User raphael - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:29:13Z http://mathoverflow.net/feeds/user/4731 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18661/pd3-groups-and-pd4-complexes PD3 groups and PD4 complexes Raphael 2010-03-18T21:29:40Z 2012-04-04T05:14:33Z <p>I am interested at the moment in what groups can occur as the fundamental group of a 4-manifold (or more generally, a 4-dimensional CW complex) with prescribed conditions on the intersection form. I have what I am hoping is a basic homotopy theory question:</p> <p>A (orientable) PD-$n$ group is a group $G$ such that the Eilenberg-Maclane space $K(G,1)$ admits "Poincare duality", i.e. there is an $n$-dimensional integer homology class in $K(G,1)$ (thought of as the "fundamental class") such that cap product with it yields an isomorphism between the corresponding cohomology and homology groups (like for closed oriented manifolds). This is more general than saying that $K(G,1)$ admits the structure of an orientable closed manifold of dimension $n$.</p> <p>Let $G$ be a PD-3 group. Is there any reason why $G$ cannot be the fundamental group of an orientable PD4 complex $X$ with vanishing second homotopy group, $\pi_2(X)=0$? </p> http://mathoverflow.net/questions/1162/atiyah-singer-index-theorem/18678#18678 Answer by Raphael for Atiyah-Singer index theorem Raphael 2010-03-18T22:58:12Z 2010-03-18T22:58:12Z <p>I have also good experience with the book of Lawson and Michelsohn. The theory of pseudo-elliptic differential operators is also well explained in the book of Wells, named "Analysis and complex geometry" (or something like that).</p> http://mathoverflow.net/questions/18661/pd3-groups-and-pd4-complexes/18680#18680 Comment by Raphael Raphael 2010-03-19T06:34:27Z 2010-03-19T06:34:27Z Thanks for your comments! I guess any group G such that $H^2(G;\mathbb{Z})$ contains an infinite cyclic group is not going to work, just because of the naturality of the cup product Paul mentioned and because $H^4(G;\mathbb{Z})=0$ by the assumption that $G$ is a $PD3$ group. I could imagine there being no restriction if $G$ has only 2-torsion in its second cohomology group, but of course this would need a slightly more careful thought. http://mathoverflow.net/questions/18661/pd3-groups-and-pd4-complexes Comment by Raphael Raphael 2010-03-18T22:55:49Z 2010-03-18T22:55:49Z Thanks for your interest. Yes, I knew that. I have given a rough definition of what a Poincare-duality group is.