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紧支撑光滑函数
首先考虑克莱因瓶如何嵌入到$\mathbb{R}^4$里,
这直接导致我们需要如下的工具.
在$\mathbb{R}^n$上, 取$\varphi(x)=\begin{cases}
e^{\frac{1}{|x|^2-1} },&|x|<1\\
0,&|x|\ge 1.
\end{cases},$
令$\alpha(x)=\frac{1}{\int_{\mathbb{R}^n} \varphi dx}\varphi(x).$
那么$\alpha(x)\in C_c^\infty(\mathbb{R}^n),$
$\operatorname{supp}\alpha=B_1,$ 满足$\int_{\mathbb{R}^n}\alpha dx=1.$
取$\alpha_\varepsilon=\frac{1}{\varepsilon^n}\alpha(\frac{x}{\varepsilon})\in C_c^\infty(\mathbb{R}^n),$
则$\operatorname{supp}\alpha_\varepsilon=B_\varepsilon,$
也满足$\int_{\mathbb{R}^n}\alpha_\varepsilon dx=1.$
定义卷积运算$f\ast g:=\int_{\mathbb{R}^n}f(y)g(x-y)dy.$ 那么对任意开集$E,$
$f=1_E\ast \alpha_\varepsilon=\int_E \alpha_\varepsilon(x-y)dy,$
满足如下性质:
$f\in [0,1];$
$\,\forall\,x$满足$d(x,E^c)< \varepsilon,$ $f(x)=1;$
$\operatorname{supp}f=\overline{O_\varepsilon(E)};$
$f\in C_c^\infty(\mathbb{R}^n).$
那么对流形上坐标卡$(U,\varphi),$ 倘若有开集$V\Subset W\Subset U,$
则$\varphi(V)\Subset \varphi(W)\Subset \varphi(U).$
记$\delta=\min(d(\overline{\varphi(V)},\varphi(W)^c),d(\overline{\varphi(W)},\varphi(U)^c))>0,$
那么取$f=(1_{\varphi(W)}\ast \alpha_{\frac{\delta}{2} })\circ \varphi,$
满足$f\in [0,1],$ $f|_{V}=1,$ 且$f\in C_c^\infty(U).$
紧流形的嵌入
基本思路是对每个坐标卡, 把坐标邻域分别嵌入到各自的$\mathbb{R}^m$里,
再留一维用来标识即可.
定理 1.1. 设$M$为$m$维紧致光滑流形, 则存在正整数$n,$ 以及光滑映射$\varphi:M\rightarrow \mathbb{R}^n,$ 使得$(M,\varphi)$是$\mathbb{R}^n$的正则子流形.
证: 取$M$的坐标图册$\{(U_i,\varphi_i;u_i^j)\}_{i=1}^k,$
由于是紧流形, 可取到有限坐标图册.
取$M\setminus\bigcup_{j=2}^k U_j\subset V_1\Subset W_1\Subset U_1,$
归纳地取$M\setminus(\bigcup_{j=1}^{i-1}V_j\cup\bigcup_{j=i+1}^k U_j)\subset V_i\Subset W_i\Subset U_i,$
得到加细$\{V_i\}$覆盖$M,$ 且对每个$V_i$由前面的说明,
存在$f_i\in C_c^\infty(U_i),$ $f_i|_{V_i}=1.$ 可以将$f_i$零延拓到$M$上.
在$M$上定义$n=k(m+1)$个光滑函数: $$\begin{aligned}
x_i^0&=f_i\\
x_i^j&=f_i \cdot u_{i}^j\\
\end{aligned}$$ 其中$i=1,\cdots,k,$ $j=1,\cdots,m.$
虽然每个函数只定义在$U_i$上, 但是同样可零延拓到整个$M$上.
这给出了光滑映射$\Phi:M\rightarrow \mathbb{R}^n,$
断言$(M,\Phi)$即为$\mathbb{R}^n$的正则子流形.
首先它确实是子流形: 先验证秩, 由于$\,\forall\,p\in M,$
$\,\exists\,V_i\supset O_p\ni p,$ 那么$f_i|_{O_p}\equiv 1,$
从而$\det \frac{\partial(x^{1}_i,\cdots,x^m_i)}{\partial (u_i^1,\cdots,u_i^m)}=1,$
$r(\Phi)=m.$ 因此$(M,\Phi)$是浸入子流形.
接下来若$\Phi(p)=\Phi(q),$ 那么倘若$V_i\supset p,$
$f_i(p)=1=f_i(q)\Rightarrow q\in V_i.$ 从而$\varphi_i(p)=\varphi_i(q),$
由于$\varphi_i$为同胚, $p=q.$ 因此$\Phi$是单射, $(M,\Phi)$是子流形.
由于$M$紧, 由上节课定理, $(M,\Phi)$自动是嵌入子流形,
而由另一定理又说明$(M,\Phi)$也是正则子流形.
维数论
通过这种方式得到的嵌入显然有些暴力, 多余的维数太多了. 以$S^2$为例,
至少要用两个坐标卡来描述,
此时按照上面的方法里要嵌入到$n=2\times(2+1)=6$维空间里.
倘若用标准的六个坐标卡来描述甚至要嵌到$18$维空间里.
事实上我们可以用维数论的方法将外围空间维数降低. 首先需要介绍Sard定理:
定理 1.2 (Sard定理). 设有光滑映照$f:M^m\rightarrow N^n,$ 则$N$上临界值全体零测.
临界值即为$M$上临界点的像, 临界点指使得$r(f)<n$的点,
即$f$在临界点非淹没.
定理 1.3. 若$m$维光滑流形$M$可单浸入$\mathbb{R}^K$中, $K>2m+1,$ 那么它也可以单浸入$\mathbb{R}^{K-1}$中.
证: 记单浸入为$\Phi,$
$\mathbb{R}^K$中所有过原点的超平面$P$全体为$\mathbb{R}\mathrm{P}^{K-1},$
它是一个$K-1$维光滑流形.
我们说明使得$\pi_P\circ \Phi$不是单浸入的$P$在$\mathbb{R}\mathrm{P}^{K-1}$上是零测集.
若其不是单射, 则$M$上有两点$p,q,$
使得$P$在原点法线$[v]=[\Phi(p)-\Phi(q)]$决定. 从而$[v]$落在
$$\alpha:M\times M\setminus \Delta_M\rightarrow \mathbb{R}\mathrm{P}^{K-1}, \quad(p,q)\mapsto[\Phi(p)-\Phi(q)]$$
的像集中, 其中$\Delta_M=\{(p,p)|p\in M\}.$ 那么$\alpha$是光滑映射,
$M\times M\setminus \Delta_M$为$2m$维开子流形, 由假设$2m<K-1,$
因此$\alpha$处处非淹没, 从而像集零测.
若其不是浸入, 则$\,\exists\,p\in M,$
$0\neq X_p\in T_pM$使得$(d\pi_P)_{\Phi(p)} (d\Phi)_p(X_p)\neq 0.$
而$\pi_P$本身是线性映照, $d\pi_P=\pi_P,$
因此$[(d\Phi)_p(X_p)]=[v]\neq 0.$ 从而$[v]$落在
$$\beta:TM\setminus\{(p,0)\,|\,p\in M\}\rightarrow \mathbb{R}\mathrm{P}^{K-1}, \quad(p,X_p)\mapsto[(d\Phi)_p(X_p)]$$
的像集中. $\beta$是光滑映射, 左侧同样为$2m$维开子流形,
由Sard定理同理有像集零测.
综上不是单浸入的$P$在$\mathbb{R}\mathrm{P}^{K-1}$上零测,
于是几乎处处可以取到投影将维数降低.
于是我们得到若$M$可单浸入$\mathbb{R}^K$中,
则必可单浸入$\mathbb{R}^{2m+1}$中. 事实上对不是浸入的证明,
可将$TM$更换为球面丛$SM=\{(p,X_p)\,|\,|X_p|=1\},$
切丛度量由外围空间诱导. 此时$SM$是$2m-1$维光滑流形, 维数可以再降低.
因此当$M$可嵌入$\mathbb{R}^{2m+1}$时, $M$可浸入$\mathbb{R}^{2m}.$
综上, 任意$m$维光滑紧流形可嵌入$\mathbb{R}^{2m+1}$中,
可浸入$\mathbb{R}^{2m}$中.
Whitney嵌入定理
对于非紧情形, 我们给出证明的思路.
定理 1.4. 任意非紧光滑流形$M$可单浸入到$\mathbb{R}^K$中.
我们将$M$划分为可列多个紧子流形的并$M=\cup_{i}{M_i}$,
保证$M_i\cap M_j=\varnothing,$ 若$|j-i|\ge 2.$
这样设$M_i$上有到$\mathbb{R}^{2m+1}$的嵌入$\varphi_i,$
取$M_i$小邻域的单位分解$\rho_i$, 则
$$\left(\sum_{i-odd}\rho_i \varphi_i, \sum_{i-even}\rho_i \varphi_i, f\right)$$
给出了想要的单浸入, $f$为控制函数, 提供了$M_i$的划分方法.
这样$M^m$可单浸入到$\mathbb{R}^{2m+1}$中, 不过不见得是嵌入.
不过可以改造单浸入成为嵌入, 利用如下命题,
即Lee的光滑流形导论中的命题4.22(b):
命题 1.5. 逆紧的光滑单浸入是嵌入. (像空间是局部紧$T_2$的即可, 此时逆紧连续映射是闭映射.)
将单浸入改造为逆紧的单浸入, 就得到了经典的Whitney嵌入定理.
定理 1.6 (Whitney嵌入定理). $m$维光滑流形$M$可嵌入$\mathbb{R}^{2m+1},$ 可浸入$\mathbb{R}^{2m}.$
其它结果
应用Whitney trick, 可以证明如下定理:
定理 1.7 (强Whitney嵌入定理). 任意$m\ge 2$维以上光滑流形可嵌入$\mathbb{R}^{2m}$中, 可浸入$\mathbb{R}^{2m-1}$中.
还有许多更细致的结果如下:
$m$维光滑紧定向流形可嵌入$\mathbb{R}^{2m-1}$中.
$m\neq 2^k$时, $m$维光滑流形可嵌入$\mathbb{R}^{2m-1}$中. $m=2^k$时,
$\mathbb{R}\mathrm{P}^m$不可嵌入$\mathbb{R}^{2m-1}$中.
$m$维光滑流形可浸入$\mathbb{R}^{2m-a(m)}$中,
$a(m)$为$m$二进制表示中$1$的数目, $m\ge 2.$
最后一个结果出自The Immersion Conjecture for Differentiable Manifolds,
Ralph L. Cohen. 方法源于E.H.Brown与F.P.Peterson的工作.
事实上由障碍性理论(Smale-Hirsch theory),
浸入的唯一阻碍是Stiefel-Whitney示性类,
由此可以说明对$m=2^{i_1}+\cdots+2^{i_l},$
取$M^m=\mathbb{R}\mathrm{P}^{2^{i_1} }\times\cdots\times \mathbb{R}\mathrm{P}^{2^{i_l} },$
$M^m$浸入的最低维数为$\sum_j(2i_j-1)=2m-a(m).$ 因此该结果是最佳的.
文章最后更新于 2021-11-01 17:27:27
Compactly supported smooth function
First consider how the Klein bottle is embedded in $\mathbb{R}^4$,
This directly leads us to need the following tools.
On $\mathbb{R}^n$, take $\varphi(x)=\begin{cases}
e^{\frac{1}{|x|^2-1} },&|x|<1\\
0,&|x|\ge 1.
\end{cases},$
Order$\alpha(x)=\frac{1}{\int_{\mathbb{R}^n} \varphi dx}\varphi(x).$
Then $\alpha(x)\in C_c^\infty(\mathbb{R}^n),$
$\operatorname{supp}\alpha=B_1,$ Satisfy$\int_{\mathbb{R}^n}\alpha dx=1.$
Take $\alpha_\varepsilon=\frac{1}{\varepsilon^n}\alpha(\frac{x}{\varepsilon})\in C_c^\infty(\mathbb{R}^n),$
Then $\operatorname{supp}\alpha_\varepsilon=B_\varepsilon,$
Also satisfied $\int_{\mathbb{R}^n}\alpha_\varepsilon dx=1.$
Define the convolution operation $f\ast g:=\int_{\mathbb{R}^n}f(y)g(x-y)dy.$ then for any open set $E,$
$f=1_E\ast \alpha_\varepsilon=\int_E \alpha_\varepsilon(x-y)dy,$
Satisfy the following properties:
$f\in [0,1];$
$\,\forall\,x$ satisfies $d(x,E^c)< \varepsilon,$ $f(x)=1;$
$\operatorname{supp}f=\overline{O_\varepsilon(E)};$
$f\in C_c^\infty(\mathbb{R}^n).$
Then on the manifold coordinate card $(U,\varphi),$, if there is an open set $V\Subset W\Subset U,$
Then $\varphi(V)\Subset \varphi(W)\Subset \varphi(U).$
Note$\delta=\min(d(\overline{\varphi(V)},\varphi(W)^c),d(\overline{\varphi(W)},\varphi(U)^c))>0,$
Then take $f=(1_{\varphi(W)}\ast \alpha_{\frac{\delta}{2} })\circ \varphi,$
Satisfies $f\in [0,1],$ $f|_{V}=1,$ and $f\in C_c^\infty(U).$
embedding of compact manifolds
The basic idea is to embed the coordinate neighborhoods into their respective $\mathbb{R}^m$ for each coordinate card.
Just leave one more dimension for identification.
Theorem 1.1. Assume $M$ is a $m$-dimensional compact smooth manifold, then there are positive integers $n,$ and smooth maps $\varphi:M\rightarrow \mathbb{R}^n,$ such that $(M,\varphi)$ is a regular submanifold of $\mathbb{R}^n$.
Certificate: Get the coordinate atlas $\{(U_i,\varphi_i;u_i^j)\}_{i=1}^k,$ of $M$
Since it is a compact manifold, a finite coordinate atlas can be obtained.
Take $M\setminus\bigcup_{j=2}^k U_j\subset V_1\Subset W_1\Subset U_1,$
Take inductively $M\setminus(\bigcup_{j=1}^{i-1}V_j\cup\bigcup_{j=i+1}^k U_j)\subset V_i\Subset W_i\Subset U_i,$
Obtain thinning $\{V_i\}$ covering $M,$ and for each $V_i$ according to the previous description,
There is $f_i\in C_c^\infty(U_i),$ $f_i|_{V_i}=1.$ that can extend $f_i$ zero to $M$.
Define $n=k(m+1)$ smooth functions on $M$: $$\begin{aligned}
x_i^0&=f_i\\
x_i^j&=f_i \cdot u_{i}^j\\
\end{aligned}$$ where $i=1,\cdots,k,$ $j=1,\cdots,m.$
Although each function is only defined on $U_i$, it can also be extended to the entire $M$.
This gives a smooth map $\Phi:M\rightarrow \mathbb{R}^n,$
Assert that $(M,\Phi)$ is the regular submanifold of $\mathbb{R}^n$.
First of all, it is indeed a submanifold: first verify the rank, since $\,\forall\,p\in M,$
$\,\exists\,V_i\supset O_p\ni p,$ Then $f_i|_{O_p}\equiv 1,$
Thus $\det \frac{\partial(x^{1}_i,\cdots,x^m_i)}{\partial (u_i^1,\cdots,u_i^m)}=1,$
$r(\Phi)=m.$ Therefore $(M,\Phi)$ is an immersed submanifold.
Next if $\Phi(p)=\Phi(q),$ then if $V_i\supset p,$
$f_i(p)=1=f_i(q)\Rightarrow q\in V_i.$ thus $\varphi_i(p)=\varphi_i(q),$
Since $\varphi_i$ is a homeomorphism, $p=q.$ therefore $\Phi$ is an injective, and $(M,\Phi)$ is a submanifold.
Since $M$ is compact, according to the theorem of the previous lesson, $(M,\Phi)$ is automatically an embedded submanifold,
And another theorem shows that $(M,\Phi)$ is also a regular submanifold.
Dimension theory
The embedding obtained in this way is obviously a bit violent and has too many redundant dimensions. Take $S^2$ as an example,
At least two coordinate cards must be used to describe,
At this time, it needs to be embedded into the $n=2\times(2+1)=6$ dimensional space according to the above method.
If it is described using standard six coordinate cards, it would even have to be embedded in $18$ dimensional space.
In fact, we can use dimension theory to reduce the dimension of the peripheral space. First we need to introduce Sard's theorem:
Theorem 1.2 (Sard’s theorem). Assuming smooth mapping $f:M^m\rightarrow N^n,$, then the upper critical value of $N$ is all zero.
The critical value is the image of the critical point on $M$, and the critical point refers to the point making $r(f)<n$,
That is, $f$ is not submerged at the critical point.
Theorem 1.3. If the $m$-dimensional smooth manifold $M$ can be single-immersed in $\mathbb{R}^K$, $K>2m+1,$, then it can also be single-immersed in $\mathbb{R}^{K-1}$.
Certificate: Record the single immersion as $\Phi,$
All hyperplanes $P$ in $\mathbb{R}^K$ passing through the origin are $\mathbb{R}\mathrm{P}^{K-1},$
It is a $K-1$-dimensional smooth manifold.
We show that $P$ such that $\pi_P\circ \Phi$ is not single-immersed has a zero measure set on $\mathbb{R}\mathrm{P}^{K-1}$.
If it is not injective, then there are two points $p,q,$ on $M$
So that $P$ is determined by the origin normal $[v]=[\Phi(p)-\Phi(q)]$. Thus $[v]$ falls on
$$\alpha:M\times M\setminus \Delta_M\rightarrow \mathbb{R}\mathrm{P}^{K-1}, \quad(p,q)\mapsto[\Phi(p)-\Phi(q)]$$
image concentration, where $\Delta_M=\{(p,p)|p\in M\}.$ then $\alpha$ is a smooth mapping,
$M\times M\setminus \Delta_M$ is the $2m$-dimensional open submanifold, by assuming $2m<K-1,$
Therefore $\alpha$ is not submerged everywhere, thus it is like a zero-measurement set.
If it is not immersed, then $\,\exists\,p\in M,$
$0\neq X_p\in T_pM$ makes $(d\pi_P)_{\Phi(p)} (d\Phi)_p(X_p)\neq 0.$
And $\pi_P$ itself is a linear mapping, $d\pi_P=\pi_P,$
Therefore $[(d\Phi)_p(X_p)]=[v]\neq 0.$ and thus $[v]$ fall in
$$\beta:TM\setminus\{(p,0)\,|\,p\in M\}\rightarrow \mathbb{R}\mathrm{P}^{K-1}, \quad(p,X_p)\mapsto[(d\Phi)_p(X_p)]$$
The image is concentrated. $\beta$ is a smooth mapping, and the left side is also a $2m$-dimensional open submanifold.
In the same way as Sard's theorem, there is a zero measure of the image set.
To sum up, it is not a zero test of single immersion $P$ on $\mathbb{R}\mathrm{P}^{K-1}$.
Therefore, projection can be used almost everywhere to reduce the dimensionality.
So we get that if $M$ can be immersed in $\mathbb{R}^K$ alone,
Then it must be immersed in $\mathbb{R}^{2m+1}$ alone. In fact, for the proof that it is not immersed,
$TM$ can be replaced by spherical bundle $SM=\{(p,X_p)\,|\,|X_p|=1\},$
The tangent bundle metric is induced by the peripheral space. At this time $SM$ is a $2m-1$-dimensional smooth manifold, and the dimension can be further reduced.
So when $M$ can be embedded in $\mathbb{R}^{2m+1}$, $M$ can be immersed in $\mathbb{R}^{2m}.$
In summary, any $m$-dimensional smooth compact manifold can be embedded in $\mathbb{R}^{2m+1}$,
Can be immersed in $\mathbb{R}^{2m}$.
Whitney embedding theorem
For the non-compact case, we give the idea of proof.
Theorem 1.4. Any non-compact smooth manifold $M$ can be simply immersed in $\mathbb{R}^K$.
We divide $M$ into a union $M=\cup_{i}{M_i}$ that can list multiple compact submanifolds,
Guarantee$M_i\cap M_j=\varnothing,$ if$|j-i|\ge 2.$
In this way, suppose $M_i$ has an embedded $\varphi_i,$ to $\mathbb{R}^{2m+1}$.
Taking the partition of unity $\rho_i$ of the small neighborhood $M_i$, then
$$\left(\sum_{i-odd}\rho_i \varphi_i, \sum_{i-even}\rho_i \varphi_i, f\right)$$
The desired single immersion is given, $f$ is the control function, and the division method of $M_i$ is provided.
In this way $M^m$ can be immersed in $\mathbb{R}^{2m+1}$ alone, but it is not necessarily embedded.
However, single immersion can be transformed into embedding, using the following proposition,
That is Proposition 4.22(b) in Lee’s Introduction to Smooth Manifolds:
Proposition 1.5. The proper smooth single immersion is an embedding. (As long as the space is locally compact $T_2$, the proper continuous map is a closed map.)
By transforming single immersion into proper single immersion, the classic Whitney embedding theorem is obtained.
Theorem 1.6 (Whitney embedding theorem). $m$ Dimensional smooth manifold $M$ Embedding $\mathbb{R}^{2m+1},$ Immersable $\mathbb{R}^{2m}.$
Other results
Applying the Whitney trick, the following theorem can be proved:
Theorem 1.7 (Strong Whitney embedding theorem). Any smooth manifold with dimensions above $m\ge 2$ can be embedded in $\mathbb{R}^{2m}$ and immersed in $\mathbb{R}^{2m-1}$.
There are many more detailed results as follows:
$m$-dimensional smooth compact directed manifold can be embedded in $\mathbb{R}^{2m-1}$.
When $m\neq 2^k$, the $m$-dimensional smooth manifold can be embedded in $\mathbb{R}^{2m-1}$. When $m=2^k$,
$\mathbb{R}\mathrm{P}^m$ cannot be embedded in $\mathbb{R}^{2m-1}$.
$m$-dimensional smooth manifold can be immersed in $\mathbb{R}^{2m-a(m)}$,
$a(m)$ is the number of $1$ in the binary representation of $m$, $m\ge 2.$
The last result comes from The Immersion Conjecture for Differentiable Manifolds,
Ralph L. Cohen. Method derived from the work of E.H. Brown and F.P. Peterson.
In fact, according to the Smale-Hirsch theory,
The only obstacle to immersion is the Stiefel-Whitney characteristic class,
This shows that $m=2^{i_1}+\cdots+2^{i_l},$
Take $M^m=\mathbb{R}\mathrm{P}^{2^{i_1} }\times\cdots\times \mathbb{R}\mathrm{P}^{2^{i_l} },$
The lowest dimensionality of $M^m$ immersion is $\sum_j(2i_j-1)=2m-a(m).$ so this result is optimal.
The article was last updated on 2021-11-01 17:27:27
