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对于$R$模$A,$ 具有滤列$\{F_pA\},$ 我们有分级模$G_pA=F_pA/F_{p-1}A.$
我们认为$\{F_pA\},\{G_pA\}$延展地决定了$A.$ 若滤列是有界的,
那么仅有$\{G_pA\}$便可延展地恢复出$\{F_pA\},$ 进而决定$A.$
假设有$(C_\ast ,\partial)$是一个滤列链复形, 我们定义了
$$
E^0_{p,q}:=G_pC_{p+q}=F_pC_{p+q}/F_{p-1}C_{p+q},
$$
其上有由$\partial$自然诱导的$\partial_0,$
$(E^0,\partial_0)$是一个链复形. 可以理解为$E^0=\oplus_{p,q}E_{p,q}^0,$
来将双分级降低为单分级.
接下来$E_{p+q}^1:=H_{p+q}(E^0_{p,q}),$ $(E^1,\partial_1)$是链复形.
$\partial_1$由$\partial_0$经过图追踪得到.
谱序列
递归地, 我们定义$E^r_{p,q}:=H_{p+q}(E^r_{p,q}),$
$\partial_r$由$\partial_{r-1}$图追踪得到,
$$
\partial_r:E^r_{p,q}\rightarrow E^{r}_{p-r,q+r-1},\quad E^r_{p,q}:=\frac{\left\{x\in F_pC_{p+q}\mid \partial x\in F_{p-r}C_{p+q-1}\right\}}{F_{p-1}C_{p+q}+\partial (F_{p+r-1}C_{p+q+1})}.
$$
称$r$为谱序列的页数, 称$E^r$为$H_\ast (C_\ast )$的$r$阶逼近.
由这样的递归定义可以得到:
引理 1. 令$(F_pC_\ast ,\partial)$为滤列复形, 有谱序列$E^r_{p,q}.$ 那么, $\partial_r$是良定的, 且$\partial_r^2=0.$ $E^{r+1}$是$(E^r,\partial_r)$的同调, $E^1_{p,q}=H_{p+q}(G_pC_\ast ).$ 若$C_i$上的滤列对每个$i$是有界的, 则$\,\forall\,p,q,$ 当$r$充分大时, $E^r_{p,q}=G_pH_{p+q}(C_\ast ).$
定义 2. 一个谱序列由如下要素组成: $R$模$E^r_{p,q},$ 微分$\partial_r:E_{p,q}^r\rightarrow E^r_{p-r,q+r-1},$ 使得$E^{r+1}$为$E^r$的同调.
定义 3. 称谱序列是收敛的, 若$\,\forall\,p,q,$ 当$r$充分大时, $\partial_r=0,$ 从而,
$$
E^\infty_{p,q}:=E^r_{p,q}=E^{r+1}_{p,q}=\cdots
$$
由引理, 我们有命题:
命题 4. 若$(F_pC_\ast ,\partial)$是一个滤列复形, 那么存在谱序列$(E^r_{p,q},\partial_r),$ 使得$E^1_{p,q}=H_{p+q}(G_pC_\ast ).$ 若滤列在$C_i$上是有界的, 那么谱序列收敛,
$$
E^\infty_{p,q}=G_pH_{p+q}(C_\ast ).
$$
例子
对于CW复形$X,$ 定义$F_pC_\ast (X)=C_\ast (X^p),$ 为$C_\ast (X)$上的滤列.
命题 5. 由CW复形产生的谱序列, $E^r$在第二页后不变, 有:
$$
E^0_{p,q}=C_\ast (X^p,X^{p-1}),
$$
$$
E^1_{p,q}=H_{p+q}(X^p,X^{p-1})=
\left\{
\begin{aligned}
&C_p^{\operatorname{cell} }(X), && q=0\\
&0 &&q\neq 0
\end{aligned}\right.
$$
$$
E^\infty_{p,q}=E^2_{p,q}=
\left\{
\begin{aligned}
&H_p(X), &&q=0\\
&0, &&q\neq 0
\end{aligned}\right.
$$
假设$X$是有限维的, 那么滤列有界,
因此命题说明了$E^\infty_{p,q}=G_pH_{p+q}(C_\ast (X))=E^2_{p,q}.$
记$p+q=i,$ 则
$$
G_pH_i(X)=\left\{
\begin{aligned}
&H_i^{\operatorname{cell} }(X), &&p=i\\
&0, &&p\neq i
\end{aligned}\right.
$$
$$
H_i(X)=G_iH_i(X)=H_i^{\operatorname{cell} }(X).
$$
这就说明了奇异同调与胞腔复形同调的同构关系.
文章最后更新于 2022-09-20 17:39:06
review
For $R$ module $A,$ with filtration $\{F_pA\},$ we have hierarchical module $G_pA=F_pA/F_{p-1}A.$
We believe that $\{F_pA\},\{G_pA\}$ extendsly determines $A.$ if the filter list is bounded,
Then only $\{G_pA\}$ can be extended to recover $\{F_pA\},$ and then decide $A.$
Suppose $(C_\ast ,\partial)$ is a filter chain complex, we define
$$
E^0_{p,q}:=G_pC_{p+q}=F_pC_{p+q}/F_{p-1}C_{p+q},
$$
There is $\partial_0,$ naturally induced by $\partial$ on it
$(E^0,\partial_0)$ is a chain complex. It can be understood as $E^0=\oplus_{p,q}E_{p,q}^0,$
to reduce double classification to single classification.
Next $E_{p+q}^1:=H_{p+q}(E^0_{p,q}),$ $(E^1,\partial_1)$ is the chain complex.
$\partial_1$ is obtained from $\partial_0$ through graph tracing.
spectral sequence
Recursively, we define $E^r_{p,q}:=H_{p+q}(E^r_{p,q}),$
$\partial_r$ is obtained by tracking the $\partial_{r-1}$ graph,
$$
\partial_r:E^r_{p,q}\rightarrow E^{r}_{p-r,q+r-1},\quad E^r_{p,q}:=\frac{\left\{x\in F_pC_{p+q}\mid \partial x\in F_{p-r}C_{p+q-1}\right\}}{F_{p-1}C_{p+q}+\partial (F_{p+r-1}C_{p+q+1})}.
$$
Call $r$ a spectral sequence Number of pages, call $E^r$ the $r$-order approximation of $H_\ast (C_\ast )$.
From this recursive definition we can get:
Lemma 1. Let $(F_pC_\ast ,\partial)$ be the filter sequence complex, and the spectral sequence $E^r_{p,q}.$ then $\partial_r$ is well-determined, and $\partial_r^2=0.$ $E^{r+1}$ is the homology of $(E^r,\partial_r)$, $E^1_{p,q}=H_{p+q}(G_pC_\ast ).$ If the filter sequence on $C_i$ is bounded for each $i$, then $\,\forall\,p,q,$ When $r$ is sufficiently large, $E^r_{p,q}=G_pH_{p+q}(C_\ast ).$
Definition 2. A spectral sequence consists of the following elements: $R$ modulo $E^r_{p,q},$ differential $\partial_r:E_{p,q}^r\rightarrow E^r_{p-r,q+r-1},$ such that $E^{r+1}$ is the homology of $E^r$.
Definition 3. The spectrum sequence is said to be convergent, if $\,\forall\,p,q,$ when $r$ is sufficiently large, $\partial_r=0,$ thus,
$$
E^\infty_{p,q}:=E^r_{p,q}=E^{r+1}_{p,q}=\cdots
$$
By the lemma, we have the proposition:
Proposition 4. If $(F_pC_\ast ,\partial)$ is a filter sequence complex, then there is a spectral sequence $(E^r_{p,q},\partial_r),$ such that $E^1_{p,q}=H_{p+q}(G_pC_\ast ).$ If the filter sequence is bounded on $C_i$, then the spectral sequence converges,
$$
E^\infty_{p,q}=G_pH_{p+q}(C_\ast ).
$$
Example
For the CW complex $X,$, define $F_pC_\ast (X)=C_\ast (X^p),$ as the filtration on $C_\ast (X)$.
Proposition 5. The spectral sequence generated by the CW complex, $E^r$ does not change after the second page, is:
$$
E^0_{p,q}=C_\ast (X^p,X^{p-1}),
$$
$$
E^1_{p,q}=H_{p+q}(X^p,X^{p-1})=
\left\{
\begin{aligned}
&C_p^{\operatorname{cell} }(X), && q=0\\
&0 &&q\neq 0
\end{aligned}\right.
$$
$$
E^\infty_{p,q}=E^2_{p,q}=
\left\{
\begin{aligned}
&H_p(X), &&q=0\\
&0, &&q\neq 0
\end{aligned}\right.
$$
Assuming that $X$ is finite-dimensional, then the filtration is bounded,
Therefore the proposition states $E^\infty_{p,q}=G_pH_{p+q}(C_\ast (X))=E^2_{p,q}.$
Note $p+q=i,$ then
$$
G_pH_i(X)=\left\{
\begin{aligned}
&H_i^{\operatorname{cell} }(X), &&p=i\\
&0, &&p\neq i
\end{aligned}\right.
$$
$$
H_i(X)=G_iH_i(X)=H_i^{\operatorname{cell} }(X).
$$
This illustrates the isomorphic relationship between singular homology and cellular complex homology.
The article was last updated on 2022-09-20 17:39:06
