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

A connection on a vector bundle is a complex-linear map $D:{\Omega }^0(E)\to {\Omega }^1(E)$ satisfying the Leibniz rule $D(fs)=df\otimes s+fDs$. It is compatible with the Hermitian metric on a holomorphic vector bundle $h$ whenand compatible with the holomorphic structure when its type-$(0,1)$ part is the Dolbeault partial connection:

In a unitary frame the metric matrix is $I$. If $A$ is the connection one-form, metric compatibility gives $0=dI=A^{\ast }+A$, sothe matrix is Skew-Hermitian. In a holomorphic local frame, every frame vector is annihilated by ${\bar{\partial }}_E$, and compatibility with the holomorphic structure gives

In a holomorphic local frame $e$, let $H=(h(e_i,e_j))$ and write $De=eA$. Holomorphic compatibility forces $A^{0,1}=0$, while metric compatibility forces the local formula for the Chern connection on a vector bundleThis proves uniqueness. The formula transforms by the connection one-form law under a holomorphic frame change, so the local definitions glue and satisfy both conditions. This proves existence of the unique Chern connection.

Let $D_1$ and $D_2$ each be a connection on a vector bundle. Their difference is tensorial, so $D_1=D_2+a$ with $a\in {\Omega }^1(\mathrm{End}E)$. Extend $D_2$ to the endomorphism bundle connection. Expanding $(D_2+a{)}^2$ with the supplied graded Leibniz rule gives the curvature difference formula

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 $D_2^{0,1}={\bar{\partial }}_{\mathrm{End}E}$, proving the curvature difference of two Chern connections