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On a connected Lie group, a left-invariant differential 1-form is closed exactly when it is bi-invariant. Connectedness is essential: on $O(2)$ every left-invariant 1-form is closed because its Lie algebra is abelian, while reflection conjugation negates every nonzero one.

Choose a nonzero element of ${\Lambda }^n{T}_e^{\ast }G$ for an $n$-dimensional Lie group and transport it by inverse left translations. This gives a smooth nowhere-zero left-invariant differential form of top degree, hence an orientation. The identity $L_{(hg{)}^{-1}}{L}_h=L_{g^{-1}}$ proves invariance. The space of invariant top forms is one-dimensional, since evaluation at the identity identifies it with ${\Lambda }^n{T}_e^{\ast }G$.

The Maurer-Cartan form of the translation-dilation group of the plane has these coefficients in its dilation–translation basis, with $a=1,2$. Left multiplication rescales each coordinate differential and $\rho$ equally, proving they are left-invariant differential forms. Their dual left-invariant vector fields are $\rho {\partial }_{\rho }$ and $\rho {\partial }_{x^a}$. The Maurer-Cartan equation is $d{\lambda }^0=0$, $d{\lambda }^a=-{\lambda }^0\wedge {\lambda }^a$. The tensor ${\sum }_j{\lambda }^j\otimes {\lambda }^j$ is a positive definite left-invariant metric.

In the faithful matrices $g=\left(\begin{array}{cc}a& b\\ 0& 1\end{array}\right)$ with $a>0$, the Maurer-Cartan form has entries $da/a$ and $db/a$. Thus the left-invariant differential forms ${\sigma }^D=da/a$, ${\sigma }^T=db/a$ obey $d{\sigma }^D=0$ and $d{\sigma }^T=-{\sigma }^D\wedge {\sigma }^T$. The Lie algebra generators satisfy $[D,T]=T$.

Use the index convention $[e_a,e_c]=C_a{}^b{}_c{e}_b$. For the left- and right-invariant coframes, the Maurer-Cartan equations areTheir dual invariant vector fields satisfyFor example, evaluating $d{\lambda }^b$ on two dual fields gives $-{\lambda }^b([L_a,L_c])$, since its invariant coefficient functions are constant. The opposite right-invariant bracket gives the other sign. The mixed bracket vanishes because left and right translations commute.

For a semisimple Lie algebra, the Killing form $B(X,Y)=\mathrm{Tr}(\mathrm{ad}X\mathrm{ad}Y)$ is nondegenerate. It is adjoint invariant: using $\mathrm{ad}[Z,X]=[\mathrm{ad}Z,\mathrm{ad}X]$ and cyclicity of trace givesGroup conjugation preserves the trace as well, so this is invariant under the full group adjoint action. Translation supplies a bi-invariant pseudo-Riemannian metric.

Lower the indicated structure index by $C_{abc}=B_{be}{C}_a{}^e{}_c$. ThenThis changes sign under exchange of $a,c$ by the bracket and under exchange of $a,b$ by invariance. Hence $C_{abc}=C_{[abc]}$.

Let $\eta (X,Y,Z)=B(X,[Y,Z])$ at the identity and extend it as a left-invariant differential form. Its value is adjoint invariant, so the extended form is also right-invariant. The three-form in the question is $-\eta$ with the index-lowering convention just specified, an irrelevant overall sign for these properties. To prove closedness rather than only invariance, apply the exterior derivative on four left-invariant fields. Its six bracket terms combine by invariance intoThe final equality is the Jacobi identity. Therefore the Cartan three-formis both bi-invariant and closed.

Differentiate the product with the first factor close to the identity. A small displacement of its $x^{\prime }$ coordinate adds to $x$; a small $y^{\prime }$ displacement adds to $y$; and a small $z^{\prime }$ displacement changes $(x,y,z)$ by $(-x,y,1)$ times that displacement. The infinitesimal left actions therefore give the right-invariant vector fieldsFor right multiplication, regard $(x^{\prime },y^{\prime },z^{\prime })$ as the fixed point and the second factor as infinitesimal. Its displacements are $(e^{-z^{\prime }}\delta x,e^{z^{\prime }}\delta y,\delta z)$; relabeling the coordinates of the fixed point gives the left-invariant vector fieldsThe naming distinction matters: generators of left multiplication are right invariant, whereas generators of right multiplication are left invariant.

All mixed Lie brackets of vector fields vanish. For example, $R_3(e^{-z})=-e^{-z}$ cancels $L_1(-x)=-e^{-z}$ in $[L_1,R_3]$, and $R_3(e^z)=e^z$ cancels $L_2(y)=e^z$ in $[L_2,R_3]$. The brackets with $R_1,R_2$ are zero because the coefficients of the $L_i$ have no $x$ or $y$ dependence; $[L_3,R_3]=0$ because the coefficients of $R_3$ have no $z$ dependence. HenceConceptually, the corresponding flows commute because $(ag)b=a(gb)$, by associativity of the Lie group multiplication.

The dual coframe iswhich satisfies ${\theta }^i(L_j)={\delta }_j^i$. Under left multiplication by $(x_0,y_0,z_0)$, the transformed coordinates are $X=x_0+e^{-z_0}x$, $Y=y_0+e^{z_0}y$, $Z=z_0+z$. Therefore $e^{Z}dX=e^{z}dx$, $e^{-Z}dY=e^{-z}dy$, and $dZ=dz$. Each one-form is a left-invariant differential form, so their sum of squares is the left-invariant metricThis is the standard metric of the Sol Lie group. As a sign check, $[L_3,L_1]=-L_1$, $[L_3,L_2]=L_2$, while $d{\theta }^1={\theta }^3\wedge {\theta }^1$ and $d{\theta }^2=-{\theta }^3\wedge {\theta }^2$, in agreement with the Maurer-Cartan equation.

Let $e$ be the identity. For a left-invariant differential form $\omega$ and a left-invariant vector field $X$, one hasThe value is therefore independent of $g$, even if the Lie group is disconnected. The same calculation applies to another left-invariant field $Y$.

The standard exterior derivative of a one-form evaluated on vector fields isBoth differentiated functions are constant in the present situation. ThusThis identity is a direct consequence of left invariance and the stated general formula for the exterior derivative.

The Maurer-Cartan form is $g^{-1}dg={\lambda }^j{X}_j$. Direct matrix multiplication givesThese are left-invariant differential forms: for a constant group element $a$, $(ag{)}^{-1}d(ag)=g^{-1}dg$. Equivalently, left translation sends $(\rho ,x)$ to $(a\rho ,b+ax)$, so all three coordinate differentials and the denominator acquire the same factor $a>0$.

The left-invariant coframe of the translation-dilation group is pointwise linearly independent because $\rho >0$. Consequentlyis positive definite and unchanged under every left translation. It therefore defines the required left-invariant metric, with precisely the coordinate expression obtained from this sum. The notation $d{\rho }^2$ in that metric means $d\rho \otimes d\rho$, not the differential of ${\rho }^2$.

Taking the exterior derivative of the left-invariant differential forms yieldsFor example, $d({\rho }^{-1}d{x}^1)=-{\rho }^{-2}d\rho \wedge d{x}^1$. With the antisymmetric structure constants of a Lie algebra found above, the two terms with indices $(0,1)$ and $(1,0)$ combine asThe same computation holds for superscript $2$, and all superscript-$0$ terms vanish. Hence the Maurer-Cartan equation isThe factor $1/2$ avoids counting each antisymmetric pair twice. If coefficients were not required to be antisymmetric in $j,k$, their symmetric parts would be invisible to the wedge product of differential forms; the standard structure constants of a Lie algebra fix this ambiguity.

For a Matrix Lie group $G$, the left Maurer-Cartan form identifies tangent vectors with the Lie algebra by translating them to the identity:It is a Lie algebra-valued differential form of degree one. For a fixed $h\in G$, replacing $g$ by $hg$ gives $(hg{)}^{-1}d(hg)=g^{-1}dg$, so it is a left-invariant differential form.

Differentiate $g^{-1}g=I$ to get $d(g^{-1})=-g^{-1}(dg)g^{-1}$. The exterior derivative then givesHere the exterior product of matrix-valued differential forms includes matrix multiplication; the order of the matrices matters. Hence the Maurer-Cartan equation is

Write $\rho ={\sigma }^{\alpha }{T}_{\alpha }$. Antisymmetry of the exterior product impliesComparison of the Lie algebra components yields the Maurer-Cartan equation in a Lie-algebra basis:Thus the canonical antisymmetric choice is $f^{\gamma }{}_{\alpha \beta }=-\frac{1}{2}{c}^{\gamma }{}_{\alpha \beta }$. The factor one half is required because the sum includes both ordered pairs $(\alpha ,\beta )$ and $(\beta ,\alpha )$. If only $\alpha <\beta$ is summed, its coefficient is $-c^{\gamma }{}_{\alpha \beta }$. Strictly, the equality of differential forms determines only the antisymmetric part of $f$: one may add any tensor symmetric in $\alpha ,\beta$ without changing it. This is the symmetric-part ambiguity in Maurer-Cartan coefficients.

A faithful matrix representation of the orientation-preserving real affine group isIts action on $(x,1{)}^T$ has first component $ax+b$. The group operation and inverse areDifferent transformations have different matrix entries, proving faithfulness. With the Lie algebra basisthe left Maurer-Cartan form isThe requested left-invariant differential forms therefore form the Maurer-Cartan coframe of the real affine group:For a direct invariance check, left translation by $(a_0,b_0)$ sends $(a,b)$ to $(a_{0}a,b_0+a_{0}b)$, and the pullbacks of the two displayed differential forms are unchanged. Their dual left-invariant vector fields are $a{\partial }_a$ and $a{\partial }_b$, whose bracket is $a{\partial }_b$, in agreement with $[D,T]=T$.

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

Suppose first that the left-invariant 1-form $\alpha$ is closed. The scalar function $\alpha (l_{\xi })$ is constant for every $\xi$, because both the form and vector field are left-invariant. Cartan's magic formula therefore givesPart b says that $g\mapsto R_g^{\ast }\alpha$ is locally constant. Since $G$ is connected, it is constant, and its value at the identity is $\alpha$. Hence $R_g^{\ast }\alpha =\alpha$ for every $g$, so $\alpha$ is bi-invariant.

Conversely, if $\alpha$ is bi-invariant, then all these Lie derivatives vanish. Cartan's formula and the constancy of $\alpha (l_{\xi })$ give ${\iota }_{l_{\xi }}d\alpha =0$. The left-invariant vector fields span every tangent space, so $d\alpha =0$.

Connectedness cannot be omitted. The orthogonal group $O(2)$ has an abelian Lie algebra, so every left-invariant 1-form is closed by the Maurer-Cartan equation in a Lie-algebra basis. Conjugation by a reflection acts as $-1$ on its one-dimensional Lie algebra, so a nonzero left-invariant 1-form is not right-invariant. This is the standard obstruction recorded by the closed left-invariant 1-form criterion.