KAMEKOS HOMOMORPHISM AND THE ALGEBRAIC TRANSFER

Abstract. Let Pk be the graded polynomial algebra F2x1, x2, . . . , xk, with the degree of each xi being 1, regarded as a module over the mod2 Steenrod algebra A, and let GLk be the general linear group over the prime field F2. We study the algebraic transfer constructed by Singer 20, using the technique of the hit problem. This transfer is a homomorphism from the homology of the mod2 Steenrod algebra, TorA k,k+n (F2, F2), to the subspace of F2⊗APk consisting of all the GLkinvariant classes of degree n. In this paper, we develop Hưng’s result in 11 on the relation between the algebraic transfer and Kameko’s homomorphism. Using this result, we show that Singer’s conjecture for the algebraic transfer is true in the case k = 5 and the degree 7.2 s − 5 with s an arbitrary positive integer.

ALGEBRAIC TRANSFER

NGUYỄN SUM†AND NGUYỄN KH ĂC TÍN´ ‡

Abstract Let P kbe the graded polynomial algebra F 2[x1, x2, , x k], with

the degree of each x ibeing 1, regarded as a module over the mod-2 Steenrod

algebra A, and let GL k be the general linear group over the prime field F 2

We study the algebraic transfer constructed by Singer [20], using the technique

of the hit problem This transfer is a homomorphism from the homology of

the mod-2 Steenrod algebra, TorAk,k+n(F 2, F2 ), to the subspace of F 2 ⊗ AP k

consisting of all the GL k -invariant classes of degree n.

In this paper, we develop Hưng’s result in [11] on the relation between the

algebraic transfer and Kameko’s homomorphism Using this result, we show

that Singer’s conjecture for the algebraic transfer is true in the case k = 5 and

the degree 7.2 s − 5 with s an arbitrary positive integer.

coho-Being the cohomology of a space, P k is a module over the mod-2 Steenrod

al-gebra, A The action of A on P k is determined by the elementary properties of

the Steenrod squares Sq i and subject to the Cartan formula (see Steenrod andEpstein [22])

Let GL k be the general linear group over the field F2 This group acts naturally

on P k by matrix substitution Since the two actions of A and GL k upon P kcommute

with each other, there is an inherited action of GL k on F2⊗AP k

Denote by (P k)n the subspace of P k consisting of all the homogeneous

polyno-mials of degree n in P k and by (F2⊗AP k)n the subspace of F2⊗AP k consisting

of all the classes represented by the elements in (P k)n In [20], Singer defined thealgebraic transfer, which is a homomorphism

ϕ k : TorAk,k+n(F2, F2) −→ (F2⊗AP k)GL k

n

from the homology of the Steenrod algebra to the subspace of (F2⊗AP k)nconsisting

of all the GL k-invariant classes It is a useful tool in describing the homology groups

of the Steenrod algebra, TorAk,k+n(F2, F2) The Singer algebraic transfer was studied

by many authors (see Boardman [2], Bruner-Hà-Hưng [3], Hà [9], Hưng [10, 11],

12000 Mathematics Subject Classification Primary 55S10; 55S05, 55T15.

2Keywords and phrases: Steenrod squares, hit problem, algebraic transfer.

Chơn-Hà [6, 7, 8], Minami [15], Nam [16], Hưng-Quỳnh [12], Quỳnh [19], the firstnamed author [24] and others).

Singer showed in [20] that ϕ k is an isomorphism for k = 1, 2 Boardman showed

in [2] that ϕ3is also an isomorphism However, for any k > 4, ϕ k is not a phism in infinitely many degrees (see Singer [20], Hưng [11].) Singer made thefollowing conjecture

monomor-Conjecture 1.1 (Singer [20]) The algebraic transfer ϕ k is an epimorphism for any k > 0.

The conjecture is true for k 6 3 Based on the results in [23, 25], it can be

verified for k = 4 We hope that it is also true in this case.

In this paper, we develop Hưng’s result in [11] on the relation between thealgebraic transfer and Kameko’s homomorphism fSq0∗ : F2⊗AP k −→ F2⊗AP k This homomorphism is an GL k-homomorphism induced by the F2-linear map, alsodenoted by fSq0∗: P k → P k, given by

Sq0∗ and fSq0∗Sq 2t+1 = 0 for any non-negative integer t.

From the result of Carlisle and Wood [4] on the boundedness conjecture, we can

see that, for any non-negative integer d, there exists a non-negative integer t such

The following is one of our main results

Theorem 1.2 Let d be an arbitrary non-negative integer Then

(fSq0∗)s−t : (F2⊗AP k)k(2 s−1)+2s d−→ (F2⊗AP k)k(2 t−1)+2t d

is an isomorphism of GL k -modules for every s > t if and only if t > t(k, d).

It is easy to see that t(k, d) 6 k − 2 for every d and k > 2 So, one gets the

Corollary 1.3 shows that the number t = k − 2 commonly serves for every degree

d In [11], Hưng predicted that t = k − 2 is the minimum number for this purpose and proved it for k = 5 It is easy to see that for d = 2 k − k + 1, we have t(k, d) = k − 2 So, his prediction is true for all k > 2.

An application of Theorem 1.2 is the following theorem

Theorem 1.4 Singer’s conjecture is true for k = 5 and in degree 7.2 s − 5 with s

an arbitrary positive integer.

For d = 2, we have t(5, 2) = 2 and 5(2 s− 1) + 2s d = 7.2 s− 5 So, by Theorem1.2,

(fSq0∗)s−2: (F2⊗AP5)7.2 s−5−→ (F2⊗AP5)23

is an isomorphism of GL5-modules for every s> 2 Hence, by an explicit tation of (F2⊗AP5)7.2 s−5and (F2⊗AP5)GL5

compu-7.2 s−5for s = 1, 2, one gets the following.

Theorem 1.5 Let m = 7.2 s − 5 with s a positive integer Then

i) dim(F2⊗AP5)m = 191 for s = 1 and dim(F2⊗AP5)m = 1248 for any s > 2.

ii) (F2⊗AP5)GL5

m = 0 for any s > 1.

The second part of this theorem has been proved by Singer [20] for s = 1 In

[11], Hưng computed the dimensions of (F2⊗AP5)mand (F2⊗AP5)GL5

m for s> 2.However, his result for dim(F2⊗AP5)mis not true

From the results of Tangora [26], Lin [14] and Chen [5],

Ext5,7.2A s(F2, F2) =

(

hP h1i, if s = 1,

hh s g s−1 i, if s > 2, and h s g s−1 6= 0, where h s denote the Adams element in Ext1,2A s(F2, F2), P is the Adams periodicity operator [1] and g s−1 ∈ Ext4,2As+2+2s+1(F2, F2) for s > 2 Bypassing to the dual, one gets

TorA5,7.2 s(F2, F2) =

(

h(P h1)∗i, if s = 1, h(h s g s−1)∗i, if s > 2.

Hence, by Theorem 1.5(ii), the homomorphism

− 5 with s a positive integer Then dim(F2⊗AP5)n=

432 for s = 1 and dim(F2⊗AP5)n = 1171 for any s > 2.

This theorem is proved in [24] for s = 1 In [11], Hưng study the dimension of

this space by using computer calculation However, his result is also not true for

s > 2.

From Tangora [26], Lin [14] and Chen [5], we have

Ext5,10.2A s(F2, F2) =

(

hh4h4, h1d0i, if s = 1,

hh s d s−1 i, if s > 2, and h s d s−1 6= 0, where d s−1 ∈ Ext4,2As+3+2s(F2, F2) for s> 1 By passing to thedual, we obtain

TorA5,10.2 s(F2, F2) =

(

h(h4h4)∗, (h1d0)∗i, if s = 1, h(h s d s−1)∗i, if s > 2.

In [24], we have proved that dim(F2⊗AP5)GL5

This paper is organized as follows In Section 2, we recall some needed

infor-mation on the admissible monomials in P k, Singer’s criterion on the hit monomialsand Kameko’s homomorphism Theorems 1.2, 1.5 and 1.6 are respectively proved

in Sections 3-5 Finally, in the appendix we list all the admissible monomials of

Let α i (a) denote the i-th coefficient in dyadic expansion of a non-negative integer

a That means a = α0(a)20+ α1(a)21+ α2(a)22+ , for α i (a) = 0 or 1 with i> 0

16j6k α i−1 (ν j (x)) = deg XJi−1 (x) , i > 1 The sequence ω(x) is called the weight vector of x.

Let ω = (ω1, ω2, , ω i , ) be a sequence of non-negative integers The sequence

ω is called the weight vector if ω i = 0 for i  0.

The sets of all the weight vectors and the sigma vectors are given the left cographical order

lexi-For a weight vector ω, we define deg ω = P

i>02i−1 ω i Denote by P k (ω) the subspace of P k spanned by all monomials y such that deg y = deg ω, ω(y) 6 ω, and by P k−(ω) the subspace of P k spanned by all monomials y ∈ P k (ω) such that ω(y) < ω.

Definition 2.3 Let ω be a weight vector and f, g two polynomials of the same

degree in P k

i) f ≡ g if and only if f − g ∈ A+P k If f ≡ 0 then f is called hit.

ii) f ≡ ω g if and only if f − g ∈ A+P k + P k−(ω).

Obviously, the relations ≡ and ≡ω are equivalence ones Denote by QP k (ω) the quotient of P k (ω) by the equivalence relation ≡ ω Then, we have

QP k (ω) = P k (ω)/((A+P k ∩ P k (ω)) + P k−(ω)).

For a polynomial f ∈ P k , we denote by [f ] the class in F2⊗AP k represented by

f If ω is a weight vector and f ∈ P k (ω), then denote by [f ] ω the class in QP k (ω) represented by f Denote by |S| the cardinal of a set S.

It is easy to see that

QP k (ω) ∼ = QP k ω := h{[x] ∈ QP k : x is admissible and ω(x) = ω}i.

We note that the weight vector of a monomial is invariant under the permutation

of the generators x i , hence QP k (ω) has an action of the symmetric group Σ k.For 16 i 6 k, define the A-homomorphism g i : P k → P k, which is determined

by g i (x i ) = x i+1 , g i (x i+1 ) = x i , g i (x j ) = x j for j 6= i, i + 1, 1 6 i < k, and

g k (x1) = x1+ x2, g k (x j ) = x j for j > 1 Note that the general linear group GL k is

generated by the matrices associated with g i , 1 6 i 6 k, and the symmetric group

Σk is generated by the ones associated with g i , 1 6 i < k So, a homogeneous polynomial f ∈ P k is an GL k -invariant if and only if g i (f ) ≡ f for 1 6 i 6 k If

g i (f ) ≡ f for 1 6 i < k, then f is an Σ k-invariant

We have the following

Lemma 2.4 If x is a monomial in P k , then g k (x) ∈ P k (ω(x)).

Proof If ν1(x) = 0, then x = g k (x) and ω(g k (x)) = ω(x) Suppose ν1(x) > 0 and

Then, g k (x) is a sum of monomials of the form

where 06 c 6 b If c = 0, then ¯ y = x and ω(¯ y) = ω(x) Suppose c > 0.

If 2 ∈ Jt uj (x) for all j, 1 6 j 6 c, then ω(¯ y) = ω(x) and ¯ y ∈ P k (ω(x)) Suppose there is an index j such that 2 6∈ J t uj (x) Let j0 be the smallest index such that

Hence ω(¯ y) < ω(x) and ¯ y ∈ P k (ω(x)) The lemma is proved. 

Lemma 2.4 implies that if ω is a weight vector and x ∈ P k (ω), then g k (x) ∈

P k (ω) Moreover, QP k (ω) is an GL k-module

Definition 2.5 Let x, y be monomials of the same degree in P k We say that

x < y if and only if one of the following holds:

i) ω(x) < ω(y);

ii) ω(x) = ω(y) and σ(x) < σ(y).

Definition 2.6 A monomial x is said to be inadmissible if there exist monomials

y1, y2, , y m such that y t < x for t = 1, 2, , m and x −Pm

t=1 y t∈ A+P k

A monomial x is said to be admissible if it is not inadmissible.

Obviously, the set of all the admissible monomials of degree n in P k is a minimal

set of A-generators for P k in degree n.

Theorem 2.7 (See Kameko [13]) Let x, y, w be monomials in P k such that ω i (x) =

0 for i > r > 0, ω s (w) 6= 0 and ω i (w) = 0 for i > s > 0.

i) If w is inadmissible, then xw2r is also inadmissible.

ii) If w is strictly inadmissible, then wy2s is also strictly inadmissible.

Now, we recall a result of Singer [21] on the hit monomials in P k

Definition 2.8 A monomial z in P k is called a spike if ν j (z) = 2 d j − 1 for d j a

non-negative integer and j = 1, 2, , k If z is a spike with d1> d2> > d r−1>

d r > 0 and d j = 0 for j > r, then it is called the minimal spike.

For a positive integer n, by µ(n) one means the smallest number r for which it

is possible to write n =P

16i6r(2d i − 1), where d i > 0 In [21], Singer showed that

if µ(n) 6 k, then there exists uniquely a minimal spike of degree n in P k

Lemma 2.9 (See [18]) All the spikes in P k are admissible and their weight vectors are weakly decreasing Furthermore, if a weight vector ω is weakly decreasing and

ω16 k, then there is a spike z in P k such that ω(z) = ω.

The following is a criterion for the hit monomials in P k

Theorem 2.10 (See Singer [21]) Suppose x ∈ P k is a monomial of degree n, where µ(n) 6 k Let z be the minimal spike of degree n If ω(x) < ω(z), then x is hit.

This result implies a result of Wood, which original is a conjecture of son [17]

Peter-Theorem 2.11 (See Wood [27]) If µ(n) > k, then (F2⊗AP k)n = 0.

Denote by

(fSq0∗)(k,m): (F2⊗AP k)2m+k−→ (F2⊗AP k)m

Kameko’s homomorphism fSq0∗: (F2⊗AP k) −→ (F2⊗AP k in degree 2m + k.

Theorem 2.12 (See Kameko [13]) Let m be a positive integer If µ(2m + k) = k,

Denote Nk =(i; I); I = (i1, i2, , i r ), 1 6 i < i1< < i r 6 k, 0 6 r < k

Definition 2.14 Let (i; I) ∈ N k , let r = `(I) be the length of I, and let u be an

integer with 16 u 6 r A monomial x ∈ P k−1 is said to be u-compatible with (i; I)

if all of the following hold:

i) ν i1−1(x) = ν i2−1(x) = = ν i (u−1)−1(x) = 2 r− 1,

ii) ν i u−1(x) > 2 r− 1,

iii) α r−t (ν i u−1(x)) = 1, ∀t, 1 6 t 6 u,

iv) α r−t (ν i t−1(x)) = 1, ∀t, u < t 6 r.

Clearly, a monomial x can be u-compatible with a given (i; I) ∈ N k for at most

one value of u By convention, x is 1-compatible with (i; ∅).

For 16 i 6 k, define the homomorphism f i : P k−1 → P k of algebras by tuting

(x2i r−1f i (x))/x (I,u) , if there exists u such that

x is u-compatible with (i, I),

Then we have an F2-linear map φ (i;I) : P k−1 → P k In particular, φ (i;∅) = f i

Definition 2.16 For any (i; I) ∈ N k , we define the homomorphism p (i;I) : P k →

Then, p (i;I) is a homomorphism of A-modules In particular, for I = ∅, p (i;∅) (x i) = 0

and p (i;I) (f i (y)) = y for any y ∈ P k−1

Lemma 2.17 (See [18]) If x is a monomial in P k , then p (i;I) (x) ∈ P k−1 (ω(x)) Lemma 2.17 implies that if ω is a weight vector and x ∈ P k (ω), then p (i;I) (x) ∈

P k−1 (ω) Moreover, p (i;I) passes to a homomorphism from QP k (ω) to QP k−1 (ω) For a subset B ⊂ P k , we denote [B] = {[f ] : f ∈ B} If B ⊂ P k (ω), then we set [B] ω = {[f ] ω : f ∈ B}.

It is easy to see that if B is a minimal set of generators for A-module P k−1 in

degree n, then Φ0(B) is a minimal set of generators for A-module P k0 in degree n

We need the following

Lemma 3.1 Let n be a positive integer Then, µ(n) = s if and only if there exists

uniquely a sequence of integers d1> d2> > d s−1 > d s > 0 such that

Proof Assume that µ(n) = s Set β(n) = min{u ∈ N : α(n + u) 6 u} We prove µ(n) = β(n).

Suppose β(n) = t Then α(n + t) = r 6 t and n = 2 c1+ 2c2+ + 2 c r − t, where

c1> c2> > c r> 0

If c r 6 t − r then

α(n + t − 1) = α(2 c1+ 2c2+ + 2 c r−1+ 2c r − 1) = r − 1 + c r 6 t − 1 Hence β(n) 6 t − 1 This contradicts the fact that β(n) = t So, c r > t − r.

If r = t then c t = c r > t − r = 0 Set d i = c i , i = 1, 2, , t We obtain

Suppose s > 2 By the inductive hypothesis, µ(n + 1 − 2 d1) = s − 1.

It is well-known that there exists uniquely an integer d such that 2 d

Suppose that m is a positive integer and µ(2m + k) = s < k By Lemma 3.1, there exists a sequence of integers d1 > d2 > > d s−1 > d s > 0 such that 2m + k =Ps

i=1(2d i − 1) Set z = x21d1−1x22d2−1 x2s ds−1∈ (P k)2m+k Since z is a spike and s < k, we have [z] 6= 0 and (f Sq0∗)(k,m) ([z]) = 0 So, one gets the following.

Corollary 3.3 Let m be a positive integer If µ(2m + k) < k, then

(F2⊗AP k)k(2 s−1)+2s d = 0, (F2⊗AP k)d = 0.

So, the theorem holds

Assume that q 6 k Using Theorem 2.12, Corollaries 3.2 and 3.3, we see that

If q + r > k, then r + t − (k − q − 1) = q + r − k + 1 + t > 0 for any t > 0 = t(k, d).

By Lemma 3.1, µ(2m + k) = k So, the theorem is true.

If q +r < k, then from Lemma 3.1 and the relation (3.2), we see that µ(2m+k) =

k if and only if t > k − q − r = t(k, d) The theorem is completely proved. 

4 Proof of Theorem 1.5

In this section we prove Theorem 1.5 by explicitly determining all admissible

monomials of degree 7.2 s − 5 in P5 Using this results, we prove that for all s> 1,(F2⊗AP5)GL5

7.2 s−5= 0 Recall that by Theorem 1.2,

(fSq0)s−2: (F2⊗AP5)7.2 s−5−→ (F2⊗AP k)23

is an isomorphism of GL5-modules for every s> 2 So, we need only to prove the

x = x i x j x ` y2 with y1 an admissible monomial of degree 3 in P5 It is easy to see

that either ω(y1) = (1, 1) or ω(y1) = (3, 0) The lemma is proved. Using this lemma, we see that

Ker(fSq0∗)(5,2)= (F2⊗AP50)9

M

QP5+(3, 1, 1)MQP5+(3, 3).

From a result in [24], we have dim(F2⊗AP0)9= 160

Proposition 4.1.2 B+5(3, 1, 1) is the set of the monomials x1x2

j x4

` x u x v such that (1, j, `, u, v) is a permutation of (1, 2, 3, 4, 5), j < ` and u < v.

Proof Let x be an admissible monomial in P5+ and ω(x) = (3, 1, 1) Then x =

x i x2

j x4

` x u x v with (i, j, `, u, v) a permutation of (1, 2, 3, 4, 5) and i < u < v If ` < j,

then using the Cartan formula, we have

x = x i x4j x2` x u x v + Sq2(x i x2j x2` x u x v ) mod(P5−(3, 1, 1)).

So, x is inadmissible, hence j < ` If j = 1, then

x = x2i x j x4` x u x v + x i x j x4` x2u x v + x i x j x4` x u x2v + Sq1(x i x j x4` x u x v ).

Hence, x is inadmissible, so 1 < j Since i < u < v and j < `, we obtain i = 1.

By a direct computation, we see that the set

{[x1x2j x4` x u x v ] : 1 < j < `, 1 < u < v}

is linearly independent in F2⊗AP5 The proposition follows. 

By a similar argument as given in the proof of Proposition 4.1.2, we get thefollowing

Proposition 4.1.3 B5+(3, 3) is the set of the monomials x1x2j x2` x u x3v such that (1, j, `, u, v) is a permutation of (1, 2, 3, 4, 5), and j < `.

It is easy to see that |B5(2)| = 10, |B+5(3, 1, 1)| = 6 and |B+5(3, 3)| = 15 Hence, the first part of Theorem 1.5 is proved for s = 1.

By an easy computation, one gets

i) h[Σ5(u i)]iΣ5 = h[p(u i )]i, i = 1, 2, 3.

ii) h[Σ5(u4)]iΣ5 = h[p1], [p2]i, where

p1= x1x2x3x24x45+ x1x2x23x4x45+ x1x2x23x44x5

+ x1x22x3x4x45+ x1x22x3x44x5+ x1x22x43x4x5,

p2= x1x2x3x24x45+ x1x2x23x4x45+ x1x2x23x44x5+ x1x2x23x24x35+ x1x2x23x34x25+ x1x2x33x24x25+ x1x22x3x24x35+ x1x22x3x34x25+ x1x22x23x4x35+ x1x22x23x34x5

computation From Propositions 4.1.2 and 4.1.3, we see that dimh[Σ (u )]i = 21

with a basis consisting of the classes represented by the following monomials:

By using (4.3), (4.4) and (4.5), we have

g4(f ) + f ≡ (γ1+ γ3)(v2+ v3) + (γ4+ γ5)(v4+ v5)

+ (γ7+ γ8)(v7+ v8+ v10+ v11+ v12+ v13) ≡ 0.

From this one gets

γ1+ γ3= γ4+ γ5= γ7+ γ8= 0. (4.6)

Now, we prove the second part of Theorem 1.5 for s = 1.

Let f ∈ (P5)9 such that [f ] ∈ (F2⊗AP5)GL5

9 Since [f ] ∈ (F2⊗AP5)Σ5

9 , usingLemmas 4.1.5 and 4.1.6, we have

f ≡ γ1p(u1) + γ2p(u2) + γ3p(u3) + γ4p1+ γ5p2+ γ6p3,

with γ j ∈ F2 By computing g5(f ) + f in terms of the admissible monomials, we

obtain

g5(f ) + f ≡ γ1x2x4x75+ γ2x32x33x34+ γ3x2x34x55+ (γ4+ γ5+ γ6)x32x3x4x45

+ γ5x2x3x24x55+ (γ5+ γ6)x2x3x4x65+ other terms ≡ 0.

This relation implies γ j = 0 for 16 j 6 6 Theorem 1.5 is proved for s = 1.

4.2 The admissible monomials of degree 10 in P5.

To prove Theorem 1.5 for s = 2, we need to determine all the admissible mials of degree 10 in P5

mono-Lemma 4.2.1 If x is an admissible monomial of degree 10 in P5, then ω(x) is one of the following sequences: (2, 2, 1), (2, 4), (4, 1, 1), (4, 3).

Proof Observe that z = x7x3 is the minimal spike of degree 10 in P5 and ω(z) = (2, 2, 1) Since [x] 6= 0, by Theorem 2.10, either ω1(x) = 2 or ω1(x) = 4. If

ω1(x) = 2, then x = x i x j y2 with y a monomial of degree 4 in P5 and i < j Since

x is admissible, by Theorem 2.7, y is admissible and y ∈ P0 Using a result in

[25], one gets either ω(y) = (2, 1) or ω(y) = (4, 0) If ω1(x) = 4, then x = X j y2

with y1 a monomial of degree 3 in P5 Since y1 is admissible, we see that either

ω(y1) = (1, 1) or ω(y1) = (3, 0) The lemma is proved. From this lemma and a result in [25], we have

iii) B5+(4, 1, 1) is the set of the following monomials:

From the this proposition and a result in [25], we get dim(F2⊗AP5)10= 280.

The proposition is proved by using Theorems 2.7, 2.10 and the following

Lemma 4.2.3 The following monomials are strictly inadmissible:

x2

j x ` x t x3

u x3, j < ` < t.

Here (j, `, t, u, v) is a permutation of (1, 2, 3, 4, 5).

The proof of this lemma is straightforward

Proof of Proposition 4.2.2 We prove the first part of the proposition The others

can be proved by a similar computation We denote

d1= x1x2x2x2x4, d2= x1x2x2x4x2, d3= x1x2x3x2x4,

d4= x1x2x3x4x2, d5= x1x2x4x4x2

Let x be an admissible monomial of degree 10 in P5 such that ω(x) = (2, 2, 1) Then x = x i x j y2 with 16 i < j 6 5 and y a monomial of degree 4 in P5 Since x

is admissible, according to Theorem 2.7, we have y ∈ B5(4)

By a direct computation we that for all y ∈ B5(4), such that x i x j y26= d u , ∀u, 1 6

u 6 5, there is a monomial w which is given in Lemma 4.2.3(i) such that x i x j y2=

wz2r with a monomial z ∈ P5, and r = max{t ∈ Z : ω t (w) > 0} By Theorem 2.7,

x i x j y2 is inadmissible Since x = x i x j y2 is admissible, one gets x = d u for suitable

u.

We now prove the set {[d u ], 1 6 u 6 5} is linearly independent in F2⊗AP5

Suppose that S = γ1d1+ γ2d2+ γ3d3+ γ4d4+ γ5d5≡ 0 with γ u∈ F2 By a simplecomputation using Theorem 2.10, we have

p(1;2)(S) ≡ γ3[x31x2x23x44] + γ4[x31x2x43x2] + γ5[x31x42x3x24] ≡ 0,

p(2;5)(S) ≡ (γ1+ γ3)[x1x2x23x64] + γ1[x1x22x3x64] + γ2[x1x22x43x34] ≡ 0 These relations imply γ u = 0 for all u The first part of the proposition is proved.



we need to compute Ker(fSq0∗)(5,9).

Lemma 4.3.1 If x is an admissible monomial of degree 23 in P5 and [x] ∈

Ker(fSq0∗)(5,9) , then ω(x) is one of the following sequences:

(3, 2, 2, 1), (3, 2, 4), (3, 4, 1, 1), (3, 4, 3).

Proof Note that z = x15

1 x7x3 is the minimal spike of degree 23 in P5 and ω(z) = (3, 2, 2, 1) Since [x] 6= 0, by Theorem 2.10, either ω1(x) = 3 or ω1(x) = 5 If

ω1(x) = 5, then x = X∅y2 with y a monomial of degree 9 in P5 Since x is admissible, by Theorem 2.7, y is admissible Hence, (f Sq0∗)(5,9) ([x]) = [y] 6= 0 This contradicts the fact that [x] ∈ Ker(f Sq0∗)(5,9) , so ω1(x) = 3 Then, we have

x = x i x j x ` y2with y1an admissible monomial of degree 10 in P5 Now, the lemma

Using Lemma 4.3.1 and a result in [25], we get

result in [25], we easily obtain dim(F2⊗AP0)23= 635

Proposition 4.3.2 B+5(ω(1)) = Φ(B4+(23))∪C, where C is the set of the following monomials:

The following lemma is proved by a direct computation

Lemma 4.3.3 The following monomials are strictly inadmissible:

i) x2

j x ` x t x3

u , j < ` < t; x2

j x ` x t x u x2, j < ` < t < u; x1x2x2x4x5 Here (j, `, t, u, v) is a permutation of (1, 2, 3, 4, 5).

ii) f i(¯x), 1 6 i 6 5, where ¯ x is one of the following monomials:

x31x122 x3x74, x31x122 x73x4, x31x122 x33x54, x31x42x93x74,

x31x52x93x64, x31x52x83x74, x71x82x33x54.

Let z ∈ B5(2, 2, 1) such that x j x ` x t z2∈ P+

5 By a direct computation using the

results in Subsection 4.2, we see that if x j x ` x t z2∈ Φ(B / 4+(23)) ∪ C, then there is a monomial w which is given in Lemma 4.3.3 such that x j x ` x t z2= wz12u with suitable

monomial z1∈ P5, and u = max{j ∈ Z : ω j (w) > 0} By Theorem 2.7, x j x ` x t z2 is

inadmissible Since x = x j x ` x t y2 and x is admissible, one gets x ∈ Φ(B4+(23)) ∪ C This implies B5+(ω(1)) ⊂ Φ(B+4(23)) ∪ C.

We now prove the set [Φ(B4+(23)) ∪ C] is linearly independent in (F2⊗AP5)23.Suppose there is a linear relation

4(23)) ∪ C We explicitly compute p (i;j)(S)

in terms of the admissible monomials in P4 From the relation p (i;j)(S) ≡ 0 with

16 i < j 6 5, one gets γ t = 0 for all t / ∈ J , where J = {1, 15, 18, 82, 98, 99, 113,

116, 121, 122, 160, 161, 168, 170, 173,199, 201, 202, 211, 214, 219, 220} and γ t = γ1, for all t ∈ J So, S = γ1p ≡ 0 with p =P

t∈J d t Consider the homomorphism

g4: P5→ P5 By computing g4(p) in terms of d t’s one gets

g4(S) ≡ γ1(d19+ other terms) ≡ 0.

Since 19 / ∈ J , we obtain γ1 = 0 Hence γ t = 0 for 1 6 t 6 293 The proposition

Proposition 4.3.4 B5(ω(4)) = B5+(ω(4)) = ∅ That means QP5(ω(4)) = 0.

Proof Let x be an admissible monomial in P5+ such that ω(x) = ω(4) Then

x = x j x ` x t y2 with y ∈ B5(2, 4) By a direct computation using Theorem 2.7, Proposition 4.2.2 and Lemma 4.3.3, we see that x is a permutation of one of the monomials: x31x42x43x54x75, x31x42x53x54x65 A simple computation shows:

x31x42x43x54x75= Sq1(x31x2x23x94x75+ x31x2x23x54x115 ) + Sq2(x51x22x23x54x75+ x51x2x23x64x75)

+ Sq4(x31x22x23x54x75+ x31x2x23x64x75) mod(P5−(B5(ω(4)))).

This relation implies [x3x4x4x5x7]ω(4) = 0 By a similar computation, we have

[x3x4x5x5x6]ω(4) = 0 The proposition is proved 

Proposition 4.3.5 B5(ω(2)) = B5+(ω(2)) is the set of the following monomials:

The proof of this lemma is straightforward

Proof of Proposition 4.3.5 Let x be an admissible monomial in P5+such that ω(x) =

ω(2) Then x = x j x ` x t y2 with y ∈ B5(4, 1, 1).

Let z ∈ B5(4, 1, 1) such that x j x ` x t z2 ∈ P5+ Denote by D the set of all the

monomials as given in this proposition By a direct computation using Proposition

4.2.2, we see that if x j x ` x t z2 ∈ D, then there is a monomial w which is given in / Lemma 4.3.6 such that x j x ` x t z2 = wz12u with suitable monomial z1 ∈ P5, and

u = max{j ∈ Z : ω j (w) > 0} By Theorem 2.7, x j x ` x t z2 is inadmissible Since

x = x j x ` x t y2with y ∈ B5(4, 1, 1) and x is admissible, one gets x ∈ D This implies

B5+(ω(2)) ⊂ D.

We now prove the set [D] is linearly independent in (F2⊗AP5)23 Suppose there

is a linear relation S = P398

γ t d t ≡ 0, where γ t ∈ F2 and d t = d 23,t ∈ D We

explicitly compute p (i;j) (S) in terms of the admissible monomials in P4 From the

relation p (i;j)(S) ≡ 0 with 16 i < j 6 5, one gets γ t= 0 for all 2946 t 6 398.

To prove [D] ω(2) is a basis of QP5(ω(2)), we need to show that [x] ω(2) 6= 0 for

all x ∈ D Observe that if x ∈ D then x is a permutation of one of the following

monomials:

a1= x1x2x2x3x155 , a2= x1x2x2x7x115 , a3= x1x2x3x6x115 , a4= x1x3x3x6x105 ,

a5= x1x2x3x7x105 , a6= x1x2x3x3x145 , a7= x3x3x133 x2x2, a8= x3x5x2x2x115 ,

a9= x3x7x9x2x2, a10= x3x3x5x2x105 .

It is easy to see that [a5]ω(2) = [σ1a3]ω(2), [a7]ω(2) = [σ2a6]ω(2), [a8]ω(2) = [σ3a3]ω(2),

[a9]ω(2) = [σ4a3]ω(2), [a10]ω(2) = [σ5a4]ω(2), where σ u are suitable permutations in

Σ5

By a similar argument as given in the proof of Proposition 4.3.2, we can prove

that the set [B5+(ω(1)) ∪ {a1, a2, a3, a4, a6}] is linearly independent in the space(F2⊗AP5)23 This fact shows that [a v]ω(2) 6= 0 for all v So, [x] ω(2) 6= 0 for all

Proposition 4.3.7 B5(ω(3)) = B5+(ω(3)) is the set of the following monomials:

The following lemma is proved by a direct computation

Lemma 4.3.8 The following monomials are strictly inadmissible:

x j x6` x3t x6u x7v , j < ` < t; x j x2` x6t x7u x7v Here (j, `, t, u, v) is a permutation of (1, 2, 3, 4, 5).

Proof of Proposition 4.3.7 Let x be an admissible monomial in P5+such that ω(x) =

ω(3) Then x = x j x ` x t y2 with y ∈ B5(4, 3).

Let z ∈ B5(4, 3) such that x j x ` x t z2 ∈ P5+ Denote by E the set of all the

monomials as given in proposition 4.3.7 By a direct computation using Proposition

4.2.2, we see that if x j x ` x t z2 ∈ E, then there is a monomial w which is given in / Lemma 4.3.8 such that x j x ` x t z2 = wz12u with suitable monomial z1 ∈ P5, and

u = max{j ∈ Z : ω j (w) > 0} By Theorem 2.7, x j x ` x t z2 is inadmissible Since

x = x j x ` x t y2 with y ∈ B5(4, 3) and x is admissible, one gets x ∈ E This implies

p (i;j)(S) ≡ 0 with 16 i < j 6 5, one gets γ t = 0 for all t.

To prove [E] ω(3) is a basis of QP5(ω(3)), we need to show that [x] ω(3) 6= 0 for all

x ∈ E.

It is easy to see that if x ∈ E then x is a permutation of one of the following

monomials:

b1= x1x32x63x64x75, b2= x31x32x53x64x65, b3= x21x32x53x64x75.

We have [b3]ω(3)= [b1]ω(3) By a similar argument as given in the proof of

Proposi-tion 4.3.2, we can prove that the set [B5+(3, 2, 2, 1)∪B5(3, 4, 1, 1)∪{b1, b2}] is linearlyindependent in the space (F2⊗AP5)23 This implies [b1]ω 6= 0 and [b2]ω(3) 6= 0,

hence [x] ω(3) 6= 0 for all x ∈ E The proposition follows. Now we compute (F2⊗AP5)GL5

t=1 γ t u t with γ t ∈ F2 and [f ] ∈ h[Σ5(b2)]ω(3)iΣ5 By a directcomputation, we have

Proof of Proposition 4.3.9 Let f ∈ P5(ω(3)) such that [f ] ω(3) ∈ QP5(ω(3))GL5

Since [f ] ω(3) ∈ QP5(ω(3))Σ5, using Lemmas 4.3.10, and 4.3.11, we have f ≡ ω(3)γp(b1) with γ ∈ F2 By computing g5(f ) + f in terms of the admissible monomials,

we obtain

g5(f ) + f ≡ ω(3) γb1+ other terms ≡ω(3)0.

Proposition 4.3.12 QP (ω )GL5= 0.