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Let $X$ be a compact complex manifold and $L\to X$ a holomorphic line bundle (without any a priori assumption on its positivity).

Suppose that for each $x,y\in X$, with $x\ne y$, there exists a $k_0\in\mathbb N$ which depends on $x$ and $y$, such that for all $k\in\mathbb N$, $k\ge k_0$ there exists a global holomorphic section $\sigma_k\in H^0(X,L^{\otimes k})$ such that $\sigma_k(x)=0$ and $\sigma_k(y)\ne 0$.

Is it then true that this property is uniform with respect to couples of distinct points of $X$?

In other words, is it then true that there exists a $k_1\in\mathbb N$ such that for all $k\in\mathbb N$, $k\ge k_1$, and for all $x,y\in X$, with $x\ne y$, one can find a global holomorphic section $\tau_k\in H^0(X,L^{\otimes k})$ such that $\tau_k(x)=0$ and $\tau_k(y)\ne 0$?

Thanks in advance!

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    $\begingroup$ Btw, in your final formula, should $\tau_k(y)=0$ be instead $\tau_k(y) \neq 0$? $\endgroup$
    – M P
    Commented Dec 14, 2011 at 11:23
  • $\begingroup$ right, I edit my question then! thanks $\endgroup$
    – diverietti
    Commented Dec 14, 2011 at 12:52

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I think this is true.

The condition implies that for some $n$ the sections of $L^{\otimes n}$ are base point free, so $L^{\otimes n}$ is obtained by pulling back $\mathcal O(1)$ along a map $X \to \mathbb P^N$ for some $N$. This map must be finite, because otherwise there would be a positive dimensional connected subspace of $X$ along which $L^{\otimes n}$ would be trivial, and then its point could not be separated by any power of $L$. By GAGA $X$ is projective, and then $L^{\otimes n}$ is ample (it is a standard result that a pullback of an ample line bundle along a finite map is ample). But this implies that $L$ is ample, so all sufficiently large powers of $L$ must be very ample.

[Addendum]: I guess I am using three facts: that the image $Y$ of $X$ in $\mathbb P^N$ is closed, hence projective, by Chow's theorem; that given a finite map $X \to Y$, if $Y$ is algebraic so is $X$, by GAGA; that the the pullback of an ample line bundle along a finite map is still ample (which implies that $X$ is projective).

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  • $\begingroup$ Dear Angelo, thank you very much for your answer! As long as I can see, it seems to work perfectly. Two comments are in order: 1) Which "kind" of GAGA are you invoking here? Does one need $X$ to be already projective in order to say that a line bundle on $X$ pullback of an ample by a finite map is ample? 2) Do you think there is a more "elementary" proof of this fact? By elementary I mean just using, for example, combining sections of different powers of $L$ to obtain such an uniformity (I don't know if I am clear enough to make you understand what I mean by elementary here...) $\endgroup$
    – diverietti
    Commented Dec 14, 2011 at 14:16
  • $\begingroup$ I didn't say that in the original question, but I was really guessing if there was a "down-to-earth proof" of this fact... $\endgroup$
    – diverietti
    Commented Dec 14, 2011 at 14:17
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    $\begingroup$ I have no idea how to give a direct elementary proof, sorry. $\endgroup$
    – Angelo
    Commented Dec 14, 2011 at 14:38
  • $\begingroup$ If the assumption is that for any $x,y \in X$ there is a global section such that it vanishes at $x$ and doesn't vanish at $y$, OR vice versa, then this system is not necessarily ample. $\endgroup$
    – Parsa
    Commented Dec 16, 2011 at 10:33
  • $\begingroup$ So, where is Angelo's answer incorrect? I cannot see! $\endgroup$
    – diverietti
    Commented Dec 18, 2011 at 19:28
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Maybe I am making a mistake, but you could try the following. First, for each $k$ the set $U_k$ of points $x$ for which there is a section of $L^k$ not vanishing at $x$ is open. Moreover, the union of all the $U_k$ is all of $X$, by assumption (case of 0-dimensional $X$ is easy!). Thus, by compactness, there is a $k_0$ such that the line bundle $L^{k_0}$ is base point free. Now use $L^{k_0}$ to give a map to projective space and reduce to the projective case. If you care about the "all greater multiples" business, you can build it in this argument, I believe.

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  • $\begingroup$ I don't think this can work so plainly, even just for the case of (asymptotic) base point freeness! First, I really need the property for all enough large multiples. Second, I don't think that your argument could straightforwardly be adapted to this: what you show is that if $U_{k_1},\dots,U_{k_N}$ gives a finite covering, then $L^{k_0}$, where $k_0:=\operatorname{lcm}(k_1,\dots,k_N)$, works (and hence its multiples, too). Do I understand well? $\endgroup$
    – diverietti
    Commented Dec 14, 2011 at 13:41

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