ON BEHAVIOR OF THE ALGEBRAIC TRANSFER
ROBERT R. BRUNER, L^E M. H`A AND NGUY^E~N H. V. HU_.NG
Dedicated to Professor Hu`ynh M`ui on the occasion of his sixtieth birthday
Abstract.Let Trk : F2 PHi(BVk) ! Extk,k+iA(F2, F2) be the algebraic
GLk
transfer, which is defined by W. Singer as an algebraic version of the ge*
*o-
metrical transfer trk : iS*((BVk)+) ! iS*(S0). It has been shown that the
algebraic transfer is highly nontrivial and, more precisely, that Trk is *
*an iso-
morphism for k = 1, 2, 3. However, Singer showed that Tr5 is not an epi-
morphism. In this paper,swe prove that Tr4 does not detect the non zero
element gs 2 Ext4,12.2A(F2, F2) for every s 1. As a consequence, the lo-
calized (Sq0)-1Tr4 given by inverting the squaring operation Sq0 is not an
epimorphism. This gives a negative answer to a prediction by Minami.
1.Introduction and statement of results
The subject of the present paper is the algebraic transfer
Trk : F2 PHi(BVk) ! Extk,k+iA(F2, F2),
GLk
which is defined by W. Singer as an algebraic version of the geometrical transf*
*er
trk : ßS*((BVk)+) ! ßS*(S0) to the stable homotopy groups of spheres. Here
Vk denotes a k-dimensional F2-vector space, PH*(BVk) is the primitive part
consisting of all elements in H*(BVk), which are annihilated by every positive-
degree operation in the mod 2 Steenrod algebra, A. Throughout the paper, the
homology is taken with coefficients in F2.
It has been proved that Trk is an isomorphism for k = 1, 2 by Singer [13]
and for k = 3 by Boardman [1]. These data together with the fact that Tr =
k 0Trk is an algebra homomorphism (see [13]) show that Trk is highly nontrivia*
*l.
Therefore, the algebraic transfer is considered to be a useful tool for studyin*
*g the
mysterious cohomology of the Steenrod algebra, Ext*,*A(F2, F2). In [13], Singer*
* also
gave computations to show that Tr4 is an isomorphism up to a range of internal
degrees. However, he proved that Tr5 is not an epimorphism.
Based on these data, we are particularly interested in the behavior of the fo*
*urth
algebraic transfer. The following theorem is the main result of this paper.
s
Theorem 1.1. Tr4 does not detect the non zero element gs 2 Ext4,12.2A(F2, F2)
for every s 1.
___________
1The third named author was supported in part by the Vietnam National Researc*
*h Program,
Grant N0140801.
22000 Mathematics Subject Classification. Primary 55P47, 55Q45, 55S10, 55T15.
3Key words and phrases. Adams spectral sequences, Steenrod algebra, Invariant*
* theory, Alge-
braic transfer.
1
2 ROBERT R. BRUNER, L^E M. H`A AND NGUY^E~N H. V. HU_.NG
The reader is referred to May [10] for the generator g1 and to Lin [7] or [8] f*
*or the
generators gs.
As a consequence, we get a negative answer to a prediction by Minami [12].
Corollary 1.2. The localization of the fourth algebraic transfer
(Sq0)-1Tr4 : (Sq0)-1F2 PHi(BV4) ! (Sq0)-1Ext4,4+iA(F2, F2)
GL4
given by inverting Sq0 is not an epimorphism.
It is well-known (see [9]) that there are squaring operations Sqi (i 0) act*
*ing
on the cohomology of the Steenrod algebra, which share most of the properties
with Sqi on the cohomology of spaces. However, Sq0 is not the identity. We refer
to Section 2 for the precise meaning of the operation Sq0 on the domain of the
algebraic transfer.
We next explain the idea of the proof of Theorem 1.1.
Let Pk := H*(BVk) be the polynomial algebra of k variables, each of degree 1.
Then, the domain of Trk, F2 PH*(BVk), is dual to (F2 Pk)GLk. In order to
GLk A
prove Theorem 1.1, it suffices to show that (F2 P4)GL412.2s-4= 0, for every s *
* 1.
A
Direct calculation of (F2 P4)12.2s-4is difficult, as P4 in degree 12 . 2s- 4*
* is
A
a huge F2-vector space, e.g. its dimension is 1771 for s = 1. To compute it, we
observe that the iterated dual squaring operation
(Sq0*)s : (F2 P4)12.2s-4! (F2 Pk)8
A A
is an isomorphism of GL4-modules for any s 1. This isomorphism is obtained by
applying repeatedly the following proposition.
Proposition 1.3. Let k and r be positive intergers. Suppose that each monomial
xi11.x.i.kkof Pk in degree 2r + k with at least one exponent it even is hit. Th*
*en
Sq0*: (F2 Pk)2r+k! (F2 Pk)r
A A
is an isomorphism of GLk-modules.
Here, as usual, we say that a polynomial Q in Pk is hit if it is A-decomposable.
Further, we show that (F2 P4)8 is an F2-vector space of dimension 55. Then,
A
by investigating a specific basis of it, we prove that (F2 P4)GL48= 0. As a co*
*nse-
A
quence, we get (F2 P4)GL412.2s-4= 0 for every s 1.
A
The reader who does not wish to follow the invariant theory computation above
may be satisfied by the following weaker theorem, and then would not need to re*
*ad
the paper's last 3 sections.
Theorem 1.4. Tr4 is not an isomorphism.
This theorem is proved by observing that, on the one hand,
(F2 P4)GL420~=(F2 P4)GL48,
A A
and on the other hand,
Ext4,4+20A(F2, F2) = F2. g1 6~=Ext4,4+8A(F2, F2) = 0.
ON BEHAVIOR OF THE ALGEBRAIC TRANSFER 3
The paper is divided into six sections and organized as follows. Section 2 st*
*arts
with a recollection of the squaring operation and ends with a proof of the iso-
morphism (F2 P4)12.2s-4~=(F2 Pk)8. Theorem 1.4 is proved in Section 3. We
A A
compute (F2 P4)8 and its GL4-invariants in Section 4. We prove Theorem 1.1 in
A
Section 5. Finally, in Section 6, we describe the GL4-module structure of (F2 *
*P4)8.
A
Acknowledgment: The research was in progress during the second named author's
visit to the IHES (France) and the third named author's visit to Wayne State
University, Detroit (Michigan) in the academic year 2001-2002.
The third named author is grateful to Daniel Frohardt, David Handel, Lowell
Hansen, John Klein, Charles McGibbon, Claude Schochet and all colleagues at the
Department of Mathematics, Wayne State University for their hospitality and for
the warm working atmosphere.
The authors express their hearty thanks to Tr^a`n N. Nam for helpful discussi*
*on.
2. A sufficient condition for the squaring operation to be an
isomorphism
This section starts with a recollection of Kameko's squaring operation
Sq0 : F2 PH*(BVk) ! F2 PH*(BVk).
GLk GLk
The most important property of Kameko's Sq0 is that it commutes with the classi*
*cal
Sq0 on Ext*A(F2, F2) (defined in [9]) through the algebraic transfer (see [1], *
*[12]).
This squaring operation is constructed as follows.
As well known, H*(BVk) is the polynomial algebra, Pk := F2[x1, ..., xk], on k
generators x1, ..., xk, each of degree 1. By dualizing,
H*(BVk) = (a1, . .,.ak)
is the divided power algebra generated by a1, . .,.ak, each of degree 1, where *
*ai
is dual to xi 2 H1(BVk). Here the duality is taken with respect to the basis of
H*(BVk) consisting of all monomials in x1, . .,.xk.
In [6] Kameko defined a homomorphism
fSq0: H*(BVk) ! H*(BVk),
a(i1)1.a.(.ik)k7!a(2i1+1)1.a.(.2ik+1)k,
where a(i1)1.a.(.ik)kis dual to xi11.x.i.kk. The following lemma is well known.*
* We
give a proof to make the paper self-contained.
0
Lemma 2.1. fSq is a GLk-homomorphism.
0
Proof.We use the explanation of fSqby Crabb and Hubbuck [3], which does not
depend on the chosen basis of H*(BVk). The element a(Vk) = a1. .a.kis nothing
but the image of the generator of k(Vk) under the (skew) symmetrization map
k(Vk) ! Hk(BVk) = k(Vk) = ( Vk___._.-.Vkz____")Sk,
k times
where the symmetric group Sk acts on Vk . . .Vk by permutations of the factors.
Let c : H*(BVk) ! H*(BVk) be the degree-halving epimorphism, which is dual to
4 ROBERT R. BRUNER, L^E M. H`A AND NGUY^E~N H. V. HU_.NG
the Frobenius monomorphism F : H*(BVk) ! H*(BVk) defined by F(x) = x2 for
any x. We have
fSq0(c(y)) = a(Vk)y,
for y 2 H*(BVk). To prove that this is well defined we 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 2 H*(BVk). Here < ., .P> denotes the dual pairing between H*(BVk) and
H*(BVk). So, if we write y = a(i1)1.a.(.ik)k, then there is at least one itwh*
*ich is
odd in each term of the sum. Therefore,
X (i ) (i )
a(Vk)y = a1. .a.k( a1 1. .a.kk) = 0,
0
because ata(it)t= 0 for any odd it. So, fSqis well defined.
0
As c is a GLk-epimorphism, the map fSqis a GLk-homomorphism.
The lemma is proved.
Further, it is easy to see that cSq2t+1*= 0, cSq2t*= Sqt*c. So we have
0 2t 0 0 t
Sq2t+1*fSq= 0, Sq* fSq= fSqSq*.
0
(See [4] for an explicit proof.) Therefore, fSqmaps PH*(BVk) to itself.
Kameko's Sq0 is defined by
0
Sq0 = 1 fSq : F2 PH*(BVk) ! F2 PH*(BVk).
GLk GLk GLk
0 0
The dual homomorphism fSq*: Pk ! Pk of fSqis obviously given by
( j1-1_ jk-1_
fSq0*(xj11.x.j.kk) = x12 . .x.2k ,j1, ..., jk odd,
0, otherwise.
Hence
0 _____
Ker(fSq*) = Even,
_____ i i
where Evendenotes the vector subspace of Pk spanned by all monomials x11. .x.kk
with at least one exponent iteven.
0
Let s : Pk ! Pk be a left inverse of fSq*defined as follows:
s(xi11.x.i.kk) = x2i1+11.x.2.ik+1k.
It should be noted that s does not commute with the doubling map on A, that is,
in general
Sq2ts 6= sSqt.
However, in one particular circumstance we have the following.
Lemma 2.2. Under the hypothesis of Proposition 1.3, the map
_s: (F
2 Pk)r! (F2 Pk)2r+k
A_ A
s[X] = [sX]
is a well-defined linear map.
ON BEHAVIOR OF THE ALGEBRAIC TRANSFER 5
Proof.We start with an observation that
_____
Im(Sq2ts - sSqt) Even.
We prove this by showing equivalently that
fSq0*(Sq2ts - sSqt) = 0.
Indeed,
fSq0*(Sq2ts - sSqt)=fSq0*Sq2ts - fSq0*sSqt
0 0 t
= SqtfSq*s - fSq*sSq
= Sqt. id - id . Sqt
= 0.
_____
As a consequence, s maps (A+Pk)r to (A+Pk + Even)2r+k. Here and in what
follows, A+ denotes the submodule of A consisting of all positive degree operat*
*ions.
Further, by the hypothesis of Proposition 2.3, we have
_____ +
(A+Pk + Even)2r+k (A Pk)2r+k.
Hence, s maps (A+Pk)r to (A+Pk)2r+k. So the map _sis well-defined. Then it is a
linear map, as s is.
The lemma is proved.
The following proposition is also numbered as Proposition 1.3
Proposition 2.3. Let k and r be positive intergers. Suppose that each monomial
xi11.x.i.kkof Pk in degree 2r + k with at least one exponent it even is hit. Th*
*en
Sq0*: (F2 Pk)2r+k! (F2 Pk)r
A A
is an isomorphism of GLk-modules.
0
Proof.On the one hand, we have Sq0*_s= id(F2 Pk)r. Indeed, from fSq*s = idPk, *
*it
A
implies
0
Sq0*_s[X] = Sq0*[sX] = [fSq*sX] = [X],
for any X in degree r of Pk.
On the other hand, we have _sSq0*= id(F2 Pk)2r+k. Indeed, by the hypoth-
A
esis, any monomial with at least one even exponent represents the 0 class in
(F2 Pk)2r+k, so we need only to check on the classes of monomials with all exp*
*o-
A
nents odd. We have
_sSq0 2i1+1 2ik+1 _ i1 ik
*[x1 . .x.k ]= s[x1 . .x.k]
= [s(xi11.x.i.kk)]
= [x2i1+11.x.2.ik+1k],
for any x2i1+11.x.2.ik+1kin degree 2r + k of Pk.
Combining the above two equalities, Sq0*_s= id(F2 Pk)rand _sSq0*= id(F2 Pk)*
*2r+k,
A A
we see that Sq0*: (F2 Pk)2r+k ! (F2 Pk)r is an isomorphism with inverse
_ A A
s: (F2 Pk)r ! (F2 Pk)2r+k.
A A
The proposition is proved.
6 ROBERT R. BRUNER, L^E M. H`A AND NGUY^E~N H. V. HU_.NG
The target of this section is the following.
Lemma 2.4. For every positive integer s,
(Sq0*)s : (F2 P4)12.2s-4! (F2 P4)8
A A
is an isomorphism of GL4-modules.
Proof.By using Proposition 2.3 repeatedly, it suffices to show that any monomial
of P4 in degree m = 12 . 2s- 4 with at least one even exponent is hit. Since m
is even, the number of even exponents in such a monomial must be either 2 or 4.
If all exponents of the monomial are even, then it is hit by Sq1. Hence we need
only to consider the case of a monomial R with exactly two even exponents (and
so exactly two odd exponents). We can write, up to a permutation of variables,
R = x1x2Q2, where Q is a monomial in degree 6 . 2s- 3.
Let Ø be the anti-homomorphism in the Steenrod algebra. The so-called Ø-trick,
which is known to Brown and Peterson in the mid-sixties, states that
uSqi(v) Ø(Sqi)(u)v mod A+M,
for u, v in any A-algebra M. (See also Wood [14].) In our case, it claims that
s-3
R = x1x2Sq6.2 (Q)
is hit if and only if Ø(Sq6.2s-3)(x1x2)Q is. We will show Ø(Sq6.2s-3)(x1x2) = 0*
* for
any s > 0.
As A is a commutative coalgebra, Ø is a homomorphism of coalgebras (see [11,
Proposition 8.6]). Then we have the Cartan formula
X
Ø(Sqn)(uv) = Ø(Sqi)(u)Ø(Sqj)(v).
i+j=n
Furthermore, it is shown by Brown and Peterson in [2] that
æ 2p p
Ø(Sqi)(xj) = xj0 ifio=t2h-e1rforwsome,pise,
for xj in degree 1. So, in order to prove Ø(Sq6.2s-3)(x1x2) = 0 we need only to
show that 6 . 2s- 3 can not be written in the form
6 . 2s- 3 = (2a - 1) + (2b- 1)
with a b. Indeed, if we have this equality, then b = 0 as the left hand size *
*is odd.
So 6 . 2s- 3 = 2a- 1, or equivalently 3 . 2s = 2a-1+ 1. As s > 0, the left hand*
* side
is even, hence a - 1 = 0. It implies 3 . 2s = 2. This equality has no solution *
*s.
The lemma is proved.
3. The fourth algebraic transfer is not an isomorphism
The target of this section is to prove the following theorem, which is also n*
*um-
bered as Theorem 1.4.
Theorem 3.1.
Tr4 : F2 PHi(BV4) ! Ext4,4+iA(F2, F2)
GL4
is not an isomorphism.
Proof.For any r, we have a commutative diagram
ON BEHAVIOR OF THE ALGEBRAIC TRANSFER 7
(F2 PHi(BV4))r _________-Tr4Ext4,4+r(F , F )
GL4 A 2 2
| |
| 0 | 0
|Sq |Sq
| |
|? |?
(F2 PHi(BV4))2r+4________-Tr4Ext4,8+2r(F , F ) ,
GL4 A 2 2
where the first vertical arrow is the Kameko Sq0 and the second vertical one is*
* the
classical Sq0.
The dual statement of Lemma 2.4 for s = 2 claims that
Sq0 : (F2 PHi(BV4))8 ! (F2 PHi(BV4))20
GL4 GL4
is an isomorphism. On the other hand, it is known (May [10]) that
Ext4,4+8A(F2, F2) = 0 6~=Ext4,4+20A(F2, F2) = F2. g1.
This implies that Tr4 is not an isomorphism. The theorem is proved.
Remark 3.2. This proof does not show whether Tr4 fails to be a monomorphism
or fails to be an epimorphism. We will see that actually Tr4 is not an epimorph*
*ism
in Section 5 below.
4.GL4-invariants of the indecomposables of P4 in degree 8
From now on, let us write x = x1, y = x2, z = x3 and t = x4 and denote the
monomial xaybzctd by (a, b, c, d) for abbreviation.
Proposition 4.1. (F2 P4)8 is an F2-vector space of dimension 55 with a basis
A
consisting of the classes represented by the following monomials:
(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),
(C)(6, 1, 1, 0), (6, 1, 0, 1), (6, 0, 1, 1), (1, 6, 1, 0), (1, 6, 0, 1), (1,*
* 1, 6, 0),
(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), (5, 0, 3, 0), (5, 0, 0, 3), (0, 5, 3, 0), (0, 5, 0, 3), (0,*
* 0, 5, 3),
(E)(5, 2, 1, 0), (5, 2, 0, 1), (5, 0, 2, 1), (2, 5, 1, 0), (2, 5, 0, 1), (2,*
* 1, 5, 0),
(2, 1, 0, 5), (2, 0, 5, 1), (2, 0, 1, 5), (0, 5, 2, 1), (0, 2, 5, 1), (0,*
* 2, 1, 5),
(F)(5, 1, 1, 1), (1, 5, 1, 1), (1, 1, 5, 1), (1, 1, 1, 5),
(G) (4, 2, 1, 1), (4, 1, 2, 1), (1, 4, 2, 1).
The proposition is proved by combining a couple of lemmas.
Lemma 4.2. (F2 P4)8 is generated by the 55 elements listed in Proposition 4.1.
A
8 ROBERT R. BRUNER, L^E M. H`A AND NGUY^E~N H. V. HU_.NG
Proof.It is easy to see that every monomial (a, b, c, d) with a, b, c, d all ev*
*en is hit
(more precisely by Sq1).
The only monomials (a, b, c, d) in degree 8 with at least one of a, b, c, d o*
*dd are
the following up to permutations of the variables:
(7, 1, 0, 0), (3, 3, 1, 1), (6, 1, 1, 0), (5, 3, 0, 0), (5, 2, 1, 0), (5, *
*1, 1, 1), (4, 2, 1, 1),
(4, 3, 1, 0), (3, 3, 2, 0), (3, 2, 2, 1).
The last 3 monomials and their permutations are expressed in terms of the fir*
*st
7 monomials and their permutations as follows:
(4, 3, 1,=0) (2, 5, 1, 0) + Sq4(1, 2, 1, 0) + Sq2(2, 3, 1, 0),
(3, 3, 2,=0) (5, 2, 1, 0) + (2, 5, 1, 0) + Sq4(2, 1, 1, 0) + Sq4(1, 2, 1, 0)
+Sq2(3, 2, 1, 0) + Sq2(2, 3, 1, 0) + Sq1(3, 3, 1, 0),
(3, 2, 2,=1) (5, 1, 1, 1) + (4, 2, 1, 1) + (4, 1, 2, 1)
+Sq2(3, 1, 1, 1) + Sq1(4, 1, 1, 1) + Sq1(3, 2, 1, 1) + Sq1(3, 1, *
*2, 1).
Hence, (F2 P4)8 is generated by the following 7 monomials and their permuta-
A
tions:
(7, 1, 0, 0), (3, 3, 1, 1), (6, 1, 1, 0), (5, 3, 0, 0), (5, 2, 1, 0), (5, 1*
*, 1, 1), (4, 2, 1, 1).
By the family of a monomial (a, b, c, d) we mean the set of all monomials whi*
*ch
are obtained from (a, b, c, d) by permutations of the variables.
The monomials in the 7 families above which are not in Proposition 4.1 can be
expressed in terms of the 55 elements listed there as follows. (We give only one
expression from each symmetry class.)
(3, 5, 0,=0) (5, 3, 0, 0) + Sq4(2, 2, 0, 0) + Sq2(3, 3, 0, 0),
(5, 1, 2,=0) (6, 1, 1, 0) + (5, 2, 1, 0) + Sq1(5, 1, 1, 0),
(4, 1, 1,=2) (4, 2, 1, 1) + (4, 1, 2, 1) + Sq1(4, 1, 1, 1),
(2, 4, 1,=1) (4, 2, 1, 1) + Sq4(1, 1, 1, 1) + Sq2(2, 2, 1, 1),
(2, 1, 1,=4) (4, 2, 1, 1) + (4, 1, 2, 1)
+Sq4(1, 1, 1, 1) + Sq2(2, 1, 1, 2) + Sq1(4, 1, 1, 1),
(1, 4, 1,=2) (4, 2, 1, 1) + (1, 4, 2, 1)
+Sq4(1, 1, 1, 1) + Sq2(2, 2, 1, 1) + Sq1(1, 4, 1, 1),
(1, 2, 1,=4) (4, 2, 1, 1) + (1, 4, 2, 1)
+Sq2(2, 2, 1, 1) + Sq2(1, 2, 1, 2) + Sq1(1, 4, 1, 1),
(1, 1, 4,=2) (4, 1, 2, 1) + (1, 4, 2, 1)
+Sq2(2, 1, 2, 1) + Sq2(1, 2, 2, 1) + Sq1(1, 1, 4, 1),
(1, 1, 2,=4) (4, 1, 2, 1) + (1, 4, 2, 1) + Sq4(1, 1, 1, 1)
+Sq2(2, 1, 2, 1) + Sq2(1, 2, 2, 1) + Sq2(1, 1, 2, 2) + Sq1(1, 1, *
*4, 1).
The lemma is proved.
Lemma 4.3. The 55 elements listed in Proposition 4.1 are linearly independent in
(F2 P4)8.
A
ON BEHAVIOR OF THE ALGEBRAIC TRANSFER 9
Proof.We will use an equivalence relation defined by saying that, for two polyn*
*o-
mials P and Q, P is equivalent to Q, denoted by P ~ Q, if P - Q is hit.
If X is one of the letters from A to G, let Xi be the i-th element in family X
according to the order listed in Proposition 4.1. (This is the lexicographical *
*order
in each family.)
Suppose there is a linear relation between the 55 elements listed there
12X X6 12X X6 12X X4 X3
aiAi+ biBi+ ciCi+ diDi+ eiEi+ fiFi+ giGi= 0,
i=1 i=1 i=1 i=1 i=1 i=1 i=1
where ai, bi, ci, di, ei, fi, gi 2 F2. We need to show all these coefficients a*
*re zero.
The proof is divided into 4 steps.
Step 1. We call a monomial a spike if each of its exponents is of the form 2n -*
* 1
for some n. It is well known that spikes do not appear in thenexpression of SqiY
for any i positive and any monomial Y , since the powers x2 -1are not hit in the
one variable case. Hence, the coefficient of any spike is zero in every linear *
*relation
in F2 Pk.
A
Among the 55 elements of Proposition 4.1, the classes of families A and B are
spikes. So ai= bj= 0, for every i and j. Then, we get
X12 X6 12X X4 X3
ciCi+ diDi+ eiEi+ fiFi+ giGi= 0.
i=1 i=1 i=1 i=1 i=1
Step 2. Consider the homomorphism F2 P4 ! F2 P2 induced by the projection
A A
P4 ! P4=(z, t) ~=P2. Under this homomorphism, the image of the above linear
relation is d1(5, 3) = 0.
In order to show d1 = 0, we need to prove that (5, 3) is non zero in F2 P2. *
*The
A
linear transformation x 7! x, y 7! x+y sents (5,3) to (8, 0)+(7, 1)+(6, 2)+(5, *
*3) ~
(7, 1) + (5, 3). As the action of the Steenrod algebra commutes with linear map*
*s, if
(5, 3) is hit then so is (7, 1) + (5, 3). But it is impossible, because (7, 1) *
*is a spike.
Hence, (5, 3) 6= 0 in F2 P2 and d1 = 0.
A
Similarly, using all the projections of P4 to its quotients by the ideals gen*
*erated
by each pair of the four variables, we get di= 0 for every i. So we get
12X 12X X4 X3
ciCi+ eiEi+ fiFi+ giGi= 0.
i=1 i=1 i=1 i=1
Step 3. Consider the homomorphism F2 P4 ! F2 P3 induced by the projection
A A
P4 ! P4=(t) ~=P3. Under this homomorphism, the linear relation above is sent to
c1(6, 1, 1) + c4(1, 6, 1) + c6(1, 1, 6) + e1(5, 2, 1) + e4(2, 5, 1) + e6(2, *
*1, 5) = 0.
Applying the linear map x 7! x, y 7! x, z 7! y to this relation, we obtain
(c1+ c4+ e1+ e4)(7, 1) + c6(2, 6) +=e6(3, 5)
(c1+ c4+ e1+ e4)(7, 1) + e6(3,=5)0.
Since (7, 1) is a spike, (c1+ c4+ e1+ e4) = 0, hence e6(3, 5) = 0. As for (5, 3*
*), we
can show (3, 5) 6= 0 2 F2 P2 and get e6 = 0.
A
10 ROBERT R. BRUNER, L^E M. H`A AND NGUY^E~N H. V. HU_.NG
By similar arguments, we have e1 = e4 = e6 = 0. The equality (c1+c4+e1+e4) =
0 shows c1+ c4 = 0 or c1 = c4. By similar arguments, c1 = c4 = c6. We denote
this common coefficient by c and get
c{(6, 1, 1) + (1, 6, 1) + (1, 1, 6)} = 0.
We prove c = 0 by showing (6, 1, 1) + (1, 6, 1) + (1, 1, 6) 6= 0. Suppose the c*
*ontrary,
that (6, 1, 1)+(1, 6, 1)+(1, 1, 6) is hit. Then, by the unstable property of th*
*e action
of A on the polynomial algebra, we have
(6, 1, 1) + (1, 6, 1) + (1, 1, 6) = Sq1(P) + Sq2(Q) + Sq4(R),
for some polynomials P, Q, R. By the degree information, Sq4(R) = R2 and this
element is hit by Sq1. Therefore, it suffices to suppose (6, 1, 1)+(1, 6, 1)+(1*
*, 1, 6) =
Sq1(P) + Sq2(Q).
Let Sq2Sq2Sq2 act on the both sides of this equality. The right hand side is *
*sent
to zero, as Sq2Sq2Sq2 annihilates Sq1 and Sq2. On the other hand,
Sq2Sq2Sq2{(6, 1, 1) + (1, 6, 1) + (1, 1, 6)} = (8, 4, 2) + symmetries6= 0.
This is a contradiction. So, it implies (6, 1, 1) + (1, 6, 1) + (1, 1, 6) 6= 0 *
*and c = 0.
We get
X4 X3
fiFi+ giGi= 0.
i=1 i=1
Step 4. Apply the linear map x 7! x, y 7! y, z 7! y, t 7! y to the above equali*
*ty and
we have
f1(5, 3) + (f2+ f3+ f4+ g3)(1, 7) + (g1+ g2)(4,=4)
f1(5, 3) + (f2+ f3+ f4+ g3)(1,=7)0.
As (7, 1) is a spike, we obtain (f2+f3+f4+g3) = 0 and f1(5, 3) = 0. As (5, 3) 6*
*= 0,
it yields f1 = 0.
P Next, applyPthe linear map x 7! x, y 7! y, z 7! x, t 7! x to the equality
3
i6=1fiFi+ i=1giGi= 0 and we have
f2(3, 5) + (f3+ f4+ g2)(7, 1) + g1(6, 2)=+ g3(4, 4)
f2(3, 5) + (f3+ f4+ g2)(7,=1)0.
As (7, 1) is a spike, we get (f3+ f4+ g2) = 0 and f2(3, 5) = 0. Since (3, 5) 6=*
* 0, it
implies f2 = 0.
Similarly,Papply the linear map x 7! x, y 7! x, z 7! y, t 7! x to the equality
f3F3+ f4F4+ 3i=1giGi= 0 and we have
f3(3, 5) + (f4+ g1)(7, 1) + (g2+ g3)(6,=2)
f3(3, 5) + (f4+ g1)(7,=1)0.
As (7, 1) is a spike, we get f4+ g1 = 0 and then f3 = 0.
Finally,Papply the linear map x 7! x, y 7! x, z 7! x, t 7! y to the equality
f4F4+ 3i=1giGi= 0 and we have
f4(3, 5) + (g1+ g2+ g3)(7,=1)
f4(3, 5) + (g1+ g2+ g3)(7,=1)0.
As (7, 1) is a spike, we get g1+ g2+ g3 = 0 and then f4 = 0.
ON BEHAVIOR OF THE ALGEBRAIC TRANSFER 11
Substituting f1 = f2 = f3 = f4 = 0 into the equations (f2+ f3+ f4+ g3) = 0,
(f3+ f4+ g2) = 0, f4+ g1 = 0, we get g1 = g2 = g3 = 0.
We have shown that all coefficients of an arbitrary linear relation between t*
*he
55 elements listed in Proposition 4.1 are zero. The lemma follows.
Combining Lemmas 4.2 - 4.3, we get Proposition 4.1.
Proposition 4.4. (F2 P4)GL48= 0.
A
Proof.If X is one of the letters A, B, C, D, E, F, G, let L(X) be the vector su*
*bspace
of (F2 P4)8spanned by the elements of family X in Proposition 4.1. Let Sk deno*
*te
A
the symmetric subgroup of GLk. According to the relations listed in the proof
of Lemma 4.2, L(A), L(B), L(C), L(D), L(F), L(G) are S4-submodules. The
subspace L(E) is not an S4-submodule. However, the sum
L(C, E) = L(C) L(E)
is. We have a decomposition of S4-modules
(F2 P4)8 = L(A) L(B) L(C, E) L(D) L(F) L(G).
A
Let ff be an arbitrary GL4-invariant in (F2 P4)8. It can uniquely be written*
* in
A
the form
ff = ffA + ffB + ffC,E+ ffD + ffF + ffG,
where ffX 2 L(X) for X 2 {A, B, D, F, G}, and ffC,E2 L(C, E). Each term of this
sum is S4-invariant.
Note that if a linear combination of elements in a family is S4-invariant, th*
*en all
of its coefficients are equal, because each element in the family can be obtain*
*ed from
any other by a suitable permutation. Let sX denote the sum of all the elements
in the family X listed in Proposition 4.1. Then, we have ffA = asA, ffB = bsB,
ffD = dsD , ffF = fsF, ffG = gsG, and ffC,E= csC+esE, where a, b, c, d, e, f, g*
* 2 F2.
Let p be the transposition given by p(x) = y, p(y) = x, p(z) = z, p(t) = t. I*
*t 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).
Further, the 10 elements different from (2, 1, 0, 5) and (2, 1, 5, 0) in family*
* E are
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 family C is full, in th*
*e sense
that it contains all the variable permutations of a monomial, we have p(sC) = s*
*C.
Hence, we get
p(ffC,E) = p(csC + esE) = csC + esE + e(1, 1, 0, 6) + e(1, 1, 6, 0).
As ffC,E is S4-invariant, e(1, 1, 0, 6) + e(1, 1, 6, 0) = 0. So e = 0, because *
*the two
elements are linearly independent by Lemma 4.3. We obtain
ff = ffA + ffB + ffC + ffD + ffF + ffG,
where ffC = ffC,E= csC.
12 ROBERT R. BRUNER, L^E M. H`A AND NGUY^E~N H. V. HU_.NG
Let us now consider the transvection ' given by '(x) = x, '(y) = y, '(z) = z,
'(t) = x + t. A routine computation shows
'(sA) = sA + (7, 1, 0, 0) + (7, 0, 1, 0) + (7, 0, 0, 1) + (1, 7, 0, 0) + (1*
*, 0, 7, 0)
+(6, 1, 0, 1) + (6, 0, 1, 1) + (1, 1, 0, 6) + (1, 0, 1, 6),
'(sB) = sB + (6, 1, 1, 0) + (1, 6, 1, 0) + (1, 1, 6, 0) + (2, 5, 1, 0) + (2*
*, 1, 5, 0)
+(5, 1, 1, 1) + (1, 5, 1, 1) + (1, 1, 5, 1) + (4, 2, 1, 1) + (4, 1,*
* 2, 1)
+(3, 3, 1, 1) + (3, 1, 3, 1),
'(sC) = sC + (6, 1, 1, 0) + (1, 6, 1, 0) + (1, 1, 6, 0),
'(sD )= sD + (7, 0, 0, 1) + (1, 6, 0, 1) + (1, 0, 6, 1) + (5, 3, 0, 0) + (5*
*, 0, 3, 0),
'(sF) = sF + (2, 5, 1, 0) + (2, 1, 5, 0) + (5, 1, 1, 1) + (4, 2, 1, 1) + (4*
*, 1, 2, 1),
'(sG) = sG + (6, 1, 1, 0).
Let rX = '(sX )-sX where X is one of the letters A, B, C, D, F, G. The equali*
*ty
'(ff) = ff is rewritten as
'(asA + bsB + csC + dsD + fsF + gsG) = asA + bsB + csC + dsD + fsF + gsG,
or equivalently
arA + brB + crC + drD + frF + grG = 0.
In this linear combination, rB and rD are the only terms containing (3, 3, 1,*
* 1)
in family B and (5, 3, 0, 0) in family D respectively. From Lemma 4.3, we get
b = d = 0, and therefore arA + crC + frF + grG = 0.
In the new linear combination, as rA, rC and rF are the only terms containing
(7, 1, 0, 0) in family A, (1, 6, 1, 0) in family C and (4, 2, 1, 1) in family F*
* respectively,
we have a = c = f = 0. As a consequence, grG = 0, so we finally get g = 0.
In summary, we have shown that every GL4-invariant ff in (F2 P4)8 equals
A
zero. The proposition is proved.
5. The fourth algebraic transfer is not an epimorphism
The goal of this paper is to prove the following theorem, which is also numbe*
*red
as Therem 1.1.
Theorem 5.1.
Tr4 : F2 PHi(BV4) ! Ext4,4+iA(F2, F2)
GL4 s
does not detect the non zero elements gs 2 Ext4,12.2A(F2, F2) for every s 1.
Proof.Combining Lemma 2.4 and Proposition 4.4 we get
(F2 P4)GL412.2s-4= 0,
A
for every non negative integer s.
On the other hand, it is well known that Ext4,24A(F2, F2) is spanned by thesg*
*ener-
ator g1 (see May [10]). Further, gs = (Sq0)s-1(g1) is non zero in Ext4,12.2A(F2*
*, F2)
(see Lin [7] and also [8]).
As F2 PH12.2s-4(BV4) is dual to (F2 P4)GL412.2s-4,
GL4 A
s
Tr4 : F2 PH12.2s-4(BV4) ! Ext4,12.2A(F2, F2)
GL4
ON BEHAVIOR OF THE ALGEBRAIC TRANSFER 13
does not detect the generator gs, for every non negative integer s.
The theorem is proved.
As a consequence, we get a negative answer to a prediction by Minami [12].
(This corollary is also numbered as Corollary 1.2.)
Corollary 5.2. The localization of the fourth algebraic transfer
(Sq0)-1Tr4 : (Sq0)-1F2 PHi(BV4) ! (Sq0)-1Ext4,4+iA(F2, F2)
GL4
given by inverting Sq0 is not an epimorphism.
Proof.Indeed, it does not detect the non zero element g, which is represented by
the family (gs)s>0with gs = (Sq0)s-1(g1). The corollary follows.
Remark 5.3. Our result does not affect Singer's conjecture that the k-th algebr*
*aic
transfer is a monomorphism for every k. (See [13].)
6.Final Remark: GL4-module structure
Boardman's study of the 3 variable problem shows that the GLk module struc-
ture of F2 Pk may be a useful tool. In this vein we close with a description o*
*f the
A
module (F2 P4)8 as a GL4-module. From the öM dular Atlas" [5] we find that
A
there are 8 irreducible modules for GL4 in characteristic 2, of dimensions 1, 4*
*, 4,
6, 14, 20, 20, and 64. With a little calculation we find the following descript*
*ion of
them:
1:the trivial module F2
N: the natural module F42,
N*: the dual of the natural module,
:the alternating square of N or N*,
S:the nontrivial constituent of N N*, which has composition factors 1, S,*
* 1,
T: a constituent of N , which has composition factors N* and T,
T*:a constituent of N* , which has composition factors N and T*,
St:the Steinberg module.
Using a "meataxe" program written in MAGMA, together with a MAGMA pro-
gram to compute Brauer characters, we have found that (F2 P4)8 is an extension
A
0 -! N* T -! (F2 P4)8 -! M -! 0,
A
where the 25-dimensional module M is an extension
0 -! 1 -! M -! N S -! 0.
The corresponding lattice of submodules of (F2 P4)8 is shown in Figure 1. We
A
name the submodules by their dimension, using a prime to distinguish the two
submodules of dimension 30. We label the edges by the corresponding quotient
module. In it, intersections are shown, but sums are omitted for clarity. That *
*is,
the intersection of the submodules 300and 35 is the submodule 24, but the sum
of 300 and 35 (a submodule of dimension 41) is not shown. The two extensions
above can be seen in the lattice, in the sense that, for example, the submodule*
* of
dimension 24 is the direct sum of the submodules of dimensions 4 and 20, since
14 ROBERT R. BRUNER, L^E M. H`A AND NGUY^E~N H. V. HU_.NG
55OO
OOOO
OOOO
OOOOO
O
"49??
N """ ??S?
""" ???
"" ?
M 45? 35
??? """
S?? N""
??? """
"
"31??
""" ??1?
""" ???
"" ?
300TTTT 25? 30
TTTT ??? """
TTTTT 1?? ""
TTTTTTT ??? """
TTTT "
"24??
T"""" ??N*??
"""" ???
4?? 20"
??N*? T """
?? ""
?? """
0
Figure 1. Some GL4-submodules of (F2 P4)8.
A
their intersection is trivial. Further, the quotient of 55 by 24 is the direct *
*sum of
the quotients of 30' by 24 and of 49 by 24.
The generators for these submodules are provided by the same computer program
used to find this decomposition and are listed below. When all the monomials in*
* one
of the seven families listed in Proposition 4.1 appear, we simply write the nam*
*e of
the family, so that, for example, all the monomials in family A are in the subm*
*odule
of dimension 20. Also, recall the element
sG = (4, 2, 1, 1) + (4, 1, 2, 1) + (1, 4, 2, 1)
used in the proof of Proposition 4.4. Finally note that elements which form bas*
*es
for the subquotients can be read off by comparing these lists of generators. For
example, the quotient of the module 30 by the submodule 24 is , and the elemen*
*ts
of family D generate it.
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, 1, 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, 1, 1, 6).
ON BEHAVIOR OF THE ALGEBRAIC TRANSFER 15
24:(A) and (C).
25:(A), (C), and sG.
30:(A), (C), and (D).
300:(A), (C) and (5, 1, 1, 1) + (1, 5, 1, 1) + sG + (3, 3, 1, 1),
(5, 1, 1, 1)+(1, 1, 5, 1)+sG+(3, 1, 3, 1), (5, 1, 1, 1)+(1, 1, 1, 5)+sG+(*
*3, 1, 1, 3),
(1, 5, 1, 1)+(1, 1, 5, 1)+sG+(1, 3, 3, 1), (1, 5, 1, 1)+(1, 1, 1, 5)+sG+(*
*1, 3, 1, 3),
(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),
(2, 1, 5, 0) + (2, 0, 5, 1) + (0, 2, 5, 1) + (1, 1, 5, 1),
(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).
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16 ROBERT R. BRUNER, L^E M. H`A AND NGUY^E~N H. V. HU_.NG
DEPARTMENT OF MATHEMATICS, WAYNE STATE UNIVERSITY
656 W. KIRBY STREET, DETROIT, MI 48202 (USA)
ROBERT R. BRUNER: rrb@math.wayne.edu
NGUY^E~N H. V. HU_.NG: nhvhung@math.wayne.edu
IHES, F-91440, BURE-SUR-YVETTE, FRANCE
L^E M. H`A: lha@ihes.fr