Let $(M, \omega)$ be a compact Kähler manifold and let $\lambda_1$ denote the first eigenvalue of the Laplace operator $\Delta$. It is a well known fact that there exists an eigenfunction $u$ such that $\Delta u = - \lambda_1 u$.
I am working through the following computation on page 65 of Gang Tian's Canonical Metrics in Kähler Geometry text.
At a point, we write the metric as $g_{i \overline{j}} = \delta_{ij}$. Then we see that \begin{eqnarray*} u_{i j} u_{\overline{i} \overline{j}} &=& (u_{i j} u_{\overline{i}})_{\overline{j}}- u_{i j \overline{j}} u_{\overline{i}}, \\ &=& \text{div}(u_{ij} u_{\overline{i}}) - u_{ji\overline{j}} u_{\overline{i}} \\ &=& \text{div}(u_{ij} u_{\overline{i}}) - u_{j \overline{j} i} u_i - R_{\overline{s} j i \overline{j}} u_s u_{\overline{i}} \end{eqnarray*}
Here $$u_{i \overline{j}} : = \frac{\partial^2 u}{\partial z_i \partial \overline{z}_j}.$$
Can someone explain to me why $u_{ij \overline{j}} = u_{j i \overline{j}}$, but the mixing of holomorphic and anti-holomorphic partial derivatives picks up a curvature term? Any help with justifying the last line of the computation of $u_{ij} u_{\overline{i} \overline{j}}$ would be very much appreciated.
Thanks in advance.
In Riemannian manifold, $\nabla_Z\nabla_X\nabla_Y f-\nabla_Z \nabla_Y\nabla _Xf=0$ so that we have it :
If $T = \nabla\nabla f$ where $T_{ij}=T(E_i,E_j)$, where $E_i$ is normal coordinate fields, and $S = \nabla T$, then \begin{align*} & S(E_i,E_j,E_k)-S(E_i,E_k,E_j)\\ &= E_i( T_{jk} - T_{kj} ) - T_{\alpha k}\Gamma_{ij}^\alpha - T_{j\alpha}\Gamma_{ik}^\alpha + T_{\alpha j}\Gamma_{ik}^\alpha +T_{k\alpha } \Gamma_{ij}^\alpha\\&=_{{\rm normal}}\ E_i( T_{jk} - T_{kj} )\\&=0\end{align*} by the following :
\begin{align*} & T_{ij} - T_{ji} \\&= E_iE_jf -E_jE_i f - \Gamma_{ij}^\alpha E_\alpha f +\Gamma_{ji}^\alpha E_\alpha f \\&=_{{\rm normal}}\ E_iE_jf -E_jE_i f \\&=0 \end{align*}
If $U=S-S'$ where $S'(X,Y,Z):=S(X,Z,Y)$, then $U=0$ for any normal coordinate.