Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is partly inspired by answers to the question: Question about Hodge number . Is there a family of compact complex manifolds, where the general fibres are Kähler, but for which $E_1$ degeneration of the Hodge to de Rham spectral sequence fails at the special fibre? Or, even better, such that the special fibre has nonclosed holomorphic forms?

I feel like I should know the answer, but somehow I don't. All the examples I know where the spectral sequence doesn't degenerate are nilmanifolds*, so they aren't even homotopic to Kähler manifolds by standard rational homotopy theoretic obstructions (e.g. they aren't formal). Also the famous Hironaka example [Ann. Math 1962] won't work either, because the special fibre is an algebraic variety, so the spectral sequence will degenerate (by an argument that can found in Deligne [Théorème de Lefschetz...]). Obviously, I haven't thought about this deeply enough, but perhaps someone else has**.


*I was bit sloppy yesterday, since the examples I have in mind include solvmanifolds. However, there are still topological obstructions to these being Kähler due to Nori and myself.

** From the answers, I gather that the work of Popovici suggests that there may be no counterexample.

share|cite|improve this question
Regarding this question, have a look at the very last sentence of this paper of Demailly-Paun where they say that they are not sure what to expect from these limiting manifolds, and they suggest that perhaps they are always bimeromorphic to Kahler, at least assuming the $E_1$ degeneration holds –  YangMills Nov 17 '11 at 2:02

3 Answers 3

up vote 10 down vote accepted

This is known, for projective (even Moishezon) manifolds as shown by Dan Popovici in his paper For general Kaehler manifold, this is conjectured. Popovici has proved that a property of "strong Gauduchon" is preserved in limits and (I think) there are no example of strong Gauduchon manifold without Hodge decomposition.

share|cite|improve this answer
So the evidence seems to point to my question having a negative answer. Actually, this would make me happier. –  Donu Arapura Oct 19 '10 at 22:00

If any example exists, then the general fibre of the family cannot be projective.

In fact, Dan Popovici ["Limits of projective manifolds under holomorphic deformations", arXiv.09102032] recently proved the following

Theorem. Let $\pi \colon \mathcal{X} \to \Delta$ be a complex analytic family of compact complex manifolds such that the fibre $X_t:=\pi^{-1}(t)$ is projective for all $t \neq 0$. Then $X_0:=\pi^{-1}(0)$ is Moishezon.

Since Moishezon manifolds admit a projective algebraic modification, it follows that their Hodge-Frolicher spectral sequence degenerates at $E_1$. In particular, Hodge decomposition holds for $X_0$. Notice that in this case $X_0$ is Kähler if and only if it is projective.

share|cite|improve this answer
Interesting, thanks. –  Donu Arapura Oct 19 '10 at 21:58
The paper of Popovici contained a mistake, so this theorem remains a conjecture. See here –  YangMills Jan 26 '13 at 2:33

Just as a comment.

In general, given a family of sG (i.e., strongly-Gauduchon) manifolds, the limit need not to be still sG: an example was provided by Ceballos, Otal, Ugarte, and Villacampa, (see also It may be possible that the sG property is preserved to the limit assuming stronger conditions on the fibres: however, the published version of the paper by Popovici does not prove this fact, as noticed also by YangMills.

On the other hand, an example of a family of (non-Kähler) manifolds satisfying the Hodge decomposition (namely, the $\partial\overline{\partial}$-Lemma) and whose limit does not, is studied in

Note that there are several examples of sG manifolds non-satisfying the Hodge decomposition: for example, every nilmanifold admitting a balanced or sG metric.

share|cite|improve this answer

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.