14. The Euler Class

If $E\stackrel{\pi }{\to }M$ is a vector bundle of rank $1$, $⟨,⟩$ a metric on $E$, then every metric compatible connection $\nabla$ is flat.

Now take $E$ of rank $k=2$. $\nabla$ is connection on $E$ compatible with a metric $⟨,⟩$. Let $s_1$, $s_2$ be a local orthogonal frame with respect to $⟨,⟩$.$$\nabla s_i=\sum _{j=1}^2{\omega }_{ij}\otimes s_j\text{with}{\omega }_{ij}\text{skew-symmetric}\left(\begin{array}{cc}0& -{\omega }_{12}\\ -{\omega }_{21}& 0\end{array}\right)$$Then$${\Omega }_{ij}=d{\omega }_{ij}-\sum _{l=1}^2{\omega }_{il}\wedge {\omega }_{lj}$$${\Omega }_{ij}$ is also skew-symmetric.$${\Omega }_{12}=d{\omega }_{12}-\sum _{l=1}^2{\omega }_{1l}\wedge {\omega }_{l2}=d{\omega }_{12}\Rightarrow d{\Omega }_{12}=0$$We assume now that $E$ is oriented and $s_1$, $s_2$ are positive with respect to this orientation. Let $s_1^{\prime }$, $s_2^{\prime }$ be another local orthogonal frame, which is also positive oriented.$$s_i^{\prime }=\sum _{j=1}^2{g}_{ij}{s}_j$$$s_i$, $s_i^{\prime }$ are defined on $E{|}_U$, $g_{ij}\in {\mathcal{C}}^{\infty }(U)$. $g\in SO(2)=S^1$ at every point.$$\begin{array}{rl}g& =\left(\begin{array}{cc}\cos (f(x))& -\sin (f(x))\\ \sin (f(x))& \cos (f(x))\end{array}\right){\Omega }^{\prime }\\ & =g\Omega g^{-1}\\ & =\left(\begin{array}{cc}\cos (f(x))& -\sin (f(x))\\ \sin (f(x))& \cos (f(x))\end{array}\right)\left(\begin{array}{cc}0& {\Omega }_{12}\\ -{\Omega }_{21}& 0\end{array}\right)\left(\begin{array}{cc}\cos (f(x))& \sin (f(x))\\ -\sin (f(x))& \cos (f(x))\end{array}\right)\\ & =\left(\begin{array}{cc}0& {\Omega }_{12}\\ -{\Omega }_{12}^{\prime }& 0\end{array}\right)=\Omega \end{array}$$The last equality is because $SO(2)$ is abelian. This shows that $\Omega$ and therefore ${\Omega }_{12}$ is independent of the choice of oriented orthogonal frames ${\mathbf{s}}_1$, ${\mathbf{s}}_2$.

${\Omega }_{12}$ is a globally well-defined closed $2$-form.

定义 14.0.1. $e(E):=-\frac{1}{2\pi }[{\Omega }_{12}]\in H_{dR}^2(M)$.

命题 14.0.2.  

(1)	$e(\overline{E})=-e(E)$, $\overline{E}$ is the vector bundle with opposite orientation.
(2)	If $E$ admits a section $s$, which is nowhere zero, then $e(E)=0$ [without loss of generality $⟨s,s⟩=1$. Then $0=2⟨s,\nabla s⟩$, so $⟨s,\nabla s⟩=0$. Take $s_1=s$. There is a unique $s_2$, s.t. $s_1$, $s_2$ are orthogonal and oriented. Globally ${\Omega }_{12}=d{\omega }_{12}$, so $[\Omega ]=0\in H_{dR}^2(M)$.]
(3)	The Euler class is independent of the choice of $\nabla$ (compatible with a fixed $⟨,⟩$). [Let ${\nabla }^0$, ${\nabla }^1$ be two different connections compatible with $⟨,⟩$. Then ${\nabla }^1-{\nabla }^0=A\in {\Omega }^1(\stackrel{\mathrm{rank}=1}{\overbrace{Skew-End(E)}})$, with respect to a local orthogonal frame $s_1$, $s_2$: $$\left(\begin{array}{cc}0& {\omega }_{12}^1\\ -{\omega }_{12}^1& 0\end{array}\right)-\left(\begin{array}{cc}0& {\omega }_{12}^0\\ -{\omega }_{12}^0& 0\end{array}\right)=\left(\begin{array}{cc}0& a\\ -a& 0\end{array}\right)$$$a\in {\Omega }^1(M)$ is a globally well-defined $1$-form, where $a$ has trivial gauge transformation. $${\omega }_{12}^1={\omega }_{12}^0+a\Rightarrow {\Omega }_{12}^1={\Omega }_{12}^0+da\Rightarrow [{\Omega }_{12}^1]=[{\Omega }_{12}^0]\in H_{dR}^2(M)]$$$E$ of rank $k$ is trivial if and only if $\exists s_1,\dots ,s_k$ sections, which are everywhere linear independent. $E$ oriented of rank $k$ is trivial if and only if $\exists s_1,\dots ,s_{k-1}$ which are everywhere linear independent.
(4)	$e(E)$ is independent of the choice of metric. [Sketch of proof: $E\times [0,1]\stackrel{\pi \in \mathrm{id}}{\to }M\times [0,1]$ as a $\mathbb{E}$ vector bundle on $M\times [0,1]$. On ${\mathbb{E}}_{(x,t)}$, we consider the metric $(1-t)⟨-,-{⟩}_x^0+t⟨-,-{⟩}_x^1=⟨⟨-,-⟩{⟩}_{x,t}$. This is a metric on $\mathbb{E}$, which restricts to $\mathbb{E}{|}_{M\times \{0\}}=E$ as $⟨,{⟩}^0$ and to $\mathbb{E}{|}_{M\times \{1\}}$ as $⟨,{⟩}^1$. Let $\nabla \nabla$ be a connection on $\mathbb{E}$ compatible with $⟨⟨,⟩⟩$. From its curvature, we determine $e(\mathbb{E})\in H_{dR}^2(M\times [0,1])$. Let $i_0,i_1:M\hookrightarrow M\times [0,1]$, where $i_0(x)=(x,0)$ and $i_1(x)=(x,1)$. Then $e(E,⟨,{⟩}^0)=i_0^{\ast }e(\mathbb{E},⟨⟨,⟩⟩)$. Similarly, $e(E,⟨,{⟩}^1)=i_1^{\ast }e(\mathbb{E},⟨⟨,⟩⟩)$. $\Rightarrow$ $e(E,⟨,{⟩}^0)=e(E,⟨,{⟩}^1)$, because $i_0^{\ast }=\mathrm{Id}=i_1^{\ast }$. By Poincaré lemma, $i_0^{\ast }$ and $i_1^{\ast }$ are homotopic maps induce the same $H_{dR}$. ]

例 14.0.3. $M=S^2$. Take two copies of ${\mathbb{R}}^2\times {\mathbb{R}}^2$. With the standard $⟨,⟩$ on the second factor. And standard flat connection compatible with $⟨,⟩$. Take $\begin{array}{rl}\psi :({\mathbb{R}}^2\setminus \{0\})\times {\mathbb{R}}^2& \to ({\mathbb{R}}^2\setminus \{0\})\times {\mathbb{R}}^2\\ (x,v)& \mapsto \left(-\frac{x}{\Vert x{\Vert }^2},g(x)v\right)\end{array}$ where $g:{\mathbb{R}}^2\setminus \{0\}\to SO(2)$. $X_1={\mathbb{R}}^2\times {\mathbb{R}}^2$, $X_2={\mathbb{R}}^2\times {\mathbb{R}}^2$ are identified via $\psi$ to get an oriented rank $2$ vector bundle $E\to S^2$, with a metric.

Let ${\nabla }^0$ be the given flat connection on $E{|}_{S^2\setminus \{N\}}$, coming from $X_1$.

Let ${\nabla }^1$ be the given flat connection on $E{|}_{S^2\setminus \{S\}}$, coming from $X_2$.
Choose a smooth partition of unity $\rho$, $1-\rho$ subordinate to the covering of $S^2$ by $S^2\setminus \{N\}$ and $S^2\setminus \{S\}$. Write $S^2\setminus \{N,S\}=S^1\times \mathbb{R}$. $$\begin{array}{rl}\rho :S^1\times \mathbb{R}& \to \mathbb{R}\\ (\phi ,t)& \mapsto \rho (t)\end{array}$$$\rho$ extends to a smooth function on $S^2$. Define $\nabla =\rho {\nabla }^1+(1-\rho ){\nabla }^0$. This is a metric connection on $E\to S^2$. Over $S^2\setminus \{N,S\}$, consider the frame which is parallel for ${\nabla }^0$ given by the standard basis for ${\mathbb{R}}^2$. With respect to this frame the connection matrix for $\nabla$ is that for ${\nabla }^1$ scaled by $\rho$.

Let $s_1^{\prime }$, $s_2^{\prime }$ be the parallel frame for ${\nabla }^1$ coming from $X_2$. In this frame, ${\nabla }^1$ has zero connection matrix. $$\begin{array}{rlrlr}& & 0={\omega }^{\prime }=dg{g}^{-1}+g\omega g^{-1}& & \\ \Rightarrow & & g\omega g^{-1}=-dg{g}^{-1}& & \\ \Rightarrow & & \omega =-g^{-1}dg& & \end{array}$$$${\omega }_{12}=-\sum _{i=1}^2{g}^{1i}d{g}_{i2}$$Take $\begin{array}{rl}g:S^1\times \mathbb{R}& \to SO(2)\\ (\phi ,t)& \mapsto \left(\begin{array}{cc}\cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{array}\right)\end{array}$, we could also take $g={\mathrm{e}}^{\mathrm{i}n\phi }$. $$\begin{array}{rlrlr}\Rightarrow & & {\omega }_{12}=-g^{11}d{g}_{12}-g^{12}d{g}_{22}=d\phi & & \end{array}$$In the frame $s_1$, $s_2$, $\nabla$ is represented by $\left(\begin{array}{cc}0& \rho d\phi \\ -\rho d\phi & 0\end{array}\right)$ $${\Omega }_{12}=d(\rho d\phi )=d\rho \wedge \underset{\text{Not really exact on}{S}^1}{\underbrace{d\phi }}=\frac{d\rho }{dt}dt\wedge d\phi$$$$\underset{S^2}{\int }{\Omega }_{12}=\underset{S^1\times \mathbb{R}}{\int }{\Omega }_{12}=-\left({\int }_{-\infty }^{+\infty }dt\frac{d\rho }{dt}\right)\left(\underset{S^1}{\int }d\phi \right)=-(\rho (\infty )-\rho (-\infty ))\cdot 2\pi =-2\pi \ne 0$$with $g={\mathrm{e}}^{\mathrm{i}n\phi }:\underset{S^2}{\int }{\Omega }_{12}=-2\pi n$, $g$ is called clutching map. We can do this for general $S^n$.

If $E$, $F$ are oriented preserving isomorphism, then $e(E)=e(F)$.

If $M$ is oriented, $n$-dimensional, then$$\underset{M}{\int }:H_c^n(M)\to \mathbb{R}$$is well defined and surjective.

$M$ is an oriented $2$-dimensional manifold, compact without boundary, then$$\underset{M}{\int }:H_{dR}^2(M)\to \mathbb{R}$$If $M$ is connected, this is an isomorphism.

定义 14.0.4. $\chi (E):=\underset{M}{\int }e(E)$ is the Euler number of $E$.

Let $M$ be oriented $2$-dimensional manifold, and $⟨,⟩$ a Riemannian metric.

How do we determine $e(TM)$?

Let $X_1$, $X_2$ be a local orthogonal frame for $(TM,⟨,⟩)$, s.t. $(X_1,X_2)$ is positive oriented.$$K(T_{p}M)=⟨R(X_1,X_2)X_2,X_1⟩=⟨{\nabla }_{X_1}{\nabla }_{X_2}{X}_2-{\nabla }_{X_2}{\nabla }_{X_1}{X}_2-{\nabla }_{[X_1,X_2]}{X}_2,X_1⟩$$where ${\nabla }_{X_i}$, $i=1,2$ is the Levi-Civita connection.$$\begin{array}{rl}\nabla X_1& ={\omega }_{12}\otimes X_2\\ \nabla X_2& =-{\omega }_{12}\otimes X_1\end{array}\Rightarrow \begin{array}{rl}{\nabla }_{X_2}{X}_2& =-{\omega }_{12}(X_2)X_1\\ {\nabla }_{X_1}{X}_1& =-{\omega }_{12}(X_1)X_1\\ {\nabla }_{[X_1,X_2]}{X}_2& =-{\omega }_{12}([X_1,X_2])X_1\end{array}$$Therefore,$$\begin{array}{rl}K(T_{p}M)& =⟨{\nabla }_{X_1}(-{\omega }_{12}(X_2)X_1)-{\nabla }_{X_2}(-{\omega }_{12}(X_1)X_1)+{\omega }_{12}⟨([X_1,X_2])X_1,X_1⟩\\ & =⟨-L_{X_1}{\omega }_{12}(X_2)\cdot X_1-\cancel{{\omega }_{12}(X_2){\nabla }_{X_1}{X}_1}+L_{X_2}{\omega }_{12}(X_1)\cdot X_1\\ & +\cancel{{\omega }_{12}(X_1){\nabla }_{X_2}{X}_1}+{\omega }_{12}([X_1,X_2])X_1,X_1⟩\\ & =-L_{X_1}{\omega }_{12}(X_2)+L_{X_2}{\omega }_{12}(X_1)+{\omega }_{12}([X_1,X_2])\\ & =-(d{\omega }_{12})(X_1,X_2)\\ & =-{\Omega }_{12}(X_1,X_2)\end{array}$$[$M$ $n$-dimensional oriented, $⟨,⟩$ on $TM$ $\Rightarrow$ $\exists !dvol\in {\Omega }^n(M)$ with $dvol(X_1,\dots ,X_n)=1$ for any oriented orthonormal basis $X_1,\dots ,X_n$ of $T_{p}M$. $dvol=X_1^{\ast }\wedge \cdots \wedge X_n^{\ast }$.]

定理 14.0.5 (Gauss Bonnet Theorem). On an oriented $2$-dimensional manifold with a metric, the equation $K(T_{p}M)=-{\Omega }_{12}(X_1,X_2)$ is equivalent to ${\Omega }_{12}=-K\cdot dvol$. The Euler number of $TM\stackrel{\pi }{\to }M$ is$$\boxed{\chi (TM)=-\frac{1}{2\pi }\underset{M}{\int }{\Omega }_{12}=\frac{1}{2\pi }\underset{M}{\int }K\cdot dvol}$$where $\chi (TM)$ is the Euler character of $M$.

例 14.0.6. $M=S^2(R)$ is the $2$-sphere of radius $R$ in ${\mathbb{R}}^3$. $$\chi (T{S}^2)=\frac{1}{2\pi }\underset{S^2(R)}{\int }K\cdot dvol=\frac{1}{2\pi R^2}\cdot vol(S^2(R))=\frac{4\pi R^2}{2\pi R^2}=2$$

例 14.0.7. Suppose the $2$-manifold $M$ admits a vector field without zeroes. Then $$\chi (TM)=0=\frac{1}{2\pi }\underset{M}{\int }K\cdot dvol$$

推论 14.0.8 (Hedgehog/Hairy Ball Theorem). $S^2$ does not admit a vector field without zeroes.

例 14.0.9. $M=T^2$.

$M$ an oriented $2$-dimensional manifold, $X\in \mathfrak{X}(M)$ a vector field with isolated zeroes.

$M$ connected, compact $\Rightarrow$ $X$ has finitely many zeroes $p_1,\dots ,p_k$.

Choose disjoint open neighborhoods $U_1,\dots ,U_k$ of $p_1,\dots ,p_k$, with each $U_i$ diffeomorphic to a disc of radius $2$ in ${\mathbb{R}}^2$ and $V_i=D(1)$ in.

We equip $M$ with a Riemannian metric, which restricts to each $U_i$ as the flat metric of $U_i\subset {\mathbb{R}}^2$.

On $M\setminus \{p_1,\dots ,p_k\}$, define $X_1=\frac{X}{\Vert X\Vert }$ with respect to our metric.

Complete $X_1$ to an oriented orthonormal basis $\{X_1,X_2\}$ on $M\setminus \{p_1,\dots ,p_k\}$.$$\begin{array}{rl}\chi (TM)& =\frac{1}{2\pi }\underset{M}{\int }K\cdot dvol\\ & \stackrel{\text{flat around poles}}{=}\frac{1}{2\pi }\underset{M\setminus (\bigcup V_i)}{\int }K\cdot dvol\\ & =-\frac{1}{2\pi }\underset{M\setminus (\bigcup V_i)}{\int }{\Omega }_{12}\\ & \stackrel{\text{Stokes}}{=}-\frac{1}{2\pi }\underset{\partial (M\setminus (\cup V_i))}{\int }{\omega }_{12}\\ & =\frac{1}{2\pi }\sum _{i=1}^n\underset{\partial V_i}{\int }{\omega }_{12}\end{array}$$

Claim 14.0.10. $X_1^{\ast }d\theta ={\omega }_{12}$.

证明. Notice that

$$(X_1^{\ast }d\theta )(Y)=d\theta ((D{X}_1)Y)\stackrel{\text{flat connection}}{=}d\theta ({\nabla }_Y{X}_1)=d\theta ({\omega }_{12}(Y)X_2)={\omega }_{12}Y$$

$\square$

$$\begin{array}{rlrlr}\Rightarrow & & \chi (TM)=\frac{1}{2\pi }\sum _{i=1}^n\underset{\partial V_i}{\int }{X}_1^{\ast }d\theta =\frac{1}{2\pi }\sum _{i=1}^n\mathrm{deg}\left(X_1{|}_{\partial V_i}\right)\underset{S^1}{\int }d\theta =\sum _{i=1}^n\underset{=\mathrm{Index}(X_1,p_i)}{\underbrace{\mathrm{deg}\left(X_1{|}_{\partial V_i}\right)}}& & \end{array}$$This is called the Poincaré–Hopf Theorem.