Lecture Notes in Mathematics An informal series of special lectures, seminars and reports on mathematical topics Edited by A. Dold, Heidelberg and B. Eckmann, Z(Jrich
3 d. Frank Adams Department of Mathematics University of Manchester
Stable Homotopy Theory Lectures delivered at the University of California at Berkeley 1961. Notes by A.T. Vasquez
1964
SpringerVerlag  Berlin. G~ttingen 9 Heidelberg N e w York
National Science Foundation Grant 10 700
Alle Rechte, insbesondere das der Obersetzung in fremde Sprachen, vorbehalten. Ohne ausdr~cldiche Genehmigung des VerlaRes ist es anch nlcht gestattet, dieses Buch oder Telle daraus auf photomechanlschem WeRe (Photokopie, Mikrokopie) oder ~tufandere Art zu vervlelf~ItIRen. @ by SprlnRerVerlag Berlin 9 G6ttlngen 9Heidelberg 1964. Library of Congress Catalog Card Number 648035 Printed in Germany. Titel NR. 7523. Druck : Beltz, Weinhelm
TABLE OF CONTENTS Lecture
Page
1.
Introduction
2.
Primary
. . . . . . . . . . . . . . . .
opera tlons.
EilenbergMacLane
(Steenrod spaces,
work on the Steemrod 3.
Stable homotopy
theorY.
and properties
1
Squares,
Milnor's
algebra.) . . . .
. . .
(Construction
of a category
of
stable objects.) . . . . . . . . . . . . . . 4.
ApplicationsLofhomo!ogical to
stable hom0t0py
(Spectral 5.
Theorems
sequences,
Comments
algebra
theory. etc.) . . . . . . . . .
in hgmological
o n .prospective
38
algebra . . . .
58
applications
of 5, work in progress, Bibliography
22
of periodicit Y and
approximation 6.
4
etc . . . . . . . .
. . . . . . . . . . . . . . . . . . .
69 74
I)
Introductlon Before I get down to the business
llke to offer a little motivation.
of exposition,
I'd
I want to show that there
are one or two places in homotopy theory where we strongly suspect that there is something systematic
going on, but
where we are not yet sure what the system is. The first question concerns
the stable Jhomomorphlsm.
I recall that this is a homomorphism j: ~r(S#) ~ ~S (sn), n large. r = ~r+n It is of interest to the differential Bott, we know that ~r(SO) r = I
~r(S0) = z~
topologlsts.
Since
is periodic with period 8:
2
3
4
8
6
7
8
0
Z
0
0
0
Z
Z2
. 1 g
Z2~
On the other hand, ~rS is not known, but we can nevertheless ask about the behavior of J. prove:
The differential
topologlsts
2
Theorem:
If r = 4k  i, so that ~r(S0) m Z, then J(~r(SO))
= Z m where m is a multiple of the denominator of ~m/4k (B~ being in the k th Bernoulli number.) Conjecture:
The above result is best possible,
i.e.
J(~r(SO)) = Z m where m is exactly this denominator. Status of con~ectu~e: ~pn~ecture L
No proof in sight.
If r = 8k or 8k + l, so that
~r(SO) = Z2, then J(~r(S0)) = Z 2. Status of conjecture:
Probably provable, but this is
work in progress.
The second question is somewhat related to the first; it concerns vector fields on spheres.
We know that S n admits
a continuous field of nonzero tangent vectors if and only if n is odd.
We also know that if n = 1,3,7 then S n is
parallelizable:
that is, S n admits n continuous tangent
vector fields which are linearly independent at every point. The question is then: number, r(n),
for each n, what is the maximum
such that S n admits r(n) continuous tangent
vector fields that are linearly independent at every point? This is a very classical problem in the theory of fibre bundles.
The best positive result is due to Hurwitz, Radon
and Eckmann who construct a certain number of vector fields by algebraic methods.
The number,
~(n), of fields which they
construct is always one of the numbers for which ~r(SO) is not zero
(0,1,3,7,8,9,11 .... ).
n + 1 = (2t +l)2V: v by one increases
To determine which, write
then p(n) depends only on v and increasing p(n) to the next allowable value.
3
Conjecture:
This result is best possible:
i.e.
p(n) = r(n). Status of conjecture:
This has been confirmed by Toda
for v< Ii. It seems best to consider separately p(n) = 8k  I, 8k, 8k + i, 8k + 3.
the cases in which
The most favourable
appears to be that in which p(n) = 8k + 3.
case
I have a line of
investigation
which gives hope of proving that the result is
best possible
in this case.
Now, I. M. James has shown that if S qI admits rfields,
then S 2qI admits r + i fields.
proposition
Therefore
the
that p(n) = r(n) when p(n) = 8k + 3 would imply
that r(n) ~ p(n) + I in the other three cases.
This would
seem to show that the result is in sight in these cases also: either one can try to refine the inference based on James' result or one can try to adapt the proof of the case p(n) = 8k + 3 to the case p(n) = 8k + i.
4
2)
Primary o_2erations It is good general philosophy that if you want to show
that a geometrical construction is possible, you go ahead and perform it; but if you wsnt to show that a proposed geometric construction is impossible, you have to find a topological invariant which shows the impossibility.
Among topological
invariants we meet first the homology and cohomology groups, with their additive and multiplicative structure.
Afte that
we meet cohomology operations, such as the celebrated Steenrod square.
I recall that this is a homomorphism
Sqi: Hn(x,Y;Z2 ) , H n+i(X,Y;Z2) defined for each palr (X,Y) and for all nonnegative integers i and n.
(Hn is to be interpreted as singular cohomology.)
The Steenrod square enjoys the following properties: l)
Naturality:
if f:(~,~) ~ (X,Y) is a map, then
f*(Sqiu) = sqlf*u. 2)
Stability:
if 5: Hn(Y;Z2) ~ Hn+I(x,Y;Z2)
is the
coboundary homomorphism of the pair (X,Y), then
Sqi( u) = (Sqlu) 3)
Properties for small values of i. i) ii)
sqOu = u sqlu = Bu
where 6 is
the Bockstein coboundary associated with the exact sequence 0 ~ Z 2 ~ Z 4 ~ Z 2 ~ O.
5
4)
Properties for small values of n. i) il)
Ifn=i
Sqiu  u 2
if n < i
Sqiu  0.
5) Cartan formula:
sqi(u.v)
z
(SqJu).(Sqkv)
J+k=i 6)
Adam relations:
if i < 2J then
SqiSq j k~
Z kk~,l sqksq ~ = i+J k>2~
where the kk, ~ are certain binomial coefficients which one finds in Adam's paper
[1].
References for these properties are found in Serre[2].
These
properties are certainly sufficient to characterize the Steenrod squares axiomatically;
as a matter of fact, it is
sufficient to take fewer properties, namely l, 2, and 4(i). Perhaps one word about Steenrod's definition is in order.
One begins by recalling that the cupproduct of
cohomology classes satisfies 7)
u.v = (  l ~ q v . u
where
u r HP(X;Z) and v ~ Hq(x;z). However the cupproduct of cochains does not satisfy this rule.
One way of proving this rule is to construct, more or
less explicitly,
a chain homotopy:
to every pair of cochains,
x, y, one assigns a cochain, usually written X ~ l x, so that
(X ly) = xy  (l)PSx
6 if x and y are cocycles of dimension p and q respectively. Therefore
if x is a mod 2 cocycle of dimension m
5 ( X ~ l x) = xx • xx = 0
mod 2.
We define sqnlx = [ X ~ l X ] ,
the mod 2 cohomology class of the cocycle x ~ i x. definition
generalized
Steenrod's
this procedure.
The notion of a primary operation is a bit more general.
Suppose given n,m,G,H where n,m are nonnegatlve
integers and G and H are abelian groups. operation of type
Then a primary
(n,m,G,H) would be a f~uctlon
Hm(X,Y;H) defined for each pair
(X,Y) and natural with respect to
mappings of such pairs. Similarly, degree i.
we define a stable prima~y operation of
This is a sequence of functions: ~n: Hn(x'Y;G) > Hn+I(X'Y;H)
defined for each n and each pair (X,Y) so that each function Sn is natural and Sn+l 5 = 65n where 5 is the coboundary homomorphism
of the pair (X,Y).
From what we have assumed it
can be shown that each function Sn is necessarily
a
homomorphism. Now let's take G = H = Z 2. operations
Then the stable primary
form a set A, which is actually a graded algebra
because two such operations obvious fashion. structure of A?"
can be added or composed in the
One should obviously ask,
"What is the
7 Theorem 1.
(Serre)
A is generated by the Steenrod
squares Sq i . (For this reason, A is usually called the Steenrod algebra, and the elements a ~ A are called Steenrod operationsO More precisely,
A has a Z2basis consisting of the
operations Sq
ilsqi2
...Sq
it
where il,...,i t take all values such that
Ir ~ 2ir+l
(i ~ r < t)
and
i t > O.
The empty product is to be admitted and interpreted as the identity operation. (The restriction i r ~ 2it+ 1 is obviously sensible in view of property 6) listed above.)
There is an analogous
theorem in which Z 2 is replaced by Zp. Remark:
The products Sq
admissible monomials.
iI
...Sq
it
considered above are called
It is comparatively elementary to show
that they are linearly independent operations. n take X = X RP  1
For example,
a Cartesian product of n copies of real
(infinite dimensional) be the generators
projective
spaces:
in the separate factors
let x i g Hl(Rp~;z2) (i = l,...,n),
so
that H*(X;Z2) is a polynomial algebra generated by Xl,...,x n. Then Serre and Thom have shown that the admissible monomlals of a given dimension d take linearly independent values on
8 the class x = Xl.X2....x n r Hn(x;z2 ) if n is sufficiently large compared to d. The computation of Sq
iI
... 'Sq
it
on the class x is
l~educed by the Caftan formula to the computation of other Iterated operations on the Xl'S themselves.
Properties 3(1),
4(I) and (ll) imply that SqOx I = Xl, sqlxi = x~, and SqJx i = 0 for j > I.
The Cartan formula than.allows us to
compute iterated operations on the xi's.
The details are
omitted. The substance,
then, of Theorem 1 is that the
admissible monomials span A. EilenbergMacLane
This is proved by using
spaces.
I recall that a space K is called an EilenbergMacLane space of type (~,n)wrltten K e K(~,n)If and only if ~r (K)
/ ~
if r = n
[0
otherwise.
It follows, by the Hurewlcz Isomorphism Theorem that Hr(K ) = 0 for r < n and Hn(X ) = ~.
(if n > l)
Hence Hn(K;~) =
Hom (~,~), and Hn(K;~) contains an element b n, the fundamental class, corresponding to the identity homomorphlsm from ~ to w. Concerning such spaces K, we have Lemma I.
Let (X,Y) be "good" pair (e.g. homotopy equivalent
to a CWpair.)
Let Map (X,Y;K,ko) denote the set of homotopy
classes of mappings from the pair (X,Y) to the pair (K, ko), k 0 being a point of K.
Then this set is in onetoone
9 correspondence with
Hn(x,Y;~).
by assigning to each class,
The correspondence is given
If], of maps the element f*b n.
This lemma is proved by obstruction theory and is classical, see e. g. [33. Lemma 2.
There is a one to one correspondence between coho
mclogy operations $, as defined above, and elements Cm of Hm(G,n;H).
The correspondence is given by $ > $(bn).
The
notation Hm(G,n;H) means the cohomology groups (coefficients H) of an EilenbergMacLane space of type (G,n), this depends only on G, n and H.
b n is the fundamental class in Hn(G,n;G).
This lemma follows from the first rather easily for "nice " pairs.
But a general pair can be replaced by a CW
pair without affecting the singular cohomology. There is a similar corollary for stable operations. In order to state it, I need to recall that if K e K(G,n) then its space of loops, QK, is an EilenbergMacLane space of type (G,n1).
The suspension c: Hm(K) >(qK) is defined
as follows: Let K denote the space of paths in Ko ~:
Then we have
(LK,GK) > (K, pt), the map that assigns to each path its
endpoint.
The map ~ is the composition:
v 5 Hm_l Rm(K) ~ ' q
)which holds
for p ~ P) B r corresponds to Br and Z r corresponds to ~ r for p + r < P. Proof:
Again by induction.
make it trivial.
For r = 1 our conventions
For r = 2 it is also clear.
The inductive
step is made by inspecting the following diagram in which p
(Zr_i/Br_l)pr, q+rI = E pr,q+ri r
(%1/%1)pr, q+r1m_>Epr, q+rI
P'q Er
mono p > E ,q/np,q i~ri
q
>re~176 Np 'q/~rlP'q
dr> Ep,
Returning to the main line of argument, we now consider the group E~ 'q where p + q = P
q ~ 1.
is preserved by f and so is Zq. mapped isomorphically
by f.
(Zq/Zq+1 )P'q Therefore
(Bq+i/Bq)p+l'O
(for 1 < q < P).
By the lemma Bp (= Zq+l)
Therefore
(Zq/Zq+l)P'q is
But
" > (Bq+I/Bq)p+I ,0 d q+l is mapped isomorphically by f
Now E P+l'O has the composition series
0 = BICB 2C...CBp+
1 = Z 1 = E P+l'O.
We have Just shown that
1S all the successive quotients are mapped isomorphically by f, therefore E~ +l'0
is mapped isomorphically by f.
This
completes the induction and proves the theorem. We can now give the proof of theorem 3.
The E 2 term
of the spectral sequence of the fibering F > E > B is E~ "q = H P ( B ) ~ H q ( F ) ~means
where Z 2 coefficients are understood and
tensor with respect to Z 2.
finitely generated.) E~'0~E~
(By assumption Hq(F) is
In other words, the cupproduct
'q gives an isomorphism.
We will now construct
another spectral sequence and apply the comparison theorem. The condition on E~ 'q is satisfied because E is a contractible space. We first construct E~r'q(i) as follows. dimension of b i be t i.
Let E~'q(1) have a basis consisting
of elements B~i~ E2(i~ ti'0 differential d
Let the
mt,,t.i ~iB~ir ~2 1 l(i).
Define the
r so that d r = 0 for 2 < r < t i and
oq +I
ti
Then ~r+l ll) has a basis consisting of one element 1 We set d r = 0 for r > t iWe now define ~ by
E** = N*'* r
r
N** (1)
(2) 
.....
With the understanding that this is to be interpreted as the
16 direct limit of the finite tensor products.
We define
d
r
by the usual formula. Note that
H(E r ) = H(E r ( 1 ) @ . . . )
r
=
=
*,*(i)  Er+ 1
=
Er+ 1
We will now define a map assumption classes
> Er
(2)
"
about the relation of the classes
drf i  0
for r < t i
drf i = [b i]
for r = t i
construct
@r
..
Because of our fi
m mi Ej ~ J @ . . . b i ~Tj j r
m i
g k mk j'x~fK b k ) ..... f]j
mj [bj]
g mk ... fkk [bk }
It is immediate that the maps @r commute with the To apply the comparison that if @ 2 : ~ , 0
p < P.
to the
by setting
fii[bi ]
then
. . .
b.l , we have
We therefore r @r(~i
@ r: Er
Er+l
(2)) @
theorem we need only check
_> E~,0 is an isomorphim for
@2: E~'q > E~'q
dr 9
is an isomorphism
p ~ P
for all
This is immediate from the follwoing:
q
and
17 cupproduct
9E.p,O
(~) E~zz,q
m
> ~.2p,q
The comparison theorem then implies that H*(B;Z2)
is an isomorphism.
92: E;,O __> E~,0
Therefore H*(B;Z2) is a
polynomial algebra generated by the b i.
This completes
the proof. Remark:
In the above theorem, the coefficients need
not be Z2, an analogous theorem is valid for coefficients in any commutative ring with ideotity. I now wish to turn to Milnor's work [6].
Milnor
remarks that the Steenrod algebra is in fact a Hopf algebra. I recall that a Hopf algebra is a graded algebra which is provided with a diagonal homomorphism ~: A m >
A~A.
(of algebras)
In our case the diagonal
defined by the Caftanformula 5) so that
z
sq j 
k.
@
is going to be
@(Sq i) =
In general, for any element a ~ A, there is
j+k=i a unique element
Z a' ~ a " r I!
r
a(u.v) = z a r' (u)'ar(V)
r
e A~A
We define
such that ~(a)
~, E a' ,f'~a"
r
better add a word about how this is proven.
r~.~
r ~
I'd
r
It is pretty
clear that there is such a formula when u and v have some fixed dlmensionssay p and q, because it is sufficient to
18 examine the case where u and v are the fundamental classes in K(Z2,p ) x K(Z2,q).
After that, one has to see that the
formula is independent of the p and q.
We omit the details.
We ought to check that @ is a homomorphism,
but this
Just amounts to saying that the two ways of computing (a.b)(u.v) are the same. associative,
We ought to check that ~ is
but this Just amounts to saying that the two
ways of computing a(uvw) are the same. counit.
Similarly ~ has a
Thus A is a Hopf algebra.
With any Hopf algebra A over a field K, you can associate the vector space dual:
A*t = HomK(A t "K) " Assuming that A is flnitley generated in each dimension,
the
structure maps
[email protected]>
A
A> A x ~ A
(where @ denotes multiplication in the algebra A) transpose to give A* ~ Hence A
A*
y
between two such stable objects is
a sequence of ~aps: fn: Xn
~>
such that
Yn
fn ~ XnI ~ SI = fnI • I. We can compose maps in the obvious fashion. A homotopy
h: f ^ g
between two such maps is a se
quence of homotopies h n:
(I ~ X n)2n'2 ___>
base points fixed and commuting with
Yn
keeping W~S I
in the obvious
fashion. This is equivalent to defining homotopy in terms of an object
"I y X"
(I • X n)n
defined as follows: ~n2
= (I x Xn/l ~ X
)
x o = vertex of
Xn
O
Whenever I want to apply notions from the general theory of categories, the word morphism is to be interpreted as a homo
27 topy class of mappings~
But we allow ourselves
to keep the
notion of maps so that we m a y speak of inclusion maps, etc. Example of a stable object. rq
dimension
The stable sphere of
,
r.
We have
X
n
= S n+r
for
n > r + 2
X n = pt. otherwise. We have to ass~nne that Warning. are available
r > O.
Since spheres of positive dimension only
in this category it is not always possible to
desuspend an object.
This is a great blemish from the ha~ets
point of view. W i t h thla category I wish to do three things. I)
To justify it by showing that at least some pheno
mena of classical stable homotopy theory go over into this category~ 2)
To make it familiar, by showing that some of the
familiar theorems for spaces go over into this category. 3)
To lay the foundations for the next lecture b y
obtaining those properties of the category which I require. We w i s h to show t~hat this category does allow us to consider some of the phenomena which are considered in classical stable homotopy theory.
28 Theorem 1.
If
K,L are CW
complexes with one
vertex and positive dimensional cells for n < r < 2n  2 then there exist stable objects
X, Y such that
one of the same homotopy type as
K
r [ n, and similarly for
L.
have these properties, correspondence with
Y
and
then
and
Xn
for
Xr+ 1 = X r ~ S i
Furthermore,
Map(X, Y)
is
X, Y
if
is in oneone
Map (K, L).
This follows from the classical
suspension theorems,
and I wish to say no more about it.
[The notes for the remainder of this lecture have been revised in order to reorganize the proofs.] Both in stable and in unstable homotopy theory we may take the maps
f: X > Y , divide them into homotopy classes,
and so form a set Hap
(X,Y).
This set we make into a group
(in favourable cases), and such groups figure in certain exact sequences.
It is here that a certain basic difference
between stable and unstable homotopy theory arises. unstable homotopy theory we take groups Map(X,Y), we try to make exact sequences by varying is a pair
X 9
In
and first
What we need
(X1,X2)that is, an inclusion map with the
homotopy extension property. sequences by varying
Y .
Secondly,
we try to make exact
In this case we need a fibering,
that is, a projection map with the homotopy lifting property. In cohomology the pair gives an exact sequence~ fibering gives a spectral sequence.
the
In stable homotopy
theol,y the distinction disappears: exact sequence sequences,
UfCX.
and
In order to construct
suppose given a map
complexes. Y
of spaces.
we have Just one sort of
f: X > Y
such exact
between stable
Then we can construct a new stable complex (Here
Y UfCX
CX
is intended to suggest
is intended
attached to
Y
to suggest
" Y
by means of the map f.")
"the cone on
X ",
with the cone
CX
The definition
is
(Y U fCX) n = (Yn ~Ifn I ~ X n )2n2 (Here
I = [0,I] , with basepolnt As indicated above,
two exact sequences.
0 .)
this construction
gives rise to
We will prove this below,
first to consider the special case in which "point"
p
(that is,
is the "constant map" stable complex
P ~yCX
in our category. induces a map Lemma I.
S:
Yn
is a point for all
y 9
We write
, and regard
SX
is a n ) and
f
for the resulting
this as the "suspension"
It is clear that a map
f: X 1 > X 2
Sf: SX 1 > SX 2 , and similarly Map
Y
but we have
(X,Y) > Map (SX, SY)
for homotoples.
is a onetoone
correspondence 9 Remark 1.
The hare would always arrange matters
that this lemma would be a triviality 9 details,
so
With the present
it seems to need proof.
Remark 2.
Our proof will involve desuspension.
Suppose given a CWpair
K,L
of dimension at most
(2n2)
,
30 and a space map
Y
which is (nl)connected;
f: S I g ~ K   >
flSI%~L
SI~%Y
, and a deformation
into a suspended map l ~ g
over
K .
In fact, the map
map
f: K > ~ ( s l s y )
flL
into a map
to extend
g:
h,g
over
~r(Q(SI~Y),Y)
= 0
h
(for some
Then we can extend the deformation g
suppose given a
h
f
of g: L> Y .)
and the desuspenslon
is equivalent to a
; we are given a deformation L > Y C q ( s l S y )
h
, and we are asked
K ; this is trivially possible, for
of
since
r < 2n .
Similar remarks apply, when we are given a map f: S I ~ K
~~S1 > s l a y
a suspension given on
by
2(sl
l~%g
SI#~L y
~$S 1 , and asked to deform it into
#~l , or to extend a deformation already
#~S 1 .
One has only to replace
Q(SI:~Y)
sl).
Remark 3.
The effect of the definition of
(SX) n = (SI.~Xn)2n2 = S I ~ x ~ Remark 4.
n3 = S I 9 ~ X n _ I ~ S
SX
is that
1 .
We shall deduce lemma 1 from the following
lemma. Lemma 2. map a map
Suppose given a pair
K,L
of stable complexes,
f: SK > SY
(consisting of
fn:
g: L > Y
(consisting of
gnl: LnI > Yni
a sequence
h = {h n]
hn_l: l ~ g n _ l ~ l
a
(SK)n > (SY)n) 9 )
and
of homotopies
~ fnlSl~Ln_l~Sl:
SI~Ln_I~SI>SI~Yn_
I~S 1
51 such that the and
h
hn
commute
can be extended
~S I
with
from
to
L
Then the maps
.
g
so as to preserve
K
these properties. We proceed
Proof. the maps
and
gr
r < n  1 .
hr
by induction
have been extended
1 @ gn ~ 1 .
over
Ln
.
We are a l s o g l v e n
S1 c
obtain
a deformation
This
a certain
By the data,
Kn_l%~
By remark
fn+l"9 S I = ~ K n
We are given
and we wish to construct map
Kn
(by applying
S: Map
Suppose K
(X,Y) > Map
X ,
L
lemma provides
S: Map
over
We thus
over
of lemma
(SX, SY)
i.
Kn 9
We first prove
is an epimorphism.
f: SX > SY ; we apply
g: X> Y
it is necessary (X,Y) > Map Suppose
g and
h
such that to prove
lemma 2, talcing trivial.
that
(SX, SY)
given
two maps
fl,f2:
be the stable
, and
complex
g: L > Y by using
"I x X" using
The
Sg ~ f 9
lemma 2 again.
f: SK > SY
over
L n u (Kn_ 1 ~ S I ) .
that Sf I ~ Sf 2 ; we will apply
we define
of it into a
the deformation
can be extended
to be a "point",
a map
Secondly,
is monomorphic.
for
> S I ~ : Yn ~ S I
hn_ 1 .)
defined
that
the induction.
given a map
to be
~SI
K
deformation
to
Suppose
over
we are given
@S 1
We now turn to the proof that
n .
the d e f o r m a t i o n
compatibly
2, the d e f o r m a t i o n
completes
over
fl
and
the homotopy
L
X > Y We take
such K
to be its two end;
f2 ' and we define Sf I ~ Sf 2
.
The
to
32 homotopies
hn
can be taken stationary
provides an extension of homotopy
fl ~ f2
g
over
on
L .
Lemma 2
"I x X" , that is, a
in our category.
This completes
the
proof of lemma I. For
r > 2
the sets Map (srx, sry)
form abelian groups,
and the product is independent
"suspension coordinate" product.
in
srx
of which
is used to define the
(In fact, the arguments which one usually uses
for spaces apply, because the constructions suspension coordinates operation
~S 1
involve
"on the left," and commute with the
"on the right" used in defining our category.
It is illuminating
to recall that in a category where
direct sums and direct products
always exist and coincide,
the "sum of two morphlsms" can be defined purely in terms of composition.
(Given two morphisms
f,g: X > Y , one
considers f~g X
> Y x Y
>Y
Y
i~i > Y .)
In our category direct sums and direct products do exist and coincide; define
given two stable complexes
X v Y
by
(XVY)n
= XnVY n 9
X
and
Y , one can
33 This
is a stable
direct
sum
complex which fulfils
X v Y
and a direct product
as soon as it is equipped The proof performed
deformed r < 2n
that
into
.
with
X v Y
by imbedding
constructions
X x Y
the obvious
is a direct
XnV
can easily
Yn
in
~r(Xn
"
maps.)
may be The desired
X n x Yn ' and then
x Yn,Xn v Yn) = 0
for
are omitted.
To sum up, we have made ou~ category category,
(at least,
product
in
both for a
structure
X n x Yn
be performed
X n V Y n , since
Details
the axioms
and we are entitled
into an abellan
to use the following
definition.
Definition: ~r(X,Y)
for any
n
assumed
that
= Map
such that
n > O,
above.
to obtain
We recall
constructed
a stable
M = Y
; we have
UfCX
Theorem
""
(It is not
2.
sequences
f: X > Y
Y U fCX ; we now write
obvious
>
the two exact
that given any map
complex
~ Y
i)
n + r > 0 .
r [ 0 .)
We will now proceed mentioned
(sn+rx, sny)
maps
q M
>
SX
9
The. sequences
~r(V,M)>
~r_l(V,X)>...
are exact. To prove this, we follow Puppe's method. Lemma 3 .
The sequence f* j* Map (X,U) Y&> (Y~fCX)>
is equivalent
J2
(Y'~,cx) j ~   >
. . .
(up to signs) to the sequence
(b) X ~ > Y J > M q>SX S__f>SY S 4 S~ > ... (In sequence two terms as
(a) each term is constructed from the previous Y~fCX
is constructed from
X f> Y .)
It is clear that lemmas 3 and 4 suffice to prove the exactness of the sequence
(i) in theorem 2:
lemma 4 has a similar application to sequence
moreover, (il) in
theorem 2. The proof of lemma 4 is unaltered from the usual case. It is sufficient to consider the first four maps in the
35 sequence.
The required constructions
can be performed
"on the left" and commute with the operation
4~S1 used
in defining our category. Lemma
5.
The sequence
q. Map (sw,M)
(sf).
>
(sw,sx)
..> Map (sw,sY)
is exact. This lemma show that if we map (a) of lemma 4, we have exactness (Y OfCX) U j C Y
9
also.
into the sequence
at the fourth term
Since all subsequent
"fourth terms," we have exactness
SW
terms are also
at all subsequent
This still assumes that the "testspace"
terms
SW
is
suspended at least once; but even so, it proves part
(Jl) of
theorem 2. Proof of lemma 5. such that
(Sf),k = 0 .
sentative map for g: W
>
X
W g>
X f>
where
CW
k
k r Map
(SW, SX)
By lemma l, we can take a repre
which is of the form
Sg, where
is a map such that fg ~ 0 ; that is, Y
can be factored through
is the "cone on W" and
can now construct a map qm  Sg 9
Suppose given a map
m: SW
In fact, we decompose
C+W = " [ 0 , 8 9
and C W .
On
h
W
i>
CW
C+W
Y ,
is a "homotopy."
> M = YUfCX SW
h>
such that
into two "cones" we define
m
by
We
36 taking
Cg: C+W
.
> CX ; on C W
we define
m
to be
h .
This completes the proof. In this category we have homology and cohomology
theories defined in the obvious way.
We have i.
..._> Hm+n(Xn)
0 > Hm+n+l(X n
S I)
> Hm+n+l(Xn+l)__>...;
these groups all become equal after a while, and we define the limit to be
Hm(X) , where
X = [Xn} .
the homology maps induced by morphisms.
Similarly for
We can define the
boundary maps for a pair because our version of suspension has been chosen to commute with
3 .
Similar remarks apply
to cohomology. We now turn to the question of EilenbergMacLane objects in this category. Theorem 3.
Suppose that
F = t~O Ft
graded module over the Steenrod algebra many generators in each dimension. complex
(i) (ii)
K
A
is a free with finitely
Then there exists stable
such that: H*(K; Z 2) m F
as an Amodule, and
~r(X,K) a HomA(F,H*(XIZ2))
for each stable complex (The symbol lower the degree by
Hom~ r .
X 9 denotes the set of Amaps that It is understood that the
isomorphism in (ii) is induced by assigning to the map
57 f: sn+rx
 > SnK
the associated cohomology homomorphism,
composed with appropriate suspension isomorphism.) Proof.
We can construct a stable EilenbergMacLane
complex of type (Z2,n) by following J. H. C. Whitehead's procedure, attaching "stable cells" to kill "stable groups". By forming the one point union we can arrange it that Amodule.
H*(K;Z2)
The clause about
K = V Ki i
of such objects,
is the required free
~r(X,K)
is Just the
reexpresslon in a new guise of lemma 1 of the last lectureor of the corresponding assertion for maps into a Cartesian proudct of EilenbergMacLane spaces.
38 4)
Applications
I ought of h o m o l o g i c a l modules i.e.
of H o m o l o g i c a l
to b e g i n by running algebra
over a graded
M =
Z
Mt
o N
is no A
39 which lower degree by
t,
as u~ual, maps induced by
We define
i.e
f(Cs) u ~ Nu. t
We have,
d,
Ext As't(M' N) =
Ker d*/Im d*
in
Horn At (Cs, N)
The notation is justified since any two resolutions are ehain equivalent,
and therefore the groups
of
M
N)
Ext~'t(M,
depend only on the objects displayed 9 Next we go back to stable homotopy theory. Rite
H*(X)
for
H*(X, Z2)
recall that ar~y map
" H~(X).
to abbreviate notation. We @. Y induces an Amap f . H*(Y) 7
f: X  7
So we get a function:
Nap (X, Y) 7 Horn[(ff :'(Y),H (X)).
This function is moreover a homomcrphism.
It is in general
neither a monomorphism nor an eplmorphism. pute terms.
Map (X, Y)
We will
In order to com
by homological methods we need further
A general formulatTon is the following. Theorem 1.
Under suitable conditions
exists a spectral sequence whose
H* (X))
Ext~'t(H*(Y),
E 2s, t
on
X, Y
there
term is
and which converges to
2 ~ r (X' Y)"
The details are as follows: i) quotient of
If
G G
is an abelian group, then
2G
means the
by the subgroup of elements of odd order.
It is clear that eleEents of odd order play no part in our studies which are confined to the prime 2. il) series for
The elements in 2~r(X,
Y)
E s't oo
which give a composition
are those for which
t  s = r.
That
40 is, the "total degree" in th~ spoctral sequence is The filtration of
211~(X, Y)
tSe
is a decreasing one, so, for
example, osr
27rr (X, Y)/F I __ Eoo E l,r+l
iii) ts
by
etc.
The differential
dr
raises
s
by
r
and decreases
I~ It is perhaps desirable to give an example of this
spectral sequence. is then
Let us take
Ext~'t(z2, Z 2)
for brevity).
X = Y = S~
I recall that
~i
2
(which is rechristened H I' *(A)
with the space of primitive elements in element
The
term
Hs't(A)
can be identified A .
thus gives us a generator
h i.
The primitive In
H''~(A)
we can define cup products and this allows us to write down part of a basis for
Hs't(A).
(
h4 O
I
3
h3
2
h2
!
3
hoh3
h h3
0
q
O
h
2 h2
ho h2 JL
.

hob 3

h3
h2
O
S
0
I 0
! ts
2
3
4
5
5
7
>
b!s_nks are to be interpreted as
0
groups for
s < 4 ts < 7
41 The differentials in this part of the table are all zero, yielding a result in good agreement with the known values of
~7r (S ~
so):
r
r =
0
I
~
3
4
8
6
V
Z
Z2
Z~
Z8
0
0
Z2
Z16
Returning to the theorem, we must state suitable conditions on
X
and
Y.
We may distinguish two halves to our
work: a)
Setting up the spectral sequence.
This is more
,p
or less formal.
I shall assume that
H~(Y)
is finitely
generated in each dimension because I'll have to assume it later anyway.
It is possible, however, that we might be able
to eliminate this restriction for this part of the work. b)
Proving the convergence of the spectral sequence.
Existing proofs require the following conditions. (I) say
X, the object mapped, must be finitolydimensional,
Xr+ 1 = SX r (II)
H*(Y)
for
r > N,
some
N.
must be finitely generated in each dimension,
as assumed above. It is, perhaps 9 an interesting exercise for the experts to try to reformulate the theorem so as to relax these conditions.
Two changes are fairly obvious.
" " E x t ~ 't (H"(Y), H~'(X) )
by
~. LS ~x~ A
9t
You can replace
(H.(X), H.(Y))
which
I believe behaves better with respect to limits; and you can redefine
2G
by replacing "subgroup of elements of odd
42 order" by "subgroup of elements high powers of 2."
divisible by arbitrarily
These changes however do not suffice to
overcome
certain obvious counterexamples.
suppose
Y
(For example,
has only one integral homology group which is
the group of rationals mod 1.) if you replace
the coefficient
integers or the reals meal l~ to the case
I have no idea what happens field
Z2(or Zp)
(The case
Zp
by the
is analogous
Z2.)
Just for variety, however,
I want to give a simple
and explicit proof of convergence, even more restrictive
which works under conditions
ths~a I have already stated.
That is~
I shall assurae: (III) E
H*(Y)
generated by
H*(Y;
Z)
of order
is a free module over the exterior algebra Sq 1.
This is equivalent
has no elements of oo order, 9f
to supposing that
and all its elements
are actually of order 2.
This evidently excludes must give one or two examples
the case
X = Y = S~ ,
so I
to show that it does not ex
clude all cases of interest. Ex. l~ Ep2t/RP 2u .
Y
is the stable object corresponding
This example is relevant
to the vector field
problem. Ex. 2.
Set up an exact sequence
S ~ Ar > K(Z, O) ~,~ > N = K(Z, so that
llr(S~
S~
m
~Ur+l(SO ' M)
to
O) uf C S~
r
>
0
43
and
Ext~'t(z~,
Z 2) ~ E x t ~ "l't for
Condition
(H*(M), Z 2)
~  s > 0
III is satisfied by
and
s > O.
M.
Setting up the spectral sequence Supoose given an object ~ H~(y) the
< @
Co ~ d
Y
and a sequence
CI Ms fs+l, Ks+l fo" MI > Ko In general,
has a realization:
Ms
such that
such that
induces the map
induces the map
such
S
M i = Y
M s = K s UfsCMs. I d: C s + i ~ > Cs,
@: C O > H " (Y).
I don't assert that a sequence of order 2 but if the sequence
is a resolution,
then
Viz.
We can choose an EilenbergMacLane that
obJect~ K
Cs
2)
it does.
I mean
H*(K o) = C o
~: C O > H" (Y).
and a map
fo: Y > Ko
object
Ko
such
inducing
This follows from the last theorem of
44 last lecture.
We for~ the "quotient"
Mo = K o UfoCT.
We look at the exact sequence
< _ H t (y) H*(Y) > 0
d: CI> Ker @ = H*(M o).
is exact we have a ma~
We can find
KI,
an Eilenberg
8,.
MacLane object such that such that H*(~)
f "i = d.
H (KI) = C I
As above, if we form
= Z I = Ker dlC I .
Ms
fs
and maps
Ps > Ks
0 <mZs_ I ~
= K I UflC M o
fs: MsI "> Ks
realizes
> Ms
C s T. s,
the following diagram is homotopy
commutative. fs Ms1
Js > Ks ~
~
sI
i
qs~
gs
s
sI m

SM
Ms
Ms
Sms. I
s
~ > SMsI
If we have such a realization of a ladder,
then we
shall obtain induced maps of all our exact sequences, hence a map of sp ctral sequences. duced by the maps two resolutions,
Xs
"
On
E1
and
the map is in
If we asstune that we started with
then the induced map on
E2
is given by m.m
the induced map of
Ext.
In particular,
we see that our spectral sequence
if we take
Y = Y,
is defined up to a canonical
i somorphi sm. Le~na I.
If
{C s}
is a resolution,
then such a ladder can
be realized. The proof is by induction over msl:
MsI > MsI "
object we can construct
Since
Ks
s.
Suppose given
is an EilenbergMacLane
gs : K s ~ > Ks
such that
ks~ = gs "
47
~,'e now wish to show that Since
~s
gsfs = ~s msI (up to homotopy).
is an EilenbergMacLane
object, it is sufficient
to show that the induced cohomology maps are equal.
Since
{C s}
is a
is a resolution,
fsl: H (Ms. 1 ) > H (Ks. 1 )
monomorphism and it thus suffices to show that Jsl
s gs = JsI m _I f
that
dX s = ks.ld Given that
.
This follows from the assumption
(using the inductive hypothesis). gsfs = ~sms.l
(up to homotopy), the
whole of the diagram required by condition 3 follows by an obvious geometric argument.
I now wish to consider the convergence spectral sequence.
of this
By recalling the theory of exact couples,
one writes down a portion of the ( qr = rfold iteration of the map @
r th
derived couple.
q..)
I ! TM q~ : Trt+ r (X, Ms+r) > ~t (X, M s) J (defined by Es,t. r+l"
J.
inSo the second group)
...... a sub .... quotient~ of Vrt(X , K s ) F (defined by
~Im q~ : W"t (X, Ms. I)
f.
from the first group)
> srt_r(X, Ms_r_ I)
Q (defined by
q.)
Im qr. :. ~rt+l(X . . . ' M s ) > ~'tr+l (X, M s  r )
48 If
is large compared with
r
and domain of
~s(X' ~t (X, M s)
consists of elements of odd order. Since the spectral
sequence is an invariant,
it is sufficient
to do this for a favorably chosen resolution. I recall the hypothesis that terior algebra
E
H*(Y)
generated b y
Sq 1.
is free over the exWe prove below that
this allows us to find a resolution such th2t for
t < 2s + 2.
Hence
At this point
Ht(Ms; Z 2) = 0
for
Z
s,t
= 0
t < 2s + 2.
and (by Serre~s rood C theorems) Vzt(S~
Ms )
is an odd torsion group for
t < 2s + 2
Hence
?Wt(VS ~ ; N s)
is an odd torsion group for
where
V S i~ is a one point u n i o n of cooies of
t ( 2s + 2
S~.
If
X
is finite dimensional, we deduce by exact sequence arguments that 7rt(X, M s ) where
c
is a constant depending on
7Ft+r(X, Ms+ r )
t < 2s + 2  C,
is an odd torsion group for X.
Therefore
is an odd torsion group for
t + r < 2s + ~r + 2  c~
for a given
s
and
t ,
this is
49 true for s u f f i c i e n t l y
Ler~na
Existence
3 9
Remark
I9
large
r.
of a "nice" resolution 9
M
is free over
homology
of the module
operator
is
M
E
if and only if the
w i t h respect
to
Sq I
as b o u n d a r y
O.
Remark
2.
In an exact
sequence
!
TI
0> M if two of the te~ms
> M
~> M
are free
  > 0,
over
E,
then t h e
third is
also. Proof:
Remark
R e m a r k 3. 0 ~ M
Co
then each
If
~"
Zs
I and the exact h o m o l o g y M
o..
Now we take
Pl KI
> MI = KI
K2
}~ ~
>
~fl CKo ql > S Ko
'
'
~
0 0 0
and construct
fgt>
K2
M2
@
f Mn.l
n>
@n Kn .__.>
Hn
qn .... > SMn. I
If we look at this, we see that it is the same sort of thing we previously called a "realization," because we kept a little spare generality in hand for this purpose. to fix up the details, that for
M.I
I have to define
is a "point";
s > n.)
realization.
I also have
Mo Ks
(Strictly,
to be
Ko
so
to be a "point"
We can look at the spectral sequence of this The first term is
Horn A* (Co, H*(X))
d* Horn A* (C1, H * (X))
d* @
*
'
Horn A (Cn' H~(X)) written
C
s
for
H"(K s ).
module on generators
ci
where I have
Now suppose that (i = l, ..., m):
C
Aface
is an
Then an
A
m~o
~3
f: C 9 >H*(X) f(c i) e H*(x). d*
is determined by giving the elements With this interpretation, each homomorphism
may be interpreted as a primary operation, from
variables to
m' variables.
Consequently each
d2
m may be
regarded as a function from the kernel of one primary operation to the cokernel of another primary operation. ing to offer you the differential
dn
I am go
as defining an opera
tion of the n th kind, and I ought to verify that this agrees with onets notion of the usual procedure with universal examples.
Our procedure is given by the following diagram: (q~) nI
i !
\ F = f.
/ ~
VTt+nl(X' Kn)
E~
J = identity (subgroup of
~t(X, K o)
That is, you realize a cohomology mtuple by a map from stx
into
Ko:
you llft this, if you can, to a ~ap,
the universal example,
Mn_l:
as giving y~1 a mTtuple in of this
m'tuple by
~
in
you now regard
~ , into
r _. :Cn>H" ~ (Mn _l )
H*(Mn_ I)
and you take the image
H'(X).
This is precisely oneTs
ordinary notion of the procedure for defining an operation by means of a universal example.
One cormnent is called for;
I have supposed given the realization consisting of the Mts S
and the
K 's. S
element of geometry;
This supposition involves an irreducible for
n _> 3, not every chain complex
5~ d
C
m>
Cn.l   > C 1 > C o
n
K s 's
can be r e a l i z e d by developed
~d
that
if ~
then
~ 9
C s! ">
Similarly, of
~
...
>
process.
He proves,
C~
"> C o! '
~
I
and if
!
> ~ C s . I > ...  > C O 9
he shows that the S p a n i e r  W h i t e h e a d
from
for
Cr> ... > C I > C O
is a s s o c i a t e d w i t h a chain c o m p l e x
constructed
is associ
is a s s o c i a t e d w i t h C~ = Cs
C r > o.o > C I ~
~ ,
Cn> ... > C o .
is a s s o c i a t e d w i t h
9 is a s s o c i a t e d w i t h
CO =C's '
He has d e f i n e d
the n o t i o n that an operation,
ated w i t h a chain c o m p l e x
and
C.R.F. Maund2r has
the t h e o r y in this direction.
axiomatically
example,
M s !s.
C r > ... > C O
dual,
c ~
,
CCo* CCl*...>cC r
by a well
defined algebraic
55 Appendix to Lecture 4 The following table gives a Z 2 basis for IIS't(A) in the range of (s,t) indicated.
The following differentials
are known: d2h 4 = hoh ~ d3(hoh4) = hog d3(h~h4)
=
h~g
d3(h2h4) = h2g d~(hoh2h4) = hoh2g d3(h~h2h 4) = h~h2g. The notation Px implies that this element corresponds under a periodicity isomporphlsm to the element x.
56 A
4~
r~
I
C~
o~!~ c~ I ~ 1
~ L
L~
A
57 A edrI
A
o
v
b,
@
L ! J
c%p
i
I, m
k.O
If')
~
C
i,
, ~:
....
"~J
~ t !
~ o . . . .~_l_~ ~_ . . . . . . . . . .
1
1
II
I
1
i
A
A 0,1
A
C!.' , ~ 0
I.~O o ~
c
/i;~i
,
. I
,
I
.
.
.
.
4 .......
[
ii LI;;~47 ia~____~ iic.
o
~
.
.
.
.
0.1 i
s8 5)
~eorems
of p e_riodicity and approximation in homological
algebra. Let us begin by contrasting the spectral sequence I have developed with the classical method of killing homotopy groups, as aDpJied to the calculation of stable homotopy groups. Both depend on a knowledge of the stable EilenbcrgMacLane groups
Hn+q(~1", n; G) (n ~ q)
of them is an algorithm.
for some
7r and
G .
Neither
By an algorithm I would mean a
procedure that comes provided with a guarantee that you can always co~pute any required group by doing a finite amount of work following the instructions blindly.
In the case of the
method of killing homotopy groups, you have no idea how far you can get before you run up against some ambiguity and don't know how to settle it.
In the case of the spectral sequence,
the situation is clearer: the groups
Ext~'t (H*(Y), H*
(x) )
are recursively computable up to any given dimension; what is left to one's intelligence is finding the differentials in the spectral sequence, and the group extensions at the end of it. This account would be perfectly satisfying to a mathematical logician: an algorithm is given for computing Ext~'t(H*(Y), H*(X)); none is given for computing
dr.
The practical mathematician, however, is forced to admit that the intelligence of mathe~aticlans is an asset at least as reliable as their willingness to do large amounts of tedious mechanical work.
The blstory of the subject shows, in fact,
59 that whenever a chance has arisen to show that a differential dr
is nonzero,
the experts have fallen on it with shouts
of joy  "Here is an interesting phenomenon! chance
to do some nice, clean researchJ"
the problem in short order.
 and they have solved
On the other hand,
the calcu
Ext~ 't groups is necessary not only for this
lation of spectral
Here is a
sequence,
operations
but also for the study of cohomology
of the n th kind: each such group can be calculated
by a large awount of tedious mechanical work:
but the process
finds few people willing to take it on. In this situation, what we want is theorems which tell us the value of the the
Ext~ 't groups.
Now it is a fact that
Ext~ 't groups enjoy a certain limited amount of periodi
city, and I would like to approach this topic
in historical
order. First recall that last time I wrote down a basis for Ext~ 't (Z 2 ,
4
h4
3
h3
2
h2
Z2)
for
small
s
and
t~
0
o2h2
0
h2h3o 2 h 2
hoh 2
0
hoh 3 1
s
1
h 0
@
hI
h 2
h3
3
7
1
t

S
>
r
6o It was implied that
~v A
of
0 < t  s ~ 7.
s
in the range
(Z2, Z 2) = 0
theorem, which is proved in Theorem i.
for larger values
This is actually a
[ 7 ].
There is a numerical function,
f(s),
such that: (i)
Ext~'t(z2,
(ii)
f(s) ~ 2s
(iii)
f(s + s')
(iv)
Z 2) = 0
for
s < t < f(s)
f(s) + Z(s')
f(O) = O, f(1) = 2, f(2) = 4, f(3) = 6, f(4) = II
The published proof of this theorem is by induction, the induction involves other than
Z2 .
Ext~ 't (M, Z 2)
for
and
Amodules
We consider the exterior algebra
E
M
gener
ated by Sq I, so that we have an injection i: E
~A
.
This induces i * : Ext~ 't (M, Z 2) ~ E x ~ 't (M, Z 2) (For example,
if
M = Z2,
then
nomial algebra with
ho
M
E, then
is a module over
operator on M where
H*
and
Ext Es't(z2, Z 2)
as its generator. Sql: M M
Ext~'t(M,
denotes the homology
is a poly
In general,
if
is a boundary
Z 2) ~ Ht'S(M)
for
s  0,
with respect to sql).
What one proves, then is the following.
61 Theorem 2. same function
~ppose
Ext~ 't (M, Z 2) >
is an isomorphism for
a matter of fact,
some correspondence
function
f(s)
is As
w i t h Liulevicius
f(s)
= 12n  I
that for
is actually true.
volved me in extended calculations which
f(4n)
with the
Ext,' t (M; Z 2)
value of
This conjecture
that the best possible
;
the map
I also conjecture
the best possible
= 3. 2n  1.
t <m
t < m + f(s).
In the same paper,
f(~n)
for
f(s) as in theorem I,
i*:
s = 2n(n _> 2)
Mt = 0
in
stroD~ly suggested
is given by
(for n > O)
f(4n + I) = 12n + 2 f(4n + 2) = 12n + 4 f(4n + 3) = 12n + 6. This is actually true,
so that the function
the
diagram is periodic with period 4 in
s
"edge" of the and w i t h period
therefore
E2 12
in
t.
f(s)
The period in
w h i c h gives
ts
is
8, a n d this strongly rem imds
us of Bott's re
As a matter of fact more is true.
~Not only is the
sults.
"edge" of the
E2
diagram periodic,
but the groups nes~
the edge are periodic:
i.e. in a n e i g h b o r h o o d
llne
Hs't(A) ~
t = 3s,
we have
More still is true.
No
of the
HS+4't+12(A).
In a bigger neighborhood,
Nk,
62 of the line period
HS't(A)" are periodic with
t = 3s, the ~ o u p s
4 9 2k
in
neighborhoods,
s,
Nk,
4s ~ g(s) ~ 6s.
12 9 2k
is the area
(Possibly
in
t.
The union of these
t < g(s)
where
g(s) = 2f(s), but I cannot
give the exact value until I have refined my methods a little.) Again, these periodicity theorems should not be E x t As,t(z 2' Z2)
restricted to the case of E xt As,t (M, Z2)
with
is free over
E,
We should deal
We deal with the case in which
the exterior algebra generated by
M
sql:
Theorem 2 showsthat:*~hls is indispensable in the general case. Although this condition is not satisfied by the module periodicity rGsults for
E x t ~ ' t ( z 2, Z 2)
Z2,
can be deduced
from the following formula. E xt As't(z2, Z2 ) m E:xt~,t(z2 ' Z9 ) + E X t A s'l't(l(A)/A Here
I(A)/A Sq I
is a free left module over
Well, now~ let us see some details. Ar
will denote the algebra generated by
when AO
r
= E
is finite;
GO
will denote
and that
Ar,
that
Lt = 0
for
L
In what fo~loWS,r Sq I, Sq 2, ..., Sq 2 Note that
E x t ~ 't (L, Z 2)
are zero if
L
is a
is free qua left module over t < ~ .
Theorem 3. (Vanishing).
r
A .
E.
For our first results, we assur~e that
left module over Ao ,
A
SqI,Z~
Ar
and
Ters,t(Z2. L)
t < ~ + T(s)
where
T
is the
63 n~merical function defined by T(4k)
= 12k
T(4k + l) = 12k + 2 T(4k + 2) = 12k + 4 T(4k + 3) = 12k + 7 Theorem 4.
(Aporoxlmation).
The maps Ar
i.: TorA~s,t(Z2, L) ~
i*: E x t ~ t (L, Z 2) ~   E X v A~s't(L, r
and if
Tors,t (Z2, L)
0 ~
p
~ r,
s _~ I
and
Z2)
are isomorphisms
t ~ ~ + T(sl)
+ 2 p+l
I will not give complete proofs, but I will try to give some of the ideas. a) Ext B
It is not too laborious
where
B
to compute Tor
is a small subalgebra
of
A.
B
and
For example,
suppose we consider the case of Theorem 3 in which r = 1 (so
B = AI ,
Sq 2)
and let
tion of
Ao
a finite algebra generated by L = A o. over
AI ,
Sq I
and
Then we can make an explicit resoluand we can see that theorem 3
is
true. b)
If
theorem 3 is true in the special case
(some fixed value) whatever
L
and
L = Ao,
then it is true for
r = R r = R
is.
In fact, if we are given theorem 3 for the A r module A
O 9
then by exact sequences we can obtain theorem 3 for
64 any
Armadule
Ao,
theorem 3 for ~uy
then by exact sequences we can obtain Atmodule which can be written as a
finite extension of modules
isomorphic
to
A o.
This is
sufficient. At this stage we have obtained theorem 3 for the case
r
=
I.
o)
Theorem 4 tends to support theorem 3.
In fact, if
we know that AI i.: Tor s,t(Z2,
A L) > Torsrt(Zg,L)
AI TOrs,t(Z2,
then
and that d)
L) = 0,
is an isomorphism,
A Torsrt(Z2,
L) = 0 .
Theorem 3 tends to support theorem 4.
we consider the Map
Ar~ApL> L
and define
K
In fact, to be its
kernal, so that 0
,
> Ar~ A L
,>K,
is sn exact sequence.
9 L
> 0
Then we have the following diagram
TorAPt (Z2, L)
A
A
The vertical map
A
Ar
is an isomorphism by a standard result on
changing rings, which is in CartanEilenberg
[ 8 ] for the
ungraded case.
then
for
Also, if L t = 0 for
t < ~ + 2 P+I .
t < ~,
Hence theorem 3 implies that
Kt = 0
65 A Torsrt(Z2,K )
A Torsrl,t(Z2,K )
and
t < ~ + 2 p+I + T(sl)
.
are zero for
This implies that
i.
is an
isomorphism in the same range. Of course, in order to apply theorem 3, it is necssary to prove that
K
is free over
A O, and this is one of the
places where we rely on a firm grasp of the structure of
A.
Given these ideas, it is possible to prove theorems 3 and 4 simultaneously by induction over the dimensions.
The
details are somewhat tricky, and I will not try to rehearse them here.
The inference
the inference
(d) goes smoothly enough; but in
(c), the conclusion of theorem 4 does not apply
to the entire range of dimensions which we wish to consider. It is therefore necessary to preserve not only the conclusion of theorem 4 from a previous stage of induction, but also the ;method of proof used in (d). Theorem 5. ~r
in
(Periodicity)
Ext~t(z2Jz2)
for
There exists an element
S ~ 2r ,
t = 3.2 r
(r _> 2)
with the following properties. (i) The map
x> X~r: Ext
is an isomorphism when and
U(s)
L
is
s,
 Ls+2r,t+3 2 4 t(L,Z2)> ~x~ A (~) r ~r A0free and
is a numerical function such that
(li) ~*(~r) = (~r_l)2 maps for different
r
t ~ U(s) + 4s < U(s) < 6s.
(This says that the periodicity
are compatible.)
66 (lli) In a certain range, where i * : Ext Art(L, Z2) <  
EXtA't(L, Z2)
is an isomorphism, the periodicity isomorphism on the left is transported by x
i*
2r 9 > <x,h 0 ,hr+l> Remarks.
to the Massey product operation on the right.
I will try to make this plausible starting
from the end and working forwards. The Massey product xy = 0
and yz = 0 .
<x,y,z>
is defined only when 2r The fact that xh 0 is zero when
lies in a suitable range is guaranteed by theorem 5. fact that
2r h 0 hr+ 1 = 0
(for
X
The
4 ~ 2) was previously known and
was proved by introducing Steenrod squaring operations into H*(A)~
However, it can be deduced from theorems 3 and 4. I have next to recall that
H*(A)
can be defined as
the cohomology of a suitable ring of cochains, by using the barconstructlon.
In fact,
h2rhr+l
is the cohomology class
determined by 2r+l
[ ll ll
..
I li l
] 9
2rt~mes Therefore we have a formula 2r+l
8c = [~ll~ll
"'"
I~ll~l
]
"
67 Now consider fornlula. i* c
i: A r > A
We have
defines a
t = 52 r .
and apply
to the above
2r+l i*(~ 1 )  0 , whence
class,
~r
in
One checks that
HS'ttAr t ) ~r
.6(i'c) = 0 for
s
=
has the property
and
2r (lli).
is actually well defined by the above description. We now begin an argument like the former one. Step (a).
The homomorphism
x> x ( i * ~ ) :
.s+4,t+12,. , mX~AI (AO,~2)
_
Ext~t(~,Z2>
is an isomorphism for
s > 0.
Proof by explicit computation. Step (b).
The homomorphism
x> x ( i * ~ ) :
Ext~[t(L,Z2)>
is an isomorphism for Proof:
s > 0
if
L
ExtS+r't+12(L,Z2)
is Aofree.
by taking successive extensions of AlmOdules
isomorphic to
A0 .
(Since the homomorphlsm
is natural we can use arguments based on the
x>
Five Lemma. ) Step (c).
It is now clear that
ExtS't(L, Z2) Ar
is periodic
in the small range where it is isomorphic to EXtAlt(L,Z 2) 9 by induction.
We now extend this result up the dimensions We form the exact sequence
68 0
4> K~ > Ar ~ A I L   >
L
> 0 9
EXtAr ( A r ~ A I L; Z 2 ) m
ExtAI(L;Z2)
and this is periodic by step (b).
Lt = 0
t < 1 , then
for
Kt = 0
use the inductive hypothesis on
for
Also, if
t < Lt4; so we can
K .
I remark that the reason this proof does not give the best value of the function calculations over possible value for
U(s)
is that I started with
A 2 , one could perhaps extract a best U(s).
69 6)
Co~nents qn prosRectlve
a
[email protected] of S), w o r k
in
p_rosre ss, etc. Once agin,
I would like to hang out a large sign
saying "
[email protected] in Progress."
My first remark
however is a theorem. Remark I.
The theorems of the previous
lecture allow
one to put an explicit upper bound on the order of elements in
2~r(S0,sO).
In fact, we have filtered
that the composition
2Vr(SO,s0)
quotients are vector spaces over
so Z2 ,
and we have put explicit upper bounds on the length of the composition elements
series.
For large r, the bound on the order of
is appriximately
2(1/2 r) ; the previous best value,
due to I. M. James, was approximately Questlon I. new cohomology
for large
r.
I've remarked that as soon as you define
operations you are entitled to some dividlend
in the way of calculation and results. operations
2r
Stable cohomology
of the n th kind are associated with free chain
complexes over A.
The work of the last lecture leads one to
consider a lot of chain complexes over A which are periodic; the f ~ d a m e n t a l
one is
x>XSq0,1 ...>A In more familiar notation
x_> xsqO, 1 >A
> A>
Sq 0"I = Sq 3 + Sq2Sq I .
7o One can certainly construct cohomoiogy operations corresponding to the fmudamental chain complex written down above; the proof relies on Botts' work.
It is also possible
to construct cohomology operations corresponding to a number of other periodic chain complexes; but the general situation is not clear. Question 2.
Behavior of the Jhomomorphism.
calculate the groups the line
Ext~ 't (Z2,Z2)
One may
in a neighborhood of
t = 3s , and it is plausible to conjecture that
certain of these represent the image of the Jhomomorphism in dimensions
8k, 8k + l, 8k + 3.
Question 3. Ext~'t(z2,z 2) .
Consider the spectral sequence
Consider the differentials which arrive in a
neighborhood of periodicity
Nk, and originate (i) in a
neighborhood of periodicity
N~
the region of nonperiodicity. periodicity?
with
~ ~ k , or (il) from
Do these differentials show
(I think it is implausible to suppose that they
show periodicity with as small a period as that which obtains in
Nk. ) Let us make the question stronger.
subgroups of
2~r(SO,s 0)
Can one find
which display periodicity?
The
first periodicity operation should be x
>
<x,16~,a>
(where the bracket is a Toda bracket, and
L,a are generators
71 for the Ostem and 7stem)~
Further periodicity operations
might be
x ~>
<x,2f~,gk+l>
where gk+l is an element of the greatest possible order in the
(2 k+l  l) stem, and
2f
is the order of
gk+l
"
The whole of this question is highly speculative. It is quite obscure how one could ever isolate the relevant sugroups directly, without bringing in the machinery introduced above. Perhaps one can isolate the essence of questions 1 and 3 in a further question. Question 4.
What geometric phenomena can one find
which show a periodicity and which on passing to algebra give the sort of periodicity encountered in the last lecture? The question is wide open. Now I want to talk a little about the vector field question.
It is classical that one can reduce this question
to studying the homotopy theory of real projective
spaces;
let us recall how this is done. One may define a map Rpni , ,
as follows.
,> so(n)
Choose a base point e in
S n1
Y ~ S n1 , assign the following rotation:
. To each point first reflect
S n1 in the hyperplane perpendicular to e, then reflect
S n1
72, in the hyperplane perpendicular to
y .
Since
y
and its
antipode give the same rotation, we have defined a map RP nI ....> S0(n)
9
By attention to detail, you can make the following diagram commutative. RP nI
.
>
Rpn I/Rp nr2
S0(n)
,
>
S0(n )/S0 (nr1)
, R~n_i/Rpn_ 2
sn_l degree 1 >
=
S0 (n)/S0(n1)
=
S nI
It follows that if we can construct a lifting Rpn1/Rp nr2 _ S n1 Conversely,
then the required rfield exists.
if the dimensions
n
and
r
are favourably
disposed, then Rpn1/Rp nrR is equivalent to up to the required dimension, lifting
k
so that the existence of the
is necessary and sufficient for the existence of
an rfield on
S n1 .
We have to decide, therefore, whether
the topdimensional homology class in spherical.
S0(n)/S0(nr1)
Rpn1/Rp nr2
It is sufficient to show that it is not spherical
after suspension. Let us examine the spectral sequence Ext,' * (H* (RpnI/Rp nr2) ,Z 2 ) ~ 2~.S (Rpni/Rpnr2)
.
73
S
E t 't(H*(RPn'I/RPn'r'2);z2)
I
~h2
h~r I
h
ts Let us assume
to
to generators whether
~>
that we have n = 2 m , r = 8k+4
the top and bottom correspond
~
cohomology
classes
x 2m'l , x 2m8k5 h,h'
in
in
Ext 0
RP

Then
RpnI/Rpnr2
.
These
Let us calculate
this is zero or not depends
class mod 8 of 2m8k
in
.
5, and Sq4x 5
correspond Sq4x 2m8k5
only on the congruence =
0, so Sq 4 x 2m8h5
= 0
.
This gives a class hh 2 in Ext s't for s = I, ts = 2 m  8h  2. By p e r i o d i c i t y
we get a class h "
in Ext s't for
S = 4k + I, t  s = 2 m  2.
answer anyway,
Question
5.
is yes.
If the a n s w e r
= 0 for r < 4k + i? is no,
so we don't need to worry ~uestiqn
that the answer Question answer
Is drh'
is yes.
6.
Is h''
Probably
then h' is not spherical
about
this question.
= drX for r < 4k +l?
One hopes
is no. 7.
Is d4k+ik'
= h''?
the
One hopes
that the