Chapter 4 Calculus and theory of surfaces Flashcards
(77 cards)
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.
