Machine-translated from Chinese.
定向与Thom类
回忆对$V$为秩$n$实向量空间, $V$上的定向为
$\operatorname{Iso}(\mathbb{R}^n,V)\cong GL_n(\mathbb{R})$
上连通分支的选取(线性代数). 也等价于 $H^n(V|0;\mathbb{Z})$
生成元的选取(代数拓扑). 也等价于
$$
\Lambda^n V\setminus\{0\}\cong \Lambda^n\mathbb{R}^n\setminus\{0\}\xrightarrow{\det}\mathbb{R}\setminus\{0\}
$$
连通分支的选取(微分流形).
对于秩$n$实向量丛,
定义$\xi$的定向为考虑外代数丛$\Lambda^n F\rightarrow E(\Lambda^n \xi)\rightarrow B,$
$\Lambda^n\xi$上一个处处不消失的截面.
当然我们也有其它的等价定义.
$\xi$的一个定向对每个纤维$F$给出了一个$H^\ast (F|0;\mathbb{Z})$的定向,
满足局部相容性.
定理 1. 若$\xi$为秩$n$定向实向量丛, 那么存在唯一一个类$u\in H^n(E|B;\mathbb{Z}),$ 限制在每个纤维上给出$H^n(F|0;\mathbb{Z})$的生成元.
证明思路: 我们采用MV方法来处理. 首先证明平凡丛$\xi$的情况,
接下来考虑$B=U_1\cup U_2$的情况, 这样能给出$B$紧的证明.
最后考虑$B$的穷竭序列即可.
这样的类$u=u(\xi)$称为Thom类. 若考虑$\mathbb{Z}_2$系数,
那么无需预先给出$\xi$的定向.
定理 2 (Thom同构定理). 若$\xi$为秩$n$定向向量丛, 有Thom类$u.$ 那么$H^i(B;\mathbb{Z})\cong H^{n+i}(E|B;\mathbb{Z}),$ $\alpha\mapsto \pi^\ast \alpha \cup u.$
应用Leray-Hirsch定理到
$$
(F,F_0)\rightarrow (E,E_0)\rightarrow B.
$$
那么首先$H^n(F|0)=\mathbb{Z}$自由生成, 其次由上一个定理, $i^\ast $是满射.
这样$H^\ast (E|0)\cong H^\ast (B)\{u\}$就给出了结论.
欧拉类
对定向丛$\xi$定义欧拉类: $e(\xi)\in H^n(B;\mathbb{Z}),$ 使得
$$
H^n(B;\mathbb{Z})\xrightarrow{\pi^\ast :\cong} H^n(E;\mathbb{Z})\leftarrow H^n(E|B;\mathbb{Z})
$$
将$e(\xi)$与$u(\xi)$打到同一个元.
它满足以下性质: 若$B\rightarrow B'$由保定向丛映射覆盖,
那么$e(\xi)(=e(f^\ast \xi'))=f^\ast e(\xi'),$ 即满足自然性; 若$\xi$定向改变,
则欧拉类改变符号;
$$
e(\xi\oplus\eta)=e(\xi)\cup e(\eta),\quad e(\xi\times \eta)=e(\xi)\times e(\eta).
$$
若$\xi$有一个无处消失截面, 那么$e(\xi)=0.$
因为这将给出$\xi=\varepsilon^1\oplus\varepsilon^\perp,$
$e(\xi)=e(\varepsilon^1)\cup e(\varepsilon^\perp)=0.$
定理 3. 对光滑定向紧连通流形$M^n,$ $e(\tau_M)=\chi(M)\cdot PD([\ast ]),$ 或者说$\left<{}e(\tau_M),[M]\right>=\chi(M).$
由此可以推出Poincaré-Hopf定理: $\chi(M)\neq 0$ 则不存在无处消失向量场.
定理 4. 若$\xi$是同上假设的$M$上的光滑定向向量丛, 则$e(\xi)=PD([Z_s]),$ $Z_s$为$s$的零截面, 需要和$M$(默认零截面)横截相交.
取$\xi=\tau_M$即推出前一定理.
若$\xi$是$B$上的秩$n$定向向量丛,
$H^n(B;\mathbb{Z})\rightarrow H^n(B;\mathbb{Z}_2)$的自然映射就是$e(\xi)\mapsto w_n(\xi).$
文章最后更新于 2022-10-26 21:15:00
Orientation and Thom Class
Recall that pair $V$ is a real vector space of rank $n$, and the orientation on $V$ is
$\operatorname{Iso}(\mathbb{R}^n,V)\cong GL_n(\mathbb{R})$
Selection of upper connected components (linear algebra). Also equivalent to $H^n(V|0;\mathbb{Z})$
Selection of generators (algebraic topology). Also equivalent to
$$
\Lambda^n V\setminus\{0\}\cong \Lambda^n\mathbb{R}^n\setminus\{0\}\xrightarrow{\det}\mathbb{R}\setminus\{0\}
$$
Selection of connected components (differential manifold).
For a real vector bundle of rank $n$,
Define the orientation of $\xi$ by considering the exterior algebra bundle $\Lambda^n F\rightarrow E(\Lambda^n \xi)\rightarrow B,$
$\Lambda^n\xi$ The previous section does not disappear everywhere.
Of course we also have other equivalent definitions.
An orientation of $\xi$ gives an orientation of $H^\ast (F|0;\mathbb{Z})$ for each fiber $F$,
Meet local compatibility.
Theorem 1. If $\xi$ is a directed real vector bundle of rank $n$, then there is a unique generator of class $u\in H^n(E|B;\mathbb{Z}),$ restricted to give $H^n(F|0;\mathbb{Z})$ on each fiber.
Proof idea: We use the MV method to deal with it. First prove the case of trivial bundle $\xi$,
Next, consider the case of $B=U_1\cup U_2$, which can give a tight proof of $B$.
Finally, consider the exhaustive sequence of $B$.
Such a class $u=u(\xi)$ is called the Thom class. If the $\mathbb{Z}_2$ coefficient is considered,
Then there is no need to give the orientation of $\xi$ in advance.
Theorem 2 (Thom’s isomorphism theorem). If $\xi$ is a directed vector bundle of rank $n$ and has Thom class $u.$, then $H^i(B;\mathbb{Z})\cong H^{n+i}(E|B;\mathbb{Z}),$ $\alpha\mapsto \pi^\ast \alpha \cup u.$
Apply the Leray-Hirsch theorem to
$$
(F,F_0)\rightarrow (E,E_0)\rightarrow B.
$$
Then first $H^n(F|0)=\mathbb{Z}$ is generated freely, and secondly, according to the previous theorem, $i^\ast $ is surjective.
In this way $H^\ast (E|0)\cong H^\ast (B)\{u\}$ gives the conclusion.
Euler class
Define the Euler class for the directed bundle $\xi$: $e(\xi)\in H^n(B;\mathbb{Z}),$ such that
$$
H^n(B;\mathbb{Z})\xrightarrow{\pi^\ast :\cong} H^n(E;\mathbb{Z})\leftarrow H^n(E|B;\mathbb{Z})
$$
Move $e(\xi)$ and $u(\xi)$ to the same element.
It satisfies the following properties: If $B\rightarrow B'$ is covered by a direction-preserving bundle map,
Then $e(\xi)(=e(f^\ast \xi'))=f^\ast e(\xi'),$ satisfies naturalness; if $\xi$ changes orientation,
Then the Euler class changes sign;
$$
e(\xi\oplus\eta)=e(\xi)\cup e(\eta),\quad e(\xi\times \eta)=e(\xi)\times e(\eta).
$$
If $\xi$ has a nowhere-vanishing section, then $e(\xi)=0.$
since this will give $\xi=\varepsilon^1\oplus\varepsilon^\perp,$
$e(\xi)=e(\varepsilon^1)\cup e(\varepsilon^\perp)=0.$
Theorem 3. For smoothly oriented compactly connected manifolds $M^n,$ $e(\tau_M)=\chi(M)\cdot PD([\ast ]),$ or $\left<{}e(\tau_M),[M]\right>=\chi(M).$
From this we can derive the Poincaré-Hopf theorem: $\chi(M)\neq 0$ then there is no vector field that disappears nowhere.
Theorem 4. If $\xi$ is a smooth oriented vector bundle on $M$ assumed as above, then $e(\xi)=PD([Z_s]),$ $Z_s$ is the zero section of $s$ and needs to intersect with $M$ (default zero section).
Taking $\xi=\tau_M$ leads to the previous theorem.
If $\xi$ is a rank $n$ directed vector bundle on $B$,
The natural mapping of $H^n(B;\mathbb{Z})\rightarrow H^n(B;\mathbb{Z}_2)$ is $e(\xi)\mapsto w_n(\xi).$
The article was last updated on 2022-10-26 21:15:00
