How does pseudoconvexity restrict the topology? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:43:48Z http://mathoverflow.net/feeds/question/97266 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97266/how-does-pseudoconvexity-restrict-the-topology How does pseudoconvexity restrict the topology? Steven Gubkin 2012-05-17T23:33:31Z 2012-05-18T00:33:33Z <p>A domain of holomorphy in $\mathbb{C}^n$ has vanishing de rham cohomology in real dimensions greater than $n$ - half of it's cohomology is missing. Are there any other restrictions? If I give you a finite dimensional graded ring, can you find a domain of holomorphy having that ring as its cohomology ring?</p> http://mathoverflow.net/questions/97266/how-does-pseudoconvexity-restrict-the-topology/97269#97269 Answer by Daniele Zuddas for How does pseudoconvexity restrict the topology? Daniele Zuddas 2012-05-18T00:33:33Z 2012-05-18T00:33:33Z <p>A theorem of <a href="http://www.worldscinet.com/ijm/01/0101/S0129167X90000034.html" rel="nofollow">Eliashberg</a> implies that an open subset of $\Bbb C^n$, $n \neq 2$, is isotopic to a Stein domain (hence to a domain of holomorphy) if and only if it admits a handlebody structure with all handles of index $\leq n$. For $n = 2$ the theorem still holds with few modifications regarding the framing of 2-handles. Is not properly the statement of Eliashberg that implies this, but the proof of the theorem itself.</p> <p>So the answer of the last question is yes if you can build handles of index $\leq n$ (and dimension $n$) so that the resulting manifold has the prescribed cohomology ring (of course not any such handlebody smoothly embeds in $\Bbb C^n$). The problem is related to embedding CW complexes of real dimension $n$ in $\Bbb C^n$, realizing the graded ring (I mean smoothly embedding each cell, then pick a regular neighborhood).</p>