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A symplectomorphism is a diffeomorphism, so it preserves the genus of a surface. It also preserves the orientation determined by the symplectic form and, by integration of a pullback of a differential form, preserves the symplectic area.

Conversely, equal genus of a surface gives a diffeomorphism $f:{\Sigma }_0\to {\Sigma }_1$. We can choose $f$ to preserve the orientations determined by ${\omega }_0,{\omega }_1$: if necessary, compose with an orientation-reversing diffeomorphism of the standard closed orientable surface of that genus of a surface, obtained by reflection of a symmetric handle model. On ${\Sigma }_0$ put $\eta =f^{\ast }{\omega }_1$. Both $\eta$ and ${\omega }_0$ are positive volume forms for the same orientation, and their integrals agree. By top-degree de Rham cohomology, $[\eta ]=[{\omega }_0]$.

The family ${\omega }_t=(1-t){\omega }_0+t\eta$ consists of positive volume forms, hence symplectic forms in dimension two, with constant de Rham cohomology class. Moser theorem supplies $\psi$ with ${\psi }^{\ast }\eta ={\omega }_0$. ConsequentlyTherefore the genus of a surface and symplectic area classify compact connected symplectic surfaces.

The exterior derivative is the real-linear degree-one map $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ characterized by $df(X)=Xf$ on smooth functions, the graded Leibniz ruleand $d^2=0$. In local coordinates $(x^1,\dots ,x^m)$, write $\alpha ={\sum }_I{\alpha }_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$ using multi-index notation. Since $d{x}^i=d(x^i)$ and hence $d(d{x}^i)=0$, the defining rules forceThis proves local uniqueness, and the coordinate formulas agree on overlaps because the same rules are preserved by the pullback of a differential form. They also directly define an operator satisfying all three rules, proving existence.

An exact differential form is a form $\omega =d\eta$. Let $\pi :S^{2n}\to {\mathbb{RP}}^{2n}$ be the antipodal double covering map and let $a(x)=-x$. For a $2n$-form $\omega$ on real projective space, $a^{\ast }{\pi }^{\ast }\omega ={\pi }^{\ast }\omega$, while the mapping degree of the antipodal map is $-1$. ThereforeBy the stated criterion, ${\pi }^{\ast }\omega =d\eta$. The invariant primitive under a finite group actionstill satisfies $d\bar{\eta }={\pi }^{\ast }\omega$ and descends to a form $\beta$ on ${\mathbb{RP}}^{2n}$. Since pullback through a covering is injective on differential forms, ${\pi }^{\ast }(d\beta -\omega )=0$ implies $d\beta =\omega$. Thus every $2n$-form on ${\mathbb{RP}}^{2n}$ is exact; this is the top-degree differential forms on even-dimensional real projective space are exact result.

The $k$th de Rham cohomology isthe closed differential forms modulo the exact ones. For the product, let $p:M\times S^1\to M$ be projection and choose a closed one-form $\nu$ on the circle with ${\int }_{S^1}\nu =1$. Averaging differential forms over the circle is cochain-homotopic to the identity, so every class has a rotation-invariant representative; if such a representative is exact, averaging a primitive gives an invariant primitive. Every invariant $k$-form has a unique decomposition $p^{\ast }\alpha +p^{\ast }\beta \wedge \nu$, andClosedness and exactness are therefore componentwise. Hence the mapis a well-defined bijection $H_{\mathrm{dR}}^k(M)\oplus H_{\mathrm{dR}}^{k-1}(M)\to H_{\mathrm{dR}}^k(M\times S^1)$, proving the de Rham cohomology of a product with a circle formula.

The Hodge decomposition theorem for compact Kähler manifolds gives a unique harmonic differential form representative of every complex de Rham cohomology class, with a decomposition into harmonic pure-type components. ConsequentlyEquivalently, the Dolbeault Hodge decomposition on a compact Hermitian manifold is

The integral of an exact differential form over the closed curve $S^1$ is zero by Stokes theorem, so the map is well defined on de Rham cohomology. It is surjective because the angular form $d\theta$ has integral $2\pi$. If a closed one-form $\alpha$ has zero integral, defineThe zero period makes this definition $2\pi$-periodic, and $dF=\alpha$. Thus the kernel is zero andis an isomorphism.

For $h=z^{-k}f$,The scalar-valued form $h^{-1}dh$ is closed, andBy the period isomorphism from part (b), $[\alpha ]=0$ in complexified de Rham cohomology. Hence there is a smooth complex-valued function $\phi$ on $U_0\cap U_1$ with $\alpha =d\phi$.

The de Rham cohomology of $X$ isBecause pullback commutes with $d$, it sends closed forms to closed forms and exact forms to exact forms. Thus a smooth map induces

Let $H:X\times I\to Y$ be a smooth homotopy and write $H_t(x)=H(x,t)$. If $V={\partial }_t$, Cartan's magic formula givesIntegrating defines a degree-minus-one operator $K$ satisfyingFor closed $\omega$, the difference is exact, so smoothly homotopic maps induce the same map on de Rham cohomology.

If $F:X\to Y$ is a homotopy equivalence with inverse up to homotopy $G$, functoriality and homotopy invariance giveHence $F^{\ast }$ is an isomorphism.

Moser's trick states that if $M$ is compact and ${\omega }_t$, $0\le t\le 1$, is a smooth family of symplectic forms whose de Rham cohomology class is independent of $t$, then there is an isotopy ${\varphi }_t$ withSince $[{\dot{\omega }}_t]=0$, choose a smooth family of one-forms ${\sigma }_t$ with ${\dot{\omega }}_t=d{\sigma }_t$. Nondegeneracy of ${\omega }_t$ uniquely determines a vector field $X_t$ byCompactness makes its flow ${\varphi }_t$ exist for the whole interval. Cartan's magic formula and $d{\omega }_t=0$ givewhich proves the theorem.

Smooth degree-$d$ hypersurfaces form the complement of the discriminant in the projective space of degree-$d$ homogeneous polynomials. This complement is path connected, so $X$ and $X^{\prime }$ lie in a smooth one-parameter family. The Ehresmann fibration theorem identifies the fibers smoothly. Under such an identification, the restrictions of the Fubini-Study form form a family ${\omega }_t$ whose cohomology class is the fixed restricted hyperplane class. Moser's trick therefore gives the symplectic equivalence of smooth projective hypersurfaces.

It remains to construct the finite subgroup for one convenient hypersurface. On the Fermat hypersurfacethe group $({\mu }_d{)}^{n+1}$ acts by diagonal coordinate multiplication. It preserves both $X_F$ and the Fubini-Study form. The kernel of its projective action is the diagonal subgroup ${\mu }_d$, so the effective Fermat hypersurface diagonal symmetry group isConjugating this action by a symplectomorphism $X_F\to X$ gives the required subgroup of $\mathrm{Symp}(X,{\omega }_{FS}{|}_X)$.

The de Rham cohomology isThe Poincaré lemma says that every closed positive-degree differential form is locally exact, and is exact on every star-shaped open subset of Euclidean space.

Let $n>1$ and let $\alpha$ be a closed one-form on $M$. The hypothesis gives $\alpha =dg$ on $M\setminus B$. Choose a slightly larger coordinate ball $B^{\prime }$ around $B$. The Poincare lemma gives $\alpha =dh$ on $B^{\prime }$. The annulus $B^{\prime }\setminus B$ is connected when $n>1$, so $d(g-h)=0$ there and $g-h$ is constant. Adjusting $h$ by this constant makes $g$ and $h$ agree on the overlap, and they glue to a global primitive of $\alpha$. Hence $H^1(M)=0$. For $n=1$ the claim fails: remove a closed proper interval from $S^1$. Its complement is an interval and has vanishing first de Rham cohomology, whereas $H^1(S^1)\cong \mathbb{R}$.

Finally choose a nowhere-vanishing $n$-form $\omega$ on $S^n$ and a coordinate $t$ on the finite interval $I$. Every $(n+1)$-form on $S^n\times I$ is $a(x,t)\omega \wedge dt$. Fix $t_0\in I$ and putSince $d_{S^n}A\wedge \omega =0$ for dimensional reasons,Every top-degree form is exact, so $H^{n+1}(S^n\times I)=0$ without using the de Rham theorem.

An orientation selects the positive ordered bases in each tangent space. On an oriented $n$-dimensional Riemannian manifold, the Riemannian volume form ${\omega }_g$ is the unique smooth $n$-form satisfyingfor every positively oriented orthonormal frame. In positively oriented local coordinates,

The metric induces an inner product on the bundle ${\Lambda }^p{T}^{\ast }M$ of $p$-forms. The Hodge star operator is the unique linear mapsuch thatfor all $p$-forms $\alpha ,\beta$. With the codifferential $\delta$, the Laplace-Beltrami operator on differential forms is

The Hodge decomposition theorem says that on a compact oriented Riemannian manifold,an $L^2$-orthogonal direct sum, where ${\mathcal{H}}^p(M)=\ker \Delta$ is the finite-dimensional space of harmonic $p$-forms. Every de Rham cohomology class has exactly one harmonic representative.

Every smooth closed differential form of positive degree is locally an exact differential form. On a coordinate ball centered at zero, put $F_t(x)=tx$ and let $R=x^i{\partial }_i$ be the radial vector field. Define the degree-lowering operatorFor a $p$-form with $p>0$, the integrand has a factor $t^{p-1}$ in its coordinate coefficients and is smooth at zero. Cartan's magic formula and differentiation of the pullback of a differential form give ${\partial }_t(F_t^{\ast }\alpha )=t^{-1}{F}_t^{\ast }(d{\iota }_R\alpha +{\iota }_{R}d\alpha )$. Integrating from zero to one gives $dK\alpha +Kd\alpha =\alpha$, since $F_0^{\ast }\alpha =0$ in positive degree. If $d\alpha =0$, take $\beta =K\alpha$. The local potential of a closed differential one-form gives the explicit degree-one construction. Global exactness can fail through nonzero periods, measured by de Rham cohomology.

A polydisc is a product of open discs in $\mathbb{C}$. It is a contractible complex manifold, with vanishing positive-degree de Rham cohomology and vanishing Dolbeault cohomology in positive antiholomorphic degree.

For a real vector bundle $E$, its Pontryagin classes are the even-index Chern classes of its complexification with the displayed alternating signs. They lie in integral cohomology of degrees $4k$ and obey a total-class product formula under direct sums. If a metric connection on a vector bundle has curvature matrix $R$, its first real representative is $p_1(E)=-\mathrm{tr}(R\wedge R)/(8{\pi }^2)$. The Bianchi identity makes this representative closed, and Chern-Weil connection transgression makes its de Rham cohomology class independent of the connection. This normalization gives degree-four A-hat form $-p_1/24$.

On a nonempty compact Kähler manifold of complex dimension $n$, the Kähler form is closed and ${\int }_X{\omega }^n>0$. If ${\omega }^k=d\eta$ for $1\le k\le n$, then ${\omega }^n=d(\eta \wedge {\omega }^{n-k})$; Stokes theorem contradicts the positive integral. Thus the de Rham cohomology is nonzero in every even degree through $2n$.

On a compact oriented smooth manifold $N$ of dimension four without boundary, set $Q([\alpha ],[\beta ])={\int }_N\alpha \wedge \beta$ for closed real two-forms. The Stokes theorem makes the expression independent of representatives, and the even degrees make it symmetric. It is the real de Rham cohomology version of the intersection form. The Hodge decomposition theorem and Hodge star operator express it as $Q(h,k)=(h,\ast k{)}_{L^2}$ on harmonic representatives.

A solid torus is the product of a circle with a closed disk. Its boundary is the torus $S^1\times S^1$. It deformation retracts onto its central circle, so its degree-one de Rham cohomology is one dimensional and injects into that of its boundary.

On a nonempty compact $2n$-dimensional symplectic manifold without boundary, with $n\ge 1$, the class $[\omega {]}^n$ in top-degree de Rham cohomology is nonzero, since its integral in the symplectic orientation is positive. In particular $\omega$ cannot be an exact differential form: if $\omega =d\alpha$, then ${\omega }^n=d(\alpha \wedge {\omega }^{n-1})$ and the Generalized Stokes theorem would make that integral zero. Thus vanishing second de Rham cohomology obstructs a closed positive-dimensional symplectic manifold.

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.

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.

The difference of two connections on a vector bundle is a global endomorphism-valued one-form. Taking the trace of the curvature difference formula cancels both the commutator terms and the trace of the square of that one-form. Consequently the trace curvatures represent the same de Rham cohomology class.