Differential Geometry
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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 4 1 Smooth Manifolds 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. 5 1 Smooth Manifolds 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 .