Subject: RE: two more on Hopf inv Date: Thu, 19 Feb 2004 18:52:28 +0800 From: "Wu Jie" Subject: Re: two more on Hopf inv Date: Wed, 18 Feb 2004 22:37:40 -0600 From: Bill Richter Jie, that's an interesting post. But I say you're wrong here: On the other hand, it seems that the (left) inverse of \Omega\eta can be given by the Ganea-Hopf invariants: I proved this is false, so we have a contradiction! That is, spurred on by Brayton's question, I showed that the composite Omega S^3 -Omega eta--> Omega S^3 -H_2--> Omega S^2 is not the identity, for any map H_2 which is degree 1 on the 2 cell, including your map. Use pi_5 Omega S^3 = Z/4, where nu' |---> -nu'. However, I don't yet see an error in your proof, so we have an interesting contradiction to resolve. We should scrutinize your Top+Apps paper, which sounds interesting anyway. Some thoughts: I think your map is given by looping the pinch map on S^2, and then projecting onto the 3rd piece of the decomposition that Ganea liked O(A vee B) = O(A) x O(B) x O(O A * O B) , and using the map f: O S^2 = J(S^1) ---> S^1 given by the H-space mult on S^1 (which as you say is a loop map). So your map is the composite O S^2 -O(comult)--> O(S^2 vee S^2) -proj--> O(O S^2 * O S^2) -O(f * f)--> O(S^1 * S^1) = O S^3 The composite to O(O S^2 * O S^2) is the total Hilton-Hopf inv, which Barratt long ago how related to the James-Hopf invs. What does f * f do to the Hilton-Hopf invs? We might as well suspend to O S^2 -your map--> O (S^3) -O(E)--> O^2 S^4 since my counterexample nu' is detected after a suspension, so we'll have the suspended Hilton-Hopf inv, which Boardman & Steer related to the suspended James Hopf invs. (Actually Boardman & Steer use more suspensions, but Michael pointed out that one suspension suffices for their calculations.) So must deal with 2 copies of S(f): S O S^2 ---> S^2 Now if instead of S(f), we had sigma: S O S^2 ---> S^2, adjoint to the identity, then we'd be done. Your composite would Boardman & Steer's lambda_2, which is where my nu' counterexample comes from. Now the Hopf inv decomposition of S O S^2 is dual to John Moore's decomposition given by the Hopf constructions, for k >= 1: S^{k+1} ---> S (S^1 x ... x S^1) -S(mult)--> S O S^2 but pushing to S^2 by your map f just gives products of eta. Since eta^4 = 0, I guess that S(f) is adjoint to the selfmap of O(S^2) S(f) = 1 + O(eta) H_2 + O(eta^2) H_3 + O(eta^3) H_4 (Hmm, on S^2, eta^4 \ne 0, but I think that comes out in the wash.) Actually we only need John Moore's splitting of S O S^2, since the Hilton Hopf invs will map into these Hopf constructed S^{k+1}. So we have a chance a James-Hopf invariant calculation of your map. And we can look at your map your way too, whether it's simplicial or Ganea-theoretic. Sounds like fun! (Perhaps we should take the discussion off Don Davis's list, and report back when we're done.) -- Best, Bill ______________________________________________________________ To: "Bill Richter" , CC: , , , , , Dear Bill, It sounds pretty fun. It's a good idea to take the discussion off Don's list before we clean up the possible contradictions. I am back to think about Brayton's comments about different Hopf invariants. Consider the following maps: \Omega \Sigma X ----H_6---> \Omega \Sigma X^{\smash 6} \Omega \Sigma X ----H_6---> \Omega \Sigma X^{\smash 6}--\Omega \Sigma (23)--> \Omega \Sigma X^{(6)}, where H_n is the James-Hopf invariant and (23) is the permutation by exchanging coordinats 2 and 3. The above two maps are much different either after suspensions or after looping because the homotopy colomit by iterating the map \Omega \Sigma X ----H_6---> \Omega \Sigma X^{\smash 6}---\Omega W_6---> \Omega \Sigma X is contractible localized at 2 by computing the homology, where W_6 is the Whitehead product, and the homotopy colimit by iterating the map \Omega \Sigma X ----H_6---> \Omega \Sigma X^{\smash 6}--\Omega \Sigma (23)--> \Omega \Sigma X^{(6)}---\Omega W_6---> \Omega \Sigma X is \Omega\Sigma L'_6(X) localized at 2, where the mod 2 homology of L'_6(X) is the Lie elements L_6(\bar H_*(X)) modulo [L_3(\bar H_*(X)),L_3(\bar H_*(X))]. For mentioning this example, I intend to say that if we are allowed to consider the Hopf invariants like: \Omega \Sigma X---H_n ----> \Omega \Sigma X^{\smash n} ---functorial equivalences---> \Omega\Sigma X^{\smash n}, would we get the left inverse of \Omega\eta? By writing (the combinatorial Ganea theoretic version of) the total Hilton-Hopf invariants back to the James-Hopf invariants, it involves many of "functorial self equivalences" of \Omega\Sigma X^{\smash n}. (These self equivalences come from collecting differences by writing one type of words (as product elements) into another type. ) Since we know that \Omega S^3--\Omega\eta ->\Omega S^2 --H_2--> \Omega S^3 is a homotopy equivalence, the left inverse of \Omega\eta is given by \Omega S^2---H_2---> \Omega S^3 --\phi--> \Omega S^3, where \phi is a self homotopy equivalence of the individual space \Omega S^3. The question may be whether we can choose \phi to be a functorial homotopy equivalence restricting to the particular case \Omega S^3. By using total Hopf invariants, there seems a chance to have the functorial version of \phi by consider the canonical retraction from \Omega\Sigma G to \Omega \Sigma (G\smash G), as I menstioned in the previous message. I do not have Boardman & Steer's paper on hand and I will have to refresh the uniqueness theorem of Boardman & Steer. In the sense of Boardman-Steer, is the group of functorial self equivalences allowed to act on the Hopf invariants? (I forget their definitions.) I mean whether Boardman-Steer allows H_n to compose with (functorial) self equivalences? If so, I get some confusions (namely possible contradictions come out). If not, possibly both arguments are correct in Brayton's sense that this is because of "different" H_2. Best Regards, Jie