DSpace at VNU: The cohomology of the Steenrod algebra and representations of the general linear groups

Let Trk be the algebraic transfer that maps from the coinvariants of certain GLk-representations to the cohomology of the Steenrod algebra.. The second main theorem is that the elements

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TRANSACTIONS OF THE

AMERICAN MATHEMATICAL SOCIETY

Volume 357, Number 10, Pages 4065-4089

S 0002-9947(05)03889-4

Article electronically published on May 20, 2005

NGUYEN H V HUNG Dedicated to Professor Nguye^n Htru Anh on the occasion of his sixtieth birthday

ABSTRACT Let Trk be the algebraic transfer that maps from the coinvariants

of certain GLk-representations to the cohomology of the Steenrod algebra

This transfer was defined by W Singer as an algebraic version of the geo-

metrical transfer trk?: 7rS((BVk)+) - irS(S?) It has been shown that the

algebraic transfer is highly nontrivial, more precisely, that Trk is an isomor-

phism for k = 1, 2, 3 and that Tr = Qk Trk is a homomorphism of algebras

In this paper, we first recognize the phenomenon that if we start from any

degree d and apply Sq? repeatedly at most (k- 2) times, then we get into

the region in which all the iterated squaring operations are isomorphisms on

the coinvariants of the GLk-representations As a consequence, every finite

Sq?-family in the coinvariants has at most (k- 2) nonzero elements Two

applications are exploited

The first main theorem is that Trk is not an isomorphism for k > 5

Furthermore, for every k > 5, there are infinitely many degrees in which Trk

is not an isomorphism We also show that if Tre detects a nonzero element in

certain degrees of Ker(Sq?), then it is not a monomorphism and further, for

each k > ?, Trk is not a monomorphism in infinitely many degrees

The second main theorem is that the elements of any Sq?-family in the

cohomology of the Steenrod algebra, except at most its first (k - 2) elements,

are either all detected or all not detected by Trk, for every k Applications of

this study to the cases k = 4 and 5 show that Tr4 does not detect the three

families g, D3 and p', and that Tr5 does not detect the family {hn+lgnl n >

1}

1 INTRODUCTION AND STATEMENT OF RESULTS

There have been several efforts, implicit or explicit, to analyze the Steenrod alge- bra by using modular representations of the general linear groups (See Muii [22, 23, 24], Madsen-Milgram [19], Adams-Gunawardena-Miller [3], Priddy-Wilkerson [27], Peterson [25], Wood [32], Singer [28], Priddy [26], Kuhn [15] and others.) In partic- ular, one of the most direct attempts in studying the cohomology of the Steenrod algebra by means of modular representations of the general linear groups was the surprising work [28] by W Singer, which introduced a homomorphism, the so-called

Received by the editors November 13, 2003

2000 Mathematics Subject Classification Primary 55P47, 55Q45, 55S10, 55T15

Key words and phrases Adams spectral sequences, Steenrod algebra, modular representations, invariant theory

This work was supported in part by the National Research Program, Grant No 140 804

)2005 by Nguyen H V Hrung, Nguyen H V Khue and Nguyen My Trang

4065

each other, there are inherited actions of GLk on F2 0H*(BVk) and PH,(BVk)

A

In [28], W Singer defined the algebraic transfer

Trk : F2 0 PHd(BVk) - Extk k+d ( F2)

GLk

as an algebraic version of the geometrical transfer trk 7rs((BVk)+) -* 7rS(S) to

the stable homotopy groups of spheres

It has been proved that Trk is an isomorphism for k = 1,2 by Singer [28] and for k = 3 by Boardman [4] Among other things, these data together with the fact that Tr = (k Trk is an algebra honmomorphism (see [28]) show that Trk is highly nontrivial Therefore, the algebraic transfer is expected to be a useful tool in the study of the mysterious cohomology of the Steenrod algebra, ExtI* (F2, F2)

Directly calculating the value of Trk on any nonzero element is difficult (see [28], [4], [11]) In this paper, our main idea is to exploit the relationship between the algebraic transfer and the squaring operation Sq? It is well known (see [18]) that there are squaring operations Sqi (i > 0) acting on the cohomology of the Steenrod algebra that share most of the properties with Sqi on the cohomology of spaces However, Sq? is not the identity On the other hand, there is an analogous squaring operation Sq?, the Kamneko one, acting on the domain of the algebraic transfer and commuting with the classical Sq? on Extk(F2, F2) through the algebraic transfer

We refer to Section 2 for its precise meaning

The key point is that the behaviors of the two squaring operations do not agree

in infinitely many certain degrees, called k-spikes A k-spike degree is a number

that can be written as (2n1 - 1) + + (2nk - 1), but cannot be written as a sum

of less than k terms of the form (2n - 1) (See a discussion of this notion after Definition 3.1.) The following result is originally due to Kameko [13]: If m is a k-spike, then

Sq : PH,(BVk) r,,- 2 -> PH, (BVk)m

0

is an isomorphism of GLk-modules, where Sq is certain GLk-hornomorphism such

that Sq? = 1 Sq (See Section 2 for an explanation of Sq )

GLk

We recognize two phenomena on thle universality and the stability of k-spikes:

First, if we start from any degree d that can be written as (2n - 1) + + (2nk - 1),

and apply the function 6k with 6k(d) = 2d + k repeatedly at most (k - 1) times, then we get a k-spike; second, k-spikes are mapped by 6k to k-spikes Therefore,

STEENROD ALGEBRA AND GENERAL LINEAR GROUPS

From the result of Carlisle and WBood [8] on the boundedness conjecture, one can see that, for any degree d, there exists t such that

-_

(Sq )i : PH,(BVk)2td+(2,-l)k - PH*(BVk)2ld+(2-1)k

is an isomorphism of GLk-modules for every i > t However, this result does not

t k- 2 commonly serves for every degree d It will be pointed out in Remark 6.5 that t = k- 2 is the minimum number for this purpose

An inductive property of k-spikes, which will also play a key role in the paper,

is that if m is a k-spike, then (2n - 1 + m) is a (k + 1)-spike for n big enough Two applications of the study will be exploited in this paper The first applica- tion is the following theorem, which is one of the paper's main results

Theorem 1.2 Trk is not an isomorphism for k > 5 Furthermore, for every

k > 5, there are infinitely many degrees in which Trk is not an isomorphism

That Tr5 is not an isomorphism in degree 9 is due to Singer [28]

In order to prove this theorem, using the notion of k-spike, we introduce the concept of a critical element in Extk (F2, F2) in such a way that if d is the stem of

a critical element, then Trk is not an isomorphism either in degree d or in degree 2d + k Further, we show that if x is critical, then so is hnX for n big enough Our inductive procedure starts with the initial critical element Ph2 for k = 5

Combining Theorem 1.2 and the results by Singer [28], Boardman [4] and Bruner- Ha-Htrng [7], we get

Corollary 1.3 (i) Trk is an isomorphism for k = 1, 2 and 3

(ii) Trk is not an isomorphism for k > 4

(iii) For k = 4 and for each k > 5, there are infinitely many degrees in which Trk is not an isomorphism

Remarkably, we do not know whether the algebraic transfer fails to be a mono- morphism or fails to be an epimorphism for k > 5 Therefore, Singer's conjecture

is still open

Conjecture 1.4 ([28]) Trk is a monomorphism for every k

The following theorem is related to this conjecture

Theorem 1.5 If Tre detects a critical element, then it is not a monomorphism and further, for each k > ?, there are infinitely many degrees in which Trk is not a monomorphism

A family {a| i > 0} of elements in Extk (F2,IF2) (or in IF2 PH,(BVk))

GLk

is called a Sq?-family if ai = (Sq?)'(ao) for every i > 0 Recall that, if a C

Extt (F2, F2), then t- k is called the stem of a, and denoted by Stem(a) The root degree of a is the maximum nonnegative integer r such that Stem(a) can be written

in the form Stem(a) - 2rd + (2' - 1)k, for some nonnegative integer d

The second application of our study is the following theorem, which is also one

of the paper's main results

Theorem 1.6 Let {aiI i > 0} be an Sq?-family in Extk(F2,F2) and let r be the root degree of ao If Trk detects an for some n > max{k-r-2, 0}, then it detects ai for every i > n and detects aj modulo Ker(Sq?)n-j for max{k - r - 2, 0}< j < n

4067

NGUYEN H V HUNG

An Sq0-family is called finite if it has only finitely many nonzero elements The existence of finite Sq?-families in Ext (IF2,IF2) is well known, and that of finite Sq?-families in F2 0 PH (BVk) will be shown in Section 9

GLk

The following is a consequence of Theorem 1.1 and Theorem 1.6

(i) If Tr4 detects bn for some n > 1, then it detects bi for every i > 1

(ii) If Tr4 detects bn for some n > 0, then it detects bi for every i > 0

Based on this event, we prove the following theorem by showing that Tr4 does

detect D3(0)

Conjecture 1.10 Tr4 is a monomorphism that detects all elements in Ext4(F2,F2) except the ones in the three Sq?-families {gil i > 1}, {D3(i)l i > 0} and {p'i i > 0}

The following theorem would complete our knowledge in Corollary 1.3 on whether Tr5 is not an isomorphism in infinitely many degrees

Theorem 1.11 If h,n+lg is nonzero, then it is not detected by Tr5

It has been claimed by Lin [16] that hn+lgn is nonzero for every n > 1

The paper is divided into nine sections and organized as follows Section 2 is a recollection of the Kameko squaring operation In Section 3, we explain the notion

of k-spike and then study the Kameko squaring and its iterated operations in k- spike degrees Section 4 deals with an inductive way of producing k-spikes, which plays a key role in the proofs of Theorems 1.1, 1.2, 1.5 and 1.6 In Section 5, based

on the concept of critical element, we prove Theorems 1.2 and 1.5 Section 6 is devoted to the proofs of Theorems 1.1 and 1.6 Sections 7 and 8 are applications

to the study of the fourth and the fifth algebraic transfers Final remarks and conjectures are given in Section 9

4068

STEENROD ALGEBRA AND GENERAL LINEAR GROUPS

2 PRELIMINARIES ON THE SQUARING OPERATION

To make the paper self-contained, this section is a recollection of the Kameko squaring operation Sq? on F2 0 PH,(BVk) The most important property of the

GLk

Kameko Sq? is that it commutes with the classical Sq? on Ext, (F2, F2) (defined in [18]) through the algebraic transfer (see [4], [21])

This squaring operation is constructed as follows

As is well known, H*(BVk) is the polynomial algebra, Pk := F2[x, ,xk], on k generators x, xk, each of degree 1 By dualizing,

H*(BVk) = (a, .,ak)

is the divided power algebra generated by a1, , ak, each of degree 1, where ai

is dual to xi E H1(BVk) Here the duality is taken with respect to the basis of H*(BVk) consisting of all monomials in xl, ,Xk

In [13] Kameko defined a homomorphism

See e.g [7] for a proof Further, there are two well-known relations,

2t+1Sq =0, q2tq = Sq Sqt

0 See [10] for an explicit proof Therefore, Sq maps PH,(BVk) to itself

The Kameko Sq? is defined by

0 0 Sq? = 1 0 Sq : F2 ( PH*(BVk) -* F2 PH,(BVk)

The dual homomorphism Sq : Pk -? Pk of Sq is obviously given by

Jl - k-1 Sq(x'2 *

)= 3 X1 "Xk 2 , li,.-.,Jk odd,

Hence

Ker(Sq.) = Even, where Even denotes the vector subspace of Pk spanned by all monomials x~1 .x.k

with at least one exponent it even

The following lemma is more or less obvious

Lemma 2.2 ([7]) Let k and d be positive integers Suppose that each monomial

x * xi of Pk in degree 2d + k with at least one exponent it even is hit Then

Sq : (F2 0Pk)2d+k (F2 0Pk)d

is an isomorphism of GLk-modules

4069

NGUYEN H V HUNG

Here, as usual, a polynomial is called hit if it is A-decomposable in Pk

A proof of this lemma is sketched as follows -0

Let s: Pk -> Pk be a right inverse of Sq, defined by

5(2ii ^ik ) 2i 1+1 2ik+l S(X * Xk ) = XI Xk

It should be noted that s does not commute with the doubling map on A, that is,

in general,

Sq2s 7 sSqt

However, Im(Sq2ts- sSqt) C Even

Let A+ denote the ideal of A consisting of all positive degree operations Under the hypothesis of the lemma, we have

(A+Pk + Even)2d+k C (A+Pk)2d+k

Therefore, the map

Therefore, Sq, is an isomorphism in degree 2d + k

3 THE ITERATED SQUARING OPERATIONS IN k-SPIKE DEGREES

The following notion, which is originally due to Kraines [14], formulates some special degrees that we will mainly be interested in

Definition 3.1 A natural number m is called a k-spike if

(a) n = (2n - 1) + + (2nk - 1) with n, ., nk > 0, and

(b) m cannot be written as a sum of less than k terms of the form (2' - 1) Note that k-spike is our terminology Other authors write /,(m) = k to say in is

a k-spike (See e.g Wood [33, Definition 4.4].)

One easily checks e.g that 20 is a 4-spike, 27 is a 5-spike and 58 is a 6-spike Let a(m) denote the number of ones in the dyadic expansion of m The following two lemmas are more or less obvious, but useful later

Lemma 3.2 Condition (a) in Definition 3.1 is equivalent to

a(rn + k) < k < rn, In _ k (mod 2)

Proof Suppose In = (2'n - 1) + + (2nk 1) with nl, ., nk > 0 Then

m > k = (21 - 1)+ .+ (2 - 1) (k terms)

In addition, from 7n + k = 2T' + + 2n with 71, , nk > 0, it implies

a(n + k) < k and n- k (mod 2)

The equality a(mr + k) = k occurs if and only if n., , nk are different from each

STEENROD ALGEBRA AND GENERAL LINEAR GROUPS where m, ., mi > 0, as m + k is even

If at least one exponent mj > 1, then we write (m + k) as a sum of (i + 1) terms

of 2-powers as follows:

m+k 2m + -2mJ-1 +2m7-l + + 2m

This procedure can be continued if at least one of the exponents mi, ., mj - 1, mj-

1, , mi is bigger than 1 After each step, the number of terms in the sum increases

by 1 The procedure stops only in the case when the sum becomes m +k = 2+ +2 with the number of terms (m + k)/2 > 2k/2 = k In particular, we reached at some step a sum of exactly k terms

n + k = 2n1 + - + 2n

with n1, , nk > 0, or equivalently

m =(2 1 -1) + + (2nk -1)

The following lemma helps to recognize k-spikes

Lemma 3.3 A natural number m is a k-spike if and only if

(i) a(m + k) < k < m, m - k (mod 2), and

(ii) a(m + i) > i for 1 < i < k

Proof From Lemma 3.2, if m satisfies (i), then m = (2 - 1) +.* + (2k - 1) with

nl, ,nk > 0 Also by Lemma 3.2, if m satisfies (ii), then it cannot be written as

a sum of less than k terms of the form (2n - 1)

So, if m satisfies (i) and (ii), then it is a k-spike

Conversely, suppose m is a k-spike Then (i) holds by Lemma 3.2

It suffices to show (ii) Suppose to the contrary that ca(m + i) < i for some i

with 1 < i < k We then have (m + i) < i < k < m Let us consider the two

By Lemma 4.3 below, we have

a(m + (i- 1)) = a(n + i) - 1 < i- 1

As m i- 1 (mod 2), we apply Lemma 3.2 again to see that m can be written as

a sum of (i- 1) terms of the form (2n - 1) This is also a contradiction

Combining the two cases, we see that if m is a k-spike, then (i) and (ii) hold

The following proposition is originally due to Kameko [13] We give a proof of

it to make the paper self-contained

Proposition 3.4 If m is a k-spike, then

NGUYEN H V HUNG

Proof By using Lemma 2.2, it suffices to show that any monomial R of Pk in degree

m with at least one even exponent is hit Such a monomial R can be written, up

to a permutation of variables, in the form

R =x .1.iQ2,

with 0 < i < k, where Q is a monomial in degree (m - i)/2

If i = 0, then R = Q2 is simply in the image of Sql (It is also in the image of

Sq , as R = Q2 = Sqn Q.) So, it suffices to consider the case 0 < i < k

Let X be the anti-homomorphism in the Steenrod algebra The so-called x-trick, which was known to Brown and Peterson in the mid-sixties, states that

uSqn(v) - X(Sqn)(u)v mod A+M, for u, v in any A-algebra M (See also Wood [32].) In our case, it claims that

R = xl' XiQ2 = xl xiSq 2 Q

is hit if and only if x(Sq 2 )(xl * * xi)Q is We will show x(Sq 2 )(xz * * * xi) = 0

As A is a commutative coalgebra, X is a homomorphism of coalgebras (see [20, Proposition 8.6]) Then we have the Cartan formula

So, in order to prove x(Sq~2 ~)(x * xi) 0 we need only to show that m2

cannot be written in the form

m = (2e 1) + + (2 - 1) 2

with ?1, , /i > 0 This equation is equivalent to

m - (21+1- 1) + (2eg+1- 1)

Since 0 < i < k, this equality contradicts the hypothesis that m is a k-spike The

The following lemma is the base for an iterated application of Proposition 3.4 Lemma 3.5 If m is a k-spike, then so is (2m + k)

Proof (a) From the definition of k-spike,

m= (2l - 1) + .+ (2nk-1),

for nl, , nk > 0 It implies that

2m + k = (2nl+l - 1) + + (2nk+1 - 1)

So, 2m + k satisfies the first condition in the definition of k-spike

(b) Also by this definition, we have

a(m+ k - j) > k- j,

4072

STEENROD ALGEBRA AND GENERAL LINEAR GROUPS

for 1 < j < k Hence

a(2m + k + (k - 2j)) = a(2(m + k-j))

= a(m + k - j) > k-j > k - 2j, (2m + + (k-2j + 1)) = a(2(m + k-j) + 1)

= c(2(m + k - j)) + 1 (by Lemma 4.3)

= a(m k- j)+ 1

> (k-j) + > -2j+1

Note that each i satisfying 1 < i < k can be written either in the form i = k- 2j

(for 1 < j < k21) or in the form i = k- 2j + 1 (for 1 < < k) So, the above two inequalities show that

a(2 + k+ i) > i, for 1 < i < k Thus, 2m + k satisfies the second condition in Definition 3.1

Remark 3.6 The converse of Lemma 3.5 is false For instance, 27 is a 5-spike, whereas 11 = (27 - 5)/2 is not

Proposition 3.7 If m is a k-spike, then

(Sq)i+1 : PH*(BVk)m-k " PH*(BVk)2m+(2i- 1)k

is an isomorphism of GLk-modules for every i > 0

Proof If m is a k-spike, then by the dual of Proposition 3.4, we have an isomorphism

Corollary 3.8 If m is a k-spike, then

(Sq0)i+ : (F2 0 PH*(BVk))m (IF2 ( PH*(BVk))2Tm+(2i-1)k

Lemma 4.1 If m is a k-spike, then (2' -1 + m) is a (k + 1)-spike for every n with 2 > m + k- 1

4073

Thus a(2n - 1 + 2P) 1 + p > 1 = a(a)

Suppose inductively that the lemma is valid for a(a) = t We now consider the case a(a) = t + 1 > 1 That is,

The following lemma is an obvious observation

Lemma 4.3 If e is an even number, then

a(e + 1) ae) + 1

Proof of Lemma 4.1 (a) Since m = (2nl - 1) + + (2nk - 1), we get

(2n - = 1) + (2 - 1) + (2 - 1) + + (2n - 1)

So the first condition in Definition 3.1 holds for (2n' - 1 + m)

(b) If 1< i < k, then 2n > n + k -1> m + i By Lemma 4.2, we have

-(2n - 1 + m + i) > a(m + i) > i

The last inequality conies from the hypothesis that m is a k-spike

Finally, we need to show Ca(2L- 1 + nm + k) > k Recall that, as m is a k-spike, then r - k (mod 2) Hence, e = (2n - 1) + m + (k- 1) is even By Lemma 4.3,

The last inequality comes from the fact that mn is a k-spike

In summary, the second condition in Definition 3.1 holds for (2n - 1 + m) Combining parts (a) and (b), we see that (2 - 1 + 7n) is a (k + 1)-spike The

4074

STEENROD ALGEBRA AND GENERAL LINEAR GROUPS Remark 4.4 Lemma 4.1 cannot be improved in the meaning that the hypothesis 2n+1 > m + k- 1 does not imply (2' - 1 + m) to be a (k + 1)-spike This is the

case of k = 5, m = 27 and 2n = 16, because 15 + 27 = 42 is not a 6-spike

The following corollary is a key point in the proof of Lemma 6.3 and therefore

in the proofs of Theorems 1.1 and 1.6

Corollary 4.5 2k - k is a k-spike for every k > 0

Proof We prove this by induction on k The corollary holds trivially for k = 1 Suppose indlctively that 2k - k is a k-spike Then, as 2k > (2k - k) + k- 1, applying Lemma 4.2 to the case n = k and m = 2k - k, we have

2k1 _ (k + 1) (2k ) + (2k k)

5 THE ALGEBRAIC TRANSFER IS NOT AN ISOMORPHISM FOR k > 4

We first briefly recall the definition of the algebraic transfer Let P1 be the submodule of F2[xl,x 1] spanned by all powers x1 with i > -1 The usual A- action on P1 = F2[xl] is canonically extended to an A-action on F2[xI, x1 1] (see Adams [2], Wilkerson [31]) P1 is an A-submodule of F2[xl,xl1] The inclusion P1 C P1 gives rise to a short exact sequence of A-modules:

O - P1 - P1 - E- F2 -> 0

Let e1 be the corresponding element in Ext(E-1F2, P) Singer set ek el

* e1 E Ext(E-kF2, Pk) (k times) Then, he defined Tr : Tor A(F2, -kF2)

TorA (F2, Pk) F2 ?Pk by Tr(z) =eknz Its image is a submodule of (F2 ?Pk)GLk

The k-th algebraic transfer is defined to be the dual of Trk

We will need to apply the following theorem by D Davis [9]

Let hn be the nonzero element in Ext 2 (F2F2F)

Theorem 5.1 ([9]) If x is a nonzero element in Ext+k d(F2, 2) with 4 < d < 23, then hnx : 0 for every n > 2j + 1

The following concept plays a key role in this section

Definition 5.2 A nonzero element x C Ext,(F2, F2) is called critical if

(a) Sq?(x) = 0, and

(b) 2Stem(x) + k is a k-spike

Note that, by Lemma 3.5, if Stem(x) is a k-spike, then so is 2Stem(x) + k Lemma 5.3 If x C Ext(F2,F2F) is critical, then so is hnx for every n with 2n > max{4d2, d + k}, where d - Stem(x)

Proof First, we show that if x is critical, then Stem(x) > 0 Indeed, suppose to

the contrary that Stem(x) - 0; then x = hk As x is critical, Sq?(x) = Sq?(ho)

hk = 0 This implies that k > 4, as h1, h , h all are nonzero, whereas h4 = 0 However, 2Stem(x) + k k is not a k-spike for k > 4, because it can be written as

a sum k = 3 + 1 + + 1 of (k- 2) terms of the form (2' - 1) This contradicts the definition of a critical element

4075

Let j be the smallest positive integer such that 2J > d Then, the smallest positive integer i with 2i > d2 should be either 2j or 2j- 1 From the hypothesis 2n > 4d2, it implies that 2n-2 > d2 Hence, we get n - 2 > i > 2j - 1, or equivalently, n > 2j + 1

Remark 5.4 (a) Suppose hnx ? 0 although 2n < 4(Stem(x))2 If x is critical

and 2n > Stem(x) + k, then hnx is also critical

(b) There is no critical element for k < 4, as Sq? is a monomorphism in positive stems of ExtA(F2, F2) for k < 4

Proposition 5.5 (i) For k = 5, there is at least one number, which is the

stem of a critical element

(ii) For each k > 5, there are infinitely many numbers, which are stems of critical elements

Proof For k = 5, Ph2 E Ext5l16(F2,IF2) is critical Indeed, it is well known (see

e.g Tangora [30]) that Ext532 (F2, F2) , so we get

Sq?(Ph2) = 0

Further, by Lemma 3.3, 2Stem(Ph2) + 5 = 27 is a 5-spike

We can start the inductive argument of Lemma 5.3 with the initial critical ele-

The following theorem is also numbered as Theorem 1.2 in the Introduction Theorem 5.6 Trk is not an isomorphism for k > 5 Furthermore, for every

k > 5, there are infinitely many degrees in which Trk is not an isomorphism

4076