Jucys–Murphy elements and a presentation for partition algebras

Abstract

We give a new presentation for the partition algebras. This presentation was discovered in the course of establishing an inductive formula for the partition algebra Jucys–Murphy elements defined by Halverson and Ram (Eur. J. Comb. 26:869–921, 2005). Using Schur–Weyl duality we show that our recursive formula and the original definition of Jucys–Murphy elements given by Halverson and Ram are equivalent. The new presentation and inductive formula for the partition algebra Jucys–Murphy elements given in this paper are used to construct the seminormal representations for the partition algebras in a separate paper.

1 Introduction

The partition algebras A _k (n), for k,n∈ℤ_⩾0, are a family of algebras defined in the work of Martin and Jones in [4–6] in connection with the Potts model and higher dimensional statistical mechanics. By [4], the partition algebra A _k (n) is in Schur–Weyl duality with the symmetric group $\mathfrak{S}_{n}$ acting diagonally on the k-fold tensor product V ^⊗k of its n-dimensional permutation representation V. In [7], Martin defined the partition algebras $A_{k+\frac {1}{2}}(n)$, for k,n∈ℤ_⩾0, as the centralisers of the subgroup $\mathfrak{S}_{n-1}\subseteq\mathfrak {S}_{n}$ acting on V ^⊗k. Including the algebras $A_{k+\frac {1}{2}}(n)$ in the tower

(1.1)

allowed for the simultaneous analysis of the whole tower of algebras using the Jones Basic construction by Martin [7] and Halverson and Ram [3].

Halverson and Ram [3] and East [1] have given a presentation for the partition algebras in terms of Coxeter generators for the symmetric group and certain contractions. Halverson and Ram [3] used Schur–Weyl duality to show that certain diagrammatically defined elements in the partition algebras A _k (n) and $A_{k+\frac {1}{2}}(n)$ play an analogous role to the classical Jucys–Murphy elements in the group algebra of the symmetric group $\mathfrak{S}_{k}$.

The Jucys–Murphy elements in the group algebra of the symmetric group respect the inclusions $\mathfrak{S}_{k-1}\subseteq\mathfrak{S}_{k}$ and are simultaneously diagonalisable in any irreducible representation of the symmetric group. The seminormal representations of the symmetric group are the irreducible matrix representations with respect to a basis on which the Jucys–Murphy elements act diagonally. These may be constructed inductively (see [8], for example).

This paper provides a new presentation for the partition algebras. This presentation was discovered in the course of establishing an inductive formula for the Jucys–Murphy elements for partition algebras given by Halverson and Ram. In a separate paper, we use this new presentation to construct seminormal representations for the partition algebras [2], solving a problem highlighted in the introduction to the paper of Halverson and Ram [3].

In Sect. 2 we recall the presentation of the partition algebras given by Halverson and Ram [3] and East [1]. In Sect. 3 we state a recursion defining a family of operators

(1.2)

and establish that the operators (1.2) form a commuting family with properties analogous to the Jucys–Murphy elements that arise in the representation theory of the symmetric group. Simultaneously, we establish some basic commutativity results for certain operators denoted

(1.3)

which arose in the recursive definition of the Jucys–Murphy elements (1.2). In Sect. 4 we show that the elements of (1.3) are involutions which are related to the Coxeter generators for the symmetric group in a precise way. Using the relation between the involutions (1.3) and the Coxeter generators for the symmetric group, and the properties established in Sect. 3, we derive a new presentation for the partition algebras. In Sect. 5 we provide formulae for the actions of the Jucys–Murphy elements (1.2) and the involutions (1.3) on tensor space. Using Schur–Weyl duality, we demonstrate that the recursive definition of Jucys–Murphy elements given in Sect. 3 is equivalent to the definition of Jucys–Murphy elements given by Halverson and Ram [3].

2 The partition algebras

In this section we follow the exposition by Halverson and Ram in [3]. For k=1,2,… , let

Any element ρ∈A _k may be represented as a graph with k vertices in the top row, labelled from left to right, by 1,2,…,k and k vertices in the bottom row, labelled, from left to right by 1′,2′,…,k′, with vertex i joined to vertex j if i and j belong to the same block of ρ. The representation of a partition by a diagram is not unique; for example the partition

can be represented by the diagrams

If ρ ₁,ρ ₂∈A _k , then the composition ρ ₁∘ρ ₂ is the partition obtained by placing ρ ₁ above ρ ₂ and identifying each vertex in the bottom row of ρ ₁ with the corresponding vertex in the top row of ρ ₂ and deleting any components of the resulting diagram which contains only elements from the middle row. The composition product makes A _k into an associative monoid with identity

Let z be an indeterminant and R=ℤ[z]. The partition algebra $\mathcal{A}_{k}(z)$ is the R-module freely generated by A _k , equipped with the product

where ℓ is the number of blocks removed from the middle row in constructing the composition ρ ₁∘ρ ₂. Let $\mathcal {A}_{k-\frac {1}{2}}(z)$ denote the subalgebra of $\mathcal{A}_{k}(z)$ generated by $A_{k-\frac {1}{2}}$. A presentation for $\mathcal{A}_{k}(z)$ has been given by Halverson and Ram [3] and East [1].

Theorem 2.1

(Theorem 1.11 of [3])

If k=1,2,… , then the partition algebra $\mathcal{A}_{k}(z)$ is the unital associative R-algebra presented by the generators

and the relations

• (1) (Coxeter relations)

  • (i) $s_{i}^{2}=1$, for i=1,…,k−1.

  • (ii) s _i s _j =s _j s _i , if j≠i+1.

  • (iii) s _i s _i+1 s _i =s _i+1 s _i s _i+1, for i=1,…,k−2.

• (2) (Idempotent relations)

  • (i) $p_{i}^{2}=z p_{i}$, for i=1,…,k.

  • (ii) $p_{i+\frac {1}{2}}^{2}=p_{i+\frac {1}{2}}$, for i=1,…,k−1.

  • (iii) $s_{i}p_{i+\frac {1}{2}}=p_{i+\frac {1}{2}}s_{i}=p_{i+\frac {1}{2}}$, for i=1,…,k−1.

  • (iv) s _i p _i p _i+1=p _i p _i+1 s _i =p _i p _i+1, for i=1,…,k−1.

• (3) (Commutation relations)

  • (i) p _i p _j =p _j p _i , for i=1,…k and j=1,…,k.

  • (ii) $p_{i+\frac {1}{2}}p_{j+\frac {1}{2}}=p_{j+\frac {1}{2}}p_{i+\frac {1}{2}}$, for i=1,…k−1 and j=1,…,k−1.

  • (iii) $p_{i}p_{j+\frac {1}{2}}=p_{j+\frac {1}{2}}p_{i}$, for j≠i,i+1.

  • (iv) s _i p _j =p _j s _i , for j≠i,i+1.

  • (v) $s_{i}p_{j+\frac {1}{2}}=p_{j+\frac {1}{2}}s_{i}$, for j≠i−1,i+1.

  • (vi) s _i p _i s _i =p _i+1, for i=1,…,k−1.

  • (vii) $s_{i}p_{i-\frac {1}{2}}s_{i}=s_{i-1}p_{i+\frac {1}{2}}s_{i-1}$, for i=2,…,k−1.

• (4) (Contraction relations)

  • (i) $p_{i+\frac {1}{2}}p_{j}p_{i+\frac {1}{2}}=p_{i+\frac {1}{2}}$, for j=i,i+1.

  • (ii) $p_{i}p_{j-\frac {1}{2}}p_{i}=p_{i}$, for j=i,i+1.

The above relations also imply that

In Theorem 2.1, the following identifications have been made:

and

We also let $\mathfrak{S}_{k}$ denote the symmetric group on k letters which is generated by s ₁,…,s _k−1. If $u\in\mathfrak {S}_{k}\subset\mathcal{A}_{k}(z)$, and $\rho\in\mathcal{A}_{k}(z)$, we will sometimes write ρ ^u=uρu ⁻¹. Let $*:\mathcal{A}_{k}(z)\to \mathcal{A}_{k}(z)$ denote the algebra anti-involution which, given ρ∈A _k , for i=1,…,k, interchanges i and i′ in ρ. Then ∗ reflects each element of the diagram basis for $\mathcal {A}_{k}(z)$ in the horizontal axis, and satisfies

and

Restricting the map ∗ from $\mathcal{A}_{k}(z)$ to $\mathcal {A}_{k-\frac {1}{2}}(z)$, gives an algebra anti-involution of $\mathcal {A}_{k-\frac {1}{2}}(z)$ which we also denote by ∗.

3 Jucys–Murphy elements

In this section we recursively define a family of Jucys–Murphy elements in $\mathcal{A}_{k}(z)$ and $\mathcal{A}_{k+\frac {1}{2}}(z)$. It will be shown in Sect. 5 that the recursive formula given below is equivalent to the combinatorial definition of Jucys–Murphy elements given by Halverson and Ram [3].

Let (σ _i :i=1,2,…) and (L _i :i=0,1,…) be given by

and, for i=1,2,… ,

(3.1)

where, for i=2,3,… ,

(3.2)

Define $(\sigma_{i+\frac {1}{2}}:i=1,2,\ldots)$ and $(L_{i+\frac {1}{2}}:i=0,1,\ldots)$ by

and, for i=1,2,… ,

(3.3)

where, for i=2,3,… ,

(3.4)

Rewriting the last summand in (3.4) as

the expression (3.4) becomes

(3.5)

Using induction, it follows that if i=0,1,… , then $\sigma_{i+\frac {1}{2}}\in\mathcal{A}_{i+\frac {1}{2}}(z)$, and $L_{i+\frac {1}{2}}\in\mathcal {A}_{i+\frac {1}{2}}(z)$. Observe that if i=0,1,… , then (L _i )^∗=L _i and (σ _i+1)^∗=σ _i+1. The fact that $(L_{i+\frac {1}{2}})^{*}=L_{i+\frac {1}{2}}$ and $(\sigma_{i+\frac {1}{2}})^{*}=\sigma_{i+\frac {1}{2}}$ will be shown in Proposition 3.3.

Example 3.1

In terms of the diagram presentation for $\mathcal{A}_{3}(z)$, we have

and

Using the above expressions for L ₂ and σ ₃, we can write L ₃ as

The relations given in the next proposition are fundamental for subsequent calculations.

Proposition 3.2

For i=1,2,… , the following statements hold:

• (1) $\sigma_{i+1}p_{i+\frac {1}{2}}=p_{i+\frac {1}{2}}$,

• (2) $s_{i+1}\sigma_{i+1}p_{i+\frac {3}{2}}=p_{i+\frac {1}{2}}s_{i+1}\sigma_{i+1}$,

• (3) $\sigma_{i+1}p_{i}p_{i+\frac {1}{2}}=s_{i}L_{i}p_{i+\frac {1}{2}}$,

• (4) $\sigma_{i+1}p_{i+1}p_{i+\frac {1}{2}}=L_{i}p_{i+\frac {1}{2}}$,

• (5) $p_{i+\frac {1}{2}}L_{i}p_{i+\frac {1}{2}}=p_{i+\frac {1}{2}}$.

Proof

(1) We consider each of the terms on the right hand side of the definition

beginning with the argument by induction:

From the fact that $p_{i+\frac {1}{2}}$ commutes with L _i−1, $s_{i}p_{i-\frac {1}{2}}L_{i-1}s_{i}p_{i-\frac {1}{2}}s_{i}p_{i+\frac {1}{2}}=p_{i-\frac {1}{2}}\,{\cdot} L_{i-1}p_{i-\frac {1}{2}}p_{i+\frac {1}{2}}$ and $p_{i-\frac {1}{2}}L_{i-1}s_{i}p_{i-\frac {1}{2}}p_{i+\frac {1}{2}}=p_{i-\frac {1}{2}}L_{i-1}p_{i-\frac {1}{2}}p_{i+\frac {1}{2}}$, while

and, from the relation $p_{i-\frac {1}{2}}p_{i}p_{i+\frac {1}{2}}s_{i-1}=p_{i-\frac {1}{2}}p_{i}s_{i-1}s_{i}p_{i-\frac {1}{2}}s_{i}$, we obtain

Substituting the terms obtained above into the definition of $\sigma_{i+1}p_{i+\frac {1}{2}}$, we observe that all terms vanish except for $p_{i+\frac {1}{2}}$, which completes the proof of (1).

(2) The definition (3.2) gives

Now consider each of the terms on the right hand side of the above equality. Firstly,

Next, use the fact that $s_{i}p_{i+\frac {3}{2}}s_{i}$ and $p_{i-\frac {1}{2}}$ commute to observe that

and that

Since $p_{i+\frac {3}{2}}$ commutes with $\mathcal{A}_{i+\frac {1}{2}}(z)$, we see that

For the last term, we use the fact that $(s_{i}p_{i+\frac {3}{2}}s_{i})$ and $p_{i-\frac {1}{2}}$ commute to see that

Putting the above together, we have

which completes the proof of (2).

(3) We consider the terms on the right hand side of the equality

beginning with

For the second term, we have

and for the third,

Using the relation $p_{i}p_{i-\frac {1}{2}}p_{i}=p_{i}$, we see that the fourth term satisfies

while, for the fifth term,

Combining the above terms, we obtain

which completes the proof of (3).

(4) We consider each of the terms on the right hand side of the expression

Firstly,

For the second term,

For the third term,

For the fourth term, we use the relation $s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}}p_{i+1}p_{i+\frac {1}{2}}=s_{i}p_{i-\frac {1}{2}}p_{i-1}\,{\cdot} p_{i}p_{i-\frac {1}{2}}p_{i+\frac {1}{2}}$ in

For the final term, we use the relation $p_{i-\frac {1}{2}}p_{i}p_{i+\frac {1}{2}}s_{i-1}=p_{i-\frac {1}{2}}s_{i}p_{i-1}p_{i-\frac {1}{2}}s_{i}$ in

Putting the above together,

which proves (4).

(5) Parts (1) and (3) give

as required. □

Proposition 3.3

If i=1,2,… , then

• (1) $(\sigma_{i+\frac {1}{2}})^{*}=\sigma_{i+\frac {1}{2}}$,

• (2) $(L_{i+\frac {1}{2}})^{*}=L_{i+\frac {1}{2}}$.

Proof

(1) We show that the summand $p_{i-\frac {1}{2}}L_{i-1}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}}$ in (3.4) is fixed under the ∗ anti-involution on $\mathcal{A}_{i+\frac {1}{2}}(z)$, using Proposition 3.2, as follows:

(2) Given that $(\sigma_{i+\frac {1}{2}})^{*}=\sigma_{i+\frac {1}{2}}$, it suffices to show that the summand $p_{i+\frac {1}{2}}L_{i}p_{i}p_{i+\frac {1}{2}}$ in (3.3) is fixed under the ∗ anti-involution on $\mathcal {A}_{i+\frac {1}{2}}(z)$. Using Proposition 3.2,

gives the required result. □

Proposition 3.4

If i=1,2,… , then $\sigma_{i+\frac {1}{2}}s_{i}=s_{i}\sigma_{i+\frac {1}{2}}=\sigma_{i+1}$.

Proof

After checking the case i=1, the statement follows from induction and the equality

which was established in the proof of Proposition 3.3. □

The following observation is made for later reference.

Lemma 3.5

If i=1,2,… , then

Proof

On the one hand,

and on the other,

as required. □

We are now in a position to prove the first commutativity result of this paper.

Theorem 3.6

The elements σ _i+1 and $\sigma_{i+\frac {1}{2}}$ satisfy the following commutativity relations:

• (1) $\sigma_{i+1}p_{i-\frac {1}{2}}=p_{i-\frac {1}{2}}\sigma_{i+1}=p_{i-\frac {1}{2}}L_{i-1}s_{i}p_{i-\frac {1}{2}}$ for i=2,3,… .

• (2) σ _i+1 p _i−1=p _i−1 σ _i+1=s _i−1 σ _i p _i+1 s _i s _i−1 for i=2,3,… .

• (3) $\sigma_{i+1}p_{i-\frac {3}{2}}=p_{i-\frac {3}{2}}\sigma_{i+1}$ for i=3,4,… .

• (4) σ _i+1 s _i−2=s _i−2 σ _i+1 for i=3,4,… .

• (5) σ _i+1 p _i−2=p _i−2 σ _i+1 for i=3,4,… .

• (6) $\sigma_{i+\frac {1}{2}}p_{i-1}=p_{i-1}\sigma_{i+\frac {1}{2}}$ for i=2,3,… .

• (7) $\sigma_{i+\frac {1}{2}}p_{i-\frac {3}{2}}=p_{i-\frac {3}{2}}\sigma_{i+\frac {1}{2}}$ for i=3,4,… .

• (8) $\sigma_{i+\frac {1}{2}}s_{i-2}=s_{i-2}\sigma_{i+\frac {1}{2}}$ for i=3,4,… .

• (9) $\sigma_{i+\frac {1}{2}}p_{i-2}=p_{i-2}\sigma_{i+\frac {1}{2}}$ for i=3,4,… .

Proof

(1) Using Lemma 3.5,

as required.

(2) We first show that

• (i) $s_{i}p_{i-\frac {1}{2}}L_{i-1}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}}p_{i-1} =s_{i}p_{i-\frac {1}{2}}L_{i-1}s_{i}p_{i-\frac {1}{2}}s_{i}p_{i-1}$,

• (ii) $p_{i-\frac {1}{2}}p_{i}p_{i+\frac {1}{2}}s_{i-1}L_{i-1}p_{i-\frac {1}{2}}s_{i}p_{i-1} =p_{i-\frac {1}{2}}L_{i-1}s_{i}p_{i-\frac {1}{2}}p_{i-1}$,

• (iii) s _i−1 s _i σ _i s _i s _i−1 p _i−1=s _i−1 s _i σ _i p _i+1 s _i s _i−1=p _i−1 s _i−1 s _i σ _i s _i s _i−1.

The left hand side of (i) gives

which is the right hand side of (i). Using the relation $p_{i+\frac {1}{2}}s_{i-1}s_{i}=s_{i-1}s_{i}p_{i-\frac {1}{2}}$, the left hand side of (ii) gives

as required. The item (iii) follows from the relation s _i−1 s _i p _i+1=p _i−1 s _i−1 s _i . Next, using the definition (3.2),

Since the right hand side of the last expression is manifestly fixed under the ∗ anti-involution on $\mathcal{A}_{i+1}(z)$, the proof of (2) is complete.

(3) We first show that

• (i) $s_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}L_{i-2}s_{i-2}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}} =s_{i}p_{i-\frac {1}{2}}s_{i-2}L_{i-2}s_{i-2}s_{i-1}p_{i+\frac {1}{2}}\,{\cdot} p_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}$,

• (ii) $s_{i}p_{i-\frac {1}{2}}s_{i-2}L_{i-2}p_{i-\frac {3}{2}}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}= s_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}L_{i-2}p_{i-1}p_{i-\frac {3}{2}}\,{\cdot} s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}$,

• (iii) $s_{i}p_{i-\frac {1}{2}}\sigma_{i-1}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}} =s_{i-1}s_{i}p_{i-\frac {3}{2}}L_{i-2}s_{i-1}p_{i-\frac {3}{2}}s_{i}s_{i-1}p_{i-\frac {3}{2}}$,

• (iv) $p_{i-\frac {1}{2}}p_{i}p_{i+\frac {1}{2}}s_{i-1}p_{i-\frac {3}{2}}L_{i-2}s_{i-2}p_{i-\frac {1}{2}}s_{i}p_{i-\frac {3}{2}} =p_{i-\frac {1}{2}}p_{i}p_{i+\frac {1}{2}}s_{i-1}\sigma_{i-1}p_{i-\frac {1}{2}}\,{\cdot} s_{i}p_{i-\frac {3}{2}}$,

• (v) $p_{i-\frac {1}{2}}p_{i}p_{i+\frac {1}{2}}s_{i-1}s_{i-2}L_{i-2}p_{i-\frac {3}{2}}p_{i-\frac {1}{2}}s_{i}p_{i-\frac {3}{2}} =p_{i-\frac {1}{2}}p_{i}p_{i+\frac {1}{2}}s_{i-1}s_{i-2}L_{i-2}s_{i-2}\,{\cdot} p_{i-\frac {1}{2}}s_{i}p_{i-\frac {3}{2}}$,

• (vi) $p_{i-\frac {1}{2}}p_{i}p_{i+\frac {1}{2}}s_{i-1}p_{i-\frac {3}{2}}L_{i-2}p_{i-1}p_{i-\frac {3}{2}}p_{i-\frac {1}{2}}s_{i}p_{i-\frac {3}{2}} =s_{i-1}s_{i}s_{i-1}p_{i-\frac {3}{2}}L_{i-2}s_{i-1}\,{\cdot} p_{i-\frac {3}{2}}s_{i-1}s_{i}s_{i-1}p_{i-\frac {3}{2}}$,

• (vii) $p_{i-\frac {1}{2}}s_{i-2}L_{i-2}p_{i-\frac {3}{2}}s_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}} =p_{i-\frac {1}{2}}s_{i-2}L_{i-2}s_{i-2}s_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}$,

• (viii) $p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}L_{i-2}s_{i-2}s_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}} =p_{i-\frac {1}{2}}\sigma_{i-1}s_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}$,

• (ix) $s_{i}p_{i-\frac {1}{2}}s_{i-2}L_{i-2}s_{i-2}s_{i}p_{i-\frac {1}{2}}s_{i}p_{i-\frac {3}{2}} =s_{i}p_{i-\frac {1}{2}}s_{i-2}L_{i-2}p_{i-\frac {3}{2}}s_{i}p_{i-\frac {1}{2}}s_{i}p_{i-\frac {3}{2}}$,

• (x) $s_{i}p_{i-\frac {1}{2}}\sigma_{i-1}s_{i}p_{i-\frac {1}{2}}s_{i}p_{i-\frac {3}{2}} =s_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}L_{i-2}s_{i-2}s_{i}p_{i-\frac {1}{2}}s_{i}p_{i-\frac {3}{2}}$.

• (xi) $s_{i-1}s_{i}s_{i-1}p_{i-\frac {3}{2}}L_{i-2}s_{i-2}p_{i-\frac {1}{2}}p_{i-1}p_{i-\frac {3}{2}} s_{i}s_{i-1}p_{i-\frac {3}{2}}= s_{i-1}\sigma_{i-\frac {3}{2}}s_{i-1}s_{i}p_{i-\frac {3}{2}}\,{\cdot} p_{i-\frac {1}{2}}$.

• (xii) $s_{i-1}s_{i}p_{i-\frac {3}{2}}p_{i-1}p_{i-\frac {1}{2}}s_{i-2}L_{i-2}p_{i-\frac {3}{2}} s_{i-1}s_{i}s_{i-1}p_{i-\frac {3}{2}}=p_{i-\frac {3}{2}}p_{i-\frac {1}{2}}s_{i}s_{i-1}\sigma_{i-\frac {3}{2}}\,{\cdot} s_{i-1}$.

For the left hand side of (i),

and for the right hand side of (i),

For the left hand side of (ii),

and for the right hand side of (ii),

For the left hand side of (iii),

and for the right hand side of (iii),

For the left hand side of (iv),

and, for the right hand side of (iv),

For the left hand side of (v),

and, for the right hand side of (v),

For the left hand side of (vi),

and, for the right hand side of (vi),

For the left hand side of (vii)

and for the right hand side of (vii)

For the left hand side of (viii),

and for the right hand side of (viii),

For the left hand side of (ix),

and for the right hand side of (ix),

For the left hand side of (x),

and for the right hand side of (x),

Starting with the left hand side of (xi),

as required. Considering the left hand side of (xii),

as required. Now, after substituting the expressions

and

into the definition (3.2), and simplifying the resulting expression using items (i)–(x), we obtain the equality

in which the two terms with negative coefficients survive from the expansion of $s_{i-1}s_{i}\sigma_{i}s_{i}s_{i-1}p_{i-\frac {3}{2}}$. By items (xi) and (xii), the two terms with negative coefficients in the last expression are interchanged by the ∗ map on $\mathcal {A}_{i+1}(z)$. Since

the right hand side of the above expression for $\sigma_{i+1}p_{i-\frac {3}{2}}$ is fixed under the ∗ anti-involution on $\mathcal{A}_{i+1}(z)$. This completes the proof of (3).

(4) We show that after substituting the expression

into the definition (3.2), conjugation by s _i−2 permutes the summands of σ _i+1 as follows:

• (i) $(s_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}L_{i-2}s_{i-2}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}})^{s_{i-2}} =s_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}L_{i-2}s_{i-2}s_{i-1}p_{i+\frac {1}{2}}\,{\cdot} p_{i}p_{i-\frac {1}{2}}$,

• (ii) $(s_{i}p_{i-\frac {1}{2}}s_{i-2}L_{i-2}p_{i-\frac {3}{2}}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}})^{s_{i-2}} =p_{i-\frac {1}{2}}p_{i}p_{i+\frac {1}{2}}s_{i-1}p_{i-\frac {3}{2}}L_{i-2}\,{\cdot} s_{i-2}p_{i-\frac {1}{2}}s_{i}$,

• (iii) $(s_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}L_{i-2}p_{i-1}p_{i-\frac {3}{2}}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}})^{s_{i-2}} =s_{i}p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}L_{i-2}s_{i-2}s_{i}\,{\cdot} p_{i-\frac {1}{2}}s_{i}$,

• (iv) $(s_{i}p_{i-\frac {1}{2}}s_{i-2}L_{i-2}s_{i-2}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}})^{s_{i-2}} =s_{i-1}s_{i}s_{i-1}p_{i-\frac {3}{2}}L_{i-2}s_{i-2}p_{i-\frac {1}{2}}\,{\cdot} p_{i-1}p_{i-\frac {3}{2}}s_{i}s_{i-1}$,

• (v) $(s_{i}p_{i-\frac {1}{2}}\sigma_{i-1}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}})^{s_{i-2}}= p_{i-\frac {1}{2}}s_{i-2}L_{i-2}p_{i-\frac {3}{2}}s_{i}p_{i-\frac {1}{2}}$,

• (vi) $(p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}L_{i-2}p_{i-1}p_{i-\frac {3}{2}}s_{i}p_{i-\frac {1}{2}})^{s_{i-2}} =p_{i-\frac {1}{2}}p_{i-\frac {3}{2}}L_{i-2}p_{i-1}p_{i-\frac {3}{2}}s_{i}p_{i-\frac {1}{2}}$,

• (vii) $(s_{i}p_{i-\frac {1}{2}}s_{i-2}L_{i-2}s_{i-2}s_{i}p_{i-\frac {1}{2}}s_{i})^{s_{i-2}} =s_{i-1}s_{i}s_{i-1}p_{i-\frac {3}{2}}L_{i-2}s_{i-1}p_{i-\frac {3}{2}}s_{i-1}s_{i}s_{i-1}$,

• (viii) $(s_{i}p_{i-\frac {1}{2}}\sigma_{i-1}s_{i}p_{i-\frac {1}{2}}s_{i})^{s_{i-2}} =p_{i-\frac {1}{2}}\sigma_{i-1}s_{i}p_{i-\frac {1}{2}}$,

• (ix) $(s_{i-1}s_{i}p_{i-\frac {3}{2}}L_{i-2}s_{i-1}p_{i-\frac {3}{2}}s_{i}s_{i-1})^{s_{i-2}} =p_{i-\frac {1}{2}}s_{i-2}L_{i-2}s_{i-2}s_{i}p_{i-\frac {1}{2}}$,

• (x) $(s_{i-1}s_{i}s_{i-2}s_{i-1}\sigma_{i-1}s_{i-1}s_{i-2}s_{i}s_{i-1})^{s_{i-2}} =s_{i-1}s_{i}s_{i-2}s_{i-1}\sigma_{i-1}s_{i-1}s_{i-2}s_{i}s_{i-1}$.

The item (i) follows from the relation $s_{i-2}p_{i-\frac {3}{2}}=p_{i-\frac {3}{2}}$ and

For the left hand side of (ii),

and, for the right hand side of (ii),

where the last equality follows from $s_{i-2}p_{i-\frac {1}{2}}\sigma_{i-1}=\sigma_{i-1}p_{i-\frac {1}{2}}s_{i-2}$.

For the left hand side of (iii),

and for the right hand side of (iii),

From the left hand side of (iv), we obtain

which is identical to the right hand side of (iv).

From the left hand side of (v), we obtain

which is identical to the right hand side of (v).

The item (vi) follows immediately from the relation $s_{i-2}p_{i-\frac {3}{2}}=p_{i-\frac {3}{2}}s_{i-2}=p_{i-\frac {3}{2}}$.

From the left hand side of (vii), we obtain

which is identical to the right hand side of (vii).

From the left hand side of (viii),

which is identical to the right hand side of (viii).

From the left hand side of (ix),

which is identical to the right hand side of (ix).

The equality (x) follows from the fact that s _i−2 s _i−1 s _i s _i−2 s _i−1=s _i−1 s _i s _i−2 s _i−1 s _i .

(5) By (2) and (4), σ _i+1 commutes with 〈s _i−2,p _i−1〉, and so with p _i−2=s _i−2 p _i−1 s _i−2.

(6) By Proposition 3.4, and (2),

(7)–(9) Can be proved using the same argument as part (6). □

Theorem 3.7

The elements L _i+1 and $L_{i+\frac {1}{2}}$ satisfy the following commutation relations:

• (1) $L_{i+1}p_{i+\frac {1}{2}}=p_{i+\frac {1}{2}}L_{i+1}=p_{i+\frac {1}{2}}L_{i}p_{i+1}p_{i+\frac {1}{2}}$, for i=1,2,… .

• (2) L _i+1 p _i =p _i L _i+1, for i=1,2,… .

• (3) $L_{i+1}p_{i-\frac {1}{2}}=p_{i-\frac {1}{2}}L_{i+1}$, for i=2,3,… .

• (4) L _i+1 s _i−1=s _i−1 L _i+1, for i=2,3,… .

• (5) L _i+1 p _i−1=p _i−1 L _i+1, for i=2,3… .

• (6) $L_{i+\frac {1}{2}}p_{i}=p_{i}L_{i+\frac {1}{2}}$, for i=1,2,… .

• (7) $L_{i+\frac {1}{2}}p_{i-\frac {1}{2}}=p_{i-\frac {1}{2}}L_{i+\frac {1}{2}}$, for i=2,3,… .

• (8) $L_{i+\frac {1}{2}}s_{i-1}=s_{i-1}L_{i+\frac {1}{2}}$, for i=2,3,… .

• (9) $L_{i+\frac {1}{2}}p_{i-1}=p_{i-1}L_{i+\frac {1}{2}}$, for i=2,3… .

Proof

(1) Using Proposition 3.2,

as required.

(2) From Proposition 3.2, we obtain $\sigma_{i+1}p_{i}=\sigma_{i+1}p_{i}p_{i+\frac {1}{2}}p_{i}=s_{i}L_{i}p_{i+\frac {1}{2}}p_{i}$. Thus

and the statement now follows from the fact that the right hand side of the above expression is fixed under the ∗ anti-involution on $\mathcal {A}_{i+1}(z)$.

(3) We first show that

• (i) $p_{i+\frac {1}{2}}L_{i}s_{i}p_{i-\frac {1}{2}}=p_{i+\frac {1}{2}}\sigma_{i}s_{i}$,

• (ii) $s_{i}L_{i}p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}=s_{i}p_{i-\frac {1}{2}}L_{i-1}p_{i}p_{i-\frac {1}{2}}p_{i+\frac {1}{2}}$,

• (iii) $p_{i+\frac {1}{2}}L_{i}p_{i+1}p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}= p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}p_{i}p_{i-\frac {1}{2}}p_{i+\frac {1}{2}}$,

• (iv) $\sigma_{i+1}p_{i-\frac {1}{2}}=p_{i-\frac {1}{2}}\sigma_{i+1}$.

(i) The definition (3.1) gives

where the last equality follows from Proposition 3.2.

(ii) The definition (3.1) gives

since $p_{i-\frac {1}{2}}L_{i-1}p_{i-\frac {1}{2}}=p_{i-\frac {1}{2}}$.

(iii) The definition (3.1) gives

(iv) was demonstrated in Theorem 3.6.

Now, using

we obtain

Since the right hand side of the last expression is fixed under the ∗ anti-involution on $\mathcal{A}_{i+1}(z)$, the proof of (3) is complete.

(4) We show that after substituting the expression

into the definition (3.1), conjugation by s _i−1 permutes the summands of L _i+1 as follows:

• (i) $s_{i-1}p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}s_{i-1}s_{i}s_{i-1} =p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}s_{i-1}s_{i}$,

• (ii) $s_{i-1}p_{i+\frac {1}{2}}s_{i-1}L_{i-1}p_{i-\frac {1}{2}}s_{i}s_{i-1} =s_{i}p_{i-\frac {1}{2}}L_{i-1}s_{i-1}p_{i+\frac {1}{2}}$,

• (iii) $s_{i-1}p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}p_{i}p_{i-\frac {1}{2}}s_{i}s_{i-1} =p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}s_{i-1}p_{i+1}p_{i+\frac {1}{2}}$,

• (iv) $s_{i-1}p_{i+\frac {1}{2}}s_{i-1}L_{i-1}s_{i-1}s_{i}s_{i-1} =s_{i}p_{i-\frac {1}{2}}L_{i-1}s_{i-1}s_{i}$,

• (v) $s_{i-1}p_{i+\frac {1}{2}}\sigma_{i}s_{i}s_{i-1} =s_{i}p_{i-\frac {1}{2}}L_{i-1}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}}$,

• (vi) $s_{i-1}p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}p_{i}p_{i-\frac {1}{2}}p_{i+1}p_{i+\frac {1}{2}}s_{i-1} =p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}p_{i}p_{i-\frac {1}{2}}p_{i+1}p_{i+\frac {1}{2}}$,

• (vii) $s_{i-1}p_{i+\frac {1}{2}}s_{i-1}L_{i-1}s_{i-1}p_{i+1}p_{i+\frac {1}{2}}s_{i-1} =s_{i}p_{i-\frac {1}{2}}L_{i-1}p_{i}p_{i-\frac {1}{2}}s_{i}$,

• (viii) $s_{i-1}p_{i+\frac {1}{2}}\sigma_{i}p_{i+1}p_{i+\frac {1}{2}}s_{i-1} =s_{i}p_{i-\frac {1}{2}}L_{i-1}s_{i}p_{i-\frac {1}{2}}s_{i}$,

• (ix) s _i−1 s _i s _i−1 L _i−1 s _i−1 s _i s _i−1=s _i s _i−1 L _i−1 s _i−1 s _i ,

• (x) $s_{i-1}p_{i-\frac {1}{2}}L_{i-1}s_{i}p_{i-\frac {1}{2}}s_{i-1} =p_{i-\frac {1}{2}}L_{i-1}s_{i}p_{i-\frac {1}{2}}$.

For item (i),

For item (ii),

For item (iii),

For item (iv),

For the left hand side of item (v),

and for the right hand side of item (v),

The item (vi) follows from the relation $s_{i-1}p_{i-\frac {1}{2}}=p_{i-\frac {1}{2}}s_{i-1}=p_{i-\frac {1}{2}}$. For item (vii),

For item (viii),

(ix) follows from the Coxeter relations, and (x) from the relation $s_{i-1}p_{i-\frac {1}{2}}=p_{i-\frac {1}{2}}s_{i-1}=p_{i-\frac {1}{2}}$.

(5) By parts (2) and (4), L _i+1 commutes with 〈p _i ,s _i−1〉, and so with p _i−1=s _i−1 p _i s _i−1.

(6) From item (3) of Proposition 3.2, $\sigma_{i+\frac {1}{2}}p_{i}=s_{i}\sigma_{i+1}p_{i}p_{i+\frac {1}{2}}p_{i}=L_{i}p_{i+\frac {1}{2}}p_{i}$, and

Since $s_{i}L_{i-\frac {1}{2}}p_{i+1}s_{i}=s_{i}p_{i+1}L_{i-\frac {1}{2}}s_{i}$, this completes the proof of (6).

(7) We show that

(3.6)

From the definition (3.3),

(3.7)

Using part (1) the first three summands in the right hand side of (3.7) are transformed as

and

Next,

Substituting each of the above into (3.7), and using part (1) of Theorem 3.6 gives (3.6). The right hand side of (3.6) being fixed under the ∗ anti-involution on $\mathcal {A}_{i+\frac {1}{2}}(z)$, the proof of (7) is complete.

(8) We show that, after substituting the expression

into the definition (3.3), conjugation by s _i−1 permutes the summands of $L_{i+\frac {1}{2}}$ as follows:

• (i) $s_{i-1}(p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}s_{i-1})s_{i-1} =p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}p_{i}p_{i-\frac {1}{2}}p_{i}p_{i+\frac {1}{2}}$,

• (ii) $s_{i-1}p_{i+\frac {1}{2}}s_{i-1}L_{i-1}p_{i-\frac {1}{2}}s_{i-1} =s_{i}p_{i-\frac {1}{2}}L_{i-1}s_{i}p_{i-\frac {1}{2}}$,

• (iii) $s_{i-1}p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}p_{i}p_{i-\frac {1}{2}}s_{i-1} =p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}p_{i}p_{i-\frac {1}{2}}$,

• (iv) $s_{i-1}(p_{i+\frac {1}{2}}s_{i-1}L_{i-1}s_{i-1})s_{i-1} =s_{i}p_{i-\frac {1}{2}}L_{i-1}s_{i}$,

• (v) $s_{i-1}p_{i+\frac {1}{2}}\sigma_{i}s_{i-1}=\sigma_{i}p_{i+\frac {1}{2}}$,

• (vi) $s_{i-1}p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}s_{i-1}p_{i}p_{i+\frac {1}{2}}s_{i-1} =p_{i+\frac {1}{2}}p_{i-\frac {1}{2}}L_{i-1}s_{i-1}p_{i}p_{i+\frac {1}{2}}$,

• (vii) $s_{i-1}p_{i+\frac {1}{2}}s_{i-1}L_{i-1}p_{i-\frac {1}{2}}p_{i}p_{i+\frac {1}{2}}s_{i-1} =s_{i}p_{i-\frac {1}{2}}p_{i}p_{i+\frac {1}{2}}s_{i-1}L_{i-1}p_{i-\frac {1}{2}}s_{i}$,

• (viii) $s_{i-1}p_{i+\frac {1}{2}}s_{i-1}L_{i-1}s_{i-1}p_{i}p_{i+\frac {1}{2}}s_{i-1} =s_{i}p_{i-\frac {1}{2}}L_{i-1}p_{i-1}p_{i-\frac {1}{2}}s_{i}$,

• (ix) $s_{i-1}p_{i-\frac {1}{2}}L_{i-1}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}}s_{i-1} =p_{i-\frac {1}{2}}L_{i-1}s_{i-1}p_{i+\frac {1}{2}}p_{i}p_{i-\frac {1}{2}}$.

The left hand side of (i) is

and the right hand side of (i) is

The left hand side of (ii) is

which is the same as the right hand side of (ii). The left hand side of (iii) is

which is the same as the right hand side of (iii). The left hand side of (iv) is

which is the same as the right hand side of (iv). The left hand side of (v) is

which is the same as the right hand side of (v). The left hand side of (vi) is

which is the same as the right hand side of (vi). The left hand side of (vii) is

which is the same as the right hand side of (vii). The left hand side of (viii) is

which is the same as the right hand side of (viii). Since the statement (ix) is evident from the relation $s_{i-1}p_{i-\frac {1}{2}}=p_{i-\frac {1}{2}}s_{i-1}=p_{i-\frac {1}{2}}$, the proof of part (8) is complete.

(9) By parts (6) and (8), $L_{i+\frac {1}{2}}$ commutes with 〈p _i ,s _i−1〉, and so with p _i−1=s _i−1 p _i s _i−1. □

Theorem 3.8

If i=1,2,… , then

• (1) L _i+1 commutes with $\mathcal{A}_{i+\frac {1}{2}}(z)$, and σ _i+1 commutes with $\mathcal{A}_{i-\frac {1}{2}}(z)$,

• (2) $L_{i+\frac {1}{2}}$ commutes with $\mathcal{A}_{i}(z)$, and $\sigma_{i+\frac {1}{2}}$ commutes with $\mathcal{A}_{i-1}(z)$.

Consequently, the family of elements $(L_{i},L_{i+\frac {1}{2}}\,:\,i=0,1,\ldots )$ is pairwise commutative.

Proof

(1) Observe that L ₁ commutes with $\mathcal{A}_{1-\frac {1}{2}}(z)$ and L ₂ commutes with $\mathcal{A}_{1+\frac {1}{2}}(z)$, while σ ₂ commutes with $\mathcal{A}_{1-\frac {1}{2}}(z)$ and σ ₃ commutes with $\mathcal{A}_{2-\frac {1}{2}}(z)$. Since

it suffices to show that

• (i) L _i+1 commutes with $\mathcal{A}_{i-\frac {3}{2}}(z)$, if i⩾2, and

• (ii) σ _i+1 commutes with $\mathcal{A}_{i-\frac{5}{2}}(z)$, if i⩾3.

If i⩾2, then induction on i shows that L _i+1 commutes with $\mathcal{A}_{i-2}(z)$, while, if i⩾3, the fact that L _i+1 commutes with $p_{i-\frac {3}{2}}$ follows from induction on i, and the fact that σ _i+1 commutes with $p_{i-\frac {3}{2}}$. Similarly, if i⩾3, then induction on i shows that σ _i+1 commutes with $\mathcal{A}_{i-3}(z)$, and that, if i⩾4, then σ _i+1 commutes with $p_{i-\frac{5}{2}}$.

(2) Observe that $L_{0+\frac {1}{2}}$ commutes with $\mathcal {A}_{0}(z)$ and $L_{1+\frac {1}{2}}$ commutes with $\mathcal{A}_{1}(z)$, while $\sigma_{1+\frac {1}{2}}$ commutes with $\mathcal{A}_{0}(z)$ and $\sigma_{2+\frac {1}{2}}$ commutes with $\mathcal{A}_{1}(z)$. Since

it suffices to show that

• (i) $L_{i+\frac {1}{2}}$ commutes with $\mathcal{A}_{i-\frac {3}{2}}(z)$, if i⩾2, and

• (ii) $\sigma_{i+\frac {1}{2}}$ commutes with $\mathcal{A}_{i-\frac {5}{2}}(z)$, if i⩾3.

If i⩾2, then induction on i shows that $L_{i+\frac {1}{2}}$ commutes with $\mathcal{A}_{i-2}(z)$, while, if i⩾3, the fact that $L_{i+\frac {1}{2}}$ commutes with $p_{i-\frac {3}{2}}$ follows from induction on i, and the fact that $\sigma_{i+\frac {1}{2}}$ commutes with $p_{i-\frac {3}{2}}$. Similarly, if i⩾3, then induction on i shows that $\sigma_{i+\frac {1}{2}}$ commutes with $\mathcal{A}_{i-3}(z)$, and that, if i⩾4, then $\sigma_{i+\frac {1}{2}}$ commutes with $p_{i-\frac{5}{2}}$. □

Proposition 3.9

For i=1,2,… , the following statements hold:

• (1) $(L_{i+\frac {1}{2}}+L_{i+1})p_{i+1}=p_{i+1}(L_{i+\frac {1}{2}}+L_{i+1})=z p_{i+1}$,

• (2) $(L_{i}+L_{i+\frac {1}{2}})p_{i+\frac {1}{2}}=p_{i+\frac {1}{2}}(L_{i}+L_{i+\frac {1}{2}})=z p_{i+\frac {1}{2}}$,

• (3) $(L_{i-\frac {1}{2}}+L_{i}+L_{i+\frac {1}{2}}+L_{i+1})s_{i}=s_{i}(L_{i-\frac {1}{2}}+L_{i}+L_{i+\frac {1}{2}}+L_{i+1})$.

Proof

(1) It suffices to observe that

• (i) $L_{i}p_{i+\frac {1}{2}}p_{i+1}=\sigma_{i+1}p_{i+1}$,

• (ii) $p_{i+\frac {1}{2}}L_{i}p_{i+1}=p_{i+\frac {1}{2}}L_{i}p_{i+1}p_{i+\frac {1}{2}}p_{i+1}$,

• (iii) $p_{i+\frac {1}{2}}L_{i}p_{i}p_{i+\frac {1}{2}}p_{i+1}=p_{i+\frac {1}{2}}L_{i}s_{i}p_{i+1}$,

• (iv) $s_{i}L_{i-\frac {1}{2}}s_{i}p_{i+1}+s_{i}L_{i}s_{i}p_{i+1}=z p_{i+1}$,

• (v) $\sigma_{i+\frac {1}{2}}p_{i+1}=s_{i}L_{i}p_{i+\frac {1}{2}}p_{i+1}$.

With the exception of (iv), each of the statements above is evident from the defining relations or from what we have already shown. Since $(L_{\frac {1}{2}}+L_{1})p_{1}=zp_{1}$,

(3.8)

gives (iv) by induction.

(2) Using the expression (3.8), we have

which yields

(3) Since $\sigma_{i+1}s_{i}=\sigma_{i+\frac {1}{2}}=s_{i}\sigma_{i+1}$, we have

as required. □

Theorem 3.10

If k=0,1,… , then

• (1) the element $z_{k+\frac {1}{2}}=L_{\frac {1}{2}}+L_{1}+L_{1+\frac {1}{2}}+\cdots +L_{k+\frac {1}{2}}$ is central in $\mathcal{A}_{k+\frac {1}{2}}(z)$,

• (2) the element $z_{k+1}=L_{\frac {1}{2}}+L_{1}+L_{1+\frac {1}{2}}+\cdots+L_{k+1}$ is central in $\mathcal{A}_{k+1}(z)$.

Proof

(1) We first show that $z_{k+\frac {1}{2}}$ commutes with p ₁,…,p _k and $p_{1+\frac {1}{2}},\ldots,p_{k+\frac {1}{2}}$. If i=1,…,k, then p _i commutes with z _i−1 and with $L_{i+\frac {1}{2}}+L_{i+1}+\cdots+L_{k+\frac {1}{2}}$. Since p _i also commutes with $L_{i-\frac {1}{2}}+L_{i}$, it follows that p _i commutes with $z_{k+\frac {1}{2}}$. Similarly, since $p_{i+\frac {1}{2}}$ commutes with $z_{i-\frac {1}{2}}$, $L_{i}+L_{i+\frac {1}{2}}$, and with $L_{i+1}+L_{i+\frac {3}{2}}+\cdots + L_{k+\frac {1}{2}}$, it follows that $p_{i+\frac {1}{2}}$ commutes with $z_{k+\frac {1}{2}}$. Next, suppose that i=1,…,k−1. Since s _i commutes with z _i−1, $L_{i-\frac {1}{2}}+L_{i}+L_{i+\frac {1}{2}}+L_{i+1}$, and with $L_{i+\frac {3}{2}}+L_{i+2}+\cdots+L_{k+\frac {1}{2}}$, it is evident that s _i commutes with $z_{k+\frac {1}{2}}$. As $\mathcal{A}_{k+\frac {1}{2}}(z)$ is generated by p ₁,…,p _k , $p_{1+\frac {1}{2}},\ldots,p_{k+\frac {1}{2}}$, and s ₁,…,s _k−1, we conclude that $z_{k+\frac {1}{2}}$ is central in $\mathcal{A}_{k+\frac {1}{2}}(z)$.

(2) Given that $z_{k+\frac {1}{2}}$ is central in $\mathcal {A}_{k+\frac {1}{2}}(z)$, it suffices to observe that s _k commutes with z _k+1. □

Proposition 3.11

For i=1,2,… , the following statements hold:

• (1) $(L_{i}+L_{i+\frac {1}{2}}+L_{i+1})\sigma_{i+1}=\sigma_{i+1}(L_{i}+L_{i+\frac {1}{2}}+L_{i+1})$,

• (2) $(L_{i-\frac {1}{2}}+L_{i}+L_{i+\frac {1}{2}})\sigma_{i+\frac {1}{2}}=\sigma_{i+\frac {1}{2}}(L_{i-\frac {1}{2}}+L_{i}+L_{i+\frac {1}{2}})$.

Proof

(1) Since σ _i+1 commutes with

and σ _i+1 also commutes with $z_{i-1}+L_{i-\frac {1}{2}}\in\mathcal {A}_{i-\frac {1}{2}}(z)$, it follows that σ _i+1 commutes with $L_{i}+L_{i+\frac {1}{2}}+L_{i+1}$.

(2) Given that $\sigma_{1+\frac {1}{2}}=1$, we may suppose that i=2,3,… . Since $\sigma_{i+\frac {1}{2}}$ commutes with

and $\sigma_{i+\frac {1}{2}}$ also commutes with $z_{i-\frac {3}{2}}+L_{i-1}\in \mathcal{A}_{i-1}(z)$, it follows that $\sigma_{i+\frac {1}{2}}$ commutes with $L_{i-\frac {1}{2}}+L_{i}+L_{i+\frac {1}{2}}$. □

4 A presentation for partition algebras

In this section, we rewrite the presentation for $\mathcal{A}_{k}(z)$ given by [3] in Theorem 2.1 in terms of the elements $\sigma_{i},\sigma_{i+\frac {1}{2}},p_{i},p_{i+\frac {1}{2}}$.

Theorem 4.1

If k=1,2,… , then $\mathcal{A}_{k}(z)$ is the unital associative algebra presented by the generators

and the relations:

• (1) (Involutions)

  • (a) $\sigma_{i+\frac {1}{2}}^{2}=1$, for i=2,…,k−1.

  • (b) $\sigma_{i+1}^{2}=1$, for i=1,…,k−1.

• (2) (Braid-like relations)

  • (a) $\sigma_{i+1}\sigma_{j+\frac {1}{2}}=\sigma_{j+\frac {1}{2}}\sigma_{i+1}$, if j≠i+1.

  • (b) σ _i σ _j =σ _j σ _i , if j≠i+1.

  • (c) $\sigma_{i+\frac {1}{2}}\sigma_{j+\frac {1}{2}}=\sigma_{j+\frac {1}{2}}\sigma_{i+\frac {1}{2}}$, if j≠i+1.

  • (d) s _i s _i+1 s _i =s _i+1 s _i s _i+1, for i=1,…,k−2, where

    

    are the Coxeter generators for $\mathfrak{S}_{k}$.

• (3) (Idempotent relations)

  • (a) $p_{i}^{2}=zp_{i}$, for i=1,…,k.

  • (b) $p_{i+\frac {1}{2}}^{2}=p_{i+\frac {1}{2}}^{2}$, for i=1,…,k−1.

  • (c) $\sigma_{i+1}p_{i+\frac {1}{2}}=p_{i+\frac {1}{2}}\sigma_{i+1}=p_{i+\frac {1}{2}}$, for i=1,…,k−1.

  • (d) $\sigma_{i+\frac {1}{2}}p_{i+\frac {1}{2}}=p_{i+\frac {1}{2}}\sigma_{i+\frac {1}{2}}=p_{i+\frac {1}{2}}$, for i=1,…,k−1.

  • (e) $\sigma_{i+\frac {1}{2}}p_{i}p_{i+1}=\sigma_{i+1}p_{i}p_{i+1}$, for i=1,…,k−1.

  • (f) $p_{i}p_{i+1}\sigma_{i+\frac {1}{2}}=p_{i}p_{i+1}\sigma_{i+1}$, for i=1,…,k−1.

• (4) (Commutation relations)

  • (a) p _i p _j =p _j p _i , for i,j=1,…,k.

  • (b) $p_{i+\frac {1}{2}}p_{j+\frac {1}{2}}=p_{j+\frac {1}{2}}p_{i+\frac {1}{2}}$, for i,j=1,…,k−1.

  • (c) $p_{i+\frac {1}{2}}p_{j}=p_{j}p_{i+\frac {1}{2}}$, for j≠i,i+1.

  • (d) σ _i p _j =p _j σ _i if j≠i−1,i.

  • (e) $\sigma_{i}p_{j+\frac {1}{2}}=p_{j+\frac {1}{2}}\sigma_{i}$, if j≠i.

  • (f) $\sigma_{i+\frac {1}{2}}p_{j}=p_{j}\sigma_{i+\frac {1}{2}}$, if j≠i,i+1.

  • (g) $\sigma_{i+\frac {1}{2}}p_{j+\frac {1}{2}}=p_{j+\frac {1}{2}}\sigma_{i+\frac {1}{2}}$, if j≠i−1.

  • (h) $\sigma_{i+\frac {1}{2}}p_{i}\sigma_{i+\frac {1}{2}}=\sigma_{i+1}p_{i+1}\sigma_{i+1}$, for i=1,…,k−1.

  • (i) $\sigma_{i+\frac {1}{2}}p_{i-\frac {1}{2}}\sigma_{i+\frac {1}{2}}=\sigma_{i}p_{i+\frac {1}{2}}\sigma_{i}$, for i=2,…,k−1.

• (5) (Contraction relations)

  • (a) $p_{i+\frac {1}{2}}p_{j}p_{i+\frac {1}{2}}=p_{i+\frac {1}{2}}$, for j=i,i+1.

  • (b) $p_{i}p_{j-\frac {1}{2}}p_{i}=p_{i}$, for j=i,i+1.

Proof

We first show that the relations given in the statement above are a consequence the presentation given by [3] in Theorem 2.1.

(1a), (1b) follow from Proposition 4.2.

(2a) To see that σ _i+1 and $\sigma_{i+\frac {1}{2}}$ commute, we use Propositions 3.4 and 4.2 to obtain

(4.1)

where s _i is a Coxeter generator for $\mathfrak{S}_{k}\subset\mathcal {A}_{k}(z)$. Theorem 3.8 shows that σ _j commutes with $\mathcal{A}_{j-\frac {3}{2}}(z)$, and hence that σ _j commutes with $\sigma_{1+\frac {1}{2}},\ldots,\sigma_{j-\frac {3}{2}}$. From Theorem 3.8, $\sigma_{i+\frac {1}{2}}$ commutes with $\mathcal{A}_{i-1}(z)$, and hence with σ ₂,…,σ _i−1. In general, however, σ _i and $\sigma_{i+\frac {1}{2}}$ do not commute.

(2b), (2c) Theorem 3.8 shows that σ _i+1 commutes with σ ₂,…,σ _i−1 and that $\sigma_{i+\frac {1}{2}}$ commutes with $\sigma_{1+\frac {1}{2}},\ldots,\sigma_{i-\frac {3}{2}}$.

(2d) Follows from (4.1), whereby for j=1,…,k−1, each product $\sigma_{j+\frac {1}{2}}\sigma_{j+1}=s_{j}$ is a Coxeter generator for $\mathfrak{S}_{k}\subset\mathcal{A}_{k}(z)$.

(3a), (3b) Are included in the set of relations given by [3].

(3c), (3d) That $\sigma_{i+1}p_{i+\frac {1}{2}}=p_{i+\frac {1}{2}}\sigma_{i+1}=p_{i+\frac {1}{2}}$ is given in Proposition 3.2. Proposition 3.4 shows that $\sigma_{i+\frac {1}{2}}p_{i+\frac {1}{2}}=s_{i}\sigma_{i+1}p_{i+\frac {1}{2}}=s_{i}p_{i+\frac {1}{2}}=p_{i+\frac {1}{2}}$.

(3e), (3f) Propositions 3.4 and 4.2 show that

(4a)–(4c) are included in the set of relations given by [3].

(4d), (4e) By Theorem 3.8, σ _i commutes with $\mathcal{A}_{i-\frac {3}{2}}(z)$, and hence with p ₁,…p _i−2 and with $p_{1+\frac {1}{2}},\ldots,p_{i-\frac {3}{2}}$. Proposition 3.2 shows that σ _i commutes with $p_{i-\frac {1}{2}}$.

(4f), (4g) By Theorem 3.8, $\sigma_{i+\frac {1}{2}}$ commutes with $\mathcal{A}_{i-1}(z)$, and hence with p ₁,…p _i−1 and with $p_{1+\frac {1}{2}},\ldots,p_{i-\frac {3}{2}}$. From (3d), it follows that $\sigma_{i+\frac {1}{2}}$ commutes with $p_{i+\frac {1}{2}}$.

(4h) Propositions 3.4 and 4.2 show that

(4i) Proposition 3.2 shows that $p_{i-\frac {1}{2}}s_{i}\sigma_{i}=s_{i}\sigma_{i}p_{i+\frac {1}{2}}$. Propositions 3.4 and 4.2, together with the fact that σ _i+1 commutes with $p_{i-\frac {1}{2}}$ give

Multiplying both sides of the last expression by $\sigma_{i}\sigma_{i+\frac {1}{2}}$ on the left, and using Proposition 4.2 once more shows that

as required.

(5a), (5b) are included in the set of relations given by [3].

Next, we derive the relations given by [3] in Theorem 2.1 from the relations (1a)–(5b) above.

(1i) By the relations (1a), (1b) and (2a),

Thus, writing $s_{i}=\sigma_{i+\frac {1}{2}}\sigma_{i+1}$, for i=1,…,k−1, we recover (1i).

(1ii) If j≠i+1, then, by (2b) and (2c),

as required.

(1iii) is equivalent to (2d) with $s_{i}=\sigma_{i+\frac {1}{2}}\sigma_{i+1}$, for i=1,…,k−1.

(2i), (2ii) are identical to the relations (3a) and (3b).

(2iii) With $s_{i}=\sigma_{i+1}\sigma_{i+\frac {1}{2}}$, the relations (3c) and (3d) give

and

as required.

(2iv) With $s_{i}=\sigma_{i+1}\sigma_{i+\frac {1}{2}}$, the relations (2a), (3e) and (3f) give

as required.

(3i)–(3iii) are identical to the relations (4a)–(4c).

(3iv) If j≠i,i+1, then the relations 4d and 4f give

as required.

(3v) If j≠i−1,i+1, then the relations (4e) and (4f) give

as required.

(3vi) From the relations (1b) and (4h),

as required.

(3vii) From the relations (4e), (4g) and (1a), (1b),

as required.

(4i), (4ii) are identical to the relations (5a) and (5b). □

Proposition 4.2

If i=0,1,… , then $(\sigma_{i+\frac {1}{2}})^{2}=1$ and (σ _i+1)²=1.

Proof

Given that $\sigma_{i+\frac {1}{2}}s_{i}=s_{i}\sigma_{i+\frac {1}{2}}=\sigma_{i+1}$, we obtain

It therefore suffices to show that $\sigma_{i+1}^{2}=1$. By definition σ ₁=1, so we proceed by induction. After taking the square of the right hand side of the definition (3.2), the proposition will follow from the relations

(4.2)

(4.3)

(4.4)

(4.5)

(4.6)

(4.7)

(4.8)

(4.9)

(4.10)

(4.11)

(4.12)

(4.13)

(4.14)

From the left hand side of (4.2), using the fact that $p_{i-\frac {1}{2}}L_{i-1}p_{i-\frac {1}{2}}=p_{i-\frac {1}{2}}$, together with $p_{i+\frac {1}{2}}s_{i-1}p_{i+\frac {1}{2}}=p_{i-\frac {1}{2}}p_{i+\frac {1}{2}}$, we obtain

Similarly, from the right hand side of (4.2), we obtain

which demonstrates (4.2). Now consider the left hand side of (4.3)

From the right hand side of (4.3), we obtain

Now,

(4.15)

which completes the proof of (4.3). From the left hand side of (4.4),

The right hand side of (4.4) gives

which demonstrates (4.4). The left hand side of (4.5) gives

while the right hand side of (4.5) gives

which demonstrates (4.5). The statement (4.6) is equivalent to (4.2) which has already been verified. The right hand side of (4.7) leads to

while

as required. Since the statement (4.8) is equivalent to (4.7), we consider (4.9). Using the relation $p_{i-\frac {1}{2}}s_{i}p_{i-\frac {1}{2}}=p_{i-\frac {1}{2}}p_{i+\frac {1}{2}}$,

On the other hand,

The left hand side of (4.10) is given by (4.15), and the right hand side by

as required. Considering the left hand side of (4.11),

which, by (4.15), is equal to the right hand side of (4.11). Since (4.9) is equivalent to (4.13), while (4.14) is equivalent to (4.4), the proof of the proposition is complete. □

We record for later reference further consequences of the presentation given in Theorem 2.1.

Proposition 4.3

For i=1,2,… , the following statements hold:

• (1) p _i+1 σ _i+1 p _i+1=L _i p _i+1,

• (2) $p_{i+1}\sigma_{i+\frac {1}{2}}p_{i+1}=(z-L_{i-\frac {1}{2}})p_{i+1}$,

• (3) $p_{i+\frac {3}{2}}\sigma_{i+1}p_{i+\frac {3}{2}}=p_{i+\frac {1}{2}}p_{i+\frac {3}{2}}$.

Proof

(1) From Proposition 3.2, we obtain $p_{i+\frac {1}{2}}p_{i+1}\sigma_{i+1}=p_{i+\frac {1}{2}}L_{i}$. Thus

as required.

(2) We first compute

Now observe that $p_{2}\sigma_{1+\frac {1}{2}}p_{2}=(z-L_{\frac {1}{2}})p_{2}$ and hence, by induction,

(3) Observe that $p_{2+\frac {1}{2}}\sigma_{2}p_{2+\frac {1}{2}}=p_{2+\frac {1}{2}}s_{1}p_{2+\frac {1}{2}}=p_{2+\frac {1}{2}}p_{1+\frac {1}{2}}$ and, by induction,

Therefore, using the definition (3.2) and the fact that $p_{i-\frac {1}{2}}L_{i-1}p_{i-\frac {1}{2}}=p_{i-\frac {1}{2}}$,

as required. □

The next statement gives an alternative recursion for the family $(L_{i+\frac {1}{2}}:i=0,1\ldots)$ for use in Sect. 5.

Theorem 4.4

If i=1,2,… , then

Proof

By Propositions 3.2, Proposition 3.4 and Proposition 4.3,

Substituting the above expression into the definition of $L_{i+\frac {1}{2}}$ given in (3.3) provides the required statement. □

5 Schur–Weyl duality

In this section we use Schur–Weyl duality to show that the family $(L_{i+\frac {1}{2}},L_{i+1}:i=1,2,\ldots)$, defined above, and the Jucys–Murphy elements given by Halverson and Ram [3] are in fact equal.

Let n=1,2,… , and V be a vector space over ℂ with basis v ₁,…,v _n . If r=1,2,… , the tensor product

and is equipped, via the inclusion $\mathfrak{S}_{n}\subset GL_{n}(\mathbb {C})$, with the diagonal $\mathfrak{S}_{n}$-action

Let $A_{r}(n)=\mathcal{A}_{r}(z)\otimes_{\mathbb{Z}[z]}\mathbb{C}$, where z acts on ℂ as multiplication by n. The action of A _r (n) on V ^⊗r is given (Sect. 3 of [3]) by

and for k=1,…,r−1,

and for k=1,…,r,

The ${A}_{r+\frac {1}{2}}(n)$-action on V ^⊗r is obtained in Sect. 3 of [3] from the action of A _r+1(n) on V ^⊗r+1 by restricting to the subspace V ^⊗r⊗v _n and identifying V ^⊗r with V ^⊗r⊗v _n . The next statement asserts that $\mathfrak{S}_{n}$ and A _r (n) act as commuting operators on V ^⊗r.

Theorem 5.1

(Theorem 3.22 of [3])

Let n,r∈ℤ_⩾0. Let $S_{n}^{\lambda}$ denote the irreducible $\mathfrak{S}_{n}$-module indexed by λ.

• (1) As $(\mathbb{C}\mathfrak{S}_{n},A_{r}(n))$-bimodules

  

  where $\lambda\in\hat{A}_{r}(n)$ is an indexing set for the irreducible A _r (n)-modules, and the vector spaces ${A}_{r}^{\lambda}(n)$, for $\lambda\in\hat{A}_{r}(n)$, are irreducible A _r (n)-modules.

• (2) As $(\mathbb{C}\mathfrak{S}_{n-1},A_{r+\frac {1}{2}}(n))$-bimodules

  

  where $\lambda\in\hat{A}_{r+\frac {1}{2}}(n)$ is an indexing set for the irreducible ${A}_{r+\frac {1}{2}}(n)$-modules, and the vector spaces ${A}_{r+\frac {1}{2}}^{\lambda}(n)$, for $\lambda\in\hat{A}_{r+\frac {1}{2}}(n)$, are irreducible ${A}_{r+\frac {1}{2}}(n)$-modules.

By Theorem 3.6 of [3], the homomorphism $A_{r}(n)\to \operatorname{End}_{\mathfrak{S}_{n}}(V^{\otimes r})$ in Theorem 5.1 is an isomorphism whenever n⩾2r.

If 1⩽i,j⩽n, let $s_{i,j}\in\mathfrak{S}_{n}$ denote the transposition which interchanges i and j. The next statement gives the action of the group $\langle\sigma_{i+\frac {1}{2}},\sigma_{i+1} \,|\, i=1,\ldots,r-1\rangle$ on V ^⊗r.

Proposition 5.2

If k=1,2,… , and $v_{i_{1}}\otimes\cdots\otimes v_{i_{k+1}}\in V^{\otimes k+1}$, then

(5.1)

and

(5.2)

Proof

The proposition is true when k=1. If k=2,3,… , observe that the linear endomorphisms defined, for $v_{i_{1}}\otimes\cdots\otimes v_{i_{k+1}}\in V^{\otimes k+1}$, by

and

commute with the diagonal action of $\mathfrak{S}_{n}$ on V ^⊗k+1. Thus, by Theorem 5.1, $\theta_{k+\frac {1}{2}}$ and θ _k+1, lie in the image of the map $A_{k+1}(n)\mapsto \operatorname{End}_{\mathfrak {S}_{n}}(V^{\otimes k+1})$. Observe also that the action of $\theta_{k+\frac {1}{2}}$ on V ^⊗k+1 commutes with the action of A _k−1(n) on V ^⊗k+1, and the action of θ _k+1 on V ^⊗k+1 commutes with the action of $A_{k-\frac {1}{2}}(n)$ on V ^⊗k+1. Since A _k+1(n) is generated by $A_{k-\frac {1}{2}}(n)$ together with $\langle\sigma_{k+\frac {1}{2}},\sigma_{k+1},p_{k},p_{k+\frac {1}{2}}\rangle$, to show that the map $A_{k+1}(n)\to \operatorname{End}_{\mathfrak {S}_{n}}(V^{\otimes k+1})$ sends

it now suffices to verify that, as operators on V ^⊗k+1,

• (i) $\theta_{k+\frac {1}{2}}^{2}=\theta_{k+1}^{2}=1$,

• (ii) $\theta_{k+\frac {1}{2}}\theta_{k+1}=\theta_{k+1}\theta_{k+\frac {1}{2}}=s_{k}$,

• (iii) $\theta_{k+\frac {1}{2}}p_{k+\frac {1}{2}}=p_{k+\frac {1}{2}}\theta_{k+\frac {1}{2}}=p_{k+\frac {1}{2}}$,

• (iv) $\theta_{k+\frac {1}{2}}p_{k}p_{k+1}=\theta_{k+1}p_{k}p_{k+1}$,

• (v) $p_{k}p_{k+1}\theta_{k+\frac {1}{2}}=p_{k}p_{k+1}\theta_{k+1}$,

• (vi) $\theta_{k+\frac {1}{2}}p_{k}\theta_{k+\frac {1}{2}}=\theta_{k+1}p_{k+1}\theta_{k+1}$,

• (vii) $\theta_{k+\frac {1}{2}}p_{k-\frac {1}{2}}\theta_{k+\frac {1}{2}}=\sigma_{k}p_{k+\frac {1}{2}}\sigma_{k}$,

where s _k in item (ii) acts on V ^⊗k by place permutation. Since each of (i)–(vii) can be verified by inspection, the proof of the proposition is complete. □

Proposition 5.3

If k=1,2,… , and $v_{i_{1}}\otimes \cdots\otimes v_{i_{k}}\in V^{\otimes k}$ then

(5.3)

and

(5.4)

Proof

Identify V ^⊗k as the subspace $V^{\otimes k}\otimes\sum_{j=1}^{n} v_{j}\subseteq V^{\otimes k+1}$. Then

By Proposition 4.3, as operators on V ^⊗k,

and the statement (5.3) follows. Next,

By Proposition 4.3, as operators on V ^⊗k,

which yields (5.4). □

As in Sect. 3 of [3], let κ _n be the central element which is the class sum corresponding to the conjugacy class of transpositions in $\mathbb{C}\mathfrak{S}_{n}$,

For ℓ=1,…,n, we also define

so that κ _n,n =κ _n−1.

Proposition 5.4

Let n=dim(V) and r∈ℤ_>0. If $z_{r-\frac {1}{2}}\in A_{r-\frac {1}{2}}(n)$ and z _r ∈A _r (n) are the central elements defined by Theorem 3.10, and $v_{i_{1}}\otimes \cdots\otimes v_{i_{r}}\in V^{\otimes r}$, then

(5.5)

and, if r=2,3,… , then

(5.6)

Proof

The proof is by induction on r. Let i=1,…,n. Since z ₁=p ₁,

which verifies (5.5) when r=1. Now observe that if $v_{i_{1}}\otimes\cdots\otimes v_{i_{k}}\in V^{\otimes r}$, then the diagonal action of κ _n and $\kappa_{n,i_{r}}$ on V ^⊗k allows us to write

and

Assuming that (5.5) holds for r=1,…,k−1 , we obtain

which verifies (5.6) for r=k. Next, since (5.6) holds for r=2,3,…,k , we obtain

which verifies (5.5) for r=k. □

Let Z _k ∈A _k (n) and $Z_{k+\frac {1}{2}}\in A_{k+\frac {1}{2}}(n)$ denote the central element defined by Halverson and Ram in Sect. 3 of [3]. Then the Jucys–Murphy elements of [3] are given by

Theorem 5.5

If k=1,2,… , then $M_{k+\frac {1}{2}}=L_{k+\frac {1}{2}}$ and M _k+1=L _k+1 as elements of A _k+1(n).

Proof

By Theorem 3.6 of [3], the homomorphism $A_{k}(n)\to \operatorname{End}_{\mathfrak{S}_{n}}(V^{\otimes k})$ is an isomorphism whenever n⩾2k. Since the coefficients in the expansions of $Z_{k},Z_{k+\frac {1}{2}}$ and $z_{k},z_{k+\frac {1}{2}}$, in terms of the diagram basis for A _k+1(n), are polynomials in n, and the map $A_{k}(n)\to \operatorname{End}_{\mathfrak {S}_{n}}(V^{\otimes k})$ is an isomorphism for infinitely many values of n, to prove the theorem, it suffices to compare the action of z _k and Z _k (resp. $z_{k+\frac {1}{2}}$ and $Z_{k+\frac {1}{2}}$) on V ^⊗k for an arbitrary choice of n. Identifying V ^⊗k with the subspace V ^⊗k⊗v _n ⊆V ^⊗k+1, and κ _n−1 with κ _n,n , then Theorem 3.35 of [3] states that, as operators on V ^⊗k,

(5.7)

Comparing (5.7) with the action of z _k and $z_{k+\frac {1}{2}}$ on V ^⊗k in (5.5) and (5.6) completes the proof. □

Remark 5.6

Whereas $L_{\frac {1}{2}}=0$ and L ₁=p ₁, in [3], the first three Jucys–Murphy elements are $M_{0}=M_{\frac {1}{2}}=1$, and M ₁=p ₁−1. Thus, although z ₁=Z ₁ as elements of A ₁(n), Theorem 5.5 cannot be extended to the case k=0.