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Let $s_1,\ldots, s_k$ be linearly independent global holomorphic sections of a holomorphic line bundle $E$ over a compact algebraic manifold $X$, with volume form $\Omega$.

For $m$ large, let $\{e_1,\ldots, e_p\}$ be a basis of $H^0(X,L^m)$, where $L$ is ample. Write $\|e\|^2 := |e_1|^2 + \cdots + |e_p|^2$ and $\|s\|^2 := |s_1|^2 + \cdots + |s_k|^2$. If $t$ is a complex linear combination $\sum_{i,j} \lambda_{ij} e_is_j$, then $\int_X \frac{|t|^2}{\|e\|^2 \|s\|^2} \Omega$ is finite, being bounded above by $\sum_{i,j} |\lambda_{ij}|^2 vol_\Omega X$, up to some constant.

Conversely, for a global holomorphic section $t$ of $L^mE$ over $X$, does the finiteness of
$$\int_X \frac{|t|^2}{\|e\|^2 \|s\|^2} \Omega$$ imply that $t$ is a complex linear combination $\sum_{1\le i\le p,1\le j\le k} \lambda_{ij} e_is_j$?

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    $\begingroup$ It might. I think you want to calculate the multiplier ideal sheaf associated to the singular Hermitian metric $h = h' e^{-\log(\|e\|^2 \|s\|^2)}$, where $h'$ is any smooth metric; see Chapter 5 of Demailly's www-fourier.ujf-grenoble.fr/~demailly/manuscripts/analmeth.pdf What we hope is that the multiplier ideal sheaf coincides with the subsheaf generated by the sections $s_j$, which we should be able to calculate locally. $\endgroup$ Apr 7, 2014 at 16:52

2 Answers 2

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to your first question :

$\int_X \frac{|t|^2}{\|e\|^2 \|s\|^2} \Omega$ is finite, being bounded above by $\sum_{i,j} |\lambda_{ij}|^2 vol_\Omega X$

i have found one paper on google which prove this theorem :

  1. http://arxiv.org/pdf/1211.2948v3.pdf (definition 3,4 )

and the inverse problem maybe not true , you can refer to this paper :

  1. http://www.math.uh.edu/~shanyuji/2012/Geometry/n-20.pdf

since, $\lambda_{ij}=\partial\partial^{'}\log h$

so, $\int_X \frac{|t|^2}{\|e\|^2 \|s\|^2} \Omega\le{\sum{(\partial\partial^{'}\log h)^2}}$$\cdot{vol\Omega}$

because, $\Vert{e}\Vert^2\cdot$$(\frac{1}{2\Vert{e}\Vert^2}+$${\log\Vert{e}\Vert{^{2}}})\ge$$1$$\Longrightarrow$$n\log n \ge1/2,$ with $n\ge1$

which contradict with the definition of the base $a$ of the logarithm :

$\frac{1}{h^2Ina}=\int\int{Inh}=Inh+h^{2}/2$

so, the inverse question holds only under the case that the manifold $X$ satisfy convexity !

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I do not know if I have well-understood your notations. I think you mean that there is smooth metric on $E$ and positive metric on $L$. Moreover, the volume form $\Omega$ is equivalent to the Hausdorff measure of some smooth Riemannian metric on $X$.

Under such assumptions, then the function $\|e\|$ you have defined is just proportional to the so-called Bergman kernel of $L^m$. As long as $m$ is large enough, it is bounded away from $0$. So when $k=1$ your condition simply says $$ \int_{X}\frac{|t|^2}{|s|^2}\Omega<\infty. $$ Thus $t$ must vanish along the divisor of $s$ at least the same times as $s$ does. So we get a decomposition $$ \mathrm{Div}\, t=\mathrm{Div}\, s+\alpha, $$ where $\alpha$ is in the linear system $|L^m|$. This means $s$ divides $t$.

The general case where $k>1$ is slightly more complicated. The sections $t$ so that the integral is finite can be described locally using the multiplier ideal sheaf of $\log\|s\|$. You may find the full result on Page 37 of Demailly's book 'analytic methods in algebraic geometry'.

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