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# Refining the grading of irreducible Lie colour algebra representations

Mitchell Ryan  School of Mathematics and Physics, University of Queensland,
St. Lucia, QLD 4072, Australia<EMAIL_ADDRESS>
###### Abstract.

We define the refining extension of representations for Lie colour algebras
which simultaneously extends the representations and refines the grading to a
larger grading group. We show that the refining extension provides a general
method for deriving all finite-dimensional irreducible Lie colour algebra
representations from those for Lie superalgebras. This method yields a
bijection between equivalence classes but, despite this, Lie colour algebras
maintain a non-trivial representation theory distinct from that of Lie
superalgebras. We expect this result to be useful for the many applications of
Lie colour algebras that make use of irreducible representations. In addition,
we show that the refining extension provides a natural way to construct the
Hilbert space realisation for colour algebra quantum-mechanical models by
giving an example that has previously appeared in the literature.

###### Key words and phrases:

Color Lie (super)algebras, Graded Lie (super)algebras, irreducible
representations

###### 2020 Mathematics Subject Classification:

17B75, 17B70, 17B10

##  1\. Introduction

Lie colour algebras are a generalisation of Lie superalgebras to grading by
any abelian group $\Gamma$ (rather than just $\mathbb{Z}_{2}$) and were
introduced by Rittenberg and Wyler [1, 2] in 1978 (however, see also [3]). Lie
colour algebras initially found applications in (for example): de Sitter
spaces [4, 5, 6], quasispin [7], strings [8] and extensions of Poincaré
algebras [9, 10]. There has been a recent increase in activity surrounding
applications of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded Lie colour
algebras. This recent activity has mainly been focussed on parastatistics [11,
12, 13, 14, 15, 16, 17, 18, 19] and
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded quantum mechanics [20, 21, 22, 23,
24, 25, 26, 27, 28, 29, 30].

Many Lie colour algebra applications make use of irreducible representations
to construct various models. A classification of such representations would
therefore be useful. Indeed, in [31], the authors conclude their paper with a
desire for a classification of all irreducible representations for the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-supersymmetry algebra with
$\mathcal{N}\geq 2$. Such a classification motivates the results of this
paper.

Various classifications of the algebras themselves under different assumptions
have already been performed [32, 33, 34, 35]. Moreover, the representation
theories of the colour versions of the following algebras have been examined:
$\mathfrak{gl}$, $\mathfrak{sl}$, $\mathfrak{osp}$ [36, 37], the Lie algebra
for the group of plane motions [38], the Heisenberg Lie algebra [39], super
Schrödinger algebras [40] and a supersymmetry algebra [31].

One of the most powerful techniques in the study of Lie colour algebras is
that of discolouration, which provides a bijection between Lie colour algebras
and $\Gamma$-graded Lie superalgebras. This bijection preserves much of the
important structure, such as subalgebras, ideals and subrepresentations.
Discolouration, in the specific case of $\mathbb{Z}_{2}^{n}$-gradings, was
introduced concurrently with the algebras themselves [1] and was soon
generalised [41] to grading by any finitely generated abelian group.

Discolouration was reformulated in [42], using so-called Klein operators lying
outside the universal enveloping algebra. It has since been shown that these
Klein operators can be expressed in an algebraic extension of the universal
enveloping algebras for a colour version of $\mathfrak{gl}$ [43] or for the
non-coloured supersymmetric $\mathfrak{gl}(m|n)$ [44] and that Klein operators
appear in the $q\to-1$ limit of $U_{q}(\mathfrak{sl}(3))$ [45]. This version
of discolouration has already been applied to the study of the representation
theory. For example, in [44] it is shown how the covariant representations of
$\mathfrak{gl}(m|n)$ can be lifted to the colour version of $\mathfrak{gl}$.

Despite widespread awareness of the results of [41], discolouration has rarely
been utilised in the recent literature surrounding applications of
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded Lie colour algebras. Perhaps a
partial explanation lies with the following nuance: the discolouration of a
Lie colour algebra representation is a $\Gamma$-graded representation, whose
representation theory can be quite different from the more familiar
$\mathbb{Z}_{2}$-graded representation theory of Lie superalgebras. For
example, an irreducible $\Gamma$-graded representation may become reducible
when considering only the $\mathbb{Z}_{2}$-grade.

To address this nuance, we introduce the notion of a refining extension which
creates a $\Gamma$-graded representation as a vector space direct sum of
multiple copies of a $\Gamma/H$-graded representation (for some finite simple
subgroup $H\leq\Gamma$). The refining extension provides a natural
correspondence between $\Gamma/H$\- and $\Gamma$-graded representations.

The refining extension is a natural structure to consider in the study of Lie
colour algebras. For example, it can be used to construct the Hilbert space
realisation for a $\Gamma$-graded supersymmetric quantum mechanical model. As
an example, we show how the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded
Hilbert space realisation in [22] can be constructed from a modified version
of Witten’s model [46] using the refining extension.

In some sense, the refining extension also preserves irreducibility: for a
finite-dimensional irreducible $\Gamma/H$-graded module $V$, either $V$ or its
refining extension is an irreducible $\Gamma$-graded module. This provides a
method to derive every irreducible finite-dimensional $\Gamma$-graded
representation from the corresponding $\mathbb{Z}_{2}$-graded representations.
This derivation is more complicated than discolouration, and allows for
interesting differences in the representation theories. In particular, we
obtain bijections $\mathcal{F}_{1}$ and $\mathcal{F}_{2}$ that map between
equivalence classes (with equivalence relations stronger than isomorphism) of
irreducible representations, as per the following diagram.

$\mathbb{Z}_{2}$-graded superalgebra irreps$\Gamma$-graded superalgebra
irreps$\Gamma$-graded colour algebra irrepsungraded colour algebra
irreps$\mathcal{F}_{1}$recolourationdiscolouration$\mathcal{F}_{2}$

These bijections are provided in Theorem 5.14. The irreducible
$\mathbb{Z}_{2}$-graded superalgebra representations have been well-studied,
and these bijections will allow us to carry across these results to Lie colour
algebras.

The paper is organised as follows. In Section 2, we review some known results
for Lie colour algebras, and results analogous to those for Lie
(super)algebras. We then introduce the notion of a refining extension in
Section 3. As an example, Section 4 uses the refining extension to construct
the Hilbert space of the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded
supersymmetric system of [22]. In Section 5, we show how the refining
extension can be used to derive the finite-dimensional irreducible
$\Gamma$-graded modules from the $\Gamma/H$-graded modules. We end with an
example of deriving the finite-dimensional irreducible modules of a colour
version of $\mathfrak{sl}_{2}$ (Section 6) and compare with [47].

##  2\. Lie colour algebras and their representations

A Lie colour algebra generalises a Lie superalgebra to grading by some
(additive) abelian group $\Gamma$ (instead of just $\mathbb{Z}_{2}$). For the
formal definition of a Lie colour algebra, we refer the reader to [41]. As an
illustrative example: if $A=\bigoplus_{\alpha\in\Gamma}A_{\alpha}$ is an
associative algebra such that $A_{\alpha}A_{\beta}\subseteq A_{\alpha+\beta}$
then $A$ can be given the structure of a Lie colour algebra with bracket

$\cbrak{x}{y}=xy-\varepsilon(\alpha,\beta)yx,\qquad x\in A_{\alpha},\,y\in
A_{\beta}.$

Here, $\varepsilon\colon\mathfrak{g}\times\mathfrak{g}\to\mathbb{F}$ (where
$\mathbb{F}$ is the field of scalars) is called a commutation factor [41] (or
antisymmetric bicharacter [47], or phase function [43]). That this generalizes
a Lie superalgebra is apparent: a Lie superalgebra is simply a Lie colour
algebra graded by the group $\Gamma=\mathbb{Z}_{2}$ with commutation factor
$\varepsilon(\alpha,\beta)=(-1)^{\alpha\cdot\beta}$. Note that, in general, a
commutation factor must satisfy certain conditions (see [41]).

We can define notions such as homomorphisms, subalgebras and representations
similarly to superalgebras. However, to retain information about the
$\Gamma$-grading, we require some maps (such as homomorphisms, intertwiners,
and inclusion maps of subalgebras and $\Gamma$-graded submodules) to be
homogeneous of degree $0$, as per the following definition.

######  Definition 2.1.

Let $V=\bigoplus_{\gamma\in\Gamma}V_{\gamma}$ and
$W=\bigoplus_{\gamma\in\Gamma}W_{\gamma}$ be vector spaces over $\mathbb{F}$.
A linear map $f\colon V\to W$ is homogeneous of degree $\gamma\in\Gamma$ if
$f(V_{\alpha})\subseteq W_{\alpha+\gamma}$ for all $\alpha\in\Gamma$. (In
particular, $f$ is a homogeneous element of the colour algebra
$\mathfrak{gl}(V,\varepsilon)$, see [41].)

###  2.1. Representations

Although representations can be defined similarly to those for superalgebras,
we make an important distinction between graded and ungraded representations.

######  Definition 2.2.

An ungraded representation of a Lie colour algebra $\mathfrak{g}$ on a vector
space $V$ is a linear map $\rho\colon\mathfrak{g}\to\operatorname{End}(V)$
satisfying
$\rho(\cbrak{x}{y})=\rho(x)\rho(y)-\varepsilon(\alpha,\beta)\rho(y)\rho(x)$
for all $x\in\mathfrak{g}_{\alpha},y\in\mathfrak{g}_{\beta}$.

A $\Gamma$-graded representation of $\mathfrak{g}$ on a $\Gamma$-graded vector
space $V$ is an ungraded representation that also satisfies
$\rho(\mathfrak{g}_{\alpha})V_{\gamma}\subseteq V_{\alpha+\gamma}$.

In either case, we say that $V$ is a $\mathfrak{g}$-module given by $\rho$ (or
simply a $\mathfrak{g}$-module).

Obviously, we do not require maps related to ungraded representations (such as
intertwiners and submodules) to be homogeneous of degree $0$. We only require
this condition of the corresponding maps for graded representations.

Recall that a $\mathfrak{g}$-module is called completely reducible if it can
be written as a direct sum of irreducible submodules. Weyl’s Theorem will not
hold for a general Lie colour algebra (indeed, Weyl’s Theorem does not hold
for Lie superalgebras [48]). As such, we need to manually determine when a
$\mathfrak{g}$-module is completely reducible. The following propositions will
be of great help.

######  Proposition 2.3.

Consider a ($\Gamma$-graded or ungraded) $\mathfrak{g}$-module $V$ that can be
written as $V=\sum_{i\in I}V_{i}$ for irreducible $\mathfrak{g}$-modules
$V_{i}$ ($i\in I$). Then, there exists some subset $J\subseteq I$ such that
$V=\bigoplus_{i\in J}V_{i}$.

######  Proposition 2.4.

Submodules and quotient modules of a ($\Gamma$-graded or ungraded) completely
reducible $\mathfrak{g}$-module are completely reducible.

These propositions are well-known facts from the module theory of rings (for
the proofs in the case of rings, see e.g. [49, Chapter XVII, Lemma 2.1 and
Proposition 2.2]). The proofs of these propositions for Lie colour algebra
modules are exactly the same as for modules over rings.

###  2.2. Graded quotient modules

Let $V=\bigoplus_{\gamma\in\Gamma}V_{\gamma}$ be a $\Gamma$-graded
$\mathfrak{g}$-module given by the representation $\rho$ and
$U=\bigoplus_{\gamma\in\Gamma}U_{\gamma}$ a $\Gamma$-graded submodule of $V$
(so that $U_{\gamma}\subseteq V_{\gamma}$).

######  Definition 2.5.

The graded quotient module of $V$ by $U$, denoted $V/U$, is a direct sum of
quotient vector spaces $\bigoplus_{\gamma\in\Gamma}V_{\gamma}/U_{\gamma}$ with
$\mathfrak{g}$-module structure given by
$\pi\colon\mathfrak{g}\to\operatorname{End}(V/U)$ defined by

$\pi(x)(v+U_{\gamma})=\rho(x)v+U_{\alpha+\gamma}\qquad\text{for}~{}x\in\mathfrak{g}_{\alpha},\,v\in
V_{\gamma}.$

The proof that the above representation is well defined is similar to that for
Lie superalgebras. The only additional property to check is whether $\pi$ is
homogeneous of degree $0$, which is easily verified.

The three isomorphism theorems hold for $\mathfrak{g}$-modules. Again, the
proofs are similar to superalgebras, needing only to verify that all maps are
homogeneous of degree $0$.

###  2.3. Discolouration

Scheunert [41] gave a bijection between Lie colour algebras and graded Lie
superalgebras and between their representations. This bijection has many nice
properties; for example, it preserves subalgebras, ideals and
subrepresentations. To compute the image of a colour algebra under this
bijection, we deform the bracket using a multiplier.

Given a Lie colour algebra $\mathfrak{g}$ and a multiplier
$\sigma\colon\Gamma\times\Gamma\to\mathbb{F}^{\times}$ (for the definition,
see [41]; also called a multiplicative $2$-cocycle [34]), we can define a new
bracket $\cbrak{\cdot}{\cdot}^{\sigma}$ by

$\cbrak{x}{y}^{\sigma}=\sigma(\alpha,\beta)\cbrak{x}{y},\qquad
x\in\mathfrak{g}_{\alpha},y\in\mathfrak{g}_{\beta}.$

Assuming that $\Gamma$ is finitely generated, we can choose this multiplier
$\sigma$ so that $\mathfrak{g}$ equipped with $\cbrak{\cdot}{\cdot}^{\sigma}$
forms a Lie superalgebra [41]. We denote this Lie superalgebra
$\mathfrak{g}^{\sigma}$. The map $\mathfrak{g}\mapsto\mathfrak{g}^{\sigma}$ is
a bijection between the Lie colour algebras with commutation factor
$\varepsilon$ and the $\Gamma$-graded Lie superalgebras [41]. As such, we call
$\mathfrak{g}^{\sigma}$ a discolouration of $\mathfrak{g}$. Discolouration is
an invertible procedure, and we will call
$(\mathfrak{g}^{\sigma})^{1/\sigma}=\mathfrak{g}$ the recolouration of
$\mathfrak{g}^{\sigma}$.

Note that there is an alternative method of discolouration which works by
adding new elements, called Klein operators, to the universal enveloping
algebra (see [42, 43, 44, 45]). However, for the remainder of this paper we
will use the original discolouration process of [41] described in this
section.

It is also possible to discolour representations. Given a graded
representation $\rho\colon\mathfrak{g}\to\mathfrak{gl}(V,\varepsilon)$ and a
discolouring multiplier
$\sigma\colon\Gamma\times\Gamma\to\mathbb{F}^{\times}$, we can define the
discolouration of $\rho$ to be the graded representation $\rho^{\sigma}$ of
$\mathfrak{g}^{\sigma}$ given by

$\rho^{\sigma}(x)v=\sigma(\alpha,\gamma)\rho(x)v,\qquad\text{for}~{}x\in\mathfrak{g}_{\alpha},v\in
V_{\gamma}.$

Similarly, the map $\rho\mapsto\rho^{\sigma}$ is a bijection between the
graded representations of $\mathfrak{g}$ and the graded representations of the
corresponding discoloured $\Gamma$-graded Lie superalgebra [41]. Thus, the
representation theory of Lie colour algebras can be completely derived from
that of $\Gamma$-graded Lie superalgebras.

The $\mathbb{Z}_{2}$-graded irreducible representations of Lie superalgebras
have been well studied. However, these results cannot be immediately applied
to find irreducible $\Gamma$-graded representations. For instance, it might
not be possible to find a $\Gamma$-grading for a $\mathbb{Z}_{2}$-graded
representation. On the other hand, there might be an irreducible
$\Gamma$-graded representation that becomes reducible when considered as a
$\mathbb{Z}_{2}$-graded representation.

##  3\. Refining extension

Consider a module graded by some group. If we wish to increase the size of the
group, we may not be able to find a grading for the module with this new
group. To find a module graded by this larger group, we will allow ourselves
to increase the size of the module itself as well. With this idea in mind, we
introduce the notion of a refining extension.

Let $\mathfrak{g}=\bigoplus_{\gamma\in\Gamma}\mathfrak{g}_{\gamma}$ be a
$\Gamma$-graded Lie colour algebra with commutation factor $\varepsilon$ and
let $H$ be a subgroup of the (additive) abelian group $\Gamma$. Notice that
$\mathfrak{g}$ has a natural $\Gamma/H$ grading, given by
$\mathfrak{g}=\bigoplus_{\Lambda\in\Gamma/H}\mathfrak{g}^{\Gamma/H}_{\Lambda}$
where
$\mathfrak{g}^{\Gamma/H}_{\Lambda}=\bigoplus_{\alpha\in\Lambda}\mathfrak{g}_{\alpha}$.
Recall [41] that $\mathfrak{g}$ also has a natural $\mathbb{Z}_{2}$-grading

$\mathfrak{g}^{0}=\bigoplus_{\gamma\in\Gamma_{0}}\mathfrak{g}_{\gamma},\qquad\mathfrak{g}^{1}=\bigoplus_{\gamma\in\Gamma_{1}}\mathfrak{g}_{\gamma}.$

where $\Gamma_{0}=\\{\gamma\in\Gamma\mid\varepsilon(\gamma,\gamma)=1\\}$ and
$\Gamma_{1}=\Gamma\setminus\Gamma_{0}$. If we choose $H\leq\Gamma_{0}$, then
$\Gamma/H$-grading respects the $\mathbb{Z}_{2}$-grading, i.e.

$\mathfrak{g}^{0}=\bigoplus_{\Lambda\in\Gamma_{0}/H}\mathfrak{g}^{\Gamma/H}_{\Lambda},\qquad\mathfrak{g}^{1}=\bigoplus_{\Lambda\in\Gamma_{1}/H}\mathfrak{g}^{\Gamma/H}_{\Lambda}.$

This is useful if we want to prevent $H$ from interfering with discolouration.

######  Definition 3.1.

Let $V=\bigoplus_{\Lambda\in\Gamma/H}V_{\Lambda}$ be a $\Gamma/H$-graded
$\mathfrak{g}$-module given by $\rho$. Construct a vector space
$V^{H}=\bigoplus_{\gamma\in\Gamma}V^{H}_{\gamma}$, where $V^{H}_{\gamma}$ is
isomorphic (as a vector space) to $V_{\gamma+H}$. Let $\varphi_{\gamma}\colon
V^{H}_{\gamma}\to V_{\gamma+H}$ be such a vector space isomorphism. The
refining extension of $V$ by $H$ is the $\mathfrak{g}$-module $V^{H}$ given by
the representation $\rho^{H}$ defined by

$\rho^{H}(x)v=\varphi_{\alpha+\gamma}^{-1}(\rho(x)\varphi_{\gamma}(v)),\qquad\text{for}~{}x\in\mathfrak{g}_{\alpha},\,v\in
V^{H}_{\gamma}.$

Intuitively, given a sector $V_{\Lambda}$ of $V$, we add a copy of
$V_{\Lambda}$ to $V^{H}$ for every element of the coset $\Lambda$. Since
$\Gamma$ is simply the union of all cosets, this gives $V^{H}$ a natural
$\Gamma$ grading. We then define the action of $\rho^{H}$ on each copy of
$V_{\Lambda}$ to be the same as $\rho$, being careful to respect the new
$\Gamma$-grading.

######  Remark 3.2.

If $\Gamma=K\times H$ for some group $K$, then we could equivalently define
the refining extension to be $V\otimes\mathbb{F}[H]$ (where $\mathbb{F}[H]$ is
the group algebra of $H$), with representation given by $\rho^{H}(x)(v\otimes
h)=\rho(x)v\otimes(h+\eta)$ for $x\in\mathfrak{g}_{(\kappa,\eta)}$ and
$v\otimes h\in V\otimes\mathbb{F}[H]$. This is similar to the approach used in
[34] to classify the simple $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour
Lie superalgebras. However, with 3.1, we do not need to make any additional
assumptions on the structure of the group $\Gamma$.

As well as a $\Gamma$-grading, $V^{H}$ has a natural $\Gamma/H$-grading given
by

$V^{H}=\bigoplus_{\Lambda\in\Gamma/H}\left(\bigoplus_{\gamma\in\Lambda}V_{\gamma}^{H}\right).$

In this sense, the $\Gamma$-grading of $V^{H}$ is a refinement of the
$\Gamma/H$-grading.

######  Proposition 3.3.

The $\mathfrak{g}$-module $V$ is a $\Gamma/H$-graded submodule of $V^{H}$
(equivalently, $V^{H}$ is an extension of $V$).

###### Proof.

Define a map $\Phi\colon V\to V^{H}$ by

$\Phi(w)=\sum_{\gamma\in\Lambda}\varphi_{\gamma}^{-1}(w)\qquad\text{for}~{}w\in
V_{\Lambda},$

extending to inhomogeneous elements by linearity. Linearity of $\Phi$ for
homogeneous elements follows from linearity of $\varphi_{\gamma}^{-1}$. For
injectivity, if $\Phi(w)=0$ then $\varphi_{\gamma}^{-1}(w)=0$ for all
$\gamma\in\Lambda$ (because each $\varphi_{\gamma}^{-1}(w)$ is in a different
sector of the $\Gamma$-grade), so $w=0$ by injectivity of
$\varphi_{\gamma}^{-1}$. Thus, $\ker\Phi=0$ and $\Phi$ is injective. The map
$\Phi$ is homogeneous of degree $0+H$ by definition. Finally, $\Phi$ is an
intertwiner: for $w\in V_{\Lambda}$ and $x\in\mathfrak{g}_{\alpha}$,

$\displaystyle\Phi(\rho(x)w)$
$\displaystyle=\sum_{\alpha+\gamma\in\alpha+\Lambda}\varphi_{\alpha+\gamma}^{-1}(\rho(x)w)$
$\displaystyle=\sum_{\alpha+\gamma\in\alpha+\Lambda}\varphi_{\alpha+\gamma}^{-1}(\rho(x)\varphi_{\gamma}\varphi_{\gamma}^{-1}(w))$
$\displaystyle=\sum_{\gamma\in\Lambda}\rho^{H}(x)\varphi_{\gamma}^{-1}(w)$
$\displaystyle=\rho^{H}(x)\Phi(w).\qed$

If $V$ can be given a $\Gamma$-grading, then we can strengthen the above
result:

######  Proposition 3.4.

If $V$ can be given a $\Gamma$-grading that respects the $\Gamma/H$-grading,
then $V$ is a $\Gamma$-graded submodule of $V^{H}$.

###### Proof.

Let $V=\bigoplus_{\gamma\in\Gamma}W_{\gamma}$ be the $\Gamma$-grading, with
$V_{\Lambda}=\bigoplus_{\gamma\in\Lambda}W_{\gamma}$ for all
$\Lambda\in\Gamma/H$ (i.e. the $\Gamma$-grading respects the $\Gamma/H$
grading). Define a map $\Phi\colon V\to V^{H}$ by
$\Phi(w)=\varphi_{\gamma}^{-1}(w)$ for $w\in W_{\gamma}$ and extending to
inhomogeneous elements by linearity. Linearity and injectivity of $\Phi$
follow immediately from linearity and injectivity of $\varphi_{\gamma}^{-1}$.
The map $\Phi$ is homogeneous of degree $0$ by definition. And $\Phi$ is an
intertwiner: for $w\in W_{\gamma}$ and $x\in\mathfrak{g}_{\alpha}$,

$\Phi(\rho(x)w)=\varphi_{\alpha+\gamma}^{-1}(\rho(x)w)=\varphi_{\alpha+\gamma}^{-1}(\rho(x)\varphi_{\gamma}\varphi_{\gamma}^{-1}(w))=\rho^{H}(x)\Phi(w).\qed$

The above results show that a refining extension is indeed an extension of the
original module and has a refined grading (i.e. grading by a larger group). As
we will soon see, refining extensions appear naturally in the applications and
representation theory of Lie colour algebras.

##  4\. Refining extension and quantum mechanical systems

A common way to obtain a $\Gamma$-graded colour supersymmetric quantum
mechanical system is by modifying an existing supersymmetric system. This is
the approach taken in (for example) [22], where the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded supersymmetric system presented in
[22, Section III] can be obtained by introducing an additional central charge
to Witten’s model [46, Section 6]. In [22], the Hilbert space realisation for
this $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded system was constructed
independently from that of Witten’s model. In this section, we will
demonstrate that this $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded Hilbert
space realisation can be constructed from Witten’s realisation using a
refining extension. In this way, the refining extension has already appeared
in the literature and is a natural structure to consider in the study of Lie
colour algebras.

Witten considered [46, Section 6] a superalgebra with even sector spanned by a
Hamiltonian $H$ and odd sector spanned by two supercharges $Q_{1}$ and $Q_{2}$
satisfying

$\commutator{Q_{i}}{H}=0\qquad\anticommutator{Q_{i}}{Q_{j}}=\delta_{ij}H$

for $i,j=1,2$. This algebra was realized as operators acting on two component
Pauli spinors in the Hilbert space
$\mathscr{H}^{\textup{W}}=L^{2}(\mathbb{R})\otimes\mathbb{C}^{2}$ via

$\displaystyle H$
$\displaystyle=\frac{1}{2}\left(p^{2}+W^{2}(x)+\hbar\sigma_{3}\derivative{W}{x}\right),$
$\displaystyle Q_{1}$ $\displaystyle=\frac{1}{2}(\sigma_{1}p+\sigma_{2}W(x)),$
$\displaystyle Q_{2}$ $\displaystyle=\frac{1}{2}(\sigma_{2}p-\sigma_{1}W(x)).$

where $\sigma_{i}$ ($i=1,2,3$) are the Pauli matrices,
$W\in\operatorname{C}^{\infty}(\mathbb{R})$ and $p=-i\hbar\derivative*{x}$ is
the momentum operator. This Hilbert space representation is a
$\mathbb{Z}_{2}$-graded representation of the superalgebra, with grading given
by $\mathbb{C}^{2}=\mathbb{C}_{0}\oplus\mathbb{C}_{1}$ (that is, the first
component of the spinor has an even grade and the second component has an odd
grade).

By adding a central charge,

$Z=-\frac{i}{2}\left(\sigma_{3}(p^{2}+W^{2}(x))+\hbar\derivative{W}{x}\right)$

the original superalgebra becomes a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebra (without
requiring recolouration) with sectors spanned by

$00\text{-sector: }H\qquad 01\text{-sector: }Q_{1}\qquad 11\text{-sector:
}Z\qquad 10\text{-sector: }Q_{2}$

and satisfying the following relations

$\commutator{Q_{2}}{Q_{1}}=Z\qquad\anticommutator{Q_{i}}{Q_{i}}=H$

while all other Lie colour brackets vanish. We can easily verify that
$\mathscr{H}^{\textup{W}}$ becomes an ungraded module of the Lie colour
algebra.

Taking the refining extension $(\mathscr{H}^{\textup{W}})^{\mathbb{Z}_{2}}$
turns $\mathscr{H}^{\textup{W}}$ into a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded module. Here, we use
$\mathbb{Z}_{2}\cong\\{00,11\\}\leq\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. Take
vector spaces $\mathscr{H}_{00},\mathscr{H}_{11}$ isomorphic to the even
sector $\mathscr{H}^{\textup{W}}_{0}$ and vector spaces
$\mathscr{H}_{01},\mathscr{H}_{10}$ isomorphic to the odd sector
$\mathscr{H}^{\textup{W}}_{1}$ with isomorphisms
$\varphi_{\gamma}\colon\mathscr{H}_{\gamma}\to\mathscr{H}_{\gamma+\\{00,11\\}}$
for $\gamma\in\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. Then, the refining
extension is
$(\mathscr{H}^{\textup{W}})^{\mathbb{Z}_{2}}=\mathscr{H}_{00}\oplus\mathscr{H}_{01}\oplus\mathscr{H}_{11}\oplus\mathscr{H}_{10}=L^{2}(\mathbb{R})\otimes\mathbb{C}^{4}$
(note the order of the sectors) with representation

$\rho^{\mathbb{Z}_{2}}(x)v=\varphi_{\alpha+\xi}^{-1}(\rho(x)\varphi_{\xi}(v)),\qquad\text{for}~{}x\in\mathfrak{g}_{\alpha},\,v\in\mathscr{H}_{\xi}.$

Using the above equality, we can explicitly compute the representation on
four-component spinors:

$\displaystyle\rho^{\mathbb{Z}_{2}}(H)$ $\displaystyle=\begin{pmatrix}H&0\\\
0&H\end{pmatrix},$ $\displaystyle\rho^{\mathbb{Z}_{2}}(Q_{1})$
$\displaystyle=\begin{pmatrix}Q_{1}&0\\\ 0&Q_{1}\end{pmatrix},$
$\displaystyle\rho^{\mathbb{Z}_{2}}(Q_{2})$
$\displaystyle=\begin{pmatrix}0&Q_{2}\\\ Q_{2}&0\end{pmatrix},$
$\displaystyle\rho^{\mathbb{Z}_{2}}(Z)$ $\displaystyle=\begin{pmatrix}0&Z\\\
Z&0\end{pmatrix}.$

Setting
$H_{00}=2\rho^{\mathbb{Z}_{2}}(H),\,Q_{01}=\rho^{\mathbb{Z}_{2}}(Q_{1}),\,Q_{10}=\rho^{\mathbb{Z}_{2}}(Q_{2})$
and $Z_{11}=2\rho^{\mathbb{Z}_{2}}(Z)$, we obtain the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-supersymmetry model in [22, Section III]
(with $m=1/2$).

The above example demonstrates that the refining extension provides a natural
method for constructing the Hilbert space of a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded supersymmetry model from a
$\mathbb{Z}_{2}$-graded supersymmetry model. The refining extension can be
applied to other colour supersymmetric models constructed in a similar way.

##  5\. Refining extension and preserving irreducibility

In this section, we prove that every finite-dimensional irreducible
$\Gamma$-graded module can be constructed from the $\Gamma/H$-graded modules
using the refining extension. In combination with discolouration, this will
allow us to derive all such irreducible $\Gamma$-graded colour algebra
representations from the irreducible $\mathbb{Z}_{2}$-graded superalgebra
representations.

The proof proceeds as follows. We first prove that every finite-dimensional
irreducible $\Gamma$-graded module appears as a quotient of a refining
extension (Theorem 5.2). The refining extension is completely reducible both
as a $\Gamma/H$\- and $\Gamma$-graded module (Theorems 5.9 and 5.11
respectively) and we explicitly provide the direct sum decomposition by
defining notions of $\Gamma/H$-representative twists and $H$-parity shifts
(respectively). Comparing these direct sum decompositions limits the possible
irreducible $\Gamma$-graded modules to either $\Gamma/H$-graded modules or
their refining extensions. We use these results to construct a bijection
between $\Gamma/H$\- and $\Gamma$-graded modules (Theorem 5.14) up to
equivalence defined by the irreducible direct summands.

###  5.1. Quotient modules of the refining extension

Suppose we are given a finite-dimensional irreducible $\Gamma/H$-graded
$\mathfrak{g}$-module $V$. To construct an irreducible $\Gamma$-graded
$\mathfrak{g}$-module, we can simply take the refining extension $V^{H}$ and
quotient out by a maximal $\Gamma$-graded submodule. Since $V^{H}$ may have
many maximal submodules, this procedure may yield many nonisomorphic
$\Gamma$-graded modules for each $\Gamma/H$-graded module. Regardless, we will
show that every irreducible $\Gamma$-graded $\mathfrak{g}$-module can be
constructed in this way.

But first, we prove the following lemma which is a restatement of the First
Isomorphism Theorem in a form that is useful for refining extensions.

######  Lemma 5.1.

Let $V=\bigoplus_{\Lambda\in\Gamma/H}V_{\Lambda}$ be a $\Gamma/H$-graded
$\mathfrak{g}$-module given by $\rho$ and
$W=\bigoplus_{\gamma\in\Gamma}W_{\gamma}$ be a $\Gamma$-graded
$\mathfrak{g}$-module given by $\pi$. Suppose there are linear maps
$f_{\gamma}\colon V_{\gamma+H}\to W_{\gamma}$ that satisfy
$f_{\alpha+\gamma}(\rho(x)v)=\pi(x)f_{\gamma}(v)$ for all
$x\in\mathfrak{g}_{\alpha}$, $v\in V_{\gamma+H}$. Let $V^{H}$ be the refining
extension of $V$ by $H$ and define $f_{\gamma}^{H}=f_{\gamma}\varphi_{\gamma}$
where $\varphi_{\gamma}\colon V_{\gamma}^{H}\to V_{\gamma+H}$ are the
isomorphisms in the construction of $V^{H}$. Then,

  1. (i)

$K=\bigoplus_{\gamma\in\Gamma}\ker f_{\gamma}^{H}$ is a submodule of $V^{H}$;

  2. (ii)

$I=\bigoplus_{\gamma\in\Gamma}\operatorname{im}f_{\gamma}^{H}$ is a submodule
of $W$; and

  3. (iii)

$V^{H}/K\cong I$.

###### Proof.

Define $f\colon V^{H}\to W$ by

$f(v)=f^{H}_{\gamma}(v)$

for $v\in V^{H}_{\gamma}$ and extending to inhomogeneous elements by
linearity. It is easy to see that $f$ is an intertwiner, $\ker f=K$ and
$\operatorname{im}f=I$. The result follows immediately by applying the First
Isomorphism Theorem to $f$. ∎

######  Theorem 5.2.

Let $\mathfrak{g}=\bigoplus_{\gamma\in\Gamma}\mathfrak{g}_{\gamma}$ be a Lie
colour algebra. Let $W=\bigoplus_{\gamma\in\Gamma}W_{\gamma}$ be a finite-
dimensional irreducible $\Gamma$-graded $\mathfrak{g}$-module. Then, for
$H\leq\Gamma$, there exists an irreducible $\Gamma/H$-graded
$\mathfrak{g}$-module $V$ such that $W$ is a graded quotient of the refining
extension of $V$ by $H$.

###### Proof.

Note that $W$ has a natural $\Gamma/H$-grading given by

$\bigoplus_{\Lambda\in\Gamma/H}\left(\bigoplus_{\gamma\in\Lambda}W_{\gamma}\right).$

Let $V$ be an irreducible $\Gamma/H$-graded submodule of $W$. Such a $V$
exists by strong induction on the dimension of $W$.

Let $\rho\colon\mathfrak{g}\to\operatorname{End}(W)$ be the representation
corresponding to $W$ ($\rho$ also provides a submodule structure for $V$).

Take $v\in V_{\Lambda}$ and write

(1) $v=\sum_{\gamma\in\Lambda}w_{\gamma}$

for $w_{\gamma}\in W_{\gamma}$. For $\gamma\in\Lambda$, let $f_{\gamma}\colon
V_{\Lambda}\to W_{\gamma}$ be the projection of $V_{\Lambda}$ onto
$W_{\gamma}$,

$f_{\gamma}(v)=w_{\gamma}.$

Note that $f_{\gamma}$ is well-defined because the linear combination in (1)
is unique due to the direct sum
$V_{\Lambda}\subseteq\bigoplus_{\gamma\in\Lambda}W_{\gamma}$. Additionally, it
is easy to see that $f_{\gamma}$ is linear.

For $x\in\mathfrak{g}_{\alpha}$, we have
$\rho(x)v=\sum_{\gamma\in\Lambda}\rho(x)f_{\gamma}(v)$. Since
$\rho(x)f_{\gamma}(v)\in W_{\alpha+\gamma}$ and such a linear combination of
homogeneous elements is unique, we have that
$f_{\alpha+\gamma}(\rho(x)v)=\rho(x)f_{\gamma}(v)$ by the definition of
$f_{\alpha+\gamma}$.

Therefore, by 5.1, $I\cong V^{H}/K$ for
$I=\bigoplus_{\gamma\in\Gamma}\operatorname{im}f_{\gamma}^{H}$ and
$K=\bigoplus_{\gamma\in\Gamma}\ker f_{\gamma}^{H}$. But $I$ is a
$\Gamma$-graded is a submodule of $W$, so $I=0$ or $I=W$ by irreducibility of
$W$. However, since $V\neq 0$, there is some nonzero sector $V_{\Lambda}$ of
$V$ (with $\Lambda\in\Gamma/H$). Any non-zero element $v\in V_{\Lambda}$ can
be written as $v=\sum_{\gamma\in\Lambda}f_{\gamma}(v)$, so $f_{\gamma}(v)\neq
0$ for at least one $\gamma\in\Gamma$. That is,
$\operatorname{im}f_{\gamma}\neq 0$, so $I\neq 0$. We conclude that $I=W$;
hence $W\cong V^{H}/K$ as claimed. ∎

Theorem 5.2 provides us with a general method of obtaining the irreducible
$\Gamma$-graded modules from the $\Gamma/H$-graded ones; namely find all
maximal submodules and take quotients. However, this theorem does not provide
much information about the actual structure of the $\Gamma$-graded modules.

###  5.2. $\Gamma/H$-grading: Refining extension is completely reducible

To aid with the structural analysis of the refining extension, from now on we
will assume that $H$ is finite. Without loss of generality, we will also
assume that $H$ is a simple group. If not, we can take a maximal subgroup
$M\leq H$ (so $H/M$ simple) and work with $(\Gamma/M)/(H/M)$ instead of
$\Gamma/H$ (note that these two quotient groups are isomorphic). Theorem 5.2
will yield $\Gamma/M$-graded modules. We can then iterate until the maximal
submodule is also simple. Since $H$ is abelian (because $\Gamma$ is), finite
and simple, we must have that $H\cong\mathbb{Z}_{p}$ for some prime $p$.

We will show that the refining extension is a completely reducible
$\Gamma/H$-graded module and determine the structure of the direct sum
decomposition. The direct summands of the refining extension will restrict
which quotients are possible, and hence which irreducible $\Gamma$-graded
representations are possible by Theorem 5.2. Given that $V^{H}$ was
constructed from $p=\absolutevalue{H}$ copies of $V$, we might expect that the
composition factors are merely $p$ copies of $V$. This is not necessarily the
case; the composition factors are (what we will call) the
$\Gamma/H$-representative twists of $V$. The structure of a
$\Gamma/H$-representative twist is almost identical to that of $V$, with the
representation on each sector simply multiplied by a root of unity. To ensure
that all roots of unity exist and have nice properties, we will assume for the
rest of this paper that we are working over the field $\mathbb{F}=\mathbb{C}$.

####  5.2.1. $\Gamma/H$-representative twist

Let $G$ be a set containing exactly one representative from each coset in
$\Gamma/H$. Note that $G$ does not necessarily form a group. Let
$V=\bigoplus_{\Lambda\in\Gamma/H}V_{\Lambda}$ be a $\Gamma/H$-graded
$\mathfrak{g}$-module given by the representation $\rho$ and let $\zeta$ be a
$p$-th root of unity. Define the $\Gamma/H$-representative twist to be a new
module $V^{\zeta}$ with representation $\rho^{\zeta}$ via the following: set
$V^{\zeta}=\bigoplus_{\Lambda\in\Gamma/H}V^{\zeta}_{\Lambda}$ where
$V^{\zeta}_{\Lambda}=V_{\Lambda}$ (as vector spaces) and, for arbitrary
$x\in\mathfrak{g}_{\alpha}$ and $v\in V_{\Lambda}$, define $\rho^{\zeta}$ by

$\rho^{\zeta}(x)v=\zeta^{h}\rho(x)v$

where $h\in H$ is an element such that $\alpha+g-h\in G$ for $g\in\Lambda\cap
G$. Here, we interpret $h$ as an integer in $\mathbb{Z}_{p}\cong H$. Note that
$g$ and $h$ are both unique because there is exactly one representative in $G$
from each coset. We will now prove that $V^{\zeta}$ is indeed a representation
and is independent of the choice of $G$.

######  Proposition 5.3.

$V^{\zeta}$ is a $\Gamma/H$-graded $\mathfrak{g}$-module.

###### Proof.

Both linearity properties of $\rho^{\zeta}$ follow immediately from that of
$\rho$. Obviously $V^{\zeta}$ has the same grading as $V$.

Let $x\in\mathfrak{g}_{\alpha},\,y\in\mathfrak{g}_{\beta}$ and $v\in
V_{\Lambda}$ be arbitrary. Let $g$ be the unique representative of $\Lambda$
that is also in $G$. We claim that
$\rho^{\zeta}(x)\rho^{\zeta}(y)v=\zeta^{h}\rho(x)\rho(y)v$ where $h\in
H,\,\alpha+\beta+g-h\in G$. Indeed, take $\eta\in H$ such that
$\beta+g-\eta\in G$. Then $\rho^{\zeta}(y)v=\zeta^{\eta}\rho(y)v$. Noting that
$\alpha+\beta+g-\eta=(\alpha+\beta+g-h)+(h-\eta))$ and $\alpha+\beta+g-h\in
G$,

$\rho^{\zeta}(x)\rho^{\zeta}(y)v=\rho^{\zeta}(x)\zeta^{\eta}\rho(y)v=\zeta^{h-\eta}\zeta^{\eta}\rho(x)\rho(y)v=\zeta^{h}\rho(x)\rho(y)v$

as claimed.

Now, $\rho^{\zeta}$ is a representation:

$\rho^{\zeta}(\cbrak{x}{y})v=\zeta^{h}(\rho(x)\rho(y)-\varepsilon(\alpha,\beta)\rho(y)\rho(x))v=(\rho^{\zeta}(x)\rho^{\zeta}(y)-\varepsilon(\alpha,\beta)\rho^{\zeta}(y)\rho^{\zeta}(x))v$

by the above claim. ∎

######  Proposition 5.4.

The $\Gamma/H$-representative twist is independent of the choice of
representatives. More precisely, let $G,\,\widetilde{G}$ be two sets which
each contain exactly one representative from each coset of $\Gamma/H$. Let
$V^{\zeta},\,\widetilde{V}^{\zeta}$ be the corresponding
$\Gamma/H$-representative twists of $V$. Then $V^{\zeta}$ and
$\widetilde{V}^{\zeta}$ are isomorphic as $\Gamma/H$-graded modules.

###### Proof.

Define a map $\Phi\colon V^{\zeta}\to\widetilde{V}^{\zeta}$ on each sector
$V^{\zeta}_{\Lambda}$ by

$\Phi(v)=\zeta^{g-\widetilde{g}}v$

for $v\in V^{\zeta}_{\Lambda},\,g\in G\cap\Lambda$ and
$\widetilde{g}\in\widetilde{G}\cap\Lambda$. This map is clearly linear,
bijective and homogeneous of degree $0$. Moreover, this map is an intertwiner:
let $x\in\mathfrak{g}_{\alpha}$ and take $h,\widetilde{h}\in H$ such that
$\alpha+g-h\in G$ and $\alpha+\widetilde{g}-\widetilde{h}\in\widetilde{G}$.
Then,

$\Phi(\rho^{\zeta}(x)v)=\zeta^{\alpha+g-h-\alpha-\widetilde{g}+\widetilde{h}}\zeta^{h}\rho(x)v=\zeta^{\widetilde{h}}\rho(x)\zeta^{g-\widetilde{g}}v=\widetilde{\rho}^{\zeta}(x)\Phi(v).\qed$

The $\Gamma/H$-representative twist has many important properties. For
example, we can easily see that $(V^{\zeta})^{\xi}=V^{\zeta\xi}$. We now prove
three more such important properties.

######  Proposition 5.5.

If $V$ is irreducible then $V^{\zeta}$ is irreducible.

###### Proof.

If $U$ is a submodule of $V^{\zeta}$ then
$\rho^{\zeta}(\mathfrak{g})U\subseteq U$. Since the action of $\rho$ only
differs from $\rho^{\zeta}$ by a scalar (at most), we have that
$\rho(\mathfrak{g})U\subseteq U$. Hence, $U$ is also submodule of $V$. If $V$
is irreducible then, as a vector space, $U=0$ or $U=V=V^{\zeta}$. ∎

######  Proposition 5.6.

There is an equality of $\mathfrak{g}$-modules:
$(V^{H})^{\zeta}=(V^{\zeta})^{H}$.

###### Proof.

Clearly $(V^{H})^{\zeta}$ and $(V^{\zeta})^{H}$ are equal as vector spaces. We
just need to check the module structure. Let $\rho^{H\zeta}$ be the
representation for $(V^{H})^{\zeta}$ and $\rho^{\zeta H}$ be the
representation for $(V^{\zeta})^{H}$. Let $v\in V^{H}_{\gamma}$ and
$x\in\mathfrak{g}_{\alpha}$. Choose $g$ to be the unique representative of
$\gamma+H$ that is also in $G$ and choose $h\in H$ such that $\alpha+g-h\in
G$. Then,

$\rho^{H\zeta}(x)v=\zeta^{h}\rho^{H}(x)v=\varphi_{\alpha+\gamma}^{-1}\zeta^{h}\rho(x)\varphi_{\gamma}v=\varphi_{\alpha+\gamma}^{-1}\rho^{\zeta}(x)\varphi_{\gamma}v=\rho^{\zeta
H}(x)v.\qed$

######  Proposition 5.7.

If $V$ can be given a $\Gamma$-grading that respects the $\Gamma/H$-grading,
then $V^{\zeta}\cong V$.

###### Proof.

Let $V=\bigoplus_{\gamma\in\Gamma}W_{\gamma}$ be the $\Gamma$-grading, with
$V_{\Lambda}=\bigoplus_{\gamma\in\Lambda}W_{\gamma}$ for all
$\Lambda\in\Gamma/H$ (i.e. the $\Gamma$-grading respects the $\Gamma/H$
grading). Take $v\in W_{\gamma}$ and define $\Phi\colon V\to V^{\zeta}$ by

$\Phi(v)=\zeta^{\eta}v$

for $\eta\in H$ and $\gamma-\eta\in G$. Obviously, $\Phi$ is linear, bijective
and homogeneous of degree $0$. Additionally, $\Phi$ is an intertwiner: let
$x\in\mathfrak{g}_{\alpha}$ and take $h\in H$ such that
$\alpha+(\gamma-\eta)-h\in G$. Then, noting that $\rho(x)v\in
W_{\alpha+\gamma}$,

$\Phi(\rho(x)v)=\zeta^{\eta+h}\rho(x)v=\zeta^{h}\rho(x)\zeta^{\eta}v=\rho^{\zeta}(x)\Phi(v)\qed$

It is clear from the definition and the above propositions that the
$\Gamma/H$-representative twist has a very similar module structure to $V$.
This will be useful when examining the structure of $V^{H}$.

####  5.2.2. Direct sum decomposition

######  Lemma 5.8.

$(V^{\zeta})^{H}\cong V^{H}$ and so $V^{\zeta}$ is a $\Gamma/H$-graded
submodule of $V^{H}$.

###### Proof.

By 5.6, $(V^{\zeta})^{H}\cong(V^{H})^{\zeta}$. Since $(V^{H})^{\zeta}$ has a
$\Gamma$-grade, we have that $(V^{H})^{\zeta}\cong V^{H}$ by 5.7. By 3.3, we
then have that $V^{\zeta}$ is a $\Gamma/H$-graded submodule of
$(V^{\zeta})^{H}\cong V^{H}$. ∎

######  Theorem 5.9.

Let $H\cong\mathbb{Z}_{p}$ be a finite simple group and
$V=\bigoplus_{\Lambda\in\Gamma/H}V_{\Lambda}$ be a $\Gamma/H$-graded module of
a Lie colour algebra $\mathfrak{g}$ over $\mathbb{C}$. As a $\Gamma/H$-graded
module, the refining extension $V^{H}$ is isomorphic to

$\bigoplus_{k\in H}V^{(\zeta^{k})},$

where $\zeta\neq 1$ is a $p$-th root of unity

###### Proof.

By 5.8, we have that $V^{\zeta^{k}}$ is a $\Gamma/H$-graded submodule of
$V^{H}$. By following the proofs of 5.8 and related results, we find the
embedding $\Phi_{k}\colon V^{\zeta^{k}}\hookrightarrow V^{H}$ is given by

$\Phi_{k}(v)=\sum_{h\in H}(\zeta^{k})^{-h}\varphi_{g+h}^{-1}(v)$

for $v\in V_{\Lambda}$ and $g\in\Lambda\cap G$.

Obviously, $\sum_{k\in H}\Phi_{k}(V^{\zeta^{k}})\subseteq V^{H}$. We claim
that this sum is direct: $\bigoplus_{k\in H}\Phi_{k}(V^{\zeta^{k}})\subseteq
V^{H}$. We will show that for any nonzero $v_{k}\in V^{\zeta^{k}}$ the set
$\\{\Phi_{k}(v_{k})\mid k\in H\\}$ is linearly independent. This will prove
the claim. Take arbitrary nonzero vectors $v_{k}\in V^{\zeta^{k}}_{\Lambda}$
and consider the equation

$\sum_{k\in H}c_{k}\Phi_{k}(v_{k})=0$

for constants $c_{k}\in\mathbb{F}=\mathbb{C}$. By decomposing
$\Phi_{k}(v_{k})$ into each graded sector of $V^{H}$, we find that

$\sum_{k\in H}c_{k}(\zeta^{k})^{-h}\varphi_{g+h}^{-1}(v_{k})=0$

for each $h\in H$. And applying $\varphi_{g+h}$ to both sides, we have

$\sum_{k\in H}c_{k}(\zeta^{k})^{-h}v_{k}=0$

for each $h\in H$. Let $\\{e_{k}\mid k=0,\ldots,p-1\\}$ be the standard basis
vectors for $\mathbb{C}^{p}$ and consider the element

$\sum_{h\in H}\sum_{k\in H}\zeta^{-kh}e_{h}\otimes
c_{k}v_{k}\in\mathbb{C}^{p}\otimes V$

which is equal to $0$. Define a matrix $A=(A_{hk})_{h,k=0}^{p-1}$ over
$\mathbb{C}$ with entries $A_{hk}=\zeta^{-hk}$. We find that

$(A\otimes\operatorname{id})\left(\sum_{k=0}^{p-1}e_{k}\otimes
c_{k}v_{k}\right)=\sum_{h\in H}\sum_{k\in H}\zeta^{-kh}e_{h}\otimes
c_{k}v_{k}=0$

We can easily verify that $A$ is invertible (with
$(A^{-1})_{hk}=(1/p)\zeta^{hk}$). So, applying
$(A^{-1}\otimes\operatorname{id})$ to both sides of the above equality, we
find that

$\sum_{k=0}^{p-1}e_{k}\otimes c_{k}v_{k}=0.$

Since $\\{e_{k}\\}$ is linearly independent and $v_{k}\neq 0$, we must have
that $c_{k}=0$ for all $k\in H$. Therefore, we find that
$\\{\Phi_{k}(v_{k})\mid k\in H\\}$ is linearly independent for any $v_{k}\in
V^{\zeta^{k}}_{\Lambda}$. It is not difficult to extend this argument to any
$v_{k}\in V^{\zeta^{k}}$ by first projecting onto the sectors of
$V^{\zeta^{k}}$. We have thus proven the claim and $\bigoplus_{k\in
H}\Phi_{k}(V^{\zeta^{k}})\subseteq V^{H}$.

Finally, we will show that $\bigoplus_{k\in H}\Phi_{k}(V^{\zeta^{k}})=V^{H}$.
Take arbitrary $w\in V^{H}_{\gamma}$ and set $v=\varphi_{\gamma}(w)\in
V_{\gamma+H}$. Choose $\eta\in H$ so that $\gamma-\eta\in G$. Then,

$w=\frac{1}{p}\sum_{k\in H}(\zeta^{k})^{\eta}\Phi_{k}(v).$

That is, any element of $V^{H}_{\gamma}$ can be written as an element of
$\bigoplus_{k\in H}\Phi_{k}(V^{\zeta^{k}}_{\gamma+H})$. Extending to
inhomogeneous elements by linearity proves $\bigoplus_{k\in
H}\Phi_{k}(V^{\zeta^{k}})=V^{H}$. ∎

Note that the above result tells us that we could have equivalently defined
the refining extension as a direct sum of $\Gamma/H$-representative twists and
then provided a $\Gamma$-grading.

Moreover, if we assume that $V$ is irreducible, then Theorem 5.9 shows that
$V^{H}$ is completely reducible as a $\Gamma/H$-graded module and gives a
direct sum decomposition. In combination with Theorem 5.2, this restricts the
possible structures of the irreducible $\Gamma$-graded modules.

###  5.3. $\Gamma$-grading: refining extension is completely reducible

We will now show that the refining extension is completely reducible as a
$\Gamma$-graded module. Similar to $\Gamma/H$-graded case, the $\Gamma$-graded
irreducible direct summands all have a similar structure: they are all (what
we will call) $H$-parity shifts of the same irreducible $\Gamma$-graded
module. The structure of a $H$-parity shift is nearly identical to the
original module: a $H$-parity shift is obtained by simply applying a circular
shift to each of the sectors.

Comparing the $\Gamma$\- and $\Gamma/H$-graded irreducible direct summands
gives us more information about the structure of the refining extension. This
will allow us to conclude that the only possible $\Gamma$-graded direct
summands of $V^{H}$ are $V$ with a $\Gamma$-grading or $V^{H}$ itself.

####  5.3.1. Direct sum decomposition

Take some $h\in H$ and a $\Gamma$-graded module $W$ with representation
$\rho$. We can define the $H$-parity shift of $W$ by $h$ as a new
$\Gamma$-graded module $W^{+h}=\bigoplus_{\gamma\in\Gamma}W^{+h}_{\gamma}$,
where $W^{+h}_{\gamma}=W_{\gamma+h}$ with representation $\rho^{+h}=\rho$.
Obviously, a $H$-parity shift of $W$ is a $\mathfrak{g}$-module. We can easily
check that the $\Gamma$-grading is still respected:
$\rho^{+h}(x)W^{+h}_{\gamma}\subseteq W^{+h}_{\alpha+\gamma}$ for
$x\in\mathfrak{g}_{\alpha}$.

Following this definition, we make three immediate observations: A $H$-parity
shift might yield a module isomorphic to the original — indeed, any $H$-parity
shift of $V^{H}$ is isomorphic to $V^{H}$. If $W$ is irreducible then $W^{+h}$
is irreducible. And finally, $W$ and $W^{+h}$ are isomorphic as
$\Gamma/H$-graded modules.

######  Lemma 5.10.

Let $V$ be a $\Gamma/H$-graded module and $W$ be a $\Gamma$-graded submodule
of $V^{H}$. Then $W^{+h}$ can be embedded in $V^{H}$.

###### Proof.

Define a map $\Phi\colon W^{+h}\to V^{H}$ by

$\Phi(w)=\varphi_{\gamma}^{-1}\varphi_{\gamma+h}w$

for $w\in W^{+h}_{\gamma}$ and extending to inhomogeneous elements by
linearity. We claim that this map is a monomorphism. The map $\Phi$ is
obviously injective, linear and homogeneous of degree $0$. Now, for
$x\in\mathfrak{g}_{\alpha},\,w\in W^{+h}_{\gamma}$, we have

$\displaystyle\rho^{H}(x)w$
$\displaystyle=\varphi_{\alpha+\gamma+h}^{-1}\rho(x)\varphi_{\gamma+h}(w)$
$\displaystyle=\varphi_{\alpha+\gamma+h}^{-1}\varphi_{\alpha+\gamma}\varphi_{\alpha+\gamma}^{-1}\rho(x)\varphi_{\gamma}\varphi_{\gamma}^{-1}\varphi_{\gamma+h}(w)$
$\displaystyle=\Phi^{-1}\rho^{H}(x)\Phi(w).\qed$

In what follows, we will identify $W^{+h}$ with its embedding in $V^{H}$.

######  Theorem 5.11.

Let $H\leq\Gamma$ be a finite simple group and
$V=\bigoplus_{\Lambda\in\Gamma/H}V_{\Lambda}$ be a finite-dimensional
irreducible $\Gamma/H$-graded module of a Lie colour algebra $\mathfrak{g}$
over $\mathbb{C}$. Then, the refining extension $V^{H}$ is completely
reducible as a $\Gamma$-graded module with

$V^{H}=\bigoplus_{h\in K}W^{+h}$

for any irreducible $\Gamma$-graded module $W\subseteq V^{H}$ and some subset
$K\subseteq H$.

###### Proof.

Take any irreducible $\Gamma$-graded module $W\subseteq V^{H}$. Such a module
exists because $V^{H}$ is finite-dimensional.

Define a map $\Phi\colon W\to V$ by

$\Phi(w)=\varphi_{\gamma}(w)$

for $w\in W_{\gamma}\subseteq V^{H}_{\gamma}$ and extending to inhomogeneous
elements by linearity. We claim that $\Phi$ is $\Gamma/H$-graded intertwiner.
That $\Phi$ is linear and homogeneous of degree $0$ is obvious. For
$x\in\mathfrak{g}_{\alpha}$, we have

$\rho(x)\Phi(w)=\rho(x)\varphi_{\gamma}(w)=\varphi_{\alpha+\gamma}\varphi_{\alpha+\gamma}^{-1}\rho(x)\varphi_{\gamma}(w)=\Phi(\rho^{H}(x)w)$

which proves the claim.

We now know that $\Phi(W)$ is a submodule of $V$. But since $W\neq 0$, one of
the sectors $W_{\gamma^{\prime}}$ of $W$ must be nonzero. And because
$\varphi_{\gamma^{\prime}}$ is a vector space isomorphism,
$\Phi(W_{\gamma^{\prime}})\neq 0$. Since $V$ is irreducible, we must have that
$\Phi(W)=V$.

Now, consider the module $\Sigma_{h\in H}W^{+h}\subseteq V^{H}$. In
particular, consider the $\gamma$ sector $\sum_{h\in H}W^{+h}_{\gamma}$ for
some $\gamma\in\Gamma$. Applying $\varphi_{\gamma}$, we find that

$\varphi_{\gamma}\left(\sum_{h\in H}W^{+h}_{\gamma}\right)=\sum_{h\in
H}\varphi_{\gamma+h}(W_{\gamma+h})=\Phi\left(\bigoplus_{h\in
H}W_{\gamma+h}\right)=V_{\gamma+H}=\varphi_{\gamma}(V^{H}_{\gamma})$

(recalling the embedding in 5.10). Therefore, $\sum_{h\in H}W^{+h}=V^{H}$.

All that remains to show is that we can take a subset of $H$ to obtain a
direct sum. But this follows from 2.3. ∎

######  Corollary 5.12.

Let $V$ be irreducible. Every irreducible $\Gamma$-graded submodule of $V^{H}$
is isomorphic as a $\Gamma/H$-graded submodule to either $V^{H}$ or $V$.

###### Proof.

Take $W$ an irreducible $\Gamma$-graded submodule of $V^{H}$. By Theorem 5.11,

$V^{H}=\bigoplus_{h\in K}W^{+h}.$

Since $V^{H}$ is a completely reducible $\Gamma/H$-graded module, we know that
$W$ is a completely reducible $\Gamma/H$-graded module with direct summands
contained in those of $V^{H}$. By Theorem 5.9, $W$ is a direct sum of some
$\Gamma/H$-representative twists of $V$. Let $n$ be the number of
$\Gamma/H$-graded direct summands in $W$.

Then, the number of direct summands in $V^{H}$ as a $\Gamma/H$-graded module
is equal to $n\absolutevalue{K}$. But, from Theorem 5.9, the number of direct
summands is equal to $p$. Therefore, $n=1$ or $n=p$. If $n=p$ then $W\cong
V^{H}$. If $n=1$ then $W$ is isomorphic to $V^{\zeta}$ for some $p$-th root of
unity $\zeta$. But then $W$ gives $V^{\zeta}$ a $\Gamma$-grading, so $W\cong
V^{\zeta}\cong V$ by 5.7. ∎

In combination with Theorem 5.2, the above corollary tells us that every
$\Gamma$-graded module appears as either an irreducible $\Gamma/H$-graded
module or the refining extension of one. This result will greatly aid in the
search for irreducible Lie colour algebra representations. One difficulty
remains: in the former case, finding a $\Gamma$-grade for a $\Gamma/H$-graded
module can be difficult in general. To aid with this, we prove a uniqueness
result for the $\Gamma$-grade.

####  5.3.2. Uniqueness of the $\Gamma$-grade

It is possible for two modules to be non-isomorphic as $\Gamma$-graded modules
but isomorphic as $\Gamma/H$-graded modules. This is important to keep in mind
when trying to find a $\Gamma$-grading for an irreducible $\Gamma/H$-graded
module; there might be multiple _different_ $\Gamma$-gradings, all of which
must be considered. However, the following proposition gives a uniqueness
result for the $\Gamma$-grade.

######  Proposition 5.13.

If $V$ is a finite-dimensional irreducible $\Gamma/H$-graded
$\mathfrak{g}$-module, then any two $\Gamma$-gradings for $V$ are equivalent
up to $H$-parity shift.

More precisely, this proposition tells us that if
$V=\bigoplus_{\gamma\in\Gamma}V_{\gamma}$ and
$V=\bigoplus_{\gamma\in\Gamma}\widetilde{V}_{\gamma}$ are two
$\Gamma$-gradings for $V$ which both respect the $\Gamma/H$ grading (i.e. for
each $\Lambda\in\Gamma/H$ the original $\Lambda$-sector is
$\bigoplus_{\gamma\in\Lambda}V_{\gamma}=\bigoplus_{\gamma\in\Lambda}\widetilde{V}_{\gamma}$)
then there exists $\eta\in H$ such that
$V^{+\eta}_{\gamma}=\widetilde{V}_{\gamma}$.

###### Proof.

Both $V=\bigoplus_{\gamma\in\Gamma}V_{\gamma}$ and
$V=\bigoplus_{\gamma\in\Gamma}\widetilde{V}_{\gamma}$ appear as irreducible
$\Gamma$-graded submodules of $V^{H}$ by 3.4. Since all irreducible direct
summands of $V^{H}$ are $H$-parity shifts of $V$ by Theorem 5.11, the result
follows. ∎

###  5.4. The bijection

Let $\mathfrak{g}=\bigoplus_{\gamma\in\Gamma}\mathfrak{g}_{\gamma}$ be a Lie
colour algebra and $H\leq\Gamma$ be a finite simple group. It is easily
verified that $\Gamma/H$-representative twists and $H$-parity shifts both
define equivalence relations. Let
$\mathsf{Irr}_{\Gamma/H}^{\textup{tw}}(\mathfrak{g})$ be the collection of
equivalence classes of finite-dimensional irreducible $\Gamma/H$-graded
$\mathfrak{g}$-modules, where two $\Gamma/H$-graded modules are equivalent up
to $\Gamma/H$-representative twist. Let
$\mathsf{Irr}_{\Gamma}^{H\textup{p}}(\mathfrak{g})$ be the collection of
equivalence classes of finite-dimensional irreducible $\Gamma$-graded
$\mathfrak{g}$-modules, where two $\Gamma$-graded modules are equivalent up to
$H$-parity shift.

######  Theorem 5.14.

Define a function
$\mathcal{F}\colon\mathsf{Irr}_{\Gamma/H}^{\textup{tw}}\to\mathsf{Irr}_{\Gamma}^{H\textup{p}}$
by

$\mathcal{F}(V)=\begin{cases}V&\text{if $V$ is $\Gamma$-gradable}\\\
V^{H}&\text{otherwise}\end{cases}$

where $V$ and $V^{H}$ are representatives of the equivalence classes. Then
$\mathcal{F}$ is well-defined (does not depend on the choice of representative
$V$) and bijective.

###### Proof.

The $\Gamma$-gradability remains unchanged after applying a
$\Gamma/H$-representative twist or an isomorphism. And any $\Gamma$-grading of
$V$ is unique up to $H$-parity shift (5.13). Observe that
$\mathcal{F}(V^{\zeta})=(\mathcal{F}(V))^{\zeta}=\mathcal{F}(V)$, since
$\Gamma/H$-representative twists are isomorphic for $\Gamma$-graded modules
(5.7). We can hence deduce that $\mathcal{F}$ is well defined.

For injectivity, assume that $\mathcal{F}(V)=\mathcal{F}(\widetilde{V})$.
Using Theorem 5.9, both $\mathcal{F}(V)$ and $\mathcal{F}(\widetilde{V})$ are
completely reducible as $\Gamma/H$-graded modules with direct summands
contained among the $\Gamma/H$-representative twists. In particular, $V$
appears as a direct summand of $\mathcal{F}(\widetilde{V})$, so we must have
$V\cong\widetilde{V}^{\zeta}$ for some $p$-th root of unity $\zeta$. That is,
$V$ and $\widetilde{V}$ are equivalent in
$\mathsf{Irr}_{\Gamma/H}^{\textup{tw}}$ and thus $\mathcal{F}$ is injective.

For surjectivity, take an arbitrary finite dimensional irreducible
$\Gamma$-graded module $W$. By Theorem 5.2 and Theorem 5.11, this module
appears as a submodule of the refining extension of some irreducible
$\Gamma/H$-graded module $V$. By 5.12, the only $\Gamma$-graded submodules of
$V^{H}$ are either $V^{H}$ itself or $V$ with a suitable $\Gamma$-grading. If
$V$ can be given a $\Gamma$-grading then it appears as a $\Gamma$-graded
submodule of $V^{H}$ (by 3.4) making $V^{H}$ reducible. Thus, there is only
one option (up to $H$-parity shift) for what $W$ could have been. This only
option is $\mathcal{F}(V)$ and so $\mathcal{F}$ is surjective. ∎

In general, it may be difficult to determine which case of $\mathcal{F}$
applies. However, there are multiple options we can test:

  * •

Try to find a $\Gamma$-grading for $V$.

  * •

Try to find $\Gamma$-graded proper submodules of $V^{H}$ (all of which will
provide a $\Gamma$-grading for $V$) or else prove that $V^{H}$ is irreducible
(and hence $V$ will not be $\Gamma$-gradable).

  * •

Check whether the $\Gamma/H$-representative twists of $V$ are isomorphic. If
not, then $V$ cannot be $\Gamma$-graded (by 5.7).

Taking $\Gamma=\mathbb{Z}_{2}\times\mathbb{Z}_{2},\,H=\\{00,11\\}$, the above
theorem (in combination with discolouration/recolouration) gives a way to
construct the irreducible modules of the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebras from the
$\mathbb{Z}_{2}$-graded Lie superalgebras. More generally, we can use iterated
applications of the theorem to produce all of the irreducible $\Gamma$-graded
modules for any finite abelian group $\Gamma$. Theorem 5.14 also gives us a
bijection between the graded and ungraded modules of the colour algebra.

##  6\. Example: colour $\mathfrak{sl}_{2}$

As an example, we will determine the finite-dimensional irreducible
representations of $\mathfrak{sl}_{2}^{\mathrm{c}}$ over $\mathbb{C}$. The
algebra $\mathfrak{sl}_{2}^{\mathrm{c}}$ is a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded Lie colour algebra with
commutation factor
$\varepsilon(\alpha_{1}\alpha_{2},\beta_{1}\beta_{2})=(-1)^{\alpha_{1}\beta_{2}-\alpha_{2}\beta_{1}}$
and sectors spanned by the following elements:

$\displaystyle 00\text{-sector:}$ $\displaystyle~{}0$ $\displaystyle
10\text{-sector:}$ $\displaystyle~{}a_{1}$ $\displaystyle 01\text{-sector:}$
$\displaystyle~{}a_{2}$ $\displaystyle 11\text{-sector:}$
$\displaystyle~{}a_{3}$

and equipped with a bracket $\cbrak{\cdot}{\cdot}$ which satisfies

$\displaystyle\cbrak{a_{1}}{a_{2}}$ $\displaystyle=a_{3}$
$\displaystyle\cbrak{a_{2}}{a_{3}}$ $\displaystyle=a_{1}$
$\displaystyle\cbrak{a_{3}}{a_{1}}$ $\displaystyle=a_{2}.$

Using the multiplier
$\sigma(\alpha_{1}\alpha_{2},\beta_{1}\beta_{2})=(-1)^{\alpha_{2}\beta_{1}}$,
$\mathfrak{sl}_{2}^{\mathrm{c}}$ discolours to a Lie algebra with Lie bracket
$\cbrak{\cdot}{\cdot}^{\sigma}=\commutator{\cdot}{\cdot}$ given by

$\displaystyle\commutator{a_{1}}{a_{2}}$ $\displaystyle=a_{3}$
$\displaystyle\commutator{a_{2}}{a_{3}}$ $\displaystyle=-a_{1}$
$\displaystyle\commutator{a_{3}}{a_{1}}$ $\displaystyle=-a_{2}.$

These commutation relations are satisfied by the following realisation

(2) $\displaystyle a_{1}$ $\displaystyle=\frac{i}{2}(e-f)$ $\displaystyle
a_{2}$ $\displaystyle=-\frac{1}{2}(e+f)$ $\displaystyle a_{3}$
$\displaystyle=-\frac{i}{2}h.$

in terms of a standard basis $\\{h,e,f\\}$ for $\mathfrak{sl}_{2}$,
($\commutator{h}{e}=2e,\,\commutator{h}{f}=-2f,\,\commutator{e}{f}=h$).
Consequently, the discolouration $(\mathfrak{sl}_{2}^{\mathrm{c}})^{\sigma}$
is isomorphic to $\mathfrak{sl}_{2}$ (hence the notation).

The Lie colour algebra $\mathfrak{sl}_{2}^{\mathrm{c}}$ has previously
appeared in [47], in which the authors found all the ungraded irreducible
representations for $\mathfrak{sl}_{2}^{\mathrm{c}}$ by embedding the
universal enveloping algebra $U(\mathfrak{sl}_{2}^{\mathrm{c}})$ in
$M_{2}(U(\mathfrak{sl}_{2}))$ (the space of $2\times 2$ matrices with entries
in $U(\mathfrak{sl}_{2})$). The approach that we will take via refining
extensions is very similar, though we will not work at the level of universal
enveloping algebras. Additionally, our approach is undertaken in the context
of the more general preceding sections.

Our strategy is as follows:

  1. (1)

find all of the ungraded irreducible representations for $\mathfrak{sl}_{2}$;

  2. (2)

use the techniques from the preceding sections to find the
$\mathbb{Z}_{2}$-graded irreducible representations for $\mathfrak{sl}_{2}$;

  3. (3)

similarly, find the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded irreducible
representations for $\mathfrak{sl}_{2}$;

  4. (4)

recolour to obtain the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded irreducible
representations for $\mathfrak{sl}_{2}^{\mathrm{c}}$.

  5. (5)

Find the ungraded irreducible representations for
$\mathfrak{sl}_{2}^{\mathrm{c}}$ as subrepresentations

###  6.1. Irreducible representations for $\mathfrak{sl}_{2}$

Recall that $\mathfrak{sl}_{2}$ has a unique irreducible representation
$V_{\lambda}$ of dimension $\lambda+1$ for every highest weight
$\lambda\in\mathbb{Z}_{\geq 0}$.We will choose a basis
$\\{v_{0},v_{1},\ldots,v_{\lambda}\\}$ for $V_{\lambda}$ such that

$\displaystyle hv_{j}$ $\displaystyle=(\lambda-2j)v_{j}$ $\displaystyle
ev_{j}$ $\displaystyle=(\lambda-j+1)v_{j-1}$ $\displaystyle fv_{j}$
$\displaystyle=(j+1)v_{j+1}$

with the convention $v_{-1}=v_{\lambda+1}=0$. In particular, note that $v_{0}$
is the highest weight vector. (In the above line and what follows, we have
omitted the explicit representation map
$\rho\colon\mathfrak{sl}_{2}\to\operatorname{End}(V_{\lambda})$.) Using (2),
we find that

$\displaystyle a_{1}v_{j}$
$\displaystyle=\frac{i}{2}((\lambda-j+1)v_{j-1}-(j+1)v_{j+1})$ $\displaystyle
a_{2}v_{j}$ $\displaystyle=-\frac{1}{2}((\lambda-j+1)v_{j-1}+(j+1)v_{j+1})$
$\displaystyle a_{3}v_{j}$ $\displaystyle=-\frac{i}{2}(\lambda-2j)v_{j}.$

###  6.2. $\mathbb{Z}_{2}$-graded irreducible representations

We wish to find the $\mathbb{Z}_{2}$-graded representations. So, we choose
$\Gamma_{1}=H_{1}=\mathbb{Z}_{2}\times\mathbb{Z}_{2}/\\{00,11\\}\cong\mathbb{Z}_{2}$.
Under this group and the original $\mathfrak{sl}_{2}^{\mathrm{c}}$ grading,
the sectors for $\mathfrak{sl}_{2}$ are spanned by the following elements:

$\displaystyle 0\text{-sector:}$ $\displaystyle~{}a_{3}$ $\displaystyle
1\text{-sector:}~{}a_{1},a_{2}.$

The ungraded module $V_{\lambda}$ can be given a $\Gamma_{1}$-grading:
$V_{\lambda,0}^{\mathrm{E}}\oplus V_{\lambda,1}^{\mathrm{E}}$ where

$\displaystyle V_{\lambda,0}^{\mathrm{E}}$
$\displaystyle=\operatorname{span}\\{v_{j}\mid j~{}\text{even}\\}$
$\displaystyle V_{\lambda,1}^{\mathrm{E}}$
$\displaystyle=\operatorname{span}\\{v_{j}\mid j~{}\text{odd}\\}.$

We can easily check that $a_{1}V_{\lambda,\gamma}^{\mathrm{E}}\subseteq
V_{\lambda,\gamma+1}^{\mathrm{E}},\,a_{2}V_{\lambda,\gamma}^{\mathrm{E}}\subseteq
V_{\lambda,\gamma+1}^{\mathrm{E}}$ and
$a_{3}V_{\lambda,\gamma}^{\mathrm{E}}\subseteq
V_{\lambda,\gamma}^{\mathrm{E}}$ for $\gamma\in\mathbb{Z}_{2}$. Call this
$\Gamma_{1}$-graded module $V_{\lambda}^{\mathrm{E}}$.

Since $V_{\lambda}$ is an irreducible ungraded module, $V_{\lambda}^{E}$ is an
irreducible $\Gamma_{1}$-graded module. By shifting the $H_{1}$-parity of
$V_{\lambda}^{\mathrm{E}}$, we get another $\Gamma_{1}$-graded irreducible
module $V_{\lambda}^{\mathrm{O}}=V_{\lambda,0}^{\mathrm{O}}\oplus
V_{\lambda,1}^{\mathrm{O}}$ where

$\displaystyle V_{\lambda,0}^{\mathrm{O}}$
$\displaystyle=\operatorname{span}\\{v_{j}\mid j~{}\text{odd}\\}$
$\displaystyle V_{\lambda,1}^{\mathrm{O}}$
$\displaystyle=\operatorname{span}\\{v_{j}\mid j~{}\text{even}\\}.$

By 5.13, we do not need to search for any other gradings. And by 3.4, we do
not need to check the refining extension at all. Thus, the finite-dimensional
irreducible $\Gamma_{1}$-graded $\mathfrak{sl}_{2}$-modules are
$V_{\lambda}^{\mathrm{E}}$ and $V_{\lambda}^{\mathrm{O}}$ for
$\lambda\in\mathbb{Z}_{\geq 0}$.

###  6.3. $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded irreducible
representations

Now we wish to find the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded
representations. So, we choose $\Gamma_{2}=\mathbb{Z}_{2}\times\mathbb{Z}_{2}$
and $H_{2}=\\{00,11\\}$. We will now find the $\Gamma_{2}$-gradings, with the
aid of the following lemma.

######  Lemma 6.1.

Let $V_{\lambda}=\bigoplus_{\gamma\in\Gamma_{2}}W_{\gamma}$ be a
$\Gamma_{2}$-grading for $V_{\lambda}$. Then
$v_{0}=\sum_{\gamma\in\Gamma_{2}}v_{0,\gamma}$ where $v_{0,\gamma}\in
W_{\gamma}$ is some linear combination of $v_{0}$ and $v_{\lambda}$ (possibly
equal to zero).

###### Proof.

Since $h=2ia_{3}$ is homogeneous of degree $11$, we find that
$h^{2}W_{\gamma}\subseteq W_{\gamma}$. If we project $v_{0}$ onto each sector
and write $v_{0}=\sum_{\gamma\in\Gamma_{2}}v_{0,\gamma}$, for $v_{0,\gamma}\in
W_{\gamma}$ then we find that

$\sum_{\gamma\in\Gamma_{2}}h^{2}v_{0,\gamma}=h^{2}(v_{0})=\sum_{\gamma\in\Gamma_{2}}\lambda^{2}v_{0,\gamma}.$

In particular, $h^{2}v_{0,\gamma}=\lambda^{2}v_{0,\gamma}$ for every
$\gamma\in\Gamma_{2}$. Using our knowledge of the eigenvectors of $h$, we find
that $v_{0,\gamma}$ must be a linear combination of $v_{0}$ and $v_{\lambda}$.
∎

######  Remark 6.2.

A similar result holds for
$v_{\lambda}=\sum_{\gamma\in\Gamma_{2}}v_{\lambda,\gamma}$.

6.1 tells us that at least one sector of any $\Gamma_{2}$-grading must contain
a nonzero linear combination of $v_{0}$ and $v_{\lambda}$. Our strategy is to
then apply $a_{1}$ and $a_{2}$ to this linear combination to help deduce the
structure of a $\Gamma_{2}$-grading or prove that one cannot exist.

####  6.3.1. The $\lambda$ even case

If $\lambda$ is even, then both $v_{0}$ and $v_{\lambda}$ are in
$V_{\lambda,0}^{\mathrm{E}}$. Therefore, using 6.1, we find that the
$00$-sector or $11$-sector must contain a nonzero linear combination of
$v_{0}$ and $v_{\lambda}$. We choose (after some trial and error)
$v_{0}+v_{\lambda}$ to be in the $00$-sector and $v_{0}-v_{\lambda}$ to be in
the $11$-sector. With this choice, we can find a $\Gamma_{2}$-grading for
$V_{\lambda}^{\mathrm{E}}$:

(3) $\displaystyle V_{\lambda,00}^{\mathrm{E}+}$
$\displaystyle=\operatorname{span}\\{v_{j}+v_{\lambda-j}\mid
j~{}\text{even}\\},$ $\displaystyle V_{\lambda,10}^{\mathrm{E}+}$
$\displaystyle=\operatorname{span}\\{v_{j}-v_{\lambda-j}\mid
j~{}\text{odd}\\},$ $\displaystyle V_{\lambda,01}^{\mathrm{E}+}$
$\displaystyle=\operatorname{span}\\{v_{j}+v_{\lambda-j}\mid
j~{}\text{odd}\\},$ $\displaystyle V_{\lambda,11}^{\mathrm{E}+}$
$\displaystyle=\operatorname{span}\\{v_{j}-v_{\lambda-j}\mid
j~{}\text{even}\\}.$

Denote this $\Gamma_{2}$-graded module by $V_{\lambda}^{\mathrm{E}+}$. We
calculate

$\displaystyle a_{1}(v_{j}\pm v_{\lambda-j})$
$\displaystyle=\frac{i}{2}((\lambda-j+1)(v_{j-1}\mp
v_{\lambda-(j-1)})-(j+1)(v_{j+1}\mp v_{\lambda-(j+1)})$ $\displaystyle
a_{2}(v_{j}\pm v_{\lambda-j})$
$\displaystyle=-\frac{1}{2}((\lambda-j+1)(v_{j-1}\pm
v_{\lambda-(j-1)})+(j+1)(v_{j+1}\pm v_{\lambda-(j+1)})$ $\displaystyle
a_{3}(v_{j}\pm v_{\lambda-j})$
$\displaystyle=-\frac{i}{2}((\lambda-2j)(v_{j}\mp v_{\lambda-j}))$

from which we can easily verify that
$a_{1}V_{\lambda,\gamma}^{\mathrm{E}+}\subseteq
V_{\lambda,\gamma+10}^{\mathrm{E}+}$,
$a_{2}V_{\lambda,\gamma}^{\mathrm{E}+}\subseteq
V_{\lambda,\gamma+01}^{\mathrm{E}+}$ and
$a_{3}V_{\lambda,\gamma}^{\mathrm{E}+}\subseteq
V_{\lambda,\gamma+11}^{\mathrm{E}+}$ for each $\gamma\in\Gamma_{2}$. Since
$V_{\lambda}^{\mathrm{E}}$ is irreducible, so is $V_{\lambda}^{\mathrm{E}+}$.
By shifting the $H_{2}$-parity of $V_{\lambda}^{\mathrm{E}+}$, we get another
$\Gamma_{2}$-graded irreducible module $V_{\lambda}^{\mathrm{E}-}$, which can
be obtained from $V_{\lambda}^{\mathrm{E}+}$ by swapping the $00$\- with the
$11$-sector and the $10$\- with the $01$-sector. Note that
$V_{\lambda}^{\mathrm{E}+}$ and $V_{\lambda}^{\mathrm{E}-}$ are not
isomorphic, because $v_{\lambda/2}=(1/2)(v_{\lambda/2}+v_{\lambda-\lambda/2})$
(the unique vector with weight $0$) will be in different sectors.

Performing similar computations with $V_{\lambda}^{\mathrm{O}}$, we find two
more irreducible $\Gamma_{2}$-graded modules $V_{\lambda}^{\mathrm{O}+}$ and
$V_{\lambda}^{\mathrm{O}-}$ (these can be obtained from
$V_{\lambda}^{\mathrm{E}+}$ by swapping the $00$\- with the $01$-sector and
the $10$\- with the $11$-sector to get $V_{\lambda}^{\mathrm{O}+}$; and by
swapping the $00$\- with the $10$-sector and the $01$\- with the $11$-sector
to get $V_{\lambda}^{\mathrm{O}-}$). By 5.13 and Theorem 5.14 we do not need
to search for any more irreducible $\Gamma_{2}$-graded modules in the case
where $\lambda$ is even.

####  6.3.2. The $\lambda$ odd case

######  Lemma 6.3.

If $\lambda$ is odd, then $V_{\lambda}^{\mathrm{E}}$ cannot be given a
$\Gamma_{2}$-grading.

###### Proof.

For contradiction, assume that we can give $V_{\lambda}^{\mathrm{E}}$ a
$\Gamma_{2}$-grading,
$V_{\lambda}^{\mathrm{E}}=\bigoplus_{\gamma\in\Gamma_{2}}V_{\lambda,\gamma}^{\mathrm{E}}$.
From the structure of $V_{\lambda}^{\mathrm{E}}$, we know that $v_{0}\in
V_{\lambda,0}^{\mathrm{E}}=V_{\lambda,00}^{\mathrm{E}}\oplus
V_{\lambda,11}^{\mathrm{E}}$. Therefore, we can write
$v_{0}=v_{0,00}+v_{0,11}$ for $v_{0,00}\in
V_{\lambda,00}^{\mathrm{E}},\,v_{0,11}\in V_{\lambda,11}^{\mathrm{E}}$. By
6.1, both $v_{0,00}$ and $v_{0,11}$ are linear combinations of $v_{0}$ and
$v_{\lambda}$. But since $v_{\lambda}\in
V_{\lambda,1}^{\mathrm{E}}=V_{\lambda,10}^{\mathrm{E}}\oplus
V_{\lambda,01}^{\mathrm{E}}$, we must have that both $v_{0,00}$ and $v_{0,11}$
are scalar multiples of $v_{0}$. And since they sum to $v_{0}\neq 0$, at least
one of them must be nonzero. That is, $v_{0}\in V_{\lambda,00}^{\mathrm{E}}$
or $v_{0}\in V_{\lambda,11}^{\mathrm{E}}$. An analogous argument shows that
$v_{\lambda}\in V_{\lambda,10}$ or $v_{\lambda}\in V_{\lambda,01}$.

If $v_{0}\in V_{\lambda,00}^{\mathrm{E}}$ and $v_{\lambda}\in
V_{\lambda,01}^{\mathrm{E}}$ then applying $a_{1}$ we find

$\displaystyle v_{1}$ $\displaystyle=2ia_{1}v_{0}\in
V_{\lambda,00}^{\mathrm{E}}\oplus V_{\lambda,10}^{\mathrm{E}}$ $\displaystyle
v_{2}$ $\displaystyle=ia_{1}v_{1}+\frac{\lambda}{2}v_{0}\in
V_{\lambda,00}^{\mathrm{E}}\oplus V_{\lambda,10}^{\mathrm{E}}$ $\displaystyle\
\vdots$ $\displaystyle v_{j+1}$
$\displaystyle=\frac{2i}{j+1}a_{1}v_{j}+\frac{\lambda-j+1}{j+1}v_{j-1}\in
V_{\lambda,00}^{\mathrm{E}}\oplus V_{\lambda,10}^{\mathrm{E}}.$

Proceeding inductively, we find that $V_{\lambda}^{\mathrm{E}}\subseteq
V_{\lambda,00}^{\mathrm{E}}\oplus V_{\lambda,10}^{\mathrm{E}}$. But, by
applying $a_{1}$ to $v_{\lambda}$ in a similar way, we find that
$V_{\lambda}^{\mathrm{E}}\subseteq V_{\lambda,01}^{\mathrm{E}}\oplus
V_{\lambda,11}^{\mathrm{E}}$, a contradiction! We similarly arrive at a
contradiction if we assume $v_{0}\in V_{\lambda,11}^{\mathrm{E}}$ and
$v_{\lambda}\in V_{\lambda,10}^{\mathrm{E}}$; $v_{0}\in
V_{\lambda,00}^{\mathrm{E}}$ and $v_{\lambda}\in V_{\lambda,10}^{\mathrm{E}}$;
or $v_{0}\in V_{\lambda,11}$ and $v_{\lambda}\in V_{\lambda,01}$. (Note that
in some of these cases we need to use $a_{2}$ instead of $a_{1}$.) ∎

We have just shown that $V_{\lambda}^{\mathrm{E}}$ cannot be given a
$\Gamma_{2}$-grading if $\lambda$ is odd. By 5.12, the refining extension
$V_{\lambda}^{\mathrm{E}H_{2}}$ must therefore be an irreducible
$\Gamma_{2}$-graded representation. If we take a basis
$\\{v_{\alpha,j}\mid\alpha\in
H_{2}\cong\mathbb{Z}_{2},\,j=0,\ldots,\lambda\\}$ for
$V_{\lambda}^{\mathrm{E}H_{2}}$, where

(4) $\displaystyle V^{\mathrm{E}H_{2}}_{\lambda,00}$
$\displaystyle=\operatorname{span}\\{v_{0,j}\mid j~{}\text{even}\\},$
$\displaystyle V^{\mathrm{E}H_{2}}_{\lambda,10}$
$\displaystyle=\operatorname{span}\\{v_{1,j}\mid j~{}\text{odd}\\},$
$\displaystyle V^{\mathrm{E}H_{2}}_{\lambda,01}$
$\displaystyle=\operatorname{span}\\{v_{0,j}\mid j~{}\text{odd}\\},$
$\displaystyle V^{\mathrm{E}H_{2}}_{\lambda,11}$
$\displaystyle=\operatorname{span}\\{v_{1,j}\mid j~{}\text{even}\\},$

then we can compute the action as

$\displaystyle a_{1}v_{\alpha,j}$
$\displaystyle=\frac{i}{2}((\lambda-j+1)v_{\alpha+1,j-1}-(j+1)v_{\alpha+1,j+1})$
$\displaystyle a_{2}v_{\alpha,j}$
$\displaystyle=-\frac{1}{2}((\lambda-j+1)v_{\alpha,j-1}+(j+1)v_{\alpha,j+1})$
$\displaystyle a_{3}v_{\alpha,j}$
$\displaystyle=-\frac{i}{2}(\lambda-2j)v_{\alpha+1,j}.$

By similar reasoning, we find that $V_{\lambda}^{\mathrm{O}H_{2}}$ is an
irreducible $\Gamma_{2}$-graded representation. We can obtain
$V_{\lambda}^{\mathrm{O}H_{2}}$ from $V_{\lambda}^{\mathrm{E}H_{2}}$ by
swapping the $00$\- with the $01$-sector and the $10$\- with the $11$-sector.
Note that if we shift the $H_{2}$-parity of either of these modules, we get
the same module. By Theorem 5.14, the only irreducible $\Gamma_{2}$-graded
modules for the case where $\lambda$ is odd are
$V_{\lambda}^{\mathrm{E}H_{2}}$ and $V_{\lambda}^{\mathrm{O}H_{2}}$.

###  6.4. Recolouring

To recolour $\mathfrak{sl}_{2}$ to $\mathfrak{sl}_{2}^{\mathrm{c}}$ we use the
multiplier
$\sigma(\alpha_{1}\alpha_{2},\beta_{1}\beta_{2})=(-1)^{\alpha_{2}\beta_{1}}$.
To obtain the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded irreducible modules
for $\mathfrak{sl}_{2}^{\mathrm{c}}$, we use this same multiplier for the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded irreducible modules for
$\mathfrak{sl}_{2}$.

######  Theorem 6.4.

The only finite-dimensional $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded
irreducible modules for $\mathfrak{sl}_{2}^{\mathrm{c}}$ are:

  1. (i)

For each even number $\lambda$, the four modules which are equivalent to
$(V_{\lambda}^{\mathrm{E}+})^{\sigma}$ up to
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-parity shift. The sectors of
$(V_{\lambda}^{\mathrm{E}+})^{\sigma}$ are given as in (3) and the action is
given by

$\displaystyle a_{1}(v_{j}\pm v_{\lambda-j})$
$\displaystyle=\frac{i}{2}((\lambda-j+1)(v_{j-1}\mp
v_{\lambda-(j-1)})-(j+1)(v_{j+1}\mp v_{\lambda-(j+1)})$ $\displaystyle
a_{2}(v_{j}\pm v_{\lambda-j})$
$\displaystyle=\mp\frac{1}{2}((\lambda-j+1)(v_{j-1}\pm
v_{\lambda-(j-1)})+(j+1)(v_{j+1}\pm v_{\lambda-(j+1)})$ $\displaystyle
a_{3}(v_{j}\pm v_{\lambda-j})$
$\displaystyle=\mp\frac{i}{2}((\lambda-2j)(v_{j}\mp v_{\lambda-j})).$

The dimension of these modules is $\lambda+1$.

  2. (ii)

For each odd number $\lambda$, the two modules which are equivalent to
$(V_{\lambda}^{\mathrm{E}H_{2}})^{\sigma}$ up to $\\{00,01\\}$-parity shift.
The sectors of $(V_{\lambda}^{\mathrm{E}H_{2}})^{\sigma}$ are given as in (4)
and the action is given by

$\displaystyle a_{1}v_{\alpha,j}$
$\displaystyle=\frac{i}{2}((\lambda-j+1)v_{\alpha+1,j-1}-(j+1)v_{\alpha+1,j+1})$
$\displaystyle a_{2}v_{\alpha,j}$
$\displaystyle=-\frac{(-1)^{\alpha}}{2}((\lambda-j+1)v_{\alpha,j-1}+(j+1)v_{\alpha,j+1})$
$\displaystyle a_{3}v_{\alpha,j}$
$\displaystyle=-\frac{i(-1)^{\alpha}}{2}(\lambda-2j)v_{\alpha+1,j}.$

The dimension of these modules is $2(\lambda+1)$.

###### Proof.

Use the results from the previous sections and the facts that
$\sigma(10,\beta)=1$ for all $\beta\in\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and

$\sigma(01,\beta)=\sigma(11,\beta)=\begin{cases}-1&\text{if}~{}\beta=10,11\\\
1&\text{if}~{}\beta=00,01.\end{cases}\qed$

Note that, for each $\lambda\in\mathbb{Z}_{\geq 0}$, there is a unique
equivalence class of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded irreducible
$\mathfrak{sl}_{2}^{\mathrm{c}}$-modules. This corresponds exactly to the
unique irreducible $\mathfrak{sl}_{2}$-modules, demonstrating the existence of
the bijection in Theorem 5.14.

###  6.5. Ungraded irreducible representations for
$\mathfrak{sl}_{2}^{\mathrm{c}}$

######  Lemma 6.5.

The ungraded irreducible representations for $\mathfrak{sl}_{2}^{\mathrm{c}}$
all appear as ungraded subrepresentations of the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded irreducible representations in
Theorem 6.4.

###### Proof.

If we knew the ungraded irreducible representations for
$\mathfrak{sl}_{2}^{\mathrm{c}}$, then we could use the same process in the
preceding sections to derive the same
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded irreducible modules. By applying
Theorem 5.14 twice, these irreducible
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded modules are either irreducible as
ungraded modules, a refining extension or a refining extension of a refining
extension. In any case, the original ungraded irreducible modules must appear
as ungraded submodules of the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded
modules. ∎

Note that $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded modules which are
equivalent up to $H$-parity shift (for some $H\leq\mathbb{Z}_{2}$) are
isomorphic as ungraded modules.

######  Lemma 6.6.

If $\lambda$ is even, then $(V_{\lambda}^{E+})^{\sigma}$ is irreducible as an
ungraded module.

###### Proof.

Choose a new basis $\\{u_{j}\mid j=0,\ldots,\lambda\\}$ where
$u_{j}=(v_{j}+v_{\lambda-j})+i(-1)^{j}(v_{j}-v_{\lambda-j})$. Note that this
is not linearly dependent since
$u_{\lambda-j}=(v_{j}+v_{\lambda-j})-i(-1)^{j}(v_{j}-v_{\lambda-j})$. The
action of $\mathfrak{sl}_{2}^{\mathrm{c}}$ is then

(5) $\displaystyle a_{1}u_{j}$
$\displaystyle=\frac{(-1)^{j+1}}{2}((\lambda-j+1)u_{j-1}-(j+1)u_{j+1})$
$\displaystyle a_{2}u_{j}$
$\displaystyle=-\frac{1}{2}((\lambda-j+1)u_{j-1}+(j+1)u_{j+1})$ $\displaystyle
a_{3}u_{j}$ $\displaystyle=\frac{(-1)^{j+1}}{2}(\lambda-2j)u_{j}$

with the convention that $u_{-1}=u_{\lambda+1}=0$. In particular,

$\displaystyle(a_{2}+a_{1})u_{j}$
$\displaystyle=\begin{cases}-(\lambda-j+1)u_{j-1}&\text{if $j$ even}\\\
-(j+1)u_{j+1}&\text{if $j$ odd}.\end{cases}$ $\displaystyle(a_{2}-a_{1})u_{j}$
$\displaystyle=\begin{cases}-(j+1)u_{j+1}&\text{if $j$ even}\\\
-(\lambda-j+1)u_{j-1}&\text{if $j$ odd}.\\\ \end{cases}$

So, if we take an arbitrary element $\sum_{j}c_{j}u_{j}$ of
$(V_{\lambda}^{E+})^{\sigma}$, then we can alternate applying $(a_{2}+a_{1})$
and $(a_{2}-a_{1})$ (i.e. we apply $((a_{2}-a_{1})(a_{2}+a_{1}))^{k}$ or
$(a_{2}-a_{1})((a_{2}+a_{1})(a_{2}-a_{1}))^{k}$ for some non-negative integer
$k$) to raise/lower the basis elements of $\sum_{j}c_{j}u_{j}$. Eventually,
this procedure will yield a scalar multiple of either $u_{0}$ or
$u_{\lambda}$, from which we can generate the entire module. Since the entire
module can be generated from a single vector,
$(V_{\lambda}^{\mathrm{E}+})^{\sigma}$ is irreducible. ∎

Now we consider the case when $\lambda$ is odd. Then
$(V_{\lambda}^{\mathrm{E}H_{2}})^{\sigma}=U_{\lambda}^{++}\oplus
U_{\lambda}^{+-}\oplus U_{\lambda}^{-+}\oplus U_{\lambda}^{--}$ where

$\displaystyle U_{\lambda}^{\zeta\xi}$
$\displaystyle=\operatorname{span}\\{u^{\zeta\xi}_{j}\mid
j=0,1,\ldots,(\lambda-1)/2\\}$ $\displaystyle u^{\zeta\xi}_{j}$
$\displaystyle=v_{0,j}+\zeta i(-1)^{j}v_{1,j}+\xi v_{0,\lambda-j}-\zeta\xi
i(-1)^{j}v_{1,\lambda-j}$

for $\zeta,\xi\in\\{-1,+1\\}$. The action of $\mathfrak{sl}_{2}^{\mathrm{c}}$
is given by

(6) $\displaystyle a_{1}u^{\zeta\xi}_{j}$
$\displaystyle=-\frac{\zeta}{2}(-1)^{j}((\lambda-j+1)u^{\zeta\xi}_{j-1}-(j+1)u^{\zeta\xi}_{j+1})$
$\displaystyle\left(j<\frac{\lambda-1}{2}\right)$ $\displaystyle
a_{1}u^{\zeta\xi}_{(\lambda-1)/2}$
$\displaystyle=-\frac{\zeta}{2}(-1)^{(\lambda-1)/2}\left(\frac{\lambda+3}{2}u^{\zeta\xi}_{(\lambda-3)/2}-\xi\frac{\lambda+1}{2}u^{\zeta\xi}_{(\lambda-1)/2}\right)$
$\displaystyle a_{2}u^{\zeta\xi}_{j}$
$\displaystyle=-\frac{1}{2}((\lambda-j+1)u^{\zeta\xi}_{j-1}+(j+1)u^{\zeta\xi}_{j+1})$
$\displaystyle\left(j<\frac{\lambda-1}{2}\right)$ $\displaystyle
a_{2}u^{\zeta\xi}_{(\lambda-1)/2}$
$\displaystyle=-\frac{1}{2}\left(\frac{\lambda+3}{2}u^{\zeta\xi}_{(\lambda-3)/2}+\xi\frac{\lambda+1}{2}u^{\zeta\xi}_{(\lambda-1)/2}\right)$
$\displaystyle a_{3}u^{\zeta\xi}_{j}$
$\displaystyle=-\frac{\zeta}{2}(-1)^{j}(\lambda-2j)u^{\zeta\xi}_{j}$

with the convention $u_{-1}=1$. (Note that $u^{\zeta\xi}_{j}$ was chosen so
that it was an eigenvector of $a_{3}$.) Clearly $U_{\lambda}^{\zeta\xi}$ gives
rise to four non-isomorphic modules, since the action of $a_{1}$, $a_{2}$ and
$a_{3}$ on $u^{\zeta\xi}_{(\lambda-1)/2}$ (the unique eigenvector for $a_{3}$
which has eigenvalue with absolute value $1/2$) is different for different
values of $\zeta,\xi$.

######  Lemma 6.7.

$U_{\lambda}^{\zeta\xi}$ is irreducible as an ungraded module.

###### Proof.

Observe that, for $j<(\lambda-1)/2$,

$\displaystyle(a_{2}+a_{1})u^{\zeta\xi}_{j}$
$\displaystyle=\begin{cases}-(\lambda-j+1)u^{\zeta\xi}_{j-1}&\text{if}~{}\zeta(-1)^{j}=1\\\
-(j+1)u^{\zeta\xi}_{j+1}&\text{if}~{}\zeta(-1)^{j}=-1\end{cases}$
$\displaystyle(a_{2}+a_{1})u^{\zeta\xi}_{(\lambda-1)/2}$
$\displaystyle=\begin{cases}-\frac{\lambda+3}{2}u^{\zeta\xi}_{(\lambda-3)/2}&\text{if}~{}\zeta(-1)^{(\lambda-1)/2}=1\\\
-\xi\frac{\lambda+1}{2}u^{\zeta\xi}_{(\lambda-1)/2}&\text{if}~{}\zeta(-1)^{(\lambda-1)/2}=-1\end{cases}$
$\displaystyle(a_{2}-a_{1})u^{\zeta\xi}_{j}$
$\displaystyle=\begin{cases}-(j+1)u^{\zeta\xi}_{j+1}&\text{if}~{}\zeta(-1)^{j}=1\\\
-(\lambda-j+1)u^{\zeta\xi}_{j-1}&\text{if}~{}\zeta(-1)^{j}=-1\end{cases}$
$\displaystyle(a_{2}-a_{1})u^{\zeta\xi}_{(\lambda-1)/2}$
$\displaystyle=\begin{cases}-\xi\frac{\lambda+1}{2}u^{\zeta\xi}_{(\lambda-1)/2}&\text{if}~{}\zeta(-1)^{(\lambda-1)/2}=1\\\
-\frac{\lambda+3}{2}u^{\zeta\xi}_{(\lambda-3)/2}&\text{if}~{}\zeta(-1)^{(\lambda-1)/2}=-1\end{cases}$

Take an arbitrary element $\sum_{j}c_{j}u_{j}$ of $U_{\lambda}^{\zeta\xi}$.
Similar to the proof of 6.6 we alternate applying $a_{2}+a_{1}$ and
$a_{2}-a_{1}$. In this case, we are able to obtain a scalar multiple of
$u_{0}$ from $\sum_{j}c_{j}u_{j}$, from which we can generate the entire
module. ∎

In summary, we have proven the following theorem:

######  Theorem 6.8.

The only finite-dimensional irreducible ungraded modules for
$\mathfrak{sl}_{2}^{\mathrm{c}}$ are

  1. (i)

For each even number $\lambda$, the module
$(V_{\lambda}^{\mathrm{E}+})^{\sigma}$ with action given in (5). The dimension
of these modules is $\lambda+1$.

  2. (ii)

For each odd number $\lambda$ and each $\zeta,\xi\in\\{-1,+1\\}$, the module
$U_{\lambda}^{\zeta\xi}$ with action given in (6). The dimension of these
modules is $(\lambda+1)/2$.

######  Remark 6.9.

It is clear that the modules $U_{\lambda}^{+\xi}$ and $U_{\lambda}^{-\xi}$ are
$\Gamma/H$-representative twists of each other, where
$\Gamma=\mathbb{Z}_{2}\times\mathbb{Z}_{2}/\\{00,01\\}$ and $H=\Gamma$. If we
instead choose
$\Gamma=\mathbb{Z}_{2}\times\mathbb{Z}_{2}/\\{00,11\\},\,H=\Gamma$, then we
find that the $\Gamma/H$-representative twist of $U_{\lambda}^{\zeta+}$ is
isomorphic to $U_{\lambda}^{\zeta-}$, with isomorphism given by
$u^{\zeta+}_{j}\mapsto(-1)^{j}u_{j}^{\zeta-}$.

The above remark shows that $U^{\zeta\xi}$ are all equivalent up to
representative twist for different values of $\zeta,\xi$. Consequently, there
is a unique equivalence class in Theorem 6.8 for each
$\lambda\in\mathbb{Z}_{\geq 0}$. This corresponds exactly to the equivalence
classes for $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded
$\mathfrak{sl}_{2}^{\mathrm{c}}$-modules in Theorem 6.4, again demonstrating
the existence of the bijection in Theorem 5.14.

######  Remark 6.10.

Although the ungraded $\mathfrak{sl}_{2}^{\mathrm{c}}$ representations were
derived from the ungraded $\mathfrak{sl}_{2}$ representations, they have a
remarkably different structure. This shows that, despite the procedure
outlined in this paper, colour algebras still have an interesting
representation theory.

### Acknowledgements

The author would like to thank Phil Isaac for his invaluable guidance, and for
taking the time to suggest improvements to the manuscript.

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