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When the Poisson equation for differential forms has one solution ${\alpha }_0$, all solutions are its translates by the kernel of a linear map of the Hodge Laplacian, namely the harmonic differential forms. The Hodge decomposition theorem identifies this translation vector space canonically with de Rham cohomology. There is no preferred affine origin until an extra normalization is chosen; orthogonality to harmonic forms picks the Green-operator solution.

On a compact oriented Riemannian manifold without boundary, a smooth $(n-p)$-form $\chi$ annihilates all harmonic $p$-forms under wedge integration exactly when its harmonic projection vanishes. Indeed ${\int }_M\phi \wedge \chi =⟨\ast \phi ,\chi {⟩}_{L^2}$, and the Hodge star maps ${\mathcal{H}}^p$ isomorphically onto ${\mathcal{H}}^{n-p}$. By the Hodge decomposition theorem, this annihilator isThus both exact and coexact additions are allowed; a harmonic representative is unique. The potentials of these additions need not be unique, and out-of-range degree spaces are zero. For a concrete coexact ambiguity, take the torus $(\mathbb{R}/2\pi \mathbb{Z}{)}^2$ with metric $d{x}^2+d{y}^2$. The form $\chi =\delta (\sin ydx\wedge dy)=\cos ydx$ annihilates every harmonic one-form by orthogonality, yet $d\chi =\sin ydx\wedge dy$ is nonzero. Thus an annihilating representative need not even be closed, let alone exact.

On a closed connected Riemannian manifold with nonnegative Ricci curvature, harmonic one-forms are parallel under nonnegative Ricci curvature. Evaluation at one point injects their space into the cotangent space. The Hodge decomposition theorem identifies its dimension with the first Betti number, giving $b_1(M)\le \dim M$. For example, a genus-two closed orientable surface times a circle has dimension three and first Betti number five, so no Riemannian metric on that product has nonnegative Ricci curvature.

On a closed oriented Riemannian manifold, a smooth isometric action of a connected Lie group fixes every harmonic differential form. Each group element is connected to the identity through a smooth path, so its action is smoothly homotopic to the identity and fixes de Rham cohomology. Isometries preserve the Hodge Laplacian; uniqueness of the harmonic representative, from the Hodge decomposition theorem, then makes each form itself fixed.

A harmonic one-form is a differential one-form $\alpha$ satisfying $\Delta \alpha =0$ for the Hodge Laplacian. On a closed manifold, the Hodge decomposition theorem identifies these forms with degree-one de Rham cohomology. The Bochner-Weitzenbock formula for one-forms implies that harmonic one-forms are parallel under nonnegative Ricci curvature.

Hodge theory connects de Rham cohomology with harmonic differential forms through the Hodge Laplacian on a compact Riemannian manifold. The Hodge decomposition theorem supplies a unique harmonic representative of each cohomology class. On a compact Kähler manifold, the decomposition additionally respects the complex bidegrees of differential forms.

On an oriented Riemannian $n$-manifold, the metric volume form is the unique top form taking value one on every positively oriented orthonormal tangent basis. In a positive coordinate chart it isFor another positive coordinate chart $y$, write $A=\partial x/\partial y$. Then $g_y=A^{\mathsf{T}}{g}_{x}A$, so $\sqrt{\det g_y}=\det A\sqrt{\det g_x}$, since $\det A>0$. The wedge of coordinate differentials acquires the same determinant factor. The two local expressions agree, proving coordinate independence and smooth global existence. Positivity of the chosen charts is what fixes the sign of the volume form.

The metric induces the inner product on exterior powers of a cotangent space. Define the Hodge star operator $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ byfor forms of the same degree, and apply it fiberwise to smooth forms. In an oriented orthonormal coframe, if $I$ is an increasing $p$-index and $I^c$ its ordered complement, $\ast e^I=ϵ(I,I^c)e^{I^c}$. Exchanging the two blocks has sign $(-1{)}^{p(n-p)}$, so

Use the real $L^2$ inner product $(\alpha ,\beta )={\int }_M\alpha \wedge \ast \beta$. For a $p$-form $\alpha$ and a $(p+1)$-form $\beta$, the graded Leibniz rule and the given boundaryless Stokes identity giveThe codifferential acting on $\beta$ has sign $(-1{)}^{n(p+2)+1}$. Applying the star once more and using its square on the $(n-p)$-form $d\ast \beta$ givesThe exponent reduction uses parity, since $p^2\equiv p(\mathrm{mod}2)$. Combining the two displayed identities proves $(d\alpha ,\beta )=(\alpha ,\delta \beta )$. Symmetry of the real inner product supplies preciselyThus Hodge integration by parts in arbitrary degree identifies $\delta$ as the formal adjoint of $d$. The displayed bilinear identity extends to complex differential forms by complex linearity. For positive norm identities on complex forms, use the Hermitian inner product with conjugation instead.

With the nonnegative convention, the Hodge Laplacian isIt is the form-valued extension of the Laplace-Beltrami operator, whose action on functions is $-\mathrm{div}\mathrm{grad}$. A harmonic differential form is a smooth form in $\ker \Delta$. The adjoint identity givesIf $\Delta \alpha =0$, both norms vanish, and smoothness impliesThe closed compact hypotheses are essential for the absence of boundary terms and this global energy conclusion.

The Hodge decomposition theorem states that on a closed oriented Riemannian manifoldis an $L^2$-orthogonal direct sum, with finite-dimensional harmonic space ${\mathcal{H}}^p$. Every smooth form has unique harmonic, exact and coexact components, although primitives of the latter two need not be unique. Equivalently every de Rham cohomology class has a unique harmonic representative. An exact harmonic form belongs to both the harmonic and exact summands; their orthogonality makes its squared norm zero. Therefore $\boxed{\alpha \text{ exact and harmonic}⟹\alpha =0}$. The same conclusion follows directly from $\alpha =d\eta$ and $\Vert \alpha {\Vert }^2=(\delta \alpha ,\eta )=0$.

The exterior derivative is the real-linear degree-one operator $d:{\Omega }^k\to {\Omega }^{k+1}$ characterized by its usual derivative on functions, the graded Leibniz ruleand $d^2=0$. In local coordinates,These rules determine it uniquely, and the coordinate formula shows its existence and invariance under coordinate changes. It also commutes with pullback of a differential form.

On the unit ball let $V=x^i{\partial }_i$ be the radial vector field, and let $F_t(x)=tx$. For $k\ge 1$ define the radial homotopy operator byHere the vectors are identified using the ambient Euclidean coordinates. Equivalently $h_k\omega ={\int }_0^1{t}^{-1}{F}_t^{\ast }({\iota }_V\omega )dt$, where ${\iota }_V$ is the interior product. The apparent factor $t^{-1}$ is harmless: the coordinate coefficients of its integrand contain $t^{k-1}$, so the integral is smooth at $t=0$ and depends smoothly on $x$. This is a linear map to ${\Omega }^{k-1}$; set $h_0=0$.

For $t>0$, differentiation of the coordinate pullback and Cartan's magic formula giveFor completeness, Cartan’s identity follows by checking it on functions and coordinate one-forms: both sides give $V(f)$ on $f$ and $d(V(x^i))$ on $d{x}^i$. Both are degree-zero derivations of the exterior algebra, so the checks imply the identity on every local form. Integrating from zero to one provesFor positive $k$, the pullback by the constant map is zero, so this is the required identity. If $d\omega =0$ and $k\ge 1$, it follows that $\omega =d(h_k\omega )$. A chart about any point contains such a ball, proving the Poincaré lemma. In degree zero the formula is instead $h_{1}df=f-f(0)$: a constant function cannot be the image of the uncorrected homotopy identity. This degree-zero correction to radial homotopy specifies the necessary positive-degree convention in the printed request.

To treat the sphere, cover $S^2$ by the complements of its north and south poles. Each is diffeomorphic to ${\mathbb{R}}^2$, where the same star-shaped homotopy applies. A closed differential one-form $\alpha$ therefore has local primitives $f_N$ and $f_S$. On their overlap $d(f_N-f_S)=0$. This overlap is connected, so the difference is a constant $c$. Replacing $f_N$ by $f_N-c$ makes the primitives agree, and they glue smoothly to a global function. ThusThis proves directly that closed one-forms on the two-sphere are exact.

For the Hodge decomposition theorem, assume a compact oriented Riemannian manifold without boundary. The Hodge star operator is characterized by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$ and obeys $\ast \ast \alpha =(-1{)}^{k(n-k)}\alpha$. The formal $L^2$ adjoint of $d$ on $k$-forms is $\delta =(-1{)}^{n(k+1)+1}\ast d\ast$. Define the Hodge Laplacian $\Delta =d\delta +\delta d$ and the space ${\mathcal{H}}^k=\ker \Delta$ of harmonic differential forms. Integration by parts, with no boundary term, givesHence a form is harmonic exactly when it is both closed and coclosed. The Hodge decomposition theorem states that ${\mathcal{H}}^k$ is finite-dimensional and that there is the orthogonal decompositionIf a closed form has decomposition $\alpha =h+d\beta +\delta \gamma$, then $⟨\alpha ,\delta \gamma ⟩=⟨d\alpha ,\gamma ⟩=0$. Orthogonality consequently forces $\delta \gamma =0$, so its de Rham cohomology class is represented by $h$. An exact harmonic form $h=d\beta$ has $\Vert h{\Vert }^2=⟨\delta h,\beta ⟩=0$, proving uniqueness. Thus

Every top-degree form is $\omega =f{\mathrm{vol}}_g$, and $d\omega =0$ for dimensional reasons. Its Hodge star is $f$, so the formula for $\delta$ shows that $\delta \omega =0$ exactly when $df=0$. On a connected manifold this says that $f$ is constant. Therefore the harmonic top-degree forms on a closed oriented manifold form the spaceAny such smooth manifold admits a Riemannian metric; the assertion is valid for every choice of that metric.

Use the standard boundaryless-manifold convention, so $M$ is a closed manifold. Its orientation and Riemannian metric determine a unique positive Riemannian volume form ${\mu }_g$ taking value one on every positively oriented orthonormal frame. In an oriented atlas it isFor the coordinate invariance of the Riemannian volume form, on an overlap put $J=\partial y/\partial x$. Then $g_x=J^T{g}_{y}J$ and $d{y}^1\wedge \cdots \wedge d{y}^n=(\det J)d{x}^1\wedge \cdots \wedge d{x}^n$. The oriented transition has $\det J>0$, so the square root of the metric determinant and the coordinate volume transform together. The local forms therefore agree and define ${\mu }_g$ globally.

For real $r$-forms the Hodge star operator is defined byThe pointwise wedge pairing is nondegenerate, so this uniquely defines a smooth map $\ast :{\Omega }^r(M)\to {\Omega }^{n-r}(M)$. On an oriented orthonormal coframe, it sends a basis wedge to the signed complementary wedge. Applying it twice gives

For Hodge star eigenvalues on graded differential forms, extend $\ast$ complex-linearly to the direct sum of all form degrees. This interpretation matters: on a fixed degree it is an endomorphism only in middle degree. In positive odd dimension, $r(n-r)$ is even for every $r$, so ${\ast }^2=I$ and only $\pm 1$ can occur. Both occur, with eigenforms $1+{\mu }_g$ and $1-{\mu }_g$.

In positive even dimension, ${\ast }^2$ is $+I$ on even-degree forms and $-I$ on odd-degree forms. Hence ${\ast }^4=I$ and the only possible eigenvalues are $\pm 1,\pm i$. The first two occur as above. For the latter, choose a smooth local orthonormal frame and multiply its first dual differential one-form by a nonzero smooth bump function supported in the chart. This gives a form $\eta$ for which $\eta$ and $\ast \eta$ are independent. For $\lambda =\pm i$,Thus all four occur. Consequently, for $n\ge 1$,On degree $m$ in dimension $2m$, the Hodge star has eigenvalues $\pm 1$ for even $m$ and $\pm i$ for odd $m$; signed complementary middle-degree wedges show both occur for $m\ge 1$. In zero dimensions it is multiplication by the signed unit volume at each point, so only the signs actually present occur.

For Hodge integration by parts in arbitrary degree, let $\alpha \in {\Omega }^{r-1}(M)$ and $\beta \in {\Omega }^r(M)$. The graded Leibniz rule and Stokes theorem on the closed manifold giveUsing the given codifferential and the square of the Hodge star,Thereforewhich proves that $\delta$ is the formal adjoint of $d$. The complex version inserts conjugation in the usual Hermitian inner product.

The Hodge decomposition theorem gives an $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^k(M)=\ker \Delta$ is finite dimensional and consists of the harmonic differential forms. Each de Rham cohomology class has a unique harmonic representative. Equivalently, with harmonic projection $P_{\mathcal{H}}$ there is a smooth Green operator of the Hodge Laplacian $G$ satisfying $\Delta G=G\Delta =I-P_{\mathcal{H}}$.

The solvability condition for the Hodge Poisson equation isNecessity follows from $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. Conversely the orthogonality condition gives $P_{\mathcal{H}}\beta =0$, so ${\alpha }_0=G\beta$ solves the Poisson equation for differential forms.

The nonnegativity of the Hodge Laplacian identityidentifies its kernel with closed and coclosed forms. Thus the affine space of solutions of the Hodge Poisson equation isFor completeness, if a closed form is decomposed as $\omega =h+d\xi +\delta \zeta$, then $d\delta \zeta =0$, and $\Vert \delta \zeta {\Vert }^2=⟨\zeta ,d\delta \zeta ⟩=0$. Hence its cohomology class is represented by $h$. An exact harmonic form $h=d\xi$ must vanish, since $\Vert h{\Vert }^2=⟨\delta h,\xi ⟩=0$. This proves the canonical isomorphism ${\mathcal{H}}^k(M)\cong H_{\mathrm{dR}}^k(M)$, so the solution set is an affine space whose translation vector space is isomorphic to the requested de Rham cohomology.

Start with any unitary connection ${\nabla }_0$. The Hodge decomposition theorem gives a unique real harmonic differential form $\eta$ representing $[i{F}_{{\nabla }_0}]$. Since they have the same de Rham cohomology class, $\eta -i{F}_{{\nabla }_0}=d\beta$ for a smooth real one-form $\beta$. Set $\nabla ={\nabla }_0-i\beta$. The vector-bundle curvature difference formula givesThis proves existence of a harmonic-curvature unitary line connection for any metric and any complex line bundle.

Two such unitary connections have the same vector-bundle curvature because harmonic representatives are unique. Their difference is therefore $i\gamma$ with $d\gamma =0$. When $b_1(X)=0$, write $\gamma =d\phi$ globally. The bundle gauge transformation $u=e^{i\phi }$ changes $\nabla$ to $u^{-1}\nabla u=\nabla +id\phi$. Thus the harmonic-curvature connection is unique up to gauge. On a general base, closed forms modulo the integral-period forms coming from circle-valued gauge transformations leave the torus $H^1(X;\mathbb{R})/(2\pi H^1(X;\mathbb{Z}))$ of possible gauge classes; the hypothesis removes this freedom.

Let $M$ be a compact Riemannian manifold without boundary. Let $\delta =d^{\ast }$ be the formal $L^2$ adjoint operator of the exterior derivative, and let ${\Delta }_H=d\delta +\delta d$ be the Hodge Laplacian. Write ${\mathcal{H}}^r(M)=\ker {\Delta }_H\cap {\Omega }^r(M)$ for the smooth harmonic differential forms. The Hodge decomposition theorem states that this space is finite-dimensional andan $L^2$-orthogonal direct sum. Thus every smooth $r$-form is uniquely the sum of a harmonic, an exact and a coexact form; its exact and coexact components are unique, though the primitives used to write those components need not be unique. AlsoThe isomorphism sends a harmonic form to its de Rham cohomology class. For example, a closed coexact component $\delta \gamma$ must vanish because $\Vert \delta \gamma {\Vert }^2=⟨\gamma ,d\delta \gamma ⟩=0$; an exact harmonic form also vanishes by orthogonality. These facts explain uniqueness of the harmonic representative of a cohomology class.

On an oriented $n$-manifold the codifferential is $\delta =(-1{)}^{n(r+1)+1}\ast d\ast$ on $r$-forms. The general formulation above also holds on nonorientable Riemannian manifolds: use the globally defined Riemannian volume density for the $L^2$ inner product and its formal adjoint, rather than require a global ordinary volume form. The closed-manifold hypothesis is essential unless suitable boundary conditions are added.

By the Hodge decomposition theorem, the first Betti number is $b_1(M)=\dim {\mathcal{H}}^1(M)$. Part (b) makes every element of this space a parallel differential form. For any fixed $p\in M$, consider the linear evaluation mapIt is injective. A parallel form with ${\omega }_p=0$ remains zero under parallel transport along any path from $p$; connectedness of the smooth manifold makes every point reachable by such a path. Equivalently, the derivative of the squared norm of a parallel form is zero in every direction, so its norm is constant on the connected manifold. Either argument gives $\omega =0$ globally. Since $\dim T_p^{\ast }M=n$, the dimension bound for harmonic one-forms under nonnegative Ricci curvature gives

For an explicit obstruction in dimension three, take $N={\Sigma }_2\times S^1$, where ${\Sigma }_2$ is a closed orientable surface of genus two. It is a compact connected smooth three-manifold. A cell decomposition of ${\Sigma }_2$ has one vertex, four one-cells and one two-cell attached by a product of commutators. Its cellular boundary into the one-cells is zero, because each generator has total exponent zero in that attaching word; the boundary of every one-cell is also zero. Therefore $H^1({\Sigma }_2;\mathbb{R})={\mathbb{R}}^4$. The Künneth theorem in degree one and $H^1(S^1;\mathbb{R})=\mathbb{R}$ giveConsequentlyThis is the Betti-number obstruction to nonnegative Ricci curvature. It rules out every Riemannian metric on that underlying manifold, not merely a particular product metric.

Let $g$ be the real Riemannian metric underlying the Hermitian metric, and let $J$ be the complex structure. The fundamental form of a Hermitian manifold isThe identity $g(Ju,Jv)=g(u,v)$ implies $g(Ju,v)=-g(u,Jv)$, so this is an alternating real two-form. Moreover $\omega (Ju,Jv)=\omega (u,v)$. A two-form with this invariance has type $(1,1)$: on two vectors of type $(1,0)$, applying $J$ to both multiplies its value by $i^2=-1$, forcing that value to vanish; the same argument applies to type $(0,1)$. Hence $\omega$ is a real $(1,1)$-form.

Use the complex-linear convention for the complex Hodge star operator, extending the real star. The complex orientation and metric define $dV={\omega }^n/n!$, andHere the Hermitian inner product is linear in the first argument. With this convention, on the even-dimensional underlying real manifold the formal adjoints are $d^{\ast }=-\ast d\ast$, ${\bar{\partial }}^{\ast }=-\ast \partial \ast$ and ${\partial }^{\ast }=-\ast \bar{\partial }\ast$. A conjugate-linear star is another convention, but then the adjoint formulas must change accordingly.

The Hodge Laplacian and Dolbeault Laplacian areAlso write ${\Delta }_{\partial }=\partial {\partial }^{\ast }+{\partial }^{\ast }\partial$. They are nonnegative, since for exampleThe requested Dolbeault Hodge decomposition on a compact Hermitian manifold is a compactness statement: on a compact Hermitian manifold without boundary,The harmonic space is finite-dimensional, and taking the harmonic representative induces an isomorphism ${\mathcal{H}}_{\bar{\partial }}^{p,q}\cong H_{\bar{\partial }}^{p,q}(X)$. Compactness is essential to this formulation; it is not asserted on an arbitrary noncompact Hermitian manifold.

Now suppose $X$ is compact and Kähler, so $d\omega =0$. The permitted Kähler Laplacian identity is ${\Delta }_d=2{\Delta }_{\bar{\partial }}=2{\Delta }_{\partial }$, making the three harmonic spaces identical in each bidegree. A $d$-closed pure-type form has $\partial \alpha =\bar{\partial }\alpha =0$, since these two terms have different types. We will construct a Green operator potential for ddbar exactness, rather than invoking the ddbar lemma as the conclusion to be proved.

Let $G$ be the inverse of ${\Delta }_{\bar{\partial }}$ on the orthogonal complement of its harmonic forms, zero on the harmonic space. It is twice the Green operator of the Hodge Laplacian. The Hodge decomposition theorem and elliptic inversion give ${\Delta }_{\bar{\partial }}G=I-H$, with $H$ harmonic projection. It preserves bidegrees. The operators $\partial$ and $\bar{\partial }$ commute with ${\Delta }_{\bar{\partial }}$: the first follows from ${\Delta }_{\bar{\partial }}={\Delta }_{\partial }$ and ${\partial }^2=0$, and the second directly from ${\bar{\partial }}^2=0$. They and their formal adjoints annihilate harmonic forms, so they preserve the orthogonal complement of the harmonic space. Consequently they commute with $G$, by uniqueness of the inverse on the harmonic complement. Thus $u=G\alpha$ and $v=G^2\alpha$ are killed by both differential operators.

The assumed orthogonality makes $H\alpha =Hu=0$. Applying the two Laplacians givesWe also need the mixed anticommutator, which follows from the permitted Laplacian identity rather than an extra unproved lemma. Expand ${\Delta }_d$ using $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$. Since ${\Delta }_d={\Delta }_{\partial }+{\Delta }_{\bar{\partial }}$, its mixed terms sum to zero. Their two bidegrees $(1,-1)$ and $(-1,1)$ are distinct, so each vanishes separately, in particular ${\bar{\partial }}^{\ast }\partial +\partial {\bar{\partial }}^{\ast }=0$. HenceWe have obtained the explicit answerThe positive assumptions on $p,q$ ensure these degrees are available. Conversely such a form is orthogonal to harmonic forms by integration by parts, so the same proof also explains the harmonic orthogonality criterion for ddbar exactness.

For injectivity of the Lefschetz operator on a Hermitian manifold on real one-forms, work at one point in a metric-orthonormal coframe $(e^1,f^1,\dots ,e^n,f^n)$ with $\omega ={\sum }_j{e}^j\wedge f^j$. If $a\ne 0$, a unitary change of coframe makes $a=c{e}^1$, $c\ne 0$. Thensince the displayed basis three-forms are linearly independent. Therefore $L(a)=a\wedge \omega$ has zero kernel. Injectivity on complex one-forms follows by taking real and imaginary parts, since $\omega$ is real. This pointwise proof of injectivity of powers of the Lefschetz operator for degree one does not require the hard Lefschetz theorem.

Closedness of $\omega$ gives $\bar{\partial }L=L\bar{\partial }$. Using the permitted Kähler identities, specifically $[{\bar{\partial }}^{\ast },L]=i\partial$, givesMultiplying by two proves $[{\Delta }_d,L]=0$. In particular the Lefschetz operator preserves harmonic forms and the injective map on one-forms restricts to an injection ${\mathcal{H}}_d^1\hookrightarrow {\mathcal{H}}_d^3$. The de Rham Hodge theorem identifies these real harmonic spaces with the corresponding real cohomology groups. Taking dimensions yields

The specified bidegree is the conjugate-linear Hodge star, not the complex-linear extension of the real star. If the pointwise Hermitian inner product is linear in its first argument, it is defined byIt is the real Hodge star operator extended complex-linearly and composed with complex conjugation; ${\ast }^2=(-1{)}^{p+q}$. On a compact Hermitian manifold,is a positive definite Hermitian inner product on smooth forms. The formal adjoint is ${\bar{\partial }}^{\ast }=-\ast \bar{\partial }\ast$ in this convention; the final star in this identity is present in the PDF but is corrupted in the TeX aid.

State the relevant Hodge theorem as follows. On a compact manifold without boundary, the Dolbeault Laplacian ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$ is elliptic, self-adjoint and nonnegative. Its harmonic kernel ${\mathcal{H}}_{\bar{\partial }}^{p,q}$ is finite-dimensional, and smooth forms split orthogonally as this kernel plus the image of ${\Delta }_{\bar{\partial }}$. There are the orthogonal projection $H$ and a smooth Green operator $G$ withFor any form $u$, expand the Laplacian term to obtainHarmonic forms are both closed and coclosed, since $({\Delta }_{\bar{\partial }}v,v)=\Vert \bar{\partial }v{\Vert }^2+\Vert {\bar{\partial }}^{\ast }v{\Vert }^2$. They are orthogonal to both other terms; the two images are orthogonal because $(\bar{\partial }a,{\bar{\partial }}^{\ast }b)=({\bar{\partial }}^{2}a,b)=0$. This proves the Dolbeault Hodge decompositionThe closed forms are exactly the first two summands: if $\bar{\partial }u=0$, then $u$ is orthogonal to the adjoint image, and conversely the first two summands are closed. It follows that every Dolbeault cohomology class has a unique harmonic representative. Negative or out-of-range bidegrees mean zero spaces. Compactness and the absence of boundary are necessary for this stated global decomposition.

Now assume a Kähler metric. Set $L=\omega \wedge -$ and $\Lambda =L^{\ast }$. The permitted Kähler identities, stated with signs, arewith ordinary commutators because $\Lambda$ has even degree. Thus ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$, andby ${\partial }^2=0$. The companion identity gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$ in the same way.

Define ${\Delta }_d=d{d}^{\ast }+d^{\ast }d$ and ${\Delta }_{\partial }=\partial {\partial }^{\ast }+{\partial }^{\ast }\partial$. Expanding $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$ makes the two mixed anticommutators vanish, so ${\Delta }_d={\Delta }_{\partial }+{\Delta }_{\bar{\partial }}$. To prove equality of the latter two, substitute the same commutator formulas and use $\partial \bar{\partial }=-\bar{\partial }\partial$:The displayed operators are identical. Therefore the Kähler Laplacian identity is

Let $\eta$ be a $\bar{\partial }$-exact form. It is closed and orthogonal to harmonic forms. The Dolbeault Hodge decomposition, or the Green identity with $\alpha =G\eta$, therefore givesFor the Green construction, $\bar{\partial }\alpha =0$ follows since $G$ commutes with $\bar{\partial }$; equivalently the coexact component of a closed form vanishes. This yields exactly the stated factorization.

If also $\partial \eta =0$, put $\beta ={\bar{\partial }}^{\ast }\alpha$ and $\xi =\partial \beta$. The identities already proved giveThus $\xi$ is harmonic for ${\Delta }_{\bar{\partial }}$ and hence for ${\Delta }_{\partial }$. But it is $\partial$-exact, so it is orthogonal to the harmonic summand in the Hodge decomposition theorem. Consequently $\xi =0$, proving that $\beta ={\bar{\partial }}^{\ast }\alpha$ is $\partial$-closed.

Furthermore $\beta$ is orthogonal to the common harmonic space, since $(\beta ,h)=(\alpha ,\bar{\partial }h)=0$. Its $\partial$-Hodge decomposition then has neither a harmonic term nor a coexact term: closedness kills the coexact term. Therefore $\beta =\partial \chi$ for a form $\chi \in A^{p-1,q-1}$. It follows thatThis proves the requested ddbar lemma, including the intermediate harmonicity and orthogonality steps. The hypotheses $p,q\ge 1$ ensure that the potential has a valid bidegree.

Finally $\eta ={\omega }_2-{\omega }_1$ is a real closed $(1,1)$ form and is $d$-exact. Integration by parts makes it orthogonal to every $d$-harmonic form. By the Kähler Laplacian identity it is also orthogonal to the Dolbeault harmonic space; because it is $\bar{\partial }$-closed, the Dolbeault Hodge decomposition makes it $\bar{\partial }$-exact. Apply the proved ddbar lemma to get $\eta =\partial \bar{\partial }\varphi$ for a smooth complex-valued function. Reality gives $\eta =-\partial \bar{\partial }\overline{\varphi }$, so averaging yieldsWith $f=\mathrm{Im}\varphi$,The potential is defined up to a real constant on each connected component. This is a statement about cohomologous Kähler forms, not about an arbitrary pair of such forms.

For a closed manifold $M$ with an oriented Riemannian metric, define $\delta$ to be the formal $L^2$ adjoint operator of the exterior derivative, equivalently $\delta =(-1{)}^{n(q+1)+1}\ast d\ast$ on $q$-forms. The Hodge Laplacian is ${\Delta }_q=d\delta +\delta d$. The Hodge decomposition theorem states that its space ${\mathcal{H}}^q=\ker {\Delta }_q$ of harmonic differential forms is finite-dimensional and smooth, and thatThe last isomorphism sends a harmonic differential form to its de Rham cohomology class: every class has exactly one harmonic representative. Out-of-range degree spaces are zero.

We first justify the analytic facts needed in the proof. In local coordinates and a smooth frame for the form bundle, the principal symbol of ${\Delta }_q$ is $g^{ij}{\xi }_i{\xi }_{j}I$. This follows from the exterior/contraction identities for $d$ and $\delta$. Freezing the leading coefficients, the Fourier transform estimate for the constant positive quadratic form gives, for each nonnegative integer $s$,Here is how to pass to variable coefficients. On a sufficiently small coordinate ball the difference of leading coefficients has arbitrarily small supremum norm. After differentiating $s$ times, its top-order contribution is bounded by $\epsilon \Vert u{\Vert }_{H^{s+2}}$; derivatives of the coefficients and first-order terms contribute $C_s\Vert u{\Vert }_{H^{s+1}}$. A partition of unity gives the same types of terms from commutators with the cutoffs. Absorb the epsilon term and use Sobolev interpolation to absorb $\Vert u{\Vert }_{H^{s+1}}$ into a small multiple of $\Vert u{\Vert }_{H^{s+2}}$ plus $C\Vert u{\Vert }_2$. This proves the estimate.

The first-order energy form $Q(u,u)=\Vert du{\Vert }_2^2+\Vert \delta u{\Vert }_2^2$ satisfies $Q(u,u)+c\Vert u{\Vert }_2^2\ge c^{\prime }\Vert u{\Vert }_{H^1}^2$ for a sufficiently large $c>0$. This follows from the same frozen-symbol argument at first order, or by integrating the principal part and absorbing the lower-order terms. The Lax-Milgram theorem thus constructs a bounded inverse $R=({\Delta }_q+c{)}^{-1}:L^2\to H^1$ in the weak sense. To obtain $H^2$ regularity of this weak solution, use a local difference quotient ${\delta }_{h}u$ and test the weak equation with $-{\delta }_{-h}({\chi }^2{\delta }_{h}u)$. Positivity of $g^{ij}$ bounds $\Vert \chi {\delta }_h\nabla u{\Vert }_2^2$; derivatives of $\chi$ and difference quotients of the smooth coefficients give terms bounded by $C\Vert u{\Vert }_{H^1}\Vert \chi {\delta }_h\nabla u{\Vert }_2$ and $C\Vert u{\Vert }_{H^1}^2$. The forcing contributes at most $C\Vert f{\Vert }_2(\Vert \chi {\delta }_h\nabla u{\Vert }_2+\Vert u{\Vert }_{H^1})$. Absorption gives a bound independent of $h$, so the difference-quotient criterion yields $u\in H^2$. Differentiating the equation and repeating gives the higher estimates and smoothness for smooth forcing. Thus the use of elliptic regularity here has been justified rather than assumed.

The inverse $R$ is compact as an $L^2$ operator by the Rellich-Kondrachov compactness theorem, and self-adjoint by symmetry of the weak energy form. It is positive and injective with dense range. The spectral theorem for compact self-adjoint operators now gives an orthonormal eigenbasis of ${\Delta }_q$, with nonnegative eigenvalues tending to infinity. Each eigenfunction is smooth by the estimates above. In particular ${\mathcal{H}}^q$ is finite-dimensional and there is a positive gap above its zero eigenvalue.

Let $H_q$ be the orthogonal projection onto ${\mathcal{H}}^q$. Define the Green operator of the Hodge Laplacian by multiplying a positive-eigenvalue coefficient by $1/\lambda$ and sending the zero coefficients to zero. The gap makes this bounded on $L^2$, and the elliptic estimate shows it maps $H^s$ to $H^{s+2}$; hence it sends smooth forms to smooth forms. It obeysFor a smooth $q$-form $\alpha$, expand the left side to obtainThese three summands are mutually orthogonal. The energy identityshows that a harmonic differential form has both $dh=0$ and $\delta h=0$, making it orthogonal to the exact and coexact summands by integration by parts. Also $⟨d\beta ,\delta \gamma ⟩=⟨d^2\beta ,\gamma ⟩=0$. This proves the claimed direct sum and its uniqueness.

Finally let $d\alpha =0$ and write $u=G_q\alpha$. We have $d{\Delta }_{q}u={\Delta }_{q+1}du$ because $d^2=0$. Since $d(\alpha -H_q\alpha )=0$, $du$ is harmonic. It is also exact, so its orthogonality to harmonic forms gives $du=0$. The decomposition therefore reduces to $\alpha =H_q\alpha +d\delta u$: every closed differential form is cohomologous to a harmonic one. If a harmonic $h$ is exact, $h=d\beta$, then $\Vert h{\Vert }_2^2=⟨\delta h,\beta ⟩=0$. Hence its class is zero only when $h=0$, establishing the isomorphism with de Rham cohomology.

The Riemannian volume form of an oriented $n$-dimensional Riemannian manifold is the unique positive $n$-form taking value one on every positively oriented orthonormal frame. In positive coordinates,The metric induces an inner product on exterior powers of the cotangent bundle, with wedges of distinct orthonormal covectors forming an orthonormal basis. The Hodge star is the pointwise linear map $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ specified byIt is an isometry and satisfies $\ast \ast \alpha =(-1{)}^{p(n-p)}\alpha$. With the positive-sign convention, the Hodge Laplacian on differential forms isHere $d$ is the exterior derivative and $\delta$ its formal adjoint; $\delta$ on functions is zero. On functions this is the Laplace-Beltrami operatorThe convention $\mathrm{div}\mathrm{grad}$ gives the negative of this operator. All subsequent Laplacians use the nonnegative convention above.