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A closed differential form satisfies $d\alpha =0$; an exact differential form has $\alpha =d\beta$. The identity $d^2=0$ makes every exact form closed. Global failure of the converse is measured by de Rham cohomology; the Poincaré lemma gives the local converse in positive degrees.

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

A nondegenerate Poisson bivector inverts to a closed differential form of degree two. Differentiating $S=P^{-1}$ gives ${\partial }_{d}S=-S({\partial }_{d}P)S$. Contracting the coordinate Jacobi condition for a Poisson bivector with three copies of $S$ yields ${\partial }_i{S}_{jk}+{\partial }_j{S}_{ki}+{\partial }_k{S}_{ij}=0$, which is precisely closure of the resulting symplectic form. An overall sign in the inverse convention does not change closure.

A differential form is coclosed when its codifferential vanishes. On a closed Riemannian manifold, a form is a harmonic differential form exactly when it is both a closed differential form and coclosed: the identity $⟨\Delta \alpha ,\alpha ⟩=\Vert d\alpha {\Vert }^2+\Vert \delta \alpha {\Vert }^2$ proves the converse as well as necessity.

Translation by a differential one-form $\beta$ is $(x,\xi )\mapsto (x,\xi +{\beta }_x)$. It changes the canonical one-form on a cotangent bundle by ${\pi }^{\ast }\beta$. A closed differential form gives a symplectomorphism; an exact differential form gives a Hamiltonian diffeomorphism when compact support is not required.

The displayed one-form on the unit round sphere has zero integral on every closed geodesic. Indeed, the antipodal map pulls it back to its negative, and the second half of an oriented great circle is the antipodal image of the first. Nevertheless $d\theta =2x_{3}d{x}_3\wedge d{x}_1$ is nonzero on an open set, so the one-form is not a closed differential form and cannot be an exact differential form. Thus a curvature-free extension of one-form injectivity fails.

Restriction of differential forms to the finitely many open and closed components gives an isomorphism of cochain complexes ${\Omega }^{\ast }(⨆M_j)\cong {⨁}_j{\Omega }^{\ast }(M_j)$. Piecewise forms glue because there are no overlaps, and the exterior derivative acts componentwise. Taking the quotient of closed differential forms by exact differential forms proves the displayed de Rham cohomology isomorphism. For an infinite disjoint union, forms and cohomology give products instead.

A differential form is a smooth field of alternating multilinear maps that can be integrated over an appropriately oriented manifold of matching dimension. A one-form measures a directed displacement; a two-form measures an oriented area element. The point is to express what is being accumulated along a curve or over a surface without tying the answer to one set of coordinates. This makes line integrals, surface integrals, and the integral theorems of vector calculus parts of one construction.

For a concrete one-form, take $\alpha =xdy$ in the plane. Travel from $(0,0)$ to $(1,1)$ along the bottom and then the right edge of the unit square. The horizontal leg contributes zero because $dy=0$; the vertical leg has $x=1$ and contributes ${\int }_0^{1}dy=1$. Taking the left and then the top edge instead gives zero: the vertical leg has $x=0$, and the horizontal leg again has $dy=0$. The two paths have the same endpoints but different integrals.

Now replace $\alpha$ with the exact differential $\beta =d(xy)=ydx+xdy$. Both paths give $1$, because integrating an exact differential along a curve gives the endpoint difference of its potential. Here that difference is $xy{|}_{(1,1)}-xy{|}_{(0,0)}=1$. This is the fundamental theorem of calculus applied to a function along a parametric curve, and is the basic reason conservative vector fields have path-independent work.

The exterior derivative records the local obstruction in the first example: $d(xdy)=dx\wedge dy$. The Generalized Stokes theorem turns the difference between the two path integrals into the integral of this two-form over the square, whose oriented area is $1$. Interior edges cancel when adjoining pieces are added; only the outer boundary survives. That cancellation connects a local calculation with a global measurement.

More precisely, a degree-$k$ differential form assigns to each point $p$ a multilinear scalar-valued function of $k$ tangent vectors at $p$. Interchanging two arguments changes its sign. A one-form is therefore a covector field, while a two-form vanishes on a pair of parallel vectors. Degree zero simply means a smooth function. The space ${\Omega }^k(M)$ is a vector space; the wedge product of differential forms combines degrees, and the exterior derivative raises degree by one.

The usefulness goes beyond bookkeeping. Pullbacks of a differential form supply the right transformation law automatically under a coordinate change. The Generalized Stokes theorem unifies the fundamental theorem of calculus, Green theorem, Stokes theorem, and divergence theorem. Meanwhile, the distinction between closed differential forms and exact differential forms detects holes through de Rham cohomology. Reyer Sjamaar develops these connections in sections 5.4 and 9.3 of his university notes. ^[ref].

The exterior derivative is the unique graded derivation $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$ that agrees with the differential on functions and satisfies $d^2=0$.

On a smooth planar one-form $\alpha =Pdx+Qdy$, the product rule and antisymmetry of the wedge product of differential forms giveFor $\alpha =xdy$, the coefficient is $1$: there is a nonzero integral around the boundary of every positively oriented rectangle. For $\beta =d(xy)$, the coefficient is ${\partial }_{x}x-{\partial }_{y}y=0$. More generally, the identity $d^2=0$ makes every exact differential form a closed differential form. It expresses cancellation between mixed partial derivatives locally; it does not imply that every closed differential form has a global potential.

Let $\alpha$ be a real closed differential form on a Riemann surface, of finite positive conformal energy $E=\int |\alpha {|}^{2}dA$. If its period on a loop homotopy class $\gamma$ is $p$, then $\lambda (\gamma ,X)\ge p^2/E$. Indeed the conformal metric $\rho =|\alpha |$ has area $E$ and each loop in the class has length at least $|\int \alpha |=|p|$. Energy is independent of the auxiliary smooth conformal metric used to compute the pointwise norm: the inverse scaling of the squared norm cancels the area scaling.

A closed differential form of degree one on the two-sphere has primitives on its north-punctured and south-punctured charts by the Poincaré lemma. The overlap is connected, so their difference is constant. Adjust one constant and glue the primitives. Thus every closed one-form is an exact differential form, proving the first de Rham cohomology vanishes.

An equivalence class of continuous maps under homotopy, with any endpoint or boundary restrictions specified. Unbased loop classes are free homotopy classes; periods of a closed differential form are constant on such classes.

A compactly supported top-degree differential form on an oriented smooth manifold is integrated using orientation-preserving coordinates and a partition of unity. The coordinate change of variables formula makes the result independent of the choices. More generally, a form is integrated over a lower-dimensional oriented submanifold by its pullback. The Generalized Stokes theorem states ${\int }_{M}d\beta ={\int }_{\partial M}\beta$ with the induced boundary orientation. Integrating a closed differential form over a closed cycle depends only on its homology class, since its integral over a boundary vanishes.

A continuously differentiable vector field on an open subset of three-dimensional Euclidean space is irrotational when its curl is zero. Under the Euclidean identification $\alpha =X_{i}d{x}^i$, this says that the differential one-form $\alpha$ is a closed differential form. The Poincaré lemma gives a potential of a conservative vector field locally. A global potential additionally requires zero line integrals around every closed curve; simply connected domains ensures this. On the punctured plane extended in the third direction, $X=(-y,x,0)/(x^2+y^2)$ has zero curl but integral $2\pi$ around a circle, so it is locally conservative and has no single-valued global potential.

For a non-null Killing vector field $V$ of a Levi-Civita connection, hypersurface orthogonality implies that $\eta =V^{♭}/F$, with $F=g(V,V)$, is a closed differential form. The Killing equation and $V^{♭}\wedge d{V}^{♭}=0$ give $2F{\nabla }_{\mu }{V}_{\nu }=V_{\nu }{\partial }_{\mu }F-V_{\mu }{\partial }_{\nu }F$, whence $d\eta =0$. The Poincaré lemma gives $\eta =d\varphi$ locally. For a timelike $V$, $\varphi$ supplies local static time coordinates. This does not assert global exactness.

For an exact differential $Md\theta +Nd\varphi =dF$, equality of mixed partial derivatives requires $M_{\varphi }=N_{\theta }$. Here the two partial derivatives areThey are not identically equal, so the differential form is not a closed differential form and hence is not an exact differential. Equality along special lines such as $\theta =0$ does not supply a potential on a two-dimensional open neighborhood, because every such neighborhood contains points where the two partial derivatives differ.

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.

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.

For the graph embedding $\iota (q)=(q,p(q))$, pull back the symplectic form in Darboux coordinates:The coordinate two-forms are linearly independent. The graph has dimension $n$, so it is a Lagrangian submanifold exactly when this pullback vanishes, equivalently when its displayed coefficient matrix is symmetric.

Let ${\theta }_L={\sum }_i{p}_i(q)d{q}^i$. Then $d{\theta }_L=-{\iota }^{\ast }\omega$, so the same condition says that ${\theta }_L$ is a closed differential form. On a sufficiently small contractible coordinate neighborhood the Poincaré lemma supplies $S$ with ${\theta }_L=dS$. ConsequentlyOne explicit primitive on a star-shaped neighborhood of $q_0$ is $S(q)=S(q_0)+{\int }_0^1{p}_i(q_0+t(q-q_0))(q^i-q_0^i)dt$. Differentiate it and use ${\partial }_j{p}_i={\partial }_i{p}_j$ to recognize the derivative of $t{p}_j(q_0+t(q-q_0))$; the result is ${\partial }_{j}S=p_j(q)$. Conversely $p=dS$ has symmetric mixed derivatives and gives a Lagrangian graph. This construction requires the local projection to the $q$ coordinates to be nonsingular. A general Lagrangian submanifold need not be such a graph in a preassigned chart, and a closed one-form on a larger domain need not be globally exact.