Second derivative of Kähler potential.

Question

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.

Answer 1

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.