Timeline for rationalization of classifying spaces

6 events

when toggle format	what		by	license	comment
Feb 1, 2016 at 16:07	vote	accept	Ulrich Pennig
Apr 5, 2014 at 20:29	comment	added	Ben Wieland		Commutativity shows that the answer to the second question in the first bullet point is "No," but does not address the first question in that bullet point. I think your second argument actually shows that finite dimensional rational $H$-spaces are commutative, so it really is the same reason. An infinite dimensional non-commutative example is given by $\Omega S^{2n}$. Its classifying space $S^{2n}$ is not an $H$-space, let alone a product of EM spaces, not even after rationalization.
Apr 5, 2014 at 9:48	history	edited	Mark Grant	CC BY-SA 3.0	fixed spelling of McCleary
Apr 5, 2014 at 9:15	comment	added	Mark Grant		Two nilpotent spaces with finite Betti numbers are rationally homotopy equivalent if and only if they have isomorphic minimal models. So the above argument should work, as long as there are only finitely many $y_i$ in each even degree.
Apr 4, 2014 at 19:57	comment	added	Ulrich Pennig		Suppose I have a group, such that $H^*(BG; \mathbb{Q}) \cong \mathbb{Q}[y_1, y_2, \dots]$ with infinitely many $y_i$'s in even degrees. Would this argument still work?
Apr 4, 2014 at 13:10	history	answered	Mark Grant	CC BY-SA 3.0