DSpace at VNU: On the behavior of the algebraic transfer

On the Behavior of the Algebraic Transfer
Author(s): Robert R. Bruner, Lê M. Hà and Nguyễn H. V. Hung
Source: Transactions of the American Mathematical Society, Vol. 357, No. 2 (Feb., 2005), pp. 473487
Published by: American Mathematical Society
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TRANSACTIONS
OF THE
SOCIETY
AMERICAN
MATHEMATICAL
Volume 357, Number 2, Pages 473-487
S 0002-9947(04)03661-X
Article electronically published on May 28, 2004

OF THE ALGEBRAIC

ON THE BEHAVIOR

TRANSFER

ROBERT R. BRUNER, LE M. HA, AND NGUYEN H. V. HUNG
Dedicated to Professor Huynh Mui on the occasion of his sixtiethbirthday
ABSTRACT.

Let Trk : F2

0

GLk

PH

(BVk)

-> Ext

k+ (F2F2)F

be the alge-

braic transfer,which is defined by W. Singer as an algebraic version of the
geometrical transfertrk : 7rS((BVk)+) -+ 7rS(S?). It has been shown that
the algebraic transferis highly nontrivial and, more precisely,that Trk is
an isomorphismfor k = 1,2,3. However, Singer showed that Tr5 is not an
epimorphism. In this paper, we prove that Tr4 does not detect the nonzero
element gs C E xt4122 (F2, F2) for every s > 1. As a consequence, the localized (Sq?)-lTr4 given by invertingthe squaring operation Sq? is not an

epimorphism. This gives a negative answer to a predictionby Minami.

1. INTRODUCTION

AND STATEMENT OF RESULTS

The subject of the presentpaper is the algebraictransfer
Trk : F2 (

GLk

PHi(BVk)

-3 Extk+i (F2,F2),

whichis definedby W. Singeras an algebraicversionof the geometricaltransfer

trk
7rs(S?) to the stable homotopy groups of spheres. Here
7rrs((BVk)+)
denotes
a
k-dimensional
F2-vectorspace, and PH*(BVk) is the primitivepart
Vk

consistingof all elementsin H (BVk) that are annihilatedby everypositive-degree
operationin the mod 2 Steenrodalgebra,A. Throughoutthe paper, the homology
in F2.
is takenwithcoefficients

It has been proved that Trk is an isomorphismfor k = 1,2 by Singer [14]
and fork = 3 by Boardman [1]. These data togetherwith the fact that Tr =
(see [14]) showthatTrk is highlynontrivial.
Dk>0Trk is an algebrahomomorphism
to be a usefultool forstudyingthe
transfer
is
considered
the
Therefore, algebraic
mysteriouscohomology of the Steenrod algebra, Ext*5 (F2, F2). In [14], Singer also
gave computationsto show that Tr4 is an isomorphismup to a range of internal

degrees.However,he provedthat Tr5 is not an epimorphism.
Based on thesedata, we are particularlyinterestedin the behaviorof the fourth
theoremis the main resultof this paper.
algebraictransfer.The following
Received by the editors June 18, 2003.
2000 Mathematics Subject Classification. Primary 55P47, 55Q45, 55S10, 55T15.
Key words and phrases. Adams spectral sequences, Steenrod algebra, invarianttheory,algebraic transfer.
The third author was supported in part by the Vietnam National Research Program, Grant
N0140801. The computer calculations herein were done on equipment supplied by NSF grant
DMS-0079743.
02004

American Mathematical

473

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Society


474

ROBERT

LE M. HA, AND NGUYEN

R. BRUNER,

H. V. HUNG

Theorem 1.1. For each s > 1, thenonzeroelementgs C Ext,12 2 (F2,F2) is not
in the image ofTr4.
The readeris referred
to May [11] forthe generatorgl and to Lin [8] or [9] for
the generatorsgs.
As a consequence,we get a negativeanswerto a predictionby Minami [13].
Corollary 1.2. The localizationof thefourthalgebraictransfer
(Sq)-Tr4

: (Sq?)-1F2

-

0 PH*(BV4)

GL4

(Sq?)-lExt4A4+

(F2, F2)

givenbyinverting
Sq? is not an epimorphism.
It is well known(see [10]) that thereare squaringoperationsSq' (i > 0) acting
on the cohomologyof the Steenrodalgebra, whichshare most of the properties
with Sqi on the cohomologyof spaces. However,Sq? is not the identity.We refer
to Section 2 forthe precise meaningof the operationSq? on the domain of the
algebraictransfer.
We next explainthe idea of the proofof Theorem1.1.
Let Pk '= H*(BVk) be the polynomialalgebra of k variables,each of degree1.
Then, the domain of Trk, F2 0 PH,(BVk), is dual to (F2 ( Pk)GLk. In orderto
A

GLk

proveTheorem1.1, it sufficesto showthat (F2 P4)12L_4 = 0, foreverys > 1.
A
Direct calculationof (F2 0P4)12.2s-4 is difficult,
as P4 in degree 12 25 - 4 is
A
a huge F2-vectorspace, e.g. its dimensionis 1771 fors = 1. To computeit, we
observethat the iterateddual squaringoperation
(SqO)

: (F2 0P4)l2.2A

-4

(F2

A

P4)8

is an isomorphism
of GL4-modulesforany s > 1. This isomorphism
is obtainedby
the
applyingrepeatedly followingproposition.
Proposition 1.3. Let k and r be positiveintegers.Suppose thateach monomial
Xl .*. xk of Pk in degree2r + k withat least one exponent
it even is hit. Then
Sq

: (F2 0 Pk)2r+k -t (F2 0Pk)r
A
A

is an isomorphismofGLk-modules.
Here, as usual, we say that a polynomialQ in Pk is hitifit is A-decomposable.
Further,we show that (F2 OP4)8 is an F2-vectorspace of dimension55. Then,
A

a specificbasis of it, we provethat (F2 pP4)GL4 = 0. As a conseby investigating
A

quence, we get (F2 P4)G2L._4 = 0 foreverys > 1.
A
The readerwho does not wishto followthe invarianttheorycomputationabove
weakertheorem,and thenwouldnot need to read
may be satisfiedby the following
the paper's last 3 sections.
Theorem 1.4. Tr4 is not an isomorphism.
This theoremis provedby observingthat,on the one hand,
(IF2

A

P4)2L4

(F2

P4)GL
A

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THE BEHAVIOR

OF THE ALGEBRAIC

TRANSFER

475

and on the otherhand,
20(F

0
F) = F2 -91 Ext44+8 (F, F
The paper is dividedintosix sectionsand organizedas follows.Section2 starts
with a recollectionof the squaring operationand ends with a proofof the isoExt4

- (F2 OPk)8.

morphism (F2 OP4)12.2s-4
A

Theorem 1.4 is proved in Section 3. We

A

compute(F2 0P4)8 and its GL4-invariantsin Section4. We proveTheorem1.1 in
A
Section5. Finally,in Section6, we describethe GL4-modulestructureof (F2 (P4)8.
A

2. A SUFFICIENT

CONDITION FOR THE SQUARING OPERATION
TO BE AN ISOMORPHISM

This sectionstartswitha recollectionof Kameko's squaringoperation

Sq? : F2 ? PH*(BVk) -> F2
GLk

GLk

PH*(BVk).

The mostimportantpropertyofKameko'sSq? is thatit commuteswiththeclassical
Sq? on Ext((F2, F2) (definedin [10]) throughthe algebraictransfer(see [1], [13]).
This squaringoperationis constructedas follows.
As is well known,H*(BVk) is the polynomialalgebra,Pk := F2[xl,... ,xk], on
k generatorsx1,..., Xk,each of degree1. By dualizing,
H,(BVk)

- r(a,..

., ak)

is the divided power algebra generatedby al,..., ak, each of degree 1, whereai
is dual to xi C H1(BVk). Here the dualityis taken with respectto the basis of
H*(BVk)

consisting of all monomials in x1,... ,Xk.

In [6] and [7] Kameko defineda homomorphism
Sq:

-

H, (BVk)

(ii

al

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

. ak
...

* * * a a1 ak
is dual to x1
1
xk
k . The following lemma is well known. We
to
a
make
self-contained.
the
give proof
paper

where a(i1

Lemma 2.1. Sq? is a GLk-homomorphism.
Proof. We use the explanation of Sq? by Crabb and Hubbuck [3], which does not
depend on the chosen basis of H*(BVk). The element a(Vk) = a ... ak is nothing

but the image of the generatorof Ak(k) underthe (skew) symmetrization

map
Ak(Vk)

-- Hk(Bk)

= rk()

= (V k

...

vk)Sk,

k times

where the symmetricgroup Sk acts on Vk ) ***0Vk by permutations of the factors.

Let c: H,(BVk) -- H*(BVk) be the degree-halving
whichis dual to
epimorphism,
the Frobenius monomorphism F : H*(BVk)

any x. We have

Sq?(c(y))

for y E H*(BVk).

-*

H*(BVk)

defined by F(x) = x2 for

= a(k)y,

To prove that this is well definedwe need to show that if

c(y) = 0, then a(Vk)y = 0. Indeed, c(y) = 0 implies (c(y),x) = (y,x2) = 0 for
every x e H*(BVk). Here (., ) denotes the dual pairing between H*(BVk) and

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ROBERT R. BRUNER, LE M. HA, AND NGUYEN H. V. HUNG

476

a(ik, thenthereis at least one it whichis
H*(BVk). So, ifwe writey = Eai)
odd in each termof the sum. Therefore,
a(i) ... a(ik)) = 0,

a(Vk)y = a ... ak(

because atait) = 0 forany odd it. So, Sq? is well defined.
As c is a GLk-epimorphism,
the map SqO is a GLk-homomorphism.
The lemmais proved.

O

Further,it is easy to see that cSq2t+l = O, cSq2t = Sqtc. So we have
Sq,2t+lSqo = 0, Sq2tSq0

Sq?S0qt.

(See [4] foran explicitproof.) Therefore,Sq? maps PH,(BVk) to itself.
Kameko's Sq? is definedby
Sq? = 1 0 Sq?: F2 0 PH,(BVk) - F2 ( PH*(BVk).
GLk

GLk

GLk

The dual homomorphism
Sq : Pk -+ Pk of Sq? is obviouslygivenby
Sq(Igl...

*tXlfk
)

?

31-1

3k-1

,Jk odd,
otherwise.

jl,...

X

xi'1

0,

Hence

Ker(Sq? : Pk -* Pk) = Even,
whereEven denotesthe vectorsubspace of Pk spannedby all monomialsx1l
withat least one exponentit even.
Let s: Pk -i Pk be a rightinverseof Sq? definedas follows:
...

s(ixl

"V l

ik
Xk;

xk

2ik+l

=271+1

....Xk

XI

It should be notedthat s does not commutewiththe doublingmap on A, that is,
in general
Sq2ts

sSqt.

However,in one particularcircumstancewe have the following.
Lemma 2.2. Underthe hypothesis
of Proposition1.3, the map
s : (F2 OPk)r
A

s[X]

-*

(F2 0Pk)2r+k,
A

=

[sX]

is a well-defined

linearmap.
Proof. We startwithan observationthat
Im(Sq2ts - sSqt) C Even.
We provethis by showingequivalentlythat
SqO(Sq2ts - sSqt) = O.

Indeed,
SqO(Sq2ts

-

sSqt)

=

SqSq2ts

=

Sqtqs

=

Sqt id - id Sqt

=

0.

- SqsSqt

- Sq?sSqt

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THE BEHAVIOR OF THE ALGEBRAIC TRANSFER

As a consequence, s maps (A+Pk)r to (A+Pk + Even)2r+k.

477

Here and in what

follows,A+ denotesthe submoduleofA consistingofall positivedegreeoperations.
Further,by the hypothesisof Proposition2.3, we have
(A+Pk

Hence, s maps (A+Pk)r

to

linearmap, as s is.
The lemma is proved.

C (A+Pk)2r+k.

+ Even)2r+k

So the map s is well defined. Then it is a

(A+Pk)2r+k.

O

The followingpropositionis also numberedas Proposition1.3.
Proposition 2.3. Let k and r be positiveintegers.Suppose that each monomial
Xi

...

xk

of Pk in degree 2r + k with at least one exponent it even is hit. Then
-

Sq* : (F2?Pk)2r+k
A

(F2 0Pk)r
A

is an isomorphismofGLk-modules.
Proof. On the one hand, we have Sqs =- id(2
followsthat

Sq s[X] = Sq[sX]

gk)
A

[Ss

. Indeed,fromSq?s = idpk, it
= [X],

X]

forany X in degreer of Pk.

On the other hand, we have sSq? = id(lF2?k)2r+k

Indeed, by the hypoth-

A

esis, any monomial with at least one even exponent representsthe 0 class in
(F2 0Pk)2r+k, so we need onlyto checkon the classes of monomialswithall expoA
nentsodd. We have

-

[s (xi

[x 2i+l

...

xik)]

...2ik+l]

~2ik+1r?kfk]
~~2i1+1
for any x21
... x2ik+1 in degree 2r + k of Pk.
Combining the two equalities, SqOs = id(F2 Pk)r and sSq
we see that SqO : (F2 (Pk)2r+k
(F2 OPk)r
A

A
(IF20Pk) 2r+k.
A

-*

A

= id(l

(Pk)2r+

A

(F2 OPk)r is an isomorphismwith inverse~s
A

The propositionis proved.

D

The targetof this sectionis the following.
Lemma 2.4. For everypositiveintegers,
(Sqo)s : (F2 0 P4)12.2--4 --(]F2
A

is an isomorphismofGL4-modules.

A

P4)8

Proof. By usingProposition2.3 repeatedly,it sufficesto show that any monomial
of P4 in degreem = 12 *2s - 4 with at least one even exponentis hit. Since m
is even,the numberof even exponentsin such a monomialmust be either2 or 4.
If all exponentsof the monomialare even, then it is hit by Sq1. Hence we need
only to considerthe case of a monomialR with exactlytwo even exponents(and
so exactlytwo odd exponents).Wood proves([15]) that if a(m + oo(R)) > ao(R)

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ROBERT

478

R. BRUNER,

LE M. HA, AND NGUYEN

H. V. HUNG

thenR is hit,whereao(R) is the numberofodd exponentsin the monomialR, and
a(n) is the numberof ones in the binaryexpansionof n. We have ao(R) = 2 and
a(m + aoQ(R))= a(12 25 - 2) = s + 2, so Wood's criterionis met,and R is hit.
D
The lemmais proved.
3. THE FOURTH ALGEBRAIC TRANSFER IS NOT AN ISOMORPHISM

The targetof this sectionis to provethe followingtheorem,whichis also numbered as Theorem1.4.
Theorem 3.1.
Tr4 : F2 (

GL4

is not an isomorphism.

PHi(BV4)

-- Ext 4+i (F2,
F2)

Proof. For any r, we have a commutativediagram
(IF2X PHi(BV4))r

Tr4

Ext 44+r(

GL422

SqO

(F2

F)

SqO

( PHi(BV4))2r+4

Tr4

Ext4,8+2r(F2,
(F F2)

GL4

wherethe firstverticalarrowis the Kameko Sq? and the secondverticalone is the
classical Sq?.
The dual statementof Lemma 2.4 fors = 2 claims that
Sq?: (F2 0 PHi(BV4))8
GL4

-

(F2 (

GL4

PHi(BV4))20

is an isomorphism.On the otherhand, it is known(May [11]) that
= 0 2 Ext4,4+20(2, IF2)=F2
Ext44+8
~A (F2 IF) ~A

9gi

This impliesthat Tr4 is not an isomorphism.The theoremis proved.

]

or
Remark3.2. This proofdoes not showwhetherTr4 failsto be a monomorphism
failsto be an epimorphism.We will see that actuallyTr4 is not an epimorphism
in Section 5 below.
4. GL4-INVARIANTS

OF THE INDECOMPOSABLES

OF P4 IN DEGREE 8

From now on, let us writex = x1, y -= 2, z = x3 and t = x4 and denotethe
monomialxaybzctdby (a, b,c, d) forabbreviation.
Proposition 4.1. (F2?P4)8 is an F2-vectorspace of dimension55 witha basis
A
consistingof the classes represented

bythefollowingmonomials:
(A) (7,1,0,0), (7,0,1,0), (7,0,0,1), (1,7,0,0), (1,0,7,0), (1,0,0,7),
(0, 7,1,0), (0,7, 0,1), (0, 1,7,0), (0,1, 0,7), (0, 0,7,1), (0,0, 1,7),
(B) (3, 3,1, 1), (3, 1,3, 1), (3, 1,1, 3), (1, 3, 3, 1), (1, 3,1, 3), (1, 1,3, 3),

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THE BEHAVIOR

OF THE ALGEBRAIC

TRANSFER47

479

(C) (6,1,1,0),(6,1,0,1),(6,0,11,1)1(1,6,1,0),(1,6,0,1)1(1,11,60),)
(1,1,0,6), (1,0,6,1), (1,0,1,6), (0,6,1,~1), (0, 1,6,1), (0,1,1,6),
(D) (5, 3,0,0),1(5, 0,3,0), (5, 0,0,3), (0,15,3,0), (0, 5,~0,3), (0, 0,5,3),
(E) (5, 2,1,10),(5, 2,0,l1),(5, 0,2,1),~(2, 5,1,10),(2, 5,0,1), (2,1,15,0),
(2,1,~0,~5),(2, 0,5,1), (2, 0,1,15),(0, 5,2, 1), (0, 2,5,1), (0, 2,1,15),
(F) (5, 1,1,1), (1, 5,1,1),(1,1,5,1), (1,1,1, 5),

The propositionis provedby combinling
a couple of lemmas.
Lemma

4.2. (F2

OP4)8

A

is generatedbythe55 elements listed in Proposition 4. 1.

Proof. It is easy to see that everymonomial(a, b,c, d) with a, b,c, d all even is hit
(morepreciselyby Sq1).
The only monomials(a, b,c, d) in degree8 with at least one of a, b,c, d odd are
the followingup to permutationsof the variables:
(7, 1,0,~0),1
(3,3,1,1), (6, 1,1,10),1
(5,3,0,0), (5,2,1,0), (5,1,1,
I~1), (4, 2,1,11),
(4, 3,1,~0),1
(3, 3,2,0), (3, 2,2,1).
The last 3 monomialsand theirpermutationsare expressedin termsof the first
7 monomialsand theirpermutationsas follows:
(4,3,1,0)

=

(3, 3,2,0)

=

(2,5,1,0)?Sq 4(1, 2,110)?+Sq2(2,3,1,0),
(5,2,1,0)?+(2,5, 1,0) +Sq 4(2,1,1, 10)?+Sq4(1,2,1,0)

?Sq 2 (3,2,1,~0) + Sq2(2, 3,1,10)+ Sq1(3, 3,1,0),

(3,2,2, 1) = (5,1,1,1) + (4,2,1,1) + (4,11,2,1)
+Sq 2 (3,1,1,11) ? Sq'(4, 1,1,11)+ Sql(3,221, 1) ? Sq1(3,1,12,1).
Hence, (F2 0&P4)8 is generatedby the following7 monomialsand theirpermutaA

tions:

By the familyof a monomial(a, b,c, d) we mean the set of all monomialswhich
are obtainedfrom(a, b,c, d) by permutationsof the variables.
The monomialsin the 7 familiesabove whichare not in Proposition4.1 can be
expressedin termsof the 55 elementslisted thereas follows. (We give only one
expressionfromeach symmetry
class.)
(3, 5,0,~0)

-

(5,1,2,0)
(4,1,1,2)
(2,4, 1,1)
(2,1,1,4)

-

(5,3, 0,0) +Sq 4(2,2, 0,0)?+Sq 2(31310,0),
(6,1 1,10)?+(5,2, 1,0)?+Sq1(5, 1,110),
(4,2, 1,1)-i- (4,1,2, 1)?+Sq1 (4,1,1, 1),
(4,2,1 1)?+Sq4(1,11,11)-i-Sq 2(2 2,1,11),
(4,2,1,1) +(4,1,2,1)
+Sq4(1, 1,1,~1) + Sq2(2, 1,1,2) + Sql(4, 1,1,1),

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480

ROBERT

(1,4,1,2)

=

(1,2,1,4)

=

(1,1,4,2)

=

(1,1,2,4)

=

R. BRUNER,

LE M. HA, AND NGUYEN

H. V. HUNG

(4,2,1,1)+(1,4,2,1)
+Sq4(, 1,1, 1) +Sq2(,2,2,1,1)+ Sq1(1,4,1,1),
(4,2,1,1)+ (1,4,2,1)
+Sq2(2,2, 1, 1) + Sq2(1, 2, 1,2) + Sq(1,4, 1, 1),
(4,1,2,1)+(1,4,2,1)
+Sq2(2,1,2,1) + Sq2(1, 2,2,1) + Sq(1, 1,4,1),
(4,1,2,1) + (1,42,,1) +Sq4(1,1,1,1)
+Sq2(2,1,2,1) + Sq2(1, 2,2,1) + Sq2(1,1,2,2) + Sql(1, 1,4,1).
D

The lemma is proved.

in
Lemma 4.3. The 55 elementslistedin Proposition4.1 are linearlyindependent
(F2 ?P4)8.
A

Proof. We will use an equivalencerelationdefinedby sayingthat, fortwo polynomials P and Q, P is equivalentto Q, denotedby P ~ Q, ifP - Q is hit.
If X is one of the lettersfromA to G, let Xi be the i-th elementin familyX
accordingto the orderlistedin Proposition4.1. (This is the lexicographicalorder
in each family.)
Suppose thereis a linearrelationbetweenthe 55 elementslistedthere,
12

12

6

Za,A,A + ybiBz
i=l

+

i=

i=

ciCi +

i=

4

12

6

diDi +

E
i=

eiEi
i=

3

+fiFi +

i=l

giGi = 0,

are
where a,, b,,ci, di,e%,f, gi E F2. We need to show that all these coefficients
zero. The proofis dividedinto 4 steps.
Step 1. We call a monomiala spikeif each of its exponentsis of the form2n - 1
forsome n. It is well knownthat spikesdo not appear in the expressionof SqiY
forany i positiveand any monomialY, since the powersx2 -1 are not hit in the
of any spikeis zero in everylinearrelation
one variablecase. Hence,the coefficient
in F2O Pk.
A
Amongthe 55 elementsof Proposition4.1, the classes of familiesA and B are
spikes. So ai = bj = 0, foreveryi and j. Then, we get
12
i=l

12

6

ciCi +,

diD +
i=l

eiEi +
2fiFi

i=l

4

3

giGi = 0.

+
i=l

i=l

F2 OP4 -* F2 ?P2 inducedby the projection
Step 2. Considerthe homomorphism
A
A
the image of the above linear
P4 - P4/(z, t) - P2. Under this homomorphism,
relationis d1(5, 3) = 0.
In orderto showthat d1 = 0, we need to provethat (5,3) is nonzeroin F2 ?P2.
A
x F- x, y - x + y sends (5, 3) to (8, 0) + (7, 1) + (6, 2) +
The lineartransformation
(5, 3) ~ (7, 1) + (5, 3). As the action of the Steenrodalgebracommuteswithlinear
maps, if (5, 3) is hit thenso is (7, 1) + (5, 3). But it is impossible,because (7, 1) is
a spike. Hence, (5,3) 0 in F2 ?P2 and dl = 0.
A

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THE

BEHAVIOR

OF

THE

ALGEBRAIC

481

TRANSFER

Similarly,usingall the projectionsof P4 to its quotientsby the ideals generated
by each pair of the fourvariables,we get dai -0 foreveryi. So we get
12

12

i 1

3

4

c +Z~~~~~~~~~~eiEi

? ZfiFi
ZeiCi

Step 3. Considerthe homomorphism
IF2OP4
P4

~~j=0
=i1

i1

ii

--

0.

F2 OP3 inducedby the projection
A

A

-- P4/(t) N P3. Underthis homomorphism,
the linearrelationabove is sent to
ci(6, 1,1) + c4(1,6, 1) + C6(1,1,6) + ei(5,2, 1) + e4(2,5, 1) + e6(2,1, 5)z 0.
Applying the linear map x H-? x, y t-* x, z

H-+

y to this relation, we obtain

(ci + C4+ e1 + e4)(7,1)+ c6(2,6) + e6(3,5)
= (cl C4+ el + e4)(7, 1) + e6(3,5) = 0.
Since (7, 1) is a spike,(cl + c4 ? e1 + e4) = 0, hence e6(3, 5) = 0. As for(5, 3), we
can showthat (3,5) $y0 E F2 OP2 and get e6 = 0.
A

=
By similar arguments,we have eI = e4 = e6 = 0. The equality (cl+c4+eI+e4)
0 showsthat cl + C4 = 0 or c1 = C4. By similararguments,cl = c4 = c6. We denote

this commoncoefficient
by c and get

c{(6, 1, 1) + (1.6,1) ? (1. 11,6)}= 0.
0. Suppose the
We provethat c - 0 by showingthat (6,1, 1) +-(1,6, 1)?(1, 1, 6) 0
is
the
unstable
that
hit.
Then, by
propertyof
contrary,
(6,1,1) + (1, 6,1) + (1, 1,6)
the action of A on the polynomialalgebra,we have
(6,1,1)?(1,6,1)+
(1,1,6) = Sql(P)+ Sq2(Q)+ Sq4(R),

f and this
forsome polynomialsP, Q, R. By the degreeinformation,
R2
Sq4(R)
to assume (6, 1, 1) + (1,6,1) + (1, 1,6) elementis hit by Sq1. Therefore,
it suffices
?
Sq1(P)
Sq2(Q).
Let Sq2Sq2Sq2 act on the both sides ofthisequality.The righthand side is sent
to zero,as Sq2 Sq2 Sq2 annihilatesSq1 and Sq2. On the otherhand,
Sq2Sq2Sq2 {(6, 1,1) + (1,6, 1) + (1,1, 6)}
This is a contradiction.So, it implies(6, 1, 1)
We get
fjFj +
i l

Step 4. Apply the linear map x

fi(5,3) +

(f2

?

(8,4,2) ? symmetries
Z 0.
=
0
and c = 0.

(1, 6,1) + (1, 1,6)

3

4

and we have

=

H-4

x, y

gjG
i Gl
~-+

0.

y,z H y,t

f3 + fh + 93)(1,

F-*

7) + (g, +

y to the above equality,
92)(4, 4)

= fi(5,3) + (f2 + f3+ f4 + 93)(1, 7) = 0.
As (7, 1) is a spike,we obtain (f2+ f3+ f4+ 93) - Oand fi(5, 3) = 0. As (5, 3)#f0,
it yieldsfi = 0.
Next, apply the linear map x F-* x, y H-+ y, z 4 x, t H-4 x to the equality
Z,f1f7F,+ E3 I1iGi = 0, and we have
f2(3,5)

+ (fh + f4 + 92)(7, 1) + gi(6, 2) + 93(4,4)

- f2(3,5) + (f3 + f4 + 92)(7, 1) = 0.

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ROBERT

482

LE M. HA, AND NGUYEN

R. BRUNER,

As (7,1) is a spike, we get (f3 + f4 +
implies f2 = 0.

92)

H. V. HUNG

= 0 and f2(3, 5) = 0. Since (3, 5) : 0, it

Similarly,apply the linear map x -* x, y H- x, z
1+
f4F4 +
f3F3 f
gGi = 0, and we have

-

y,t

-*

x to the equality

f3(3,5) + (f4 + gi)(7, 1) + (92 + 93)(6, 2)
= f3(3,5) + (f4+ gi)(7, 1) =0.
As (7,1) is a spike,we get f4 +

91

= 0 and thenf3 = 0.

Finally, apply the linear map x i- x,y i- x,z i- x,t
f4F4 + E3i= gGi - 0, and we have

) y to the equality

f4(3, 5) + (g9 + 92 + 93)(7, 1)
= f4(3,5) + (9 + 92 + 93)(7,1) = 0.
As (7,1) is a spike, we get 91 + 92 + 93 = 0 and then f4 = 0.
Substituting fi = f2 = f3 = f4 = 0 into the equations (f2 + f3 + f4 + 93) = 0,
(f3 + f4 + 92) = 0, f4 + 91 = 0, we get 91 = 92 = g3 = 0.

of an arbitrarylinearrelationbetweenthe
We have shownthat all coefficients
D
55 elementslistedin Proposition4.1 are zero. The lemmafollows.
CombiningLemmas 4.2-4.3, we get Proposition4.1.
Proposition

4.4. (F2 ?P4)GL4
A

= 0.

Proof. If X is one ofthe lettersA, B, C, D, E, F, G, let L(X) be thevectorsubspace
of (IF2 P4)8 spannedbythe elementsoffamilyX in Proposition4.1. Let Sk denote
A
the symmetricsubgroupof GLk. Accordingto the relationslisted in the proof
of Lemma 4.2, L(A), ?(B), L(C), L(D), L(F), L(G) are S4-submodules. The
subspace L(E) is not an S4-submodule.However,the sum
C(C, E) = L(C) ( (E)
is. We have a decompositionof S4-modules
(F2 ?P4)8 - L(A)
LC(B)f ?(C, E)
A

12(D)

?C(F)

?2(G).

Let a be an arbitraryGL4-invariantin (F2 OP4)8. It can uniquelybe writtenin
the form

A

a = aA + COB+ aC,E + aD + aOF + acG,

whereax E ?(X) forX E {A, B, D, F, G}, and aC,E E )(C, E). Each termofthis
sum is S4-invariant.
thenall
Note that ifa linearcombinationofelementsin a familyis S4-invariant,
are equal, because each elementin thefamilycan be obtainedfrom
ofits coefficients
any otherby a suitable permutation.Let sx denote the sum of all the elements
in the family X listed in Proposition 4.1. Then, we have aA = asA, aB = bsB,
where a, b,c, d, e, f, g E F2.
aD = dsD, aF = fSF, aG = 9SG, and ac,E = csc+esE,

Let p be the transpositiongivenby p(x) = y, p(y) = x, p(z) = z, p(t) = t. It is
easy to see that
p(2,1,0,5) = (1,2,0,5)
(2,1,0,5)+(1,1,0,6),
=
p(2,1,5,0)

(1,2,5,0)=(2,1,5,0)+(1,1,6,0).

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THE

BEHAVIOR

OF

THE

ALGEBRAIC

TRANSFER

483

from(2, 1,0, 5) and (2, 1,5,0) in familyE are
Further,the 10 elementsdifferent
divided into 5 pairs with p acting on each pair by twisting. So, p(sE) = SE +
(1,1,0, 6) ? (1,1,6, 0). On the other hand, as the familyC is full,in the sense
that it containsall the variablepermutationsof a monomial,we have p(sc) = sc.
Hence,we get
- p(cSC

p(ac,E)

+ eSE)

-

cSC + eSE + e(1, 1,O, 6) + e(1, 1, 6, 0).

As aC,E is S4-invariant, e(l, 1, 0, 6) ? e(1, 1, 6, 0) = 0. So e = 0, because the two
elementsare linearlyindependentby Lemma 4.3. We obtain
a = aA + aB + aC + aD + aF + aG,

whereac = ac,E - csc.

Let us now consider the transvection co given by yv(x) - x, yO(y) = y, 0p(z) = x,

cp(t)= x + t. A routinecomputationshows
=
A ? (7,1,0,0) + (7,0,10) -- (7,0,0,1) + (1,7,0,0) + (1,0,7,0)
(P(SA)
?(6, 1,0,1) ? (6, 0,1,1) + (1,1,0, 6) ? (1,0,1,6),
SB + (6,1,1,0) + (1,6,1,0) ? (1,1,6,60)- (2, 5,1,0) + (2,1, 5, 0)
=
cp(SB)
?(5, 1,1,1) + (1,5,1,11)+ (1, 1,5,1) ? (4,2,1,) 1? (4, 1,2,1)
?(3,3,1,1) + (3,1,3, 1),
co(sc) = sc + (6, 1,1,0) + (1,6,1,0) + (1, 1,6,0),
= SD + (7, 0, 0,1) ? (1,6, 0,1) ? (1,0, 6,1) + (5,3, 0,0) + (5,0,3, 0),
(O(SD)
? (2, 5,1,0) ? (2,1,5,0) ? (5,,1,11) + (4, 2,1,1) ? (4,1,2,1),

SO(SF)

=

SF

cp(SG)

=

Sc + (6, 1, 1,0).

Let rx = yp(sx)- sx whereX is one ofthe lettersA, B, C, D, F, G. The equality
a is rewrittenas
co(a)=
cp(aSA + bSB + cSc + dSD?

or equivalently

f SF + gSG)

= aSA + bSB + cSC + dSD + fSF + gSG,

arA + brB + crc + drD + frF + grG = 0.

In this linearcombination,rB and rD are the only termscontaining(3,3, 1, 1)
in familyB and (5,3,0,0) in familyD respectively.From Lemma 4.3, we get
b - d - 0, and therefore
arA ? crC + frF + grG = 0.
In the new linearcombination,as rA, rc and rF are the only termscontaining
(7, 1,0,0) in familyA, (1, 6, 1,0) in familyC and (4, 2, 1, 1) in familyF respectively,

we have a = c = f = 0. As a consequence,grG - 0, so we finallyget g - 0.
In summary,we have shown that everyGL4-invarianta in (F2 0P4)8 equals
A
D
zero. The propositionis proved.
5.

THE

FOURTH

ALGEBRAIC

TRANSFER

IS NUT

AN EPIMORPHISM

The goal ofthispaper is to provethe following
theorem,whichis also numbered
as Therein1.1.
Theorem 5.1. For each s > 1,
Tr4 :F2 0 PH (BV4)
GL4

--

ExtA4 (F2,F2)

does not detectthe nonzeroelementg, G Ext 412-2' (F2, F2).

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ROBERT

484

R. BRUNER,

LE M. Hk, AND NGUYEN

H. V. HUNG

Proof. CombiningLemma 2.4 and Proposition4.4 we get
(F2

A

= 0,

)P4)G24

foreverynonnegativeintegers.
On the otherhand, it is wellknownthat Ext424(IF2,F2) is spannedby the generator gl (see May [11]). Further,gs = (Sq0)S-l(gl) is nonzeroin Ext4l12'2 (F2,F2)
(see Lin [8] and also [9]).
As F2 0 PH12.2--4(BV4)
GL4

Tr4:

JF2(

GL4

is dual to (F2 ?P4)2GL
A

PH12.25--4(BV4)

-

Ext

4i
,122 (IF2,

2)

does not detectthe generatorgs, foreverynonnegativeintegers.
The theoremis proved.

D

As a consequence,we get a negativeanswerto a predictionby Minami [13].
(This corollaryis also numberedas Corollary1.2.)
Corollary 5.2. The localizationof thefourthalgebraictransfer
(Sq?)-1Tr4:

(Sq?)-1F2

0 PH*(BV4)

--

GL4

(Sq?)-1Ext4j4+*(F2,IF2)

givenbyinvertingSq? is not an epimorphism.
Proof. Indeed, it does not detect the nonzeroelementg, whichis representedby
D
the family(gs)s>o withgs = (SqO)s-1(gi). The corollaryfollows.
Remark5.3. Our resultdoes not affectSinger'sconjecturethat the k-thalgebraic
foreveryk. (See [14].)
transferis a monomorphism
6. FINAL REMARK: GL4-MODULE

STRUCTURE

Boardman's studyof the 3 variableproblemshowsthat the GLk modulestructureof F2 0 Pk may be a usefultool. In thisveinwe close witha descriptionofthe
A

module (F2?P4)8
A

as a GL4-module. From the "Modular Atlas" [5] we findthat

thereare 8 irreduciblemodulesforGL4 in characteristic2, of dimensions1, 4, 4,
6, 14, 20, 20, and 64. With a littlecalculationwe findthe following
descriptionof
them:
1: the trivialmoduleF2,
N: the naturalmoduleF4,
N*: the dual of the naturalmodule,
A: the alternatingsquare of N or N*,
S: the nontrivialconstituentof N ( N*, whichhas compositionfactors1,S, 1,
T: a constituentof N 0 A, whichhas compositionfactorsN* and T,
T*: a constituentof N* 0 A, whichhas compositionfactorsN and T*,
St: the Steinbergmodule.
Using a "meataxe" programwrittenin MAGMA, togetherwitha MAGMA programto computeBrauer characters,we have foundthat (F2 P4)8 is an extension
A

A

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THE BEHAVIOR

OF THE ALGEBRAIC

TRANSFER

485

A

49
N
M

S

45

35

31
31

/

30'

25

30

~1 \

A

24

4

20
T

0

FIGURE 1. Some GL4-submodules of (F2

A

P4)8s

wherethe 25-dimensionalmodule M is an extension
O - >iA

- -M
--

N

S --0.

The correspondinglattice of submodulesof (F2 OP4)8 is shownin Figure 1. We
A
name the submodulesby theirdimension,using a prime to distinguishthe two
submodulesof dimension30. We label the edges by the correspondingquotient
module. In it, intersections
are shown,but sums are omittedforclarity.That is,
the intersectionof the submodules30' and 35 is the submodule24, but the sum
of 30' and 35 (a submoduleof dimension41) is not shown. The two extensions
above can be seen in the lattice,in the sense that, forexample,the submoduleof

dimension24 is the directsum of the submodulesof dimensions4 and 20, since
theirintersectionis trivial. Further,the quotientof 55 by 24 is the directsum of
the quotientsof 30' by 24 and of 49 by 24.
The generatorsforthesesubmodulesare providedbythesame computerprogram
used to findthisdecompositionand are listedbelow. When all themonomialsin one
of the seven familieslistedin Proposition4.1 appear, we simplywritethe name of
thefamily,
so that,forexample,all the monomialsin familyA are in the submodule

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486

ROBERT R. BRUNER, LE M. HA, AND NGUYEN H. V. HUNG

of dimension20. Also, recallthe element
SG = (4, 2, 1, 1) + (4, 1,2, 1)+ (1, 4, 2, 1)

used in the proofof Proposition4.4. Finally note that elementswhichformbases
forthe subquotientscan be read offby comparingthese lists of generators.For
example,the quotientofthe module30 by the submodule24 is A, and the elements
of familyD generateit.
4: (6, 1,1,0)+ (1,6,1,0) + (1,1,6,0), (6,1,0,1) + (1,6,0,1) + (1,1,0,6),
(6,0,1,1) + (1,0,6,1)+ (1,0,1,6), (0,6,1,1)+ (0,1,6,1)+ (0,1,1,6).
20: (A), (6,11,1,0)+ (1,1,6,0), (6,1,0,1)+ (1, 1,0,6), (6,0,1,1) + (1,0,1,6),
(1,6,1,0) + (1,1,6,0), (1,6,0,1)+ (1,1,0,6), (1,0,6,1)+ (1,0,1,6),
(0,6,1,1) + (0,1,1,6), (0,1,6,1)+ (0,11,1,6).
24: (A) and (C).

25: (A), (C), and sG.

30: (A), (C), and (D).
30': (A), (C) and (5,1,1,1) + (1,5,1,1) + SG+(3,3,1,1),
1, 3, 1), (5, 1,1, 1)+(1, 111, 5)+G+(3,
(5,1,1, 1)+(1, 1,5, 1)+sG+(3,
1,1,3),
3, 3, 1), (1, 5, 1, 1)+(1, 1, 1, 5)+SG+(1, 3,1,3),
(1, 5, 1, 1)+(1, 1, 5, 1)+SG+(1,
(1,1, 5,1) + (1,1,1,5)+ sG + (1,1,3,3).
31: (A),(C), (D) and sG.
35: (A), (C), (D), SG and
(5,2,1,0) + (5,2,0,1)+ (5,0,2,1)+ (5,1,1,1),
(2,5,1,0)+(2,5,0,1)+
(0,5,2,1)+(1,5,1,1),
1),
(2,1,5,0) + (2,0,5,1)+(0,2,5,1)+(1,1,5,
(2,1,0,5) + (2,0,1,5)+ (0,2,1,5)+ (1,1,1,5).
45: (A), (C), (D), (E) and (G).
49: (A), (C), (D), (E), (F) and (G).
ACKNOWLEDGMENT

The researchwas in progressduringthe secondnamed author'svisitto theIHES
(France) and the thirdnamed author's visit to Wayne State University,Detroit
(Michigan)in the academic year2001-2002.
The second named authorthanksLionel Schwartzforhis supportand encouragement.
The thirdnamed author is gratefulto Daniel Frohardt,David Handel, Lowell
Hansen,JohnKlein, Charles McGibbon,Claude Schochetand all colleaguesat the
fortheirhospitalityand for
Departmentof Mathematics,WayneState University,

the warmworkingatmosphere.
The authorsexpresstheirheartythanksto Tran N. Nam forhelpfuldiscussions.
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DEPARTMENTOF MATHEMATICS,
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MICHIGAN48202
E-mail address: rrbOmath.wayne.edu
UNIVERSITEDE LILLE I, UFR DE MATHEMATIQUES,
UMR 8524, 59655 VILLENEUVED'ASCQ
CEDEX, FRANCE
E-mail address:
DEPARTMENTOF MATHEMATICS,
VIETNAMNATIONALUNIVERSITY,334 NGUYENTRAI STREET,
HANOI,VIETNAM
E-mail address: nhvhung@vnu.
edu.vn

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