CS计算机代考程序代写 L2_Surfaces

L2_Surfaces

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L2: Surfaces

Hao Su

Machine Learning meets Geometry

Our Focus Today: Surface

http://web.mit.edu/manoli/crust/www/slides/piggy.jpg

Agenda

• Parameterized Surface

• Manifold

• Differential Map

• Curvature

• Principal Curvature

Lots of (sloppy) math!

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Parameterized Surface

Parametrized Surface

Image from Wikipedia

A parameterized surface is a map from a two-
dimensional region into U ⊂ ℝ2 ℝn

f : U → ℝn

The set of points is called the image of the parameterization.f(U)

Example
• Example: We can express a saddle as a

parameterized surface:

U

1

1
f

U := {(u, v) ∈ ℝ2 : u2 + v2 ≤ 1}
f(u, v) = [u, v, u2 − v2]T

• Smoothly “interpolate” between a set of points Pi

Application: Bezier Surface, Spline Surface

Application: Bezier Surface, Spline Surface
Widely used in design industry (e.g., car modeling)

(Differentiable) Manifold

Smoothness as a Local Property
• Things that can be discovered by local

observation: point + neighborhood

Local Smoothness
• Things that can be discovered by local

observation: point + neighborhood

differentiable
1-1 mapping

• Things that can be discovered by local
observation: point + neighborhood

Local to Global

continuous
1-1 mapping

u

v

Tangents, normals,
curvatures, curve
angles, distances

Tangent Plane
• One can attach to every point a tangent plane
• Intuitively, it contains the possible directions in which

one can tangentially pass through .

p Tp

p

p
Tp(ℝ

3)

Differential Map

Differential of a Surface
• Relate the movement of point in the domain and on

the surface

X f(U)U

Dfp(X )

f

p

fp

Differential of a Surface

Total differential: df =
∂f
∂u

du +
∂f
∂v

dvTotal differential:

X
f(U)

U

Df (X)

f

If point moves along vector by , the
movement of is:

p ∈ ℝ2 X = [u, v]T ϵ
fp

Δfp ≈
∂f
∂u

(ϵu) +
∂f
∂v

(ϵv) = ϵ [ ∂f∂u ,
∂f
∂v ] [uv]

Δf ≈
∂f
∂u

Δu +
∂f
∂v

Δv

Differential of a Surface

Total differential: df =
∂f
∂u

du +
∂f
∂v

dvTotal differential:

X
f(U)

U

Df (X)

f

If point moves with velocity by , the
movement of is:

p ∈ ℝ2 X = [u, v]T ϵ
fp

Δfp ≈
∂f
∂u

(ϵu) +
∂f
∂v

(ϵv) = ϵ [ ∂f∂u ,
∂f
∂v ] [uv]

Dfp := [ ∂f∂u , ∂f∂v ] ∈ ℝ3×2

Differential of a Surface

Total differential: df =
∂f
∂u

du +
∂f
∂v

dvTotal differential:

X
f(U)

U

Df (X)

f

If point moves with velocity by , the
movement of is:

p ∈ ℝ2 X = [u, v]T ϵ
fp

Δfp ≈
∂f
∂u

(ϵu) +
∂f
∂v

(ϵv) = ϵ [ ∂f∂u ,
∂f
∂v ] [uv] = ϵ[Dfp]X

Dfp := [ ∂f∂u , ∂f∂v ] ∈ ℝ3×2 : differential (Jacobian) a linear map.
Dfp

Differential of a Surface

Total differential: df =
∂f
∂u

du +
∂f
∂v

dvTotal differential:

X
f(U)

U

Df (X)

f

If point moves with velocity by , the
movement of is:

p ∈ ℝ2 X = [u, v]T ϵ
fp

Δfp ≈
∂f
∂u

(ϵu) +
∂f
∂v

(ϵv) = ϵ [ ∂f∂u ,
∂f
∂v ] [uv] = ϵ[Dfp]X

velocity in 2D domainDfp := [ ∂f∂u , ∂f∂v ] ∈ ℝ3×2

Differential of a Surface

df =
∂f
∂u

du +
∂f
∂v

dvTotal differential:

X
f(U)

U

Df (X)

f

If point moves with velocity by , the
movement of is:

p ∈ ℝ2 X = [u, v]T ϵ
fp

Δfp ≈
∂f
∂u

(ϵu) +
∂f
∂v

(ϵv) = ϵ [ ∂f∂u ,
∂f
∂v ] [uv] = ϵ[Dfp]X

velocity in 3D space

velocity in 2D domainDfp := [ ∂f∂u , ∂f∂v ] ∈ ℝ3×2

Differential of a Surface

𝚍f =
∂f
∂u

𝚍u +
∂f
∂v

𝚍vTotal differential:

X
f(U)

U

Df (X)

f

If point moves along vector by , the
movement of is:

p ∈ ℝ2 X = [u, v]T ϵ
fp

Δfp ≈
∂f
∂u

(ϵu) +
∂f
∂v

(ϵv) = ϵ [ ∂f∂u ,
∂f
∂v ] [uv] = ϵ[Dfp]X

Intuitively, the differential of a parameterized surface
tells us how tangent vectors on the domain get mapped
to tangent vectors in space:

Tangent Plane

is a vector in 3D tangent plane[ ∂f∂u ,
∂f
∂v ] [uv]

These vectors don’t have to be orthogonal

Δfp ≈
∂f
∂u

(ϵu) +
∂f
∂v

(ϵv) = ϵ [ ∂f∂u ,
∂f
∂v ] [uv]

Tangent plane at point is
spanned by

f(u, v)

fu =
∂f
∂u

, fv =
∂f
∂v

fu fv

Df(0,0)(X )

An Example

f(u, v) = [u, v, u2 − v2]T

Dfp =
∂f1/∂u ∂f1/∂v
∂f2/∂u ∂f2/∂v
∂f3/∂u ∂f3/∂v

= [
1 0
0 1

2u −2v]

Df(0,0)(X )

An Example

f(u, v) = [u, v, u2 − v2]T

Dfp =
∂f1/∂u ∂f1/∂v
∂f2/∂u ∂f2/∂v
∂f3/∂u ∂f3/∂v

= [
1 0
0 1

2u −2v]

Df(0,0)(X )

An Example

f(u, v) = [u, v, u2 − v2]T

Dfp =
∂f1/∂u ∂f1/∂v
∂f2/∂u ∂f2/∂v
∂f3/∂u ∂f3/∂v

= [
1 0
0 1

2u −2v]
X :=

3
4

[1, − 1]T

Df(X) =
3
4

[1, − 1,2(u + v)]T

e.g., at u = v = 0 : Df(X) = [
3
4

, −
3
4

,0]T

Df(0,0)(X )

An Example

f(u, v) = [u, v, u2 − v2]T

Dfp =
∂f1/∂u ∂f1/∂v
∂f2/∂u ∂f2/∂v
∂f3/∂u ∂f3/∂v

= [
1 0
0 1

2u −2v]
X :=

3
4

[1, − 1]T

Df(X) =
3
4

[1, − 1,2(u + v)]T

e.g., at u = v = 0 : Df(X) = [
3
4

, −
3
4

,0]T

Df(0,0)(X )

An Example

f(u, v) = [u, v, u2 − v2]T

Dfp =
∂f1/∂u ∂f1/∂v
∂f2/∂u ∂f2/∂v
∂f3/∂u ∂f3/∂v

= [
1 0
0 1

2u −2v]

at , tangent space is spanned by and . u = v = 1 [
1
0
2] [

0
1

−2]
e.g., at u = v = 0 : Df(X) = [

3
4

, −
3
4

,0]T

X :=
3
4

[1, − 1]T

Df(X) =
3
4

[1, − 1,2(u + v)]T

Df(0,0)(X )

An Example

f(u, v) = [u, v, u2 − v2]T

Dfp =
∂f1/∂u ∂f1/∂v
∂f2/∂u ∂f2/∂v
∂f3/∂u ∂f3/∂v

= [
1 0
0 1

2u −2v]

at , tangent space is spanned by and . u = v = 1 [
1
0
2] [

0
1

−2]
e.g., at u = v = 0 : Df(X) = [

3
4

, −
3
4

,0]T

X :=
3
4

[1, − 1]T

Df(X) =
3
4

[1, − 1,2(u + v)]T

Summary of Differential Map
• Tells us the velocity of point in 3D when the parameter

changes in 2D

• Maps a vector in the tangent space of the domain to
the tangent space of the surface

• Allows us to construct the bases of tangent plane

• Is a linear map

Dfp : Tp(ℝ
2) → Tf(p)(ℝ

3)

Curvature

Goal
Quantify how a surface bends.

Curvature of CurvesRecall:

Theorem:
Curvature and torsion determine geometry

of a curve up to rigid motion.

35

The Binormal Vector

For points s, s.t. !(s) ≠ 0, the
binormal vector B(s) is defined
as:

B(s) = T(s) ” N(s)

The binormal vector defines the
osculating plane

T

N

B
N

T

B

T

N

N

T

Can curvature/torsion of
a curve help us

understand surfaces?

Curves: Change of Normal
Describes Curve Bending

http://mathworld.wolfram.com/images/eps-gif/UnitSphere_800.gif

Surfaces: Change of Normal
Describes Surface Bending

q

Surface Normals

Surface normal:

fu :=
∂f
∂u

, fv :=
∂f
∂v

N(u, v) =
fu × fv

∥fu × fv∥

also as a function of N u, v

fu fv

Consider a nonstandard parameterization of the cylinder
(sheared along ):z

Example

Df =
−sin(u) 0
cos(u) 0

1 1

N =
−sin(u)
cos(u)

1
× [

0
0
1] =

cos(u)
sin(u)

0

f(u, v) := [cos(u), sin(u), u + v]T

Df (X2) Df (X1)
N

http://mathworld.wolfram.com/images/eps-gif/UnitSphere_800.gif

Measure the Change of Normal
Assume moves along a curve parameterized by arc-
length: , and the normal is with unit norm

q γ
q = γ(s) N(s)

q

http://mathworld.wolfram.com/images/eps-gif/UnitSphere_800.gif

Measure the Change of Normal
Assume moves along a curve parameterized by arc-
length: , and the normal is with unit norm

q γ
q = γ(s) N(s)

0 ≡
d
ds

⟨N(s), N(s)⟩ = 2⟨ ·N(s), N(s)⟩

Local change of normal is
always in the tangent plane!

·N(s) ⊥ N(s)
q

Differential of Normal

dN =
∂N
∂u

du +
∂N
∂v

dvTotal differential:

X U

f

If point moves with velocity by , the
movement of is:

p ∈ ℝ2 X = [u, v]T ϵ
Np

ΔNp =
∂N
∂u

(ϵu) +
∂N
∂v

(ϵv) = ϵ [ ∂N∂u ,
∂N
∂v ] [uv] = ϵ[DNp]X

DNp := [ ∂N∂u , ∂N∂v ] ∈ ℝ3×2

p
q = fp

Differential of Normal

dN =
∂N
∂u

du +
∂N
∂v

dvTotal differential:

If point moves with velocity by , the
movement of is:

p ∈ ℝ2 X = [u, v]T ϵ
Np

ΔNp =
∂N
∂u

(ϵu) +
∂N
∂v

(ϵv) = ϵ [ ∂N∂u ,
∂N
∂v ] [uv] = ϵ[DNp]X

X U

f

DNp := [ ∂N∂u , ∂N∂v ] ∈ ℝ3×2
Note: [DNp]X ∈ Tp(ℝ3)

p
q = fp

Curvature of at ⃗κ γ p
• Recall we need the arc-length parameterization and

measure the change of normal
• Recall that tangent vector under arc-length

parameterization. So we need to scale by so that:

• As moves with velocity , the tangent is

• the velocity of normal change is:

∥T∥ = 1
X μ

∥Dfp[μX]∥ = 1 ⟹ μ =
1

∥DfpX∥
p μX

Dfp[μX] =
DfpX

∥DfpX∥

DNp[μX] =
DNpX

∥DfpX∥

Curvature of at ⃗κ γ p

• The velocity of normal change is:

• We denote this quantity as in this lecture (note that
in the last lecture is a scalar, the norm of this vector)

DNp[μX] =
DNpX

∥DfpX∥

⃗κ
κ

Directional Normal Curvature

κn(X) := ⟨T, ⃗κ ⟩ =
⟨Dfp(X), DNp(X)⟩

∥Dfp(X)∥2

Df

Df

Note: is not the curvature of κn κ γ

X U

p

f

Dfp(X)

DNp(X)

N

tangent plane

Relationship to Curvature of Curves

http://www.solitaryroad.com/c335.html

Drawing by Adrian Butscher

κn := ⟨ ⃗κ , T⟩

κg := ⟨ ⃗κ , N × T⟩
(Geodesic curvature)

Consider a nonstandard parameterization of the cylinder
(sheared along ):z

Example

Df =
−sin(u) 0
cos(u) 0

1 1
f(u, v) := [cos(u), sin(u), u + v]T

Df (X2) Df (X1)
N

N =
cos(u)
sin(u)

0
DN =

−sin(u) 0
cos(u) 0

0 0

Consider a nonstandard parameterization of the cylinder
(sheared along ):z

Example

Df =
−sin(u) 0
cos(u) 0

1 1
f(u, v) := [cos(u), sin(u), u + v]T

Df (X2) Df (X1)
N

N =
cos(u)
sin(u)

0
DN =

−sin(u) 0
cos(u) 0

0 0

Consider a nonstandard parameterization of the cylinder
(sheared along ):z

Example

Df =
−sin(u) 0
cos(u) 0

1 1
f(u, v) := [cos(u), sin(u), u + v]T

Df (X2) Df (X1)
N

N =
cos(u)
sin(u)

0

X1 = [01]
κn(X1) =

⟨Df(X1), DN(X1)⟩
∥Df(X1)∥2

= 0

DN =
−sin(u) 0
cos(u) 0

0 0

X2 = [−11 ]

κn(X2) =
⟨Df(X2), DN(X2)⟩

∥Df(X2)∥2
= 1

Consider a nonstandard parameterization of the cylinder
(sheared along ):z

Example

Df =
−sin(u) 0
cos(u) 0

1 1
f(u, v) := [cos(u), sin(u), u + v]T

Df (X2) Df (X1)
N

N =
cos(u)
sin(u)

0

X1 = [01]
κn(X1) =

⟨Df(X1), DN(X1)⟩
∥Df(X1)∥2

= 0

DN =
−sin(u) 0
cos(u) 0

0 0

X2 = [−11 ]

κn(X2) =
⟨Df(X2), DN(X2)⟩

∥Df(X2)∥2
= 1

Summary of Curvature

• Curvature quantifies the bending of surfaces

• Local change of normal (differential of normal) is
always in the tangent plane

• Directional normal curvature quantifies how fast a
surface bends along a direction

Principal Curvatures

Principal Curvatures
Maximal curvature:

Minimal curvature:

κ1 = κmax = max
φ

κn(φ)

κ2 = κmin = min
φ

κn(φ)

N

X1
X2

N X1N X2

Principal Directions

min curvature max curvaturetangent plane in 3D

ϕ min

t1
t2

Principal directions:
tangent vectors
corresponding to

and φmax φmin

Euler’s Theorem: Planes of principal curvature are orthogonal
and independent of parameterization.

Principal Directions

κn(φ) = κ1 cos
2 φ + κ2 sin

2 φ, φ = angle with t1

Tangent plane with principal
directions as axes

t1

t2

φ

Principal Directions

Summary of Principal Curvatures

• The direction that bends fastest / slowest are principal
directions, which are orthogonal to each other

• The corresponding curvatures are principal curvatures