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解析流形
微分学
回忆$C^\omega$结构$[\mathcal{F}]$指坐标图册$\mathcal{F}$中的所有坐标卡$(U,\varphi)$彼此是$C^\omega$相容的,
即每个坐标变换$\psi\circ\varphi^{-1}\in C^\omega,$ 是解析的.
定义 1.1. $m$维$C^\omega$流形=$m$维拓扑流形+$C^\omega$结构\[$\mathcal{F}$\].
称$C^\omega$流形为解析流形.
在光滑流形上能够研究的微分学在解析流形上都能研究, 除此之外,
可以研究解析流形上函数的解析性, 这是由解析函数的复合仍是解析函数保证的.
只需取合适的开球, 让内函数值域落在外函数收敛域上即可.
例子
不解析的光滑流形
记$f(x)=\operatorname{sgn}(x)e^{-\frac{1}{x^2} }\in C^\infty(\mathbb{R}),$
令其图像作为$1$维拓扑流形$M.$
取$[\{(M,\operatorname{pr}_x),(M,\operatorname{pr}_y)\}]$为坐标图册,
那么这个图册就是$C^\infty$而非$C^\omega$的. 因为在$\mathbb{R}$上,
$\operatorname{pr}_y\circ \operatorname{pr}_x^{-1}$就是$f,$
是一个光滑不解析函数. 从而转移函数光滑不解析, 图册仅是$C^\infty$的,
$M$在该图册下成为一个光滑流形, 而非解析流形.
注意到其实$M\approx \mathbb{R},$ 整体欧氏, 根本就不需要转移函数,
因此可以是任意$\ast $-图册, $\ast $为对转移函数的要求.
如把$M$投影到$x$轴或$y$轴上.
但现将两个图册拼起来反而只能是$C^\infty$的了,
因为这两个图册并不解析相容.
因此$M$是在这个图册下的$C^\infty$流形,
它当然可以在其它图册下是其它微分流形, 但我们赋予它的微分结构是这样的.
我们可以选择赋予流形不同的微分结构.
$n$维球面
最典型的$S^n$当然可以是解析流形. 对$j=1,\cdots,n+1,$
定义$$U_{2j-1}=\{x\in S^n:x_j>0\}, \quad U_{2j}=\{x\in S^n:x_j<0\}.$$
对$i=1,\cdots,2n+2,$
定义$$\varphi_i:U_i\rightarrow \mathbb{R}^n, \quad \varphi_i(x)=(x_1,\cdots,\hat x_j,\cdots, x_{n+1}), \quad i=2j-1 \text{或} 2j.$$
那么对$r<s,$
$\varphi_r^{-1}(x)=(x_1,\cdots,\pm \sqrt{1-|x|^2},\cdots, x_n),$
添加项出现在$\lceil \frac{r}{2} \rceil$位置.
$\varphi_s\varphi_r^{-1}(x)=(x_1,\cdots,\pm \sqrt{1-|x|^2},\cdots,\hat x_k,\cdots, x_n),$
$s=2k-1$或$2k.$ 显然坐标变换是解析的, 因此$S^n\in C^\omega.$
射影空间
回忆开集$U_i=\{[x=(x^1,\cdots,x^{m+1})]\in M|x^i\neq 0\},$
$\varphi_i([x])=\frac{1}{x_i}(x^1,\cdots, \widehat{x^{i} },\cdots, x^{m+1}).$
记$\xi_{(i)}^k=\frac{x^k}{x^i}.$ 任取$1\le i<j\le m+1,$
那么$U_i\cap U_j$上坐标变换$\varphi_j\circ\varphi_i^{-1}$在分量上的作用为$\xi_{(i)}^j\mapsto \xi_{(j)}^i=\frac{1}{\xi_{(i)}^j},$
$\xi_{(i)}^l\mapsto \xi_{(j)}^l=\frac{\xi^l_{(i)} }{\xi_{(i)}^j},$
$l\neq i,j,$ 是$\mathbb{R}^m$上的解析映射, 因此坐标图册是$C^\omega$的,
实射影空间为解析流形.
类似地, 对复射影空间, 将$x$换为$z,$ $\xi$换为$w,$
那么坐标转移函数$\varphi_j\circ\varphi_i^{-1}:$
$w_{(i)}^j\mapsto w_{(j)}^i=\frac{1}{w_{(i)}^j},$
$w_{(i)}^l\mapsto w_{(j)}^l=\frac{w_{(i)}^l}{w_{(i)}^j},$ $l\neq i,j,$
是复解析映射, 从而复射影空间$\mathbb{C}P^n$是$n$维复流形.
Grassmann流形
在课上我们接触了Grassmann流形的定义,
接下来我们赋予它解析结构使其成为一个解析流形.
取$k$维子空间$P\subset V.$ 取$n-k$维互补子空间$Q$使得$V=P\oplus Q.$
那么任意线性映射$X:P\rightarrow Q$的图像$\Gamma(X)=\{v+Xv:v\in P\}$为一个$k$维子空间.
易见图像和$Q$平凡地相交.
反过来任意$k$维子空间$S\subset V$若与$Q$平凡地相交,
$S$总可以表示为某个线性映射$X$的图像, 考虑$S$的基即可.
记$P$到$Q$的线性映射全体为$L(P,Q).$
$G_k(V)$中与$Q$平凡相交的子空间全体为$U_Q.$
(注意与$\mathbb{R}\mathrm{P}^n$作类比.)
图像$\Gamma$可视为$L(P,Q)\rightarrow U_Q$的双射.
由于$L(P,Q)$可视为$M((n-k)\times k,\mathbb{R}),$
也可视为$\mathbb{R}^{k(n-k)}.$ 取$\varphi=\Gamma^{-1},$
那么$(U_Q,\varphi)$即可成为坐标卡.
需要注意$\varphi$同胚的说明. 直观上,
$U_Q$上的连续性即讨论子空间的连续性, 由于$U_Q$上取商拓扑,
可归结到特定基的连续性. 可以取到子空间的基形如$f_i=e_i+g_i,$
$\{e_i\}$构成$P$的标准基, $g_i\in Q,$ $i=1,\cdots,k.$
而$[g_1,\cdots,g_k]\in M((n-k)\times k,\mathbb{R})$恰恰给出了$\varphi,$
因此$\varphi$是同胚.
接下来说明坐标变换是解析的. 首先需要对$G_k(V)$有更清楚的认识.
它是$M(n\times k,\mathbb{R})$中的秩$k$矩阵全体的商空间,
每列描述了子空间的一个基. 秩$k$在说这$k$个基不能线性相关.
商空间的等价关系由$A\sim AP,$ $A\in M(n\times k,\mathbb{R}),$
$P\in GL(k)$给出. 因为交换基, 数乘基,
线性组合基后张成的还是同一个子空间.
在$U_Q\cap U_Q'$里的都是与$Q,Q'$均平凡相交的子空间. 对其中的任一元素$K,$
$\varphi(K)$将它打到以$P\oplus Q$为标准基的$k$个向量,
$\varphi'(K)$则打到$P'\oplus Q'$中. 注意这$k$个向量前$k$行是平凡的,
因此实际取出来的为后$n-k$行组成的$n-k$维向量.
由此可见坐标变换$\varphi'\circ\varphi^{-1}$不过是线性空间的坐标变换而已.
记$P'\oplus Q'$到$P\oplus Q$的过渡矩阵为$A,$
那么给定$\varphi$下的坐标$X,$ 还原为完整的$\widetilde{X}=\begin{bmatrix}
I\\X
\end{bmatrix},$ 左乘$A$即得到新的表示$A\widetilde{X}.$
由于它与$Q'$平凡相交, 取$P'\oplus Q'$标准基, 前$k$行组成的矩阵必满秩.
取其逆阵$B,$ 那么$\varphi'$下的坐标即为$A\widetilde{X}B$的后$n-k$行,
$A\widetilde{X}B$前$k$行是单位阵.
事实上李群中的乘法, 求逆运算也都是解析的, 因此就得到了坐标变换是解析的,
从而$[\{(U_Q,\varphi)\}]\in C^\omega,$ 流形有解析结构.
Hausdorff性与第二可数性由拓扑取商拓扑保证.
也可以通过分别说明两个$k$维子空间可以落在同一个坐标邻域上,
$G_k(V)$可由可列(事实有限)个$U_Q$覆盖得到.
注 1.2. 对于同胚性更严格的说明, 考虑空间$\pi^{-1}U_Q,$ 是一些秩$k$的$n\times k$矩阵. 设前$k$行表出$P$(即左乘某个置换阵), 那么$U_Q$中的每个元的原像可以取一个代表元$\begin{bmatrix} I\\A \end{bmatrix},$ 因此逆像中所有矩阵前$k$行是满秩的, 右乘一个逆阵即可给出标准型. 将下面的$n-k$行的矩阵提出, 即给出了$M((n-k)\times k,\mathbb{R})$中的对应. 这$k$列给出了$U_Q$中图像的刻画, 即指出了$k$个基. 由于$GL(k)$为李群, 乘法, 求逆运算均是连续的, 因此上面的映射$M(n\times k,\mathbb{R})\rightarrow M((n-k)\times k,\mathbb{R})$也是连续的, 从而$\varphi$是连续的. 接下来这个映射也是开映射, 因为对$M(n\times k,\mathbb{R})$的小扰动, 令原像下面$n-k$行产生小扰动即可映过去, 由商拓扑性质便可判断出$\varphi$是同胚.
研究意义
面对光滑流形, 我们可能产生这样的问题:
为什么通常我们只研究光滑流形就够了? 下面的结论给出了回答:
命题 1.3. 每个$C^r$流形都$C^r$微分同胚于一个$C^\omega$流形, 且在$C^\omega$微分同胚意义下唯一. $1\le r\le \infty.$
这里$C^r$微分同胚$f:M\rightarrow N$, 指(至少是)$C^r$流形间的$C^r$映射,
为同胚, 且逆映射也是$C^r$的. 命题中的解析性换为光滑性当然也对,
且光滑的情形的证明容易许多, 是’留作习题’程度的难度.
于是, 如果我们想在一个$C^r$流形上研究$C^s$微分学的问题,
$1\le r<s\le \infty$或$s=\omega,$
只需转换到与它$C^r$微分同胚的解析流形上去研究.
在转换过程中可以保证$C^k(1\le k\le r)$微分学性质不变.
另外, 每个$C^r$函数都可以被$C^\omega$函数逼近, 因此,
我们总可以假设函数与流形均有极好的正则性.
鸿沟
需要注意的是, 上面的命题对$r\ge 1$才成立,
$C^0$与$C^1$间存在着不可逾越的鸿沟. Whitehead证明,
所有光滑流形都有一个三角剖分(都光滑同胚于一个仿射流形,
而仿射流形有三角剖分), 然而存在不可三角剖分的$4$维拓扑流形(Micheal W.
Davis, Tadeusz Januszkiewicz). 三角剖分即为一个从单纯复形到流形的同胚.
$r\ge 1$的$C^r$流形由于同胚于某个光滑流形, 均有诱导的三角剖分.
因此存在$C^0$流形不同胚于任意$C^r$流形, $1\le r\le \infty.$
不可三角剖分的拓扑流形最低维数就是$4,$ 这种不可剖分的障碍称为
Kirby-Siebenmann 光滑性障碍.
复流形
我们知道流形都至少是$C^0$的, 更特殊一点可以是$C^r$流形,
比较好的是$C^{\infty}$流形, 更好的是$C^\omega$流形. 那有没有还要好的呢?
如果解析流形是$2m$维的, 那么由$\mathbb{R}^{2m}\approx \mathbb{C}^m,$
可以让坐标函数$\varphi:\mathbb{C}^m\rightarrow \mathbb{C}^m$是复解析的.
由于它每个分量实虚部都是解析的, 因此复解析流形是解析流形,
比它要更好一点, 简称复流形.
回忆单复变函数里由全纯函数的Cauchy积分公式:
$$f(z)=\frac{1}{2\pi i}\int_{\partial\Omega}\frac{f(\xi)}{\xi-z}d\xi,\quad z\in\Omega,$$
可以推出高阶导数公式:$$f^{(n)}(z)=\frac{n!}{2\pi i}\int_{\partial |\xi-c|=r}\frac{f(\xi)}{(\xi-z)^{n+1} }d\xi, \quad \,\forall\,z\in U(c,r)\subset\subset \Omega,$$
进而利用函数最大模控制导数推知全纯函数都是解析的.
对多复变函数中的全纯函数, 要求它每个分量实虚部都可微的基础上,
再加$m$对Cauchy-Riemann方程.
记开多圆盘$D(z,r):=\prod_{j=1}^{n}U(z_j,r_j)$, 有Osgood引理:
$$f(z)=\frac{1}{(2\pi i)^n}\int\cdots\int_{\partial D(z,r)}\frac{f(\xi)}{(\xi_1-z_1)\cdots(\xi_n-z_n)}d\xi_1\cdots d\xi_n, \,\forall\,D(z,r)\subset\subset \Omega,$$
类似地也有高阶导数公式版本, 从而在多复变函数里, 全纯函数也都是解析的.
复解析的函数要求对所有$\bar z_i$偏导为零, 此即C-R方程,
因此复解析与全纯是等价的. 由于可微的一个充分条件是$C^1,$
因此只需要对偶数维$C^1$流形附加坐标变换满足C-R方程的要求,
即有$C^1$流形是复流形.
由此也可以看出$C^1$到更好的正则性的距离确实并不远,
但$C^0$到$C^1$之间确实有着巨大的障碍.
文章最后更新于 2021-10-08 11:31:36
Analytic manifold
Differential calculus
Recall that $C^\omega$ structure $[\mathcal{F}]$ means that all coordinate cards $(U,\varphi)$ in the coordinate atlas $\mathcal{F}$ are $C^\omega$ compatible with each other,
That is, every coordinate transformation $\psi\circ\varphi^{-1}\in C^\omega,$ is analytical.
Definition 1.1. $m$-dimensional $C^\omega$ manifold = $m$-dimensional topological manifold + $C^\omega$ structure \[$\mathcal{F}$\].
Call the $C^\omega$ manifold Analytic manifold.
Differential calculus that can be studied on smooth manifolds can also be studied on analytic manifolds. In addition,
The analyticity of functions on analytic manifolds can be studied, which is guaranteed by the fact that the composition of analytic functions is still an analytic function.
Just take a suitable kickoff and let the value domain of the inner function fall on the convergence domain of the outer function.
Example
Unanalytical smooth manifold
Remember$f(x)=\operatorname{sgn}(x)e^{-\frac{1}{x^2} }\in C^\infty(\mathbb{R}),$
Let its image be a $1$-dimensional topological manifold $M.$
Take $[\{(M,\operatorname{pr}_x),(M,\operatorname{pr}_y)\}]$ as the coordinate atlas,
Then this album belongs to $C^\infty$ instead of $C^\omega$. Because on $\mathbb{R}$,
$\operatorname{pr}_y\circ \operatorname{pr}_x^{-1}$ is $f,$
is a smooth and non-analytic function. Therefore, the transfer function is smooth and non-analytic, and the atlas is only for $C^\infty$,
$M$ becomes a smooth manifold in this atlas, rather than an analytic manifold.
Note that in fact, $M\approx \mathbb{R},$ global Euclidean does not need a transfer function at all.
Therefore, it can be any $\ast $-atlas, $\ast $ is the requirement for the transfer function.
For example, project $M$ onto the $x$ axis or $y$ axis.
But now when the two albums are put together, they can only be from $C^\infty$.
Because these two albums are not analytically compatible.
Therefore $M$ is the $C^\infty$ manifold under this album,
Of course it can be other differential manifolds in other atlases, but the differential structure we give it is this.
We can choose to give the manifold a different differential structure.
$n$ Dimensional sphere
The most typical $S^n$ can of course be an analytic manifold. For $j=1,\cdots,n+1,$
Definition$$U_{2j-1}=\{x\in S^n:x_j>0\}, \quad U_{2j}=\{x\in S^n:x_j<0\}.$$
Right $i=1,\cdots,2n+2,$
Definition$$\varphi_i:U_i\rightarrow \mathbb{R}^n, \quad \varphi_i(x)=(x_1,\cdots,\hat x_j,\cdots, x_{n+1}), \quad i=2j-1 \text{或} 2j.$$
So right $r<s,$
$\varphi_r^{-1}(x)=(x_1,\cdots,\pm \sqrt{1-|x|^2},\cdots, x_n),$
The added item appears at the $\lceil \frac{r}{2} \rceil$ position.
$\varphi_s\varphi_r^{-1}(x)=(x_1,\cdots,\pm \sqrt{1-|x|^2},\cdots,\hat x_k,\cdots, x_n),$
$s=2k-1$ or $2k.$ Obviously the coordinate transformation is analytical, so $S^n\in C^\omega.$
projective space
Memories Opening Episode$U_i=\{[x=(x^1,\cdots,x^{m+1})]\in M|x^i\neq 0\},$
$\varphi_i([x])=\frac{1}{x_i}(x^1,\cdots, \widehat{x^{i} },\cdots, x^{m+1}).$
Remember $\xi_{(i)}^k=\frac{x^k}{x^i}.$ Take any $1\le i<j\le m+1,$
Then the effect of coordinate transformation $\varphi_j\circ\varphi_i^{-1}$ on the component $U_i\cap U_j$ is $\xi_{(i)}^j\mapsto \xi_{(j)}^i=\frac{1}{\xi_{(i)}^j},$
$\xi_{(i)}^l\mapsto \xi_{(j)}^l=\frac{\xi^l_{(i)} }{\xi_{(i)}^j},$
$l\neq i,j,$ is the analytical mapping on $\mathbb{R}^m$, so the coordinate atlas is $C^\omega$,
The real projective space is an analytic manifold.
Similarly, for the complex projective space, replace $x$ with $z,$ and $\xi$ with $w,$
Then the coordinate transfer function $\varphi_j\circ\varphi_i^{-1}:$
$w_{(i)}^j\mapsto w_{(j)}^i=\frac{1}{w_{(i)}^j},$
$w_{(i)}^l\mapsto w_{(j)}^l=\frac{w_{(i)}^l}{w_{(i)}^j},$ $l\neq i,j,$
is a complex analytic mapping, so the complex projective space $\mathbb{C}P^n$ is a $n$-dimensional complex manifold.
Grassmann manifold
In class we came into contact with the definition of Grassmann manifold,
Next we give it an analytic structure to make it an analytic manifold.
Take the $k$-dimensional subspace $P\subset V.$ and take the $n-k$-dimensional complementary subspace $Q$ such that $V=P\oplus Q.$
Then the image $\Gamma(X)=\{v+Xv:v\in P\}$ of any linear mapping $X:P\rightarrow Q$ is a $k$-dimensional subspace.
The easy-to-see image trivially intersects $Q$.
Conversely, if any $k$-dimensional subspace $S\subset V$ trivially intersects $Q$,
$S$ can always be expressed as an image of a certain linear mapping $X$, just consider the basis of $S$.
Let the linear mapping from $P$ to $Q$ be $L(P,Q).$
The total subspace in $G_k(V)$ that trivially intersects with $Q$ is $U_Q.$
(Note the analogy with $\mathbb{R}\mathrm{P}^n$.)
The image $\Gamma$ can be regarded as a bijection of $L(P,Q)\rightarrow U_Q$.
Since $L(P,Q)$ can be regarded as $M((n-k)\times k,\mathbb{R}),$
It can also be regarded as $\mathbb{R}^{k(n-k)}.$ and take $\varphi=\Gamma^{-1},$
Then $(U_Q,\varphi)$ can become a coordinate card.
It is necessary to pay attention to the description of $\varphi$ homeomorphism. Intuitively,
The continuity on $U_Q$ is the continuity of the subspace discussed. Due to the quotient topology on $U_Q$,
It can be attributed to the continuity of a specific basis. The basis shape of the subspace can be taken as $f_i=e_i+g_i,$
$\{e_i\}$ constitutes the standard base of $P$, $g_i\in Q,$ $i=1,\cdots,k.$
And $[g_1,\cdots,g_k]\in M((n-k)\times k,\mathbb{R})$ just gives $\varphi,$
Therefore $\varphi$ is a homeomorphism.
Next, we will explain that the coordinate transformation is analytical. First, we need to have a clearer understanding of $G_k(V)$.
It is the quotient space of all rank $k$ matrices in $M(n\times k,\mathbb{R})$,
Each column describes a basis of the subspace. Rank $k$ means that these $k$ basis cannot be linearly related.
The equivalent relationship of quotient space is given by $A\sim AP,$ $A\in M(n\times k,\mathbb{R}),$
$P\in GL(k)$ is given. Because the commutative basis, the multiplication basis,
The same subspace is formed after linear combination of the bases.
Everything in $U_Q\cap U_Q'$ is a subspace that trivially intersects with $Q,Q'$. For any element $K,$
$\varphi(K)$ Beat it to $k$ vectors with $P\oplus Q$ as the standard base,
$\varphi'(K)$ is typed into $P'\oplus Q'$. Note that the $k$ lines before the $k$ vector are ordinary,
Therefore, what is actually taken out is the $n-k$-dimensional vector composed of the last $n-k$ rows.
It can be seen that coordinate transformation $\varphi'\circ\varphi^{-1}$ is nothing more than coordinate transformation in linear space.
Denote the transition matrix from $P'\oplus Q'$ to $P\oplus Q$ as $A,$
Then the coordinates $X,$ under the given $\varphi$ are restored to the complete $\widetilde{X}=\begin{bmatrix}
I\\X
\end{bmatrix},$ and left multiplied by $A$ to obtain the new representation $A\widetilde{X}.$
Since it trivially intersects with $Q'$, taking $P'\oplus Q'$ standard basis, the matrix composed of the first $k$ rows must be of full rank.
Take its inverse matrix $B,$, then the coordinates under $\varphi'$ are the $n-k$ rows after $A\widetilde{X}B$,
The rows before $A\widetilde{X}B$ and $k$ are unit arrays.
In fact, the multiplication and inversion operations in Lie groups are also analytical, so the coordinate transformation is analytical.
Thus the $[\{(U_Q,\varphi)\}]\in C^\omega,$ manifold has an analytic structure.
Hausdorff property and second countability are guaranteed by topological quotient topology.
It can also be explained separately that two $k$-dimensional subspaces can fall on the same coordinate neighborhood,
$G_k(V)$ can be covered by a listable (limited number of facts) $U_Q$.
Note 1.2. For a more rigorous explanation of homeomorphism, consider that the space $\pi^{-1}U_Q,$ is some $n\times k$ matrix of rank $k$. Suppose the first $k$ row expresses $P$ (that is, left multiplied by a permutation matrix), then the original image of each element in $U_Q$ can take a representative element $\begin{bmatrix} I\\A \end{bmatrix},$. Therefore, the first $k$ rows of all matrices in the inverse image are of full rank, and right multiplication by an inverse matrix can give the standard form. Propose the matrix of the following $n-k$ row, That is, the correspondence in $M((n-k)\times k,\mathbb{R})$ is given. The $k$ column gives the description of the image in $U_Q$, that is, the $k$ basis is pointed out. Since $GL(k)$ is a Lie group, the multiplication and inversion operations are all continuous, so the above mapping $M(n\times k,\mathbb{R})\rightarrow M((n-k)\times k,\mathbb{R})$ is also continuous, so $\varphi$ is continuous. The next mapping is also an open mapping, because a small perturbation to $M(n\times k,\mathbb{R})$ causes a small perturbation to the $n-k$ row below the original image to reflect it. From the quotient topological properties, it can be judged that $\varphi$ is a homeomorphism.
Research significance
Facing smooth manifolds, we may have such problems:
Why is it usually enough for us to only study smooth manifolds? The following conclusion gives the answer:
Proposition 1.3. Every $C^r$ manifold is $C^r$ diffeomorphic to a $C^\omega$ manifold, and is unique in the sense of $C^\omega$ diffeomorphism. $1\le r\le \infty.$
Here $C^r$ diffeomorphism $f:M\rightarrow N$ refers to (at least) the $C^r$ mapping between $C^r$ manifolds,
is homeomorphism, and the inverse mapping is also $C^r$. Of course, it is also correct to replace the analyticity in the proposition with smoothness,
And the proof of the smooth case is much easier, and is as difficult as 'leaving it as an exercise'.
So, if we want to study the problem of $C^s$ differential calculus on a $C^r$ manifold,
$1\le r<s\le \infty$ or $s=\omega,$
Just convert it to the analytic manifold that is diffeomorphic with it $C^r$ and study it.
It can be ensured that the differential properties of $C^k(1\le k\le r)$ remain unchanged during the conversion process.
In addition, each $C^r$ function can be approximated by the $C^\omega$ function, therefore,
We can always assume that both functions and manifolds have excellent regularity.
chasm
It should be noted that the above proposition is only true for $r\ge 1$,
There is an unbridgeable gap between $C^0$ and $C^1$. Whitehead proved that,
All smooth manifolds have a triangulation (all are smooth homeomorphic to an affine manifold,
While affine manifolds have triangulation), there are $4$-dimensional topological manifolds that cannot be triangulated (Micheal W.
Davis, Tadeusz Januszkiewicz). Triangulation is a homeomorphism from a simple complex to a manifold.
$r\ge 1$ and $C^r$ manifolds have induced triangulation because they are homeomorphic to a smooth manifold.
Therefore, there exists a $C^0$ manifold that is not embryonic to any $C^r$ manifold, $1\le r\le \infty.$
The minimum dimension of a topological manifold that cannot be triangulated is $4,$. This type of obstacle that cannot be triangulated is called
Kirby-Siebenmann smoothness disorder.
complex manifold
We know that all manifolds are at least $C^0$, and more specifically, they can be $C^r$ manifolds.
The better one is the $C^{\infty}$ manifold, and the better one is the $C^\omega$ manifold. Is there anything better?
If the analytic manifold is $2m$ dimensional, then $\mathbb{R}^{2m}\approx \mathbb{C}^m,$
The coordinate function $\varphi:\mathbb{C}^m\rightarrow \mathbb{C}^m$ can be made complex analytic.
Since the real and imaginary parts of each of its components are analytic, the complex analytic manifold is an analytic manifold,
Better than that, for short complex manifold.
Recall that among functions of a single complex variable, the holomorphic function Cauchy integral formula:
$$f(z)=\frac{1}{2\pi i}\int_{\partial\Omega}\frac{f(\xi)}{\xi-z}d\xi,\quad z\in\Omega,$$
The higher-order derivative formula can be derived: $$f^{(n)}(z)=\frac{n!}{2\pi i}\int_{\partial |\xi-c|=r}\frac{f(\xi)}{(\xi-z)^{n+1} }d\xi, \quad \,\forall\,z\in U(c,r)\subset\subset \Omega,$$
Then use the maximum mode control derivative of the function to deduce All pure functions are analytic.
For a holomorphic function in a multi-variable function, it is required that the real and imaginary parts of each component are differentiable,
Add $m$ pair Cauchy-Riemann equation.
remember Open long disk $D(z,r):=\prod_{j=1}^{n}U(z_j,r_j)$, there is Osgood’s lemma:
$$f(z)=\frac{1}{(2\pi i)^n}\int\cdots\int_{\partial D(z,r)}\frac{f(\xi)}{(\xi_1-z_1)\cdots(\xi_n-z_n)}d\xi_1\cdots d\xi_n, \,\forall\,D(z,r)\subset\subset \Omega,$$
Similarly, there are versions of higher-order derivative formulas, so that in functions of multiple complex variables, holomorphic functions are also analytic.
The complex analytic function requires that the partial derivatives for all $\bar z_i$ are zero, which is the C-R equation,
Therefore Complex analysis and holomorphism are equivalent. Since a sufficient condition for differentiability is $C^1,$
Therefore, it is only necessary to add coordinate transformation to the even-dimensional $C^1$ manifold to meet the requirements of the C-R equation.
That is, the $C^1$ manifold is a complex manifold.
It can also be seen that the distance from $C^1$ to better regularity is indeed not far.
But there is indeed a huge obstacle between $C^0$ and $C^1$.
The article was last updated on 2021-10-08 11:31:36
