Braid Group Action on Projective Quantum $\mathfrak {sl}(2)$ Modules

Abstract

We define a family of braid group representations via the action of the R-matrix (of the quasitriangular extension) of the restricted quantum $\mathfrak {sl}(2)$ on a tensor power of a simple projective module. This family is an extension of the Lawrence representation specialized at roots of unity. Although the center of the braid group has finite order on the specialized Lawrence representations, this action is faithful for our extension.

Appendix: A. Proofs of Various Statements

1.1 A.1 Action of R-Matrix on Weight Modules

We show that the action of the R-matrix (2.2) on any (weight) module V of D is given by (2.13)

$$ R = q^{H \otimes H/2} \sum\limits_{n=0}^{r-1} \frac{\{1\}^{2n}}{\{n\}!} q^{n(n-1)/2} \left( E^{n} \otimes F^{n} \right). $$

Due to (2.1) we have that Eⁱk = q⁻ⁱkEⁱ and Fⁱk = qⁱkFⁱ for any $i \in \mathbb N$. Therefore (2.2) can be written as $R = C \circ \widetilde {R}$, where

$$ C = \frac{1}{4r} \sum\limits_{m,m^{\prime}=0}^{4r-1} q^{- m m^{\prime} / 2} k^{m} \otimes k^{m^{\prime}} \quad\text{and}\quad \widetilde{R} = \sum\limits_{n=0}^{r-1} \frac{\{1\}^{2n}}{\{n\}!} q^{n(n-1)/2} \left( E^{n} \otimes F^{n} \right). $$

Therefore, it remains to prove that C = q^H⊗H/2 (2.12) as operators acting on V ⊗V. Let $-2r \leq \lambda , \lambda ^{\prime } < 2r$ and let v and w be weight vectors of V, such that kv = q^λ/2v and $k w = q^{\lambda ^{\prime }/2} w$. We have that

$$ \begin{array}{@{}rcl@{}} 4r C(v \otimes w) = \sum\limits_{m,m^{\prime}=0}^{4r-1} q^{- m m^{\prime} / 2} k^{m} v \otimes k^{m^{\prime}} w &=& \sum\limits_{m,m^{\prime}=0}^{4r-1} q^{(- m m^{\prime} + m \lambda + m^{\prime} \lambda^{\prime}) /2} v \otimes w \\ &=& \sum\limits_{m=0}^{4r-1} q^{m \lambda /2} \sum\limits_{m^{\prime}=0}^{4r-1} q^{\frac{\lambda^{\prime}-m}{2}m^{\prime}} v \otimes w. \end{array} $$

Note that $q^{\frac {\lambda ^{\prime }-m}{2}}=1$ if and only if $\lambda ^{\prime }-m = 0 \mod 4r$, since q is a primitive 2r th root of unity. Since 0 ≤m ≤ 4r − 1 this happens only for one such m. Denote this value of m by $m_{0} = 4 \mu r + \lambda ^{\prime }$ for some $\mu \in \mathbb Z$. Then at the second summation (over $m^{\prime }$) for all $m^{\prime } \neq m_{0}$ the corresponding summand equals 0, since it is multiple of the sum of all $4r/\gcd (4r,\lambda ^{\prime }-m)$-roots of unity. Therefore

$$ \begin{array}{@{}rcl@{}} \sum\limits_{m,m^{\prime}=0}^{4r-1} q^{- m m^{\prime} / 2} k^{m} v \otimes k^{m^{\prime}} w &=& 4r q^{m_{0} \lambda/2} q^{(\lambda^{\prime} - m_{0}) /2} v \otimes w \\ &=& 4r q^{\lambda \lambda^{\prime} /2} v \otimes w = 4r q^{H \otimes H/2} (v \otimes w), \end{array} $$

that is, C = q^H⊗H/2 on V ⊗V.

1.2 A.2 R-Matrix on $\mathsf {V}_{r-1}^{\otimes 2}$

Applying the R-matrix (2.13) composed with the permutation operator τ on a vector u_i ⊗u_j of $\mathsf {V}_{r-1}^{\otimes 2}$ we get

$$ \begin{array}{@{}rcl@{}} (\tau \circ R)(u_{i} \otimes u_{j}) = q^{H \otimes H /2}{\sum}_{n=0}^{r-1} \frac{\{1\}^{2n}}{\{n\}!} q^{n(n-1)/2} F^{n} u_{j} \otimes E^{n} u_{i}. \end{array} $$

Note that Eⁿu_i≠ 0 if n < i + 1 and Fⁿu_j≠ 0 if n < r −j. Hence, the summation is up to ${\min \limits } (i,r-j-1)$ and all other terms are zero. Substituting the action of E and F we get

$$ \begin{aligned} (\tau \circ R)(u_{i} \otimes u_{j}) =& \sum\limits_{n=0}^{\min (i,r-j-1)} \left( {\vphantom{\prod\limits_{m=0}^{n-1}}}\frac{\{1\}^{2n}}{\{n\}!} q^{\frac{n(n-1)}{2}} \right.\\ &\left.\cdot \prod\limits_{m=0}^{n-1} [m+j+1] []r - 1 -(m+j)] q^{H \otimes H/2} (u_{j+n} \otimes u_{i-n}) \right). \end{aligned} $$

(A.1)

It holds that

$$ \frac{\{1\}^{n}}{\{n\}!} \prod\limits_{m=0}^{n-1} [m+j+1] = ([n]!)^{-1} \prod\limits_{m=1}^{n} [m+j] = \frac{[n+j]!}{[j]! [n]!} = \begin{bmatrix}{n+j}\\ {j}\end{bmatrix}. $$

(A.2)

Using the equality {1}⋅ [r − 1 − (m + j)] = −{− 1 − (m + j)} = {m + j + 1} and substituting (A.2) into (A.1) we get

$$ (\tau \circ R)(u_{i} \otimes u_{j}) = {\sum}_{n=0}^{\min (i,r-j-1)} q^{n(n-1)/2} \begin{bmatrix}{n+j} \\ {j}\end{bmatrix} {\prod}_{m=0}^{n-1} \{m+j+1\} q^{H \otimes H /2} (u_{j+n} \otimes u_{i-n}). $$

(A.3)

We set s := q^r− 1 = −q^− 1 (2.5). The action of q^H⊗H/2 on u_j+n ⊗u_i−n is as follows

$$ \begin{array}{@{}rcl@{}} q^{H \otimes H /2} (u_{j+n} \otimes u_{i-n}) &=& q^{\frac{(r - 1 -2(i-n))(r - 1 -2(j+n))}{2}} u_{j+n} \otimes u_{i-n} \\ &=& q^{\frac{(r - 1)^{2}}{2}} q^{-(r - 1) (i+j)} q^{2(i-n)(j+n)} u_{j+n} \otimes u_{i-n} \\ &=& q^{\frac{(r - 1)^{2}}{2}} s^{-(i+j)} q^{2(i-n)(j+n)} u_{j+n} \otimes u_{i-n}. \end{array} $$

The factor $q^{\frac {(r - 1)^{2}}{2}}$ is annihilated by the action of the operator R defined as in (3.1). Hence, using also (A.3), the action of R on the vector u_i ⊗u_j of $\mathsf {V}_{r-1}^{\otimes 2}$ is

$$ \begin{array}{@{}rcl@{}} \mathsf{R}(u_{i} \otimes u_{j}) = s^{-(i+j)} \sum\limits_{n=0}^{\min (i,r-j-1)} q^{2(i-n)(j+n)} q^{n(n-1)/2} \begin{bmatrix} {n+j} \\ {j} \end{bmatrix} \prod\limits_{m=0}^{n-1} \{m+j+1\} u_{j+n} \otimes u_{i-n}. \end{array} $$

1.3 A.3 Proof of Lemma 3.8

Let a_ε ∈A_n,ℓ as in (3.6) and b ∈B_n,ℓ. By the definition of the map Φ (3.9) we have that (Φ− 1)(b) = 0. Moreover, for m = 1 it holds b_ε,1 = 1 and hence, by (3.6) we get (Φ− 1)(a_ε) ∈B_n,ℓ. Therefore (Φ− 1)²(a_ε) = 0.

1.4 A.4 Proof of Lemma 3.9

Let 1 ≤ℓ < r. We first show that $E \circ \varPhi \vert _{\boldsymbol {A}_{n,\ell }} = 0$. We have that

$$ \begin{array}{@{}rcl@{}} E \circ \varPhi(a_{\boldsymbol{\varepsilon}}) &=& E \left( \sum\limits_{m=0}^{\ell} b_{\boldsymbol{\varepsilon},m} u_{0}^{\otimes j - 2} \otimes u_{m} \otimes E^{m-1} u_{\boldsymbol{\varepsilon}} \right) \\ &=& \sum\limits_{m=0}^{\ell} b_{\boldsymbol{\varepsilon},m} u_{0}^{\otimes j - 2} \otimes u_{m} \otimes E^{m} u_{\boldsymbol{\varepsilon}} + \sum\limits_{m=0}^{\ell} b_{\boldsymbol{\varepsilon},m} u_{0}^{\otimes j - 2} \otimes u_{m-1} \otimes K E^{m-1} u_{\boldsymbol{\varepsilon}}. \end{array} $$

Note that u_ε ∈V_{n−j− 1,ℓ− 1} by (3.6), therefore E^ℓu_ε = 0 and the summand for m = ℓ at the first sum is zero. Further, at the second sum for m = 0 we have that u_m− 1 = 0. Finally, KE^m− 1u_ε = s^n−j+ 1q^{− 2(ℓ−m)}E^m− 1u_ε since $E^{m-1} u_{\boldsymbol {\varepsilon }} \in \boldsymbol {V}_{n-j-1, \ell - m}$. Putting it all together, we have

$$ \begin{array}{@{}rcl@{}} E \circ \varPhi(a_{\boldsymbol{\varepsilon}}) &=& \sum\limits_{m=0}^{\ell-1} b_{\boldsymbol{\varepsilon},m} u_{0}^{\otimes j - 2} \otimes u_{m} \otimes E^{m} u_{\boldsymbol{\varepsilon}} \\ & +& \sum\limits_{m=1}^{\ell} s^{n-j+1} q^{-2(\ell-m-1)} b_{\boldsymbol{\varepsilon},m} u_{0}^{\otimes j - 2} \otimes u_{m-1} \otimes E^{m-1} u_{\boldsymbol{\varepsilon}} \\ &=& \sum\limits_{m=1}^{\ell} (b_{\boldsymbol{\varepsilon},m} + s^{n-j+1} q^{-2(\ell-m-1)} b_{\boldsymbol{\varepsilon},m+1}) u_{0}^{\otimes j - 2} \otimes u_{m} \otimes E^{m} u_{\boldsymbol{\varepsilon}}. \end{array} $$

But b_ε,m + s^n−j+ 1q^{− 2(ℓ−m− 1)}b_ε,m+ 1 = 0 and therefore, E∘Φ = 0 on A_n,ℓ. Now, by definition, Φ is the identity on B_n,ℓ. Hence, $E \circ \varPhi = 0 \oplus E \vert _{\boldsymbol {B}_{n,\ell }}$.

By Lemma 3.7 $E \vert _{\boldsymbol {B}_{n, \ell }}$ is injective for all ℓ ≥ 1 and by Lemma 3.8 the map Φ is an automorphism of V_n,ℓ. Therefore, $\ker (E \circ \varPhi ) \cap \boldsymbol {V}_{n,\ell } = \ker E \cap \boldsymbol {V}_{n,\ell } = \boldsymbol {A}_{n,\ell }$. Hence, due to Lemma 3.8 we conclude that A_n,ℓ≅W_n,ℓ.

1.5 A.5 Proof of (4.1) and (4.2)

Let w be a weight vector of P_i of weight $q^{(i + 2m^{\prime })/2}$, where $m^{\prime } \in \{-r+1, \ldots , j+1\}$. Since $E u_{0}^{\alpha } = F u_{0}^{\alpha } = 0$, by (2.13) we have that

$$ c_{\mathsf{V}^{\alpha}_{0}, \mathsf{P}_{i}} \left( u_{0}^{\alpha} \otimes w\right) = \tau \circ R \left( u_{0}^{\alpha} \otimes w\right) = \tau \left( q^{H \otimes H /2} u_{0}^{\alpha} \otimes w\right) = q^{\frac{mr (i + 2m^{\prime})}{2}} w \otimes u_{0}^{\alpha}. $$

Similarly, we get

$$ c_{\mathsf{P}_{i}, \mathsf{V}^{\alpha}_{0}} (w \otimes u_{0}^{\alpha}) = q^{\frac{mr (i + 2m^{\prime})}{2}} u_{0}^{\alpha} \otimes w. $$

After combining the two equations, we finally have

$$ c_{\mathsf{P}_{i}, \mathsf{V}^{\alpha}_{0}} \circ c_{\mathsf{V}^{\alpha}_{0}, \mathsf{P}_{i}} (u_{0}^{\alpha} \otimes w) = q^{mr(i+2m^{\prime})} u_{0}^{\alpha} \otimes w = q^{mri} u_{0}^{\alpha} \otimes w, $$

which proves (4.1).

For the calculations involving the twist operator, we use the ribbon element as given by [30] and [10], that is

$$ \theta = K^{r-1} \sum\limits_{n=0}^{r-1} \frac{\{1\}^{2n}}{\{n\}!} q^{n(n-1)/2} S(F^{n}) q^{-H^{2}/2} E^{n}, $$

(A.4)

where the operator $q^{-H^{2}/2}$ is defined as $q^{-H^{2}/2} v = q^{-\lambda ^{2} / 2} v$ for a weight vector v, where λ is the strong weight of v (2.10). Since $E u_{0}^{\alpha } = F u_{0}^{\alpha } = 0$, by (A.4) we have that

$$ \theta (u_{0}^{\alpha}) = K^{r-1} q^{-H^{2}/2} u_{0}^{\alpha} = q^{mr(r-1) - \frac{(mr)^{2}}{2}} u_{0}^{\alpha}. $$

Since the action of the twist operator $\theta _{\mathsf {V}_{0}^{\alpha }}$ is defined by the action of 𝜃^− 1, (4.2) is proved.

1.6 A.6 Proof of (4.3)

By the naturality of the twist, it holds that

$$ \theta_{\mathsf{P}^{\alpha}_{i}} = \theta_{\mathsf{V}^{\alpha}_{0} \otimes \mathsf{P}_{i}}= \left( \theta_{\mathsf{V}^{\alpha}_{0}} \otimes \theta_{\mathsf{P}_{i}} \right) c_{\mathsf{P}_{i}, \mathsf{V}^{\alpha}_{0}} \circ c_{\mathsf{V}^{\alpha}_{0}, \mathsf{P}_{i}}. $$

By (4.1) and (4.2) we get

$$ \theta_{\mathsf{V}^{\alpha}_{0} \otimes \mathsf{P}_{i}}= q^{\frac{(mr)^{2}}{2} + mr - m r^{2}} (-1)^{m i } \left( \text{Id}_{\mathsf{V}^{\alpha}_{0}} \otimes \theta_{\mathsf{P}_{i}} \right). $$

Therefore, by (2.14)

$$ \theta_{\mathsf{P}^{\alpha}_{i}} = q^{\frac{(mr)^{2}+i^{2}}{2} + mr - m r^{2} + i} (-1)^{(m+1)i} \left( \text{Id}_{\mathsf{V}^{\alpha}_{0}} \otimes I_{1,i} - (r - i - 1) \frac{\{1\}^{2}}{\{i+1\}} \text{Id}_{\mathsf{V}^{\alpha}_{0}} \otimes x_{1,i} \right). $$

Note that $\text {Id}_{\mathsf {V}^{\alpha }_{0}} \otimes I_{1,i} = I_{\alpha ,i}$ and $\text {Id}_{\mathsf {V}^{\alpha }_{0}} \otimes x_{1,i} = x_{\alpha ,i}$. Moreover, we have that

$$ q^{\frac{(mr)^{2}+i^{2}}{2}} = q^{\frac{(mr+i)^{2}}{2}} (-1)^{-mi} \quad\text{and}\quad q^{-mr^{2}} = (-1)^{-mr}. $$

Combining all the above together, we get (4.3).

1.7 A.7 Proof of (5.2)

First we prove by induction the following:

$$ F u_{0}^{\otimes m} = \sum\limits_{j=1}^{m} s^{-(j-1)} c_{j}. $$

(A.5)

For m = 1 we have that Fu₀ = u₁ = c₁. Suppose the statement holds for n. Then for n + 1, we have that

$$ \begin{array}{@{}rcl@{}} F u_{0}^{\otimes m+1} &=& K^{-1} u_{0} \otimes F u_{0}^{\otimes m} + F u_{0} \otimes u_{0}^{\otimes m} = s^{-1} \sum\limits_{j=1}^{m} s^{-(j-1)} u_{0} \otimes c_{j} + u_{1} \otimes u_{0}^{\otimes m}\\ &=& \sum\limits_{j=1}^{m} s^{-j} c_{j+1} + (-1)^{k} c_{1} = \sum\limits_{j=1}^{m+1} s^{-(j-1)} c_{j}. \end{array} $$

Now by (A.5) we have that

$$ \begin{array}{@{}rcl@{}} F \left( u_{1} \otimes u_{0}^{\otimes m}\right) &=& K^{-1} u_{1} \otimes F u_{0}^{\otimes m} + F u_{1} \otimes u_{0}^{\otimes m} \\ &=& s^{-1} q^{2} \sum\limits_{j=1}^{m} s^{-(j-1)} u_{1} \otimes c_{j} + [2]^{2} u_{2} \otimes u_{0}^{\otimes m} = q^{2} \sum\limits_{j=1}^{m} s^{-j} a_{1,j+1} + [2]^{2} b_{1} . \end{array} $$

Note that [2]≠ 0, since 2 + ℓ < r. And finally, we have that

$$ \begin{array}{@{}rcl@{}} &&F c_{i} = F \left( u_{0}^{\otimes i-1} \otimes u_{1} \otimes u_{0}^{\otimes n-i}\right) = K^{-1} u_{0}^{\otimes i-1} \otimes F \left( u_{1} \otimes u_{0}^{\otimes n-i}\right) + F u_{0}^{\otimes i-1} \otimes u_{1} \otimes u_{0}^{\otimes n-i} \\ &&= s^{-(i-1)} \left[ q^{2} \sum\limits_{j=1}^{n-i} s^{-j} u_{0}^{\otimes i-1} \otimes a_{1,j+1} + [2]^{2} u_{0}^{\otimes i -1} \otimes b_{1} \right] + \sum\limits_{j=1}^{i-1} s^{-(j-1)} c_{j} \otimes u_{1} \otimes u_{0}^{\otimes n-i} \\ &&= s^{-(i-1)} q^{2} \sum\limits_{j=1}^{n-i} s^{-j} a_{i,i+j} + \sum\limits_{j=1}^{i-1} s^{-(j-1)} a_{j,i} + [2]^{2} s^{-(i-1)} b_{i}. \end{array} $$

1.8 A.8 Proof of Lemma 5.1

We prove the statement for the matrix corresponding to σ₁ ∈B_n. Then it follows immediately for the rest of the generators of B_n since the generators of B_n are all conjugate to each other. Since p(X) is the minimal polynomial of the representation W_n,2, we have that p(σ₁)w_i,j = 0 for all 1 ≤i < j ≤n. It remains to prove the same for the additional basis vectors of N_n,2,0 and N_n,2,1. We also have that

$$ p(X) = X^{3} + \left( -1 + s^{-2} - s^{-4} q^{2}\right) X^{2} + \left( -s^{-2} + s^{-4} q^{2} - s^{-6} q^{2}\right) X + s^{-6} q^{2}. $$

We start with N_n,2,0. We denote t = s^− 3(1 −q²). By an induction on k it holds that

$$ {\sigma_{1}^{k}} b = b + t \sum\limits_{m=0}^{k-1} \left( s^{-4}q^{2}\right)^{m} w_{1,2}. $$

Therefore, the coefficient of b in p(σ₁)b is zero, since the sum of the coefficients of p(X) is zero. Moreover, the coefficient of w_1,2 in p(σ₁)b equals

$$ \begin{array}{@{}rcl@{}} t\left( 1 + s^{-4}q^{2} + s^{-8}q^{4}\right) + t\left( -1 + s^{-2} - s^{-4} q^{2}\right) \left( 1 + s^{-4}q^{2}\right) + t\left( -s^{-2} + s^{-4} q^{2} - s^{-6} q^{4}\right) = 0. \end{array} $$

Hence, the matrix for σ₁ satisfies p(X) and p(X) is the minimal polynomial for the matrix (since it is the minimal polynomial for the matrices of W_n,2 when the eigenvalues are distinct, that is when r ≥ 5).

We proceed now with N_n,2,1. Since $\sigma _{1} b^{\prime }_{j} = b_{j}^{\prime }$ for 3 ≤j < n, and since the coefficients of p(X) sum up to zero, we have that $p(\sigma _{1})b_{j}^{\prime } = 0$ for 3 ≤j < n. For j = 1, 2 we solve the following equation:

$$ {\sigma_{1}^{3}} b^{\prime}_{j} + x {\sigma_{1}^{2}} b^{\prime}_{j} + y \sigma_{1} b^{\prime}_{j} + z b^{\prime}_{j} = 0. $$

Note that

$$ \begin{array}{@{}rcl@{}} {\sigma_{1}^{3}} b^{\prime}_{2} + x {\sigma_{1}^{2}} b^{\prime}_{2} + y \sigma_{1} b^{\prime}_{2} + z b^{\prime}_{2} = 0 &\Leftrightarrow& s^{-1} {\sigma_{1}^{2}} b^{\prime}_{1} + s^{-1} x \sigma_{1} b^{\prime}_{1} + s^{-1} y \sigma_{1} b^{\prime}_{1} + z b^{\prime}_{2} = 0 \\ &\Leftrightarrow& s^{-1} {\sigma_{1}^{3}} b^{\prime}_{1} + s^{-1} x {\sigma_{1}^{2}} b^{\prime}_{1} + s^{-1} y {\sigma_{1}^{2}} b^{\prime}_{1} + s^{-1} z b^{\prime}_{1} = 0 \\ &\Leftrightarrow& {\sigma_{1}^{3}} b^{\prime}_{1} + x {\sigma_{1}^{2}} b^{\prime}_{1} + y {\sigma_{1}^{2}} b^{\prime}_{1} + z b^{\prime}_{1} = 0. \end{array} $$

So, it suffices to solve the equation for either $b^{\prime }_{1}$ or $b^{\prime }_{2}$. Calculating ${\sigma _{1}^{k}} b^{\prime }_{2}$ for k = 1, 2, 3 we find that the equation is satisfied when x = − 1 + s^− 2 −s^− 4q², y = −s^− 2 + s^− 4q² −s^− 6q² and z = s^− 6q². We provide a verification with a Mathematica program Nn2_min_poly.nb (available at [18]). So, $p(\sigma _{1})b_{j}^{\prime } = 0$ for every 1 ≤j < n and similarly as before, p(X) is the minimal polynomial for the matrix corresponding to σ₁.

1.9 A.9 Proof of Proposition 6.1

Note that the action of B_n on the submodule spanned by the vectors w_i,j is the same as the action of B_n on the vectors w_i,j of W_n,2 as in (3.12), which is isomorphic to the LKB representation. Therefore, it remains to prove that the braid group relations are satisfied on the vector b. Let 1 ≤i,j≠n − 1 such that $\lvert i -j \rvert > 1$. Then

$$ \begin{array}{@{}rcl@{}} \sigma_{j} \sigma_{i} b &=& t \sigma_{j} w_{i,i+1} + \sigma_{j} b = t \sigma_{j} w_{i,i+1} + t w_{j,j+1} + b \\ &=& t w_{i,i+1} + t \sigma_{i} w_{j,j+1} + b = t \sigma_{i} w_{j,j+1} + \sigma_{i} b = \sigma_{i} \sigma_{j} b, \end{array} $$

where at the third equality we have used the fact that $\lvert i -j \rvert > 1$. Moreover, for 1 ≤i < n − 2, by applying the formula for the action of B_n on b, we have that

$$ \begin{array}{@{}rcl@{}} \sigma_{i} \sigma_{i+1} \sigma_{i} b &=& t \left( \sigma_{i} \sigma_{i+1} w_{i,i+1} + \sigma_{i} w_{i+1,i+2} + w_{i,i+1} \right) + b, \\ \sigma_{i+1} \sigma_{i} \sigma_{i+1} b &=& t \left( \sigma_{i+1} \sigma_{i} w_{i+1,i+2} + \sigma_{i+1} w_{i,i+1} + w_{i+1,i+2} \right) + b. \end{array} $$

A simple calculation using the action of B_n on the vectors w_i,j shows that the two expressions in parentheses are both equal to s^− 1w_i,i+ 2 + w_i,i+ 1 + s^− 2w_{i+ 1,i+ 2}, which proves the existence of the B_n-representation $\widetilde {\boldsymbol {N}}_{n,2,0}$ over $\mathbb Z[q^{\pm }, s^{\pm }, t^{\pm }]$. Furthermore, the variable t does not appear in any negative powers since the action of the inverses of the braid group generators on the vector b is given by

$$ \sigma_{i}^{-1} = - s^{4} q^{-2} t w_{i,i+1} + b. $$

That is, the representation is actually defined over $\mathbb Z[q^{\pm }, s^{\pm }, t]$, which concludes the proof of the statement.

1.10 A.10 Proof of Proposition 6.8

We prove first the statement for k = 1. We first prove that for every m ≤n

$$ (\sigma_{1} {\ldots} \sigma_{m-1})(\sigma_{1} {\ldots} \sigma_{m-2}) {\ldots} \sigma_{1} b = b + t \sum\limits_{1 \leq i, j \leq m} w_{i,j}. $$

(A.6)

For m = 2, we have that σ₁b = b + tw_1,2. Suppose the statement holds for any number less than m ≤n. Then for m − 1 it holds that

$$ (\sigma_{1} {\ldots} \sigma_{m-2}) {\ldots} \sigma_{1} b = b + t \sum\limits_{1 \leq i, j \leq m-1} w_{i,j}. $$

Now, for m we have that

$$ (\sigma_{1} {\ldots} \sigma_{m-1}) (\sigma_{1} {\ldots} \sigma_{m-2}) {\ldots} \sigma_{1} b = (\sigma_{1} {\ldots} \sigma_{m-1}) b + t \sum\limits_{1 \leq i, j \leq m-1} (\sigma_{1} {\ldots} \sigma_{m-1}) w_{i,j}. $$

Using (3.12) with s = q = 1 we get that

$$ \begin{array}{@{}rcl@{}} (\sigma_{1} {\ldots} \sigma_{m-1}) w_{i,j} &=& \sigma_{1} {\ldots} \sigma_{j} w_{i,j} = \sigma_{1} {\ldots} \sigma_{j-1} w_{i,j+1} = \sigma_{1} {\ldots} \sigma_{i} w_{i,j+1} \\ &=& \sigma_{1} {\ldots} \sigma_{i-1} w_{i+1,j+1} = w_{i+1,j+1}. \end{array} $$

Now, we prove that for every $m^{\prime } <n$

$$ \sigma_{1} {\ldots} \sigma_{m}^{\prime} b = b + t \sum\limits_{2 \leq j \leq m^{\prime}+1} w_{1,j}. $$

For $m^{\prime }=2$ the statement is obvious. Supposing the statement for any number less than $m^{\prime } < n$, we have that

$$ \begin{array}{@{}rcl@{}} \sigma_{1} {\ldots} \sigma_{m^{\prime}} b &=& \sigma_{1} {\ldots} \sigma_{m^{\prime}-1} b + t \sigma_{1} {\ldots} \sigma_{m^{\prime}} w_{m^{\prime},m^{\prime}+1} = b + t \sum\limits_{2 \leq j \leq m^{\prime}} w_{1,j} + t w_{1,m^{\prime}+1} \\ &=& b + t \sum\limits_{2 \leq j \leq m^{\prime}+1} w_{1,j}, \end{array} $$

where at the second equality we use the inductive statement and (3.12). Combining all the above, it holds that

$$ \begin{array}{@{}rcl@{}} (\sigma_{1} {\ldots} \sigma_{m-1}) (\sigma_{1} {\ldots} \sigma_{m-2}) {\ldots} \sigma_{1} b &=& b + t \sum\limits_{2 \leq j \leq m} w_{1,j} + t \sum\limits_{1 \leq i, j \leq m-1} w_{i+1,j+1} \\ &=& b + t \sum\limits_{2 \leq j \leq m} w_{1,j} + t \sum\limits_{2 \leq i, j \leq m} w_{i,j}, \end{array} $$

which proves (A.6). By substituting m = n in (A.6) we get (6.8) for k = 1.

We now use induction on k to prove (6.8) for any $k \in \mathbb N_{\geq 1}$. Suppose that the statement holds for any number less than k. Then

$$ \begin{array}{@{}rcl@{}} {{\varDelta}_{n}^{k}} b = {\varDelta}_{n} {\varDelta}_{n}^{k-1} b = {\varDelta}_{n} b + (k-1) t {\varDelta}_{n} \sum\limits_{1 \leq i < j \leq n} w_{i,j}. \end{array} $$

As mentioned before, the matrices corresponding to the generators of B_n for the representation $\overline {\boldsymbol {W}}_{n,2}$ are permutation matrices. Therefore, any braid in B_n acts trivially on ${\sum }_{1 \leq i, j \leq m} w_{i,j}$. Therefore

$$ {{\varDelta}_{n}^{k}} b = {\varDelta}_{n} b + (k-1) t \sum\limits_{1 \leq i < j \leq n} w_{i,j} = b + k t \sum\limits_{1 \leq i < j \leq n} w_{i,j}, $$

and the statement is proved for every $k \in \mathbb N_{\geq 1}$. Now, it is easy to see that

$$ {\varDelta}_{n}^{-1} = b - kt \sum\limits_{1 \leq i, j \leq n} w_{i,j}. $$

By a similar inductive argument, we can prove the statement for every k < 0, which concludes the proof.

1.11 A.11 Proof of Proposition 6.2

It remains to show that the braid group relations are satisfied for the vectors $b^{\prime }_{m}$, with 1 ≤m ≤n − 1. We start with the commutation relations σ_iσ_j = σ_jσ_i for |i −j|> 1. Suppose first that i,j≠n − 1 and that m∉{i,i + 1,j,j + 1}. Then we have that $\sigma _{i} \sigma _{j} b^{\prime }_{m} = b^{\prime }_{m} = \sigma _{j} \sigma _{m} b^{\prime }_{m}$. If m = i (or without loss of generality m = j) then

$$ \sigma_{i} \sigma_{j} b^{\prime}_{i} = t w_{i,i+1} + \left( 1-s^{-2}\right) b^{\prime}_{i} + s^{-1} b^{\prime}_{i+1} = \sigma_{j} \sigma_{i} b^{\prime}_{i}. $$

Similarly if m = i + 1 (or m = j + 1)

$$ \sigma_{i} \sigma_{j} b^{\prime}_{i+1} = s^{-1} b^{\prime}_{i} = \sigma_{j} \sigma_{i} b^{\prime}_{i+1}. $$

Now, suppose i or j is equal to n − 1 (we choose without loss of generality i = n − 1 and hence j < n − 2) and 1 ≤m < n − 1. If m≠j,j + 1, we have that

$$ \sigma_{n-1} \sigma_{j} b^{\prime}_{m} = b^{\prime}_{m} - s^{n-k-1} b^{\prime}_{n-1} = \sigma_{j} \sigma_{n-1} b^{\prime}_{m}. $$

For m = j

$$ \sigma_{n-1} \sigma_{j} b^{\prime}_{j} = s^{j-n} t w_{j,j+1} + \left( 1-s^{-2}\right) b^{\prime}_{j} + s^{-1} b^{\prime}_{j+1} - s^{n-j-1} b^{\prime}_{n-1} = \sigma_{j} \sigma_{n-1} b^{\prime}_{j}. $$

For m = j + 1

$$ \sigma_{n-1} \sigma_{j} b^{\prime}_{j+1} = s^{-1} b^{\prime}_{j} - s^{-1} s^{n-j-1} b^{\prime}_{n-1} = \sigma_{j} \sigma_{n-1} b^{\prime}_{j+1}. $$

Finally, if m = n − 1

$$ \sigma_{n-1} \sigma_{j} b^{\prime}_{n-1} = s^{-1} t w_{n-1,n} - s^{-2} b^{\prime}_{n-1} = \sigma_{j} \sigma_{n-1} b^{\prime}_{n-1}. $$

Now we proceed to the braiding relations σ_iσ_i+ 1σ_i = σ_i+ 1σ_iσ_i+ 1 for 1 ≤i < n − 1. We start with the case 1 ≤i < n − 2. If m≠i,i + 1,i + 2 then the braiding relations are satisfied since $\sigma _{i} b^{\prime }_{m} = b^{\prime }_{m} = \sigma _{i+1} b^{\prime }_{k}$. For k = i we have that

$$ \begin{array}{@{}rcl@{}} \sigma_{i} \sigma_{i+1} \sigma_{i} b^{\prime}_{i} =& s^{i-n} t \overbrace{\left( \sigma_{i} \sigma_{i+1} w_{i,i+1} + \sigma_{i} w_{i+1, i+2} + (1 - s^{-2}) w_{i,i+1} \right)}^{A_{1} :=} \\ &+ (1- s^{-2}) b^{\prime}_{i} + s^{-1} (1- s^{-2}) b^{\prime}_{i+1} + s^{-2} b^{\prime}_{i+2}. \end{array} $$

On the other hand

$$ \begin{array}{@{}rcl@{}} \sigma_{i+1} \sigma_{i} \sigma_{i+1} b^{\prime}_{i} &=& s^{i-n} t \overbrace{\left( \sigma_{i+1} w_{i,i+1} + w_{i+1, i+2} \right)}^{A_{2} :=}\\ &&\!\!+ \left( 1- s^{-2}\right) b^{\prime}_{i} + s^{-1} \left( 1- s^{-2}\right) b^{\prime}_{i+1} + s^{-2} b^{\prime}_{i+2}. \end{array} $$

Note that in both expressions we have the same terms involving the vectors $b^{\prime }_{k}$ for k = 1,…,n. It remains to compare the expressions A₁ and A₂. In Appendix A.9 it was proved that

$$ \sigma_{i} \sigma_{i+1} w_{i,i+1} + \sigma_{i} w_{i+1,i+2} + w_{i,i+1} = \sigma_{i+1} \sigma_{i} w_{i+1,i+2} + \sigma_{i+1} w_{i,i+1} + w_{i+1,i+2}. $$

(A.7)

Therefore

$$ A_{1} - A_{2} = \sigma_{i+1} \sigma_{i} w_{i+1, i+2} - s^{-2} w_{i,i+1} = 0, $$

where the last equality is due to (3.12). Hence, $\sigma _{i} \sigma _{i+1} \sigma _{i} b^{\prime }_{i} = \sigma _{i+1} \sigma _{i} \sigma _{i+1} b^{\prime }_{i}$. For m = i + 1, it holds that

$$ \sigma_{i} \sigma_{i+1} \sigma_{i} b^{\prime}_{i+1} = s^{i-n-1} t w_{i+1,i+2} + s^{-1} \left( 1-s^{-2}\right) b^{\prime}_{i} + s^{-2} b^{\prime}_{i+1} = \sigma_{i+1} \sigma_{i} \sigma_{i+1} b^{\prime}_{i+1}. $$

If m = i + 2, then

$$ \sigma_{i} \sigma_{i+1} \sigma_{i} b^{\prime}_{i+2} = s^{-2} b^{\prime}_{i} = \sigma_{i+1} \sigma_{i} \sigma_{i+1} b^{\prime}_{i+2}. $$

Now we examine the case i = n − 2. If m < n − 2, then we have that

$$ \begin{array}{@{}rcl@{}} \sigma_{n-2} \sigma_{n-1} \sigma_{n-2} b^{\prime}_{n-2} &=& s^{-2} t \overbrace{\left( \sigma_{n-2} \sigma_{n-1} w_{n-2,n-1} + \sigma_{n-2} w_{n-1, n} + \left( 1 - s^{-2}\right) w_{n-2,n-1} \right)}^{B_{1} :=} \\ &&\!\!- s^{-2} b^{\prime}_{n-2} + s^{-1} \left( 1- s^{-2}\right) b^{\prime}_{n-1}. \end{array} $$

On the other hand

$$ \sigma_{n-1} \sigma_{n-2} \sigma_{n-1} b^{\prime}_{n-2} = s^{-2} t \overbrace{\left( \sigma_{n-1} w_{n-2,n-1} + w_{n-1, n} \right)}^{B_{2} :=} - s^{-2} b^{\prime}_{n-2} + s^{-1} \left( 1- s^{-2}\right) b^{\prime}_{n-1}. $$

Again, due to (A.7) we have that B₁ = B₂ (the proof is analogous to the proof of A₁ = A₂). Hence, $\sigma _{n-2} \sigma _{n-1} \sigma _{n-2} b^{\prime }_{n-2} = \sigma _{n-1} \sigma _{n-2} \sigma _{n-1} b^{\prime }_{n-2}$. And, finally, if m = n − 1

$$ \sigma_{n-2} \sigma_{n-1} \sigma_{n-2} b^{\prime}_{n-1} = s^{-3} t w_{n-2,n-1} - s^{-3} b^{\prime}_{n-2} + s^{-2} b^{\prime}_{n-1} = \sigma_{n-1} \sigma_{n-2} \sigma_{n-1} b^{\prime}_{n-1}, $$

which completes the proof.

1.12 A.12 Proof of (7.2)

Using the fact that β_i = 1 for n ≡− 2 mod r (see Section 5.4) and (5.4), we have that

$$ F c_{i} = s^{-(i-1)} q^{2} \sum\limits_{j=1}^{n-i} s^{-j} w_{i,i+j} + \sum\limits_{j=1}^{i-1} s^{-(j-1)} w_{j,i} + s^{-(i-1)} \sum\limits_{j=1}^{n} b_{j}. $$

So

$$ \begin{array}{@{}rcl@{}} F \overline{c}_{i} &= & F c_{i} - s^{n-i} F c_{n} = s^{-(i-1)} q^{2} \sum\limits_{j=1}^{n-i} s^{-j} w_{i,i+j} + \sum\limits_{j=1}^{i-1} s^{-(j-1)} w_{j,i} + s^{-(i-1)} \sum\limits_{j=1}^{n} b_{j} \\ & &- s^{n-i} \sum\limits_{j=1}^{n-1} s^{-(j-1)} w_{j,n} - s^{i-1} \sum\limits_{j=1}^{n} b_{j} \\ &=& s^{-(i-1)} q^{2} \sum\limits_{j=1}^{n-i} s^{-j} w_{i,i+j} + \sum\limits_{j=1}^{i-1} s^{-(j-1)} w_{j,i} - s^{n-i} \sum\limits_{j=1}^{n-1} s^{-(j-1)} w_{j,n}. \end{array} $$