[proofplan] The proof is local, because all four commutator identities are identities of first-order differential operators. At a fixed point we choose a unitary holomorphic coframe in which the Kähler form has its standard expression and the connection coefficients vanish. In that frame the differential operators become sums of wedge operators, contraction operators, and covariant derivatives, so the identities reduce to the elementary creation-annihilation relations in the exterior algebra. The remaining two identities follow by taking adjoints, and the Laplacian equality follows by expanding $d=\partial+\bar{\partial}$ and using the already proved commutator identities. [/proofplan]

[step:Reduce the commutators to a pointwise calculation in a unitary holomorphic frame]Fix a point $x_0 \in X$. Since the desired identities are local differential-operator identities, it is enough to verify them at $x_0$ on an arbitrary smooth form. Choose holomorphic coordinates $(z_1,\dots,z_n)$ on a neighbourhood $U \subset X$ of $x_0$ which are Kähler normal at $x_0$: the Hermitian metric satisfies $g_{j\bar{k}}(x_0)=\delta_{jk}$ and $\partial_{z_\ell}g_{j\bar{k}}(x_0)=\partial_{\bar{z}_\ell}g_{j\bar{k}}(x_0)=0$ for all indices. Let $\nabla$ denote the Chern connection on complex-valued differential forms; on a Kähler manifold this connection agrees with the Levi-Civita connection after complexification and preserves the bidegree decomposition. Define the pointwise covectors \begin{align*} \theta_j := dz_j|_{x_0} \in T_{x_0}^{1,0*}X, \qquad \bar{\theta}_j := d\bar{z}_j|_{x_0} \in T_{x_0}^{0,1*}X. \end{align*} Then $(\theta_1,\dots,\theta_n)$ is a unitary coframe of $T_{x_0}^{1,0*}X$, the connection satisfies $\nabla dz_j=0$ at $x_0$, and \begin{align*} \omega_{x_0}=i\sum_{j=1}^n \theta_j\wedge\bar{\theta}_j. \end{align*} Let \begin{align*} E_j: \Lambda^\bullet T_{x_0}^*X \otimes \mathbb{C} &\to \Lambda^{\bullet+1} T_{x_0}^*X \otimes \mathbb{C},& \alpha &\mapsto \theta_j \wedge \alpha,\\ \bar{E}_j: \Lambda^\bullet T_{x_0}^*X \otimes \mathbb{C} &\to \Lambda^{\bullet+1} T_{x_0}^*X \otimes \mathbb{C},& \alpha &\mapsto \bar{\theta}_j \wedge \alpha \end{align*} be wedge operators on the full complex exterior algebra. Let $I_j$ and $\bar{I}_j$ denote their pointwise Hermitian adjoints, equivalently contraction by the unit vectors dual to $\theta_j$ and $\bar{\theta}_j$, respectively. At $x_0$ we have \begin{align*} L &= i\sum_{j=1}^n E_j\bar{E}_j, & \Lambda &= -i\sum_{j=1}^n \bar{I}_j I_j. \end{align*} Let $Z_j$ be the local $(1,0)$ coordinate vector field $\partial/\partial z_j$, and let $\bar{Z}_j$ be its complex conjugate. At $x_0$, because the connection coefficients in the Kähler [normal coordinates](/theorems/2713) vanish and $\nabla$ preserves type, \begin{align*} \partial &= \sum_{j=1}^n E_j\nabla_{Z_j}, & \bar{\partial} &= \sum_{j=1}^n \bar{E}_j\nabla_{\bar{Z}_j},\\ \partial^* &= -\sum_{j=1}^n I_j\nabla_{\bar{Z}_j}, & \bar{\partial}^* &= -\sum_{j=1}^n \bar{I}_j\nabla_{Z_j}. \end{align*}[/step]

[guided]We prove the operator identities at one fixed point $x_0 \in X$. This is enough because both sides of each Kähler identity are differential operators; if two such operators agree on every form at every point, then they are the same operator. The Kähler condition allows us to choose holomorphic coordinates $(z_1,\dots,z_n)$ on a neighbourhood $U \subset X$ of $x_0$ which are normal for the Kähler metric at $x_0$. This means $g_{j\bar{k}}(x_0)=\delta_{jk}$ and $\partial_{z_\ell}g_{j\bar{k}}(x_0)=\partial_{\bar{z}_\ell}g_{j\bar{k}}(x_0)=0$ for all indices. Let $\nabla$ denote the Chern connection on complex-valued forms; because the metric is Kähler, this is the complexified Levi-Civita connection and it preserves $(p,q)$-type. Define \begin{align*} \theta_j := dz_j|_{x_0} \in T_{x_0}^{1,0*}X, \qquad \bar{\theta}_j := d\bar{z}_j|_{x_0} \in T_{x_0}^{0,1*}X. \end{align*} Then $(\theta_1,\dots,\theta_n)$ is unitary at the single point $x_0$, the connection coefficients vanish at $x_0$, and \begin{align*} \omega_{x_0}=i\sum_{j=1}^n \theta_j \wedge \bar{\theta}_j. \end{align*} The vanishing of the connection coefficients means that, at $x_0$, differentiating the components of a form produces the whole first-order part of $\partial$ and $\bar{\partial}$; no extra connection-coefficient terms appear. Define the wedge operators \begin{align*} E_j: \Lambda^\bullet T_{x_0}^*X \otimes \mathbb{C} &\to \Lambda^{\bullet+1} T_{x_0}^*X \otimes \mathbb{C},& \alpha &\mapsto \theta_j \wedge \alpha,\\ \bar{E}_j: \Lambda^\bullet T_{x_0}^*X \otimes \mathbb{C} &\to \Lambda^{\bullet+1} T_{x_0}^*X \otimes \mathbb{C},& \alpha &\mapsto \bar{\theta}_j \wedge \alpha. \end{align*} Let $I_j:=E_j^*$ and $\bar{I}_j:=\bar{E}_j^*$ be their Hermitian adjoints. These are contraction operators. Since $L$ means wedge multiplication by $\omega$, the normal expression for $\omega$ gives \begin{align*} L = i\sum_{j=1}^n E_j\bar{E}_j. \end{align*} Taking pointwise adjoints gives \begin{align*} \Lambda = L^* = -i\sum_{j=1}^n \bar{I}_j I_j, \end{align*} because $(E_j\bar{E}_j)^*=\bar{I}_j I_j$ and the complex conjugate of $i$ is $-i$. Finally, if $Z_j$ is the unitary $(1,0)$ vector field dual to $\theta_j$, then at $x_0$ the Dolbeault operators and their formal adjoints have the local expressions \begin{align*} \partial &= \sum_{j=1}^n E_j\nabla_{Z_j}, & \bar{\partial} &= \sum_{j=1}^n \bar{E}_j\nabla_{\bar{Z}_j},\\ \partial^* &= -\sum_{j=1}^n I_j\nabla_{\bar{Z}_j}, & \bar{\partial}^* &= -\sum_{j=1}^n \bar{I}_j\nabla_{Z_j}. \end{align*} The minus signs in the adjoint formulas come from [integration by parts](/theorems/2098) in a normal frame; at the chosen point no lower-order connection term remains. Thus the proof has been reduced to computing the commutators of $\Lambda$ with the wedge operators $E_j$ and $\bar{E}_j$.[/guided]

[step:Compute the exterior algebra commutators with $\Lambda$]The wedge and contraction operators satisfy the canonical anticommutation relations \begin{align*} I_jE_k+E_kI_j &= \delta_{jk}\operatorname{id},& \bar{I}_j\bar{E}_k+\bar{E}_k\bar{I}_j &= \delta_{jk}\operatorname{id}, \end{align*} and all mixed wedge-contraction pairs anticommute. Using \begin{align*} \Lambda=-i\sum_{j=1}^n \bar{I}_jI_j, \end{align*} we compute for each $k \in \{1,\dots,n\}$: \begin{align*} [\Lambda,E_k] &= -i\sum_{j=1}^n [\bar{I}_jI_j,E_k] = -i\bar{I}_k, \\ [\Lambda,\bar{E}_k] &= -i\sum_{j=1}^n [\bar{I}_jI_j,\bar{E}_k] = iI_k. \end{align*}[/step]

[guided]The whole proof now rests on the elementary algebra of wedge and contraction. The operators $E_j$ and $I_j$ are adjoint creation and annihilation operators for the $(1,0)$ basis element $\theta_j$, while $\bar{E}_j$ and $\bar{I}_j$ are the corresponding operators for $\bar{\theta}_j$. They satisfy \begin{align*} I_jE_k+E_kI_j &= \delta_{jk}\operatorname{id},& \bar{I}_j\bar{E}_k+\bar{E}_k\bar{I}_j &= \delta_{jk}\operatorname{id}. \end{align*} Mixed operators, such as $I_j$ with $\bar{E}_k$ and $\bar{I}_j$ with $E_k$, anticommute because they act on distinct exterior generators. First compute $[\Lambda,E_k]$. Since \begin{align*} \Lambda=-i\sum_{j=1}^n \bar{I}_jI_j, \end{align*} we need $[\bar{I}_jI_j,E_k]$. If $j\ne k$, both $I_j$ and $\bar{I}_j$ anticommute past $E_k$, and the two sign changes cancel, so the commutator is zero. If $j=k$, then \begin{align*} \bar{I}_kI_kE_k &= \bar{I}_k(\operatorname{id}-E_kI_k)\\ &= \bar{I}_k-\bar{I}_kE_kI_k\\ &= \bar{I}_k+E_k\bar{I}_kI_k. \end{align*} Therefore \begin{align*} [\bar{I}_kI_k,E_k] = \bar{I}_kI_kE_k-E_k\bar{I}_kI_k = \bar{I}_k. \end{align*} Multiplying by $-i$ gives \begin{align*} [\Lambda,E_k]=-i\bar{I}_k. \end{align*} The calculation for $\bar{E}_k$ is analogous but the sign changes in the other order. For $j=k$, \begin{align*} \bar{I}_kI_k\bar{E}_k &= -\bar{I}_k\bar{E}_kI_k\\ &= -(\operatorname{id}-\bar{E}_k\bar{I}_k)I_k\\ &= -I_k+\bar{E}_k\bar{I}_kI_k. \end{align*} Hence \begin{align*} [\bar{I}_kI_k,\bar{E}_k] = -I_k. \end{align*} After multiplying by $-i$, we obtain \begin{align*} [\Lambda,\bar{E}_k]=iI_k. \end{align*} These two algebraic identities are exactly the pointwise input needed for the Kähler identities.[/guided]

[step:Derive the two commutator identities involving $\Lambda$] At $x_0$, the algebraic commutators commute with the scalar covariant derivatives $\nabla_{Z_j}$ and $\nabla_{\bar{Z}_j}$ acting on the coefficient functions. No derivative of $\Lambda$ appears: since $\nabla\omega=0$ for the Kähler connection, the induced connection on exterior powers satisfies $\nabla L=0$, and taking Hermitian adjoints gives $\nabla\Lambda=0$. Therefore the commutator of the zero-order operator $\Lambda$ with the first-order expression $E_j\nabla_{Z_j}$ is exactly $[\Lambda,E_j]\nabla_{Z_j}$ at $x_0$. Thus \begin{align*} [\Lambda,\partial] &= \sum_{j=1}^n [\Lambda,E_j]\nabla_{Z_j}\\ &= -i\sum_{j=1}^n \bar{I}_j\nabla_{Z_j}\\ &= i\bar{\partial}^*. \end{align*} Similarly, \begin{align*} [\Lambda,\bar{\partial}] &= \sum_{j=1}^n [\Lambda,\bar{E}_j]\nabla_{\bar{Z}_j}\\ &= i\sum_{j=1}^n I_j\nabla_{\bar{Z}_j}\\ &= -i\partial^*. \end{align*} Since $x_0$ was arbitrary, these identities hold globally: \begin{align*} [\Lambda,\partial]=i\bar{\partial}^*, \qquad [\Lambda,\bar{\partial}]=-i\partial^*. \end{align*} [/step]

[step:Take adjoints to obtain the two commutator identities involving $L$] Taking formal adjoints of \begin{align*} [\Lambda,\partial]=i\bar{\partial}^* \end{align*} gives \begin{align*} (\Lambda\partial-\partial\Lambda)^* = (i\bar{\partial}^*)^*. \end{align*} Since $\Lambda^*=L$ and $(\bar{\partial}^*)^*=\bar{\partial}$, this becomes \begin{align*} \partial^*L-L\partial^*=-i\bar{\partial}. \end{align*} Equivalently, \begin{align*} [L,\partial^*]=i\bar{\partial}. \end{align*} Taking adjoints of \begin{align*} [\Lambda,\bar{\partial}]=-i\partial^* \end{align*} gives \begin{align*} \bar{\partial}^*L-L\bar{\partial}^*=i\partial, \end{align*} hence \begin{align*} [L,\bar{\partial}^*]=-\i\partial. \end{align*} [/step]

[step:Use the commutator identities to compare the Dolbeault Laplacians]From \begin{align*} [\Lambda,\bar{\partial}]=-i\partial^* \end{align*} we have \begin{align*} \partial^*=i[\Lambda,\bar{\partial}]. \end{align*} Thus \begin{align*} \Delta_{\partial} &= \partial\partial^*+\partial^*\partial\\ &= i\partial[\Lambda,\bar{\partial}]+i[\Lambda,\bar{\partial}]\partial. \end{align*} Expanding the commutators and using $\partial\bar{\partial}+\bar{\partial}\partial=0$, we get \begin{align*} \Delta_{\partial} &= i\partial\Lambda\bar{\partial} -i\partial\bar{\partial}\Lambda +i\Lambda\bar{\partial}\partial -i\bar{\partial}\Lambda\partial\\ &= i\partial\Lambda\bar{\partial} +i\bar{\partial}\partial\Lambda -i\Lambda\partial\bar{\partial} -i\bar{\partial}\Lambda\partial. \end{align*} Similarly, from \begin{align*} [\Lambda,\partial]=i\bar{\partial}^* \end{align*} we have \begin{align*} \bar{\partial}^*=-i[\Lambda,\partial]. \end{align*} Therefore \begin{align*} \Delta_{\bar{\partial}} &= \bar{\partial}\bar{\partial}^*+\bar{\partial}^*\bar{\partial}\\ &= -i\bar{\partial}[\Lambda,\partial]-i[\Lambda,\partial]\bar{\partial}\\ &= -i\bar{\partial}\Lambda\partial +i\bar{\partial}\partial\Lambda -i\Lambda\partial\bar{\partial} +i\partial\Lambda\bar{\partial}. \end{align*} The two final displayed expressions are identical term by term, so \begin{align*} \Delta_{\partial}=\Delta_{\bar{\partial}}. \end{align*}[/step]

[guided]The goal is to prove that the two Dolbeault Laplacians are equal using only the Kähler identities already established and the basic Dolbeault relation \begin{align*} \partial\bar{\partial}+\bar{\partial}\partial=0. \end{align*} Start with $\Delta_{\partial}$. The identity \begin{align*} [\Lambda,\bar{\partial}]=-i\partial^* \end{align*} is equivalent to \begin{align*} \partial^*=i[\Lambda,\bar{\partial}]. \end{align*} Substituting this into the definition of $\Delta_{\partial}$ gives \begin{align*} \Delta_{\partial} &= \partial\partial^*+\partial^*\partial\\ &= i\partial[\Lambda,\bar{\partial}]+i[\Lambda,\bar{\partial}]\partial. \end{align*} Now expand the commutators: \begin{align*} \Delta_{\partial} &= i\partial(\Lambda\bar{\partial}-\bar{\partial}\Lambda) +i(\Lambda\bar{\partial}-\bar{\partial}\Lambda)\partial\\ &= i\partial\Lambda\bar{\partial} -i\partial\bar{\partial}\Lambda +i\Lambda\bar{\partial}\partial -i\bar{\partial}\Lambda\partial. \end{align*} The relation $\partial\bar{\partial}=-\bar{\partial}\partial$ changes the middle two terms: \begin{align*} -i\partial\bar{\partial}\Lambda = i\bar{\partial}\partial\Lambda, \qquad i\Lambda\bar{\partial}\partial = -i\Lambda\partial\bar{\partial}. \end{align*} Hence \begin{align*} \Delta_{\partial} = i\partial\Lambda\bar{\partial} +i\bar{\partial}\partial\Lambda -i\Lambda\partial\bar{\partial} -i\bar{\partial}\Lambda\partial. \end{align*} Now do the same computation for $\Delta_{\bar{\partial}}$. The identity \begin{align*} [\Lambda,\partial]=i\bar{\partial}^* \end{align*} is equivalent to \begin{align*} \bar{\partial}^*=-i[\Lambda,\partial]. \end{align*} Therefore \begin{align*} \Delta_{\bar{\partial}} &= \bar{\partial}\bar{\partial}^*+\bar{\partial}^*\bar{\partial}\\ &= -i\bar{\partial}[\Lambda,\partial]-i[\Lambda,\partial]\bar{\partial}\\ &= -i\bar{\partial}(\Lambda\partial-\partial\Lambda) -i(\Lambda\partial-\partial\Lambda)\bar{\partial}\\ &= -i\bar{\partial}\Lambda\partial +i\bar{\partial}\partial\Lambda -i\Lambda\partial\bar{\partial} +i\partial\Lambda\bar{\partial}. \end{align*} This is the same expression obtained for $\Delta_{\partial}$, only with the terms written in a different order. Thus \begin{align*} \Delta_{\partial}=\Delta_{\bar{\partial}}. \end{align*}[/guided]

[step:Expand the de Rham Laplacian and cancel the mixed terms] Since \begin{align*} d=\partial+\bar{\partial}, \qquad d^*=\partial^*+\bar{\partial}^*, \end{align*} we have \begin{align*} \Delta_d &= dd^*+d^*d\\ &= \Delta_{\partial}+\Delta_{\bar{\partial}} + \partial\bar{\partial}^* +\bar{\partial}^*\partial + \bar{\partial}\partial^* +\partial^*\bar{\partial}. \end{align*} It remains to show that the two mixed anticommutators vanish. Using $\bar{\partial}^*=-i[\Lambda,\partial]$, we compute \begin{align*} \partial\bar{\partial}^*+\bar{\partial}^*\partial &= -i\partial[\Lambda,\partial]-i[\Lambda,\partial]\partial\\ &= -i\partial\Lambda\partial +i\partial^2\Lambda -i\Lambda\partial^2 +i\partial\Lambda\partial\\ &=0, \end{align*} because $\partial^2=0$. Using $\partial^*=i[\Lambda,\bar{\partial}]$, we also get \begin{align*} \bar{\partial}\partial^*+\partial^*\bar{\partial} &= i\bar{\partial}[\Lambda,\bar{\partial}] +i[\Lambda,\bar{\partial}]\bar{\partial}\\ &= i\bar{\partial}\Lambda\bar{\partial} -i\bar{\partial}^2\Lambda +i\Lambda\bar{\partial}^2 -i\bar{\partial}\Lambda\bar{\partial}\\ &=0, \end{align*} because $\bar{\partial}^2=0$. Therefore \begin{align*} \Delta_d = \Delta_{\partial}+\Delta_{\bar{\partial}}. \end{align*} Together with $\Delta_{\partial}=\Delta_{\bar{\partial}}$, this gives \begin{align*} \Delta_d=2\Delta_{\partial}=2\Delta_{\bar{\partial}}. \end{align*} This completes the proof. [/step]