# Modified Ariki-Koike algebra and Yokonuma-Hecke like relations

Myungho Kim, Sungsoon Kim  Department of Mathematics, Kyung Hee University,
Seoul 02447, Korea<EMAIL_ADDRESS>Dṕartement de Mathématiques, CNRS UMR 7352
-UPJV, 33 rue St. Leu, Amiens, FRANCE<EMAIL_ADDRESS>
###### Abstract.

We find new presentations of the modified Ariki-Koike algebra (known also as
Shoji’s algebra) $\mathcal{H}_{n,r}$ over an integral domain $R$ associated
with a set of parameters $q,u_{1},\ldots,u_{r}$ in $R$. It turns out that the
algebra $\mathcal{H}_{n,r}$ has a set of generators $t_{1},\ldots,t_{n}$ and
$g_{1},\ldots g_{n-1}$ subject to a set of defining relations similar to the
relations of Yokonuma-Hecke algebra. We also obtain a presentation of
$\mathcal{H}_{n,r}$ which is independent to the choice of $u_{1},\ldots
u_{r}$. Hence the algebras associated with the parameters $q,u_{1},\ldots
u_{r}$ and $q,u^{\prime}_{1},\ldots u^{\prime}_{r}$ are isomorphic even in the
case that $(u_{1},\ldots u_{r})$ and $(u^{\prime}_{1},\ldots u^{\prime}_{r})$
are different. As applications of the presentations, we find an explicit trace
form on the algebra $\mathcal{H}_{n,r}$ which is symmetrising provided the
parameters $u_{1},\ldots,u_{r}$ are invertible in $R$. We also show that the
symmetric group $\mathfrak{S}(r)$ acts on the algebra $\mathcal{H}_{n,r}$, and
find a basis and a set of generators of the fixed subalgebra
$\mathcal{H}_{n,r}^{\mathfrak{S}(r)}$.

The research of M. Kim was supported by the National Research Foundation of
Korea (NRF) Grant funded by the Korea government(MSIP) (NRF-
NRF-2022R1F1A1076214 and NRF-2020R1A5A1016126).

S. Kim is grateful of K.I.A.S. for the host during this work as well as l’IEA-
France for partial supports (2023).

###### Contents

  1. 1 Introduction
  2. 2 Modified Ariki-Koike algebras
    1. 2.1 The subalgebra $R[t_{1},\ldots,t_{n}]$
    2. 2.2 Actions of symmetric groups on $R[t_{1},\ldots,t_{n}]$
  3. 3 Yokonuma-Hecke like presentation
  4. 4 Symmetrizing trace form
  5. 5 Symmetric group action on $\mathcal{H}_{n,r}$ and the fixed subalgebra $\mathcal{H}_{n,r}^{\mathfrak{S}(r)}$

##  1\. Introduction

In the article [7], Shoji introduced a new presentation of the generic Ariki-
Koike algebra of the group $G(r,1,n)$, a complex reflection group consisting
of $n\times n$ monomial matrices whose non-zero entries are $r$-th roots of
unity, while the product of all non-zero entries is a $r$-th root of unity.
Since this algebra is defined over the ring
${\mathbb{Z}}[q,q^{-1},u_{1},\ldots,u_{r}][\Delta^{-1}]$, where
$\Delta=\prod_{1\leq i<j\leq r}(u_{i}-u_{j})$ and $q,u_{1},\ldots,u_{r}$ are
indeterminates, it can be specialized to an algebra over an integral domain
$R$, i.e., the $R$-algebra with the same generators and relations with respect
to the parameters $q,u_{1},\ldots,u_{r}$ in $R$, provided that $q$ and
$\prod_{1\leq i<j\leq r}(u_{i}-u_{j})$ are invertible in $R$. Its definition,
as given in Definition 2.1, is with the generators $\\{T_{i},t_{j}\,|\,1\leq
i\leq n-1,\ 1\leq j\leq n\\}$ with the defining relations involving the
parameters $q,u_{1},\ldots,u_{r}$. Here $T_{1},\ldots,T_{n-1}$ are the
generators of the Iwahori-Hecke algebra of the symmetric group
$\mathfrak{S}(n)$ and $t_{1},\ldots,t_{n}$ satisfy a common relation of degree
$r$.

This algebra with new presentation is called the _Modified Ariki-Koike
algebra_ , or _Shoji algebra_ , and we denote it by
$\mathcal{H}_{n,r}(R,q,u_{1},\ldots,u_{r})$ or simply by $\mathcal{H}_{n,r}$
if there is no risk of confusion. The algebra $\mathcal{H}_{n,r}$ has the
following $R$-basis ([7, Theorem 3.7])

$\displaystyle\left\\{t_{1}^{c_{1}}\cdots t_{n}^{c_{n}}T_{w}\mid
w\in\mathfrak{S}(n),\ 0\leq c_{1},\ldots,c_{n}\leq r-1\right\\}$ (1.1)

thus $\mathcal{H}_{n,r}$ is a free $R$-module of rank $\mid
G(n,1,r)\mid=r^{n}n!$.

One advantage of this new presentation of the generic Ariki-Koike algebra of
$G(r,1,n)$ is that, using the fact that the generators $\\{t_{j}\,|\,1\leq
j\leq n-1\\}$ are symmetric in the defining relations, one can embed
$\mathcal{H}_{n,r}\otimes\mathcal{H}_{m,r}$ into $\mathcal{H}_{m+n,r}$ as in
the case of the group algebras. It follows that all the irreducible
representations of $\mathcal{H}_{n,r}$ can be constructed as induced modules
from certain subalgebras in an analogous way to the group case ([7]). In [8],
the specialized algebra $\mathcal{H}_{n,r}(R,q,u_{1},\ldots,u_{r})$ is studied
in the connection with the Ariki-Koike algebra over $R$ with the same
parameters $q,u_{1},\ldots,u_{r}$ (see [8, Section 1]). Even they are not
isomorphic in general, there is an interesting algebra homomorphism from the
latter to the former, which allowed the authors in [8] to produce some
connections between thier representations. For example, an estimate of the
decomposition numbers for the Ariki-Koike algebra of $G(r,1,n)$ is obtained in
terms of the decomposition numbers for the modified Ariki-Koike algebra.

In this paper, we obtain new presentations of the modified Ariki-Koike algebra
$\mathcal{H}_{n,r}$ and study their applications. In short, these new
presentations are quite similar to those of the _Yokonuma-Hecke algebra_ (of
type A) $\mathcal{Y}_{n,r}$ and therefore some properties analogous to those
we had in the case of $\mathcal{Y}_{n,r}$ can be developed for
$\mathcal{H}_{n,r}$. The Yokonouma-Hecke algebra $\mathcal{Y}_{n,r}$ is a
deformation of the group algebra of $G(r,1,n)$. Since it is also a quotient of
the group algebra of the framed braid group ${\mathbb{Z}}\wreath B_{n}$ (see
[1]) it has been studied in connection with with knot theory (see for example
the introduction of [4] and references therein).

As the first step, we take an $R$-basis

$\left\\{b_{k_{1},\ldots,k_{n}}:=\displaystyle\prod_{1\leq i\leq
n}\prod_{1\leq j\leq r,\,j\neq
k_{i}}\dfrac{t_{i}-u_{j}}{u_{k_{i}}-u_{j}}\,|\,(k_{1},\ldots,k_{n})\in[1,r]^{n}\right\\}$

of the subalgebra $R[t_{1},\ldots,t_{n}]$ generated by the mutually commuting
generators $t_{1},\ldots,t_{n}$. This basis forms a complete system of
orthogonal idempotents of the subalgebra $R[t_{1},\ldots,t_{n}]$ (Lemma 2.4).
Define

$g_{i}:=T_{i}-(q-q^{-1})\sum_{k_{i}<k_{i+1}}b_{k_{1},\ldots,k_{n}}\quad(1\leq
i\leq n-1).$

Then we have $g_{i}t_{i}=t_{i+1}g_{i}$, $g_{i}t_{j}=t_{j}g_{i}$ $(|i-j|\geq
2)$ and $g_{i}$’s satisfy the braid relations. With these new elements, using
(1.1), we obtain the following new $R$-bases of $\mathcal{H}_{n,r}$ :

$\displaystyle\left\\{t_{1}^{c_{1}}\cdots t_{n}^{c_{n}}g_{w}\mid
w\in\mathfrak{S}(n),\ 0\leq c_{i}\leq r-1\right\\}\quad\text{and}$ (1.2)
$\displaystyle\left\\{b_{k_{1},\ldots,k_{n}}g_{w}\mid(k_{1},\ldots,k_{n})\in[1,r]^{n},\
w\in\mathfrak{S}(n)\right\\}.$ (1.3)

Finally these bases allow us to obtain two new presentations of
$\mathcal{H}_{n,r}$: See Theorem 3.11, and Theorem 3.14, respectively. We call
the presentation in Theorem 3.11 with generators $g_{i}$’s and $t_{j}$’s the
_Yokonuma-Hecke like presentation_ of $H_{n,r}$, since it is similar to a
standard presentation of the Yokonuma-Hecke algebra $\mathcal{Y}_{n,r}$ (see
for example [4]). Indeed this Yokonuma-Hecke like presentation is motivated by
[3, Theorem 13], for which the authors proved that the Yokonuma-Hecke algerba
$\mathcal{Y}_{n,r}$ is isomorphic to $\mathcal{H}_{n,r}$ over
${\mathbb{C}}[q,q^{-1}]$ with the choice $u_{k}=exp(2\pi ik/r)$ (see Remark
3.12). In this way, the modified Ariki-Koike algebra $\mathcal{H}_{n,r}$ can
be considered as a generalization of the Yokonuma-Hecke algebra. As an
application of the Yokonuma-Hecke like presentation, in Section 4, we
construct a trace form on $\mathcal{H}_{n,r}$. Recall that an $R$-linear map
$f:\mathcal{A}\to R$ from an $R$-algerba $\mathcal{A}$ to its base ring $R$ is
called a _trace form_ if $f(xy)=f(yx)\,\,\text{for}\,\,x,y\in\mathcal{A}.$ A
trace form $f$ is called _symmetrizing_ if the bilinear form
$\mathcal{A}\times\mathcal{A}\to R$ given by $(x,y)\mapsto f(xy)$ is non-
degenerate. The trace form $\tau:\mathcal{H}_{n,r}\longrightarrow R$ is given
by

$\displaystyle\tau(t_{1}^{c_{1}}\cdots
t_{n}^{c_{n}}g_{w})=\begin{cases}1&\text{if}\ \
w=\operatorname{id}_{\mathfrak{S}(n)},c_{1}=\cdots=c_{n}=0\\\ 0&\text{
otherwise }\end{cases}$ (1.4)

for $0\leq c_{1},\ldots,c_{r}\leq r-1$, $w\in\mathfrak{S}(n)$. Note that this
formula of $\tau$ is exactly the same as the one in [4, Proposition 10] for
the Yokonuma-Hecke algebra $\mathcal{Y}_{n,r}$. Since we have the Yokonuma-
Hecke like presentation, the same proof as in [4, Proposition 10] works for
$H_{n,r}$ to show that $\tau$ is a trace form (Corollary 4.2). We further show
that $\tau$ is symmetrizing if $\sigma_{r}:=u_{1}\cdots u_{r}$ is invertible
in $R$ (Corollary 4.6) by constructing an explicit dual basis to the basis
in(1.2).

The other new presentation of $\mathcal{H}_{n,r}$ in Theorem 3.14 is
interesting at first glance because it is independent of the parameters
$u_{1},\ldots,u_{r}$. That is, the algebra $\mathcal{H}_{n,r}$ is generated by
$g_{i}$’s and $b_{k_{1},\ldots,k_{n}}$’s with defining relations in which the
parameters $u_{1},\ldots,u_{r}$ do not appear. Hence the algebras
$H_{n,r}(R,q,u_{1},\ldots u_{r})$ are isomorphic no matter what values we
choose for $u_{1},\ldots,u_{r}$ (Corollary 3.14). This property is quite
unexpected for the authors of the present paper at the beginning of the work.

Note that there is a natural action of the symmetric group $\mathfrak{S}(r)$
on $R[t_{1},\ldots,t_{n}]$ given by
${}^{\sigma}b_{k_{1},\ldots,k_{n}}:=b_{\sigma(k_{1}),\ldots,\sigma(k_{n})}$
for $\sigma\in\mathfrak{S}(r)$. Then we show that $\mathfrak{S}(r)$ acts on
the algebra $\mathcal{H}_{n,r}$ by the presentation in Theorem 3.14. In [5,
6], the authors introduce a basis $\\{E_{\chi}\\}$ of the commutative
subalgebra ${\mathbb{C}}[q,q^{-1}][t_{1},\ldots,t_{n}]$ of the Yokonuma-Hekce
algebra $\mathcal{Y}_{n,r}$ and show that $\mathfrak{S}(r)$ acts on the basis
and the algebra $\mathcal{Y}_{n,r}$. Indeed the basis
$\\{b_{k_{1},\ldots,k_{n}}\\}$ and the presentation in Theorem 3.14 can be
regarded as a generalization of $\\{E_{\chi}\\}$ and the presentation in [6,
Section 2.2] so that one can consider the fixed subalgebra
$\mathcal{H}_{n,r}^{\mathfrak{S}(r)}$ of the modified Ariki-Koike algebra as
it was done in [6] for the Yokonuma-Hecke algebras. Similary as in [6], we
obtain an $R$-basis and a set of generators of the fixed subalgebra. In the
case of the Yokonuma Hecke algebra with $n\geq r$, the fixed subalgebra is
known as the _the algebra of braids and ties_ and denoted by $BT_{n}$ (see [6,
Section 4 ] and refereces therein).

Lastly, this paper is organized as follows. In Section 2, we recall the
definition of the modified Ariki-Koike algebra $\mathcal{H}_{n,r}$. We
intorduce and study the basis elements $b_{k_{1},\ldots,k_{n}}$ of the
commutative subalgebra of $\mathcal{H}_{n,r}$ generated by
$t_{1},\ldots,t_{n}$. In Section 3, we show various relations among the
elements $g_{j}$’s and others. Then we obtain the bases (1.2), (1.3), and the
new presentations of $\mathcal{H}_{n,r}$. In Section 4, we define the trace
form (1.4) and show that it is non-degenerate if $u_{i}$’s are all invertible
in $R$. In Section 5, we studied the fixed subalgebra of the action of
$\mathfrak{S}(r)$ on $\mathcal{H}_{n,r}$.

Acknowledgments. We thank Toshiaki Shoji for a valuable communication.

##  2\. Modified Ariki-Koike algebras

Let $R$ be an integral domain and $R^{\times}$ be the group of invertible
elements in $R$.

We take $q,u_{1},\ldots,u_{r}\in R$ such that

$\displaystyle q\in
R^{\times}\quad\text{and}\quad\Delta:=\prod_{i>j}(u_{i}-u_{j})\in R^{\times}.$
(2.1)

######  Definition 2.1.

The _modified Ariki-Koike algebra (also called the Shoji’s algebra)_
$\mathcal{H}_{n,r}=\mathcal{H}_{n,r}(R,q,u_{1},\ldots,u_{r})$ is an
associative $R$-algebra generated by

$t_{1},\ldots,t_{n},T_{1},\ldots,T_{n-1}$

with the defining relations

$\displaystyle(T_{i}-q)(T_{i}+q^{-1})=0\qquad(1\leq i\leq n-1)$ (2.2)
$\displaystyle(t_{i}-u_{1})\cdots(t_{i}-u_{r})=0\qquad(1\leq i\leq n)$ (2.3)
$\displaystyle T_{i}T_{i+1}T_{i}=T_{i+1}T_{i}T_{i+1}\qquad(1\leq i\leq n-2)$
(2.4) $\displaystyle T_{i}T_{j}=T_{j}T_{i}\qquad(|i-j|\geq 2)$ (2.5)
$\displaystyle t_{i}t_{j}=t_{j}t_{i}\qquad(1\leq i,j\leq n)$ (2.6)
$\displaystyle T_{j}t_{k}=t_{k}T_{j}\qquad(k\neq j,j+1)$ (2.7) $\displaystyle
T_{j-1}t_{j}=t_{j-1}T_{j-1}+\Delta^{-2}\displaystyle\sum_{1\leq
c_{1}<c_{2}\leq
r}(u_{c_{2}}-u_{c_{1}})(q-q^{-1})F_{c_{1}}(t_{j-1})F_{c_{2}}(t_{j})$ (2.8)
$\displaystyle T_{j-1}t_{j-1}=t_{j}T_{j-1}-\Delta^{-2}\displaystyle\sum_{1\leq
c_{1}<c_{2}\leq
r}(u_{c_{2}}-u_{c_{1}})(q-q^{-1})F_{c_{1}}(t_{j-1})F_{c_{2}}(t_{j}),$ (2.9)

where $\Delta=\displaystyle\prod_{i>j}(u_{i}-u_{j})$, and $F_{i}(X)\in R[X]$
is the polynomial uniquely determined by the conditions that $\deg(F_{i})=r-1$
and $F_{c}(u_{c^{\prime}})=\delta_{c,c^{\prime}}\Delta$ for $1\leq
c,c^{\prime}\leq r$.

If we write $F_{i}(X)=\sum_{j=1}^{r}h_{i,j}X^{j-1}$, then the matrix
$H(u_{1},\ldots,u_{r}):=(h_{i,j})_{1\leq i,j\leq r}$ is given by
$H(u_{1},\ldots,u_{r})=\Delta V(u_{1},\ldots,u_{r})^{-1}$, the adjugate matrix
of the Vandermonde matrix $V(u_{1},\ldots,u_{r})$ where

$\displaystyle V(u_{1},\ldots,u_{r})_{i,j}=u_{j}^{i-1}\text{ for}\quad 1\leq
i,j\leq r.$ (2.10)

Note that

$\displaystyle t_{i}^{r}=\sum_{k=0}^{r-1}(-1)^{r-k+1}\sigma_{r-k}\,t_{i}^{k}$
(2.11)

for each $1\leq i\leq n$, where $\sigma_{r}$ denotes the $r$-th elementary
symmetric polynomial in $u_{1},\ldots,u_{r}$.

######  Remark 2.2.

  1. (1)

Note that the algebra in [7] is the case in Definition 2.1 when
$R={\mathbb{Z}}[q,q^{{}^{-1},}u_{1},\ldots,u_{r},\Delta]\subset\mathbb{Q}(q,u_{1},\ldots,u_{r})$,
where $q,u_{1}\ldots,u_{r}$ are indeterminates and
$\mathbb{Q}(q,u_{1},\ldots,u_{r})$ denotes the ring of rational functions in
them. In this paper, we consider the specializations of the one in [7]
following [8, Section 1.2].

  2. (2)

The generators $t_{i}$ and $T_{j-1}$ correspond to $\xi_{i}$ and $a_{j}$ in
[7], respectively.

Let $\mathfrak{S}(n)$ be the symmetric group of $n$-letters. The following
fundamental result is proved in [7].

######  Theorem 2.3.

[7, Theorem 3.7] The algebra $\mathcal{H}_{n,r}$ has an $R$-basis

$\displaystyle\left\\{t_{1}^{c_{1}}\cdots t_{n}^{c_{n}}T_{w}\mid
w\in\mathfrak{S}(n),\ 0\leq c_{1},\ldots,c_{n}\leq r-1\right\\}.$ (2.12)

It follows that the subalgebra generated by $T_{1},\ldots T_{n-1}$ is
isomorphic to the Iwahori-Hecke algebra of the symmetric group
$\mathfrak{S}(n)$. The subalgebra $R[t_{1},\ldots,t_{n}]$ generated by
$t_{1},\ldots,t_{n}$ is isomorphic to the quotient ring $R[X_{1},\ldots
X_{n}]/(\left\\{\prod_{j=1}^{r}(X_{i}-u_{j})\mid 1\leq i\leq n\right\\})$ of
the polynomial ring $R[X_{1},\ldots X_{n}]$. Note that $R[t_{1},\ldots,t_{n}]$
is a free $R$-module with a basis $\left\\{t_{1}^{c_{1}}\cdots
t_{n}^{c_{n}}\mid 0\leq c_{1},\ldots,c_{n}\leq r-1\right\\}$.

###  2.1. The subalgebra $R[t_{1},\ldots,t_{n}]$

Set

$\displaystyle[1,r]:=\left\\{a\in{\mathbb{Z}}\mid 1\leq a\leq
r\right\\}\quad\text{and}\quad[1,r]^{n}=\overbrace{[1,r]\times\cdots\times[1,r]}^{n-\text{times}}.$

For each $(k_{1},\ldots k_{n})\in[1,r]^{n}$, define

$\displaystyle\tilde{b}_{k_{1},\ldots,k_{n}}:=\displaystyle\prod_{1\leq i\leq
n}\prod_{1\leq j\leq r,\,j\neq k_{i}}\dfrac{X_{i}-u_{j}}{u_{k_{i}}-u_{j}}\in
R[X_{1},\ldots,X_{n}]\quad\text{and}$ $\displaystyle
b_{k_{1},\ldots,k_{n}}:=\displaystyle\prod_{1\leq i\leq n}\prod_{1\leq j\leq
r,\,j\neq k_{i}}\dfrac{t_{i}-u_{j}}{u_{k_{i}}-u_{j}}\in
R[t_{1},\ldots,t_{n}].$

In other words, the element $\tilde{b}_{k_{1},\ldots,k_{n}}$ is the polynomial
in $X_{1},\ldots,X_{n}$ of degree $r-1$ for each $X_{i}$ such that
$\tilde{b}_{k_{1},\ldots,k_{n}}(u_{j_{1}},\ldots,u_{j_{n}})=\delta_{(k_{1},\ldots,k_{n}),(j_{1},\ldots,j_{n})}$
for any $(j_{1},\ldots,j_{n})\in[1,r]^{n}$ and $b_{k_{1},\ldots,k_{n}}$ is the
image of $\tilde{b}_{k_{1},\ldots,k_{n}}$ in the quotient ring
$R[t_{1},\ldots,t_{n}]$.

######  Lemma 2.4.

The set

$\left\\{b_{k_{1},\ldots,k_{n}}\in
R[t_{1},\ldots,t_{n}]\mid(k_{1},\ldots,k_{n})\in[1,r]^{n}\right\\}$

forms an $R$-basis of $R[t_{1},\ldots,t_{n}]$. It is a complete system of
orthogonal idempotents of $R[t_{1},\ldots,t_{n}]$.

###### Proof.

Note that

$\displaystyle t_{i}\prod_{1\leq j\leq r,\,j\neq
c}(t_{i}-u_{j})=u_{c}\prod_{1\leq j\leq r,\,j\neq c}(t_{i}-u_{j})$

for any $1\leq i\leq n$, $1\leq c\leq r$. Hence we have
$(t_{i}-u_{k_{i}})b_{k_{1},\ldots,k_{n}}=0$ so that

$\displaystyle f(t_{1},\ldots,t_{n})b_{k_{1},\ldots,k_{n}}=f(u_{k_{1}},\ldots
u_{k_{n}})b_{k_{1},\ldots,k_{n}}$ (2.13)

for any $f\in R[X_{1},\ldots,X_{n}]$ and $(k_{1},\ldots,k_{n})\in[1,r]^{n}$.
It follows that $b_{k_{1},\ldots,k_{n}}$’s are idempotents and orthogonal to
each other. Thus they are linearly independent over $R$ and hence form an
$R$-basis of $R[t_{1},\ldots,t_{n}]$, which is a free $R$-module of rank
$r^{n}$. It follows that

$\displaystyle
f(t_{1},\ldots,t_{n})=\sum_{(k_{1},\ldots,k_{n})\in[1,r]^{n}}f(u_{k_{1}},\ldots,u_{k_{n}})b_{k_{1},\ldots,k_{n}}$

for any polynomial $f(X_{1},\ldots,X_{n})$. In particular we have
$1=\sum_{(k_{1},\ldots,k_{n})\in[1,r]^{n}}b_{k_{1},\ldots,k_{n}}$, as deisred.
∎

Set

$\displaystyle
B_{j-1}:=-\Delta^{-2}\displaystyle\sum_{c_{1}<c_{2}}(u_{c_{2}}-u_{c_{1}})(q-q^{-1})F_{c_{1}}(t_{j-1})F_{c_{2}}(t_{j})$

so that

$\displaystyle T_{j-1}t_{j}=t_{j-1}T_{j-1}-B_{j-1}\quad\text{and}\quad$
$\displaystyle T_{j-1}t_{j-1}=t_{j}T_{j-1}+B_{j-1}.$

Let
$\widetilde{B}_{j-1}:=-\Delta^{-2}\displaystyle\sum_{c_{1}<c_{2}}(u_{c_{2}}-u_{c_{1}})(q-q^{-1})F_{c_{1}}(X_{j-1})F_{c_{2}}(X_{j}).$
Then

$\displaystyle\widetilde{B}_{j-1}(u_{k_{1}},\ldots,u_{k_{n}})=-\Delta^{-2}\displaystyle\sum_{c_{1}<c_{2}}(u_{c_{2}}-u_{c_{1}})(q-q^{-1})\delta_{c_{1}=k_{j-1},c_{2}=k_{j}}\Delta^{2}$
$\displaystyle=\begin{cases}(q-q^{-1})(u_{k_{j-1}}-u_{k_{j}})&\text{if}\
k_{j-1}<k_{j},\\\ 0&\text{otherwise}.\end{cases}$

It follows that

$\displaystyle B_{j-1}=(q-q^{-1})\sum_{(k_{1},\ldots,k_{n})\in[1,r]^{n},\
k_{j-1}<k_{j}}(u_{k_{j-1}}-u_{k_{j}})b_{k_{1},\ldots,k_{n}}.$ (2.14)

######  Lemma 2.5.

For $1\leq p\leq n-1$, we have

$\displaystyle
T_{p}b_{k_{1},\ldots,k_{p},k_{p+1},\ldots,k_{n}}-b_{k_{1},\ldots,k_{p+1},k_{p},\ldots,k_{n}}T_{p}$
$\displaystyle=(q-q^{-1})\left(\delta(k_{p}<k_{p+1})b_{k_{1},\ldots,k_{p},k_{p+1},\ldots,k_{n}}-\delta(k_{p}>k_{p+1})b_{k_{1},\ldots,k_{p+1},k_{p},\ldots,k_{n}}\right).$

###### Proof.

First, we will prove when $n=2$ and $p=1$ case. Note that

$\displaystyle
T_{1}(t_{1}-u)(t_{2}-u^{\prime})=(t_{2}-u)(t_{1}-u^{\prime})T_{1}+B_{1}(u-u^{\prime})$
(2.15)

for any $u,u^{\prime}\in R$.

Hence we have

$\displaystyle T_{1}b_{x,y}=\left(\prod_{1\leq j\leq r,\,j\neq
x}\dfrac{t_{1}-u_{j}}{u_{x}-u_{j}}\right)\left(\prod_{1\leq j\leq r,\,j\neq
y}\dfrac{t_{2}-u_{j}}{u_{y}-u_{j}}\right)$
$\displaystyle=T_{1}\,\left(\prod_{1\leq j\leq r,\,j\neq
x,y}\left(\dfrac{t_{1}-u_{j}}{u_{x}-u_{j}}\right)\left(\dfrac{t_{2}-u_{j}}{u_{y}-u_{j}}\right)\right)\left(\left(\dfrac{t_{1}-u_{y}}{u_{x}-u_{y}}\right)\left(\dfrac{t_{2}-u_{x}}{u_{y}-u_{x}}\right)\right)^{\delta(x\neq
y)}$ $\displaystyle=\left(\prod_{1\leq j\leq r,\,j\neq
x,y}\left(\dfrac{t_{1}-u_{j}}{u_{x}-u_{j}}\right)\left(\dfrac{t_{2}-u_{j}}{u_{y}-u_{j}}\right)\right)\,T_{1}\,\left(\left(\dfrac{t_{1}-u_{y}}{u_{x}-u_{y}})\right)\left(\dfrac{t_{2}-u_{x}}{u_{y}-u_{x}}\right)\right)^{\delta(x\neq
y)}$ $\displaystyle=\begin{cases}\left(\displaystyle\prod_{1\leq j\leq
r,\,j\neq
x,y}\left(\dfrac{t_{1}-u_{j}}{u_{x}-u_{j}}\right)\left(\dfrac{t_{2}-u_{j}}{u_{y}-u_{j}}\right)\right)\left(\left(\dfrac{t_{2}-u_{y}}{u_{x}-u_{y}}\right)\left(\dfrac{t_{1}-u_{x}}{u_{y}-u_{x}}\right)\,T_{1}+\dfrac{B_{1}}{u_{x}-u_{y}}\right)&\text{if
}\ x\neq y\\\ \left(\displaystyle\prod_{1\leq j\leq r,\,j\neq
x}\left(\dfrac{t_{1}-u_{j}}{u_{x}-u_{j}}\right)\left(\dfrac{t_{2}-u_{j}}{u_{y}-u_{j}}\right)\right)T_{1}&\text{if
}\ x=y.\end{cases}$

On the other hand, we have

$\displaystyle b_{y,x}T_{1}=\left(\prod_{1\leq j\leq r,\,j\neq
y}\dfrac{t_{1}-u_{j}}{u_{y}-u_{j}}\right)\left(\prod_{1\leq j\leq r,\,j\neq
x}\dfrac{t_{2}-u_{j}}{u_{x}-u_{j}}\right)\,T_{1}$
$\displaystyle=\left(\prod_{1\leq j\leq r,\,j\neq
x,y}\left(\dfrac{t_{1}-u_{j}}{u_{y}-u_{j}}\right)\left(\dfrac{t_{2}-u_{j}}{u_{x}-u_{j}}\right)\right)\left((\dfrac{t_{1}-u_{x}}{u_{y}-u_{x}})(\dfrac{t_{2}-u_{y}}{u_{x}-u_{y}})\right)^{\delta(x\neq
y)}\,T_{1}.$

Hence

$\displaystyle T_{1}\,b_{xy}-b_{y,x}\,T_{1}=\delta(x\neq
y)\left(\displaystyle\prod_{1\leq j\leq r,\,j\neq
x,y}\left(\dfrac{t_{1}-u_{j}}{u_{x}-u_{j}}\right)\left(\dfrac{t_{2}-u_{j}}{u_{y}-u_{j}}\right)\right)\dfrac{B_{1}}{u_{x}-u_{y}}$

Set

$\displaystyle G(X_{1},X_{2}):=\displaystyle\prod_{1\leq j\leq r,\,j\neq
x,y}\left(\dfrac{X_{1}-u_{j}}{u_{x}-u_{j}}\right)\left(\dfrac{X_{2}-u_{j}}{u_{y}-u_{j}}\right)\in
R[X_{1},X_{2}].$

Then we have

$\displaystyle\begin{cases}G(u_{a},u_{b})=0&\text{if }\ a\neq x,y\ \text{or}\
b\neq x,y,\\\ G(u_{x},u_{y})=G(u_{y},u_{x})=1.\end{cases}$

Hence

$\displaystyle G(t_{1},t_{2})=a_{x,x}b_{x,x}+a_{y,y}b_{y,y}+b_{x,y}+b_{y,x}$

for some $a_{x,x},a_{y,y}\in R$. Thus

$\displaystyle G(t_{1},t_{2})\dfrac{B_{1}}{u_{x}-u_{y}}$
$\displaystyle=G(t_{1},t_{2})\left(\dfrac{q-q^{-1}}{u_{x}-u_{y}}\right)\left(\sum_{1\leq\alpha<\beta\leq
r}(u_{\alpha}-u_{\beta})b_{\alpha,\beta}\right)$
$\displaystyle=\left(\dfrac{q-q^{-1}}{u_{x}-u_{y}}\right)\left(\delta(x<y)(u_{x}-u_{y})b_{x,y}+\delta(x>y)(u_{y}-u_{x})b_{y,x}\right)$
$\displaystyle=(q-q^{-1})\left(\delta(x<y)b_{x,y}-\delta(x>y)b_{y,x}\right),$

as desired.

Now assume that $n\geq 2$ and $1\leq p\leq n-1$. Note that

$\prod_{i\neq p,p+1}\left(\prod_{1\leq j\leq r,\ k_{i}\neq
j}\dfrac{t_{i}-u_{j}}{u_{k_{i}}-u_{j}}\right)$

commutes with $T_{p}$. Hence by the above calculation for $n=2,p=1$ case, we
have

$\displaystyle
T_{p}b_{k_{1},\ldots,k_{p},k_{p+1},\ldots,k_{n}}-b_{k_{1},\ldots,k_{p+1},k_{p},\ldots,k_{n}}T_{p}$
$\displaystyle=(q-q^{-1})\left(\delta(k_{p}<k_{p+1})b_{k_{1},\ldots,k_{p},k_{p+1},\ldots,k_{n}}-\delta(k_{p}>k_{p+1})b_{k_{1},\ldots,k_{p+1},k_{p},\ldots,k_{n}}\right)\prod_{i\neq
p,p+1}\left(\prod_{1\leq j\leq r,\ k_{i}\neq
j}\dfrac{t_{i}-u_{j}}{u_{k_{i}}-u_{j}}\right).$

Since

$\dfrac{t_{i}-u_{j}}{u_{k_{i}}-u_{j}}b_{k_{1},\ldots,k_{p},k_{p+1},\ldots,k_{n}}=b_{k_{1},\ldots,k_{p},k_{p+1},\ldots,k_{n}}\quad\text{and}\quad\dfrac{t_{i}-u_{j}}{u_{k_{i}}-u_{j}}b_{k_{1},\ldots,k_{p+1},k_{p},\ldots,k_{n}}=b_{k_{1},\ldots,k_{p+1},k_{p},\ldots,k_{n}}$

for any $i\neq k_{p},k_{p+1}$ and $1\leq j\leq r$ with $k_{i}\neq j$, we get
the desired result. ∎

###  2.2. Actions of symmetric groups on $R[t_{1},\ldots,t_{n}]$

Note that the symmetric group $\mathfrak{S}(n)$ acts on
$R[t_{1},\ldots,t_{n}]$ as $R$-algebra automorphisms by

$w.t_{i}:=t_{w(i)}\qquad\text{ for $1\leq i\leq n$, $w\in\mathfrak{S}(n)$ }.$

It also acts on the set $[1,r]^{n}$ by the place permutations
$w.(k_{1},\ldots,k_{n}):=(k_{w^{-1}(1)},\ldots,k_{w^{-1}(n)})$. Then we have

$\displaystyle
w.b_{(k_{1},\ldots,k_{n})}=b_{w.(k_{1},\ldots,k_{n})}\qquad\text{for}\
w\in\mathfrak{S}(n),\ (k_{1},\ldots,k_{n})\in[1,r]^{n}.$ (2.16)

On the other hand, the symmetric group $\mathfrak{S}(r)$ acts on $[1,r]^{n}$
by

${}^{\sigma}(k_{1},\ldots,k_{n}):=(\sigma(k_{1}),\ldots,\sigma(k_{n}))\qquad\text{
for $1\leq k_{i}\leq r$, $\sigma\in\mathfrak{S}(r)$},$

which induces an action on the basis given by

${}^{\sigma}b_{k_{1},\ldots,k_{n}}:=b_{\sigma(k_{1}),\ldots,\sigma(k_{n})}.$
(2.17)

Since

${{}^{\sigma}b_{k_{1},\ldots,k_{n}}}{{}^{\sigma}b_{k^{\prime}_{1},\ldots,k^{\prime}_{n}}}=\delta_{(k_{1},\ldots,k_{n}),(k^{\prime}_{1},\ldots,k^{\prime}_{n})}{{}^{\sigma}b_{k_{1},\ldots,k_{n}}},$

the $R$-linear map $\sigma$ extends to an $R$-algebra automorphism and hence
the group $\mathfrak{S}(r)$ acts on $R[t_{1},\ldots,t_{n}]$.

Note that the acions of $\mathfrak{S}(n)$ and $\mathfrak{S}(r)$ on $[1,r]^{n}$
are commuting to each other.

######  Remark 2.6.

Let $OP_{r}(n)$ be the set of sequences $(I_{1},\ldots,I_{r})$ in subsets of
$[1,n]$ such that $I_{i}\cap I_{j}=\emptyset$ for $i\neq j$ and
$\displaystyle\bigsqcup_{1\leq i\leq r}I_{i}=[1,n]$. Then there is a bijection
between the sets $[1,r]^{n}$ and $OP_{r}(n)$ given by

$\displaystyle\Psi:[1,r]^{n}$ $\displaystyle\to$ $\displaystyle OP_{r}(n)$
$\displaystyle f$ $\displaystyle\mapsto$
$\displaystyle(f^{-1}(1),\ldots,f^{-1}(r)),$

where we identify the set $[1,r]^{n}$ with the set of functions from $[1,n]$
to $[1,r]$. Then the action of $\mathfrak{S}(n)\times\mathfrak{S}(r)$ on
$[1,r]^{n}$ is nothing but

$\displaystyle w.(^{\sigma}f)=^{\sigma}(w.f)=\sigma\circ f\circ w^{-1}\qquad
w\in\mathfrak{S}(n),\quad\sigma\in\mathfrak{S}(r),$

and $\Psi$ commutes with the action of $\mathfrak{S}(n)\times\mathfrak{S}(r)$
in [6, Section 2.2], in which $\mathfrak{S}(r)$ acts on $OP_{r}(n)$ as place
permutations. Thus the set of orbits of $[1,r]^{n}$ under the action of
$\mathfrak{S}(r)$ is in bijection with the set of partitions of $n$ at most
$r$-many parts. More precisely, we have

$\displaystyle\hskip-20.00003pt(\ell_{1},\ldots,\ell_{n})\in[k_{1},\ldots,k_{n}]\
\quad\text{if and olny if}\quad(\ell_{i}=\ell_{j}\Leftrightarrow
k_{i}=k_{j}\quad\text{for all}\ 1\leq i,j\leq n),$ (2.18)

where $[k_{1},\ldots,k_{n}]$ denote the orbit of $(k_{1},\ldots,k_{n})$ under
the action of $\mathfrak{S}(r)$.

Let $s_{p}(1\leq p\leq n-1)$ is the simple transposition in the symmetric
group $\mathfrak{S}(n)$ permuting $p$ and $p+1$.

######  Lemma 2.7.

Let $a\in R[t_{1},\ldots,t_{n}]$ and $1\leq p\leq n-1$. If $s_{p}.a=a$, then
we have

$\displaystyle T_{p}a=aT_{p}.$

###### Proof.

Write

$a=\sum_{(k_{1},\ldots,k_{n})\in[1,r]^{n}}a_{(k_{1},\ldots,k_{n})}b_{k_{1},\ldots,k_{n}}$

for some $a_{(k_{1},\ldots,k_{n})}\in R$. Since $s_{i}.a=a$, we have

$a_{(k_{1},k_{p},k_{p+1},\ldots,k_{n})}=a_{(k_{1},\ldots,k_{p+1},k_{p},\ldots,k_{n})}$

for all $(k_{1},\ldots,k_{n})\in[1,r]^{n}$. Then by Lemma 2.5, we have

$\displaystyle(q-q^{-1})^{-1}(T_{p}a-s_{p}.a\,T_{p})$
$\displaystyle=\sum_{k_{p}<k_{p+1}}a_{(k_{1},\ldots,k_{n})}b_{k_{1},\ldots,k_{n}}-\sum_{k_{p+1}<k_{p}}a_{(k_{1},\ldots,k_{n})}b_{s_{p}.(k_{1},\ldots,k_{n})}$
$\displaystyle=\sum_{k_{p}<k_{p+1}}a_{(k_{1},\ldots,k_{n})}b_{k_{1},\ldots,k_{n}}-\sum_{k_{p}<k_{p+1}}a_{s_{p}.(k_{1},\ldots,k_{n})}b_{k_{1},\ldots,k_{n}}=0,$

as desired. ∎

##  3\. Yokonuma-Hecke like presentation

For each $1\leq i,j\leq n$, define

$\displaystyle
B^{\prime}_{i,j}:=-(q-q^{-1})\sum_{k_{i}<k_{j}}b_{k_{1},\ldots,k_{n}}.$ (3.1)

Note that

$\displaystyle-(t_{i}-t_{i+1})B_{i,i+1}^{\prime}=B_{i}\quad\text{for}\ 1\leq
i\leq n-1.$ (3.2)

We also define for each $1\leq i\leq n-1$

$\displaystyle e_{i}:=\sum_{k_{i}=k_{i+1}}b_{k_{1},\ldots,k_{n}}.$ (3.3)

For each $1\leq i\leq n-1$ we define

$\displaystyle g_{i}:=T_{i}+B^{\prime}_{i,i+1}.$ (3.4)

######  Remark 3.1.

The above element $g_{i}$ is denoted by $S_{j+1}$ in [7] . It corresponds to
$\mathbb{G}_{i}$ in [3, Definition 5].

######  Example 3.2.

Let $n=3$, $r=2$. Then the group $G(2,1,3)$ is the Weyl group of type $B_{3}$.
We have

$\begin{cases}&b_{111}=\dfrac{t_{1}-u_{2}}{u_{1}-u_{2}}\cdot\dfrac{t_{2}-u_{2}}{u_{1}-u_{2}}\cdot\dfrac{t_{3}-u_{2}}{u_{1}-u_{2}},\qquad
b_{121}=\dfrac{t_{1}-u_{2}}{u_{1}-u_{2}}\cdot\dfrac{t_{2}-u_{1}}{u_{2}-u_{1}}\cdot\dfrac{t_{3}-u_{2}}{u_{1}-u_{2}},\\\
&b_{112}=\dfrac{t_{1}-u_{2}}{u_{1}-u_{2}}\cdot\dfrac{t_{2}-u_{2}}{u_{1}-u_{2}}\cdot\dfrac{t_{3}-u_{1}}{u_{2}-u_{1}},\qquad
b_{122}=\dfrac{t_{1}-u_{2}}{u_{1}-u_{2}}\cdot\dfrac{t_{2}-u_{1}}{u_{2}-u_{1}}\cdot\dfrac{t_{3}-u_{1}}{u_{2}-u_{1}},\\\
&b_{211}=\dfrac{t_{1}-u_{1}}{u_{2}-u_{1}}\cdot\dfrac{t_{2}-u_{2}}{u_{1}-u_{2}}\cdot\dfrac{t_{3}-u_{2}}{u_{1}-u_{2}},\qquad
b_{221}=\dfrac{t_{1}-u_{1}}{u_{2}-u_{1}}\cdot\dfrac{t_{2}-u_{1}}{u_{2}-u_{1}}\cdot\dfrac{t_{3}-u_{2}}{u_{1}-u_{2}},\\\
&b_{212}=\dfrac{t_{1}-u_{1}}{u_{2}-u_{1}}\cdot\dfrac{t_{2}-u_{2}}{u_{1}-u_{2}}\cdot\dfrac{t_{3}-u_{1}}{u_{2}-u_{1}},\qquad
b_{222}=\dfrac{t_{1}-u_{1}}{u_{2}-u_{1}}\cdot\dfrac{t_{2}-u_{1}}{u_{2}-u_{1}}\cdot\dfrac{t_{3}-u_{1}}{u_{2}-u_{1}},\end{cases}$
$\begin{cases}&B^{\prime}_{12}=-(q-q^{-1})\\{b_{121}+b_{122}\\},\\\
&B^{\prime}_{13}=-(q-q^{-1})\\{b_{112}+b_{122}\\},\\\
&B^{\prime}_{23}=-(q-q^{-1})\\{b_{112}+b_{212}\\},\end{cases}$

and

$\begin{cases}&e_{1}=b_{111}+b_{112}+b_{221}+b_{222},\qquad
e_{2}=b_{111}+b_{211}+b_{122}+b_{222},\\\ &g_{1}=T_{1}+B^{\prime}_{12},\qquad
g_{2}=T_{2}+B^{\prime}_{23}.\end{cases}$

######  Proposition 3.3.

For $1\leq i\leq n$, $1\leq j\leq n-1$, we have

$\displaystyle g_{j}t_{i}=t_{s_{i}(j)}g_{j}.$ (3.5)

Hence for any $f\in R[t_{1},\ldots,t_{n}],\ 1\leq j\leq n-1$, we have

$\displaystyle g_{j}f=(s_{j}.f)g_{j}.$ (3.6)

###### Proof.

We recall the following relations from Definition 2.1:

$\displaystyle T_{j}t_{k}=t_{k}T_{j}\ (k\neq j,j+1),\qquad
T_{j}t_{j+1}=t_{j}T_{j}-B_{j},\quad\text{and}\quad
T_{j}t_{j}=t_{j+1}T_{j}+B_{j}.$

For the case $i\neq j,j+1$, there is nothing to prove since $s_{j}(i)=i$.

For $i=j+1$, by using the formula (3.2), we get

$\displaystyle
g_{j}t_{j+1}=(T_{j}+B^{\prime}_{j,j+1})t_{j+1}=t_{j}T_{j}-B_{j}+B^{\prime}_{j,j+1}t_{j+1}=t_{j}T_{j}+(t_{j}-t_{j+1})B_{j,j+1}^{\prime}+B^{\prime}_{j,j+1}t_{j+1}$
$\displaystyle=t_{j}(T_{j}+B_{j,j+1}^{\prime})=t_{j}g_{j}=t_{s_{i}(j)}g_{j}$

Similarly, for the case $i=j$, we obtain

$\displaystyle
g_{j}t_{j}=(T_{j}+B^{\prime}_{j,j+1})t_{j}=t_{j+1}T_{j}+B_{j}+B^{\prime}_{j,j+1}t_{j}=t_{j+1}T_{j}-(t_{j}-t_{j+1})B_{j,j+1}^{\prime}+B^{\prime}_{j,j+1}t_{j}$
$\displaystyle=t_{j+1}(T_{j}+B_{j,j+1}^{\prime})=t_{s_{i}(j)}g_{j},$

as desired. ∎

######  Lemma 3.4.

We have

$\displaystyle g_{k}B^{\prime}_{i,j}=B^{\prime}_{s_{k}(i),s_{k}(j)}g_{k},$
(3.7) $\displaystyle
T_{k}B^{\prime}_{i,j}=B^{\prime}_{s_{k}(i),s_{k}(j)}T_{k}+(B^{\prime}_{s_{k}(i),s_{k}(j)}-B^{\prime}_{i,j})B^{\prime}_{k,k+1}$
(3.8)

for $1\leq k\leq n-1$, $1\leq i,j\leq n$.

###### Proof.

Note that for any $1\leq i,j,\leq n$, $w\in\mathfrak{S}(n)$, we have

$\displaystyle
w.B^{\prime}_{i,j}=(q-q^{-1})\sum_{k_{i}<k_{j}}b_{(k_{w^{-1}(1)},\ldots,k_{w^{-1}(n)})}=(q-q^{-1})\sum_{k_{w(i)}<k_{w(j)}}b_{(k_{1},\ldots,k_{n})}=B^{\prime}_{w(i),w(j)}.$
(3.9)

Thus Proposition 3.3 implies (3.7). The seconde formula results immediately
from the first one. ∎

######  Proposition 3.5.

For $1\leq i\leq n-1$, we have

$\displaystyle g_{i}^{2}=1+(q-q^{-1})e_{i}g_{i}.$ (3.10)

###### Proof.

We have

$g_{i}^{2}=(T_{i}+B^{\prime}_{i,i+1})^{2}=T_{i}^{2}+T_{i}B^{\prime}_{i,i+1}+B^{\prime}_{i,i+1}T_{i}+{B^{\prime}_{i,i+1}}^{2}$
$=\left((q-q^{-1})T_{i}+1\right)+\left(B^{\prime}_{i+1,i}T_{i}+(B^{\prime}_{i+1,i}-B^{\prime}_{i,i+1})B^{\prime}_{i,i+1}\right)+B^{\prime}_{i,i+1}T_{i}+{B^{\prime}_{i,i+1}}^{2},$

by the quadratic relation on $T_{i}$ and Lemma 3.4. Then by the fact that
$\sum b_{k_{1}\cdots k_{n}}=1$ and the definition of $B^{\prime}_{i,i+1}$, it
becomes

$g_{i}^{2}=(q-q^{-1})\left(\sum b_{k_{1}\cdots
k_{n}}-\sum_{k_{i+1}<k_{i}}b_{k_{1}\cdots
k_{n}}-\sum_{k_{i}<k_{i+1}}b_{k_{1}\cdots
k_{n}}\right)T_{i}+1+B^{\prime}_{i+1,i}B^{\prime}_{i,i+1}.$

We know that $B^{\prime}_{i+1,i}B^{\prime}_{i,i+1}=0$, thus we obtain the
desired result

$g_{i}^{2}=(q-q^{-1})e_{i}\,T_{i}+1=(q-q^{-1})e_{i}g_{i}+1,$

by using $e_{i}T_{i}=e_{i}(g_{i}-B^{\prime}_{i,i+1})=e_{i}g_{i}$ because
$e_{i}B^{\prime}_{i,i+1}=0.$ ∎

The below relations can be verified in the faithful representation of
$\mathcal{H}_{n,r}$ constructed by Shoji (see [10, Lemma 3.7]). We include
here a direct proof using the defining relations.

######  Proposition 3.6.

The following relations hold.

  1. (1)

$g_{i}g_{j}=g_{j}g_{i}$ for $|i-j|>1$,

  2. (2)

$g_{i}g_{i+1}g_{i}=g_{i+1}g_{i}g_{i+1}$ for $1\leq i\leq n-1$.

###### Proof.

(1) If $|i-j|>1$, we know that $T_{i}T_{j}=T_{j}T_{i}$. Then the difference
$g_{i}g_{j}-g_{j}g_{i}$ becomes

$g_{i}g_{j}-g_{j}g_{i}=T_{i}B^{\prime}_{j,j+1}+B^{\prime}_{i,i+1}T_{j}-T_{j}B^{\prime}_{i,i+1}-B^{\prime}_{j,j+1}T_{i}=0,$

where the last equality follows from Lemma 3.4.

(2) We have

$g_{i}g_{i+1}g_{i}=g_{i}g_{i+1}(T_{i}+B^{\prime}_{i,i+1})=g_{i}g_{i+1}T_{i}+B^{\prime}_{i+1,i+2}g_{i}g_{i+1}$

and

$g_{i+1}g_{i}g_{i+1}=(T_{i+1}+B^{\prime}_{i+1,i+2})g_{i}g_{i+1}=T_{i+1}g_{i}g_{i+1}+B^{\prime}_{i+1,i+2}g_{i}g_{i+1}.$

Thus it is enough to show that

$\displaystyle g_{i}g_{i+1}T_{i}=T_{i+1}g_{i}g_{i+1}.$ (3.11)

By (2.2) and (3.8), we get

$\displaystyle
g_{i}g_{i+1}T_{i}=T_{i}T_{i+1}T_{i}+B^{\prime}_{i,i+1}T_{i+1}T_{i}+\left((q-q^{-1})B^{\prime}_{i,i+2}+B^{\prime}_{i,i+1}B^{\prime}_{i,i+2}\right)T_{i}+B^{\prime}_{i,i+2}.$
(3.12)

On the other hand we have

$\displaystyle T_{i+1}g_{i}g_{i+1}$
$\displaystyle=T_{i+1}(T_{i}+B^{\prime}_{i,i+1})(T_{i+1}+B^{\prime}_{i+1,i+2})$
$\displaystyle=T_{i+1}T_{i}T_{i+1}+T_{i+1}B^{\prime}_{i,i+1}T_{i+1}+T_{i+1}T_{i}B^{\prime}_{i+1,i+2}+T_{i+1}B^{\prime}_{i,i+1}B^{\prime}_{i+1,i+2}.$

For the second term, we get

$\displaystyle
T_{i+1}B^{\prime}_{i,i+1}T_{i+1}=\left((q-q^{-1})B^{\prime}_{i,i+2}+(B^{\prime}_{i,i+2}-B^{\prime}_{i,i+1})B^{\prime}_{i+1,i+2}\right)T_{i+1}+B_{i,i+2}.$

For the last two terms, we have

$\displaystyle
T_{i+1}T_{i}B^{\prime}_{i+1,i+2}+T_{i+1}B^{\prime}_{i,i+1}B^{\prime}_{i+1,i+2}$
$\displaystyle=\left(T_{i+1}B^{\prime}_{i,i+2}T_{i}+T_{i+1}B^{\prime}_{i,i+2}B^{\prime}_{i,i+1}-T_{i+1}B^{\prime}_{i+1,i+2}B^{\prime}_{i,i+1}\right)+T_{i+1}B^{\prime}_{i,i+1}B^{\prime}_{i+1,i+2}$
$\displaystyle=T_{i+1}B^{\prime}_{i,i+2}T_{i}+T_{i+1}B^{\prime}_{i,i+2}B^{\prime}_{i,i+1}$
$\displaystyle=\left(B^{\prime}_{i,i+1}T_{i+1}T_{i}+((B^{\prime}_{i,i+1}-B^{\prime}_{i,i+2})B^{\prime}_{i+1,i+2})T_{i}\right)+T_{i+1}B^{\prime}_{i,i+2}B^{\prime}_{i,i+1}.$

Since
$s_{i+1}.(B^{\prime}_{i,i+2}B^{\prime}_{i,i+1})=B^{\prime}_{i,i+2}B^{\prime}_{i,i+1}$,
by Lemma 2.7 we have

$\displaystyle
T_{i+1}B^{\prime}_{i,i+2}B^{\prime}_{i,i+1}=B^{\prime}_{i,i+2}B^{\prime}_{i,i+1}T_{i+1}.$

Summing up, we obtain

$\displaystyle
T_{i+1}g_{i}g_{i+1}=T_{i+1}T_{i}T_{i+1}+B^{\prime}_{i,i+1}T_{i+1}T_{i}+A\,T_{i+1}+C\,T_{i}+B^{\prime}_{i,i+2},$
(3.13)

where

$\displaystyle
A=(q-q^{-1})B^{\prime}_{i,i+2}-B^{\prime}_{i,i+1}B^{\prime}_{i+1,i+2}+B^{\prime}_{i,i+2}B^{\prime}_{i+1,i+2}+B^{\prime}_{i,i+2}B^{\prime}_{i,i+1}\quad\text{and},$
$\displaystyle C=(B^{\prime}_{i,i+1}-B^{\prime}_{i,i+2})B^{\prime}_{i+1,i+2}.$

Note that if $A=0$, then
$C=(q-q^{-1})B^{\prime}_{i,i+2}+B^{\prime}_{i,i+1}B^{\prime}_{i,i+2}$ and
hence by comparing (3.12) and (3.13), we obtain
$g_{i}g_{i+1}T_{i}=T_{i+1}g_{i}g_{i+1}$.

Since

$\displaystyle(q-q^{-1})B^{\prime}_{i,i+2}+B^{\prime}_{i,i+2}B^{\prime}_{i+1,i+2}+B^{\prime}_{i,i+2}B^{\prime}_{i,i+1}$
$\displaystyle=(q-q^{-1})^{2}(-\sum_{k_{i}<k_{i+2}}b_{k_{1}\cdots
k_{n}}+\sum_{k_{i}<k_{i+2},\,k_{i+1}<k_{i+2}}b_{k_{1}\cdots
k_{n}}+\sum_{k_{i}<k_{i+2},\,k_{i}<k_{i+1}}b_{k_{1}\cdots k_{n}})$
$\displaystyle=-(q-q^{-1})^{2}\sum_{k_{i}<k_{i+2},\,k_{i}<k_{i+1},\,k_{i+1}<k_{i+2}}b_{k_{1}\cdots
k_{n}}$
$\displaystyle=-(q-q^{-1})^{2}\sum_{k_{i}<k_{i+1}<k_{i+2}}b_{k_{1}\cdots
k_{n}}=-B^{\prime}_{i,i+1}B^{\prime}_{i+1,i+2},$

we obtain $A=0$, as deisred. ∎

Hence for each $w\in\mathfrak{S}(n)$, the element

$\displaystyle g_{w}:=g_{i_{1}}\cdots g_{i_{l}}$ (3.14)

is well-defined, i.e. independent of the choice of the reduced expression of
$w$, where $s_{i_{1}}\cdots s_{i_{l}}$ is a reduced expression of $w$.

The lemma below follows immediately from Proposition 3.5.

######  Lemma 3.7.

For $w\in\mathfrak{S}(n)$, we have

$g_{w}g_{s_{i}}=\begin{cases}g_{ws_{i}}\,\,\,\text{if}\,\,\,\ell(ws_{i})>\ell(w)\\\
g_{ws_{i}}+(q-q^{-1})g_{w}e_{i}\,\,\,\text{if}\,\,\,\ell(ws_{i})<\ell(w).\end{cases}$

and

$g_{s_{i}}g_{w}=\begin{cases}g_{s_{i}w}\,\,\,\text{if}\,\,\,\ell(s_{i}w)>\ell(w)\\\
g_{s_{i}w}+(q-q^{-1})e_{i}g_{w}\,\,\,\text{if}\,\,\,\ell(s_{i}w)<\ell(w).\end{cases}$

######  Lemma 3.8.

For $w\in\mathfrak{S}(n)$, we have

$g_{w}\in T_{w}+\sum_{w^{\prime}<w}R[t_{1},\ldots,t_{n}]T_{w^{\prime}},$

where $<$ denotes the Bruhat order on $\mathfrak{S}(n)$.

###### Proof.

We prove by induction on the length of $w$. Suppose that the claim holds for
$w\in\mathfrak{S}(n)$ and assume that $\ell(s_{i}w)>\ell(w)$. We will verify
the claim for $s_{i}w$. By the assumption, there is
$A\in\sum_{w^{\prime}<w}R_{1}[t_{1},\ldots,t_{n}]T_{w^{\prime}}$ such that

$g_{s_{i}w}=g_{i}g_{w}=g_{i}(T_{w}+A)=(T_{i}+B^{\prime}_{i,i+1})(T_{w}+A)=T_{s_{i}w}+B^{\prime}_{i,i+1}T_{w}+B^{\prime}_{i,i+1}A+T_{i}A.$

Since $w<s_{i}w$, it is enough to show that

$T_{i}A\in\sum_{v<s_{i}w}R[t_{1},\ldots,t_{n}]T_{v}.$

Indeed we have

$\displaystyle T_{i}A$
$\displaystyle\in\sum_{w^{\prime}<w}T_{i}R[t_{1},\ldots,t_{n}]T_{w^{\prime}}$
$\displaystyle\subset\sum_{w^{\prime}<w}R[t_{1},\ldots,t_{n}]T_{i}T_{w^{\prime}}+\sum_{w^{\prime}<w}R[t_{1},\ldots,t_{n}]T_{w^{\prime}}$
$\displaystyle\subset\sum_{w^{\prime}<w}R[t_{1},\ldots,t_{n}]T_{s_{i}w^{\prime}}+\sum_{w^{\prime}<w}R[t_{1},\ldots,t_{n}]T_{w^{\prime}}$

by Lemma 2.5 and Lemma 3.7. Notice that if $\ell(s_{i}w)>\ell(w)$, then
$w^{\prime}<w$ implies $s_{i}w^{\prime}<s_{i}w$, which completes the proof. ∎

By Theorem 2.3 and Lemma 3.8, we get

######  Theorem 3.9.

The algebra $\mathcal{H}_{n,r}$ has a $R$-basis given as follows:

$B:=\left\\{t_{1}^{c_{1}}\cdots t_{n}^{c_{n}}g_{w}\mid w\in\mathfrak{S}(n),\
0\leq c_{i}\leq r-1\right\\}.$

######  Corollary 3.10.

The set

$\displaystyle{\mathbf{B}}:=\left\\{b_{k_{1},\ldots,k_{n}}g_{w}\mid(k_{1},\ldots,k_{n})\in[1,r]^{n},\
w\in\mathfrak{S}(n)\right\\}$ (3.15)

forms an $R$-basis of $\mathcal{H}_{n,r}$.

The following is one of the main theorems of this paper.

######  Theorem 3.11.

The $R$-algebra $\mathcal{H}_{n,r}$ has the following presentation.

Generators: $t_{1},\ldots,t_{n}$ and $g_{1},\ldots,g_{n-1}$

Relations:

  1. (1)

$(t_{i}-u_{1})\cdots(t_{i}-u_{r})=0$

  2. (2)

$t_{i}t_{j}=t_{j}t_{i}$

  3. (3)

$g_{j}t_{i}=t_{s_{j}(i)}g_{j}$

  4. (4)

$g_{i}g_{j}=g_{j}g_{i}$ if $|i-j|>1$

  5. (5)

$g_{i}g_{i+1}g_{i}=g_{i+1}g_{i}g_{i+1}$

  6. (6)

$g_{i}^{2}=1+(q-q^{-1})e_{i}g_{i}$,

where $e_{i}:=\sum_{k_{i}=k_{i+1}}b_{k_{1},\ldots,k_{n}}$ and
$b_{k_{1},\ldots,k_{n}}:=\displaystyle\prod_{1\leq i\leq n}\prod_{1\leq j\leq
r,\,j\neq k_{i}}\dfrac{t_{i}-u_{j}}{u_{k_{i}}-u_{j}}$.

###### Proof.

Since $g_{i}=T_{i}+B^{\prime}_{i,i+1}$, the set
$\\{t_{1},\ldots,t_{n},g_{1},\ldots,g_{n-1}\\}$ generates $\mathcal{H}_{n,r}$.
The relations (3)-(6) were shown through this section.

Now let $\mathcal{H}_{n,r}^{\prime}$ be the $R$-algebra defined by the
presentation above. Then there is a surjective $R$-algebra homomorphism
$\psi:\mathcal{H}_{n,r}^{\prime}\to\mathcal{H}_{n,r}$. Let $B^{\prime}$ be the
subset of $\mathcal{H}_{n,r}^{\prime}$ analogus to $B$ in Theorem 3.9. By the
relations, any elements in $\mathcal{H}_{n,r}^{\prime}$ can be written as an
$R$-linear combination of elements in $B^{\prime}$. Moreover $\psi$ maps
$B^{\prime}$ to $B$, which is linearly independent over $R$. Hence
$B^{\prime}$ is $R$-basis of $\mathcal{H}_{n,r}^{\prime}$ and $\psi$ is an
isomorphism, as desired. ∎

Note that $e_{i}$ commutes with $g_{i}$. Hence there exists an $R$-algebra
anti-invoultion which fixes the generators $g_{j}$’s and $t_{i}$’s.

######  Remark 3.12.

If we take $R={\mathbb{C}}[q,q^{-1}]$ and
$u_{k}:=e^{\frac{2\pi\sqrt{-1}k}{r}}$ $(1\leq k\leq r)$, then the algebra
_defined by_ the above presentation is called the Yokonuma-Hecke algebra (see
[1], [4]). Note that under this choice of base ring and parameters, we have

$\displaystyle e_{i}=\dfrac{1}{r}\sum_{s=0}^{r-1}t_{i}^{s}t_{i}^{r-s}.$

In [3, Theorem 13], it is shown that the algebra
$\mathcal{H}_{n,r}({\mathbb{C}}[q,q^{-1}],q,e^{\frac{2\pi\sqrt{-1}k}{r}}\,(1\leq
k\leq r))$ is isomorphic to the Yokonuma-Hecke algebra, which can be
understood as a special case of Theorem 3.11. Note that the isomorphism in
this case can be obatined by comparing the results of Lusztig in [2] and
Jacon-Poulain d’Andecy in [5] on the structure of Yokonuma-Hekce algebras and
the ones of Sawada-Shoji in [8] and Hu-Stoll in [9] on $\mathcal{H}_{n,r}$, as
explained in [3, Section 3.1].

######  Remark 3.13.

It is known that (see, [8, (8.3.2)]) if

$\displaystyle\text{(separation condition) }\qquad q^{2k}u_{i}-u_{j}\in
R^{\times}\quad\text{for }\ -n<k<n,\,i\neq j,$ (3.16)

then the algebra $\mathcal{H}_{n,r}$ is isomorphic to the Ariki-Koike algebra
associated with the same parameter (see [8, (8.3.2)], and see [8, Section 1.1]
for the definition of Ariki-Koike algebras over $R$ associated with
$(q,u_{1},\ldots,u_{r})$). Hence the theorem above also provides a new
presentation of the Ariki-Koike algerba with the condition (3.16) on the
parametres, which includes the generic Ariki-Koike algebra.

######  Theorem 3.14.

There is a presentation of $\mathcal{H}_{n,r}$ with the generators

$g_{1},\ldots,g_{n-1}\qquad\text{and}\quad
b_{k_{1},\ldots,k_{n}}\quad\text{for}\ (k_{1},\ldots,k_{n})\in[1,r]^{n}$

subjet to the relations

  1. (1)

$g_{i}g_{j}=g_{j}g_{i}$ if $|i-j|>1$,

  2. (2)

$g_{i}g_{i+1}g_{i}=g_{i+1}g_{i}g_{i+1}$,

  3. (3)

$g_{i}^{2}=1+(q-q^{-1})e_{i}g_{i}$,

  4. (4)

$b_{k_{1},\ldots,k_{n}}b_{k^{\prime}_{1},\ldots,k^{\prime}_{n}}=\delta_{(k_{1},\ldots,k_{n}),(k^{\prime}_{1},\ldots,k^{\prime}_{n})}b_{k_{1},\ldots,k_{n}}$,

  5. (5)

$g_{i}b_{k_{1},\ldots,k_{n}}=b_{s_{i}.(k_{1},\ldots,k_{n})}g_{i}$

  6. (6)

$\sum b_{k_{1},\ldots,k_{n}}=1$,

where $e_{i}:=\sum_{k_{i}=k_{i+1}}b_{k_{1},\ldots,k_{n}}$.

###### Proof.

Let $\mathcal{H}_{n,r}^{\prime}$ be the $R$-algebra defined by the above
presentation. Then there is a surjective $R$-algbera homomorphism
$\phi:\mathcal{H}_{n,r}^{\prime}\to\mathcal{H}_{n,r}$ assigning $g_{i}\mapsto
g_{i}$ and $b_{k_{1},\ldots,k_{n}}\mapsto b_{k_{1},\ldots,k_{n}}$. Since

$\displaystyle t_{i}=\sum u_{k_{i}}b_{k_{1},\ldots,k_{n}}\qquad(1\leq i\leq
n)\quad\text{in}\ \mathcal{H}_{n,r},$ (3.17)

the homomorphism $\phi$ is surjective.

Let ${\mathbf{B}}^{\prime}$ be the subset of $\mathcal{H}_{n,r}^{\prime}$
analogus to ${\mathbf{B}}$, the basis in (3.15). Then by the relations
(1)–(6), any element in $\mathcal{H}_{n,r}^{\prime}$ can be written as an
$R$-linear combination of ${\mathbf{B}}^{\prime}$. In particular, the relation
(3) together with (6) implies that $g_{i}^{2}$ can be written as an $R$-linear
combination of ${\mathbf{B}}^{\prime}$. Moreover $\phi$ maps
${\mathbf{B}}^{\prime}$ to ${\mathbf{B}}$, which is linearly independent over
$R$. Hence ${\mathbf{B}}^{\prime}$ is a basis of $\mathcal{H}_{n,r}^{\prime}$
and $\phi$ is an isomorphism, as desired. ∎

Note that in the presentation of the theorem above, the parameters
$u_{1},\ldots,u_{r}$ do not show up. Hence we have

######  Corollary 3.15.

Let $R$ be an integral domain and $q\in R^{\times}$. Assume that
$(u_{1},\ldots,u_{r})$ and $(\tilde{u}_{1},\ldots,\tilde{u}_{r})$ be
$r$-tuples of elements in $R$ such that

$\displaystyle\Delta=\prod_{i>j}(u_{i}-u_{j})\in
R^{\times}\quad\text{and}\quad\tilde{\Delta}:=\prod_{i>j}(\tilde{u}_{i}-\tilde{u}_{j})\in
R^{\times}.$ (3.18)

Then there is an isomorphism

$\displaystyle\mathcal{H}_{n,r}(R,q,\tilde{u}_{1},\ldots,\tilde{u}_{r})\mathop{\xrightarrow[\raisebox{1.29167pt}[0.0pt][1.29167pt]{$\scriptstyle{}$}]{{\raisebox{-2.58334pt}[0.0pt][-2.58334pt]{$\mspace{2.0mu}\sim\mspace{2.0mu}$}}}}\mathcal{H}_{n,r}(R,q,u_{1},\ldots,u_{r})$

of $R$-algebras which assigns

$\displaystyle\tilde{T}_{j}\mapsto
T_{j},\quad\text{and}\quad\tilde{t}_{i}\mapsto\sum_{(k_{1},\ldots,k_{n})\in[1,r]^{n}}\tilde{u}_{k_{i}}b_{k_{1},\ldots,k_{n}}=\sum_{j=0}^{r-1}a_{j}t_{i}^{j},$

where $\tilde{T}_{j}$, $\tilde{t}_{i}$ denote the generators of
$\mathcal{H}_{n,r}(R,q,\tilde{u}_{1},\ldots,\tilde{u}_{r})$ of the
presentation in Definition 2.1, and $a_{0},\ldots,a_{r-1}\in R$ are given by
the equation

$\displaystyle\begin{pmatrix}1&u_{1}&u_{1}^{2}&\cdots&u_{1}^{r-1}\\\
1&u_{2}&u_{2}^{2}&\cdots&u_{2}^{r-1}\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\
1&u_{r}&u_{r}^{2}&\cdots&u_{r}^{r-1}\end{pmatrix}\begin{pmatrix}a_{0}\\\
a_{1}\\\ \vdots\\\ a_{r-1}\end{pmatrix}=\begin{pmatrix}\tilde{u}_{1}\\\
\tilde{u}_{2}\\\ \vdots\\\ \tilde{u}_{r}\end{pmatrix}.$

We remark here that the above matrix is the transpose of the Vandermonde
matrix $V(u_{1},\ldots,u_{r})$ in (2.10).

###### Proof.

By Theorem 3.14, there is an $R$-algebra isomorphism from
$\mathcal{H}_{n,r}(R,q,\tilde{u}_{1},\ldots,\tilde{u}_{r})$ to
$\mathcal{H}_{n,r}(R,q,u_{1},\ldots,u_{r})$ which matches the generators in
the presentation in Theorem 3.14. By (3.4) and (3.17), the image of
$\tilde{t}_{i}$ is the same with
$\sum\tilde{u}_{k_{i}}b_{k_{1},\ldots,k_{n}}$. Since

$\displaystyle\left(\sum_{k=0}^{r-1}a_{j}t_{i}^{j}\right)b_{k_{1},\ldots,k_{n}}=\left(\sum_{j=0}^{r-1}a_{j}u_{k_{i}}^{j}\right)b_{k_{1},\ldots,k_{n}}=\tilde{u}_{k_{i}}b_{k_{1},\ldots,k_{n}}\quad\text{for
any}\quad(k_{1},\ldots k_{n})\in[1,r]^{n},$

we have
$\sum\tilde{u}_{k_{i}}b_{k_{1},\ldots,k_{n}}=\sum_{j=0}^{r-1}a_{j}t_{i}^{j}$,
as desired. ∎

##  4\. Symmetrizing trace form

Define an $R$-linear map $\tau:\mathcal{H}_{n,r}\longrightarrow R$ by

$\displaystyle\tau(t_{1}^{c_{1}}\cdots
t_{n}^{c_{n}}g_{w})=\begin{cases}1&\text{if}\ \
w=\operatorname{id}_{\mathfrak{S}(n)},c_{1}=\cdots=c_{n}=0\\\ 0&\text{
otherwise }\end{cases}$ (4.1)

for $0\leq c_{1},\ldots,c_{r}\leq r-1$, $w\in\mathfrak{S}(n)$. Note that for
any polynomial $f(X)\in R[X]$ in one variable and any $1\leq i\leq n$, we have

$\displaystyle\tau(f(t_{i}))=\tau(\tilde{f}(t_{i}))=a,$ (4.2)

where $\tilde{f}(X)$ is the polynomial with degree $<r$ congruent to $f(X)$
modulo $(X-u_{1})\cdots(X-u_{n})$, and $a$ is the constant term of
$\tilde{f}(X)$. Note also that if $f_{1}(X),\ldots,f_{n}(X)\in R[X]$, then

$\displaystyle\tau(f_{1}(t_{1})\cdots
f_{n}(t_{n}))=\tau(\tilde{f}_{1}(t_{1})\cdots\tilde{f}_{n}(t_{n}))=a_{1}\cdots
a_{n}=\tau(f_{1}(t_{1}))\cdots\tau(f_{n}(t_{n})),$ (4.3)

where $a_{i}$ denotes the constant term of $\tilde{f}_{i}(X)$.

The proof of the below lemma is identical to that of [4, Proposition 10]), but
we include it here for reader’s convenience.

######  Lemma 4.1.

([4, Proposition 10]) Let $w,w^{\prime}\in\mathfrak{S}(n)$. Then

$\tau(g_{w}g_{w^{\prime}})=\delta(w^{\prime}=w^{-1}).$

It is equivalent to saying that for any $A\in R[t_{1},\ldots,t_{n}]$,

$\tau(Ag_{w}g_{w^{\prime}})=\begin{cases}\tau(A)&\text{if}\
w^{-1}=w^{\prime}\\\ 0&\text{if}\ w^{-1}\neq w^{\prime}\end{cases}$

###### Proof.

Since $\\{t_{1}^{c_{1}}\cdots t_{n}^{c_{n}}g_{w}\\}$ is an $R$-basis of
$\mathcal{H}_{n,r}$, the second statement follows from the first, by the
definition of $\tau$.

When $w=\operatorname{id}$, there is nothing to prove. Assume that
$\ell(w)\geq 1$. Then there is $1\leq i\leq n-1$ such that
$\ell(ws_{i})<\ell(w)$. Then

$g_{w}=g_{(ws_{i})s_{i}}=g_{ws_{i}}g_{s_{i}}.$

(Case 1) $\ell(s_{i}w^{\prime})>\ell(w^{\prime})$: We have

$g_{s_{i}}g_{w^{\prime}}=g_{s_{i}w^{\prime}}$

On the other hand, $\ell(ws_{i})<\ell(w)$ implies that
$\ell(s_{i}w^{-1})<\ell(w^{-1})$ so that $w^{\prime}\neq w^{-1}$ and hence
$s_{i}w^{-1}=(ws_{i})^{-1}\neq s_{i}w^{\prime}$. By induction on $\ell(w)$, we
have

$\tau(g_{w}g_{w^{\prime}})=\tau(g_{ws_{i}}g_{s_{i}}g_{w^{\prime}})=\tau(g_{ws_{i}}g_{s_{i}w^{\prime}})=0.$

(Case 2) $\ell(s_{i}w^{\prime})<\ell(w^{\prime})$: We have

$g_{s_{i}}g_{w^{\prime}}=g_{s_{i}w^{\prime}}+(q-q^{-1})e_{i}g_{w^{\prime}}$

and hence

$\tau(g_{w}g_{w}^{\prime})=\tau(g_{ws_{i}}g_{s_{i}w^{\prime}}+(q-q^{-1})g_{ws_{i}}e_{i}g_{w^{\prime}})=\tau(g_{ws_{i}}g_{s_{i}w^{\prime}})+(q-q^{-1})\tau(g_{ws_{i}}e_{i}g_{w^{\prime}})$

On the other hand, $\ell(s_{i}w^{\prime})<\ell(w^{\prime})$ implies that
$\ell((w^{\prime})^{-1}s_{i})<\ell((w^{\prime})^{-1})$ so that
$(w^{\prime})^{-1}\neq w{s_{i}}$. Since $g_{ws_{i}}e_{i}=Ag_{ws_{i}}$ for some
$A\in R_{1}[t_{1},\ldots,t_{n}]$, we have

$\tau(g_{ws_{i}}e_{i}g_{w^{\prime}})=\tau(Ag_{ws_{i}}g_{w^{\prime}})=0$

by the induction hypothesis.

If $w^{-1}\neq w^{\prime}$, then $(ws_{i})^{-1}\neq s_{i}w^{\prime}$, and by
induction,

$\tau(g_{ws_{i}}g_{s_{i}w^{\prime}})=0$

If $w^{-1}=w^{\prime}$, then $(ws_{i})^{-1}=s_{i}w^{\prime}$, and by
induction,

$\tau(g_{ws_{i}}g_{s_{i}w^{\prime}})=1,$

as desired. ∎

Recall that an $R$-linear map $f:\mathcal{A}\to R$ from an $R$-algerba
$\mathcal{A}$ to its base ring $R$ is called a _trace form_ if

$\displaystyle f(xy)=f(yx)\quad\text{for}\ x,y\in\mathcal{A}.$

A trace form $f$ is called _symmetrizing_ if the bilinear form

$\displaystyle\mathcal{A}\times\mathcal{A}\to R\qquad\text{given
by}\quad(x,y)\mapsto f(xy)$

is nondegenerate.

######  Corollary 4.2.

$\tau$ is a trace form on $\mathcal{H}_{n,r}$.

###### Proof.

Because $\left\\{t_{1}^{c_{1}}\cdots t_{n}^{c_{n}}g_{w}\mid 0\leq c_{i}\leq
r-1,\,w\in\mathfrak{S}(n)\right\\}$ and $\left\\{g_{w}t_{1}^{c_{1}}\cdots
t_{n}^{c_{n}}\mid 0\leq c_{i}\leq r-1,\,w\in\mathfrak{S}(n)\right\\}$ are
$R$-basis of $\mathcal{H}_{n,r}$, it is enough to show that

$\tau(t_{1}^{c_{1}}\cdots t_{n}^{c_{n}}g_{w}g_{w^{\prime}}t_{1}^{d_{1}}\cdots\
t_{n}^{d_{n}})=\tau(g_{w^{\prime}}t_{1}^{d_{1}}\cdots\
t_{n}^{d_{n}}t_{1}^{c_{1}}\cdots t_{n}^{c_{n}}g_{w})$

for all $w,w^{\prime}\in\mathfrak{S}(n)$ and $0\leq c_{i},d_{i}\leq r-1$. We
have

$\tau(t_{1}^{c_{1}}\cdots t_{n}^{c_{n}}g_{w}g_{w^{\prime}}t_{1}^{d_{1}}\cdots\
t_{n}^{d_{n}})=\delta(w^{\prime}=w^{-1})\tau(t_{1}^{c_{1}+d_{1}}\cdots
t_{n}^{c_{n}+d_{n}})$

and

$\tau(g_{w^{\prime}}t_{1}^{d_{1}}\cdots\ t_{n}^{d_{n}}t_{1}^{c_{1}}\cdots
t_{n}^{c_{n}}g_{w})=\tau(t_{w^{\prime}(1)}^{c_{1}+d_{1}}\cdots\
t_{w^{\prime}(n)}^{c_{n}+d_{n}}g_{w^{\prime}}g_{w})=\delta(w^{\prime}=w^{-1})\tau(t_{w^{\prime}(1)}^{d_{1}+c_{1}}\cdots\
t_{w^{\prime}(n)}^{c_{n}+d_{n}})$

Hence it amounts to show that

$\tau(t_{1}^{p_{1}}\cdots t_{n}^{p_{n}})=\tau(t_{v(1)}^{p_{1}}\cdots
t_{v(n)}^{p_{n}})$

for any $0\leq p_{i}\leq 2r-2$ and any $v\in\mathfrak{S}(n)$. Indeed, it
follows from (4.2) and (4.3). ∎

######  Remark 4.3.

The trace form $\tau$ in the case of the Yokonuma-Hekce algebra appeared in
[4]. It is known to be the same with the trace form obtained from an
isomrophism between Yokonuma-Hecke algebra and a matrix algebra over Hecke
algebras of symetric groups ([5]).

######  Lemma 4.4.

For $0\leq s\leq r-1$ and $\leq i\leq n$, we have

$\displaystyle\tau(t_{i}^{r+s})=\tau(t_{i}^{r})h_{s}=(-1)^{r+1}(u_{1}\cdots
u_{r})h_{s}=(-1)^{r+1}\sigma_{r}h_{s}$ (4.4)

where $h_{s}=h_{s}(u_{1},\ldots,u_{r})$ denotes the $s$-th complete
homogeneous symmetric polynomial in $u_{1},\ldots,u_{r}$.

###### Proof.

Note that for any polynomial $f(X)$ in single variable, the value
$\tau(f(t_{i}))$ is independent to $i$. Hence in the proof we will denote $t$
as a representative of $t_{i}$’s. Note that
$\tau(t^{r})=(-1)^{r+1}(u_{1}\cdots u_{r})$ by the defining relation.

We will proceed by induction on $s$. When $s=0$, it is trivial.

Assume that $s\geq 1$. From (2.11), we have

$\displaystyle t^{r+s}=\sum_{k=0}^{r-1}(-1)^{r-k+1}\sigma_{r-k}\,t^{k+s},$

where $\sigma_{k}=\sigma_{k}(u_{1},\ldots,u_{r})$ is the $k$-th elementary
symmetric polynomial in $u_{1},\ldots,u_{r}$. It follows that

$\displaystyle\tau(t^{r+s})$
$\displaystyle=\sum_{k=0}^{r-1}(-1)^{r-k+1}\sigma_{r-k}\,\tau(t^{k+s})=\sum_{k=r-s}^{r-1}(-1)^{r-k+1}\sigma_{r-k}\,\tau(t^{k+s})$
$\displaystyle=\sum_{k=r-s}^{r-1}(-1)^{r-k+1}\sigma_{r-k}\,\tau(t^{r})h_{k+s-r}=\tau(t^{r})\sum_{j=1}^{s}(-1)^{j+1}\sigma_{j}\,h_{s-j}=\tau(t^{r})h_{s}$

where the third equality comes from the induction hypothesis, and the last
equality follows from the well-known Newton’s identity below.

$\displaystyle\sum_{j=0}^{s}(-1)^{j}\sigma_{j}\,h_{s-j}=\begin{cases}1&\text{if}\
s=0\\\ 0&\text{if}\ s>0.\end{cases}$ (4.5)

∎

######  Lemma 4.5.

For $0\leq c,d\leq r-1$ and $1\leq i\leq n$, we have

$\displaystyle\tau\left(t_{i}^{c}\left(\sum_{j=0}^{r-d-1}(-1)^{j}\sigma_{j}t_{i}^{r-d-j}\right)\right)=\delta(c=d)(-1)^{r+1}\sigma_{r}.$

###### Proof.

We have

$\displaystyle\tau\left(t_{i}^{c}\left(\sum_{j=0}^{r-d-1}(-1)^{j}\sigma_{j}t_{i}^{r-d-j}\right)\right)=\tau\left(\sum_{j=0}^{r-d-1}(-1)^{j}\sigma_{j}t_{i}^{r+c-d-j}\right)$
$\displaystyle=\sum_{j=0}^{r-d-1}(-1)^{j}\sigma_{j}\tau(t_{i}^{r+c-d-j})=\sum_{j=0}^{c-d}(-1)^{j}\sigma_{j}\tau(t_{i}^{r+c-d-j})$
$\displaystyle=(-1)^{r+1}\sigma_{r}\sum_{j=0}^{c-d}(-1)^{j}\sigma_{j}h_{c-d-j}=(-1)^{r+1}\sigma_{r}\delta(c=d),$

as desired. ∎

Assume that $\sigma_{r}=u_{1}u_{2}\cdots u_{r}\in R^{\times}$. For each $1\leq
i\leq n$ and $1\leq c\leq r-1$, set

$\displaystyle(t_{i}^{c})^{\vee}:=\dfrac{(-1)^{r+1}}{\sigma_{r}}\sum_{0\leq
j\leq r-c-1}(-1)^{j}\sigma_{j}t_{i}^{r-c-j}.$

Then for each $w\in\mathfrak{S}(n)$ and $(c_{1},\ldots,c_{n})\in[0,r-1]^{n}]$,
we set

$\displaystyle\left(t_{1}^{c_{1}}\cdots
t_{n}^{c_{n}}g_{w}\right)^{\vee}:=g_{w^{-1}}\prod_{i:c_{i}\neq
0}(t_{i}^{c_{i}})^{\vee}.$ (4.6)

######  Proposition 4.6.

Assume that $u_{1},\ldots,u_{r}\in R^{\times}$. For
$(c_{1},\ldots,c_{n})\in[0,r-1]^{n}$ and $w\in\mathfrak{S}(n)$, we have

$\displaystyle\tau\left(t_{1}^{c_{1}}\cdots
t_{n}^{c_{n}}g_{w}(t_{1}^{d_{1}}\cdots
t_{n}^{d_{n}}g_{u})^{\vee}\right)=\delta(\text{$w=u$ and $c_{i}=d_{i}$ for all
$i$}).$

###### Proof.

We have

$\displaystyle\tau\left(t_{1}^{c_{1}}\cdots
t_{n}^{c_{n}}g_{w}\,(t_{1}^{d_{1}}\cdots
t_{n}^{d_{n}}g_{u})^{\vee}\right)=\tau\left(t_{1}^{c_{1}}\cdots
t_{n}^{c_{n}}g_{w}\,g_{u^{-1}}(t_{1}^{d_{1}})^{\vee}\cdots(t_{n}^{d_{n}})^{\vee}\right)$
$\displaystyle=\delta(w=u)\tau\left(t_{1}^{c_{1}}\cdots
t_{n}^{c_{n}}(t_{1}^{d_{1}})^{\vee}\cdots(t_{n}^{d_{n}})^{\vee}\right)=\delta(\text{$w=u$
and $c_{i}=d_{i}$ for all $i$}).$

where the last equality follows from (4.3) and Lemma 4.5. ∎

######  Corollary 4.7.

The trace form $\tau$ is symmetrizing if $\sigma_{r}:=u_{1}\cdots u_{r}$ is
invertible in $R$.

##  5\. Symmetric group action on $\mathcal{H}_{n,r}$ and the fixed subalgebra
$\mathcal{H}_{n,r}^{\mathfrak{S}(r)}$

Recall that there is an action of the symmetric group $\mathfrak{S}(r)$ on the
subalgebra $R[t_{1},\ldots,t_{n}]$ given by

${}^{\sigma}b_{k_{1},\ldots,k_{n}}:=b_{\sigma(k_{1}),\ldots,\sigma(k_{n})}\quad\text{for
$\sigma$ in $\mathfrak{S}(r)$}.$

For each $\sigma\in\mathfrak{S}(r)$ we obtain an $R$-linear endomorphism on
$\mathcal{H}_{n,r}$, denoted by $\sigma$ again, by setting

${}^{\sigma}g_{w}:=g_{w}\quad\text{for $w$ in $\mathfrak{S}(n)$}.$

The following is a direct consequence of Theorem 3.14.

######  Proposition 5.1 (cf. [Proposition 2.2 in [6]).

The map $\sigma$ on $\mathcal{H}_{n,r}$ is an $R$-algebra automorphism so that
the group $\mathfrak{S}(r)$ acts on $\mathcal{H}_{n,r}$ by $R$-algerba
automorphisms.

For each $1\leq i,j\leq n$, define

$\displaystyle
e_{i,j}:=\sum_{(k_{1},\ldots,k_{n})\in[1,r]^{n},\,k_{i}=k_{j}}b_{k_{1},\ldots,k_{n}}.$

Then $e_{i,i+1}=e_{i}$ for $1\leq i\leq n-1$ and ${}^{\sigma}e_{i,j}=e_{i,j}$
for all $\sigma\in\mathfrak{S}(r)$ and $1\leq i,j\leq n$.

######  Lemma 5.2.

For $1\leq i<j\leq n-1$, we have

$\displaystyle g_{j-1}\cdots g_{i+1}e_{i}g_{i+1}^{-1}\cdots
g^{-1}_{j-1}=e_{i,j}.$

###### Proof.

We have

$g_{i+1}e_{i}=\sum_{k_{i}=k_{i+1}}g_{i+1}b_{k_{1},\ldots,k_{n}}=\sum_{k_{i}=k_{i+2}}b_{k_{1},\ldots,k_{n}}g_{i+1}=e_{i,i+1}g_{i+1}.$

By induction on $j-i$, we obtain the assertion. ∎

Let $[k_{1},\ldots,k_{n}]$ denote the orbit of $(k_{1},\ldots,k_{n})$ under
the action of $\mathfrak{S}(r)$ and define

$\displaystyle
b_{[k_{1},\ldots,k_{n}]}:=\sum_{(\ell_{1},\ldots,\ell_{n})\in{[k_{1},\ldots,k_{n}]}}b_{\ell_{1},\ldots,\ell_{n}}=\sum_{\stackrel{{\scriptstyle(\ell_{1},\ldots,\ell_{n})\in[1,r]^{n}}}{{\text{such
that}\,\ell_{i}=\ell_{j}\,\text{if and only
if}\,k_{i}=k_{j}}}}b_{\ell_{1},\ldots,\ell_{n}}.$

######  Lemma 5.3 (cf. Lemma 4.2 in [6]).

For each $(k_{1},\ldots,k_{n})\in[1,r]^{n}$, we have

$\displaystyle b_{[k_{1},\ldots,k_{n}]}=\left(\prod_{1\leq i<j<\leq
n,k_{i}=k_{j}}e_{i,j}\right)\left(\prod_{1\leq i<j<\leq n,k_{i}\neq
k_{j}}(1-e_{i,j})\right).$

###### Proof.

We have

$\displaystyle\prod_{k_{i}\neq k_{j}}(1-e_{i,j})$
$\displaystyle=\sum_{(\ell_{1},\ldots,\ell_{n})\in[1,r]^{n}}\left(\prod_{k_{i}\neq
k_{j}}(1-e_{i,j})\right)b_{\ell_{1},\ldots,\ell_{n}}$
$\displaystyle=\sum_{(\ell_{1},\ldots,\ell_{n})\in[1,r]^{n}}\left(\prod_{k_{i}\neq
k_{j}}(1-e_{i,j})b_{\ell_{1},\ldots,\ell_{n}}\right)$
$\displaystyle=\sum_{(\ell_{1},\ldots,\ell_{n})\in[1,r]^{n}}\left(\prod_{k_{i}\neq
k_{j}}(1-\delta(\ell_{i}=\ell_{j}))b_{\ell_{1},\ldots,\ell_{n}}\right)$
$\displaystyle=\sum_{(\ell_{1},\ldots,\ell_{n})\in[1,r]^{n}}\left(\prod_{k_{i}\neq
k_{j}}(\delta(\ell_{i}\neq\ell_{j}))b_{\ell_{1},\ldots,\ell_{n}}\right)$
$\displaystyle=\sum_{\stackrel{{\scriptstyle(\ell_{1},\ldots,\ell_{n})\in[1,r]^{n}}}{{\text{such
that}\,\ell_{i}\neq\ell_{j}\,\text{if}\,k_{i}\neq
k_{j}}}}b_{\ell_{1},\ldots,\ell_{n}}.$

Similarly we obtain

$\displaystyle\prod_{k_{i}=k_{j}}e_{i,j}=\sum_{\stackrel{{\scriptstyle(\ell_{1},\ldots,\ell_{n})\in[1,r]^{n}}}{{\text{such
that}\,\ell_{i}=\ell_{j}\,\text{if}\,k_{i}=k_{j}}}}b_{\ell_{1},\ldots,\ell_{n}},$

and hence we have

$\displaystyle\left(\prod_{k_{i}=k_{j}}e_{i,j}\right)\left(\prod_{k_{i}\neq
k_{j}}(1-e_{i,j})\right)=\sum_{\stackrel{{\scriptstyle(\ell_{1},\ldots,\ell_{n})\in[1,r]^{n}}}{{\text{such
that}\,\ell_{i}=\ell_{j}\,\text{if and only
if}\,k_{i}=k_{j}}}}b_{\ell_{1},\ldots,\ell_{n}}.$

Hence the assertion follows from (2.18). ∎

Let $\mathcal{H}_{n,r}^{\mathfrak{S}(r)}$ be the fixed subalgebra of
$\mathcal{H}_{n,r}$ under the action of $\mathfrak{S}(r)$.

######  Proposition 5.4.

  1. (1)

The set

$\displaystyle\left\\{b_{[k_{1},\ldots,k_{n}]}g_{w}\mid[k_{1},\ldots,k_{n}]\in\mathfrak{S}(r)\backslash[1,r]^{n},\
w\in\mathfrak{S}(n)\right\\}$

forms an $R$-basis of the subalgebra $\mathcal{H}_{n,r}^{\mathfrak{S}(r)}$.

  2. (2)

The subalgebra of $\mathcal{H}_{n,r}$ generated by
$g_{1},\ldots,g_{n-1},e_{1},\ldots,e_{n-1}$ is equal to the subalgebra
$\mathcal{H}_{n,r}^{\mathfrak{S}(r)}$.

###### Proof.

The first assertion is immediate from the definitions.

By definition the elements $g_{1},\ldots,g_{n-1},e_{1},\ldots,e_{n-1}$ belong
to $\mathcal{H}_{n,r}^{\mathfrak{S}(r)}$. The second assertion follows from
(1) together with Lemma 5.2 and Lemma 5.3. ∎

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