Because the two <Chern connections> have the same $(0,1)$ part, $a=D_1-D_2$ has type $(1,0)$. The <curvature difference formula> has only $(2,0)$ and $(1,1)$ parts. Both Chern curvatures have type $(1,1)$, so the total $(2,0)$ part vanishes. The remaining part is obtained from $D_2^{0,1}=\bar\partial_{\operatorname{End}E}$, proving the <curvature difference of two Chern connections>
$$
\boxed{F_{D_1}-F_{D_2}=\bar\partial_{\operatorname{End}E}a.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 118 / 4 / e / Solution

{title2=$F_{D_1}-F_{D_2}=\bar\partial a$}

Let $D_1$ and $D_2$ be <Chern connections> on the same <holomorphic vector bundle>, possibly for different <Hermitian metrics>, and put $a=D_1-D_2$. Then $a$ has type $(1,0)$ and
$$
F_{D_1}-F_{D_2}=\bar\partial_{\operatorname{End}E}a.
$$
Indeed, the <curvature difference formula> has only types $(2,0)$ and $(1,1)$; both Chern curvatures have type $(1,1)$, so the $(2,0)$ part vanishes and the remaining part is $D_2^{0,1}a=\bar\partial_{\operatorname{End}E}a$.

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 Incoming links: Curvature difference of two Chern connections