# Representations of the orthosymplectic Yangian

A. I. Molev

###### Abstract

We give a complete description of the finite-dimensional irreducible
representations of the Yangian associated with the orthosymplectic Lie
superalgebra $\mathfrak{osp}_{1|2}$. The representations are parameterized by
monic polynomials in one variable, they are classified in terms of highest
weights. We give explicit constructions of a family of elementary modules of
the Yangian and show that a wide class of irreducible representations can be
produced by taking tensor products of the elementary modules.

##  1 Introduction

It is well-known that the classification of finite-dimensional irreducible
representations of the orthosymplectic Lie superalgebras
$\mathfrak{osp}_{M|2n}$ is more complicated than that of its general linear
counterpart $\mathfrak{gl}_{M|N}$; see e.g. books [6] and [15]. It is
therefore not surprising that such a disparity extends to the respective
Yangians.

The Yangian for $\mathfrak{gl}_{M|N}$ was introduced by Nazarov [16] and its
finite-dimensional irreducible representations were classified by Zhang [19].
The orthosymplectic Yangians ${\rm Y}(\mathfrak{osp}_{M|2n})$ were introduced
by Arnaudon et al. [1], and the general classification problem still remains
open. Our goal in this paper is to describe finite-dimensional irreducible
representations of the Yangian ${\rm Y}(\mathfrak{osp}_{1|2})$. One can expect
that this simplest case will be instrumental in solving the general
classification problem. In particular, an extension to $\mathfrak{osp}_{1|2n}$
appears to be straightforward.

According to [1], the $R$-matrix originated in [18] leads to the $RTT$-type
definition of the extended Yangian ${\rm X}(\mathfrak{osp}_{M|2n})$. The
Yangian ${\rm Y}(\mathfrak{osp}_{M|2n})$ can be regarded as the quotient of
the extended Yangian by the ideal generated by elements of the center,
implying that the finite-dimensional irreducible representations of these two
algebras are essentially the same. An explicit description of the center and
the Hopf algebra structure were given in [1]. In the subsequent work [2], a
Drinfeld-type presentation of the Yangian ${\rm Y}(\mathfrak{osp}_{1|2})$ was
produced, the double Yangian was constructed and its universal $R$-matrix was
calculated in an explicit form. Applications of the orthosymplectic Yangians
to spin chain models were discussed in [3]. More recently, linear and
quadratic evaluations in the Yangian ${\rm Y}(\mathfrak{osp}_{M|2n})$ were
investigated in [8] and [10].

By analogy with the $R$-matrix form of its orthogonal and symplectic
counterparts, every finite-dimensional irreducible representation of the
orthosymplectic Yangian is a highest weight representation; cf. [4]. For the
extended Yangian ${\rm X}(\mathfrak{osp}_{1|2})$, this means that the
representations are determined by the triples
$(\lambda_{1}(u),\lambda_{2}(u),\lambda_{3}(u))$ of formal series
$\lambda_{i}(u)\in 1+u^{-1}\mathbb{C}\hskip 1.0pt[[u^{-1}]]$ which should
satisfy the consistency condition

$\lambda_{1}(u)\lambda_{3}(u+1/2)=\lambda_{2}(u)\lambda_{2}(u+1/2).$ (1.1)

The key step in the classification is to find the conditions on such triples
for the corresponding representation to be finite-dimensional. Here we need a
super-extension of the approach originally used by Tarasov [17] and which had
already been adapted and applied to the twisted Yangians associated with the
classical Lie algebras; see [14, Ch. 3 & 4]. This leads to the following
result.

######  Main Theorem.

The finite-dimensional irreducible representation of the algebra ${\rm
X}(\mathfrak{osp}_{1|2})$ associated with the highest weight
$(\lambda_{1}(u),\lambda_{2}(u),\lambda_{3}(u))$ is finite-dimensional if and
only if

$\frac{\lambda_{2}(u)}{\lambda_{1}(u)}=\frac{P(u+1)}{P(u)}$ (1.2)

for some monic polynomial $P(u)$ in $u$. Hence, the finite-dimensional
irreducible representations of the Yangian ${\rm Y}(\mathfrak{osp}_{1|2})$ are
parameterized by monic polynomials $P(u)$.

The parametrization turns out to be the same as for the representations of the
Yangian ${\rm Y}(\mathfrak{sl}_{2})$ given in [7], and we will call $P(u)$ the
Drinfeld polynomial of the representation. The appearance of the consistency
conditions (1.1) is best explained by the Gauss decomposition of the generator
matrix for ${\rm X}(\mathfrak{osp}_{1|2})$ and can be compared with the
similar conditions for the Yangian ${\rm Y}(\mathfrak{o}_{3})$; see [12], [13,
Sec. 5.3], which were derived in a different way in [4]. However, unlike the
case of the Yangian ${\rm Y}(\mathfrak{sl}_{2})$ (or ${\rm
Y}(\mathfrak{o}_{3})$), there is no epimorphism ${\rm
X}(\mathfrak{osp}_{1|2})\to{\rm U}(\mathfrak{osp}_{1|2})$, so that in general,
representations of $\mathfrak{osp}_{1|2}$ do not extend to the Yangian.

An essential step in the proof of the Main Theorem, which will be completed in
Sec. 4.4, is the analysis of the elementary modules $L(\alpha,\beta)$ over
${\rm X}(\mathfrak{osp}_{1|2})$ associated with the highest weights of the
form

$\lambda_{1}(u)=\frac{u+\alpha}{u+\beta},\qquad\lambda_{2}(u)=1,\qquad\lambda_{3}(u)=\frac{u+\beta-1/2}{u+\alpha-1/2}.$
(1.3)

The corresponding small Verma module $M(\alpha,\beta)$ turns out to be
irreducible if and only if $\beta-\alpha$ and $\beta-\alpha+1/2$ are not
nonnegative integers. The elementary modules $L(\alpha,\beta)$ are the
irreducible quotients of $M(\alpha,\beta)$ and so they split into three
families, according to these conditions. The module $L(\alpha,\beta)$ is
finite-dimensional if and only if $\beta-\alpha\in\mathbb{Z}\hskip 1.0pt_{+}$.
In this case, when regarded as an $\mathfrak{osp}_{1|2}$-module,
$L(\alpha,\beta)$ decomposes into the direct sum

$L(\alpha,\beta)\cong\bigoplus_{p=0}^{\lfloor\frac{\beta-\alpha}{2}\rfloor}V(\beta-\alpha-2\hskip
1.0ptp),$

where $V(\mu)$ denotes the $2\mu+1$-dimensional $\mathfrak{osp}_{1|2}$-module
with the highest weight $\mu\in\mathbb{Z}\hskip 1.0pt_{+}$. In particular,

$\dim L(\alpha,\beta)=\binom{\beta-\alpha+2}{2}.$

We construct a basis of each small Verma module $M(\alpha,\beta)$ and give
explicit formulas for the action of the generators of ${\rm
X}(\mathfrak{osp}_{1|2})$. This leads to a corresponding description of all
elementary modules. We show that, up to twisting by a multiplication
automorphism of ${\rm X}(\mathfrak{osp}_{1|2})$, every finite-dimensional
irreducible representation of this algebra is isomorphic to the quotient of
the submodule of the tensor product module of the form

$L(\alpha_{1},\beta_{1})\otimes\dots\otimes L(\alpha_{k},\beta_{k}),$ (1.4)

generated by the tensor product of the highest vectors. The final step is to
investigate irreducibility conditions for such tensor products.

In the case of the Yangian ${\rm Y}(\mathfrak{sl}_{2})$, an irreducibility
criterion for tensor products of evaluation modules was given by Chari and
Pressley [5]; see also [14, Ch. 3]. Such tensor products exhaust all finite-
dimensional irreducible ${\rm Y}(\mathfrak{sl}_{2})$-modules. This property
turns out not to extend to representations of the Yangian for
$\mathfrak{osp}_{1|2}$; see Example 4.21 below. A wide class of irreducible
modules can still be constructed explicitly via tensor products of the form
(1.4); see Theorem 4.17.

##  2 Sign conventions, definitions and preliminaries

Consider the three-dimensional space $\mathbb{C}\hskip 1.0pt^{1|2}$ over
$\mathbb{C}\hskip 1.0pt$ with basis elements $e_{1},e_{2},e_{3}$ where we
assume a $\mathbb{Z}\hskip 1.0pt_{2}$-gradation defined by setting that the
vector $e_{2}$ is even and the vectors $e_{1}$ and $e_{3}$ are odd. It will be
convenient to use the involution on the set $\\{1,2,3\\}$ defined by $i\mapsto
i^{\prime}=4-i$. The endomorphism algebra ${\rm{End}\,}\mathbb{C}\hskip
1.0pt^{1|2}$ then gets a $\mathbb{Z}\hskip 1.0pt_{2}$-gradation with the
parity of the matrix unit $e_{ij}$ found by $i+j\mod 2$.

We will consider $3\times 3$ matrices with entries in superalgebras. All such
matrices will be even so that the $(i,j)$ entry will have the parity $i+j\mod
2$. We want to posit that the product of two matrices with entries in a
superalgebra $\mathcal{A}$ is calculated in the standard way (without any
additional signs). Accordingly, the algebra of even matrices over
$\mathcal{A}$ will be identified with the tensor product algebra
${\rm{End}\,}\mathbb{C}\hskip 1.0pt^{1|2}\otimes\mathcal{A}$. A matrix
$A=[a_{ij}]$ will be regarded as the element

$A=\sum_{i,j=1}^{3}e_{ij}\otimes a_{ij}(-1)^{i\hskip
1.0ptj+j}\in{\rm{End}\,}\mathbb{C}\hskip 1.0pt^{1|2}\otimes\mathcal{A}.$

We will use the involutive matrix super-transposition $t$ defined by
$(A^{t})_{ij}=A_{j^{\prime}i^{\prime}}(-1)^{ij+j}\hskip
1.0pt\theta_{i}\theta_{j}$, where we set

$\theta_{1}=\theta_{2}=1\qquad\text{and}\qquad\theta_{3}=-1.$

This super-transposition is associated with the bilinear form on the space
$\mathbb{C}\hskip 1.0pt^{1|2}$ defined by the anti-diagonal matrix
$G=[\delta_{ij^{\prime}}\hskip 1.0pt\theta_{i}]$. We will also regard $t$ as
the linear map

$t:{\rm{End}\,}\mathbb{C}\hskip 1.0pt^{1|2}\to{\rm{End}\,}\mathbb{C}\hskip
1.0pt^{1|2},\qquad e_{ij}\mapsto e_{j^{\prime}i^{\prime}}(-1)^{ij+i}\hskip
1.0pt\theta_{i}\theta_{j}.$ (2.1)

In the case of multiple tensor products of the endomorphism algebras, we will
indicate by $t_{a}$ the map (2.1) acting on the $a$-th copy of
${\rm{End}\,}\mathbb{C}\hskip 1.0pt^{1|2}$.

A standard basis of the general linear Lie superalgebra $\mathfrak{gl}_{1|2}$
is formed by elements $E_{ij}$ of the parity $i+j\mod 2$ for $1\leqslant
i,j\leqslant 3$ with the commutation relations

$[E_{ij},E_{kl}]=\delta_{kj}\,E_{i\hskip 1.0ptl}-\delta_{i\hskip
1.0ptl}\,E_{kj}(-1)^{(i+j)(k+l)}.$

We will regard the orthosymplectic Lie superalgebra $\mathfrak{osp}_{1|2}$
associated with the bilinear from defined by $G$ as the subalgebra of
$\mathfrak{gl}_{1|2}$ spanned by the elements

$F_{ij}=E_{ij}-E_{j^{\prime}i^{\prime}}(-1)^{i\hskip
1.0ptj+i}\,\theta_{i}\theta_{j}.$

For any given $\mu\in\mathbb{C}\hskip 1.0pt$ we will denote by $V(\mu)$ the
irreducible highest weight module over $\mathfrak{osp}_{1|2}$ generated by a
nonzero vector $\xi$ such that

$F_{11}\xi=\mu\,\xi\qquad\text{and}\qquad F_{12}\hskip 1.0pt\xi=0.$

The module $V(\mu)$ is finite-dimensional if and only if
$\mu\in\mathbb{Z}\hskip 1.0pt_{+}$. In that case, $\dim V(\mu)=2\mu+1$.

Introduce the permutation operator $P$ as the element

$P=\sum_{i,j=1}^{3}e_{ij}\otimes e_{ji}(-1)^{j}\in{\rm{End}\,}\mathbb{C}\hskip
1.0pt^{1|2}\otimes{\rm{End}\,}\mathbb{C}\hskip 1.0pt^{1|2}$

and set

$Q=P^{\hskip 1.0ptt_{1}}=P^{\hskip 1.0ptt_{2}}=\sum_{i,j=1}^{3}e_{ij}\otimes
e_{i^{\prime}j^{\prime}}(-1)^{ij}\,\theta_{i}\theta_{j}\in{\rm{End}\,}\mathbb{C}\hskip
1.0pt^{1|2}\otimes{\rm{End}\,}\mathbb{C}\hskip 1.0pt^{1|2}.$

The rational function in $u$ given by

$R(u)=1-\frac{P}{u}+\frac{Q}{u-\kappa},\qquad\kappa=-3/2,$

is an $R$-matrix, it satisfies the Yang–Baxter equation as originally found in
[18]. The $R$-matrices produced in that paper are known to extend to the
Brauer algebra so that the Yang–Baxter equation can be verified by taking a
suitable Brauer algebra representation in tensor products of the
$\mathbb{Z}\hskip 1.0pt_{2}$-graded spaces; cf. [8], [11].

Following [1], define the extended Yangian ${\rm X}(\mathfrak{osp}_{1|2})$ as
an associative superalgebra with generators $t_{ij}^{(r)}$ of parity $i+j\mod
2$, where $1\leqslant i,j\leqslant 3$ and $r=1,2,\dots$, satisfying certain
quadratic relations. In order to write them down, introduce the formal series

$t_{ij}(u)=\delta_{ij}+\sum_{r=1}^{\infty}t_{ij}^{(r)}\,u^{-r}\in{\rm
X}(\mathfrak{osp}_{1|2})[[u^{-1}]]$ (2.2)

and combine them into the $3\times 3$ matrix $T(u)=[t_{ij}(u)]$ so that

$T(u)=\sum_{i,j=1}^{3}e_{ij}\otimes t_{ij}(u)(-1)^{i\hskip
1.0ptj+j}\in{\rm{End}\,}\mathbb{C}\hskip 1.0pt^{1|2}\otimes{\rm
X}(\mathfrak{osp}_{1|2})[[u^{-1}]].$

Consider the algebra ${\rm{End}\,}\mathbb{C}\hskip
1.0pt^{1|2}\otimes{\rm{End}\,}\mathbb{C}\hskip 1.0pt^{1|2}\otimes{\rm
X}(\mathfrak{osp}_{1|2})[[u^{-1}]]$ and introduce its elements $T_{1}(u)$ and
$T_{2}(u)$ by

$T_{1}(u)=\sum_{i,j=1}^{3}e_{ij}\otimes 1\otimes t_{ij}(u)(-1)^{i\hskip
1.0ptj+j},\qquad T_{2}(u)=\sum_{i,j=1}^{3}1\otimes e_{ij}\otimes
t_{ij}(u)(-1)^{i\hskip 1.0ptj+j}.$

The defining relations for the superalgebra ${\rm X}(\mathfrak{osp}_{1|2})$
can then be written in the form of the $RTT$-relation

$R(u-v)\,T_{1}(u)\,T_{2}(v)=T_{2}(v)\,T_{1}(u)\,R(u-v).$ (2.3)

As shown in [1], the product $T(u)\,T^{\hskip 1.0ptt}(u-\kappa)$ is a scalar
matrix with

$T(u-\kappa)\,T^{\hskip 1.0ptt}(u)=c(u)1,$ (2.4)

where $c(u)$ is a series in $u^{-1}$. All its coefficients belong to the
center ${\rm ZX}(\mathfrak{osp}_{1|2})$ of ${\rm X}(\mathfrak{osp}_{1|2})$ and
generate the center.

The Yangian ${\rm Y}(\mathfrak{osp}_{1|2})$ is defined as the subalgebra of
${\rm X}(\mathfrak{osp}_{1|2})$ which consists of the elements stable under
the automorphisms

$t_{ij}(u)\mapsto f(u)\,t_{ij}(u)$ (2.5)

for all series $f(u)\in 1+u^{-1}\mathbb{C}\hskip 1.0pt[[u^{-1}]]$. We have the
tensor product decomposition

${\rm X}(\mathfrak{osp}_{1|2})={\rm ZX}(\mathfrak{osp}_{1|2})\otimes{\rm
Y}(\mathfrak{osp}_{1|2}).$ (2.6)

The Yangian ${\rm Y}(\mathfrak{osp}_{1|2})$ can be equivalently defined as the
quotient of ${\rm X}(\mathfrak{osp}_{1|2})$ by the relation

$T(u-\kappa)\,T^{\hskip 1.0ptt}(u)=1.$

We will also use a more explicit form of the defining relations (2.3) written
in terms of the series (2.2) as follows:

$\displaystyle[\hskip 1.0ptt_{ij}(u),t_{kl}(v)]$
$\displaystyle=\frac{1}{u-v}\big{(}t_{kj}(u)\,t_{il}(v)-t_{kj}(v)\,t_{il}(u)\big{)}(-1)^{i\hskip
1.0ptj+i\hskip 1.0ptk+j\hskip 1.0ptk}$
$\displaystyle-\frac{1}{u-v-\kappa}\Big{(}\delta_{ki^{\hskip
1.0pt\prime}}\sum_{p=1}^{3}\,t_{pj}(u)\,t_{p^{\prime}l}(v)(-1)^{i+i\hskip
1.0ptj+j\hskip 1.0ptp}\,\theta_{i}\theta_{p}$ (2.7)
$\displaystyle\qquad\qquad\qquad{}-\delta_{lj^{\hskip
1.0pt\prime}}\sum_{p=1}^{3}\,t_{k\hskip
1.0ptp^{\prime}}(v)\,t_{ip}(u)(-1)^{j+p+i\hskip 1.0ptk+j\hskip 1.0ptk+i\hskip
1.0ptp}\,\theta_{j}\theta_{p}\Big{)}.$

The coefficients of the series $t_{11}(u),t_{12}(u),t_{21}(u)$ and $c(u)$
generate the algebra ${\rm X}(\mathfrak{osp}_{1|2})$. The mapping
$t_{ij}(u)\mapsto t_{ij}(-u)$ defines an anti-automorphism of ${\rm
X}(\mathfrak{osp}_{1|2})$, while each of the mappings

$t_{ij}(u)\mapsto t_{ij}(u+a),\quad a\in\mathbb{C}\hskip 1.0pt,$ (2.8)

and $t_{ij}(u)\mapsto t_{i^{\prime}j^{\prime}}(u)\,\theta_{i}\theta_{j}$
defines an automorphism. Consider their composition to define the anti-
automorphism

$\omega:t_{ij}(u)\mapsto
t_{i^{\prime}j^{\prime}}(-u+1/2)\,\theta_{i}\theta_{j}.$ (2.9)

The universal enveloping algebra ${\rm U}(\mathfrak{osp}_{1|2})$ can be
regarded as a subalgebra of ${\rm X}(\mathfrak{osp}_{1|2})$ via the embedding

$F_{ij}\mapsto\frac{1}{2}\big{(}t_{ij}^{(1)}-t_{j^{\prime}i^{\prime}}^{(1)}(-1)^{j+ij}\,\theta_{i}\theta_{j}\big{)}(-1)^{i}.$
(2.10)

This fact relies on the Poincaré–Birkhoff–Witt theorem for the orthosymplectic
Yangian which was essentially pointed out in [1] and [2], as the associated
graded algebra for ${\rm Y}(\mathfrak{osp}_{1|2})$ is isomorphic to ${\rm
U}(\mathfrak{osp}_{1|2}[u])$. It states that given any total ordering on the
set of generators $t_{ij}^{(r)}$ with $i+j\leqslant 4$ and $r\geqslant 1$, the
ordered monomials in the generators with the powers of odd generators not
exceeding $1$, form a basis of ${\rm X}(\mathfrak{osp}_{1|2})$. A detailed
proof can be carried over by adapting the arguments of [4, Sec. 3] to the
super case in a straightforward way with the use of the vector representation
recalled below in (4.20).

The extended Yangian ${\rm X}(\mathfrak{osp}_{1|2})$ is a Hopf algebra with
the coproduct defined by

$\Delta:t_{ij}(u)\mapsto\sum_{k=1}^{3}t_{ik}(u)\otimes t_{kj}(u).$ (2.11)

For the image of the series $c(u)$ we have $\Delta:c(u)\mapsto c(u)\otimes
c(u)$ and so the Yangian ${\rm Y}(\mathfrak{osp}_{1|2})$ inherits the Hopf
algebra structure from ${\rm X}(\mathfrak{osp}_{1|2})$.

##  3 Gaussian generators

A Drinfeld-type presentation of the Yangian for $\mathfrak{osp}_{1|2}$ was
given in [2] with the use of the Gauss decomposition of the matrix $T(u)$. We
will be using some calculations produced therein and derive consistency
relations for the Gaussian generators.

Apply the Gauss decomposition to the generator matrix $T(u)$ for ${\rm
X}(\mathfrak{osp}_{1|2})$,

$T(u)=F(u)\,H(u)\,E(u),$ (3.1)

where $F(u)$, $H(u)$ and $E(u)$ are uniquely determined matrices of the form

$F(u)=\begin{bmatrix}1&0&0\,\\\ f_{21}(u)&1&0\\\
f_{31}(u)&f_{32}(u)&1\end{bmatrix},\qquad
E(u)=\begin{bmatrix}\,1&e_{12}(u)&e_{13}(u)\,\\\ \,0&1&e_{23}(u)\\\
0&0&1\end{bmatrix},$

and $H(u)={\rm diag}\,\big{[}h_{1}(u),h_{2}(u),h_{3}(u)\big{]}$. Explicit
formulas for the entries of the matrices $F(u)$, $H(u)$ and $E(u)$ can be
written with the use of the Gelfand–Retakh quasideterminants [9]; cf. [13,
Sec. 4]. In particular, we have

$h_{1}(u)=t_{11}(u),\qquad h_{2}(u)=\begin{vmatrix}t_{11}(u)&t_{12}(u)\\\
t_{21}(u)&\boxed{t_{22}(u)}\end{vmatrix},\qquad
h_{3}(u)=\begin{vmatrix}t_{11}(u)&t_{12}(u)&t_{13}(u)\\\\[1.99997pt]
t_{21}(u)&t_{22}(u)&t_{23}(u)\\\
t_{31}(u)&t_{32}(u)&\boxed{t_{33}(u)}\end{vmatrix},$

whereas

$e_{12}(u)=h_{1}(u)^{-1}\,t_{12}(u),\qquad
e_{23}(u)=h_{2}(u)^{-1}\,\begin{vmatrix}t_{11}(u)&t_{13}(u)\\\
t_{21}(u)&\boxed{t_{23}(u)}\end{vmatrix},$

and

$f_{21}(u)=t_{21}(u)\,h_{1}(u)^{-1},\qquad
f_{32}(u)=\begin{vmatrix}t_{11}(u)&t_{12}(u)\\\
t_{31}(u)&\boxed{t_{32}(u)}\end{vmatrix}\,h_{2}(u)^{-1}.$

######  Proposition 3.1.

The following relations for the Gaussian generators hold:

$e_{12}(u)=-e_{23}(u+1/2),\qquad f_{21}(u)=f_{32}(u+1/2),$ (3.2)

and

$h_{1}(u)\hskip 1.0pth_{3}(u+1/2)=h_{2}(u)\hskip 1.0pth_{2}(u+1/2).$ (3.3)

Moreover,

$c(u)=h_{1}(u)\hskip 1.0pth_{1}(u+1)^{-1}\hskip 1.0pth_{2}(u+1)\hskip
1.0pth_{2}(u+3/2).$ (3.4)

###### Proof.

The argument is quite similar to the proof of the corresponding relations for
the Gaussian generators of ${\rm Y}(\mathfrak{o}_{3})$ given in [12]; see also
[13, Sec. 5.3]. We will outline a few key steps.

By inverting the matrices on both sides of (3.1), we get

$T(u)^{-1}=E(u)^{-1}\hskip 1.0ptH(u)^{-1}\hskip 1.0ptF(u)^{-1}.$

On the other hand, relation (2.4) implies $T^{\hskip 1.0ptt}(u)=c(u)\hskip
1.0ptT(u-\kappa)^{-1}$. Hence, by equating the $(i,j)$ entries with $i,j=2,3$
in this matrix relation, we derive

$\displaystyle h_{1}(u)$ $\displaystyle=c(u)\hskip 1.0pth_{3}(u-\kappa)^{-1},$
$\displaystyle h_{1}(u)\hskip 1.0pte_{12}(u)$ $\displaystyle=-c(u)\hskip
1.0pte_{23}(u-\kappa)\hskip 1.0pth_{3}(u-\kappa)^{-1},$ (3.5) $\displaystyle
f_{21}(u)\hskip 1.0pth_{1}(u)$ $\displaystyle=c(u)\hskip
1.0pth_{3}(u-\kappa)^{-1}\hskip 1.0ptf_{32}(u-\kappa),$

and

$\displaystyle h_{2}(u)+f_{21}(u)\hskip 1.0pt$ $\displaystyle h_{1}(u)\hskip
1.0pte_{12}(u)$
$\displaystyle=c(u)\big{(}h_{2}(u-\kappa)^{-1}+e_{23}(u-\kappa)\hskip
1.0pth_{3}(u-\kappa)^{-1}\hskip 1.0ptf_{32}(u-\kappa)\big{)}.$ (3.6)

Calculating as in [2] and [12], we verify that the coefficients of the series
$h_{1}(u),h_{2}(u)$ and $h_{3}(u)$ pairwise commute. Furthermore, we get

$h_{1}(u)\hskip 1.0pte_{12}(u)=e_{12}(u+1)\hskip
1.0pth_{1}(u)\qquad\text{and}\qquad h_{1}(u)\hskip
1.0ptf_{21}(u+1)=f_{21}(u)\hskip 1.0pth_{1}(u)$

which together with relations (3.5) imply the first two desired identities,
where we replaced $\kappa$ by its value $-3/2$. They imply that relation (3.6)
can be written in the form

$h_{2}(u)-c(u)\hskip
1.0pth_{2}(u-\kappa)^{-1}=-\big{[}e_{12}(u+1),f_{21}(u)\big{]}\hskip
1.0pth_{1}(u).$ (3.7)

As a final step, use one more relation between the Gaussian generators,

$\big{[}e_{12}(u),f_{21}(v)\big{]}=\frac{h_{1}(u)^{-1}\hskip
1.0pth_{2}(u)-h_{1}(v)^{-1}\hskip 1.0pth_{2}(v)}{u-v},$

so that eliminating $c(u)$ from (3.7) we come to (3.3). Relation (3.4) follows
by eliminating $h_{3}(u)$ from the first relation in (3.5) with the use of
(3.3). ∎

Observe that the coefficients of the series $e_{12}(u)$ and $f_{21}(u)$ are
stable under all automorphisms (2.5) and so belong to the subalgebra ${\rm
Y}(\mathfrak{osp}_{1|2})$ of ${\rm X}(\mathfrak{osp}_{1|2})$. Together with
the coefficients of the series $h(u)=h_{1}(u)^{-1}h_{2}(u)$ they generate the
Yangian ${\rm Y}(\mathfrak{osp}_{1|2})$, and the defining relations for these
generators are given in [2] in a slightly different setting.

##  4 Yangian representations

###  4.1 Highest weight representations

A representation $V$ of the algebra ${\rm X}(\mathfrak{osp}_{1|2})$ is called
a highest weight representation if there exists a nonzero vector $\xi\in V$
such that $V$ is generated by $\xi$,

$\displaystyle t_{ij}(u)\,\xi$ $\displaystyle=0\qquad$
$\displaystyle\text{for}\quad 1\leqslant i<j\leqslant 3,\qquad\text{and}$
$\displaystyle t_{ii}(u)\,\xi$ $\displaystyle=\lambda_{i}(u)\,\xi\qquad$
$\displaystyle\text{for}\quad i=1,2,3,$ (4.1)

for some formal series

$\lambda_{i}(u)\in 1+u^{-1}\mathbb{C}\hskip 1.0pt[[u^{-1}]].$ (4.2)

The vector $\xi$ is called the highest vector of $V$ and the triple
$\lambda(u)=(\lambda_{1}(u),\lambda_{2}(u),\lambda_{3}(u))$ is the highest
weight of $V$.

The quasideterminant formulas for the Gaussian generators $h_{i}(u)$ given in
Sec. 3 imply that the conditions (4.1) in the above definition can be replaced
with

$h_{i}(u)\,\xi=\lambda_{i}(u)\,\xi\qquad\text{for}\quad i=1,2,3.$

Hence, relation (3.3) of Proposition 3.1 implies the consistency condition
(1.1) for the components $\lambda_{i}(u)$ of the highest weight.

Given any triple $\lambda(u)=(\lambda_{1}(u),\lambda_{2}(u),\lambda_{3}(u))$
of formal series of the form (4.2) safisfying the consistency condition (1.1),
define the Verma module $M(\lambda(u))$ as the quotient of ${\rm
X}(\mathfrak{osp}_{1|2})$ by the left ideal generated by all coefficients of
the series $t_{ij}(u)$ with $1\leqslant i<j\leqslant 3$, and
$t_{ii}(u)-\lambda_{i}(u)$ for $i=1,2,3$. The Poincaré–Birkhoff–Witt theorem
implies that the Verma module $M(\lambda(u))$ is nonzero and we denote by
$L(\lambda(u))$ its irreducible quotient.

######  Remark 4.1.

In terms of the Drinfeld presentation of the Yangian ${\rm
Y}(\mathfrak{osp}_{1|2})$ given in [2], the highest vector conditions take the
form

$e_{12}(u)\,\xi=0\qquad\text{and}\qquad h(u)\,\xi=\mu(u)\,\xi$

for a series $\mu(u)\in 1+u^{-1}\mathbb{C}\hskip 1.0pt[[u^{-1}]]$. The Main
Theorem implies that the irreducible highest weight representation of ${\rm
Y}(\mathfrak{osp}_{1|2})$ associated with $\mu(u)$ is finite-dimensional if
and only if

$\mu(u)=\frac{P(u+1)}{P(u)}$

for some monic polynomial $P(u)$ in $u$. ∎

######  Proposition 4.2.

Every finite-dimensional irreducible representation of the algebra ${\rm
X}(\mathfrak{osp}_{1|2})$ is isomorphic to $L(\lambda(u))$ for a certain
highest weight $\lambda(u)$ satisfying (1.1).

###### Proof.

The argument is a straightforward adaptation of the proof of [4, Thm 5.1] to
the super case, which amounts to taking care of additional signs in the
calculations. ∎

To prove the Main Theorem, we only need to determine which of the modules
$L(\lambda(u))$ are finite-dimensional. The first step is to establish some
necessary conditions.

######  Proposition 4.3.

If the module $L(\lambda(u))$ is finite-dimensional, then

$\frac{\lambda_{1}(u)}{\lambda_{2}(u)}=\frac{(u+\alpha_{1})\dots(u+\alpha_{k})}{(u+\beta_{1})\dots(u+\beta_{k})}$

for $k\in\mathbb{Z}\hskip 1.0pt_{+}$ and certain complex numbers
$\alpha_{i},\beta_{i}$.

###### Proof.

We follow the proof of a similar property for the Yangian ${\rm
Y}(\mathfrak{gl}_{2})$; see [14, Prop. 3.3.1]. Note that due to (1.1), the
series $\lambda_{3}(u)$ is uniquely determined by $\lambda_{1}(u)$ and
$\lambda_{2}(u)$, and so we can parameterize the highest weights by arbitrary
pairs of series $\lambda(u)=(\lambda_{1}(u),\lambda_{2}(u))$, omitting
$\lambda_{3}(u)$. By twisting the action of the extended Yangian ${\rm
X}(\mathfrak{osp}_{1|2})$ on $L(\lambda(u))$ by the automorphism (2.5) with
$f(u)=\lambda_{2}(u)^{-1}$, we get an ${\rm X}(\mathfrak{osp}_{1|2})$-module
isomorphic to $L(\mu(u),1)$ with $\mu(u)=\lambda_{1}(u)/\lambda_{2}(u)$. Let
$\xi$ denote the highest vector of $L(\mu(u),1)$. Since this representation is
finite-dimensional, the vectors $t_{21}^{(i)}\xi\in L(\mu(u),1)$ with
$i\geqslant 1$ are linearly dependent,

$\sum_{i=1}^{m}c_{i}\,t_{21}^{(i)}\xi=0$

with $c_{i}\in\mathbb{C}\hskip 1.0pt$, assuming $c_{m}\neq 0$. Apply the
operators $t_{12}^{(r)}$ for all $r\geqslant 1$ to the linear combination on
the left hand side and take the coefficient of $\xi$. Since $t_{12}(u)\hskip
1.0pt\xi=0$, we get from the defining relations (2.7) that

$t_{12}(u)\hskip 1.0ptt_{21}(v)\hskip
1.0pt\xi=\frac{1}{u-v}\big{(}t_{22}(u)\,t_{11}(v)-t_{22}(v)\,t_{11}(u)\big{)}\hskip
1.0pt\xi=-\frac{\mu(u)-\mu(v)}{u-v}\hskip 1.0pt\xi.$

Hence, writing

$\mu(u)=1+\mu^{(1)}u^{-1}+\mu^{(2)}u^{-2}+\dots,\qquad\mu^{(i)}\in\mathbb{C}\hskip
1.0pt,$

we derive

$t_{12}^{(r)}\,t_{21}^{(i)}\xi=\mu^{(r+i-1)}\hskip 1.0pt\xi.$

Therefore, for all $r\geqslant 1$ we have the relations

$\sum_{i=1}^{m}c_{i}\hskip 1.0pt\mu^{(r+i-1)}=0.$

They imply

$\mu(u)\hskip 1.0pt(c_{1}+c_{2}\hskip 1.0ptu+\dots+c_{m}\hskip
1.0ptu^{m-1})=(b_{1}+b_{2}\hskip 1.0ptu+\dots+b_{m}\hskip 1.0ptu^{m-1})$

so that $\mu(u)$ can be written as a rational function in $u$, as required. ∎

We will use the name elementary module for the module $L(\lambda(u))$ with

$\lambda_{1}(u)=\frac{u+\alpha}{u+\beta}\qquad\text{and}\qquad\lambda_{2}(u)=1$
(4.3)

and denote it by $L(\alpha,\beta)$.

The Hopf algebra structure on the extended Yangian ${\rm
X}(\mathfrak{osp}_{1|2})$ allows us to regard tensor products of the form

$L=L(\alpha_{1},\beta_{1})\otimes\dots\otimes L(\alpha_{k},\beta_{k})$ (4.4)

as ${\rm X}(\mathfrak{osp}_{1|2})$-modules. Let $\xi^{(i)}$ denote the highest
vector of $L(\alpha_{i},\beta_{i})$.

######  Proposition 4.4.

The ${\rm X}(\mathfrak{osp}_{1|2})$-module $L(\lambda(u))$ with

$\lambda_{1}(u)=\frac{(u+\alpha_{1})\dots(u+\alpha_{k})}{(u+\beta_{1})\dots(u+\beta_{k})}\qquad\text{and}\qquad\lambda_{2}(u)=1$
(4.5)

is isomorphic to the irreducible quotient of the submodule of $L$, generated
by the tensor product of the highest vectors
$\xi^{(1)}\otimes\dots\otimes\xi^{(k)}$.

###### Proof.

The coproduct formula (2.11) implies that the cyclic span ${\rm
X}(\mathfrak{osp}_{1|2})(\xi^{(1)}\otimes\dots\otimes\xi^{(k)})$ is a highest
weight module with the highest weight $(\lambda_{1}(u),\lambda_{2}(u))$ which
implies the claim. ∎

We will need to find the conditions for the elementary modules to be finite-
dimensional and establish some sufficient conditions for the module $L$ in
(4.4) to be irreducible.

###  4.2 Small Verma modules

Note that by twisting the action of the extended Yangian on a highest weight
module with the highest weight (4.3) by the shift automorphism (2.8) with
$a=-\beta$, we get the corresponding module whose highest weight is found by
shifting $\alpha\mapsto\alpha-\beta$ and $\beta\mapsto 0$. We will now assume
that $\beta=0$. Let $\alpha\in\mathbb{C}\hskip 1.0pt$ and consider the Verma
module $M(\lambda(u))$ with

$\lambda_{1}(u)=\frac{u+\alpha}{u},\qquad\lambda_{2}(u)=1,\qquad\lambda_{3}(u)=\frac{u-1/2}{u+\alpha-1/2}.$
(4.6)

Let $K$ be the submodule of $M(\lambda(u))$ generated by all vectors of the
form

$t_{21}^{(r)}\hskip 1.0pt\xi\quad\text{for}\quad r\geqslant
2\qquad\text{and}\qquad\big{(}t_{31}^{(r)}+(\alpha-1/2)\hskip
1.0ptt_{31}^{(r-1)}\big{)}\hskip 1.0pt\xi\quad\text{for}\quad r\geqslant 3,$
(4.7)

where $\xi$ denotes the highest vector of the Verma module. Introduce the
small Verma module $M(\alpha)$ as the quotient $M(\lambda(u))/K$. We will keep
the notation $\xi$ for the image of the highest vector of the Verma module in
the quotient. More general small Verma modules of the form $M(\alpha,\beta)$
corresponding to the highest weights (4.3) are then obtained by twisting the
modules $M(\alpha)$ by suitable automorphisms (2.8).

######  Proposition 4.5.

The module $M(\alpha)$ is spanned by the vectors

$t_{31}^{(2)\hskip 1.0ptr}\hskip 1.0ptt_{21}^{(1)\hskip 1.0pts}\,\xi,\qquad
r,s\in\mathbb{Z}\hskip 1.0pt_{+}.$ (4.8)

###### Proof.

By the Poincaré–Birkhoff–Witt theorem for the extended Yangian, the Verma
module $M(\lambda(u))$ has the basis

$t_{31}^{(k_{1})}\dots t_{31}^{(k_{p})}\hskip 1.0ptt_{21}^{(l_{1})}\dots
t_{21}^{(l_{q})}\hskip 1.0pt\xi,$ (4.9)

where $k_{1}\geqslant\dots\geqslant k_{p}\geqslant 1$ and
$l_{1}>\dots>l_{q}\geqslant 1$. Hence, the induction on the length of the
monomial in (4.9) reduces the argument to the verification of the property
that the span of the vectors (4.8) is stable under the action of the
generators $t_{31}^{(k)}$ and $t_{21}^{(l)}$.

The defining relations (2.7) imply that $[t_{31}^{(k)},t_{31}^{(m)}]=0$ and
$[t_{31}^{(k)},t_{21}^{(1)}]=0$ for all $k,m$. Therefore, for $k\geqslant 2$
in $M(\lambda(u))$ we have

$t_{31}^{(k)}\hskip 1.0ptt_{31}^{(2)\hskip 1.0ptr}\hskip
1.0ptt_{21}^{(1)\hskip
1.0pts}\,\xi\equiv(-\alpha+1/2)^{k-2}\,t_{31}^{(2)\hskip 1.0ptr+1}\hskip
1.0ptt_{21}^{(1)\hskip 1.0pts}\,\xi\mod K.$

The property is also clear for $k=1$ because $t_{31}^{(1)}=2\hskip
1.0ptt_{21}^{(1)\hskip 1.0pt2}$. Furthermore, since

$[t_{21}^{(l)},t_{31}^{(2)}]=t_{21}^{(1)}\hskip
1.0ptt_{31}^{(l)}-t_{31}^{(1)}\hskip 1.0ptt_{21}^{(l)}$

and $[t_{21}^{(l)},t_{21}^{(1)}]=t_{31}^{(l)}$, the property for the
generators $t_{21}^{(l)}$ easily follows too. ∎

We will regard $M(\alpha)$ as an $\mathfrak{osp}_{1|2}$-module via the
embedding (2.10). We get the weight space decomposition

$M(\alpha)=\bigoplus_{p=0}^{\infty}M(\alpha)_{-\alpha-p},$

where we define the weight subspaces of an arbitrary
$\mathfrak{osp}_{1|2}$-module $V$ by

$V_{\gamma}=\\{v\in V\ |\ F_{11}\hskip 1.0ptv=\gamma\hskip 1.0ptv\\}.$ (4.10)

Proposition 4.5 implies that

$\dim M(\alpha)_{-\alpha-p}\leqslant\lfloor p/2\rfloor+1.$ (4.11)

For all values $i,j\in\\{1,2,3\\}$ set $T_{ij}(u)=u\hskip
1.0pt(u+\alpha-1/2)\,t_{ij}(u)$. We will regard the coefficients of these
Laurent series in $u$ as operators in $M(\alpha)$.

######  Proposition 4.6.

All operators $T_{ij}(u)$ on the small Verma module $M(\alpha)$ are
polynomials in $u$.

###### Proof.

Calculating modulo $K$, we get

$t_{21}(u)\hskip 1.0pt\xi=u^{-1}\hskip 1.0ptt_{21}^{(1)}\hskip
1.0pt\xi\qquad\text{and}\qquad t_{31}(u)\hskip 1.0pt\xi=\big{(}u^{-1}\hskip
1.0ptt_{31}^{(1)}+\frac{1}{u(u+\alpha-1/2)}t_{31}^{(2)}\big{)}\hskip 1.0pt\xi$

so that the claim holds for the action of the operators $T_{21}(u)$ and
$T_{31}(u)$ on $\xi$. By acting on the vectors (4.8) of the spanning set, we
note that the operator $T_{31}(u)$ commutes with $t_{31}^{(2)}$ and
$t_{21}^{(1)}$, while for the operator $T_{21}(u)$ we have the relations

$[T_{21}(u),t_{31}^{(2)}]=t_{21}^{(1)}\hskip 1.0ptT_{21}(u)-t_{31}^{(1)}\hskip
1.0ptT_{31}(u)\qquad\text{and}\qquad[T_{21}(u),t_{21}^{(1)}]=T_{31}(u).$

Hence the property for the operators $T_{21}(u)$ and $T_{31}(u)$ follows by an
obvious induction.

As a next step, consider the relations for the series $T_{11}(u)$ implied by
(2.7):

$[T_{11}(u),t_{21}^{(1)}]=T_{21}(u),\qquad[T_{11}(u),t_{31}^{(1)}]=2\hskip
1.0ptT_{31}(u)$

and

$[T_{11}(u),t_{31}^{(2)}]=T_{31}(u)\big{(}2\hskip
1.0ptu+1/2+t_{11}^{(1)}\big{)}-2\hskip 1.0ptt_{31}^{(1)}\hskip
1.0ptT_{11}(u)-t_{21}^{(1)}\hskip 1.0ptT_{21}(u).$

Together with the relation

$T_{11}(u)\,\xi=(u+\alpha-1/2)(u+\alpha)\,\xi$ (4.12)

they imply the claim for the operator $T_{11}(u)$. For the remaining operators
the property follows from the relations

$[t_{12}^{(1)},T_{21}(u)]=T_{11}(u)-T_{22}(u),\qquad[t_{21}^{(1)},T_{22}(u)]=T_{32}(u)-T_{21}(u)$

and

$[t_{12}^{(1)},T_{11}(u)]=T_{12}(u),\qquad[t_{23}^{(1)},T_{32}(u)]=T_{33}(u)-T_{22}(u),\qquad[t_{23}^{(1)},T_{33}(u)]=-T_{23}(u),$

which are consequences of (2.7). ∎

For any $r,s\in\mathbb{Z}\hskip 1.0pt_{+}$ introduce vectors of the small
Verma module $M(\alpha)$ by setting

$\xi_{rs}=T_{21}(-\alpha-r+3/2)\dots T_{21}(-\alpha-1/2)\hskip
1.0ptT_{21}(-\alpha+1/2)\\\\[3.00003pt] {}\times T_{21}(-\alpha-s+1)\dots
T_{21}(-\alpha-1)\hskip 1.0ptT_{21}(-\alpha)\,\xi.$ (4.13)

We would like to show that under certain additional conditions the vectors
$\xi_{rs}$ form a basis of $M(\alpha)$; see Theorem 4.10 and Corollary 4.11
below. This will require a few lemmas where the action of the operators
$T_{ij}(u)$ on these vectors is calculated.

######  Lemma 4.7.

In the module $M(\alpha)$ we have

$T_{11}(u)\,\xi_{rs}=(u+\alpha+r-1/2)(u+\alpha+s)\,\xi_{rs}.$

###### Proof.

The formula holds for $\xi_{00}=\xi$ by (4.12). The defining relations (2.7)
give

$T_{11}(u)\hskip 1.0ptT_{21}(v)=\frac{u-v+1}{u-v}\,T_{21}(v)\hskip
1.0ptT_{11}(u)-\frac{1}{u-v}\,T_{21}(u)\hskip 1.0ptT_{11}(v),$

which implies the desired formula by an obvious induction. ∎

######  Lemma 4.8.

In the module $M(\alpha)$ for all $r\leqslant s+1$ we have

$T_{21}(u)\,\xi_{rs}=\frac{(-1)^{r+1}\hskip
1.0pt(s-r+1)(2u+2\alpha+2r-1)}{(s+1)(2s-2r+1)}\,\xi_{r,s+1}+\frac{2\hskip
1.0pt(u+\alpha+s)}{2s-2r+1}\,\xi_{r+1,s}.$

###### Proof.

By the definition of the vectors $\xi_{rs}$ we have
$T_{21}(-\alpha-r+1/2)\hskip 1.0pt\xi_{rs}=\xi_{r+1,s}$. Next we point out the
following relation for generators of ${\rm X}(\mathfrak{osp}_{1|2})$:

$(u-v-1/2)\,t_{21}(u)\hskip 1.0ptt_{21}(v)+(u-v+1/2)\,t_{21}(v)\hskip
1.0ptt_{21}(u)=t_{31}(v)\hskip 1.0ptt_{11}(u)-t_{31}(u)\hskip 1.0ptt_{11}(v).$

It is derived by calculating the commutators $[t_{21}(u),t_{21}(v)]$ and
$[t_{11}(u),t_{31}(v)]$ by (2.7) and eliminating the term $t_{11}(u)\hskip
1.0ptt_{31}(v)$. By Lemma 4.7 we have $T_{11}(u)\hskip 1.0pt\xi_{rs}=0$ for
$u=-\alpha-r+1/2$ and $u=-\alpha-s$. Hence, we come to the relation

$(r-s-1)\,T_{21}(-\alpha-s)\hskip
1.0ptT_{21}(-\alpha-r+1/2)\,\xi_{rs}=-(r-s)\,T_{21}(-\alpha-r+1/2)\hskip
1.0ptT_{21}(-\alpha-s)\,\xi_{rs}.$

Since $T_{21}(-\alpha-s)\,\xi_{0\hskip 1.0pts}=\xi_{0,s+1}$, applying the
relation repeatedly, we get the formula

$T_{21}(-\alpha-s)\,\xi_{rs}=\frac{(-1)^{r}\hskip
1.0pt(s-r+1)}{s+1}\,\xi_{r,s+1}$ (4.14)

which is valid for all $r\leqslant s+1$. Finally, using the Lagrange
interpolation formula

$T_{21}(u)=\frac{u+\alpha+r-1/2}{r-s-1/2}\,T_{21}(-\alpha-s)-\frac{u+\alpha+s}{r-s-1/2}\,T_{21}(-\alpha-r+1/2).$

we get the relation in the lemma. ∎

######  Lemma 4.9.

In the module $M(\alpha)$ for all $r\leqslant s$ we have

$\displaystyle T_{12}(u)\,\xi_{rs}$ $\displaystyle=-\frac{r\hskip
1.0pt(s-r+1)(2\hskip 1.0pt\alpha+2r-3)(u+\alpha+s)}{2\hskip
1.0pt(2s-2r+1)}\,\xi_{r-1,s}$ $\displaystyle+\frac{(-1)^{r+1}\hskip
1.0pts\hskip 1.0pt(2s+1)(\alpha+s-1)(2u+2\alpha+2r-1)}{4\hskip
1.0pt(2s-2r+1)}\,\xi_{r,s-1}.$

###### Proof.

By Proposition 4.6, the operator $T_{12}(u)$ is a polynomial in $u$ of degree
one. As in the proof of Lemma 4.8, it will be sufficient to calculate the
action of the operator for two different values $u=-\alpha-r+1/2$ and
$u=-\alpha-s$, and then apply the Lagrange interpolation formula.

Recall from Sec. 3 that the coefficients of the series $h_{1}(u)$ and
$h_{2}(u)$ pairwise commute. Set $d(u)=h_{1}(u)\hskip 1.0pth_{2}(u+1)$. Using
the defining relations (2.7), we can also write this series in the form

$d(u)=t_{22}(u)\hskip 1.0ptt_{11}(u+1)+t_{12}(u)\hskip 1.0ptt_{21}(u+1).$

The coefficients of the series $c(u)$ act by scalar multiplication in the
small Verma module. The scalars are found from (3.4) and given by

$c(u)\mapsto\frac{(u+1)\hskip 1.0pt(u+\alpha)}{u\hskip 1.0pt(u+\alpha+1)}.$
(4.15)

On the other hand, by Lemma 4.7, the coefficients of the series
$h_{1}(u)=t_{11}(u)$ act on each vector $\xi_{rs}$ as multiplications by
scalars depending on $r$ and $s$. Hence the same property holds for the
coefficients of $d(u)$ whose action is uniquely determined by the relation

$d(u)\hskip 1.0ptd(u+1/2)=c(u)\hskip 1.0pth_{1}(u+1/2)\hskip 1.0pth_{1}(u+1)$

implied by (3.4). Therefore, the action is found by

$d(u)\mapsto\frac{(u+1/2)\hskip 1.0pt(u+\alpha)}{u\hskip
1.0pt(u+\alpha+1/2)}\,h_{1}(u+1/2).$

For the corresponding polynomial operator

$D(u)=T_{22}(u)\hskip 1.0ptT_{11}(u+1)+T_{12}(u)\hskip 1.0ptT_{21}(u+1)$
(4.16)

we then have

$D(u)=(u+1)(u+\alpha-1/2)\,T_{11}(u+1/2).$ (4.17)

For any $r,s\in\mathbb{Z}\hskip 1.0pt_{+}$ we find from (4.16) by applying
Lemma 4.7 that

$\displaystyle D(-\alpha-r-1/2)\,\xi_{rs}$
$\displaystyle=T_{12}(-\alpha-r-1/2)\hskip
1.0ptT_{21}(-\alpha-r-1/2)\,\xi_{rs}$
$\displaystyle=T_{12}(-\alpha-r-1/2)\,\xi_{r+1,s}.$

Hence using (4.17) and replacing $r$ by $r-1$ we find

$T_{12}(-\alpha-r+1/2)\,\xi_{rs}=-\frac{1}{4}\,r(s-r+1)(2\hskip
1.0pt\alpha+2r-3)\,\xi_{r-1,s},$

which holds for $r\geqslant 1$. To extend this formula to the case $r=0$ use
Lemma 4.7 and relations

$[T_{12}(u)\hskip 1.0ptT_{21}(v)]=\frac{1}{u-v}\,\big{(}T_{22}(u)\hskip
1.0ptT_{11}(v)-T_{22}(v)\hskip 1.0ptT_{11}(u)\big{)}$ (4.18)

implied by (2.7) to derive by induction on $s$ that
$T_{12}(-\alpha+1/2)\,\xi_{0\hskip 1.0pts}=0$.

Similarly, taking $u=-\alpha-s-1$ in (4.16) and (4.17), we get by using (4.14)
that

$T_{12}(-\alpha-s)\,\xi_{rs}=\frac{1}{4}\,(-1)^{r}\hskip 1.0pts\hskip
1.0pt(2s+1)(\alpha+s-1)\,\xi_{r,s-1},$

which holds for $r<s$. This formula extends to the case $r=s$ by applying
relation (4.18) and taking into account Lemma 4.7. ∎

######  Theorem 4.10.

Suppose that $-\alpha\notin\mathbb{Z}\hskip 1.0pt_{+}$ and
$-\alpha+1/2\notin\mathbb{Z}\hskip 1.0pt_{+}$. Then the ${\rm
X}(\mathfrak{osp}_{1|2})$-module $M(\alpha)$ is irreducible. Moreover, the
vectors $\xi_{rs}$ with $r\leqslant s$ form a basis of $M(\alpha)$ and
$\xi_{rs}=0$ for $r>s$.

###### Proof.

We start by showing that all vectors $\xi_{rs}$ with $0\leqslant r\leqslant s$
are nonzero in $M(\alpha)$. The conditions on $\alpha$ and Lemma 4.9 imply
that it is sufficient to verify that $\xi\neq 0$; the vector $\xi_{rs}$ would
then also have to be nonzero, because the application of suitable operators
$T_{12}(v)$ to $\xi_{rs}$ gives the vector $\xi$ with a nonzero coefficient.

The relation $\xi=0$ in $M(\alpha)$ would mean that $\xi$, as an element of
the Verma module $M(\lambda(u))$ with the highest weight given in (4.6),
belongs to the submodule $K$. That is, $\xi$ is a linear combination of
vectors of the form

$x_{r}\hskip 1.0ptt_{21}^{(r)}\hskip 1.0pt\xi\quad\text{for}\quad r\geqslant
2\qquad\text{and}\qquad y_{r}\hskip
1.0pt\big{(}t_{31}^{(r)}+(\alpha-1/2)\hskip 1.0ptt_{31}^{(r-1)}\big{)}\hskip
1.0pt\xi\quad\text{for}\quad r\geqslant 3,$

with $x_{r},y_{r}\in{\rm X}(\mathfrak{osp}_{1|2})$. The elements $x_{r}$ and
$y_{r}$ must have the respective $\mathfrak{osp}_{1|2}$-weights $1$ and $2$ as
eigenvectors of the operator $F_{11}$. Write these elements as linear
combinations of the vectors of the Poincaré–Birkhoff–Witt basis of ${\rm
X}(\mathfrak{osp}_{1|2})$ by using any ordering on the generators consistent
with the increasing $\mathfrak{osp}_{1|2}$-weights. The right-most generators
occurring in each basis monomial will have positive
$\mathfrak{osp}_{1|2}$-weights. On the other hand, calculating in the Verma
module $M(\lambda(u))$ we find

$\displaystyle t_{12}(u)\hskip
1.0pt\big{(}t_{21}(v)-v^{-1}t_{21}^{(1)}\big{)}\,\xi{}$
$\displaystyle=\frac{1}{u-v}\,\big{(}t_{22}(u)\hskip
1.0ptt_{11}(v)-t_{22}(v)\hskip 1.0ptt_{11}(u)\big{)}\,\xi$
$\displaystyle-v^{-1}\,\big{(}t_{11}(u)-t_{22}(u)\big{)}\,\xi=0,$

as the coefficient of $\xi$ equals

$\frac{1}{u-v}\,\big{(}\frac{v+\alpha}{v}-\frac{u+\alpha}{u}\big{)}-\alpha\hskip
1.0ptu^{-1}v^{-1}=0.$

Now combine the second family of generators of the submodule $K$ given in
(4.7) into the generating series

$t_{31}(v)-v^{-1}\hskip
1.0ptt_{31}^{(1)}-\frac{1}{v(v+\alpha-1/2)}\,t_{31}^{(2)}$

which can be written as the anti-commutator of $t_{21}^{(1)}$ with the series

$t_{21}(v)-v^{-1}\hskip
1.0ptt_{21}^{(1)}-\frac{1}{v(v+\alpha-1/2)}t_{21}^{(2)}$

whose coefficients are also generators of $K$. Working first with one part of
the anti-commutator and using the previous calculation we get

$t_{12}(u)\hskip 1.0ptt_{21}^{(1)}\hskip 1.0pt\big{(}t_{21}(v)-v^{-1}\hskip
1.0ptt_{21}^{(1)}\big{)}\,\xi=\big{(}t_{11}(u)-t_{22}(u)\big{)}\big{(}t_{21}(v)-v^{-1}\hskip
1.0ptt_{21}^{(1)}\big{)}\,\xi.$

By the previous argument, the coefficients of this series vanish under the
action of the coefficients of the series $t_{12}(w)$. Turning to the second
part of the anti-commutator, we find that the expression

$t_{12}(u)\hskip 1.0pt\big{(}t_{21}(v)-v^{-1}\hskip
1.0ptt_{21}^{(1)}-\frac{1}{v(v+\alpha-1/2)}\hskip
1.0ptt_{21}^{(2)}\big{)}\hskip 1.0ptt_{21}^{(1)}\,\xi$

equals

$-\big{(}t_{21}(v)-v^{-1}\hskip
1.0ptt_{21}^{(1)}-\frac{1}{v(v+\alpha-1/2)}t_{21}^{(2)}\big{)}\hskip
1.0ptt_{12}(u)\hskip 1.0ptt_{21}^{(1)}\,\xi$ (4.19)

plus

$\displaystyle\frac{1}{u-v}\,\big{(}t_{22}(u)\hskip 1.0ptt_{11}(v)$
$\displaystyle-t_{22}(v)\hskip 1.0ptt_{11}(u)\big{)}\hskip
1.0ptt_{21}^{(1)}\,\xi-v^{-1}\,\big{(}t_{11}(u)-t_{22}(u)\big{)}\hskip
1.0ptt_{21}^{(1)}\,\xi$ $\displaystyle-\frac{1}{v(v+\alpha-1/2)}\hskip
1.0pt\big{(}(u+t_{22}^{(1)})\hskip
1.0ptt_{11}(u)-t_{22}(u)(u+t_{11}^{(1)})\big{)}\hskip 1.0ptt_{21}^{(1)}\,\xi.$

The expression (4.19) vanishes under the action of the coefficients of the
series $t_{12}(w)$, so we only need to transform the second expression. We
will do this modulo terms of the form $x_{r}t_{21}^{(r)}\hskip 1.0pt\xi$ with
$r\geqslant 2$ which were already considered above. Note the commutators

$[t_{11}(u),t_{21}^{(1)}]=t_{21}(u),\qquad[t_{22}(u),t_{21}^{(1)}]=t_{21}(u)-t_{32}(u).$

Using the second relation in (3.2) and writing the Gaussian generators in
terms of the $t_{ij}(u)$, we find

$t_{21}(u)\hskip 1.0ptt_{22}(u+1/2)\,\xi=t_{32}(u)\hskip
1.0ptt_{11}(u+1/2)\,\xi.$

Since $t_{21}(u)\hskip 1.0pt\xi\equiv u^{-1}\hskip 1.0ptt_{21}^{(1)}\,\xi$, we
derive that $t_{32}(u)\hskip 1.0pt\xi\equiv(u+\alpha-1/2)^{-1}\hskip
1.0ptt_{21}^{(1)}\,\xi$. Therefore, the expression in question is then
simplified by using relations

$t_{11}(u)\hskip 1.0ptt_{21}^{(1)}\,\xi\equiv u^{-1}\hskip
1.0ptt_{21}^{(1)}\,\xi\qquad\text{and}\qquad t_{22}(u)\hskip
1.0ptt_{21}^{(1)}\,\xi\equiv\frac{u^{2}+(\alpha-1/2)(u+1)}{u(u+\alpha-1/2)}\hskip
1.0ptt_{21}^{(1)}\,\xi$

and thus verifying that it reduces to zero. This completes the proof that
$\xi\not\equiv 0\mod K$.

As a next step, observe that since the vectors $\xi_{rs}$ with $0\leqslant
r\leqslant s$ are nonzero in $M(\alpha)$, they are eigenvectors for the
operator $T_{11}(u)$, whose eigenvalues are distinct as polynomials in $u$.
Hence they are linearly independent. The number of those vectors of the
$\mathfrak{osp}_{1|2}$-weight $-\alpha-p$ equals $\lfloor p/2\rfloor+1$, which
together with the inequality (4.11) proves that they form a basis of the
weight space $M(\alpha)_{-\alpha-p}$. Thus, all vectors $\xi_{rs}$ with
$0\leqslant r\leqslant s$ form a basis of $M(\alpha)$. Any vector $\xi_{rs}$
with $r>s$ cannot be nonzero, because otherwise it would be an eigenvector for
the operator $T_{11}(u)$ whose eigenvalue does not occur among those of the
vectors in $M(\alpha)$.

Finally, we prove the irreducibility of $M(\alpha)$. As we noted in the
beginning of the proof, the application of suitable operators $T_{12}(v)$ to
an arbitrary basis vector $\xi_{rs}$ yields the highest vector $\xi$ with a
nonzero coefficient. This implies that any nonzero submodule of $M(\alpha)$
must contain $\xi$ and so coincide with $M(\alpha)$. ∎

######  Corollary 4.11.

For any $\alpha\in\mathbb{C}\hskip 1.0pt$ the vectors $\xi_{rs}$ with
$0\leqslant r\leqslant s$ form a basis of $M(\alpha)$.

###### Proof.

Consider the vector space $\widetilde{M}(\alpha)$ with basis elements
$\widetilde{\xi}_{rs}$ labelled by $r,s\in\mathbb{Z}\hskip 1.0pt_{+}$ with
$0\leqslant r\leqslant s$. Define the action of the generators $t_{11}^{(r)}$,
$t_{21}^{(r)}$ and $t_{12}^{(r)}$ of $X(\mathfrak{osp}_{1|2})$ in
$\widetilde{M}(\alpha)$ by using the formulas of Lemmas 4.7, 4.8 and 4.9,
where the vectors $\xi_{rs}$ with $r\leqslant s$ are respectively replaced
with $\widetilde{\xi}_{rs}$, while all vectors $\xi_{rs}$ with $r>s$ are
replaced by $0$. Also, let the coefficients of the series $c(u)$ act in
$\widetilde{M}(\alpha)$ by scalar multiplication defined by (4.15). By Theorem
4.10, this assignment endows the space $\widetilde{M}(\alpha)$ with a
$X(\mathfrak{osp}_{1|2})$-module structure for all
$-\alpha\notin\mathbb{Z}\hskip 1.0pt_{+}$ and
$-\alpha+1/2\notin\mathbb{Z}\hskip 1.0pt_{+}$. Since the matrix elements of
the generators in the basis depend polynomially on $\alpha$, the same formulas
define a representation of $X(\mathfrak{osp}_{1|2})$ in
$\widetilde{M}(\alpha)$ for all values of $\alpha$ by continuity.

The formulas for the action of the generators in the basis
$\widetilde{\xi}_{rs}$ show that for any $\alpha\in\mathbb{C}\hskip 1.0pt$
there is an $X(\mathfrak{osp}_{1|2})$-module epimorphism
$\pi:M(\lambda(u))\to\widetilde{M}(\alpha)$ defined by
$\xi\mapsto\widetilde{\xi}_{00}$, where the highest weight $\lambda(u)$ of the
Verma module is given by (4.6). Moreover, the submodule $K$ of $M(\lambda(u))$
is contained in the kernel of $\pi$ which gives rise to an epimorphism
$\bar{\pi}:M(\alpha)\to\widetilde{M}(\alpha)$ with
$\xi_{rs}\mapsto\widetilde{\xi}_{rs}$. By taking into account the dimensions
of the respective $\mathfrak{osp}_{1|2}$-weight components, we conclude from
(4.11) that $\bar{\pi}$ is an isomorphism. ∎

As was pointed out in the proof of Corollary 4.11, for any
$\alpha\in\mathbb{C}\hskip 1.0pt$ the vectors (4.8) form a basis of
$M(\alpha)$, and (4.11) is in fact an equality:

$\dim M(\alpha)_{-\alpha-p}=\lfloor p/2\rfloor+1.$

###  4.3 Elementary modules

The elementary modules $L(\alpha)$ are defined as the irreducible quotients of
$M(\alpha)$. We would like to describe the structure of $L(\alpha)$ for the
values of $\alpha$ which do not satisfy the assumptions of Theorem 4.10; that
is, $-\alpha\in\mathbb{Z}\hskip 1.0pt_{+}$ or $-\alpha+1/2\in\mathbb{Z}\hskip
1.0pt_{+}$.

######  Proposition 4.12.

Suppose that $-\alpha=k\in\mathbb{Z}\hskip 1.0pt_{+}$. The linear span $J$ of
all basis vectors $\xi_{rs}$ of $M(-k)$ with $s>k$ is an ${\rm
X}(\mathfrak{osp}_{1|2})$-submodule. The module $L(-k)$ is isomorphic to the
quotient $M(-k)/J$, and the vectors $\xi_{rs}\mod J$ with $0\leqslant
r\leqslant s\leqslant k$ form its basis.

###### Proof.

The formula of Lemma 4.9 gives

$T_{12}(u)\,\xi_{r,k+1}=\frac{1}{2}\,r\hskip 1.0pt(k-r+2)(u+1)\,\xi_{r-1,k+1}$

for all $r\leqslant k+1$. This implies that the subspace $J$ of $M(-k)$ is
invariant under the action of ${\rm X}(\mathfrak{osp}_{1|2})$. Furthermore,
the formula of Lemma 4.9 also shows that the quotient $M(-k)/J$ is irreducible
and hence isomorphic to $L(-k)$. ∎

######  Proposition 4.13.

Suppose that $-\alpha+1/2=k\in\mathbb{Z}\hskip 1.0pt_{+}$. The linear span $I$
of all basis vectors $\xi_{rs}$ of $M(-k+1/2)$ with $r>k$ is an ${\rm
X}(\mathfrak{osp}_{1|2})$-submodule. The module $L(-k+1/2)$ is isomorphic to
the quotient $M(-k+1/2)/I$, and the vectors $\xi_{rs}\mod I$ with $0\leqslant
r\leqslant k$ form its basis.

###### Proof.

The formula of Lemma 4.9 now gives

$T_{12}(u)\,\xi_{k+1,s}=\frac{(-1)^{k}}{4}\hskip 1.0pts\hskip
1.0pt(2s+1)(u+1)\,\xi_{k+1,s-1}$

for all $s\geqslant k+1$. Recalling that $\xi_{rs}=0$ for $r>s$ we conclude
that the subspace $I$ of $M(-k+1/2)$ is invariant under the action of ${\rm
X}(\mathfrak{osp}_{1|2})$. Furthermore, Lemma 4.9 implies that the quotient
$M(-k+1/2)/I$ is irreducible and hence isomorphic to $L(-k+1/2)$. ∎

######  Corollary 4.14.

We have the following criteria.

  1. 1.

The ${\rm X}(\mathfrak{osp}_{1|2})$-module $M(\alpha)$ is irreducible if and
only if $-\alpha\notin\mathbb{Z}\hskip 1.0pt_{+}$ and
$-\alpha+1/2\notin\mathbb{Z}\hskip 1.0pt_{+}$.

  2. 2.

The ${\rm X}(\mathfrak{osp}_{1|2})$-module $L(\alpha)$ is finite-dimensional
if and only if $-\alpha=k\in\mathbb{Z}\hskip 1.0pt_{+}$. Moreover,

$\dim L(-k)=\binom{k+2}{2}.$

###### Proof.

All parts are immediate from Theorem 4.10 and Propositions 4.12 and 4.13. ∎

As the above description of the elementary modules shows, they admit bases
formed by $\mathfrak{osp}_{1|2}$-weight vectors. Accordingly, we can define
their characters by using formal exponents of a variable $q$ and using the
definition (4.10) of $\mathfrak{osp}_{1|2}$-weight subspaces. Namely, we set

${\rm{ch}\,}V=\sum_{\gamma}\dim V_{-\gamma}\,q^{\gamma}.$

In particular, the character of the irreducible highest weight module $V(\mu)$
over $\mathfrak{osp}_{1|2}$ is found by

${\rm{ch}\,}V(\mu)=\frac{\ \
q^{-\mu}}{1-q}\qquad\text{and}\qquad{\rm{ch}\,}V(\mu)=\frac{q^{-\mu}-q^{\hskip
1.0pt\mu+1}}{1-q}$

for $\mu\notin\mathbb{Z}\hskip 1.0pt_{+}$ and $\mu\in\mathbb{Z}\hskip
1.0pt_{+}$, respectively.

######  Corollary 4.15.

  1. 1.

The character of $M(\alpha)$ is given by

${\rm{ch}\,}M(\alpha)=\frac{q^{\alpha}}{(1-q)(1-q^{2})}.$

  2. 2.

For $-\alpha=k\in\mathbb{Z}\hskip 1.0pt_{+}$ we have

${\rm{ch}\,}L(-k)=q^{-k}\,\frac{(1-q^{k+1})(1-q^{k+2})}{(1-q)(1-q^{2})}.$

  3. 3.

For $-\alpha+1/2=k\in\mathbb{Z}\hskip 1.0pt_{+}$ we have

${\rm{ch}\,}L(-k+1/2)=q^{-k+1/2}\,\frac{1-q^{2k+2}}{(1-q)(1-q^{2})}.$

###### Proof.

The formulas follow by evaluating the dimensions of the weight subspaces. ∎

In terms of the characters of the $\mathfrak{osp}_{1|2}$-modules, we can write
the above formulas as

${\rm{ch}\,}L(-k)=\sum_{p=0}^{\lfloor k/2\rfloor}{\rm{ch}\,}V(k-2p)$

and

${\rm{ch}\,}L(-k+1/2)=\sum_{p=0}^{k}{\rm{ch}\,}V(k-1/2-2p).$

Finite-dimensional modules over the Lie superalgebras $\mathfrak{osp}_{1|2n}$
are known to be completely reducible; see e.g. [6, Sec. 2.2.5]. The formulas
for the action of the generator $F_{12}$ of $\mathfrak{osp}_{1|2}$ in the
basis $\xi_{rs}$ of $L(-k)$ show that there are singular vectors of the
weights $k$, $k-2$, etc., to imply the direct sum decomposition

$L(-k)\cong\bigoplus_{p=0}^{\lfloor k/2\rfloor}V(k-2\hskip 1.0ptp).$

######  Corollary 4.16.

The restriction of the module $L(\alpha)$ to the Lie superalgebra
$\mathfrak{osp}_{1|2}$ is irreducible if and only if $\alpha=0,-1$ or $1/2$. ∎

Corollary 4.16 shows that the $\mathfrak{osp}_{1|2}$-modules $V(0)$, $V(1)$
and $V(-1/2)$ can be extended to ${\rm X}(\mathfrak{osp}_{1|2})$. The Yangian
action on the three-dimensional vector representation $V(1)=\mathbb{C}\hskip
1.0pt^{1|2}$ arises from the replacement of $T(u)$ in the $RTT$-relation (2.3)
by a transposed $R$-matrix $R(u)$ which satisfies the Yang–Baxter equation;
cf. [2]. It is given explicitly by setting

$t_{ij}(u)\mapsto\delta_{ij}+u^{-1}\hskip
1.0pte_{ij}(-1)^{i}-(u+\kappa)^{-1}\hskip
1.0pte_{j^{\prime}i^{\prime}}(-1)^{ij}\,\theta_{i}\theta_{j}$ (4.20)

and is isomorphic to $L(-1)$.

###  4.4 Tensor product modules

We will now use the results of the previous sections to complete the proof of
the Main Theorem.

Recall that the elementary modules of the form $L(\alpha,\beta)$ and small
Verma modules $M(\alpha,\beta)$ are associated with the highest weights of the
form (1.3). They can be obtained by twisting the respective modules
$L(\alpha)$ and $M(\alpha)$ with the shift automorphisms (2.8). Corollary 4.14
(2) implies that the module $L(\alpha,\beta)$ is finite-dimensional if and
only if $\beta-\alpha\in\mathbb{Z}\hskip 1.0pt_{+}$.

For the highest weight of the form (4.5), the existence of a monic polynomial
$P(u)$ satisfying (1.2) is equivalent to the condition that the parameters
$\beta_{1},\dots,\beta_{k}$ can be renumbered in such a way that all
differences $\beta_{i}-\alpha_{i}$ with $i=1,\dots,k$ belong to
$\mathbb{Z}\hskip 1.0pt_{+}$. If this condition holds, then the tensor product
module (4.4) is finite-dimensional and so is its irreducible subquotient
$L(\lambda(u))$. This thus proves that the conditions of the Main Theorem are
sufficient for the irreducible highest weight module to be finite-dimensional.
In the rest of this section, we will show that the conditions are also
necessary.

By the results of Sec. 4.2, each small Verma module $M(\alpha,\beta)$ has the
basis $\xi_{rs}$ parameterized by $r,s\in\mathbb{Z}\hskip 1.0pt_{+}$ with
$r\leqslant s$ and the generators of the extended Yangian ${\rm
X}(\mathfrak{osp}_{1|2})$ act by the rules implied by Lemmas 4.7, 4.8 and 4.9.
Namely, for all $i,j\in\\{1,2,3\\}$ we now introduce the operators
$T_{ij}(u)=(u+\alpha-1/2)\hskip 1.0pt(u+\beta)\,t_{ij}(u)$, and the formulas
take the following form, where the vectors $\xi_{rs}$ with $r>s$ are equal to
zero:

$T_{11}(u)\,\xi_{rs}=(u+\alpha+r-1/2)(u+\alpha+s)\,\xi_{rs}$

together with

$T_{21}(u)\,\xi_{rs}=\frac{(-1)^{r+1}\hskip
1.0pt(s-r+1)(2u+2\alpha+2r-1)}{(s+1)(2s-2r+1)}\,\xi_{r,s+1}+\frac{2\hskip
1.0pt(u+\alpha+s)}{2s-2r+1}\,\xi_{r+1,s}$

and

$\displaystyle T_{12}(u)\,\xi_{rs}$ $\displaystyle=-\frac{r\hskip
1.0pt(s-r+1)(2\hskip 1.0pt\alpha-2\hskip 1.0pt\beta+2r-3)(u+\alpha+s)}{2\hskip
1.0pt(2s-2r+1)}\,\xi_{r-1,s}$ $\displaystyle+\frac{(-1)^{r+1}\hskip
1.0pts\hskip 1.0pt(2s+1)(\alpha-\beta+s-1)(2u+2\alpha+2r-1)}{4\hskip
1.0pt(2s-2r+1)}\,\xi_{r,s-1}.$

The coefficients of the series $c(u)$ act on $M(\alpha,\beta)$ by scalar
multiplication, with the scalars found from (3.4) and given by

$c(u)\mapsto\frac{(u+\alpha)\hskip 1.0pt(u+\beta+1)}{(u+\alpha+1)\hskip
1.0pt(u+\beta)}.$

By Corollary 4.14 (1), the ${\rm X}(\mathfrak{osp}_{1|2})$-module
$M(\alpha,\beta)$ is irreducible if and only if
$\beta-\alpha\notin\mathbb{Z}\hskip 1.0pt_{+}$ and
$\beta-\alpha+1/2\notin\mathbb{Z}\hskip 1.0pt_{+}$. In the cases where
$M(\alpha,\beta)$ is reducible, the above formulas for the action of
$T_{ij}(u)$ extend to the irreducible quotients $L(\alpha,\beta)$ with the
assumption that the vectors $\xi_{rs}$ belonging to the maximal proper
submodule of $M(\alpha,\beta)$ are understood as equal to zero.

Our argument will rely on certain sufficient conditions for the tensor product
of the form (4.4) to be irreducible as an ${\rm
X}(\mathfrak{osp}_{1|2})$-module. To state the conditions we will use a
notation involving multisets of complex numbers $\\{z_{1},\dots,z_{l}\\}$. For
such a multiset we will write $\\{z_{1},\dots,z_{l}\\}_{+}$ to denote the
multiset formed by all elements $z_{i}$ which belong to $\mathbb{Z}\hskip
1.0pt_{+}$.

######  Theorem 4.17.

Suppose that for each $h=1,\dots,k-1$ the following holds:

  1. 1.

If the multiset $\\{\beta_{h}-\alpha_{i},\ \beta_{i}-\alpha_{h}\ |\
i=h,\dots,k\\}_{+}$ is not empty, then $\beta_{h}-\alpha_{h}$ is a minimal
element of the multiset $\\{\beta_{h}-\alpha_{i},\ \beta_{i}-\alpha_{h},\
\beta_{h}-\alpha_{i}+1/2,\ \beta_{i}-\alpha_{h}+1/2\ |\ i=h,\dots,k\\}_{+}$.

  2. 2.

If the multiset $\\{\beta_{h}-\alpha_{i},\ \beta_{i}-\alpha_{h}\ |\
i=h,\dots,k\\}_{+}$ is empty and the multiset
$\\{\beta_{h}-\alpha_{i}+1/2,\ \beta_{i}-\alpha_{h}+1/2\ |\
i=h,\dots,k\\}_{+}$ is not empty, then $\beta_{h}-\alpha_{h}+1/2$ is a minimal
element of this multiset.

Then the ${\rm X}(\mathfrak{osp}_{1|2})$-module $L$ defined in (4.4) is
irreducible.

###### Proof.

We let $\xi_{rs}^{(l)}$ denote the basis vectors of the module
$L(\alpha_{l},\beta_{l})$ with the highest vector $\xi^{(l)}$. Proposition 4.6
implies that all operators

$T_{ij}(u)=\prod_{l=1}^{k}(u+\alpha_{l}-1/2)\hskip
1.0pt(u+\beta_{l})\,t_{ij}(u)$

acting in the module $L$ are polynomials in $u$.

As a first step, we will show by induction on $k$ that any vector $\zeta\in L$
satisfying the condition $T_{12}(u)\hskip 1.0pt\zeta=0$ is proportional to
$\xi^{(1)}\otimes\dots\otimes\xi^{(k)}$. The case $k=1$ is clear so we suppose
that $k\geqslant 2$. We may assume that such a vector $\zeta$ is an
$\mathfrak{osp}_{1|2}$-weight vector and write

$\zeta=\sum_{r,s}\xi_{rs}^{(1)}\otimes\zeta_{rs},\qquad\zeta_{rs}\in
L(\alpha_{2},\beta_{2})\otimes\dots\otimes L(\alpha_{k},\beta_{k}).$

The sum is finite and taken over the pairs $r\leqslant s$ with the condition
that the $\xi_{rs}^{(1)}$ are basis vectors of $L(\alpha_{1},\beta_{1})$. Let
$p$ be the maximal sum $r+s$ for which there are nonzero elements $\zeta_{rs}$
in the expression. By taking the coefficient of $\xi^{(1)}_{rs}$ with $r+s=p$
in the relation $T_{12}(u)\hskip 1.0pt\zeta=0$, we get $T_{12}(u)\hskip
1.0pt\zeta_{rs}=0$. By the induction hypothesis, $\zeta_{rs}$ is proportional
to the vector $\xi^{\prime}=\xi^{(2)}\otimes\dots\otimes\xi^{(k)}$.
Furthermore, the defining relations (2.7) give

$T_{12}(u)\hskip 1.0ptT_{11}(v)=\frac{u-v-1}{u-v}\,T_{11}(v)\hskip
1.0ptT_{12}(u)+\frac{1}{u-v}\,T_{11}(u)\hskip 1.0ptT_{12}(v).$

Hence, for any value of $v$, the vector $T_{11}(v)\hskip 1.0pt\zeta$ is also
annihilated by the operator $T_{12}(u)$. Note that the basis vectors
$\xi_{rs}^{(1)}$ are eigenvectors for the operator $T_{11}(v)$ with distinct
eigenvalues, as polynomials in $v$. This implies that by taking a suitable
value of $v$, we can find a linear combination of the vectors
$T_{11}(v)^{m}\hskip 1.0pt\zeta$ with $m=0,1,\dots$ to get an
$\mathfrak{osp}_{1|2}$-weight vector $\zeta$ of the form

$\zeta=\xi_{r_{0}s_{0}}^{(1)}\otimes\xi^{\prime}+\sum_{r+s<p}\xi_{rs}^{(1)}\otimes\zeta_{rs},$
(4.21)

with $r_{0}+s_{0}=p$ such that $T_{12}(u)\hskip 1.0pt\zeta=0$.

Next we will show that the condition $r_{0}<s_{0}$ is impossible in such a
vector. Indeed, if this condition holds, consider the coefficient of the
vector $\xi_{r_{0},s_{0}-1}^{(1)}\otimes\xi^{\prime}$ in the relation
$T_{12}(u)\hskip 1.0pt\zeta=0$. This coefficient can only arise from the terms

$T_{12}(u)\,\xi_{r_{0},s_{0}}^{(1)}\otimes T_{22}(u)\,\xi^{\prime}\pm
T_{11}(u)\,\xi_{r_{0},s_{0}-1}^{(1)}\otimes T_{12}(u)\,\zeta_{r_{0},s_{0}-1}$

with the sign depending on the parity of the vector
$\xi_{r_{0},s_{0}-1}^{(1)}$. The $\mathfrak{osp}_{1|2}$-weight condition
implies that

$\zeta_{r_{0},s_{0}-1}=\sum_{l=2}^{k}c_{l}\,\xi^{(2)}\otimes\dots\otimes\xi_{01}^{(l)}\otimes\dots\otimes\xi^{(k)}$

for some constants $c_{l}\in\mathbb{C}\hskip 1.0pt$. We have

$T_{12}(u)\,\zeta_{r_{0},s_{0}-1}=\sum_{l=2}^{k}\pm\hskip
1.0ptc_{l}\,T_{11}(u)\,\xi^{(2)}\otimes\dots\otimes
T_{12}(u)\,\xi_{01}^{(l)}\otimes\dots\otimes T_{22}(u)\,\xi^{(k)}.$

By using the formulas for the action of the operators $T_{ij}(u)$ and equating
the coefficient in question to zero, we get

$b\hskip 1.0pt(u+\alpha_{1}+r_{0}-1/2)\hskip
1.0pt\prod_{i=2}^{k}(u+\alpha_{i}-1/2)(u+\beta_{i})\\\
+(u+\alpha_{1}+r_{0}-1/2)(u+\alpha_{1}+s_{0}-1)\sum_{l=2}^{k}b_{l}\,\prod_{i=2}^{l-1}(u+\alpha_{i}-1/2)(u+\alpha_{i})\\\
{}\times(u+\alpha_{l}-1/2)\prod_{i=l+1}^{k}(u+\alpha_{i}-1/2)(u+\beta_{i})=0,$
(4.22)

where $b_{l}$ are some constants, while $b$ is a nonzero constant because of
the condition $s_{0}\leqslant\beta_{1}-\alpha_{1}$ in the case
$\beta_{1}-\alpha_{1}\in\mathbb{Z}\hskip 1.0pt_{+}$ implied by Proposition
4.12. By cancelling the common factors and setting $u=-\alpha_{1}-s_{0}+1$ we
get

$\prod_{i=2}^{k}(\beta_{i}-\alpha_{1}-s_{0}+1)=0.$

It follows from this relation that the multiset $\\{\beta_{i}-\alpha_{1}\ |\
i=1,\dots,k\\}_{+}$ is not empty, because
$\beta_{i}-\alpha_{1}=s_{0}-1\in\mathbb{Z}\hskip 1.0pt_{+}$ for some
$i\in\\{2,\dots,k\\}$. By assumption (1) of the theorem, we have
$\beta_{1}-\alpha_{1}\in\mathbb{Z}\hskip 1.0pt_{+}$ and
$\beta_{1}-\alpha_{1}\leqslant\beta_{i}-\alpha_{1}$. However, this makes a
contradiction, as by Proposition 4.12 we must have
$s_{0}\leqslant\beta_{1}-\alpha_{1}$.

Excluding the condition $r_{0}<s_{0}$ in (4.21), we show next that the
condition $r_{0}=s_{0}\geqslant 1$ is impossible either. If this condition
holds, consider the coefficient of the vector
$\xi_{r_{0}-1,r_{0}}^{(1)}\otimes\xi^{\prime}$ in the relation
$T_{12}(u)\hskip 1.0pt\zeta=0$. This coefficient can only arise from the terms

$T_{12}(u)\,\xi_{r_{0},r_{0}}^{(1)}\otimes T_{22}(u)\,\xi^{\prime}\pm
T_{11}(u)\,\xi_{r_{0}-1,r_{0}}^{(1)}\otimes T_{12}(u)\,\zeta_{r_{0}-1,r_{0}}.$

By the $\mathfrak{osp}_{1|2}$-weight condition,

$\zeta_{r_{0}-1,r_{0}}=\sum_{l=2}^{k}c_{l}\,\xi^{(2)}\otimes\dots\otimes\xi_{01}^{(l)}\otimes\dots\otimes\xi^{(k)}$

for some constants $c_{l}\in\mathbb{C}\hskip 1.0pt$. Calculating as in the
previous case, we now come to the relation

$b\hskip 1.0pt(u+\alpha_{1}+r_{0})\hskip
1.0pt\prod_{i=2}^{k}(u+\alpha_{i}-1/2)(u+\beta_{i})\\\
+(u+\alpha_{1}+r_{0}-3/2)(u+\alpha_{1}+r_{0})\sum_{l=2}^{k}b_{l}\,\prod_{i=2}^{l-1}(u+\alpha_{i}-1/2)(u+\alpha_{i})\\\
{}\times(u+\alpha_{l}-1/2)\prod_{i=l+1}^{k}(u+\alpha_{i}-1/2)(u+\beta_{i})=0,$
(4.23)

where $b_{l}$ are some constants, while $b$ is a nonzero constant. The latter
property holds because of the condition
$r_{0}\leqslant\beta_{1}-\alpha_{1}+1/2$ in the case
$\beta_{1}-\alpha_{1}+1/2\in\mathbb{Z}\hskip 1.0pt_{+}$ implied by Proposition
4.13. Cancel the common factors and set $u=-\alpha_{1}-r_{0}+3/2$ to get

$\prod_{i=2}^{k}(\beta_{i}-\alpha_{1}-r_{0}+3/2)=0.$

This means that for some $i\in\\{2,\dots,k\\}$ we have
$\beta_{i}-\alpha_{1}+1/2=r_{0}-1\in\mathbb{Z}\hskip 1.0pt_{+}$. If the
multiset $\\{\beta_{1}-\alpha_{j},\ \beta_{j}-\alpha_{1}\ |\
j=1,\dots,k\\}_{+}$ is not empty, then by assumption (1) of the theorem, we
have $\beta_{1}-\alpha_{1}\in\mathbb{Z}\hskip 1.0pt_{+}$ and
$\beta_{1}-\alpha_{1}\leqslant\beta_{i}-\alpha_{1}+1/2$. This is impossible
because by Proposition 4.12 we must have $r_{0}\leqslant\beta_{1}-\alpha_{1}$.
Hence assumption (2) of the theorem for $h=1$ should apply, and we have
$\beta_{1}-\alpha_{1}+1/2\in\mathbb{Z}\hskip 1.0pt_{+}$ together with the
inequality

$\beta_{1}-\alpha_{1}+1/2\leqslant\beta_{i}-\alpha_{1}+1/2.$

This makes a contradiction, as by Proposition 4.13 we must have
$r_{0}\leqslant\beta_{1}-\alpha_{1}+1/2$.

We have thus showed that any vector $\zeta\in L$ with $T_{12}(u)\hskip
1.0pt\zeta=0$ is proportional to $\xi^{(1)}\otimes\xi^{\prime}$. By looking at
the set of $\mathfrak{osp}_{1|2}$-weights of any nonzero submodule of $L$ we
derive that such a submodule must contain a nonzero vector $\zeta$ with
$T_{12}(u)\hskip 1.0pt\zeta=0$, and so contain the vector
$\xi^{(1)}\otimes\xi^{\prime}$. It remains to prove this vector is cyclic in
$L$.

Consider the vector space $L^{*}$ dual to $L$ which is spanned by all linear
maps $\sigma:L\to\mathbb{C}\hskip 1.0pt$ satisfying the condition that the
linear span of the vectors $\eta\in L$ such that $\sigma(\eta)\neq 0$, is
finite-dimensional. Equip $L^{*}$ with an ${\rm
X}(\mathfrak{osp}_{1|2})$-module structure by setting

$(x\,\sigma)(\eta)=\sigma(\omega(x)\,\eta)\quad\text{for}\quad x\in{\rm
X}(\mathfrak{osp}_{1|2})\quad\text{and}\quad\sigma\in L^{*},\ \eta\in L,$
(4.24)

where $\omega$ is the anti-automorphism of the algebra ${\rm
X}(\mathfrak{osp}_{1|2})$ defined in (2.9). It is easy to verify that $L^{*}$
is isomorphic to the tensor product module

$L(-\beta_{1},-\alpha_{1})\otimes\dots\otimes L(-\beta_{k},-\alpha_{k}).$
(4.25)

Moreover, the highest vector of the module $L(-\beta_{i},-\alpha_{i})$ can be
identified with the dual basis vector $\xi^{(i)\hskip 1.0pt*}$. Suppose now
that the submodule $N={\rm
X}(\mathfrak{osp}_{1|2})(\xi^{(1)}\otimes\dots\otimes\xi^{(k)})$ of $L$ is
proper and consider its annihilator

${\rm{Ann}\,}N=\\{\rho\in L^{*}\ |\ \rho(\eta)=0\quad\text{for
all}\quad\eta\in N\\}.$ (4.26)

Then ${\rm{Ann}\,}N$ is a nonzero submodule of $L^{*}$, which does not contain
the vector $\xi^{(1)\hskip 1.0pt*}\otimes\dots\otimes\xi^{(k)\hskip 1.0pt*}$.
However, this contradicts the claim verified in the first part of the proof,
because the conditions on the parameters $\alpha_{i}$ and $\beta_{i}$ stated
in the theorem will remain satisfied after we replace each $\alpha_{i}$ by
$-\beta_{i}$ and each $\beta_{i}$ by $-\alpha_{i}$. ∎

######  Proposition 4.18.

Suppose that the ${\rm X}(\mathfrak{osp}_{1|2})$-module $L(\lambda(u))$ with
the highest weight (4.5) is finite-dimensional. Then for any nonnegative
integers $l_{1},\dots,l_{k}$ and $m_{1},\dots,m_{k}$ the module
$L(\lambda^{+}(u))$ with the highest weight

$\lambda^{+}_{1}(u)=\frac{(u+\alpha_{1}-l_{1})\dots(u+\alpha_{k}-l_{k})}{(u+\beta_{1}+m_{1})\dots(u+\beta_{k}+m_{k})}\qquad\text{and}\qquad\lambda^{+}_{2}(u)=1$
(4.27)

is also finite-dimensional.

###### Proof.

The highest weight module $L(\lambda^{+}(u))$ is isomorphic to an irreducible
subquotient of the finite-dimensional module

$L(\lambda(u))\otimes L(\alpha_{1}-l_{1},\alpha_{1})\otimes\dots\otimes
L(\alpha_{k}-l_{k},\alpha_{k})\otimes
L(\beta_{1},\beta_{1}+m_{1})\otimes\dots\otimes L(\beta_{k},\beta_{k}+m_{k})$

and hence is finite-dimensional. ∎

We now return to proving the Main Theorem. Let the irreducible highest weight
module $L(\lambda(u))$ with the highest weight (4.5) be finite-dimensional. To
argue by contradiction, suppose that it is impossible to renumber the
parameters $\beta_{1},\dots,\beta_{k}$ in such a way that all differences
$\beta_{i}-\alpha_{i}$ with $i=1,\dots,k$ belong to $\mathbb{Z}\hskip
1.0pt_{+}$. By Proposition 4.18, all modules $L(\lambda^{+}(u))$ with the
highest weight of the form (4.27) are also finite-dimensional. It is possible
to choose nonnegative integers $l_{i}$ and $m_{i}$ to ensure that the
assumptions of Theorem 4.17 are satisfied by the shifted parameters
$\alpha^{\prime}_{i}=\alpha_{i}-l_{i}$ and
$\beta^{\prime}_{i}=\beta_{i}+m_{i}$, after a possible renumbering. This can
be done by induction, beginning with the multiset

$\\{\beta_{1}-\alpha_{i},\ \beta_{i}-\alpha_{1}\ |\ i=1,\dots,k\\}$

and renumbering the parameters $\alpha_{i}$ and $\beta_{i}$, if necessary, to
ensure that $\beta_{1}-\alpha_{1}$ is a minimal element of the multiset

$\\{\beta_{1}-\alpha_{i},\ \beta_{i}-\alpha_{1}\ |\ i=1,\dots,k\\}_{+}$ (4.28)

if it is nonempty. Then assumption (1) of the theorem for $h=1$ is achieved by
suitable shifts $\alpha_{i}\mapsto\alpha_{i}-l_{i}$ and
$\beta_{i}\mapsto\beta_{i}+m_{i}$ for $i=2,\dots,k$. If the multiset (4.28) is
empty, then assumption (2) for $h=1$ is achieved by a suitable renumbering of
the parameters $\alpha_{i}$ and $\beta_{i}$. Then we continue in the same way
to consider the multisets for $h=2$, etc. As a result, by Theorem 4.17, the
module $L(\lambda^{+}(u))$ is isomorphic to the tensor product of the
corresponding elementary modules. Since it is finite-dimensional, all new
differences $\beta^{\prime}_{i}-\alpha^{\prime}_{i}$ must be nonnegative
integers due to Corollary 4.14 (2).

This argument implies, that all the differences $\beta_{i}-\alpha_{i}$ of the
original parameters may be assumed to be integers. Moreover, we can apply some
shifts as given in Proposition 4.18, to further suppose that
$\beta_{i}-\alpha_{i}\in\mathbb{Z}\hskip 1.0pt_{+}$ for $i=1,\dots,k-1$, while
$\alpha_{k}-\beta_{k}\in 1+\mathbb{Z}\hskip 1.0pt_{+}$, and that it is
impossible to renumber the parameters to make all the differences
$\beta_{i}-\alpha_{i}$ nonnegative integers.

Now consider all the parameters $\alpha_{i}$ and $\beta_{i}$ which belong to
the $\mathbb{Z}\hskip 1.0pt$-coset in $\mathbb{C}\hskip 1.0pt$ containing
$\alpha_{k}$ and $\beta_{k}$. Renumbering them, if necessary, suppose that
they correspond to $i=d+1,\dots,k$ for some $d\in\\{0,1,\dots,k-1\\}$. After a
further renumbering to satisfy the assumptions of Theorem 4.17, we obtain that
the ${\rm X}(\mathfrak{osp}_{1|2})$-module

$L^{(2)}=L(\alpha_{d+1},\beta_{d+1})\otimes\dots\otimes
L(\alpha_{k},\beta_{k})$

is irreducible. Similarly, by applying suitable shifts of Proposition 4.18 to
the remaining parameters $\alpha_{i},\beta_{i}$ with $i=1,\dots,d$, and
possible relabelling, we may assume that they satisfy the assumptions of
Theorem 4.17 and so the ${\rm X}(\mathfrak{osp}_{1|2})$-module

$L^{(1)}=L(\alpha_{1},\beta_{1})\otimes\dots\otimes L(\alpha_{d},\beta_{d})$

is also irreducible. If the tensor product $L=L^{(1)}\otimes L^{(2)}$ turns
out to be irreducible, then we arrive at a contradiction, because the module
$L(\alpha_{k},\beta_{k})$ is infinite-dimensional. So we will suppose that $L$
is not irreducible and denote by $\mu$ the $\mathfrak{osp}_{1|2}$-weight of
the vector $\xi^{(1)}\otimes\dots\otimes\xi^{(k)}$. Consider the multiset

$\\{\beta_{i}-\alpha_{j}+1/2\ |\ 1\leqslant i\leqslant d,\quad d+1\leqslant
j\leqslant k\\}_{+}$ (4.29)

and let $p_{0}$ denote its minimal element, if the multiset is nonempty, or
set $p_{0}=+\infty$ otherwise.

######  Lemma 4.19.

The $\mathfrak{osp}_{1|2}$-weight component $N_{\mu-p}$ of the cyclic span

$N={\rm X}(\mathfrak{osp}_{1|2})(\xi^{(1)}\otimes\dots\otimes\xi^{(k)})$

coincides with $L_{\mu-p}$ for all $0\leqslant p\leqslant 2\hskip 1.0ptp_{0}$.

###### Proof.

Equip $L^{*}$ with an ${\rm X}(\mathfrak{osp}_{1|2})$-module structure by
using (4.24). By considering the annihilator ${\rm{Ann}\,}N$, as defined by
(4.26), it will be sufficient to show that any vector $\zeta\in L^{*}$ of the
$\mathfrak{osp}_{1|2}$-weight $\mu-p$ with the property $t_{12}(u)\hskip
1.0pt\zeta=0$ is proportional to the vector $\xi^{(1)\hskip
1.0pt*}\otimes\dots\otimes\xi^{(k)\hskip 1.0pt*}$. As before, we will identify
$L^{*}$ with the tensor product module (4.25) and denote by
$\check{\xi}^{(i)}$ the highest vector of the elementary module
$L(-\beta_{i},-\alpha_{i})$. We will now follow the first part of the proof of
Theorem 4.17 to derive by a reverse induction on $l\in\\{1,\dots,k\\}$,
beginning with $l=k$, that any vector

$\zeta^{(l)}\in L(-\beta_{l},-\alpha_{l})\otimes\dots\otimes
L(-\beta_{k},-\alpha_{k})$

of the $\mathfrak{osp}_{1|2}$-weight $\mu-p$ with the property
$t_{12}(u)\hskip 1.0pt\zeta^{(l)}=0$ is proportional to
$\check{\xi}^{(l)}\otimes\dots\otimes\check{\xi}^{(k)}$. This is clear for the
values $l=d+1,\dots,k$, because the assumptions of Theorem 4.17 are satisfied
by the corresponding parameters.

Now suppose that $l\in\\{1,\dots,d\\}$ and repeat the argument of the first
part of the proof of Theorem 4.17 to come to the expression

$\zeta^{(l)}=\check{\xi}_{r_{0}s_{0}}^{(l)}\otimes\xi^{\prime}+\sum_{r+s<p}\check{\xi}_{rs}^{(l)}\otimes\zeta_{rs},$

analogous to (4.21), where $r_{0}+s_{0}=p$ and
$\xi^{\prime}=\check{\xi}^{(l+1)}\otimes\dots\otimes\check{\xi}^{(k)}$.
Arguing as in that proof, we find that the condition $r_{0}<s_{0}$ is
impossible, leading to the only possibility that $r_{0}=s_{0}\geqslant 1$. In
this case, with our conditions of the parameters, we must have $p=2r_{0}$ and

$\beta_{l}-\alpha_{j}+1/2=r_{0}-1$ (4.30)

for some $d+1\leqslant j\leqslant k$. Since $r_{0}-1\in\mathbb{Z}\hskip
1.0pt_{+}$, relation (4.30) implies that $p_{0}$ has a finite value and
$r_{0}>p_{0}$. This makes a contradiction, because $p=2r_{0}\leqslant 2p_{0}$
by the assumption, thus completing the proof of the lemma. ∎

For any $s\in\mathbb{Z}\hskip 1.0pt_{+}$ set
$\eta_{s}=\xi^{(1)}\otimes\dots\otimes\xi^{(k-1)}\otimes\xi^{(k)}_{0\hskip
1.0pts}\in L(\alpha_{1},\beta_{1})\otimes\dots\otimes
L(\alpha_{k},\beta_{k})$.

######  Lemma 4.20.

In the tensor product module, for any $s\in\mathbb{Z}\hskip 1.0pt_{+}$ we have

$\displaystyle T_{21}(-\alpha_{k}-s)\hskip 1.0pt\eta_{s}{}$
$\displaystyle=\prod_{i=1}^{k-1}(\beta_{i}-\alpha_{k}-s)(\alpha_{i}-\alpha_{k}-s-1/2)\,\eta_{s+1},$
(4.31) $\displaystyle T_{12}(u)\hskip 1.0pt\eta_{s}$
$\displaystyle=\frac{s}{2}\,(\beta_{k}-\alpha_{k}-s+1)\,\prod_{i=1}^{k-1}(u+\alpha_{i}-1/2)(u+\alpha_{i})\,\eta_{s-1},$
and $\displaystyle T_{11}(u)\hskip 1.0pt\eta_{s}$
$\displaystyle=(u+\alpha_{k}-1/2)(u+\alpha_{k}+s)\,\prod_{i=1}^{k-1}(u+\alpha_{i}-1/2)(u+\alpha_{i})\,\eta_{s}.$

###### Proof.

All relations are immediate from the coproduct rule (2.11) and the formulas
for the action of the generators of the extended Yangian in the basis
$\xi_{rs}$ of the elementary module $L(\alpha,\beta)$, which were recalled in
the beginning of this section. In particular, for (4.31) we take into account
the relations $T_{11}(-\alpha_{k}-s)\,\xi^{(k)}_{0\hskip 1.0pts}=0$ and
$T_{21}(-\alpha_{k}-s)\,\xi^{(k)}_{0\hskip 1.0pts}=\xi^{(k)}_{0,s+1}$ in
$L(\alpha_{k},\beta_{k})$. ∎

Observe that the numerical coefficient on the right hand side of (4.31) is
nonzero for any values of $s$ outside the multisets

$\\{\beta_{i}-\alpha_{k}\ |\
i=d+1,\dots,k-1\\}_{+}\quad\text{and}\quad\\{\alpha_{i}-\alpha_{k}-1/2\ |\
i=1,\dots,d\\}_{+}.$ (4.32)

On the other hand, recalling that $p_{0}$ is the minimal element of the
multiset (4.29) when it is nonempty, note that we can use the shifts of the
parameters $\alpha_{i},\beta_{i}$ with $i=1,\dots,d$ as in Proposition 4.18 to
keep the assumptions of Theorem 4.17 satisfied. The module $L^{(1)}$ with the
shifted parameters remains irreducible, while we can make the value of $p_{0}$
arbitrarily large. It will be sufficient to make $p_{0}$ large enough for the
elements of both multisets in (4.32) not to exceed $2p_{0}$, noting that the
elements of the second multiset can only decrease after the shifts
$\alpha_{i}\mapsto\alpha_{i}-l_{i}$ for $i=1,\dots,d$.

The $\mathfrak{osp}_{1|2}$-weight of the vector $\eta_{s}$ equals $\mu-s$, and
hence, by Lemma 4.19, all vectors $\eta_{s}$ with $s\leqslant 2p_{0}$ belong
to the cyclic span $N={\rm X}(\mathfrak{osp}_{1|2})\hskip 1.0pt\eta_{0}$. This
property extends to all values $s\in\mathbb{Z}\hskip 1.0pt_{+}$ by relation
(4.31) of Lemma 4.20, because the numerical coefficient of $\eta_{s+1}$ does
not vanish for $s>2p_{0}$. The remaining two relations of Lemma 4.20 imply
that the images of the vectors $\eta_{s}$ in the irreducible quotient
$L(\lambda(u))$ of $N$ are linearly independent. Hence, $L(\lambda(u))$ is
infinite-dimensional, as it contains an infinite family of linearly
independent vectors. This contradiction completes the proof of the first part
of the Main Theorem. The second part concerning representations of the Yangian
${\rm Y}(\mathfrak{osp}_{1|2})$ is immediate from the decomposition (2.6); cf.
[4, Sec. 5.3].

Comparing the irreducibility conditions with those for the evaluation modules
over the Yangian ${\rm Y}(\mathfrak{gl}_{2})$ (see e.g. [14, Sec. 3.3]), note
that unlike that case, it is not possible, in general, to renumber the
parameters of the given highest weight (4.5) to satisfy the assumptions of
Theorem 4.17. In fact, not every module $L(\lambda(u))$ is isomorphic to a
tensor product module of the form (4.4), as illustrated by the following
example.

######  Example 4.21.

To describe the ${\rm X}(\mathfrak{osp}_{1|2})$-module $L(\lambda(u))$ with

$\lambda_{1}(u)=\frac{(u-1)(u-5/2)}{u\hskip
1.0pt(u-3/2)},\qquad\lambda_{2}(u)=1,$

consider the tensor product $L=L(-1,0)\otimes L(-5/2,-3/2)$ of two three-
dimensional modules. Note that its parameters do not satisfy the assumptions
of Theorem 4.17. The module $L$ turns out to have a proper submodule $K$ which
is generated by the vector

$\zeta=\xi_{11}^{(1)}\otimes\xi^{(2)}+3\,\xi_{01}^{(1)}\otimes\xi_{01}^{(2)}-\xi^{(1)}\otimes\xi_{11}^{(2)}.$

The submodule $K$ is one-dimensional, isomorphic to a highest weight module
$L(\mu(u))$ with the components

$\mu_{1}(u)=\mu_{2}(u)=\frac{(u-1/2)(u-5/2)}{(u-3/2)^{2}}.$

The module $L(\lambda(u))$ is isomorphic to the quotient $L/K$ with $\dim
L(\lambda(u))=8$ and so does not admit a tensor product decomposition of the
form (4.4). ∎

To conclude, we note that by analysing submodules of reducible modules
$M(\alpha,\beta)$, we can obtain explicit constructions of some modules
$L(\lambda(u))$ beyond the elementary modules. In particular, for any
$k\in\mathbb{Z}\hskip 1.0pt_{+}$ the submodule of $M(-k)$ generated by the
vector $\xi_{0,k+1}$ is isomorphic to the highest weight module
$L(\lambda(u))$ with

$\lambda_{1}(u)=\frac{u+1}{u}\qquad\text{and}\qquad\lambda_{2}(u)=\frac{(u+1/2)(u-k-1)}{u(u-k-1/2)}.$

The vectors $\xi_{rs}$ with $r\leqslant s$ and $s>k$ form its basis, and the
action of the generators is described in Sec. 4.2. The character of
$L(\lambda(u))$, as defined in Sec. 4.3, is found by

${\rm{ch}\,}L(\lambda(u))=\frac{q+q^{2}-q^{k+3}}{(1-q)(1-q^{2})}.$

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