Subject: Re: two more on Hopf inv Date: Thu, 19 Feb 2004 13:00:42 -0600 From: Bill Richter [Don, please post this today, I have serious retractions to make.] I twice posted a false claim, and there is indeed a map F= H_2 + \Omega(\eta_3\eta_4)\circ H_4 : \Omega S^2 -> \Omega S^3 which is a left inverse of Omega eta, after one loop: O^2 S^3 -O^2 eta--> O^2 S^2 -O H--> O^2 S^3 is homotopic to the identity, at least 2 locally. So Brayton was on the right track, and Jie Wu may well have a proof also. Neil Strickland independently came up with F, and realized it worked for my nu', and also noted my error. Neil wrote me: I think you sketched an argument based on the fact that the higher Hopf invariants of \nu' are trivial. However, it's really the higher Hopf invariants of \eta_2\circ\nu' that are relevant in your argument. Here's a sketch proof that the Brayton-inspired F indeed works: By using the retraction O S^3 >-O E--> O^2 S^4 we can turn H_2 into Boardman Steer's lambda_2. So Boardman Steer's composition formula implies that the composite O S^3 -O eta--> O S^2 -H_2--> O S^3 is the identity plus the composite O S^3 -H_2--> O S^2 -O eta^2--> O S^3 So now F O(eta) = (H + O(eta^2) H^2) O(eta) = H O(eta) + O(eta^2) H^2 O(eta) = 1 + O(eta^2) H + O(eta^2) H ( 1 + O(eta^2) H ) as a selfmap of O S^3. But now after a loop, we can distribute the H across the +, but the 2nd term is zero: since eta^2: S^5 ---> S^3 is a suspension, the naturality of the James Hopf invariant says that. H O(eta^2) = O(eta^4) H = 0 \qed