Let $\mathcal{F}$ be a sheaf on a topological space $X$. An open cover $\mathcal{U} = \{U_i\}_{i\in I}$ of $X$ is called acyclic for $\mathcal{F}$ if for all $i_0, \dots, i_p \in I$, $H^q(U_{i_0}\cap\dots\cap U_{i_q}, \mathcal{F}) = 0$ for every $q \geq 1$. Not every sheaf admits an acyclic cover as is shown here.

Let $X$ be a complex manifold and consider the sheaf $\mathcal{O}$ of holomorphic functions on $X$. Does $X$ always admit an open cover which is acyclic for $\mathcal{O}$? What about for the sheaf $\mathcal{O}^*$ of nowhere zero holomorphic functions on $X$?

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    $\begingroup$ Michael, for starters, it seems like because of the Dolbeault Lemma (for the case of $\mathscr O$), you only need to consider an open cover by polydisks, as then intersections will likewise be polydisks. [This would be the analogue of geodesically convex coverings of a Riemannian manifold.] I think you ought to be able to deduce that it will work for $\mathscr O^*$ as well, but I haven't thought this through. $\endgroup$ – Ted Shifrin Apr 14 '15 at 1:56
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    $\begingroup$ I don't think the intersection of polydiscs is necessarily a polydisc. For example, $(D(0, 1)\times D(0, 1))\cap(D(1, 1)\times D(0, 1))$ is not a polydisc. Is the intersection of polydiscs necessarily biholomorphic to a polydisc? If so, then as $H^q(\cdot, \mathcal{O}) = H^{0,q}_{\bar{\partial}}(\cdot)$ and the latter is a biholomorphism invariant, we can conclude that a covering by polydiscs is an acyclic cover for $\mathcal{O}$. $\endgroup$ – Michael Albanese Apr 14 '15 at 3:33
  • $\begingroup$ Michael: I've given away my Gunning and Rossi (which is where I would have looked). But in the proof of Lemma 5.3.9 of Neeman's Algebraic and Analytic Geometry he argues that the intersection of generalized polydisks is a generalized polydisk; a generalized polydisk is the preimage of a polydisk under a polynomial mapping, so we're not quite there, but ... $\endgroup$ – Ted Shifrin Apr 14 '15 at 12:42
  • $\begingroup$ I haven't been able to find anything in Gunning and Rossi, but I am not familiar with the book so I might be looking in the wrong place. I have read the proof of Lemma 5.3.9, but I have no intuition about generalised polydiscs, in particular, their biholomorphism type. Was your ellipsis an indication that I should be able to arrive at a conclusion from your remarks or just a trailing thought? $\endgroup$ – Michael Albanese Apr 16 '15 at 4:14
  • $\begingroup$ Dear Michael, the intersection of two polydiscs is most certainly not isomorphic to a polydisc: it needn't even be connected! You can check that already in $\mathbb C$, where any simply connected open subset distinct from $\mathbb C$ is isomorphic to a (poly)disc, according to Riemann's uniformization theorem. However the intersection of two polydiscs is Stein: see my answer. $\endgroup$ – Georges Elencwajg Apr 17 '15 at 9:04

If $X$ is a complex space, not necessarily a manifold nor even reduced, any cover $\mathcal U=(U_i)_{i\in I}$ by open subsets $U$ which are Stein is acyclic for any coherent sheaf.
Indeed any finite intersection $U=U_1\cap\dots\cap U_n $ of sets $U\in \mathcal U$ is Stein : this is trivial if you define "Stein" as holomorphically convex and holomorphically separable.
But is it legitimate to define "Stein" that way?
Sure! That's what the brothers Ludger and Burchard Kaup do in their extremely pedagogical book Holomorphic functions of Several Variables
One may say that the basic theory of Stein spaces consists of proving (in about 100 pages!) the equivalence of several definitions of Stein spaces, in particular the one that says that a Stein space is a complex space for which theorem B holds!

1) A complex space $Y$ is said to be holomorphically convex if, given a compact subset $K\subset Y$, its holomorphic hull $h(K)$ is also compact.
That hull is defined as the set of $y\in Y$ such that for all $f\in \mathcal O(Y)$ we have $\mid f(y)\mid\leq \sup_{k\in K} \mid f(k)\mid$
A (non-trivially!) equivalent definition of holomorphically convex is:
Given a discrete closed subset $D\subset Y$, there exists a holomorphic function $h\in \mathcal O(Y)$ such that $\sup_{d\in D} \mid h(d)\mid=\infty$.

2) A complex space $Y$ is said to be holomorphically separable if, given $y\neq y'\in Y$, there exists $g \in \mathcal O(Y)$ with $g(y)\neq g(y')$ .

3) Very roughly one might divide complex analysis into two parts:
$\bullet$ The kingdom of partial differential equations with its emphasis on plurisubharmonicity, pseudoconvexity, $L^2$-estimates, $\bar {\partial}$-operator, $\dots$ The basic book here is Hörmander's.
$\bullet \bullet$ The kingdom of geometry and algebra with its emphasis on sheaves, homological and commutative algebra, non-reduced spaces, algebraic topology, $\dots$
The Kaups's book and the above answer are written in the spirit of this second point of view (I used holomorphic convexity instead of pseudoconvexity) , which is more adapted to the study of the general non reduced complex spaces introduced by Grauert under the influence of Grothendieck.
The definitive reference on Stein spaces written by Grauert-Remmert, the masters themselves, Theory of Stein spaces doesn't even mention plurisubharmonicity and pseudoconvexity (although, needless to say, these are extremely useful tools in many contexts).

4) How do I know this? Because in my youth I betrayed the first Kingdom for the second and then I betrayed the second for the Scheming Empire of Algebraic Geometry...

  • $\begingroup$ So a covering by Stein manifolds is acyclic for any coherent sheaf by Cartan's Theorem B, in particular for the sheaf $\mathcal{O}$. What about $\mathcal{O}^*$? Is it a coherent sheaf? $\endgroup$ – Michael Albanese Apr 18 '15 at 21:32
  • $\begingroup$ No, it isn't, so that a covering by Stein open sets has no reason to be acyclic for $\mathcal O^*$, unfortunately. $\endgroup$ – Georges Elencwajg Apr 18 '15 at 23:14

OK, so what you want is H. Cartan's Theorem B, which states that whenever $X$ is a closed analytic subvariety of a polydisk and $\mathscr F$ is a coherent sheaf, then $H^i(X,\mathscr F) = 0$ for $i>0$. This should be in Gunning and Rossi. :)

Trying again (darn, I'm rusty on this stuff): Any polydisk is a simple example of a domain of holomorphy in $\Bbb C^n$. The preimage under an analytic map of a domain of holomorphy is again a domain of holomorphy (Theorem 2.5.14 of Hörmander's Introduction to Complex Analysis in Several Variables), and intersections of domains of holomorphy are again domains of holomorphy (Corollary 2.5.7). Moreover, the big theorem (Theorem 4.2.8) tells us that domains of holomorphy are equivalent to pseudoconvex domains. And the $\overline\partial$ problem is always solvable in pseudoconvex domains. [I'm sure there are more basic ways to get this ...]

Perhaps an easier way, avoiding polydisks, is to put a hermitian metric on the complex manifold and (as in the smooth manifold case), cover with sufficiently small geodesic balls. It should be the case that these are pseudoconvex, and then we're back in business, intersections being again pseudoconvex.

  • $\begingroup$ Sorry for my ignorance, but why is the intersection of polydiscs a closed subvariety of a polydisc? $\endgroup$ – Michael Albanese Apr 16 '15 at 14:22
  • $\begingroup$ Dear Michael, I'm not sure this is true but it doesn't matter: the intersection of two polydiscs is Stein and Theorem B holds for all Stein manifolds. $\endgroup$ – Georges Elencwajg Apr 17 '15 at 9:25

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