2.1. Garside groups and their representations

By [Reference Dehornoy and Paris9], the groups $G(n,m)$ are Garside groups. We will not recall all the theory of Garside monoids and groups here, but only some properties that we will use later on. For the interested reader, more on Garside structures can be found in [Reference Dehornoy, Digne, Krammer, Godelle and Michel8], [Reference Dehornoy and Paris9]; as explained in Remark 2.2 below, the fact that $G(n,m)$ is a Garside group is in fact not needed in this article, but our approach rather reproves some important properties that $G(n,m)$ admits as a Garside group.

Roughly speaking, saying that $G(n,m)$ is a Garside group means that $G(n,m)$ is the group of fractions of the monoid $\mathcal {M}(n,m)$ defined by the same presentation (which we will call a Garside presentation), that the latter is cancellative and has good divisibility properties, and that there exists an element $\Delta $ (the Garside element) whose set of left- and right-divisors coincide, and form a finite generating set $\mathrm {Div}(\Delta )$ of $\mathcal {M}(n,m)$ (and hence of $G(n,m)$ ), called the set of simple elements of $\mathcal {M}(n,m)$ . In our case, the Garside element is given by $X^m = Y^n$ and is central (a Garside element always has a central power, and since Garside groups are torsion-free, a Garside group in particular has a nontrivial center). In this case, it is even a generator of the center, as shown by Schreier [Reference Schreier22].

In a Garside group G attached to a Garside monoid $\mathcal {M}$ , every element $g\in G$ can be written in the form $x^{-1}y$ , where $x, y\in \mathcal {M}$ . Garside groups have a solvable word problem and every element admits canonical left- and right-normal forms, whose factors are elements of the set $\mathrm {Div}(\Delta )$ . In fact, given any element $x\in \mathcal {M}\backslash \{1\}$ , the good divisibility properties of $\mathcal {M}$ ensure that there is a unique element $x_1$ in $\mathrm {Div}(\Delta )\backslash \{1\}$ , maximal for left-divisibility, that can be factored out from x. We thus have $x=x_1 x'$ for some (by cancellativity uniquely-defined) element $x'\in \mathcal {M}$ , and one can go on factoring $x'$ . The properties of Garside monoids guarantee that this procedure terminates, yielding a canonical decomposition $x=x_1 x_2 \dots x_k$ with $x_i \in \mathrm {Div}(\Delta ) \backslash \{1\}$ , the left-Garside normal form of x. Factoring out from the right yields the right-Garside normal form. In our case, we have $\mathrm {Div}(\Delta )=\{1, X, X^2, \dots , X^{m-1}, Y, Y^2, \dots , Y^{n-1}, \Delta = X^m = Y^n\}$ .

We shall construct two-dimensional representations of $G(n,m)$ and show that they are faithful using the following procedure, which is the philosophy used by Krammer to show that Artin’s n-strand braid groups are linear [Reference Krammer16]:

1. 1. To show that an action of a Garside group G (with Garside monoid $\mathcal {M}$ ) on a set X is faithful, since every element of G can be written under the form $x^{-1} y$ with $x, y\in \mathcal {M}$ , it suffices to show that the monoid $\mathcal {M}$ acts faithfully.

2. 2. To show that the action of $\mathcal {M}$ is faithful, one can try to reconstruct the Garside normal form of every element $m\in \mathcal {M}$ from its corresponding automorphism of X.

This approach has been successfully applied to show that several (linear or categorical) actions of braid groups or more generally spherical-type Artin–Tits groups are faithful (see, for instance, [Reference Brav and Thomas4], [Reference Digne10], [Reference Jensen15], [Reference Lehrer and Xi17], [Reference Licata and Queffelec18]). Since Krammer’s approach is Garside-theoretic, it is natural to wonder whether it can be applied for other families of Garside groups. To the best of our knowledge, this has not been investigated before, and we shall do it below for the family of groups $G(n,m)$ , which is among the easiest families of Garside groups distinct from spherical-type Artin–Tits groups.

2.2. Properties of the groups under consideration

As already pointed out, the groups $G(n,m)$ form one of the easiest families of Garside groups. Let us point out some additional features:

• • If $n=2$ and m is odd, then $G(n,m)$ is isomorphic to the Artin group of dihedral type $I_2(m)$ .

• • If $(n,m)=(3,4)$ , respectively, $(3,5)$ , then $G(n,m)$ is isomorphic to the complex braid group of the exceptional complex reflection group $G_{12}$ , respectively, $G_{22}$ (see [Reference Bannai1]).

• • If $n, m$ are coprime, which in particular covers all the examples above, then $G(n,m)$ is the knot group of the torus knot $T_{n,m}$ . Since every torus knot (except the unknot) is of this form, it covers all the torus knots apart from the knot group of the unknot, which is isomorphic to $\mathbb {Z}$ .

Since $G(2,3)$ is isomorphic to the three-strand braid group $B_3$ , the groups $G(n,m)$ can be considered as generalizations of $B_3$ . Note that the presentation $\langle X, Y \ \vert \ X^3 = Y^2 \rangle $ is not the standard Artin presentation of $B_3$ . The standard Artin presentation of $B_3$ can in fact be generalized when n and m are coprime as we now recall, and this suggests to investigate whether properties of $B_3$ can actually be generalized to the groups of the form $G(n,m)$ with n and m coprime. In a previous work [Reference Gobet12], we introduced and classified analogs of the complex reflection groups appearing as quotients of $B_3$ when adding torsion on the generators of the standard presentation, and studied Garside structures on $G(n,m)$ analogous to those that are known for $B_3$ [Reference Gobet11]. Note that Haladjian recently generalized some of these results to a larger family of Garside groups, in such a way that they include all the complex braid groups of complex reflection groups of rank two [Reference Haladjian13], [Reference Haladjian14]. In this article, we continue this program by taking a more representation-theoretic road: we will see that for most groups $G(n,m)$ with n and m coprime, we can construct a representation generalizing the Burau representation of $B_3$ and show that it is faithful using the philosophy recalled in Section 2.1. This will be done by proving Proposition 1.1—which in fact will allow us to produce faithful representations for all the groups of the form $G(n,m)$ , that is, not only when n and m are coprime (see §4.2 below).

2.3. Case where the parameters are coprime

Let $n,m \geq 2$ with n and m coprime. The group $G(n,m)$ admits the presentation

(2.1) $$ \begin{align}\langle \ x_1, x_2, \dots, x_n \ \vert\ \underbrace{x_1 x_2 \dots}_{m~\text{factors}} = \underbrace{x_2 x_3\dots}_{m~\text{factors}} = \dots = \underbrace{x_n x_1 \dots}_{m~\text{factors}} \ \rangle,\end{align} $$

where indices are taken modulo n if $n<m$ . Since $G(n,m)\cong G(m,n)$ , it also admits the presentation

(2.2) $$ \begin{align}\langle \ y_1, y_2, \dots, y_m \ \vert\ \underbrace{y_1 y_2 \dots}_{n~\text{factors}} = \underbrace{y_2 y_3\dots}_{n~\text{factors}} = \dots = \underbrace{y_m y_1 \dots}_{n~\text{factors}} \ \rangle.\end{align} $$

For $B_3 \cong G(2,3)$ (more generally for the Artin group of dihedral type $I_2(m)$ with m odd, which is isomorphic to $G(2,m)$ ), Presentation (2.1) is nothing but the classical Artin–Tits group presentation, while Presentation (2.2) is nothing but its Birman–Ko–Lee or dual presentation. An explicit isomorphism between $G(n,m)$ and the group with Presentation (2.1) is given by $X \mapsto x_1 x_2 \dots x_n$ , $Y \mapsto x_1 x_2 \dots x_m$ . We refer to [Reference Gobet11, Section 3] for more details and proofs of these facts. The above two presentations are also Garside presentations.

In fact, since n and m are coprime, all the $x_i$ ’s are conjugate in $G(n,m)$ . It thus suffices to add the relation $x_1^k=1$ . These groups behave like reflection groups in some sense, and the corresponding torus knot groups like their braid groups, generalizing the quotient $B_3 \twoheadrightarrow S_3$ (see [Reference Gobet12]).

3.1. Preliminary observations

Let $n,m \geq 2$ . Recall from Section 2.1 that the presentation $\langle X, Y \ \vert \ X^m = Y^n \rangle $ defines a Garside structure on $G(n,m)$ . Denote $\mathcal {M}=\mathcal {M}(n,m)$ the corresponding Garside monoid with $\Delta = X^m = Y^n$ . Let $u\in \mathcal {M}$ be an element such that $u\neq 1$ and u does not admit $\Delta = X^m = Y^n$ as a left- (equivalently right-) divisor and denote by $\mathcal {M}_{\Delta \not \leq }$ the set of such elements of $\mathcal {M}$ . Then u can be uniquely written in the form

$$ \begin{align*}Y^{n_1} X^{m_1} Y^{n_2} X^{m_2} \dots Y^{n_k} X^{m_k},\end{align*} $$

where $k\geq 1$ , $1 \leq m_i \leq m-1$ for all $i=1, \ldots , k-1$ , $0 \leq m_k \leq m-1$ , $1 \leq n_i \leq n - 1$ for all $i=2, \dots , k$ , $0 \leq n_1 \leq n-1$ . We denote by $L(u)$ the first letter of the above word, and by $R(u)$ its last letter. Every factor $X^{m_i}$ , $Y^{n_i}$ is a simple element (in fact, their concatenation as above yields both the left- and right-Garside normal form of u); we call such factors the Garside factors of u. Let $\ell (u)$ denote the Garside length of u, that is,

$$ \begin{align*}\ell(u)= 2(k-1) + \mathbf{{1}}_{n_1 \neq 0} + \mathbf{{1}}_{m_k \neq 0}.\end{align*} $$

We denote by $\ell _w(u)$ its weighted length, that is, the integer $Mn+Nm$ , where $M=\sum m_i$ and $N= \sum n_i$ . For later use, for an arbitrary $x\in \mathcal {M}$ , one can define $\ell _w(x)$ by choosing a word for x, counting the number M (resp. N) of occurrences of X (resp. of Y), and setting $\ell _w(x) = Mn+ Nm$ ; the relation $X^m = Y^n$ guarantees that it is well-defined. Given a Laurent polynomial $P\in \mathbb {C}[t^{\pm 1}, q^{\pm 1}]$ , denote by $d_q(P)$ its degree in q. By convention, the polynomial $P=0$ has degree $-\infty $ .

To show that the representation $X \mapsto M_X$ , $Y \mapsto M_Y$ from Proposition 1.1 is faithful, we first establish the following.

We symbolize the four situations occurring in Proposition 3.1 in Figure 1.

Corollary 3.2. The elements $L(u)$ and $R(u)$ can be read off from $M_u$ . The integer $\ell (u)$ can be read off from $M_u$ .

Figure 1

Reading the first and last letters of the word from the matrix.

Proof of Proposition 3.1

The proof will be by induction on $\ell (u)$ . If $\ell (u)=1$ , then $u=X^k$ for some $1\leq k \leq m-1$ or $u=Y^k$ for some $1\leq k \leq n-1$ , hence we either lie in Situation $1$ or in Situation $4$ , and the result holds by assumption on the matrices $M_X^k$ , $M_Y^k$ .

Hence, assume that $\ell (u) \geq 1$ and let $\alpha $ be a Garside factor such that, denoting $v= \alpha u$ , we have $\ell (v)= \ell (u)+1$ . Assume that the result holds by induction on u. Let $M_u=\begin {pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end {pmatrix}$ . If $\alpha =X^k$ for some $1 \leq k \leq m-1$ , then

$$ \begin{align*}M_v=M_X^k M_u = t^{nk} \begin{pmatrix} c_1(k) a_{11} + \underbrace{c_2(k)}_{\neq 0} a_{21} & c_1(k) a_{12} + \underbrace{c_2(k)}_{\neq 0} a_{22} \\ c_3(k) a_{11} + c_4(k) a_{21} & c_3(k) a_{12} + c_4(k) a_{22} \end{pmatrix},\end{align*} $$

where the $c_i(k)$ ’s are complex numbers.

If $\alpha =Y^k$ for some $1\leq k \leq n-1$ , then

$$ \begin{align*}M_v=M_Y^k M_u = t^{km} \begin{pmatrix} \underbrace{P_1(k)}_{\mathrm{deg}\leq 0} a_{11} + \underbrace{P_2(k)}_{\mathrm{deg}\leq 0} a_{21} & \underbrace{P_1(k)}_{\mathrm{deg}\leq 0} a_{12} + \underbrace{P_2(k)}_{\mathrm{deg}\leq 0} a_{22} \\ \underbrace{P_3(k)}_{ \mathrm{deg} = 1} a_{11} + \underbrace{P_4(k)}_{\mathrm{deg}\leq 0} a_{21} & \underbrace{P_3(k)}_{ \mathrm{deg} = 1} a_{12} + \underbrace{P_4(k)}_{\mathrm{deg}\leq 0} a_{22} \end{pmatrix},\end{align*} $$

where the $P_i(k)$ ’s are Laurent polynomials in q.

Assume that u lies in Situation 1. This means that the first Garside factor of u is a nontrivial power of Y, hence that $\alpha =X^k$ for some $1\leq k \leq m-1$ . We have $n_q(\alpha u)= n_q(u)$ . By induction, the maximal degree of a coefficient of $M_u$ is $n_q(u)$ and is achieved in $a_{21}$ and only in this coefficient. The above calculation for $M_v=M_X^k M_u=(b_{ij})_{1\leq i,j \leq 2}$ then yields that the maximal degree of a coefficient in $M_v$ is still $n_q(u)$ and achieved in $b_{11}$ and possibly in $b_{21}$ as well, but not in the other coefficients. Since v lands in Situation $3$ , this shows the statement in this case.

Assume that u lies in Situation 2. The first Garside factor of u is as in the previous case, we can thus argue similarly. We have $n_q(\alpha u)= n_q(u)$ and by induction, the maximal degree of a coefficient of $M_u$ is equal to $n_q(u)$ and is achieved in $a_{22}$ and possibly in $a_{21}$ , but not in the other coefficients. The above calculation for $M_v=M_X^k M_u=(b_{ij})_{1\leq i,j \leq 2}$ then yields that the maximal degree of a coefficient in $M_v$ is still $n_q(u)$ and achieved in $b_{12}$ and possibly in any other coefficient of $M_v$ as well. Since v lands in Situation $4$ , this shows the statement in this case.

Assume that u lies in Situation 3. This means that the first Garside factor of u is a nontrivial power of X, hence that $\alpha =Y^k$ for some $1\leq k \leq n-1$ . We have $n_q(\alpha u)= n_q(u)+1$ . By induction, the maximal degree of a coefficient of $M_u$ is $n_q(u)$ and is achieved in $a_{11}$ and possibly in $a_{21}$ , but not in the other coefficients. The above formula for $M_v=M_Y^k M_u=(b_{ij})_{1\leq i,j \leq 2}$ then yields that the maximal degree of a coefficient in $M_v$ is $n_q(u)+1$ and is achieved in $b_{21}$ and only in $b_{21}$ . Since v lands in Situation $1$ , this shows the statement in this case.

Finally, assume that u lies in Situation 4. The first Garside factor of u is as in the previous case, we can thus argue similarly. We have $n_q(\alpha u) = n_q(u)+1$ . By induction, the maximal degree of a coefficient of $M_u$ is $n_q(u)$ and is achieved in $a_{12}$ and possibly in any other coefficient. The above formula for $M_v=M_Y^k M_u=(b_{ij})_{1\leq i,j \leq 2}$ then yields that the maximal degree of a coefficient in $M_v$ is $n_q(u)+1$ and is achieved in $b_{22}$ and possibly in $b_{21}$ as well, but not in the other coefficients. Since v lands in Situation $2$ , this shows the statement in this case.

We now have all the tools at our disposal to show the faithfulness.

3.2. Proof of Proposition 1.1

3.3. Recovering the Garside normal form from the representation

Assume again that $M_X$ and $M_Y$ satisfy the assumptions of Proposition 1.1. The aim of this section is to show that we can recover the Garside normal form of any element u of $\mathcal {M}$ from its matrix, that we still denote $M_u$ .

Consider an element $u\in \mathcal {M}$ . We can write it in the form $u' \Delta ^k$ Footnote ¹ , with $u'\in \mathcal {M}_{\Delta \not \leq }\cup \{1\}$ . In our representation, we then have

$$ \begin{align*}M_u=M_{u'} M_\Delta^k.\end{align*} $$

Now since $\Delta =X^m=Y^n$ is central in $G(n,m)$ , the matrix $M_\Delta =t^{nm}\begin {pmatrix}a & b \\ c & d \end {pmatrix}$ , with $a,b,c,d\in \mathbb {C}$ has to commute with both $M_X$ and $M_Y$ . Writing $P_i=P_i(1)$ and $c_i=c_i(1)$ for simplicity, this implies that

$$ \begin{align*}\begin{pmatrix} a c_1 + b c_3 & a c_2 + b c_4 \\ c c_1 + d c_3 & c c_2 + d c_4 \end{pmatrix} &=\begin{pmatrix} a c_1 + c c_2 & b c_1 + d c_2 \\ a c_3 + c c_4 & b c_3 + d c_4 \end{pmatrix},\\ \begin{pmatrix} a P_1 + b P_3 & a P_2 + b P_4 \\ c P_1 + d P_3 & c P_2 + d P_4 \end{pmatrix} &=\begin{pmatrix} a P_1 + c P_2 & b P_1 + d P_2 \\ a P_3 + c P_4 & b P_3 + d P_4 \end{pmatrix},\end{align*} $$

yielding

$$ \begin{align*}\begin{pmatrix} b c_3 & a c_2 + b c_4 \\ c c_1 + d c_3 & c c_2 \end{pmatrix} =\begin{pmatrix} c c_2 & b c_1 + d c_2 \\ a c_3 + c c_4 & b c_3 \end{pmatrix}\end{align*} $$

and

$$ \begin{align*}\begin{pmatrix} b P_3 & a P_2 + b P_4 \\ c P_1 + d P_3 & c P_2 \end{pmatrix} =\begin{pmatrix} c P_2 & b P_1 + d P_2 \\ a P_3 + c P_4 & b P_3 \end{pmatrix}.\end{align*} $$

Since $\deg (P_3)=1$ while $\deg (P_2) \leq 0$ , this last equality of matrices forces to have $b=0$ , yielding $c P_2=0$ and $a=d$ . The first equality becomes

$$ \begin{align*}\begin{pmatrix} 0 & a c_2 \\ c c_1 + a c_3 & c c_2 \end{pmatrix} =\begin{pmatrix} c c_2 & a c_2 \\ a c_3 + c c_4 & 0 \end{pmatrix},\end{align*} $$

yielding $c=0$ since $c_2 \neq 0$ . The matrix $M_\Delta $ is thus the diagonal matrix $t^{nm}\begin {pmatrix} a & 0 \\ 0 & a \end {pmatrix}$ .

It follows that the matrix $M_u$ has the same degrees of coefficients (in q) as $M_{u'}$ . One can thus with Proposition 3.1 recover $L(u')$ from $M_u$ , and inductively recover the Garside normal form of $u'$ (recall that the Garside length is also recovered): we consider the matrix $M_{L(u')}^{-1} M_{u}$ and iterate the procedure until reaching a diagonal matrix with the same coefficients on the diagonal (note that, since by convention $\deg (0)= -\infty $ , Situations $1$ – $4$ cannot produce such a matrix). This matrix is then $M_\Delta ^k$ , and one recovers the exponent by looking at the power of t of any diagonal coefficient (which has to be $t^{knm}$ ). This allows one to recover the (right) Garside normal form $x_1 x_2 \dots x_\ell \Delta ^k$ of u. The $x_i$ ’s are just the maximal powers of X or Y that are obtained during the procedure.

Note that the algorithm still works in the representation specialized at $q=t$ : in this case, the matrix $M_u$ does not have the same degrees of coefficients in q as $M_{u'}$ if $k>0$ , but all coefficients are just multiplied by the same power of q (which is $q^{knm}$ ), hence one can still apply Proposition 3.1 until reaching a diagonal matrix with the same coefficients on the diagonal.

4.1. Odd dihedral Burau representation

Let $m=2$ and $n=2\ell +1 \geq 3$ be odd. Then $G(n,m)$ is isomorphic to the Artin group of dihedral type $I_2(n)$ ; its classical Artin presentation is Presentation (2.2) in this case, which we shall write

$$ \begin{align*}\langle \sigma_1, \sigma_2 \ \vert \ \underbrace{\sigma_1 \sigma_2 \dots}_{n~\text{factors}} = \underbrace{\sigma_1 \sigma_2 \dots}_{n~\text{factors}} \rangle.\end{align*} $$

The generator Y corresponds to $\sigma _1 \sigma _2$ while X corresponds to $\underbrace {\sigma _1 \sigma _2 \dots }_{n~\text {factors}}$ . In the classical (reduced) Burau representation $\rho _1 : G(n,m) \longrightarrow \mathrm {GL}_2(\mathbb {R}[q^{\pm 1}])$ , one has

$$ \begin{align*}\rho_1(Y)= q^2 \begin{pmatrix} -1 & -q^{-1} \\ c q & c-1\end{pmatrix}, \ \rho_1(X) = q^n \begin{pmatrix} 0 & - \frac{\zeta_n^\ell - \zeta_n^{-\ell}}{\zeta_n - \zeta_n^{-1}} \\ c \frac{\zeta_n^{\ell+1} - \zeta_n^{-\ell-1}}{\zeta_n - \zeta_n^{-1}} & 0 \end{pmatrix},\end{align*} $$

where $c = 4 \cos ^2(\pi / n)$ (see, for instance, [Reference Lehrer and Xi17, (3.2), (3.3.1), and Proposition 4.1]). To apply Proposition 1.1 with the matrices $M_X:=\rho _1(X)$ and $M_Y:=\rho _1(Y)$ , one needs to calculate the powers of $M_Y$ , which is done in [Reference Lehrer and Xi17, Lemma 3.3]: for $k \geq 0$ , setting $[\zeta _n]_k:=\frac {\zeta _n^k - \zeta _n^{-k}}{\zeta _n - \zeta _n^{-1}}$ , we have

$$ \begin{align*}M_Y^k = q^{2k} \begin{pmatrix} -[\zeta_n]_k - [\zeta_n]_{k-1} & -q^{-1} [\zeta_n]_k \\ cq [\zeta_n]_k & [\zeta_n]_k + [\zeta_n]_{k+1} \end{pmatrix}.\end{align*} $$

Since $m=2$ , the only power of $M_X$ to consider is $M_X$ itself, while for $M_Y,$ it is $M_Y^k$ for all $1\leq k \leq n-1$ . Since n is odd, we have $c\neq 0$ , $\zeta _n^\ell \neq \zeta _n^{-\ell }$ , and $[\zeta _n]_k \neq 0$ for all $1\leq k \leq n-1$ , hence the conditions of Proposition 1.1 (with $q=t$ ) are fulfilled. This yields a new proof that the Burau representation of odd dihedral Artin groups is faithful. This will be reobtained in Section 4.3 below (see Example 4.10) by viewing the odd dihedral Burau representation as a particular case of a representation built for a larger family of groups inside the family of groups of the form $G(n,m)$ .

4.2. A faithful representation for arbitrary groups of the form $X^m=Y^n$

Let $n,m\geq 2$ be two integers. We shall construct a faithful representation $\rho _2 : G(n,m) \longrightarrow \mathrm {GL}_2(\mathbb {C}[t^{\pm 1}, q^{\pm 1}])$ . Note that if $\ell \geq 2$ is an integer, for all $k\geq 0,$ we have

$$ \begin{align*}\begin{pmatrix} \zeta_\ell & 1 \\ 0 & \zeta_{\ell}^{-1} \end{pmatrix}^k =\begin{pmatrix} \zeta_\ell^k & [\zeta_\ell]_k \\ 0 & \zeta_\ell^{-k} \end{pmatrix}, \ \begin{pmatrix} \zeta_\ell & 0 \\ 1 & \zeta_{\ell}^{-1} \end{pmatrix}^k =\begin{pmatrix} \zeta_\ell^k & 0 \\ [\zeta_\ell]_k & \zeta_\ell^{-k} \end{pmatrix}.\end{align*} $$

Let $\rho _2(X) := t^n \begin {pmatrix} \zeta _{2m} & 1 \\ 0 & \zeta _{2m}^{-1} \end {pmatrix}$ . For all $k \geq 0,$ we thus have

$$ \begin{align*}\rho_2(X)^k = t^{nk} \begin{pmatrix}\zeta_{2m}^k & [\zeta_{2m}]_k \\ 0 & \zeta_{2m}^{-k} \end{pmatrix}.\end{align*} $$

Let $\rho _2(Y):=t^m \begin {pmatrix} \zeta _{2n} & 0 \\ q & \zeta _{2n}^{-1} \end {pmatrix}$ . Noting that $\rho _2(Y) = P (\rho _2(Y))_{q=1} P^{-1}$ , where $P=\begin {pmatrix} 1 & 0 \\ 0 & q \end {pmatrix}$ , we get that for all $k\geq 0$ ,

$$ \begin{align*}\rho_2(Y)^k = t^{mk} \begin{pmatrix} \zeta_{2n}^k & 0 \\ q[\zeta_{2n}]_k & \zeta_{2n}^{-k} \end{pmatrix}.\end{align*} $$

Setting $M_X=\rho _2(X)$ and $M_Y=\rho _2(Y),$ we then have:

• • $M_Y^n = - t^{nm} I_2=M_X^m.$

• • For all $1 \leq k \leq m-1$ (resp. for all $1\leq k \leq n-1$ ), the matrix $M_X^k$ (resp. $M_Y^k$ ) satisfies assumption $2$ (resp. $3$ ) of Proposition 1.1 (observe that $[\zeta _{2\ell }]_k=0$ if and only if $\ell $ divides k).

By Proposition 1.1, we thus get that $\rho _2$ is faithful.

4.3. A Burau representation for some torus knot groups

In this section, we shall assume that $n, m\geq 2$ are coprime and that m is not divisible by $3$ .

Let $a, b\in \mathbb {Z}$ be such that $an-bm=1$ . Consider the matrices

$$ \begin{align*}M_X= (-1)^b i^{n+1} t^n \begin{pmatrix} U^{\prime} & V \\ V & -U \end{pmatrix}, \ M_Y= (-1)^a i^m t^m \begin{pmatrix} A_1 [\boldsymbol{\mu}^b]_1 + \mu_1 & q^{-1} V [\boldsymbol{\lambda}]_a [\boldsymbol{\mu}^b]_1 \\ - \frac{A_1 A_2}{ q^{-1} V [\boldsymbol{\lambda}]_a} [\boldsymbol{\mu}^b]_1 & \mu_1 - A_2 [\boldsymbol{\mu}^b]_1 \end{pmatrix},\end{align*} $$

where

• • $U= \zeta _{4m}^{3m-2} + \zeta _{4m}^{3m+2} = \zeta _{4m}^{3m-2} - (\zeta _{4m}^{3m-2})^{-1}$ .

• • $U^{\prime } = U + \lambda _1 + \lambda _2$ , where $\lambda _1= \zeta _{4m}^{3m-6}$ and $\lambda _2= \zeta _{4m}^{3m+6}$ . Note that $\lambda _2 = - \lambda _1^{-1}$ .

• • $V= -1 - \zeta _{m} - \zeta _{m}^{-1}$ if $m> 2$ , and $-1$ otherwise. Since $m\neq 3,$ we have $V \neq 0$ . Note that $U^{\prime } U + V^2 = 1$ .

• • $\mu _1= \zeta _n$ and $\mu _2=\zeta _n^{-1}$ .

• • For $k\in \mathbb {Z}$ , $[\boldsymbol {\lambda }]_k=\frac {\lambda _2^k - \lambda _1^k}{\lambda _2 - \lambda _1}$ , and $[\boldsymbol {\mu }^b]_k= \frac {\mu _2^k - \mu _1^k}{\mu _2^b - \mu _1^b}$ . Note that $[\boldsymbol {\lambda }]_0=0$ , $[\boldsymbol {\lambda }]_1=1=[\boldsymbol {\lambda }]_{-1}$ , and $[\boldsymbol {\lambda }]_k=0$ if and only if m divides $3k$ , if and only if m divides k since we assumed that $3$ does not divide m. In particular, $[\boldsymbol {\lambda }]_a\neq 0$ since $(a,m)=1$ . Similarly, we have $[\boldsymbol {\mu }^b]_k=0$ if and only if n divides $2k$ .

• • $A_k= q^{-1} U^{\prime } [\boldsymbol {\lambda }]_a + q^{-1} [\boldsymbol {\lambda }]_{a-1} - \mu _k^b$ (where $k\in \{1,2\}$ ).

Proposition 4.3. Let $k\in \mathbb {Z}$ . We have

$$ \begin{align*} M_X^k &= (-1)^{bk} i^{nk+k} t^{nk}\begin{pmatrix} U^{\prime} [\boldsymbol{\lambda}]_k + [\boldsymbol{\lambda}]_{k-1} & V [\boldsymbol{\lambda}]_k \\ V [\boldsymbol{\lambda}]_k & -U^{\prime} [\boldsymbol{\lambda}]_k + [\boldsymbol{\lambda}]_{k+1}\end{pmatrix},\\ M_Y^k&= (-1)^{ak} i^{mk} t^{mk} \begin{pmatrix} A_1 [\boldsymbol{\mu}^b]_{k} + \mu_1^{k} & q^{-1} V [\boldsymbol{\lambda}]_a [\boldsymbol{\mu}^b]_{k} \\ - \frac{A_1 A_2}{ q^{-1} V [\boldsymbol{\lambda}]_a} [\boldsymbol{\mu}^b]_{k} & \mu_1^{k} - A_2 [\boldsymbol{\mu}^b]_{k} \end{pmatrix}. \end{align*} $$

Proof. We have $[\boldsymbol \lambda ]_m=0$ , $[\boldsymbol \lambda ]_{m-1}=(-1)^{m+1} i^m = [\boldsymbol \lambda ]_{m+1}$ , and $[\boldsymbol \mu ^b]_n=0$ , thus using Proposition 4.3, we get

$$ \begin{align*} M_X^m &= (-1)^{bm} i^{nm+m} t^{nm} \begin{pmatrix} (-1)^{m+1} i^m & 0 \\ 0 & (-1)^{m+1} i^m \end{pmatrix} = (-1)^{bm+1} i^{nm+m} (-i)^m t^{nm} I_2 \\ &= (-1)^{an} i^{nm} t^{nm} I_2 = M_Y^n.\\[-34pt] \end{align*} $$

We now apply Proposition 1.1 to the above-constructed representation, under the further assumption that n is odd.

For $k\in \mathbb {Z,}$ we have $[\boldsymbol {\lambda }]_{-k}=(-1)^{k+1} [\boldsymbol {\lambda }]_k$ . Using Proposition 4.3, we thus have

$$ \begin{align*} M_X^{-a}&= (-1)^{ab} i^{-an-a} t^{-an} \begin{pmatrix} U^{\prime} (-1)^{a+1} [\boldsymbol{\lambda}]_a +(-1)^a [\boldsymbol{\lambda}]_{a+1} & V (-1)^{a+1} [\boldsymbol{\lambda}]_a \\ V (-1)^{a+1} [\boldsymbol{\lambda}]_a & -U^{\prime} (-1)^{a+1} [\boldsymbol{\lambda}]_a +(-1)^a [\boldsymbol{\lambda}]_{a-1}\end{pmatrix},\end{align*} $$

and setting $A=q^{-1} U^{\prime } [\boldsymbol {\lambda }]_a + q^{-1} [\boldsymbol {\lambda }]_{a-1}$ , we have

$$ \begin{align*} M_Y^b &=(-1)^{ab} i^{bm} t^{bm} \begin{pmatrix} A & q^{-1} V [\boldsymbol{\lambda}]_a \\ - q \frac{( A - \zeta_n^b)(A - \zeta_n^{-b})}{V [\boldsymbol\lambda]_a} & \zeta_n^b + \zeta_n^{-b} - A \end{pmatrix}. \end{align*} $$

We calculate $M_Y^b M_X^{-a}$ using that $[\boldsymbol \lambda ]_a^2 - [\boldsymbol \lambda ]_{a-1} [\boldsymbol \lambda ]_{a+1}=(-1)^{a+1}$ , $[\boldsymbol \lambda ]_{a} (\lambda _1 +\lambda _2) + [\boldsymbol \lambda ]_{a-1} - [\boldsymbol \lambda ]_{a+1}=0$ and $V^2 = 1 - U^{\prime 2} +(\lambda _1+\lambda _2)U^{\prime }$ . We write

$$ \begin{align*}M_Y^b M_X^{-a} = i^{-a-1} t^{-1}\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix},\end{align*} $$

and explicitly determine $a_{11}, a_{12}$ , and $a_{22}$ , which will be enough for our purposes. We have

$$ \begin{align*} a_{11} &= A (U^{\prime} (-1)^{a+1} [\boldsymbol{\lambda}]_a + (-1)^a [\boldsymbol{\lambda}]_{a+1}) + q^{-1} V^2 (-1)^{a+1}[\boldsymbol{\lambda}]_a^2 \\ &=q^{-1} U^{\prime2} (-1)^{a+1} [\boldsymbol{\lambda}]_a^2 + q^{-1} U^{\prime} (-1)^{a+1} [\boldsymbol{\lambda}]_a [\boldsymbol{\lambda}]_{a-1} + q^{-1} U^{\prime} (-1)^a[\boldsymbol{\lambda}]_a [\boldsymbol{\lambda}]_{a+1} + q^{-1} (-1)^a [\boldsymbol{\lambda}]_{a-1} [\boldsymbol{\lambda}]_{a+1} \\ &~~~+q^{-1} (1 - U^{\prime2} + (\lambda_1 + \lambda_2) U^{\prime}) (-1)^{a+1}[\boldsymbol{\lambda}]_a^2 \\ &= q^{-1} U^{\prime} (-1)^{a+1} [\boldsymbol{\lambda}]_{a} ( \underbrace{[\boldsymbol\lambda]_{a} (\lambda_1 +\lambda_2) + [\boldsymbol\lambda]_{a-1} - [\boldsymbol\lambda]_{a+1}}_{=0}) + q^{-1}(-1)^{a+1} (\underbrace{[\boldsymbol\lambda]_{a}^2 - [\boldsymbol\lambda]_{a-1}[\boldsymbol\lambda]_{a+1}}_{=(-1)^{a+1}}) = q^{-1},\\ a_{12} &= A V (-1)^{a+1} [\boldsymbol\lambda]_{a} + q^{-1} V [\boldsymbol\lambda]_{a} (\underbrace{- U^{\prime} (-1)^{a+1} [\boldsymbol\lambda]_{a} + (-1)^a [\boldsymbol\lambda]_{a-1}}_{= (-1)^a q A})=0, \\ a_{22} &= - q (A - \zeta_n^b)(A - \zeta_n^{-b})(-1)^{a+1} + (\zeta_n^b + \zeta_n^{-b}-A)(\underbrace{- U^{\prime}(-1)^{a+1} [\boldsymbol\lambda]_{a}+ (-1)^a [\boldsymbol\lambda]_{a-1}}_{{= (-1)^a q A}}) \\ &= (-1)^a q. \end{align*} $$

We thus get that $M_Y^b M_X^{-a}$ is of the form

(4.1) $$ \begin{align} i^{-a-1} t^{-1}\begin{pmatrix} q^{-1} & 0 \\ \star & (-1)^a q \end{pmatrix}. \end{align} $$

This yields the following lemma.

Recall the 2-toric reflection group from Definition 2.4.

We now study some particular cases of the representation that we just constructed.

Example 4.10. Let $n \geq 3$ be odd and assume that $m=2$ . In this case, $G(n,2)$ is isomorphic to the Artin group of dihedral type $I_2(n)$ . We have $U=U^{\prime }=0$ and $V=-1$ . Denoting $n=2n' + 1,$ we have $a=1$ and $b=n'$ . This yields the matrices

$$ \begin{align*} M_X = t^n \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \ M_Y &= t^2 \begin{pmatrix} - \zeta_n^{n'} \frac{\zeta_n^{-1}- \zeta_n}{ \zeta_n^{-n'} - \zeta_n^{n'}} + \zeta_n & -q^{-1} \frac{\zeta_n^{-1}- \zeta_n}{\zeta_n^{-n'} - \zeta_n^{n'}} \\ q \frac{\zeta_n^{-1}- \zeta_n}{\zeta_n^{-n'} - \zeta_n^{n'}} & \zeta_n + \zeta_n^{-n'}\frac{\zeta_n^{-1}- \zeta_n}{\zeta_n^{-n'} - \zeta_n^{n'}} \end{pmatrix}\\ &= t^2 \begin{pmatrix}\frac{\zeta_n^{-(n'-1)} - \zeta_n^{n'-1}}{\zeta_n^{-n'} - \zeta_n^{n'}} & -q^{-1} \frac{\zeta_n^{-1}- \zeta_n}{\zeta_n^{-n'} - \zeta_n^{n'}} \\q \frac{\zeta_n^{-1}- \zeta_n}{\zeta_n^{-n'} - \zeta_n^{n'}} & \frac{\zeta_n^{-(n'+1)}- \zeta_n^{n'+1}}{\zeta_n^{-n'} - \zeta_n^{n'}} \end{pmatrix}.\end{align*} $$

Conjugating these two matrices by the matrix $P=\begin {pmatrix} 1 & 0 \\ 0 & \frac {\zeta _n^{-1}- \zeta _n}{\zeta _n^{n'} - \zeta _n^{-n'}} \end {pmatrix}$ yields after simplification the two matrices

$$ \begin{align*}M_X'= t^n \begin{pmatrix} 0 & \frac{\zeta_n^{n'}- \zeta_n^{-n'}}{\zeta_n^{-1} - \zeta_n} \\ \frac{\zeta_n^{-1} - \zeta_n}{ \zeta_n^{n'} - \zeta_n^{-n'}} & 0 \end{pmatrix}, \ M_Y' = t^2 \begin{pmatrix} \zeta_n + 1 + \zeta_n^{-1} & q^{-1} \\ -q (\zeta_n + 2 + \zeta_n^{-1}) & -1 \end{pmatrix}.\end{align*} $$

Recall that, in the notation of Section 4.1, the isomorphism between $G(n,2)$ and the dihedral Artin group of type $I_2(n)$ sends Y to $\sigma _1 \sigma _2$ and X to $\underbrace {\sigma _1 \sigma _2 \dots }_{n~\text {factors}}$ . Replacing q with $q^{-1}$ and specializing t to q yields the matrices for the lift of $w_0$ and $st$ of the reduced Burau representation of $I_2(n)$ , which can also be found in [Reference Lehrer and Xi17, (3.2), (3.3.1), and Proposition 4.1].Footnote ² We thus get that, when the group is $G(n,2)$ , the representation $\rho _3$ is isomorphic to the reduced Burau representation of the Artin group of type $I_2(n)$ .

Example 4.11 (Burau representation for $G_{12}$ )

The previous example shows that the representation $\rho _3$ is not new when $m=2$ . The “smallest” case for which our representation is new is the case $n=3, m=4$ . In this case, the group $G(3,4)$ is isomorphic to the complex braid group of the exceptional complex reflection group $G_{12}$ . We have $U=- \sqrt {-2}$ , $U^{\prime }=0$ , $V=-1$ , $a=b=-1$ , $[\boldsymbol {\lambda }]_{-2}= U$ , $[\boldsymbol {\lambda }]_{-1}=1$ , and $[\boldsymbol {\mu }^{-1}]_{1}=-1$ . This yields the matrices

$$ \begin{align*}M_X= t^3 \begin{pmatrix} 0 & 1 \\ 1 & U \end{pmatrix}, M_Y = t^4 \begin{pmatrix} 1+q^{-1} U & -q^{-1} \\ - 2q^{-1} + q + U & -q^{-1}U\end{pmatrix}.\end{align*} $$

Specializing q and t to $1$ yields matrices generating the complex reflection group $G_{12}$ . One way to check this is as follows: the group $G_{12}$ has order $48$ , with a reflection presentation given by Presentation (2.1) with $n=3$ , $m=4$ , and the additional relations $x_i^2=1$ for all i. The $x_i$ ’s are reflections and there is only one conjugacy class of reflections. Using Remark 2.5, we get that an alternative presentation of $G_{12}$ is given by $X^4 = Y^3$ and $(Y^{-1} X)^2=1$ , with $Y^{-1} X$ sent to some $x_i$ . Since $M_X$ and $M_Y$ satisfy the first relation (hence so do their specializations) and $M_Y^{-1} M_X = M_Y^b M_X^{-a}$ has the form (4.1) hence specializes to a matrix of order $2$ , we get that $\langle (M_X)_{t=q=1}, (M_Y)_{t=q=1} \rangle $ is a quotient of $G_{12}$ . But one checks using a computer that this group has order $48$ , we thus obtain an isomorphism and the conjugacy class of the $x_i$ ’s is the same as the conjugacy class of $M_Y^{-1} M_X$ which is indeed a complex reflection.

Example 4.12 (Burau representation for $G_{22}$ )

Consider the case $n=3, m=5$ . In this case, $G(3,5)$ is isomorphic to the complex braid group of the exceptional complex reflection group $G_{22}$ . We have $U= \zeta _{20}^{13}+\zeta _{20}^{17}$ , $U^{\prime }=-i$ , $V=\zeta _5^2+\zeta _5^3$ , $a=2, b=1$ , $[\boldsymbol {\lambda }]_{2}= \zeta _{20}^9 + \zeta _{20}$ , and $[\boldsymbol {\lambda }]_1={[\boldsymbol {\mu }^1]}_1=1$ . Since a is even, following Remark 4.9, we replace q by $iq$ in the defining matrices, yielding the matrices

$$ \begin{align*}(M_X)_{q \mapsto iq} &= -t^3 \begin{pmatrix} -i & \zeta_5^2 + \zeta_5^3 \\ \zeta_5^2 + \zeta_5^3 & -(\zeta_{20}^{13} + \zeta_{20}^{17})\end{pmatrix}, \\ (M_Y)_{q \mapsto iq} &= i t^5 \begin{pmatrix} - q^{-1} (\zeta_{20}^9 + \zeta_{20}) - i q^{-1} & - iq^{-1} (\zeta_5^2+\zeta_5^3)(\zeta_{20}^9 + \zeta_{20}) \\ \frac{(- q^{-1} (\zeta_{20}^9 + \zeta_{20}) - i q^{-1} - \zeta_3)(- q^{-1} (\zeta_{20}^9 + \zeta_{20}) - i q^{-1} - \zeta_3^{-1})}{iq^{-1}(\zeta_5^2+\zeta_5^3)(\zeta_{20}^9 + \zeta_{20})} &-1 + q^{-1} (\zeta_{20}^9 + \zeta_{20}) + i q^{-1} \end{pmatrix}. \end{align*} $$

As for Example 4.11, specializing q and t to $1$ yields matrices generating the complex reflection group $G_{22}$ : this can be checked in the same way as for $G_{12}$ .

4.4. Unitarizability

If a is odd, then the representation constructed in Section 4.3 is unitarizable: setting $J= \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix}$ , it is straightforward to check that ${ }^t\!{\overline {M_X}} J M_X = J$ and ${ }^t\!{\overline {M_Y}} J M_Y=J$ , where for $P\in \mathbb {C}[t^{\pm 1}, q^{\pm 1}]$ , we set $\overline {P(q,t)}= \overline {P}(-q, t^{-1})$ . Note that in this case, since $\overline {\lambda _j} = - \lambda _k$ (where $\{j,k\} = \{1,2\}$ ), $\overline {U^{\prime }} = - U^{\prime }$ , $\overline { [\boldsymbol {\lambda }]_a} = [\boldsymbol {\lambda }]_a$ , and $\overline { [\boldsymbol {\lambda }]_{a-1}} = - [\boldsymbol {\lambda }]_{a-1} $ , we have $\overline {A_j}= A_k$ (where $\{j,k\}=\{1,2\}$ ).

If a is even, then we again replace q by $iq$ in the matrices from Section 4.3. For $P\in \mathbb {C}[t^{\pm 1}, q^{\pm 1}],$ we again set $\overline {P(q,t)}= \overline {P}(-q, t^{-1})$ . Since $\overline {\lambda _j} = - \lambda _k$ (where $\{j,k\} = \{1,2\}$ ), we still have $\overline {U^{\prime }} = - U^{\prime }$ , but since a is even, we get $\overline { [\boldsymbol {\lambda }]_a} = -[\boldsymbol {\lambda }]_a$ and $\overline { [\boldsymbol {\lambda }]_{a-1}} = [\boldsymbol {\lambda }]_{a-1} $ . This again yields $\overline {(A_j)_{q \mapsto iq}}= (A_k)_{q \mapsto iq}$ (where $\{j,k\}=\{1,2\}$ ). Then the same matrix J as above yields unitarizability.