# Non-Weight modules over affine Nappi-Witten Lie algebras

Priyanshu Chakraborty  Priyanshu Chakraborty:School of Mathematical Sciences,
Ministry of Education Key Laboratory of Mathematics and Engineering
Applications and Shanghai Key Laboratory of PMMP, East China Normal
University, No. 500 Dongchuan Rd., Shanghai 200241, China.
<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Santanu Tantubay  S.
Tantubay: Department of Mathematics, The Institute of Mathematical Sciences, A
CI of Homi Bhabha National Institute, IV Cross Road, CIT Campus Taramani
Chennai 600 113 Tamil Nadu, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this paper, we study the representation theory of affine Nappi-Witten Lie
algebra $\widehat{H_{4}}$ corresponding to the Nappi-Witten Lie algebra
$H_{4}$. We completely classify all Cartan-free modules of rank one for the
Nappi-Witten Lie algebra $H_{4}$. With the help of Cartan free $H_{4}$ modules
we classify all Cartan-free modules of rank one over affine Nappi Witten Lie
algebra. We also give a necessary and sufficient condition for these modules
to be irreducible. Finally as an application we classify Cartan free modules
of rank one for affine-Virasoro Nappi-Witten Lie algebras.

###### Key words and phrases:

affine Kac-Moody Algebras, Nappi-Witten Lie algebra, affine Nappi-Witten Lie
algebra

###### 2020 Mathematics Subject Classification:

17B68; 17B67

##  1\. Introduction

There are numerous uses for two-dimensional conformal field theories (CFTs) in
mathematics and physics. A significant category of CFTs include Wess-Zumino-
Novikov-Witten (WZNW) models [12]. Nappi and Witten showed in [13] that a WZNW
model is based on a central extension of the two-dimensional Euclidean group
that describes the homogeneous four-dimensional space-time corresponding to a
gravitational plane wave. The corresponding Lie algebra $H_{4}$ is known as
Nappi Witten Lie algebra. This Lie algebra is neither abelian nor semisimple.
Nappi-Witten Lie algebra has non-degenerate symmetric bilinear form which is
$H_{4}$ invariant, so like affine Kac-Moody Lie algebra, one can affinize
Nappi-Witten Lie algebra. The central extension of loop of $H_{4}$ is known as
affine Nappi-Witten Lie algebra. Now one naturally add the Virasoro Lie
algebra with affine Nappi-Witten Lie algebra and construct the afiine-Virasoro
Nappi-Witten Lie algebra.

The study of the representation theory of the Nappi-witten Lie algebra $H_{4}$
was started in [17]. The representation theory of affine Nappi-witten Lie
algebra $\widehat{H}_{4}$ was studied by [20]. Later a systematic study of
representations of $\widehat{H}_{4}$ was initiated in [14], where the authors
studied the structures of generalized Verma modules and gave vertex operator
algebra constructions. In [19], the authors classified the simple restricted
modules for the affine Nappi-Witten algebra. The category of weight modules
with finite dimensional weight spaces over the Nappi-Witten vertex operator
algebra was studied in [18], here the simple modules were classified and the
characters of these simple mod ules were computed. The imaginary Verma modules
and irreducible weight modules with finite weight spaces for the twisted
affine Nappi-witten Lie algebra were studied in [15] and [16] respectively.

Non-weight simple modules for any Lie algebras are less understood. A class of
non-weight modules for finite-dimensional simple Lie algebra of type $A_{l}$
is studied by J. Nilsson in [5], which are known as Cartan-free modules. In
[6], the author showed that Cartan-free modules exist only for type
$A_{l}(l\geq 1)$ and type $C_{l}(l\geq 2)$. For Witt algebra such
classification was done in [7]. For other classes of Lie algebras, towards
classification of Cartan free modules one can see [1, 8, 7, 9, 3]. It should
mention that in [2] authors studied Cartan free module affine Kac-Moody Lie
algebras, which encourage us to study this type of modules for affine Nappi-
Witten Lie algebras. Recently non-weight modules for affine-Virasoro Nappi-
Witten Lie algebras has been studied in [11].

In the present paper, we study representation of affine Nappi-Witten Lie
algebra. In Section 2, we recall definitions of Nappi-Witten Lie algebra,
affine Nappi-Witten Lie algebra as well as Nappi-Witten type Virasoro algebra.
In Section 3, we study Cartan-free representation of Nappi-Witten Lie
algebras. We classify all possible Cartan-free modules for $H_{4}$ and find
out the irreducibility condition of this type of modules. In Section 4, we
consider Cartan-free modules over affine Nappi-WItten Lie algebras and find
the conditions of irreduciblity of these modules. We show that most of all
such modules are coming from Cartan-free modules of Nappi-Witten Lie algebras.
Finally as an application of our results we recover the Cartan free modules
for affine-Virasoro Nappi-Witten Lie algebras which was obtained in [11].

##  2\. Notations and Preliminaries

Throughout this paper, $\mathbb{Z}$, $\mathbb{C}$, and $\mathbb{C}^{*}$ denote
the sets of integers, complex numbers, and nonzero complex numbers,
respectively. For a Lie algebra $\mathfrak{a}$, we denote the universal
enveloping algebra of $\mathfrak{a}$ as $\mathcal{U}(\mathfrak{a})$. All the
vector spaces, algebras, and tensor products are over $\mathbb{C}$, unless it
is specified.

###  2.1. Nappi-Witten Lie algebras

The Nappi-Witten Lie algebra $H_{4}$ is a four-dimensional vector space

$H_{4}=\mathbb{C}p\oplus\mathbb{C}q\oplus\mathbb{C}r\oplus\mathbb{C}s$

together with the Lie brackets

$[p,q]=r,\ [s,p]=p,\;[s,q]=-q,[r,.]=0.$

Let $\mathfrak{h}=\mathbb{C}s$ be a Cartan subalgebra of $H_{4}$. In the next
Section, we will describe a class of modules for $H_{4}$ with respect to this
Cartan subalgebra.
Let $(,)$ be a symmetric bilinear form on $H_{4}$ defined by

$(p,q)=1,\;(r,s)=1,\;\text{otherwise},(\;,\;)=0.$

It is easy to see that $(\;,\;)$ is a non-degenerate $H_{4}$-invariant
symmetric bilinear form on $H_{4}$.

###  2.2. Affine Nappi-Witten Lie algebras.

Let $\mathbb{C}[t^{\pm 1}]$ be the Laurent polynomial ring over $\mathbb{C}$.
Now we consider the space

$\widehat{H_{4}}=H_{4}\otimes\mathbb{C}[t^{\pm
1}]\oplus\mathbb{C}\mathbf{k}\oplus\mathbb{C}d$

with the Lie brackets

$[h_{1}\otimes t^{m},h_{2}\otimes t^{n}]=[h_{1},h_{2}]\otimes
t^{m+n}+m(h_{1},h_{2})\delta_{m+n,0}\mathbf{k},\;[\widehat{H_{4}},\mathbf{k}]=0,\;[d,h_{1}\otimes
t^{m}]=mh_{1}\otimes t^{m},$

where $h_{1},h_{2}\in H_{4},\;m,n\in\mathbb{Z}$.
Throughout the paper, we will denote the element $h_{1}\otimes t^{n}$ by
$h_{1}(n)$.

###  2.3. Virasoro Lie algebras

By definition, the Virasoro algebra
$Vir:=\mathbb{C}\\{d_{m},\mathbf{k}:m\in\mathbb{Z}\\}$, with the brackets

$[d_{m},d_{n}]=(n-m)d_{m+n}+\delta_{m+n,0}\frac{m^{3}-m}{12}\mathbf{k},\;[d_{m},\mathbf{k}=0],$

for all $m,n\in\mathbb{Z}$.

###  2.4. Affine-Virasoro Nappi-Witten Lie algebras

We define the affine-Virasoro Nappi-Witten algebra as follows:
The underlying vector space is given by

$\overline{H_{4}}=H_{4}\otimes\mathbb{C}[t^{\pm
1}]\oplus\mathbb{C}\mathbf{k}\;\displaystyle{\bigoplus_{m\in\mathbb{Z}}}\mathbb{C}d_{m},$

where the Lie brackets are given by:

$[h_{1}\otimes t^{m},h_{2}\otimes t^{n}]=[h_{1},h_{2}]\otimes
t^{m+n}+m(h_{1},h_{2})\delta_{m+n,0}\mathbf{k},\;[\overline{H_{4}},\mathbf{k}]=0,\;[d_{m},h_{1}\otimes
t^{n}]=nh_{1}\otimes t^{m+n},$
$[d_{m},d_{n}]=(n-m)d_{m+n}+\delta_{m+n,0}\frac{m^{3}-m}{12}\mathbf{k}.$

### 2.5.

Let $\mathfrak{a}$ be any Lie algebra over $\mathbb{C}$ and $\mathfrak{b}$ be
any finite-dimensional abelian subalgebra of $\mathfrak{a}$. Denote
$\mathcal{M}(\mathfrak{a},\mathfrak{b})$ by the full subcategory of
$U(\mathfrak{a})$-modules consisting of objects whose restriction to
$U(\mathfrak{b})$ is a free module of rank $1$, i.e.,

$\mathcal{M}(\mathfrak{a},\mathfrak{b})=\\{M\in
U(\mathfrak{a})-\hbox{Mod}|\hbox{Res}_{U(\mathfrak{b})}^{U(\mathfrak{a})}M\cong_{U(\mathfrak{b})}U(\mathfrak{b})\\}.$

In this paper, we will study this subcategory for Nappi-Witten, affine Nappi-
Witten, affine-Virasoro Nappi-Witten Lie algebras.

##  3\. Cartan free modules over Nappi-Witten Lie algebras

In this Section, we will discuss Cartan free modules of rank one over Nappi-
Witten Lie algebra. Recall that we consider a Cartan subalgebra
$\mathfrak{h}=\mathbb{C}s$ for $H_{4}$. Then $\mathcal{U}(\mathfrak{h})$ is
the polynomial algebra $\mathbb{C}[s]$. We define an automorphism
$\tau:\mathcal{U}(\mathfrak{h})\rightarrow\mathcal{U}(\mathfrak{h})$ by
$\tau(s)=s-1$.

######  Lemma 3.1.

Let $x\in\mathbb{C}[s]$ be any non-constant polynomial. Then
$\text{deg}(\tau(x)-x)=\text{deg}(x)-1$.

###### Proof.

It is easy to prove. ∎

Now we define certain class of modules for $H_{4}$ on the polynomial algebra
$\mathbb{C}[s]$. For any $g\in\mathbb{C}[s]$, we define
$M_{(g,0)}=\mathbb{C}[s]$ with the actions:

$\begin{cases}p.x=\tau(x)g,\\\ q.x=0,\\\ r.x=0,\\\ s.x=sx,\end{cases}$ (3.1)

where $x\in\mathbb{C}[s]$.

Similarly we define $M_{(0,g)}=\mathbb{C}[s]$ with the actions:

$\begin{cases}p.x=0,\\\ q.x=\tau^{-1}(x)g,\\\ r.x=0,\\\ s.x=sx,\end{cases}$
(3.2)

where $x\in\mathbb{C}[s]$.
Now suppose $h(s)=a_{1}s+a_{2}$ be a non-zero one-degree polynomial in
$\mathbb{C}[s]$ and $b\in\mathbb{C}^{*}$. We define $M_{(h,b)}=\mathbb{C}[s]$
with the actions:

$\begin{cases}p.x=\tau(x)h(s),\\\ q.x=\tau^{-1}(x)b,\\\ r.x=-a_{1}b,\\\
s.x=sx,\end{cases}$ (3.3)

where $x\in\mathbb{C}[s]$.
Similarly we define $M_{(b,h)}=\mathbb{C}[s]$ with the actions:

$\begin{cases}p.x=\tau(x)b,\\\ q.x=\tau^{-1}(x)h(s),\\\ r.x=-a_{1}b,\\\
s.x=sx,\end{cases}$ (3.4)

where $x\in\mathbb{C}[s]$.

Let $a,b$ be two non-zero constant and define $M_{(a,b)}=\mathbb{C}[s]$ with
the actions:

$\begin{cases}p.x=\tau(x)a,\\\ q.x=\tau^{-1}(x)b,\\\ r.x=0,\\\
s.x=sx,\end{cases}$ (3.5)

where $x\in\mathbb{C}[s]$.
Define one another class of module for $H_{4}$ on $M_{0}=\mathbb{C}[s]$ by the
action given by:

$\begin{cases}p.x=q.x=r.x=0,\\\ s.x=sx,\end{cases}$ (3.6)

where $x\in\mathbb{C}[s]$.

It is easy to see, with the above actions $M_{(g,0)}$, $M_{(0,g)}$,
$M_{(h,b)}$, $M_{(b,h)}$, $M_{(a,b)}$ and $M_{0}$ becomes $H_{4}$-module,
which are Cartan free modules of rank one.

Let us define a linear map $\eta:H_{4}\to h_{4}$ by sending

$p\mapsto-q,\hskip 28.45274ptq\mapsto p,\hskip 28.45274ptr\mapsto r,\,\hskip
28.45274pts\mapsto-s.$

It is easy to see that $\eta$ is an automorphism of $H_{4}$. Now we twist the
modules $M_{(g,0)}$ and $M_{(h,b)}$ by the above automorphsim and denote the
twisted modules by $M^{t}_{(g,0)}$ and $M^{t}_{(h,b)}$ respectively. Then we
have the isomorphism of modules $M^{t}_{(g(s),0)}\cong M_{(0,g(-s))}$ and
$M^{t}_{(h(s),b)}\cong M_{(-b,h(-s))}$ under the map $f(s)\mapsto f(-s)$ (in
both cases).

Now we are going to prove that these are the all $U(h)$ free modules for
$H_{4}$. The following Lemma easily follows from the Lie brackets of $H_{4}.$

######  Lemma 3.2.

Let $M\in\mathcal{M}(H_{4},\mathfrak{h})$. Then for any $x\in M$ we have:

$\begin{cases}p.x=\tau(x)p.1,\\\ q.x=\tau^{-1}(x)q.1,\\\ r.x=xr.1,\\\
s.x=sx\end{cases}$ (3.7)

######  Lemma 3.3.

Let $M\in\mathcal{M}(H_{4},\mathfrak{h})$.

  1. (1)

If $p.1=0$ or $q.1=0$ then $r.1=0$. In these cases $M$ will be isomorphic to
$M_{(g,0)}$ or $M_{(0,g)}$ for some $g\in\mathbb{C}[s]$.

  2. (2)

If $p.1=q.1=0$, then $M$ will be isomorphic to $M_{0}.$

###### Proof.

There is nothing to prove for (2). Suppose $p.1=0$. Then we have
$r.1=p.q.1-q.p.1=\tau(q.1)p.1-q.p.1=0$. Similarly, we prove the other part. ∎

######  Lemma 3.4.

Let $M\in\mathcal{M}(H_{4},\mathfrak{h})$. If $p.1\neq 0$ or $q.1\neq 0$, then
we have $r.1\in\mathbb{C}$,

###### Proof.

Suppose $p.1\neq 0$. Now $[p,r].1=0$, which will give us
$(\tau(r.1)-r.1)p.1=0$. This will imply that $r\in\mathbb{C}$ by Lemma 3.1.
Similarly, we can prove the other part of the Lemma. ∎

From the above three Lemmas, we see that now we need to only consider the case
$p.1\neq 0$ and $q.1\neq 0$. From Lemma 3.4 we have $r.1\in\mathbb{C}$.

######  Proposition 3.5.

Let $M\in\mathcal{M}(H_{4},\mathfrak{h})$ with $p.1$ and $q.1$ are non-zero.
Then $M\cong M_{(h,b)}$ or $M_{(b,h)}$ for some
$h(s)=a_{1}s+a_{2}\in\mathbb{C}[s]$ and $b\in\mathbb{C}^{*}$

###### Proof.

We know
$r.1=p.q.1-q.p.1=\tau(q.1)p.1-\tau^{-1}(p.1)q.1=\tau(\tau^{-1}(p.1)q.1)-\tau^{-}1(p.1)q.1$.
By assumption $\tau^{-1}(p.1)q.1$ is a non-zero polynomial in $\mathbb{C}[s]$,
therefore by Lemma 3.1, we have either degree of $\tau^{-1}(p.1)q.1$ is one or
$p.1$ and $q.1$ both non-zero constant.
Case I : Let $p.1$ and $q.1$ are constant. It is clear that in this case
$M\cong M_{a,b}$.
Case II: Let both of $p.1$ and $q.1$ are not constant. This will imply that
the possibility of $\text{deg}(p.1,q.1)$ is $(1,0)$ or $(0,1)$. If
$\text{deg}(p.1,q.1)=(1,0)$, then assume $p.1=a_{1}s+a_{2}$ and $q.1=b$ for
some $a_{1},b\in\mathbb{C}^{*}$ and $a_{2}\in\mathbb{C}$. Then
$r.1=p.q.1-q.p.1=b(a_{1}s+a_{2})-q.(a_{1}s+a_{2})=b(a_{1}s+a_{2})-b(a_{1}(s+1)+a_{2})=-a_{1}b$.
So $M\cong M_{(h,b)}$. Similarly, we can prove if $\text{deg}(p.1,q.1)=(0,1)$,
then $M\cong M_{(b,h)}$.

∎

######  Theorem 3.6.

$\mathcal{M}(H_{4},\mathfrak{h})=\\{M_{(g,0)},M_{(0,g)},M_{(h,b)},M_{(b,h)},M_{(a,b)},M_{0}:g,h\in\mathbb{C}[s],\;a,b\in\mathbb{C}^{*},\;\text{deg}(h)=1\\}$

###### Proof.

Follows from Lemma 3.3 and Proposition 3.5. ∎

Now we are interested to find the conditions when the modules of
$\mathcal{M}(H_{4},\mathfrak{h})$ are irreducible.

######  Theorem 3.7.

  1. (1)

The modules $M_{(g,0)}$ and $M_{(0,g)}$ are irreducible iff $g$ is a non-zero
constant polynomial.

  2. (2)

The modules $M_{(h,b)}$, $M_{(b,h)}$ and $M_{(a,b)}$ are irreducible.

###### Proof.

Let $g$ be a non-zero constant polynomial and $W$ be a non-zero submodule of
$M_{(g,0)}.$ Note that due to the action of $s$, it is sufficient to prove
that $1\in W$. Let $f(s)=\displaystyle{\sum_{i=0}^{k}}a_{i}s^{k-i}$ be a non-
zero polynomial of degree $k$ in $W$. Then $p.f(s)=f(s-1)g\in W.$ Now consider
the vector

$\displaystyle f_{1}(s)=\frac{1}{g}p.f(s)-f(s)$
$\displaystyle=\displaystyle{\sum_{i=0}^{k}}a_{i}(s-1)^{k-i}-\displaystyle{\sum_{i=0}^{k}}a_{i}s^{k-i}$
$\displaystyle=-ka_{0}s^{k-1}\;(mod\;\displaystyle{\bigoplus_{i=2}^{k}}\mathbb{C}s^{k-i}).$

This means $f_{1}(s)$ is a non-zero polynomial of degree less than $k$ in $W$.
Continuing this process we have $1\in W.$ This proves that $M_{(g,0)}$ is
irreducible.
On the other hand if $g$ is a polynomial of degree greater equal to 1, then
the ideal generated by $g$ is a proper submodule of $M_{(g,0)},$ Hence
$M_{(g,0)}$ is reducible. Similar proof works for $M_{(0,g)}.$
To prove (2) consider a non-zero submodule $W_{1}$ of $M_{(h,b)}.$ Let $f(s)$
be a degree $k$ non-zero polynomial in $W_{1}.$ Now consider the vector
$\frac{1}{b}q.f(s)-f(s)$ and proceed similarly like (1) to conclude $1\in
W_{1}.$ This proves that $M_{(h,b)}$ is irreducible. Similarly $M_{(b,h)}$ is
irreducible.

∎

##  4\. Cartan Free modules over affine Nappi-Witten Lie algebras

In this section, we will study the categories
$\mathcal{M}(\widehat{H_{4}},\widehat{\mathfrak{h}})$ for the Lie algebras
$\widehat{H_{4}}$.

Suppose $\mathbb{C}[s,d]$ be the polynomial algebra over two indeterminant $s$
and $d$. Let us consider $\tau,\;\sigma$ be two automorphism of
$\mathbb{C}[s,d]$ defined by

$\tau(s)=s-1,\;\tau(d)=d,\;\sigma(s)=s,\;\sigma(d)=d-1.$

We see that $\tau$ and $\sigma$ are two commuting automorphisms. Suppose $M$
is a Cartan free module over the Nappi-Witten Lie algebra $H_{4}$. For
$\alpha\in\mathbb{C}^{*}$ and a sequence of complex numbers
$\mathbf{\beta}=\\{\beta_{i}:i\in\mathbb{Z},\;\beta_{0}=0\\}$, let
$\tilde{M}(\alpha,\beta)=M\otimes\mathbb{C}[d]$ with action of
$\widehat{H_{4}}$ given by:

$\begin{cases}p\otimes t^{k}.x(s,d)=\alpha^{k}\tau\sigma^{k}(x(s,d))p.1\\\
q\otimes t^{k}.x(s,d)=\alpha^{k}\tau^{-1}\sigma^{k}(x(s,d))q.1\\\ r\otimes
t^{k}.x(s,d)=\alpha^{k}\sigma^{k}(x(s,d))r.1\\\ s\otimes
t^{k}.(s,d)=\sigma^{k}(x(s,d))(\alpha^{k}s+\beta_{k})\\\ K.(x(s,d))=0\\\
d.x(s,d)=dx(s,d),\end{cases}$ (4.1)

where $x(s,d)\in\mathbb{C}[s,d]$.

For a sequence of functions
$\mathbf{f}=\\{f_{k}(s):f_{k}(s)\in\mathbb{C}[s],\;f_{0}(s)=s\\}$, define a
$\widehat{H}_{4}$-module structure on $\tilde{M}(\mathbf{f})=\mathbb{C}[s,d]$
with the action of $\widehat{H_{4}}$ given by:

$\begin{cases}s\otimes t^{k}.x(s,d)=\sigma^{k}(x(s,d))f_{k}(s)\\\
d.x(s,d)=dx(s,d),\\\ p\otimes t^{k}.x(s,d)=q\otimes t^{k}.x(s,d)=r\otimes
t^{k}.x(s,d)=K.(x(s,d))=0\par\end{cases}$ (4.2)

######  Proposition 4.1.

Under actions of (4.1) and (4.2), $\tilde{M}(\alpha,\beta)$ and
$\tilde{M}(\mathbf{f})$ forms $\widehat{H}_{4}$-module.

###### Proof.

Let $k,l\in\mathbb{Z}$ and $x(s,d)\in\mathbb{C}[s,d]$. Now we compute all
possible relations to prove that actions in (4.1) and (4.2) are module
actions.

$\displaystyle(i).\,\,[p\otimes t^{k},q\otimes t^{l}].x(s,d)$
$\displaystyle=(p\otimes t^{k}.q\otimes t^{l}-q\otimes t^{l}.p\otimes
t^{k}).x(s,d)$ (4.3) $\displaystyle=\alpha^{l}p\otimes
t^{k}.x(s+1,d-l)q.1-\alpha^{k}q\otimes t^{l}.x(s-1,d-k)p.1$ (4.4)
$\displaystyle=\alpha^{k+l}(x(s,d-l-k)\tau(q.1)p.1-x(s,d-l-k)\tau^{-1}(p.1)q.1)$
(4.5) $\displaystyle=\alpha^{k+l}x(s,d-l-k)(\tau(q.1)p.1-\tau^{-1}(p.1)q.1)$
(4.6) $\displaystyle=\alpha^{k+l}x(s,d-l-k)r.1$ (4.7) $\displaystyle=r\otimes
t^{k+l}.x(s,d)$ (4.8)

$\displaystyle(ii).\,\,[p\otimes t^{k},r\otimes t^{l}].x(s,d)$
$\displaystyle=(p\otimes t^{k}.r\otimes t^{l}-r\otimes t^{l}.p\otimes
t^{k}).x(s,d)$ (4.10) $\displaystyle=\alpha^{l}p\otimes
t^{k}.x(s,d-l)r.1-\alpha^{k}r\otimes t^{l}.x(s-1,d-k)p.1$ (4.11)
$\displaystyle=\alpha^{k+l}(x(s-1,d-l-k)\tau(r.1)p.1-x(s-1,d-l-k)p.1r.1)$
(4.12) $\displaystyle=0,$ (4.13)

since $\tau(r.1)=r.1$.

$\displaystyle(iii).\,[p\otimes t^{k},s\otimes t^{l}].x(s,d)$
$\displaystyle=(p\otimes t^{k}.s\otimes t^{l}-s\otimes t^{l}.p\otimes
t^{k}).x(s,d)$ (4.15) $\displaystyle=p\otimes
t^{k}.x(s,d-l)(\alpha^{l}s.1+\beta_{l})-\alpha^{k}s\otimes
t^{l}.x(s-1,d-k)p.1$ (4.16)
$\displaystyle=\alpha^{k}x(s-1,d-l-k)[\tau(\alpha^{l}s.1+\beta_{l})p.1-\sigma(p.1)(\alpha^{l}s.1+\beta_{l})]$
(4.17)
$\displaystyle=\alpha^{k}x(s-1,d-l-k)[(\alpha^{l}(s-1)+\beta_{l})p.1-(\alpha^{l}s+\beta_{l})p.1)$
(4.18) $\displaystyle=-\alpha^{k+l}x(s-1,d-l-k)p.1=-p\otimes t^{k+l}.x(s,d)$
(4.19)

$\displaystyle(iv).\,\,[q\otimes t^{k},r\otimes t^{l}]$
$\displaystyle=(q\otimes t^{k}.r\otimes t^{l}-r\otimes t^{l}.q\otimes
t^{k}).x(s,d)$ (4.21) $\displaystyle=\alpha^{l}q\otimes
t^{k}.x(s,d-l)r.1-\alpha^{k}r\otimes t^{l}.x(s+1,d-k)q.1$ (4.22)
$\displaystyle=\alpha^{k+l}(x(s+1,d-l-k)\tau^{-1}(r.1)q.1-x(s+1,d-l-k)q.1r.1)$
(4.23) $\displaystyle=0=[q\otimes t^{k},r\otimes t^{l}].x(s,d),$ (4.24)

since $\tau^{-1}(r.1)=r.1$.

$\displaystyle(v).\,\,[q\otimes t^{k},s\otimes t^{l}]$
$\displaystyle=(q\otimes t^{k}.s\otimes t^{l}-s\otimes t^{l}.q\otimes
t^{k}).x(s,d)$ (4.26) $\displaystyle=q\otimes
t^{k}.x(s,d-l)(\alpha^{l}s.1+\beta_{l})-\alpha^{k}s\otimes
t^{l}.x(s+1,d-k)q.1$ (4.27)
$\displaystyle=\alpha^{k}x(s+1,d-l-k)[\tau(\alpha^{l}s.1+\beta_{l})p.1-\sigma(q.1)(\alpha^{l}s.1+\beta_{l})]$
(4.28)
$\displaystyle=\alpha^{k}x(s+1,d-l-k)[(\alpha^{l}(s+1)+\beta_{l})q.1-(\alpha^{l}s+\beta_{l})q.1)$
(4.29) $\displaystyle=\alpha^{k+l}x(s+1,d-l-k)q.1=q\otimes t^{k+l}.x(s,d)$
(4.30)

For the action (4.1) other relations are easy to check. Now we check the
module relations for the actions of (4.2).

$\displaystyle(s\otimes t^{k}.s\otimes t^{l}-s\otimes t^{l}.s\otimes
t^{k}).x(s,d)$
$\displaystyle=x(s,d-k-l)f_{k}(s)f_{l}(s)-x(s,d-k-l)f_{k}(s)f_{l}(s)$ (4.32)
$\displaystyle=0=[s\otimes t^{k},s\otimes t^{l}].x(s,d)$ (4.33)

Again,

$\displaystyle[d,s\otimes t^{l}].x(s,d)$ $\displaystyle=(d.s\otimes
t^{l}-s\otimes t^{l}.d).x(s,d)$ (4.34)
$\displaystyle=dx(s,d-l)f_{l}(s)-(d-l)x(s,d-l)f_{l}(s)=lx(s,d-l)f_{l}(s)$
(4.35) $\displaystyle=ls\otimes t^{l}.x(s,d).$ (4.36)

It is easy to see that, due to the zero actions of $p\otimes t^{k},q\otimes
t^{k},r\otimes t^{k}$ in (4.2) all other relations satisfies trivially. This
completes the proof.

∎

######  Theorem 4.2.

$\mathcal{M}(\widehat{H_{4}},\widehat{\mathfrak{h}})=\\{\tilde{M}(\alpha,\mathbf{\beta}),\;\tilde{M}(\mathbf{f}):\alpha\in\mathbb{C}^{*},\;\mathbf{\beta}\in(\mathbb{C})^{\mathbb{N}},\;\mathbf{f}\in(\mathbb{C}[s])^{\mathbb{N}},M\in\mathcal{M}(H_{4},\mathfrak{h}),\;\beta_{0}=0,f_{0}=s\\},$
here $A^{\mathbb{N}}$ denote the set of all functions from $\mathbb{Z}$ to
$A.$

Suppose $M$ is a Cartan-free module over $\widehat{H_{4}}$, then as vector
space $M=\mathbb{C}[s,d]$. Now we assume $p\otimes
t^{k}.1=g_{k}(s,d),\;q\otimes t^{k}.1=h_{k}(s,d),\;s\otimes
t^{k}=f_{k}(s,d)\in\mathbb{C}[s,d]$, where $k\in\mathbb{Z}$. Note that if
$g_{k}(s,d)$ (or $h_{k}(s,d)$ ) is zero for some $k\in\mathbb{Z}$, then
$g_{0}(s,d)$ (or $h_{0}(s,d)$) is also zero polynomial.

######  Lemma 4.3.

Let $f_{k}(s,d)$ be as above. If for some $k\in\mathbb{Z}$, $f_{k}$ is a non-
zero polynomial, then we will have $deg_{d}(f_{k})=0$.

###### Proof.

We see that for all $k\neq 0,$ $0=[s\otimes t^{k},s\otimes
t^{-k}].1=f_{k}(s,d)f_{-k}(s,d-k)-f_{k}(s,d+k)f_{-k}(s,d)$, hence
$H(s,d)=H(s,d+k)$, where $H(s,d)=f_{k}(s,d)f_{-k}(s,d-k)$. So we get
$deg_{d}(f_{k})=deg_{d}(f_{-k})=deg_{d}(H)=0$. ∎

######  Lemma 4.4.

Suppose $g_{0}$ (or $h_{0}$) is a non-zero polynomial, then $deg_{d}(f_{k})=0$
and $deg_{s}(f_{k})\leq 1$ for all $k\in\mathbb{Z}$.

###### Proof.

We will prove for the case $g_{0}\neq 0$. Similarly one prove for $h_{0}\neq
0$.
We see that $f_{k}\neq 0$ for all $k\in\mathbb{Z}$, otherwise
$g_{0}=p.1=[s\otimes t^{k},p\otimes t^{-k}].1=0$. Now the first part of the
statement follows from Lemma 4.3.
We have $g_{k}(s,d)=[s\otimes
t^{k},p].1=g_{0}(s,d-k)f_{k}(s)-f_{k}(s-1)g_{0}(s,d)$ and putting this value
in the equation $0=[p,p\otimes t^{k}].1$, we get:

$0=g_{k}(s-1,d)g_{0}(s,d)-g_{k}(s,d)g_{0}(s-1,d-k)$
$=\\{g_{0}(s-1,d-k)f_{k}(s-1)-f_{k}(s-2)g_{0}(s-1,d)\\}g_{0}(s,d)-$
$\\{g_{0}(s,d-k)f_{k}(s)-f_{k}(s-1)g_{0}(s,d)\\}g_{0}(s-1,d-k).$

This will give us

$2f_{k}(s-1)g_{0}(s-1,d-k)g_{0}(s,d)=f_{k}(s)g_{0}(s-1,d-k)g_{0}(s,d-k)+f_{k}(s-2)g_{0}(s,d)g_{0}(s-1,d).$

Let $g_{0}(s,d)=\displaystyle{\sum_{i=0}^{m}}a_{i}(s)d^{i}.$ Now putting the
value of $g_{0}(s,d)$ in the above equation we have,
$2f_{k}(s-1)\displaystyle{\sum_{i=0}^{m}}a_{i}(s-1)(d-k)^{i}\displaystyle{\sum_{i=0}^{m}}a_{i}(s)d^{i}=f_{k}(s)\displaystyle{\sum_{i=0}^{m}}a_{i}(s-1)(d-k)^{i}\displaystyle{\sum_{i=0}^{m}}a_{i}(s)(d-k)^{i}$

$+f_{k}(s-2)\displaystyle{\sum_{i=0}^{m}}a_{i}(s-1)d^{i}\displaystyle{\sum_{i=0}^{m}}a_{i}(s-1)d^{i}.$

Now comparing the coefficient of $d^{2m}$ in the above equation we will get

$2f_{k}(s-1)=f_{k}(s)-f_{k}(s-2)$

and hence $deg_{s}(f_{k})\leq 1$. ∎

Similarly, we can prove that if $q.1\neq 0$, then $deg_{d}(f_{k})=0$ and
$deg_{s}(f_{k})\leq 1$. So in both cases, we assume
$f_{k}=\alpha_{k}s+\beta_{k}$, where $\alpha_{k},\beta_{k}\in\mathbb{C}$. Note
that $\alpha_{0}=1$ and $\beta_{0}=0$. We assume $\alpha_{1}=\alpha$ and
$\beta_{1}=\beta$.

######  Lemma 4.5.

Let $g_{k}(s,d),\;h_{k}(s,d),\;f_{k}$ be as above. If $g_{0}$ or $h_{0}$ is
non-zero polynomial, then we have $\alpha_{-1}=\alpha^{-1}$ and

  1. (1)

$deg_{d}(g_{0}(s,d))=0$ if $g_{0}\neq 0,$

  2. (2)

$deg_{d}(h_{0}(s,d))=0$ if $h_{0}\neq 0.$

.

###### Proof.

Here we will prove for $g_{0}\neq 0$, one proves the other case similarly. We
assume $g_{0}(s,d)=\displaystyle{\sum_{i=0}^{m}}a_{i}(s)d^{i}$. We need to
prove $m=0$.
From the relation $[s\otimes t,p].1=p\otimes t.1$ we say that

$g_{1}(s,d)=(\alpha s+\beta)(g_{0}(s,d-1)-g_{0}(s,d))+\alpha g_{0}(s,d).$
(4.37)

Again from the relation $[s\otimes t^{-1},p\otimes t].1=p.1$ and the equation
(4.37) we will have

$\begin{split}(\alpha_{-1}s+\beta_{-1})\\{(\alpha
s+\beta)(g_{0}(s,d)-g_{0}(s,d+1))+\alpha g_{0}(s,d+1)\\}-\\\
(\alpha_{-1}(s-1)+\beta_{-1})\\{(\alpha
s+\beta)(g_{0}(s,d-1)-g_{0}(s,d))+\alpha g_{0}(s,d)\\}=g_{0}(s,d).\end{split}$
(4.38)

Note that coefficient of $d^{m}$ in both $g_{0}(s,d)-g_{0}(s,d+1)$ and
$g_{0}(s,d-1)-g_{0}(s,d)$ are zero. Therefore comparing the coefficients of
$d^{m}$ on both sides of the equation (4.38), we get

$\alpha(\alpha_{-1}s+\beta_{-1})a_{m}(s)-\alpha(\alpha_{-1}(s-1)+\beta_{-1})a_{m}(s)=a_{m}(s)$

Hence we have $\alpha_{-1}=\alpha^{-1}$.
Now from the relation $[s\otimes t^{-1},p].1=p\otimes t^{-1}.1$ we have

$g_{-1}(s,d)=(\alpha_{-1}s+\beta_{-1})(g_{0}(s,d+1)-g_{0}(s,d))+\alpha_{-1}g_{0}(s,d).$
(4.39)

Now from the relation $[p\otimes t,p\otimes t^{-1}].1=0$, we get

$g_{-1}(s-1,d-1)g_{1}(s,d)=g_{-1}(s,d)g_{1}(s-1,d+1)$ (4.40)

Now we compute the coefficient of $d^{2m-1}$ on both side of equation (4.40)
with the help of equation (4.39) and equation (4.37). Now

$g_{-1}(s-1,d-1)g_{1}(s,d)$

$=\\{(\alpha_{-1}(s-1)+\beta_{-1})(g_{0}(s-1,d)-g_{0}(s-1,d-1))+\alpha_{-1}g_{0}(s-1,d-1)\\}\times\\{(\alpha
s+\beta)(g_{0}(s,d-1)-g_{0}(s,d))+\alpha g_{0}(s,d)\\}$

$=\\{(\alpha_{-1}(s-1)+\beta_{-1})(ma_{m}(s-1)d^{m-1}+O(d^{m-2}))+\alpha_{-1}(a_{m}(s-1)d^{m}-ma_{m}(s-1)d^{m-1}+a_{m-1}(s-1)d^{m-1}+O(d^{m-2})\\}\times\\{(\alpha
s+\beta)(-ma_{m}(s)d^{m-1}+O(d^{m-2}))+\alpha(a_{m}(s)d^{m}+a_{m-1}(s)d^{m-1}+O(d^{m-2})\\}$

$=\\{\alpha_{-1}a_{m}(s-1)d^{m}+[ma_{m}(s-1)(\alpha_{-1}(s-1)+\beta_{-1})-\alpha_{-1}ma_{m}(s-1)+\alpha_{-1}a_{m-1}(s-1)]d^{m-1}+O(d^{m-2})\\}\times\\{\alpha
a_{m}(s)d^{m}+[-(\alpha s+\beta)ma_{m}(s)+a_{m-1}(s)]d^{m-1}+O(d^{m-2})\\}$

From the above we have coefficient of $d^{2m-1}$ of
$g_{-1}(s-1,d-1)g_{1}(s,d)$ equal to

$a_{-1}a_{m}(s-1)[\alpha a_{m-1}(s)-m(\alpha s+\beta)a_{m}(s)]+\alpha
a_{m}(s)[ma_{m}(s-1)(\alpha_{-1}(s-1)+\beta_{-1})-\alpha_{-1}ma_{m}(s-1)+\alpha_{-1}a_{m-1}(s-1)].$

In a similar way we have coefficient of $d^{2m-1}$ of
$g_{-1}(s,d)g_{1}(s-1,d+1)$ equal to

$\alpha
a_{m}(s-1)[\alpha_{-1}a_{m-1}(s)+m(\alpha_{-1}s+\beta_{-1})a_{m}(s)]+\alpha_{-1}a_{m}(s)[m\alpha
a_{m}(s-1)+\alpha a_{m-1}(s-1)-m(\alpha(s-1)+\beta)a_{m}(s-1)].$

Now equating both coefficient of $d^{2m-1}$ we get

$\alpha_{-1}a_{m}(s-1)\\{\alpha a_{m-1}(s)-m(\alpha s+\beta)a_{m}(s)\\}+\alpha
a_{m}(s)\\{ma_{m}(s-1)(\alpha_{-1}(s-1)+\beta_{-1})-\alpha_{-1}(ma_{m}(s-1)-a_{m-1}(s-1)\\}$
$=\alpha
a_{m}(s-1)\\{\alpha_{-1}a_{m-1}(s)+m(\alpha_{-1}s+\beta_{-1})a_{m}(s)\\}+\alpha_{-1}a_{m}(s)\\{m\alpha
a_{m}(s-1)+\alpha a_{m-1}(s-1)-m(\alpha(s-1)+\beta)a_{m}(s-1)\\}.$
$\implies 4m\alpha\alpha_{-1}a_{m}(s)a_{m}(s-1)=0\implies m=0$. This completes
the proof.

∎

######  Proposition 4.6.

Let $M\in\mathcal{M}(\widehat{H_{4}},\hat{\mathfrak{h}})$, then $p\otimes
t^{k}.1=\alpha^{k}p.1$, $q\otimes t^{k}.1=\alpha^{k}q.1,$ $r\otimes
t^{k}.1=\alpha^{k}r.1$ for all $k\in\mathbb{Z}$.

###### Proof.

From $p\otimes t^{k}=[s\otimes t^{k},p]$ we see that if $p.1=0$, then
$p\otimes t^{k}=0$ for all $k\in\mathbb{Z}$. Now we assume $p.1\neq 0$. From
Lemma 4.5, we know that $\deg_{d}(g_{0})=0$. Now $p\otimes t.1=[s\otimes
t,p].1=(\alpha s+\beta)g_{0}(s)-g_{0}(s)(\alpha(s-1)+\beta)=\alpha
g_{0}(s)=\alpha p.1$. Also we know, $p\otimes t^{k+1}=[s\otimes t,p\otimes
t^{k}]$. Now use induction on $k$ for positive integer $k$. Similarly, we
prove for negative integers with the property $\alpha_{1}=\alpha^{-1}$. One
proves for $q\otimes t^{k}$ similarly. Now $r\otimes t^{k}.1=[p\otimes
t^{k},q].1=[p,q\otimes t^{k}].1=\alpha^{k}r.1$. ∎

Proof of theorem 4.2 :

###### Proof.

From Lemma 4.5 we have, $p.1$ and $q.1$ are in $\mathbb{C}[s]$. This implies
that $\mathbb{C}[s]$ is a rank one $U(H_{4})$ submodule of $\mathbb{C}[s,d].$
Therefore we get the actions of $p.1,q.1$ and $r.1$ from Theorem 3.6.
Case I:  Let $p.1\neq 0$ ( similar case will arise if $q.1\neq 0$). Then we
use the relation $[s\otimes t^{k},p].1=p\otimes t^{k}.1$ to conclude that
$\alpha_{k}=\alpha^{k}.$
Subcase (i):  Let $p.1=a$ and $q.1=b$, $a,b\in\mathbb{C}^{*}.$ Then $r.1=0$
and,

$\displaystyle r.1+K.1=[p\otimes t,q\otimes t^{-1}]=0$ (4.41)
$\displaystyle\implies K.1=0.$ (4.42)

Subcase (ii):  Let $p.1=a_{1}s+a_{2}$ and $q.1=b$,
$a_{1},b\in\mathbb{C}^{*},a_{2}\in\mathbb{C}.$ Then $r.1=-a_{1}b$ and

$\displaystyle K.1+r.1=-a_{1}b\alpha_{1}\alpha_{-1}$ (4.43)
$\displaystyle\implies K.1=0.$ (4.44)

Subcase (iii):  Let $p.1=g(s)$ and $q.1=0$. Then clearly we have $K.1=0.$
Therefore in Case I we have $M\cong\tilde{M}(\alpha,\mathbf{\beta})$ with the
help of Proposition 4.6.
Case II: Let $p.1,q.1=0.$ Then clearly $M\cong\tilde{M}({\mathbf{f}})$, with
the help of Lemma 4.3. This completes the proof.

∎

For notational convenience we denote the Cartan free rank one modules for
$\widehat{H_{4}}$ as $\widehat{M}_{(g,0,\alpha,\beta)}$,
$\widehat{M}_{(0,g,\alpha,\beta)}$, $\widehat{M}_{(h,b,\alpha,\beta)}$,
$\widehat{M}_{(b,h,\alpha,\beta)}$ and $\widehat{M}_{(a,b,\alpha,\beta)}$ when
corresponding Cartan free rank one $H_{4}$ modules are $M_{(g,0)}$,
$M_{(0,g)}$, $M_{(h,b)}$, $M_{(b,h)}$ and $M_{(a,b)}$ respectively.

######  Theorem 4.7.

  1. (1)

$\widehat{M}_{(g,0,\alpha,\beta)}\cong\widehat{M}_{(g_{1},0,\alpha_{1},\beta_{1})}$
iff $g=g_{1},\alpha=\alpha_{1},\beta=\beta_{1}$

  2. (2)

$\widehat{M}_{(0,g,\alpha,\beta)}\cong\widehat{M}_{(0,g_{1},\alpha_{1},\beta_{1})}$
iff $g=g_{1},\alpha=\alpha_{1},\beta=\beta_{1}$

  3. (3)

$\widehat{M}_{(h,b,\alpha,\beta)}\cong\widehat{M}_{(h_{1},b_{1},\alpha_{1},\beta_{1})}$
iff $h=h_{1},\alpha=\alpha_{1},\beta=\beta_{1},b=b_{1}$

  4. (4)

$\widehat{M}_{(b,h,\alpha,\beta)}\cong\widehat{M}_{(b_{1},h_{1},\alpha_{1},\beta_{1})}$
iff $h=h_{1},\alpha=\alpha_{1},\beta=\beta_{1},b=b_{1}$

  5. (5)

$\widehat{M}_{(a,b,\alpha,\beta)}\cong\widehat{M}_{(a_{1},b_{1},\alpha_{1},\beta_{1})}$
iff $a=a_{1},\alpha=\alpha_{1},\beta=\beta_{1},b=b_{1}$

###### Proof.

We prove (3) and all other follows with the simialr proof. Let
$\phi:\widehat{M}_{(h,b,\alpha,\beta)}\to\widehat{M}_{(h_{1},b_{1},\alpha_{1},\beta_{1})}$
be an isomorphism. Then we can see that $\phi(1)$ must be a non-zero scalar.
Now we use the relation $\phi(x.1)=x.\phi(1)$, for $x=p,q$. This gives us
$h=h_{1}$ and $b=b_{1}.$ Then using the relation $\phi(x\otimes
t^{k}.1)=x\otimes t^{k}.\phi(1)$ for $x=p,s$ and $k\neq 0$ we will get that
$\alpha=\alpha_{1}$ and $\beta=\beta_{1}.$ This completes the proof. ∎

######  Theorem 4.8.

  1. (1)

$\widehat{M}_{(h,b,\alpha,\beta)}$, $\widehat{M}_{(b,h,\alpha,\beta)}$ and
$\widehat{M}_{(a,b,\alpha,be)}$ are irreducible $\widehat{H}_{4}$ module.

  2. (2)

$\widehat{H}_{4}$-modules $\widehat{M}_{(g,0,\alpha,\beta)}$ and
$\widehat{M}_{(0,g,\alpha,\beta)}$ are irreducible iff $g$ is constant.

###### Proof.

(1). We prove it for $\widehat{M}_{(h,b,\alpha,\beta)}$ (similar proof work
for other two modules). Let $W$ be a non-zero submodule of
$\widehat{M}_{h,b,\alpha,\beta}$ and $w\in W$ be a non-zero element of $W$ of
smallest degree in $d.$ Note that $deg_{d}(q\otimes t.w-f_{1}q.w)<deg_{d}(w)$
and $q\otimes t.w-f_{1}q.w$ is a non-zero element of $W$, hence
$deg_{d}(w)=0$, i.e $w\in M_{h,b}.$ Now observe that $U(H_{4})w$ is a non-zero
submodule of $M_{h,b}$, so $M_{h,b}\subseteq W.$ Now using the action of $d$
we have $W=M_{h,b}\otimes\mathbb{C}[d].$ This completes the proof of (1).
(2) Note that if $g$ is constant the proof of (1) will run through and hence
$\widehat{M}_{(g,0,\alpha,\beta)}$ is irreducible. On the other hand, if $g$
is non-constant, then corresponding to a factor $s-\alpha$ of $g$, the ideal
generated by $s-\alpha$ in $\mathbb{C}[s,d]$ forms a proper submodule of
$\widehat{M}_{(g,0,\alpha,\beta)}$. ∎

Now as an application of Theorem 4.2 we recover the Cartan free modules for
Affine-Virasoro Nappi-Witten Lie algebras, which was obtained in [11]. For
this we need the help of Theorem 3.2 of [10]. We state it here. Let $Vir(0,0)$
be the Lie algebra with basis $\\{d_{n},W_{n},K:n\in\mathbb{Z}\\}$ with
bracket relations:

$[d_{n},d_{m}]=(n-m)d_{m+n}+\delta_{m+n,0}\frac{m^{3}-m}{12}K$
$[d_{n},W_{m}]=mW_{m+n}$ $[K,.]=0$

for all $m,n\in\mathbb{Z}.$

######  Theorem 4.9.

Let $M$ be a free $U(\mathbb{C}d_{0}\oplus\mathbb{C}W_{0})$-module of rank 1
for $Vir(0,0)$. Then $M\cong M(\lambda,f)$, where $\lambda\in\mathbb{C}^{*}$
and $f$ is a polynomial in $W_{0}.$ Actions of elements of $Vir(0,0)$ on
$U(\mathbb{C}d_{0}\oplus\mathbb{C}W_{0})$ are given by

$\displaystyle
d_{m}.x(d_{0},W_{0})=\lambda^{m}(d_{0}+mf(W_{0}))x(d_{0}-m,W_{0})$ (4.45)
$\displaystyle W_{m}.x(d_{0},W_{0})=\lambda^{m}W_{0}x(d_{0}-m,W_{0})$ (4.46)
$\displaystyle K.x(s,d)=0,$ (4.47)

for all $m\in\mathbb{Z}$, $x(d_{0},W_{0})\in
U(\mathbb{C}d_{0}\oplus\mathbb{C}W_{0})$.

######  Theorem 4.10.

Let $M$ be a $U(\mathbb{C}d_{0}\oplus\mathbb{C}s)$ free module of rank one for
$\overline{H_{4}}$. Then they are the classes defined in Theorem 3.2 of [11].

###### Proof.

Case I: Let $p.1=g(s)\neq 0$ (similar method works if $q.1\neq 0$).
Note that we have $d_{n}.f(s,d)=f(s,d-n)d_{n}.1$. Let $d_{n}.1=f_{n}(s,d)$.
Now consider $[d_{n},p].1=0,$ which implies that
$g(s)f_{n}(s,d)=g(s)f_{n}(s-1,d).$ Since $g(s)\neq 0$, we have
$f_{n}(s,d)=f_{n}(s-1,d)$, this forces that $deg_{s}(f_{n}(s,d))=0.$ Let
$d_{n}.1=f_{n}(d).$ For all $n\in\mathbb{Z}\setminus\\{0\\}$ consider the
relation,

$\displaystyle\alpha_{1}^{n+1}g(s)=p\otimes t^{n+1}.1$
$\displaystyle=[d_{n},p\otimes t].1$
$\displaystyle=d_{n}(\alpha_{1}g(s))-p\otimes t.(f_{n}(d))$
$\displaystyle=\alpha_{1}g(s)f_{n}(d)-\alpha_{1}f_{n}(d-1)g(s)$
$\displaystyle=\alpha_{1}g(s)(f_{n}(d)-f_{n}(d-1)).$

The above relation implies that $f_{n}(d)-f_{n}(d-1)=\alpha_{1}^{n}$ and hence
$f_{n}(d)$ is a one degree polynomial. Therefore we have
$f_{n}(d)=\alpha_{1}^{n}d+\mu_{n}$ for some constant $\mu_{n},$ for all
$n\in\mathbb{Z}.$ Note that $\mu_{0}=0$. We know that $s\otimes
t^{k}.1=\alpha_{1}^{k}s+\beta_{k}$ for all $k\in\mathbb{Z}$ and $\beta_{0}=0.$
Claim : $\beta_{k}=0$ for all $k\in\mathbb{Z}.$ Consider the relation

$\displaystyle ks\otimes t^{k+n}.1$ $\displaystyle=[d_{n},s\otimes t^{k}].1$
$\displaystyle=k\alpha_{1}^{n}(\alpha_{1}^{k}s+\beta_{k}).$

Now comparing both side we have, $\beta_{k+n}=\alpha_{1}^{n}\beta_{k}$ for all
$k\neq 0$, for all $n\in\mathbb{Z}$. Now putting $n=-k$ and using the fact
$\beta_{0}=0$ we have $\beta_{k}=0$ for all $k\in\mathbb{Z}$.

Now we consider the relation $[(n-m)d_{m+n}.1=[d_{m},d_{n}].1$, this implies
that

$\displaystyle(n-m)d_{m+n}.1=(\alpha_{1}^{n}(d-m)+\mu_{n})(\alpha_{1}^{m}d+\mu_{m})-(\alpha_{1}^{m}(d-n)+\mu_{m})(\alpha_{1}^{n}d+\mu_{n})$
(4.48)

Putting $m=-n\neq 0,$ in equation (4.48) and equating constant term we have,
$\mu_{n}=-\alpha_{1}^{2n}\mu_{-n}$, for all $n\neq 0.$
Claim:  $\mu_{n}=(n-1)\alpha_{1}^{n-2}\mu_{2}-(n-2)\alpha_{1}^{n-1}\mu_{1},$
for all $n>2.$ It is easy to verify from the equation (4.48) with $m=1,n=2$
that $\mu_{3}$ satisfy the property of the claim. Assume the claim is true for
$n=k.$ Now comparing the constant terms in equation (4.48) with $m=1,n=k$, we
have,

$\displaystyle(k-1)\mu_{k+1}$
$\displaystyle=-\alpha_{1}^{k}\mu_{1}+k\alpha_{1}\mu_{k}$
$\displaystyle=-\alpha_{1}^{k}\mu_{1}+k\alpha_{1}[(k-1)\alpha_{1}^{k-2}\mu_{2}-(k-2)\alpha_{1}^{k-1}\mu_{1}]$
$\displaystyle=k(k-1)\alpha_{1}^{k-1}\mu_{2}-(k-1)^{2}\alpha_{1}^{k}\mu_{1}.$

Therefore the claim is proved by induction principal. Now putting $m=-1,n=3$
in equation (4.48) and comparing the constant terms we have,

$\displaystyle 4\mu_{2}$
$\displaystyle=\mu_{-1}(\mu_{3}+\alpha_{1}^{3})-\mu_{3}(\mu_{-1}-3\alpha_{1}^{-1})$
$\displaystyle=\mu_{-1}\alpha_{1}^{3}+3\mu_{3}\alpha_{1}^{-1}$
$\displaystyle=-\alpha_{1}\mu_{1}+3\alpha_{1}^{-1}(2\alpha_{1}\mu_{2}-\alpha_{1}^{2}\mu_{1}).$

From the above equation implies that $\mu_{2}=2\alpha_{1}\mu_{1}.$ Hence we
have $\mu_{n}=n\alpha_{1}^{n-1}\mu_{1}=n\alpha_{1}^{n}\lambda$, where
$\mu_{1}=\alpha_{1}\lambda,$ for some $\lambda\in\mathbb{C}$. Then we get
$\mu_{-n}=-n\alpha_{1}^{-n}\lambda$ for all $n<0$. Hence for all
$n\in\mathbb{Z}$ we have $\mu_{n}=n\alpha_{1}^{n}\lambda,$ for some
$\lambda\in\mathbb{C}.$ This proves that actions of $d_{m}$ and $s\otimes
t^{m}$ are identical with the actions of [11] Theorem 3.2. This completes the
proof for case I with the help of Theorem 4.2.
Case II: Let $p.1=q.1=0.$ In this case $M$ will be free module for the Lie
algebra $span\\{s\otimes t^{k},d_{k},K:k\in\mathbb{Z}\\}\cong Vir(0,0).$ Now
the result follows from Theorem 4.9. Hence completes the proof.

∎

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# Multiple Magnetorotons and Spectral Sum Rules in Fractional Quantum Hall
Systems

Dung Xuan Nguyen  Brown Theoretical Physics Center and Department of Physics,
Brown University, 182 Hope Street, Providence, Rhode Island 02912, USA  F. D.
M. Haldane  Department of Physics, Princeton University, Princeton, New Jersey
08544, USA  E. H. Rezayi  Physics Department, California State University Los
Angeles, Los Angeles, California 90032, USA  Dam Thanh Son  Kadanoff Center
for Theoretical Physics, University of Chicago, Chicago, Illinois 60637, USA
Kun Yang  NHMFL and Department of Physics, Florida State University,
Tallahassee, Florida 32306, USA

###### Abstract

We study numerically the the charge neutral excitations (magnetorotons) in
fractional quantum Hall systems, concentrating on the two Jain states near
quarter filling, $\nu=2/7$ and $\nu=2/9$, and the $\nu=1/4$ Fermi-liquid state
itself. In contrast to the $\nu=1/3$ states and the Jain states near half
filling, on each of the two Jain states $\nu=2/7$ and $\nu=2/9$ the graviton
spectral densities show two, instead of one, magnetoroton peaks. The
magnetorotons have spin 2 and have opposite chiralities in the $\nu=2/7$ state
and the same chirality in the $\nu=2/9$ state. We also provide a numerical
verification of a sum rule relating the guiding center spin $\bar{s}$ with the
spectral densities of the stress tensor.

Introduction.—The exploration of topological phases of matter started with the
discovery of the fractional quantum Hall (FQH) effect [1, 2, 3]. Under a
strong magnetic field, electrons in two dimensions form strongly correlated
quantum Hall systems. In the lowest Landau level (LLL) limit, the kinetic
energy of the electrons is constant, and the two-dimensional electron system
is driven to numerous exotic topological phases depending on the filling
fraction and the effective interactions. The topological characteristics of
the FQH ground state and its charged excitations can be understood using the
wavefunction approach, pioneered by Laughlin [2] and further developed by many
other authors [4, 5, 6, 7, 8]. The idea of the composite fermion (CF) by Jain
[9, 10] provides an explanation of numerous gapped FQH states observed in
experiment [11, 12] and an intuitive construction of model wavefunctions, and
suggests the abelian and non-abelian braiding statistic of quasi-particles and
quasi-holes [13, 14]. Inspired by Jain’s intuitive picture, a field theory
description of FQH states was developed (the Halperin-Lee-Read, or HLR,
theory), based on the idea of flux attachment [15], which predicts a gapless
Fermi-liquid state [16] at half filling, which has been confirmed
experimentally [17].

Recently, modifications to the HLR theory has been suggested to make it
consistent with the symmetries of a single Landau level. Particle-hole
symmetry, which has been long an issue [18, 19], is restored in the Dirac CF
theory for FQH states near half filling [20]. The dipole coupling of the CF to
the electric field takes care of the consistency of the theory with
diffeomorphism [21, 22]. In combination, these two modifications lead to
response functions consistent with all known symmetries (see, e.g., Ref.
[23]).

An important feature of gapped quantum Hall states is the existence of the
neutral magnetoroton mode, first suggested by Girvin, MacDonald and Platzman
(GMP) [24] and later observed in experiment [25, 26, 27]. Though GMP
originally introduced the magnetoroton as a charge density wave, recent works
[28, 29, 30], employing the lowest Landau level symmetries, suggest that
magnetoroton has spin 2 and thus can be considered as a massive “emergent
graviton” in FQH systems. The magnetoroton has been studied from many
different perspectives: by constructing the wavefunction [31, 32] (which
conforms with the spin-2 structure), as an excitation of the CF crossing
$\Lambda$ levels [33], by exact diagonalization [34, 35, 36, 37, 38, 39], or
within the Dirac CF theory [30, 21, 40], where it is interpreted as the shear
deformation of the composite Fermi surface. In the latter studies, the
chirality of the magnetoroton is determined by the direction of the residual
magnetic field seen by the Dirac CFs, and the positions of the minima of the
magnetoroton dispersion relation match the experimental results [25, 26, 27]
rather well.

Very recently, the Dirac CF theory has been generalized to Jain states near
$\nu=\frac{1}{4}$ [41, 42, 43]. The situation with the magnetoroton there
seems to be very different from that near $\nu=\frac{1}{2}$: it is necessary
[43] to postulate a “Haldane mode,” i.e., an extra high-energy magnetoroton
(or multiple magnetorotons), in addition to the low-energy magnetoroton that
emerges from the dynamics of the CFs. The additional magnetoroton(s), which
contribute to the projected static structure factor (pSSF), are crucial for
the Haldane bound [44]

$S_{4}\geq\frac{|\bar{s}|}{4}\,,$ (1)

where $S_{4}$ is the coefficient of the leading $Q^{4}$ 111Here we define
$Q=q\ell_{B}$ with $q$ being the momentum and $\ell_{B}$ the magnetic length.
term in the pSSF, and $\bar{s}$ is the guiding center spin. On the LLL
$\bar{s}=\frac{1}{2}(\mathcal{S}-1)$ where $\mathcal{S}$ is the Wen-Zee shift.
The extra magnetoroton was heuristically suggested to arise from the
microscopic structure of the CF. While the electric dipole moment of the CF is
constrained by its momentum [21, 43], a higher moment deformation of its shape
could in principle generate a spin-2 mode, which is the high-energy
magnetoroton.

In this Letter we investigate the magnetoroton excitations, guided by the FQH
spectral sum rules [46, 47, 43] relating the chiral graviton spectral
functions with $\bar{s}$ and $S_{4}$. These sum rules constrain the spectral
densities of the $\nu=p/(2np\pm 1)$ states and the Fermi-liquid-like state at
$\nu=1/4$, and suggest the chiralities of the magnetorotons there. We
calculate the spectral densities numerically, from which we read out the
chirality of the magnetorotons and verify the sum rules.

Graviton spectral sum rules.—In the LLL limit, when the interacting energy is
much smaller than the cyclotron energy, one can obtain the exact sum rules
involving the spectral densities of the stress tensor [46, 47]. In the complex
coordinate $z=x+iy$, the two components of the traceless part of the stress
tensor, $T_{zz}=\frac{1}{4}\left(T_{xx}-T_{yy}-2iT_{xy}\right)$ and
$T_{\bar{z}\bar{z}}=\frac{1}{4}\left(T_{xx}-T_{yy}+2iT_{xy}\right)$, can be
used to define two spectral densities [46]

$\displaystyle I_{-}(\omega)$ $\displaystyle=\frac{1}{N_{e}}\sum_{n}|\langle
n|\\!\int\\!d\mathbf{x}\,T_{zz}|0\rangle|^{2}\delta(\omega-E_{n}),$ (2)
$\displaystyle I_{+}(\omega)$ $\displaystyle=\frac{1}{N_{e}}\sum_{n}|\langle
n|\\!\int\\!d\mathbf{x}\,T_{\bar{z}\bar{z}}|0\rangle|^{2}\delta(\omega-
E_{n}),$ (3)

where $N_{e}$ is the total number of electrons, $|0\rangle$ is the ground
state, the sum is taken over all excited states $|n\rangle$ in the lowest
Landau level, and $E_{n}$ is the energy of the state $|n\rangle$ relative to
the ground state. Physically, (2) and (3) are the densities of spin-2 states
with opposite chiralities at frequency $\omega$, and as such they depend on
the microscopic details of the FQH problem. The expressions for the integrals
of $T_{zz}$ and $T_{\bar{z}\bar{z}}$ over space in terms of the LLL operators
have been derived in Ref. [48]. We expect $I_{-}(\omega)$ and $I_{+}(\omega)$
to vanish at frequencies below the energy gap; we also expect them to rapidly
go to 0 at energies much larger than the energy scale set by the Coulomb
interaction. Using the $U(1)$ charge conservation and the LLL limit of
momentum conservation, one can obtain the following exact sum rules [46, 47,
43] 222The upper limit $\infty$ here means a frequency which is much higher
than the Coulomb gap but also smaller the the cyclotron energy.

$\displaystyle\int_{0}^{\infty}\frac{d\omega}{\omega^{2}}\left[I_{-}(\omega)+I_{+}(\omega)\right]$
$\displaystyle=S_{4},$ (4)
$\displaystyle\int_{0}^{\infty}\frac{d\omega}{\omega^{2}}\left[I_{-}(\omega)-I_{+}(\omega)\right]$
$\displaystyle=\frac{\bar{s}}{4}~{}\left(=\frac{\mathcal{S}-1}{8}~{}\text{on
LLL}\right).$ (5)

For derivations of the sum rules see Refs. [50, 47, 43]. Both sum rules do not
rely on a microscopic details and can be applied for fractional quantum Hall
states in the single Landau level limit, where Landau-level mixing is ignored.
By definition $I_{\pm}(\omega)$ are non-negative, therefore the sum rules
imply the Haldane bound (1). This bound is saturated if an only if the FQH
state is chiral, i.e., when one of the spectral densities vanishes identically
[i.e., $I_{-}(\omega)=0$ or $I_{+}(\omega)=0$].

General Jain states— The Wen-Zee shift $\mathcal{S}$ of the general Jain state
has been found previously [51, 52, 43]

$\displaystyle\nu_{+}=\frac{p}{2np+1},$
$\displaystyle\quad\mathcal{S}_{+}=p+2n,$ (6)
$\displaystyle\nu_{-}=\frac{p}{2np-1},$
$\displaystyle\quad\mathcal{S}_{-}=-p+2n.$ (7)

The subscript index $+$ ($-$) corresponds to the residual magnetic field seen
by CFs being in the same (opposite) direction as the applied magnetic field
[21, 43]. The direction of the residual magnetic field determines the
chirality of the low-energy magnetoroton, the one induced by the deformation
of composite Fermi surface. If the residual magnetic field is in the same
(opposite) direction of external field, the low-energy magnetoroton has
negative (positive) chirality. Consequently, we expect a low energy peak in
$I_{-}(\omega)$ ($I_{+}(\omega)$) of state $\nu_{+}$ ($\nu_{-}$).

In Ref. [43] a Dirac CF model of the $\nu_{\pm}$ states were presented. The
model is supposed to be reliable at large $p$ and its result for the spectral
densities can be summarized as

$\begin{split}\nu_{+}:\quad&\frac{I_{-}(\omega)}{\omega^{2}}=\frac{p+1}{8}\delta(\omega-\omega_{L})+\frac{n-1}{4}\delta(\omega-\omega_{H}),\\\
&I_{+}(\omega)=0,\end{split}$ (8)
$\begin{split}\nu_{-}:\quad&\frac{I_{-}(\omega)}{\omega^{2}}=\frac{n-1}{4}\delta(\omega-\omega_{H}),\phantom{+\frac{0+0}{8}\delta(\omega-\omega_{L})}\\\
&\frac{I_{+}(\omega)}{\omega^{2}}=\frac{p-1}{8}\delta(\omega-\omega_{L}),\end{split}$
(9)

where $\omega_{L}$ and $\omega_{H}$ are the energies of the low- and high-
energy magnetorotons, respectively. Note that the delta-function
$\delta(\omega-\omega_{H})$ may be broadened by the decay of the high-energy
magnetoroton or splits into several peaks. One can introduce the integrated
spectral densities

$\mathcal{I}_{\mp}=\int_{0}^{\infty}\frac{d\omega}{\omega^{2}}I_{\mp}(\omega).$
(10)

The prediction of Ref. [43] reads

$\displaystyle\nu_{+}:\quad\mathcal{I}_{-}=\frac{p+2n-1}{8},\quad\mathcal{I}_{+}=0,$
(11)
$\displaystyle\nu_{-}:\quad\mathcal{I}_{-}=\frac{n-1}{4},\quad\mathcal{I}_{+}=\frac{p-1}{8}.$
(12)

and from the sum rule (4) one finds the $S_{4}$ coefficient of the $\nu_{\pm}$
states:

$\displaystyle S_{4}(\nu_{+})=\frac{p+2n-1}{8}\,,\quad
S_{4}(\nu_{-})=\frac{p+2n-3}{8}.$ (13)

Some remarks are in order. (i) For $n=1$ (near half filling), there is no
high-energy magnetoroton, and both $\nu_{\pm}$ states are chiral. (ii) For
$n\neq 1$, only the $\nu_{+}$ state is chiral, with $S_{4}$ saturating the
Haldane bound, while the $\nu_{-}$ is not chiral. (iii) Strictly speaking, the
formulas presented above are obtained in the large $p$ limit, so the
application of these formulas for the case of, say, $p=2$ should be taken with
a grain of salt. On the other hand, one may expect that the qualitative
statements about the chirality of the magnetoroton modes are robust.

The Fermi-liquid states.—In the Fermi-liquid state with $\nu=1/2n$ the CFs are
in zero emergent magnetic field and form a Fermi liquid, whose excitations do
not contribute to the sum rule (5). The only contribution to the the sum rule
(5) is from the high-energy magnetoroton (the Haldane mode). Thus we find

$\mathcal{I}_{-}-\mathcal{I}_{+}=\frac{n-1}{4}\,,$ (14)

Note that since the state is ungapped, the notion of $S_{4}$ does not apply.
Naively we can associate $n-1$ with the guiding center spin for of the Fermi-
liquid state. This, in turn, can be explained if one thinks of the CF at
$\nu=1/2n$ as a CF at $\nu=1/2$ state with $2(n-1)$ flux quanta attached. The
CF at $\nu=1/2$ has no spin, while each attached flux quanta increases the
spin by $\frac{1}{2}$.

Looking at Eqs. (8), (9), and (14), we notice that the contribution of the
Haldane mode to the guiding center spin of states near $1/2n$ is universal. In
the Fermi-liquid state $\nu=1/2$, the spectral densities $I_{-}(\omega)$ and
$I_{+}(\omega)$ should be identical due to the particle-hole symmetry,
therefore $\mathcal{I}_{-}-\mathcal{I}_{+}=0$. (The same should be valid for
the PH-Pfaffian state [20].)

Numerical results—The graviton spectral function has been investigated on both
boson and fermion FQHE states which include Moore-Read states [53] as well as
Laughlin states in Refs. [54, 55]. In this Letter we present some results on
the guiding center spin $\bar{s}$, the coefficient $S_{4}$ of the $Q^{4}$ term
in the pSSF, and show the FQHE graviton spectral functions.

Figure 1: Graphical representation of $\bar{s}$ and $S_{4}$ obtained from the
sum rules at electron filling factor of 2/5 for both the Coulomb and the hard-
core potentials. The latter is just a Haldane pseudo-potential with its value
set to 1. We see that $\bar{s}/4$ and $S_{4}$ are mostly identical, which is
expected for a chiral state.

We find good agreement with the theoretical predictions of Eqs. (4) and (5).
We will present some of the results in this Letter, delegating to the
Supplement Material some others not germane to the main topic of the Letter.
To obtain the correct sum rules for the Coulomb interactions it is necessary
to use the complete stress tensor given in the Supplement Material [56] to
evaluate the spectral densities $I_{-}(\omega)$ and $I_{+}(\omega)$.

In Figs. 1 and 2 we present the results for the sum rules for the Jain states
at fillings 2/5, 2/7, and the Fermi liquid state 1/4. We use the left vertical
axis to represent $\bar{s}/4$, while the right vertical axis represents
$S_{4}$. As predicted by Haldane [44] only chiral states saturate the $S_{4}$
bound. Generic states such as the ones obtained from the Coulomb interaction
exceed this bound [46, 43]. The numerical results of $\bar{s}$ and $S_{4}$ of
Jain states converge to the theoretical predictions in Eqs. (13), (6) and (7).
We also numerically verify the Wen-Zee shift of the Fermi liquid state
$\nu=1/4$ predicted by Eq. (14). The numerical results presented in Figs. 1
and 2 are highly non-trivial, they are the first numerical check of the exact
graviton spectral sum rules in the LLL limit.

Figure 2: Same as in Fig. 1 but at electron filling factors $\nu=2/7$ and
$\nu=1/4$ for the Coulomb potential. The results for the 2/7 for $\bar{s}$ are
given by the filled symbols and their values are on the left vertical axis
(LVA). The RVA gives results for the $S_{4}$ of 2/7 (open symbols) and for
$\bar{s}/4$ of $\nu=1/4$ (open symbols with a central dot). The long lines
show the exact values of $\bar{s}/4$ on both vertical axes. The short dotted
line gives the same but for 1/4 state.  Figure 3: The graviton spectral
function for the filling factor $\nu=2/5$ and pure Coulomb potential. The
positive chirality total strength of the spectral function is much less than
the negative chirality. As a result we have normalized the spectrum of both
chiralities by the total weight of the gravitons with negative chirality. The
quasi-chiral nature can be attributed to dearth of quasi-holes of the chiral
parent, which is 1/3 for the hierarchy [57]. Similarly, in Jain’s CF approach,
the spectrum is nearly chiral if there is not any opposing flux (to the
attached ones) from the filled LLs.  Figure 4: Same as in Fig. 3 but for
$\nu=2/7$ filling and with the modified Coulomb. A clear separation of
predominantly positive chirality is seen on the left. On the right is the
spectrum for the negative chirality, which appears to contain the main weight
of total spectral density. Here we have the opposite case from 2/5. Hierarchy
involves quasiholes and filled LLs in the Jain construction have opposite
B-field to the attached fluxes.

Figures 3–6 present the FQH graviton spectral functions for $\nu=2/5$, $2/7$,
$1/4$, and $2/9$. For $\nu=2/5$, we use just the Coulomb potential since we
want to compare it to the spectrum obtained with the hard-core potential, but
for the remaining three fractions we calculate the spectral functions using a
modified version of Coulomb. The two forms give the same qualitative results
for the distributions of the graviton weights, particularly in instances where
there are two distinct energy sectors.

All the theoretical predictions on graviton’s chirality for the general Jain
states from Dirac CF model of Ref. [43] are confirmed numerically. For $n=1$,
the chirality of gravitons are determined by the residue magnetic field seen
by the CFs, therefore the graviton of $\nu=2/5$ has negative chirality as
showed in Fig. 3. With $n=2$ the chirality of low energy graviton is also
determined by the residue magnetic field, and the chirality of the high energy
graviton is universal for all Jain states near $1/4$. The predictions are
confirmed in Figs. 4 and 6.

Interestingly, the graviton spectral functions of Fermi-liquid state $\nu=1/4$
in Fig. 5 shows the high energy graviton with expected chirality. From both
the shift sum rule and the spectral densities, we see that the Haldane mode is
universal for all FQH states near $1/4$: it does not care if there is a
composite Fermi surface or if there is a residue magnetic field. Fig. 5 also
shows the low-energy excitations of both chirality with equal weight. We
expect that the spectral densities of the Fermi-liquid state $\nu=1/2$ share
the same feature with the low energy spectral densities of $\nu=1/4$ with
$I_{-}(\omega)$ and $I_{+}(\omega)$ being similar, the only difference is the
non-appearance of the Haldane mode.

Figure 5: The spectral function but for $\nu=1/4$ composite Fermi liquid for
the modified Coulomb potential. Note that the low energy spectrum has an equal
weight for both chiralities with our normalization. This means the
contribution to $\bar{s}$ is solely due to the Haldane mode.  Figure 6: The
spectral function but for $\nu=2/9$ with the modified Coulomb potential. The
spectrum is nearly completely chiral as would be expected since it is to the
right of the 1/5 in the hierarchy or involves filled LLs with no opposing flux
in the Jain’s construction as in 2/5 filling. Two well separated sectors with
negative chirality can be seen clearly. Here the size is too small for us to
get a meaningful $\bar{s}$ and $S_{4}$. However, the qualitative feature of
having two separated regions of nontrivial graviton spectral density will
persist for larger sizes.

Conclusion.—In this Letter, we compute numerically the graviton spectral
densities for the $\nu=2/7$, $2/9$, and $1/4$ states. For the first two
states, we observe in the spectral densities two magnetoroton peaks, one at
low energy with chirality that depends on the residual magnetic field acting
on the CFs (and thus are opposite for the $\nu=2/7$ and $2/9$ states), and one
at high energy with the same chirality for the two states. The higher-energy
magnetoroton is also observed in a spectral density of the $\nu=1/4$ state,
and this magnetoroton has approximately the same energy in all three filling
fractions. The result is consistent with the two-magnetoroton model of FQH
states near $1/4$ proposed in Ref. [43]. In addition, we have verified the FQH
graviton spectral sum rules for the two Jain states and the Fermi liquid state
$\nu=1/4$.

We hope that our results will motivate the experimental exploration of the
magnetoroton spectrum of the FQH states near $\nu=1/4$, as well as more detail
study of the Haldane mode which may reveal its nature.

This work is supported, in part, by the US Department of Energy, Basic Energy
Sciences grant DE-SC0002140 (FDMH, EHR, KY), U.S. DOE grant No. DE-
FG02-13ER41958 (DTS), a Simons Investigator grant and by the Simons
Collaboration on Ultra-Quantum Matter, which is a grant from the Simons
Foundation (651440, DTS). KY’s work was performed at the National High
Magnetic Field Laboratory, which is supported by National Science Foundation
Cooperative Agreement No. DMR-1644779, and the State of Florida. DXN was
supported by Brown Theoretical Physics Center.

Note: While this manuscript is being prepared, we became aware of the work
[58] that also discusses the extra graviton mode in the FQH states $\nu=2/7$
and $\nu=2/9$. We thank Ajit C. Balram, Zhao Liu, Andrey Gromov, and Zlatko
Papić for sharing their manuscript with us before publication.

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— Supplementary Material —
Multiple Magnetorotons and Spectral Sum Rules in Fractional Quantum Hall
Systems
Dung Xuan Nguyen, E. H. Rezayi, Dam Thanh Son, and Kun Yang

##  S1 Stress tensor operator on a general Landau level

In order to verify numerically the spectral sum rules, we need to use the
single Landau level limit of the holomorphic components of the stress-tensor
operator. The zero momentum stress tensor projected to a general Landau level
$N$ is given in Ref. [48],

$\int\\!d\mathbf{x}\,T_{ij}(\mathbf{x})=\frac{1}{2}\sum_{\mathbf{q}}\frac{q_{i}q_{j}}{q}\frac{\partial}{\partial
q}\left\\{e^{-\frac{q^{2}\ell_{B}^{2}}{2}}\left[L_{N}\left(\frac{q^{2}\ell_{B}^{2}}{2}\right)\right]^{2}V(q)\right\\}\bar{\rho}(\mathbf{q})\bar{\rho}(-\mathbf{q}).$
(S1)

with $L_{N}(x)$ being the Laguerre polynomial and $V(q)$ the Fourier
transformed of the two-body interaction,

$V(\mathbf{q})=\int
d^{2}\mathbf{x}\,V(\mathbf{x})e^{i\mathbf{q}\cdot\mathbf{x}}.$ (S2)

Since we assume the interaction is rotationally invariant, we expect that
$V(\mathbf{q})$ only depends on

$q=\sqrt{\mathbf{q}\cdot\mathbf{q}}.$ (S3)

In the complex coordinate, we define the complex momentum as

$q_{z}=\frac{1}{2}(q_{x}+iq_{y}),\quad
q_{\bar{z}}=\frac{1}{2}\left(q_{x}-iq_{y}\right).$ (S4)

Using Eq. (S1), we can write the holomorphic components of the stress tensor
at zero momentum as

$\int\\!d\mathbf{x}\,T_{zz}(\mathbf{x})=\frac{1}{2}\sum_{\mathbf{q}}\frac{q_{z}q_{z}}{q}\frac{\partial}{\partial
q}\left\\{e^{-\frac{q^{2}\ell_{B}^{2}}{2}}\left[L_{N}\left(\frac{q^{2}\ell_{B}^{2}}{2}\right)\right]^{2}V(q)\right\\}\bar{\rho}(\mathbf{q})\bar{\rho}(-\mathbf{q}),$
(S5)
$\int\\!d\mathbf{x}\,T_{\bar{z}\bar{z}}(\mathbf{x})=\frac{1}{2}\sum_{\mathbf{q}}\frac{q_{\bar{z}}q_{\bar{z}}}{q}\frac{\partial}{\partial
q}\left\\{e^{-\frac{q^{2}\ell_{B}^{2}}{2}}\left[L_{N}\left(\frac{q^{2}\ell_{B}^{2}}{2}\right)\right]^{2}V(q)\right\\}\bar{\rho}(\mathbf{q})\bar{\rho}(-\mathbf{q}).$
(S6)

In numerical calculations, we use the explicit form of the stress tensor in
Eqs. (S5), (S6) and the ground state from exact diagonalization to obtain the
spectral functions $I_{-}(\omega)$, $I_{+}(\omega)$.

##  S2 Numerical procedure

Our numerical studies are based on the evaluation of two stress tensor
spectral densities

$\displaystyle I_{-}(\omega)$ $\displaystyle=\frac{1}{N_{e}}\sum_{n}|\langle
n|\\!\int\\!d\mathbf{x}\,T_{zz}|0\rangle|^{2}\delta(\omega-E_{n}),$ (S7)
$\displaystyle I_{+}(\omega)$ $\displaystyle=\frac{1}{N_{e}}\sum_{n}|\langle
n|\\!\int\\!d\mathbf{x}\,T_{\bar{z}\bar{z}}|0\rangle|^{2}\delta(\omega-
E_{n}),$ (S8)

where the explicit expression of the full stress tensor at zero momentum is
given in Eqs. (S5) and (S6). Note that the expressions of the stress-tensor
components (S5) and (S6) differs from the ones used in Refs. [54, 55, 59, 60].
As explained in Ref. [48], the expressions in Ref. [54, 55, 59, 60] includes
only the “kinetic part” but omits the “potential part” of the stress tensor.

With the short-range pseudo-potential, the two expressions for the stress
tensor give the same numerical result, since one can show that the
contribution to zero momentum stress tensor from interaction annihilates the
ground state [47]. However, to obtain the correct sum rules (4), (5) for the
Coulomb interactions, it is necessary to use the complete stress tensor (S5)
and (S6) to evaluate the spectral densities $I_{-}(\omega)$ and
$I_{+}(\omega)$. The end result is that the LLL Coulomb potential is modified
by adding to $1/Q$ its cube. The first Haldane pseudo-potential is unaffected
for both bosons and fermions. We’ve performed numerical calculation of both
versions in this work.

The computer resources required for these types of calculations are much more
demanding than those in obtaining the Hall viscosity [61] from the Berry
curvature. The gain is the calculation of $S_{4}$ and the spectral functions
of Jain (or hierarchy)states, which are new. In this Letter and in the
Supplement Material, we present the numerical results of the spectral
densities $I_{-}(\omega)$ and $I_{+}(\omega)$ as well as the integrated
spectral densities $\mathcal{I}_{-}$ and $\mathcal{I}_{+}$.

##  S3 Supplemental Figures

In this supplemental section, we show some more numerical results of the shift
sum rule for $\nu=1/3$ and $\nu=2/3$ in Fig. S1. The chiralities of graviton
for these states come out as expected with negative for $\nu=1/3$ and positive
for $\nu=2/3$. We see that the sum rules are again confirmed by numerical
calculations.

(a) $\nu=1/3$ (b) $\nu=2/3$

Figure S1: (a) Graphical representation of $\bar{s}$ and $S_{4}$ obtained from
the sum rules at electron filling factor of $\nu=1/3$ for both the Coulomb and
the hard-core potentials. We see that $\bar{s}/4$ and $S_{4}$ are mostly
identical, which is expected for a chiral state. (b) Graphical representation
of $\bar{s}$ obtained from the sum rules at electron filling factor of
$\nu=2/3$ for hard-core potentials.

We also include the graviton’s spectral function for $\nu=2/5$, but for the
hard-core ($v_{1}=1$) and the modified Coulomb potentials in Fig. S2.

Figure S2: The graviton spectral functions for the filling factor $\nu=2/5$
with the hard-core ($v_{1}=1$) and the modified Coulomb potentials. We compare
these because they produce consistent values of $\bar{s}$ and $S_{4}$ in Fig
1. The spectrum on the left of 0.25 (in units of
$e^{2}/(4\pi\epsilon_{0}\ell_{B})$ is that of the modified Coulomb
interactions and the spectrum to the right is for hard-core potential with
$v_{1}=1$. The energies scales are different but the relative size of peaks
appear to be mirrored in both cases. The modified Coulomb potential has the
same scale and peak positions as the pure Coulomb potential in Fig 3. The
graviton with the negative chirality dominates both spectral functions as
expected.