= Solution
The unheaded definitions use the <complex structure> $J$ on the real tangent bundle. Compatibility means $g(Ju,Jv)=g(u,v)$. Its <fundamental Hermitian form> is $\omega(u,v)=g(Ju,v)$; the compatibility identity makes this real, alternating and of type $(1,1)$. The metric is a <Kähler metric> precisely when $d\omega=0$. In complex dimension one a real three-form is zero, so every compatible metric is a <Kähler metric>.
Now assume closedness and prove the <Kähler normal holomorphic coordinates> condition. Start with holomorphic coordinates $w$ centered at the point and make a complex-linear change so that the positive Hermitian coefficient matrix satisfies $h(0)=I$. Closedness of the $(1,1)$ form gives
$$
\partial_{w_k}h_{i\bar j}=\partial_{w_i}h_{k\bar j}.
$$
Define $a_{i\bar j,k}=\partial_{w_k}h_{i\bar j}(0)$; it is symmetric in $i,k$. Choose new coordinates implicitly by
$$
w_j=z_j-\frac12\sum_{i,k}a_{i\bar j,k}z_iz_k.
$$
The derivative of this <holomorphic map> at zero is the identity, so the <holomorphic inverse function theorem> makes $z$ valid local coordinates. In the new coordinates,
$$
h'_{i\bar j}(z)=\sum_{a,b}h_{a\bar b}(w(z))
\frac{\partial w_a}{\partial z_i}\overline{\frac{\partial w_b}{\partial z_j}}.
$$
At zero, $h'=I$, and differentiating gives
$$
\partial_{z_k}h'_{i\bar j}(0)
=a_{i\bar j,k}+\partial_{z_i}\partial_{z_k}w_j(0)=0.
$$
Hermitian symmetry makes all antiholomorphic first derivatives zero as well. <Taylor's theorem> for a <smooth function> now yields
$$
\boxed{h'_{i\bar j}=\delta_{ij}+O(|z|^2).}
$$
Thus condition (a) implies condition (b), with the exact factor $i/2$ in the fundamental-form convention retained.
Back to article page
