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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-118/4/e/solution