Second derivative of Kähler potential. Does the following second covariant (in terms of Kähler geometry) derivative of Kähler potential vanish?
\begin{equation}
K_{ij}\equiv\nabla_i\nabla_j K=0,
\end{equation}
\begin{equation}
K_{i^*j^*}\equiv\nabla_{i^*}\nabla_{j^*} K=0?
\end{equation}
Indices represent complex coordinates ($A_i$ and $\bar{A}_i$) on Kähler manifold.
 A: The answer is in general No. Take e.g. the Fubini-Study Kaehler potential 
$$K~=~\ln D, \qquad D~=~1+ Q,\qquad Q~=~\sum_{k=1}^n z^k \bar{z}^k, \tag{1}$$
with Hermitian metric 
$$g_{\imath\bar{\jmath}} ~=~ \partial_{\imath} \bar{\partial}_{\bar{\jmath}}K~=~ \frac{\delta_{\imath\bar{\jmath}}}{D}-\frac{\bar{z}^{\imath}z^{\bar{\jmath}}}{D^2}~=~ D^{-1}\left(\delta_{\imath\bar{\jmath}}-\frac{\bar{z}^{\imath}z^{\bar{\jmath}}}{D}\right), \tag{2} $$
and inverse metric
$$ g^{\bar{\imath}\jmath}~=~D(\delta^{\bar{\imath}\jmath}+\bar{z}^{\bar{\imath}}z^{\jmath} ) ,  \tag{3}$$
and Hermitian Christoffel symbols
$$ \Gamma_{\imath\jmath}^{\ell} ~=~ \partial_{\imath}g_{\jmath\bar{k}}~g^{\bar{k}\ell}~=~-\frac{\bar{z}^{\imath}\delta_{\jmath}^{\ell}}{D}-\frac{\bar{z}^{\jmath}\delta_{\imath}^{\ell}}{D}. \tag{4}$$ 
The covariant derivative of the Kaehler potential is
$$ \nabla_{\ell}K~=~\partial_{\ell}K ~=~\frac{\bar{z}^{\ell}}{D}.\tag{5} $$
Now calculate the sought-for quantity
$$ \nabla_{\imath}\nabla_{\jmath}K 
~=~ \nabla_{\imath}\partial_{\jmath}K
~=~\partial_{\imath}\partial_{\jmath}K
-\Gamma_{\imath\jmath}^{\ell}~\partial_{\ell}K
~=~\frac{\bar{z}^{\imath}\bar{z}^{\jmath}}{D^2}
~\neq~0, \tag{6}$$
which does not vanish.
