[proofplan] The proof is local on $X$. On a sufficiently small [open set](/page/Open%20Set), choose a holomorphic frame of $E$ and write an $E$-valued form in components relative to that frame. In such a frame the bundle Dolbeault operator acts by applying the scalar Dolbeault operator to each coefficient form. A coordinate computation gives the scalar identity $\bar\partial^2=0$, hence the componentwise expression for $\bar\partial_E^2$ vanishes locally, and local vanishing implies the global identity. [/proofplan] [step:Write the $E$-valued form in a local holomorphic frame] Let $\alpha \in A^{p,q}(X,E)$. Since the identity to be proved is local, fix a point $x_0 \in X$ and choose an open neighbourhood $U \subset X$ of $x_0$ on which $E$ admits a holomorphic frame. Denote this frame by $e_1,\dots,e_r$, where $r$ is the rank of $E$. The restriction $\alpha|_U$ has a unique expression \begin{align*} \alpha|_U = \sum_{i=1}^{r} \alpha_i \otimes e_i \end{align*} with scalar forms $\alpha_i \in A^{p,q}(U)$ for $i \in \{1,\dots,r\}$. By the defining local formula for the Dolbeault operator associated to a holomorphic vector bundle, \begin{align*} \bar\partial_E(\alpha|_U) = \sum_{i=1}^{r} \bar\partial \alpha_i \otimes e_i. \end{align*} Here $\bar\partial: A^{p,q}(U) \to A^{p,q+1}(U)$ is the scalar Dolbeault operator on the complex manifold $U$. [/step] [step:Verify that the scalar Dolbeault operator squares to zero in coordinates] We recall the local scalar computation. Choose a holomorphic coordinate chart $(V,\varphi)$ with $V \subset U$ and holomorphic coordinates $z_1,\dots,z_n$ on $V$. For each $i \in \{1,\dots,r\}$, write the scalar form $\alpha_i|_V$ as \begin{align*} \alpha_i|_V = \sum_{|I|=p,\, |J|=q} a_{i,I,J}\, dz_I \wedge d\bar z_J, \end{align*} where $I$ and $J$ range over strictly increasing multi-indices, each coefficient is a smooth map $a_{i,I,J}: V \to \mathbb{C}$, and $dz_I \wedge d\bar z_J$ denotes the corresponding coordinate basis form. For a fixed coefficient $a_{i,I,J}$, the scalar Dolbeault operator differentiates only in the anti-holomorphic coordinate directions. Thus $\bar\partial^2$ contributes second derivatives of the form \begin{align*} \frac{\partial^2 a_{i,I,J}}{\partial \bar z_k \partial \bar z_\ell}\, d\bar z_\ell \wedge d\bar z_k \wedge dz_I \wedge d\bar z_J \end{align*} summed over $k,\ell \in \{1,\dots,n\}$, up to the fixed sign convention determined by the placement of the $d\bar z_k$ factors. The terms with $k=\ell$ vanish because $d\bar z_k \wedge d\bar z_k=0$. For $k \neq \ell$, the term indexed by $(k,\ell)$ cancels the term indexed by $(\ell,k)$ because smoothness of $a_{i,I,J}$ gives equality of mixed partial derivatives, while \begin{align*} d\bar z_\ell \wedge d\bar z_k = - d\bar z_k \wedge d\bar z_\ell. \end{align*} Therefore $\bar\partial^2 \alpha_i = 0$ on $V$. Since these coordinate neighbourhoods cover $U$, $\bar\partial^2 \alpha_i = 0$ on $U$. [guided] We prove the scalar identity locally because $\bar\partial_E$ becomes componentwise once a holomorphic frame is chosen. Fix a coordinate neighbourhood $V \subset U$ with holomorphic coordinates $z_1,\dots,z_n$. Each scalar coefficient form $\alpha_i|_V$ can be expanded uniquely as \begin{align*} \alpha_i|_V = \sum_{|I|=p,\, |J|=q} a_{i,I,J}\, dz_I \wedge d\bar z_J, \end{align*} where $a_{i,I,J}: V \to \mathbb{C}$ is smooth. The operator $\bar\partial$ differentiates the coefficient functions in the anti-holomorphic directions and wedges in one anti-holomorphic differential. Applying $\bar\partial$ twice therefore produces sums involving the second anti-holomorphic partial derivatives \begin{align*} \frac{\partial^2 a_{i,I,J}}{\partial \bar z_k \partial \bar z_\ell} \end{align*} together with a wedge factor containing $d\bar z_\ell \wedge d\bar z_k$. There are two possible ways such a term can vanish. If $k=\ell$, the wedge product contains $d\bar z_k \wedge d\bar z_k$, which is zero by alternatingness of the exterior product. If $k \neq \ell$, compare the contribution from the ordered pair $(k,\ell)$ with the contribution from $(\ell,k)$. The coefficient functions are smooth, so the mixed partial derivatives agree: \begin{align*} \frac{\partial^2 a_{i,I,J}}{\partial \bar z_k \partial \bar z_\ell} = \frac{\partial^2 a_{i,I,J}}{\partial \bar z_\ell \partial \bar z_k}. \end{align*} The corresponding wedge factors have opposite signs because \begin{align*} d\bar z_\ell \wedge d\bar z_k = - d\bar z_k \wedge d\bar z_\ell. \end{align*} Thus the two terms cancel. Since every term in the coordinate expansion either vanishes by repeated wedge factors or cancels with its transposed partner, we get $\bar\partial^2 \alpha_i = 0$ on $V$. The coordinate neighbourhoods $V$ cover $U$, so $\bar\partial^2 \alpha_i = 0$ on all of $U$. [/guided] [/step] [step:Apply the componentwise formula a second time] Using the local formula for $\bar\partial_E$ in the holomorphic frame $e_1,\dots,e_r$, we compute on $U$: \begin{align*} \bar\partial_E^2(\alpha|_U) = \bar\partial_E\left(\sum_{i=1}^{r} \bar\partial \alpha_i \otimes e_i\right). \end{align*} Applying the same componentwise formula to the scalar forms $\bar\partial\alpha_i \in A^{p,q+1}(U)$ gives \begin{align*} \bar\partial_E^2(\alpha|_U) = \sum_{i=1}^{r} \bar\partial^2 \alpha_i \otimes e_i. \end{align*} By the scalar computation from the previous step, each term $\bar\partial^2\alpha_i$ is zero. Hence \begin{align*} \bar\partial_E^2(\alpha|_U)=0. \end{align*} [/step] [step:Globalize the local vanishing and identify the cochain complex] The point $x_0 \in X$ was arbitrary, and for each such point we found a neighbourhood $U$ on which $\bar\partial_E^2(\alpha|_U)=0$. Therefore the global form $\bar\partial_E^2\alpha \in A^{p,q+2}(X,E)$ restricts to zero on an open cover of $X$, so \begin{align*} \bar\partial_E^2\alpha = 0 \end{align*} on $X$. Since $\alpha \in A^{p,q}(X,E)$ was arbitrary, the composition \begin{align*} A^{p,q}(X,E) \xrightarrow{\bar\partial_E} A^{p,q+1}(X,E) \xrightarrow{\bar\partial_E} A^{p,q+2}(X,E) \end{align*} is zero for every $q$. Hence, for each fixed $p$, the graded space $A^{p,\bullet}(X,E)$ equipped with the differential $\bar\partial_E$ is a cochain complex. [/step]