Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We call a compact complex manifold Moisezon manifold, if its dimension coincides with the algebraic dimension, i.e. it has as many algebraically independent meromorphical functions as its complex dimension. Boris Moisezon himself gave a proof to the following theorem: Let $X$ be a Moisezon mainfold, then for $X$ to be projective it is necessary and sufficient to be Kähler.

It was told that it could also be formulated as a criterion for projectivity:

A compact complex manifold is projective if and only if it is kähler and moisezon.

I didn't find any complete proof to the second formulation. Does someone know where to find it? Or at least how it works?

Edit: I already found two versions for the implication "projective=>moisezon" (necessity to be kähler is in fact clear), but i am also interested in alternatives, especially for the other one. On the one hand Huybrechts, Complex Geometry and on the other hand Wells, Moisezon spaces and the kodaira embeddingtheorem.

Next Edit: I don't want this to look like a jeaopardy question but, after postponing the problem I stumbled upon this:

There is a different proof by Thomas Peternell, given in "Algebraicity Criteria for Compact Complex Manifolds", Math. Ann. 275, 653-672 (1986). Theorem 1.4. states a slight variation of the theorem of Moisezon which is indeed equivalent. More precisely it states, that if there exist a real $(1,1)$-form $\omega$ and a real $2$-form $\varphi$ on a Moisezon manifold $X$ such that $\omega$ is positive definite, $d(\omega-\varphi) = 0$ and $\int_C \varphi = 0$ for all curves $C\subset X$, then $X$ is projective.

share|cite|improve this question

Are you asking why a projective complex variety has "enough" meromorphic functions to be Moisezon?

If so, the answer is that if $X$ is projective of dimension $d$, then the field of rational functions on $X$ (all of which are meromorphic --- and conversely, although we won't need this latter fact) is of transcendence dimension $d$, i.e. contains (exactly) $d$ algebraically independent elements. This is part of the basic dimension theory of algebraic varieties, and I'm pretty sure that it is discussed in Chapter I of Hartshorne's book (and in many other places too).

share|cite|improve this answer
Of corse it is - how could i miss this. How to make yourself a fool....Thank you! – Ben Jul 21 '11 at 1:05
I shouldn't have accepted this, because an alternative for the even more interesting implication is still missing...I'm sorry for toggling – Ben Jul 21 '11 at 6:46
@Ben: Dear Ben, No problem. But which implication is "the even more interesting one"? (Sorry, I've just gotten myself confused about what it is you're trying to find out.) Best wishes, – Matt E Jul 21 '11 at 7:36
As you pointed, the "projective=>moisezon"-one is easy...Its the converse that i didn't find any proof of except the original (translated) by Moisezon himself...To prevent more confusion: I'm looking for another proof of "moisezon and kähler => projective". I thought of something like ~constructing a Hodge class... – Ben Jul 21 '11 at 8:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.