††thanks: Dedicated to Prof. Daciberg L. Goncalves for his seventies birthday
and to Prof. Bill Singer for his 76th birthday.

# Structure of the space of $GL_{4}(\mathbb{Z}_{2})$-coinvariants
$\mathbb{Z}_{2}\otimes_{GL_{4}(\mathbb{Z}_{2})}PH_{*}(\mathbb{Z}_{2}^{4},\mathbb{Z}_{2})$
in some generic degrees
and its application

Đặng Võ Phúc  Faculty of Education Studies, University of Khanh Hoa, Viet nam
<EMAIL_ADDRESS>
###### Abstract.

Let $A$ denote the Steenrod algebra at the prime 2 and let $k=\mathbb{Z}_{2}.$
An open problem of homotopy theory is to determine a minimal set of
$A$-generators for the polynomial ring
$P_{q}=k[x_{1},\ldots,x_{q}]=H^{*}(k^{q},k)$ on $q$ generators
$x_{1},\ldots,x_{q}$ with $|x_{i}|=1.$ Equivalently, one can write down
explicitly a basis for the graded vector space $Q^{\otimes
q}:=k\otimes_{A}P_{q}$ in each non-negative degree $n.$ This is the content of
the classical ”hit problem” in literature [30]. Based on this problem, we are
interested in the $q$-th cohomological transfer $Tr_{q}^{A}$ of William M.
Singer [39], which is one of the useful tools for describing mod-2 cohomology
of the algebra $A.$ This transfer is a linear map from the space of
$GL_{q}(k)$-coinvariant $k\otimes_{GL_{q}(k)}P((P_{q})_{n}^{*})$ of
$Q^{\otimes q}$ to the $k$-cohomology group of the Steenrod algebra, ${\rm
Ext}_{A}^{q,q+n}(k,k).$ Here $GL_{q}(k)$ is the general linear group of degree
$q$ over the field $k,$ and $P((P_{q})_{n}^{*})$ is the primitive part of
$(P_{q})^{*}_{n}$ under the action of $A.$ Singer conjectured that
$Tr_{q}^{A}$ is a monomorphism, but this remains unanswered for all $q\geq 4.$
The present paper is to devoted to the investigation of this conjecture for
the rank 4 case. More specifically, basing the techniques of the hit problem
of four variables, we explicitly determine the structure of
$k\otimes_{GL_{4}(k)}P((P_{4})_{n}^{*})$ in some generic degrees $n.$ Applying
these results and a representation of $Tr_{4}^{A}$ over the lambda algebra, we
notice that Singer’s conjecture is true for the rank 4 transfer in those
degrees $n$. Also, we give some conjectures on the dimensions of
$k\otimes_{GL_{q}(k)}((P_{4})_{n}^{*})$ for the remaining degrees $n.$ As a
consequence, Singer’s conjecture holds for $Tr_{4}^{A}.$ This study and our
previous results [36, 37] have been provided a panorama of the behavior of the
fourth cohomological transfer.

###### keywords:

Adams spectral sequences; Primary cohomology operations; Steenrod algebra;
Lambda algebra; Hit problem; Actions of groups on commutative rings; Algebraic
transfer

###### 2010 Mathematics Subject Classification:

55Q45, 13A50, 55S10, 55S05, 55T15, 55R12

###### Contents

  1. 1 Introduction
  2. 2 Outline of main results
  3. 3 Final remarks
  4. 4 Proofs of main results
    1. 4.1 Proof of Theorem 2.2
    2. 4.2 Proof of Theorem 2.5
    3. 4.3 Proof of Theorem 2.7
    4. 4.4 Proof of Theorem 2.9
    5. 4.5 Proof of Theorem 2.11

##  1\. Introduction

We will adopt the following notations and conventions throughout the paper. We
work at the prime 2 and write $k:=\mathbb{Z}_{2}$ for the field of two
elements. Let us denote by $A$ the classical, singly-graded Steenrod algebra
over $k,$ which is the ring of stable operations on cohomology with $k$
coefficients generated by the Steenrod squares $Sq^{t}$ for $t\geq 0$ modulo
the Adem relations. Let $P_{q}=k[x_{1},\ldots,x_{q}]=H^{*}(k^{q},k)$ be the
polynomial ring on generators $x_{i}$ of degree $1,$ which may be considered
as the symmetric power algebra on $(k^{q})^{*}$ and as an unstable left
$A$-module. Here $k^{q}$ denotes a rank $q$ elementary abelian 2-group, which
views as a $q$-dimensional $k$-vector space. The algebra $A$ acts upon $P_{q}$
by use the Cartan formula $Sq^{t}(fg)=\sum_{i+j=t}Sq^{i}(f)Sq^{j}(g)$ for
arbitrary $f,\,g\in P_{q},$ and the rule that $Sq^{t}(x_{j})$ is $x_{j}$ if
$t=0,$ $x_{j}^{2}$ if $t=1,$ and is 0 otherwise. The algebra $P_{q}$ is
equipped with the usual structure of a right module over the general linear
group $GL_{q}(k)$ by means of substitution of variables as follows: For each
$\sigma=(\sigma_{i,j})\in GL_{q}(k)$ and any $f(x_{1},\ldots,x_{q}):=f\in
P_{q},$ one defines

$(f\sigma)(x_{1},\ldots,x_{q})=f(x_{1}\sigma,\ldots,x_{q}\sigma)=f(\sum_{1\leq
i\leq q}x_{i}\sigma_{i,1},\ldots,\sum_{1\leq i\leq q}x_{i}\sigma_{i,q}).$

Consider the tensor product $Q^{\otimes q}:=k\otimes_{A}P_{q}$. We notice that
the mapping $k\to End(Q^{\otimes q})$ is a ring homomorphism in which
$End(Q^{\otimes q})$ is the ring of endomorphism of $Q^{\otimes q}$ (as
abelian group). So, $Q^{\otimes q}$ has the structure of a $k$-module (or,
equivalently, a $k$-vector space). Writing $(P_{q})_{n}=H^{n}(k^{q},k)$ for
the $k$-subspace of $P_{q}$ generated by the homogeneous polynomials of the
non-negative degree $n$ in $P_{q}.$ Then, we have that $(P_{q})_{n}$ is a
right $kGL_{q}(k)$-module for every $n\geq 0$ and that the graded space
$\\{(P_{q})^{*}_{n}\\}_{n\geq 0},$ where $(P_{q})^{*}_{n}$ denotes the dual of
$(P_{q})_{n},$ is the divided power algebra
$\Gamma(a_{1}^{(1)},\ldots,a_{q}^{(1)})$ generated by
$a_{1}^{(1)},\ldots,a_{q}^{(1)},$ in which $a_{i}^{(1)}$ is the linear dual to
$x_{i}$. Moreover, this $\Gamma(a_{1}^{(1)},\ldots,a_{q}^{(1)})$ is a
bicommutative Hopf algebra with the vector space basis $a_{1}^{(i_{1})}\ldots
a_{q}^{(i_{q})},\,i_{r}\geq 0,$ for all $1\leq r\leq q,$ with multiplication

$\prod_{1\leq r\leq q}a_{r}^{(i_{r})}\prod_{1\leq r\leq
q}a_{r}^{(j_{r})}=\prod_{1\leq r\leq q}\binom{i_{r}+j_{r}}{i_{r}}\prod_{1\leq
r\leq q}a_{r}^{(i_{r}+j_{r})}.$

The right $A$-module structure of this algebra is described by the rule

$(a^{(n)}_{i})Sq^{t}=\binom{n-t}{t}a^{(n-t)}_{i}=Sq_{*}^{t}(a^{(n)}_{i})$

and Cartan’s formula for Steenrod squares on cross products. The group
$GL_{q}(k)$ acts on $\Gamma(a_{1}^{(1)},\ldots,a_{q}^{(1)})$ by the formula

$\sigma(\prod_{1\leq r\leq q}a_{r}^{(i_{r})})=(\sum_{1\leq i\leq
q}\sigma_{i,1}a_{i}^{(1)})^{(i_{1})}(\sum_{1\leq i\leq
q}\sigma_{i,2}a_{i}^{(1)})^{(i_{2})}\ldots(\sum_{1\leq i\leq
q}\sigma_{i,q}a_{i}^{(1)})^{(i_{q})},$

for all $\sigma=(\sigma_{i,j})\in GL_{q}(k).$ So,
$\Gamma(a_{1}^{(1)},\ldots,a_{q}^{(1)})$ is a left $kGL_{q}(k)$-module. Let us
denote by $Q_{n}^{\otimes q}:=(k\otimes_{A}P_{q})_{n}$ the $k$-vector subspace
of $Q^{\otimes q}$ consisting of all the classes represented by the elements
in $(P_{q})_{n}.$ The action of $GL_{q}(k)$ commutes with that of the Steenrod
squares $Sq^{t}$ on $P_{q}$, (i.e., $Sq^{t}:(P_{q})_{n-t}\to(P_{q})_{n}$ is a
map of $kGL_{q}(k)$-modules, and $Sq^{t}(f)\sigma=Sq^{t}(f\sigma)$ for every
$f\in(P_{q})_{n-t},\,\sigma\in GL_{q}(k)$) and therefore there exists an
action of $GL_{q}(k)$ on $Q^{\otimes q}_{n}.$

Let us now recall that an interesting problem of homotopy theory is to
determine the set of homotopy classes $[\mathbb{S}^{n+q},\mathbb{S}^{n}]$ of
continuous based map between spheres. It is known that for $n+q>0,$ these sets
have a natural group structure and they are abelian when $n+q>1.$ The
Freudenthal suspension theorem in [7] showed a relationship between the groups
$[\mathbb{S}^{n+q},\mathbb{S}^{n}]$ for fixed $q$ and varying $n.$ The
supension map induces a sequence:

$\cdots\longrightarrow[\mathbb{S}^{n-1+q},\mathbb{S}^{n-1}]\longrightarrow[\mathbb{S}^{n+q},\mathbb{S}^{n}]\longrightarrow[\mathbb{S}^{n+1+q},\mathbb{S}^{n+1}]\longrightarrow\cdots$

of group homomorphisms, and when $n+q>1,$ these homomorphisms are
isomorphisms. Then, the stable value $[\mathbb{S}^{n+q},\mathbb{S}^{n}]$ for
$n$ sufficiently large is known as the $q$-th stable homotopy group of
spheres, $\pi_{q}.$ The cohomology of $A$ with $k$-coefficients, ${\rm
Ext}_{A}(k,k)=\\{{\rm Ext}^{q,t}_{A}(k,k)\\}_{(q,t)\in\mathbb{Z}^{2},\,q\geq
0,\,t\geq 0}$ is an object of much interest in Algebraic topology. It features
prominently in homotopy theory as the $E_{2}$-page of the Adams spectral
sequence (the ASS) $E_{2}^{q,t}={\rm Ext}^{q,t}_{A}(k,k)$ for the computation
of the groups $\pi_{q}.$ (It is to be noted that a spectral sequence consists
of a sequence of intermediate dual chain complexes called pages
$E_{0},E_{1},E_{2},\ldots,$ with differentials denoted by
$d_{0},d_{1},d_{2},\ldots,$ such that $E_{t+1}$ is the cohomology of $E_{t}.$)
Although the bigraded algebra ${\rm Ext}_{A}(k,k)$ has been intensively
studied by many authors like Adams [1], Adem [2], Wall [47], Wang [48], Lin
[22], Chen [8]), its structure remains largely mysterious. Up to now, it
remains open for every cohomological degree $q>5.$ The May spectral sequence
[24] is the best way to compute ${\rm Ext}_{A}(k,k)$ by hand. Another tool
that is also quite efficient to study the cohomology of $A$ is the Singer
cohomological ”transfer” [39], which is a linear transformation

$Tr_{q}^{A}:k\otimes_{GL_{q}(k)}P((P_{q})_{n}^{*})\to{\rm Ext}_{A}^{\dim
k^{q},\dim k^{q}+n}(k,k)={\rm Ext}_{A}^{q,q+n}(k,k),$

where
$P((P_{q})_{n}^{*}):=\langle\\{\theta\in(P_{q})_{n}^{*}:\,(\theta)Sq^{i}=0,\,\mbox{for
all $i>0$}\\}\rangle=(Q_{n}^{\otimes q})^{*},$ the space of primitive homology
classes as a representation of $GL_{q}(k)$ for all $n$ and the coinvariant
$k\otimes_{GL_{q}(k)}P((P_{q})_{n}^{*})$ is isomorphic as an $k$-vector space
to $(Q_{n}^{\otimes q})^{GL_{q}(k)},$ the subspace of $GL_{q}(k)$-invariants
of $Q^{\otimes q}.$ The Singer transfer has been studying for a long time: see
Boardman [4], Chơn and Hà [9], Crossley [10], Hà [15], Hưng [18], Hưng-Quỳnh
[19], Minami [26], the present writer [31, 33, 34, 35, 36, 37], Sum [44, 46],
and others. By the works of Singer himself [39] and Boardman [4], $Tr_{q}^{A}$
is known to be an isomorphism for $q\leq 3.$ Moreover, Singer [39] shows that
the ”total” transfer $Tr^{A}:=\\{Tr^{A}_{q}\\}_{q\geq
0}:\\{k\otimes_{GL_{q}(k)}P((P_{q})_{n}^{*})\\}_{q\geq
0}\longrightarrow\\{{\rm Ext}_{A}^{q,q+n}(k,k)\\}_{q\geq 0}$ is a homomorphism
of bigraded algebras with respect to the product by concatenation in the
domain and the usual Yoneda product for the Ext group. These works
demonstrated that the cohomological transfer is highly nontrivial.
Furthermore, we knew that this transfer homomorphism is induced over the
$E_{2}$-page of the ASS by the geometrical transfer map
$\Sigma^{\infty}(B(k^{q})_{+})\longrightarrow\Sigma^{\infty}(\mathbb{S}^{0})$
between the suspension spectrum in stable homotopy category (see also Mitchell
[28]). Here $\mathbb{S}^{0}=\\{0,1\\}$ with $0$ as base-point. The work of
Minami [27] tells us that these transfers are closely related to the problem
of finding permanent cycles in the ASS. In the second cohomology groups of
$A$, by Mahowald [23] and Lin-Mahowald [21], the classes $h_{1}h_{j}$ for
$j\geq 3$ and $h_{j}^{2}$ for $0\leq j\leq 5,$ are known to be the permanent
cycles in the ASS. In 2016, Hill, Hopkins, and Ravenel [16] claimed that when
$j\geq 7,$ the class $h_{j}^{2}$ is not a permanent cycle in the ASS. The case
$j=6$ has not been solved and has the status of a hypothesis by Snaith [41].
Most recently, Snaith’s conjecture can be established by Akhmet’ev [3]. The
question of whether these $h_{j}^{2}$ are the permanent cycles in the ASS or
not is called Kervaire invariant problem in literature [6]. This is one of the
oldest unresolved issues in Differential and Algebraic topology. Return to
Singer’s transfer, in [39], Singer sets up the following conjecture, which has
not been proven or refuted for $q\geq 4.$

######  Conjecture 1.1.

The transfer homomorphism $Tr_{q}^{A}$ is a monomorphism.

The present study is to devoted to the investigation of this conjecture for
$q=4.$ Beyond the above methods, the mod two lambda algebra $\Lambda$ of
Bousfield et al. [5] can also be used as an useful tool to describe mysterious
Ext groups. One can view $\Lambda$ as the $E_{1}$-term of the classical Adams
spectral sequence converging to the 2-component of the stable homotopy groups
of spheres. For the reader’s convenience, let us note again that $\Lambda$ is
an associative differential bigraded algebra with generators
$\lambda_{n}\in\Lambda^{1,n}$ ($n\geq 0$) and the Adem relations

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt\lambda_{i}\lambda_{2i+n+1}&=\sum_{j\geq
0}\binom{n-j-1}{j}\lambda_{i+n-j}\lambda_{2i+1+j}\\\ &\quad\quad(i\geq 0,\
n\geq 0)\end{array}$ (1)

with differential

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0ptd(\lambda_{n-1})&=\sum_{j\geq
1}\binom{n-j-1}{j}\lambda_{n-j-1}\lambda_{j-1}\\\ &\quad\quad\quad(n\geq
1),\end{array}$ (2)

where $d(\lambda_{0})=0.$ We refer to [48] for the relations (1) and [5, 38]
for that the differential in (2) is a well-defined endomorphism of $\Lambda.$
According to [48], there is a unique differential algebra endomorphism
$\theta:\Lambda\to\Lambda$ with $\theta(\lambda_{n})=\lambda_{2n+1}.$ This
map, which is one-to-one, induces the classical $Sq^{0}$, an endomorphism of
${\rm Ext}_{A}(k,k)$. Indeed, Palmieri [29] explained that as follows: It is
well-known that the graded dual $A^{*}$ of $A$ is a commutative Hopf algebra
over $k$, and as an algebra over $k,$ $A^{*}\cong k[\xi_{1},\xi_{2},\ldots],$
the polynomial algebra on generators $\xi_{i}$ of degree $2^{i}-1.$ Thence,
the Frobenius map of $A^{*}$ ($A^{*}\longrightarrow
A^{*},\,\xi\longmapsto\xi^{2}$) is a Hopf algebra map. Due to Priddy [38],
$\lambda_{n}$ is represented in the cobar construction by $[\xi_{1}^{n+1}].$
Due to May [25], $Sq^{0}$ is induced by the dual to the Frobenius map and
therefore, $Sq^{0}$ takes $[\xi_{1}^{n+1}]$ to $[\xi_{1}^{2(n+1)}].$ Thus, it
agrees with endomorphism induced by $\theta.$ Alternatively, people have shown
that $Sq^{0}$ commutes with the so-called Kameko $Sq^{0}$ through the
cohomological transfer $Tr_{q}^{A}$ (see also [4], [26] for discussions
therein). Now for non-negative integers $j_{1},\ldots,j_{q}$, a monomial
$\prod_{1\leq s\leq q}\lambda_{j_{s}}$ in $\Lambda$ is said to be an
admissible of length q if $j_{s}\leq 2j_{s+1}$ for all $1\leq s\leq q-1.$ The
admissible monomials form an additive basis of $\Lambda$ by [5, 38]. We denote
by $\Lambda^{q,n}$ the $k$-vector subspace of $\Lambda$ generated by all the
admissible monomials of length $q.$ In [9], Chơn and Hà gave an interesting
linear transformation $\psi_{q}:(P_{q})_{n}^{*}\longrightarrow\Lambda^{q,n},$
which is determined by $\psi_{q}(\prod_{1\leq s\leq
q}a_{s}^{(j_{s})})=\lambda_{j_{q}}$ if $q=1,$ while $\psi_{q}(\prod_{1\leq
s\leq q}a_{s}^{(j_{s})})=\sum_{k\geq
j_{q}}\psi_{q-1}(Sq_{*}^{k-j_{q}}(\prod_{1\leq s\leq
q-1}a_{s}^{(j_{s})}))\lambda_{k}$ if $q>1,$ for any $\prod_{1\leq s\leq
q}a_{s}^{(j_{s})}$ in $(P_{q})_{n}^{*}.$ This map can be considered as the
$E_{1}$-level of the Singer transfer; further the authors have stated the
following, which is a dual version of the one in Hưng [17].

######  Theorem 1.2.

With the notation chosen, if $\zeta\in P((P_{q})_{n}^{*}),$ then
$\psi_{q}(\zeta)$ is a cycle in $\Lambda$ and is a representative of
$Tr_{q}^{A}([\zeta]).$

Our motivation to write up this work is to describe the structure of the
coinvariants $k\otimes_{GL_{q}(k)}P((P_{k})_{n}^{*})$ and also to investigate
Conjecture 1.1 in the rank 4 case. One of the major difficulties involved in
studying Singer’s conjecture is that we do not determine a representation of
$GL_{q}(k)$ (or a basis of $k\otimes_{GL_{q}(k)}P((P_{q})_{n}^{*})$). So, our
current goal is to make some progress towards that problem when $q=4$. More
precisely, by using the techniques in the hit problem of four variables (see
the works of Sum [42, 43]), we explicitly determine the structure of
$k\otimes_{GL_{4}(k)}P((P_{4})_{n}^{*})$ in some generic degrees $n.$ Applying
these results and the representation in the algebra $\Lambda$ of the fourth
transfer, we show that Conjecture 1.1 holds for $q=4$ and degrees given. We
also give some conjectures on the dimensions of
$k\otimes_{GL_{q}(k)}((P_{4})_{n}^{*})$ for the remaining degrees $n.$ As a
consequence, Singer’s conjecture for the rank 4 transfer has been favored.
From these results and our previous works [36, 37], we obtain a complete
picture of the behavior of $Tr_{4}^{A}$. Our method is completely different
from that of Singer [39], Boardman [4], Chơn-Hà [9], Hà [15], Hưng [18], Hưng-
Quỳnh [19] in the study of the domain of the cohomological transfer.

##  2\. Outline of main results

To motivate the statement of our main result, we review the hit problem of
Frank Peterson [30] and some known results on the graded space ${\rm
Ext}_{A}^{q,*}(k,k)$ for $q\leq 4.$ Firstly, we say that a polynomial
$f\in(P_{q})_{n}$ is ”hit” (or $A$-decomposable), if it satisfies a hit
equation, $f=\sum_{t>0}Sq^{t}(f_{t})$ for some $Sq^{t}\in A$ and
$f_{t}\in(P_{q})_{n-t}.$ Said differently, $f$ belongs to
$kGL_{q}(k)$-submodule,
$\widehat{A}(P_{q})_{n}:=(P_{q})_{n}\cap\widehat{A}P_{q}=(P_{q})_{n}\cap\sum_{t>0}{\rm
Im}(Sq^{t}),$ where $\widehat{A}$ is the kernel of the epimorphism of graded
$k$-algebras $A\to k$ given by $Sq^{0}\longmapsto 1$ and $Sq^{i}\longmapsto 0$
for all $i>0.$ For instance, $x^{2}\in(P_{1})_{2}$ is hit since
$x^{2}=Sq^{1}(x).$ In general, it is not easy to determine whether a
polynomial is hit, except for special cases given in Theorem 4.5 (see section
four). Solving the hit problem is to determine a minimal set of $A$-generators
for the polynomial ring $P_{q}$ in each degree $n.$ Equivalently, when $k$ has
the trivial $A$ action, one can write down explicitly a minimal generating set
for the ”cohit” $k$-module,

${\rm Tor}^{A}_{0,n}(k,P_{q})=Q_{n}^{\otimes
q}\cong(P_{q})_{n}/\widehat{A}(P_{q})_{n},\ q\geq 1,\ n\geq 0.$

The structure of this space, which is closely related to some classical
problems in homotopy theory (e.g., cobordism theory of manifolds, modular
representation theory of the general linear group, stable homotopy type of
classifying spaces of finite groups), has explicitly been determined for
$q\leq 4$ and all degrees $n$: see the works of Peterson [30], Kameko [20] and
Sum [42, 43]. Due to Minami [26], the hit problem also plays an important role
for studying permanent cycles in the ASS. Recently, the problem for $q\geq 5$
in some certain degrees is known by Sum [45, 46] and the present author [31,
32, 33, 34, 35, 37]. By these, it seems likely that an explicit description of
$Q_{n}^{\otimes q}$ for general $q$ will appear in the near future. Hit
problems for the symmetric polynomials have been investigated by the works of
Janfada and Wood (see [11, 12, 13, 14]). The following Kameko maps [20] are
often used in studying the hit problem: For each $n\geq 0,$ the down Kameko
map $\overline{Sq}^{0}:(P_{q})_{2n+q}\to(P_{q})_{n}$ is a surjective linear
map defined on monomials by $\overline{Sq}^{0}(f)=g$ if $f=\prod_{1\leq i\leq
q}x_{i}g^{2}$ and $\overline{Sq}^{0}(f)=0$ otherwise. This map sends hit
polynomials to hit polynomials. The up Kameko map
$\varphi:(P_{q})_{n}\to(P_{q})_{2n+q}$ is an injective linear map defined on
monomials by $\varphi(g)=\prod_{1\leq i\leq q}x_{i}g^{2}.$ In general, this
map does not send hit polynomials to hit polynomials. However, let
$\mu(n)=\mbox{min}\big{\\{}*\in\mathbb{N}:\ \alpha(n+*)\leq*\big{\\}},$ where
the $\alpha$ function counts the number of ones in the binary expansion of its
argument. Then, if $\mu(2n+q)=q$, then $\overline{Sq}^{0}$ induces an
isomorphism of $kGL_{q}(k)$-modules $\overline{Sq}^{0}:Q^{\otimes q}_{2n+q}\to
Q^{\otimes q}_{n}$ with inverse $\varphi.$ In [30], Peterson conjectured that
$Q^{\otimes q}_{n}$ is trivial if and only if $\mu(n)>q.$ (His motivation for
this was to prove that if $\mathcal{M}$ is a smooth manifold of dimension $n$
such that all products of length $q$ of Stiefel-Whiney classes of its nomal
bundle vanish, then either $\alpha(n)\leq q$ or $\mathcal{M}$ is cobordant to
zero.) A number of researchers have investigated Peterson’s conjecture, but at
last it was solved by Wood [49]. Additionally, in [40], Singer proved another
result which generalizes the Peterson conjecture and identifies a new class of
monomials in $\widehat{A}(P_{q})_{n}.$ So, from these facts, it is sufficient
to solve the hit problem in degrees $n$ satisfying $\mu(n)<q.$ By this
condition, a careful but straightforward computation shows that $n$ is of the
”generic” form:

$n=r(2^{s}-1)+v.2^{s},$ (3)

where $0\leq\mu(v)<r<q$ and $s\geq 0$. Thus, we only need to study the
behavior of the algebraic transfer at these degrees. Next, we provide a brief
information on the graded space ${\rm Ext}_{A}^{q,*}(k,k)$ for $q\leq 4.$

######  Theorem 2.1 (see Adams [1], Adem [2], Wall [47], Wang [48], Lin [22]).

The following hold:

  1. i)

${\rm Ext}_{A}^{1,*}(k,k)$ is generated by $h_{i}$ for $i\geq 0;$

  2. ii)

${\rm Ext}_{A}^{2,*}(k,k)$ is generated by $h_{i}h_{j}$ for $j\geq i\geq 0$
and $j\neq i+1;$

  3. iii)

${\rm Ext}_{A}^{3,*}(k,k)$ is generated by $h_{i}h_{j}h_{\ell},\,c_{t}$ for
$t\geq 0;$ $\ell\geq j\geq i\geq 0,$ and subject only to the relations
$h_{i}h_{i+1}=0,\ h_{i}h_{i+2}^{2}=0$ and $h_{i}^{3}=h^{2}_{i-1}h_{i+1};$

  4. iv)

${\rm Ext}_{A}^{4,*}(k,k)$ is generated by
$h_{i}h_{j}h_{\ell}h_{m},\,h_{u}c_{v},\,d_{t},\,e_{t},\,f_{t},\,g_{t+1},\,p_{t},\,D_{3}(t),\,p^{\prime}_{t}$
for $m\geq\ell\geq j\geq i\geq 0$, $u,\ v,\ t\geq 0$, and subject to the
relations in iii) together with
$h^{2}_{i}h^{2}_{i+3}=0,\,h_{v-1}c_{v}=0,\,h_{v}c_{v}=0,\,h_{v+2}c_{v}=0$ and
$h_{v+3}c_{v}=0.$

We are now in a position to formulate the main results of the paper. For this
let us study Conjecture 1.1 in the rank $4$ case and the generic degrees of
the form (3) where

$(r,v)=\left\\{\begin{array}[]{ll}(3,2^{t+1}-1)&\ \mbox{for $t\neq
3$},\\\\[2.84526pt] (2,2^{t}-1)&\ \mbox{for $t=1$ and $t\geq
5$},\\\\[2.84526pt] \end{array}\right.$

We refer the reader to some remarks in section three for explaining why we
need to calculate these degrees. We first consider the pair
$(r,v)=(3,2^{t+1}-1)$ for $t\neq 3.$ Thence, one has the following cases.

The case $t=1$. Based upon an admissible basis of the $k$-vector space
$Q^{\otimes 4}_{3(2^{s}-1)+3.2^{s}}$ in [43], we find that

######  Theorem 2.2.

For a positive integer $s$, then

$\dim
k\otimes_{GL_{4}(k)}P((P_{4})_{3(2^{s}-1)+3.2^{s}}^{*})=\left\\{\begin{array}[]{ll}0&\mbox{if
$s=2$},\\\ 1&\mbox{if $s\neq 2$}.\end{array}\right.$

Let us sketch the proof of the theorem. Firstly, by Sum [43], the dimensions
of the $k$-vector spaces $Q^{\otimes 4}_{3(2^{s}-1)+3.2^{s}}$ are computed as
follows:

$\dim Q^{\otimes
4}_{3(2^{s}-1)+3.2^{s}}=\left\\{\begin{array}[]{ll}46&\mbox{if $s=1$},\\\
94&\mbox{if $s=2$},\\\ 105&\mbox{if $s\geq 3$}.\end{array}\right.$

Moreover, they have an admissible monomial basis, which are given in [42].
Now, thanks to these results, for $s\in\\{1,2\\},$ a direct computation
indicates that

$k\otimes_{GL_{4}(k)}P((P_{4})_{3(2^{s}-1)+3.2^{s}}^{*})=\left\\{\begin{array}[]{ll}\langle[\zeta_{1}]\rangle&\mbox{if
$s=1$},\\\ 0&\mbox{if $s=2$},\\\ \end{array}\right.$

where
$\zeta_{1}=a_{1}^{(1)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(2)}+a_{1}^{(1)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(1)}+a_{1}^{(1)}a_{2}^{(5)}a_{3}^{(2)}a_{4}^{(1)}+a_{1}^{(1)}a_{2}^{(6)}a_{3}^{(1)}a_{4}^{(1)}$
belongs to $P((P_{4})_{3(2^{1}-1)+3.2^{1}}^{*}).$ By the unstable condition,
to verify that $\zeta_{1}$ is $\widehat{A}$-annihilated, we only need to
consider the effects of $Sq^{1}$ and $Sq^{2}$.

For $s\geq 3,$ based on the basis of $Q^{\otimes 4}_{3(2^{s}-1)+3.2^{s}},$ we
obtain

$\dim k\otimes_{GL_{4}(k)}P((P_{4})_{3(2^{s}-1)+3.2^{s}}^{*})\leq 1.$ (4)

On the other side, it is easy to check that the elements
$\zeta_{s}=a_{1}^{(0)}a_{2}^{(2^{s+1}-1)}a_{3}^{(2^{s+1}-1)}a_{4}^{(2^{s+1}-1)}$
in $(P_{4})^{*}_{3(2^{s}-1)+3.2^{s}}$ are $\widehat{A}$-annihilated. Moreover,
notice that $\lambda_{2^{s}-1}\in\Lambda^{1,2^{s}-1}$ are a cycle in the
algebra $\Lambda$ and $h_{s}=[\lambda_{2^{s}-1}]\in{\rm
Ext}_{A}^{1,2^{s}}(k,k).$ Then, since $\zeta_{s}\in
P((P_{4})_{3(2^{s}-1)+3.2^{s}}^{*})$, by Theorem 1.2, deduce that the cycles
$\psi_{4}(\zeta_{s})=\lambda_{0}\lambda_{2^{s+1}-1}^{3}$ in $\Lambda$ are
representative of the non-zero elements $h_{0}h_{s+1}^{3}\in{\rm
Ext}_{A}^{4,6.2^{s}+1}(k,k).$ These lead to $h_{0}h_{s+1}^{3}\in{\rm
Im}(Tr_{4}^{A}).$ Invoking Theorem 2.1, it follows that

${\rm Ext}_{A}^{4,6.2^{s}+1}(k,k)\\\ =\left\\{\begin{array}[]{ll}\langle
h_{0}h_{2}^{3},h_{1}c_{0}\rangle=\langle h_{1}c_{0}\rangle&\mbox{if $s=1$},\\\
\langle h_{0}h_{s+1}^{3}\rangle&\mbox{if $s\geq 2$},\end{array}\right.$ (5)

with $h_{0}h_{s+1}^{3}=0$ for $s=2$ and
$h_{0}h_{s+1}^{3}=h_{0}h_{s}^{2}h_{s+2}\neq 0$ for $s\geq 3.$ These data and
the inequality (4) imply that the coinvariants space
$k\otimes_{GL_{4}(k)}P((P_{4})_{3(2^{s}-1)+3.2^{s}}^{*})$ is $1$-dimensional.
Moreover, by a direct computation using the monomial bases of $Q^{\otimes
4}_{3(2^{s}-1)+3.2^{s}},$ one can obtain that the coinvariants
$k\otimes_{GL_{4}(k)}P((P_{4})_{3(2^{s}-1)+3.2^{s}}^{*})$ are generated by the
classes $[\zeta_{s}],$ for all $s\geq 3.$

######  Remark 2.3.

Clearly, $\lambda_{3}^{2}\lambda_{2}$ is a representative of the non-zero
element $c_{0}\in{\rm Ext}_{A}^{3,11}(k,k).$ Then, since $\zeta_{1}$ is
$\widehat{A}$-annihilated, by a direct computation using the representation of
$Tr_{4}^{A}$ over $\Lambda$ and Theorem 1.2, we deduce that

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt\psi_{4}(a_{1}^{(1)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(2)})&=\lambda_{1}\lambda_{3}^{2}\lambda_{2}+\lambda_{1}\lambda_{3}\lambda_{4}\lambda_{1}+\lambda_{1}\lambda_{4}\lambda_{3}\lambda_{1},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\psi_{4}(a_{1}^{(1)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(1)})&=\lambda_{1}\lambda_{3}\lambda_{4}\lambda_{1}+\lambda_{1}\lambda_{4}\lambda_{3}\lambda_{1}+\lambda_{1}\lambda_{5}\lambda_{2}\lambda_{1},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\psi_{4}(a_{1}^{(1)}a_{2}^{(5)}a_{3}^{(2)}a_{4}^{(1)})&=\lambda_{1}\lambda_{5}\lambda_{2}\lambda_{1}+\lambda_{1}\lambda_{6}\lambda_{1}^{2},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\psi_{4}(a_{1}^{(1)}a_{2}^{(6)}a_{3}^{(1)}a_{4}^{(1)})&=\lambda_{1}\lambda_{6}\lambda_{1}^{2},\end{array}$

and therefore the cycle
$\psi_{4}(\zeta_{1})=\lambda_{1}\lambda_{3}^{2}\lambda_{2}$ in $\Lambda^{4,9}$
is a representative of the element $h_{1}c_{0}\in{\rm Ext}_{A}^{4,13}(k,k).$
This fact and the equality (5) lead to the non-zero element $h_{1}c_{0}$ being
in the image of $Tr_{4}^{A}.$

Combining Remark 2.3 with Theorem 2.2 and the equality (5), it may be
concluded that

######  Corollary 2.4.

The Singer transfer

$Tr_{4}^{A}:k\otimes_{GL_{4}(k)}P((P_{4})_{6.2^{s}-3}^{*})\to{\rm
Ext}_{A}^{4,6.2^{s}+1}(k,k)$

is an isomorphism for every positive integer $s.$

The case $t=2$. The structure of the coinvariant spaces
$k\otimes_{GL_{4}(k)}P((P_{4})_{3(2^{s}-1)+7.2^{s}}^{*})$ are given as
follows.

######  Theorem 2.5.

With a positive integer $s,$ we have

$k\otimes_{GL_{4}(k)}P((P_{4})_{3(2^{s}-1)+7.2^{s}}^{*})=\left\\{\begin{array}[]{ll}\langle[\zeta]\rangle&\mbox{if
$s=1$},\\\ 0&\mbox{if $s>1$},\\\ \end{array}\right.$

where $\zeta$ is the following sum:

$\begin{array}[]{ll}&a_{1}^{(5)}a_{2}^{(5)}a_{3}^{(5)}a_{4}^{(2)}+a_{1}^{(5)}a_{2}^{(5)}a_{3}^{(6)}a_{4}^{(1)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(8)}a_{4}^{(1)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(8)}a_{4}^{(1)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(3)}a_{2}^{(6)}a_{3}^{(7)}a_{4}^{(1)}\\\
&+a_{1}^{(5)}a_{2}^{(7)}a_{3}^{(4)}a_{4}^{(1)}+a_{1}^{(7)}a_{2}^{(5)}a_{3}^{(4)}a_{4}^{(1)}+a_{1}^{(3)}a_{2}^{(9)}a_{3}^{(4)}a_{4}^{(1)}+a_{1}^{(9)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(1)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(3)}a_{2}^{(9)}a_{3}^{(3)}a_{4}^{(2)}\\\
&+a_{1}^{(9)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(2)}+a_{1}^{(5)}a_{2}^{(9)}a_{3}^{(2)}a_{4}^{(1)}+a_{1}^{(9)}a_{2}^{(5)}a_{3}^{(2)}a_{4}^{(1)}+a_{1}^{(5)}a_{2}^{(10)}a_{3}^{(1)}a_{4}^{(1)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(9)}a_{2}^{(6)}a_{3}^{(1)}a_{4}^{(1)}\\\
&+a_{1}^{(3)}a_{2}^{(11)}a_{3}^{(2)}a_{4}^{(1)}+a_{1}^{(11)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(1)}+a_{1}^{(5)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(4)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(4)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(3)}a_{2}^{(5)}a_{3}^{(5)}a_{4}^{(4)}\\\
&+a_{1}^{(3)}a_{2}^{(12)}a_{3}^{(1)}a_{4}^{(1)}+a_{1}^{(11)}a_{2}^{(4)}a_{3}^{(1)}a_{4}^{(1)}+a_{1}^{(7)}a_{2}^{(8)}a_{3}^{(1)}a_{4}^{(1)}+a_{1}^{(7)}a_{2}^{(7)}a_{3}^{(1)}a_{4}^{(2)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(13)}a_{2}^{(2)}a_{3}^{(1)}a_{4}^{(1)}\\\
&+a_{1}^{(14)}a_{2}^{(1)}a_{3}^{(1)}a_{4}^{(1)}+a_{1}^{(6)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(3)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(3)}+a_{1}^{(3)}a_{2}^{(6)}a_{3}^{(5)}a_{4}^{(3)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(6)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(5)}\\\
&+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(5)}+a_{1}^{(3)}a_{2}^{(6)}a_{3}^{(3)}a_{4}^{(5)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(6)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(6)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(6)}\\\
&+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(8)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(7)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(2)}a_{4}^{(7)}+a_{1}^{(3)}a_{2}^{(6)}a_{3}^{(1)}a_{4}^{(7)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(9)}a_{4}^{(2)}\\\
&+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(10)}a_{4}^{(1)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(7)}a_{4}^{(2)}+a_{1}^{(5)}a_{2}^{(7)}a_{3}^{(3)}a_{4}^{(2)}+a_{1}^{(7)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(2)}.\end{array}$

To check that $\zeta$ is $\widehat{A}$-annihilated, we need only to compute
the actions of $Sq^{i}$ for $i\in\\{1,2,4\\}$ because of the unstable
condition. The proof of the theorem is based upon the admissible bases for the
$k$-vector spaces $Q^{\otimes 4}_{3(2^{s}-1)+7.2^{s}}$ in [42].

It is apparently that the non-zero element $e_{0}$ in ${\rm
Ext}_{A}^{4,21}(k,k)$ is represented by the cycle

$\overline{e}_{0}:=\lambda_{3}^{3}\lambda_{8}+\lambda_{3}\lambda_{5}^{2}\lambda_{4}+\lambda_{3}^{2}\lambda_{7}\lambda_{4}+\lambda_{7}\lambda_{5}\lambda_{3}\lambda_{2}+\lambda_{3}^{2}\lambda_{5}\lambda_{6}$

in $\Lambda^{4,17}.$ Then, since $\zeta\in
P((P_{4})_{3(2^{1}-1)+7.2^{1}}^{*}),$ by a direct computation using the
differential (2) and the representation of $Tr_{4}^{A}$ over $\Lambda,$ we
conclude that

$\psi_{4}(\zeta)=\overline{e}_{0}+d(\lambda_{3}\lambda_{5}\lambda_{10}+\lambda_{3}\lambda_{12}\lambda_{3}+\lambda_{4}\lambda_{7}^{2}+\lambda_{0}\lambda_{11}\lambda_{7})$

is a cycle in $\Lambda^{4,3(2^{1}-1)+7.2^{1}},$ from which one gets

$Tr_{4}^{A}([\zeta])=[\psi_{4}(\zeta)]=[\overline{e}_{0}]=e_{0}.$ (6)

On the other side, using Theorem 2.1, we may deduce that

${\rm Ext}_{A}^{4,10.2^{s}+1}(k,k)\\\ =\left\\{\begin{array}[]{ll}\langle
h_{0}h_{2}h_{3}^{2},e_{0}\rangle=\langle e_{0}\rangle&\mbox{if $s=1$},\\\
\langle h_{0}h_{s+1}h_{s+2}^{2}\rangle=0&\mbox{if $s>1$}.\end{array}\right.$
(7)

Taking Theorem 2.5, together with the equalities (6) and (7), the readers can
easily see that

######  Corollary 2.6.

$Tr_{4}^{A}$ is an isomorphism in the internal degree $10.2^{s}-3$ for any
$s\geq 1.$

The cases $t\geq 4$. By Sum [42], for each $t\geq 4,$ the dimension of the
$k$-vector spaces $Q^{\otimes 4}_{3(2^{s}-1)+2^{s}(2^{t+1}-1)}$ are determined
as follows:

$\dim Q^{\otimes
4}_{3(2^{s}-1)+2^{s}(2^{t+1}-1)}=\left\\{\begin{array}[]{ll}\vskip 6.0pt plus
2.0pt minus 2.0pt150&\mbox{if $s=1$},\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt195&\mbox{if $s=2$},\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt210&\mbox{if
$s\geq 3$},\end{array}\right.$

Thanks to these results, by direct calculations, we obtain the following.

######  Theorem 2.7.

Let $s$ and $t$ be positive integers such that $t\geq 4.$ Then,

$\dim
k\otimes_{GL_{4}(k)}P((P_{4})_{3(2^{s}-1)+2^{s}(2^{t+1}-1)}^{*})=\left\\{\begin{array}[]{ll}1&\mbox{if
$s=1,2$},\\\ 2&\mbox{if $s\geq 3$}.\end{array}\right.$

Moreover,

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt&k\otimes_{GL_{4}(k)}P((P_{4})_{3(2^{s}-1)+2^{s}(2^{t+1}-1)}^{*})\\\
&=\left\\{\begin{array}[]{ll}\langle[a_{1}^{(0)}a_{2}^{(2^{s+1}-1)}a_{3}^{(2^{s+t}-1)}a_{4}^{(2^{s+t}-1)}]\rangle&\mbox{if
$s=1,2$},\\\\[2.84526pt]
\langle[a_{1}^{(0)}a_{2}^{(2^{s+1}-1)}a_{3}^{(2^{s+t}-1)}a_{4}^{(2^{s+t}-1)}],[a_{1}^{(0)}a_{2}^{(2^{s}-1)}a_{3}^{(2^{s}-1)}a_{4}^{(2^{s+t+1}-1)}]\rangle&\mbox{if
$s\geq 3$}.\end{array}\right.\end{array}$

Following Theorem 2.1, one gets

${\rm Ext}_{A}^{4,2^{s+t+1}+2^{s+1}+1}(k,k)=\langle
h_{0}h_{s+1}h_{s+t}^{2},\,h_{0}h_{s}^{2}h_{s+t+1}\rangle.$

It should be noted that if $t=3,$ then $h_{0}h_{s+1}h_{s+t}^{2}=0.$ If $t=1$
and $s>2,$ then $h_{0}h_{s+1}h_{s+t}^{2}=h_{0}h_{s}^{2}h_{s+t+1}$. When $t=2$
and $s>1,$ we have $h_{0}h_{s+1}h_{s+t}^{2}=0$ and
$h_{0}h_{s}^{2}h_{s+t+1}=0.$ In the case in which $s\in\\{1,2\\}$, we have
$h_{0}h_{s}^{2}h_{s+t+1}=0.$ On the other hand, we observe that the elements
$\lambda_{2^{i}-1}\in\Lambda^{1,2^{i}-1}$ are representative of the non-zero
elements $h_{i}\in{\rm Ext}_{A}^{1,2^{i}}(k,k)$ for
$i=0,\,s,\,s+1,,s+t,\,s+t+1$ ($t\geq 4$). So, since
$a_{1}^{(0)}a_{2}^{(2^{s+1}-1)}a_{3}^{(2^{s+t}-1)}a_{4}^{(2^{s+t}-1)}$ and
$a_{1}^{(0)}a_{2}^{(2^{s}-1)}a_{3}^{(2^{s}-1)}a_{4}^{(2^{s+t+1}-1)}$ are
$\widehat{A}$-annihilated, by Theorem 1.2, we claim that the cycles

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt\lambda_{0}\lambda_{2^{s+1}-1}\lambda_{2^{s+t}-1}^{2}&=\psi_{4}(a_{1}^{(0)}a_{2}^{(2^{s+1}-1)}a_{3}^{(2^{s+t}-1)}a_{4}^{(2^{s+t}-1)}),\\\
\lambda_{0}\lambda^{2}_{2^{s}-1}\lambda_{2^{s+t+1}-1}&=\psi_{4}(a_{1}^{(0)}a_{2}^{(2^{s}-1)}a_{3}^{(2^{s}-1)}a_{4}^{(2^{s+t+1}-1)})\end{array}$

in $\Lambda^{4,3(2^{s}-1)+2^{s}(2^{t+1}-1)}$ are representative of the non-
zero elements $h_{0}h_{s+1}h_{s+t}^{2}$ and $h_{0}h_{s}^{2}h_{s+t+1},$
respectively and so, the following is a direct consequence from these data and
Theorem 2.7.

######  Corollary 2.8.

The fourth transfer is an isomorphism when acting on the coinvariant spaces
$k\otimes_{GL_{4}(k)}P((P_{4})_{2^{s+t+1}+2^{s+1}-3}^{*})$ for all $s>0$ and
$t>3.$

Next, consider the pair $(r,v)=(2,2^{t}-1)$ with $t=1,$ we remark that

$Q^{\otimes 4}_{2(2^{s}-1)+2^{s}}\cong{\rm
Ker}[\overline{Sq}^{0}]_{2(2^{s}-1)+2^{s}}\bigoplus Q^{\otimes
4}_{2^{s-1}+2^{s}-3},$

because the Kameko homomorphism

$[\overline{Sq}^{0}]_{2(2^{s}-1)+2^{s}}:=\overline{Sq}^{0}:Q^{\otimes
4}_{2(2^{s}-1)+2^{s}}\to Q^{\otimes 4}_{2^{s-1}+2^{s}-3}$

is an epimorphism of $kGL_{4}(k)$-modules, from which we get

$\begin{array}[]{ll}\dim
k\otimes_{GL_{4}(k)}P((P_{4})_{2(2^{s}-1)+2^{s}}^{*})&\leq\dim({\rm
Ker}[\overline{Sq}^{0}]_{2(2^{s}-1)+2^{s}})^{GL_{4}(k)}\\\ &\qquad+\dim
k\otimes_{GL_{4}(k)}P((P_{4})_{2^{s-1}+2^{s}-3}^{*}).\end{array}$ (8)

The following assertion is based on a monomial basis of ${\rm
Ker}[\overline{Sq}^{0}]_{2(2^{s}-1)+2^{s}}$ in [42].

######  Theorem 2.9.

The subspaces of $GL_{4}(k)$-invariants $({\rm
Ker}[\overline{Sq}^{0}]_{2(2^{s}-1)+2^{s}})^{GL_{4}(k)}$ are trivial for all
$s>0.$

This theorem together with the inequality (8) give the following results:
For $s=1,$ evidently $Q^{\otimes 4}_{2^{1-1}+2^{1}-3}\cong k,$ and so

$k\otimes_{GL_{4}(k)}P((P_{4})_{2^{1-1}+2^{1}-3}^{*})=\langle[1]\rangle.$ (9)

Suppose that $[f]\in k\otimes_{GL_{4}(k)}P((P_{4})_{2(2^{1}-1)+2^{1}}^{*}).$
Then, $[f]$ is dual to $[\widetilde{f}]\in(Q^{\otimes
4}_{2(2^{1}-1)+2^{1}})^{GL_{4}(k)}.$ Since the Kameko map
$[\overline{Sq}^{0}]_{2(2^{1}-1)+2^{1}}$ is an epimorphism, the dual of
$[\overline{Sq}^{0}]_{2(2^{1}-1)+2^{1}}([\widetilde{f}])$ belongs to
$k\otimes_{GL_{4}(k)}P((P_{4})_{2(2^{1}-1)+2^{1}}^{*}).$ So, due to Theorem
2.9 and the equality (9), $[f]$ is dual to
$(\gamma[\varphi(1)]+[\widetilde{f}^{\prime}]),$ where $\gamma\in k,$
$[\widetilde{f}^{\prime}]\in{\rm Ker}[\overline{Sq}^{0}]_{2(2^{1}-1)+2^{1}}$
and the up Kameko map $\varphi$ is determined by

$\begin{array}[]{ll}\varphi:k&\longrightarrow(P_{4})_{2(2^{1}-1)+2^{1}}\\\ \ \
\ \ u&\longmapsto\left\\{\begin{array}[]{ll}0&\mbox{if $u=0$},\\\ \prod_{1\leq
i\leq 4}x_{i}&\mbox{if $u=1$}.\end{array}\right.\end{array}$

By straightforward computations using Lemma 4.4.1(i) in section four and an
admissible monomial basis of $Q^{\otimes 4}_{2(2^{1}-1)+2^{1}},$ show that
$[\widetilde{f}]=0,$ and therefore $[f]=([\widetilde{f}])^{*}=0.$ This means
that $k\otimes_{GL_{4}(k)}P((P_{4})_{2(2^{1}-1)+2^{1}}^{*})$ is trivial.
For $s\in\\{2,4\\},$ combining Theorems 2.2, 2.9 with the inequality (8) and
the fact that the invariant space $(Q^{\otimes
4}_{2^{2-1}+2^{2}-3})^{GL_{4}(k)}$ is trivial (see Sum [44]), it may be
concluded that the coinvariant spaces
$k\otimes_{GL_{4}(k)}P((P_{4})_{2(2^{s}-1)+2^{s}}^{*})$ are trivial, too.
For $s\not\in\\{1,2,4\\},$ the following inequality is immediate from Theorems
2.2 and 2.9 and the inequality (8):

$\dim k\otimes_{GL_{4}(k)}P((P_{4})_{2(2^{s}-1)+2^{s}}^{*})\leq 1.$ (10)

On the other side, we observe that $\lambda_{2^{s}-1}\in\Lambda^{1,2^{s}-1}$
and
$\lambda_{7}^{2}\lambda_{5}=Sq^{0}(\lambda_{3}^{2}\lambda_{2})\in\Lambda^{3,19}$
are cycles in $\Lambda,$ and are representative of $h_{s}\in{\rm
Ext}_{A}^{1,2^{s}}(k,k)$ and $c_{1}=Sq^{0}(c_{0})\in{\rm Ext}_{A}^{3,22}(k,k)$
respectively. Moreover, it is routine to verify that the elements

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt\zeta_{3}&=(a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(7)}a_{4}^{(5)}+a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(9)}a_{4}^{(3)}+a_{1}^{(3)}a_{2}^{(11)}a_{3}^{(5)}a_{4}^{(3)}+a_{1}^{(3)}a_{2}^{(13)}a_{3}^{(3)}a_{4}^{(3)})\in(P_{4})^{*}_{2(2^{3}-1)+2^{3}},\\\
\zeta_{s}&=a_{1}^{(1)}a_{2}^{(2^{s-1}-1)}a_{3}^{(2^{s-1}-1)}a_{4}^{(2^{s+1}-1)}\in(P_{4})^{*}_{2(2^{s}-1)+2^{s}},\
\mbox{for $s\geq 5$}\end{array}$

are $\widehat{A}$-annihilated. So, by Theorem 1.2, it implies that the cycles
$\psi_{4}(\zeta_{3})=\lambda_{3}\lambda_{7}^{2}\lambda_{5}$ and
$\psi_{4}(\zeta_{s})=\lambda_{1}\lambda_{2^{s-1}-1}^{2}\lambda_{2^{s+1}-1}$ in
$\Lambda$ are representative of the elements $h_{2}c_{1}\in{\rm
Ext}_{A}^{4,3.2^{3}+2}(k,k)$ and $h_{1}h_{s-1}^{2}h_{s+1}\in{\rm
Ext}_{A}^{4,3.2^{s}+2}(k,k),$ respectively. It should be noted that with the
$\widehat{A}$-annihilated elements $a^{(2^{s}-1)}\in(P_{1})_{2^{s}-1}^{*}$ and

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt\widehat{\zeta}&=(a_{1}^{(7)}a_{2}^{(7)}a_{3}^{(5)}+a_{1}^{(7)}a_{2}^{(9)}a_{3}^{(3)}+a_{1}^{(11)}a_{2}^{(5)}a_{3}^{(3)}+a_{1}^{(13)}a_{2}^{(3)}a_{3}^{(3)})\in(P_{3})_{19}^{*},\end{array}$

we have $h_{s}=Tr_{1}^{A}([a_{1}^{(2^{s}-1)}])$ and
$c_{1}=Sq^{0}(c_{0})=Tr^{A}_{3}([\widehat{\zeta}])$ (since the classical
$Sq^{0}$ commutes with the Kameko $Sq^{0}$ via the rank 3 algebraic transfer).
These arguments and the inequality (10) imply that

$\dim k\otimes_{GL_{4}(k)}P((P_{4})_{2(2^{s}-1)+2^{s}}^{*})=1,\ \mbox{for
$s\not\in\\{1,2,4\\}$},$ (11)

and that the cohomological transfer is an epimorphism in the bidegree
$(4,3.2^{s}+2).$ Moreover, according to Theorem 2.1, we easily get

${\rm Ext}_{A}^{4,3.2^{s}+2}(k,k)\\\ =\left\\{\begin{array}[]{ll}\langle
h_{1}h_{3}^{3},h_{2}c_{1}\rangle=\langle h_{2}c_{1}\rangle&\mbox{if $s=3$},\\\
0&\mbox{if $s\in\\{1,2,4\\}$},\\\ \langle h_{1}h_{s}^{3}\rangle&\mbox{if
$s>4$},\\\ \end{array}\right.$ (12)

where $h_{1}h_{s}^{3}=h_{1}h_{s-1}^{2}h_{s+1}\neq 0,$ from which, by (11), we
have immediately

######  Corollary 2.10.

The Singer algebraic transfer is an isomorphism in bidegree $(4,3.2^{s}+2)$
for every positive integer $s.$

Finally, we consider the pair $(r,v)=(2,2^{t}-1)$ for $t\geq 5$ and obtain the
following theorem.

######  Theorem 2.11.

Let $s$ and $t$ be positive integers such that $t\geq 5.$ Then,

$k\otimes_{GL_{4}(k)}P((P_{4})_{2(2^{s}-1)+2^{s}(2^{t}-1)}^{*})\\\
=\left\\{\begin{array}[]{ll}\langle[\zeta_{s,\,t}]\rangle&\mbox{if
$s=1,\,2$},\\\\[2.84526pt]
\langle[\zeta_{s,\,t}],[\widetilde{\zeta}_{s,\,t}]\rangle&\mbox{if
$s=3,\,4$},\\\\[2.84526pt]
\langle[\zeta_{s,\,t}],[\widetilde{\zeta}_{s,\,t}],[\widehat{\zeta}_{s,\,t}]\rangle&\mbox{if
$s\geq 5$},\end{array}\right.$

where

$\begin{array}[]{lll}\vskip 6.0pt plus 2.0pt minus
2.0pt\zeta_{s,\,t}&:=a_{1}^{(1)}a_{2}^{(2^{s}-1)}a_{3}^{(2^{s+t-1}-1)}a_{4}^{(2^{s+t-1}-1)},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\widetilde{\zeta}_{s,\,t}&:=a_{1}^{(0)}a_{2}^{(0)}a_{3}^{(2^{s}-1)}a_{4}^{(2^{s+t}-1)},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\widehat{\zeta}_{s,\,t}&:=a_{1}^{(1)}a_{2}^{(2^{s-1}-1)}a_{3}^{(2^{s-1}-1)}a_{4}^{(2^{s+t}-1)}.\end{array}$

The theorem indicates that the elements $\zeta_{s,\,t},$
$\widetilde{\zeta}_{s,\,t}$ and $\widehat{\zeta}_{s,\,t}$ belong to
$P((P_{4})_{2(2^{s}-1)+2^{s}(2^{t}-1)}^{*}).$ So, by Theorem 1.2,
$\psi_{4}(\zeta_{s,\,t})$ are cycles in $\Lambda;$ moreover and for each
$t\geq 5,$ using the representation in $\Lambda$ of $Tr_{4}^{A},$ it may be
concluded that

$\begin{array}[]{ll}Tr_{4}^{A}([\zeta_{s,\,t}])&=[\psi_{4}(\zeta_{s,\,t})]=[\lambda_{1}\lambda_{2^{s}-1}\lambda^{2}_{2^{s+t-1}-1}]\\\
&=h_{1}h_{s}h_{s+t-1}^{2}\in{\rm Ext}_{A}^{4,2^{s+t}+2^{s}+2}(k,k),\ \mbox{for
$s\geq 1,\ s\neq 2$},\\\\[2.84526pt]
Tr_{4}^{A}([\widetilde{\zeta}_{s,\,t}])&=[\psi_{4}(\widetilde{\zeta}_{s,\,t})]=[\lambda_{0}^{2}\lambda_{2^{s}-1}\lambda_{2^{s+t}-1}]\\\
&=h_{0}^{2}h_{s}h_{s+t}\in{\rm Ext}_{A}^{4,2^{s+t}+2^{s}+2}(k,k),\ \mbox{for
$s\geq 2$},\\\\[2.84526pt]
Tr_{4}^{A}([\widehat{\zeta}_{s,\,t}])&=[\psi_{4}(\widehat{\zeta}_{s,\,t})]=[\lambda_{1}\lambda^{2}_{2^{s-1}-1}\lambda_{2^{s+t}-1}]\\\
&=h_{1}h^{2}_{s-1}h_{s+t}\in{\rm Ext}_{A}^{4,2^{s+t}+2^{s}+2}(k,k),\ \mbox{for
$s\geq 5$}.\end{array}$

On the other side, using Theorem 2.1, one has that

${\rm Ext}_{A}^{4,2^{s+t}+2^{s}+2}(k,k)=\left\\{\begin{array}[]{ll}\langle
h_{1}^{2}h_{7}^{2},D_{3}(2)\rangle&\mbox{if $s=1$ and $t=7$},\\\\[2.84526pt]
\langle h_{1}^{2}h_{t}^{2}\rangle&\mbox{if $s=1$ and $t\geq 5,\,t\neq
7$},\\\\[2.84526pt] \langle h_{0}^{2}h_{2}h_{t+2}\rangle=\langle
h_{1}^{3}h_{t+2}\rangle&\mbox{if $s=2$ and $t\geq 5$},\\\\[2.84526pt] \langle
h_{1}h_{s}h_{s+t-1}^{2},h_{0}^{2}h_{s}h_{s+t},h_{1}h_{s-1}^{2}h_{s+t}\rangle&\mbox{if
$s\geq 3$ and $t\geq 5$},\end{array}\right.$

where $h_{1}h_{s-1}^{2}h_{s+t}=0$ if $s=3,\,4,$ and $t\geq 5.$ Combining these
data with a fact in [18] that the rank 4 transfer does not detect the element
$D_{3}(2)$ in $Sq^{0}$-family $\\{D_{3}(s):\,s\geq 0\\}$ in ${\rm
Ext}_{A}^{4,2^{s}+2^{s+6}}(k,k)$ for all $s\geq 0,$ we are forced to conclude
that

######  Corollary 2.12.

The transfer homomorphism

$Tr_{4}^{A}:k\otimes_{GL_{4}(k)}P((P_{4})_{2(2^{s}-1)+2^{s}(2^{t}-1)}^{*})\to{\rm
Ext}_{A}^{4,2^{s+t}+2^{s}+2}(k,k)$

is not an epimorphism if $s=1$ and $t=7$ and is an isomorphism if $s\geq 1$
and $t\geq 5,\,t\neq 7.$

Moreover, the following remark is useful.

######  Remark 2.13.

Let us consider the generic degrees of the form (3) with $r=4$ and $v=61,$
(i.e, $n=2^{s+6}+2^{s}-4.$) Then, it is easily verified that $\mu(n)=4$ for
all $s>2.$ This leads to the iterated Kameko map
$(\overline{Sq}^{0})^{s-2}:Q^{\otimes 4}_{n}\to Q^{\otimes
4}_{2^{2+6}+2^{2}-4}$ being an isomorphism of $kGL_{4}(k)$-modules for any
$s\geq 2.$ So, by Theorem 2.11 and the results in Sum [44], deduce that the
coinvariant space $k\otimes_{GL_{4}(k)}P((P_{4})_{n}^{*})$ is trivial if $s=0$
and has dimension $1$ if $s\geq 1.$ On the other hand, from Theorem 2.1, it
implies that

${\rm Ext}_{A}^{4,4+n}(k,k)=\left\\{\begin{array}[]{ll}\langle
D_{3}(0)\rangle&\mbox{if $s=0$},\\\ \langle
D_{3}(s),h_{s-1}^{2}h_{s+5}^{2}\rangle&\mbox{if $s\geq 1$}.\end{array}\right.$

Moreover, following Hưng [18], the elements $D_{3}(s)$ are not detected by
$Tr_{4}^{A},$ and therefore the fourth algebraic transfer
$Tr_{4}^{A}:k\otimes_{GL_{4}(k)}P((P_{4})_{2^{s+6}+2^{s}-4}^{*})\to{\rm
Ext}_{A}^{4,2^{s+6}+2^{s}}(k,k)$ is a monomorphism, but not an isomorphism for
every non-negative integer $s.$

Thus, observing from Corollaries 2.4, 2.6, 2.8, 2.10 and 2.12, we can assert
that

######  Corollary 2.14.

Let $s$ and $t$ be two positive integers. Then, Singer’s conjecture for the
rank 4 transfer holds in degree $2^{s+t+1}+2^{s+1}-3$ for $t\neq 3$ and degree
$2^{s+t}+2^{s}-2$ for $t=1$ and arbitrary $t\geq 5.$

##  3\. Final remarks

We conclude this article with some remarks. It may need to be recalled that
$Q^{\otimes 4}_{n}$ is trivial if $\mu(n)>4$ and that $Q^{\otimes 4}_{n}\cong
Q^{\otimes 4}_{(n-4)/2}$ if $\mu(n)=4.$ This means that we need only to study
the cohit module $Q^{\otimes 4}_{n}$ in degrees $n$ with $\mu(n)<4$ (or the
generic degrees $n$ of the form (3)). So, since the domain of the fourth
Singer transfer is dual to the invariant $(Q^{\otimes 4}_{n})^{GL_{4}(k)},$ it
suffices to verify Singer’s conjecture for $Tr_{4}^{A}$ in the following
degrees $n$:

$\begin{array}[]{lll}i)&n&=2^{s+1}-\ell,\ \ell\in\\{1,2,3\\},\\\\[2.84526pt]
ii)&n&=2^{s+t+1}+2^{s+1}-3,\\\\[2.84526pt]
iii)&n&=2^{s+t}+2^{s}-2,\\\\[2.84526pt]
iv)&n&=2^{s+t+u}+2^{s+t}+2^{s}-3,\end{array}$

whenever $s,\,t,\,u$ are the positive integers (see also [43]). It is easy to
check that the above degrees can be rewritten as (3). For example, consider
i), we have $n=\ell(2^{s}-1)+(2-\ell).2^{s}$ with $\ell\leq 2$ and
$\mu(2-\ell)<\ell.$ When $\ell=3$ and any $s\geq 1,$ we may rewrite
$2^{s+1}-3$ as $2^{s+2}-3$ with $s\geq 0$ and so $n=3(2^{s}-1)+1.2^{s}$ where
$\mu(1)=1<3<4.$

The case i) is known by Sum [44]. It is remarkable that, by dualizing and
Sum’s work [44], in this case when the pairs $(\ell,s)=(3,5)$ and $(2,6),$ we
deduce that $k\otimes_{GL_{4}(k)}P((P_{4})_{2^{5+1}-3}^{*})$ is trivial and
that $k\otimes_{GL_{4}(k)}P((P_{4})_{2^{6+1}-2}^{*})$ is 1-dimensional. On the
other hand, following Theorem 2.1 and Remark 2.13, we have seen that ${\rm
Ext}_{A}^{4,2^{5+1}+1}(k,k)=\langle D_{3}(0)\rangle$ and ${\rm
Ext}_{A}^{4,2^{6+1}+2}(k,k)=\langle h_{0}^{2}h_{6}^{2},D_{3}(1)\rangle,$ where
$h_{0}^{2}h_{6}^{2}=[\psi_{4}(a_{1}^{(0)}a_{2}^{(0)}a_{3}^{(63)}a_{4}^{(63)})]=Tr_{4}^{A}([a_{1}^{(0)}a_{2}^{(0)}a_{3}^{(63)}a_{4}^{(63)}])\in{\rm
Im}(Tr_{4}^{A})$ and $D_{3}(s)\not\in{\rm Im}(Tr_{4}^{A})$ for every non-
negative integer $s.$ Combining these data, we may state that $Tr_{4}^{A}$ is
not an isomorphism when acting on the coinvariants
$k\otimes_{GL_{4}(k)}P((P_{4})_{2^{5+1}-3}^{*})$ and
$k\otimes_{GL_{4}(k)}P((P_{4})_{2^{6+1}-2}^{*}).$

The case iii) where $t=2,4$ and the case iv) have been investigated by us in
[36, 37]. The results have been computed in this paper for the case ii) with
$t\neq 3$ and the case iii) with $t=1$ and $t\geq 5.$ Thus, Singer’s
conjecture is undetermined for the cases ii) and iii) where $t=3$. We now will
discuss these remaining cases.

Firstly, let us consider the case ii) in which $t=3.$ Naturally, we are
tempted to propose the following in viewpoint of Theorems 2.2, 2.5, 2.7, 2.9,
2.11 and our previous works [36, 37].

######  Conjecture 3.1.

For a positive integer $s,$ the coinvariant
$k\otimes_{GL_{4}(k)}P((P_{4})_{2^{s+4}+2^{s+1}-3}^{*})$ is trivial if $s=2$
and has dimension $1$ if $s\neq 2.$

On the other hand, by Theorem 2.1, we have

${\rm Ext}_{A}^{4,2^{s+4}+2^{s+1}+1}(k,k)\\\
=\left\\{\begin{array}[]{ll}\langle p_{0}\rangle&\mbox{if
$s=1$},\\\\[2.84526pt] \langle p^{\prime}_{0}\rangle&\mbox{if
$s=2$},\\\\[2.84526pt] \langle
h_{0}h_{s+1}h_{s+3}^{2},h_{0}h_{s}^{2}h_{s+4}\rangle=\langle
h_{0}h_{s}^{2}h_{s+4}\rangle&\mbox{if $s\geq 3$}.\end{array}\right.$

Moreover, it is straightforward to see that the elements
$a_{1}^{(0)}a_{2}^{(2^{s}-1)}a_{3}^{(2^{s}-1)}a_{4}^{(2^{s+4}-1)}$ belong to
${\rm Ext}_{A}^{0,2^{s+4}+2^{s+1}-3}(k,P_{4}).$ So, following Theorem 1.2, the
cycles

$\lambda_{0}\lambda_{2^{s}-1}^{2}\lambda_{2^{s+4}-1}=\psi_{4}(a_{1}^{(0)}a_{2}^{(2^{s}-1)}a_{3}^{(2^{s}-1)}a_{4}^{(2^{s+4}-1)})$

in $\Lambda$ are representative of the non-zero elements
$h_{0}h_{s}^{2}h_{s+4}$ for any $s\geq 3$. This fact and the previous results
by Hưng-Quỳnh [19] show that the elements $p_{0}$ and $h_{0}h_{s}^{2}h_{s+4}$
are in the image of $Tr_{4}^{A},$ except $p^{\prime}_{0}.$ And therefore, by
Conjecture 3.1, it follows that

######  Corollary 3.2.

The transfer homomorphism

$Tr_{4}^{A}:k\otimes_{GL_{4}(k)}P((P_{4})_{2^{s+4}+2^{s+1}-3}^{*})\to{\rm
Ext}_{A}^{4,2^{s+4}+2^{s+1}+1}(k,k)$

is an isomorphism for $s\neq 2$, but it is not an epimorphism for $s=2.$

Next, we discuss the case iii) with $t=3.$ Because Kameko’s squaring operation

$[\overline{Sq}^{0}]_{2^{s+3}+2^{s}-2}:=\overline{Sq}^{0}:Q^{\otimes
4}_{2^{s+3}+2^{s}-2}\to Q^{\otimes 4}_{2^{s+2}+2^{s-1}-3}$

is an epimorphism of $kGL_{4}(k)$-modules, we have an isomorphism

$Q^{\otimes 4}_{2^{s+3}+2^{s}-2}\cong{\rm
Ker}[\overline{Sq}^{0}]_{2^{s+3}+2^{s}-2}\bigoplus Q^{\otimes
4}_{2^{s+2}+2^{s-1}-3}.$ (13)

Following Sum [44] and Conjecture 3.1, the invariant spaces $(Q^{\otimes
4}_{2^{s+2}+2^{s-1}-3})^{GL_{4}}$ are trivial if $s=1,\,4$ and are
1-dimensional otherwise. On the other side, by Theorem 2.1, it implies that

${\rm Ext}_{A}^{4,2^{s+3}+2^{s}+2}(k,k)=\left\\{\begin{array}[]{ll}\langle
h_{1}^{2}h_{3}^{2}\rangle=0&\mbox{if $s=1$},\\\\[2.84526pt] \langle
h_{0}^{2}h_{2}h_{5},h_{1}h_{2}h_{4}^{2}\rangle=\langle
h_{1}^{3}h_{5}\rangle&\mbox{if $s=2$},\\\\[2.84526pt] \langle
h_{0}^{2}h_{3}h_{6},h_{1}h_{2}^{2}h_{6},h_{1}h_{3}h_{5}^{2},p_{1}\rangle=\langle
h_{0}^{2}h_{3}h_{6},p_{1}\rangle&\mbox{if $s=3$},\\\\[2.84526pt] \langle
h_{0}^{2}h_{4}h_{7},h_{1}h_{3}^{2}h_{7},h_{1}h_{4}h_{6}^{2},p^{\prime}_{1}\rangle=\langle
h_{0}^{2}h_{4}h_{7},p^{\prime}_{1}\rangle&\mbox{if $s=4$},\\\\[2.84526pt]
\langle
h_{0}^{2}h_{s}h_{s+3},h_{1}h_{s-1}^{2}h_{s+3},h_{1}h_{s}h_{s+2}^{2}\rangle=\langle
h_{0}^{2}h_{s}h_{s+3},h_{1}h_{s-1}^{2}h_{s+3}\rangle&\mbox{if $s\geq
5$}.\end{array}\right.$

Taking these data, we make the following additional conjecture.

######  Conjecture 3.3.

The invariant spaces $({\rm
Ker}[\overline{Sq}^{0}]_{2^{s+3}+2^{s}-2})^{GL_{4}}$ are trivial if $s=1,\,2$
and have dimension $1$ if $s\geq 3.$

As is well known, $p_{1}\in{\rm Im}(Tr_{4}^{A})$ (see Hưng-Quỳnh [19]) and
$p_{1}^{\prime}\not\in{\rm Im}(Tr_{4}^{A})$ (see Hưng [18]). These, together
with the equality (13) and Conjecture 3.3, tempt us to propose:

######  Conjecture 3.4.

For a positive integer $s,$ then

$\dim
k\otimes_{GL_{4}(k)}P((P_{4})_{2^{s+3}+2^{s}-2}^{*})=\left\\{\begin{array}[]{ll}0&\mbox{if
$s=1$},\\\\[2.84526pt] 1&\mbox{if $s=2$},\\\\[2.84526pt] 2&\mbox{if
$s=3$},\\\\[2.84526pt] 1&\mbox{if $s=4$},\\\\[2.84526pt] 2&\mbox{if $s\geq
5$}.\end{array}\right.$

So, the following corollary is immediate.

######  Corollary 3.5.

The cohomological transfer

$Tr_{4}^{A}:k\otimes_{GL_{4}(k)}P((P_{4})_{2^{s+3}+2^{s}-2}^{*})\to{\rm
Ext}_{A}^{4,2^{s+3}+2^{s}+2}(k,k)$

is an isomorphism for $s\neq 4$, but it is not an epimorphism for $s=4.$

Additionally, one has the following important observation: Consider the
generic degree $2^{s+6}+2^{s+3}+2^{s}-4$ with $s$ an arbitrary non-negative
integer. It is not hard to check that $\mu(2^{s+6}+2^{s+3}+2^{s}-4)=4$ for all
$s>1$ and so by iteration of the Kameko homomorphism, we deduce that the map
$(\overline{Sq}^{0})^{s-1}:Q^{\otimes 4}_{2^{s+6}+2^{s+3}+2^{s}-4}\to
Q^{\otimes 4}_{2^{1+6}+2^{1+3}+2^{1}-4}$ is an isomorphism of
$kGL_{4}(k)$-modules, for arbitrary $s\geq 1.$ Note that $Q^{\otimes
4}_{2^{1+6}+2^{1+3}+2^{1}-4}=Q^{\otimes 4}_{2^{4+3}+2^{4}-2}.$ Then, due to a
previous result in Sum [44] and Conjecture 3.1, it can be easily seen that the
coinvariant $k\otimes_{GL_{4}(k)}P((P_{4})_{2^{s+6}+2^{s+3}+2^{s}-4}^{*})$ is
trivial if $s=0$ and is 1-dimensional if $s\geq 1.$ On the other side,
clearly, Theorem 2.1 implies that

${\rm
Ext}_{A}^{4,2^{s+6}+2^{s+3}+2^{s}}(k,k)=\left\\{\begin{array}[]{ll}\langle
p^{\prime}_{0}\rangle&\mbox{if $s=0$},\\\ \langle
p^{\prime}_{s},h_{s-1}^{2}h_{s+3}h_{s+6}\rangle&\mbox{if $s\geq
1$},\end{array}\right.$

and moreover, the non-zero elements $p^{\prime}_{s}$ are not detected by
$Tr_{4}^{A}$ (see Hưng [18]). Therefore, the fourth transfer
$Tr_{4}^{A}:k\otimes_{GL_{4}(k)}P((P_{4})_{2^{s+6}+2^{s+3}+2^{s}-4}^{*})\to{\rm
Ext}_{A}^{4,2^{s+6}+2^{s+3}+2^{s}}(k,k)$ is a monomorphism, but not an
isomorphism for any $s\geq 0.$

Finally, we believe that Conjectures 3.1, 3.3 and 3.4 are true and
verification will be soon communicated. Taking Corollaries 2.14, 3.2 and 3.5,
together with the work in Sum [44] and our previous results [36, 37], it may
be concluded that

######  Corollary 3.6.

Conjecture 1.1 holds for $q=4.$

######  Acknowledgments.

This work has been supported in part by the NAFOSTED of Viet Nam under grant
no. 101.04-2017.05.

##  4\. Proofs of main results

The whole section is devoted to prove Theorems 2.2, 2.5, 2.7, 2.9, and 2.11.
Firstly, to make the paper self-contained, let us briefly review Singer’s
algebraic transfer and the relevant notions.

Singer’s algebraic transfer. Consider the polynomial ring of one variable,
$P_{1}=k[x_{1}].$ The canonical $A$-action on $P_{1}$ is extended to an
$A$-action on $k[x_{1},x_{1}^{-1}],$ the ring of finite Laurent series. Then,
$\overline{\mathscr{P}}=\langle\\{x_{1}^{t}|\ t\geq-1\\}\rangle$ is
$A$-submodule of $k[x_{1},x_{1}^{-1}].$ One has a short-exact sequence:

$0\to
P_{1}\xrightarrow{j}\overline{\mathscr{P}}\xrightarrow{\pi}\Sigma^{-1}k,$ (14)

where $j$ is the inclusion and $\pi$ is given by $\pi(x_{1}^{t})=0$ if
$t\neq-1$ and $\pi(x_{1}^{-1})=1.$ Denote by $e_{1}$ the element in ${\rm
Ext}_{A}^{1}(\Sigma^{-1}k,P_{1}),$ which is represented by the cocyle
associated to (14). For each $i\geq 1,$ the short-exact sequence

$0\to P_{i+1}\cong
P_{i}\otimes_{k}P_{1}\xrightarrow{1\otimes_{k}j}P_{i}\otimes_{k}\overline{\mathscr{P}}\xrightarrow{1\otimes_{k}\pi}\Sigma^{-1}P_{i},$
(15)

determines a class $(e_{1}\times P_{i})\in{\rm
Ext}_{A}^{1}(\Sigma^{-1}P_{i},P_{i+1}).$ Then, using the cross product and
Yoneda product, we have the element

$e_{q}=(e_{1}\times P_{q-1})\circ(e_{1}\times
P_{q-2})\circ\cdots\circ(e_{1}\times P_{1})\circ e_{1}\in{\rm
Ext}_{A}^{q}(\Sigma^{-q}k,P_{q}).$

Let $\Delta(e_{1}\times P_{i}):{\rm Tor}_{q-i}^{A}(k,\Sigma^{-q}P_{i})\to{\rm
Tor}_{q-i-1}^{A}(k,P_{i+1})$ be the connecting homomorphism associated to
(15). Then, we have a composition of the connecting homomorphisms

$\overline{\varphi}^{A}_{q}=\Delta(e_{1}\times P_{q-1})\circ\Delta(e_{1}\times
P_{q-2})\circ\cdots\circ\Delta(e_{1}\times P_{1})\circ\Delta(e_{1})$

from ${\rm Tor}_{q}^{A}(k,\Sigma^{-q}k)$ to ${\rm
Tor}_{0}^{A}(k,P_{q})=k\otimes_{A}P_{q}$, determined by
$\overline{\varphi}_{q}(z)=e_{q}\cap z$ for any $z\in{\rm
Tor}_{q}^{A}(k,\Sigma^{-q}k).$ Here $\cap$ denotes the cap product in homology
with $k$-coefficients. The image of $\overline{\varphi}_{q}$ is a submodule of
the invariant space $(k\otimes_{A}P_{q})^{GL_{q}(k)}.$ Hence,
$\overline{\varphi}^{A}_{q}$ induces homomorphism

$\varphi_{q}^{A}:{\rm
Tor}_{q}^{A}(k,\Sigma^{-q}k)\to(k\otimes_{A}P_{q})^{GL_{q}(k)}.$

Since the supension $\Sigma^{-q}$ induces an isomorphism ${\rm
Tor}_{q,n}^{A}(k,\Sigma^{-q}k)\cong{\rm Tor}_{q,q+n}^{A}(k,k),$ we have the
homomorphism $\varphi_{q}^{A}:{\rm
Tor}_{q,q+n}^{A}(k,k)\to(k\otimes_{A}P_{q})_{n}^{GL_{q}(k)}.$ So its dual

$Tr_{q}^{A}:k\otimes_{GL_{q}(k)}P((P_{q})_{n}^{*})\to{\rm
Ext}_{A}^{q,q+n}(k,k)$

is called the cohomological transfer (or algebraic transfer).

Some definitions. A sequence of non-negative integers
$\omega=(\omega_{1},\omega_{2},\ldots,\omega_{i},\ldots)$ is called a weight
vector, if $\omega_{i}=0$ for $i\gg 0.$ One defines $\deg(\omega)=\sum_{i\geq
1}2^{i-1}\omega_{i},$ For a natural number $n,$ let us denote by
$\alpha_{j}(n)$ the $j$-th coefficients in dyadic expansion of $n,$ from which
we have $\alpha(n)=\sum_{j\geq 0}\alpha_{j}(n)$ and $n=\sum_{j\geq
0}\alpha_{j}(n)2^{j},$ where $\alpha_{j}(n)\in\\{0,1\\},\ j=0,1,\ldots.$ Let
$x=\prod_{1\leq i\leq 4}x_{i}^{a_{i}}$ be a monomial in $P_{4},$ define two
sequences associated with $x$ by
$\omega(x):=(\omega_{1}(x),\omega_{2}(x),\ldots,\omega_{j}(x),\ldots)$ and
$(a_{1},a_{2},\ldots,a_{4}),$ where $\omega_{j}(x)=\sum_{1\leq i\leq
4}\alpha_{j-1}(a_{i}),$ for $j\geq 1.$ These sequences are respectively called
the weight vector and the exponent vector of $x.$ By convention, the sets of
all the weight vectors and the exponent vectors are given the left
lexicographical order.

######  Definition 4.1.

Assume that $x=\prod_{1\leq i\leq 4}x_{i}^{a_{i}}$ and $y=\prod_{1\leq i\leq
4}x_{i}^{b_{i}}$ are the monomials in $(P_{4})_{n}.$ We write $a,\,b$ for the
exponent vectors of $x$ and $y,$ respectively. We say that $x<y$ if and only
if one of the following holds:

  1. (i)

$\omega(x)<\omega(y);$

  2. (ii)

$\omega(x)=\omega(y)$ and $a<b.$

Let us recall that the algebra $A$ is generated by Steenrod squares $Sq^{i}$
for $i\geq 0$ and that the action of $A$ on the polynomial ring
$P_{4}=k[x_{1},x_{2},x_{3},x_{4}]$ is determined by
$Sq^{i}(x_{i}^{n})=\binom{n}{i}x^{n+i}$ for every $x_{i}\in P_{4}$ and
Cartan’s formula $Sq^{i}(fg)=\sum_{m+n=i}Sq^{i}(f)Sq^{i}(g),$ for all the
homogeneous polynomials $f,\,g\in P_{4}.$ One should note that
$\binom{n}{i}=0$ if $i>n.$

######  Definition 4.2.

For a weight vector $\omega$ of degree $n,$ we denote two subspaces associated
with $\omega$:

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt(P_{4})_{n}^{\omega}&=\langle\\{x\in(P_{4})_{n}|\,\deg(x)=\deg(\omega),\
\omega(x)\leq\omega\\}\rangle,\\\
(P_{4})_{n}^{<\omega}&=\langle\\{x\in(P_{4})_{n}|\,\deg(x)=\deg(\omega),\
\omega(x)<\omega\\}\rangle.\end{array}$

For homogeneous polynomials $f$ and $g$ in $(P_{4})_{n}.$ We define the
equivalence relations ”$\equiv$” and ”$\equiv_{\omega}$” on $(P_{4})_{n}$:

  1. (i)

$f\equiv g$ if and only if $(f-g)\in\widehat{A}(P_{4})_{n}.$

  2. (ii)

$f\equiv_{\omega}g$ if and only if $f,\,g\in(P_{4})_{n}^{\omega}$ and
$(f-g)\in((\widehat{A}(P_{4})_{n}\cap(P_{4})_{n}^{\omega})+(P_{4})_{n}^{<\omega}).$
Specifically, if $f\equiv 0$ (resp. $f\equiv_{\omega}0$), then $f$ is called
hit (resp. $\omega$-hit).

From the equivalence relation ”$\equiv_{\omega}$”, we have the $k$-quotient
space:

$(Q_{n}^{\otimes
4})^{\omega}:=(P_{4})_{n}^{\omega}/(((P_{4})_{n}^{\omega}\cap\widehat{A}(P_{4})_{n})+(P_{4})_{n}^{<\omega}).$

According to Sum [46], this space has the structure of an $GL_{4}$-module.
Furthermore, by the definitions above, it is straightforward to show that

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus 2.0pt\dim Q^{\otimes
4}_{n}&=\sum_{\deg(\omega)=n}\dim(Q_{n}^{\otimes 4})^{\omega},\\\
\dim(Q^{\otimes
4}_{n})^{GL_{4}(k)}&\leq\sum_{\deg(\omega)=n}\dim((Q_{n}^{\otimes
4})^{\omega})^{GL_{4}(k)}.\end{array}$

Let $\underline{(P_{4})_{n}}$ and $\overline{(P_{4})_{n}}$ denote the
$k$-subspaces of $(P_{4})_{n}$ spanned all the monomials $\prod_{1\leq i\leq
4}x_{i}^{a_{i}}$ such that $\prod_{1\leq i\leq 4}a_{i}=0,$ and $\prod_{1\leq
i\leq 4}a_{i}>0,$ respectively. We put

$\underline{Q^{\otimes 4}_{n}}=k\otimes_{A}\underline{(P_{4})_{n}},\
\mbox{and}\ \overline{Q^{\otimes 4}_{n}}=k\otimes_{A}\overline{(P_{4})_{n}}.$

Then, we have

$Q^{\otimes 4}_{n}\cong\underline{Q^{\otimes
4}_{n}}\bigoplus\overline{Q^{\otimes 4}_{n}}.$

######  Definition 4.3.

A monomial $x\in(P_{4})_{n}$ is said to be inadmissible if there exist
monomials $y_{1},y_{2},\ldots,y_{k}$ in $(P_{4})_{n}$ such that $y_{j}<x$ for
$1\leq j\leq k$ and $x\equiv\sum_{1\leq j\leq k}y_{j}.$ Then, $x$ is said to
be admissible if it is not inadmissible.

With the above definitions on hand, it is easily seen that $Q^{\otimes 4}_{n}$
is a $k$-vector space with a basis consisting of all the classes represent by
the admissible monomials in $(P_{4})_{n}.$

######  Definition 4.4.

A monomial $z=x_{1}^{v_{1}}\ldots x_{4}^{v_{4}}$ in $(P_{4})_{n}$ is called a
spike if every exponent $v_{j}$ is of the form $2^{\xi_{j}}-1.$ In particular,
if the exponents $\xi_{j}$ can be arranged to satisfy
$\xi_{1}>\xi_{2}>\ldots>\xi_{m-1}\geq\xi_{m}\geq 1$ where only the last two
smallest exponents can be equal, and $\xi_{r}=0$ for $m<r\leq 4,$ then $z$ is
called a minimal spike.

According to Singer [39], we have the following technical result on a hit
monomial.

######  Theorem 4.5.

For each $q\geq 1$ and degree $n\geq 0,$ suppose that the monomial $x$ belongs
to $(P_{q})_{n}$ and $\mu(n)\leq q.$ Let $z$ be a minimal spike in
$(P_{q})_{n}.$ Therefrom, if $\omega(x)<\omega(z),$ then $x$ is hit.

######  Definition 4.6.

For each $1\leq j\leq 4,$ one defines $k$-homomorphism

$\sigma_{j}:k^{4}\cong\langle x_{1},\ldots,x_{4}\rangle\to k^{4}\cong\langle
x_{1},\ldots,x_{4}\rangle,$

which is given by

$\left\\{\begin{array}[]{ll}\sigma_{j}(x_{j})&=x_{j+1},\\\
\sigma_{j}(x_{j+1})&=x_{j},\\\ \sigma_{j}(x_{t})&=x_{t},\end{array}\right.$

for $t\not\in\\{j,j+1\\},\ j=1,2,3,$ and $\sigma_{4}(x_{1})=x_{1}+x_{2},\
\sigma_{4}(x_{t})=x_{t}$ for $2\leq t\leq 4.$

The following remark is useful.

######  Remark 4.7.

We denote by $\Sigma_{4}$ the symmetric group of rank $4.$ Then, $\Sigma_{4}$
is generated by the ones associated with $\sigma_{j},\ 1\leq j\leq 3.$ For
each permutation in $\Sigma_{4}$, consider corresponding permutation matrix;
these form a group of matrices isomorphic to $\Sigma_{4}.$ So, $GL_{4}(k)\cong
GL(k^{4})$ is generated by the matrices associated with $\sigma_{j},\ 1\leq
j\leq 4.$ Let $x=\prod_{1\leq i\leq 4}x_{i}^{a_{i}}$ be an monomial in
$(P_{4})_{n}.$ Then, the weight vector $\omega$ of $x$ is invariant under the
permutation of the generators $x_{i},\ i=1,2,\ldots,4.$ Hence $(Q^{\otimes
4}_{n})^{\omega}$ also has a $\Sigma_{4}$-module structure.

For a polynomial $f\in(P_{4})_{n},$ we denote by $[f]$ the classes in
$Q^{\otimes 4}_{n}$ represented by $f.$ If $\omega$ is a weight vector and
$f\in(P_{4})_{n}^{\omega},$ then denote by $[f]_{\omega}$ the classes in
$(Q^{\otimes 4}_{n})^{\omega}$ represented by $f.$ By the above remark and
$\sigma_{j}$ induces a homomorphism of $A$-algebras which is also denoted by
$\sigma_{j}:P_{4}\to P_{4},$ a class $[f]_{\omega}\in(Q^{\otimes
4}_{n})^{\omega}$ is an $GL_{4}(k)$-invariant if and only if
$\sigma_{j}(f)\equiv_{\omega}f$ for $1\leq j\leq 4.$ It is an
$\Sigma_{4}$-invariant if and only if $\sigma_{j}(f)\equiv_{\omega}f$ for
$1\leq j\leq 3.$

In what follows, for any monomials $v_{1},v_{2},\ldots,v_{s}\in(P_{4})_{n}$
and for a subgroup $G$ of $GL_{4}(k),$ we denote by
$G(v_{1};v_{2};\ldots,v_{s})$ the $G$-submodule of $(Q^{\otimes
4}_{n})^{\omega}$ generated by the set $\\{[v_{i}]_{\omega}:\,1\leq i\leq
s\\}.$ Note that if $\omega$ is a weight vector of a minimal spike, then
$[v_{i}]_{\omega}=[v_{i}]$ for all $i.$

###  4.1. Proof of Theorem 2.2

For convenience, put $n_{s}:=3(2^{s}-1)+3.2^{s},$ according to Sum [43], we
have

$\dim Q^{\otimes 4}_{n_{s}}=\left\\{\begin{array}[]{ll}46&\mbox{if $s=1$},\\\
94&\mbox{if $s=2$},\\\ 105&\mbox{if $s\geq 3$}.\end{array}\right.$

Recall that $Q^{\otimes 4}_{n_{s}}\cong\underline{Q^{\otimes
4}_{n_{s}}}\bigoplus\overline{Q^{\otimes 4}_{n_{s}}}.$ By Sum [43], a monomial
basis for $\underline{Q^{\otimes 4}_{n_{s}}}$ is the set consisting of all the
classes represented monomials $a_{s,i}$ which are determined as follows:

$a_{s,\,1}=x_{2}^{2^{s}-1}x_{3}^{2^{s}-1}x_{4}^{2^{s+2}-1}$, |  $a_{s,\,2}=x_{2}^{2^{s}-1}x_{3}^{2^{s+2}-1}x_{4}^{2^{s}-1}$, |  $a_{s,\,3}=x_{2}^{2^{s+2}-1}x_{3}^{2^{s}-1}x_{4}^{2^{s}-1}$,
---|---|---
$a_{s,\,4}=x_{1}^{2^{s}-1}x_{3}^{2^{s}-1}x_{4}^{2^{s+2}-1}$, |  $a_{s,\,5}=x_{1}^{2^{s}-1}x_{3}^{2^{s+2}-1}x_{4}^{2^{s}-1}$, |  $a_{s,\,6}=x_{1}^{2^{s}-1}x_{2}^{2^{s}-1}x_{4}^{2^{s+2}-1}$,
$a_{s,\,7}=x_{1}^{2^{s}-1}x_{2}^{2^{s}-1}x_{3}^{2^{s+2}-1}$, |  $a_{s,\,8}=x_{1}^{2^{s}-1}x_{2}^{2^{s+2}-1}x_{4}^{2^{s}-1}$, |  $a_{s,\,9}=x_{1}^{2^{s}-1}x_{2}^{2^{s+2}-1}x_{3}^{2^{s}-1}$,
$a_{s,\,10}=x_{1}^{2^{s+2}-1}x_{3}^{2^{s}-1}x_{4}^{2^{s}-1}$, |  $a_{s,\,11}=x_{1}^{2^{s+2}-1}x_{2}^{2^{s}-1}x_{4}^{2^{s}-1}$, |  $a_{s,\,12}=x_{1}^{2^{s+2}-1}x_{2}^{2^{s}-1}x_{3}^{2^{s}-1}$,
$a_{s,\,13}=x_{2}^{2^{s}-1}x_{3}^{2^{s+1}-1}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,14}=x_{2}^{2^{s+1}-1}x_{3}^{2^{s}-1}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,15}=x_{2}^{2^{s+1}-1}x_{3}^{3.2^{s}-1}x_{4}^{2^{s}-1}$,
$a_{s,\,16}=x_{1}^{2^{s}-1}x_{3}^{2^{s+1}-1}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,17}=x_{1}^{2^{s}-1}x_{2}^{2^{s+1}-1}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,18}=x_{1}^{2^{s}-1}x_{2}^{2^{s+1}-1}x_{3}^{3.2^{s}-1}$,
$a_{s,\,19}=x_{1}^{2^{s+1}-1}x_{3}^{2^{s}-1}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,20}=x_{1}^{2^{s+1}-1}x_{3}^{3.2^{s}-1}x_{4}^{2^{s}-1}$, |  $a_{s,\,21}=x_{1}^{2^{s+1}-1}x_{2}^{2^{s}-1}x_{4}^{3.2^{s}-1}$,
$a_{s,\,22}=x_{1}^{2^{s+1}-1}x_{2}^{2^{s}-1}x_{3}^{3.2^{s}-1}$, |  $a_{s,\,23}=x_{1}^{2^{s+1}-1}x_{2}^{3.2^{s}-1}x_{4}^{2^{s}-1}$, |  $a_{s,\,24}=x_{1}^{2^{s+1}-1}x_{2}^{3.2^{s}-1}x_{3}^{2^{s}-1}$,
$a_{s,\,25}=x_{2}^{2^{s+1}-1}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+1}-1}$, |  $a_{s,\,26}=x_{1}^{2^{s+1}-1}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+1}-1}$, |  $a_{s,\,27}=x_{1}^{2^{s+1}-1}x_{2}^{2^{s+1}-1}x_{4}^{2^{s+1}-1}$,
$a_{s,\,28}=x_{1}^{2^{s+1}-1}x_{2}^{2^{s+1}-1}x_{3}^{2^{s+1}-1}$. |  |

Thanks to these, a direct computation shows that

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt\Sigma_{4}(a_{s,\,1})&=\langle\\{[a_{s,\,j}]:\,1\leq j\leq
12\\}\rangle,\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\Sigma_{4}(a_{s,\,13})&=\langle\\{[a_{s,\,j}]:\,13\leq j\leq
24\\}\rangle,\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\Sigma_{4}(a_{s,\,25})&=\langle\\{[a_{s,\,j}]:\,25\leq j\leq
28\\}\rangle.\end{array}$

These imply that there is a direct summand decomposition of the
$\Sigma_{4}$-modules:

$\underline{Q^{\otimes
4}_{n_{s}}}=\Sigma_{4}(a_{s,\,1})\bigoplus\Sigma_{4}(a_{s,\,13})\bigoplus\Sigma_{4}(a_{s,\,25}).$

In the following, it will play a key role in the proof of the theorem.

######  Lemma 4.1.1.

We have that

  * i)

$\Sigma_{4}(a_{s,\,1})^{\Sigma_{4}}=\langle[q_{s,\,1}]\rangle,$ with
$q_{s,\,1}=\sum_{1\leq j\leq 12}a_{s,\,j};$

  * ii)

$\Sigma_{4}(a_{s,\,13})^{\Sigma_{4}}=\langle[q_{s,\,2}]\rangle,$ with
$q_{s,\,1}=\sum_{13\leq j\leq 24}a_{s,\,j};$

  * iii)

$\Sigma_{4}(a_{s,\,25})^{\Sigma_{4}}=\langle[q_{s,\,3}]\rangle,$ with
$q_{s,\,1}=\sum_{25\leq j\leq 28}a_{s,\,j}.$

###### Proof.

We frist prove Part $i).$ Since the set $\\{[a_{s,\,j}]:\,1\leq j\leq 12\\}$
is a basis of $\Sigma_{4}(a_{s,\,1}),$ if
$[f]\in\Sigma_{4}(a_{s,\,1})^{\Sigma_{4}},$ then

$\begin{array}[]{ll}f&\equiv\big{(}\gamma_{1}a_{s,\,1}+\gamma_{2}a_{s,\,2}+\gamma_{3}a_{s,\,3}+\gamma_{4}a_{s,\,4}+\gamma_{5}a_{s,\,5}+\gamma_{6}a_{s,\,6}\\\
&\quad\gamma_{7}a_{s,\,7}+\gamma_{8}a_{s,\,8}+\gamma_{9}a_{s,\,9}+\gamma_{10}a_{s,\,10}+\gamma_{11}a_{s,\,11}+\gamma_{12}a_{s,\,12}\big{)},\end{array}$
(16)

in which $\gamma_{j}\in k$ for every $j.$ Acting the homomorphisms
$\sigma_{i}:(P_{4})_{n_{s}}\to(P_{4})_{n_{s}},\,i=1,2,3,$ on both sides of
(16), we get

$\begin{array}[]{ll}\sigma_{1}(f)&\equiv\big{(}\gamma_{4}a_{s,\,1}+\gamma_{5}a_{s,\,2}+\gamma_{10}a_{s,\,3}+\gamma_{1}a_{s,\,4}+\gamma_{2}a_{s,\,5}+\gamma_{6}a_{s,\,6}+\gamma_{7}a_{s,\,7}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\gamma_{11}a_{s,\,8}+\gamma_{12}a_{s,\,9}+\gamma_{3}a_{s,\,10}+\gamma_{8}a_{s,\,11}+\gamma_{9}a_{s,\,12}\big{)};\\\
\sigma_{2}(f)&\equiv\big{(}\gamma_{1}a_{s,\,1}+\gamma_{3}a_{s,\,2}+\gamma_{2}a_{s,\,3}+\gamma_{6}a_{s,\,4}+\gamma_{8}a_{s,\,5}+\gamma_{4}a_{s,\,6}+\gamma_{9}a_{s,\,7}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\gamma_{5}a_{s,\,8}+\gamma_{7}a_{s,\,9}+\gamma_{11}a_{s,\,10}+\gamma_{10}a_{s,\,11}+\gamma_{12}a_{s,\,12}\big{)}.\\\
\sigma_{3}(f)&\equiv\big{(}\gamma_{2}a_{s,\,1}+\gamma_{1}a_{s,\,2}+\gamma_{3}a_{s,\,3}+\gamma_{5}a_{s,\,4}+\gamma_{4}a_{s,\,5}+\gamma_{7}a_{s,\,6}+\gamma_{6}a_{s,\,7}\\\
&\quad+\gamma_{9}a_{s,\,8}+\gamma_{8}a_{s,\,9}+\gamma_{10}a_{s,\,10}+\gamma_{12}a_{s,\,11}+\gamma_{11}a_{s,\,12}\big{)}.\end{array}$
(17)

Since $[f]\in\Sigma_{4}(a_{s,\,1})^{\Sigma_{4}},$ $\sigma_{i}(f)\equiv f$ for
$1\leq i\leq 3.$ Combining this and the equalities (16) and (17), we find that

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt(\gamma_{1}+\gamma_{4})a_{s,\,1}+(\gamma_{2}+\gamma_{5})a_{s,\,2}+(\gamma_{3}+\gamma_{10})a_{s,\,3}+(\gamma_{8}+\gamma_{11})a_{s,\,8}+(\gamma_{9}+\gamma_{12})a_{s,\,9}&\equiv
0,\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt(\gamma_{2}+\gamma_{3})a_{s,\,2}+(\gamma_{4}+\gamma_{6})a_{s,\,4}+(\gamma_{5}+\gamma_{8})a_{s,\,5}+(\gamma_{7}+\gamma_{9})a_{s,\,7}+(\gamma_{10}+\gamma_{11})a_{s,\,10}&\equiv
0,\\\
(\gamma_{1}+\gamma_{2})a_{s,\,1}+(\gamma_{4}+\gamma_{5})a_{s,\,4}+(\gamma_{6}+\gamma_{7})a_{s,\,6}+(\gamma_{8}+\gamma_{9})a_{s,\,8}+(\gamma_{11}+\gamma_{12})a_{s,\,11}&\equiv
0.\end{array}$

This implies that $\gamma_{1}=\gamma_{j}$ for all $j,\,2\leq j\leq 12.$ By a
similar computation, we also get Part $iii).$

We now prove Part $ii).$ We see that a basis for $\Sigma_{4}(a_{s,\,13})$ is
the set $\\{[a_{s,\,j}]:\,13\leq j\leq 24\\}.$ Then, assume that
$[g]\in\Sigma_{4}(a_{s,\,13})^{\Sigma_{4}},$ then we have

$g\equiv\sum_{13\leq j\leq 24}\beta_{j}a_{s,\,j},$ (18)

with $\beta_{j}\in k,\,j=13,\ldots,24.$ Acting the homomorphisms
$\sigma_{i}:(P_{4})_{n_{s}}\to(P_{4})_{n_{s}},\,1\leq i\leq 3,$ on both sides
of (18), we get

$\begin{array}[]{ll}\sigma_{1}(g)&\equiv\big{(}\beta_{16}a_{s,\,13}+\beta_{19}a_{s,\,14}+\beta_{20}a_{s,\,15}+\beta_{13}a_{s,\,16}+\beta_{21}a_{s,\,17}\\\
&\quad+\beta_{22}a_{s,\,18}+\beta_{14}a_{s,\,19}+\beta_{15}a_{s,\,20}+\beta_{17}a_{s,\,21}+\beta_{18}a_{s,\,22}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\beta_{23}x_{1}^{3.2^{s}-1}x_{2}^{2^{s+1}-1}x_{4}^{2^{s}-1}+\beta_{24}x_{1}^{3.2^{s}-1}x_{2}^{2^{s+1}-1}x_{3}^{2^{s}-1}\big{)};\\\
\sigma_{2}(g)&\equiv\big{(}\beta_{14}a_{s,\,13}+\beta_{13}a_{s,\,14}+\beta_{17}a_{s,\,16}+\beta_{16}a_{s,\,17}+\beta_{21}a_{s,\,19}\\\
&\quad+\beta_{23}a_{s,\,20}+\beta_{19}a_{s,\,21}+\beta_{24}a_{s,\,22}+\beta_{20}a_{s,\,23}+\beta_{22}a_{s,\,24}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\beta_{15}x_{2}^{3.2^{s}-1}x_{3}^{2^{s+1}-1}x_{4}^{2^{s}-1}+\beta_{18}x_{1}^{2^{s}-1}x_{2}^{3.2^{s}-1}x_{3}^{2^{s+1}-1}\big{)};\\\
\sigma_{3}(g)&\equiv\big{(}\beta_{15}a_{s,\,14}+\beta_{14}a_{s,\,15}+\beta_{18}a_{s,\,17}+\beta_{17}a_{s,\,18}+\beta_{20}a_{s,\,19}\\\
&\quad+\beta_{19}a_{s,\,20}+\beta_{22}a_{s,\,21}+\beta_{21}a_{s,\,22}+\beta_{24}a_{s,\,23}+\beta_{23}a_{s,\,24}\\\
&\quad+\beta_{13}x_{2}^{2^{s}-1}x_{3}^{3.2^{s}-1}x_{4}^{2^{s+1}-1}+\beta_{16}x_{1}^{2^{s}-1}x_{3}^{3.2^{s}-1}x_{4}^{2^{s+1}-1}\big{)}.\end{array}$
(19)

Using Cartan’s formula and Theorem 4.5, we have

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}^{3.2^{s}-1}x_{2}^{2^{s+1}-1}x_{4}^{2^{s}-1}&=Sq^{2}(x_{1}^{3.2^{s}-3}x_{2}^{2^{s+1}-1}x_{4}^{2^{s}-1})+a_{s,\,23}\mod(\widehat{A}(P_{4})_{n_{s}});\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}^{3.2^{s}-1}x_{2}^{2^{s+1}-1}x_{3}^{2^{s}-1}&=Sq^{2}(x_{1}^{3.2^{s}-3}x_{2}^{2^{s+1}-1}x_{3}^{2^{s}-1})+a_{s,\,24}\mod(\widehat{A}(P_{4})_{n_{s}});\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{2}^{3.2^{s}-1}x_{3}^{2^{s+1}-1}x_{4}^{2^{s}-1}&=Sq^{2}(x_{2}^{3.2^{s}-3}x_{3}^{2^{s+1}-1}x_{4}^{2^{s}-1})+a_{s,\,15}\mod(\widehat{A}(P_{4})_{n_{s}});\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}^{2^{s}-1}x_{2}^{3.2^{s}-1}x_{3}^{2^{s+1}-1}&=Sq^{2}(x_{1}^{2^{s}-1}x_{2}^{3.2^{s}-3}x_{3}^{2^{s+1}-1})+a_{s,\,18}\mod(\widehat{A}(P_{4})_{n_{s}});\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{2}^{2^{s}-1}x_{3}^{3.2^{s}-1}x_{4}^{2^{s+1}-1}&=Sq^{2}(x_{2}^{2^{s}-1}x_{3}^{3.2^{s}-3}x_{4}^{2^{s+1}-1})+a_{s,\,13}\mod(\widehat{A}(P_{4})_{n_{s}});\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}^{2^{s}-1}x_{3}^{3.2^{s}-1}x_{4}^{2^{s+1}-1}&=Sq^{2}(x_{1}^{2^{s}-1}x_{3}^{3.2^{s}-3}x_{4}^{2^{s+1}-1})+a_{s,\,16}\mod(\widehat{A}(P_{4})_{n_{s}}).\end{array}$
(20)

Then, using the equalities (18), (19) and (20) and the relations
$\sigma_{i}(g)\equiv g,$ for $i\in\\{1,2,3\\},$ deduce that
$\beta_{13}=\beta_{j}$ for all $j,\,14\leq j\leq 24.$ The lemma is completely
proved. ∎

###### Proof of Theorem 2.2.

According to Sum [43], a basis for $\overline{Q^{\otimes 4}_{n_{s}}}$ is the
set consisting of all the classes represented monomials $a_{s,i}$ which are
determined as follows:

For $s\geq 1,$

$a_{s,\,29}=x_{1}x_{2}^{2^{s}-1}x_{3}^{2^{s}-1}x_{4}^{2^{s+2}-2}$, |  $a_{s,\,30}=x_{1}x_{2}^{2^{s}-1}x_{3}^{2^{s+2}-2}x_{4}^{2^{s}-1}$, |  $a_{s,\,31}=x_{1}x_{2}^{2^{s+2}-2}x_{3}^{2^{s}-1}x_{4}^{2^{s}-1}$,
---|---|---
$a_{s,\,32}=x_{1}x_{2}^{2^{s}-1}x_{3}^{2^{s+1}-2}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,33}=x_{1}x_{2}^{2^{s+1}-2}x_{3}^{2^{s}-1}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,34}=x_{1}x_{2}^{2^{s+1}-2}x_{3}^{3.2^{s}-1}x_{4}^{2^{s}-1}$,
$a_{s,\,35}=x_{1}x_{2}^{2^{s+1}-2}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+1}-1}$, |  $a_{s,\,36}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{2^{s+1}-2}x_{4}^{2^{s+1}-1}$, |  $a_{s,\,37}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+1}-2}$,
$a_{s,\,38}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{2^{s+1}-2}x_{4}^{2^{s+1}-1}$, |  $a_{s,\,39}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+1}-2}$, |  $a_{s,\,40}=x_{1}^{2^{s+1}-1}x_{2}^{2^{s+1}-1}x_{3}x_{4}^{2^{s+1}-2}$.

For $s=1,$

$a_{1,\,41}=x_{1}x_{2}x_{3}^{3}x_{4}^{4}$, |  $a_{1,\,42}=x_{1}x_{2}^{3}x_{3}x_{4}^{4}$, |  $a_{1,\,43}=x_{1}x_{2}^{3}x_{3}^{4}x_{4}$,
---|---|---
$a_{1,\,44}=x_{1}^{3}x_{2}x_{3}x_{4}^{4}$, |  $a_{1,\,45}=x_{1}^{3}x_{2}x_{3}^{4}x_{4}$, |  $a_{1,\,46}=x_{1}^{3}x_{2}^{4}x_{3}x_{4}$.

For $s\geq 2,$

$a_{s,\,41}=x_{1}x_{2}^{2^{s}-2}x_{3}^{2^{s}-1}x_{4}^{2^{s+2}-1}$, |  $a_{s,\,42}=x_{1}x_{2}^{2^{s}-2}x_{3}^{2^{s+2}-1}x_{4}^{2^{s}-1}$, |  $a_{s,\,43}=x_{1}x_{2}^{2^{s}-1}x_{3}^{2^{s}-2}x_{4}^{2^{s+2}-1}$, |
---|---|---|---
$a_{s,\,44}=x_{1}x_{2}^{2^{s}-1}x_{3}^{2^{s+2}-1}x_{4}^{2^{s}-2}$, |  $a_{s,\,45}=x_{1}x_{2}^{2^{s+2}-1}x_{3}^{2^{s}-2}x_{4}^{2^{s}-1}$, |  $a_{s,\,46}=x_{1}x_{2}^{2^{s+2}-1}x_{3}^{2^{s}-1}x_{4}^{2^{s}-2}$, |
$a_{s,\,47}=x_{1}^{2^{s}-1}x_{2}x_{3}^{2^{s}-2}x_{4}^{2^{s+2}-1}$, |  $a_{s,\,48}=x_{1}^{2^{s}-1}x_{2}x_{3}^{2^{s+2}-1}x_{4}^{2^{s}-2}$, |  $a_{s,\,49}=x_{1}^{2^{s}-1}x_{2}^{2^{s+2}-1}x_{3}x_{4}^{2^{s}-2}$, |
$a_{s,\,50}=x_{1}^{2^{s+2}-1}x_{2}x_{3}^{2^{s}-2}x_{4}^{2^{s}-1}$, |  $a_{s,\,51}=x_{1}^{2^{s+2}-1}x_{2}x_{3}^{2^{s}-1}x_{4}^{2^{s}-2}$, |  $a_{s,\,52}=x_{1}^{2^{s+2}-1}x_{2}^{2^{s}-1}x_{3}x_{4}^{2^{s}-2}$, |
$a_{s,\,53}=x_{1}x_{2}^{2^{s}-2}x_{3}^{2^{s+1}-1}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,54}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{2^{s}-2}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,55}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{3.2^{s}-1}x_{4}^{2^{s}-2}$, |
$a_{s,\,56}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{2^{s}-2}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,57}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{3.2^{s}-1}x_{4}^{2^{s}-2}$, |  $a_{s,\,58}=x_{1}^{2^{s+1}-1}x_{2}^{3.2^{s}-1}x_{3}x_{4}^{2^{s}-2}$, |
$a_{s,\,59}=x_{1}^{2^{s}-1}x_{2}x_{3}^{2^{s}-1}x_{4}^{2^{s+2}-2}$, |  $a_{s,\,60}=x_{1}^{2^{s}-1}x_{2}x_{3}^{2^{s+2}-2}x_{4}^{2^{s}-1}$, |  $a_{s,\,61}=x_{1}^{2^{s}-1}x_{2}^{2^{s}-1}x_{3}x_{4}^{2^{s+2}-2}$, |
$a_{s,\,62}=x_{1}^{2^{s}-1}x_{2}x_{3}^{2^{s+1}-2}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,63}=x_{1}x_{2}^{2^{s}-1}x_{3}^{2^{s+1}-1}x_{4}^{3.2^{s}-2}$, |  $a_{s,\,64}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{2^{s}-1}x_{4}^{3.2^{s}-2}$, |
$a_{s,\,65}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{3.2^{s}-2}x_{4}^{2^{s}-1}$, |  $a_{s,\,66}=x_{1}^{2^{s}-1}x_{2}x_{3}^{2^{s+1}-1}x_{4}^{3.2^{s}-2}$, |  $a_{s,\,67}=x_{1}^{2^{s}-1}x_{2}^{2^{s+1}-1}x_{3}x_{4}^{3.2^{s}-2}$, |
$a_{s,\,68}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{2^{s}-1}x_{4}^{3.2^{s}-2}$, |  $a_{s,\,69}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{3.2^{s}-2}x_{4}^{2^{s}-1}$, |  $a_{s,\,70}=x_{1}^{2^{s+1}-1}x_{2}^{2^{s}-1}x_{3}x_{4}^{3.2^{s}-2}$, |
$a_{s,\,71}=x_{1}^{3}x_{2}^{2^{s}-1}x_{3}^{2^{s+2}-3}x_{4}^{2^{s}-2}$, |  $a_{s,\,72}=x_{1}^{3}x_{2}^{2^{s+2}-3}x_{3}^{2^{s}-2}x_{4}^{2^{s}-1}$, |  $a_{s,\,73}=x_{1}^{3}x_{2}^{2^{s+2}-3}x_{3}^{2^{s}-1}x_{4}^{2^{s}-2}$, |
$a_{s,\,74}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{2^{s}-2}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,75}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{3.2^{s}-1}x_{4}^{2^{s}-2}$, |  $a_{s,\,76}=x_{1}^{3}x_{2}^{2^{s+1}-1}x_{3}^{3.2^{s}-3}x_{4}^{2^{s}-2}$, |
$a_{s,\,77}=x_{1}^{2^{s+1}-1}x_{2}^{3}x_{3}^{3.2^{s}-3}x_{4}^{2^{s}-2}$, |  $a_{s,\,78}=x_{1}^{3}x_{2}^{2^{s}-1}x_{3}^{2^{s+1}-3}x_{4}^{3.2^{s}-2}$, |  $a_{s,\,79}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{2^{s}-1}x_{4}^{3.2^{s}-2}$, |
$a_{s,\,80}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{3.2^{s}-2}x_{4}^{2^{s}-1}$, |  $a_{s,\,81}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{2^{s+1}-2}x_{4}^{2^{s+1}-1}$, |  $a_{s,\,82}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+1}-2}$, |
$a_{s,\,83}=x_{1}^{3}x_{2}^{2^{s+1}-1}x_{3}^{2^{s+1}-3}x_{4}^{2^{s+1}-2}$, |  $a_{s,\,84}=x_{1}^{2^{s+1}-1}x_{2}^{3}x_{3}^{2^{s+1}-3}x_{4}^{2^{s+1}-2}$. |  |

For $s=2,$

$a_{2,\,85}=x_{1}^{3}x_{2}^{3}x_{3}^{3}x_{4}^{12}$, |  $a_{2,\,86}=x_{1}^{3}x_{2}^{3}x_{3}^{12}x_{4}^{3}$, |  $a_{2,\,87}=x_{1}^{7}x_{2}^{9}x_{3}^{2}x_{4}^{3}$,
---|---|---
$a_{2,\,88}=x_{1}^{7}x_{2}^{9}x_{3}^{3}x_{4}^{2}$, |  $a_{2,\,89}=x_{1}^{3}x_{2}^{3}x_{3}^{4}x_{4}^{11}$, |  $a_{2,\,90}=x_{1}^{3}x_{2}^{3}x_{3}^{7}x_{4}^{8}$,
$a_{2,\,91}=x_{1}^{3}x_{2}^{7}x_{3}^{3}x_{4}^{8}$, |  $a_{2,\,92}=x_{1}^{3}x_{2}^{7}x_{3}^{8}x_{4}^{3}$, |  $a_{2,\,93}=x_{1}^{7}x_{2}^{3}x_{3}^{3}x_{4}^{8}$,
$a_{2,\,94}=x_{1}^{7}x_{2}^{3}x_{3}^{8}x_{4}^{3}$. |  |

For $s\geq 3,$

$a_{s,\,85}=x_{1}^{3}x_{2}^{2^{s}-3}x_{3}^{2^{s}-2}x_{4}^{2^{s+2}-1}$, |  $a_{s,\,86}=x_{1}^{3}x_{2}^{2^{s}-3}x_{3}^{2^{s+2}-1}x_{4}^{2^{s}-2}$, |  $a_{s,\,87}=x_{1}^{3}x_{2}^{2^{s+2}-1}x_{3}^{2^{s}-3}x_{4}^{2^{s}-2}$,
---|---|---
$a_{s,\,88}=x_{1}^{2^{s+2}-1}x_{2}^{3}x_{3}^{2^{s}-3}x_{4}^{2^{s}-2}$, |  $a_{s,\,89}=x_{1}^{3}x_{2}^{2^{s}-3}x_{3}^{2^{s}-1}x_{4}^{2^{s+2}-2}$, |  $a_{s,\,90}=x_{1}^{3}x_{2}^{2^{s}-3}x_{3}^{2^{s+2}-2}x_{4}^{2^{s}-1}$,
$a_{s,\,91}=x_{1}^{3}x_{2}^{2^{s}-1}x_{3}^{2^{s}-3}x_{4}^{2^{s+2}-2}$, |  $a_{s,\,92}=x_{1}^{2^{s}-1}x_{2}^{3}x_{3}^{2^{s}-3}x_{4}^{2^{s+2}-2}$, |  $a_{s,\,93}=x_{1}^{2^{s}-1}x_{2}^{3}x_{3}^{2^{s+2}-3}x_{4}^{2^{s}-2}$,
$a_{s,\,94}=x_{1}^{3}x_{2}^{2^{s}-3}x_{3}^{2^{s+1}-2}x_{4}^{3.2^{s}-1}$, |  $a_{s,\,95}=x_{1}^{3}x_{2}^{2^{s}-3}x_{3}^{2^{s+1}-1}x_{4}^{3.2^{s}-2}$, |  $a_{s,\,96}=x_{1}^{3}x_{2}^{2^{s+1}-1}x_{3}^{2^{s}-3}x_{4}^{3.2^{s}-2}$,
$a_{s,\,97}=x_{1}^{2^{s+1}-1}x_{2}^{3}x_{3}^{2^{s}-3}x_{4}^{3.2^{s}-2}$, |  $a_{s,\,98}=x_{1}^{2^{s}-1}x_{2}^{3}x_{3}^{2^{s+1}-3}x_{4}^{3.2^{s}-2}$, |  $a_{s,\,99}=x_{1}^{7}x_{2}^{2^{s+2}-5}x_{3}^{2^{s}-3}x_{4}^{2^{s}-2}$,
$a_{s,\,100}=x_{1}^{7}x_{2}^{2^{s+1}-5}x_{3}^{2^{s}-3}x_{4}^{3.2^{s}-2}$, |  $a_{s,\,101}=x_{1}^{7}x_{2}^{2^{s+1}-5}x_{3}^{3.2^{s}-3}x_{4}^{2^{s}-2}$, |  $a_{s,\,102}=x_{1}^{7}x_{2}^{2^{s+1}-5}x_{3}^{2^{s+1}-3}x_{4}^{2^{s+1}-2}$.

For $s=3,$

$a_{3,\,103}=x_{1}^{7}x_{2}^{7}x_{3}^{7}x_{4}^{24}$, |  $a_{3,\,104}=x_{1}^{7}x_{2}^{7}x_{3}^{9}x_{4}^{22}$, |  $a_{3,\,105}=x_{1}^{7}x_{2}^{7}x_{3}^{25}x_{4}^{6}$.
---|---|---

For $s\geq 4,$

$a_{s,\,103}=x_{1}^{7}x_{2}^{2^{s}-5}x_{3}^{2^{s}-3}x_{4}^{2^{s+2}-2}$, |  $a_{s,\,104}=x_{1}^{7}x_{2}^{2^{s}-5}x_{3}^{2^{s+1}-3}x_{4}^{3.2^{s}-2}$, |  $a_{s,\,105}=x_{1}^{7}x_{2}^{2^{s}-5}x_{3}^{2^{s+2}-3}x_{4}^{2^{s}-2}$.
---|---|---

We now consider the following cases.

Case $s=1$. We have a direct summand decomposition of the
$\Sigma_{4}$-modules:

$\overline{Q^{\otimes
4}_{n_{1}}}=\Sigma_{4}(a_{1,\,29},a_{1,\,32})\bigoplus\Sigma_{4}(a_{1,\,35}),$

where
$\Sigma_{4}(a_{1,\,29};a_{1,\,32})=\langle\\{[a_{1,\,j}]:\,j=29,30,\ldots,34,41,42,\ldots,46\\}\rangle,$
and

$\Sigma_{4}(a_{1,\,35})=\langle\\{[a_{1,\,j}]:\,35\leq j\leq 40\\}\rangle.$

######  Lemma 4.1.2.

The following statements are true:

  * i)

$\Sigma_{4}(a_{1,\,29};a_{1,\,32})^{\Sigma_{4}}=\langle[q_{1,\,4}]\rangle,$
with $q_{1,\,4}=\sum_{29\leq j\leq 31}a_{1,\,j}+\sum_{44\leq j\leq
46}a_{1,\,j};$

  * ii)

$\Sigma_{4}(a_{1,\,35})^{\Sigma_{4}}=0.$

Combining Lemmas 4.1.1 and 4.1.2 gives

######  Corollary 4.1.3.

The space of $\Sigma_{4}$-invariants $(Q^{\otimes 4}_{n_{1}})^{\Sigma_{4}}$ is
generated by the classes $[q_{1,\,t}]$ for all $t,\,1\leq t\leq 4.$

###### Proof of Lemma 4.1.2.

The proof of $ii)$ is straightforward. We prove $i)$ in detail. Suppose that
$[g]\in\Sigma_{4}(a_{1,\,29};a_{1,\,32})^{\Sigma_{4}}.$ Then, we have

$g\equiv\big{(}\sum_{29\leq j\leq 34}\gamma_{j}a_{1,\,j}+\sum_{41\leq j\leq
46}\gamma_{j}a_{1,\,j}\big{)},$ (21)

with $\gamma_{j}\in k.$ Acting the homomorphisms
$\sigma_{i}:(P_{4})_{n_{1}}\to(P_{4})_{n_{1}},\,1\leq i\leq 3,$ on both sides
of (21), we get

$\begin{array}[]{ll}\sigma_{1}(g)&\equiv\big{(}\gamma_{29}a_{1,\,29}+\gamma_{30}a_{1,\,30}+\gamma_{32}a_{1,\,32}+\gamma_{41}a_{1,\,41}+\gamma_{44}a_{1,\,42}\\\
&\quad+\gamma_{45}a_{1,\,43}+\gamma_{42}a_{1,\,44}+\gamma_{43}a_{1,\,45}+\gamma_{31}x_{1}^{6}x_{2}x_{3}x_{4}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\gamma_{33}x_{1}^{2}x_{2}x_{3}x_{4}^{5}+\gamma_{34}x_{1}^{2}x_{2}x_{3}^{5}x_{4}+\gamma_{46}x_{1}^{4}x_{2}^{3}x_{3}x_{4}\big{)}\\\
\sigma_{2}(g)&\equiv\big{(}\gamma_{29}a_{1,\,29}+\gamma_{31}a_{1,\,30}+\gamma_{30}a_{1,\,31}+\gamma_{33}a_{1,\,32}+\gamma_{42}a_{1,\,41}+\gamma_{41}a_{1,\,42}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\gamma_{44}a_{1,\,44}+\gamma_{46}a_{1,\,45}+\gamma_{45}a_{1,\,46}+\gamma_{34}x_{1}x_{2}^{5}x_{3}^{2}x_{4}+\gamma_{43}x_{1}x_{2}^{4}x_{3}^{3}x_{4}\big{)}\\\
\sigma_{3}(g)&\equiv\big{(}\gamma_{30}a_{1,\,29}+\gamma_{29}a_{1,\,30}+\gamma_{31}a_{1,\,31}+\gamma_{34}a_{1,\,43}+\gamma_{33}a_{1,\,34}+\gamma_{43}a_{1,\,42}+\gamma_{42}a_{1,\,43}\\\
&\quad+\gamma_{45}a_{1,\,44}+\gamma_{44}a_{1,\,45}+\gamma_{46}a_{1,\,46}+\gamma_{32}x_{1}x_{2}x_{3}^{5}x_{4}^{2}+\gamma_{41}x_{1}x_{2}x_{3}^{4}x_{4}^{3}\big{)}.\end{array}$
(22)

Using the Cartan formula and Theorem 4.5, we have

$\begin{array}[]{ll}x_{1}^{6}x_{2}x_{3}x_{4}&=Sq^{1}(x_{1}^{5}x_{2}x_{3}x_{4})+Sq^{2}(x_{1}^{3}x_{2}^{2}x_{3}x_{4}+x_{1}^{3}x_{2}x_{3}^{2}x_{4}+x_{1}^{3}x_{2}x_{3}x_{4}^{2})\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+a_{1,\,44}+a_{1,\,45}+a_{1,\,46}\mod(\widehat{A}(P_{4})_{n_{1}});\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}^{2}x_{2}x_{3}x_{4}^{5}&=Sq^{1}(x_{1}x_{2}x_{3}x_{4}^{5})+a_{1,\,29}+a_{1,\,32}+a_{1,\,33}\mod(\widehat{A}(P_{4})_{n_{1}});\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}^{2}x_{2}x_{3}^{5}x_{4}&=Sq^{1}(x_{1}x_{2}x_{3}^{5}x_{4})+Sq^{2}(x_{1}x_{2}x_{3}^{3}x_{4}^{2})+a_{1,\,30}+a_{1,\,34}+a_{1,\,41}\mod(\widehat{A}(P_{4})_{n_{1}});\\\
x_{1}^{4}x_{2}^{3}x_{3}x_{4}&=Sq^{1}(x_{1}x_{2}^{5}x_{3}x_{4})+Sq^{2}(x_{1}x_{2}^{3}x_{3}x_{4}^{2}+x_{1}x_{2}^{3}x_{3}^{2}x_{4}+x_{1}^{2}x_{2}^{3}x_{3}x_{4})\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+a_{1,\,31}+a_{1,\,42}+a_{1,\,43}\mod(\widehat{A}(P_{4})_{n_{1}});\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}x_{2}^{5}x_{3}^{2}x_{4}&=Sq^{2}(x_{1}x_{2}^{3}x_{3}^{2}x_{4})+a_{1,\,43}\mod(\widehat{A}(P_{4})_{n_{1}});\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}x_{2}^{4}x_{3}^{3}x_{4}&=Sq^{2}(x_{1}x_{2}^{2}x_{3}^{3}x_{4})+a_{1,\,34}\mod(\widehat{A}(P_{4})_{n_{1}});\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}x_{2}x_{3}^{5}x_{4}^{2}&=Sq^{2}(x_{1}x_{2}x_{3}^{3}x_{4}^{2})+a_{1,\,41}\mod(\widehat{A}(P_{4})_{n_{1}});\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}x_{2}x_{3}^{4}x_{4}^{3}&=Sq^{2}(x_{1}x_{2}x_{3}^{2}x_{4}^{3})+a_{1,\,32}\mod(\widehat{A}(P_{4})_{n_{1}}).\end{array}$
(23)

Combining (21), (22) and (23), we get

$\begin{array}[]{ll}(\sigma_{1}(g)+g)&\equiv\big{(}\gamma_{33}(a_{1,\,29}+a_{1,\,32})+\gamma_{34}(a_{1,\,30}+a_{1,\,41})+(\gamma_{31}+\gamma_{46})a_{1,\,31}+(\gamma_{42}+\gamma_{44}+\gamma_{46})a_{1,\,42}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+(\gamma_{43}+\gamma_{45}+\gamma_{46})a_{1,\,43}+(\gamma_{31}+\gamma_{42}+\gamma_{44})a_{1,\,44}+(\gamma_{31}+\gamma_{43}+\gamma_{45})a_{1,\,45}\big{)},\\\
(\sigma_{2}(g)+g)&\equiv\big{(}(\gamma_{30}+\gamma_{31})a_{1,\,30}+(\gamma_{32}+\gamma_{33})a_{1,\,32}+(\gamma_{34}+\gamma_{43})a_{1,\,34}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+(\gamma_{41}+\gamma_{42})a_{1,\,41}+(\gamma_{45}+\gamma_{46})a_{1,\,45}\big{)}\\\
(\sigma_{3}(g)+g)&\equiv\big{(}(\gamma_{29}+\gamma_{30})a_{1,\,29}+(\gamma_{32}+\gamma_{41})a_{1,\,32}+(\gamma_{33}+\gamma_{34})a_{1,\,33}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+(\gamma_{42}+\gamma_{43})a_{1,\,42}+(\gamma_{44}+\gamma_{45})a_{1,\,44}\big{)}.\end{array}$

From these equalities and the relations $\sigma_{j}(g)\equiv g,\,1\leq j\leq
3,$ we obtain $\gamma_{j}=0$ for $j\in\\{32,33,34,41,42,43\\},$ and
$\gamma_{29}=\gamma_{30}=\gamma_{31}=\gamma_{44}=\gamma_{45}=\gamma_{46}.$
This completes the proof of the lemma. ∎

The following assertion is useful.

######  Proposition 4.1.4.

We have $(Q^{\otimes 4}_{n_{1}})^{GL_{4}(k)}=\langle[q_{1,\,4}]\rangle.$

###### Proof.

Let $h\in(P_{4})_{n_{1}}$ such that $[h]\in(Q^{\otimes
4}_{n_{1}})^{GL_{4}(k)}.$ Since $\Sigma_{4}\subset GL_{4}(k),$
$[h]\in(Q^{\otimes 4}_{n_{1}})^{\Sigma_{4}}$ and therefore by Corollary 4.1.3,
we have $h\equiv\sum_{1\leq t\leq 4}m_{t}q_{1,\,t},$ where $m_{t}\in k$ for
every $t.$ Since $[h]\in(Q^{\otimes 4}_{n_{1}})^{GL_{4}(k)},$
$(\sigma_{4}(h)+h)\equiv 0$ with
$\sigma_{4}:(P_{4})_{n_{1}}\to(P_{4})_{n_{1}}.$ From Theorem 4.5 and a direct
computation, we obtain

$(\sigma_{4}(h)+h)\equiv(m_{1}a_{1,\,1}+(m_{1}+m_{2})a_{1,\,8}+(m_{2}+m_{3})a_{1,\,17}+\mbox{other
terms})\equiv 0.$

and therefore $h\equiv q_{1,\,4}.$ The proposition is proved. ∎

Now, we consider the element

$\zeta=(a_{1}^{(1)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(2)}+a_{1}^{(1)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(1)}+a_{1}^{(1)}a_{2}^{(5)}a_{3}^{(2)}a_{4}^{(1)}+a_{1}^{(1)}a_{2}^{(6)}a_{3}^{(1)}a_{4}^{(1)})\in(P_{4})_{3(2^{1}-1)+3.2^{1}}^{*}.$

Then, it is $\widehat{A}$-annihilated. Indeed, by the unstable condition, we
need only to consider the effects of $Sq^{1}$ and $Sq^{2}$. By a direct
computation, we get

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt(\zeta)Sq^{1}&=a_{1}^{(1)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(1)}+a_{1}^{(1)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(1)}+a_{1}^{(1)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(1)}+a_{1}^{(1)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(1)}=0,\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt(\zeta)Sq^{2}&=0+a_{1}^{(1)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(1)}+a_{1}^{(1)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(1)}+0=0.\end{array}$

On the other side, since $\langle[q_{1,\,4}],[\zeta]\rangle=1,$ by Proposition
4.1.4, implies that

$k\otimes_{GL_{4}(k)}P((P_{4})_{n_{1}}^{*})=((Q^{\otimes
4}_{n_{1}})^{GL_{4}(k)})^{*}=\langle([q_{1,\,4}])^{*}\rangle=\langle[\zeta]\rangle.$

Case $s=2$. We first have the following remark.

######  Remark 4.1.5.

Based on the bases of $\underline{Q^{\otimes 4}_{n_{s}}}$ and
$\overline{Q^{\otimes 4}_{n_{s}}}$ above, the weight vector of $a_{s,\,j}$ is
one of the following sequences:

$\omega_{(s,1)}:=\underset{\mbox{{$s$ times of
$3$}}}{\underbrace{(3,3,\ldots,3}},1,1),\ \
\omega_{(s,2)}:=\underset{\mbox{{$s+1$ times of
$3$}}}{\underbrace{(3,3,\ldots,3)}}.$

Hence, we have the following isomorphisms:

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus 2.0ptQ^{\otimes
4}_{n_{s}}&\cong(Q^{\otimes 4}_{n_{s}})^{\omega_{(s,1)}}\bigoplus(Q^{\otimes
4}_{n_{s}})^{\omega_{(s,2)}},\\\ (Q^{\otimes
4}_{n_{s}})^{\omega_{(s,j)}}&\cong\underline{(Q^{\otimes
4}_{n_{s}})}^{\omega_{(s,j)}}\bigoplus\overline{(Q^{\otimes
4}_{n_{s}})}^{\omega_{(s,j)}},\ j=1,\,2,\end{array}$

where

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus 2.0pt\underline{(Q^{\otimes
4}_{n_{s}})}^{\omega_{(s,j)}}&:=\langle\\{[x=\prod_{1\leq i\leq
4}x_{i}^{a_{i}}]\in Q^{\otimes 4}_{n_{s}}:\,\prod_{1\leq i\leq
4}a_{i}=0,\omega(x)=\omega_{(s,j)}\\}\rangle,\\\ \overline{(Q^{\otimes
4}_{n_{s}})}^{\omega_{(s,j)}}&:=\langle\\{[x=\prod_{1\leq i\leq
4}x_{i}^{a_{i}}]\in Q^{\otimes 4}_{n_{s}}:\,\prod_{1\leq i\leq
4}a_{i}>0,\omega(x)=\omega_{(s,j)}\\}\rangle.\end{array}$

Then, for $s=2,$ we have $Q^{\otimes 4}_{n_{2}}\cong(Q^{\otimes
4}_{n_{2}})^{\omega_{(2,1)}}\bigoplus(Q^{\otimes 4}_{n_{2}})^{\omega_{(2,2)}}$
and the following.

######  Proposition 4.1.6.

The spaces of invariants $((Q^{\otimes
4}_{n_{2}})^{\omega_{(2,1)}})^{GL_{4}(k)}$ and $((Q^{\otimes
4}_{n_{2}})^{\omega_{(2,2)}})^{GL_{4}(k)}$ are trivial.

###### Proof.

We first prove $((Q^{\otimes 4}_{n_{2}})^{\omega_{(2,1)}})^{GL_{4}(k)}=0.$ By
Definition 4.4, the monomial
$a_{s,\,12}=x_{1}^{2^{s+2}-1}x_{2}^{2^{s}-1}x_{3}^{2^{s}-1}$ is minimal spike
in $(P_{4})_{n_{s}}$ and $\omega(a_{s,\,12})=\omega_{(s,1)}$ for any $s\geq
1.$ Hence, $[a_{s,\,j}]_{\omega_{(s,1)}}=[a_{s,\,j}].$ Recall that
$(Q^{\otimes 4}_{n_{2}})^{\omega_{(2,1)}}\cong\underline{(Q^{\otimes
4}_{n_{2}})}^{\omega_{(2,1)}}\bigoplus\overline{(Q^{\otimes
4}_{n_{2}})}^{\omega_{(2,1)}}.$ Then, by Lemma 4.1.1, we get
$\underline{(Q^{\otimes
4}_{n_{2}})}^{\omega_{(2,1)}}=\Sigma_{4}(a_{2,\,1})\bigoplus\Sigma_{4}(a_{2,\,13}),$
and

$(\underline{(Q^{\otimes
4}_{n_{2}})}^{\omega_{(2,1)}})^{\Sigma_{4}}=\langle[q_{2,\,1}],[q_{2,\,2}]\rangle.$
(24)

Using an admissible basis of $\overline{Q^{\otimes 4}_{n_{2}}}$ above, we have
a direct summand decomposition of the $\Sigma_{4}$-modules:

$\overline{(Q^{\otimes
4}_{n_{2}})}^{\omega_{(2,1)}}=\Sigma_{4}(a_{2,\,41})\bigoplus\Sigma_{4}(a_{2,\,53})\bigoplus\mathbb{V},$

where

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt&\Sigma_{4}(a_{2,\,41})=\langle\\{[a_{2,\,j}]:\,41\leq j\leq
52\\}\rangle,\ \Sigma_{4}(a_{2,\,53})=\langle\\{[a_{2,\,j}]:\,53\leq j\leq
58\\}\rangle,\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt&\mathbb{V}=\langle\\{[a_{2,\,j}]:\,j=29,\ldots,34,59,\ldots,80,85,\ldots,94\\}\rangle.\end{array}$

Then, by a similar technique as in the proof of Lemma 4.1.2, we find that

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt&\Sigma_{4}(a_{2,\,41})^{\Sigma_{4}}=\langle[\widehat{q_{2,\,1}}]\rangle\
\mbox{with}\ \widehat{q_{2,\,1}}:=\sum_{41\leq j\leq 52}a_{2,\,j};\\\ \vskip
6.0pt plus 2.0pt minus
2.0pt&\Sigma_{4}(a_{2,\,53})^{\Sigma_{4}}=\langle[\widehat{q_{2,\,2}}]\rangle\
\mbox{with}\ \widehat{q_{2,\,2}}:=\sum_{53\leq j\leq 58}a_{2,\,j};\\\ \vskip
6.0pt plus 2.0pt minus 2.0pt&\mathbb{V}^{\Sigma_{4}}=0.\end{array}$ (25)

Thus, from the equalities (24) and (25), we get

$((Q^{\otimes
4}_{n_{2}})^{\omega_{(2,1)}})^{\Sigma_{4}}=\langle[q_{2,\,1}],[q_{2,\,2}],[\widehat{q_{2,\,1}}],[\widehat{q_{2,\,2}}]\rangle.$
(26)

Now, let $h\in(P_{4})_{n_{2}}$ such that $[h]\in((Q^{\otimes
4}_{n_{2}})^{\omega_{(2,1)}})^{GL_{4}(k)}.$ Since $[h]\in((Q^{\otimes
4}_{n_{2}})^{\omega_{(2,1)}})^{\Sigma_{4}},$ by the equality (26), we have

$h\equiv(\xi_{1}q_{2,\,1}+\xi_{2}q_{2,\,2}+\xi_{3}\widehat{q_{2,\,1}}+\xi_{4}\widehat{q_{2,\,2}}),$
(27)

where $\xi_{t}\in k$ for $1\leq t\leq 4.$ Acting the homomorphism
$\sigma_{4}:(P_{4})_{n_{2}}\to(P_{4})_{n_{2}}$ on both sides of (27), and
using Theorem 4.5 with the relation $(\sigma_{4}(h)+h)\equiv 0,$ we obtain

$(\sigma_{4}(h)+h)\equiv((\xi_{1}+\xi_{3})a_{2,\,1}+\xi_{1}a_{2,\,3}+(\xi_{1}+\xi_{2})a_{2,\,8}+(\xi_{2}+\xi_{4})a_{2,\,13}+\mbox{other
terms})\equiv 0.$

The last equality implies $\xi_{1}=\xi_{2}=\xi_{3}=\xi_{4}.$ This means that
$((Q^{\otimes 4}_{n_{2}})^{\omega_{(2,1)}})^{GL_{4}(k)}=0.$

Next, we have $(Q^{\otimes
4}_{n_{2}})^{\omega_{(2,2)}}\cong\underline{(Q^{\otimes
4}_{n_{2}})}^{\omega_{(2,2)}}\bigoplus\overline{(Q^{\otimes
4}_{n_{2}})}^{\omega_{(2,2)}}.$ According to Lemma 4.1.1, we get

$(\underline{(Q^{\otimes
4}_{n_{2}})}^{\omega_{(2,2)}})^{\Sigma_{4}}=\langle[q_{2,\,3}]_{\omega_{(2,2)}}\rangle.$

A basis of $\overline{(Q^{\otimes 4}_{n_{2}})}^{\omega_{(2,2)}}$ is the set
$\\{[a_{2,\,j}]_{\omega_{(2,2)}}:\,j=35,36,\ldots,40,81,\ldots,84\\}$ and we
have a direct summand decomposition of the $\Sigma_{4}$-modules:

$\overline{(Q^{\otimes
4}_{n_{2}})}^{\omega_{(2,2)}}=\Sigma_{4}(a_{2,\,35})\bigoplus\Sigma_{4}(a_{2,\,81}),$

where
$\Sigma_{4}(a_{2,\,35})=\langle\\{[[a_{2,\,j}]_{\omega_{(2,2)}}:\,j=35,\ldots,40]\\}\rangle$
and
$\Sigma_{4}(a_{2,\,81})=\langle\\{[[a_{2,\,j}]_{\omega_{(2,2)}}:\,j=81,\ldots,84]\\}\rangle.$
By using Theorem 4.5 and a similar computation as above, we also obtain

$\Sigma_{4}(a_{2,\,35})^{\Sigma_{4}}=\langle[\widehat{q_{2,\,3}}]_{\omega_{(2,\,2)}}\rangle,\
\
\Sigma_{4}(a_{2,\,81})^{\Sigma_{4}}=\langle[\widehat{q_{2,\,4}}]_{\omega_{(2,\,2)}}\rangle,$

where $\widehat{q_{2,\,3}}=\sum_{35\leq j\leq 40}a_{2,\,j}$ and
$\widehat{q_{2,\,4}}=\sum_{81\leq j\leq 84}a_{2,\,j}.$ Then,

$((Q^{\otimes
4}_{n_{2}})^{\omega_{(2,2)}})^{\Sigma_{4}}=\langle[q_{2,\,3}]_{\omega_{(2,\,2)}},\langle[\widehat{q_{2,\,3}}]_{\omega_{(2,\,2)}},\langle[\widehat{q_{2,\,4}}]_{\omega_{(2,\,2)}}\rangle.$

Let $[g]_{{}_{\omega_{(2,\,2)}}}\in((Q^{\otimes
4}_{n_{2}})^{\omega_{(2,2)}})^{GL_{4}(k)},$ we have
$g\equiv_{{}_{\omega_{(2,\,2)}}}(\ell_{1}q_{2,\,3}+\ell_{2}\widehat{q_{2,\,3}}+\ell_{3}\widehat{q_{2,\,4}})$
with $\ell_{i}\in k$ for every $i.$ By a direct computation using the relation
$(\sigma_{4}(g)+g)\equiv_{\omega_{(2,\,2)}}0,$ we obtain

$(\sigma_{4}(g)+g)\equiv_{{}_{\omega_{(2,\,2)}}}((\ell_{1}+\ell_{2})a_{2,\,25}+(\ell_{2}+\ell_{3})a_{2,\,36}+\ell_{3}a_{2,\,83}+\mbox{other
terms})\equiv_{\omega_{(2,\,2)}}0.$

This means that $\ell_{1}=\ell_{2}=\ell_{3}=0.$ The proposition follows. ∎

Now, since $\dim(Q^{\otimes 4}_{n_{2}})^{GL_{4}(k)}\leq\dim((Q^{\otimes
4}_{n_{2}})^{\omega_{(2,1)}})^{GL_{4}(k)}+\dim((Q^{\otimes
4}_{n_{2}})^{\omega_{(2,2)}})^{GL_{4}(k)},$ by Proposition 4.1.6, we get

$k\otimes_{GL_{4}(k)}P((P_{4})_{n_{2}}^{*})=((Q^{\otimes
4}_{n_{2}})^{GL_{4}(k)})^{*}=0.$

Case $s\geq 3$. For simplicity, we prove the cases $s\geq 4.$ Calculations of
the case $s=3$ use similar techniques. Recall that

$Q^{\otimes 4}_{n_{s}}\cong(Q^{\otimes
4}_{n_{s}})^{\omega_{(s,1)}}\bigoplus(Q^{\otimes
4}_{n_{s}})^{\omega_{(s,2)}},$

where $\dim(Q^{\otimes 4}_{n_{s}})^{\omega_{(s,1)}}=90$ and $\dim(Q^{\otimes
4}_{n_{s}})^{\omega_{(s,2)}}=15$ for all $s\geq 4.$

######  Proposition 4.1.7.

The following hold:

  * i)

$((Q^{\otimes 4}_{n_{s}})^{\omega_{(s,1)}})^{GL_{4}(k)}$ is trivial.

  * ii)

$((Q^{\otimes
4}_{n_{s}})^{\omega_{(s,2)}})^{GL_{4}(k)}=\langle[\zeta_{s}^{*}]_{\omega_{(s,2)}}\rangle,$
where

$\zeta_{s}^{*}=\sum_{25\leq j\leq 28}a_{s,\,j}+\sum_{35\leq j\leq
40}a_{s,\,j}+\sum_{81\leq j\leq 84}a_{s,\,j}+a_{s,\,102}$

###### Outline of the proof.

We will give the sketch of proof of ii). The proofs of i) use similar idea.

We have an isomorphism of the $k$-vector spaces

$(Q^{\otimes 4}_{n_{s}})^{\omega_{(s,2)}}\cong\underline{(Q^{\otimes
4}_{n_{s}}})^{\omega_{(s,2)}}\bigoplus\overline{(Q^{\otimes
4}_{n_{s}}})^{\omega_{(s,2)}},$

where

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus 2.0pt\underline{(Q^{\otimes
4}_{n_{s}}})^{\omega_{(s,2)}}&=\langle\\{[a_{s,\,j}]_{\omega_{(s,2)}}:\,25\leq
j\leq 28\\}\rangle,\\\ \overline{(Q^{\otimes
4}_{n_{s}}})^{\omega_{(s,2)}}&=\langle\\{[a_{s,\,j}]_{\omega_{(s,2)}}:\,j=35,\ldots,40,81,\ldots,84,102\\}\rangle.\end{array}$

Then, by Lemma 4.1.1,

$(\underline{(Q^{\otimes
4}_{n_{s}}})^{\omega_{(s,2)}})^{\Sigma_{4}}=\langle[q_{s,\,3}]_{\omega_{(s,2)}}\rangle.$

Now, if $[f]\in(\overline{(Q^{\otimes
4}_{n_{s}}})^{\omega_{(s,2)}})^{\Sigma_{4}},$ then we have

$f\equiv_{\omega_{(s,2)}}\big{(}\sum_{35\leq j\leq
40}\gamma_{j}a_{s,\,j}+\sum_{81\leq j\leq
84}\gamma_{j}a_{s,\,j}+\gamma_{102}a_{s,\,102}\big{)}$

in which $\gamma_{j}\in k.$ So, by a simple computation using the relations
$\sigma_{i}(f)\equiv_{\omega_{(s,2)}}f$ for $i\in\\{1,2,3\\},$ where
$\sigma_{i}:(P_{4})_{n_{s}}\to(P_{4})_{n_{s}},$ we get
$\gamma_{35}=\gamma_{36}=\cdots=\gamma_{40},$ and
$\gamma_{81}=\gamma_{82}=\gamma_{83}=\gamma_{84}.$ This leads to

$(\overline{(Q^{\otimes
4}_{n_{s}}})^{\omega_{(s,2)}})^{\Sigma_{4}}=\langle[q_{s,\,4}]_{\omega_{(s,2)}},\,[q_{s,\,5}]_{\omega_{(s,5)}},[a_{s,\,102}]_{\omega_{(s,2)}}\rangle,$

with $q_{s,\,4}:=\sum_{35\leq j\leq 40}a_{s,\,j}$ and $q_{s,\,5}:=\sum_{81\leq
j\leq 84}a_{s,\,j}.$

Let $h\in(P_{4})_{n_{s}}$ such that $[h]\in((Q^{\otimes
4}_{n_{s}})^{\omega_{(s,2)}})^{GL_{4}(k)}.$ Since $\Sigma_{4}\subset
GL_{4}(k),$ from the above calculations, we have

$h\equiv_{\omega_{(s,2)}}\big{(}\beta_{1}q_{s,\,3}+\beta_{2}q_{s,\,4}+\beta_{3}q_{s,\,5}+\beta_{4}a_{s,\,102}\big{)},$

with $\beta_{i}\in k$ for every $i.$ Direct computing from the relation
$(\sigma_{4}(h)+h)\equiv 0,$ we obtain

$\begin{array}[]{ll}(\sigma_{4}(h)+h)&\equiv_{\omega_{(s,2)}}\big{(}(\beta_{1}+\beta_{2})a_{s,\,25}+(\beta_{2}+\beta_{3})(a_{s,\,36}+a_{s,\,37})+(\beta_{3}+\beta_{4})a_{s,\,83}\big{)}\equiv
0,\end{array}$

and this is apparently that $\beta_{1}=\beta_{2}=\beta_{3}=\beta_{4}.$ The
proposition is proved. ∎

The following inequality is immediate from Proposition 4.1.7.

$\dim k\otimes_{GL_{4}(k)}P((P_{4})_{n_{s}}^{*})\leq\dim((Q^{\otimes
4}_{n_{s}})^{\omega_{(s,1)}})^{GL_{4}(k)}+\dim((Q^{\otimes
4}_{n_{s}})^{\omega_{(s,s)}})^{GL_{4}(k)}=1.$

On the other hand, as indicated in the introduction, we can verify that the
element

$\zeta_{s}=a_{1}^{(0)}a_{2}^{(2^{s+1}-1)}a_{3}^{(2^{s+1}-1)}a_{4}^{(2^{s+1}-1)}\in(P_{4})^{*}_{3(2^{s}-1)+3.2^{s}}\in
P((P_{4})_{n_{s}}^{*})$

and that the cycle $\psi_{4}(\zeta_{s})=\lambda_{0}\lambda_{s+1}^{3}$ in
$\Lambda$ is a representative of the non-zero element $h_{0}h_{s+1}^{3}\in{\rm
Ext}_{A}^{4,6.2^{s}+1}(k,k).$ Hence, $h_{0}h_{s+1}^{3}$ is in the image of
$Tr_{4}^{A}.$ Moreover, by Theorem 2.1, we have ${\rm
Ext}_{A}^{4,6.2^{s}+1}(k,k)=\langle h_{0}h_{s+1}^{3}\rangle$ with
$h_{0}h_{s+1}^{3}=h_{0}h_{s}^{2}h_{s+2}\neq 0$ for any $s\geq 3.$ Combining
these data gives $\dim k\otimes_{GL_{4}(k)}P((P_{4})_{n_{s}}^{*})=1.$
Moreover, by a simple computation, we find that $(Q^{\otimes
4}_{n_{s}})^{GL_{4}(k)}=\langle[\zeta_{s}^{*}]\rangle.$ From this and the fact
that $\langle[\zeta_{s}^{*}],[\zeta_{s}]\rangle=1,$ it may be concluded that
$k\otimes_{GL_{4}(k)}P((P_{4})_{n_{s}}^{*})=\langle[\zeta_{s}]\rangle.$ The
theorem is proved. ∎

###  4.2. Proof of Theorem 2.5

Let $n_{s}:=3(2^{s}-1)+7.2^{s},$ in order to prove the theorem, we first have
the following.

######  Proposition 4.2.1.

For a positive integer $s,$ the invariant spaces $(Q^{\otimes
4}_{n_{s}})^{GL_{4}(k)}$ are determined as follows:

$(Q^{\otimes
4}_{n_{s}})^{GL_{4}(k)}=\left\\{\begin{array}[]{ll}\langle[\widetilde{\zeta}]\rangle&\mbox{for
$s=1$},\\\ 0&\mbox{otherwise},\end{array}\right.$

where

$\begin{array}[]{ll}\widetilde{\zeta}&=x_{1}x_{2}x_{3}x_{4}^{14}+x_{1}x_{2}x_{3}^{14}x_{4}+x_{1}x_{2}^{3}x_{3}x_{4}^{12}+\vskip
6.0pt plus 2.0pt minus 2.0ptx_{1}x_{2}^{3}x_{3}^{12}x_{4}\\\
&\quad+x_{1}^{3}x_{2}x_{3}x_{4}^{12}+x_{1}^{3}x_{2}x_{3}^{12}x_{4}+x_{1}^{3}x_{2}^{5}x_{3}x_{4}^{8}+x_{1}^{3}x_{2}^{5}x_{3}^{8}x_{4}.\end{array}$

###### Proof.

The proof of the proposition is based on an admissible monomial basis of the
$k$-vector space $Q^{\otimes 4}_{n_{s}}.$ We have an isomorphism of $k$-vector
spaces: $Q^{\otimes 4}_{n_{s}}\cong\underline{Q^{\otimes
4}_{n_{s}}}\bigoplus\overline{Q^{\otimes 4}_{n_{s}}}.$ According to Sum [42],
$\underline{Q^{\otimes 4}_{n_{s}}}=\langle\\{[b_{s,\,j}]:\,1\leq j\leq
40\\}\rangle,$ for all $s\geq 1,$ where the admissible monomials $b_{s,\,j}$
are determined as follows:

$b_{s,\,1}=x_{2}^{2^{s+1}-1}x_{3}^{2^{s+2}-1}x_{4}^{2^{s+2}-1}$, |  $b_{s,\,2}=x_{2}^{2^{s+2}-1}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+2}-1}$, |  $b_{s,\,3}=x_{2}^{2^{s+2}-1}x_{3}^{2^{s+2}-1}x_{4}^{2^{s+1}-1}$,
---|---|---
$b_{s,\,4}=x_{1}^{2^{s+1}-1}x_{3}^{2^{s+2}-1}x_{4}^{2^{s+2}-1}$, |  $b_{s,\,5}=x_{1}^{2^{s+2}-1}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+2}-1}$, |  $b_{s,\,6}=x_{1}^{2^{s+2}-1}x_{3}^{2^{s+2}-1}x_{4}^{2^{s+1}-1}$,
$b_{s,\,7}=x_{1}^{2^{s+1}-1}x_{2}^{2^{s+2}-1}x_{4}^{2^{s+2}-1}$, |  $b_{s,\,8}=x_{1}^{2^{s+2}-1}x_{2}^{2^{s+1}-1}x_{4}^{2^{s+2}-1}$, |  $b_{s,\,9}=x_{1}^{2^{s+2}-1}x_{2}^{2^{s+2}-1}x_{4}^{2^{s+1}-1}$,
$b_{s,\,10}=x_{1}^{2^{s+1}-1}x_{2}^{2^{s+2}-1}x_{3}^{2^{s+2}-1}$, |  $b_{s,\,11}=x_{1}^{2^{s+2}-1}x_{2}^{2^{s+1}-1}x_{3}^{2^{s+2}-1}$, |  $b_{s,\,12}=x_{1}^{2^{s+2}-1}x_{2}^{2^{s+2}-1}x_{3}^{2^{s+1}-1}$,
$b_{s,\,13}=x_{1}^{2^{s}-1}x_{2}^{2^{s}-1}x_{3}^{2^{s+3}-1}$, |  $b_{s,\,14}=x_{1}^{2^{s}-1}x_{2}^{2^{s+3}-1}x_{3}^{2^{s+3}-1}$, |  $b_{s,\,15}=x_{1}^{2^{s+3}-1}x_{2}^{2^{s}-1}x_{3}^{2^{s}-1}$,
$b_{s,\,16}=x_{2}^{2^{s}-1}x_{3}^{2^{s+3}-1}$, |  $b_{s,\,17}=x_{2}^{2^{s+3}-1}x_{3}^{2^{s+3}-1}$, |  $b_{s,\,18}=x_{2}^{2^{s}-1}x_{3}^{2^{s}-1}$,
$b_{s,\,19}=x_{1}^{2^{s}-1}x_{3}^{2^{s+3}-1}$, |  $b_{s,\,20}=x_{1}^{2^{s+3}-1}x_{3}^{2^{s+3}-1}$, |  $b_{s,\,21}=x_{1}^{2^{s}-1}x_{3}^{2^{s}-1}$,
$b_{s,\,22}=x_{1}^{2^{s}-1}x_{2}^{2^{s+3}-1}$, |  $b_{s,\,23}=x_{1}^{2^{s+3}-1}x_{2}^{2^{s+3}-1}$, |  $b_{s,\,24}=x_{1}^{2^{s}-1}x_{2}^{2^{s}-1}$,
$b_{s,\,25}=x_{1}^{2^{s}-1}x_{2}^{2^{s+1}-1}x_{3}^{7.2^{s}-1}$, |  $b_{s,\,26}=x_{1}^{2^{s+1}-1}x_{2}^{2^{s}-1}x_{3}^{7.2^{s}-1}$, |  $b_{s,\,27}=x_{1}^{2^{s+1}-1}x_{2}^{7.2^{s}-1}x_{3}^{2^{s}-1}$,
$b_{s,\,28}=x_{2}^{2^{s+1}-1}x_{3}^{7.2^{s}-1}$, |  $b_{s,\,29}=x_{2}^{2^{s}-1}x_{3}^{7.2^{s}-1}$, |  $b_{s,\,30}=x_{2}^{7.2^{s}-1}x_{3}^{2^{s}-1}$,
$b_{s,\,31}=x_{1}^{2^{s+1}-1}x_{3}^{7.2^{s}-1}$, |  $b_{s,\,32}=x_{1}^{2^{s}-1}x_{3}^{7.2^{s}-1}$, |  $b_{s,\,33}=x_{1}^{7.2^{s}-1}x_{3}^{2^{s}-1}$,
$b_{s,\,34}=x_{1}^{2^{s+1}-1}x_{2}^{7.2^{s}-1}$, |  $b_{s,\,35}=x_{1}^{2^{s}-1}x_{2}^{7.2^{s}-1}$, |  $b_{s,\,36}=x_{1}^{7.2^{s}-1}x_{2}^{2^{s}-1}$,
$b_{s,\,37}=x_{1}^{2^{s+1}-1}x_{2}^{3.2^{s}-1}x_{3}^{5.2^{s}-1}$, |  $b_{s,\,38}=x_{2}^{3.2^{s}-1}x_{3}^{5.2^{s}-1}$, |  $b_{s,\,39}=x_{1}^{3.2^{s}-1}x_{3}^{5.2^{s}-1}$,
$b_{s,\,40}=x_{1}^{3.2^{s}-1}x_{2}^{5.2^{s}-1}$. |  |

Then, we have a direct summand decomposition of $\Sigma_{4}$-modules:

$\underline{Q^{\otimes
4}_{n_{s}}}=\Sigma_{4}(b_{s,\,1})\bigoplus\Sigma_{4}(b_{s,\,13})\bigoplus\Sigma_{4}(b_{s,\,25})\bigoplus\Sigma_{4}(b_{s,\,37}),$

where

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt\Sigma_{4}(b_{s,\,1})&=\langle\\{[b_{s,\,j}]:\,1\leq j\leq 12\\}\rangle;\
\ \ \Sigma_{4}(b_{s,\,13})=\langle\\{[b_{s,\,j}]:\,13\leq j\leq
24\\}\rangle\\\ \Sigma_{4}(b_{s,\,25})&=\langle\\{[b_{s,\,j}]:\,25\leq j\leq
36\\}\rangle;\ \ \ \Sigma_{4}(b_{s,\,37})=\langle\\{[b_{s,\,j}]:\,37\leq j\leq
40\\}\rangle.\end{array}$

Denote the bases of $\Sigma_{4}(b_{s,\,j})$ by the sets
$[\mathscr{B}(b_{s,\,j})]$ for $j\in\\{1,13,25,37\\},$ where
$\mathscr{B}(b_{s,\,j})=\\{b_{s,\,j}\\}.$ Suppose that $f\in(P_{4})_{n_{s}}$
such that $[f]\in\Sigma_{4}(b_{s,\,j})^{\Sigma_{4}}$. Then,
$f\equiv\sum_{x\in\mathscr{B}(b_{s,\,j})}\gamma_{x}.x$ in which $\gamma_{x}\in
k.$ By a direct computation, we can see that the action of $\Sigma_{4}$ on
$\underline{Q^{\otimes 4}_{n_{s}}}$ induces the one of it on
$[\mathscr{B}(b_{s,\,j})]$ and this action is transitive. So,
$\gamma_{x}=\gamma_{x^{\prime}}=\gamma\in k$ for all
$x,\,x^{\prime}\in\mathscr{B}(b_{s,\,j}),$ and therefore, the spaces of
$\Sigma_{4}$-invariants $\Sigma_{4}(b_{s,\,j})^{\Sigma_{4}}$ are determined as
follows.

######  Lemma 4.2.2.

We find that

  * i)

$\Sigma_{4}(b_{s,\,1})^{\Sigma_{4}}=\langle[q_{s,\,1}]\rangle$ with
$q_{s,\,1}:=\sum_{1\leq j\leq 12}b_{s,\,j};$

  * ii)

$\Sigma_{4}(b_{s,\,13})^{\Sigma_{4}}=\langle[q_{s,\,2}]\rangle$ with
$q_{s,\,2}:=\sum_{13\leq j\leq 24}b_{s,\,j};$

  * iii)

$\Sigma_{4}(b_{s,\,25})^{\Sigma_{4}}=\langle[q_{s,\,3}]\rangle$ with
$q_{s,\,3}:=\sum_{25\leq j\leq 36}b_{s,\,j};$

  * iv)

$\Sigma_{4}(b_{s,\,37})^{\Sigma_{4}}=\langle[q_{s,\,4}]\rangle$ with
$q_{s,\,4}:=\sum_{37\leq j\leq 40}b_{s,\,j}.$

By Sum [42], the admissible bases of the spaces $\overline{Q^{\otimes
4}_{n_{s}}}$ are represented by the following admissible monomials:

For $s\geq 1,$

$b_{s,\,41}=x_{1}x_{2}^{2^{s}-1}x_{3}^{2^{s}-1}x_{4}^{2^{s+3}-2}$, |  $b_{s,\,42}=x_{1}x_{2}^{2^{s}-1}x_{3}^{2^{s+3}-2}x_{4}^{2^{s}-1}$,
---|---
$b_{s,\,43}=x_{1}x_{2}^{2^{s+3}-2}x_{3}^{2^{s}-1}x_{4}^{2^{s}-1}$, |  $b_{s,\,44}=x_{1}x_{2}^{2^{s}-1}x_{3}^{2^{s+1}-2}x_{4}^{7.2^{s}-1}$,
$b_{s,\,45}=x_{1}x_{2}^{2^{s+1}-2}x_{3}^{2^{s}-1}x_{4}^{7.2^{s}-1}$, |  $b_{s,\,46}=x_{1}x_{2}^{2^{s+1}-2}x_{3}^{7.2^{s}-1}x_{4}^{2^{s}-1}$,
$b_{s,\,47}=x_{1}x_{2}^{2^{s+1}-2}x_{3}^{3.2^{s}-1}x_{4}^{5.2^{s}-1}$, |  $b_{s,\,48}=x_{1}x_{2}^{2^{s+1}-2}x_{3}^{2^{s+2}-1}x_{4}^{2^{s+2}-1}$,
$b_{s,\,49}=x_{1}x_{2}^{2^{s+2}-1}x_{3}^{2^{s+1}-2}x_{4}^{2^{s+2}-1}$, |  $b_{s,\,50}=x_{1}x_{2}^{2^{s+2}-1}x_{3}^{2^{s+2}-1}x_{4}^{2^{s+1}-2}$,
$b_{s,\,51}=x_{1}^{2^{s+2}-1}x_{2}x_{3}^{2^{s+1}-2}x_{4}^{2^{s+2}-1}$, |  $b_{s,\,52}=x_{1}^{2^{s+2}-1}x_{2}x_{3}^{2^{s+2}-1}x_{4}^{2^{s+1}-2}$,
$b_{s,\,53}=x_{1}^{2^{s+2}-1}x_{2}^{2^{s+2}-1}x_{3}x_{4}^{2^{s+1}-2}$, |  $b_{s,\,54}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+2}-1}$,

$b_{s,\,55}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{2^{s+2}-1}x_{4}^{2^{s+2}-2}$, |  $b_{s,\,56}=x_{1}x_{2}^{2^{s+2}-2}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+2}-1}$,
---|---
$b_{s,\,57}=x_{1}x_{2}^{2^{s+2}-2}x_{3}^{2^{s+2}-1}x_{4}^{2^{s+1}-1}$, |  $b_{s,\,58}=x_{1}x_{2}^{2^{s+2}-1}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+2}-2}$,
$b_{s,\,59}=x_{1}x_{2}^{2^{s+2}-1}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+1}-1}$, |  $b_{s,\,60}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+2}-1}$,
$b_{s,\,61}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{2^{s+2}-1}x_{4}^{2^{s+2}-2}$, |  $b_{s,\,62}=x_{1}^{2^{s+1}-1}x_{2}^{2^{s+2}-1}x_{3}x_{4}^{2^{s+2}-2}$,
$b_{s,\,63}=x_{1}^{2^{s+2}-1}x_{2}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+2}-2}$, |  $b_{s,\,64}=x_{1}^{2^{s+2}-1}x_{2}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+1}-1}$,
$b_{s,\,65}=x_{1}^{2^{s+2}-1}x_{2}^{2^{s+1}-1}x_{3}x_{4}^{2^{s+2}-2}$, |  $b_{s,\,66}=x_{1}^{3}x_{2}^{2^{s+2}-3}x_{3}^{2^{s+1}-2}x_{4}^{2^{s+2}-1}$,
$b_{s,\,67}=x_{1}^{3}x_{2}^{2^{s+2}-3}x_{3}^{2^{s+2}-1}x_{4}^{2^{s+1}-2}$, |  $b_{s,\,68}=x_{1}^{3}x_{2}^{2^{s+2}-1}x_{3}^{2^{s+2}-3}x_{4}^{2^{s+1}-2}$,
$b_{s,\,69}=x_{1}^{2^{s+2}-1}x_{2}^{3}x_{3}^{2^{s+2}-3}x_{4}^{2^{s+1}-2}$, |  $b_{s,\,70}=x_{1}^{3}x_{2}^{2^{s+1}-1}x_{3}^{2^{s+2}-3}x_{4}^{2^{s+2}-2}$,
$b_{s,\,71}=x_{1}^{3}x_{2}^{2^{s+2}-3}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+2}-2}$, |  $b_{s,\,72}=x_{1}^{3}x_{2}^{2^{s+2}-3}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+1}-1}$.

For $s=1,$

$b_{1,\,73}=x_{1}^{3}x_{2}^{3}x_{3}^{4}x_{4}^{7}$, |  $b_{1,\,74}=x_{1}^{3}x_{2}^{3}x_{3}^{7}x_{4}^{4}$, |  $b_{1,\,75}=x_{1}^{3}x_{2}^{7}x_{3}^{3}x_{4}^{4}$,
---|---|---
$b_{1,\,76}=x_{1}^{7}x_{2}^{3}x_{3}^{3}x_{4}^{4}$, |  $b_{1,\,77}=x_{1}x_{2}x_{3}^{3}x_{4}^{12}$, |  $b_{1,\,78}=x_{1}x_{2}^{3}x_{3}x_{4}^{12}$,
$b_{1,\,79}=x_{1}x_{2}^{3}x_{3}^{12}x_{4}$, |  $b_{1,\,80}=x_{1}^{3}x_{2}x_{3}x_{4}^{12}$, |  $b_{1,\,81}=x_{1}^{3}x_{2}x_{3}^{12}x_{4}$,
$b_{1,\,82}=x_{1}x_{2}^{3}x_{3}^{4}x_{4}^{9}$, |  $b_{1,\,83}=x_{1}^{3}x_{2}x_{3}^{4}x_{4}^{9}$, |  $b_{1,\,84}=x_{1}x_{2}^{3}x_{3}^{5}x_{4}^{8}$,
$b_{1,\,85}=x_{1}^{3}x_{2}x_{3}^{5}x_{4}^{8}$, |  $b_{1,\,86}=x_{1}^{3}x_{2}^{5}x_{3}x_{4}^{8}$, |  $b_{1,\,87}=x_{1}^{3}x_{2}^{5}x_{3}^{8}x_{4}$.

For $s\geq 2,$

$b_{s,\,73}=x_{1}^{2^{s+1}-1}x_{2}^{3}x_{3}^{2^{s+2}-3}x_{4}^{2^{s+2}-2}$, |  $b_{s,\,74}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+2}-1}$,
---|---
$b_{s,\,75}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{2^{s+2}-1}x_{4}^{2^{s+2}-2}$, |  $b_{s,\,76}=x_{1}^{3}x_{2}^{2^{s+2}-1}x_{3}^{2^{s+1}-3}x_{4}^{2^{s+2}-2}$,
$b_{s,\,77}=x_{1}^{2^{s+2}-1}x_{2}^{3}x_{3}^{2^{s+1}-3}x_{4}^{2^{s+2}-2}$, |  $b_{s,\,78}=x_{1}^{7}x_{2}^{2^{s+2}-5}x_{3}^{2^{s+1}-3}x_{4}^{2^{s+2}-2}$,
$b_{s,\,79}=x_{1}^{7}x_{2}^{2^{s+2}-5}x_{3}^{2^{s+2}-3}x_{4}^{2^{s+1}-2}$, |  $b_{s,\,80}=x_{1}^{7}x_{2}^{2^{s+1}-5}x_{3}^{2^{s+2}-3}x_{4}^{2^{s+2}-2}$,
$b_{s,\,81}=x_{1}x_{2}^{2^{s}-2}x_{3}^{2^{s}-1}x_{4}^{2^{s+3}-1}$, |  $b_{s,\,82}=x_{1}x_{2}^{2^{s}-2}x_{3}^{2^{s+3}-1}x_{4}^{2^{s}-1}$,
$b_{s,\,83}=x_{1}x_{2}^{2^{s}-1}x_{3}^{2^{s}-2}x_{4}^{2^{s+3}-1}$, |  $b_{s,\,84}=x_{1}x_{2}^{2^{s}-1}x_{3}^{2^{s+3}-1}x_{4}^{2^{s}-2}$,
$b_{s,\,85}=x_{1}x_{2}^{2^{s+3}-1}x_{3}^{2^{s}-2}x_{4}^{2^{s}-1}$, |  $b_{s,\,86}=x_{1}x_{2}^{2^{s+3}-1}x_{3}^{2^{s}-1}x_{4}^{2^{s}-2}$,
$b_{s,\,87}=x_{1}^{2^{s}-1}x_{2}x_{3}^{2^{s}-2}x_{4}^{2^{s+3}-1}$, |  $b_{s,\,88}=x_{1}^{2^{s}-1}x_{2}x_{3}^{2^{s+3}-1}x_{4}^{2^{s}-2}$,
$b_{s,\,89}=x_{1}^{2^{s}-1}x_{2}^{2^{s+3}-1}x_{3}x_{4}^{2^{s}-2}$, |  $b_{s,\,90}=x_{1}^{2^{s+3}-1}x_{2}x_{3}^{2^{s}-2}x_{4}^{2^{s}-1}$,
$b_{s,\,91}=x_{1}^{2^{s+3}-1}x_{2}x_{3}^{2^{s}-1}x_{4}^{2^{s}-2}$, |  $b_{s,\,92}=x_{1}^{2^{s+3}-1}x_{2}^{2^{s}-1}x_{3}x_{4}^{2^{s}-2}$,
$b_{s,\,93}=x_{1}^{2^{s}-1}x_{2}x_{3}^{2^{s}-1}x_{4}^{2^{s+3}-2}$, |  $b_{s,\,94}=x_{1}^{2^{s}-1}x_{2}x_{3}^{2^{s+3}-2}x_{4}^{2^{s}-1}$,
$b_{s,\,95}=x_{1}^{2^{s}-1}x_{2}^{2^{s}-1}x_{3}x_{4}^{2^{s+3}-2}$, |  $b_{s,\,96}=x_{1}x_{2}^{2^{s}-2}x_{3}^{2^{s+1}-1}x_{4}^{7.2^{s}-1}$,
$b_{s,\,97}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{2^{s}-2}x_{4}^{7.2^{s}-1}$, |  $b_{s,\,98}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{7.2^{s}-1}x_{4}^{2^{s}-2}$,
$b_{s,\,99}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{2^{s}-2}x_{4}^{7.2^{s}-1}$, |  $b_{s,\,100}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{7.2^{s}-1}x_{4}^{2^{s}-2}$,
$b_{s,\,101}=x_{1}^{2^{s+1}-1}x_{2}^{7.2^{s}-1}x_{3}x_{4}^{2^{s}-2}$, |  $b_{s,\,102}=x_{1}^{2^{s}-1}x_{2}x_{3}^{2^{s+1}-2}x_{4}^{7.2^{s}-1}$,
$b_{s,\,103}=x_{1}x_{2}^{2^{s}-1}x_{3}^{2^{s+1}-1}x_{4}^{7.2^{s}-2}$, |  $b_{s,\,104}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{2^{s}-1}x_{4}^{7.2^{s}-2}$,
$b_{s,\,105}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{7.2^{s}-2}x_{4}^{2^{s}-1}$, |  $b_{s,\,106}=x_{1}^{2^{s}-1}x_{2}x_{3}^{2^{s+1}-1}x_{4}^{7.2^{s}-2}$,
$b_{s,\,107}=x_{1}^{2^{s}-1}x_{2}^{2^{s+1}-1}x_{3}x_{4}^{7.2^{s}-2}$, |  $b_{s,\,108}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{2^{s}-1}x_{4}^{7.2^{s}-2}$,
$b_{s,\,109}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{7.2^{s}-2}x_{4}^{2^{s}-1}$, |  $b_{s,\,110}=x_{1}^{2^{s+1}-1}x_{2}^{2^{s}-1}x_{3}x_{4}^{7.2^{s}-2}$,
$b_{s,\,111}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{3.2^{s}-2}x_{4}^{5.2^{s}-1}$, |  $b_{s,\,112}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{3.2^{s}-2}x_{4}^{5.2^{s}-1}$,
$b_{s,\,113}=x_{1}x_{2}^{2^{s+1}-1}x_{3}^{3.2^{s}-1}x_{4}^{5.2^{s}-2}$, |  $b_{s,\,114}=x_{1}^{2^{s+1}-1}x_{2}x_{3}^{3.2^{s}-1}x_{4}^{5.2^{s}-2}$,
$b_{s,\,115}=x_{1}^{2^{s+1}-1}x_{2}^{3.2^{s}-1}x_{3}x_{4}^{5.2^{s}-2}$, |  $b_{s,\,116}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{2^{s}-2}x_{4}^{7.2^{s}-1}$,
$b_{s,\,117}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{7.2^{s}-1}x_{4}^{2^{s}-2}$, |  $b_{s,\,118}=x_{1}^{3}x_{2}^{2^{s+1}-1}x_{3}^{7.2^{s}-3}x_{4}^{2^{s}-2}$,
$b_{s,\,119}=x_{1}^{2^{s+1}-1}x_{2}^{3}x_{3}^{7.2^{s}-3}x_{4}^{2^{s}-2}$, |  $b_{s,\,120}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{3.2^{s}-2}x_{4}^{5.2^{s}-1}$,
$b_{s,\,121}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{3.2^{s}-1}x_{4}^{5.2^{s}-2}$, |  $b_{s,\,122}=x_{1}^{3}x_{2}^{2^{s+1}-1}x_{3}^{3.2^{s}-3}x_{4}^{5.2^{s}-2}$,
$b_{s,\,123}=x_{1}^{2^{s+1}-1}x_{2}^{3}x_{3}^{3.2^{s}-3}x_{4}^{5.2^{s}-2}$, |  $b_{s,\,124}=x_{1}^{3}x_{2}^{2^{s}-1}x_{3}^{2^{s+3}-3}x_{4}^{2^{s}-2}$,
$b_{s,\,125}=x_{1}^{3}x_{2}^{2^{s+3}-3}x_{3}^{2^{s}-2}x_{4}^{2^{s}-1}$, |  $b_{s,\,126}=x_{1}^{3}x_{2}^{2^{s+3}-3}x_{3}^{2^{s}-1}x_{4}^{2^{s}-2}$,
$b_{s,\,127}=x_{1}^{3}x_{2}^{2^{s}-1}x_{3}^{2^{s+1}-3}x_{4}^{7.2^{s}-2}$, |  $b_{s,\,128}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{2^{s}-1}x_{4}^{7.2^{s}-2}$.

For $s=2,$

$b_{2,\,129}=x_{1}^{3}x_{2}^{3}x_{3}^{3}x_{4}^{28}$, |  $b_{2,\,130}=x_{1}^{3}x_{2}^{3}x_{3}^{28}x_{4}^{3}$, |  $b_{2,\,131}=x_{1}^{3}x_{2}^{3}x_{3}^{4}x_{4}^{27}$,
---|---|---
$b_{2,\,132}=x_{1}^{3}x_{2}^{3}x_{3}^{7}x_{4}^{24}$, |  $b_{2,\,133}=x_{1}^{3}x_{2}^{7}x_{3}^{3}x_{4}^{24}$, |  $b_{2,\,134}=x_{1}^{7}x_{2}^{3}x_{3}^{3}x_{4}^{24}$,
$b_{2,\,135}=x_{1}^{7}x_{2}^{7}x_{3}^{9}x_{4}^{14}$. |  |

For $s\geq 3,$

$b_{s,\,129}=x_{1}^{3}x_{2}^{2^{s+1}-3}x_{3}^{7.2^{s}-2}x_{4}^{2^{s}-1}$, |  $b_{s,\,130}=x_{1}^{2^{s}-1}x_{2}^{3}x_{3}^{2^{s+1}-3}x_{4}^{7.2^{s}-2}$,
---|---
$b_{s,\,131}=x_{1}^{2^{s}-1}x_{2}^{3}x_{3}^{2^{s+3}-3}x_{4}^{2^{s}-2}$, |  $b_{s,\,132}=x_{1}^{3}x_{2}^{2^{s}-3}x_{3}^{2^{s}-2}x_{4}^{2^{s+3}-1}$,
$b_{s,\,133}=x_{1}^{3}x_{2}^{2^{s}-3}x_{3}^{2^{s+3}-1}x_{4}^{2^{s}-2}$, |  $b_{s,\,134}=x_{1}^{3}x_{2}^{2^{s+3}-1}x_{3}^{2^{s}-3}x_{4}^{2^{s}-2}$,
$b_{s,\,135}=x_{1}^{2^{s+3}-1}x_{2}^{3}x_{3}^{2^{s}-3}x_{4}^{2^{s}-2}$, |  $b_{s,\,136}=x_{1}^{3}x_{2}^{2^{s}-3}x_{3}^{2^{s}-1}x_{4}^{2^{s+3}-2}$,
$b_{s,\,137}=x_{1}^{3}x_{2}^{2^{s}-3}x_{3}^{2^{s+3}-2}x_{4}^{2^{s}-1}$, |  $b_{s,\,138}=x_{1}^{3}x_{2}^{2^{s}-1}x_{3}^{2^{s}-3}x_{4}^{2^{s+3}-2}$,
$b_{s,\,139}=x_{1}^{2^{s}-1}x_{2}^{3}x_{3}^{2^{s}-3}x_{4}^{2^{s+3}-2}$, |  $b_{s,\,140}=x_{1}^{3}x_{2}^{2^{s}-3}x_{3}^{2^{s+1}-2}x_{4}^{7.2^{s}-1}$,
$b_{s,\,141}=x_{1}^{3}x_{2}^{2^{s}-3}x_{3}^{2^{s+1}-1}x_{4}^{7.2^{s}-2}$, |  $b_{s,\,142}=x_{1}^{3}x_{2}^{2^{s+1}-1}x_{3}^{2^{s}-3}x_{4}^{7.2^{s}-2}$,
$b_{s,\,143}=x_{1}^{2^{s+1}-1}x_{2}^{3}x_{3}^{2^{s}-3}x_{4}^{7.2^{s}-2}$, |  $b_{s,\,144}=x_{1}^{7}x_{2}^{2^{s+3}-5}x_{3}^{2^{s}-3}x_{4}^{2^{s}-2}$,
$b_{s,\,145}=x_{1}^{7}x_{2}^{2^{s+1}-5}x_{3}^{2^{s}-3}x_{4}^{7.2^{s}-2}$, |  $b_{s,\,146}=x_{1}^{7}x_{2}^{2^{s+1}-5}x_{3}^{3.2^{s}-3}x_{4}^{5.2^{s}-2}$,
$b_{s,\,147}=x_{1}^{7}x_{2}^{2^{s+1}-5}x_{3}^{7.2^{s}-3}x_{4}^{2^{s}-2}$. |

For $s=3,$

$b_{3,\,148}=x_{1}^{7}x_{2}^{7}x_{3}^{7}x_{4}^{56}$, |  $b_{3,\,149}=x_{1}^{7}x_{2}^{7}x_{3}^{9}x_{4}^{54}$, |  $b_{3,\,150}=x_{1}^{7}x_{2}^{7}x_{3}^{57}x_{4}^{6}$.
---|---|---

For $s\geq 4,$

$b_{s,\,148}=x_{1}^{7}x_{2}^{2^{s}-5}x_{3}^{2^{s}-3}x_{4}^{2^{s+3}-2}$, |  $b_{s,\,149}=x_{1}^{7}x_{2}^{2^{s}-5}x_{3}^{2^{s+1}-3}x_{4}^{7.2^{s}-2}$, |  $b_{s,\,150}=x_{1}^{7}x_{2}^{2^{s}-5}x_{3}^{2^{s+3}-3}x_{4}^{2^{s}-2}$.
---|---|---

######  Remark 4.2.3.

Based upon the bases of the spaces $Q^{\otimes 4}_{n_{s}}$ above,
$\omega(b_{s,\,j})$ is one of the following sequences:

$\omega_{(s,1)}:=\underset{\mbox{{$(s+1)$ times of
$3$}}}{\underbrace{(3,3,\ldots,3}},2),\ \ \omega_{(s,2)}:=\underset{\mbox{{$s$
times of $3$}}}{\underbrace{(3,3,\ldots,3}},1,1,1).$

Moreover, since $\omega_{(s,2)}$ is the weight vector of the minimal spike
$b_{s,\,15},$ $[x]_{\omega_{(s,2)}}=[x]$ for all monomials $x$ in
$(P_{4})_{n_{s}}.$

Case $s=1$. From the admissible basis of $\overline{Q^{\otimes 4}_{n_{1}}}$
above, we have a direct summand decomposition of $\Sigma_{4}$-modules:
$\overline{Q^{\otimes 4}_{n_{1}}}=\mathcal{M}_{1}\bigoplus\mathcal{M}_{2},$
where
$\mathcal{M}_{1}=\langle\\{[b_{1,\,j}]:\,j=41,\ldots,47,77,\ldots,87\\}\rangle$
and $\mathcal{M}_{2}=\langle\\{[b_{1,\,j}]:\,j=48,\ldots,76\\}\rangle.$

######  Lemma 4.2.4.

The following assertions are true:

  * i)

$\mathcal{M}_{1}^{\Sigma_{4}}=\langle[\widetilde{\zeta}]\rangle$ with
$\widetilde{\zeta}=b_{1,\,41}+b_{1,\,42}+b_{1,\,78}+b_{1,\,79}+b_{1,\,80}+b_{1,\,81}+b_{1,\,86}+b_{1,\,87};$

  * ii)

$\mathcal{M}_{2}^{\Sigma_{4}}=0.$

###### Proof.

We prove i) in detail. By a similar technique, we also get ii). Suppose that
$f\in(P_{4})_{n_{1}}$ such that $[f]\in\mathcal{M}_{1}^{\Sigma_{4}}.$ Then, we
have

$f\equiv\big{(}\sum_{41\leq j\leq 47}\gamma_{j}b_{1,\,j}+\sum_{77\leq j\leq
87}\gamma_{j}b_{1,\,j}\big{)}.$

Direct calculating $\sigma_{i}(f)$ in terms of
$b_{1,\,j},\,j\in\\{41,\ldots,47,77,\ldots,87\\}$ modulo
($\widehat{A}(P_{4})_{n_{1}}$) and using the relations
$(\sigma_{i}(f)+f)\equiv 0,$ for $i\in\\{1,2,3\\},$ we get

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt(\sigma_{1}(f)+f)&\equiv\big{(}\gamma_{45}b_{1,\,41}+\gamma_{45}b_{1,\,46}+(\gamma_{45}+\gamma_{47})b_{1,\,44}+(\gamma_{46}+\gamma_{47})b_{1,\,77}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+(\gamma_{43}+\gamma_{78}+\gamma_{80})b_{1,\,78}+(\gamma_{43}+\gamma_{79}+\gamma_{81})b_{1,\,79}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+(\gamma_{82}+\gamma_{83})b_{1,\,82}+(\gamma_{84}+\gamma_{85})b_{1,\,84}+\gamma_{43}b_{1,\,86}+\mbox{others
terms}\big{)}\equiv 0,\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt(\sigma_{2}(f)+f)&\equiv\big{(}\gamma_{43}b_{1,\,41}+(\gamma_{42}+\gamma_{43}+\gamma_{87})b_{1,\,42}+(\gamma_{42}+\gamma_{43}+\gamma_{81})b_{1,\,43}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+(\gamma_{44}+\gamma_{45}+\gamma_{83})b_{1,\,44}+(\gamma_{46}+\gamma_{79}+\gamma_{87})b_{1,\,46}+(\gamma_{47}+\gamma_{82})b_{1,\,47}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+(\gamma_{77}+\gamma_{78}+\gamma_{87})b_{1,\,77}+(\gamma_{77}+\gamma_{78}+\gamma_{81})b_{1,\,78}+(\gamma_{46}+\gamma_{79}+\gamma_{81})b_{1,\,79}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\gamma_{83}b_{1,\,80}+(\gamma_{81}+\gamma_{87})b_{1,\,81}+(\gamma_{81}+\gamma_{87})b_{1,\,81}+(\gamma_{85}+\gamma_{86}+\gamma_{87})b_{1,\,85}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+(\gamma_{81}+\gamma_{85}+\gamma_{86})b_{1,\,86}+\mbox{others
terms}\big{)}\equiv 0,\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt(\sigma_{3}(f)+f)&\equiv\big{(}(\gamma_{41}+\gamma_{42})b_{1,\,41}+(\gamma_{44}+\gamma_{47})b_{1,\,44}+(\gamma_{45}+\gamma_{46})b_{1,\,45}+(\gamma_{78}+\gamma_{79})b_{1,\,78}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+(\gamma_{80}+\gamma_{81})b_{1,\,80}+(\gamma_{82}+\gamma_{84})b_{1,\,82}+(\gamma_{83}+\gamma_{85})b_{1,\,83}\\\
&\quad+(\gamma_{86}+\gamma_{87})b_{1,\,86}+\mbox{others terms}\big{)}\equiv
0.\end{array}$

These equalities imply that
$\gamma_{41}=\gamma_{42}=\gamma_{78}=\cdots=\gamma_{81}=\gamma_{86}=\gamma_{87}$
and $\gamma_{j}=0$ with $j\not\in\\{41,42,78,79,80,81,86,87\\}.$ The lemma
follows. ∎

Now , assume that $[h]\in(Q^{\otimes 4}_{n_{1}})^{GL_{4}(k)},$ then since
$\Sigma_{4}\subset GL_{4}(k),$ by Lemmas 4.2.2 and 4.2.4, we have

$h\equiv\big{(}\beta_{1}q_{2,2}+\beta_{2}q_{2,3}+\beta_{3}q_{2,4}+\beta_{4}\widetilde{\zeta}\big{)},$

where $\beta_{i}\in k$ for every $i.$ Computing $\sigma_{4}(h)$ in terms of
$b_{1,\,j},\,1\leq i\leq 87$ modulo ($\widehat{A}(P_{4})_{n_{1}}$) and using
the relation $(\sigma_{4}(f)+f)\equiv 0,$ we obtain

$(\sigma_{4}(f)+f)\equiv\big{(}\beta_{1}b_{1,\,13}+(\beta_{1}+\beta_{2})b_{1,\,20}+(\beta_{2}+\beta_{3})b_{1,\,31}+\mbox{others
terms}\big{)}\equiv 0.$

This leads to $\beta_{1}=\beta_{2}=\beta_{3}=0.$

Case $s\geq 2.$ It is sufficient to show that the spaces $((Q^{\otimes
4}_{n_{s}})^{\omega_{(s,1)}})^{GL_{4}(k)}$ and $((Q^{\otimes
4}_{n_{s}})^{\omega_{(s,2)}})^{GL_{4}(k)}$ are trivial for every $s\geq 2.$
Indeed, we prove this statement for the invariant $((Q^{\otimes
4}_{n_{s}})^{\omega_{(s,1)}})^{GL_{4}(k)}$ with $s\geq 4.$ The other cases can
be proved by similar computations.

We have $(Q^{\otimes 4}_{n_{s}})^{\omega_{(s,1)}}\cong\underline{(Q^{\otimes
4}_{n_{s}})}^{\omega_{(s,1)}}\bigoplus\overline{(Q^{\otimes
4}_{n_{s}})}^{\omega_{(s,1)}},$ where

$\underline{(Q^{\otimes
4}_{n_{s}})}^{\omega_{(s,1)}}=\langle\\{[b_{s,j}]_{\omega_{(s,1)}}:\,1\leq
j\leq 12\\}\rangle,\ \mbox{and}\ \overline{(Q^{\otimes
4}_{n_{s}})}^{\omega_{(s,1)}}=\langle\\{[b_{s,j}]_{\omega_{(s,1)}}:\,48\leq
j\leq 80\\}\rangle.$

Then, it is easy to check that

$\overline{(Q^{\otimes
4}_{n_{s}})}^{\omega_{(s,1)}}\cong\Sigma_{4}(b_{s,\,48})\bigoplus\Sigma_{4}(b_{s,\,54},b_{s,\,66})\bigoplus\mathcal{M},$

where

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt\Sigma_{4}(b_{s,\,48})&=\langle\\{[b_{s,\,j}]_{\omega_{(s,1)}}:\,j=48,\ldots,53\\}\rangle,\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\Sigma_{4}(b_{s,\,54},b_{s,\,66})&=\langle\\{[b_{s,\,j}]_{\omega_{(s,1)}}:\,j=54,\ldots,69,74,75,76,77\\}\rangle,\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\mathcal{M}&=\langle\\{[b_{s,\,j}]_{\omega_{(s,1)}}:\,j=70,71,72,73,78,79,80\\}\rangle.\end{array}$

######  Lemma 4.2.5.

The following hold:

$\Sigma_{4}(b_{s,\,48})^{\Sigma_{4}}=\langle[\widehat{q_{s,\,2}}]_{\omega_{(s,1)}}\rangle$
with $\widehat{q_{s,\,2}}:=\sum_{48\leq j\leq 53}b_{s,\,j}.$

$\Sigma_{4}(b_{s,\,54},b_{s,\,66})^{\Sigma_{4}}=\langle[\widehat{q_{s,\,3}}]_{\omega_{(s,1)}}\rangle$
with $\widehat{q_{s,\,3}}:=\sum_{54\leq j\leq 69}b_{s,\,j}+\sum_{74\leq j\leq
77}b_{s,\,j}.$

$\mathcal{M}^{\Sigma_{4}}=\langle[\widehat{q_{s,\,4}}]_{\omega_{(s,1)}},[\widehat{q_{s,\,5}}]_{\omega_{(s,1)}}\rangle$
with

$\widehat{q_{s,\,4}}:=\sum_{70\leq j\leq 73}b_{s,\,j}+b_{s,\,78}+b_{s,\,79},\
\ \widehat{q_{s,\,5}}:=\sum_{70\leq j\leq 73}b_{s,\,j}+b_{s,\,80}.$

###### Proof.

We prove Part iii) in detail. The others parts can be computed by a similar
idea. Computing the general case is very technical. Indeed, we have the set
$\\{[b_{s,\,j}]_{\omega_{(s,1)}}:\,j=70,71,72,73,78,79,80\\},$ which is an
admissible basis of $\mathcal{M}.$ Suppose that
$[f]\in\mathcal{M}^{\Sigma_{4}},$ then

$f\equiv_{\omega_{(s,1)}}\big{(}\sum_{70\leq j\leq
73}\gamma_{j}b_{s,\,j}+\sum_{78\leq j\leq 80}\gamma_{j}b_{s,\,j}\big{)},$

with $\gamma_{j}\in k$ for every $j.$ Using the homomorphisms
$\sigma_{i}:(P_{4})_{n_{s}}\to(P_{4})_{n_{s}},\,i=1,2,3,$ we have

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt\sigma_{1}(f)&\equiv_{\omega_{(s,1)}}\big{(}\gamma_{73}b_{s,\,70}+\gamma_{70}b_{s,\,73}+\gamma_{71}x_{1}^{2^{s+2}-3}x_{2}^{3}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+2}-2}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\gamma_{72}x_{1}^{2^{s+2}-3}x_{2}^{3}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+1}-1}+\gamma_{78}x_{1}^{2^{s+2}-5}x_{2}^{7}x_{3}^{2^{s+1}-3}x_{4}^{2^{s+2}-2}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\gamma_{79}x_{1}^{2^{s+2}-5}x_{2}^{7}x_{3}^{2^{s+2}-3}x_{4}^{2^{s+1}-2}+\gamma_{80}x_{1}^{2^{s+1}-5}x_{2}^{7}x_{3}^{2^{s+2}-3}x_{4}^{2^{s+2}-2}\big{)},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\sigma_{2}(f)&\equiv_{\omega_{(s,1)}}\big{(}\gamma_{71}b_{s,\,70}+\gamma_{70}b_{s,\,71}+\gamma_{72}x_{1}^{3}x_{2}^{2^{s+2}-2}x_{3}^{2^{s+2}-3}x_{4}^{2^{s+1}-1}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\gamma_{73}x_{1}^{2^{s+1}-1}x_{2}^{2^{s+2}-3}x_{3}^{3}x_{4}^{2^{s+2}-2}+\gamma_{78}x_{1}^{7}x_{2}^{2^{s+1}-3}x_{3}^{2^{s+2}-5}x_{4}^{2^{s+2}-2}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\gamma_{79}x_{1}^{7}x_{2}^{2^{s+2}-3}x_{3}^{2^{s+2}-5}x_{4}^{2^{s+1}-2}+\gamma_{80}x_{1}^{7}x_{2}^{2^{s+2}-3}x_{3}^{2^{s+1}-5}x_{4}^{2^{s+2}-2}\big{)},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\sigma_{3}(f)&\equiv_{\omega_{(s,1)}}\big{(}\gamma_{72}b_{s,\,71}+\gamma_{71}b_{s,\,72}+\gamma_{70}x_{1}^{3}x_{2}^{2^{s+1}-1}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+2}-3}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\gamma_{73}x_{1}^{2^{s+1}-1}x_{2}^{3}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+2}-3}+\gamma_{78}x_{1}^{7}x_{2}^{2^{s+2}-5}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+1}-3}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+\gamma_{79}x_{1}^{7}x_{2}^{2^{s+2}-5}x_{3}^{2^{s+1}-2}x_{4}^{2^{s+2}-3}+\gamma_{80}x_{1}^{7}x_{2}^{2^{s+1}-5}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+2}-3}\big{)}.\end{array}$

Using the actions of the Steenrod squares $Sq^{i}$ for $i\in\\{1,2,4,8\\}$ and
the Cartan formula, we get

$\begin{array}[]{ll}x_{1}^{3}x_{2}^{2^{s+1}-1}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+2}-3}&\equiv_{\omega_{(s,1)}}b_{s,\,70},\bigg{|}\
x_{1}^{3}x_{2}^{2^{s+2}-2}x_{3}^{2^{s+2}-3}x_{4}^{2^{s+1}-1}\equiv_{\omega_{(s,1)}}b_{s,\,72},\\\
x_{1}^{7}x_{2}^{2^{s+1}-5}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+2}-3}&\equiv_{\omega_{(s,1)}}b_{s,\,80},\bigg{|}\
x_{1}^{7}x_{2}^{2^{s+1}-3}x_{3}^{2^{s+2}-5}x_{4}^{2^{s+2}-2}\equiv_{\omega_{(s,1)}}b_{s,\,80},\\\
x_{1}^{7}x_{2}^{2^{s+2}-5}x_{3}^{2^{s+1}-2}x_{4}^{2^{s+2}-3}&\equiv_{\omega_{(s,1)}}b_{s,\,78},\bigg{|}\
x_{1}^{7}x_{2}^{2^{s+2}-5}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+1}-3}\equiv_{\omega_{(s,1)}}b_{s,\,79},\\\
x_{1}^{7}x_{2}^{2^{s+2}-3}x_{3}^{2^{s+1}-5}x_{4}^{2^{s+2}-2}&\equiv_{\omega_{(s,1)}}b_{s,\,78},\bigg{|}\
x_{1}^{7}x_{2}^{2^{s+2}-3}x_{3}^{2^{s+2}-5}x_{4}^{2^{s+1}-2}\equiv_{\omega_{(s,1)}}b_{s,\,79},\\\
x_{1}^{2^{s+1}-5}x_{2}^{7}x_{3}^{2^{s+2}-3}x_{4}^{2^{s+2}-2}&\equiv_{\omega_{(s,1)}}b_{s,\,80},\bigg{|}\
x_{1}^{2^{s+1}-1}x_{2}^{3}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+2}-3}\equiv_{\omega_{(s,1)}}b_{s,\,73},\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}^{2^{s+1}-1}x_{2}^{2^{s+2}-3}x_{3}^{3}x_{4}^{2^{s+2}-2}&\equiv_{\omega_{(s,1)}}\big{(}b_{s,\,73}+b_{s,\,78}+b_{s,\,80}\big{)},\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}^{2^{s+2}-5}x_{2}^{7}x_{3}^{2^{s+1}-3}x_{4}^{2^{s+2}-2}&\equiv_{\omega_{(s,1)}}\big{(}b_{s,\,78}+b_{s,\,80}\big{)},\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}^{2^{s+2}-5}x_{2}^{7}x_{3}^{2^{s+2}-3}x_{4}^{2^{s+1}-2}&\equiv_{\omega_{(s,1)}}\big{(}b_{s,\,79}+b_{s,\,80}\big{)},\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}^{2^{s+2}-3}x_{2}^{3}x_{3}^{2^{s+1}-1}x_{4}^{2^{s+2}-2}&\equiv_{\omega_{(s,1)}}\big{(}b_{s,\,71}+b_{s,\,80}\big{)},\\\
\vskip 6.0pt plus 2.0pt minus
2.0ptx_{1}^{2^{s+2}-3}x_{2}^{3}x_{3}^{2^{s+2}-2}x_{4}^{2^{s+1}-1}&\equiv_{\omega_{(s,1)}}\big{(}b_{s,\,72}+b_{s,\,80}\big{)}.\end{array}$

Combining the above computations and the relations
$\sigma_{i}(f)\equiv_{\omega_{(s,\,1)}}f$ for $1\leq i\leq 3,$ we obtain
$\gamma_{78}=\gamma_{79}$ and $\gamma_{j}+\gamma_{79}+\gamma_{80}=0,$ for
$70\leq j\leq 73.$ The lemma is proved. ∎

Now, let $[g]\in(Q^{\otimes 4}_{n_{s}})^{\omega_{(s,1)}})^{GL_{4}(k)},$ by
Lemmas 4.2.2 and 4.2.5, we have

$g\equiv_{\omega_{(s,1)}}\big{(}\rho_{1}q_{s,\,1}+\rho_{2}\widehat{q_{s,\,2}}+\rho_{3}\widehat{q_{s,\,3}}+\rho_{4}\widehat{q_{s,\,4}}+\rho_{5}\widehat{q_{s,\,5}}\big{)},$

with $\rho_{i}\in k$ for all $i,\,1\leq i\leq 5.$ Direct computing from the
relation $(\sigma_{4}(g)+g)\equiv_{\omega_{(s,1)}}0,$ we get

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt(\sigma_{4}(g)+g)&\equiv_{\omega_{(s,1)}}\big{(}(\rho_{1}+\rho_{2})b_{s,\,1}+(\rho_{1}+\rho_{3})b_{s,\,2}+\rho_{1}b_{s,\,7}+(\rho_{4}+\rho_{5})b_{s,\,58}\\\
&\quad\quad\quad+\rho_{4}b_{s,\,76}+\mbox{others
terms}\big{)}\equiv_{\omega_{(s,1)}}0,\end{array}$

and therefore $\rho_{i}=0$ for all $i.$ This completes the proof of the
proposition. ∎

Now, we consider the element $\zeta\in(P_{4})^{*}_{n_{1}}$, which is the
following sum:
$\begin{array}[]{ll}&a_{1}^{(5)}a_{2}^{(5)}a_{3}^{(5)}a_{4}^{(2)}+a_{1}^{(5)}a_{2}^{(5)}a_{3}^{(6)}a_{4}^{(1)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(8)}a_{4}^{(1)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(8)}a_{4}^{(1)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(3)}a_{2}^{(6)}a_{3}^{(7)}a_{4}^{(1)}\\\
&+a_{1}^{(5)}a_{2}^{(7)}a_{3}^{(4)}a_{4}^{(1)}+a_{1}^{(7)}a_{2}^{(5)}a_{3}^{(4)}a_{4}^{(1)}+a_{1}^{(3)}a_{2}^{(9)}a_{3}^{(4)}a_{4}^{(1)}+a_{1}^{(9)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(1)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(3)}a_{2}^{(9)}a_{3}^{(3)}a_{4}^{(2)}\\\
&+a_{1}^{(9)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(2)}+a_{1}^{(5)}a_{2}^{(9)}a_{3}^{(2)}a_{4}^{(1)}+a_{1}^{(9)}a_{2}^{(5)}a_{3}^{(2)}a_{4}^{(1)}+a_{1}^{(5)}a_{2}^{(10)}a_{3}^{(1)}a_{4}^{(1)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(9)}a_{2}^{(6)}a_{3}^{(1)}a_{4}^{(1)}\\\
&+a_{1}^{(3)}a_{2}^{(11)}a_{3}^{(2)}a_{4}^{(1)}+a_{1}^{(11)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(1)}+a_{1}^{(5)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(4)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(4)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(3)}a_{2}^{(5)}a_{3}^{(5)}a_{4}^{(4)}\\\
&+a_{1}^{(3)}a_{2}^{(12)}a_{3}^{(1)}a_{4}^{(1)}+a_{1}^{(11)}a_{2}^{(4)}a_{3}^{(1)}a_{4}^{(1)}+a_{1}^{(7)}a_{2}^{(8)}a_{3}^{(1)}a_{4}^{(1)}+a_{1}^{(7)}a_{2}^{(7)}a_{3}^{(1)}a_{4}^{(2)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(13)}a_{2}^{(2)}a_{3}^{(1)}a_{4}^{(1)}\\\
&+a_{1}^{(14)}a_{2}^{(1)}a_{3}^{(1)}a_{4}^{(1)}+a_{1}^{(6)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(3)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(3)}+a_{1}^{(3)}a_{2}^{(6)}a_{3}^{(5)}a_{4}^{(3)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(6)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(5)}\\\
&+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(5)}+a_{1}^{(3)}a_{2}^{(6)}a_{3}^{(3)}a_{4}^{(5)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(6)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(6)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(6)}\\\
&+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(8)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(7)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(2)}a_{4}^{(7)}+a_{1}^{(3)}a_{2}^{(6)}a_{3}^{(1)}a_{4}^{(7)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(9)}a_{4}^{(2)}\\\
&+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(10)}a_{4}^{(1)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(7)}a_{4}^{(2)}+a_{1}^{(5)}a_{2}^{(7)}a_{3}^{(3)}a_{4}^{(2)}+a_{1}^{(7)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(2)}.\end{array}$
Then, $\zeta$ is $\widehat{A}$-annihilated (or $\zeta\in
P((P_{4})^{*}_{n_{1}})$). Indeed, notice that by the unstable condition, we
need only to compute the (right) action of $Sq^{i}$ for $i=1,2,4.$ It is easy
to check that $(\zeta)Sq^{i}=0$ for $i=1,2.$ A direct calculation shows that
$(\zeta)Sq^{4}$ is the following sum:
$\begin{array}[]{ll}&a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(2)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(2)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(2)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(1)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(3)}a_{2}^{(5)}a_{3}^{(4)}a_{4}^{(1)}\\\
&+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(1)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(1)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(1)}+a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(2)}a_{4}^{(1)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(7)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(1)}\\\
&+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(4)}a_{4}^{(1)}+a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(2)}a_{4}^{(1)}+a_{1}^{(5)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(1)}+a_{1}^{(7)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(1)}+\vskip
6.0pt plus 2.0pt minus 2.0pta_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(2)}\\\