Machine-translated from Chinese.
Mayer-Vietoris序列是代数拓扑中经典的结论, 对于计算同调群有着莫大的帮助.
它当然也存在于de Rham复形中.
上同调MV序列
复形的短正合列
在代数拓扑中我们知道, 若存在复形的短正合列
$$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0,$$
则由zig-zag引理(图追踪), 可以构造长正合列
$$...\rightarrow H^p(A)\rightarrow H^p(B)\rightarrow H^p(C)\rightarrow H^{p+\varepsilon}(A)\rightarrow ...$$
其中$\varepsilon=\pm 1$, 取决于考虑上同调还是下同调. 因此为了得到MV序列,
我们只需构造出复形的短正合列即可.
设$U,V$为流形$M$的开子集, $M=U\cup V$, $U\cap V$是开的, 其上有到$U$,
$V$的包含映射:
$$\partial_0:U\cap V\rightarrow V, \quad \partial_1:U\cap V\rightarrow U,$$
下标的数字指遗忘第几个位置, 用于在之后考虑更一般的MV序列(可列多个开集).
记$U\sqcup V$为两者做无交并, 也称为拓扑和(因为在同一空间中看,
它们当然是有交的). 注意不是笛卡尔积.
于是有包含映射$U,V\rightarrow M$诱导的映射$U\sqcup V\rightarrow M$,
从而我们有映射列
$$M=U\cup V\leftarrow U\sqcup V\mathop{\leftleftarrows}\limits_{\partial_1}^{\partial_0} U\cap V,$$
这里双箭头指选取其中一个映射, 即事实上我们有两个映射列.
应用反变函子$\Omega^\ast$, 做拉回映射得到
$$\Omega^\ast (M)\rightarrow \Omega^\ast (U)\oplus \Omega^\ast (V)\mathop{\rightrightarrows}\limits_{\partial_1^\ast }^{\partial_0^\ast }\Omega^\ast (U\cap V),$$
上面每个拉回都等价于对微分形式的限制. 为了得到正合列,
我们需适当改造该映射列:
$$0\rightarrow\Omega^\ast (M)\xrightarrow{r}\Omega^\ast (U)\oplus\Omega^\ast (V)\xrightarrow{\delta}\Omega^\ast (U\cap V)\rightarrow 0,$$
其中$r(\omega)=(\omega|_U, \omega|_V)$,
$\delta=\partial_0^\ast -\partial_1^\ast$,
即$\delta(\omega,\tau)=\tau|_{U\cap V}-\omega|_{U\cap V}$. 一般的,
若定义域明了, 方便起见也可简记为$\tau-\omega$.
我们需要验证上述复形列正合.
证 $r$即为将微分形式分别限制在$U$和$V$上, 很自然它是单射;
由$\delta$的特殊构造, 容易看到在中间项也正合, 只需证$\delta$满.
$\,\forall\,\omega\in \Omega^q(U\cap V)$,
取从属于$\{U,V\}$的单位分解$\{\rho_U,\rho_V\}$, 在$U$上定义
$$\rho_V\omega=
\begin{cases}
\rho_V|_{U\cap V}\cdot\omega, \quad & \text{on}\quad U\cap V\\
0, &\text{on} \quad U-U\cap V
\end{cases},$$
这里$\rho_V$定义域是$V$, 在$U$上自然地零延拓至0.
$\omega$定义域是$U\cap V$,
因此此时$\rho_V\omega$扩充到$U$上也可以给定定义, 且是光滑的.
在$V$上类似定义$\rho_U\omega$. 取$\xi=(-\rho_V\omega, \rho_U\omega)$,
则$\delta\xi=\omega$. 从而复形列正合.
Mayer-Vietoris序列
由zig-zag引理, 短正合列诱导了长正合列
$$\cdots\rightarrow H^q(M)\xrightarrow{r}H^q(U)\oplus H^q(V)\xrightarrow{\delta}H^q(U\cap V)\xrightarrow{d^\ast }H^{q+1}(M)\rightarrow \cdots$$
这里连接同态$d^\ast$的定义由zig-zag引理过程给出: 利用图追踪,
$\omega\Rightarrow \xi=(-\rho_V\omega,\rho_U\omega)$,
再推出$r\eta=d\xi$. $r$就是限制映射, 因此$\eta=
\begin{cases}
d(-\rho_V\omega) &\quad\text{on}\quad U\\
d(\rho_U\omega) &\quad\text{on}\quad V
\end{cases}$.
在$U\cap V$处, 由$\rho_U+\rho_V=1$, 与$d\omega=0$, 知定义合法.
于是$d^\ast [\omega]=[\eta]$.
注意$\rho_U \omega, \rho_V\omega$ 均不是整体定义在$M$上的,
不一定有$\eta=d\tau$, $\tau\in\Omega^\ast (M)$,
因此$\eta$并不一定是恰当的.
紧支集MV序列
设$f:N\rightarrow M$为光滑映射,
但$f^\ast :\Omega^\ast _c(M)\rightarrow \Omega^\ast _c(N)$一般并不成立,
无法确保将紧支撑的微分形式仍然拉回到紧支撑的微分形式,
如取$f$为投射$(x_1,x_2)\mapsto x_1$. 不过光滑的逆紧映射就可以了,
因为其定义就是满足\”紧集的逆像仍是紧集\”的映射.
一般的, 我们更多考虑的是包含映射: 若有$j: U\rightarrow M$,
定义$j_\ast : \Omega^\ast _c(U)\rightarrow \Omega^\ast _c(M)$,
$j_\ast \omega$为将$\omega$零延拓到$M-U$上.
注意这时映射的星号标在下方.
若只考虑开集间的包含映射, $\Omega^\ast _c$成为共变函子. 类似地,
对这个函子也有MV序列,
由包含映射列$U\cap V\mathop \rightrightarrows \limits_{j_U}^{j_V} U\sqcup V\rightarrow M$诱导.
这里双箭头指分别映到$U,V$上. 由此构造出复形的短正合列:
$$0\rightarrow \Omega^\ast _c(U\cap V)\xrightarrow{\delta} \Omega^\ast _c(U)\oplus \Omega^\ast _c(V)\xrightarrow{s} \Omega^\ast _c(M)\rightarrow 0,$$
其中$\delta\omega=(-j_\ast \omega,j_\ast \omega)$,
$s(\omega,\tau)=j_\ast \omega+j_\ast \tau$.
为了证明正合性, 同理只需证明$s$满,
类似地取$\{\rho_U,\rho_V\}$为单位分解,
构造$\rho_U\omega,\rho_V\omega$分别是$U$,$V$上的微分形式就可以了.
由于$\omega$在$M$上有紧支撑, $\rho_U$支集在$U$中,
可证$\rho_U\omega$在$U$上有紧支撑, $\rho_V\omega$同理.
进而由zig-zag引理导出长正合列:
$$\cdots\rightarrow H^q_c(U\cap V)\xrightarrow{\delta} H^q_c(U)\oplus H^q_c(V)\xrightarrow{s} H^q_c(M) \xrightarrow{d_\ast }H^{q+1}_c(U\cap V)\rightarrow \cdots$$
同理可判断连接同态.
图追踪找到$\eta=d(\rho_V\omega)|_{U\cap V}=-d(\rho_U\omega)|_{U\cap V}$,
$d_\ast [\omega]=[\eta]$.
同样的在紧上同调中, $\eta$并不一定恰当.
因为无论是$\rho_U\omega$还是$\rho_V\omega$, 虽然这时有了整体定义,
但限制在$U\cap V$中不一定具有紧支集.
$H^1(S^1)\simeq \mathbb{R}$
看一个简单的例子. 由于$S^1$紧, $H_c^\ast (S^1)=H^\ast (S^1)$.
这时用上同调, 紧上同调的MV序列都可以容易地证出这个命题,
只需将$S^1$切成两半即可.
或者用局部坐标$\theta$, $d\theta$为$S^1$上$1$-形式.
$1$-形式$\omega$恰当时,
$\omega=dF=ad\theta \Leftrightarrow \int_{S^1}\omega=0.$ 因此
$B^1(S^1)=\ker \int_{S^1}.$
注意到$\int_{S^1}:\Omega^1(S^1)\rightarrow \mathbb{R}.$
从而$H^1(S^1)=Z^1(S^1)/B^1(S^1)=\Omega^1(S^1)/\ker \int_{S^1}\cong \mathbb{R}$.
文章最后更新于 2021-10-09 12:32:30
The Mayer-Vietoris sequence is a classic conclusion in algebraic topology and is of great help in calculating homology groups.
It certainly exists in de Rham complexes as well.
Cohomological MV sequence
Short exact sequence of complex
In algebraic topology we know that if there is a short exact sequence of complex
$$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0,$$
Then according to the zig-zag lemma (graph tracking), a long exact sequence can be constructed
$$...\rightarrow H^p(A)\rightarrow H^p(B)\rightarrow H^p(C)\rightarrow H^{p+\varepsilon}(A)\rightarrow ...$$
Among them, $\varepsilon=\pm 1$ depends on whether we consider cohomology or hypohomology. Therefore, in order to obtain the MV sequence,
We only need to construct the short exact sequence of the complex.
Let $U,V$ be the open subset of the manifold $M$, $M=U\cup V$, $U\cap V$ are open, and $U$,
Inclusion mapping for $V$:
$$\partial_0:U\cap V\rightarrow V, \quad \partial_1:U\cap V\rightarrow U,$$
The number in the subscript refers to the forgotten position, which is used to consider more general MV sequences later (multiple open sets can be listed).
Let $U\sqcup V$ be the intersection-free union of the two, also called the topological sum (because when viewed in the same space,
Of course they intersect). Note that it is not a Cartesian product.
Then there is a mapping $U\sqcup V\rightarrow M$ induced by the inclusion mapping $U,V\rightarrow M$,
Thus we have mapped columns
$$M=U\cup V\leftarrow U\sqcup V\mathop{\leftleftarrows}\limits_{\partial_1}^{\partial_0} U\cap V,$$
The double arrow here points to selecting one of the mappings, that is, in fact we have two mapping columns.
Apply the inverse functor $\Omega^\ast$ and do pullback mapping to get
$$\Omega^\ast (M)\rightarrow \Omega^\ast (U)\oplus \Omega^\ast (V)\mathop{\rightrightarrows}\limits_{\partial_1^\ast }^{\partial_0^\ast }\Omega^\ast (U\cap V),$$
Each pullback above is equivalent to a restriction on the differential form. To get the exact series,
We need to appropriately transform the mapping column:
$$0\rightarrow\Omega^\ast (M)\xrightarrow{r}\Omega^\ast (U)\oplus\Omega^\ast (V)\xrightarrow{\delta}\Omega^\ast (U\cap V)\rightarrow 0,$$
Among them $r(\omega)=(\omega|_U, \omega|_V)$,
$\delta=\partial_0^\ast -\partial_1^\ast$,
That is $\delta(\omega,\tau)=\tau|_{U\cap V}-\omega|_{U\cap V}$. Generally,
If the definition domain is clear, it can also be abbreviated as $\tau-\omega$ for convenience.
We need to verify that the above complex columns are exact.
certificate $r$ means to limit the differential forms to $U$ and $V$ respectively, so naturally it is an injective;
From the special structure of $\delta$, it is easy to see that the middle term is also consistent, and we only need to prove that $\delta$ is satisfied.
$\,\forall\,\omega\in \Omega^q(U\cap V)$,
Take the partition of unity $\{\rho_U,\rho_V\}$ belonging to $\{U,V\}$ and define it on $U$
$$\rho_V\omega=
\begin{cases}
\rho_V|_{U\cap V}\cdot\omega, \quad & \text{on}\quad U\cap V\\
0, &\text{on} \quad U-U\cap V
\end{cases},$$
Here the domain of $\rho_V$ is $V$, and zero naturally extends to 0 on $U$.
The domain of $\omega$ is $U\cap V$,
Therefore, the expansion of $\rho_V\omega$ to $U$ can also be defined at this time, and it is smooth.
Define $\rho_U\omega$ similarly on $V$. Take $\xi=(-\rho_V\omega, \rho_U\omega)$,
Then $\delta\xi=\omega$. Therefore, the complex sequence is exact.
Mayer-Vietoris sequence
According to the zig-zag lemma, the short exact sequence induces the long exact sequence
$$\cdots\rightarrow H^q(M)\xrightarrow{r}H^q(U)\oplus H^q(V)\xrightarrow{\delta}H^q(U\cap V)\xrightarrow{d^\ast }H^{q+1}(M)\rightarrow \cdots$$
Here the definition of connection homomorphism $d^\ast$ is given by the zig-zag lemma process: using graph tracking,
$\omega\Rightarrow \xi=(-\rho_V\omega,\rho_U\omega)$,
Then derive $r\eta=d\xi$. $r$ is the restriction mapping, so $\eta=
\begin{cases}
d(-\rho_V\omega) &\quad\text{on}\quad U\\
d(\rho_U\omega) &\quad\text{on}\quad V
\end{cases}$.
At $U\cap V$, from $\rho_U+\rho_V=1$, and $d\omega=0$, the definition is legal.
So $d^\ast [\omega]=[\eta]$.
Note that $\rho_U \omega, \rho_V\omega$ is not entirely defined on $M$.
There may not necessarily be $\eta=d\tau$, $\tau\in\Omega^\ast (M)$,
Therefore $\eta$ is not necessarily appropriate.
Compactly supported MV sequence
Let $f:N\rightarrow M$ be a smooth mapping,
But $f^\ast :\Omega^\ast _c(M)\rightarrow \Omega^\ast _c(N)$ is generally not true,
There is no guarantee that the differential form of the tight support will still be pulled back to the differential form of the tight support,
For example, take $f$ as the projection $(x_1,x_2)\mapsto x_1$. However, a smooth proper mapping will suffice.
Because its definition is a mapping that satisfies "the inverse image of a compact set is still a compact set".
Generally, what we consider more is the inclusion mapping: if there is $j: U\rightarrow M$,
Definition$j_\ast : \Omega^\ast _c(U)\rightarrow \Omega^\ast _c(M)$,
$j_\ast \omega$ is general $\omega$ zero extension Go to $M-U$.
Note that the mapped asterisk is marked below.
If only the inclusion mapping between open sets is considered, $\Omega^\ast _c$ becomes a covariant functor. Similarly,
There is also a MV sequence for this functor,
Induced by the containing mapped column $U\cap V\mathop \rightrightarrows \limits_{j_U}^{j_V} U\sqcup V\rightarrow M$.
Here the double arrows point to $U,V$ respectively. From this, the short exact sequence of the complex is constructed:
$$0\rightarrow \Omega^\ast _c(U\cap V)\xrightarrow{\delta} \Omega^\ast _c(U)\oplus \Omega^\ast _c(V)\xrightarrow{s} \Omega^\ast _c(M)\rightarrow 0,$$
Among them $\delta\omega=(-j_\ast \omega,j_\ast \omega)$,
$s(\omega,\tau)=j_\ast \omega+j_\ast \tau$.
In order to prove the legitimacy, we only need to prove that $s$ is full,
Similarly, take $\{\rho_U,\rho_V\}$ as the partition of unity,
Just construct $\rho_U\omega,\rho_V\omega$ as the differential forms on $U$ and $V$ respectively.
Since $\omega$ has a tight support on $M$, $\rho_U$ is supported on $U$,
It can be proved that $\rho_U\omega$ is tightly supported on $U$, and the same is true for $\rho_V\omega$.
Then the long exact sequence is derived from the zig-zag lemma:
$$\cdots\rightarrow H^q_c(U\cap V)\xrightarrow{\delta} H^q_c(U)\oplus H^q_c(V)\xrightarrow{s} H^q_c(M) \xrightarrow{d_\ast }H^{q+1}_c(U\cap V)\rightarrow \cdots$$
In the same way, connection homomorphism can be judged.
Graph tracking found $\eta=d(\rho_V\omega)|_{U\cap V}=-d(\rho_U\omega)|_{U\cap V}$,
$d_\ast [\omega]=[\eta]$.
Similarly, in tight cohomology, $\eta$ is not necessarily appropriate.
Because whether it is $\rho_U\omega$ or $\rho_V\omega$, although there is an overall definition at this time,
But the restriction does not necessarily have compact support in $U\cap V$.
$H^1(S^1)\simeq \mathbb{R}$
Let's look at a simple example. Since $S^1$ is tight, $H_c^\ast (S^1)=H^\ast (S^1)$.
At this time, this proposition can be easily proved using cohomology and compact cohomology MV sequences.
Just cut the $S^1$ in half.
Or use local coordinates $\theta$, $d\theta$ as $S^1$ on $1$-form.
$1$-When the form $\omega$ is appropriate,
$\omega=dF=ad\theta \Leftrightarrow \int_{S^1}\omega=0.$ Therefore
$B^1(S^1)=\ker \int_{S^1}.$
Notice$\int_{S^1}:\Omega^1(S^1)\rightarrow \mathbb{R}.$
Thus $H^1(S^1)=Z^1(S^1)/B^1(S^1)=\Omega^1(S^1)/\ker \int_{S^1}\cong \mathbb{R}$.
The article was last updated on 2021-10-09 12:32:30
