User raphael - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-21T16:29:13Zhttp://mathoverflow.net/feeds/user/4731http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/18661/pd3-groups-and-pd4-complexesPD3 groups and PD4 complexesRaphael2010-03-18T21:29:40Z2012-04-04T05:14:33Z
<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#18678Answer by Raphael for Atiyah-Singer index theoremRaphael2010-03-18T22:58:12Z2010-03-18T22:58:12Z<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#18680Comment by RaphaelRaphael2010-03-19T06:34:27Z2010-03-19T06:34:27ZThanks 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-complexesComment by RaphaelRaphael2010-03-18T22:55:49Z2010-03-18T22:55:49ZThanks for your interest. Yes, I knew that. I have given a rough definition of what a Poincare-duality group is.