Machine-translated from Chinese.
基本概念
对于秩$n$实向量丛
$$
\mathbb{R}^n\rightarrow E(\xi)\xrightarrow{\pi}B(\xi),
$$
定义Stiefel-Whitney classes为:
$$
\omega_i(\xi)\in H^i(B(\xi);\mathbb{Z}_2),\quad i=0,1,2,\cdots
$$
满足如下公理:
$\omega_0(\xi)=1\in H^0(B(\xi);\mathbb{Z}_2)$; $\omega_i(\xi)=0,$
$\,\forall\,i>n.$
(自然性)对于$f:B(\xi)\rightarrow B(\eta)$由丛映射覆盖($\xi \cong f^\ast \eta$),
$\omega_i(\xi)=(\omega_i(f^\ast \eta)=)f^\ast \omega_i(\eta).$
(Whitney乘积公式) 若$\xi,\eta$为底空间相同的实向量丛,
那么$\omega_k(\xi\oplus\eta)=\sum_i \omega_i(\xi)\cup \omega_{k-i}(\eta).$
对于$\mathbb{R}\mathrm{P}^1$上的经典线丛$\gamma_1^1,$
$\omega_1(\gamma_1^1)\neq 0.$
我们后面将看到这样的示性类存在唯一.
性质
一些不加证明的简单的性质如下.
若$\xi\cong \eta,$ 则$\omega_i(\xi)=\omega_i(\eta).$
若向量丛$\varepsilon^n$平凡, 则$\omega_i(\varepsilon)=0,$
$\,\forall\,i>0.$
若向量丛$\varepsilon^n$平凡,
则$\omega_i(\xi\oplus\varepsilon^n)=\omega_i(\xi).$
命题 1. 若$\xi$是欧氏秩$n$实向量丛, 有无处消失截面, 则$\omega_n(\xi)=0.$
若$\xi$有无处消失截面, 则$\xi$包含一个平凡秩$1$子向量丛$\varepsilon.$
由欧氏性, $\xi=\varepsilon\oplus \varepsilon^{\perp},$
从而$\omega_n(\xi)=\omega_n(\varepsilon^\perp)=0.$
最后一个等号是因为$\varepsilon^\perp$是秩$n-1$的.
因此我们可以将Stiefel-Whitney classes视为一种阻碍, 若最高维示性类非零,
则$\xi$上一定不存在无处消失截面. 类似地, 我们还有:
命题 2. 若$\xi$为欧氏秩$n$实向量丛, 有$k$个处处线性无关截面, 则$\omega_n(\xi)=\cdots=\omega_{n-k+1}(\xi)=0.$
总SW示性类
定义总SW示性类$\xi$为
$$
\omega(\xi):=1+\omega_1(\xi)+\omega_2(\xi)+\cdots \in H^\ast (B(\xi);\mathbb{Z}_2)\subset H^\pi(B(\xi);\mathbb{Z}_2).
$$
最后一个包含关系指直和包含于直积.
引理 3. $\{\omega:\omega\text{起始系数为$1$}\}\subset H^\pi(B;\mathbb{Z}_2)$ 组成了在杯积下不变的交换子群.
只需证每个$\omega$存在逆. 记$\omega=1+\omega_1+\cdots,$
$\overline\omega=1+\overline\omega_1+\cdots,$
为了让$\omega\overline\omega=1,$ 我们需要$\,\forall\,k>0,$
$$
\sum_{i}\omega_i\overline\omega_{k-i}=0.
$$
故取如下元素作为逆即可.
$$
\overline\omega_k=\omega_1\overline\omega_{k-1}+\cdots+\omega_k.
$$
命题 4. 若$\xi,\eta$都是$B$上的向量丛, 使得$\xi\oplus\eta$是平凡的, 那么$\omega(\xi)=\overline\omega(\eta).$
证明只需回忆:
$$
\omega(\xi\oplus\eta)=\omega(\xi)\cup\omega(\eta)=1.
$$
推论 5 (Whitney对偶). 若$M\subset \mathbb{R}^N$是一个光滑子流形, $\tau$为切丛, $\nu$为法丛, 则$\omega(\nu)=\overline\omega(\tau).$
只需注意到:
$$
\tau\oplus\nu\cong T\mathbb{R}^N|_M=\varepsilon.
$$
计算SW类
对于标准球面$S^n,$ 法丛平凡, 因此切丛的SW类平凡.
需要注意到偶数维球面上切丛非平凡,
因此SW类最高维为零也不见得有处处非零截面.
考虑$\gamma_n^1$为$\mathbb{R}\mathrm{P}^n$上的经典线丛, 我们有:
$$
\omega(\gamma_n^1)=1+a,\quad a\neq 0\in H^1(\mathbb{R}\mathrm{P}^n;\mathbb{Z}_2).
$$
考虑$\mathbb{R}^2\hookrightarrow \mathbb{R}^{n+1},$
诱导了$\mathbb{R}\mathrm{P}^1\hookrightarrow \mathbb{R}\mathrm{P}^n,$
由丛映射覆盖. 这就得到了结论.
令$\mathbb{R}^{n+1}$具标准度量,
$\gamma^\perp$为$\mathbb{R}\mathrm{P}^n$上的向量丛,
每根纤维是经典线丛纤维的正交补.
那么$\gamma^\perp\oplus \gamma_n^1=\varepsilon,$ 从而
$$
\omega(\gamma^\perp)=\overline\omega(\gamma_n^1)=(1+a)^{-1}=1+a+\cdots+a^n.
$$
特别地, $\omega_i(\gamma^\perp)=a^i\neq 0,$
$a^i\in H^i(\mathbb{R}\mathrm{P}^n).$
SW类的应用
可除代数
定理 6 (Stiefel). 设$\mathbb{R}^n$有双线性乘法$p:\mathbb{R}^n\times\mathbb{R}^n\rightarrow \mathbb{R}^n,$ 没有零因子, 则$n$是$2$的幂次.
我们分为如下两部分来证明:
1. 假设乘法存在, 那么切丛$\tau_{\mathbb{R}\mathrm{P}^{n-1} }$平凡.
2. 若$\tau_{\mathbb{R}\mathrm{P}^{n-1} }$平凡, 则$n$为$2$的幂次.
引理 7. $\mathbb{R}\mathrm{P}^n$上的切丛有同构$\tau_{\mathbb{R}\mathrm{P}^n}\cong \operatorname{Hom}(\gamma_n^1,\gamma^\perp).$
构造
$$
E(\tau_{\mathbb{R}\mathrm{P}^n})=\{\pm(x,v)\in S^n\times \mathbb{R}^{n+1}||x|=1,x\cdot v=0.\}
$$
那么每个对$(x,v)$决定了一个函数$f\in \operatorname{Hom}(L,L^\perp),$
$L:=[x]\cong \mathbb{R}^1,$ $f(\lambda x)=\lambda v.$
验证发现$(x,v)\mapsto f$就给出了引理中的同构.
命题 8. $\tau_{\mathbb{R}\mathrm{P}^n}\oplus \varepsilon^1\cong \gamma_n^1\oplus \cdots\oplus \gamma_n^1,$ $n+1$个.
$\tau\oplus \varepsilon^1\cong \operatorname{Hom}(\gamma_n^1,\gamma^\perp)\oplus \operatorname{Hom}(\gamma_n^1,\gamma_n^1)\cong \operatorname{Hom}(\gamma_n^1,\gamma^\perp\oplus \gamma_n^1)=\operatorname{Hom}(\gamma_n^1,\varepsilon)^{\oplus (n+1)}\cong (\gamma_n^1)^{\oplus(n+1)}$
文章最后更新于 2022-10-11 15:12:27
Basic concepts
For a real vector bundle of rank $n$
$$
\mathbb{R}^n\rightarrow E(\xi)\xrightarrow{\pi}B(\xi),
$$
Define Stiefel-Whitney classes as:
$$
\omega_i(\xi)\in H^i(B(\xi);\mathbb{Z}_2),\quad i=0,1,2,\cdots
$$
Satisfy the following axioms:
$\omega_0(\xi)=1\in H^0(B(\xi);\mathbb{Z}_2)$; $\omega_i(\xi)=0,$
$\,\forall\,i>n.$
(Natural) For $f:B(\xi)\rightarrow B(\eta)$ covered by the bundle map ($\xi \cong f^\ast \eta$),
$\omega_i(\xi)=(\omega_i(f^\ast \eta)=)f^\ast \omega_i(\eta).$
(Whitney product formula) If $\xi,\eta$ is a real vector bundle with the same base space,
Then $\omega_k(\xi\oplus\eta)=\sum_i \omega_i(\xi)\cup \omega_{k-i}(\eta).$
For the classical line bundle $\gamma_1^1,$ on $\mathbb{R}\mathrm{P}^1$
$\omega_1(\gamma_1^1)\neq 0.$
We will see later that such characteristic classes exist uniquely.
nature
Some simple properties without proof are as follows.
If $\xi\cong \eta,$ then $\omega_i(\xi)=\omega_i(\eta).$
If the vector bundle $\varepsilon^n$ is trivial, then $\omega_i(\varepsilon)=0,$
$\,\forall\,i>0.$
If the vector bundle $\varepsilon^n$ is trivial,
Then $\omega_i(\xi\oplus\varepsilon^n)=\omega_i(\xi).$
Proposition 1. If $\xi$ is a real vector bundle of Euclidean rank $n$ and has a nowhere-vanishing section, then $\omega_n(\xi)=0.$
If $\xi$ has a nowhere-vanishing section, then $\xi$ contains a trivial rank $1$ subvector bundle $\varepsilon.$
By Euclidean properties, $\xi=\varepsilon\oplus \varepsilon^{\perp},$
Thus $\omega_n(\xi)=\omega_n(\varepsilon^\perp)=0.$
The last equal sign is because $\varepsilon^\perp$ is of rank $n-1$.
Therefore we can regard Stiefel-Whitney classes as an obstacle. If the highest dimensional representation class is non-zero,
Then there must be no nowhere-vanishing section on $\xi$. Similarly, we also have:
Proposition 2. If $\xi$ is a real vector bundle of Euclidean rank $n$ and there are $k$ linearly independent sections everywhere, then $\omega_n(\xi)=\cdots=\omega_{n-k+1}(\xi)=0.$
Total SW Indicative Category
Define the total SW characteristic class $\xi$ as
$$
\omega(\xi):=1+\omega_1(\xi)+\omega_2(\xi)+\cdots \in H^\ast (B(\xi);\mathbb{Z}_2)\subset H^\pi(B(\xi);\mathbb{Z}_2).
$$
The last inclusion relation means that direct sums are included in direct products.
Lemma 3. $\{\omega:\omega\text{起始系数为$1$}\}\subset H^\pi(B;\mathbb{Z}_2)$ forms a commutative subgroup that is invariant under the cup product.
We only need to prove that every $\omega$ has an inverse. Let $\omega=1+\omega_1+\cdots,$
$\overline\omega=1+\overline\omega_1+\cdots,$
To get $\omega\overline\omega=1,$ we need $\,\forall\,k>0,$
$$
\sum_{i}\omega_i\overline\omega_{k-i}=0.
$$
Therefore, just take the following elements as the inverse.
$$
\overline\omega_k=\omega_1\overline\omega_{k-1}+\cdots+\omega_k.
$$
Proposition 4. If $\xi,\eta$ are all vector bundles on $B$, making $\xi\oplus\eta$ trivial, then $\omega(\xi)=\overline\omega(\eta).$
To prove, just recall:
$$
\omega(\xi\oplus\eta)=\omega(\xi)\cup\omega(\eta)=1.
$$
Corollary 5 (Whitney dual). If $M\subset \mathbb{R}^N$ is a smooth submanifold, $\tau$ is the tangent bundle, and $\nu$ is the normal bundle, then $\omega(\nu)=\overline\omega(\tau).$
Just note:
$$
\tau\oplus\nu\cong T\mathbb{R}^N|_M=\varepsilon.
$$
Calculate SW class
For the standard sphere $S^n,$, the normal bundle is trivial, so the SW class of the tangent bundle is trivial.
It should be noted that the tangent bundle on an even-dimensional sphere is non-trivial,
Therefore, even if the highest dimension of the SW class is zero, it does not necessarily mean that there are non-zero sections everywhere.
Considering $\gamma_n^1$ as the classical line bundle on $\mathbb{R}\mathrm{P}^n$, we have:
$$
\omega(\gamma_n^1)=1+a,\quad a\neq 0\in H^1(\mathbb{R}\mathrm{P}^n;\mathbb{Z}_2).
$$
Consider $\mathbb{R}^2\hookrightarrow \mathbb{R}^{n+1},$
Induced $\mathbb{R}\mathrm{P}^1\hookrightarrow \mathbb{R}\mathrm{P}^n,$
Covered by a bundle map. This leads to the conclusion.
Let $\mathbb{R}^{n+1}$ have a standard metric,
$\gamma^\perp$ is the vector bundle on $\mathbb{R}\mathrm{P}^n$,
Each fiber is the orthogonal complement of a classical line bundle fiber.
Then $\gamma^\perp\oplus \gamma_n^1=\varepsilon,$ thus
$$
\omega(\gamma^\perp)=\overline\omega(\gamma_n^1)=(1+a)^{-1}=1+a+\cdots+a^n.
$$
In particular, $\omega_i(\gamma^\perp)=a^i\neq 0,$
$a^i\in H^i(\mathbb{R}\mathrm{P}^n).$
SW type applications
divisible algebra
Theorem 6 (Stiefel). Assume $\mathbb{R}^n$ has bilinear multiplication $p:\mathbb{R}^n\times\mathbb{R}^n\rightarrow \mathbb{R}^n,$ without zero factors, then $n$ is the power of $2$.
We divide it into two parts to prove:
1. Assuming multiplication exists, then the tangent bundle $\tau_{\mathbb{R}\mathrm{P}^{n-1} }$ is trivial.
2. If $\tau_{\mathbb{R}\mathrm{P}^{n-1} }$ is trivial, then $n$ is the power of $2$.
Lemma 7. The tangent bundle on $\mathbb{R}\mathrm{P}^n$ has isomorphism $\tau_{\mathbb{R}\mathrm{P}^n}\cong \operatorname{Hom}(\gamma_n^1,\gamma^\perp).$
structure
$$
E(\tau_{\mathbb{R}\mathrm{P}^n})=\{\pm(x,v)\in S^n\times \mathbb{R}^{n+1}||x|=1,x\cdot v=0.\}
$$
Then each pair $(x,v)$ determines a function $f\in \operatorname{Hom}(L,L^\perp),$
$L:=[x]\cong \mathbb{R}^1,$ $f(\lambda x)=\lambda v.$
It is verified that $(x,v)\mapsto f$ gives the isomorphism in the lemma.
Proposition 8. $\tau_{\mathbb{R}\mathrm{P}^n}\oplus \varepsilon^1\cong \gamma_n^1\oplus \cdots\oplus \gamma_n^1,$ $n+1$ pieces.
$\tau\oplus \varepsilon^1\cong \operatorname{Hom}(\gamma_n^1,\gamma^\perp)\oplus \operatorname{Hom}(\gamma_n^1,\gamma_n^1)\cong \operatorname{Hom}(\gamma_n^1,\gamma^\perp\oplus \gamma_n^1)=\operatorname{Hom}(\gamma_n^1,\varepsilon)^{\oplus (n+1)}\cong (\gamma_n^1)^{\oplus(n+1)}$
The article was last updated on 2022-10-11 15:12:27
