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de Rham
上同调是流形的最重要的微分同胚不变量，而不变量是我们很期待的东西，用不变量可以快速区分/分类流形。
$\mathbb{R}^n$ de Rham 上同调
微分形式
在$\mathbb{R}^n$中坐标为$(x_1,...,x_n)$, 记自变量微分$dx_1,...,dx_n$,
引入记号$\wedge$(wedge), 满足$dx_i\wedge dx_i=0$,
$dx_i\wedge dx_j=-dx_j\wedge dx_i$, $i\neq j$.
记$\Omega^q=\operatorname{span}_R\{dx_{i_1}\wedge\cdots\wedge dx_{i_q}|{}i_{1}<...<{}i_{ {}q}\},$
$\Omega^\ast=\oplus_{q=0}^n \Omega^q$, 易见其维数为$2^n$.
记光滑$q-$形式全体为$\Omega^q(\mathbb{R}^n)=C^\infty(\mathbb{R}^n)\otimes_\mathbb{R}\Omega^q$,
光滑形式全体为$\Omega^\ast(\mathbb{R}^n)=\oplus_{q=0}^n \Omega^q(\mathbb{R}^n)$.
于是任意光滑形式$\omega\in \Omega^\ast(\mathbb{R}^n)$可以按型分解为$\omega=\sum_{q=0}^{n} \omega_q$,
$\omega_q\in \Omega^q(\mathbb{R}^n)$.
$\omega_q=\sum_{I:{}i_{1}<...<{}i_{ {}q} }f_Idx_I$, $I$称为多重指标。
算子
外积算子$\wedge$:(可以看成原来记号的扩充定义)
$\tau=\sum_{I:{}i_{1}<...<{}i_{ {}p} }f_Idx_I\in \Omega^p(\mathbb{R}^n)$,
$\omega=\sum_{J:{}j_{1}<...<{}j_{ {}q} }g_Jdx_J\in \Omega^q(\mathbb{R}^n)$,
则
$\tau\wedge\omega=\sum_{I,J}f_Ig_Jdx_I\wedge dx_J=\sum_{K:{}k_{1}<...<{}k_{ {}p+q} }h_Kdx_K$.
其中$h_K$可由$f_I,g_J$表出。
命题 1.1. $\tau\wedge \omega = (-1)^{pq}\omega \wedge \tau$
外微分算子$d:\Omega^q(\mathbb{R}^n)\rightarrow \Omega^{q+1}(\mathbb{R}^n)$,
$\omega=\sum_I f_I dx_I$,
$d\omega=\sum_I df_I\wedge dx_I=\sum_{i,I} \frac{\partial {}f_I}{\partial {}x_i}dx_i\wedge dx_I$,
是$C^\infty$上微分算子的自然延拓.
命题 1.2. $\tau\in \Omega^p(\mathbb{R}^n)$, 则$d(\tau\wedge \omega)=d\tau \wedge \omega + (-1)^p \tau\wedge d\omega,$ $d^2=0$.
de Rham 复形
由于$d^2=0$, $\{\Omega^\ast(\mathbb{R}^n),d\}$形成$\mathbb{R}^n$的de
Rham复形：
$$\Omega^0(\mathbb{R}^n)\xrightarrow{d}\cdots\xrightarrow{d}\Omega^n(\mathbb{R}^n).$$
可以对$q<0$及$q>n$添加$\Omega^q(\mathbb{R}^n)=0$.
定义$Z^q(\mathbb{R}^n)=\{\omega\in\Omega^q(\mathbb{R}^n)|d\omega=0\},$
其中元素称为闭的$q$-形式.
定义$B^q(\mathbb{R}^n)=\{d\tau | \tau\in\Omega^{q-1}(\mathbb{R}^n)\},$
其中元素称为恰当的$q$-形式. 特别地,
规定$Z^n(\mathbb{R}^n)=\Omega^n(\mathbb{R}^n),$ $B^0(\mathbb{R}^n)=0.$
由于$d^2=0$, $B^q(\mathbb{R}^n)\subset Z^q(\mathbb{R}^n).$
定义$\mathbb{R}^n$的第$q$个de
Rham上同调为$H^q_{dR}(\mathbb{R}^n)=Z^q(\mathbb{R}^n)/B^q(\mathbb{R}^n).$
不引起歧义时, 通常隐去下标.
闭$q$-形式$\omega$所在上同调类记为$[\omega].$
类似地, 对$\mathbb{R}^n$开集$U$上的de Rham复形,
只需在一开始将光滑函数定义域设置为$U$即可.
我们有上同调的Poincaré引理: $H^q(\mathbb{R}^n)=\begin{cases}
\mathbb{R}\quad&q=0\\
0&q\neq 0
\end{cases}.$
紧上同调
记$\Omega^\ast_c(\mathbb{R}^n)=C^\infty_c(\mathbb{R}^n)\otimes_\mathbb{R}\Omega^\ast,$
即将光滑函数改为具有紧支集的光滑函数. 类似地, 可以将其按型分解.
由于光滑函数偏导支集闭包含于原来的紧支集, 它也是有紧支撑的,
因此$\{\Omega^\ast_c(\mathbb{R}^n),d\}$是一个复形, 称为具紧支集的de
Rham复形.
类似地,
定义$Z^q_c(\mathbb{R}^n)=\{\omega\in\Omega_c^q(\mathbb{R}^n)|d\omega=0\},$
$B^q_c(\mathbb{R}^n)=\{d\tau | \tau\in\Omega_c^{q-1}(\mathbb{R}^n)\},$
$H^q_{c}(\mathbb{R}^n)=Z^q_c(\mathbb{R}^n)/B^q_c(\mathbb{R}^n).$
注意$Z_c^q(\mathbb{R}^n)=\Omega^q_c(\mathbb{R}^n)\cap Z^q(\mathbb{R}^n),$
但一般$B^q_c(\mathbb{R}^n)\neq B^q(\mathbb{R}^n)\cap \Omega_c^q(\mathbb{R}^n).$
即不是$d\tau$有紧支集而是$d\tau$中$\tau$有紧支集.
我们有紧上同调的Poincaré引理: $H^q_c(\mathbb{R}^n)=\begin{cases}
\mathbb{R}\quad&q=n\\
0&q\neq n
\end{cases}.$
注意这说明了紧上同调不是同伦不变量. 类似地,
我们可对一般的$U\subset \mathbb{R}^n$定义$H_c^\ast(U)$.
微分复形
称向量空间直和$C=\oplus_{q\in\mathbb{Z} } C^q$为微分复形,
若有同态$d:C\rightarrow C,$ 满足$d:C^q\rightarrow C^{q+1},$ $d^2=0.$
可以由$d$对复形定义上同调. de Rham复形即为微分复形.
与代数拓扑相同, 我们可定义复形间的链映射及其诱导同态, 复形的短正合列,
以及至关重要的zig-zag引理(图追踪定义同调的长正合列).
文章最后更新于 2021-09-19 18:59:49
de Rham
Cohomology is the most important diffeomorphic invariant of manifolds, and invariants are something we are looking forward to. Invariants can be used to quickly distinguish/classify manifolds.
$\mathbb{R}^n$ de Rham cohomology
differential form
In $\mathbb{R}^n$, the coordinate is $(x_1,...,x_n)$, and the independent variable differential $dx_1,...,dx_n$ is recorded.
Introduce the symbol $\wedge$(wedge), satisfy $dx_i\wedge dx_i=0$,
$dx_i\wedge dx_j=-dx_j\wedge dx_i$, $i\neq j$.
Remember$\Omega^q=\operatorname{span}_R\{dx_{i_1}\wedge\cdots\wedge dx_{i_q}|{}i_{1}<...<{}i_{ {}q}\},$
$\Omega^\ast=\oplus_{q=0}^n \Omega^q$, it is easy to see that its dimension is $2^n$.
Let the whole smooth $q-$ form be $\Omega^q(\mathbb{R}^n)=C^\infty(\mathbb{R}^n)\otimes_\mathbb{R}\Omega^q$,
The smooth form is generally $\Omega^\ast(\mathbb{R}^n)=\oplus_{q=0}^n \Omega^q(\mathbb{R}^n)$.
So any smooth form $\omega\in \Omega^\ast(\mathbb{R}^n)$ can Break down by type is $\omega=\sum_{q=0}^{n} \omega_q$,
$\omega_q\in \Omega^q(\mathbb{R}^n)$.
$\omega_q=\sum_{I:{}i_{1}<...<{}i_{ {}q} }f_Idx_I$, $I$ are called Multiple indicators。
operator
External product operator $\wedge$: (can be seen as an expanded definition of the original symbol)
$\tau=\sum_{I:{}i_{1}<...<{}i_{ {}p} }f_Idx_I\in \Omega^p(\mathbb{R}^n)$,
$\omega=\sum_{J:{}j_{1}<...<{}j_{ {}q} }g_Jdx_J\in \Omega^q(\mathbb{R}^n)$,
rule
$\tau\wedge\omega=\sum_{I,J}f_Ig_Jdx_I\wedge dx_J=\sum_{K:{}k_{1}<...<{}k_{ {}p+q} }h_Kdx_K$.
Among them $h_K$ can be expressed by $f_I,g_J$.
Proposition 1.1. $\tau\wedge \omega = (-1)^{pq}\omega \wedge \tau$
Exterior derivative operator $d:\Omega^q(\mathbb{R}^n)\rightarrow \Omega^{q+1}(\mathbb{R}^n)$,
$\omega=\sum_I f_I dx_I$,
$d\omega=\sum_I df_I\wedge dx_I=\sum_{i,I} \frac{\partial {}f_I}{\partial {}x_i}dx_i\wedge dx_I$,
It is a natural continuation of the differential operator on $C^\infty$.
Proposition 1.2. $\tau\in \Omega^p(\mathbb{R}^n)$, then $d(\tau\wedge \omega)=d\tau \wedge \omega + (-1)^p \tau\wedge d\omega,$ $d^2=0$.
de Rham complex
Since $d^2=0$, $\{\Omega^\ast(\mathbb{R}^n),d\}$ form $\mathbb{R}^n$ de
Rham complex:
$$\Omega^0(\mathbb{R}^n)\xrightarrow{d}\cdots\xrightarrow{d}\Omega^n(\mathbb{R}^n).$$
$\Omega^q(\mathbb{R}^n)=0$ can be added to $q<0$ and $q>n$.
Definition$Z^q(\mathbb{R}^n)=\{\omega\in\Omega^q(\mathbb{R}^n)|d\omega=0\},$
where the elements are called closed $q$-Form.
Definition$B^q(\mathbb{R}^n)=\{d\tau | \tau\in\Omega^{q-1}(\mathbb{R}^n)\},$
where the elements are called appropriate $q$-Form. In particular,
Regulations $Z^n(\mathbb{R}^n)=\Omega^n(\mathbb{R}^n),$ $B^0(\mathbb{R}^n)=0.$
Due to $d^2=0$, $B^q(\mathbb{R}^n)\subset Z^q(\mathbb{R}^n).$
Define the $q$th de of $\mathbb{R}^n$
Rham cohomology is $H^q_{dR}(\mathbb{R}^n)=Z^q(\mathbb{R}^n)/B^q(\mathbb{R}^n).$
When no ambiguity is caused, subscripts are usually omitted.
The cohomology class where the closed $q$-form $\omega$ is located is recorded as $[\omega].$
Similarly, for the de Rham complex on $\mathbb{R}^n$ open set $U$,
Just set the smooth function domain to $U$ at the beginning.
we have Cohomological Poincaré's Lemma: $H^q(\mathbb{R}^n)=\begin{cases}
\mathbb{R}\quad&q=0\\
0&q\neq 0
\end{cases}.$
cohomology
Remember$\Omega^\ast_c(\mathbb{R}^n)=C^\infty_c(\mathbb{R}^n)\otimes_\mathbb{R}\Omega^\ast,$
That is to say, the smooth function is changed into a smooth function with compact support set. Similarly, it can be decomposed according to the shape.
Since the set of partial derivative supports of smooth functions is contained in the original compact support set, it also has compact support,
Therefore $\{\Omega^\ast_c(\mathbb{R}^n),d\}$ is a complex, called with compact support de
Rham complex.
Similarly,
Definition$Z^q_c(\mathbb{R}^n)=\{\omega\in\Omega_c^q(\mathbb{R}^n)|d\omega=0\},$
$B^q_c(\mathbb{R}^n)=\{d\tau | \tau\in\Omega_c^{q-1}(\mathbb{R}^n)\},$
$H^q_{c}(\mathbb{R}^n)=Z^q_c(\mathbb{R}^n)/B^q_c(\mathbb{R}^n).$
Note$Z_c^q(\mathbb{R}^n)=\Omega^q_c(\mathbb{R}^n)\cap Z^q(\mathbb{R}^n),$
But generally $B^q_c(\mathbb{R}^n)\neq B^q(\mathbb{R}^n)\cap \Omega_c^q(\mathbb{R}^n).$
That is, it is not that $d\tau$ has a compact support but $\tau$ in $d\tau$ has a compact support.
we have Poincaré's Lemma for Compact Cohomology: $H^q_c(\mathbb{R}^n)=\begin{cases}
\mathbb{R}\quad&q=n\\
0&q\neq n
\end{cases}.$
Note that this shows Cohomology is not homotopy invariant.Similarly,
We can define $H_c^\ast(U)$ for general $U\subset \mathbb{R}^n$.
differential complex
The vector space direct sum $C=\oplus_{q\in\mathbb{Z} } C^q$ is called differential complex,
If there is a homomorphism $d:C\rightarrow C,$ that satisfies $d:C^q\rightarrow C^{q+1},$ $d^2=0.$
Cohomology can be defined for a complex by $d$. A de Rham complex is a differential complex.
As in algebraic topology, we can define chain mappings between complexes and their induced homomorphisms, short exact sequences of complexes,
And the crucial zig-zag lemma (graph tracing defines long exact sequences of homology).
The article was last updated on 2021-09-19 18:59:49
