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Let $\omega ={\theta }^{-1}(\alpha )$. The preceding invariant formula for the exterior derivative identifies the expression in the question withIts vanishing for every pair of global vector fields is equivalent to $d\omega =0$: global fields with arbitrary prescribed tangent values are available by bump-function extension. Thus it is exactly the condition that $\omega$ be a closed differential one-form.

By definition, the first de Rham cohomology is closed one-forms modulo exact differential forms. The hypothesis $H_{\mathrm{dR}}^1(M)=0$ makes every closed one-form exact. Hence $d\omega =0$ gives a globally smooth function $g$ with $\omega =dg$, andConversely, if $\alpha (X)=X(g)$, thenby the definition of the Lie bracket of vector fields. Equivalently, $d(dg)=0$. The bracket compatibility condition is necessary and, when $H_{\mathrm{dR}}^1(M)=0$, sufficient for a global potential. The potential is unique up to a constant on each connected component, since the difference of two potentials has zero differential.

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.

Treat the printed $e^{\mu }$ as an orthonormal coframe, with frame metric $\eta =\mathrm{diag}(-1,1,1,1)$. Work on a patch $z\ne 0$ and $F=1-\alpha z^3>0$, so the given real coframe exists. Define $f=\sqrt{F}$ and $A=z{f}^{\prime }-f$. The exterior derivatives areThe connection 1-forms satisfying Cartan's first structure equation and metric compatibility areAll other forms vanish. With a Lorentzian frame, it is the lowered forms ${\omega }_{ab}={\eta }_{ac}{\omega }^c{}_b$ that are antisymmetric; the two mixed time–space forms are equal. These displayed forms solve the torsion-free structure equation, and uniqueness of the Levi-Civita connection identifies them as the required connection.

Apply Cartan's second structure equation. For example,The remaining derivatives give ${\Theta }^1{}_3=Af{e}^1\wedge e^3$ and ${\Theta }^1{}_2=-f^2{e}^1\wedge e^2$, with analogous forms for index 2. Put $q=\alpha z^3$ and $B=1+q/2$. From $f^2=1-q$,Therefore all six independent curvature 2-forms areThe other six are fixed by ${\Theta }^b{}_a=-{\eta }_{aa}{\eta }_{bb}{\Theta }^a{}_b$ for $a\ne b$, with no sum: equal for time–space pairs and opposite for spatial pairs. All diagonal forms are zero. The Lorentzian connection-form antisymmetry is essential to these signs.

The exterior derivative is an $\mathbb{R}$-linear map $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$, agrees with the differential $df$ of a smooth function, satisfies the graded Leibniz rule $d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^p\alpha \wedge d\beta$ for $\alpha \in {\Omega }^p(M)$, and has $d^2=0$. These properties determine it locally and hence globally. To see locality directly, if a differential form $\alpha$ vanishes near $x$, choose a smooth bump function $\chi$ equal to one near $x$ and supported where $\alpha =0$. The identity $d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $(d\alpha {)}_x=0$. Thus global forms can be computed using local extensions in a manifold chart.

In coordinates write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. Since $d(d{x}^i)=d^2{x}^i=0$, the graded Leibniz rule forcesThis is the uniqueness of the exterior derivative from its axioms. The coordinate formula also establishes existence: it has the stated properties, and the chain rule shows that coordinate changes give the same operator.

The de Rham cohomology is the real quotient vector spaceThus a cohomology class records a closed differential form modulo an exact differential form. In degree zero there are no exact forms; closed functions are locally constant. The Poincaré lemma says that every closed form of positive degree on a star-shaped open subset of ${\mathbb{R}}^n$ is exact, and hence that positive-degree closed forms are locally exact on a smooth manifold.

For the first de Rham cohomology of the two-sphere, let $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with ${\mathbb{R}}^2$, so a closed one-form $\alpha$ has primitives $f_U,f_V$. On the connected overlap $U\cap V$, the derivative of $f_U-f_V$ is zero, so this difference is a constant. Subtracting that constant from $f_U$ makes the primitives agree. They glue to a global smooth primitive. Therefore $H_{\mathrm{dR}}^1(S^2)=0$.

Let $q:S^2\to {\mathbb{RP}}^2$ be the double covering map and $a$ the antipodal map. The pullback of a differential form identifies forms downstairs with $a$-invariant forms upstairs. Averaging $(\eta +a^{\ast }\eta )/2$ commutes with $d$. If an invariant form is exact upstairs, averaging its primitive proves it exact downstairs. Conversely an invariant cohomology class has an invariant representative by the same averaging. This proves the de Rham cohomology of a finite quotient identification $H_{\mathrm{dR}}^p({\mathbb{RP}}^2)\cong H_{\mathrm{dR}}^p(S^2{)}^{a^{\ast }}$. In degree zero the sphere is connected and $a^{\ast }$ fixes constants; degree one is zero; in degree two the granted action is multiplication by $-1$, whose invariant real subspace is zero. Forms of degree greater than two vanish. HenceThis is real de Rham cohomology; it does not detect the integral two-torsion of the real projective plane.

Let $P=({\omega }^{ab})$ be a smooth antisymmetric matrix field. It defines a Poisson bivector throughThis Poisson bracket is bilinear, antisymmetric and a derivation in each argument. It defines a Poisson manifold when it also satisfies the Jacobi identity. Apply that identity in the form $\{\{f,g\},h\}+\{\{g,h\},f\}+\{\{h,f\},g\}=0$ to the coordinate functions. Since $\{x^a,x^b\}={\omega }^{ab}$, it givesThis is the coordinate Jacobi condition for a Poisson bivector. It is also sufficient: in the Jacobi identity for arbitrary functions the terms containing second derivatives cancel by antisymmetry, leaving $J^{abc}({\partial }_{a}f)({\partial }_{b}g)({\partial }_{c}h)$. Thus the coordinate condition captures the whole obstruction.

Now suppose the Poisson bivector is nondegenerate. Write $S=P^{-1}$, so $S_{ab}{\omega }^{bc}={\delta }_a^c$, and define the 2-formAn overall minus sign in identifying the symplectic form depends on the convention for Hamiltonian vector fields; it does not affect the closure argument. Differentiating the inverse matrix givesContract the coordinate Jacobi identity with $S_{ia}{S}_{jb}{S}_{kc}$. Antisymmetry gives ${\omega }^{dc}{S}_{kc}=-{\delta }_k^d$, and similarly for the other terms, henceThe cyclic expression is precisely the coefficient of the exterior derivative $d\Omega$. ThereforeSince $S$ is antisymmetric and nondegenerate, $\Omega$ is a symplectic form. This proves the closure of the inverse of a nondegenerate Poisson bivector. The matrix entries ${\omega }^{ab}$ are scalar functions; the codomain in the source's matrix description should be read as the space of antisymmetric matrices, rather than a vector-valued individual entry.

A Chern number turns local curvature data into a global integer. Let $E\to M$ be a complex vector bundle over a compact oriented manifold without boundary, with a unitary connection represented locally by an anti-Hermitian matrix-valued one-form $A$. Its curvature form of a connection is $F=dA+A\wedge A$. The Chern-Weil theory representative of the total Chern class isThe coefficients are closed differential forms. They represent the images of integral Chern classes in de Rham cohomology. On an oriented $2r$-manifold, a product $c_{i_1}\cdots c_{i_s}$ with $i_1+\cdots +i_s=r$ pairs with the fundamental class to give a Chern number, an integer independent of the unitary connection.

For example,For an $SU(N)$ bundle with $N\ge 2$, $\mathrm{Tr}F=0$. On an oriented four-manifold the Second Chern number is thenusing the fundamental matrix trace and the stated anti-Hermitian convention. The trace-product correction is required for a general $U(N)$ bundle; it cannot simply be omitted. The sign of a physical instanton number also depends on the trace and orientation convention, so these conventions must accompany the formula.

The local origin of closure is the Bianchi identity $D_{A}F=0$ and the cyclic matrix trace: $d\mathrm{Tr}(F\wedge F)=0$. Independence of the connection one-form follows more concretely by varying a family $A_t$. Since ${\dot{F}}_t=D_{A_t}{\dot{A}}_t$, the Chern-Weil connection transgression isIts integral on a closed four-manifold is zero by the Generalized Stokes theorem. This proves connection independence of $q$; the integrality is the global Chern class statement, not merely a consequence of the local formula.

The local primitive of the Second Chern form is the Chern-Simons three-formHere we continue to use $SU(N)$, so $c_2=\mathrm{Tr}(F\wedge F)/(8{\pi }^2)$. The cubic coefficient is forced by the exterior derivative. In the graded cyclic matrix trace, $d\mathrm{Tr}(A^3)=3\mathrm{Tr}(dA\wedge A^2)$ and $\mathrm{Tr}(A^4)=0$, the latter because cycling one degree-one factor past the other three changes its sign. ThereforeMatrix-valued differential forms require both matrix order and the graded signs; treating all factors as commuting scalars would lose this derivation.

The Chern-Simons three-form depends on a local trivialization and is not itself gauge-invariant. For the Yang-Mills gauge transformation convention $A^g=g^{-1}Ag+g^{-1}dg$, set $u=g^{-1}dg$. The gauge change of the Chern-Simons three-form isThe last term is closed by the Maurer-Cartan equation. On a closed three-manifold its integral is an integer with the fundamental $SU(N)$ normalization. Consequently the Chern-Simons integral is naturally defined modulo integers, while its exponential $\exp (2\pi i\ell \int Y)$ is invariant under large Yang-Mills gauge transformations for integer level $\ell$.

This also explains why a nonzero Chern number is compatible with $dY=c_2$: $Y$ need not be a globally defined three-form. On $S^4$, trivialize over two hemispheres and let $A_N=A_S^g$ on their common equator $\Sigma =S^3$, oriented as the boundary of the northern hemisphere. The Generalized Stokes theorem and the gauge-change formula giveFor $SU(2)\cong S^3$, this is the degree of the transition map, with compatible group orientation; for $SU(N)$ it is the corresponding integer in ${\pi }_3(SU(N))$. Thus the Second Chern number measures the obstruction to choosing one trivialization over the whole four-sphere.

A simpler First Chern class example is a line bundle over $S^2$. Write $A=-ia$ and choose local real potentialsThey have common curvature $da=(k/2)\sin \theta d\theta \wedge d\varphi$, so ${\int }_{S^2}{c}_1=(2\pi {)}^{-1}\int da=k$. Their difference $a_N-a_S=kd\varphi$ corresponds to the transition function $e^{-ik\varphi }$, which is single-valued exactly when $k\in \mathbb{Z}$. This illustrates how the global integer arises from patching, rather than from an arbitrary flux normalization.

In physics, these constructions distinguish topological sectors of Yang-Mills instantons and relate four-dimensional characteristic densities to three-dimensional boundary actions. The Chern-Simons three-form itself gives a metric-independent gauge action in three dimensions. On a closed manifold its first variation isso its classical equation is $F=0$. The common thread is that a local expression in the connection one-form records global topology: curvature produces the invariant Chern number, while its local Chern-Simons three-form primitive retains gauge and boundary information.

Put $E_{\lambda }=e^{-i\beta (\lambda z-\bar{z}/\lambda )}$. The Wirtinger derivatives give $(E_{\lambda }{)}_z=-i\beta \lambda E_{\lambda }$ and $(E_{\lambda }{)}_{\bar{z}}=i\beta E_{\lambda }/\lambda$. Write $W=Adz+Bd\bar{z}$, whereUsing the exterior derivative, $dW=(B_z-A_{\bar{z}})dz\wedge d\bar{z}$. The terms involving $u_z$ and $u_{\bar{z}}$ cancel, leavingBecause $E_{\lambda }$ never vanishes for $\lambda \ne 0$, $dW=0$ if and only if $u_{z\bar{z}}-{\beta }^{2}u=0$. In real coordinates this is the modified Helmholtz equation with mass parameter $2\beta$, not $\beta$. In fact any one nonzero spectral parameter suffices for the equivalence; the full family supplies many independent boundary tests.

A tangent vector at $p$ is a derivation at a point: a real-linear map $v:C_p^{\infty }(M)\to \mathbb{R}$ on germs of smooth functions satisfyingAddition and scalar multiplication preserve this identity, defining the tangent space $T_{p}M$ as a vector space. A derivation annihilates constants. In a manifold chart $x=(x^1,\dots ,x^n)$ centered near $p$, the local factorization $f-f(p)={\sum }_i(x^i-x^i(p))f_i$, with $f_i(p)={\partial }_{i}f(p)$, gives $v(f)={\sum }_{i}v(x^i){\partial }_{i}f(p)$. The factorization follows by integrating the first derivatives along a straight segment in a sufficiently small coordinate ball. ThereforeConversely these coordinate derivatives satisfy the derivation identity, so they really give a basis rather than merely a spanning family.

The cotangent space is the dual vector space $T_p^{\ast }M=(T_{p}M{)}^{\ast }$. Its basis $d{x}^i{|}_p$ is characterized by $d{x}^i({\partial }_j)={\delta }_j^i$. If $\alpha =a_{i}d{x}^i=b_{j}d{y}^j$, the chain rule gives the cotangent coordinate transitionThe cotangent bundle is the disjoint union $T^{\ast }M={\coprod }_{p\in M}{T}_p^{\ast }M$, with projection $\pi (p,\alpha )=p$. For each base manifold chart $(U,x)$, defineIts inverse sends $(z,a)$ to $(x^{-1}(z),a_{i}d{x}^i{|}_{x^{-1}(z)})$. On overlaps the transition is $(x,a)\mapsto (y(x),b)$, with the displayed fibre-linear transformation. Its coefficients and inverse are smooth, so these maps give a smooth atlas in dimension $2n$. Give the total space the topology obtained by transporting the product topology through these charts. Different base points are separated by disjoint base neighborhoods; distinct points over the same base point are separated within one bundle chart. A countable base smooth atlas and countable product bases give second countability. Thus the total space is a Hausdorff, second-countable smooth manifold. The same charts are vector bundle trivializations, since $\pi$ is product projection and their fibre changes are invertible linear maps.

The intrinsic canonical one-form on a cotangent bundle isIn the bundle chart it is $\lambda ={\sum }_i{a}_{i}d{x}^i$. This definition is coordinate independent, hence its exterior derivative is the globally defined smooth differential two-formThis sign follows the fibre-first order in this problem; the equally common position-first symplectic form is $-d\lambda$. The volume formnever vanishes. A smooth manifold is orientable exactly when it admits a nowhere-zero top-degree form; its positive ordered bases determine a consistent orientation. Consequently $T^{\ast }M$ is an orientable smooth manifold, even when $M$ is not. This is cotangent bundle orientation. In dimension zero the same conclusion uses the nowhere-zero zero-form $1$.

Use real differential forms and the usual convention that a smooth manifold has no boundary unless one is specified. On a noncompact manifold the formal-adjoint identity below uses compactly supported forms. On a compact manifold it has no boundary contribution; if a boundary were allowed, extra boundary conditions would be necessary.

The Riemannian metric gives the dual inner product on $T_x^{\ast }M$ and hence the inner product on exterior powers of a cotangent space:Extend this bilinearly. Equivalently, the increasing wedge products of an orthonormal coframe are an orthonormal basis of ${\Lambda }^p{T}_x^{\ast }M$. The Hodge star operator is the unique map $\ast :{\Lambda }^p{T}_x^{\ast }M\to {\Lambda }^{n-p}{T}_x^{\ast }M$ satisfying $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$. If $e^1,\dots ,e^n$ is a positively oriented orthonormal coframe, $\ast e^I=\epsilon (I,I^c)e^{I^c}$. Moving the $p$ factors of $I$ across the $n-p$ factors of its complement changes the sign by $(-1{)}^{p(n-p)}$, soThe basis formula also proves that $\ast$ is an isometry.

For $\beta \in {\Omega }^{p-1}(M)$ and $\alpha \in {\Omega }^p(M)$ with compactly supported product, the Stokes theorem and the Leibniz rule for the exterior derivative giveThus $(d\beta ,\alpha {)}_{L^2}=(-1{)}^p{\int }_M\beta \wedge d\ast \alpha$. The formal adjoint operator $\delta$ satisfies $\ast \delta \alpha =(-1{)}^{p}d\ast \alpha$. Applying the inverse star and its square on degree $p-1$ gives the codifferentialThe exponent is congruent to $p+(p-1)(n-p+1)$ modulo two, so this convention is consistent in every degree. Define the Hodge Laplacian by $\Delta =d\delta +\delta d$ and a harmonic differential form by $\Delta \alpha =0$. On a compact manifold,Both nonnegative terms vanish for a harmonic form, giving $d\alpha =\delta \alpha =0$. Conversely these two equations imply harmonicity.

Now let $n=4$. The printed eigenvalue relation omits the subscript on its right-hand side; the intended equation is $\ast {\alpha }_{\pm }=\pm {\alpha }_{\pm }$. On two-forms ${\ast }^2=1$, soThese are respectively self-dual two-forms and anti-self-dual two-forms, and sum to $\alpha$. Their uniqueness follows because the $+1$ and $-1$ eigenspaces have zero intersection. Star is an orthogonal involution here, hence is self-adjoint and its opposite eigenspaces are orthogonal.

For closed two-forms define $Q([\alpha ],[\beta ])={\int }_N\alpha \wedge \beta$. Adding $d\xi$ to $\alpha$ changes the integral by ${\int }_{N}d(\xi \wedge \beta )=0$; adding an exact form to $\beta$ has the same effect. This is the real de Rham intersection form in dimension four. The wedge product of differential forms in degrees two and two is symmetric because $(-1{)}^{2\cdot 2}=1$; hence $Q$ is a symmetric bilinear form on $H_{\mathrm{dR}}^2(N;\mathbb{R})$.

In four dimensions $\delta =-\ast d\ast$ in every degree. If $h$ is a harmonic two-form, $\delta h=0$ gives $d\ast h=0$, and $dh=0$ gives $\delta \ast h=-\ast dh=0$. Thus $\ast h$ is harmonic, and ${\mathcal{H}}^2={\mathcal{H}}^{+}\oplus {\mathcal{H}}^{-}$ is the harmonic star-eigenspace decomposition. Use the permitted unique harmonic representative for each de Rham cohomology class. For harmonic $h,k$,The mixed terms vanish by orthogonality. The first restriction is positive definite and the second negative definite. In particular a nonzero harmonic $h$ has $Q(h,\ast h)=\Vert h{\Vert }_{L^2}^2>0$, proving nondegeneracy. Hence the signature of the intersection form from harmonic duality isThe finite dimensions and the harmonic description also follow from the Hodge decomposition theorem.

For completeness, the Hodge decomposition theorem on a compact oriented Riemannian manifold without boundary states the finite-dimensional harmonic space and the $L^2$-orthogonal direct sumThe three summand components are unique, although their potentials need not be. Given an exact three-form $\beta =d\omega$ on $N$, decompose its smooth two-form potential as $\omega =h+d\xi +\delta \psi$, with $\psi \in {\Omega }^3$. Since $dh=0$ and $d^2=0$, $\beta =d\delta \psi$. Put $a=\delta \psi$. In dimension four, $a=-\ast d\ast \psi$, and ${\ast }^2=1$ on two-forms gives $\ast a=-d\ast \psi$, so $d\ast a=0$. ConsequentlyThis constructs a self-dual primitive of an exact three-form. The sum contains no factor $1/2$: it is twice the self-dual projection of the coexact potential. Using the unscaled projection alone would give $d\sigma =\beta /2$. This proves the final unheaded request.

We use complex-valued smooth differential forms. The sheaf of smooth differential forms is ${\mathcal{A}}^k(U)=\Gamma (U,{\bigwedge }^k(T^{\ast }X\otimes \mathbb{C}))$, with the usual restriction maps. Dualizing the type decomposition of the complexified tangent bundle and taking exterior powers decomposes this bundle into the summandsTheir smooth sections form the sheaf of differential forms of type (p, q) ${\mathcal{A}}^{p,q}$. Locally a section is a sum of $f_{I,J}d{z}_I\wedge d{\bar{z}}_J$ with $|I|=p$, $|J|=q$ and smooth coefficients. Holomorphic transition maps preserve types, so the local decompositions agree globally. HenceThe exterior derivative splits as $d=\partial +\bar{\partial }$, with bidegrees $(1,0)$ and $(0,1)$. Its square being zero gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$. The Dolbeault cohomology is thereforeForms in negative or out-of-range bidegrees are understood to be zero.

Use the standard meaning of static spacetime: a stationary spacetime chart whose time-space components also vanish. The printed static definition needs to inherit the timelike requirement $g_{00}<0$; without it, the requested timelike conclusion does not follow.

The preceding argument shows that $V={\partial }_0$ is a timelike Killing vector field. Its metric-dual one-form is $v=V_{\mu }d{x}^{\mu }=g_{00}d{x}^0$. Its exterior derivative satisfiesFor a torsion-free connection, antisymmetrizing the covariant derivative of a one-form gives its exterior derivative. Consequently the Killing vector has hypersurface orthogonality:Geometrically, the hypersurfaces $x^0=\mathrm{constant}$ are orthogonal to $V$.

For the spherical null expansions in double-null coordinates, the cross-sectional area of a symmetry sphere is $A=4\pi r^2$. For the two affine null geodesic congruences, the definition of null expansion therefore givesThere is no extra derivative of $f$ in these formulas: the null expansion differentiates the transverse area, not the normalization of the tangent. Equivalently, the projected metric is $q_{AB}=r^2{\gamma }_{AB}$ and its optical tensor is $(k(r)/r)q_{AB}$, or $(\ell (r)/r)q_{AB}$ for the other congruence. It is purely trace, so both null shears vanish.

The null twist vanishes because the covector tangents are gradients of $-U$ and $-V$, and hence their exterior derivatives vanish. Each congruence is orthogonal to a family of null hypersurfaces. This also identifies the Frobenius theorem reason for zero null twist.

For the Komar integral of a five-dimensional Reissner--Nordstrom metric, use the one-form metrically dual to the stationary Killing vector field: $k^{♭}=-f(r)dt$. Its exterior derivative is $d{k}^{♭}=f^{\prime }(r)dt\wedge dr$. The stated orientation gives volume formBecause the squared norm of $dt\wedge dr$ is $g^{tt}{g}^{rr}=-1$, the defining wedge identity for the Hodge star operator givesThus the Komar integral over the positively oriented angular three-sphere issince the unit three-sphere has volume $2{\pi }^2$. Write $s=r_{+}^2+r_{-}^2$, $q=r_{+}^2{r}_{-}^2$. Then $f=1-s/r^2+q/r^4$ and $r^3{f}^{\prime }=2s-4q/r^2$. With the normalization specified for the Komar mass,The orientation makes this positive. The problem's coefficient $1/(16\pi )$ must be retained: it is not the usual dimension-adjusted five-dimensional ADM-normalized Komar coefficient. With that conventional normalization the ADM mass would instead be $3\pi s/8$.

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.