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秩定理
定理 1.1 (秩定理). 设光滑映照$F:M^m\rightarrow N^n$在$p$点邻域内秩为$k,$ 则$\,\exists\,$分别含$p,q=F(p)$的坐标系$(U,\varphi),(V,\psi),$ 使得$p,q$坐标为零且局部表示$\widehat{F}$为投影.
证: 先取定一个坐标系使得$p,q$坐标为零, 记$\widetilde{U}=\varphi(U),$
$\widetilde{V}=\varphi(V).$ 考虑$k=m\le n$的情形.
取局部表示$\widehat{F}$的Jacobi矩阵$D\widehat{F},$
不妨设其前$m$行在原点邻域是非退化的. 取
$$I_\varepsilon=\{(x^{m+1},\cdots,x^n)| |x^\gamma|\le \varepsilon, m+1\le \gamma\le n\}\subset \mathbb{R}^{n-m},$$
$$\widetilde{\psi}:\widetilde{U}\times I_\varepsilon\rightarrow \widetilde{V}$$
$$\begin{aligned}
\widetilde{\psi}^i(t)&=f^i(t^1,\cdots,t^m) & 1\le i\le m;\\
\widetilde{\psi}^i(t)&=t^i+f^i(t^1,\cdots,t^m) & m+1\le i\le n.
\end{aligned}$$
那么容易说明$\widetilde{\psi}$的Jacobi矩阵非退化. 由反函数定理,
不妨假设$\widetilde{\psi}$本身为微分同胚,
那么当然$\widetilde{\psi}^{-1}$也是.
考虑$\widetilde{\psi}^{-1}\circ\widehat{F},$
它是$(U,\varphi),(V,\widetilde{\psi}\circ\psi)$的局部表示.
它的Jacobi阵前$m$行为单位阵, 后$n-m$行为零. 结合其将原点映为原点的性质,
该局部表示即为投影. 从$\widetilde{\psi}^{-1}$的原始定义出发同样可以说明.
对$k=n$的情形, 构造合适的$\widetilde{\varphi}$即可. 对一般的情形,
同时改造两侧的坐标系即可给出证明.
特别地, 对该定理, 可以要求$\varphi(U)=C_\varepsilon^m(0),$
$\psi(V)=C_\varepsilon^n(0).$
$C_\varepsilon^r(0)$为$r$维$\varepsilon$边长的立方体.
取$0\le k\le m,$ 对$a\in \varphi(U),$
取$S=\{q\in U|x^i(q)=a^i, k+1\le i\le m\}.$ 赋予其诱导拓扑,
使之成为$k$维拓扑流形.
取$\widetilde{\varphi}=T\circ\varphi|_S:S\rightarrow T\circ\varphi(S)\subset \mathbb{R}^k$为同胚,
$T$为投影. 那么其有坐标系$(S,\widetilde{\varphi}),$
成为$k$维$C^\infty$流形, 称为$(U,\varphi)$的$k$维切片.
因此秩定理说明, $F(U)$是$V$内的$k$维切片, 在$U,V$像取立方体时.
子流形
回忆曲线: $C:(a,b)\rightarrow \mathbb{R}^3,$ $C'(t)\neq 0,$
即$r(C(t))=1;$
曲面: $F:\Omega\subset \mathbb{R}^2\rightarrow \mathbb{R}^3,$
$(u,v)\mapsto F(u,v).$ $F_u,F_v$线性无关, 即$r(F)=2.$
一般地, 设$F:M^m\rightarrow N^n$为$C^\infty$映照. 若$F$每点秩为$m,$
则称$(M,F)$为$N$中的浸入子流形. 如果$F$又是单射, 则称之为子流形.
浸入子流形在局部考虑都是子流形. 包含映射当然是秩$m$的,
因此$M$中开子集都是开子流形.
注意(浸入)子流形指$(M,F)$对, 而非像.
事实上有一个子流形与非子流形的浸入子流形像一致.
文章最后更新于 2021-10-20 16:23:00
rank theorem
Theorem 1.1 (rank theorem). Assume that the rank of the smooth mapping $F:M^m\rightarrow N^n$ in the neighborhood of $p$ is $k,$, then $\,\exists\,$ respectively contains the coordinate system $(U,\varphi),(V,\psi),$ of $p,q=F(p)$, so that the coordinates of $p,q$ are zero and the local representation $\widehat{F}$ is a projection.
Certificate: First determine a coordinate system so that the $p,q$ coordinate is zero, denoted as $\widetilde{U}=\varphi(U),$
$\widetilde{V}=\varphi(V).$ Consider the situation of $k=m\le n$.
Get the Jacobi matrix $D\widehat{F},$ of the local representation $\widehat{F}$
It may be assumed that the previous $m$ rows are non-degenerate in the neighborhood of the origin. Take
$$I_\varepsilon=\{(x^{m+1},\cdots,x^n)| |x^\gamma|\le \varepsilon, m+1\le \gamma\le n\}\subset \mathbb{R}^{n-m},$$
$$\widetilde{\psi}:\widetilde{U}\times I_\varepsilon\rightarrow \widetilde{V}$$
$$\begin{aligned}
\widetilde{\psi}^i(t)&=f^i(t^1,\cdots,t^m) & 1\le i\le m;\\
\widetilde{\psi}^i(t)&=t^i+f^i(t^1,\cdots,t^m) & m+1\le i\le n.
\end{aligned}$$
It is then easy to show that the Jacobi matrix of $\widetilde{\psi}$ is non-degenerate. According to the inverse function theorem,
Let’s assume that $\widetilde{\psi}$ itself is a diffeomorphism,
Then of course $\widetilde{\psi}^{-1}$ is too.
Consider $\widetilde{\psi}^{-1}\circ\widehat{F},$
It is a local representation of $(U,\varphi),(V,\widetilde{\psi}\circ\psi)$.
The front $m$ row of its Jacobi matrix is a unit matrix, and the back $n-m$ row is zero. Combined with its property of mapping the origin to the origin,
This local representation is the projection. It can also be explained starting from the original definition of $\widetilde{\psi}^{-1}$.
For the case of $k=n$, just construct the appropriate $\widetilde{\varphi}$. For the general case,
The proof can be given by transforming the coordinate systems on both sides at the same time.
In particular, for this theorem, it can be required that $\varphi(U)=C_\varepsilon^m(0),$
$\psi(V)=C_\varepsilon^n(0).$
$C_\varepsilon^r(0)$ is a cube with $r$ dimension and $\varepsilon$ side length.
Take $0\le k\le m,$ versus $a\in \varphi(U),$
Take $S=\{q\in U|x^i(q)=a^i, k+1\le i\le m\}.$ and give it the induced topology,
Make it a $k$-dimensional topological manifold.
Take $\widetilde{\varphi}=T\circ\varphi|_S:S\rightarrow T\circ\varphi(S)\subset \mathbb{R}^k$ as homeomorphism,
$T$ is the projection. Then it has the coordinate system $(S,\widetilde{\varphi}),$
It becomes a $k$-dimensional $C^\infty$ manifold, which is called a $k$-dimensional slice of $(U,\varphi)$.
Therefore, the rank theorem states that $F(U)$ is a $k$-dimensional slice within $V$, when $U,V$ is like a cube.
submanifold
Recall curve: $C:(a,b)\rightarrow \mathbb{R}^3,$ $C'(t)\neq 0,$
That is $r(C(t))=1;$
Surface: $F:\Omega\subset \mathbb{R}^2\rightarrow \mathbb{R}^3,$
$(u,v)\mapsto F(u,v).$ $F_u,F_v$ is linearly independent, that is, $r(F)=2.$
Generally, let $F:M^m\rightarrow N^n$ be the mapping of $C^\infty$. If $F$, the rank of each point is $m,$
Then $(M,F)$ is called $N$ in immersed submanifold. If $F$ is injective again, it is called submanifold.
Immersed submanifolds are all submanifolds when considered locally. Of course, the inclusion map is of rank $m$,
Therefore, the open subsets in $M$ are all open submanifolds.
Note that (immersed) submanifolds refer to $(M,F)$ pairs, not images.
There is actually a submanifold that is consistent with the immersed submanifold image of the non-submanifold.
The article was last updated on 2021-10-20 16:23:00
