Machine-translated from Chinese.
引子
固定$F$为拓扑空间, 取$G=\operatorname{Homeo}(F).$ (紧开拓扑.) 我们有
$$
\{F\rightarrow E\rightarrow B\}/\cong\leftrightarrow \{G\rightarrow P\rightarrow B\}/\cong \leftrightarrow [B,BG]
$$
$$
\xi\mapsto (\operatorname{Homeo}(\xi_b,F)\rightarrow P\rightarrow B),
$$
像称为标架丛. 反过来, 我们有:
$$
(F\rightarrow P\times_G F\rightarrow B)\leftarrow (G\rightarrow P\rightarrow B).
$$
光滑流形上的向量丛
圆丛
设$F=M^n$为(可定向)光滑流形, $G=\operatorname{Diffeo}^+(M^n)$保持定向.
那么
$$
\{\text{(oriented) $M^n$-bundle}\}\leftrightarrow [B,BG].
$$
若$F=S^1,$
$U(1)=SO(2)\hookrightarrow \operatorname{Diffeo}^+(S^1)$为同伦等价, 那么
$$
B\operatorname{Diffeo}^+(S^1)=BU(1)=BSO(2)=\mathbb{C}\mathrm{P}^\infty=K(\mathbb{Z},2).
$$
于是我们有
$$
\{B\text{上定向$S^1$丛}\}/\cong\leftrightarrow [B,K(\mathbb{Z},2)]\cong H^2(B;\mathbb{Z}),
$$
$$
\xi\mapsto e(\xi).
$$
注 1. 每个可定向$S^1$丛都是一个复线丛的球丛. Euler类是一个完全不变量.
推论 2. 每个定向圆丛平凡若其有一个截面.
球丛
对一般的球丛, $M^n=S^n,$
$SO(n+1)\hookrightarrow \operatorname{Diffeo}^+(S^n)$为一个弱同伦等价,
当$n=2,3$时. 它一定不是弱同伦等价, 若$n\ge 4.$
注 3. $S^n$-丛理论和$n+1$维实向量丛理论是一致的.
环丛
若$M^n=T^2=S^1\times S^1,$
$$
\{B\text{上}T^2\text{丛}\}\leftrightarrow [B,B\operatorname{Diffeo}(T^2)],
$$
$$
\{B\text{上}T^2\text{丛, 有截面}\}\leftrightarrow [B,B\operatorname{Diffeo}(T^2,\ast )],
$$
$\operatorname{Diffeo}(T^2,\ast )$表示有不动点.
命题 4. $SL(2,\mathbb{Z})\hookrightarrow \operatorname{Diffeo}^+(T^2,\ast )$为一个同伦等价.
推论 5. $B\operatorname{Diffeo}^+(T^2,\ast )=BSL(2,\mathbb{Z}).$
命题 6. $\,\exists\,\operatorname{Diffeo}^+(T^2,\ast )\rightarrow \operatorname{Diffeo}^+(T^2)\rightarrow T^2.$
$$
B\operatorname{Diffeo}^+(T^2,\ast )=BSL(2,\mathbb{Z})\rightarrow BSL(2,\mathbb{R})\cong BSO(2).
$$
给出了
$$
\{B\text{上定向}T^2\text{丛具截面}\}\xrightarrow{\phi} \{B\text{上定向秩$2$丛}\}.
$$
(是每个截面处取切平面)
曲面丛
考虑一般的$M^n=\Sigma_g,$ $g\ge 2.$ 那么
$$
\{B\text{上光滑$\Sigma_g$丛}\}/\cong\leftrightarrow [B,B\operatorname{Diffeo}(\Sigma_g)].
$$
Earle-Eells(1967):
单位元所在的分支$\operatorname{Diffeo}_0(\Sigma_g)$可缩.
定义$\Sigma_g$的映射类群为离散群
$$
\operatorname{Mod}(\Sigma_g):=\operatorname{Diff}^+(\Sigma_g)/\operatorname{Diffeo}_0(\Sigma_g)=\pi_0(\operatorname{Diff}^+).
$$
因此,
$$
B\operatorname{Diff}^+(\Sigma_g)=B\operatorname{Mod}(\Sigma_g).
$$
$$
H^\ast (B\operatorname{Mod}(\Sigma_g))=\{\text{曲面丛示性类}\}
$$
定理 7 (Harer,1985). $H^k(B\operatorname{Mod}(\Sigma_g);\mathbb{Z})$稳定, 当$g\rightarrow\infty.$
定理 8 (Madsen-Weiss,2006). $H^\ast (B\operatorname{Mod}(\Sigma_\infty);\mathbb{Q})\rightarrow \mathbb{Q}[\kappa_1,\kappa_2,\cdots],$ 当$g\rightarrow\infty,$ $|\kappa_i|=2i.$
定理 9 (Harer-Zagier,1986). $\chi(B\operatorname{Mod}(\Sigma_g))=\frac{\zeta(1-2g)}{2-2g}\sim (-1)^g\frac{(2g-1)!}{(2-2g)2^{2g-1}\pi^{2g} }.$
注 10. $\chi(B\operatorname{Mod}(\Sigma_g))$超指数增长, 但稳定群的示性数仅是多项式增长.
这称为”暗物质问题”.
Pontrjagin类
回忆线性代数理论,
$n$维实向量空间可以复化为$n$维复线性空间$V\otimes_\mathbb{R}\mathbb{C}=\{x+iy|x,y\in V\}.$
我们有复向量空间的同构
$$
V\otimes_\mathbb{R}\mathbb{C}\rightarrow\overline{V\otimes_\mathbb{R}\mathbb{C} },\quad x+iy\mapsto x-iy.
$$
对于向量丛, 设$\xi$为秩$n$实向量丛,
$\xi\otimes_\mathbb{R}\mathbb{C}$为秩$n$复向量丛,
$\xi\otimes_\mathbb{R}\mathbb{C}\cong \overline{\xi\otimes_\mathbb{R}\mathbb{C} }$为复向量丛的同构.
$\,\forall\,k,$
$$
c_k(\xi\otimes \mathbb{C})=c_k(\overline{\xi\otimes \mathbb{C} })=(-1)^kc_k(\xi\otimes \mathbb{C}).
$$
因此$k$为奇数时, 总有$2c_k(\xi\otimes\mathbb{C})=0.$
定义第$i$个Pontrjagin示性类为
$$
p_i(\xi):=(-1)^ic_{2i}(\xi\otimes \mathbb{C})\in H^{4i}(B;\mathbb{Z}).
$$
显然$p_i(\xi)=0,$ $\,\forall\,i>\frac{\dim \xi}{2}.$
记$p(\xi)=1+p_1(\xi)+\cdots+p_{[\frac{n}{2}]}(\xi).$
$$
p(\xi\oplus \eta)=p(\xi)\cup p(\eta)+2 \text{torsions},
$$
$$
2[p(\xi\oplus\eta)-p(\xi)\cup p(\eta)]=0.
$$
文章最后更新于 2022-11-23 15:12:50
Introduction
Fix $F$ as the topological space, take $G=\operatorname{Homeo}(F).$ (compact open topology.) We have
$$
\{F\rightarrow E\rightarrow B\}/\cong\leftrightarrow \{G\rightarrow P\rightarrow B\}/\cong \leftrightarrow [B,BG]
$$
$$
\xi\mapsto (\operatorname{Homeo}(\xi_b,F)\rightarrow P\rightarrow B),
$$
The image is called a frame bundle. In turn, we have:
$$
(F\rightarrow P\times_G F\rightarrow B)\leftarrow (G\rightarrow P\rightarrow B).
$$
Vector bundles on smooth manifolds
Round bush
Let $F=M^n$ be a (orientable) smooth manifold, and $G=\operatorname{Diffeo}^+(M^n)$ remain oriented.
Then
$$
\{\text{(oriented) $M^n$-bundle}\}\leftrightarrow [B,BG].
$$
If $F=S^1,$
$U(1)=SO(2)\hookrightarrow \operatorname{Diffeo}^+(S^1)$ is homotopy equivalent, then
$$
B\operatorname{Diffeo}^+(S^1)=BU(1)=BSO(2)=\mathbb{C}\mathrm{P}^\infty=K(\mathbb{Z},2).
$$
So we have
$$
\{B\text{上定向$S^1$丛}\}/\cong\leftrightarrow [B,K(\mathbb{Z},2)]\cong H^2(B;\mathbb{Z}),
$$
$$
\xi\mapsto e(\xi).
$$
Note 1. Every orientable $S^1$ bundle is a spherical bundle of polylinear bundles. The Euler class is a complete invariant.
Corollary 2. Every oriented circular bundle is trivial if it has one cross section.
ball cluster
For general ball clusters, $M^n=S^n,$
$SO(n+1)\hookrightarrow \operatorname{Diffeo}^+(S^n)$ is a weak homotopy equivalent,
When $n=2,3$. It must not be weakly homotopy equivalent, if $n\ge 4.$
Note 3. $S^n$-bundle theory is consistent with $n+1$-dimensional real vector bundle theory.
ring cluster
If $M^n=T^2=S^1\times S^1,$
$$
\{B\text{上}T^2\text{丛}\}\leftrightarrow [B,B\operatorname{Diffeo}(T^2)],
$$
$$
\{B\text{上}T^2\text{丛, 有截面}\}\leftrightarrow [B,B\operatorname{Diffeo}(T^2,\ast )],
$$
$\operatorname{Diffeo}(T^2,\ast )$ means there is a fixed point.
Proposition 4. $SL(2,\mathbb{Z})\hookrightarrow \operatorname{Diffeo}^+(T^2,\ast )$ is a homotopy equivalent.
Corollary 5. $B\operatorname{Diffeo}^+(T^2,\ast )=BSL(2,\mathbb{Z}).$
Proposition 6. $\,\exists\,\operatorname{Diffeo}^+(T^2,\ast )\rightarrow \operatorname{Diffeo}^+(T^2)\rightarrow T^2.$
$$
B\operatorname{Diffeo}^+(T^2,\ast )=BSL(2,\mathbb{Z})\rightarrow BSL(2,\mathbb{R})\cong BSO(2).
$$
given
$$
\{B\text{上定向}T^2\text{丛具截面}\}\xrightarrow{\phi} \{B\text{上定向秩$2$丛}\}.
$$
(The tangent plane is taken at each section)
Surface bundle
Consider the general $M^n=\Sigma_g,$ $g\ge 2.$ then
$$
\{B\text{上光滑$\Sigma_g$丛}\}/\cong\leftrightarrow [B,B\operatorname{Diffeo}(\Sigma_g)].
$$
Earle-Eells(1967):
The branch $\operatorname{Diffeo}_0(\Sigma_g)$ where the unit element is located can be contracted.
Definition $\Sigma_g$ Mapping taxa is a discrete group
$$
\operatorname{Mod}(\Sigma_g):=\operatorname{Diff}^+(\Sigma_g)/\operatorname{Diffeo}_0(\Sigma_g)=\pi_0(\operatorname{Diff}^+).
$$
Therefore,
$$
B\operatorname{Diff}^+(\Sigma_g)=B\operatorname{Mod}(\Sigma_g).
$$
$$
H^\ast (B\operatorname{Mod}(\Sigma_g))=\{\text{曲面丛示性类}\}
$$
Theorem 7 (Harer,1985). $H^k(B\operatorname{Mod}(\Sigma_g);\mathbb{Z})$ is stable, when $g\rightarrow\infty.$
Theorem 8 (Madsen-Weiss,2006). $H^\ast (B\operatorname{Mod}(\Sigma_\infty);\mathbb{Q})\rightarrow \mathbb{Q}[\kappa_1,\kappa_2,\cdots],$ when $g\rightarrow\infty,$ $|\kappa_i|=2i.$
Theorem 9 (Harer-Zagier,1986). $\chi(B\operatorname{Mod}(\Sigma_g))=\frac{\zeta(1-2g)}{2-2g}\sim (-1)^g\frac{(2g-1)!}{(2-2g)2^{2g-1}\pi^{2g} }.$
Note 10. $\chi(B\operatorname{Mod}(\Sigma_g))$ grows superexponentially, but the indicator number of the stable group only grows polynomially.
This is called the "dark matter problem."
Pontrjagin class
Recalling linear algebra theory,
$n$-dimensional real vector space can be complexed into $n$-dimensional complex linear space $V\otimes_\mathbb{R}\mathbb{C}=\{x+iy|x,y\in V\}.$
We have the isomorphism of complex vector spaces
$$
V\otimes_\mathbb{R}\mathbb{C}\rightarrow\overline{V\otimes_\mathbb{R}\mathbb{C} },\quad x+iy\mapsto x-iy.
$$
For vector bundles, let $\xi$ be a real vector bundle of rank $n$,
$\xi\otimes_\mathbb{R}\mathbb{C}$ is a complex vector bundle of rank $n$,
$\xi\otimes_\mathbb{R}\mathbb{C}\cong \overline{\xi\otimes_\mathbb{R}\mathbb{C} }$ is the isomorphism of the complex vector bundle.
$\,\forall\,k,$
$$
c_k(\xi\otimes \mathbb{C})=c_k(\overline{\xi\otimes \mathbb{C} })=(-1)^kc_k(\xi\otimes \mathbb{C}).
$$
Therefore, when $k$ is an odd number, there is always $2c_k(\xi\otimes\mathbb{C})=0.$
Define the $i$th Pontrjagin characteristic class as
$$
p_i(\xi):=(-1)^ic_{2i}(\xi\otimes \mathbb{C})\in H^{4i}(B;\mathbb{Z}).
$$
Obviously $p_i(\xi)=0,$ $\,\forall\,i>\frac{\dim \xi}{2}.$
Note$p(\xi)=1+p_1(\xi)+\cdots+p_{[\frac{n}{2}]}(\xi).$
$$
p(\xi\oplus \eta)=p(\xi)\cup p(\eta)+2 \text{torsions},
$$
$$
2[p(\xi\oplus\eta)-p(\xi)\cup p(\eta)]=0.
$$
The article was last updated on 2022-11-23 15:12:50
