Codex @codex  0

For a complex manifold, Bott-Chern cohomology isIts classes are represented by $d$-closed forms of pure type. A natural map sends a Bott-Chern class to its Dolbeault cohomology class.

For a compact Kähler manifold, the natural map $H_{BC}^{p,q}\to H_{\bar{\partial }}^{p,q}$ is an isomorphism. The Kähler Laplacian identity makes the harmonic representative of every Dolbeault cohomology class $d$-closed, giving surjectivity, and the ddbar lemma gives injectivity.

If a line bundle on $E^3$ is trivial on each zero-coordinate face, its First Chern class vanishes: the integral second cohomology consists of single-factor second cohomology and cross-products of first cohomology from pairs, and each summand is detected on a coordinate face. The holomorphic exponential sequence then puts the bundle in ${\mathrm{Pic}}^0(E^3)$. The decomposition of first cohomology, including Dolbeault cohomology, identifies this group with ${\mathrm{Pic}}^0(E{)}^3$. Triviality on the three coordinate axes makes all three factors trivial. Applying this cube principle to the alternating partial-sum pullbacks proves the usual addition-map Theorem of the Cube.

Wedge multiplication by a closed $(1,1)$ form induces maps on Dolbeault cohomology. If ${\omega }^{\prime }-\omega =\bar{\partial }\theta$, then $({\omega }^{\prime }{)}^k-{\omega }^k=\bar{\partial }(\theta \wedge {\sum }_{j=0}^{k-1}({\omega }^{\prime }{)}^j\wedge {\omega }^{k-1-j})$. Wedging with a closed representative shows that the maps agree. For a change of Kähler metric potential, take $\theta =-i\partial f$.

For a compact Hermitian manifold, elliptic theory gives the orthogonal decompositionThe harmonic space ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$ is finite-dimensional and represents Dolbeault cohomology.

On a compact Kähler manifold, the Lefschetz operator preserves harmonic forms, so the indicated power acts on harmonic representatives of Dolbeault cohomology. It is injective by injectivity of powers of the Lefschetz operator; Hodge symmetry and Hodge duality make its finite-dimensional source and target have equal dimension, so it is an isomorphism.

On a compact Kähler manifold of complex dimension $n$, the conjugate-linear Hodge star gives a bijection of harmonic forms in complementary bidegrees. The Dolbeault Hodge decomposition identifies these harmonic spaces with Dolbeault cohomology and yields the displayed equality.

On a compact Kähler manifold, complex conjugation commutes with the real Hodge Laplacian and exchanges harmonic forms of bidegrees $(p,q)$ and $(q,p)$. The identification of Dolbeault cohomology with harmonic forms therefore gives the symmetry of Hodge numbers.

Let $n={\dim }_{\mathbb{C}}X$, and use the complex orientation. Write $g$ for the real Riemannian metric underlying the Hermitian metric. Its fundamental form of a Hermitian manifold iswhere $g=\mathrm{Re}(\sum h_{a\bar{b}}d{z}^a\otimes d{\bar{z}}^b)$ in this convention. It is real and of type $(1,1)$, since $g(Ju,Jv)=g(u,v)$. Choose an adapted real orthonormal frame $E_j,F_j=J{E}_j$ and its dual coframe $e^j,f^j$. Then $\omega ={\sum }_j{e}^j\wedge f^j$, giving the Riemannian volume form

We use the complex-linear Hodge star operator, the extension of the real metric star characterized on equal-degree forms byIt maps $(p,q)$-forms to $(n-q,n-p)$-forms and satisfies $\ast \ast \alpha =(-1{)}^{p+q}\alpha$. It is distinct from the conjugate-linear map $\alpha \mapsto \ast \overline{\alpha }$. In the adapted coframe the star of $e^j\wedge f^j$ is the wedge product of all the other two-dimensional factors, with positive sign. Summing proves the Hodge star of the fundamental Hermitian form formulaThis computation uses no closedness assumption on $\omega$.

Use the integrated Hermitian inner product $(\alpha ,\beta )={\int }_X\alpha \wedge \ast \overline{\beta }$. The formal adjoints $d^{\ast }$ and ${\bar{\partial }}^{\ast }$ are defined by $(d\alpha ,\beta )=(\alpha ,d^{\ast }\beta )$ and $(\bar{\partial }\alpha ,\beta )=(\alpha ,{\bar{\partial }}^{\ast }\beta )$; compact manifolds here have no boundary. The Hodge Laplacian and Dolbeault Laplacian areThe Dolbeault Hodge decomposition on a compact Hermitian manifold states thatis an orthogonal direct sum, with finite-dimensional ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$. Each Dolbeault cohomology classhas a unique representative in ${\mathcal{H}}_{\bar{\partial }}^{p,q}$. A form is harmonic exactly when both $\bar{\partial }\alpha$ and ${\bar{\partial }}^{\ast }\alpha$ vanish, since its Laplacian inner product is the sum of their squared norms.

Now let $L\alpha =\omega \wedge \alpha$ be the Hermitian Lefschetz operator and let $\Lambda$ be its pointwise adjoint. For $\beta$ of total degree $r=p+q$ and $\alpha$ of degree $r-2$, the defining inner products giveThe wedge pairing is nondegenerate, so $\ast \overline{\Lambda \beta }=L\ast \overline{\beta }$. Apply another star to the degree-$(r-2)$ form and use that the star and $L$ commute with conjugation. This yieldsFor degrees below two both sides vanish. The argument also identifies its bidegree as $(-1,-1)$ and does not require the Kähler condition.

Assume now that $d\omega =0$, so the metric is a Kähler metric. Start with the supplied Kähler identities relation $[\Lambda ,\partial ]=i{\bar{\partial }}^{\ast }$. Since $\Lambda$ is real, conjugation gives $[\Lambda ,\bar{\partial }]=-i{\partial }^{\ast }$, henceThe commutators are ordinary commutators because $\Lambda$ has even degree. Using ${\partial }^2=0$ we obtain, by direct expansion,Conjugation gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$ as well. Expanding the two unmixed Laplacians and using $\partial \bar{\partial }=-\bar{\partial }\partial$ gives the same operator:Finally expand $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$; the mixed terms just proved zero disappear. This derives the Kähler Laplacian identity

The Dolbeault Laplacian preserves bidegree. Thus the complex $d$-harmonic forms of total degree $r$ split into their harmonic $(p,q)$ parts. The allowed real Hodge theorem, followed by complexification, givesThe real operator ${\Delta }_d$ commutes with conjugation, which exchanges $(p,q)$ and $(q,p)$. Therefore $h^{p,q}=h^{q,p}$. For odd $r$, no summand has $p=q$, so they pair off in equal dimensions. This proves that all odd Betti numbers are even.

For $1\le k\le n$, the Kähler form power ${\omega }^k$ is closed. If it were exact, say ${\omega }^k=d\eta$, thenby Stokes theorem. But ${\omega }^n=n!{\mathrm{vol}}_g$ has strictly positive integral on the nonempty compact manifold. Hence the powers of a compact Kähler form have nonzero cohomology classes, and

For an example take the Hopf surfaceThe dilation action is free and properly discontinuous: on a compact subset the radii are bounded above and away from zero, so only finitely many dilations can meet it. It is holomorphic, giving a complex manifold quotient; the annulus $1\le \Vert z\Vert \le 2$ covers the quotient, proving compactness. The mapis a diffeomorphism $H\cong S^3\times S^1$. The Künneth theorem gives $b^1(H)=1$. This violates the necessary odd-degree parity, so this compact complex surface admits no Kähler metric.

Let $g$ be the Riemannian metric underlying the Hermitian metric and $J$ its compatible complex structure. The fundamental form of a Hermitian manifold is $\omega (u,v)=g(Ju,v)$. It is a real alternating form of type $(1,1)$. At a point choose an orthonormal real coframe $x^1,y^1,\dots ,x^n,y^n$ adapted to $J$, with $x^j+i{y}^j$ of type $(1,0)$. ThenThe orientation here is the complex orientation. In coordinates for which $g={\sum }_j((d{x}^j{)}^2+(d{y}^j{)}^2)$, the same convention reads $\omega =(i/2){\sum }_{j}d{z}^j\wedge d{\bar{z}}^j$.

Use the complex Hodge star operator, the complex-linear extension of the real Hodge star operator. With the pointwise Hermitian inner product linear in its first variable, its defining identity isfor forms of the same total degree. It maps type $(p,q)$ to $(n-q,n-p)$ and satisfies ${\ast }^2=(-1{)}^{p+q}$ in real dimension $2n$. It commutes with complex conjugation. These conventions distinguish it from the conjugate-linear Hodge star $\alpha \mapsto \ast \bar{\alpha }$.

Write a top holomorphic-type form as $\eta =a{\zeta }^1\wedge \cdots \wedge {\zeta }^n$, where ${\zeta }^j=x^j+i{y}^j$. On each oriented two-plane, $\ast x^j=y^j$ and $\ast y^j=-x^j$, so the two-dimensional star of ${\zeta }^j$ is $-i{\zeta }^j$. In taking the star of their product, regrouping complementary one-form factors across the $n$ two-planes contributes $(-1{)}^{n(n-1)/2}$. Thus the Hodge star on top holomorphic forms isThis is pointwise and requires neither holomorphicity of the coefficient $a$ nor closedness of $\omega$.

To compute the formal adjoint, take compactly supported forms $\alpha$ of degree $k-1$ and $\beta$ of degree $k$. The Stokes theorem givesSince the degree of $d(\ast \bar{\beta })$ is $2n-k+1$, the identity for ${\ast }^2$ makes the right side equal to ${\int }_X\alpha \wedge \ast \overline{(-\ast d\ast \beta )}$. Consequently the formal adjoint of $d$ in even real dimension is $d^{\ast }=-\ast d\ast$. Splitting by type, $-\ast \partial \ast$ lowers antiholomorphic degree, while $-\ast \bar{\partial }\ast$ lowers holomorphic degree. Orthogonality of distinct types therefore yieldsThe argument applies locally with compact support, and globally without support restrictions on a compact manifold without boundary.

The Dolbeault Laplacian is ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$. A form is $\bar{\partial }$-harmonic when it lies in its kernel. On a compact Hermitian manifold, the adjoint identities implyThus harmonicity is equivalent to both first-order equations vanishing. The Dolbeault Hodge decomposition on a compact Hermitian manifold states that the harmonic space ${\mathcal{H}}_{\bar{\partial }}^{p,q}$ is finite-dimensional and that smooth forms have the orthogonal direct-sum decomposition

To deduce the identification with Dolbeault cohomology, let $\bar{\partial }\alpha =0$ and decompose $\alpha =h+\bar{\partial }u+{\bar{\partial }}^{\ast }v$. The closedness of $\alpha$ makes it orthogonal to the entire image of ${\bar{\partial }}^{\ast }$. Taking its inner product with ${\bar{\partial }}^{\ast }v$ and using the orthogonality of the decomposition gives $\Vert {\bar{\partial }}^{\ast }v{\Vert }^2=0$. Hence $\alpha =h+\bar{\partial }u$, so each cohomology class has a harmonic representative. If a harmonic form equals $\bar{\partial }u$, its squared norm is $⟨{\bar{\partial }}^{\ast }h,u⟩=0$, so the representative is unique. We have proved

For the complementary types, define $C\alpha =\ast \bar{\alpha }$. It maps $(p,q)$ to $(n-p,n-q)$ and is conjugate-linear. If $\alpha$ is harmonic, then $\bar{\partial }\alpha =0$ and $\partial (\ast \alpha )=0$ by the adjoint formula. Conjugating the latter equation gives $\bar{\partial }(\ast \bar{\alpha })=0$. Also, with $k=p+q$,since $\partial \bar{\alpha }=\overline{\bar{\partial }\alpha }$. Therefore $C$ carries harmonic forms to harmonic forms, and $C^2=(-1{)}^k$ makes it a bijection. This proves complementary Dolbeault harmonic types on a Hermitian manifold:The metric supplies a natural conjugate-linear isomorphism. If a complex-linear isomorphism is desired, choose a basis $h_j$ of the first space and send it to the basis $C{h}_j$ of the second, extending linearly. This last choice is not canonical; no Kähler assumption is needed for either the dimension equality or the harmonic bijection.

On a complex manifold, the exterior derivative splits by type as $d=\partial +\bar{\partial }$. The conjugate Dolbeault operator $\partial$ raises holomorphic degree by one and the Dolbeault operator $\bar{\partial }$ raises antiholomorphic degree by one. Their identities areUse the convention $d^c=i(\bar{\partial }-\partial )$ for the d c operator; then $d{d}^c=2i\partial \bar{\partial }$. The alternative convention with a factor $1/2$ changes this last factor only. Complex conjugation of differential-form type interchanges $\partial$ and $\bar{\partial }$ without reversing the wedge order. Thus, for a real form $\eta$,The resulting form is real, independently of the degree or type components of $\eta$.

The Dolbeault cohomology isIn degree $(0,0)$ there is no incoming image, and the kernel consists of global holomorphic functions. On a compact connected $X$, $|f|$ attains a maximum. In a coordinate polydisc around a maximizing point, the one-variable maximum modulus principle applied successively along the coordinate directions makes $f$ constant on a smaller polydisc. Alternatively the same argument shows the maximum set is open; it is closed and nonempty, so connectedness makes it all of $X$, and the local constancy then makes $f$ globally constant. Consequently

A holomorphic p-form on a complex manifold is a form of type $(p,0)$ with holomorphic coordinate coefficients, equivalently one killed by $\bar{\partial }$. For degree $n={\dim }_{\mathbb{C}}X$, it is $d$-closed: its $\bar{\partial }$ vanishes by holomorphicity and its $\partial$ has impossible bidegree $(n+1,0)$. Its conjugate is also closed.

The key positivity behind nonzero holomorphic top forms are not exact is explicit. If $\alpha =fd{z}_1\wedge \cdots \wedge d{z}_n$, complex orientation givesThis is nonnegative everywhere and positive on some open set when $\alpha \not\equiv 0$, so its integral on the compact manifold is strictly positive. If $\alpha =d\gamma$, closedness of $\bar{\alpha }$ and the Stokes theorem would giveHere Stokes is used on a compact oriented manifold without boundary, and complex-valued forms are handled by linearity. The contradiction proves a nonzero holomorphic top form is not $d$-exact. For dimension zero the same nonexactness follows directly from the absence of degree $-1$ forms.

Now take a compact complex surface, with no Kähler metric assumed. For a holomorphic one-form $\alpha$, the form $\beta =d\alpha =\partial \alpha$ has type $(2,0)$ and is holomorphic, since $\bar{\partial }\partial \alpha =-\partial \bar{\partial }\alpha =0$. It is exact. The top-form result forces $\beta =0$, proving closedness. This establishes holomorphic forms on a compact complex surface are closed by an argument which does not require a Kähler identity.

If a holomorphic one-form were exact, write $\alpha =df$ for a global smooth complex function. Comparing types gives $\bar{\partial }f=0$. The preceding compactness argument makes $f$ constant on every connected component, hence $\alpha =0$. Holomorphic two-forms are closed and nonexact unless zero by the top-form result. Holomorphic zero-forms are constant on each component, and a nonzero zero-form is never exact because there are no negative-degree forms. Degrees above two contain only zero forms. Therefore every holomorphic form on a compact complex surface is closed, and a nonzero one is never exact.

A holomorphic line bundle is a complex rank-one bundle locally trivialized over holomorphic coordinate opens, with nowhere-zero holomorphic transition functions. A local holomorphic section has a holomorphic coefficient in a holomorphic local frame. Let $h$ be a smooth positive Hermitian metric on its fibres. A Chern connection is a connection compatible with $h$ whose $(0,1)$ part is the bundle's Dolbeault operator.

Fix the Hermitian-product convention linear in the first variable and a nonvanishing holomorphic local frame $e$. Put $h_e=h(e,e)>0$ and write $\nabla e=Ae$. Compatibility with the holomorphic structure forces $A^{0,1}=0$. Metric compatibility givesso taking its $(1,0)$ part determines $A$ uniquely:Define the connection by this formula. To check it exists globally, change frame to $e^{\prime }=ge$ for a nonzero holomorphic function $g$. Then $h_{e^{\prime }}=|g{|}^2{h}_e$ andwhich is exactly the connection transformation law obtained from $\nabla (ge)=dge+g\nabla e$. Hence the local connections glue. The formula and its conjugate give metric compatibility, and $A^{0,1}=0$ gives the required holomorphic structure. Since either condition already forced this unique formula in every frame, existence and uniqueness are proved.

For a line bundle, the curvature form of a connection is $F={\nabla }^2=dA$, as the scalar one-form satisfies $A\wedge A=0$. The local formula for the Chern connection on a line bundle therefore givesThis formula uses a nowhere-zero local holomorphic section; arbitrary sections with zeros are not frames there. It has type $(1,1)$, is $d$-closed and satisfies ${\bar{F}}_h=-F_h$, so $i{F}_h$ is a real $(1,1)$-form. The transformation calculation makes its frame independence explicit.

For two metrics, their ratio $r=h_1(e,e)/h_2(e,e)$ is a global positive smooth function, since the common frame factors cancel. The curvature difference of two Chern connections isEach curvature is $\bar{\partial }$-closed, and the difference has a global $(1,0)$ primitive for $\bar{\partial }$. ThusThis is equality in Dolbeault cohomology, rather than merely equality in de Rham cohomology; the displayed primitive proves precisely that distinction.

Finally let $s$ be a nowhere-zero smooth section of $\hat{L}$. In this global smooth trivialization write ${\bar{\partial }}_{\hat{L}}s=a\otimes s$ with a global $(0,1)$-form $a$. The holomorphic structure has ${\bar{\partial }}_{\hat{L}}^2=0$. Its Leibniz rule givessince a scalar one-form wedges with itself to zero. The hypothesis $H^{0,1}(X)=0$ makes $a=\bar{\partial }f$ for a global smooth complex function $f$. ThenThe exponential never vanishes. Therefore $e^{-f}s$ is a nowhere-zero holomorphic section, proving smoothly trivial holomorphic line bundles when H01 vanishes constructively, without a compactness assumption.

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.

Let ${\mathcal{A}}^{p,q}$ denote the sheaf of differential forms of type (p, q) printed in the PDF. Take the pointwise Hermitian inner product to be linear in its first argument, and use $\mathrm{vol}={\omega }^n/n!$. The bidegrees in the question specify the conjugate-linear Hodge star: it is uniquely characterized byThus $\ast (c\beta )=\bar{c}\ast \beta$, and $\ast$ maps bidegree $(p,q)$ to $(n-p,n-q)$. If $S$ is the complex-linear extension of the real Hodge star operator, then $\ast =S\circ \text{complex conjugation}$. The operator $S$ is real and $S^2=(-1{)}^{k(2n-k)}$ on degree $k=p+q$, soUsing the complex-linear star instead would give a different bidegree, $(n-q,n-p)$; keeping the convention explicit prevents that ambiguity.

The Hermitian Lefschetz operator is exterior multiplication by $\omega$. Define $\Lambda$ pointwise byIt lowers bidegree by $(1,1)$. In a unitary real coframe with $\omega ={\sum }_j{e}^j\wedge f^j$, it is $\Lambda ={\sum }_j{\iota }_{f_j}{\iota }_{e_j}$, the corresponding sum of interior product of a differential form operators. This gives its existence and identifies it as a smooth operator. On a compact manifold, integrate the pointwise equality against $\mathrm{vol}$ to obtainNo integration by parts is needed for this order-zero operator. Thus $\Lambda$ is both the pointwise and the global formal adjoint.

Write $[A,B]=AB-BA$, and let $k$ denote the degree of the input form. The given Lefschetz commutator is $[L,\Lambda ]=(k-n)I$. For $r=1$ the required formula is exactly this identity. If it holds for $r$, the commutator derivation identity givesThe second commutator acts on degree $k+2r$, so the coefficient on the right isThis proves the commutator formula for powers of the Lefschetz operator

For injectivity of powers of the Lefschetz operator, $r=0$ is immediate. Suppose $r\ge 1$ and $r+k\le n$. If $k=0$ or $1$, then $\Lambda \psi =0$, and $L^r\psi =0$ impliesThe scalar is nonzero, since $k+r\le n$ makes $k-n+r-1\le -1$. Repeating with the smaller power shows $\psi =0$. For $k\ge 2$, induct on $r+k$, treating all $r=0$ cases as already established. The same commutator calculation yields, with $c=r(k-n+r-1)\ne 0$,The induction hypothesis applies to $L^{r-1}$ on degree $k$, since $(r-1)+k<r+k$ and $(r-1)+k\le n$. Hence $\psi =L\alpha$ with $\alpha =c^{-1}\Lambda \psi$ of degree $k-2$. Moreover $L^{r+1}\alpha =L^r\psi =0$. Since $(r+1)+(k-2)=r+k-1$, induction also makes $L^{r+1}$ injective on that degree, so $\alpha =0$ and $\psi =0$. All operators preserve restriction to open sets; the argument applies on every open set. Therefore $L^r$ is an injective morphism of sheaves whenever $r+k\le n$.

On a Kähler manifold, the Kähler identities, with the positive form convention used above, areTaking formal adjoints conjugates the scalar and reverses the order in the commutator. Since ${\Lambda }^{\ast }=L$, this givesTo derive the Kähler Laplacian identity, put $P=\partial$ and $B=\bar{\partial }$, so $P^2=B^2=0$ and $PB+BP=0$. Write $\{A,B\}=AB+BA$ for an anticommutator. The Kähler identities say $B^{\ast }=-i[\Lambda ,P]$ and $P^{\ast }=i[\Lambda ,B]$. Thereforebecause expansion leaves only terms containing $P^2$ or $B^2$. Furthermore,Expanding these expressions and using $PB=-BP$ makes them equal. Since $d=P+B$ and $d^{\ast }=P^{\ast }+B^{\ast }$, the mixed anticommutators already vanish, and consequentlyClosedness of the Kähler form implies $[L,P]=[L,B]=0$. Using the adjoint identity above,The equality of the three Laplacians therefore proves each Laplacian commutes with $L$.

For a compact Kähler manifold, the Dolbeault Hodge decomposition is the orthogonal decompositionHere harmonicity for the Dolbeault Laplacian and for the Hodge Laplacian agrees by the preceding identity. If $\bar{\partial }\alpha =0$ and $\alpha =h+\bar{\partial }u+{\bar{\partial }}^{\ast }v$, then $\bar{\partial }{\bar{\partial }}^{\ast }v=0$. Taking its inner product with $v$ gives $\Vert {\bar{\partial }}^{\ast }v{\Vert }^2=0$. Thus $\alpha$ is cohomologous to $h$. Conversely, a harmonic form is $\bar{\partial }$-closed, and a harmonic exact form $h=\bar{\partial }u$ satisfies $\Vert h{\Vert }^2=({\bar{\partial }}^{\ast }h,u)=0$. This proves existence and uniqueness of the harmonic representative and henceThese spaces are finite dimensional by the ellipticity of the Dolbeault Laplacian on the compact manifold.

Complex conjugation commutes with the real operator ${\Delta }_d$ and interchanges the bidegrees $(p,q)$ and $(q,p)$. It is therefore a conjugate-linear bijection of the harmonic spaces, proving Hodge symmetry. The real Hodge star operator commutes with ${\Delta }_d$, as does complex conjugation; hence their composition, our conjugate-linear Hodge star, sends harmonic $(p,q)$-forms bijectively to harmonic $(n-p,n-q)$-forms. Its square is the nonzero scalar established above. This proves Hodge duality, and gives

For the final hard Lefschetz isomorphism on Dolbeault cohomology, take $k=p+q\le n$ and $r=n-k$. This restriction is necessary to make the displayed power nonnegative. The Hermitian Lefschetz operator raises bidegree by $(1,1)$, and the Lefschetz operator preserves harmonic forms; thereforeis well defined. It is injective by injectivity of powers of the Lefschetz operator, since $r+k=n$. Hodge symmetry and Hodge duality give $h^{n-q,n-p}=h^{q,p}=h^{p,q}$, so it is a bijection between finite-dimensional spaces of equal dimension. Because $[L,\bar{\partial }]=0$, the map on harmonic representatives agrees with exterior multiplication by $[\omega {]}^{n-k}$ on Dolbeault cohomology. We concludeFor $p+q>n$, the corresponding valid statement is the inverse of the positive-power isomorphism from bidegree $(n-q,n-p)$ to $(p,q)$; a negative exterior-multiplication power is not defined.

Let ${\mathcal{A}}^{p,q}(X)$ be the space of smooth complex differential forms of type $(p,q)$. The Dolbeault operator satisfies ${\bar{\partial }}^2=0$, and the Dolbeault cohomology isFor $q=0$ the denominator is zero; forms outside the dimension range are zero.

A sequence of sheaves is an exact sequence of sheaves precisely when its sequence of stalks at every point is exact. Here this means the first map is injective on stalks, the last is surjective on stalks, and the image equals the kernel at the middle stalk. In particular, a section of the last sheaf need only have local lifts; surjectivity on all global sections is not required.

The associated long exact sequence in sheaf cohomology of Čech cohomology, understood in the direct limit over open covers, isThe degree-zero groups are global sections. The connecting map is obtained by locally lifting a Čech cocycle to the middle sheaf and taking its Čech coboundary, which takes values in the first sheaf. Different lifts change it by a Čech coboundary.