3
_ as an upper triangular matrix
Jonathan Barker and Victor Snaith
Abstract
In the 2-local stable homotopy category the group of left-bu-module
automorphisms of bu^bo which induce the identity on mod 2 homology
is isomorphic to the group of infinite upper triangular matrices with
entries in the 2-adic integers. We identify the conjugacy class of the
matrix corresponding to 1 ^ _3, where _3 is the Adams operation.
1 Introduction
Let bu and bo denote the stable homotopy spectra representing 2-adically com-
pleted unitary and orthogonal connective K-theory respectively. The main
result of [7] is the existence of an isomorphism of groups
~= 0
: U1 Z2 -! Aut left-bu-mod(bu ^ bo).
Here U1 Z2 is the group of upper triangular matrices with coefficients in the
2-adic integers and Aut0left-bu-mod(bu^bo) denotes the group of left bu-module
automorphisms of bu ^ bo in the stable homotopy category of 2-local spectra,
which induce the identity on mod 2 singular homology. The details of this
isomorphism are recapitulated in x2.
This isomorphism is defined up to inner automorphisms of U1 Z2. Given
an important automorphism in Aut 0left-bu-mod(bu ^ bo) one is led to ask what
is its conjugacy class in U1 Z2. By far the most important such automorphism
is 1 ^ _3, where _3 : bo -! bo denotes the Adams operation.
The following is our main result, which proved by combining the discussion
of x3.5 with Theorem 4.2.
Theorem 1.1.
Under the isomorphism the automorphism 1 ^ _3 corresponds to an
1
element in the conjugacy class of the matrix
0 1
1 1 0 0 0 . . .
B C
B C
B C
B 0 9 1 0 0 . . .C
B C
B C
B 2 C
B 0 0 9 1 0 . . .C.
B C
B C
B 3 C
B 0 0 0 9 1 . . .C
B C
@ A
.. . . . . .
. .. .. .. .. ..
By techniques which are described in ([1] pp.338-360) and reiterated in x3,
Theorem 1.1 reduces to calculating the effect on 1 ^ _3 on ss*(bu ^ bo) modulo
torsion. The difficulty arises because, in order to identify -1(1 ^ _3) one
must compute the map on homotopy modulo torsion in terms of an unknown
2-adic basis defined in terms of the Mahowald splitting of bu ^ bo (see x2 and
x3). On the other hand a very convenient 2-adic basis is defined in [3] and the
crucial fact is that 1 ^ _3 acts on the second basis by the matrix of Theorem
1.1. This fact was pointed out to one of us (VPS) by Francis Clarke in 2001
and led to the confident prediction appearing as a footnote in ([7] p.1273).
Verifying the prediction has proved a little more difficult than first imagined!
Once one has Theorem 1.1 a number of homotopy problems become merely
a matter of matrix algebra. In x5 we give an example concerning the maps
1^(_3-1)(_3-9) . .(._3-9n-1) where we prove a vanishing result (Theorem
5.4) which is closely related to the main theorem of [5], as explained in Remark
5.5. In subsequent papers we shall give further applications to homotopy
theory and to algebraic K-theory.
2 2-adic homotopy of bu ^ bo
2.1. Let bu and bo denote the stable homotopy spectra representing 2-adically
completed unitary and orthogonal connective K-theory respectively. We shall
begin by recalling the 2-local homotopy decomposition of bu ^ bo which is one
of a number of similar results which were discovered by Mark Mahowald in
the 1970's. These results may be proved in several ways [1], [4] and [5]. For
notational reasons we shall refer to the proof, a mild modification of ([1]
pp.190-196), which appears in ([7] x2).
Consider the second loopspace of the 3-sphere, 2S3. There exists a model
for 2S3 which is filtered by finite complexes ([2] , [6])
[
S1 = F1 F2 F3 . . . 2S3 = Fk
k 1
2
and there is a stable homotopy equivalence, an example of the so-called Snaith
splitting, of the form
2S3 ' _k 1Fk=Fk-1.
There is a 2-local homotopy equivalence of left-bu-module spectra (see [7]
Theorem 2.3(ii)) of the form
^L: _k 0bu ^ (F4k=F4k-1) -'! bu ^ bo.
The important fact about this homotopy equivalence is that its induced map
on mod 2 homology is a specific isomorphism which is described in ([7] x2.2)
From this decomposition we obtain left-bu-module spectrum maps of the
form
'k,l: bu ^ (F4k=F4k-1) -! bu ^ (F4l=F4l-1)
where 'k,k= 1, 'k,l= 0 if l > k and, as explained in ([7] x3.1), 'k,lis defined
up to multiplication by a 2-adic unit when k > l.
Consider the ring of left bu-module endomorphisms of degree zero in the
stable homotopy category of spectra [1], which we shall denote by
End left-bu-mod(bu ^ bo). The group of units in this ring will be denoted
by Aut left-bu-mod(bu ^ bo), the group of homotopy classes of left bu-module
homotopy equvialences and let Aut 0left-bu-mod(bu ^ bo) denote the subgroup
of left bu-module homotopy equivalences which induce the identity map on
H*(bu ^ bo; Z=2).
Let U1 Z2 denote the group of infinite, invertible upper triangular matrices
with entries in the 2-adic integers. That is, X = (Xi,j) 2 U1 Z2 if Xi,j2 Z2
for each pair of integers 0 i, j and Xi,j= 0 if j > i and Xi,iis a 2-adic
unit. This upper trianglular group is not equal to the direct limit lim!nUnZ2
of the finite upper triangular groups. The main result of [7] is the existence
of an isomorphism of groups
~= 0
: U1 Z2 -! Aut left-bu-mod(bu ^ bo).
By the Mahowald decomposition of bu ^ bo the existence of _ is equivalent to
an isomorphism of the form
~= 0
: U1 Z2 -! Aut left-bu-mod(_k 0bu ^ (F4k=F4k-1)).
If we choose 'k,lto satisfy 'k,l= 'l+1,l'l+2,l+1. .'.k,k-1for all k - l 2 the*
*n,
for X 2 U1 Z2, we define ([7] x3.2)
X
(X-1 ) = Xl,k'k,l: bu ^ (_k 0F4k=F4k-1) -! bu ^ (_k 0F4k=F4k-1).
l k
The ambiguity in the definition of the 'k,l's implies that is defined up to
conjugation by a diagonal matrix in U1 Z2.
3
2.2. Bases for ss*(bu^bo)_Z2Torsion
Let Gs,tdenote the 2-adic homotopy group modulo torsion
sss(bu ^ F4t=F4t-1) Z2
Gs,t= _______________________
Torsion
so
sss(bu ^ F4t=F4t-1) Z2 ss*(bu ^ bo) Z2
G*,*= s,t_______________________~= ________________.
Torsion Torsion
From [1] or [7] 8
< Z2 ifs even, s 4t,
Gs,t~=
: 0 otherwise
and if "Gs,tdenotes sss(bu ^ F4t=F4t-1) Z2 then "Gs,t~=Gs,t Ws,twhere Ws,t
is a finite, elementary abelian 2-group.
In [3] a Z2-basis is given for G*,*consisting of elements lying in the subr*
*ing
Z2[u=2, v2=4] of Q2[u=2, v2=4]. One starts with the elements
_ !
v2 - 9i-1u2
c4k = ki=1 ___________ , k = 1, 2, . . .
9k - 9i-1
and "rationalises" them, after the manner of ([1] p.358), to obtain elements of
Z2[u=2, v2=4]. In order to describe this basis we shall require a few well-known
preparatory results about 2-adic valuations.
Proposition 2.3.
n n+3
For any integer n 0, 92 - 1 = 2 (2s + 1) for some s 2 Z.
Proof
We prove this by induction on n, starting with 9 - 1 = 23. Assuming the
result is true for n, we have
(n+1) 2n 2n
92 - 1 = (9 - 1)(9 + 1)
n 2n
= (92 - 1)(9 - 1 + 2)
= 2n+3(2s + 1)(2n+3(2s + 1) + 2)
= 2n+4(2s + 1) (2n+2(2s + 1) + 1)
________-z_______"
odd
as required.
Proposition 2.4.
For any integer l 0, 9l- 1 = 2 2(l)+3(2s + 1) for some s 2 Z, where 2(l)
denotes the 2-adic valuation of l.
4
Proof
Write l = 2e1 + 2e2 + . .+.2ek with 0 e1 < e2 < . . .< ek so that
2(l) = e1. Then, by Proposition 2.3,
e1+2e2+...+22k
9l- 1 = 92 - 1
= ((2s1 + 1)2e1+3+ 1)) . .(.(2sk + 1)2ek+3+ 1)) - 1
(2s1 + 1)2e1+3 (modulo 2e1+4)
= 2e1+3(2t + 1)
as required.
Proposition 2.5. Q
For any integer l 1, li=1(9l- 9i-1) = 2 2(l!)+3l(2s + 1) for some s 2 Z.
Proof
By Proposition 2.4 we have
Q l l i-1 Q l l-i+1 i-1
i=1 (9 - 9 ) = i=1(9 - 1)9
Q l
= i=1 2 2(l-i+1)+3(2ti+ 1)9i-1
= (2t + 1)2 2(l!)+3l,
as required.
Proposition 2.6. For any integer l 0, 2 2(l!)+3l= 24l-ff(l)where ff(l) is
equal to the number of 1's in the dyadic expansion of l.
Proof
Write l = 2e1+2e2+. .+.2ek with 0 e1 < e2 < . .<.ek so that ff(l) = k.
9l- 1 = 2 2(l)+3(2s + 1) for some s 2 Z, where 2(l) denotes the 2-adic
valuation of l. Then
2(l!) = 2ff1-1 + 2ff2-1 + . . .+ 2ffk-1
+ 2ff1-2 + 2ff2-2 + . . .+ 2ffk-2
.. .
. ..
+ 1 + 2ff2-ff1+ . . .+ 2ffk-ff1
+ 1 + . . .+ 2ffk-ff2
+ 1
because the first row counts the multiples of 2 less than or equal to l, the
second row counts the multiples of 4, the third row counts multiples of 8 and
so on. Adding by columns we obtain
2(l!) = 2ff1- 1 + 2ff2- 1 + . .+.2ffk- 1 = l - k
5
which implies that 23l+ 2(l!)= 23l+l-ff(l)= 24l-ff(l), as required.
2.7. Bases continued
2-9i-1u2
Consider the elements c4k = ki=1v_______9k-9i-1, introduced in x2.2, for a*
* par-
ticular k = 1, 2, . ...For completeness write c0 = 1 so that c4k 2 Q2[u=2, v2=4*
*].
Since the degree of the numerator of c4k is 2k, Proposition 2.6 implies that
v2 - 9i-1u2
f4k = 24k-ff(k)-2kc4k = 22k-ff(k) ki=1___________
9k - 9i-1
lies in Z2[u=2, v2=4] but 24k-ff(k)-2k-1c4k =2Z2[u=2, v2=4]. Similarly (u=2)f4k*
* =
24k-ff(k)-2k-1uc4k 2 Z2[u=2, v2=4] but 24k-ff(k)-2k-2uc4k =2Z2[u=2, v2=4] and so
on. This process is the "rationalisation yoga" referred to in x2.2. One forms
ujc4k and then multiplies by the smallest positive power of 2 to obtain an
element of Z2[u=2, v2=4].
By Proposition 2.6, starting with f4l= 24l-ff(l)-2lc4lthis process produces
the following set of elements of Z2[u=2, v2=4]
f4l, (u=2)f4l, (u=2)2f4l, . . ., (u=2)2l-ff(l)f4l,
u(u=2)2l-ff(l)f4l, u2(u=2)2l-ff(l)f4l, u3(u=2)2l-ff(l)f4l, . ...
As explained in ([1] p.352 et seq), the Hurewicz homorphism defines an
injection of graded groups of the form
ss*(bu_^_bo)___Z2_ 2
-! Q2[u=2, v =4]
Torsion
which, by the main theorem of [3], induces an isomorphism between ss*(bu^bo)_Z2*
*Torsion
and the free graded Z2-module whose basis consists of the elements of
Z2[u=2, v2=4] listed above for l = 0, 1, 2, 3, . ...
From this list we shall be particularly interested in the elements whose
degree is a multiple of 4. Therefore denote by g4m,4l2 Z2[u=2, v2=4] for l m
the element produced from f4lin degree 4m. Hence, for m l, g4m,4lis given
by the formula
8
>< u2m-4l+ff(l)u2l-ff(l)f4l_if4l - ff(l) 2m,
22l-ff(l)
g4m,4l=
>: u2(m-l)f4l_
22(m-l) if 4l - ff(l) > 2m.
Lemma 2.8.
In the notation of x2.2, let denote the projection
ss*(bu ^ bo) Z2
: ________________~= G*,*-! G*,k= m Gm,k.
Torsion
Then (g4k,4i) = 0 for all i < k.
6
Proof
Since Gm,k is torsion free it suffices to show that (g4k,4i) vanishes in
G*,k Q2. When i < k, by definition
ss4i(bu ^ bo) Z2 ss4k(bu ^ bo) Z2
g4k,4i2 u2k-2i_________________ Q2 _________________ Q2.
Torsion Torsion
However projects onto s sss(bu^F4k=F4k-1)_Z2Torsionand commutes with multi-
plication by u so the result follows from the fact that the homotopy of
bu ^ F4k=F4k-1 is trivial in degrees less than 4k (see [7] x3).
2.9. Recall from x2.2 that G4k,k~=Z2 for k = 0, 1, 2, 3, . .s.o we may choose
a generator z4k for this group as a module over the 2-adic integers (with the
convention that z0 = f0 = 1). Let "z4kbe any choice of an element in the
2-adic homotopy group "G4k,k~=G4k,k W4k,kwhose first coordinate is z4k.
Lemma 2.10.
Let B denote the exterior subalgebra of Z=2 Steenrod algebra generated by
Sq1 and Sq0,1. In the collapsed Adams spectral sequence (see [1] or [7])
Es,t2~=Exts,tB(H*(F4k=F4k-1; Z=2), Z=2)
=) sst-s(bu ^ (F4k=F4k-1)) Z2
the homotopy class "z4kis represented either in E0,4k2or E1.4k+12.
Proof
Recall from x2.2 that ss4k(bu ^ (F4k=F4k-1)) Z2 = G"4k,k~=Z2 W4k,k.
The following behaviour of the filtration coming from the spectral sequence
is well-known, being explained in [1]. The group "G4k,khas a filtration
. . .F i . .F.2 F 1 F 0= "G4k,k
with F i=F i+1~=Ei,4k+i2and 2F i F i+1. Also 2 . W4k,k= 0, every non-trivial
element of W4k,kbeing represented in E0,4k2. Furthermore for i = 1, 2, 3, . . .
we have 2F i= F i+1and F 1~=Z2.
Now suppose that "z4kis represented in Ej,4k+j2for j 2 then "z4k2
F j. From the multiplicative structure of the spectral sequence there exists a
generator ^z4kof F 1such that 2j^z4kgenerates F j+1and therefore 2jflz^4k= 2"z4k
for some 2-adic integer fl. Hence 2(2j-1flz^4k-z"4k) = 0 and so 2j-1flz^4k-z"4k2
W4k,kwhich implies the contradiction that the generator z4k is divisible by 2
in G4k,k.
Theorem 2.11.
7
In the notation of x2.7 and x2.9
ss4k(bu ^ bo) Z2
z4k = ki=02fi(k,i)~4k,4ig4k,4i2 _________________
Torsion
with ~s,t2 Z2, ~4k,4k2 Z*2and
8
< 4(k - i) - ff(k) + ff(i)if4i - ff(i) > 2k,
fi(k, i) =
: 2k - ff(k) if4i - ff(i) 2k.
Proof
From [3], as explained in x2.7, a Z2-module basis for G4k,*is given by
{g4k,4l}0 l k. Hence there is a relation of the form
z4k = ~4k,4kg4k,4k+ "~4k,4(k-1)g4k,4(k-1)+ . .+."~4k,0g4k,0
where "~4k,4iand ~4k,4k are 2-adic integers. Applying the projection
: G4k,*-! G4k,kwe see that z4k = (z4k) = ~4k,4k (g4k,4k), by Lemma 2.8.
Hence, if ~4k,4kis not a 2-adic unit, then z4k would be divisible by 2 in G4k,k
and this is impossible since z4k is a generator, by definition.
Multiplying the relation
z4k = ~4k,4k (g4k,4k) = ~4k,4k (f4k) 2 G4k,k.
by (u=2)2k-ff(k)we obtain (u=2)2k-ff(k)z4k = ~4k,4k ((u=2)2k-ff(k)f4k), which
lies in G8k-2ff(k),k, by the discussion of x2.7. Therefore, in G8k-2ff(k),k Q2*
* we
have the relation
k-1X
(u=2)2k-ff(k)z4k = (u=2)2k-ff(k)f4k+ "~4k,4i(u=2)2k-ff(k)g4k,4i.
i=0
Since the left hand side of the equation lies in G8k-2ff(k),k, the Q2 coefficie*
*nts
must all be 2-adic integers once we re-write the right hand side in terms of
the basis of x2.7.
For i = 0, 1, . .,.k - 1
8 u2k-ff(k)+2k-4i+ff(i)+2i-ff(i)
< _____________________22k-ff(k)+2i-ff(i)f4iif4i - ff*
*(i) 2k,
(u=2)2k-ff(k)g4k,4i=
: u2k-ff(k)+2k-2i_
22k-ff(k)+2k-2if4i if4i - ff(i) > 2k
8 u4k-2i-ff(k)
< ____________22k+2i-ff(k)-ff(i)f4iif4i - ff(i) 2k,
=
: u4k-2i-ff(k)_
24k-2i-ff(k)f4i if 4i - ff(i) > 2k.
8
Now we shall write (u=2)2k-ff(k)g4k,4ias a power of 2 times a generator derived
from f4i in x2.7 (since we did not define any generators called g4k+2,4ithe
generator in question will be g8k-2ff(k),4ionly when ff(k) is even).
Assume that 4i - ff(i) 2k so that 2i - ff(i) 4k - 2i - ff(k) and
__u4k-2i-ff(k)_ 1 4k-4i-ff(k)+ff(i) 2i-ff(i)
f4i= ________u (u=2) f4i
22k+2i-ff(k)-ff(i) 22k-ff(k)
which implies that "~4k,4iis divisible by 22k-ff(k)in the 2-adic integers, as
required.
Finally assume that 4i-ff(i) > 2k. We have 2i-ff(i) 4k-2i-ff(k) also.
To see this observe that ff(i) + ff(k - i) - ff(k) 0 because, by Proposition
k
2.6, this equals the 2-adic valuation of the binomial coefficient (i). Therefore
ff(k) - ff(i) ff(k - i) k - i < 4(k - i).
Then, as before,
u4k-2i-ff(k)_ 1 4k-4i-ff(k)+ff(i) 2i-ff(i)
f4i= ______________u (u=2) f4i
24k-2i-ff(k) 24k-4i-ff(k)+ff(i)
which implies that "~4k,4iis divisible by 24k-4i-ff(k)+ff(i)in the 2-adic integ*
*ers,
as required.
Theorem 2.12.
(i) In the collapsed Adams spectral sequence and the notation of Lemma
2.10 "z4kmay be chosen to be represented in E0,4k2.
(ii) In fact, "z4kmay be taken to be the smash product of the unit j of the
bu-spectrum with the inclusion of the bottom cell jk into F4k=F4k-1
j^jk
S0 ^ S4k -! bu ^ F4k=F4k-1.
Proof
For part (i), suppose that "z4kis represented in E1,4k+12. By Lemma 2.10
we must show that this leads to a contradiction. From [7] we know that on the
s = 1 line the non-trivial groups are precisely E1,4k+12, E1,4k+32, . .,.E1,8k+*
*2-2ff(k)2
which are all of order two. From the multiplicative structure of the spectral
sequence, if a homotopy class w is represented Ej,4k+2j-12and Ej,4k+2j+12is
non-zero then there is a homotopy class w0 represented in Ej,4k+2j+12such
that 2w0 = uw. Applied to "z4kthis implies that the homotopy element
u2k-ff(k)+1"z4kis divisible by 22k-ff(k)+1. Hence u2k-ff(k)+1z4k is divisible *
*by
22k-ff(k)+1in G*,*, which contradicts the proof of Theorem 2.11.
For part (ii) consider the Adams spectral sequence
Es,t2= Exts,tB(H*(F4k=F4k-1; Z=2), Z=2) =) sst-s(bu ^ F4k=F4k-1) Z2.
9
We have an isomorphism
Ht(F4k=F4k-1; Z=2)
E0,t2= Hom(____________________________________________________, Z=2).
Sq1Ht-1(F4k=F4k-1; Z=2) + Sq0,1Ht-3(F4k=F4k-1; Z=2)
The discussion of the homology groups H*(F4k=F4k-1; Z=2) given in ([1] p.341;
see also x3.1) shows that E0,4k2~=Z=2 generated by the Hurewicz image of
j ^ jk. Therefore the generator of E0,4k2represents j ^ jk. Since there is only
one non-zero element in E0,4k2it must also represent "z4k, by part (i), which
completes the proof.
3 The Matrix
3.1. Consider the left-bu-module spectrum map of x2.1
'k,l: bu ^ (F4k=F4k-1) -! bu ^ (F4l=F4l-1)
when l > k. This map is determined up to homotopy by its restriction, via
the unit of bu, to (F4k=F4k-1). By S-duality this restriction is equivalent to a
map of the form
S0 -! D(F4k=F4k-1) ^ bu ^ (F4l=F4l-1),
which DX denotes the S-dual of X. Maps of this form are studied by means
of the (collapsed) Adams spectral sequence (see [7] x3.1), where B is as in
Lemma 2.10,
Es,t2= Exts,tB(H*(D(F4k=F4k-1); Z=2) H*(F4l=F4l-1; Z=2), Z=2)
=) sst-s(D(F4k=F4k-1) ^ (F4l=F4l-1) ^ bu) Z2.
Recall from ([1] p.332) that a is the (invertible) B-module given by
Z=2 in degree a, -a = Hom( a, Z=2) and I is the augmentation ideal,
I = ker(ffl : B - ! Z=2). Hence, if b > 0, I-b = Hom(Ib, Z=2), where Ib is
the b-fold tensor product of I. These duality identifications may be verified
using the criteria of ([1] p.334 Theorem 16.3) for identifying aIb.
In ([1] p.341) it is shown that the B-module given by
H-* (D(F4k=F4k-1); Z=2) ~=H*(F4k=F4k-1; Z=2)
r-1+1 2r-1-1 r
is stably equivalent to 2 I when 0 < 4k = 2 . Therefore
r-1+1)1-2r-1
H*(D(F4k=F4k-1); Z=2) is stably equivalent to -(2 I when 0 <
4k = 2r. If k is not a power of two we may write 4k = 2r1+ 2r2+ . .+.2rt
with 2 r1 < r2 < . .<.rt. In this case
H*(F4k=F4k-1; Z=2) ~= rtj=r1H*(F2j=F2j-1; Z=2)
10
which is stably equivalent to 2k+ff(k)I2k-ff(k), where ff(k) equals the num-
ber of 1's in the dyadic expansion of k, as in Proposition 2.6. Similarly,
H*(D(F4k=F4k-1); Z=2) is stably equivalent to -2k-ff(k)Iff(k)-2k. From this,
for all s > 0, one easily deduces a canonical isomorphism ([7]p.1267) of the
form
Es,t2 ~=Exts,tB( 2l-2k+ff(l)-ff(k)I2l-2k-ff(l)+ff(k), Z=2)
~=Exts+2l-2k-ff(l)+ff(k),t-2l+2k-ff(l)+ff(k)B(Z=2, Z=2).
Also there is an algebra isomorphism of the form Ext*,*B(Z=2, Z=2) ~=Z=2[a, b]
where a 2 Ext1,1B, b 2 Ext1,3B. As explained in ([7] p.1270) ik,lis represented
in
E4(k-l)+ff(l)-ff(k),4(k-l)+ff(l)-ff(k)2~=Ext2k-2l,6k-6l)B(Z=2, Z=2) ~=Z=2 = **.
Proposition 3.2.
For l < k, in the notation of x2, the homomorphism
('k,l)* : G4k,k-! G4k,l
satisfies ('k,l)*(z4k) = ~4k,4l22k-2l-ff(k)+ff(l)u2k-2lz4lfor some 2-adic unit *
*~4k,4l.
Proof
Let "z4k2 "G4k,kbe as in x2.9 so that, proved in a similar manner to Lemma
2.10, 2"z4kis represented in E1,4k+12in the spectral sequence
Es,t2= Exts,tB(H*(F4k=F4k-1; Z=2), Z=2)
=) sst-s(bu ^ (F4k=F4k-1)) Z2.
where, from x3.1, we have
E1,4k+12~=Ext1+2k-ff(k),4k+1-2k-ff(k)B(Z=2, Z=2) ~=Z=2 = .
The multiplicative pairing between these spectral sequences shows that
('k,l)*(2"z4k) 2 "G4k,lis represented in the spectral sequence
Es,t2= Exts,tB(H*(F4l=F4l-1; Z=2), Z=2)
=) sst-s(bu ^ (F4l=F4l-1)) Z2.
by the generator of E1+4k-4l-ff(k)+ff(l),1+8k-4l-ff(k)+ff(l)2because a2k+1-ff(k*
*)b2k-2l
is the generator of
E1+4k-4l-ff(k)+ff(l),1+8k-4l-ff(k)+ff(l)2~=Ext1+4k-2l-ff(k),1+8k-6l-ff(k)B(Z=*
*2, Z=2).
11
Since multiplication by a and b in the spectral sequence corresponds to
multiplication by 2 and u respectively on homotopy groups we have the fol-
lowing table of representatives in ss*(bu ^ (F4l=F4l-1)) Z2.
_homotopy_element___||representative||dimension___
________2z4l________||_a2l-ff(l)+1_||_____4l______
_____(u=2)(2z4l)_____||_a2l-ff(l)b____||4l_+_2____
____(u=2)2(2z4l)____||a2l-ff(l)-1b2___||4l_+_4____
.. . .
__________._________||_____..______||_____..______
__(u=2)2l-ff(l)(2z4l)__||ab2l-ff(l)_||8l_-_2ff(l)_
__u(u=2)2l-ff(l)(2z4l)_b||2l-ff(l)+1||8l_-_2ff(l)_+ 2
_u2(u=2)2l-ff(l)(2z4l)_||b2l-ff(l)+28||l_-_2ff(l)_+ 4
.. . .
__________._________||_____..______||_____..______
Therefore there are two cases for ('k,l)*(2"z4k). If 2k-2l 2l-ff(l)+1 then
b2k-2lrepresents u2k-2l-(2l-ff(l))(u=2)2l-ff(l)"z4l= u2k-4l+ff(l))(u=2)2l-ff(l)*
*"z4land,
up to multiplication by 2-adic units, ('k,l)*(2"z4k) is equal *
* to
21+2k-ff(k)u2k-4l+ff(l))(u=2)2l-ff(l)"z4l, as required. On the other han*
*d, if
2k - 2l 2l - ff(l) then a2l-ff(l)+1-(2k-2l)b2k-2l= a4l-2k-ff(l)+1b2k-2lrepres*
*ents
(u=2)2k-2l(2"z4l) which shows that, up to 2-adic units, ('k,l)*(2"z4k) is equal*
* to
21+2k-ff(k)-(4l-2k-ff(l)+1)(u=2)2k-2l(2"z4l) = 24k-ff(k)-4l+ff(l)(u=2)2k-2l(2"z*
*4l), as re-
quired.
Proposition 3.3.
Let _3 : bo -! bo denote the Adams operation, as usual. Then, in the
notation of x2.7,
8
>> 9kg4k,4k+ 9k-12 2(k)+3g4k,4k-4ifk 3,
>>
>>
>> 2 3
< 9 g8,8+ 9 . 2 g8,4 ifk = 2,
(1 ^ _3)*(g4k,4k) =
>>
>> 9g4,4+ 2g4,0 ifk = 1,
>>
>>
:
g0,0 ifk = 0.
12
Proof
The map (1^_3)* fixes u, multiplies v by 9 and is multiplicative. Therefore
2 i-1 2
(1 ^ _3)*(c4k) = ki=19v_-9___u_9k-9i-1
(9v2-9ku2+9ku2-u2) k (v2-9i-2u2)
= 9k-1 ___________________i=2______ k k i-1
i=1(9 -9 )
(9v2-9ku2) k (v2-9i-2u2) (9ku2-u2) k (v2-9i-2u2)
= 9k-1 ___________i=2_______ k+ 9kk-1i-1_______i=2_______kki-1
i=1(9 -9 ) i=1(9 -9 )
(v2-9k-1u2) k (v2-9i-2u2) (9ku2-u2) k (v2-9i-2u2)
= 9k ____________i=2_______ k+ 9kk-1i-1_______i=2_______kki-1
i=1(9 -9 ) i=1(9 -9 )
u2 k-1(v2-9i-1u2)
= 9kc4k+ 9k-1(9k - 1) ____i=1_________(9k-1) k-1 k i-1
i=1(9 -9 )
= 9kc4k+ 9k-1u2c4k-4.
Hence, for k 1, we have
(1 ^ _3)*(f4k) = 22k-ff(k)(1 ^ _3)*(c4k)
= 22k-ff(k)9kc4k+ 9k-1u222k-ff(k)-2k+2+ff(k-1)+2k-2-ff(k-1)c4k-4
= 9kf4k+ 9k-1u222-ff(k)+ff(k-1)f4k-4
= 9kf4k+ 9k-1u22 2(k)+1f4k-4,
which yields the result, by the formulae of x2.7.
Proposition 3.4.
When k > l
8 l l-1
>> 9 g4k,4l+ 9 g4k,4l-4 if4l - ff(l) 2k,
>>
>>
>> l l-1 4l-ff(l)-2k
< 9 g4k,4l+ 9 2 g4k,4l-4if4l - ff(l) - 2(l) - 3
(1^_3)*(g4k,4l) = 2k < 4l - ff(l),
>>
>>
>> l l-1 3+ (k)
>> 9 g4k,4l+ 9 2 2 g4k,4l-4 if2k < 4l - ff(l) - 2(l) - 3
:
< 4l - ff(l)
13
Proof
Suppose that 4l - ff(l) 2k then, by Proposition 3.3 (proof),
(1 ^ _3)*(g4k,4l)
u2l-ff(l)f
= (1 ^ _3)*(u2k-4l+ff(l)_______224ll-ff(l))
u2l-ff(l)(9lf +9l-1u22 2(l)+1f )
= u2k-4l+ff(l)____________4l___________22l-4l-4ff(l)
u2l-ff(l)u22 2(l)+1f
= 9lg4k,4l+ 9l-1u2k-4l+ff(l)________________22l4l-4-ff(l)
u2l+2-ff(l)2 2(l)+1f
= 9lg4k,4l+ 9l-1u2k-4l+ff(l)________________22l4l-4-ff(l).
Then, since 2(l) = 1 + ff(l - 1) - ff(l),
4(l - 1) - ff(l - 1) = 4l - ff(l) + ff(l) - ff(l - 1) - 4 = 4l - ff(l) - 3 - 2*
*(l) < 2k
so that 2l-2-ff(l-1)
g4k,4l-4 = u2k-4l+4+ff(l-1)u________f4l-4_22l-2-ff(l-1)
u2l+2-ff(l)f
= u2k-4l+ff(l)_____________4l-4_22l-2-ff(l)+ff(l)-ff(*
*l-1)
u2l+2-ff(l)f
= u2k-4l+ff(l)___________4l-422l-ff(l)- 2(l)-1
so that, for 0 < l < k suppose that 4l - ff(l) 2k,
(1 ^ _3)*(g4k,4l) = 9lg4k,4l+ 9l-1g4k,4l-4.
Similarly, for 0 < l < k if 4l - ff(l) > 2k then, by Proposition 3.3 (proof*
*),
(1 ^ _3)*(g4k,4l)
2(k-l)f4l
= (1 ^ _3)*( u______22(k-l))
2(k-l)(9lf4l+9l-1u22 2(k)+1f4k-4)
= u________________________22(k-l)
u2k-2l+22 2(k)+1f
= 9lg4k,4l+ 9l-1 _________________4k-422(k-l).
This situation splits into two cases given by
(i) 4l - ff(l) - 2(l) - 3 2k < 4l - ff(l) or
(ii) 2k < 4l - ff(l) - 2(l) - 3 < 4l - ff(l).
14
In case (i) 4l - 4 - ff(l - 1) = 4l - ff(l) - 2(l) - 3 2k and so again we
have 2l-2-ff(l-1)
g4k,4l-4 = u2k-4l+4+ff(l-1)u________f4l-4_22l-2-ff(l-1)
2k-2l+2f4l-4
= _u__________22l-1- 2(l)-ff(l)
2k-2l+221+ 2(l)f4l-4
= u_______________22k-2l+4l-ff(l)-2k
so that
(1 ^ _3)*(g4k,4l) = 9lg4k,4l+ 9l-124l-ff(l)-2kg4k,4l-4.
In case (ii)
u2k-2l+2f
g4k,4l-4 = _________4k-4_22(k-l+2)
so that
(1 ^ _3)*(g4k,4l) = 9lg4k,4l+ 9l-123+ 2(k)g4k,4l-4.
3.5. In the notation of x2.1, suppose that A 2 U1 Z2 satisfies
(A-1) = [1 ^ _3] 2 Aut 0left-bu-mod(bu ^ bo).
Therefore, by definition of and the formula of Theorem 2.11
P 3
l kAl,k('k,l)*(z4k)= (1 ^ _ )*(z4k)
= ki=02fi(k,i)~4k,4i(1 ^ _3)*(g4k,4i).
On the other hand
P
l kAl,k('k,l)*(z4k)
P
= Ak,kz4k+ l< u2k-2lu2l-4i+ff(i)u_____f4iif4i - ff(i) 2l,
22i-ff(i)
u2k-2lg4l,4i=
>: u2(l-i)f4i
u2k-2l _______22(l-i) if4i - ff(i) > 2l.
8 2k-2i
>< u____f4i_if4i - ff(i) 2l,
22i-ff(i)
=
>: u2k-2if4i_
22l-2i if4i - ff(i) > 2l
15
while 8
>< u2k-4i+ff(i)u2i-ff(i)f4i_if4i - ff(i) 2k,
22i-ff(i)
g4k,4i=
>: u2(k-i)f4i_
22(k-i) if4i - ff(i) > 2k.
From these formulae we find that
8
>> g4k,4i if4i - ff(i) 2l 2k,
>>
<
u2k-2lg4l,4i= 24i-ff(i)-2lg4k,4iif2l < 4i - ff(i) 2k,
>>
>>
: 22k-2lg
4k,4i if2l < 2k < 4i - ff(i).
Now let us calculate Al,k.
When k = 0 we have z0 = (1 ^ _3)*(z0) = A0,0('0,0)*(z0) = A0,0z0 so that
A0,0= 1.
When k = 1 we have
P 3
l 1Al,1('1,l)*(z4)= (1 ^ _ )*(z4)
= ~4,4(1 ^ _3)*(g4,4) + 2~4,0(1 ^ _3)*(g4,0)
= ~4,4(9g4,4+ 2g4,0) + 2~4,0g4,0
and
P
l 1Al,k('1,l)*(z4)= A1,1z4 + A0,1~1,02g4,0
= A1,1(2~4,0g4,0+ ~4,4g4,4) + A0,1~1,02g4,0
which implies that A1,1= 9 and A0,1= ~-11,0(~4,4- 8~4,0) so that A0,12 Z*2.
When k = 2 we have
P 3
l 2Al,2('2,l)*(z8)= (1 ^ _ )*(z8)
= (1 ^ _3)*(~8,8g8,8+ 23~8,4g8,4+ 23~8,0g8,0)
= ~8,8(92g8,8+ 9 . 23g8,4) + 23~8,4(9g8,4+ g8,0) + 23~8,0g*
*8,0
16
and
P
l 2Al,2('2,l)*(z8)= A2,2z8 + A1,2('2,1)*(z8) + A0,2('2,0)*(z8)
= A2,2(~8,8g8,8+ 23~8,4g8,4+ 23~8,0g8,0)
+A1,2(~8,422u2z4) + A0,2(~8,023u4z0)
= A2,2(~8,8g8,8+ 23~8,4g8,4+ 23~8,0g8,0)
+A1,2~8,422(2~4,0g8,0+ ~4,4u2g4,4) + A0,2~8,023g8,0
= A2,2(~8,8g8,8+ 23~8,4g8,4+ 23~8,0g8,0)
+A1,2~8,422(2~4,0g8,0+ ~4,42g8,4) + A0,2~8,023g8,0.
Therefore we obtain
~8,8(92g8,8+ 9 . 23g8,4) + 23~8,4(9g8,4+ g8,0) + 23~8,0g8,0
= A2,2(~8,8g8,8+ 23~8,4g8,4+ 23~8,0g8,0)
+A1,2~8,422(2~4,0g8,0+ ~4,42g8,4) + A0,2~8,023g8,0
which yields
92 = A2,2,
~8,8. 9 + ~8,4(9 - 92) = A1,2~8,4~4,4,
~8,4+ ~8,0(1 - 92) = A1,2~8,4~4,0+ A0,2~8,0.
Hence A1,22 Z*2.
Now assume that k 3 and consider the relation derived above
ki=02fi(k,i)~4k,4i(1 ^ _3)*(g4k,4i)
= Ak,k ki=02fi(k,i)~4k,4ig4k,4i
P
+ l n
and the other entries are given by the formula
X
(Xn)s,s+t= A(k1)A(k2) . .A.(kt)
1 k1**