《微分拓扑》复习笔记(3)-带边流形 "Differential Topology" Review Notes (3) — Manifolds with Boundary

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带边流形

$n$维拓扑带边流形是满足第二可数公理的Hausdorff空间, 且每点有一个邻域同胚于$H^n$的一个开集.

定理 1 (边界的拓扑不变性). $M$是带边拓扑流形, 则$M$上的点不能既是边界点又是内点. 于是$\partial M\cap \operatorname{Int}M=\varnothing,$ $\partial M\cup \operatorname{Int}M=M.$

定理 2 (Brouwer区域不变性定理). 设$U\subset \mathbb{R}^n$为$\mathbb{R}^n$中开集, $f:U\rightarrow \mathbb{R}^n$为单的连续映射, 则$f(U)$是$\mathbb{R}^n$中开集.

定理 3 (边界的微分不变性). 设$M$为光滑带边流形, $p\in M.$ 若存在$M$的光滑坐标卡$(U,\phi)$使$\phi(U)\subset H^n,$ $\phi(p)\in \partial H^n,$ 则任意含$p$的坐标卡都如此.

光滑情形利用逆映射定理即可.

命题 4. 若$M$为$n$维带边流形, 则$\operatorname{Int}M$是$n$维流形, $\partial M$是$(n-1)$维无边流形.

命题 5. 无边流形$M$和带边流形$N$的乘积是带边流形. 并且, $\partial(M\times N)=M\times \partial N,$ $\dim (M\times N)=\dim M+\dim N.$

引理 6. 设 $M$ 是 $m$ 维光滑无边流形, 设 $0$ 是光滑函数 $f:M\rightarrow \mathbb{R}$ 的正则值. 则 $M_0:=\{p\in M|f(p)\ge 0\}$ 是 $m$ 维带边流形, 且 $\partial M_0=f^{-1}(0).$

$\{p\in M|f(p)>0\}$为开集, 因此是$m$维开子流形, 只需确定边界. 由正则原像定理, $f^{-1}(0)$为$m-1$维嵌入子流形, 有坐标卡$(U,\phi)$使得$U\cap f^{-1}(0)$为切片.

由$f$为淹没, $\frac{\partial {}f}{\partial {}x_m}(p)\neq 0,$ $\,\forall\,p\in f^{-1}(0).$ 不妨设其为正, 那么$\,\forall\,q\in U_p,$ $q\in M_0$当且仅当$x_m(q)\ge 0.$ 于是$U_0:=U\cap M_0$为边界坐标卡.

带边版本定理

定理 7 (正则原像定理带边版本). 设$M$是$m$维光滑带边流形, $N$是$n$维光滑无边流形. 设$f:M\rightarrow N$是光滑映射. 若$q\in N$为$f,\partial f:=f|_{\partial M}$的正则值, 那么$S:=f^{-1}(q)$是$M$的$(m-n)$维子流形, $\partial S=S\cap \partial M.$

只需给出边界处的坐标系, 考虑$M=H^m,$ $N=\mathbb{R}^n,$ $q=0.$ 设$p\in f^{-1}(0)\cap \partial H^m.$ 由$f$光滑, 存在$p$处邻域$U\subset \mathbb{R}^m,$ $F:U\rightarrow \mathbb{R}^n$使得$F|_{U\cap H^m}=f|_{U\cap H^m}.$

由$0$为$\partial f$的正则值, $d(\partial f)_p=dF_p|_{T_x\partial M}$为满射. 设$dF_p|_{\operatorname{span}\{\frac{\partial {} }{\partial {}x_1},\cdots,\frac{\partial {} }{\partial {}x_n}\} }$为同构. 现取$U$上的坐标映射(类似淹没典范映射证明. 不能直接用典范表示的原因是确保边界$\{x_m=0\}$不动).

$$ \Phi:U\rightarrow \mathbb{R}^m,\quad \Phi(x)=(F(x),x_{n+1},\cdots,x_m). $$

此时 $$ d\Phi_p=\begin{bmatrix} dF_p|_{\operatorname{span}\{\frac{\partial {} }{\partial {}x_1},\cdots,\frac{\partial {} }{\partial {}x_n}\} }&0\\ *&I \end{bmatrix}, $$ 为同构, 因此由反函数定理, $\Phi$为局部微分同胚, 不妨设$\Phi:U\rightarrow V$为微分同胚.

此时$\Phi:U\cap S\mapsto V\cap (\{0\}\times H^{m-n}).$ 这就给出了$p$处的子流形边界坐标卡.

定理 8 (横截原像定理的带边版本). 设$M$为光滑带边流形, $N$为光滑无边流形. $f:M\rightarrow N$为光滑映射. 设$Z\subset N$为$N$中的无边子流形, 且$f\pitchfork Z,$ $\partial f\pitchfork Z,$ 则$S:=f^{-1}(Z)$为$M$中的带边子流形, 且$\partial (f^{-1}(Z))=f^{-1}(Z)\cap \partial M,$ $\operatorname{codim}_M f^{-1}(Z)=\operatorname{codim}_N Z.$

只需考虑边界处的坐标卡. 考虑$M=H^m,$ $N=\mathbb{R}^n,$ $Z=\mathbb{R}^{\dim Z}.$ 设$p\in f^{-1}(Z)\cap \partial H^m.$ 同上题取光滑映射$F.$

取投影$\pi:\mathbb{R}^n\rightarrow \mathbb{R}^{\operatorname{codim}Z},$ $Z=\pi^{-1}(0).$ 断言$0$为$\pi\circ F$的正则值. 只需证

$$ d\pi\circ dF_p T_xM=\mathbb{R}^{\operatorname{codim}Z}. $$

而这是因为

$$ dF_p T_xM+T_{F(x)}Z=\mathbb{R}^n,\quad d\pi (T_{F(x)}Z)=0. $$

同理$0$为$\pi\circ \partial f$的正则值. 从而由正则原像定理带边版本, $f^{-1}\circ \pi^{-1}(0)=f^{-1}(Z)=S$为$M$的$(m-\operatorname{codim}Z)$维子流形, $\partial S=S\cap \partial M,$ $\operatorname{codim}_M f^{-1}(Z)=\operatorname{codim}_N Z.$

定理 9 (带边流形的Sard定理). 设$f:M\rightarrow N$为带边流形$M$到无边流形$N$的光滑映射. $f,\partial f$的公共正则值集在$N$中稠密.

只需说明$f,\partial f$的临界值集零测. $f|_{\operatorname{Int}M},$ $\partial f$已有说明. 只需讨论$p\in \partial M$为$f$的临界点的情形. 然而此时它必是$\partial f$的临界点, 已有讨论. 这就给出了结论.

文章最后更新于 2023-06-11 15:06:16

Manifolds with Boundary

$n$-dimensional topological edge manifold is a Hausdorff space that satisfies the second countability axiom, And each point has a neighborhood homeomorphic to an open set of $H^n$.

Theorem 1 (Topological invariance of boundaries). $M$ is a topological manifold with edges, so the points on $M$ cannot be both boundary points and interior points. So $\partial M\cap \operatorname{Int}M=\varnothing,$ $\partial M\cup \operatorname{Int}M=M.$

Theorem 2 (Brouwer's zone invariance theorem). Assume $U\subset \mathbb{R}^n$ is the open set in $\mathbb{R}^n$, $f:U\rightarrow \mathbb{R}^n$ is a single continuous map, then $f(U)$ is the open set in $\mathbb{R}^n$.

Theorem 3 (Differential invariance of boundaries). Let $M$ be a smooth edged manifold, $p\in M.$. If there is a smooth coordinate card $(U,\phi)$ of $M$ such as $\phi(U)\subset H^n,$ $\phi(p)\in \partial H^n,$, then any coordinate card containing $p$ will be the same.

In the smooth case, just use the inverse mapping theorem.

Proposition 4. If $M$ is a $n$-dimensional edge manifold, then $\operatorname{Int}M$ is a $n$-dimensional manifold, and $\partial M$ is a $(n-1)$-dimensional edgeless manifold.

Proposition 5. The product of the edgeless manifold $M$ and the edged manifold $N$ is an edged manifold. And, $\partial(M\times N)=M\times \partial N,$ $\dim (M\times N)=\dim M+\dim N.$

Lemma 6. Let $M$ be a $m$-dimensional smooth edgeless manifold, and let $0$ be the regular value of the smooth function $f:M\rightarrow \mathbb{R}$. Then $M_0:=\{p\in M|f(p)\ge 0\}$ is a $m$-dimensional edged manifold, and $\partial M_0=f^{-1}(0).$

$\{p\in M|f(p)>0\}$ is an open set, so it is a $m$-dimensional open submanifold, and only the boundaries need to be determined. According to the regular primitive image theorem, $f^{-1}(0)$ is the $m-1$-dimensional embedded submanifold, There is a coordinate card $(U,\phi)$ that makes $U\cap f^{-1}(0)$ a slice.

From $f$ to submerged, $\frac{\partial {}f}{\partial {}x_m}(p)\neq 0,$ $\,\forall\,p\in f^{-1}(0).$ Let’s assume it is positive, then $\,\forall\,q\in U_p,$ $q\in M_0$ If and only if $x_m(q)\ge 0.$ then $U_0:=U\cap M_0$ is the boundary coordinate card.

The edged version of the theorem

Theorem 7 (Edged version of the regular prime image theorem). Let $M$ be a $m$-dimensional smooth edged manifold, $N$ be a $n$-dimensional smooth edgeless manifold. Let $f:M\rightarrow N$ be a smooth map. If $q\in N$ is the regular value of $f,\partial f:=f|_{\partial M}$, then $S:=f^{-1}(q)$ is the $(m-n)$-dimensional submanifold of $M$, $\partial S=S\cap \partial M.$

Just give the coordinate system at the boundary, consider $M=H^m,$ $N=\mathbb{R}^n,$ $q=0.$ Assume $p\in f^{-1}(0)\cap \partial H^m.$ is smoothed by $f$, There is $p$ neighborhood $U\subset \mathbb{R}^m,$ $F:U\rightarrow \mathbb{R}^n$ makes $F|_{U\cap H^m}=f|_{U\cap H^m}.$

From $0$ to the regular value of $\partial f$, $d(\partial f)_p=dF_p|_{T_x\partial M}$ is a surjection. Let $dF_p|_{\operatorname{span}\{\frac{\partial {} }{\partial {}x_1},\cdots,\frac{\partial {} }{\partial {}x_n}\} }$ be isomorphism. Now take the coordinate mapping on $U$ (similar to the submerged canonical mapping proof. The reason why it cannot be directly represented by canonical is to ensure that the boundary $\{x_m=0\}$ does not move).

$$ \Phi:U\rightarrow \mathbb{R}^m,\quad \Phi(x)=(F(x),x_{n+1},\cdots,x_m). $$

At this time $$ d\Phi_p=\begin{bmatrix} dF_p|_{\operatorname{span}\{\frac{\partial {} }{\partial {}x_1},\cdots,\frac{\partial {} }{\partial {}x_n}\} }&0\\ *&I \end{bmatrix}, $$ is isomorphism, so by the inverse function theorem, $\Phi$ is a local diffeomorphism. Let $\Phi:U\rightarrow V$ be a diffeomorphism.

At this time $\Phi:U\cap S\mapsto V\cap (\{0\}\times H^{m-n}).$ This gives the submanifold boundary coordinate card at $p$.

Theorem 8 (The edged version of the transversal preimage theorem). Let $M$ be a smooth edged manifold, $N$ be a smooth edgeless manifold. $f:M\rightarrow N$ be a smooth map. Let $Z\subset N$ be the edgeless submanifold in $N$, and $f\pitchfork Z,$ $\partial f\pitchfork Z,$ then $S:=f^{-1}(Z)$ be the edged submanifold in $M$, and $\partial (f^{-1}(Z))=f^{-1}(Z)\cap \partial M,$ $\operatorname{codim}_M f^{-1}(Z)=\operatorname{codim}_N Z.$

Just consider the coordinate cards at the boundaries. Consider $M=H^m,$ $N=\mathbb{R}^n,$ $Z=\mathbb{R}^{\dim Z}.$ Set $p\in f^{-1}(Z)\cap \partial H^m.$ Same as above to get smooth mapping $F.$

Get projection$\pi:\mathbb{R}^n\rightarrow \mathbb{R}^{\operatorname{codim}Z},$ $Z=\pi^{-1}(0).$ Assert that $0$ is the regular value of $\pi\circ F$. Just prove

$$ d\pi\circ dF_p T_xM=\mathbb{R}^{\operatorname{codim}Z}. $$

And this is because

$$ dF_p T_xM+T_{F(x)}Z=\mathbb{R}^n,\quad d\pi (T_{F(x)}Z)=0. $$

In the same way $0$ is the canonical value of $\pi\circ \partial f$. Therefore, from the edged version of the canonical primitive image theorem, $f^{-1}\circ \pi^{-1}(0)=f^{-1}(Z)=S$ is the $(m-\operatorname{codim}Z)$-dimensional submanifold of $M$, $\partial S=S\cap \partial M,$ $\operatorname{codim}_M f^{-1}(Z)=\operatorname{codim}_N Z.$

Theorem 9 (Sard’s theorem for manifolds with edges). Let $f:M\rightarrow N$ be a smooth mapping from the edged manifold $M$ to the edgeless manifold $N$. The set of common regular values of $f,\partial f$ is dense in $N$.

Just state that $f,\partial f$ is a zero measure of the set of critical values. $f|_{\operatorname{Int}M},$ $\partial f$ has been explained. We only need to discuss the situation where $p\in \partial M$ is the critical point of $f$. However, it must be the critical point of $\partial f$ at this time, which has been discussed. This gives the conclusion.

The article was last updated on 2023-06-11 15:06:16