Representations of the necklace braid group ${{\mathcal {N}}{\mathcal {B}}}_n$ of dimension 4 ($n=2,3,4$)

Abstract

We consider the irreducible representations each of dimension 2 of the necklace braid group ${\mathcal {N}}{\mathcal {B}}_n$ ($n=2,3,4$). We then consider the tensor product of the representations of ${\mathcal {N}}{\mathcal {B}}_n$ ($n=2,3,4$) and determine necessary and sufficient condition under which the constructed representations are irreducible. Finally, we determine conditions under which the irreducible representations of ${\mathcal {N}}{\mathcal {B}}_n$ ($n=2,3,4$) of degree 2 are unitary relative to a hermitian positive definite matrix.

1 Introduction

Bullivant et al. [4] studied representations of the necklace braid group ${\mathcal {N}}{\mathcal {B}}_n$, especially those obtained as extensions of representations of the braid group $B_n$ and the loop braid group $LB_n$ (see [3, 5]). They showed that any irreducible $B_n$ representation extends to ${\mathcal {N}}{\mathcal {B}}_n$ in a standard way. Moreover, they proved that any local representation of $B_n$, coming from a braided vector space, can be extended to ${\mathcal {N}}{\mathcal {B}}_n$.

A link $\mathcal {L}_n=K_0\cup K_1\cup \cdots \cup K_n$ is called a necklace if:

• $K_0$ is an Euclidean circle of center O and of radius 1

• each $K_i$ is an Euclidean circle whose center $O_i$ belongs to $K_0$ and radius $r_i$ such that $0<r_i<\frac{1}{2}$ for $0<i\leqslant n$

• the plane of each $K_i$ is the one containing the line $(OO_i)$ and perpendicular to the plane of $K_0$.

• If $O_i=O_j$ then $r_i\ne r_j$

Fig. 1

Necklace $\mathcal {L}_n$

The motion group ${\mathcal {N}}{\mathcal {B}}_n$, the necklace braid group, as described in [1] is identified with the fundamental group of the configuration space $\mathcal {L}_n$. The group ${\mathcal {N}}{\mathcal {B}}_n$ is generated by the elements $\sigma _1,\ldots ,\sigma _n$ and $\tau$ where $\sigma _i$ is the motion, up to homotopy, of passing the i th circle through the $(i+1)$th circle, while $\tau$ corresponds to shifting each circle one position in the counterclockwise direction.

In Sect. 2, we consider the irreducible representations defined on the necklace braid group ${{\mathcal {N}}{\mathcal {B}}}_n$ ($n=2,3,4$) each of dimension 2. In Sect. 3, we construct the tensor product of representations on ${{\mathcal {N}}{\mathcal {B}}}_n$. Then, we discuss the irreducibility of the tensor product of the representations of ${\mathcal {N}}{\mathcal {B}}_n$ ($n=2,3,4$). Theorem 3.5 gives necessary and sufficient condition for the irreducibility of tensor product of the representations of ${\mathcal {N}}{\mathcal {B}}_4$. Theorem 3.17 provides necessary and sufficient conditions under which the tensor product of the representations of ${\mathcal {N}}{\mathcal {B}}_3$ is irreducible. Theorem 3.18 gives necessary and sufficient condition for the irreducibility of the tensor product of the representations of ${\mathcal {N}}{\mathcal {B}}_2$. In Sect. 4, we prove that the irreducible representations of dimension 2 of ${\mathcal {N}}{\mathcal {B}}_n$ are unitary relative to hermitian positive definite matrix in the case $n=3,4$ (see Propositions 4.4, 4.5 and Proposition 4.6).

Definition 1.1

[1] The necklace braid group ${\mathcal {N}}{\mathcal {B}}_n$ is identified with the fundamental group of the configuration space of $\mathcal {L}_n$.

The following theorem gives a presentation of the necklace braid group ${\mathcal {N}}{\mathcal {B}}_n$ by generators and relations.

Theorem 1.2

[1] The necklace braid group ${{\mathcal {N}}{\mathcal {B}}}_n$ has a presentation by generators $\sigma _1, \ldots ,\sigma _n,\tau$ and relations as follows:

• (B1) $\sigma _i \sigma _{i+1} \sigma _i = \sigma _{i+1} \sigma _i \sigma _{i+1}$ for $1\leqslant i\leqslant n$;

• (B2) $\sigma _i \sigma _j = \sigma _j \sigma _i \text { for } |i-j|\ne 1$;

• (N1) $\tau \sigma _i\tau ^{-1}=\sigma _{i+1}$ for $1\leqslant i \leqslant n~(\text {mod } n)$;

• (N2) $\tau ^{2n}=1$

Here, indices are taken modulo n, with $\sigma _{n+1}=\sigma _1$ and $\sigma _0=\sigma _n$. The relations (B1) and (B2) are those for the braid group $B_n$ (see [2]).

2 Irreducible representations of dimension 2 of ${\mathcal {N}}{\mathcal {B}}_n$ ($n=2,3,4$)

Consider the necklace braid group ${\mathcal {N}}{\mathcal {B}}_n$ for $n=2,3,4$. We have the following proposition (see [4]).

Proposition 2.1

Any irreducible representation of dimension 2 of ${\mathcal {N}}{\mathcal {B}}_n (n=2,3,4)$ is isomorphic to the representation $\rho =\rho (T,t,a,c,d)$ that is defined by:

$$\begin{aligned} \rho (\tau )=\left( \begin{matrix} Tt&{}0\\ 0&{}t \end{matrix}\right) \text { and } \rho (\sigma _i)=\left( \begin{matrix} a&{}T^{i-1}\\ cT^{1-i}&{}d \end{matrix}\right) , \end{aligned}$$

with the following conditions:

n	T	Conditions
2	$T=-1$	$c=a^2-ad+d^2$, $c\ne 0$ and $a\ne d$
3	$T=e^{\pm i2\pi /3}$	$a=\omega d$, $c\ne 0$, $c\ne \omega d^2$ and $\omega =e^{\pm i\pi /3}$
4	$T=-1$	$c=a^2-ad+d^2$, $c\ne 0$ and $a\ne d$
	$T=\pm i$	$c=-d^2$, $a=d\ne 0$

Here, t is any $(2n)^{th}$ root of unity.

Proof

Let $(\rho ,V)$ be an irreducible representation of dimension 2 of ${\mathcal {N}}{\mathcal {B}}_n$ (for $n=2,3,4$). From the fact that $\tau$ has order 2n, we may choose a basis for V such that $\rho (\tau ) = \left( \begin{matrix} t_1&{}0\\ 0&{}t_2 \end{matrix}\right)$, where $t_1$ and $t_2$ are (2n)th roots of unity. Since $\rho$ is irreducible representation, it follows that $t_1\ne t_2$. Similarly we have that $\rho (\sigma _1)$ is neither upper nor lower triangular, because (1, 0) or (0, 1) would generate an invariant subspace. Due to rescaling, we may assume that $\rho (\sigma _1) = \left( \begin{matrix} a&{}1\\ c&{}d \end{matrix}\right)$. Since $\rho (\sigma _1)$ is neither diagonal nor triangular, we have that $c\ne 0$. To require invertibility, we assume that $ad\ne c$. Using $\rho (\sigma _1)=\left( \begin{matrix} a&{}1\\ c&{}d \end{matrix}\right)$ and the condition (N1): $\tau \sigma _i\tau ^{-1}=\sigma _{i+1}$, we get

$$\begin{aligned} \rho (\sigma _i)=\left( \begin{matrix} a&{}T^{i-1}\\ cT^{1-i}&{}d \end{matrix}\right) , \end{aligned}$$

where $T=t_1t_2^{-1}$. Note that $T\ne 1$ because $t_1\ne t_2$. Set $t_2=t$ then $t_1=Tt$ and $\rho (\tau )$ has the form $\left( \begin{matrix} Tt&{}0\\ 0&{}t \end{matrix}\right)$. Since $\sigma _{n+1}=\sigma _1$, we have $T^{n}=T^{-n}=1$. Thus, T is a primitive nth root of unity.

Now, we check the conditions (B1): $\sigma _i\sigma _{i+1}\sigma _i=\sigma _{i+1}\sigma _i\sigma _{i+1}$. By Lemma 1.2 in [4], it is sufficient to check (B1) for $i=1$ (i.e. $\sigma _1\sigma _2\sigma _1=\sigma _2\sigma _1\sigma _2$ ).

By direct calculations, we get

$$\begin{aligned} \rho (\sigma _1)\rho (\sigma _2)\rho (\sigma _1)=\rho (\sigma _2)\rho (\sigma _1) \rho (\sigma _2)\Leftrightarrow (a^2-ad+d^2)T+c(1+T+T^2)=0. \end{aligned}$$

For $n=2$, we have that $T=-1$ and $c=a^2-ad+d^2$. Under the assumption that $ad\ne c$, we have $a\ne d$.

For $n=3$, we have that $T=e^{\pm 2\pi i/3}$. Then, $a=\omega d$ where $\omega =e^{\pm \pi i/3}$. Under the assumption that $ad\ne c$, we have $c\ne \omega d^2$.

For $n=4$, we have that $T=-1, \pm i$.

• If $T=-1$, then $c=a^2-ad+d^2$ with $a\ne d$.

• If $T=\pm i$, then $c=-a^2+ad-d^2$.

It remains to verify the condition (B2): $\sigma _i\sigma _j=\sigma _j\sigma _i$ for $|i-j|>1$ when $n=4$. By Lemma 1.2 in [4] we just check (B2) for $i=1$ and $j=3$.

If $T=-1$, then it is clear that $\rho (\sigma _1)\rho (\sigma _3)=\rho (\sigma _3)\rho (\sigma _1)$.

If $T=\pm i$, then

$$\begin{aligned} \rho (\sigma _1)\rho (\sigma _3)=\rho (\sigma _3)\rho (\sigma _1)&\Longleftrightarrow \begin{pmatrix} a&{}1\\ c&{}d \end{pmatrix} \begin{pmatrix} a&{}-1\\ -c&{}d \end{pmatrix} = \begin{pmatrix} a&{}-1\\ -c&{}d \end{pmatrix} \begin{pmatrix} a&{}1\\ c&{}d \end{pmatrix}\\&\Longleftrightarrow \begin{pmatrix} a^2-c&{}-a+d\\ ca-dc&{}-c+d^2 \end{pmatrix} = \begin{pmatrix} a^2-c&{}a-d\\ -ca+dc&{}-c+d^2 \end{pmatrix}\\&\Longleftrightarrow a=d \end{aligned}$$

It follows that $c=a^2+ad-d^2=-d^2$. $\square$

3 Representations of dimension 4 of ${\mathcal {N}}{\mathcal {B}}_n$ ($n=2,3,4$)

Consider two irreducible representations $\rho _1=\rho (T_1,t_1,a_1,c_1,d_1)$ and $\rho _2=\rho (T_2,t_2,a_2,c_2,d_2)$ of ${\mathcal {N}}{\mathcal {B}}_n$ ($n=2,3,4$) each of dimension 2.

$$\begin{aligned} \rho _1(\tau )=\left( \begin{matrix} T_1t_1&{}0\\ 0&{}t_1 \end{matrix}\right) \text { and } \rho _1(\sigma _i)=\left( \begin{matrix} a_1&{}T_1^{i-1}\\ c_1T_1^{1-i}&{}d_1 \end{matrix}\right) ; \\ \rho _2(\tau )=\left( \begin{matrix} T_2t_2&{}0\\ 0&{}t_2 \end{matrix}\right) \text { and } \rho _2(\sigma _i)=\left( \begin{matrix} a_2&{}T_2^{i-1}\\ c_2T_1^{1-i}&{}d_2 \end{matrix}\right) . \end{aligned}$$

Definition 3.1

Consider the tensor product $\rho _1\otimes \rho _2$ given by $\rho _1\otimes \rho _2(\alpha )=\rho _1(\alpha )\otimes \rho _2(\alpha ),$ where $\alpha$ is a generator of ${\mathcal {N}}{\mathcal {B}}_n$ ($n=2,3,4$). We get the following matrices for the generators $\tau$ and $\sigma _1$.

$$\begin{aligned} \rho (\tau )=t_1t_2\left( \begin{matrix} T_1T_2&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad T_1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad T_2&{}\quad 0\\ 0&{}0&{}0&{}1 \end{matrix}\right) ~~\text { and }~~ \rho (\sigma _1)=\left( \begin{matrix} a_1a_2&{}\quad a_1&{}\quad a_2&{}\quad 1\\ a_1c_2&{}\quad a_1d_2&{}\quad c_2&{}\quad d_2\\ c_1a_2&{}\quad c_1&{}\quad d_1a_2&{}\quad d_1\\ c_1c_2&{}\quad c_1d_2&{}\quad d_1c_2&{}\quad d_1d_2 \end{matrix}\right) . \end{aligned}$$

3.1 Representations of ${\mathcal {N}}{\mathcal {B}}_4$

In this section, we study the irreducibility of the representation $\rho$ of the necklace braid group ${\mathcal {N}}{\mathcal {B}}_4$. Actually, we have four cases: (3.1.1) $T_1=T_2=-1$, (3.1.2) $T_1=T_2=\pm i$, (3.1.3) $T_1=-T_2=\pm i$ and (3.1.4) $T_1=-1, T_2=\pm i$.

In what follows, suppose that $a_1a_2d_1d_2\ne 0$.

3.1.1 Case $T_1=T_2=-1$

Direct computations show that the representation $\rho$ is given by

$$\begin{aligned} \rho (\tau )= t_1t_2\left( \begin{matrix} 1&{}0&{}0&{}0\\ 0&{}-1&{}0&{}0\\ 0&{}0&{}-1&{}0\\ 0&{}0&{}0&{}1 \end{matrix}\right) , ~ \rho (\sigma _1)=\rho (\sigma _3)=\left( \begin{matrix} a_1a_2&{}a_1&{}a_2&{}1\\ a_1c_2&{}a_1d_2&{}c_2&{}d_2\\ c_1a_2&{}c_1&{}d_1a_2&{}d_1\\ c_1c_2&{}c_1d_2&{}d_1c_2&{}d_1d_2 \end{matrix}\right) \end{aligned}$$

and

$$\begin{aligned} \rho (\sigma _2)=\rho (\sigma _4)=\left( \begin{matrix} a_1a_2&{}-a_1&{}-a_2&{}1\\ -a_1c_2&{}a_1d_2&{}c_2&{}-d_2\\ -c_1a_2&{}c_1&{}d_1a_2&{}-d_1\\ c_1c_2&{}-c_1d_2&{}-d_1c_2&{}d_1d_2 \end{matrix}\right) . \end{aligned}$$

The eigenvalues of $\rho (\tau )$ are $\lambda _1=t_1t_2$ and $\lambda _2=-t_1t_2$. Both are of multiplicities 2. The corresponding eigenvectors are $\alpha _1e_1+\alpha _4e_4=(\alpha _1,0,0,\alpha _4)$ for $\lambda _1=t_1t_2$ and $\alpha _2e_2+\alpha _3e_3=(0,\alpha _2,\alpha _3,0)$ for $\lambda _2=-t_1t_2$, where $\{\alpha _1,\alpha _2,\alpha _3,\alpha _4\}\subset {\mathbb {C}}$.

We now determine conditions under which the representation $\rho$ is irreducible.

Proposition 3.2

The representation $\rho :{{\mathcal {N}}{\mathcal {B}}}_4\rightarrow GL(4,{\mathbb {C}})$ has no non-trivial proper invariant subspaces of dimension 1 if and only if

$$\begin{aligned} a_1a_2\ne d_1d_2~\text { and }~a_1d_2\ne a_2d_1. \end{aligned}$$

Proof

The subspaces of dimension 1 that are invariant under $\rho (\tau )$ are those spanned by one of the following vectors: $e_1,~e_2,~e_3,~e_4,~e_1+ye_4~\text { and }e_2+x e_3~~\text { for }x\ne 0$ and $y\ne 0$.

$\rho (\sigma _1)(e_1)=\left( \begin{matrix} a_1a_2\\ a_1c_2\\ c_1a_2\\ c_1c_2 \end{matrix}\right) \not \in \langle e_1\rangle$ because, $c_1c_2\ne 0$. So, $\langle e_1\rangle$ is not invariant.

$\rho (\sigma _1)(e_2)=\left( \begin{matrix} a_1\\ a_1d_2\\ c_1\\ c_1d_2 \end{matrix}\right) \not \in \langle e_2\rangle$ because, $c_1\ne 0$. So, $\langle e_2\rangle$ is not invariant.

$\rho (\sigma _1)(e_3)=\left( \begin{matrix} a_2\\ c_2\\ d_1a_2\\ d_1c_2 \end{matrix}\right) \not \in \langle e_3\rangle$ because, $c_2\ne 0$. Thus, $\langle e_3\rangle$ is not invariant.

$\rho (\sigma _1)(e_4)=\left( \begin{matrix} 1\\ d_2\\ d_1\\ d_1d_2 \end{matrix}\right) \not \in \langle e_4\rangle$. So, $\langle e_4\rangle$ is not invariant.

Now consider the subspaces of the form $\langle e_1+ye_4\rangle$ with $y\ne 0$

Since $\rho (\sigma _i)(e_1+ye_4)=\left( \begin{matrix} a_1a_2+y\\ \pm a_1c_2\pm d_2y\\ \pm c_1a_2\pm d_1y\\ c_1c_2+d_1d_2y \end{matrix}\right)$, it follows that $\langle e_1+ye_4\rangle$ is invariant under $\rho (\sigma _i)$ for $1\leqslant i\leqslant n.$ It follows that $\left( \begin{matrix} a_1a_2+y\\ \pm a_1c_2\pm d_2y\\ \pm c_1a_2\pm d_1y\\ c_1c_2+d_1d_2y \end{matrix}\right) = \left( \begin{matrix} \alpha \\ 0\\ 0\\ \alpha y \end{matrix}\right)$ for some $\alpha \in {\mathbb {C}}\setminus \{0\}$.

This is equivalent to the system:

$$\begin{aligned}&\left\{ \begin{array}{l} (a_1a_2+y)y= c_1c_2+d_1d_2y \\ a_1c_2+d_2y=0\\ c_1a_2+d_1y=0 \end{array}\right. \\ \Longleftrightarrow&\left\{ \begin{array}{l} y^2+(a_1a_2-d_1d_2)y-c_1c_2=0\\ y=-\dfrac{a_1c_2}{d_2}\\ y=-\dfrac{c_1a_2}{d_1} \end{array}\right. \\ \Longleftrightarrow&\left\{ \begin{array}{l} \dfrac{a_1^2c_2^2}{d_2^2}-\dfrac{a_1^2a_2c_2}{d_2}+d_1a_1c_2-c_1c_2=0\\ -\dfrac{a_1c_2}{d_2}= -\dfrac{c_1a_2}{d_1} \end{array}\right. \\ \Longleftrightarrow&\left\{ \begin{array}{l} a_1^2c_2-a_1^2a_2d_2+d_1a_1d_2^2-c_1d_2^2=0\\ a_1d_1c_2=a_2d_2c_1 \end{array}\right. \\ \Longleftrightarrow&\left\{ \begin{array}{l} a_1^2(c_2-a_2d_2)-d_2^2(c_1-a_1d_1)=0\\ a_1d_1(a_2^2-a_2d_2+d_2^2)=a_2d_2(a_1^2-a_1d_1+d_1^2) \end{array}\right. \\ \Longleftrightarrow&\left\{ \begin{array}{l} a_1^2(a_2-d_2)^2-d_2^2(a_1-d_1)^2=0\\ a_1d_1a_2^2-a_1d_1a_2d_2+a_1d_1d_2^2=a_2d_2a_1^2-a_2d_2a_1d_1+a_2d_2d_1^2 \end{array}\right. \\\ \Longleftrightarrow&\left\{ \begin{array}{l} a_1^2(a_2-d_2)^2-d_2^2(a_1-d_1)^2=0\\ a_1d_1a_2^2-2a_1d_1a_2d_2+a_1d_1d_2^2=a_2d_2a_1^2-2a_2d_2a_1d_1+a_2d_2d_1^2 \end{array}\right. \\ \Longleftrightarrow&\left\{ \begin{array}{l} a_1^2(a_2-d_2)^2-d_2^2(a_1-d_1)^2=0\\ a_1d_1(a_2^2-2a_2d_2+d_2^2)=a_2d_2(a_1^2-2a_1d_1+d_1^2) \end{array}\right. \\ \Longleftrightarrow&\left\{ \begin{array}{l} a_1^2(a_2-d_2)^2=d_2^2(a_1-d_1)^2\\ a_1d_1(a_2-d_2)^2=a_2d_2(a_1-d_1)^2 \end{array}\right. \\ \Longleftrightarrow&\left\{ \begin{array}{l} \dfrac{a_1^2}{d_2^2}=\dfrac{(a_1-d_1)^2}{(a_2-d_2)^2}\\ \dfrac{a_1d_1}{a_2d_2}=\dfrac{(a_1-d_1)^2}{(a_2-d_2)^2} \end{array}\right. \\ \Longleftrightarrow&\dfrac{a_1}{d_2}=\dfrac{d_1}{a_2}\\ \Longleftrightarrow&a_1a_2=d_1d_2. \end{aligned}$$

We consider the subspaces of the form $\langle e_2+xe_3\rangle$ with $x\ne 0$.

$\rho (\sigma _j)(e_2+xe_3)=\left( \begin{matrix} \pm a_1\pm a_2x\\ a_1d_2+c_2x\\ c_1+d_1a_2x\\ \pm c_1d_2\pm d_1c_2x \end{matrix}\right)$ for $1\leqslant j\leqslant 4$.

If we assume that the subspace $\langle e_2+xe_3\rangle$ is invariant, then $\left( \begin{matrix} a_1+a_2x\\ a_1d_2+c_2x\\ c_1+d_1a_2x\\ c_1d_2+d_1c_2x \end{matrix}\right) = \left( \begin{matrix} 0\\ \alpha \\ \alpha x\\ 0 \end{matrix}\right)$ for some $\alpha \in {\mathbb {C}}^*$. This is equivalent to:

$$\begin{aligned}&\left\{ \begin{array}{l} a_1+a_2x=0,\\ c_1d_2+d_1c_2x=0, \\ a_1d_2+c_2x=\alpha \text { and } \\ c_1+d_1a_2x=\alpha x \end{array} \right. \\ \Longleftrightarrow&x=-\dfrac{a_1}{a_2},~x=-\dfrac{c_1d_2}{d_1c_2},~\text { and }~ c_2x^2+(a_1d_2-d_1a_2)x-c_1=0\\ \Longleftrightarrow&c_2\dfrac{a_1^2}{a_2^2}-\dfrac{a_1^2d_2}{a_2}+d_1a_1-c_1=0 ~\text { and }~ a_1c_2d_1=a_2c_1d_2\\ \Longleftrightarrow&c_2a_1^2-a_1^2d_2a_2+d_1a_1a_2^2-c_1a_2^2=0 ~\text { and }~ a_1d_1(a_2^2-a_2d_2+d_2^2)=a_2d_2(a_1^2-a_1d_1+d_1^2)\\ \Longleftrightarrow&a_1^2(c_2-d_2a_2)=a_2^2(-d_1a_1+c_1) ~\text { and }~ a_1d_1(a_2^2-2a_2d_2+d_2^2)=a_2d_2(a_1^2-2a_1d_1+d_1^2)\\ \Longleftrightarrow&a_1^2(a_2-d_2)^2=a_2^2(a_1-d_1)^2 ~\text { and }~ a_1d_1(a_2-d_2)^2=a_2d_2(a_1-d_1)^2\\ \Longleftrightarrow&\dfrac{a_1^2}{a_2^2}=\dfrac{(a_1-d_1)^2}{(a_2-d_2)^2}=\dfrac{a_1d_1}{a_2d_2} \\\ \Longleftrightarrow&\dfrac{a_1}{a_2}=\dfrac{d_1}{d_2}\\ \Longleftrightarrow&a_1d_2=a_2d_1. \end{aligned}$$

$\square$

Proposition 3.3

The representation $\rho :{{\mathcal {N}}{\mathcal {B}}}_4\rightarrow GL(4,{\mathbb {C}})$ has no non-trivial proper invariant subspaces of dimension 2 if and only if

$$\begin{aligned} a_1a_2\ne d_1d_2\text { and }a_1d_2\ne a_2d_1. \end{aligned}$$

Proof

The subspaces of dimension 2 that are possibly invariant are those spanned by the following subsets of vectors: $\{e_1,he_2+je_3\},~\{e_4,re_2+se_3\},~\{e_2,ke_1+me_4\},~\{e_3,pe_1+qe_4\},~\{e_1+xe_4,e_2+ye_3\}$, where $\{h,j,r,s,k,m,p,q,x,y\}\subset {\mathbb {C}}$.

$\rho (\sigma _1)(e_1)=\left( \begin{matrix} a_1a_2\\ a_1c_2\\ c_1a_2\\ c_1c_2 \end{matrix}\right) \not \in \langle e_1,he_2+je_3\rangle$ because $c_1c_2\ne 0$. So $\langle e_1,he_2+je_3\rangle$ are not invariant subspaces for all $h,j\in {\mathbb {C}}$.

$\rho (\sigma _1)(e_4)=\left( \begin{matrix} 1\\ d_2\\ d_1\\ d_1d_2 \end{matrix}\right) \not \in \langle e_4,re_2+se_3\rangle$. So, $\langle e_4,re_2+se_3\rangle$ are not invariant subspaces for all $r,s\in {\mathbb {C}}$.

$\rho (\sigma _1)(e_2)=\left( \begin{matrix} a_1\\ a_1d_2\\ c_1\\ c_1d_2 \end{matrix}\right) \not \in \langle e_2,ke_1+me_4\rangle$ because $c_1\ne 0$. So, the subspaces $\langle e_2,ke_1+me_4\rangle$ are not invariant for all $k,m\in {\mathbb {C}}$.

$\rho (\sigma _1)(e_3)=\left( \begin{matrix} a_2\\ c_2\\ d_1a_2\\ d_1c_2 \end{matrix}\right) \not \in \langle e_3,pe_1+qe_4\rangle$ because $c_2\ne 0$. So, the subspaces $\langle e_3,pe_1+qe_4\rangle$ are not invariant for all $p,q\in {\mathbb {C}}$.

Now consider the subspaces of the form $\langle e_1+xe_4,e_2+ye_3\rangle$ where $x,y\in {\mathbb {C}}^*.$ Put $u=e_1+xe_4$ and $v=e_2+ye_3$ then

$\rho (\sigma _j)(u)=\left( \begin{matrix} a_1a_2+x\\ \pm a_1c_2\pm d_2x\\ \pm c_1a_2\pm d_1x\\ c_1c_2+d_1d_2x \end{matrix}\right)$ and $\rho (\sigma _j)(v)=\left( \begin{matrix} \pm a_1\pm a_2y\\ a_1d_2+c_2y\\ c_1+d_1a_2y\\ \pm c_1d_2\pm d_1c_2y \end{matrix}\right) .$

The subspace $\langle u,v\rangle$ is invariant if and only if $\rho (\sigma _j)(u)=\alpha u+\alpha 'v$ and $\rho (\sigma _j)(v)=\beta u+\beta 'v$ for some $\alpha , \alpha ', \beta , \beta '\in {\mathbb {C}}$.

This is equivalent to $\left\{ \begin{matrix} a_1a_2+x=\alpha \\ \pm (a_1c_2+d_2x)=\alpha '\\ \pm (c_1a_2+d_1x)=\alpha 'y\\ c_1c_2+d_1d_2x=\alpha x \end{matrix}\right.$ and $\left\{ \begin{matrix} \pm (a_1+a_2y)=\beta \\ a_1d_2+c_2y=\beta '\\ c_1+d_1a_2y=\beta 'y\\ \pm (c_1d_2+d_1c_2y)=\beta x \end{matrix}\right. .$

By eliminating $\alpha ,\alpha ', \beta$ and $\beta '$, we get the following system.

$$\begin{aligned}&c_1c_2+d_1d_2x=a_1a_2x+x^2,\end{aligned}$$

(3.1a)

$$\begin{aligned}&c_1a_2+d_1x=a_1c_2y+d_2xy,\end{aligned}$$

(3.1b)

$$\begin{aligned}&c_1d_2+d_1c_2y=a_1x+a_2xy,\end{aligned}$$

(3.1c)

$$\begin{aligned}&c_1+d_1a_2y=a_1d_2y+c_2y^2. \end{aligned}$$

(3.1d)

The Eqs. (3.1a) and (3.1d) lead to four solutions:

$$\begin{aligned} x=\dfrac{-a_1a_2+d_1d_2\pm \sqrt{4c_1c_+(a_1a_2-d_1d_2)^2}}{2};\quad y=\dfrac{a_2d_1-a_1d_2\pm \sqrt{4c_1c_+(a_2d_1-a_1d_2)^2}}{2c_2}. \end{aligned}$$

Substitute each of these solutions into the Eqs. (3.1b) and (3.1c) then, using Mathematica, we get the following relations:

$(a_2=d_2),~~~(a_1=d_1),~~~(d_1=e^{\pm \pi i/3}a_1), ~~~ (d_1d_2=a_1a_2),~~~ (d_1a_2=a_1d_2),~~~ (a_1=a_2=0),~~~(a_1=d_2=0),~~~(d_1=a_2=0),~~~(d_1=d_2=0),~~~(d_1=a_1,a_2=0)~~~(d_1=a_1,d_2=0),~~~(d_1=e^{\pm \pi i/3}a_1,a_2=0)$ or $(d_1=e^{\pm \pi i/3}a_1,d_2=0)$

But we have $a_1\ne d_1,~~ a_2\ne d_2,~~ a_1a_2\ne d_1d_2,~~ a_1d_2\ne a_2d_1~$ and $a_1d_1a_2d_2\ne 0$. So, the only 2 relations left are $d_1=e^{\pm \pi i/3}a_1$ which lead to $c_1=a_1^2-a_1d_1+d_1^2=0$. This contradicts the fact that $c_1\ne 0$.

Therefore $\langle u,v\rangle$ is not invariant under $\rho (\sigma _j)$ for $1\leqslant j\leqslant 4$. $\square$

Proposition 3.4

$\rho :{{\mathcal {N}}{\mathcal {B}}}_4\rightarrow GL(4,{\mathbb {C}})$ has no non-trivial proper invariant subspaces of dimension 3 if and only if

$$\begin{aligned} a_1a_2\ne d_1d_2\text { and }a_1d_2\ne a_2d_1. \end{aligned}$$

Proof

The subspaces of dimension 3 that are possibly invariant are those spanned by the following subsets of vectors: $\{e_1,e_4,e_3\},~\{e_1,e_4,e_2\},~\{e_1,e_2,e_3\},~\{e_4,e_2,e_3\},~\{e_1,e_4,e_2+xe_3\},~\{e_1+ye_4,e_2,e_3\}$, where $x,y\in {\mathbb {C}}^*$.

$\rho (\sigma _1)(e_3)=\left( \begin{matrix} a_2\\ c_2\\ d_1a_2\\ d_1c_2 \end{matrix}\right) \not \in \langle e_1,e_4,e_3\rangle$ since $c_2\ne 0$. Thus, $\langle e_1,e_4,e_3\rangle$ is not invariant.

$\rho (\sigma _1)(e_2)=\left( \begin{matrix} a_1\\ a_1d_2\\ c_1\\ c_1d_2 \end{matrix}\right) \not \in \langle e_1,e_4,e_2\rangle$ since $c_1\ne 0$. So, $\langle e_1,e_4,e_2\rangle$ is not invariant

$\rho (\sigma _1)(e_1)=\left( \begin{matrix} a_1a_2\\ a_1c_2\\ c_1a_2\\ c_1c_2 \end{matrix}\right) \not \in \langle e_1,e_2,e_3\rangle$ since $c_1c_2\ne 0$. So, $\langle e_1,e_2,e_3\rangle$ is not invariant.

$\rho (\sigma _1)(e_4)=\left( \begin{matrix} 1\\ d_2\\ d_1\\ d_1d_2 \end{matrix}\right) \not \in \langle e_4,e_2,e_3\rangle$. So, $\langle e_4,e_2,e_3\rangle$ is not invariant.

Now, consider the subspace $\langle e_1,e_4,e_2+xe_3\rangle$ with $x\ne 0$.

Note that $\rho (\sigma _j)(e_1)=\left( \begin{matrix} a_1a_2\\ \pm a_1c_2\\ \pm c_1a_2\\ c_1c_2 \end{matrix}\right)$, $\rho (\sigma _j)(e_4)=\left( \begin{matrix} 1\\ \pm d_2\\ \pm d_1\\ d_1d_2 \end{matrix}\right)$ and $\rho (\sigma _j)(e_2+xe_3)=\left( \begin{matrix} \pm a_1\pm a_2x\\ a_1d_2+c_2x\\ c_1+d_1a_2x\\ \pm c_1d_2\pm d_1c_2x \end{matrix}\right)$.

Assume that the subspace $\langle e_1,e_4,e_2+xe_3\rangle$ is invariant. It follows that $\{\rho (\sigma _1)(e_1),\rho (\sigma _1)(e_4), \rho (\sigma _1)(e_2+xe_3)\}\subset \langle e_1,e_4,e_2+xe_3\rangle$. This is equivalent to the following system.

$$\begin{aligned}&c_1a_2=a_1c_2x, \end{aligned}$$

(3.2a)

$$\begin{aligned}&d_1=d_2x, \end{aligned}$$

(3.2b)

$$\begin{aligned}&c_2x^2+(a_1d_2-d_1a_2)x-c_1=0. \end{aligned}$$

(3.2c)

We determine conditions for which the system above is consistent.

From Eqs. (3.2a) and (3.2b), we get $x=\dfrac{c_1a_2}{a_1c_2}=\dfrac{d_1}{d_2}$ then substitute it into the Eq. (3.2c) to obtain $c_2\dfrac{d_1^2}{d_2^2}+a_1d_1-\dfrac{d_1^2a_2}{d_2}-c_1=0.$

$\Longrightarrow a_2d_2c_1=a_1d_1c_2\text { and }c_2d_1^2-a_2d_2d_1^2=d_2^2c_1-d_2^2a_1d_1$

$\Longrightarrow a_2d_2(a_1^2-a_1d_1+d_1^2)=a_1d_1(a_2^2-a_2d_2+d_2^2) ~ \text { and } ~ d_1^2(c_2-a_2d_2)=d_2^2(c_1-a_1d_1)$

$\Longrightarrow a_2d_2(a_1^2-2a_1d_1+d_1^2)=a_1d_1(a_2^2-2a_2d_2+d_2^2) ~ \text { and } ~ d_1^2(a_2-d_2)^2=d_2^2(a_1-d_1)^2$

$\Longrightarrow \dfrac{a_2d_2}{a_1d_1}=\dfrac{(a_2-d_2)^2}{(a_1-d_1)^2}=\dfrac{d_2^2}{d_1^2}$

$\Longrightarrow \dfrac{a_2}{a_1}=\dfrac{d_2}{d_1}$

$\Longrightarrow a_2d_1=a_1d_2$.

This gives a contradiction.

Now, consider the subspace $\langle e_1+ye_4,e_2,e_3\rangle$ with $y\ne 0$. Note that $\rho (\sigma _j)(e_2)=\left( \begin{matrix} \pm a_1\\ a_1d_2\\ c_1\\ \pm c_1d_2 \end{matrix}\right)$,  $\rho (\sigma _j)(e_3)=\left( \begin{matrix} \pm a_2\\ c_2\\ d_1a_2\\ \pm d_1c_2 \end{matrix}\right)$  and  $\rho (\sigma _j)(e_1+ye_4)=\left( \begin{matrix} a_1a_2+y\\ \pm a_1c_2\pm d_2y\\ \pm c_1a_2\pm d_1y\\ c_1c_2+d_1d_2y \end{matrix}\right)$.

Assume that the subspace $\langle e_1+ye_4,e_2,e_3\rangle$ is invariant. It follows that the vectors $\rho (\sigma _i)(e_2),~~\rho (\sigma _i)(e_3)$ and $\rho (\sigma _i)(e_1+ye_4)$ are linear combinations of $e_1+ye_4,~e_2,$ and $e_3$.

$\Longrightarrow \dfrac{c_1d_2}{a_1}=\dfrac{d_1c_2}{a_2}=y$ and $y^2+(a_1a_2-d_1d_2)y-c_1c_2=0$

$\Longrightarrow a_2d_2c_1=a_1d_1c_2 ~ \text { and } ~ \dfrac{c_1^2d_2^2}{a_1^2}+a_2d_2c_1-\dfrac{c_1d_1d_2^2}{a_1}-c_1c_2=0$

$\Longrightarrow a_2d_2(a_1^2-a_1d_1+d_1^2)=a_1d_1(a_2^2-a_2d_2+d_2^2) ~ \text { and } ~ c_1d_2^2+a_2d_2a^2_1-d_1d_2^2a_1-c_2a_1^2=0$

$\Longrightarrow a_2d_2(a_1^2-2a_1d_1+d_1^2)=a_1d_1(a_2^2-2a_2d_2+d_2^2) ~ \text { and } ~ d_2^2(c_1-d_1a_1)=a_1^2(c_2-a_2d_2)$

$\Longrightarrow a_2d_2(a_1-d_1)^2=a_1d_1(a_2-d_2)^2 ~ \text { and } ~ d_2^2(a_1-d_1)^2=a_1^2(a_2-d_2)^2$

$\Longrightarrow \dfrac{a_2d_2}{a_1d_1}=\dfrac{d_2^2}{a_1^2}=\dfrac{(a_2-d_2)^2}{(a_1-d_1)^2}$

$\Longrightarrow \dfrac{a_2}{d_1}=\dfrac{d_2}{a_1}$.

This gives a contradiction. $\square$

We, thus, get our main theorem.

Theorem 3.5

The representation $\rho :{\mathcal {N}}{\mathcal {B}}_4\rightarrow GL(4,{\mathbb {C}})$ is irreducible if and only if $a_1a_2\ne d_1d_2$ and $a_1d_2\ne a_2d_1$.

3.1.2 Case $T_1=T_2=\pm i$

In the case $T_1=T_2=i$, the representation $\rho$ of ${\mathcal {N}}{\mathcal {B}}_4$ is given by

$$\begin{aligned} \rho (\tau )= & {} t_1t_2\left( \begin{matrix} -1&{}0&{}0&{}0\\ 0&{}i&{}0&{}0\\ 0&{}0&{}i&{}0\\ 0&{}0&{}0&{}1 \end{matrix}\right) , ~ \rho (\sigma _1)=\left( \begin{matrix} d_1d_2&{}d_1&{}d_2&{}1\\ -d_1d^2_2&{}d_1d_2&{}-d^2_2&{}d_2\\ -d^2_1d_2&{}-d_1^2&{}d_1d_2&{}d_1\\ d^2_1d^2_2&{}-d^2_1d_2&{}-d_1d^2_2&{}d_1d_2 \end{matrix}\right) , \\ \rho (\sigma _2)= & {} \left( \begin{matrix} d_1d_2&{} id_1&{} id_s2&{}-1\\ id_1d^2_2&{}d_1d_2&{}-d^2_2&{} id_2\\ id^2_1d_2&{}-d_1^2&{}d_1d_2&{} id_1\\ -d^2_1d^2_2&{} id^2_1d_2&{} id_1d^2_2&{}d_1d_2 \end{matrix}\right) , ~ \rho (\sigma _3)=\left( \begin{matrix} d_1d_2&{}-d_1&{}-d_2&{}1\\ d_1d^2_2&{}d_1d_2&{}-d^2_2&{}-d_2\\ d^2_1d_2&{}-d_1^2&{}d_1d_2&{}-d_1\\ d^2_1d^2_2&{}d^2_1d_2&{}d_1d^2_2&{}d_1d_2 \end{matrix}\right) , \\ \rho (\sigma _4)= & {} \left( \begin{matrix} d_1d_2&{}- id_1&{}- id_2&{}-1\\ id_1d^2_2&{}d_1d_2&{}-d^2_2&{}- id_2\\ id^2_1d_2&{}-d_1^2&{}d_1d_2&{}- id_1\\ -d^2_1d^2_2&{}- id^2_1d_2&{}- id_1d^2_2&{}d_1d_2 \end{matrix}\right) . \end{aligned}$$

Likewise, we write the matrices for $T_1=T_2=-i.$

Proposition 3.6

The representation $\rho$ is reducible.

Proof

Let S be the subspace spanned by the vector $v=d_2e_2-d_1e_3$. Note that $e_2$ and $e_3$ are eigen-vectors of $\rho (\tau )$ corresponding the same eigenvalue. So, S is invariant under $\rho (\tau )$. It is easy to show that $\rho (\sigma _j)(v)=2d_1d_2v$ for $j=1,2,3,4$. This completes the proof. $\square$

3.1.3 Case $T_1=-T_2=\pm i$

In the case $T_1=-T_2=i$, the representation $\rho$ of ${\mathcal {N}}{\mathcal {B}}_4$ is given by

$$\begin{aligned} \rho (\tau )= & {} t_1t_2\left( \begin{matrix} 1&{}0&{}0&{}0\\ 0&{} i&{}0&{}0\\ 0&{}0&{}- i&{}0\\ 0&{}0&{}0&{}1 \end{matrix}\right) , ~ \rho (\sigma _1)=\left( \begin{matrix} d_1d_2&{}d_1&{}d_2&{}1\\ -d_1d^2_2&{}d_1d_2&{}-d^2_2&{}d_2\\ -d^2_1d_2&{}-d_1^2&{}d_1d_2&{}d_1\\ d^2_1d^2_2&{}-d^2_1d_2&{}-d_1d^2_2&{}d_1d_2 \end{matrix}\right) ,\\ \rho (\sigma _2)= & {} \left( \begin{matrix} d_1d_2&{}- id_1&{} id_2&{}1\\ - id_1d^2_2&{}d_1d_2&{}d^2_2&{} id_2\\ id^2_1d_2&{}d_1^2&{}d_1d_2&{}- id_1\\ d^2_1d^2_2&{} id^2_1d_2&{}- id_1d^2_2&{}d_1d_2\end{matrix}\right) , ~ \rho (\sigma _3)=\left( \begin{matrix} d_1d_2&{}-d_1&{}-d_2&{}1\\ d_1d^2_2&{}d_1d_2&{}-d^2_2&{}-d_2\\ d^2_1d_2&{}-d_1^2&{}d_1d_2&{}-d_1\\ d^2_1d^2_2&{}d^2_1d_2&{}d_1d^2_2&{}d_1d_2 \end{matrix}\right) , \\ \rho (\sigma _4)= & {} \left( \begin{matrix} d_1d_2&{} id_1&{}- id_2&{}1\\ id_1d^2_2&{}d_1d_2&{}d^2_2&{}- id_2\\ - id^2_1d_2&{}d_1^2&{}d_1d_2&{} id_1\\ d^2_1d^2_2&{}- id^2_1d_2&{} id_1d^2_2&{}d_1d_2 \end{matrix}\right) . \end{aligned}$$

Likewise, we write the matrices for $T_1=-T_2=-i.$

Proposition 3.7

The representation $\rho$ is reducible.

Proof

Let S be the subspace spanned by the vector $v=e_1+d_1d_2e_4$. The subspace S is invariant under $\rho (\tau )$ because, $e_1$ and $e_4$ are eigenvectors of $\rho (\tau )$ corresponding to the same eigenvalue. By direct calculations we get $\rho (\sigma _j)(v)=2d_1d_2v\in S$ for $j=1,2,3,4$. Hence, S is invariant under $\rho (\sigma _j)$. $\square$

3.1.4 Case $T_1=-1,T_2=\pm i$

Without loss of generality, assume that $T_1=-1$ and $T_2=i$. Then, we have $T_1T_2=-i$ and

$$\begin{aligned} \rho (\tau )= t_1t_2\left( \begin{matrix} - i&{}0&{}0&{}0\\ 0&{}-1&{}0&{}0\\ 0&{}0&{} i&{}0\\ 0&{}0&{}0&{}1 \end{matrix}\right) , ~ \rho (\sigma _1)=\left( \begin{matrix} a_1d_2&{}a_1&{}d_2&{}1\\ -a_1d^2_2&{}a_1d_2&{}-d^2_2&{}d_2\\ c_1d_2&{}c_1&{}d_1d_2&{}d_1\\ -c_1d^2_2&{}c_1d_2&{}-d_1d^2_2&{}d_1d_2 \end{matrix}\right) \end{aligned}$$

Likewise, we write the matrices for $T_1=-1, T_2=-i.$

Note that the vectors $e_j$ ($j=1,2,3,4$) are eigenvectors of $\rho (\tau )$ corresponding to four distinct eigenvalues (each of multiplicity 1).

Proposition 3.8

The representation $\rho$ is irreducible.

Proof

We have $\rho (\sigma _1)(e_1)=\left( \begin{matrix} a_1d_2\\ -a_1d_2^2\\ c_1d_2\\ -c_1d_2^2 \end{matrix}\right)$, $\rho (\sigma _1)(e_2)=\left( \begin{matrix} a_1\\ a_1d_2\\ c_1\\ c_1d_2 \end{matrix}\right)$, $\rho (\sigma _1)(e_3)=\left( \begin{matrix} d_2\\ -d_2^2\\ d_1d_2\\ -d_1d_2^2 \end{matrix}\right)$ and $\rho (\sigma _1)(e_4)=\left( \begin{matrix} 1\\ d_2\\ d_1\\ d_1d_2 \end{matrix}\right)$. Note that $a_1\ne 0$, $d_1\ne 0$, $c_1\ne 0$ and $d_2\ne 0$. This implies that none of the proper subspaces spanned by the vectors $e_j$ are invariant under $\rho (\sigma _1)$. Likewise, we show that there are no invariant subspaces of dimensions 2 and 3. Hence $\rho$ is irreducible. $\square$

3.2 Representation of ${\mathcal {N}}{\mathcal {B}}_3$

Consider two irreducible representations of dimension 2 of the necklace braid group ${\mathcal {N}}{\mathcal {B}}_3$ $\rho _1=\rho (T_1,t_1,\omega _1,c_1,d_1)$ and $\rho _2=\rho (T_2,t_2,\omega _2,c_2,d_2)$ (see Proposition 2.1).

Recall that $T_1=e^{\pm i2\pi /3},T_2=e^{\pm i2\pi /3},\omega _1=e^{\pm i\pi /3}$ and $\omega _2=e^{\pm i\pi /3}$.

Let $\rho$ be the representation of ${\mathcal {N}}{\mathcal {B}}_3$, which is defined in Definition 3.1. We have

$$\begin{aligned} \rho (\tau )= t_1t_2\left( \begin{matrix} T_1T_2&{}0&{}0&{}0\\ 0&{}T_1&{}0&{}0\\ 0&{}0&{}T_2&{}0\\ 0&{}0&{}0&{}1 \end{matrix}\right) \end{aligned}$$

and

$$\begin{aligned} \rho (\sigma _j)= \left( \begin{matrix} \omega _1\omega _2d_1d_2&{}\omega _1d_1T_2^{j-1}&{}\omega _2d_2T_1^{j-1}&{}(T_1T_2)^{j-1}\\ \omega _1d_1c_2T_2^{1-j}&{}\omega _1d_1d_2&{}c_2T_1^{j-1}T_2^{1-j}&{}d_2T_1^{j-1}\\ \omega _2c_1d_2T_1^{1-j}&{}c_1T_1^{1-j}T_2^{j-1}&{}\omega _2d_2d_1&{}d_1T_2^{j-1}\\ c_1c_2(T_1T_2)^{1-j}&{}c_1d_2T_1^{1-j}&{}d_1c_2T_2^{1-j}&{}d_1d_2 \end{matrix}\right) . \end{aligned}$$

for $j=1,2,3$. We have two cases (1) $T_1=e^{2\pi i/3},T_2=e^{-2\pi i/3}$ and (2) $T_1=T_2=e^{\pm 2\pi i/3}$.

We determine a necessary and sufficient condition under which $\rho$ is an irreducible representation of ${\mathcal {N}}{\mathcal {B}}_3$ of dimension 4.

In what follows, suppose that $d_1d_2\ne 0$.

3.2.1 Case $T_1=e^{2\pi i/3}, T_2=e^{-2\pi i/3}$

In this case, we have

$$\begin{aligned} \rho (\tau )= t_1t_2\left( \begin{matrix} 1&{}0&{}0&{}0\\ 0&{}e^{2\pi i/3}&{}0&{}0\\ 0&{}0&{}e^{-2\pi i/3}&{}0\\ 0&{}0&{}0&{}1 \end{matrix}\right) . \end{aligned}$$

Then, the eigenvalues of $\rho (\tau )$ are $\lambda _1=t_1t_2$ (of multiplicity 2) and $\lambda _2=e^{2\pi i/3}t_1t_2$ and $\lambda _3=e^{-2\pi i/3}t_1t_2$. The corresponding eigenvectors are $x_1e_1+x_4e_4$ for $\lambda _1$ and $e_2$ and $e_3$ for $\lambda _2$ and $\lambda _3$, respectively, where $\{x_1,x_4\}\subset {\mathbb {C}}^*$.

Lemma 3.9

If $T_1=e^{2\pi i/3}$, $T_2=e^{-2\pi i/3},\omega _1=e^{\pm i\pi /3}$ and $\omega _2=e^{\pm i\pi /3}$, then the representation $\rho$ of ${\mathcal {N}}{\mathcal {B}}_3$ has no invariant proper subspaces of dimension 1 if and only if

$$\begin{aligned} \omega _1\omega _2\ne 1 \text { or }\omega _1d_1^2c_2\ne \omega _2d_2^2c_1. \end{aligned}$$

Proof

By direct calculations, we check that each of the subspaces spanned by the vectors $e_j$ ($j=1,2,3$) are not invariant under $\rho (\sigma _1)$.

It remains to consider the subspace S spanned by the vector $v=e_1+xe_4$ with $x\in {\mathbb {C}}^*$. Since $T_1T_2=1$, it follows that

$$\begin{aligned} \rho (\sigma _j)(v)=\left( \begin{matrix} \omega _1\omega _2d_1d_2+x\\ \omega _1d_1c_2T_2^{1-j}+d_2T_1^{j-1}x\\ \omega _2c_1d_2T_1^{1-j}+d_1T_2^{j-1}x\\ c_1c_2+d_1d_2x \end{matrix}\right) = \left( \begin{matrix} \omega _1\omega _2d_1d_2+x\\ (\omega _1d_1c_2+d_2x)T_1^{j-1}\\ (\omega _2c_1d_2+d_1x)T_2^{j-1}\\ c_1c_2+d_1d_2x \end{matrix}\right) . \end{aligned}$$

So

$$\begin{aligned} \rho (\sigma _j)(v)\in S&\Leftrightarrow \rho (\sigma _j)(v)=\lambda v \text { for some }\lambda \in {\mathbb {C}} \\&\Leftrightarrow x=\frac{-\omega _1d_1c_2}{d_2}=\frac{-\omega _2c_1d_2}{d_1} \text { and } x^2+(\omega _1\omega _2-1)d_1d_2x-c_1c_2=0\\&\Leftrightarrow \omega _1d_1^2c_2=\omega _2d_2^2c_1\text { and } (\omega _1\omega _2-1)(c_2-\omega _2d_2^2)=0\\&\Leftrightarrow \omega _1d_1^2c_2=\omega _2d_2^2c_1\text { and } \omega _1\omega _2-1=0. \end{aligned}$$

Therefore the subspace S, that is spanned by $v=e_1-\frac{\omega _1d_1c_2}{d_2}e_4$, is invariant under $\rho (\sigma _j)$ if $\omega _1\omega _2=1$ and $\omega _1d_1^2c_2= \omega _2d_2^2c_1$. This gives a contradiction. $\square$

Lemma 3.10

If $T_1=e^{2\pi i/3}$, $T_2=e^{-2\pi i/3},\omega _1=e^{\pm i\pi /3}$ and $\omega _2=e^{\pm i\pi /3}$ then the representation $\rho$ of ${\mathcal {N}}{\mathcal {B}}_3$ has no invariant proper subspaces of dimension 2.

Proof

The possible two-dimensional invariant subspaces candidates to study are the following:

• 1. $\langle e_i,e_j\rangle$ for $i\ne j$

• 2. $\langle e_2,e_1+xe_4\rangle$ for $x\ne 0$

• 3. $\langle e_3,e_1+ye_4\rangle$ for $y\ne 0$

Since $c_1\ne 0, c_2\ne 0$, $d_1\ne 0$ and $d_2\ne 0$, it follows that none of the subspaces mentioned above is invariant under $\rho (\sigma _1)$. $\square$

Lemma 3.11

If $T_1=e^{2\pi i/3}$, $T_2=e^{-2\pi i/3}, \omega _1=e^{\pm i\pi /3}$ and $\omega _2=e^{\pm i\pi /3}$ then the representation $\rho$ of ${\mathcal {N}}{\mathcal {B}}_3$ has no invariant proper subspaces of dimension 3 if and only if

$$\begin{aligned} \omega _1\omega _2\ne 1 \text { or }\omega _1d_1^2c_2\ne \omega _2d_2^2c_1. \end{aligned}$$

Proof

The possible invariant three-dimensional invariant subspaces to study are:

• 1. $\langle e_i,e_j,e_k\rangle$ where i, j, k are pairwise distinct

• 2. $\langle e_2,e_3,e_1+xe_4\rangle$ for $x\ne 0$

By direct computations, it is easy to check that all the subspaces in case 1 are not invariant.

In case 2, we let S be a subspace of the form $\langle e_2,e_3,e_1+xe_4\rangle$ with $x\ne 0$.

$$\begin{aligned} \rho (\sigma _j)(e_2)=\left( \begin{matrix} \omega _1d_1T_2^{j-1}\\ \omega _1d_1d_2\\ c_1T_1^{1-j}T_2^{j-1}\\ c_1d_2T_1^{1-j} \end{matrix}\right) , ~\rho (\sigma _j)(e_3)=\left( \begin{matrix} \omega _2d_2T_1^{j-1}\\ c_2T_1^{j-1}T_2^{1-j}\\ \omega _2d_1d_2\\ d_1c_2T_2^{1-j} \end{matrix}\right) \end{aligned}$$

and

$$\begin{aligned} \rho (\sigma _j)(e_1+xe_4)=\left( \begin{matrix} \omega _1\omega _2d_1d_2+x\\ \omega _1d_1c_2T_2^{1-j}+d_2T_1^{j-1}x\\ \omega _2c_1d_2T_1^{1-j}+d_1T_2^{j-1}x\\ c_1c_2+d_1d_2x \end{matrix}\right) , \end{aligned}$$

Therefore

$$\begin{aligned} \rho (\sigma _j)(e_2)\in S\Leftrightarrow x=\dfrac{c_1d_2}{\omega _1d_1} \end{aligned}$$

and

$$\begin{aligned}&\rho (\sigma _j)(e_3)\in S\Leftrightarrow x=\dfrac{d_1c_2}{\omega _2d_2},\\&\quad \rho (\sigma _j)(e_1+xe_4)\in S&\Leftrightarrow \left( \begin{matrix} \omega _1\omega _2d_1d_2+x\\ \omega _1d_1c_2T_2^{1-j}+d_2T_1^{j-1}x\\ \omega _2c_1d_2T_1^{1-j}+d_1T_2^{j-1}x\\ c_1c_2+d_1d_2x \end{matrix}\right) = \left( \begin{matrix} \lambda \\ \alpha \\ \beta \\ \lambda x \end{matrix}\right) \text { for some } \alpha , \beta , \gamma \in {\mathbb {C}}. \end{aligned}$$

This is equivalent to

$$\begin{aligned} x^2+(\omega _1\omega _2d_1d_2-d_1d_2)x-c_1c_2=0. \end{aligned}$$

Therefore, S is invariant under $\rho (\sigma _j)$ ($j=1,2,3$) if and only if

$$\begin{aligned} x=\dfrac{c_1d_2}{\omega _1d_1}=\dfrac{c_2d_1}{\omega _2d_2} \end{aligned}$$

(3.3)

and

$$\begin{aligned} c_1^2d_2^2+\omega _1d_1^2c_1d_2^2(\omega _1\omega _2-1)-\omega _1d_1^2c_1c_2=0. \end{aligned}$$

(3.4)

The Eqs.  (3.3) and (3.4) imply that

$$\begin{aligned} \omega _1d_1^2c_2=\omega _2d_2^2c_1 \end{aligned}$$

and

$$\begin{aligned} (\omega _1\omega _2-1)(\omega _2d_2^2-c_2)=0. \end{aligned}$$

Since $\omega _2d_2^2\ne c_2$, it follows that

$$\begin{aligned} \omega _1\omega _2=1 \end{aligned}$$

Therefore the subspace S, which is spanned by $v=e_1-\frac{c_1d_2}{\omega _1d_1}e_4$, is invariant under $\rho (\sigma _j)$ if $\omega _1\omega _2=1$ and $\omega _1d_1^2c_2= \omega _2d_2^2c_1$. This gives a contradiction. $\square$

Now, we have the following proposition.

Proposition 3.12

If $T_1=e^{i2\pi /3}$, $T_2=e^{-i2\pi /3}, \omega _1=e^{\pm i\pi /3}$ and $\omega _2=e^{\pm i\pi /3}$ then $\rho$ is irreducible representation of ${\mathcal {N}}{\mathcal {B}}_3$ if and only if $\omega _1\omega _2\ne 1$ or $\omega _2d_2^2c_1\ne \omega _1d_1^2c_2$.

3.2.2 Case ($T_1=T_2=e^{2\pi i/3}$) and ($T_1=T_2=e^{-2\pi i/3}$)

In this case, we have $T_1=T_2$. Then,

$$\begin{aligned} \rho (\tau )= t_1t_2\left( \begin{matrix} e^{\mp i2\pi /3}&{}0&{}0&{}0\\ 0&{}e^{\pm i2\pi /3}&{}0&{}0\\ 0&{}0&{}e^{\pm i2\pi /3}&{}0\\ 0&{}0&{}0&{}1 \end{matrix}\right) \end{aligned}$$

and

$$\begin{aligned} \rho (\sigma _j)= \left( \begin{matrix} \omega _1\omega _2d_1d_2&{}\omega _1d_1T_1^{j-1}&{}\omega _2d_2T_1^{j-1}&{}(T_1)^{2j-2}\\ \omega _1d_1c_2T_1^{1-j}&{}\omega _1d_1d_2&{}c_2&{}d_2T_1^{j-1}\\ \omega _2c_1d_2T_1^{1-j}&{}c_1&{}\omega _2d_2d_1&{}d_1T_1^{j-1}\\ c_1c_2(T_1)^{2-2j}&{}c_1d_2T_1^{1-j}&{}d_1c_2T_1^{1-j}&{}d_1d_2 \end{matrix}\right) \end{aligned}$$

for $j=1,2,3$. The eigenvalues of $\rho (\tau )$ are then $\lambda _2=t_1t_2e^{\pm i2\pi /3}$ (of multiplicity 2) and $\lambda _1=e^{\mp i2\pi /3}t_1t_2$ and $\lambda _3=t_1t_2$. The corresponding eigenvectors are $xe_2+ye_3$ for $\lambda _2$ and $e_1$ and $e_4$ for $\lambda _1$ and $\lambda _3$, respectively.

Lemma 3.13

If $T_1=T_2=e^{\pm 2\pi i/3},\omega _1=e^{\pm i\pi /3}$ and $\omega _2=e^{\pm i\pi /3}$ then the representation $\rho$ of ${\mathcal {N}}{\mathcal {B}}_3$ has no invariant proper subspaces of dimension 1 if and only if $\omega _1\ne \omega _2$ or $\omega _1d_1^2c_2\ne \omega _2d_2^2c_1$.

Proof

By direct calculations, we check that each of the subspaces spanned by the vectors $e_j$ are not invariant under $\rho (\sigma _1)$.

It remains to consider the subspace S spanned by the vector $v=e_2+xe_3$ with $x\in {\mathbb {C}}^*$. Since $T_1=T_2$, it follows that

$$\begin{aligned} \rho (\sigma _j)(v)=\left( \begin{matrix} \omega _1d_1T_1^{j-1}+\omega _2d_2T_1^{j-1}x\\ \omega _1d_1d_2+c_2x\\ c_1+\omega _2d_2d_1x\\ c_1d_2T_1^{1-j}+d_1c_2T_1^{1-j}x \end{matrix}\right) = \left( \begin{matrix} (\omega _1d_1+\omega _2d_2x)T_1^{j-1}\\ \omega _1d_1d_2+c_2x\\ c_1+\omega _2d_2d_1x\\ (c_1d_2+d_1c_2x)T_1^{1-j} \end{matrix}\right) . \end{aligned}$$

So

$$\begin{aligned} \rho (\sigma _j)(v)\in S&\Leftrightarrow \rho (\sigma _j)(v)=\lambda v \text { for some }\lambda \in {\mathbb {C}} \\&\Leftrightarrow x=\frac{-\omega _1d_1}{\omega _2d_2}=\frac{-c_1d_2}{d_1c_2} \text { and } c_2 x^2+(\omega _1-\omega _2)d_1d_2x-c_1=0\\&\Leftrightarrow \omega _1d_1^2c_2=\omega _2d_2^2c_1\text { and } c_2\omega _1^2d_1^2+(\omega _2-\omega _1)d_1^2d_2^2\omega _1\omega _2-c_1\omega _2^2d_2^2=0\\&\Leftrightarrow \omega _1d_1^2c_2=\omega _2d_2^2c_1\text { and } (\omega _1-\omega _2)(c_2-\omega _2d_2^2)=0\\&\Leftrightarrow \omega _1d_1^2c_2=\omega _2d_2^2c_1\text { and } \omega _1=\omega _2\text { because } c_2\ne \omega _2d_2^2 \end{aligned}$$

Therefore the subspace S, which is spanned by $v=e_2-\frac{\omega _1d_1}{\omega _2d_2}e_3$, is invariant under $\rho (\sigma _j)$ if $\omega _1=\omega _2$ and $\omega _1d_1^2c_2= \omega _2d_2^2c_1$. This gives a contradiction. $\square$

Lemma 3.14

If $T_1=T_2=e^{\pm 2\pi i/3}$ and $\omega _1=e^{\pm i\pi /3}$ and $\omega _2=e^{\pm i\pi /3}$, then the representation $\rho$ of ${\mathcal {N}}{\mathcal {B}}_3$ has no invariant proper subspaces of dimension 2.

Proof

The possible two-dimensional invariant subspaces candidates to study are the following:

• 1. $\langle e_i,e_j\rangle$ for $i\ne j$

• 2. $\langle e_1,e_2+xe_3\rangle$ for $x\ne 0$

• 3. $\langle e_4,e_2+ye_3\rangle$ for $y\ne 0$

Since $c_1\ne 0, c_2\ne 0$, $d_1\ne 0$ and $d_2\ne 0$, it follows that none of the subspaces mentioned above is invariant under $\rho (\sigma _1).$ $\square$

Lemma 3.15

If $T_1=T_2=e^{\pm 2\pi i/3}$, $\omega _1=e^{\pm i\pi /3}$ and $\omega _2=e^{\pm i\pi /3}$, then the representation $\rho$ of ${\mathcal {N}}{\mathcal {B}}_3$ has no invariant proper subspaces of dimension 3 if and only if $\omega _1\ne \omega _2$ or $\omega _1d_1^2c_2\ne \omega _2d_2^2c_1$.

Proof

The possible invariant three-dimensional invariant subspaces to study are:

• 1. $\langle e_i,e_j,e_k\rangle$ where i, j, k are pairwise distinct

• 2. $\langle e_1,e_4,e_2+xe_3\rangle$ for $x\ne 0$

By direct computations, it is easy to check that all the subspaces in case 1 are not invariant under $\rho (\sigma _1)$.

In case 2, we let S be a subspace of the form $\langle e_1,e_4,e_2+xe_3\rangle$ with $x\ne 0$.

$$\begin{aligned} \rho (\sigma _j)(e_1)=\left( \begin{matrix} \omega _1\omega _2d_1d_2\\ \omega _1d_1c_2T_1^{1-j}\\ \omega _2c_1d_2T_1^{1-j}T_2^{j-1}\\ c_1c_2T_1^{2-2j} \end{matrix}\right) , ~\rho (\sigma _j)(e_4)=\left( \begin{matrix} T_1^{2j-2}\\ d_2T_1^{j-1}\\ d_1T_1^{j-1}\\ d_1d_2 \end{matrix}\right) \end{aligned}$$

and

$$\begin{aligned} \rho (\sigma _j)(e_2+xe_3)=\left( \begin{matrix} \omega _1d_1T_1^{j-1}+\omega _2d_2T_1^{j-1}x\\ \omega _1d_1d_2+c_2x\\ c_1+\omega _2d_2d_1x\\ c_1d_2T_1^{1-j}+d_1c_2T_1^{1-j}x \end{matrix}\right) . \end{aligned}$$

Therefore, we get

$$\begin{aligned} \rho (\sigma _j)(e_1)\in S\Leftrightarrow x=\dfrac{\omega _2c_1d_2}{\omega _1d_1c_2} \end{aligned}$$

and

$$\begin{aligned}&\rho (\sigma _j)(e_4)\in S\Leftrightarrow x=\dfrac{d_1}{d_2},\\&\quad rho(\sigma _j)(e_2+xe_3)\in S \Leftrightarrow \begin{pmatrix} \omega _1d_1T_1^{j-1}+\omega _2d_2T_1^{j-1}x\\ \omega _1d_1d_2+c_2x\\ c_1+\omega _2d_2d_1x\\ c_1d_2T_1^{1-j}+d_1c_2T_1^{1-j}x \end{pmatrix}= \left( \begin{matrix} \alpha \\ \lambda \\ \lambda x\\ \beta \end{matrix}\right) \text { for some } \alpha , \beta , \gamma \in {\mathbb {C}}. \end{aligned}$$

This is equivalent to

$$\begin{aligned} c_2x^2+(\omega _1-\omega _2)d_1d_2x-c_1=0. \end{aligned}$$

Therefore, S is invariant under $\rho (\sigma _j)$ ($j=1,2,3$) if and only if:

$$\begin{aligned} x=\dfrac{d_1}{d_2}=\dfrac{\omega _2c_1d_2}{\omega _1d_1c_2} \end{aligned}$$

(3.5)

and

$$\begin{aligned} c_2d_1^2+(\omega _1-\omega _2)d_1^2d_2^2-c_1d_2^2=0. \end{aligned}$$

(3.6)

The Eqs. (3.5) and (3.6) imply that

$$\begin{aligned} \omega _1d_1^2c_2=\omega _2d_2^2c_1 \end{aligned}$$

and

$$\begin{aligned} (\omega _1-\omega _2)(\omega _1d_1^2-c_1)=0. \end{aligned}$$

Since $\omega _1d_1^2\ne c_1$, it follows that $\omega _1=\omega _2$, which gives a contradiction. Therefore, the subspace S, that is spanned by $v=e_2-\frac{d_1}{d_2}e_3$, is invariant under $\rho (\sigma _j)$ if $\omega _1=\omega _2$ and $\omega _1d_1^2c_2= \omega _2d_2^2c_1$. $\square$

Proposition 3.16

If $T_1=T_2=e^{\pm 2\pi i/3}$, $\omega _1=e^{\pm i\pi /3}$ and $\omega _2=e^{\pm i\pi /3}$ then $\rho$ is an irreducible representation of ${\mathcal {N}}{\mathcal {B}}_3$ if and only if $\omega _1\ne \omega _2$ or $\omega _2d_2^2c_1\ne \omega _1d_1^2c_2$.

Proof

The proof follows directly from Lemmas 3.13, 3.14 and 3.15. $\square$

Theorem 3.17

The representation $\rho$ of ${\mathcal {N}}{\mathcal {B}}_3$ is irreducible if and only if one of the following conditions hold.

• 1. $T_1=e^{2\pi i/3}, T_2=e^{-2\pi i/3}$ and $\omega _1\omega _2\ne 1$

• 2. $T_1=e^{2\pi i/3}, T_2=e^{-2\pi i/3}$ and $\omega _1d_1^2c_2\ne \omega _2d_2^2c_1$

• 3. $T_1=T_2=e^{\pm 2\pi i/3}$ and $\omega _1\ne \omega _2$

• 4. $T_1=T_2=e^{\pm 2\pi i/3}$ and $\omega _1d_1^2c_2\ne \omega _2d_2^2c_1$

Proof

The proof follows directly from Propositions 3.12 and 3.16$\square$

3.3 Representation of ${\mathcal {N}}{\mathcal {B}}_2$

In this section, we consider the representation $\rho$ given by Definition 3.1. In this case, we have $T_1=T_2=-1$. A similar work to that done for ${\mathcal {N}}{\mathcal {B}}_4$ (Theorem 3.5), we get the following theorem.

Theorem 3.18

The representation $\rho$ is irreducible if and only if $a_1a_2\ne d_1d_2$ and $a_1d_2\ne a_2d_1$.

4 Unitary representations of ${\mathcal {N}}{\mathcal {B}}_n$ of dimension 2 ($n=2,3,4$)

Definition 4.1

A square matrix A is unitary relative to a matrix M if $AMA^*=M$. Here, $A^*$ is the conjugate transpose of A.

Definition 4.2

A representation $\rho :G\rightarrow GL(n,{\mathbb {C}})$ of a group G is called unitary if $\rho (g)$ is unitary for every element $g\in G$.

We consider the irreducible representations of ${\mathcal {N}}{\mathcal {B}}_n$, given by Proposition 2.1. Then, we determine a necessary and sufficient condition under which these representations are unitary relative to some hermitian positive definite matrix.

Recall that

$$\begin{aligned} \rho (\tau )=\left( \begin{matrix} Tt&{}0\\ 0&{}t \end{matrix}\right) , ~ \rho (\sigma _j)=\begin{pmatrix} a&{}T^{j-1}\\ cT^{1-j}&{}d \end{pmatrix}, \end{aligned}$$

where t is a (2n)th root of unity and $T\in \{-1,i,-i,e^{i2\pi /3},e^{-i2\pi /3}\}$.

Lemma 4.3

Let $\rho$ be a complex irreducible representation of ${\mathcal {N}}{\mathcal {B}}_n$ ($n=2,3,4$), given by Proposition 2.1. If $\rho$ is unitary relative to a matrix $M=\left( \begin{matrix} x&{}y\\ z&{}u \end{matrix}\right)$, then $y=z=0$ and

$$\begin{aligned} \left\{ \begin{array}{l} (|a|^2-1)x+u=0\\ |c|^2x+(|d|^2-1)u=0\\ a{\bar{c}}x+{\bar{d}}u=0 \end{array}\right. . \end{aligned}$$

Proof

Since $\rho$ is unitary relative to a matrix $M=\left( \begin{matrix} x&{}y\\ z&{}u \end{matrix}\right)$, we have $\rho (\tau )M\rho (\tau )^*=M$. Thus, we get

$$\begin{aligned} \left( \begin{matrix} Tt&{}0\\ 0&{}t \end{matrix}\right) \left( \begin{matrix} x&{}y\\ z&{}u \end{matrix}\right) \left( \begin{matrix} {\bar{T}}{\bar{t}}&{}0\\ 0&{}{\bar{t}} \end{matrix}\right) =\left( \begin{matrix} x&{}y\\ z&{}u \end{matrix}\right) . \end{aligned}$$

Then, we get

$$\begin{aligned} \left( \begin{matrix} x&{}Ty\\ {\bar{T}}z&{}u \end{matrix}\right) = \begin{pmatrix} x&{}y\\ z&{}u \end{pmatrix}. \end{aligned}$$

It follows that

$$\begin{aligned} Ty=y \text { and } {\bar{T}}z=z. \end{aligned}$$

Therefore, $y=z=0$ because $T\ne 1$.

We also have $\rho (\sigma _j)M\rho (\sigma _j)^*=M$ ($j=1,2,3,4$).

Then, we get

$$\begin{aligned} \begin{pmatrix} a&{}T^{j-1}\\ cT^{1-j}&{}d \end{pmatrix} \begin{pmatrix} x&{}0\\ 0&{}u \end{pmatrix} \begin{pmatrix} {\bar{a}}&{}{\bar{c}}{\bar{T}}^{1-j}\\ {\bar{T}}^{j-1}&{}{\bar{d}} \end{pmatrix} =\begin{pmatrix} x&{}0\\ 0&{}u \end{pmatrix}. \end{aligned}$$

So we have

$$\begin{aligned} \begin{pmatrix} |a|^2x+u&{}a{\bar{c}}{\bar{T}}^{1-j}x+{\bar{d}}T^{j-1}u\\ c{\bar{a}}T^{1-j}x+d{\bar{T}}^{j-1}u&{}|c|^2x+|d|^2u \end{pmatrix} =\begin{pmatrix} x&{}0\\ 0&{}u \end{pmatrix}. \end{aligned}$$

Since ${\bar{T}}=T^{-1}$, it follows that

$$\begin{aligned} \begin{pmatrix} |a|^2x+u&{}(a{\bar{c}}x+{\bar{d}}u)T^{j-1}\\ (c{\bar{a}}x+du)T^{1-j}&{}|c|^2x+|d|^2u \end{pmatrix} =\begin{pmatrix} x&{}0\\ 0&{}u \end{pmatrix}. \end{aligned}$$

Then we get

$$\begin{aligned} \left\{ \begin{array}{l} (|a|^2-1)x+u=0\\ |c|^2x+(|d|^2-1)u=0\\ a{\bar{c}}x+{\bar{d}}u=0 \end{array}\right. . \end{aligned}$$

$\square$

Proposition 4.4

Let $\rho :{{\mathcal {N}}{\mathcal {B}}}_4\rightarrow GL(2,{\mathbb {C}})$ be the irreducible representation of ${{\mathcal {N}}{\mathcal {B}}}_4$ given in Proposition 2.1$(T=-1)$. Then, $\rho$ is unitary relative to a hermitian positive definite matrix M if and only if

$$\begin{aligned} |a|<1 \end{aligned}$$

and

$$\begin{aligned} a=d+e^{i\theta }, \left\{ \begin{array}{ll} d=\delta '-\frac{1}{2}i&{}\text { if }\theta =\frac{\pi }{2}\text { mod }2\pi \\ d=\delta '+\frac{1}{2}i&{}\text { if }\theta =-\frac{\pi }{2}\text { mod }2\pi \\ d=-\frac{1}{2}\sec \theta -(\tan \theta ) \delta +i\delta ,&{} \text { if }\theta \ne \pm \frac{\pi }{2} \text { mod }2\pi \end{array}\right. , \end{aligned}$$

where $\theta ,\delta ',\delta \in {\mathbb {R}}$

Proof

Recall that

$$\begin{aligned} \rho (\tau )=\left( \begin{matrix} -t&{}0\\ 0&{}t \end{matrix}\right) ~\text {and }~ \rho (\sigma _j)=\left( \begin{matrix} a&{}(-1)^{j-1}\\ c(-1)^{1-j}&{}d \end{matrix}\right) , \end{aligned}$$

where t is an eighth root of unity and $c=a^2-ad+d^2$ such that $a\ne d$ and $c\ne 0$. Suppose that $\rho$ is unitary relative to matrix M. By Lemma 4.3, we have $M=\left( \begin{matrix} x&{}0\\ 0&{}u \end{matrix}\right)$ and

$$\begin{aligned} \left\{ \begin{array}{l} (|a|^2-1)x+u=0\\ |c|^2x+(|d|^2-1)u=0\\ a{\bar{c}}x+{\bar{d}}u=0 \end{array}\right. . \end{aligned}$$

To have a non-trivial solution of the first two equations of the system above, the determinant of the system is zero and so

$$\begin{aligned} (|a|^2-1)(|d|^2-1)=|c|^2. \end{aligned}$$

(4.1)

The first equation yields

$$\begin{aligned} u=(1-|a|^2)x. \end{aligned}$$

(4.2)

From the third equation of the last system, we conclude that $ad\ne 0$ and

$$\begin{aligned} x=-\dfrac{{\bar{d}}u}{a{\bar{c}}}. \end{aligned}$$

(4.3)

The Eq. (4.2) and the Eq. (4.3) imply that

$$\begin{aligned} a{\bar{c}}=(|a|^2-1){\bar{d}}. \end{aligned}$$

(4.4)

Using the fact $c=a^2-ad+d^2$ and the Eqs.  (4.1) and (4.4), we get

$$\begin{aligned} {\bar{d}}=-a({\bar{a}}-{\bar{d}})^2\text { and } {\bar{d}}(a-d)^2=-a. \end{aligned}$$

Thus, we get

$$\begin{aligned} |a-d|^2=1\text { and }\frac{a}{{\bar{d}}}=-(a-d)^2. \end{aligned}$$

Hence, there exists $\theta \in {\mathbb {R}}$ such that

$$\begin{aligned} a-d=e^{i\theta }\text { and } a=-{\bar{d}}e^{2i\theta }. \end{aligned}$$

Therefore we get

$$\begin{aligned} a=d+e^{i\theta }\text { and }{\bar{d}}e^{2i\theta }+e^{i\theta }+d=0. \end{aligned}$$

Set $d=\delta '+i\delta$ where $\delta ',\delta \in {\mathbb {R}}$. Then

$$\begin{aligned} (1+e^{2i\theta })\delta '+i\delta (1-e^{2i\theta })+e^{i\theta }=0. \end{aligned}$$

(4.5)

To solve the Eq. (4.5), consider two cases: (1) $1+e^{2i\theta }=0$, (2) $1+e^{2i\theta }\ne 0$.

If $1+e^{2i\theta }=0$ then $d=\delta '\mp \dfrac{i}{2}$ and $a=\delta '\pm \dfrac{i}{2}$.

If $1+e^{2i\theta }\ne 0$ then $d=-\dfrac{1}{2}\sec \theta -(\tan \theta )\delta +\delta i$.

Therefore, we have

$$\begin{aligned} a=d+e^{i\theta }\text { and } \left\{ \begin{array}{ll} d=\delta '-\frac{1}{2}i&{}\text { if }\theta =\frac{\pi }{2}\text { mod }2\pi \\ d=\delta '+\frac{1}{2}i&{}\text { if }\theta =-\frac{\pi }{2}\text { mod }2\pi \\ d=-\frac{1}{2}\sec \theta -(\tan \theta ) \delta +i\delta ,&{} \text { if }\theta \ne \pm \frac{\pi }{2} \text { mod }2\pi \end{array}\right. , \end{aligned}$$

where $\theta ,\delta ',\delta \in {\mathbb {R}}$.

The condition $|a|<1$ follows directly from the Eq. (4.2) and the fact that M is positive definite. The matrix M becomes $\begin{pmatrix} 1&{}0\\ 0&{}1-|a|^2 \end{pmatrix}$ up to a constant. $\square$

Proposition 4.5

Let $\rho :{{\mathcal {N}}{\mathcal {B}}}_4\rightarrow GL(2,{\mathbb {C}})$ be the irreducible representation of ${{\mathcal {N}}{\mathcal {B}}}_4$ given by Proposition 2.1$(T=\pm i)$. The representation $\rho$ is unitary relative to a hermitian positive definite matrix M if and only if

$$\begin{aligned} |d|^2=\dfrac{1}{2}. \end{aligned}$$

Proof

Recall that

$$\begin{aligned} \rho (\tau )=\left( \begin{matrix} \pm it&{}0\\ 0&{}t \end{matrix}\right) ~\text {and }~ \rho (\sigma _j)=\left( \begin{matrix} d&{}(\pm i)^{j-1}\\ -d^2(\pm i)^{1-j}&{}d \end{matrix}\right) , \end{aligned}$$

where t is an eighth root of unity and $d\ne 0$.

Suppose that $\rho$ is unitary relative to hermitian positive definite matrix M. Then, by Lemma 4.3, we have $M=\left( \begin{matrix} x&{}0\\ 0&{}u \end{matrix}\right)$ and

$$\begin{aligned} \left\{ \begin{array}{l} (|a|^2-1)x+u=0\\ |c|^2x+(|d|^2-1)u=0\\ a{\bar{c}}x+{\bar{d}}u=0 \end{array}\right. . \end{aligned}$$

Since $a=d$ and $c=-d^2$, it follows that

$$\begin{aligned} \left\{ \begin{array}{l} (|d|^2-1)x+u=0\\ |d|^4x+(|d|^2-1)u=0\\ -d{\bar{d}}^2x+{\bar{d}}u=0 \end{array}\right. . \end{aligned}$$

To have a non-trivial solution of the first two equations of the last system, the determinant of the system is zero and so

$$\begin{aligned} (|d|^2-1)(|d|^2-1)=|d|^4. \end{aligned}$$

The first and third equations of the last system imply that

$$\begin{aligned} x=\frac{u}{d{\bar{d}}}=\frac{u}{1-|d|^2}. \end{aligned}$$

Then, we get

$$\begin{aligned} |d|^2=\frac{1}{2} \end{aligned}$$

and the matrix M becomes $\begin{pmatrix} 2&{}0\\ 0&{}1 \end{pmatrix}$ up to a constant. $\square$

Proposition 4.6

Let $\rho :{{\mathcal {N}}{\mathcal {B}}}_3\rightarrow GL(2,{\mathbb {C}})$ be the irreducible representation of ${{\mathcal {N}}{\mathcal {B}}}_4$ that is defined, in Proposition 2.1$(T=e^{\pm 2\pi i/3})$. The representation $\rho$ is unitary relative to a hermitian positive definite matrix M if and only if $|d|<1$ and

$$\begin{aligned} c=\dfrac{\omega d(|d|^2-1)}{{\bar{d}}}. \end{aligned}$$

Proof

Recall that

$$\begin{aligned} \rho (\tau )=\left( \begin{matrix} e^{\pm 2\pi i/3}t&{}0\\ 0&{}t \end{matrix}\right) ~\text {and }~ \rho (\sigma _j)=\left( \begin{matrix} \omega d&{}e^{\pm (j-1)2\pi i/3}\\ e^{\pm (1-j)2\pi i/3}c&{}d \end{matrix}\right) , \end{aligned}$$

where t is a sixth root of unity and $\omega =e^{\pm i\pi /3}$ such that $c\ne 0$.

Suppose that $\rho$ is unitary relative to a matrix M. Then, by Lemma 4.3, we have $M=\left( \begin{matrix} x&{}0\\ 0&{}u \end{matrix}\right)$ and

$$\begin{aligned} \left\{ \begin{array}{l} (|a|^2-1)x+u=0\\ |c|^2x+(|d|^2-1)u=0\\ a{\bar{c}}x+{\bar{d}}u=0 \end{array}\right. . \end{aligned}$$

Since $a=\omega d$, it follows that

$$\begin{aligned} \left\{ \begin{array}{l} (|d|^2-1)x+u=0\\ |c|^2x+(|d|^2-1)u=0\\ \omega d{\bar{c}}x+{\bar{d}}u=0 \end{array}\right. . \end{aligned}$$

So

$$\begin{aligned} x=-\frac{{\bar{d}}u}{\omega d{\bar{c}}}=\frac{u}{1-|d|^2} \end{aligned}$$

(4.6)

Since $\omega ^{-1}={\bar{\omega }}$, it follows that

$$\begin{aligned} c=\frac{\omega d(|d|^2-1)}{{\bar{d}}}. \end{aligned}$$

The condition $|d|<1$ follows from the Eq. (4.6) and the fact that M is a positive definite. The matrix M becomes $\left( \begin{matrix} 1&{}0\\ 0&{}1-|d|^2 \end{matrix}\right)$ up to a constant. $\square$