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L is a holomorphic line bundle on a compact complex manifold X. The Kodaira dimension of L is defined as the maximal dimension of the image of the map associated to the powers $ mL(m \in N)$. I want to prove the asymptotic estimate

$$ h^0 (X,mL) \leq O(m^{k(L)})$$

I heard that it is an easy consequence of the Schwarz lemma. Maybe it is a similar argument used by Siegel to prove the theorem "the transcendental degree of the meromorphic function field of a compact complex manifold is not bigger than the dimension of the manifold". But I'm afraid to deal with meromorphic mappings and singularities in the image.So I cannot complete the argument myself.

Can somebody tell me how the argument goes?

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3 Answers 3

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Hi,

An enlightening and very elementary proof of this fact can be found in the very complete book of X. Ma and G. Marinescu "Holomorphic Morse inequalities and Bergman kernels".

You will find this in Chapter 2. Their approach is exactly what you are looking for (only elementary complex analysis in several variables and a slightly modified Schwarz inequality).

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  • $\begingroup$ Yes,that is the proof I'm looking for.Thank you. $\endgroup$
    – Jun Li
    Commented Jun 29, 2011 at 7:43
  • $\begingroup$ Ok! I am glad I gave you the answer you wanted... $\endgroup$
    – diverietti
    Commented Jun 29, 2011 at 14:53
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Actually, it is possible to prove the following statement. Set

$$\mathbb{N}(L, X):=\{a > 0 \ | \ h^0(X, aL) \geq 1 \}$$

and let $d$ be the largest common divisor of the elements of $\mathbb{N}(L, X)$.

Then there exist two positive integers $\alpha$, $\beta$ such that for $m$ large enough one has

$$\alpha m^{k(L)}\leq h^0(X, mdL) \leq \beta m^{k(L)}.$$

For a proof, see [Ueno, Classification theory of algebraic varieties and compact complex manifolds, Lecture Notes in Mathematics 439, Theorem 8.1 p. 86].

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  • $\begingroup$ This does not seem to work if $L$ is a non-trivial line bundle of finite order. Perhaps the statement needs to be changed slightly? $\endgroup$
    – naf
    Commented Jun 19, 2011 at 10:18
  • $\begingroup$ Now it should be ok (this is actually the statement that one finds in Ueno's book). Thank you. $\endgroup$ Commented Jun 19, 2011 at 11:29
  • $\begingroup$ Thank you for the reference.I find many topics interesting in this book. $\endgroup$
    – Jun Li
    Commented Jun 20, 2011 at 3:35
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For an answer in terms of Bergman kernels associated with line bundles (a little bit of functional analysis is involved), see Theorem 4.2.3 in:

Berndtsson, Bo An introduction to things $\overline\partial$. Analytic and algebraic geometry, 7–76, IAS/Park City Math. Ser., 17, Amer. Math. Soc., Providence, RI, 2010

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  • $\begingroup$ Unfortunately I can't find this book in our library.Thanks for the reference. $\endgroup$
    – Jun Li
    Commented Jun 29, 2011 at 7:37

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