Chapter 4: Metric quantities

Question 1

Lemma 4.4.1: dot product wrt basis

Answer 1

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

Question 2

DEF 4.1.2

first fundamental form

Answer 2

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

Question 3

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

Answer 3

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

for all (u, v) ∈ R^2

Question 4

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)

Answer 4

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

Question 5

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

Answer 5

[x’(t)^2 + z’(t)^2 0 ]

[0 x(t)^2 ]

Question 6

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

FFF=

Answer 6

[1 0]

[0 1]

Question 7

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

Answer 7

[1 0 ]

[0 cos^2(t)]

Question 8

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.

Answer 8

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

Question 9

image on surface

Answer 9

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

Question 10

LEMMA 4.2.1: length of γ˜ = σ ◦ γ

Answer 10

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.

Question 11

DEF 4.2.2: local isometry

Answer 11

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

Question 12

THM 4.2.3

local isometry if

Answer 12

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

Question 13

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

Answer 13

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•)

Question 14

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

Answer 14

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

Question 15

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

Answer 15

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

Question 16

isometry

Answer 16

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

Question 17

local isometry

Answer 17

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

Question 18

angle between two non-zero vectors

u, v ∈ R^3

Answer 18

cos θ = (u · v) / [||u|| ||v||] =

u · v)/(√[u · u] √[v · v]

Question 19

Lemma 4.3.1.
two tangent
vectors

The angle θ between them is given by

Answer 19

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

Question 20

DEF 4.3.2 conformal

Answer 20

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

Question 21

THM 4.3.3

conformal if

Answer 21

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

Question 22

when are locally isometric parametrizations conformal

Answer 22

Any locally isometric parameterization is conformal, here

λ =1

Question 23

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?

Answer 23

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.

Question 24

Lemma 4.4.1. area of vectors

Answer 24

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

Question 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.

Answer 25

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)

Question 26

DEF 4.4.2 surface area of a parametrized surface

Answer 26

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 .

Question 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

Answer 27

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)

Question 28

Definition 4.4.4

equiareal

Answer 28

.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)

Question 29

Theorem 4.4.5

equiareal thm

equiareal IF

Answer 29

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

Question 30

equiareal and isometric?

Answer 30

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

Question 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

Answer 31

dots of normal and basis

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]
……

[(1/(1-z^2) 0 ]
[ 0 1-z^2]

which has determinant 1 so by 4.4.5 this is a equiarial