《代数拓扑2》笔记(18)-分类空间 "Algebraic Topology 2" Notes (18) — Classifying Spaces

Machine-translated from Chinese.

主(Principal)$G$-丛

$G$为拓扑群. 一个主$G$-丛是一个局部平凡的纤维丛$\pi:P\rightarrow B,$ 在总空间上有一个连续的右作用$P\times G\rightarrow P,$ 使得$G$在每根纤维上的作用是自由可迁的. (保持纤维.)

注 1. $\,\forall\,b\in B,$ $G\cong \pi^{-1}(b),$ $g\mapsto x_0\cdot g.$ 因此可写$G\rightarrow P\rightarrow B.$ 局部平凡化意味着可以写$B\cong P/G.$ 每个右作用也给出左作用$g\cdot x:=x\cdot g^{-1}.$

$G$为离散群时, 主$G$-丛就是一个正则$G$覆盖. 设$\xi$为秩$n$实向量丛, $\mathbb{R}^n\cong \xi_b\rightarrow E\rightarrow B.$ 考虑$n$-标架丛:

$$ F(\xi):F(\xi_b)\rightarrow F(E)\rightarrow B, $$

$$ F(\xi_b):=\{(v_1,\cdots,v_n)|v_i\in \xi_b \text{ 线性无关}\}\cong \operatorname{Iso}(\mathbb{R}^n,\xi_b). $$

那么这个标架丛就是一个主$GL_n(\mathbb{R})$丛.

注意到我们有$B$上秩$n$实向量丛到主$GL_n(\mathbb{R})$丛的双射, 逆映射由$E:=P\times_{GL_n(\mathbb{R})} \mathbb{R}^n\xrightarrow{p}P/GL_n(\mathbb{R})=B$给出.

设$\xi$有一个欧氏度量, $OF(\xi_b):=\{(v_1,\cdots,v_n)|\text{标准正交}\}=V_n(\xi_b)\cong \operatorname{Isometry(\mathbb{R}^n,\xi_b)},$ 是一个主$O_n$丛. 类似地, 我们也有具度量秩$n$实向量丛到主$O_n$丛的双射.

把实数改为复数也有类似的结论. 另外, 取$B=G_n(\mathbb{R}^{n+k}),$ $P=V_n(\mathbb{R}^{n+k}),$ $V_n(\mathbb{R}^{n+k})\rightarrow G_n(\mathbb{R}^{n+k}),$ $(v_1,\cdots,v_n)\mapsto \operatorname{span}\{v_1,\cdots,v_n\}$是一个主$O_n$丛. 这个丛就是$OF(\gamma_k^n).$ 回忆$G_n(\mathbb{R}^{n+k})\cong O_{n+k}/O_n\times O_k,$ $V_n(\mathbb{R}^{n+k})=O_{n+k}/O_k,$ 纤维为$O_n.$

取秩$1$的情况, 就是$S^0\cong O(1)\rightarrow S(\mathbb{R}^{n+1})\rightarrow \mathbb{R}\mathrm{P}^n,$ $S^1\cong U(1)\rightarrow S(\mathbb{C}^{n+1})\rightarrow \mathbb{C}\mathrm{P}^n,$ 类似地还有四元数的情形, 纤维记作$S_p(1)\cong SU(2)\cong S^3.$

一个反例是$\mathbb{Q}\rightarrow \mathbb{R}\rightarrow \mathbb{R}/\mathbb{Q},$ 不是一个主$\mathbb{Q}$丛. 因为此时底空间拓扑平凡, 局部平凡要求了整体平凡, 给出矛盾. 但当$G$为李群, $H$为其闭子群时, 有$H\rightarrow G\rightarrow G/H$为主$H$丛, 底空间称为齐性空间.

$G$光滑自由作用在光滑流形$M$上, $M\rightarrow M/G$可能不是一个主$G$丛, 如$M=T^2,$ $G=\mathbb{R},$ 非有理地作用在$M$上, 使得轨道在$M$上是稠密的. 不过当$G$紧的时候它是主$G$丛.

命题 2. 一个主$G$丛是平凡的当且仅当它有一个截面.

注 3. 这对一般的纤维丛当然是错误的. $S^n$可平行化当且仅当$n=0,1,3,$ 但$S^n$有非平凡向量场当且仅当$\chi(S^n)=0,$ 当且仅当$n$为奇数.

设$P\rightarrow B$有一个截面, 定义$B\times G\rightarrow P,$ $(b,g)\mapsto s(b)\cdot g$即可. 验证它是一个同胚.

分类空间

我们称$BG$是拓扑群$G$的一个分类空间, 若存在主$G$丛$EG\xrightarrow{\pi}BG,$ 使得$EG$是弱可缩的, 即$\pi_i=0,$ $\,\forall\,i.$

定理 4. 对任意CW复形$B,$ 有映射$[B,BG]\rightarrow Bun_G(B),$ 后者为$B$上主$G$丛, $\phi\mapsto \phi^\ast (EG)$为一个双射.

满射: 任意给定p.G.b (principle $G$-bundle), 有$EG\rightarrow P\times_G EG\rightarrow P/G=B.$ 截面存在因为$ob\in H^{i+1}(B;\pi_i(EG))=0.$ $s:P/G\rightarrow P\times_G EG,$ $p\mapsto (p,\widetilde\phi(p))=(p\cdot g,g\cdot \widetilde\phi(p)).$ $p=p\cdot g\mapsto (p\cdot g,\widetilde\phi(p\cdot g)).$ 因此我们有映射$\widetilde{\phi}:P\rightarrow EG,$ 为$G$-等变映射. 于是可以定义$\phi:P/G\rightarrow EG/G,$ 验证这是一个拉回即可.

单射: 设$\psi:f_0^\ast EG\cong f_1^\ast EG.$ 每个$f_0,f_1$给出$EG\rightarrow P\times_G EG\rightarrow P/G=B$的截面$s_0,s_1.$ 由阻碍理论, 它们是同伦的.

注 5. $BG$可以取为CW复形. $G\rightarrow EG\rightarrow BG$给出$\pi_i(BG)\cong \pi_{i-1}(G).$ $BG$是CW同伦等价意义下唯一的, 称为$G$的分类空间. Milnor证明了对任意拓扑群$G$, $BG$存在.

Milnor的构造方法是Join. 给两个拓扑空间$X,Y,$

$$ X\ast Y:=X\times[0,1]\times Y/(x,0,y_1)\sim (x,0,y_2),(x_1,1,y)\sim (x_2,1,y). $$

一些例子是$S^0\ast S^0=S^1,$ $S^n\ast S^m=S^{n+m-1}.$

一个事实是Join会增加连通性, $X^{\ast n}$是$(n-2)$连通的. 令$n\rightarrow\infty,$ 就有它弱可缩. $EG:=\bigcup_{n=1}^\infty G^{\ast n}.$ 显然$G$在$G^{\ast n},EG$上有一个自然的作用, 且是自由的. 定义$BG=EG/G$即可.

文章最后更新于 2022-11-17 11:35:13

Principal $G$-Bundles

$G$ is a topological group. A principal $G$-bundle is a locally trivial fiber bundle $\pi:P\rightarrow B,$ There is a continuous right action on the total space $P\times G\rightarrow P,$ Make the effect of $G$ on each fiber freely transitive. (Keep the fiber.)

Note 1. $\,\forall\,b\in B,$ $G\cong \pi^{-1}(b),$ $g\mapsto x_0\cdot g.$ So one can write $G\rightarrow P\rightarrow B.$ Local trivialization means one can write $B\cong P/G.$ Every right action also gives a left action $g\cdot x:=x\cdot g^{-1}.$

When $G$ is a discrete group, the main $G$-bundle is a regular $G$ cover. Let $\xi$ be a real vector bundle of rank $n$, $\mathbb{R}^n\cong \xi_b\rightarrow E\rightarrow B.$ Consider $n$-frame bundle:

$$ F(\xi):F(\xi_b)\rightarrow F(E)\rightarrow B, $$

$$ F(\xi_b):=\{(v_1,\cdots,v_n)|v_i\in \xi_b \text{ 线性无关}\}\cong \operatorname{Iso}(\mathbb{R}^n,\xi_b). $$

Then this frame bundle is a main $GL_n(\mathbb{R})$ bundle.

Notice that we have a bijection from the real vector bundle of rank $n$ on $B$ to the main $GL_n(\mathbb{R})$ bundle, The inverse mapping is given by $E:=P\times_{GL_n(\mathbb{R})} \mathbb{R}^n\xrightarrow{p}P/GL_n(\mathbb{R})=B$.

Assume $\xi$ has a Euclidean metric, $OF(\xi_b):=\{(v_1,\cdots,v_n)|\text{标准正交}\}=V_n(\xi_b)\cong \operatorname{Isometry(\mathbb{R}^n,\xi_b)},$ is a principal $O_n$ bundle. Similarly, we also have a bijection from a real vector bundle of metric rank $n$ to a principal $O_n$ bundle.

There is a similar conclusion when changing real numbers to complex numbers. In addition, take $B=G_n(\mathbb{R}^{n+k}),$ $P=V_n(\mathbb{R}^{n+k}),$ $V_n(\mathbb{R}^{n+k})\rightarrow G_n(\mathbb{R}^{n+k}),$ $(v_1,\cdots,v_n)\mapsto \operatorname{span}\{v_1,\cdots,v_n\}$ is a master $O_n$ plex. This cluster is $OF(\gamma_k^n).$ Memories$G_n(\mathbb{R}^{n+k})\cong O_{n+k}/O_n\times O_k,$ $V_n(\mathbb{R}^{n+k})=O_{n+k}/O_k,$ fiber is $O_n.$

Taking rank $1$, It’s $S^0\cong O(1)\rightarrow S(\mathbb{R}^{n+1})\rightarrow \mathbb{R}\mathrm{P}^n,$ $S^1\cong U(1)\rightarrow S(\mathbb{C}^{n+1})\rightarrow \mathbb{C}\mathrm{P}^n,$ Similarly, in the case of quaternions, the fiber is denoted as $S_p(1)\cong SU(2)\cong S^3.$

A counterexample is $\mathbb{Q}\rightarrow \mathbb{R}\rightarrow \mathbb{R}/\mathbb{Q},$ is not a main $\mathbb{Q}$ bundle. Because the base space topology is trivial at this time, Local triviality requires global triviality, giving a contradiction. But when $G$ is a Lie group and $H$ is its closed subgroup, There is $H\rightarrow G\rightarrow G/H$ as the main $H$ bundle, and the base space is called a homogeneous space.

$G$ smooth free action on smooth manifold $M$, $M\rightarrow M/G$ may not be a main $G$ bundle, For example, $M=T^2,$ $G=\mathbb{R},$ acts on $M$ irrationally, making the orbit on $M$ dense. But when $G$ is tight, it is the main $G$ cluster.

Proposition 2. A principal $G$ bundle is trivial if and only if it has a section.

Note 3. This is of course wrong for general fiber bundles. $S^n$ is parallelizable if and only if $n=0,1,3,$ but $S^n$ has a nontrivial vector field if and only if $\chi(S^n)=0,$ if and only if $n$ is odd.

Assume $P\rightarrow B$ has a section, define $B\times G\rightarrow P,$ $(b,g)\mapsto s(b)\cdot g$ is enough. Verify that it is a homeomorphism.

classification space

We call $BG$ a classification space of the topological group $G$, If there is a main $G$ bundle $EG\xrightarrow{\pi}BG,$ such that $EG$ is weakly shrinkable, that is, $\pi_i=0,$ $\,\forall\,i.$

Theorem 4. For any CW complex $B,$, there is a mapping $[B,BG]\rightarrow Bun_G(B),$, the latter is the main $G$ bundle on $B$, and $\phi\mapsto \phi^\ast (EG)$ is a bijection.

Surjection: Given any p.G.b (principle $G$-bundle), Yes $EG\rightarrow P\times_G EG\rightarrow P/G=B.$ The section exists because $ob\in H^{i+1}(B;\pi_i(EG))=0.$ $s:P/G\rightarrow P\times_G EG,$ $p\mapsto (p,\widetilde\phi(p))=(p\cdot g,g\cdot \widetilde\phi(p)).$ $p=p\cdot g\mapsto (p\cdot g,\widetilde\phi(p\cdot g)).$ Therefore we have the mapping $\widetilde{\phi}:P\rightarrow EG,$ as $G$-equivariant mapping. So we can define $\phi:P/G\rightarrow EG/G,$ Just verify that this is a pullback.

Injection: Let $\psi:f_0^\ast EG\cong f_1^\ast EG.$ Each $f_0,f_1$ gives the section $s_0,s_1.$ of $EG\rightarrow P\times_G EG\rightarrow P/G=B$ According to the obstruction theory, they are homotopic.

Note 5. $BG$ can be taken as a CW complex. $G\rightarrow EG\rightarrow BG$ gives $\pi_i(BG)\cong \pi_{i-1}(G).$ $BG$, which is the only classification space in the sense of CW homotopy equivalence, called $G$. Milnor proved that for any topological group $G$, $BG$ exists.

Milnor's construction method is Join. Given two topological spaces $X,Y,$

$$ X\ast Y:=X\times[0,1]\times Y/(x,0,y_1)\sim (x,0,y_2),(x_1,1,y)\sim (x_2,1,y). $$

Some examples are $S^0\ast S^0=S^1,$ $S^n\ast S^m=S^{n+m-1}.$

One fact is that Join will increase connectivity, $X^{\ast n}$ is connected to $(n-2)$. Let $n\rightarrow\infty,$ have it weakly shrinkable. $EG:=\bigcup_{n=1}^\infty G^{\ast n}.$ Obviously $G$ has a natural effect on $G^{\ast n},EG$ and is free. Just define $BG=EG/G$.

The article was last updated on 2022-11-17 11:35:13