Machine-translated from Chinese.
定义
对$\pi:E\rightarrow M$秩$n+1$向量丛, 给定度量$\left<{}-,-\right>$,
可定义$r^2(v)=\left<{}v,v\right>$.
置$S(E)_x=\{v\in E_x|r(v)=1\}\simeq S^n$, $x\in M$;
$S(E)=\prod_{x\in M}S(E)_x$, 则$S(E)$是$M$上纤维为$S^n$的纤维丛.
它的结构群是$O(n+1)$.
把纤维为球面$S^n$的纤维丛称为$n$维球面丛或$S^n$丛.
结构群是微分同胚群$\operatorname{Diff}(S^n)$.
它一般并不一定能约化到$O(n+1)$.
事实上能约化到$O(n+1)$当且仅当该球面丛由向量丛构造(上一段).
设$\pi:E\rightarrow M$为$n$维向量丛, 根据Leray-Hirsch定理,
若在$E$上存在闭的整体$n$-形式, 它到每条纤维的限制生成纤维的上同调,
则$E$的上同调是$H^\ast (E)=H^\ast (M)\otimes H^\ast (S^n)$.
若结构群是$SO(2)$, 则其上有整体角形式$\psi$与欧拉类$d\psi=-\pi^\ast e$.
若欧拉类为零, 则整体角形式闭, 满足上述条件.
一般的, 考虑结构群为$\operatorname{Diff}(S^n)$或$O(n+1)$的球面丛.
我们研究可定向性与欧拉类.
可定向性
假定$M$是连通的, 有好覆盖$\{U_\alpha\}$.
取$\pi:E\rightarrow M$为$n$维球面丛,
$E|_{U_\alpha}\approx U_\alpha\times S^n$,
$H^n(E|_{U_\alpha})=\mathbb{R}$. 设$[\sigma_\alpha]$为生成元. 先验地,
$[\sigma_\alpha]_{E|_{U_{\alpha\beta} } }=\pm[\sigma_\beta]_{E|_{U_{\alpha\beta} } }$.
若存在定向平凡化,
使得$[\sigma_\alpha]_{E|_{U_{\alpha\beta} } }=[\sigma_\beta]_{E|_{U_{\alpha\beta} } }$,
则称$E$是可定向的. 它等价于对每条纤维$E_x$,
选取$H^n(E_x)$生成元$[\sigma_x]$, 满足局部相容性条件:
$M$上每点有邻域$U$, 与$H^n(E|_U)$生成元$[\sigma_U]$,
使得$\,\forall\,x\in U$, $[\sigma_U]|_{E_x}=[\sigma_x]$.
相容性条件等价于$\sigma_\beta-\sigma_\alpha=d\sigma_{\alpha\beta}$.
这是讨论欧拉类的出发点.
对于定向球面丛, 有两组相容生成元, 每组生成元称为球面丛的一个定向.
给定定向的球面丛称为定向球面丛. 流形$M$上$S^0$丛是$M$的二重覆盖.
如果$M$是连通的, 该丛可定向当且仅当它有两个连通分支.
固定一个定向, 取$[\sigma]\in H^n(S^n)$为生成元. 若$g\in SO(n+1)$,
则$g(S^n)=S^n$且定向相同.
$g^\ast [\sigma]=\sigma$当且仅当正交变换$g$的行列式为正.
现在研究向量丛$E$可定向性与$S(E)$可定向性.
命题 1. 向量丛$E$可定向当且仅当球面丛$S(E)$可定向.
证: 给定$E$上度量,
给定保内积的局部平凡化$\{(U_\alpha,\phi_\alpha)\}$.
$g_{\alpha\beta}$约化到$O(n+1)$选取,
诱导了$S(E)$的局部平凡化$\{(U_\alpha,\varphi_\alpha)\}$.
$S(E)$的结构群也是$O(n+1)$.
取投射$\rho_\alpha:U_\alpha\times S^n\rightarrow S^n$,
利用$\rho_\alpha^\ast ,\varphi_\alpha^\ast $拉回$H^n(S^n)$生成元$[\sigma]$,
给出$H^n(S(E)|_{U_\alpha})$生成元$[\sigma_\alpha]_x$.
可以发现,
$[\sigma_\alpha]_x=[\sigma_\beta]_x\Leftrightarrow [\sigma]=g_{\alpha\beta}(x)^\ast [\sigma]$.
从而命题得证.
注 2. $SO(1)=\{1\}$, 因此连通流形$M$上的线丛$\pi:L\rightarrow M$可定向当且仅当它是平凡的. 此时球面丛$S(L)$有两个连通分支. ($S^0=\{\pm 1\}$)
若$\{g_{\alpha\beta}\}$为$E$的转移函数,
则$\{\det g_{\alpha\beta}\}$为行列式丛的转移函数. 因此有:
命题 3. 向量丛$E$可定向当且仅当行列式丛$\det E$可定向.
由于$0$-球面丛$S(\det E)$是$M$的二重覆盖,
$E$可定向当且仅当$S(\det E)$不连通.
由于单连通流形不可能有重数大于$1$的连通覆盖空间, 从而有:
命题 4. 单连通流形上每个向量丛可定向.
以及:
推论 5. 单连通流形可定向.
欧拉类
$\sigma_\beta-\sigma_\alpha=d\sigma_{\alpha\beta}$,
定义了$0$-上链$\sigma^{0,n}=(\sigma_\alpha,\sigma_\beta,...)$,
与$1$-上链$\sigma^{1,n-1}=(\sigma_{\alpha\beta},\sigma_{\alpha\gamma},...)$.
利用$\{U_\alpha\}$为好覆盖, 以及Künneth公式, 可继续沿对角线往右下方走,
得到欧拉类$[\varepsilon]\in H^{n+1}(\mathcal{U},\mathbb{R})$,
这对应了$[e]\in H^{n+1}(M)$.
若$E$是定向向量丛, 零截面补$E^0$与定向球面丛有相同伦型.
$E$欧拉类定义为$E^0$欧拉类. $E$欧拉类也可以通过度量成为$S(E)$的欧拉类.
命题 6. 给定定向, 欧拉类与$\sigma^{j,n-j}$选取无关.
凭借下一行的元素即可, 找$\tau\in K^{n-1}$, $D\tau=\bar\sigma-\sigma$,
这将推得$\delta \tau=\bar\varepsilon-\varepsilon$,
这里$\tau$跑到了$C^n(\pi^{-1}\mathcal{U},\mathbb{R})$.
$[\bar\varepsilon]=[\varepsilon]\in H^{n+1}(M)$.
命题 7. 欧拉类$e(E)$与好覆盖选取无关.
取$\mathcal{W}$为$\mathcal{U},\mathcal{V}$的公共加细好覆盖,
则由Čech同构,
$[\varepsilon_\mathcal{U}]=[\varepsilon_\mathcal{W}]=[\varepsilon_\mathcal{V}]$.
若欧拉类$e(E)\in H^{n+1}(M)$为零, 则$\varepsilon=\delta \tau$为上边缘,
那么$\delta (\sigma^{n,0}+i\tau)=-i\varepsilon+i\varepsilon=0$,
从而可用$\sigma^{n,0}+i\tau$代替$\sigma^{n,0}$,
这样$\delta \sigma^{n,0}=0$, $D\sigma=0$. 由Collating公式,
这对应了$E$上的整体闭$n$-形式$\eta$,
使得$[\eta|_{E{|_{U_\alpha} } }]=[\sigma_\alpha]$,
限制在每条纤维上为生成元.
总结一下, 我们从球面丛$E$可定向, 给出了欧拉类. 当欧拉类为零时,
推出了整体定义的$\eta$. 反过来, 若有一个整体定义的闭形式$\eta$,
限制在每个纤维上为生成元, 那么取$[\sigma]_x=[\eta|_x]$,
$E$当然是可定向的. 此时$\sigma_\alpha=\sigma_\beta=\sigma_x$,
$d\sigma_{\alpha\beta}=0$, 从而推出欧拉类为零.
综上,
球面丛$E$可定向且欧拉类为零当且仅当$E$上存在一个限制在每条纤维上是生成元的整体闭形式.
欧拉类用来衡量定向向量丛的扭曲程度:
命题 8. 若定向球面丛$E$有截面, 则它的欧拉类为零.
证: 若$E$有截面$s$, 注意到$-\pi^\ast \varepsilon=D\sigma$,
作用$s^\ast $有$-\varepsilon=Ds^\ast \sigma$, 因此欧拉类为零($D$-上边缘,
$[-\varepsilon]_\delta=[Ds^\ast \sigma]_D=0=[e]_d$).
注 9. 之前也有$-i\varepsilon=D\sigma$, 为什么欧拉类可能非零? 因为$\pi^{-1}\mathcal{U}$并不是一个好覆盖, 没有$H^\ast (\pi^{-1}\mathcal{U},\mathbb{R})\cong H^\ast (K)=H^\ast (\pi^{-1}\mathcal{U},\Omega^\ast )$. 但经过$s^\ast $拉回, $s^\ast \sigma$所在的双复形就是基于好覆盖的了. 所以这一部分的内容可以看成两层双复形, 其间由$\pi^\ast $和$\pi^\ast $逆映射(不一定是$s^\ast $)联系. 上层是$\pi^{-1}\mathcal{U}$的双复形, 是纤维层; 下层是$\mathcal{U}$的双复形, 是流形层. 从纤维层左上方由定向类出发, 往右下方跑到$[\pi^\ast \varepsilon]$. 因为利用好覆盖的性质, $\pi^\ast $在Čech上同调上是同构(可缩空间上的向量丛都是平凡丛, 再利用连通性说明). 这对应的是$[\varepsilon]$, 落在下层. 然后往左上方跑到$[e]$, 这就是欧拉类了.
整体角形式
设$\pi:E\rightarrow M$为定向$S^n$丛, 问是否存在$\psi\in \Omega^n(E)$,
使得$[\psi|_{E_x}]\in H^n(E_x)$是生成元, $d\psi=-\pi^\ast e$.
由Collating公式, $e=(-1)^{n+1}(D''K)^{n+1}\varepsilon$.
$D\sigma=-\pi^\ast \varepsilon$,
则$\psi=f(\sigma)=\sum_{i=0}^n(-1)^i(D''K)^i\sigma^{i,n-i}+(-1)^nK(D''K)^n(-\pi^\ast \varepsilon)$.
这里因为$\{\pi^\ast \rho\}$是$\pi^{-1}\mathcal{U}$上的单位分解,
可以把流形层上的同伦算子照搬到纤维层上.
可以看出, $d\psi=(-1)^n dK(D''K)^n (\pi^\ast \varepsilon)=-\pi^\ast e$.
同时由于同伦算子$K$中的单位分解为$\pi^\ast $拉回的, 限制在$E_x$上不动,
因此$\psi$所有单项中前面有$K$的项, 限制在纤维上都为零. 从而在纤维上,
$[\psi|_{E_x}]=[\sigma^{0,n}|_{E_x}]\in H^n(E_x)$.
因此$\psi$恰为整体角形式.
注 10. 设$\{U_\alpha\}$为$M$开覆盖, 局部平凡化$n$维球面丛$E$. 则上面的$e,\psi$满足$\operatorname{supp}d\psi\subset \pi^{-1}(U_{\alpha_0\cdots\alpha_n})$, $\operatorname{supp}e\subset \cup U_{\alpha_0\cdots \alpha_n}$.
这自然是因为$\varepsilon$只在$\cup U_{\alpha_0\cdots\alpha_{n+1} }$上局部常值,
每次经过$K$后确保支集落在一个$U_\alpha$内. 由 Collating 公式,
经过$n+1$次$K$后, 欧拉类$e$支集落在$\cup U_{\alpha_0\cdots\alpha_n}$上.
由$d\psi=-\pi^\ast e$得到另一结果.
文章最后更新于 2021-11-21 15:56:09
definition
For $\pi:E\rightarrow M$ rank $n+1$ vector bundle, given metric $\left<{}-,-\right>$,
Definable$r^2(v)=\left<{}v,v\right>$.
Set$S(E)_x=\{v\in E_x|r(v)=1\}\simeq S^n$, $x\in M$;
$S(E)=\prod_{x\in M}S(E)_x$, then $S(E)$ is the fiber bundle of $S^n$ on $M$.
Its structural group is $O(n+1)$.
The fiber bundle whose fibers are spherical $S^n$ is called $n$-dimensional spherical bundle or $S^n$ bundle.
The structural group is a diffeomorphism group $\operatorname{Diff}(S^n)$.
It generally does not necessarily reduce to $O(n+1)$.
In fact, it can be reduced to $O(n+1)$ if and only if the spherical bundle is constructed from a vector bundle (previous paragraph).
Let $\pi:E\rightarrow M$ be a $n$-dimensional vector bundle, according to the Leray-Hirsch theorem,
If there is a closed global $n$-form on $E$, its restriction to each fiber generates fiber cohomology,
Then the cohomology of $E$ is $H^\ast (E)=H^\ast (M)\otimes H^\ast (S^n)$.
If the structural group is $SO(2)$, then there are global angular forms $\psi$ and Euler classes $d\psi=-\pi^\ast e$ on it.
If the Euler class is zero, then the overall angular form is closed and satisfies the above conditions.
Generally, consider the spherical bundle whose structural group is $\operatorname{Diff}(S^n)$ or $O(n+1)$.
We study orientability and Euler classes.
Orientability
Assume $M$ is connected and has good coverage $\{U_\alpha\}$.
Let $\pi:E\rightarrow M$ be the $n$-dimensional spherical bundle,
$E|_{U_\alpha}\approx U_\alpha\times S^n$,
$H^n(E|_{U_\alpha})=\mathbb{R}$. Let $[\sigma_\alpha]$ be the generator. A priori,
$[\sigma_\alpha]_{E|_{U_{\alpha\beta} } }=\pm[\sigma_\beta]_{E|_{U_{\alpha\beta} } }$.
If there is directed trivialization,
So that $[\sigma_\alpha]_{E|_{U_{\alpha\beta} } }=[\sigma_\beta]_{E|_{U_{\alpha\beta} } }$,
Then $E$ is said to be orientable. It is equivalent to for each fiber $E_x$,
Select $H^n(E_x)$ generator $[\sigma_x]$ to satisfy the local compatibility condition:
Each point on $M$ has a neighborhood $U$, which generates element $[\sigma_U]$ with $H^n(E|_U)$,
Make $\,\forall\,x\in U$, $[\sigma_U]|_{E_x}=[\sigma_x]$.
The compatibility condition is equivalent to $\sigma_\beta-\sigma_\alpha=d\sigma_{\alpha\beta}$.
This is the starting point for discussing Euler classes.
For a directed spherical bundle, there are two sets of compatible generators, and each set of generators is called an orientation of the spherical bundle.
A spherical bundle with a given orientation is called an oriented spherical bundle. The $S^0$ bundle on the manifold $M$ is a double cover of $M$.
If $M$ is connected, the bundle is directional if and only if it has two connected components.
Fix an orientation and take $[\sigma]\in H^n(S^n)$ as the generator. If $g\in SO(n+1)$,
Then $g(S^n)=S^n$ has the same orientation.
$g^\ast [\sigma]=\sigma$ If and only if the determinant of the orthogonal transformation $g$ is positive.
Now we study the $E$ orientability and $S(E)$ orientability of the vector bundle.
Proposition 1. The vector bundle $E$ is orientable if and only if the spherical bundle $S(E)$ is orientable.
Certificate: Given the upper metric $E$,
Local trivialization of given inner-preserving products $\{(U_\alpha,\phi_\alpha)\}$.
$g_{\alpha\beta}$ reduces to $O(n+1)$ selection,
Induces the local trivialization $\{(U_\alpha,\varphi_\alpha)\}$ of $S(E)$.
The structural group of $S(E)$ is also $O(n+1)$.
Take projection $\rho_\alpha:U_\alpha\times S^n\rightarrow S^n$,
Use $\rho_\alpha^\ast ,\varphi_\alpha^\ast $ to pull back $H^n(S^n)$ to generate element $[\sigma]$,
Give $H^n(S(E)|_{U_\alpha})$ generator $[\sigma_\alpha]_x$.
can be found,
$[\sigma_\alpha]_x=[\sigma_\beta]_x\Leftrightarrow [\sigma]=g_{\alpha\beta}(x)^\ast [\sigma]$.
Thus the proposition is proved.
Note 2. $SO(1)=\{1\}$, so the line bundle $\pi:L\rightarrow M$ on the connected manifold $M$ can be oriented if and only if it is trivial. At this time, the spherical bundle $S(L)$ has two connected components. ($S^0=\{\pm 1\}$)
If $\{g_{\alpha\beta}\}$ is the transfer function of $E$,
Then $\{\det g_{\alpha\beta}\}$ is the transfer function of the determinant bundle. Therefore, there is:
Proposition 3. The vector bundle $E$ is orientable if and only if the determinant bundle $\det E$ is orientable.
Since $0$-spherical bundle $S(\det E)$ is a double cover of $M$,
$E$ can be directed if and only if $S(\det E)$ is not connected.
Since a simply connected manifold cannot have a connected covering space with a multiplicity greater than $1$, we have:
Proposition 4. Every vector bundle on a simply connected manifold can be oriented.
and:
Corollary 5. Simply connected manifolds can be oriented.
Euler class
$\sigma_\beta-\sigma_\alpha=d\sigma_{\alpha\beta}$,
Defined $0$-on-chain $\sigma^{0,n}=(\sigma_\alpha,\sigma_\beta,...)$,
With $1$ - wind up $\sigma^{1,n-1}=(\sigma_{\alpha\beta},\sigma_{\alpha\gamma},...)$.
Using $\{U_\alpha\}$ for good coverage and the Künneth formula, you can continue to go diagonally to the lower right,
Obtain Euler class $[\varepsilon]\in H^{n+1}(\mathcal{U},\mathbb{R})$,
This corresponds to $[e]\in H^{n+1}(M)$.
If $E$ is a directed vector bundle, the zero-section complement $E^0$ has the same type as the directed spherical bundle.
$E$ Euler class is defined as $E^0$ Euler class. $E$ Euler class can also become the Euler class of $S(E)$ through metric.
Proposition 6. Given an orientation, the Euler class is independent of the choice of $\sigma^{j,n-j}$.
Just rely on the elements of the next row to find $\tau\in K^{n-1}$, $D\tau=\bar\sigma-\sigma$,
This will lead to $\delta \tau=\bar\varepsilon-\varepsilon$,
Here $\tau$ ran to $C^n(\pi^{-1}\mathcal{U},\mathbb{R})$.
$[\bar\varepsilon]=[\varepsilon]\in H^{n+1}(M)$.
Proposition 7. The Euler class $e(E)$ has nothing to do with good coverage selection.
Take $\mathcal{W}$ as the public plus fine coverage of $\mathcal{U},\mathcal{V}$,
Then Čech is isomorphic,
$[\varepsilon_\mathcal{U}]=[\varepsilon_\mathcal{W}]=[\varepsilon_\mathcal{V}]$.
If the Euler class $e(E)\in H^{n+1}(M)$ is zero, then $\varepsilon=\delta \tau$ is the upper edge,
Then $\delta (\sigma^{n,0}+i\tau)=-i\varepsilon+i\varepsilon=0$,
Therefore, $\sigma^{n,0}+i\tau$ can be used instead of $\sigma^{n,0}$,
In this way $\delta \sigma^{n,0}=0$, $D\sigma=0$. According to the Collating formula,
This corresponds to the global closed $n$-form $\eta$ on $E$,
So that $[\eta|_{E{|_{U_\alpha} } }]=[\sigma_\alpha]$,
Restricted to generators on each fiber.
To summarize, we can orientate from the spherical bundle $E$, giving the Euler class. When the Euler class is zero,
The overall definition $\eta$ is derived. On the contrary, if there is a closed form $\eta$ of the overall definition,
Restrict each fiber to be a generator, then take $[\sigma]_x=[\eta|_x]$,
$E$ is of course orientable. At this time $\sigma_\alpha=\sigma_\beta=\sigma_x$,
$d\sigma_{\alpha\beta}=0$, it follows that the Euler class is zero.
To sum up,
The spherical bundle $E$ is orientable and the Euler class is zero if and only if there is a limit on $E$ that is the global closed form of the generator on each fiber.
The Euler class is used to measure the degree of distortion of the directed vector bundle:
Proposition 8. If the directed spherical bundle $E$ has a cross section, its Euler class is zero.
Certificate: If $E$ has section $s$, note $-\pi^\ast \varepsilon=D\sigma$,
The effect $s^\ast $ has $-\varepsilon=Ds^\ast \sigma$, so the Euler class is zero ($D$-upper edge,
$[-\varepsilon]_\delta=[Ds^\ast \sigma]_D=0=[e]_d$).
Note 9. There was $-i\varepsilon=D\sigma$ before, so why might the Euler class be non-zero? Because $\pi^{-1}\mathcal{U}$ is not a good cover, there is no $H^\ast (\pi^{-1}\mathcal{U},\mathbb{R})\cong H^\ast (K)=H^\ast (\pi^{-1}\mathcal{U},\Omega^\ast )$. But after $s^\ast $ is pulled back, the bicomplex where $s^\ast \sigma$ is located is based on a good cover. Therefore, the content of this part can be regarded as two layers of double complexes, which are connected by the inverse mapping of $\pi^\ast $ and $\pi^\ast $ (not necessarily $s^\ast $). The upper layer is the double complex of $\pi^{-1}\mathcal{U}$, which is the fiber layer; the lower layer is the double complex of $\mathcal{U}$, which is the manifold layer. Starting from the orientation class at the upper left of the fiber layer, run to $[\pi^\ast \varepsilon]$ at the lower right. Because of the good use of the properties of coverage, $\pi^\ast $ is isomorphic on the Čech cohomology (the vector bundles on the shrinkable space are all trivial bundles, and then the connectivity is used to illustrate). This corresponds to $[\varepsilon]$, which falls on the lower level. Then run to the upper left to $[e]$, which is the Euler class.
Global Angular Form
Let $\pi:E\rightarrow M$ be the directed $S^n$ bundle, and ask whether $\psi\in \Omega^n(E)$ exists,
Let $[\psi|_{E_x}]\in H^n(E_x)$ be the generator, $d\psi=-\pi^\ast e$.
By Collating formula, $e=(-1)^{n+1}(D''K)^{n+1}\varepsilon$.
$D\sigma=-\pi^\ast \varepsilon$,
Then $\psi=f(\sigma)=\sum_{i=0}^n(-1)^i(D''K)^i\sigma^{i,n-i}+(-1)^nK(D''K)^n(-\pi^\ast \varepsilon)$.
Here, because $\{\pi^\ast \rho\}$ is the partition of unity on $\pi^{-1}\mathcal{U}$,
The homotopy operator on the manifold layer can be copied to the fiber layer.
It can be seen that $d\psi=(-1)^n dK(D''K)^n (\pi^\ast \varepsilon)=-\pi^\ast e$.
At the same time, since the units in the homotopy operator $K$ are decomposed into $\pi^\ast $ and pulled back, they are restricted to $E_x$.
Therefore, all single terms of $\psi$ preceded by $K$ are restricted to zero on the fiber. Thus, on the fiber,
$[\psi|_{E_x}]=[\sigma^{0,n}|_{E_x}]\in H^n(E_x)$.
Therefore $\psi$ is exactly the overall angular form.
Note 10. Let $\{U_\alpha\}$ be the open cover of $M$, and locally trivialize the $n$-dimensional spherical bundle $E$. Then the above $e,\psi$ satisfies $\operatorname{supp}d\psi\subset \pi^{-1}(U_{\alpha_0\cdots\alpha_n})$, $\operatorname{supp}e\subset \cup U_{\alpha_0\cdots \alpha_n}$.
This is naturally because $\varepsilon$ is only locally constant on $\cup U_{\alpha_0\cdots\alpha_{n+1} }$,
After each pass through $K$, ensure that the support set falls within one $U_\alpha$. According to the Collating formula,
After $n+1$ times $K$, the Euler class $e$ support falls on $\cup U_{\alpha_0\cdots\alpha_n}$.
Another result is obtained from $d\psi=-\pi^\ast e$.
The article was last updated on 2021-11-21 15:56:09
