Machine-translated from Chinese.
映射的微分学
映射
$M^m$是$C^\infty$流形, $f:M\rightarrow \mathbb{R},$ $p\in M.$
若$\,\exists\,$含$p$点允许坐标系$(U,\varphi)$(即$\in [\mathcal{F}]$)使$f\circ \varphi^{-1}$在$\varphi(p)$的某个邻域内是$C^\infty$的,
则称$f$在$p$点$C^\infty.$ 若$f$点点光滑, 则称$f\in C^\infty(M).$
允许坐标系限制在更小的开子集上显然仍是允许坐标系,
因此上述定义等价于存在某个允许坐标系使得$f\circ \varphi^{-1}$在整个$\varphi(U)$上是$C^\infty$的.
由于坐标变换是光滑的, 上述概念与允许坐标系选择无关. 类似地,
可以定义$f$在$p$点是$C^r$的, 或$f\in C^r(M).$
映照
设$M^m,N^n$是光滑流形, $F:M\rightarrow N,$ $p\in M.$
若存在含$p,q=F(p)$的允许坐标系$(U,\varphi),(V,\psi),$ $F(U)\subset V,$
使得局部表示$\widehat{F}:=\psi\circ F\circ\varphi^{-1}$在$\varphi(p)$某邻域是$C^\infty$的,
则称$F$在$p$点是$C^\infty$的. 若$F$点点$C^\infty,$
则称$F$是光滑映照, 记为$F\in C^\infty(M,N).$
注 1.1. 若$F(U)\not\subset V,$ 则可取$\widetilde{U}=F^{-1}(V)\cap U.$
同理上述概念与允许坐标系选取无关.
微分同胚
如果$(U,\varphi)$是$M$的允许坐标系,
则$\varphi:U\rightarrow \varphi(U)$及$\varphi^{-1}:\varphi(U)\rightarrow U$依定义显然是光滑的,
再取同样的允许坐标系$(U,\varphi)$说明即可. 注意$U$从$M$上诱导光滑结构,
为开子流形, 而非嵌入到外围空间的子流形(微分结构不同),
如$\widetilde{\varphi}:U\rightarrow \varphi(U)\hookrightarrow \mathbb{R}^k$不见得光滑.
易见光滑映照的复合仍是光滑映照.
设$F:M\rightarrow N$是光滑流形间的双射. 如果$F,F^{-1}$是光滑的,
则称$F:M\rightarrow N$是$C^\infty$微分同胚.
两个微分同胚的光滑流形视为一样的.
例如球极投影$F:S^2\setminus\{p\}\rightarrow\mathbb{R}^2$, scaling
$B^n(1)\rightarrow \mathbb{R}^n$都是光滑同胚.
微分结构问题
一个基本问题是, 给定拓扑流形$M^m,$ 其上是否一定存在$C^\infty$结构?
存在的话是否(在$C^\infty$同胚意义下)唯一? 若$M$有两个微分结构,
问是否有光滑同胚$(M,\mathcal{F}_1)\approx (M,\mathcal{F}_2).$
注意同胚的选取不见得取恒等映射$\mathrm{id},$
因此不能简简单单只验证$\mathcal{F}_1,\mathcal{F}_2$是否$C^\infty$相容.
已知结果:
$\mathbb{R}^m,$ $m\neq 4$上存在唯一的$C^\infty$结构;
$\mathbb{R}^4$上存在无穷多个互不等价的$C^\infty$结构;
$S^7$-Milnor怪球上存在28种互不等价的$C^\infty$结构;
维数$m\le 3$的拓扑流形上存在唯一的$C^\infty$结构;
$\,\forall\,m>3,$ 都存在$m$维拓扑流形, 其上不存在$C^\infty$结构.
下面假设拓扑流形都给定了$C^\infty$结构.
切空间
给定光滑流形$M^m,$ 我们尝试内蕴地推广切向量, 切平面的概念.
将曲线$C:(a,b)\rightarrow M^m$视为光滑流形间的映照,
我们可以定义光滑曲线的概念. 设$C(t_0)=p,$
用$C_p^\infty(M)$表示$M$上在$p$处光滑的函数全体,
我们称映照(方向导数算子)$X:C_p^\infty(M)\rightarrow \mathbb{R},$
$f\mapsto \frac{d}{dt}[f(c(t))]|_{t=t_0}$为$C(t)$在$p$处切向量,
即求方向导数的运算过程蕴含了该方向本身.
上述”切向量”$X$满足$X(af+bg)=aX(f)+bX(g),$
$X(f\cdot g)=X(f)g(p)+f(p)X(g).$
取$\widetilde{T}_pM=\{\widetilde{X}:C_p^\infty\rightarrow \mathbb{R}\text{满足(1),(2)}\}.$
$(\widetilde{X}_1+\widetilde{X}_2)f:=\widetilde{X}_1(f)+\widetilde{X}_2(f),$
$(x\widetilde{X})(f):=a\widetilde{X}(f).$
容易验证$\widetilde{X}_1+\widetilde{X}_2,$
$a\widetilde{X}$满足$(1),(2),$ 即$\widetilde{T}_pM$是向量空间.
$C_p^\infty(M)$和$C^\infty(M)$不同, 后者为向量空间, 但前者不是,
因为零元并不唯一. 原因是前者的加法仅在公共光滑邻域上做加法.
因此$X$也暂时不是线性映射.
我们引入等价关系来改变这一状况:
记$f\sim g\Leftrightarrow \,\exists\,p$点小邻域$W,$ 使得$f|_W=g|_W.$
称$[f]$为$C^\infty$函数芽,
$C_p^\infty(M)/_\sim$为$C^\infty$函数芽空间$\mathcal{F}_p^\infty.$
$X$可以诱导其上的映射,
仍记为$X:\mathcal{F}_p^\infty\rightarrow \mathbb{R},$ $[f]\mapsto Xf.$
它也满足类似前面$(1),(2)$的性质.
此时$\mathcal{F}_p^\infty$确实是向量空间, 因此$X$成为线性映射.
文章最后更新于 2021-09-26 16:39:45
Differential Calculus of Maps
mapping
$M^m$ is the $C^\infty$ manifold, $f:M\rightarrow \mathbb{R},$ $p\in M.$
If $\,\exists\,$ contains $p$ points, the coordinate system $(U,\varphi)$ (i.e. $\in [\mathcal{F}]$) allows $f\circ \varphi^{-1}$ to be $C^\infty$ in a neighborhood of $\varphi(p)$,
Then it is called $f$ at $p$ point $C^\infty.$. If $f$ point is smooth, it is called $f\in C^\infty(M).$
Allowing coordinate systems to be restricted to smaller open subsets is obviously still allowing coordinate systems,
Therefore, the above definition is equivalent to the existence of a certain allowed coordinate system such that $f\circ \varphi^{-1}$ is $C^\infty$ over the entire $\varphi(U)$.
Since coordinate transformations are smooth, the above concepts are irrelevant to allowing coordinate system selection. Similarly,
You can define $f$ to be $C^r$ at $p$ point, or $f\in C^r(M).$
reflect
Let $M^m,N^n$ be a smooth manifold, $F:M\rightarrow N,$ $p\in M.$
If there is an allowed coordinate system $(U,\varphi),(V,\psi),$ $F(U)\subset V,$ containing $p,q=F(p)$
make local representation $\widehat{F}:=\psi\circ F\circ\varphi^{-1}$ is $C^\infty$ in a certain neighborhood of $\varphi(p)$,
Then $F$ is said to be $C^\infty$ at point $p$. If $F$ points to $C^\infty,$
Then $F$ is said to be smooth reflection, recorded as $F\in C^\infty(M,N).$
Note 1.1. If $F(U)\not\subset V,$ then $\widetilde{U}=F^{-1}(V)\cap U.$ is available
In the same way, the above concepts have nothing to do with the selection of allowed coordinate systems.
diffeomorphism
If $(U,\varphi)$ is an allowed coordinate system of $M$,
Then $\varphi:U\rightarrow \varphi(U)$ and $\varphi^{-1}:\varphi(U)\rightarrow U$ are obviously smooth by definition,
Just take the same allowed coordinate system $(U,\varphi)$ for explanation. Note that $U$ induces a smooth structure from $M$,
is an open submanifold, rather than a submanifold embedded in the peripheral space (the differential structure is different),
For example, $\widetilde{\varphi}:U\rightarrow \varphi(U)\hookrightarrow \mathbb{R}^k$ may not be smooth.
It is easy to see that the composite of smooth reflections is still smooth reflections.
Let $F:M\rightarrow N$ be a bijection between smooth manifolds. If $F,F^{-1}$ is smooth,
Then $F:M\rightarrow N$ is said to be $C^\infty$ diffeomorphism.
Two diffeomorphic smooth manifolds are considered to be the same.
For example, spherical projection $F:S^2\setminus\{p\}\rightarrow\mathbb{R}^2$, scaling
$B^n(1)\rightarrow \mathbb{R}^n$ are all smooth homeomorphisms.
Differential structure problem
A basic question is, given a topological manifold $M^m,$, is there necessarily a $C^\infty$ structure on it?
If it exists, is it unique (in the sense of $C^\infty$ homeomorphism)? If $M$ has two differential structures,
Ask whether there is a smooth homeomorphism$(M,\mathcal{F}_1)\approx (M,\mathcal{F}_2).$
Note that the selection of homeomorphisms does not necessarily lead to identity mapping $\mathrm{id},$
Therefore, we cannot simply verify whether $\mathcal{F}_1,\mathcal{F}_2$ is compatible with $C^\infty$.
Known results:
There is a unique $C^\infty$ structure on $\mathbb{R}^m,$ $m\neq 4$;
There are infinitely many non-equivalent $C^\infty$ structures on $\mathbb{R}^4$;
$S^7$-There are 28 mutually non-equivalent $C^\infty$ structures on Milnor's strange ball;
There is a unique $C^\infty$ structure on the topological manifold of dimension $m\le 3$;
There is a $m$-dimensional topological manifold for $\,\forall\,m>3,$, and there is no $C^\infty$ structure on it.
It is assumed below that the topological manifold is given a $C^\infty$ structure.
Tangent Space
Given a smooth manifold $M^m,$ we try implicitly Generalize the concepts of tangent vectors and tangent planes.
Treat curve $C:(a,b)\rightarrow M^m$ as a mapping between smooth manifolds,
We can define the concept of smooth curve. Let $C(t_0)=p,$
Use $C_p^\infty(M)$ to represent the entire function on $M$ that is smooth at $p$,
We call it reflection ( Directional derivative operator)$X:C_p^\infty(M)\rightarrow \mathbb{R},$
$f\mapsto \frac{d}{dt}[f(c(t))]|_{t=t_0}$ is $C(t)$ at $p$ tangent vector,
That is, the operation process of finding the directional derivative contains the direction itself.
The above "tangent vector" $X$ satisfies $X(af+bg)=aX(f)+bX(g),$
$X(f\cdot g)=X(f)g(p)+f(p)X(g).$
Take $\widetilde{T}_pM=\{\widetilde{X}:C_p^\infty\rightarrow \mathbb{R}\text{满足(1),(2)}\}.$
$(\widetilde{X}_1+\widetilde{X}_2)f:=\widetilde{X}_1(f)+\widetilde{X}_2(f),$
$(x\widetilde{X})(f):=a\widetilde{X}(f).$
Easy to verify$\widetilde{X}_1+\widetilde{X}_2,$
$a\widetilde{X}$ satisfies $(1),(2),$, that is, $\widetilde{T}_pM$ is a vector space.
$C_p^\infty(M)$ is different from $C^\infty(M)$. The latter is a vector space, but the former is not.
because Zero yuan is not the only one.The reason is that the addition of the former only adds to the common smooth neighborhood.
Therefore $X$ is not a linear mapping for the time being.
We introduce the equivalence relationship to change this situation:
Note $f\sim g\Leftrightarrow \,\exists\,p$ point small neighborhood $W,$ such that $f|_W=g|_W.$
Call $[f]$ $C^\infty$ function bud,
$C_p^\infty(M)/_\sim$ is $C^\infty$ function bud space $\mathcal{F}_p^\infty.$
$X$ can induce the mapping on it,
Still recorded as $X:\mathcal{F}_p^\infty\rightarrow \mathbb{R},$ $[f]\mapsto Xf.$
It also satisfies properties similar to the previous $(1),(2)$.
At this time $\mathcal{F}_p^\infty$ is indeed a vector space, so $X$ becomes a linear mapping.
The article was last updated on 2021-09-26 16:39:45
