[proofplan] We first use the [type decomposition of complex differential forms](/theorems/7008) to split the curvature, since $F_\nabla$ is an $\operatorname{End}(E)$-valued two-form. The identity for $(\nabla^{0,1})^2$ is local, so we compute in a local frame where $\nabla = d + A$ and split the matrix-valued one-form $A$ into its $(1,0)$ and $(0,1)$ parts. Extracting the $(0,2)$ component of $dA + A \wedge A$ gives $\bar{\partial}A^{0,1} + A^{0,1} \wedge A^{0,1}$, which is exactly the square of $\bar{\partial}+A^{0,1}$ on local section vectors. [/proofplan]

[step:Decompose the curvature by form type]Since $\nabla$ is a complex-linear connection on the smooth complex vector bundle $E \to X$, its curvature is the $\operatorname{End}(E)$-valued two-form \begin{align*} F_\nabla := \nabla\circ\nabla \in \Omega^2(X;\operatorname{End}(E)). \end{align*} The complex cotangent bundle decomposes as $T^*X \otimes_{\mathbb{R}} \mathbb{C} = (T^{1,0}X)^* \oplus (T^{0,1}X)^*$, hence \begin{align*} \Omega^2(X;\operatorname{End}(E)) = \Omega^{2,0}(X;\operatorname{End}(E)) \oplus \Omega^{1,1}(X;\operatorname{End}(E)) \oplus \Omega^{0,2}(X;\operatorname{End}(E)). \end{align*} Therefore $F_\nabla$ has a unique decomposition \begin{align*} F_\nabla = F_{\nabla}^{2,0} + F_{\nabla}^{1,1} + F_{\nabla}^{0,2} \end{align*} with $F_{\nabla}^{p,q} \in \Omega^{p,q}(X;\operatorname{End}(E))$.[/step]

[guided]The first assertion is purely a type-decomposition statement. A connection $\nabla$ on $E$ extends to $E$-valued forms, and its square is not another first-order operator: it is tensorial. More precisely, the curvature is the $\operatorname{End}(E)$-valued two-form \begin{align*} F_\nabla := \nabla\circ\nabla \in \Omega^2(X;\operatorname{End}(E)). \end{align*} Because $X$ is a complex manifold, the complexified cotangent bundle splits into holomorphic and anti-holomorphic cotangent directions: \begin{align*} T^*X \otimes_{\mathbb{R}} \mathbb{C} = (T^{1,0}X)^* \oplus (T^{0,1}X)^*. \end{align*} Taking exterior powers gives the decomposition of complex two-forms into types $(2,0)$, $(1,1)$, and $(0,2)$. Tensoring with $\operatorname{End}(E)$ preserves this direct-sum decomposition, so \begin{align*} \Omega^2(X;\operatorname{End}(E)) = \Omega^{2,0}(X;\operatorname{End}(E)) \oplus \Omega^{1,1}(X;\operatorname{End}(E)) \oplus \Omega^{0,2}(X;\operatorname{End}(E)). \end{align*} Since this is a direct sum, every element has a unique expression as a sum of components in the three summands. Applying this to $F_\nabla$ gives \begin{align*} F_\nabla = F_{\nabla}^{2,0} + F_{\nabla}^{1,1} + F_{\nabla}^{0,2}. \end{align*}[/guided]

[step:Write the connection locally and isolate its Dolbeault component] Let $U \subset X$ be an [open set](/page/Open%20Set) over which $E$ has a smooth complex frame $e = (e_1,\dots,e_r)$. A section $s \in \Gamma(E|_U)$ is represented by a smooth map \begin{align*} u: U \to \mathbb{C}^r \end{align*} through $s = \sum_{a=1}^r u_a e_a$. There is a matrix-valued one-form \begin{align*} A \in \Omega^1(U;\operatorname{End}(\mathbb{C}^r)) \end{align*} such that \begin{align*} \nabla s = e(du + Au). \end{align*} Here $e(v)$ denotes the section $\sum_{a=1}^r v_a e_a$ for a $\mathbb{C}^r$-valued form $v$. Decompose \begin{align*} A = A^{1,0} + A^{0,1} \end{align*} with $A^{1,0} \in \Omega^{1,0}(U;\operatorname{End}(\mathbb{C}^r))$ and $A^{0,1} \in \Omega^{0,1}(U;\operatorname{End}(\mathbb{C}^r))$. Since $d = \partial + \bar{\partial}$ on complex-valued forms, the local expression for the $(0,1)$ part of the connection is \begin{align*} \nabla^{0,1}s = e(\bar{\partial}u + A^{0,1}u). \end{align*} [/step]

[step:Compute the local formula for the $(0,2)$ curvature component] In the same local frame, the curvature is represented by the matrix-valued two-form \begin{align*} F_A := dA + A \wedge A \in \Omega^2(U;\operatorname{End}(\mathbb{C}^r)). \end{align*} The [exterior derivative](/theorems/1525) splits as $d = \partial + \bar{\partial}$, and the type decomposition of $A$ gives \begin{align*} dA = \partial A^{1,0} + \bar{\partial}A^{1,0} + \partial A^{0,1} + \bar{\partial}A^{0,1}. \end{align*} Among these terms, the only one of type $(0,2)$ is $\bar{\partial}A^{0,1}$. Similarly, \begin{align*} A \wedge A = A^{1,0}\wedge A^{1,0} + A^{1,0}\wedge A^{0,1} + A^{0,1}\wedge A^{1,0} + A^{0,1}\wedge A^{0,1}. \end{align*} The only term of type $(0,2)$ is $A^{0,1}\wedge A^{0,1}$. Hence the local representative of $F_{\nabla}^{0,2}$ is \begin{align*} F_A^{0,2} = \bar{\partial}A^{0,1} + A^{0,1}\wedge A^{0,1}. \end{align*} [/step]

[step:Extend $\nabla^{0,1}$ by the graded Leibniz rule] For $q \geq 0$, define $\nabla^{0,1}$ on decomposable $E$-valued forms by \begin{align*} \nabla^{0,1}(\alpha \otimes s) := \bar{\partial}\alpha \otimes s + (-1)^q \alpha \wedge \nabla^{0,1}s \end{align*} for $\alpha \in \Omega^{0,q}(X)$ and $s \in \Gamma(E)$, and extend $\mathbb{C}$-linearly. This gives a map \begin{align*} \nabla^{0,1}: \Omega^{0,q}(X;E) \to \Omega^{0,q+1}(X;E). \end{align*} The formula is local in $X$ and compatible with changes of frame because it is the $(0,1)$ part of the usual connection Leibniz rule for $E$-valued forms. Uniqueness follows because every local $E$-valued $(0,q)$-form is a finite sum of decomposable forms $\alpha \otimes e_a$ in a local frame, and the displayed rule fixes the value on each such summand. [/step]

[step:Compare the square of $\nabla^{0,1}$ with the local curvature formula]Let $s \in \Gamma(E|_U)$ be represented by the smooth map $u: U \to \mathbb{C}^r$. Using the local formula $\nabla^{0,1} = \bar{\partial} + A^{0,1}$ on $\mathbb{C}^r$-valued forms, we compute \begin{align*} (\nabla^{0,1})^2s = e\big((\bar{\partial} + A^{0,1}\wedge)(\bar{\partial}u + A^{0,1}u)\big). \end{align*} Expanding by the graded Leibniz rule, \begin{align*} (\bar{\partial} + A^{0,1}\wedge)(\bar{\partial}u + A^{0,1}u) = \bar{\partial}^2u + \bar{\partial}(A^{0,1}u) + A^{0,1}\wedge \bar{\partial}u + A^{0,1}\wedge A^{0,1}u. \end{align*} Since $u$ is a $\mathbb{C}^r$-valued zero-form, $\bar{\partial}^2u = 0$. The graded Leibniz rule for $\bar{\partial}$ applied to the product of the degree-one matrix-valued form $A^{0,1}$ and the degree-zero vector-valued form $u$ gives \begin{align*} \bar{\partial}(A^{0,1}u) = (\bar{\partial}A^{0,1})u - A^{0,1}\wedge \bar{\partial}u. \end{align*} Substituting this identity cancels the two mixed terms and leaves \begin{align*} (\bar{\partial} + A^{0,1}\wedge)(\bar{\partial}u + A^{0,1}u) = (\bar{\partial}A^{0,1} + A^{0,1}\wedge A^{0,1})u. \end{align*} By the previous computation, the matrix-valued form in parentheses is $F_A^{0,2}$. Therefore \begin{align*} (\nabla^{0,1})^2s = e(F_A^{0,2}u) = F_{\nabla}^{0,2}s. \end{align*} This identity holds on every local frame domain $U$, so it holds globally on $X$.[/step]

[guided]We now verify the key identity by doing the full local calculation. Fix an open set $U \subset X$ with a smooth complex frame $e = (e_1,\dots,e_r)$ for $E|_U$. A section $s \in \Gamma(E|_U)$ is represented by a smooth map \begin{align*} u: U \to \mathbb{C}^r \end{align*} through $s = \sum_{a=1}^r u_a e_a$. In this frame, the $(0,1)$ part of the connection is the operator \begin{align*} \bar{\partial} + A^{0,1}\wedge \end{align*} acting on $\mathbb{C}^r$-valued forms, where $A^{0,1} \in \Omega^{0,1}(U;\operatorname{End}(\mathbb{C}^r))$ is the $(0,1)$ component of the local connection matrix. Apply this operator twice to the section vector $u$. First, \begin{align*} \nabla^{0,1}s = e(\bar{\partial}u + A^{0,1}u). \end{align*} Applying $\nabla^{0,1}$ again gives \begin{align*} (\nabla^{0,1})^2s = e\big((\bar{\partial} + A^{0,1}\wedge)(\bar{\partial}u + A^{0,1}u)\big). \end{align*} Now expand the right-hand side. The terms are \begin{align*} (\bar{\partial} + A^{0,1}\wedge)(\bar{\partial}u + A^{0,1}u) = \bar{\partial}^2u + \bar{\partial}(A^{0,1}u) + A^{0,1}\wedge \bar{\partial}u + A^{0,1}\wedge A^{0,1}u. \end{align*} The first term vanishes because $\bar{\partial}^2 = 0$ on smooth vector-valued functions, component by component: \begin{align*} \bar{\partial}^2u = 0. \end{align*} The important point is the sign in the next term. The object $A^{0,1}$ is a matrix-valued form of degree one, while $u$ is a vector-valued form of degree zero. The graded Leibniz rule for $\bar{\partial}$ therefore gives \begin{align*} \bar{\partial}(A^{0,1}u) = (\bar{\partial}A^{0,1})u - A^{0,1}\wedge \bar{\partial}u. \end{align*} The minus sign appears because $\bar{\partial}$ passes across the degree-one form $A^{0,1}$. This is exactly why the mixed terms cancel. Substituting the displayed identity into the expansion gives \begin{align*} (\bar{\partial} + A^{0,1}\wedge)(\bar{\partial}u + A^{0,1}u) = (\bar{\partial}A^{0,1})u - A^{0,1}\wedge \bar{\partial}u + A^{0,1}\wedge \bar{\partial}u + A^{0,1}\wedge A^{0,1}u. \end{align*} Canceling the two mixed terms yields \begin{align*} (\bar{\partial} + A^{0,1}\wedge)(\bar{\partial}u + A^{0,1}u) = (\bar{\partial}A^{0,1} + A^{0,1}\wedge A^{0,1})u. \end{align*} The expression $\bar{\partial}A^{0,1} + A^{0,1}\wedge A^{0,1}$ is not an accidental formula: it is exactly the $(0,2)$ component of the curvature matrix $dA + A \wedge A$. Thus the local representative of $F_{\nabla}^{0,2}$ is \begin{align*} F_A^{0,2} = \bar{\partial}A^{0,1} + A^{0,1}\wedge A^{0,1}. \end{align*} Therefore \begin{align*} (\nabla^{0,1})^2s = e(F_A^{0,2}u) = F_{\nabla}^{0,2}s. \end{align*} Because both sides are globally defined $E$-valued $(0,2)$-forms and the equality holds in every local frame, the equality holds globally on $X$.[/guided]