Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 118 / 3 / b / Solution

Fix the convention that the Hermitian metric is conjugate-linear in its first argument. In a holomorphic local frame $e=(e_1,\dots ,e_r)$, let $H_{ij}=h(e_i,e_j)$, so $h(ev,ew)=v^{†}Hw$. Write $De=e\theta$, or equivalently $D(ev)=e(dv+\theta v)$.

Compatibility with the holomorphic vector bundle structure means $D^{0,1}={\bar{\partial }}_E$. Since the frame is holomorphic, this forces ${\theta }^{0,1}=0$. Metric compatibility meansor, in the chosen frame, $dH={\theta }^{†}H+H\theta$. Conjugate transpose here also conjugates the one-form factors. Taking the $(1,0)$ component gives $\partial H=H\theta$, and henceThis proves uniqueness and suggests existence. The proposed $\theta$ has type $(1,0)$, and its conjugate transpose obeys ${\theta }^{†}H=\bar{\partial }H$, since $H=H^{†}$. Therefore $dH={\theta }^{†}H+H\theta$, proving metric compatibility as well.

Under a change of holomorphic local frame $e^{\prime }=eG$, we have $H^{\prime }=G^{†}HG$ and $\partial G^{†}=0$. A direct computation yieldsBecause $G$ is holomorphic, $\partial G=dG$; this is exactly the transformation rule for a connection one-form. The locally defined connections therefore glue to a global connection on a vector bundle. The Chern connection exists uniquely, and its local formula for the Chern connection on a vector bundle is