12. de Rham Cohomology

$$H_{dR}^{\ast }(M)=\underset{k=0}{\overset{n}{⨁}}{H}_{dR}^k(M)$$

定义 12.0.1. $H_c^k(M):=\frac{\ker (d:{\Omega }_0^k(M)\to {\Omega }_0^{k+1}(M))}{\mathrm{im}(d:{\Omega }_0^{k-1}(M)\to {\Omega }_0^k(M))}$ the de Rham cohomology of $M$ with compact support, where $\ker (d:{\Omega }_0^k(M)\to {\Omega }_0^{k+1}(M))$ is the closed form and $\mathrm{im}(d:{\Omega }_0^{k-1}(M)\to {\Omega }_0^k(M))$ is the exact form.

例 12.0.2. $H_c^0(M)=$locally constant functions with compact support. If $M$ is connected, then $$H_c^0(M)=\{\begin{array}{rl}\mathbb{R}& M\text{compact}\\ 0& M\text{non-compact}\end{array}$$

$M=\mathbb{R}$, $k=1$$$H_c^1(\mathbb{R})=\frac{{\Omega }_c^1(\mathbb{R})}{\mathrm{im}(d:{\Omega }_c^0(\mathbb{R})\to {\Omega }_c^1(\mathbb{R}))}$$${\Omega }_c^1(\mathbb{R})\ni \omega =fdt$, $f\in {\mathcal{C}}_0^{\infty }(\mathbb{R})$. Before$$F(x)={\int }_{-\infty }^{x}f(t)dt,\omega =dF$$But $F$ does not have compact support, if $\underset{-\infty }{\overset{+\infty }{\int }}f(t)dt=c\ne 0$.

If $c=0$, then $F\in {\Omega }_0^0(\mathbb{R})$ and $dF=\omega$, then $[\omega ]=0\in H_c^1(\mathbb{R})$. If $c\ne 0$, then still $dF=\omega$, but $F\notin {\Omega }_0^0(\mathbb{R})$.

Suppose $G\in {\mathcal{C}}_0^{\infty }(\mathbb{R})$ and $dG=\omega$. Then $d(F-G)=0$. $\Rightarrow$ $F-G=d$ is constant, for $x\ll 0$: $G(x)=d$ and for $x\gg 0$: $G(x)=-d+c$.

Since $G$ has compact support $\Rightarrow$ $d=0$ and $d=c$. This leads to a contradiction since $c\ne 0$.

$\Rightarrow$ $\omega \notin \mathrm{im}(d:{\Omega }_0^0(\mathbb{R})\to {\Omega }_0^1(\mathbb{R}))$.

$\Rightarrow$ $H_c^1(\mathbb{R})\ne 0$.

定理 12.0.3. If $M$ is a smooth $n$-dimensional oriented manifold without boundary, then $$\underset{M}{\int }:H_c^n(M)\to \mathbb{R}$$is well-defined and surjective.

证明. If $[{\omega }^{\prime }]=[\omega ]\in H_c^n(M)$, then ${\omega }^{\prime }=\omega +d\alpha$, with $\alpha \in {\Omega }_0^{n-1}(M)$. $$\underset{M}{\int }{\omega }^{\prime }=\underset{M}{\int }\omega +\underset{M}{\int }d\alpha \stackrel{\text{Stokes}}{=}\underset{M}{\int }\omega +\underset{\partial M}{\int }\alpha =\underset{M}{\int }\omega$$Let $\rho ⩾0$ be a bump function, with support in a chart: $\omega =\rho d{x}_1\wedge \cdots \wedge d{x}_n$. $$\underset{M}{\int }\omega =\underset{U}{\int }\rho d{x}_1\wedge \cdots \wedge d{x}_n={\int }_{-\infty }^{+\infty }\cdots {\int }_{-\infty }^{+\infty }\rho d{x}_1\cdots d{x}_n>0$$By linearity, surjective follows.

$\square$

例 12.0.4. $M=\mathbb{R}$ $$\underset{\mathbb{R}}{\int }:H_c^1(\mathbb{R})\to \mathbb{R}$$

Claim 12.0.5. This is an isomorphism.

证明. $fdt=\alpha \in {\Omega }_0^1(\mathbb{R})$, where $f\in {\Omega }_0^0(\mathbb{R})$. $$\underset{\mathbb{R}}{\int }\alpha =\underset{\mathbb{R}}{\int }fdt=c$$If $[\alpha ]\in \ker \left(\underset{\mathbb{R}}{\int }\right)$ $$\begin{array}{rlrlrl}\iff & & c& =0& & \\ \iff & & \alpha & =dF\text{with}F\in {\Omega }_0^0(\mathbb{R})& & \\ \iff & & [\alpha ]& =0& & \end{array}$$The instance of Poincaré duality.

$M=\mathbb{R}$	$H_{dR}^k(\mathbb{R})$	$H_c^k(\mathbb{R})$
$k=0$	$\mathbb{R}$	$0$
$k=1$	$0$	$\mathbb{R}$

$\square$

$H_c^{\ast }(M)=\underset{k=0}{\overset{n}{⨁}}{H}_c^k(M)$ is an algebra with $\wedge$ induced by wedge product on forms.$$H_{dR}^k(M)\times H_c^l(M)\to H_c^{k+l}(M)$$because a wedge product has compact support if one of the factors does.

Suppose $f:M\to N$ is a smooth map between smooth manifolds. The pullback $f^{\ast }:{\Omega }^k(N)\to {\Omega }^k(M)$ commutes with $d$. In particular, if $d\omega =0$, then $d{f}^{\ast }\omega =f^{\ast }d\omega =0$ and if $\omega =d\alpha$, $f^{\ast }\omega =d{f}^{\ast }\alpha$.

$\Rightarrow$ $\begin{array}{rl}[t]f^{\ast }:H_{dR}^k(N)& \to H_{dR}^k(M)\\ [\omega ]& \mapsto [f^{\ast }\omega ]\end{array}$ is well-defined, linear. This $f^{\ast }$ induces an algebra homomorphism.$$f^{\ast }:H_{dR}(N)\to H_{dR}(M)$$

例 12.0.6. $M=\overline{B_1(0)}\subset {\mathbb{R}}^n$.

Claim 12.0.7. There is no smooth map $n:M\to \partial M$, $r{|}_{\partial M}=\mathrm{Id}{|}_{\partial M}$.

证明. Assume there is such an $r$, then This leads to a contradiction.

$\square$