Chapter 2 Global theory of curves Flashcards

Question

DEFN normal component of acceleration

Answer 1

γₙ''(t) found by [γ''(t)- [γ''(t) *γ'(t)]\[|γ'(t)|²] γ'(t) ]

Answer 2

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

Answer 3

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'

Answer 4

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??

Answer 5

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

Answer 6

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

Answer 7

=1 =0 as if USC velocity orthogonal to acceleration

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

DEFORMATIONS OF A CIRCLE e.g point end at same place

Answer 12

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

Answer 13

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.

Answer 14

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

Answer 15

are bounded, may be periodic

Answer 16

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 17

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

Answer 18

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

Answer 19

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

Answer 20

*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

Answer 21

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)

Answer 22

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.

Answer 23

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.

Answer 24

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"

Answer 25

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

Answer 26

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

Answer 27

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)

Answer 28

REGULAR CLOSED PC? has extension γ¬(t) = (cos(2pi*t, sin4pi*t) t in R γ_0(t)=2pi(-sin(2pi*t), 2cos(4pi*t)) periodic function in [0,o.5] will vanish when both sin cos =0 but =/0

Answer 29

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)

Answer 30

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

Answer 31

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.

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

space of all paths joining points

Answer 37

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

Answer 38

*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

Answer 39

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

Answer 40

Γ _1(t)_overline = overlineγ_1(t) periodic extension same as (cos 2pi*t, sin 2pi*t) 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

Answer 41

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)cos2pi*t, sin 2pi*t) is a smooth 1-perioduc funct deriv ert t=/0 checked >= 4pi^2(s^2+c^2) = 4pi^2 >0 2) kth derivatives wrt t of F(𝜏,t) = ((1+𝜏) ∂ᵏ/∂tᵏ cos2pi*t, ∂ᵏ/∂tᵏ sin2pi*t) 1+𝜏 depends on tau continuously (polynomial of first order)

Answer 42

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

Answer 43

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 2pi*t] [sin 2pi*t] dep continuously on tai rank 1 rotation fraction tau/pi if tau=1 angle is pi [-1 0] [0 -1]

Answer 44

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 45

(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

Answer 46

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

Answer 47

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 48

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 .

Answer 49

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

Answer 50

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

Answer 51

v(t) = γ'(t)/ |γ'(t)|

Answer 52

uses components of unit tangent vector n(t) := (−v2(t), v1(t)). like rotation 90 degrees

Answer 53

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

Answer 54

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

Answer 55

t |κ(t)| = |k(t)|.

Answer 56

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

Answer 57

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.

Answer 58

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

Answer 59

v(t)= (Cos(θ(t), sinθ(t))

Answer 60

r(γ) := (1/2π)(θ(1) − θ(0)) ∈ Z is called the rotation index of γ. #times unit tangent vector v(t) winds around the unit circle S^1 anticlockwise counting positively angle is from 0 anticlockwise

Answer 61

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(4pi*t)) v(t)= (-sin(2pi*t, 2cos(4pi*t) =(cos(θ(t), sin(θ(t)) cos(2pi*t + (pi/2)) = cos(pi/2)cos(2pi*t) - sin(pi/2)sin(2pi*t) =-sin2pi*t .....difficult analytically starts clockiwise turns anti but never makes full #turns so 0

Answer 62

rotation index = ℓ θ(t) = (pi/2) + 2*pi*ℓ*t

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

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)).

Answer 68

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

Answer 69

curves arent RH to each other

Answer 70

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

Answer 71

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

Answer 72

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

Answer 73

complicated drawing is actually regularly homotopic to circle traversed once clockwise

Answer 74

when we have closed curves any closed curve can be parametrised to USC

Answer 75

subset in R^2 open subset bounded by a SIMPLE REGULAR CLOSED PC image of surrounds it as a boundary

Answer 76

restriction part injective with no self intersection until 0,1 restrict to [0,1) so we can use

Answer 77

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)

Answer 78

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.

Answer 79

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.

Answer 80

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.

Answer 81

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

Answer 82

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

Answer 83

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).

Answer 84

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.

Answer 85

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

Answer 86

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].

Answer 87

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

Answer 88

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

Answer 89

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}.

Answer 90

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

Answer 91

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)

Answer 92

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

Answer 93

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.

Answer 94

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

Chapter 2 Global theory of curves Flashcards

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