Exact differential form ($\alpha =d\beta$)

A differential form is exact when it is the exterior derivative of a form of one lower degree. Since $d^2=0$, every exact form is closed.

The $p$th de Rham cohomology group is the quotient of closed $p$-forms by exact $p$-forms.

For a holomorphic submersion, relative differential forms are forms modulo those containing a base differential. The relative exterior derivative gives a complex whose fibrewise hypercohomology is complex cohomology for a smooth proper family. Its filtration by degree defines the Hodge filtration; retaining exactly one base differential in the absolute complex constructs the Gauss–Manin connection.

There are no exact forms in degree zero. A smooth function has zero exterior derivative precisely when it is locally constant, so $H_{\mathrm{dR}}^0(M)$ is the space of locally constant functions. On a connected nonempty smooth manifold, it is $\mathbb{R}$. An overlapping cover by two connected open subsets therefore does not give the direct-sum formula for de Rham cohomology of a finite disjoint union: restriction sends a constant to equal constants on both pieces.

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 complex of smooth differential forms resolves the constant complex sheaf by the Poincaré lemma and consists of fine sheaves. Its global cohomology therefore agrees with sheaf cohomology of that constant sheaf. A good-cover Čech-de Rham double complex gives an explicit comparison, used in Čech-de Rham curvature descent.

For a compact connected oriented boundaryless $n$-dimensional smooth manifold, choose a Riemannian metric. Harmonic functions are constant because $⟨f,\Delta f⟩=\Vert df{\Vert }^2$, and Hodge star commutes with the Hodge Laplacian. Thus the Hodge star operator identifies the constant functions with the harmonic differential forms of top degree, exactly $\mathbb{R}{\omega }_g$. The Hodge decomposition theorem identifies these with the top de Rham cohomology. The metric volume form has nonzero class by Stokes theorem. The harmonic-representative theorem is stated in Denis Auroux's Hodge theory lecture.

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.

For a free smooth action of a finite group $G$ on a smooth manifold $M$, the quotient projection inducesInvariant differential forms descend uniquely through local inverse branches of the covering map. Averaging by $|G{|}^{-1}{\sum }_{g\in G}{g}^{\ast }$ commutes with the exterior derivative, produces invariant representatives of invariant classes, and produces an invariant primitive for every invariant exact form. Hence taking invariant forms and taking invariant cohomology give the same result over the real numbers.

On a compact connected oriented $n$-dimensional smooth manifold without boundary, integration is a linear isomorphism $H_{\mathrm{dR}}^n(M)\to \mathbb{R}$. Thus two top-degree differential forms have the same de Rham cohomology class exactly when their integrals agree.

A compact Lie group has finitely many connected components, each an oriented compact manifold of dimension $n$. Componentwise integration identifies its top-degree de Rham cohomology with one real coordinate per component. For a connected group, if top-degree classes are represented by invariant forms, the invariant volume form on a Lie group spans the cohomology: invariant top forms are one-dimensional, and its positive integral proves that its class is nonzero by the Generalized Stokes theorem. A disconnected group cannot have an invariant representative of every top-degree class, since left translations permute its components. For example, $S^1\times \mathbb{Z}/2$ has two independent degree-one classes but only one independent invariant top form.

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.

Choose a closed one-form $\nu$ on $S^1$ with integral one and let $p:M\times S^1\to M$ be projection. Every rotation-invariant $k$-form is uniquelyThe exterior derivative acts componentwise. Averaging therefore proves thatis the displayed isomorphism.

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

For an open cover $X=U\cup V$, restriction of forms gives a short exact sequence of cochain complexesand hence a long exact sequence in de Rham cohomology.

Cover the sphere by complements of its two poles. Each open set is contractible, and their intersection deformation retracts to the equatorial sphere. The Poincaré lemma and homotopy invariance of de Rham cohomology make the two neighboring groups in the Mayer--Vietoris sequence for de Rham cohomology zero, so its connecting map is an isomorphism. Its inverse takes a top-degree form to the difference of its local primitives on the equator, up to the connecting-map sign convention.