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Let $X,Y$ are two projective varieties and $f:X\to Y$ is an Iitaka fibration. Consider the following singular hermitian metric $$h(\sigma,\sigma)=\left(\int_{X_y}|\sigma|^{\frac{2}{m!}}\right)^{m!}$$ where $y\in Y$ and $\sigma$ is a section of $$\frac{1}{m!}f_*\mathcal O_X(m!K_{X/Y})|_{X_y}$$

then, the curvature of hermitian metric $h$, i.e., $\sqrt{-1}\Theta_h$ is $C^\infty$? A counter-example for it is appreciated.

Under which assumption on $X,Y$ it can be $C^\infty$?

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Is there any singular hermitian metric on $$\frac{1}{m!}f_*\mathcal O_X(m!K_{X/Y})$$ such that its curvature is $C^\infty$

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  • $\begingroup$ If the curvature of a singular Hermitian metric on a holomorphic line bundle is $C^\infty$ then so is the metric itself. Indeed, we can write $h=h_0e^{-\varphi}$ where $h_0$ is a smooth Hermitian metric and $\varphi$ is quasi-psh. Your assumption is that $\Theta_{h_0}+dd^c\varphi$ is smooth. In particular, the Laplacian of $\varphi$ is smooth, so by local elliptic estimates you obtain than $\varphi$ is smooth. $\endgroup$
    – YangMills
    Commented Aug 1, 2016 at 5:27
  • $\begingroup$ Thank you. But you repeated my question again, since by definition $\Theta_h=Θ_{h_0}+dd^cφ $, ;) $\endgroup$
    – user21574
    Commented Aug 1, 2016 at 11:06
  • $\begingroup$ What do you mean repeated your question? I answered it. There is no genuinely singular metric on a line bundle whose curvature is smooth! $\endgroup$
    – YangMills
    Commented Aug 1, 2016 at 17:10
  • $\begingroup$ If $X_0$ only has canonical singularities, or if $X$ is smooth and $X_0$ only has isolated ordinary quadratic singularities, then such $L^2$ metric is continuous. See Remark 2.10. of perso.univ-rennes1.fr/christophe.mourougane/recherche/metric/… $\endgroup$
    – user21574
    Commented Feb 4, 2017 at 9:56
  • $\begingroup$ continuous is not the same as smooth! Your question here is about smooth curvature $\endgroup$
    – YangMills
    Commented Feb 6, 2017 at 4:59

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