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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.

We use complex-valued smooth differential forms. The sheaf of smooth differential forms is ${\mathcal{A}}^k(U)=\Gamma (U,{\bigwedge }^k(T^{\ast }X\otimes \mathbb{C}))$, with the usual restriction maps. Dualizing the type decomposition of the complexified tangent bundle and taking exterior powers decomposes this bundle into the summandsTheir smooth sections form the sheaf of differential forms of type (p, q) ${\mathcal{A}}^{p,q}$. Locally a section is a sum of $f_{I,J}d{z}_I\wedge d{\bar{z}}_J$ with $|I|=p$, $|J|=q$ and smooth coefficients. Holomorphic transition maps preserve types, so the local decompositions agree globally. HenceThe exterior derivative splits as $d=\partial +\bar{\partial }$, with bidegrees $(1,0)$ and $(0,1)$. Its square being zero gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$. The Dolbeault cohomology is thereforeForms in negative or out-of-range bidegrees are understood to be zero.