Chapter 2 Global theory of curves

Question 1

smooth

Answer 1

A vector funct is smooth on I if it extends to an infinitely differentiable vector function defined on open interval J containing I

e.g I=(a,b) inff diff on x
I=[a,b]-smooth funct on I extends to smooth funct on (a-e,b+e) e>0

Question 2

PARAMETRISED CURVE

Answer 2

A PARAMETRISED CURVE is a SMOOTH vector function

Question 3

REGULAR

Answer 3

if γ’(t) =/ 0 for all t in I

Question 4

VELOCITY

Answer 4

γ’

Question 5

SPEED

Answer 5

|γ’|

Question 6

e.g of parametrised curves
simple circular curve

when is it regular

Answer 6

parametrised plane curve
γ:R to R^n
γ(t) = (rcos 2πt, rsin 2πt) r>0

regular
γ’(t) =(-2πr sin2πt, 2πr cos 2πt)
|γ’(t)| =2πr>0 never 0

Question 7

e.g of parametrised curve
is it regular
γ:R to R^3
γ(t) = (t^3,t^2)

Answer 7

γ’(t) =(3t^2,2t)
=(0,0) for t=0
not regular

Question 8

ACCELERATION

Answer 8

γ’’

Question 9

LENGTH

Answer 9

L(γ) = Integral_[a,b] |γ’(t)|.dt

Question 10

e.g of parametrised curves
straight line

when is it regular

give another repara

Answer 10

γ:R to R^n
γ(t) =a+tv a,v in R^n
regular when γ’(t) =v iff v=\0

γ~(𝜏)= a+ (tan𝜏)v
𝜏 in J=(-pi/2,pi/2) is a reparameterization of γ
φ(𝜏) = tan𝜏 regular
but
𝜏^3 wouldnt be regular
-𝜏 would be traversed in opposite direction

Question 11

circular helix
is it regular?

Answer 11

γ(t) = (rcost,rsint,ht) r>0

regular
RPC:
for any h in R
γ’(t)=(-rsint,rcost,t)
if h=0 but periodic so never equal to 0

Question 12

DEF
RE-PARAMETRISATION

Answer 12

of a parametrised curve
γ:I to R^n
is another parametrised curve
γ~ : J to R^n s.t

γ~= γ∘φ
where
φ:J to I is a SMOOTH SURJECTIVE funct s.t φ’(t)>0 for any t in J (strictly increasing)

RETAINS IMAGE, DIRECTION TRAVERSED, #TIMES IF CLOSED

Question 13

γ~= γ∘φ
where
φ:J to I is a SMOOTH SURJECTIVE funct s.t φ’(t)>0 for any t in J (strictly increasing)J to I is a SMOOTH SURJECTIVE funct s.t φ’(t)>0 for any t in J (strictly increasing)

Answer 13

φ
is called the parameter transform
allows us to evaluate original PC using different input φ(t) for t in J

and is a BIJECTION
as injective

Question 14

γ~= γ∘φ

Answer 14

used to reparametrise

Question 15

consider
γ(t) = (a + t(b-a)) t in [0,1]

Answer 15

thus we can always define a PC on [0,1] from [a,b]
using the reparametrisation
φ’(t)>0

φ(t) = a +t(b-a)
φ(0)=a
φ(1)=b
can be used for smooth surjective funct

Question 16

PROPN 2.1
which re-parametrisations of regular closed curves are RPC

Answer 16

EVERY re-para of an RPC is an RPC

proof:
by chain rule
γ~’(𝜏)= (γ∘φ)’(𝜏)
= φ’(𝜏) γ’(φ(𝜏) =/0
we know
φ’(𝜏)>0 for all 𝜏 in J
γ’(t) =/0 for all t in I
thus PC γ~ is
REGULAR

Question 17

length of arc

Answer 17

Let γ : I → R
n be a PC. For a closed interval [t_0, t_1] ⊂ I the quantity
L(γ| [t_0, t_1]) =
integral_[t_0,t_1]
|γ’(t)| dt

is called the length of the arc
γ| [t_0, t_1].

Question 18

arc length for repara
prop 2.2

Answer 18

The length of an arc does not change under a re-parametrisation

Question 19

proof: prop 2.2

The length of an arc does not change under a re-parametrisation

Answer 19

Let ˜γ : J → R^n be a re-parametrisation of a PC γ : I → R^n, that is ˜γ = γ ◦ϕ, where ϕ : J → I is a function that satisfies the hypotheses of Definition II.3. Suppose that ϕ maps [τ_0, τ_1] → [t_0, t_1] bijectively. Then by the change of variables formula, we obtain
L( ˜γ| [τ0, τ1]) =
∫{τ_0,τ_1] |γ˜’(τ )| dτ =
∫{τ_0,τ_1] |γ’(ϕ(τ ))| ϕ’(τ )dτ =
∫_{t_0,_1] |γ’(t)| dt
= L(γ| [t_0, t_1]),
ϕ’(τ ) > 0 t_0 = ϕ(τ_0), t_1 = ϕ(τ_1).

Question 20

change of vars integral fromula

Answer 20

∫{a,b] f(x).dx = ∫{c,d} f(g(t__g’(t).dt

g([c,d]) to [a,b]

Question 21

example: considering PCS
γ_ℓ:R to R^2
γ_ℓ (t)=(cos2πℓ, sin2πℓ)

γ_1(t)=(cos2π, sin2π)

are they reparametrisations?

Answer 21

γ_1 =γ_ℓ ◦ϕ
ϕ(t)= ℓt
t to ℓt
but
NO: check ARC LENGTH
|γ_ℓ ‘(t)| = 2πℓ
L(γ_ℓ|[0,1]) = ∫_{0,1} 2πℓ .dt
=2πℓ

|γ_1 ‘(t)| = 2π
L(γ_1|[0,1]) = ∫_{0,1} 2π .dt
=2π
not the same

note their image looks the same!

Question 22

UNIT SPEED CURVE

Answer 22

A PC γ : I → R^n is called a unit speed curve (USC) if |γ’(t)| = 1 for any t ∈ I.

MAKES IT EASIER TO FIND LENGTH

Question 23

PROPN 2.3 USC exist?

Answer 23

Let γ : I → R^n be an RPC. Then there exists a re-parametrisation γ˜ of γ that has unit speed.

Question 24

PROOF:
Let γ : I → R^n be an RPC. Then there exists a re-parametrisation γ˜ of γ that has unit speed.

Answer 24

Proof. Pick a point t0 ∈ I and consider the arc-length function
s(t) = integral_[t_0,t] |γ’(τ )| dτ.

Clearly, s : I → R is a smooth function, and s’(t) = |γ’(t)| > 0. Denote by J the image of s; it
is an interval in R . Then s : I → J is bijective, and there is an
inverse function
ϕ = s−1: J → I.
By the inverse function theorem, ϕ is smooth,
ϕ’(τ ) =
(1/s’(ϕ(τ ))) > 0.

ϕ can be used as a parameter transformation

γ˜’(τ ) = γ’(ϕ(τ ))ϕ’(τ ) chain rule
= γ’(ϕ(τ ))/s’(ϕ(τ ))
= γ(ϕ(τ ))/ |γ’(ϕ(τ ))| ,
thus USC
|γ˜’(τ )| = 1.

Question 25

DEFN
normal component of acceleration

Answer 25

γₙ’‘(t)

found by

[γ’‘(t)- [γ’‘(t) *γ’(t)][|γ’(t)|²] γ’(t) ]

Question 26

CURVATURE VECTOR OF
γ

Answer 26

k(t)

(1/|γ’(t)|²)
[γ’‘(t)- [γ’‘(t) *γ’(t)][|γ’(t)|²] γ’(t) ]

=γₙ’‘(t) /|γ’(t)|²

RECALL IN EXAM

Question 27

diagrams for acceleration, velocity vectors

Answer 27

For a curve compute acceleration and velocity
taking projection of acceleration on velocity
v =γ’(t)/|γ’(t)|

taking orthogonal component and γ’’ prokected onto notmal

ie γ’’ projected onto γ’

Question 28

true or false
the curvature vector orthogonal to velocity?

Answer 28

true always
k(t) γ’(t)
=
(1/|γ’(t)|²)
[γ’‘(t)γ’(t)
- [γ’‘(t) γ’(t)][|γ’(t)|²] γ’(t)γ’(t) ]
=0??

Question 29

true or false reparametrisation keeps arc length

Answer 29

true, it will have same properties, if we find a phi s,t it doesnt clearly this isnt a repara

Question 30

PROP 2.4
curvature vector
how does is change under re-parametrisation?

Answer 30

The curvature vector is unchanged under a re-parametrisation, that is if
γ˜ : J → R^n is a re-parametrisation of an RPC γ : I → R^n, then

˜k(t) = k(ϕ(t)) for any t ∈ J,
where γ˜ = γ ◦ ϕ,

˜k is the curvature vector of γ˜, and k is the curvature vector of γ

Question 31

if USC
dot prod
γ’ * γ’=

γ’’ * γ’

Answer 31

=1
=0
as if USC velocity orthogonal to acceleration

Question 32

if our curve isnt USC?

Answer 32

we can always reparameterise
into a USC
preserving image
traversed in same direction
same #times
regularity
length
and curvature preserved

Question 33

EXAM EXAMPLE
given a curve check its regular:

Given 2 parametrised curves determine whether related by parametrisation or not

γ_1(t)= a+tv line in R^n
γ_2(t) = r(cos 2pit, sin2pit)
γ_3(t) = a +t^3v

Answer 33

circle
take line and wrap around circle
1 &3:
1 regular 3 not regular as γ’_3(0)=0 so not

1 and 2: straight line has curvature k(t)=0 so cant be repara of a circel as regions where curvature never equals 0

Question 34

Given 2 parametrised curves determine whether related by parametrisation or not

γ_2(t) = r(cos 2pit, sin2pit)
defined for [0,1]
γ_4(t) = r(cos 4pit, sin4pit)

Answer 34

both regular
velocity >0
γ’_4(t) = 2γ’_2(t)
length preserved?
|γ’_2| = 2pir
|γ’_4| = 4pir
length arc different so not repara
integrating over [0,1] gives double

Question 35

CLOSED CURVES

Answer 35

DEFORMATIONS OF A CIRCLE

e.g
point end at same place

Question 36

SPHERE IN R^3

Answer 36

2 parameters
2D object in R^3,R^4,..

Question 37

def T-Periodic

Answer 37

A vector-function f : R → R^n is called T-periodic (or periodic with period T > 0) if f(t + T) = f(t) for any t ∈ R.

Question 38

e.g
periodicity of
f(t) = sin(2πt)
g(x) = sin (x )
h(t) = t^2

Answer 38

f(t) = sin(2πt) 1-periodic smooth function,
g(x) = sin x 2π-periodic smooth function.

On the other hand, any unbounded continuous function h : R → R
(e.g. h(t) = t) can not be T-periodic with T > 0

Question 39

continuous funct on closed interval period?

Answer 39

are bounded, may be periodic

Question 40

PROPN 2.5 when do we have a periodic extension

Answer 40

A smooth function γ : [a, b] → R^n has a smooth (b − a)-periodic extension
IFF
all derivatives at the ends coincide
γ⁽ᵐ⁾(a) = γ⁽ᵐ⁾(b)
for any non-negative integer m.

Question 41

L5 PROOF
A smooth function γ : [a, b] → R^n has a smooth (b − a)-periodic extension
IFF
all derivatives at the ends coincide
γ⁽ᵐ⁾(a) = γ⁽ᵐ⁾(b)
for any non-negative integer m.

Answer 41

LONG SUMMARISE??
uses that ends coincide then periodic can be shown as adding f(a+T) =f(b)?

Question 42

(T₁, T₂)-equivariant

Answer 42

A real-valued function ϕ : R → R is called (T₁, T₂)-equivariant for given positive real numbers T₁& T₂ , if

ϕ(t + T₂) = ϕ(t) + T₁ for any t ∈ R.

ie
evaluated at t + T₂
same as evaluating at t and adding T₁

diagram 2T_2 for 2T_1 increments

Question 43

example
ϕ(t) = 2πt
(T₁, T₂)-equivariant

Answer 43

( 2π, 1)-equivariant
for any t in R
ϕ(t + 1) = 2π(t + 1)
= 2πt + 2π = ϕ(t) + 2π.

Question 44

how are T periodic functs and (t_1,t_2) equivariant functs linked

Answer 44

*any T-periodic function
is (0, T)-equivariant function.

• derivative of a (T_1, T_2)-equivariant function is a T_2-periodic function.

*equivariant functions often occur as integrals of periodic functions

Question 45

L5:
Let f : R → R be a T2-periodic function. Can you show that the function ϕ, defined below, is
(T1, T2)-periodic:

ϕ(t) = ∫{0,t} f(s)ds,
where
T1 =∫{0,T_2} f(s)ds

Answer 45

ϕ(t +T_2) =
∫_{0,t+T_2} f(s)ds
using change of vars
𝜏=s-T_2
d𝜏/ds =1
𝜏_1= s_1-T_2=0
𝜏_2 =s-T_2 -t = ?

= ∫{0,T_2} f(s)ds + ∫{T_2,t+T_2} f(s)ds
= T_1 + ∫_{0,t+T_2} f(s)ds
= T_1 +ϕ(t)

Question 46

PROPN 2.6 equivariant function composed with periodic function gives

Answer 46

Let ϕ : R → R be a (T1, T2)-equivariant function. Then for any T1-periodic vector-function f : R → R^n the composition
f ◦ ϕ is a T_2-periodic vector function.

proof:
For any t ∈ R
f ◦ ϕ(t + T2) = f(ϕ(t + T2)) = f(ϕ(t) + T1) = f(ϕ(t)) = f ◦ ϕ(t),

where in the second relation we used equivariance of ϕ, and in the third – the periodicity of f.

Question 47

PROP 2.7
when to we have a (T_1,T_2) equivariant extension

Answer 47

Let ϕ : [c, d] → [a, b] be a surjective smooth function. Then it has a smooth (T1, T2)-equivariant extension ϕ¯ : R → R with
T_1 = (b − a) and
T_2 = (d − c)
IFF
ϕ(c) = a,
ϕ(d) = b, and
ϕ⁽ᵐ⁾(c) = ϕ⁽ᵐ⁾(d)
for any integer m > 0.

Proof. (For MATH5113M only.) The proof follows an argument similar to the one in the proof of
Proposition II.5.

Question 48

CLOSED PARAMETRISED CURVE

Answer 48

A vector-function γ : [a, b] → R
n is called a closed parameterised curve (CPC) if it has a (b − a)-periodic extension ¯γ : R → R^n that is a smooth map.

“no distinguished start /end”

Question 49

when is a closed parametrised curve regular

Answer 49

A closed parameterised curve γ
is called regular (RCPC), if its periodic extension ¯γ is regular

Question 50

simple circular curve

Answer 50

an example of a closed regular curve
γ : [0, 1] → R^2
γ(t) = (cos(2πt),sin(2πt)).

Question 51

example:
is this a closed parametrised curve

Consider a PC γ : [0, 1] → R
2 given by γ(t) = (t^2 − t, sin 2πt)

Answer 51

γ(0) = γ(1) but not a CPC
If γ would have a smooth periodic extension, then γ’(0) = γ’(1).
but
γ’(t) = (2t − 1, 2π cos 2πt)
⇒
γ’(0) = (−1, 2π) /= (1, 2π) = γ’(1)

Question 52

γ_0(t)=(cos(2pit), sin(4pit)) t in [0,1] is this a RCPC

Answer 52

REGULAR CLOSED PC?
has extension
γ¬(t) = (cos(2pit, sin4pit) t in R
γ_0(t)=2pi(-sin(2pit), 2cos(4pit)) periodic function in [0,o.5] will vanish when both sin cos =0 but =/0

Question 53

is this a closed PC?
γ:[0,1] to R^2
γ_0(t)= (t^2-t, sin2pi*t)

Answer 53

no because if it was a CPC it would have a smooth periodic extension and
γ’(0)=γ’(1) at end points but these arent equal
γ’(0)= (-1, 2pi) =/ (1, 2,pi) = γ’(1)

Question 54

FOR A SMOOTH PERIODIC EXTENSION TO EXIST

Answer 54

MUST HAVE DERIVS PERIODIC
AND
COINCIDE AT a and b end points

Question 55

when is a CPC a reparameterization

Answer 55

A CPC ˜γ : [c, d] → R^n is called the re-parametrisation of a CPC
γ : [a, b] → R^n if there exists a surjective map
ϕ : [c, d] → [a, b] that has a (T_1, T_2)-equivariant extension
¯ϕ : R → R with
T1 = b − a and
T2 = d − c s.t
¯ϕ is smooth, ¯ϕ’(t) > 0 for any t ∈ R, and
γ¯˜ = ¯γ ◦ ϕ¯, where
γ¯˜ and ¯γ are periodic extensions of ˜γ and γ respectively.

Question 56

where does a CPC repara come from

Answer 56

We know any PC γ:[a,b] to R^n can be reparametrised to a PC defined on [0,1]
γ~(t)= γ(a+t(b-a)) t in [0,1]

if γ is a CPC then PC γ~ defined is also a CPC
can use the above repara parameter as it is a ((b-a),1) equivariant extension
with deriv >0

Question 57

SIMPLE

Answer 57

A closed parametrised curve (CPC) γ : [a, b] → R^n is called simple if the restricted
map γ| [a, b) is injective

no self intersects

Question 58

e.g
is this simple?
CPC γ` : [0, 1] → R^2
γ_ℓ(t) = (cos 2πℓt ,sin 2πℓt), where ℓ is
an integer

Answer 58

is simple if and only if ℓ = 1 or = −1.

Geometrically γ_ℓ wraps around γ_1 ` ℓ times; if ℓ > 0 it is traversed in the same direction, while
if ℓ < 0 – in the opposite.

0/w self intersects and loops

Question 59

true or false
being SIMPLE depends on the parametrisation

Answer 59

false
SIMPLE
doensnt dep on repara
as for each point there is a unique t in [a,b] meaning unique tau in [alpha,beta]

Question 60

HOMOTOPY OF CLOSED CURVES

Answer 60

space of all paths joining points

Question 61

DEF
REGULAR HOMOTOPY

Answer 61

A regular homotopy from a closed curve α : [0, 1] → R^ n to a closed curve β : [0, 1] → R ^n is a continuous map F : [0, 1] × [0, 1] → R^ n that satisfies the following properties:

(i) α(t) = F(0, t), β(t) = F(1, t) for any t ∈ [0, 1], and for any fixed τ ∈ [0, 1] the map[
t in [0, 1]
t → F(τ, t) ∈ R^n
is an RCPC.

(ii) For any integer k > 0 the derivatives (∂k/∂tkF(τ, t)) are continuous in τ

i idea of deformation
ii ensures curvature change continuously under deformation

Question 62

regular homotopy key points

Answer 62

*MAP will be a FAMILY OF CLOSED CURVES
each value of tai
smooth derivatives

deformations change
alpha(t) F(0,t)
beta(t) F(1,t)

F(tau,t) some other closed curve

derivative property ensures curve geometry affected, acceleration and velocity affect curvature

Question 63

in exam will need to know

Answer 63

definitions of curvature

regular homotopy
stating these

given two curves alpha and beta, asked to prove RH by checking conditions and give formula for F

Question 64

EXAM EXAMPLE
Γ _1(t)=(cos 2πt,sin 2πt)
t in [0.5,3/2]

Answer 64

Γ _1(t)_overline = overlineγ_1(t)

periodic extension same as
(cos 2pit, sin 2pit)

so
φ_overline : R to R
φ_overline (t) = t -0.5
φ(t) =t -0.5 for [0.5,3/2]
φ_overline (t+1) = φ_overline (t) +1
LHS: t +1 -0.5 = t -0.5+1 = RHS
γ:[a,b] to R^n is a CPC
overlineγ :R to R^n a (b-a) periodic extension

φ:[0,1] to [a,b]
φ(t)=a+t)b-a)
φ’(t)=b-a>0
φ(t+1) = a+(t+1)(b-a) = overline_φ(t) + (b-a)
and thusφ overline is (b-a,1) equivariant and this we have
γ~ using this as a repara parameter is a repara of γ

note arc lengths are the same for γ and overline_γ for a closed curve

HOMOTOPY CAN CHANGE GEOMETRY OF THE CURVE

Question 65

EXAM EXAMPLE
check that the simple circle
α(t) = (cos 2πt,sin 2πt)
is regularly homotopic to the ellipse β(t) = (2 cos 2πt,sin 2πt), where t ∈ [0, 1].

Answer 65

2) kth derivatives wrt t of F(𝜏,t)

F(τ, t) = ((1 + τ ) cos 2πt,sin 2πt), where τ, t ∈ [0, 1],
CONDITIONS
1) F(0,t)= α(t) & F(1,t)= β(t)

fix 𝜏 in [0,1]
Show F(𝜏,t) is an RCPC
F_𝜏: t to F(𝜏,t):
check if periodic extension is regular
OVERLINE_F_𝜏: R to R^2
F‾_𝜏(t) = ((1+𝜏)cos2pit, sin 2pit)
is a smooth 1-perioduc funct
deriv ert t=/0 checked >= 4pi^2(s^2+c^2) = 4pi^2 >0

((1+𝜏) ∂ᵏ/∂tᵏ cos2pit, ∂ᵏ/∂tᵏ sin2pit)
1+𝜏 depends on tau continuously (polynomial of first order)

Question 66

deforming by homotopy methods

Answer 66

we can deform for example by horizontal lines or rays from the origin

check velocity o this extension has no 0’s ie never equal to 0

a periodic is easier to check for intervals

Question 67

For example consider
Γ _1(s)=(cos 2πs,sin 2πs)
t in [0.5,3/2]
same image as [
Γ _2(t)=(-cos 2πt,-sin 2πt)
t in [0,1]

homotopy?

Answer 67

same image as
Γ _2(t)=(-cos 2πt,-sin 2πt)
t in [0,1]

by repara change of vars t=s-0.5

F(tau,t)=
[cos pi𝜏, -sin pi𝜏]
[sin pi𝜏 cos pi𝜏] *
[cos 2pit]
[sin 2pit]

dep continuously on tai

rank 1

rotation fraction tau/pi
if tau=1 angle is pi
[-1 0]
[0 -1]

Question 68

closed curves for integer vales of ℓ
gamma_ℓ
are these regularly homotopic

Answer 68

no

Question 69

recall in exam regularly homotopic property PROP 2.8

Answer 69

The relation of being regularly homotopic is an equivalence relation on the set of closed regular curves

(1) reflexivity α ∼ α, RH to itself
(2) symmetry α ∼ β =⇒ β ∼ α,
(3) transitivity α ∼ β and β ∼ γ =⇒ α ∼ γ.

Question 70

proof
The relation of being regularly homotopic is an equivalence relation on the set of closed regular curves

(1) reflexivity α ∼ α, RH to itself
(2) symmetry α ∼ β =⇒ β ∼ α,
(3) transitivity α ∼ β and β ∼ γ =⇒ α ∼ γ.

Answer 70

(1) For any RCPC α : [0, 1] → R^n we define F(τ, t) = α(t). satisfied.
(2) Let α and β : [0, 1] → R
n be two RCPC for exists a regular homotopy from α to β, F : [0, 1] × [0, 1] → R^n ,F(0, t) = α(t), F(1, t) = β(t). Then the map
G(τ, t) := F(1 − τ, t), where τ, t ∈ [0, 1]
defines a regular homotopy from β to α, that is G(0, t) = β(t) and G(1, t) = α(t). The conditions
in Definition II.18 for G follow from similar conditions for F. For example, for any fixed τ the map
t 7−→ G(τ, t) is an RCPC, since G(τ, t) = F(1 − τ, t) and t 7−→ F(1 − τ, t) is an RCPC. We also see
that for any k > 0 the derivative
∂^k/∂tk G(τ, t) = ∂ k/ ∂tk F(1 − τ, t)
is continuous in τ , as a composition of continuous maps.

(3) Suppose that α is regularly homotopic to β via a regular homotopy F, and β is regularly homotopic to γ via G. Consider the map H : [0, 1] × [0, 1] → R^n
, H(τ, t) =
{F(2τ, t), τ ∈ [0, 1/2];
{G(2τ − 1, t), τ ∈ [1/2, 1].
H satisfies condition a regular
homotopy from α to γ

Question 71

PROP 2.9
Every RCPC is regularly homotopic to

Answer 71

Every RCPC is regularly homotopic to an RCPC of unit length.

proof: uses map
F:[0,1] x [0,1’ to R^m
F(tau,t) = (1-tau + tau/L) gamma (t)
and verifies conditions

Question 72

Proposition II.10. Let γ˜ : [0, 1] → R^n be a re-parametrisation of an RCPC γ : [0, 1] → R^n. Then

Answer 72

Proposition II.10. Let γ˜ : [0, 1] → R^
n be a re-parametrisation of an RCPC γ : [0, 1] → R^n. Then γ
and γ˜ are regularly homotopic.

Question 73

PROOF
Proposition II.10. Let γ˜ : [0, 1] → R^
n be a re-parametrisation of an RCPC γ : [0, 1] → R^n. Then γ
and γ˜ are regularly homotopic.

Answer 73

Since ˜γ is a re-parametrisation of γ there exists a smooth function ¯ϕ : R → R such that:
* ϕ¯(0) = 0, ¯ϕ(1) = 1, and ¯ϕ’(t) > 0 for any t ∈ R;
* ϕ¯(t + 1) = ¯ϕ(t) + 1 for any t ∈ R;
* γ¯˜ = ¯γ ◦ ϕ¯, where γ¯˜ and ¯γ are 1-periodic extensions of ˜γ and γ respectively.
We define a map F¯ : [0, 1] × R → R
n by setting
F¯(τ, t) = ¯γ(τϕ¯(t) + (1 − τ )t).
We claim that for any τ ∈ [0, 1] the map F¯_τ : t 7→ F¯(τ, t) is smooth, 1-periodic, and has non-vanishing
derivative. Indeed, it is smooth as a composition of smooth functions. To verify 1-periodicity, we
write
F¯_τ (t + 1) = ¯γ(τϕ¯(t + 1) + (1 − τ )(t + 1)) = ¯γ(τϕ¯(t) + τ + (1 − τ )t + (1 − τ )) =
γ¯(τϕ¯(t) + (1 − τ )t + 1) = ¯γ(τϕ¯(t) + (1 − τ )t) = F¯τ (t).
Finally, computing the derivative we obtain
F¯’τ (t) = ∂/∂tF¯(τ, t) = ¯γ’(τϕ¯(t) + (1 − τ )t)(τϕ¯’(t) + (1 − τ )).

Since ¯γ’ /= 0, we obtain that for any τ ∈ [0, 1] the derivative F¯’ τ(t) 6= 0. These properties show that
for any τ ∈ [0, 1] the restriction of F¯
τ to [0, 1] defines an RCPC.
Now we define the homotopy F between γ and ˜γ by setting
F(τ, t) := γ(τϕ(t) + (1 − τ )t).
Clearly, F(0, t) = γ(t) and F(1, t) = γ(ϕ(t)) = ˜γ(t). We verify that the conditions of Definition II.18
hold. For any fixed τ ∈ [0, 1] the RPC F¯_τ is an 1-periodic extension of t → F(τ, t), and hence,
the latter is an RCPC. To verify the second condition of Definition II.18 we should show that the
derivative ∂^kF¯(τ, t)/∂t^k
is continuous in τ for any t ∈ R. The latter is a consequence of the fact that
F¯(τ, t) is defined as a composition of the maps whose all derivatives are continuous in τ .

Question 74

to save time:
notes from now on only thms and statements
examples in written notes!

Answer 74

:(

Question 75

Corollary II.11. Every RCPC γ is regularly homotopic to

Answer 75

Every RCPC γ is regularly homotopic to an RCPC of constant speed L, where L is the length of γ

Question 76

Corollary II.12. Every RCPC γ is regularly homotopic to

Answer 76

Corollary II.12. Every RCPC γ is regularly homotopic to an RCPC of unit speed.

Proof. The statement is direct consequence of Proposition II.9 and Corollary II.11

Question 77

UNIT TANGENT VECTOR

Answer 77

v(t) = γ’(t)/ |γ’(t)|

Question 78

unit normal vector

Answer 78

uses components of unit tangent vector
n(t) := (−v2(t), v1(t)).

like rotation 90 degrees

Question 79

unit normal and unit tangent vector

Answer 79

vectors v(t) and n(t) are orthogonal and form a positively oriented basis of R

Question 80

signed curvature

Answer 80

k(t) be a curvature vector to γ. As we know, it is orthogonal to v(t), and since γ is a plane curve, we conclude that
k(t) = κ(t)n(t)

for some real-valued function κ(t). The function κ(t) is called
the signed curvature of γ

Question 81

signed curvature and curvature

Answer 81

t |κ(t)| = |k(t)|.

Question 82

Geometrically the signed curvature function κ(t)

Answer 82

measures the rate of change of a tangent line direction: when κ > 0 the curve is bending towards the
unit normal, while when κ < 0 it is bending away from the normal.

diagram of smother version of a kink
___B__
| |
—-A-| |——–

signed curvature positive A
curve bends towards unit normal
points up ish
signed curvature negative B
curve bends away from unit normal
points upish

Question 83

left for readin: THM 2.13 Fundamental thm of plane curves

Answer 83

Given a smooth real-valued function
κ : [a, b] → R, a point t0 ∈ [a, b], and vectors γ0, v0 ∈ R^2
such that |v_0| = 1, there exists a unique unit speed parametrised curve γ : [a, b] → R^2 whose signed curvature equals κ(t) and γ(t0) = γ0,
γ’(t_0) = v_0.

Question 84

Lemma
θ function

Answer 84

Let γ : [a, b] → R^2
be a regular parametrised curve. Then there exists a smooth function
θ : [a, b] → R such that the unit tangent vector v(t) satisfies the relation
v(t) = (cos θ(t),sin θ(t)).

If θ1 and θ2 are two such functions, then they differ only by an integer multiple of 2π, that is

θ1(t) = θ2(t) + 2πm, where m ∈ Z is a constant. In particular, the quantity

θ(b) − θ(a) uniquely determined by PC

Question 85

angle parametrisation for unit tangent vector

Answer 85

v(t)= (Cos(θ(t), sinθ(t))

Question 86

ROTATION INDEX for an RCPC

Answer 86

times unit tangent vector v(t) winds around the unit circle S^1

r(γ) :=
(1/2π)(θ(1) − θ(0)) ∈ Z
is called the rotation index of γ.

anticlockwise counting positively

angle is from 0 anticlockwise

Question 87

e.g draw plane RCPC given
γ1(t) = (cos(2πt),sin(2πt)) and
γ2(t) = (cos(2πt),sin(4πt)), where t ∈ [0, 1].

Compute (analytically
or visually) how many times the unit tangent vector winds around the unit circle S^1
for each curve.

Answer 87

circle anticlockwise
curvature 1
derivative used in formula for rotation index =1
v(t)=(1/2pi)(deriv)
used to find theta functs

figure 8
curvature =0 never makes full turn

unit tangent vector:
γ’(t)=(-2pisin(2pit), 4picos(4pit))
v(t)= (-sin(2pit, 2cos(4pi*t)
=(cos(θ(t), sin(θ(t))

cos(2pit + (pi/2))
= cos(pi/2)cos(2pit) - sin(pi/2)sin(2pit)
=-sin2pit
…..difficult

analytically starts clockiwise turns anti but never makes full #turns so 0

Question 88

θ for γ _ℓ(t)

Answer 88

rotation index = ℓ

θ(t) = (pi/2) + 2piℓ*t

Question 89

PROP 2.15 rotation index geoemtric formula

Answer 89

Let γ : [0, 1] → R^2 be a plane RCPC. Then its rotation index satisfies the relation
r(γ) = (1/2π)
integral_{0,1} κ(t)|γ(t)| dt,
where κ is the signed curvature of γ

proof: we know smooth theta exists s.t we can write the unit tangent vector.
our first deriv will be multiple of this( modulus of first deriv) then subbin into formula

Question 90

PROP 2.16 regularly homotopic does rotation index change

Answer 90

Regularly homotopic RCPCs have the same rotation index

proof: defining a regular homotopy the rotation index defined as a formula for any curve in the homotopy takes integer values so must be constant by IVT

Question 91

important thme
WHITNEY GRAUSTEIN

Answer 91

Two regular closed plane curves are regularly homotopic if
and only if they have the same rotation index

Question 92

thm 2.18 Hopf

Answer 92

Let γ be a simple regular closed plane curve. Then its rotation index r(γ) equals either 1 or −1.

Question 93

corollary 2.19 for simple regular closed plane curves regularly homotopic to

Answer 93

Any simple regular closed plane curve is regularly homotopic either to the standard circle
[0, 1]
t → (cos(2πt),sin(2πt))
or to the reversed standard circle
[0, 1]
t→ (cos(2πt), − sin(2πt)).

Question 94

Corollary II.20. For any simple plane RCPC

integral_[0,1] |κ(t)| |γ’(t)| dt > 2π

Answer 94

Corollary II.20. For any simple plane RCPC γ : [0, 1] → R^2 the following relation holds
integral_[0,1] |κ(t)| |γ’(t)| dt > 2π;

the equality occurs
IFF
κ(t) does not change sign
(that is, κ(t) > 0 everywhere or κ(t)<=0 everywhere).

proof in notes

Question 95

if rotation index differs

Answer 95

curves arent RH to each other

Question 96

circular curve cant be deformed to curve of length 2pi st
signed curvature (t) >1

Answer 96

changing curvature changes length everywhere

1=r(gamma)=r(gamma~) if deformed
same length
contradiction as r(gamma~) = fromula using signed curvatiure > 1

Question 97

if we have a closed curve with RI ℓ then its homotopic to

Answer 97

curve with RI ℓ

so given a curve we cab either
find the integral formula fro RI
or guess and prove by constructng a regular homotopy
applying whitney graustein thm

Question 98

TOPOLOGICAL DEGREE METHOD

Answer 98

for RI counts #pre-images with signs
1) fix direction curve traversed
tangent traverses anticlockwise for t in [0,1]

2) find points with same curvature
signs will show curvature

drawn a sketch
chosen all points with same tangent/direction upwards
curvature at 90 degrees anticlockwise to this drawn fixed unit bector

signed curvature
positive if points into a curve
negative if points out of a curve
added
in exam we can do this or cumpute by constructing RH

Question 99

cos 2pi*t, -sin2pit)

Answer 99

complicated drawing is actually regularly homotopic to circle traversed once clockwise

Question 100

L5:

when does a periodic extension exist?

Answer 100

when we have closed curves

any closed curve can be parametrised to USC

Question 101

L5:
Bounded domain Ω

Answer 101

subset in R^2
open subset bounded by a SIMPLE REGULAR CLOSED PC

image of surrounds it as a boundary

Question 102

an example
γ[0,1] to R^2 simple

Answer 102

restriction part injective with no self intersection until 0,1

restrict to [0,1) so we can use

Question 103

common qs L5

Answer 103

among all domains in R^3 of a given area A find boundary with least length

among all simple regular closed curves of given length L find a curve that encloses a domain of greatest area

find a region of greatest area bounded by a straight line aand a curve of a fixed lengthwhose ends lie on the line (answer disk part of circle)

Question 104

ISOPERIMETRIC INEQUALITY

Answer 104

Let Ω ⊂ R^2
be a domain bounded by a simple
regular closed parametrised curve γ : [0, 1] → R^2
. Let A be the area of Ω, and L be the
length of γ. Then the inequality
4πA <= L^2 holds, and the equality is achieved if and only if
Ω is a disk.

Question 105

Lemma II.2 (Green’s formula).

Answer 105

Let P(x, y) and Q(x, y) be two smooth real-valued functions
defined on a domain Ω. Then the following relation holds:
∫∫_Ω [∂P/∂x + ∂Q/∂y]dxdy =

∫_γ (P dy − Qdx),

where the right hand-side above is the anti-clockwise integral along the boundary curve
γ = ∂Ω.

Question 106

example

Answer 106

Let Ω ⊂ R^2 be a domain bounded by a simple regular closed unit speed curve γ : [0, L] → R^2
; that is |γ’(t)| = 1 for any t ∈ [0, 1], and hence, L equals to the length of γ
L(γ) =
∫[0,L] |γ’(t)| dt =
∫[0,L] 1dt = L

Let n(t) be a unit normal vector to γ such that
det(γ’(t), n(t)) = 1 for any t ∈ [0, L]. In
other words, if γ’(t) has coordinates
(x’(t), y’(t)), then
n(t) = (−y’(t), x’(t)).
We claim that
the area A of the domain Ω satisfies the following relation:
2A = −∫[0.L] (γ(t) · n(t))dt.
Indeed, choosing P(x, y) = x and Q(x, y) = y, by Green’s formula we obtain
2A =
2∫∫Ωdxdy =
∫∫_Ω (∂P/∂x + ∂Q/∂y) dxdy =
∫_γ (xdy − ydx) =
∫[0,L] (xy’ − yx’)(t)dt =
− ∫[0,L] (γ(t) · n(t))dt.

Question 107

WIRTINGERS INEQUALITY

Answer 107

Let f : R → R be a smooth L-periodic function such
that ∫_[0,L] f(t)dt = 0.

Then the following inequality holds
∫[0,L] f^2(t)dt <=
(L^2/4π^2)
∫[0,L] (f’(t))^dt;
the equality occurs if and only if f(t) = a cos(2πt/L) + b sin(2πt/L) for some a, b ∈ R.

REMEMBER THE NAME?
integral over period =0

tranlation left

Question 108

curvature

total curvature

Answer 108

For a given RCPC γ : [0, 1] → R
n the quantity
µ(γ) =
integral_{0,1} |k(t)| |γ’(t)| dt,
where k(t) is the curvature vector, is called the total curvature of γ

Question 109

propn 2.4 for total curvature

Answer 109

Proposition II.4. Let γ˜ : [0, 1] → R
n be a re-parametrisation of an RCPC γ : [0, 1] → R^n.
Then the total curvatures of γ˜ and γ are equal, µ(˜γ) = µ(γ).

Question 110

Theorem II.5 (Fenchel-Borsuk).

Answer 110

Let γ : [0, 1] → Rn be a simple regular closed parametrised
curve. Then its total curvature µ(γ) is at least 2π, and it equals 2π if and only if the curve lies in an affine 2-plane and its signed curvature does not change sign.

The statement above is a special property of closed curves. Simple examples show that the total curvature is not a topological invariant, that is it changes under the deformations of a curve.

Question 111

isotopy

Answer 111

An isotopy of R^n is a continuous map Φ : [0, 1] × R^n → R^n such that for any fixed τ ∈ [0, 1] the map R^n x→ Φ(τ, x) ∈ R^ n is a homeomorphism, that is bijective,
continuous, and the inverse map is also continuous

Question 112

ambient isotopic l5

Answer 112

Definition II.3. Two simple RCPCs α : [0, 1] → Rn and β : [0, 1] → R
n are called ambient
isotopic if there exists an isotopy of Rn such that:
* Φ(0, x) = x for any x ∈ Rn;
* Φ(1, α(t)) = β(t) for any t ∈ [0, 1].

Question 113

knotted

Answer 113

A simple RCPC is called unknotted if it is ambient isotopic to a plane
circle. Otherwise, it is called knotted.

Question 114

Theorem II.6 (F´ary-Milnor).

Answer 114

Let γ be a knotted simple RCPC. Then its total curvature
is at least 4π, that is µ(γ) > 4π.

Question 115

note l5

Answer 115

Tangent indicatrix. Let γ : [0, 1] → R
n be an RCPC, and v(t) = γ’(t)/ |γ’(t)| be its
unit tangent vector. In the sequel we use the notation S^{n−1}
for a unit (n − 1)-dimensional
sphere centred at the origin in the Euclidean space R^n

S^{n−1} = {(x1, . . . , xn) ∈ R^n
: x^2_1 + . . . + x^2_n = 1}.

Question 116

tangent indicatrix

Answer 116

For a given RCPC γ : [0, 1] → R
n the parametrised curve
Γ : [0, 1] → S^{n−1} ⊂ R^n
Γ(t) = v(t) = γ’(t)/|γ’(t)|
is called the tangent indicatrix of γ

Question 117

Proposition II.7. The total curvature and tangent indicatrix

Answer 117

Proposition II.7. The total curvature of an RCPC γ : [0, 1] → R^n is equal to the length of the tangent indicatrix Γ of γ, that is L(Γ) = µ(γ)

Question 118

great circle

Answer 118

A great circle, or equator, in a unit sphere S
n−1
is an RCPC γ : [0, 1] → R
n
of the form
γ(t) = cos(2πt)e1 + sin(2πt)e2,
where e1 and e2 are orthonormal vectors

Question 119

Fact II.1. For any two different points p and q in a sphere S
n−1
there exists a great circle
γ that passes through p and q. Moreover, if p and q are not anti-podal, p /= −q, then such
a great circle is unique.

Answer 119

Suppose that given points p and q ∈ S
n−1
lie in a great circle γ. Then they divide it
into two arcs γ1 and γ2, and we define the spherical distance between p and q by setting
distS(p, q) = min{L(γ1), L(γ2)}.
Note that distS(p, q) 6 π for any points p and q, and the equality occurs if and only if p
and q are anti-podal, that is p = −q. Besides, distS(p, q) = 0 if and only if p = q. Note
also that distS(p, q) = π/2 if and only if the points p and q, viewed as vectors in R
n, are
orthogonal.

Question 120

Fact II.2. Let γ : [0, 1] → S
n−1 ⊂ R
n be a PC such that γ(0) = p and γ(1) = q. Then the
spherical distance is not greater that the length of γ, that is distS(p, q) 6 L(γ). Moreover,
the equality is achieved if and only if γ is an arc of a great circl

Question 121

Exercise II.7. Check that the distance function distS(p, q) satisfies the triangle inequality

Answer 121

distS(p, q) 6 distS(p, z) + distS(z, q) for any points p, q, and z ∈ S
n−1