University of Cambridge
Mathematics Tripos
Part II Differential Geometry
Lent, 2018
Lectures by M. Dafermos
Notes by Qiangru Kuang Contents
Contents
1 1 Smooth Manifolds
1 Smooth Manifolds
1.1 Definitions
Definition (Smooth). Let 푈 ⊆ R푛 be an open set and 푓 ∶ 푈 → R푚. 푓 is said to be smooth if 휕|훼|푓 휕푥훼 exists for all multi-indices 훼.
Notation. A multi-index 훼 = (훼1, … , 훼푛) is a tuple of non-negative integers |훼| = ∑푛 훼 and 푖=1 푖.
Definition (Smooth). Let 푋 ⊆ R푛 be a subset. 푓 ∶ 푋 → R푚 is smooth if for all 푥 ∈ 푋, there exists 푈 ⊆ R푛, 푥 ∈ 푈 such that there exists smooth ̃ 푚 푓 ∶ 푈 → R extending 푓|푋∩푈. i.e. the following diagram commutes
푓 ̃ 푈 R푚 휄 푓 푈 ∩ 푋
Remark. 1. This is a local property.
2. 푋 is a topological space with subspace topology induced from R푛. 푓 is smooth implies that 푓 is continuous.
Definition (Diffeomorphism). Let 푋 ⊆ R푛 and 푌 ⊆ R푚. A map 푓 ∶ 푋 → 푌 is a diffeomorphism if 푓 is smooth and bijective and its inverse 푓−1 ∶ 푌 → 푋 is also smooth. Remark. 1. Diffeomorphism implies homeomorphism (with repsect to the subspace topology). 2. Diffeomorphism is an equivalence relation.
Definition (Manifold). A 푘-dimensional manifold is a set 푋 ⊆ R푁 such that for all 푥 ∈ 푋 there exists 푉 ⊆ 푋 open, 푥 ∈ 푉 such that 푉 is diffeomorphic to an open subset 푈 ⊆ R푘, i.e. 푋 is locally diffeomorphic to R푘.
Remark. The diffeomorphism 휑 ∶ 푈 → 푉 is called a local parameterisation and its inverse 휑−1 ∶ 푉 → 푈 is called coordinate charts. More specifically, let 푘 푥푖 ∶ R → R be the projection of the 푖th coordinate and we have local coordinate −1 푥푖 ∘ 휑 ∶ 푉 → R. However, sometimes we abuse the notation and just write 푥푖 ∶ 푉 → R for the above map.
2 1 Smooth Manifolds
Notation. Write dim 푋 = 푘 if 푋 is a 푘-dimensional manifold. Example.
id 1. R푁 is an 푁-dim manifold. For example, 휑 ∶ R푁 → R푁.
2. An open subset 푉 ⊆ R푁 is an 푁-dim manifold, for example by taking 휑 to be the restriction id |푉. 3. In general, if 푋 is a manifold and 푉 ⊆ 푋 is an open subset then 푉 is also a manifold.
푛 푛+1 2 2 4. The 푛-sphere, 푆 ⊆ R = {푥1 + ⋯ + 푥푛+1 = 1} is an 푛-dim manifold. 푛 Suppose 푥 ∈ 푆 lies in the upper half hyperplane define by 푥푛+1 > 0. Take
푛 푛 2 푛 푉 = 푆 ∩ {푥푛+1 > 0}, 푈 = {(푥1, … , 푥푛) ∶ ∑ 푥푖 < 1} = 퐵(ퟎ, 1) ⊆ R 푖=1 and define
휑 ∶ 푈 → 푉
2 2 (푥1, … , 푥푛) ↦ (푥1, … , 푥푛, +√1 − (푥1 + ⋯ + 푥푛))
Let’s do a reality check: the map consists of algebraic operations so is smooth. It is easy to check that the image lies in 푉. Its inverese is the projection onto the first 푛 coordinates, which is the restriction of a smooth function so also smooth.
Similarly, if 푥 lies in {푥푛+1 < 0}, change + to − in the last coordinate of 휑 will do. This accounts for “most” points, with the exception of those on the equator. However, there is nothing special about 푥푛+1 so we may repeat for the other coordinates. As
푛+1 푛 푛 푛 푆 = ⋃ ((푆 ∩ {푥푖 > 0}) ∪ (푆 ∩ {푥푖 < 0})) 푖=1
(which in English says that at least one coordinate is non-zero), we have covered 푆푛 so done.
Exercise. Suppose 푋 ⊆ R푛 and 푌 ⊆ R푚 are manifolds. Show that 푋 × 푌 ⊆ R푛 × R푚 = R푛+푚 is a manifold of dimension dim 푋 + dim 푌.
Definition (Submanifold). Let 푋 ⊆ R푁 be a manifold. A manifold 푌 ⊆ R푁 such that 푌 ⊆ 푋 is a submanifold of 푋.
Remark. By definition, all manifolds are submanifolds of R푁 for some 푁. Note that we have yet shown that the dimension of a manifold is well-defined but we will do so in a while. Assuming so, we define
3 1 Smooth Manifolds
Definition (Codimension). The codimension of 푌 in 푋 is
codim푋(푌 ) = dim 푋 − dim 푌 .
1.2 Tangent space
Definition (Differential). Let 푈 ⊆ R푛 and 푓 ∶ 푈 → R푚 be a smooth map. The differential of 푓 at 푥 ∈ 푈, 푑푓푥 is a linear map
푛 푚 푑푓푥 ∶ R → R 푓(푥 + 푡ℎ) − 푓(푥) ℎ ↦ lim 푡→0 푡
Explicitly, 푑푓푥 is represented by the matrix 푓 휕푓 1 ( ) where 푓 = ⎜⎛ ⋮ ⎟⎞ . 휕푥푗 ⎝푓푚⎠ It then follows that ℎ 휕푓 ⎛ 1 ⎞ 푑푓푥(ℎ) = ( ) ⎜ ⋮ ⎟ . 휕푥푗 ⎝ℎ푛⎠
Proposition 1.1 (Chain rule). Let 푓 ∶ 푈 → 푉 , 푔 ∶ 푉 → R푝 be smooth maps where 푈 ⊆ R푛, 푉 ⊆ R푚. Let 푥 ∈ 푈 and 푓(푥) ∈ 푉. Then
푑(푔 ∘ 푓)푥 = 푑푔푓(푥) ∘ 푑푓푥.
The aim of this section is to define differentials of maps between manifolds. Before that we have to find where differential lives. Certainly in R푛 it is a linear map living in a linear space. Let 푋 ⊆ R푁 be a 푘-dim manifold and 푥 ∈ 푋. Let 휑 be a local parameteri- sation around 푥. Wlog assume 휑−1(푥) = 0. Note that 휑 is defined on an open subset of R푘 so we can do calculus on it (this is false for 휑−1 as 푉 ⊆ R푁 may not be open). We can define
푘 푁 푑휑0 ∶ R → R .
Definition (Tangent space). The tangent space of 푋 at 푥, denoted 푇푥푋, is 푘 푁 푑휑0(R ), a subspace of R .
For this to be a good definition, we need to show that it is independent of 휑 푘 and dim 푑휑0(R ) = 푘. Suppose 휑 ∶ 푈 → 푉 and ̃휑∶ 푈̃ → 푉̃ are parameterisations around 푥, again assumming wlog 휑(0) = 푥 = ̃휑(0). By taking intersection we can assume 푉 = 푉.̃ Write 휑 = ̃휑∘ (⏟ ̃휑−1 ∘ 휑) . 푈→푈̃
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The term in the parenthesis is a map between open subsets of R푘 so we can do calculus on it. By chain rule
−1 푑휑0 = 푑 ̃휑0 ∘ 푑( ̃휑 ∘ 휑)0.
Note that 휑−1 ∘ ̃휑 is the inverse of ̃휑−1 ∘ 휑 so by applying chain rule to id, we get
−1 −1 id = 푑(휑 ∘ ̃휑)0 ∘ 푑( ̃휑 ∘ 휑)0
so both of them are invertible. Thus
푘 푘 푑휑0(R ) = 푑 ̃휑0(R ).
푘 Next we want to show dim 푑휑0(R ) = 푘. Certainly linear algebra tells us that it is at most 푘. By definition there exists a smooth 휓 ∶ 푊 → R푘, where 푊 ⊆ R푁 is an open subset containing 푥 such that
−1 휓|푋∩푊 = 휑 .
Assume wlog 푉 ⊆ 푋 ∩ 푊. Then
id푈 = 휓 ∘ 휑
so by chain rule id = 푑휓푥 ∘ 푑휑0. 푘 Thus 푑휑0(R ) is 푘-dimensional. Remark. The functions 휑−1 ∘ ̃휑 are called transition functions and will appear later in the course.
Corollary 1.2. The dimension of a manifold is well-defined.
Exercise. If 푋 ⊆ R푁 is a manifold and 푌 is a submanifold of 푋 then
푇푦푌 ≤ 푇푦푋
for all 푦 ∈ 푌. In particular codim푋(푌 ) ≥ 0. Example.
푁 푁 1. 푇푥R = R . 푁 푁 2. Let 푋 be an open subset of R . Then 푇푥푋 = R . 3. Let 푁 푋 = {(푥1, 푥2, … , 푥푘, 0, … , 0)} ⊆ R which can be seen as the image of the embedding 휑 ∶ R푘 → R푁. This is a 푘-dim manifold. This is a linear map so 푑휑푥 = 휑. Thus 푇푥푋 = 푋. In general, this holds if 휑 is an injective linear map.
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푛 푛+1 푛 4. Let 푆 ⊆ R . Given a point 푥 ∈ 푆 with 푥푛+1 > 0, use the parameteri- sation
휑 ∶ 퐵(ퟎ, 1) → R푛+1
2 2 (푥1, … , 푥푛) ↦ (푥1, … , 푥푛, +√1 − 푥1 − ⋯ − 푥푛)
Then 휕휑 휕휑 푑휑 = { ,…, }. im (푥1,…,푥푛) span 휕푥1 휕푥푛 We have 휕휑 푥 = (0, … , 1, … , 0, − 푖 ) 휕푥푖 푥푛+1 and we can verify that for all 푖, 휕휑 ⋅ (푥1, … , 푥푛, 푥푛+1) = 0 휕푥푖
푛 푛+1 so 푇푥푆 is {푣 ∶ 푣 ⋅ 푥 = 0} ⊆ R . Now we go on to define differential of maps between manifolds. Let 푓 ∶ 푋 → 푌 be a smooth map of manifolds where 푋 ⊆ R푛, 푌 ⊆ R푚. Let 휑 ∶ 푈 → 푉 and ̃휑∶ 푈̃ → 푉̃ be local parameterisations around 푥 and 푓(푥), where 푈 ⊆ R푘, 푈̃ ⊆ Rℓ. wlog we assume 휑(0) = 푥, ̃휑(0)= 푓(푥). We also assume 푓(푉 ) ⊆ 푉.̃ We can thus define a map ̃휑−1 ∘ 푓 ∘ 휑 ∶ 푈 → 푈̃ and do calculus on it. As 푇푥푋 = im 푑휑0 and 푑휑0 is an injective map, we can define the inverse of its restriction to the image, denoted by
−1 푘 (푑휑0) ∶ 푇푥푋 → R .
Definition (Differential). The differential map of 푓 at 푥 ∈ 푋, denoted 푑푓푥, is a linear map 푑푓푥 ∶ 푇푥푋 → 푇푓(푥)푌, defined as the composition
−1 −1 푑 ̃휑0 ∘ 푑( ̃휑 ∘ 푓 ∘ 휑)0 ∘ (푑휑0) .
푑푓0 푇푥푋 푇푓(푥)푌
푑휑0 푑 ̃휑0
푘 ℓ R −1 R 푑( ̃휑 ∘푓∘휑)0
Now as everything else we defined in differential geometry, we have to check that it is independent of parameterisation. It will be a horrible job, and one that you would want to do only (at most) once in your lifetime. Let 휓, 휓̃ be another pair of parameterisations. Define 휑 in terms of transition maps 휑 = 휓 ∘ (휓−1 ∘ 휑). By chain rule, −1 푑휑0 = 푑휓0 ∘ 푑(휓 ∘ 휑)0
6 1 Smooth Manifolds
so −1 −1 −1 −1 −1 −1 (푑휑0) = (푑(휓 ∘ 휑)0) ∘ (푑휓0) = 푑(휑 ∘ 휓)0 ∘ (푑휓0) where the last equality comes from applying chain rule to (휑−1 ∘ 휓) ∘ (휓−1 ∘ 휑) = id . Thus
−1 −1 푑푓푥 = 푑 ̃휑0 ∘ 푑( ̃휑 ∘ 푓 ∘ 휑)0 ∘ (푑휑0) −1 −1 −1 = 푑 ̃휑0 ∘ 푑( ̃휑 ∘ 푓 ∘ 휑)0 ∘ 푑(휑 ∘ 휓)0 ∘ (푑휓0) −1 −1 = 푑 ̃휑0 ∘ 푑( ̃휑 ∘ 푓 ∘ 휓)0 ∘ (푑휑0) = ⋯ ̃ −1̃ −1 = 푑휓0 ∘ 푑(휓 ∘ 푓 ∘ 휓)0 ∘ (푑휓0) where the omitted lines are similar and are left as an exercise.
Proposition 1.3 (Chain rule). Let 푋, 푌 , 푍 be manifolds, 푓 ∶ 푋 → 푌 , 푔 ∶ 푌 → 푍 smooth. Then for all 푥 ∈ 푋,
푑(푔 ∘ 푓)푥 = 푑푔푓(푥) ∘ 푑푓푥.
Proof. Tedious exercise using transition maps and chain rule on maps between Euclidean spaces.
Theorem 1.4 (Inverse Function Theorem). Let 푓 ∶ 푋 → 푌 be a smooth map between manifolds. If 푑푓푥 ∶ 푇푥푋 → 푇푓(푥)푌 is an isomorphism (which implies that in particular dim 푋 = dim 푌), then 푓 is a local diffeomorphism. In other words, there exists an open subset 푉 ⊆ 푋 such that 푓|푉 ∶ 푉 → 푌 is a diffeomorphism onto its image.
Proof. This is easy using Inverse Function Theorem from analysis. Let 휑, ̃휑 be −1 local parameterisations around 푥 and 푓(푥) respectively. Then 푑( ̃휑 ∘ 푓 ∘ 휑)0 is surjection R푘 → R푘 so ̃휑−1 ∘ 푓 ∘ 휑 is a local diffeomorphism of Euclidean spaces. Thus 푓 is a local diffeomorphism of manifolds. Exercise. Suppose 푌 is a submanifold of 푋 and 푓 ∶ 푋 → 푍 is a smooth map between manifolds. Then for all 푦 ∈ 푌, show 푑(푓| ) = 푑푓 | . 푌 푦 푦 푇푦푌 Exercise. Let 푓∶푋→푌,푔∶푋→푍 be smooth maps between manifolds. We can define map (푓, 푔) ∶ 푋 → 푌 × 푍 푥 ↦ (푓(푥), 푔(푥))
Show that this map is smooth and for all 푥 ∈ 푋,
푑(푓, 푔)|푥 = (푑푓푥, 푑푔푥). Thus if differential is thought as a matrix, the map to the product spacehas differential made of blocks of matrices.
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1.3 Pre-image theorem We defined manifolds in terms of charts but in reality, the pre-image theorem actually gives the most convenient way to define a manifold. First define some terminologies:
Definition (Critial point, critial value, regular value). Let 푓 ∶ 푋 → 푌 be a smooth map between manifolds.
• A critical point of 푓 is a point 푥 such that 푑푓푥 ∶ 푇푥푋 → 푇푓(푥)푌 fails to be surjective. We denote the set of all critical points 퐶. • A critical value of 푓 is a point 푦 ∈ 푌 such that there exists 푥 ∈ 푋 critical and 푓(푥) = 푦, i.e. the set of critical values is 푓(퐶) ⊆ 푌. • 푦 ∈ 푌 is a regular value if 푦 is not a critical value, i.e. 푦 ∈ 푌 \ 푓(퐶).
Remark. 1. If dim 푋 < dim 푌 then 퐶 = 푋. 2. If 푓−1(푦) = ∅, i.e. 푦 ∉ 푓(푋) then 푦 is a regular value.
Theorem 1.5 (Pre-image theorem). Let 푓 ∶ 푋 → 푌 be a smooth map between manifolds. Given 푦 a regular value of 푓, 푓−1(푦) ⊆ 푋 is a submanifold of 푋 with codimension equal to the dimension of 푌, i.e.
dim 푋 − dim(푓−1(푦)) = dim 푌 .
Remark. Suppose 푋 is non-empty (which we may add to our definition of manifolds). Then this theorem gives a proof that codim ≥ 0 for this particular case. Proof. Let 푥 ∈ 푓−1(푦). Suppose 푋, 푌 ⊆ R푁 and 푞 = dim 푋 − dim 푌. By surjectivity and rank-nullity, dim ker 푑푓푥 = 푔. It is an exercise in linear algebra 푁 푞 to show that there exists a linear map 푇 ∶ R → R such that ker 푇 ∩ker 푑푓푥 = 0. Now define
퐹 ∶ 푋 → 푌 × R푞 푥 ↦ (푓(푥), 푇 (푥))
Then by the exercises above 퐹 is smooth with differential 푑퐹푥 = (푑푓푥, 푑푇푥) and 푑(푇 | ) = 푑푇 | = 푇 | 푑퐹 = (푑푓 , 푇 ) 푋 푥 푇푥푋 푇푥푋 so 푥 푥 which is an isomorphism. Thus by Inverse Function Theorem 퐹 is a diffeomorphism around 푥. There exists an open neighbourhood of 푥 푉 ⊆ 푋 such that
퐹 |푉 ∶ 푉 → 푓(푉 ) × 푇⏟ (푉 ) 푞 =푈⊆R is a diffeomorphism. Then
−1 휑 = (퐹 |푉) |{푦}×푈 ∶ 푈 → 푉 is a local parameterisation of 푓−1(푦).
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Corollary 1.6. Suppose dim 푋 = dim 푌 and 푦 is a regular value of some 푓 ∶ 푋 → 푌. Then 푓−1(푦) is a manifold of dimension 0.
If 푍 ⊆ R푁 is a manifold of dimension 0, then it is a collection of discrete points, i.e. for each 푧 ∈ 푍, there exists an open neighbourhood of 푧 whose intersection with 푍 is {푧}. Thus if 푋 is in addition compact, then 푓−1(푦) is a finite set of points.
Theorem 1.7 (Stack of records theorem). Under the above assumptions, there exist open neighbourhoods 푊푖 of 푥푖’s such that
푛 푛 −1 푓 (⋂ 푓(푊푖) \ 푓(푋 \ ⋃ 푊푖)) 푖=1 푖=1
is a disjoint union of open neighbourhoods of 푥푖’s and the restriction of 푓 on each of which is a diffeomorphism.
−1 Given a regular value 푦, 푇푥푓 (푦) = ker 푑푓푥. See example sheet 1. Example.
1. Let 푆푛 ⊆ R푛+1 and
푓 ∶ 푆푛 → R 2 2 (푥1, … , 푥푛+1) ↦ 푥1 + ⋯ + 푥푛+1
which is smooth. We have
푑푓 = (2푥1, … , 2푥푛+1)
which is surjective everywhere on 푆푛. Thus by pre-image theorem 푓−1(1) ⊆ R푛+1 is a manifold of dimension 1, which is precisely 푆푛. 2. Let 푂(푛) be the set of 푛 × 푛 orthogonal matrices, i.e. matrices 퐴 such 푇 푛2 that 퐴퐴 = 퐼. 푂(푛) can be seen as a subset of ℳ푛(R) = R . Claim that 푛(푛−1) 푂(푛) is a manifold of dimension 2 .
Let 푆(푛) ⊆ ℳ푛(R) be the subset of symmetric matrices. This is a subspace 푛2 푛(푛+1) of R so clearly a submanifold of dimension 2 . Consider that map
푓 ∶ ℳ푛 → 푆(푛) 퐴 ↦ 퐴퐴푇
which is a smooth map. As 푓−1(퐼) = 푂(푛), if 퐼 is a regular value of 푓 then 푛(푛−1) 푂(푛) is a manifold of dimension 2 . (퐴 + 푠퐻)(퐴 + 푠퐻)푇 − 퐴퐴푇 푑푓퐴(퐻) = lim 푠→0 푠 = 퐴퐻푇 + 퐻퐴푇
퐶퐴 Given 퐶 ∈ 푆(푛), 퐻 = 2 satisfies 푑푓퐴(퐻) = 퐶 so 푑푓퐴 is surjective.
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Remark. In addition to being a manifold, 푂(푛) is a group under matrix multi- plication. The group operations are smooth. We call it a Lie group. In this case, objects that preserve geometric objects (i.e. transformations) are themselves geometric objects. It is also worthing pointing out that 푂(푛) is not connected, as seen from the continuous map det ∶ 푂(푛) → {±1}. The connected component of identity is called special orthogonal group
푆푂(푛) = {퐴 ∈ 푂(푛) ∶ det 퐴 = 1}.
Definition. A set 푆 ⊆ R푛 is said to be measure 0 if for all 휀 > 0 there exists a countable collection of cubes 퐶푚 such that 푆 ⊆ ⋃ 퐶푚 and ∑ volume 퐶푚 < 휀.
Definition. Let 푋 be an 푛-dim manfiold and 푆 ⊆ 푋. 푆 has measure 0 if for all local parameterisations 휑 ∶ 푈 → 푉 ⊆ 푋, 휑−1(푉 ∩ 푆) ⊆ R푛 is of measure 0.
This is well-defined since we can show any manifold can be parameterised by a chart of countably many parameterisations and the countable union of measure 0 subsets in R푛 has measure 0. Remark. 1. An open subset 푉 ⊆ 푋 is not of measure 0. 2. Suppose 푌 ⊆ 푋 is a submanifold of 푋 and dim 푌 < dim 푌. Then 푌 is of measure 0 in 푋.
Theorem 1.8 (Sard’s Theorem). Let 푓 ∶ 푋 → 푌 be a smooth map between manifolds. Then the set of critical values of 푓 is of measure 0 in 푌.
Corollary 1.9. Regular values are dense.
1.4 Transversality Suppose 푓 ∶ 푋 → 푌 is a smooth map between manifolds and 푍 ⊆ 푌 is a submanifold.
Definition (Transversality). 푓 is transversal to 푍, denoted 푓 푍, if for all 푥 ∈ 푓−1(푍), im 푑푓푥 + 푇푓(푥)푍 = 푇푓(푥)푌 .
Remark. 1. If 푓(푋) ∩ 푍 = ∅ then 푓 is vacuously transversal to 푍. 2. This is a generalisation of regular value: if 푍 = {푦} then 푓 푍 if and only if 푦 is a regular value of 푓 since the tangent space of a dimension 0 manifold is 0.
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Example. Let 푋 = R, 푌 = R2, 푓(푡) = (0, 푡) and 푍 ⊆ 푌 is the 푥-axis. Then 0 푤 2 푑푓0 = (1) and 푇(푤,0)푍 = span{(0)} so Im 푑푓0 + 푇(푤,0)푍 = R so 푓 푍.
2 1 Example. Let 푓(푡) = (푡, 푡 ). Then 푑푓0 = (0) so
2 Im 푑푓0 + 푇(푤,0)푍 ≠ R
so 푓 is not transversal to 푍. The corresponding generalisation of pre-image theorem is
Theorem 1.10 (Transversality theorem). Suppose 푓 ∶ 푋 → 푌 is a smooth map between manifolds and 푍 ⊆ 푌 is a submanifold. If 푓 푍 then 푓−1(푍) ⊆ 푋 is a submanifold of 푋 with codimension equal to codim푌(푍).
Proof. It is enough to show that for all 푥 ∈ 푍, there exists 푉 ⊆ 푍 open such that 푓−1(푉 ) is a submanifold of 푋, i.e. we can work locally to deduce global result. This is because if {푉푖}푖∈퐼 is a cover of 푍 then
−1 −1 푓 (푍) = ⋃ 푓 (푉푖). 푖∈퐼
Given 푥 ∈ 푍, let codim푌 푍 = 푘. Claim that there exist 푉 ⊆ 푍 open containing 푥: there exists 푉̃ ⊆ 푌 open such that
⎧ℎ = 0 { 1 { ̃ ℎ2 = 0 푉 = 푍 ∩ 푉 = ⎨ {⋮ { ⎩ℎ푘 = 0
̃ ̃ −1 where 푘푖 ∶ 푉 → R and moreover 푍 ∩ 푉 = ℎ (0) and 0 is a regular value of ℎ. This is basically saying that a submanifold of codimension 푘 can be expressed as the pre-image of 0 where 0 is a regular value. This is left as an exercise. Now we prove the theorem. wlog assume 푍 = 푍 ∩푉.̃ Consider ℎ∘푓 ∶ 푋 → R푘. Its differential is 푑(ℎ ∘ 푓)푥 = 푑ℎ푓(푥) ∘ 푑푓푥. Claim that 0 is a regular value of ℎ ∘ 푓: 푥 ∈ (ℎ ∘ 푓)−1(0) if and only if 푥 ∈ 푓−1(푍). Recall that 푇푦푍 = ker 푑ℎ푦 (see example sheet). From transversality we know
Im 푑푓푥 + 푇푓(푥)푍 = 푇푓(푥)푌
so Im 푑푓푥 + ker 푑ℎ푓(푥) = 푇푓(푥)푌 .
Apply 푑ℎ푓(푥) to both sides, we get
푘 Im 푑ℎ푓(푥) ∘ 푑푓푥 = Im 푑ℎ푓(푥) = R where the last equality is by surjectivity. Therefore (ℎ ∘ 푓)−1(0) ⊆ 푋 is a submanifold of codimension 푘. So does 푓−1(푍).
11 1 Smooth Manifolds
Consider the special case where 푋 and 푍 are both submanifolds of 푌, i.e. 푓 = 휄 ∶ 푋 → 푌. Then 휄 푍 if and only if
Im 푑휄푥 + 푇휄(푥)푍 = 푇휄(푥)푌 which can be rewritten as
푇푥푋 + 푇푥푍 = 푇푥푌 .
In this case write 푋 푍 (which is clearly a symmetric relation). Example. If 푓 푍, it is not necessarily the case that 푓(푋) 푍. Consider 푋 = R, 푌 = R2, 푓(푡) = (0, 푡2) and 푍 ⊆ 푌 is the 푥-axis. Then 푓−1(푍) ∩ 푋 is a submanifold of 푋 of dimension 0.
Corollary 1.11. If 휄−1(푍) ⊆ 푋, then 푋 ∩ 푍 is a submanifold of 푌 with codimension codim푌(푋 ∩ 푍) = codim푌 푋 + codim푌 푍.
1.5 Manifolds with boundary 푛 푛 Notation. H = {(푥1, … , 푥푛) ∶ 푥푖 ≥ 0} ⊆ R .
Definition (Manifold with boundary). 푋 ⊆ R푛 is an 푛-dimensional manifold with boundary if for all 푥 ∈ 푋 there exists 푉 ⊆ 푋 open, 푈 ⊆ H푛 open and 휑 ∶ 푈 → 푉 a diffeomorphism.
It is possible that 푈 is also open in R푛.
Definition (Boundary, interior). The boundary of 푋, denoted 휕푋, is defined to be
휕푋 = {푥 ∶ ∃휑 as above such that 푥 = 휑(푥1, … , 푥푛−1, 0)}.
푛 푛 Equivalently, 푥 ∈ 휑(푈 ∩ 휕H ) where 휕H = {(푥1, … , 푥푛−1, 0)}. The interior of 푋, denoted Int 푋, is defined to be 푋 \ 휕푋.
Remark. 1. The above definition is consistent, i.e. independent of parameterisation: if 푥 ∈ 휑(푈 ∩ 휕퐻푛) for some 휑 then it holds for all 휑 such that 푥 ∈ Im 휑.
2. H푛 is an 푛-dimensional manifold with boundary. 3. Boundary of manifolds is not the same as topological boundary, defined to be the set difference between the closure and the interior. 4. For any manifold with boundary 푋, 휕푋 and Int 푋 are manifolds without boundary of dimension dim 푋 − 1 and dim 푋 respectively. This can be seen by restricting image of 휑 to 휕H and Int H respectively. Example.
12 1 Smooth Manifolds
푛 1. The closed unit ball 퐵푛 ⊆ R is a manifold with bounday, whose boundary 푛−1 is 휕퐵푛 = 푆 . 2. Any manifold is a manifold with boundary (without boundary).
Exercise (Classification of 1-dim connected manifolds with boundary). Show that any 1-dim connected manifolds is up to diffeomorphism one of
푋 휕푋 Compactness 푆1 ∅ Yes (0, 1) ∅ No [0, 1) {0} No [0, 1] {0, 1} Yes
Table 1: 1-dim manifolds
Lemma 1.12. Let 푋 be an 푛-dim manifold (without boundary) and 푓 ∶ 푋 → R smooth. Suppose 0 is a regular value of 푓. Then {푥 ∈ 푋 ∶ 푓(푥) ≥ 0} ⊆ 푋 is a submanifold with boundary whose boundary is {푥 ∈ 푋 ∶ 푓(푥) = 0}.
Proof. Immediate from pre-image theorem. Suppose 푥 ∈ 푋 such that 푓(푥) > 0 then 푥 ∈ 푓−1((0, ∞) which is open in 푋. If 푓(푥) = 0, just proceed as in pre-image −1 theorem and consider (퐹 |푈) ∶ 퐼 × 푇 (푈) → 푈 restricted to 퐼 ∩ {푡 ≥ 0}. Example. Let
푓 ∶ R푛 → R 푛 2 (푥1, … , 푥푛) ↦ 1 − ∑ 푥푖 푖=1
Then 퐵푛 = 푓−1((0, ∞)) is a manifold with boundary. Note that with the definition of tangent space as before, forall 푥 ∈ 푋 (even those on the bounday), we have
dim 푑푇푥푋 = dim 푋.
Also for 푥 ∈ 휕푋, 푇푥휕푋 ≤ 푇푥푋. Next up is an extension of pre-image theorem
Theorem 1.13. Let 푋 be a manifold with boundary and 푓 ∶ 푋 → 푌 be a smooth map betwween manifolds. Suppose 푦 ∈ 푌 is a regular value of 푓 −1 and 푓|휕푋, then 푓 (푦) is a submanifold with boundary of 푋, with boundary 휕푓−1(푦) = 푓−1(푦) ∩ 휕푋.
In this case dim 푋 > dim 푌.
13 1 Smooth Manifolds
Proof. Note that if 푥 ∈ 푓−1 ∩ Int 푋 we can apply pre-image theorem. Thus assume 푥 ∈ 푓−1(푦) ∩ 휕푋. As an open set in H푛 with non-empty intersection with its boundary is diffeomorphic to H푛, assume wlog 푋 = H푛 and similarly 푌 = R푚, where dim 푋 = 푛, dim 푌 = 푚. Given such an 푥, there exists 푈 open in R푛 containing 푥 such that there 푚 퐹 ∶ 푈 → 퐹 | 푛 = 푓| 푦 푓 exists R smooth such that 푈∩H 푈. is a regular value of so 푛 푑푓푥 is surjective. For 푈 small enough, 푑푓푧 surjective for all 푧 ∈ 푈 ∩ H . Then 푑퐹푧 is surjective for all 푧 ∈ 푈. Then 푦 is a regular value of 푈. We want 푓−1(푦) ∩ 푈 to be a manifold with boundary. Claim that the set equal to 퐹 −1 ∩ H푛. 퐹 −1(푦) is a manifold by pre-image theorem. Introduce projection map
휋 ∶ R푛 → R
(푥1, … , 푥푛) ↦ 푥푛
−1 푛 −1 −1 Then 퐹 (푦) ∩ H = 퐹 (푦){푥 ∶ 휋(푥) ≥ 0} = {푥 ∶ (휋|푓−1(푦)) (푥) ≥ 0}. All we −1 need to do is to show 0 is a regular value of the map (휋|푓−1(푦)) . But from example sheet
푑휋| −1 = 휋| −1 = 휋| = 휋| . 푓 (푦) 푇푥퐹 (푥) ker 푑퐹푥 ker 푑푓푥
푛−1 Thus just need to show ker 푑푓푥 ≠ R = 푇푥휕푋. Done by assumption on 푓.
If 푓 푍 and 푓|휕푋 푍 Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Etiam lobortis facilisis sem. Nullam nec mi et neque pharetra sollicitudin. Praesent imperdiet mi nec ante. Donec ullamcorper, felis non sodales commodo, lectus velit ultrices augue, a dignissim nibh lectus placerat pede. Vivamus nunc nunc, molestie ut, ultricies vel, semper in, velit. Ut porttitor. Praesent in sapien. Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Duis fringilla tristique neque. Sed interdum libero ut metus. Pellentesque placerat. Nam rutrum augue a leo. Morbi sed elit sit amet ante lobortis sollicitudin. Praesent blandit blandit mauris. Praesent lectus tellus, aliquet aliquam, luctus a, egestas a, turpis. Mauris lacinia lorem sit amet ipsum. Nunc quis urna dictum turpis accumsan semper.
1.6 Degree mod 2 Let 푋 be a manifold and 푌 be a manifold with boundary.
Definition (Smooth homotopy). Two smooth maps 푓, 푔 ∶ 푋 → 푌 are said to be smoothly homotopic if there exists a smooth map 퐹 ∶ [0, 1] × 푋 → 푌 such that 퐹 |{0}×푋 = 푓 and 퐹 |{1}×푋 = 푔.
Notation. For 푡 ∈ [0, 1], 푓푡 = 퐹 |{푡}×푋. It is a restriction of a smooth map so is smooth.
Remark. 푓 smoothly homeomorphic to 푔 is an equivalence relation. See example sheet. This is slightly less trivial than its topological counterpart since arbitrary reparameterisation is not necessarily smooth.
14 1 Smooth Manifolds
Definition (Smooth isotopy). If 푓 and 푔 above are smoothly homotopic diffeomorphisms and 푓푡 is a diffeomrophism for all 푡 ∈ [0, 1] then we say that 퐹 is a smooth isotopy and 푓 and 푔 are smoothly isotopic.
Remark. This is also an equivalence relation.
Lemma 1.14 (Homogeneity lemma). Let 푋 be a connected manifold, 푦, 푧 ∈ 푋. Then there exists a map 휑 ∶ 푋 → 푋 smoothly isotopic to the identity such that 휑(푦) = 푧.
This says that in differential topology, there is no “particular” point ina connected component of a manifold. Sketch of proof. Say 푦 and 푧 isotopic if there exists such a 휑. This defines an equivalence relation ∼ on 푋. It then suffices, by connectivity, to show that the equivalence classes are open. Given 푦 ∈ 푋, must find an open neighbourhood 푉 of 푦 such that for all 푧 ∈ 푉, 푧 ∼ 푦. Suffices to show given 푧 ∈ 퐵(0, 1) ⊆ R푛, there exists 퐹 ∶ R푛 × [0, 1] → R푛 푓 = 푛 푓 (0) = 푧 smooth isotopy of diffeomorphisms such that 0 idR and 1 with 푓 | 푛 푡 R \퐵(0,2) (by considering a chart covering ???, we can pushforward the map and extend it by identity outside the ball to ensure smoothness). One way to do this is to use cut-off function and ODE theory.
Lemma 1.15. Let 푋 be a compact manifold, 푌 a manifold and dim 푋 = dim 푌. Let 푓, 푔 ∶ 푋 → 푌 be smoothly homotopic. Let 푦 be a regular value of both 푓 and 푔. Then
#푓−1(푦) = #푔−1(푦) (mod 2).
Proof. Let 퐹 ∶ 푋 ×[0, 1] → 푌 be the smooth homotopy between 푓 and 푔. Assume first 푦 is a regular value of 퐹. Note that 휕(푋 × [0, 1]) = {0} × 푋 ∪ {1} × 푋
so denote 퐹 |휕(푋×[0,1]) = 푓 ∪ 푔. Then by generalised pre-image theorem, 퐹 −1(푦) is a manifold with boundary, whose boundary is
휕퐹 −1(푦) = 퐹 −1 ∩ 휕(푋 × [0, 1]) = ( ⋃ {0} × {푥}) ∪ ( ⋃ {1} × {푥}) . 푥∈푓−1(푦) 푥∈푔−1(푦)
By assumption 퐹 −1(푦) is a compact 1-dim manifold, which by classification above is a union of disjoint circles or closed intervales. But both of these have even number of boundary points so
#휕퐹 −1(푦) = #푓−1(푦) + #푔−1(푦) = 0 (mod 2).
Suppose instead 푦 is not a regular value of 퐹. Consider an open neighbourhood 푉 ⊆ 푌 of 푦. By Sard’s Theorem the sets of critical values of 푓, 푔 and 퐹 are all
15 1 Smooth Manifolds
of measure 0, so is the union of them. Then there exist ̃푦∈ 푉 that is a regular value of 푓, 푔, 퐹. But by stack of records theorem there exists 푉 containing 푦 such 푛 that 푓−1(푉 ) = ⋃ 푈 such that 푓| ∶ 푈 → 푉 is a diffeomorphism. 푖=1 푖 푈푖 푖
Theorem 1.16 (Degree mod 2). If 푦 and 푧 are regular values of 푓 then
#푓−1(푦) = 푓−1(푧) (mod ()2).
Proof. Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Etiam lobortis facilisis sem. Nullam nec mi et neque pharetra sollicitudin. Praesent imperdiet mi nec ante. Donec ullamcorper, felis non sodales commodo, lectus velit ultrices augue, a dignissim nibh lectus placerat pede. Vivamus nunc nunc, molestie ut, ultricies vel, semper in, velit. Ut porttitor. Praesent in sapien. Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Duis fringilla tristique neque. Sed interdum libero ut metus. Pellentesque placerat. Nam rutrum augue a leo. Morbi sed elit sit amet ante lobortis sollicitudin. Praesent blandit blandit mauris. Praesent lectus tellus, aliquet aliquam, luctus a, egestas a, turpis. Mauris lacinia lorem sit amet ipsum. Nunc quis urna dictum turpis accumsan semper. Putting everyting together, under the assumptions above, we have 푓−1(푦) does not depend on the choice of regular value 푦. Moreover if 푔 is smoothly homotopic to 푓 then 푔−1(푧) = 푓−1(푦) (mod 2) for any regular value 푧 of 푔.
Definition (Degree mod 2). Suppose 푋 is a compact manifold, 푌 a connected manifold and dim 푋 = dim 푌. Let 푓 ∶ 푋 → 푌 be a smooth map. Then the −1 degree mod 2 of 푓, denoted deg2(푓), is defiend to be #푓 (푦) (mod 2) for some regular value 푦 of 푓.
By Sard’s theorem, there exists a regular value and by theorems above, this is well-defined. Moreover, this only depends on the homotopy classof 푓. Example. 푛 1. Let 푋, 푌 = 푆 , 푓 ∶ 푋 → 푌 be a constant map. Then deg2 푓 = 0 since we can find 푦 ∈ 푌 \ Im 푓 and #푓−1(푦) = 0.
푛 2. Let 푋, 푌 = 푆 , id ∶ 푋 → 푌 be the identity. Then deg2 id = 1. Thus 푓 and id are not smoothly homotopic.
Corollary 1.17. There does not exist a smooth retraction 퐵푛 → 휕퐵푛 = 푛−1 푛 푛 푛 푛 푆 , i.e. a map 푓 ∶ 퐵 → 휕퐵 such that 푓|휕퐵푛 ∶ 휕퐵 → 휕퐵 is the identity.
Proof. Suppose not. Define 퐹 ∶ 푆푛−1 × [0, 1] → 푆푛−1 (푥, 푡) ↦ 푓(푡푥) whose smoothness is left as an exercise. Then 퐹 is an homotopy of 푓 and id. Absurd.
16 1 Smooth Manifolds
Corollary 1.18 (Brouwer’s fixed point theorem). Let 푓 ∶ 퐵푛 → 퐵푛 be 푛 smooth. Then there exists 푥0 ∈ 퐵 such that 푓(푥0) = 푥0.
Proof. Suppose 푓 is smooth without fixed point. Then the ray from 푓(푥) to 푥 intersects 휕퐵푛 at precisely one point, say 푔(푥). Claim that 푔 is smooth (left as an exercise). If 푥 ∈ 휕퐵푛 then 푔(푥) = 푥. Then 푔 is a smooth retraction. Absurd.
Corollary 1.19 (Brouwer’s fixed point theorem). Same true with continuous assumption.
Proof. Weierstrass approximation theorem.
1.6.1 Intersection number mod 2 Let 푋 be a compact manifold, 푌 be a manifold, 푍 ⊆ 푌 a closed subset. Suppose 푓 ∶ 푋 → 푌 is smooth and 푓 푍 and dim 푋 + dim 푍 = dim 푌.
Definition. The intersection number mod 2 is
−1 퐼2(푓, 푍) = #푓 (푍) (mod 2).
Claim that if 푓 and 푔 are smoothly homotopic and 푔 푍 then 퐼2(푓, 푍) = 퐼2(푔, 푍). In the speical case 푋 ⊆ 푌, we have 푋 푍 and denote it 퐼2(푋, 푍). Example. Let 푌 be a torus. Let 푋 and 푍 be the two “perpendicular circles” on 푌. Then 퐼2(푋, 푍) = 1. It turns out we can still define intersection number without the assumption 푋 푍.
dim 푌 Example. Suppose dim 푌 is even and dim 푋 is 2 . Note that 푋 is not transversal to 푋. Let 푌 be the Möbius band and 푋 a central circle, 퐼2(푋, 푋) = 1.
1.7 Abstract manifolds
Definition (Smooth manifold). A smooth 푛-dimensional manifold is a Hausdorff, second-countable topological space 푋 together with a collection of 푛 maps 휑 ∶ 푈훼 → 푉훼 where 푈훼 ⊆ R , 푉훼 ⊆ 푋, called charts, with the following properties:
1. 휑훼 ∶ 푈훼 → 푉훼 are homeomorphisms, ⋃ 푉 = 푋 2. 훼 훼 , −1 휑 ∘ 휑 | −1 훼, 훽 3. 훽 훼 휑훼 (푈훼∩푉훽) are diffeomorphisms for all , 4. lastly, it is conventional to require the collection of charts to be maximal, i.e. if {휑훼}훼 ∪ { ̃휑} satisfies 1, 2, 3 then ̃휑= 휑훼 for some 훼.
17 1 Smooth Manifolds
Note that Hausdorffness and second-countability are hereditary properties. We can now define smooth maps 푋 → R푚 and smooth maps between manifolds, similar as before. Also compatibility condition requires us only to check smoothness on an open cover of 푋. What about differentials and tangent spaces? Since our manifold 푋 is no longer sitting inside an Euclidean space, the differential of a chart is not very meaningful. This is point where the abstract theory diverges from what we have learned. In III Differential Geometry you will learn to do it properly using tangent bundles.
Definition (Immersion, submersion, embedding). A smooth map 푓 ∶ 푋 → 푌 between manifolds is an immersion if 푑푓푥 is injective for all 푥 ∈ 푋, a submersion if 푑푓푥 is surjective for all 푥 ∈ 푋, and an embedding if it is an immersion and a homeomorphism onto its image.
Theorem 1.20 (Whitney embedding theorem). Let 푋 be a smooth manifold. Then there exists an embedding 푓 ∶ 푋 → R2푛+1.
Thus the abstract formulation of manifold is not any more general than the one we begin with. In fact, Whitney proved the better upper bound R2푛, which is in fact a sharp bound.
18 2 Geometry
2 Geometry
Definition (Regular curve). Let 푋 be a manifold. A regular curve in 푋 is a smooth immersion 훾 ∶ 퐼 → 푋 where 퐼 ⊆ R is an interval, i.e. 푑훾푡(1) ≠ 0.
Notation. We write ̇훾(푡) = 푑훾푡.
Exercise. Suppose 푋 = R3 and 훾(푡) = (푥(푡), 푦(푡), 푧(푡)), show that ̇훾(푡) = ( ̇푥(푡), ̇푦(푡), ̇푧(푡)).
3 Definition (Arc-length). Let 훾 ∶ 퐼 → R is a regular curve and 푡0 ∈ 퐼. The arc-length of 훾 (measured from 푡0) is
푡 푠(푡) = ∫ | ̇훾(푡′)|푑푡′
푡0
where | ⋅ | is the Euclidean norm.
Arc-length is invariant to reparameterisation: a reparameterisation of 훾 is a map ̂훾(푡)̂ such that 푡 = 푡(푡)̂ is a diffeomorphism, i.e. the following diagram commutes: 훾 퐼 R3
̂훾 퐼 ̂ 푑푡 > 0 푡 = 푡(푡̂ ) Wlog suppose 푑푡̂ . Let 0 0 , then
푡 푡 푑훾 푡̂ 푑훾 푠(푡) = ∫ | ̂훾(푡′)|푑푡′ = ∫ ∣ ∣ 푑푡′ = ∫ ∣ ∣ 푑푡′ = 푠(푡)̂ ′ ′ 푑푡 ̂ 푑푡̂ 푡0 푡0 푡0
Note that 푠′(푡) = | ̂훾(푡)| ≠ 0 by Fundmental Theorem of Calculus. Thus by Inverse Function Theorem and connectivity 푠(푡) is a diffeomorphism, i.e. can be inverted to 푡(푠). Thus we can reparameterise any regular curve by 푠. We will write 훾(푠) for 훾(푡(푠)). Missed a lecture Frenet equations
푡̇ 0 푘 0 푡 ⎜⎛ ̇푛⎟⎞ = ⎜⎛−푘 0 −휏⎟⎞ ⎜⎛푛⎟⎞ ⎝푏̇⎠ ⎝ 0 휏 0 ⎠ ⎝푏⎠ The curvature 푘 and torsion 휏 is invariant under isometries.
Theorem 2.1 (Fundamental theorem of curves in R3). Given smooth func- tion 푘 > 0 and 휏(푠), 푠 ∈ 퐼, there exists a regular curve 훼 ∶ 퐼 → R3 such that 푠 is the arclength, 푘(푠) is the curvature and 휏(푠) is the torision of 훼. Moreover, given 푘 = 푘,̃ 휏 =, ̃휏 then there exists an isometry of R3 such
19 2 Geometry
that ̃훾= 푅 ∘ 훾 + 푏. Proof. Let 푏 = ̃훾(0) − 훾(0) and we can assume ̃훾(0) = 훾(0) = 0. Given two orthonormal basis {푡(0), 푠(0), 푛(0)} and {푡(0),̃ ̃푠(0), ̃푛(0)} in the same orientation, it is an exercise in Euclidean geometry to show that there exists an element 푅 ∈ SO(3) sending the first triplet to the second. To avoid excessive notation, replace 훾 with 푅 ∘ 훾. Then 훾(푠) and ̃훾(푠) satisfy the Frenet equations, which is a linear ODE, with the same initial condition. Thus by uniqueness of solutions of ODE 훾 =. ̃훾 Geometric intuition: curvature measures the turning of the tangent, 푘 the magnitude and 푛 measures the direction. Torsion measures the non-planiness of a curve.
Proposition 2.2. Given a regular curve 훾(푠), 휏 = 0 if and only if 훾 lies in a plane.
Proof. Suppose 휏 = 0. Then 푏̇ = 0 so (푡 ∧ 푛)′ = 0. Let Π be the plane spanned by 푡 and 푛 and so Π(푠) = Π0 is constant. Consider the plane 훾(0) + Π, claim 훾(퐼) ⊆ 훾(0) + Π0: this is equivalent to claiming that 훾(푠) − 훾(0) ∈ Π for all 푠. But
푠 푠 훾(푠) − 훾(0) = ∫ 훾′(푢)푑푢 = ∫ 푡(푢)푑푢 ∈ Π. 0 0 The other direction is left as an exercise.
Definition (Plane curve). A regular curve is said to be a plane curve if its image lies in a plane Π ⊆ R3.
Wlog we will consider all plane curves as maps 훾 ∶ 퐼 → R2. For regular plane curves, there is an alternative definition of the curvature:
Definition (Signed curvature). Let 훾(푠) be a plane curve in R2 parame- terised by arc-length 푠. 푡 = ̇훾(푠) is the tangent. Pick 푛(푠) so that 푡, 푛 has the usual orientation (exercise: exhibit such 푛). Then 푡(푠)̇ ⋅ 푡(푠) = 0 so define signed curvature 푘(푠) to be such that
푡(푠)̇ = 푘(푠)푛(푠)
where 푘(푠) ∈ R, i.e. not necessarily non-vanishing.
Note the subtleties in the order of terms being defined: in general regular curves we defined 푘(푠) and 푛(푠) in terms of 푡(푠)̇ but for plane curves, we fixed 푛(푠) first (using orientation of orthonormal basis) and then define 푘(푠) to be the signed magitude of 푡(푠)̇ . Exercise. Define the Frenet formulae for plane curves:
푡(푠)̇ = 푘(푠)푛(푠) ̇푛(푠)= −푘(푠)푡(푠)
20 2 Geometry
2.1 Isoperimetric inequality in the plane The is the first global result we meet in differential geometry. Let 훾 be a smooth simple closed curve in R2, i.e. 훾 ∶ 푆1 → R2 is a smooth embedding. Thanks to
Theorem 2.3 (Jordan curve theorem). 훾(푆1) is the boundary of a bounded, simply-connected open set in R2. we let Ω be the bouned open set (alternatively assume Ω is a 2-manifold with boundary in R2 with boundary diffeomorphic to 푆1). Note. We want to define 훼(R) = 훼(푆1). Let 훼 ∶ R → R2 be a regular curve parameterised by arc-length. Let 퐿 be the first 푠 > 0 such that 훼(퐿) = 훼(0). Then 훼 be periodic with period 퐿. 퐿 is the arc-length of 훾.
Theorem 2.4 (Isoperimetric inequality).
퐿2 ≥ 4휋 area(Ω)
with equality if and only if 훾 is a circle.
Remark. It suffices to assume 훼 is 퐶1. Write 훼(푠) = (푥(푠), 푦(푠)) where 푥(푠) and 푦(푠) are 퐶1 functions. Let 퐕 ∶ R2 → R2 be the identity and think of it as a vector field. By divergence theorem applied to 퐕 in Ω,
2 area(Ω) = ∫ 훁 ⋅ 퐕 Ω 퐿 = ∫ 퐕 ⋅ (−푛)푑푠 0 퐿 ≤ ∫ ‖퐕(푠)‖푑푠 0 1/2 퐿 퐿 ≤ (∫ ‖퐕(푠)‖2푑푠 ∫ 12푑푠) 0 0
Missed a lecture
2.2 Riemannian metric something something Riemannian metric. See printed notes.
Definition (Isometry). Let 푋 and 푌 be (Riemannian) manifolds with Rie- mannian metrics 푔푝 and 푞̃푔 respectively. We say 푋 and 푌 are isometric if there exists 푓 ∶ 푋 → 푌 smooth such that
푑푓푥 ∶ 푇푥푋 → 푇푓(푥)푌
21 2 Geometry
is an isometry of inner product spaces, i.e. for all 푣, 푤 ∈ 푇푥푋,
푔푥(푣, 푤) =푓(푥) ̃푔 (푑푓푥(푣), 푑푓푥(푤)).
Furthermore 푑푓푥 is invertible.
Note that conventions vary. If 푑푓푥 is not required to be invertible then 푓 is usually called an isometric embedding.
Definition (Local isometry). Two Riemannian manifolds 푋 and 푌 are locally isometric around 푝 ∈ 푋 if there exist 푉 ∋ 푝 open in 푋 and 푊 open in 푌 such that 푓 ∶ 푉 → 푊 is an isometry.
Question.
1. When are two surfaces 푋 and 푌 (say embedded in R3) locally isometric? Globally isometric?
2. When is 푋 locally isometric to the plane? (intrinsic geometry) 3. When do 푋 and 푌 differ by a Euclidean isometry, i.e. 푌 = 푅(푋) + 푏 where 푅 ∈ SO(3)? (extrinsic geometry) Let 푋 be a Riemannian 2-manifold. Let 휑 ∶ 푈 → 푉 be a local parameterisation. Define
퐸(푢, 푣) = 푔(휑푢, 휑푢) > 0
퐹 (푢, 푣) = 푔(휑푢, 휑푣)
퐺(푢, 푣) = 푔(휑푣, 휑푣) > 0
2 퐸 퐹 with 퐸퐺 − 퐹 > 0 and 퐸, 퐹 , 퐺 smooth. Claim that ( 퐹 퐺 ) defines a Riemannian metric. A converse question is, given such 퐸, 퐹 , 퐺, does there exist a local param- eterisation 휑 ∶ 푈 → 푉 such that 퐸, 퐹 , 퐺 are coefficients of 퐼푝? The answer is positive for analytic functions, but in more general settings this is still an open problem. Thus it is possible that local Riemannian geometry is richer than Euclidean geometry. First fundamental form
Example (Embedded torus).
Exercise. Do for general surfaces of revolution.
Definition. Let 푋 be a surface in R3. Let 휑 ∶ 푈 → 푉 be a local parameteri- sation. Let Ω ⊆ 푉 be an open set with smooth boundary. Then the area of Ω is
∫ |휑푢 × 휑푣|푑푢푑푣. 휑−1(Ω)
Claim that the definition is independent of parameterisation. Use chain rule for change of variable.
22 2 Geometry
Exercise. √ area Ω = ∫ 퐸퐺 − 퐹 2푑푢푑푣 휑−1(Ω) so the area depends only on the first fundamental form. Thus an isometry preserves area. In the other words, area is an intrinsic geometric property.
Remark. Using partition of unity, we can drop the restriction that Ω lies in the range of a local parameterisation. In fact there exists a Borel measure on 푋, denoted 푑퐴, such that area(Ω) = ∫ 푑퐴. Ω Subsequently, we can define the integral of any smooth (actually measurable) map with respect to the measure. Note that 퐸 ⊆ 푋 has measure zero as defined before if and only if 퐸 has measure 0 with respect to 푑퐴.
Example. The area of the embedded torus 푋 define before is
2휋 2휋 area(푋) = ∫ ∫ 푟(푎 + 푟 cos 푢)푑푢푑푣 = 4휋2푎푟. 0 0
2.3 Gauss map
Let 푋 be a surface in R3. Let 휑 ∶ 푈 → 푉 be a local parameterisation. Given 푝 ∈ 푋, 휑푢, 휑푣 form a basis for 푇푝푋. Define a map
푓 ∶ 푉 → 푆2 −1 (휑푢 × 휑푣)(휑 (푝)) 푝 ↦ −1 |(휑푢 × 휑푣)(휑 (푝))| which is a smooth map such that 푓(푝) ⟂ 푇푝푋. Note that the map with such property is not unique, for example taking it negative value.
Definition (Gauss map). Let 푋 be a surface in R3.A Gauss map of 푋 is a 2 smooth map 푁 ∶ 푋 → 푆 such that 푁(푝) ⟂ 푇푝푋 for all 푝 ∈ 푋.
Note. Let 푋 be connected. If there exists a Gauss map then there exists precisely 2 Gauss maps 푁, −푁. Moreover 휑 × 휑 푁 = ± 푢 푣 |휑푢 × 휑푣| in a local chart.
Definition (Orientability). If there exists a Gauss map, we say 푋 is ori- entable. (푋, 푁) is called an oriented surface.
Remark.
23 2 Geometry
1. There exists non-orientable surface, such as the Möbius band. 2. The above definition is an extrinsic one. It is possible to characterise orientability intrinsically using transition maps.
Example. The sphere is orientable, with 푁 = ± id. Let 푋 be an orientable surface and 푁 ∶ 푋 → 푆2 a Gauss map. Consider 2 the differential 푑푁푝 ∶ 푇푝푋 → 푇푁(푝)푆 . Since 푁(푝) ⟂ 푇푝푋 by definition, and 2 푁(푝) ⟂ 푇푁(푝)푆 by the observation in the example above, we can identify 2 푇푝푋 = 푇푁(푝)푆 . Thus we can treat 푑푁푝 ∶ 푇푝푋 → 푇푝푋 as an endormorphism of an inner product space. Next time we will show that it is self-adjoint with respect to the first fundamental form. Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Etiam lobortis facilisis sem. Nullam nec mi et neque pharetra sollicitudin. Praesent imperdiet mi nec ante. Donec ullamcorper, felis non sodales commodo, lectus velit ultrices augue, a dignissim nibh lectus placerat pede. Vivamus nunc nunc, molestie ut, ultricies vel, semper in, velit. Ut porttitor. Praesent in sapien. Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Duis fringilla tristique neque. Sed interdum libero ut metus. Pellentesque placerat. Nam rutrum augue a leo. Morbi sed elit sit amet ante lobortis sollicitudin. Praesent blandit blandit mauris. Praesent lectus tellus, aliquet aliquam, luctus a, egestas a, turpis. Mauris lacinia lorem sit amet ipsum. Nunc quis urna dictum turpis accumsan semper.
2.4 Theorema Egregium
Since {휑푢, 휑푣, 푁} form an orthonormal basis, we can express 휑푢푢, 휑푢푣 etc as unique linear combinations of them. But what are the indices? Recall that
푒 = −⟨휑푢, 푁푢⟩ = ⟨휑푢푢, 푁⟩
and 푁 is orthogonal to 휑푈, 휑푣 so we have
1 2 휑푢푢(푢, 푣) = Γ11(푢, 푣)휑푢 + Γ11휑푣 + 푒푁
and etc.
푘 Definition. Γ푖푗 are called the Christoffel symbols.
Remark. 1. The Christoffel symbols are symmetric with repsect to lower indices, i.e. 푘 푘 Γ푖푗 = Γ푗푖.
푘 2. Γ푖푗 = 퐴(퐸, 퐹 , 퐺, 휕퐸, … , 휕퐺) where 퐴 is a universal function. Take the inner product of the above relations with 휑푢 and recoginising ⟨휑푢, 휑푢⟩ = 퐸, we get 1 1 Γ1 퐸 + Γ2퐹 = ⟨휑 , 휑 ⟩ = 휕 ⟨휑 , 휑 ⟩ = 휕 퐸 11 1 푢푢 푢 2 푢 푢 푢 2 푢 1 Γ1 퐹 + Γ2퐺 = ⟨휑 , 휑 ⟩ = 휕 ⟨휑 , 휑 ⟩ − ⟨휑 , 휑 ⟩ = 휕 퐹 − 휕 퐸 11 1 푢푢 푣 푢 푢 푣 푢 푣푢 푢 2 푣
24 2 Geometry
Write it in matrix form,
1 1 퐸 퐹 Γ11 2 휕푢퐸 ( )( 2 ) = ( 1 ) 퐹 퐺 Γ11 휕푢퐹 − 2 휕푣퐸
푘 We can repeat for the other Γ푖푗’s. Let’s go on and continue differentiating. Differentiate * with respect to 푣 and ** with respect to 푢, and by symmetry of partial derivatives,
1 2 1 2 휕푣Γ11휑푢 + 휕푣Γ11휑푣 + (휕푣푒)푁 + Γ11휑푢푣 + Γ11휑푣푣 + 푒푁푣 1 2 1 2 =휕푢Γ12휑푢 + 휕푢Γ12휑푣 + (휕푢푓)푁 + Γ12휑푢푢 + Γ12휑푣푢 + 푓푁푢
Expand this in our favourite basis {휑푢, 휑푣, 푁}, we get three equations, each for a basis element. The first three components on each side are already inthis form. Look up in * and **, we get
2 휕푣Γ11
Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Etiam lobortis facilisis sem. Nullam nec mi et neque pharetra sollicitudin. Praesent imperdiet mi nec ante. Donec ullamcorper, felis non sodales commodo, lectus velit ultrices augue, a dignissim nibh lectus placerat pede. Vivamus nunc nunc, molestie ut, ultricies vel, semper in, velit. Ut porttitor. Praesent in sapien. Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Duis fringilla tristique neque. Sed interdum libero ut metus. Pellentesque placerat. Nam rutrum augue a leo. Morbi sed elit sit amet ante lobortis sollicitudin. Praesent blandit blandit mauris. Praesent lectus tellus, aliquet aliquam, luctus a, egestas a, turpis. Mauris lacinia lorem sit amet ipsum. Nunc quis urna dictum turpis accumsan semper. And after some algebra we find that 퐾 can be expressed purely in terms of the first fundamental form and its derivative.
25 3 Critical points of length and curve
3 Critical points of length and curve
Suppose 푋 ⊆ R3 is a surface and 푝, 푞 ∈ 푋. Let Ω(푝, 푞) be the set of all regular paths from 푝 to 푞. Then the length could be seen as a functional on Ω(푝, 푞). Since length depends only on the first fundamental form, i.e. an intrinsic geometric quantity, it turns out that the curves minimising the length functional are also special and intrinsic to the surface. They are called geodesic. Similarly, area is also a functional on the space of all surfaces. The area- miniming surfaces are called minimal surfaces and eventually we can derive an equation depending on the mean curvature.
26 Index arc-length, 19 isometry, 21, 22 isoperimetric inequality, 21 boundary, 12 Brouwer’s fixed point theorem, 17 manifold, 2 abstract smooth manifold, 17 codimension, 4 manifold with boundary, 12 critical point, 8 submanifold, 3 critical value, 8 orientability, 23 degree mod 2, 16 diffeomorphism, 2 plane curve, 20 differential, 4, 6 regular value, 8 embedding, 18 signed curvature, 20 Fundamental theorem of curves in smooth, 2 R3, 19 smooth homotopy, 14 smooth isotopy, 15 Gauss map, 23 submersion, 18 homogeneity lemma, 15 tangent space, 4 transversality, 10 immersion, 18 transversality theorem, 11 interior, 12 inverse function theorem, 7 Whitney embedding theorem, 18
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