Let $f:X\setminus D\to Y$ be a smooth family of Kahler manifolds, where $D=\{\sigma=0\}$ is a divisor on $X$. Taking a local coordinate $(s_1,...,s_d)$ of $Y$ and a local coordinate $(z_1,...,z_n)$ of a fiber of $f$, $(z_1,...,z_n,\sigma, s_2,...,s_d)$ forms a local coordinate of $X\setminus D$ such that under this coordinate, the holomorphic mapping $f$ is locally given by $f(z_1,...,z_n,\sigma, s_2,...,s_d) = (\sigma, s_2,...,s_d)$

Now take $\omega$ be a $(1,1)$-form on $X'=X\setminus D$. We can write:

$$\omega=\sqrt{-1}\left(\omega_{z_i\bar z_j}dz_i\wedge d\bar z_j+\omega_{z_i\bar s_j}dz_i\wedge d\bar s_j+\omega_{s_i\bar z_j}ds_i\wedge d\bar z_j+\omega_{s_i\bar s_j}ds_i\wedge d\bar s_j+\omega_{\sigma\bar \sigma}d\sigma\wedge d\bar \sigma+\omega_{\sigma\bar z_j}d\sigma\wedge d\bar z_j+\omega_{z_i\bar \sigma}dz_i\wedge d\bar \sigma+\omega_{s_i\bar \sigma}ds_i\wedge d\bar \sigma+\omega_{\sigma\bar s_j}d\sigma\wedge d\bar s_j\right)$$

Is the $\omega$-horizontal lift of type $(1, 0)$ of the tangent vector $\partial/\partial s_j$ equal to:

$$\partial/\partial s_j-\omega_{i\bar \beta}\omega^{\bar \beta\alpha}\partial/\partial z_i?$$ Can someone give a proof? See page 4 http://arxiv.org/pdf/1006.2966.pdf

What is the horizontal lift of $\partial/\partial \sigma$?