By definition, $\sigma$ is the orthogonal projection of $\partial_{w_{1}}$ onto $(TH)^{\perp}$, so we have along $H$,

(3.299)

$$a_{1}^{-1}\partial_{y}=\partial_{w_{1}}+\sum_{j=2}^{n-1}b_{j}\partial_{w_{j}},$$

where $a_{1}=|\sigma|^{-1}>0$ and $b_{j}$ are smooth functions on $H$. Now write

(3.300)

$$\partial_{y}=\sum_{j=1}^{n-1}\frac{\partial w_{j}}{\partial y}\partial_{w_{j}}+\sum_{j=1}^{n-1}\frac{\partial\bar{w}_{j}}{\partial{y}}\partial_{\bar{w}_{j}}.$$

then we get that along $H$,

(3.301)

$\displaystyle\frac{\partial\bar{w}_{j}}{\partial y}=0,\ j\geq 1,$

which in particular implies

(3.302)

$$\frac{\partial w_{j}}{\partial\bar{y}}=\overline{\frac{\partial\bar{w}_{j}}{\partial y}}=0,\ j\geq 1.$$

Now by the definition of the normal exponential map, we have at $p$,

(3.303)

$$\nabla_{\partial_{y}}\partial_{y}=\nabla_{\partial_{\bar{y}}}\partial_{y}=\nabla_{\partial_{\bar{y}}}\partial_{\bar{y}}=0.$$

Using the Kähler condition we have

(3.304)

$$\nabla_{\partial_{w_{j}}}{\partial_{\bar{w}_{k}}}=\nabla_{\partial_{\bar{w}_{j}}}\partial_{w_{k}}=0,\ \ j,k\geq 1.$$

Then by (3.300) we get

(3.305)

$$\frac{\partial^{2}w_{j}}{\partial y\partial\bar{y}}=\frac{\partial^{2}w_{j}}{\partial\bar{y}^{2}}=0,\ \ \ j\geq 1.$$

Therefore, the conclusion follows.

∎