1 Introduction

Let Vk be an elementary abelian 2-group of rank k and let BVk be the classifying space of Vk. Then,

$$ P_{k}:= H^{\ast}(BV_{k}) \cong \mathbb F_{2}[x_{1},x_{2},{\ldots} ,x_{k}], $$

a polynomial algebra in k generators x1,x2,…,xk, each of degree 1. Here the cohomology is taken with coefficients in the prime field \(\mathbb F_{2}\) of two elements.

Being the cohomology of a topological space, Pk is a module over the mod-2 Steenrod algebra, \(\mathcal A\). The action of \(\mathcal A\) on Pk is determined by the elementary properties of the Steenrod squares Sqi and subject to the Cartan formula (see Steenrod and Epstein [27]).

A polynomial g in Pk is called hit if it can be written as a finite sum \(g = {\sum }_{j > 0}Sq^{j}(g_{j})\) for suitable polynomials gjPk. That means g belongs to \(\mathcal {A}^{+}P_{k}\), where \(\mathcal {A}^{+}\) is the augmentation ideal of \(\mathcal A\).

The hit problem is to find a minimal generating set for Pk regarded as a module over the mod-2 Steenrod algebra. Equivalently, we want to find a vector space basis for \(\mathbb F_{2} \otimes _{\mathcal A} P_{k}\) in each degree d. Such a basis may be represented by a list of monomials of degree d.

The hit problem was first studied by Peterson [18], Wood [39], Singer [25], and Priddy [21], who showed its relation to several classical problems in the homotopy theory. Then, this problem was investigated by Carlisle and Wood [3], Crabb and Hubbuck [7], Janfada and Wood [11], Kameko [12], Mothebe [15], Nam [16], Phúc and Sum [19, 20], Silverman [23], Silverman and Singer [24], Singer [26], Walker and Wood [35,36,37], Wood [39, 40] and others.

The vector space \(\mathbb F_{2} \otimes _{\mathcal A} P_{k}\) was explicitly calculated by Peterson [18] for k = 1,2, by Kameko [12] for k = 3, and by us [28, 30] for k = 4. Recently, the hit problem and it’s applications to representations of general linear groups have been presented in the books of Walker and Wood [37, 38].

Let GLk be the general linear group over the field \(\mathbb F_{2}\). Since Vk is an \(\mathbb F_{2}\)-vector space of dimension k, this group acts naturally on Vk and therefore on the cohomology Pk of BVk. The two actions of \(\mathcal A\) and GLk upon Pk commute with each other. Hence, there is an inherited action of GLk on \(\mathbb F_{2}\otimes _{\mathcal A}P_{k}\).

For a non-negative integer d, denote by (Pk)d the subspace of Pk consisting of all the homogeneous polynomials of degree d in Pk and by \((\mathbb F_{2}\otimes _{\mathcal A}P_{k})_{d}\) the subspace of \(\mathbb F_{2}\otimes _{\mathcal A}P_{k}\) consisting of all the classes represented by the elements in (Pk)d. In [25], Singer defined the algebraic transfer, which is a homomorphism

$$ \varphi_{k} :\quad\text{Tor}^{\mathcal A}_{k,k+d} (\mathbb F_{2},\mathbb F_{2}) \longrightarrow (\mathbb F_{2}\otimes_{\mathcal A}P_{k})_{d}^{GL_{k}} $$

from the homology of the Steenrod algebra, \(\text {Tor}^{\mathcal A}_{k,k+d} (\mathbb F_{2},\mathbb F_{2})\), to the subspace of \((\mathbb F_{2}{\otimes }_{\mathcal A}P_{k})_{d}\) consisting of all the GLk-invariant classes. The Singer algebraic transfer is a useful tool in describing the homology groups of the Steenrod algebra. It was studied by many authors (see Boardman [1], Bruner–Hà–Hưng [2], Chơn–Hà [5, 6], Hà [8], Hưng [9], Minami [14], Nam [17], Hưng–Quỳnh [10], Quỳnh [22], the present author [29] and others).

It was shown that the algebraic transfer is an isomorphism for k = 1,2 by Singer in [25] and for k = 3 by Boardman in [1]. However, for any k ≥ 4, φk is not a monomorphism in infinitely many degrees (see Singer [25], Hưng [9].) Singer made the following conjecture.

Conjecture 1 (Singer [25])

The algebraic transfer φk is an epimorphism for any k ≥ 0.

The conjecture is true for k ≤ 3. Based on the results in [28, 30], we are currently trying to verify the conjecture for the case k = 4. However, for k ≥ 5, the conjecture is still open.

There is a classical operator, known as Kameko’s squaring operation

$$ \widetilde{Sq}^{0}_{\ast}:\quad (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{2d+k} \longrightarrow (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{d}, $$

which is induced by an \(\mathbb F_{2}\)-linear map ϕ : PkPk given by

$$ \phi(x) = \begin{cases} y&\text{ if }x=x_{1}x_{2}{\ldots} x_{k}y^{2},\\ 0& \text{ otherwise,} \end{cases} $$

for any monomial xPk. Note that ϕ is not an \(\mathcal A\)-homomorphism. However, ϕSq2i = Sqiϕ and ϕSq2i+ 1 = 0 for any non-negative integer i. Note that since \(\widetilde {Sq}^{0}_{*}\) is an epimorphism of GLk-modules, it induces a homomorphism which is also denoted by \(\widetilde {Sq}^{0}_{\ast }: (\mathbb F_{2} \otimes _{\mathcal A} P_{k})^{GL_{k}}_{2d+k} \longrightarrow (\mathbb F_{2} \otimes _{\mathcal A} P_{k})^{GL_{k}}_{d}\). It was recognized by Boardman [1] for k = 3 and by Minami [14] for general k that Kameko’s squaring operation commutes with the dual of the classical squaring operation on the cohomology of the Steenrod algebra, \(Sq^{0}: \text {Ext}_{\mathcal A}^{k,d+k}(\mathbb F_{2},\mathbb F_{2}) \to \text {Ext}_{\mathcal A}^{k,2d+2k}(\mathbb F_{2},\mathbb F_{2})\), through the Singer algebraic transfer. This means that the following diagram is commutative:

For a positive integer n, by μ(n) one means the smallest number r for which it is possible to write \(n = {\sum }_{1\leq i\leq r}(2^{u_{i}}-1),\) where ui > 0.

Theorem 1 (Kameko 12)

Let d be a non-negative integer. If μ(2d + k) = k, then

$$ \widetilde{Sq}^{0}_{\ast}:\quad (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{2d+k}\longrightarrow (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{d} $$

is an isomorphism of GLk-modules.

From the result of Carlisle and Wood [3] on the boundedness conjecture, Hưng observes in [9] that, for any degree d, there exists a non-negative integer t such that

$$ (\widetilde{Sq}^{0}_{\ast})^{s-t}:\quad (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{k(2^{s}-1) + 2^{s}d} \longrightarrow (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{k(2^{t}-1) + 2^{t}d} $$

is an isomorphism of GLk-modules for every st. However, this result does not confirm how large t should be.

Denote by α(n) the number of ones in the dyadic expansion of a positive integer n and by ζ(n) the greatest integer u such that n is divisible by 2u. That means n = 2ζ(n)m with m an odd integer. We set

$$ t(k,d) = \max\{0,k- \alpha(d+k) -\zeta(d+k)\}. $$

The following is one of our main results.

Theorem 2

Let d be an arbitrary non-negative integer. Then

$$ (\widetilde{Sq}^{0}_{\ast})^{s-t}:\quad (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{k(2^{s}-1) + 2^{s}d} ~\longrightarrow~ (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{k(2^{t}-1) + 2^{t}d} $$

is an isomorphism of GLk-modules for every st if and only if tt(k,d).

It is easy to see that t(k,d) ≤ k − 2 for every d and k ≥ 2. Hence, one gets the following.

Corollary 1 (Hưng [9])

Let d be an arbitrary non-negative integer. If k ≥ 2, then

$$ (\widetilde{Sq}^{0}_{\ast})^{s-k+2}:\quad (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{k(2^{s}-1) + 2^{s}d}~\longrightarrow~(\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{k(2^{k-2}-1) + 2^{k-2}d} $$

is an isomorphism of GLk-modules for every sk − 2.

Corollary 1 shows that the number t = k − 2 commonly serves for every degree d. Hưng stated in [9] 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 = 2kk + 1, we have t(k,d) = k − 2. So, his statement is true for all k ≥ 2.

An application of Theorem 2 is the case k = 5 and d = 0. Then, we have t(5,0) = 3. So, Theorem 2 implies that

$$ (\widetilde{Sq}^{0}_{\ast})^{s-3}:\quad (\mathbb F_{2} \otimes_{\mathcal A} P_{5})_{5(2^{s}-1)}~\longrightarrow~(\mathbb F_{2} \otimes_{\mathcal A} P_{5})_{35} $$

is an isomorphism of GL5-modules for every s ≥ 3. Hence, by computing the space \((\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{5(2^{s}-1)}\) for s = 1,2,3, we obtain the following.

Theorem 3

For s an arbitrary positive integer, we have

$$ \dim(\mathbb F_{2}{\otimes}_{\mathcal A}P_{5})_{5(2^{s}-1)} = \begin{cases} 46 &\text{ if $s=1$},\\ 432 &\text{ if $s=2$},\\ 1117 &\text{ if $s \geq 3$}. \end{cases} $$

This theorem has been proved in [29] for s = 2. In [9], Hưng also proved this theorem for s = 2,3 by using a computer program of Shpectorov written in GAP. However, the detailed proof was unpublished at the time of the writing.

Theorem 3 is proved by determining the admissible monomials of degree 5(2s − 1) in P5. The computations are based on some results in [12, 25, 30] on the admissible monomials and the hit monomials (see Section 2).

For s > 3, we have 5(2s − 1) = 5(2s− 3 − 1) + 35 ⋅ 2s− 3, μ(35) = 3 and α(35 + μ(35)) = μ(35). Hence, by combining Theorem 3 and [30, Theorem 1.3], we get

Corollary 2

Let s be a positive integer. If s ≥ 8, then

$$ \dim (\mathbb F_{2} \otimes_{\mathcal A} P_{6})_{5(2^{s}-1)} = (2^{6}-1)\dim (\mathbb F_{2} \otimes_{\mathcal A} P_{5})_{35} = 70371. $$

To verify Conjecture 1 for k = 5 and the degree d = 5(2s − 1), we need the following.

Theorem 4

For any positive integer s, we have

$$ \dim(\mathbb F_{2} \otimes_{\mathcal A} P_{5})_{5(2^{s}-1)}^{GL_{5}} = \begin{cases} 0 &\text{ if } s=1,\\ 2 &\text{ if } s=2,\\ 1 &\text{ if } s \geq 3. \end{cases} $$

This result confirms the one of Hưng in [9] for s = 2,3, which was also proved by using a computer calculation.

From the results of Chen [4], Lin [13] and Tangora [32], we have

$$ \text{Tor}_{5,5\cdot 2^{s}}^{\mathcal A}(\mathbb F_{2},\mathbb F_{2}) =\begin{cases} 0 &\text{ if } s=1,\\ \langle ({h_{0}^{4}}h_{4})^{\ast},(h_{1}d_{0})^{\ast}\rangle &\text{ if } s=2,\\ \langle (h_{s-1}d_{s-2})^{\ast}\rangle &\text{ if } s \geq 3, \end{cases} $$

and hs− 1ds− 2≠ 0, where hs− 1 denote the Adams element in \(\text {Ext}_{\mathcal A}^{1,2^{s-1}}(\mathbb F_{2}, \mathbb F_{2})\) and \(d_{s-2} \in \text {Ext}_{\mathcal A}^{4,2^{s+2}+2^{s-1}}(\mathbb F_{2}, \mathbb F_{2})\) for s ≥ 2. In [29], we have proved that

$$ \varphi_{5}:\quad \text{Tor}_{5,20}^{\mathcal A}(\mathbb F_{2},\mathbb F_{2}) ~\longrightarrow~ (\mathbb F_{2}{\otimes}_{\mathcal A}P_{5})_{15}^{GL_{5}} $$

is an isomorphism. Combining the results of Hà [8] and Singer [25], we have φ5((hs− 1ds− 2))≠ 0. Hence, Theorem 4 implies that

$$ \varphi_{5}:\quad \text{Tor}_{5,5\cdot 2^{s}}^{\mathcal A}(\mathbb F_{2},\mathbb F_{2}) ~\longrightarrow~ (\mathbb F_{2}{\otimes}_{\mathcal A}P_{5})_{5(2^{s}-1)}^{GL_{5}} $$

is also an isomorphism for s ≥ 3. So, we get the following.

Corollary 3

Singer’s conjecture is true for k = 5 and the degree 5(2s − 1), with s an arbitrary positive integer.

For d = 2, we have t(5,2) = 2 and 5(2s − 1) + 2sd = 7 ⋅ 2s − 5. Hence, by Theorem 2, \((\widetilde {Sq}^{0}_{\ast })^{s-2}: (\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{7\cdot 2^{s}-5} \longrightarrow (\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{23}\) is an isomorphism of GL5-modules for every s ≥ 2. So, by an explicit computation of \((\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{7\cdot 2^{s}-5}^{GL_{5}}\) for s = 1,2, Tín proved in [34] the following.

Theorem 5 (Tín 34)

Let s be a positive integer. Then, \((\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{7\cdot 2^{s}-5}^{GL_{5}} = 0\).

The theorem has been proved by Singer [25] for s = 1. In [9], Hưng also proved this theorem for s = 2 by using a computer calculation. However, the detailed proof was also unpublished.

From the results of Tangora [32], Lin [13], and Chen [4], we can see that for any s ≥ 1, \(\dim \text {Tor}_{5,7\cdot 2^{s}}^{\mathcal A}(\mathbb F_{2},\mathbb F_{2}) = 1\). Hence, by Theorem 5, the homomorphism φ5 is an epimorphism in the degree 7 ⋅ 2s − 5. However, it is not a monomorphism. This result confirms the one of Hưng.

Corollary 4 (Hưng [9])

There are infinitely many degrees in which φ5 is not a monomorphism.

This paper is organized as follows. In Section 2, we recall some needed information on the admissible monomials in Pk and Singer’s criterion on the hit monomials. Theorem 2 is proved in Section 3. In Section 4, we explicitly determine a system of \(\mathcal A\)-generators for P5 in degree 5(2s − 1). Theorem 4 is proved in Section 5 by using the results in Section 4. The admissible monomials of the degree 5(2s − 1) in P5 are given in Section 6 of the online version [31].

Theorems 2 and 5 have already been announced in [33].

2 Preliminaries

In this section, we recall some needed information from Kameko [12] and Singer [26], which will be used in the next sections.

Notation 1

Let αi(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 αi(a) = 0 or 1 with i ≥ 0.

For a set of integers \(\mathbb J = \{j_{1},j_{2},{\ldots } , j_{s}\}\) with 1 ≤ juk, 1 ≤ us, we define the monomial \(X_{\mathbb J} \in P_{k}\) by setting \(X_{\mathbb J} = {\prod }_{j\ne j_{u}, \forall u}x_{j} = x_{1}{\cdots } \hat x_{j_{1}}{\cdots } \hat x_{j_{s}}{\cdots } x_{k}\).

Let \(x=x_{1}^{\nu _{1}(x)}x_{2}^{\nu _{2}(x)}{\cdots } x_{k}^{\nu _{k}(x)} \in P_{k}\). Set

$$ \mathbb J_{t}(x) = \{j : 1 \leq j \leq k,~\alpha_{t}(\nu_{j}(x)) =0\} $$

for t ≥ 0. Then, we have \(x = {\prod }_{t\geq 0}X_{\mathbb J_{t}(x)}^{2^{t}}\).

Definition 1

For a monomial x in Pk, define two sequences associated with x by

$$ \omega(x)=(\omega_{1}(x),\omega_{2}(x),{\ldots} , \omega_{i}(x), \ldots),\quad \sigma(x) = (\nu_{1}(x),\nu_{2}(x),{\ldots} ,\nu_{k}(x)), $$

where \(\omega _{i}(x) = {\sum }_{1\leq j \leq k} \alpha _{i-1}(\nu _{j}(x))= \text {deg}X_{\mathbb J_{i-1}(x)},~i \geq 1\). The sequence ω(x) is called the weight vector of x.

The sequence ω = (ω1,ω2,…,ωi,…) of non-negative integers is called the weight vector if ωi = 0 for i ≫ 0.

The sets of the weight vectors and the exponent vectors are given the left lexicographical order.

For a weight vector ω, we define \(\text {deg}\omega = {\sum }_{i > 0}2^{i-1}\omega _{i}\). Denote by Pk(ω) the subspace of Pk spanned by all monomials y such that degy = degω, ω(y) ≤ ω, and by \(P_{k}^{-}(\omega )\) the subspace of Pk spanned by all monomials yPk(ω) such that ω(y) < ω.

Definition 2

Let ω be a weight vector and f, g two polynomials of the same degree in Pk.

  1. i)

    fg if and only if \(f + g \in \mathcal A^{+}P_{k}\). If f ≡ 0 then f is called hit.

  2. ii)

    fωg if and only if \(f + g \in \mathcal A^{+}P_{k}+P_{k}^{-}(\omega )\).

Obviously, the relations ≡ and ≡ω are equivalence ones. Denote by QPk(ω) the quotient of Pk(ω) by the equivalence relation ≡ω. Then, we have

$$ QP_{k}(\omega)= P_{k}(\omega)/ \big((\mathcal A^{+}P_{k}\cap P_{k}(\omega))+P_{k}^{-}(\omega)\big). $$

For a polynomial fPk, we denote by [f] the class in \(\mathbb F_{2}\otimes _{\mathcal A}P_{k} = P_{k}/\mathcal A^{+}P_{k}\) represented by f. If ω is a weight vector, then denote by [f]ω the class in the space \(P_{k}/(\mathcal A^{+}P_{k}+P_{k}^{-}(\omega ))\) represented by f.

Denote by |S| the cardinal of a set S. If S is a subset of a vector space, then we denote by 〈S〉 the subspace spanned by S.

It is easy to see that

$$ QP_{k}(\omega) \cong QP_{k}^{\omega} := \langle\{[x] \in QP_{k} : x \text{ is admissible and } \omega(x) = \omega\}\rangle. $$

So, we get

$$ (\mathbb F_{2}{\otimes}_{\mathcal A}P_{k})_{n} = \bigoplus_{\text{deg}\omega = n}QP_{k}^{\omega} \cong \bigoplus_{\text{deg}\omega = n}QP_{k}(\omega). $$

Hence, we can identify the vector space QPk(ω) with \(QP_{k}^{\omega } \subset QP_{k}\).

We note that the weight vector of a monomial is invariant under the permutation of the generators xi, hence QPk(ω) is an Σk-module, where ΣkGLk is the symmetric group.

Note that Vk≅〈x1,x2,…,xk〉⊂ Pk. For 1 ≤ ik, define the \(\mathbb F_{2}\)-linear map ρi : VkVk, which is determined by ρi(xi) = xi+ 1,ρi(xi+ 1) = xi, ρi(xj) = xj for ji,i + 1, 1 ≤ i < k, and ρk(x1) = x1 + x2, ρk(xj) = xj for j > 1. The general linear group GLkGL(Vk) is generated by ρi, 1 ≤ ik, and the symmetric group Σk is generated by ρi, 1 ≤ i < k. The \(\mathbb F_{2}\)-linear map ρi induces a homomorphism of \(\mathcal A\)-algebras which is also denoted by ρi : PkPk. So, an element [f]ωQPk(ω) is an GLk-invariant if and only if ρi(f) ≡ωf for 1 ≤ ik. It is an Σk-invariant if and only if ρi(f) ≡ωf for 1 ≤ i < k.

Proposition 1

For any weight vector ω, the space QPk(ω) is an GLk-module.

Proof

Since QPk(ω) is an Σk-module, we prove the proposition by proving the fact that if x is a monomial in Pk, then \(\rho _{k}(x) \in P_{k}(\omega (x))\).

If ν1(x) = 0, then x = ρk(x) and ω(ρk(x)) = ω(x). Suppose ν1(x) > 0 and \(\nu _{1}(x) = 2^{t_{1}} + {\cdots } + 2^{t_{b}}\), where \(0 \leq t_{1} < {\cdots } < t_{b}\), b ≥ 1.

Since \(x = {\prod }_{t\geq 0}X_{\mathbb J_{t}(x)}^{2^{t}} \in P_{k}\) and ρk is a homomorphism of algebras, we have

$$ \rho_{k}(x) = \prod\limits_{t\geq 0}\left( \rho_{k}(X_{\mathbb J_{t}(x)})\right)^{2^{t}}= \left( \prod\limits_{u=1}^{b}\left( (x_{1}+x_{2})X_{\mathbb J_{t_{u}}(x)\cup 1}\right)^{2^{t_{u}}} \right)\left( \prod\limits_{t \ne t_{1}, t_{2},\ldots, t_{b}}X_{\mathbb J_{t}(x)}^{2^{t}}\right). $$

We have \((x_{1}+x_{2})X_{\mathbb J_{t_{u}}(x)\cup 1} = X_{\mathbb J_{t_{u}}(x)} + x_{2}X_{\mathbb J_{t_{u}}(x)\cup 1}\), hence ρk(x) is a sum of monomials of the form

$$ \bar y = \left( \prod\limits_{j=1}^{c}\left( x_{2}X_{\mathbb J_{t_{u_{j}}}(x)\cup 1}\right)^{2^{t_{u_{j}}}} \right)\left( \prod\limits_{t \ne t_{u_{1}}, \ldots, t_{u_{c}}}X_{\mathbb J_{t}(x)}^{2^{t}}\right), $$

where 0 ≤ cb. If c = 0, then \(\bar y = x\) and \(\omega (\bar y) = \omega (x)\). Suppose that c > 0. If \(2 \in \mathbb J_{t_{u_{j}}}(x)\) for all j, 1 ≤ jc, then \(\omega (\bar y) = \omega (x)\) and \(\bar y \in P_{k}(\omega (x))\). Assume that there is an index j such that \(2 \not \in \mathbb J_{t_{u_{j}}}(x)\). Let j0 be the smallest index such that \(2\not \in \mathbb J_{t_{u_{j_{0}}}}(x)\). Then, we have

$$ \omega_{i}(\bar y) = \begin{cases} \omega_{i}(x) &\text{ if } i \leq t_{u_{j_{0}}},\\ \omega_{i}(x)-2 &\text{ if } i = t_{u_{j_{0}}}+1. \end{cases} $$

Hence, \(\omega (\bar y) < \omega (x)\) and \(\bar y \in P_{k}(\omega (x))\). The proposition is proved. □

Definition 3

Let x,y be monomials of the same degree in Pk. We say that x < y if and only if one of the following holds:

  1. i)

    ω(x) < ω(y);

  2. ii)

    ω(x) = ω(y) and σ(x) < σ(y).

Definition 4

A monomial x is said to be inadmissible if there exist monomials \(y_{1},y_{2},\ldots , y_{t}\) such that yj < x for j = 1,2,…,t and \(x + {\sum }_{j=1}^{t}y_{j} \in \mathcal A^{+}P_{k}\). 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 Pk is a minimal set of \(\mathcal {A}\)-generators for Pk in degree n.

Definition 5

A monomial x in Pk is said to be strictly inadmissible if and only if there exist monomials \(y_{1},y_{2},\ldots , y_{t}\) such that yj < x, for j = 1,2,…,t and

$$ x = \sum\limits_{j=1}^{t} y_{j} + \sum\limits_{u=1}^{2^{s}-1}Sq^{u}(q_{u}) $$

with \(s = \max \limits \{i : \omega _{i}(x) > 0\}\) and suitable polynomials quPk.

It is easy to see that if x is strictly inadmissible, then it is inadmissible.

Theorem 6 (See Kameko 12)

Let x,y,w be monomials in Pk such that ωi(x) = 0 for i > r > 0, ωs(w)≠ 0 and ωi(w) = 0 for i > s > 0.

  1. i)

    If w is inadmissible, then \(xw^{2^{r}}\) is also inadmissible.

  2. ii)

    If w is strictly inadmissible, then \(wy^{2^{s}}\) is also strictly inadmissible.

Now, we recall a result of Singer [26] on the hit monomials in Pk.

Definition 6

A monomial z in Pk is called a spike if \(\nu _{j}(z)=2^{t_{j}}-1\) for tj a non-negative integer and j = 1,2,…,k. If z is a spike with \(t_{1}>t_{2}>{\cdots } >t_{r-1}\geq t_{r}>0\) and tj = 0 for j > r, then it is called the minimal spike.

In [26], Singer showed that if μ(n) ≤ k, then there exists uniquely a minimal spike of degree n in Pk.

The following is a criterion for the hit monomials in Pk.

Theorem 7 (See Singer 26)

Suppose xPk is a monomial of degree n, where μ(n) ≤ k. Let z be the minimal spike of degree n. If ω(x) < ω(z), then x is hit.

This result implies the one of Wood, which originally is a conjecture of Peterson [18].

Theorem 8 (See Wood 39)

If μ(n) > k, then \((\mathbb F_{2} \otimes _{\mathcal A}P_{k})_{n} = 0\).

Now, we recall some notations and definitions in [30], which will be used in the next sections. We set

$$ \begin{array}{@{}rcl@{}} {P_{k}^{0}} &=&\langle\{x=x_{1}^{a_{1}}x_{2}^{a_{2}}{\cdots} x_{k}^{a_{k}}~:~ a_{1}a_{2}{\cdots} a_{k}=0\}\rangle,\\ P_{k}^{+} &=& \langle\{x=x_{1}^{a_{1}}x_{2}^{a_{2}}{\cdots} x_{k}^{a_{k}} ~:~ a_{1}a_{2}{\cdots} a_{k}>0\}\rangle. \end{array} $$

It is easy to see that \({P_{k}^{0}}\) and \(P_{k}^{+}\) are the \(\mathcal {A}\)-submodules of Pk. Furthermore, we have the following.

Proposition 2

We have a direct summand decomposition of the \(\mathbb F_{2}\)-vector spaces \(\mathbb F_{2}\otimes _{\mathcal A}P_{k} =Q{P_{k}^{0}} \oplus QP_{k}^{+}\). Here \(Q{P_{k}^{0}} = \mathbb F_{2}\otimes _{\mathcal A}{P_{k}^{0}}\) and \(QP_{k}^{+} = \mathbb F_{2}\otimes _{\mathcal A}P_{k}^{+}\).

Definition 7

For 1 ≤ ik, define the homomorphism \(f_{i}: P_{k-1} \to P_{k}\) of algebras by substituting

$$ f_{i}(x_{j}) = \begin{cases} x_{j}&\text{ if } 1 \leq j <i,\\ x_{j+1} &\text{ if } i \leq j <k. \end{cases} $$

Obviously, we have the following.

Proposition 3

If B is a minimal set of generators for \(\mathcal A\)-module Pk− 1 in degree n, then \(f(B):=\bigcup _{1 \leq i \leq k}f_{i}(B)\) is also a minimal set of generators for \(\mathcal A\)-module \({P_{k}^{0}}\) in degree n.

Definition 8

For any 1 ≤ i < jk, we define the homomorphism \(p_{(i;j)}: P_{k} \to P_{k-1}\) of algebras by substituting

$$ p_{(i;j)}(x_{u}) =\begin{cases} x_{u} &\text{ if } 1 \leq u < i,\\ x_{j-1} &\text{ if } u = i,\\ x_{u-1}&\text{ if } i< u \leq k. \end{cases} $$

Then, p(i;j) is a homomorphism of \(\mathcal A\)-modules. In particular, \(p_{(i;j)}(f_{i}(y)) = y\) for any yPk− 1.

Lemma 1 (See 19)

Let x be a monomial in Pk. Then, p(i;j)(x) ∈ Pk− 1(ω(x)).

Lemma 1 implies that if ω is a weight vector and xPk(ω), then p(i;j)(x) ∈ Pk− 1(ω). Moreover, p(i;j) passes to a homomorphism from QPk(ω) to QPk− 1(ω).

For a subset BPk and a weight vector ω, we denote [B] = {[f] : fB} and [B]ω = {[f]ω : fB}. From Theorem 7, we see that if ω is the weight vector of a minimal spike in Pk, then [B]ω = [B].

From now on, we denote by Bk(n) the set of all admissible monomials of degree n in Pk, \({B_{k}^{0}}(n) = B_{k}(n)\cap {P_{k}^{0}}\), \(B_{k}^{+}(n) = B_{k}(n)\cap P_{k}^{+}\). For a weight vector ω of degree n, we set \(B_{k}(\omega ) = B_{k}(n)\cap P_{k}(\omega )\), \(B_{k}^{+}(\omega ) = B_{k}^{+}(n)\cap P_{k}(\omega )\). Then, [Bk(ω)]ω and \([B_{k}^{+}(\omega )]_{\omega }\), are respectively the bases of the \(\mathbb F_{2}\)-vector spaces QPk(ω) and \(QP_{k}^{+}(\omega ) := QP_{k}(\omega )\cap QP_{k}^{+}\).

3 Proof of Theorem 2

To make the paper self-contained, we give here a proof for the following lemma, which is an elementary property of the μ-function.

Lemma 2

Let n be a positive integer. Then, μ(n) = s if and only if there exists uniquely a sequence of integers \(v_{1} > v_{2} >{\cdots } > v_{s-1}\geq v_{s}>0\) such that

$$ n = 2^{v_{1}} + 2^{v_{2}} + {\cdots} + 2^{v_{s-1}} + 2^{v_{s}} - s= \sum\limits_{i=1}^{s}(2^{v_{i}}-1). $$
(1)

Proof

Assume that μ(n) = s. Set \(\upbeta (n) = \min \limits \{u\in \mathbb N: \alpha (n+u) \leq u\}\). We prove μ(n) =β (n).

Suppose β(n) = t. Then α(n + t) = rt, and \(n = 2^{c_{1}} + 2^{c_{2}} + {\cdots } +2^{c_{r}} -t\), where \(c_{1}> c_{2} > {\cdots } > c_{r} \geq 0\).

If crtr then

$$ \alpha(n+t-1) = \alpha(2^{c_{1}} + 2^{c_{2}} + {\cdots} +2^{c_{r-1}}+ 2^{c_{r}}-1) = r-1 + c_{r} \leq t-1. $$

Hence β(n) ≤ t − 1. This contradicts the fact that β(n) = t. So, cr > tr.

If r = t then ct = cr > tr = 0. Set \(v_{i}= c_{i}\), i = 1,2,…,t. We obtain

$$ n = 2^{v_{1}} + 2^{v_{2}} + {\cdots} + 2^{v_{t-1}} + 2^{v_{t}} - t = {\sum}_{i=1}^{t}(2^{v_{i}}-1), $$

where v1 > v2 > ⋯ > vt− 2 > vt− 1 > vt > 0. Hence, μ(n) ≤ t =β (n).

Suppose r < t. Obviously,

$$ 2^{c_{r}} = 2^{c_{r}-1} + {\cdots} + 2^{c_{r} -t+r+1}+2^{c_{r} -t+r}+2^{c_{r} -t+r}. $$

Set

$$ \begin{array}{@{}rcl@{}} v_{i} &=& c_{i}, \quad i=1,2,\ldots, r-1,\\ v_{r+\ell} &=&c_{r}-\ell- 1> 0, \quad \ell = 0,1, {\ldots} ,t-r-2,\\ v_{t-1}&=& v_{t} = c_{r} -t+r > 0. \end{array} $$

Then, we get

$$ n = 2^{v_{1}} + 2^{v_{2}} + {\cdots} + 2^{v_{t-1}} + 2^{v_{t}} - t = {\sum}_{i=1}^{t}(2^{v_{i}}-1), $$

with \(v_{1}>v_{2} > {\cdots } >v_{t-2} > v_{t-1} = v_{t}>0\). Hence μ(n) ≤ t =β (n).

Since μ(n) = s, \(n = {\sum }_{i=1}^{s}(2^{h_{i}}-1)\) with hi positive integers. Then, \(\alpha (n+s) = \alpha ({\sum }_{i=1}^{s}2^{h_{i}})\leq s\). So, we get μ(n) = s ≥β (n). Hence, t =β (n) = μ(n) = s. Thus, n is of the form (1).

Now, assume that n is of the form (1). Then, μ(n) ≤ s. We prove μ(n) = s by induction on s.

If s = 1, then μ(n) = 1, since μ(n) > 0. If s = 2, then \(\alpha (n+1) = \alpha (2^{v_{1}} + 2^{v_{2}}-1) = 1 + v_{2} >1\). Hence, μ(n) =β (n) ≥ 2. So, μ(n) = 2.

Suppose s > 2. By the inductive hypothesis, \(\mu (n+1-2^{v_{1}}) = s-1\).

It is well-known that there exists uniquely an integer d such that \(2^{d} \leq n+1 < 2^{d+1}\). Since \(v_{1}>v_{2}>{\cdots } > v_{s-1} \geq v_{s}> 0\), we have

$$ 2^{v_{1}} \leq n+1 < 2^{v_{1}}+ 2^{v_{1}-1}+{\cdots} + 2^{v_{1}-s+2}+2^{v_{1}-s+2} = 2^{v_{1}+1}. $$

So, we get v1 = d. Set μ(n) = ts. There exists u1 > u2 > ⋯ > ut− 1ut > 0 such that \(n = 2^{u_{1}}+2^{u_{2}} + {\cdots } + 2^{u_{t}}-t\). Then, u1 = d = v1 and \(\alpha (n+1-2^{d} +t-1) \leq t-1\). Hence, \(t-1 \geq \upbeta (n+1-2^{d}) = \mu (n+1-2^{d}) = s-1\). This implies ts and μ(n) = t = s.

By induction on i, we get ui = vi for 1 ≤ is. The lemma is proved. □

From this lemma we easily obtain the following.

Corollary 5 (See Kameko [12])

Let n,k be positive integers. Then

  1. i)

    μ(n) > k if and only if α(n + k) > k.

  2. ii)

    If n > μ(n), then nμ(n) is even and \(\mu \big (\frac {n-\mu (n)}2\big ) \leq \mu (n)\).

  3. iii)

    μ(2n + μ(n)) = μ(n).

Suppose that d is a non-negative integer such that μ(2d + k) = s < k. By Lemma 2, there exists a sequence of integers \(v_{1} > v_{2} >{\cdots } > v_{s-1}\geq v_{s}>0\) such that \(2d+k = {\sum }_{i=1}^{s}(2^{v_{i}}-1)\). Set \(z = x_{1}^{2^{v_{1}}-1}x_{2}^{2^{v_{2}}-1}{\cdots } x_{s}^{2^{v_{s}}-1} \in (P_{k})_{2d+k}\). Since z is a spike and s < k, we have [z]≠ 0 and \(\widetilde {Sq}^{0}_{\ast }([z]) = 0\). So, one gets the following.

Corollary 6

Let d be an arbitrary non-negative integer. If μ(2d + k) < k, then

$$ (\widetilde{Sq}^{0}_{\ast})_{(k,d)}:= \widetilde{Sq}^{0}_{\ast} :~ (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{2d+k}~\longrightarrow~ (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{d} $$

is not a monomorphism.

Now, we are ready to prove Theorem 2.

Proof Proof of Theorem 2

Set q = α(d + k), r = ζ(d + k) and \(m = k(2^{t}-1) + 2^{t}d\). From the proof of Lemma 2 and Corollary 5, we see that if q > k, then

$$ \mu(k(2^{s}-1) + 2^{s}d) \geq \mu(d)>k $$

for any s ≥ 0 = t(k,d). By Theorem 8,

$$ (\mathbb F_{2} \otimes_{\mathcal A}P_{k})_{k(2^{s}-1) + 2^{s}d} =0,\quad (\mathbb F_{2} \otimes_{\mathcal A}P_{k})_{d} =0. $$

So, the theorem holds.

Assume that qk. Using Theorem 1, Corollaries 5 and 6, we see that the homomorphism

$$ (\widetilde{Sq}^{0}_{\ast})^{s-t}:\quad (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{k(2^{s}-1) + 2^{s}d} ~\longrightarrow~ (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{k(2^{t}-1) + 2^{t}d} $$

is an isomorphism of GLk-modules for every st if and only if μ(2m + k) = k.

Since α(d + k) = q and ζ(d + k) = r, there exists a sequence of integers \(c_{1} > c_{2} > {\cdots } > c_{q-1} > c_{q} = r\geq 0\) such that

$$ d+k = 2^{c_{1}} + 2^{c_{2}} + {\cdots} + 2^{c_{q}}. $$

If q = k, then

$$ \begin{array}{@{}rcl@{}} 2m +k &=& k(2^{t+1}-1) + 2^{t+1}d = k(2^{t+1}-1) + 2^{t+1}(2^{c_{1}} + 2^{c_{2}} + {\cdots} + 2^{c_{k}}-k)\\ &=& 2^{c_{1}+t+1} + 2^{c_{2}+t+1} +{\cdots} + 2^{c_{k}+t+1} - k. \end{array} $$

By Lemma 2, μ(2m + k) = k for any t ≥ 0 = t(k,d). Hence, the theorem holds.

Suppose that q < k. Then, we have

$$ \begin{array}{@{}rcl@{}} 2m +k &=& 2^{c_{1}+t+1} + 2^{c_{2}+t+1} +{\cdots} + 2^{c_{q-1}+t+1} + 2^{r+t+1} - k\\ &=& 2^{c_{1}+t+1} + 2^{c_{2}+t+1} +{\cdots} + 2^{c_{q-1}+t+1}\\ && + 2^{r+t}+ 2^{r+t-1} + {\cdots} + 2^{r+t- (k-q-1)} + 2^{r+t- (k-q-1)} - k. \end{array} $$
(2)

If q + rk, then r + t − (kq − 1) = q + rk + 1 + t > 0 for any t ≥ 0 = t(k,d). By Lemma 2, μ(2m + k) = k. So, the theorem is true.

If q + r < k, then from Lemma 2 and the relation (2), we see that μ(2m + k) = k if and only if tkqr = t(k,d). The theorem is completely proved.□

4 \(\mathcal A\)-Generators for P 5 in Degree 5(2s − 1)

In this section, we prove Theorem 3 by determining the admissible monomials of P5 in degree 5(2s − 1). The computations are technical and complicated. So, we present some main calculations. We refer the reader to the online version [31] for intermediate calculations. From Theorem 2, we see that

$$ (\widetilde{Sq}^{0}_{\ast})^{s-3}:\quad (\mathbb F_{2} \otimes_{\mathcal A} P_{5})_{5(2^{s}-1)}~ \longrightarrow~ (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{35} $$

is an isomorphism of GL5-modules for every s ≥ 3. So, we need only to determine \(\mathcal A\)-generators for P5 in degree 5(2s − 1) for s = 1,2,3.

4.1 The Cases s = 1,2

By using a result in [30], we can easily obtain the following.

Proposition 4

There exist exactly 46 admissible monomials of degree 5 in P5. Consequently \(\dim (\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{5} = 46\).

For s = 2, we have 5(2s − 1) = 15. The space \((\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{15}\) has been computed in [29].

Proposition 5 (See 29)

There exist exactly 432 admissible monomials of degree 15 in P5. Consequently \(\dim (\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{15} = 432.\)

The admissible monomials of degrees 5 and 15 are explicitly determined as in Sections 6.1 and 6.3 of the online version [31].

4.2 The Admissible Monomials of Degree 16 in P 5

To determine the admissible monomials of P5 in degree 5(2s − 1) for s = 3, we need to determine the admissible monomials of degree 16 in P5.

Lemma 3

If x is an admissible monomial of degree 16 in P5, then ω(x) is one of the following sequences:

$$ (2,1,1,1),\quad (2,1,3),\quad (2,3,2),\quad (4,2,2), \quad (4,4,1). $$

Proof

Observe that \(z = x_{1}^{15}x_{2}\) is the minimal spike of degree 16 in P5 and ω(z) = (2,1,1,1). Since [x]≠ 0, by Theorem 7, either ω1(x) = 2 or ω1(x) = 2. If ω1(x) = 2, then \(x = x_{i}x_{j}y^{2}\) with y a monomial of degree 7 in P5. Since x is admissible, by Theorem 6, y is admissible. A routine computation shows that either ω(y) = (1,1,1) or ω(y) = (1,3) or ω(y) = (3,2). If ω1(x) = 4, then \(x = X_{j}{y_{1}^{2}}\) with y1 an admissible monomial of degree 6 in P5. It is easy to see that either ω(y1) = (2,2) or ω(y1) = (4,1). The lemma is proved. □

We have \((\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{16} = (Q{P_{5}^{0}})_{16}\bigoplus (QP_{5}^{+})_{16}\). From a result in [30] and Proposition 3, we easily obtain \(\dim (Q{P_{5}^{0}})_{16} = 255\).

Proposition 6

\((QP_{5}^{+})_{16}\) is the \(\mathbb F_{2}\)-vector space of dimension 188 with a basis consisting of all the classes represented by the monomials \(a_{t} = a_{16,t}\), 1 ≤ t ≤ 188, which are determined as in Section 6.4 of the online version [31].

Proof

From Lemma 3, we have

$$ B_{5}^{+}(16) = B_{5}^{+}(2,1,1,1)\cup B_{5}^{+}(2,1,3)\cup B_{5}^{+}(2,3,2)\cup B_{5}^{+}(4,2,2)\cup B_{5}^{+}(4,4,1). $$

We prove \(|B_{5}^{+}(2,1,1,1)| = 4\), \(|B_{5}^{+}(2,1,3)| = 5\), \(|B_{5}^{+}(2,3,2)| = 20\), \(|B_{5}^{+}(4,2,2)| = 110\) and \(|B_{5}^{+}(4,4,1)| = 49\). For simplicity, we prove \(|B_{5}^{+}(2,3,2)| = 20\) by showing that \(B_{5}^{+}(2,3,2) = \{a_{t} = a_{16,t}: 10 \leq t \leq 29\}\). The others can be proved by the similar computations.

Let x be an admissible monomial in \(P_{5}^{+}\) such that ω(x) = (2,3,2) := ω. Then \(x = x_{j}x_{\ell } y^{2}\) with 1 ≤ j < ≤ 5 and y a monomial of degree 7 in P5. Since x is admissible, by Theorem 6, yB5(3,2). Let zB5(3,2) such that \(x_{j}x_{\ell } z^{2} \in P_{5}^{+}\). By a direct computation, we see that if \(x_{j}x_{\ell } z^{2}\ne a_{t}\), for all t, 10 ≤ t ≤ 29, then there is a strictly inadmissible monomial w which is given in Lemma 4.2.3 of [31] such that \(x_{j}x_{\ell } z^{2}= wz_{1}^{2^{u}}\) with suitable monomial z1P5, and \(u = \max \limits \{j \in \mathbb Z : \omega _{j}(w) >0\}\). By Theorem 6, \(x_{j}x_{\ell } z^{2}\) is inadmissible. Since \(x = x_{j}x_{\ell } y^{2}\) with yB5(3,2) and x admissible, one obtain x = at for some t, 10 ≤ t ≤ 29. Hence, \(QP_{5}^{+}(\omega )\) is spanned by the set {[at] : 10 ≤ t ≤ 29}.

We now prove the set {[at] : 10 ≤ t ≤ 29} is linearly independent in QP5(ω). Suppose there is a linear relation \(\mathcal S = {\sum }_{t = 10}^{29}\gamma _{t}a_{t} \equiv _{\omega } 0\), where \(\gamma _{t} \in \mathbb F_{2}\). By a direct computation from the relations \(p_{(i,;j)}(\mathcal S) \equiv _{\omega } 0\) with 1 ≤ i < j ≤ 5, we obtain γt = 0 for all t, 10 ≤ t ≤ 29. The proof is completed. □

By combining the above results, one gets the following.

Corollary 7

There exist exactly 443 admissible monomials of degree 16 in P5. Consequently \(\dim (\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{16} = 443.\)

4.3 The Case s = 3

For s = 3, we have 5(2s − 1) = 35. Since Kameko’s squaring operation

$$ (\widetilde{Sq}^{0}_{\ast})_{(5,15)}:\quad (\mathbb F_{2} \otimes_{\mathcal A} P_{5})_{35} ~\longrightarrow~ (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{15} $$

is an epimorphism, we have \((\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{35} \cong \text {Ker}(\widetilde {Sq}^{0}_{\ast })_{(5,15)} \bigoplus (\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{15}\). Hence, we need only to compute \(\text {Ker}(\widetilde {Sq}^{0}_{\ast })_{(5,15)}\).

Lemma 4

If x is an admissible monomial of degree 35 in P5 and \([x] \in \text {Ker}(\widetilde {Sq}^{0}_{\ast })_{(5,15)}\), then ω(x) is one of the following sequences:

$$ \begin{array}{@{}rcl@{}} \omega_{(1)} &=& (3,2,1,1,1),\quad \omega_{(2)} = (3,2,1,3),\quad \omega_{(3)} = (3,2,3,2),\\ \omega_{(4)} &=& (3,4,2,2), \quad \omega_{(5)} = (3,4,4,1). \end{array} $$

Proof

Note that \(z = x_{1}^{31}{x_{2}^{3}}x_{3}\) is the minimal spike of degree 35 in P5 and ω(z) = (3,2,1,1,1). Since [x]≠ 0, by Theorem 7, either ω1(x) = 3 or ω1(x) = 5. If ω1(x) = 5, then \(x = X_{\emptyset } y^{2}\) with y a monomial of degree 15 in P5. Since x is admissible, by Theorem 6, y is admissible. Hence, \((\widetilde {Sq}^{0}_{\ast })_{(5,15)}([x]) = [y] \ne 0\). This contradicts the fact that \([x] \in \text {Ker}(\widetilde {Sq}^{0}_{\ast })_{(5,15)}\), so ω1(x) = 3. Then, we have \(x = x_{i}x_{j}x_{\ell } {y_{1}^{2}}\) with y1 an admissible monomial of degree 16 in P5. Now, the lemma follows from Lemma 3. □

From Lemma 4, we obtain

$$ \text{Ker}(\widetilde{Sq}^{0}_{\ast})_{(5,15)} \cong \bigoplus_{j=1}^{5} QP_{5}(\omega_{(j)}). $$

Proposition 7

There exist exactly 620 admissible monomials in P5 such that their weight vectors are ω(1). Consequently \(\dim QP_{5}(\omega _{(1)})= 620\).

Proof

Using a result in [30] and Proposition 3 we obtain \(QP_{5}(\omega _{(1)}) = (Q{P_{5}^{0}})_{35}\oplus QP_{5}^{+}(\omega _{(1)})\) and \(\dim (Q{P_{5}^{0}})_{35} = 460\). We prove \(B_{5}^{+}(\omega _{(1)}) = \{a_{t} = a_{35,t}: 1 \leq t \leq 160\}\), as given in Section 6.5 of the online version [31].

Let x be an admissible monomial such that ω(x) = ω(1). Then, \(x = x_{j}x_{\ell } x_{t}y^{2}\) with 1 ≤ j < < t ≤ 5 and yB5(2,1,1,1).

Let zB5(2,1,1,1) such that \(x_{j}x_{\ell } x_{t}z^{2} \in P_{5}^{+}\). By a direct computation using the results in Section 4.2, we see that if \(x_{j}x_{\ell } x_{t}z^{2} \ne a_{t}\), ∀t, 1 ≤ t ≤ 160, then there is a strictly inadmissible monomial w which is given in one of Lemmas 4.3.3–4.3.5 of [31] such that \(x_{j}x_{\ell } x_{t}z^{2}= wz_{1}^{2^{u}}\) with suitable monomial z1P5, and \(u = \max \limits \{j \in \mathbb Z : \omega _{j}(w) >0\}\). By Theorem 6, xjxxtz2 is inadmissible. Since \(x = x_{j}x_{\ell } x_{t}y^{2}\) and x is admissible, one gets x = at for some t, 1 ≤ t ≤ 160. This implies \(B_{5}^{+}(\omega _{(1)}) \subset \{a_{t} : 1 \leq t \leq 160\}\).

We now prove the set {[at] : 1 ≤ t ≤ 160} is linearly independent in \((\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{35}\). Suppose there is a linear relation \(\mathcal S = {\sum }_{t = 1}^{160}\gamma _{t}a_{t} \equiv 0\), where \(\gamma _{t} \in \mathbb F_{2}\). We explicitly compute \(p_{(i;j)}(\mathcal S)\) in terms of the admissible monomials in \(P_{4} (\text {mod}(\mathcal A^{+}P_{4})\)). By a direct computation from the relations \(p_{(i;j)}(\mathcal S) \equiv 0\) with 1 ≤ i < j ≤ 5, we obtain γt = 0 for 1 ≤ t ≤ 160. The proposition follows. □

By an analogous argument as the previous one we get the following propositions.

Proposition 8

We have QP5(ω(2)) = 0 and QP5(ω(3)) = 0.

Consider the monomials at = a35,t, 161 ≤ t ≤ 225 as given in Section 6.5 of the online version [31].

Proposition 9

  1. i)

    The space \(QP_{5}(\omega _{(4)})\) is an GL5-module generated by the class \([a_{203}]_{\omega _{(4)}}\) and \(B_{5}(\omega _{(4)}) = \{a_{t} : 161 \leq t \leq 210\}\).

  2. ii)

    The space \(QP_{5}(\omega _{(5)})\) is an GL5-module generated by the class \([a_{225}]_{\omega _{(5)}}\) and \(B_{5}(\omega _{(5)}) = \{a_{t} : 211 \leq t \leq 225\}\).

By combining the above results we obtain Theorem 3.

5 Proof of Theorem 4

In this section, we prove Theorem 4 by using the results in Section 4. From Theorem 2, we see that the homomorphism

$$ (\widetilde{Sq}^{0}_{\ast})^{s-3}:\quad (\mathbb F_{2} \otimes_{\mathcal A} P_{5})_{5(2^{s}-1)}^{GL_{5}} ~\longrightarrow~ (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{35}^{GL_{5}} $$

is an isomorphism for every s ≥ 3. So, we need only to prove the theorem for s = 1,2,3.

5.1 The Cases s = 1,2

For s = 1, we have 5(2s − 1) = 5.

Proposition 10

\((\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{5}^{GL_{5}} = 0\).

Proof

Denote by \(a_{t} = a_{5,t}\), 1 ≤ t ≤ 46, the admissible monomials of degree 5 in P5 (see [31, Section 6.1]). Suppose \(f \in (P_{5})_{5}\) such that \([f] \in (\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{5}^{GL_{5}}\). Then, \(f \equiv {\sum }_{t=1}^{46}\gamma _{t}a_{t}\) with \(\gamma _{t} \in \mathbb F_{2}\) and ρi(f) + f ≡ 0 for i = 1,2,3,4,5. By computing directly from the relations ρi(f) + f ≡ 0 for i = 1,2,3,4, we easily obtain \(\gamma _{t} = \gamma _{1}\), 1 ≤ t ≤ 30, and γt = 0 for 31 ≤ t ≤ 45. Now, by computing ρ5(f) + f in terms of the admissible monomials, we get

$$ \rho_{5}(f) +f \equiv \gamma_{1}a_{4} + \gamma_{46}a_{31} + \text{ other terms} \equiv 0. $$

The last equality implies γ1 = γ46 = 0. The proposition follows. □

For s = 2, the theorem has been proved in [29]. We have

Proposition 11 (See 29)

\((\mathbb F_{2} \otimes _{\mathcal A} P_{5})_{15}^{GL_{5}}\) is an \(\mathbb F_{2}\)-vector space of dimension 2 with a basis consisting of the 2 classes represented by the polynomials p and q, which are determined as in Section 6.6 of [31].

5.2 The Case s = 3

For s = 3, 5(2s − 1) = 35. From the results in Section 4, we have

$$ \text{Ker}(\widetilde{Sq}^{0}_{\ast})_{(5,15)} = \bigoplus_{1\leq j \leq 5} QP_{5}(\omega_{(j)}). $$

Since QP5(ω(2)) = 0 and QP5(ω(3)) = 0, we need only to compute \(QP_{5}(\omega _{(j)})^{GL_{5}}\) for j = 1,4,5.

Proposition 12

\(QP_{5}(\omega _{(1)})^{GL_{5}} = 0\).

Proof

We denote the admissible monomials in \((Q{P_{5}^{0}})_{35}\) by bt, 1 ≤ t ≤ 460, as given in [31, Section 6.5]. By a direct computation, we have

$$ QP_{5}(\omega_{(1)})^{{\Sigma}_{5}} = \langle\{[p_{j}] : 1 \leq j \leq 9\}\rangle, $$

where \(p_{1} ={\sum }_{t=1}^{60}b_{t}\), \(p_{2}={\sum }_{t=61}^{90}b_{t}\), \(p_{3}={\sum }_{t=61}^{130}b_{t}\), \(p_{4}= {\sum }_{t=181}^{250}b_{t} + {\sum }_{t=371}^{415}b_{t}\), \(p_{5}= {\sum }_{t=251}^{265}b_{t} + {\sum }_{t=271}^{395}b_{t}\), \(p_{6} = {\sum }_{t=266}^{375}b_{t} + {\sum }_{t=396}^{415}b_{t}\), \(p_{7} = {\sum }_{t=11}^{31}a_{t} + {\sum }_{t=38}^{51}a_{t} + {\sum }_{t=66}^{85}a_{t}\), \(p_{8} = {\sum }_{t=1}^{10}a_{t} + {\sum }_{t=52}^{85}a_{t}\), \(p_{9} = {\sum }_{t=32}^{85}a_{t}\).

Let \(f \in P_{5}(\omega _{(1)})\) such that \([f] \in QP_{5}(\omega _{(1)})^{GL_{5}}\). Then, \(f \equiv {\sum }_{j=1}^{9}\gamma _{j}p_{j}\) with \(\gamma _{j} \in \mathbb F_{2}\). By a direct computation using Theorem 7, we have

$$ \begin{array}{@{}rcl@{}} \rho_{5}(f) + f &\equiv& \gamma_{1}b_{7} + \gamma_{2}b_{48} + \gamma_{3}b_{36} + \gamma_{4}b_{172} + \gamma_{5}b_{232} + (\gamma_{1} + \gamma_{5} + \gamma_{6})b_{21}\\ && + \gamma_{7}b_{206} + \gamma_{8}a_{29} + \gamma_{9}b_{416} + \text{ other terms} \equiv 0. \end{array} $$

The last equality implies γt = 0 for 1 ≤ t ≤ 9. The proposition is proved. □

By a same argument as the previous one we get the following.

Proposition 13

\(QP_{5}(\omega _{(4)})^{GL_{5}} = 0\) and \(QP_{5}(\omega _{(5)})^{GL_{5}} = 0\).

Combining the above results, we easily obtain the following.

Corollary 8

\(\text {Ker}(\widetilde {Sq}^{0}_{\ast })_{(5,15)}^{GL_{5}} = 0\).

Proof of Theorem 4 for s = 3

Let \(f \in (P_{5})_{35}\) such that \([f] \in (\mathbb F_{2}\otimes P_{5})_{35}^{GL_{5}}\). Since Kameko’s squaring operation

$$ (\widetilde{Sq}^{0}_{\ast})_{(5,15)}:\quad (\mathbb F_{2} \otimes_{\mathcal A} P_{5})_{35} ~\longrightarrow~ (\mathbb F_{2} \otimes_{\mathcal A} P_{k})_{15} $$

is an epimorphism of GL5-modules, \((\widetilde {Sq}^{0}_{\ast })_{(5,15)}([f]) \in (\mathbb F_{2}\otimes P_{5})_{15}^{GL_{5}}\). By Proposition 11, \((\widetilde {Sq}^{0}_{\ast })_{(5,15)}([f]) = \lambda _{1} [p] + \lambda _{2}[q]\) with \(\lambda _{1}, \lambda _{2} \in \mathbb F_{2}\). Hence, we have

$$ f \equiv \lambda_{1} \psi(p) + \lambda_{2}\psi(q) + \bar f, $$

where \(\bar f \in (P_{5})_{35}\) such that \([\bar f] \in \text {Ker}(\widetilde {Sq}^{0}_{\ast })_{(5,15)}\), and \(\psi : P_{5} \to P_{5}\) is the \(\mathbb F_{2}\)-linear map determined by \(\psi (y) = x_{1}x_{2}x_{3}x_{4}x_{5}y^{2}\) for any yP5.

Now, we prove that if [f]≠ 0, then λ2 = 1. Suppose the contrary, that λ2 = 0. By a direct computation, we have

$$ \begin{array}{@{}rcl@{}} \rho_{1}(\psi(p)) + \psi(p) &\equiv& 0, \end{array} $$
(3)
$$ \begin{array}{@{}rcl@{}} \rho_{2}(\psi(p)) + \psi(p) &\equiv& a_{17} + a_{18} + a_{20} + a_{21} + a_{104} + a_{105} + a_{106} + a_{107} + a_{108}\\ && + a_{109} + a_{113} + a_{114} + a_{115} + a_{116} + a_{123} + a_{124}, \end{array} $$
(4)
$$ \begin{array}{@{}rcl@{}} \rho_{3}(\psi(p)) + \psi(p) &\equiv& a_{11} + a_{18} + a_{35} + a_{37} \\ && + a_{52} + a_{54} + a_{91} + a_{93} + a_{95} + a_{105}, \end{array} $$
(5)
$$ \begin{array}{@{}rcl@{}} \rho_{4}(\psi(p)) + \psi(p) &\equiv& a_{1} + a_{4} + a_{11} + a_{12} + a_{15}\\ && + a_{16} + a_{32} + a_{33} + a_{54} + a_{55} + a_{87}. \end{array} $$
(6)

From the relations (3)–(6), we have \(\rho _{i}(\bar f) + \bar f \equiv _{\omega _{(t)}} \rho _{i}(f) + f \equiv _{\omega _{(t)}} 0\), for i = 1,2,3,4,5 and t = 4,5. From this and Proposition 13, we get \([\bar f]_{\omega _{(t)}} \in QP_{5}(\omega _{(t)})^{GL_{5}} = 0\). By combining this and the facts that QP5(ω(2)) = 0 and QP5(ω(3)) = 0 we get \(\bar f \equiv f^{\prime } \in QP_{5}(\omega _{(1)})\). Now, by a direct computation from the relations ρi(f) + f ≡ 0, for i = 1,2,3,4, and using Proposition 7, we get λ1 = 0. Hence, by Proposition 12, \([f] = [f^{\prime }] \in QP_{5}(\omega _{(1)})^{GL_{5}} = 0\). This contradicts the hypothesis [f]≠ 0. Hence, λ2 = 1 and \(f \equiv \lambda _{1}\psi (p) + \psi (q) + \bar f\).

Suppose that \(f^{\ast } = \lambda \psi (p) + \psi (q) + \bar f^{\ast }\) with \(\lambda \in \mathbb F_{2}\), \([\bar f^{\ast }] \in \text {Ker}(\widetilde {Sq}^{0}_{\ast })_{(5,15)}\) and \([f^{\ast }] \in (\mathbb F_{2}\otimes P_{5})_{35}^{GL_{5}}\). Then, \((\lambda _{1} + \lambda )\psi (p) + \bar f + \bar f^{\ast } = f + f^{\ast }\) and

$$ [f+f^{\ast}] = [f] + [f^{\ast}] \in (\mathbb F_{2}\otimes P_{5})_{35}^{GL_{5}}. $$

By the same argument as the previous one we obtain λ + λ1 = 0. So, [f + f] ∈Ker \((\widetilde {Sq}^{0}_{\ast })_{(5,15)}^{GL_{5}} = 0\). This implies \([f] = [f^{\ast }]\). Thus, we have proved that \(\dim (\mathbb F_{2}\otimes P_{5})_{35}^{GL_{5}} \leq 1\).

In [25], Singer showed that the Adams elements hi are in the image of \(\varphi _{1}^{\ast }\). Hà showed in [8] that the elements di are in the image of \(\varphi _{4}^{\ast }\). Since \(\varphi ^{\ast } = \{\varphi _{k}^{\ast }: k \geq 0\}\) is a homomorphism of graded algebras, we see that the element \(h_{2}d_{1}\) is in the image of \(\varphi _{5}^{\ast }\), hence \(\varphi _{5}((h_{2}d_{1})^{\ast }) \ne 0\). This implies \(\dim (\mathbb F_{2}\otimes P_{5})_{35}^{GL_{5}} \geq 1\). Theorem 4 is completely proved.□