# Review of the Rham Cohomology Theory

Let $M$ be a smooth closed (i.e., compact and without boundary) manifold. Let $TM$ and $T^\ast M$ denote the tangent and cotangent (vector) bundle of $M$, respectively. Denote $\Lambda^\cdot(T^\ast M)$ the complex exterior algebra bundle of $T^\ast M$ and $\Omega^\cdot(M,\C)=\Gamma(\Lambda^\cdot(T^\ast M))$ be the space of smooth sections of $\Lambda^\cdot(T^\ast M)$. Particularly, for any integer $p$ with $0\leq p\leq n=\dim M$, we denote by
$\Omega^p(M)=\Gamma(\Lambda^p(T^\ast M))$
be the space of smooth $p$-forms over $M$.

Between the $\Omega^\cdot(M,\C)$, we have an operator $\rm d$, i.e., the exterior differential operator, which maps a $p$-form to a $p+1$-form and with the property $\rm d^2=0$. Then, we obtain a complex chain $(\Omega^\cdot(M,\C),\rd)$

Define the $p$-th (complex coefficient) de Rham cohomology of $M$ as
$H^p_{dR}(M,\C)=\frac{\ker\rd|_{\Omega^p(M,\C)}} {\rd(\Omega^{p-1}(M,\C))},$
and the total de Rham cohomology of $M$ as
$H^\ast_{dR}(M;\C)=\bigoplus_{p=0}^{\dim M}H_{dR}^p(M;\C).$
Particularly, a form $\omega$ is called closed if $\rd\omega=0$; called exact if there exists an $\eta$ such that $\rd\eta=\omega$.

Excise 1. For general $p$ the $p$-th de Rham cohomology may be hard to compute, but prove the following two special cases
• $H_{dR}^0(M;\C)$ can be written as the direct sum of $\C$ with multiplicity be its connected components, i.e.,
$H_{dR}^0(M;\C)=\bigoplus_{\textrm{connected components}}\C.$
• Suppose that $M$ be connected, and define an functional
$\int_M\mathpunct{:}\Omega^n(M;\C)\to\C,$
by $\omega\to\int_M\omega$, then prove that
$\ker\int_M=\rd(\Omega^{n-1}(M;\C)),$
from which we can compute the $n$-th de Rham cohomology $H_{dR}^n(M;\C)$. (cf. [1])

Apparently $H_{dR}^\ast(M;\C)$ is a vector space, since for any $\omega$, $\omega’\in\Omega^\ast(M;\C)$ (maybe one is $k$ form and other is $k’$ form), then we can verify the following equation holds
$[a\omega]=a[\omega],\quad[\omega+\omega’]=[\omega]+[\omega’],$
where $a$ is a constant function on $M$. With a litter more effort, we can prove that $H_{dR}^\ast(M;\C)$ is a ring. In fact, define
$[\omega]\wedge[\omega’]=[\omega\wedge\omega’],$
then for any two differential forms $\eta,\eta’$ on $M$, we have (note that $\rd\omega=0=\rd\omega’$)
\begin{align*}
(\omega+\rd\eta)\wedge(\omega’+\rd\eta’)
&=\omega\wedge\omega’+\omega\wedge\rd\eta’
+\rd\eta\wedge\omega’+\rd\eta\wedge\rd\eta’\\
&=\omega\wedge\omega’+\rd((-1)^{|\omega|}\omega\wedge\eta’
+\eta\wedge\omega’+\eta\wedge\rd\eta’),
\end{align*}
which means
$[\omega]\wedge[\omega’]=[\omega\wedge\omega’].$
Moreover, $H^\ast_{dR}(M;\C)$ is superexchange, i.e.,
$[\omega]\wedge[\eta] =(-1)^{|\omega|\cdot|\eta|}[\eta]\wedge[\omega].$

When $\dim M=n=4m$, consider $H_{dR}^{2m}(M;\C)$, then for any $[\omega]$, $[\eta]\in H_{dR}^{2m}(M;\C)$ we have
$[\omega]\wedge[\eta]=[\eta]\wedge[\omega].$
Define a bi-linear operator $< \cdot,\cdot>$ on $H_{dR}^{2m}(M;\C)$ as
$< [\omega],[\eta]>=\int_M\omega\wedge\eta.$
we can prove that $< \cdot,\cdot>$ is a non-singular quadric on $H_{dR}^{2m}(M;\C)$ and its signature is a topological invariants, called the characteristic number of $M$.

Now, we state the de Rham theory without proof (for the proof cf. [1]).

Theorem 1. If $M$ is a smooth closed orientable manifold, then for any integer $p$ with $0\leq p\leq\dim M$, we have
1. $\dim H_{dR}^p(M;\C)<+\infty$;
2. $H_{dR}^p(M;\C)$ is canonically isomorphic to $H^p_{sing}(M;\C)$, the $p$-th singular cohomology of $M$.
Remark 1. The above Theorem shows that algebraic topology can be represented by differential forms, and Chern-Weil theory claims that the characteristic classes of vector bundle (which is a cohomology class) can be represented by de Rham cohomology class.