[proofplan] The proof is local, because a first-order differential operator and its formal adjoint have coefficients determined pointwise once the metric, volume form, and bundle metric are fixed. In normal holomorphic coordinates and an $h$-unitary frame with vanishing connection matrix at $x$, the operators $\bar\partial_E$ and $\partial_E$ have the same principal form at $x$ as the flat Euclidean operators $\sum_j d\bar z_j \wedge \partial_{\bar z_j}$ and $\sum_j dz_j \wedge \partial_{z_j}$. [Integration by parts](/theorems/2098) identifies the adjoint of $\nabla_{\partial/\partial \bar z_j}$ with $-\nabla_{\partial/\partial z_j}$ at $x$, and the pointwise adjoint of wedge multiplication by $d\bar z_j$ is contraction by $\partial/\partial \bar z_j$, with the analogous statement for $dz_j$. [/proofplan] [step:Reduce the computation to the flat normal expression at the chosen point] Let $(U,z_1,\dots,z_n)$ be the chosen holomorphic coordinate neighbourhood of $x$, and let $(e_\alpha)_{\alpha=1}^r$ be the chosen local holomorphic frame for $E|_U$. Write \begin{align*} \frac{\partial}{\partial z_j} &:= \partial_{z_j},& \frac{\partial}{\partial \bar z_j} &:= \partial_{\bar z_j}. \end{align*} At $x$, the hypotheses give an orthonormal coframe \begin{align*} dz_1,\dots,dz_n,d\bar z_1,\dots,d\bar z_n \end{align*} for the complexified cotangent space, and an $h$-orthonormal frame $(e_\alpha(x))_{\alpha=1}^r$ for $E_x$. Since the Chern connection matrix vanishes at $x$, for every smooth $E$-valued form $v$ defined near $x$, \begin{align*} \nabla_{\partial_{z_j}}v(x) &= \partial_{z_j}v(x),& \nabla_{\partial_{\bar z_j}}v(x) &= \partial_{\bar z_j}v(x) \end{align*} when the coefficients of $v$ are written in the frame $(e_\alpha)$ and the coordinate coframe. Thus, at $x$, \begin{align*} \bar\partial_E v &= \sum_{j=1}^n d\bar z_j \wedge \nabla_{\partial_{\bar z_j}}v,\\ \partial_E v &= \sum_{j=1}^n dz_j \wedge \nabla_{\partial_{z_j}}v. \end{align*} [guided] The point of the normal coordinate and normal frame assumptions is to remove all zero-order correction terms at the single point $x$. Let $(U,z_1,\dots,z_n)$ be the chosen holomorphic coordinate neighbourhood, and let $(e_\alpha)_{\alpha=1}^r$ be the chosen local holomorphic frame for $E|_U$. We abbreviate the coordinate vector fields by \begin{align*} \partial_{z_j} &:= \frac{\partial}{\partial z_j},& \partial_{\bar z_j} &:= \frac{\partial}{\partial \bar z_j}. \end{align*} At $x$, the normality of the Kähler metric says that the coordinate coframe is orthonormal to first order: \begin{align*} g_{j\bar k}(x)=\delta_{jk}, \qquad \partial_{z_\ell}g_{j\bar k}(x)=\partial_{\bar z_\ell}g_{j\bar k}(x)=0. \end{align*} The frame condition says that $(e_\alpha(x))_{\alpha=1}^r$ is $h$-orthonormal and that the connection matrix of the Chern connection vanishes at $x$. Therefore, if a smooth $E$-valued form $v$ is written in this local frame and coordinate coframe, covariant differentiation at $x$ acts on the coefficient functions only: \begin{align*} \nabla_{\partial_{z_j}}v(x) &= \partial_{z_j}v(x),& \nabla_{\partial_{\bar z_j}}v(x) &= \partial_{\bar z_j}v(x). \end{align*} Consequently the first-order parts of $\bar\partial_E$ and $\partial_E$ at $x$ are exactly the flat model expressions: \begin{align*} \bar\partial_E v &= \sum_{j=1}^n d\bar z_j \wedge \nabla_{\partial_{\bar z_j}}v,\\ \partial_E v &= \sum_{j=1}^n dz_j \wedge \nabla_{\partial_{z_j}}v. \end{align*} All possible correction terms from the metric, the volume density, and the bundle connection vanish at $x$ by the normality hypotheses. [/guided] [/step] [step:Identify the pointwise adjoints of wedge multiplication] For each $j \in \{1,\dots,n\}$, define the wedge operators on complexified exterior powers at $x$ by \begin{align*} L_{\bar j}: \Lambda^\bullet T_x^*X \otimes E_x &\to \Lambda^{\bullet+1}T_x^*X \otimes E_x,\\ \eta &\mapsto d\bar z_j \wedge \eta,\\ L_j: \Lambda^\bullet T_x^*X \otimes E_x &\to \Lambda^{\bullet+1}T_x^*X \otimes E_x,\\ \eta &\mapsto dz_j \wedge \eta. \end{align*} With respect to the pointwise Hermitian inner product induced by $\omega$ and $h$ at $x$, their adjoints are \begin{align*} L_{\bar j}^* &= \iota_{\partial_{\bar z_j}},& L_j^* &= \iota_{\partial_{z_j}}. \end{align*} Indeed, the monomials \begin{align*} dz_I \wedge d\bar z_J \otimes e_\alpha \end{align*} form an orthonormal basis at $x$, where $I,J \subset \{1,\dots,n\}$ are strictly increasing multi-indices and $\alpha \in \{1,\dots,r\}$. Wedge multiplication by $d\bar z_j$ sends a basis element either to $0$ if $j \in J$, or to the corresponding basis element with $j$ inserted into $J$, with the exterior sign determined by the insertion. Contraction by $\partial_{\bar z_j}$ reverses exactly this operation with the same sign. Hence $L_{\bar j}^*=\iota_{\partial_{\bar z_j}}$. The proof for $L_j^*=\iota_{\partial_{z_j}}$ is identical with $dz_j$ in place of $d\bar z_j$. [/step] [step:Compute the adjoint of the coefficient derivative by integration by parts] Let $\phi$ and $\psi$ be compactly supported smooth scalar functions on the coordinate neighbourhood $U$. At the point $x$, the first-order part of the formal adjoint is determined by the flat density because the first derivatives of the metric coefficients vanish at $x$. In the flat model, using the Hermitian convention that the $L^2$ inner product is linear in the first argument, \begin{align*} \int_U (\partial_{\bar z_j}\phi)\,\overline{\psi}\,d\mathcal{L}^{2n} &= -\int_U \phi\,\partial_{\bar z_j}\overline{\psi}\,d\mathcal{L}^{2n} \\ &= -\int_U \phi\,\overline{\partial_{z_j}\psi}\,d\mathcal{L}^{2n}. \end{align*} Thus the formal adjoint of $\partial_{\bar z_j}$ is $-\partial_{z_j}$ at $x$. Replacing scalar coefficients by coefficients of $E$-valued forms in the normal unitary frame gives \begin{align*} (\nabla_{\partial_{\bar z_j}})^* &= -\nabla_{\partial_{z_j}} \end{align*} at $x$. Similarly, \begin{align*} (\nabla_{\partial_{z_j}})^* &= -\nabla_{\partial_{\bar z_j}} \end{align*} at $x$. [guided] We now compute the adjoint of the directional derivative. Let $\phi,\psi:U\to \mathbb{C}$ be compactly supported smooth scalar functions. In the flat model with Lebesgue measure $\mathcal{L}^{2n}$ and with the Hermitian inner product linear in the first argument, [integration by parts](/theorems/210) gives \begin{align*} \int_U (\partial_{\bar z_j}\phi)\,\overline{\psi}\,d\mathcal{L}^{2n} &= -\int_U \phi\,\partial_{\bar z_j}\overline{\psi}\,d\mathcal{L}^{2n}. \end{align*} The boundary term vanishes because $\phi$ and $\psi$ are compactly supported in $U$. Since complex conjugation interchanges $\partial_{\bar z_j}$ and $\partial_{z_j}$, \begin{align*} \partial_{\bar z_j}\overline{\psi} = \overline{\partial_{z_j}\psi}. \end{align*} Therefore \begin{align*} \int_U (\partial_{\bar z_j}\phi)\,\overline{\psi}\,d\mathcal{L}^{2n} &= -\int_U \phi\,\overline{\partial_{z_j}\psi}\,d\mathcal{L}^{2n}. \end{align*} This proves that the formal adjoint of $\partial_{\bar z_j}$ is $-\partial_{z_j}$ in the flat model. Why does this flat computation give the value at $x$ on the manifold? The formal adjoint is a first-order local operator. Its first-order coefficient is determined by the pointwise metric and bundle metric, and its zero-order correction terms come from derivatives of the metric density and the connection matrix. In our chosen coordinates and frame, the first derivatives of the metric vanish at $x$, and the Chern connection matrix vanishes at $x$. Hence the same computation yields \begin{align*} (\nabla_{\partial_{\bar z_j}})^* &= -\nabla_{\partial_{z_j}} \end{align*} at $x$. Repeating the same integration by parts with $\partial_{z_j}$ gives \begin{align*} (\nabla_{\partial_{z_j}})^* &= -\nabla_{\partial_{\bar z_j}} \end{align*} at $x$. [/guided] [/step] [step:Combine the wedge and derivative adjoints for $\bar\partial_E^*$] At $x$, write \begin{align*} \bar\partial_E = \sum_{j=1}^n L_{\bar j}\nabla_{\partial_{\bar z_j}}. \end{align*} The operator $L_{\bar j}$ is algebraic, and $\nabla_{\partial_{\bar z_j}}$ is the first-order coefficient derivative. Taking the formal adjoint term by term gives \begin{align*} \bar\partial_E^* &= \sum_{j=1}^n \left(L_{\bar j}\nabla_{\partial_{\bar z_j}}\right)^* \\ &= \sum_{j=1}^n (\nabla_{\partial_{\bar z_j}})^* L_{\bar j}^* \\ &= -\sum_{j=1}^n \nabla_{\partial_{z_j}}\iota_{\partial_{\bar z_j}}. \end{align*} The contraction operator $\iota_{\partial_{\bar z_j}}$ is parallel at $x$ in the chosen [normal coordinates](/theorems/2713), so it commutes with $\nabla_{\partial_{z_j}}$ at $x$. Therefore, for every smooth $E$-valued form $u$, \begin{align*} \bar\partial_E^*u(x) &= -\sum_{j=1}^n \iota_{\partial_{\bar z_j}}\nabla_{\partial_{z_j}}u(x). \end{align*} [/step] [step:Combine the wedge and derivative adjoints for $\partial_E^*$] At $x$, write \begin{align*} \partial_E = \sum_{j=1}^n L_j\nabla_{\partial_{z_j}}. \end{align*} Taking the formal adjoint term by term and using the identities already proved gives \begin{align*} \partial_E^* &= \sum_{j=1}^n \left(L_j\nabla_{\partial_{z_j}}\right)^* \\ &= \sum_{j=1}^n (\nabla_{\partial_{z_j}})^*L_j^* \\ &= -\sum_{j=1}^n \nabla_{\partial_{\bar z_j}}\iota_{\partial_{z_j}}. \end{align*} The contraction operator $\iota_{\partial_{z_j}}$ is parallel at $x$ in the chosen normal coordinates, so it commutes with $\nabla_{\partial_{\bar z_j}}$ at $x$. Hence, for every smooth $E$-valued form $u$, \begin{align*} \partial_E^*u(x) &= -\sum_{j=1}^n \iota_{\partial_{z_j}}\nabla_{\partial_{\bar z_j}}u(x). \end{align*} Together with the formula for $\bar\partial_E^*$, this proves the stated pointwise identities. [/step]