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Are there non-compact complex manifolds that a) Don't embed in C^n (holomorphically) and b) Cannot be covered by a finite number of coordinate open sets? If b) can be satisfied, then I think so can a) be by taking a product with a compact complex manifold. If one takes a Riemann surface of infinite genus, one does not have a "good" finite open cover, but I allow non-contractible open covers as well. Apologies in advance for this elementary question.

flag	asked Mar 20 2011 at 17:46 Vamsi 1,317●4●12

A simple example of (b) would be a disjoint union of countably-many copies of $\mathbb C$. Perhaps you want connected and/or other restrictions? – Ryan Budney Mar 20 2011 at 17:51

Yeah I want it to be connected and second-countable. – Vamsi Mar 20 2011 at 18:58

If there a simple example of a smooth manifold that does not satisfy b)? If you give such an example that might help to construct a complex manifold with this property. There are many surface of infinite genus, and the simplest one that I can imagine seem to satisfy b) even if the charts are assumed to be connected an contractible. – Dmitri Apr 5 2011 at 12:47

If you have a good finite open cover, the betti numbers would be finite (I think) by a Mayer-Vietoris argument. – Vamsi Apr 8 2011 at 17:27

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If you don't put restrictions on the topology/geometry/etc of the open sets, then b) has a simple answer: every complex manifold has an open cover by one open set: the whole manifold :-P – David Roberts Jul 21 2011 at 5:37

Ricky Demer · Accepted Answer · 2012-07-09 10:01:04Z UTC

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Fornaess and Stout proved that EVERY complex manifold (connected and second countable) can be covered by finitely many open subsets biholomorphic to a polydisc (Lemma II.1 in MR0470251). They even have an explicit bound on the size of the cover in terms of the dimension of the manifold. Further results of a similar flavour are contained in their papers MR0435441 and MR0662439.

edited Jul 9 at 10:01

Ricky Demer

answered Jul 21 2011 at 5:02

Finnur Larusson
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Welcome to MO, Finnur! – David Roberts Jul 21 2011 at 5:33

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The paper Finnur quotes does prove that there is a finite covering by open sets biholomorphic to polydisks. But you can't choose them to be relatively compact. Actually lemma II.1 proves that you can choose a covering by relatively compact open sets, biholomorphic to polydisks, with a bound on the number of them that can intersect nontrivially. You can cover the complex plane by 3 open sets biholomorphic to polydisks, but they can't be chosen relatively compact. – Ben McKay Feb 16 2012 at 9:33

Thanks for the correction, Ben, and apologies for citing the wrong result. It is Theorem VI.1 in Fornaess and Stout's paper that I should have cited. And the words "relatively compact" should certainly be deleted. – Finnur Larusson Apr 12 2012 at 4:57

Lucas Kaufmann · Answer 2 · 2012-05-04 20:33:58Z UTC

If $\widetilde X$ is a compact complex manifold of dimension $\geq 2$ and $x \in \widetilde X$ then $X = \widetilde X - \lbrace x \rbrace$ is a non-compact manifold that cannot be holomorphically embedded in $\mathbb C^N$. This is because, by Hartogs' Theorem, we have $\mathcal O(X) = \mathcal O(\widetilde X)$ and therefore global holomorphic functions on $X$ are constant, which is not the case for complex submanifolds of $\mathbb C^N$.

Georges Elencwajg · Answer 3 · 2011-03-21 12:12:08Z UTC

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Blow-up $\mathbb C^2$ simultaneously at all points of $\mathbb Z \times \{0\}$.

Part a) is evident for the blown-up manifold $X$ since it contains one-dimensional compact submanifolds. As for part b), I am guessing (but cannot prove) that the huge first Chern class of the line bundle $\mathcal O_X(D)$ associated to the exceptional divisor $D$ prevents the existence of a finite number of holomorphic or even differentiable charts.

edited Mar 21 2011 at 12:12

answered Mar 20 2011 at 21:56

Georges Elencwajg
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I am probably being very daft here, but why does the first Chern class being huge prevent the existence of a finite number of charts? (is it similar to the argument that goes into proving that for compact manifolds, the first Chern class is the Poincare dual of the fundamental class of the divisor? If there are finitely many charts, we are allowed to integrate using a partition of unity and so on...) – Vamsi Mar 21 2011 at 2:25

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Dear Vamsi, you are not being daft at all! On the contrary, it is all my fault: I thought that I could find a proof of my guess by somehow adapting the proofs that $\mathbb P^n $ cannot be covered by $n$ open affines and that the tautological bundle on $\mathbb P^n (\mathbb R)$ cannot be trivialized topologically by $n$ open subsets but I failed. I have edited my post to emphasize its status as guess. I hope one of the incredibly competent users here will help us settle the question. – Georges Elencwajg Mar 21 2011 at 12:26

Stefan Waldmann · Answer 4 · 2011-03-21 08:15:27Z UTC

It depends probably a bit on the notion of "chart" but if you allow your charts to have countably infinitely many connected components then it is a consequence of dimension theory that a $n$ dimensional manifold can always be covered by $n+1$ (?) charts even in such a way that the connected components of arbitrary intersections are either empty or contractible. This is not quite a "good" cover, but comes very close. In particular, vector bundles trivialize locally over such a finite atlas. You can find this in e.g. Well's book of complex manifolds. I don't have a copy here, so I can't tell you the precise page but you surely will find it.

Al-burcas · Answer 5 · 2011-07-27 18:49:44Z UTC

I suppose most (which?) complex $n$-manifolds can be classified by a degree $p\in\mathbb{N}$, that of holomorphic completeness. The simplest way I see it is as the dimension $p-1$ of the compact factor on a product of a compact complex manifold by $\mathbb{C}^{n-p+1}$. I see it also as the maximal dimension (+1) a compact complex submanifold can attain... Stein manifolds=holomorphic 1-complete=holomor. embeddable in some $\mathbb{C}^n$.

Andreotti, Cartan, Grauert, Remmert, Stein worked a lot on the notion in the middle of XXth century. A consequence is that every coherent sheaf's $i$-cohomology vanishes for $i\geq p$. But how could we deduce the number of charts from cohomology? For which sheaf?

An example of a complex manifold without a finite open cover

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