MATH-214 spring 2021 的课堂小 exercise
《把量子计算当成李群转转转》
莫名其妙的交叉:量子计算、逻辑、范畴、一般拓扑、层理论、流形、李群
# 一堆定义
# Basis of topology
B ⊂ P ( X ) \mathcal{B}\subset \mathcal{P}(X) B ⊂ P ( X ) is called a basis for the topology on X X X if for every A ⊂ X A\subset X A ⊂ X ,
A is open ⟺ A is a union of elemens in B ⟺ ∀ p ∈ A . ∃ B ∈ B . s . t . p ∈ B ⊂ A \begin{aligned}
A\text{ is open}&\iff A\text{ is a union of elemens in }\mathcal{B}\\
&\iff \forall p\in A.\;\exists B\in\mathcal{B}.\;s.t.\;p\in B\subset A
\end{aligned}
A is open ⟺ A is a union of elemens in B ⟺ ∀ p ∈ A . ∃ B ∈ B . s . t . p ∈ B ⊂ A
# Neighborhood
For point p ∈ W p\in W p ∈ W in topological space X X X , W W W is called a neighborhood of p p p if there exists an open set such that p ∈ U ⊂ W ⊂ X p\in U\subset W\subset X p ∈ U ⊂ W ⊂ X .
# Hausdorff
Topological space X X X is Hausdorff if for any p , q ∈ X p,q\in X p , q ∈ X and p ≠ q p\neq q p = q , exists open subsets U , V ⊂ X U,V\subset X U , V ⊂ X such that p ∈ U , q ∈ V , U ∩ V = ∅ p\in U,q\in V,U\cap V=\emptyset p ∈ U , q ∈ V , U ∩ V = ∅ .
# Compact
K ⊂ X K\subset X K ⊂ X is called compact if every open cover K ⊂ ⋃ i ∈ I U i K\subset\bigcup_{i\in I}U_i K ⊂ ⋃ i ∈ I U i (in topogical space X X X ) has a finite subcover K ⊂ ⋃ i ∈ F U i K\subset\bigcup_{i\in F}U_i K ⊂ ⋃ i ∈ F U i , ∣ F ∣ < ∞ |F|<\infty ∣ F ∣ < ∞ .
# Second countable
Topological space X X X is called second countable if it has a coutable basis.
# Locally Euclidean
Topological space X X X is called n n n -dimensional locally Euclidean if for any point p ∈ X p\in X p ∈ X , there is an open subset p ∈ U ⊂ X p\in U\subset X p ∈ U ⊂ X such that U U U is homeomorphic to an open set in R n \mathbb{R}^n R n .
With out loss of generality, the definition could be modified as U ≈ B ( 0 , 1 ) ∈ R n U\approx B(0,1)\in\mathbb{R}^n U ≈ B ( 0 , 1 ) ∈ R n . (See exercise below)
# Topological manifold
A topological space M M M is called n n n -dimensional topological manifold if M M M is
locally n n n -dimensional Euclidean.
Hausdorff.
second countable.
# Coordinate chart
A (coordinate) chart on n n n -dimensional manifold M M M is a pair ( U , φ ) (U,\varphi) ( U , φ ) where U ⊂ M U\subset M U ⊂ M is an open set and φ : U → U ~ \varphi:U\rightarrow \tilde{U} φ : U → U ~ is a homeomorphism to an open subset in R n \mathbb{R}^n R n . i.e.,
U ~ : = φ ( U ) ⊂ open R n \tilde{U}:=\varphi(U)\mathop{\subset}_{\text{open}}\mathbb{R}^n
U ~ : = φ ( U ) ⊂ open R n
# (Path) connectivity (component)
Topological space X X X is connected if the only subset that are both open and closed are ∅ , X \emptyset,X ∅ , X .
Topological space X X X is path-connected if ∀ p , q ∈ X \forall p,q\in X ∀ p , q ∈ X , there is a continuous path r : [ 0 , 1 ] → X r:[0,1]\rightarrow X r : [ 0 , 1 ] → X with r ( 0 ) = p , r ( 1 ) = q r(0)=p,r(1)=q r ( 0 ) = p , r ( 1 ) = q .
Remark : In manifold M M M , M M M is connected if and only if it’s path connected.
While in topological space X X X , path-connected is a stronger requirement.
The component of X X X is the maximal path-connected subsets of X X X .
# Interior
The interior of a subset S ⊂ X S\subset X S ⊂ X in topological space X X X is the union of all subsets of S S S that are open in X X X . i.e.,
Int S = ⋃ A ⊂ S , A is open in X A \text{Int}S=\bigcup_{A\subset S,A\text{ is open in }X}A
Int S = A ⊂ S , A is open in X ⋃ A
Remark : Int S \text{Int}S Int S is the largest open subset of S S S .
# Exhaustion by compact subsets
An exhaustion by compact subsets is an increasing sequence K 1 ⊂ K 2 ⊂ . . . ⊂ X K_1\subset K_2\subset...\subset X K 1 ⊂ K 2 ⊂ . . . ⊂ X such that:
∀ i . K i \forall i.\;K_i ∀ i . K i is compact.
K i ⊂ Int K i + 1 K_i\subset\text{Int}K_{i+1} K i ⊂ Int K i + 1 .
⋃ i = 1 ∞ K i = X \bigcup_{i=1}^{\infty}K_i=X ⋃ i = 1 ∞ K i = X .
Remark : Since X X X is open, thus X = ⋃ i = 1 ∞ Int K i X=\bigcup_{i=1}^{\infty}\text{Int}K_i X = ⋃ i = 1 ∞ Int K i .
# Limit points
The limit points for a subset A ⊂ X A\subset X A ⊂ X in topological space X X X are points p p p such that, for any open subset U U U in X X X ,
p ∈ U ⊂ X ⟹ ∃ q ∈ U . q ≠ p and q ∈ A p\in U\subset X\implies \exists q\in U.\;q\neq p\text{ and }q\in A
p ∈ U ⊂ X ⟹ ∃ q ∈ U . q = p and q ∈ A
Remark : An interior point is not necessarily a limit point, i.e. p ∈ A ⟹ p p\in A\cancel{\implies} p p ∈ A ⟹ p is a limit point for A A A .
The discrete set X = { a , b , c } X=\{a,b,c\} X = { a , b , c } with topology P ( A ) \mathcal{P}(A) P ( A ) . The interior point a ∈ { a } a\in \{a\} a ∈ { a } is not a limit point of { a } \{a\} { a } .
# Closure
The closure of a subset A ⊂ X A\subset X A ⊂ X in topological space X X X is all interior points together with limit points of A A A , which is denoted as A ˉ = A ∪ lim A \bar{A}=A\cup \text{lim}A A ˉ = A ∪ lim A .
Remark : closure are closed sets.
we prove that X − A ˉ X-\bar{A} X − A ˉ is open.
Given any point p ∈ X − A ˉ p\in X-\bar{A} p ∈ X − A ˉ , it’s not a limit point of A A A , which means, exists open subset p ∈ U p ⊂ X p\in U_p\subset X p ∈ U p ⊂ X such that U p ∩ A = ∅ U_p\cap A=\emptyset U p ∩ A = ∅ .
Then ⋃ p ∈ X − A ˉ U p \bigcup_{p\in X-\bar{A}}U_p ⋃ p ∈ X − A ˉ U p is a open cover of X − A ˉ X-\bar{A} X − A ˉ and does not intersect with A A A .
Thus X − A ˉ = ⋃ p ∈ X − A ˉ U p X-\bar{A}=\bigcup_{p\in X-\bar{A}}U_p X − A ˉ = ⋃ p ∈ X − A ˉ U p is open.
# Precompact
Subset A ⊂ X A\subset X A ⊂ X in topological space X X X is precompact if its closure A ˉ \bar{A} A ˉ is compact in X X X .
# (Locally finite) cover
U ⊂ P ( X ) \mathcal{U}\subset \mathcal{P}(X) U ⊂ P ( X ) is called a cover of X X X if X = ⋃ u ∈ U u X=\bigcup_{u\in\mathcal{U}}u X = ⋃ u ∈ U u .
Moreover, the cover is called locally finite if for any point p ∈ X p\in X p ∈ X , there exists a neighborhood p ∈ W ⊂ X p\in W\subset X p ∈ W ⊂ X such that W W W only intersects finitely many u ∈ U u\in\mathcal{U} u ∈ U .
# Refinement
Cover V ⊂ P ( X ) \mathcal{V}\subset\mathcal{P}(X) V ⊂ P ( X ) is called a refinement of cover U ⊂ P ( X ) \mathcal{U}\subset\mathcal{P}(X) U ⊂ P ( X ) if ∀ v ∈ V . ∃ u ∈ U . v ⊂ u \forall v\in\mathcal{V}.\;\exists u\in\mathcal{U}.\;v\subset u ∀ v ∈ V . ∃ u ∈ U . v ⊂ u .
# Paracompact
Topological space X X X is called paracompact if every open cover has a locally finite refinement.
Remark : Every topological manifold is paracompact.
# Linear map
A linear map (also called a linear mapping , linear transformation , vector space homomorphism , or in some contexts linear function ) is a mapping V → W V\to W V → W between two vector spaces that preserves the operations of vector addition and scalar multiplication.
# Smooth (component function, partial derivative)
If U U U and V V V are open subsets of Euclidean spaces R n \mathbb{R}^n R n and R m \mathbb{R}^m R m , respectively, then a function F : U → V F:U\rightarrow V F : U → V is said to be smooth (or C ∞ C^\infty C ∞ , or infinitely differentiable) if each of its component functions has continuous partial derivatives of all orders at any point in U U U .
Notation: Suppose u → ∈ R n \overrightarrow{u}\in\mathbb{R}^n u ∈ R n and F ( u → ) = ( F 1 ( u → ) , . . . , F m ( m → ) ) ∈ R m F(\overrightarrow{u})=(F_1(\overrightarrow u),...,F_m(\overrightarrow{m}))\in\mathbb{R}^m F ( u ) = ( F 1 ( u ) , . . . , F m ( m ) ) ∈ R m , then each F 1 , . . , F m F_1,..,F_m F 1 , . . , F m is a component function of F F F .
Then F : R n → R m F:\mathbb{R}^n\to\mathbb{R}^m F : R n → R m is in C 1 C^1 C 1 means that,
∂ F i ∂ u j = lim h → 0 F i ( u 1 , . . . , u j + h , . . . , u n ) − F ( u 1 , . . . , u j , . . . , u n ) h \begin{aligned}
& \frac{\partial F_i}{\partial u_j}=\lim_{h\to 0}\frac{F_i(u_1,...,u_j+h,...,u_n)-F(u_1,...,u_j,...,u_n)}{h}
\end{aligned}
∂ u j ∂ F i = h → 0 lim h F i ( u 1 , . . . , u j + h , . . . , u n ) − F ( u 1 , . . . , u j , . . . , u n )
is continuous for any i = 1 , . . . , m j = 1 , . . . , n i=1,...,m\quad j=1,...,n i = 1 , . . . , m j = 1 , . . . , n , and for any u → ∈ U \overrightarrow{u}\in U u ∈ U .
Besides, F ∈ C 2 F\in C^2 F ∈ C 2 means
∂ 2 F i ∂ u j ∂ u k \frac{\partial^2 F_i}{\partial u_j\partial u_k}
∂ u j ∂ u k ∂ 2 F i
is continuous for i = 1 , . . . , m j = 1 , . . . , n k = 1 , . . . , n i=1,...,m\quad j=1,...,n\quad k=1,...,n i = 1 , . . . , m j = 1 , . . . , n k = 1 , . . . , n .
And so on. F F F is smooth is defined as F ∈ C ∞ F\in C^{\infty} F ∈ C ∞ .
# Diffeomorphism
If smooth function F F F is bijective and has a smooth inverse map, then it is called a diffeomorphism .
# Smoothly compatible
We say two coordinate charts ( U , φ ) , ( V , ψ ) (U,\varphi),(V,\psi) ( U , φ ) , ( V , ψ ) are smoothly compatible if both transition maps:
ψ ∘ φ − 1 : φ ( U ∩ V ) → ψ ( U ∩ V ) φ ∘ ψ − 1 : ψ ( U ∩ V ) → φ ( U ∩ V ) \psi\circ\varphi^{-1}:\varphi(U\cap V)\rightarrow \psi(U\cap V)\\
\varphi\circ\psi^{-1}:\psi(U\cap V)\to \varphi(U\cap V)
ψ ∘ φ − 1 : φ ( U ∩ V ) → ψ ( U ∩ V ) φ ∘ ψ − 1 : ψ ( U ∩ V ) → φ ( U ∩ V )
are smooth.
# (Smooth) atlas
An atlas A \mathcal{A} A of M M M is a collection of charts such that M = ⋃ ( U , φ ) ∈ A U M=\bigcup_{(U,\varphi)\in\mathcal{A}}U M = ⋃ ( U , φ ) ∈ A U .
An atlas is called smooth if all charts in it are smoothly compatible.
An atlas A \mathcal{A} A is maximal if there is no other atlas A ′ \mathcal{A}' A ′ such that A ⊊ A ′ \mathcal{A}\subsetneq\mathcal{A}' A ⊊ A ′ .
Remark : Given any smooth atlas A \mathcal{A} A , we could construct the maximal smooth atlas as:
A ‾ = { ( U , ψ ) ∣ ( U , ψ ) is smooth compatible with all ( V , φ ) ∈ A } \overline{\mathcal{A}}=\{(U,\psi)\mid (U,\psi)\text{ is smooth compatible with all }(V,\varphi)\in \mathcal{A}\}
A = { ( U , ψ ) ∣ ( U , ψ ) is smooth compatible with all ( V , φ ) ∈ A }
and the maximal construction is unique. That is, if every chart in A \mathcal{A} A is smooth compatible with every chart in A ′ \mathcal{A}' A ′ , then A ‾ = A ′ ‾ \overline{\mathcal{A}}=\overline{\mathcal{A}'} A = A ′ .
# Smooth structure
A maximal smooth altlas on a topological manifold M M M is called a smooth structure on M M M .
# Smooth manifold
A smooth manifold is a pair ( M n , A ) (M^n,\mathcal{A}) ( M n , A ) , where M n M^n M n is a n n n -dimensional topological manifold and A \mathcal{A} A is a smooth structure.
For short, we write M = ( M n , A ) M=(M^n,\mathcal{A}) M = ( M n , A ) , and say that chart ( U , φ ) (U,\varphi) ( U , φ ) is smooth if ( U , φ ) ∈ A (U,\varphi)\in\mathcal{A} ( U , φ ) ∈ A . (Just like open sets in topological space)
Remark : “Smooth structure” here is a closure (see maximal atlas), which means if ( U , φ ) (U,\varphi) ( U , φ ) is smooth compatible with all charts in A \mathcal{A} A , then ( U , φ ) ∈ A (U,\varphi)\in\mathcal{A} ( U , φ ) ∈ A .
In words, if chart is smooth compatible with all smooth charts, then it’s smooth.
# Lecture 1
Exercise . If X = Y = R X=Y=\mathbb{R} X = Y = R , verify that the continuous map defined by topological space is equivalent to ε − δ \varepsilon-\delta ε − δ language.
一方面,
考虑f : R → R f:\mathbb{R}\rightarrow\mathbb{R} f : R → R 是一个 topological continuous function。那么意味着
∀ ( a , b ) ⊂ R . f − 1 ( a , b ) = { x ∣ a < f ( x ) < b } is open \forall (a,b)\subset\mathbb{R}.\;f^{-1}(a,b)=\{x\mid a<f(x)<b\}\text{ is open}
∀ ( a , b ) ⊂ R . f − 1 ( a , b ) = { x ∣ a < f ( x ) < b } is open
所以,对于任意 point x ∈ R x\in\mathbb{R} x ∈ R 和ε > 0 \varepsilon>0 ε > 0 ,有( f ( x ) − ε , f ( x ) + ε ) (f(x)-\varepsilon,f(x)+\varepsilon) ( f ( x ) − ε , f ( x ) + ε ) 是R \mathbb{R} R 的一个 open set。因此存在
f − 1 ( f ( x ) − ε , f ( x ) + ε ) = ⋃ i ∈ I ( a i , b i ) ⊂ R f^{-1}(f(x)-\varepsilon,f(x)+\varepsilon)=\bigcup_{i\in I}(a_i,b_i)\subset \mathbb{R}
f − 1 ( f ( x ) − ε , f ( x ) + ε ) = i ∈ I ⋃ ( a i , b i ) ⊂ R
即f − 1 ( f ( x ) − ε , f ( x ) + ε ) f^{-1}(f(x)-\varepsilon,f(x)+\varepsilon) f − 1 ( f ( x ) − ε , f ( x ) + ε ) 是一个 open set。又因为不难验证x ∈ f − 1 ( f ( x ) − ε , f ( x ) + ε ) x\in f^{-1}(f(x)-\varepsilon,f(x)+\varepsilon) x ∈ f − 1 ( f ( x ) − ε , f ( x ) + ε ) ,所以∃ i . x ∈ ( a i , b i ) \exists i.\;x\in(a_i,b_i) ∃ i . x ∈ ( a i , b i ) 。
所以我们取对应的( a i , b i ) (a_i,b_i) ( a i , b i ) 满足x ∈ ( a i , b i ) x\in (a_i,b_i) x ∈ ( a i , b i ) 。那么存在一个δ = max ( b i − x , x − a ) \delta=\max(b_i-x,x-a) δ = max ( b i − x , x − a ) ,使得有:
∀ x ′ ∈ ( x − δ , x + δ ) . f ( x ′ ) ∈ ( f ( x ) − ε , f ( x ) + ε ) \forall x'\in (x-\delta,x+\delta).\;f(x')\in (f(x)-\varepsilon,f(x)+\varepsilon)
∀ x ′ ∈ ( x − δ , x + δ ) . f ( x ′ ) ∈ ( f ( x ) − ε , f ( x ) + ε )
即对应了ε − δ \varepsilon-\delta ε − δ 语言:
∀ ε > 0. ∃ δ > 0. ∀ x ′ . ∣ x − x ′ ∣ < δ ⟹ ∣ f ( x ) − f ( x ′ ) ∣ < ε \forall \varepsilon>0.\;\exists \delta>0.\;\forall x'.\;|x-x'|<\delta\implies |f(x)-f(x')|<\varepsilon
∀ ε > 0 . ∃ δ > 0 . ∀ x ′ . ∣ x − x ′ ∣ < δ ⟹ ∣ f ( x ) − f ( x ′ ) ∣ < ε
另一方面,
如果有ε − δ \varepsilon-\delta ε − δ 语言,我们证明:任给一个 open set ⋃ i ∈ I ( a i , b i ) ⊂ R \bigcup_{i\in I}(a_i,b_i)\subset\mathbb{R} ⋃ i ∈ I ( a i , b i ) ⊂ R ,有f − 1 ( ⋃ i ∈ I ( a i , b i ) ) f^{-1}\left (\bigcup_{i\in I}(a_i,b_i)\right ) f − 1 ( ⋃ i ∈ I ( a i , b i ) ) 也是 open 的。
注意到:
f − 1 ( ⋃ i ∈ I ( a i , b i ) ) = { x ∈ R ∣ ∃ i . a i < f ( x ) < b i } = ⋃ i ∈ I f − 1 ( a i , b i ) f^{-1}\left (\bigcup_{i\in I}(a_i,b_i)\right )=\{x\in\mathbb{R}\mid \exists i.\;a_i<f(x)<b_i\}=\bigcup_{i\in I}f^{-1}(a_i,b_i)
f − 1 ( i ∈ I ⋃ ( a i , b i ) ) = { x ∈ R ∣ ∃ i . a i < f ( x ) < b i } = i ∈ I ⋃ f − 1 ( a i , b i )
因此我们证明,每个f − 1 ( a i , b i ) f^{-1}(a_i,b_i) f − 1 ( a i , b i ) 都是 open 的,那么根据 open set 的 axiom,它们的 union 也是 open 的。
任给一个( a , b ) (a,b) ( a , b ) ,根据实数的连续性,可以知道( a , b ) = ⋃ y ∈ ( a , b ) ( y − ε y , y + ε y ) (a,b)=\bigcup_{y\in(a,b)}(y-\varepsilon_y,y+\varepsilon_y) ( a , b ) = ⋃ y ∈ ( a , b ) ( y − ε y , y + ε y ) 。即开区间( a , b ) (a,b) ( a , b ) 可以由每个开区间中的元素的邻域覆盖。
那么∀ x ∈ f − 1 ( a , b ) \forall x\in f^{-1}(a,b) ∀ x ∈ f − 1 ( a , b ) ,我们取y : = f ( x ) ∈ ( a , b ) y:=f(x)\in (a,b) y : = f ( x ) ∈ ( a , b ) ,那么根据ε − δ \varepsilon-\delta ε − δ 语言,存在一个δ x \delta_x δ x 使得:
( x − δ x , x + δ x ) ⊂ f − 1 ( y − ε y , y + ε y ) ⊂ f − 1 ( a , b ) (x-\delta_x,x+\delta_x)\subset f^{-1}(y-\varepsilon_y,y+\varepsilon_y)\subset f^{-1}(a,b)
( x − δ x , x + δ x ) ⊂ f − 1 ( y − ε y , y + ε y ) ⊂ f − 1 ( a , b )
那么有,对于∀ x ∈ f − 1 ( a , b ) . ∃ δ x . ( x − δ x , x + δ x ) ⊂ f − 1 ( a , b ) \forall x\in f^{-1}(a,b).\; \exists \delta_x.\;(x-\delta_x,x+\delta_x)\subset f^{-1}(a,b) ∀ x ∈ f − 1 ( a , b ) . ∃ δ x . ( x − δ x , x + δ x ) ⊂ f − 1 ( a , b ) 。即对于f − 1 ( a , b ) f^{-1}(a,b) f − 1 ( a , b ) 中的每个元素,都存在一个开邻域。那么
f − 1 ( a , b ) = ⋃ x ∈ f − 1 ( a , b ) ( x − δ x , x + δ x ) f^{-1}(a,b)=\bigcup_{x\in f^{-1}(a,b)}(x-\delta_x,x+\delta_x)
f − 1 ( a , b ) = x ∈ f − 1 ( a , b ) ⋃ ( x − δ x , x + δ x )
也是 open 的。
□ \square □ .
Exercise . If f : X → Z f:X\rightarrow Z f : X → Z between two topological spaces is continuous, then for any sub topological space ( Y , O ( Y ) ) (Y,\mathcal{O}(Y)) ( Y , O ( Y ) ) with
Y ⊂ X , O ( Y ) = { Y ∩ A ∣ A ∈ O ( X ) } Y\subset X,\;\mathcal{O}(Y)=\{Y\cap A\mid A\in\mathcal{O}(X)\}
Y ⊂ X , O ( Y ) = { Y ∩ A ∣ A ∈ O ( X ) }
we have the restriction map f ∣ Y : Y → Z f|_Y:Y\rightarrow Z f ∣ Y : Y → Z being continuous.
不难发现,对于任意Z Z Z 中 open set A A A ,有:
f − 1 ( A ) ∩ Y = { x ∈ Y ∣ f ( x ) ∈ A } = { x ∈ Y ∣ f ∣ Y ( x ) ∈ A } = f ∣ Y − 1 ( A ) \begin{aligned}
f^{-1}(A)\cap Y&=\{x\in Y\mid f(x)\in A\}\\
&=\{x\in Y\mid f|_Y(x)\in A\}\\
&=f|_Y^{-1}(A)
\end{aligned}
f − 1 ( A ) ∩ Y = { x ∈ Y ∣ f ( x ) ∈ A } = { x ∈ Y ∣ f ∣ Y ( x ) ∈ A } = f ∣ Y − 1 ( A )
因为f f f continuous,既然f − 1 ( A ) ∈ O ( X ) f^{-1}(A)\in\mathcal{O}(X) f − 1 ( A ) ∈ O ( X ) ,那么f ∣ Y − 1 ( A ) = f − 1 ( A ) ∩ Y ∈ O ( Y ) f|_Y^{-1}(A)=f^{-1}(A)\cap Y\in\mathcal{O}(Y) f ∣ Y − 1 ( A ) = f − 1 ( A ) ∩ Y ∈ O ( Y ) 。故f ∣ Y f|_Y f ∣ Y continuous。
□ \square □ .
Exercise . The subspace topology ( Y , O ( Y ) ) (Y,\mathcal{O}(Y)) ( Y , O ( Y ) ) with Y ⊂ X , O ( Y ) = { Y ∩ A ∣ A ∈ O ( X ) } Y\subset X,\;\mathcal{O}(Y)=\{Y\cap A\mid A\in\mathcal{O}(X)\} Y ⊂ X , O ( Y ) = { Y ∩ A ∣ A ∈ O ( X ) } is the coarest topology on Y Y Y (i.e. fewest subsets) such that injection i : Y ↪ X i:Y\hookrightarrow X i : Y ↪ X is continuous.
注意到对于任意X X X 的 open set A ∈ O ( X ) A\in\mathcal{O}(X) A ∈ O ( X ) ,有
i − 1 ( A ) = { y ∈ Y ∣ i ( y ) ∈ A } = { y ∈ Y ∣ y ∈ A } = Y ∩ A \begin{aligned}
i^{-1}(A)&=\{y\in Y\mid i(y)\in A\}\\
&=\{y\in Y\mid y\in A\}\\
&=Y\cap A
\end{aligned}
i − 1 ( A ) = { y ∈ Y ∣ i ( y ) ∈ A } = { y ∈ Y ∣ y ∈ A } = Y ∩ A
因此要求i i i continuous,至少要所有的i − 1 ( A ) = Y ∩ A i^{-1}(A)=Y\cap A i − 1 ( A ) = Y ∩ A 都在O ( Y ) \mathcal{O}(Y) O ( Y ) 中。不难验证,O ( Y ) = { Y ∩ A ∣ A ∈ O ( X ) } \mathcal{O}(Y)=\{Y\cap A\mid A\in\mathcal{O}(X)\} O ( Y ) = { Y ∩ A ∣ A ∈ O ( X ) } 也确实满足 openset axioms,所以它就是最小的一个。
□ \square □ .
Theorem . 对于 Euclidean space R m , R n \mathbb{R}^m,\mathbb{R}^n R m , R n ,如果有非空 open set ∅ ≠ U ∈ O ( R m ) , ∅ ≠ V ∈ O ( R n ) \emptyset\neq U\in\mathcal{O}(\mathbb{R}^m),\emptyset\neq V\in\mathcal{O}(\mathbb{R}^n) ∅ = U ∈ O ( R m ) , ∅ = V ∈ O ( R n ) 同胚 (homemorphism,即存在 continuous 的 bijection,记U ≈ V U\approx V U ≈ V ),那么有n = m n=m n = m 。
证明略(beyond course scope)
Exercise . If X X X is locally Euclidean at point p ∈ X p\in X p ∈ X , i.e. there exists an open neighborhood p ∈ U ⊂ X p\in U\subset X p ∈ U ⊂ X that is homemorphic to some open set in R n \mathbb{R}^n R n , then we can always take the open set as B ( 0 , 1 ) ⊂ R n B(0,1)\subset \mathbb{R}^n B ( 0 , 1 ) ⊂ R n .
注:B ( 0 , 1 ) B(0,1) B ( 0 , 1 ) 就是R n \mathbb{R}^n R n 中的原点附近半径为 1 的球形邻域。
假如说对于 open set p ∈ U ⊂ X p\in U\subset X p ∈ U ⊂ X 有φ ( U ) = ⋃ i ∈ I B ( a i , r i ) \varphi(U)=\bigcup_{i\in I}B(a_i,r_i) φ ( U ) = ⋃ i ∈ I B ( a i , r i ) 是R n \mathbb{R}^n R n 中的 openset。
那么不难发现,存在一个i i i 使得φ ( p ) ∈ B ( a i , r i ) \varphi(p)\in B(a_i,r_i) φ ( p ) ∈ B ( a i , r i ) ,不妨记φ ( p ) ∈ B ( a 0 , r 0 ) \varphi(p)\in B(a_0,r_0) φ ( p ) ∈ B ( a 0 , r 0 ) 。
因为φ , φ − 1 \varphi,\varphi^{-1} φ , φ − 1 都是 continuous 的,故φ − 1 ( B ( a 0 , r 0 ) ) \varphi^{-1}(B(a_0,r_0)) φ − 1 ( B ( a 0 , r 0 ) ) 是X X X 中的一个 open set。
那么不失一般性,我们不妨取p ∈ φ − 1 ( B ( a 0 , r 0 ) ) ⊂ X p\in \varphi^{-1}(B(a_0,r_0))\subset X p ∈ φ − 1 ( B ( a 0 , r 0 ) ) ⊂ X 代替p ∈ U ⊂ X p\in U\subset X p ∈ U ⊂ X ,此时φ − 1 ( B ( a 0 , r 0 ) ) ≈ B ( a 0 , r 0 ) ≈ B ( 0 , 1 ) \varphi^{-1}(B(a_0,r_0))\approx B(a_0,r_0)\approx B(0,1) φ − 1 ( B ( a 0 , r 0 ) ) ≈ B ( a 0 , r 0 ) ≈ B ( 0 , 1 ) 。
因此X X X 是 locally Euclidean 的,当且仅当对于任意 point p ∈ X p\in X p ∈ X ,存在一个 open neighborhood p ∈ U ⊂ X p\in U\subset X p ∈ U ⊂ X 使得U ≈ B ( 0 , 1 ) U\approx B(0,1) U ≈ B ( 0 , 1 ) 。
□ \square □ .
Exercise . If there is a continous function f : X → R f:X\rightarrow\mathbb{R} f : X → R such that ∀ p , q ∈ X . p ≠ q ⟹ f ( p ) ≠ f ( q ) \forall p,q\in X.\;p\neq q\implies f(p)\neq f(q) ∀ p , q ∈ X . p = q ⟹ f ( p ) = f ( q ) , then X X X is Hausdorff space.
很简单,对于p ≠ q ∈ X p\neq q\in X p = q ∈ X ,因为f ( p ) ≠ f ( q ) f(p)\neq f(q) f ( p ) = f ( q ) ,所以存在两个不相交的 intervals 满足:
f ( p ) ∈ ( a , b ) f ( q ) ∈ ( c , d ) ( a , b ) ∩ ( c , d ) = ∅ f(p)\in (a,b)\quad f(q)\in(c,d)\quad (a,b)\cap (c,d)=\emptyset
f ( p ) ∈ ( a , b ) f ( q ) ∈ ( c , d ) ( a , b ) ∩ ( c , d ) = ∅
注意到f f f 是 continous,因此f − 1 ( a , b ) , f − 1 ( c , d ) f^{-1}(a,b),f^{-1}(c,d) f − 1 ( a , b ) , f − 1 ( c , d ) 都是X X X 中的 open set,且p ∈ f − 1 ( a , b ) , q ∈ f − 1 ( c , d ) p\in f^{-1}(a,b),q\in f^{-1}(c,d) p ∈ f − 1 ( a , b ) , q ∈ f − 1 ( c , d ) 。
不难发现f − 1 ( a , b ) ∩ f − 1 ( c , d ) = ∅ f^{-1}(a,b)\cap f^{-1}(c,d)=\emptyset f − 1 ( a , b ) ∩ f − 1 ( c , d ) = ∅ 。因为如果不为空,就∃ x . f ( x ) ∈ ( a , b ) , f ( x ) ∈ ( c , d ) \exists x.\;f(x)\in (a,b),f(x)\in (c,d) ∃ x . f ( x ) ∈ ( a , b ) , f ( x ) ∈ ( c , d ) ,那么( a , b ) ∩ ( c , d ) ≠ ∅ (a,b)\cap (c,d)\neq\emptyset ( a , b ) ∩ ( c , d ) = ∅ 矛盾。
故X X X 是 Hausdorff 空间。
□ \square □ .
Exercise . If X X X is Hausdorff space, and Y ⊂ X Y\subset X Y ⊂ X is a sub topological space, then Y Y Y is Hausdorff.
简单的,对于任意p ≠ q ∈ Y p\neq q\in Y p = q ∈ Y ,存在U , V ∈ O ( X ) U,V\in\mathcal{O}(X) U , V ∈ O ( X ) 使得p ∈ U , q ∈ V , U ∩ V = ∅ p\in U,q\in V,U\cap V=\emptyset p ∈ U , q ∈ V , U ∩ V = ∅ 。
那么故就有U ∩ Y , V ∩ Y U\cap Y,V\cap Y U ∩ Y , V ∩ Y 满足
p ∈ U ∩ Y q ∈ V ∩ Y ( U ∩ Y ) ∩ ( V ∩ Y ) = ∅ p\in U\cap Y\;q\in V\cap Y\;(U\cap Y)\cap(V\cap Y)=\emptyset
p ∈ U ∩ Y q ∈ V ∩ Y ( U ∩ Y ) ∩ ( V ∩ Y ) = ∅
故Y Y Y 也是 Hausdorff。
□ \square □ .
Exercise . If X X X is Hausdorff space, and subset K ⊂ X K\subset X K ⊂ X is compact, then K K K is closed. i.e., X − K X-K X − K is open.
对于任意 point x ∈ X − K x\in X-K x ∈ X − K ,因为X X X 是 Hausdorff 的,那么对于任意y ∈ K y\in K y ∈ K 都有 open neighborhood U y , V y U_y,V_y U y , V y 使得
x ∈ U y y ∈ V y U y ∩ V y = ∅ x\in U_y\quad y\in V_y\quad U_y\cap V_y=\emptyset
x ∈ U y y ∈ V y U y ∩ V y = ∅
注意到,⋃ y ∈ K V y \bigcup _{y\in K}V_y ⋃ y ∈ K V y 是K K K 的一个 cover,因为K K K 是 compact 的,因此存在一个 finite cover K ⊂ ⋃ y ∈ F V y K\subset \bigcup_{y\in F}V_y K ⊂ ⋃ y ∈ F V y 。
那么⋂ y ∈ F U y \bigcap_{y\in F}U_y ⋂ y ∈ F U y 就是x x x 的一个 open neighborhood,而且和K K K disjoint。不妨记为A x A_x A x 。
因此⋃ x ∈ X − K A x \bigcup_{x\in X-K}A_x ⋃ x ∈ X − K A x 就是一个X − K X-K X − K 的 cover,而且和K K K 不相交。故X − K = ⋃ x ∈ X − K A x X-K=\bigcup_{x\in X-K}A_x X − K = ⋃ x ∈ X − K A x 是一个 open set。
□ \square □ .
(注意:单纯的把每个x , y x,y x , y 对应的U y U_y U y 并起来,虽然也是X − K X-K X − K 的一个 cover,但无法保证和K K K 不相交。)
Exercise . If φ : X → Y \varphi:X\rightarrow Y φ : X → Y is continous, and K ⊂ X K\subset X K ⊂ X is compact, then φ ( K ) \varphi(K) φ ( K ) is compact in Y Y Y .
不妨考虑φ ( K ) \varphi(K) φ ( K ) 的一个 open cover,即φ ( K ) ⊂ ⋃ i ∈ I U i \varphi(K)\subset \bigcup_{i\in I}U_i φ ( K ) ⊂ ⋃ i ∈ I U i 。
那么不难发现,有K ⊂ ⋃ i ∈ I φ − 1 ( U i ) K\subset \bigcup_{i\in I}\varphi^{-1}(U_i) K ⊂ ⋃ i ∈ I φ − 1 ( U i ) 。那么因为K K K compact,所以K ⊂ ⋃ i ∈ F φ − 1 ( U i ) K\subset \bigcup_{i\in F}\varphi^{-1}(U_i) K ⊂ ⋃ i ∈ F φ − 1 ( U i ) 。
因此φ ( K ) ⊂ ⋃ i ∈ F U i \varphi(K)\subset \bigcup_{i\in F}U_i φ ( K ) ⊂ ⋃ i ∈ F U i 。故φ ( K ) \varphi(K) φ ( K ) compact。
□ \square □ .
Exercise . If φ : X → Y \varphi:X\rightarrow Y φ : X → Y is a continuous bijection between compact space X X X and Hausdorff space Y Y Y , then φ \varphi φ is homeomorphism.
我们需要证明φ − 1 : Y → X \varphi^{-1}:Y\rightarrow X φ − 1 : Y → X 是 continuous 的,即要证明对于任意 open set U ⊂ X U\subset X U ⊂ X ,有φ ( U ) \varphi(U) φ ( U ) open in Y Y Y 。
也就等价为,对于任意 closed set U ⊂ X U\subset X U ⊂ X ,有φ ( U ) \varphi(U) φ ( U ) closed in Y Y Y 。
我们首先证明,a closed subset of compact set is compact。
对于 closed set U ⊂ X U\subset X U ⊂ X 和一个 open cover U ⊂ ⋃ i ∈ I U i U\subset\bigcup_{i\in I}U_i U ⊂ ⋃ i ∈ I U i ,那么X − U X-U X − U 是 open 的,那么X ⊂ ( X − U ) ∪ ⋃ i ∈ I U i X\subset (X-U)\cup\bigcup_{i\in I}U_i X ⊂ ( X − U ) ∪ ⋃ i ∈ I U i 。
因为X X X 是 compact,因此X ⊂ ( X − U ) ∪ ⋃ i ∈ F U i X\subset (X-U)\cup\bigcup_{i\in F}U_i X ⊂ ( X − U ) ∪ ⋃ i ∈ F U i 。
因此U ⊂ ⋃ i ∈ F U i U\subset \bigcup_{i\in F}U_i U ⊂ ⋃ i ∈ F U i (因为否则如果有u ∈ U , u ∉ ⋃ i ∈ F U i u\in U,u\notin \bigcup_{i\in F}U_i u ∈ U , u ∈ / ⋃ i ∈ F U i ,那么就有u ∈ X u\in X u ∈ X 但是u ∉ ( X − U ) ∪ ⋃ i ∈ F U i u\notin (X-U)\cup\bigcup_{i\in F}U_i u ∈ / ( X − U ) ∪ ⋃ i ∈ F U i ,矛盾)
因此U U U 是 compact 的。
所以有对于 closed set U ⊂ X U\subset X U ⊂ X ,有U U U 也是 compact 的。
再根据上个 exercise,有φ ( U ) \varphi(U) φ ( U ) is compact in Y Y Y 。
再根据上上个 exercise,有φ ( U ) \varphi(U) φ ( U ) is closed in Y Y Y 。
□ \square □ .
Exercise . If M M M is 0-dimensional topological manifold, then M M M is a countable set equipped with discrete topology.
0-dimensional topological manifold 意思就是对于M M M 中的每一个 point p p p ,都存在一个开邻域p ∈ U ⊂ M p\in U\subset M p ∈ U ⊂ M 同胚于一个点,即R 0 \mathbb{R}^0 R 0 。
注意到,对于任意 point p p p ,上述存在的开邻域中一定只存在p p p 一个 point。
因为如果对于某个U U U 含有两个不同的 point p ≠ q p\neq q p = q 且U U U 同胚于R 0 \mathbb{R}^0 R 0 ,那么因为M M M 是 Hausdorff 的,就存在U 1 , U 2 U_1,U_2 U 1 , U 2 使得p ∈ U 1 , q ∈ U 2 , U 1 ∩ U 2 = ∅ , U 1 ∪ U 2 ⊂ U p\in U_1,q\in U_2,U_1\cap U_2=\empty,U_1\cup U_2\subset U p ∈ U 1 , q ∈ U 2 , U 1 ∩ U 2 = ∅ , U 1 ∪ U 2 ⊂ U 。不难发现此时不可能有U ≈ R 0 U\approx \mathbb{R}^0 U ≈ R 0 。
因此对于M M M 中的每个点,都有存在一个 open set 只包含那一个点。故根据 open set axiom,O ( M ) = P ( M ) \mathcal{O}(M)=\mathcal{P}(M) O ( M ) = P ( M ) 。
那么显然地,最小的M M M 的 topological basis 就是B = { { x } : x ∈ M } \mathcal{B}=\{\{x\}: x\in M\} B = { { x } : x ∈ M } 。因为每个{ x } \{x\} { x } 都是 open set,而且它们无法写成其他任何集合的 union,故它们必须在 basis 里。
所以这告诉我们M M M 的 topological basis 应该是B \mathcal{B} B 再添加一些M M M 的 subset。
假如M M M is uncoutable,那么B \mathcal{B} B 也是 uncoutable 的。那么就违背了M M M 作为 manifold 是 second coutable 的条件。
故M M M 是 countable 的。
□ \square □ .
Exercise . If M M M is a n n n -dimensional topological manifold and M ′ ⊂ M M'\subset M M ′ ⊂ M is a open subset, then M ′ M' M ′ with sub topology is n n n -dimensional manifold.
实际上就想证明,manifold 要求的三个 property 都是可以被 open sub topology 保持的。
locally Euclidean。对于任意 point p p p ,存在一个 open set p ∈ U ⊂ M p\in U\subset M p ∈ U ⊂ M 使得U U U 同胚于R n \mathbb{R}^n R n 中的某个 open set,假设这个同胚是 continous function f , f − 1 f,f^{-1} f , f − 1 。
那么考虑M ′ M' M ′ 中,U ∩ M ′ U\cap M' U ∩ M ′ 是一个 open set in M M M ,那么f ( U ∩ M ′ ) f(U\cap M') f ( U ∩ M ′ ) 也是R n \mathbb{R}^n R n 中的 open set。(因为f − 1 f^{-1} f − 1 continous)
因此M ′ M' M ′ 中,对于任意一个 point p p p ,都存在 open set p ∈ U ∩ M ′ p\in U\cap M' p ∈ U ∩ M ′ ,且U ∩ M ′ U\cap M' U ∩ M ′ 同胚于R n \mathbb{R}^n R n 中的 open set f ( U ∩ M ′ ) f(U\cap M') f ( U ∩ M ′ ) 。
Hausdorff。这个更容易,对于M ′ M' M ′ 中不同两点p ≠ q p\neq q p = q ,那么因为M M M 是 Hausdorff 的,就存在U , V U,V U , V 使得
p ∈ U q ∈ V U ∩ V = ∅ p\in U\quad q\in V\quad U\cap V=\emptyset
p ∈ U q ∈ V U ∩ V = ∅
那么不难发现在M ′ M' M ′ 中有
p ∈ U ∩ M ′ q ∈ V ∩ M ′ U ∩ M ′ ∩ V ∩ M ′ = ∅ p\in U\cap M'\quad q\in V\cap M'\quad U\cap M'\cap V\cap M'=\emptyset
p ∈ U ∩ M ′ q ∈ V ∩ M ′ U ∩ M ′ ∩ V ∩ M ′ = ∅
故M ′ M' M ′ 也是 Hausdorff 的。
second coutable。这个也容易,对于M M M 的一个 countable basis B \mathcal{B} B ,那么下面就是一个M ′ M' M ′ 的 countable basis:
B ′ = { B ∩ M ′ ∣ B ∈ B } \mathcal{B}'=\{B\cap M'\mid B\in\mathcal{B}\}
B ′ = { B ∩ M ′ ∣ B ∈ B }
□ \square □ .
注意,2 和 3 都不需要要求M ′ M' M ′ 是 open 的,所以更弱,有时候只需要 check 1 即可。
譬如我们想 check S n − 1 = { ( x 1 , . . . , x n ) ∈ R n ∣ ∑ x i 2 = 1 } S^{n-1}=\{(x_1,...,x_n)\in\mathbb{R}^n\mid \sum x_i^2=1\} S n − 1 = { ( x 1 , . . . , x n ) ∈ R n ∣ ∑ x i 2 = 1 } 是 manifold 时,因为S n − 1 S^{n-1} S n − 1 是R n \mathbb{R}^n R n 的 sub topology,所以它已经是 Hausdorff 和 second countable 了。
但是由于S n − 1 S^{n-1} S n − 1 并不是一个R n \mathbb{R}^n R n 中的 open set,所以还需要验证S n − 1 S^{n-1} S n − 1 是 local Euclidean 的。
# Lecture 2
Exercise . Consider projective space R P n : = ( R n + 1 − { 0 } ) / ∼ \mathbb{R}P^n:=(\mathbb{R}^{n+1}-\{0\})/\sim R P n : = ( R n + 1 − { 0 } ) / ∼ where x → ∼ y → \overrightarrow{x}\sim\overrightarrow{y} x ∼ y ⟺ x → = λ y → \iff \overrightarrow{x}=\lambda\overrightarrow{y} ⟺ x = λ y for some λ ≠ 0 \lambda\neq 0 λ = 0 , (Intuitively, R P n \mathbb{R}P^n R P n is set of lines crossing the origin) and projection mapping π : R n − { 0 } → R P n \pi:\mathbb{R}^n-\{0\}\rightarrow\mathbb{R}P^n π : R n − { 0 } → R P n .
Then the topology on R P n \mathbb{R}P^n R P n could be defined as quotient topology: A ⊂ R P n A\subset\mathbb{R}P^n A ⊂ R P n is open if and only if π − 1 ( A ) \pi^{-1}(A) π − 1 ( A ) is open in R n − { 0 } \mathbb{R}^n-\{0\} R n − { 0 } .
Show that another quotient topology: A ⊂ R P n A\subset \mathbb{R}P^n A ⊂ R P n is open if and only if π ∣ S n − 1 ( A ) \pi|_{S^n}^{-1}(A) π ∣ S n − 1 ( A ) is open in S n S^n S n , is just the same as the quotien topology above.
我们证明,对于任意 subset A ⊂ R P n A\subset \mathbb{R}P^n A ⊂ R P n ,有π − 1 ( A ) \pi^{-1}(A) π − 1 ( A ) is open in R n + 1 − { 0 } \mathbb{R}^{n+1}-\{0\} R n + 1 − { 0 } 当且仅当π ∣ S n − 1 ( A ) \pi|_{S^n}^{-1}(A) π ∣ S n − 1 ( A ) is open in S n S^n S n 。
(直观地,可以把A A A 想成一些穿过原点的直线的集合)
注意到
π − 1 ( A ) = { x → ∈ R n + 1 − { 0 } ∣ [ x → ] ∈ A } π ∣ S n − 1 ( A ) = { x → ∈ S n ∣ [ x → ] ∈ A } \pi^{-1}(A)=\{\overrightarrow{x}\in\mathbb{R}^{n+1}-\{0\}\mid [\overrightarrow{x}]\in A\}\\
\pi|_{S^n}^{-1}(A)=\{\overrightarrow{x}\in S^n\mid [\overrightarrow{x}]\in A\}
π − 1 ( A ) = { x ∈ R n + 1 − { 0 } ∣ [ x ] ∈ A } π ∣ S n − 1 ( A ) = { x ∈ S n ∣ [ x ] ∈ A }
其中[ x → ] [\overrightarrow{x}] [ x ] 就是将点x → \overrightarrow{x} x 对应到它的等价类,即一条直线。那么不难发现
π − 1 ( A ) ∩ S n = π ∣ S n − 1 ( A ) \pi^{-1}(A)\cap S_n=\pi|_{S^n}^{-1}(A)
π − 1 ( A ) ∩ S n = π ∣ S n − 1 ( A )
又因为S n S_n S n 作为R n + 1 − { 0 } \mathbb{R}^{n+1}-\{0\} R n + 1 − { 0 } 的 sub topology,就有:
π ∣ S n − 1 ( A ) is open ⟺ π − 1 ( A ) ∩ S n is open ⟺ π − 1 ( A ) is open \pi|_{S^n}^{-1}(A)\text{ is open}\iff \pi^{-1}(A)\cap S^n\text{ is open}\iff \pi^{-1}(A)\text{ is open}
π ∣ S n − 1 ( A ) is open ⟺ π − 1 ( A ) ∩ S n is open ⟺ π − 1 ( A ) is open
所以证毕,这两个是相同的 quotient topology。
□ \square □ .
Exercise . R P n \mathbb{R}P^n R P n is Hausdorff and second coutable.
我们记S + n = S n / ∼ S^n_+=S^n/\sim S + n = S n / ∼ ,其中,x → ∼ y → ⟺ x → = ± y → \overrightarrow{x}\sim\overrightarrow{y}\iff \overrightarrow{x}=\pm\overrightarrow{y} x ∼ y ⟺ x = ± y 。在三维中可以想象把球削掉了一半。
那么其实很显然地,R P n ≈ S + n \mathbb{R}P^n\approx S^n_+ R P n ≈ S + n ,即π ∣ S + n \pi|_{S^n_+} π ∣ S + n 就是一个连续的双射。
因为S + n S^n_+ S + n 是S n S^n S n 的一个 subset,虽然它并不 open,但足以保持 Hausdorff 和 second countable 了。
□ \square □ .
Exercise . Suppose X X X is a locally path-connected space, prove that each component of X X X is open.
注意,locally path-connected 就是说,对于任意 point p ∈ X p\in X p ∈ X ,存在一个 open set p ∈ U ⊂ X p\in U\subset X p ∈ U ⊂ X 使得U U U 是 path-connected。而且 locally path-connected 无法推出X X X 是 path-connected 的。(直观想象,越靠近 component 边界的点,存在的 open set 越小)
那么对于任意一个 component(即 maximal path-connected subsets)U U U ,我们证明它是 open 的。
对于每一个点p ∈ U p\in U p ∈ U ,都存在一个 open subset p ∈ U p p\in U_p p ∈ U p 使得U p U_p U p 是 path-connected 的。显然U p ⊂ U U_p\subset U U p ⊂ U ,因为根据 maximal 的要求,U U U 需要包含所有和p p p path-connected 的点。
那么显然U = ⋃ p ∈ U U p U=\bigcup_{p\in U}U_p U = ⋃ p ∈ U U p ,是 open 的。
□ \square □ .
# Lecture 3
Exercise . Given smooth manifold M M M , prove that if chart ( U , φ ) (U,\varphi) ( U , φ ) is smooth, and U ′ ⊂ U U'\subset U U ′ ⊂ U is open, then ( U ′ , φ ∣ U ′ ) (U',\varphi|_{U'}) ( U ′ , φ ∣ U ′ ) is smooth.
根据 smooth structure 中 maximal 的构造(见前述 (smooth) atlas 中的 remark),实际上我们就是要证明,对于任意一个 chart ( V , ψ ) (V,\psi) ( V , ψ ) ,如果( U , φ ) (U,\varphi) ( U , φ ) is compatible with ( V , ψ ) (V,\psi) ( V , ψ ) ,那么( U ′ , φ ∣ U ′ ) (U',\varphi|_{U'}) ( U ′ , φ ∣ U ′ ) is compatible with ( V , ψ ) (V,\psi) ( V , ψ ) 。
这其实是显然的,也就是ψ ∘ φ − 1 \psi\circ\varphi^{-1} ψ ∘ φ − 1 是 smooth 的话,那么( ψ ∘ φ − 1 ) ∣ φ ( U ′ ∩ V ) (\psi\circ\varphi^{-1})|_{\varphi(U'\cap V)} ( ψ ∘ φ − 1 ) ∣ φ ( U ′ ∩ V ) 这个 restricted function 也是 smooth 的。
而且注意到,( ψ ∘ φ − 1 ) ∣ φ ( U ′ ∩ V ) = ψ ∘ ( φ ∣ U ′ ) − 1 (\psi\circ\varphi^{-1})|_{\varphi(U'\cap V)}=\psi\circ(\varphi|_{U'})^{-1} ( ψ ∘ φ − 1 ) ∣ φ ( U ′ ∩ V ) = ψ ∘ ( φ ∣ U ′ ) − 1 。故ψ ∘ φ ∣ U ′ − 1 \psi\circ\varphi|_{U'}^{-1} ψ ∘ φ ∣ U ′ − 1 也是 smooth,故( U ′ , φ ∣ U ′ ) (U',\varphi|_{U'}) ( U ′ , φ ∣ U ′ ) 和( V , ψ ) (V,\psi) ( V , ψ ) 也 smooth compatible。
另一方向的 smooth 同理。
□ \square □ .
Exercise . Given smooth manifold M M M , prove that if chart ( U , φ ) (U,\varphi) ( U , φ ) is smooth and χ : φ ( U ) ↦ χ ( φ ( U ) ) ⊂ R n \chi:\varphi(U)\mapsto \chi(\varphi(U))\subset\mathbb{R}^n χ : φ ( U ) ↦ χ ( φ ( U ) ) ⊂ R n is a diffeomorphism, then ( U , χ ∘ φ ) (U,\chi\circ\varphi) ( U , χ ∘ φ ) is smooth.
类似上一个 exercise,我们证明对于任意 chart ( V , ψ ) (V,\psi) ( V , ψ ) ,如果( U , φ ) (U,\varphi) ( U , φ ) is smooth compatible with ( V , ψ ) (V,\psi) ( V , ψ ) ,那么( U , χ ∘ φ ) (U,\chi\circ\varphi) ( U , χ ∘ φ ) is smooth compatible with ( V , ψ ) (V,\psi) ( V , ψ ) 。
首先,( U , χ ∘ φ ) (U,\chi\circ\varphi) ( U , χ ∘ φ ) 的确也是一个 chart。因为χ ∘ φ \chi\circ \varphi χ ∘ φ 是两个 homeomorphism 的复合,故也是 homeomorphism。
其次,我们已知ψ ∘ φ − 1 : φ ( U ∩ V ) → ψ ( U ∩ V ) \psi\circ\varphi^{-1}:\varphi(U\cap V)\to\psi(U\cap V) ψ ∘ φ − 1 : φ ( U ∩ V ) → ψ ( U ∩ V ) 和φ ∘ ψ − 1 : ψ ( U ∩ V ) → φ ( U ∩ V ) \varphi\circ\psi^{-1}:\psi(U\cap V)\to\varphi(U\cap V) φ ∘ ψ − 1 : ψ ( U ∩ V ) → φ ( U ∩ V ) 是 smooth 的,那么我们要证明:
ψ ∘ φ − 1 ∘ χ − 1 : χ ( φ ( U ∩ V ) ) → ψ ( U ∩ V ) χ ∘ φ ∘ ψ − 1 : ψ ( U ∩ V ) → χ ( φ ( U ∩ V ) ) \psi\circ\varphi^{-1}\circ\chi^{-1}:\chi(\varphi(U\cap V))\to \psi(U\cap V)\\
\chi\circ\varphi\circ\psi^{-1}:\psi(U\cap V)\to\chi(\varphi(U\cap V))
ψ ∘ φ − 1 ∘ χ − 1 : χ ( φ ( U ∩ V ) ) → ψ ( U ∩ V ) χ ∘ φ ∘ ψ − 1 : ψ ( U ∩ V ) → χ ( φ ( U ∩ V ) )
也是 smooth 的。不难发现,利用 composition of differentiable functions is differentiable 就直接得出了。
ψ ∘ φ − 1 ∘ χ − 1 = ( ψ ∘ φ − 1 ) ∘ χ − 1 \psi\circ\varphi^{-1}\circ\chi^{-1}=(\psi\circ\varphi^{-1})\circ\chi^{-1} ψ ∘ φ − 1 ∘ χ − 1 = ( ψ ∘ φ − 1 ) ∘ χ − 1 就是两个 differentiable function 的复合,故为 smooth。
□ \square □ .
Exercise . Given smooth manifold M M M , if U ⊂ M U\subset M U ⊂ M is open, and φ : U → R n \varphi:U\to \mathbb{R}^n φ : U → R n is injective and has the following property: for any p ∈ U p\in U p ∈ U there is an open neighborhood p ∈ U p ⊂ U p\in U_p\subset U p ∈ U p ⊂ U such that ( U p , φ ∣ U p ) (U_p,\varphi|_{U_p}) ( U p , φ ∣ U p ) is smooth.
Prove that ( U , φ ) (U,\varphi) ( U , φ ) is smooth.
这个定理实际上说明的是 smooth 的确是一个 local property。
首先,我们证明( U , φ ) (U,\varphi) ( U , φ ) 的确是一个 chart。也就是说,φ \varphi φ 是一个 homeomorphism。这是显然的,因为我们考虑的φ \varphi φ 的值域就是φ ( U ) \varphi(U) φ ( U ) ,并且φ \varphi φ 是一个单射。
其次,我们证明,对于任意一个 chart ( V , ψ ) (V,\psi) ( V , ψ ) ,如果∀ p ∈ U , ( U p , φ ∣ U p ) \forall p\in U,(U_p,\varphi|_{U_p}) ∀ p ∈ U , ( U p , φ ∣ U p ) is smooth compatible with ( V , ψ ) (V,\psi) ( V , ψ ) ,那么有( U , φ ) (U,\varphi) ( U , φ ) 也 smooth compatible with ( V , ψ ) (V,\psi) ( V , ψ ) 。
不难想象,因为ψ ∘ φ ∣ U p − 1 \psi\circ\varphi|_{U_p}^{-1} ψ ∘ φ ∣ U p − 1 在每一个p ∈ φ ( U p ∩ V ) p\in \varphi(U_p\cap V) p ∈ φ ( U p ∩ V ) 都是 infinitely differentiable 的,且
φ ( U ∩ V ) = φ ( ⋃ p ∈ U U p ∩ V ) = ⋃ p ∈ U φ ( U p ∩ V ) \varphi(U\cap V)=\varphi\left (\bigcup_{p\in U}U_p\cap V\right )=\bigcup_{p\in U}\varphi(U_p\cap V)
φ ( U ∩ V ) = φ ⎝ ⎛ p ∈ U ⋃ U p ∩ V ⎠ ⎞ = p ∈ U ⋃ φ ( U p ∩ V )
后一个等号是因为φ \varphi φ 是 injective。那么对于任意p ∈ φ ( U ∩ V ) p\in \varphi(U\cap V) p ∈ φ ( U ∩ V ) ,都有p ∈ φ ( U p ∩ V ) p\in\varphi(U_p\cap V) p ∈ φ ( U p ∩ V ) 且ψ ∘ φ ∣ U p − 1 \psi\circ\varphi|_{U_p}^{-1} ψ ∘ φ ∣ U p − 1 在点p p p 处是无限可导的。
那么ψ ∘ φ − 1 \psi\circ\varphi^{-1} ψ ∘ φ − 1 在点p p p 处也是无限可导,也就推出ψ ∘ φ − 1 \psi\circ\varphi^{-1} ψ ∘ φ − 1 是 smooth 的了。
总之就是根据定义,smooth 的含义就是在每一个点都无限可导,即它是一个 local property。
□ \square □ .