1
$\begingroup$

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 $\partial/\partial \sigma$?

$\endgroup$
  • $\begingroup$ I don't see a reason why $\sigma$ should be a horizontal coordinate. Do you assume that $D$ is a union of fibers of $f$, so that in fact $D=(\sigma\circ f)^{-1}(0)$? And do you want $\omega$ to be Kähler, or at least nondegenerate? $\endgroup$ – Sebastian Goette Feb 17 '16 at 18:54
  • $\begingroup$ $D$ is a simple normal crossing divisor with conical or cusp singularities. The restriction of $\omega $ on each fibre $X_s\setminus D_s$ is Kahler $\endgroup$ – user21574 Feb 17 '16 at 18:59
  • $\begingroup$ It seems there is a typo in your formula for the $\omega$-horizontal lift of $\partial s_j$. Namely, the super-index $\alpha$ is neither contracted nor related to $s_j$, isnt' it ? $\endgroup$ – Holonomia Feb 18 '16 at 12:50
  • $\begingroup$ $\omega^{\bar \beta\alpha} $ is just $\omega^{\bar z_i z_j}$, sorry for typo $\endgroup$ – user21574 Feb 18 '16 at 13:35
  • $\begingroup$ At page 5 of the paper you refered is defined "the horizontal lift" of $\partial s_j$. Can you indicated where is the definition of $\omega$-horizontal lift ? Or they are the same thing? $\endgroup$ – Holonomia Feb 18 '16 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy