Subject: Toda's result
Date: Tue, 16 Jul 2002 04:43:59 -0400 (EDT)
From: Robert Bruner
To: dmd1@lehigh.edu
Dear Don,
After reading Doug's letter, I thought people might be interested
in the following. I will trust your judgement whether to send it
out or not. I sent this to Zhou last February, but have not had
any response.
Regarding Toda's result, that \alpha_1 \beta_1^p = 0:
in addition to Toda's proof by 'composition methods' there
is also a proof by 'smash product methods'. This can be
found in Springer Lecture Notes in Mathematics, Vol. 1176,
Cor 1.15 (p. 138). The proof is on p. 167, though you
have to work your way back through several other results to
extract the whole proof.
Here is the idea. In what follows, I will use spectra, but the whole
argument can be done in a very elementary fashion on the space level
by simply suspending enough.
Let G be either the cyclic group of order p or the symmetric group
\Sigma_p and let E be a contractible free G space. Define
D(X) = (E+ \smash X^(p))/G
where ^(p) is the p-fold smash product and + denotes addition of a
disjoint basepoint, when X is a space, and the spectrum level analog as
in V. 1176 when X is a spectrum. The spectrum level functor is an
analog in the sense that it commutes with the suspension spectrum
functor, \Sigma^\infty : Top --> Spectra.
We can see easily that when G is cyclic, then
D(S^n) = \Sigma^n L_{n(p-1)},
where L is the infinite lens space S^\infty/G and L_k = L/L^(k-1), the
quotient of L by its (k-1)-skeleton. The action of the Steenrod
algebra here is well known. When G= \Sigma_p, D(S^n) localized at (p)
splits from the corresponding D_{cyclic}(S^n), and is a complex
containing cells in dimensions n or n-1 modulo 2(p-1). (Thm 2.9 p.
146)
There is a map D(S) --> S extending the product S^(p) ---> S,
where S is the sphere spectrum. (On the space level, one gets
D^i(S^k) --> S^kp extending the identity map of S^kp, where D^i
is defined using S^i in place of S^\infty, and k is congruent to
0 modulo a power of p depending on i.)
So, given x: S^n --> S, we get D(S^n) --> D(S) --> S extending x^p on
the bottom cell S^{np}. If a \in \pi_k(D(S^n)), we can compose with it
to get an element
a^*(x) : S^k --> D(S^n) --> D(S) --> S.
Using the diagonal map E --> E x E, we get a natural tranformation
D(X \smash Y) ---> D(X) \smash D(Y)
and from this we can compute a Cartan formula for these operations a^*.
Suppose n is even. Then
(*) D(S^n) = S^{pn} \wedge S^{pn+2p-3} \union e^{pn+2p-2}
\union higher cells
where the np+2p-2 cell is attached by p to the cell just below it, and to
the bottom cell by -k \alpha, where n = 2k.
The inclusion of the pn + 2p-3 cell gives an operation I shall call B,
since it is the second operation, the first being the p^th power.
It is easy to use the diagonal map E --> E x E to compute that the Cartan
formula for B is
B(xy) = x^p B(y) + B(x) y^p + c \alpha_1 x^p y^p
for some constant c, if x and y are both in even stems. Now, apply
this to 0 = p \beta_1. We get
0 = B(0)
= B(p \beta_1)
= p^p B(\beta_1) + B(p) \beta_1^p + 0
= 0 + \alpha_1 \beta_1^p + 0
and hence \alpha_1 \beta_1^p = 0. Here I am using B(p) = \alpha_1,
and the relation pB(x) = k \alpha_1 x^p, which comes from the attaching
map of the pn+2p-2 cell in (*) so that
p^p B(x) = p^{p-1} k \alpha_1 x^p = 0,
since p \alpha_1 = 0.
So, to demolish Toda's result, \alpha_1 \beta_1^p = 0, this argument
has to be demolished as well. I think we can assert with confidence
that Toda got it right.
Best Regards,
Bob Bruner