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For the vector-bundle curvature matrix $R$ of a metric connection on a vector bundle on a real oriented vector bundle, the A-hat form is the even differential form defined by the displayed finite-degree power series, with square root having constant term one. The expression is independent of the choice of orthonormal frame. Equivalently, for formal paired curvature Chern roots $\pm x_j$, it is ${\prod }_j(x_j/2)/\sinh (x_j/2)$. The Bianchi identity makes it closed, and changing the connection changes it by an exact differential form via Chern-Weil connection transgression. Its degree-four part is $-p_1/24$, where $p_1=-\mathrm{tr}(R\wedge R)/(8{\pi }^2)$ is the first Pontryagin class form.

A variation of the connection one-form has $\dot{F}=D_A\dot{A}$. The Bianchi identity and graded cyclic matrix trace imply ${\partial }_t\mathrm{Tr}(F\wedge F)=2d\mathrm{Tr}(\dot{A}\wedge F)$. Thus the Chern-Weil theory four-form changes by an exact differential form, and its integral on a closed four-cycle is connection-independent. For a general unitary bundle, include the corresponding trace-product term in the Second Chern form.

The Chern-Weil homomorphism turns a polynomial in curvature into a topological cohomology class. For a principal bundle with Lie group $G$, a degree-$k$ polynomial $P$ on its Lie algebra that is invariant under the adjoint action determineswhere $F$ is the curvature form of any principal connection. The Bianchi identity makes $P(F)$ a closed differential form. Changing the principal connection changes it by an exact differential form, so the de Rham cohomology class depends on the principal bundle, not that choice. Products of polynomials become cup products.^[ref]

For example, the trace-square polynomial produces $\mathrm{tr}(F\wedge F)$, underlying the first Pontryagin class. This is the bridge between the locally measured curvature and a global topological invariant.

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.

A differential one-form is closed when its exterior derivative vanishes. In coordinates, $\eta ={\eta }_{i}d{x}^i$ is closed exactly when ${\partial }_i{\eta }_j={\partial }_j{\eta }_i$. The Poincaré lemma makes it locally an exact differential form, but global exactness can fail through nonzero integrals around closed curves.

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.

The de Rham complex of a smooth manifold $M$ is the cochain complex of smooth differential forms with exterior derivative:Here $n=\dim M$ and ${\Omega }^0(M)=C^{\infty }(M)$. The identity $d^2=0$ makes every exact differential form closed. The quotient of closed $k$-forms by exact $k$-forms is the de Rham cohomology group $H_{\mathrm{dR}}^k(M)$.^[ref]

In Euclidean space the Poincaré lemma makes this complex exact in positive degrees. On a general manifold it is locally exact in small coordinate balls, but global exactness can fail. For the unit circle, the restriction of $\omega =-ydx+xdy$ has integral $2\pi$ around one positive turn. It is closed, since the circle has no nonzero two-forms. It cannot equal $df$ for a globally defined smooth function on the circle: the integral of a derivative around a closed loop is zero.

The complex therefore separates local differential identities from global obstructions. In degree zero, $df=0$ means that $f$ is locally constant; in degree one, periods around loops can detect a failure to possess a global potential.

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].

For real $a,b\ne 0$, this differential one-form on the punctured plane is the pullback of a differential form of the nonexact angular one-form under the invertible linear transformation $(X,Y)=(ax,by)$. It is a closed differential form for every such pair. On a region admitting a continuous complex argument, it has the local potential $f=\mathrm{Arg}(ax+iby)$. In a chart with $x\ne 0$, this can be written $f=\arctan (by/(ax))+C$.

There is no globally single-valued potential on the whole punctured plane. On the closed path $x=\cos t/a$, $y=\sin t/b$, $0\le t\le 2\pi$, substitution gives ${\omega }_{a,b}=dt$, hence its line integral is $2\pi \ne 0$. This contradicts the zero integral around every closed path required of an exact differential form. Around a positively oriented once-winding path the integral is $2\pi \mathrm{sgn}(ab)$. The local equation ${\omega }_{a,b}=0$ gives radial rays; away from a vertical ray its solutions as graphs are $y=Cx$ on intervals not crossing the excluded origin.

Let $M$ be a compact oriented Riemannian manifold of dimension four, without boundary. An exact two-form $\beta =d\alpha$ that is an anti-self-dual differential form must vanish. Indeed the Generalized Stokes theorem and the Hodge star operator identity $\ast \beta =-\beta$ giveThe boundaryless hypothesis is essential. On a Euclidean four-ball, the nonzero constant two-form $d{x}^1\wedge d{x}^2-d{x}^3\wedge d{x}^4$ is both an exact differential form and an anti-self-dual differential form; the omitted boundary integral need not vanish there.

An equation $P(x,y)+Q(x,y)y^{\prime }=0$ is exact if $Pdx+Qdy$ is an exact differential form: there is a potential $F$ with $F_x=P$ and $F_y=Q$. Its solution curves are levels $F=C$. For continuously differentiable coefficients, $P_y=Q_x$ suffices locally and globally on a simply connected domain. Global topology matters; a closed form on a punctured domain need not be exact.

An exact symplectic manifold has a symplectic form that is an exact differential form, written here $\omega =-d\lambda$. A positive-dimensional example without boundary cannot be compact, by the Generalized Stokes theorem applied to the top wedge product of differential forms.

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.

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.

On a star-shaped planar domain containing zero, let $\eta$ be a smooth closed differential one-form and set $f(x)={\int }_{[0,x]}\eta$. Green theorem on the triangle with vertices $0,x,x+h$ gives $f(x+h)-f(x)={\int }_{[x,x+h]}\eta$. Dividing by a displacement and taking its limit proves $df=\eta$. Thus every such form is an exact differential form, a degree-one version of the Poincaré lemma which uses only the planar integral theorem.

For a differential one-form $Pdx+Qdy$, seek a nonzero integrating factor $\mu (x)$. Expanding $(\mu P{)}_y=(\mu Q{)}_x$ gives the displayed equation where $Q\ne 0$. Such a factor exists locally when $(P_y-Q_x)/Q$ depends only on $x$, and is $\mu =\exp \int [(P_y-Q_x)/Q]dx$. Likewise, a factor depending only on $y$ obeys ${\mu }^{\prime }/\mu =(Q_x-P_y)/P$ where $P\ne 0$. The resulting product is an exact differential form on a sufficiently small regular domain; singular lines of the factor need separate treatment.

For the differential one-form $fdy+gdx$, a nonzero integrating factor $\mu$ makes $\mu fdy+\mu gdx$ an exact differential form locally. If it equals $dF$, then $F_y=\mu f$, $F_x=\mu g$, and equality of mixed partial derivatives gives the displayed compatibility condition. Conversely this condition is local sufficiency for smooth coefficients on a small rectangle; global exactness additionally depends on the domain. Zeros of the multiplying factor can change the solution set, so equivalence is asserted only where the factor is nonzero.

Let a finite group $G$ act smoothly on a smooth manifold, and suppose that a $G$-invariant differential form $\omega$ is an exact differential form, say $\omega =d\eta$. Averaging gives the invariant primitiveIf the action is free, invariant forms descend uniquely through the resulting covering map.