Machine-translated from Chinese.
概述
参考书目:
Hutchings, Introduction to spectral sequences;
Hatcher, AT Chapter 5;
Ramos, Spectral sequences via examples.
动机
给定纤维丛
$$
F\rightarrow E\rightarrow B,
$$
我们希望能够利用谱序列计算$H_\ast (E),$ 利用$H_\ast (F)$与$H_\ast (B)$的信息.
回忆对于CW复形对$(X,A)$, 有
$$
A\rightarrow X\rightarrow X/A,
$$
它也称为余纤维(cofibration)(商纤维, 商丛?).
其上我们有zig-zag引理得到长正合列来计算同调群, 但对于基本群,
没有太多明显的工具.
而对于纤维丛, 我们可以通过长正合列来计算基本群,
而同调群的计算则仰仗于谱序列.
谱序列
对于链复形$(C_\ast ,\partial),$ 取$F_0C_\ast \subset C_\ast $为子复形,
我们有短正合列:
$$
0\rightarrow F_0C_\ast \rightarrow C_\ast \rightarrow C_\ast /F_0C^\ast \rightarrow 0.
$$
这诱导了长正合列
$$
\cdots\rightarrow H_i(F_0C^\ast )\rightarrow H_i(C_\ast )\rightarrow H_i(C_\ast /F_{0}C^\ast )\xrightarrow{\delta_i}H_{i-1}(F_0C_\ast )\rightarrow\cdots.
$$
回忆$\delta$的定义为$\alpha\mapsto [\partial x],$ $[x]=\alpha.$
假设我们希望从$H_\ast (F_0C_\ast ),H_\ast (C_\ast /F_0C_\ast )$计算$H_\ast (C_\ast ),$
我们将长正合列拆为短正合列,
$$
0\rightarrow \operatorname{coker}(\delta_{i+1})\rightarrow H_i(C_\ast )\rightarrow \ker(\delta_i)\rightarrow 0.
$$
总结一下, 我们为了计算$H_i(C_\ast ),$ 我们需要:
计算$H_i(F_0C_\ast ),H_i(C_\ast /F_0C_\ast );$
考虑仅有两项的复形$H_{i}(C_\ast /F_0C_\ast )\xrightarrow{\delta_i}H_{i-1}(F_0C_\ast ),$
它的同调是:
$$
G_0H_\ast =\operatorname{coker}\delta,\quad G_1H_\ast =\ker\delta.
$$
- 我们得到了短正合列:
$$
0\rightarrow G_0H_\ast \rightarrow H_\ast (C_\ast )\rightarrow G_1H_\ast \rightarrow 0.
$$
注意短正合列不一定可分裂, 因此用代数术语来描述, 我们说$H_\ast (C_\ast )$是
延展地(\”up to extensions\”) 决定的.
滤列
一个滤列$R$模是一个$R$模$A,$ 有一列递增子模
$$
\cdots \subset F_p A\subset F_{p+1}A\subset \cdots,\quad p\in \mathbb{Z},
$$
使得$\bigcup_p F_pA=A,$ $\bigcap_pF_pA=0.$ 这被称为滤列.
称滤列是有界的, 若非平凡滤子有限, 即
$$
\cdots =0=F_p A\subset \cdots \subset F_{q} A=A=\cdots.
$$
那么定义它的联系分级模为
$$
G_pA:=F_pA/F_{p-1}A.
$$
我们认为$\{F_p A\},$ $\{G_pA\}$归纳地延展地决定$A.$
$$
0\rightarrow F_{p-1}A\rightarrow F_pA\rightarrow G_pA\rightarrow 0.
$$
注意延展是平凡的, 若$R$是一个域.
一个滤列链复形是一个链复形$(C_\ast ,\partial),$
以及关于$C_i$的滤列$\{F_pC_i\},$
使得$\partial(F_pC_i)\subset F_pC_{i-1}.$ 可以验证, $\,\forall\,p,$
$(G_pC_\ast ,\partial)$是一个链复形, 可以考虑其上的同调$H_i(G_pC_\ast )$.
一个$C_\ast $上的滤列诱导了$H_\ast (C_\ast )$上的滤列:
$$
F_pH_i(C_\ast ):=\{\alpha\in H_i(C_\ast )|\,\exists\,x\in F_pC_\ast , [x]=\alpha\}.
$$
接下来我们有$G_pH_i(C_\ast ).$ 一个自然的问题是它与$H_i(G_pC_\ast )$有什么联系?
回忆前面有
$$
0\subset F_0C_\ast \subset F_1C_\ast =C_\ast ,
$$
则$G_pH_\ast (C_\ast )$是$H_\ast (G_1C_\ast )\xrightarrow{\delta} H_{\ast -1}(G_0C_\ast )$的同调.
此时$H_i(G_pC_\ast )$基本决定了$G_pH_i(C_\ast ).$
当滤列有更多非平凡项时, 为了考虑它们之间的关系,
我们希望能考虑一些连续的逼近$E^1,E^2,\cdots$收敛到想要的结果,
类似于Taylor展开. 这就称为谱序列.
假设$(F_pC_\ast ,\partial)$是一个滤列复形, 定义:
$$
E^0_{p,q}:=G_pC_{p+q}=F_pC_{p+q}/F_{p-1}C_{p+q},
$$
那么$E$是双分级的,
$p$称为滤度(filtration degree), $p+q$是同调度(homology degree),
$q$称为补度(complementary degree).
那么$(E_0,\partial_0)$是一个链复形,
$\partial_0:E_{p,q}^0\rightarrow E^0_{p,q-1}.$
定义$E^1_{p,q}:=H_{p+q}(G_pC_\ast ),$ 它是$H_\ast (C_\ast )$的一阶逼近.
如下定义$\partial_1:E^1_{p,q}\rightarrow E^1_{p-1,q}:$
$\,\forall\,\alpha\in E^1_{p,q}=H_{p+q}(G_pC_\ast ),$ 取$x\in F_pC_{p+q},$
$[x]=\alpha,$ $\partial x\in F_{p-1}C_{p+q-1}.$
记$\partial_1\alpha=[\partial x].$
文章最后更新于 2022-09-15 10:54:54
Overview
Bibliography:
Hutchings, Introduction to spectral sequences;
Hatcher, AT Chapter 5;
Ramos, Spectral sequences via examples.
motivation
Given fiber bundle
$$
F\rightarrow E\rightarrow B,
$$
We hope to use the spectral sequence to calculate $H_\ast (E),$ and use the information of $H_\ast (F)$ and $H_\ast (B)$.
Recall that for the CW complex pair $(X,A)$, we have
$$
A\rightarrow X\rightarrow X/A,
$$
it is also called remaining fiber (cofibration)(Shang Fiber, Shang Cong?).
On top of it, we have the zig-zag lemma to get the long exact sequence to calculate the homology group, but for the fundamental group,
There aren't many obvious tools.
For fiber bundles, we can calculate the fundamental group through long exact sequences,
The calculation of homology groups relies on spectral sequences.
spectral sequence
For the chain complex $(C_\ast ,\partial),$, take $F_0C_\ast \subset C_\ast $ as the subcomplex,
We have short exact sequences:
$$
0\rightarrow F_0C_\ast \rightarrow C_\ast \rightarrow C_\ast /F_0C^\ast \rightarrow 0.
$$
This induces a long positive column
$$
\cdots\rightarrow H_i(F_0C^\ast )\rightarrow H_i(C_\ast )\rightarrow H_i(C_\ast /F_{0}C^\ast )\xrightarrow{\delta_i}H_{i-1}(F_0C_\ast )\rightarrow\cdots.
$$
Recall $\delta$ is defined as $\alpha\mapsto [\partial x],$ $[x]=\alpha.$
Suppose we wish to calculate $H_\ast (C_\ast ),$ from $H_\ast (F_0C_\ast ),H_\ast (C_\ast /F_0C_\ast )$
We split the long exact sequence into short exact sequences,
$$
0\rightarrow \operatorname{coker}(\delta_{i+1})\rightarrow H_i(C_\ast )\rightarrow \ker(\delta_i)\rightarrow 0.
$$
To summarize, in order to calculate $H_i(C_\ast ),$ we need:
Calculate$H_i(F_0C_\ast ),H_i(C_\ast /F_0C_\ast );$
Consider the complex $H_{i}(C_\ast /F_0C_\ast )\xrightarrow{\delta_i}H_{i-1}(F_0C_\ast ),$ with only two terms
Its homology is:
$$
G_0H_\ast =\operatorname{coker}\delta,\quad G_1H_\ast =\ker\delta.
$$
- We get the short exact sequence:
$$
0\rightarrow G_0H_\ast \rightarrow H_\ast (C_\ast )\rightarrow G_1H_\ast \rightarrow 0.
$$
Note that the short exact sequence is not necessarily divisible, so to describe it in algebraic terms, we say that $H_\ast (C_\ast )$ is
Extended (\"up to extensions\") determined.
filtration
A filtration $R$ module is a $R$ module $A,$ with a column of increasing submodules.
$$
\cdots \subset F_p A\subset F_{p+1}A\subset \cdots,\quad p\in \mathbb{Z},
$$
such that $\bigcup_p F_pA=A,$ $\bigcap_pF_pA=0.$ this is called filtration.
The filtration is said to be Bounded , if the non-trivial filter is finite, that is
$$
\cdots =0=F_p A\subset \cdots \subset F_{q} A=A=\cdots.
$$
Then define it Contact Grading Model for
$$
G_pA:=F_pA/F_{p-1}A.
$$
We think $\{F_p A\},$ $\{G_pA\}$ inductively extendedly Decide$A.$
$$
0\rightarrow F_{p-1}A\rightarrow F_pA\rightarrow G_pA\rightarrow 0.
$$
Note that the extension is trivial if $R$ is a domain.
one filter chain complex is a chain complex $(C_\ast ,\partial),$
And the filtration $\{F_pC_i\},$ about $C_i$
Make $\partial(F_pC_i)\subset F_pC_{i-1}.$ verifiable, $\,\forall\,p,$
$(G_pC_\ast ,\partial)$ is a chain complex, and the homology $H_i(G_pC_\ast )$ on it can be considered.
A filtration on $C_\ast $ induces a filtration on $H_\ast (C_\ast )$:
$$
F_pH_i(C_\ast ):=\{\alpha\in H_i(C_\ast )|\,\exists\,x\in F_pC_\ast , [x]=\alpha\}.
$$
Next we have $G_pH_i(C_\ast ).$ and a natural question is how is it related to $H_i(G_pC_\ast )$?
Memories were there before
$$
0\subset F_0C_\ast \subset F_1C_\ast =C_\ast ,
$$
Then $G_pH_\ast (C_\ast )$ is the homology of $H_\ast (G_1C_\ast )\xrightarrow{\delta} H_{\ast -1}(G_0C_\ast )$.
At this point $H_i(G_pC_\ast )$ basically determines $G_pH_i(C_\ast ).$
When the filtration has more non-trivial items, in order to consider the relationship between them,
We hope to consider some continuous approximations $E^1,E^2,\cdots$ to converge to the desired result,
Similar to Taylor expansion. This is called spectral sequence.
Suppose $(F_pC_\ast ,\partial)$ is a filter complex, defined as:
$$
E^0_{p,q}:=G_pC_{p+q}=F_pC_{p+q}/F_{p-1}C_{p+q},
$$
Then $E$ is double hierarchical,
$p$ is called filtration degree, $p+q$ is homology degree,
$q$ is called complementary degree.
Then $(E_0,\partial_0)$ is a chain complex,
$\partial_0:E_{p,q}^0\rightarrow E^0_{p,q-1}.$
Define $E^1_{p,q}:=H_{p+q}(G_pC_\ast ),$ which is a first-order approximation of $H_\ast (C_\ast )$.
Define as follows $\partial_1:E^1_{p,q}\rightarrow E^1_{p-1,q}:$
$\,\forall\,\alpha\in E^1_{p,q}=H_{p+q}(G_pC_\ast ),$ takes $x\in F_pC_{p+q},$
$[x]=\alpha,$ $\partial x\in F_{p-1}C_{p+q-1}.$
Note$\partial_1\alpha=[\partial x].$
The article was last updated on 2022-09-15 10:54:54
