Cylinder algebras

Tammo tom Dieck

Preliminary version, February 9, 2009^{∗}

Cylinder algebras generalize algebras of Birman-Wenzl [??] [??] and Murakami [??]. Geometrically, they are related to tangles and ribbons in the cylinder. For generic parameters, we define algebras ℨ_{n}(k) of height k which are quotients of the group algebra of the braid group ZB_{n}. Therefore we call these algebras the algebras with n strings. This article assembles some foundational material for the general investigation of these algebras.

In the first part, we present a thorough discussion of the skein relations. This is based on the algebras with 2 strings.

In the second part, we describe a combinatorial model with the aid of coloured balanced Brauer graphs for certain specializations of the algebras.

1. The algebras of height ∞

We define the cylinder algebras ℨ(∞) = ℨ_{2}(∞) on two strings and of height ∞. The algebras depend on a choice of parameters in the ground ring. We suppress the dependence on the parameters in the notation.

We need some notation. Let 𝔎 be a commutative ring. We fix a family

(1.1) P = (λ, q, A_{j} | j ∈ ZZ)

of parameters in the ring 𝔎. We write A_{j} = a_{j} + ε_{j} where ε_{j} = 1 if j is even and ε_{j} = 0 otherwise. We assume that λ and q are invertible and set δ = q − q^{−1}. We call

(1.2) R_{0} := λ − λ^{−1} + δa_{0} = 0

(1.3)  R_{n} := λa_{n} − λ^{−1}a_{−n} − δ ∑_{j=1}^{n−1} a_{j}a_{j−n} + δε_{n} \frac{n}{2} = 0,  n ≥ 1.

the basic relations between the parameters. We see from (1.3) that the a_{n}, n < 0 can be computed from the remaining parameters. If δ = 0, then a_{j} = a_{−j}.

(1.4) Definition. The algebra ℨ(∞) is the associative unital 𝔎-algebra generated by invertible elements X, Y and another element E with the following relations (1) X − X^{−1} = δ(1 − E)

∗z.tex

(2) EX = λE = XE

(3) XY XY = Y XY X

(4) EXY XY = E = XY XY E

(5) EY^{j}E = A_{j}E, j ∈ ZZ. The parameters are assumed to satisfy the basic relations (1.2) and (1.3). ♡

We make some remarks concerning this definition.

We multiply (1) by X^{2} and λX, use (2), and eliminate E. This gives the relation

(1.5) (X − λ)(X − q)(X + q^{−1}) = 0

as a consequence of (1) and (2).

(1.6) Remark. Suppose the ring 𝔎 carries an involution φ (called conjugation, also denoted x ↦ x^{∗}) such that

φ(λ) = λ^{−1}, φ(q) = q^{−1}, φ(A_{j}) = A_{−j}.

We can certainly define an involution with these properties on the ring

ZZ[λ^{±}, q^{±}, A_{j} | j ∈ ZZ].

We use this to apply φ formally to any integral polynomial in the parameters. The involution φ can be extended to a conjugate-linear involution φ on ℨ(∞) such that

φ(X) = X^{−1}, φ(Y) = Y^{−1}, φ(E) = E.

The algebra always has a 𝔎-linear antiinvolution ι which fixes X, Y, E. ♡

(1.7) Remark. The generic ring for the definition of ℨ(∞) is

ZZ[λ^{±}, q^{±}, A_{j} | j ≥ 0] ≅ ZZ[λ^{±}, q^{±}, A_{j} | j ∈ ZZ]/(R_{0}, R_{1}, …).

Since φ(R_{n}) = −R_{−n}, this ring carries an involution as in (1.6). This involution is complicated in terms of the A_{j}, j ≥ 0 alone. ♡

(1.8) Remark. Suppose δ ∈ 𝔎 is invertible. Then

E = 1 − δ^{−1}(X − X^{−1}).

We see that ℨ(∞) is generated by invertible elements X and Y. The algebra is therefore a quotient of the group algebra of the braid group ZB_{2} = ⟨ X, Y | XY XY = Y XY X ⟩. ♡

The basic relations are justified by the next result.

(1.9) Proposition. Suppose X, Y, E are elements in a 𝔎-algebra which satisfy (1.4.1) − (1.4.5). Suppose the annihilator ideal of E is zero. Then the basic relations hold.

T. tom Dieck 1. The algebras of height ∞ 3

A further justification will be given later when we construct a certain module for ℨ(∞). A proof of (1.9) is based on some consequences of the defining relations (1.4).

(1.10) Proposition. Suppose X, Y, E are elements in a 𝔎-algebra which satisfy (1.4.1) − (1.47.5). Then the following relations hold (n ≥ 1):

(1) EY^{n}X = EY^{n−1}XY^{−1} − δEY^{n−2} + δA_{n−1}EY^{−1}

(2) EY^{−n}X = EY^{−n+1}XY + δEY^{−n} − δA_{−n}E

(3) EY^{n}X = λ^{−1}EY^{−n} − δE ∑_{j=1}^{n−1}(Y^{−n+2j} − A_{n−j}Y^{−j})

(4) EY^{−n}X = λEY^{n} + δE ∑_{j=0}^{n−1}(Y^{−n+2j} − A_{−n+j}Y^{j}). There are analogous relations for left multiplication (formally apply the antiinvolution ι).

Proof. Note that, by (1.4.3) and (1.4.4), XY XY is contained in the center. Relation (1.4.4) yields EY^{n}XY X = EY XY XY^{n−1} = EY^{n−1} and hence the first line in

EY^{n}X = EY^{n−1}X^{−1}Y^{−1}
= EY^{n−1}(X + δ(E − 1))Y^{−1}
= EY^{n−1}XY^{−1} + δA_{n−1}EY^{−1} − δEY^{n−2}.

(2) Relation (1.4.4) yields EY^{−n}X^{−1} = EY^{−n+1}XY. Now eliminate X^{−1} with (1.4.1).

(3) and (4) follow by induction from (1) and (2). □

Proof of (1.9). We multiply (1.4.1) by E and use (1.4.2). This gives (λ − λ^{−1} + δa_{0})E = 0 and hence (1.2). We multiply (1.10) from the right with E and obtain 0 = λA_{n} − λ^{−1}A_{−n} + δ(∑_{j=1}^{n−1} A_{2j−n} − A_{n−j}A_{−j}). A rewriting in terms of the a_{j} gives the result for n ≥ 1. □

(1.11) Corollary. The ideal ⟨ E ⟩ of ℨ(∞) generated by E is spanned over 𝔎 by the elements Y^{i}EY^{j} (i, j ∈ ZZ). The right ideal [E ⟩ generated by E is spanned by the elements EY^{j} (j ∈ ZZ). Similarly for the left ideal ⟨ E] generated by E.□

(1.12) Corollary. Under the hypothesis of (1.10), right multiplication by X and Y XY maps

I(n) = Span{EY^{−n}, EY^{−n+1}, …, EY^{n−1}, EY^{n}}

into itself. □

Suppose the (EY^{j} | j ∈ ZZ) are linearly independent over 𝔎. We can express the right multiplication by X, Y, E in terms of by matrices. We now formally construct this right module by specifying matrices.

Let X̃ denote the following IN × IN-matrix with entries in 𝔎. The first row is

(λ, 0, −δa_{−1}, −δ, −δa_{−2}, 0, −δa_{−3}, −δ, −δa_{−4}, …).

The remaining part of the matrix, except the first row and column, is decomposed into 2 × 2-blocks. As such it is an upper triangular block matrix; beginning with the diagonal term, all rows are equal to Λ, α_{1}, α_{2}, … with

Λ = ( \begin{matrix}0 & λ \\ λ^{−1} & δ \end{matrix} ) , α_{j} = δ ( \begin{matrix}−ε_{j} & −a_{−j} \\ a_{j} & ε_{j} \end{matrix} ) .

Let Ỹ be the shift matrix with 1’s below the diagonal and zeros otherwise. Finally, let Ẽ be the matrix with first row

(A_{0}, A_{1}, A_{−1}, A_{2}, A_{−2}, …).

If we order the EY^{n} according to the exponents (0, 1, −1, 2, −2, 3, −3, …), then X̃, Ỹ, and Ẽ describe the right multiplication by X, Y, and E, respectively. For X, this follows from (1.10); the cases Y and E are clear. Since we work with right ideals, Ỹ X̃ gives the action of XY. The matrix Ỹ X̃ is an upper triangular matrix of 2 × 2-blocks. The rows are, beginning with the diagonal term,

( \begin{matrix}0 & λ^{−1} & δ & δa_{1} & 0 & δa_{2} & δ & δa_{3} & ⋯ \\ λ & 0 & −δa_{−1} & −δ & −δa_{−2} & 0 & −δa_{−3} & −δ & ⋯ \end{matrix} ) .

The square (Ỹ X̃)^{2} is therefore again a matrix of the same shape. The block on the diagonal is the unit matrix. The block j positions to the right is the diagonal matrix with diagonal (δR_{j}, δR_{j}). If the relations R_{j} = 0 hold, then we can compute X̃^{−1} as Ỹ X̃Ỹ. The result is the following matrix. The first row is

(λ^{−1}, δa_{1}, 0, δa_{2}, δ, δa_{3}, 0, …).

The rest is again 2 × 2 upper triangular; the rows are as for X̃, except that the diagonal block is replaced by its inverse. These data and results are used in the following construction.

(1.13) The basic module. Suppose the basic relations (1.2) and (1.3) hold. We construct a right ℨ(∞)-module with basis the symbols (ey^{j} | j ∈ ZZ). We order the ey^{j} as we did for the EY^{j} and let X, Y, and E act through X̃, Ỹ, and Ẽ. We have to verify the relations (1.4). By construction, Y XY X acts as identity. Hence (3) and (4) hold. From statements of the last paragraph we see that (1) holds. Also (2) and (5) are immediate from the definitions. The assignment ey^{j} ↦ EY^{j} is now a ℨ-linear epimorphism onto the right ideal generated by E. We shall see in a moment that this is an isomorphism. ♡

(1.14) Theorem. The algebra ℨ(∞) is additively a free 𝔎-module with basis

Y^{i}XY^{j}, Y^{i}XY^{j}X^{−1}, Y^{i}EY^{j} (i, j ∈ ZZ).

Proof. It is known that the factor algebra by the ideal generated by E is a free 𝔎-module with basis consisting of the first two families in (1.14). This is shown in [??] by a reduction to Ariki-Koike algebras [??]. Another proof in [??] reduces

T. tom Dieck 2. The algebras of height k 5

it to Hecke algebras of affine root systems. We sketch here an independent proof along the lines of Hecke algebra theory (see e. g. [??]). We construct a right module over A = ℨ(∞)/⟨ E ⟩ in the free 𝔎-module with basis y(i, j), z(i, j) (such that the assignement y(i, j) ↦ Y^{i}XY^{j}, z(i, j) ↦ Y^{i}XY^{j}X^{−1} is a morphism of 𝔄-modules. We have to define the action of X and Y on the basis elements and verify the relations X − X^{−1} = δ and XY XY = Y XY X. We define

z(i, j)X = y(i, j)
z(i, j)Y = z(i + 1, j) + δy(i + j, 1) − δy(i, j + 1)
y(i, j)X = z(i, j) + δy(i, j)
y(i, j)Y = y(i, j + 1)

We leave the verification of the relations to the reader. If we apply a linear combination of the Y^{i}XY^{j}, Y^{i}XY^{j}X^{−1} to z(0, 0), we obtain the analogous linear combination of the y(i, j), z(i, j). Hence we see that the elements in question are linearlx independent.

This shown, it remains to verify that the Y^{i}EY^{j} are linearly independent. For this purpose we let them act on the basic module (1.13). We have

(ey^{a})(∑_{i,j} λ_{ij}Y^{i}EY^{j}) = ∑_{i,j} λ_{ij}A_{i+a}ey^{j}.

If c = ∑_{i,j} λ_{ij}Y^{i}EY^{j} = 0, we can assume that i, j are taken from the interval 1 ≤ i, j ≤ n. If the determinant of the n × n-matrix (u, v) ↦ A_{u+v} is not a zero divisor in 𝔎, then (ey^{a})c = 0 implies λ_{ij} = 0. The hypothesis on the determinant certainly holds in the generic ring. Hence in this case, ℨ(∞) has a basis as indicated. Over any other ring, the algebra is obtained by a change of scalars from the generic case, hence has the same basis. □

(1.15) Remark. If we work over a field such that all the determinants of the matrices (u, v) ↦ A_{u+v} are non-zero, then the basic module is simple. This holds for the quotient field of the generic ring. Thus, in this generic case, we have an infinite-dimensional irreducible representation of the braid group ZB_{2}. ♡

2. The algebras of height k

The algebras ℨ_{2}(k) for 2 ≤ k < ∞ are defined as for k = ∞. We only require that Y satisfies an equations of degree k. Fix, in addition to (1.1), parameters

(2.1) q_{0} = 1, q_{1}, …, q_{k}

in the ground ring 𝔎. We assume that q_{k} is invertible.

(2.2) Definition. The algebra ℨ(k) = ℨ_{2}(k) on two strings and of height k is the associative unital 𝔎-algebra generated by invertible elements X, Y and another element E with the relations

(1) X − X^{−1} = δ(1 − E)

(2) EX = λE = XE

(3) XY XY = Y XY X

(4) EXY XY = E = XY XY E

(5) EY^{j}E = A_{j}E, j ∈ ZZ

(6) ∑_{j=0}^{k} q_{j}Y^{k−j} = 0. This is a tentative definition. The final definition will add basic relations for the parameters (1.1) and (2.1). ♡

It takes some time to define and justify the basic relations for (2.2). To begin with we combine (2.2.5) and (2.2.6) as follows. Suppose we have (A_{j} | j ∈ ZZ) such that for all l ∈ ZZ

(2.3)  ∑_{j=0}^{k} q_{j}A_{l+k−j} = 0.

Given A_{l+i}, 0 ≤ i < k, l fixed, we can define A_{j} for all j ∈ ZZ uniquely such that (2.3) holds. We often do this by fixing l suitably. If (5) holds for l ≤ j ≤ l + k, then it holds for all j ∈ ZZ. Thus we assume (2.3) and (5) for all j ∈ ZZ in the sequel.

Before we discuss and justify these relations, we consider the examples k = 2, 3 in the next section.

Occcasionally it is convenient to split (2.2) into linear factors. This uses parameters p_{j} ∈ 𝔎. We set

(2.4)  ∏_{j=1}^{k}(Y − p_{j}) = ∑_{j=0}^{k} q_{j}Y^{k−j} =: Q(Y).

Hence (−1)^{j}q_{j} is the j-th elementary symmetric function in the p_{i}. Since q_{k} is invertible, all p_{i} are invertible. If we assume that Δ = ∏_{i<j}(p_{i} − p_{j}) is invertible in 𝔎, then the algebra ℨ_{1}(k) = 𝔎[Y]/(Q(Y)) has the pairwise orthogonal idempotents

(2.5) Y_{j} = ∏_{i,i≠j} \frac{Y − p_{i}}{p_{j} − p_{i}}

which are also a 𝔎-basis. In terms of the involution, we require p_{j}^{∗} = p_{j}^{−1}. This is then compatible with the relation for q_{j}^{∗} above.

3. The examples for k = 2 and k = 3.

The case k = 2. We assume given an algebra (2.2) such that the elements E and EY are linearly independent. We deduce from this the basic relations. The right ideal generated by E is spanned by these elements. This follows from

EY X = λ^{−1}EY^{−1} = −λ^{−1}q_{2}^{−1}q_{1}E − λ^{−1}q_{2}^{−1}EY.

T. tom Dieck 3. The examples for k = 2 and k = 3. 7

The right multiplications by X, Y, E have, with respect to E, EY, the matrices

X = ( \begin{matrix}λ & −λ^{−1}q_{2}^{−1}q_{1} \\ 0 & −λ^{−1}q_{2}^{−1} \end{matrix} ) , Y = ( \begin{matrix}0 & −q_{2} \\ 1 & −q_{1} \end{matrix} ) , E = ( \begin{matrix}A_{0} & A_{1} \\ 0 & 0 \end{matrix} ) .

These matrices satisfy Y XY X = XY XY = 1_{2}, and the relations (2.2) are satisfied if and only if (with μ = λq_{2})

(3.1) \begin{matrix}R_{0} := λ − λ^{−1} + δa_{0} = 0 \\ R_{1} := λa_{1} − λ^{−1}a_{−1} = 0 \\ S_{0} := μ − μ^{−1} − δ = 0 \\ S_{1} := δλa_{1} + q_{1} − q_{1}^{∗} = 0. \end{matrix}

As to the second relation, we remark that a_{−1} is defined by A_{1}+q_{1}A_{0}+q_{2}A_{−1} = 0. Therefore this relation is equivalent to

(3.2) (1 + λμ)A_{1} + q_{1}A_{0} = 0.

We could define the algebra over the generic ring

ZZ[λ^{±}, q^{±}, A_{0}, A_{1}, q_{1}, q_{2}]/(R_{0}, R_{1}, S_{0}, S_{1}).

Since the second relation has the form (λq_{2} − q)(λq_{2} + q^{−1}) = 0, this is not an integral domain. In an integral domain, we must have λq_{2} = q or λq_{2} = −q^{−1}. These two cases are essentially isomorphic. We prefer the relation

(3.3) λq_{2} = −q^{−1},

because it is compatible with λ = q = 1 and Y^{2} = 1.

If δ is not a zero divisor, then (3.2) is a consequence of R_{0}, S_{0}, S_{1}, and, similarly, R_{0} is a consequence of R_{1}, S_{0}, S_{1}. Under the conditions (3.1) we can, as in (1.13), construct a right ℨ(2) module with basis ey^{0}, ey^{1}, where X, Y, E act through the matrices above.

Under the conditions (3.1) and (3.3), the generic ring would be

ZZ[λ^{±}, q^{±}, A_{0}, A_{1}, q_{1}]/(R_{0}, R_{1}, S_{1}).

Is this an integral domain?

The case k = 3. We assume that EY^{−1}, E, EY are linearly independent. Then these elements span the right ideal generated by E, and the matrices of X, Y, and E in this basis are

X = ( \begin{matrix}δ & 0 & λ^{−1} \\ −δA_{−1} & λ & 0 \\ λ & 0 & 0 \end{matrix} ) Y = ( \begin{matrix}0 & 0 & −q_{3} \\ 1 & 0 & −q_{2} \\ 0 & 1 & −q_{1} \end{matrix} ) E = ( \begin{matrix}0 & 0 & 0 \\ A_{−1} & A_{0} & A_{1} \\ 0 & 0 & 0 \end{matrix} ) .

A computation with these matrices shows the relation Y XY X = 1_{3}, and (2.2) is satisfied if an only if the following relations hold (with μ = λq_{3})

(3.4) \begin{matrix}R_{0} := λ − λ^{−1} + δa_{0} = 0 \\ R_{1} := λa_{1} − λ^{−1}a_{−1} = 0 \\ S_{0} := μ^{2} − 1 = 0 \\ S_{1} := δλa_{1} + δλq_{3} + q_{1} − q_{1}^{∗} = 0. \end{matrix}

Again, in an integral domain we have a choice for λq_{3}. We prefer the choice

(3.5) λq_{3} = −1,

since it is compatible with the specialization λ = q = 1 and Y^{3} = 1. Under the relations (3.4), we can define a free module of rank three by letting X, Y, E act through the matrices above. (We think formally of the basis EY^{−1}, E, EY.)

Later we discuss conditions on the parameters under which the ideal generated by E splits off as an algebra factor. In the case k = 2, this is the condition that

(p_{1}A_{0} + A_{1})(p_{2}A_{0} + A_{1}) = q_{2}(A_{0} − λ^{2}A_{1}^{2})

be invertible in 𝔎.

4. The basic relations for k < ∞

In this section we investigate the basic relations for algebras of height k. We write k = 2n or k = 2n + 1, as the case may be. We set μ = λq_{k} and call it the basic parameter. We set S_{0} = μ^{2} − 1 for k odd and S_{0} = μ − μ^{−1} − δ for k even. We use for 1 ≤ j ≤ n the following polynomial in the parameters (the dependance on k will not be denoted):

(4.1) S_{j} = δλ(∑_{ν=0}^{j−1} q_{ν}a_{j−ν} + ∑_{ν=0}^{j−1} ε_{k+j−ν}q_{k−ν}) + q_{j} − q_{j}^{∗}.

We call S_{j}^{∗} the value of S_{j} under the formal involution (1.6).

(4.2) Theorem. Let an algebra as in (2.2) be given. Suppose E, EY, …, EY^{k−1} are linearly independent over 𝔎. Then the relations R_{j} = 0 and S_{j} = 0 hold for 0 ≤ j ≤ n.

Proof. We know that the elements (EY^{j} | 0 ≤ j ≤ k − 1) span the right ideal generated by E. We write the matrices for the right multiplication by X, Y, E in terms of this basis and state the conditions under which the relations (2.2) hold.

We begin with the case k = 2n + 1. We assume that the elements EY^{−n}, …, EY^{n} are linearly independent. They are then a basis for the right ideal generated by E. We compute the matrices for the right multiplication by X, Y, and E in this basis. For simplicity, we denote the matrices again by the same symbol. The matrices for Y and E are clear. The matrix for X is the upperleft k × k-block of the matrix X̃ in section one, if we order the basis according to the exponents (0, 1, −1, …, n, −n). The matrix Y X has the following structure:

T. tom Dieck 4. The basic relations for k < ∞ 9

Take the upper-left k × k-block of Ỹ X̃ in section one and add to the last column the transpose of

−λ(q_{n−1}, q_{n}, q_{n+2}, …, q_{2n}, q_{1}, q_{2n+1} + 1).

The relation EX = XE = λE holds if and only if the relations R_{j} = 0 hold for 0 ≤ j ≤ n hold. If these R_{j} = 0, then, be the results of section one, (Y X)^{2} differs from the unit matrix formally only in the last column. A computation shows that the formal structure of the last column is, now written with exponents in the natural order,

(R_{1} − S_{1}, …, R_{n} − S_{n}, −λμS_{n}^{∗}, …, λμS_{1}^{∗}, μ^{2}).

Hence (Y X)^{2} = 1_{k} if and only if μ^{2} = 1 and S_{j} = S_{j}^{∗} = 0 for 1 ≤ j ≤ n. We shall see later that the relations S_{j}^{∗} = 0 are implied by the other relations.

The case k = 2n is handled similarly. We work with the basis EY^{1−n}, …, EY^{n} for the right ideal generated by E. The matrix Y X is the upper-left k × kblock of the corresponding matrix in section one. Under the assumptions R_{j} = 0 for 1 ≤ j ≤ n we have therefore (Y X)^{2} = 1_{k}. In this case, the relation X(X + δ(E − 1)) = 1_{k} is the problem. Again, only the last column can differ from the identity, if we assume that the R_{j} = 0. This last cloumn is, written in the natural order of the exponents,

(−λ^{−1}μ^{−1}S_{1}, …, −λ^{−1}μ^{−1}S_{n−1}, −S_{n}^{∗}, …, −S_{1}^{∗}, −μ^{−1}(δ + μ^{−1})).

This is the last column of the unit matrix if and only if S_{j} = S_{j}^{∗} = 0. Again we do not need the S_{j}^{∗}. □

We now discuss the basic relations. We introduce the formal power series in an indeterminate t

(4.3)  A(t) = 1 + δλ ∑_{j=1}^{∞} A_{j}t^{j},  a(t) = 1 + δλ ∑_{j=1}^{∞} a_{j}t^{j}.

They are related by

(4.4) A(t) = a(t) + δλ\frac{t^{2}}{1 − t^{2}} .

We set A_{j}^{∗} = A_{−j}, a_{j}^{∗} = a_{−j}, and a^{∗}(t) = 1 − δλ^{−1} ∑_{j=1}^{∞} a_{j}^{∗}t^{j}. The identity

(4.5) a(t)a^{∗}(t) = \frac{(1 − q^{−2}t^{2})(1 − q^{2}t^{2})}{(1 − t^{2})^{2}}

is equivalent to δR_{j} = 0 for j ≥ 1.

Under the hypothesis (2.3), the power series A(t), a(t), A^{∗}(t), and a^{∗}(t) are rational functions. We set q(t) = ∑_{j=0}^{k} q_{j}t^{j}, q_{j}^{∗} = q_{k}^{−1}q_{k−j}, and q^{∗}(t) = ∑_{j=0}^{k} q_{j}^{∗}t^{j}. We extend the formal involution (1.6) by φ(q_{j}) = q_{j}^{∗}. We write

r(t) = q(t) + δλρ(t) = q(t) + δλ ∑_{j=1}^{k} ρ_{j}t^{j}, with ρ_{j} = ∑_{ν=0}^{j−1} q_{ν}A_{j−ν}.

We obtain r^{∗}(t) and ρ^{∗}(t) by applying the involution to the coefficients. With these data we have

(4.6)  A(t) = \frac{r(t)}{q(t)},  A^{∗}(t) = \frac{r^{∗}(t)}{q^{∗}(t)}

and, of course, still (4.4) and its conjugate. The identities of power series

a(t) = \frac{q^{∗}(t)}{q(t)} · \frac{1 − μδt − t^{2}}{1 − t^{2}} , k odd

a(t) = \frac{q^{∗}(t)}{q(t)} · \frac{1 − μ^{2}t^{2}}{1 − t^{2}} , k even

are equivalent to

S(t) := δλ(ρ(t) − s(t)) + q(t) − q^{∗}(t) = 0

with

s(t) = t^{2} \frac{q(t) − q_{k}q^{∗}(t)}{1 − t^{2}} k even

s(t) = \frac{tq(t) − q_{k}q^{∗}(t)}{1 − t^{2}} k odd.

(Note that the fractions which appear in s(t) are polynomials.) For a proof, multiply the power series relations by q(t) and rewrite the result. A small computation shows that if we write S(t) = ∑_{j=1}^{k−1} S_{j}t^{j}, then the S_{j} for 1 ≤ j ≤ n are the polynomials (4.1).

The S(t) have a symmetry property. In its proof we use the following identities.

(4.7) t^{k}q(t^{−1}) = q_{k}q^{∗}(t).

(4.8) t^{k}s(t^{−1}) = \{ \begin{matrix}t^{−2}s(t) & k even \\ t^{−1}s(t) & k odd. \end{matrix}

(4.9) s^{∗}(t) = \{ \begin{matrix}−q_{k}^{−1}s(t) & k even \\ −q_{k}^{−1}t^{−1}s(t) − q^{∗}(t) & k odd. \end{matrix}

The first one is an immediate consequence of the definition and the other two are verified by using (4.7) in the definition of s(t). The following identity is a direct consequence of the definition and (2.3).

(4.10) δλρ(t) = −δμt^{k}ρ^{∗}(t^{−1}) − δμA_{0}t^{k}q^{∗}(t^{−1}).

Here is the symmetry of the S(t):

(4.11) Proposition. Assume R_{0} = 0 and S_{0} = 0. Then

t^{k}S(t^{−1}) = λμS^{∗}(t).

T. tom Dieck 5. The splitting 11

Proof. One uses the assumption on A_{0}, the quadratic equation and (4.7) – (4.10) in order to verify the identity in question. □

The symmetry property is now used to show that R_{j} = 0 and S_{j} = 0 for 1 ≤ j ≤ n imply S_{j}^{∗} = 0 in the same range. Proof:

(4.12) The basic module. From the last proof we also see that, under the basic relations, the matrices X and Y just defined satisfy Y XY X = 1 and X − X^{−1} = δ(1 − E), in both cases. We can therefore define a right module of the algebra with 𝔎-basis ey^{−n}, …, ey^{n} for k = 2n + 1 and ey^{1−n}, …, ey^{n} for k = 2n such that the right multiplication by X and Y is given by the matrices with the same name of the previous proof. The action of E should be clear. The assignment ey^{j} ↦ EY^{j} is now a ℨ(k)-linear map. ♡

We give a heuristic argument for the general form of the relations. We have

a(t) = \frac{r(t)}{q(t)} − \frac{δλt^{2}}{1 − t^{2}}

and similarly for a^{∗}(t). The basic equation (4.5) thus leads to

(1−q^{−2}t^{2})(1−q^{2}t^{2})q(t)q^{∗}(t) = (r(t)(1−t^{2})−δλt^{2}q(t)(r^{∗}(t)(1−t^{2})+δλ^{−1}t^{2}q^{∗}(t)).

In a sufficiently generic situation, q(t) and q^{∗}(t) are different irreducible polynomials. So q^{∗}(t) must divide one of the factors on the right. It cannot divide the second factor, because then it would divide r^{∗}(t), and this would imply that A^{∗}(t) is a polynomial. Therefore

a(t) = \frac{q^{∗}(t)}{q(t)} · \frac{b(t)}{1 − t^{2}}

where b(t) is a quadratic factor of (1 − q^{−2}t^{2})(1 − q^{2}t^{2}).

We finally show that the relations above imply R_{t} = 0 for all t ∈ ZZ. Proof:

5. The splitting

In this section, we discuss conditions under which the ideal ⟨ E ⟩ splits off as an algebra. This amounts to finding a central idempotent which is the unit element of ⟨ E ⟩. We consider the case k < ∞ and use the k × k-matrix A = (A_{i+j} | 0 ≤ i, j ≤ k − 1).

(5.1) Proposition. Suppose A is invertible and B = (B_{ij}) an inverse. Then ε = ∑_{i,j} B_{ij}Y^{i}EY^{j} is a unit element of ⟨ E ⟩. If (Y^{i}EY^{j} | 0 ≤ i, j ≤ k − 1) is a 𝔎-basis of ⟨ E ⟩ and ⟨ E ⟩ has a unit element, then A is invertible and ε is the unit element.

Proof. Let ε = ∑_{i,j} B_{ij}Y^{i}EY^{j} be an arbitrary element of ⟨ E ⟩. Then

εY^{a}EY^{b} = ∑_{i,j} B_{ij}A_{j+a}Y^{i}EY^{b}.

If ε is a unit element and (Y^{a}EY^{b}) a basis, then ∑_{j} B_{ij}A_{j+a} = δ_{ia}, hence B is an inverse of A. The same computation shows that ε, constructed from an inverse B of A, is a unit element. □

Suppose from now on that the basic relations are satisfied. Then we have the fundamental right ℨ-module M at our disposal (4.12). We have an antihomomorphism of algebras

r: ⟨ E ⟩ → Hom_{𝔎}(M, M)

which maps x to the right multiplication r_{x} by x.

(5.2) Proposition. Suppose A is invertible. Then r is an isomorphism and (Y^{i}EY^{j} | 0 ≤ i, j ≤ k − 1) is a basis of ⟨ E ⟩.

Proof. Let r_{x} = 0. The right ideal [E ⟩ is a surjective image of M. Hence r_{x} is zero in [E ⟩. The right module ⟨ E ⟩ is the sum of the isomorphic right ideals Y^{i}[E ⟩. Hence right multiplication by x is zero in ⟨ E ⟩. Since this algebra has a unit, x = 0, and r is injective.

Let x = ∑_{i,j} C_{ij}Y^{i}EY^{j}. Then r_{x} has, with respect to ey^{j}, the matrix AC. Hence, if A is invertible, we can realize any matrix by some r_{x}. Therefore r is surjective too. This also shows the linear independence of the Y^{i}EY^{j}. □

If we work over a field 𝔎, the preceding results yield:

(5.3) Proposition. The algebra ℨ(k) is semi-simple if and only if ℨ/⟨ E ⟩ is semi-simple and A invertible. □

(5.4) Example. In the case k = 2, we have

| \begin{matrix}A_{0} & A_{1} \\ A_{1} & A_{2} \end{matrix} | = −(p_{1}A_{0} + A_{1})(p_{2}A_{0} + A_{1}).

In the case δ = 0, Y^{2} = 1, this is non-zero iff A_{0} ≠ ±A_{1}. If δ is invertible this is non-zero iff

(q − q^{−1}) − (λ − λ^{−1}) ≠ ±(p_{1} − p_{1}^{−1} + p_{2} − p_{2}^{−1}).

(By the way, the algebras should always be free as a 𝔎-module, if we only assume the basic equations.) ♡

6. Quantum algebra specializations

The generic ring for the definition of the algebra ℨ(k) is obtained by treating the parameters as indeterminates and adding the relations R_{0}, …, R_{n}, S_{0}, …, S_{n}. This ring is not very tractable. Instead, one looks for specializations into rings which should fulfill the following requirements:

(1) Integral domain

(2) Allows the specialization q = 1

(3) Allows specialization to the Brauer algebras defined below

T. tom Dieck 7. Coloured Brauer graphs 13

(4) Allows enough specializations so as to capture all information of the generic case

(5) Works for all representations via quantum algebra, e. g. R-matrices and

four braid relations. We suggest that ℜ = ZZ[q^{±}, μ_{1}^{±}, …, μ_{n}^{±}] is a suitable ring (k = 2n, 2n + 1). This is based on the following choice of parameters:

Case k = 2n + 1.

λ = q^{−t}, q_{k} = −q^{t}
p_{2j−1} = μ_{j}q^{r(j)}, p_{2j} = μ_{j}^{−1}q^{r(j)}, 1 ≤ j ≤ n,
p_{2n+1} = q^{r(2n+1)}

with t = r(2n + 1) + 2 ∑_{j=1}^{n} r(j). Under these conditions, the coefficients of the polynomial q(t) − q^{∗}(t) are formally divisible by δ. The relations S_{j} therefore allow to express the a_{j} as elements of ℜ, because q_{j}^{∗} − q_{j} is divisible by δ. The specialization to the Brauer algebra is obtained by q = 1 and {μ_{j}, μ_{j}^{−1} | 1 ≤ j ≤ n} are the n-th roots of unity different from 1.

Case k = 2n + 2.

λ = q^{−t−1}, q_{k} = −q^{t}
p_{2j−1} = μ_{j}q^{r(j)}, p_{2j} = μ_{j}^{−1}q^{r(j)}, 1 ≤ j ≤ n,
p_{2n+1} = q^{r(2n+1)}, p_{2n+2} = −q^{r(2n+2)}

with t = r(2n + 1) + r(2n + 2) + ∑_{j=1}^{n} r(j). The specialization to the Brauer algebra is obtained by q = 1 and {μ_{j}, μ_{j}^{−1} | 1 ≤ j ≤ n} are the 1 + n-th roots of unity different from ±1.

Since the exponents r(j) can be chosen arbitrarily (subject to the requirement about the value of λq_{k}), we have for each of the parameters p_{j} an infinite number of specializations.

7. Coloured Brauer graphs

Let k, l ∈ IN_{0} = {0, 1, 2, …} be natural numbers. A (k, l)-Brauer-graph is a free involution σ on the set

P (k, l) = {1, …, k} × 0 ∪ {1, …, l} × 1.

Our aim is to construct certain categories based on Brauer graphs. For this purpose we fix the following data:

(1) A commutative ring 𝔎.

(2) A group G.

(3) A function z: G → 𝔎 with the properties z(g) = z(g^{−1}) and z(ghg^{−1}) = z(h) for g, h ∈ G.

We call z the cycle character. (The property z(g) = z(g^{−1}) holds for characters of selfdual finite dimensional representations of G.)

A G-coloured (k, l)-Brauer-graph is a pair (σ, v) which consists of a (k, l)- Brauer-graph σ and a mapping v: P (k, l) → G satisfying the balance relation v(σx) = v(x)^{−1}.

(7.1) Example. let M be an n-manifold and D ⊂ M an n-disk. Then a path w: I → M with w(0), w(1) ∈ D defines uniquely an element of the fundamental group π = π_{1}(M) with base point in D. Choose a set B_{k} ⊂ D with k points. A set of paths [0, 1] → M × [0, 1] with total boundary point set B_{k} × 0 ∪ B_{l} × 1 defines a π-coloured (k, l)-Brauer graph σ as the involution which maps the first point of a path to its last point. Its colouring is given by sending a path to the element of the fundamental group it defines. ♡

We construct a 𝔎-category T (G, z). The objects are the natural numbers n ∈ IN_{0}. The morphism 𝔎-module Mor_{T (G,z)}(k, l) is the free 𝔎-module on the set of G-coloured (k, l)-graphs.

By definition of a 𝔎-category, the composition

Mor(l, m) × Mor(k, l) → Mor(k, m), (β, α) ↦ β ∘ α

has to be a 𝔎-bilinear map. Thus it suffices to specify the composition

(τ, w) ∘ (σ, v)

of two coloured Brauer graphs. The definition of the composition uses the cycle character z.

Let |σ, v| denote the one-dimensional CW-complex with set of zero-cells P (k, l) and a one-cell for each orbit {x, σx} of σ with boundary set {x, σx}. Picking x in the orbit {x, σx} of σ amounts to the choice of an orientation of the one-cell, directed from x to σx. We define an oriented 1-cochain by mapping the oriented 1-cell {x, σx} to v(x). This assigment is called the colouring of |σ, v|.

Let (τ, w) be an (l, m)-graph. We define a one-complex

|τ, w| ∪ |σ, v|

by identifying the 0-cell (a, 1) ∈ |σ, v| with the 0-cell (a, 0) ∈ |τ, w| for 1 ≤ a ≤ l. The path components of this complex are either homeomorphic to the unit interval I or to the circle S^{1}. We orient the path components of this complex. Each oriented 1-cell of a path component is then a 1-cell of |σ, v| or of |τ, w| and has an assigned colour in G. We assign to an interval component the value g_{r} ⋯ g_{2}g_{1} if g_{1}, …, g_{r} are the colours in consecutive order with respect to the chosen orientation. Similarly, to a circle component we assign the colour g_{r} ⋯ g_{1} by starting at some vertex and then running around cyclically according to the chosen orientation. The colour of an oriented circle component is in general not well defined; but the conjugacy class of the colour is independent of the choice of the initial vertex.

T. tom Dieck 8. An example 15

The interval components of |τ, w| ∪ |σ, v| with the colours just defined form a G-coloured (k, m)-graph

|τ, w| ∧ |σ, v|.

By assumption about z, the value z(g) = z(S) of a g-valued oriented circle component S does not depend on the orientation and the choice of the initial vertex. We now define the composition in our category by

(τ, w) ∘ (σ, v) = ∏_{S circle} z(S) |τ, w| ∧ |σ, v|.

The product is taken over the circle components. It is immediate from the construction, that composition is associative. The identity of the object k is given by the involution σ(a, 0) = (a, 1) and colour the neutral element e ∈ G on each orbit of σ. If k = 0, we have to consider the free involution of the empty set; there are no orbits, hence no colours.

Remark. The category T (G, z) may be called the semi-oriented or balanced category. There are analogous unoriented categories and categories of oriented Brauer graphs. In the latter case, the objects are functions {1, 2, …, k} → {±1}. Can one use character theory to describe the dependence of T (G, z) on z? ♡

8. An example

In order to get a feeling for the definition of T (G, z), we describe the special case of the endomorphism algebra of the object 2. Suppose k + l = 2n. There are 1 · 3 · 5 · . . . · 2n − 1 = (2n − 1)!! (k, l)-Brauer-graphs. Therefore, if G is finite, Mor(k, l) is a free 𝔎-module of rank |G|^{n}(2n − 1)!. In particular, this is the rank of the endomorphism algebra of the object n in T (G, z). As usual, we visualize a (k, l)-Brauer-graph σ by drawing an arc from x to σx in the strip IR × [0, 1]. If the graph is coloured, we orient the arcs and write the corresponding colour next to it.

The endomorphism algebra T_{2} of the object 2 thus has the following 𝔎-basis.

F igure

The composition of these morphisms is given by

I(a, b)I(c, d) = I(ac, bd)
I(a, b)X(c, d) = X(ac, bd)
X(a, b)I(c, d) = X(ad, bc)
X(a, b)X(c, d) = I(ad, bc)
I(a, b)E(c, d) = E(c, bda^{−1})
E(a, b)I(c, d) = E(d^{−1}ac, b)
X(a, b)E(c, d) = E(c, bd^{−1}a^{−1})
E(a, b)X(c, d) = E(c^{−1}a^{−1}d, b)
E(a, b)E(c, d) = z(a^{−1}d)E(c, b).

We describe this algebra by generators and relations.

(8.1) Definition. Let U_{2} denote the associative unital 𝔎-algebra gerated by X, E, Y ∈ G with relations

(1) X^{2} = 1

(2) EX = XE = 1

(3) The elements of G multiply according to the group multiplication

(4) Y XZX = XZXY, for Y, Z ∈ G

(5) EY XZX = EZ^{−1}Y, for Y, Z ∈ G

(6) XZXY E = Y Z^{−1}E, for Y, Z ∈ G

(7) EY E = z(Y)E, for Y ∈ G. (The relation (4) is a generalized four braid relation.) ♡

(8.2) Proposition. The assignment X ↦ X(1, 1), Y ↦ I(Y, 1), E ↦ E(1, 1) yields an isomorphism ψ: U_{2} → T_{2}.

Proof. It is easily verified that the assignment is compatible with the relations of U_{2} and therefore yields a well-defined homomorphism ψ of algebras.

The relations

I(Y, 1)X(1, 1)I(Z, 1) = X(Y, Z)
I(Y, 1)X(1, 1)I(Z, 1)X(1, 1) = I(Y, Z)
I(Z^{−1}, 1)E(1, 1)I(Y, 1) = E(Y, Z)

show that ψ is surjective.

Injectivity follows if we show that U_{2} is generated as a 𝔎-module by elements of the form Y XZ, Y XZX, Y EZ for Y, Z ∈ G. One first verifies from the relations that the elements of the form Y EZ span a twosided ideal ⟨ E ⟩. We factor out this ideal. The resulting algebra has generators X, Y ∈ G with relations X^{2} = 1, Y XZX = XZXY. It is easily seen that this algebra is spanned by elements of the form Y XZ and Y XZX. It is also the group algebra 𝔎Γ of the semi-direct product

1 → G × G → Γ → ZZ/2 → 1

where ZZ/2 is generated by X and interchanges the factors of G. □

(8.3) Remark. If the group G is given by a presentation, then we can use this presentation in the presentation of the algebra U_{2}. Thus, if G = ⟨ Y | Y^{k} = 1 ⟩, then U_{2} is generated by X, Y, E with relations X^{2} = 1, Y^{k} = 1, XY XY = Y XY X, EX = E = XE, EY^{i}E = z(Y^{i})E, EXY XY = E = XY XY E. This is a special case of our algebras ℨ(k), namely for the parameters λ = q = 1, z(Y^{i}) = A_{i}. ♡