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.

This post imported from StackExchange Mathematics at 2016-06-23 20:24 (UTC), posted by SE-user Qmechanic