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In their book "Compact complex surfaces", W.P. Barth, K. Hulek, C.A.M. Peters and A. Van de Ven refer to the following theorem:

Let $X$ be a compact complex space and $L$ a holomorphic line bundle on $X$. Then $L$ is ample if and only if the folowing holds: given any irreducible analytic subset $Y$ of strictly positive dimension on $X$, there exist an $n=n(Y)$ such that $L^{\otimes n} |_Y$ has a section which has at least one zero, but does not vanish identically.

They don't give a proof in their book. Instead they refer to Grauert's original paper Über Modifikationen und exzeptionelle analytische Mengen which was published in Math. Ann. 1962.

I do not know German. So my question is: Can I find the proof of this theorem somewhere else? Or instead, some comments on the idea of proof will also be very helpful.

Finally, this paper of Grauert is among one of the papers I want to read with greatest enthusiasm. Is there a translation? Or can I find some books or papers which give an explanation of the results of this paper?

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2 Answers 2

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You can find a proof in Kleiman's famous paper Toward a Numerical Theory of Ampleness, Theorem 1 page 317.

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  • $\begingroup$ Kleiman states that his proof is for complete algebraic schemes. Does it work verbatim for compact analytic spaces? $\endgroup$
    – Ben McKay
    Sep 21, 2011 at 8:55
  • $\begingroup$ I had a (quick) look at the proof and it seems to me that it should extend without problems to the compact analytic setting. $\endgroup$ Sep 21, 2011 at 9:16
  • $\begingroup$ The reference is very helpful.Thank you. $\endgroup$
    – Jun Li
    Sep 21, 2011 at 12:02
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    $\begingroup$ even if you think you don't read german, you might try Grauert's paper, in pretty clear german, from middle page 347 to top 349. he uses induction, finds X1, the zero set in X of a section of a power of F, then applies induction to the restriction of F to X1. If Kleiman uses the same argument you might read them together. Grauert uses Remmert's reduction theorem which allows one to blow down compact subvarieties of a holom convex space to a Stein space, with the structure sheaf also pushing down. gdz.sub.uni-goettingen.de/en/dms/load/img $\endgroup$
    – roy smith
    Sep 22, 2011 at 20:26
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    $\begingroup$ verschwinden = vanishes, schnittfla"che = section, keine = no, Menge = set, Teilmenge = subset, Garbe = sheaf, Geradenbundel = line bundle, Jede = any, fortsetzen =(?) extends, and you can guess: exakte Sequenz = .... $\endgroup$
    – roy smith
    Sep 22, 2011 at 20:35
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Suppose that $H^0(L^{\otimes n})|_ Y$ is non-zero for all $Y$ and $n$ sufficiently big, and has a section which vanishes somewhere on $Y$. Then it follows that the base set of $L$ is trivial: indeed, $L$ has a non-zero section on any complex subvariety, which includes the base set. This implies that the natural map $P_n:\; X\rightarrow {\mathbb P}(H^0(X, L^{\otimes n}))$ is holomorphic, for $n$ sufficiently big. Also from this assumption it follows that $P_n$ does not map any irreducible, positive-dimensional subvariety to a point (again, for $n$ sufficiently big). This implies that $P_n$ is a finite, proper map to a projective variety, hence $X$ is a ramified covering of a projective variety. A ramified covering of a projective variety is projective, which can be seen from vanishing of cohomology of powers of $L$ (a finite map is acyclic on coherent sheaves, hence the cohomology of $L^{\otimes n}$ on $X$ are the same as cohomology of ${\cal O}(1)$ on its image).

However, this works only when $H^0(L^{\otimes n} |_Y)\neq 0$; the implication $H^0(L^{\otimes n} |_Y)\neq 0$ $\Rightarrow$ $H^0(L^{\otimes n})|_Y\neq 0$ is not that easy.

If $Y$ does not lie in a zero divisor of $L^{\otimes n}$, we are done, otherwise we replace $Y$ with the zero divisor of $L^{\otimes n}$. Using induction on dimension, we may already assume that the restriction $L|_Y$ is ample. Consider the exact sequence $$ 0\rightarrow H^0(L^{\otimes k}) \rightarrow H^0(L^{\otimes n+k}) \stackrel r \rightarrow H^0(Y, L^{\otimes n+k} |_Y) \rightarrow H^1(L^{\otimes k}) \rightarrow H^1(L^{\otimes n+k}) \rightarrow 0. $$ The arrow $r$ of this sequence is the restriction map; we need to prove that $r$ does not vanish. If it vanishes, we have $\dim H^0(Y, L^{\otimes n+k}|_Y) \leq \dim H^1(L^{\otimes k})$ for all $n\gg 0$ and $k$. The last term of this exact sequence actually implies that $\dim H^1(L^{\otimes n+k})\leq \dim H^1(L^{\otimes k})$, for all $k$, and all $n\gg 0$, hence $H^1(L^{\otimes n+k})$ is bounded by a universal constant. This implies that $\dim H^0(Y, L^{\otimes n+k} |_Y)$ is also bounded, whenever $r=0$, which is impossible, because $L|_Y$ is ample.

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  • $\begingroup$ How do you know you can lift sections of $L^{\otimes n}|_Y$ to sections of $L^{\otimes n}$? How do you find a uniform value of $n$? $\endgroup$ Nov 27, 2021 at 12:33
  • $\begingroup$ Sorry! I misread the statement of the theorem, and my argument works when there are many sections of $L^n$ restricted to $Y$. However, $H^0(L|_Y)\neq 0$ implies $H^0(L^2)|_Y\neq 0$. I will edit the answer to add this argument. $\endgroup$ Nov 27, 2021 at 17:35
  • $\begingroup$ as for the uniformity of the number, it was assumed in the original question: "given any irreducible analytic subset Y of strictly positive dimension on X, there exist an n=n(Y) such that L⊗n|Y has a section which has at least one zero, but does not vanish identically.", otherwise it would take some extra work $\endgroup$ Nov 27, 2021 at 17:46

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