Uniformization theorem in higher dimensions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:49:08Z http://mathoverflow.net/feeds/question/3519 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3519/uniformization-theorem-in-higher-dimensions Uniformization theorem in higher dimensions Jonah Sinick 2009-10-31T07:42:39Z 2012-01-07T02:09:14Z <p>Let \$M\$ be a 4-manifold with a complex structure.</p> <p>Does there exist a finite list of simply connected complex 4-manifolds \$M_1, ... , M_n\$ such that M is the quotient of some \$M_i\$ by the action of a group acting discretely on \$M\$?</p> <p>This would be an analog of the Poincare-Koebe uniformization theorem in (real) dimension 2. People who I've asked this question to think that it's unlikely that there is such a list, but haven't been able to offer an argument or a reference.</p> http://mathoverflow.net/questions/3519/uniformization-theorem-in-higher-dimensions/3523#3523 Answer by Georges Elencwajg for Uniformization theorem in higher dimensions Georges Elencwajg 2009-10-31T09:43:22Z 2009-10-31T09:43:22Z <p>There exist infinitely many holomorphically non-isomorphic complex structures on the unit ball of R^4 (or more generally R^2n): this is a beautiful theorem of Burns, Shnider, Wells ( Inventiones Math. 46, p. 237-253 ,1978) . Since these complex manifolds are simply connected (even contractible) they are their own covering spaces and this proves that the set (hum, hum) of universal covers of complex manifolds of dimension 2 is infinite. As a historical aside, Poincaré was first in noticing that the unit ball and the bidisk (DXD) in C^2 were not holomorphically isomorphic. Here is the reference to Kang-Tae's nice survey on the subject.</p> <p><a href="http://mathnet.kaist.ac.kr/mathnet/kms_tex/100248.pdf" rel="nofollow">http://mathnet.kaist.ac.kr/mathnet/kms_tex/100248.pdf</a></p> http://mathoverflow.net/questions/3519/uniformization-theorem-in-higher-dimensions/3530#3530 Answer by Pete L. Clark for Uniformization theorem in higher dimensions Pete L. Clark 2009-10-31T11:02:43Z 2009-11-04T15:46:44Z <p>Given any smooth complex surface, you can blow up any point on the surface:</p> <p><a href="http://en.wikipedia.org/wiki/Blowing_up" rel="nofollow">http://en.wikipedia.org/wiki/Blowing_up</a></p> <p>This gives you another smooth, complex surface with isomorphic fundamental group but with the second Betti number increased by 1. (There is, in some sense, essentially one more curve on the new surface than there was on the old one, called the "exceptional divisor".) </p> <p>Starting with any one simply connected smooth complex surface (e.g. the projective plane) and repeatedly blowing up shows that your question has a negative answer.</p> http://mathoverflow.net/questions/3519/uniformization-theorem-in-higher-dimensions/3541#3541 Answer by engelbrekt for Uniformization theorem in higher dimensions engelbrekt 2009-10-31T13:41:10Z 2009-10-31T13:41:10Z <p>As a complement to Georges Elencwajg's answer, there are lots of different simply connected hypersurfaces in projective space. The second volume of Shafarevich's introduction to algebraic geometry has a discussion of just this question of uniformization of complex manifolds.</p>