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On a star-shaped planar domain containing zero, let $\eta$ be a smooth closed differential one-form and set $f(x)={\int }_{[0,x]}\eta$. Green theorem on the triangle with vertices $0,x,x+h$ gives $f(x+h)-f(x)={\int }_{[x,x+h]}\eta$. Dividing by a displacement and taking its limit proves $df=\eta$. Thus every such form is an exact differential form, a degree-one version of the Poincare lemma which uses only the planar integral theorem.

In a manifold chart, write a smooth differential form as $\eta ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$, with $I$ increasing. Define its exterior derivative byTo check that this definition is global, change coordinates to $y$. The chain rule expresses $d{x}^i$ as ${\sum }_j({\partial }_j{x}^i)d{y}^j$. Differentiating it gives zero because the second partial derivatives of $x^i$ are symmetric and $d{y}^k\wedge d{y}^j$ is antisymmetric. The coordinate definition satisfies the graded Leibniz rule, so differentiating the transformed expression for $\eta$ leaves exactly ${\sum }_{I}d{a}_I\wedge d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. This agrees with the original formula and proves independence of the manifold chart. The same cancellation of mixed derivatives gives $d^2=0$.

The de Rham cohomology is thereforethe vector space of closed differential forms modulo exact differential forms.

Here is a Green-theorem proof of exactness on a disc. Identify the disc with a round disc centred at zero, and let $\eta =Pdx+Qdy$ be a closed differential form. Set $f(x)={\int }_{[0,x]}\eta$. For $x$ and $x+h$ in the disc, Green theorem on the oriented triangle with vertices $0,x,x+h$ gives zero boundary integral, since ${\partial }_{x}Q-{\partial }_{y}P=0$. HenceTaking $h=\epsilon v$ and dividing by $\epsilon$ yields $d{f}_x(v)={\eta }_x(v)$. The radial integral defining $f$ is smooth, so $\eta =df$ andDegenerate triangles follow by continuity. The same proof works on any star-shaped plane domain, including the whole plane.

Integration requires an orientation: take $M$ to be an oriented compact $n$-dimensional smooth manifold without boundary. The Generalized Stokes theorem says ${\int }_{M}d\beta =0$ for every smooth $(n-1)$-form $\beta$. Every $n$-form is closed by dimension, and changing it by an exact differential form leaves its integral unchanged. Thusis a well-defined linear map $H_{\mathrm{dR}}^n(M;\mathbb{R})\to \mathbb{R}$. Compactness alone does not supply the orientation omitted from the printed integration statement.

For an $n$-dimensional Lie group, choose a nonzero $\lambda \in {\Lambda }^n{T}_e^{\ast }G$ and define the invariant volume form on a Lie group byEach $d{L}_{g^{-1}}$ is an isomorphism, so $\omega$ is nowhere zero and defines an orientation. The composition identity $L_{(hg{)}^{-1}}\circ L_h=L_{g^{-1}}$ gives $L_h^{\ast }\omega =\omega$. Moreover, every left-invariant differential form of degree $n$ is determined by its value at $e$, so the space of such forms is one-dimensional.

Under the supplied invariant-representative assumption, all top-degree de Rham cohomology classes are multiples of $[\omega ]$. For compact $G$, orient it so $\omega$ is positive; then ${\int }_G\omega >0$, whereas the Generalized Stokes theorem makes every exact top form have integral zero. Therefore $[\omega ]\ne 0$ andunder that assumption. For actual compact Lie groups, this argument applies when $G$ is connected. The supplied assumption fails for disconnected groups: on $S^1\times \mathbb{Z}/2$, the two components have independent degree-one classes, while globally left-invariant top forms form only a one-dimensional space. In general the top-degree de Rham cohomology of a compact Lie group is ${\mathbb{R}}^c$, where $c$ is its number of connected components.

For $S^2$, use $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with the plane. The disc argument gives $\eta {|}_U=d{f}_U$ and $\eta {|}_V=d{f}_V$ for any closed differential one-form $\eta$. The overlap $U\cap V$ is connected, so $d(f_U-f_V)=0$ makes the difference a constant. Adjust one potential by that constant; they glue to a global smooth function $f$ with $df=\eta$. ThusFinally, $S^3$ is a compact connected Lie group: the identificationidentifies it smoothly with the special unitary group $SU(2)$. Matrix multiplication and inversion supply its smooth group operations. The preceding invariant-volume argument therefore gives

For $u_{z\bar{z}}={\beta }^{2}u$, pull back its spectral closed differential one-form along each straight side $z=m_j+s{h}_j$. Under counterclockwise traversal and outward normal derivatives $q_j$, the side density is $i{e}^{-i\beta (\lambda z-\bar{z}/\lambda )}[|h_j|q_j+\beta (\lambda h_j+{\bar{h}}_j/\lambda )g_j]$, where $g_j$ is the Dirichlet boundary data. The Generalized Stokes theorem gives the displayed global relation. Clockwise traversal reverses the normal-derivative coefficient; all orientations must be changed consistently.