Rational Group Cohomology - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T10:12:01Zhttp://mathoverflow.net/feeds/question/5057http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/5057/rational-group-cohomologyRational Group CohomologyJames Griffin2009-11-11T15:12:38Z2012-02-16T09:07:57Z
<p>This is a general question about group cohomology. I'm interested in the case when the coefficients are the rational numbers and hence I suppose when my groups are infinite. The question splits into two:</p>
<p>1) Are there any favoured examples that you would recommend a look at? (Recommended references would be just as welcome.)</p>
<p>And the main question:</p>
<p>2) What sort of functors on the category of groups leave the rational cohomology unchanged? In particular is there a projection onto a special subcategory of groups that is in some way the right category to study?</p>
<p>I have a feeling that someone with a good knowledge of rational homotopy theory would be able to answer this question with relative ease.</p>
http://mathoverflow.net/questions/5057/rational-group-cohomology/5064#5064Answer by Danny Calegari for Rational Group CohomologyDanny Calegari2009-11-11T15:58:00Z2009-11-11T15:58:00Z<p>Stallings showed that if $f:\Gamma \to \Delta$ is a homomorphism between finitely presented groups where <img src="http://latex.mathoverflow.net/png?%24f%5Fi%3AH%5Fi%28%5CGamma%2CQ%29%20%5Cto%20H%5Fi%28%5CDelta%2CQ%29%24" alt="$f\sb i:H\sb i(\Gamma,Q) \to H\sb i(\Delta,Q)$" title="" /> is bijective for $i\le 1$ and surjective for $i=2$ then $f\otimes Q: \Gamma \otimes Q \to \Delta \otimes Q$ is an isomorphism between the $Q$-unipotent completions. So "taking the $Q$-unipotent completion" is a functor with some interesting properties with respect to rational group (co)homology. I'm not sure if this is the kind of thing you're after . . .</p>
<p>Stallings' paper is </p>
<p>MR0175956 (31 #232)
Stallings, John
Homology and central series of groups.
J. Algebra 2 1965 170--181. </p>
http://mathoverflow.net/questions/5057/rational-group-cohomology/88613#88613Answer by Mark Grant for Rational Group CohomologyMark Grant2012-02-16T09:07:57Z2012-02-16T09:07:57Z<p>One class of groups whose rational cohomology is particularly easy to handle are the finitely generated torsion-free nilpotent groups. This is because they admit refined Postnikov systems which can be localised, leading to a recipe for computing the Sullivan minimal model of cochains in terms of their central series. This means one can also read off cup and Massey products quite easily.</p>
<p>A good reference (with a crash course in rational homotopy theory included) is the paper</p>
<p>Oprea, John <em>The category of nilmanifolds.</em> Enseign. Math. (2) 38 (1992), no. 1-2, 27–40.</p>