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

BP(X)\mathcal{B}\subset \mathcal{P}(X) is called a basis for the topology on XX if for every AXA\subset X,

A is open    A is a union of elemens in B    pA.  BB.  s.t.  pBA\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}

# Neighborhood

For point pWp\in W in topological space XX, WW is called a neighborhood of pp if there exists an open set such that pUWXp\in U\subset W\subset X.

# Hausdorff

Topological space XX is Hausdorff if for any p,qXp,q\in X and pqp\neq q, exists open subsets U,VXU,V\subset X such that pU,qV,UV=p\in U,q\in V,U\cap V=\emptyset.

# Compact

KXK\subset X is called compact if every open cover KiIUiK\subset\bigcup_{i\in I}U_i (in topogical space XX) has a finite subcover KiFUiK\subset\bigcup_{i\in F}U_i, F<|F|<\infty.

# Second countable

Topological space XX is called second countable if it has a coutable basis.

# Locally Euclidean

Topological space XX is called nn-dimensional locally Euclidean if for any point pXp\in X, there is an open subset pUXp\in U\subset X such that UU is homeomorphic to an open set in Rn\mathbb{R}^n.

With out loss of generality, the definition could be modified as UB(0,1)RnU\approx B(0,1)\in\mathbb{R}^n. (See exercise below)

# Topological manifold

A topological space MM is called nn-dimensional topological manifold if MM is

  • locally nn-dimensional Euclidean.
  • Hausdorff.
  • second countable.

# Coordinate chart

A (coordinate) chart on nn-dimensional manifold MM is a pair (U,φ)(U,\varphi) where UMU\subset M is an open set and φ:UU~\varphi:U\rightarrow \tilde{U} is a homeomorphism to an open subset in Rn\mathbb{R}^n. i.e.,

U~:=φ(U)openRn\tilde{U}:=\varphi(U)\mathop{\subset}_{\text{open}}\mathbb{R}^n

# (Path) connectivity (component)

Topological space XX is connected if the only subset that are both open and closed are ,X\emptyset,X.

Topological space XX is path-connected if p,qX\forall p,q\in X, there is a continuous path r:[0,1]Xr:[0,1]\rightarrow X with r(0)=p,r(1)=qr(0)=p,r(1)=q.

Remark: In manifold MM, MM is connected if and only if it’s path connected.

While in topological space XX, path-connected is a stronger requirement.

The component of XX is the maximal path-connected subsets of XX.

# Interior

The interior of a subset SXS\subset X in topological space XX is the union of all subsets of SS that are open in XX. i.e.,

IntS=AS,A is open in XA\text{Int}S=\bigcup_{A\subset S,A\text{ is open in }X}A

Remark: IntS\text{Int}S is the largest open subset of SS.

# Exhaustion by compact subsets

An exhaustion by compact subsets is an increasing sequence K1K2...XK_1\subset K_2\subset...\subset X such that:

  • i.  Ki\forall i.\;K_i is compact.
  • KiIntKi+1K_i\subset\text{Int}K_{i+1}.
  • i=1Ki=X\bigcup_{i=1}^{\infty}K_i=X.

Remark: Since XX is open, thus X=i=1IntKiX=\bigcup_{i=1}^{\infty}\text{Int}K_i.

# Limit points

The limit points for a subset AXA\subset X in topological space XX are points pp such that, for any open subset UU in XX,

pUX    qU.  qp and qAp\in U\subset X\implies \exists q\in U.\;q\neq p\text{ and }q\in A

Remark: An interior point is not necessarily a limit point, i.e. pA    pp\in A\cancel{\implies} p is a limit point for AA.

The discrete set X={a,b,c}X=\{a,b,c\} with topology P(A)\mathcal{P}(A). The interior point a{a}a\in \{a\} is not a limit point of {a}\{a\}.

# Closure

The closure of a subset AXA\subset X in topological space XX is all interior points together with limit points of AA, which is denoted as Aˉ=AlimA\bar{A}=A\cup \text{lim}A.

Remark: closure are closed sets.

we prove that XAˉX-\bar{A} is open.

Given any point pXAˉp\in X-\bar{A}, it’s not a limit point of AA, which means, exists open subset pUpXp\in U_p\subset X such that UpA=U_p\cap A=\emptyset.

Then pXAˉUp\bigcup_{p\in X-\bar{A}}U_p is a open cover of XAˉX-\bar{A} and does not intersect with AA.

Thus XAˉ=pXAˉUpX-\bar{A}=\bigcup_{p\in X-\bar{A}}U_p is open.

# Precompact

Subset AXA\subset X in topological space XX is precompact if its closure Aˉ\bar{A} is compact in XX.

# (Locally finite) cover

UP(X)\mathcal{U}\subset \mathcal{P}(X) is called a cover of XX if X=uUuX=\bigcup_{u\in\mathcal{U}}u.

Moreover, the cover is called locally finite if for any point pXp\in X, there exists a neighborhood pWXp\in W\subset X such that WW only intersects finitely many uUu\in\mathcal{U}.

# Refinement

Cover VP(X)\mathcal{V}\subset\mathcal{P}(X) is called a refinement of cover UP(X)\mathcal{U}\subset\mathcal{P}(X) if vV.  uU.  vu\forall v\in\mathcal{V}.\;\exists u\in\mathcal{U}.\;v\subset u.

# Paracompact

Topological space XX 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 VWV\to W between two vector spaces that preserves the operations of vector addition and scalar multiplication.

# Smooth (component function, partial derivative)

If UU and VV are open subsets of Euclidean spaces Rn\mathbb{R}^n and Rm\mathbb{R}^m, respectively, then a function F:UVF:U\rightarrow V is said to be smooth (or CC^\infty, or infinitely differentiable) if each of its component functions has continuous partial derivatives of all orders at any point in UU.

Notation: Suppose uRn\overrightarrow{u}\in\mathbb{R}^n and F(u)=(F1(u),...,Fm(m))RmF(\overrightarrow{u})=(F_1(\overrightarrow u),...,F_m(\overrightarrow{m}))\in\mathbb{R}^m, then each F1,..,FmF_1,..,F_m is a component function of FF.

Then F:RnRmF:\mathbb{R}^n\to\mathbb{R}^m is in C1C^1 means that,

Fiuj=limh0Fi(u1,...,uj+h,...,un)F(u1,...,uj,...,un)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}

is continuous for any i=1,...,mj=1,...,ni=1,...,m\quad j=1,...,n, and for any uU\overrightarrow{u}\in U.

Besides, FC2F\in C^2 means

2Fiujuk\frac{\partial^2 F_i}{\partial u_j\partial u_k}

is continuous for i=1,...,mj=1,...,nk=1,...,ni=1,...,m\quad j=1,...,n\quad k=1,...,n.

And so on. FF is smooth is defined as FCF\in C^{\infty}.

# Diffeomorphism

If smooth function FF 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) are smoothly compatible if both transition maps:

ψφ1:φ(UV)ψ(UV)φψ1:ψ(UV)φ(UV)\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)

are smooth.

# (Smooth) atlas

An atlas A\mathcal{A} of MM is a collection of charts such that M=(U,φ)AUM=\bigcup_{(U,\varphi)\in\mathcal{A}}U.

An atlas is called smooth if all charts in it are smoothly compatible.

An atlas A\mathcal{A} is maximal if there is no other atlas A\mathcal{A}' such that AA\mathcal{A}\subsetneq\mathcal{A}'.

Remark: Given any smooth atlas A\mathcal{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}\}

and the maximal construction is unique. That is, if every chart in A\mathcal{A} is smooth compatible with every chart in A\mathcal{A}', then A=A\overline{\mathcal{A}}=\overline{\mathcal{A}'}.

# Smooth structure

A maximal smooth altlas on a topological manifold MM is called a smooth structure on MM.

# Smooth manifold

A smooth manifold is a pair (Mn,A)(M^n,\mathcal{A}), where MnM^n is a nn-dimensional topological manifold and A\mathcal{A} is a smooth structure.

For short, we write M=(Mn,A)M=(M^n,\mathcal{A}), and say that chart (U,φ)(U,\varphi) is smooth if (U,φ)A(U,\varphi)\in\mathcal{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) is smooth compatible with all charts in A\mathcal{A}, then (U,φ)A(U,\varphi)\in\mathcal{A}.

In words, if chart is smooth compatible with all smooth charts, then it’s smooth.

# Lecture 1

Exercise. If X=Y=RX=Y=\mathbb{R}, verify that the continuous map defined by topological space is equivalent to εδ\varepsilon-\delta language.

一方面,

考虑f:RRf:\mathbb{R}\rightarrow\mathbb{R} 是一个 topological continuous function。那么意味着

(a,b)R.  f1(a,b)={xa<f(x)<b} is open\forall (a,b)\subset\mathbb{R}.\;f^{-1}(a,b)=\{x\mid a<f(x)<b\}\text{ is open}

所以,对于任意 point xRx\in\mathbb{R}ε>0\varepsilon>0,有(f(x)ε,f(x)+ε)(f(x)-\varepsilon,f(x)+\varepsilon)R\mathbb{R} 的一个 open set。因此存在

f1(f(x)ε,f(x)+ε)=iI(ai,bi)Rf^{-1}(f(x)-\varepsilon,f(x)+\varepsilon)=\bigcup_{i\in I}(a_i,b_i)\subset \mathbb{R}

f1(f(x)ε,f(x)+ε)f^{-1}(f(x)-\varepsilon,f(x)+\varepsilon) 是一个 open set。又因为不难验证xf1(f(x)ε,f(x)+ε)x\in f^{-1}(f(x)-\varepsilon,f(x)+\varepsilon),所以i.  x(ai,bi)\exists i.\;x\in(a_i,b_i)

所以我们取对应的(ai,bi)(a_i,b_i) 满足x(ai,bi)x\in (a_i,b_i)。那么存在一个δ=max(bix,xa)\delta=\max(b_i-x,x-a),使得有:

x(xδ,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)

即对应了εδ\varepsilon-\delta 语言:

ε>0.  δ>0.  x.  xx<δ    f(x)f(x)<ε\forall \varepsilon>0.\;\exists \delta>0.\;\forall x'.\;|x-x'|<\delta\implies |f(x)-f(x')|<\varepsilon

另一方面,

如果有εδ\varepsilon-\delta 语言,我们证明:任给一个 open set iI(ai,bi)R\bigcup_{i\in I}(a_i,b_i)\subset\mathbb{R},有f1(iI(ai,bi))f^{-1}\left (\bigcup_{i\in I}(a_i,b_i)\right ) 也是 open 的。

注意到:

f1(iI(ai,bi))={xRi.  ai<f(x)<bi}=iIf1(ai,bi)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)

因此我们证明,每个f1(ai,bi)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),根据实数的连续性,可以知道(a,b)=y(a,b)(yεy,y+εy)(a,b)=\bigcup_{y\in(a,b)}(y-\varepsilon_y,y+\varepsilon_y)。即开区间(a,b)(a,b) 可以由每个开区间中的元素的邻域覆盖。

那么xf1(a,b)\forall x\in f^{-1}(a,b),我们取y:=f(x)(a,b)y:=f(x)\in (a,b),那么根据εδ\varepsilon-\delta 语言,存在一个δx\delta_x 使得:

(xδx,x+δx)f1(yεy,y+εy)f1(a,b)(x-\delta_x,x+\delta_x)\subset f^{-1}(y-\varepsilon_y,y+\varepsilon_y)\subset f^{-1}(a,b)

那么有,对于xf1(a,b).  δx.  (xδx,x+δx)f1(a,b)\forall x\in f^{-1}(a,b).\; \exists \delta_x.\;(x-\delta_x,x+\delta_x)\subset f^{-1}(a,b)。即对于f1(a,b)f^{-1}(a,b) 中的每个元素,都存在一个开邻域。那么

f1(a,b)=xf1(a,b)(xδx,x+δx)f^{-1}(a,b)=\bigcup_{x\in f^{-1}(a,b)}(x-\delta_x,x+\delta_x)

也是 open 的。

\square.

Exercise. If f:XZf:X\rightarrow Z between two topological spaces is continuous, then for any sub topological space (Y,O(Y))(Y,\mathcal{O}(Y)) with

YX,  O(Y)={YAAO(X)}Y\subset X,\;\mathcal{O}(Y)=\{Y\cap A\mid A\in\mathcal{O}(X)\}

we have the restriction map fY:YZf|_Y:Y\rightarrow Z being continuous.

不难发现,对于任意ZZ 中 open set AA,有:

f1(A)Y={xYf(x)A}={xYfY(x)A}=fY1(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}

因为ff continuous,既然f1(A)O(X)f^{-1}(A)\in\mathcal{O}(X),那么fY1(A)=f1(A)YO(Y)f|_Y^{-1}(A)=f^{-1}(A)\cap Y\in\mathcal{O}(Y)。故fYf|_Y continuous。

\square.

Exercise. The subspace topology (Y,O(Y))(Y,\mathcal{O}(Y)) with YX,  O(Y)={YAAO(X)}Y\subset X,\;\mathcal{O}(Y)=\{Y\cap A\mid A\in\mathcal{O}(X)\} is the coarest topology on YY (i.e. fewest subsets) such that injection i:YXi:Y\hookrightarrow X is continuous.

注意到对于任意XX 的 open set AO(X)A\in\mathcal{O}(X),有

i1(A)={yYi(y)A}={yYyA}=YA\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}

因此要求ii continuous,至少要所有的i1(A)=YAi^{-1}(A)=Y\cap A 都在O(Y)\mathcal{O}(Y) 中。不难验证,O(Y)={YAAO(X)}\mathcal{O}(Y)=\{Y\cap A\mid A\in\mathcal{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 Rm,Rn\mathbb{R}^m,\mathbb{R}^n,如果有非空 open set UO(Rm),VO(Rn)\emptyset\neq U\in\mathcal{O}(\mathbb{R}^m),\emptyset\neq V\in\mathcal{O}(\mathbb{R}^n) 同胚 (homemorphism,即存在 continuous 的 bijection,记UVU\approx V),那么有n=mn=m

证明略(beyond course scope)

Exercise. If XX is locally Euclidean at point pXp\in X, i.e. there exists an open neighborhood pUXp\in U\subset X that is homemorphic to some open set in Rn\mathbb{R}^n, then we can always take the open set as B(0,1)RnB(0,1)\subset \mathbb{R}^n.

注:B(0,1)B(0,1) 就是Rn\mathbb{R}^n 中的原点附近半径为 1 的球形邻域。

假如说对于 open set pUXp\in U\subset Xφ(U)=iIB(ai,ri)\varphi(U)=\bigcup_{i\in I}B(a_i,r_i)Rn\mathbb{R}^n 中的 openset。

那么不难发现,存在一个ii 使得φ(p)B(ai,ri)\varphi(p)\in B(a_i,r_i),不妨记φ(p)B(a0,r0)\varphi(p)\in B(a_0,r_0)

因为φ,φ1\varphi,\varphi^{-1} 都是 continuous 的,故φ1(B(a0,r0))\varphi^{-1}(B(a_0,r_0))XX 中的一个 open set。

那么不失一般性,我们不妨取pφ1(B(a0,r0))Xp\in \varphi^{-1}(B(a_0,r_0))\subset X 代替pUXp\in U\subset X,此时φ1(B(a0,r0))B(a0,r0)B(0,1)\varphi^{-1}(B(a_0,r_0))\approx B(a_0,r_0)\approx B(0,1)

因此XX 是 locally Euclidean 的,当且仅当对于任意 point pXp\in X,存在一个 open neighborhood pUXp\in U\subset X 使得UB(0,1)U\approx B(0,1)

\square.

Exercise. If there is a continous function f:XRf:X\rightarrow\mathbb{R} such that p,qX.  pq    f(p)f(q)\forall p,q\in X.\;p\neq q\implies f(p)\neq f(q), then XX is Hausdorff space.

很简单,对于pqXp\neq q\in X,因为f(p)f(q)f(p)\neq 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

注意到ff 是 continous,因此f1(a,b),f1(c,d)f^{-1}(a,b),f^{-1}(c,d) 都是XX 中的 open set,且pf1(a,b),qf1(c,d)p\in f^{-1}(a,b),q\in f^{-1}(c,d)

不难发现f1(a,b)f1(c,d)=f^{-1}(a,b)\cap f^{-1}(c,d)=\emptyset。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),那么(a,b)(c,d)(a,b)\cap (c,d)\neq\emptyset 矛盾。

XX 是 Hausdorff 空间。

\square.

Exercise. If XX is Hausdorff space, and YXY\subset X is a sub topological space, then YY is Hausdorff.

简单的,对于任意pqYp\neq q\in Y,存在U,VO(X)U,V\in\mathcal{O}(X) 使得pU,qV,UV=p\in U,q\in V,U\cap V=\emptyset

那么故就有UY,VYU\cap Y,V\cap Y 满足

pUY  qVY  (UY)(VY)=p\in U\cap Y\;q\in V\cap Y\;(U\cap Y)\cap(V\cap Y)=\emptyset

YY 也是 Hausdorff。

\square.

Exercise. If XX is Hausdorff space, and subset KXK\subset X is compact, then KK is closed. i.e., XKX-K is open.

对于任意 point xXKx\in X-K,因为XX 是 Hausdorff 的,那么对于任意yKy\in K 都有 open neighborhood Uy,VyU_y,V_y 使得

xUyyVyUyVy=x\in U_y\quad y\in V_y\quad U_y\cap V_y=\emptyset

注意到,yKVy\bigcup _{y\in K}V_yKK 的一个 cover,因为KK 是 compact 的,因此存在一个 finite cover KyFVyK\subset \bigcup_{y\in F}V_y

那么yFUy\bigcap_{y\in F}U_y 就是xx 的一个 open neighborhood,而且和KKdisjoint。不妨记为AxA_x

因此xXKAx\bigcup_{x\in X-K}A_x 就是一个XKX-K 的 cover,而且和KK 不相交。故XK=xXKAxX-K=\bigcup_{x\in X-K}A_x 是一个 open set。

\square.

(注意:单纯的把每个x,yx,y 对应的UyU_y 并起来,虽然也是XKX-K 的一个 cover,但无法保证和KK 不相交。)

Exercise. If φ:XY\varphi:X\rightarrow Y is continous, and KXK\subset X is compact, then φ(K)\varphi(K) is compact in YY.

不妨考虑φ(K)\varphi(K) 的一个 open cover,即φ(K)iIUi\varphi(K)\subset \bigcup_{i\in I}U_i

那么不难发现,有KiIφ1(Ui)K\subset \bigcup_{i\in I}\varphi^{-1}(U_i)。那么因为KK compact,所以KiFφ1(Ui)K\subset \bigcup_{i\in F}\varphi^{-1}(U_i)

因此φ(K)iFUi\varphi(K)\subset \bigcup_{i\in F}U_i。故φ(K)\varphi(K) compact。

\square.

Exercise. If φ:XY\varphi:X\rightarrow Y is a continuous bijection between compact space XX and Hausdorff space YY, then φ\varphi is homeomorphism.

我们需要证明φ1:YX\varphi^{-1}:Y\rightarrow X 是 continuous 的,即要证明对于任意 open set UXU\subset X,有φ(U)\varphi(U) open in YY

也就等价为,对于任意 closed set UXU\subset X,有φ(U)\varphi(U) closed in YY

我们首先证明,a closed subset of compact set is compact。

对于 closed set UXU\subset X 和一个 open cover UiIUiU\subset\bigcup_{i\in I}U_i,那么XUX-U 是 open 的,那么X(XU)iIUiX\subset (X-U)\cup\bigcup_{i\in I}U_i

因为XX 是 compact,因此X(XU)iFUiX\subset (X-U)\cup\bigcup_{i\in F}U_i

因此UiFUiU\subset \bigcup_{i\in F}U_i(因为否则如果有uU,uiFUiu\in U,u\notin \bigcup_{i\in F}U_i,那么就有uXu\in X 但是u(XU)iFUiu\notin (X-U)\cup\bigcup_{i\in F}U_i,矛盾)

因此UU 是 compact 的。

所以有对于 closed set UXU\subset X,有UU 也是 compact 的。

再根据上个 exercise,有φ(U)\varphi(U) is compact in YY

再根据上上个 exercise,有φ(U)\varphi(U) is closed in YY

\square.

Exercise. If MM is 0-dimensional topological manifold, then MM is a countable set equipped with discrete topology.

0-dimensional topological manifold 意思就是对于MM 中的每一个 point pp,都存在一个开邻域pUMp\in U\subset M 同胚于一个点,即R0\mathbb{R}^0

注意到,对于任意 point pp,上述存在的开邻域中一定只存在pp 一个 point。

因为如果对于某个UU 含有两个不同的 point pqp\neq qUU 同胚于R0\mathbb{R}^0,那么因为MM 是 Hausdorff 的,就存在U1,U2U_1,U_2 使得pU1,qU2,U1U2=,U1U2Up\in U_1,q\in U_2,U_1\cap U_2=\empty,U_1\cup U_2\subset U。不难发现此时不可能有UR0U\approx \mathbb{R}^0

因此对于MM 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)

那么显然地,最小的MM 的 topological basis 就是B={{x}:xM}\mathcal{B}=\{\{x\}: x\in M\}。因为每个{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MM 的 topological basis 应该是B\mathcal{B} 再添加一些MM 的 subset。

假如MM is uncoutable,那么B\mathcal{B} 也是 uncoutable 的。那么就违背了MM 作为 manifold 是 second coutable 的条件。

MM 是 countable 的。

\square.

Exercise. If MM is a nn-dimensional topological manifold and MMM'\subset M is a open subset, then MM' with sub topology is nn-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。

  1. locally Euclidean。对于任意 point pp,存在一个 open set pUMp\in U\subset M 使得UU 同胚于Rn\mathbb{R}^n 中的某个 open set,假设这个同胚是 continous function f,f1f,f^{-1}

    那么考虑MM' 中,UMU\cap M' 是一个 open set in MM,那么f(UM)f(U\cap M') 也是Rn\mathbb{R}^n 中的 open set。(因为f1f^{-1}continous)

    因此MM' 中,对于任意一个 point pp,都存在 open set pUMp\in U\cap M',且UMU\cap M' 同胚于Rn\mathbb{R}^n 中的 open set f(UM)f(U\cap M')

  2. Hausdorff。这个更容易,对于MM' 中不同两点pqp\neq q,那么因为MM 是 Hausdorff 的,就存在U,VU,V 使得

    pUqVUV=p\in U\quad q\in V\quad U\cap V=\emptyset

    那么不难发现在MM' 中有

    pUMqVMUMVM=p\in U\cap M'\quad q\in V\cap M'\quad U\cap M'\cap V\cap M'=\emptyset

    MM' 也是 Hausdorff 的。

  3. second coutable。这个也容易,对于MM 的一个 countable basis B\mathcal{B},那么下面就是一个MM' 的 countable basis:

    B={BMBB}\mathcal{B}'=\{B\cap M'\mid B\in\mathcal{B}\}

\square.

注意,2 和 3 都不需要要求MM' 是 open 的,所以更弱,有时候只需要 check 1 即可。

譬如我们想 check Sn1={(x1,...,xn)Rnxi2=1}S^{n-1}=\{(x_1,...,x_n)\in\mathbb{R}^n\mid \sum x_i^2=1\} 是 manifold 时,因为Sn1S^{n-1}Rn\mathbb{R}^n 的 sub topology,所以它已经是 Hausdorff 和 second countable 了。

但是由于Sn1S^{n-1} 并不是一个Rn\mathbb{R}^n 中的 open set,所以还需要验证Sn1S^{n-1} 是 local Euclidean 的。

# Lecture 2

Exercise. Consider projective space RPn:=(Rn+1{0})/\mathbb{R}P^n:=(\mathbb{R}^{n+1}-\{0\})/\sim where xy\overrightarrow{x}\sim\overrightarrow{y}     x=λy\iff \overrightarrow{x}=\lambda\overrightarrow{y} for some λ0\lambda\neq 0, (Intuitively, RPn\mathbb{R}P^n is set of lines crossing the origin) and projection mapping π:Rn{0}RPn\pi:\mathbb{R}^n-\{0\}\rightarrow\mathbb{R}P^n.

Then the topology on RPn\mathbb{R}P^n could be defined as quotient topology: ARPnA\subset\mathbb{R}P^n is open if and only if π1(A)\pi^{-1}(A) is open in Rn{0}\mathbb{R}^n-\{0\}.

Show that another quotient topology: ARPnA\subset \mathbb{R}P^n is open if and only if πSn1(A)\pi|_{S^n}^{-1}(A) is open in SnS^n, is just the same as the quotien topology above.

我们证明,对于任意 subset ARPnA\subset \mathbb{R}P^n,有π1(A)\pi^{-1}(A) is open in Rn+1{0}\mathbb{R}^{n+1}-\{0\} 当且仅当πSn1(A)\pi|_{S^n}^{-1}(A) is open in SnS^n

(直观地,可以把AA 想成一些穿过原点的直线的集合)

注意到

π1(A)={xRn+1{0}[x]A}πSn1(A)={xSn[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\}

其中[x][\overrightarrow{x}] 就是将点x\overrightarrow{x} 对应到它的等价类,即一条直线。那么不难发现

π1(A)Sn=πSn1(A)\pi^{-1}(A)\cap S_n=\pi|_{S^n}^{-1}(A)

又因为SnS_n 作为Rn+1{0}\mathbb{R}^{n+1}-\{0\} 的 sub topology,就有:

πSn1(A) is open    π1(A)Sn is open    π1(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}

所以证毕,这两个是相同的 quotient topology。

\square.

Exercise. RPn\mathbb{R}P^n is Hausdorff and second coutable.

我们记S+n=Sn/S^n_+=S^n/\sim,其中,xy    x=±y\overrightarrow{x}\sim\overrightarrow{y}\iff \overrightarrow{x}=\pm\overrightarrow{y}。在三维中可以想象把球削掉了一半。

那么其实很显然地,RPnS+n\mathbb{R}P^n\approx S^n_+,即πS+n\pi|_{S^n_+} 就是一个连续的双射。

因为S+nS^n_+SnS^n 的一个 subset,虽然它并不 open,但足以保持 Hausdorff 和 second countable 了。

\square.

Exercise. Suppose XX is a locally path-connected space, prove that each component of XX is open.

注意,locally path-connected 就是说,对于任意 point pXp\in X,存在一个 open set pUXp\in U\subset X 使得UU 是 path-connected。而且 locally path-connected 无法推出XX 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)UU,我们证明它是 open 的。

对于每一个点pUp\in U,都存在一个 open subset pUpp\in U_p 使得UpU_p 是 path-connected 的。显然UpUU_p\subset U,因为根据 maximal 的要求,UU 需要包含所有和pp path-connected 的点。

那么显然U=pUUpU=\bigcup_{p\in U}U_p,是 open 的。

\square.

# Lecture 3

Exercise. Given smooth manifold MM, prove that if chart (U,φ)(U,\varphi) is smooth, and UUU'\subset U is open, then (U,φU)(U',\varphi|_{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),如果(U,φ)(U,\varphi) is compatible with (V,ψ)(V,\psi),那么(U,φU)(U',\varphi|_{U'}) is compatible with (V,ψ)(V,\psi)

这其实是显然的,也就是ψφ1\psi\circ\varphi^{-1} 是 smooth 的话,那么(ψφ1)φ(UV)(\psi\circ\varphi^{-1})|_{\varphi(U'\cap V)} 这个 restricted function 也是 smooth 的。

而且注意到,(ψφ1)φ(UV)=ψ(φU)1(\psi\circ\varphi^{-1})|_{\varphi(U'\cap V)}=\psi\circ(\varphi|_{U'})^{-1}。故ψφU1\psi\circ\varphi|_{U'}^{-1} 也是 smooth,故(U,φU)(U',\varphi|_{U'})(V,ψ)(V,\psi) 也 smooth compatible。

另一方向的 smooth 同理。

\square.

Exercise. Given smooth manifold MM, prove that if chart (U,φ)(U,\varphi) is smooth and χ:φ(U)χ(φ(U))Rn\chi:\varphi(U)\mapsto \chi(\varphi(U))\subset\mathbb{R}^n is a diffeomorphism, then (U,χφ)(U,\chi\circ\varphi) is smooth.

类似上一个 exercise,我们证明对于任意 chart (V,ψ)(V,\psi),如果(U,φ)(U,\varphi) is smooth compatible with (V,ψ)(V,\psi),那么(U,χφ)(U,\chi\circ\varphi) is smooth compatible with (V,ψ)(V,\psi)

首先,(U,χφ)(U,\chi\circ\varphi) 的确也是一个 chart。因为χφ\chi\circ \varphi 是两个 homeomorphism 的复合,故也是 homeomorphism。

其次,我们已知ψφ1:φ(UV)ψ(UV)\psi\circ\varphi^{-1}:\varphi(U\cap V)\to\psi(U\cap V)φψ1:ψ(UV)φ(UV)\varphi\circ\psi^{-1}:\psi(U\cap V)\to\varphi(U\cap V) 是 smooth 的,那么我们要证明:

ψφ1χ1:χ(φ(UV))ψ(UV)χφψ1:ψ(UV)χ(φ(UV))\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))

也是 smooth 的。不难发现,利用 composition of differentiable functions is differentiable 就直接得出了。

ψφ1χ1=(ψφ1)χ1\psi\circ\varphi^{-1}\circ\chi^{-1}=(\psi\circ\varphi^{-1})\circ\chi^{-1} 就是两个 differentiable function 的复合,故为 smooth。

\square.

Exercise. Given smooth manifold MM, if UMU\subset M is open, and φ:URn\varphi:U\to \mathbb{R}^n is injective and has the following property: for any pUp\in U there is an open neighborhood pUpUp\in U_p\subset U such that (Up,φUp)(U_p,\varphi|_{U_p}) is smooth.

Prove that (U,φ)(U,\varphi) 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) 的确是一个 chart。也就是说,φ\varphi 是一个 homeomorphism。这是显然的,因为我们考虑的φ\varphi 的值域就是φ(U)\varphi(U),并且φ\varphi 是一个单射。

其次,我们证明,对于任意一个 chart (V,ψ)(V,\psi),如果pU,(Up,φUp)\forall p\in U,(U_p,\varphi|_{U_p}) is smooth compatible with (V,ψ)(V,\psi),那么有(U,φ)(U,\varphi) 也 smooth compatible with (V,ψ)(V,\psi)

不难想象,因为ψφUp1\psi\circ\varphi|_{U_p}^{-1} 在每一个pφ(UpV)p\in \varphi(U_p\cap V) 都是 infinitely differentiable 的,且

φ(UV)=φ(pUUpV)=pUφ(UpV)\varphi(U\cap V)=\varphi\left (\bigcup_{p\in U}U_p\cap V\right )=\bigcup_{p\in U}\varphi(U_p\cap V)

后一个等号是因为φ\varphi 是 injective。那么对于任意pφ(UV)p\in \varphi(U\cap V),都有pφ(UpV)p\in\varphi(U_p\cap V)ψφUp1\psi\circ\varphi|_{U_p}^{-1} 在点pp 处是无限可导的。

那么ψφ1\psi\circ\varphi^{-1} 在点pp 处也是无限可导,也就推出ψφ1\psi\circ\varphi^{-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.