Machine-translated from Chinese.
回顾
回忆若$G$是一个拓扑群, 有分类空间$BG,$ 使得有$G$主丛:
$$
G\rightarrow EG\rightarrow BG,
$$
即$G$在$EG$上有一个依纤维自由可迁的作用. $EG$弱可缩.
命题 1. 设有拓扑群间连续映射$f:H\rightarrow G,$ 那么它诱导空间上的连续映射$Bf:BH\rightarrow BG,$ 同伦意义下唯一.
注 2. $G\mapsto BG$是一个拓扑群到CW复形同伦范畴的函子.
考虑一个$G$主丛:
$$
G\rightarrow EH\times_HG\rightarrow EH/H=BH.
$$
那么分类映射$\phi:BH\rightarrow BG$就是诱导的映射, 同伦意义下唯一.
推论 3. 若$f:H\rightarrow G$还是一个同伦等价, 那么$Bf:BH\rightarrow BG$也是一个同伦等价.
$f_\ast :\pi_{i-1}(H)\cong \pi_{i-1}(G),$
这就诱导了$Bf_\ast :\pi_i(BH)\cong \pi_i(BG).$
弱同伦等价在CW复形上就是同伦等价.
Gram-Schmidt正交化$O(n)\hookrightarrow GL_n(\mathbb{R})$是一个同伦等价.
另一种观点来看,
$GL_n(\mathbb{R})/O_n=\{\left<{}-,-\right>_{\mathbb{R}^n}\}=\{S\in M_n(\mathbb{R})|S>0\}.$
由于后者为凸锥, 可缩.
因此, $BO_n=BGL_n(\mathbb{R}).$ 类似地, $BU_n=BGL_n(\mathbb{C}).$
考虑$(\mathbb{R}^n,+),$ 那么$B(\mathbb{R}^n)=B(\ast )=\ast .$
因此任意$\mathbb{R}^n$主丛都是平凡的.
BG的显式模型
若$G$是离散的, 那么$BG=K(G,1).$ 如$E\mathbb{Z}=\mathbb{R},$
$B\mathbb{Z}=\mathbb{R}/\mathbb{Z}=S^1.$ $EU(1)=S^\infty,$
$BU(1)=\mathbb{C}\mathrm{P}^\infty.$
$O(1)=\{\pm 1\}\subset U(1),$ $EO(1)=EU(1)=S^\infty,$
$BO(1)=\mathbb{R}\mathrm{P}^\infty=K(\mathbb{Z}_2,1).$ 一般地,
考虑$\mu_n=\{\lambda|\lambda^n=1\}\subset U(1)\subset \mathbb{C},$
$E\mu_n=EU(1)=S^\infty,$ $B\mu_n=S^\infty/\mu_n=K(\mu_n,1)=$ lens space
with dimension $\infty.$
回忆有$O_n\rightarrow V_n(\mathbb{R}^{n+k})\rightarrow G_n(\mathbb{R}^{n+k}),$
令$k\rightarrow \infty,$ 即有$V_n:=V_n(\mathbb{R}^\infty)=EO_n,$
$G_n:=G_n(\mathbb{R}^\infty)=BO_n.$
之前提过$GL_n(\mathbb{R})$收缩到$O_n,$ $\,\forall\,B$为CW复形,
$[B,BO_n]$有到$B$上具度量秩$n$向量丛间的双射,
$[B,BGL_n(\mathbb{R})]$有到$B$上秩$n$向量丛的双射.
所以后者之间也有一个双射.
类似地, 对$BU_n=BGL_n(\mathbb{C})=G_n(\mathbb{C}^\infty)$也有类似的理论.
结构群
对向量丛$\xi,$ 它的定向性, 无处消失截面存在性,
是否具度量等许多问题可以用一致的方法来变为讨论结构群改变的问题.
命题 4. 设$G$是一个李群, $H\subset G$为一个闭子群, (此时$H\rightarrow G\rightarrow G/H$是一个$H$主丛.) 若$G\rightarrow P\rightarrow B$为CW复形上的$G$主丛. 那么以下等价:
$P$由$H$主丛诱导, 即$\,\exists\,H\rightarrow Q\rightarrow B,$
使得$P\cong Q\times_HG.$
$G/H\rightarrow P\times_G(G/H)\rightarrow B$有一个截面.
$P$的分类映射可以同伦唯一地提升到$BH.$
给一个$B$上秩$n$实向量丛$\xi,$ 问给定$H\rightarrow GL_n(\mathbb{R}),$
何时三结论成立?
举例来看, 若$H=O_n,$ 那么提升总存在; 若$H=SO_n,$
那么提升存在当且仅当$\xi$可定向. 我们有双重覆盖
$$
O_n/SO_n\rightarrow BSO_n\rightarrow BO_n\leftarrow B,
$$
提升的阻碍在$H^1(B;\pi_0(\pm 1))$中, 恰是$w_1(\xi).$
于是$\xi$可定向当且仅当$w_1(\xi)=0.$
$H=Spin(n),$ 回忆$\pi_1(SO_n)\cong \pi_1(SO_3)=\mathbb{Z}_2,$
$$
SO(n-1)\rightarrow SO(n)\rightarrow S^{n-1},
$$
$Spin(n)$定义为$SO_n$的万有覆盖.
$$
0\rightarrow \mathbb{Z}_2\rightarrow Spin(n)\rightarrow SO_n\rightarrow 0,
$$
这诱导了
$$
B\mathbb{Z}_2\rightarrow BSpin(n)\rightarrow BSO_n\leftarrow B.
$$
提升存在当且仅当$\xi$上有一个Spin结构,
第一阻碍也是唯一阻碍在$H^2(B;\pi_1(B\mathbb{Z}_2))=H^2(B;\mathbb{Z}_2)$中,
为$w_2(\xi).$
命题 5. 对一个可定向的$\xi,$ $\xi$有一个spin结构当且仅当$w_2(\xi)=0.$
分裂原理($\mathbb{C}$)
$\omega$为一个复向量丛,
$\omega$可分裂为线丛的和当且仅当分类映射$B\rightarrow BU_n$可提升到$B(U_1)^{\times n}$上.
$(U_1)^{\times n}\hookrightarrow U_n,$ 嵌入到对角阵中.
考虑$G=U_n,$ $H=T^n=(U_1)^{\times n},$ 有纤维化
$$
U_n/T_n\rightarrow BT^n\rightarrow BU_n
$$
分裂原理可写为:
$$
H^\ast (BU_n;\mathbb{Z})\hookrightarrow H^\ast (BT^n;\mathbb{Z}),\quad \mathbb{Z}[c_1,\cdots,c_n]\rightarrow \mathbb{Z}[x_1,\cdots, x_n],\quad c_i\mapsto \sigma_i(x_1,\cdots,x_n).
$$
$$
U_n/T_n=\{(L_1,\cdots,L_n)|L_i\subset \mathbb{C}^n\text{ 为正交直线}\}.
$$
文章最后更新于 2022-11-18 17:11:11
review
Recall that if $G$ is a topological group, there is a classification space $BG,$ such that there is a main cluster $G$:
$$
G\rightarrow EG\rightarrow BG,
$$
That is, $G$ has a freely movable fiber-dependent function on $EG$. $EG$ is weakly contractible.
Proposition 1. Given a continuous mapping between topological groups $f:H\rightarrow G,$, then it induces a spatially continuous mapping $Bf:BH\rightarrow BG,$ that is unique in the sense of homotopy.
Note 2. $G\mapsto BG$ is a functor from a topological group to the CW complex homotopy category.
Consider a $G$ master bundle:
$$
G\rightarrow EH\times_HG\rightarrow EH/H=BH.
$$
Then the classification map $\phi:BH\rightarrow BG$ is an induced map, unique in the sense of homotopy.
Corollary 3. If $f:H\rightarrow G$ is still a homotopy equivalent, then $Bf:BH\rightarrow BG$ is also a homotopy equivalent.
$f_\ast :\pi_{i-1}(H)\cong \pi_{i-1}(G),$
This induces $Bf_\ast :\pi_i(BH)\cong \pi_i(BG).$
Weak homotopy equivalence is homotopy equivalence on the CW complex.
Gram-Schmidt orthogonalization $O(n)\hookrightarrow GL_n(\mathbb{R})$ is a homotopy equivalent.
From another point of view,
$GL_n(\mathbb{R})/O_n=\{\left<{}-,-\right>_{\mathbb{R}^n}\}=\{S\in M_n(\mathbb{R})|S>0\}.$
Since the latter is a convex cone, it can be contracted.
Therefore, $BO_n=BGL_n(\mathbb{R}).$ similarly, $BU_n=BGL_n(\mathbb{C}).$
Consider $(\mathbb{R}^n,+),$ then $B(\mathbb{R}^n)=B(\ast )=\ast .$
Therefore any $\mathbb{R}^n$ principal bundle is trivial.
Explicit model of BG
If $G$ is discrete, then $BG=K(G,1).$ is like $E\mathbb{Z}=\mathbb{R},$
$B\mathbb{Z}=\mathbb{R}/\mathbb{Z}=S^1.$ $EU(1)=S^\infty,$
$BU(1)=\mathbb{C}\mathrm{P}^\infty.$
$O(1)=\{\pm 1\}\subset U(1),$ $EO(1)=EU(1)=S^\infty,$
$BO(1)=\mathbb{R}\mathrm{P}^\infty=K(\mathbb{Z}_2,1).$ Generally,
Consider $\mu_n=\{\lambda|\lambda^n=1\}\subset U(1)\subset \mathbb{C},$
$E\mu_n=EU(1)=S^\infty,$ $B\mu_n=S^\infty/\mu_n=K(\mu_n,1)=$ lens space
with dimension $\infty.$
There are memories$O_n\rightarrow V_n(\mathbb{R}^{n+k})\rightarrow G_n(\mathbb{R}^{n+k}),$
Let $k\rightarrow \infty,$ be $V_n:=V_n(\mathbb{R}^\infty)=EO_n,$
$G_n:=G_n(\mathbb{R}^\infty)=BO_n.$
As mentioned before, $GL_n(\mathbb{R})$ shrinks to $O_n,$ and $\,\forall\,B$ is a CW complex.
$[B,BO_n]$ has a bijection between vector bundles with metric rank $n$ on $B$,
$[B,BGL_n(\mathbb{R})]$ has a bijection to the vector bundle of rank $n$ on $B$.
So there is also a bijection between the latter.
Similarly, there is a similar theory for $BU_n=BGL_n(\mathbb{C})=G_n(\mathbb{C}^\infty)$.
structural group
For the vector bundle $\xi,$, its directionality, the existence of the nowhere-vanishing section,
Many questions, such as whether they are metric or not, can be turned into questions about changes in structural groups using a consistent approach.
Proposition 4. Suppose $G$ is a Lie group, $H\subset G$ is a closed subgroup, (at this time $H\rightarrow G\rightarrow G/H$ is a $H$ principal bundle.) If $G\rightarrow P\rightarrow B$ is the $G$ principal bundle on the CW complex. Then the following is equivalent:
$P$ is induced by the $H$ main cluster, that is, $\,\exists\,H\rightarrow Q\rightarrow B,$
Make$P\cong Q\times_HG.$
$G/H\rightarrow P\times_G(G/H)\rightarrow B$ has a section.
The classification map of $P$ can be homotopically uniquely promoted to $BH.$
Given a real vector bundle $\xi,$ of rank $n$ on $B$, ask the given $H\rightarrow GL_n(\mathbb{R}),$
When are the three conclusions established?
For example, if $H=O_n,$ then promotion always exists; if $H=SO_n,$
Then promotion exists if and only if $\xi$ is directional. We have double coverage
$$
O_n/SO_n\rightarrow BSO_n\rightarrow BO_n\leftarrow B,
$$
The obstacle to improvement is in $H^1(B;\pi_0(\pm 1))$, which is exactly $w_1(\xi).$
Then $\xi$ can be oriented if and only if $w_1(\xi)=0.$
$H=Spin(n),$ Memories$\pi_1(SO_n)\cong \pi_1(SO_3)=\mathbb{Z}_2,$
$$
SO(n-1)\rightarrow SO(n)\rightarrow S^{n-1},
$$
$Spin(n)$ is defined as the universal coverage of $SO_n$.
$$
0\rightarrow \mathbb{Z}_2\rightarrow Spin(n)\rightarrow SO_n\rightarrow 0,
$$
This induced
$$
B\mathbb{Z}_2\rightarrow BSpin(n)\rightarrow BSO_n\leftarrow B.
$$
Promotion exists if and only if there is a Spin structure on $\xi$,
The first and only obstacle is in $H^2(B;\pi_1(B\mathbb{Z}_2))=H^2(B;\mathbb{Z}_2)$,
for $w_2(\xi).$
Proposition 5. There is a spin structure for an orientable $\xi,$ $\xi$ if and only if $w_2(\xi)=0.$
Split Principle ($\mathbb{C}$)
$\omega$ is a complex vector bundle,
$\omega$ can be split into the sum of line bundles if and only if the classification map $B\rightarrow BU_n$ can be promoted to $B(U_1)^{\times n}$.
$(U_1)^{\times n}\hookrightarrow U_n,$ is embedded in a diagonal matrix.
Consider $G=U_n,$ $H=T^n=(U_1)^{\times n},$ fibrosis
$$
U_n/T_n\rightarrow BT^n\rightarrow BU_n
$$
The splitting principle can be written as:
$$
H^\ast (BU_n;\mathbb{Z})\hookrightarrow H^\ast (BT^n;\mathbb{Z}),\quad \mathbb{Z}[c_1,\cdots,c_n]\rightarrow \mathbb{Z}[x_1,\cdots, x_n],\quad c_i\mapsto \sigma_i(x_1,\cdots,x_n).
$$
$$
U_n/T_n=\{(L_1,\cdots,L_n)|L_i\subset \mathbb{C}^n\text{ 为正交直线}\}.
$$
The article was last updated on 2022-11-18 17:11:11
