[proofplan] The identities are local first-order operator identities, so it suffices to verify them at an arbitrary point in unitary holomorphic coordinates and a unitary holomorphic frame for $E$ in which the Chern connection matrix vanishes at that point. In such a frame the bundle-valued operators have the same pointwise first-order form as the scalar Dolbeault operators, with coefficients in the fixed Hermitian [vector space](/page/Vector%20Space) $E_x$. The proof reduces to the finite-dimensional commutator relations between wedging and contracting by $dz_j$ and $d\bar z_j$, together with the local formulas for the formal adjoints. The two identities involving $L$ then follow by taking formal adjoints of the two identities involving $\Lambda$. [/proofplan] [step:Choose normal holomorphic data at one point] Fix a point $x_0 \in X$. Let $r=\operatorname{rank}_{\mathbb C}E$ denote the complex rank of the holomorphic vector bundle $E$. By the Kähler normal-coordinate construction, choose holomorphic coordinates \begin{align*} z=(z_1,\dots,z_n): U \to z(U) \subset \mathbb C^n \end{align*} on a neighbourhood $U$ of $x_0$ such that, at $x_0$, \begin{align*} \omega = i\sum_{j=1}^n dz_j \wedge d\bar z_j, \end{align*} and the first derivatives of the Kähler metric coefficients vanish. By applying a holomorphic gauge normalized at $x_0$, choose a holomorphic frame \begin{align*} e=(e_1,\dots,e_r) \end{align*} for $E|_U$ such that $h_{\alpha\bar\beta}(x_0)=\delta_{\alpha\beta}$ and the Chern connection one-form vanishes at $x_0$. Let $Z_j$ denote the local vector field $\partial/\partial z_j$, and let $\bar Z_j$ denote $\partial/\partial\bar z_j$. Define the wedge operators \begin{align*} \varepsilon_j: \Lambda^{*,*}T_{x_0}^*X \otimes E_{x_0} &\to \Lambda^{*+1,*}T_{x_0}^*X \otimes E_{x_0}, & \varepsilon_j \alpha &= dz_j \wedge \alpha,\\ \bar\varepsilon_j: \Lambda^{*,*}T_{x_0}^*X \otimes E_{x_0} &\to \Lambda^{*,*+1}T_{x_0}^*X \otimes E_{x_0}, & \bar\varepsilon_j \alpha &= d\bar z_j \wedge \alpha. \end{align*} Let $\iota_j$ and $\bar\iota_j$ be their pointwise Hermitian adjoints, respectively; equivalently, $\iota_j$ contracts with $Z_j$ and $\bar\iota_j$ contracts with $\bar Z_j$ at $x_0$. In these coordinates and this frame, the Lefschetz operators at $x_0$ are \begin{align*} L &= i\sum_{j=1}^n \varepsilon_j\bar\varepsilon_j, & \Lambda &= -i\sum_{j=1}^n \bar\iota_j\iota_j. \end{align*} [guided] We localize at a single point because both sides of the desired identities are first-order differential operators, and equality of first-order operators can be checked pointwise after choosing coordinates and frames adapted to the metrics. Fix $x_0 \in X$. Let $r=\operatorname{rank}_{\mathbb C}E$ denote the complex rank of $E$. Since $(X,\omega)$ is Kähler, the Kähler normal-coordinate construction gives holomorphic coordinates \begin{align*} z=(z_1,\dots,z_n): U \to z(U) \subset \mathbb C^n \end{align*} centered at $x_0$ such that, at $x_0$, \begin{align*} \omega = i\sum_{j=1}^n dz_j \wedge d\bar z_j, \end{align*} and the first derivatives of the Kähler metric coefficients vanish. The vanishing of the first derivatives is what removes lower-order Christoffel terms from the adjoint formulas at $x_0$. Next apply a holomorphic gauge normalized at $x_0$ to choose a holomorphic frame \begin{align*} e=(e_1,\dots,e_r) \end{align*} for $E|_U$ satisfying $h_{\alpha\bar\beta}(x_0)=\delta_{\alpha\beta}$ and such that the Chern connection one-form is zero at $x_0$. This is the bundle analogue of choosing [normal coordinates](/theorems/2713): at the point $x_0$, the connection differentiates the coefficient functions without adding a zeroth-order matrix term. Let $Z_j=\partial/\partial z_j$ and $\bar Z_j=\partial/\partial\bar z_j$. We define the pointwise wedge maps \begin{align*} \varepsilon_j: \Lambda^{*,*}T_{x_0}^*X \otimes E_{x_0} &\to \Lambda^{*+1,*}T_{x_0}^*X \otimes E_{x_0}, & \varepsilon_j \alpha &= dz_j \wedge \alpha,\\ \bar\varepsilon_j: \Lambda^{*,*}T_{x_0}^*X \otimes E_{x_0} &\to \Lambda^{*,*+1}T_{x_0}^*X \otimes E_{x_0}, & \bar\varepsilon_j \alpha &= d\bar z_j \wedge \alpha. \end{align*} Their pointwise Hermitian adjoints are denoted by $\iota_j$ and $\bar\iota_j$. Thus $\iota_j$ contracts by the vector $Z_j$, and $\bar\iota_j$ contracts by the vector $\bar Z_j$. Because $\omega=i\sum_j dz_j\wedge d\bar z_j$ at $x_0$, wedging by $\omega$ gives \begin{align*} L = i\sum_{j=1}^n \varepsilon_j\bar\varepsilon_j. \end{align*} Taking the pointwise Hermitian adjoint and using $(AB)^*=B^*A^*$ and $\overline{i}=-i$ gives \begin{align*} \Lambda=L^*=-i\sum_{j=1}^n \bar\iota_j\iota_j. \end{align*} [/guided] [/step] [step:Compute the algebraic commutators with wedge operators] Let \begin{align*} I: \Lambda^{*,*}T_{x_0}^*X \otimes E_{x_0} &\to \Lambda^{*,*}T_{x_0}^*X \otimes E_{x_0} \end{align*} denote the identity map. The exterior algebra relations at $x_0$ are \begin{align*} \iota_j\varepsilon_k+\varepsilon_k\iota_j &= \delta_{jk}I, & \bar\iota_j\bar\varepsilon_k+\bar\varepsilon_k\bar\iota_j &= \delta_{jk}I,\\ \iota_j\bar\varepsilon_k+\bar\varepsilon_k\iota_j &=0, & \bar\iota_j\varepsilon_k+\varepsilon_k\bar\iota_j &=0. \end{align*} Using these relations and the expression for $\Lambda$, we obtain, for each $k$, \begin{align*} [\Lambda,\varepsilon_k] &= \left[-i\sum_{j=1}^n \bar\iota_j\iota_j,\varepsilon_k\right]\\ &= -i\bar\iota_k, \end{align*} and similarly \begin{align*} [\Lambda,\bar\varepsilon_k] &= \left[-i\sum_{j=1}^n \bar\iota_j\iota_j,\bar\varepsilon_k\right]\\ &= i\iota_k. \end{align*} [guided] The differential part of the [Kähler identities](/theorems/3853) will come from the coefficient derivatives. The signs and factors of $i$ come entirely from the finite-dimensional exterior algebra at one point. Let \begin{align*} I: \Lambda^{*,*}T_{x_0}^*X \otimes E_{x_0} &\to \Lambda^{*,*}T_{x_0}^*X \otimes E_{x_0} \end{align*} denote the identity map. The wedge and contraction operators satisfy \begin{align*} \iota_j\varepsilon_k+\varepsilon_k\iota_j &= \delta_{jk}I, & \bar\iota_j\bar\varepsilon_k+\bar\varepsilon_k\bar\iota_j &= \delta_{jk}I,\\ \iota_j\bar\varepsilon_k+\bar\varepsilon_k\iota_j &=0, & \bar\iota_j\varepsilon_k+\varepsilon_k\bar\iota_j &=0. \end{align*} These are the usual creation-annihilation relations: contraction by a vector dual to the one-form being wedged gives one identity term, while operators of different holomorphic type anticommute. We compute $[\Lambda,\varepsilon_k]$. Since \begin{align*} \Lambda=-i\sum_{j=1}^n \bar\iota_j\iota_j, \end{align*} all terms with $j\ne k$ commute with $\varepsilon_k$ after two anticommutations. The only nonzero contribution is the $j=k$ term: \begin{align*} [\Lambda,\varepsilon_k] &= -i\bar\iota_k\iota_k\varepsilon_k +i\varepsilon_k\bar\iota_k\iota_k\\ &= -i\bar\iota_k(I-\varepsilon_k\iota_k) +i\varepsilon_k\bar\iota_k\iota_k\\ &= -i\bar\iota_k +i\bar\iota_k\varepsilon_k\iota_k +i\varepsilon_k\bar\iota_k\iota_k. \end{align*} Using $\varepsilon_k\bar\iota_k=-\bar\iota_k\varepsilon_k$, the last two terms cancel, so \begin{align*} [\Lambda,\varepsilon_k]=-i\bar\iota_k. \end{align*} The computation for $[\Lambda,\bar\varepsilon_k]$ is the conjugate calculation. The only nonzero contribution again comes from the $j=k$ term: \begin{align*} [\Lambda,\bar\varepsilon_k] &= -i\bar\iota_k\iota_k\bar\varepsilon_k +i\bar\varepsilon_k\bar\iota_k\iota_k\\ &= i\bar\iota_k\bar\varepsilon_k\iota_k +i\bar\varepsilon_k\bar\iota_k\iota_k\\ &= i(I-\bar\varepsilon_k\bar\iota_k)\iota_k +i\bar\varepsilon_k\bar\iota_k\iota_k\\ &= i\iota_k. \end{align*} [/guided] [/step] [step:Write the Dolbeault operators and their adjoints at the chosen point] Let $\nabla^E$ denote the connection induced on $\Lambda^{*,*}T^*X\otimes E$ by the Levi-Civita connection of $\omega$ on $T^*X$ and the Chern connection of $(E,h)$ on $E$. For a local vector field $V$ on $X$, write $\nabla_V^E$ for covariant differentiation in the direction $V$. At $x_0$, because the Chern connection matrix and the first derivatives of the Kähler metric vanish in the chosen data, the operators on compactly supported $E$-valued forms have the pointwise formulas \begin{align*} \partial_E &= \sum_{j=1}^n \varepsilon_j\nabla_{Z_j}^E, & \bar\partial_E &= \sum_{j=1}^n \bar\varepsilon_j\nabla_{\bar Z_j}^E,\\ \partial_E^* &= -\sum_{j=1}^n \iota_j\nabla_{\bar Z_j}^E, & \bar\partial_E^* &= -\sum_{j=1}^n \bar\iota_j\nabla_{Z_j}^E. \end{align*} At $x_0$ these formulas have no additional zeroth-order metric or connection terms. Moreover, since $(X,\omega)$ is Kähler and the Chern connection of $(E,h)$ is Hermitian, the induced connection on $\Lambda^{*,*}T^*X\otimes E$ preserves $L$ and $\Lambda$. Thus, at $x_0$, \begin{align*} \nabla_{Z_j}^E(\Lambda u)&=\Lambda\nabla_{Z_j}^E u, & \nabla_{\bar Z_j}^E(\Lambda u)&=\Lambda\nabla_{\bar Z_j}^E u \end{align*} for every local smooth $E$-valued form $u$. [guided] We now translate the algebra into differential operators. Let $\nabla^E$ denote the connection induced on $\Lambda^{*,*}T^*X\otimes E$ by the Levi-Civita connection of $\omega$ on $T^*X$ and the Chern connection of $(E,h)$ on $E$. For a local vector field $V$ on $X$, write $\nabla_V^E$ for covariant differentiation in the direction $V$. In the chosen holomorphic coordinates and holomorphic frame, the Chern connection decomposes into its $(1,0)$ and $(0,1)$ parts, namely $\partial_E$ and $\bar\partial_E$. At $x_0$, the connection matrix vanishes, so the first-order parts are exactly wedging by $dz_j$ or $d\bar z_j$ followed by covariant differentiation in the corresponding complex direction: \begin{align*} \partial_E &= \sum_{j=1}^n \varepsilon_j\nabla_{Z_j}^E, & \bar\partial_E &= \sum_{j=1}^n \bar\varepsilon_j\nabla_{\bar Z_j}^E. \end{align*} The formal adjoints are computed with respect to the $L^2$ inner product induced by $\omega$ and $h$. Compact support removes boundary terms. Since the first derivatives of the metric coefficients and the Chern connection matrix vanish at $x_0$, [integration by parts](/theorems/2098) gives the pointwise adjoint formulas \begin{align*} \partial_E^* &= -\sum_{j=1}^n \iota_j\nabla_{\bar Z_j}^E, & \bar\partial_E^* &= -\sum_{j=1}^n \bar\iota_j\nabla_{Z_j}^E. \end{align*} The change from $Z_j$ to $\bar Z_j$ in the adjoint of $\partial_E$ is the complex analogue of the fact that the formal adjoint conjugates the coefficient derivative. Finally, the commutator with $\Lambda$ contains no derivative of $\Lambda$. The reason is geometric: the Kähler condition gives $\nabla\omega=0$, hence the induced connection preserves $L=\omega\wedge(\cdot)$ and its pointwise adjoint $\Lambda$. The Hermitian property of the Chern connection ensures the same preservation after tensoring with $E$. Therefore \begin{align*} \nabla_{Z_j}^E(\Lambda u)&=\Lambda\nabla_{Z_j}^E u, & \nabla_{\bar Z_j}^E(\Lambda u)&=\Lambda\nabla_{\bar Z_j}^E u \end{align*} for every local smooth $E$-valued form $u$ at $x_0$. [/guided] [/step] [step:Derive the two commutators involving $\Lambda$] Let $u$ be a compactly supported smooth $E$-valued form. At $x_0$, \begin{align*} [\Lambda,\partial_E]u &= \sum_{j=1}^n [\Lambda,\varepsilon_j]\nabla_{Z_j}^E u\\ &= -i\sum_{j=1}^n \bar\iota_j\nabla_{Z_j}^E u\\ &= i\bar\partial_E^*u. \end{align*} Similarly, \begin{align*} [\Lambda,\bar\partial_E]u &= \sum_{j=1}^n [\Lambda,\bar\varepsilon_j]\nabla_{\bar Z_j}^E u\\ &= i\sum_{j=1}^n \iota_j\nabla_{\bar Z_j}^E u\\ &= -i\partial_E^*u. \end{align*} Since $x_0$ was arbitrary, these identities hold on all of $X$: \begin{align*} [\Lambda,\partial_E]&=i\bar\partial_E^*, & [\Lambda,\bar\partial_E]&=-i\partial_E^*. \end{align*} [guided] We now combine the algebraic commutators with the differential formulas. Let $u$ be a compactly supported smooth $E$-valued form. Because $\Lambda$ is parallel for the induced connection, the commutator with $\partial_E$ acts only on the wedge operator $\varepsilon_j$, not on the derivative $\nabla_{Z_j}^E$. Thus, at $x_0$, \begin{align*} [\Lambda,\partial_E]u &= \Lambda\left(\sum_{j=1}^n \varepsilon_j\nabla_{Z_j}^E u\right) - \sum_{j=1}^n \varepsilon_j\nabla_{Z_j}^E(\Lambda u)\\ &= \sum_{j=1}^n \left(\Lambda\varepsilon_j-\varepsilon_j\Lambda\right)\nabla_{Z_j}^E u\\ &= \sum_{j=1}^n [\Lambda,\varepsilon_j]\nabla_{Z_j}^E u. \end{align*} Using the algebraic identity $[\Lambda,\varepsilon_j]=-i\bar\iota_j$, we obtain \begin{align*} [\Lambda,\partial_E]u &= -i\sum_{j=1}^n \bar\iota_j\nabla_{Z_j}^E u. \end{align*} By the local adjoint formula \begin{align*} \bar\partial_E^*=-\sum_{j=1}^n \bar\iota_j\nabla_{Z_j}^E, \end{align*} this is exactly \begin{align*} [\Lambda,\partial_E]u=i\bar\partial_E^*u. \end{align*} The same computation for $\bar\partial_E$ uses $\bar\varepsilon_j$ in place of $\varepsilon_j$: \begin{align*} [\Lambda,\bar\partial_E]u &= \sum_{j=1}^n [\Lambda,\bar\varepsilon_j]\nabla_{\bar Z_j}^E u\\ &= i\sum_{j=1}^n \iota_j\nabla_{\bar Z_j}^E u. \end{align*} Since \begin{align*} \partial_E^*=-\sum_{j=1}^n \iota_j\nabla_{\bar Z_j}^E, \end{align*} we get \begin{align*} [\Lambda,\bar\partial_E]u=-i\partial_E^*u. \end{align*} The point $x_0$ was arbitrary, and both sides are globally defined differential operators. Therefore \begin{align*} [\Lambda,\partial_E]&=i\bar\partial_E^*, & [\Lambda,\bar\partial_E]&=-i\partial_E^* \end{align*} on $\mathcal A_c^{*,*}(X,E)$. [/guided] [/step] [step:Take formal adjoints to obtain the two commutators involving $L$] Since $L^*=\Lambda$ and $\Lambda^*=L$, taking formal adjoints of \begin{align*} [\Lambda,\partial_E]=i\bar\partial_E^* \end{align*} gives \begin{align*} [\Lambda,\partial_E]^* &= (\Lambda\partial_E-\partial_E\Lambda)^*\\ &= \partial_E^*L-L\partial_E^*\\ &= -[L,\partial_E^*], \end{align*} while \begin{align*} (i\bar\partial_E^*)^*=-i\bar\partial_E. \end{align*} Hence \begin{align*} [L,\partial_E^*]=i\bar\partial_E. \end{align*} Taking formal adjoints of \begin{align*} [\Lambda,\bar\partial_E]=-i\partial_E^* \end{align*} gives \begin{align*} [\Lambda,\bar\partial_E]^* &= \bar\partial_E^*L-L\bar\partial_E^*\\ &= -[L,\bar\partial_E^*], \end{align*} and \begin{align*} (-i\partial_E^*)^*=i\partial_E. \end{align*} Therefore \begin{align*} [L,\bar\partial_E^*]=-i\partial_E. \end{align*} Combining the four identities proves the theorem. [/step]