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好覆盖组合计算de Rham上同调
设$\mathcal{U}=\{U_\alpha\}$为$M$的开覆盖, 则$\mathcal{U}$的 nerve
$N(\mathcal{U})$为一个单纯复形: 对每个$U_\alpha$赋予一个顶点$\alpha$,
若$U_{\alpha\beta...\gamma}$非空, 则添加$[\alpha,\beta,...,\gamma]$单形.
$\delta :C^\ast (\mathcal{U},\mathbb{R})\rightarrow C^{\ast +1}(\mathcal{U},\mathbb{R})$事实上即为复形层面的上同调.
于是这提供了另一种求上同调的方法:
即考虑求$H^\ast (\mathcal{U},\mathbb{R})\cong H^\ast (M)$.
显式同构例子
$H^2(M)\cong H^2(\mathcal{U},\mathbb{R})$为几何上的重要例子.
这个同构具体的给出, 顺着双复形沿对角线$(K^2)$跑即可,
需适当添加下一行对角线$(K^1)$上的元素.
即$[\omega]_d=[\omega]_D=[c]_D=[c]_\delta$, 其中存在$\eta\in K^1$,
$D\eta=c-\omega$.
反过来的跑, 需要利用上节课程中的同伦算子$K$.
$(Kc)_{\beta\alpha}=\sum \rho_\gamma c_{\gamma\beta\alpha}$. 从而显式地,
我们有对应于$c$的$((D''K)^2c)_\alpha=\sum d\rho_\gamma\wedge d\rho_\beta \cdot c_{\gamma\beta\alpha}\in \Omega^2(U_\alpha)$,
它在$M$上整体定义.
显式同构
引理 1.1. $\delta(D'' K)^i=(D''K)^i\delta-(D''K)^{i-1}D''$
计算即可. 由此得到:
命题 1.2. $\eta\in C^n(\mathcal{U},\mathbb{R})$, $\delta \eta=0$, 则它对应的闭$n$形式$\omega=(-1)^n(D''K)^n\eta$.
由于$\Omega^\ast (M)$与$C^\ast (\mathcal{U},\Omega^\ast )$上同调同构,
其间应有链映射$f$, 与$r$在上同调意义下互为逆映射.
我们先来定义$f$. 设$\alpha=\sum_{i=0}^n\alpha_i$,
$\alpha_i\in C^i(\mathcal{U},\Omega^{n-i})$. 对每个$\alpha_i$,
定义$f(\alpha_i)=(1-K\delta)(-D''K)^i\alpha_i$,
则$\delta f(\alpha_i)=0$, $f(\alpha)=\sum_{i=0}^n f(\alpha_i)$整体定义.
引理 1.3. $D\alpha=\beta=\sum_{i=0}^{n+1}\beta_i$, $f(\alpha)=\sum_{i=0}^n(-D''K)^i\alpha_i-\sum_{i=1}^{n+1}K(-D''K)^{i-1}\beta_i$.
利用之前的引理即可. 接下来我们有:
引理 1.4. $f$是链映射, $f\circ r=\mathrm{id}$.
前项利用先前的公式验证$df=fD$即可,
后项注意$r\alpha$只有$\alpha_0$项且整体定义.
还有一个命题需要证明:
引理 1.5. $1-r\circ f=DL+LD$, 同伦算子$L\alpha=\sum_{p=0}^{n-1}(L\alpha)_p$, $(L\alpha)_p=\sum_{i=p+1}^nK(-D''K)^{i-(p+1)}\alpha_i$.
事实上这个同伦算子很容易想到, 就是对于每一个$n-1$-上链的分量,
把$n$-上链每一个分量能打到该位置的都往上打即可. 从而, 我们有:
命题 1.6 (Collating 公式). $f:C^\ast (\mathcal{U},\Omega^\ast )\cong \Omega^\ast (M)$为同构, 逆映射就是$r$.
文章最后更新于 2021-10-25 14:18:02
Good covering combination calculation de Rham cohomology
Assume $\mathcal{U}=\{U_\alpha\}$ is the open cover of $M$, then the nerve of $\mathcal{U}$
$N(\mathcal{U})$ is a simplicial complex: assign a vertex $\alpha$ to each $U_\alpha$,
If $U_{\alpha\beta...\gamma}$ is not empty, add $[\alpha,\beta,...,\gamma]$ simplex.
$\delta :C^\ast (\mathcal{U},\mathbb{R})\rightarrow C^{\ast +1}(\mathcal{U},\mathbb{R})$ is actually cohomology at the complex level.
So this provides another way to find cohomology:
That is, consider finding $H^\ast (\mathcal{U},\mathbb{R})\cong H^\ast (M)$.
Explicit isomorphism example
$H^2(M)\cong H^2(\mathcal{U},\mathbb{R})$ is an important example in geometry.
This isomorphism is given concretely, just run along the diagonal $(K^2)$ along the double complex,
The elements on the next row of diagonals $(K^1)$ need to be added appropriately.
That is $[\omega]_d=[\omega]_D=[c]_D=[c]_\delta$, where there is $\eta\in K^1$,
$D\eta=c-\omega$.
To run in reverse, you need to use the homotopy operator $K$ in the previous lesson.
$(Kc)_{\beta\alpha}=\sum \rho_\gamma c_{\gamma\beta\alpha}$. Thus explicitly,
We have $((D''K)^2c)_\alpha=\sum d\rho_\gamma\wedge d\rho_\beta \cdot c_{\gamma\beta\alpha}\in \Omega^2(U_\alpha)$ corresponding to $c$,
It is defined overall on $M$.
explicit isomorphism
Lemma 1.1. $\delta(D'' K)^i=(D''K)^i\delta-(D''K)^{i-1}D''$
Just calculate. From this we get:
Proposition 1.2. $\eta\in C^n(\mathcal{U},\mathbb{R})$, $\delta \eta=0$, then its corresponding closed $n$ form $\omega=(-1)^n(D''K)^n\eta$.
Since $\Omega^\ast (M)$ and $C^\ast (\mathcal{U},\Omega^\ast )$ are cohomologically isomorphic,
There should be a chain mapping $f$, which is an inverse mapping to $r$ in the sense of cohomology.
Let’s define $f$ first. Let $\alpha=\sum_{i=0}^n\alpha_i$,
$\alpha_i\in C^i(\mathcal{U},\Omega^{n-i})$. For each $\alpha_i$,
Definition$f(\alpha_i)=(1-K\delta)(-D''K)^i\alpha_i$,
Then $\delta f(\alpha_i)=0$, $f(\alpha)=\sum_{i=0}^n f(\alpha_i)$ are defined as a whole.
Lemma 1.3. $D\alpha=\beta=\sum_{i=0}^{n+1}\beta_i$, $f(\alpha)=\sum_{i=0}^n(-D''K)^i\alpha_i-\sum_{i=1}^{n+1}K(-D''K)^{i-1}\beta_i$.
Just use the previous lemma. Next we have:
Lemma 1.4. $f$ is the chain mapping, $f\circ r=\mathrm{id}$.
The previous term can be verified using the previous formula $df=fD$,
Note that the latter term $r\alpha$ only has $\alpha_0$ term and is defined as a whole.
There is another proposition that needs to be proved:
Lemma 1.5. $1-r\circ f=DL+LD$, homotopy operators $L\alpha=\sum_{p=0}^{n-1}(L\alpha)_p$, $(L\alpha)_p=\sum_{i=p+1}^nK(-D''K)^{i-(p+1)}\alpha_i$.
In fact, this homotopy operator is easy to think of, that is, for each $n-1$-on-chain component,
Just move up every component of the $n$-winding chain that can reach the position. Thus, we have:
Proposition 1.6 (Collating formula). $f:C^\ast (\mathcal{U},\Omega^\ast )\cong \Omega^\ast (M)$ is isomorphism, and the inverse mapping is $r$.
The article was last updated on 2021-10-25 14:18:02
