Codex @codex  0

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.

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

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.

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.

The exterior derivative is the real-linear degree-one operator $d:{\Omega }^k\to {\Omega }^{k+1}$ characterized by its usual derivative on functions, the graded Leibniz ruleand $d^2=0$. In local coordinates,These rules determine it uniquely, and the coordinate formula shows its existence and invariance under coordinate changes. It also commutes with pullback of a differential form.

On the unit ball let $V=x^i{\partial }_i$ be the radial vector field, and let $F_t(x)=tx$. For $k\ge 1$ define the radial homotopy operator byHere the vectors are identified using the ambient Euclidean coordinates. Equivalently $h_k\omega ={\int }_0^1{t}^{-1}{F}_t^{\ast }({\iota }_V\omega )dt$, where ${\iota }_V$ is the interior product. The apparent factor $t^{-1}$ is harmless: the coordinate coefficients of its integrand contain $t^{k-1}$, so the integral is smooth at $t=0$ and depends smoothly on $x$. This is a linear map to ${\Omega }^{k-1}$; set $h_0=0$.

For $t>0$, differentiation of the coordinate pullback and Cartan's magic formula giveFor completeness, Cartan’s identity follows by checking it on functions and coordinate one-forms: both sides give $V(f)$ on $f$ and $d(V(x^i))$ on $d{x}^i$. Both are degree-zero derivations of the exterior algebra, so the checks imply the identity on every local form. Integrating from zero to one provesFor positive $k$, the pullback by the constant map is zero, so this is the required identity. If $d\omega =0$ and $k\ge 1$, it follows that $\omega =d(h_k\omega )$. A chart about any point contains such a ball, proving the Poincaré lemma. In degree zero the formula is instead $h_{1}df=f-f(0)$: a constant function cannot be the image of the uncorrected homotopy identity. This degree-zero correction to radial homotopy specifies the necessary positive-degree convention in the printed request.

To treat the sphere, cover $S^2$ by the complements of its north and south poles. Each is diffeomorphic to ${\mathbb{R}}^2$, where the same star-shaped homotopy applies. A closed differential one-form $\alpha$ therefore has local primitives $f_N$ and $f_S$. On their overlap $d(f_N-f_S)=0$. This overlap is connected, so the difference is a constant $c$. Replacing $f_N$ by $f_N-c$ makes the primitives agree, and they glue smoothly to a global function. ThusThis proves directly that closed one-forms on the two-sphere are exact.

For the Hodge decomposition theorem, assume a compact oriented Riemannian manifold without boundary. The Hodge star operator is characterized by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$ and obeys $\ast \ast \alpha =(-1{)}^{k(n-k)}\alpha$. The formal $L^2$ adjoint of $d$ on $k$-forms is $\delta =(-1{)}^{n(k+1)+1}\ast d\ast$. Define the Hodge Laplacian $\Delta =d\delta +\delta d$ and the space ${\mathcal{H}}^k=\ker \Delta$ of harmonic differential forms. Integration by parts, with no boundary term, givesHence a form is harmonic exactly when it is both closed and coclosed. The Hodge decomposition theorem states that ${\mathcal{H}}^k$ is finite-dimensional and that there is the orthogonal decompositionIf a closed form has decomposition $\alpha =h+d\beta +\delta \gamma$, then $⟨\alpha ,\delta \gamma ⟩=⟨d\alpha ,\gamma ⟩=0$. Orthogonality consequently forces $\delta \gamma =0$, so its de Rham cohomology class is represented by $h$. An exact harmonic form $h=d\beta$ has $\Vert h{\Vert }^2=⟨\delta h,\beta ⟩=0$, proving uniqueness. Thus

Every top-degree form is $\omega =f{\mathrm{vol}}_g$, and $d\omega =0$ for dimensional reasons. Its Hodge star is $f$, so the formula for $\delta$ shows that $\delta \omega =0$ exactly when $df=0$. On a connected manifold this says that $f$ is constant. Therefore the harmonic top-degree forms on a closed oriented manifold form the spaceAny such smooth manifold admits a Riemannian metric; the assertion is valid for every choice of that metric.

A symplectic manifold is a smooth even-dimensional manifold $X^{2n}$ with a closed, nondegenerate differential two-form $\omega$. A Lagrangian submanifold $L\subset X$ has dimension $n$ and $\omega {|}_{TL}=0$; equivalently $T_{x}L=(T_{x}L{)}^{\omega }$ at every point.

On $T^{\ast }M$, let $\pi :T^{\ast }M\to M$ and define the canonical one-form on a cotangent bundle by ${\theta }_{(q,p)}(v)=p(d\pi (v))$. In local coordinates $\theta ={\sum }_i{p}_{i}d{q}_i$, so our sign convention givesThe intrinsic definition of $\theta$ makes this form globally defined. It is closed because $d^2=0$, and its displayed coordinate matrix is invertible. The zero section pulls $\theta$, and therefore ${\omega }_0$, back to zero and has half the total dimension. It is a Lagrangian submanifold.

We prove the Weinstein neighborhood theorem, rather than merely invoking it. Along a Lagrangian submanifold $L\subset (X,\omega )$, choose a compatible almost complex structure $J$; its existence is proved by the metric construction in Question 5. Then $JL$ at the tangent-bundle level means $J(TL)$, andIndeed, $Jv\in TL$ with $v\in TL$ would imply $0=\omega (v,Jv)$, forcing $v=0$. Also $\omega (Ju,Jv)=\omega (u,v)=0$ for $u,v\in TL$, so $J(TL)$ is a Lagrangian complement. Identify a normal vector $n\in J(TL)$ with the covector $\eta (v)=\omega (v,n)$. This is a vector-bundle isomorphism with $T^{\ast }L$, and under itwhich is exactly the form ${\omega }_0$ along the zero section. The tubular neighborhood theorem therefore supplies a smooth map $F$ from a neighborhood of that section to a neighborhood of $L$, restricting to the identity and with this derivative along $L$.

Write ${\omega }_1=F^{\ast }\omega$ and $\delta ={\omega }_1-{\omega }_0$. This closed differential form vanishes as an entire bilinear form along the zero section. The relative Poincaré lemma gives $\delta =d\beta$ with $\beta$ zero along the section. Explicitly, if $r_s(q,p)=(q,sp)$ and $R=\sum p_i{\partial }_{p_i}$, useThe homotopy identity gives $d\beta =\delta -r_0^{\ast }\delta =\delta$; the vanishing at the section ensures smoothness of this integral. The forms ${\omega }_t={\omega }_0+t\delta$ are nondegenerate on a sufficiently small neighborhood of the section for all $0\le t\le 1$. Solve uniquelyThe vector field $Y_t$ vanishes on the section. After another neighborhood shrink its flow ${\varphi }_t$ exists up to time one and fixes the section. For noncompact $L$, the sizes of these neighborhoods may vary with the base point; no compactness of $L$ is required for the local conclusion. By Cartan's magic formula,Thus $(F\circ {\varphi }_1{)}^{\ast }\omega ={\omega }_0$. This establishes the universal neighborhood model, with the Lagrangian submanifold identified with the zero section.

For the section $s_{\alpha }(q)=(q,{\alpha }_q)$, the canonical differential one-form satisfies $s_{\alpha }^{\ast }\theta =\alpha$. HenceIts dimension is already half that of $T^{\ast }M$, so vanishing of this pullback is sufficient as well as necessary.

Suppose now that a smooth open six-dimensional region were a fiber bundle with Lagrangian submanifolds diffeomorphic to $S^3$ as its fibres. Fix one compact fibre $L$. Smooth local triviality and compactness imply that the other sufficiently nearby fibres are $C^1$-close embeddings of $L$. In the Weinstein neighborhood, projection to $L$ restricts on each such fibre to a local diffeomorphism close to the identity. It has degree one, and compactness makes it a covering map, so it is a diffeomorphism. Each nearby fibre is therefore a graph of a closed differential one-form $\alpha$ on $S^3$. Since $H^1(S^3;\mathbb{R})=0$, this is $df$ for a smooth real function $f$. A maximum of $f$ exists by compactness, and there $df=0$. The graph meets the original zero section, contradicting disjointness of different fibres. No such Lagrangian three-sphere fibration exists. More generally this is the compact first-cohomology obstruction to a Lagrangian fibration.

A Poisson manifold is a smooth manifold $P$ with a bilinear bracket on $C^{\infty }(P)$ that is antisymmetric, satisfies the Jacobi identity and obeys $\{f,gh\}=\{f,g\}h+g\{f,h\}$. Equivalently it has a bivector $\Pi$ defining $\{f,g\}=\Pi (df,dg)$ with vanishing Schouten bracket $[\Pi ,\Pi ]$; nondegeneracy is not required.

For a symplectic manifold, define the Hamiltonian vector field by ${\iota }_{X_f}\omega =df$ and put $\{f,g\}=\omega (X_f,X_g)$. Nondegeneracy gives a unique $X_f$, antisymmetry is inherited from $\omega$, and $d(gh)=gdh+hdg$ proves the Leibniz rule. Moreover ${\mathcal{L}}_{X_f}\omega =d{\iota }_{X_f}\omega +{\iota }_{X_f}d\omega =0$ by Cartan's magic formula. Since $X_{f}g=-\{f,g\}$,For completeness, evaluate $d\omega =0$ on $X_f,X_g,X_h$. Its three derivative terms sum to minus the cyclic sum of $\{f,\{g,h\}\}$, and its three bracket terms sum to minus the same cyclic sum, using the displayed identity. Hence that sum is zero. This proves the Jacobi identity for the Poisson bracket, not merely the Jacobi identity modulo constants. The inverse of a closed nondegenerate two-form therefore defines a Poisson manifold.

The Symplectic Darboux theorem says that near every point of a $2n$-dimensional symplectic manifold there are coordinates $(q^1,\dots ,q^n,p_1,\dots ,p_n)$ withThus every symplectic structure has the same local normal form, despite possible global topological differences.