An example of a complex manifold without a finite open cover - MathOverflow

An example of a complex manifold without a finite open cover

2011-03-20T17:46:11Z

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.

Answer by Georges Elencwajg for An example of a complex manifold without a finite open cover

2011-03-20T21:56:58Z

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.

Answer by Stefan Waldmann for An example of a complex manifold without a finite open cover

2011-03-21T08:15:27Z

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.

Answer by Finnur Larusson for An example of a complex manifold without a finite open cover

2011-07-21T05:02:12Z

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.

Answer by Al-burcas for An example of a complex manifold without a finite open cover

2011-07-27T18:49:44Z

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?

Answer by Lucas Kaufmann for An example of a complex manifold without a finite open cover

2012-05-04T20:33:58Z

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 $\mathbb C^N$.