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

The Cohomology of the Steenrod Algebra and Representations of the General Linear Groups
Author(s): Nguyễn H. V. Hung
Source: Transactions of the American Mathematical Society, Vol. 357, No. 10 (Oct., 2005), pp.
4065-4089
Published by: American Mathematical Society
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OF THE
TRANSACTIONS
AMERICAN
MATHEMATICAL
SOCIETY
Volume 357, Number 10, Pages 4065-4089
S 0002-9947(05)03889-4
Article electronically published on May 20, 2005

THE COHOMOLOGY

REPRESENTATIONS

ALGEBRA AND
OF THE STEENROD
OF THE GENERAL LINEAR GROUPS
NGUYEN H. V. HUNG

Dedicated to Professor Nguye^nHtruAnh on the occasion of his sixtiethbirthday
ABSTRACT. Let Trk be the algebraic transferthat maps fromthe coinvariants
of certain GLk-representationsto the cohomology of the Steenrod algebra.
This transferwas defined by W. Singer as an algebraic version of the geometrical transfertrk?: 7rS((BVk)+) - irS(S?). It has been shown that the
algebraic transferis highlynontrivial,more precisely,that Trk is an isomorphism fork = 1, 2, 3 and that Tr = Qk Trk is a homomorphismof algebras.
In this paper, we firstrecognize the phenomenonthat if we start fromany
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 isomorphismson
the coinvariantsof 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 firstmain theorem is that Trk is not an isomorphism for k > 5.
Furthermore,forevery k > 5, there are infinitelymany 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 monomorphismand further,for
each k > ?, Trk is not a monomorphismin infinitelymany degrees.
The second main theorem is that the elements of any Sq?-family in the
cohomologyof the Steenrod algebra, except at most its first(k - 2) elements,
are eitherall detected or all not detected by Trk, forevery k. Applications of
this study to the cases k = 4 and 5 show that Tr4 does not detect the three
familiesg, D3 and p', and that Tr5 does not detect the family{hn+lgnl n >
1}.

1. INTRODUCTION

AND STATEMENT OF RESULTS

Therehave been severalefforts,
implicitor explicit,to analyzethe Steenrodalgeofthe generallineargroups. (See Muii[22,23,
bra byusingmodularrepresentations
Adams-Gunawardena-Miller
[27],
[3],Priddy-Wilkerson
24], Madsen-Milgram[19],
Peterson[25],Wood [32],Singer[28],Priddy[26],Kuhn [15]and others.) In particular, one of the most directattemptsin studyingthe cohomologyof the Steenrod
of the generallineargroupswas the
algebra by means of modularrepresentations
the so-called
a homomorphism,
work
W.
which
introduced
surprising
Singer,
[28]by
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,
invarianttheory.
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

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NGUYEN

4066

H. V. HUNG

of the
algebraictransfer,
mappingfromthe coinvariantsof certainrepresentations
linear
to
the
of
the
Steenrod
algebra.
general
group
cohomology
Let Vk denote a k-dimensionalF2-vectorspace, and let PH,(BVk) denotethe
primitivesubspace consistingof all elementsin H,(BVk) that are annihilatedby
everypositive-degreeoperationin the mod 2 Steenrodalgebra, A. Throughout
in F2. The generallineargroup
the paper, the homologyis taken withcoefficients
on

on the homologyand cohomolacts
and
therefore
:=
GLk
regularly Vk
GL(Vk)
of
the
two
actions
of
and
Since
A
GLk upon H*(BVk) commutewith
ogy BVk.
each other,thereare inheritedactionsof GLk on F2 0H*(BVk) and PH,(BVk).
A
In [28],W. Singerdefinedthe algebraictransfer
Trk : F2 0 PHd(BVk)
GLk

- Extk k+d(

as an algebraic version of the geometrical transfertrk

F2)

7rs((BVk)+)

-* 7rS(S)

to

the stable homotopygroupsof spheres.
It has been provedthat Trk is an isomorphismfork = 1,2 by Singer[28] and
fork = 3 by Boardman [4]. Amongotherthings,thesedata togetherwiththe fact
that Tr = (k Trk is an algebra honmomorphism
(see [28]) showthat Trk is highly
nontrivial.Therefore,the algebraictransferis expectedto be a usefultool in the
studyof the mysteriouscohomologyof the Steenrodalgebra,ExtI* (F2, F2).
Directlycalculatingthe value ofTrk on any nonzeroelementis difficult
(see [28],
[4], [11]). In this paper, our main idea is to exploitthe relationshipbetweenthe
algebraictransferand the squaringoperationSq?. It is well known(see [18]) that
thereare squaringoperationsSqi (i > 0) actingon the cohomologyofthe Steenrod
algebra that share most of the propertieswith Sqi on the cohomologyof spaces.
However,Sq? is not the identity.On the otherhand,thereis an analogoussquaring
and
one, actingon the domainofthe algebraictransfer
operationSq?, the Kamneko
transfer.
commutingwiththe classical Sq? on Extk(F2, F2) throughthe algebraic
We referto Section2 forits precisemeaning.
The keypointis that the behaviorsof the twosquaringoperationsdo not agree
in infinitely
manycertaindegrees,called k-spikes. A k-spikedegreeis a number
that can be written as (2n1 - 1) + ... + (2nk - 1), but cannot be written as a sum

of less than k termsof the form(2n - 1). (See a discussionof this notionafter
Definition3.1.) The followingresultis originallydue to Kameko [13]: If m is a
k-spike,then
Sq

: PH,(BVk) r,,2

->

PH, (BVk)m

-- 0

such
is an isomorphism
of GLk-modules,whereSq is certainGLk-hornomorphism
that Sq? = 1

GLk

Sq . (See Section 2 for an explanation of Sq .)

We recognizetwo phenomenaon thleuniversalityand the stabilityof k-spikes:

First, if we start fromany degree d that can be writtenas (2n - 1) +...+

(2nk- 1),

and apply the function6k with 6k(d) = 2d + k repeatedlyat most (k - 1) times,
then we get a k-spike;second, k-spikesare mapped by 6k to k-spikes.Therefore,

we have
Theorem

1.1. Let d be an arbitrarynonnegative integer. Then
(Sq

)i-k+2

: PH*(BVk)2k-2d+(2^-2-1)k

-

PH*(BVk)2id+(21-)k

is an isomorphism of GLk-modules for every i > k- 2.

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STEENROD

ALGEBRA

AND GENERAL

LINEAR

GROUPS

4067

Fromthe resultofCarlisleand WBood[8]on the boundednessconjecture,one can
see that,forany degreed, thereexistst such that
---_

(Sq )i

: PH,(BVk)2td+(2,-l)k

- PH*(BVk)2ld+(2-1)k

is an isomorphismof GLk-modulesforeveryi > t. However,this resultdoes not
confirmhow large t should be. Theorem 1.1 shows that a rathersmall number
t k- 2 commonlyservesforeverydegreed. It will be pointedout in Remark6.5
that t = k- 2 is the minimumnumberforthis purpose.
An inductivepropertyof k-spikes,whichwill also play a key role in the paper,
is that ifm is a k-spike,then (2n - 1 + m) is a (k + 1)-spikeforn big enough.
Two applicationsof the studywill be exploitedin this paper. The firstapplication is the followingtheorem,whichis 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 infinitelymany degrees in which Trk is not an isomorphism.

That Tr5 is not an isomorphismin degree9 is due to Singer[28].
In orderto prove this theorem,using the notion of k-spike,we introducethe
conceptof a criticalelementin Extk (F2, F2) in such a way that ifd is the stem of
a criticalelement,then Trk is not an isomorphismeitherin degreed or in degree
2d + k. Further,we showthat ifx is critical,thenso is hnXforn big enough. Our
inductiveprocedurestartswiththe initialcriticalelementPh2 fork = 5.
CombiningTheorem1.2 and theresultsby Singer[28],Boardman[4]and BrunerHa-Htrng[7],we get
Corollary 1.3.

(i) Trk is an isomorphism for k = 1, 2 and 3.
is
not
an
isomorphism for k > 4.
(ii) Trk
k
For
4
and
=
for each k > 5, there are infinitelymany degrees in which
(iii)
is
not
an
Trk
isomorphism.

Remarkably,we do not knowwhetherthe algebraictransferfailsto be a monomorphismor failsto be an epimorphismfork > 5. Therefore,Singer'sconjecture
is stillopen.
foreveryk.
Conjecture 1.4 ([28]). Trk is a monomorphism
The followingtheoremis relatedto this conjecture.
Theorem 1.5. If Tre detects a critical element, then it is not a monomorphism
and further,for each k > ?, there are infinitelymany degrees in which Trk is not a
monomorphism.

A family{a|

i > 0} of elementsin Extk (F2,IF2)

is called a Sq?-family if ai = (Sq?)'(ao)

(or in IF2

for every i > 0.

GLk

PH,(BVk))

Recall that, if a C

Extt (F2, F2), thent- k is called the stemof a, and denotedby Stem(a). The root
degreeofa is the maximumnonnegativeintegerr such that Stem(a) can be written
in the formStem(a) - 2rd+ (2' - 1)k, forsome nonnegativeintegerd.
The second applicationof our studyis the followingtheorem,whichis 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,
then
it detects ai
0},
>
n
and
r
i

detects
modulo
2, 0}< j < n.
for every
Ker(Sq?)n-j for max{k
aj

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NGUYEN

4068

H. V. HUNG

An Sq0-familyis called finiteifit has onlyfinitely
manynonzeroelements.The
existenceof finiteSq?-familiesin Ext (IF2,IF2)is well known,and that of finite
Sq?-familiesin F2 0 PH (BVk) will be shownin Section9.
GLk
The followingis a consequenceof Theorem1.1 and Theorem1.6.
Corollary 1.7.
(i) EveryfiniteSq?-familyin F2 0 PH*(BVk) has at most
GLk
(k - 2) nonzeroelements.
thenit does not detectany elementof a finite
(ii) If Trk is a monomorphism,
Sq?-familyin Extk(F2,F2) withat least (k - 1) nonzeroelements.

The followingis an applicationof Theorem1.6 intothe investigation
of Tr4.
Proposition 1.8. Let {bil i > 0} and {bi i > 0} be theSq?-familiesin ExtA(F2,IF2)
withbo one of the usual five elementsdo,eo,Po,D3(O),pO, and bo one of the usual
two elementsfo,g.
(i) If Tr4 detectsbn for some n > 1, thenit detectsbi for everyi > 1.
(ii) If Tr4 detectsbnfor some n > 0, thenit detectsbi for everyi > 0.
Based on this event,we provethe followingtheoremby showingthat Tr4 does

not detect gl, D3(1), p'.

Theorem 1.9. Tr4 does not detectany elementin the threeSq?-families{gil i >
1}, {D3(i)l i > 0} and {p' i > 0}.
This theoremgives furthernegativeinformation
on Minami's ([21]) conjecture
that the localization of the algebraic transfergiven by invertingSq? is an isomorphism. The firstnegative answer to this conjecturewas given in BrunerHa-Hirng [7] by showingthat the elementin (Sq?)-1Ext (F2,F2) representedby
the family{gil i > 1} is not detected by (Sqo)-lTr4. From Theorem 1.9, the
two elementsin (Sq)-lExt (IF2, F2) representedrespectivelyby the two families
{D3(i)l i > 0} and {pi[ i > 0} are also not detectedby (Sq?)-1Tr4.
Recently,T. N. Nam informedthe author about his claim that Tr4 does not
detectD3(0).
Conjecture 1.10. Tr4 is a monomorphismthat detects all elements in
Ext4(F2,F2) except the ones in the threeSq?-families{gil i > 1}, {D3(i)l i > 0}
and {p'i i > 0}.
The following
theoremwouldcompleteourknowledgein Corollary1.3 on whether
Tr5 is not an isomorphismin infinitely
manydegrees.
Theorem 1.11. If h,n+lg is nonzero,thenit is not detectedbyTr5.
It has been claimedby Lin [16] that hn+lgn is nonzeroforeveryn > 1.

The paper is dividedinto nine sectionsand organizedas follows.Section 2 is a
recollectionof the Kameko squaringoperation.In Section3, we explainthe notion
of k-spikeand then study the Kameko squaringand its iteratedoperationsin kspike degrees. Section4 deals withan inductiveway of producingk-spikes,which
plays a keyrolein the proofsof Theorems1.1, 1.2, 1.5 and 1.6. In Section5, based
on the concept of criticalelement,we prove Theorems1.2 and 1.5. Section 6 is
devotedto the proofsof Theorems1.1 and 1.6. Sections7 and 8 are applications
to the study of the fourthand the fifthalgebraic transfers.Final remarksand
conjecturesare givenin Section 9.

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2.

AND GENERAL

ALGEBRA

STEENROD

ON THE

PRELIMINARIES

LINEAR

GROUPS

4069

OPERATION

SQUARING

this sectionis a recollectionof the Kameko
To make the paper self-contained,
squaringoperationSq? on F2 0 PH,(BVk). The most importantpropertyof the
GLk

Kameko Sq? is that it commuteswiththe classical Sq? on Ext, (F2, F2) (definedin
[18]) throughthe algebraictransfer(see [4], [21]).
This squaringoperationis constructedas follows.
As is well known,H*(BVk) is the polynomialalgebra,Pk := F2[x, ...,xk], on k
generatorsx, ...xk, each of degree1. By dualizing,
H*(BVk) =

(a,..

.,ak)

is the divided power algebra generatedby a1,..., ak, each of degree 1, where ai
is dual to xi E H1(BVk). Here the dualityis taken with respectto the basis of
H*(BVk) consistingof all monomialsin xl,... ,Xk.
In [13] Kameko defineda homomorphism
Sq:
a

ai

where a(il

ik)
*

-

H*(BVk)
(i)

..

.

H*(BVk),
... ak(2ik+l)

a (2i1+1)

(ik)

ak

k

is dual to x1

x.
i.

The followinglemma is well known.

-0

Lemma 2.1. Sq is a homomorphism
ofGLk-modules.
See e.g. [7] fora proof.Further,thereare two well-known
relations,
--0

See [10] for an explicit proof. Therefore, Sq

maps PH,(BVk)

The Kameko Sq? is definedby
0--0

Sq? = 1 0 Sq
GLk

: F2 (

GLk

-0

The dual homomorphism Sq
.

Sq(x'2

*

)=

PH*(BVk)

: Pk

3

= Sq Sqt.

q2tq

=0,

2t+1Sq

-?

~0

Pk of Sq

Jl

X1

O 0,

-

k-1

"Xk

F2

-*

GLk

to itself.

PH,(BVk).

is obviously given by

2 ,

li,.-.,Jk odd,

otherwise.

Hence
Ker(Sq.) = Even,
whereEven denotesthe vectorsubspace ofPk spannedby all monomialsx~1.x.k
withat least one exponentit even.
The followinglemmais moreor less obvious.

Lemma 2.2 ([7]). Let k and d be positiveintegers.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
A

--

(F2 0Pk)d
A

is an isomorphism of GLk-modules.

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4070

NGUYEN

H. V. HUNG

Here, as usual, a polynomialis called hitifit is A-decomposablein Pk.

A proofof this lemmais sketchedas follows.
Let s: Pk

-> Pk

-0

be a right inverse of Sq, defined by
5(2ii
S(X

*

2i
XI

^ik
Xk )) =

1+1 ...

2ik+l
Xk

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

in general,

However, Im(Sq2ts-

Sq2s 7 sSqt.
sSqt) C Even.

Let A+ denotethe ideal of A consistingof all positivedegreeoperations.Under
the hypothesisof the lemma,we have
+ Even)2d+k

(A+Pk

C (A+Pk)2d+k.

Therefore,the map
S:(F2

A

Pk)d

-

(F2 ?Pk)2d+k

s[X]

=

[sX]

A

is a well-defined
linearmap. Further,it is the inverseof
-0

Sq

: (F2 ?Pk)2d+k
A

-- (F2 ?Pk)d.
A

0

Therefore,Sq, is an isomorphismin degree2d + k.
3. THE ITERATED

SQUARING OPERATIONS

IN k-SPIKE DEGREES

The followingnotion,whichis originallydue to Kraines [14], formulatessome
special degreesthat we will mainlybe interestedin.
Definition 3.1. A naturalnumberm is called a k-spikeif
(a) n = (2n - 1) + ... + (2nk 1) with n, ..., nk > 0, and
m
cannot
be
written
as

a
sum
of less than k termsof the form(2' - 1).
(b)
Note that k-spikeis our terminology.
Otherauthorswrite/,(m)= k to say in is
a k-spike.(See e.g. Wood [33, Definition4.4].)
One easily checkse.g. that 20 is a 4-spike,27 is a 5-spikeand 58 is a 6-spike.
Let a(m) denotethe numberofones in the dyadicexpansionofm. The following
two lemmasare moreor less obvious,but usefullater.
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, from7n+ 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 differentfrom each

other.
Conversely,suppose that a(m + k) < k < Im and m = k (mod 2). Let i

a(m + k). Then we have
m + k = 2"1 +

. + 2"2,

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STEENROD

ALGEBRA

AND GENERAL

LINEAR

GROUPS

4071

where m, ..., mi > 0, as m + k is even.

If at least one exponentmj > 1, thenwe write(m + k) as a sum of (i + 1) terms
of 2-powersas 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 biggerthan 1. Aftereach step,the numberoftermsin the sum increases
by 1. The procedurestopsonlyin the case whenthesum becomesm+k = 2+ --+2
withthe numberofterms(m + k)/2 > 2k/2= k. In particular,we reachedat some
step a sum of exactlyk terms
n + k = 2n1 + - . + 2n
with n1, ..., nk > 0, or equivalently
m =(2

1 -1)

+

+ (2nk -1).

D

The lemmais proved.
The followinglemmahelps to recognizek-spikes.
Lemma 3.3. A naturalnumberm is a k-spikeif and onlyif
(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, ifm satisfies(ii), then it cannot be writtenas
a sum of less than k termsof the form(2n - 1).
So, ifm satisfies(i) and (ii), then it is a k-spike.
suppose m is a k-spike.Then (i) holds by Lemma 3.2.
Conversely,
It sufficesto show (ii). Suppose to the contrarythat ca(m+ i) < i forsome i
with 1 < i < k. We then have
cases.

(m + i) < i < k < m. Let us consider the two

Case 1: m - i (mod 2). Then, by Lemma 3.2, we get m = (2n - 1) +..
(2n - 1) with nl, ..., ni > 0. This contradicts to the definitionof a k-spike.

+

Case 2: m- i - (mod 2). It impliesi > 1. Indeed, if i = 1, combiningthe
hypothesisa(m + 1) < 1 with the fact m + 1 is odd, we get m + 1 = 1. This
contradictsthe hypothesisthat m is a naturalnumber.
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 writtenas
a sum of (i- 1) termsof the form(2n - 1). This is also a contradiction.
Combiningthe two cases, we see that if m is a k-spike,then (i) and (ii) hold.
D
The lemmafollows.
The followingpropositionis originallydue to Kameko [13]. We give a proofof
it to make the paper self-contained.

Proposition 3.4. If m is a k-spike,then
--

Sq, : (F2 ?Pk)m A

(F2

A

Pk) m-2

is an isomorphismofGLk-modules.

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NGUYEN H. V. HUNG

4072

to showthatanymonomialR ofPk in degree
Proof. By usingLemma 2.2, it suffices
m with at least one even exponentis hit. Such a monomialR can be written,up
to a permutationof 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 simplyin the image of Sql. (It is also in the image of
Sq , as R = Q2 = Sqn Q.) So, it sufficesto considerthe case 0 < i < k.
Let X be the anti-homomorphism
in the Steenrodalgebra. The so-calledx-trick,
whichwas knownto Brownand Petersonin the mid-sixties,states that
uSqn(v)

- X(Sqn)(u)v

mod A+M,

foru, v in any A-algebraM. (See also Wood [32].) In our case, it claims that
R = xl' .
is hit if and only if x(Sq

XiQ2 = xl...

xiSq

2

Q

2 )(xl ** xi)Q is. We will show x(Sq

2 )(xz * **xi) = 0.

As A is a commutativecoalgebra,X is a homomorphism
of coalgebras(see [20,

Then
we
have
the Cartan formula
Proposition8.6]).
E

=

x(Sqn)(uv)

i+j =n

x(Sqi)(u)x(Sqj)(v)

it is shownby Brownand Petersonin [5] that
Furthermore,
X(Sqn')(x)
)(
X(q

-

2,

\i

0,

forxj in degree1.

So, in order to prove x(Sq~2 ~)(x

cannot be writtenin the form
m
2

= (2e

if n =2q - l for some q,
otherwise,
*.. xi)

0 we need only to show that m2

1) +... +(2

- 1)

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

m - (21+1-

1) + ...

(2eg+1- 1).

Since 0 < i < k, this equalitycontradictsthe hypothesisthat m is a k-spike.The
D
propositionis completelyproved.

The followinglemmais the base foran iteratedapplicationof Proposition3.4.
Lemma 3.5. If m is a k-spike,thenso is (2m + k).
Proof. (a) Fromthe definitionof k-spike,
m=

(2l

- 1) + ...+

(2nk-1),

for nl, .., nk > 0. It implies that

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

+ (2nk+1- 1).

of k-spike.
So, 2m + k satisfiesthe firstconditionin the definition
Also
we
have
this
definition,
by
(b)
a(m+

k - j) > k- j,

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4073

for1 < j < k. Hence
a(2m + k + (k - 2j))
(2m +

+ (k-2j

+ 1))

=

a(2(m + k-j))
= a(m + k - j) > k-j > k - 2j,
= 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
(for1 < j < k21) or in the formi = k- 2j + 1 (for1 < < k). So, the above two

inequalitiesshowthat

a(2 + k+ i) > i,
for1 < i < k. Thus, 2m + k satisfiesthe second conditionin Definition3.1.
Combiningparts (a) and (b), we see that 2m + k is a k-spike.

D

Remark 3.6. The converseof Lemma 3.5 is false. For instance,27 is a 5-spike,
whereas11 = (27 - 5)/2 is not.
Proposition 3.7. If m is a k-spike,then
: PH*(BVk)m-k

(Sq)i+1

"

PH*(BVk)2m+(2i-

1)k

is an isomorphismofGLk-modulesfor everyi > 0.
Proof. Ifm is a k-spike,thenbythedual ofProposition3.4, we have an isomorphism
of GLk-modules
--

Sq

-+ PH*(BVk)m.

: PH*(BVk)m-

On the otherhand, fromLemma 3.5, ifm is a k-spike,thenso is 2im + (2i - l)k
foreveryi > 0. Hence, repeatedlyapplyingthe dual of Proposition3.4, we get an
isomorphismof GLk-modules
(Sq )i+ : PH*(BVk)m-k

PH*(BVk)2,m+(2i-1)k.

D

The propositionis proved.
Corollary 3.8. If m is a k-spike,then
(Sq0)i+

: (F2 0 PH*(BVk))m

GLk

2

--

(IF2

(

GLk

PH*(BVk))2Tm+(2i-1)k

is an isomorphism
for everyi > 0.
4.

RECOGNITION

OF k-SPIKES

In this section,we introducean inductiveway of producingk-spikes,whichwill
play a key role in the proofsof Theorems 1.1, 1.2, 1.5 and 1.6 in the next two
sections.
Lemma 4.1. If m is a k-spike,then(2' -1 + m) is a (k + 1)-spikefor everyn
with2 > m + k- 1.

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NGUYEN

4074

To provethis lemma,we need the followingtwo lemmas.
Lemma 4.2. If 2n > a, then
(2n _-1 + a) > ca(a).
Proof. The proofproceedsby inductionon a(a). If a(a) = 1, then a is a powerof
2, say a= 2P < 2n. We have
2 -1 +2 = 2n + (2p - 1) = 2n + (2P-l +
+ 2?).
Thus a(2n - 1 + 2P) 1 + p > 1 = a(a).
Suppose inductivelythat the lemmais valid fora(a) = t. We now considerthe
case a(a) = t + 1 > 1. That is,
a = 2't+1

+

2t

+ ** +

2T71

with nt+l > nt > ..

> ni.

Set b 2nt + .. + 2nl < 2nt+,; then a = 2nt+l + b, and c(b) = t. From 2n > a, it
implies 2n > 2nt+l. Therefore, we obtain
'
ac(2 - 1 + a) = a(2 + 2nt+ -1 + b)

= 1 + (2nt+ 1 + b)
> 1 + ca(b) (by the inductivehypothesis)
+ t = a(a).

D

The lemma is proved.
The followinglemmais an obviousobservation.
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

(2 - 1)
- 1) +

+ (2n - 1).

So the firstconditionin Definition3.1 holds for(2n'- 1 + m).
(b) If 1< i < k, then2n > n + k -1> m + i. By Lemma 4.2, we have
-(2n - 1 + m + i) > a(m + i) > i.
The last inequalityconies fromthe hypothesisthat m is a k-spike.
+ k) > k. Recall that,as m is a k-spike,
Finally,we need to show Ca(2L- 1 + nm
then r - k (mod 2). Hence, e = (2n - 1) + m + (k- 1) is even. By Lemma 4.3,
we have
a(27-1 -+ mr+ k) = a(2n - + m + (k - 1) + 1)
=

a(2'- 1 +rr+ (k- 1)) + 1.
4.2 to the case 2" > m + k- 1, we get
Now, applyingLelrmma
0(2

- 1 +m+(k-1))

+ 1 >
>

a(m+(k-1))+
=k.
(k-1)+1

1

The last inequalitycomes fromthe factthat mnis a k-spike.
In summary,the second conditionin Definition3.1 holds for(2n - 1 + m).
Combiningparts (a) and (b), we see that (2 - 1 + 7n) is a (k + 1)-spike. The
D
lemma is proved.

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4075

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 followingcorollaryis a keypoint in the proofof Lemma 6.3 and therefore

in the proofsof Theorems1.1 and 1.6.
Corollary 4.5. 2k - k is a k-spikefor everyk > 0.
Proof.We provethis by inductionon k. The corollaryholds triviallyfork = 1.
Suppose indlctivelythat 2k - k is a k-spike. Then, as 2k > (2k - k) + k- 1,
applyingLemma 4.2 to the case n = k and m = 2k - k, we have
2k1

_ (k + 1)

(2k

) + (2k

k)

D

to be a (k + 1)-spike. The corollary follows.
5. THE ALGEBRAIC TRANSFER IS NOT AN ISOMORPHISM

FOR k > 4

We firstbrieflyrecall the definitionof the algebraic transfer. Let P1 be the
submoduleof F2[xl,x 1] spanned by all powersx1 with i > -1. The usual Aaction on P1 = F2[xl] is canonicallyextendedto an A-action on F2[xI, x1 1] (see
Adams [2], Wilkerson[31]). P1 is an A-submoduleof F2[xl,xl1]. The inclusion
P1 C P1 givesrise to a shortexact sequence of A-modules:

O - P1 - P1 E-

F2 -> 0.

Let e1 be the corresponding element in Ext(E-1F2,
Singer set ek
P).
el
* . e1 E Ext(E-kF2,
Tr
:
he
defined
Tor
A(F2, -kF2)
Pk) (k times). Then,
TorA (F2, Pk) F2 ?Pk by Tr(z) =eknz.
Its image is a submodule of (F2 ?Pk)GLk.
A

The k-th algebraic transferis defined to be the dual of Trk.

A

We will need to apply the followingtheoremby D. Davis [9].
Let hnbe the nonzeroelementin Ext 2 (F2F2F).

Theorem 5.1 ([9]). If x is a nonzeroelementin Ext+k d(F2, 2) with4 < d < 23,
thenhnx : 0 for everyn > 2j + 1.
The followingconceptplays a keyrole 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,thenso is 2Stem(x) + k.
Lemma 5.3. If x C Ext(F2,F2F) is critical,then so is hnx for everyn with
2n > max{4d2,d + k}, whered - 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 impliesthat k > 4, as h1,h ,h all are nonzero,whereas h4 = 0.

However,2Stem(x) + k k is not a k-spikefork > 4, because it can be writtenas
a sum k = 3 + 1 +
+ 1 of (k- 2) termsof the form(2' - 1). This contradicts
of a criticalelement.
the definition

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4076

H. V. HUNG

Now we have Stem(x) > 0. Combiningthe factthat Sq? is a monomorphism
in
<
positivestemsof ExtA(F2, F2) fork 4, and that x is critical,we get k > 4. As x

is a nonzeroelementof positivestemin ExtA (F2, F2) withk > 4, by the vanishing
line theorem(see [1]) we have Stem(x) > 7. Therefore,
x satisfiesthe hypothesisof
Theorem5.1 that d= Stem(x) > 4.
Let j be the smallest positive integersuch that 2J > d. Then, the smallest
positiveintegeri with2i > d2 shouldbe either2j or 2j- 1. Fromthe hypothesis
2n > 4d2, it impliesthat 2n-2 > d2. Hence, we get n - 2 > i > 2j - 1, or
n > 2j + 1.
equivalently,
Therefore,by Theorem5.1, hnx i 0 if27 > 4d2.
As SqO is a homomorphism
of algebras,we have
Sq?(hnx) = Sq?(hn)Sq?(x)

= Sq?(hn)?

= 0.

Since x is critical,m := 2d + k is a k-spike.We need to showthat 2Stem(hnx)+
(k + 1) is a (k + 1)-spike.We have
Stem(hnx) = 2n- 1 + Stem(x) = 2n - 1 + d.
A routinecalculationshows
2Stem(hnx)+ (k + 1)

=

2(2 -1 + d) + (k + 1)
2n+

- 2 + (2d + k) + 1 = 2+

- 1 + m.

By Lemma 4.1, thisnumberis a (k + 1)-spikeforeveryn with2n+1 > m + k - 1 =
2(d + k) - 1, or equivalently2n > d + k.
In summary,hnx is criticalforeveryn with
2n > max{4d2,d + k}.

The lemmais proved.

D

Remark5.4.
(a) Suppose hnx ? 0 although2n < 4(Stem(x))2. If x is critical
and 2n > Stem(x) + k, thenhnxis also critical.
in positive
(b) Thereis no criticalelementfork < 4, as Sq? is a monomorphism
stemsof ExtA(F2, F2) fork < 4.
Proposition 5.5.
(i) For k = 5, thereis at least one number,whichis the
stem of a criticalelement.
many numbers,whichare stems of
(ii) For each k > 5, thereare infinitely
criticalelements.
Proof. For k = 5, Ph2 E Ext5l16(F2,IF2)
e.g. Tangora [30]) that Ext532 (F2, F2)

is critical. Indeed, it is well known (see

, so we get

Sq?(Ph2) = 0.
Further,by Lemma 3.3, 2Stem(Ph2) + 5 = 27 is a 5-spike.
We can startthe inductiveargumentof Lemma 5.3 withthe initialcriticaleleD
mentPh2. The propositionfollows.
The followingtheoremis also numberedas Theorem1.2 in the Introduction.
Theorem 5.6. Trk is not an isomorphismfor k > 5. Furthermore,
for every
k > 5, thereare infinitely
manydegreesin whichTrk is not an isomorphism.

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ALGEBRA

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4077

Proof. In orderto prove the theorem,by means of Proposition5.5 it sufficesto
show that Trk is not an isomorphismeitherin degreed or in degree2d + k, where
d denotesthe stemof a criticalelementx E ExtA(F2,F2).

We considerthe followingtwo cases:
Case 1: x is not in the image of Trk.
Then, Trk is not an epimorphismin degreed.
Case 2: x = Trk(y) for some y E F2 0 PH,(BVk).
GLk

Fromx 5 0, it impliesthat y 570. Accordingto Boardman [4, Thm 6.9 and Cor
6.12] and Minami [21, Thm 4.4], we have a commutativediagram
(F2 0

GLk

Trk

PH(BVk))d

Extk,k+d(2

Sqo

2)

Sq0
v

?

Trk

(F2 ? PH*(BVk))2d+k

GLk

Extk2(kd)

)

wherethe leftverticalarrowis the Kameko Sq? and the rightverticalone is the
classical squaringoperation.
As m = 2d+ k is a k-spike,by Corollary3.8 the Kameko Sq? is an isomorphism.
So, fromy :: 0, we have
z = Sq(y)

$ 0.

of the diagram,we get
Now, by the commutativity
Trk(z) = Trk(Sq?(y))

= Sq?(Trk(y)) ==

0.

in degree 2d + k. The theoremis
This means that Trk is not a monomorphism
D
completelyproved.
Remark5.7. (a) We can showthat F2 0 PH*(BV5)11 = 0. It impliesthat Ph2 is
GL5
not detectedby Tr5.
(b) By Lemma 5.3, hnPh2 is criticalforeveryn > 9, as Stem(Ph2) + 5 <

4(Stem(Ph2))2

= 4 112 = 484 < 29 = 512. Also, by Remark 5.4, hnPh2 is critical

forn = 4, 5,6, as it is nonzero(see [6]) and 24 > Stem(Ph2) + 5 = 16. R. Bruner
privatelyclaimed h7Ph2 ~ 0. It seems likelythat h8Ph2 7 0. If so, by the same
argument,thesetwo elementsare also critical.
The following
corollaryis also numberedas Corollary1.3 in the Introduction.
Corollary 5.8.
(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 infinitelymany degrees in which
Trk is not an isomorphism.

This resultis due to Singer[28] fork = 1,2, to Boardman [4] fork = 3, and to
Bruner-Ha-Hirng[7] fork = 4. That Tr5 is not an isomorphismin degree9 is also
due to Singer[28]. The remainingpart is shownby Theorem5.6.
Our knowledge'sgap on whetherTr5 is not an isomorphismin infinitely
many
degreeswill be studiedin Section8.

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NGUYEN

4078

The followingtheoremis also numberedas Theorem1.5 in the Introduction.
Theorem 5.9. If Trt detectsa criticalelement,thenit is not a monomorphism,
and further,
manydegreesin whichTrk is not a
for each k > e, thereare infinitely
monomorphism.
Proof. The proofproceedsby inductionon k > ?.
For k = ?, suppose Tre detects a criticalelementxe E ExtA(F2,F2). Then,
in degree
by Case 2 in the proofof Theorem 5.6, Tre is not a monomorphism
2Stem(xe) + f.
By means of this argument,it sufficesto show that if Trk detects a critical
elementXk,then Trk+l detectsinfinitely
manycriticalelements,whose stemsare
different
fromeach other.
Fromthe hypothesis,xk = Trk(yk)forsome YkE F2 0 PH*(BVk). WithambiGLk

guityof notation,let hn also denotethe elementin F2?
under Tr1 is the usual hn E Extl(IF2, F2). As Tr -

algebras (see [28]), we have

GL1

PH,(BV1), whoseimage

Trk is a homomorphism of

(k

Trk+l(hnyk) = Trl(hn)Trk(yk) = hnxk.
By Lemma 5.3, the element hnXk is critical forevery n with 2n > max{4d2, d + k}.
By the firstpart of the theorem,since Trk+l detectsthe criticalelementhnXk,
in degree2Stem(hnxk)+ (k + 1) foreveryn with2n >
it is not a monomorphism

in infinitely
max{4d2,d + k}. Thus, Trk+l is not a monomorphism
manydegrees.
D
The theoremfollows.
6. THE STABILITY OF THE ITERATED

SQUARING OPERATIONS

The followingtheorem,whichis also numberedas Theorem1.1 in the Introduction,showsthat SqO is eventuallyisomorphicon F2 0 PH (BVk). More precisely,
GLk

it claims that ifwe startfromany degreed of this module,and apply SqO repeatedly at most (k- 2) times,then we get into the region,in whichall the iterated
squaringoperationsare isomorphisms.
6.1. Let d be an arbitrarynonnegative integer. Then

Theorem

)i

(Sq

+2

-

PH*(BVk)2k-2d+(2k-2-_)k

PH*(BVk)2,d+(22-1)k

is an isomorphism of GLk-modules for every i > k- 2.
In the theorem, for k = 1 we take the convention that 21-2d + (21-2

Let us denote
(Sq?)

1(F2

0
GLk

PH*(BVk))d

-lim..

.

Sq(

i

(F2 0

GLk

PH*(BVk))2d+(2,-1)k

1)k = d.
Sq(

The followingcorollaryis an immediateconsequenceof Theorem6.1.
Corollary 6.2. Let d be an arbitrarynonnegative integer. Then,
(i) The following iterated operation is an isomorphism for every i > k - 2:
(SqO)i-k+2

F2 (

GLk

PH,(BVk)2k-2d+(2k-2-)k

-

F2 0

GLk

PH*(BVk)21d+(2-l)k.

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GROUPS

4079

(ii)
(Sq?)-

(F2 (

GLk

(F2 0
PH,(BVk))2k-2d+(2k-2_1)k.

PH,(BVk))d

GLk

(iii) If d = 2k-2d' + (2k-2 - 1)k for some nonnegative integer d', then

(Sq0)- (F2 0 PH*(BVk))d - (F2 0 PH*(BVk))d.
GLk

GLk

In orderto proveTheorem6.1, we need the followinglemma. Let Sk denotethe
functiongivenby 6k(d) = 2d + k.
Lemma 6.3. If d is a nonnegativeintegerwitha(d + k) < k, then 5k-l(d)
2k-ld + (2k-l - l)k is a k-spike.

Proof. The lemmaholdstriviallyfork = 1. Indeed,fromthe hypothesisa(d + 1) <
1 it implies that d = 2 - 1 for some n. Then 0?(d) - d- 2n - 1 is an 1-spike.
We now consider the case of k > 2. First, we observe that k < 2k- d +
k (mod 2) and
(2k- - 1)k

a(2k- d + (2k-1 - l)k + k) = a(2k-l(d + k)) = a(d + k) < k.
By Lemma3.2, k-l(d) = 2k-ld+(2k-~-ll)k satisfiescondition(a) ofDefinition3.1.
So, in orderto provethe lemma,it sufficesto showthat
-l)k+i)

a(2k-ld+(2k-1

>i

for1 < i < k.

We now work modulo 2k-1. First, we have

2kld

+ (2k-1 - 1)k _ (2k-1 - l)k (mod 2k-1).

Let k = 2n + + + 2n be the dyadic expansion of k with nt > ... > nl. We get
(2k-1 - 1)k = 2k-1(2nt + ..

+ 2n2) + (2k-l+nl

-

*
(2nt +

+ 2n1)).

Thus
(2k-1 - l)k

-

2k-l+nl

-

-

.
+ 2T1) (mod 2k-l)
(2nt +

..
2k 1 - (2nt +
+ 2n1) (mod 2k-1)

-

2k-1-

k(mod2k-l),

where2k-1 - k > 0 because k > 2.
As a consequence,we get
2k-ld + (2k-1 - 1)k + i - 2k 1 - k + i (mod 2k-1)

for1 < i < k. Since k > 2 and d > 0 we have
2k-ld + (2k-1 - l)k + i > (2k-1 - 1)2 + 1 > 2-1

From this inequality it implies that, in the dyadic expansion of 2k-ld +

(2k-1- )k + i, there is at least one nonzero term 2n with n > k- 1. On the other
hand, as 2k-1 - k + i < 2k-1 for 1 < i < k, the dyadic expansion of 2k-1 - k + i is
just a combination of the 2-powers 20, 21,..., 2k-2. Therefore, in order to prove

a(2k-ld+ (2k-1

)k

i) > i

for1 < i < k, we need onlyto showthat

a(2

- k +i)

>i.

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4080

FromCorollary4.5, 2k-1 - (k- 1) is a (k- 1)-spike. Then we have
a(2k-1 - (k - 1) + j) > j
for1 < j < k-1. Set i = j + 1; we get
k +i) >i

a(2k-1-

for2 < i < k. In addition,it is obviousthat
> 1.

k+1)

a(2k-1-

In summary,we have shownthat
a(2k-1

> i

-k+i)

for1 < i < k. The lemmais proved.

O

Remark6.4. (a) Lemma 6.3 cannot be improvedin the meaningthat the number

6~-2(d) = 2k-2d + (2k-2 - l)k is not a k-spike in general.

Indeed, takingd = 2t + 1 - k witht big enoughso that d > 0, we have
a(2k-2d + (2k-2 - l)k + (k - 1)) = a(2t+k-2

+ (2k-2 - 1)) = k- 1.

By Lemma 3.3, 2k-2d + (2k-2 - 1)k is not a k-spike.

(b) However,a numbercould be a k-spikealthoughit is not of the form k-1(d)
foranynonnegativeintegerd. For instance,thisis thecase ofthe following
numbers
with k = 4:
Stem(fl) = 40,

Stem(e2) = 80,

Stem(D3(2)) = 256, Stem(p) = 288,

Stem(p2) = 144,

wheree2,fl, P2,D3(2), p' are the usual elementsin Ext4 (F2,F2). This observation
will be helpfulin the proofof Proposition7.2 below.
Proofof Theorem6.1. Accordingto Wood's theorem[32] (it was originallyPeter-

son's conjecture), the primitive part PH,(BVk)

is concentrated in the degrees d

witha(d + k) < k. This facttogetherwiththe equality

+ k) = a(2i(d + k)) = a(d + k)

a(6'(d)

show that, ifa(d + k) > k, then the domainand the targetof the homomorphism
in the theoremare both zero.
If a(d + k) < k, then the theoremis an immediateconsequenceof Lemma 6.3
D
and Proposition3.7. The theoremis proved.
Remark 6.5. Let k = 5 and d = 0. As 65-2(0) = 35, Theorem 6.1 claims that
(Sq?)i-3

: PH

(BV5)35

-

PH*(BV5)5(2-_1)

is an isomorphismof GL5-modulesfori > 3. In the finalsectionwe will see that
Sq? : F2 0 PH*(BV5)l5
GL5

- F2 0
PH*(BV5)35
GL5

is not a monomorphism.This showsthat Theorem6.1 cannotbe improvedin the
meaningthat (k - 2) is, in general,the minimumtimesthat we must repeatedly
apply SqO to get into "theisomorphism
region"ofthe iteratedsquaringoperations.
A family {ail i > 0} of elements in ExtA(F,F2F2)

is called an Sq?-family if

ai = (Sq?)'(ao) foreveryi > 0. An Sq?-familyin F2 0 PH*(BVk) is similarly
GLk
defined.

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4081

Definition 6.6. Let ao cExtk(F2, F2). The root degreeof ao is the maximum
nonnegativeintegerr such that Stem(ao) can be writtenin the form
Stem(ao) = 5(d) = 2d + (2' - 1)k,
forsome nonnegativeintegerd.
The followingtheoremis also numberedas Theorem1.6 in the Introduction.
Theorem 6.7. Let {ail i > 0} be an Sq?-familyin Ext(F2,F2,) and let r be the
rootdegreeofao. If Trk detectsan forsome n > max{k-r-2, 0}, thenit detectsai
for everyi > n and detectsaj moduloKer(Sq?)n-j formax{k - r - 2, 0} < j < n.
Proof.It is easy to see that
ca(Stem(ai) + k) = a(2(Stem(ao) + k)) - a(Stem(ao) + k).
Suppose a(Stem(ao) + k) > k; thenwe have a(Stem(ai) + k) > k foreveryi > 0.
By Wood's theorem[32] (it was originallyPeterson'sconjecture),PH*(BVk)t = 0
in any degreet withac(t+ k) > k. So, all elementsofthe family{ai i > 0} are not
detectedby Trk.
Now we considerthe case wherea(Stem(ao) + k) < k. We observethat
a(Stem(ao)

+ k) = a(2r(d + k)) = a(d + k) < k.

Set q = max{k - r - 2, 0}, and we have
+l (Stem(ao)) = 6++l(d).

Stem(aq+l) -=

Note that
q+ r

1 = max{k - r - 2,0}+

r+1

> (k-r-2)

+ r1

=k-1.

So, by Lemmas 6.3 and 3.5, Stem(aq+l) is a k-spike.
Accordingto Theorem6.1, ifc = Stem(aq), then
: PH*(BVk)c

(Sq)i-

-)

PH*(BVk)2i-qc+(2i-q-

)k

is an isomorphismof GLk-modulesforeveryi > q.
Suppose Trk detects an with n > q, that is, an = Trk(an) for some an in
As the squaring
F2 0 PH*(BVk). If i > n, then we set ai = (SqO)i-n(a).
GLk

we have
operationscommutewitheach otherthroughthe algebraictransfer,
ai

(an)=

=

(Sq)i-n

=

Trk(SqO)i-

(Sq)i-nTrk(an)

(an) = Trk(ai).

Thus, ai is detectedby Trk foreveryi > n.
Next we considerj withmax{k - r- 2, 0} < j < n. Then we set
aj = [(Sq?)n-j]-l(n).

This makes sense, as it is shownabove that (Sqo)n-j is isomorphicin the degree
of aj. Again, as the squaringoperationscommutewith each other throughthe
we have
algebraictransfer,
(Sq?)n-iTrk((a j)

= Trk(Sq?)n-j(aj)
= Trk(an)
= an = (Sq)n-j (aj).

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4082

NGUYEN

H. V. HUNG

As a consequence,we get
Trk((aj) = aj (mod Ker(Sq?)n-j).

This means that Trk detectsaj moduloKer(Sq)n~-j. The theoremis proved. D
Remark6.8. (a) Underthe hypothesisof Theorem6.7, let
a' = Trk(Sq?)'-'((a)
foreveryi > max{k - r - 2, 0} no matterwhetheri > n or i < n. Then we get
a new Sq?-family{al i > maxk - r - 2,0}}, whose everyelementis detectedby
Trk and
ifi > n,
,_
ai,
if i < n.
ai l ai (mod Ker(Sq?)n-i),
The new Sq?-familyis called the adjustmentof the originalone.
(b) Theorem6.7 is still valid and can be shownby the same proofifwe replace
max{k - r- 2, 0} by any numberq suchthat Stem(aq+i) is a k-spike.This remark
will be usefulin the proofof Proposition7.2 forthe case k = 4.
Corollary 6.9. Let {ail i > 0} be an Sq?-familyin ExtA(F2,IF2) and let r be
the rootdegreeof ao. Suppose the classical Sq? is a monomorphism
in the stems
of the elements{ail i > max{k - r - 2,0}}. If Trk detectsan for some n >
max{k - r - 2, 0}, thenit detectsai for everyi > max{k - r - 2, 0}.

An Sq?-familyis calledfiniteifit has onlyfinitely
manynonzeroelements,infinite
if all of its elementsare nonzero. The followingis also numberedas Corollary1.7
in the Introduction.
Corollary 6.10.

(i) EveryfiniteSq?-familyin F2 0 PH,(BVk) has at most
GLk

(k - 2) nonzeroelements.
thenit does not detectany elementof a finite
Trk is a monomorphism,
If
(ii)
at least (k - 1) nonzeroelements.
in
with
Sq?-family ExtkA(F2,
F2)
Proof. (i) Suppose that {ail i > 0} is an Sq?-familyin F2 0 PH,(BVk) withat
GLk
least (k - 1) nonzeroelements. Then ao, a, ...,ak-2 are its first(k - 1) nonzero

elements. Set d = deg(ao); then deg(ak-2) = 2k-2d + (2k-2 - 1)k. Therefore,by

Corollary6.2,
(Sq)i-+2

: F2

0( PH,(BVk)2k-2d+(2k-2_l)k -- F2

GLk

GLk

PH,(BVk)2td+(2i-l)k

is an isomorphismforeveryi > k- 2. Therefore,fromak-2 7~ 0 it impliesthat
ai = (Sq0)i-k+2(ak-2) is nonzeroforeveryi > k - 2. Thus, the Sq?-familyis
infinite.
(ii) Let ao, al,..., ak-2 be the last (k - 1) nonzeroelementsof the givenfinite
Sq?-familyin Ext(F2, IF2). As ak-2 is the last nonzeroelementin the Sq?-family,

we have Sq?(ak_2)

= 0. Set d = Stem(ao); then by Lemma 6.3, 2Stem(ak_2) + k

2k-ld + (2k-1 - l)k is a k-spike.So, ak-2 is critical.
Suppose to the contrarythat Trk detectssome (nonzero) elementin the Sq?family. Then, as the squaringoperationscommutewith each otherthroughthe
algebraictransfer,Trk also detectsthe criticalelementak-2. Accordingto Theorem5.9, thiscontradictsthehypothesisthatTrk is a monomorphism.
The corollary
is proved.
D

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STEENROD

7.

ON

ALGEBRA

BEHAVIOR

AND GENERAL

OF THE

LINEAR

ALGEBRAIC

FOURTH

GROUPS

4083

TRANSFER

This sectionis an applicationof the previoussectionintothe studyof Tr4. We
referto [30], [6], [17] foran explanationof the generatorsof ExtA (F2, F2).
It has been known(see [17]) that the graded module ExtA(F2, F2) is generated
by hihjhehm,hicj, di,ei, fi,gi+ pi, D3(i), p and subject to the relations
=
hihi+l O,

hih+2 =0,

hi = h2-hi+l,

fori j-1,j,j
+2, j 3.
0, hicj = O
hh+3
The followingis also numberedas Conjecture1.10 in the Introduction.
that detectsall elementsin Ext (F2, F2)
Conjecture 7.1. Tr4 is a monomorphism
in
the
the
ones
three
Sq?-families
except
{gi\i > 1}, {D3(i)l i > 0} and {p[ i > 0}.
That Tr4 does not detectthe family{gij i > 1} is due to Bruner-Ha-Hurng[7].
the authorabout his claim that Tr4 does not detect
Recently,T. N. Nam informed
the elementD3(0).
The followingproposition,which is also numberedas Proposition 1.8 in the
Introduction,is an attemptto preparefora proofof Conjecture7.1.
Proposition 7.2. Let {biI i > 0} and {bi i > 0} betheSq?-familiesin Ext\(F2, F2)
withbo one of the usual fiveelementsdo,eo,po,D3(0),p, and bo one of the usual
two elementsfo,gi.
(i) If Tr4 detectsbnfor some n > 1, thenit detectsbi for everyi > 1.

(ii) If Tr4 detectsbnfor some n > O, thenit detectsbi for everyi > 0.
Proof. Althoughthe stemsof b2 and bl cannot be writtenas 63(d) forsome nonnegativeintegerd (except forb2 = d2 and bl = 92), it is easy to check by using
Lemma 3.3 that theyall are 4-spikes.
Followingpart (b) of Remark 6.8, we can show this propositionby the same
as Sq? is a monomorargumentas givenin the proofofTheorem6.7. Furthermore,
phism in positive stems of ExtA(F2,F2) (see e.g. [17]), the propositionhas the
as in Corollary6.9.
strongformulation
D
The propositionis proved.
to show that:
By means of Proposition7.2, to proveConjecture7.1 it suffices
(1) Tr4 detectsdo,eo,fo,Po;
(2) Tr4 does not detect gl, D3 (1),p;

and

(3) Tr4 is a monomorphism.
The followingtheoremis also numberedas Theorem1.9 in the Introduction.
Theorem 7.3. Tr4 does not detectany elementin the threeSq?-families{gi i >
1}, {D3(i)I i > 0} and {p'l i > 0}.
Outlineof theproof.First, we show that F2 0 PH,(BV4) is zero in degree 20.
GL4

Therefore,Tr4 does not detect gl of stem 20 and, by Proposition7.2, does not
detectany elementin the Sq?-family{gil i > 1}. (Note again that this part of the
theoremis due to Bruner-Ha-Hirng[7].)
Second, as the stemsof D3(1) and pl are respectively126 and 142, we focusto
the GL4-modulePH.(BV4) in degrees126 and 142. By routinecomputations,we
showthat PH, (BV4) has dimension80 and 285 in degrees126 and 142,respectively,

and furtherthat F2 0 PH.(BV4) is of dimension1 in these two degrees.
GL4

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NGUYEN H. V. HUNG

4084

of
Note that,as Tr1 detectsthe family{hnl n > O} (see [28]),the homomorphism
=
>
the
n
Tr
detects
the
O}.
family{hnl
algebras
subalgebrageneratedby
kTrk
sends the two generatorsof its domainin degrees126 and 142 to
So, Tr4 definitely
the nonzeroelementsh2h2 and h2h4h7, respectively.Therefore,the two elements
stems 126,142 are not detectedby Tr4.
D3(1) and p' of,respectively,

D
The theoremis provedby combiningthis fact and Proposition7.2.
8. AN OBSERVATION ON THE FIFTH ALGEBRAIC TRANSFER

From Corollary5.8, the followingconjecturenaturallycomes up.
Conjecture 8.1. There are infinitely
manydegreesin whichTr5 is not an isomorphism.
The factthat gn is not detectedby Tr4 and that Tr = )k Trk is a homomorphism of algebras do not implythat hign is not detectedby Tr5. For instance,
hog1= h2eo and h1gl = h2fo are presumablydetectedby Tr5, as eo and fo are
expectedlydetectedby Tr4.
The purposeof this sectionis to provethe following,
whichis also numberedas
Theorem1.11 in the Introduction.
Theorem 8.2. If hn+lgn is nonzero,thenit is not detectedbyTr5.
of algebras,
Outlineof theproof.We firstobservethat,as Sq? is a homomorphism
>
{hn+lgnl n 1} is an Sq?-family,that is,
hn+lgn,
(Sqo) n-(h2gl)=
>
foreveryn 1.
Next, using Lemma 3.3 we easily show that Stem(h2g1)= 23 is not a 5-spike,
but 65(23) = 2 . 23 + 5 = 51 is. So, by Proposition3.7,
(Sq ) : PH,(BV5)23

-

PH,(BVk)2i.23+(2i-1)5

is an isomorphismof GL5-modulesforeveryi > 0.
In addition,a routinecomputationshowsthat PH,(BV5) is of dimension1245
in degree23, and furtherthat
F2 0 PH*(BV5)23 = 0.
GL5

As a consequence,we get
F2

(

GL5

PH*(BV5)2i.23+(2,-1)5 = 0,

foreveryi > 0. So, the domain of Tr5 is zero in the degreethat equals
Stem(hn+lgn) = 2n-1 23 + (2n-1 - 1)5

foreveryn > 1.
Therefore,if hn+lgn is nonzero,thenit is not detectedby Tr5. The theoremis
?
proved.
Corollary 8.3. If hn+lgn is nonzerofor everyn > 1, then thereare infinitely
many degrees,namelythe degreesof hn+lgn for n > 1, in whichTr5 is not an
epimorphism.
The corollary'shypothesisis claimedto be true by Lin [16]. So, Conjecture8.1
is established.

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STEENROD ALGEBRA AND GENERAL LINEAR GROUPS

4085

of algebras,
Remark 8.4. As h3g2= h5gl (see [30]) and Sq? is a homomorphism
Theorem8.2 also showsthat ifhn+4gnis nonzero,then it is not detectedby Tr5.
Whichelementsin Ext(5 F2,F2) are detectedby Tr5?
This questioncan be partiallyansweredby usingthe factthat Tr = Ok Trk is
on elementsdetectedby Trk for
and the information
an algebra homomorphism
k < 4. For instance,h3D3(0) = hod2(see [6]) is presumablydetectedby Tr5, as ho
is detectedby Tr1 and d2 is expectedlydetectedby Tr4 (see Conjecture7.1).
Based on Theorem6.7 and concretecalculations,the followingconjecturepresentssome "new" families,whichare expectedlydetectedby Tr5.
Conjecture 8.5. Tr5 detectseveryelementin the Sq?-familiesinitiatedby the
classes n, x, hog2,D1, H1, hlD3 (0), h2D3(0), Q3, h4D3(0), h6gl,hog3ofstems31, 37,
44, 52, 62, 62, 64, 67, 76, 83, 92, respectively.
Conjectures8.5 and 7.1, togetherwiththe factthat Tr = Dk Trk is an algebra
homomorphism,
predictthat Tr5 detects all Sq?-familiesinitiatedby the classes
of stems < 125, exceptpossiblythe threefamilies,whichare respectivelyinitiated
by Phl, Ph2 and hop'. Since Sq?(Phl) = h2gl, everyelementof the Sq?-family
initiatedby Phl is not detected by Tr5 (see [28] for Phl and Theorem 8.2 for
hn+lg,). It has been knownthat Tr5 does not detectthe Sq?-familyofexactlyone
nonzeroelement{Ph2} (see Remark5.7). We have no predictionon whetherthe
Sq?-familyinitiatedby hop' of stem 69 is detectedor not.
9. FINAL REMARKS

or
Remark 9.1. We still do not know whetherTrk fails to be a monomorphism
fails to be an epimorphismfork > 5. If Singer's Conjecture 1.4 that Trk is a
foreveryk is true,thenby Theorem1.5 the algebraictransferdoes
monomorphism
not detectthe kernelof Sq? in k-spikedegrees.
This leads us to the studyof the kernelof Sq? in F2 ? PH*(BVk). The map
GLk

--0

Sq : PH*(BVk) -- PH*(BVk)
is obviouslyinjective. Taking this observationtogetherwith Corollary3.8 into
account,one would expect that the Kameko map
Sq? = 1

GLk

0-~0

Sq:

F2 ? PH,(BVk)
GLk

-+ F2

GLk

PH*(BVk)

is also a monomorphism.
However,this is false. Indeed, PH,(BV5) has dimension
whileF2 0 PH (BV5) has dimen432 and 1117 in degrees15 and 35, respectively,
GL5

sion 2 and 1 in degrees15 and 35, respectively.We obtainedthis claim by usinga
computerprogramof S. Shpectorovwrittenin GAP.
As in theproofofTheorem5.9, let hnalso denotetheelementin F2 0 PH, (BV),
GL1

Tr1 is the usual hn C Ext4(F2, F2). In the folwhoseimageunderthe isomorphism
lowingremark,we willuse the productof ODk(F2 0 PH,(BVk)) definedby Singer
in [28] and his result that Tr =

is a homomorphism
of algebras.

Dk Trk

k(F2

GLk

0 PH*(BVk))

GLk

- Ext(F2,

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2)


NGUYEN H. V. HUNG

4086

Remark 9.2.

(a) Let t5 =

Tr5 (t) - h4h4 7 0.
(b) If tk E F2 0 PH,(BVk)
GLk

h4h4

(F2 0 PH,(BV5))15.
GL5

Then Sq?(t5) = 0 and

is a positivedegreeelementwith Sq?(tk) = 0 and

Trk(tk) - 0, then Sq?(hntk) = 0 and Trk+l(hntk)
2n > 4(deg(tk))2.

$/0 for every n with

"Proofof Remark9.2 ". Part (a) of this proofproceedsundertwo hypotheses:
(i) F2 0 PH,(BV5) has respectivelydimension2 and 1 in degrees15 and 35.
GL5

(This is knownby a computercalculationas writtenabove.)
(ii) do E Im(Tr4). (This is a part of Conjecture1.10. When this paper was
beingrevised,Le M. Ha privatelyinformedthe authorthat he provedthis
claim.)
It would be betterto writea directproofin the framework
of invarianttheory
forthe factSq0(h4h4) = 0 in F2 0 PH*(BV5).
GL5

(a) As is well known,Ext 5+15(F2,JF2) = Span{h4h4,h1do}. Combiningthe
fact that Tr = (k Trk is an algebra homomorphism
with the one that hn is in
the image of Tr1 foreveryn, and do is in the image of Tr4, we conclude that
hoh4 and h1do are both in the image of Tr5. On the otherhand, the domainof
Tr5 has dimension2 in degree 15. So, Tr5 is an isomorphismin degree 15. As
F2 0 PH*(BV5) has respectivelydimension2 and 1 in degrees15 and 35, there
GL5

existsa nonzeroelementt5 C F2 ( PH*(BV5) in degree15 such that Sq?(t5) = 0.
GL5

Since Tr5 is an isomorphism
in degree15, Tr5(t5) $ 0.

Next, we show that Tr5(ts) = h4h4. Indeed, we suppose to the contrarythat

Tr5(t5) = Ah4h4 + hldo, for some A

IF2. Then, as Sq?(hO)

-

we have
Ext, (F2, F2) and Sq? is an algebra homomorphism,
Sq?Tr(ts)

= ASq?(h4h4) + Sq?(hdo)

h4

0 in

= Sq?(hldo) = h2d.

Since Tr5 commuteswiththe squaringoperations,we get
Tr5Sq?(t5) = Sq?Tr5(t5) = h2dl : 0.

This contradictsthe above conclusionthat Sq?(t5) = O. Therefore,
Tr (ts) + Ah4h4+ h1do
forany A C F2. Combiningthiswiththe factthat Trs(ts) / 0 in Span{h h4,hldo},
we get Trs(t5) hh4.
With ambiguity of notation, we also have Tr5(h4h4)

is an isomorphismin degree15, we obtain t5 = h4h4.

we have
(b) As Sq? is an algebrahomomorphism,
Sqo(hntk) = Sqo(hn)Sqo(tk)

= h4h4 = Tr5(t5). As Tr5

- 0.

On the otherhand, as Tr = (k Trk is also an algebrahomomorphism,
we get
Trk+l (htk)

Trl (hn)Trk(tk)

= hnTrk(tk).

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STEENROD

ALGEBRA

AND GENERAL

LINEAR

GROUPS

4087

As shownin the proofof Lemma 5.3, a consequenceof Davis' Theorem5.1 claims
that,ifTrk(tk) 7 0, then hnTrk(tk) / 0 foreveryn with
2n > 4(Stem(Trk(tk)))2 = 4(deg(tk))2.

[2

The remarkis proved.
As an immediateconsequence,we have
Corollary

9.3.

(i) Ker(Sq?) n (F2 0 PH,(BVk))
GLk

is nonzero for k = 5 and

has an infinitedimension for k > 5.
k = 5, Trk detects a nonzero element in the kernel of Sq?, and for each
For
(ii)

k > 5, it detectsinfinitely
manynonzeroelementsin thiskernel.

It has been known(see [28], [4]) that Sq? is injectiveon F2 0 PH*(BVk) for
GLk

k < 3.

in positivedegreesofF2 0 PH* (BV4).
Conjecture 9.4. Sq? is a monomorphism
GL4

In otherwords,Sq? is a monomorphism
in positivedegreesof F2 0 PH*(BVk) if
GLk

and only if k < 4.

The followingis an analogue of Corollary6.2 and is relatedto Corollary6.10.
Conjecture 9.5 (Sq? is eventuallyisomorphicon the Ext groups). Let Im(Sq0)i
denotethe image of (Sq?)i on Ext(lF2, F2). There is a numbert dependingon k
such that
Im(SqO)t

(Sq)i-t

-

Im(Sq?)i

is an isomorphismforeveryi > t.
In otherwords,Ker(Sq0)i = Ker(Sq?)t on Ext(lF2,F2) foreveryi > t. As a
consequence,any finiteSq?-familyin Ext (F2, F2) has at mostt nonzeroelements.
Is the conjecturetruefort = k- 2?
An observationon the knowngeneratorsof the Ext groupssupportsthe above
conjecturewitht muchsmallerthan k - 2.

It also leads us to the questionon whetherSqO is an isomorphismon
Im(Sq?)t C F2 0 PH*(BVk)
GLk

forsome t < k- 2. (This questionhas an affirmative
answergivenby Corollary6.2
fort = k - 2.)
ACKNOWLEDGMENT

The researchwas in progressduringmyvisitto WayneState University,
Detroit
(Michigan) in the academic year 2002-2003. I would like to expressmy warmest
thanksto LowellHansen and all his colleaguesat the Departmentof Mathematics
fortheirhospitalityand forthe wonderful
workingatmosphere.In particular,I am
to
Robert
Daniel
grateful
Bruner,
Frohardt,Kay Magaard and SergeyShpectorov
forfruitful
conversationson the Ext groupsand modularrepresentations.

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4088

NGUYEN H. V. HUNG
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