The Unstable Adams Spectral Sequence for
TwoStage Towers
Kathryn Lesh 1
Department of Mathematics, University of Toledo, Toledo, Ohio 436063390 USA
_____________________________________________________________________________
Abstract
Let KP denote a generalized mod 2 EilenbergMacLane space and let Y be the fiber
of a map X ! KP to which the MasseyPeterson theorem applies. We study the
relationship of the mod 2 unstable Adams spectral sequence (UASS) for X and for
Y . Given conditions on X, we split the E2term for Y , and we use a primary le*
*vel
calculation to compute d2 for Y up to an error term. If the UASS for X collapses
at E2 (for example, if X is an EilenbergMacLane space), the UASS for Y collaps*
*es
at E3, and we have the entire UASS for Y . We also give examples and address a
conjecture of Bousfield on the UASS for the Lie group SO.
Key words: unstable Adams spectral sequence, unstable Ext groups, SO
1991 MSC: 55T15, 55Q52, 55R45
_____________________________________________________________________________
1 Introduction
Let KP denote a mod 2 generalized EilenbergMacLane space. In this paper we
study the mod 2 unstable Adams spectral sequence (UASS) of a space Y with
polynomial cohomology which is obtained as the fiber of a map between simply
connected spaces X ! KP to which the MasseyPeterson theorem applies.
By using specially constructed Adams resolutions, we study the relationship
between the UASS for X and for Y . Given certain conditions on X, we give
a splitting of the E2term of the UASS for Y , and we show how to use a
primary level calculation to compute almost complete information about the
d2 differentials in the UASS for Y . In the case that the UASS for X collapses *
*at
E2, the UASS for Y collapses at E3, and we have almost complete information
about it.
_______
1 Email: klesh@uoft02.utoledo.edu
Preprint submitted to Topology and Its Applications 15 May 1998
To state our results more precisely, we make some definitions. Throughout the
paper, we take p = 2 and cohomology with mod 2 coefficients. Let U_ denote
the category of unstable modules over the mod 2 Steenrod algebra A . If M
is an object in U_, let U(M) denote the free unstable A algebra generated by
M, as described by [5]. If P is a free unstable A module, we write KP for the
EilenbergMacLane space with H*KP ~= U(P ).
We call a map X ! KP MasseyPeterson if H*X ~= U(M), H*X is polyno
mial of finite type, X and KP are simply connected, and H*KP ! H*X is
induced by a map f : P ! M. We think of the topological map as realizing
f. If X ! KP is MasseyPeterson, with fiber Y , then the MasseyPeterson
theorem [4] tells us that H*Y ~= U(N), where there is a short exact sequence
0 ! cok(f) ! N ! ker(f) ! 0:
In general, this fundamental sequence does not split as A modules. However,
the following theorem states conditions under which it does split at the level
of Ext .
Theorem 1.1 Let X be a simply connected space with polynomial mod 2 co
homology H*X ~=U(M), and let X ! KP be MasseyPeterson, inducing
f : P ! M in cohomology. Let Y be the fiber of X ! KP , and suppose
H*Y ~= U(N) is polynomial. If X has E2 = E3 in its unstable Adams spectral
sequence, then there is a splitting
Ext *U_(N; tF2 )~=Ext *U_(cok (f); tF2 )Ext*U_( ker(f) ; tF2 :)
In general, the E2term of an UASS is an Ext group taken over the category
K__of unstable algebras over the Steenrod algebra. Because H*Y ~= U(N),
Ext *K_(H*Y ; H*St~)=Ext *U_(N; tF2 );
and hence Theorem 1.1 describes the E2term of the UASS for Y . Many differ
entials in the UASS for Y can be described in this case; we make a precise stat*
*e
ment in Theorem 4.5. Roughly speaking, the d2 differentials that go between
the two summands of Ext *U_(N; tF2 )can be computed by a primary level
calculation involving an algebraic mapping cone. In fact, if X has E2 = E1 in
its UASS, then the UASS for Y can be computed almost completely by this
calculation (Corollary 4.7).
The motivation for studying this situation was provided by a conjecture of
Bousfield from the 1970s about the E2term of the UASS for the Lie group
SO. We describe this conjecture in Section 5, along with examples of the
application of Theorem 1.1 and Theorem 4.5.
Our strategy for studying the fibration Y ! X ! KP where H*X ~= U(M)
2
and the fibration is induced by f : P ! M is to build a resolution of M using
P and resolutions of cok(f) and ker(f). When realized, this resolution gives
an Adams tower for X which can be manipulated to give an Adams tower for
Y . Comparison of the two towers gives information about the differentials in
the UASS for Y .
The organization of the paper is as follows. In Section 2 we construct a spe
cial chain complex resolving M and use it to construct an Adams tower for
X and a tower with inverse limit Y which are closely related. In Section 3,
we construct an Adams tower for Y by "rearranging" the tower for X. In
Section 4, we study the homotopy spectral sequences of X and Y to prove
Theorem 1.1, and we state and prove further results on differentials in the
UASS for Y (Theorem 4.5 and Corollary 4.7). In Section 5, we give examples
of the application of Theorem 1.1 and Theorem 4.5.
The notation set up in the introduction will be carried through the paper.
2 Adams tower for X
In this section we construct an Adams tower for X and use it to obtain a
tower whose inverse limit is the 2completion of Y . The tower for Y will not
be an Adams tower, but we show in Section 3 that its kinvariants can be
"rearranged" in a controllable way to give an Adams tower for Y . We will
retain enough information through the rearrangement that we can relate the
homotopy spectral sequences of the two towers, thus obtaining infomation
about the UASS for Y from that of X.
We will need to make frequent use of the algebraic looping functor on U_, and
so we review the basic facts here. (See also [4].) The functor : U_ ! U_ is
right adjoint to the suspension functor : U_ ! U_. The module M is the
largest unstable quotient of the desuspension of M:
M (1M)=(1Sq 0M);
where Sq 0x = Sqxx. We write k for the kfold iterate of . The functor
is not exact, but only its first derived functor, which we denote 11, can be
nonzero. The module 11M can be expressed as a regrading of the kernel of
Sq 0on M. In particular, if Sq0 acts freely on M, then 11M = 0. We write kj
for the jth derived functor of k and note that there is a composite functor
spectral sequence (the Singer spectral sequence) sitjM ) s+ti+jM which
allows us to calculate derived functors of k inductively. For any unstable
module M, kjM = 0 for j > k.
Now we turn to the construction of an Adams tower for X. Recall that H*X ~=
3
U(M) and that Y is the fiber of a MasseyPeterson map X ! KP which
realizes f : P ! M. We know that H*Y ~=U(N) where N is given by an
extension
0 ! cok(f) ! N ! ker(f) ! 0:
We will construct an Adams tower for X that can be related to a tower for Y
by using a particular resolution of M. There is a short exact sequence
0 ! P= ker(f) ! M ! cok(f) ! 0;
and so we note that both M and N are given by extensions which involve
cok(f) and ker(f). Let C* and D* be minimal projective resolutions of cok(f)
and ker(f), respectively. Later in the paper we will need the following lemma.
Lemma 2.1 D* is a projective resolution of ker(f).
Proof. The kth homology group of the complex D* is 1kker(f). If k > 1,
then the functor 1kis identically zero. On the other hand, Sq0 acts freely on
ker(f) because ker(f) P , which is itself Sq0free. Hence 11ker(f) = 0.
We construct a projective resolution of M as follows. Since D* is a projective
resolution of ker(f) and P is projective, we can make a resolution of P= ker(f)
by shifting Di to homological dimension i + 1 and putting P into homological
dimension 0. Let B* be this chain complex,
. .!.D1 ! D0 ! P;
which resolves P= ker(f). Augment B* by f : P ! M to obtain HiB* = 0 for
i 0 and H1B* = cok(f). Let Shift(C*) be the augmented chain complex ob
tained by shifting C* down one homological degree, that is, Shift(C*)i = Ci+1,
and let the augmentation be the differential C1 ! C0. Then HiShift(C*) = 0
for i 0 and H1Shift(C*) = cok (f). It is easy to construct a chain map
g : Shift(C*) ! B*,
. .C.3   ! C2  ! C1 ffl!C0
?? ? ? ?
?y ??y ??y ??y
. .D.1   ! D0  ! P ffl!M;
which is an isomorphism on all homology groups. In dimension 1, we define
g by a lift C0 ! M of C0 ! cok(f). Then we note that C1 ! C0 ! M lifts
to P , since C1 ! C0 ! M ! cok(f) is zero. The rest of the map g follows
from projectivity of C* and acyclicity of D*.
4
Let X* be the algebraic mapping cone. Hence X0 = C0 P , Xi = Ci Di1
for i 1, X* has an augmentation to M, and the differential is defined by
d = (dC ; g dD ):
D2 _____//D1____//D0____//PPPPP>>>>>>
___ ___ """ PP''
____ ___ """ nM77
_ __ "" nnnnn
C3 _____//C2____//C1____//C0
The augmented complex X* is acyclic because g is a homology isomorphism.
Hence X* ! M is a projective resolution, and it can be realized to give
an Adams tower {Xk} for X. (See, for example, [2, Section 3].) We note the
following facts about the tower {Xk}, which follow from its being a realization
of the projective resolution X* ! M where H*X ~= U(M):
(1) limkXk ' X^2.
(2) The kinvariants Xk ! Kk(Ck+1 Dk) are MasseyPeterson maps.
(3) The composite of the inclusion of a fiber with a kinvariant
Kk(Ck Dk1) ! Xk ! Kk(Ck+1 Dk)
realizes the looped down differential of X*.
(4) X0 ' KC0 x KP .
To construct a chain complex modeling Y , we define E* as the quotient of X*
by P in dimension 0: Ei = Xi for i 1 and E0 = X0=P . If we regard P as
a chain complex concentrated in degree 0, there is a short exact sequence of
chain complexes
0 ! P ! X* ! E* ! 0
which can be regarded as modeling the fibration Y ! X ! KP . We realize
this situation by defining a new tower as follows. Define a map of {Xk} to
the constant tower {KP } by Xk ! X0 ! KP , where the second map is
projection, and let {Ek} be the tower of fibers. The following lemma gives the
essential facts about {Ek}.
Lemma 2.2
(1) limkEk ' Y2^.
(2) The kinvariants Ek ! Kk(Ck+1 Dk) are MasseyPeterson maps.
(3) The composites
Kk(Ck Dk1) ! Ek ! Kk(Ck+1 Dk)
realize the looped down differential of E*.
5
(4) The commutative diagram of kinvariants produced by the map of towers
{Ek} ! {Xk} realizes the quotient map of chain complexes X* ! E*
appropriately looped down.
Proof. For the first item, we note that since Ek is the fiber of the map Xk !
KP , we have that limkEk is the fiber of the inverse limit map limkXk ! KP .
In view of the fact that KP is its own 2completion, since it is a mod 2
EilenbergMacLane space, the map between inverse limits is the 2completion
of X ! KP . Since all the spaces are simply connected, the fiber lemma for
pcompletion tells us that the fiber of the map between completions is the
completion of the fiber, that is,
limEk ' Fiber(limXk ! KP )
k k
' Fiber(X^2! KP )
' [Fiber (X ! KP )]^2
' Y2^
To relate the tower {Xk} and the tower {Ek}, we study the following commu
tative diagram:
Ek+1   ! Ek   ! Kk(Ck+1 Dk)
?? ? ?
?y ??y =??y
Xk+1   ! Xk   ! Kk(Ck+1 Dk)
?? ? ?
?y ??y ??y
KP  =! KP   ! *
The upper right hand square gives the kinvariants, and hence the third part
of the lemma follows immediately from the corresponding fact for {Xk} and
the following commuting diagram of kinvariants and inclusions of fibers:
Kk(Ck Dk1)  ! Ek   ! Kk(Ck+1 Dk)
?? ?? ??
=?y ?y =?y
Kk(Ck Dk1)  ! Xk   ! Kk(Ck+1 Dk):
Likewise the fourth part of the lemma follows from noting that at the bottom
6
of the towers we have
E0 ' KC0  ! K(C1 D0)
?? ?
?y =??y
X0 ' K(P C0)  ! K(C1 D0)
?? ??
ss?y ?y
KP  ! *:
To prove the second part of the lemma, we proceed inductively. Consider the
map of fibrations arising from the first kinvariant:
E1  ! E0 ' KC0  ! K(C1 D0)
?? ? ?
?y ??y ??y
X1  ! X0 ' K(P C0)  ! K(C1 D0):
Each of the kinvariants is clearly a MasseyPeterson map, and so the coho
mologies of E1 and X1 are "very nice" in the sense of MasseyPeterson, say
H*E1 ~= U(N1) and H*X1 ~= U(M1). The fundamental sequence is natural,
giving us a ladder of short exact sequences:
0   ! cok (f)  ! N1  ! ker(C1 D0 ! C0)  ! 0
x? x x
?? ??? ???
0   ! M  ! M1  ! ker(C1 D0 ! C0 P )  ! 0:
In order to show that E1 ! K(C2 D1) is MasseyPeterson, we need to
know that H*E1 is polynomial. It suffices to check that Sq 0 acts freely on
the two ends of the short exact sequence for N1. Certainly Sq0 acts freely on
ker(C1 D0 ! C0), since this is a submodule of the projective (C1 D0).
Further, by the assumption that H*Y ~= U(N) is polynomial and the fact that
cok(f) N, Sq0 acts freely on cok(f). It follows easily that E1 ! K(C2D1)
is MasseyPeterson.
We need to examine what happens in cohomology for the fibration E2 ! E1 !
K(C2 D1), and then the rest of the induction will follow. In particular,
we need to know cok((C2 D1) ! N1) in order to understand the cokernel
side of the fundamental sequence for H*E2. Consider the following commuting
diagram, where the rows are MasseyPeterson fibrations:
Y  ! X   ! KP
?? ? ?
?y ??y ??y
E1  ! E0 ' KC0   ! K(C1 D0):
(Commutativity of the right hand square follows from the fact that the com
7
posite C1D0 ! C0P ! M is zero because it is the square of the differential
in the chain complex X*.) We obtain a commuting ladder of fundamental se
quences for H*Y and H*E1:
0  ! cok(f)   ! N  ! ker(f)  ! 0
x? x x
?? ??? ???
0  ! cok(f)   ! N1  ! ker(C1 D0 ! C0)  ! 0:
Now the map E1 ! K(C2D1) induces a map (C2D1) ! N1 which goes
into the ker(C1 D0 ! C0) part of N1; specifically, it maps to ker(C1
D0 ! C0 P ). Therefore cok((C2 D1) ! N1) is an extension of cok(f) by
cok [ ker(C1 D0 ! C0 P ) ! ker(C1 D0 ! C0)];
which is ker(f), and the ladder of exact sequences above shows that this
extension is N.
It is now easy to see that im (H*Ek ! H*Ek+1) ~= U(N) for k > 1, and
since the part of H*Ek+1 coming from the fiber is also polynomial, we have
that H*Ek+1 is polynomial, as required. The induction now follows without
difficulty.
3 Adams tower for Y
In this section, we take the tower {Ek} constructed in Section 2 and "rear
range" it into an Adams tower for Y . We do this by taking the kinvariants,
which are products, and separating them in such a way as to give a tower
{E0k} which is interleaved with the tower {Ek}. In Section 4, we will study the
homotopy spectral sequence of the tower {E0k} and show that this tower is an
Adams tower for Y .
Recall that Ek+1 is the fiber of a MasseyPeterson map Ek ! Kk(Ck+1Dk).
The tower {E0k} is defined as follows.
Definition 3.1 Let E0kbe the homotopy fiber of Ek ! KkDk. Define a map
Ek+1 ! E0kby the commutative ladder
Ek+1  ! Ek  ! Kk(Ck+1 Dk)
?? ? ?
?y =??y ss??y
E0k  ! Ek  ! KkDk
The map E0k+1! E0kis defined as the composite E0k+1! Ek+1 ! E0k.
8
Corollary 3.2 limkEk ' limkE0k.
Our first task is to understand the kinvariants for the tower {E0k}, so that
we will later be able to analyze the homotopy spectral sequence. We need
the following lemma, which allows us to combine kinvariants of successive
fibrations.
Lemma 3.3 Suppose given a tower of principal fibrations
W3


fflfflg
W2 _____//K2==___
____
p ________h
fflfflf_____
W1 _____//K1
and suppose that g factors up to homotopy as g = hp. Then the homotopy
fiber of (f; h) : W1 ! K1 x K2 is homotopy equivalent to W3.
Proof. The commutative square
p
W2   ! W1
?? ?
?y f??y
*   ! K1
is a homotopy pullback square, hence remains so when we cross the bottom
row with K2 to obtain the square
p
W2   ! W1
?? ??
g?y (f;h)?y
(*;id)
K2   ! K1 x K2:
Since the fibers of the vertical maps are homotopy equivalent and the fiber of
g is W3, the lemma follows.
We want to apply this lemma to find a kinvariant that will get us from E0kto
E0k+1, and as a preliminary, we find a kinvariant to go from E0kto Ek+1.
Lemma 3.4 The homotopy fiber of the composite E0k! Ek ! KkCk+1 is
Ek+1.
9
Proof. Apply the same reasoning as in the previous lemma, crossing the
bottom row of the homotopy pullback square
E0k  ! Ek
?? ?
?y ??y
*  ! KkDk;
with KkCk+1. Then the fiber of the new vertical maps will be Ek+1.
To show there is a kinvariant to go from E0kto E0k+1and to compute it, we
look at the following tower of three spaces:
E0k+1
??
?y
Ek+1  ! Kk+1Dk+1
??
?y
E0k  ! KkCk+1:
We next show that the second kinvariant factors through E0kand apply
Lemma 3.3.
Lemma 3.5 The map Ek+1 ! Kk+1Dk+1 factors through Ek+1 ! E0kup to
homotopy.
Proof. Since we are mapping into an EilenbergMacLane space, the statement
amounts to a factoring on the level of cohomology. To explain the algebraic
fact that makes this work, consider a part of the tower {Ek}:
Kk+1(Ck+1 Dk)  ! Ek+1  ! Kk+1(Ck+2 Dk+1)
??
?y
Ek  ! Kk(Ck+1 Dk):
The composite Kk+1(Ck+1 Dk) ! Ek+1 ! Kk+1(Ck+2 Dk+1) is a
realization of the looped down chain complex differential of the complex E*.
The critical fact we need for our proof is that the composite
Dk+1 ,! Ck+2 Dk+1 ! Ck+1 Dk ! Ck+1
is zero. Heuristically, this means that putting in Kk+1Ck+1 is not necessary
in order to put in Kk+1Dk+1; once we have put in Kk+1Dk (to obtain E0k),
we can then put in Kk+1Dk+1.
10
We compare the fundamental sequences for the cohomology of Ek+1 and E0k
arising from the fibrations
Ek+1  ! Ek  ! Kk(Ck+1 Dk)
?? ? ?
?y =??y ss??y
E0k  ! Ek  ! KkDk:
Let H*Ek ~= U(Nk), let H*E0k ~= U(N0k), and note that the cokernel of
k(Ck+1Dk) ! Nk is N. The ladder of fundamental sequences corresponding
to the above diagram is
0 ! cok (kDk ! Nk) ! N0k ! ker(kDk ! Nk) ! 0
?? ? ?
y ?y ?y
0 ! N ! Nk+1 ! ker(k(Ck+1 Dk) ! Nk) ! 0:
The left vertical arrow is an epimorphism because N ~= cok[k(Ck+1 Dk) !
Nk], and we are mapping to it from the cokernel of kDk ! Nk.
Now we want to consider the map k+1Dk+1 ! Nk+1 and show that it lifts to
N0k. The composite
k+1Dk+1 ! Nk+1 ! ker(k(Ck+1 Dk) ! Nk)
is zero into the factor k(Ck+1), because of the critical algebraic fact mention*
*ed
above. Hence k+1Dk+1 lifts to ker(kDk ! Nk). Then a standard diagram
chase using the projectivity of k+1Dk+1 and the fact that cok(kDk ! Nk) !
N is an epimorphism shows that k+1Dk+1 ! Nk+1 lifts to N0k.
Corollary 3.6 E0k+1is the fiber of a MasseyPeterson map
E0k! KkCk+1 x Kk+1Dk+1:
Note that, in particular, E00' KC0 x KD0.
4 Homotopy Spectral Sequences
In this section we study the homotopy spectral sequence for the tower {E0k}
constructed in Section 3. By examining d1 we show that {E0k} is an Adams
tower for Y . Because the homotopy spectral sequence of {Ek} is almost the
same as that of {Xk}, that is, the UASS for X, we can compare the UASS
11
for X and for Y using the two towers {Ek} and {E0k}. We deduce relation
ships between the differentials to establish Theorem 1.1 and the related results
Theorem 4.5 and Corollary 4.7.
To study d1 of the homotopy spectral sequence of {E0k}. we need to examine
the composite
KkCk x Kk+1Dk ! E0k! KkCk+1 x Kk+1Dk+1:
The following lemma relates these maps to the corresponding kinvariants and
inclusions of fibers for {Ek}.
Lemma 4.1 The following diagram of kinvariants and inclusions of fibers
commutes:
Kk+1Ck+1 x Kk+1Dk  ! Ek+1  ! Kk+1Ck+2 x Kk+1Dk+1
?? ?? ??
(*;ss)?y ?y (*;ss)?y
KkCk x Kk+1Dk  ! E0k  ! KkCk+1 x Kk+1Dk+1
?? ?? ??
(ss;*)?y ?y (ss;*)?y
KkCk x KkDk1  ! Ek  ! KkCk+1 x KkDk
Proof. The upper left square commutes by the commutative ladder which
defines the map Ek+1 ! E0k. (See Definition 3.1.) The upper right square
commutes by Lemma 3.5 and the fact that Ek+1 ! E0k! KkCk+1 is a
fibration. The lower right square commutes because E0k! KkCk+1 is defined
as the composite E0k! Ek ! KkCk+1 and because E0k! Ek ! KkDk is
a fibration.
For the lower left square, we expand the 2 x 2 square
Ek  ! E0k1
?? ?
?y ??y
KkDk  ! *
to a 3 x 3 square:
KkCk x Kk+1Dk  ! E0k  ! E0k1
?? ? ?
?y ??y =??y
KkCk  ! Ek  ! E0k1
?? ?? ??
*?y ?y ?y
KkDk =! KkDk  ! *:
12
Because the composite KkCk ! Ek ! KkDk is known to be null, the map
KkCk x Kk+1Dk ! KkCk is homotopic to projection.
Proposition 4.2 {E0k} is an Adams tower for Y .
Proof. We already know that limkE0k' limkEk ' Y2^. We must show that
(1) Y ! E0kgives an epimorphism H*E0k! H*Y ;
(2) ker(H*E0k! H*E0k+1) ~=ker(H*E0k! H*Y ).
Consider the commutative ladder of fundamental sequences at the end of the
proof of Lemma 3.5. The map Ek+1 ! E0kinduces a map N0k! Nk+1 which
is onto N. Since Y ! Ek+1 induces an isomorphism from N Nk+1 to
N H*Y ~= U(N), the desired epimorphism H*E0k! H*Y follows. To prove
the second item, we note that Y ! E0kfactors as Y ! E0k+1! Ek+1 ! E0k.
Since ker(H*E0k! H*Ek+1) ~= ker(H*E0k! H*Y ) the desired isomorphism
follows.
Now we wish to calculate the homotopy spectral sequence of {E0k}, which is
the UASS for Y . The d1 differential comes from the composite
KkCk x Kk+1Dk ! E0k! KkCk+1 x Kk+1Dk+1:
The components ss*KkCk ! ss*KkCk+1 and ss*Kk+1Dk ! ss*Kk+1Dk+1
are zero because C* and D* are minimal resolutions. Further, by Lemma 4.1
Kk+1Dk ! E0k! KkCk+1 is actually null because it factors through E0k!
Ek and Kk+1Dk ! E0k! Ek is a fibration. Hence the only possible nonzero
part of d1 takes ss*KkCk ! ss*Kk+1Dk+1.
Remark 4.3 Note the contrast between the d1 differentials in the homotopy
spectral sequences for {Ek} and {E0k}. In the homotopy spectral sequence for
{Ek}, d1 comes from the composite
KkCk x KkDk1 ! Ek ! KkCk+1 x KkDk:
The only possible nonzero component of d1 is ss*KkDk1 ! ss*KkCk+1,
corresponding to the part of the chain complex differential of E* taking Ck+1 !
Dk1.
Before giving the proof of Theorem 1.1, we need the following preliminary
lemma.
Lemma 4.4 Let x 2 ss*KkCk be a boundary in the homotopy spectral se
quence of {Ek}. Then x is in the kernel of d1 of the homotopy spectral sequence
of {E0k}.
13
Proof. We use the diagram of Lemma 4.1. Let y be the image of x in ss*E0k.
If x is a boundary in the homotopy spectral sequence of {Ek}, then x goes
to 0 in ss*Ek, and so y does also. The long exact sequence for the fibration
Kk+1Dk ! E0k! Ek gives a preimage for y, call it z, in ss*Kk+1Dk.
However, d1(z) is certainly zero, and therefore since x and z have the same
image in ss*E0k, we have d1(x) = 0 also.
We are now ready to prove our Ext splitting result, which we reproduce here
for the convenience of the reader.
Theorem 1.1 Let X be a simply connected space with polynomial mod 2
cohomology H*X ~= U(M), and let X ! KP be MasseyPeterson, inducing
f : P ! M in cohomology. Let Y be the fiber of X ! KP , and suppose
H*Y ~= U(N) is polynomial. If X has E2 = E3 in its unstable Adams spectral
sequence, then there is a splitting
Ext *U_(N; tF2 )~=Ext *U_(cok (f); tF2 )Ext*U_( ker(f) ; tF2 :)
Proof. In view of the remarks preceding the statement of the theorem in
the introduction, and the fact that D* ! ker(f) and C* ! ker(f) are
resolutions, we have only to show that d1 in the homotopy spectral sequence
of {E0k} is zero. Assuming that C* and D* are chosen to be minimal resolutions,
this will follow from showing that ss*KkCk ! ss*Kk+1Dk+1 is zero.
We will relate this part of d1 to the UASS for X using the diagram of
Lemma 4.1. To do this, we must discuss the relationship of the homotopy
spectral sequences for {Ek} and {Xk}, the latter being the UASS for X. We
claim that they are almost the same. To be more precise, note the map of
towers {Ek} ! {Xk} is the identity on the kinvariants except at the very
bottom, where E0 ! X0 is the inclusion KC0 ! KC0 x KP . Hence any
nonzero differential in {Ek} gives rise to a corresponding nonzero differential
in {Xk}, and any zero differential in {Ek} gives rise to a corresponding zero
differential in {Xk}. In fact, there are only two differences between the ho
motopy spectral sequences for {Ek} and {Xk}. The first is that classes in the
homotopy spectral sequence of {Xk} which are boundaries of ss*KP will sur
vive to E1 in the homotopy spectral sequence of {Ek}. These correspond to
the kernel of ss*Y ! ss*X. The second is that infinite cycles in the homotopy
spectral sequence of {Xk} which are in ss*KP will not appear in the homotopy
spectral sequence of {Ek}. These correspond to the cokernel of ss*Y ! ss*X.
To complete the proof of the theorem, we claim that the diagram of Lemma 4.1
shows that a nonzero map ss*KkCk ! ss*Kk+1Dk+1 would produce a
nonzero d2 in the homotopy spectral sequence of {Ek}, and hence in the UASS
14
of X. To prove this, we use the interplay between the homotopy spectral se
quences for the towers {Ek} and {E0k}. If x 2 ss*KkCk has a nonzero image
in ss*Kk+1Dk+1, then x is not in the kernel of d1 for the homotopy spectral
sequence of {E0k}. Hence by Lemma 4.4, we know that x is not a boundary
in the homotopy spectral sequence for {Ek}. However, this means that x sur
vives to E2 in the homotopy spectral sequence for {Ek}, because all elements
of ss*KkCk are in the kernel of d1. On the other hand, since E0k+1is the fiber
of Ek+1 ! Kk+1Dk+1, x does not lift even to E0k+1, let alone to Ek+2, and
so x must support a d2 differential.
Therefore if X has E2 = E3 in its unstable Adams spectral sequence, that is,
there are no nonzero d2's, it must be that ss*KkCk ! ss*Kk+1Dk+1 is zero,
completing the proof of the theorem.
Lastly, we prove a theorem about d2 in the UASS for Y .
Theorem 4.5 Given the hypotheses of Theorem 1.1, let x 2 ss*KkDk1 and
let ffi : ss*KkDk1 ! ss*KkCk+1 be given by d1 in the homotopy spectral
sequence of {Ek}. In the unstable Adams spectral sequence of Y , d2(x) =
ffi(x) y where y 2 ss*Kk+1Dk+1. If x 2 ker(ffi), then y = 0.
Remark 4.6 If we write Hom t(A; B) for morphisms A ! tB in the category
of unstable A modules, then ffi is the only nonzero part of the differential in
the cochain complex Hom *(E*; F2), where E* is the chain complex constructed
in Section 2. Hence this theorem says that d2's in the unstable Adams spectral
sequence for Y can be determined up to an error term y by the primary level
algebraic calculation of E*. Furthermore, if an element x looks like a cycle
under d2 (ffi(x) = 0), then it is a cycle (the error term vanishes).
Proof of Theorem 4.5 Again we examine the diagram in Lemma 4.1, this
time extending it downward by a row:
KkCk x Kk+1Dk  ! E0k  ! KkCk+1 x Kk+1Dk+1
?? ?? ??
(ss;*)?y ?y (ss;*)?y
KkCk x KkDk1  ! Ek  ! KkCk+1 x KkDk
?? ?? ??
(*;ss)?y ?y (*;ss)?y
Kk1Ck1 x KkDk1  ! E0k1  ! Kk1Ck x KkDk:
The first part of the theorem follows from a diagram chase starting in the lower
left hand corner. The second part follows once we note that if x 2 ker(ffi), th*
*en
x lifts to Ek+1, and therefore also to Ek+2 (there are no nonzero d2 differenti*
*als
15
in the homotopy spectral sequence for {Ek}). Therefore x must go to zero in
ss*Kk+1Dk+1 or there would be an obstruction to lifting.
Corollary 4.7 If E2 = E1 for X, then E3 = E1 for Y .
Essentially, the corollary says that if E2 = E1 for X, then the UASS for Y
is almost completely determined by the primary level calculation of the chain
complex E*. The only piece of information missing is the potentially nonzero
term y of Theorem 4.5.
Proof. If E2 = E1 for X, then once a class in homotopy lifts from Ek to
Ek+1, it lifts all the way up the tower to Y2^. If a class lifts from E0k1to E*
*0k+1,
it has lifted from Ek to Ek+1 in the process, hence lifts all the way to Y2^.
5 Examples
In this section, we discuss two examples of Theorem 1.1 and Theorem 4.5. In
the first example, we show how the theory works itself out in a very easy case,
by considering the unstable Adams spectral sequence for Y = K(Z=4; n). In
the second example, which motivated looking for a theorem along the lines
of Theorem 1.1, we discuss a conjecture of Bousfield on the unstable Adams
spectral sequence for the Lie group SO, and we give a result related to the
first stage of the conjecture.
5.1 The UASS for K(Z=4; n)
As an example where Theorems 1.1 and 4.5 can be easily and completely
worked out, we let X = K(Z=2; n), KP = K(Z=2; n + 1), and f : X ! KP
be Sq1. Hence the space Y whose UASS is being described by the theorems is
Y = K(Z=4; n). Of course, we already know everything there is to know about
the UASS in this case, so it is simply an exercise in following the definitions
of the previous sections and seeing where they lead, particularly regarding the
differentials.
We write F (n)_for the free unstable A module on a generator of dimension n,
and_we write F (n) F_(n)=Sq 1. Then M = F (n), P = F (n + 1), cok(f) =
F (n), and ker(f) ~=F (n + 2). The resolution C* ! cok(f) has Ci = F (n + i),
and the resolution D* ! ker(f) has Di = F (n + i + 2), while in both cases all
the differentials are Sq1.
16
If we lift C0 i cok(f) to M, we get the identity map F (n) ! F (n). Hence
the chain map g : Shift(C*) ! B*,
. .C.3   ! C2  ! C1 ffl!C0
?? ? ? ?
?y ??y ??y ??y
. .D.1   ! D0  ! P ffl!M;
becomes
Sq1 Sq1 ffl
. . .F (n + 3) ! F (n + 2)  ! F (n + 1)  ! F (n)
?? ?? ?? ??
= ?y =?y =?y =?y
Sq1 Sq1 ffl
. . .F (n + 3) ! F (n + 2)  ! F (n + 1)  ! F (n):
The complex E*, which we realize to obtain Y , is the mapping cone, without
augmentation and factored out by P :
D2 ____//_D1___//_D0>>
__ __>>
_____ ____
_ __
C3 _____//C2____//_C1___//_C0:
In our example, the complex E* becomes
F (n + 4)_____//F8(n8+ 3)___//_F (n + 2)
qqq qq88q
qqqqqq qqqqq
qq qqq
F (n + 3)_____//F (n + 2)___//_F (n + 1)__//_F (n);
where the slant maps are the identities, since they come from the vertical
maps in g*. To obtain the Adams resolution for Y , we "pull" D* down one
homological dimension (looping as we do so since we are beginning in a lower
homological dimension and the topological dimension must remain the same):
D3 _____//D2 _____//D15_____//D05kk55k
k k k kk
k k k
k k k k k
C3_______//C2______//C1_____//_C0:
The dashed slant arrows indicate that we are remembering where d1 used to
go, because it will become d2 in the UASS. In our example, we get
F (n + 5) _____//F (n + 4)_____//F3(n3+g3)_____//F (n + 2)
g g g g g g g33
gg g g g g g
gg g g g gg
F (n + 3) ______//_F (n +_2)____//F (n + 1)______//F (n):
17
*
Finally, to get the E2term of the UASS for Y , we compute Hom (; F2):
n+4F2 oo___n+3F2 oo___n+2F2i oo___n+1F2i
i i i i i i i i
ttiii ttiii
n+3F2 oo___n+2F2 oo___n+1F2 oo___nF2:_
Rewriting this in the usual s versus (t  s) format, and including the d2's
indicated by the slant arrows, we obtain the UASS for Y = K(Z=4; n):

_ r rd
 AK 
_ rA rd
 A
 AK 
_ rAArd
 A
 AAKA
_ r rd
 AKA
_ rAArd
1  A
  
________________________________rrdA
n n + 1
The UASS for K(Z=4; n).
5.2 The UASS for SO
Theorem 1.1 and Theorem 4.5 came up as part of an effort to prove a conjec
ture of Bousfield about the E2term_of the unstable Adams spectral sequence
* 1 *
for the Lie group SO. Let P1 = H RP , and recall that H SO ~= U(P1 ).
Hence in the UASS for SO, we have Es;t2~=Ext sU_(P1 ; tF2 .)Writing the
elements of P1 as xi, we filter P1 by letting Pn be the elements xi with at
most n nonzero digits in the dyadic expansion of i. There is an associated
spectral sequence converging to Ext sU_(P1 ; tF2 ,)and Bousfield's conjecture
is that this spectral sequence collapses, giving
Ext sU_(P1 ; tF2 )~=nExt sU_(Pn=Pn1 ; tF2 :)
For an algebraic exploration of the conjecture and of the conjectured workings
of the UASS of SO, see [1].
It is a suggestion of Mahowald that one try to relate Bousfield's conjecture to
the Postnikov tower for SO, which consists of spaces with polynomial coho
mology of the form U() and kinvariants which are MasseyPeterson maps.
In Figure 1 we display the beginning of_the_tower._Note that the composites
knin run cyclically through the list Sq 3, Sq 5, Sq 2, Sq 2. If we let Xn be the
nth space in the Postnikov tower, then im (H*Xn ! H*SO) ~= U(Pn). That
is, the filtration of H*SO given by the Postnikov tower is the same as that
18
..
.
K(Z=2; 8)  i4!X4 k4! K(Z=2; 10)
??
?y
K(Z; 7)  i3!X3 k3! K(Z=2; 9)
??
?y
K(Z; 3)  i2!X2 k2! K(Z; 8)
??
?y
K(Z=2; 1)  i1!X1 k1! K(Z; 4)
??
?y
*
Fig. 1. The Postnikov tower for SO
given by dyadic expansion. Hence the Postnikov tower gives some geometric
significance to the dyadic filtration of P1 . The theorems in this paper are a
first attempt to study the conjecture of Bousfield by comparing neighboring
terms in the Postnikov tower of SO.
It is an easy matter to split off the first two filtrations.
Proposition 5.1 Ext sU_(P1 ; tF2 )is isomorphic to the direct sum
Ext sU_(P1 ; tF2 E)xtsU_(P2=P1 ; tF2 )ExtsU_(P1 =P2 ; tF2: )
Proof. First, we claim that the boundary map in the long exact sequence for
Ext arising from the short exact sequence
0 ! P1 ! P1 ! P1 =P1 ! 0
is zero. To prove this, it is sufficient to show that the map
Ext sU_(P1 ; tF2 !)Ext sU_(P1 ; tF2 )
is an epimorphism. Since P1 ~=F (1), a free module in the category of unstable
A modules, we have Ext sU_(P1 ; tF2 =)0 when s > 0, so surjectivity is clear
for s > 0. For s = 0, we need only note that P1 ,! P1 takes the generator of
P1 to a generator of P1 .
To finish the proof of the proposition, we will show that the boundary map in
19
Ext associated to the short exact sequence
0 ! P2=P1 ! P1 =P1 ! P1 =P2 ! 0
is zero. Again, it is sufficient to show that
Ext sU_(P1 =P1 ; tF2 !)Ext sU_(P2=P1 ; tF2 )
__ s
is an epimorphism. Since P2=P1 ~=F (3), we know that Ext U_(P2=P1 ; tF2 )=
0 unless t  s = 3, so we certainly have an epimorphism for t  s 6= 3. On the
other hand, P1 =P2 has its first class in dimension 7, so Ext sU_(P1 =P2 ; tF2 *
*=)
0 for t  s < 7. Therefore we have the required epimorphism for t  s = 3
also.
Because of the preceding proposition, to prove Bousfield's conjecture it is
sufficient to prove a splitting for Ext sU_(P1 =P2 ; tF2. )Note that U(P1 =P2) *
*~=
H*SO<7>, where SO <7> denotes the 6connected cover of SO. The first stage
of a splitting of Ext sU_(P1 =P2 ; tF2 w)ould be
Ext sU_(P4=P2 ; tF2 )~=ExtsU_(P4=P3 ; tF2 )ExtsU_(P3=P2 ; tF2 :)
We can use Theorems 1.1 and 4.5 to offer a step in this direction. The first
two stages of the Postnikov tower for SO <7> are
K(Z=2; 8)  ! SO[7; 8]
??
?y
Sq2
K(Z; 7)  ! K(Z=2; 9);
and this is a situation to which we can apply Theorems 1.1 and 4.5.
We note the following facts, which follow directly from [3], [6], and the Masse*
*y
Peterson Theorem:
__ 2
(1) Let f : F (9) ! F (7) be Sq . Then cok(f) ~=P3=P2, and so H*SO[7; 8] ~=
U(N) where there is a short exact sequence
0 ! P3=P2 ! N ! ker(f) ! 0:
(2) There is a short exact sequence
0 ! ! ker(f) ! P4=P3 ! 0:
(3) There is a short exact sequence
0 ! ! N ! P4=P2 ! 0:
20
As an immediate corollary of these facts and Theorem 1.1, we have the fol
lowing.
Proposition 5.2
Ext sU_(N; tF2 )~=Ext sU_(P3=P2 ; tF2 )ExtsU_( ker(f) ; tF2 :)
It further follows from the facts enumerated above that there is a filtration
0 = N0 N1 N2 N3 = N
with the property that
N1=N0 ~= P3=P2
N2=N1 ~=
N3=N2 ~= P4=P3:
There is an associated spectral sequence converging to Ext sU_(N; tF2 ), and
the preceding proposition tells us that all elements of Ext sU_(P3=P2 ; tF2 a)re
infinite cycles. Another result necessary for the first level of a splitting re*
*sult
would be the following.
Conjecture 5.3 d1 : Ext sU_(; tF2 !)Ext s+1U_(P4=P3 ; tF2 i)s zero.
Corollary to Conjecture 5.3
Ext sU_(P4=P2 ; tF2 )~=ExtsU_(P4=P3 ; tF2 )ExtsU_(P3=P2 ; tF2 :)
Proof. Conjecture 5.3 would tell us that
Ext sU_(N; tF2 )~=Ext sU_(P3=P2 ; tF2 )ExtsU_(; tF2 )
Ext sU_(P4=P3 ; tF2 :)
However, it is also possible to filter N by
0 = N00 N01 N02 N03= N
with the property that
N01=N00~=
N02=N01~= P3=P2
N03=N02~= P4=P3
21
and the spectral sequence for Ext arising from this filtration would also col
lapse. Therefore the inclusion ! N would give an epimorphism
Ext sU_(N; tF2 )! Ext sU_(; tF2 ;)
and hence the short exact sequence
0 ! ! N ! P4=P2 ! 0
would split on Ext. The corollary would then follow.
Even in the absence of Conjecture 5.3, however, we can apply Theorem 4.5 and
Corollary 4.7 to get the following chart for the UASS of SO [7; 8]. The plain
dots come from the summand Ext sU_(P3=P2 ; tF2 .)The circled dots come
from the summand Ext sU_( ker(f) ; tF2 .)No circled dot can survive to be
an element of homotopy, which is found only in dimensions 7 and 8, so since
E3 = E1 , all circled dots support d2 differentials. Further, the error term y *
*of
Theorem 4.5 is always zero, because y must be a d2 cycle and in this situation
there are no such cycles, as we have just discussed.


10 _ r r rd
   
_ r r rd
   
_ r r rd
   
_ r r rd
   
_ r r rd r
   
_ r r rd r r
5    
    AAK
_ r r rd r r Ar rd
     A%
_ r r rd r r rrrdd%
  AK   
_ r r A rd r rd rd rd
  AK  AK  
_ r r A Ard A rdrd rd
  
___________________________________________________________*
*rdrArdA
5 10 15
The UASS for SO [7; 8]
Acknowledgement
I am grateful to Mark Mahowald for many conversations about SO and to
Don Davis for introducing me to Bousfield's conjecture.
22
References
[1]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 (199*
*2)
4153.
[2]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.
[3]J. Long, Two contributions to the homotopy theory of Hspaces, Princeton
University thesis (1979).
[4]W. S. Massey and F. P. Peterson, The mod 2 cohomology structure of certain
fiber spaces, Mem. Amer. Math. Soc. Number 74, Providence, 1967.
[5]N. E. Steenrod and D. B. A. Epstein, Cohomology Operations, (Ann. Math.
Studies Number 50, Princeton University Press, Princeton, NJ, 1962).
[6]R. Stong, Determination of H*(BO(k; 1)) and H*(BU(k; 1)), Trans. Amer.
Math. Soc. 107 (1963) 526544.
23