Chapter 4: Metric quantities Flashcards

1
Q

Lemma 4.4.1: dot product wrt basis

A

Given a smooth regular surface σ : U → R^3, the tangent space
corresponding to (u, v) has a natural basis σ_u(u, v) and σ_v(u, v). The dot product
can be written, with respect to this basis, as follows.
(a_1σ_u(u, v) + b_1σ_v(u, v)) · (a_2σ_u(u, v) + b_2σ_v(u, v))

=
(a_1, b_1)

[σ_u(u, v) · σ_u(u, v) σu(u, v) · σv(u, v)]
[σu(u, v) · σv(u, v) σv(u, v) · σv(u, v)]
[a_2]
[b_2]

1x2 * 2x2 * 2x1
1x2 * FFF I *2x1
σ_u and σ_v form a basis expressed as linear combo

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2
Q

DEF 4.1.2

first fundamental form

A

Suppose σ : U → R3
is a regular parametrized surface. For each (u, v) ∈ U we have an inner product I(u,v) on R^2, known as the first
fundamental form, which has matrix

[E F]
[F G]

=
[E(u, v) F(u, v)]
[F(u, v) G(u, v)]
:=
[σu(u, v) · σu(u, v) σu(u, v) · σv(u, v)]
[σu(u, v) · σv(u, v) σv(u, v) · σv(u, v)]
where this defines the three functions E, F, G : U → R.

  • FFF of σ is the dot product of R^3 restricted to tangent space at (u,v) wrt natural basis σ_u,σ_v
  • FFF changes for different parameterization of the same surface
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3
Q

EXAMPLE: The plane R^2

can be seen as a surface in R^3 and parametrized by σ : R^2 → R^3, (u, v) → (u, v, 0). This has first fundamental form

A
[E(u, v) F(u, v)]
[F(u, v) G(u, v)]
\:=
[1 0]
[0 1]

for all (u, v) ∈ R^2

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4
Q

EXAMPLE: Compute FFF of the plane R^2 seen as a surface in R^3, but parametrized by σ : R^2 → R^3
(u, v) → (2u + v, −u + 3v, 0)

A

σ_u=(2,-1,0) σ_v=(1,3,0)
σ_u . σ_u =4+1=5 etc

[E(u, v) F(u, v)]
[F(u, v) G(u, v)]
\:=
[5 -1]
[-1 10]

for all (u, v) ∈ R^2

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5
Q

EXAMPLE: For parametrised curve
γ: ]α, β[ → R^3
t→ (x(t), 0, z(t))
we obtain the parametrized surface of revolution
σ : ]α, β[ × R → R^3
σ(t, θ) = (x(t) cos θ, x(t) sin θ, z(t)).
The first fundamental form is

A

[x’(t)^2 + z’(t)^2 0 ]

[0 x(t)^2 ]

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6
Q

EXAMPLE: cylinder x(t)=1 z(t)=t for all t

FFF=

A

[1 0]

[0 1]

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7
Q

EXAMPLE:
For the sphere S^2
in Example 3.3.6, x(t) = cos(t) and z(t) = sin(t). So for all (t, θ) the matrix of the first fundamental form is

A

[1 0 ]

[0 cos^2(t)]

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8
Q

EXAMPLE: The catenoid is the surface of rotation of the curve γ(t) := (cosh(t), 0, t) for t ∈ R. Compute the first fundamental form of the catenoid.

A
[cosh^2t      0     ]
[0           cos^2(t)]
= cosh^2t
[1 0]
[0 1]
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9
Q

image on surface

A

For a parametrized surface σ : U→ R^3, a parametrized curve γ: ]α, β[ → U
will have an image γ˜ = σ ◦ γ on the surface. We will see that the first fundamental
form encodes how much the parametrization σ distorts distances.

diagram mapping real line to open subset in R^2, by γto surface in R^3,by σ
]α, β[ on R

line on subset of R^2
by γ
U
to
surface in R^3
by σ
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10
Q

LEMMA 4.2.1: length of γ˜ = σ ◦ γ

A

For a parametrized surface σ:U→ R^3, a parametrized curve γ: ]α, β[ → U and any a, b with α < a ≤ b < β, the length of γ˜ = σ ◦ γ between a and b is
∫ _[a,b] √ [ I_γ(t)(γ˙(t), γ˙(t))]dt.

*we only need tangent vectors to γ and FFF along γ to compute the length of the
image of γ in S.

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11
Q

DEF 4.2.2: local isometry

A

A parametrization σ : U → R^3
is a local isometry if it takes any curve in U to a curve of the same length in the surface

ie alter shape doesn’t change arc lengths

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12
Q

THM 4.2.3

local isometry if

A

A parametrization σ : U → R^3
of surface is a local isometry if
and only if its first fundamental form I is represented by the identity matrix

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13
Q

proof of LEMMA 4.2.1: length of γ˜ = σ ◦ γ

A

proof: we know the arc length of γ˜ from a to b is
∫ _[a,b] ||γ˜•||.dτ
chain rule: γ˜ = σ ◦ γ
γ(t)= (u(t),v(t))
dγ˜/dt = (du/dt)(∂σ/∂t) + (dv/dt)(∂σ/∂t) = u’(t)σu +v’(t)σv
by 4.1.1

(γ˜^•)•(γ˜^•) =
(u’(t) v’(t))
[E F] [u’(t)]
[F G] [v’(t)]

but (γ•)(t)= (u’(t), v’(t))
ie (γ˜^•)•(γ˜^•) = I_γ(t) (γ•,γ•)

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14
Q

proof of

A parametrization σ : U → R^3
of surface is a local isometry if
and only if its first fundamental form I is represented by the identity matrix

A

The arc length functs are
s_γ(t)= ∫ _[t_0,t] ||γ˜•(τ)||.dτ=
∫ _[t_0,t] √ [ γ˙(τ) • γ˙(τ)]dτ by 4.2.1

and
s_γ~(t)= ∫ _t_0,t.dτ

so σ is a local isometry if and only if these two arc lengths equal for all curves γ: ]α, β[ → U and all t in ]α, β[

CONVERSE: if I is represented by the identity matrix then
I_{γ(t)} (γ˙(t),γ˙(t)) = γ˙(t)• γ˙(t) so s_γ (t)=s_γ~ (t) for all γ and t. Then √[γ˙(t)• γ˙(t)]= √[I_{γ(t)} (γ˙(t),γ˙(t))]

but by 3.4.6
every X ∈T_(u,v) σ is of the form γ˙(t) for some curve γ

so X•X=I_(u,v) (X,X) for all (u,v) ∈U X∈T_(u,v) σ. We want to show I_(u,v)(X,Y)= X•Y for all (u,v)∈U and X•Y∈T_(u,v)σ

Here we use a standard fact on inner products in vector spaces:
(FFF is inner prod)
*if g is an inner product on the vector space V then
g( x, y) = 0.5(g(x+y, x+y) - g(x,x) -g(y,y)) for all x,y in V

Now I_(u,v) (X,Y)= 0.5(I_(u,v) (X+Y,X+Y) -I_(u,v) (X,X) - I_(u,v) (Y,Y))

=0.5( (X+Y)•(X+Y)- X•X - Y•Y)
=X•Y
since •is inner product so I is represented by the identity Matrix

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15
Q

EXAMPLE 4.2.4 the cylinder (u, v) 7→ (cos(u), sin(u), v), the matrix of the first fundamental
form of is constant and equal to the identity matrix

is this an isometry

A

By the theorem, this
parametrization is a local isometry. This is a fancy way of saying that a flat
sheet of paper can be rolled into a cylinder without tearing or crumpling
the paper.

rectangle in R^2 rolled up to R^3 cylinder

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16
Q

isometry

A

map that preserves distances between points e.g translations and rotations

17
Q

local isometry

A

local isometry is only guaranteed to
preserve distances between points that are close enough, but not distances
between all points. In Example 4.2.4, the distance between (0,0) and (3π/2, 0)
on R2 is 3π/2, but the distance between their images Φ(0, 0) and Φ(3π/2, 0) on C is π/2

18
Q

angle between two non-zero vectors

u, v ∈ R^3

A

cos θ = (u · v) / [||u|| ||v||] =

u · v)/(√[u · u] √[v · v]

19
Q

Lemma 4.3.1.
two tangent
vectors

The angle θ between them is given by

A

For a parametrized surface σ : U → R^3, consider two tangent
vectors X,Y ∈ T(u0,v0)σ at a point P = σ(u0, v0):

X = aσ_u(u0, v0) + bσ_v(u0, v0),
Y = cσu(u0, v0) + dσv(u0, v0).
The angle θ between them is given by

cos θ = (X · Y) /(√[X · X] √[Y · Y]) =
(I_(u_0,v_0)((a,b)(c,d)) /
(√[I_(u_0,v_0)(a,b)(a,b)] √[I_(u_0,v_0)(c,d)(c,d)])

shows that the first fundamental form determines angles between tangent vectors
on a surface.
*tangent vectors with respect to a basis at point

20
Q

DEF 4.3.2 conformal

A

A parametrized surface σ : U → R^3 is conformal if for any
two curves γ1 and γ2 in U that intersect at a point (u, v) ∈ U, the angle of intersection of γ1 and γ2 at (u, v) is
equal to

the angle of intersection of

γ˜1 = σ ◦ γ1 and γ˜2 = σ ◦ γ2 at σ(u, v)

σ is conformal if it preserves angles

21
Q

THM 4.3.3

conformal if

A

A parametrized surface σ : U → R^3
is conformal if and only if
its first fundamental form is

 [1 0 ] λ   [0 1 ] for some FUNCTION λ: U →R
22
Q

when are locally isometric parametrizations conformal

A

Any locally isometric parameterization is conformal, here

λ =1

23
Q

The inverse stereographic projection

σ : R^2 → S^2\ {(0, 0, 1)},

σ(u, v) = ([2u/(u^2+v^2+1)],[2v/(u^2+v^2+1)],[(u^2+v^2−1)/(u^2+v^2+1)]

is it conformal?

A

yes,

 [1 0 ] λ   [0 1 ]

for FUNCTION λ: U →R

λ is..

Therefore σ^−1
can be used (and is indeed often used) to produce maps of the Earth that correctly represent all angles

*angles on map ≈ angles on earth, bearings and navigation. Distorts relative areas.

24
Q

Lemma 4.4.1. area of vectors

A

Two vectors u and v span a parallelogram of area

area of ||gram= base x height = ||u||·||v||·sinθ =√[||u||^2·||v||^2·sin^2(θ)] =
√[||u||^2·||v||^2·(1- cos^2(θ)] =
√[(u · u)(v · v) − (u · v)^2]

As dot involves cos

25
approximating : Let σ : U → R3 be a parametrized surface. Given a region R ⊂ U we want to make sense of the area of σ(R). Decompose R into very small rectangles of side ∆u and ∆v.
Then the image of the rectangle with corner (u0, v0) is roughly a parallelogram with sides σu(u0, v0)∆u and σv(u0, v0)∆v. So its area is approximately √[(σu(u0, v0) · σu(u0, v0))(σv(u0, v0) · σv(u0, v0)) − (σu(u0, v0) · σv(u0, v0))2∆u∆v] i.e. √[E(u0, v0)G(u0, v0) − F(u0, v0)^2] ∆u∆v. (det of FFF) *approximating by Taylors thm vector between vertices diagrams: small rectangle sides ∆u and ∆v vertices (u_0,v_0) , (u_0+ ∆u, v_0) , (u_0,v_0+∆v), (u_0+∆u,v_0+∆v) to region R in R^2 squares to surface in R^3 in parallelograms with vertices σ(u_0,v_0) , σ(u_0+ ∆u, v_0) , σ(u_0,v_0+∆v), σ(u_0+∆u,v_0+∆v)
26
DEF 4.4.2 surface area of a parametrized surface
Let σ : U → R3 be a parametrized surface, and let R ⊂ U be a region. Then the surface area of σ(R) is A_σ(R) = ∫ ∫_R √[EG − F^2] du dv .
27
EXAMPLE: The catenoid is the surface of revolution of the catenary (i.e. cosh) curve; it has parametrization σ : R^2 → R^3, σ(t, θ) = (cosh t cos θ, cosh t sin θ, t). The area on between the latitudes t = 0 and t = 1
area= π(1 + (1/2)sinh 2)= 8.838 . . . . FFF is cosh^2(t)* [1 0] [0 1] so √[EG − F^2] = √[cosh^2(t)] A_σ(R) = ∫ ∫_R √[EG − F^2] du dv =A_σ(R) = ∫_[t=0,1] ∫_[θ=0, 2π] [cosh^2(t)] .dt dθ= (∫_[t=0,1]cosh^2(t) .dt)(∫_[θ=0, 2π] .dθ)= 2π[0.5t + (1/4)sinh2t ]_t=0 to 1 = π(1 + (1/2)sinh 2)
28
Definition 4.4.4 equiareal
.A parametrized σ : U → R^3 surface is equiareal if it takes any region in U to a region of the same area in σ(U)
29
Theorem 4.4.5 equiareal thm equiareal IF
A parametrized surface σ : U → R^3 is equiareal if and only if its first fundamental form has determinant 1. *from 4.2.3 and 4.3.3
30
equiareal and isometric?
Any local isometry preserves surface areas. by 4.2.3 local isometry has FFF I_σ = [1 0] [0 1] so det=1 and by 4.4.3 this preserves surface area
31
Example 4.4.7 (Cylindrical coordinates). The inverse cylindrical projection of the sphere (minus North and South Poles) is the parametrization corresponding to the revolution of the parametrized semicircle γ: ]1, −1[ → R^3, t → (√[1 − t^2], 0, t). Writing z for t, the parametrization is σ : R × ]1, −1[ → R^3, (θ, z) → (√[1 − z^2] cos θ,√[1 − z^2] sin θ, z). is this equiareal
This is an equiareal parametrization. We can state this another way. This gives a map of the Earth that correctly represents all areas! area preserving property of this radial projection from the sphere to the cylinder and formula for surface area of a sphere A= 4 pi r^2 FFF for the surface of Revolution: by 4.1.5 [x'(t)^2 +z'(t)^2 0 ] [ 0 x(t)^2 ] where z=t and x=√[1-z^2] ...... dots of normal and basis = [(1/(1-z^2) 0 ] [ 0 1-z^2] which has determinant 1 so by 4.4.5 this is a equiarial