e) → IR

"

be an are length parameterizationof some curve C

recall Its ) = F' is ) and I Cs) =Fcs ) are unit

orthonormal vectors in IR"

so they span a plane in IR"

P = { a Flat b Fcs ) I a. b E IR)

translate P so thatit godsthroughBcs , =p

Ppc= { peso ) ta TT so ) t bit Csd I a. bEIR )

this is called the osculating plane to C at p = I Go)

.

i :c

note . Ppc contains the tangent line Tpc

exercise Convince yourself that Ppc is the plane that C

comes closest to lying in atp

I later we will see precisely when C lies in Ppc )

Recall given 3 points I ,I

, xj in IR"

that do not lie on a line

then they① determine a unique plane

PII ,I .

= I ,t span I Ii - I , ,

I ,- I ,

}

② deter min a unique circle Ctx , ,IT in PCI , E. Is )

" """" '

Facts : let F ! Lo, e) →IR" be a regular parameterization of C

suppose so C- [ oil ] Sf. K Cso ) ¥0

I ) for points s, , Sa

,s ,

C- Co,e ] sufficiently close to So

§ Is.)

, fish , f- Cs,) do not lie on a line

I ) the osculatingplane is

Pasok=

, .

!Ys,

→PCB 's

. ? Goal , ptsd )

HI ) The limit im CC fist , fish , ptsd )4.5.55 ' So

is a circle Cpcso ,in PBisnt

it is called the osculating circle and can be

parameterized byIN = pics , + ¥ ,

N' Iso) t IT, , Iint ) Icsd + lost ) Is;]

so the circle has radius ¥ ,

~ .

osculating circle

note i ) osculating circle is tangent to C at B

( has "

order 2"

contact with C )

2) if K is close to 0 then C is almost straightif K is large then C curves a lot

.