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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 $[D,T]=T$.

The product of the magnetic field with the velocity in this question is the three-dimensional cross product. Fix ${ϵ}_{123}=1$ and write $K^i{}_j={ϵ}^i{}_{kj}{B}^k$, so $K\mathbf{v}=\mathbf{B}\times \mathbf{v}$. In coordinates $y^0=t$, $y^i=x^i$ on ${\mathbb{R}}^3\times \mathbb{R}$, prescribe an affine connection by the following Christoffel symbols:with all other Christoffel symbols zero. These are smooth globally in the given Cartesian coordinates. The lower-index symmetry makes this a torsion-free connection. It is the geometrization of the Lorentz force by an affine connection; no condition on the curl of $\mathbf{E}$ or the divergence of $\mathbf{B}$ is required for this construction.

An affinely parametrized geodesic, with primes denoting differentiation with respect to $\lambda$, satisfiesOn the branch $t^{\prime }=c\ne 0$, use $t$ itself as an affine parameter. Dividing the spatial equation by $c^2$ givesConversely, each solution of this equation makes $\lambda =t$, $y(t)=(t,\mathbf{x}(t))$ an affinely parametrized geodesic of the constructed affine connection. Therefore its image is also an unparametrized geodesic. A general change of parameter adds a term proportional to the tangent in the geodesic equation, leaving the curve unchanged. The branch $t^{\prime }=0$ is not a trajectory with time as parameter.

For the metric realization, assume $\mathbf{E}=-\nabla U$ and $\nabla \cdot \mathbf{B}=0$. Introduce the differential formThen $dF=(\nabla \cdot \mathbf{B})d{x}^1\wedge d{x}^2\wedge d{x}^3=0$. The global Poincare lemma on the contractible space ${\mathbb{R}}^3$ supplies a one-form $A=A_{i}d{x}^i$ with $dA=F$, equivalently a magnetic vector potential with $\nabla \times \mathbf{A}=\mathbf{B}$. An explicit radial-gauge potential for a divergence-free magnetic field isThis is the radial homotopy formula for a closed two-form and is smooth even at the origin. For a constant magnetic field it gives $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.

Consider the Eisenhart-Duval lift with the Lorentzian metricThe independent one-forms $d{x}^1,d{x}^2,d{x}^3,dt,du-2A_{i}d{x}^i-Udt$ exhibit three positive directions and a two-dimensional block with one positive and one negative direction. Thus the metric is nondegenerate, of signature $(4,1)$. Its coefficients do not depend on $u$, and $g_{uu}=0$, so $\xi ={\partial }_u$ is a null Killing vector field. Indeed $g_{Au}$ are constant, so ${\Gamma }^A{}_{Bu}=0$ for the Levi-Civita connection and $\xi$ is parallel.

The geodesic Lagrangian for an affine parameter $\lambda$ isThe cyclic coordinate $u$ gives the conserved quantity $p_u=\partial L/\partial u^{\prime }=t^{\prime }$. Work at a nonzero value of $p_u$ and rescale the affine parameter to set $t^{\prime }=1$. This is the essential step in the null Kaluza-Klein reduction of a stationary force: one fixes the momentum along the null isometry and projects its geodesics, rather than dividing by $g_{uu}$.

The spatial Euler-Lagrange equations, before setting $t^{\prime }=1$, areSince $t^{\prime \prime }=0$ and $F_{ij}{v}^j=-(\mathbf{B}\times \mathbf{v}{)}_i$, their reduction isEquivalently, after taking $t$ as the affine parameter, the term $\dot{u}$ is a total derivative and the reduced Lagrangian is $L_{\mathrm{red}}=\frac{1}{2}|\dot{\mathbf{x}}{|}^2-2\mathbf{A}\cdot \dot{\mathbf{x}}-U$. Its Euler-Lagrange equations give the same sign and factor two.

There is also an explicit converse using null geodesics. For any physical trajectory, setThis makes its five-dimensional tangent null. The conserved momentum of the cyclic coordinate $t$ becomesThe quantity in parentheses is conserved, because its derivative is $\dot{\mathbf{x}}\cdot (2\mathbf{B}\times \dot{\mathbf{x}})=0$. Hence the $t$ equation, as well as the spatial and $u$ Euler-Lagrange equations, is satisfied. Every trajectory admits a null geodesic lift with $p_u=1$, and every such lift projects to the required trajectory.

Finally, a time-independent gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ is absorbed by $u\mapsto u+2\chi$. The one-form $du-2A_{i}d{x}^i-Udt$ and the five-dimensional metric are unchanged. These gauge transformations of an Eisenhart-Duval lift alter the reduced Lagrangian only by the total derivative $-2d\chi /dt$.

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

Fix a reference path ${\gamma }_{\ast }$ from $L_0$ to $L_1$. By path connectedness, a path $\gamma$ can be joined to it by a smooth homotopy strip $u:[0,1{]}_s\times [0,1{]}_t\to M$ withSmooth homotopies may be used by the relative smoothing theorem; piecewise smooth homotopies give the same integrals. Orient the square by $ds\wedge dt$ and define the Lagrangian path-area functional byHere $\mathcal{A}$ denotes the functional called $A$ in the question. The minus sign is the convention for which downward gradient flow is $J$-holomorphic on the standard oriented strip.

For relative symplectic-area independence, glue two choices, with one orientation reversed, along their common reference and final paths. The resulting oriented surface is a relative $2$-cycle $Z$ in $(M,L_0\cup L_1)$. Since $H_2(M,L_0\cup L_1)=0$, it bounds a relative $3$-chain: as an absolute chain, $Z=\partial B+C$ for a chain $C$ supported in $L_0\cup L_1$. The symplectic form is closed, and it restricts to zero on each Lagrangian submanifold. Using piecewise smooth relative chains, Stokes theorem givesEquivalently, integration of the closed symplectic form defines the zero pairing on the zero relative homology group. The two strips therefore define the same value of $\mathcal{A}$.

Choosing a different reference path changes every value by the negative area of one fixed connecting strip, independent of $\gamma$. Changing the initially assigned reference value also adds a constant. Path connectedness ensures that this is one global constant, rather than an independent constant on each component.

For the first variation, let $\xi (t)$ be a variation vector field along $\gamma$, with $\xi (0)\in T_{\gamma (0)}{L}_0$ and $\xi (1)\in T_{\gamma (1)}{L}_1$. Differentiate the defining strip integral. Stokes theorem, or the variation formula for the integral of a closed differential form, reduces it to the final edge; the side edges give zero because their variations and tangent vectors lie in the Lagrangian submanifolds. With the chosen sign,If this vanishes for every admissible $\xi$, it in particular vanishes for every field supported in $(0,1)$. Nondegeneracy of the symplectic form, together with the fundamental lemma of the calculus of variations, forces $\dot{\gamma }(t)=0$ in the interior and hence everywhere by smoothness. Conversely, a constant path makes the displayed integral zero. The constant must lie in both endpoint submanifolds. ThusNo transverse-intersection assumption is required; the critical set can be empty or have positive dimension.

A compatible almost complex structure is a smooth bundle map $J:TM\to TM$ such thatThese conditions makea Riemannian metric. Its symmetry follows from the $J$-invariance and skew symmetry of $\omega$, and its positivity is the final compatibility condition. It also satisfies $g(Jv,w)=\omega (v,w)$.

Use the formal $L^2$ metric on the path space,The variation formula becomesThus a formal downward flow $s\mapsto u(s,\cdot )$ satisfiesPut the standard complex structure $j$ on the strip, with $j{\partial }_s={\partial }_t$. The J-holomorphic curve equation $du\circ j=J\circ du$ says $u_t=J{u}_s$, which is equivalent to the displayed flow equation because $J^2=-I$. The converse is identical: a smooth J-holomorphic curve on the strip with the specified Lagrangian boundary conditions gives a formal downward trajectory of $\mathcal{A}$.

As a sign check, along such a strip,The drop in $\mathcal{A}$ therefore equals the positive symplectic area swept out by the strip. The word formal matters: the $L^2$ calculation gives the interior equation and its Lagrangian boundary conditions; it does not assert that arbitrary smooth initial paths produce a well-posed ordinary flow on the smooth path space. In particular, the expression $J\dot{\gamma }$ need not satisfy the endpoint tangent restrictions for an arbitrary initial path. The claimed correspondence concerns smooth solutions of the strip equation.

Orient both spheres in the standard way and normalize their area forms to total area $4\pi$. The degree of a map between oriented manifolds can be obtained in two ways. For a regular value $y$, use the degree as a sum of local degrees:Each inverse image is isolated by the inverse function theorem, and compactness makes the set finite. The determinant is computed in consistently oriented local coordinates. A second method is spherical degree by area pullback:where the last formula represents $g$ as a unit vector in ${\mathbb{R}}^3$. The pullback of a differential form already contains the signed Jacobian determinant; no extra $\sin \theta$ is to be inserted in that last coordinate expression.

To relate the methods, replace $\omega$ by a smooth top-degree differential form with the same total integral supported in a small neighbourhood of a regular value. Two such top-degree forms with equal integral differ by an exact differential form on $S^2$, by its top-degree de Rham cohomology. Their pullbacks therefore have the same integral by Stokes theorem. Over the chosen neighbourhood, $g$ splits into local inverse branches; the change of variables formula makes the contribution of each branch its orientation sign times $4\pi$. Their sum is precisely the first formula. Thus the area integral is an integer and agrees with the signed inverse-image count.

For a nonconstant rational map, first use common-factor reduction of a rational map so $p$ and $q$ are coprime. Write $k=\max (\mathrm{deg}p,\mathrm{deg}q)$ for these reduced polynomials. A generic finite target value $w$ has inverse images at the roots of $p-wq$: avoiding exceptional values makes its degree $k$ and its roots simple. The fundamental theorem of algebra supplies $k$ roots. A holomorphic map has positive real Jacobian determinant $|R^{\prime }{|}^2$ at a regular point, so every local sign is $+1$. HenceThe source leaves coprimality implicit. In an unreduced representation the answer is $\max (\mathrm{deg}p,\mathrm{deg}q)-\mathrm{deg}\gcd (p,q)$, including degree zero for a constant reduced map. For example $(z^2-1)/(z-1)$ extends to $z+1$ and has degree one, although the unreduced maximum degree is two. Exceptional inverse images at infinity or multiple roots do not change the degree of a rational map of the Riemann sphere.

For the rational map approximation for Skyrmions, use stereographic projection $z=\tan (\theta /2)e^{i\phi }$ and the unit target vectorCombine this rational map with a radial profile to form a special unitary group field:where $\mathbf{\sigma }$ are the Pauli matrices. The endpoint values make $U(0)=-1$ independent of angle and $U(\infty )=1$. Appropriate radial behaviour gives an admissible finite-energy field configuration. With $L_i=U^{†}{\partial }_{i}U$, choose the topological baryon number in the Skyrme model conventionSeparating the radial and angular factors givesThus the degree of a rational map of the Riemann sphere supplies the Skyrmion charge.

In conventional dimensionless massless Skyrme model units, its static energy reduces towith the angular Jacobian of a rational mapThe Cauchy-Schwarz inequality gives $\mathcal{I}\ge k^2$. These formulas follow from the radial strain $|f^{\prime }|$ and the two equal angular strains $\sin f\sqrt{J_R}/r$: the quadratic energy sums their squares and the quartic Skyrme term sums their pairwise products of squares. Minimize the angular integral in the rational map approximation over degree-$k$ maps, then minimize the remaining radial energy with the stated endpoints. This replaces a three-dimensional field minimization by finitely many map coefficients and an ordinary differential equation for $f$.

The method constructs a charge-$k$ variational approximation, with topology built in and with rotational symmetry of a rational map translated into combined spatial and isospin rotations. It is efficient for identifying shapes and providing initial data for unrestricted numerical relaxation. Its restrictions are equally concrete: it uses one radial profile and a holomorphic angular map independent of radius, so it cannot represent arbitrary radial-angular correlations, separated clusters, or all deformations. Apart from the degree-one Skyrmion hedgehog ansatz, it generally does not solve the full field equation exactly. Massive-pion terms can be included in the radial functional but do not remove these restrictions, and multi-shell or unrestricted fields may be needed for larger charges. Approximate energy minima and a final collective-coordinate quantization are distinct steps.

Pullback of a differential form commutes with the exterior derivative, so $f^{\ast }$ gives a linear map on de Rham cohomology. Functoriality and $f\circ f=\mathrm{id}$ imply $(f^{\ast }{)}^2=I$. Over the real numbers, define the complementary projectionsThey obey $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$. Every cohomology class therefore decomposes asThe first term is fixed and the second negated by $f^{\ast }$. Their intersection is zero, since a class in both equals its own negative. ConsequentlyThis is the eigenspace decomposition of a linear involution. The factor $1/2$ uses the real coefficient field; a corresponding argument over characteristic two would fail. At the level of differential forms, the same averaging projections give representatives of either parity whenever the class has that parity.

First work at the level of differential forms. The fiber description implies $\pi \circ f=\pi$, so every pullback ${\pi }^{\ast }\beta$ is $f$-invariant. Since $\pi$ is a surjective local diffeomorphism, its pullback on forms is injective: at any $y\in N$, choose a lift $x$ and use the inverse of $d{\pi }_x$ to evaluate the form on arbitrary tangent vectors at $y$.

Conversely let $\alpha$ be an $f$-invariant form on $M$. For a local inverse $s:V\to M$ of $\pi$, define $\beta {|}_V=s^{\ast }\alpha$. If two inverses select the same lift they agree locally; if they select different lifts, the fiber condition identifies the second with $f$ composed with the first. Locally these alternatives are stable, using disjoint inverse neighborhoods at distinct lifts. The equality $f^{\ast }\alpha =\alpha$ therefore makes the local definitions agree. They glue to a unique smooth form $\beta$ on $N$ with ${\pi }^{\ast }\beta =\alpha$. Thusis an isomorphism of complexes, with the exterior derivative on both sides.

It remains to compare the cohomology of the invariant complex with invariant de Rham cohomology classes. Put $P=(I+f^{\ast })/2$ on forms. If $f^{\ast }[\alpha ]=[\alpha ]$ and $d\alpha =0$, then $P\alpha$ is invariant, closed and represents the same class, because its class is the average of two equal classes. If an invariant form is exact, $\alpha =d\eta$, then $\alpha =d(P\eta )$ with an invariant primitive. This proves both surjectivity and injectivity on cohomology. For $p=0$ there are no negative-degree primitives, and the same conclusion follows directly from closed functions.

Combining the two steps givesThis is the de Rham cohomology of a finite quotient argument specialized to an involution. It does not require orientation. The hypotheses do not explicitly exclude $f=\mathrm{id}$; then all fibers are singletons, $\pi$ is a bijective local diffeomorphism and the conclusion is the ordinary diffeomorphism invariance. If $f$ has one fixed point, local injectivity of $\pi$ makes its fixed set open as well as closed; connectedness of $M$ forces this identity case. Otherwise the fibers have two points and $\pi$ is the usual double covering.

Define the Lie derivative of a differential form by ${\mathcal{L}}_X\alpha =(d/dt{)}_{t=0}{\varphi }_t^{\ast }\alpha$, using the local flow. The pullback of a differential form respects the wedge product of differential forms and commutes with the exterior derivative. Therefore ${\mathcal{L}}_X$ is a derivation of an algebra of degree zero and commutes with $d$.

Let $D_X={\iota }_{X}d+d{\iota }_X$, where ${\iota }_X$ is the interior product of a differential form. The graded product rules for $d$ and ${\iota }_X$ show that their anticommutator $D_X$ is also a degree-zero derivation of an algebra: the two mixed terms cancel. Since $d^2=0$, it commutes with $d$. On a function $f$, ${\iota }_{X}f=0$ and $D_{X}f={\iota }_{X}df=Xf={\mathcal{L}}_{X}f$. Hence $D_X(df)=d(Xf)={\mathcal{L}}_X(df)$.

In a manifold chart, every differential form is a sum of functions times products $d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. The two derivations agree on its generators $f$ and $d{x}^i$, and therefore agree on every such form. This proves Cartan's magic formula, including the case $p=0$:

There is an orientation convention in this statement: the two volume forms must be positive for one fixed orientation, as is usual in Moser theorem for volume forms. Without that convention, the stated necessity is false. For example, on an oriented two-sphere, take a positive volume form ${\beta }_0$ and an orientation-reversing diffeomorphism $r$, and set ${\beta }_1=r^{\ast }{\beta }_0$. Then $(r^{-1}{)}^{\ast }{\beta }_1={\beta }_0$ but $\int {\beta }_1=-\int {\beta }_0$. We prove the intended result with a common orientation.

If ${\varphi }^{\ast }{\beta }_1={\beta }_0$, positivity forces $\varphi$ to preserve the orientation, and integration of the pullback of a differential form gives equality of the integrals.

For sufficiency, top-degree de Rham cohomology gives a differential form $\alpha$ of degree $n-1$ with ${\beta }_1-{\beta }_0=d\alpha$. Put ${\beta }_t=(1-t){\beta }_0+t{\beta }_1$. Positivity makes every ${\beta }_t$ a volume form. The interior product of a differential form gives an linear isomorphismThus there is a unique smooth time-dependent vector field $X_t$ satisfying ${\iota }_{X_t}{\beta }_t=-\alpha$. Since $M$ is compact, its local flow exists throughout $0\le t\le 1$ and defines a smooth isotopy ${\varphi }_t$ starting at the identity. By Cartan's magic formula, and since a top-degree differential form has zero exterior derivative,HenceThe zero-dimensional case, a single point, is immediate from equality of the integrals.

An orientable smooth manifold is a smooth $n$-manifold that admits a smoothly varying orientation of its tangent spaces. Equivalently, it has an atlas whose coordinate-transition Jacobian determinants are positive, or a nowhere-vanishing smooth top-degree differential form.

The equivalent formulations of orientability of a smooth manifold are:

A positive chart orients its coordinate frame. Conversely, smoothly oriented frames determine local positive coordinate volume forms. A partition of unity subordinate to an oriented atlas glues these positive forms: at each point they are positive multiples of one another, so their weighted sum cannot vanish. Finally, a nowhere-zero top form $\omega$ declares $(v_1,\dots ,v_n)$ positive exactly when $\omega (v_1,\dots ,v_n)>0$. These constructions are inverse at the level of orientations.

For a smooth divergence-free field on ${\mathbb{R}}^3$, the differential form $F={\iota }_{\mathbf{B}}(d{x}^1\wedge d{x}^2\wedge d{x}^3)$ is closed. The radial homotopy formula gives a global magnetic vector potential $\mathbf{A}(\mathbf{x})={\int }_0^{1}s\mathbf{B}(s\mathbf{x})\times \mathbf{x}ds$ with curl $\mathbf{B}$. For constant $\mathbf{B}$, this reduces to $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.

On a star-shaped domain, let ${\rho }_t(x)=tx$ and $R=\sum x_j{\partial }_{x_j}$. On positive-degree differential forms, $H\alpha ={\int }_0^1{\rho }_t^{\ast }({\iota }_R\alpha )dt/t$ satisfies $dH+Hd=\mathrm{id}-{\rho }_0^{\ast }$. It gives a primitive of every closed positive-degree form. On a star-shaped domain in a complex vector space, contraction with the two type components of $R$ shows that $H$ sends type $(p,q)$ into $(p-1,q)\oplus (p,q-1)$. There is no primitive assertion for a nonzero constant function.

A differential form $\alpha$ on a Lie group is right-invariant if $R_g^{\ast }\alpha =\alpha$ for every right translation $R_g(p)=pg$. A right Maurer-Cartan form is represented in a matrix group by $dg{g}^{-1}$, whereas the left version is $g^{-1}dg$. Its components supply a right-invariant coframe.

On a smooth manifold, ${\mathcal{A}}^k(U)$ is the space of smooth complex-valued degree-$k$ differential forms on $U$. Restriction of forms supplies the sheaf maps. Smooth partitions of unity make this a fine sheaf.

The top-degree differential form ${\omega }^n/n!$ on a $2n$-dimensional symplectic manifold. Every canonical transformation preserves this form. For $n>1$, preservation of this volume alone does not imply preservation of the symplectic form.

A smooth family $Y_t$ assigns a tangent vector at each space-time point. Its flow maps solve ${\partial }_t{\rho }_t=Y_t\circ {\rho }_t$, with ${\rho }_0$ the identity. For a varying differential form, differentiation gives ${\partial }_t({\rho }_t^{\ast }{\alpha }_t)={\rho }_t^{\ast }({\dot{\alpha }}_t+{\mathcal{L}}_{Y_t}{\alpha }_t)$.

On a compact connected oriented $n$-dimensional smooth manifold without boundary, integration is a linear isomorphism $H_{\mathrm{dR}}^n(M)\to \mathbb{R}$. Thus two top-degree differential forms have the same de Rham cohomology class exactly when their integrals agree.

An $\mathbb{R}$-linear degree-one graded derivation of the exterior algebra of smooth differential forms that equals the differential on functions and squares to zero is unique. A smooth bump function and the graded Leibniz rule first establish locality. Then $d(d{x}^i)=0$, and applying the graded Leibniz rule to $\alpha ={\sum }_I{a}_{I}d{x}^I$ gives the displayed coordinate formula. The chain rule makes that formula coordinate-independent.