Subject: Hopf invariant
Date: Thu, 12 Feb 2004 23:09:03 -0500
From: Tom Goodwillie <tomg@math.brown.edu>

>From: "Neil Strickland" <N.P.Strickland@sheffield.ac.uk>
>
>Let \eta : S^3 --> S^2 be the Hopf map, and let
>H : \Omega S^2 --> \Omega S^3 be the Hopf invariant.
>The map
>
>                \Omega^3\eta
>  \Omega^3 S^3 ---------------> \Omega^3 S^2
>
>is an equivalence, and the map
>
>                \Omega^2 H
>  \Omega^3 S^2 ---------------> \Omega^3 S^3
>
>is a 2-local equivalence.  Can anyone tell me if the
>composite is 2-locally homotopic to the identity?
>
>Neil
>_______________________________________________________________

I don't know, but the map

                          \Omega^2 H
  \Omega^3 S^2 ---------------> \Omega^3 S^3

is an equivalence, not just 2-locally. In fact, any map

                              H
\Omega S^2 ---------------> \Omega S^3

that induces an iso on \pi_2 must induces iso on all higher \pi_n.
The reason is that the composed map

                      \Omega\eta                             H
\Omega S^3 ---------------->\Omega S^2 -------------> \Omega S^3

is an equivalence. And the reason for that is that any map

                     \Omega S^3 -----------> \Omega S^3

inducing iso on \pi_2 must be an equivalence. And the reason for that
is that the integral cohomology ring of \Omega ^3 is such that a
self-map must be iso if it is iso on H^2.

  -- Tom Goodwillie
______________________________________________________________

Subject: Re: four postings
Date: Thu, 12 Feb 2004 22:26:37 -0600
From: Bill Richter <richter@math.northwestern.edu>

Neil, the 2-local composite is not the identity:

Omega^3 S^3 -Omega^3 eta--> Omega^3 S^2 -Omega^2 H--> Omega^3 S^3

You can see this in the homotopy group pi_3.  All you need is
Boardman-Steer's composition formula for the suspended Hopf invariant.

Pick a generator
              nu' in pi_6 S^3 = Z/4 >---> pi_7 S^4 = Z/8
I'll show that

H(eta . nu') \ne nu' in pi_6 S^3

and this will be a counterexample to your conjecture.

There are 2 generators for pi_6 S^3, and
                      nu' + eta^3  in pi_6 S^3
is the other choice of generator, since
                          eta^3  = 2 nu',
and
                    H(nu') = eta  2 in pi_6 S^5

Toda proves all of this in his book, but it's due to Barratt, who gave
the first example of a Toda bracket, nu' = {eta, 2iota, eta} and
computed its Hopf invariant.  Toda's book is essentially taking this
one example of Barratt to the max.  I like this counterexample here,
because Mahowald always says that you can't tell these 2 choices for
nu' apart, as the indeterminacy of the Toda bracket is eta^3.  But
even if you can't tell 'em apart, they're not equal!  Continuing,

it's suffices to prove this after a suspension, and that means

lambda_2(eta . nu') \ne nu' in pi_7 S^4,

using Boardman-Steer's suspended Hopf invariant.  But by
Boardman-Steer's  composition formula (Thm. 3.16, which has sign
errors for  higher Hopf invariants),

lambda_2(eta . nu')  = lambda_2(eta) . nu'

             + (eta smash eta) . lambda_2(nu')

             = nu' + eta^3 in pi_7(S^4)
\qed

BTW I liked Clarence's point, all Whitehead products for spheres are
at the prime 2.

Fred Cohen makes a good point in his Seattle notes lectures on
unstable homotopy, that at the prime 2, only the Whitehead square
w_n = [iota, iota] is nonzero.

That is, [[iota, iota], iota] = 0 by the Jacobi identity, which says
that either 3 or 1 kills it, depending on the parity.  So therefore
the 4-fold and higher Whitehead products are zero as well:

[[iota, iota], ... iota]] = 0

Fred disposes of other Whitehead products, such as

[w_n, w_n] = 0

One way to see that is by naturality and that that composition with
w_n factors through the double suspension.  So at the prime 2 for
spheres, the Barcus-Barratt theorem drastically simplifies:

[f, g] = [iota, iota] . H(f) . H(f)

So we only have to worry about compositions [iota, iota] . alpha.