# Categorification of based modules over the complex representation ring of
$S_{4}$

Wenxia Wu  School of Mathematics and Information Science, Guangzhou
University, Waihuan Road West 230, Guangzhou 510006, China<EMAIL_ADDRESS>and  Yunnan Li  School of Mathematics and Information Science, Guangzhou
University, Waihuan Road West 230, Guangzhou 510006, China<EMAIL_ADDRESS>
###### Abstract.

The complex representation rings of finite groups are the fundamental class of
fusion rings, categorified by the corresponding fusion categories of complex
representations. The category of $\mathbb{Z}_{+}$-modules of finite rank over
such a representation ring is also semisimple.

In this paper, we classify the irreducible based modules of rank up to 5 over
the complex representation ring $r(S_{4})$ of the symmetric group $S_{4}$.
Totally 16 inequivalent irreducible based modules are obtained. Based on such
a classification result, we further discuss the categorification of based
modules over $r(S_{4})$ by module categories over the complex representation
category ${\rm Rep}(S_{4})$ of $S_{4}$ arisen from projective representations
of certain subgroups of $S_{4}$.

###### Key words and phrases:

$\mathbb{Z}_{+}$-module, $\mathbb{Z}_{+}$-ring, representation ring, module
category, symmetric group

2020 Mathematics Subject Classification. 13C05, 18M20, 19A22.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 $\mathbb{Z}_{+}$-rings and $\mathbb{Z}_{+}$-modules
    2. 2.2 Based rings and based modules
  3. 3 Irreducible based modules over $r(S_{4})$
    1. 3.1 Irreducible based modules of rank $\leq 3$ over $r(S_{4})$
    2. 3.2 Irreducible based modules of rank $4,\,5$ over $r(S_{4})$
  4. 4 Categorified based modules by module categories over ${\rm Rep}(S_{4})$
    1. 4.1 The module categories over ${\rm Rep}(S_{4})$ with trivial twists
    2. 4.2 The module categories over ${\rm Rep}(S_{4})$ with nontrivial twists

##  1\. Introduction

Tensor categories should be thought as counterparts of rings in the world of
categories, i.e., the categorification of groups and rings [1, 10, 11, 16, 17,
23]. They are ubiquitous in noncommutative algebra and representation theory.
Tensor categories were introduced by Bénabou [2] in 1963 and Mac Lane in [24]
as “categories with multiplication”, and its related theories are now widely
used in many fields of mathematics, including algebraic geometry, algebraic
topology, number theory, operator algebraic theory, etc. [3, 13, 36, 5]. The
theory of tensor categories is also seen as a sequel developed from that of
Hopf algebras and their representation theory [4, 15]. As an important
invariant in the theory of tensor categories, the concept of
$\mathbb{Z}_{+}$-ring can be traced back to Lusztig’s work [30] in 1987. Later
in [12, 32], the notion of $\mathbb{Z}_{+}$-module over $\mathbb{Z}_{+}$-ring
was introduced. Module categories over multitensor categories were first
considered in [16, 18], and then the notion of indecomposable module category
was introduced in [32]. As a categorification of irreducible
$\mathbb{Z}_{+}$-modules, it is interesting to classify indecomposable exact
module categories over a given tensor category. In this process, it is often
necessary to first classify all irreducible $\mathbb{Z}_{+}$-modules over the
Grothendieck ring of a given tensor category.

Typical examples of $\mathbb{Z}_{+}$-rings are the representation rings of
Hopf algebras [8, 21, 29, 39, 40, 41] and the Grothendieck rings of tensor
categories [6, 7, 31, 33]. It is natural to consider the classification of all
irreducible $\mathbb{Z}_{+}$-modules over them. For example, Etingof and
Khovanov classified irreducible $\mathbb{Z}_{+}$-modules over the group ring
$\mathbb{Z}G$, and showed that indecomposable $\mathbb{Z}_{+}$-modules over
the representation ring of ${\rm SU}(2)$, under certain conditions, correspond
to affine and infinite Dynkin diagrams [12]. Also, there appears a lot of
related research in the context of near-group fusion categories. For instance,
Tambara and Yamagami classified semisimple tensor categories with fusion rules
of self-duality for finite abelian groups. Evans, Gannon and Izumi have
contributed to the classification of the near-group $C^{*}$-categories [19,
22]. Li, Yuan and Zhao in [42] studied irreducible $\mathbb{Z}_{+}$-modules of
near-group fusion ring $K\left(\mathbb{Z}_{3},3\right)$ and so on.

In this paper, we will explore the problem of classifying irreducible based
modules of rank up to 5 over the complex representation ring $r(S_{4})$, and
then discuss their categorification. In contrast with the representation ring
of $S_{3}$, $r(S_{n})$ is no longer a near-group fusion ring when $n>3$, and
the classification of irreducible $\mathbb{Z}_{+}$-modules over general
$r(S_{n})$ seems to be a hopeless task. In fact, the fusion rule of $r(S_{n})$
is already a highly nontrivial open problem in combinatorics, namely counting
the multiplicities of irreducible components of the tensor product of any two
irreducible complex representations of $S_{n}$ (so called the Kronecker
coefficients).

The paper is organized as follows. In Section 2, we recall some basic
definitions and propositions. In Section 3, we discuss the irreducible based
modules of rank up to 5 over $r(S_{4})$ and give the classification of all
these based modules (Propositions 3.1–3.5). In Section 4, we first show that
any $\mathbb{Z}_{+}$-module over the representation ring $r(G)$ of a finite
group $G$ categorified by a module category over the representation category
${\rm Rep}(G)$ should be a based module (Theorem 4.3), and then determine
which irreducible based modules over $r(S_{4})$ can be categorified then
(Theorem 4.18).

##  2\. Preliminaries

Throughout the paper, all rings are assumed to be associative with unit. Let
$\mathbb{Z}_{+}$ denote the set of nonnegative integers. First we recall the
definitions of $\mathbb{Z}_{+}$-rings and $\mathbb{Z}_{+}$-modules. For more
details about these concepts, readers can refer to [14, 32].

###  2.1. $\mathbb{Z}_{+}$-rings and $\mathbb{Z}_{+}$-modules

######  Definition 2.1.

Let $A$ be a ring which is free as a $\mathbb{Z}$-module.

  1. (i)

A $\mathbb{Z}_{+}$-basis of $A$ is a basis $B=\left\\{b_{i}\right\\}_{i\in I}$
such that $b_{i}b_{j}=\sum_{k\in I}c_{ij}^{k}b_{k}$, where
$c_{ij}^{k}\in\mathbb{Z}_{+}$.

  2. (ii)

A $\mathbb{Z}_{+}$-ring is a ring with a fixed $\mathbb{Z}_{+}$-basis and with
unit 1 being a non-negative linear combination of the basis elments.

  3. (iii)

A $\mathbb{Z}_{+}$-ring is unital if the unit $1$ is one of its basis
elements.

######  Definition 2.2.

Let $A$ be a $\mathbb{Z}_{+}$-ring with basis $\left\\{b_{i}\right\\}_{i\in
I}$. A $\mathbb{Z}_{+}$-module over $A$ is an $A$-module $M$ with a fixed
$\mathbb{Z}$-basis $\left\\{m_{l}\right\\}_{l\in L}$ such that all the
structure constants $a_{il}^{k}$ (defined by the equality
$b_{i}m_{l}=\sum_{k}a_{il}^{k}m_{k}$) are non-negative integers.

A $\mathbb{Z}_{+}$-module has the following equivalent definition referring to
[14, Section 3.4].

######  Definition 2.3.

Let $A$ be a $\mathbb{Z}_{+}$-ring with basis $\left\\{b_{i}\right\\}_{i\in
I}$. A $\mathbb{Z}_{+}$-module $M$ over $A$ means an assignment where each
basis $b_{i}$ in $A$ is in one-to-one correspondence with a non-negative
integer square matrix $M_{i}$ such that $M$ forms a representation of $A$:
$M_{i}M_{j}=\sum_{k\in I}c_{ij}^{k}M_{k},\forall i,j,k\in I$, and also the
unit of $A$ corresponds to the identity matrix. The rank of a
$\mathbb{Z}_{+}$-module $M$ is equal to the order of the matrix $M_{i}$.

######  Definition 2.4.

(i) Two $\mathbb{Z}_{+}$-modules $M_{1},\,M_{2}$ over $A$ with bases
${\left\\{m_{i}^{1}\right\\}}_{i\in L_{1}},{\left\\{m_{j}^{2}\right\\}}_{j\in
L_{2}}$ are equivalent if and only if there exists a bijection $\phi:L_{1}\to
L_{2}$ such that the induced $\mathbb{Z}$-linear map $\tilde{\phi}$ of abelian
groups $M_{1},\,M_{2}$ defined by $\tilde{\phi}(m_{i}^{1})=m_{{\phi}(i)}^{2}$
is an isomorphism of $A$-modules. In other words, for $a\in A$, let
$a_{M_{1}}$ and $a_{M_{2}}$ be the matrices respect to the bases
${\left\\{m_{i}^{1}\right\\}}_{i\in L_{1}}$ and
${\left\\{m_{j}^{2}\right\\}}_{j\in L_{2}}$, respectively. Then two
$\mathbb{Z}_{+}$-modules $M_{1}$, $M_{2}$ of rank $n$ are equivalent if and
only if there exists an $n\times n$ permutation matrix $P$ such that
$a_{M_{2}}=Pa_{M_{1}}P^{-1}$,$\forall a\in A$.

  1. (ii)

The direct sum of two $\mathbb{Z}_{+}$-modules $M_{1},\,M_{2}$ over $A$ is the
module $M_{1}\oplus M_{2}$ over $A$ whose basis is the union of the bases of
$M_{1}$ and $M_{2}$.

  2. (iii)

A $\mathbb{Z}_{+}$-module $M$ over $A$ is indecomposable if it is not
equivalent to a nontrivial direct sum of $\mathbb{Z}_{+}$-modules.

  3. (iv)

A $\mathbb{Z}_{+}$-submodule of a $\mathbb{Z}_{+}$-module $M$ over $A$ with
basis $\left\\{m_{l}\right\\}_{l\in L}$ is a subset $J\subset L$ such that
abelian subgroup of $M$ generated by ${\left\\{m_{j}\right\\}}_{j\in J}$ is an
$A$-submodule.

  4. (v)

A $\mathbb{Z}_{+}$-module $M$ over $A$ is irreducible if any
$\mathbb{Z}_{+}$-submodule of $M$ is 0 or $M$. In other words, the
$\mathbb{Z}$-span of any proper subset of the basis of $M$ is not an
$A$-submodule.

###  2.2. Based rings and based modules

Let $A$ be a $\mathbb{Z}_{+}$-ring with basis $\left\\{b_{i}\right\\}_{i\in
I}$, and let $I_{0}$ be the set of $i\in I$ such that $b_{i}$ occurs in the
decomposition of 1. Let $\tau:A\to\mathbb{Z}$ denote the group homomorphism
defined by

$\tau(b_{i})=\begin{cases}1&\text{ if }\ i\in I_{0},\\\ 0&\text{ if }\ i\notin
I_{0}.\end{cases}$

######  Definition 2.5.

A $\mathbb{Z}_{+}$-ring with basis $\left\\{b_{i}\right\\}_{i\in I}$ is called
a based ring if there exists an involution $i\mapsto i^{*}$ of $I$ such that
the induced map

$a=\sum_{i\in I}a_{i}b_{i}\mapsto{}a^{*}=\sum_{i\in
I}a_{i}b_{i^{*}},a_{i}\in\mathbb{Z},$

is an anti-involution of the ring $A$, and

$\tau(b_{i}b_{j})=\begin{cases}1&\text{ if }\ i=j^{*},\\\ 0&\text{ if }\ i\neq
j^{*}.\end{cases}$

A fusion ring is a unital based ring of finite rank.

######  Definition 2.6.

A based module over a based ring $A$ with basis $\left\\{b_{i}\right\\}_{i\in
I}$ is a $\mathbb{Z}_{+}$-module $M$ with basis $\left\\{m_{l}\right\\}_{l\in
L}$ over $A$ such that $a_{il}^{k}=a_{i^{\ast}k}^{l}$, where $a_{il}^{k}$ are
defined as in Definition 2.2.

Let $A$ be a unital $\mathbb{Z}_{+}$-ring of finite rank with basis
$\left\\{b_{i}\right\\}_{i\in I}$ and let $M$ be a $\mathbb{Z}_{+}$-module
over $A$ with $\mathbb{Z}$-basis $\left\\{m_{l}\right\\}_{l\in L}$. Take
$b=\sum_{i\in I}b_{i}$. For any fixed $m_{l_{0}}$, the
$\mathbb{Z}_{+}$-submodule of $M$ generated by $m_{l_{0}}$ is the
$\mathbb{Z}$-span of $\left\\{m_{k}\right\\}_{k\in Y}$, where the set $Y$
consists of $k\in L$ such that $m_{k}$ is a summand of $bm_{l_{0}}$. Also, we
need the following facts.

######  Proposition 2.7 ([14, Prop. 3.4.6]).

Let $A$ be a based ring of finite rank over $\mathbb{Z}$. Then there exist
only finitely many irreducible $\mathbb{Z}_{+}$-module over $A$.

######  Proposition 2.8 ([14, Exercise 3.4.3. (i)]).

A $\mathbb{Z}_{+}$-module over a based ring is irreducible if and only if it
is indecomposable.

As a result, any $\mathbb{Z}_{+}$-module of finite rank over a fusion ring is
completely reducible, and then only irreducible $\mathbb{Z}_{+}$-modules need
to be classified.

In general, the rank of an irreducible $\mathbb{Z}_{+}$-module over a fusion
ring $A$ may be larger than the rank of $A$; e.g. $A=r(D_{5})$ for the
dihedral group $D_{5}$ ([25, Remark 1]). In this paper, we explore which
irreducible based modules over $r(S_{4})$ can be categorified by
indecomposable exact module categories over the representation category ${\rm
Rep}(S_{4})$. Since all these module categories are of rank not greater than
5, we only deal with based modules of rank up to 5 correspondingly.

##  3\. Irreducible based modules over $r(S_{4})$

In this section, we will classify the irreducible based modules over the
complex representation ring $r(S_{4})$ of $S_{4}$ up to equivalence.
$r(S_{4})$ is a commutative fusion ring having a $\mathbb{Z}_{+}$-basis
$\left\\{1,V_{\psi},V_{\rho_{1}},V_{\rho_{2}},V_{\rho_{3}}\right\\}$ with the
following fusion rule.

(1) $\begin{array}[]{l}V_{\psi}^{2}=1,\quad
V_{\psi}V_{\rho_{1}}=V_{\rho_{1}},\quad
V_{\psi}V_{\rho_{2}}=V_{\rho_{3}},\quad
V_{\rho_{1}}^{2}=1+V_{\psi}+V_{\rho_{1}},\\\\[5.0pt]
V_{\rho_{1}}V_{\rho_{2}}=V_{\rho_{2}}+V_{\rho_{3}},\quad
V_{\rho_{2}}^{2}=1+V_{\rho_{1}}+V_{\rho_{2}}+V_{\rho_{3}},\end{array}$

where $1$, $V_{\psi}$, $V_{\rho_{1}}$ denote the trivial representation, sign
representation and 2-dimensional irreducible representation respectively,
while $V_{\rho_{2}}$ stands for the 3-dimensional standard representation and
$V_{\rho_{3}}$ denotes its conjugate representation, then we have

|  $(1)$ |  $(12)$ |  $(123)$ |  $(1234)$ |  $(12)(34)$
---|---|---|---|---|---
$\chi_{1}$ |  $1$ |  $1$ |  $1$ |  $1$ |  $1$
$\chi_{\psi}$ |  $1$ |  $-1$ |  $1$ |  $-1$ |  $1$
$\chi_{\rho_{1}}$ |  $2$ |  $0$ |  $-1$ |  $0$ |  $2$
$\chi_{\rho_{2}}$ |  $3$ |  $1$ |  $0$ |  $-1$ |  $-1$
$\chi_{\rho_{3}}$ |  $3$ |  $-1$ |  $0$ |  $1$ |  $-1$

Table 1. The complex character table of $S_{4}$

Let $M$ be a based module of $r(S_{4})$ with the basis
$\left\\{m_{l}\right\\}_{l\in L}$. Let $T,\,Q,\,U$ and $W$ be the matrices
representing the action of
$V_{\psi},\,V_{\rho_{1}},\,V_{\rho_{2}},\,V_{\rho_{3}}$ on $M$ respectively.
They are all symmetric matrices with nonnegative integer entries by Definition
2.6. Let $E$ be the identity matrix. By the fusion rule of $r(S_{4})$, we have

(2) $\displaystyle T^{2}=E,$ (3) $\displaystyle TQ=QT=Q,$ (4) $\displaystyle
TU=UT=W,$ (5) $\displaystyle Q^{2}=E+T+Q,$ (6) $\displaystyle QU=U+TU,$ (7)
$\displaystyle U^{2}=E+Q+U+TU.$

In particular, since $T^{2}=E$ and $T$ has nonnegative integer entries, we
know that $T$ is a symmetric permutation matrix.

Convention. Let $P_{n}$ be the group of $n\times n$ permutation matrices.
Since there is naturally a group isomorphism between $S_{n}$ and $P_{n}$, we
will use the cycle notation of permutations to represent permutation matrices.

###  3.1. Irreducible based modules of rank $\leq 3$ over $r(S_{4})$

We define a $\mathbb{Z}_{+}$-module $M_{1,1}$ of rank 1 over $r(S_{4})$ by
letting

(8) $V_{\psi}\mapsto 1,\quad V_{\rho_{1}}\mapsto 2,\quad V_{\rho_{2}}\mapsto
3,\quad V_{\rho_{3}}\mapsto 3.$

######  Proposition 3.1.

Any irreducible based module of rank $1$ over $r(S_{4})$ is equivalent to
$M_{1,1}$.

###### Proof.

Note that any integral fusion ring $A$ has the unique character ${\rm
FPdim}:A\to\mathbb{Z}$ which takes non-negative values on the
$\mathbb{Z}_{+}$-basis, so there exists a unique $\mathbb{Z}_{+}$-module $M$
of rank 1 over it. Clearly, such $M$ is a based module. Now this argument is
available for the situation $A=r(S_{4})$. ∎

Next we consider irreducible based modules of rank $2,3$. According to the
fusion rule of $r(S_{4})$ given in (1), it is sufficient to only list the
representation matrices of $V_{\psi}$, $V_{\rho_{1}}$ and $V_{\rho_{2}}$
acting on them. For simplicity, we choose to present our result for the cases
of small rank 2 and 3 directly, and then analyze the cases of higher rank 4
and 5 with details.

######  Proposition 3.2.

Let $M$ be an irreducible based module of rank $2$ over $r(S_{4})$. Then $M$
is equivalent to one of the based modules $M_{2,i},\,1\leq i\leq 3$, listed in
TABLE 2.

|  $V_{\psi}$ |  $V_{\rho_{1}}$ |  $V_{\rho_{2}}$
---|---|---|---
$M_{2,1}$ |  $\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$ |  $\begin{pmatrix}2&0\\\ 0&2\end{pmatrix}$ |  $\begin{pmatrix}1&2\\\ 2&1\end{pmatrix}$
$M_{2,2}$ |  $\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$ |  $\begin{pmatrix}1&1\\\ 1&1\end{pmatrix}$ |  $\begin{pmatrix}2&1\\\ 1&2\end{pmatrix}$
$M_{2,3}$ |  $\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$ |  $\begin{pmatrix}1&1\\\ 1&1\end{pmatrix}$ |  $\begin{pmatrix}1&2\\\ 2&1\end{pmatrix}$

Table 2. Inequivalent irreducible based modules of rank 2 over $r(S_{4})$

######  Proposition 3.3.

Let $M$ be an irreducible based module of rank $3$ over $r(S_{4})$. Then $M$
is equivalent to one of the based modules $M_{3,i},\,1\leq i\leq 3$, listed in
TABLE 3.

|  $V_{\psi}$ |  $V_{\rho_{1}}$ |  $V_{\rho_{2}}$
---|---|---|---
$M_{3,1}$ |  $\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&1\\\ 1&0&1\\\ 1&1&0\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&1\\\ 1&1&1\\\ 1&1&1\\\ \end{pmatrix}$
$M_{3,2}$ |  $\begin{pmatrix}0&1&0\\\ 1&0&0\\\ 0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&0&1\\\ 0&0&1\\\ 1&1&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&0&1\\\ 0&1&1\\\ 1&1&2\\\ \end{pmatrix}$
$M_{3,3}$ |  $\begin{pmatrix}0&1&0\\\ 1&0&0\\\ 0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&0&1\\\ 0&0&1\\\ 1&1&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&1\\\ 1&0&1\\\ 1&1&2\\\ \end{pmatrix}$

Table 3. Inequivalent irreducible based modules of rank 3 over $r(S_{4})$

###  3.2. Irreducible based modules of rank $4,\,5$ over $r(S_{4})$

######  Proposition 3.4.

Let $M$ be an irreducible based module of rank $4$ over $r(S_{4})$. Then $M$
is equivalent to one of the based modules $M_{4,i},\,1\leq i\leq 7$, listed in
TABLE 4.

|  $V_{\psi}$ |  $V_{\rho_{1}}$ |  $V_{\rho_{2}}$
---|---|---|---
$M_{4,1}$ |  $\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&0&1&1\\\ 0&2&0&0\\\ 1&0&0&1\\\ 1&0&1&0\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&0&0\\\ 1&2&1&1\\\ 0&1&0&0\\\ 0&1&0&0\\\ \end{pmatrix}$
$M_{4,2}$ |  $\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}2&0&0&0\\\ 0&2&0&0\\\ 0&0&2&0\\\ 0&0&0&2\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\ 1&1&0&1\\\ 1&1&1&0\\\ \end{pmatrix}$
$M_{4,3}$ |  $\begin{pmatrix}0&1&0&0\\\ 1&0&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&0&0\\\ 1&1&0&0\\\ 0&0&2&0\\\ 0&0&0&2\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&0&1&1\\\ 0&1&1&1\\\ 1&1&0&1\\\ 1&1&1&0\\\ \end{pmatrix}$
$M_{4,4}$ |  $\begin{pmatrix}0&1&0&0\\\ 1&0&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&0&0\\\ 1&1&0&0\\\ 0&0&2&0\\\ 0&0&0&2\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\ 1&1&0&1\\\ 1&1&1&0\\\ \end{pmatrix}$
$M_{4,5}$ |  $\begin{pmatrix}0&1&0&0\\\ 1&0&0&0\\\ 0&0&0&1\\\ 0&0&1&0\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&0&0\\\ 1&1&0&0\\\ 0&0&1&1\\\ 0&0&1&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\ 1&1&0&1\\\ 1&1&1&0\\\ \end{pmatrix}$
$M_{4,6}$ |  $\begin{pmatrix}0&1&0&0\\\ 1&0&0&0\\\ 0&0&0&1\\\ 0&0&1&0\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&0&0\\\ 1&1&0&0\\\ 0&0&1&1\\\ 0&0&1&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\ 1&1&1&0\\\ 1&1&0&1\\\ \end{pmatrix}$
$M_{4,7}$ |  $\begin{pmatrix}0&1&0&0\\\ 1&0&0&0\\\ 0&0&0&1\\\ 0&0&1&0\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&0&0\\\ 1&1&0&0\\\ 0&0&1&1\\\ 0&0&1&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&0&1&1\\\ 0&1&1&1\\\ 1&1&1&0\\\ 1&1&0&1\\\ \end{pmatrix}$

Table 4. Inequivalent irreducible based modules of rank 4 over $r(S_{4})$

###### Proof.

Let $M$ be a based module of rank 4 over $r(S_{4})$, with the action of
$r(S_{4})$ on it given by

$V_{\psi}\mapsto T,\quad V_{\rho_{1}}\mapsto Q=(a_{ij})_{1\leq i,j\leq
4},\quad V_{\rho_{2}}\mapsto U=(b_{ij})_{1\leq i,j\leq 4},\quad
V_{\rho_{3}}\mapsto W=TU,$

where $a_{ij}=a_{ji}$, $b_{ij}=b_{ji}$.

The symmetric group $S_{4}$ has two conjugacy classes of permutations of order
2. One conjugacy class of 6 permutations includes $(12)$, and the other one of
3 permutations includes $(12)(34)$. As previously seen, $T$ is the unit or an
element of order 2 in $P_{4}$, so we have 10 candidates for $T$, and each of
them is conjugate to one of the following 3 matrices.

$T_{1}=E_{4},\quad T_{2}=(12),\quad T_{3}=(12)(34).$

Hence, for the based module $M$ determined by the pair $(T,Q,U)$, there exists
a $4\times 4$ permutation matrix $P$ such that $T^{\prime}=PTP^{-1}$ is one of
the above $T_{r}$’s $(1\leq r\leq 3)$. Correspondingly, let
$Q^{\prime}=PQP^{-1}$, $U^{\prime}=PUP^{-1}$. Then we get a based module
$M^{\prime}$ determined by the pair $(T^{\prime},Q^{\prime},U^{\prime})$ and
equivalent to $M$ as based modules by Definition (v) (i). So we have reduced
the proof to the situation when $T=T_{r}$.

Case 1 $T=T_{1}=E_{4}$. Since $Q$ satisfies Eq. (5), we obtain the following
system of integer equations:

$\left\\{\begin{array}[]{l}a_{11}^{2}+a_{12}^{2}+a_{13}^{2}+a_{14}^{2}=2+a_{11},\\\
a_{11}a_{12}+a_{12}a_{22}+a_{13}a_{23}+a_{14}a_{24}=a_{12},\\\
a_{11}a_{13}+a_{12}a_{23}+a_{13}a_{33}+a_{14}a_{34}=a_{13},\\\
a_{11}a_{14}+a_{12}a_{24}+a_{13}a_{34}+a_{14}a_{44}=a_{14},\\\
a_{12}^{2}+a_{22}^{2}+a_{23}^{2}+a_{24}^{2}=2+a_{22},\\\
a_{12}a_{13}+a_{22}a_{23}+a_{23}a_{33}+a_{24}a_{34}=a_{23},\\\
a_{12}a_{14}+a_{22}a_{24}+a_{23}a_{34}+a_{24}a_{44}=a_{24},\\\
a_{13}^{2}+a_{23}^{2}+a_{33}^{2}+a_{34}^{2}=2+a_{33},\\\
a_{13}a_{14}+a_{23}a_{24}+a_{33}a_{34}+a_{34}a_{44}=a_{34},\\\
a_{14}^{2}+a_{24}^{2}+a_{34}^{2}+a_{44}^{2}=2+a_{44}.\end{array}\right.$

We use Matlab to figure out all the solutions of $Q$ as follows.

$\begin{array}[]{l}Q_{1}=\begin{pmatrix}0&0&1&1\\\ 0&2&0&0\\\ 1&0&0&1\\\
1&0&1&0\end{pmatrix},\quad Q_{2}=\begin{pmatrix}0&1&0&1\\\ 1&0&0&1\\\
0&0&2&0\\\ 1&1&0&0\end{pmatrix},\quad Q_{3}=\begin{pmatrix}0&1&1&0\\\
1&0&1&0\\\ 1&1&0&0\\\ 0&0&0&2\end{pmatrix},\\\\[20.00003pt]
Q_{4}=\begin{pmatrix}2&0&0&0\\\ 0&0&1&1\\\ 0&1&0&1\\\
0&1&1&0\end{pmatrix},\quad Q_{5}=\begin{pmatrix}2&0&0&0\\\ 0&2&0&0\\\
0&0&2&0\\\ 0&0&0&2\end{pmatrix}.\end{array}$

Next we calculate $U$ after taking $Q$ as one $Q_{k}$ $\left(1\leq k\leq
5\right)$. Case 1.1 $Q=Q_{1}$.

Since $U$ satisfies Eq. (6), we get

$\begin{split}b_{12}&=b_{23}=b_{24},\\\
b_{11}&=b_{13}=b_{14}=b_{33}=b_{34}=b_{44}.\end{split}$

Then by Eq. (7), we have

$\left\\{\begin{array}[]{l}3b_{11}^{2}+b_{12}^{2}=2b_{11}+1,\\\
3b_{11}b_{12}+b_{12}b_{22}=2b_{12},\\\
3b_{12}^{2}+b_{22}^{2}=2b_{22}+3.\end{array}\right.$

The solutions of $U$ given by Matlab are as follows.

$U_{1}=\begin{pmatrix}0&1&0&0\\\ 1&2&1&1\\\ 0&1&0&0\\\
0&1&0&0\end{pmatrix},\quad U_{2}=\begin{pmatrix}1&0&1&1\\\ 0&3&0&0\\\
1&0&1&1\\\ 1&0&1&1\end{pmatrix}.$

It is easy to check that the based module determined by
$\left(T_{1},Q_{1},U_{1}\right)$ is an irreducible based module denoted as
$M_{4,1}$, while the based module determined by
$\left(T_{1},Q_{1},U_{2}\right)$ is reducible.

Note that there exists a permutation matrix $P=(14)(23)$ such that
$PQ_{1}P^{-1}=Q_{2}$. Let $U^{\prime}_{1}=PU_{1}P^{-1}$. There is an
irreducible based module $N^{\prime}$ determined by the pair
$\big{(}T_{1},Q_{2},U^{\prime}_{1}\big{)}$ and equivalent to $M_{4,1}$ by
Definition (v) (i). Conversely, any irreducible based module with
representation matrices $T_{1}$ and $Q_{2}$ is equivalent to $M_{4,1}$. The
same analysis tells us that irreducible based modules with representation
matrices $T_{1}$ and $Q_{3}$ (or $Q_{4}$) are also equivalent to $M_{4,1}$.
Case 1.2 $Q=Q_{5}$.

Since $U$ satisfies Eqs. (6) and (7), we get a system of integer equations as
follows:

$\left\\{\begin{array}[]{l}b_{11}^{2}+b_{12}^{2}+b_{13}^{2}+b_{14}^{2}=2b_{11}+3,\\\
b_{11}b_{12}+b_{12}b_{22}+b_{13}b_{23}+b_{14}b_{24}=2b_{12},\\\
b_{11}b_{13}+b_{12}b_{23}+b_{13}b_{33}+b_{14}b_{34}=2b_{13},\\\
b_{11}b_{14}+b_{12}b_{24}+b_{13}b_{34}+b_{14}b_{44}=2b_{14},\\\
b_{12}^{2}+b_{22}^{2}+b_{23}^{2}+b_{24}^{2}=2b_{22}+3,\\\
b_{12}b_{13}+b_{22}b_{23}+b_{23}b_{33}+b_{24}b_{34}=2b_{23},\\\
b_{12}b_{14}+b_{22}b_{24}+b_{23}b_{34}+b_{24}b_{44}=2b_{24},\\\
b_{13}^{2}+b_{23}^{2}+b_{33}^{2}+b_{34}^{2}=2b_{33}+3,\\\
b_{13}b_{14}+b_{23}b_{24}+b_{33}b_{34}+b_{34}b_{44}=2b_{34},\\\
b_{14}^{2}+b_{24}^{2}+b_{34}^{2}+b_{44}^{2}=2b_{44}+3.\\\ \end{array}\right.$

Thus, the solutions of $U$ by Matlab are as follows:

$\begin{array}[]{l}U_{1}=\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\ 1&1&0&1\\\
1&1&1&0\end{pmatrix},\quad U_{2}=\begin{pmatrix}1&0&0&2\\\ 0&1&2&0\\\
0&2&1&0\\\ 2&0&0&1\end{pmatrix},\quad U_{3}=\begin{pmatrix}1&0&2&0\\\
0&1&0&2\\\ 2&0&1&0\\\ 0&2&0&1\end{pmatrix},\quad
U_{4}=\begin{pmatrix}1&2&0&0\\\ 2&1&0&0\\\ 0&0&1&2\\\
0&0&2&1\end{pmatrix},\\\\[20.00003pt] U_{5}=\begin{pmatrix}1&0&2&0\\\
0&3&0&0\\\ 2&0&1&0\\\ 0&0&0&3\end{pmatrix},\quad
U_{6}=\begin{pmatrix}1&0&0&2\\\ 0&3&0&0\\\ 0&0&3&0\\\
2&0&0&1\end{pmatrix},\quad U_{7}=\begin{pmatrix}1&2&0&0\\\ 2&1&0&0\\\
0&0&3&0\\\ 0&0&0&3\end{pmatrix},\quad U_{8}=\begin{pmatrix}3&0&0&0\\\
0&3&0&0\\\ 0&0&1&2\\\ 0&0&2&1\end{pmatrix},\\\\[20.00003pt]
U_{9}=\begin{pmatrix}3&0&0&0\\\ 0&1&2&0\\\ 0&2&1&0\\\
0&0&0&3\end{pmatrix},\quad U_{10}=\begin{pmatrix}3&0&0&0\\\ 0&1&0&2\\\
0&0&3&0\\\ 0&2&0&1\end{pmatrix},\quad U_{11}=\begin{pmatrix}3&0&0&0\\\
0&3&0&0\\\ 0&0&3&0\\\ 0&0&0&3\end{pmatrix}.\end{array}$

Since $T_{1}$ and $Q_{5}$ are diagonal and the solutions $U_{t}$ $\left(2\leq
t\leq 11\right)$ are block diagonal with at least two blocks, only the based
module determined by $(T_{1},Q_{5},U_{1})$ is irreducible, denoted as
$M_{4,2}$.

Case 2 $T=T_{2}=(12)$. Since $Q$ satisfies Eq. (3), we get

$Q=\begin{pmatrix}a_{11}&a_{11}&a_{13}&a_{14}\\\
a_{11}&a_{11}&a_{13}&a_{14}\\\ a_{13}&a_{13}&a_{33}&a_{34}\\\
a_{14}&a_{14}&a_{34}&a_{44}\end{pmatrix}.$

Since $Q$ also satisfies Eq. (5), we have the following system of integer
equations:

$\left\\{\begin{array}[]{l}2a_{11}^{2}+a_{13}^{2}+a_{14}^{2}=a_{11}+1,\\\
2a_{11}a_{13}+a_{13}a_{33}+a_{14}a_{34}=a_{13},\\\
2a_{11}a_{14}+a_{13}a_{34}+a_{14}a_{44}=a_{14},\\\
2a_{13}^{2}+a_{33}^{2}+a_{34}^{2}=a_{33}+2,\\\
2a_{13}a_{14}+a_{33}a_{34}+a_{34}a_{44}=a_{34},\\\
2a_{14}^{2}+a_{34}^{2}+a_{44}^{2}=a_{44}+2.\\\ \end{array}\right.$

Hence, the solutions of $Q$ by Matlab are as follows:

$\begin{array}[]{l}Q_{1}=\begin{pmatrix}0&0&0&1\\\ 0&0&0&1\\\ 0&0&2&0\\\
1&1&0&1\end{pmatrix},\quad Q_{2}=\begin{pmatrix}0&0&1&0\\\ 0&0&1&0\\\
1&1&1&0\\\ 0&0&0&2\end{pmatrix},\quad Q_{3}=\begin{pmatrix}1&1&0&0\\\
1&1&0&0\\\ 0&0&2&0\\\ 0&0&0&2\end{pmatrix}.\end{array}$

Since $U$ satisfies Eq. (4), we get

$U=\begin{pmatrix}b_{11}&b_{12}&b_{13}&b_{14}\\\
b_{12}&b_{11}&b_{13}&b_{14}\\\ b_{13}&b_{13}&b_{33}&b_{34}\\\
b_{14}&b_{14}&b_{34}&b_{44}\end{pmatrix}.$

Next we calculate $U$ after taking $Q$ as one $Q_{k}$ $\left(1\leq k\leq
3\right)$. Case 2.1 $Q=Q_{1}$.

Since $U$ satisfies Eqs. (6) and (7), the solutions of $U$ given by Matlab are
as follows.

$\begin{array}[]{l}U_{1}=\begin{pmatrix}0&1&0&1\\\ 1&0&0&1\\\ 0&0&3&0\\\
1&1&0&2\end{pmatrix},\quad U_{2}=\begin{pmatrix}1&0&0&1\\\ 0&1&0&1\\\
0&0&3&0\\\ 1&1&0&2\end{pmatrix}.\end{array}$

Since $T_{2}$, $Q_{1}$ and all the solutions $U_{t}$ for $t=1,2$ are block
diagonal with at least two blocks, the based modules determined by each pair
$(T_{2},Q_{1},U_{t})$ are reducible.

Note that there exists a permutation matrix $P=(12)(34)$ such that
$PQ_{1}P^{-1}=Q_{2}$. Let $U^{\prime}_{t}=PU_{t}P^{-1}$. Then each based
module $N_{t}$ determined by the pair
$\left(T_{2},Q_{2},U^{\prime}_{t}\right)$ is reducible. Namely, any based
module with representation matrices $T_{2}$ and $Q_{2}$ is reducible.

Case 2.2 $Q=Q_{3}$.

Since $U$ satisfies Eqs. (6) and (7), we have

$\begin{array}[]{l}U_{1}=\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\ 1&1&0&1\\\
1&1&1&0\end{pmatrix},\quad U_{2}=\begin{pmatrix}1&0&1&1\\\ 0&1&1&1\\\
1&1&0&1\\\ 1&1&1&0\end{pmatrix},\quad U_{3}=\begin{pmatrix}1&2&0&0\\\
2&1&0&0\\\ 0&0&1&2\\\ 0&0&2&1\end{pmatrix},\quad
U_{4}=\begin{pmatrix}1&2&0&0\\\ 2&1&0&0\\\ 0&0&3&0\\\
0&0&0&3\end{pmatrix},\\\\[20.00003pt] U_{5}=\begin{pmatrix}2&1&0&0\\\
1&2&0&0\\\ 0&0&1&2\\\ 0&0&2&1\end{pmatrix},\quad
U_{6}=\begin{pmatrix}2&1&0&0\\\ 1&2&0&0\\\ 0&0&3&0\\\
0&0&0&3\end{pmatrix}.\end{array}$

Since $T_{2}$, $Q_{3}$ and the solutions $U_{s}$ $(3\leq s\leq 6)$ are block
diagonal with at least two blocks, only the based module determined by
$(T_{2},Q_{3},U_{1})$ and $(T_{2},Q_{3},U_{2})$ are irreducible, denoted as
$M_{4,3}$ and $M_{4,4}$ respectively. It is easy to check that $M_{4,3}$ and
$M_{4,4}$ are inequivalent based modules.

Case 3 $T=T_{3}=(12)(34)$. Since $Q$ satisfies Eq. (3), we get

$Q=\begin{pmatrix}a_{11}&a_{11}&a_{13}&a_{13}\\\
a_{11}&a_{11}&a_{13}&a_{13}\\\ a_{13}&a_{13}&a_{33}&a_{33}\\\
a_{13}&a_{13}&a_{33}&a_{33}\end{pmatrix}.$

Then by Eq. (5), we have the following system of integer equations:

$\left\\{\begin{array}[]{l}2a_{11}^{2}+2a_{13}^{2}=a_{11}+1,\\\
2a_{11}a_{13}+2a_{13}a_{33}=a_{13},\\\
2a_{13}^{2}+2a_{33}^{2}=a_{33}+1.\end{array}\right.$

$Q$ has the following unique solution.

$Q_{1}=\begin{pmatrix}1&1&0&0\\\ 1&1&0&0\\\ 0&0&1&1\\\ 0&0&1&1\end{pmatrix}.$

Since $U$ satisfies Eq. (4), we get

$U=\begin{pmatrix}b_{11}&b_{12}&b_{13}&b_{14}\\\
b_{12}&b_{11}&b_{14}&b_{13}\\\ b_{13}&b_{14}&b_{33}&b_{34}\\\
b_{14}&b_{13}&b_{34}&b_{33}\end{pmatrix}.$

Since $U$ also satisfies Eqs. (6) and (7), we obtain the solutions of $U$ by
Matlab as follows:

$\begin{array}[]{l}U_{1}=\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\ 1&1&0&1\\\
1&1&1&0\end{pmatrix},\quad U_{2}=\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\
1&1&1&0\\\ 1&1&0&1\end{pmatrix},\quad U_{3}=\begin{pmatrix}1&0&1&1\\\
0&1&1&1\\\ 1&1&1&0\\\ 1&1&0&1\end{pmatrix},\quad
U_{4}=\begin{pmatrix}1&0&1&1\\\ 0&1&1&1\\\ 1&1&0&1\\\
1&1&1&0\end{pmatrix},\\\\[20.00003pt] U_{5}=\begin{pmatrix}1&2&0&0\\\
2&1&0&0\\\ 0&0&1&2\\\ 0&0&2&1\end{pmatrix},\quad
U_{6}=\begin{pmatrix}1&2&0&0\\\ 2&1&0&0\\\ 0&0&2&1\\\
0&0&1&2\end{pmatrix},\quad U_{7}=\begin{pmatrix}2&1&0&0\\\ 1&2&0&0\\\
0&0&1&2\\\ 0&0&2&1\end{pmatrix},\quad U_{8}=\begin{pmatrix}2&1&0&0\\\
1&2&0&0\\\ 0&0&2&1\\\ 0&0&1&2\end{pmatrix}.\end{array}$

Clearly, $T_{3}$, $Q_{1}$ and the solutions $U_{s}$ are block diagonal with at
least two blocks, but the based module determined by the pair
$\left(T_{3},Q_{1},U_{t}\right)$ is irreducible, denoted as $M_{4,s}$, where
$5\leq s\leq 8,1\leq t\leq 4$. Define the $\mathbb{Z}$-module isomorphism
$\phi:M_{4,6}\to M_{4,8}$ by

$\phi(v_{1}^{1})=v_{4}^{2},\quad\phi(v_{2}^{1})=v_{3}^{2},\quad\phi(v_{3}^{1})=v_{2}^{2},\quad\phi(v_{4}^{1})=v_{1}^{2}.$

It is easy to see that $M_{4,6}$ is equivalent to $M_{4,8}$ as based modules
over $r(S_{4})$ under $\phi$. Then we can check that
$\left\\{M_{4,s}\right\\}_{5\leq s\leq 7}$ are inequivalent irreducible based
modules. ∎

Finally, we construct two based modules $M_{5,i}\ \left(i=1,2\right)$ over
$r(S_{4})$ with the actions of $r(S_{4})$ on them presented in TABLE 5.

|  $V_{\psi}$ |  $V_{\rho_{1}}$ |  $V_{\rho_{2}}$
---|---|---|---
$M_{5,1}$ |  $\begin{pmatrix}0&1&0&0&0\\\ 1&0&0&0&0\\\ 0&0&0&1&0\\\ 0&0&1&0&0\\\ 0&0&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&0&0&0&1\\\ 0&0&0&0&1\\\ 0&0&1&1&0\\\ 0&0&1&1&0\\\ 1&1&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&0&0&1&0\\\ 0&0&1&0&0\\\ 0&1&1&1&1\\\ 1&0&1&1&1\\\ 0&0&1&1&0\\\ \end{pmatrix}$
$M_{5,2}$ |  $\begin{pmatrix}0&1&0&0&0\\\ 1&0&0&0&0\\\ 0&0&0&1&0\\\ 0&0&1&0&0\\\ 0&0&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&0&0&0\\\ 1&1&0&0&0\\\ 0&0&1&1&0\\\ 0&0&1&1&0\\\ 0&0&0&0&2\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&0&0&1&1\\\ 0&0&1&0&1\\\ 0&1&0&0&1\\\ 1&0&0&0&1\\\ 1&1&1&1&1\\\ \end{pmatrix}$

Table 5. Inequivalent irreducible based modules of rank 5 over $r(S_{4})$

######  Proposition 3.5.

Let $M$ be an irreducible based module of rank $5$ over $r(S_{4})$. Then $M$
is equivalent to one of the based modules $M_{5,i}\,(i=1,2)$, listed in TABLE
5.

###### Proof.

Let $M$ be a based module of rank 5 over $r(S_{4})$, with the action of
$r(S_{4})$ on it given by

$V_{\psi}\mapsto T,\quad V_{\rho_{1}}\mapsto Q=(a_{ij})_{1\leq i,j\leq
5},\quad V_{\rho_{2}}\mapsto U=(b_{ij})_{1\leq i,j\leq 5},\quad
V_{\rho_{3}}\mapsto W=TU,$

where $a_{ij}=a_{ji}$, $b_{ij}=b_{ji}$.

First by the similar argument applied in the case of rank 4, we only need to
deal with one of the following 3 cases for $T$.

$T_{1}=E_{5},\quad T_{2}=(12),\quad T_{3}=(12)(34).$

Case 1 $T=T_{1}=E_{5}$. There are 11 solutions of $Q$ satisfying Eq. (5), but
only two conjugacy classes by permutation matrices with their representatives
given as follows.

$Q_{1}=\begin{pmatrix}0&1&1&0&0\\\ 1&0&1&0&0\\\ 1&1&0&0&0\\\ 0&0&0&2&0\\\
0&0&0&0&2\end{pmatrix},\quad Q_{2}=\begin{pmatrix}2&0&0&0&0\\\ 0&2&0&0&0\\\
0&0&2&0&0\\\ 0&0&0&2&0\\\ 0&0&0&0&2\end{pmatrix}.$

Next we calculate $U$ after taking $Q$ as one $Q_{k}$ $\left(k=1,2\right)$.
Case 1.1 $Q=Q_{1}$.

There are 4 solutions of $U$ satisfying Eqs. (6) and (7) as follows.

$U_{1}=\begin{pmatrix}0&0&0&1&0\\\ 0&0&0&1&0\\\ 0&0&0&1&0\\\ 1&1&1&2&0\\\
0&0&0&0&3\end{pmatrix},\ \ U_{2}=\begin{pmatrix}0&0&0&0&1\\\ 0&0&0&0&1\\\
0&0&0&0&1\\\ 0&0&0&3&0\\\ 1&1&1&0&2\end{pmatrix},\ \
U_{3}=\begin{pmatrix}1&1&1&0&0\\\ 1&1&1&0&0\\\ 1&1&1&0&0\\\ 0&0&0&3&0\\\
0&0&0&0&3\\\ \end{pmatrix},\ \ U_{4}=\begin{pmatrix}1&1&1&0&0\\\ 1&1&1&0&0\\\
1&1&1&0&0\\\ 0&0&0&1&2\\\ 0&0&0&2&1\end{pmatrix}.$

Case 1.2 $Q=Q_{2}$.

There are 31 solutions of $U$ satisfying Eqs. (6) and (7), but only 4
conjugacy classes by permutation matrices and their representatives as
follows.

$U_{1}=\begin{pmatrix}0&1&1&1&0\\\ 1&0&1&1&0\\\ 1&1&0&1&0\\\ 1&1&1&0&0\\\
0&0&0&0&3\end{pmatrix},\ \ U_{2}=\begin{pmatrix}1&0&0&2&0\\\ 0&1&2&0&0\\\
0&2&1&0&0\\\ 2&0&0&1&0\\\ 0&0&0&0&3\end{pmatrix},\ \
U_{3}=\begin{pmatrix}1&2&0&0&0\\\ 2&1&0&0&0\\\ 0&0&3&0&0\\\ 0&0&0&3&0\\\
0&0&0&0&3\\\ \end{pmatrix},\ \ U_{4}=\begin{pmatrix}3&0&0&0&0\\\ 0&3&0&0&0\\\
0&0&3&0&0\\\ 0&0&0&3&0\\\ 0&0&0&0&3\end{pmatrix}.$

Each pair $(T_{1},Q_{k},U_{r})$ above determines a based module, but not
irreducible for $1\leq r\leq 4$. Case 2 $T=T_{2}=(12)$.

There are 5 solutions of $Q$ satisfying Eq. (5), but only 3 conjugacy classes
with the following representatives.

$Q_{1}=\begin{pmatrix}0&0&0&0&1\\\ 0&0&0&0&1\\\ 0&0&2&0&0\\\ 0&0&0&2&0\\\
1&1&0&0&1\end{pmatrix},\quad Q_{2}=\begin{pmatrix}1&1&0&0&0\\\ 1&1&0&0&0\\\
0&0&0&1&1\\\ 0&0&1&0&1\\\ 0&0&1&1&0\end{pmatrix},\quad
Q_{3}=\begin{pmatrix}1&1&0&0&0\\\ 1&1&0&0&0\\\ 0&0&2&0&0\\\ 0&0&0&2&0\\\
0&0&0&0&2\end{pmatrix}.$

Next we calculate $U$ after choosing $Q$. Case 2.1 $Q=Q_{1}$.

There are 4 solutions of $U$ satisfying Eqs. (6) and (7) as follows.

$U_{1}=\begin{pmatrix}0&1&0&0&1\\\ 1&0&0&0&1\\\ 0&0&1&2&0\\\ 0&0&2&1&0\\\
1&1&0&0&2\end{pmatrix},\ \ U_{2}=\begin{pmatrix}0&1&0&0&1\\\ 1&0&0&0&1\\\
0&0&3&0&0\\\ 0&0&0&3&0\\\ 1&1&0&0&2\end{pmatrix},\ \
U_{3}=\begin{pmatrix}1&0&0&0&1\\\ 0&1&0&0&1\\\ 0&0&1&2&0\\\ 0&0&2&1&0\\\
1&1&0&0&2\\\ \end{pmatrix},\ \ U_{4}=\begin{pmatrix}1&0&0&0&1\\\ 0&1&0&0&1\\\
0&0&3&0&0\\\ 0&0&0&3&0\\\ 1&1&0&0&2\end{pmatrix}.$

Case 2.2 $Q=Q_{2}$.

There are 2 solutions of $U$ satisfying Eqs. (6) and (7) as follows.

$U_{1}=\begin{pmatrix}1&2&0&0&0\\\ 2&1&0&0&0\\\ 0&0&1&1&1\\\ 0&0&1&1&1\\\
0&0&1&1&1\end{pmatrix},\ \ U_{2}=\begin{pmatrix}2&1&0&0&0\\\ 1&2&0&0&0\\\
0&0&1&1&1\\\ 0&0&1&1&1\\\ 0&0&1&1&1\end{pmatrix}.$

Case 2.3 $Q=Q_{3}$. There are 14 solutions of $U$ satisfying Eqs. (6) and (7),
but only 6 conjugacy classes by permutation matrices and their representatives
as follows.

$\begin{array}[]{l}U_{1}=\begin{pmatrix}0&1&1&1&0\\\ 1&0&1&1&0\\\ 1&1&0&1&0\\\
1&1&1&0&0\\\ 0&0&0&0&3\end{pmatrix},\quad U_{2}=\begin{pmatrix}1&0&1&1&0\\\
0&1&1&1&0\\\ 1&1&0&1&0\\\ 1&1&1&0&0\\\ 0&0&0&0&3\end{pmatrix},\quad
U_{3}=\begin{pmatrix}1&2&0&0&0\\\ 2&1&0&0&0\\\ 0&0&1&2&0\\\ 0&0&2&1&0\\\
0&0&0&0&3\end{pmatrix},\\\\[25.00003pt] U_{4}=\begin{pmatrix}2&1&0&0&0\\\
1&2&0&0&0\\\ 0&0&1&2&0\\\ 0&0&2&1&0\\\ 0&0&0&0&3\end{pmatrix},\quad
U_{5}=\begin{pmatrix}1&2&0&0&0\\\ 2&1&0&0&0\\\ 0&0&3&0&0\\\ 0&0&0&3&0\\\
0&0&0&0&3\\\ \end{pmatrix},\quad U_{6}=\begin{pmatrix}2&1&0&0&0\\\
1&2&0&0&0\\\ 0&0&3&0&0\\\ 0&0&0&3&0\\\ 0&0&0&0&3\end{pmatrix}.\end{array}$

Through analysis, all based modules derived from Case 2 are reducible.

Case 3 $T=T_{3}=(12)(34)$. There are 3 solutions of $Q$ satisfying Eq. (5) as
follows.

$Q_{1}=\begin{pmatrix}0&0&0&0&1\\\ 0&0&0&0&1\\\ 0&0&1&1&0\\\ 0&0&1&1&0\\\
1&1&0&0&1\end{pmatrix},\quad Q_{2}=\begin{pmatrix}1&1&0&0&0\\\ 1&1&0&0&0\\\
0&0&1&1&0\\\ 0&0&1&1&0\\\ 0&0&0&0&2\end{pmatrix},\quad
Q_{3}=\begin{pmatrix}1&1&0&0&0\\\ 1&1&0&0&0\\\ 0&0&0&0&1\\\ 0&0&0&0&1\\\
0&0&1&1&1\end{pmatrix}.$

Next we calculate $U$ after fixing $Q$. Case 3.1 $Q=Q_{1}$.

There are 6 solutions of $U$ satisfying Eqs. (6) and (7) as follows.

$\begin{array}[]{l}U_{1}=\begin{pmatrix}0&0&0&1&0\\\ 0&0&1&0&0\\\ 0&1&1&1&1\\\
1&0&1&1&1\\\ 0&0&1&1&0\end{pmatrix},\quad U_{2}=\begin{pmatrix}0&0&1&0&0\\\
0&0&0&1&0\\\ 1&0&1&1&1\\\ 0&1&1&1&1\\\ 0&0&1&1&0\end{pmatrix},\quad
U_{3}=\begin{pmatrix}0&1&0&0&1\\\ 1&0&0&0&1\\\ 0&0&1&2&0\\\ 0&0&2&1&0\\\
1&1&0&0&2\end{pmatrix},\\\\[25.00003pt] U_{4}=\begin{pmatrix}0&1&0&0&1\\\
1&0&0&0&1\\\ 0&0&2&1&0\\\ 0&0&1&2&0\\\ 1&1&0&0&2\end{pmatrix},\quad
U_{5}=\begin{pmatrix}1&0&0&0&1\\\ 0&1&0&0&1\\\ 0&0&1&2&0\\\ 0&0&2&1&0\\\
1&1&0&0&2\\\ \end{pmatrix},\quad U_{6}=\begin{pmatrix}1&0&0&0&1\\\
0&1&0&0&1\\\ 0&0&2&1&0\\\ 0&0&1&2&0\\\ 1&1&0&0&2\end{pmatrix}.\end{array}$

Each pair $(T_{3},Q_{1},U_{r})$ $\left(1\leq r\leq 6\right)$ above determines
a based module, but only the based modules with representation matrices
$U_{1}$ and $U_{2}$ are irreducible. Denote $M_{5,1}$ and $M^{\prime}_{5,1}$
these two irreducible based modules with the corresponding $\mathbb{Z}$-basis
$\left\\{v_{1}^{k},v_{2}^{k},v_{3}^{k},v_{4}^{k},v_{5}^{k}\right\\}$ for
$k=1,2$ respectively. Define the $\mathbb{Z}$-module isomorphism
$\phi:M_{5,1}\to M^{\prime}_{5,1}$ by

$\phi(v_{s}^{1})=v_{s}^{2},\quad\phi(v_{3}^{1})=v_{4}^{2},\quad\phi(v_{4}^{1})=v_{3}^{2},\quad
s=1,2,5.$

Then it is easy to see that $M_{5,1}$ is equivalent to $M^{\prime}_{5,1}$ as
based modules over $r(S_{4})$ under $\phi$.

Case 3.2 $Q=Q_{2}$.

There are 10 solutions of $U$ satisfying Eqs. (6) and (7), but only 7
conjugacy classes with their representatives given as follows.

$\begin{array}[]{l}U_{1}=\begin{pmatrix}0&0&0&1&1\\\ 0&0&1&0&1\\\ 0&1&0&0&1\\\
1&0&0&0&1\\\ 1&1&1&1&1\end{pmatrix},\ \ U_{2}=\begin{pmatrix}0&1&1&1&0\\\
1&0&1&1&0\\\ 1&1&0&1&0\\\ 1&1&1&0&0\\\ 0&0&0&0&3\\\ \end{pmatrix},\ \
U_{3}=\begin{pmatrix}0&1&1&1&0\\\ 1&0&1&1&0\\\ 1&1&1&0&0\\\ 1&1&0&1&0\\\
0&0&0&0&3\\\ \end{pmatrix},\ \ U_{4}=\begin{pmatrix}1&0&1&1&0\\\ 0&1&1&1&0\\\
1&1&1&0&0\\\ 1&1&0&1&0\\\ 0&0&0&0&3\\\ \end{pmatrix},\\\\[25.00003pt]
U_{5}=\begin{pmatrix}1&2&0&0&0\\\ 2&1&0&0&0\\\ 0&0&1&2&0\\\ 0&0&2&1&0\\\
0&0&0&0&3\end{pmatrix},\ \ U_{6}=\begin{pmatrix}1&2&0&0&0\\\ 2&1&0&0&0\\\
0&0&2&1&0\\\ 0&0&1&2&0\\\ 0&0&0&0&3\end{pmatrix},\ \
U_{7}=\begin{pmatrix}2&1&0&0&0\\\ 1&2&0&0&0\\\ 0&0&2&1&0\\\ 0&0&1&2&0\\\
0&0&0&0&3\end{pmatrix}.\end{array}$

Each pair $(T_{3},Q_{2},U_{t})$ $\left(2\leq t\leq 7\right)$ above determines
a based module, but only the based module with representation matrix $U_{1}$
is irreducible. We denote it by $M_{5,2}$.

Also, the based modules obtained by taking $Q=Q_{3}$ are equivalent to the
based module $M_{5,1}$ found in Case 3.1. ∎

##  4\. Categorified based modules by module categories over ${\rm
Rep}(S_{4})$

In this section, we will apply the knowledge of module categories over the
complex representation category of a finite group to find which based modules
over $r(S_{4})$ can be categorified by module categories over the
representation category ${\rm Rep}(S_{4})$ of $S_{4}$. For the details about
module categories over tensor categories, see e.g. [14, Chapter 7].

First we recall the required result for the upcoming discussion. For any
finite group $G$, the second cohomology group $H^{2}(G,{\mathbb{C}}^{*})$ is
known to be a finite abelian group called the Schur multiplier and classifies
central extensions of $G$. The notion of universal central extension of a
finite group was first investigated by Schur in [38].

Let ${\rm Rep}(G,\alpha)$ denote the semisimple abelian category of projective
representations of $G$ with the multiplier $\alpha\in
Z^{2}(G,{\mathbb{C}}^{*})$. Equivalently, ${\rm Rep}(G,\alpha)$ is the
representation category of the twisted group algebra ${\mathbb{C}}G_{\alpha}$
of $G$ with multiplication

$g\cdot_{\alpha}h=\alpha(g,h)gh,\quad g,h\in G.$

In particular, ${\rm Rep}(G,\alpha)={\rm Rep}(G)$ when taking $\alpha=1$.

Let $\alpha\in Z^{2}(G,{\mathbb{C}}^{*})$ represent an element of order $d$ in
$H^{2}(G,{\mathbb{C}}^{*})$. Define

${\rm Rep}^{\alpha}(G)=\bigoplus_{j=0}^{d-1}{\rm Rep}(G,\alpha^{j}).$

According to the result in [9], we know that ${\rm Rep}^{\alpha}(G)$ becomes a
fusion category with the tensor product of two projective representations in
${\rm Rep}(G,\alpha^{i})$ and ${\rm Rep}(G,\alpha^{j})$ respectively lying in
${\rm Rep}(G,\alpha^{i+j})$, and the dual object in ${\rm Rep}(G,\alpha^{i})$
lying in ${\rm Rep}(G,\alpha^{d-i})$. Correspondingly, we have the fusion ring

(9) $r^{\alpha}(G)=\bigoplus_{j=0}^{d-1}r(G,\alpha^{j}).$

Now let $H$ be a subgroup of $G$ and $\alpha\in Z^{2}(H,{\mathbb{C}}^{*})$.
The category ${\rm Rep}(H,\alpha)$ is a module category over ${\rm Rep}(G)$ by
applying the restriction functor ${\rm Res}_{H}^{G}:{\rm Rep}(G)\to{\rm
Rep}(H)$.

######  Theorem 4.1 ([32, Theorem 3.2]).

The indecomposable exact module categories over the representation category
${\rm Rep}(G)$ are of the form ${\rm Rep}(H,\alpha)$ and are classified by
conjugacy classes of pairs $(H,[\alpha])$.

Consequently by [14, Prop. 7.7.2], we know that

######  Proposition 4.2.

The Grothendieck group

$r(H,\alpha)={\rm Gr}({\rm Rep}(H,\alpha))$

is an irreducible $\mathbb{Z}_{+}$-module over $r(G)$.

Next we show that any $\mathbb{Z}_{+}$-module over the complex representation
ring $r(G)$ of a finite group $G$ categorified by this way is a based module.

######  Theorem 4.3.

Let $G$ be a finite group, $H$ a subgroup of $G$ and $\alpha\in
Z^{2}(H,{\mathbb{C}}^{*})$. The $\mathbb{Z}_{+}$-module $r(H,\alpha)$ over
$r(G)$ is a based module.

###### Proof.

Let $\left\\{{\psi}_{i}\right\\}_{i\in I}$ be the $\mathbb{Z}_{+}$-basis of
$r(G)$. Take $r^{\alpha}(H)$ defined in Eq. (9) as a $\mathbb{Z}_{+}$-module
over $r(G)$ with the $\mathbb{Z}$-basis $\left\\{{\chi}_{k}\right\\}_{k\in J}$
such that

$\psi_{i}.\chi_{k}=\sum_{l}a_{ik}^{l}\chi_{l},\quad
a_{ik}^{l}\in\mathbb{Z}_{+}.$

On the other hand, we write the fusion rule of the fusion ring $r^{\alpha}(H)$
as follows,

$\chi_{i}\chi_{j}=\sum_{k=1}^{s}n_{ij}^{k}\chi_{k},\quad
n_{ij}^{k}\in\mathbb{Z}_{+}.$

Since the number $n_{ij}^{k^{*}}$ is invariant under cyclic permutations of
$i,j,k$, we have

$n_{ij}^{k}=n_{k^{*}i}^{j^{*}}=n_{i^{*}k}^{j}.$

By the restriction rule, we interpret $r(G)$ as a subring of $r^{\alpha}(H)$
and write down

$\psi_{i}=\sum_{j}r_{ij}\chi_{j},\quad r_{ij}\in\mathbb{Z}_{+}.$

Then

$\psi_{i}.\chi_{k}=\sum_{j}r_{ij}\chi_{j}\chi_{k}=\sum_{j,\,l}r_{ij}n_{jk}^{l}\chi_{l}.$

By comparing the coefficients, we see that

$a_{ik}^{l}=\sum_{j}r_{ij}n_{jk}^{l}=\sum_{j}r_{ij}n_{j^{*}l}^{k}=\sum_{j}r_{i^{*}j^{*}}n_{j^{*}l}^{k}=\sum_{j}r_{i^{*}j}n_{jl}^{k}=a_{i^{*}l}^{k},$

so $r^{\alpha}(H)$ is a based module over $r(G)$, and $r(H,\alpha)$ is clearly
a based submodule of $r^{\alpha}(H)$. Equivalently, any
$\mathbb{Z}_{+}$-module over $r(G)$ categorified by a module category ${\rm
Rep}(H,\alpha)$ over ${\rm Rep}(G)$ must be a based module. ∎

By Theorem 4.3, we only need to focus on those inequivalent irreducible based
modules $M_{i,j}$ over $r(S_{4})$ collected in Section 3 , each of which is
possibly categorified by a module category ${\rm Rep}(H,\alpha)$ for some
$H<S_{4}$ and $\alpha\in Z^{2}(H,{\mathbb{C}}^{*})$.

All the non-isomorphic subgroups of the symmetric group $S_{4}$ are as
follows:

  1. (i)

the symmetric group $S_{3}$;

  2. (ii)

the cyclic groups $\mathbb{Z}_{i},\,1\leq i\leq 4$;

  3. (iii)

the Klein 4-group $K_{4}$;

  4. (iv)

the alternating group $A_{4}$;

  5. (v)

the dihedral group $D_{4}$;

  6. (vi)

the symmetric group $S_{4}$ itself.

Correspondingly, the Schur multipliers we concern here are given as follows
(see e.g. [28]),

$H^{2}(\mathbb{Z}_{n},{\mathbb{C}}^{*})\cong
H^{2}(S_{3},{\mathbb{C}}^{*})\cong 0,\ n\geq 1,\quad\
H^{2}(K_{4},{\mathbb{C}}^{*})\cong H^{2}(D_{4},{\mathbb{C}}^{*})\cong
H^{2}(A_{4},{\mathbb{C}}^{*})\cong
H^{2}(S_{4},{\mathbb{C}}^{*})\cong\mathbb{Z}_{2}.$

As a result, we only need to consider the following two situations.

(1) Module category ${\rm Rep}(H)$ for any subgroup $H<S_{4}$;

(2) Module category ${\rm Rep}(H,\alpha)$ for any subgroup $H<S_{4}$ and
nontrivial twist $\alpha\in Z^{2}(H,{\mathbb{C}}^{*})$.

###  4.1. The module categories over ${\rm Rep}(S_{4})$ with trivial twists

i First we consider the representation category ${\rm Rep}(S_{3})$ as a module
category over ${\rm Rep}(S_{4})$.

######  Theorem 4.4.

$r(S_{3})={\rm Gr}({\rm Rep}(S_{3}))$ is an irreducible based module over
$r(S_{4})={\rm Gr}({\rm Rep}(S_{4}))$ equivalent to the based module $M_{3,2}$
in TABLE 3.

###### Proof.

According to the branching rule of symmetric groups (see e.g. [37, Theorem
2.8.3]), we have the following restriction rules.

$\displaystyle{\rm Res}_{S_{3}}^{S_{4}}\left(1\right)=1,\quad{\rm
Res}_{S_{3}}^{S_{4}}\left(V_{\psi}\right)=\chi,\quad{\rm
Res}_{S_{3}}^{S_{4}}\left(V_{\rho_{1}}\right)=V,$ $\displaystyle{\rm
Res}_{S_{3}}^{S_{4}}\left(V_{\rho_{2}}\right)=1+V,\quad{\rm
Res}_{S_{3}}^{S_{4}}\left(V_{\rho_{3}}\right)=\chi+V,$

where $\chi$ and $V$ denote the sign representation and the standard
representation in ${\rm Rep}({S_{3}})$ respectively. Hence we get the
representation matrices of basis elements of $r(S_{4})$ acting on $r({S_{3}})$
as follows.

$\displaystyle 1\mapsto E_{3},\quad V_{\psi}\mapsto\begin{pmatrix}0&1&0\\\
1&0&0\\\ 0&0&1\end{pmatrix},\quad V_{\rho_{1}}\mapsto\begin{pmatrix}0&0&1\\\
0&0&1\\\ 1&1&1\end{pmatrix},\quad V_{\rho_{2}}\mapsto\begin{pmatrix}1&0&1\\\
0&1&1\\\ 1&1&2\end{pmatrix},\quad V_{\rho_{3}}\mapsto\begin{pmatrix}0&1&1\\\
1&0&1\\\ 1&1&2\end{pmatrix}.$

We see that $r(S_{3})$ is an irreducible based module $M_{3,2}$ according to
TABLE 3. In other words, the based module $M_{3,2}$ can be categorified by the
module category ${\rm Rep}({S_{3}})$ over ${\rm Rep}(S_{4})$. ∎

######  Remark 4.5.

Since the roles of the standard representation and its dual in $r(S_{4})$ are
symmetric, we can exchange the notations $V_{\rho_{2}}$ and $V_{\rho_{3}}$ for
them to get the following restriction rules instead.

${\rm Res}_{S_{3}}^{S_{4}}\left(V_{\rho_{2}}\right)=\chi+V,\quad{\rm
Res}_{S_{3}}^{S_{4}}\left(V_{\rho_{3}}\right)=1+V.$

Therefore, we get another action of $r(S_{4})$ on $r({S_{3}})$ such that
$r(S_{3})$ is an irreducible based module over $r(S_{4})$ equivalent to the
based module $M_{3,3}$ according to TABLE 3. In other words, the based module
$M_{3,3}$ can also be categorified by the module category ${\rm Rep}({S_{3}})$
over ${\rm Rep}(S_{4})$.

ii Secondly, we consider ${\rm Rep}(\mathbb{Z}_{4})$ as a module category over
${\rm Rep}(S_{4})$.

######  Theorem 4.6.

$r(\mathbb{Z}_{4})={\rm Gr}({\rm Rep}(\mathbb{Z}_{4}))$ is an irreducible
based module over $r(S_{4})$ equivalent to the based module $M_{4,5}$ in TABLE
4.

###### Proof.

Let $\mathbb{Z}_{4}=\left\\{1,g,g^{2},g^{3}\right\\}$ be the cyclic group of
order $4$, and it has four non-isomorphic 1-dimensional irreducible
representations, denoted by $U_{i},\,i=0,1,2,3$. Let $U_{0}=1$ represent the
trivial representation and

$\chi_{U_{1}}(g)=\sqrt{-1},\quad\chi_{U_{2}}(g)=-1,\quad\chi_{U_{3}}(g)=-\sqrt{-1}.$

On the other hand, we consider $\mathbb{Z}_{4}$ as the subgroup of $S_{4}$
generated by $g=(1234)$. Then by the character table of $S_{4}$ (TABLE 1), we
have

$\displaystyle\chi_{\psi}(g^{i})=(-1)^{i},\quad\chi_{\rho_{1}}(g^{i})=1+(-1)^{i},\quad\chi_{\rho_{2}}(g^{i})=(-1)^{i}+(\sqrt{-1})^{i}+(-\sqrt{-1})^{i},$
$\displaystyle\chi_{\rho_{3}}(g^{i})=1+(\sqrt{-1})^{i}+(-\sqrt{-1})^{i}.$

So the restriction rule of $r(S_{4})$ on $r(\mathbb{Z}_{4})$ is given as
follows.

$\displaystyle{\rm Res}_{\mathbb{Z}_{4}}^{S_{4}}\left(1\right)=1,\quad{\rm
Res}_{\mathbb{Z}_{4}}^{S_{4}}\left(V_{\psi}\right)=U_{2},\quad{\rm
Res}_{\mathbb{Z}_{4}}^{S_{4}}\left(V_{\rho_{1}}\right)=1+U_{2},$
$\displaystyle{\rm
Res}_{\mathbb{Z}_{4}}^{S_{4}}\left(V_{\rho_{2}}\right)=U_{1}+U_{2}+U_{3},\quad{\rm
Res}_{\mathbb{Z}_{4}}^{S_{4}}\left(V_{\rho_{3}}\right)=1+U_{1}+U_{3}.$

Then we get the representation matrices of basis elements of $r(S_{4})$ acting
on $r(\mathbb{Z}_{4})$ as follows.

$\displaystyle 1\mapsto E_{4},\quad V_{\psi}\mapsto\begin{pmatrix}0&0&1&0\\\
0&0&0&1\\\ 1&0&0&0\\\ 0&1&0&0\end{pmatrix},\quad
V_{\rho_{1}}\mapsto\begin{pmatrix}1&0&1&0\\\ 0&1&0&1\\\ 1&0&1&0\\\
0&1&0&1\end{pmatrix},$ $\displaystyle
V_{\rho_{2}}\mapsto\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\ 1&1&0&1\\\
1&1&1&0\end{pmatrix},\quad V_{\rho_{3}}\mapsto\begin{pmatrix}1&1&0&1\\\
1&1&1&0\\\ 0&1&1&1\\\ 1&0&1&1\end{pmatrix}.$

Let $\left\\{w_{i}\right\\}_{1\leq i\leq 4}$ be the stated $\mathbb{Z}$-basis
of $M_{4,5}$, and define a $\mathbb{Z}$-linear map $\varphi:M_{4,5}\to
r(\mathbb{Z}_{4})$ by

$\varphi(w_{1})=U_{3},\quad\varphi(w_{2})=U_{1},\quad\varphi(w_{3})=U_{2},\quad\varphi(w_{4})=1.$

Then it is easy to check that $\varphi$ is an isomorphism of
$r(S_{4})$-modules, so $M_{4,5}\cong r(\mathbb{Z}_{4})$ as based modules by
Definition (v) (i). In other words, based module $M_{4,5}$ can be categorified
by the module category ${\rm Rep}(\mathbb{Z}_{4})$ over ${\rm Rep}(S_{4})$. ∎

######  Remark 4.7.

By the same argument as in Remark 4.5, $V_{\rho_{2}}$ and $V_{\rho_{3}}$ can
be required to satisfy the following restriction rules instead.

${\rm
Res}_{\mathbb{Z}_{4}}^{S_{4}}\left(V_{\rho_{2}}\right)=1+U_{1}+U_{3},\quad{\rm
Res}_{\mathbb{Z}_{4}}^{S_{4}}\left(V_{\rho_{3}}\right)=U_{1}+U_{2}+U_{3}.$

Therefore, we get another action of $r(S_{4})$ on $r({\mathbb{Z}_{4}})$ such
that $r(\mathbb{Z}_{4})$ is an irreducible based module over $r(S_{4})$
equivalent to the based module $M_{4,7}$ according to TABLE 4. In other words,
the based module $M_{4,7}$ can also be categorified by the module category
${\rm Rep}(\mathbb{Z}_{4})$ over ${\rm Rep}(S_{4})$.

Also, one can similarly check that the module category ${\rm
Rep}(\mathbb{Z}_{2})$ over ${\rm Rep}(S_{4})$ categorifies the based modules
$M_{2,2}$ and $M_{2,3}$, while ${\rm Rep}(\mathbb{Z}_{3})$ over ${\rm
Rep}(S_{4})$ categorifies the based module $M_{3,1}$.

iii Now we consider ${\rm Rep}(K_{4})$ as a module category over ${\rm
Rep}(S_{4})$.

######  Theorem 4.8.

$r(K_{4})={\rm Gr}({\rm Rep}(K_{4}))$ is an irreducible based module over
$r(S_{4})$ equivalent to the based module $M_{4,7}$ in TABLE 4.

###### Proof.

We consider $K_{4}$ as the subgroup of $S_{4}$ generated by $(12)$ and $(34)$,
and it has four non-isomorphic 1-dimensional irreducible representations
$Y_{0}=1$ and $Y_{1},Y_{2},Y_{3}$ such that

$\displaystyle\chi_{Y_{1}}((12))=-1,\quad\chi_{Y_{1}}((34))=1;\quad\chi_{Y_{2}}((12))=1,\quad\chi_{Y_{2}}((34))=-1;$
$\displaystyle\chi_{Y_{3}}((12))=-1,\quad\chi_{Y_{3}}((34))=-1.$

On the other hand, by the character table of $S_{4}$ (TABLE 1), we have

$\displaystyle\chi_{\psi}((12))=\chi_{\psi}((34))=-1,\quad\chi_{\psi}((12)(34))=1;$
$\displaystyle\chi_{\rho_{1}}((12))=\chi_{\rho_{1}}((34))=0,\quad\chi_{\rho_{1}}((12)(34))=2;$
$\displaystyle\chi_{\rho_{2}}((12))=\chi_{\rho_{2}}((34))=1,\quad\chi_{\rho_{2}}((12)(34))=-1;$
$\displaystyle\chi_{\rho_{3}}((12))=\chi_{\rho_{3}}((34))=-1,\quad\chi_{\rho_{3}}((12)(34))=-1.$

So we have the following restriction rules.

$\displaystyle{\rm Res}_{K_{4}}^{S_{4}}\left(1\right)=1,\quad{\rm
Res}_{K_{4}}^{S_{4}}\left(V_{\psi}\right)=Y_{3},\quad{\rm
Res}_{K_{4}}^{S_{4}}\left(V_{\rho_{1}}\right)=1+Y_{3},$ $\displaystyle{\rm
Res}_{K_{4}}^{S_{4}}\left(V_{\rho_{2}}\right)=1+Y_{1}+Y_{2},\quad{\rm
Res}_{K_{4}}^{S_{4}}\left(V_{\rho_{3}}\right)=Y_{1}+Y_{2}+Y_{3}.$

Then, we get the representation matrices of basis elements of $r(S_{4})$
acting on $r(K_{4})$ as follows.

$\displaystyle 1\mapsto E_{4},\quad V_{\psi}\mapsto\begin{pmatrix}0&0&0&1\\\
0&0&1&0\\\ 0&1&0&0\\\ 1&0&0&0\end{pmatrix},\quad
V_{\rho_{1}}\mapsto\begin{pmatrix}1&0&0&1\\\ 0&1&1&0\\\ 0&1&1&0\\\
1&0&0&1\end{pmatrix},$ $\displaystyle
V_{\rho_{2}}\mapsto\begin{pmatrix}1&1&1&0\\\ 1&1&0&1\\\ 1&0&1&1\\\
0&1&1&1\end{pmatrix},\quad V_{\rho_{3}}\mapsto\begin{pmatrix}0&1&1&1\\\
1&0&1&1\\\ 1&1&0&1\\\ 1&1&1&0\end{pmatrix}.$

Let $\left\\{w_{i}\right\\}_{1\leq i\leq 4}$ be the stated $\mathbb{Z}$-basis
of $M_{4,7}$ listed in TABLE 4, then

$w_{1}\mapsto Y_{2},\quad w_{2}\mapsto Y_{1},\quad w_{3}\mapsto 1,\quad
w_{4}\mapsto Y_{3},$

defines an equivalence of $\mathbb{Z}_{+}$-modules between $M_{4,7}$ and
$r(K_{4})$. In other words, the irreducible based module $M_{4,7}$ can be
categorified by the module category ${\rm Rep}(K_{4})$ over ${\rm
Rep}(S_{4})$. ∎

######  Remark 4.9.

In a manner analogous to the argument in Remark 4.5, it follows that the
irreducible based module $M_{4,5}$ can also be categorified by the module
category ${\rm Rep}(K_{4})$ over ${\rm Rep}(S_{4})$.

iv We consider ${\rm Rep}(A_{4})$ as a module category over ${\rm
Rep}(S_{4})$.

######  Theorem 4.10.

$r(A_{4})={\rm Gr}({\rm Rep}(A_{4}))$ is an irreducible based module over
$r(S_{4})$ equivalent to the based module $M_{4,1}$ in TABLE 4.

###### Proof.

We know that $A_{4}$ has three non-isomorphic 1-dimensional irreducible
representations and one 3-dimensional irreducible representation, denoted by
$N_{0},N_{1},N_{2}$ and $N_{3}$ respectively, where $N_{0}=1$ represents the
trivial representation, and

$\displaystyle\chi_{N_{1}}((123))=\omega,\quad\chi_{N_{1}}((12)(34))=1;\quad\chi_{N_{2}}((123))=\omega^{2},\quad\chi_{N_{2}}((12)(34))=1;$
$\displaystyle\chi_{N_{3}}((123))=\chi_{N_{3}}((132))=0,\quad\chi_{N_{3}}((12)(34))=-1,\quad\omega=\frac{-1+\sqrt{-3}}{2}.$

On the other hand, the character table of $S_{4}$ (TABLE 1) tells us that

$\displaystyle\chi_{\psi}((123))=1,\quad\chi_{\psi}((12)(34))=1;\quad\chi_{\rho_{1}}((123))=-1,\quad\chi_{\rho_{1}}((12)(34))=2;$
$\displaystyle\chi_{\rho_{2}}((123))=0,\quad\chi_{\rho_{2}}((12)(34))=-1;\quad\chi_{\rho_{3}}((123))=0,\quad\chi_{\rho_{3}}((12)(34))=-1.$

So we have the following restriction rules.

$\displaystyle{\rm Res}_{A_{4}}^{S_{4}}\left(1\right)={\rm
Res}_{A_{4}}^{S_{4}}\left(V_{\psi}\right)=1,\quad{\rm
Res}_{A_{4}}^{S_{4}}\left(V_{\rho_{1}}\right)=N_{1}+N_{2},\quad{\rm
Res}_{A_{4}}^{S_{4}}\left(V_{\rho_{2}}\right)={\rm
Res}_{A_{4}}^{S_{4}}\left(V_{\rho_{3}}\right)=N_{3}.$

Hence, we get the representation matrices of basis elements of $r(S_{4})$
acting on $r(A_{4})$ as follows.

$\displaystyle 1\mapsto E_{4},\quad V_{\psi}\mapsto E_{4},\quad
V_{\rho_{1}}\mapsto\begin{pmatrix}0&1&1&0\\\ 1&0&1&0\\\ 1&1&0&0\\\
0&0&0&2\end{pmatrix},\quad
V_{\rho_{2}},V_{\rho_{3}}\mapsto\begin{pmatrix}0&0&0&1\\\ 0&0&0&1\\\
0&0&0&1\\\ 1&1&1&2\end{pmatrix}.$

Then $r(A_{4})$ is an irreducible based module over $r(S_{4})$ equivalent to
$M_{4,1}$ listed in TABLE 4. In other words, the irreducible based module
$M_{4,1}$ can be categorified by the module category ${\rm Rep}(A_{4})$ over
${\rm Rep}(S_{4})$. ∎

v Next we consider ${\rm Rep}(D_{4})$ as a module category over ${\rm
Rep}(S_{4})$.

######  Theorem 4.11.

$r(D_{4})={\rm Gr}({\rm Rep}(D_{4}))$ is an irreducible based module over
$r(S_{4})$ equivalent to the based module $M_{5,2}$ in TABLE 5.

###### Proof.

The dihedral group $D_{4}=\langle r,s\,|\,r^{4}=s^{2}=(rs)^{2}=1\rangle$ has
four 1-dimensional irreducible representations and one 2-dimensional
irreducible representation up to isomorphism, denoted by
$W_{0},W_{1},W_{2},W_{3}$ and $W_{4}$ respectively. Let $W_{0}=1$ stand for
the trivial representation and

$\displaystyle\chi_{W_{1}}(r)=1,\quad\chi_{W_{1}}(s)=-1;\quad\chi_{W_{2}}(r)=-1,\quad\chi_{W_{2}}(s)=1;$
$\displaystyle\chi_{W_{3}}(r)=-1,\quad\chi_{W_{3}}(s)=-1;\quad\chi_{W_{4}}(r)=\chi_{W_{4}}(s)=\chi_{W_{4}}(rs)=0.$

On the other hand, we consider $D_{4}$ as the subgroup of $S_{4}$ by taking
$r=(1234)$ and $s=(12)(34)$. Then $rs=(13)$. By the character table of $S_{4}$
(TABLE 1), we have

$\displaystyle\chi_{\psi}((1234))=-1,\quad\chi_{\psi}((12)(34))=1,\quad\chi_{\psi}((13))=-1;$
$\displaystyle\chi_{\rho_{1}}((1234))=0,\quad\chi_{\rho_{1}}((12)(34))=2,\quad\chi_{\rho_{1}}((13))=0;$
$\displaystyle\chi_{\rho_{2}}((1234))=-1,\quad\chi_{\rho_{2}}((12)(34))=-1,\quad\chi_{\rho_{2}}((13))=1;$
$\displaystyle\chi_{\rho_{3}}((1234))=1,\quad\chi_{\rho_{3}}((12)(34))=-1,\quad\chi_{\rho_{3}}((13))=-1.$

So we have the following restriction rules.

$\displaystyle{\rm Res}_{D_{4}}^{S_{4}}\left(1\right)=1,\quad{\rm
Res}_{D_{4}}^{S_{4}}\left(V_{\psi}\right)=W_{2},\quad{\rm
Res}_{D_{4}}^{S_{4}}\left(V_{\rho_{1}}\right)=1+W_{2},$ $\displaystyle{\rm
Res}_{D_{4}}^{S_{4}}\left(V_{\rho_{2}}\right)=W_{3}+W_{4},\quad{\rm
Res}_{D_{4}}^{S_{4}}\left(V_{\rho_{3}}\right)=W_{1}+W_{4}.$

Then, we get the representation matrices of basis elements of $r(S_{4})$
acting on $r(D_{4})$ as follows.

$\displaystyle 1\mapsto E_{5},\quad V_{\psi}\mapsto\begin{pmatrix}0&0&1&0&0\\\
0&0&0&1&0\\\ 1&0&0&0&0\\\ 0&1&0&0&0\\\ 0&0&0&0&1\end{pmatrix},\quad
V_{\rho_{1}}\mapsto\begin{pmatrix}1&0&1&0&0\\\ 0&1&0&1&0\\\ 1&0&1&0&0\\\
0&1&0&1&0\\\ 0&0&0&0&2\end{pmatrix},$ $\displaystyle
V_{\rho_{2}}\mapsto\begin{pmatrix}0&0&0&1&1\\\ 0&0&1&0&1\\\ 0&1&0&0&1\\\
1&0&0&0&1\\\ 1&1&1&1&1\end{pmatrix},\quad
V_{\rho_{3}}\mapsto\begin{pmatrix}0&1&0&0&1\\\ 1&0&0&0&1\\\ 0&0&0&1&1\\\
0&0&1&0&1\\\ 1&1&1&1&1\end{pmatrix}.$

Then $r(D_{4})$ is an irreducible based module over $r(S_{4})$ equivalent to
$M_{5,2}$ listed in TABLE 5. In other words, the irreducible based module
$M_{5,2}$ can be categorified by the module category ${\rm Rep}(D_{4})$ over
${\rm Rep}(S_{4})$. ∎

vi Finally, we consider ${\rm Rep}(S_{4})$ as a module category over itself.

######  Theorem 4.12.

The regular $\mathbb{Z}_{+}$-module $r(S_{4})$ over itself is equivalent to
the irreducible based module $M_{5,1}$ in TABLE 5.

###### Proof.

Let $r(S_{4})$ be the regular $\mathbb{Z}_{+}$-module over itself with the
$\mathbb{Z}$-basis
$\left\\{1,V_{\psi},V_{\rho_{1}},V_{\rho_{2}},V_{\rho_{3}}\right\\}$, and the
action of $r(S_{4})$ on it is given as follows.

$\displaystyle 1\mapsto E_{5},\quad V_{\psi}\mapsto\begin{pmatrix}0&1&0&0&0\\\
1&0&0&0&0\\\ 0&0&1&0&0\\\ 0&0&0&0&1\\\ 0&0&0&1&0\end{pmatrix},\quad
V_{\rho_{1}}\mapsto\begin{pmatrix}0&0&1&0&0\\\ 0&0&1&0&0\\\ 1&1&1&0&0\\\
0&0&0&1&1\\\ 0&0&0&1&1\end{pmatrix},$ $\displaystyle
V_{\rho_{2}}\mapsto\begin{pmatrix}0&0&0&1&0\\\ 0&0&0&0&1\\\ 0&0&0&1&1\\\
1&0&1&1&1\\\ 0&1&1&1&1\end{pmatrix},\quad
V_{\rho_{3}}\mapsto\begin{pmatrix}0&0&0&0&1\\\ 0&0&0&1&0\\\ 0&0&0&1&1\\\
0&1&1&1&1\\\ 1&0&1&1&1\end{pmatrix}.$

Then the regular $\mathbb{Z}_{+}$-module $r(S_{4})$ over itself is equivalent
to $M_{5,1}$ listed in TABLE 5. In other words, the irreducible based module
$M_{5,1}$ over $r(S_{4})$ can be categorified by the module category ${\rm
Rep}(S_{4})$ over itself. ∎

######  Remark 4.13.

Following the argument presented in Remark 4.5, if we exchange the notations
$V_{\rho_{2}}$ and $V_{\rho_{3}}$ with their restriction rules given in the
proof of Theorem 4.11 and Theorem 4.12, we see that $r(D_{4})$ and $r(S_{4})$
are still equivalent to $M_{5,2}$ and $M_{5,1}$ respectively.

###  4.2. The module categories over ${\rm Rep}(S_{4})$ with nontrivial twists

Lastly, we consider the module category ${\rm Rep}(H,\alpha)$ over ${\rm
Rep}(S_{4})$, where $H$ is a subgroup of $S_{4}$ with $\alpha$ representing
the unique nontrivial cohomological class in $H^{2}(H,{\mathbb{C}}^{*})$. All
non-isomorphic irreducible projective representations of $H$ with the
multiplier $\alpha$ form a $\mathbb{Z}$-basis of $r(H,\alpha)$, whose
cardinality is the number of $\alpha$-regular conjugacy classes by [26,
Theorem 6.1.1].

Firstly, we consider the twisted group algebra of $K_{4}$. There is only one
irreducible projective representation with respect to $\alpha$ up to
isomorphism, see e.g. [35, Appendix D.1]. Hence, $r(K_{4},\alpha)$ is a based
module of rank $1$ over $r(S_{4})$ equivalent to $M_{1,1}$ defined in (8).
Namely, the based module $M_{1,1}$ can also be categorified by ${\rm
Rep}(K_{4},\alpha)$.

Secondly, we consider the twisted group algebra of $D_{4}$.

######  Theorem 4.14.

$r(D_{4},\alpha)={\rm Gr}({\rm Rep}(D_{4},\alpha))$ is an irreducible based
module over $r(S_{4})$ equivalent to the based module $M_{2,3}$ in TABLE 2.

###### Proof.

Let $D_{4}=\langle r,s\,|\,r^{4}=s^{2}=(rs)^{2}=1\rangle$. Let $\alpha\in
Z^{2}(D_{4},{\mathbb{C}}^{*})$ be the 2-cocycle defined by

$\alpha(r^{i}s^{j},r^{i^{\prime}}s^{j^{\prime}})=(\sqrt{-1})^{ji^{\prime}}.$

Here
$i,i^{\prime}\in{\left\\{0,1,2,3\right\\}},\,j,j^{\prime}\in\left\\{{0,1}\right\\}$
. As shown in [27, Section 3.7], this is a unitary 2-cocycle representing the
unique non-trivial cohomological class in $H^{2}(D_{4},{\mathbb{C}}^{*})$.
According to [9, Section 3], there exist two (2-dimensional) non-isomorphic
irreducible projective representations of $D_{4}$ with respect to $\alpha$,
which are given by

$\displaystyle\pi_{l}:D_{4}$
$\displaystyle\rightarrow\mathrm{GL}_{2}(\mathbb{C})$ $\displaystyle
r^{i}s^{j}$ $\displaystyle\mapsto A_{l}^{i}B^{j},$

where $A_{l}=\begin{pmatrix}(\sqrt{-1})^{l}&0\\\
0&(\sqrt{-1})^{1-l}\end{pmatrix},B=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$,
$l=1,2$. Also, for irreducible representations $W_{0},W_{1},W_{2},W_{3}$ and
$W_{4}$ of $D_{4}$ mentioned in the proof of Theorem 4.11, we have

$W_{0}\otimes\pi_{l}=W_{1}\otimes\pi_{l}=\pi_{l},\quad
W_{2}\otimes\pi_{l}=W_{3}\otimes\pi_{l}=\pi_{3-l},\quad
W_{4}\otimes\pi_{l}=\pi_{1}+\pi_{2}.$

Next, using the previous restriction rule of $r(S_{4})$ on $r(D_{4})$, we get
the representation matrices of basis elements of $r(S_{4})$ acting on
$r(D_{4},\alpha)$ as follows.

$\displaystyle 1\mapsto E_{2},\quad V_{\psi}\mapsto\begin{pmatrix}0&1\\\
1&0\end{pmatrix},\quad V_{\rho_{1}}\mapsto\begin{pmatrix}1&1\\\
1&1\end{pmatrix},\quad V_{\rho_{2}}\mapsto\begin{pmatrix}1&2\\\
2&1\end{pmatrix},\quad V_{\rho_{3}}\mapsto\begin{pmatrix}2&1\\\
1&2\end{pmatrix}.$

Then $r(D_{4},\alpha)$ is an irreducible based module over $r(S_{4})$
equivalent to $M_{2,3}$ listed in TABLE 2. In other words, the irreducible
based module $M_{2,3}$ can be categorified by the module category ${\rm
Rep}(D_{4},\alpha)$ over ${\rm Rep}(S_{4})$. ∎

######  Remark 4.15.

As discussed in Remark 4.5, it follows that the irreducible based module
$M_{2,2}$ can also be categorified by the module category ${\rm
Rep}(D_{4},\alpha)$ over ${\rm Rep}(S_{4})$.

Next we consider the twisted group algebras of $A_{4}$ and $S_{4}$. By [26,
Theorem 6.1.1], $A_{4}$ has three (2-dimensional) non-isomorphic irreducible
projective representations, denoted as $V_{\gamma_{1}},V_{\gamma_{2}}$ and
$V_{\gamma_{3}}$ respectively. Similarly, $S_{4}$ has two (2-dimensional) non-
isomorphic irreducible projective representations $V_{\xi_{1}},V_{\xi_{2}}$
and one (4-dimensional) irreducible projective representation $V_{\xi_{3}}$.
We give the character table for projective representations of $A_{4}$ and
$S_{4}$ as follows, where primes are used to differentiate between the two
classes splitting from a single conjugacy class of $A_{4}$ in its double cover
$\tilde{A}_{4}$, and the same applies to $S_{4}$; subscripts distinguish
between the two classes splitting from the conjugacy classes
$(31)^{{}^{\prime}}$ and $(31)^{{}^{\prime\prime}}$ in the double cover
$\tilde{S}_{4}$ of $S_{4}$ respectively. For more details, see [20, Chapter
4].

|  $(1^{4})^{{}^{\prime}}$ |  $(1^{4})^{{}^{\prime\prime}}$ |  $(2^{2})$ |  $(31)^{{}^{\prime}}_{1}$ |  $(31)^{{}^{\prime\prime}}_{1}$ |  $(31)^{{}^{\prime}}_{2}$ |  $(31)^{{}^{\prime\prime}}_{2}$
---|---|---|---|---|---|---|---
$\chi_{\gamma_{1}}$ |  $2$ |  $-2$ |  $0$ |  $1$ |  $-1$ |  $1$ |  $-1$
$\chi_{\gamma_{2}}$ |  $2$ |  $-2$ |  $0$ |  $\omega$ |  $-\omega$ |  $\omega^{2}$ |  $-\omega^{2}$
$\chi_{\gamma_{3}}$ |  $2$ |  $-2$ |  $0$ |  $\omega^{2}$ |  $-\omega^{2}$ |  $\omega$ |  $-\omega$

Table 6. The character table for irreducible projective representations of
$A_{4}$

|  $(1^{4})^{{}^{\prime}}$ |  $(1^{4})^{{}^{\prime\prime}}$ |  $(21^{2})$ |  $(2^{2})$ |  $(31)^{{}^{\prime}}$ |  $(31)^{{}^{\prime\prime}}$ |  $(4)^{{}^{\prime}}$ |  $(4)^{{}^{\prime\prime}}$
---|---|---|---|---|---|---|---|---
$\chi_{\xi_{1}}$ |  $2$ |  $-2$ |  $0$ |  $0$ |  $1$ |  $-1$ |  $\sqrt{2}$ |  $-\sqrt{2}$
$\chi_{\xi_{2}}$ |  $2$ |  $-2$ |  $0$ |  $0$ |  $1$ |  $-1$ |  $-\sqrt{2}$ |  $\sqrt{2}$
$\chi_{\xi_{3}}$ |  $4$ |  $-4$ |  $0$ |  $0$ |  $-1$ |  $1$ |  $0$ |  $0$

Table 7. The character table for irreducible projective representations of
$S_{4}$

In TABLE 6 we denote $\omega=e^{2\pi\sqrt{-1}/3}=\frac{-1+\sqrt{-3}}{2}$. Then
we have the following theorems.

######  Theorem 4.16.

$r(A_{4},\alpha)={\rm Gr}({\rm Rep}(A_{4},\alpha))$ is an irreducible based
module over $r(S_{4})$ equivalent to the based module $M_{3,1}$ in TABLE 3.

###### Proof.

For the irreducible representations $N_{0},N_{1},N_{2}$ and $N_{3}$ of $A_{4}$
mentioned in the proof of Theorem 4.10, we obtain the following tensor product
rule in $r^{\alpha}(A_{4})$ by computing the values of products of characters,

$\displaystyle N_{0}\otimes V_{\gamma_{i}}=V_{\gamma_{i}},\quad N_{1}\otimes
V_{\gamma_{j}}=V_{\gamma_{j+1}},\quad N_{1}\otimes
V_{\gamma_{3}}=V_{\gamma_{1}};$ $\displaystyle N_{2}\otimes
V_{\gamma_{1}}=V_{\gamma_{3}},\quad N_{2}\otimes
V_{\gamma_{2}}=V_{\gamma_{1}},\quad N_{2}\otimes
V_{\gamma_{3}}=V_{\gamma_{2}};$ $\displaystyle N_{3}\otimes
V_{\gamma_{i}}=V_{\gamma_{1}}+V_{\gamma_{2}}+V_{\gamma_{3}},$

where $i=1,2,3$, $j=1,2$. Next, by combining this with the previous
restriction rule of $r(S_{4})$ on $r(A_{4})$, we obtain

$1,V_{\psi}\mapsto E_{3},\quad V_{\rho_{1}}\mapsto\begin{pmatrix}0&1&1\\\
1&0&1\\\ 1&1&0\end{pmatrix},\quad
V_{\rho_{2}},V_{\rho_{3}}\mapsto\begin{pmatrix}1&1&1\\\ 1&1&1\\\
1&1&1\end{pmatrix}.$

Then $r(A_{4},\alpha)$ is an irreducible based module over $r(S_{4})$
equivalent to $M_{3,1}$ listed in TABLE 3. In other words, the irreducible
based module $M_{3,1}$ can be categorified by ${\rm Rep}(A_{4},\alpha)$. ∎

######  Theorem 4.17.

$r(S_{4},\alpha)={\rm Gr}({\rm Rep}(S_{4},\alpha))$ is an irreducible based
module over $r(S_{4})$ equivalent to the based module $M_{3,3}$ in TABLE 3.

###### Proof.

Let $\alpha$ be a nontrivial 2-cocycle in $Z^{2}(S_{4},{\mathbb{C}}^{*})$ (see
e.g. [34, Section 3.2.4]). By checking products of characters, we get the
following tensor product rule in $r^{\alpha}(S_{4})$.

$\displaystyle 1\otimes V_{\xi_{i}}=V_{\xi_{i}},\quad V_{\psi}\otimes
V_{\xi_{j}}=V_{\xi_{3-j}},\quad V_{\psi}\otimes V_{\xi_{3}}=V_{\xi_{3}};$
$\displaystyle V_{\rho_{1}}\otimes V_{\xi_{j}}=V_{\xi_{3}},\quad
V_{\rho_{1}}\otimes V_{\xi_{3}}=V_{\xi_{1}}+V_{\xi_{2}}+V_{\xi_{3}};$
$\displaystyle V_{\rho_{2}}\otimes V_{\xi_{j}}=V_{\xi_{3-j}}+V_{\xi_{3}},\quad
V_{\rho_{2}}\otimes V_{\xi_{3}}=V_{\rho_{3}}\otimes
V_{\xi_{3}}=V_{\xi_{1}}+V_{\xi_{2}}+2V_{\xi_{3}};$ $\displaystyle
V_{\rho_{3}}\otimes V_{\xi_{j}}=V_{\xi_{j}}+V_{\xi_{3}};$

where $i=1,2,3$, $j=1,2$. Thus, we get

$1\mapsto E_{3},\quad V_{\psi}\mapsto\begin{pmatrix}0&1&0\\\ 1&0&0\\\
0&0&1\end{pmatrix},\quad V_{\rho_{1}}\mapsto\begin{pmatrix}0&0&1\\\ 0&0&1\\\
1&1&1\end{pmatrix},\quad V_{\rho_{2}}\mapsto\begin{pmatrix}0&1&1\\\ 1&0&1\\\
1&1&2\end{pmatrix},\quad V_{\rho_{3}}\mapsto\begin{pmatrix}1&0&1\\\ 0&1&1\\\
1&1&2\end{pmatrix}.$

Then $r(S_{4},\alpha)$ is an irreducible based module over $r(S_{4})$
equivalent to $M_{3,3}$ listed in TABLE 3. In other words, the irreducible
based module $M_{3,3}$ over $r(S_{4})$ can be categorified by ${\rm
Rep}(S_{4},\alpha)$.∎

In summary, we have the following classification theorem.

######  Theorem 4.18.

The inequivalent irreducible based modules over $r(S_{4})$ are $M_{1,1}$,
${\left\\{M_{2,i}\right\\}}_{i=1,2,3}$,
${\left\\{M_{3,j}\right\\}}_{j=1,2,3}$, ${\left\\{M_{4,s}\right\\}}_{1\leq
s\leq 7}$ and ${\left\\{M_{5,t}\right\\}}_{t=1,2}$, among which $M_{1,1}$,
${\left\\{M_{2,i}\right\\}}_{i=2,3}$, ${\left\\{M_{3,j}\right\\}}_{j=1,2,3}$,
${\left\\{M_{4,s}\right\\}}_{s=1,5,7}$ and
${\left\\{M_{5,t}\right\\}}_{t=1,2}$ can be categorified by module categories
over ${\rm Rep}(S_{4})$; see TABLE 8.

|  |  $V_{\psi}$ |  $V_{\rho_{1}}$ |  $V_{\rho_{2}}$ |  categorification
---|---|---|---|---|---
Rank 1 |  $M_{1,1}$ |  $1$ |  $2$ |  $3$ |  ${\rm Rep}(\mathbb{Z}_{1})$, ${\rm Rep}(K_{4},\alpha)$
Rank 2 |  $M_{2,1}$ |  $\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$ |  $\begin{pmatrix}2&0\\\ 0&2\end{pmatrix}$ |  $\begin{pmatrix}1&2\\\ 2&1\end{pmatrix}$ |  No
|  $M_{2,2}$ |  $\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$ |  $\begin{pmatrix}1&1\\\ 1&1\end{pmatrix}$ |  $\begin{pmatrix}2&1\\\ 1&2\end{pmatrix}$ |  ${\rm Rep}(\mathbb{Z}_{2})$, ${\rm Rep}(D_{4},\alpha)$
|  $M_{2,3}$ |  $\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$ |  $\begin{pmatrix}1&1\\\ 1&1\end{pmatrix}$ |  $\begin{pmatrix}1&2\\\ 2&1\end{pmatrix}$ |  ${\rm Rep}(\mathbb{Z}_{2})$, ${\rm Rep}(D_{4},\alpha)$
Rank 3 |  $M_{3,1}$ |  $\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&1\\\ 1&0&1\\\ 1&1&0\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&1\\\ 1&1&1\\\ 1&1&1\\\ \end{pmatrix}$ |  ${\rm Rep}(\mathbb{Z}_{3})$, ${\rm Rep}(A_{4},\alpha)$
|  $M_{3,2}$ |  $\begin{pmatrix}0&1&0\\\ 1&0&0\\\ 0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&0&1\\\ 0&0&1\\\ 1&1&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&0&1\\\ 0&1&1\\\ 1&1&2\\\ \end{pmatrix}$ |  ${\rm Rep}(S_{3})$
|  $M_{3,3}$ |  $\begin{pmatrix}0&1&0\\\ 1&0&0\\\ 0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&0&1\\\ 0&0&1\\\ 1&1&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&1\\\ 1&0&1\\\ 1&1&2\\\ \end{pmatrix}$ |  ${\rm Rep}(S_{3})$, ${\rm Rep}(S_{4},\alpha)$
Rank 4 |  $M_{4,1}$ |  $\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&0&1&1\\\ 0&2&0&0\\\ 1&0&0&1\\\ 1&0&1&0\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&0&0\\\ 1&2&1&1\\\ 0&1&0&0\\\ 0&1&0&0\\\ \end{pmatrix}$ |  ${\rm Rep}(A_{4})$
|  $M_{4,2}$ |  $\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}2&0&0&0\\\ 0&2&0&0\\\ 0&0&2&0\\\ 0&0&0&2\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\ 1&1&0&1\\\ 1&1&1&0\\\ \end{pmatrix}$ |  No
|  $M_{4,3}$ |  $\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&0&0\\\ 1&1&0&0\\\ 0&0&2&0\\\ 0&0&0&2\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\ 1&1&0&1\\\ 1&1&1&0\\\ \end{pmatrix}$ |  No
|  $M_{4,4}$ |  $\begin{pmatrix}0&1&0&0\\\ 1&0&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&0&0\\\ 1&1&0&0\\\ 0&0&2&0\\\ 0&0&0&2\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&0&1&1\\\ 0&1&1&1\\\ 1&1&0&1\\\ 1&1&1&0\\\ \end{pmatrix}$ |  No
|  $M_{4,5}$ |  $\begin{pmatrix}0&1&0&0\\\ 1&0&0&0\\\ 0&0&0&1\\\ 0&0&1&0\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&0&0\\\ 1&1&0&0\\\ 0&0&1&1\\\ 0&0&1&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\ 1&1&0&1\\\ 1&1&1&0\\\ \end{pmatrix}$ |  ${\rm Rep}(\mathbb{Z}_{4}),\,{\rm Rep}(K_{4})$
|  $M_{4,6}$ |  $\begin{pmatrix}0&1&0&0\\\ 1&0&0&0\\\ 0&0&0&1\\\ 0&0&1&0\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&0&0\\\ 1&1&0&0\\\ 0&0&1&1\\\ 0&0&1&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&1&1&1\\\ 1&0&1&1\\\ 1&1&1&0\\\ 1&1&0&1\\\ \end{pmatrix}$ |  No
|  $M_{4,7}$ |  $\begin{pmatrix}0&1&0&0\\\ 1&0&0&0\\\ 0&0&0&1\\\ 0&0&1&0\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&0&0\\\ 1&1&0&0\\\ 0&0&1&1\\\ 0&0&1&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&0&1&1\\\ 0&1&1&1\\\ 1&1&1&0\\\ 1&1&0&1\\\ \end{pmatrix}$ |  ${\rm Rep}(\mathbb{Z}_{4}),\,{\rm Rep}(K_{4})$
Rank 5 |  $M_{5,1}$ |  $\begin{pmatrix}0&1&0&0&0\\\ 1&0&0&0&0\\\ 0&0&0&1&0\\\ 0&0&1&0&0\\\ 0&0&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&0&0&0&1\\\ 0&0&0&0&1\\\ 0&0&1&1&0\\\ 0&0&1&1&0\\\ 1&1&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&0&0&1&0\\\ 0&0&1&0&0\\\ 0&1&1&1&1\\\ 1&0&1&1&1\\\ 0&0&1&1&0\\\ \end{pmatrix}$ |  ${\rm Rep}(S_{4})$
|  $M_{5,2}$ |  $\begin{pmatrix}0&1&0&0&0\\\ 1&0&0&0&0\\\ 0&0&0&1&0\\\ 0&0&1&0&0\\\ 0&0&0&0&1\\\ \end{pmatrix}$ |  $\begin{pmatrix}1&1&0&0&0\\\ 1&1&0&0&0\\\ 0&0&1&1&0\\\ 0&0&1&1&0\\\ 0&0&0&0&2\\\ \end{pmatrix}$ |  $\begin{pmatrix}0&0&0&1&1\\\ 0&0&1&0&1\\\ 0&1&0&0&1\\\ 1&0&0&0&1\\\ 1&1&1&1&1\\\ \end{pmatrix}$ |  ${\rm Rep}(D_{4})$

Table 8. Inequivalent irreducible based modules over $r(S_{4})$

Open problem. Classifying irreducible $\mathbb{Z}_{+}$-modules over $r(S_{4})$
is more challenging, especially for high-rank cases, and it might be
considered in a sequel of this paper. Additionally, the existence of any
irreducible based module of rank $\geq$ 6 over $r(S_{4})$ warrants further
investigation.

Acknowledgments. We would like to thank Zhiqiang Yu for helpful discussion.
This work is supported by Natural Science Foundation of China (12071094,
12171155) and Guangdong Basic and Applied Basic Research Foundation
(2022A1515010357).

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