Chapter 5 Geometry of regular surfaces Flashcards
intrinsic geometry
intrinsic geometry, that is the study of geometric properties
that are seen by the inhabitants of a surface only.
e.g if we bend a surface the length of curve wont change from point of observers geometry wont change but the shape of surface changes
dot product
in R^n is a positive definite symmetric bilinear form on R^n
dot prod refers to notion of chosen basis
for any scalar product B on R^n there exists basis
e_1‾,….,e_n‾ s,t for any vectors u and v in R^n
symmetric bilinear form,
Let V be a finite-dimensional vector space. A map B : V × V → R is called a symmetric bilinear form, if it satisfies the following properties
(i) B(α1u1 + α2u2, v) = α1B(u__1, v) + α2B(u2, v) for any vectors u1, u2, and v ∈ V , and any real
numbers α1 and α2;
(ii) B(u, v) = B(v, u) for any vectors u and v ∈ V
positive definite
A symmetric bilinear form B on a vector space V is called positive definite (or a scalar product), if B(u, u) > 0 for any u ∈ V , and B(u, u) = 0 if and only if u = 0
matrix of the basis
by property 2: symmetric b_ij=b_ji for any i and j
A MATRIX CAN REPRESENT BILINEAR FORMS
First Fundamental Form G
(RESTRICTION to the tangent space of the dot prod from euclidean spae R^n)
on a regular surface Σ^n ⊂ R^m is a family of scalar products G_p : T_pΣ^n×T_pΣ^n → R (that is a family of positive definite symmetric bilinear forms), where
p ranges over Σ^n, such that
G_p(X, Y ) = X · Y for any X, Y ∈ T_pΣ^n and any p ∈ Σ^n
——————————–
Let P_n be space of polynomials
f:R to R whose degree not greater than n
is this a positive definite symmetric bilinear form on P_n:
B(f,g) =∫_{0,1} f(t)g(t).dt
We claim that the scalar product B is
uniquely determined by the matrix bij = B(ei
, ej ), where i, j = 1, . . . , n. Indeed, for any vectors u and v ∈ V by properties (i) and (ii)
B(u,v) = sum_i,j b_ij u_i v_j
Thus, we conclude that the value B(u, v) can be computed from the knowledge of the matrix (bij ), and the claim holds.
checking linear wrt first var
non neg integral is non neg
measuring length of vector in Bilinear form
distance and angles
given vector in V
length in sense of given bilinear form
they can also imply cs
b(u,v) <= sqrt (B(u,u) B(v,v))
so we can do everything that we can in euclidean space
CS bilinear form
bilinear triangle inequality
interested inpath length?
We are concerned with computing
the lengths of curves on surfaces Σ^n ⊂ R^m
Let γ : [a, b] → R^m be a PC such that γ(t) ∈ Σ
n for any t ∈ [a, b]. As we know, the length of γ is given by the formula
L(γ) = integral_[a,b] |γ’(t)| dt.
Since γ lies in Σn, the velocity vector γ’(t) lies in the tangent space Tγ(t)Σ^n; due Definition IV.15
(to show γ’(t) lies in tangent space sufficient to construct generating curve γ~ :(-e,e) to surface s,t γ~(0)=p and γ~’(0)=γ’(t)
we are interested in the geometry of the surface ie being able to compute lengths of paths that lie in surface
and hence we only need to know values |X| for |X| in the tangent space for all p in the surface
(speeds lengths for tangent vectors AND THATS ALL WE NEED TO KNOW)
Consider the space P_2 of polynomials f : R → R whose degree is not greater than two. Pick a basis in this space and compute the matrix of the form given by formula corresponding to it
what is dimension of this space?
what is dimension of this space?
basis consists of 3 polynomials so it is 3
basis
(1,t,t^2)
compute
matrix of
B(f,g) = integral_[0,a] f(t)g(t).dt in this basis
bij = B(ei, ej )
integrating
1, t, t^2
t, t^2, t^3
t^2, t^3, t^4
integrating over [0,1]
[ 1 1/2 1/3]
[1/2 1/3 1/4]
[1/3 1/4 1/5]
on 6 components
MATRICES AND COMPUTATION
for some chart (U,phi,V) around p in Σ^n
we have a CANONICAL BASIS
X_i(p) = D_x(ϕ(e_i)) = ∂ϕ/∂xi(x),
of tangent space and we may compute corresponding matrix in this chart
we obtain
gij (ϕ(x)) = G_ϕ(x) (Xi, Xj ) = ∂ϕ/∂xi(x) ·∂ϕ/∂xj
(x)
if p lies in this chart then
D_x(ϕ) :R^n to tangent space
is a linear isometry
in maps the standard basis in R^n to the tangent space spanned by above
THE FIRST FUNDAMENTAL FORM
for a given chart the first fundamental form
G_p for a given point on surface and basis
Using the First Fundamental Form we can now re-write the formula for the length of a curve on
Σn in the following fashion:
L(γ) = integral_[a,b]
SQRT(Gγ(t)(γ’(t), γ’(t))dt.
so how would we compute the arc length? of a curve
if you are inhabitant of the surface and you dont know the formula, directions and lengths of tose directions not tangent to surface
gij (ϕ(x)) = G_ϕ(x) (Xi, Xj ) = ∂ϕ/∂xi(x) ·∂ϕ/∂xj
(x) we know
could write coordinates on surface
ϕ-1 . γ(t) = (γ¹(t),…, γⁿ(t))
speed of tangent vector: in terms of first fundamental form speed would be
|γ’(t)|_R^n = sqrt(G_γ(t)( γ’(t), γ’(t))
so using this into integral of:
sqrt[sum_ij (gij(γ¹(t),…,γⁿ(t))) dγ^i/dt (t)dγ^j/dt(t) ].dt
FORMULA FOR THE LENGTH FOR __SURFACE
has no idea what R^n is
Let Σn ⊂ R^{n+1} be a graph of a smooth function f : Ω → R, that is the set
Σ^n = {(x, f(x)) : x ∈ Ω}.
using charts
we have only one chart (Ω, ϕ, Ω×R), where the map
ϕ : Ω → Ω×R is given by
ϕ(x) = (x, f(x)).
(cylinder Ω×R
Using relations
X_i(p) = D_x(ϕ(e_i)) = ∂ϕ/∂xi(x),
=(0,…,1,0,..,0,∂f/∂x_i), for ith place
we obtain the following formula for the matrix of the first fundamental form:
g_ij (ϕ(x)) =
∂ϕ/∂xi(x) ·∂ϕ/∂xj(x)
=X_i . X_i
= δ_ij +∂f/∂xi(x) ∂f/∂xj(x),
where i, j = 1, . . . , n, and
δij is the Kronecker delta. δij = 1 iff i = j,
in this e.g if f(x) =c constant the surface is just the plane itself, so becomes identity matrix as the euclidean matrix
e.g f(x,y)=x^3+y^3 decide on curve (t,0)
inhabitants see, but if we live in R^3 for us the curv will be gamma(t)=(t,0, t^3 +0)
g_ij (ϕ(x)) =
= δ_ij +∂f/∂xi(x) ∂f/∂xj(x),
…9x^2 etc
matrix representation
The family of matrices (gij ) ◦ ϕ defined by relation (V.7) is called the matrix
representation of the first fundamental form in a given coordinate chart (U, ϕ, V ) on a regular surface Σ^n ⊂ R^m.
(not discussed)
e.g
Unit sphere S^n ⊂ R^{n+1). Consider a unit sphere S^n ⊂ R^{n+1}
Hyperplace subset viewed as graph of 0 function
why is first fundamental form useful?
INNER GEOMETRIC
geometric properties/quantaties of a regular surface that are determined by the first fundamental form only are called intrinsic or INNER GEOMETRIC
INTRINSIC DISTANCE FUNCTION
an example of an intrinsic quality
example
instrinsic distance function on unit sphere
SPHERICAL DISTANCE
PROP 5.1 inequality for intrinsic distance
it has constant speed
then using CS inequality and corollary 2.22
example: on a plane the instrisic distance coincides with euclidean distance
the inequality might be strict only for a curved shape
LOCAL ISOMETRY BETWEEN SURFACES
when is a local isometry a GLOBAL ISOMETRY
A local isometry is called a global isometry if it is a bijective map between surfaces
REMARKS ABOUT ISOMETRIES
relationships
first fundamental forms
example
for an ORTHOGONAL TRANSFORMATION
map restricted is a GLOBAL ISOMETRY OF THE UNIT SPHERE
example: PLANE AND CYLINDER REGULAR 2 DIM SURFACES
SHOW THERE IS A LOCAL ISOMETRY BETWEEN THE TWO
PROPN 5.2 local isometry instrinsic distance functions inequality
corollary 5.3
GLOBAL ISOMETRY and intrinsic distance
COROLLARY 5.4
global isometry
Euclidean isometry
did he go over these?
Corollary V.4. Let Φ : R^n → R^n be a smooth map.
Then Φ is a global isometry of R
n in the sense of Definition V.13
if and only if
it is a Euclidean isometry in the sense of Definition I.20.
hypersurface
regular surfaces whose dimension is one less than the dimension of the ambient
space
GAUSS MAP
Is applied to a hyperspace
units nomal field
diagram: VECTORS coming out at different points on surface p N_p and q N_q
these form our unit normal field
EXAMPLE of GAUSS MAP
of the unit sphere
further on example of S^n in R^{n+1)
sphere
given this surface this example will have tangent space
thus giving the normal fields on p
with different orientations
N(p) =p is a unit normal field
the Gauss map coincides with identity
N(p)-p is another unit normal field corresponding gauss map is called antipodal map
SHAPE OPERATOR
propn shape operator is self adjoint
eigenvalues always real
self-adjoint linear operator
SHAPE OPERATOR PROPERTIES
PRINCIPLE CURVATURES
MEAN CURVATURES
GAUSS CURVATURES
UNIT SPHERE
shape operator
principle curvatures
mean
gauss curvature
differwntial N_1 =-id
as k_n(p)=-1
D_p N_2=id
HYPERPLANE
shape operator
principle curvatures
mean
gauss curvature
Can you see that principal curvatures in Definition V.19 depend on
an orientation on Σn? More precisely, can you see that principal curvatures change sign when a unit normal field N is replaced by −N?
reversed orientation changes sign of principle curvatures
SECOND FUNDAMENTAL FORM
COROLLARY 5.6
second fundamental form for tangent vectors XY
for a regular hypersurface oriented by unit normal N
COROLLARY 5.7 proving the PROP 5.5
PROPN 5.8 relationship between
(b_ij)
gauss curvature and g_ij
useful?
SUMMARY properties of sec fund form
applying our useful relations:
EXAMPLE cylinder
CYLINDER EXAMPLE
notes
cylinder example working out partial derivs
spent all lecture…
gauss curvature:
plane and cylinder locally isometric
their curvatures related?
Theorem V.9 (Theorema Egregium)
summary
EXAMPLE:
using
Theorem V.9 (Theorema Egregium) on eucludean plane and sphere
thm
gauss curvature vanishes
then
rotation index visually:
with signed curvature
l5: PRE SECTION I MISSED
L5 prop 5.1:
LIE BRACKET and properties
IT DOESNT DEPEND ON GEOMETRY OF SURFACE AND IF IS DIFFEOMORPHISM…
LIE BRACKET properties
lie bracket example from lecture
C(surface^n)
CURVATURE TENSOR) level 5
measures how far the separation
CURVATURE TENSOR PROPERTIES
prop 5.2
e.g multiplied by function factors out….
R is an intrinsic geometric quantity
corollary 5.3 another curvature tensor
missed
example: curvature tensor for n dimensional euclidean space covered by a chart
missed
PROPN 5.4
local isometry and relationships missed
PROP 5.5
THE CURVATURE TENSOR OF A REGULAR N DIM SURFACE SATISFIES (l5)
missed
PROPN 5.6
GAUSS EQUATIONS
will this be on exam?
gone through with proof
gauss equation from lecture
USES PROPN 5.6!!
Tangent vectors will give 0
PROP 5.7 shape operator satisfies equations
GAUSS EQUATION
propn 5.8
(Theorema Egregium)
level 5 e.g summary 1
for notes
computing these is hard: THIS THM IS INTERESTING FOR DIMENSION 2
COROLLARY 5.9
LOCAL ISOMETRY AND GAUSS CURVATURE
gone through in lecture and proven
EXAMPLE
