Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there new classification of Kahler manifolds of complex dimension n and new results for necessary and sufficient conditions for a manifold being Kahler? I know if redactivity of Lie algebra on vector bundels and some theorems like this. But I am looking for other results.

share|improve this question
    
I don't think Kähler differentials are relevant here. They are named after Erich Kähler, not for their relationship to Kähler geometry. See en.wikipedia.org/wiki/Kähler_differential –  Michael Albanese Jun 14 '12 at 13:04
    
Do you know Simpson's The construction problem in Kähler geometry? It does not seem to be on the arXiv, here is a link: citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.67.743. It's basically a great list of open problems, with a vast bibliography - not brand new, but the best I know. –  Balazs Jul 9 '12 at 21:30

1 Answer 1

up vote 11 down vote accepted

I thought this had already been answered here on MO but my searches didn't turn anything up. It might be good to have a general purpose answer here somewhere.

I'll restrict myself to compact manifolds, since the question is already complicated enough there!

In complex dimension one every manifold is Kahler, simply because the exterior derivative of any 2-form is zero, so every hermitian metric is Kahler.

In complex dimension two a surface is Kahler if and only if its first Betti number is even. This is known since the 80's by classification of surfaces and hard results of Siu, but modern proofs by Buchdahl and Lamari that do not make use of classification of surfaces appeared in 1999.

The condition that the odd Betti numbers of the manifold be even for it to be Kahler is necessary by the Hodge decomposition theorem. Interestingly this is sufficient in dimension two, implying that the Kahler condition is topological there. This fails in higher dimensions, for there is an example due to Hironaka of Kahler manifolds of dimension 3 that can be deformed to a non-Kahler manifold. Since the underlying smooth manifolds of these complex manifolds are all diffeomorphic, a sufficient condition for being a Kahler manifold cannot be read off the topological or smooth structure of a given complex manifold.

The list of necessary conditions that a manifold must satisfy to be Kahler is long and growing. First among these are the multiple conditions that the Hodge decomposition and the hard Lefschetz theorem impose on its cohomology ring. The properties of this ring are not yet well understood, considering how recently Voisin constructed examples demonstrating that there are Kahler manifolds whose homotopy type differs from that of projective manifolds.

Another active area of research is on the fundamental groups of Kahler manifolds. I can't say I know much about this, but just to give an idea then examples of Serre show that any finite group can be the fundamental group of a Kahler manifold (projective surface, even). Finer results are available under additional hypotheses, for example Paun has shown that the fundamental group of a Kahler manifold with nef anticanonical bundle has polynomial growth.

One can say more about necessary conditions for a manifold to be Kahler (strictly speaking any theorem that begins "Let $X$ be a compact Kahler manifold..." can be considered one!), but by contrast almost nothing is known about sufficient conditions in higher dimensions. With the examples of Hironaka in mind such conditions are likely to be extremely subtle, if they exist at all.

share|improve this answer
1  
Did Hironaka's result provide the first example of a non-Kahler compact complex manifold with odd Betti numbers even? –  Michael Albanese Jun 14 '12 at 12:35
3  
I don't know if it had been remarked that this could happen before Hironaka, but there are certainly easier examples of such things. For a cheap one, take $X = S \times S$ to be the product of two Hopf surfaces $S$. Then $X$ is certainly compact non-Kahler, but its interesting Betti numbers are $b_1 = 2$, $b_2 = 1$, $b_3 = 2$ and $b_4 = 4$. In particular, the odd $b_k$ are even. –  Gunnar Þór Magnússon Jun 14 '12 at 14:23
1  
That's a very nice example, thanks. –  Michael Albanese Jun 14 '12 at 14:35
1  
Are there nice results in low dimension? For example, is there a nice list of fairly computable conditions in complex dimension 3? –  Steven Gubkin Jun 14 '12 at 17:50

Your Answer

 
discard

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.