Machine-translated from Chinese.
万有丛
一个秩$n$实向量丛$\gamma^n$称为万有丛, 若其满足如下性质:
$\,\forall\,$ 仿紧底空间$B,$
任意以$B$为底空间的秩$n$向量丛$\xi,$
有一个丛映射$\xi\rightarrow \gamma^n.$
$\,\forall\,f,g:\xi\rightarrow\gamma^n$为丛映射,
它们在丛映射意义下同伦.
因此, 任意秩$n$向量丛$\xi$都是$\gamma^n$的拉回.
万有丛是在丛同伦等价的意义下唯一的.
$B$上秩$n$向量丛的同构类和连续映射$f:B\rightarrow B(\gamma^n)$的同伦类间有一个一一对应.
万有丛的构造
$\,\forall\,n,k\in \mathbb{N},$ 定义Grassman流形
$$
G_n(\mathbb{R}^{n+k}):=\{H|\text{$H$为$\mathbb{R}^{n+k}$中的$n$维子空间}\}
$$
$GL_{n+k}(\mathbb{R})$可以可迁地作用在$G_n(\mathbb{R}^{n+k}),$
因此$G_n(\mathbb{R}^{n+k})\cong GL_{n+k}(\mathbb{R})/\operatorname{Stab}_H.$
设$H=\{e_1,\cdots,e_n\},$ 则
$$
\operatorname{Stab}_H=\begin{bmatrix}
*&*\\
0&*
\end{bmatrix}\subset GL_{n+k},
$$
为$GL_{n+k}$的闭子群.
因此$G_n(\mathbb{R}^{n+k})$为$nk$维的光滑流形.
事实上, 它还是紧的.
这是因为$O_{n+k}\subset GL_{n+k}$就可以可迁地作用在$G_n(\mathbb{R}^{n+k})$上了.
显然可以看出$G_n(\mathbb{R}^{n+k})\cong G_k(\mathbb{R}^{n+k}),$
$H\mapsto H^\perp,$ 且$G_1(\mathbb{R}^{1+k})=\mathbb{R}\mathrm{P}^k.$
定义$G_n(\mathbb{R}^{n+k})$上的典范丛$\gamma_k^n$为
$$
E(\gamma_k^n):=\{(H,v)|H\in G_n(\mathbb{R}^{n+k}),v\in H\},
$$
构造$\mathbb{R}^\infty,$ 那么我们有
$$
G_n(\mathbb{R}^n)\subset G_n(\mathbb{R}^{n+1})\subset\cdots\subset G_n(\mathbb{R}^\infty)=\bigcup_i G_n(\mathbb{R}^{n+i}).
$$
定义$\gamma^n$为$G_n(\mathbb{R}^\infty)$上的典范丛.
那么事实上它就是满足前述要求的万有丛.
定理 1. $G_n(\mathbb{R}^\infty)$上的典范从$\gamma^n$为秩$n$万有丛.
引理 2. 设有仿紧集$B$上的秩$n$丛$\xi,$ 存在线性映射$\widehat f:E(\xi)\rightarrow \mathbb{R}^\infty,$ 在每个纤维上是单射.
取局部平凡化$\{U_i\}$覆盖$B.$ 由仿紧性,
我们可以构造可数覆盖且是局部有限的.
取从属于$\{U_i\}$的单位分解$\{\lambda_i\},$
使得$W_i\Subset V_i\Subset U_i,$ $\lambda_i\in C_0^\infty(V_i),$
$\lambda_i|_{W_i}=1.$ $\sum \lambda_i(x)=1.$
由于$\xi|_{U_i}$是平凡的, 取$h_i:\pi^{-1}U_i\cong \mathbb{R}^n.$
定义连续映射$h_i':E(\xi)\rightarrow \mathbb{R}^n,$
$h_i'(e)=\lambda_i(\pi(e))h_i(e).$
定义$\widehat f:E(\xi)\rightarrow \mathbb{R}^\infty,$
$\widehat f(e):=(h_i'(e))^\infty_{i=0}\in \mathbb{R}^\infty.$
这就是满足要求的映射了.
我们回到定理的证明: 考虑映射$f:E(\xi)\rightarrow E(\gamma^n),$
$$
e\mapsto (\widehat f(F_{\pi(e)}(\xi)),\widehat f(e)),
$$
这就是一个丛映射. 接下来对于任意两个丛映射$f,g:\xi\rightarrow \gamma^n,$
每个都给出$\widehat{f},\widehat g:E(\xi)\rightarrow \mathbb{R}^\infty.$
定义同伦$\widehat h_t(e)=(1-t)\widehat f(e)+t\widehat g(e),$
$\widehat h_t$是单射, 若$\widehat f(e),\widehat{g} (e)$是线性无关的.
定义$d_{2}:e_i\mapsto e_{2i},$ $d_{1}:e_i\mapsto e_{2i+1},$ 则
$$
f\simeq d_1\circ f\simeq d_2\circ g\simeq g.
$$
示性类
一个秩$n$向量丛上的示性类是对$\xi$自然地赋予$c(\xi)\in H^\ast (B(\xi);R),$
使得$c(f^\ast \xi)=f^\ast c(\xi).$ 因此我们只需考虑万有丛上的示性类即可.
考虑计算$H^\ast (G_n(\mathbb{R}^\infty);R).$ 取$R=\mathbb{Z}_2.$
断言其同构于$\mathbb{Z}_2[w_1,w_2,\cdots,w_n].$
考虑$G_n(\mathbb{R}^\infty)$上的CW结构. 固定
$$
\mathbb{R}^0\subset \mathbb{R}^1\subset \cdots \subset \mathbb{R}^m.
$$
任意$X\in G_n(\mathbb{R}^m)$给出一个序列:
$$
0\le \dim (X\cap \mathbb{R}^1)\le\cdots\le \dim(X\cap \mathbb{R}^m)=n.
$$
定义Schubert符号$\sigma=(\sigma_1,\cdots,\sigma_n)$记录了序列中跳跃的部分,
$1\le \sigma_1<\cdots<\sigma_n\le m.$
$$
e(\sigma):=\{X\in G_n(\mathbb{R}^m)|\dim (X\cap \mathbb{R}^{\sigma_i})=i,\:\dim(X\cap \mathbb{R}^{\sigma_i-1})=i-1\}.
$$
断言$\{e(\sigma)\}$组成了$G_n(\mathbb{R}^m)$的CW结构.
令$H^k\subset \mathbb{R}^k$为$x_k>0$的上半平面,
引理 3. 每个$X\in e(\sigma)$有唯一一个正交基$(x_1,\cdots,x_n)\in H^{\sigma_1}\times\cdots\times H^{\sigma_n}.$
只需注意到下面的序列, 由线性空间的基本知识即得.
$$
\dim(X\cap \mathbb{R}^\sigma_1)=1,\quad \dim(X\cap \mathbb{R}^\sigma_2)=2,\quad \cdots.
$$
$$
e'(\sigma):=\{(x_1,\cdots,x_n)|\{x_i\}\text{正交, }x_i\in H^{\sigma_i}\}.
$$
对$n$做归纳, 即可得到$\overline e'(\sigma)\cong$闭球.
文章最后更新于 2022-10-14 16:39:02
Universal bundle
A real vector bundle $\gamma^n$ of rank $n$ is called Universal bundle, if it satisfies the following properties:
$\,\forall\,$ Paracompact base space $B,$
Any rank $n$ vector bundle $\xi,$ of space with $B$ as the base
There is a plex map $\xi\rightarrow \gamma^n.$
$\,\forall\,f,g:\xi\rightarrow\gamma^n$ is the bundle map,
They are homotopic in the sense of bundle mapping.
Therefore, any vector bundle $\xi$ of rank $n$ is a pullback of $\gamma^n$.
The universal bundle is unique in the sense of bundle homotopy equivalence.
There is a one-to-one correspondence between the isomorphism class of $B$ upper-rank $n$ vector bundles and the homotopy class of continuous maps $f:B\rightarrow B(\gamma^n)$.
The structure of the universal bundle
$\,\forall\,n,k\in \mathbb{N},$ Define Grassman manifold
$$
G_n(\mathbb{R}^{n+k}):=\{H|\text{$H$为$\mathbb{R}^{n+k}$中的$n$维子空间}\}
$$
$GL_{n+k}(\mathbb{R})$ can be relocated to $G_n(\mathbb{R}^{n+k}),$
Therefore $G_n(\mathbb{R}^{n+k})\cong GL_{n+k}(\mathbb{R})/\operatorname{Stab}_H.$
Assume $H=\{e_1,\cdots,e_n\},$ then
$$
\operatorname{Stab}_H=\begin{bmatrix}
*&*\\
0&*
\end{bmatrix}\subset GL_{n+k},
$$
is the closed subgroup of $GL_{n+k}$.
Therefore $G_n(\mathbb{R}^{n+k})$ is a smooth manifold of $nk$ dimension.
In fact, it's still tight.
This is because $O_{n+k}\subset GL_{n+k}$ can be moved to $G_n(\mathbb{R}^{n+k})$.
It is obvious that $G_n(\mathbb{R}^{n+k})\cong G_k(\mathbb{R}^{n+k}),$
$H\mapsto H^\perp,$ and $G_1(\mathbb{R}^{1+k})=\mathbb{R}\mathrm{P}^k.$
Define the canonical bundle $\gamma_k^n$ on $G_n(\mathbb{R}^{n+k})$ as
$$
E(\gamma_k^n):=\{(H,v)|H\in G_n(\mathbb{R}^{n+k}),v\in H\},
$$
Construct$\mathbb{R}^\infty,$ then we have
$$
G_n(\mathbb{R}^n)\subset G_n(\mathbb{R}^{n+1})\subset\cdots\subset G_n(\mathbb{R}^\infty)=\bigcup_i G_n(\mathbb{R}^{n+i}).
$$
Define $\gamma^n$ as the canonical bundle on $G_n(\mathbb{R}^\infty)$.
So in fact it is the universal bundle that meets the above requirements.
Theorem 1. The canon on $G_n(\mathbb{R}^\infty)$ from $\gamma^n$ is a universal bundle of rank $n$.
Lemma 2. Suppose there is a linear map $\widehat f:E(\xi)\rightarrow \mathbb{R}^\infty,$ of rank $n$ bundle $\xi,$ on the paracompact set $B$ that is injective on each fiber.
Take the local trivialization $\{U_i\}$ to cover $B.$ due to paracompactness,
We can construct countable covers that are locally finite.
Take the partition of unity $\{\lambda_i\},$ belonging to $\{U_i\}$
Make$W_i\Subset V_i\Subset U_i,$ $\lambda_i\in C_0^\infty(V_i),$
$\lambda_i|_{W_i}=1.$ $\sum \lambda_i(x)=1.$
Since $\xi|_{U_i}$ is trivial, take $h_i:\pi^{-1}U_i\cong \mathbb{R}^n.$
Define continuous mapping $h_i':E(\xi)\rightarrow \mathbb{R}^n,$
$h_i'(e)=\lambda_i(\pi(e))h_i(e).$
Definition$\widehat f:E(\xi)\rightarrow \mathbb{R}^\infty,$
$\widehat f(e):=(h_i'(e))^\infty_{i=0}\in \mathbb{R}^\infty.$
This is the mapping that meets the requirements.
We return to the proof of the theorem: consider the mapping $f:E(\xi)\rightarrow E(\gamma^n),$
$$
e\mapsto (\widehat f(F_{\pi(e)}(\xi)),\widehat f(e)),
$$
This is a bundle map. Next, for any two bundle maps $f,g:\xi\rightarrow \gamma^n,$
Each is given $\widehat{f},\widehat g:E(\xi)\rightarrow \mathbb{R}^\infty.$
Define homotopy$\widehat h_t(e)=(1-t)\widehat f(e)+t\widehat g(e),$
$\widehat h_t$ is injective, if $\widehat f(e),\widehat{g} (e)$ is linearly independent.
Define $d_{2}:e_i\mapsto e_{2i},$ $d_{1}:e_i\mapsto e_{2i+1},$ then
$$
f\simeq d_1\circ f\simeq d_2\circ g\simeq g.
$$
Characteristic class
The characteristic class on a vector bundle of rank $n$ is naturally assigned to $c(\xi)\in H^\ast (B(\xi);R),$ for $\xi$
So that $c(f^\ast \xi)=f^\ast c(\xi).$ we only need to consider the characteristic classes on the universal bundle.
Consider calculating $H^\ast (G_n(\mathbb{R}^\infty);R).$ and taking $R=\mathbb{Z}_2.$
Assert that it is isomorphic to $\mathbb{Z}_2[w_1,w_2,\cdots,w_n].$
Consider the CW structure on $G_n(\mathbb{R}^\infty)$. Fixed
$$
\mathbb{R}^0\subset \mathbb{R}^1\subset \cdots \subset \mathbb{R}^m.
$$
Any $X\in G_n(\mathbb{R}^m)$ gives a sequence:
$$
0\le \dim (X\cap \mathbb{R}^1)\le\cdots\le \dim(X\cap \mathbb{R}^m)=n.
$$
Define Schubert symbol $\sigma=(\sigma_1,\cdots,\sigma_n)$ to record the jump part in the sequence,
$1\le \sigma_1<\cdots<\sigma_n\le m.$
$$
e(\sigma):=\{X\in G_n(\mathbb{R}^m)|\dim (X\cap \mathbb{R}^{\sigma_i})=i,\:\dim(X\cap \mathbb{R}^{\sigma_i-1})=i-1\}.
$$
Assert that $\{e(\sigma)\}$ forms the CW structure of $G_n(\mathbb{R}^m)$.
Let $H^k\subset \mathbb{R}^k$ be the upper half plane of $x_k>0$,
Lemma 3. Each $X\in e(\sigma)$ has a unique orthonormal basis $(x_1,\cdots,x_n)\in H^{\sigma_1}\times\cdots\times H^{\sigma_n}.$
Just note the following sequence, which is obtained from the basic knowledge of linear spaces.
$$
\dim(X\cap \mathbb{R}^\sigma_1)=1,\quad \dim(X\cap \mathbb{R}^\sigma_2)=2,\quad \cdots.
$$
$$
e'(\sigma):=\{(x_1,\cdots,x_n)|\{x_i\}\text{正交, }x_i\in H^{\sigma_i}\}.
$$
By induction on $n$, we can get the closed ball $\overline e'(\sigma)\cong$.
The article was last updated on 2022-10-14 16:39:02
