Machine-translated from Chinese.
整体角形式
在定向流形$M$上, 最高次形式是正的若它在$M$的定向类中.
取$\sigma\in H^{n-1}(S^{n-1})$为生成元,
$p:\mathbb{R}^n-\{0\}\rightarrow S^{n-1}$为形变收缩,
规定$\sigma$在$S^{n-1}$上是正的当且仅当$dr\wedge p^\ast \sigma$在$\mathbb{R}^n-\{0\}$上是正的.
设$\sigma$是$S^{k-1}$上正形式, 且$\int_{S^{k-1} }\sigma=1$,
则$\psi=p^\ast \sigma$称为$\mathbb{R}^k-0$上的角形式.
$[\psi]\in H^{k-1}(\mathbb{R}^k-\{0\})$为生成元.
设$\rho(r)\in C^\infty[0,+\infty)$, $\rho(r)=\begin{cases}
-1&r\le \delta_1\\
0&r\ge \delta_2
\end{cases}$, $\delta_1<\delta_2$, 中间部分光滑连接.
$d\rho=\rho'(r)dr$是一个bump-1形式.
这时, $[d\rho\wedge \psi]\in H_c^k(\mathbb{R}^k)$, 挖去$B_\varepsilon$,
利用$\rho$性质与Stokes定理可证$\int_{\mathbb{R}^k}d\rho\wedge \psi=1$.
即$[d\rho\wedge \psi]$为生成元.
考虑$M$上秩$k$定向向量丛$\pi :E\rightarrow M$, $E^0=E-s(E)$,
$s$为零截面. 设$\left<{},\right>$为度量, 对每条纤维上的向量$v$,
定义$r(v)=\sqrt{\left<{}v,v\right>}$. 那么$r$在$E$上有意义,
$\rho(r)$可视为定义在$E$上的函数.
称$E^0$的一个$k-1$-形式$\psi$为整体角形式,
若$\psi$限制在每条纤维$E_x^0\simeq \mathbb{R}^k-\{0\}$上是角形式,
并且$d\psi$能自然延拓成$E$的一个$k$-形式.
若$E^0$上存在整体角形式$\psi$,
则$E$的Thom类为$\Phi=d\rho\wedge\psi+\rho d\psi$.
这是因为限制在纤维上第二项消失, 而第一项积分为$1$.
平面丛欧拉类的显示表示
假定$E$是定向平面丛, 即秩为$2$. 取坐标图册与定向平凡化,
在$U_\alpha$上有坐标$(x_1^\alpha,...,x_n^\alpha)$.
利用定向平凡化, 定义$E^0|_{E_\alpha}$的极坐标$r_\alpha$,
$\theta_\alpha$, 从而有坐标$\pi^\ast x^\alpha, r_\alpha, \theta_\alpha$.
假定$E$上度量保内积, 即转移函数约化到$SO(2)$, 则$r_\alpha=r_\beta$,
我们定义了整体的$r$. 事实上这就是先前的$r(v)=\sqrt{\left<{}v,v\right>}$.
但$\theta_\alpha,\theta_\beta$相差一个旋转,
可定义$\varphi_{\alpha\beta}$,
使得$\theta_\beta=\theta_\alpha+\pi^\ast \varphi_{\alpha\beta}$.
在$\mod 2\pi$意义下,
$\varphi_{\alpha\beta}$满足cocycle条件以及可被唯一定义.
因此若采用de Rham上同调, 则$\{d\varphi_{\alpha\beta}\}$满足cocycle条件.
定义$\xi_\alpha=\frac{1}{2\pi} \sum_\gamma \rho_\gamma d\varphi_{\gamma\alpha}$,
则在$U_{\alpha\beta}$上,
$\frac{1}{2\pi}d\varphi_{\alpha\beta}=\xi_\beta-\xi_\alpha$.
从而, $d\xi_\alpha=d\xi_\beta$, 可以拼成整体定义$2$-形式$e$:
$e|_{U_\alpha}=d\xi_\alpha$. 显然$de=0$.
断言$[e]\in H^2(M)$与$\{\xi_\alpha\}$选取无关,
称之为平面丛$E$的欧拉类, 也记作$e(E)$.
选取无关是因为倘若有另一个$\tilde \xi_\alpha$满足要求,
那么$\tilde \xi_\alpha-\xi_\alpha$将满足相容性, 可以整体定义为$\eta$.
那么$d\eta=d\tilde\xi_\alpha-d\xi_\alpha$, 从而在上同调层面它们是相等的.
若给$E$一个定向平凡化, 假设约化到了$SO(2)=SU(1)$,
则$\varphi_{\alpha\beta}=-\frac{1}{i}\log g_{\alpha\beta}$, 进一步
$e|_{U_\alpha}=d\xi_\alpha=-\frac{1}{2\pi i}\sum_\gamma d(\rho_\gamma \cdot d\log g_{\gamma\alpha})$,
称为平面丛欧拉类的显示表示公式.
当$E$为乘积丛时, 转移函数为常值矩阵, 从而欧拉类为零.
它反映出了定向向量丛$E$的扭曲程度. 由表示公式, 可以证明欧拉类的函子性:
命题 1.1. $f:N\rightarrow M$为光滑映射, $\pi:E\rightarrow M$为秩2定向向量丛, 则$e(f^{-1}E)=f^\ast e(E)\in H^2(N)$.
平面丛Thom类的显示表示
$\frac{d\theta_\alpha}{2\pi}-\pi^\ast \xi_\alpha$定义了$E^0$上的$1$-形式$\psi$.
$\psi$限制在每条纤维上$E_x^0$上是$\frac{d\theta_\alpha}{2\pi}$,
为$E_x^0$的角形式; 同时$d\psi=-\pi^\ast e$在$E$上有定义, $\psi$是整体角形式.
于是, $E$的Thom类为$\Phi=d\rho(r)\wedge \psi-\rho(r)\pi^\ast e$.
若$s:M\rightarrow E$为零截面, 则$s^\ast \Phi=e$.
即零截面把Thom类拉回为欧拉类.
利用欧拉类的显式表示,
即得$\Phi=d\rho(r)\wedge \frac{d\theta_\alpha}{2\pi}+\frac{1}{2\pi i}d(\rho(r)\pi^\ast \sum_\gamma\rho_\gamma d\log g_{\gamma\alpha})$.
前一项形式非常直观, 而后一项是因为向量丛扭曲引起的修正项.
文章最后更新于 2021-10-12 10:25:18
Global Angular Form
On the oriented manifold $M$, the highest-order form is True If it is in the orientation class of $M$.
Take $\sigma\in H^{n-1}(S^{n-1})$ as the generator,
$p:\mathbb{R}^n-\{0\}\rightarrow S^{n-1}$ is deformation shrinkage,
It is stipulated that $\sigma$ is positive on $S^{n-1}$ if and only if $dr\wedge p^\ast \sigma$ is positive on $\mathbb{R}^n-\{0\}$.
Let $\sigma$ be the upper positive form of $S^{k-1}$, and $\int_{S^{k-1} }\sigma=1$,
Then $\psi=p^\ast \sigma$ is called on $\mathbb{R}^k-0$ Angular form.
$[\psi]\in H^{k-1}(\mathbb{R}^k-\{0\})$ is the generator.
Assume $\rho(r)\in C^\infty[0,+\infty)$, $\rho(r)=\begin{cases}
-1&r\le \delta_1\\
0&r\ge \delta_2
\end{cases}$, $\delta_1<\delta_2$, the middle part is smoothly connected.
$d\rho=\rho'(r)dr$ is a bump-1 form.
At this time, $[d\rho\wedge \psi]\in H_c^k(\mathbb{R}^k)$, dig out $B_\varepsilon$,
Using the $\rho$ property and Stokes' theorem, we can prove $\int_{\mathbb{R}^k}d\rho\wedge \psi=1$.
That is, $[d\rho\wedge \psi]$ is the generator.
Consider $M$ upper rank $k$ directed vector bundle $\pi :E\rightarrow M$, $E^0=E-s(E)$,
$s$ is the zero section. Let $\left<{},\right>$ be the metric, for the vector $v$ on each fiber,
Define $r(v)=\sqrt{\left<{}v,v\right>}$. Then $r$ makes sense on $E$,
$\rho(r)$ can be regarded as a function defined on $E$.
A $k-1$-form $\psi$ of $E^0$ is called Global Angular Form,
If $\psi$ is restricted to the angular form on each fiber $E_x^0\simeq \mathbb{R}^k-\{0\}$,
And $d\psi$ can be naturally extended into a $k$-form of $E$.
If there is a global angular form $\psi$ on $E^0$,
Then the Thom class of $E$ is $\Phi=d\rho\wedge\psi+\rho d\psi$.
This is because the second term disappears when restricted to the fiber, and the first term integrates as $1$.
Explicit representation of Euler classes for planar bundles
Assume that $E$ is an oriented planar bundle, that is, the rank is $2$. Taking the coordinate atlas and oriented trivialization,
There are coordinates $(x_1^\alpha,...,x_n^\alpha)$ on $U_\alpha$.
Using directional trivialization, define the polar coordinates $r_\alpha$ of $E^0|_{E_\alpha}$,
$\theta_\alpha$, thus there are coordinates $\pi^\ast x^\alpha, r_\alpha, \theta_\alpha$.
Assume that $E$ is a metric-preserving inner product, that is, the transfer function is reduced to $SO(2)$, then $r_\alpha=r_\beta$,
We defined the overall $r$. In fact this was the previous $r(v)=\sqrt{\left<{}v,v\right>}$.
But $\theta_\alpha,\theta_\beta$ differs by one rotation,
Definable $\varphi_{\alpha\beta}$,
Make $\theta_\beta=\theta_\alpha+\pi^\ast \varphi_{\alpha\beta}$.
In the sense of $\mod 2\pi$,
$\varphi_{\alpha\beta}$ Meets cocycle conditions and can be uniquely defined.
Therefore, if de Rham cohomology is adopted, $\{d\varphi_{\alpha\beta}\}$ satisfies the cocycle condition.
Definition$\xi_\alpha=\frac{1}{2\pi} \sum_\gamma \rho_\gamma d\varphi_{\gamma\alpha}$,
Then on $U_{\alpha\beta}$,
$\frac{1}{2\pi}d\varphi_{\alpha\beta}=\xi_\beta-\xi_\alpha$.
Thus, $d\xi_\alpha=d\xi_\beta$, can be put together into the overall definition $2$-form $e$:
$e|_{U_\alpha}=d\xi_\alpha$. Obviously $de=0$.
Assert that $[e]\in H^2(M)$ has nothing to do with $\{\xi_\alpha\}$ selection,
Called the planar bundle $E$ Euler class, also denoted as $e(E)$.
The selection is irrelevant because if there is another $\tilde \xi_\alpha$ that meets the requirements,
Then $\tilde \xi_\alpha-\xi_\alpha$ will satisfy the compatibility and can be defined as $\eta$ as a whole.
Then $d\eta=d\tilde\xi_\alpha-d\xi_\alpha$, thus they are equal at the cohomology level.
If $E$ is given a directional trivialization, assuming it is reduced to $SO(2)=SU(1)$,
Then $\varphi_{\alpha\beta}=-\frac{1}{i}\log g_{\alpha\beta}$, further
$e|_{U_\alpha}=d\xi_\alpha=-\frac{1}{2\pi i}\sum_\gamma d(\rho_\gamma \cdot d\log g_{\gamma\alpha})$,
called Explicit representation formulas for Euler classes of plane bundles.
When $E$ is a product bundle, the transfer function is a constant matrix, so the Euler class is zero.
It reflects the degree of distortion of the directed vector bundle $E$. From the expression formula, the functor property of the Euler class can be proved:
Proposition 1.1. $f:N\rightarrow M$ is a smooth mapping, $\pi:E\rightarrow M$ is a rank-2 directed vector bundle, then $e(f^{-1}E)=f^\ast e(E)\in H^2(N)$.
Display representation of planar bundle Thom class
$\frac{d\theta_\alpha}{2\pi}-\pi^\ast \xi_\alpha$ defines $1$-form $\psi$ on $E^0$.
$\psi$ is limited to $E_x^0$ on each fiber and is $\frac{d\theta_\alpha}{2\pi}$,
It is the angular form of $E_x^0$; at the same time, $d\psi=-\pi^\ast e$ is defined on $E$, and $\psi$ is the overall angular form.
Therefore, the Thom class of $E$ is $\Phi=d\rho(r)\wedge \psi-\rho(r)\pi^\ast e$.
If $s:M\rightarrow E$ is zero section, then $s^\ast \Phi=e$.
That is, zero cross section pulls the Thom class back to the Euler class.
Using explicit representations of Euler classes,
That is, $\Phi=d\rho(r)\wedge \frac{d\theta_\alpha}{2\pi}+\frac{1}{2\pi i}d(\rho(r)\pi^\ast \sum_\gamma\rho_\gamma d\log g_{\gamma\alpha})$ is obtained.
The form of the former term is very intuitive, while the latter term is a correction term caused by the distortion of the vector bundle.
The article was last updated on 2021-10-12 10:25:18
