[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 $F_{\nabla}^{p,q} \in \Omega^{p,q}(X;\operatorname{End}(E))$.
[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]
[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 $X$.
[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]
[/step]
