Weights of Markov traces and generic degrees

Meinolf Geck, Lacrimioara Iancu and Gunter Malle

Abstract In this paper we prove the second author’s conjecture about weights of Markov traces for Ariki–Koike algebras. As an application, we obtain a proof of the third author’s conjecture on the generic degrees of these algebras. This yields explicit combinatorial formulas for the generic degrees which are important for several applications, for example, the construction of unipotent degrees for “spetses”.

1 Introduction and main results

Let n, e ⩾ 1 and W_{n}^{(e)} be the subgroup of GL_{n}(ℂ) consisting of all matrices whose non-zero coefficients are eth roots of unity and where there is precisely one nonzero coefficient in each row and each column. By [1, 2.1], W_{n}^{(e)} has a presentation with generators t, s_{1}, …, s_{n−1} where the defining relations are t^{e} = 1, s_{1}^{2} = ⋯ = s_{n−1}^{2} = 1 and the homogeneous relations

\{ \begin{matrix}ts_{1}ts_{1} & = & s_{1}ts_{1}t, & ts_{i} & = & s_{i}t for i ⩾ 2, \\ s_{i}s_{i+1}s_{i} & = & s_{i+1}s_{i}s_{i+1} for i ⩾ 1, & s_{i}s_{j} & = & s_{j}s_{i} if |i − j| ⩾ 2. \end{matrix}

The group W_{n}^{(e)} can also be seen as the wreath product C_{e} ≀ 𝔖_{n} where C_{e} is the cyclic group of order e (generated by t) and 𝔖_{n} is the symmetric group on n letters (generated by s_{1}, …, s_{n−1}). Following Ariki–Koike [1] (see also [3]) we associate with W_{n}^{(e)} an algebra H_{n}^{(e)} over the ring A = ℚ[q^{±1}, u_{1}, …, u_{e}], where q, u_{1}, …, u_{e} are indeterminates. The algebra H_{n}^{(e)} is defined by a presentation with generators T, S_{1}, …, S_{n−1} where the defining relations are

(T − u_{1})(T − u_{2}) ⋯ (T − u_{e}) = 0,
S_{i}^{2} = q 1 + (q − 1)S_{i} for 1 ⩽ i ⩽ n − 1,

and homogeneous relations analogous to those above; here, 1 denotes the identity element of H_{n}^{(e)}. It is shown in [1, Theorem 3.10] that H_{n}^{(e)} is free as an A-module, of rank equal to |W_{n}^{(e)}| = e^{n}n!. Standard bases of H_{n}^{(e)} are described in [2]; see also (4.2) below.

Now let K = ℚ(q, u_{1}, …, u_{e}) be the field of fractions of A and Ĥ_{n}^{(e)} be the algebra obtained by extending scalars from A to K. Ariki and Koike [1] show that Ĥ_{n}^{(e)} is split semisimple and that the simple Ĥ_{n}^{(e)}-modules have a natural parametrization by the set Λ_{n}^{(e)} consisting of all e-tuples of partitions α := (α_{1}, …, α_{e}) such that |α_{1}| + ⋯ + |α_{e}| = n. We shall write

Irr(Ĥ_{n}^{(e)}) = {χ_{α} | α ∈ Λ_{n}^{(e)}}

for the set of irreducible characters of Ĥ_{n}^{(e)}. This paper is concerned with a class of trace functions on Ĥ_{n}^{(e)} which were introduced by Ocneanu/Jones [8] for e = 1, by Geck and Lambropoulou [5] for e = 2 and by Lambropoulou [9] for e ⩾ 3. These trace functions are defined inductively, using the chain of subgroups W_{1}^{(e)} ⊂ W_{2}^{(e)} ⊂ ⋯ ⊂ W_{n}^{(e)} and the corresponding chain of subalgebras Ĥ_{1}^{(e)} ⊂ Ĥ_{2}^{(e)} ⊂ ⋯ ⊂ Ĥ_{n}^{(e)}, where W_{m}^{(e)} (resp., Ĥ_{m}^{(e)}) is generated by t, s_{1}, …, s_{m−1} (resp., T, S_{1}, …, S_{m−1}) for 1 ⩽ m ⩽ n. (We also set W_{0}^{(e)} = {1} and Ĥ_{0}^{(e)} = K1 by convention.) The precise definition is as follows.

Definition 1.1 Let ζ, η_{1}, …, η_{e−1} ∈ K and τ: Ĥ_{n}^{(e)} → K be a trace function, i.e., τ is K-linear and we have τ(hh′) = τ(h′h) for all h, h′ ∈ Ĥ_{n}^{(e)}. We say that τ is a Markov trace with parameters ζ, η_{1}, …, η_{e−1} if τ(1) = 1 and

τ(hS_{m}) = ζ τ(h)  and  τ(h(T′_{m})^{k}) = η_{k} τ(h)

for any 1 ⩽ k ⩽ e − 1 and any h ∈ Ĥ_{m}^{(e)} with 0 ⩽ m < n, where we set S_{0} := ζ and

T′_{m} = S_{m}S_{m−1} ⋯ S_{1}T S_{1}^{−1} ⋯ S_{m−1}^{−1}S_{m}^{−1}.

By [9, §4], given ζ, η_{1}, …, η_{e−1} ∈ K, a Markov trace is uniquely determined by the above conditions. The unique elements ω_{α}(ζ, η_{1}, …, η_{e−1}) ∈ K such that

τ = ∑_{α∈Λ_{n}^{(e)}} ω_{α}(ζ, η_{1}, …, η_{e−1}) χ_{α}

are called the weights of τ. (Note that Irr(Ĥ_{n}^{(e)}) is a basis for the vector space of trace functions on Ĥ_{n}^{(e)}.)

The aim of this paper is to prove Conjecture 3.3 in [7], which yields explicit combinatorial formulae for the weights of Markov traces on Ĥ_{n}^{(e)}. At the same time, this also proves the existence of Markov traces for any given choice of ζ, η_{1}, …, η_{e−1} ∈ K. (This existence proof is independent from that in [9].) To state the main result, we have to introduce some notation.

1.2. For any e-tuple \underline{S} = (S_{1}, …, S_{e}) of finite subsets S_{k} ⊂ ℕ_{0} with |S_{1}| = r + 1, |S_{2}| = ⋯ = |S_{e}| = r, we define as in [11, 2.19] (but we omit a factor equal to (q − 1)^{n}):

D\underline{S}(q; u_{1}, …, u_{e})

:= \frac{(−1)^{\binom{e}{2}\binom{r}{2}+n(e−1)} · ∏_{1⩽k⩽l⩽e} ∏_{{(λ,λ′)∈S_{k}×S_{l} \atop λ>λ′ if k = l}} (q^{λ}u_{k} − q^{λ′}u_{l}) · ∏_{k=1}^{e} u_{k}^{n}}{q^{f(r,e)} · ∏_{k,l=1}^{e} ∏_{λ∈S_{k}} ∏_{h=1}^{λ}(q^{h}u_{k}−u_{l}) · ∏_{1⩽k<l⩽e}(u_{k}−u_{l})^{r}} ,

where f(r, e) = \binom{e(r−1)+1}{2} + \binom{e(r−2)+1}{2} + · · ·.

Now let α ∈ Λ_{n}^{(e)}. As in [11, 2.13], we associate with α = (α_{1}, …, α_{e}) an etuple \underline{S} = (S_{1}, …, S_{e}) of finite subsets S_{k} ⊂ ℕ_{0} with |S_{1}|−1 = |S_{2}| = ⋯ = |S_{e}|, as follows. Writing α_{1} = (α_{1,1} ⩾ … ⩾ α_{1,r+1} ⩾ 0) and α_{k} = (α_{k,1} ⩾ … ⩾ α_{k,r} ⩾ 0) for 2 ⩽ k ⩽ e, for some large r, we set

S_{1} := {α_{1,i} + r − i + 1 | 1 ⩽ i ⩽ r + 1},
S_{k} := {α_{k,i} + r − i | 1 ⩽ i ⩽ r}  (2 ⩽ k ⩽ e).

Finally, we define D_{α} := D\underline{S}(q; u_{1}q^{−1}, u_{2}, …, u_{e}) ∈ K. As pointed out in the remarks following [11, 2.19], D_{α} does not depend on the choice of r.

Theorem 1.3 Let ζ, η_{1}, …, η_{e−1} ∈ K. For any α ∈ Λ_{n}^{(e)}, we define:

R_{α}(ζ,η_{1}, …, η_{e−1}) := ∏_{k=1}^{e} ∏_{x∈α_{k}}[ζ(1 − q^{c(x)}) ∏_{{l=1 \atop l≠k}}^{e}(u_{k}q^{c(x)} − u_{l})

+(1 − q)(∑_{i=1}^{e−1}(u_{k}^{e−i−1}q^{(e−i)c(x)} ∑_{j=1}^{i}(−1)^{i−j}η_{j}σ_{i−j}) + (−1)^{e−1} ∏_{{l=1 \atop l≠k}}^{e} u_{l})],

where σ_{i} denotes the i-th elementary symmetric function in u_{1}, …, u_{e} and c(x) denotes the content of x ∈ α_{k} (see Remark 1.4 below). Then

τ_{ζ,η_{1},... ,η_{e−1}} := ∑_{α∈Λ_{n}^{(e)}}(−1)^{ne}(∏_{k=1}^{e} u_{k}^{|α_{k}|−n})D_{α} R_{α}(ζ, η_{1}, …, η_{e−1}) χ_{α}

is a Markov trace on Ĥ_{n}^{(e)} with parameters ζ, η_{1}, …, η_{e−1}.

The proof consists of two steps. First, we need a generalization of Orellana’s results [14] to the present context; this will be discussed in Section 2. This will yield Theorem 1.3 for special choices of the parameters. The general case will then be settled in Section 3, using the methods developed in [7].

The above result contains as special cases the known formulae for e = 1, due to Ocneanu (see [15]), and for e = 2, due to Orellana [14] and Iancu [7]. However, our proof of Theorem 1.3 requires that the case e = 1 is known in advance.

1.4. The case e = 1. The algebra Ĥ_{n}^{(1)} is the generic Iwahori–Hecke algebra associated with the symmetric group 𝔖_{n}, where the generators s_{1}, …, s_{n−1} correspond to the basic transpositions (1, 2), …, (n−1, n). In this case, a Markov trace τ only depends on the parameter ζ. Setting e = 1 in Theorem 1.3, we obtain the weight formula

ω_{α}(ζ) = q^{n(α)} ∏_{x∈α} \frac{q − 1 + (q^{c(x)} − 1)ζ}{q^{h(x)} − 1} .

Here, we identify α ⊢ n with its diagram, i.e., the set of all (i, j) ∈ ℕ × ℕ such that 1 ⩽ i ⩽ l and 1 ⩽ j ⩽ α_{i}, where α_{1} ⩾ … ⩾ α_{l} > 0 are the non-zero parts of α. Using this convention, c(x) = j − i denotes the content and h(x) denotes the hook length of x ∈ α (see [10, Ex. I.1.1]) and n(α) = ∑_{i=1}^{l}(i − 1)α_{i}. A proof of the above formula can be found in [15, §3].

Corollary 1.5 Let τ be the Markov trace on Ĥ_{n}^{(e)} with parameters ζ = η_{1} = ⋯ = η_{e−1} = 0. Then τ is the canonical symmetrizing trace τ_{0} introduced in [2, §2B]. Setting ζ = η_{1} = ⋯ = η_{e−1} = 0 in the formula in Theorem 1.3, we find that

τ_{0} = ∑_{α∈Λ_{n}^{(e)}}(q − 1)^{n}D_{α} χ_{α}.

The above formula for the weights of the canonical symmetrizing trace τ_{0} was conjectured by the third author in [11, 2.20]. The proof will be given in Lemma 4.3. See [3] for the relevance of that trace and its weights in the representation theory of finite reductive groups.

In the final section we verify that the elements (q − 1)^{n}D_{α} satisfy a certain palindromicity property. By a result from [4] this implies that τ_{0} can also be characterized as the unique trace form on H_{n}^{(e)} satisfying certain natural conditions.

2 Markov traces and specializations

As in [14], we begin by establishing a weight formula for Markov traces on Ĥ_{n}^{(e)} for special choices of the parameters. This is based on the following connection between weights of Markov traces in type A and Schur functions; see [15].

2.1. Let r ⩾ 1 and consider a partition ν with at most r parts. Writing ν = (ν_{1} ⩾ ⋯ ν_{r} ⩾ 0) we define, as in [14, §1], the normalized Schur function

s_{ν,r}(q) := q^{n(ν)}(\frac{1 − q}{1 − q^{r}} )^{|ν|} · ∏_{1⩽i<j⩽r} \frac{1 − q^{ν_{i}−ν_{j}+j−i}}{1 − q^{j−i}} .

Using the identities in [10, I.1], we find that

s_{ν,r}(q) = q^{n(ν)}(\frac{1 − q}{1 − q^{r}} )^{|ν|} · ∏_{x∈ν} \frac{1 − q^{r+c(x)}}{1 − q^{h(x)}} ,

where c(x) denotes the content and h(x) denotes the hook length of x ∈ ν. A straightforward computation and a comparison with the weight formula for type A in (1.4) shows that

ω_{ν}(ζ(r)) = s_{ν,r}(q) where ζ(r) := q^{r} \frac{1 − q}{1 − q^{r}} .

Thus, the weights in type A for a Markov trace with parameter of form ζ(r) are given by normalized Schur functions.

2.2. Now we consider non-negative integers r_{1}, …, r_{e} and m_{1}, …, m_{e} such that the following hold:

\{ \begin{matrix}r_{i} & ⩾ & n & for i = 1, …, e, \\ m_{i} − m_{i+1} & ⩾ & n & for i = 1, …, e − 1, \\ m_{e} & = & 0. \end{matrix} (∗)

Then we set

r := r_{1} + ⋯ + r_{e}  and  f := r_{1}m_{1} + ⋯ + r_{e}m_{e}.

We associate with these data the partition λ of f which has r_{1} parts equal to m_{1}, r_{2} parts equal to m_{2} and so on. Thus, λ has precisely r − r_{e} non-zero parts.

Now let α = (α_{1}, …, α_{e}) ∈ Λ_{n}^{(e)}. We define a corresponding partition μ_{α} ⊢ n + f as follows. Let α = (α_{1}, …, α_{e}) and write

α_{k} = (α_{k,1} ⩾ α_{k,2} ⩾ ⋯ ⩾ α_{k,r_{k}} ⩾ 0) for 1 ⩽ k ⩽ e.

(This is always possible since |α_{k}| ⩽ n ⩽ r_{k}.) Then μ_{α} has parts

m_{1} + α_{1,1} ⩾ m_{1} + α_{1,2} ⩾ ⋯ ⩾ m_{1} + α_{1,r_{1}}
⩾ m_{2} + α_{2,1} ⩾ m_{2} + α_{2,2} ⩾ ⋯ ⩾ m_{2} + α_{2,r_{2}}
⋮
⩾ m_{e} + α_{e,1} ⩾ m_{e} + α_{e,2} ⩾ ⋯ ⩾ m_{e} + α_{e,r_{e}}.

(The inequalities hold since |α_{k}| ⩽ n and thanks to conditions (∗).) Thus, μ_{α} is obtained by adding to the first r_{1} parts of λ (which are all equal to m_{1}) the parts of α_{1}, to the next r_{2} parts of λ (which are all equal to m_{2}) the parts of α_{2} and so on. Note that μ_{α} has at most r parts.

Let θ: A → ℚ(q) be the ring homomorphism defined by θ(q) = q and

θ(u_{k}) = −q^{r_{k}+···+r_{e−1}+m_{k}} for 1 ⩽ k ⩽ e.

Note that we have θ(u_{e}) = −1. Let H̃_{n}^{(e)} = ℚ(q) ⊗_{A} H_{n}^{(e)} be the corresponding specialized Ariki–Koike algebra. For any α ∈ Λ_{n}^{(e)}, we have χ_{α}(h) ∈ A for all h ∈ H_{n}^{(e)}, since A is integrally closed in K. So we have induced trace functions χ̃_{α} on H̃_{n}^{(e)}, for any α ∈ Λ_{n}^{(e)}. The following result generalizes [14, Theorem 5.5].

Proposition 2.3 The trace function τ̃: H̃_{n}^{(e)} → ℚ(q) defined by

τ̃ := ∑_{α∈Λ_{n}^{(e)}} \frac{s_{μ_{α},r}(q)}{s_{λ,r}(q)} χ̃_{α}

is a Markov trace, i.e., we have τ̃(1) = 1 and there exist ζ, η̃_{1}, …, η̃_{e−1} ∈ ℚ(q) such that

τ̃(hS_{m}) = ζ τ̃(h)  and  τ̃(h(T′_{m})^{k}) = η̃_{k} τ̃(h)

for any 1 ⩽ k ⩽ e − 1 and any h ∈ H̃_{m}^{(e)} with 0 ⩽ m < n. We have ζ = q^{r}(1 − q)/(1 − q^{r}).

Proof. This is analogous to that in [14]. The whole argument is based on the construction of a (non-unital) algebra homomorphism from H̃_{n}^{(e)} to the Iwahori–Hecke algebra H_{n+f}(q) of the symmetric group 𝔖_{n+f}. We define Δ_{f}^{2} = (S_{f−1} ⋯ S_{1})^{f} and Δ_{f+1}^{2} = (S_{f} ⋯ S_{1})^{f+1}, as in [14, 1.3]; these are central elements in H_{f}(q) and H_{f+1}(q), respectively. Let z_{λ} be the scalar by which Δ_{f}^{2} acts on the simple H_{f}(q)-module labelled by λ. Now there are precisely e partitions of f + 1 which can be obtained by adding a box to the diagram of λ; we can add such a box in the row r_{1} + ⋯ + r_{k−1} + 1, for 1 ⩽ k ⩽ e. Denote the corresponding partition of f + 1 by λ^{(k)} and let z_{k} be the scalar by which Δ_{f+1}^{2} acts on the simple H_{f+1}(q)-module labelled by λ^{(k)}. Using [14, Lemma 3.1], we find that

z_{k} = z_{λ} q^{f+m_{k}−r_{1}−···−r_{k−1}}.

Furthermore, let p_{λ} be a primitive idempotent affording the simple H_{f}(q)-module labelled by λ. Then there exists a unique algebra homomorphism ρ_{f}: H̃_{n}^{(e)} → H_{n+f}(q) such that

ρ_{f}(1) = p_{λ}, ρ_{f}(T) = −\frac{z_{λ}}{z_{λ^{(e)}}} Δ_{f}^{−2}Δ_{f+1}^{2}, ρ_{f}(S_{i}) = p_{λ}S_{f+i} for 1 ⩽ i ⩽ n − 1.

As in [14, Lemma 3.2], this is proved by checking the defining relations. Now consider the Markov trace τ_{ζ} on H_{n+f}(q) with parameter ζ = q^{r}(1 − q)/(1 − q^{r}). Restricting this trace to H_{f}(q) and using the formula in (2.1), we see that s_{λ,r}(q) = τ_{ζ}(p_{λ}). Furthermore, by the branching rule, the element p_{λ} acts as 0 in an irreducible representation of H_{n+f}(q) unless that representation is labelled by a partition of form μ_{α} for some α ∈ Λ_{n}^{(e)}. Thus, we have

τ̃(h) = \frac{1}{s_{λ,r}(q)} τ_{ζ}(ρ_{f}(h)) for all h ∈ H̃_{n}^{(e)}.

This already shows that τ̃(1) = 1 and that τ̃(hS_{m}) = ζ τ̃(h) for any h ∈ H̃_{m}^{(e)} with 0 ⩽ m < n. The remaining assertion is proved in exactly the same way as in [14, Lemma 5.4]. We omit further details. □

Theorem 2.4 Let \underline{r} = (r_{1}, …, r_{e}) ∈ ℕ^{e} (with r_{i} ⩾ n for all i) and define, for any α ∈ Λ_{n}^{(e)},

ω_{α,\underline{r}}(q, u_{1}, …, u_{e}) =q^{n(α)}(\frac{1 − q}{1 − q^{r}} )^{n} · ∏_{k=1}^{e} ∏_{1⩽i<j⩽r_{k}} \frac{1 − q^{α_{k,i}−α_{k,j}+j−i}}{1 − q^{j−i}}

× ∏_{1⩽k<l⩽e} ∏_{i=1}^{r_{k}} ∏_{j=1}^{r_{l}} \frac{u_{l}q^{α_{l,j}−j} − u_{k}q^{α_{k,i}−i}}{u_{l}q^{−j} − u_{k}q^{−i}}

where n(α) = ∑_{k=1}^{e} n(α_{k}) and r = r_{1} + ⋯ + r_{e}. Then the trace function

τ\underline{r} := ∑_{α∈Λ_{n}^{(e)}} ω_{α,\underline{r}}(q, u_{1}, …, u_{e}) χ_{α}

is a Markov trace on Ĥ_{n}^{(e)} with parameters ζ = ζ(\underline{r}) = q^{r}(1 − q)/(1 − q^{r}) and

η_{k} = η_{k}(\underline{r}) = ∑_{i=1}^{e} u_{i}^{k} \frac{1 − q^{r_{i}}}{1 − q^{r}} ∏_{{j=1 \atop j≠i}}^{e} \frac{u_{i}q^{r_{j}} − u_{j}}{u_{i} − u_{j}} for 1 ⩽ k ⩽ e − 1.

Proof. This is also analogous to that in [14]. First we consider a specialization θ: A → ℚ(q) as above, for any choice of m_{1}, …, m_{e} satisfying the conditions in (2.2). Then we have

θ(ω_{α,\underline{r}}(q, u_{1}, …, u_{e})) = \frac{s_{μ_{α},r}(q)}{s_{λ,r}(q)} ∈ ℚ(q). (∗∗)

Note that θ(u_{i}) ≠ θ(u_{j}) for i ≠ j, so we may apply θ to ω_{α,\underline{r}}(q, u_{1}, …, u_{e}). Formula (∗∗) is proved as follows. Using the defining formula and noting that |μ_{α}| = |λ| + n, we have

\frac{s_{μ_{α},r}(q)}{s_{λ,r}(q)} = q^{n(μ_{α})−n(λ)}(\frac{1 − q}{1 − q^{r}} )^{n} ∏_{1⩽i<j⩽r} \frac{1 − q^{μ_{i}−μ_{j}+j−i}}{1 − q^{λ_{i}−λ_{j}+j−i}} ,

where we write μ_{α} = (μ_{1} ⩾ ⋯ ⩾ μ_{r} ⩾ 0) and λ = (λ_{1} ⩾ ⋯ ⩾ λ_{r} ⩾ 0). Let us first compare n(μ_{α}) and n(λ). By definition, we have

n(μ_{α}) = \underbrace{∑_{k=1}^{e} ∑_{i=1}^{r_{k}}(r_{1} + ⋯ + r_{k−1} + i − 1)m_{k}}_{=n(λ)} + ∑_{k=1}^{e} ∑_{i=1}^{r_{k}}(r_{1} + ⋯ + r_{k−1} + i − 1)α_{k,i}

= n(λ) + n(α) + ∑_{k=1}^{e}(r_{1} + ⋯ + r_{k−1})|α_{k}|.

To compute the remaining terms in the above expression for ω_{α,\underline{r}}(q), we split the product over all pairs (i, j) with 1 ⩽ i < j ⩽ r as follows. We have a partition {1, …, r} = I_{1} ∪⋯∪I_{e}, where I_{1} consists of the first r_{1} numbers, I_{2} of the next r_{2} numbers and so on. Thus, the desired product can be computed by first considering all pairs (i, j) where i, j ∈ I_{k}, for fixed 1 ⩽ k ⩽ e, and then all pairs (i, j) where i ∈ I_{k} and j ∈ I_{l} with 1 ⩽ k < l ⩽ e. A straightforward computation yields the desired formula (∗∗) for ω_{α,\underline{r}}(q). Note that, in the case e = 2, we obtain precisely the formula in [14, §5, Eq. 12].

Thus, we have seen that the trace function τ\underline{r} specializes to the Markov trace τ̃ in Proposition 2.3. The value for η̃_{k} is obtained by evaluating τ̃ on T^{k} for 1 ⩽ k ⩽ e − 1, as in the proof of [14, Theorem 5.5]. Since T^{k} lies in H̃_{1}^{(e)}, we have the identity

η̃_{k} = τ̃(T^{k}) = ∑_{α∈Λ_{1}^{(e)}} θ(ω_{α,\underline{r}}(q, u_{1}, …, u_{e})) χ̃_{α}(T_{k}).

Now Λ_{1}^{(e)} consists of all the e-tuples α in which one component is the partition [1] and all other components are the empty partition; if the partition [1] occurs in the ith component, the value of the corresponding irreducible character on T^{k} is given by u_{i}^{k}. Inserting the weight formula yields that

η̃_{k} = θ(∑_{i=1}^{e} u_{i}^{k} \frac{1 − q^{r_{i}}}{1 − q^{r}} ∏_{{j=1 \atop j≠i}}^{e} \frac{u_{i}q^{r_{j}} − u_{j}}{u_{i} − u_{j}} ).

Since this works for all possible choices of m_{1}, …, m_{e}, we conclude that τ\underline{r} itself must be a Markov trace on Ĥ_{n}^{(e)}, with parameters ζ, η_{1}, …, η_{e−1} as claimed. Indeed, first note that it is enough to consider the case where we set u_{e} = −1. (The general case is obtained from that special case by replacing the generator T by a suitable scalar multiple of T.) Next, the fact that τ\underline{r} is a Markov trace as claimed is equivalent to the validity of certain polynomial identities in the variables u_{1}, …, u_{e−1} (as in the proof of [7, Lemma 2.3]). If θ runs over the specializations corresponding to all possible choices of m_{1}, …, m_{e}, we see that the set of the corresponding values (θ(u_{1}), …, θ(u_{e−1})) is a Zariski dense subset of ℚ(q)^{e−1}. Thus, the above mentioned polynomial identities hold for all values of the variables in a Zariski dense subset. Hence they must hold without specialization. □

Orellana and Wenzl informed us that they independently obtained similar generalizations of the results in [14].

3 Proof of Theorem 1.3

We will derive Theorem 1.3 from Theorem 2.4, using the strategy in [7, §2]. For this purpose we define, for any \underline{r} = (r_{1}, …, r_{e}) ∈ ℕ^{e},

ζ(\underline{r}) := q^{r} \frac{1 − q}{1 − q^{r}}  where r = r_{1} + ⋯ + r_{e},
η_{k}(\underline{r}) := ∑_{i=1}^{e} u_{i}^{k} \frac{1 − q^{r_{i}}}{1 − q^{r}} ∏_{{j=1 \atop j≠i}}^{e} \frac{u_{i}q^{r_{j}} − u_{j}}{u_{i} − u_{j}}  for 1 ⩽ k ⩽ e − 1.

As in [7, §2], it is then enough to prove the following two statements.

3.1. For any α ∈ Λ_{n}^{(e)} and any \underline{r} = (r_{1}, …, r_{e}) with r_{i} ⩾ n, we have

(−1)^{ne}(∏_{k=1}^{e} u_{k}^{|α_{k}|−n}) D_{α} R_{α}(ζ(\underline{r}), η_{1}(\underline{r}), …, η_{e−1}(\underline{r})) = ω_{α,\underline{r}}(q, u_{1}, …, u_{e}).

3.2. The set

\{(ζ(\underline{r}), η_{1}(\underline{r}), …, η_{e−1}(\underline{r})) | \underline{r} = (r_{1}, …, r_{e}) with r_{i} ⩾ n\}

is a Zariski dense subset of K^{e}.

We shall need the following result.

Lemma 3.3 Let \underline{r} = (r_{1}, …, r_{e}) ∈ ℕ^{e} and r = r_{1} + ⋯ + r_{e}. Then we have

η_{k}(\underline{r}) = \frac{q^{r}}{1 − q^{r}} ∑_{i=0}^{k}(−1)^{i+1} h_{k−i}(u_{1}, …, u_{e}) σ_{i}(u_{1}q^{−r_{1}}, …, u_{e}q^{−r_{e}})

for any 1 ⩽ k ⩽ e − 1, where h_{k−i} denotes the complete symmetric function of degree k − i and σ_{i} denotes the elementary symmetric function of degree i.

Proof. To simplify notation we define

τ_{j} = σ_{j}(u_{1}q^{−r_{1}}, …, u_{e}q^{−r_{e}}),

for 0 ⩽ j ⩽ e and reserve the symbol σ_{j} for σ_{j}(u_{1}, …, u_{e}). We shall use the following identities:

(1)  ∑_{i=1}^{e} \frac{u_{i}^{e−1+t}}{∏_{{j=1 \atop j≠i}}^{e}(u_{i} − u_{j})} = h_{t}, 0 ⩽ t ⩽ e−1 ;

(2)  ∑_{i=1}^{e} \frac{u_{i}^{e−1−t}}{∏_{{j=1 \atop j≠i}}^{e}(u_{i} − u_{j})} = 0, 1 ⩽ t ⩽ e−1 .

(These follow from equations (3.1) and (3.9) in [10, I.3].) Now for any 1 ⩽ i ⩽ e we have:

(1 − q^{r_{i}}) ∏_{{j=1 \atop j≠i}}^{e}(u_{i}q^{r_{j}} − u_{j}) = q^{r}u_{i}^{−1}(q^{−r_{i}}u_{i} − u_{i}) ∏_{{j=1 \atop j≠i}}^{e}(u_{i} − u_{j}q^{−r_{j}})

= −q^{r}u_{i}^{−1} ∑_{j=0}^{e}(−1)^{j}u_{i}^{e−j}τ_{j} = q^{r} ∑_{j=0}^{e}(−1)^{j+1}u_{i}^{e−1−j}τ_{j}

and hence, for 1 ⩽ k ⩽ e − 1, we obtain:

\frac{1 − q^{r}}{q^{r}} η_{k}(\underline{r}) = ∑_{i=1}^{e} \frac{u_{i}^{k} ∑_{j=0}^{e}(−1)^{j+1}u_{i}^{e−1−j}τ_{j}}{∏_{{l=1 \atop l≠i}}^{e}(u_{i} − u_{l})}

= ∑_{j=0}^{e}((−1)^{j+1}τ_{j} ∑_{i=1}^{e} \frac{u_{i}^{e−1+k−j}}{∏_{{l=1 \atop l≠i}}^{e}(u_{i} − u_{l})}) =^{(1),(2)} ∑_{j=0}^{k}(−1)^{j+1}τ_{j}h_{k−j} ,

which completes the proof. □

3.4. Proof of (3.1) By using Lemma 3.3 above and induction one proves, for any 1 ⩽ i ⩽ e − 1, the following identity:

∑_{j=1}^{i}(−1)^{i−j}η_{j}(\underline{r})σ_{i−j} = \frac{(−1)^{i+1}q^{r}}{1 − q^{r}} (τ_{i} − σ_{i})

Then the expression for R_{α} becomes:

R_{α}(\underline{r}) := R_{α}(ζ(\underline{r}), η_{1}(\underline{r}), …, η_{e−1}(\underline{r})) = (\frac{1 − q}{1 − q^{r}} )^{n}

× ∏_{k=1}^{e} ∏_{x∈α_{k}}[q^{r}(1 − q^{c(x)}) ∏_{{l=1 \atop l≠k}}^{e}(u_{k}q^{c(x)} − u_{l}) + (−1)^{e−1} ∏_{{l=1 \atop l≠k}}^{e} u_{l}

+ q^{r} ∑_{i=1}^{e−1}(−1)^{i+1}u_{k}^{e−1−i}q^{(e−i)c(x)}τ_{i} + q^{r} ∑_{i=1}^{e}(−1)^{i}u_{k}^{e−1−i}q^{(e−i)c(x)}σ_{i}] .

Weights of Markov traces 10 Now note that we have

q^{r}(1 − q^{c(x)}) ∏_{{l=1 \atop l≠k}}^{e}(u_{k}q^{c(x)} − u_{l}) = −q^{r}u_{k}^{−1} ∏_{l=1}^{e}(u_{k}q^{c(x)} − u_{l})

= −q^{r}u_{k}^{−1} ∑_{j=0}^{e}(−1)^{j}u_{k}^{e−j}q^{c(x)(e−j)}σ_{j}.

So we obtain the equality

q^{r}(1 − q^{c(x)}) ∏_{{l=1 \atop l≠k}}^{e}(u_{k}q^{c(x)} − u_{l})

+ q^{r} ∑_{i=1}^{e}(−1)^{i}u_{k}^{e−1−i}q^{(e−i)c(x)}σ_{i} = −q^{r}u_{k}^{e−1}q^{ec(x)} ,

which yields

R_{α}(\underline{r}) = (\frac{1−q}{1−q^{r}} )^{n} ∏_{k=1}^{e} ∏_{x∈α_{k}}[q^{r} ∑_{i=0}^{e}(−1)^{i+1}u_{k}^{e−1−i}q^{(e−i)c(x)}τ_{i}].

The final step in computing R_{α}(\underline{r}) is to observe that we have

q^{r} ∑_{i=0}^{e}(−1)^{i+1}u_{k}^{e−1−i}q^{(e−i)c(x)}τ_{i} = −u_{k}^{−1}q^{r} ∑_{i=0}^{e}(−1)^{i}u_{k}^{e−i}q^{(e−i)c(x)}τ_{i}

= −u_{k}^{−1}q^{r} ∏_{l=1}^{e}(u_{k}q^{c(x)} − u_{l}q^{−r_{l}}) = (1 − q^{r_{k}+c(x)}) ∏_{{l=1 \atop l≠k}}^{e}(u_{k}^{r_{l}+c(x)} − u_{l})

and so we get

R_{α}(\underline{r}) = (\frac{1 − q}{1 − q^{r}} )^{n} ∏_{k=1}^{e} ∏_{x∈α_{k}}[(1 − q^{r_{k}+c(x)}) ∏_{{l=1 \atop l≠k}}^{e}(u_{k}^{r_{l}+c(x)} − u_{l})]. (3)

As for D_{α}, following the definition in (1.2), we get

D_{α} = (−1)^{\binom{r}{2}\binom{e}{2}+n(e−1)} · q^{−n} · q^{−f(r,e)} · ∏_{k=1}^{e} u_{k}^{n}

× \frac{∏_{1⩽i<j⩽r+1} u_{1}(q^{α_{1,i}+r−i} − q^{α_{1,j}+r−j}) · ∏_{k=2}^{e} ∏_{1⩽i<j⩽r} u_{k}(q^{α_{k,i}+r−i} − q^{α_{k,j}+r−j})}{∏_{i=1}^{r+1} ∏_{j=1}^{α_{1,i}+r+1−i} u_{1}q^{−1}(q^{j} − 1) · ∏_{k=2}^{e} ∏_{i=1}^{r} ∏_{j=1}^{α_{k,i}+r−i} u_{k}(q^{j} − 1)}

× \frac{∏_{k=2}^{e} ∏_{i=1}^{r+1} ∏_{j=1}^{r}(u_{1}q^{α_{1,i}+r−i} − u_{k}q^{α_{k,j}+r−j}) ∏_{k=2}^{e}(u_{1}q^{−1} − u_{k})^{−r}}{∏_{k=2}^{e} ∏_{i=1}^{r+1} ∏_{j=1}^{α_{1,i}+r+1−i}(u_{1}q^{j−1} − u_{k}) · ∏_{k=2}^{e} ∏_{i=1}^{r} ∏_{j=1}^{α_{k,i}+r−i}(u_{k}q^{j} − u_{1}q^{−1})}

× \frac{∏_{2⩽k<l⩽e} ∏_{i,j=1}^{r}(u_{k}q^{α_{k,i}+r−i} − u_{l}q^{α_{l,j}+r−j})}{∏_{{k,l=2 \atop k≠l}}^{e} ∏_{i=1}^{r} ∏_{j=1}^{α_{k,i}+r−i}(u_{k}q^{j} − u_{l}) · ∏_{2⩽k<l⩽e}(u_{k} − u_{l})^{r}}

In order to find an equivalent expression for D_{α} which would suit our purpose, we shall use the following equalities, valid for 2 ⩽ k ⩽ e:

∏_{i=1}^{r} ∏_{j=1}^{α_{k,i}+r−i} u_{k}(q^{j−1} − 1)

= (−1)^{|α_{k}|} · u_{k}^{|α_{k}|} · u\binom{r}{2}_{k} · ∏_{1⩽i<j⩽r}(q^{j−i} − 1) · ∏_{x∈α_{k}}(1 − q^{r+c(x)}) ,

∏_{1⩽i<j⩽r}u_{k}(q^{α_{k,i}+r−i} − q^{α_{k,j}+r−j})

= u\binom{r}{2}_{k} · q^{n(α_{k})} · q^{r(r−1)(r−2)/6} · ∏_{1⩽i<j⩽r}(q^{α_{k,i}−α_{k,j}+j−i} − 1)

and similar ones corresponding to k = 1. We shall also make use of the following identity, valid for any partition α and any integers r, s ⩾ l(α) (see [10, I.1])

\frac{∏_{1⩽i<j⩽r} \frac{1 − q^{α_{i}−α_{j}+j−i}}{1 − q^{j−i}}}{∏_{x∈α}(1 − q^{r+c(x)})} = \frac{∏_{1⩽i<j⩽s} \frac{1 − q^{α_{i}−α_{j}+j−i}}{1 − q^{j−i}}}{∏_{x∈α}(1 − q^{s+c(x)})}

In this way we obtain a simpler expression:

D_{α} = (−1)^{ne} · q^{n(α)} · ∏_{k=1}^{e} u_{k}^{n−|α_{k}|} · \frac{∏_{k=1}^{e} ∏_{1⩽i<j⩽r_{k}} \frac{1 − q^{α_{k,i}−α_{k,j}+j−i}}{1 − q^{j−i}}}{∏_{k=1}^{e} ∏_{x∈α_{k}}(1 − q^{r_{k}+c(x)})}

× \frac{∏_{k=2}^{e} ∏_{i=1}^{r+1} ∏_{j=1}^{r} \frac{u_{1}q^{α_{1,i}−i} − u_{k}q^{α_{k,j}−j}}{u_{1}q^{−i} − u_{k}q^{−j}}}{∏_{k=2}^{e} ∏_{x∈α_{1}}(u_{1}q^{r+c(x)} − u_{k}) · ∏_{k=2}^{e} ∏_{x∈α_{k}}(u_{k}q^{r+1+c(x)} − u_{1})}

× \frac{∏_{2⩽k<l⩽e} ∏_{i,j=1}^{r} \frac{u_{k}q^{α_{k,i}−i} − u_{l}q^{α_{l,j}−j}}{u_{k}q^{−i} − u_{l}q^{−j}}}{∏_{{k,l=2 \atop k≠l}}^{e} ∏_{x∈α_{k}}(u_{k}q^{r+c(x)} − u_{l})}

Finally, we use the following identity, valid for any indeterminates u, v, any partition α and any integer r ⩾ l(α) (see [7, 2.3b] and [10, I.1]),

∏_{x∈α}(uq^{r+c(x)} − v) = ∏_{i=1}^{r} \frac{φ_{α_{i}+r−i}(q)}{φ_{r−i}(q)} ,

where φ_{k}(q) = (uq − v)(uq^{2} − v) ⋯ (uq^{k} − v), and we obtain, for 2 ⩽ k < l ⩽ e

∏_{i,j=1}^{r} \frac{u_{k}q^{α_{k,i}−i} − u_{l}q^{α_{l,j}−j}}{u_{k}q^{−i} − u_{l}q^{−j}} = ∏_{i=1}^{r_{k}} ∏_{j=1}^{r_{l}} \frac{u_{k}q^{α_{k,i}−i} − u_{l}q^{α_{l,j}−j}}{u_{k}q^{−i} − u_{l}q^{−j}}

× ∏_{x∈α_{k}} \frac{u_{k}q^{r+c(x)} − u_{l}}{u_{k}q^{r_{l}+c(x)} − u_{l}} · ∏_{x∈α_{l}} \frac{u_{k} − u_{l}q^{r+c(x)}}{u_{k} − u_{l}q^{r_{k}+c(x)}} ,

and a similar equality holds for k = 1, l ∈ {2, …, e}. Combining (3) and all the above computations for D_{α}, we obtain:

D_{α}R_{α}(ζ(\underline{r}), η_{1}(\underline{r}), …, η_{e−1}(\underline{r})) = (−1)^{ne} · ω_{α,r}(\underline{r}) · ∏_{k=1}^{e} u_{k}^{n−|α_{k}|}

and the proof of (3.1) is now complete.

3.5. Proof of (3.2) We let A be the K-algebra of all functions from the set \{\underline{r} = (r_{1}, …, r_{e}) ∈ ℕ^{e} | r_{i} ⩾ n\} to K (with pointwise defined operations). Then to say that the set in (3.2) is Zariski dense is the same as to say that the functions ζ, η_{1}, …, η_{e−1} ∈ A are algebraically independent. In order to prove that these functions indeed are algebraically independent, we proceed as follows. We begin with the functions x_{1}, …, x_{e} ∈ A where

x_{i}(\underline{r}) = u_{i}q^{−r_{i}} for 1 ⩽ i ⩽ e.

These functions are certainly algebraically independent, since \underline{r} runs over all etuples of positive integers ⩾ n. Then, by [10, I.2.4], the elementary symmetric functions σ_{i}(x_{1}, …, x_{e}) ∈ A (1 ⩽ i ⩽ e) are also algebraically independent. Now note the following general facts. Assume that α_{1}, …, α_{m} ∈ A are algebraically independent. Then the following hold.

(i) If α_{1} is invertible in A, then α_{1}^{−1}, α_{2}, …, α_{m} are algebraically independent, and so are α_{1}, α_{1}α_{2}, …, α_{1}α_{m};

(ii) If (b_{ij}) ∈ GL_{m}(K), then β_{1}, …, β_{m} ∈ A are algebraically independent, where β_{j} = ∑_{i=1}^{m} b_{ij}α_{i}.

(iii) If c_{1}, …, c_{m} ∈ K, then α_{1} + c_{1}, …, α_{m} + c_{m} are algebraically independent.

We will now show that ζ, η_{1}, …, η_{e−1} are obtained from the elementary symmetric functions σ_{i}(x_{1}, …, x_{e}) by applying the above three rules. First define functions y′_{1}, …, y′_{e−1} ∈ A by

y′_{k} = ∑_{i=0}^{k} \underbrace{(−1)^{i+1}h_{k−i}(u_{1}, …, u_{e})}_{∈K} σ_{i}(x_{1}, …, x_{e}) for 1 ⩽ k ⩽ e − 1.

Note that y′_{k} = (−1)^{k−1}σ_{k}(x_{1}, …, x_{e})+ K-linear combination of σ_{i}(x_{1}, …, x_{e}) with 0 ⩽ i < k (where σ_{0} = 1). Hence, using (ii) and (iii), the functions σ_{e}(x_{1}, …, x_{e}), y′_{1}, …, y′_{e−1} are also algebraically independent. Now observe that

σ_{e}(x_{1}, …, x_{e})(\underline{r}) = x_{1}(\underline{r}) ⋯ x_{e}(\underline{r}) = u_{1} ⋯ u_{e}q^{−r}.

Thus, q^{−r}, y′_{1}, …, y′_{e−1} are algebraically independent. Then so are the functions q^{−r} − 1, y′_{1}, …, y′_{e−1}, using (iii). Since q^{r} ≠ 1 for all r considered, the function q^{−r} − 1 is invertible in A; we have

ζ′(\underline{r}) := (q^{−r} − 1)^{−1} = \frac{q^{r}}{1 − q^{r}} .

Hence, using (i), we see that ζ′, ζ′y′_{1}, …, ζ′y′_{e−1} are algebraically independent. Hence ζ, η_{1}, …, η_{e−1} are algebraically independent, as required. This completes the proof of (3.2).

4 The canonical symmetrizing trace

We now specialize the result of Theorem 1.3 to obtain the weights of a particular very important Markov trace on Ĥ_{n}^{(e)}. The existence of this trace function was proved by Bremke and Malle [2] who constructed it by giving its values on a suitable basis of Ĥ_{n}^{(e)}. The values of the corresponding weights had been previously conjectured by the third author in [11].

We start with an easy lemma:

Lemma 4.1 Assume that we have a Markov trace τ: Ĥ_{n}^{(e)} → K with parameters ζ, η_{1}, …, η_{e−1} ∈ K where ζ = 0. Then, for any 0 ⩽ m < n and h ∈ Ĥ_{m}^{(e)}, we have

τ(hT_{m}^{(k)}) = q^{m}η_{k}τ(h) for all 1 ⩽ k ⩽ e − 1,

where we set T_{m}^{(k)} = S_{m} ⋯ S_{1}T^{k}S_{1} ⋯ S_{m}.

Proof. Let 0 ⩽ m ⩽ n − 1, h ∈ Ĥ_{m}^{(e)} and 1 ⩽ k ⩽ e − 1. Furthermore, we set

T̃_{i} := S_{m} ⋯ S_{1}T^{k}S_{1}^{−1} ⋯ S_{i}^{−1}S_{i+1} ⋯ S_{m} for 0 ⩽ i ⩽ m.

Thus, we have T̃_{0} = T_{m}^{(k)} and T̃_{m} = (T′_{m})^{k}. Now we show that

τ(hT̃_{i}) = q^{m−i}η_{k}τ(h) for i = 0, …, m.

We do this by downward induction on i. If i = m, then T̃_{m} = (T′_{m})^{k} and the required relation holds by the definition in (1.1). Now assume that i < m. Using the quadratic relation S_{i+1}^{2} = q1 + (q − 1)S_{i+1}, we can write T̃_{i} = q T̃_{i+1} + (q − 1)x where

x := S_{m} ⋯ S_{1}T^{k}S_{1}^{−1} ⋯ S_{i}^{−1}S_{i+2} ⋯ S_{m}
= (S_{m} ⋯ S_{i+2}S_{i+1}S_{i+2} ⋯ S_{m})S_{i} ⋯ S_{1}T^{k}S_{1}^{−1} ⋯ S_{i}^{−1}.

Now the term in brackets equals S_{i+1} ⋯ S_{m−1}S_{m}S_{m−1} ⋯ S_{i+1}. Thus, we can write hx in the form hx_{1}S_{m}x_{2} where x_{1}, x_{2} ∈ Ĥ_{m}^{(e)}. Using the fact that ζ = 0, we obtain that

τ(hx) = τ(hx_{1}S_{m}x_{2}) = τ(x_{2}hx_{1}S_{m}) = ζτ(x_{2}hx_{1}) = 0.

It follows that τ(T̃_{i}) = q τ(T̃_{i+1}), and we are done by induction. Thus, the assertion follows if we take i = 0. □

We now proceed to show that the trace introduced in [2] is a Markov trace with all parameters equal to 0. For this, we recall the construction in [2].

4.2. For 1 ⩽ k ⩽ n, a ∈ ℤ, we define the following elements in H_{n}^{(e)}:

T_{k,a} := \begin{cases}S_{k−1} ⋯ S_{1}T^{a} & for a ≠ 0, \\ 1 & for a = 0. \end{cases}

A short calculation [2, Lemma 2.3] shows that these elements satisfy

(a) S_{i}T_{k,a} = \begin{cases}T_{k,a}S_{i} & if i > k or a = 0, \\ T_{k+1,a} & if i = k, a ≠ 0, \\ T_{k,a}S_{i+1} & if i < k − 1, a ≠ 0. \end{cases}

Note that the Iwahori-Hecke algebra of the symmetric group 𝔖_{n} ≅ W_{n}^{(1)} is isomorphic to H_{n}^{(1)} and is embedded naturally into H_{n}^{(e)} as the subalgebra generated by S_{1}, … S_{n−1}. Thus it makes sense to speak of elements T_{w} ∈ H_{n}^{(e)} for w ∈ 𝔖_{n}: take a reduced expression of w and form the corresponding product of standard generators of H_{n}^{(1)} ⊆ H_{n}^{(e)}. By Matsumoto’s lemma [6, Lemma 4.4.3] the resulting element does not depend on the choice of reduced expression.

It was proved in [2, Lemma 1.5] that the set

B := {T_{1,a_{1}} ⋯ T_{n,a_{n}}T_{w} | 0 ⩽ a_{i} ⩽ e − 1, w ∈ 𝔖_{n} = W_{n}^{(1)}} (b)

forms a basis of H_{n}^{(e)}, and that the function τ_{0} : H_{n}^{(e)} → A defined by linearly extending τ_{0}(1) = 1, τ_{0}(U) = 0 for U ∈ B \ {1}, is a trace function [2, Prop. 2.7].

Lemma 4.3 The Markov trace τ: Ĥ_{n}^{(e)} → K with parameters ζ = η_{1} = … = η_{e−1} = 0 equals the canonical trace τ_{0} defined above, i.e., we have

τ(1) = 1 and τ(U) = 0 for U ∈ B \ {1}.

Proof. By the definition of a Markov trace τ(1) = 1, so we have to show that τ vanishes on all elements U ∈ B \ {1}. For U ∈ B \ {1} we write m(U) for the smallest integer m such that U ∈ H_{m}^{(e)}. In particular 1 ⩽ m(U) ⩽ n. The proof will now proceed by induction on m(U). If m(U) = 1 then U = T^{k} for some 1 ⩽ k ⩽ e − 1, and by the second defining equation of a Markov trace

τ(U) = τ(T^{k}) = η_{k}τ(1) = 0.

Now assume that m > 1 and τ(U) = 0 has been shown for all U ∈ B \ {1} with m(U) < m. Let U ∈ B \ {1} with m(U) = m. Then U = T_{1,a_{1}} ⋯ T_{m,a_{m}}T_{w} with w ∈ 𝔖_{m}, and S_{m−1} occurs at least once in U. If it occurs precisely once then U = U_{1}S_{m−1}U_{2} with U_{1}, U_{2} ∈ H_{m−1}^{(e)}. In particular

τ(U) = τ(U_{1}S_{m−1}U_{2}) = τ(U_{2}U_{1}S_{m−1}) = ζτ(U_{2}U_{1}) = 0

as required. Hence we may now assume that S_{m−1} occurs at least twice in U. In particular U = T_{1,a_{1}} ⋯ T_{m,a_{m}}T_{w} and w ∈ 𝔖_{m} \ 𝔖_{m−1}. Clearly the set

{1, s_{m−1}, s_{m−2}s_{m−1}, …, s_{1} ⋯ s_{m−1}}

forms a system of left coset representatives of 𝔖_{m−1} in 𝔖_{m}. This implies that any element w ∈ 𝔖_{m} \ 𝔖_{m−1} has a reduced expression of the form s_{j}s_{j+1} ⋯ s_{m−1}w′ with w′ ∈ 𝔖_{m−1}, for some 1 ⩽ j ⩽ m − 1. In particular we may assume that S_{m−1} occurs only once in T_{w} and so we must have a_{m} ≠ 0 in U. First assume that j > 1. Then by (4.2)(a) we have

U = T_{1,a_{1}} ⋯ T_{m,a_{m}}S_{j} ⋯ S_{m−1}T_{w′} = T_{1,a_{1}} ⋯ T_{m−1,a_{m−1}}S_{j−1} ⋯ S_{m−2}T_{m,a_{m}}T_{w′}.

Hence U can be written in a form such that S_{m−1} occurs only once, and we are done by the previous step.

Finally if j = 1 then

U = T_{1,a_{1}} ⋯ T_{m,a_{m}}S_{1} ⋯ S_{m−1}T_{w′} = T_{1,a_{1}} ⋯ T_{m−1,a_{m−1}}T_{m}^{(a_{m})}T_{w′}

and thus

τ(U) = τ(T_{w′}T_{1,a_{1}} ⋯ T_{m−1,a_{m−1}}T_{m}^{(a_{m})}) = η_{a_{m}}τ(T_{w′}T_{1,a_{1}} ⋯ T_{m−1,a_{m−1}}) = 0

by Lemma 4.1. This completes the induction step and hence the proof. □

4.4. For a reduced expression w (with respect to the standard generators of W_{n}^{(e)}) of an element w ∈ W_{n}^{(e)} let T_{w} denote the corresponding product of standard generators of H_{n}^{(e)}. In contrast to the situation for Iwahori-Hecke algebras, T_{w} may well depend on the choice of reduced expression for w. Nevertheless it was shown in [2, Th. 2.8] that the trace form τ_{0} satisfies the following property:

τ_{0}(1) = 1 and τ_{0}(T_{w}) = 0 for any reduced expression w of 1 ≠ w ∈ W_{n}^{(e)}.

4.5. We note that the proof of conjecture [11, 2.20] has several useful consequences. First, by [11, 5.13] and [12, 3.9], the generic degrees for the cyclotomic algebra attached to any imprimitive complex reflection group with respect to the canonical trace form constructed in [2] are now known.

Secondly, by [11, 3.16 and 6.11], it implies the validity of conjecture (dHV6) in [3] for classical groups. That is, all Φ_{d}-blocks of classical groups of Lie type satisfy the so-called generalized d-Howlett-Lehrer-Lusztig theories: the degrees of the unipotent characters in a Φ_{d}-block b are obtained by a suitable specialization from the generic degrees of the cyclotomic algebra attached to the relative Weyl group of b.

Thirdly, [11, 3.14 and 6.10] shows that this generalized d-Howlett-Lehrer-Lusz-tig theory is also valid for the Φ-blocks of unipotent degrees of the complex reflection groups W_{n}^{(e)} = G(e, 1, n) and G(e, e, n) constructed in [11].

Fourthly, the precise form of the generic degrees attached to the canonical trace form on H_{n}^{(e)} is used in [13, Prop. 8.1] to define the notion of spetsiality. It entails the fact that imprimitive complex reflection groups are spetsial if and only if they have unipotent degrees.

5 A palindromicity property

We end by proving that the Schur elements of H_{n}^{(e)} satisfy a certain palindromicity property (see Theorem 5.2). By a result of Broué, Malle and Michel [4] this implies that the Markov trace τ_{0} considered in the previous section is the unique trace form on H_{n}^{(e)} enjoying certain natural properties.

We first need to compute some auxiliary invariants. The irreducible characters of Ĥ_{n}^{(e)} and of all semisimple specializations of H_{n}^{(e)} are parametrized by multipartitions α = (α_{1}, …, α_{e}) of n (see Section 1). We will interpret χ_{α} ∈ Irr(Ĥ_{n}^{(e)}) also as character of any such specialization, in particular of the reflection group W_{n}^{(e)}. As such, it is obtained by inducing a suitable irreducible character from the Young subgroup W_{α}^{(e)} = C_{e}^{n}.𝔖_{α} to W_{n}^{(e)} (see for example [11, (2.3)]). We write m_{j}(α) (1 ⩽ j ⩽ e) for the multiplicity of u_{j} as eigenvalue of T in a matrix representation affording χ_{α}, and similarly m_{0}(α) for the multiplicity of q as eigenvalue of S_{1}. Note that this is the same as the dimension of the exp(2πij/e)-eigenspace of t, resp. the dimension of the fixed space of s_{1} in a representation of W_{n}^{(e)} affording χ_{α}.

Lemma 5.1 Let α = (α_{1}, …, α_{e}), α_{i} ⊢ n_{i}, ∑ n_{i} = n. Then we have

m_{e−j+1}(α) = \frac{(n − 1)!}{|𝔖_{α}|}n_{j} ∏_{i=1}^{e} χ_{α_{i}}(1) (1 ⩽ j ⩽ e),

m_{0}(α) = \frac{1}{2}χ_{α}(1) + \frac{(n − 2)!χ_{α}(1)}{n!} ∑_{i=1}^{e} \binom{n_{i}}{2}\frac{χ_{α_{i}}(s_{1})}{χ_{α_{i}}(1)}.

For the proof, we determine the multiplicity of ζ^{e−j+1} as an eigenvalue of t, respectively of 1 as an eigenvalue of s_{1}, in a representation of W_{n}^{(e)} affording χ_{α}. But this is straightforward from the description of χ_{α} as induced character and hence omitted.

Let π denote the central element π := (T S_{1} ⋯ S_{n−1})^{en} of Ĥ_{n}^{(e)} and for χ_{α} ∈ Irr(Ĥ_{n}^{(e)}) write Ω_{α} for the corresponding central character. Finally let

S_{α}(q, u_{1}, …, u_{e}) := ((q − 1)^{n}D_{α}(q; u_{1}q^{−1}, u_{2}, …, u_{e}))^{−1}

be the Schur element of χ_{α} on Ĥ_{n}^{(e)} with respect to τ_{0}. Thus S_{α}(q, u_{1}, …, u_{e}) is a rational function in the indeterminates q, u_{1}, …, u_{e}.

Theorem 5.2 For all α ⊢ n we have

Ω_{α}(π)S_{α}(q^{−1}, u_{1}^{−1}, …, u_{e}^{−1}) = τ_{0}(π)S_{α}(q, u_{1}, …, u_{e}).

Proof. It is clear by inspection of the defining formula for D_{α} that the stated formula is true up to a sign and a factor which is a monomial in q, u_{1}, …, u_{e}. By [4, 2.1.2(b)] we have

τ_{0}(π) = (−1)^{n(e−1)}q^{e\binom{n}{2}} ∏_{i=1}^{e} u_{i}^{n}.

Furthermore, by [4, p. 169]

Ω_{α}(π) = q^{l_{0}} ∏_{i=1}^{e} u_{i}^{l_{i}},

where

l_{0} = en(n − 1)m_{0}(α)/χ_{α}(1) = e\binom{n}{2} + e ∑_{i=1}^{e} \binom{n_{i}}{2}\frac{χ_{α_{i}}(s_{1})}{χ_{α_{i}}(1)}

and

l_{i} = enm_{e−i+1}(α)/χ_{α}(1) = en_{i}

by Lemma 5.1.

Thus it remains to determine the effect of replacing the indeterminates by their inverses in S_{α}. First note that when replacing u_{i} by u_{i}^{−1} in D_{α} we obtain a factor u_{i}^{n−en_{i}}. Assume that α is represented by an e-symbol \underline{S} = (S_{1}, …, S_{e}) as in Section 1.2 with |S_{1}| − 1 = |S_{2}| = … = |S_{e}| = r. Then upon replacing q by q^{−1} we obtain a factor q^{h} with

h = ern + \frac{1}{6}er(r + 1)(er − e + 3) − e ∑_{λ∈\underline{S}} \binom{λ + 1}{2} + en_{1}.

Now by Lemma 5.3 this is equal to

h = −e ∑_{i=1}^{e} \binom{n_{i}}{2}\frac{χ_{α_{i}}(s_{1})}{χ_{α_{i}}(1)}.

Collecting the above results shows that the asserted formula is true up to sign. The correctness of the sign is now seen by specializing all terms to the reflection group W_{n}^{(e)}. □

The following statement on the values of irreducible characters of symmetric groups on transpositions was needed in the proof of the preceding theorem:

Lemma 5.3 Let n ⩾ 2, s ∈ 𝔖_{n} a transposition, α ⊢ n a partition with corresponding β-set S of length |S| = r. Then

\binom{n}{2}\frac{χ_{α}(s)}{χ_{α}(1)} = ∑_{λ∈S} \binom{λ + 1}{2} − nr − \binom{r + 1}{3}.

Proof. According to [4, p. 176] the Iwahori-Hecke algebra of 𝔖_{n} has a trace form satisfying [4, 2.1]. Thus by [4, Lemma 2.7] the assertion of Theorem 5.2 is true for e = 1. In particular n(n − 1)m_{0}(α)/χ_{α}(1) equals ∑_{λ∈S} \binom{λ+1}{2} − nr − \binom{r+1}{3} up to a summand independent of α and r. The value of this summand can easily be found by evaluating both sides at α = (n). □

Note that by [4, Lemma 2.7] the above Theorem 5.2 implies that τ_{0} satisfies [4, 2.1 1(c)]. In particular, together with the other results in [4, Section 2] this shows that τ_{0} is the trace form uniquely determined by the conditions stated in [4, 2.1(1)].

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Weights of Markov traces 19

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[15] H. Wenzl, Hecke algebras of type A_{n} and subfactors, Invent. Math. 92 (1988), 349–383.

M.G.: Institut Girard Desargues, bat. 101, Université Lyon 1, 43 bd du 11 novembre 1918, F–69622 Villeurbanne cedex, France

E-mail address: geck@desargues.univ-lyon1.fr

L.I.: Institut Girard Desargues, bat. 101, Université Lyon 1, 43 bd du 11 novembre 1918, F–69622 Villeurbanne cedex, France and Facultatea de Stiinte, Universitatea Baia Mare, Victoriei 76, 4800 Baia Mare, Romania

E-mail address: iancu@desargues.univ-lyon1.fr

G.M.: FB Mathematik/Informatik, Universität Kassel, Heinrich-Plett-Str. 40, D–34132 Kassel, Germany.

E-mail address: malle@mathematik.uni-kassel.de