Chern connection

Remarks

Write $\nabla = \nabla^{1,0} + \nabla^{0,1}$ for the splitting by type. In a local holomorphic frame $e_1, \dots, e_r$, let $H$ be the matrix with entries $H_{ij} = h(e_j, e_i)$. Then the connection matrix is $\theta = H^{-1} \partial H$ in a suitable index convention, and the curvature is $\Theta = \bar\partial \theta$, a $(1,1)$-form with values in $\operatorname{End} E$.

Example: let $L$ be a holomorphic line bundle with local holomorphic frame $e$, and let $H = h(e, e) > 0$. Put $\nabla e = \partial \log H \otimes e$. Clause 2 holds because $\bar\partial_E e = 0$ and $\partial \log H$ has type $(1,0)$. Clause 1 holds because $h(\nabla e, e) + h(e, \nabla e) = (\partial \log H + \bar\partial \log H) H = dH$. The curvature is $\Theta = d \partial \log H = -\partial \bar\partial \log H$.

Non-example: on the trivial line bundle over $\C$ with the standard metric, let $\nabla s = ds + i s\, dx$, where $x = \operatorname{Re} z$. Since $i\, dx$ is purely imaginary, clause 1 holds. But $i\, dx = \tfrac{i}{2}(dz + d\bar z)$, so $\nabla^{0,1} = \bar\partial + \tfrac{i}{2} d\bar z \ne \bar\partial$, and clause 2 fails.