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Let $S^1$ act on $M\times S^1$ by rotation of the second factor. Averaging a differential form over this action is a cochain map. Every rotation is homotopic to the identity through smooth rotations, and integrating Cartan's magic formula along them gives a cochain homotopy between the averaging map and the identity. Consequently every de Rham cohomology class has a rotation-invariant representative, and an invariant exact form has an invariant primitive obtained by averaging any primitive.

Suppose a nonempty compact $(n+1)$-manifold with a volume form has $n$ pointwise independent volume-preserving vector fields that commute and are tangent to its boundary. If restriction $H_{\mathrm{dR}}^1(M)\to H_{\mathrm{dR}}^1(\partial M)$ is injective, the boundary must have at least two connected components.

Indeed $\eta ={\iota }_{X_1}\cdots {\iota }_{X_n}\omega$ is nowhere zero. Cartan's magic formula and commutation show $d\eta =0$, while boundary tangency gives $j^{\ast }\eta =0$. Injectivity in de Rham cohomology makes $\eta =df$. On a connected boundary $f$ is constant; a maximum or minimum differing from that boundary value would be an interior critical point, contradicting $df\ne 0$. Empty boundary is also impossible.

The Cartan product formula says that total Steenrod squares and total Steenrod reduced powers preserve the cup product:It is named after Henri Cartan and is distinct from Cartan's magic formula for the Lie derivative of a differential form, named after Élie Cartan.

Suppose $V$ is a Killing vector field, $F$ is an electromagnetic field tensor satisfying $dF=0$ and ${\mathcal{L}}_{V}F=0$, and a particle of nonzero mass $m$ obeys the Lorentz force equation ${\nabla }_{u}u=(q/m)F^a{}_b{u}^b$. Cartan's magic formula gives $d({\iota }_{V}F)=0$, so the local potential of a closed differential one-form proof gives $d\Phi ={\iota }_{V}F$ locally. The Killing equation eliminates $u^a{u}^b{\nabla }_a{V}_b$, and the force term gives $d(V\cdot u)/d\tau =(q/m)d\Phi /d\tau$. Therefore $mV\cdot u-q\Phi$ is locally conserved. The sign assumes contraction in the first argument of $F$. A global potential additionally requires the closed one-form to have zero periods; invariance alone is insufficient without the homogeneous Maxwell equations.

For $F=dA$ and an embedding variation $V$, the pullback Lie derivative and Cartan's magic formula give the displayed variation. Consequently the interior force from the Wess-Zumino brane coupling depends only on the gauge-invariant field strength. A change $A\mapsto A+d\Lambda$ changes the action only on the boundary, by the Generalized Stokes theorem; boundaries require the appropriate fixed boundary data or additional boundary coupling.

A generalized calibration bounds the full energy of a supersymmetric brane, including its potential or gauge couplings. For a membrane, compatible spinor bilinears obey $d\Omega ={\iota }_{K}F$. In a stationary gauge ${\mathcal{L}}_{K}A=0$, Cartan's magic formula proves $d(\Omega +{\iota }_{K}A)=0$. The closed charge form then supplies the energy bound. This extends ordinary calibration from volume alone to charged branes, as developed in Topological charges for branes in M-theory.

For a symplectic left Lie group action, use actual fundamental fields ${\xi }_P$ and ${\iota }_{X_f}\omega =df$. They satisfy $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$. Cartan's magic formula then gives ${\iota }_{[\xi ,\eta {]}_P}\omega =d[\omega ({\xi }_P,{\eta }_P)]$. Thus every commutator generator has a globally exact contraction. A semisimple Lie algebra is perfect, so every generator has a global Hamiltonian. The separate semisimple moment-map equivariance result removes the constant two-cocycle obstruction and gives an equivariant moment map for a connected group.

If ${\iota }_{X_{H_t}}\omega =-d{H}_t$ and ${\varphi }_t$ is its flow, then Cartan's magic formula givesThus every defined time slice of a Hamiltonian isotopy is a symplectomorphism.

A Hamiltonian isotopy is a smooth path of symplectomorphisms ${\varphi }_t$, with ${\varphi }_0=\mathrm{id}$, generated by a smooth time-dependent Hamiltonian function. In the convention ${\iota }_{X_{H_t}}\omega =-d{H}_t$, it satisfiesThe flow must exist throughout the chosen time interval. This is automatic on a closed symplectic manifold; on an open one, uniformly compactly supported Hamiltonians also suffice. The Hamiltonian flow preserves the symplectic form by Cartan's magic formula.

Its time-one map is a Hamiltonian diffeomorphism, an element of the group of Hamiltonian diffeomorphisms. A path and its endpoint are different data: a full rotation of the symplectic plane gives a nonconstant Hamiltonian path whose endpoint is the identity. Distinct paths can therefore represent the same Hamiltonian diffeomorphism.

On a symplectic manifold, use ${\iota }_{X_f}\omega =-df$ and $\{f,g\}=\omega (X_f,X_g)=X_f(g)$ for the Poisson bracket. Cartan's magic formula gives ${\mathcal{L}}_{X_f}\omega =0$. Taking the interior product of a differential form with $[X_f,X_g]$ then yields $-d\{f,g\}$, proving the homomorphism. Its kernel consists of locally constant functions, since $X_f=0$ precisely when $df=0$. On a connected manifold these are the constants. The quotient of smooth functions by this kernel is therefore the Lie algebra of Hamiltonian vector fields. A different contraction sign requires a compatible bracket sign.

Smoothly homotopic maps induce the same map on de Rham cohomology. Integrating Cartan's magic formula along a homotopy gives a cochain homotopy

Let $X$ be a smooth vector field with local flow ${\varphi }_t$. The Lie derivative of a differential form isPulling back before taking the derivative compares forms in the same cotangent space. The Lie derivative of a differential form preserves degree and obeys the Leibniz rule for the wedge product of differential forms.

Cartan's magic formula expresses it using two intrinsic operations:where $d$ is the exterior derivative and ${\iota }_X$ is the interior product. In particular, it commutes with the exterior derivative.

On the real line, $X=x{\partial }_x$ has local flow ${\varphi }_t(x)=e^{t}x$. Hence ${\varphi }_t^{\ast }dx=e^{t}dx$ and ${\mathcal{L}}_{X}dx=dx$. The vector field vanishes at zero, but its differential still stretches the one-form there. For a volume form $\omega$, the equation ${\mathcal{L}}_X\omega =0$ defines a volume-preserving vector field.

For a smooth vector field $X$, the Lie derivative is the tensor derivation determined by ${\mathcal{L}}_{X}f=X(f)$ and ${\mathcal{L}}_{X}Y=[X,Y]$. The identity $[X,fY]=f[X,Y]+X(f)Y$ permits its unique extension. On a differential one-form,On differential forms it agrees with the usual Lie derivative of a differential form and Cartan's magic formula.

For an exact symplectic manifold with $\omega =d\lambda$, the Liouville vector field is defined by ${\iota }_Z\omega =\lambda$. Cartan's magic formula gives ${\mathcal{L}}_Z\omega =\omega$. On a cotangent bundle using $\omega =-d\alpha$, take $\lambda =-\alpha$; then $Z={\sum }_i{p}_i{\partial }_{p_i}$ generates the cotangent fiber-dilation flow.

Let $M$ be a closed smooth manifold, and let ${\omega }_t$ be a smooth family of symplectic forms with constant de Rham cohomology class. Choose a smooth family of differential one-forms ${\sigma }_t$ with ${\dot{\omega }}_t=d{\sigma }_t$. Nondegeneracy uniquely solvesCompactness without boundary gives its flow ${\varphi }_t$ throughout $0\le t\le 1$. By Cartan's magic formula,so ${\varphi }_t^{\ast }{\omega }_t={\omega }_0$. This converts a constant-cohomology deformation into an actual isotopy of forms by diffeomorphisms.^[ref]

On a manifold with boundary one additionally needs a boundary-preserving flow, for instance by choosing $X_t$ tangent to the boundary. On a noncompact manifold the flow may fail to exist for the full interval. For example, on ${\mathbb{R}}^2$ the positive forms $dx\wedge dy$ and $e^{-(x^2+y^2)}dx\wedge dy$ are cohomologous, and their affine interpolation stays symplectic. They cannot be symplectomorphic, because their total symplectic volumes are respectively infinite and $\pi$.

On a compact connected smooth manifold without boundary, two positive volume forms for the same orientation are related by an orientation-preserving diffeomorphism exactly when their total integrals agree. A positive interpolation and Cartan's magic formula give the smooth isotopy realizing the equality.

Use the Hamiltonian vector field convention ${\iota }_{X_{H_t}}\omega =-d{H}_t$. A smooth time-dependent Hamiltonian function $H:[0,1]\times M\to \mathbb{R}$ with flow defined over the whole interval generates the Hamiltonian isotopy determined byThe group of Hamiltonian diffeomorphisms isThese maps are symplectomorphisms: by Cartan's magic formula, ${\mathcal{L}}_{X_{H_t}}\omega =d{\iota }_{X_{H_t}}\omega =0$. Thus $d({\varphi }_H^t{)}^{\ast }\omega /dt=0$.

The term “group” is justified explicitly. If ${\varphi }_t,{\psi }_t$ are generated by $H_t,K_t$, their product path is generated by $H_t+K_t\circ {\varphi }_t^{-1}$, since the pushforward of $X_{K_t}$ by a symplectomorphism is $X_{K_t\circ {\varphi }_t^{-1}}$. The inverse path is generated by $-H_t\circ {\varphi }_t$. The zero Hamiltonian gives the identity.

On a closed symplectic manifold, all these flows exist on the finite time interval. On an open manifold one often means the compactly supported Hamiltonian diffeomorphism group ${\mathrm{Ham}}_c$, using Hamiltonians with compact support in $[0,1]\times M$. We use that subgroup for the finite-energy definitions on Euclidean space below. The unrestricted complete-flow group is needed for the global cotangent translations in Question 4; those translations are generally not compactly supported.

Write the exact differential form as $\alpha =du$ for a smooth real-valued function $u$ on $N$. On $T^{\ast }N$ take the Hamiltonian function $H=u\circ \pi$. With $\omega ={\sum }_{j}d{q}_j\wedge d{p}_j$ and ${\iota }_{X_H}\omega =-dH$, the Hamiltonian vector field isIts complete flow is explicit:The base point stays fixed, so this flow exists for all real $t$, even if $u$ is unbounded on a noncompact base. Its time-one map is exactly $f$, proving that the cotangent fiber translation is a Hamiltonian diffeomorphism in the unrestricted complete-flow group.

This construction also justifies the hint for this family, without treating exactness as a general substitute for a Hamiltonian isotopy. For any Hamiltonian isotopy ${\psi }_t$ on this cotangent bundle, Cartan's magic formula, $d\theta =-\omega$ and ${\iota }_{X_{H_t}}\omega =-d{H}_t$ giveIntegrating proves that ${\psi }_1^{\ast }\theta -\theta$ is exact. For the present translation this difference is ${\pi }^{\ast }\alpha$, which is exact if and only if $\alpha$ is exact: one direction is pullback of $du$, and the other is pullback by the zero section. Necessity therefore follows from the differential identity, while sufficiency follows from the complete flow just constructed.

Compact support is a separate issue. If $du$ is nonzero at some $q$, the map moves every point of the entire unbounded fiber over $q$, so it is not a member of the compactly supported Hamiltonian diffeomorphism group. For example $N=\mathbb{R}$, $u(q)=q$ gives $(q,p)\mapsto (q,p+1)$. Thus the claim about this global translation uses the full Hamiltonian group, not ${\mathrm{Ham}}_c$; its flow can be cut off to agree on a prescribed compact swept set, but such a cutoff does not give the same map on the entire cotangent bundle.

The exterior derivative is a linear map $d:{\Omega }^k(M)\to {\Omega }^{k+1}(M)$, agrees with the ordinary differential on functions, satisfies $d^2=0$, and obeys the graded Leibniz ruleIn local coordinates it differentiates the coefficient functions:Antisymmetry cancels the second partial derivatives, proving $d^2=0$ locally. These properties determine the coordinate-independent derivative.

For any smooth $\varphi :N\to M$, the pullback of a differential form is defined byThere is no requirement that $\varphi$ be invertible or an immersion. Pullback preserves wedge products and sends $df$ to $d(f\circ \varphi )$ by the chain rule. Locally every form is a sum of $fd{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k}$; applying the preceding identities and $d^2{\varphi }^i=0$ proves

Describe the brane by an embedding $X:\Sigma \to M$, with nondegenerate induced worldvolume metric $h=X^{\ast }g$. Locally the Poincaré lemma gives $F=dA$ because $F$ is a closed differential form. For a timelike worldvolume, the geometric area and Wess-Zumino brane coupling give the action$T$ is the brane tension and $q$ the charge in this normalization. The geometric term is the higher-dimensional version of the Nambu–Goto action. Both integrals are invariant under changes of worldvolume coordinates.

The additive gauge transformation is $A\mapsto A+d\Lambda$. Its change of action isThe last step is the Generalized Stokes theorem. It vanishes for a closed worldvolume, or with suitable boundary restrictions. An open worldvolume can instead have a compensating boundary coupling, as in gauge invariance of an open-brane coupling.

More directly, for a variation vector $V$ of the embedding, Cartan's magic formula givesThus the interior force depends on $F$, not the gauge potential. Define the trace of the second fundamental form byThe area variation is $-\int \sqrt{|h|}g(V,K)$ plus a boundary term. For compactly supported interior variations the resulting equations areThey exhibit the field-strength force on a charged brane and are gauge invariant.

Two global qualifications are necessary. First, replacing $A$ by $d\Lambda$ alone, as printed, gives $d{A}^{\prime }=0$ and would erase a nonzero $F$; the intended transformation must be the additive shift above. Second, closedness of $F$ only guarantees a potential locally. If a global potential does not exist and the worldvolume bounds a filling $B$, its charged term can be defined as $q{\int }_{B}F$. Different fillings differ by a period of $F$; independence of the quantum phase requires $q{\int }_{C}F\in 2\pi \hslash \mathbb{Z}$ for the corresponding closed cycles. Without the required global potential or patching/filling data, the given closed form still defines the local variational equations, but not automatically a single globally defined real-valued action.

Use ${\iota }_{X_f}\omega =-df$ and $\{f,g\}=X_{g}f$, so in the stated real coordinates $\{q^i,p_j\}={\delta }_j^i$ and $\dot{f}=\{f,H\}$. For $\xi \in \mathfrak{g}$, let ${\xi }_P$ be the actual fundamental vector field of the left action, obtained by differentiating $\exp (t\xi )\cdot x$ at zero. Since the action preserves $\omega$, Cartan's magic formula gives $d({\iota }_{{\xi }_P}\omega )=0$.

A global component Hamiltonian ${\mu }_{\xi }$ requires this closed one-form to be exact, withChoose these functions linearly in $\xi$. This defines the moment map, with equivariance when the constants have the required normalization. A symplectic action alone need not admit it: translation in a periodic coordinate of a symplectic two-torus contracts $\omega$ to a closed one-form with nonzero period. A sufficient existence condition is $H_{\mathrm{dR}}^1(P)=0$; more generally one assumes the action is weakly Hamiltonian and resolves its moment-map equivariance obstruction.

Because the group also preserves $H$,The component vector fields commute with the Hamiltonian flow: in this convention $[X_f,X_g]=-X_{\{f,g\}}$, hence $[X_{{\mu }_{\xi }},X_H]=0$. These are both conserved observables and infinitesimal symmetries of the dynamics.

For a left action, fundamental vector fields obey $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$. Together with the Hamiltonian-field identity, this makeshave zero Hamiltonian vector field. It is therefore constant on each connected component of $P$. One sufficient condition for $c=0$ is an equivariant moment map, $\mu (gx)={\mathrm{Ad}}_g^{\ast }\mu (x)$, where $⟨{\mathrm{Ad}}_g^{\ast }\ell ,\xi ⟩=⟨\ell ,{\mathrm{Ad}}_{g^{-1}}\xi ⟩$. Differentiating this condition along ${\eta }_P$ gives directlyA useful concrete sufficient condition is a common fixed point $x_0$ and the normalization ${\mu }_{\xi }(x_0)=0$. The constant $c$ is zero there because all fundamental fields vanish, and hence zero everywhere on connected $P$. This is fixed-point normalization removes the moment-map cocycle. If the action has an ineffective Lie-algebra kernel, the map to observables can have that kernel; equality of brackets alone is not a claim of a faithful representation.

For the oscillator, the printed complex coordinates give $id{z}_i\wedge d{\bar{z}}_i=d{p}_i\wedge d{q}^i$ and $\{z_i,{\bar{z}}_j\}=i{\delta }_{ij}$. For an anti-Hermitian matrix $\xi \in \mathfrak{u}(3)$, the fundamental field is $\delta z=\xi z$. Direct contraction yieldsConsequently the unitary moment map of an isotropic oscillator in our declared convention isIt is real because $\xi$ is anti-Hermitian. Identifying $\mathfrak{u}(3{)}^{\ast }$ with Hermitian matrices by $⟨M,\xi ⟩=-i\mathrm{Tr}(M\xi )$, this is simply $\mu (z)=z{z}^{†}$. Unitary invariance of $H=z^{†}z$ proves conservation of all nine real matrix components. Under $z\mapsto Uz$, the matrix transforms as $z{z}^{†}\mapsto Uz{z}^{†}{U}^{-1}$, exactly the coadjoint transformation in this pairing. The origin is the required fixed point.

The bracket can also be checked without an equivariance theorem. For $f=-i\bar{z}\xi z$ and $g=-i\bar{z}\eta z$, useDifferentiation gives $\{{\mu }_{\xi },{\mu }_{\eta }\}=-i{z}^{†}[\xi ,\eta ]z={\mu }_{[\xi ,\eta ]}$. The central generator $\xi =iI$ has ${\mu }_{\xi }=H$, while a real antisymmetric generator $\xi v=a\times v$ hasThus the special orthogonal group subgroup has the usual angular momentum components $L=q\times p$, with $\{L_i,L_j\}={\epsilon }_{ijk}{L}_k$. The remaining unitary generators supply the other conserved quadratic observables, with their signs fixed by the stated choice of complex coordinates and Hamiltonian convention.