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对于向量丛$\xi:$ $\mathbb{R}^n\rightarrow E\rightarrow B,$
$\omega_i(\xi)\in H^i(B;\mathbb{Z}_2),$
$\omega(\xi):=\sum_i \omega_i(\xi).$
我们有例子$\omega(\tau_{S^n})=1,$ $\omega(\gamma_n^1)=1+a,$
$\omega(\gamma^\perp)=\overline\omega(\gamma_n^1)=(1+a)^{-1}=1+a+\cdots+a^n.$
对于$\omega(\tau_{\mathbb{R}\mathrm{P}^n}),$ 我们有引理:
引理 1. $\tau_{\mathbb{R}\mathrm{P}^n}=\operatorname{Hom}(\gamma_n^1,\gamma^\perp)$
命题 2. $\tau_{\mathbb{R}\mathrm{P}^n}\oplus\varepsilon^1\cong \gamma_n^1\oplus\cdots\oplus \gamma_n^1$ $(n+1)$个.
$$
\tau_{\mathbb{R}\mathrm{P}^n}\oplus\varepsilon^1=\operatorname{Hom}(\gamma_n^1,\gamma^\perp)\oplus \operatorname{Hom}(\gamma_n^1,\gamma_n^1)=\operatorname{Hom}(\gamma_n^1,\gamma^\perp\oplus \gamma_n^1=\varepsilon^{n+1})=\operatorname{Hom}(\gamma_n^1,\varepsilon^1)^{n+1}=(\gamma_n^1)^{n+1}
$$
这样我们得到,
$\omega(\tau_{\mathbb{R}\mathrm{P}^n})=\omega(\gamma_n^1)^{n+1}=(1+a)^{n+1}.$
但注意$a^{n+1}=0.$
推论 3 (Stiefel). $\omega(\tau_{\mathbb{R}\mathrm{P}^n})=1$ 当且仅当 $(n+1)=2^k.$
可除代数
定理 4 (Stiefel). 假设$\mathbb{R}^n$有一个双线性乘法$p:\mathbb{R}^n\times \mathbb{R}^n\rightarrow\mathbb{R}^n,$ 没有零因子, 那么$n$为$2$的幂次.
只需证明若$p$存在, 那么$\tau_{\mathbb{R}\mathrm{P}^{n-1} }$是平凡的.
$\,\forall\,z\neq 0,$
$y\mapsto p(y,z)$给出了$\mathbb{R}^n$到$\mathbb{R}^n$的线性同构.
取一组基$\{e_i\},$ 那么存在同构$v_i:\mathbb{R}^n\cong \mathbb{R}^n,$
使得$v_i(p(y,e_1))=p(y,e_i).$
对任意$L\subset \mathbb{R}^n,$ 定义$\overline v_i:L\rightarrow L^\perp,$
$\overline v_i(x):=pv_i(x),$ $p$为$\mathbb{R}^n$到$L^\perp$的投影.
这样$\overline v_i$给出了一个$\operatorname{Hom}(\gamma_{n-1}^1,\gamma^\perp)\cong \tau_{\mathbb{R}\mathrm{P}^{n-1} }$的一个截面.
验证发现$\{\overline v_i\}_{i\neq 1}$给出了一组标架(线性无关),
故$\tau_{\mathbb{R}\mathrm{P}^{n-1} }$是平凡的. 有前面的推论立即得到结论.
SW示性类的其它应用
SW示性类阻碍浸入
命题 5. 若$M^n$可以浸入到$\mathbb{R}^{n+k}$中, 则$\overline\omega_i(\tau_M)=0,$ $\,\forall\,i>k.$
设有一个浸入, 那么$\tau_M\oplus \nu_M=\varepsilon^{n+k},$
$\omega(\nu)=\overline\omega(\tau).$ 然而$\dim\nu=k,$
这就说明了$\overline\omega_i(\tau)=\omega_i(\nu)=0,$ $\,\forall\,i>k.$
命题 6. 若$n=2^r,$ $\mathbb{R}\mathrm{P}^n$可浸入$\mathbb{R}^{n+k},$ 则$k\ge n-1.$
$\omega(\tau_{\mathbb{R}\mathrm{P}^n})=(1+a)^{n+1}=(1+a)(1+a^n)=1+a+a^n.$
那么$\overline\omega(\tau_{\mathbb{R}\mathrm{P}^n})=1+a+\cdots+a^{n-1}.$
由命题, $\overline\omega_i(\tau_{\mathbb{R}\mathrm{P}^n})=0,$
$\,\forall\,i>k,$ 故$k\ge n-1.$
配边
考虑底空间$B=M^n$为闭流形, $\omega_i(M):=\omega_i(\tau_M).$ 考虑单项式
$$
\omega_1(M)^{r_1}\cdots\omega_n(M)^{r_n}\in H^D(M,\mathbb{Z}_2),\quad D=\sum_{i=1}^n ir_i.
$$
若$D=n,$ 考虑$[M]\in H_n(M),$ 我们就得到了一个数字
$$
\left<{}\omega_1(M)^{r_1}\cdots\omega_n(M)^{r_n},[M]\right>\in \mathbb{Z}_2,
$$
称为Stiefel-Whitney数, 与单项$\omega_1^{r_1}\cdots \omega_n^{r_n}$相关.
定理 7 (Pontrjagin). 若$M^n$是$(n+1)$维紧流形$V^{n+1}$的边界, 那么所有$M$的SW数都是零.
$M=\partial V,$
有$H_{n+1}(V,\partial V)\xrightarrow{\partial} H_n(\partial V),$
$\partial[V]=[M].$ 对偶地, 我们有
$$
\left<{}\delta \alpha,[V]\right>=\left<{}\alpha,\partial[V]\right>=\left<{}\alpha,[M]\right>.
$$
考虑$\tau_V|_M=i^\ast \tau_V,$ $i:M=\partial V\hookrightarrow V.$
注意到$\nu_M^V$是平凡的, 那么$i^\ast \tau_V=\tau_M\oplus\nu_M,$
$i^\ast \omega_k(\tau_V)=\omega_k(i^\ast \tau_V)=\omega_k(\tau_M)$
考虑长正合列
$$
H^n(V)\rightarrow H^n(M)\rightarrow H^{n+1}(V,M),
$$
$M$的SW数为:
$$
\left<{}\omega_1(M)^{r_1}\cdots\omega_n(M)^{r_n},[M]\right>=\left<{}\delta(\bullet),[V]\right>=\left<{}\delta i^\ast (\omega_1(V)^{r_1}\cdots\omega_n(V)^{r_n}),[V]\right>=0.
$$
定理 8 (Thom). 逆命题成立.
对于两个$n$维光滑闭流形$M_1,M_2,$ 称他们属于相同的配边类(cobordism
class), 若$M_1\sqcup M_2$是一个光滑紧$(n+1)$维流形的边界.
推论 9. $M_1,M_2$属于相同的配边类当且仅当它们的SW数都相同.
文章最后更新于 2022-10-12 15:23:48
review
For the vector bundle $\xi:$ $\mathbb{R}^n\rightarrow E\rightarrow B,$
$\omega_i(\xi)\in H^i(B;\mathbb{Z}_2),$
$\omega(\xi):=\sum_i \omega_i(\xi).$
We have examples $\omega(\tau_{S^n})=1,$ $\omega(\gamma_n^1)=1+a,$
$\omega(\gamma^\perp)=\overline\omega(\gamma_n^1)=(1+a)^{-1}=1+a+\cdots+a^n.$
For $\omega(\tau_{\mathbb{R}\mathrm{P}^n}),$ we have the lemma:
Lemma 1. $\tau_{\mathbb{R}\mathrm{P}^n}=\operatorname{Hom}(\gamma_n^1,\gamma^\perp)$
Proposition 2. $\tau_{\mathbb{R}\mathrm{P}^n}\oplus\varepsilon^1\cong \gamma_n^1\oplus\cdots\oplus \gamma_n^1$ $(n+1)$ pieces.
$$
\tau_{\mathbb{R}\mathrm{P}^n}\oplus\varepsilon^1=\operatorname{Hom}(\gamma_n^1,\gamma^\perp)\oplus \operatorname{Hom}(\gamma_n^1,\gamma_n^1)=\operatorname{Hom}(\gamma_n^1,\gamma^\perp\oplus \gamma_n^1=\varepsilon^{n+1})=\operatorname{Hom}(\gamma_n^1,\varepsilon^1)^{n+1}=(\gamma_n^1)^{n+1}
$$
This way we get,
$\omega(\tau_{\mathbb{R}\mathrm{P}^n})=\omega(\gamma_n^1)^{n+1}=(1+a)^{n+1}.$
But pay attention $a^{n+1}=0.$
Corollary 3 (Stiefel). $\omega(\tau_{\mathbb{R}\mathrm{P}^n})=1$ if and only if $(n+1)=2^k.$
divisible algebra
Theorem 4 (Stiefel). Suppose $\mathbb{R}^n$ has a bilinear multiplication $p:\mathbb{R}^n\times \mathbb{R}^n\rightarrow\mathbb{R}^n,$ with no zero factors, then $n$ is the power of $2$.
Just prove that if $p$ exists, then $\tau_{\mathbb{R}\mathrm{P}^{n-1} }$ is trivial.
$\,\forall\,z\neq 0,$
$y\mapsto p(y,z)$ gives the linear isomorphism of $\mathbb{R}^n$ to $\mathbb{R}^n$.
Take a set of bases $\{e_i\},$ then there is an isomorphism $v_i:\mathbb{R}^n\cong \mathbb{R}^n,$
Make$v_i(p(y,e_1))=p(y,e_i).$
Define $\overline v_i:L\rightarrow L^\perp,$ for any $L\subset \mathbb{R}^n,$
$\overline v_i(x):=pv_i(x),$ $p$ is the projection from $\mathbb{R}^n$ to $L^\perp$.
Thus $\overline v_i$ gives a cross-section of $\operatorname{Hom}(\gamma_{n-1}^1,\gamma^\perp)\cong \tau_{\mathbb{R}\mathrm{P}^{n-1} }$.
Verification found that $\{\overline v_i\}_{i\neq 1}$ gives a set of frames (linearly independent),
Therefore $\tau_{\mathbb{R}\mathrm{P}^{n-1} }$ is trivial. With the previous inference, the conclusion can be obtained immediately.
Other applications of the SW sexual category
SW Characteristic class hinders immersion
Proposition 5. If $M^n$ can be immersed in $\mathbb{R}^{n+k}$, then $\overline\omega_i(\tau_M)=0,$ $\,\forall\,i>k.$
Suppose there is an immersion, then $\tau_M\oplus \nu_M=\varepsilon^{n+k},$
$\omega(\nu)=\overline\omega(\tau).$ However $\dim\nu=k,$
This explains $\overline\omega_i(\tau)=\omega_i(\nu)=0,$ $\,\forall\,i>k.$
Proposition 6. If $n=2^r,$ $\mathbb{R}\mathrm{P}^n$ can be immersed in $\mathbb{R}^{n+k},$ then $k\ge n-1.$
$\omega(\tau_{\mathbb{R}\mathrm{P}^n})=(1+a)^{n+1}=(1+a)(1+a^n)=1+a+a^n.$
Then $\overline\omega(\tau_{\mathbb{R}\mathrm{P}^n})=1+a+\cdots+a^{n-1}.$
By proposition, $\overline\omega_i(\tau_{\mathbb{R}\mathrm{P}^n})=0,$
$\,\forall\,i>k,$ Therefore $k\ge n-1.$
edging
Consider the base space $B=M^n$ as a closed manifold, $\omega_i(M):=\omega_i(\tau_M).$ consider a monomial
$$
\omega_1(M)^{r_1}\cdots\omega_n(M)^{r_n}\in H^D(M,\mathbb{Z}_2),\quad D=\sum_{i=1}^n ir_i.
$$
If $D=n,$ considers $[M]\in H_n(M),$ we get a number
$$
\left<{}\omega_1(M)^{r_1}\cdots\omega_n(M)^{r_n},[M]\right>\in \mathbb{Z}_2,
$$
Called the Stiefel-Whitney number, it is related to the single term $\omega_1^{r_1}\cdots \omega_n^{r_n}$.
Theorem 7 (Pontrjagin). If $M^n$ is the boundary of $(n+1)$-dimensional compact manifold $V^{n+1}$, then the SW numbers of all $M$ are zero.
$M=\partial V,$
Yes $H_{n+1}(V,\partial V)\xrightarrow{\partial} H_n(\partial V),$
$\partial[V]=[M].$ Dually, we have
$$
\left<{}\delta \alpha,[V]\right>=\left<{}\alpha,\partial[V]\right>=\left<{}\alpha,[M]\right>.
$$
Consider$\tau_V|_M=i^\ast \tau_V,$ $i:M=\partial V\hookrightarrow V.$
Note that $\nu_M^V$ is trivial, then $i^\ast \tau_V=\tau_M\oplus\nu_M,$
$i^\ast \omega_k(\tau_V)=\omega_k(i^\ast \tau_V)=\omega_k(\tau_M)$
Consider long and straight columns
$$
H^n(V)\rightarrow H^n(M)\rightarrow H^{n+1}(V,M),
$$
The SW number of $M$ is:
$$
\left<{}\omega_1(M)^{r_1}\cdots\omega_n(M)^{r_n},[M]\right>=\left<{}\delta(\bullet),[V]\right>=\left<{}\delta i^\ast (\omega_1(V)^{r_1}\cdots\omega_n(V)^{r_n}),[V]\right>=0.
$$
Theorem 8 (Thom). The converse proposition holds.
For two $n$-dimensional smooth closed manifolds $M_1,M_2,$, they are said to belong to the same cobordism
class), if $M_1\sqcup M_2$ is the boundary of a smooth compact $(n+1)$-dimensional manifold.
Corollary 9. $M_1,M_2$ belongs to the same side class if and only if their SW numbers are the same.
The article was last updated on 2022-10-12 15:23:48
