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Created: 2026-03-06 07:59:30

Updated: 2026-03-06 08:17:13

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6.1 Stokes' theorem

Let the standard $n$ -simplex $\bar{\sigma}_{r}$ given by

$\begin{equation} \bar{\sigma}_{r} = \left\{(x^{1},\dots,x^{r})\in \mathbb{R}^{r}\mid x^{\mu}\geq 0, \sum_{\mu=1}^{r}x^{\mu}\leq 1\right\} \end{equation}$

or $\bar{\sigma}_{r}=(p_{0}p_{1}\dots p_{r})$ where $p_{0}=(0,\dots,0)$ , $p_{1}=(1,0,\dots,0)$ , $p_{r}=(0,\dots,1)$
An $r$ -form $\omega$ in $\mathbb{R}^{r}$ is written as

$\begin{equation} \omega = a(x) dx^{1}\wedge dx^{2}\wedge \dots \wedge dx^{r} \end{equation}$

Define the integration of $\omega$ over $\bar{\sigma}_{r}$ by

$\begin{equation} \int _{\bar{\sigma}_{r}} \, \omega \equiv \int _{\bar{\sigma}_{r}} \, a(x) dx^{1}dx^{2}\dots dx^{r} \end{equation}$

We may define $r$ -chain, $r$ -cycle and $r$ -boundary in a manifold $M$ . We have the nilpotent map $\partial:C_{r}(M)\to C_{r-1}(M)$ corresponding to geometrical boundary of $s_{r}$ with an induced orientation.

We can define the integration of a $r$ -form $\omega$ over an $r$ -chain. The integration on $r$ -simplex $s_{r}$ of $M$ is defined by

$\begin{equation} \int _{s_{r}}\omega =\int _{\bar{\sigma}_{r}}f^{*} \, \omega \end{equation}$

where $f:\bar{\sigma}_{r}\to M$ is a smooth map s.t. $s_{r}=f(\bar{\sigma}_{r})$ . Since $f^{*}\omega$ is an $r$ -form in $\mathbb{R}^{r}$ , the RHS is usual r-fold integral. For general $r$ -chain $c=\sum_{i}a_{i}s_{r,i}\in C_{r}(M)$ define

$\begin{equation} \int _{c} \, \omega=\sum_{i}a_{i}\int _{s_{r,i}} \, \omega \end{equation}$

Let $\omega\in \Omega ^{r-1}(M)$ and $c\in C_{r}(M)$ . Then

$\begin{equation} \int _{c} \, d\omega = \int _{\partial c} \, \omega \end{equation}$

6.2 de Rham cohomology groups

Let $M$ be an $m$ -dimensional differentiable manifold. The set of closed $r$ -forms is called the $r$ th cocycle group $Z^{r}(M)$ , the set of exact $r$ -forms is called $r$ th boundary group $B^{r}(M)$ , since $d^{2}=0$ , $B^{r}(M)\subset Z^{r}(M)$ .

$\begin{equation} \begin{aligned} Z^{r}(M) = \left\{\omega\in \Omega ^{r}(M)\mid d\omega=0\right\} \\ B^{r}(M) = \left\{\omega \in \Omega ^{r}(M)\mid \omega=d\psi,\exists \psi\in \Omega ^{r-1}(M)\right\} \end{aligned} \end{equation}$

If $\omega\in Z^{r}(M),\psi\in Z^{s}(M)$ , then $\omega \wedge \psi\in Z^{r+s}(M)$
If $\omega\in Z^{r}(M)$ , $\psi\in B^{s}(M)$ , $\omega \wedge \psi\in B^{r+s}(M)$
If $\omega\in B^{r}(M),\psi\in B^{s}(M),\omega \wedge \psi\in B^{r+s}(M)$
The $r$ th de Rham cohomology group is defined as $H^{r}(M;\mathbb{R})\equiv Z^{r}(M) / B^{r}(M)$
Let $\omega\in Z^{r}(M)$ . Then $[\omega]\in H^{r}(M)$ is the equivalence class $\{\omega'\in Z^{r}(M)|\omega'=\omega+d\psi,\psi\in\Omega ^{r-1}(M)\}$ . Two forms which differ by an exact form are called cohomologous. We will see that $H^{r}(M)$ is isomorphic to $H_{r}(M)$ .

$r=0$ , $B^{0}(M)$ has no meaning. Define $\Omega ^{-1}(M)=\emptyset$ , $B^{0}(M)=0$ , $H^{0}(M)=Z^{0}(M)=\{f\in \Omega ^{0}(M)=\mathcal{F}(M)\mid df=0\}$ . If $M$ connected, $df=0$ iff $f$ is constant over $M$ . Hence $H^{0}(M) \cong \mathbb{R}$ . If $M$ has $n$ conected components, $H^{0}(M)\cong \mathbb{R}^{n}$ .

Let $M=\mathbb{R}$ . $H^{0}(\mathbb{R})=\mathbb{R}$ . Let $\omega=fdx,d\omega=0$ ; Let $\omega=fdx=d(\int _{0}^{x}f(x) \, dx)$ , so any form is closed as well as exact, $H^{1}(\mathbb{R})=\Omega ^{1}(\mathbb{R})/\Omega ^{1}(\mathbb{R})=\{0\}$

Since $S^{1}$ connected, $H^{0}(S^{1})=\mathbb{R}$ . Let $\omega=f(\theta)d\theta \in\Omega ^{1}(S^{1})$ , is it possible to write $\omega=dF$ for some function $F$ ? Let $F(\theta)=\int _{0}^{\theta}f(\theta') \, d\theta'$ , since $F(0)=F(2\pi)$ , $\int _{0}^{2\pi}f(\theta') \, d\theta'=0$ . If we define a map $\lambda:\omega=fd\theta\mapsto \int _{0}^{2\pi} \, f(\theta')d\theta'$ , then $B^{1}(S^{1})$ is identified with $\mathrm{ker}\lambda$ .

$\begin{equation} H^{1}(S^{1})= \Omega ^{1}(S^{1}) / \mathrm{ker}\lambda = \mathrm{Im}\lambda = \mathbb{R} \end{equation}$

6.2.2 Duality of $H_{r}(M)$ and $H^{r}(M)$ ; de Rham's theorem

The cohomology group is a dual space of the homology group. The duality is provided by Stroke's theorem. Define inner product of $r$ -form and $r$ -chain in $M$ . Let $C_{r}(M)$ be the chain group of $M$ . Take $c\in C_{r}(M),\omega\in\Omega ^{r}(M),1\leq r\leq m$ . Define inner product $C_{r}(M)\times\Omega ^{r}(M)\to \mathbb{R}$ :

$\begin{equation} c,\omega \mapsto (c,\omega) = \int _{c} \, \omega \end{equation}$

Clearly the map is linear in both $c$ and $\omega$ :

$\begin{equation} \begin{aligned} (c_{1}+c_{2},\omega) & = \int _{c_{1}+c_{2}} \, \omega = \int _{c_{1}} \,\omega + \int _{c_{2}} \, \omega \\ (c,\omega_{1}+\omega_{2}) & = \int _{c} \, \omega_{1}+\omega_{2} = \int _{c} \, \omega_{1} + \int _{c} \, \omega_{2} \end{aligned} \end{equation}$

The Stroke's theorem takes the form $(c,d\omega)=(\partial c,\omega)$

The inner product naturally induces an inner product $\lambda$ between the elements of $H_{r}(M)$ and $H^{r}(M)$ . Now we show that $H_{r}(M)$ is dual of $H^{r}(M)$ . Let $[c]\in H_{r}(M),[\omega]\in H^{r}(M)$ , define inner product $\Lambda:H_{r}(M)\times H^{r}(M)\to \mathbb{R}$ by

$\begin{equation} \Lambda([c],[\omega]) \equiv (c,\omega) = \int _{c} \, \omega \end{equation}$

This is well defined since it is independent of the choice of representatives:

$\begin{equation} (c+\partial c',\omega)= (c,\omega) +(c',d\omega) = (c,\omega) \end{equation}$

where $d\omega=0$ has been used. Similarly, for $\omega+d\psi,\psi\in\Omega ^{r-1}(M)$ ,

$\begin{equation} (c,\omega+d\psi) = (c,\omega) + (\partial c,\psi) = (c,\omega) \end{equation}$

(de Rham's Theorem) If $M$ is a compact manifold, $H_{r}(M)$ and $H^{r}(M)$ are finite dimensional. The map $\Lambda$ is bilinear and non-degenerate. Thus $H^{r}(M)$ is dual vector space of $H_{r}(M)$

We accept Theorem 6.2 without proof which is highly nontrivial.

$(c,\omega)$ vanishes if $\omega$ is exact or if $c$ is a boundary.
Corollary 6.1: Let $M$ be a compact manifold and $k$ be the $r$ th Betti number. Let $c_{1},c_{2},\dots,c_{k}$ be properly chosen elements of $Z_{r}(M)$ s.t. $[c_{i}]\neq[c_{j}]$ .

A closed $r$ -form $\psi$ is exact iff $\int _{c_{i}} \psi \,=0$ , $1\leq i\leq k$
For any set of real numbers $b_{1},b_{2},\dots,b_{k}$ , $\exists$ closed $r$ -form $\omega$ s.t. $\int _{c_{i}}\omega b_{i}$ , $1\leq i\leq k$

de Rham's theorem states that bilinear form $\Lambda([c],[\omega])$ is non-degenerate. Hence if $\Lambda([c_{i}],\cdot)$ is a linear map acting on $H^{r}(M)$ , the ker $\Lambda$ consists of trivial element, the cohomology class of exact forms. $\psi$ is an exact form.
We may choose dual basis $[\omega_{i}]$ of $H^{r}(M)$ s.t. $\Lambda([c_{i}],[\omega_{j}])=\delta_{ij}$ , define $\omega=\sum_{i=1}^{k} b_{i}\omega_{i}$ completes the proof.

$\begin{equation} \begin{aligned} H^{0}(M)\cong H_{0}(M) \cong \underbrace{ \mathbb{R}\oplus \dots \oplus \mathbb{R} }_{ n } & \text{ if M has n connected components} \\ H^{1}(S^{1})\cong H_{1}(S^{1})\cong \mathbb{R} & \end{aligned} \end{equation}$

and the betti number satisfies

$\begin{equation} b^{r}(M)\equiv \text{dim}H^{r}(M)= \text{dim}H_{r}(M) = b_{r}(M) \end{equation}$

so Euler characteristic is

$\begin{equation} \chi(M) = \sum_{r=1}^{m}(-1)^{r}b^{r}(M) \end{equation}$

The LHS is purely topological while RHS is given by analytic condition ( $d\omega=0$ is a set of PDEs)

Summary:

$\begin{equation} \begin{aligned} \leftarrow C_{r-1}(M) \overset{ \partial_{r} }{ \leftarrow } C_{r}(M) \overset{ \partial_{r+1} }{ \leftarrow } C_{r+1}(M) \leftarrow \\ \to \Omega ^{r-1}(M) \overset{ d_{r} }{ \to } \Omega ^{r}(M) \overset{ d_{r+1} }{ \to } \Omega ^{r+1}(M) \to \\ H_{r}(M) = Z_{r}(M) / B_{r}(M) = \mathrm{ker} \partial_{r} / \mathrm{im} \partial_{r+1} \\ H^{r}(M) = Z^{r}(M) / B^{r}(M) = \mathrm{ker} d_{r+1} / \text{im}\, d_{r} \end{aligned} \end{equation}$

6.3 Poincare's lemma

If a coordinate neighbourhood $U$ of a manifold $M$ is contractible to a point $p_{0}\in M$ , any closed $r$ -form on $U$ is also exact.

$U$ is smoothly contractible, to $p_{0}$ , so $\exists$ smooth map $F:U\times I\to U$ s.t.

$\begin{equation} F(x,0)=x,F(x,1)=p_{0} \text{ for }x\in U \end{equation}$

Let $\eta\in\Omega ^{r}(U\times I)$ ,

$\begin{equation} \eta= a_{i_{1}\dots i_{r}}(x,t) dx^{i_{1}}\wedge\dots \wedge dx^{i_{r}} + b_{j_{1}\dots j_{r-1}}(x,t) dt \wedge dx^{j_{1}}\wedge\dots \wedge dx^{j_{r-1}} \end{equation}$

where $x$ is coordinate of $U$ , $t$ of $i$ . Define $P:\Omega ^{r}(U\times I)\to\Omega ^{r-1}(U)$ by

$\begin{equation} P\eta \equiv \left( \int _{0}^{1} \, ds\ b_{j_{1}\dots j_{r-1}}(x,s) \right) dx^{j_{1}}\wedge\dots \wedge dx^{j_{r-1}} \end{equation}$

define $f_{t}:U\to U\times I$ by $f_{t}(x)=(x,t)$ . The pulback of first term of $\eta$ is an element of $\Omega ^{r}(U)$ ,

$\begin{equation} f_{t}^{*} \eta = a_{i_{1}\dots i_{r}}(x,t) dx^{i_{1}}\wedge\dots \wedge dx^{i_{r}}\in \Omega ^{r}(U) \end{equation}$

Now prove the following identity:

$\begin{equation} d(P\eta) + P(d \eta) = f_{1}^{*}\eta-f_{0}^{*}\eta \end{equation}$

LHS:

$\begin{equation} \begin{aligned} \mathrm{d} \left(P \eta\right) & = \mathrm{d}\left( \int _{0}^{1} \, ds\ b_{j_{1}\dots j_{r-1}} \right) dx^{j_{1}}\wedge\dots \wedge dx^{j_{r-1}} \\ &= \int _{0}^{1} \, ds \left( \frac{ \partial b_{j_{1}\dots j_{r-1}} }{ \partial x^{j_{r}} } \right) dx^{j_{r}}\wedge dx^{j_{1}}\wedge\dots \wedge dx^{j_{r-1}} \\ P\mathrm{d}\eta & = P\bigg{[} \left(\frac{ \partial a_{i_{1}\dots i_{r}} }{ \partial x^{i_{r+1} } }\right) dx^{i_{r+1}}\wedge dx^{i_{1}}\dots \wedge dx^{i_{r}} + \\ & \left(\frac{ \partial a_{i_{1}\dots i_{r}} }{ \partial t }\right) dt \wedge dx^{i_{1}}\wedge\dots \wedge dx^{i_{r}} +\\ & \left( \frac{ \partial b_{j_{1}\dots j_{r-1}} }{ \partial x^{j_{r}} } \right) dx^{j_{r}}\wedge dt \wedge dx^{j_{1}}\wedge\dots \wedge dx^{j_{r-1}} \bigg{ ]} \\ & = \left[\int _{0}^{1} \, ds \left(\frac{ \partial a_{i_{1}\dots i_{r}} }{ \partial s } \right) \right] dx^{i_{1}}\wedge\dots \wedge dx^{i_{r}} \\ & - \left[ \int _{0}^{1} \, ds \left(\frac{ \partial b_{j_{1}\dots j_{r-1}} }{ \partial x^{j_{r}} }\right) \right] dx^{j_{r}}\wedge dx^{j_{1}}\wedge\dots \wedge dx^{j_{r-1}} \end{aligned} \end{equation}$

$\begin{equation} \begin{aligned} \mathrm{d}(P \eta) + P(\mathrm{d} \eta) & = \left[\int _{0}^{1} \, ds \left( \frac{ \partial a_{i_{1}\dots i_{r}} }{ \partial s } \right) \right] dx^{i_{1}}\wedge \dots \wedge dx^{i^{r}} \\ & = [ a_{i_{1}\dots i_{r}}(x,1)-a_{i_{1}\dots i_{r}}(x,0 )] dx^{i_{1}}\wedge\dots \wedge dx^{i_{r}} \\ & = f_{1}^{*}\eta - f_{0}^{*}\eta \end{aligned} \end{equation}$

Poincare's lemma readily follows. Let $\omega$ be closed $r$ form on a contractible chart $U$ . We will show that $\omega$ can be written as an exact form,

$\begin{equation} \omega= d(-PF^{*}\omega) \end{equation}$

$F$ being the smooth contraction map. In fact, if $\eta$ is replaced by $F^{*}\omega\in\Omega ^{r}(U\times I)$ , we have

$\begin{equation} dPF^{*}\omega + PdF^{*}\omega = f_{1}^{*}\circ F^{*}\omega - f_{0}^{*} \circ F^{*}\omega = (F\circ f_{1})^{*} \omega - (F\circ f_{0})^{*}\omega \end{equation}$

Clearly $F\circ f_{1}:U\to U$ is a constant map $x\mapsto p_{0}$ , hence $(F\circ f_{1})^{*}=0$ . $F\circ f_{0}=id_{U}$ , hence RHS is simply $-\omega$ . The second term of LHS vanishes since $\omega$ is closed; $dF^{*}\omega=F^{*}d\omega=0$ . Finally, $\omega=-\mathrm{d}PF^{*}\omega$ , which proves the theorem.

e.g. since $\mathbb{R}^{n}$ contractible, $H^{r}(\mathbb{R}^{n})=0,1\leq r\leq n$ . However note that $H^{0}(\mathbb{R}^{n})=\mathbb{R}$ .

6.4 Structure of de Rham cohomology groups

Let $M$ be $m$ -dim compact manifold, $\omega\in H^{r}(M),\eta\in H^{m-r}(M)$ . Note that $\omega \wedge \eta$ is volume element, define inner product: $H^{r}(M)\times H^{m-r}\to \mathbb{R}$ by

$\begin{equation} \left<\omega,\eta\right> \equiv \int _{M} \, \omega \wedge \eta \end{equation}$

The inner product is bilinear but not singular: if $\omega\neq 0$ or $\eta\neq 0$ , $\left<\omega,\eta\right>$ cannot vanish identically. Thus the inner product defines duality of $H^{r}(M)$ and $H^{m-r}(M)$ :

$\begin{equation} H^{r}(M)\cong H^{m-r}(M) \end{equation}$

called Poincare duality ,and $b_{r}=b_{m-r}$ . It follows that in odd dim space, Eular characteristic vanishes:

$\begin{equation} \chi(M) = \sum(-)^{r}b_{r}=0 \end{equation}$

6.4.2 Cohomology rings

Let $[\omega]\in H^{q}(M)$ , $[\eta]\in H^{r}(M)$ , define the product by

$\begin{equation} [\omega]\wedge[\eta]=[\omega \wedge \eta] \end{equation}$

$\omega \wedge \eta$ is closed, so $[\omega \wedge \eta]\in H^{q+r}(M)$ . $\left[\omega \wedge \eta\right]$ is also independent of representatives of $\left[\omega\right]$ and $[\eta]$ : if we take $\omega'=\omega+\mathrm{d}\psi$ ,

$\begin{equation} [\omega']\wedge[\eta]= [(\omega+\mathrm{d}\psi)\wedge \eta] = [\omega \wedge \eta + \mathrm{d}(\psi \wedge \eta)] = [\omega \wedge \eta] \end{equation}$

so the $\wedge$ is well defined. The cohomology ring is defined by direct sum

$\begin{equation} H^{*}(M) = \bigoplus_{r=1}^{m} H^{r}(M) \end{equation}$

the product is provided by $\wedge:H^{*}(M)\times H^{*}(M)\to H^{*}(M)$ , the addition is the formal sum of two elements of $H^{*}(M)$ . One of the superiorities of cohomology groups over homology groups resides here: Products of chains are not well defined, homology groups cannot have ring structure.

6.4.3 The Kunneth formula

Let $M$ be a product