Subject: RE: resp re Hopf inv Date: Tue, 17 Feb 2004 08:59:39 -0000 From: "Neil Strickland" Incidentally, Bill's argument desuspends. Corollary III.6.3 in Baues's "Commutator calculus" book says that in \pi_*(S^*) we have H(b \circ a) = H(b) \circ a + (b # b) \circ H(a), provided that H is defined by the James-Hopf procedure with the left lexicographic ordering. Taking b = \eta_2 gives H(\eta_2\circ a) = a + \eta_3\circ\eta_4\circ H(a), which gives plenty of examples with H(\eta_2\circ a) \neq a. Neil _______________________________________________________________ Subject: Re: resp re Hopf inv Date: Mon, 16 Feb 2004 22:24:57 -0600 From: Bill Richter Neil asked if the composite was 2-locally homotopic to 1: Omega^3 S^3 -Omega^3 eta--> Omega^3 S^2 -Omega^2 H--> Omega^3 S^3 I showed it's not, as nu' in pi_6(S^3) = Z/4 goes to -nu'. This is surprising, as Neil's conjecture is true in the lambda algebra. Brayton responded: This certainly depends on what you mean by the Hopf invariant. Even the James Hopf invariant is dependent on a choice of lexicographic ordering, of which there are at least 2. No, Boardman & Steer showed the question is largely independent of the definition of the Hopf invariant, as my proof only used lambda_2. Boardman & Steer showed that for all proposed definitions of the Hopf invariants H_n, their suspensions lambda_n = Sigma^{n-1} H_n are equal, and characterized by the Cartan formula. Noch einmals: lambda_2(eta . nu') = lambda_2(eta) . nu' + (eta smash eta) . lambda_2(nu') = nu' + eta^3 = - nu' in pi_7(S^4) So Neil's (very reasonable!) conjecture is false for any of the choices for H_2 that Boardman & Steer considered. As a variation on the second Hopf invariant(H_2), for example, we could add the composition of \Omega\eta^2 with H_4, the fourth James Hopf invariant. This would add a term to the composition, which in homotopy would add \eta^3 composed with H_4. Hmm, that's a meaning of H_2 that Boardman & Steer didn't consider. You mean to take any map H_2: Omega S^2 ---> Omega S^3 which is degree one on the 2-cell. The James splitting shows there lots of such maps, all sums of the sort you consider. But my proof shows that Neil's conjecture is false for any such H_2! Because nu' doesn't have any higher Hopf invariants. nu' could have a third Hopf invariant, but (1 \pm (12) + (123)) lambda_3 = 0, so lambda_3 is killed by either 1 or 3, hence lambda_3(nu') = 0. More on the James splitting: The adjoint of our map H_2 above is map H_2^*: Sigma Omega S^2 ---> S^3 and the James splitting is an equivalence Sigma Omega S^2 ---> vee_{k > 1} S^k given by the sum of the H_n, for any choice of H_n's, so we might as well take a standard choice, say left-lexicographical James-Hopf. So H_2 is determined by a collection of maps f_{k-1}: S^k ---> S^3, for k > 0, and any such H_2 you want is then sum_{k=0}^infty Omega(f_k) H_k where these RHS H_k's are the standard ones. As Fred likes to say, the condition on our H_2 is that the diagram homotopy commutes: Omega S^2 -H_2--> Omega S^3 ^ ^ | | | | S^1 x S^1 -pinch--> S^2 That means that f_1 = 0 and f_2 = 1, and the higher f_k are arbitrary, and any such So by taking f_4 = eta^2, all other f_k = 0, we get your example. I don't have an example off hand where this perturbation is not zero, but suspect that such examples exist. I think so, Brayton. Michael used to say that 2-locally, H_4 is pretty much H_2 . H_2, i.e. the composite Omega S^2 -H_2--> Omega S^3 -H_2--> Omega S^5 Now Neil's conjecture is pretty much true, so let's say that the first H_2 of our H_4 just removes the eta. We want alpha in pi_k S^3 with eta^2 H_2(alpha) \ne 0 pi_k S^3 The Lambda Curtis table say 2233 <--- 911 11233 <--- 8111 That indicates there's an element alpha in pi_{10} S^3 with eta^2 H_2(alpha) = eta^2 epsilon \ne 0 in pi_{10} S^3 and furthermore, H_2(alpha) = epsilon and Sigma^6 alpha = w_9 eta^2. We can check this in Toda's tables, or ask Mark Mahowald. Speaking of which, I should've said that my calculation of lambda_2(eta nu') = -nu' is an example of the correctness of Mark position, that we can't tell the generators in nu' apart. I wrote as if I was refuting Mark, but I backed him up instead: We can of course choose a canonical generator nu', as it's in the image of J, the image of the double cover S^3 --->> SO(3). Mark's point must be that this isn't helpful. Just take any generator g, and so nu' = {g, -g}. And that's exactly what I did, and and showed that g went to -g. But Mark does that all the time: a Toda bracket has an indeterminacy, so at first your element is fudged, but you play around and end up with an element of the indeterminacy, which you have to compute. So the indeterminacy of nu' is {0, eta^3}, and I showed that Neil's example led to the nonzero element eta^3. That's the drill! -- Best, Bill