Chapter 4 Calculus and theory of surfaces Flashcards
DEFN
REGULAR n-DIMENSIONAL SURFACE
need to state for exam
A subset Σ ⊂ R^m is called a regular n-dimensional surface, where n ≤m is a positive integer, if for any point p ∈ Σ there exists an open neighbourhood V of p ∈ R^m and an open set U ⊂ R^n together with a smooth map φ : U → R^m such that:
(i) φ(U) = Σ∩V and the map φ : U → V is a continuous embedding
( φ : U → Σ ∩ V is bijective, and p_k → p if and only if φ(pk) → φ(p));
(ii) rank D_xφ = n for any x ∈ U
diagram for n reg surface
surface Σ
point p on surface with NHD Σ∩V
which is mapped to by φ on U
φ: U to V open containing point p
we can use coords in euclidean space
euclidean coords (x_1,…,x_n) on v are local coords around p
trivial example
(R^m, Id, R^m) identity map
triple (U, φ, V )
local parametrisation of a surface
surface Σ around a point p
local patch p
triple (U, φ, V ) that satisfies the hypotheses for REGULAR n-DIMENSIONAL SURFACE is called the local parametrisation of a surface Σ around a point p,
also called chart around p
chart around p
triple (U, φ, V ) that satisfies the hypotheses for REGULAR n-DIMENSIONAL SURFACE is called the local parametrisation of a surface Σ around a point p,
also called local patch
coordinate nhd
set Σ ∩ V is called the coordinate neighbourhood of a point p ∈ Σ.
local coordinates
The Euclidean coordinates
(x_1, . . . , x_n) on U are called local coordinates around a point p ∈ Σ.
summary
in order to prove surface is regular its sufficient to construct a #charts s.t cover whole image with corresponding V’s
charts cover surface
continuous embedding hypothesis
IN EXAM
construction of charts
for simple surface
construct a chart
and briefly explain why its a chart
simple example:
with RCPC
drawing closed
γ:[0,1] to R^n
Γ=γ([0,1]) subset in R^n (image)
The image φ(v)=H (H intersection V)
we give the chart describing
{γ(t): t in [0,1]}{γ(0)}
(part of γ not γ(0))
u=(0,1)
φ=γ|_[0,1]
V=R^n{{γ(0)}
all due to gamma being a simple RCPC
we have chart (u_1,φ_1,v_1)
now if we want to cover {γ(0)} we need to produce another chart (u_2,φ_2,v_2)
u_1=(1/2,3/2)
γ‾|_[1/2,3/2]
v_2=R^n{{γ(1/2)}
two charts sufficiently cover whole image
Example IV.2 (Linear subspaces). is it an n dimensional regular surface?
linear subspace
H⊂ R^m of dimension n
H is an n-dimensional regular surface:
To construct a chart:
choose orthonormal basis
h1, . . . , h_n in H
define a map
φ : R^n → R^m
φ(x1, . . . , xn) = x_1h_1 + · · · + x_nh_n ∈ H ⊂ R^m
(for which we can easily compute coeffs)
linear map
(because injective its a linear map so kernel trivial means nullity 0, could use RNthm)
or range equals H,
hence, its rank equals n = dim H.
by prev e.g for any x ∈ R^n we have Dxφ = φ,
rank Dxφ =n
Claim triple (R^n, φ, R^m) is a chart for H.
remains to check φ is
a continuous embedding.
injective and continuous
(φ linear map and injective, bc for 2 different n tuples will give different vector in V as basis means unique composition )
checking
φ((x_k)_1, . . . ,(x_k)_n) → φ(x_1, . . . , x_n) ⇒ (x_k)i → xi as k → +∞ for any i = 1, . . . , n,
for any sequence (xk) and a point x in R^n. Since a basis h_1, . . . , h_n is orthonormal, by the map
used to define φ, we obtain
(x_k)_i = h_i· φ((x_k)_1, . . . ,(x_k)_n) → h_i· φ(x_1, . . . , x_n) = x_i as k → +∞
for any i = 1, . . . , n, and conclude that implication holds.
PROP 4.1
image for a_________ is a 1 dimensional surface
Let γ : [0, 1] → R^n be a simple regular closed parametrised curve (RCPC). Then its image Γ = γ([0, 1]) ⊂ R^ n is a regular 1-dimensional surface in R^n
Example IV.3 (Graphs of functions). Let f : U ⊂ R^n → R be a smooth function, where U ⊂ R^n is
an open set. We claim that its graph
Γ = {(x_1, . . . , x_n, f(x_1, . . . , xn)) : (x1, . . . , xn) ∈ U} ⊂ R^{n+1
is a regular n-dimensional surface
diagram:
open set U represents R^n in plane (e.g x-y plane)
Γ gives
this open set moved up, ie representing R^{n+1}
Note set
V = U × R = {(x1, . . . , xn, x_{n+1}) : (x1, . . . , xn) ∈ U}
is an open subset of R^{n+1}
triple (U, φ, V) is a chart on Γ:
φ : U
(x_1, . . . , x_n) → (x_1, . . . , x_n, f(x_1, . . . , x_n)) ∈ Γ ∩ V ⊂ V,
map φ is smooth, bijective onto Γ ∩ V
continuous embedding:
since continuous we only need to check for any p ∈ U and any sequence pk ∈ U
such that φ(pk) → φ(p), we have pk → p as k → +∞
(pk, f(pk)) = φ(pk) → φ(p) = (p, f(p)) =⇒ pk → p.
checking second part:
For any p ∈ U the matrix of the
differential D_pφ’s rank:
matrix of differential has form
J_p φ =
[1 0 . . . 0]
[0 1 . . . 0]
[. . . . . . . . ]
[0 0 . . . 1]
[∂f/∂x_1|_p ∂f/∂x_2|_p. . .∂f/∂x_n|_p]
first n rows linearly indep
CLEAR ARGUMENT NEEDED HERE
b.c first cols are we must have
thus maximal rank =n
thus the triple is a chart
(example uses cyclinder?)
unit sphere S^n ⊂ R^{n+1} is a regular n dimensional surface
S^n =
{(x_1,..,x_{n+1}) ∈ R^{n+1} : sum_{i=1, n+1} x^2_i =1}
⊂R^{n+1}
construct chart/s for this
(stereographic projection, pole is removed (0,0,1)
BREAK INTO TO HEMISPHERES
to show its a reg n dim surface sufficient to construct collection of charts
(Uᵢ±,φᵢ±, Vᵢ±) I=1,…,n+1
s.t any point p∈ Sⁿ
belongs to either Sⁿ∩Vᵢ+ or Sⁿ∩Vᵢ- for some i
we have collection of 2(n+1) charts
sets Sⁿ∩Vᵢ± form a covering of the sphere Sⁿ
set Vᵢ± to be half spaces
Vᵢ+ = {(y_1,.., yₙ₊₁ ∈Rⁿ⁺¹ : yᵢ>0}
Vᵢ- = {(y_1,.., yₙ₊₁ ∈Rⁿ⁺¹ : yᵢ<0}
so intersections Sⁿ∩Vᵢ± are hemispheres
all sets Uᵢ± are set to be unit n dim ball
Uᵢ±=Bⁿ = {(x₁,..,xₙ)∈ Rⁿ: sum_{i=1,n} xᵢ²<1 }
i =1,…,n+1
so maps defined by
φᵢ±: Uᵢ± toSⁿ∩Vᵢ±
φᵢ+(x₁,..,xₙ) = (x₁,..,xᵢ₋₁, sqrt( 1 - Σⁿⱼ₌₁ xⱼ²), xᵢ ,xₙ )
φᵢ-(x₁,..,xₙ)= (x₁,..,xᵢ₋₁, -sqrt( 1 - Σⁿⱼ₌₁ xⱼ²), xᵢ ,xₙ )
we can then show satisfy the hypothesis of a chart
Self-Check Question IV.2. In Example IV.4 the hemi-spheres S^n ∩ V_i+and S^n ∩ V_i−
i in {1, . . . , n + 1}, cover the whole sphere. In other words, any point p ∈ S^n belongs
either toS^n ∩ V_i+ or S^n ∩ V_i−
for some i
explain
if p in S^n
if lies in sphere there exists i s.t x_i not equal to 0
0/w length 0
origin =0
so if x_i >0 then p in V_i+
if x_i<0 then p in V_i -
Thus every p in S^n belongs to either chart
for a 2 dim sphere in 3 dimensions how many open hemispheres will we need to cover
6 open hemispheres required
equator not covered in this example
upper lower in x_1
in x_2 in x_3
level set
smooth map Φ : R^m → R^ℓ by its level set we mean the pre-image
Φ−1(y) := {x ∈ R^m : Φ(x) = y},
where y ∈ R^ℓ
Theorem IV.2 (Regular Value Theorem
smooth map Φ : R^m → R^ℓ
where m > `ℓ
Suppose that for some y ∈ R^ℓ
the pre-image Φ−1 (y) does not contain singular points. Then it is a
regular (m − ℓ )-dimensional surface in R^m
proof:
need to show for any point in preimage to construct chart s.t point is in intersection of preimage and V
use constant rank thm….
WHY do we use the regular value thm
gives method for producing surface s as preimages under maps
if we can check preimage has no singular points we have a regular surface
WE DONT NEED TO CONSTRUCT CHARTS JUST USE THM
Example IV.5 (Unit sphere S^n ⊂ R^{n+1} revisited).
differential
n dimensional surface?
Consider the following function
Φ : R^{n+1} → R, Φ(x1, . . . , xn+1) =
Σⁿ⁺¹ⱼ₌₁ xⱼ²
level set Φ−1(1) is the unit sphere S^n ⊂ Rⁿ⁺¹
differential DxΦ is 1 × n-matrix
J_xΦ =
(∂Φ/∂x_1, . . . ,∂Φ/∂x_{n+1})
= (2x_1, . . . , 2x_{n+1}),
rank D_xΦ =1 for any x ∈ R^{n+1}{0}.
Theorem IV.2: conclude unit sphere S^n = Φ−1(1) is a regular n-dimensional surface in Rⁿ⁺¹
Example IV.6 (Hyperboloids.).
consider
Φ : Rⁿ⁺¹ → R, Φ(x₁, . . . , xₙ₊₁) = x₁²+x₂²+…+xₙ²-xₙ₊₁²
show its a regular n dimensional surface
range is whole real line R
differential DxΦ
1 x (n-1) matrix
JxΦ =
(∂Φ/∂x_1, . . . , ∂Φ/∂x_{n+1})
= (2x₁, . . . , 2xₙ, −2xₙ₊₁),
rank D_xΦ =1 for any x ∈ R^{n+1}{0}.
( can be 1 or 0
X singular IFF rank J_x <1 IFF rank =0 IFF jacobian =0 iff X_i=0 for all i=1,..,n} thus the set of singular points consists only of {0} not in S^n)
TAKE CARE WITH 3 CASES
for real c /= 0:
preimage Φ−1 (c) does not contain singular points, and by the Regular Value Theorem it is a regular n-dimensional surface.
For c > 0 the pre-image Φ−1(c) is a hyperboloid of one sheet,
take hyperbola in plane spanned by x_1 and x_2 coords and rotate
c < 0 it is a hyperboloid of two sheets.
pre-image Φ−1(0) is a cone, is not a regular surface CANNOT APPLY RVT FOR THIS CASE
only first two cases are reg surfaces of dim n
there will be no chart for singularity 0
Example IV.7 (Special Linear group SL(2)).
view the 4-dimensional Euclidean space R^4 as
the set of 2 × 2-matrices, by identifying the 4-tuple (x_1, x_2, x_3, x_4) with the matrix
[x1 x2]
[x3 x4]
Recall that the Special Linear group SL(2) is a subset consisting of those matrices whose determinant equals one. We claim that it is a 3-dimensional regular surface in R^4 .
by considering
Φ : R^4 → R, Φ(x1, x2, x3, x4) = x_1x_4 − x_2x_3.
is SL(2) a 3 dimensional regular surface?
The level set Φ−1(1) is precisely SL(2).
The differential DxΦ is given by the 1 × 4-matrix
JxΦ = (x4, −x3, −x2, x1),
rank DxΦ equals one for any x ∈ R^4{0}.
pre-image Φ−1(1) does not contain
singular points, and by the Regular Value theorem we conclude that SL(2) is indeed a 3-dimensional regular surface
RVT IN EXAM
need to show the regular value thm in exam
apply this theorem
to a SPHERE?
main example, exercises
every exam has a question like this
GIVEN SURFACE
you need to represent the surface as a preimage of function
calculate the regular points and apply the theorem
writing preimages from surfaces is the main part of module
LEMMA 4.3
if there exists a chart
a map is smooth iff
Let Σⁿ ⊂ Rᵐ be a regular n-dimensional surface, and
Φ : W ⊂ R^ → Rᵐ a map defined on an open subset W ⊂ R^ and takes values in Σⁿ.
Suppose that there exists a chart
(U, ϕ, V ) for
Σⁿ such that
Φ(W) ⊂ Σⁿ∩V .
Then the map Φ is smooth
IFF
the map ϕ⁻¹◦Φ : W → U ⊂ Rⁿ is smooth.
in principle any graph is a preimage
locally we can put a graph
(x_1,..,x_n, f(x_1,..,x_n))
phi = x_{n-1} - f(x_1,..,x_n)
inverse of phi (0) is image
there will be an exam q on
RVT
(e.g whitneys umbrella which is the singular set for which RVT doesnt apply singular point @ origin preimage of
becomes regular as we restrict arb close to 0_
proof discussion:
Then the map Φ is smooth
IFF
the map ϕ⁻¹◦Φ : W → U ⊂ Rⁿ is smooth.
converse trivial we look at composition not charts here its a consequence of chain rule
composition of smooth maps is a smooth map
implication: non trivial
we use IVT
consider surface, chart, point q s.t phi maps open set U containing point p to NHD of q, V
choose q in phi(w) in intersection of surface and V
q=phi(p)
can find 3d nhd of p
phi is smooth map inverse defined…
proof continues finding diffeomorphism etc long
Corollary IV.4
two charts
Let Σⁿ ⊂ Rᵐ be a regular n-dimensional surface, and (U₁, ϕ₁, V₁) and (U₂, ϕ₂, V₂)
be two charts such that the set
W := Σⁿ ∩ V₁ ∩ V₂ is non-empty. Then the map ϕ₂⁻¹◦ ϕ₁ : ϕ₁⁻¹(W) ⊂ Rⁿ → ϕ₂⁻¹(W) ⊂ Rⁿ is smooth
transition functions
from corollary 4,4
maps ϕ₂⁻¹◦ ϕ₁
transition functions between charts
maps have smooth inverse maps ϕ₁⁻¹◦ ϕ₂
thus are diffeomorphisms
Example IV.8 (Transition functions on a unit sphere S^n ⊂ R^n+1)
Prev constructed charts on unit sphere (Uᵢ±,φᵢ±, Vᵢ±)
write the transition functs for the charts
(Uᵢ+,φᵢ+, Vᵢ+) and (U+_j,φ+_j, V+_j)
for some values i<j explicitly
first set the quarter sphere:
W=S^n ∩Vᵢ+∩V+_j
W= {(x₁,…,xₙ₊₁)∈S^n xᵢ>0,xⱼ>0}
transition function
(φ+_j)−1 ◦ (φ+_i) is defined on the set
(φ+_i)−1 (W) = {(x1, . . . , xn) ∈ B^n: x_j > 0},
and takes values in the set
(φ+_j)−1(W) = {(x1, . . . , xn) ∈ B^n: x_i > 0}.
Using the explicit formulae for φ+_iand φ+_j
in Example IV.4, we obtain
(φ+_j)−1◦ (φ+_i) : {B^n: x_j > 0}
(x1, . . . , xn) →(x1, . . . , xi−1, sqrt(1 −1 - Σxₖ²), xᵢ,…,xⱼ₋₁^,…, xₙ } ∈ {B^n: x_i > 0},
where the hat ˆ over xj−1 says that this coordinate is omitted.
In particular, we also see that the
transition map (φ+_j)−1 ◦ (φ+)i) is smooth; this conclusion agrees with Corollary IV.4.
Lemma IV.5
EQUIVALENT HYPOTHESIS FOR REGULAR N DIMENSIONAL
Let Σ^n ⊂ R^m be a regular n-dimensional surface, and let q ∈ Σ^n be an arbitrary
point. Then for a map Φ : Σn → R^ℓ
the following hypotheses are equivalent:
(i) there exists an open neighbourhood V˜ ⊂ R
m of a point q such that the map Φ : Σn ∩ V˜ → R^ℓ extends to a smooth map Φ : ˜ V˜ → R^ℓ;
(ii) for any chart (U, ϕ, V ) such that q ∈ V the map Φ ◦ ϕ : U → R^ℓ is smooth in a neighbourhood
of ϕ−1(q);
(iii) there exists a chart (U, ϕ, V ) such that q ∈ V and the map Φ ◦ ϕ : U → R^ℓ
is smooth in a neighbourhood of ϕ−1(q).
a map is called smooth if…
Let Σ₁ⁿ-¹ ⊂ Rᵐ-¹ and Σ₂ⁿ-² ⊂ Rᵐ-² be two regular surfaces of dimensions n₁ and n₂ respectively.
A map Φ : Σ₁ⁿ-¹ → Σ₂ⁿ-² is called smooth at a point p ∈ Σ₁ⁿ-¹ if for some (and hence any)
local charts
(U₁, ϕ₁, V₁) on Σ₁ⁿ-¹
and
(U₂, ϕ₂, V₂) on Σ₂ⁿ-²
such that p ∈ V_1 and
Φ(Σ₁ⁿ-¹ ∩ V_1) ⊂ Σ₂ⁿ-² ∩ V_2
the map
ϕ₂−1◦ Φ ◦ ϕ₁ : U_1 ⊂ Rⁿ-¹ → U₂ ⊂ Rⁿ-²
is smooth at ϕ₁−1(p)
Example IV.10
notion of diffeomorphism
consider funct
Φ : R^3 → R,
Φ(x1, x2, x3) = x_1x_2.
We may restrict it to a unit sphere S^2 ⊂ R^3 and view it as a map S^2 → R
Since the real line R is
an 1-dimensional regular surface with one chart (R,Id, R), the local representation of Φ : S
2 → R is reduced to the map Φ ◦ φ±_i: U±_i → R, where we use the charts (U±i, φ±i, V ±i), i = 1, 2, 3, on a sphere S^2
constructed in Example IV.4. For instance, in the chart (U+_1, φ+1, V +1) the local representation of Φ is the map Φ ◦ φ+1: B^2 →R,
Φ ◦ φ+_1(z_1, z_2) = z_1SQRT(1 − z_1^2 − z_2 ^2)
where B^2 ⊂ R^2 is a unit ball; in the chart (U_3−, φ_3−, V_3 −) the local representation of Φ is the map Φ ◦ φ−_3: B^2 → R, Φ ◦ φ−_3 (z1, z2) = z1z2.
DIFFEOMORPHISM BETWEEN SURFACES DEFN
Let Σ₁ⁿ-¹ ⊂ Rᵐ-¹ and Σ₂ⁿ-² ⊂ Rᵐ-² be two regular surfaces of dimensions n₁ and n₂ respectively.
A map Φ : Σ₁ⁿ-¹ → Σ₂ⁿ-²
is called a diffeomorphism
if Φ is smooth, bijective, and the
inverse map
Φ-1Σ₂ⁿ-² →Φ : Σ₁ⁿ-¹ is smooth
surfaces which there exists a diffeomorphism are called diffeomorphic
EXAMPLE
orthogonal transformations of unit sphere
Let A: R{n+1) to R^ {n+1} be an orthogonal transformation
since preserves lengths
|Av|=|v| for any v in R^{n+1}
concluding that restriction A|_{S^n} is a map of unit sphere S^n ⊂ R^{n+1} to itself
claiming its a smooth map:
A|_S^n : S^n to S^n
Indeed, if (U1, ϕ1, V1) is a chart
on S^n, then the map A ◦ ϕ1 : U1 → R^n+1 is smooth as a composition of smooth maps. If (U2, ϕ2, V2) is another chart on S^ n such that A(V1) ⊂ V2, then by Lemma IV.3 the map ϕ−1_2◦ A ◦ ϕ_1 is smooth.
Thus, the condition in Definition IV.9 is verified. Moreover, note that for any A ∈ O(n + 1) the map
A|Sn : Sn → S n is actually a diffeomorphism in the sense of Definition IV.11. Indeed, the restriction
A−1|_Sn : Sn → Sn of the inverse map to A ∈ O(n + 1) is the inverse map to A|Sn : S^n → S^n, and is smooth by the reasoning above
e.g consider n dimensional ellipsoid
e.g smooth funct
TANGENT VECTOR
TANGENT SPACE
propn 4.6
n DIM SURFACE
TANGENT SPACE
bases for tangent space
UNIT SPHERE TANGENT SPACE EXAMPLE
special linear group tangent space
dimensions
VERY USEFUL: HOW TO WORK OUT TANGENT SPACES with RVT
DIFFERENTIAL TO MAPS BETWEEN SURFACES
proposition for differential using with PC
worked example similar to exercises
continue worked example to give the matrix of D_p relative to bases of tangent space
note
orthogonal trasform restructing A to unit sphere
TANGENT VECTOR FIELD
EXAMPLE: velocity vector field
rotation vector field example
remark
example normal vector field
normal vector field
example curvature vector field
mobius strip
propn special linear group
ORIENTABLE
path connected
regular level sets
propn orientable and path connected
example:path connected
level 5 determinant property
propn det
corollary for non degenerate matrix det/=0 (Level 5)
little summary of matrices and groups
level 5 orthogonal projection recap
level 5 tangent vector field
xample coord vector field L5
level5 covariant derivative
level 5 propn 4.6
vector function dep on
lewl 5 propn 5.7 properties of covariant derivative
CHRISTOFFEL SYMBOLS
covariant derivative using christoffel
EXAMPLE working out christoffel symbols
properties of christoffel symbols
