A CONJECTURE ON THE UNSTABLE ADAMS SPECTRAL
SEQUENCES FOR SO AND U
KATHRYN LESH
Abstract.In this paper we give a systematic account of a conjecture sug
gested by Mark Mahowald on the unstable Adams spectral sequences for the
groups SO and U. The conjecture is related to a conjecture of Bousfield o*
*n a
splitting of the E2term and to an algebraic spectral sequence constructe*
*d by
Bousfield and Davis. In this paper, we construct and realize topologicall*
*y a
chain complex which is conjectured to contain in its differential the str*
*ucture
of the unstable Adams spectral sequence for SO. A filtration of this chain
complex gives rise to a spectral sequence that is conjectured to be the u*
*n
stable Adams spectral sequence for SO. If the conjecture is correct, then*
* it
means that the entire unstable Adams spectral sequence for SO is available
from a primary level calculation. We predict the unstable Adams filtratio*
*n of
the homotopy elements of SO based on the conjecture, and we give an exam
ple of how the chain complex predicts the differentials of the unstable A*
*dams
spectral sequence. Our results are also applicable to the analogous situa*
*tion
for the group U.
1.Introduction
In this paper, we consider the unstable Adams spectral sequence (UASS) of the
group SO at the prime 2. In particular, we give a systematic account of a conje*
*cture
suggested by Mark Mahowald concerning the calculation of the differentials in t*
*his
spectral sequence. We give a geometric realization of the conjecture in the for*
*m of a
tower with the 2completion of SO as inverse limit. Our tower comes equipped wi*
*th
a map from the destabilization of the stable Adams tower for the infinite deloo*
*ping
of SO. We use this map and theorems of Bousfield on h0towers in unstable Ext
to predict the Adams filtrations of the unstable homotopy of SO. Our results are
equally valid for the group U, and thus differentials and unstable filtrations *
*can be
predicted for this group as well. We note that, of course, the homotopy of SO a*
*nd
U is well known by Bott periodicity, and that what is of interest is the workin*
*gs of
the UASS, not the end result.
Before we describe our results and conjectures, we establish some notation. We
work entirely at the prime 2, all cohomology will be taken with mod 2 coefficie*
*nts,
and all spaces will be taken to be completed at 2 as appropriate. Let A be the *
*mod
2 Steenrod algebra, let U_be the category of unstable Amodules, and let K_be
the category of unstable Aalgebras. There is a functor U : U_! K_, described by
Massey and Peterson [MP ], which takes the free unstable Aalgebra on an unsta*
*ble
Amodule. This functor is left adjoint to the forgetful functor from unstable A
algebras to unstable Amodules.
___________
Date: July 22, 2001.
1991 Mathematics Subject Classification. 55T15, 55Q52, 55U99.
1
2 KATHRYN LESH
In general, the unstable Adams spectral sequence for a space X has the form
Es,t2= ExtsK_(H*X , H*St) ) ßtsX,
where Ext is a derived functor in the nonabelian category K_. However, for a sp*
*ace
X with the property that H*X ~=U(N) for some N 2 U_, the unstable Adams
spectral sequence has the form
Es,t2= ExtsU_(N, tF2) ) ßtsX.
We will follow the stable notation and write Exts,tU_(N, F2) for ExtsU_(N, tF2*
*).
We will be discussing the unstable Adams spectral sequence for the special or
thogonal group SO and indicating_modifications*to be made to the discussion for
the unitary group U. Let M1 = H RP1 , with nonzero elements xiin dimension i
and Aaction Sqjxi= ijxi+j; then H*SO ~=U(M1 ). Hence the unstable Adams
spectral sequence for SO takes the form
Exts,tU_(M1 , F2) ) ßtsSO.
Let ff(i) be the number of ones in the dyadic expansion of i, and filter M1 by
Mn = {xiff(i) n}. This filtration leads to a spectral sequence converging to*
* the
E2term of the UASS:
Ext*,*U_(Mn=Mn1 , F2) ) Ext*,*U_(M1 , F2).
It is a conjecture of Bousfield from the 1970s that this spectral sequence coll*
*apses,
giving
Ext*,*U_(M1 , F2) ~= n Ext*,*U_(Mn=Mn1 , F2).
A similar conjecture for the_E2term*of the UASS for the group U arises from the
fact that if we take M1 = H CP1+, then H*U ~=U(M1 ), and in this case also,
M1 can be filtered by dyadic expansion of the dimension of the elements.
In this paper, we use the destabilization of the stable Adams resolution of t*
*he
connective so spectrum to construct a chain complex whose constituent parts are
minimal resolutions of the filtration quotients Mn=Mn1. When realized topo
logically using the MasseyPeterson theorem [MP ], this chain complex gives a
tower of spaces whose inverse limit is SO (2completed), and whose homotopy
spectral sequence collapses at E2. The E1term of the homotopy spectral sequence
is Ext*,*U_(Mn=Mn1 , F2), a very large vector space, while E2 = E1 is the as*
*soci
ated graded to ß*SO, a rather small vector space (ßiSO ~=Z for i 3 mod 4, and
Z=2 for i 0 or 1 mod 8). Hence the spectral sequence has a very complicated d*
*1,
which is, however, completely calculated by the calculation of the chain comple*
*x, a
primary level calculation. The conjecture suggested by Mahowald (Conjecture 5.1)
is that in a certain precise sense, this d1 differential contains all the diffe*
*rentials in
the UASS. Because the tower comes equipped with a map from a modified Post
nikov tower for SO, it is possible to use theorems of Bousfield on unstable Ext
to predict where the homotopy of SO is represented, and this, in turn, allows a
prediction of the unstable Adams filtration of those elements. It is the hope *
*of
the author that in the future it will be possible to manipulate this tower by an
elaboration of methods of [Lesh] to prove Conjecture 5.1. Extensive knowledge of
differentials in the UASS for SO would allow the computation of differentials in
other unstable Adams spectral sequences by naturality. For example, it should be
possible to recover a form of Hopf invariant one from the model's calculation o*
*f the
UASS for SO.
UASS FOR SO AND U 3
The splitting conjecture of Bousfield was discussed and an algebraic model for
the UASS for U and SO constructed in [BD]. However, the model was considered
strictly on an algebraic level and was not realized topologically. Although the
author believes that the spectral sequence of [BD] is the same as that of the *
*current
work, the advantages of the model described here seem to be the following. Firs*
*t,
the construction of the model is essentially formal, and very similar to a stan*
*dard
construction of homological algebra to obtain the spectral sequence converging *
*to
the derived functors of a composite functor. All of the differentials can be ca*
*lculated
by a primary level calculation that is a strictly mechanical process. Second, t*
*he
model comes equipped with a topological realization. It seems that in order to
prove that the model actually does give the UASS, it will be necessary to have *
*such
a realization.
The rest of this paper is organized as follows. In Section 2, we give some ba*
*ck
ground on the stable and destabilized Postnikov towers of so, as well as some
algebraic preliminaries. In Section 3, we construct a tower of spaces and an a*
*s
sociated chain complex that models the UASS for SO. In Section 4, we study
the homotopical properties of the tower. Finally, in Section 5 we use theorems *
*of
Bousfield to predict the unstable Adams filtration of elements of ß*SO, we give
a counterexample to a conjecture of [BD], we draw some conclusions about what
may be necessary to prove Conjecture 5.1, and we give an example of a different*
*ial
in the UASS that is predicted by our methods.
2.Preliminaries
In this section, we review algebraic properties of the category of unstable A
modules, we recall the MasseyPeterson theorem, and we consider the cohomology
of the stages of the destabilized Adams tower of so.
We begin by reviewing properties of the algebraic looping functor : U_! U_
and its iterates. (See also [MP ].) The functor : U_! U_is the left adjoint *
*to the
suspension functor : U_! U_. Given an unstable Amodule M, the module M
can be calculated as the largest unstable quotient of the desuspension of M:
M ( 1M)=( 1Sq0M),
where Sq0x = Sqxx. The functor is not exact, but it can have at most one
nonzero derived functor, which we denote 11. The module 11M can be expressed
as a regrading of the kernel of Sq0 on M. In particular, if Sq0 acts freely on *
*M,
then 11M = 0. We write n for the nfold iterate of , and we write njfor the*
* jth
derived functor of n. There is a composite functor spectral sequence (the Sing*
*er
spectral sequence) si tjM ) s+ti+jM which allows us to calculate derived func*
*tors
of n inductively. For any unstable module M, njM = 0 for j > k.
We will also need the following routine lemma.
Lemma 2.1. Let g : N1 ! N2 be a map of unstable Amodules. If im(g) is Sq0
free, then the natural map ker(g) ! ker( g) is a monomorphism. If in addition
N2 is Sq0free, then there is a short exact sequence
0 ! ker(g) ! ker( g) ! 11cok(g) ! 0.
Proof.The map g factors as N1 ! im(g) ! N2, and since is right exact,
N1 ! im(g) is an epimorphism. Thus there is a short exact sequence
(2.1) 0 ! ker[ N1 ! im(g)] ! ker( g) ! ker[ im(g) ! N2] ! 0.
4 KATHRYN LESH
To calculate the lefthand term, observe that the short exact sequence
0 ! ker(g) ! N1 ! im(g) ! 0
gives rise to an exact sequence
11im(g) ! ker(g) ! N1 ! im(g) ! 0.
Thus if 11im(g) = 0, then ker[ N1 ! im(g)] ~= ker(g), proving that ker(g)
injects into ker( g).
Consider the righthand term of equation (2.1). The short exact sequence
0 ! im(g) ! N2 ! cok(g) ! 0
gives rise to a long exact sequence
0 ! 11im(g) ! 11N2 ! 11cok(g) ! im(g) ! N2 ! cok(g) ! 0.
If 11N2 = 0, then ker[ im(g) ! N2] ~= 11cok(g), concluding the proof of the
lemma.
Remark 2.2. Suppose that M is an unstable Amodule, that N1and N2are unstable
projective Amodules, and that we are given a map M ! ker(N1 ! N2). Then
we can consider the composition
M ! ker(N1 ! N2)
,! ker( N1 ! N2)
,! N1,
and so ker[M ! ker(N1 ! N2)] = ker(M ! N1). We will use this remark
repeatedly in Section 3.
Going in the opposite direction from looping, we define a "deloopingö n free
modules. Given a free unstable Amodule P, we write BP for the free unstable A
module whose generators are one dimension higher than those of P. Thus, BF(n) =
F(n + 1), and BP ~=P. Note that "delooping" is not a functor on U_, because
given a map g : P1 ! P0, there is no canonical choice of map Bg : BP1 ! BP0 with
Bg = g. In most cases where we will use this notation, P will itself be an ite*
*rated
looping, and BP will simply mean one fewer loops: P = iN and BP = i1N.
We remind the reader of the content of the MasseyPeterson theorem, which
we will need to use repeatedly. Essentially, this theorem says that under favo*
*r
able conditions, the Serre spectral sequence for a fibration behaves much like *
*the
long exact sequence in cohomology for a stable cofibration. Note that if we wri*
*te
F(n) for the free unstable Amodule on a single generator in dimension n, then
H*K(Z=2, n) ~=U(F(n)). Therefore, if P is a free unstable Amodule, we write
KP for the EilenbergMacLane space with H*KP ~=U(P).
Definition 2.3. We call a map X ! KP MasseyPeterson if the following hold.
(1)There is an unstable Amodule M with H*X ~=U(M)
(2)There is a map f : P ! M that induces the map on cohomology. That is,
H*KP ! H*X is U(f).
(3)im(H*KP ! H*X) is contained in a polynomial subalgebra of H*X.
(4)X is simple and of finite type.
We think of the topological map X ! KP as realizing f, and by abuse of
notation we call the topological map f as well. If Y is the homotopy fiber of a
MasseyPeterson map f : X ! KP, then the MasseyPeterson theorem says that
UASS FOR SO AND U 5
H*Y ~=U(N), where there is a short exact sequence (the fundamental sequence of
f)
0 ! cok(f) ! N ! ker(f) ! 0.
The short exact sequence does not, in general, split as Amodules, although U(N)
is split as an algebra as the tensor product of U(cok(f)) and U( ker(f)).
We begin our discussion of SO by describing the stable Postnikov tower of so,
which is very_close to its stable Adams resolution.1 We know H*so ~= A =A Sq3,
and letting A = A =A Sq1, the stable Postnikov tower of so realizes the acyclic
complex of stable Amodules
__ 11 9__ 4__
(2.2) . .!. 13A ! A ! A ! A ! A
where each_term_is_monogenic and the differentials_run cyclically through the l*
*ist
Sq2, Sq2, Sq3, Sq5. Only the fact that A is not projective keeps this chain com*
*plex
from being the Adams resolution. Next we destabilize the stable Postnikov tower
for the spectrum so by taking the zero space of the infinite loop spectrum at e*
*ach
level of the tower. We obtain the unstable Postnikov tower for SO, a tower of
spaces {Xn} (Figure 1) with very nice cohomological properties_summarized_in the
following lemma. (Recall that Mn is the nth filtration of M H*RP1 by dyadic
expansion.)
Lemma 2.4. [Long]
(1)holimnXn ' SO.
(2)kn is a MasseyPeterson map.
(3)ker(H*Xn ! H*Xn+1) = ker(H*Xn ! H*SO).
(4)im(H*Xn ! H*Xn+1) ~=im(H*Xn ! H*SO) ~=U(Mn).
However, we will be interested in the destabilization, not of the Postnikov t*
*ower
for so, but of the Adams tower. The only difference this introduces is that ins*
*tead
of having only one homotopy group in each dimension, we have to introduce the
copies of the integers one Z=2 at a time (building up the completion Z^2). To do
this, take a projective resolution of each term in (2.2), take the total comple*
*x, and
destabilize. The realization of this projective chain complex will have the for*
*m of
Figure 2. An exercise in homological algebra shows that the tower has the same
cohomological properties as those of the Postnikov tower which were summarized
in Lemma 2.4:
Lemma 2.5.
(1)holimnYn ' SO^2.
(2)There is an unstable Amodule Zn with H*Yn ~=U(Zn).
(3)ker(H*Yn ! H*Yn+1) = ker(H*Yn ! H*SO).
(4)im(H*Yn ! H*Yn+1) ~=im(H*Yn ! H*SO) ~=U(Mn).
Remark 2.6.
(1)For the reader interested in carrying out this calculation, we note that *
*the
issues are the same as those laid out in detail in the proofs of Proposit*
*ion 4.3
and Proposition 4.1.
___________
1An appropriate reference for the remainder of the section is [Long].
6 KATHRYN LESH
SO
??
y
..
.
K(F(8)) ! X4 ! K(F(10))
??
y
__
K(F (7))! X3 ! K(F(9))
??
y
__ __
K(F (3))! X2 ! K(F (8))
??
y
__
K(F(1)) ! X1 ! K(F (4))
??
y
* ! K(F(2))
Figure 1. The Postnikov tower for SO
SO^2
??
y
..
.
K(F(8) F(7) F(3))i4!Y4k4!K(F(10) F(8) F(4))
??
y
K(F(7) F(3))i3!Y3 k3!K(F(9) F(8) F(4))
??
y
K(F(3)) i2!Y2 k2! K(F(8) F(4))
??
y
K(F(1)) i1!Y1 k1! K(F(4))
??
y
* ! K(F(2))
Figure 2. The destabilized Adams tower for SO
UASS FOR SO AND U 7
(2)Let Pn be the unstable projective such that KPn is the homotopy fiber of
Yn ! Yn1. Thus P1 = F(1), P2 = F(3), P3 = F(7) F(3), etc. Then it
is a consequence of Lemma 2.5(4) that
__ker(BPn_!_Pn1)_~M =M .
im(BPn+1 ! Pn) = n n1
3.A chain complex model for the UASS
In this section, we use {Yn}, the destabilized Adams tower of so, to construc*
*t a
tower {En} that also has SO^2as its inverse limit, but that involves in its ki*
*nvariants
the unstable resolutions of the filtration quotients Mn=Mn1. The tower {En} wi*
*ll
come equipped with a map {Yn} ! {En}, which will allow us to calculate where
the homotopy of SO is represented in the homotopy spectral sequence of {En}.
This in turn will allow us in Section 5 to make predictions about unstable Adams
filtrations in the homotopy of SO.
We need a considerable amount of notation. Choose a minimal projective U_
resolution Dn*of Mn=Mn1. The tower we are going to build will have the form
En+1
#
K(Dn0 Dn11 . . . n1D1n1)E!n! KB(Dn+10 Dn1 . . . nD1n).
Note that Dn*will make its first appearance at the nth stage of the tower. Beca*
*use
the module Dniappears in the tower as iDni, we avoid excessive loops in our
notation by letting Cni= iDniand BCni= i1Dni. We write Ln = ni=1Cini,
and our tower will have the form
En+1
#
KLn ! En ! KBLn+1.
We define the following filtration, along with similar filtrations of BLn and *
*Ln:
FjLn = ni=jCini.
Thus Cn0= Fn F(n1) ... F1Ln = Ln.
The tower of spaces {En} that we construct in this section has the following
properties. Recall from Lemma 2.5 that Zn is the unstable Amodule such that
H*Yn ~=U(Zn), and from Remark 2.6 that Pn is the unstable projective such that
Yn is the homotopy fiber of a map Yn1 ! BPn.
(1)There exists an unstable Amodule Fn1 such that H*En1 ~=U(Fn1),
and En is the homotopy fiber of a MasseyPeterson map En1 ! KBLn.
(2)There are commuting diagrams of MasseyPeterson maps
KPn1 ! Yn1 ! KBPn
?? ? ?
y ?y ?y
KLn1 ! En1 ! KBLn
8 KATHRYN LESH
induced by commuting diagrams of unstable Amodules
BLn ! Fn1! Ln1
? ? ?
hn?y ?y hn1?y
BPn ! Zn1! Pn1.
(3)ker(BLn ! Fn1) = ker(BLn ! Ln1).
(4)cok(BLn ! Fn1) ! cok(BPn ! Zn1) is an isomorphism.
(5)Algebraic properties of the map fn described in detail below.
Property (3) is analogous to Lemma 2.5(3), and both say that the kinvariants do
not kill any cohomology that comes from lower down in the tower. Property (4) is
related to Lemma 2.5(4), and arranges for the towers {En} and {Yn} to give the
same filtration of H*SO.
To describe the last set of properties we recall from the MasseyPeterson the*
*orem
that if En1 is the fiber of a MasseyPeterson map En2 ! KBLn1, then the
fundamental sequence for En1 is
0 ! cok(BLn1 ! Fn2) ! Fn1 ! ker(BLn1 ! Fn2) ! 0,
where the righthand term is the contribution of the fiber, KLn1, to H*En1. The
next space, En, will be the fiber of a MasseyPeterson map En1 ! KBLn, and
our last requirement is on the composition of the kinvariants, KLn1 ! En1 !
KBLn. Let fn denote the composite BLn ! Fn1 ! ker(BLn1 ! Fn2). The
final requirement on the tower {En} is detailed below.
(5)fn has the following algebraic properties:
(a)fn is filtration preserving.
(b)For 1 j n, on Fj=F(j+1)the map E0(fn) is the map
BCjnj! ker(BCjnj1! Cjnj2).
that comes from looping down the differential in the resolution Dj*!
Mj=Mj1.
(c)E0(ker(fn)) ~=ker(E0(fn)).
We will use Remark 2.2 freely throughout this section. In particular, Remark *
*2.2
together with requirement (5) tell us that the associated graded of ker(fn) is
Fj=F(j+1)(ker(fn)) ~=ker(BCjnj! Cjnj1).
The construction of {En} is inductive. For the first stage we observe that P1*
* =
L1 = C10, and we define L1 ! P1 to be the identity map. Thus Y1 = KP1 = KL1 =
E1, and the theorem is certainly true in this case. Observe that P1 = Z1 = F1 =*
* L1.
At the next stage, L2 = C20 C11; we want a commuting diagram
BL2 ! L1 = F1 B(C20 C11)! C10
? ? ? ?
h2?y =?y i.e. h2?y = ?y
BP2 ! P1 = Z1. BP2 ! P1.
We define BC20! C10to be zero, and BC11! C10by the differential for C1*. The
composite BC11! C10= P1 ! cok(BP2 ! P1) ~=M1 is zero because BC11! C10!
M1 begins a resolution, and so the composite BC11! C10! P1 factors through
BP2. We use this factoring to define h2 : BL2 ! BP2 on the factor BC11. To defi*
*ne
UASS FOR SO AND U 9
h2 on the factor BC20, choose a class x2 2 ker(BP2 ! P1) that, when looped, giv*
*es
the generator of ker(BP2 ! P1)= im(BP3 ! P2) ~=M2=M1. This gives us the
desired commuting diagram above. Looking at the topological realization,
Y1 ! KBP2
?? ?
y ?y
E1 ! KBL2,
the properties required for E1 ! KBL2 are easily verified by inspection, and we
take homotopy fibers in the diagram to obtain the space E2together a map Y2 ! E2
and maps of fundamental sequences
0 ! cok(BL2 ! L1) ! F2 ! ker(BL2 ! L1) ! 0
# # #
0 ! cok(BP2 ! P1)! Z2 ! ker(BP2 ! P1) ! 0.
For an inductive hypothesis, we assume that for i n we have defined spaces
Ei and maps fi satisfying the required conditions, and we seek to define En+1.
Thus we have maps BPn+1 ! Zn and Fn ! Zn induced by Yn ! KBPn+1 and
Yn ! En, respectively. We need to define a commuting diagram
BLn+1 ! Fn
? ?
hn+1?y ?y
BPn+1 ! Zn
and verify that when we realize it by a diagram of spaces
Yn ! KBPn+1
?? ?
y ?y
En ! KBLn+1,
taking horizontal fibers gives rise to a space En+1 and a map Yn+1 ! En+1 that
satisfies the inductive hypotheses.
Consider the ladder of fundamental sequences for Yn and En:
0 ! cok(BLn ! Fn1) ! Fn ! ker(BLn ! Fn1) ! 0
# # #
0 ! cok(BPn ! Zn1) ! Zn ! ker(BPn ! Zn1) ! 0.
We know by Lemma 2.5 that ker(BPn ! Zn1) = ker(BPn ! Pn1), and by
the inductive hypothesis, ker(BLn ! Fn1) = ker(BLn ! Ln1), with the
map between them induced by hn. Our strategy is to construct a commuting
diagram
BLn+1 ! ker(BLn ! Ln1)
? ?
(3.1) hn+1?y hn?y
BPn+1 ! ker(BPn ! Pn1)
This will give a map of BLn+1into the righthand term of the fundamental sequen*
*ce
above, and then we will lift to Fn using projectivity of BLn+1. We will make the
construction in such a way that hn induces an isomorphism between the cokernel*
* of
10 KATHRYN LESH
BLn+1 ! ker(BLn ! Ln1) and the cokernel of BPn+1 ! ker(BPn ! Pn1),
which we know to be Mn=Mn1. This will lead to condition (4) for the tower {En}.
To construct diagram (3.1), we compute ker(BLn ! Ln1). From inductive
hypothesis (5), we know the associated graded of ker(BLn ! Ln1), and since
commutes with cokernels, we know that ker(BLn ! Ln1) has associated graded
Fj=F(j+1)~= ker[BCjnj! ker(BCjnj1! Cjnj2)]
= ker(BCjnj! Cjnj1).
We first define a filtration preserving map gn+1 : BLn+1 ! ker(BLn ! Ln1) as
follows. On the lowest filtration, F(n+1)= BCn+10, let gn+1 be zero. In filtra*
*tion
(j), let gn+1 : BCjnj+1! ker(BLn ! Ln1) lift the natural map
BCjnj+1! ker(BCjnj! Cjnj1)
= Fj=F(j+1)( ker(BLn ! Ln1))
to Fj( ker(BLn ! Ln1)). Note that Fn( ker(BLn ! Ln1)) = Cn0splits off
from ker(BLn ! Ln1), and hence we can take gn+1 : n1j=1BCjn+1j! Cn0to
be zero, and the only factor on which gn+1 : BLn+1 ! Cn0is nonzero is BCn1.
Lemma 3.1. gn+1 is filtration preserving and cok(gn+1) ~=Mn=Mn1.
Proof.gn+1 is filtration preserving by its construction. To calculate the coker*
*nel,
we first consider the cokernel on the level of the associated graded. For j 1*
*, in
filtration Fj=F(j+1)we have
BCjnj+1! ker(BCjnj! Cjnj1),
that is,
njDjnj+1! ker( nj1Djnj! nj1Djnj1).
Because Dj*! Mj=Mj1is a resolution, for j < n the homology at the middle of the
three term sequence nj1Djnj+1! nj1Djnj! nj1Djnj1calculates
nj1njMj=Mj1, which we know is zero since n  j > n  j  1. Hence the map
nj1Djnj+1! ker( nj1Djnj! nj1Djnj1)
is a surjection. Looping preserves surjections, and hence
BCjnj+1! ker(BCjnj! Cjnj1)
is a surjection.
Thus the cokernel of E0(gn+1) is zero on Fj=F(j+1)for j < n. Consider j = n:
on Fn we have defined gn+1 to be the differential BCn1! Cn0, whose cokernel is
Mn=Mn1. Since we have taken gn+1 to be zero from higher filtrations into Fn,
we find that cok(gn+1) ~=Mn=Mn1 as desired.
Recall that the cokernel of BPn+1 ! ker(BPn ! Pn1) is Mn=Mn1 (Re
mark 2.6). To get diagram (3.1), we must have a map fn+1 : BLn+1 ! ker(BLn !
Ln1) whose cokernel is Mn=Mn1and whose composition with hn factors through
BPn+1. So far, we have a map gn+1 : BLn+1 ! ker(BLn ! Ln1) whose coker
nel is Mn=Mn1, but the composition of gn+1 with hn does not necessarily factor
UASS FOR SO AND U 11
through BPn+1. To adjust gn+1, consider the composite
n1j=1BCjnj+1,! BLn+1gn+1! ker(BLn ! Ln1)
hn! ker(BP
n ! Pn1)
! Mn=Mn1.
Choose a lift of the composite across the epimorphism Cn0! Mn=Mn1. We
define fn+1 : BLn+1 ! ker(BLn ! Ln1) as the sum of gn+1 with the lift
n1j=1BCjnj+1! Cn0. Observe that fn+1 is the same as gn+1 on the factors
BCn+10and BCn1of BLn+1, and further, the adjustment added to gn+1 to obtain
fn+1 strictly lowers filtration; thus fn+1 and gn+1 induce the same map on the
associated graded. By construction, hn O fn+1 : nj=1BCjnj+1! ker(BPn !
Pn1) composes to zero in Mn=Mn1, and so hn O fn+1 factors through BPn+1.
We define hn+1 : BLn+1 ! BPn+1 to be the sum of this factoring with a map
BCn+10! BPn+1 that hits a class xn+1 whose looping generates ker(BPn+1 !
Pn)= im(BPn+2 ! Pn+1) ~=Mn+1=Mn.
Lemma 3.2. The commuting diagram
BLn+1 fn+1! ker(BLn ! Ln1)
? ?
hn+1?y hn?y
BPn+1 dn+1! ker(BPn ! Pn1)
induces an isomorphism
cok(fn+1) ~=cok(dn+1).
Proof.By the construction of hn : BLn ! BPn at the previous stage, ker(BLn !
Ln1) ! cok(dn+1) ~=Mn=Mn1 is an epimorphism. On the other hand, the
cokernel of E0(fn+1) is Mn=Mn1 in filtration n and zero in higher filtrations,
and so hn induces an isomorphism cok(fn+1) ~=cok(dn+1).
Corollary 3.3. E0(kerfn+1) ~=ker(E0(fn+1)).
Proof.The result follows from the proof of the preceding lemma, since we establ*
*ish
that E0(cokfn+1) ~=cok(E0(fn+1)).
Now we are ready to define the kinvariant that takes us from En to En+1. Let
kn+1 be a lift of fn+1 across the epimorphism Fn ! ker(BLn ! Ln1) that
comes from the fundamental sequence for En.
Lemma 3.4. kn+1 can be chosen to give a commuting diagram
BLn+1 kn+1!Fn
? ?
hn+1?y ?y
BPn+1 ! Zn
Proof.The choice of the lift kn+1can be adjusted if necessary by a routine diag*
*ram
chase. Use the ladder of fundamental sequences
0 ! cok(BLn ! Fn1) ! Fn ! ker(BLn ! Fn1) ! 0
# # #
0 ! cok(BPn ! Zn1) ! Zn ! ker(BPn ! Zn1) ! 0.
12 KATHRYN LESH
in which the left vertical arrow is an isomorphism by induction, and the commut*
*ing
diagram
BLn+1 fn+1! ker(BLn ! Ln1)= ker(BLn ! Fn1)
? ?
hn+1?y hn?y
BPn+1 ! ker(BPn ! Pn1)= ker(BPn ! Zn1).
We now begin verification of the inductive hypotheses. Let
Yn ! KBPn+1
? ?
(3.2) ?y ?y
En ! KBLn+1
be a homotopy commutative diagram of spaces that realizes the commutative di
agram of Lemma 3.4, let En+1 be the homotopy fiber of En ! KBLn+1, and
let Yn+1 ! En+1 be the map between the homotopy fibers. By construction,
En ! KBLn+1 is a MasseyPeterson map, because the image of BLn+1 ! Fn
injects to ker(BLn ! Ln1) Ln, and thus is Sq0free. The commuting
square (3.2)is a map between MasseyPeterson maps by construction, and thus
we get the first two inductive hypotheses immediately.
Lemma 3.5. ker(kn+1) = ker(fn+1).
Proof.fn+1 is the top composite in the commuting diagram
BLn+1 kn+1!Fn! ker(BLn ! Ln1)
? ? ?
hn+1?y ?y hn+1?y
BPn+1 ! Zn ! ker(BPn ! Pn1).
Certainly ker(kn+1) ker(fn+1). Suppose x 2 ker(fn+1); then
hn+1(x)2 ker[BPn+1 ! ker(BPn ! Pn1)]
= ker[BPn+1 ! Zn]
by Lemma 2.5. Thus kn+1(x) 2 ker(Fn ! Zn). However, by inductive hypothesis
(4) and the ladder of fundamental sequences for Yn and En, ker[Fn ! ker(BLn !
Ln1)] ~=ker[Zn ! ker(BPn ! Pn1)]. Thus kn+1(x) = 0, which establishes the
lemma.
Lemma 3.6. cok(BLn+1 ! Fn) ~=cok(BPn+1 ! Zn).
Proof.Apply the Snake Lemma to the ladder of short exact sequences
0 ! 0 ! BPn+1 ! BPn+1 ! 0
# # #
0 ! cok(BPn ! Zn1) ! Zn ! ker(BPn ! Pn1)! 0.
Because ker(BPn+1 ! Zn) ~=ker[BPn+1 ! ker(BPn ! Pn1)] by Lemma 2.5,
the cokernels of the vertical maps form a short exact sequence . Apply the same
UASS FOR SO AND U 13
reasoning with BLn+1 and the fundamental sequence for En to get a commuting
ladder of short exact sequences
0 ! cok(BLn ! Fn1) ! cok(BLn+1 ! Fn) ! cok(fn+1) ! 0
# # #
0 ! cok(BPn ! Zn1) ! cok(BPn+1 ! Zn) ! cok(dn+1) ! 0.
The leftmost column is an isomorphism by the inductive hypothesis and the right
hand column is an isomorphism by Lemma 3.2, which implies the desired conclu
sion.
Corollary 3.7. The natural map lim!nFn ! lim!nZn is an isomorphism.
Proof.Consider
Fn ! Zn
?? ?
y ?y
Fn+1 ! Zn+1
?? ?
y ?y
lim!nFn!lim!nZn
By the preceding lemma, im(Fn ! Fn+1) ~=im(Zn ! Zn+1), and by Lemma 2.5,
im(Zn ! Zn+1) ~=im(Zn ! Zn+j) for j > 1. The corollary follows.
4.Homotopical properties of {En}
In this section we give the homotopical and homological properties of the tow*
*er
{En}. We prove that it has inverse limit SO^2and that its homotopy spectral
sequence collapses at the E2term. Notation is continued from Section 3.
Proposition 4.1. The map of towers {Yn} ! {En} induces a homotopy equiva
lence on the homotopy inverse limits.
Proof.We already know from Corollary 3.7 that the map of towers induces an
isomorphism lim!nH*En ! lim!nH*Yn. Although cohomology is not in general
wellrelated to inverse limits, an application of [Lannes, Lemme 3.2.3] tells u*
*s that
in our situation,
H* holimYn~=lim!H*Yn
n n
and H* holimEn~=lim!H*En.
n n
The essential ingredients that allow the use of Lannes's lemma are:
(1)For all n, the spaces Yn and En are connected and have mod 2 cohomology
that is finite in each dimension.
(2)The towers of groups {ß1Yn} and {ß1En} are constant.
(3)The towers of groups {H1Yn} and {H1En} are constant.
The proposition then follows by observing that holimnYn and holimnEn are 2
complete (each is built from mod 2 EilenbergMacLane spaces by fibrations) and
that the map between them is a mod 2 cohomology isomorphism.
14 KATHRYN LESH
Corollary 4.2. holimnEn ' SO^2.
Our next goal is Corollary 4.5, in which we prove that the homotopy spectral
sequence of {En} collapses at the E2term. This follows by using a homological
argument to show that the map {Yn} ! {En} induces an isomorphism at E2 of
the homotopy spectral sequences, and then observing that the homotopy spectral
sequence of {Yn} does in fact collapse at E2. The following proposition performs
the main technical calculation.
Proposition 4.3. The following ladder gives a homology isomorphism at the mid
dle term:
BLn+1 ! Ln ! Ln1
?? ? ?
y ?y ?y
BPn+1 ! Pn ! Pn1.
Proof.The proof is inductive. For n = 1, we take P0 = L0 = 0 and the result is
easily established by direct calculation. Suppose that the lemma is true for
BLn ! Ln1 ! Ln2
?? ? ?
y ?y ?y
BPn ! Pn1 ! Pn2
and consider the next stage. By Lemma 3.2, we already know that
__ker(BLn_!_Ln1)_~__ker(BPn_!_Pn1)_.
im(BLn+1 ! Ln) = im(BPn+1 ! Pn)
Let iL : ker(BLn_! Ln1) ! ker(Ln ! Ln1) be the natural map ker(fn) !
ker( fn), let iL be the induced map on cokernels, and consider the diagram of
exact sequences
BLn+1 fn+1! ker(BLn ! Ln1)!_ker(BLn!Ln1)_im(BLn+1!Ln)!0
? ? ?
=?y iL?y __iL?y
BLn+1 ! ker(Ln ! Ln1) ! ker(Ln!_Ln1)_im(BLn+1!Ln)!0.
__ __
By Lemma 2.1 and the Snake Lemma, iL and iLare monomorphisms and cok(iL) ~=
cok(iL) ~= 11cok(BLn ! Ln1). The same argument with iP : ker(BPn_!
Pn1)_!_ker(Pn ! Pn1) and the corresponding_map of cokernels, iP, shows
that iP is a monomorphism and cok(iP) ~= 11cok(BPn ! Pn1). Consider the
diagram
_ker(BLn!Ln1)_~= _ker(BPn!Pn1)_
im(BLn+1!Ln)!? im(BPn+1!Pn)?
__iL?y __iP?y
ker(Ln!_Ln1)_ ker(Pn!_Pn1)_
im(BLn+1!Ln)! im(BPn+1!Pn).
__ __
We already know that the top row is an isomorphism. Since iLand iPare monomor
phisms, the lemma will be established_by the Five_Lemma if we prove that the
diagram induces an isomorphism cok(iL) ! cok(iP). Thus we must show that
11cok(BLn ! Ln1) ~= 11cok(BPn ! Pn1).
UASS FOR SO AND U 15
The three term sequence BLn ! Ln1 ! Ln2 gives us a short exact sequence
0 ! ker(Ln1_!__Ln2)_im(BL! _____Ln1_____! ______Ln1______! 0.
n !iLn1)m(BLn ! Ln1)ker(Ln1 ! Ln2)
The middle term is cok(BLn ! Ln1), and the right hand term is Sq0free, because
it injects into Ln2, which is itself Sq0free. This argument and a similar o*
*ne
applied to BPn ! Pn1 ! Pn2 give us
~ ~
11cok(BLn ! Ln1)~= 11ker(Ln1_!__Ln2)_im(BL
~ ~ n ! Ln1)
11cok(BPn ! Pn1)~= 11ker(Pn1_!__Pn2)_im(BP,
n ! Pn1)
and these are isomorphic by the inductive hypothesis, completing the proof of t*
*he
lemma.
Corollary 4.4. The commuting ladder
BLn+1 ! Ln ! Ln1
?? ? ?
y ?y ?y
BPn+1 ! Pn ! Pn1
induces an isomorphism on H* HomU_(, tF2) for all t at the middle term.
Proof.It is sufficient to prove that in the commuting ladder
BLn+1 ! Ln ! Ln1 ! 2Ln2 . . .! n1L1
?? ? ? ? ?
y ?y ?y ?y ?y
BPn+1 ! Pn ! Pn1 ! 2Pn2 . . .! n1P1,
the map between the upper and lower rows is an isomorphism on homology up to
and including Ln ! Pn. The proof is by induction, beginning with
BL2 ! L1 ! 0
?? ? ?
y =?y ?y
BP2 ! P1 ! 0
In the case of SO, BL2 ! BP2 is an equality. In the case of U, we observe
BL2 = BC11 BC20= BP2 BC20where the BP2 summand maps to BP2 by
the identity and BC20maps to L1 by the zero map. Thus we have a base for the
induction in the case of U also.
Suppose that
BLn ! Ln1 ! 1Ln2...! n2L1
?? ? ? ?
y ?y ?y ?y
BPn ! Pn1 ! 1Pn2...! n2P1
16 KATHRYN LESH
induces an isomorphism on homology up to Ln1 ! Pn1. Applying to both
complexes, we find that
Ln ! Ln1 ! 2Ln2...! n1L1
?? ? ? ?
y ?y ?y ?y
Pn ! Pn1 ! 2Pn2...! n1P1
is an isomorphism on homology up to Ln1 ! Pn1, and joining this with the
result of Lemma 4.3, we find that
BLn+1 ! Ln ! Ln1 ! 2Ln2 . . .! n1L1
?? ? ? ? ?
y ?y ?y ?y ?y
BPn+1 ! Pn ! Pn1 ! 2Pn2 . . .! n1P1
is an isomorphism on homology up to and including Ln ! Pn. The lemma follows.
Corollary 4.5. The homotopy spectral sequence of {En} collapses at E2.
Proof.By Corollary 4.4, the map {Yn} ! {En} induces a map of homotopy spec
tral sequences which is an isomorphism on the E2term. Since the homotopy spec
tral sequence of {Yn} has no further differentials (in fact, it collapses at E1*
*), the
homotopy spectral sequence of {En} collapses at E2.
5.A Model for the UASS, and Some Predictions and Reflections
In the preceding sections, we used the resolutions of the filtration quotients
Mn=Mn1 to construct a complicated tower {En} that involves those resolutions,
converges to SO^2, and has a homotopy spectral sequence that collapses at E2. T*
*he
tower {En} realizes the chain complex L*, where the notation L*is to be interpr*
*eted
as BLn+1 ! Ln ! Ln1at the nth level. The differential of the chain complex L*
gives rise to the only nonzero differential in the homotopy spectral sequence o*
*f {En},
since the E1term is Hom U_(Ln, *F2) at level n, and En,t2~=En,t1(Corollary 4.*
*5).
In this section, we describe how the complex L* gives a model for the unstable
Adams spectral sequences of SO and U, we make some predictions based on the
model, and we discuss some related work of Bousfield and Davis [BD].
5.1. A Model for the UASS.
The conjecture suggested by Mahowald is, loosely, that the differential of the
chain complex L* contains all the information of the unstable Adams spectral
sequence, including all of its many nonzero differentials. We already know that
H*[Hom U_(L*, t tF2)] is the associated graded for the filtration of ß*SO^2by *
*the
destabilized Adams tower (Corollary 4.4 and Corollary 4.5). The assertion is th*
*at
it is possible to produce the UASS from the complex Hom U_(L*, *F2) by a com
bination of filtering and regrading.
To describe the proposed model, let L* be the cochain complex of graded vector
spaces defined by
(Ln)j= HomU_(Ln, jF2),
UASS FOR SO AND U 17
and use the differential BLn+1 ! Ln and adjointness to define d : (Ln)j !
(Ln+1)j1by
Hom U_(Ln, jF2)H!omU_(BLn+1, jF2)
~=Hom U_( BLn+1, j1F2)
~=Hom U_(Ln+1, j1F2).
We filter Ln by
(FsLn)j= HomU_( ni=sCnii, jF2).
We have F0 F1 F2 . .,.and comparing to the construction of BLn+1 ! Ln
in Section 3, it is easy to check that the differential on L* is filtrationpre*
*serving.
Thus the filtration gives rise to a spectral sequence that converges to H*L*, a*
*nd
we grade it as
Es,t1= n HomU_(Cns, tsF2).
Recall that the abutment, H*L*, is the associated graded to ß*SO^2. Also, Cns=
sDns, and hence by the adjointness of and , we have
Es,t1= n HomU_(Dns, tF2).
The d1differential is induced by differential in the resolution Dn*! Mn=Mn1, *
*and
thus the spectral sequence becomes
Es,t2= n ExtsU_(Mn=Mn1 , tF2) ) ß*SO^2.
Conjecture 5.1. The spectral sequence Es,trdefined above is the UASS for SO.
If Conjecture 5.1 is correct, then it has the consequence that all of the dif*
*ferentials
in the unstable Adams spectral sequence can be computed from the primary level
calculation of the complex L*. In principle, this could be done indefinitely fa*
*r out
by computer.
Corollary to Conjecture 5.1 .
Exts,tU_(M1 , F2) ~= n Exts,tU_(Mn=Mn1 , F2).
Proof.The E2term of the UASS for SO is given by Es,t2~=ExtsU_(M1 , tF2), and
hence if Conjecture 5.1 is correct, these two must be isomorphic.
In fact, there is a general spectral sequence that is very close to the spect*
*ral se
quence of Conjecture 5.1, namely the Grothendieck spectral sequence for the cal*
*cu
lation of the derived functors ExtsA( A=Sq3 , tF2). Let D be the destabilizati*
*on
functor from the category of (stable) Amodules to U_, the category of unstable
Amodules. (This functor is often denoted 1 .) Because tF2 is an unstable
Amodule, any map to tF2 from a stable Amodule factors through the desta
bilization. Hence the functor Hom A(, tF2) can be written as the composition
Hom U_(, tF2) O D(), giving rise to a composite functor spectral sequence
ExtsrU_(Dr , tF2) =) ExtsA( , tF2).
In the case of A=Sq3, ExtsA( A=Sq3 , tF2) actually gives the associated graded
to the stable homotopy, because there are no differentials in the stable Adams
spectral sequence for infinite delooping of SO. Thus the Grothendieck spectral
sequence gives a spectral sequence starting from an unstable Ext and converging
to ß*SO.
18 KATHRYN LESH
The Grothendieck spectral sequence is very closely related to the spectral se
quence we have constructed, but it is not quite the same. In particular, let
X = A=Sq3, so that we are considering the case of SO. Then it can be shown that
Mn+1=Mn ~=Dn nX, the ingredients being found in Lemma 2.5, Lemma 2.1, and
the proof of Proposition 4.3, and our construction gives a spectral sequence
ExtsrU_(Dr rX , tF2) =) ExtsA(X, tF2).
However, the situation for the group U is a little different, the difference_be*
*ing*
caused by the fact that while H*SO is the free unstable Aalgebra_on*H RP1 ,
which is Sq0free, H*U is the free unstable Aalgebra on H CP1+, which_is not.
In fact, contrary to the assertion of [BD, Proposition 4.1], if X ~= A = 1, wh*
*ere
1 is the subalgebra of A generated by the Milnor primitives Q0 and Q1, then
Dn nX is not Mn+1=Mn Z=2 but a much larger module. The problem lies
not in the spectral sequence constructed in the proof of the proposition, but i*
*n its
assumption that the homology being converged to is Mn+1=Mn.
However, a small variation can repair the problem. Let X be an Amodule, and
let C* be a stable resolution of X. For n 1, define
D0rX = __ker(D_Cr_!_D_Cr1)_im(DC.
r+1! DCr)
Using methods similar to those of Proposition 4.3, one can show that the defini*
*tion
of D0rX is independent of the resolution used, and that the modules D0rX and DrX
are different exactly_when Dr1 X is not Sq0free. If we let X = A=Sq3 (in the
case of SO) or X = A =Sq3 (in the case of U), then for both SO and U,
D0n nX ~=Mn+1=Mn,
__*
where the modules_Mn=Mn1 are the filtration quotients of H RP1 (in the case
of SO) or H *CP1+ (in the case of U). The construction of the previous section
gives, for a general Amodule X, two spectral sequences, depending on whether we
use D0ror Dr:
(5.1) ExtsrU_(D0r rX, tF2)=) ExtsA(X, tF2)
(5.2) ExtsrU_(Dr rX , tF2)=) ExtsA(X, tF2).
(The spectral sequence of Conjecture 5.1 is (5.1).) These spectral sequences ca*
*n be
given a construction almost exactly like that of the Grothendieck spectral sequ*
*ence.
Conjecture 5.1 observes that because the stable Adams spectral sequences for SO
and U collapse, the target of spectral sequence (5.1) is actually the associate*
*d graded
to the homotopy of the space. Since the E2term is closely related to the homol*
*ogy
of the space, because D0r rX is the associated graded for the cohomology of SO
(or U), this variation of the Grothedieck spectral sequence could actually be t*
*he
unstable Adams spectral sequence.
5.2. Predictions.
Next we discuss some predictions that arise from Conjecture 5.1 and some empiri*
*cal
data that support the conjecture. The main tool in making these predictions is a
vanishing theorem of Bousfield [B , Theorem 2.6] that describes the location of*
* h0
towers in unstable Ext by giving values of t  s where towers occur, though not
the value of s in which they begin. Application of Bousfield's theorem gives us*
* the
UASS FOR SO AND U 19
following proposition. Recall that ff(n) denotes the number of ones in the dyad*
*ic
expansion of n.
Proposition 5.2.
__*
(1)For M = H RP1 :
(a)The h0towers of ExtsU_(M, tF2) are found in stem degrees satisfying
(t  s) 3 mod 4, and there is exactly one h0tower in each such
dimension.
(b)The h0towers of ExtsU_(Mn=Mn1 , tF2) are found in stem degrees
satisfying (t  s) 3 mod 4 and ff(t  s) = n, and there is exactly *
*one
h0tower_in*each such dimension.
(2)For M = H CP1+,
(a)The h0towers of ExtsU_(M, tF2) are found in stem degrees satisfying
(t  s) 1 mod 2, and there is exactly one h0tower in each such
dimension.
(b)The h0towers of ExtsU_(Mn=Mn1 , tF2) are found in stem degrees
satisfying (t  s) 1 mod 2 and ff(t  s) = n, and there is exactly *
*one
h0tower in each such dimension.
Proof.An easy calculation with [B , Theorem 2.6].
Remark 5.3. Proposition 5.2 says that Corollary to Conjecture 5.1 is correct at*
* least
at the level of h0towers, since ExtsU_(M, tF2) and n ExtsU_(Mn=Mn1 , tF2) *
*have
exactly the same towers.
Bousfield's theorem also gives a vanishing line above which Ext is zero except
for h0towers. To describe his theorem as it applies to our situation, we defin*
*e a
function OE(m) for positive integers m as follows. Suppose that m = 8k + i where
i < 8. Then
(1)OE(m) = 4k + i for i = 0, 1, 2, 3;
(2)OE(m) = 4k + 3 for i = 4, 5, 6;
(3)OE(m) = 4k + 4 for i = 7.
We specialize Bousfield's theorem to our situation as follows.
Theorem 5.4 ([B , Theorem 2.6]). Let N be an unstable Amodule such that Ni= 0
for i < c, where c 5. Then ExtsU_(N, tF2) is free over F2[h0] for s > OE(ts*
*c).
This gives a vanishing line of slope 1=2 in the UASS.
We are going to use Theorem 5.4 to predict the unstable Adams filtrations of
the elements of ß*SO and ß*U. From the map of towers {Yn} ! {En}, the maps
KPn+1 ! KLn+1 induce on homotopy a map
(5.3) ExtnA( A=Sq3 , tF2) ! nr=1Extnr+1U_(Mr=Mr1 , tr+1F2),
and this map commutes with the action of h0. All of the elements on the left
represent homotopy, and since the righthand side is the E2term for the spectr*
*al
sequence of Conjecture 5.1, the map tells us where the homotopy is represented *
*in
this spectral sequence, which predicts the unstable Adams filtration of ß*SO.
Consider first the case of SO. Suppose k 3 mod 4; if k 3 mod 8, define
n = (k  1)=2, and if k 7 mod 8, define n = (k  3)=2. Then ßkSO ~=Z,
represented by an h0tower in Ext*A( A=Sq3 , *+kF2) beginning in filtration s *
*= n.
On the right side of (5.3), the only term with an h0tower in dimension k is r *
*= ff(k)
20 KATHRYN LESH
(Proposition 5.2), and so the part of (5.3)that carries the bottom element of t*
*he
h0tower is
ExtnA( A=Sq3 , tF2) ! Extnff(k)+1U_(Mff(k)=Mff(k)1, tff(k)+1F2).
Thus we obtain the following prediction.
Conjecture 5.5. If ßkSO is torsion free, then the unstable Adams filtration of
ßkSO is ff(k)  1 less than the stable Adams filtration of the corresponding st*
*em.
By exactly the same reasoning we obtain the same prediction for the case of U,
where all the homotopy is torsion free.
Conjecture 5.6. The unstable Adams filtration of ßkU is ff(k)  1 less than the
stable Adams filtration of the corresponding stem.
Next, we predict the unstable Adams filtration of the torsion elements of ß*S*
*O,
namely ßkSO ~=Z=2 for k 0 or 1 mod 8. Consider first the case k 0 mod 8,
and let n = (1=2)k  1. Then ßkSO is represented in ExtnA( A=Sq3 , n+kF2). As
before, we predict the unstable Adams filtration by considering the image of th*
*is
element under the map of (5.3):
ExtnA( A=Sq3 , n+kF2) ! nr=1Extnr+1U_(Mr=Mr1 , n+kr+1F2).
Using Theorem 5.4, we will prove that only the r = 3 summand has h0torsion
elements in high enough filtration to be in the image of this map. We already k*
*now
that M1 has exactly one torsion element in Ext for s = 0 and nothing else, and
M2=M1 has exactly one h0tower in Ext for k = 3, and nothing else. Suppose that
r 4, and note that Mr=Mr1begins in dimension 2r 1. To use Theorem 5.4 to
show that Extnr+1U_(Mr=Mr1 , n+kr+1F2) has no h0torsion elements, we must
show that
(n  r  1) > OE[(n + k  r  1)  (n  r  1)  (2r 1)],
a task that is easily accomplished using k 0 mod 8 and n = (1=2)k  1. An
almost identical calculation leads to the same conclusion if k 1 mod 8. This
leaves the r = 3 summand as the only one where the torsion elements can go,
and since r = 3 causes a filtration drop of 2 from the stable Ext, we arrive at*
* the
following prediction.
Conjecture 5.7. If ßkSO ~=Z=2 is represented in filtration n in the stable Adams
spectral sequence, then it has filtration n2 in the unstable Adams spectral se*
*quence.
Remark 5.8. The author has verified the preceding conjectures as to filtration *
*for
ß*SO up to approximately ß50, using charts of unstable Extprovided by R. Bruner*
*'s
computer calculations. Likewise the author has verified the Corollary to Conjec
ture 5.1 for SO in the same range.
We close this discussion by giving an example of the calculation of a differe*
*ntial
in the spectral sequence modelling the UASS for SO In the Figure 3, we exhibit
part of the UASS for SO. We will show how to use the spectral sequence of
Conjecture 5.1 to predict the first differential in the UASS for SO, which goes
from (s, t  s) = (0, 15) to (s, t  s) = (2, 14). (This differential propagate*
*s to give
differentials connecting the two lightning flashes, but we will deal only with *
*the
first differential.)
UASS FOR SO AND U 21


10 _ b r r rd
  
_ b r r rd
  
_ b r r rd
  
_ b r r rd
  
_ b r r rd r
  
_ b r r rdr r
5   
     
_ b r r r rd r r
     BB 
_ b r r r rrd r rd
   
_ b r r r rd rdrd
  AK 
_ b r r A rdrd rd
 
______________________________________________________b*
*brrdA
5 10 15
Figure 3. The E2term of the UASS for SO
Elements represented by open circles arise from M1 and M2=M1. El
ements represented by black dots arise from M3=M2. Elements repre
sented by circled dots arise from M4=M3.
In order to do this, we will have to calculate the first few stages of the co*
*mplex
L*. In particular, we will be looking at the commuting three term sequences
BL5 ! L4 ! L3
? ? ?
(5.4) h5?y h4?y 2h3?y
BP5 ! P4 ! P3,
which is detailed in Table 1. We need the result that Mn=Mn1 ~= F(2n 
1)=Sq1, Sq2, ..., Sq2n2[Massey], and we remind the reader that in the diagram
above, the top row involves resolutions of Mn=Mn1 for n = 1, 2, 3, 4, and 5,
where the resolution of Mn=Mn1 is looped down 4  n times. When n = 1,
M1 ~=F(1) is a projective,_and has a resolution of length 1. Hence C1i= 0 for
i > 0. Further, M2=M1 ~=F(3) is almost projective. Its projective resolution is
. .!.F(5) ! F(4) ! F(3) (each map given by Sq1), and so all the elements
contributed lie in t  s = 3. It turns out that this resolution does not intera*
*ct with
any of the other parts of L*, corresponding to the fact that no differentials i*
*n the
UASS for SO involve t  s = 3.
In Table 1, we provide all the summands of each of the terms in (5.4)and show
the horizontal maps between them. In the commuting square
L4! L3
? ?
h4?y 2h3?y
P4! P3,
2h3 is the identity, and h4 is the map is the identity on the summands F(3),
F(7), and F(8). To describe h4 on the summand F(15) of L4, we recall that
22 KATHRYN LESH
________BL5_________________!__________________L4_____!_______L3__
C2*: F(4) Sq1'3 F(3) Sq1'2 F(2)
8 1
< F(8) Sq '7 F(7) Sq1'
C3*: : F(10) Sq2'8+ Sq3'7 ____F(8) Sq2'6 F(6)
F(15) Sq7'8+ Sq4,2,1'8+ Sq6,2'7+ _'15_ 6
8
< F(16) Sq1'15
C4*: : F(17) Sq2'15 F(15)
F(19) Sq4'15
C5*: F(32)
_____________________________________________________________________
________BP5__________________!_________________P4______!______P3__
F(4) Sq1'3 F(3) Sq1'2 F(2)
F(8) Sq1'7 F(7) Sq1'6 F(6)
________F(10)____________Sq2'8+_Sq3'7__________F(8)____Sq2'6________
Table 1. The chain complexes of Section 3
'15 2 L4 must hit an element of P4 that represents an A module generator of
the homology of the three term sequence BP5 ! P4 ! P3, and the element in
question is Sq7'8+ Sq4,2,1'8+ Sq6,2'7 2 P4.
Now for the differential, which is predicted by the construction of the map B*
*L5 !
L4. It comes about because the map BL5 ! L4 must be defined in such a way
that the composite BL5 ! L4 ! P4 lifts across BP5 ! P4. Since there are no
interactions between the filtrations in the map L4 ! L3, the map BL5 ! L4
can be constructed simply by using the differentials within the resolutions Cn**
*, and
then making adjustments as needed to ensure the required lifting. In terms of t*
*he
construction of Section 3, this is saying that the map g5 is just the sum of the
differentials in the individual resolutions.
No corrections need to be made until we reach F(15) BL5. At this point, if *
*no
adjustments were made, the composite BL5 ! L4 ! P4 would take the generator
'152 BL5 to Sq7'8 + Sq4,2,1'8 + Sq6,2'7 2 P4. Since this element generates the
homology at P4, it certainly does not lift to BP5. Thus we add '152 L4to the im*
*age
of '152 BL5 (boxed for emphasis in the table). This gives a differential between
adjoining filtrations in L*, which translates to the prediction of the nonzero *
*d2
differential taking (s, t  s) = (0, 15) to (s, t  s) = (2, 14) in the UASS of*
* SO.
UASS FOR SO AND U 23
5.3. Relation to [BD]. Bousfield and Davis make a much more general conjecture
than our Conjecture 5.1 in [BD]. Suppose given a diagram of unstable Amodules
F1 F2 F3
?? ? ?
yf0 ?yf1 ?yf2
X0 p0!X1 p1!X2 p2!. ..!X
? ?
i1?y i2?y
F1 F2
satisfying the following conditions.
(1)Fn ! Xn1 ! Xn ! Fn ! Xn1 is exact.
(2)Fn is a direct sum of F(m)'s and/or F0(m)'s (where F(m) is a free unstable
A module on a generator of dimension m and F0(m) = F(m)=Sq1).
(3)(infn)* : ExtsU_( Fn , tF2) ! ExtsU_(Fn+1, tF2) is the zero map.
(4)ker(Xn ! X) = ker(Xn ! Xn+1).
(5)X ~=lim!n(Xn).
Let Mn = im(Xn ! X).
Conjecture 5.9 ([BD, Conjecture 5.1]).
ExtsU_(X, tF2) ~= n ExtsU_(Mn=Mn1 , tF2).
However, this conjecture is false, as shown by the counterexample that follow*
*s.
Consider the following tower, whose kinvariants are described below.
K(Z=2, 10)i4!Y4
??
y
K(Z=2, 8)i3!Y3k3!K(Z=2, 10)
??
y
K(Z=2, 8)i2!Y2k2!K(Z=2, 9)
??
y
K(Z, 7)i1!Y1k1!K(Z=2, 9)
??
y
* ! K(Z, 8)
Let H*Yi= U(Zi). The first kinvariant is k1 = Sq2'7 and the second is k2 = 0.
For the_third,_let x10 be a class in Z2 with (i2)*(x10) = Sq2'9 2 ker(Sq2 :
F(9) ! F(7)), and let x010denote its image in Z3. Let x8 be a class in Z3 with
(i3)*(x8) = '8, the fundamental class. Then the third kinvariant is defined by
k3 = x010+ Sq2x8.
24 KATHRYN LESH
We consider Bousfield and Davis's conjecture for this situation, where the di*
*a
gram is given by
__
F(8) F(9) F(9) F(10)
?? ? ? ?
y Sq2'7?y 0?y x010+Sq2x8?y
__ p1 p2 p3
0 ! F(7)! Z2 ! Z3 ! Z4 = X
? ? ? ?
i1?y i2?y ?y ?y
__
F(7) F(8) F(8) F(9).
In particular, we consider Ext0, so that we are really looking at Amodule gene*
*ra
tors. We find that Ext0has nonzero groups only in the following dimensions.
(1)Ext0U_(M1 , tF2) = Z=2 if t = 7.
(2)Ext0U_(M2=M1 , tF2) = Z=2 if t = 10 or 15.
(3)Ext0U_(M3=M2 , tF2) = Z=2 if t = 8.
(4)Ext0U_(M4=M3 , tF2) = Z=2 if t = 12 or 31.
(5)Ext0U_(X, tF2) = Z=2 if t = 7, 8, 12, 15 and 31.
In particular, Ext0U_(X, tF2) has no nonzero class for t = 10. In fact, M3=M2 *
*~=
F(8)=Sq2, and in the spectral sequence for Ext*U_(X, tF2) arising from the fil*
*tration
of X, there is a nonzero differential
Ext0U_(M2=M1 , 10F2) ! Ext1U_(M3=M2 , 10F2).
In effect, what we have done in this example is to introduce a generator in M2
(namely x10, corresponding to Sq2'8) and then to equate it with a Steenrod oper*
*a
tion on another class at a later stage, thus eliminating it from the list of ge*
*nerators.
However, it is possible to revise Conjecture 5.9 to deal with this problem. T*
*he
salient feature that distinguishes the situation for SO and U from the example
above is that there is a stable resolution in the background. In other words, *
*in
the case of the_tower_{Yn} defined in Section 2, the tower realizes a destabili*
*zed
resolution of A =Sq3, whereas in the counterexample above, it realizes the uns*
*table
complex
__ Sq2 0 Sq2
F(7) F(9) F(10)  F(12),
which is certainly not the destabilization of a resolution. To reflect this, we*
* refine
Bousfield and Davis's conjecture as follows.
Conjecture_5.10._Conjecture_5.9_is_true if we add the hypothesis that there exi*
*st
Amodules_Fn and maps dn: Fn+1! Fn satisfying the following_conditions:
(1)F nis_the sum of copies of A and A=Sq1, and nDF n~=Fn.
(2) nD(dn)_=_in O fn.
(3)(F *, d*) is a chain complex whose only nonzero homology group occurs in
the lowest homological dimension.
References
[B] A. K. Bousfield, A vanishing theorem for the unstable Adams spectral sequ*
*ence, Topology
9 (1970) 337  344.
[BD] A. K. Bousfield and D. M. Davis, On the unstable Adams spectral sequence *
*for SO and
U, and splittings of unstable Ext groups, Bol. Soc. Mat. Mex. 37 (1992) 4*
*153.
UASS FOR SO AND U 25
[HM] J. R. Harper and H. R. Miller, Looping MasseyPeterson towers, in: S. M. *
*Salamon, B.
Steer, W. A. Sutherland, eds., Advances in Homotopy Theory (London Math. *
*Soc. Lec.
Notes 139, Cambridge University Press, Cambridge, 1989) 6986.
[Lannes]J. Lannes, Sur les espaces fonctionels dont la source est le classifian*
*t d'un pgroupe
ab`elien 'elementaire, Publ. Math. Inst. Hautes Etudes Sci. 75 (1992), 13*
*5  244.
[Lesh]K. Lesh, The unstable Adams spectral sequence for twostage towers, Topol*
*ogy Appl.
101 (2000), no. 2, 161180.
[Long]J. Long, Two contributions to the homotopy theory of Hspaces, Princeton *
*University
thesis (1979).
[Massey]W. S. Massey, unpublished manuscript, 1978.
[MP] W. S. Massey and F. P. Peterson, The mod 2 cohomology structure of certai*
*n fiber
spaces, Mem. Amer. Math. Soc. Number 74, Providence, 1967.
[SE] N. E. Steenrod and D. B. A. Epstein, Cohomology Operations, (Ann. Math. S*
*tudies
Number 50, Princeton University Press, Princeton, NJ, 1962).
[S] R. Stong, Determination of H*(BO(k, 1)) and H*(BU(k, 1)), Trans. Amer. Ma*
*th. Soc.
107 (1963) 526544.
Department of Mathematics, Union College, Schenectady, NY 12308
Email address: klesh@member.ams.org