[proofplan] We prove all identities locally in holomorphic frames, where a bundle-valued form is a finite sum of scalar forms multiplied by holomorphic frame vectors. In such frames, the Dolbeault operator acts only on the scalar coefficient forms, so the tensor, pullback, dual, and exterior-power formulas reduce to the corresponding scalar rules for $\bar{\partial}$. The only remaining point is that the local definitions glue: changes of holomorphic frame have holomorphic transition matrices, whose scalar $\bar{\partial}$ vanishes. This also gives uniqueness, because a bundle-valued Dolbeault operator is determined locally by its action on coefficients in holomorphic frames. [/proofplan]

[step:Write the Dolbeault operator in holomorphic frames]Let $U\subset X$ be an [open set](/page/Open%20Set) over which $E$ has a holomorphic frame $e_1,\dots,e_m$, where each $e_i:U\to E|_U$ is a holomorphic local section and $m=\operatorname{rank}E$. Every element $\alpha\in A^{p,q}(U,E)$ has a unique expression \begin{align*} \alpha=\sum_{i=1}^m \alpha_i\otimes e_i \end{align*} with scalar forms $\alpha_i\in A^{p,q}(U)$. By the local definition of the Dolbeault operator on a holomorphic vector bundle, \begin{align*} \bar{\partial}_E\alpha=\sum_{i=1}^m \bar{\partial}\alpha_i\otimes e_i. \end{align*} This formula is independent of the chosen holomorphic frame. Indeed, if $e'_1,\dots,e'_m$ is another holomorphic frame on $U$ and $g:U\to GL(m,\mathbb{C})$ is the holomorphic transition map defined by \begin{align*} e'_j=\sum_{i=1}^m g_{ij}e_i, \end{align*} then $g^{-1}:U\to GL(m,\mathbb{C})$ is holomorphic as well. If $\alpha=\sum_i\alpha_i\otimes e_i=\sum_j\alpha'_j\otimes e'_j$, then the coefficient vector $\alpha'=(\alpha'_1,\dots,\alpha'_m)$ satisfies $\alpha'=g^{-1}\alpha$ in matrix notation. Since every entry of $g^{-1}$ is holomorphic, $\bar{\partial}(g^{-1}_{ji})=0$ for all $i,j$, and the scalar Leibniz rule gives \begin{align*} \bar{\partial}\alpha'_j=\sum_{i=1}^m g^{-1}_{ji}\bar{\partial}\alpha_i. \end{align*} Thus the coefficient formula transforms by the same transition matrix, so the local expressions glue to a globally defined operator.[/step]

[guided]The key local fact is that a holomorphic frame removes the bundle part of the Dolbeault operator. Let $U\subset X$ be an open set and let $e_1,\dots,e_m$ be a holomorphic frame for $E|_U$. This means each $e_i:U\to E|_U$ is a holomorphic local section and every vector in each fibre of $E|_U$ is written uniquely in this frame. Therefore every bundle-valued form $\alpha\in A^{p,q}(U,E)$ has a unique decomposition \begin{align*} \alpha=\sum_{i=1}^m \alpha_i\otimes e_i \end{align*} where each $\alpha_i\in A^{p,q}(U)$ is an ordinary scalar differential form. The Dolbeault operator on $E$ is defined in this frame by applying the scalar Dolbeault operator to the coefficients: \begin{align*} \bar{\partial}_E\alpha=\sum_{i=1}^m \bar{\partial}\alpha_i\otimes e_i. \end{align*} Why is this not dependent on the frame? Suppose $e'_1,\dots,e'_m$ is another holomorphic frame on the same open set. There is a holomorphic transition map $g:U\to GL(m,\mathbb{C})$ such that \begin{align*} e'_j=\sum_{i=1}^m g_{ij}e_i. \end{align*} Since $g$ is holomorphic and takes values in invertible matrices, $g^{-1}:U\to GL(m,\mathbb{C})$ is holomorphic. Writing the same form in both frames, \begin{align*} \alpha=\sum_{i=1}^m \alpha_i\otimes e_i=\sum_{j=1}^m \alpha'_j\otimes e'_j, \end{align*} the coefficient vectors satisfy $\alpha'=g^{-1}\alpha$. Thus, for each $j$, \begin{align*} \alpha'_j=\sum_{i=1}^m g^{-1}_{ji}\alpha_i. \end{align*} Applying the scalar Leibniz rule for $\bar{\partial}$ gives \begin{align*} \bar{\partial}\alpha'_j=\sum_{i=1}^m \bar{\partial}(g^{-1}_{ji}\alpha_i). \end{align*} Because $g^{-1}_{ji}$ is a [holomorphic function](/page/Holomorphic%20Function), $\bar{\partial}(g^{-1}_{ji})=0$. Hence \begin{align*} \bar{\partial}\alpha'_j=\sum_{i=1}^m g^{-1}_{ji}\bar{\partial}\alpha_i. \end{align*} So the transformed coefficients of $\bar{\partial}_E\alpha$ are exactly the transition transform of the old coefficients. This proves that the coefficient formula defines a global operator and explains why holomorphic frames are essential: the transition functions contribute no extra $\bar{\partial}$ terms.[/guided]

[step:Reduce the tensor product identity to the scalar Leibniz rule] Let $U\subset X$ be an open set over which $E$ has holomorphic frame $e_1,\dots,e_m$ and $F$ has holomorphic frame $f_1,\dots,f_n$. Write \begin{align*} \alpha=\sum_{i=1}^m \alpha_i\otimes e_i \end{align*} and \begin{align*} \beta=\sum_{j=1}^n \beta_j\otimes f_j \end{align*} with $\alpha_i\in A^{p,q}(U)$ and $\beta_j\in A^{r,s}(U)$. The tensors $e_i\otimes f_j$ form a holomorphic frame for $(E\otimes F)|_U$, and \begin{align*} \alpha\wedge\beta=\sum_{i=1}^m\sum_{j=1}^n(\alpha_i\wedge\beta_j)\otimes(e_i\otimes f_j). \end{align*} Since $\alpha_i$ has total degree $p+q$, the scalar Dolbeault Leibniz rule gives \begin{align*} \bar{\partial}(\alpha_i\wedge\beta_j)=\bar{\partial}\alpha_i\wedge\beta_j+(-1)^{p+q}\alpha_i\wedge\bar{\partial}\beta_j. \end{align*} Applying the local coefficient formula for $\bar{\partial}_{E\otimes F}$, \begin{align*} \bar{\partial}_{E\otimes F}(\alpha\wedge\beta)=\bar{\partial}_E\alpha\wedge\beta+(-1)^{p+q}\alpha\wedge\bar{\partial}_F\beta. \end{align*} For sections $s\in C^\infty(X,E)$ and $t\in C^\infty(X,F)$, the form degree of $s$ is $0$, so the same formula gives \begin{align*} \bar{\partial}_{E\otimes F}(s\otimes t)=(\bar{\partial}_E s)\otimes t+s\otimes(\bar{\partial}_F t). \end{align*} [/step]

[step:Prove holomorphic pullback commutes with the Dolbeault operator] Let $V\subset X$ be an open set with holomorphic frame $e_1,\dots,e_m$ for $E|_V$, and let $W=f^{-1}(V)\subset Y$. The pulled-back sections $f^*e_1,\dots,f^*e_m$ form a holomorphic frame for $(f^*E)|_W$. For $\alpha\in A^{p,q}(V,E)$, write \begin{align*} \alpha=\sum_{i=1}^m \alpha_i\otimes e_i \end{align*} with $\alpha_i\in A^{p,q}(V)$. Then \begin{align*} f^*\alpha=\sum_{i=1}^m f^*\alpha_i\otimes f^*e_i. \end{align*} Because $f:Y\to X$ is holomorphic, scalar pullback preserves type and satisfies \begin{align*} \bar{\partial}(f^*\alpha_i)=f^*(\bar{\partial}\alpha_i) \end{align*} for every scalar form $\alpha_i\in A^{p,q}(V)$. Therefore \begin{align*} \bar{\partial}_{f^*E}(f^*\alpha)=\sum_{i=1}^m f^*(\bar{\partial}\alpha_i)\otimes f^*e_i=f^*(\bar{\partial}_E\alpha). \end{align*} These local identities agree on overlaps by the frame-independence proved above, hence the identity holds globally on $Y$. [/step]

[step:Characterize the dual Dolbeault operator by graded evaluation] Let $U\subset X$ be an open set with holomorphic frame $e_1,\dots,e_m$ for $E|_U$, and let $e^1,\dots,e^m$ be the dual holomorphic frame for $E^*|_U$, characterized by $e^i(e_j)=\delta_{ij}$. For $\lambda\in A^{p,q}(U,E^*)$, write \begin{align*} \lambda=\sum_{i=1}^m \lambda_i\otimes e^i \end{align*} with $\lambda_i\in A^{p,q}(U)$, and define locally \begin{align*} \bar{\partial}_{E^*}\lambda=\sum_{i=1}^m \bar{\partial}\lambda_i\otimes e^i. \end{align*} This definition is independent of the frame. Indeed, if $e'^1,\dots,e'^m$ is the dual frame corresponding to another holomorphic frame $e'_1,\dots,e'_m$, then the transition matrix for $e'^1,\dots,e'^m$ is obtained from the holomorphic inverse transpose of the transition matrix for $e_1,\dots,e_m$. Its entries are holomorphic, so applying scalar $\bar{\partial}$ to the transformed coefficients produces no transition-derivative terms. Let $s\in C^\infty(U,E)$ and write \begin{align*} s=\sum_{i=1}^m s_i e_i \end{align*} with smooth functions $s_i:U\to\mathbb{C}$. The graded evaluation convention is \begin{align*} \lambda(s)=\sum_{i=1}^m \lambda_i s_i \end{align*} and \begin{align*} \lambda(\bar{\partial}_E s)=\sum_{i=1}^m \lambda_i\wedge\bar{\partial}s_i. \end{align*} Using the scalar Leibniz rule and the fact that $\lambda_i$ has total degree $p+q$, \begin{align*} \bar{\partial}\bigl(\lambda(s)\bigr)=\sum_{i=1}^m \bar{\partial}(\lambda_i s_i)=\sum_{i=1}^m(\bar{\partial}\lambda_i)s_i+(-1)^{p+q}\sum_{i=1}^m\lambda_i\wedge\bar{\partial}s_i. \end{align*} Thus \begin{align*} \bar{\partial}\bigl(\lambda(s)\bigr)=\bigl(\bar{\partial}_{E^*}\lambda\bigr)(s)+(-1)^{p+q}\lambda(\bar{\partial}_E s). \end{align*} Conversely, if an operator $D:A^{p,q}(X,E^*)\to A^{p,q+1}(X,E^*)$ satisfies this identity for every smooth section $s$, then evaluating against the local frame sections $e_j$ gives \begin{align*} (D\lambda)(e_j)=\bar{\partial}(\lambda(e_j))=\bar{\partial}\lambda_j. \end{align*} Hence $D\lambda=\sum_j\bar{\partial}\lambda_j\otimes e^j=\bar{\partial}_{E^*}\lambda$, proving uniqueness. [/step]

[step:Descend the tensor product operator to exterior powers] First let $k\geq 1$. On an open set $U\subset X$ with holomorphic frame $e_1,\dots,e_m$ for $E|_U$, the wedge tensors \begin{align*} e_{i_1}\wedge\cdots\wedge e_{i_k} \end{align*} with $1\leq i_1<\cdots<i_k\leq m$ form a holomorphic frame for $\Lambda^kE|_U$. Therefore a form $\gamma\in A^{p,q}(U,\Lambda^kE)$ has a unique expression \begin{align*} \gamma=\sum_{1\leq i_1<\cdots<i_k\leq m}\gamma_{i_1,\dots,i_k}\otimes(e_{i_1}\wedge\cdots\wedge e_{i_k}) \end{align*} with scalar forms $\gamma_{i_1,\dots,i_k}\in A^{p,q}(U)$, and the Dolbeault operator is locally \begin{align*} \bar{\partial}_{\Lambda^kE}\gamma=\sum_{1\leq i_1<\cdots<i_k\leq m}\bar{\partial}\gamma_{i_1,\dots,i_k}\otimes(e_{i_1}\wedge\cdots\wedge e_{i_k}). \end{align*} Holomorphic frame changes on $\Lambda^kE$ are obtained by applying $\Lambda^k$ to the holomorphic transition matrices of $E$, so the same frame-independence argument shows that these local formulas glue. The alternating quotient map \begin{align*} \operatorname{Alt}_k:E^{\otimes k}\to\Lambda^kE \end{align*} is the holomorphic bundle morphism normalized by \begin{align*} \operatorname{Alt}_k(s_1\otimes\cdots\otimes s_k)=s_1\wedge\cdots\wedge s_k. \end{align*} In holomorphic frames, the local coefficient formula for $\bar{\partial}_{E^{\otimes k}}$ applies scalar $\bar{\partial}$ to coefficients and leaves the frame tensors fixed. The kernel of $\operatorname{Alt}_k$ is locally generated by the usual alternating relations among these frame tensors, so this coefficientwise operator preserves $\ker(\operatorname{Alt}_k)$ and therefore descends to the quotient $\Lambda^kE$. The tensor-product Leibniz rule for degree-zero sections gives \begin{align*} \bar{\partial}_{E^{\otimes k}}(s_1\otimes\cdots\otimes s_k)=\sum_{j=1}^k s_1\otimes\cdots\otimes\bar{\partial}_E s_j\otimes\cdots\otimes s_k. \end{align*} Applying $\operatorname{Alt}_k$ gives \begin{align*} \bar{\partial}_{\Lambda^kE}(s_1\wedge\cdots\wedge s_k)=\sum_{j=1}^k s_1\wedge\cdots\wedge\bar{\partial}_E s_j\wedge\cdots\wedge s_k. \end{align*} For $k=0$, $\Lambda^0E=X\times\mathbb{C}$, so the induced operator is the scalar Dolbeault operator. [/step]

[step:Conclude uniqueness and compatibility with contractions] The [tensor product](/page/Tensor%20Product), dual, and exterior-power constructions above were defined in holomorphic frames by applying scalar $\bar{\partial}$ to coefficient forms, and the preceding steps show that these definitions are invariant under holomorphic transition functions. Hence the identities hold globally. Uniqueness follows locally. In a holomorphic frame, compatibility with tensor products forces the coefficient formula on $E\otimes F$, compatibility with evaluation forces the coefficient formula on $E^*$, and compatibility with the alternating quotient forces the coefficient formula on $\Lambda^kE$. Since local holomorphic frames cover $X$ and the local formulas agree on overlaps, there is at most one global operator with these compatibilities. Finally, contractions are compositions of tensor products with evaluation pairings between a bundle and its dual. The tensor-product Leibniz rule and the dual evaluation identity therefore imply the corresponding graded Leibniz identities for every contraction. This proves the asserted natural compatibility of Dolbeault operators with pullback, tensor products, dual bundles, exterior powers, evaluation, contraction, and alternating quotients. [/step]