Representations of the orthosymplectic Yangian

We give a complete description of the finite-dimensional irreducible representations of the Yangian associated with the orthosymplectic Lie superalgebra $\frak{osp}_{1|2}$. 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.


Introduction
It is well-known that the classification of finite-dimensional irreducible representations of the orthosymplectic Lie superalgebras osp M |2n is more complicated than that of its general linear counterpart 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 gl M |N was introduced by Nazarov [16] and its finite-dimensional irreducible representations were classified by Zhang [19].The orthosymplectic Yangians Y(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 Y(osp 1|2 ).One can expect that this simplest case will be instrumental in solving the general classification problem.In particular, an extension to osp 1|2n appears to be straightforward.
According to [1], the R-matrix originated in [18] leads to the RT T -type definition of the extended Yangian X(osp M |2n ).The Yangian Y(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 finitedimensional 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 Y(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 Y(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 X(osp 1|2 ), this means that the representations are determined by the triples (λ 1 (u), λ 2 (u), λ 3 (u)) of formal series λ i (u) ∈ 1 + u −1 C[[u −1 ]] which should satisfy the consistency condition λ 1 (u)λ 3 (u + 1/2) = λ 2 (u)λ 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 X(osp 1|2 ) associated with the highest weight (λ 1 (u), λ 2 (u), λ 3 (u)) is finite-dimensional if and only if for some monic polynomial P (u) in u.Hence, the finite-dimensional irreducible representations of the Yangian Y(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 Y(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 X(osp 1|2 ) and can be compared with the similar conditions for the Yangian Y(o 3 ); see [12], [13,Sec. 5.3], which were derived in a different way in [4].However, unlike the case of the Yangian Y(sl 2 ) (or Y(o 3 )), there is no epimorphism X(osp 1|2 ) → U(osp 1|2 ), so that in general, representations of 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(α, β) over X(osp 1|2 ) associated with the highest weights of the form The corresponding small Verma module M(α, β) turns out to be irreducible if and only if β − α and β−α+1/2 are not nonnegative integers.The elementary modules L(α, β) are the irreducible quotients of M(α, β) and so they split into three families, according to these conditions.The module L(α, β) is finite-dimensional if and only if β − α ∈ Z + .In this case, when regarded as an osp 1|2 -module, L(α, β) decomposes into the direct sum where V (µ) denotes the 2µ + 1-dimensional osp 1|2 -module with the highest weight µ ∈ Z + .In particular, We construct a basis of each small Verma module M(α, β) and give explicit formulas for the action of the generators of X(osp 1|2 ).This leads to a corresponding description of all elementary modules.We show that, up to twisting by a multiplication automorphism of X(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 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 Y(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 Y(sl 2 )-modules.This property turns out not to extend to representations of the Yangian for 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.

Sign conventions, definitions and preliminaries
Consider the three-dimensional space C 1|2 over C with basis elements e 1 , e 2 , e 3 where we assume a Z 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 → i ′ = 4 − i.The endomorphism algebra End C 1|2 then gets a Z 2 -gradation with the parity of the matrix unit e ij found by i + j mod 2.
We will consider 3 × 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 A is calculated in the standard way (without any additional signs).Accordingly, the algebra of even matrices over A will be identified with the tensor product algebra End C 1|2 ⊗ A. A matrix A = [a ij ] will be regarded as the element We will use the involutive matrix super-transposition t defined by This super-transposition is associated with the bilinear form on the space C 1|2 defined by the anti-diagonal matrix G = [δ ij ′ θ i ].We will also regard t as the linear map 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 End C 1|2 .
A standard basis of the general linear Lie superalgebra gl 1|2 is formed by elements E ij of the parity i + j mod 2 for 1 i, j 3 with the commutation relations We will regard the orthosymplectic Lie superalgebra osp 1|2 associated with the bilinear from defined by G as the subalgebra of gl 1|2 spanned by the elements For any given µ ∈ C we will denote by V (µ) the irreducible highest weight module over osp 1|2 generated by a nonzero vector ξ such that Introduce the permutation operator P as the element and set The rational function in u given by 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 Z 2 -graded spaces; cf.[8], [11].
Following [1], define the extended Yangian X(osp 1|2 ) as an associative superalgebra with generators t (r) ij of parity i + j mod 2, where 1 i, j 3 and r = 1, 2, . . ., satisfying certain quadratic relations.In order to write them down, introduce the formal series and combine them into the 3 × 3 matrix T (u) = [t ij (u)] so that

Consider the algebra End
] and introduce its elements T 1 (u) and T 2 (u) by The defining relations for the superalgebra X(osp 1|2 ) can then be written in the form of the As shown in [1], the product T (u) T t (u − κ) is a scalar matrix with where c(u) is a series in u −1 .All its coefficients belong to the center ZX(osp 1|2 ) of X(osp 1|2 ) and generate the center.
The Yangian Y(osp 1|2 ) is defined as the subalgebra of X(osp 1|2 ) which consists of the elements stable under the automorphisms for all series f (u) We have the tensor product decomposition The Yangian Y(osp 1|2 ) can be equivalently defined as the quotient of X(osp 1|2 ) by the relation We will also use a more explicit form of the defining relations (2.3) written in terms of the series (2.2) as follows: The coefficients of the series t 11 (u), t 12 (u), t 21 (u) and c(u) generate the algebra X(osp 1|2 ).The mapping t ij (u) → t ij (−u) defines an anti-automorphism of X(osp 1|2 ), while each of the mappings and t ij (u) → t i ′ j ′ (u) θ i θ j defines an automorphism.Consider their composition to define the anti-automorphism ω : (2.9) The universal enveloping algebra U(osp 1|2 ) can be regarded as a subalgebra of X(osp 1|2 ) via the embedding 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 Y(osp 1|2 ) is isomorphic to U(osp 1|2 [u]).It states that given any total ordering on the set of generators t (r) ij with i + j 4 and r 1, the ordered monomials in the generators with the powers of odd generators not exceeding 1, form a basis of X(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 (4.19).
The extended Yangian X(osp 1|2 ) is a Hopf algebra with the coproduct defined by For the image of the series c(u) we have ∆ : c(u) → c(u) ⊗ c(u) and so the Yangian Y(osp 1|2 ) inherits the Hopf algebra structure from X(osp 1|2 ).

Gaussian generators
A Drinfeld-type presentation of the Yangian for 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 X(osp 1|2 ), where F (u), H(u) and E(u) are uniquely determined matrices of the form and H(u) = diag h 1 (u), h 2 (u), h 3 (u) .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 Proposition 3.1.The following relations for the Gaussian generators hold: ) Proof.The argument is quite similar to the proof of the corresponding relations for the Gaussian generators of Y(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 On the other hand, relation (2.4) implies T t (u) = c(u)T (u − κ) −1 .Hence, by equating the (i, j) entries with i, j = 2, 3 in this matrix relation, we derive ) and 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 which together with relations (3.5) imply the first two desired identities, where we replaced κ by its value −3/2.They imply that relation (3.6) can be written in the form As a final step, use one more relation between the Gaussian generators, 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 Y(osp 1|2 ) of X(osp 1|2 ).Together with the coefficients of the series h(u) = h 1 (u) −1 h 2 (u) they generate the Yangian Y(osp 1|2 ), and the defining relations for these generators are given in [2] in a slightly different setting.

Yangian representations 4.1 Highest weight representations
for some formal series The vector ξ is called the highest vector of V and the triple λ 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 Hence, relation (3.3) of Proposition 3.1 implies the consistency condition (1.1) for the components λ i (u) of the highest weight.Given any triple λ(u) = (λ 1 (u), λ 2 (u), λ 3 (u)) of formal series of the form (4.2) safisfying the consistency condition (1.1), define the Verma module M(λ(u)) as the quotient of X(osp 1|2 ) by the left ideal generated by all coefficients of the series t ij (u) with 1 i < j 3, and t ii (u) − λ i (u) for i = 1, 2, 3.The Poincaré-Birkhoff-Witt theorem implies that the Verma module M(λ(u)) is nonzero and we denote by L(λ(u)) its irreducible quotient.Remark 4.1.In terms of the Drinfeld presentation of the Yangian Y(osp 1|2 ) given in [2], the highest vector conditions take the form The Main Theorem implies that the irreducible highest weight representation of Y(osp 1|2 ) associated with µ(u) is finite-dimensional if and only if for some monic polynomial P (u) in u.
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(λ(u)) are finite-dimensional.The first step is to establish some necessary conditions.
for k ∈ Z + and certain complex numbers α i , β i .
Proof.We follow the proof of a similar property for the Yangian Y(gl 2 ); see [14,Prop. 3.3.1].
Note that due to (1.1), the series λ 3 (u) is uniquely determined by λ 1 (u) and λ 2 (u), and so we can parameterize the highest weights by arbitrary pairs of series λ(u) = (λ 1 (u), λ 2 (u)), omitting λ 3 (u).By twisting the action of the extended Yangian X(osp 1|2 ) on L(λ(u)) by the automorphism (2.5) with f (u) = λ 2 (u) −1 , we get an X(osp 1|2 )-module isomorphic to L(µ(u), 1) with µ(u) = λ 1 (u)/λ 2 (u).Let ξ denote the highest vector of L(µ(u), 1).Since this representation is finite-dimensional, the vectors t 12 for all r 1 to the linear combination on the left hand side and take the coefficient of ξ.Since t 12 (u)ξ = 0, we get from the defining relations (2.7) that Hence, writing Therefore, for all r 1 we have the relations They imply so that µ(u) can be written as a rational function in u, as required.
We will use the name elementary module for the module L(λ(u)) with and denote it by L(α, β).
The Hopf algebra structure on the extended Yangian X(osp 1|2 ) allows us to regard tensor products of the form as X(osp 1|2 )-modules.Let ξ (i) denote the highest vector of L(α i , β i ).
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.

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 = −β, we get the corresponding module whose highest weight is found by shifting α → α − β and β → 0. We will now assume that β = 0. Let α ∈ C and consider the Verma module M(λ(u)) with where 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  We will regard M(α) as an osp 1|2 -module via the embedding (2.10).We get the weight space decomposition where we define the weight subspaces of an arbitrary osp 1|2 -module V by For all values i, j ∈ {1, 2, 3} set T ij (u) = u(u + α − 1/2) t ij (u).We will regard the coefficients of these Laurent series in u as operators in M(α).Proposition 4.6.All operators T ij (u) on the small Verma module M(α) are polynomials in u.
Proof.Calculating modulo K, we get 31 ξ so that the claim holds for the action of the operators T 21 (u) and T 31 (u) on ξ.By acting on the vectors (4.8) of the spanning set, we note that the operator T 31 (u) commutes with t (2) 31 and t (1) 21 , while for the operator T 21 (u) we have the relations 21 ] = 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): 21 T 21 (u).Together with the relation they imply the claim for the operator T 11 (u).For the remaining operators the property follows from the relations 23 , T 33 (u)] = −T 23 (u), which are consequences of (2.7).
For any r, s ∈ Z + introduce vectors of the small Verma module M(α) by setting We would like to show that under certain additional conditions the vectors ξ rs form a basis of M(α); 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(α) we have Proof.The formula holds for ξ 00 = ξ by (4.12).The defining relations (2.7) give which implies the desired formula by an obvious induction.
Lemma 4.8.In the module M(α) for all r s + 1 we have Proof.By the definition of the vectors ξ rs we have T 21 (−α − r + 1/2)ξ rs = ξ r+1,s .Next we point out the following relation for generators of X(osp 1|2 ): Since T 21 (−α − s) ξ 0 s = ξ 0,s+1 , applying the relation repeatedly, we get the formula which is valid for all r s + 1.Finally, using the Lagrange interpolation formula we get the relation in the lemma.
Lemma 4.9.In the module M(α) for all r s we have 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 = −α − r + 1/2 and u = −α − 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)h 2 (u + 1).Using the defining relations (2.7), we can also write this series in the form d(u) = t 22 (u)t 11 (u + 1) + t 12 (u)t 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 On the other hand, by Lemma 4.7, the coefficients of the series h 1 (u) = t 11 (u) act on each vector ξ 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 implied by (3.4).Therefore, the action is found by For the corresponding polynomial operator we then have For any r, s ∈ Z + we find from (4.15) by applying Lemma 4.7 that Hence using (4.16) and replacing r by r − 1 we find which holds for r 1.To extend this formula to the case r = 0 use Lemma 4.7 and relations implied by (2.7) to derive by induction on s that T 12 (−α + 1/2) ξ 0 s = 0.
Similarly, taking u = −α − s − 1 in (4.15) and (4.16), we get by using (4.13) that which holds for r < s.This formula extends to the case r = s by applying relation (4.17) and taking into account Lemma 4.7.
Proof.We start by showing that all vectors ξ rs with 0 r s are nonzero in M(α).The conditions on α and Lemma 4.9 imply that it is sufficient to verify that ξ = 0; the vector ξ rs would then also have to be nonzero, because the application of suitable operators T 12 (v) to ξ rs gives the vector ξ with a nonzero coefficient.
The relation ξ = 0 in M(α) would mean that ξ, as an element of the Verma module M(λ(u)) with the highest weight given in (4.6), belongs to the submodule K.That is, ξ is a linear combination of vectors of the form ξ for r 3, with x r , y r ∈ X(osp 1|2 ).The elements x r and y r must have the respective 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 X(osp 1|2 ) by using any ordering on the generators consistent with the increasing osp 1|2 -weights.The right-most generators occurring in each basis monomial will have positive osp 1|2 -weights.On the other hand, calculating in the Verma module M(λ(u)) we find as the coefficient of ξ equals Now combine the second family of generators of the submodule K given in (4.7) into the generating series which can be written as the anti-commutator of t (1) 21 with the series whose coefficients are also generators of K. Working first with one part of the anti-commutator and using the previous calculation we get 21 ξ.
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 21 ξ and thus verifying that it reduces to zero.This completes the proof that ξ ≡ 0 mod K.As a next step, observe that since the vectors ξ rs with 0 r s are nonzero in M(α), 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 osp 1|2 -weight −α − p equals ⌊p/2⌋ + 1, which together with the inequality (4.11) proves that they form a basis of the weight space M(α) −α−p .Thus, all vectors ξ rs with 0 r s form a basis of M(α).Any vector ξ 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(α).
Finally, we prove the irreducibility of M(α).As we noted in the beginning of the proof, the application of suitable operators T 12 (v) to an arbitrary basis vector ξ rs yields the highest vector ξ with a nonzero coefficient.This implies that any nonzero submodule of M(α) must contain ξ and so coincide with M(α).Proof.Consider the vector space M(α) with basis elements ξ rs labelled by r, s ∈ Z + with 0 r s.Define the action of the generators t 12 of X(osp 1|2 ) in M (α) by using the formulas of Lemmas 4.7, 4.8 and 4.9, where the vectors ξ rs with r s are respectively replaced with ξ rs , while all vectors ξ rs with r > s are replaced by 0. Also, let the coefficients of the series c(u) act in M (α) by scalar multiplication defined by (4.14).By Theorem 4.10, this assignment endows the space M (α) with a X(osp 1|2 )-module structure for all −α / ∈ Z + and −α + 1/2 / ∈ Z + .Since the matrix elements of the generators in the basis depend polynomially on α, the same formulas define a representation of X(osp 1|2 ) in M (α) for all values of α by continuity.
The formulas for the action of the generators in the basis ξ rs show that for any α ∈ C there is an X(osp 1|2 )-module epimorphism π : M(λ(u)) → M (α) defined by ξ → ξ 00 , where the highest weight λ(u) of the Verma module is given by (4.6).Moreover, the submodule K of M(λ(u)) is contained in the kernel of π which gives rise to an epimorphism π : M(α) → M (α) with ξ rs → ξ rs .By taking into account the dimensions of the respective osp 1|2 -weight components, we conclude from (4.11) that π is an isomorphism.
As was pointed out in the proof of Corollary 4.11, for any α ∈ C the vectors (4.8) form a basis of M(α), and (4.11) is in fact an equality: dim M(α) −α−p = ⌊p/2⌋ + 1.

Elementary modules
The elementary modules L(α) are defined as the irreducible quotients of M(α).We would like to describe the structure of L(α) for the values of α which do not satisfy the assumptions of Theorem 4.10; that is, −α ∈ Z + or −α + 1/2 ∈ Z + .Proposition 4.12.Suppose that −α = k ∈ Z + .The linear span J of all basis vectors ξ rs of M(−k) with s > k is an X(osp 1|2 )-submodule.The module L(−k) is isomorphic to the quotient M(−k)/J, and the vectors ξ rs mod J with 0 r s k form its basis.
Proof.The formula of Lemma 4.9 gives for all r k + 1.This implies that the subspace J of M(−k) is invariant under the action of X(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).for all s k + 1. Recalling that ξ rs = 0 for r > s we conclude that the subspace I of M(−k + 1/2) is invariant under the action of X(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.

The X(osp
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 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 osp 1|2 -weight subspaces.Namely, we set In particular, the character of the irreducible highest weight module V (µ) over osp 1|2 is found by for µ / ∈ Z + and µ ∈ Z + , respectively.

The character of M(α) is given by
.
Proof.The formulas follow by evaluating the dimensions of the weight subspaces.
In terms of the characters of the osp 1|2 -modules, we can write the above formulas as Finite-dimensional modules over the Lie superalgebras 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 osp 1|2 in the basis ξ rs of L(−k) show that there are singular vectors of the weights k, k − 2, etc., to imply the direct sum decomposition Corollary 4.16 shows that the osp 1|2 -modules V (0), V (1) and V (−1/2) can be extended to X(osp 1|2 ).The Yangian action on the three-dimensional vector representation V (1) = C 1|2 arises from the replacement of T (u) in the RT T -relation (2.3) by a transposed R-matrix R(u) which satisfies the Yang-Baxter equation; cf.[2].It is given explicitly by setting and is isomorphic to L(−1).

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(α, β) and small Verma modules M(α, β) are associated with the highest weights of the form (1.3).They can be obtained by twisting the respective modules L(α) and M(α) with the shift automorphisms (2.8).Corollary 4.14 (2) implies that the module L(α, β) is finite-dimensional if and only if β − α ∈ Z + .
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 β 1 , . . ., β k can be renumbered in such a way that all differences β i − α i with i = 1, . . ., k belong to Z + .If this condition holds, then the tensor product module (4.4) is finite-dimensional and so is its irreducible subquotient L(λ(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(α, β) has the basis ξ rs parameterized by r, s ∈ Z + with r s and the generators of the extended Yangian X(osp 1|2 ) act by the rules implied by Lemmas 4.7, 4.8 and 4.9.Namely, for all i, j ∈ {1, 2, 3} we now introduce the operators T ij (u) = (u + α − 1/2)(u + β) t ij (u), and the formulas take the following form, where the vectors ξ rs with r > s are equal to zero: The coefficients of the series c(u) act on M(α, β) by scalar multiplication, with the scalars found from (3.4) and given by By Corollary 4.14(1), the In the cases where M(α, β) is reducible, the above formulas for the action of T ij (u) extend to the irreducible quotients L(α, β) with the assumption that the vectors ξ rs belonging to the maximal proper submodule of M(α, β) 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 X(osp 1|2 )-module.To state the conditions we will use a notation involving multisets of complex numbers {z 1 , . . ., z l }.For such a multiset we will write {z 1 , . . ., z l } + to denote the multiset formed by all elements z i which belong to Z + .Theorem 4.17.Suppose that for each h = 1, . . ., k − 1 the following holds:
Proof.We let ξ (l) rs denote the basis vectors of the module L(α l , β l ) with the highest vector ξ (l) .Proposition 4.6 implies that all operators acting in the module L are polynomials in u.
As a first step, we will show by induction on k that any vector ζ ∈ L satisfying the condition T 12 (u)ζ = 0 is proportional to ξ (1) ⊗ . . .⊗ ξ (k) .The case k = 1 is clear so we suppose that k 2. We may assume that such a vector ζ is an osp 1|2 -weight vector and write The sum is finite and taken over the pairs r s with the condition that the ξ (1)  rs are basis vectors of L(α 1 , β 1 ).Let p be the maximal sum r + s for which there are nonzero elements ζ rs in the expression.By taking the coefficient of ξ (1)  rs with r + s = p in the relation T 12 (u)ζ = 0, we get T 12 (u)ζ rs = 0.By the induction hypothesis, ζ rs is proportional to the vector ξ ′ = ξ (2) ⊗. ..⊗ξ (k) .Furthermore, the defining relations (2.7) give Hence, for any value of v, the vector T 11 (v)ζ is also annihilated by the operator T 12 (u).Note that the basis vectors ξ (1)  rs 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 ζ with m = 0, 1, . . . to get an osp 1|2 -weight vector ζ of the form with r 0 + s 0 = p such that T 12 (u)ζ = 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 ξ with the sign depending on the parity of the vector ξ (1) r 0 ,s 0 −1 .The osp 1|2 -weight condition implies that for some constants c l ∈ C. We have By using the formulas for the action of the operators T ij (u) and equating the coefficient in question to zero, we get where b l are some constants, while b is a nonzero constant because of the condition s 0 β 1 − α 1 in the case β 1 − α 1 ∈ Z + implied by Proposition 4.12.By cancelling the common factors and setting u = −α 1 − s 0 + 1 we get k i=2 It follows from this relation that the multiset {β i − α 1 | i = 1, . . ., k} + is not empty, because β i − α 1 = s 0 − 1 ∈ Z + for some i ∈ {2, . . ., k}.By assumption (1) of the theorem, we have However, this makes a contradiction, as by Proposition 4.12 we must have s 0 β 1 − α 1 .
Excluding the condition r 0 < s 0 in (4.20), we show next that the condition r 0 = s 0 1 is impossible either.If this condition holds, consider the coefficient of the vector ξ (1) r 0 −1,r 0 ⊗ ξ ′ in the relation T 12 (u)ζ = 0.This coefficient can only arise from the terms T 12 (u) ξ (1)  r 0 ,r 0 ⊗ T 22 (u) ξ ′ ± T 11 (u) ξ (1) By the osp 1|2 -weight condition, for some constants c l ∈ C. Calculating as in the previous case, we now come to the relation where b l are some constants, while b is a nonzero constant.The latter property holds because of the condition r 0 β 1 − α 1 + 1/2 in the case This means that for some i ∈ {2, . . ., k} we have . ., k} + is not empty, then by assumption (1) of the theorem, we have . This is impossible because by Proposition 4.12 we must have r 0 β 1 − α 1 .Hence assumption (2) of the theorem for h = 1 should apply, and we have β 1 − α 1 + 1/2 ∈ Z + together with the inequality This makes a contradiction, as by Proposition 4.13 we must have r 0 β 1 − α 1 + 1/2.
We have thus showed that any vector ζ ∈ L with T 12 (u)ζ = 0 is proportional to ξ (1) ⊗ ξ ′ .By looking at the set of osp 1|2 -weights of any nonzero submodule of L we derive that such a submodule must contain a nonzero vector ζ with T 12 (u)ζ = 0, and so contain the vector ξ (1) ⊗ξ ′ .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 σ : L → C satisfying the condition that the linear span of the vectors η ∈ L such that σ(η) = 0, is finitedimensional.Equip L * with an X(osp 1|2 )-module structure by setting where ω is the anti-automorphism of the algebra X(osp 1|2 ) defined in (2.9).It is easy to verify that L * is isomorphic to the tensor product module Moreover, the highest vector of the module L(−β i , −α i ) can be identified with the dual basis vector ξ (i) * .Suppose now that the submodule N = X(osp 1|2 )(ξ (1) ⊗ . . .⊗ ξ (k) ) of L is proper and consider its annihilator Then Ann N is a nonzero submodule of L * , which does not contain the vector ξ (1) * ⊗ . . .⊗ ξ (k) * .However, this contradicts the claim verified in the first part of the proof, because the conditions on the parameters α i and β i stated in the theorem will remain satisfied after we replace each α i by −β i and each β i by −α i .
Proof.The highest weight module L(λ + (u)) is isomorphic to an irreducible subquotient of the finite-dimensional module and hence is finite-dimensional.
We now return to proving the Main Theorem.Let the irreducible highest weight module L(λ(u)) with the highest weight (4.5) be finite-dimensional.To argue by contradiction, suppose that it is impossible to renumber the parameters β 1 , . . ., β k in such a way that all differences β i − α i with i = 1, . . ., k belong to Z + .By Proposition 4.18, all modules L(λ + (u)) with the highest weight of the form (4.24) 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 α ′ i = α i − l i and β ′ i = β i + m i , after a possible renumbering.This can be done by induction, beginning with the multiset This argument implies, that all the differences β i − α 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 and that it is impossible to renumber the parameters to make all the differences β i − α i nonnegative integers.Now consider all the parameters α i and β i which belong to the Z-coset in C containing α k and β k .Renumbering them, if necessary, suppose that they correspond to i = d + 1, . . ., k for some d ∈ {0, 1, . . ., k − 1}.After a further renumbering to satisfy the assumptions of Theorem 4.17, we obtain that the X(osp 1|2 )-module is irreducible.Similarly, by applying suitable shifts of Proposition 4.18 to the remaining parameters α i , β i with i = 1, . . ., d, and possible relabelling, we may assume that they satisfy the assumptions of Theorem 4.17 and so the X(osp 1|2 )-module is also irreducible.If the tensor product L = L (1) ⊗L (2) turns out to be irreducible, then we arrive at a contradiction, because the module L(α k , β k ) is infinite-dimensional.So we will suppose that L is not irreducible and denote by µ the osp 1|2 -weight of the vector ξ (1) ⊗ . . .⊗ ξ (k) coincides with L µ−p for all 0 p 2p 0 .
Proof.Equip L * with an X(osp 1|2 )-module structure by using (4.21).By considering the annihilator Ann N, as defined by (4.23), it will be sufficient to show that any vector ζ ∈ L * of the osp 1|2 -weight µ − p with the property t 12 (u)ζ = 0 is proportional to the vector ξ (1) * ⊗ . . .⊗ ξ (k) * .As before, we will identify L * with the tensor product module (4.22) and denote by ξ(i) the highest vector of the elementary module L(−β i , −α i ).We will now follow the first part of the proof of Theorem 4.17 to derive by a reverse induction on l ∈ {1, . . ., k}, beginning with l = k, that any vector of the osp 1|2 -weight µ−p with the property t 12 (u)ζ (l) = 0 is proportional to ξ(l) ⊗. ..⊗ ξ(k) .This is clear for the values l = d + 1, . . ., k, because the assumptions of Theorem 4.17 are satisfied by the corresponding parameters.Now suppose that l ∈ {1, . . ., d} and repeat the argument of the first part of the proof of Theorem 4.17 to come to the expression analogous to (4.20), where r 0 + s 0 = p and ξ ′ = ξ(l+1) ⊗ . . .⊗ ξ(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 1.In this case, with our conditions of the parameters, we must have p = 2r 0 and β l − α j + 1/2 = r 0 − 1 (4.27) for some d + 1 j k.Since r 0 − 1 ∈ Z + , relation (4.27) implies that p 0 has a finite value and r 0 > p 0 .This makes a contradiction, because p = 2r 0 2p 0 by the assumption, thus completing the proof of the lemma.
For any s ∈ Z + set η s = ξ (1) ⊗ . . .⊗ ξ (k−1) ⊗ ξ (k) 0 s ∈ L(α 1 , β 1 ) ⊗ . . .⊗ L(α k , β k ).Lemma 4.20.In the tensor product module, for any s ∈ Z + we have Observe that the numerical coefficient on the right hand side of (4.28) is nonzero for any values of s outside the multisets On the other hand, recalling that p 0 is the minimal element of the multiset (4.26) when it is nonempty, note that we can use the shifts of the parameters α i , β i with i = 1, . . ., 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 (4.29) not to exceed 2p 0 , noting that the elements of the second multiset can only decrease after the shifts α i → α i − l i for i = 1, . . ., d.
The osp 1|2 -weight of the vector η s equals µ − s, and hence, by Lemma 4.19, all vectors η s with s 2p 0 belong to the cyclic span N = X(osp 1|2 )η 0 .This property extends to all values s ∈ Z + by relation (4.28) of Lemma 4.20, because the numerical coefficient of η s+1 does not vanish for s > 2p 0 .The remaining two relations of Lemma 4.20 imply that the images of the vectors η s in the irreducible quotient L(λ(u)) of N are linearly independent.Hence, L(λ(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 Y(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 Y(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(λ(u)) is isomorphic to a tensor product module of the form (4.4), as illustrated by the following example.01 − ξ (1) ⊗ ξ (2) 11 .
The submodule K is one-dimensional, isomorphic to a highest weight module L(µ(u)) with the components The module L(λ(u)) is isomorphic to the quotient L/K with dim L(λ(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(α, β), we can obtain explicit constructions of some modules L(λ(u)) beyond the elementary modules.In particular, for any k ∈ Z + the submodule of M(−k) generated by the vector ξ 0,k+1 is isomorphic to the highest weight module L(λ(u)) with The vectors ξ rs with r s and s > k form its basis, and the action of the generators is described in Sec.4.2.The character of L(λ(u)), as defined in Sec.4.3, is found by ch L(λ(u)) = q + q 2 − q k+3 (1 − q)(1 − q 2 ) .

Corollary 4 . 11 .
For any α ∈ C the vectors ξ rs with 0 r s form a basis of M(α).
We will keep the notation ξ for the image of the highest vector of the Verma module in the quotient.More general small Verma modules of the form M(α, β) corresponding to the highest weights (4.3) are then obtained by twisting the modules M(α) by suitable automorphisms (2.8).
Proof.By the Poincaré-Birkhoff-Witt theorem for the extended Yangian, the Verma module M(λ(u)) has the basis t(k 1 )31 . . .t The expression(4.18)vanishesunder 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 ] = 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 findt 21 (u)t 22 (u + 1/2) ξ = t 32 (u)t 11 (u + 1/2) ξ.Since t 21 (u)ξ ≡ u −1 t (1)21 ξ, we derive that t 32 (u)ξ ≡ (u + α − 1/2) −1 t (2)and renumbering the parameters α i and β i , if necessary, to ensure thatβ 1 − α 1 is a minimal element of the multiset {β 1 − α i , β i − α 1 | i = 1, ..., k} + (4.25)if it is nonempty.Then assumption (1) of the theorem for h = 1 is achieved by suitable shiftsα i → α i − l i and β i → β i + m i for i = 2, ..., k.If the multiset (4.25) is empty, then assumption (2) for h = 1 is achieved by a suitable renumbering of the parameters α i and β 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(λ + (u)) is isomorphic to the tensor product of the corresponding elementary modules.Since it is finite-dimensional, all new differences β ′ i − α ′ i must be nonnegative integers due to Corollary 4.14(2).
denote its minimal element, if the multiset is nonempty, or set p 0 = +∞ otherwise.