An example of a complex manifold without a finite open cover - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T00:04:20Z http://mathoverflow.net/feeds/question/58988 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58988/an-example-of-a-complex-manifold-without-a-finite-open-cover An example of a complex manifold without a finite open cover Vamsi 2011-03-20T17:46:11Z 2012-07-09T10:01:04Z <p>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.</p> http://mathoverflow.net/questions/58988/an-example-of-a-complex-manifold-without-a-finite-open-cover/58999#58999 Answer by Georges Elencwajg for An example of a complex manifold without a finite open cover Georges Elencwajg 2011-03-20T21:56:58Z 2011-03-21T12:12:08Z <blockquote> <p>Blow-up $\mathbb C^2$ simultaneously at all points of $\mathbb Z \times \{0\}$. </p> </blockquote> <p>Part a) is evident for the blown-up manifold $X$ since it contains one-dimensional compact submanifolds. As for part b), I am <em>guessing</em> (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.</p> http://mathoverflow.net/questions/58988/an-example-of-a-complex-manifold-without-a-finite-open-cover/59044#59044 Answer by Stefan Waldmann for An example of a complex manifold without a finite open cover Stefan Waldmann 2011-03-21T08:15:27Z 2011-03-21T08:15:27Z <p>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.</p> http://mathoverflow.net/questions/58988/an-example-of-a-complex-manifold-without-a-finite-open-cover/70877#70877 Answer by Finnur Larusson for An example of a complex manifold without a finite open cover Finnur Larusson 2011-07-21T05:02:12Z 2012-07-09T10:01:04Z <p>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.</p> http://mathoverflow.net/questions/58988/an-example-of-a-complex-manifold-without-a-finite-open-cover/71426#71426 Answer by Al-burcas for An example of a complex manifold without a finite open cover Al-burcas 2011-07-27T18:49:44Z 2011-07-27T18:49:44Z <p>I suppose most (which?) complex $n$-manifolds can be classified by a degree $p\in\mathbb{N}$, that of <em>holomorphic completeness</em>. 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 $\mathbb{C}^n$.</p> <p>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?</p> http://mathoverflow.net/questions/58988/an-example-of-a-complex-manifold-without-a-finite-open-cover/96021#96021 Answer by Lucas Kaufmann for An example of a complex manifold without a finite open cover Lucas Kaufmann 2012-05-04T20:33:58Z 2012-05-04T20:33:58Z <p>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 $\mathbb C^N$.</p>