That Stein manifolds have all $(p,q), p \geq 0, q \geq 1$ vanishing Dolbeault cohomology groups is more or less standard. I am a little bit confused about the reverse implication: whether the vanishing of all Dolbeault cohomologies ($p \geq 0, q \geq 1$) implies Steinness? I could not find a reference, yet I noticed the following:

In the recent (and quite authoritative) book by Forstnerič ''Stein Manifolds and Holomorphic Mappings'' Theorem 2.4.6 says:

On any Stein manifold $X$ the Dolbeault cohomology groups vanish: $H^{p,q}_{\bar\partial}(X) = 0$ for all $p \geq 0, q \geq 1$.

So the reverse is not even mentioned.

In the older book by Grauert and Remmert ''Theory of Stein spaces'' on page 81 one reads:

The assumptions of Theorem 5 are always satisfied by Stein manifolds. It is not known if there exist non-Stein manifolds of this type (i.e. $H^q(X,\Omega^p)=0, p \geq 0, q \geq 1$)

So I assume this was open at least as of 2004 (?)

On the other hand a recent paper by Chakrabarti and Shaw entitled ''The $L^2$-cohomology of a bounded smooth Stein domain is not necessarily Hausdorff'' (on arXiv, the author's website says that it is to appear in Math. Ann.) begins with:

For each bidegree $(p,q)$, with $p\geq0,q >0$, the Dolbeault Cohomology group $H^{(p,q)}(\Omega)$ of a Stein manifold $\Omega$ in degree $(p,q)$ vanishes, and indeed this property characterizes Stein manifolds among complex manifolds.

The provide references for that fact are the books of Hörmander and Gunning-Rossi (where I couldn't find it).

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    $\begingroup$ A complex manifold X is Stein iff quasi-coherent sheaves have no higher cohomology. This is Cartan's Theorem B, attributed to Serre. There exists a version for complex spaces. $\endgroup$ Mar 2, 2015 at 22:48

2 Answers 2


The reference to the books of Hörmander and Gunning-Rossi only seems to cover a special case of this: a domain $X$ in $\mathbb{C}^n$ is a Stein manifold if and only if $H^i(X,\mathcal{O}_X)=0$ for all $i$ with $1\leq i\leq n-1$. Another possible reference is Theorem 63.7 in

  • L. and B. Kaup: Holomorphic functions of several variables. De Gruyter, 1983.

A similar result is stated (with similar references to Hörmander and Gunning-Rossi) for open subsets in Stein manifolds in

  • M.G. Eastwood and G. Vigna Suria. Cohomologically complete and pseudoconvex domains. Comment. Math. Helvetici 55 (1980), 413-426.

These results both assume that the complex manifold sits inside a Stein manifold to start with. For an absolute statement, you need the vanishing of cohomology for all coherent analytic sheaves, as mentioned in Peter Dalakov's comment. (However, unless I am missing something, this does not seem obviously equivalent to vanishing of Dolbeault cohomology.)

  • $\begingroup$ so, to sum up, without additional assumptions the statement is open, and probably not true, yet an example is missing. am I right? $\endgroup$
    – hassan
    Mar 30, 2015 at 22:03
  • $\begingroup$ Well, I would not go that far. There is a conspicuous absence of statements like "Dolbeault cohomology vanishing implies Steinness" in the literature. For Riemann surfaces you get this implication from the Behnke-Stein theorem. $\endgroup$ Mar 31, 2015 at 10:58
  • $\begingroup$ Another thing to note is: vanishing of Dolbeault cohomology gives a vanishing of de Rham cohomology in dimension $\geq n$, when $n$ is the complex dimension via the Frölicher spectral sequence. Then the theorem of Eliashberg says that the given complex structure can be homotoped to a complex structure which is Stein (complex dimension $>2$ required here). This, however, changes the complex structure. $\endgroup$ Mar 31, 2015 at 11:02
  • $\begingroup$ Matthias, Eliashberg requires an exhausting Morse function without critical points of index $>n$. How is existence of such a function implied by vanishing de Rham cohomology? (E.g. Couldn't there be high degree torsion in singular cohomology?) $\endgroup$
    – Tim Perutz
    Apr 8, 2015 at 14:35
  • $\begingroup$ @TimPerutz: you are right, there could be torsion in singular cohomology above degree $n$. I don't know any examples where this happens, but I also don't know of an argument ruling this out... $\endgroup$ Apr 10, 2015 at 11:14

Sorry for the misleading remark in the paper. Indeed we were thinking about domains in Stein manifolds, and were careless in making the side remark, which of course has nothing to do with the main topic of the paper. Thanks for pointing out!


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