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上次课我们对定向实向量丛$\xi^n,$
定义了Thom类$u(\xi)\in H^n(E,E_0;\mathbb{Z}).$
这给出了Euler类$e(\xi)\in H^n(B;\mathbb{Z}),$ 当$B=M^n$光滑时,
$$
e(\tau_M)=\chi(M)\cdot PD([\ast ]).
$$
一般的,
对$\xi$截面$s$与$M$横截相交的零截面$Z_s,$ 有
$$
e(\xi)=PD(Z_s),\quad Z_s=\{s=0\}
$$
横截性理论
设$X$为紧定向$n$维流形, 那么有Poincaré对偶:
$$
PD_X:H_i(X)\rightarrow H^{n-i}(X), \alpha\cap [M]\leftarrow \alpha.
$$
考虑横截积:
$$
H_i(X)\times H_j(X)\rightarrow H_{i+j-n}(X),\quad (a,b)\mapsto a\cdot b,
$$
$$
PD_X(a\cdot b)=PD_X(b)\cup PD_X(a).
$$
定理 1. 设$X$还是光滑的, $A,B$为横截相交闭光滑定向子流形. 那么$[A]_X\cdot [B]_X=[A\cap B]_X,$ 其中$[A]_X=i_\ast [A].$
给定$A\subset X,$ 定义$\tau_A^X:=PD([A]_X)\in H^{x-a}(X;\mathbb{Z}),$
称为$A$在$X$中的Thom类. 那么
$$
[A]_X=\tau_A^X\cap [X].
$$
记$\nu_A^X$为$A$在$X$中法丛, 我们有管状邻域定理:
存在$A$在$X$中的管状邻域$N$, 满足:
$$
E(\nu_A^X)\cong N.
$$
这样,
$$
u(\nu_A^X)\in H^{x-a}(E(\nu_A^X),E_0(\nu_A^X))\cong H^{x-a}(N,N\setminus A)\cong H^{x-a}(X,X\setminus A)\rightarrow H^{x-a}(X)\ni \tau_A^X.
$$
我们将说明$\nu_A^X$的Thom类打到$A$在$X$中的Thom类, 这也是命名后者的原因.
我们希望说明,
$$
\nu_{B\cap A}^A\cong \nu_B^X|_{B\cap A}.
$$
我们依纤维来检查: $\,\forall\,x\in A\cap B,$ $T_pA+T_pB=T_pX,$ 那么
$$
T(i_A^X):T_pA/T_p(B\cap A)\cong T_pX/T_pB.
$$
因此, $\tau^A_{B\cap A}=(i_A^X)^\ast \tau_B^X.$ 现在就可以证明定理了:
$$
\begin{aligned}
[B\cap A]_X&=(i_A^X)_*(i^A_{B\cap A})_*[B\cap A]\\
&=(i_A^X)_*(\tau_{B\cap A}^A\cap [A])\\
&=(i_A^X)_*((i_A^X)^*\tau_B^X\cap [A])\\
&=\tau_B^X\cap ((i_A^X)_*[A])\\
&=\tau_B^X\cap (\tau_A^X\cap [X])\\
&=(\tau_B^X\cup\tau_A^X)\cap [X]\\
&=PD^{-1}(PD([B]_X)\cup PD([A]_X))\\
&=[B]_X\cdot [A]_X.
\end{aligned}
$$
Euler类
回忆对定向丛$\xi,$ 我们给出Euler类:
$$
H^n(E,E_0)\rightarrow H^n(E)\leftarrow H^n(B), \quad u(\xi)\mapsto u(\xi)|_E\leftarrow e(\xi).
$$
定理 2. 若$B$为闭定向光滑流形, $e(\xi)=PD([Z_s]).$ $Z_s=\{b\in B|s(b)=0\},$ $s$为$\xi$截面, 与零截面横截相交.
记$Z:=Z_s.$ 考虑$\nu_Z^B\cong \xi|_Z.$ $E(\nu_Z^B)\cong N\subset B.$
$$
H^n(E,E_0)\rightarrow H^n(N,N\setminus Z)\rightarrow H^n(B,B\setminus Z)\rightarrow H^n(B).
$$
$$
u(\xi)\mapsto u(\nu_Z^B)\mapsto \tau_Z^B=PD([Z]).
$$
$$
H^n(B)\rightarrow H^n(E)\leftarrow H^n(E,E_0)\rightarrow H^n(N,N\setminus Z)\rightarrow H^n(B),
$$
$$
e(\xi)\mapsto u(\xi)\mapsto u(\nu_Z^B)\mapsto \tau_Z^B.
$$
这样就看到了
$e(\xi)=\tau_Z^B.$
定理 3. $B=M^n$为闭定向流形, $\left<{}e(\tau_M),[M]\right>=\chi(M).$ 这与Poincaré-Hopf定理等价.
任取一个横截相交截面, 那么:
$$
\begin{aligned}
\left<{}e(\tau_M),[M]\right>=\left<{}PD_M[Z_s],[M]\right>=\left<{}PD_M[M],[Z_s]\right>=\sum_{p\in Z_s} \text{index of $s$ at $p$}=\chi(M)
\end{aligned}
$$
指标由$\det(\nabla s:T_pM\rightarrow T_pM)$的符号决定.
一个应用是, $\chi(S^n)=2,$ 若$n$为偶数. 这样$e(\tau_{S^n})=2,$
这与$w_n(\tau_{S^n})=0$对应. 那么$\tau_{S^n}$非平凡,
且$\tau_{S^n}$不包含任何真子丛$\eta^m.$
这是因为$e(\tau_{S^n})=e(\eta)\cup e(\eta^\perp).$
横截性理论的应用
杯积
回忆$H^\ast (\mathbb{C}\mathrm{P}^n;\mathbb{Z})$中的杯积. 取
$$
\mathbb{C}\mathrm{P}^{n-1}\approx h_i=\{x_i=0\}\subset \mathbb{C}\mathrm{P}^n,
$$
$$
[h_i]\in H_{2n-2}(\mathbb{C}\mathrm{P}^n).
$$
由横截性理论,
$$
[h_1]\cdots[h_n]=[h_1\cap\cdots\cap h_n]=[\ast ]\in H_0(\mathbb{C}\mathrm{P}^n).
$$
令$\tau_i:=PD[h_i]\in H^2(\mathbb{C}\mathrm{P}^n),$
$\tau_1\cup\cdots\cup \tau_n=PD[\ast ].$
断言$\tau_1=\cdots=\tau_n=\tau,$ 称为超(射影)平面类.
这样$H^\ast (\mathbb{C}\mathrm{P}^n)=\mathbb{Z}[\tau]/\tau^{n+1}=0.$
Bezout’s 定理
假设$F_1(x,y,z),$ $F_2(x,y,z)$为齐次多项式, $\deg F_i=d_i.$
$$
V_i:=\{[x:y:z]\in \mathbb{C}\mathrm{P}^2|F_i=0\}\subset \mathbb{C}\mathrm{P}^2,
$$
在代数几何中称为$\deg=d_i$的代数曲线.
定理 4. 若$V_1,V_2$横截相交, 那么$[V_1\cap V_2]_{\mathbb{C}\mathrm{P}^2}=d_1d_2[\ast ]_{\mathbb{C}\mathrm{P}^2}.$
简单来看, 就是因为$PD[V_1]=d_1\cdot \tau,$ $PD[V_2]=d_2\cdot \tau.$
文章最后更新于 2022-10-28 16:41:58
review
In the last lesson we studied the directed real vector bundle $\xi^n,$
The Thom class $u(\xi)\in H^n(E,E_0;\mathbb{Z}).$ is defined
This gives the Euler class $e(\xi)\in H^n(B;\mathbb{Z}),$ when $B=M^n$ is smooth,
$$
e(\tau_M)=\chi(M)\cdot PD([\ast ]).
$$
general,
For the zero section $Z_s,$ where $\xi$ section $s$ crosses $M$, we have
$$
e(\xi)=PD(Z_s),\quad Z_s=\{s=0\}
$$
transversality theory
Assume $X$ is a compactly oriented $n$-dimensional manifold, then there is Poincaré duality:
$$
PD_X:H_i(X)\rightarrow H^{n-i}(X), \alpha\cap [M]\leftarrow \alpha.
$$
Consider the cross-sectional product:
$$
H_i(X)\times H_j(X)\rightarrow H_{i+j-n}(X),\quad (a,b)\mapsto a\cdot b,
$$
$$
PD_X(a\cdot b)=PD_X(b)\cup PD_X(a).
$$
Theorem 1. Assume $X$ is still smooth and $A,B$ is a transversally intersecting closed smooth directed submanifold. Then $[A]_X\cdot [B]_X=[A\cap B]_X,$ where $[A]_X=i_\ast [A].$
Given $A\subset X,$ define $\tau_A^X:=PD([A]_X)\in H^{x-a}(X;\mathbb{Z}),$
Call $A$ the Thom class in $X$. Then
$$
[A]_X=\tau_A^X\cap [X].
$$
Let $\nu_A^X$ be the normal bundle in $A$ in $X$, we have the tubular neighborhood theorem:
There is a tubular neighborhood $N$ where $A$ is in $X$, satisfying:
$$
E(\nu_A^X)\cong N.
$$
In this way,
$$
u(\nu_A^X)\in H^{x-a}(E(\nu_A^X),E_0(\nu_A^X))\cong H^{x-a}(N,N\setminus A)\cong H^{x-a}(X,X\setminus A)\rightarrow H^{x-a}(X)\ni \tau_A^X.
$$
We will describe the Thom class in $\nu_A^X$ as the Thom class in $A$ in $X$, which is why the latter is named.
We wish to explain,
$$
\nu_{B\cap A}^A\cong \nu_B^X|_{B\cap A}.
$$
Let’s check by fiber: $\,\forall\,x\in A\cap B,$ $T_pA+T_pB=T_pX,$ Then
$$
T(i_A^X):T_pA/T_p(B\cap A)\cong T_pX/T_pB.
$$
Therefore, $\tau^A_{B\cap A}=(i_A^X)^\ast \tau_B^X.$ can now prove the theorem:
$$
\begin{aligned}
[B\cap A]_X&=(i_A^X)_*(i^A_{B\cap A})_*[B\cap A]\\
&=(i_A^X)_*(\tau_{B\cap A}^A\cap [A])\\
&=(i_A^X)_*((i_A^X)^*\tau_B^X\cap [A])\\
&=\tau_B^X\cap ((i_A^X)_*[A])\\
&=\tau_B^X\cap (\tau_A^X\cap [X])\\
&=(\tau_B^X\cup\tau_A^X)\cap [X]\\
&=PD^{-1}(PD([B]_X)\cup PD([A]_X))\\
&=[B]_X\cdot [A]_X.
\end{aligned}
$$
Euler class
Recall that for directed bundles $\xi,$ we give the Euler class:
$$
H^n(E,E_0)\rightarrow H^n(E)\leftarrow H^n(B), \quad u(\xi)\mapsto u(\xi)|_E\leftarrow e(\xi).
$$
Theorem 2. If $B$ is a closed directional smooth manifold, $e(\xi)=PD([Z_s]).$ $Z_s=\{b\in B|s(b)=0\},$ $s$ is the $\xi$ section, intersecting with the zero section.
Remember $Z:=Z_s.$ Consider $\nu_Z^B\cong \xi|_Z.$ $E(\nu_Z^B)\cong N\subset B.$
$$
H^n(E,E_0)\rightarrow H^n(N,N\setminus Z)\rightarrow H^n(B,B\setminus Z)\rightarrow H^n(B).
$$
$$
u(\xi)\mapsto u(\nu_Z^B)\mapsto \tau_Z^B=PD([Z]).
$$
$$
H^n(B)\rightarrow H^n(E)\leftarrow H^n(E,E_0)\rightarrow H^n(N,N\setminus Z)\rightarrow H^n(B),
$$
$$
e(\xi)\mapsto u(\xi)\mapsto u(\nu_Z^B)\mapsto \tau_Z^B.
$$
That's how you see it
$e(\xi)=\tau_Z^B.$
Theorem 3. $B=M^n$ is a closed directional manifold, $\left<{}e(\tau_M),[M]\right>=\chi(M).$ is equivalent to the Poincaré-Hopf theorem.
Take any cross-sectional intersection section, then:
$$
\begin{aligned}
\left<{}e(\tau_M),[M]\right>=\left<{}PD_M[Z_s],[M]\right>=\left<{}PD_M[M],[Z_s]\right>=\sum_{p\in Z_s} \text{index of $s$ at $p$}=\chi(M)
\end{aligned}
$$
The indicator is determined by the sign of $\det(\nabla s:T_pM\rightarrow T_pM)$.
One application is, $\chi(S^n)=2,$ if $n$ is an even number. Thus $e(\tau_{S^n})=2,$
This corresponds to $w_n(\tau_{S^n})=0$. Then $\tau_{S^n}$ is non-trivial,
And $\tau_{S^n}$ does not contain any true subbundle $\eta^m.$
This is because $e(\tau_{S^n})=e(\eta)\cup e(\eta^\perp).$
Application of transversality theory
cup accumulation
Recall the cup product in $H^\ast (\mathbb{C}\mathrm{P}^n;\mathbb{Z})$. Take
$$
\mathbb{C}\mathrm{P}^{n-1}\approx h_i=\{x_i=0\}\subset \mathbb{C}\mathrm{P}^n,
$$
$$
[h_i]\in H_{2n-2}(\mathbb{C}\mathrm{P}^n).
$$
According to the transversality theory,
$$
[h_1]\cdots[h_n]=[h_1\cap\cdots\cap h_n]=[\ast ]\in H_0(\mathbb{C}\mathrm{P}^n).
$$
Let$\tau_i:=PD[h_i]\in H^2(\mathbb{C}\mathrm{P}^n),$
$\tau_1\cup\cdots\cup \tau_n=PD[\ast ].$
The assertion $\tau_1=\cdots=\tau_n=\tau,$ is called a super (projective) plane class.
Like this$H^\ast (\mathbb{C}\mathrm{P}^n)=\mathbb{Z}[\tau]/\tau^{n+1}=0.$
Bezout’s theorem
Assume $F_1(x,y,z),$ $F_2(x,y,z)$ is a homogeneous polynomial, $\deg F_i=d_i.$
$$
V_i:=\{[x:y:z]\in \mathbb{C}\mathrm{P}^2|F_i=0\}\subset \mathbb{C}\mathrm{P}^2,
$$
The algebraic curve called $\deg=d_i$ in algebraic geometry.
Theorem 4. If $V_1,V_2$ intersects transversely, then $[V_1\cap V_2]_{\mathbb{C}\mathrm{P}^2}=d_1d_2[\ast ]_{\mathbb{C}\mathrm{P}^2}.$
To put it simply, it is because $PD[V_1]=d_1\cdot \tau,$ $PD[V_2]=d_2\cdot \tau.$
The article was last updated on 2022-10-28 16:41:58
