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de Rham上同调的Poincaré引理
本节我们希望证明,
$H^\ast (\mathbb{R}^n\times \mathbb{R}^1)\cong H^\ast (\mathbb{R}^n)$.
我们记$\pi:\mathbb{R}^n\times \mathbb{R}^1\rightarrow \mathbb{R}^n$为投射,
$s:\mathbb{R}^n\rightarrow \mathbb{R}^n\times \mathbb{R}^1$为零截面,
即$\pi(x,t)=x$, $s(x)=(x,0)$, $x\in \mathbb{R}^n$.
我们有$\pi\circ s=\mathrm{id}\Rightarrow s^\ast \circ\pi^\ast =\mathrm{id}.$
但是$s\circ\pi \neq \mathrm{id}$,
不过确实有$\pi^\ast \circ s^\ast =\mathrm{id}$.
为了证明这一点, 类似代数拓扑中引入$D$,
我们引入同伦算子$K:\Omega^\ast (\mathbb{R}^n\times \mathbb{R}^1)\rightarrow \Omega^{\ast -1}(\mathbb{R}^n\times \mathbb{R}^1).$
若$\mathrm{id}-\pi^\ast \circ s^\ast =\pm (dK \pm Kd)$,
则称$K$为同伦算子,
记$\pi^\ast \circ s^\ast \simeq \mathrm{id}$链同伦.
链同伦的两个映射作用到上同调群中的元素$\omega$上时, 因为$d\omega=0$,
只剩$\pm dK\omega$项, 是恰当形式. 因此两个映射在同调意义下相同,
从而$\pi^\ast \circ s^\ast =id$.
将$\Omega^\ast (\mathbb{R}^n\times \mathbb{R}^1)$中的元素线性拆分为$\pi^\ast \phi\cdot f(x,t)$(无$dt$项),
和$\pi^\ast \phi\wedge f(x,t)dt$(含$dt$项).
$\phi\in \Omega^\ast (\mathbb{R}^n)$.
现令$K:\pi^\ast \phi\cdot f(x,t)\mapsto 0, \pi^\ast \phi\wedge f(x,t)dt\mapsto \pi^\ast \phi\cdot \int_0^tf(x,u)du$.
引理 1.1. $K$为同伦算子.
计算$(\mathrm{id}-\pi^\ast s^\ast )$和$dK-Kd$作用在$\omega$上的值,
由线性性分别考虑上述两种情况,
得到$\mathrm{id}-\pi^\ast s^\ast =(-1)^{q-1}(dK-Kd)$即可.
式子比较长但计算比较简单, 需要注意符号.
从而, 我们有:
命题 1.2. $\pi^\ast$, $s^\ast$互为同构, $H^\ast (\mathbb{R}^n\times \mathbb{R}^1)\cong H^\ast (\mathbb{R}^n)$.
以及推论:
推论 1.3. $H^q(\mathbb{R}^n)=H^q(\ast )= \begin{cases} \mathbb{R}, \quad &q=0\\ 0, &q\neq 0 \end{cases}$, 这里$\ast =\mathbb{R}^0$, 即单点空间.
更一般的, 我们有如下定理:
定理 1.4 (de Rham 上同调同伦公理). $M,N$为光滑流形, $f,g$是光滑同伦的两个从$M$到$N$的光滑映射, 则$f^\ast =g^\ast$.
证 取$F:M\times I\rightarrow N$ 为光滑同伦,
$\,\forall\,\omega\in \Omega^q(N)$,
$F^\ast \omega\in \Omega^q(M\times I)$.
局部将$F^\ast \omega$分解为$F^\ast \omega=\omega_1+\omega_2\wedge dt$,
$\omega_1,\omega_2$分别是$q,q-1$形式.
仿照之前的构造方法, 构造$K$把$\omega$映为$0$,
把$\omega_2$中的函数项关于$t$积分. 这时再利用$F(x,0)=f(x)$,
$F(x,1)=g(x)$, 利用简单的Leibnitz法则证明下式即可:
$$dK(F^\ast \omega)\pm Kd(F^\ast \omega)=g^\ast \omega-f^\ast \omega.$$
注意$dF^\ast =F^\ast d$, 因此上式成立意味着$KF^\ast$是同伦算子,
$g^\ast \simeq f^\ast$, 在同调群意义下相等.
推论 1.5. 同伦等价的两个流形有相同的de Rham上同调.
特别地, 有:
推论 1.6. 若$A$为$M$的形变收缩, 则$A$与$M$有相同的de Rham上同调.
紧上同调的Poincaré引理
类似地, 对于紧上同调, 我们希望证明:
$H^\ast _c(\mathbb{R}^n\times \mathbb{R}^1)\cong H^{\ast -1}_c(\mathbb{R}^n).$
此时$\pi^\ast$不再能保持紧支集, 我们需要考虑一个新的映射$\pi_\ast$.
$\,\forall\,\omega\in \Omega^\ast _c(M\times \mathbb{R}^1)$,
将其按是否含$dt$项分为两种形式的线性组合,
即$\pi^\ast \phi\cdot f(x,t)$与$\pi^\ast \phi\wedge f(x,t)dt$,
其中$\phi\in \Omega^\ast (M)$,
$f(x,t)\in C_c^\infty (M\times \mathbb{R}^1)$.
我们定义$\pi_\ast :\phi\cdot f(x,t)\mapsto 0$,
$\phi\wedge f(x,t)dt\mapsto \phi\cdot\int_{\mathbb{R}^1}f(x,t)dt$.
经过简单的验证发现$d\pi_\ast =\pi_\ast d$是链映射,
从而可以诱导$\pi_\ast : H^q_c(M\times \mathbb{R}^1)\rightarrow H^{q-1}_c(M)$.
我们接下来找它的逆映射$e_\ast$.
定义$e_\ast :\Omega^q_c(M)\rightarrow \Omega_c^{q+1}(M\times \mathbb{R}^1)$,
$e_\ast (\phi)=\pi^\ast \phi \wedge e$,
其中$e=e(t)dt\in \Omega_c^1(\mathbb{R}^1)$,
$\int_{\mathbb{R}^1}e(t)dt=1$(如 bump 1-形式). 容易验证它也是链映射,
诱导了紧上同调群上的映射.
接下来就有如下命题:
命题 1.7. $\pi_\ast \circ e_\ast =\mathrm{id}$, $e_\ast \circ \pi_\ast \simeq \mathrm{id}$
证 第一部分显然, 主要需要构造同伦算子$K$连接$e_\ast$与$\pi_\ast$.
定义$K:\pi^\ast \phi\cdot f(x,t)\mapsto 0$,
$\pi^\ast \phi\wedge f(x,t)dt\mapsto \pi^\ast \phi\cdot \int_{-\infty}^{t}f(x,u)du-\pi^\ast \phi\cdot A(t)\cdot \int_{\mathbb{R}^1}f(x,t)dt$,
其中$A(t)=\int_{-\infty}^t e(u)du$.
验证$id-e_\ast \circ \pi_\ast =(-1)^{q-1}(dK-Kd)$即可, 注意利用紧性,
很多积分积出来为$0$.
从而得到下面的命题:
命题 1.8. $\pi_\ast ,e_\ast$互为逆映射, $H_c^\ast (M\times \mathbb{R}^1)\cong H_c^{\ast -1}(M)$
以及推论:
推论 1.9. $H^q(\mathbb{R}^n)=\begin{cases} \mathbb{R}, \quad &q=n\\ 0, &q\neq n \end{cases}$.
通过$0$-形式的生成元$1$,
可以用$e_\ast$依次找到$n$-形式生成元$e(x_1)\cdots e(x_n){} dx_{1}\wedge\cdots\wedge{} dx_{ {}n}$,
是$\mathbb{R}^n$上的一个bump $n$-形式, 紧支集可任意小.
逆紧映射的度
接下来是紧支集Poincaré引理的应用, 给出了逆紧映射的光滑不变量.
若$f:\mathbb{R}^n\rightarrow \mathbb{R}^n$光滑逆紧,
则有拉回$f^\ast :\Omega^\ast _c(\mathbb{R}^n)\rightarrow \Omega^\ast _c(\mathbb{R}^n)$,
诱导$f^\ast :H^\ast _c(\mathbb{R}^n)\rightarrow H^\ast _c(\mathbb{R}^n)$有定义.
取$[\alpha]\in H^n_c(\mathbb{R}^n)$为生成元,
$\int_{\mathbb{R}^n}\alpha =1$, $f^\ast [\alpha]=c[\alpha]$,
$c$称为$f$的度, 记作$\deg f$.
易见$\deg f=\int_{\mathbb{R}^n} f^\ast \alpha$. 令人意外的,
$\deg f\in \mathbb{Z}$.
证明
首先给几个微分几何中的定义:
若$f:\mathbb{R}^m\rightarrow \mathbb{R}^n$为光滑映射,
则其诱导了$f_{\ast p}:T_p\mathbb{R}^m\rightarrow T_{f(p)}\mathbb{R}^n$为切映射.
若其非满, 则称$p$为临界点, $f(p)$为临界值.
$\mathbb{R}^n$中其余点都是正则值.
命题 1.10. 若$f:\mathbb{R}^n\rightarrow \mathbb{R}^n$非满, 则$\deg f=0$.
证 $f$逆紧, 从而$f(\mathbb{R}^n)$闭,
$\mathbb{R}^n-f(\mathbb{R}^n)$开.
存在开集$U\subset \mathbb{R}^n-f(\mathbb{R}^n)$.
令生成元$[\alpha]$支集在$U$中, 则$f^\ast \alpha=0$, $\deg f=0$.
之后首先先承认一个定理:
定理 1.11 (Sard 定理). 对任意正整数$m,n$, 临界值集是$\mathbb{R}^n$零测集.
接下来证明主命题: $\deg f\in \mathbb{Z}$.
证 设$f:\mathbb{R}^n\rightarrow \mathbb{R}^n$满, 由Sard定理,
$f(\mathbb{R}^n)$中几乎处处为正则值. 设$q\in f(\mathbb{R}^n)$为正则值.
$\,\forall\,p\in f^{-1}(q)$, $f_{\ast p}$是满线性映射, 从而为同构.
由反函数定理, $f$在$p$点附近微分同胚. 于是$f^{-1}(q)$为离散点集.
又$f$逆紧, $q$点逆像紧, 从而$f^{-1}(q)$为有限点集,
记为$f^{-1}(q)=\{p_1,...,p_N\}$.
对每个$p_i$, 取$U_i\approx f(U_i)$, 使得$U_i\cap U_j=\varnothing$.
取$\alpha \in \Omega_c^n(\mathbb{R}^n)$支集在$\bigcap_i f(U_i)$里.
从而$\operatorname{supp}f^\ast \alpha\subset \bigcup U_i$.
$\int_{U_i}f^\ast \alpha=\pm \int_{f(U_i)} \alpha=\pm 1$,
符号由微分同胚Jacobi正负号决定.
于是$\int_{\mathbb{R}^n} f^\ast \alpha=\sum_{f^{-1}(q)}\pm 1\in \mathbb{Z}$.
文章最后更新于 2021-09-20 16:39:03
de Rham's cohomological Poincaré lemma
In this section we hope to prove,
$H^\ast (\mathbb{R}^n\times \mathbb{R}^1)\cong H^\ast (\mathbb{R}^n)$.
We record $\pi:\mathbb{R}^n\times \mathbb{R}^1\rightarrow \mathbb{R}^n$ as projection,
$s:\mathbb{R}^n\rightarrow \mathbb{R}^n\times \mathbb{R}^1$ is zero section,
That is $\pi(x,t)=x$, $s(x)=(x,0)$, $x\in \mathbb{R}^n$.
We have $\pi\circ s=\mathrm{id}\Rightarrow s^\ast \circ\pi^\ast =\mathrm{id}.$
But $s\circ\pi \neq \mathrm{id}$,
But there is indeed $\pi^\ast \circ s^\ast =\mathrm{id}$.
In order to prove this point, similar to the introduction of $D$ in algebraic topology,
We introduce homotopy operator $K:\Omega^\ast (\mathbb{R}^n\times \mathbb{R}^1)\rightarrow \Omega^{\ast -1}(\mathbb{R}^n\times \mathbb{R}^1).$
If $\mathrm{id}-\pi^\ast \circ s^\ast =\pm (dK \pm Kd)$,
Then $K$ is called a homotopy operator,
Note$\pi^\ast \circ s^\ast \simeq \mathrm{id}$ chain homotopy.
When two mappings of chain homotopy act on the element $\omega$ in the cohomology group, because $d\omega=0$,
Only the $\pm dK\omega$ term is left, which is the appropriate form. Therefore, the two mappings are the same in the homology sense,
Thus $\pi^\ast \circ s^\ast =id$.
Linearly split the elements in $\Omega^\ast (\mathbb{R}^n\times \mathbb{R}^1)$ into $\pi^\ast \phi\cdot f(x,t)$ (no $dt$ items),
and $\pi^\ast \phi\wedge f(x,t)dt$ (including $dt$ items).
$\phi\in \Omega^\ast (\mathbb{R}^n)$.
Now order$K:\pi^\ast \phi\cdot f(x,t)\mapsto 0, \pi^\ast \phi\wedge f(x,t)dt\mapsto \pi^\ast \phi\cdot \int_0^tf(x,u)du$.
Lemma 1.1. $K$ is the homotopy operator.
Calculate the value of $(\mathrm{id}-\pi^\ast s^\ast )$ and $dK-Kd$ acting on $\omega$,
Consider the above two situations respectively based on linearity,
Just get $\mathrm{id}-\pi^\ast s^\ast =(-1)^{q-1}(dK-Kd)$.
The formula is relatively long but the calculation is relatively simple, so you need to pay attention to the symbols.
Thus, we have:
Proposition 1.2. $\pi^\ast$, $s^\ast$ are isomorphic to each other, $H^\ast (\mathbb{R}^n\times \mathbb{R}^1)\cong H^\ast (\mathbb{R}^n)$.
and corollary:
Corollary 1.3. $H^q(\mathbb{R}^n)=H^q(\ast )= \begin{cases} \mathbb{R}, \quad &q=0\\ 0, &q\neq 0 \end{cases}$, here $\ast =\mathbb{R}^0$, is a single point space.
More generally, we have the following theorem:
Theorem 1.4 (de Rham's cohomology and homotopy axiom). $M,N$ is a smooth manifold, $f,g$ is two smooth homotopy smooth mappings from $M$ to $N$, then $f^\ast =g^\ast$.
certificate Take $F:M\times I\rightarrow N$ as smooth homotopy,
$\,\forall\,\omega\in \Omega^q(N)$,
$F^\ast \omega\in \Omega^q(M\times I)$.
Partially decompose $F^\ast \omega$ into $F^\ast \omega=\omega_1+\omega_2\wedge dt$,
$\omega_1,\omega_2$ is in the form of $q,q-1$ respectively.
Following the previous construction method, construct $K$ and map $\omega$ to $0$,
Integrate the function terms in $\omega_2$ with respect to $t$. Then use $F(x,0)=f(x)$,
$F(x,1)=g(x)$, just use the simple Leibnitz rule to prove the following formula:
$$dK(F^\ast \omega)\pm Kd(F^\ast \omega)=g^\ast \omega-f^\ast \omega.$$
Pay attention to $dF^\ast =F^\ast d$, so the above formula is true, which means $KF^\ast$ is a homotopy operator,
$g^\ast \simeq f^\ast$, equal in the sense of a homology group.
Corollary 1.5. Two manifolds that are homotopically equivalent have the same de Rham cohomology.
In particular, there are:
Corollary 1.6. If $A$ is the deformation contraction of $M$, then $A$ and $M$ have the same de Rham cohomology.
Poincaré's Lemma for Compact Cohomology
Similarly, for tight cohomology, we wish to prove:
$H^\ast _c(\mathbb{R}^n\times \mathbb{R}^1)\cong H^{\ast -1}_c(\mathbb{R}^n).$
At this time $\pi^\ast$ can no longer maintain a compact support, and we need to consider a new mapping $\pi_\ast$.
$\,\forall\,\omega\in \Omega^\ast _c(M\times \mathbb{R}^1)$,
Divide it into two forms of linear combinations according to whether it contains the $dt$ term,
That is $\pi^\ast \phi\cdot f(x,t)$ and $\pi^\ast \phi\wedge f(x,t)dt$,
Among them $\phi\in \Omega^\ast (M)$,
$f(x,t)\in C_c^\infty (M\times \mathbb{R}^1)$.
We define $\pi_\ast :\phi\cdot f(x,t)\mapsto 0$,
$\phi\wedge f(x,t)dt\mapsto \phi\cdot\int_{\mathbb{R}^1}f(x,t)dt$.
After simple verification, it was found that $d\pi_\ast =\pi_\ast d$ is a chain mapping,
This can induce $\pi_\ast : H^q_c(M\times \mathbb{R}^1)\rightarrow H^{q-1}_c(M)$.
We next find its inverse mapping $e_\ast$.
Definition$e_\ast :\Omega^q_c(M)\rightarrow \Omega_c^{q+1}(M\times \mathbb{R}^1)$,
$e_\ast (\phi)=\pi^\ast \phi \wedge e$,
Among them $e=e(t)dt\in \Omega_c^1(\mathbb{R}^1)$,
$\int_{\mathbb{R}^1}e(t)dt=1$ (as bump 1-form). Easy to verify it is also a chain map,
Induces mappings on tight cohomology groups.
What follows is the following proposition:
Proposition 1.7. $\pi_\ast \circ e_\ast =\mathrm{id}$, $e_\ast \circ \pi_\ast \simeq \mathrm{id}$
certificate Obviously, in the first part, we mainly need to construct the homotopy operator $K$ to connect $e_\ast$ and $\pi_\ast$.
Definition$K:\pi^\ast \phi\cdot f(x,t)\mapsto 0$,
$\pi^\ast \phi\wedge f(x,t)dt\mapsto \pi^\ast \phi\cdot \int_{-\infty}^{t}f(x,u)du-\pi^\ast \phi\cdot A(t)\cdot \int_{\mathbb{R}^1}f(x,t)dt$,
Among them $A(t)=\int_{-\infty}^t e(u)du$.
Just verify $id-e_\ast \circ \pi_\ast =(-1)^{q-1}(dK-Kd)$, pay attention to the use of compactness,
The product of many integrals is $0$.
This leads to the following proposition:
Proposition 1.8. $\pi_\ast ,e_\ast$ are inverse mappings to each other, $H_c^\ast (M\times \mathbb{R}^1)\cong H_c^{\ast -1}(M)$
and corollary:
Corollary 1.9. $H^q(\mathbb{R}^n)=\begin{cases} \mathbb{R}, \quad &q=n\\ 0, &q\neq n \end{cases}$.
Through the generator $1$ of the form $0$,
You can use $e_\ast$ to find $n$-form generator $e(x_1)\cdots e(x_n){} dx_{1}\wedge\cdots\wedge{} dx_{ {}n}$ in sequence,
is a bump $n$-form on $\mathbb{R}^n$, and the compact support can be arbitrarily small.
degree of proper mapping
Next is the application of Poincaré's lemma on compact supported sets, which gives smooth invariants of proper maps.
If $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is smooth and inversely compact,
Then there is pullback $f^\ast :\Omega^\ast _c(\mathbb{R}^n)\rightarrow \Omega^\ast _c(\mathbb{R}^n)$,
Induction$f^\ast :H^\ast _c(\mathbb{R}^n)\rightarrow H^\ast _c(\mathbb{R}^n)$ is defined.
Take $[\alpha]\in H^n_c(\mathbb{R}^n)$ as the generator,
$\int_{\mathbb{R}^n}\alpha =1$, $f^\ast [\alpha]=c[\alpha]$,
$c$ is called the degree of $f$, denoted as $\deg f$.
Yi Jian$\deg f=\int_{\mathbb{R}^n} f^\ast \alpha$. Surprisingly,
$\deg f\in \mathbb{Z}$.
prove
First, give some definitions in differential geometry:
If $f:\mathbb{R}^m\rightarrow \mathbb{R}^n$ is a smooth mapping,
Then it induces $f_{\ast p}:T_p\mathbb{R}^m\rightarrow T_{f(p)}\mathbb{R}^n$ to be a tangent map.
If it is not full, $p$ is called the critical point and $f(p)$ is the critical value.
The remaining points in $\mathbb{R}^n$ are regular values.
Proposition 1.10. If $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is not full, then $\deg f=0$.
certificate $f$ is inversely tight, so $f(\mathbb{R}^n)$ is closed,
$\mathbb{R}^n-f(\mathbb{R}^n)$Open.
There is an open set $U\subset \mathbb{R}^n-f(\mathbb{R}^n)$.
Let the support of generator $[\alpha]$ be in $U$, then $f^\ast \alpha=0$, $\deg f=0$.
Then first admit a theorem:
Theorem 1.11 (Sard's theorem). For any positive integer $m,n$, the critical value set is $\mathbb{R}^n$ the zero test set.
Next, prove the main proposition: $\deg f\in \mathbb{Z}$.
certificate Assume $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is full, according to Sard's theorem,
Almost everywhere in $f(\mathbb{R}^n)$ there are regular values. Let $q\in f(\mathbb{R}^n)$ be the regular value.
$\,\forall\,p\in f^{-1}(q)$, $f_{\ast p}$ are full linear mappings, thus isomorphism.
According to the inverse function theorem, $f$ is diffeomorphic near the point $p$. So $f^{-1}(q)$ is a discrete point set.
Also $f$ is inversely compact, $q$ is inversely compact, so $f^{-1}(q)$ is a finite point set,
Marked as $f^{-1}(q)=\{p_1,...,p_N\}$.
For each $p_i$, take $U_i\approx f(U_i)$ such that $U_i\cap U_j=\varnothing$.
Take the support $\alpha \in \Omega_c^n(\mathbb{R}^n)$ in $\bigcap_i f(U_i)$.
Thus $\operatorname{supp}f^\ast \alpha\subset \bigcup U_i$.
$\int_{U_i}f^\ast \alpha=\pm \int_{f(U_i)} \alpha=\pm 1$,
The sign is determined by the diffeomorphism Jacobi sign.
So $\int_{\mathbb{R}^n} f^\ast \alpha=\sum_{f^{-1}(q)}\pm 1\in \mathbb{Z}$.
The article was last updated on 2021-09-20 16:39:03
