# Skew Symmetric Extended Affine Lie algebras

S. Eswara Rao and Priyanshu Chakraborty  S. Eswara Rao: School of Mathematics,
Tata Institute of Fundamental Research, Homi Bhaba Road, Colaba, Mumbai
400005, India<EMAIL_ADDRESS><EMAIL_ADDRESS>Priyanshu
Chakraborty: School of Mathematical Sciences, East China Normal University,
Shanghai, 200241, China<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

For any skew symmetric matrix over complex numbers, we introduce an EALA and
it is called Skew Symmetric Extended Affine Lie Algebra (SSEALA). This way we
get a large class of EALAs and most often they are non-isomorphic. In this
paper we study irreducible integrable modules for SSEALA with finite
dimensional weight spaces. We classify all such modules in the level zero case
with non degenerate skew symmetric matrix.

###### Key words and phrases:

Toroidal Lie algebras, Extended affine Lie algebras.

###### 2010 Mathematics Subject Classification:

17B67, 17B66

Notations:

  * •

Throughout this paper we will work over the base field $\mathbb{C}$.

  * •

Let $\mathbb{C},\mathbb{R},\mathbb{Z}$ denote the set of complex numbers, set
of real numbers and set of integers respectively.

  * •

Let $\mathbb{Z}_{+}$ and $\mathbb{N}$ denote set of non-negative integers and
set of positive integers respectively. For $n\in\mathbb{N}$,
$\mathbb{C}^{n}=\\{(x_{1},\ldots,x_{n}):x_{i}\in\mathbb{C},1\leq i\leq n\\}$
and $\mathbb{R}^{n},\mathbb{Z}^{n},\mathbb{N}^{n}$ and $\mathbb{Z}_{+}^{n}$
are defined similarly.

  * •

Elements of $\mathbb{C}^{n},\mathbb{R}^{n}$ and $\mathbb{Z}^{n}$ are written
in boldface.

  * •

For any Lie algebra $\mathfrak{g}$, $U(\mathfrak{g})$ will denote universal
enveloping algebra of $\mathfrak{g}$.

  * •

Let $(\cdot|\cdot)$ denote the standard inner product on $\mathbb{C}^{n}$.

  * •

For any matrix $B$, $B^{T}$ denote the transpose matrix of $B.$

##  1\. Introduction

In this paper we introduce a class of Extended Affine Lie Algebras (EALA)
which we call Skew Symmetric Extended Affine Lie Algebras ( for short SSEALA).
For any skew symmetric matrix $B$ $(B^{T}=-B)$ over complex numbers we define
an EALA $\tau_{B}$ which we call Skew Symmetric Extended Affine Lie Algebra
(SSEALA). The first author in [16] introduced Hamiltonian Extended Affine Lie
Algebra (HEALA) and a Contact Extended Affine Lie Algebra (KEALA). They both
are SSEALA and the underlying skew symmetric matrix is non-degenerate for
HEALA (see Example 3.1) and degenerate for KEALA (see Example 3.2).

EALAs have been extensively studied in the last two decades, see [1, 2, 13,
14, 3] and references there in. There authors mainly study the structure of an
EALA and able to give a definite shape to the core of EALA (see 2.1 for the
definition of Core of an EALA). Later several authors studied the
representations of EALAs, see ([16, 23, 30, 5, 6, 27, 28]). In fact the first
author of the current paper has classified irreducible integrable modules for
HEALA (both level zero and non-zero level) in [16].

In this paper after introducing SSEALA we attempt to study integrable modules
for these EALAs assuming the skew symmetric matrix to be non-degenerate (In
the degenerate case the problem is more challenging and our methods does not
work in this case).

As seen in the previous cases like Toroidal Extended Affine Lie Algebra
(TEALA) in [23] and Hamiltonian Extended Affine Lie Algebra (HEALA) in [16]
the problem of classifying irreducible integrable modules for $\tau_{B}$ gets
reduced to classification of Jet modules (see Definition 4.1) for the
derivation algebra (denoted by $H_{B}$, see Remark 3.1). Our arguments on the
SSEALA reduced the classification problem to the case of Jet modules (see
Definition 4.1) of the derivation algebra (see Remark 3.1) of HEALA denoted by
$H_{N}$. The problem for this case has been solved by J.Talboom, [29]. Our
classification problem works only when the core of EALA (see 2.1) acts non
trivially on the modules. Otherwise the classification problem gets reduced to
the ordinary module of derivation algebras (see Remark 3.1) which is open.
Even in the known cases like S-type (TEALA) and H-type (HEALA) the problems
are open.

We will now give details of each section. In Section 2 we recall the
definitions of Toroidal Lie algebra, Full Toroidal Lie Algebra, Toroidal
Extended Affine Lie Algebra and Hamiltonian Extended Affine Lie Algebra. We
note that the root system of all the algebras are same but with different
multiplicities. We also recall the definition of EALA and provide some
examples.

In Section 3 we introduce the main object Skew Symmetric Extended Affine Lie
Algebra (SSEALA) and prove some standard properties (Proposition 3.1). In
Example 3.1 and Example 3.2 we explain how HEALA and KEALA are SSEALA by
exhibiting the corresponding skew symmetric matrices. Let $J$ be the matrix
for the HEALA (see Example 3.1) and $B$ be any non-degenerate skew symmetric
matrix. It is a classical fact that there exists a matrix $A$ in
$GL(N,\mathbb{C})$ such that $A^{T}BA=J$. Suppose $A$ is an integral matrix in
$GL(N,\mathbb{Z})$ then one can see that $\tau_{B}\simeq\tau_{J}$ (Proposition
3.2), where $\tau_{J}$ ia actually HEALA.

Unfortunately for any $B$ (for example when $B$ has irrational entries or
complex entries) we cannot find such an integral matrix $A$. Thus the matrices
$B$ and $J$ are related but the corresponding EALAs need not be isomorphic.

As noted above the classification of integrable modules for $\tau_{B}$ gets
reduced to the classification of Jet modules (see Definition 4.1) of
derivation algebra denoted by $H_{B}$ (see Remark 3.1), which are actually
modules for $\widetilde{H}_{B}\ltimes A$ with finite dimensional weight
spaces. The Section 4 is our main technical section where we reduce the
classification of irreducible modules for $\widetilde{H}_{B}\ltimes A$ to the
classification of finite dimensional irreducible modules for a certain infinte
dimensional Lie algebra denoted by $T$. $T$ contains decreasing sequence of
cofinite ideals (see Lemma 4.4). We prove in Lemma 4.7 that on any irreducible
finite dimensional module for $T$, some cofinite ideal acts trivially and
hence the finite dimensional module is actually a module for finite
dimensional Lie algebra. In Lemmas 4.9,4.10 and 4.11 we prove that the finite
dimensional Lie algebra can be taken as $\mathfrak{sp}_{2m}\oplus R$, where
$R$ is central. Thus $R$ acts as scalar on the finite dimensional module. We
believe that these scalars are trivial but could not prove it. Thus any finite
dimensional irreducible module for $T$ is actually an irreducible module for
$\mathfrak{sp}_{2m}$ with some center acting as scalars. We summarized this
results in Theorem 4.1.

In Section 5 we classify irreducible integrable modules for $\tau_{B}$ where
the center elements $K_{1},\dots,K_{N}$ acts trivially (level zero case). The
non-zero level case is open.

In Section 6 we classify irreducible integrable modules for KEALA in the non-
zero level case and the level zero case is open. We have included this as we
believe the right EALA for $K$-type is KEALA and the corresponding derivation
algebra (see Remark 3.1) is $K$-type (see [20]). Let us recall an important
class of infinite dimensional Lie algebras which are Lie algebras of Cartan
type, arises in the study of vector fields on manifold. These Lie algebras are
categorized into four series, namely the General, Special, Hamiltonian and
Contact, denoted by $W_{N},S_{N},H_{N},$ and $K_{N}$ respectively. We have
EALAs for type S (TEALA), and type H (HEALA). For type W we have full toroidal
Lie algebra which falls short of EALA. We propose KEALA for type $K_{N}$ (see
[20] for more motivation).

##  2\. Toroidal Lie algebras and Full toroidal Lie algebras

Fix a positive integer $N$. Let
$A=A_{N}=\mathbb{C}[t_{1}^{\pm{1}},t_{2}^{\pm{1}},\dots,t_{N}^{\pm{1}}]$ be
denote the Laurent polynomial ring in $N$ commuting variables. For
${\bf{r}}=(r_{1},r_{2},\dots,r_{N})\in\mathbb{Z}^{N}$ denote
$t^{\bf{r}}=t_{1}^{r_{1}}t_{2}^{r_{2}}\dots t_{N}^{r_{N}}\in A.$
Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra and
$\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$. It is well known that
$\mathfrak{g}$ has a root space decomposition with respect to $\mathfrak{h}$
given by
$\mathfrak{g}=\mathfrak{h}\displaystyle{\bigoplus_{\alpha\in\Delta}}\mathfrak{g}_{\alpha}$,
where $\Delta$ is the corresponding finite root system. Let $<.,.>$ be a
symmetric, non-degenerate and invariant bilinear form on $\mathfrak{g}$.

Let us consider the multi-loop algebra $\mathfrak{g}\otimes A$ with the usual
Lie bracket. Let

$\Omega_{A}=span\\{t^{{\bf r}}K_{i}:1\leq i\leq N,\bf{r}\in\mathbb{Z}^{N}\\},$
where $K_{i}=t_{i}^{-1}dt_{i}$, for $1\leq i\leq N$.

It is clear that $\Omega_{A}$ is $\mathbb{Z}^{N}$-graded vector space with
each component being $N$ dimensional. Let $d_{A}$ be the subspace of
$\Omega_{A}$ defined by
$d_{A}=span\\{\displaystyle{\sum_{i=1}^{N}}r_{i}t^{\bf{r}}K_{i}:\bf{r}\in\mathbb{Z}^{N}\\}$
and set ${Z}=\Omega_{A}/d_{A}$. Then $Z$ is $\mathbb{Z}^{N}$-graded, more
precisely each non-zero graded components of $Z$ are $N-1$ dimensional and
zeroth graded component is $N$ dimensional. Let
$K({\bf{u},\bf{r}})=\displaystyle{\sum_{i=1}^{N}{u_{i}{t^{\bf{r}}}K_{i}}}$ for
$\bf{u}\in\mathbb{C}^{N},\bf r\in\mathbb{Z}^{N}$. Now we define toroidal Lie
algebra. As a vector space toroidal Lie algebra is given by

$\tau=\mathfrak{g}\otimes A\oplus Z\oplus D,$

where $D$ is the degree derivation space spanned by
$\\{d_{1},d_{2},\dots,d_{N}\\}$. Before defining the brackets of the toroidal
Lie algebra we fix some notations for this paper. For convenience we set
$X({\bf{r}})=X\otimes t^{\bf r}$ and $\mathfrak{g}({\bf
r})=\mathfrak{g}\otimes\mathbb{C}t^{\bf r}$, $\bf r\in\mathbb{Z}^{N}$ and
$X\in\mathfrak{g}.$ Lie brackets of the toroidal Lie algebra are given by

  1. A1.

$[X({\bf r}),Y({\bf s})]=[X,Y]({\bf{r+s}})+<X,Y>K({\bf{r,r+s}}).$

  2. A2.

$Z$ is central in $\mathfrak{g}\otimes A.$

  3. A3.

$[d_{i},X({\bf r})]=r_{i}X({\bf r}),$ $[d_{i},K({\bf u,\bf r})]=r_{i}K({\bf
u,\bf r})$, $[d_{i},d_{j}]=0,$
for all ${\bf r,\bf s}\in\mathbb{Z}^{N},{\bf
u}\in\mathbb{C}^{N},X,Y\in\mathfrak{g},1\leq i,j\leq N.$

Let $\widetilde{\mathfrak{h}}=\mathfrak{h}\oplus Z_{0}\oplus D$, where
$Z_{0}=$ span $\\{K_{1},K_{2},\dots,K_{N}\\}$. This $\widetilde{\mathfrak{h}}$
plays the role of Cartan subalgebra for toroidal Lie algebra $\tau.$ One
should observe that the toroidal Lie algebra is the $N$ variable
generalization of affine Kac-Moody Lie algebra.
Let $DerA$ be the derivation algebra of $A.$ It is well known that $DerA$ has
a basis given by the vectors $\\{t^{\bf r}d_{i}:{\bf r}\in\mathbb{Z}^{N},1\leq
i\leq N\\},$ where $d_{i}=t_{i}\frac{\delta}{\delta t_{i}}$. Let $D({\bf u,\bf
r})=\displaystyle{\sum_{i=1}^{N}}u_{i}t^{\bf r}d_{i}$, where ${\bf
u}=(u_{1},u_{2},\dots,u_{N})\in\mathbb{C}^{N}$ and ${\bf r}\in\mathbb{Z}^{N}.$
Then $DerA$ forms a Lie algebra with respect to the bracket operation:

(2.1) $\displaystyle\left[D({\bf u,\bf r}),D({\bf v,\bf s})\right]=D({\bf
w,\bf r+\bf s}),$

where ${\bf w}=({\bf u}|{\bf s}){\bf v}-({\bf v}|{\bf r}){\bf u}$ for all
${\bf r,\bf s}\in\mathbb{Z}^{N}$, ${\bf u,\bf v}\in\mathbb{C}^{N}.$
It is known that $DerA$ admits an abelian extension on $Z$ by the following
actions:

(2.2) $\displaystyle[D({\bf u,\bf r}),D({\bf v,\bf s})]=D({\bf w,\bf r+\bf
s})-({\bf u}|{\bf s})({\bf v}|{\bf r})K({\bf{r,r+s}}),$ (2.3)
$\displaystyle[D({\bf u,\bf r}),K({\bf v,\bf s})]=({\bf u|\bf s})K({\bf v,\bf
r+\bf s})+({\bf u}|{\bf v})K({\bf{r,r+s}}),$

for more details see [22]. Moreover it is known from [25] that $DerA$ has no
non trivial central extension for $N\geq 2.$ For $N=1$ above defined abelian
extension becomes central extension. Now we are prepared to define the full
toroidal Lie algebra. As a vector space full toroidal Lie algebra is given by

$\tilde{\tau}=\mathfrak{g}\otimes A\oplus Z\oplus DerA.$

The bracket operations on $\tilde{\tau}$ are given by A1, A2, (2.2), (2.3) and
the following:

(2.4) $\displaystyle[D({\bf u,\bf r}),X({\bf s})]=({\bf u|\bf s})X({\bf r+\bf
s})$

for all ${\bf r,\bf s}\in\mathbb{Z}^{N}$, ${\bf u}\in\mathbb{C}^{N}.$
Now we define an automorphism of $\tilde{\tau}.$ Let $GL(N,\mathbb{Z})$ be the
groups of integral matrices with determinant $\pm 1$. Then there is a natural
action of $GL(N,\mathbb{Z})$ on $\mathbb{Z}^{N}.$ Let us denote this action by
$F{\bf r}$ for $F\in GL(N,\mathbb{Z})$ and ${\bf r}\in\mathbb{Z}^{N}.$ Then
each $F\in GL(N,\mathbb{Z})$ define an automorphism of $\tilde{\tau}$ by the
following map:

(2.5) $\displaystyle F.X({\bf r})=X(F{\bf r}),\hskip 14.22636ptF.K({\bf u,\bf
r})=K(F{\bf u},F{\bf r}),\hskip 14.22636ptF.D({\bf u,\bf
r})=D((F^{T})^{-1}{\bf u},F{\bf r}).$

It is easy to verify that this map defines an automorphism of $\tilde{\tau}$
(see [21] for details). We denote this automorphism of $\tilde{\tau}$ by
$\Phi_{F}$.
We now define roots and coroots of $\tilde{\tau}.$ Note that root system of
$\tilde{\tau}$ is same as the root system of toroidal Lie algebra. Therefore
we just recall root system for $\tau$ from [17]. For $1\leq i\leq N$ define
$\delta_{i},\omega_{i}\in{\widetilde{\mathfrak{h}}}^{\ast}$
(${\widetilde{\mathfrak{h}}}^{\ast}$ represents the dual of
$\widetilde{\mathfrak{h}}$ ) such that

$\delta_{i}(\mathfrak{h})=0,\hskip 14.22636pt\delta_{i}(K_{j})=0,\hskip
14.22636pt\delta_{i}(d_{j})=\delta_{ij},\hskip 14.22636pt1\leq j\leq N;$

$\omega_{i}(\mathfrak{h})=0,\hskip
14.22636pt\omega_{i}(K_{j})=\delta_{ij},\hskip
14.22636pt\omega_{i}(d_{j})=0,\hskip 14.22636pt1\leq j\leq N.$

For ${\bf{m}}=(m_{1},...,m_{N})\in\mathbb{Z}^{N}$, set $\delta_{\bf m}=$
$\displaystyle\sum_{i=1}^{N}m_{i}\delta_{i}$. Let $\pi=$
$\\{\alpha_{1},\alpha_{2},...,\alpha_{d}\\}$ be a set of simple roots of
${\Delta}$ and
$\pi^{\vee}=\\{\alpha_{1}^{\vee},\alpha_{2}^{\vee},\dots,\alpha_{d}^{\vee}\\}$
be the set of corresponding coroots of $\Delta$, where $d=$ dim
$\mathfrak{h}$. Then the set of vectors
$\\{\alpha_{1},\alpha_{2},...,\alpha_{d},\delta_{1},\delta_{2},\dots,\delta_{N},\omega_{1},\omega_{2},\dots,\omega_{N}\\}$
forms a basis for ${\widetilde{\mathfrak{h}}}^{*}$. Now we extend the bilinear
form of $\mathfrak{h}^{*}$ to a symmetric non-degenerate bilinear form on
${\widetilde{\mathfrak{h}}}^{*}$ by defining:

$<\alpha,\delta_{k}>=<\alpha,\omega_{k}>=0$, for all $\alpha\in\Delta,1\leq
k\leq N,$

$<\delta_{k},\delta_{p}>=<\omega_{k},\omega_{p}>=0,\hskip
5.69046pt<\omega_{k},\delta_{p}>=\delta_{kp},\hskip 5.69046pt1\leq k,p\leq N.$

Again $\widetilde{\mathfrak{h}}$ has a basis
$\\{\alpha_{1}^{\vee},\alpha_{2}^{\vee},\dots,\alpha_{d}^{\vee},K_{1},K_{2},\dots,K_{N},d_{1},d_{2},\dots,d_{N}\\}$
with bilinear form defined by:

$<h,K_{i}>=<h,d_{i}>=0,$ for all $h\in\mathfrak{h}$ and $1\leq i\leq N.$

$<d_{i},d_{j}>=<K_{i},K_{j}>=0$ and $<d_{i},K_{j}>=\delta_{ij}$ for $1\leq
i,j\leq N.$

It is clear that dim $\mathfrak{\widetilde{h}}$ = dim
$\mathfrak{\widetilde{h}}^{*}=2N+d$. Let $\Delta^{re}=\\{\alpha+\delta_{\bf
r}:\alpha\in\Delta,{\bf r}\in\mathbb{Z}^{N}\\}$ and
$\Delta^{im}=\\{\delta_{\bf r}:{\bf r}\in\mathbb{Z}^{N}\\}.$ Then
$\widetilde{\Delta}=\Delta^{re}\cup\Delta^{im}$ is the root system of
$\tilde{\tau}$ with respect to the Cartan subalgebra $\tilde{\mathfrak{h}}.$
Let $\bar{\lambda}$ be denote the restriction of
$\lambda\in\widetilde{\mathfrak{h}}^{*}$ to $\mathfrak{h}.$ Again any
$\mu\in\mathfrak{h}^{*}$ can be extend to $\widetilde{\mathfrak{h}}^{*}$ by
defining $\mu(d_{i})=\mu(K_{i})=0$ for $1\leq i\leq N.$ Therefore any
$\lambda\in\widetilde{\mathfrak{h}}^{*}$ can be expressed as
$\lambda=\bar{\lambda}+\displaystyle{\sum_{i=1}^{N}g_{i}\delta_{i}}+\displaystyle{\sum_{i=1}^{N}s_{i}\omega_{i}},$
see [17] for reference.
We also define $\alpha_{d+j}=-\beta+\delta_{j}$ for $1\leq j\leq N$, where
$\beta$ is the highest root $\mathfrak{g}.$ It is easy to see that the set of
vectors
$\\{\alpha_{1},\dots,\alpha_{d},\alpha_{d+1},\dots,\alpha_{d+N},\omega_{1},\dots,\omega_{N}\\}$
forms an another basis of $\widetilde{\mathfrak{h}}^{*}.$ For
$\gamma=\alpha+\delta_{\bf m}\in\Delta^{re}$, define a coroot by

$\gamma^{\vee}=\alpha^{\vee}+\frac{2}{<\alpha,\alpha>}{\displaystyle\sum_{i=1}^{N}}m_{i}K_{i}.$

For a real root $\gamma\in\Delta^{re},$ define a reflection $r_{\gamma}$ on
${\widetilde{\mathfrak{h}}}^{*}$ by

$r_{\gamma}(\lambda)=\lambda-\lambda(\gamma^{\vee})\gamma,\hskip
5.69046pt\lambda\in{\widetilde{\mathfrak{h}}}^{*}.$

Let $\Omega$ be the Weyl group generated by the reflections corresponding to
real roots of $\widetilde{\Delta}$.

######  Definition 2.1.

A module $V$ for $\tilde{\tau}$ (or over $\tau$ ) is said to be integrable if
$V$ satisfy the following properties:

  1. (1)

$V$ can be decomposed as
$V=\displaystyle{\bigoplus_{\lambda\in{\tilde{\mathfrak{h}}}^{*}}}{V_{\lambda}}$,
where $V_{\lambda}=\\{v\in V:hv=\lambda(h)v,\forall
h\in{\widetilde{\mathfrak{h}}}^{*}\\}.$

  2. (2)

$X_{\alpha}({\bf m})$ acts locally nilpotently on $V$ for all
$\alpha\in\Delta,X_{\alpha}\in{\mathfrak{g}}_{\alpha},{\bf
m}\in\mathbb{Z}^{N}.$

Let $V$ be an integrable module for $\tilde{\tau}$ (or over $\tau$) with
finite dimensional weight spaces, i.e dim $V_{\lambda}<\infty$ for all
$\lambda\in{\widetilde{\mathfrak{h}}}^{*}.$ Let
$P(V)=\\{\lambda\in{\widetilde{\mathfrak{h}}}^{*}:V_{\lambda}\neq 0\\}$ be
denote the set of all weights of $V$. Then the following are very standard (
see [17] ).

######  Lemma 2.1.

(1) $P(V)$ is $\Omega$ invariant.
(2) dim $V_{\lambda}=$ dim $V_{w\lambda}$ for all $w\in\Omega,\lambda\in
P(V).$
(3) If $\alpha\in\Delta^{re},\lambda\in P(V)$ and $\lambda(\alpha^{\vee})>0,$
then $\lambda-\alpha\in P(V)$

######  Definition 2.2.

Extended affine Lie algebra (EALA) : Let $L$ be a Lie algebra such that:

  1. EA1.

$L$ is endowed with a non-degenerate symmetric bilinear form $(.,.)$ which is
invariant (i.e $([x,y],z)=(x,[y,z]),$ for all $x,y,z\in L).$

  2. EA2.

$L$ possesses a non-trivial finite dimensional self-centralizing ad-
diagonalizable abelian subalgebra $H$.

To complete the definition of EALA, we need three more axioms. We observe some
consequence of EA2 which will be useful to define other three axioms. Note
that by property EA2 we have

$L=\displaystyle{\bigoplus_{\alpha\in H^{*}}}{L_{\alpha}}$, where
$L_{\alpha}=\\{x\in L:[h,x]=\alpha(h)x,\forall h\in H\\}.$

Let $R=\\{\alpha\in H^{*}:L_{\alpha}\neq 0\\},$ this set is called root system
of $L$ with respect to $H$. Note that $0\in R$ and $\alpha,\beta\in R$ with
$\alpha+\beta\neq 0\implies(L_{\alpha},L_{\beta})\neq 0.$ Hence we have
$R=-R$. The form restricted to $H$ is non-degenerate, thus we can induce a
form on $H^{*}$. Let

$R^{\times}=\\{\alpha\in R:(\alpha,\alpha)\neq 0\\},$ $R^{0}=\\{\alpha\in
R:(\alpha,\alpha)=0\\}.$

The elements of $R^{\times}$( respectively $R^{0})$ are called non-isotropic
(respectively isotropic) roots. It is clear that $R=R^{\times}\cup R^{0}.$

  1. EA3.

If $x_{\alpha}\in L_{\alpha}$ for $\alpha\in R^{\times}$, then $adx_{\alpha}$
act locally nilpotently on $L.$

  2. EA4.

$R$ is a discreat subset of $H^{*}$.

  3. EA5.

$R$ is an irreducible root system. This means that
(i) If $R^{\times}=R_{1}\cup R_{2}$ and $(R_{1},R_{2})=0$, then either
$R_{1}=0$ or $R_{2}=0$.
(ii) If $\sigma\in R^{0}$, then there exists a $\alpha\in R^{\times}$ such
that $\sigma+\alpha\in R.$

A Lie algebra satisfying EA1 to EA5 is called an EALA.

###  2.1. Core of EALA

The core of the EALA $L$ is defined as the subalgebra generated by
$\displaystyle{\bigcup_{\alpha\in R^{\times}}}L_{\alpha}$. It should mention
that the core of $L$ forms an ideal of $L$.
Extensive research has been done on EALA and its representation theory, for
instance see [1, 2, 13, 14, 3] and references there in. One can observe that
the full toroidal Lie algebra falls short to form an EALA, as it does not
satisfy (EA1). Now we study some known examples of EALAs.

######  Example 2.1.

Consider the Lie algebra $\mathfrak{g}\otimes
A\displaystyle{\bigoplus_{i=1}^{i=N}}\mathbb{C}K_{i}\displaystyle{\bigoplus_{i=1}^{i=N}}\mathbb{C}d_{i}$
with the Lie brackets given by

$[X({\bf r}),Y({\bf
s})]=[X,Y]({\bf{r+s}})+\delta_{\bf{r+s},0}<X,Y>\displaystyle{\sum_{i=1}^{i=N}}\mathbb{C}K_{i},$

$[K_{i},K_{j}]=[d_{i},d_{j}]=[K_{i},d_{j}]=0$,

$[d_{i},X({\bf r})]=r_{i}X({\bf r}),$

and $K_{i}$ are central, for all $X,Y\in\mathfrak{g},{\bf r,\bf
s}\in\mathbb{Z}^{N},1\leq i,j\leq N$.

It is the minimal EALA coming from the extension of a multiloop algebra.
Irreducible modules for this Lie algebra have been studied in [26], also see
[18].

######  Example 2.2.

A certain subalgebra of $\tilde{\tau}$ forms an EALA. Define
$S_{N}=span\\{D({\bf u,\bf r}):({\bf u|\bf r})=0,{\bf u}\in\mathbb{C}^{N},{\bf
r}\in\mathbb{Z}^{N}\\},$ this is a subalgebra of $DerA$. It is known to be a
simple Lie algebra of $S$ type. Let

$\tau(S_{N})=\mathfrak{g}\otimes A\oplus Z\oplus S_{N}.$

Define a bilinear form on $\tau(S_{N})$ by

$(X({\bf r}),Y({\bf s}))=<X,Y>\delta_{\bf r+\bf s,0},$ for all
$X,Y\in\mathfrak{g}$, $\bf r,s$ $\in\mathbb{Z}^{N}$,

$(D({\bf u,\bf r}),K({\bf v},{\bf s}))=\delta_{\bf r+\bf s,0}({\bf u}|{\bf
v})$ for all $\bf r,\bf s$ $\in\mathbb{Z}^{N}$, ${\bf u,\bf
v}\in\mathbb{C}^{N}$,

and all other values are zero.

It is a standard fact that $\tau(S_{N})$ is an EALA with the above bilinear
form (see [14] for general construction of EALAs). It is the largest EALA
coming from the extension of a multiloop algebra. There is a analogous notion
of twisted version of $\tau(S_{N})$ and irreducible integrable module for
these Lie algebras have been studied in [23, 30, 5, 6].

######  Example 2.3.

We define one another class of subalgebras of $\tilde{\tau}$ which forms an
EALA. We closely follow [29] to define Hamiltonian Lie algebra. Let $N=2m$ and
${\bf r}=(r_{1},\dots,r_{2m})$.
Let $\overline{\bf r}=(r_{m+1},\dots,r_{2m},-r_{1},\dots,-r_{m})$,
$H_{N}=span\\{D(\overline{\bf r},{\bf r}):0\neq{\bf r}\in\mathbb{Z}^{N}\\}$
and $\widetilde{H_{N}}=H_{N}\oplus D$. It is easy to verify that
$\widetilde{H_{N}}$ is a Lie algebra with respect to the bracket operations

$[D(\overline{\bf r},{\bf r}),D(\overline{\bf s},{\bf s})]=(\overline{\bf
r}|{\bf s})D(\overline{\bf r+\bf s},{\bf r+\bf s})$ and

$[D({\bf u},0),D(\overline{\bf r},{\bf r})]$=$({\bf u|\bf r})D(\overline{\bf
r},{\bf r})$ for all ${\bf r,\bf s}\in\mathbb{Z}^{N},{\bf
u}\in\mathbb{C}^{N}$.

The Lie algebra $\widetilde{H_{N}}$ is known as Hamiltonian Lie algebra and it
is also called as $H$ type Lie algebra.
Let $K=span\\{K({\bf u,\bf r}):({\bf u}|\overline{\bf r})=0\\}$. It is trivial
to check by (2.3) that $[H_{N},K]\subseteq K.$ Now consider the Lie algebra,
$\mathfrak{g}\otimes A\oplus Z\oplus\widetilde{H_{N}}\oplus D.$ Note that $K$
is an ideal in this Lie algebra. Let us define HEALA by
$\tau(H_{N})=\mathfrak{g}\otimes A\oplus Z/K\oplus{H_{N}}\oplus D.$ It is
proved in [16] that $\tau(H_{N})$ is an EALA and called it HEALA. Irreducible
integrable modules for HEALA have been studied in [16].

##  3\.  Skew symmetric Extended Affine Lie algebras

In this section we introduce a class of extended affine Lie algebras called
them by Skew symmetric extended affine Lie algebras (In short SSEALAs), see
[14] for general construction of EALAs. We provide some concrete examples of
SSEALA which includes few known EALAs arises from extensions of multiloop
algebras.
Let $B$ be a non zero $N\times N$ skew symmetric matrix over $\mathbb{C}$ i.e,
$B^{T}=-B$. Corresponding to $B$ we associate an EALA and called them skew
symmetric EALA (in short SSEALA). Let $h_{\bf r}=D(B{\bf r,r}),$ for ${\bf
r}\in\mathbb{Z}^{N}$, $G_{B}=\\{{\bf r}\in\mathbb{Z}^{N}:B{\bf r}=0\\},$
$H_{B}=span\\{D(B{\bf r},{\bf r}):{\bf r}\in\mathbb{Z}^{N}\\},$
$\widetilde{H_{B}}=H_{B}\oplus D.$ It is easy to see that $H_{B}$ forms a
subalgebra of $DerA$ with bracket operation $[D(B{\bf r},{\bf r}),D(B{\bf
s},{\bf s})]=(B{\bf r}|{\bf s})D(B({\bf r+\bf s}),{\bf r+\bf s}).$
For a skew symmetric matrix $B$ we associate a skew symmetric bilinear form on
$\mathbb{C}^{N}$ by $B({\bf x,\bf y})=(B{\bf x|\bf y})=-({\bf x}|B^{T}{\bf
y})$. Note that the following properties holds for the bilinear form $B$.

  1. 3.1

$(B{\bf r},{\bf r})=0$ for all ${\bf r}\in\mathbb{Z}^{N}$.

  2. 3.2

$(B{\bf r},{\bf s})=0$ for all ${\bf r}\in\mathbb{Z}^{N},$ ${\bf s}\in G_{B}$.

  3. 3.3

$(B{\bf r},{\bf s})=0$ for all ${\bf r+\bf s}\in G_{B}$.

Define $K_{B}=span\\{K({\bf u,\bf r}):(B{\bf u}|{\bf r})=0\\}$. It is trivial
to check that $[H_{B},K_{B}]\subseteq K_{B}$ (see (2.3)). Now consider the Lie
algebra, $\mathfrak{g}\otimes A\oplus Z\oplus\widetilde{H_{B}}.$ Note that
$K_{B}$ is an ideal in this Lie algebra. Let us denote
$\widetilde{Z}=Z/K_{B}\displaystyle{\bigoplus_{i=1}^{N}}\mathbb{C}K_{i}$. Let
us define SSEALA by $\tau_{B}=\mathfrak{g}\otimes
A\oplus\widetilde{Z}\oplus\widetilde{H_{B}}.$

######  Proposition 3.1.

(1) dim $(\widetilde{Z})_{\bf r}=0$ if $0\neq{\bf r}\in G_{B}$.
(2) dim $(\widetilde{Z})_{\bf r}=1$ if ${\bf r}\notin G_{B}$.
(3) dim $(\widetilde{Z})_{0}=N$.
($4$) dim $(\widetilde{H}_{B})_{\bf r}=0$ if $0\neq{\bf r}\in G_{B}$.
(5) dim $(\widetilde{H}_{B})_{\bf r}=1$ if ${\bf r}\notin G_{B}$.
(6) dim $(\widetilde{H}_{B})_{0}=N$.
(7) $\tau_{B}$ is an EALA.
(8) The core of $\tau_{B}$ is $\mathfrak{g}\otimes A\oplus\widetilde{Z}$.

###### Proof.

Note that $(1)$ and $(3)-(6)$ follows from definitions. To prove $(2)$, let
${\bf r}\notin G_{B}$ and ${\bf{u},\bf{v}}\in\mathbb{C}^{N}$ such that $(B{\bf
u}|\bf r)\neq 0$ and $(B{\bf v}|\bf r)\neq 0$. Now consider

$(B(\frac{u}{(B\bf u|\bf r)}-\frac{v}{(B\bf v|\bf r)})|r)=0.$

This implies that $K({\bf u,\bf r})=\lambda K({\bf v,\bf r})$ for some non
zero $\lambda\in\mathbb{C}$. Therefore $K(B\bf r,\bf r)$ can be taken as a
basis element for $(\widetilde{Z})_{\bf r}$ when ${\bf r}\notin G_{B}$.
To prove $(7)$, it is easy to verify that all the axioms of EALA are true
except to find a bilinear form. Define a bilinear form by

$(X({\bf r}),Y({\bf s}))=<X,Y>\delta_{\bf r+\bf s,0},$ for all
$X,Y\in\mathfrak{g}$, $\bf r,s$ $\in\mathbb{Z}^{N}$,

$(h_{\bf r},K(B{\bf s},{\bf s}))=\delta_{\bf r+\bf s,0}(B{\bf r}|B{\bf s})$
for all $\bf r,\bf s$ $\notin G_{B}$,

$(D{(\bf u},0),K{(\bf v},0))=(\bf u|\bf v)$ for all $\bf u,\bf v$
$\in\mathbb{C}^{N}$ and all other values are zero.

Note that this form desends from the form of $\tau(S_{N})$ after noting that
$(D(B{\bf r},{\bf r})|K_{B})=0$. It is easy to verify that the form is
invariant. Now to check that the form is non degenerate consider $\bf r+\bf
s=0$ and note that $(h_{\bf r},K(B{\bf s},{\bf s}))=-(B{\bf r},B{\bf r})\neq
0$ for all ${\bf r}\notin G_{B}$ and $(X({\bf r}),Y({\bf s}))=<X,Y>\neq 0$,
for $X\in\mathfrak{g}_{\alpha}$ and $Y\in\mathfrak{g}_{-\alpha}$. (8) is
standard to prove by definition of the core of EALA. This completes the proof.

∎

######  Remark 3.1.

Each SSEALA comes with a certain subalgebra of $DerA$ (see Section 2) denoted
by $\widetilde{H}_{B}$. This is called the derivation algebra of SSEALA.

######  Example 3.1.

Consider the skew symmetric matrix

$J=\begin{pmatrix}0&I_{m\times m}\\\ -I_{m\times m}&0\\\ \end{pmatrix}.$

The Lie algebra $\tau_{J}$ is a SSEALA which the HEALA considered in [16].
Note that in this example the skew symmetric bilinear form corresponding to
$J$ is non-degenerate and rank of $G_{B}$ is 0.

######  Example 3.2.

Now consider the matrix

$J_{1}=\begin{pmatrix}J&&&1\\\ &&&\vdots\\\ &&&1\\\ -1&\ldots&-1&0\\\
\end{pmatrix}$

The Lie algebra $\tau_{J_{1}}$ is another example of SSEALA. Note that the
skew symmetric bilinear form corresponding to $J_{1}$ is degenerate and rank
of $G_{J_{1}}$ is 1. This SSEALA was constructed in [16] and named as KEALA.

Now we recall a result from Goodman and Wallah which we will use later.

######  Lemma 3.1.

Let $V$ be a finite dimensional vector space of dimension $n$ and $B$ be a
non-degenerate skew symmetric bilinear form on $V$. Then $n=2m$ and there
exists a basis $v_{1},v_{2},\dots v_{2m}$ of $V$ such that
$B(v_{i},v_{j})_{1\leq i,j\leq n}=J$.

######  Remark 3.2.

Let $V=\mathbb{C}^{N}$ be a vector space with non-degenerate bilinear form
$(e_{i},e_{j})=\delta_{ij}$, $\\{e_{1},e_{2},\dots,e_{N}\\}$ is a basis of
$V$. Let $B$ be a skew symmetric matrix. Define $B({\bf x,\bf y})=(B{\bf
x}|{\bf y})$. Let $V_{B}=\\{{\bf v}\in V:B{\bf v}=0\\}.$ Then $B({\bf x},{\bf
y})=0$ if $\bf x$ or $\bf y$ $\in V_{B}$. Now consider the induced bilinear
form $\overline{B}$ on $V/V_{B}$, this $\overline{B}$ is non-degenerate. Hence
by above lemma there exists a basis ${\bf v_{1}},{\bf v_{2}},\dots,{\bf
v_{2m}}$ such that $\overline{B}({\bf v_{i},\bf v_{j}})_{1\leq i,j\leq n}=J$.
Let $k=$ dim $V_{B}$. Then we have $N=2m+k$. Also let ${\bf v_{2m+1}},{\bf
v_{2m+2}},\dots,{\bf v_{N}}$ be a basis for $V_{B}$. Then with respect to the
basis ${\bf v_{1}},{\bf v_{2}},\dots,{\bf v_{N}}$ of $V$, we have

$B^{\prime}=B({\bf v_{i},\bf v_{j}})_{1\leq i,j\leq
n}=\begin{pmatrix}J_{2m\times 2m}&0_{2m\times k}\\\ 0_{k\times 2m}&0_{k\times
k}\\\ \end{pmatrix}$

Now it is a general fact that there exists $A\in GL_{N}(\mathbb{C})$ such that
$B=A^{T}B^{\prime}A$. Unfortunately we cannot find any relation between
$\tau_{B}$ and $\tau_{B^{\prime}}$ unless $A$ is an integral matrix.

######  Proposition 3.2.

Let $B$ be a skew symmetric matrix in $\mathfrak{gl}_{N}(\mathbb{C})$ and
$A\in GL_{N}(\mathbb{Z})$ such that $B=A^{T}B^{\prime}A$ for some
$B^{\prime}$. Then $\tau_{B}\simeq\tau_{B^{\prime}}$.

###### Proof.

Consider the automorphism of full toroidal Lie algebra $\tilde{\tau}$ given by
$\Phi_{A}$ (2.5) corresponding to the matrix $A$. Note that $\Phi_{A}(D(B{\bf
r},{\bf r}))=D((A^{T})^{-1})B{\bf r},A{\bf r})=D(B^{\prime}A{\bf r},A{\bf
r}).$ Since $A$ is an invertible linear operator from the above we have
$\Phi_{A}(H_{B})=H_{B^{\prime}}.$ Also observe that $\Phi_{A}(K({\bf u},{\bf
r}))=(K(A{\bf u},A{\bf r}))$ and $(B^{\prime}A{\bf u}|A{\bf
r})=((A^{T})^{-1}B{\bf u}|A{\bf r})=(A^{T}(A^{T})^{-1}B{\bf u}|{\bf r})=(B{\bf
u}|{\bf r}).$ Hence $\Phi_{A}$ maps $K_{B}$ into $K_{B^{\prime}}$. Therefore
$\Phi_{A}$ induces an isomorphism between $\tau_{B}$ and $\tau_{B^{\prime}}$.
∎

######  Remark 3.3.

For a given skew symmetric matrix $B\in\mathfrak{gl}_{N}(\mathbb{C})$ it is
not easy to find a matrix $A\in GL_{N}(\mathbb{Z})$ such that
$B=A^{T}B^{\prime}A$ holds. Note that $A^{T}B^{\prime}A$ is obtained from
$B^{\prime}$ by simultaneous row and column operations. Now in the case when
$B=\begin{pmatrix}J&0_{2m,1}\\\ 0_{1,2m}&0\\\ \end{pmatrix}$ and
$B^{\prime}=J_{1}$ the simultaneous row and column operations induces only
integer matrices with entries in the set $\\{1,0,-1\\}$. Now instead of
providing the elementary row and column operations method we just provide the
matrix $A$, which is given by

$\begin{pmatrix}1&0&0\ldots&0&0\ldots&0&0\\\ 0&I_{m-1}&&\vdots&-I_{m-1}&&0\\\
\vdots&&&&&&\vdots\\\ 0&\ldots&0&0&0&\ldots 0&1\\\ -1&0_{m-1}&&0&I_{m-1}&&0\\\
\vdots&&&\vdots&&&\vdots\\\ -1&-1&\ldots&1&\ldots&1&-1&\\\ \end{pmatrix}_{.}$

Now it is easy to check that $AJ_{1}A^{T}=B$ and $A\in GL_{2m+1}(\mathbb{Z})$.

##  4\. Classification of irreducible modules for $\widetilde{H_{B}}\ltimes
A$.

For the rest of the paper we consider $N=2m$ (except for section 6 where we
assume N=2m+1) and $B$ as a non-degenerate skew symmetric matrix. Consider the
action of $\widetilde{H_{B}}$ on $A=A_{N}$ as

(4.1) $\displaystyle h_{\bf r}.t^{\bf s}=(B{\bf r|s})t^{\bf r+s}$ (4.2)
$\displaystyle D({\bf u},0).t^{\bf s}=({\bf u|s})t^{\bf s},$

for all ${\bf r,s}\in\mathbb{Z}^{N}$ and $\bf u$ $\in\mathbb{C}^{N}$. This
action defines a Lie algebra structure on $\widetilde{H_{B}}\ltimes A$. In
this section we classify irreducible modules for $\widetilde{H_{B}}\ltimes A$
with finite dimensional weight spaces and use this classification to classify
modules for SSEALA $\tau_{B}$.

######  Definition 4.1.

A module $V$ for $\widetilde{H}_{B}$ is said to be a jet module if $V$ can be
extended to a module for $\widetilde{H}_{B}\ltimes A$ with associative action
of $A$, in the sense that $t^{\bf r}t^{\bf s}=t^{\bf r+s}$, for all $\bf r,s$
$\in\mathbb{Z}^{N}$ and action $t^{\bf 0}$ as identity operator on $V$.

Let $V$ be an irreducible jet module for $\widetilde{H_{B}}\ltimes A$ with
finite dimensional weight spaces with respect to $H=D\oplus\mathbb{C}$. Choose
a weight $\lambda\in H^{*}$ such that $V_{\lambda}\neq 0$. Then due to
irreducibility of $V$ we have
$V=\displaystyle{\bigoplus_{m\in\mathbb{Z}^{n}}}V_{m},$ where $V_{m}=\\{v\in
V:D(u,0)v=(u|m+\alpha)v$ for all $u\in\mathbb{C}^{n}\\}$ and
$\alpha=(\lambda(d_{1}),\lambda(d_{2}),....,\lambda(d_{n})\in\mathbb{C}^{n}.$
Let $U$ denote the universal enveloping algebra of $\widetilde{H_{B}}\ltimes
A.$ Let $L$ be the two sided ideal of $U$ generated by $t^{\bf r}t^{\bf
s}-t^{\bf r+s}$ and $t^{0}-1$. Consider $U^{\prime}=U/L$ an associative
algebra. Note that $U^{\prime}$ is a $\mathbb{Z}^{N}$ graded algebra, in fact
$U^{\prime}=\displaystyle{\bigoplus_{{\bf r}\in\mathbb{Z}^{N}}}U^{\prime}_{\bf
r},$ where $U^{\prime}_{{\bf r}}=\\{X\in U^{\prime}:[D({\bf u},0),X]=({\bf
u|\bf r})X,\forall{\bf u}\mathbb{C}^{N}\\}$. Obviously $U^{\prime}_{0}$ is an
associative algebra.
Let $T({\bf r})=k({\bf-r})h_{\bf r}-D(B{\bf r},0)$, where $k({\bf r})=t^{\bf
r}$ for all $\bf r\in\mathbb{Z}^{N}$. Let $T=span\\{T({\bf r}):{\bf
r}\in\mathbb{Z}^{N}\\}$. Then it is easy to check that $[T({\bf r}),T({\bf
s})]=(B{\bf r}|{\bf s})[T({\bf r+\bf s})-T({\bf r})-T({\bf s})].$ In
particular, $T$ forms a Lie algebra with the above Lie bracket.

######  Lemma 4.1.

$U(T)=U^{\prime}_{0}$.

###### Proof.

Note that $U(T)\subseteq U_{0}^{\prime}$ is obvious. Let $X\in
U_{0}^{\prime}$. Then by PBW theorem we have $X$ is a linear combination of
monomials $k(-{\bf r_{1}})\dots k(-{\bf r_{l}})h_{\bf s_{1}}\dots h_{\bf
s_{k}}$, where ${\bf r_{i},\bf s_{i}}\in\mathbb{Z}^{N}$, $l,k\in\mathbb{N}$
and $\displaystyle{\sum_{i}}{\bf r_{i}}=\displaystyle{\sum_{i}}{\bf s_{i}}$.
Now using the fact $[h_{\bf r},k({\bf s})]=(B{\bf r}|{\bf s})k({\bf r+s})$ and
$k({\bf r})k({\bf s})=k({\bf r+s})$ we see that each monomial can be written
as sum of monomials of the form $k(-{\bf r_{1}})\dots k(-{\bf r_{l}})h_{\bf
r_{1}}\dots h_{\bf r_{l}}$. Hence $X\in U(T)$.

∎

######  Lemma 4.2.

(1) Each $V_{\bf r}$ is irreducible $T$ module.
(2) $V_{\bf r}\simeq V_{\bf s}$ as $T$-module.
(3) $k({\bf s}).V_{\bf r}=V_{\bf r+\bf s}$.

###### Proof.

Let $u,v\in V_{\bf r}$. Then by irreducibility of $V$ and weight arguments we
find $X\in U_{0}^{\prime}$ such that $X.u=v$. Hence $(1)$ follows by Lemma
4.1. Note that $k({\bf s})k(-{\bf s})=1$ implies $k({\bf s})$ is invertible,
this proves $(3)$. To prove $(2)$ define a map $f:V_{\bf r}\to V_{\bf s}$
given by $f(v)=k({\bf s-r})v$. Now

$f(T({\bf k})v)=k({\bf s-r})T({\bf k})v=T({\bf k})k({\bf s-r})v=T({\bf
k})f(v),$

since $k({\bf r})$ commutes with $T({\bf k})$. Hence $f$ is a $T$ module map.
It is easy to see that $f$ is both injective and surjective.

∎

Now it is clear that, $V\simeq V_{0}\otimes A$, where $V_{0}$ is a finite
dimensional irreducible $T$-module. Conversely given any finite dimensional
irreducible $T$-module $V_{0}$ we can define a $\widetilde{H}_{B}\ltimes
A$-module structure on $V_{0}\otimes A$. Let ${\bm{\beta}}\in\mathbb{C}^{N}$
and define

(4.3) $\displaystyle D(B{\bf r},{\bf r}).v\otimes t^{\bf k}=(B{\bf r}|{\bf
k+\bm{\beta}})v\otimes t^{\bf k+r}+(T({\bf r})v)\otimes t^{\bf k+r}.$ (4.4)
$\displaystyle t^{r}.v\otimes t^{\bf k}=v\otimes t^{k+r}.$

Now we check that the above defined action is a $\widetilde{H}_{B}\ltimes
A$-module action on $V_{0}\otimes A$. It is easy to check that $[D(B{\bf
r},{\bf r}),t^{\bf s}](v\otimes t^{\bf k})=(B{\bf r}|{\bf s})t^{\bf
r+s}(v\otimes t^{\bf k}).$ We check that $[D(B{\bf r},{\bf r}),D(B{\bf s},{\bf
s})](v\otimes t^{\bf k})=(B{\bf r}|{\bf s})D(B({\bf r+\bf s}),{\bf r+\bf
s})(v\otimes t^{\bf k}).$

Consider $[D(B{\bf r},{\bf r})D(B{\bf s},{\bf s})-D(B{\bf s},{\bf s})D(B{\bf
r},{\bf r})].v\otimes t^{\bf k}$

$=(B{\bf s}|{\bf k+\bm{\beta}})\\{(B{\bf r}|{\bf k+s+\bm{\beta}})v+T({\bf
r})v\\}\otimes t^{\bf k+r+s}+\\{(B{\bf r}|{\bf k+s+\bm{\beta}})T({\bf
s})v+T({\bf r})T({\bf s})v\\}\otimes t^{\bf k+r+s}$ $-(B{\bf r}|{\bf
k+\bm{\beta}})\\{(B{\bf s}|{\bf k+r+\bm{\beta}})v+T({\bf s})v\\}\otimes t^{\bf
k+r+s}-\\{(B{\bf s}|{\bf k+r+\bm{\beta}})T({\bf r})v+T({\bf s})T({\bf
r})v\\}\otimes t^{\bf k+r+s}$

$=\\{(B{\bf s}|{\bf k+\bm{\beta}})(B{\bf r}|{\bf k+s+\bm{\beta}})-(B{\bf
r}|{\bf k+\bm{\beta}})(B{\bf s}|{\bf k+r+\bm{\beta}})\\}v\otimes t^{\bf
k+s+r}+(B{\bf r}|{\bf s})T({\bf s})v\otimes t^{\bf k+s+r}-(B{\bf s}|{\bf
r})T({\bf r})v\otimes t^{\bf k+s+r}+[T({\bf r}),T({\bf s})]v\otimes t^{\bf
k+r+s}$

$=\\{(B{\bf s}|{\bf k+\bm{\beta}})(B{\bf r}|{\bf s})-(B{\bf r}|{\bf
k+\bm{\beta}})(B{\bf s}|{\bf r})\\}v\otimes t^{\bf k+s+r}+(B{\bf r}|{\bf
s})\\{T({\bf s})+T({\bf r})\\}v\otimes t^{\bf k+s+r}+[T({\bf r}),T({\bf
s})]v\otimes t^{\bf k+r+s}$

$=(B{\bf r}|{\bf s})(B({\bf r+s})|{\bf k+\bm{\beta}})v\otimes t^{\bf
k+r+s}+(B{\bf r}|{\bf s})T({\bf r+s})v\otimes t^{\bf k+r+s}$

$=(B{\bf r}|{\bf s})D(B({\bf r+\bf s}),{\bf r+\bf s})v\otimes t^{\bf k}.$

Therefore to classify irreducible modules for $\widetilde{H}_{B}\ltimes A$
with finite dimensional weight spaces, it is sufficient to classify finite
dimensional irreducible $T$-modules. For this we proceed similarly like [19].

######  Lemma 4.3.

( Proposition 19.1,[11]) Let $\phi:\mathfrak{L}\to\mathfrak{gl}(V)$ be a
finite dimensional irreducible representation of a Lie algebra $\mathfrak{L}$.
Then $\phi({\mathfrak{L}})$ is a reductive Lie algebra with at most one
dimensional center.

For $q\in\mathbb{N}$, define

$T_{q}({\bf s,\bf r_{1},\dots,r_{q}})=T({\bf s})-\displaystyle{\sum_{i}}T({\bf
s+r_{i}})+\displaystyle{\sum_{i<j}}T({\bf s+r_{i}+r_{j}})-\dots+(-1)^{q}T({\bf
s+r_{1}+\dots+r_{q}}).$

Let $I_{q}=span\\{T_{q}({\bf s,\bf r_{1},\dots,r_{q}}):{\bf s,\bf
r_{1},\dots,r_{q}}\in\mathbb{Z}^{N},q\in\mathbb{N}\\}$.

######  Lemma 4.4.

(1) $T_{q}({\bf s,\bf r_{1},\dots,r_{q}})=T_{q}({\bf s,\bf
r_{\sigma(1)},\dots,r_{\sigma(q)}}),$ for any permutations $\sigma$ on $q$
letters.
(2) $T_{q}({\bf s,\bf r_{1},\dots,r_{q}})=T_{q-1}({\bf s,\bf
r_{1},\dots,\widehat{r_{j}},\dots,r_{q}})-T_{q-1}({\bf s+r_{j},\bf
r_{1},\dots,\widehat{r_{j}},\dots,r_{q}})$, where $T_{q-1}({\bf s,\bf
r_{1},\dots,\widehat{r_{j}},\dots,r_{q}})$ represents that ${\bf r_{j}}$ does
not appear in this expression.
(3) $T_{q}({\bf s,\bf r_{1},\dots,r_{q}})=0$ if $\bf r_{j}=0$ for some $1\leq
j\leq q$.
($4$) $I_{q+1}\subseteq I_{q}$, for all $q\in\mathbb{N}$.
(5) $I_{1}=T$.
(6) $[T({\bf r}),T_{q}({\bf s,\bf r_{1},\dots,r_{q}})]=-(B{\bf r}|{\bf
s})T_{q+1}({\bf s,r,\bf r_{1},\dots,r_{q}})+$

$\displaystyle{\sum_{i=1}^{q}}(B{\bf r}|{\bf r_{i}})T_{q}({\bf s+r_{i},\bf
r_{1},\dots,\widehat{\bf r_{i}},\dots,r_{q},r})$

for all $q\geq 2$.
(7) $I_{q}$ is an ideal of $T$ for all $q\geq 1$.
(8) $[I_{p},I_{q}]\subseteq I_{p+q-2}$ for all $p+q\geq 3$ and
$[I_{1},I_{1}]\subseteq I_{1}$.

###### Proof.

Note that $(1),(5)$ follows from the definition. To prove $(2)$ collect all
the terms where ${\bf r_{j}}$ does not appear in the expressions of
$T_{q}({\bf s,\bf r_{1},\dots,r_{q}})$. It is easy to see that these terms
equal to $T_{q-1}({\bf s,\bf r_{1},\dots,\widehat{r_{j}},\dots,r_{q}})$. On
the other hand sum of the remaining terms of $T_{q}({\bf s,\bf
r_{1},\dots,r_{q}})$ is $-T_{q-1}({\bf s+r_{j},\bf
r_{1},\dots,\widehat{r_{j}},\dots,r_{q}})$. This proves $(2)$ and $(3),(4)$
immediately follows from $(2)$. we prove $(6)$ by induction principal. Note
that,

$[T({\bf r}),T_{2}({\bf s,r_{1},r_{2}})]=[T({\bf r}),T({\bf s})-T({\bf
s+r_{1}})-T({\bf s+r_{2}})+T({\bf s+r_{1}+r_{2}})]$

$=(B{\bf r}|{\bf s})\\{T({\bf r+s})-T({\bf r})-T({\bf
s})\\}-\displaystyle{\sum_{i=1}^{2}}(B{\bf r}|{\bf r_{i}+s})\\{T({\bf
r+s+r_{i}})-T({\bf r})-T({\bf s+r_{i}})\\}+(B{\bf r}|{\bf
s+r_{1}+r_{2}})\\{T({\bf r+s+r_{1}+r_{2}})-T({\bf r})-T({\bf
s+r_{1}+r_{2}})\\}$

$=-(B{\bf r}|{\bf s})T_{3}({\bf s,r,\bf
r_{1},\dots,r_{2}})+\displaystyle{\sum_{i=1}^{2}}(B{\bf r}|{\bf
r_{i}})T_{2}({\bf s+r_{i},\bf r_{1},\dots,\widehat{\bf
r_{i}},\dots,r_{2},r})$.
Hence the claim is true for $q=2$. Now assume that the claim is true for $q$.
Then
$[T({\bf r}),T_{q+1}({\bf s,\bf r_{1},\dots,r_{q}})]$
$=[T({\bf r}),T_{q}({\bf s,\bf r_{1},\dots,r_{q}})-T_{q}({\bf s+r_{q+1},\bf
r_{1},\dots,r_{q}})]$
$=-(B{\bf r}|{\bf s})T_{q+1}({\bf s,r,\bf
r_{1},\dots,r_{q}})+\displaystyle{\sum_{i=1}^{q}}(B{\bf r}|{\bf
r_{i}})T_{q}({\bf s+r_{i},\bf r_{1},\dots,\widehat{\bf r_{i}},\dots,r_{q},r})$
$+(B{\bf r}|{\bf s+r_{q+1}})T_{q+1}({\bf s+r_{q+1},r,\bf
r_{1},\dots,r_{q}})-\displaystyle{\sum_{i=1}^{q}}(B{\bf r}|{\bf
r_{i}})T_{q}({\bf s+r_{q+1}+r_{i},\bf r_{1},\dots,\widehat{\bf
r_{i}},\dots,r_{q},r})$
$=-(B{\bf r}|{\bf s})T_{q+2}({\bf s,r,\bf
r_{1},\dots,r_{q+1}})+\displaystyle{\sum_{i=1}^{q+1}}(B{\bf r}|{\bf
r_{i}})T_{q+1}({\bf s+r_{i},\bf r_{1},\dots,\widehat{\bf
r_{i}},\dots,r_{q+1},r}).$ Therefore by induction principal the claim is true
for all $q\geq 2$. Now using $(2,5,6)$ we have that $I_{q}$ is an ideal of
$T$. We prove (8) in the appendix. ∎

######  Lemma 4.5.

(1) $T_{q}({\bf s,\bf r_{1},\dots,r_{q}})\notin I_{q+1}$ .
(2) $\displaystyle{\bigcap_{q\geq 1}}I_{q}=\\{0\\}$.
(3) $T_{q}({\bf s,\bf r_{1},r_{2},\dots,r_{q}})+T_{q}({\bf s,\bf
n_{1},r_{2},\dots,r_{q}})=T_{q}({\bf s,\bf r_{1}+\bf
n_{1},r_{2},\dots,r_{q}})$ mod $I_{q+1}$.
($4$) $T_{q}({\bf s,-\bf r_{1},r_{2},\dots,r_{q}})=-T_{q}({\bf s-r_{1},\bf
r_{1},\dots,r_{q}})$.

###### Proof.

Define a map $\eta:T\to A$ by $T_{q}({\bf s,\bf
r_{1},\dots,r_{q}})\to\prod_{i=1}^{q}t^{\bf s}(1-t^{\bf r_{i}})$= $P_{q}({\bf
s,r_{1},\dots,r_{q}})$. Note that, $\eta(T_{q+1}({\bf s,\bf
r_{1},\dots,r_{q+1}}))$
$=\eta(T_{q}({\bf s,\bf r_{1},\dots,r_{q}})-T_{q}({\bf s+r_{1},\bf
r_{1},\dots,r_{q}}))=\prod_{i=1}^{q}t^{\bf s}(1-t^{\bf
r_{i}})-\prod_{i=1}^{q}t^{\bf s+r_{1}}(1-t^{\bf r_{i}})$
$=\prod_{i=1}^{q+1}t^{\bf s}(1-t^{\bf r_{i}})$.
This proves that the map is well defined. Now using the method of Claim 1 and
Claim 2 of Lemma 3.5 ([19]) we have $P_{q}({\bf s,r_{1},\dots,r_{q}})$ can not
be written as linear combination of $P_{q+1}({\bf s,r_{1},\dots,r_{q+1}})$,
this proves $(1)$.
To prove $(2)$ consider $J_{q}=span\\{t^{\bf r}(1-t^{\bf r_{1}})\dots(1-t^{\bf
r_{q}}):{\bf r,r_{i},\dots,r_{q}}\in\mathbb{Z}^{N}\\}.$ Clearly $J_{q}$ is an
ideal in $A$ and $J_{q}=J_{1}^{q}$. Note that $\eta(I_{q})=J_{q}$ implies that
to prove $(2)$ it is sufficient to prove that $\displaystyle{\cap_{q\geq
1}}J_{q}=0$. Let $f\in\displaystyle{\cap_{q\geq 1}}J_{q}$. Then
$f=\displaystyle{\sum_{i}^{a}c_{\bf m_{i}}t^{\bf m_{i}}}$, where ${\bf
m_{i}}=(m^{1}_{i},\dots m^{N}_{i})$. We multiply $f$ by a monomial such that
components of $\bf m_{i}$, for $1\leq i\leq a$ are non negative. Choose
$l_{1}=max\\{m_{i}^{1}:1\leq i\leq a\\}$. Let us assume that
$m^{1}_{1}=\dots=m^{1}_{s}=l_{1}$. Then
$(\frac{d}{dt_{1}})^{l_{1}}f=l_{1}!\displaystyle{\sum_{i}^{s}c_{\bf
m_{i}}t_{2}^{\bf m_{i}^{2}}\dots t_{N}^{\bf m_{i}^{N}}}$. Continuing this
process we can find $l_{2},\dots l_{N}$ with the above properties. Now using
the operator $d=(\frac{d}{dt_{1}})^{l_{1}}\dots(\frac{d}{dt_{1}})^{l_{N}}$ on
$f$ we get $df\neq 0$. Let $\displaystyle{\sum_{i}^{N}l_{i}}=L.$ Now $f\in
J_{L+1}$, hence every non zero components of $df$ has a factor of the form
$(1-t^{\bf m})$. Therefore $df(1,1,..,1)=0$, a contradiction. $(3)$ and $(4)$
follows on similar lines as $(2,3)$ of Lemma 3.5 ([19]).

∎

######  Lemma 4.6.

(1) $T_{q}({\bf s,\bf r_{1},\dots,r_{q}})=T_{q}({\bf r,\bf
r_{1},\dots,r_{q}})$ mod $I_{q+1}$.
(2) $-T_{q}({0,\bf r_{1},\dots,r_{q}})=-T_{q}({0,\bf-r_{1},\dots,r_{q}})$ mod
$I_{q+1}$.
(3) dim $I_{q}/I_{q+1}\leq N^{q}$.
($4$) $I_{q}$ is a co-finite ideal in $T$.

###### Proof.

Proof is similar to Lemma 3.6 of [19]. ∎

######  Lemma 4.7.

Let $J$ be a co-finite ideal in $T$. Then $J$ contains $I_{k}$ for some large
$k.$

###### Proof.

Consider the map $\pi:T\to T/J$. Note that the image of $I_{k}$ under $\pi$
goes to $I_{k}/{J\cap I_{k}}$. Consider the family $\\{I_{k}/J\cap I_{k}\\}$
of decreasing sequences of finite dimensional algebras. Thus there exists a
$k\in\mathbb{N}$ such that dim $I_{k}/J\cap I_{k}=$ dim $I_{k+i}/J\cap
I_{k+i}$ for all $i\geq 1$. To prove the lemma it is sufficient to prove that
$I_{k}/J\cap I_{k}=0$.
Claim: $I_{k}/J\cap I_{k}=0$. Let $0\neq x_{k}\in I_{k}/J\cap I_{k}$. Since
dim $I_{k}/J\cap I_{k}=$ dim $I_{k+i}/J\cap I_{k+i}$ and $\\{I_{k}/J\cap
I_{k}\\}$ is a family of decreasing sequences, there exists $x_{k+i}\in
I_{k+i}/J\cap I_{k+i}$ such that $x_{k}=x_{k+i}$ mod $I_{k}\cap J$, i.e
$x_{k}-x_{k+i}=0$ in $I_{k}/I_{k}\cap J.$ This means that $x_{k}-x_{k+i}=0$ in
$I_{k+1}/I_{k+1}\cap J.$ Hence there exists $y_{k+i}\in I_{k+i}\cap J$ such
that $x_{k}-x_{k+i}=y_{k+i}$, i.e $x_{k}\in I_{k+i}$ for all $i\geq 1$. This
proves that $x_{k}\in\cap_{k\geq 1}I_{k}$, a contradiction to Lemma 4.5(2).

∎

######  Proposition 4.1.

Let $V_{0}$ be a finite dimensional irreducible module for $T$. Then $I_{3}$
acts trivially on $V_{0}$.

###### Proof.

Let $\phi:T\to\mathfrak{gl}(V_{0})$ be the representation. Then by Lemma 4.7
$I_{k+1}\subseteq ker\phi$ for some large $k$. Now by Lemma 4.4(7)
$[I_{k},I_{k}]\subseteq I_{2k}$ and $I_{2k}\subseteq ker\phi$ we have
$\phi(I_{k})$ is an abelian ideal. Note that Lemma 4.3 imples that
$\phi(I_{k})$ is central and hence $\phi[T,I_{k}]$ acts trivially on $V_{0}$.
Lemma 4.4(6) gives
$[T({\bf m}),T_{k}({\bf r,s,\dots,s})]=-(B{\bf m}|{\bf r})T_{k+1}({\bf r,m,\bf
s,\dots,s})+$ $q(B{\bf m}|{\bf s})T_{k}({\bf r+s,\bf s,\dots,s,m}).$ Since
$I_{k+1}$ and $[T,I_{k}]$ contained in $ker\phi$, from the above formula we
have $T_{k}({\bf r+s,s,\dots,s,m})\in ker\phi$, when $(B{\bf m}|{\bf s})\neq
0$. Hence we have $T_{k}({\bf a,m,s,\dots,s})\in ker\phi$, when $(B{\bf
m}|{\bf s})\neq 0$ for arbitrary ${\bf a}\in\mathbb{Z}^{N}$. If $(B{\bf
m}|{\bf s})=0$, we consider $b\in\mathbb{Z}^{N}$ such that $(B{\bf b}|s)\neq
0$, this is possible as $B$ is non-degenerate. Therefore we have $T_{k}({\bf
a,b,s,\dots,s})\in ker\phi$ and $T_{k}({\bf a,m-b,s,\dots,s})\in ker\phi$, as
$(B({\bf m-b})|s)\neq 0$. Hence by Lemma 4.5(2) we have $T_{k}({\bf
a,m,s,\dots,s})\in ker\phi$ for all ${\bf a,m}\in\mathbb{Z}^{N}$. Continuing
this process we have $I_{k}\subseteq ker\phi$. Now instead of $I_{k+1}$ we
consider $I_{k}\subseteq ker\phi$ and repeat the above process to deduce
$I_{3}\subseteq ker\phi$ (see Lemma 4.4(8) why it fails for $I_{2}$). ∎

######  Lemma 4.8.

dim $T/I_{3}=$ $4m^{2}+2m$.

###### Proof.

We know that $I_{3}\subseteq I_{2}\subseteq I_{1}=T$. It is clear that dim
$T/I_{3}$= dim $T/I_{2}$ \+ dim $I_{2}/I_{3}$. Note that by Lemma 4.5(3) and
4.6(2,3) we have $T_{2}(0,\bf r,\bf s)$ is linear in both $\bf r,\bf s$ mod
$I_{3}$. Also $I_{2}$ is spanned by $T_{2}(0,\bf r,\bf s)$ mod $I_{3}$, hence
$\\{T_{2}(0,{\bf{e_{i}}},{\bf{e_{j}}}):1\leq i,j\leq 2m\\}$ forms a basis for
$I_{2}/I_{3}$. By a similar argument we have $I_{1}$ is spanned by
$T_{1}(0,\bf r)$ mod $I_{2}$ and $T_{1}(0,\bf r)$ is linear in $\bf r$ mod
$I_{2}$. Hence $\\{T_{1}(0,{\bf{e_{i}}}):1\leq i\leq 2m\\}$ forms a basis for
$I_{1}/I_{2}$. This proves the lemma. ∎

For a skew symmetric matrix $B$ define
$\mathfrak{G}_{B}=\\{X\in\mathfrak{gl}_{N}(\mathbb{C}):BX=-X^{T}B\\}$. It is
easy to check that $\mathfrak{G}_{B}$ is a Lie algebra and ${\bf r}(B{\bf
r})^{T}\in\mathfrak{G}_{B}$. Note that if $B$ can be written as
$B=A^{T}B^{\prime}A$ for some $A\in GL_{N}(\mathbb{C})$ then $B^{\prime}$ is
also a skew symmetric matrix and
$A\mathfrak{G}_{B}A^{-1}=\mathfrak{G}_{B^{\prime}}$. Let

$J=\begin{pmatrix}0&I_{m\times m}\\\ -I_{m\times m}&0\\\ \end{pmatrix}.$

Then $J$ is a skew symmetric matrix with $N=2m$. It is well known that
$\mathfrak{G}_{J}\simeq\mathfrak{sp}_{2m}$ and dimension of $\mathfrak{G}_{J}$
is $2m^{2}+m$. Consider a Lie algebra homomorphism
$\psi_{B}:T\to\mathfrak{G}_{B}$ given by

$\psi_{B}(T({\bf r}))={\bf r}(B{\bf r})^{T},$

here we treated ${\bf r}$ as column vector and hence $(B{\bf r})^{T}$ is a row
vector for all ${\bf r}\in\mathbb{Z}^{N}$.

######  Lemma 4.9.

Let $B=J$, $T_{J}=span\\{B_{\bf r}:{\bf r}\in\mathbb{Z}^{2m}\\}$, where
$B_{\bf r}=k({\bf-r})D({J\bf r,\bf r})-D(J{\bf r},0)$. Then
$\psi_{J}:T_{J}\to\mathfrak{sp}_{2m}$ is a surjective Lie algebra
homomorphism.

###### Proof.

Let $\overline{\bf r}=J{\bf r}$ and note that ${\bf r}\overline{\bf
r}^{T}\in\mathfrak{sp}_{2m}.$ For $1\leq i\leq m$, take ${\bf r=e_{i}}$, then
${\bf r}\overline{\bf r}^{T}=-{\bf e_{i}e_{m+i}}=-E_{i,m+i}$. Again for ${\bf
r=e_{m+i}}$, ${\bf r}\overline{\bf r}^{T}=-E_{m+i,i}$. It is easy to see that
${\bf r\overline{s}^{T}}+\bf s\overline{r}^{T}\in\mathfrak{sp}_{2m}$. Now we
construct a table from which it follows that $\psi_{J}$ is surjective (see
[11], Section 1.2 to find a basis of $\mathfrak{sp}_{2m}$).

${\bf r}$ | ${\bf s}$ | ${\bf r\overline{s}^{T}}+\bf s\overline{r}^{T}$
---|---|---
${\bf e_{i}}$ | ${\bf e_{j}}$ | $-(E_{i,m+j}+E_{j,m+i})$
${\bf e_{i}}$ | ${\bf e_{m+j}}$ | $-(E_{i,j}-E_{m+j,m+i})$
${\bf e_{m+i}}$ | ${\bf e_{j}}$ | $-E_{m+i,m+j}+E_{j,i}$
${\bf e_{m+i}}$ | ${\bf e_{m+j}}$ | $E_{m+i,j}+E_{m+j,i}$

In the above table $i,j$ runs from $1$ to $m$. ∎

######  Lemma 4.10.

$\psi_{B}$ is a surjective map.

###### Proof.

From Lemma 3.1 we have $B=A^{T}JA$ for some $A\in GL_{N}(\mathbb{C})$. Also we
know that $A\mathfrak{G}_{B}A^{-1}=\mathfrak{G}_{J}.$ Note that,
$\mathfrak{G}_{J}=span\\{{\bf r}(J{\bf r})^{T}:{\bf r}\in\mathbb{Z}^{N}\\}$
$=span\\{{\bf r}(J{\bf r})^{T}:{\bf r}\in\mathbb{C}^{N}\\}$ $=span\\{{A\bf
r}(J{A\bf r})^{T}:{\bf r}\in\mathbb{C}^{N}\\}$, for some fixed $A\in
GL_{N}(\mathbb{C})$. Now consider ${\bf r}(B{\bf r})^{T}={\bf r}(A^{T}JA{\bf
r})^{T}={\bf r}{\bf r}^{T}A^{T}J^{T}A={\bf r}(JA{\bf r})^{T}A=A^{-1}(A{\bf
r})(JA{\bf r})^{T}A.$ Therefore we have $span\\{{\bf r}(B{\bf r})^{T}:{\bf
r}\in\mathbb{Z}^{N}\\}=A^{-1}span\\{A{\bf r}(JA{\bf r})^{T}:{\bf
r}\in\mathbb{Z}^{N}\\}A=A^{-1}\mathfrak{G}_{J}A=\mathfrak{G}_{B}.$ This
completes the proof. ∎

######  Lemma 4.11.

$ker\psi_{B}$ is central ideal modulo $I_{3}$.

###### Proof.

It is easy to see that $ker\psi_{B}$ is an ideal and $I_{3}\subseteq
ker\psi_{B}$. Let $X=\displaystyle{\sum a_{\bf r}T({\bf r})}\in ker\psi_{B}$.
Then $\displaystyle{\sum a_{\bf r}{\bf r}(B{\bf r})^{T}}=0.$ Hence for any
${\bf k}=(k_{1},\dots,k_{N})$ we have, $\displaystyle{\sum a_{\bf r}{\bf
r}(B{\bf r})^{T}}{\bf k}=\displaystyle{\sum a_{\bf r}(B{\bf r}|{\bf k}){\bf
r}}=0$. This implies that $\displaystyle{\sum a_{\bf r}(B{\bf r}|{\bf
k}){r_{i}}}=0,$ for $1\leq i\leq N$. Now consider

$[T({\bf k}),X]=\displaystyle{\sum a_{\bf r}[T({\bf k})},T({\bf
r})]=\displaystyle{\sum a_{\bf r}}(B{\bf k}|{\bf r})[T({\bf r+\bf k})-T({\bf
k})-T({\bf r})]$

$=\displaystyle{\sum a_{\bf r}}(B{\bf k}|{\bf r})T_{2}(0,{\bf k,\bf
r})=\displaystyle{\sum a_{\bf r}}(B{\bf k}|{\bf
r})\displaystyle{\sum_{i}^{N}}r_{i}T_{2}(0,{\bf k,\bf e_{i}})=0,$ as
$\displaystyle{\sum a_{\bf r}(B{\bf r}|{\bf k}){r_{i}}}=0,$ for $1\leq i\leq
N$. Note that the 4th equality follows from Lemma 4.5(3) and the fact that
$I_{3}\subseteq ker\psi_{B}$.

∎

Note that dimension of $ker\psi_{B}/I_{3}=2m^{2}+m$, which follows from Lemma
4.8 and the fact that dimension of $\mathfrak{sp}_{2m}=2m^{2}+m$. Now
accumulating the results of Proposition 4.1 and Lemmas 4.9,4.10,4.11 we have
the following theorem.

######  Theorem 4.1.

(1) Let $V$ be an irreducible jet module for $\widetilde{H_{B}}$ with finite
dimensional weight spaces. Then $V\simeq V_{0}\otimes A$, where $V_{0}$ is an
irreducible module for $T$.
(2) Let $V_{0}$ be a finite dimensional irreducible module for $T$. Then
$V_{0}$ is an irreducible module for $\mathfrak{sp}_{2m}$. Conversely given
any finite dimensional irreducible module for $\mathfrak{sp}_{2m}$ it can be
extended to an irreducible module for $T$, where $ker\psi_{B}$ acts as
scalars.

##  5\. Level zero modules for SSEALA

In this section we classify level zero irreducible integrable modules for
$\tau_{B}$ with finite dimensional weight spaces. Consider a triangular
decomposition of $\tau_{B}$ given by
$\tau_{B}^{+}=span\\{X_{\alpha}({\bf r}):\alpha\in\Delta^{+},{\bf
r}\in\mathbb{Z}^{N}\\}$,
$\tau_{B}^{-}=span\\{X_{\alpha}({\bf r}):\alpha\in\Delta^{-},{\bf
r}\in\mathbb{Z}^{N}\\}$, and
$\tau_{B}^{0}=span\\{h({\bf r}),K(B{\bf r},{\bf r}),K({\bf u,0}),D({\bf
u,0}),D(B{\bf{r},r}):h\in\mathfrak{h},{\bf r}\in\mathbb{Z}^{N},\bf
u\in\mathbb{C}^{N}\\}$. Throughout this section fix an irreducible integrable
module $V$ for $\tau_{B}$ with finite dimensional weight spaces with respect
to $\widetilde{\mathfrak{h}}$.

######  Theorem 5.1.

Let $V$ be an irreducible integrable module for $\tau_{B}$ with finite
dimensional weight spaces with respect to $\widetilde{\mathfrak{h}}$. Suppose
all $K_{i}$ $(1\leq i\leq N)$ acting trivially on $V$. Then there exists a
weight vector $v_{0}$ such that ${\tau_{B}}^{+}.v_{0}=0$.

###### Proof.

First we apply the similar argument as in [19] to find a weight $\mu$ of $V$
such that $V_{\mu+\alpha}=0$ for all $\alpha\in\Delta^{+}$. We claim that
$V_{\mu+\alpha+\delta_{\bf m}}=0$ for all $\alpha\in\Delta^{+},{\bf
m}\in\mathbb{Z}^{N}$. If not, suppose $V_{\mu+\alpha+\delta_{\bf r}}\neq 0$
for some $\beta\in\Delta^{+},{\bf r}\in\mathbb{Z}^{N}$.
Claim : $V_{\lambda+\beta+\delta_{\bf s}}=0$ for all $\beta\in\Delta^{+},{\bf
s}\in\mathbb{Z}^{N}$, $\lambda=\mu+\alpha+\delta_{\bf r}$. Suppose this is
also not true, then $V_{\lambda+\beta+\delta_{\bf s}}\neq 0$ for some
$\beta\in\Delta^{+},{\bf s}\in\mathbb{Z}^{N}$. Since
$\alpha,\beta\in\Delta^{+}$ either $<\alpha+\beta,\alpha>>0$ or
$<\alpha+\beta,\beta>>0$. Let $<\alpha+\beta,\alpha>>0$. Then we have
$<\lambda+\beta+\delta_{\bf s},\alpha+\delta_{\bf s}+\delta_{\bf
r}>=<\mu+\alpha+\beta,\alpha>>0$. Hence by Lemma 2.1 $V_{\mu+\beta}\neq 0$, a
contradiction. This completes the proof. ∎

Let $V^{+}=\\{v\in V:\tau_{B}^{+}.v=0\\}$. Then by Theorem 5.1 we have
$V^{+}\neq 0$. It follows by some weight argument and PBW theorem that $V^{+}$
is an irreducible $\tau_{B}^{0}$-module. Note that $\mathfrak{h}$ is central
in $\tau_{B}^{0}$ and hence acts by a single linear functional (say
$\bar{\lambda}$) on $V^{+}$. By Lemma 2.1 and Theorem 5.1, $\bar{\lambda}$ is
a dominant integral weight. Now fix a weight vector $v\in V^{+}$ of weight
$\lambda$. Then all the weights of $V^{+}$ are lies in the set
$\\{\bar{\lambda}+\delta_{\bf
r}+\displaystyle{\sum_{i=1}^{N}g_{i}\delta_{i}}:{\bf r}\in\mathbb{Z}^{N}\\}$
and some ${\bf g}=(g_{1},\dots,g_{N})\in\mathbb{C}^{N}$.

######  Remark 5.1.

We assume that $\mathfrak{g}\otimes A$ acts non-trivially on $V$. It is easy
to see that this condition is equivalent to say that $\bar{\lambda}\neq 0$. If
$\mathfrak{g}\otimes A$ acts trivially on $V$, then $V$ becomes an irreducible
module for $\widetilde{H}_{B}$ and we have no classification result for
irreducible modules of $\widetilde{H}_{B}$.

######  Proposition 5.1.

(1) Weight spaces of $V^{+}$ are uniformly bounded.
(2) $\widetilde{Z}$ acts trivially on $V^{+}$.

###### Proof.

Let $\lambda$ be a weight of $V^{+}$. Then we know that all the weights are
given by $\lambda_{\bf r}=\bar{\lambda}+\delta_{\bf
r}+\displaystyle{\sum_{i=1}^{N}g_{i}\delta_{i}},$ for all ${\bf
r}\in\mathbb{Z}^{N}$. Define $\lambda^{0}=min\\{\lambda(h):\lambda(h)>0\\}$.
Note that by Remark 5.1 we get $\lambda^{0}\in\mathbb{N}.$ Now using similar
argument as in [[7], Lemma 3.1] with the fact that $K_{i}$ acting trivially we
have, for any $\lambda_{\bf r}$ there exists a $\omega_{r}\in\Omega$ such that
$\omega_{\bf r}(\lambda_{\bf r})=\lambda_{\bf s}$ and $0\leq
s_{i}<\lambda^{0}$ for $1\leq i\leq N$. Therefore Lemma 2.1 suggests that
dimensions of weight spaces of $V^{+}$ are bounded by
max$\\{dimV^{+}_{\lambda_{\bf s}}:0\leq s_{i}<\lambda^{0}\\},$ this proves
(1).
Let $M=\displaystyle{\bigoplus_{0\leq s_{i}<\lambda^{0}}V^{+}_{\lambda_{\bf
s}}}$ and $\tau_{N}=\mathfrak{g}\otimes
A\oplus\widetilde{Z}\displaystyle{\bigoplus_{i=1}^{N}}\mathbb{C}d_{i}$. Note
that $\tau_{N}$ is the quotient of $N$-variable toroidal Lie algebra. It is
also clear from the irreducibility of $V$ that any $\tau_{N}$ submodule of $V$
intersecting $V^{+}$ non-trivially is generated by elements of $V^{+}$. Hence
by Lemma 2.1 any such submodule is generated by a subset of $M$. Let
$W_{1}\supset W_{2}\supset\dots,$ be a strictly decreasing sequence of
$\tau_{N}$-submodules of $V$ intersecting $V^{+}$ non-trivially. As the
dimension of $M$ is finite, this sequence must terminate. Therefore there
exists a minimal submodule, say $V_{min}$ such that $V^{+}\cap V_{min}\neq 0$.
This $V_{min}$ may not be irreducible. But we can find an unique irreducible
quotient of $V_{min}$, say $\widetilde{V}$. Then by [17] we have
$\widetilde{Z}$ acts trivially on $\widetilde{V}$. This implies that
$\widetilde{Z}.(V^{+}\cap V_{min})=0$. Now it is easy to see that $W=\\{v\in
V^{+}:\widetilde{Z}.v=0\\}$ is a non zero $\tau_{B}^{0}$ submodule of $V^{+},$
hence $W=V^{+}$. ∎

Let $\alpha\in\Delta^{+}$ such that $\lambda(h_{\alpha})\neq 0$. Then by
results of [8] it follows that $h_{\alpha}\otimes t^{\bf r}$ acts trivially on
$V^{+}$ for all ${\bf r}\neq 0$ if and only if $\lambda(h_{\alpha})=0$.
Therefore $\lambda(h_{\alpha})\neq 0$ implies that $h_{\alpha}\otimes t^{\bf
r}$ acting non-trivially on $V^{+}$ for some ${\bf r}\neq 0$. Now it follows
from Appendix that $h_{\alpha}\otimes t^{\bf r}$ acting non-trivially on
$V^{+}$ for all ${\bf r}\neq 0$. Now we have $V^{+}$ is an irreducible module
for $\widetilde{H}_{B}\ltimes(\mathfrak{h}\otimes A)$. Here we treat
$\mathfrak{h}\otimes A$ as finitely many copies of $A$. In the appendix we
have taken only one copy of A but the arguments go through for finitely many
copies of $A$ (we need to use the fact that $\mathfrak{h}\otimes A$ is
commutative). Thus from appendix we have $\lambda(h_{\alpha})\neq 0$ on
$V^{+}$. Hence by Theorem 7.1 we find non-zero scalars
$\lambda_{\alpha},\mu_{\alpha}$ such that on $V^{+}$ it satisfy the conditions

(5.1) $\displaystyle h_{\alpha}\otimes t^{\bf r}.h_{\alpha}\otimes t^{\bf
s}=\lambda_{\alpha}h_{\alpha}\otimes t^{\bf r+s},{\bf r,s,r+s}\notin\\{0\\},$
(5.2) $\displaystyle h_{\alpha}\otimes t^{\bf r}.h_{\alpha}\otimes
t^{\bf-r}=\lambda(h_{\alpha})\mu_{\alpha},\forall{\bf r}\neq 0$

and $\lambda_{\alpha}^{2}=\mu_{\alpha}\lambda(h_{\alpha}).$ Then from Lemma
7.5 of [16] it follows that
$\lambda_{\alpha}=\mu_{\alpha}=c=\lambda(h_{\alpha})$ for all
$\alpha\in\Delta^{+}$ such that $\lambda(h_{\alpha})\neq 0$ [In Lemma 7.5 of
[16] we get the equation
$\displaystyle{\sum_{i}\lambda_{i}(h_{\alpha})a_{i}^{s}}=\lambda_{\alpha},$
where LHS is independent of basis of $V^{+}$ and RHS is independent of s. But
by the above equation both are independent of basis and s. Thus arguing as in
Lemma 7.5 of [16] we see that
$\mu_{\alpha}=\lambda_{\alpha}=\lambda(h_{\alpha})=c$].
We know from Proposition 5.1 that $\widetilde{Z}.V^{+}=0$. Now consider
$W=\\{v\in V:\widetilde{Z}.v=0\\}$ and observe that $W$ is a non-zero
$\tau_{B}^{0}$ submodule of $V$. Therefore by irreducibility of $V$,
$\widetilde{Z}$ acts trivially on $V$. Hence we have a nice description of
$V^{+}$ from Theorem 4.1, given by $V^{+}\simeq V_{N}\otimes A$, where $V_{N}$
is a finite dimensional irreducible $\mathfrak{sp}_{N-1}$ module.

######  Theorem 5.2.

Let $V$ be an irreducible integrable module for $\tau_{B}$ having finite
dimensional weight spaces with respect to $\widetilde{h}$ and
$\mathfrak{g}\otimes A$ acts non-trivially on $V$. Then there exists a finite
dimensional irreducible module $V(\mu)$ for $\mathfrak{g}$ and an finite
dimensional irreducible module $V_{N}$ for $\mathfrak{sp}_{N}$ such that
$V\simeq V(\mu)\otimes V_{N}\otimes A$. The action of $\tau_{B}$ on
$V(\mu)\otimes V_{N}\otimes A$ given by following:
$X({\bf r}).(v\otimes w\otimes t^{\bf s})=X.v\otimes w\otimes t^{\bf r+s}$,
for all $X\in\mathfrak{g}$, $\bf r,s\in\mathbb{Z}^{N},$ $v\in V(\mu),w\in
V_{N}$,
$\widetilde{Z}$ acts trivially on $V$ and action of $\widetilde{H}_{B}$ is
given by Theorem 4.1.

###### Proof.

Note that the highest weight is dominant integral for $\mathfrak{g}$, hence we
can find $V(\mu)$. Then it is easy to see that $V(\mu)\otimes V_{N}\otimes A$
is an irreducible module for $\tau_{B}$. Observe that
$\mathbb{C}v_{\mu}\otimes V_{N}\otimes A$ is the highest weight space for
$\tau_{B}$-module $V(\mu)\otimes V_{N}\otimes A$ with respect to the
considered triangular decomposition, where $v_{\mu}$ is the highest weight
vector of $V(\mu)$. Now since highest weight space of $V$ and $V(\mu)\otimes
V_{N}\otimes A$ are isomorphic as $\tau_{B}^{0}$-module. Hence $V\simeq
V(\mu)\otimes V_{N}\otimes A$. ∎

######  Remark 5.2.

Let us consider the matrix $J^{\prime}=\begin{pmatrix}J&0_{2m,1}\\\
0_{1,2m}&0\\\ \end{pmatrix}$. From Remark 3.3 we see that
$\tau_{J^{\prime}}\simeq\tau_{J_{1}}$. Though $J^{\prime}$ looks much simpler
than $J_{1}$, it does not simplify the above arguments any further for
$J_{1}$.

##  6\. Non Zero level modules for KEALA

In this section we classify non zero level irreducible integrable modules for
the KEALA constructed in [16] and obtained in this paper when $B=J_{1}$, i.e
$\tau_{J_{1}}$. We start this section with some notations for convenient. Let
$N=2m+1$ for this section. Let $\underline{\bf r}=J_{1}{\bf
r}=(r_{m+1}+r_{N},\dots,r_{2m}+r_{M},-r_{1}+r_{N},\dots,-r_{m}+r_{N},-\displaystyle{\sum_{i=1}^{2m}r_{i}})$
and $G_{J_{1}}=\\{{\bf r}\in\mathbb{Z}^{N}:J_{1}{\bf r}=0\\}$. Now consider
the triangular decomposition of $\tau_{J_{1}}$ given by

(6.1) $\displaystyle\tau_{J_{1}}^{+}=span\\{\mathfrak{g}^{+}\otimes t^{\bf
m},\mathfrak{g}\otimes t^{\bf r},K(\underline{\bf s},{\bf s}),D(\underline{\bf
s},{\bf s}):m_{N}=0,r_{N},s_{N}>0,{\bf m,r,s}\in\mathbb{Z}^{N},{\bf s}\notin
G_{J_{1}}\\}$ (6.2) $\displaystyle\tau_{J_{1}}^{0}=span\\{\mathfrak{h}\otimes
t^{\bf m},K(\underline{\bf s},{\bf s}),D(\underline{\bf s},{\bf
s}):m_{N}=0,s_{N}=0,{\bf m,s}\in\mathbb{Z}^{N},{\bf s}\notin
G_{J_{1}}\\}\oplus D,\hfill$ (6.3)
$\displaystyle\tau_{J_{1}}^{-}=span\\{\mathfrak{g}^{-}\otimes t^{\bf
m},\mathfrak{g}\otimes t^{\bf r},K(\underline{\bf s},{\bf s}),D(\underline{\bf
s},{\bf s}):m_{N}=0,r_{N},s_{N}<0,{\bf m,r,s}\in\mathbb{Z}^{N},{\bf s}\notin
G_{J_{1}}\\},$

where $\mathfrak{g}=\mathfrak{g}^{-}\oplus\mathfrak{h}\oplus\mathfrak{g}^{+}$
is the usual triangular decomposition of $\mathfrak{g}$. Let $V$ be an
irreducible integrable non zero level (i.e not all $K_{i}$ acting trivially on
$V$) module for $\tau_{J_{1}}$ with finite dimensional weight spaces with
respect to $\widetilde{\mathfrak{h}}$. Then by standard argument we can assume
that after twisting the module by an automorphism (2.5) we have only $K_{N}$
acting non trivially and all other $K_{i}$ acting trivially on $V$. Now
similar to the proof of Proposition 2.4 of [17] (also see Theorem 2.1 of [21])
we have the following theorem.

######  Theorem 6.1.

Let $V$ be an irreducible integrable non zero level module for $\tau_{J_{1}}$
with finite dimensional weight spaces with respect to
$\widetilde{\mathfrak{h}}$. Then there exists a weight $\mu$ of $V$ such that
$\tau_{J_{1}}^{+}.V_{\mu}=0$.

Let $W^{+}=\\{v\in V:\tau_{J_{1}}^{+}.v=0\\}$. Then by Theorem 6.1, $W^{+}\neq
0$. By standard argument it follows that $W^{+}$ is an irreducible module for
$\tau_{J_{1}}^{0}$ with finite dimensional weight spaces. Let $\lambda$ be a
weight of $W^{+}$. Then by irreducibility of $W^{+}$ we get weights of $W^{+}$
lies in the set $\\{\bar{\lambda}+\delta_{\bf
r}+\displaystyle{\sum_{i=1}^{N}g_{i}\delta_{i}}:{\bf r}\in\mathbb{Z}^{2m}\\}$
for some ${\bf g}=(g_{1},\dots,g_{N})\in\mathbb{C}^{N}$ and
$\bar{\lambda}=\lambda|_{\mathfrak{h}}$.

######  Remark 6.1.

We assume that $\bar{\lambda}\neq 0$. Otherwise using $\mathfrak{sl}_{2}$
theory and integrability of $V$ we can prove that $\mathfrak{h}\otimes A_{2m}$
acts trivially on $W^{+}$. Moreover, for any ${\bf r,s}\in\mathbb{Z}^{N}$ with
$s_{N}=r_{N}=0$, we have $[h({\bf s}),h({\bf r})]=<h,h>K({\bf r},{\bf r+s})$
acts trivially on $W^{+}$. Now consider $[D({\underline{\bf r}},{\bf
r}),K({\bf e_{N}},{\bf s})]=(\underline{\bf r}|{\bf s})K({\bf e_{N}},{\bf
r+s})+(\underline{\bf r}|{\bf e_{N}})K({\bf r,r+s})$, for $r_{N}=s_{N}=0$.
This implies that $K({\bf e_{N}},{\bf r+s})$ acts trivially on $W^{+}$ for all
$r_{N}=s_{N}=0$, when $(\underline{\bf r}|{\bf s})\neq 0$. Now for any ${\bf
k}\in\mathbb{Z}^{N}$ with $k_{N}=0$, it is easy to find ${\bf
r,s}\in\mathbb{Z}^{N}$ such that $r_{N}=s_{N}=0$, $(\underline{\bf r}|{\bf
s})\neq 0$ and ${\bf r+s=k}$. This proves that $W^{+}$ is an irreducible
module for $span\\{D(\underline{\bf s},{\bf s}):s_{N}=0,{\bf
s}\in\mathbb{Z}^{N},{\bf s}\notin G_{J_{1}}\\}\oplus D$. We proved in Lemma
6.1 that $span\\{D(\underline{\bf s},{\bf s}):s_{N}=0,{\bf
s}\in\mathbb{Z}^{N},{\bf s}\notin G_{J_{1}}\\}$ is isomorphic to $H_{N-1}$
(see Example 2.3 for the construction of $H_{N-1}$) as Lie algebra. Hence
$W^{+}$ is an irreducible module for the Lie algebra
$H_{N-1}\oplus\displaystyle{\bigoplus_{i=1}^{N-1}}\mathbb{C}d_{i}$, as $d_{N}$
is central in $\tau_{J_{1}}^{0}.$ But we don’t have any classification theorem
for $H_{N}\oplus D$ irreducible modules.

######  Proposition 6.1.

(1) Weight spaces of $W^{+}$ are uniformly bounded.
(2) $span\\{t^{\bf s}K_{i}:{\bf s}\in\mathbb{Z}^{N-1},1\leq i\leq
N-1\\}\cap\widetilde{Z}$ acting trivially on $W^{+}$.
(3) $t^{\bf m}K_{N}$ acts trivially on $W^{+}$ for ${\bf
m}\in\mathbb{Z}^{N}\setminus\\{0\\}$ with $m_{N}=0$.

###### Proof.

By the Remark 6.1, there exits $h\in\mathfrak{h}$ such that
$\bar{\lambda}(h)\neq 0$. Now by the fact that $\bar{\lambda}$ is dominant
integral we have $\lambda^{0}=min\\{\lambda(h):\lambda(h)>0\\}>0$. Now we can
proceed similarly like Proposition 5.1(1) to prove (1). To prove (2) we
consider $M=\displaystyle{\bigoplus_{\begin{subarray}{c}{\bf
s}\in\mathbb{Z}^{N-1}\\\ 0\leq
s_{i}<\lambda^{0}\end{subarray}}W^{+}_{\lambda_{\bf s}}}$, where $\lambda_{\bf
s}$ are the weights of $W^{+}$. Also let $\tau_{N-1}=\mathfrak{g}\otimes
A_{N-1}\oplus span\\{t^{\bf s}K_{i}:{\bf s}\in\mathbb{Z}^{N-1},1\leq i\leq
N-1\\}\cap\widetilde{Z}\oplus\displaystyle{\bigoplus_{i=1}^{N-1}}\mathbb{C}d_{i}$
be the quotient of $N-1$ variable toroidal Lie algebra. Now the proof follows
similarly like 5.1(2).
To prove (3), first note that when $r_{N}=0$ and
$\displaystyle{\sum_{i=1}^{N-1}r_{i}}=0$, then $K(e_{N},{\bf r})$ is zero in
$\tau_{J_{1}}$. Now consider the following bracket for $r_{N}=s_{N}=0$.

$[D({\underline{\bf s}},{\bf s}),K({\bf e_{N}},{\bf r})]=(\underline{\bf
s}|{\bf r})K({\bf e_{N}},{\bf r+s})+(\underline{\bf s}|{\bf e_{N}})K({\bf
s,r+s}).$

Since $K({\bf s,r+s})$ acting trivially on $W^{+}$ by (2), hence we have
$K(e_{N},{\bf r+s})$ acts trivially on $W^{+}$, when $(\underline{\bf s}|{\bf
r})\neq 0$. Now for given ${\bf m}\in\mathbb{Z}^{N}$ with $m_{N}=0$ and
$\displaystyle{\sum_{i=1}^{N-1}m_{i}}\neq 0$, it is easy to find ${\bf
r,s}\in\mathbb{Z}^{N-1}$ such that $(\underline{\bf s}|{\bf r})\neq 0$,
$\displaystyle{\sum_{i=1}^{N-1}r_{i}}=0$ and ${\bf r+s}={\bf m}$
(particularly, set ${\bf r}$ as, for $1\leq i\leq m$, $r_{i}=a$ and for
$m+1\leq i\leq 2m$, $r_{i}=-a$, $a\neq 0\in\mathbb{Z}$. Set $\bf s$ as,
$s_{i}=m_{i}-a$, for $1\leq i\leq m$ and $s_{i}=m_{i}+a$ for $m+1\leq i\leq
2m$ ). This completes the proof.

∎

######  Lemma 6.1.

Let $S=span\\{D(\underline{\bf s},{\bf s}):s_{N}=0,{\bf s}\notin
G_{J_{1}}\\}$. Then $S\simeq H_{N-1}$.

###### Proof.

Define a map $\eta:S\to H_{N-1}$ by $\eta(D(\underline{\bf s},{\bf
s}))=D(\overline{\bf s},{\bf s})$, where $\overline{\bf s}=J{\bf s}$. We check
that it is a Lie algebra homomorphism. Note that

$\eta([D(\underline{\bf r},{\bf r}),D(\underline{\bf s},{\bf
s})])=(\underline{\bf r},{\bf s})\eta(D(\underline{\bf r+s},{\bf
r+s}))=(\underline{\bf r},{\bf s})(D(\overline{\bf r+s},{\bf r+s}))$
$=(\overline{\bf r},{\bf s})(D(\overline{\bf r+s},{\bf
r+s}))=[\eta(D(\underline{\bf r},{\bf r})),\eta(D(\underline{\bf s},{\bf
s}))],$

$(\underline{\bf r},{\bf s})=(\overline{\bf r},{\bf s})$ due to the fact that
$s_{N}=0$. It is easy to see that $\eta$ is bijective. ∎

Therefore from Proposition 6.1 and Lemma 6.1 we have that $W^{+}$ is an
irreducible module for $\mathfrak{h}\otimes A_{N-1}\oplus
H_{N-1}\oplus\displaystyle{\bigoplus_{i=1}^{N-1}}\mathbb{C}d_{i}$ with
uniformly bounded weight spaces, as $K_{N},d_{N}$ are central in
$\tau_{J_{1}}^{0}$. Now we proceed similarly like SSEALA case in Section 5 and
conclude that $W^{+}$ is an irreducible module for
$A_{N-1}\rtimes(H_{N-1}\oplus\displaystyle{\bigoplus_{i=1}^{N-1}}\mathbb{C}d_{i})$.
Since $H_{N}=H_{J}$, therefore by Theorem 4.1 (also see [29] Theorem 5.2 ) we
have $W^{+}\simeq V_{N-1}\otimes A_{N-1}$, where $V_{N-1}$ is a finite
dimensional irreducible $\mathfrak{sp}_{N-1}$ module.
Let $W$ be an irreducible module for $\tau_{J_{1}}^{0}$. Now consider the
Verma module

$M(W^{+})=U(\tau_{J_{1}})\bigotimes_{\tau_{J_{1}}^{+}\oplus\tau_{J_{1}}^{0}}W,$

where $\tau_{J_{1}}^{+}$ acts trivially on $W$. Let $L(W)$ be the unique
irreducible quotient of $M(W)$. Moreover it is easy to see that if
$W_{1}\simeq W_{2}$ as $\tau_{J_{1}}^{0}$ then $L(W_{1})\simeq L(W_{2})$ as
$\tau_{J_{1}}$ module. Now since $W^{+}\simeq V_{N-1}\otimes A_{N-1}$ as
$\tau_{J_{1}}^{0}$ module, hence we have the following theorem.

######  Theorem 6.2.

Let $V$ be an irreducible integrable non zero level module for $\tau_{J_{1}}$
with finite dimensional weight spaces with respect to
$\widetilde{\mathfrak{h}}$ and $\mathfrak{h}$ acts non trivially on $V$. Then
upto a twist of an automorphism $V\simeq L(V_{N-1}\otimes A_{N-1})$, where
$V_{N-1}$ is a finite dimensional irreducible $\mathfrak{sp}_{N-1}$ module.

##  7\. appendix

In this section we prove two technical result which we have used in last three
sections. We follow the approach of [9, 16, 10] to prove Theorem 7.1. For
convenient we mention some notations here as in section 4. Recall the action
of ${H_{B}}$ on $A$ given by

(7.1) $\displaystyle h_{\bf r}.t^{\bf s}=(B{\bf r|s})t^{\bf r+s},$

for all $\bf r,s\in\mathbb{Z}^{N}$. Let $G=H_{B}\ltimes A$ be the semi direct
product of $H_{B}$ and $A$. Clearly $G=\displaystyle{\bigoplus_{\bf
r\in\mathbb{Z}^{N}}}G_{\bf r}$ is a $\mathbb{Z}^{N}$ graded Lie algebra. Let
$G^{\prime}=\displaystyle{\bigoplus_{\bf r\in\mathbb{Z}^{N}-\\{0\\}}}G_{\bf
r}$, then $G^{\prime}$ is an ideal of $G$. Also let
$A^{\prime}=\displaystyle{\bigoplus_{\bf
r\in\mathbb{Z}^{N}-\\{0\\}}}\mathbb{C}t^{\bf r}$. Let $W$ be an irreducible
module for $G$ which satisfy the following properties.

  1. (1)

$W=\displaystyle{\bigoplus_{\bf r\in\mathbb{Z}^{N}}W_{\bf r}}$ and $G_{\bf
r}.W_{\bf s}\subseteq W_{\bf r+s}$.

  2. (2)

$W$ is uniformly bounded $G$ module, i.e there exists a natural number $M$
such that dim $W_{\bf r}\leq M$ for all ${\bf r}\in\mathbb{Z}^{N}$.

######  Theorem 7.1.

Let $A^{\prime}$ acts non-trivially on $W$ and $t^{\bf 0}$ acts as a non-zero
scalar $c$ on $W$. Then there exists non-zero scalars $\lambda_{\bf r,s}$ such
that $t^{\bf r}t^{\bf s}=\lambda_{\bf r,s}t^{\bf r+s}$ on $W$. Moreover

  1. (1)

$\lambda_{\bf r,s}=\lambda$ for all ${\bf r,s}\neq 0,{\bf r+s}\neq 0.$

  2. (2)

$\lambda_{\bf r,-r}=\mu$ for all ${\bf r}\neq 0$.

  3. (3)

$\lambda_{0,\bf r}=c$ and $\lambda^{2}=\mu c$.

We prove this theorem using several lemmas. The next two lemmas follows from
Lemma 3.1 and Lemma 3.2 of [10].

######  Lemma 7.1.

Let $W=\displaystyle{\bigoplus_{\bf r\in\mathbb{Z}^{N}}W_{\bf r}}$ be an
irreducible uniformly bounded module for $A$. Then dim $W_{\bf r}\leq 1$ for
all $\bf r\in\mathbb{Z}^{N}$.

######  Lemma 7.2.

Let $g\in U(A)$ such that $g.v=0$ for some $v\in W$. Then $g$ is locally
nilpotent on $W$.

######  Lemma 7.3.

(1) Either each $t^{\bf r}$, $r\neq 0$ acts injectively on $W$ or every
$t^{\bf r}$ acts locally nilpotently on $W$.
(2) If $t^{\bf r}$ acts locally nilpotently for all $\bf r\neq 0$, then
$A^{\prime}.W=0$.

###### Proof.

Let $t^{\bf r}$ acts injectively on $W$ for some ${\bf r}\neq 0$. Let ${\bf
s}\in\mathbb{Z}^{N}$ such that $(B{\bf r}|{\bf s})\neq 0$ and $t^{\bf s}$ acts
locally nilpotently on $W$. Then using the arguments similar to Claim 1 of
[[9],Proposition 3.4 ] we have $t^{\bf s}$ acts nilpotently on $W$. Now
following arguments of Claim 3 of [[9],Proposition 3.4 ] and the fact that
$(B{\bf r}|{\bf s})\neq 0$ we get a contradiction. Hence $t^{\bf s}$ acts
injectively on $W$ for all ${\bf s}$ satisfying $(B{\bf r}|{\bf s})\neq 0$. If
$(B{\bf r}|{\bf s})=0$ then consider ${\bf s_{1}}\in\mathbb{Z}^{N}$ such that
$(B{\bf r}|{\bf s_{1}})\neq 0$ and $(B{\bf s_{1}}|{\bf s})\neq 0$. Hence by
above we have action of $t^{\bf s}$ is injective on $W$. This completes the
proof of 1. Proof of 2 follows just by changing $\bar{\bf r}$ to $B\bf r$ in
the proof of Proposition 9.5 of [16]. ∎

Proof of Theorem 7.1 : The proof runs parallel to Theorem 9.1, [16]. From
Lemma 7.3 it follows that $t^{\bf r}$ acts injectively on $W$ for all $\bf
r\neq 0$. This implies that dim $W_{\bf r}$ = dim $W_{\bf s}$ for all $\bf
r,s\in\mathbb{Z}^{N}$. Now by defining $T_{\bf r,\bf s}=t^{\bf r}t^{\bf
s}-\lambda_{\bf r,s}t^{\bf r+s}$ and following the proof of Theorem 9.1, [16]
we have

(7.2) $\displaystyle(B{\bf l}|{\bf s})\lambda_{\bf{r,s+l}}+(B{\bf l}|{\bf
r})\lambda_{\bf s,r+l}-(B{\bf l}|{\bf r+s})\lambda_{\bf r,s}=0,$

for all ${\bf l,r,s}\in\mathbb{Z}^{N}\setminus\\{0\\}.$ Now we prove some
lemmas to complete the proof of Theorem 7.1.

######  Lemma 7.4.

Let $(B{\bf l}|{\bf s})\neq 0$. Then we have

  1. (1)

$\lambda_{\bf s+l,s}=\lambda_{\bf s,s}$.

  2. (2)

$\lambda_{\bf s,l}=\lambda_{\bf s,s}=\lambda_{\bf l,l}$.

  3. (3)

$\lambda_{\bf s,s}=\lambda_{\bf s+l,s+l}=\lambda_{\bf l,l}$.

  4. (4)

$\lambda_{{\bf s+l},j{\bf l}}=\lambda_{\bf s,s}=\lambda_{\bf l,l}$ for all
$j\in\mathbb{Z}\setminus\\{0\\}$.

  5. (5)

$\lambda_{{\bf s},j{\bf s+l}}=\lambda_{\bf s,s}=\lambda_{\bf l,l}$ for all
$j\in\mathbb{Z}$.

###### Proof.

Proof of this lemma runs parallelly with Lemma 9.6 of [16], just note that
$(B{\bf s}|{\bf s})=0$. ∎

######  Lemma 7.5.

Let ${\bf r,s}\in\mathbb{Z}^{N}\setminus\\{0\\}$ be such that $(B{\bf r}|{\bf
s})=0$ and ${\bf r+s}\neq 0$. Then we have the following.

  1. (1)

$\lambda_{\bf r,r}=\lambda_{\bf s,s}=\lambda_{\bf r+s,r+s}$.

  2. (2)

$\lambda_{\bf s+r,s}=\lambda_{\bf s,s}=\lambda_{\bf r,r}=\lambda_{\bf r,s+r}$.

  3. (3)

$\lambda_{\bf r,s}=\lambda_{\bf s,s}=\lambda_{\bf r,r}$.

  4. (4)

$\lambda_{{\bf s+r,}j{\bf r}}=\lambda_{\bf s,s}=\lambda_{\bf r,r}$, for all
$j\in\mathbb{Z}\setminus\\{0\\}.$

  5. (5)

$\lambda_{{\bf s,}j{\bf s+r}}=\lambda_{\bf s,s}=\lambda_{\bf r,r}$, when
$j{\bf s}+{\bf r}\neq 0$, $j\in\mathbb{Z}$.

###### Proof.

First note that (4) and (5) follows immediately from (2) and (3). Therefore we
prove (1,2,3). Let $U_{\bf r}$ be denote the orthogonal complement of $B{\bf
r}$ for all ${\bf r}\in\mathbb{Z}^{N}$. Since $B$ is non degenerate
$\mathbb{Z}^{N}\setminus(U_{\bf r}\cup U_{\bf s}\cup U_{\bf
r+s})\cap\mathbb{Z}^{N}$ is a proper subset of $\mathbb{Z}^{N}$. Hence there
exists a ${\bf k}\in\mathbb{Z}^{N}$ such that $(B{\bf r}|{\bf k}),(B{\bf
s}|{\bf k}),(B({\bf r+s})|{\bf k})\in\mathbb{C}\setminus\\{0\\}$.
Now from Lemma 7.4 we have $\lambda_{\bf r,r}=\lambda_{\bf s,s}=\lambda_{\bf
r+s,r+s}=\lambda_{\bf k,k}$. Notice that $(B{\bf r}|{\bf s+k})\neq 0\neq(B{\bf
s}|{\bf r+k})$. Hence from Lemma 7.4 and the equality $\lambda_{\bf
r,r}=\lambda_{\bf s,s}$ we have $\lambda_{\bf r,r}=\lambda_{\bf
r,s+k}=\lambda_{\bf s,s}=\lambda_{\bf s,r+k}$. Now in equation (7.2)
substituting ${\bf l}={\bf k}$ we have $\lambda_{\bf r,s+k}=\lambda_{\bf
r,s}.$ Thus we have $\lambda_{\bf r,r}=\lambda_{\bf r,s}=\lambda_{\bf s,s}$.
Now put ${\bf s}={\bf r+s}$ in the last equality, then we have $\lambda_{\bf
r,r}=\lambda_{\bf r,r+s}=\lambda_{\bf r+s,r+s}$. Similarly, $\lambda_{\bf
s,s}=\lambda_{\bf s,r+s}=\lambda_{\bf r+s,r+s}$. This completes the proof. ∎

The next lemma completes the proof of Theorem 7.1 and it follows from verbatim
same proof of Lemma 9.9 and Lemma 9.10 of [16] using Lemma 7.4 and Lemma 7.5.

######  Lemma 7.6.

(1) $\lambda_{j{\bf s},j{\bf s}}=\lambda_{\bf s,s}$ for all
$j\in\mathbb{Z}\setminus\\{0\\}$.
(2) $\lambda_{j{\bf s},p{\bf s}}=\lambda_{\bf s,s}$ for all
$j+p\in\mathbb{Z}\setminus\\{0\\}$.
(3) $\lambda_{{\bf r},{\bf s}}=\lambda$ for all ${\bf r+s}\neq 0$ and ${\bf
r,s}\neq 0.$
($4$) $\lambda_{0,\bf s}=c,$ for all $\bf s$ $\in\mathbb{Z}^{N}$.
(5) $\lambda_{\bf r,-r}=\mu$ for all $\bf r$ $\neq 0$.
(6) $c\mu=\lambda^{2}$.

Proof of Lemma 4.4(8), i.e $[I_{p},I_{q}]\subseteq I_{p+q-2}$ for all $p+q\geq
3$ and $[I_{1},I_{1}]\subseteq I_{1}.$
Let $I\subseteq\\{1,2,\dots,p\\}$ and $J\subseteq\\{1,2,\dots,q\\}$. Let ${\bf
s_{I}}=\displaystyle{\sum_{i\in I}{\bf s_{i}}}$ and ${\bf
r_{J}}=\displaystyle{\sum_{i\in J}{\bf r_{i}}}$. Also set ${\bf s_{\Phi}}=0$
and ${\bf r_{\Phi}}=0$. Then

$T_{p}({\bf k,s_{1},\dots,s_{p}})=\displaystyle{\sum_{0\leq a\leq
p}\sum_{|I|=a}(-1)^{a}T({\bf k+s_{I}})}$ and

$T_{q}({\bf l,r_{1},\dots,r_{q}})=\displaystyle{\sum_{0\leq b\leq
q}\sum_{|J|=b}(-1)^{b}T({\bf l+r_{J}})}.$

Therefore, $[T_{p}({\bf k,s_{1},\dots,s_{p}}),T_{q}({\bf
l,r_{1},\dots,r_{q}})]$

$=\displaystyle{\sum_{\begin{subarray}{c}0\leq a\leq p\\\ 0\leq b\leq
q\end{subarray}}\sum_{\begin{subarray}{c}|I|=a\\\
|J|=b\end{subarray}}}(-1)^{a+b}[T({\bf k+s_{I}}),T({\bf l+r_{J}})]$
$=-\displaystyle{\sum_{\begin{subarray}{c}0\leq a\leq p\\\ 0\leq b\leq
q\end{subarray}}\sum_{\begin{subarray}{c}|I|=a\\\
|J|=b\end{subarray}}}(-1)^{a+b}(B({\bf k+s_{I}})|{\bf l+r_{J}})(T({\bf
k+s_{I}})+T({\bf l+r_{J}})-T({\bf k+l+s_{I}+r_{J}})).$
Claim 1: $\displaystyle{\sum_{\begin{subarray}{c}0\leq a\leq p\\\ 0\leq b\leq
q\end{subarray}}\sum_{\begin{subarray}{c}|I|=a\\\
|J|=b\end{subarray}}}(-1)^{a+b}(B({\bf k+s_{I}})|{\bf l+r_{J}})T({\bf
l+r_{J}})=0.$
Fix some $J$ such that $|J|=b$. Now to prove this claim it is sufficient to
prove that $\displaystyle{\sum_{0\leq a\leq p}\sum_{|I|=a}}(-1)^{a}(B({\bf
k+s_{I}})|{\bf l+r_{J}})=0,$ which means it is sufficient to prove that
$\displaystyle{\sum_{0\leq a\leq p}\sum_{|I|=a}}(-1)^{a}({\bf k+s_{I}})=0$.
But this sum is equal to
$\displaystyle{\sum_{0\leq a\leq p}}(-1)^{a}{{p}\choose{a}}{\bf
k}+\displaystyle{\sum_{0\leq a\leq p}\sum_{|I|=a}}(-1)^{a}{\bf s_{I}}$, and
this is easy to check to be 0.
Claim 2: $\displaystyle{\sum_{\begin{subarray}{c}0\leq a\leq p\\\ 0\leq b\leq
q\end{subarray}}\sum_{\begin{subarray}{c}|I|=a\\\
|J|=b\end{subarray}}}(-1)^{a+b}(B({\bf k+s_{I}})|{\bf l+r_{J}})T({\bf
k+s_{I}})=0.$ This follows similarly like Claim 1.
Now we evaluate $\displaystyle{\sum_{\begin{subarray}{c}0\leq a\leq p\\\ 0\leq
b\leq q\end{subarray}}\sum_{\begin{subarray}{c}|I|=a\\\
|J|=b\end{subarray}}}(-1)^{a+b}(B({\bf k+s_{I}})|{\bf l+r_{J}})T({\bf
k+l+s_{I}+r_{J}}).$ For that we write the above expression as the sum of four
terms given by

(7.3) $\displaystyle\displaystyle{\sum_{\begin{subarray}{c}0\leq a\leq p\\\
0\leq b\leq q\end{subarray}}\sum_{\begin{subarray}{c}|I|=a\\\
|J|=b\end{subarray}}}(-1)^{a+b}(B{\bf k}|{\bf l})T({\bf k+l+s_{I}+r_{J}})$
(7.4) $\displaystyle\displaystyle{\sum_{\begin{subarray}{c}0\leq a\leq p\\\
0\leq b\leq q\end{subarray}}\sum_{\begin{subarray}{c}|I|=a\\\
|J|=b\end{subarray}}}(-1)^{a+b}(B{\bf s_{I}}|{\bf l})T({\bf k+l+s_{I}+r_{J}})$
(7.5) $\displaystyle\displaystyle{\sum_{\begin{subarray}{c}0\leq a\leq p\\\
0\leq b\leq q\end{subarray}}\sum_{\begin{subarray}{c}|I|=a\\\
|J|=b\end{subarray}}}(-1)^{a+b}(B{\bf k}|{\bf r_{J}})T({\bf k+l+s_{I}+r_{J}})$
(7.6) $\displaystyle\displaystyle{\sum_{\begin{subarray}{c}0\leq a\leq p\\\
0\leq b\leq q\end{subarray}}\sum_{\begin{subarray}{c}|I|=a\\\
|J|=b\end{subarray}}}(-1)^{a+b}(B{\bf s_{I}}|{\bf r_{J}})T({\bf
k+l+s_{I}+r_{J}}).$

Now consider (6.3) which is clearly equal to $(B{\bf k}|{\bf l})T_{p+q}({\bf
k+l,s_{1},\dots,s_{p},r_{1},\dots,r_{q}}).$ Now consider the expression (6.5),
to simplify this we fix $i\in\\{1,2,\dots,q\\}$. Then consider the expression
$\displaystyle{\sum_{\begin{subarray}{c}0\leq a\leq p\\\ 1\leq b\leq
q\end{subarray}}\sum_{\begin{subarray}{c}|I|=a\\\
|J|=b\end{subarray}}}(-1)^{a+b}(B{\bf k}|{\bf r_{i}})T({\bf k+l+s_{I}+r_{J}})$
(Note that $b=0$ not occur here). This is equal to $(B{\bf k}|{\bf
r_{i}})T_{p+q-1}({\bf
k+l+r_{i},s_{1},\dots,s_{p},r_{1},\dots,\hat{r_{i}},\dots,r_{q}}).$ Hence
(6.5) is equal to $-\displaystyle{\sum_{i=1}^{q}}(B{\bf k}|{\bf
r_{i}})T_{p+q-1}({\bf
k+l+r_{i},s_{1},\dots,s_{p},r_{1},\dots,\hat{r_{i}},\dots,r_{q}}).$
In the similar manner (6.4) is equal to $-\displaystyle{\sum_{i=1}^{p}}(B{\bf
s_{i}}|{\bf l})T_{p+q-1}({\bf
k+l+s_{i},s_{1},\dots,\hat{s_{i}},\dots,s_{p},r_{1},\dots,r_{q}}).$ Now fix
$i\in\\{1,2,\dots,p\\}$, $j\in\\{1,2,\dots,q\\}$ and consider the expression
$\displaystyle{\sum_{\begin{subarray}{c}1\leq a\leq p\\\ 1\leq b\leq
q\end{subarray}}\sum_{\begin{subarray}{c}|I|=a\\\ |J|=b\\\ i\in I,j\in
J\end{subarray}}}(-1)^{a+b}(B{\bf s_{i}}|{\bf r_{j}})T({\bf k+l+s_{I}+r_{J}})$
$=(B{\bf s_{i}}|{\bf r_{j}})T_{p+q-2}({\bf
k+l+s_{i}+r_{j},s_{1},\dots,\hat{s_{i}},\dots,s_{p},r_{1},\dots,\hat{r_{j}},\dots,r_{q}}).$
Therefore (6.6) is equal to $\displaystyle{\sum_{\begin{subarray}{c}1\leq
i\leq p\\\ 1\leq j\leq q\end{subarray}}}(B{\bf s_{i}}|{\bf
r_{j}})T_{p+q-2}({\bf
k+l+s_{i}+r_{j},s_{1},\dots,\hat{s_{i}},\dots,s_{p},r_{1},\dots,\hat{r_{j}},\dots,r_{q}}).$
This argument fails when $p=q=1$. In fact in that case (6.6) is equal to
$(B{\bf s_{1}}|{\bf r_{1}})T({\bf k+l+s_{1}+r_{1}})$.

Therefore $[T_{p}({\bf k,s_{1},\dots,s_{p}}),T_{q}({\bf
l,r_{1},\dots,r_{q}})]$

$=-(B{\bf k}|{\bf l})T_{p+q}({\bf k+l,s_{1},\dots,s_{p},r_{1},\dots,r_{q}})$
$+\displaystyle{\sum_{i=1}^{q}}(B{\bf k}|{\bf r_{i}})T_{p+q-1}({\bf
k+l+r_{i},s_{1},\dots,s_{p},r_{1},\dots,\hat{r_{i}},\dots,r_{q}})$
$+\displaystyle{\sum_{i=1}^{p}}(B{\bf s_{i}}|{\bf l})T_{p+q-1}({\bf
k+l+s_{i},s_{1},\dots,\hat{s_{i}},\dots,s_{p},r_{1},\dots,r_{q}})$
$-\displaystyle{\sum_{\begin{subarray}{c}1\leq i\leq p\\\ 1\leq j\leq
q\end{subarray}}}(B{\bf s_{i}}|{\bf r_{j}})T_{p+q-2}({\bf
k+l+s_{i}+r_{j},s_{1},\dots,\hat{s_{i}},\dots,s_{p},r_{1},\dots,\hat{r_{j}},\dots,r_{q}}),$
when $(p,q)\neq(1,1)$

$=-(B{\bf k}|{\bf l})T_{2}({\bf k+l,s_{1},r_{1}})+(B{\bf k}|{\bf
r_{1}})T_{1}({\bf k+l+r_{1},s_{1}})+(B{\bf s_{1}}|{\bf l})T_{1}({\bf
k+l+s_{1},r_{1}})$
$-(B{\bf s_{1}}|{\bf r_{1}})T_{1}({\bf k+l+s_{1}+r_{1}})$, when $(p,q)=(1,1)$.

This completes the proof.

## References

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