on the cohomology of the steenrod algebraon the cohomology of the steenrod algebra 19 theory in this...

Math. Z. 116, 18-64 (1970) �9 by Springer-Verlag 1970

On the Cohomology of the Steenrod Algebra

MARTIN C. TANGORA

Table of Contents

Chapter 1. The Structure of E2 =H*(E~ . . . . . . . . . . . . . . . . . . . . . . 20 Chapter 2. The May Spectral Sequence . . . . . . . . . . . . . . . . . . . . . . . 23 Chapter 3. The Method of Calculation of E~ . . . . . . . . . . . . . . . . . . . . . 25 Chapter 4. Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Chapter 5. Differentials by the Imbedding Method . . . . . . . . . . . . . . . . . . . 48 Chapter 6. Remarks and Examples . . . . . . . . . . . . . . . . . . . . . . . . . 55 Appendix 1. Dictionary of Indecomposable Elements of E~ . . . . . . . . . . . . . . . 60 Appendix 2. Display Chart of E~ . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

Introduction

The object of this study is to develop a fast and efficient technique for calculating the cohomology of the mod 2 Steenrod algebra.

The ultimate origin of this problem is the long-standing problem of computing the homotopy groups of spheres. The connection is established by the now-famous spectral sequence of Adams ([1, 2]). If X is a space and p a prime integer, then there is an Adams spectral sequence having as its E 2

t e r m Extae(H*(X; Zp),Zp), where Ap is the m o d p Steenrod algebra, and having as Its E~ term a bi-graded algebra associated to the p-primary com- ponents of the stable homotopy groups of the space X. In the fundamental case where X is a sphere, the E z term is just EXtAp(Z2, Z2) , i.e. the cohomology of the Steenrod algebra. In this case, moreover, each term of the spectral sequence is a differential algebra, where the product structure in E 2 is the usual one in Ext ([-2, 4]), while that in E~ is obtained from the composit ion product in the stable homotopy of spheres.

To calculate the cohomology of the Steenrod algebra, one needs an appropriate resolution of Zp over Ap. This is a purely algebraic problem, and two different procedures were known, each providing an effective solution in theory, but unsuitable in practice except in very low dimensions. (I refer to the bar construction and the minimal resolution.)

We needed a new resolution, and this problem was solved by May in his dissertation [-7]. May constructs a spectral sequence which passes from the cohomology of E~ the associated graded of the Steenrod algebra, to an algebra associated to the cohomology of Ap. Since E~ is a primitively generated Hopf algebra, it is isomorphic to the universal enveloping algebra of the restricted Lie algebra of its primitive elements [8]. May develops his

On the Cohomology of the Steenrod Algebra 19

theory in this framework, and applies it to the Steenrod algebra to obtain the cohomology of E ~ in terms of generators and relations for a much larger range of dimensions than had previously been accessible.

However, even with May's results, there remained a mechanical obstacle. For the case of the prime 2 (which is the only one we will discuss here), the E 2 term of the May spectral sequence has many generators in low dimensions, and the first differential dz is non-zero on almost all of these generators. Thus the calculation of the E3 term of the May spectral sequence, though theoretically accomplished as soon a s d 2 is known on the generators, is still a laborious process, though a great improvement.

Our contribution in the present work is to introduce a computational technique which removes this obstacle, reducing the routine work by an order of magnitude. Though based on rather simple ideas (see Theorem 1.3 and Chapter 3), it is remarkably effective: one can now calculate the May spectral sequence (rood 2) through dimension 30 in one or two hours, whereas, before May's breakthrough, the latter part of this range was still unknown.

This paper is organized as follows. In Chapter 1 we recall May's results on H* (E ~ A2) in a form convenient for computation, and make a key observation on the structure of this algebra. In Chapter 2 we recall from May's thesis the basic properties of his spectral sequence. Chapter 3 presents the main ideas underlying the computational techniques. These are put to work in Chapter 4, which begins with a fairly detailed description of the calculations in low dimensions, and then proves the key differentials, often by arguments which depend only on the formal properties of the spectral sequence. Theorem 4.42 summarizes the calculations by listing the generators of the Eoo term of the May spectral sequence through dimension 70. Chapter 5 contains proofs of certain differentials using May's technique of imbedding the dual spectral sequence in the bar construction; this uses very special knowledge and language, and is largely independent of the rest of our work. Chapter 6 is devoted to the study of some examples which illustrate the difficulties which arise in the routine calculations.

The Eo~ term of the May spectral sequence is isomorphic with the E 2 term of the Adams spectral sequence as a Z2-module, but not as an algebra, because of the usual difficulty with spectral sequences: products are lost if there is a filtration shift. Various methods of recovering the multiplicative structure have been developed, and will be discussed elsewhere; we will not be concerned with this problem in this work.

Applications of our results to calculation of the mod 2 Adams spectral sequence for the stable homotopy of spheres have already been published [5]. For an important result about the cohomology of A 2 of a quite different flavor, see [6], where the present results are combined with other techniques to obtain information about part of the algebra structure of H*(A2).

The present paper is essentially the author's doctoral dissertation, presented in 1966 at Northwestern University. The author gratefully acknowledges the help of Professor Mark Mahowald, his thesis adviser, for suggesting the May 2*

20 M . C . T a n g o r a :

spectral sequence as a subject of investigation, and for his enthusiastic support and encouragement during and after the time that this research was done. Thanks are also due to Professor May for helpful suggestions, in particular for the technique used in Chapter 5.

Chapter 1. The Structure of E 2 = H* (E ~ A)

We are concerned only with the prime 2; therefore from now on we will write A instead of A 2 for the mod 2 Steenrod algebra.

In this chapter we quote May's results on the algebra H*(E~ which forms the E 2 term of his spectral sequence, and then make some observations about the structure of this algebra.

The first two theorems are proved by May in his thesis [-7].

Theorem 1.1. H*(E ~ A) is the homology of the complex 91, where 9t is the polynomial algebra over Z z with generators {Ri.j: i>0, j > l } and with the differential given by j-1

6(Ri, i)= ~ Ri, k Ri+k.i-k. k = l

Cup products in H*(E ~ A) correspond to products of representative cycles in 91.

By means of a sequence of spectral sequences, May obtains the following description of E 2 in terms of generators and relations for a certain range of dimensions.

Theorem 1.2. H*(E ~ A) is generated as an algebra over Z 2 by the following generators at least; and no other generators occur in the range t - s<165 or

in the range s < 4 : hi~ E2O, l, t ' t=2i;

bij~EzZ-2j, 2j, t, t=2i+l(2J_1);

hi(1)~E2 -2,4,t, t=9.2i ;

hi(1,3)~E2 -4,7,t, t=41.2i;

hi(1,2)~E2 -6'9"t, t=49.2i;

here i>O and j > 2. These generators are subject to the relations in Table 1.2, and to many others.

All relations in the range t - s < 165 are implied by those listed in Theorem 11.5.18 of May's thesis. Table 1.2 is simply the restriction of May's list to the range t - s _-< 80.

Table 1.1 lists the generators individually, for the range t - s____ 80. We have modified May's original notation by writing b~ (or b~,j) in place

of his b~, and similarly for Rij. This is purely for convenience; we find b023 b03 ~ b122 easier to print, and easier to read aloud, than tb ~ (b%2 tbl) 2 ~, 21 31 ~ 2 �9

The algebra E 2 is tri-graded, as will have been inferred from the notation of Theorem 1.2. We write E 2 .... t, where u is the filtration degree, v the com- plementary degree or weight, and t is associated with the degree in the Steenrod algebra. The homological degree is given by s = u + v . This is the filtration

On the Cohomology of the Steenrod Algebra

Table 1.1. Generators of E 2 for t - s < 8 0

21

t - s s Name u v t t - s s Name u v t

0 1 h o 0 1 1 1 1 h 1 0 1 2 3 1 h 2 0 1 4 4 2 bo2 - 2 4 6 7 1 h a 0 1 8 7 2 ho(1) - 2 4 9

10 2 hi2 - 2 4 12 12 2 boa - 4 6 14 15 1 h a 0 1 16 16 2 h 1 (1) - 2 4 18 22 2 b22 - 2 4 24 26 2 b13 - 4 6 28

28 2 boa - 6 8 30 31 1 h 5 0 1 32 34 2 h 2 (1) - 2 4 36 38 3 ho(1,3) - 4 7 41 46 2 b32 - 2 4 48 46 3 ho(1,2 ) - 6 9 49 54 2 b23 - 4 6 56 58 2 bla - 6 8 60 60 2 bo5 - 8 10 62 63 1 h 6 0 1 64 70 2 h 3 (1) - 2 4 72 79 3 h 1 (1, 3) - 4 7 82

Table 1.2. Relations in E 2 for t - s~80

1 2.1 2.2 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 5.1 5.2 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 8.1 8.2 8.3 8.4

10.1 10.2 11.1 11.2 12

hlhi+l=O (i~O) bl, 2 bi+2,2=hi 2 bi+l,3-.}-hi+32 bi, 3 (i~O) boa ba2 = b02 b23 + ho 2 b14 + h42 b04 hi+2 bl,2 =hi hi(l) (i->_0) hi+2 hi(l) =hi bi+l. 2 (iw~ 0) hi+ a h i ( l ) = 0 ( i>0) h~_~ hi(l) = 0 (i>__ 1) h~(1) 2 =b~, 2 bi+l, 2 q-hi+l 2 bl, a ( i>0) h~(1) hi+l(1)=O (i>O) ho(1) h2(1)= ho h4 bla ho(1) h3(1)=0 b~,2 hi+l(1)=hi+l hi+3 hi. 3 (i>=O) bi+2, 2 hi(1)=hihi+2 hi+l, 3 (i>O) b~,2h~+2(1)=hihi(1,3) (i>=O) h2 ho(1, 3)=ho h4 bl3 b32 ho(1)=h 4 ho(1, 3) h 5 ho(1, 3 )=0 h o h I (1, 3 )=0 boa h2(1)=ho ho(1, 2 ) + h 2 h4 bo4 b23 ho(1)=h 4 ho(1, 2 )+ho h2 b14 h s ho(1, 2 ) = 0 h I (1) ho(1, 3 )=h 1 h 3 ho(1, 2) hi2 ho(1, 3)=h l 2 ho(1,2)+ha bl3 ho(1) b22 ho(l, 3)= h32 ho(1, 2 ) + h o bla h2(1 ) boa h o (1, 3) = bo2 h o (1, 2) + h a boa h o (1) h o (1) h o (1, 3)= hi 2 h a boa + h a bo2 b13 h2(l) ho(1, 3)=ho ha 2 bla+h o bl 3 b32 ho(1) ho(1, 2 )=h a boa b12 + ha bo3 hi3 h2(1) ho(1, 2)=ho b14 b22 +ho bla b2a h 0 (1, 3) 2= hi 2 h32 bos + hi 2 boa b32-t-h32 b02 bla + b02 bla b32

d e g r e e i n t h e A d a m s s p e c t r a l s e q u e n c e . I t is t h e g r o u p s E 2 . . . . t w i t h f i x e d

t - s = t - u - v w h i c h a r e a s s o c i a t e d w i t h t h e s t a b l e ( t - s ) - s t e m i n t h e h o m o t o p y g r o u p s o f s p h e r e s .

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The key to our computat ional technique is the following observation on the structure of E 2.

Theorem 1.3. Let x be an element of E2=H*(E~ A) and suppose that h i x is non-zero in E 2. Then hinx is non-zero for all positive integers n, at least in the range t - s < 80.

This has been verified in the stated range simply by checking against the relations of 1.2. Of course it would be desirable to have a proof that would give more insight. I conjecture that the theorem holds in all dimensions but I do not have a proof at present.

The "stabi l i ty" under multiplication by h i which holds in E 2 according to Theorem 1.3 is in sharp contrast to the situation in H* A. In H* A we may have ho"- 1 x ~: 0, ho" x = 0 for arbitrarily large n; for example, x = hi, n = 2 i - 1.

The next two propositions follow immediately from Theorem 1.3 and show its power in analyzing the structure of E 2 .

Corollary 1.4. No non-zero element of E 2 can be a multiple of h i and at the same time a multiple of hi+l, at least in the range t - s < 8 0 .

Proof. If h i x = y = h i + l z , then h i2x=h ih i+ l z=O, so that Theorem l.3 implies y = h i x = O.

Contrast 1.4 with the behavior of H* A where one has, for example, hi 3 = ho 2 h 2 , etc.

Given x e E 2 , we say that x "survives to Er" ( 2 < r < oo) if x projects to a non-zero element in E r. When r = ~ we say simply that x "survives" or "is a survivor." If d r x = 0 for all r we say that x is a "permanent cycle". This is a necessary (but not sufficient) condition for x to be a survivor.

Corollaryl.5. Let x E E 2 and take i>0 . Then h i n x c a n n o t survive to E 3

of the M a y spectral sequence for any integer n > 4, at least in the range t - s < 80.

We state this proposit ion here to illustrate the usefulness of Theorem 1.3; for its proof we need some elementary facts about the spectral sequence, quoted from May's thesis in the next chapter. Suppose hi"x is a cycle in E 2 ;

then, since d E h i = O (Corollary 2.2), we have hi" d E x = 0 , so that h i d 2 x = 0 by Theorem 1.3. Now d E b i_ 1, 2 = hl 3 d- h i_ t 2 h i+ 1 (Theorem 2.4 (ii)) and therefore

d2(hi n - 3 b i_ l , 2 x ) - - h l " - a b i_ l , 2 d2 x - [ - x . d 2 ( h i n - 3 bi_ l , 2)

= hi n x

whenever n__>4, since h i d 2 x = O as above, and hlhi_l=O by Relation I of Theorem 1.2. Thus we have shown that if hi" x is a cycle in E 2 it must be a boundary there, and this proves the corollary.

Another corollary to Theorem 1.3 appears as Remark 3.1 in Chapter 3.

The list of relations given by May and reproduced in part in 1.2 above is not a convenient package. For example, Relations 1, 2.1, and 2.2 involve only the h i and the bij, while all other relations involve other kinds of generators; but there are relations between the h i and the bij which cannot be deduced


from 1, 2.1, and 2.2 alone. Thus h o h a b12=0, since h 3 bl2=hl ha(1 ) by 3.1 and h o times this is zero by either 1 or 3.4. Or, alternatively, h o b12=h2 ho(1 ) by 3.2 and h 3 times this is zero by either 1 or 3.3.

It is therefore necessary to develop a systematic procedure for writing E 2 in such a way that the relations are taken properly into account. This presents no difficulty and is largely a matter of taste, so we will not discuss the matter here.

Chapter 2. The May Spectral Sequence In this chapter we set forth the basic properties of May's spectral sequence.

The following theorem is proved by May in his thesis.

Theorem 2.1. There exists a spectral sequence (E,, dr), converging to E ~ H* A, and having as its E 2 term H*(E~ A). Each E r is a tri-graded algebra, and each d r

is a homomorphism dr: Er,, v, t ~ E," + . . . . . + 1, t

which is a derivation with respect to the algebra structure.

It is implicit that the products in E,+, (r>2) are induced by those in Er, and similarly for products in E~.

As usual in dealing with spectral sequences, we use the same letter for an element in E 2 and for its projection (if any) to E, (2<r__< oe).

Combining 2.1 with 1.2 we have two important observations,

Corollary 2.2. dr(hi)= 0 for every h i (i > O) and all d r (r > 2).

This follows from the fact that h/has filtration degree 0, d r raises filtration degree by r, and nothing can have positive filtration degree. (This holds without restriction on t - s . )

Corollary 2.3. d r is identically zero if r is odd.

This follows from the fact that d r raises filtration degree by r and that all generators have even filtration degree, at least in the range t - s < 165 (Theo- rem 1.2).

It will be convenient to list here the differentials which are non-zero in low dimensions.

Theorem 2.4. In the range t - s < 165, d E is given as follows:

(i) d 2 h i=0; (ii) d 2 bi, z=hi+13-Fhi z hi+z;

(i/i) d E b i j= h i + 1 bi+l , j_l+hi+jbi , j_ 1 (j_->3); (iv) d 2 h/(1)==h i hi+22;

(v) d 2 hi(l, 3)= hi(1 ) hi+42 + h i hi+ 2 h/+2(1); (vi) d 2 hi(l, 2)=hi+ 3 hi(l, 3).

This is most easily proved by the imbedding method as in May's thesis (see Chapter 5 for the application of this method to more difficult questions). All

24 M.C. Tangora:

of the above differentials which occur in the range t - s < 70 are given new proofs in Chapter 4.

There are also many higher differentials, of course, but there is no need to list them here; they will be stated and proved in the course of the calculations in Chapter 4.

It is noteworthy that E 2 becomes very large very soon, but the action ofd2, which by 2.4 is non-zero on all the algebra generators except the hi, is so de- structive that in general E 3 is a very much smaller and quite manageable algebra. To give an idea of how big these algebras actually are, we give the following table. Here by "rank ofE 2" we mean, more precisely, the rank of 2; u E 2 . . . . . . 2 s - 1

as a vector space over Z 2. This group has coordinates ( s -1 , s) on the usual display, and is "typical" of the groups for fixed ( t - s) in a certain sense (see 3.4). Similarly for "rank of E 4.''

t - s Rank ofE 2 Rank o fE 4 t - s Rank o fE 2 Rank ofE 4

10 1 0 20 3 0 30 13 2 40 25 2 50 47 1 60 83 1 70 156 8

1 0 0 3 1 0 7 2 1

15 5 2 31 15 4 63 103 10

We have listed the rank for ( t - s ) = 2 k - 1 in this table because it is here that E 4 is usually most complicated (and E r for all higher r). The values in the first column of the table are more representative.

Clearly the tedious work is done in calculating E3, and, since this is rendered completely routine by Theorems 1.2 and 2.4, it seems like a good problem to turn over to a machine, with or without the refinements developed in the next chapter.

We call attention to the fact that d r leaves t fixed (2.1), and therefore the determination of E~ for a fixed (s, t) can theoretically be done by examining only that part of E z which has the appropriate t. In practice this is not efficient because the structure given by products with h0 is a very powerful tool in computing higher differentials (Remarks 3.4, and applications in Chapter 4).

Similarly, May's theorems make it theoretically possible to calculate E~o for any given (s, t) by examining that neighborhood, without knowing the behavior of the spectral sequence in the range from zero out to the neighborhood in question; but in practice one needs the latter knowledge to avoid lengthy detours in the calculation of higher differentials. For the most efficient use of the techniques of Chapter 3, one should work forward from zero, calculating the spectral sequence for a fixed value of t - s , i.e. for a given "stem", before moving on to the next. It is true that certain questions are sometimes left open for a time, to be settled more easily by manipulations in higher dimensions. (For examples, see the proofs of Propositions 4.17, 4.18, and 4.31-4.34.) However,


the use of the "ho-stable" structure of E 2 (Chapter 3, particularly Remarks 3.4) requires preliminary knowledge of E 2 for all values of s corresponding to a fixed value of t - s, in order to settle certain questions about differentials. For example, in Proposition 4.37(a) we determine that a certain element with s = 5 and t - s = 69 is not a survivor, by considering E 2 for t - s = 69 and s very large (s > 70). The work of writing down the latter group overlaps considerably with the work of writing E 2 for other values of s (with t - s still equal to 69); thus it is usually convenient to write all groups with a fixed value of t - s at one time.

We close this chapter by stating for reference Adams' theorem on the zeroes of Ext]t(Z2, Z2) , known as "the edge theorem."

Theorem 2.5 (Adams [3]). Ex t~(Za , Z 2 ) = 0 / f ( t - s )<q(s where the func- tion q is given by q (4 k) = 8 k - 1;

q (4k+ 1 ) = 8 k + l ;

q(4k+2)=8k+2;

q(4k + 3)=8k + 3.

For our purposes, the essential feature of this theorem is the following: Ext must vanish if s > � 8 9 + 3). Actually all we need is that for a given ( t - s ) Ext vanishes for sufficiently large s.

Chapter 3. The Method of Calculation of E~

In this chapter we explain the computational technique for calculating in the May spectral sequence. The obstacle to be overcome is that there are too many generators in E2, and that d E is non-zero on all of the algebra generators except the h i, so that the calculation of d 2, though straightforward, is tedious. Thus the key to rapid calculation is to find a method for sifting out those elements of E 2 which contain the essential information, and discarding the rest. We need, in effect, a "garbage-disposal" system.

We suppose that the tri-graded algebra E 2 has been written out completely. We will always understand that E 2 (and E, as well) is displayed in the first quadrant of the plane, with ( t - s ) as the abscissa and s as the ordinate, as in Appendix 2 for E~. Thus, for example, when we say "all elements above x " we mean "all elements having the same value of ( t - s ) as does x but having greater values of s."

In the next chapter we will carry the calculations forward from the beginning, and will derive each new differential in May's spectral sequence as we come to it. In the present chapter our object is to explain the computational techniques, and particularly the criteria by which we distinguish between the "essential" elements of E 2 and those which can be ignored. For clarity we will therefore assume for the present that the differentials in Theorem 2.4 are already known. In this chapter we will be concerned only with the calculation of d 2 and the results for E 3 . The groups E 3 .... t are very much smaller than the corresponding groups of E z, as a general rule. Therefore our computational

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technique is most valuable for calculating E 3 from E 2. It is also much more efficient here than in calculating E r for r > 4; some of the difficulties which arise in the later stages of the spectral sequence will be discussed in Chapter 6.

We adopt the notat ion Er( t - s , s) for the direct sum of the groups E, T M

for which u+v=s . Thus Er(i,j ) is the group whose generators appear at the lattice point with coordinates (i, j) in the plane upon which E r is displayed.

Let x be an element of E,. We will refer to the elements hi" x, where n > 1, as the hi-multiples of x (in E~). In this language, Theorem 1.3 states that in E z if h i x is non-zero then all the hi-multiples of x are non-zero. When we write out E2, we might therefore do well to write it by columns, that is, to write first E 2 (0, s) for all s, then E 2 (1, s), etc. Then if we have an element x in E 2 ( t - - s , s)

such that h o x is non-zero, we can immediately enter all the ho-multiples of x, ho" x ~ E 2 ( t - s, s + n).

The next observations are immediate consequences of Theorem 1.3 and Corollary 2.2. Recall that Theorem 1.3 is established for t - s < 80 and is conjec- tured true for all dimensions; thus similar restrictions apply to the following.

Remark 3.1. In E2, suppose that d 2 x = y and that hix and hiy are non- zero. Then hi" x and hi" y are non-zero, and d z (hi" x) = hi" y. Thus none of these elements survives the May spectral sequence; in fact they disappear already from E 3 .

Under these circumstances we say that "the differential d 2 x is hi-stable." We will be primarily concerned with the cases i= 0, 1.

If we consider the elements hi"x and hi"y located in the display of E2 ,

and draw line segments joining hi m x to hi "+1 x for each m, similarly for y, and then indicate the differentials by arrows, we obtain a figure which resembles a ladder. When d 2 x is hi-stable, we will speak o f " the hcladder of d 2 x ," by which we mean the set {hi"x: n>0} u {hi"y: n>0}, where y = d 2 x.

The root idea of our technique is to throw out all such ladders. Certainly no elements of the ladders survive the spectral sequence; but we must verify that to ignore these elements will not have any essential effect on E 3 . We will therefore make use of the following elementary result from homological algebra.

Proposition 3.2. Let B be a chain complex and A an aeyclie subcomplex. Then the projection p: B ~ B/A induces isomorphisms p.: H,(B) ~ H,(B/A) for all n.

See, for example, MacLane 's Homology [4], Theorem 4.1 of Chapter II. We have noted in Chapter 2 that the groups E 3 .... t are just the homology

groups of the complex E 2. ' *' ~. Now observe that a given ladder intersects such a complex, if at all, in a subcomplex containing exactly two non-zero elements, namely hi" x and hi" y for the n which is appropriate to h i, x, y, and t; and this is obviously an acyclic subcomplex of Ez*' *' t. Therefore the homology of the algebra obtained from E 2 by discarding (i.e. factoring out) the ladder will be isomorphic to E 3 under the natural projection.

Thus in a very precise sense we can suppress an hi-ladder from E 2 without essentially changing E 3 .


It would be convenient to collect all the ladders into one big acyclic complex to be factored out of E z. However, different ladders may in general intersect in various ways, so that it is not true that the union of two ladders gives an aeyclic subcomplex of E z. For example, the differentials d2(hobozb13)=

h 0 h 4 bo2 b12 and d2(h 4 b022 ho(1))=ho 3 h 4 bo2 b12 are both ho-stable. If we discarded both ladders we would lose the element ho 3 bo2 ba3 + h 4 b 0 22 ho(1 ) which is actually a surviving cycle, appearing as the element s in the 30-stem; so we would be "throwing out the baby with the bath water."

We must therefore proceed by factoring out one ladder at a time. We illustrate the procedure using the above example. Arriving at the 30-stem, working up from low values of s, we find the ho-stable differential d 2 (h o b o 2 bl 3) and discard its h0-1adder. Subsequently we arrive, higher in the 30-stem, at the element h 4 b o z z ho(1), or rather at its image in the quotient complex where we are now working. In that quotient complex, this element is a cycle, since its image under d z has been factored out. Ultimately we find this element to survive the May spectral sequence, and we conclude that there is actually an element in E 2 which survives the spectral sequence and which is equal to h 4 b o 22 ho(1 ) plus other terms, possibly, which are hi-multiples of elements of lower s-degree.

We can always recover complete information. For example, when our computation shows a surviving term such as " h 4 b o 2 2 ho(1)" we compute d 2 of this term. If this d 2 is zero (and if all d r = 0, r > 2), we have found the actual cycle. If this d 2 is non-zero, we need only examine that part of E 2 which lies below and to the left of the present element and which could generate ladders that cross the present tri-grading. We test the elements in that part of E 2 to find those terms of the actual cycle which had been suppressed, as elements of ladders, in the computation. In this way the actual cycle can always be re- constructed after a finite search which is usually very quick. In the above example, only three elements of E 2 have the tri-grading which qualifies them as candidates for the missing terms of the pseudo-cycle h 4 b 0 22 h 0 (1), SO one quickly recovers the correct term ho 3 bo2 b13 , as the only one of the three which has the boundary ho 3 h 4 b o 2 b12.

Remark 3.3. By Corollary 1.4, no hi-ladder can ever intersect any h~+ 1-1adder, except possibly in the initial term.

In practice we proceed as follows. We write Ez, stem by stem, as far as we want to compute the spectral sequence, and at least one stem beyond. (Recall that by "s tem" we mean grading in ( t - s ) . ) Assuming that we know d z on the algebra generators, we begin at the lower left-hand corner of the display and proceed upward, and move to the right when the stem has been finished. Each stem represents a f ini te computation; for the grading of the generators is such that, when s > ( t - s), only elements with a factor of h o can occur, and, by 1.3, 2.5, and 3.1, everything will be ho-stable above this point. Whenever a differential d 2 is ho-stable and/or hi-stable, we factor out the corresponding ladders. By Remark 3.3 we know that the h~-ladders do not contain any ho-multiples in

28 M.C. Tangora:

later stems. Thus our successive quotient complexes always have the same homology as E2, and we obtain E 3 . In practice relatively few elements survive to E3, but also more delicate situations arise in the calculation of the later stages of the spectral sequence, so we do not discuss these matters here but defer them to the next chapter.

We next record some remarks about the use of the Adams "edge theorem" 2.5 to compute differentials.

Remarks 3.4. As noted above, for any fixed ( t - s ) , the groups E 2 ( t -s , s) stabilize as s increases; precisely, multiplication by h o gives an isomorphism E2(t-s , s )~E2(t -s , s+ 1) as soon as s>(t-s) . Since every d r in the May spectral sequence is a derivation with respect to the algebra structure of E r, and since h o is a permanent cycle, this means that Er(t-s, s) is also ho-stable for large s. We may therefore consider the group E,(t-s), defined as this stable limit, the direct limit of Er(t -s ,s ) as s ~ o o . By 2.5, Eoo(t-s ) is trivial, i.e. contains only the zero element. On the other hand, we have a natural structure of chain complex in the sequence of groups Er(t-s), ( t -s)=0, 1, 2 . . . . . where the differential is induced in the obvious way by d r. It is very often possible to deduce dr(x ) for an element x in E,( t - s , s) with small s, from the fact that all groups Eoo ( t - s ) must vanish. Many specific examples in the following chapter will show the efficacy of this technique based on the Adams theorem and the hi-stability Theorem 1.3.

Remarks 3.5. A word is in order about the multiplicative structure of H* A and of Eo~ = E ~ (H* A). The products in Eoo may be thought of as induced from the products in H* A by deleting terms which have the wrong filtration degree (or, equivalently, the wrong weight). From our point of view, this means that if two elements x, y~E 2 survive the May spectral sequence, then in H*A the product of the elements corresponding to (x) and (y) contains as a summand the element corresponding to (x y), where x y is the product in E2, but it may also contain other terms. In order to be a candidate for such a lost summand, a survivor z must have the same bi-grading (s, t) as the product x y, and must have weight less than the weight ofx y. (See [7], Remarks II.6.12.) For t - s >= 30, such ambiguities become a common occurrence. At present various methods are known for detecting the lost summands, but none are completely satisfactory.

As an example of what can happen, consider the survivors h 2 and eo 2 in the 3-stem and the 34-stem respectively. They lie in E2 ~ 1.4 and E 2- 8,16, 42, or in coordinates (3, 1) and (34, 8). Thus the product of the corresponding elements in H*A must have coordinates (37, 9) and have weight w< 1 + 16= 17. Now hoax has coordinates (37, 9) and weight 15. Therefore, although h2 eo 2 does not survive to Eoo, so that the product is zero in Eoo, it is not clear whether in H*A the product is equal to zero or to ho 4 x. Actually this product is known to be non-zero [5], but the methods of proof are necessarily different in nature from techniques based on May's spectral sequence.

In this paper we do not attempt to recover such products. All products should be interpreted as products in Eoo = E ~ (H'A).


Chapter 4. Calculations

In this chapter we indicate the calculation of E ~ ( H ' A ) for the range ( t - s ) = 70. It is impossible to give all details, of course. The method of calculating E 3 of May's spectral sequence when d 2 is known on the generators has been explained in the preceding chapter. In this chapter we explain how most differentials can be proved independently of the imbedding method of May's thesis.

The following methods of proof are known for the various differentials in May's spectral sequence.

1. The Imbedding Method. In his thesis May shows that d r can in general be obtained by imbedding the dual spectral sequence in the bar construction. Thus to ascertain whether d r x = y in May's spectral sequence, one identifies dual cycles x , and y , in the dual spectral sequence, imbeds in the bar construction, and determines whether in the bar construction d ( y . ) - ( x , ) modulo elements of deeper filtration. In theory this method is effective, but in practice, like any method using the bar construction, it is sometimes prohibitively slow. In fact this method works quickly and neatly for d 2 but is unpredictably quick or tedious for higher differentials. It also requires virtuosity of a special kind in the use of the Milnor description of the Steenrod algebra. We therefore supplement it whenever possible with an independent proof by one of the alternative methods below.

2. Manipulative Methods. By this we mean arguments based on, among other things, a) the May relations in E 2 (see Theorem 1.2), b) the fact that each d r is a derivation, c) the fact that d r d r-- 0, d) the fact that each h i is a permanent cycle (Corollary 2.2). For example, ifh i x = 0 and h i y:t:O in Er, then d, x cannot equal y, since that would lead to the contradiction

0 = dr(h i x ) = h i d, ( x )= h i y ~: O.

Another pattern of proof in this category is to use a May relation of the form x 1 x z = x 3 x 4 to compute dr(xl) when the action of d, is known on xz , x 3 and x 4 (and where the product is non-zero in Er).

3. Application o f Adams' Results on H* A. In Remarks 3.4 we indicated a method for using the "edge theorem" to calculate certain hi-stable differentials. This will be a powerful tool. We can also make use of other results of Adams such as his relations hi 3 = h i_ 12 hi+ 1 ; but this will be of minor importance in our study.

4. Application o f Known Results in Homotopy Theory. Since our problem is an algebraic problem, we prefer to give purely algebraic proofs wherever possible, but as a last resort we can occasionally make deductions about the structure of H*(A) based on topological methods, particularly, of course, on the Adams spectral sequence.

In what follows, each independent differential will be stated and proved as a proposition. Often several independent proofs will be offered ; a proof based on, say, method (3) above will be indicated thus: "Proof(3)."

30 M.C. Tangora:

We have accomplished the following program: all differentials through the 70-stem, using methods (1)-(4); all differentials through the 45-stem, without using method (1); all differentials except one, through the 45-stem, using only methods (2) and

(3). The exceptional and troublesome case is d4(boa 2) which is proved either by method (1) or by method (4) as in Proposition 4.9* below.

After each new independent differential is obtained, one must do a certain amount of routine calculation, using the techniques explained in Chapter 3 for d 2 and using obvious or analogous techniques for higher differentials. We omit this part of calculation, but we give some indications at the beginning to aid the reader in developing his understanding, and we discuss in Chapter 6 some of the tricky cases that arise.

We now begin the calculations. We do not assume that any of the differentials are known at the outset, except d r h i which is zero for dimensional reasons (Corollary 2.2); thus we will make no use of Theorem 2.4; in fact we will be proving that theorem as we proceed. A reader who wishes to understand proofs based on method (1) should consult May's thesis, Chapter II.6, for more details, as well as Chapter 5 below.

We assume that E 2 = H * ( E ~ has been written down for the range ( t - s ) < 73, by the methods discussed in Chapter 1.

In the 0-stem we have just the elements ho s for all s>0 . In the 1-stem we have only h 1. Since h I is a permanent cycle (Proposition 2.2),

ho s survives the spectral sequence for all s >0. In the 2-stem we have only ha 2, which is a permanent cycle. Thus h a survives

the spectral sequence. In the 3-stem we have ha 3, and also h0 s-a h a for all s > 1. All such elements

are cycles. Thus ha 2 survives. In the 4-stem we have hi 4, and also ho s- z bo 2 for all s > 2.

Proposition 4.1. d 2 (bo2) = h 13 + ho 2 h2 '

Proof(3). The element ho" h E c a n n o t survive to H* A for large n, by the Adams edge theorem 2.5. Since it is a cycle for every d r, it must be the image of some differential. But the only element in E 2 (4, s) for large s is ho s- 2 bo2. We must therefore have dr(ho s-2 boz)=ho s h 2 for large s; moreover, r must be 2, since d r changes the filtration degree u by r, and u is ( - 2 ) and 0 for ho" bo2 and hom h2 respectively. F rom dimensional considerations we have d 2 ( b o 2 ) : Aho 2 hE+A'ha 3 where A and A' are the coefficients (in Z2) which we must determine. The above argument implies that A = 1 ; for d 2 (ho n b02 ) = A ho n + 2 h2 '

since h o h a = 0 in E 2 . TO see that A ' = 1, we apply the edge theorem again, but with respect to multiplication by h a . The grading of the algebra generators of E 2 implies that only elements which have h o or h a as a factor can appear above the edge. By Corollary 1.4 (see also Remark 3.3), these elements may be divisible by h o or by h a but not by both. Therefore the element ha s of E(s, s) would survive to H* A unless killed by a differential on some ha-multiple; and the only candidate is d2(ha s-3 bo2 ). This shows that A ' = 1 and proves 4.1.


Another proof of type (3) may be based on Adams' result that hi 3 = ho 2 h 2 in H* A. Since these elements have the same filtration, this relation must already hold in E~ i.e. in E~ of May's spectral sequence. But the only way to obtain it is by d 2 (b02).

Our conclusion is that in the 3-stem, E 3 = E 4 . . . . . E~o, with the survivors being h2, h o h2, and ho E h2 =- hi 3.

The differentials d E (h i bo2 ) are hi-stable , i= 0, 1, and we factor out these two ladders from E 2 .

In the 5-stem nothing appears whatsoever. E 2 contains ha 5 and h 1 bo2 but these elements have been factored out as elements of the hi-ladder of d E (h a bo2 ).

Therefore there are no survivors in either the 4-stem or the 5-stem. In the 6-stem we have ho s-2 hE 2. In the 7-stem we have hoS-lh3 and

ho s- 2 ho (1). (The elements hl 7 and h 13 bo 2 have been discarded.) Note the relation h E bo2--= h o ho(1 ) (Theorem 1.2, Relation 3.1). Of course h 3 and its ho-multiples are cycles.

Proposition 4.2. d E (h o (1))= h o h22.

Proof(i). This is given for illustrative purposes in Chapter 5.

Proof(2). h o d 2 (h o (1)) = d 2 (h o h o (1)) = d 2 (h E b02 ) = h 2 d 2 (bo2) = h o 2 hE 2 using 4.1 and the relation h a h2=0. This excludes the possibility dE(ho(1))=0 , and 4.2 is the only other possibility, for dimensional reasons.

Proof(3). For large s, ho s- 2 h22 cannot survive, and it can only be killed by d E (ho s-3 ho(1)); as in the preceding proof, this implies 4.2.

Second Proof(3). Adams [2] has shown h o hE 2 = 0 in H*(A). The only way to kill this element in the May spectral sequence is by d E (ho(1)).

The differential of 4.2 is ho-stable and we factor out its ladder. Note that this differential is not hi-stable.

In the 8-stem we have h a ha, h a ho(1), and ho s-4 bo22 (s>4). Clearly h a h 3 is a permanent cycle. We have d 2 (h I h 0 (1))-- h a h o h22 = 0, and all higher differentials vanish on h 1 ho(1 ) for dimensional (or rather filtrational) reasons since u is ( - 2 ) for this element, d r raises filtration degree u by r, and there are no elements in E 2 with positive filtration degree. We have dE(bo2)2=0 since d 2 is a derivation in the Z2-algebra E 2 .

Proposition 4.3. d 4 (bo22) ~ ho 4 h a .

Proof(3). Otherwise ho" h a would survive for large n, which is impossible.

Consequently the survivors in the 7-stem are just the ho i h a for 0 _ i_< 3. In the 9-stem we have ha 2 ha, ho s-3 hE 3, hi E ho(1), and h 1 b022. All these

elements are permanent cycles for obvious reasons. Therefore the survivors in the 8-stem are h a h 3 and h I h o (1). This latter element is the first to appear which is not decomposable in terms of the hi. Following May's notation we call it Co.

We now consider the 10-, 11- and 12-stems, where two new generators appear, b12 and boa. From dimensional considerations, d2(b12)=A1 ha 2 h a +

32 M.C.Tangora:

A 2 h23 and d2(bo3)=A3 h 3 b o 2 + A 4 h a ba2 , where the A i are coefficients to be determined. In addition these three stems contain the ha-multiples of c o and bo22, and the elements bo2 ho(1), h2 4, and bo2 3, together with the appropriate h o- and ha-multiples.

Proposition 4.4. (a) d 2 (b12) = h23 + hi 2 h 3 ;

(b) dz(boa)=hl b l E + h a bo2.

Proof(2, 3). The only way to kill ho" hE 3 for large n is by d2(ho n ba2 ). There- fore A 2 = 1. The only way to kill ho ~ h a bo2 for large n is by d 2 (ho n b03); therefore A 3 = 1. Since d 2 d2(bo3)=0, we have, using 4.1,

0 = h a 3 h 3 + A 4 ha d2(b12),

so that A 4 and A t must both be 1.

Another proof using methods (2) and (3) is based on the May relation (ho (1)) 2 = bo2 ba 2 + h12 bo 3 (Theorem 1.2, Relation 4.1). We first obtain A 3 = 1 as in the preceding argument. Then the calculation of d 2 (h o (1)2) --- d 2 (bo2 b a 2 + hi 2 bo3), which is zero because d z is a derivation, leads immediately to A 4 = 1, A2= 1, and A 1 = A 3 (= 1). (One uses 4.1 and the relation h2 a boz=ho 2 h z b12 , based on Relations 3.1 and 3.2 of Theorem 1.2.)

The differentials d z (h i b12 ) and d 2 (h i bo3 ) are hi-stable, i = 0, 1, and we discard these four ladders. Observe the effect of this. The differential d 2 (bo3) appears to have only the term h3 bo2, because the term hi blz is factored out as an element of the h~-ladder of d 2 (h 1 ba2 ). The relation d 2 d2 = 0 continues to hold in the quotient complex, because h3 bo2 is a pseudo-cycle, since its boundary ha 3 h3 has been discarded as an element of the ladder of dz(h 1 b12 ). This illus- trates the operation of Proposition 3.2.

Although the work sheets for the 11- and 12-stems now show d 2 (bo3) = h 3 bo2 , we must keep books on the correct differential of b03 in the original complex E2, as stated in 4.4, because we will need to compute dz(bo3 x) for various x e E 2, and we have no right to discard the term h 1 b12 x in such calculations.

In the 9-stem, then, the survivors are hz3=h12 h3, hi z ho(1)=hac o, and hi bo22. We write Px h a for h I bo22, and in general we will write pk X for bo22k x, since multiplication by bo22 is the Adams periodicity operator pX at least in a certain range (in a neighborhood of the "edge ') .

In the 11-stem we have bo2 h o (1). The value ofd 2 on this element is computed directly from 4.1 and 4.2 using the fact that d 2 is a derivation; we obtain, using the relation ho E h 2 h 0 (1)= h 0 h22 bo2 (see Theorem 1.2, Relation 3.1), the result d E (bo2 ho(1)) = ha 3 ho(1 ) = ha 2 c o. Such calculations will henceforth be taken for granted. Multiplication by h o gives a cycle, h o b02 ho (1 )=h 2 b022 (by the same relation as above); for dimensional reasons this must be a permanent cycle. Thus the only survivor in the 10-stem is hi 2 b o 2 2 = P 1 hi 2.

In the 12-stem we have the cycle h24 with its ho-multiples, and we have b 3. 02 , from 4.1, d E (b02) 3 = ho 2 h 2 b022+ hi 3 b022. This differential is hi-stable (at least when multiplied by hi), i= 0, 1, and we discard the two ladders. Thus the sur-


vivors in the 11-stem are just ho i+1 bo2 h o (1)-- ho i h 2 b022 = p1 ho i h2 for 0-< iN 2, with p1 ho 2 hE =p1 hla.

In the 13-stem our quotient complex contains only h 2 blE with its ho-multiples; from 4.4, d 2 (h 2 bl 2)= h24. This is ho-stable and we discard its ladder. The consequence is that there are no survivors in the 12-stem or the 13-stem.

We remark that the above differential is hE-stable. However, there are enough complications in working with hi-ladders for i>= 2 to make their use of rather dubious value, at least in the range (t-s)____ 70. We will retain h2 n and h2 n- 3 bl 2 in our computations, to avoid such detours as, for example, proving that no such element (for any n) is divisible by h o .

In the 14-stem we find ho s-2 h32, obviously cycles, and ho ~-4 bo2 b12. The latter elements are apparently cycles, since the 13-stem of the quotient complex is empty. In fact boE blE is a pseudo-cycle, since dE(bo2 blE)=hl E h a bo2 by 4.1, 4.4, and a relation in E 2 . The actual cycle is readily seen to be boE blE + hi E boa , or ho(1) E. For n>_ 1, ho" b02 bl2 is actually a cycle.

In the 15-stem we have h 4, h 2 boa, hE 5, h a b02 E, b022 ho(1), and their h o- multiples; and h~ ha 2 and h 1 bo2 blE. In EE, all these elements are cycles except tha t d2 (bo2 E ho(1))=h 0 hE 2 bo22=ho 3 boE bl2 , and this d 2 is ho-stable. We are therefore faced for the first time with extensive computations in later stages of the May spectral sequence. (We remark that it is typical, at least in the known range, that higher differentials play a much more important role in the neighborhood of the 2k-stem or the 3- 2k-stem than in the intervals between.) Before talking about d 4 on such elements as h 2 boa or h 3 bo22 we should compute d 2 on the 16-stem to verify that these elements actually survive to E 4. We have, in the 16-stem, a new generator hi(l), and also b02 boa , h2 E b12 , and bo24, along with various h o- and h~-multiples.

Proposition 4.5. d 2 (h1(1)) = h 1 h32.

Proof(2). From dimensional considerations, d 2 (h1(1)) = A h I h3 E + A' ho E h 4 . Since h o h1(1)=0 (Theorem 1.2, Relation 3.4) while ho a h4~0 , we have A '=0 . Now h 1 h1(1)= h a b12 by Relation 3.1; therefore d 2 (h i h1(1))= hi E h32 from 4.4, and thus A = 1.

Proof(3). Same as for 4.2, second method-(3) proof.

From 4.1 and 4.4 we have

dE(boE boa)=hl a boa+ho E h E boa +h i b02 bl2 +ha boE 2

where the first term will not appear on our work sheet because it is in the ladder of dz(h 1 boa ). We also have, of course, d 2 (hE 2 b12)--hE 5, and this is ho-stable. Clearly bo24 is a cycle in E 2 .

We now have E 3 through the 15-stem and h 3 bo22 is non-zero. From 4.3, d4(h a bo22) = ho 4 ha 2, and this higher differential is ho-stable.

Proposition 4.6. d4(h 2 boa)= ho 2 h32.

Proof(2). We have dE(honboEboa)=hon+EhEboa+honhab022, for every n => 1. Thus the two terms on the right of this equation project to the same ele- 3 Math. Z., Bd. 116

34 M.C. Tangora:

ment o f E 4 . But d4(h 0 h a b022) = ho 5 h32. Therefore d4(ho 3 h 2 b03 ) = ho 5 h32; this implies the proposition.

This completes the work for the 14-stem, in which the survivors are h32, h o h32, and ho i do, 0 < i < 2, where (as in May's notation) d o = h o (1) 2 = bo2 bla + hi 2 b03, represented in our quotient complex by bo2 b12.

Proposition 4.7. d 8 (bo2 4) = ho s h 4.

Proof(3). Nothing else is available to kill ho" h4, n > 8.

This completes the work for the 15-stem, where the survivors are ho" h4, 0_<n_<7, and h 1 d o.

The differentials now at hand make the computations routine through the 20-stem. By now we have given a taste of the methods used, and henceforth we shall discuss only certain points of interest in the calculations, suppressing all routine details.

We come to three new algebra generators in the 22-, 26-, and 28-stems.

Proposition 4.8. (a) d 2 (bEE) = hE 2 h 4 + ha 3 ;

(b) d2(baa)=h4 blE+h 2 b22 ; (C) dz(bo4)=h4bo3+h lb13. No other terms are possible, by inspection of the tri-grading; so a proof

consists of showing that the six coefficients are non-zero.

Proof(2, 3). We must first show that d2(bo4 ) contains the term h 4 bo3 by a careful and detailed application of the technique for using Adams' edge theorem given in Remarks 3.4. It is a routine matter to write down the ho-stable group E 2 (29), as defined in 3.4. It contains seven elements. None of these can kill bo4 in E,(28). The only two which have sufficiently deep filtration to be candidates are bo3 b12 ho(1) and b023 b12 ho(1 ) (i.e. the corresponding elements in the limit group E r (29)); but we can calculate d 2 on these elements; d 2 on the first is zero, so it cannot kill b04; d 2 on the other is non-zero, so that d4 on this element cannot kill b04. Therefore hon b04 cannot be a permanent cycle, for large n. But then ho b04 cannot be a permanent cycle. (As always we are making strong use of Theorem 1.3.) However, if d2(bo,,) did not contain the term h,~ b03 , we would have d2(h 0 b04)=0, and moreover d 4 and d 6 would vanish on h 0 b04 for dimensional reasons, so that ho b04 would be a permanent cycle. This argument shows that d2(b04 ) contains the term h 4 bo3.

There are several different ways to complete the proof of 4.8 using methods (2) and (3). We indicate one. Since d 2 d 2 (b04) = 0, and since 82 (h 4 b03 ) = h~ h4 bl 2 by 4.4, dE(bo4 ) must contain the other term h 1 b13 , which proves (c) and also shows that d 2 (b13) contains the term h 4 b12. Then d 2 d2(b13)=0 implies, by a similar argument, that d 2 (b13) m u s t contain also h 2 b22 , proving (b), and also proving that d 2 (b22) m u s t contain the t e r m h22 h 4 . Finally, the fact that d 2 (b22) contains h33 is quite easy to prove (independently of the above), since h~ b22 = h 3 h1(1 ) (Relation 3.2) and d 2 (h 3 h1(1))= h 1 h33.

Proposition 4.8 completes the groundwork for calculation of d 2 up to the 33-stem, since the next algebra generator is h 2 (1) in the 34-stem. However, we


encounter some higher differentials in this range. In particular we find

dr Ah2 b122 + Bh 4 bo2 z

where we must determine the coefficients A and B.

Proposition 4.9. d 4 (bo32) = h 2 b122 + h 4 bo22.

Proof(l). It is not difficult to show by the imbedding method that both A and B are non-zero. See Chapter 5.

Proof that B = 1 by method (3): there is nothing else to kill h 4 bo2 z in the stable group E,(23).

We cannot use the ho-stable method on the coefficient A, since h 2 b122 is zero in E4(23 ) because ho 3 h 2 b122 =d2(h 0 bo2 b122). [ do not know any proof that A = 1 using only methods (2) and (3). A proof by method (4) has been found jointly by the author and Mahowald. Since it depends on further calculations in the May spectral sequence, specifically on the range 42 < t - s < 44, we post- pone it until that range has been computed, and it appears after 4.18 as Pro- position 4.9*. However, it can be verified that the result of Proposition 4.9 is not needed for the proofs of any of the other propositions of this chapter, although it is essential to many parts of Theorem 4.42.

Proposition 4.10. d s (h 3 bo32) --ho 4 h42. Proof(2). Using 4.3 and 4.9, we have

d4(bo22 boaE)=ho 4 h a b032 +h4 bo24+(A)h 2 hOE 2 b122

where A is the ambiguous coefficient in 4.9. The last term, however, is not ho-stable in E4; for ho 3 h E b022 b122 =d2(h 0 b023 bl2a). Thus, regardless of the value of A, we have ho 7 h a boa 2 =ho 3 h4 bo24 in E 8 . But 4.7 gives ds(ho a h, hOE 4) =ho 1~ h42. This implies 4.10.

There is a survivor i=hobo2 bo3 ho(1 ) in the 23-stem and a survivor s=h 4 boz 2 ho(1)+ho 3 bo2 b13 in the 30-stem. (The cycle s was used as an example in Chapter 3.)

Proposition 4.11. d 6 (bo22 i )= ho 5 s.

Proof(3). We make use of the stable groups and the Adams edge theorem. The group E 4 (29) is readily seen to be zero, and the group E 6 (30) is seen to contain only the elements arising from s and h42. Since s is a permanent cycle, something must kill it in Er(30 ). The only non-zero elements of E6(31 ) arise from hs, h a bo32, and bo22 i. We know that h 5 is a cycle and we know that h 3 bo32 is also a cycle in E6 (in any case the filtration is such as to rule out this possibility). Thus b o 2 2 i is the only thing that can kill s in Er(30 ). Since the filtration degrees are ( - 1 2 ) and ( -6 ) , the differential is d6, hence 4.11.

We remark that this is the first non-zero dr for which r is not a power of 2.

Proposition 4.12. d16 (bo28) = ho 16 h 5 .

Proof(3). This is the only way to kill h 5 in Er(31), since bo2 s is the only element of Er(32 ) for r>4 . 3*

36 M.C.Tangora:

Proposition 4.13. d 4 (h 3 b13 ) = hl 2 h42.

Proof(2). For dimensional reasons, d4(h 3 b13 ) --A h 12 h42 and we need only determine the coefficient A. We must take an excursion out to the neighborhood of b 022 ha bin. From 4.3 we have d 4 (h a b 022 bxa) = ho 4 ha 2 bl 3 + A h 12 h42 bo 22. The first term is actually zero in E4, being equal to d2(ho 2 h a boa b22 ). Now d4(hl 2 h4bo32)=hl 2 h42 b022 (regardless of the ambiguous term in 4.9). But h a b022 b13 = hi 2 h 4 b032 in E4, since the sum of these two terms is dz(bo2 b03 b13 q- b02 b04 b12+hl 2 b03 b04 ). This implies that A = 1, provided that h 3 b022 b13 is actually non-zero in E4, but this is a routine matter to verify. (The verification is independent 0f4.14 below.)

Proposition 4.14. d 2 (h 2 (1)) = h 2 h42. Proof(2). We have d2(h2(1))=Ah 2 h42+Bhl 2 h 5. Since hi a h 5 is non-zero

and h 1 h 2 (1) is zero, B = 0. We now proceed as in 4.5 or 4.2: h 22 h42 = d2(h 4 b22) = d2(h2 h2(1))=Ah2 z h42, so A = 1.

Proof(I) or Proof(3). Cf. 4.5 and 4.2.

Proposition 4.15. d 2 (h o (1, 3)) = h42 h o (1) + h o h 2 h 2 (1).

Proof(2). For dimensional reasons there are no other terms to consider. Since d 2 d z = 0, we must have both or neither. But we have a relation h 2 h o (1, 3) -- ho h4 bla; d2 (ho h4 ba3) is equal to h 2 times the sum of the two terms; hence the result.

Proposition 4.16. d 2 (h o (1, 2)) = h a h o (1, 3).

Proof(2). By Relation 8.1 of Theorem 1.2, h 1 h a ho(1 , 2)= h1(1 ) ho(1 , 3); so, by 4.5 and 4.15,

h 1 h a �9 d 2 (h o (1, 2)) = d 2 (h 1 (1) h 0 (1, 3)) = h a h 32 ho (1, 3) ~e 0,

which implies the proposition.

Proposition 4.17. d4(h o h a bo4 ) = ho 3 h 2 (1) d- h o h42 bo2.

Proof(2). Otherwise the differential must be zero. Now hoE h33 bo4= h o h a bo2 ho(1, 3) in E 4 (in the 49-stem), because the sum of these elements is da(ho2bo4b22+hoboEho(1,2)). On the other hand, hohabo2ho(1,3)= h a b022 h2(l ) by Relation 6.1. Since h 3 hE(1 ) is a surviving cycle, d4(h 3 bo22 h2(1)) = h 3 h2(1 ). ho4h3 by 4.3, and this element is non-zero; the result follows.

Observe that the manipulative proof of d4(h o h 3 bo4), a differential in the 35-stem, is based on data from much higher dimensions (4.17 uses 4.16 which refers to the element h a h a h 0 (1, 2) from the 54-stem). This is very common in proofs of this type.

Proposition 4.18. d 6 (bo22 r)= ho 6 x.

Here r=hz 2 bo3 2 and x=h 3 boz bin+h12 h 3 bo4 are surviving cycles in the 30- and 37-stems respectively.


Proof(2). In E2, bo26r--b022i 2 where i is the surviving cycle h o bo2 boa ho(1)=h 2 bo22 boa. Using 4.11, d6(bo26 r)=ho 5. i. s. In the 53-stem, hoS.i .s=ho6bo24X in E r for r>2 , since the sum of these elements is d2(ho 6 b024 boa bx3 ). N o w b026 r may be written (b024) (b022 r), so d6(b026 r)-- bo24 d6(b022 r), but we have shown d6(b026 r )= boE4(ho 6 x), and this element is non-zero in E 6 (as can be verified with the tools at hand). Since d 6 (bo24)-- 0, the only possible conclusion is 4.18.

We return to a piece of unfinished business.

Proposition 4.9*. In d4(bo32):(A) h 2 b122 + h 4 bo22, A = 1.

Proof(4). Suppose otherwise that A=0 , and compute the May spectral sequence accordingly for the 4 3 - a n d 44-stems. In the 43-stem, 9_<s_<13, one has survivors h E g2 (g_b122), ho h2 g2 p1 m=do k (k=h 2 bo2 boa b12 , m=h2 bo a b122), p1 ho m, p1 ho 2 m. In the 44-stem, one has at least, in 8 < s < 10, the survivors boa 2 g, h 0 boa2 g, and ho 2 boa 2 g=d o r. In the Adams spectral sequence Mahowald [5] has proved the differential d 3 r - -ho E k. Since d o is a permanent cycle in the Adams sequence, we have d 3 (d o r) = ho 2 d o k = P1 ho E m. But the Adams differential d E (p1 m) is non-zero [7], so that p1 m and p1 hom are zero in E a of the Adams sequence; and this leads to a contradiction. For h o boa 2 g survives to E 3 of the Adams sequence, and is a cycle there, but this contradicts the statement d a (ho E boa 2 g) ~ d 3 (d o r)-- p1 ho 2 m =~ 0. The contradiction establishes A = 1.

The above proof is due jointly to the author, who made the parallel calculation of H* A under the false assumption A = 0 and verified that the relevant Adams differentials were not affected, and to Professor Mahowald, who pointed out the contradiction involving d o r.

We have now independently proved all the differentials given by May in his thesis (Theorem II.6.13) for the range ( t - s ) ~ 4 3 . Proposition4.11 does not appear in the first printing of May's thesis because the elements i and s had not been found at that time. Moreover, Proposition 4.18 was overlooked first by May and then independently by the present author, and was first discovered by Mahowald through an apparent inconsistency in the Adams spectral sequence, and first proved by him in this way, i.e. by method (4). The author subsequently found the above proof of 4.18.

Proposition 4.19. d 6 (boa b13 + bo4 b12 ) h o (1) = ho 3 g2, g2 = b222"

Proof(3). This is a good example of a proof by means of ho-stabitity and the Adams edge; the argument is detailed but straightforward. One must write the stable groups E2(44 ) and E2(45), which contain 26 and 33 elements respectively, and also E2(43 ) and E2(46 ). Computing dE, one finds that the only survivors to E 4 are the elements arising from g2 in the 44-stem and haEh5 and (boablad-bo4b12)ho(1) in the 45-stem. Clearly haEh5 is of no relevance (it is killed by d4(h 2 h 5 boa)). Since g2 is a permanent cycle, it must be killed somehow, and 4.19 is the only possibility.

Proposition 4.20. d 4 (h E bo4 h 1(1)) = hi 2 g2.

38 M.C.Tangora:

Proof There is a possible term h t h 5 d o but this is ruled out because in E 4 hi 2 h 5 d0+0 and h 1 h z boa hi( l )=0. To establish the term hi z g2 I know no proof except by method (1). See Chapter 5.

Proposition 4.21. d 8 (bo34) = h s bo24.

Proof(3). There is no other way to kill the element arising from bo34 in Er(48 ).

May states that d 8 (b034) = h 5 b024+ h 3 b124, but h 3 b124= d 2 (b03 b122 ht (1)) , so the two versions are consistent.

Proposition 4.22. d 4 (h 2 h o (1, 2)) = h o h 32 h2 (1).

Proof(2). We have ho 2 h 2 h 0(1, 2 ) = h 0 h33 b o 4 + d 2(h 0 b04 b22), and d4(h 0 h33 bo4)=ho 3 h32 ha(1 ) by 4.17, from which the result follows.

Proposition 4.23. (a) d 2 (b32) ~ h32 h 5 + h43 ;

(b) d 2 (b23) = h 5 b22 + h 3 b32 ; (c) dz(b14)=h5 b13 +h 2 b23;

(d) d2(bos)=h5 bo4 +h 1 b14.

Proof We are in a situation entirely similar to that of Proposition 4.8 but much more complicated. A proof by method (1) is easy. Otherwise we must appeal at one point to 3.4, and then we can obtain the other details by manipulation. We denote by At, A2, B t . . . . , D 2 the eight coefficients to be determined, and indicate one method of proof by methods (2) and (3).

We first use 3.4 to show that D r = l , i.e. that d2(bo5 ) contains the term h 5 bo4. It is easy to see that otherwise bo5 gives rise to a permanent cycle in the stable spectral sequence. To see that this is impossible, consider Er(61 ). E2(61 ) contains 93 distinct non-zero elements. Eleven of them have filtration degree u = - 1 0 so that they might hit bo5 with a d2; but in each case we can already calculate d 2 (without the present proposition) and this does not occur. There are dozens of elements of E2(61 ) which have deeper filtration and therefore might hit bos with a higher differential, but only two of them survive to E4(61 ). One is killed by a d 4 from E4(62) and the other is bo24 times the element whose d 6 is given by 4.19. In this way all other possibilities are eliminated and we may conclude Dr=0.

Manipulation gives the remaining coefficients. For instance, the relation 2.2 of Theorem 1.2 may be differentiated by d2, giving A 2 = B 2 = C2=1 and A t = B t = Ca; then the relation 12 may be differentiated by d 2 to give D 4 = l and A t = D t. This completes the proof.

We remark that there seems to be no guarantee that the procedures of 4.4, 4.8, and 4.23 will continue to settle the differentials dz(bij ) for i+j>=6. The experience already obtained suggests that the edge method 3.4 must be used at least once, if a proof by methods (2, 3) is desired; but 4.23 certainly demonstrates that use of the edge becomes exceedingly delicate and tedious in higher dimensions, and may at some point cease to be effective. Of course any d 2 is always easily obtained by method (1) as in May's thesis.


Proposition4.23 completes the determination of d 2 for t - s < 6 9 and therefore the following calculation is less delicate.

Proposition 4.24. (a) dg(b132) = h 5 b122 + h 3 b222;

(b) d4(bo42)=h5 bo32+h2 b13 z.

Proof(2, 3). The stable group E~(57) is zero. Therefore the four elements of E4(56 ) cannot be permanent cycles. It follows that dg(bo42) contains the term h 5 b032. (The other term is not ho-stable. ) It follows from 4.9 that the other term must also be present and that dr 2 b132) contains the term h 2 h 5 b122. Finally, b132 gives the only element of E4(52 ) and therefore d4(b132) must contain the term h 3 b22 z, since nothing else can kill this permanent cycle in the stable 51-stem.

Proposition 4.25. d 6 (h 3 b026 b032) = ho 7 S 1 .

Here S l = h o 3 bo2 bo3(b03 b13 +b04 b12 ).

Proof(3). E,(56) is zero for r > 4 and therefore h 3 bo26 bo32 cannot give rise to a permanent cycle in E,(55). This implies 4.25 since there is no other possibility. Moreover, there is no other way to kill S 1 in E,(54), so we have a double proof.

Proposition 4.26. diE (h 4 bo212) = ho 14" h 5 bo22 i,

Proof(3). E4(61 ) contains only two non-zero elements which are easily accounted for, showing that h 5 bo22i is a permanent cycle in the ho-stable groups of the 62-stem. The stable differentials from 63 to 62 all follow from what has gone before except for the one given here, which then follows.

Proposition 4.27. d12 (h a b028 boa 2) = ho 1~ h 5 bo22 i.

Proof. d4(bo21~ b032)= ho 4 h a bo28 b032 +h4 bo212 d- ho bo29 b122 ho(1), by 4.3 and 4.9. The last term is a permanent cycle, h o p4 do eo" The result follows from 4.26.

There is an odd situation here: h 5 bo22i has order 21~ in E~ as suggested by 4.27. However, the element h 5 i only has order 8 in the 54-stem, since ho 3 h 5 i=h 0 h 3 h 5 boz4+d2(ho h 5 bo2 3 bo3), and h o h a h 5 bo24~ d8(h o h a bo34). It is for this reason that we have not written pX h5 i. See Note 3 at the end of this chapter, and also Chapter 6.

Proposition 4.28. d16 (h 4 boa 4) = ho s h52.

Proof(3). From 4.7 and 4.21, ds(bo24 boa4)=ho 8 h 4 boa4+h5 bo28. The result follows from 4.12.

The analogy between 4.28 and 4.10 is noteworthy.

Proposition 4.29. d32 (b0216) = ho 32 h 6 .

Proof(3). This is the only way to dispose of these elements in the ho-stable 63- and 64-stems.

Proposition 4.30. d4(h 2 boa bo4 hi(l))= b122 b22 h1(1 ).

40 M.C.Tangora:

As with 4.20 I know no proof of this proposition except by method (1). See Chapter 5.

In order to complete the computations for the range t-s__<48 we must establish d 6 (~) where

~b=hl b12(b03 b13 q- b04. b12)

is an element in (49, 7). A proof by method (1) would make the computations self-contained for the range t-s____ 48 and would be desirable for that reason. However, it is in the nature of the elements in question that such a proof would be of astronomical length, being perhaps an order of magnitude longer than any of the proofs indicated in Chapter 5. I have not carried through such a proof.

In order to obtain d6(q~ ) we therefore use manipulative methods which will take us out as far as the 68-stem. The result, 4.34, completes the computation for t - s < 6 4 .

Let q~' denote hobo3(bo3 b13+bo4b12)ho(1)~E2(57,9 ). The notation is suggested by the fact that ho 2 q~,_p1 ~ in E 3 .

Proposit ion 4.31. d 4 (q~') = 0.

Proof(I). See Chapter 5.

The proof of 4.31 by method (1) is very easy, in contrast to 4.34. Ironically, I know no other proof of 4.31.

A higher differential on ~b' is obtained in 4.33. Let X = h o bo2 bo3 (bo3 b13 + bo4 b12)~E 2 (54, 9).

Proposit ion 4.32. d 6 (X) = h o h s b023 b12.

Proof. There seems to be no shortcut to this key result. A very long proof by the imbedding method has been found, and serves, by its formidable length, as ample demonstration of the limitations of that method. We also have a proof by method (2) which involves an excursion to the 68-stem, as follows. It is easy to see that do X = i Y where do and i are survivors in the 14- and 23-stems respectively, and Y-=(bo3 b13 +boa . b12 ) ho(1 ). Therefore we have d6(d o X)=ho 3 b222i by 4.19. If we can show that this element equals do(h o h 5 b023 b12 ) and is non-zero in E6(67 , 14), we are through. But

ho 3 b222 i=ho # bo2 b03 b222 ho(1)

= ho 7 h 2 b03 bin 2 (Relations 3.1 and 2.1)

=ho h5 bo24 b122 +d2(ho 5 b02 boa b132)+d4(ho bo24 b132).

It is a routine matter to verify that this element is non-zero in E 6 and therefore d6(d o X)= do(h o h 5 bo23 b12 ). This implies 4.32.

Proposition 4.33. (a) d 6 (t~')= h o h 5 b022 b12 h o (1);

(b) d6(b022 b12(b03 b13 q-b04 blE))=hl h 3 h 5 bo24; (C) d6(b024 b12(bo3 bin q-b04 bl2))---h 1 h 3 h 5 b 6. 02 (d) d6(b022 ~b')--h 0 h 5 b024 b12 ho(1 ).


Proof. These results are corollary to 4.32. We first obtain (a) from the relation hzX=hoqS', since h2d6(X)=~O in E 6. Then (b) follows because hi(b022 blz(b03 b13+b04 bla))+ho 2 ~b' is a boundary in E2, namely d2(boa z bo3(b03 b13 +b04 b12)), and because ho 3 h5 p1 eo=hl 3 h5 p1 do = hi hs p2 hi h3 + do(hi2 h5 b023 b03)" Now heuristically (c) and (d) are obtained from (b) and (a) by multiplication by boa 2 (or p1); but this does not give a proof, since d2(bo22)=ho4h3@O so that this multiplication does not make sense in E 6 . However, we can readily obtain (c) and (d) as we have obtained (b). For p l h i is a survivor and thus d6(pihlbo22b12(boab13+bo4bi2))= Plhl(hlh3hsbo24); the image is non-zero, and (c) follows. But also pi hi bo22 b12(bo 3 bx 3 +bo4 b~2)=p~ ho E ~b,+dE(bo24 bo3(bo 3 bl 3 +bo4 b12) ) so that (d) follows from (c) as (b) did from (a).

Finally we have d 6 (~b) as a consequence of the above results.

Proposition 4.34. d 6 (~b) = p1 hi E h3 h5 (= hi 3 h5 do).

Proof. d 6 (p2 ~b)---hi 2 h 3 h 5 boz 6 by (c) above. This element is non-zero and therefore d 6 (~b) must be non-zero, since d6(bo24)-- 0. This proves 4.34.

Proposition 4.35. d4(b132 hi (1) + h 3 2 bo 4 b13) = h5 t. Here t= b122 hi(l)+ hi 2 boa b22 is a survivor in the 36-stem.

Proof(2). By 4.24, d4(h 2 b~32 h~(1))=h 2 h5 b122 h~(1)= h 2 hs t, and this image is non-zero in E 4. The result follows.

Proposition 4.36. d 4 (h 4 b23 ) = h22 h52.

Proof(I). We omit the proof of this proposition, stated by May in general form (cf. 4.6 and 4.13) and proved by the imbedding method.

Proposition 4.37. (a) d 4 (h 3 bo4 h E (1)) = ho 2 h 2 (1) 2 ;

(b) d6(b022 boa 2 x)=ho 6 boa 2 b222;

(C) d6(ho 2 b032 Y)=ho 5 b032 b222.

Here x = h 3 bo2 bla+h12h3bo4 is a survivor in the 37-stem, and Y is as in 4.32.

Proof(2, 3). E4(68) contains two permanent cycles, namely h2(1) 2 and boa 2 b222. E4(69 ) contains four elements. One of these, p4 x, is a boundary in E4, another, h4b23, is a permanent cycle (4.36 is not ho-stable ). But h a bo4 hE(1 ) cannot hit bo32 b222 since it has the wrong filtration. The only possibility for the stable differentials is therefore the one corresponding to (a) and (b). Now (c) follows from (b) since hoabo32 YWbo22bo32X= d2(bo22 bo32(bo3 b13 +bo4 b12)).

Proposition 4.38. d E (h 3 (1)) = h a hs 2.

Compare 4.14, 4.5, 4.2. Proof by method (1) is easy, as is proof by inference from the Adams relation h a h52---0 in H* A. Proofs by manipulation or by the edge can be dug out but hardly seem worthwhile.

42 M.C. Tangora:

Proposition 4.39. (a) d 8 (h o h32 b042) = ho s h 4 b23 ; (b) d8(ho 2 h 2 h 5 bo33)=d8(h3 h 5 bo22 bo32)=ho 8 h 4 b23.

b 2 i s Proof(3). We find two survivors to E8(70 ), and one of them, h52 02 , a permanent cycle. Therefore the other one must kill h 4 b23, and the proposition follows, using the relations d4(h 3 b022 bo42)=ho 4 h32 bo42q-h3 h 5 b022 b032 and d4(ho 2 h 2 b03 bo42)=ho 4 h a b042 +ho 2 h 2 h 5 bo33+d2(ho b03 b132 ho(1)).

Proposition 4.40. d s (ho 2 h 4 bo42) = ho 2 h52 bo22.

Proof(3). Nothing else can kill h52 b022 in Es(70), and moreover there is nothing in the ho-stable 72-stem that can kill h 4 b042 and no other way to dispose of this element in E s (71).

The element hoh4bo42 is not present in E8, since d4(hoh4bo42)= h o h 2 h4b13 by 4.24(b). But ho 2 h 2 h 4 b13 is a boundary in E 2 and thus ho 2 h 4 bo42 survives to E s (71, 7); the d s of 4.40 is then ho-stable.

Proposition 4.41. d4(h o h 3 bo2 bo4 hi3 ) = ho 4 b13 ho(1 , 3).

Proof(3). There is no other way to dispose of these elements in E4(64 ) and E4(65 ).

In the light of the fact that d4(hoh3bo4)=hoah2(1)+hoh,~2bo2 (Proposition 4.17), and that d4(h o h 3 bo2 b13) =d4(h o x)=0, one has d4(ho 2 h32 bo2 bo4 b13)=ho 5 h 3 b13 ho(1, 3) from the relations of Theorem 1.2. However, this is a fictitious calculation in the sense that what it says is d4(0 ) = 0, since these elements are boundaries in E2; thus this argument cannot be used to prove 4.41.

The differentials which we have proved in the forty-one propositions above give all the essential information about the May spectral sequence in the range t-s___70. Countless other differentials must be cranked out, but they all follow from those given above by more or less elementary arguments or routine calculations. In the actual work, the proofs of these fundamental differentials appear, with a few notable exceptions, as momentary pauses in the main task, which is to use the ideas of Chapter 3 to reduce the routine work to a manageable size. However, it is not feasible to present the routine work here, and probably not desirable in any case; so in this paper the present chapter assumes a disproportionate importance.

We collect the results into the following theorem. Relations which derive from the May relations in E 2 (Theorem 1.2) are indicated by equality signs; relations which come out of differentials in the May spectral sequence are indicated by the homology sign ~ .

Theorem 4.42. Table 4.43 lists all generators of the Z2-module Eoo = E ~ * A) in the range t-s<= 70. These generators are subject to the relations indicated (and all relations deducible from them).

We give in Appendix 1 the definitions of all indecomposable generators in terms of May's generators for E 2 . A display chart ofEo~ is given in Appendix 2.

On the C o h o m o l o g y of the Steenrod A l g e b r a

Tab le 4.43

43

t - s s Gene ra to r s Per iod ic i ty

0 L 2, ... 1, ho, ho 2, ...

1 1 h I pi i ~ 8

2 2 hi 2 pi i~=8

3 1 - 3 h2 ' ho h2 ' ho 2 h2~h l 3 p i i<=8

6 2 h22 pi, i< 8

7 1 - 4 h3 ' ho h3 ' ho 2 h3 ' ho 3 ha p2i, i<3

8 2 hi h3 pi, i < 7 3 c o pi i<7

9 3 h23 hi 2 ha pi, i<7 4 hi Co pi, i < 7

14 2 - 3 h32, hoh3 ~ 4 - 6 do , ho do , ho 2 do =p1 h22 pi, i<7

15 1 --8 h4, h o h 4 . . . . . ho 7 h 4 (see no te 1) 5 hi do pi, i < 6

16 2 h i h 4 6 hi 2 d o = c o Z ~ P 1 h 1 h 3

17 3 hi 2 h 4 4 - 6 eo ' ho eo = h 2 do; ho 2 eo; pi, i<6

ho a eo = p l h23 hi 3 do

18 2 - 4 h2h4, hoh2h4,ho2h2h4 Pih2h4, i~=6 4 f o , h o f o ~ h i e o pi, i<6

19 3 c 1

20 4 - 6 g, hog = h2 eo ; ho 2 g pi, i < 6

21 3 h33 ~h22 h 4

h z f o ~ h l g pi, i<=6

22 4 h 2 c 1

23 4 h 4 c o 5 - 6 h2 g ~ p 1 h4 ; ho h2 g (see no te 1) 7 - 1 2 i, h o i; ho 2 i ~ P 2 h 3 + p1 hi do p2, p4

24 5 h 1 h 4 c o

26 6 h22 g ~ p 1 h2 h4 pi, i~=5 7 - 9 j, ho j=h2 i = p l f o ; ho2j pi, i< 5

28 8 do 2 = p1 g

29 7 - 9 k, h o k =h2j ; ho 2 k ~ P i h l g pi, i<5

30 2 - 5 h42, ho h42, ho 2 h42, ho 3 h42 p1 h42 6 r p2, p4

7- - 11 s, h o s, ..., ho 4 s p2, p4

31 1 - 1 6 hs ,hohs , . . . , hol5 hs 3 h 1 he 2

5 n 8 - 1 0 doeo; h o d o e o = P l h2g pi, i<=4

44 M.C. Tangora :

Table 4.43 (continued)

t - s s Genera tors Periodicity

32 2 hi h5 pa 4 d~ 6 q pi, i__< 4 7 - 9 l,h ol=h 2k=dofo; ho 2 l pi, i < 4

33 3 h12 hs p1

4 p, h o p ~ h 1 d 1 7 h 1 q~h 2 r U, i<4

34 2 - 4 h2 hs,hoh2 hs,ho2 h2 h5 pl 6 h 2 n 8 -- 10 eo 2 = do g; ho %2; h02 e02 = p1 h22 g ~ p2 h2 h4 p i i < 4

35 5 h 2 d a ~ h 4 g 7 - 9 m, h o m=h 2 l=eo fo; ho 2 m~h 1%2 pi, i=<4

36 6 t

37 3 h2 2 h 5 5 - 10 x, h o x, ho 2 x ; ho 3 x = h a s ; ho 4 x, ho 5 x p2, p#

7 h 1 t ~ h 2 2/'/=)Co cl 8 eog pi, i<4 1 1 - 1 3 doi=Ptk

38 2 - 5 h a h s . . . . . ho a h a h a (see note 3) 4 e 1 6 h22 dl =cl2 ~ h 2 h4 g~p1 h42

7 - 1 0 y, hoy, ho2y~h2m=fog;hoay~hleog

39 3 hi ha hs p1 4 hs Co p i 5 h 1 e I = h 3 d 1 7 c l g = h 2 t 9 u pi, i < 3

40 4 h2 3 h 5 ~ hi 2 h a h 5 4 f l , hofl=ha P; hoEfl ~ h l 2 el 5 hi hs Co p1 8 g2 pi, i < 3 10 h i u ~ p1 q 1 1 - 1 3 doj=e o i=P II

41 3 - - 5 c2,hoc2,ho2c2 I0 z,h o z=fo i~ha 2 u ~ P lh I q pi, i<3

42 9 v P~, i < 3 12 do a = p1 eo 2

43 1 1 - 1 3 dok=eoj=gi=Plm

44 4 - 6 g2, ho g2, ho 2 g2 p2 10 d o r Pi, i < 3

45 3 - 4 h43 ~ h a 2 hs; h o h4 a

5 hi g2 = ha el 5 - 7 h 5 d o . . . . . ho 2 h5 do=P lh22 h 5 P lh5 do 9 w P~, i < 3

46 6 hi hs do p l 7 B 1 pi, i__< 3 8 N


Table 4.43 (continued)

t - s s Generators Periodicity

(46) 11 d o l=e o k = g j pi, i<3 14 i 2=p 2 r

47 5 - 6 h2g2~ha f l ;hoh2g2=ha2p 7 h12 h5 do ~ p1 hi ha h5 8 hi B1 pi, i < 2 10 eor pi, i<2 1 3 - 2 4 Q, hoQ . . . . ; h o 4 Q ~ P a h 2 g + P 4 h 4 ; . . . h o n Q

48 4 h 3 c 2 5 - 8 h5 eo ' . . . , ho 2 h5 eo p1 hs eo 7 - 9 B 2 , h 0 B 2, ho 2 B 2 ~ h l 2 B 1 P~, i < 2 14 hi Q = p 2 q~h l p1 u

49 5 - 6 h s f o , h o h s f o ~ h l h ~ e o p1 11 d o m = e o l = g k pi, i<2 14 i j = p1 z

50 4 h 5 c 1 6 C 10 g r pi, i<2

51 5 - 7 h5 g ~ h 3 g2; ho h5 g, ho 2 h5 g 8 h2B 2 P~, i<2 9 gn 12 eo 3 = d o eo g pi, i<2

52 5 D 1 6 hi h5 g ~ h 2 h5 fo 8 d 1 g 11 g l = e o m pi, i<2 14 j 2 = i k = P 1 d o r 20 - 22 p4 g = p3 do 2

53 5 h 2 h 5 c 1 7 h 2 C ~ h 1 b12 b13 hi(1 ) (cf. i) 1 0 - 1 8 x', . . . . h o a x , ~ p 2 x ; h o 4 x , ~ i s ; . . .hoax, p2 13 d o u=i r = P 1 w

54 6 G PI, i < 2 8 - 1 0 h 5 i, h o h 5 i; ho 2 h s i ~ p1 hi h5 do (see note 3) 10 R 1 p2 11 h 1 x' = p1 B1 1 1 - 1 7 Sl, hoS1, ...,ho6S1 p2 12 eo2 g = d o g 2 pi, i<2

55 7 h 1 G p1 11 gm pl 14 j k = i l = P l e o r

56 10 g t 10 Q1 p1 11 h 2 x ' = P 1 B 2 13 d o v=e o u = j r p1 16 do4=p2 gZ

57 7 - 9 Q2,hoQ2,ho2Q2 P1Q2 8--10 hsj, h o h s j ; h o 2 h s j ~ p l h l h s e o p l h s j 11 h2 RI ~ h l Q1 p1 12 eo g2 p~

46 M . C . T a n g o r a :

T a b l e 4.43 (con t inued)

t - s s G e n e r a t o r s Pe r iod ic i ty

58 6 - 8 D z . . . . . ho 2 D 2 14 k 2 = j l = i m = P l gr

59 1 0 - 1 2 B21, hoB21;ho2B21=plh2B2 13 d o w = e o v = g u = k r

60 7 B 3

8 - 9 h 5 k, h o h 5 k

9 - 1 1 B,,, h o B4, ho 2 B 4 ~ h 1 B21 1 1 - 1 4 g2' ; ho g2' = p 2 g2 12 g3

61 4 D 3

6 - 8 A; h o A = h 2 D2; ho 2 A ~ h 1B 3 6 - 7 A', how 9 - 1 3 X 1 . . . . ,ho 3 X l = h 3 S a; ho 4 X 1 11 rn 14 k l = j m = g z = d o e o r

62 2 - 9 h52 . . . . . ho 7 h52 5 h 1 D 3 5 H 1

6 h 5 n

8 Co 8 E 1

1 0 - 1 1 R, hoR 1 0 - 1 2 B 2 2 , h o B 2 2 = h 2 B 2 1 = d o B2; ho 2 B22 1 2 - 2 1 hsPXi, h o h s P l i ; h o 2 h s P l i ~ P a h 3 h s ; . . .ho9hsPl i 13 g v = e o w = l r

63 1 - 3 2 h6, h o h 6 . . . . , ho 31 h 6 3 h 1 h52

6 h 1 H i 7 C '

7 X 2 8 - 9 h 5 l, h o h 5 1

9 h 1 E 1 1 0 - 1 1 h 2 B 4 , h o h 2 B 4 ~ e o B l = h 1 B22 15 dog k = e o g j = g 2 i=Pl g m

64 2 h 1 h 6 4 hi 2 hs 2 5 h 2 D 3 6 - 8 A", h o A", ho 2 A" 7 h 2 A

8 g g 2 8 - 1 0 h3Q2,hohaQ2,ho2h3Q2~h12E1 10 ql 14 12=k m=do g r

65 3 hi 2 h 6 3 - 5 h 2 hs 2, h o h 2 h52, ho 2 h 2 h sZ~h l 3 h52 6 h 2 H x 7 h 2 h s n 7 - 9 h 3 D 2, . . . , h o 2 h3 D2 9 h 2 C o

p1 p1

p1 pX

p1 ho 2 B4

p1

p1

p l A, p1 ho A

pl

p1 h52 p1 ho h52 p1

p1

p1

O n the C o h o m o l o g y of the S t e e n r o d A l g e b r a 47

T a b l e 4.43 ( con t inued)

t - s s G e n e r a t o r s Pe r iod ic i ty

(65) 1 0 - 1 2 823 , h 0 B23 = h 2 822 , ho 2 823 = h22 B21 = h 2 d o B 2 11 hi qa = p1 Q 2 13 g w 1 3 - 1 5 R a , h o R 2 ; h o E R 2 = P l h 2 R i ~ p l h l Q 1

66 2 - - 4 h 2 h6, h 0 h 2 h6, ho 2 h 2 h 6 ~ h l 3 h 6 6 - 7 ri, ho q 7 - 8 Go,h o G o s h 2 C' 1 0 - 1 1 B5,ho, B s ~ h 2 2 B 4 ~ g B i = h i B 2 3 1 0 - 1 2 D2' , . . . , ho 2 D E' 12 m n = r t 15 g2 j=d o g 1

67 5 - 6 ni , h 0 rt 1 ~ h 2 2 D 3

5 -- "7 Q3 . . . . . ho 2 Q3 = ]72 A " 8 h i G o ~ hz 2 A 9 C"

9 - 1 2 X 3 , . . . , h o 3 X 3 = s x = h 392 ' 11 C l l 14 l m = e o g r

68 4 - 5 d2, h o d 2

6 ha Q3 7 h22 H a

7 - 8 h 3 A = h 3 A ' ; h o h 3 A 8 - - 1 2 G2i . . . . . ho4 Gzl 10 d a t

11 g B 2 ~ h 4 x ' = p 1 B 3 13 - 1 4 Gas , h o Gl l = d o R 1 15 d o Sa~P i ho 2 B 4 ~ P l h i B2i 16 eo ~ = p i g3

69 3 h22 h 6 4 - 7 p' . . . . ,ho3 p '

6 h 3 H i

8 - 9 h 2 Go,hoh 2 Go~hz 2 C' 9 h i G2i = h 3 E 1 11 h 2 B s 13 W~ 15 g2 k = eo 2 m

18 eo j2 = pa g z

70 2 - 5 h a h 6 . . . . . ho 3 h 3 h 6

4 Pl 5 h52 c o

7 h a h 3 H a 8 d i e a 10 h 2 C"=g C ~ h i 2 G21 12 h 2 Cla 14 m 2 = g 2 r

1 4 - - 1 6 eox'=PX B~z,.. . ,ho2 eox ' 1 7 - - 1 8 Ri ' ,hoRI '=P2 Ri 20 do 5 = p 2 eo 2 g

48 M.C. Tangora:

Note 1. If a = b or if a ~ b and if c is a permanent cycle, then c a = c b or c awe b respectively. If a = b then also P~ a=Pib for all i. However, P~ is not a permanent cycle, so that a ~ b does not necessarily imply U a,,~P ~ b. For example, the relation h 2 g--~ p1 h4 is derived from the differential d4(b032) = h2b122+h4bo22 (4.9). Since d4(bo22)=hogh3, we have Plhzg~PZh4+ ho 4 h 3 b032. Thus it turns out that p1 h2 g and p2 h4 are not related; in fact pi h2 g (and h o times it) will be a survivor for each i in the known range, but p2 h4 and p6 h4 are not permanent cycles, while p4 preserves not only h 4 and h 0 h 4 but the entire " tower" h 4 . . . . . ho 7 h 4.

There are no other anomalies of this type in the known range. In fact the relation hzZg~plhzh4 is perfectly regular (since hzd4(boz2)=O). The question does not arise for the relation h 2 h 4 g~p1 h42, since this element is not periodic at all.

It is easy to see exactly what is permitted. Since U corresponds to bo22~, and the . . . . i ~ only non-zero differential on this element is d4~, we have P a P b whenever the relation a ~ b is based on a differential d r with r < 4 i. All relations given above are based on d 2 (and therefore give no difficulty for any U) except h 2 g,,~p1 h4 ' discussed above, and h 3 g2 "~ h5 g, based on d4(b132) (4.24), which creates no problems since p1 ha g2 does not survive.

Note 2. Many elements containing a factor of h 5 reappear under P1; for example, h I hs, h 2 hs, h 5 Co, h s do, etc. Here we can hardly consider p1 to be a "periodici ty" operator since p2 h5 a is never a survivor for any survivor a; for 4.21 implies that if a is a permanent cycle then p2 h5 a=ds(bo34 a).

pa h3 h5 is an anomalous special case; see Proposition 6.5.

Note 3. Evidently the survivor h 5 bo22i in the 62-stem should not be regarded as Pa(h 5 i) in any meaningful sense; for ho 3 h 5 i = 0 while ho 9 h 5 bo2 2 i is still non-zero. See Chapter 6, Proposition 6.5.

Chapter 5. Differentials by the Imbedding Method In his thesis ([7], Chapter II-6) May has indicated how all differentials

in his spectral sequence may be calculated (at least in theory) using an imbedding procedure. In this chapter we will prove Propositions 4.20, 4.30, and 4.31 by this method, since we have no other proofs of these three differentials. It does not seem feasible to give an exposition of the method here. The method requires familiarity and facility with the bar construction, the shuffle product, behavior of the differential with respect to the shuffle product, and also Milnor 's form of the multiplication in the Steenrod algebra. We will present some very easy examples to give an idea of what is going on. Perhaps these examples will enable the reader to follow the calculations mechanically at least.

May's spectral sequence is a cohomology sequence (note that each differentials raises the homological degree s) and we will now be dealing also with the dual spectral sequence in homology; therefore in this chapter we will denote the May differentials by 6 r instead of by d r, reserving the latter notation


for the differentials in the dual spectral sequence. We will also revert to May's original notation and write Rj, Pj, bj instead of R~j, P~, b/j respectively.

The method works, very roughly speaking, as follows. To determine whether 6r(x)=y in the May spectral sequence, we will identify the dual cycles x* and y*, imbed the dual spectral sequence in the bar construction over A, and determine whether dr(y*)=x*.

Recall from Theorem 1.1 that E 2 = - H * E ~ A is the homology of the complex 9t which is a polynomial algebra over Z 2 with generators {Rj: i>0, j > l } . According to May ([7], II.5), representative co-cycles for the generators of E 2 may be taken from the following table:

h~ R~

b) (R~) 2

h/(1) R~R~ + I + R 3 R a i i+~

ho(1,3) R ~ o 1 a o + R 2 R 4 R3+R~ R1 + R4 R1 ho(1,2) R ~ 2 0 1 2 0 I 2 0 + R3 R4 R2 + R4 R2 R3 + R4 R~ R 2 0 1 2 0 + R s R2R2+RsR31 R12

(For h/(1, *) with i>0 , simply increase the superscripts by i throughout the above rep_resentation of h 0 (1, ,).) May has proved that the dual complex, which he calls X, is an algebra with divided powers on generators Pj, where Rj is dual to 71.(Pj). This algebra imbeds in the bar construction in the most natural way: 7. (Pj) maps to i i {P)IP)I ... [Pj} (n factors)

and the product in X goes over to the shuffle product. The element Pj in A or E ~ A (we abuse the notation by using the same symbol) is the element in the Milnor basis P(q, r2, ...) which has rj = 2 ~ and no other non-zero entries; thus, for example, hi is represented by R~ which is the dual of i_ P] - P ( 2 / ) = S q 21.

May has shown ([7], II.2) that in E~ the elements P} commute according to a very simple rule: if i_< k, then

i k - - i [Pj, Pl] -- P}+l provided that i + j = k

and otherwise the bracket is zero and the elements commute. In the Steenrod algebra, therefore, the same relations hold true modulo elements of greater weight.

As an illustration consider the differential of Proposition 4 .2 :6 2 ho(1)= h 0 h2 2. The element h 0 h2 2 ~E 2 is represented by R~ 2 E91 which dualizes to 7I(P ~ 72 (p2) and thus imbeds as

{pO}. {P~IP~}

where the asterisk (,) denotes the shuffle product. The fact that the terms have length 3 reflects the homological degree s = 3; the sum of the subscripts is the weight, also 3. Since h o h2 2 is the only element in E2 ~ 3, 9 there is no possible ambiguity when we dualize. The element h o (1) is represented by R~ R~ + R~ R~. As the dual cycle we may take either ~I(P ~ 71(P~) or ?~(pO) ya(p~)" The choice is 4 Math. Z.,Bd. 116

50 M.C. Tangora:

irrelevant since these two cycles are homologous in 32 under d(Vl(P ~ v,(PI)" yi(P~)); we have not given the formula for the differential in X since it corresponds to that in the bar construction over E ~ where

d({ P~ * {PI} * {P~})= {pO}, {e~} + {pO}, {PI}.

We note that ho(1) is the only element in E: -z'4'9. Thus to establish that 62 ho(1)= h o h:: it is sufficient to show that in the bar construction over A we

2 2 0 have d({po} �9 {p~lpl})~{p3 } �9 {p~}

modulo terms of weight greater than 4. The calculation may be written out as follows (cf. May [7], p. II-6.6):

{pO}, {p21]p~ } ~ {pO}

= {pO}

= {pO}

{p~p~}+ o 2 ~,{p2} * {[P1, P1]~ * {P(2, 2)} + {P(2, 1)} �9 {p2}

* {PI P~} + {po PI} * {P~} +( '") {pO}, {p~[p~} ~ {pO}, {p~ p~} + {pOlpX } + {p~lpO}

0 1 {pOlp~}, {p2} ~ {pO p~}, {p~} + {P3iPx } + {pO]p12} {p~lpO} + o 1 {P31Pl } = {pO}, {p~}

where we write x ~ y for d(x)=y, and where (...) denotes terms of weight greater than the weight we are interested in (4 in this example). Thus indeed d({P~ is homologous to {pO},{pl} modulo terms of weight greater than 4, which is a proof of Proposition 4.2.

The above calculation makes implicit use of the following lemmas about the behavior of the differential with respect to the shuffle product in any bar construction:

{a} * {bib} --* (a} �9 {b 2} + ([a, b]} �9 {b};

{a} * {bl c} ~ {a} * {be} + {[a, bile} + {bl [a, c]}.

With a little practice anyone can generate such lemmas as are needed in the calculations.

An essential step in the execution of the above calculation is to rewrite PI: p2 as P1 P~2 plus terms of weight greater than 3, and similarly for [pO, pzl] '

2 2 =p(2 , 2), In the first case, the Milnor recipe for multiplication [8] gives PI P1 which has weight 3, since May has shown that P(q ,r 2, ...) has weight Z Zij(iaij) where r i=Zja~j2 j is the binary expansion of r~. We can write P(2, 2) as PI p1 modulo terms of greater weight, since May has shown that

P(rl ' r2 ' . ) - p%,o,(p%,02 (pliatl(Dl]a12.." (p2),~1.. . . . . . \ a l l ~," 2 ]

modulo terms of strictly greater weight. (See Theorem II.2.2 and Lemma II.2.8 of May's thesis.) The bracket o 2 [P1, P1] is handled similarly.

The procedure thus consists of taking the candidate for the coboundary, identifying the dual cycle as it appears in the bar construction, taking the boundary there, and trying to make a clever choice of other elements in the bar


construction whose boundaries produce a homology with the dual cycle of the class whose coboundary is desired. How many such "other elements" may be needed, and how to choose them, is a matter of luck and of skill, and it is often quite surprising how long or how short the calculation may turn out to be.

There is no difficulty in obtaining t~ 2 o n the generators of E 2 by the above technique. This program was carried out by May. For higher differentials, the idea is the same, but the technical problems are greater. This is because fir changes the weight by r - 1 , so that in the bar construction over A we must keep track of terms with weight up to (r - 1) greater than the initial weight, and such terms may be very clumsy to handle.

As an illustration of a higher differential which is not much of a problem, we will prove that 64(b~ 2 contains the term hz~(bl~Z2j, which was the subject of much discussion in the last chapter (Propositions 4.9 and 4.9*). The candidate hz(b2X) 2 appears in the bar construction as {p~}4, {p2}, where we have intro- duced the abbreviation {a}" for the term {a] a]... [a} (n factors). The calculation may be written out as follows:

{p21}4, {p2} __, {p~}2 , {p~ Pi} * (p2} + {pzl} 3 * t,~lPa2, P21~,-,

=0+ {p~}3, {pO pO PI} + {p~}3, {pO pO}

{p~}a,{pO]pO}_.+O+{paz}3, {pOpO}+{p1}2, {[p02, P2J ]P3}i 0 0 0 {pO pO} + {p~}2, { p z l [ P 3 , p~]} = {p1}3,

{ p ~ } 3 , { p o l p o p ~ } _ _ _ , O + { p ~ } 3 o o , , { P i P 3 P ~ } + { p ~ } 2 , { E p o , 1 o V2]lV3 P~}

+ {pzi}2, {pol][p~, pO p]]}

o o I = {pzl}3, {pO pO p~} + {p~}2, {p3[P3 p,} {p t2}2 , {pOlpOlp~}~O+{p~}2 o o i 1 z o o 1 �9 {P3 P3 tP,} + {P2} * {P3[P3 Px} + 0 + 0

+{n2 l} o 0 1 l �9 {P3]P3I[P1, P2]} { p ~ } Z , { p O l p O p ~ } + { p ~ } 0 0 0 0

= * { P 3 [ P 3 [ P 1 P3} { e ~ } , 0 0 0 0 0 0 0 {pO~4 {P3[P3[P31PI}~O+O+{P~2}*{P31P3]P~ P~ +0+0+0+ 3J

which shows that in the dual spectral sequence d({P~} 4 �9 {p2}) is homologous to {pO}4 (modulo zero). This implies the desired result. In the above calculation we have made use of the following lemmas on the interaction of d with �9 :

{a} 4 * {b} --, {a} z * {a 2} * {b} + {a} 3 * {[a, b]}

{a} 3 * {blc} --, {a} �9 {a 2} �9 {Nc} + {a} 3 * {b c}

+ {a} z �9 {[a, b] Ic} + {a} 2 * {bl [a, c]}

{a} z * {blblc} --, {a 2} * {blblc} + {a} z * {b bit} + {a} z �9 {bib c}

+ {a} * {la, b] Iblc} + {a} * {hi la, b] Ic} + {a} * {blbl [a, c]}

{a} * {blblblc} --, {a} �9 d({blblblc})+ {la, b]lblblc} + {bl [a, b][b[c}

+ {blbl[a, b]]c} + {blb]bl[a, c]}. 4*

52 M.C.Tangora:

Henceforward such lemmas will be assumed; a more general formulation of them is given by May ([71, Chapter t.4), but the best way to learn them is to do them. We have also used the following facts about the Steenrod algebra:

1 2 0 2 0 1 (Pz) = 0, (P3) - 0, [P2, P2] = 0, [po, p~] = 0, [pO, pO] = 0, [PI, Pz x] = P(1,0,1) = pO pO p1 3, [ 2,P2~]=P(3,0,1)+P(O,I, 1)=P~176 0 0 P2P3, and [P~ p O p , i = [pO p ~ ] p ~ _ p o r p l 1 0 0 0 , - 3 L-l, P2] = 0 + P3 P1 P3 = 0. These facts are all readily obtained from the Milnor multiplication formula.

Our next calculation proves Proposition 4.31 by showing that the dual to (b~) 4 hi(1 ) is a cycle in the bar construction. This is the only candidate for ~4 (~b') and it will follow that 64(~b')=0. Although we are working in relatively high dimensions, since (@4 hi(1)eE~ lo, 2o, 66, this calculation turns out to require only two lines:

{p~}9. {p2 z} __+ {pzi}S. {[p~, p~]} = {pz~iS. {pO 2 pO}

{p2a}s, {pOipO} __+ {p21}s, {pO pO}

pi using only the facts [ z, P~] = P(O, 1, O, 1)= pO pO, rp,~2_ 0 pO t 2, - , [ 2,Pz 1]=0,and [P4 ~ P2 l] = 0. Thus 4.31, for which I know no other proof, is quite readily available by the imbedding method.

The well-known capriciousness of the Steenrod algebra is reflected in the surprising brevity of some calculations, such as the above, and in the staggering length of others, such as those to follow. The only rule of thumb known to me for predicting the length of these proofs is that they tend to be long when the superscripts are larger than the subscripts. This is actually much more than a mere rule of thumb; it reflects the fact that [Pj, P~] =0 in the Steenrod algebra (not just in E ~ A) if i < 1 and k <j.

The remaining calculations would be too long to write in detail and will only be sketched. We must prove Propositions 4.20 and 4.30 in order to complete the proof of the main theorem 4.42. These involve (~4(h2 b ~ hi(l)) and ~4(h2 b ~ b ~ hi(l)) respectively. Oddly enough, the latter calculation is shorter by an order of magnitude. According to the rule of thumb given above, this may be accounted for by the fact that the candidate for the image in 4.30 is (b~) 2 bz 2 hx(1), while that in 4.20 is h 2"b2"~2 and the latter involves a bracket 1 2 [P1, P2] which is already 1 k 21 ~

non-zero in E ~ A. We now outline the proof of Proposition 4.30. We must show that in the bar

construction d({P~} 5 �9 {p2}3) is homologous to {p2}, {pO}2, {po}2, {p2 a} , {p~} (5.1)

modulo terms of weight greater than 19. In fact d({p~}5, {p2}3)= {pza }4. {pO pO}, {pZ2}2 (5.2)

which is already of weight 18. By adding to (5.2) the boundaries of three terms, {p~}4. {pO[pO}. {p2}2

{p21}4. {pO}2. {n2}. {n~}

{p~}2. {po}2. {p2}2. {p~}. {p~}


we can replace (5.2) by the sum of three terms of weight 19. We have reached the desired weight, so that the remaining calculations are essentially in the bar construction of E ~ A, which streamlines the work considerably. Adding in the boundaries of five more terms,

{p21}2, {pO}2. {pl]p~ } , {p~}, {p~}

o o {p~}, {pO}2, {pl[P3 } , {p2}2, {p~}

(p1}2, {pO}2, {pO]pO}, (p~}2

{po}, {pO}3, {p~}, {p~}3

{pO}2, {pO}:, {Pl} * {p2}2, {p22}

everything adds out except a term (5.1) which represents h: b ~ b ~ h1(1 ). This result implies 4.30.

The above proof requires three boundaries in the first stage, to rewrite the boundary into weight 19, and then five boundaries in the second and final stage, to rewrite it in the desired form so as to be recognizable as the dual cycle to the desired class. The proof takes two-thirds of a page in the author's handwriting. By way of comparison, the proof of 4.20 which is indicated below fills seven pages, requiring 16 boundaries to push the weight to the proper level and 37 more to obtain the result in the desired form. Each boundary may contain several terms, particularly in the early stages; in fact the boundary of one of the terms of weight 10 has eleven terms. Therefore we will not give the entire proof. We will help the reader find his way to the right weight, and then leave him to fend for himself.

The candidate for 54(h 2 b~ h12(b2) 2. Now this element may be represented by (Rl)2(RZ2)4e~t, but 72(P~) 74(P2z) is not a cycle in X and so the dual cycle must contain other terms. It is a routine matter to verify that the dual cycle may be taken as a class which imbeds in the bar construction as

{p~}2, {p~}4

+ {p~}2, {p~}:, {p~}2

+ {Pi} * {P~} * {P~} * {p~}3.

When we calculate the boundary of the sum of these three terms, we obtain terms of weights 11, 12, and 13, as well as terms of greater weight than 13 which do not concern us. No terms of weight 10 remain since a cycle in X imbeds as a cycle in the bar construction over E ~ A.

By adding in the boundaries of four terms of weight 11,

{P~176 } �9 {P~}~ + {p1}2, {p2}2, {p12[g2}

+ {pOlpO } , {p~}, {p~}3

Ip2~ 3 + {P2 ~} * {P~[p21} *t 2, ,

54 M.C.Tangora:

we eliminate all terms of weight 11. By adding in the boundaries of nine terms of weight 12,

{p~}, {p~}, {p~}2, {pO[pO} + {p~}, {p3}, {p~}2, {pO[pO}

+ {p1}2, {p~}, {p3}, {p~ P~IP~} + {P}} * {P~} * {p~}2, {p~ p~[p~} 1 1 + {p~}2, {p22} , {p3}, {p~[p~} + {p~}, {pl)}, {p~}2, {pE]P3 }

+ {p}}, {p~}, {p~}2, {pO}2 + {p~}, {p3}, {p~}2, {p31}2

Ip1~.2 +{P~}*{P~}* {P~}* {P~}*( 3, ,

we eliminate all terms of weight less than 13. In the author's calculation there now remain 38 terms of weight 13 which may be transformed, by the addition of the boundaries of 37 elements of weight 13, into the single term

{p~}, {pO}:, {p~}, {p~}

which represents h E b ~ h,(1). Thus 4.20 is proved. There are innumerable choices to be made in such a calculation and I do

not claim that the proofs indicated are the shortest possible. However, considerable time and practice have wrought a great improvement in the formulation of these calculations; and, while I do not know whether they can be made shorter, I know for sure that they can be made much, much longer.

The methods of this chapter may be used to give a very short proof of a very interesting general result.

Theorem 5.1. The elements do", eo', and g" are all survivors, for arbitrary n.

Proof. Consider for example do"= ho(1) 2". The dual cycle is represented in the bar construction by {po}2,. {P21}2,. This element is obviously a cycle in the bar construction over the Steenrod algebra, since (pO)Z, ~tPl~22j, and [po, P2rl are all zero in A (not just in E ~ A). But it follows from the duality that do" can never be a coboundary in the May spectral sequence. Since do is a permanent cycle, do n is also a permanent cycle, and thus it is a survivor. The proofs for e 0 and g" are similar.

In fact this argument applies to any element in E 2 of the form

p* do* eo* g*,

where the asterisks denote arbitrary positive integers, provided only that the element is a permanent cycle. This latter proviso should be considered because P" stands for bo22" which is not a permanent cycle. However, pk gk= do2k is a permanent cycle. Also, the Adams periodicity theory implies that P" d o and P" e o are survivors. Combining these results with the relation eoZ=do g we see that the proviso is fulfilled in all cases.

It is also possible to prove 5.1 for g" by finding a representative for g in the cobar construction. Such a proof is much longer than the above, and not readily generalized, but it is more direct, since it makes no use of duality.


Chapter 6. Remarks and Examples In this chapter we collect some examples and sample calculations of special

interest, to give some flavor of the calculations in higher dimensions, and to illustrate the limitations of the methods discussed in Chapter 3.

We begin by showing how to obtain some relations in E ~ Ext which are useful in working with the Adams spectral sequence [5].

Proposition 6.1. The following relations hold in Er for all r > 4 :

(a) pi+t hi h3 = p i hi2 do ; (b) P1 h4=h2 g;

(c) h z d 1 = h 4 g; (d) h0 2 Y=fo g; (e) ho 3 x'= P 2 x.

Proof of (a). By 4.1 and 4.4 we have

d2(b02 boa)=boE(h3 boz + hi bl2)+bo3(hl 3 +ho 2 h 2)

= ho 2 h E bo3 + ha bo22 + hi do.

Multiplying by h 1 bo2 zi gives (a). We remark that we also have the relation pj+2 h3 = p j + l hl do+P;ho z i when j is even; this is obtained by multiplying the above differential by bo2 z(j+x), since i=h 2 bo22 bo3. When j is odd, Pih 3 is not a survivor.

Proof of (b). This is immediate from 4.9. For some discussion of this relation see Note 1 to Theorem 4.42.

Proof of (c). From 4.4 and 4.8,

d2 (b12 b13) = hi 2 h 3 b13 q- h 4 b122 q- h 2 d 1

(cf. the calculation in (a) above). Also from 4.8,

d2(hl h3 bo4)=hl 2 h a b13 and the result follows.

Parts (a)-(c) were known to May.

Proof of (d). By 4.6 and 4.9,

d4(h 2 b033) --- h 2 b03 (h2 g q- h 4 bo2 2) q- b032 (ho 2 h32)

=fo g+h2 h4b022 b03 +ho 2 Y.

In E 4 the second term is zero since, by 4.8,

dz (h2 boz 2 b04:)= hE h4 b022 b03.

This gives the desired relation. Note that f o g = h2 m is a true relation in E 2 .

Proof of (e). By 4.4 and 4.8,

d2(b03 b:t3 + b04 bl2)=x+h2 boa b22 q-h23 b04. (6.1)

56 M.C.Tangora:

The radical on the left appears frequently and will be referred to briefly as (B). Now h2bo22b22=ho 3 b13ho(1 ) in E 2 by Relations2.1 and 3.1 of Theorem 1.2. Thus

d 2 (bo24 (B))-- bo24 x + ho 3 bo22 bo3 b13 h o (1) + h23 bo24 bo4.

We also have h23 b022 =ho 3 b12 ho(1 ) in E 2 (Relations 3.1 and 3.2) so that the right-hand side is ho 3 bo22(B) ho(1 ) or ho 3 X ~ and we are finished.

In formula (6.1) above we have written x for h3(b02 b13 q-h12 b04 ) and this will always be taken as the canonical form for x, although (6.1) shows that x could also be defined as h 2 (bo3 b22-}-h22 bo4 ).

The survivor z in the 41-stem was overlooked by May in his first pioneering exploration of what were then considered to be very high dimensions. Since it is of typical complexity for this range we will describe it in detail.

Proposition 6.2. The element z=ho 2 bo3 2 eo~E2(41 , 10) is a surviving cycle. So also is h o Z= fo i~h l 2 u,,~ P 1 h 1 q.

Proof. Clearly d 2 z = 0. Now 4.9 gives

d 4 z=ho 2 eo(h 2 g+h4 bo2 2) but this turns out to be zero in E4 since

da(b12 do eo)= hE 3 d o eo-k ho 2 h 2 eog and

d2 (ho h2 b023 bt3)= ho 3 h22 b022 b13 + h 0 h22 b023 b22 q-h 0 h2 h4 bo23 b12

= 0 + h o 2 h 4 bo22 e o

so that d 4 z is a boundary in E 2. (If the reader has difficulty following the above calculation, he is probably not making full use of the relations in E2, in particular Relations 2.1, 3.1, 3.2, and 3.3.)Thus d 4 z=0 . Now E6(40 , 11)=0, and thus z is a permanent cycle. The fact that E6(40 , 11)=0 is not claimed as obvious; E2(40, 11) contains 25 generators, but nothing survives to E6, as can be verified either by brute force or by the techniques of Chapter 3. In order to see that z survives we must know that it does not bound. Now E 2 (42, 9) has 22 generators. The filtration degree of z is - 12 so only terms of filtration degree - 1 4 or less could hit z with a differential. There are no such terms, as may be ascertained by inspection of the 22 generators. Alter- natively, one can easily calculate the boundaries of all the elements of E, (42, 9) and verify that none of these boundaries hit z. This latter method of proof is the one which is streamlined by the technique of Chapter 3. In fact, in the quotient complex obtained by discarding all h o- and hi-ladders, E2(42, 9) contains only 4 generators, so that by this method it is very easy to check that z survives.

To check that h o z also survives, one must examine Er(42, 10). Here the technique of Chapter 3 shows its full power, for E2(42 , 10) has 38 generators, of which two (given below) have filtration degree -1 4 ; however, all but 5 belong to stable h o- or hi-ladders, which means essentially that we do not


have to check their differentials because we have already checked them (at the terms where the ladders originate). The two terms of degree - 1 4 give the relations claimed in Proposition 6.2, as follows:

dz(b022 boa 2 bl2)=b022 boa2(hl 2 h 3 q-ha 3)

= p i h lq+ho z;

dE(hi 2 bo2 boa 3) = hi 2 b032 (ho E h 2 boa-+-h a b022~ h i do)

= 0 + P lh i q + h l 2 u

since u = h 1 bo32 d o. The relation with fo i is true in E 2 by the relations of Theorem 1.2.

We remark that pi z = i j which also is true already in E2, as can be readily seen from the definitions (and Relations 3.1, 3.2).

Another element of H*(A) which is interesting from the point of view of May's spectral sequence occurs in the 54-stem.

Proposition 6.3. The element G = h l 2 b132EE2(54, 6) is a survivor; moreover G,-~ h42 bo32 ,-~ h i h4(B ) = c o h o (1, 2).

Proof. Clearly d E G = 0. By 4.24,

d 4 G=hl2(h3 b222-I- h5 b12 2) but this is zero since

hi 2 h a b222= d 2 (b22 dl), hi 2 h 5 b122= d 2 (hi h5 bo3 b12),

where note that d 1 denotes the survivor in the 32-stem (and not a differential !). Nothing in E2(53, 7) with filtration degree 0 or - 2 survives to E 6, so G is a permanent cycle. There are two generators in E 2 (55, 5) with filtration degree -10, which give the stated relations:

d2(hi bo4 bi3) = hi 2 bla 2 + hi h4(B);

d2 (h4 boa bo4)= h 1 h4(B)+ h42 b032 .

The relation with c o h o (1, 2) is immediate from Relation 11.1 of Theorem 1.2. This chameleonic element also bears two more relations after multiplication

by hi: h i G survives,

h i G = h l 3 b132 =ho 2 h 2 b132-FdE(bo2 bi32)

and finally, using 4.24(b),

h i G~ho 2 h 2 b132 =ho 2 h 5 boa2q-d4(ho 2 b042).

Thus we have six different representations for h i G.

Use of the techniques of Chapter 3 gives much insight into the periodicity of Ext, since one observes that the only elements in the quotient complex near the Adams edge are multiples of bo22 (with certain types of exceptions, such as ho n hi). We will not attempt any proofs here. However, to clarify the nature of the Adams periodicity, we will make an observation about the non-periodic nature of certain low-dimensional elements. (See also Note 2 to Theorem 4.42.)

58 M.C. Tangora:

Proposition 6.4. The following elements are always zero in E 2 for any i> O: bo22i el, b022i n, b0221 h 2 d a. The following are always zero in E 3 : b02 zi h32, bozZ~p.

Proof The first statement follows immediately from the definitions of the elements and the fact that h 2 bo2 h I (1) = 0 by Relations 1 and 5.1 (or 3.1 and 4.2) of Theorem 1.2. The second statement follows from the differentials d2(h 3 b02 bo3)=h32 b022 and dz(h 0 bo2(B))=h 0 h 3 b022 b13 :b022 p.

Remark 6.4. The following are always zero in Ext, for any i>0 : U c~, pi p, pi ho p, pi h2 n, pi h2 dl"

This is independent of the proposition; there are simply no elements in Ext of the proper bi-grading. Otherwise we might argue from the proposition, using the connection between U and bo2i; but such phenomena as that discussed in 6.5 below raise some misgivings about the latter procedure.

We next discuss an anomalous case of multiplication by bo22.

Proposition 6.5. The element h 5 i~E2(54 , 8) is a survivor; h o h 5 i and ho 2 h 5 i also survive, while h03 h 5 i is zero. On the other hand hoi h 5 bo22 i~E2(62 , 12+j) survives for all j such that 0 = j < 9 .

Proof Obviously h 5 i is a permanent cycle; to see that it survives requires an exhaustive check of E 2 (55, 7), which it is not feasible to present here. Now

d2(h5 b023 bo3)=ho 2 h5 i+h l h5 bo22 do+h 3 h 5 bo24 (6.2)

and moreover ds (h 3 bo34) = h a h 5 bo24 �9 (6.3)

Thus ho 2 h 5 i survives and is related to p1 hi h5 do , while ho 3 h 5 i does not survive.

Multiplying by bo22, we lose (6.3) since d4(h 3 b022 b034) is non-zero; however,

d8 (hi bo34 do)= h 1 h 5 bo24 d o (6.Y)

and this is essentially the same thing since

hi b034 do =(h3 bo22 +ho 2 h 2 b03 ) b034

in E s (by virtue of d E bo2 bo35). The difference is that now when we look at ho 3 h 5 bo22i the higher boundary (6.3') is no longer with us, and we have only one relation, based on bo22(6.2), between the two elements ho 3 h 5 bo22 i and ho 3 h a h 5 bo24. Thus we have a survivor h j h 5 bo22 i=ho j h a h 5 bo24 for 3 < j < 9 . Finally we have ho 1~ h 5 i=dlz(h 3 b028 b032) by Proposition 4.27. We remark that this is the first d12 to appear in May's spectral sequence.

It is natural to inquire about h 5 bo24 i. Here nothing survives because each of the three terms in bo24(6.2) is a boundary in E 8.

It is striking that the seven elements of the S 1 column are sitting over h 5 i in the 54-stem and that there are just enough of them to match the height of the column in the 62-stem. However, the filtration shift between ho a h 5 i~


E2 -8'19'65 and SIdEr -14"25'65 is in the wrong direction for ho 3 h 5 i = S 1 to be a true relation in Ext disguised in E ~ Ext by a filtration shift. Evidently there is some explanation for Proposition 6.5 which involves $1 but the May spectral sequence is not the right framework in which to find it.

Another bi-grading of peculiar interest is E2(61, 6), where two different monomial generators are survivors, but a relation between their ho-multiples has led us to choose a different basis in the statement of results (4.42).

Proposition6.6. The elements A = h 2 boa ho(1, 2) and A*=h o h 3 bo4 b13~ E2(61 , 6) are survivors, as is their sum A'. In (61, 7) all the ho-multiples survive, while in (61, 8) we have the relations ho 2 A N ho E A * ~ h 1 B3, and there is only this one survivor.

Proof It is easy to verify that A, A* and their sum are permanent cycles; in fact, there is nothing in the quotient complex that they could possibly hit. Everything survives, because nothing in E 2 (62, 5) has deeper filtration degree than these elements. In (62, 6), however, we have an element boa b04 b22 of deep filtration, and

d2(bo3 bo4 bz2)--h 0 A+ho A * + h I bz2(B )

which implies, at the next step up, that hoZA'--0. We also have, coming from (62, 7),

d 2 (b02 b03 h 0 (1, 2))= ho 2 A + h I B 3.

This proves the relations given in the proposition.

Now one normally chooses monomial generators for Eo~, but in this case this would produce an anomalous diagram, so we take A and A' as the generators for E~ (61, 6), making it clear that the groups obtained are Z 8 + Z 4 (interpreting h o as a non-trivial extension).

It is hoped that the success of our methods gives ample support to the methods of Chapter 3 for calculating in the May spectral sequence. We close by indicating some of the pitfalls that await in the path of anyone who uses the quotient complex without taking the right precautions.

Remark 6.7. Many relations in E ~ Ext are misrepresented by the quotient complex.

For example, the survivor s = h 4 b022 h o (1) + ho 3 bo2 b13 ~Er(30, 7) is represented in the quotient complex by the first term only. The survivor x = h 3 bozb13+hlZh3bo4~E,(37,5) is likewise represented by its first term only. Therefore the relation h3s=ho3x , which is obviously true in Ez, is disguised in the quotient complex where one sees only that h 3 (h 4 bo22 h o (1))= 0 =t= ho 3 x.

Remark 6.8. Many higher differentials are misrepresented by the quotient complex.

Consider for example h4bo24 ho(1)eE2(38, 11). In the quotient complex this is a cycle under d 2 because its boundary is in the ladder ofd2(h o b023 b13) =

60 M.C. Tangora:

h o h 4 b023 b12 , since

d 2 (h 4 bo2 4 h o (1))= h o h2 2 h 4 bo2*= ho 3 h 4 bo2 a b12.

Actually of course this element represents bo2 2 s. NOW if one treated the quotient complex naively one would perhaps suppose

d 4 (h 4 bo2 4 h o (1))-- d 4 (bo2 2 s) -- ho 4 h a s = ho* h 3 (h 4 bo2 2 h o (1))-- 0

where we have made the mistake indicated in Remark 6.7. Actually it is essential to restore all terms before computing higher differentials, and the correct calculation is

d4(bo2 2 s ) = h o 4 h a s=ho 4 h 3 ( h 4 b022 ho(1)q - ho 3 b02 b l 3 ) = h o 7 h 3 b02 b13 = h o 7 x

so that the differential in question is actually non-zero.

Remark 6.9. The quotient-complex method is not valid for higher differentials in May's spectral sequence.

The difficulty is that the ladder lemma 3.1 does not hold in E r for any r >2 . The ladder of a higher differential d,(x)=y can be mutilated by the appearance of a lower differential ds(z)=hfy (where s<r). For example, the ladder of d6(bo22r)=ho6x (Proposition4.18) is broken by the differential d4(bo2 2 s)= ho 7 x (see Remark 6.8 just above).

The examples in Remarks 6 .7-6 .9 are chosen for their simplicity and perspicuity. The situations arising in practice in higher dimensions are so complicated that it would hardly be worthwhile to give examples of such long and difficult arguments, which are seen after considerable effort to be incorrect.

Appendix 1 Dictionary of Indecomposable Elements of E~

Name t - s s Description

c i 8, 19, 41 3 d i 14, 32, 68 4 e i 17, 38 4 f~ 18,40 4 g~ 20, 44 4 g2' 60 11 h i 2 i - 1 1 i 23 7 j 26 7 k 29 7 l 32 7 m 35 7 n i 31, 67 5 Pl 33, 70 4 p' 69 4 q 32 6 qi 64 10

hi+l hi(1 ) ( i=0,1 , 2) hi(1)2=bl.2 hi+l. 2 +h i+ l 2 bl. s ( i=0, 1, 2) bi+l. 2 hi(1 ) ( i=0, 1) hi+2 2 bi. 3 ( i=0, 1) bl,2 2 ( i=1, 2); g=gl ho 3 bo2 2 b13 2 h i ( i=0,1, 2, 3, 4, 5, 6) ho bo2 boa ho(1) ho bo2 boa b12 ho boa b12 ho(1)=h2 boa do ho bo3 hi2 2 = h 2 boa eo h2 boa b12 2 = h 2 boa g hi+: bi, a hi+l(1 ) ( i=0,1) ; n=n o h~hi+abi+l, a ( i=0,1) ; P=Po ho h4 b23 h 1 h a bo32 h I h4 boa 4


Name t - s s Description

30,66 6 s 30 7 t 36 6 u 39 9 v 42 9 w 45 9 x 37 5 x' 53 10 y 38 6 z 41 10

A 61 6 A' 61 6 A" 64 6

hi+22 bi,32 (i=0,1); r = ~ h 4 b022 ho(1)+ho 3 b02 bla b122 hl(1)q-hl 2 b03 b22 h 1 boa 2 ho(1)2~ho 2 h z boaa+h3 b02 z boa 2 h 1 b032 b12 ho(1)~ho h2 z bo33 h i boa 2 b122~h2 a boa a ha(b02 bla+hl 2 bo4)~h2(boa b22q-h22 b04) b022 ho(1)(boa bla+b04 b12) h32 b032 ho 2 boa 2 b12 ho(1)=ho 2 boa 2 eo=h 2 boaJ

h 2 boa ho(1 , 2) A+ho h a bo4 bla h o bla ho(1, 3)

(B) - not a B 1 46 B 2 48 B21 59 B22 62 B23 65 B a 60 B 4 60 B 5 66 C 50 C' 63 C" 67 C O 62 Cll 67 D 1 52 D 2 58 D 2' 66 D a 61 E 1 62 G 54 G O 66 Gll 68 G21 68 H 1 62 N 46 Q 47 Q' 47 Q1 56 Q2 57 Q3 67 R 62 R 1 54 R 2 65 R l' 70 S 1 54 W 1 69 X 1 61 X 2 63 X a 67

survivor - denotes the radical (boa bla + bo4 b12 ) 7 hi ho(1)(B)~h4 b032 ho(1)+ho h22 boa b04 7 h o b12 (B)

10 ho(1)a(B)=do ho(1)(B ) 10 b12 ho(1) 2 (B)=e o ho(1)(B ) 10 bl22 ho(1)(B)=g ho(1)(B ) 7 h 4 ho(1)(B)~d o ho(1, 2) 9 h o boa bx2(B )

10 h22 boa bl2(B)=fo bl2(B ) 6 hz 2 bo4 hi(1 ) 7 h 2 boa hi(l) 2 9 h 2 boa , b122 h1(1) 8 h22 boa bo# hl(1)=bo3 C

11 h 2 boa 4 hi ( l )= boa 4 c 1 5 h22 ho(1, 2) 6 h o boa ho(1, 2)

10 ho 2 h32 boa 2 bo,,+bo22 D 2 4 h 4 ho(1, 2) 8 bo32 b22 hl(1)=bo32 e 1 6 hi 2 bla2 ~ h42 boa2~hl h4(B)=c o ho(1, 2) 7 Bl22 ho(1, 2 )=g ho(1, 2)

13 h o bo34 b122 = h 0 boa 4 g 8 boa 2 b222=b032 g2 5 h1(1 ) ho(1, 2) 8 h22 boa 2 hl(1)=h 2 boa n=r hi(1 )

13 h a bo24 bo32 13 Q+pl u 10 h 1 h 3 bosa=b032 q 7 h a boa 2 bl3 5 h 4 bla 2

10 h32 boa4=b032 y 10 h22 boa4=bo32 r 13 h 0 boa 4 b12 ho(1 ) 17 ho b023 b034 bl2 = boa4 pa ho do 11 h4 b022 boa 2 ho(1)+ho a b02 boa 2 bla =bo32 s 13 h23 b03 s = h 2 boa Rl~b032 w 9 h a boa2(b02 bla+hl 2 bo4)=b032 x 7 h 3 bla(b02 b la+hl 2 bo4)=bi3 x 9 h a b022 b132

62 M.C.Tangora: On the Cohomology of the Steenrod Algebra

Vertical and h0rizenta[ lines indicate multipticati0n by h 0 and hi, respectively.

Appendix 2 Display chart of E~o for t - s<70

7 6 5 4 3 2

s= l ( t - s )= 0 l 2 3 4 5 6 7 8 9 I0 II 12 13

Fig. 1

15 14 13 12 II

I0 9

15

13 P3h 2 12

h2cl nlc 0 & do e0 o cl 3

2 I

8 7 6 5 4 3 2

s=l ( t-s)= 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

Fig. 2

23 22 21 20 19 18 17 16 15 14 13 12 II I0 9 8 7 6 5 4 3 2

s = l

S r

i n dl ~ h2d I

~ / h s ~ ~" h2h ~

23

p5 2 21 20

P~h 2 17 ~ p3 16

15 14

i ~ Pieog 12

/ I T /-? u 2 v w 9 o 8

, 'te 1, ; t, 3 h2h5 { ~ d2 h~, 2 ~h~ I

( t - s ) -30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Fig. 3

31 30 29 28 27 26 25 24 23 22 21 20

P + h i ~ h 2

p 7 h l ~ ~ h 2

P S d ~ t PSe'o~j

~ k I o pig[ I~ "gip,o,r P g P'gk Pg" ,, gz

�9 eo t t~ ~ 4 g ~ f "

B~ B~t~k/fX?l

C t ~ 2 G / -~ I12 I' B3 ~'?

h5cl hsg D I h~c I s = L h3c2 D 3

( t - s ) - 46 47 48 /,9 50 51 52 53 54 55 56 57 58 59 60 61

3& 33 PSh 2 32 P o

3O

27 pS k

25 ~! 22 P~r / P3z , P dar ~ ~ {

~ 3 ~ .~ 21 , u .3 2 P v 4 3 ~,P"w .2 2 20 '~ r rg ~ re o 2 ? reag ? 19 ,Pgj {_, P'gk 2 P gl ~2.....----'~p2~sl 18 , ~,Feor , P gr , .f ra P.'R I 17 '12 } "~'fP'dov I 2 P'dow I 3 rgZ pleo w i{ I

+,~ ",,~,~'~~ ~,,~' ~ ' ~'j , t / + , ~ ~, .,I 14 , f P'(~dogr t P B2~r ds ,,, g2 r ~r,22 13

/ !

'0 F p/G z ' / f h ' 2 B 4 ~ ql B23 B, D 2 t ~11| P]~ f

8 , ~ le]

6 ,l~n ~ {&" h2H, ~', ? { {h~ pIh~ H / I ' / L / / ,2

:",~ 1 ~ I. ;,/o

62 63 64 65 66 67 68 69 70

35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 B 18 17 16 15 L4 13 12 LI I0

9

8

7

6

5 4 3 2 I

3I 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 t3 12 II I0 9 8 7 6 5 4

63

Fig. 4

Fig. 5

64 M.C. Tangora: On the Cohomology of the Steenrod Algebra

References

1. Adams, J.F.: On the structure and applications of the Steenrod algebra. Commentarii Math. Helvet. 32, 180-214 (1958).

2. - Onthenon-existenceofelementsofHopfinvariantone. Ann. ofMath. 72, 20-104 (1960). 3. - Stable homotopy theory. Lecture notes. Berlin-G6ttingen-Heidelberg: Springer 1964. 4. MacLane, S.: Homology. Berlin-G/~ttingen-Heidelberg: Springer 1963. 5. Mahowald, M. E., Tangora, M. C.: Some differentials in the Adams spectral sequence. Topology

6, 349-369 (1967). 6. - - An infinite subalgebra of ExtA(Z 2, Z2). Trans. Amer. Math. Soc. 132, 263-274 (1968). 7. May, J. P.: The cohomology of restricted Lie algebras and of Hopf algebras; application to the

Steenrod algebra. Dissertation, Princeton, 1964. 8. Milnor, J., Moore, J.C.: On the structure of Hopf algebras. Ann. of Math. 81, 211 - 264 (1965).

Dr. Martin C. Tangora Department of Mathematics University of Chicago Chicago, Illinois 60637, USA

(Received February 24, 1969)

on the cohomology of the steenrod algebraon the cohomology of the steenrod algebra 19 theory in this...

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