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示性类与障碍性理论
回顾
定理 1. 假设$F\rightarrow E\rightarrow B$为纤维丛, $F$是$n$-simple的, $B$为CW复形. 令$s$为$B^{n-1}$上的截面, 可以被延拓到$B^n$上. 那么$s$可以被延拓为$B^{n+1}$上的截面当且仅当$ob(s)\in H^{n+1}(B;\pi_n(F))$为零.
注 2. 一般地, $\pi_1B$在$\pi_nF$上可能有作用, $H^{n+1}(B;\pi_n(F))$是局部系数的. 若$F$是$(n-1)$连通的, 那么第一障碍类$ob(s)\in H^{n+1}(B,\pi_n(F))$与$s$选取无关. 此时它也是自然的, 拉回丛上的障碍类就是障碍类的拉回.
SW类作为障碍
定义Stiefel流形:
$V_k(\mathbb{R}^n):=\{(v_1,\cdots,v_k)|\mathbb{R}^n\text{中标准正交}\}\cong O(n)/O(n-k).$
给定$\xi$为秩$n$实向量丛(配备度量),
$\mathbb{R}^n\rightarrow E\rightarrow B.$ 取$V_k(\xi),$ 为$B$上纤维丛
$$
V_k(\mathbb{R}^n)\rightarrow V_k(E)\rightarrow B.
$$
注意到$\xi$有$k$个线性无关/正交截面当且仅当$V_k(\xi)$有一个截面.
我们希望说明$\xi$的SW类就是$V_k(\xi)$截面的第一阻碍类.
我们下面来计算$\pi_i(V_k(\mathbb{R}^n)).$ 先做一些准备.
定理 3 (Gysin sequence). 若$\xi$为秩$n$定向丛, 那么存在长正合列,
$$
\cdots\rightarrow H^i(B)\xrightarrow{\cup e} H^{i+n}(B)\xrightarrow{\pi^\ast } H^{i+n}(E_0)\rightarrow H^{i+1}(B)\rightarrow\cdots
$$
已知有
$$
\cdots\rightarrow H^j(E,E_0)\rightarrow H^j(E)\rightarrow H^j(E_0)\rightarrow \cdots
$$
注意到有$H^{j-n}(B)\cong H^{j-n}(E)\cong H^j(E,E_0),$
$H^j(B)\cong H^j(E).$ 由自然性即有定理中的长正合列.
注 4. 若非可定向, 那么定理在$\mod 2$意义下成立.
引理 5. $\pi_i(V_k(\mathbb{R}^n))=0,$ $\,\forall\,i<n-k.$ $\pi_{n-k}(V_k(\mathbb{R}^n))=\begin{cases} \mathbb{Z}& n-k \text{为偶数, 或}k=1\\ \mathbb{Z}_2 & \text{else} \end{cases}$
考虑纤维丛
$$
V_{k-1}(\mathbb{R}^{n-1})\rightarrow V_k(\mathbb{R}^n)\rightarrow S^{n-1},\quad (v_1,\cdots,v_k)\mapsto v_k.
$$
迭代构造, 我们有:
$$
S^{n-k}=V_1(\mathbb{R}^{n-k+1})\hookrightarrow V_2(\mathbb{R}^{n-k+2})\hookrightarrow\cdots\hookrightarrow V_k(\mathbb{R}^n)
$$
每个$\hookrightarrow$都对应了一个纤维丛. 作用$\pi_i,$
我们看到$\pi_i(-)=0$在全体空间上, 若$i<n-k.$
当$i=n-k$时, 考虑$V_2(\mathbb{R}^{n-k+2}).$ 由长正合列,
这和$\pi_{n-k}(V_k(\mathbb{R}^n))$是一样的, $k\ge 2.$ 我们有
$$
S^{n-k}\rightarrow V_2(\mathbb{R}^{n-k+2})\rightarrow S^{n-k+1}.
$$
注意到, 记$UT$为单位切丛,
$$
V_2(\mathbb{R}^{n-k+2})\cong UT(S^{n-k+1})\subset TS^{n-k+1}.
$$
考虑$S^{n-k+1}$切丛的Gysin sequence,
$E_0\cong UT(S^{n-k+1})\cong V_2(\mathbb{R}^{n-k+2}),$
$$
0\rightarrow H^{n-k}(V_2)\rightarrow H^0(S^{n-k+1})\xrightarrow{\cup e} H^{n-k+1}(S^{n-k+1})\rightarrow H^{n-k+1}(V_2)\rightarrow H^1(S^{n-k+1})=0.
$$
注意到$e=\chi(S^{n-k+1})\cdot PD(\ast ),$
我们有$\chi:\mathbb{Z}\xrightarrow{\cup e} \mathbb{Z}.$
因此若$n-k$为偶数, $\chi=0,$ $e=0,$ $H^{n-k}(V_2)=\mathbb{Z}.$
由UCT与Hurewicz定理, $\pi_{n-k}(V_2)\cong \mathbb{Z}.$
若$n-k$为奇数, $\chi=2,$ $H^{n-k}(V_2)=0,$
$H^{n-k+1}(V_2)=\mathbb{Z}_2.$ 由UCT与Hurewicz定理,
$\pi_{n-k}(V_2)=\mathbb{Z}_2.$
$k=1$时单独考虑, 由$V_1(\mathbb{R}^n)=S^{n-1}$即得.
给定$\xi$为$B$上的秩$n$实向量丛. $\,\forall\,k,$
$V_k(\xi)$为$B$上纤维丛.
定理 6. $V_k(\xi)$的第一障碍类$o_j(\xi)\in H^j(B;\pi_{j-1}(V_{n-j+1}(\mathbb{R}^n))),$ $j:=n-k+1.$ 映到$H^j(B;\mathbb{Z}_2),$ 我们有$o_j(\xi)\mapsto w_j(\xi),$ $\,\forall\,j\le n.$
回忆我们有$w_j(\xi)\neq 0$时, $V_{n-j+1}(\xi)$没有截面. 反过来不见得对:
当$n-k$为偶数时, $ o_j(\xi)\in H^j(B;\widetilde{\mathbb{Z}}), $
可能为非零偶数. 即使$n-k$为奇数, 作为第一障碍类, $o_j(\xi)=w_j(\xi)=0,$
我们也可能在之后有进一步的阻碍.
Chern类作为障碍
$V_{n-q}(\mathbb{C}^n):=\{\mathbb{C}^n\text{下的标准正交基}\},$
$\pi_i(V_{n-q}(\mathbb{C}^n))=\begin{cases}
0,&i\le 2q\\
\mathbb{Z},&i=2q+1
\end{cases}$ 给定秩$n$复向量丛$\omega,$
$V_{n-q}(\omega)$截面的第一障碍类$o_{q+1}(\omega)\in H^{2q+2}(B;\pi_{2q+1}(V_{n-q}(\mathbb{C}^n))),$
$o_{q+1}(\omega)= c_{q+1}(\omega)\in H^{2q+2}(B;\mathbb{Z}).$
这里非局部系数是由复向量丛的(定向)特殊性.
Euler类
设$\xi$为秩$n$实向量丛. $\mathbb{R}^n\rightarrow E\rightarrow B.$
考虑球丛$S(\xi):$
$$
S^{n-1}\rightarrow S(E)\rightarrow B.
$$
那么$S(\xi)$纤维$(n-2)$连通,
第一障碍类$o_n\in H^n(B;\pi_{n-1}(S^{n-1})).$
定理 7. 若$\xi$是定向的, 那么$\pi_1(B)$在$\pi_{n-1}(S^{n-1})$上的作用是平凡的. $o_n=e(\xi)\in H^{n}(B;\mathbb{Z}).$
文章最后更新于 2022-11-14 19:10:05
Characteristic Classes and Obstruction Theory
review
Theorem 1. Suppose $F\rightarrow E\rightarrow B$ is a fiber bundle, $F$ is $n$-simple, and $B$ is a CW complex. Let $s$ be the section on $B^{n-1}$, which can be extended to $B^n$. Then $s$ can be extended to the section on $B^{n+1}$ if and only if $ob(s)\in H^{n+1}(B;\pi_n(F))$ is zero.
Note 2. Generally, $\pi_1B$ may have an effect on $\pi_nF$, and $H^{n+1}(B;\pi_n(F))$ is a local coefficient. If $F$ is connected to $(n-1)$, then the first obstacle class $ob(s)\in H^{n+1}(B,\pi_n(F))$ has nothing to do with the selection of $s$. At this time, it is also natural, and the obstacle class on the pullback cluster is the pullback of the obstacle class.
SW class as barrier
Define Stiefel manifold:
$V_k(\mathbb{R}^n):=\{(v_1,\cdots,v_k)|\mathbb{R}^n\text{中标准正交}\}\cong O(n)/O(n-k).$
Given $\xi$ is a real vector bundle of rank $n$ (equipped with a metric),
$\mathbb{R}^n\rightarrow E\rightarrow B.$ Take $V_k(\xi),$ as the upper fiber bundle of $B$
$$
V_k(\mathbb{R}^n)\rightarrow V_k(E)\rightarrow B.
$$
Note that $\xi$ has $k$ linearly independent/orthogonal cross-sections if and only if $V_k(\xi)$ has one cross-section.
We hope to show that the SW class of $\xi$ is the first obstruction class of the $V_k(\xi)$ section.
Let’s calculate $\pi_i(V_k(\mathbb{R}^n)).$ and make some preparations first.
Theorem 3 (Gysin sequence). If $\xi$ is a directed bundle of rank $n$, then there is a long exact sequence,
$$
\cdots\rightarrow H^i(B)\xrightarrow{\cup e} H^{i+n}(B)\xrightarrow{\pi^\ast } H^{i+n}(E_0)\rightarrow H^{i+1}(B)\rightarrow\cdots
$$
It is known that there are
$$
\cdots\rightarrow H^j(E,E_0)\rightarrow H^j(E)\rightarrow H^j(E_0)\rightarrow \cdots
$$
Notice that there is $H^{j-n}(B)\cong H^{j-n}(E)\cong H^j(E,E_0),$
$H^j(B)\cong H^j(E).$ By nature, there is a long exact sequence in the theorem.
Note 4. If it is not orientable, then the theorem holds in the sense of $\mod 2$.
Lemma 5. $\pi_i(V_k(\mathbb{R}^n))=0,$ $\,\forall\,i<n-k.$ $\pi_{n-k}(V_k(\mathbb{R}^n))=\begin{cases} \mathbb{Z}& n-k \text{为偶数, 或}k=1\\ \mathbb{Z}_2 & \text{else} \end{cases}$
Consider fiber bundles
$$
V_{k-1}(\mathbb{R}^{n-1})\rightarrow V_k(\mathbb{R}^n)\rightarrow S^{n-1},\quad (v_1,\cdots,v_k)\mapsto v_k.
$$
Iteratively constructed, we have:
$$
S^{n-k}=V_1(\mathbb{R}^{n-k+1})\hookrightarrow V_2(\mathbb{R}^{n-k+2})\hookrightarrow\cdots\hookrightarrow V_k(\mathbb{R}^n)
$$
Each $\hookrightarrow$ corresponds to a fiber bundle. Effect $\pi_i,$
We see that $\pi_i(-)=0$ is in the entire space, if $i<n-k.$
When $i=n-k$, consider $V_2(\mathbb{R}^{n-k+2}).$ by the long exact sequence,
This is the same as $\pi_{n-k}(V_k(\mathbb{R}^n))$, $k\ge 2.$ we have
$$
S^{n-k}\rightarrow V_2(\mathbb{R}^{n-k+2})\rightarrow S^{n-k+1}.
$$
Note that $UT$ is the unit tangent bundle,
$$
V_2(\mathbb{R}^{n-k+2})\cong UT(S^{n-k+1})\subset TS^{n-k+1}.
$$
Consider the Gysin sequence of the $S^{n-k+1}$ tangent bundle,
$E_0\cong UT(S^{n-k+1})\cong V_2(\mathbb{R}^{n-k+2}),$
$$
0\rightarrow H^{n-k}(V_2)\rightarrow H^0(S^{n-k+1})\xrightarrow{\cup e} H^{n-k+1}(S^{n-k+1})\rightarrow H^{n-k+1}(V_2)\rightarrow H^1(S^{n-k+1})=0.
$$
Notice$e=\chi(S^{n-k+1})\cdot PD(\ast ),$
We have $\chi:\mathbb{Z}\xrightarrow{\cup e} \mathbb{Z}.$
Therefore, if $n-k$ is an even number, $\chi=0,$ $e=0,$ $H^{n-k}(V_2)=\mathbb{Z}.$
From UCT and Hurewicz’s theorem, $\pi_{n-k}(V_2)\cong \mathbb{Z}.$
If $n-k$ is an odd number, $\chi=2,$ $H^{n-k}(V_2)=0,$
$H^{n-k+1}(V_2)=\mathbb{Z}_2.$ According to UCT and Hurewicz theorem,
$\pi_{n-k}(V_2)=\mathbb{Z}_2.$
When $k=1$ is considered separately, it can be obtained from $V_1(\mathbb{R}^n)=S^{n-1}$.
Given $\xi$ is a real vector bundle of rank $n$ on $B$. $\,\forall\,k,$
$V_k(\xi)$ is the upper fiber plexus of $B$.
Theorem 6. The first obstacle class $o_j(\xi)\in H^j(B;\pi_{j-1}(V_{n-j+1}(\mathbb{R}^n))),$ $j:=n-k+1.$ of $V_k(\xi)$ maps to $H^j(B;\mathbb{Z}_2),$ and we have $o_j(\xi)\mapsto w_j(\xi),$ $\,\forall\,j\le n.$
Recall that when we have $w_j(\xi)\neq 0$, $V_{n-j+1}(\xi)$ has no section. The reverse is not necessarily true:
When $n-k$ is an even number, $ o_j(\xi)\in H^j(B;\widetilde{\mathbb{Z}}), $
May be a non-zero even number. Even if $n-k$ is an odd number, as the first obstacle class, $o_j(\xi)=w_j(\xi)=0,$
We may also have further obstacles later.
Chern class as barrier
$V_{n-q}(\mathbb{C}^n):=\{\mathbb{C}^n\text{下的标准正交基}\},$
$\pi_i(V_{n-q}(\mathbb{C}^n))=\begin{cases}
0,&i\le 2q\\
\mathbb{Z},&i=2q+1
\end{cases}$ Given rank $n$ complex vector bundle $\omega,$
$V_{n-q}(\omega)$ Section 1 obstacle class $o_{q+1}(\omega)\in H^{2q+2}(B;\pi_{2q+1}(V_{n-q}(\mathbb{C}^n))),$
$o_{q+1}(\omega)= c_{q+1}(\omega)\in H^{2q+2}(B;\mathbb{Z}).$
Here the non-local coefficients are caused by the (directional) specificity of complex vector bundles.
Euler class
Let $\xi$ be a real vector bundle of rank $n$. $\mathbb{R}^n\rightarrow E\rightarrow B.$
Consider ball clusters$S(\xi):$
$$
S^{n-1}\rightarrow S(E)\rightarrow B.
$$
Then $S(\xi)$ fiber $(n-2)$ is connected,
First obstacle category $o_n\in H^n(B;\pi_{n-1}(S^{n-1})).$
Theorem 7. If $\xi$ is directional, then the effect of $\pi_1(B)$ on $\pi_{n-1}(S^{n-1})$ is trivial. $o_n=e(\xi)\in H^{n}(B;\mathbb{Z}).$
The article was last updated on 2022-11-14 19:10:05
