Composition Methods in the homotopy groups
of ring spectra
Brayton Gray
1.
Progress in calculating the homotopy groups of spheres has seen two major break
throughs. The first was Toda's work, culminating in his book [11] in which the *
*EHP
sequences of James and Whitehead were used inductively; "composition methods"
were used to construct elements and evaluate homomorphisms. The second was the
Adams spectral sequence. Each method has advantages and disadvantages. Toda's
method has the advantage that unstable homotopy groups are calculated along
with stable groups. It has the disadvantage that it applies only to spheres and*
* in
particular, naturality under maps between spaces cannot be applied. The Adams
method has the advantage that much of the bookkeeping work is accomplished in
advance during the calculation of the Ext groups. The disadvantage that it does
not calculate unstable groups has been eliminated, for certain nice spaces, by *
*the
work of [3]. This work implies that for certain spaces, the unstable homotopy i*
*s as
accessible as the stable homotopy. It is the purpose of this work to examine ho*
*w,
in certain cases, the methods of Toda can be used for spaces other than spheres.
We will begin by summarizing the methods used by Toda. We will discuss the
possibility of using these methods for other spectra and work out the example of
the Moore space spectrum S0 [pr e1 for p > 3.
2.
The main tool Toda used was the EHP sequence. Localized at 2, this is a long
exact sequence for each n 1
. . .___ ssr+2 (S2n+1 ) ____P
ssr(Sn ) ____Essr+1 (Sn+1 ) ____Hssr+1 (S2n+1 ) ____Pssr1 (Sn ) _*
*___ . . .
By induction first on the stem oe = r  n and then on n, the determination of
ssr+1 (Sn+1 ) above is done when the other 4 groups are known. Compositions are
used, and formulas for P and H on compositions allow one to calculate the groups
involved. As an example, we cite the following formulas:
2 Brayton Gray
Proposition 2.1.
a) H(ff O Efi) = H(ff) O Efi.
b) H(Eff O fi) = E(ff ^ ff) O H(fi).
c) P (ff O E2fi) = P (ff) O fi.
Proposition 2.2 (BarrattHilton formula). If ff 2 ssr(Sn ) and fi 2 sss(Sm ) t*
*hen
Em ff O Erfi = (1)(rn)(sm)En fi O Esff. (In other words, on Sn+*
*m ,
fffi = (1)(rn)(sm)fiff.)
We will demonstrate how to calculate the 4 stem with these methods. We
first need knowledge of previous stems.
Proposition 2.3.
group generators relations
8
< Z ; n = 2 j
ssn+1 (Sn ) = :
Z=2 ; n > 2 j 2j
ssn+2 (Sn ) = Z=2 j2
8
>>>Z=2 ; n = 2 3
>>> j
< Z=4 ; n = 3 ! 2! = j3
ssn+3 (Sn ) = >
>>>Z Z=4 ; n = 4 ; !
>>:
Z=8 ; n 5 2 = !
Now the Hopf map j :S3 ____ S2 induces an isomorphism in homotopy:
~= 2
ssr(S3) ____ ssr(S )
when r > 2, so ss6(S2) ' Z=4 generated by the composition j!. The map
ss5(S5) _____Pss3(S2) generates the kernal of E so P () = 2j. We wish to eval
uate P
ss8(S5) ____ ss6(S2)
ss8(S5) is generated by which is not a double suspension. However 2 = ! is a
double suspension so we apply 2.1 c)
P (2) = P (!) = P () O ! = 2j O ! = j O 2 O ! = j O ! O 2 = 2(j!)
since 2 O ! = ! O 2 stably by 2.2 and hence on S3 since ss6(S3) ____ ss6+n (S3*
*+n )
is a monomorphism. But P (2) = 2(j!) implies that P () = j! since P is a
homomorphism. It follows that E(j!) = 0 and 0 _____ ss7(S3) _____ ss7(S5) is
exact. We can define two compositions in ss7(S3): !j and j. We apply 2.1 (a)
in the first case to get H(!j) = H(!) O j = j2 and 2.1 (b) in the second case
Composition Methods in the homotopy groups of ring spectra 3
to get H(j O ) = j2 O H() = j2. Thus !j = j generates ss7(S3) and has
order 2. Since has Hopf invariant 1, the EHP sequence splits when n = 3 and
ssn+1 (S4) ' ssn(S3) ssn+1 (S7); in particular ss8(S4) ' Z=2 Z=2 generated by
!j = j and j. By 2.2, on S5; !j = j! = 0, so in the sequence
ss10(S9) ____Pss8(S4) ____Ess9(S5) ____Hss9(S9)
P (j) = !j. Since ss9(S5) is finite, H = 0 and ss9(S5) = Z=2 generated by j. By
2.2, on S6 j = j = !j = 0, so E ss9(S5) = 0. Furthermore, ss10(S11) = 0 so
ss10(S6) = 0 and ssn+4 (Sn ) = 0 for n 6. The interested reader will find that*
* the
calculation of the 5 stem is just as easy. There is one other useful result whi*
*ch we
did not need here.
Proposition 2.4 (BarrattToda formula). If ff 2 ssr(Sn ) and fi 2 sss(Sm ), th*
*e dif
ference
[ff; fi] = Em1 ff O Er1 fi  (1)(rn)(sm)En1 fi O Es1 ff :Sr+s1 ____ S*
*m+n1
is equal to P EH(ff) ^ H(fi) .
Using this one can see, for example, that
j = j + j = P () on S5
!j = !j + j! = P (j) on S4(since H(!) = j):
This method would continue very successfully if the homotopy groups of
spheres were finitely generated. They clearly are not and one soon runs out of
compositions. As a partial remedy to this Toda defined "secondary compositions",
or what are commonly called Toda brackets and proved formulas similar to 2.1 for
these operations.
When localized at p > 2, there are two types of EHP sequences. Define Sn
by the formula (
2k+1 ifn = 2k + 1
Sn = Sb
S2k = Jp1 (S2k) ifn = 2k
where Jm is the mth stage of the James construction.
Then the EHP sequences are defined by fibrations:
S2n ____ES2n+1 ____HpS2np+1
S2n1 ____ES2n ____HS2np1
where Hp is the pth James Hopf invariant and H is the TodaHopf invariant ([7],
[11]). Note that in case p = 2, these both degenerate into a single fibration
Sn ____ Sn+1 ____H2S2n+1
and Sn = Sn . Composition methods have been applied for p > 2 [6] with similar
success. Proposition 2.1 generalizes directly while 2.2 requires some considera*
*tion.
We define a modified suspension oe as follows: if f :Sr ____ Sn let us write o*
*e(f)
for the adjoint of E O f :Sr _____ Sn+1 . We will write SX = X ^ S1 for the
4 Brayton Gray
classical suspension of X and if f :X ____ Y , we write Sf :SX ____ SY for the
suspension of f.
Corresponding to Proposition 2.2 (at p = 2) we have
Theorem 2.5 (p local BarratHilton Theorem). Suppose that ff 2 ssr(Sn) and fi*
* 2
sss(Sm ). Then
oem ff O oerfi = (1)(rn)(sm)oen fi O oesff
in ss*(Sn+m ).
Proof. We consider 3 cases:
a) if n = 2k +1 and m = 2`+1, the formula holds on S2k+2`+2 ; since the su*
*s
pension E :ssr+s(S2k+2`+1 ) _____ ssr+s+1 (S2k+2`+2 ) is a monomorphis*
*m,
this formula holds on S2k+2`+1 = Sm+n1 and hence on Sm+n
b) if n = 2k and m = 2` + 1, apply case a) to oeff and fi to demonstrate t*
*he
formula on S2k+2`+1 = Sm+n
c) if n = 2k and m = 2`, new arguments are needed.
Let : S2n ____ bS2nbe the inclusion. To discuss case c) we rely on
Lemma 2.6. Suppose g :Sr ____ bS2n. Then the diagram:
Sr+1 ______SgSSb2n
QQ 6
Q Q S
oe(g) Qs 
S2n+1
commutes up to homotopy.
Proof. Let Kn = S2n1 [wn e2np2 be the 2np  2 skeleton of 2S2n+1 . According
to [7], [10], there is a map fl :SKn ____ bS2nsuch that fl has a right homotopy
inverse. In particular, there is a lifting of g to a map g0: Sr ____ SKn
Sr ________gbS2n
Q Q 6
QQ0 fl
g Qs 
SKn
Write oe(fl): S2Kn _____ S2n+1 for the adjoint of E O fl :SKn _____ S2n+1 . *
*We
now show that
S2Kn ______SflSSb2n
QQ 6
Q Q S
oe(fl) Qs 
S2n+1
commutes up to homotopy. The Lemma then follows by combining these two dia
grams.
Composition Methods in the homotopy groups of ring spectra 5
Now SSb2n ' S2n+1 _S4n+1 _. ._.S2(p1)n+1; let Hi: bS2n____ S2n1____ S2n*
*i1
be the restriction for the ith JamesHopf invariant for 1 i p  1. Combining,
we get a map:
p1Y
:bS2n____ S2ni1' (S2n+1 x . .x.S2n(p1)+1) ____ (SSb2n);
i=1
where the last map is the splitting map from the loops on a product to the loops
on a wedge. Since Hi has order prime to p for 1 < i p  1,
HiO fl :SKn ____ bS2n____ S2ni1
is null homotopic if 1 < i p  1. Therefore O fl factors through S2n+1 . Since
the adjoint of is an equivalence, Sfl factors through S2n+1 . This factorizati*
*on
can be identified as oe(fl) by applying the inclusion bS2n ____ S2n+1 and tak*
*ing
adjoints.
To finish the proof of 2.5, choose a map : bS2k^ bS2`____ bS2k+2`which is
degree 1 in dimension 2k + 2`. Then consider the diagram:
S2k ^ Ss ___________________________1^oe(fi)S2k+2l

OE Z Z aeae 
 Z 1^fi 1^ ae 
^1 Z Z aeae 
? Z ae 
oe(ff)^1 b2k s Z"2k b=ae2l 
S ^ S S ^ S 
ae> Z  
aeae Z Z  
ae Z 1^fi ^1 
aeae ff^1 Z Z  
ae ff^fi Z" ? ?
Sr ^ Ss __________________________bS2k^ bS2l________Sb2k+2l
JZ Z aeae> 6 6
J Z 1^fi ae  
J Z Z aeaeff^1 1^ 
J Z ae  
J Z"r b2ael b2k 2l 
1^oe(fJi) S ^ S S ^ S 
J 6 ae> "Z 
J  aeae Z Z 
J 1^ ae Z 
J  aeae ff^1 ^1 Z Z 
J^ ae oe(fi)^1 Z 
Sr ^ S2l _________________________ S2k ^ S2l
where the lemma is applied in 4 of the 6 triangular regions. If we identify oet*
*(f)
with 1 ^ f, we introduce signs (1)(s1)(r2k)and (1)(r2k)(2l1)which combine
to give (1)rs.
6 Brayton Gray
3.
The utility of the EHP method applies when what you are studying is not a single
space, but a sequence of spaces {Xn} together with a suspension homomorphism
ssr(Xn) ____oessr+1 (Xn+1 ). If oe is induced by a map Xn ____enXn+1 , what w*
*e are
in fact dealing with is a spectrum X = {Xn; en}. The utility of an EHP approach
depends on the extent to which one has control over the fiber of en:
Fn ____ Xn ____enXn+1
From this point of view, it is clear that one can form two stably equivalent sp*
*ectra
{Xn; xn} ' {Yn; yn} with the spaces Xn and Yn vastly different. We think of the
sequence of spaces {Xn} as an unstable development of the spectrum X. We seek
a favorable unstable development of X. Ideally, one might seek an unstable deve*
*l
opment in which Fn = k(n)X`(n)for some functions k(n); `(n). This is explored
in [9] and it appears formally that the spectra V (m) of Smith and Toda are good
candidates for this. In particular, V (1) = S0 and V (0) = S0 [p e1 for p > 2.
Considerable attention to this spectrum and the related spectra S0 [pr e1 will *
*be
deferred to section 4.
To apply "composition methods," unstable classes need to be composed. This
suggests a ring structure in ssS*(X) which is associative (composition is assoc*
*iative)
and commutative (if we wish to have a BarrattHilton formula). Thus one might
begin by considering a homotopy associative homotopy commutative ring spectrum
X. For Toda brackets we may need to consider higher homotopies of associativity
ahnd commutativity. However, in order to have a successful unstable composition
theory it is first necessary to develop a stable composition theory. To do this*
* we
will find certain self maps of a ring spectrum X which correspond to elements in
the stable homotopy of X in such a way that the product in ss*X corresponds to
composition.
Suppose now that X is a ring spectrum. Using the multiplication
X ^ X ____ X we can think of X as an "Xmodule spectrum".
Definition 3.1.Call a map OE: SrX ____ X right modular if there is a commutati*
*ve
diagram:
SrX ^ X ____OE^1X ^ X
Sr  
? OE ?
SrX _________X
Note that the composition of modular maps is modular. Write Mod r(X)
for the group of homotopy classes of right modular maps ae: SrX ____ X. Then
Mod r(X) is a graded ring with unit under composition.
Theorem 3.2. If X is homotopy associative, ssS*(X) ~=Mod *(X) as graded rings.
Composition Methods in the homotopy groups of ring spectra 7
Proof. Define F :ssSr(X) ____ Mod r(X) by F (ff) = bffwhere ^ffis the composi*
*tion:
SrX ____ff^1X ^ X ____ X
to check modularity, consider the diagram:
SrX ^ X ______ff^1^1X ^ X ^_X_^1X ^ X
Sr  1^  
? ff^1 ? ?
SrX __________X ^ X _________X
Now define G: Mod r(X) ____ ssSr(X) by G(OE) = OE O Sr
r OE
Sr = Sr ^ S0 ____SSr ^ X ____ X:
Clearly G O F = Id. To see that F O G = Id, consider the diagram:
Sr = Sr ^ S0 ^ X _____1^^1Sr ^ X ^ X___OE^1X ^ X
Q Q  
QQ 1^ 
1 Qs ? OE ?
SrX __________X
where (F O G)(OE) is the upper right hand composite. Furthermore, F is a ring
homomorphism. To see this consider the following diagram where ff 2 ssSr(X) and
fi 2 ssSs(X). The top right hand composition represents F (ff O fi) while the l*
*ower
left hand composition represents F (ff) O F (fi)
Sr ^ Ss ^ X _____________ff^fi^1X ^ X ^ X
 Q Q ^ j 
 QQff^fi 1^j j ^1
 Qs +jj ?
Sr^fi X ^ X X ^ X
 j3 Q 
 ff^jj1 Q Q 
?j j f^f QQs ?
SrX ______________________X
This completes the proof.
Now suppose in addition that X is homotopy commutative. Then by 3.2,
composition is graded commutative, i.e.;
bffO Srfbi~ (1)rsbfiO Ssfbf
4.
In this section we will develop the special case of the Moore space spectrum
S0[pre1, p > 2. By the results of [9], this spectrum can be represented as {Tm *
*; oen},
where T2n = S2n+1 {pr}, the fiber of the degree pr map on S2n+1 and T2n1 is a
8 Brayton Gray
space constructed by Anick [1] and developed further in [2] and [12]. T2n1 is *
*the
total space in a fibration
S2n1 ____ T2n1 ____ S2n+1
where the connecting map 2S2n+1 _____ S2n1 has degree pr. We will write
Tn(pr) when we wish to keep the exponent in mind. There are EHP fibrations:*
T2n1 ____ET2n ____HBWn
T2n ____ET2n+1 ____HBWn+1
The spaces Tm are a homotopically simple unstable representation of the Moore
space spectrum and we seek an unstable version of modularity. To do this we
need to find a functorial extension of a map Sm _____ffTn to a "modular map"
Tm ____bffTn.
This is accomplished in 2 steps:
0
a) extend ff to a map Pm = Sm [pr em+1 ____ffTn so that it is "modular"*
* in
the appropriate sense.
b) extend ff0 to bffso that it is an H map.
The idea behind point a) is to use the fact that prss*(Tn) = 0 to define an ext*
*ension.
This has some technical complexity. Part b) is easily obtained from
Theorem 4.1 ([9, 2, 12]). a) Tn is a Homotopy commutative and homotopy
associative H space and pr O ss*(Tn) = 0.
b) Let ff0: Pm _____ X where X is a homotopy commutative and homotopy
associative H space and prssr(X) = 0. Then there is a unique H map
bff:Tm ____ X extending ff0.
Proof. The case n even was worked out in [9] and the existence but not the
uniqueness when n is odd appear in [2]. The uniqueness and H property in case n
is odd appears in [12].
Proposition 4.2. Given an H map OE: Tm ______ Tn, there is a unique H map
oeOE: Tm+1 ____ Tn+1 such that the diagram:
Tm _________OETn
E  E
? oeOE ?
Tm+1 _____Tn+1
is homotopy commutative.
______________
*Note that it is an open conjecture [8] that BWn = T2np1(p) making these seque*
*nces formally
similar to those in section 2.
Composition Methods in the homotopy groups of ring spectra 9
Proof. There is a unique H extension of the composite:
SPm ____ STm ____OESTn ____ Tn+1
to an H map oe(OE): Tm+1 ____ Tn+1 . It then follows that the diagram in ques*
*tion
commutes when restricted to Pr. Since all maps are H maps, it commutes to
homotopy by 4.1 b).
Definition 4.3.Suppose X; Y; Z are H spaces, and f :X ^ Y ____ Z. We will call
f an H map in two variables (or an H map for short) if the adjoints of f:
X ____ ZY
Y ____ ZX
are H maps with the induced H space structure on the function space.
Proposition 4.4. There is a unique H map in two variables
: Tr ^ Ts ____ Tr+s
extending the composition Pr ^ Ps _____ssPr+s _____ Tr+s where ss*(1 1) = 1 *
*in
homology.
Proof. The adjoint map Pr _____ TrPs+shas a unique extension to an H map by
4.1. This gives a map Tr ____ TrPs+sand hence Tr ^ Ps ____ Tr+s. The extension
to a map Tr ^ Ts ____ Tr+s is similar.
Clearly we have commutative diagrams:
Tr ^ Ts
 Q Q
 Q Q
 Qs
o (1)rs Tr+s
 j3
 j j
? j j
Ts ^ Tr
and a similar associativity diagram if p > 3 or r > 1.
Given spaces A; B, we will often encounter the map A ^ B ____e(A ^ B)
which is adjoint to the evaluation map. e is an H map in the second variable. In
particular we can then extend
Pr ^ Ts ______e(Pr ^ Ts) _______Tr+s
i i1
 i i i
? i i ie0
T i i i
r ^ Ts
10 Brayton Gray
so that e0 is an H map in two variables, and
0
Tr ^ Ts ____eTr+s
1^E 6 6E
 
Tr ^ Ts1 ____Tr+s1
commutes up to homotopy, and similarly in the other variable.
In order to define modularity, we require the following.
Lemma 4.5. There is a map ffl: P1 ^ Tn ____ Tn so that the diagram:
P1 ^ Tn oe______1^oeP1 ^ Tn1
6 Q Q
 Q Qffl 
 Q Qs ?
S1 ^ Tn __________evTn
commutes up to homotopy.
Proof. Since the identity map of Tn has order p, there is a lifting Tn ____ T*
*n{p}
where Tn{p} is the fiber of the pth power map:
: : :___ Tn{p} ____ Tn ____pTn
However Tn{p} is homotopy equivalent to the fiber of the map Tn _____ Tn
induced by the degree p map on S1; i.e., Tn{p} ' TnP1 as H spaces. It follows
that there is an H map ffl0: Tn ____ TnP1with the composite
0 1
Tn ____fflTnP1 ____ TnS ' Tn
homotopic to the identity. Furthermore, the restriction
0
Tn1 ____ETn ____fflTnP1
is an H map and hence is the adjoint to : P1 ^ Tn1 _____ Tn. Taking adjoints
yields the lemma.
Our next step is to define, for each homotopy class ff 2 ssr(Tn) an H map
Tr _____bffTn extending ff. It suffices to construct a map Pr _____ Tn and th*
*is is
accomplished as the composition:
* ffl
Pr = P1 ^ Sr1 ____1^dP1 ^ Tn ____ Tn
bffis then the unique H extension. Consequently we have defined a homomorphism
Fffl:ssr(Tn) _____ [Tr; Tn]H
by Fffl(ff) = bff.
Clearly we have a left inverse
G: [Tr; Tn]H ____ ssr(Tn)
Composition Methods in the homotopy groups of ring spectra 11
by restriction. We wish to determine the image of Ffflas the maps which are in
some sense "modular." We will call a map OE kmodular if the diagram:
Tk ^ Tr _____1^OETk ^ Tn
 
? oek(OE) ?
Tk+r _______ Tk+n
commutes. Clearly OE is kmodular iff the composition dk(OE):
Pk+r+1 ____rPk ^ Pr ____ Pk ^ Tr ____1^OEPk ^ Tn ____ Tk+n
is null homotopic. Now oe dk(OE) = dk+1 (OE), so if OE is kmodular, it is (k *
*+ 1)
modular.
Strong Conjecture 4.6. There is a choice of ffl so that the diagram:
P1 ^ P1 ^ Tn _________________1^fflP1 ^ Tn
^1  
? ev ?
P2 ^ Tn ______P1 ^ Tn _____Tn+1
commutes up to homotopy.
Proposition 4.7. The strong conjecture implies that bffis kmodular for each k *
* 1.
Proof. We prove that bffis 1modular by factoring d1(bff)
Pr+2 _________rP1 ^ Pr___________1^^ffP1 ^_Tn_____Tn+1
6 6 6 6
? ? *  
P3 ^ Sr1 ___P1 ^ P1 ^ Sr1 ____1^ffP1 ^ P1 ^ Tn 
Q Q   
QQ ^1 ^1 
* Qs ? 1^ff* ? 
P2 ^ Sr1 _________ P2 ^ Tn ______P1 ^ Tn
By including S1 ^ P1 ^ Tn into P1 ^ P1 ^ Tn, we see that the strong
conjecture implies the
Weak Conjecture 4.8. There is a choice of ffl so that the diagram:
S1 ^ P1 ^ Tn ___1^fflS1 ^ Tn
ev  oe
? ?
P1 ^ Tn ________ Tn+1
commutes up to homotopy.
Proposition 4.9. The weak conjecture implies that bffis modular for each k 2.
12 Brayton Gray
Proof. We factor d2(bff):
Pr+3 _____rP2^ Pr__________1^^ffP2_^_Tn_______________________________ Tn+2
 
C 6 6 "ZZ ' 7 6*
*
C   Z 
C   Z" 1^E* 
C ' 1^ffl P1 ^ STn ________P1 ^ Tn+1 
C   6 6 
C   1^Sffl ^1 
*C ?    
C P ^ P ^ Sr1____1^ff*P ^ P ^oTe_'P ^ SP ^ T___evP ^ P ^ T 
C 2 1 2 1 n 1 1 n 1 1 n
C   S 
C ^1 ^1 ^1S 
CW ? 1^ff* ? ' 1^ev Sw 
P3 ^ Sr1 ________P3 ^ Tn oe___P2 ^ S1 ^ Tn _______________ P2 ^*
* Tn
We call OE 0modular if the composition d0(OE):
* ffl
Pr+1 ____rP1 ^ Pr1 ____1^OEP1 ^ Tn ____ Tn
is null homotopic. It is easy to see that the weak conjecture implies that if O*
*E is
0modular, it is 1modular and that if OE is 0modular and G(OE) = * then OE ~ *
**. If
ffl can be chosen so that bffis 0modular, there is then an isomorphism between*
* the
0modular H maps Tr ____ Tn and ssr(Tn). An even more complicated conjecture
about ffl will imply that bffis always 0modular. We will not pursue this line.
Proposition 4.10. The weak conjecture holds when n is even.
Proof. In this case we have an Hfibration sequence:
T2n ____ T2n+1 ____ BWn;
we will prove that such an ffl exists by showing that the composition
P1 ^ T2n ____e(P1 ^ T2n) ____ T2n+1 ____ BWn+1
is null homotopic. We begin by considering the diagram:
P1 ^ T2n ____________________________T2n+1
 
 
? ?
P1 ^ 2T2n+2 ' P1 ^ (S2n+3 )P2_______evS2n+3
since both compositions are H maps of the second variable, they are homotopic
iff they are homotopic when restricted to P1 ^ P2n. This is clear. We now form a
Composition Methods in the homotopy groups of ring spectra 13
diagram which contains the loops on this diagram:
P1 ^ T2n ______ (P1 ^ T2n) ________T2n+1
   Q Q
   Q Q
   Qs
'    BWn+1
   j3
? ?  j j
P1 P2 (ev) ?j j
P1 ^ S2n+1 (P1 ^ S2n+3 ) _____2S2n+3
6
' ? 6 '
?
P1 ^ S2n+2 P2 ____P1 ^ S2n+3 P3 ______ev2S2n+3
Q Q j3
Q Qev j j
Qs j j
2S2n+2
However the composition S2n+2 ____ 2S2n+3 ____ BWn+1 is null homotopic
since S2n+2 ____ 2S2n+3 ____ BWn+1 xS4n+3 is a fibration sequence [8]. This
completes the proof.
Theorem 4.11 (BarrattHilton Theorem). Suppose that ff :Tr ________ Tn a*
*nd
fi :Ts ____ Tm are k modular where n; m k. Then
oem (ff) O oer(fi) = (1)(nr)(ms)oen (fi)oes(ff): Tr+s ____ Tm+n :
Proof. Exactly as in the case of spheres, we have (ff ^ 1) O (1 ^ fi) = ff ^ f*
*i =
(1 ^ fi) O (ff ^ 1).
Now we also have:
Tk ^ Ts _____1^fiTk ^ Tm
 
? oen(fi) ?
Tk+s _______ Tk+m
for each k, while:
Tr ^ Tk _____________ff^1Tn ^ Tk
 
 
? (1)k(nr)oe(ff) ?
Tr+k _______________ Tn+k .
It follows that the indicated equation holds when preceded*
* by
: Tr ^ Ts ____ Tr+s. However, the inclusion of Pr+s into Tr+s factors through ,
so the equation holds when restricted to Pr+s. Since both composites are H maps,
they are homotopic by the universal property.
14 Brayton Gray
Appendix A.
So far we have discussed composition behavior which mimics the behavior of the
sphere spectrum localized at 2. Here we will discuss compositions in the sphere
spectrum localized at a prime p > 2. Recall that we have:
( 2k+1
S` = Sb ; if ` = 2k + 1
S2k = Jp1 (S2k) ; if ` = 2k.
The fact that half of the spaces in this spectrum are not spheres presents a di*
*fficulty
in forming compositions. The question we deal with first is this: is there a se*
*nsible
way of forming a "composite" of maps f :Sr _____ bS2nand g :S2n _____ S` to
obtain a map g * f :Sr ____ S`? If we allow ourselves to suspend once this can*
* be
done:
Sr+1 ____oefS2n+1 ____oegS`+1
It is remarkable that the case p = 3 is easier than that of larger primes.
Proposition A.1. Suppose p = 3 and X is a space such that X is homotopy
commutative in the loop space structure. Then:
a) S` is homotopy commutative in the loop space structure.
b) Each map g :S2n ____ X extends to a map
bg:bS2n____ X
c) If bg; "g:bS2n____ X are any two extensions of g, bg~ "g.
d) The diagram:
Sb2k ________bgX
 
 
? 2 ?
2S2k+1 _____g2X
commutes up to homotopy.
Corollary A.2. Suppose p = 3. If f :Sr _____ bS2kand g :S2k _____ S` we can
define a composition
g * f :Sr ____ S`
by g * f = bgO f. This is well defined and
oe(g * f) = oe(g) O oe(f): Sr+1 ____ S`+1:
Proof. a) is well known if ` is odd and is proven in [7] in case ` is even. b)*
* is
due to Hua Feng [5]. It follows since when p = 3, Sb2k = S2k [ e4k where the
attaching map is the Whitehead product. The obstruction to defining bgis thus
the composition S4k1 ____ S2k ____ X. The adjoint of this is the composition
S4k2 ____ S2k ____ X where the first map is a commutator. To prove c), note
Composition Methods in the homotopy groups of ring spectra 15
that two extensions bgand "gagree on S2n . Let ffi :S4k _____ X be the differe*
*nce
element so that bgis homotopic to the composition:
Sb2n ____ bS2n_ S4n ____"g_ffiX
Consider the diagram of fibrations:
SK * S4n ____fl*1Sb2n* S4n
 
 
? ?
SKn _ S4n _______fl_1bS2n_ S4n_____"g_ffiX
 
 
? ?
SKn x S4n ______flx1bS2nx S4n
where Kn = S2n1 [ e2np2 is the 2np  2 skeleton of Sb2n and fl :SKn ____ bS2n
is the adjoint of the inclusion. Since X is homotopy commutative, ("g_ ffi)(fl *
*_ 1)
extends over SKn xS4n , as this space is the mapping cone of a Whitehead produc*
*t.
It follows that the composite along the top is nullhomotopic. But since fl has*
* a
right homotopy inverse, the composite:
Sb2n* S4n ____ bS2n_ S4n ____ X
is nullhomotopic. Since the right hand sequence is a fibration, we get an exte*
*nsion:
(Sb2n_ S4n ) _______(_"g_ffi)X
j3
 jj
? j j
(Sb2nx S4n )
and must be "gx ffi. Fitting these diagrams together, we get
(Sb2n_ S4n )
j3  Q
jj  QQ ("g_ffi)
j j  Q Qs
Sb2n  X
Q Q  j3
QQ  j j
1xss Qs ?j j ("g)+ffi
Sb2nx S4n
Hence bg= "g+ (ffi) O ss. But by [8, Proposition 7] ss ~ * so bg~ "g. Finally,
to prove d) observe both composites:
Sb2n ____bgX ____ X
bS2n____ S2n+1 ____gX
extend S2n ____ X ____ X, so we may apply c) replacing X by X.
16 Brayton Gray
Theorem A.3. For any p and any map g :S2m _____ S`. There is an extension
bg:bS2m ____ S` such that the diagram:
bS2m __________bgS`
 
E  E
? (oeg)* ?
S2m+1 ______ S`+1
commutes up to homotopy.
Note A.4. We make no statement about the uniqueness of bgin the general case.
Proof. In case ` is odd, S` is an H space and g can then be extended to a map
g1 : S2m1 ____ S`. The diagram clearly commutes in this case. Suppose now that
` = 2n. We first lift g to a map g0: S2m ____ SKn so that g0 ~ g. Then consid*
*er
the diagram:
0 fl
S2m _________________gSKn ________ bS2n
  
  
? 0? ?
bS2m = Jp1 (S2m )_______Jp1(g)Jp1 (SKn)_OEbS2n

  
? ? ?
J1 (S2m ) ____________J1 (SKn) ______S2n1
The proof of A3 is complete when we construct OE so that the two diagrams on the
right homotopy commute. First we must prove
Proposition A.5. Suppose f :SX ____ SY and the composite:
* Hk
X ____fSY ____ SY k
is null homotopic for each k > 1. Then there is a commutative diagram:
SX ______________fSY
 
Hk ? ?Hk
SX(k) SY (k)
 
 
? ?
kSkX(k) _________(f^...^f)kSkY (k)
Proof. Of course, if f = Sf0, this is clear by naturality of the Hopf invarian*
*t.
We will prove this using the results of BoardmanSteer [4]. They describe Hopf
invariants n :[SA; SB] _____ for each n 1. Let ev :SSX _____ SX be the
evaluation map. Then [4, 3.15] n(ev): Sn (SX) _____ Sn X(n) is the adjoint of
Composition Methods in the homotopy groups of ring spectra 17
the composite SX ____HnSX(n) ____ nSn X(n). Consequently, the diagram in
question is equivalent to the diagram:
kf
SkSX ______SSkSY
 
k(ev) ? ?k(ev)
SkX(k) _______f^...^fSkY (k)
To establish this, we apply the composition formula [4, 3.16] to the composites*
* in
the square:
SSX ______evSX
Sf  f
? ev ?
SSY _______SY
Since q(f) ~ * for each q > 1, k(f O ev) = (f ^ . .^.f) O k(ev). However
q(Sf) ~ * for each q > 1, so k(ev O Sf) = k(ev) O Skf. This establishes
the result.
We now apply this to the map oefl :S2Kn ____ S2n+1 :
S2Km ________(oeg)S2m+1
 
p ? ?p
pS2pK(p)m _____pS(2m+1)p
S2Km ' (SKm )1 and Jp1 (SKm ) maps trivially under Hp and hence under p.
Thus the composition:
Jp1 (SKm ) ____ (SKm )1 ____ S2m1 ____HpS2mp1 ____ pS(2m+1)p
is nullhomotopic. Since dim Jp1 (SK) = (2mp  1)(p  1) < 2(mp + 1)p  3,
the composite of the first three maps:
Jp1 (SKm ) ____ (SKm )1 ____ S2m1 ____HpS2mp1
is nullhomotopic. We have proven the first part of
Corollary A.6. There is a map OE: Jp1 (SKm ) ____ bS2m such that the diagram:
Jp1 (SKm ) ______OEbS2m
 
 
? (oefl) ?
(SKm )1 ________1S2m1
homotopy commutes. fi
Furthermore, OEfiSKm ~ fl
18 Brayton Gray
Proof. For the second part, we observe that the composites:
SKm ____ Jp1 (SKm ) ____OEbS2m____ES2m1
SKm ____flbS2m____ES2m1fi
are homotopic, so the difference OEfiSKm  fl factors through the fiber of E. W*
*e will
alter OE so that this difference vanishes. To do this, note that the choice of *
*OE can
be modified by any element of [Jp1 (SKm ); 2S2mp+1 ]. But the restriction:
[Jp1 (SKm ); 2S2mp+1 ] _____ [SKm ; 2S2mp+1 ]
isfontoisince SJp1 (SKm ) splits. Therefore an appropriate choice of OE yields
OEfiSKm ~ fl.
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Department of Mathematics, Statistics, and Computer Science (M/C 249), Uni
versity of Illinois at Chicago, 851 South Morgan Street, Chicago, IL 606077045
Email address: brayton@uic.edu