《代数拓扑与微分形式》笔记(11-2)-Hopf指标定理 "Algebraic Topology and Differential Forms" Notes (11-2)-Hopf Index Theorem

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欧拉数与截面的孤立奇点

假定$M$是紧定向$n$维流形, $\pi:E\rightarrow M$为定向$S^{n-1}$丛, 结构群可约化到$O(n)$. 即它是某个秩$n$向量丛的球面丛.

$e(E)\in H^n(M)$, 与欧拉数$\int_M e(E)$等同. 定义$M$的欧拉数为$\int_M e(TM)$. 欧拉数依赖于$E,M$的定向选取.

命题 1. 条件如前面所述, 则存在$\{x_1,...,x_q\}\in M$, $E$限制在$M-\{x_1,...,x_q\}$上有截面.

证: 存在秩$n$向量丛$E'$与度量$\left<{},\right>$, 使得$E=S(E')$. 设$s':M\rightarrow E'$为截面. 若$s'(x)\neq 0$, 则$s(x)=\frac{s'(x)}{\Vert s'(x)\Vert}\in E_x$. 只需证明存在$s':M\rightarrow E'$, 只有有限多个零点.

我们需要如下定理:

定理 2 (横截性定理). 设$X$紧, $Z$为$Y$的闭子流形, $f':X\rightarrow Y$为光滑映射, 则存在$f'$的充分小扰动$f:X\rightarrow Y$, 使得$f$与$Z$横截.

设$s_0$为零截面, $S_0=s_0(M)\subset E'$为闭子流形. 由横截性定理, 存在充分小扰动$f:M\rightarrow E'$, $f$与$S_0$横截. $g=\pi\circ f:M\approx M$(因为是恒等映射的充分小扰动), 定义$s':M\rightarrow E'$为$s'(x)=f(g^{-1}(x))$, 那么$\pi\circ s'=\mathrm{id}$, $s'$为截面.

由于$f$横截于$S_0$, $s'$横截于$S_0$(只跟像有关), $S=s'(M)$与$S_0$横截相交. 由于$\dim S+\dim S_0=2n=\dim E'$, 因此$\dim (S\cap S_0)=0$. 又因为$S\cap S_0$紧, 交集为有限多个点. 那么$s':M\rightarrow E'$只有有限多个零点.

注 3. 事实上, 不需要能够约化到$O(n)$, 命题也成立.

设$s$为$E|_{M-\{x_1,...,x_q\} }$的一个截面, 取开覆盖$\mathcal{U}$, 使得$U_\alpha$上有坐标图,平凡化且最多包含一个$x_i$. 由下面的引理, 还可以要求每点$x_i$有邻域$W_i$, 只包含于一个$U_\alpha$.

引理 4. 存在$\mathcal{U}$的一个加细$\mathcal{V}$, 使得每点有一个邻域$W_i$, 只与某个$V_\alpha$相交.

证: 对于每点$x$, 只需给定某个开邻域$W$包含于$U_\alpha$, 改造开覆盖, 让其他的开集减去$\bar W$即可. 有限次对每个点这样做即可.

承上, 设$x_i\in W_i\subset U_{\alpha_i}$为奇点, 取充分小开邻域$D_i\subset W_i$, $\psi_i(D_i)\subset \mathbb{R}^n$为半径为$r_i$的开圆盘. $\partial D_i\approx S^{n-1}$. 有如下映射列:

$$\partial D_i\xrightarrow{s|_{\partial D_i} }E|_{\partial D_i}\xrightarrow{i} E|_{D_i}\xrightarrow[\sim]{\phi_i} D_i\times S^{n-1}\xrightarrow{\operatorname{pr}_2}S^{n-1}.$$

记$s_i$为前两个映射的复合, 后两个映射复合记为$\rho_i$.

选取定向使得$E|_{D_i}\simeq D_i\times S^{n-1}$保定向(直和定向). 选取$\sigma\in H^{n-1}(S^{n-1})$为正生成元. 定义$s$在$x_i$的局部度数记为$l.d.(x_i):=\int_{\partial D_i} s^\ast _i\rho^\ast _i\sigma$. (l.d. means local degree)

定理 5. 条件继承上面所有讨论. 则$E$的欧拉数$\int_M e(E)=\sum_{i=1}^{q} l.d.(x_i)$.

证: 由上节课内容, $\operatorname{supp}e\subset \cup U_{\alpha_0\cdots \alpha_{n-1} }$, $e$限制在每个开邻域$W_i$上取零. 因此由Stokes定理,

$$\int_M e=\int_{M-\cup D_i} s^\ast \pi^\ast e=-\int_{M-\cup D_i} s^\ast d\psi=\sum_{i=1}^q \int_{\partial D_i} s_i^\ast \psi.$$

虽然整体角形式$\psi$不一定闭, 但限制后有

$$d\psi|_{E|_{W_i} }=-(\pi^\ast e)|_{E|_{W_i} }=-\pi^\ast (e|_{W_i})=0.$$

因此$[\psi|_{E|_{W_i} }]\in H^{n-1}(E|_{W_i})$为生成元, 因为它限制在每条纤维上是生成元. 设$[\sigma]\in H^{n-1}(S^{n-1})$为生成元, 则$[\psi|_{E|_{W_i} }]=[\tilde \rho_i^\ast \sigma]$, $\tilde \rho_i:E|_{W_i}\rightarrow W_i\times S^{n-1}\rightarrow S^{n-1}$. 因此$\psi|_{E|_{W_i} }=\tilde \rho_i^\ast \sigma +d\tau_i$, 拉回到$\partial D_i$上, $s_i^\ast \psi=s_i^\ast \rho_i^\ast \sigma +s_i^\ast d\tau_i$.

两边积分, 由Stokes定理, $\int_{\partial D_i} s_i^\ast \psi=l.d.(x_i)$. 定理得证.

上述定理可以叙述在向量丛上. 设$\pi:E\rightarrow M$为紧定向$n$维流形上秩$n$定向向量丛, $s$为$E$的具有有限多零点的截面. $s$在零点$x$的重数定义为$E$的单位球面丛$S(E)$的奇异截面$s/\Vert s\Vert$在奇异点$x$的局部度数. 定义与度量选取无关, 因为局部度数是同伦不变量.

由于$e(E)$是$M$上$n$-形式, $\int_M c\cdot e(E)=cm=m\int_P c$, $m$为$E$的欧拉数, $P$是任意点. 因此$e(E)$ Poincaré对偶于$mP$. 我们自然有如下定理:

定理 6. 设$\pi:E\rightarrow M$为紧定向$M^n$上秩$n$定向向量丛, $s$为$E$的具有限零点截面, 那么$E$的欧拉类Poincaré对偶于$s$的零点的形式线性组合, 组合系数为重数.

欧拉示性数与Hopf指标定理

我们希望证明, $M$的欧拉数就是欧拉示性数.

设$\{\omega_i\}$为$H^\ast (M)$一组基, $\{\tau_i\}$为对偶基, 即$\int_M\omega_i\wedge \tau_j=\delta_{ij}$. 设$\pi,\rho:M\times M\rightarrow M$为两个投射. 由Künneth公式, $H^\ast (M\times M)=H^\ast (M)\otimes H^\ast (M)$, 基为$\{\pi^\ast \omega_i\wedge\rho^\ast \tau_j\}$. 称对角流形$\Delta =\{(x,x)\in M\times M|x\in M\}\subset M\times M$.

引理 7. 对角流形的Poincaré对偶$\eta_\Delta =\sum (-1)^{\deg \omega_i}\pi^\ast \omega_i\wedge \rho^\ast \tau_i$.

证: 用两种方法计算$\int_\Delta \pi^\ast \tau_k\wedge\rho^\ast \omega_l$即可. $(-1)^{\deg \tau_k\cdot \deg \omega_l}\delta_{lk}=(-1)^{\deg \omega_k\cdot (\deg \omega_l+\deg \tau_k)}c_{kl}$, 设$\eta_\Delta = \sum_{i,j}c_{ij}\pi^\ast \omega_i\wedge\rho^\ast \tau_j$.

引理 8. $N_{\Delta /M\times M}\simeq T\Delta$.

证: 只需注意到$M\approx \Delta$由$x\mapsto(x,x)$给出. 从而$T\Delta \cong TM$, $T(M\times M)|_{\Delta}\cong TM\oplus TM$, 于是$N_{\Delta/M\times M}\simeq TM\oplus TM/TM\simeq TM\simeq T\Delta$.

命题 9. 对紧定向流形$M$, $\int_M e(TM)=\chi (M)$.

证: 一方面, $\int_\Delta i^\ast \eta_\Delta =\int_\Delta i^\ast \Phi(N_\Delta)=\int_\Delta e(N_\Delta)=\int_\Delta e(T\Delta)=\int_M e(TM)$.

$\int_\Delta i^\ast \eta_\Delta =\int_{M\times M}\eta_\Delta \wedge \eta_\Delta$, 回忆两个横截相交闭子流形的交的Poincaré对偶等于这两个子流形的Poincaré对偶的外积, 因此该积分也称为$\Delta$在$M\times M$的自相交数.

另一方面, $\int_\Delta i^\ast \eta_\Delta=\sum_i(-1)^{\deg \omega_i}=\sum_q(-1)^q\dim H^q(M)=\chi(M)$. 由此得证.

自然得到:

定理 10 (Hopf指标定理). 紧定向流形$M$上光滑向量场指标和(奇异点的局部度数之和)等于$M$的欧拉示性数.

文章最后更新于 2021-11-24 13:33:55

Euler Numbers and Isolated Singularities of Sections

Assume that $M$ is a compact oriented $n$-dimensional manifold, $\pi:E\rightarrow M$ is an oriented $S^{n-1}$ bundle, The structure group can be reduced to $O(n)$. That is, it is a spherical bundle of a certain rank $n$ vector bundle.

$e(E)\in H^n(M)$, and Euler number $\int_M e(E)$Equal. Define the Euler number of $M$ as $\int_M e(TM)$. The Euler number depends on the directional selection of $E,M$.

Proposition 1. If the conditions are as mentioned before, then there is $\{x_1,...,x_q\}\in M$, and $E$ is restricted to have a section on $M-\{x_1,...,x_q\}$.

Certificate: There exists a vector bundle $E'$ of rank $n$ and a metric $\left<{},\right>$, such that $E=S(E')$. Let $s':M\rightarrow E'$ be the cross section. If $s'(x)\neq 0$, Then $s(x)=\frac{s'(x)}{\Vert s'(x)\Vert}\in E_x$. It is only necessary to prove that there is $s':M\rightarrow E'$, which has only a finite number of zeros.

We need the following theorem:

Theorem 2 (Transversality theorem). Assume $X$ is compact, $Z$ is the closed submanifold of $Y$, and $f':X\rightarrow Y$ is a smooth map, then there is a sufficiently small perturbation $f:X\rightarrow Y$ of $f'$ such that $f$ and $Z$ are transverse.

Let $s_0$ be zero cross-section and $S_0=s_0(M)\subset E'$ be a closed submanifold. According to the transversality theorem, There are sufficiently small perturbations $f:M\rightarrow E'$, $f$ and $S_0$ to intersect. $g=\pi\circ f:M\approx M$ (because it is a sufficiently small perturbation of the identity map), Define $s':M\rightarrow E'$ as $s'(x)=f(g^{-1}(x))$, Then $\pi\circ s'=\mathrm{id}$, $s'$ are cross sections.

Since $f$ is transverse to $S_0$ and $s'$ is transverse to $S_0$ (only related to images), $S=s'(M)$ intersects $S_0$ transversely. Since $\dim S+\dim S_0=2n=\dim E'$, Therefore $\dim (S\cap S_0)=0$. And because $S\cap S_0$ is compact, the intersection is a finite number of points. Then $s':M\rightarrow E'$ has only a finite number of zero points.

Note 3. In fact, it does not need to be reduced to $O(n)$ for the proposition to be true.

Let $s$ be a cross-section of $E|_{M-\{x_1,...,x_q\} }$, take the covering $\mathcal{U}$, Let there be a coordinate graph on $U_\alpha$, which is trivial and contains at most one $x_i$. According to the following lemma, You can also require that each point $x_i$ has a neighborhood $W_i$, which is only contained in one $U_\alpha$.

Lemma 4. There is a thinning $\mathcal{V}$ of $\mathcal{U}$, so that each point has a neighborhood $W_i$ that only intersects a certain $V_\alpha$.

Certificate: For each point $x$, only need to give a certain open neighborhood $W$ contained in $U_\alpha$, transform the open coverage, Just subtract $\bar W$ from other open sets. This can be done for each point a finite number of times.

Following the above, let $x_i\in W_i\subset U_{\alpha_i}$ be the singular point, Take a sufficiently small open neighborhood $D_i\subset W_i$, $\psi_i(D_i)\subset \mathbb{R}^n$ is an open disk with radius $r_i$. $\partial D_i\approx S^{n-1}$. There are the following mapping columns:

$$\partial D_i\xrightarrow{s|_{\partial D_i} }E|_{\partial D_i}\xrightarrow{i} E|_{D_i}\xrightarrow[\sim]{\phi_i} D_i\times S^{n-1}\xrightarrow{\operatorname{pr}_2}S^{n-1}.$$

$s_i$ is denoted as the composite of the first two mappings, and the composite of the last two mappings is denoted as $\rho_i$.

Select the orientation so that $E|_{D_i}\simeq D_i\times S^{n-1}$ remains oriented (straight and oriented). Select $\sigma\in H^{n-1}(S^{n-1})$ as the positive generator. Define the local degree of $s$ in $x_i$ as $l.d.(x_i):=\int_{\partial D_i} s^\ast _i\rho^\ast _i\sigma$. (l.d. means local degree)

Theorem 5. The condition inherits all discussions above. Then $E$ is the Euler number $\int_M e(E)=\sum_{i=1}^{q} l.d.(x_i)$.

Certificate: From the content of the previous lesson, $\operatorname{supp}e\subset \cup U_{\alpha_0\cdots \alpha_{n-1} }$, $e$ is restricted to take zero on every open neighborhood $W_i$. Therefore by Stokes’ theorem,

$$\int_M e=\int_{M-\cup D_i} s^\ast \pi^\ast e=-\int_{M-\cup D_i} s^\ast d\psi=\sum_{i=1}^q \int_{\partial D_i} s_i^\ast \psi.$$

Although the overall angular form $\psi$ is not necessarily closed, after restriction,

$$d\psi|_{E|_{W_i} }=-(\pi^\ast e)|_{E|_{W_i} }=-\pi^\ast (e|_{W_i})=0.$$

Therefore $[\psi|_{E|_{W_i} }]\in H^{n-1}(E|_{W_i})$ is the generator, Because it is restricted to each fiber as a generator. Let $[\sigma]\in H^{n-1}(S^{n-1})$ be the generator, Then $[\psi|_{E|_{W_i} }]=[\tilde \rho_i^\ast \sigma]$, $\tilde \rho_i:E|_{W_i}\rightarrow W_i\times S^{n-1}\rightarrow S^{n-1}$. Therefore $\psi|_{E|_{W_i} }=\tilde \rho_i^\ast \sigma +d\tau_i$, Pull back to $\partial D_i$, $s_i^\ast \psi=s_i^\ast \rho_i^\ast \sigma +s_i^\ast d\tau_i$.

Integral on both sides, by Stokes' theorem, $\int_{\partial D_i} s_i^\ast \psi=l.d.(x_i)$. The theorem is proved.

The above theorem can be stated on the vector bundle. Let $\pi:E\rightarrow M$ be a rank $n$ directed vector bundle on a compactly oriented $n$ dimensional manifold, $s$ is a section of $E$ with finitely many zero points. $s$ at zero point $x$ Multiplicity It is defined as the local degree of the singular section $s/\Vert s\Vert$ of the unit spherical bundle $S(E)$ of $E$ at the singular point $x$. The definition is independent of the choice of metric, since local degree is a homotopy invariant.

Since $e(E)$ is $n$-form on $M$, $\int_M c\cdot e(E)=cm=m\int_P c$, $m$ is the Euler number of $E$, and $P$ is an arbitrary point. Therefore $e(E)$ Poincaré is dual to $mP$. We naturally have the following theorem:

Theorem 6. Assume $\pi:E\rightarrow M$ is a compactly oriented vector bundle of rank $n$ on $M^n$, $s$ is a zero-limited section of $E$, then the Euler-like Poincaré dual of $E$ is a linear combination of the zero points of $s$, and the combination coefficient is a multiplicity.

Euler's characteristic number and Hopf index theorem

We hope to prove that the Euler number of $M$ is the Euler characteristic number.

Let $\{\omega_i\}$ be a set of bases $H^\ast (M)$, $\{\tau_i\}$ be a dual base, That is $\int_M\omega_i\wedge \tau_j=\delta_{ij}$. Let $\pi,\rho:M\times M\rightarrow M$ be two projections. According to Künneth's formula, $H^\ast (M\times M)=H^\ast (M)\otimes H^\ast (M)$, The basis is $\{\pi^\ast \omega_i\wedge\rho^\ast \tau_j\}$. weigh diagonal manifold $\Delta =\{(x,x)\in M\times M|x\in M\}\subset M\times M$.

Lemma 7. Poincaré duality of diagonal manifolds$\eta_\Delta =\sum (-1)^{\deg \omega_i}\pi^\ast \omega_i\wedge \rho^\ast \tau_i$.

Certificate: There are two ways to calculate $\int_\Delta \pi^\ast \tau_k\wedge\rho^\ast \omega_l$. $(-1)^{\deg \tau_k\cdot \deg \omega_l}\delta_{lk}=(-1)^{\deg \omega_k\cdot (\deg \omega_l+\deg \tau_k)}c_{kl}$, Let $\eta_\Delta = \sum_{i,j}c_{ij}\pi^\ast \omega_i\wedge\rho^\ast \tau_j$.

Lemma 8. $N_{\Delta /M\times M}\simeq T\Delta$.

Certificate: Just note that $M\approx \Delta$ is given by $x\mapsto(x,x)$. Thus $T\Delta \cong TM$, $T(M\times M)|_{\Delta}\cong TM\oplus TM$, So $N_{\Delta/M\times M}\simeq TM\oplus TM/TM\simeq TM\simeq T\Delta$.

Proposition 9. For compactly oriented manifolds $M$, $\int_M e(TM)=\chi (M)$.

Certificate: On the one hand, $\int_\Delta i^\ast \eta_\Delta =\int_\Delta i^\ast \Phi(N_\Delta)=\int_\Delta e(N_\Delta)=\int_\Delta e(T\Delta)=\int_M e(TM)$.

$\int_\Delta i^\ast \eta_\Delta =\int_{M\times M}\eta_\Delta \wedge \eta_\Delta$, Recall that the Poincaré duality of the intersection of two transversely intersecting closed submanifolds is equal to the outer product of the Poincaré duality of the two submanifolds, Therefore this integral is also called the self-intersection number of $\Delta$ in $M\times M$.

On the other hand, $\int_\Delta i^\ast \eta_\Delta=\sum_i(-1)^{\deg \omega_i}=\sum_q(-1)^q\dim H^q(M)=\chi(M)$. This is proven.

Naturally get:

Theorem 10 (Hopf index theorem). The index sum of the smooth vector field on the compactly oriented manifold $M$ (the sum of the local degrees of singular points) is equal to the Euler characteristic number of $M$.

The article was last updated on 2021-11-24 13:33:55