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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.

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.

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].

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 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.

For a smooth Anosov flow generated by a surface thermostat on a closed oriented Riemannian surface, a smooth base function plus a smooth one-form is a smooth flow coboundary precisely when the base function is zero and the one-form is an exact differential form. This is the nontrivial degree-zero/one kernel theorem. It does not require negative curvature or preservation of Liouville volume. The converse follows directly from $F(f\circ \pi )=df(v)$. A primary proof of the kernel theorem is arxiv.org/abs/math/0601039.

On the punctured plane, this one-form is closed and locally the differential of the polar angle. Its line integral around a closed curve is $2\pi$ times the winding number. A once-winding circle therefore has nonzero integral, proving that no global single-valued angle potential exists and the form is not an exact differential form.

For a counterclockwise circle of radius $R>0$, substitute $x=R\cos t$, $y=R\sin t$. The pullback of a differential form isThe radius cancels. In coordinates, $P=-y/(x^2+y^2)$ and $Q=x/(x^2+y^2)$ satisfy ${\partial }_{x}Q={\partial }_{y}P$ away from the origin, so $d\omega =0$ there. The nonzero integral is a global obstruction to being an exact differential form, not a nonzero local exterior derivative.

Compare this with the smooth one-form $\eta =(-ydx+xdy)/2$ on the whole plane. Here $d\eta =dx\wedge dy$, and its integral around a radius-$R$ circle is $\pi R^2$, the enclosed area. The Generalized Stokes theorem applies to the full disc for $\eta$. For $\omega$, that disc crosses the excluded origin, where the one-form is undefined. On an annulus the theorem applies, but its induced inner boundary is clockwise: the two integrals $2\pi$ and $-2\pi$ cancel. This explains exactly why closedness and nonzero circulation can coexist.

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.

Write the standard complex coordinates as $z_j=q_j+i{p}_j$ and let $J$ denote multiplication by $i$. Use the standard symplectic form ${\omega }_0={\sum }_{j}d{q}_j\wedge d{p}_j$, for which ${\omega }_0(v,Jv)=|v{|}^2$. In real coordinates $s,t$ on the disk, holomorphicity and the Cauchy-Riemann equations implyTherefore the pulled-back two-form isIf this integral were zero, the continuous nonnegative integrand would vanish everywhere in the interior. The displayed Cauchy-Riemann relation would then make both first derivatives zero, forcing $h$ to be constant on the connected disk and, by continuity, on its boundary. For the nonconstant map under consideration the symplectic area is consequently strictly positive:Take the primitive $\lambda =\frac{1}{2}{\sum }_j(q_{j}d{p}_j-p_{j}d{q}_j)$ from Question 3. Let $c=h{|}_{\partial D^2}$, oriented as the boundary of the disk. Since $c$ lies in $L$, the Stokes theorem givesThe restriction $\lambda {|}_L$ is closed because $L$ is a Lagrangian submanifold. If its Liouville class of a Lagrangian submanifold were zero, it would be an exact differential form, and its integral along every closed curve in $L$ would vanish. The boundary loop above contradicts this. The Liouville class is therefore nonzero. This proves that a holomorphic disk detects a nonzero Liouville class directly by positivity and Stokes; no disk-existence theorem is being assumed beyond the disk given here.