Graded torsion-free ${\mathfrak{sl}_2(\mathbb{C})}$-modules of rank 2

TL;DR

This paper investigates the structure of graded, simple infinite-dimensional modules over the Lie algebra ${\mathfrak{sl}_2(\mathbb{C})}$, focusing on modules of rank 2 and their grading properties.

Contribution

It introduces a new perspective on grading structures of infinite-dimensional simple modules over ${\mathfrak{sl}_2(\mathbb{C})}$, extending previous finite-dimensional studies.

Findings

Characterization of graded structures on infinite-dimensional modules

Extension of grading concepts from finite to infinite dimensions

Identification of conditions for rank 2 modules to admit gradings

Abstract

In this paper we explore the possibility of endowing simple infinite-dimensional ${\mathfrak{sl}_2(\mathbb{C})}$-modules by the structure of the graded module. The gradings on finite-dimensional simple module over simple Lie algebras has been studied in [arXiv:1308.6089] and [arXiv:1601.03008].

Equations

$$V=\underset{g\in G}{⨁}{V}_g.$$

$$\Gamma :V=\underset{g\in G}{⨁}{V}_g.$$

$$A_g{A}_h\subset A_k.$$

$$x=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],h=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right],y=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right].$$

$$\Gamma :L_0=⟨h⟩,L_1=⟨x⟩,L_2=⟨y⟩.$$

$${\Gamma }^{\prime }:L_1=⟨x,y⟩,L_0=⟨h⟩.$$

$${\Gamma }_0:L_{-1}=⟨x⟩,L_0=⟨h⟩,L_1=⟨y⟩.$$

$$A_g{M}_h\subset M_{g+h}\text{mbox}forallg,h\in G.$$

$$M_{\lambda }^C:=U(I_{\lambda })/U(I_{\lambda })C,$$

$$x=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],h=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right],y=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right],$$

$$A=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]=h,B=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],C=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right].$$

$${\Gamma }_{{\mathfrak{sl}}_2}^1:{\mathfrak{sl}}_2(\mathbb{C})$$

$$\text{mbox}where{L}_0$$

$${\Gamma }_{{\mathfrak{sl}}_2}^2:{\mathfrak{sl}}_2(\mathbb{C})$$

$$\text{mbox}where{L}_{(1,0)}$$

$$c=(h+1{)}^2+4yx=h^2+1+2xy+2yx.$$

$$c$$

$$c=B^2-C^2+h^2+I=A^2+B^2-C^2+I.$$

$$\{h^k{x}^{\ell }|0\le k,\ell \le n,k+\ell \le n\}\cup \{h^k{y}^{\ell }|0\le k,\ell \le nk+\ell \le n\}.$$

$$h^k{B}^l{C}^2=h^k{B}^l(h^2+B^2-\mu )=h^k(B^l{h}^2+B^{l+2}-\mu B^l).$$

$$h^k{B}^l{h}^2=h^{k+2}{B}^l+h^k[B^l,h^2]$$

$$⟨h^k{B}^l{C}^m\mid k+l+m\le n\text{ and }m=0,1⟩.$$

$$h^k{B}^2\equiv h^k(\mu -h^2)=\mu h^k-h^{k+2},$$

$$C.f(h)$$

$$B.f(h)$$

$$xB$$

$$=\frac{1}{2}(B^2+BC+2h)$$

$$\equiv \frac{1}{2}(\mu -h^2+2h),$$

$$yB$$

$$=\frac{1}{2}(B^2-BC-2h)$$

$$\equiv \frac{1}{2}(\mu -h^2-2h).$$

$$C^2.f(h)=$$

$$=$$

$$\equiv$$

Full text

Keywords: graded Lie algebras, enveloping algebras, simple modules, graded modules

2010 Mathematics Subject Classification: Primary 17B70, Secondary 17B10, 17B20, 17B35.

The first author acknowledges support by NSERC grant # 227060-14. The second author acknowledges support by NSERC grant # 227060-14 and NSERC grant # -18.

1. Introduction

Let $G$ be a non-empty set. A vector space $V$ over a field $\mathbb{F}$ is called $G$-graded if it can be written as the direct sum of subspaces indexed by $G$, as follows:

[TABLE]

We will sometimes use Greek letters to refer to the gradings, for example,

[TABLE]

Generally, it is allowed that some of $V_{g}$ are zero subspaces. The subset $S\subset G$ consisting of those $g\in G$ for which $V_{g}\neq\{0\}$ is called the support of the grading $\Gamma$. One writes $S=\operatorname{\mathrm{Supp}}\Gamma$ or $S=\operatorname{\mathrm{Supp}}V$. The subspaces $V_{g}$ are called the * homogeneous components* of $\Gamma$. Nonzero elements in $V_{g}$ are called homogeneous of degree $g$. An $\mathbb{F}$-subspace $U$ in $V$ is called graded in $V$ (or in $\Gamma$) if $U=\bigoplus_{g\in G}U\cap V_{g}$.

Now let $\Gamma$ and $\Gamma^{\prime}:V=\bigoplus_{g^{\prime}\in G^{\prime}}V^{\prime}_{g^{\prime}}$ be two gradings on $V$ with supports $S$ and $S^{\prime}$, respectively. We say that $\Gamma$ is a refinement of $\Gamma^{\prime}$ (or $\Gamma^{\prime}$ is a coarsening of $\Gamma$), if for any $s\in S$ there exists $s^{\prime}\in S^{\prime}$ such that $V_{s}\subseteq V^{\prime}_{s^{\prime}}$. The refinement is proper if this inclusion is strict for at least one $s\in S$. A grading is called fine if it does not have proper refinements.

An $\mathbb{F}$-algebra $A$ (not necessarily associative) is said to be graded by a set $G$, or $G$-graded if $A$ is a $G$-graded vector space and for any $g,h\in G$ such that $A_{g}A_{h}\neq\{0\}$ there is $k\in G$ such that

[TABLE]

In this paper, we will always assume that $G$ is an abelian group and $k$ in Equation (2) is determined by the operation of $G$. Thus, if $G$ is written additively (as is commonly done in the papers on Lie theory), then Equation (2) becomes $A_{g}A_{h}\subset A_{g+h}$. If $G$ is written multiplicatively, then it becomes $A_{g}A_{h}\subset A_{gh}$.

Given a grading $\Gamma:A=\bigoplus_{g\in G}A_{g}$ with support $S$, the universal (abelian) group of $\Gamma$, denoted by $G^{u}$, is the abelian group given in terms of generators and defining relations as follows: $G^{u}=(S\>|\>R)$, where $gh=k$ is a relation in $R$ if $\{0\}\neq A_{g}A_{h}\subset A_{k}$. If $\Gamma$ is a group grading then the identity map $\mathrm{id}_{S}$ extends to a homomorphism of $G^{u}$ to $G$ so that $\Gamma$ can be viewed as a $G^{u}$-grading. Any coarsening $\Gamma^{\prime}:A=\bigoplus_{g^{\prime}\in G^{\prime}}A^{\prime}_{g^{\prime}}$ of $\Gamma$ is a quotient-grading of this $G^{u}$-grading of $A$ in the sense that there exists a unique homomorphism $\nu:G^{u}\to G^{\prime}$ such that $A_{g^{\prime}}=\bigoplus_{g\in\nu^{-1}(g^{\prime})}A_{g}$, for any $g^{\prime}\in G^{\prime}$.

For example, consider the standard basis of $L=\mathfrak{sl}_{2}(\mathbb{C})$

[TABLE]

and give $L$ the following grading by $G=\mathbb{Z}_{3}$:

[TABLE]

The support is $G$ itself.

The following grading by $G^{\prime}=\mathbb{Z}_{2}$ is a coarsening of $\Gamma$ :

[TABLE]

The universal group of $\Gamma$ is $\mathbb{Z}$, with

[TABLE]

Both $\Gamma$ and $\Gamma^{\prime}$ are factor-gradings of $\Gamma_{0}$; $\Gamma^{\prime}$ is a coarsening but not a factor-grading of $\Gamma$.

A left module $M$ over a $G$-graded associative algebra $A$ is called $G$-graded if $M$ is a $G$-graded vector space and

[TABLE]

(Later on, the multiplicative notation for the graded modules will be used, as well).

A left $G$-graded $A$-module $M$ is called graded-simple if $M$ has no graded submodules different from $\{0\}$ and $M$. Graded modules and graded-simple modules over a graded Lie algebra $L$ are defined in the same way.

If a Lie algebra $L$ is $G$-graded then its universal enveloping algebra $U(G)$ is also $G$-graded. Every graded $L$-module is a left $U(L)$-graded module, and vice versa. The same is true for simple graded $L$-modules and graded-simple $L$-modules.

An $\mathfrak{sl}_{2}(\mathbb{C})$-module is called torsion-free if it is torsion-free as an $\mathbb{C}[h]$-module. Every simple module is either torsion -free or torsion, also called weight module. Torsion-free $\mathfrak{sl}_{2}(\mathbb{C})$-modules has been studied in [Bav90, Blo81, MP16, Maz09, Nil15]. In 1992 Bavula constructed a family of simple torsion-free modules [Bav92]. The torsion-free $\mathfrak{sl}_{2}(\mathbb{C})$-modules of rank $1$ have been classified in [Nil15]. In [MP16], the author produced torsion-free modules of arbitrary finite rank.

The gradings of the torsion-free $\mathfrak{sl}_{2}(\mathbb{C})$-module of rank $1$ have been dealt with in [BKS18]; the main result was that the torsion-free $\mathfrak{sl}_{2}(\mathbb{C})$-modules of rank $1$ cannot be $\mathbb{Z}$- or $\mathbb{Z}_{2}^{2}$-graded. In [BKS18] one also considered the gradings of the family of modules constructed in [Bav92].

In this paper, our main goal is to construct a simple torsion-free $\mathbb{Z}_{2}^{2}$-graded module. For this, we consider the following module (see the notations in Sections 2.1 and 2.2):

[TABLE]

where $\lambda\in\mathbb{C}$. The main result in this paper that the graded module $M_{\lambda}^{C}$ is simple when $\lambda$ is not an even integer. Otherwise, the module $M_{\lambda}^{C}$ contains a proper graded simple torsion-free module of rank $2$.

2. Group gradings of $\mathfrak{sl}_{2}(\mathbb{C})$-modules

In this section we restrict our attention to the modules over the Lie algebra of the type $A_{1}$, which can be realized as $\mathfrak{sl}_{2}(\mathbb{C})$.

2.1. Group gradings of $\mathfrak{sl}_{2}(\mathbb{C})$

All group gradings on $\mathfrak{sl}_{2}(\mathbb{C})$ are well-known, see e.g [EK13]. We will use the following bases:

[TABLE]

as earlier, and

[TABLE]

Up to equivalence, there are precisely two fine gradings on $\mathfrak{sl}_{2}(\mathbb{C})$ (see [EK13, Theorem 3.55]):

• •

Cartan grading with the universal group $\mathbb{Z}$,

[TABLE]

• •

Pauli grading with the universal group $\mathbb{Z}^{2}_{2}$,

[TABLE]

Hence, up to isomorphism, any $G$-grading on $\mathfrak{sl}_{2}(\mathbb{C})$ is a coarsening of one of the two gradings: Cartan or Pauli.

Note that any grading $\Gamma$ of a Lie algebra $L$ uniquely extends to a grading $U(\Gamma)$ of its universal enveloping algebra $U(L)$. The grading $U(\Gamma)$ is a grading in the sense of associative algebras but also as $L$-modules where $U(L)$ is either a (left) regular $L$-module or an adjoint $L$-module. In the study of gradings on $\mathfrak{sl}_{2}(\mathbb{C})$-modules one often considers a $\mathbb{Z}_{2}$-coarsening of $U(\Gamma^{2}_{\mathfrak{sl}_{2}})$, in which the component of the coarsening labeled by [math] is the sum of components of the original grading labeled by $(0,0)$ and $(1,0)$ while the component labeled by $1$ is the sum of components labeled by $(0,1)$ and $(1,1)$.

2.2. Algebras $U(I_{\lambda})$

Let $c\in U(\mathfrak{sl}_{2}(\mathbb{C}))$ be the Casimir element for $\mathfrak{sl}_{2}(\mathbb{C})$. With respect to the basis $\left\{h,x,y\right\}$ of $\mathfrak{sl}_{2}(\mathbb{C})$, this element can be written as

[TABLE]

It is will-known that the center of $U({\mathfrak{sl}_{2}(\mathbb{C})})$ is the polynomial ring $\mathbb{C}[c]$. Note that $c$ is a homogeneous element of degree zero, with respect to the Cartan grading of $U(\mathfrak{sl}_{2}(\mathbb{C}))$. One can write the Casimir element with respect to the basis $\left\{h,B,C\right\}$ of $\mathfrak{sl}_{2}(\mathbb{C})$.

Namely,

[TABLE]

and so

[TABLE]

It follows that $c$ is also homogeneous, of degree (0,0), with respect to the Pauli grading of $U(\mathfrak{sl}_{2}(\mathbb{C}))$. Note that this basis of the center of the universal enveloping algebra is a very particular case of the bases given in [BM].

Let $R$ be an associative algebra (or just an associative ring), and $V$ be a left $R$-module. The annihilator of $V$, denoted by $\mathrm{Ann}_{R}(V)$, is the set of all elements $r$ in $R$ such that, for all $v$ in $V$, $r.v=0:$

$\mathrm{Ann}_{R}(V)=\left\{r\in R\mid r.v=0\text{ for all }v\in V\right\}$.

Given $\lambda\in\mathbb{C}$, let $I_{\lambda}$ be the two-side ideal of $U(\mathfrak{sl}_{2}(\mathbb{C}))$, generated by the central element $c-(\lambda+1)^{2}$.

Theorem 2.1**.**

[Maz09, Theorem 4.7]** For any simple $U({\mathfrak{sl}_{2}(\mathbb{C})})$-module $M$, there exists $\lambda\in\mathbb{C}$ such that $I_{\lambda}\subset\mathrm{Ann}_{U({\mathfrak{sl}_{2}(\mathbb{C})})}(M)$.

Clearly, if $R$ be a graded algebra and $M$ be a graded $R$-module, then $\mathrm{Ann}_{R}(M)$ is a graded ideal.

Proposition 2.2** ([BKS18]).**

The ideal $I_{\lambda}$ is both $\mathbb{Z}-$ and $\mathbb{Z}_{2}^{2}-$graded ideal.

Now for any $\lambda\in\mathbb{C}$, we write $U(I_{\lambda}):=U({\mathfrak{sl}_{2}(\mathbb{C})})/{I_{\lambda}}$. Using Proposition 2.2, it follows that $U(I_{\lambda})$   has natural $\mathbb{Z}$- and $\mathbb{Z}_{2}^{2}$-gradings induced from $L$. It is well-known (see e.g [Maz09]) that the algebra $U(I_{\lambda})$ is a free $\mathbb{C}[h]$-module with basis $\mathcal{B}_{0}=\left\{1,x,y,x^{2},y^{2},\ldots\right\}$, and so it is a vector space over $\mathbb{C}$ with basis $\mathcal{B}=\left\{1,h,h^{2},\ldots\right\}.\mathcal{B}_{0}$. Note that the basis $\mathcal{B}$ is a basis of $U(I_{\lambda})$ consisting of homogeneous element with respect to the Cartan grading by $\mathbb{Z}$.

A basis of $U(I_{\lambda})$ over $\mathbb{C}$ consisting of homogeneous element with respect to the Pauli grading by $\mathbb{Z}_{2}^{2}$ can be computed as follows. Set $\widehat{B}_{0}=\{1,B,C,BC,B^{2},B^{2}C,B^{3},B^{3}C,\ldots\}$. Then easy calculations, using the induction by the degree in $B$ and the relation $C^{2}=h^{2}+B^{2}-\lambda^{2}-2\lambda$ show that the set $\widehat{B}=\{1,h,h^{2},\ldots\}\cdot\widehat{B}_{0}$ is a $\mathbb{Z}_{2}^{2}$-homogeneous basis of $U(I_{\lambda})$.

Now let $p(t)=\frac{1}{4}((\lambda^{2}+2\lambda)-2t-t^{2})\in{\mathbb{C}}[t]$. Then, inside $U(I_{\lambda})$  for any $q(t)\in\mathbb{C}[t]$, we have the following relations:

$x^{k}q(h)=q(h-2k)x^{k}$

$y^{j}q(h)=q(h+2j)x^{j}$

If $k\geq j$ then

$x^{k}y^{j}=p(h-2k)\cdots p(h-2(k-j+1))x^{k-j}$

$y^{j}x^{k}=p(h+2(j-1))\cdots p(h)x^{k-j}$

If $j\geq k$ then

$x^{k}y^{j}=p(h-2k)\cdots p(h-2)y^{j-k}$

$y^{j}x^{k}=p(h+2(j-1))\cdots p(h+2(j-k))y^{j-k}$.

One more property that is important for us is the following.

Theorem 2.3**.**

[Maz09, Theorem 4.26]** For any non-zero left ideal $I\subset U(I_{\lambda})$, the $U(I_{\lambda})$-module $U(I_{\lambda})/I$ has finite length.

Proposition 2.4**.**

Set $\widehat{B}_{0}=\{1,B,C,BC,B^{2},B^{2}C,B^{3},B^{3}C,\ldots\}$. ​​Then $\widehat{B}=\{1,h,h^{2},\ldots\}\cdot\widehat{B}_{0}$ is a $\mathbb{Z}_{2}^{2}$-homogeneous basis of $U(I_{\lambda})$.

Proof.

Consider the natural ascending filtration $\{U^{(n)}\;|\;n=0,1,2,\ldots\}$ of $U(\mathfrak{sl}_{2}(\mathbb{C}))$, where $U^{(n)}$ is the linear span of monomials of the form $e_{1}e_{2}\cdots e_{k}$ where $e_{i}\in\mathfrak{sl}_{2}(\mathbb{C})$ for all $i$ and $k\leq n$. Using PBW Theorem, if $\left\{f_{1},f_{2},f_{3}\right\}$ be an ordered basis of $\mathfrak{sl}_{2}(\mathbb{C})$, then this filtration can just be the linear span of the standard monomials of degree less than or equal $n$, that is the monomials of the form $f_{1}^{i}f_{2}^{j}f_{3}^{l}$ where the degree of the monomial is the sum of the powers. Consider now the filtration $\{I^{(n)}\;|\;n=0,1,2,\ldots\}$ of $U(I_{\lambda})$ induced by the natural filtration of $U(\mathfrak{sl}_{2}(\mathbb{C}))$. Using the above basis $\mathcal{B}$, we can write that the basis of $I^{(n)}$ in the form of the set

[TABLE]

Now using (4), we can assume that the basis of $I^{(n)}$ is formed by some of the monomials $\{h^{k}B^{l}C^{m}\mid k+l+m\leq n\}$.

Let us prove that actually $I^{(n)}$ is spanned by the same monomials such that $m=0\text{ or }1$. Let us use induction on $n$, with an obvious basis, and the relation $C^{2}=h^{2}+B^{2}-\mu$, where $\mu=\lambda^{2}+2\lambda$, holding in $U(I_{\lambda})$. As a result, it is sufficient to deal with the elements $h^{k}B^{l}C^{2}$ with $k+\ell+2=n$. We write

[TABLE]

Since

[TABLE]

where the second term belong to $I^{(s)}$ for some $s<n$, we can use the induction step and write $h^{k}B^{l}C^{2}$ as a sum of the terms of the desired form. Now it follows that $I^{(n)}$ is spanned by the set of monomials

[TABLE]

Using the basis (6), we find that the dimension of $I^{(n)}$ equals $\sum_{i=0}^{n}(2i+1)$. Clearly, the number of elements in (7) is the same and hence a basis of $U(I_{\lambda})$ is formed by the monomials $\{h^{k}B^{\ell}\}\cup\{h^{k}B^{\ell}C\}$ where $k,\ell=0,1,2,\ldots$ ∎

2.3. $\mathfrak{sl}_{2}(\mathbb{C})$-modules: torsion-free modules

Definition 2.5**.**

Following [Maz09], an $\mathfrak{sl}_{2}(\mathbb{C})$-module $M$ is called torsion if for any $m\in M$ there exists non-zero $p(t)\in{\mathbb{C}}[t]$ such that $p(h).m=0$. In other words, $h$ has an eigenvector in $M$. We call $M$ torsion-free if $M\neq 0$ and $p(h).m\neq 0$ for all $0\neq m\in M$ and all non-zero $p(t)\in{\mathbb{C}}[t]$. In other words, $M$ is a free $\mathbb{C}[h]$-module. If the rank of this module is $n$, we say that $M$ is a torsion-free $\mathfrak{sl}_{2}(\mathbb{C})$-module of rank $n$.

Theorem 2.6**.**

[Maz09, Theorem 6.3]** A simple $\mathfrak{sl}_{2}(\mathbb{C})$-module is either a weight or a torsion-free module. $\Box$

Theorem 2.6 means that if $h$ has at least one eigenvector on $M$, then $M$ is a weight module.

As a consequence of Theorem 2.1, it is sufficient to describe simple torsion-free $U(I_{\lambda})$-modules instead of simple $U({\mathfrak{sl}_{2}(\mathbb{C})})$-modules (see e.g [Maz09]).

The classification of simple torsion free $U(I_{\lambda})$-module is given in the major paper [Blo81]. In [Bav90] the author uses a certain localization $\mathbb{A}$ of $U(I_{\lambda})$ to obtain the following.

Theorem 2.7**.**

[Bav90, Proposition 3]** Let $M$ be a simple torsion-free $U(I_{\lambda})$-module, then $M\cong U(I_{\lambda})/(U(I_{\lambda})\cap{\mathbb{A}}\alpha)$, for some $\alpha\in U(I_{\lambda})$ which is irreducible as an element of $\mathbb{A}$.

Theorem 2.8** ([BKS18]).**

Torsion free $\mathfrak{sl}_{2}(\mathbb{C})$-modules of rank $1$ cannot be $\mathbb{Z}$ or $\mathbb{Z}^{2}_{2}$-graded.

3. Simple graded $\mathfrak{sl}_{2}(\mathbb{C})$-modules of rank $2$

In this section we will study simple torsion-free $\mathfrak{sl}_{2}(\mathbb{C})$-modules of rank $2$. In particular, we will construct the first family of simple $\mathbb{Z}_{2}^{2}$-graded torsion-free $\mathfrak{sl}_{2}(\mathbb{C})$-modules of rank $2$.

Let $\lambda\in\mathbb{C}$, and consider the $U(I_{\lambda})$-module $M_{\lambda}^{C}=U(I_{\lambda})/U(I_{\lambda})C$. For $u,v\in U(\mathfrak{sl}_{2}(\mathbb{C}))$, we say that $u\equiv v$ and say that $u$ is equivalent to $v$, if and only if $u+U(I_{\lambda})C=v+U(I_{\lambda})C$. All elements of the form ${h}^{k}{B}^{l}{C}$ are equivalent to 0. Moreover ${B}^{2}\equiv\mu-{h}^{2}$ where $\mu=\lambda^{2}+2\lambda$ . Hence

[TABLE]

which implies that any element of $M_{\lambda}^{C}$ can be written as a linear combination of elements of the form ${h}^{k}{B}^{m}$ where $m=0,1$. This means that $M_{\lambda}^{C}$ is a torsion-free $\mathfrak{sl}_{2}(\mathbb{C})$-module of rank $2$, with basis $\left\{{1},{B}\right\}$ as a $\mathbb{C}[h]$-module, that is, $M_{\lambda}^{C}$ can be identified as $M_{\lambda}^{C}=\mathbb{C}[h]\oplus\mathbb{C}[h]{B}$. Note that $M_{\lambda}^{C}$ is graded module since $U(I_{\lambda})C$ is a graded left ideal.

Theorem 3.1** (Main Theorem 1).**

Let $\lambda\in\mathbb{C}\setminus 2\mathbb{Z}$, then $M_{\lambda}^{C}$ is a simple $\mathfrak{sl}_{2}(\mathbb{C})$-module .

Theorem 3.2** (Main Theorem 2).**

Let $\lambda\in 2\mathbb{Z}$, then $M_{\lambda}^{C}$ has a unique maximal (graded) submodule $N_{\lambda}^{C}$ such that $N_{\lambda}^{C}=P\oplus Q$, where $P$ and $Q$ are simple $\mathfrak{sl}_{2}(\mathbb{C})$-module of rank $1$.

Before we prove the above theorems, we will start with some necessary calculations and relations.

Let $f({h})\in\mathbb{C}[h]\subset M_{\lambda}^{C}$, then

[TABLE]

[TABLE]

Note that

[TABLE]

and

[TABLE]

Using (8), (3), and (3), we write

[TABLE]

Hence

[TABLE]

In particular, if $f(h)=h^{n}$, then

[TABLE]

and

[TABLE]

Hence

[TABLE]

All other coefficients equal zero.

One more relation that we will need refers to $CB.{h}^{n}$.

Let $f({h})\in\mathbb{C}[h]\subset M_{\lambda}^{C}$, then using (9), (3), and (3), we can write

[TABLE]

Hence

[TABLE]

In particular, if $f(h)=h^{n}$, then

[TABLE]

and

[TABLE]

Hence

[TABLE]

Lemma 3.3**.**

Let $\lambda\in\mathbb{C}\setminus 2\mathbb{Z}$. Consider a non-zero submodule $N$ of $M_{\lambda}^{C}$. If there is a non-zero element $0\neq f(h)\in N$, then $1\in N$.

Proof.

Let $0\neq v=f({h})\in N$; our claim is to show that ${1}\in N$.

We will prove this fact by induction on the degree of $f({h})$. For the base of induction, we will consider $i=1$, let $0\neq v={h}+a\in N$, where $a\in\mathbb{C}$. If $a\neq 0$ then

[TABLE]

which implies that $0\neq a\in N$. If $a=0$, then $v=h\in N$ and

[TABLE]

and since ${h}\in N$ then ${h}^{2}\in N$, hence $0\neq\mu\in N$, hence ${1}\in N$.

For the induction step, let

[TABLE]

Using (13), for $n>3$, we have

[TABLE]

For $n=2,3$ we have

[TABLE]

Let $v_{1}=f_{1}({h})=C^{2}.v+4n^{2}.v\in N$, then $f_{1}({h})$ is a polynomial of degree less than $n$. If $C^{2}.v\neq-4n^{2}.v$, then we have found a non-zero polynomial of degree less than $n$ in $N$, so induction applied to $f_{1}(h)$. Otherwise, we have

[TABLE]

which implies that $b_{n-1}=0$. For $n>3$ we have

[TABLE]

hence

[TABLE]

which implies that

[TABLE]

For $n=2$ we have

[TABLE]

For $n=3$ we have

[TABLE]

Now if we have failed to produce a non-zero polynomial of degree less than $n$ using $v_{1}$, we can use another action to get such element. Using (15), for $n>2$, we have

[TABLE]

For $n=2$

[TABLE]

Now consider the element $v_{2}=f_{2}({h})=CB.v-2(n+1)h.v$, then $f_{2}({h})$ is a polynomial of degree less than $n$. If $CB.v\neq 2(n+1)h.v$, then we have found a non-zero polynomial of degree less than $n$ in $N$, so induction applied to $f_{2}(h)$. Otherwise, for $n>2$

[TABLE]

which implies that

[TABLE]

For $n=2$ we have

[TABLE]

In fact $v_{1}$ and $v_{2}$ cannot be both zero. To see this, assume that $v_{1}$ and $v_{2}$ are both zero, then using (16) and (17), for $n>3$ we have

[TABLE]

Multiplying both side by 12, we have

[TABLE]

and so

[TABLE]

since $n>0$

[TABLE]

which implies that

[TABLE]

which is not the case.

For $n=2$ we have

[TABLE]

Then $\lambda=-4,2$ which is not the case.

For $n=3$ we have

[TABLE]

Then $\lambda=-6,4$, which is again not the case. Thus $1\in N$ ∎

Corollary 3.4**.**

Let \lambda\in 2\mathbb{Z}\, then there is a uniquely determined monic non-constant polynomial of degree $n$ (as in (19)), say $r(h)\in M_{\lambda}^{C}$, such that

[TABLE]

Corollary 3.5**.**

Let $\lambda\in 2\mathbb{Z}$, and $r(h)$ as in Corollary 3.4. Let $N$ be any non-zero submodule of $M_{\lambda}^{C}$. If $f(h)\in N$ with $f(h)\neq\gamma r(h)$ for all $\gamma\in\mathbb{C}$, then

1. (1)

If $\deg(f)\leq n$, then $N=M_{\lambda}^{C}$. 2. (2)

If $\deg(f)>n$, then either $r(h)\in N$ or $N=M_{\lambda}^{C}$ .$\qquad\Box$

Note that if $\lambda\in 2\mathbb{Z}$, then the polynomial $r(h)$ (as in (3.4)) has degree $n$, where

[TABLE]

Moreover, $\lambda\in 2\mathbb{Z}$ if and only if $\mu=4(n^{2}-n)$.

Next, we can argue in the same way as just above, to evaluate the action on the element $g(h)B\in M_{\lambda}^{C}$.

Let $g({h})B\in\mathbb{C}[h]B\subset M_{\lambda}^{C}$, then

[TABLE]

Using (21), we can write

[TABLE]

In particular, if $g(h)=h^{l}$, then

[TABLE]

Another relation that we will need refers to $CB.({h}^{l}B)$.

Let $g({h})B\in\mathbb{C}[h]B\subset M_{\lambda}^{C}$; then

[TABLE]

Using (3), (3), and (25), we can write

[TABLE]

Hence

[TABLE]

In particular, if $g(h)=h^{l}$, then

[TABLE]

Lemma 3.6**.**

Let $\lambda\in\mathbb{C}\setminus 2\mathbb{Z}$. Suppose $N$ is a non-zero submodule of $M_{\lambda}^{C}$. If there is a non-zero element $0\neq g(h)B\in N$, then $B\in N$.

Proof.

Let $\lambda\in\mathbb{C}\setminus 2\mathbb{Z}$ and assume that $N$ is a nonzero submodule of $M_{\lambda}^{C}$. If $0\neq u=g(h)B\in N$, then we are going to prove $B\in N$. We will prove this fact by induction on the degree of $g(h)$. For the base of induction, we will consider $i=1$. Let $0\neq u=({h}+a)B\in N$, where $a\in\mathbb{C}$. If $a\neq 0$ then

[TABLE]

which implies that $0\neq(12a)B\in N$. If $a=0$, then $u=hB\in N$ and

[TABLE]

and since ${h}B\in N$ then ${h}^{2}B\in N$. But $\lambda$ is not an even integer, so that $0\neq 4(8-\mu)B\in N$, hence ${B}\in N$.

For the induction step, let u=g(h)B=\big{(}{h}^{l}+b_{1}{h}^{l-1}+b_{l-2}{h}^{l-2}+\cdots+b_{0}\big{)}B\in N. Using (24), for $l>3$, we will have

[TABLE]

For $l=2,3$, we will have

[TABLE]

Let $u_{1}=f_{1}({h})B=C^{2}.u-(-4(l+1)^{2}.u)\in N$, then $f_{1}({h})$ is a polynomial of degree less than $l$. If $C^{2}.u\neq-4(l+1)^{2}.u$, then we have found a non-zero polynomial of degree less than $l$, so induction applied to $u_{1}$.

Otherwise

[TABLE]

which implies that $b_{l-1}=0$, and for $l>3$ we have

[TABLE]

which implies that

[TABLE]

For $l=2$ we have

[TABLE]

For $l=3$ we have

[TABLE]

Now if we failed to produce a non-zero element with polynomial has a degree less than $l$ using $u_{1}$, we can use another action to get such element. Using (28), for $l>2$, we have

[TABLE]

If $l=2$, we have

[TABLE]

Now we can consider an element $u_{2}=f_{2}({h})B=CB.u-2(l+2)h.u$. We have that $f_{2}({h})$ is a polynomial of degree less than $l$. If $CB.u\neq 2(l+2)h.u$, then we have found a non-zero element in $N$ with polynomial of degree less than $l$, so our induction applied to $u_{2}$.

Otherwise,

[TABLE]

which implies that

[TABLE]

For $l=2$

[TABLE]

In fact $u_{1}$ and $u_{2}$ cannot be both zero. To see this, assume that $v_{1}$ and $v_{2}$ are both zero, then using (29) and (30), for $l>3$ we have

[TABLE]

Multiplying both side by 12, we will have

[TABLE]

which implies that

[TABLE]

Since $l>0$, we have $\mu=4(l^{2}+l),$ which means that $\lambda$ is an even integer, and this is not the case.

For $l=2$ we have

[TABLE]

It follows that $\lambda=4,-6$, which is not the case. For $l=3$ we have

[TABLE]

which means that $\lambda=6,-8$, which is again not the case. Hence if $0\neq v=g(h)B\in N$ then $B\in N$. ∎

We can see from the previous proof that the equations $u_{1}=0$ and $u_{2}=0$ has unique solution when $\lambda\in 2\mathbb{Z}$. Moreover, if $r=r(h)+g(h)B$ such that $C^{2}.r=-4n^{2}r$ and $CB.r=2(n+1)hr$ is satisfied, then $g(h)$ has the degree $l=n-1$.

Corollary 3.7**.**

Let $\lambda\in 2\mathbb{Z}$, then there is a uniquely determined monic polynomial $r^{*}(h)\in M_{\lambda}^{C}$ of degree $l=n-1$ such that

[TABLE]

Note that the element $C.r(h)$ satisfies the above conditions, where $l=n-1$, and since $r^{*}(h)$ is uniquely determined, we can say that $r^{*}(h)=C.(r(h))$.

Corollary 3.8**.**

Let $\lambda\in 2\mathbb{Z}$, $r(h)$ as in Corollary 3.4, and let $N$ be any non-zero submodule of $M_{\lambda}^{C}$. If $g(h)B\in N$ with $g(h)\neq\gamma C.r(h)$ for all $\gamma\in\mathbb{C}$, then

1. (1)

If $\deg(g(h))\leq n-1$, then $N=M_{\lambda}^{C}$. 2. (2)

If $\deg(g(h))>n-1$, then either $C.r(h)\in N$ or $N=M_{\lambda}^{C}$ $\Box$

Corollary 3.9**.**

Let \lambda\in 2\mathbb{Z}\, then the elements $u_{(\alpha_{1},\alpha_{2})}=\alpha_{1}r(h)+\alpha_{2}(C.r(h))$, where $\alpha_{1},\alpha_{2}\in\mathbb{C}$, are the only elements in $M_{\lambda}^{C}$, such that

[TABLE]

where $r(h)$ and $n$ as in (3.4).$\ \qquad\qquad\qquad\qquad\qquad\qquad\Box$

Now we are ready to prove Theorem 3.1.

Proof of Theorem 3.1.

Assume that $N$ be a nonzero submodule of $M_{\lambda}^{C}$. Choose a non-zero element $v\in N$

Case 1: If $0\neq v=f({h})\in N$ then by Lemma 3.3, $1\in N$, hence $N=M_{\lambda}^{C}.$

Case 2: For the general case, let us assume that $v=f({h})+g({h})B$ with $g({h})\neq 0$. Using Case 1, we may assume that $v=\gamma+g({h})B$, where $\gamma\in\mathbb{C}$ and $g({h})\neq 0$.

We have

[TABLE]

where

[TABLE]

and $\deg(q({h}))\geq 2$. If $\gamma=0$ then $0\neq B.v=q(h)\in N$, which return us to the first case, and if $0\neq\gamma$, then

[TABLE]

which also brings us back to the first case. As a result, $M_{\lambda}^{C}$ is a simple module. ∎

Corollary 3.10**.**

Let $\lambda\in\mathbb{C}\setminus 2\mathbb{Z}$, then $M_{\lambda}^{C}$ is simple $\mathbb{Z}_{2}^{2}$-graded $\mathfrak{sl}_{2}(\mathbb{C})$-module of rank $2$. $\qquad\Box$

Now assume $\lambda\in 2\mathbb{Z}$, and consider the subspace

[TABLE]

Lemma 3.11**.**

Let $\lambda\in 2\mathbb{Z}$, then $N_{\lambda}^{C}$ is a submodule of the module $M_{\lambda}^{C}$.

Proof.

Let $u=f(h)r(h)+g(h)(C.r(h))$, it is clear the action of $h$ leaves $N_{\lambda}^{C}$ invariant. Now

[TABLE]

where

[TABLE]

and similarly for $g^{\prime}(h)$ and $g^{\prime\prime}(h)$.

Note that the second and third terms are in $N_{\lambda}^{C}$, and so it is enough to show that B.r(h)\equiv\frac{1}{2}\big{(}r(h-2)+r(h+2)\big{)}B is in $N_{\lambda}^{C}$.

Using (12), (3.4), and $\mu=4(n^{2}-n)$, we find that

[TABLE]

It follows that

[TABLE]

Similarly, using (14), (3.4), and $\mu=4(n^{2}-n)$, we get

[TABLE]

Thus

[TABLE]

Adding (38) and (40), and canceling out $2$, we get

[TABLE]

or

[TABLE]

which implies that

[TABLE]

If we replace $h$ by $h+2$ in (41) we get the relation

[TABLE]

Using (42), we obtain

[TABLE]

Similarly,

[TABLE]

As a result,

[TABLE]

Hence $N_{\lambda}^{C}$ is a submodule of $M_{\lambda}^{C}$. ∎

Remark 3.12**.**

The relation (45) leads to the following result. If the first two relations in the Corollaries (3.4), (3.7), and (3.9) hold then the last two also hold.

Now we are ready to start proving Theorem 3.2.

Proof.

Let $W$ be a non-zero submodule of $M_{\lambda}^{C}$. Choose a non-zero element $u=f(h)+g(h)B\in W$. We can apply the actions by $C^{2}$ or $CB$ and reduce the degree of the polynomial $f(h)$ until we get either a constant or $r(h)$. Hence we can reduce the cases of the element $u$, up to the scalar multiplication, to the following cases: either $u=f(h)$, $u=g(h)B$, $u=1+g(h)B$, or $u=r(h)+g(h)B$, where $g(h)\neq 0$ in Cases 2,3, and 4.

Case 1: If $0\neq u=f(h)\in W$, then by Corollary 3.5, either $W=M_{\lambda}^{C}$ or $W=N_{\lambda}^{C}$.

Case 2: Now let $u=g(h)B$ where $g(h)\neq 0$, then $B.u$ is a non-zero element in $\mathbb{C}[h]$ in $W$, see (35), which returns us to the Case 1.

Case 3: If $u=1+g(h)B$ with $g(h)\neq 0$, then $g(h)B.u-u$ is a non-zero element in $\mathbb{C}[h]$ in $W$, see (36), which also means that either $W=M_{\lambda}^{C}$ or $W=N_{\lambda}^{C}$.

Case 4: If $u=r(h)+g(h)B$ and $g(h)\neq 0$, then

[TABLE]

for some polynomial $g^{*}(h)$. If $g^{*}(h)\neq 0$, then by Case 2, either $W=M_{\lambda}^{C}$ or $W=N_{\lambda}^{C}$. Otherwise, $C^{2}u\equiv-4n^{2}u$, in this case we can try again with the element

[TABLE]

for some polynomial $g^{**}(h)$. If $g^{**}(h)\neq 0$, then by Case 2, either $W=M_{\lambda}^{C}$ or $W=N_{\lambda}^{C}$. Otherwise, $CB.u\equiv 2(n+1)hu$, which means by Corollary (3.9), that $u=r(h)+\alpha(C(r(h)))$ where $\alpha\in\mathbb{C}^{*}$. In this case, $\alpha C.u=\alpha C(r(h))-4n^{2}\alpha^{2}r(h)\in W$, which implies that $\alpha C.u+u=(4n^{2}\alpha^{2}+1)r(h)\in W$. If $(4n^{2}\alpha^{2}+1)\neq 0$, then $r(h)\in W$, which implies that $W=N_{\lambda}^{C}$. Hence, the only case remaining is $u=r(h)+\alpha(C(r(h)))$ and $(4n^{2}\alpha^{2}+1)=0$, in other words, when $u=r(h)\pm\frac{i}{2n}(C.r(h))$.

Lemma 3.13**.**

Let $\lambda\in 2\mathbb{Z}$, and $r(h)$ as in Corollary 3.4, then $P=\mathbb{C}[h](r(h)+\frac{i}{2n}(C.r(h)))$ and $Q=\mathbb{C}[h](r(h)-\frac{i}{2n}(C.r(h)))$ are submodules of $M_{\lambda}^{C}$, which are torsion-free modules of rank one.

Proof.

Using Corollary (3.4) and (45) we have,

[TABLE]

and

[TABLE]

For arbitrary elements, let $u=f(h)(r(h)+\frac{i}{2n}(C.r(h)))$, it is clear that the action by $h$ is invariant. Now

[TABLE]

where $f^{\prime}(h)$ and $f^{\prime\prime}(h)$ as in 37. Also

[TABLE]

which belongs to $P$. Hence $P$ is a submodule of $N_{\lambda}^{C}$. A similar calculation shows that $Q$ is also another submodule of $N_{\lambda}^{C}$.

∎

Hence $M_{\lambda}^{C}$ has exactly $3$ proper submodules, $N_{\lambda}^{C},P,\text{ and }Q$. Moreover, $N_{\lambda}^{C}$ is the unique maximal submodule of the module $M_{\lambda}^{C}$. It is clear that $N_{\lambda}^{C}=P+Q$; suppose that $0\neq v\in P\cap Q$. Then $v$ can be written as $v=f(h)(r(h)+\frac{i}{2n}(C.r(h)))$ for some $0\neq f(h)\in\mathbb{C}[h]$, and $V=g(h)(r(h)-\frac{i}{2n}(C.r(h)))$. It follows that $f(h)g(h)r(h)\in P\cap Q$, which means that $P\cap Q$ is either $M_{\lambda}^{C}$ or $N_{\lambda}^{C}$. This is a contradiction and so $N_{\lambda}^{C}=P\oplus Q$.

∎

Corollary 3.14**.**

Let $\lambda\in 2\mathbb{Z}$, then $N_{\lambda}^{C}$ is a $\mathbb{Z}_{2}^{2}$-graded simple $\mathfrak{sl}_{2}(\mathbb{C})$-module.

Proof.

Since $P$ and $Q$ are of rank $1$, then they are not graded submodules, see [BKS18], which are the only submodules of $N_{\lambda}^{C}$. But since $N_{\lambda}^{C}$ is the unique maximal submodule of the graded module $M_{\lambda}^{C}$, then $N_{\lambda}^{C}$ is a graded module which has no graded proper submodules. Hence $N_{\lambda}^{C}$ is graded-simple. ∎

Remark 3.15**.**

Let $\lambda\in 2\mathbb{Z}$, and consider the quotient module $V=M_{\lambda}^{C}/N_{\lambda}^{C}=\mathbb{C}[h]\oplus\mathbb{C}[h]B/\mathbb{C}[h]r(h)\oplus\mathbb{C}[h]r^{*}(h)B$, where $r^{*}(h)B=C.r(h)$. Since the polynomials $r(h),\ r^{*}(h)$ have degrees $n,\ n-1$, respectively, the module $V$ is a finite-dimensional $\mathfrak{sl}_{2}(\mathbb{C})$-module, with $\dim(V)=2n-1$, hence a weight module. Moreover, since the module $N_{\lambda}^{C}$ is maximal in $M_{\lambda}^{C}$, $V$ is simple. Using [Maz09, Theorem 3.32], we have

[TABLE]

4. $\mathbb{Z}$-gradings of torsion-free $\mathfrak{sl}_{2}(\mathbb{C})$-modules of finite rank.

In this section, we prove the following.

Theorem 4.1**.**

Any simple torsion-free $\mathfrak{sl}_{2}(\mathbb{C})$-module of finite rank is not a $\mathbb{Z}$-graded $\mathfrak{sl}_{2}(\mathbb{C})$-module.

Proof.

Let $M$ be any simple torsion-free $\mathfrak{sl}_{2}(\mathbb{C})$-module of finite rank $n$. Assume that $M$ is a $\mathbb{Z}$-graded $\mathfrak{sl}_{2}(\mathbb{C})$-module, that is, $M=\underset{i\in\mathbb{Z}}{\bigoplus}M_{i}$. Since $h$ has degree [math] in $U(\mathfrak{sl}_{2}(\mathbb{C}))$, it follows that $\mathbb{C}[h]M_{i}\subset M_{i}$ for all $i\in\mathbb{Z}$. Hence $M_{i}$ is also $\mathbb{C}[h]$-submodule of some finite rank $n_{i}\leq n$, which implies that $M=\underset{i\in I}{\bigoplus}M_{i}$, where $I=\left\{i\in\mathbb{Z}\mid M_{i}\neq\left\{0\right\}\right\}$, where $\left|I\right|=r$ for some positive integer $r\leq n$. Let $t\in I$ be maximal. Then $x.M_{t}\subset M_{t+1}=0$. For any $0\neq v\in M_{t}$, then $x.v=0$, that is $yx.v=0$. Since $M$ is simple, then the Casimir element $c$ act as a scalar on $M$. Hence $c.v=((h+1)^{2}+4yx).v=(h+1)^{2}.v=\alpha v$, for some $\alpha\in\mathbb{C}$, which implies that $M$ is a torsion module. A contradiction. ∎

Corollary 4.2**.**

Let $\lambda\in\mathbb{C}\setminus 2\mathbb{Z}$, then the module $M_{\lambda}^{C}$ is not a $\mathbb{Z}$-graded $\mathfrak{sl}_{2}(\mathbb{C})$-module.