Machine-translated from Chinese.
证明
定理 1. SW示性类存在.
只需考虑CW复形$B$上向量丛的的示性类. 不然, 考虑CW逼近$B'\rightarrow B,$
考虑拉回丛即可.
给定向量丛$\xi:\mathbb{R}^n\rightarrow E\xrightarrow{\pi} B,$
考虑投影空间丛$P(\xi):P(\mathbb{R}^n)=\mathbb{R}\mathrm{P}^{n-1}\rightarrow P(E)\xrightarrow{p(\pi)}B,$
$$
P(E)=\{(b,l)|b\in B,l\subset F_b(\xi)\text{ 为直线}\}.
$$
$P(\xi)$为纤维丛. 我们希望应用Leray-Hirsch定理到该丛上.
定理 2 (Leray-Hirsch). 设$F\xrightarrow{i} E\xrightarrow{p} B$为纤维丛. 取定某个交换环$R,$ 若$H^n(F;R)$为自由有限生成的$R$模, 存在$c_j\in H^\ast (E;R)$使得$\{i^\ast c_j\}$组成了$H^\ast (F;R)\cong R\{c_j\}_j$的一组$R$-基. 那么有如下$H^\ast (B;R)$模同构:
$$
H^\ast (B;R)\bigotimes H^\ast (F;R)\cong H^\ast (E;R), \quad \sum_{i,j}b_i\otimes i^\ast (c_j)\mapsto \sum_{i,j}p^\ast (b_i)\cup c_j.
$$
我们验证条件,
$H^i(\mathbb{R}\mathrm{P}^{n-1},\mathbb{Z}_2)=\left<{}a^i\right>$的确自由有限生成,
$a\in H^1,$ $i\le n-1.$ 我们希望说明$i^\ast $满射.
回忆$\,\exists\,\widehat f:E(\xi) \rightarrow \mathbb{R}^\infty,$
使得$\widehat f$在纤维上是线性单射. 那么考虑
$$
\mathbb{R}^n\hookrightarrow E\xrightarrow{\widehat f}\mathbb{R}^\infty,
$$
投影诱导
$$
P(\mathbb{R}^n)\cong \mathbb{R}\mathrm{P}^{n-1}\xrightarrow{i}P(E)\xrightarrow{P(\widehat{f})}\mathbb{R}\mathrm{P}^\infty.
$$
取拉回映射,
$$
H^\ast (\mathbb{R}\mathrm{P}^{n-1})\xleftarrow{i^\ast } H^\ast (P(E))\leftarrow H^\ast (\mathbb{R}\mathrm{P}^\infty).
$$
复合在低维为同构, 因此$i^\ast $满.
于是条件满足, 有$H^\ast (B)$模同构
$$
H^\ast (P(E))\cong H^\ast (B)\otimes H^\ast (\mathbb{R}\mathrm{P}^{n-1})=H^\ast (B)\{1,x,\cdots,x^{n-1}\}.
$$
存在唯一的一组$w_1,w_2,\cdots,w_n\in H^\ast (B),$
使得$x^n=\sum_{i=1}^n w_ix^{n-i}.$ 这被称为投影丛公式.
定义示性类$w_i(\xi):=w_i,$ $i=1,\cdots,n.$ 我们来验证它满足SW类公理.
$w_0(\xi)=1$ ok, 对于自然性, 考虑
$$
E'\rightarrow E\xrightarrow{\widehat f} \mathbb{R}\mathrm{P}^\infty,
$$
这拉回了
$$
H^\ast (\mathbb{R}\mathrm{P}^{n-1})\leftarrow H^\ast (P(E'))\leftarrow H^\ast (P(E))\leftarrow H^\ast (\mathbb{R}\mathrm{P}^\infty).
$$
这就得到了结论. 对于Whitney乘积公式, 定义$E_1=E(\xi),$ $E_2=E(\eta),$
$E_1\oplus E_2=E(\xi\oplus \eta),$ $\dim(\xi)=m,$ $\dim(\eta)=n.$ 考虑
$$
E_i\hookrightarrow E_1\oplus E_2\rightarrow \mathbb{R}^\infty,
$$
拉回了
$$
H^\ast (P(E_i))\leftarrow H^\ast (P(E_1\oplus E_2))\leftarrow H^\ast (\mathbb{R}\mathrm{P}^\infty).
$$
定义$u:=\sum_j w_j(\xi)x^{m-j},$ $v:=\sum_j w_j(\eta)x^{n-j}.$
断言$uv=0.$ 这样就有
$$
w_j(\xi\oplus\eta)=\sum_{r+s=j}w_r(\xi)w_s(\eta).
$$
至于断言, 考虑$P(E_1),P(E_2)\subset P(E_1\oplus E_2),$
$P(E_1)\cap P(E_2)=\varnothing.$ $U_i=P(E_1\oplus E_2)\setminus P(E_j),$
$U_i$形变收缩至$P(E_j).$
$$
H^m(P(E_1\oplus E_2),U_2)\cong H^m(P(E_1\oplus E_2),P(E_1))\rightarrow H^m(P(E_1\oplus E_2))\rightarrow H^m(P(E_1)),
$$
那么$\,\exists\,\widetilde u\mapsto u\mapsto 0.$
类似地取$\widetilde{v}.$
我们有$\widetilde u\cup \widetilde v=0\in H^{m+n}(X,U_1\cup U_2)=0.$
由自然性, 便有$uv=0.$
最后只需验证$w_1(\gamma_1^1)=0\in H^1(\mathbb{R}\mathrm{P}^1).$
$\gamma_1^1:\mathbb{R}^1\rightarrow E\rightarrow \mathbb{R}\mathrm{P}^1$投影诱导了$\mathbb{R}\mathrm{P}^0=\ast \rightarrow P(E)\rightarrow \mathbb{R}\mathrm{P}^1.$
定义$\widehat f:E\rightarrow \mathbb{R}^\infty,$ $(l,v)\mapsto v.$
$P(\widehat f):P(E)\cong \mathbb{R}\mathrm{P}^1\rightarrow \mathbb{R}\mathrm{P}^\infty,$
拉回$H^1(\mathbb{R}\mathrm{P}^1)\leftarrow H^1(\mathbb{R}\mathrm{P}^\infty),$
$\alpha\mapsto x\neq 0.$ 那么$x^1=w_1x^0,$ $x=w_1(\gamma_1^1)\neq 0.$
分裂定理
定理 3. 对于秩$n$平面丛$\xi,$ 底空间为仿紧集$B.$ 存在一个$F(E),$ $f:F(E)\rightarrow B,$ 使得: $f^\ast \xi$为$F(E)$上线丛的和, 且$f^\ast :H^\ast (B;\mathbb{Z}_2)\rightarrow H^\ast (F(E);\mathbb{Z}_2)$为单射.
对$n$归纳, 只需找到$f:F(E)\rightarrow B,$
使得$f^\ast \xi=\eta\oplus \eta^\perp,$ 且$f^\ast $为单射.
取$F(E)=P(E)\rightarrow B,$ $f=p(\pi).$
那么$f^\ast :H^\ast (B)\hookrightarrow H^\ast (P(E))\cong H^\ast (B)\{1,\cdots,x^{n-1}\}$为单射.
考虑全空间
$$
E(f^\ast \xi)=\{(b,l,v)|b\in B, l\subset F_b(\xi),v\in F_b(\xi)\}.
$$
考虑秩$1$子丛
$$
E(\eta):=\{(b,l,v)|b\in B,l\subset F_b(\xi),v\in l\}.
$$
那么$\eta$就是想找的线丛.
注 4. *重复找$n$次线丛, 那么$F(E)=\{(b,l_1,\cdots,l_n)|l_1+\cdots+l_n=F_b\xi\}.$ 那么就有标架丛
$$
GL_n(\mathbb{R})/(\mathbb{R}^\ast )^n\cong \{frames\}\rightarrow F(E)\rightarrow B,
$$
计算$w(\xi\otimes \xi').$ 第一步假设它们都是线丛的和, 那么
$$
w(\xi\otimes \xi')=w\left(\bigoplus_{i,j} L_i\otimes L_j'\right)=\prod_{i,j} w(L_i\otimes L_j').
$$
然后利用分裂定理, 设$f^\ast \xi,f^\ast \xi'$分裂(可以分步来处理,
$f=f_1\circ f_2$), 那么
$$
f^\ast (w(\xi\otimes \xi'))=w(f^\ast (\xi)\otimes f^\ast (\xi'))=F(w(f^\ast (\xi)),w(f^\ast (\eta)))=f^\ast F(w(\xi),w(\eta)).
$$
这样就可以把线丛上的计算方法套到一般丛上. 这也是分裂定理的主要用途之一.
文章最后更新于 2022-10-24 19:04:00
prove
Theorem 1. SW indicates the existence of sex type.
It is only necessary to consider the characteristic class of vector bundles on the CW complex $B$. Otherwise, consider the CW approximation $B'\rightarrow B,$
Just consider pulling back into the cluster.
Given vector bundle $\xi:\mathbb{R}^n\rightarrow E\xrightarrow{\pi} B,$
Consider the projected space bundle$P(\xi):P(\mathbb{R}^n)=\mathbb{R}\mathrm{P}^{n-1}\rightarrow P(E)\xrightarrow{p(\pi)}B,$
$$
P(E)=\{(b,l)|b\in B,l\subset F_b(\xi)\text{ 为直线}\}.
$$
$P(\xi)$ is a fiber bundle. We hope to apply the Leray-Hirsch theorem to this bundle.
Theorem 2 (Leray-Hirsch). Let $F\xrightarrow{i} E\xrightarrow{p} B$ be a fiber bundle. Let a certain commutative ring $R,$ be determined. If $H^n(F;R)$ is a free finitely generated $R$ module, $c_j\in H^\ast (E;R)$ exists such that $\{i^\ast c_j\}$ forms a set of $R$-basis of $H^\ast (F;R)\cong R\{c_j\}_j$. Then there is the following $H^\ast (B;R)$ module isomorphism:
$$
H^\ast (B;R)\bigotimes H^\ast (F;R)\cong H^\ast (E;R), \quad \sum_{i,j}b_i\otimes i^\ast (c_j)\mapsto \sum_{i,j}p^\ast (b_i)\cup c_j.
$$
We verify the conditions,
$H^i(\mathbb{R}\mathrm{P}^{n-1},\mathbb{Z}_2)=\left<{}a^i\right>$ is indeed free and finitely generated,
$a\in H^1,$ $i\le n-1.$ We wish to show that $i^\ast $ is surjective.
Memories$\,\exists\,\widehat f:E(\xi) \rightarrow \mathbb{R}^\infty,$
Let $\widehat f$ be a linear injector on the fiber. Then consider
$$
\mathbb{R}^n\hookrightarrow E\xrightarrow{\widehat f}\mathbb{R}^\infty,
$$
projection induction
$$
P(\mathbb{R}^n)\cong \mathbb{R}\mathrm{P}^{n-1}\xrightarrow{i}P(E)\xrightarrow{P(\widehat{f})}\mathbb{R}\mathrm{P}^\infty.
$$
Get the pull back mapping,
$$
H^\ast (\mathbb{R}\mathrm{P}^{n-1})\xleftarrow{i^\ast } H^\ast (P(E))\leftarrow H^\ast (\mathbb{R}\mathrm{P}^\infty).
$$
Composition is isomorphic in low dimensions, so $i^\ast $ is full.
So the conditions are met, and there is $H^\ast (B)$ modular isomorphism
$$
H^\ast (P(E))\cong H^\ast (B)\otimes H^\ast (\mathbb{R}\mathrm{P}^{n-1})=H^\ast (B)\{1,x,\cdots,x^{n-1}\}.
$$
There is a unique group $w_1,w_2,\cdots,w_n\in H^\ast (B),$
Such that $x^n=\sum_{i=1}^n w_ix^{n-i}.$ This is called the projective bundle formula.
Define the characteristic class $w_i(\xi):=w_i,$ $i=1,\cdots,n.$ and let us verify that it satisfies the SW class axiom.
$w_0(\xi)=1$ ok, for naturalness, consider
$$
E'\rightarrow E\xrightarrow{\widehat f} \mathbb{R}\mathrm{P}^\infty,
$$
This pulls back
$$
H^\ast (\mathbb{R}\mathrm{P}^{n-1})\leftarrow H^\ast (P(E'))\leftarrow H^\ast (P(E))\leftarrow H^\ast (\mathbb{R}\mathrm{P}^\infty).
$$
This leads to the conclusion. For the Whitney product formula, define $E_1=E(\xi),$ $E_2=E(\eta),$
$E_1\oplus E_2=E(\xi\oplus \eta),$ $\dim(\xi)=m,$ $\dim(\eta)=n.$ Consider
$$
E_i\hookrightarrow E_1\oplus E_2\rightarrow \mathbb{R}^\infty,
$$
Pulled back
$$
H^\ast (P(E_i))\leftarrow H^\ast (P(E_1\oplus E_2))\leftarrow H^\ast (\mathbb{R}\mathrm{P}^\infty).
$$
Definition $u:=\sum_j w_j(\xi)x^{m-j},$ $v:=\sum_j w_j(\eta)x^{n-j}.$
Assert $uv=0.$ This way we have
$$
w_j(\xi\oplus\eta)=\sum_{r+s=j}w_r(\xi)w_s(\eta).
$$
As for assertions, consider $P(E_1),P(E_2)\subset P(E_1\oplus E_2),$
$P(E_1)\cap P(E_2)=\varnothing.$ $U_i=P(E_1\oplus E_2)\setminus P(E_j),$
$U_i$ deforms and shrinks to $P(E_j).$
$$
H^m(P(E_1\oplus E_2),U_2)\cong H^m(P(E_1\oplus E_2),P(E_1))\rightarrow H^m(P(E_1\oplus E_2))\rightarrow H^m(P(E_1)),
$$
Then $\,\exists\,\widetilde u\mapsto u\mapsto 0.$
Take $\widetilde{v}.$ similarly
We have $\widetilde u\cup \widetilde v=0\in H^{m+n}(X,U_1\cup U_2)=0.$
From nature, there is $uv=0.$
Finally just verify $w_1(\gamma_1^1)=0\in H^1(\mathbb{R}\mathrm{P}^1).$
$\gamma_1^1:\mathbb{R}^1\rightarrow E\rightarrow \mathbb{R}\mathrm{P}^1$ Projection induced $\mathbb{R}\mathrm{P}^0=\ast \rightarrow P(E)\rightarrow \mathbb{R}\mathrm{P}^1.$
Definition $\widehat f:E\rightarrow \mathbb{R}^\infty,$ $(l,v)\mapsto v.$
$P(\widehat f):P(E)\cong \mathbb{R}\mathrm{P}^1\rightarrow \mathbb{R}\mathrm{P}^\infty,$
Pull back$H^1(\mathbb{R}\mathrm{P}^1)\leftarrow H^1(\mathbb{R}\mathrm{P}^\infty),$
$\alpha\mapsto x\neq 0.$ Then $x^1=w_1x^0,$ $x=w_1(\gamma_1^1)\neq 0.$
splitting theorem
Theorem 3. For the rank $n$ plane bundle $\xi,$, the base space is a paracompact set $B.$, and there is a $F(E),$ $f:F(E)\rightarrow B,$ such that: $f^\ast \xi$ is the sum of the line bundles on $F(E)$, and $f^\ast :H^\ast (B;\mathbb{Z}_2)\rightarrow H^\ast (F(E);\mathbb{Z}_2)$ is injective.
To generalize $n$, just find $f:F(E)\rightarrow B,$
Let $f^\ast \xi=\eta\oplus \eta^\perp,$ and $f^\ast $ be injective.
Take $F(E)=P(E)\rightarrow B,$ $f=p(\pi).$
Then $f^\ast :H^\ast (B)\hookrightarrow H^\ast (P(E))\cong H^\ast (B)\{1,\cdots,x^{n-1}\}$ is an injective.
Consider the whole space
$$
E(f^\ast \xi)=\{(b,l,v)|b\in B, l\subset F_b(\xi),v\in F_b(\xi)\}.
$$
Consider the rank $1$ subbundle
$$
E(\eta):=\{(b,l,v)|b\in B,l\subset F_b(\xi),v\in l\}.
$$
Then $\eta$ is the line bundle you want to find.
Note 4. *Repeatedly find $n$ secondary line cluster, then $F(E)=\{(b,l_1,\cdots,l_n)|l_1+\cdots+l_n=F_b\xi\}.$ then there is a frame cluster
$$
GL_n(\mathbb{R})/(\mathbb{R}^\ast )^n\cong \{frames\}\rightarrow F(E)\rightarrow B,
$$
The first step to calculate $w(\xi\otimes \xi').$ is assuming that they are all sums of line bundles, then
$$
w(\xi\otimes \xi')=w\left(\bigoplus_{i,j} L_i\otimes L_j'\right)=\prod_{i,j} w(L_i\otimes L_j').
$$
Then use the splitting theorem, assuming $f^\ast \xi,f^\ast \xi'$ splits (it can be processed in steps,
$f=f_1\circ f_2$), then
$$
f^\ast (w(\xi\otimes \xi'))=w(f^\ast (\xi)\otimes f^\ast (\xi'))=F(w(f^\ast (\xi)),w(f^\ast (\eta)))=f^\ast F(w(\xi),w(\eta)).
$$
In this way, the calculation method on line bundles can be applied to general bundles. This is also one of the main uses of the splitting theorem.
The article was last updated on 2022-10-24 19:04:00
