Schur–Weyl duality for tensor powers of the Burau representation

Abstract

Artin’s braid group $B_n$ is generated by $\sigma _1, \ldots , \sigma _{n-1}$ subject to the relations

$$\begin{aligned} \sigma _i \sigma _{i+1} \sigma _i = \sigma _{i+1} \sigma _i \sigma _{i+1}, \quad \sigma _i\sigma _j = \sigma _j \sigma _i \text { if } |i-j|>1. \end{aligned}$$

For complex parameters $q_1,q_2$ such that $q_1q_2 \ne 0$, the group $B_n$ acts on the vector space $\mathbf {E}= \sum _i \mathbb {C}\mathbf {e}_i$ with basis $\mathbf {e}_1, \ldots , \mathbf {e}_n$ by

$$\begin{aligned} \sigma _i \cdot \mathbf {e}_i= & {} (q_1+q_2)\mathbf {e}_i + q_1\mathbf {e}_{i+1}, \quad \sigma _i \cdot \mathbf {e}_{i+1} = -q_2\mathbf {e}_i, \\ \sigma _i \cdot \mathbf {e}_j= & {} q_1 \mathbf {e}_j \quad \text { if }\; j \ne i,i+1. \end{aligned}$$

This representation is (a slight generalization of) the Burau representation. If $q = -q_2/q_1$ is not a root of unity, we show that the algebra of all endomorphisms of $\mathbf {E}^{\otimes r}$ commuting with the $B_n$-action is generated by the place-permutation action of the symmetric group $S_r$ and the operator $p_1$, given by

$$\begin{aligned} p_1(\mathbf {e}_{j_1} \otimes \mathbf {e}_{j_2} \otimes \cdots \otimes \mathbf {e}_{j_r}) = q^{j_1-1} \, \sum _{i=1}^n \mathbf {e}_i \otimes \mathbf {e}_{j_2} \otimes \cdots \otimes \mathbf {e}_{j_r} . \end{aligned}$$

Equivalently, as a $(\mathbb {C}B_n, \mathcal {P}'_r([n]_q))$-bimodule, $\mathbf {E}^{\otimes r}$ satisfies Schur–Weyl duality, where $\mathcal {P}'_r([n]_q)$ is a certain subalgebra of the partition algebra $\mathcal {P}_r([n]_q)$ on 2r nodes with parameter $[n]_q = 1+q+\cdots + q^{n-1}$, isomorphic to the semigroup algebra of the “rook monoid” studied by W. D. Munn, L. Solomon, and others.

Appendix: Proof of Theorem 5.2

Assume that $q_1q_2 \ne 0$ and $q = -q_2/q_1$ is not a root of unity. In this appendix (which is independent of Sects. 6–10) we give an elementary proof that the Zariski closure $\overline{G}$ contains ${\text {SL}}(\mathbf {F})$, where $G = \rho (B_n)$ is the image of the reduced Burau representation $\rho : B_n \rightarrow {\text {GL}}(\mathbf {F})$. This result is needed to obtain Schur–Weyl duality for $\mathbf {F}^{\otimes k}$ (see Theorem 5.3), which is needed for the proof of the main results in Sect. 9. In case q is transcendental, the result is a special case of [30, Theorem B], obtained by different methods.

We get the following result from Lemma 3.3 by an elementary induction on k, which is left to the reader.

Lemma A.1

Let k be a positive integer. For any $i,j = 1, \ldots , n-1$ the action of $\sigma _i^k \in B_n$ on $\mathbf {f}_j \in \mathbf {F}$ is given by the rules

• (a) $\rho (\sigma _i^k) \mathbf {f}_j = q_1^k \mathbf {f}_j$ if $j \ne i-1, i, i+1$.

• (b) $\rho (\sigma _i^k) \mathbf {f}_{i-1} = q_1^k \mathbf {f}_{i-1} + q_1 \Phi _k(q_1,q_2) \mathbf {f}_i$.

• (c) $\rho (\sigma _i^k) \mathbf {f}_i = q_2^k \mathbf {f}_i$.

• (d) $\rho (\sigma _i^k) \mathbf {f}_{i+1} = -q_2 \Phi _k(q_1,q_2) \mathbf {f}_i + q_1^k\mathbf {f}_{i+1}$

where $\Phi _k(q_1,q_2) = \sum _{j=0}^{k-1} q_1^j q_2^{k-1-j} = \frac{q_1^k-q_2^k}{q_1-q_2}$.

Set $\Phi _k = \Phi _k(q_1,q_2)$ for short. For convenience of reference, the matrices of the operators $\rho (\sigma _i^k)$ with respect to the $\{\mathbf {f}_i\}$ basis are listed below:

$$\begin{aligned} \rho (\sigma _1^k)&= \begin{bmatrix} q_2^k &{}\quad -q_2\Phi _k &{}\quad 0 &{}\quad \cdots &{}\quad 0\\ 0 &{}\quad q_1^k &{}\quad 0 &{}\quad \cdots &{}\quad 0\\ 0 &{}\quad 0 &{}\quad q_1^k &{}\quad \cdots &{}\quad 0\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad q_1^k \end{bmatrix} , \quad \rho (\sigma _{n-1}^k) = \begin{bmatrix} q_1^k &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad \cdots &{}\quad q_1^k &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad \cdots &{}\quad 0 &{}\quad q_1^k &{}\quad 0 \\ 0 &{}\quad \cdots &{}\quad 0 &{}\quad q_1\Phi _k &{}\quad q_2^k \\ \end{bmatrix}, \\ \rho (\sigma _i^k)&= \begin{bmatrix} q_1^k &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0\\ \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad 0\\ 0 &{}\quad \cdots &{}\quad q_1^k &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0\\ 0 &{}\quad \cdots &{}\quad q_1\Phi _k &{}\quad q_2^k &{}\quad -q_2\Phi _k &{}\quad \cdots &{}\quad 0\\ 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad q_1^k &{}\quad \cdots &{}\quad 0\\ \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad q_1^k \end{bmatrix} \quad (1< i < n-1) . \end{aligned}$$

Henceforth, we identify ${\text {GL}}(\mathbf {F}) \cong {\text {GL}}_{n-1}(\mathbb {C})$ by means of the basis $\{\mathbf {f}_i\}$.

Recall (Remark 3.4) that $\det (\sigma _j) = q_1^{n-2}q_2$ for all j. The proof of Theorem 5.2 falls naturally into two cases, depending whether $q_1^{n-2}q_2$ is or is not a root of unity. The non-root of unity case requires the following lemma.

Lemma A.2

Suppose that $q_1^{n-2}q_2$ is not a root of unity. If $g\in G$, $0 \ne z\in \mathbb {C}$, then $zg=zI_{n-1}g\in \overline{G}$.

Proof

It is well known [24] that the “full twist” $\theta _n = \Delta _n^2 \in B_n$, where

$$\begin{aligned} \Delta _n = (\sigma _1 \cdots \sigma _{n-1})(\sigma _1 \cdots \sigma _{n-2}) \cdots (\sigma _1\sigma _2) \sigma _1, \end{aligned}$$

generates the center of the braid group $B_n$. (An easy exercise [24, Exercise 1.3.2] gives the alternate formula $\theta _n = (\sigma _1 \cdots \sigma _{n-1})^n$.) Thus, by Schur’s Lemma it follows that $\rho (\theta _n)=\lambda I_{n-1}$ for some $\lambda \in \mathbb {C}$. Now an easy calculation shows that $\lambda = (q_1^{n-2}q_2)^n$. As $\lambda$ is not a root of unity, the Zariski closure of the subgroup generated by $\rho (\theta _n)$ is $H=\{zI_{n-1}\,:\, z\ne 0\}$ and so $zI_{n-1} \in \overline{G}$ for every $0 \ne z\in \mathbb {C}$. Since $\overline{G}$ is a group, for any $g\in G$ it follows that $zg = g zI_{n-1} \in \overline{G}$. $\square$

Let $\{e_{ij}\}_{i,j=1}^{n-1}$ be the standard basis of matrix units for matrix space, defined in terms of the Kronecker delta symbols by $e_{ij} = (\delta _{ik}\delta _{jl})_{k,l=1}^{n-1}$. Set

$$\begin{aligned} E(i) = I_{n-1}-e_{ii} = \textstyle \sum _{j \ne i} e_{jj}. \end{aligned}$$

We also need the constants

$$\begin{aligned} a = \frac{q}{1+q}, \quad b = \frac{1}{1+q}. \end{aligned}$$

Notice that $a = -q_2/(q_1-q_2)$, $b = q_1/(q_1-q_2)$.

Proposition A.3

Assume that $q_1q_2 \ne 0$ and $q = -q_2/q_1$ is not a root of unity. Let $\overline{G}$ be the Zariski closure of $G = \rho (B_n)$.

• (a) If $q_1^{n-2}q_2$ is not a root of unity, then $\overline{G}$ contains the one-parameter subgroups $H_1, \ldots , H_{n-1}$ where

  $$\begin{aligned} H_i&= (1-\delta _{i,1})b(1-z_i)e_{i,i-1}+z_i e_{ii} \\&\quad +(1-\delta _{i,n-1})a(1-z_i)e_{i,i+1} +E(i) \end{aligned}$$

  for nonzero complex scalars $z_1,\ldots z_{n-1}$.

• (b) If $q_1^{n-2}q_2$ is a root of unity, then $\overline{G}$ contains the one-parameter subgroups $K_1, \ldots , K_{n-1}$ where

  $$\begin{aligned} K_i&= (1-\delta _{i,1})b(w_i-w_i^{2-n})e_{i,i-1}+w_i^{2-n} e_{ii}\\&\quad +(1-\delta _{i,n-1})a(w_i-w_i^{2-n})e_{i,i+1} + w_iE(i) \end{aligned}$$

  for nonzero complex scalars $w_1,\ldots w_{n-1}$.

Proof

(a) It follows from Lemma A.2 that $q_1^{-1} \rho (\sigma _i)\in \overline{G}$ for all $i=1,\ldots ,n-1$. Hence, $(q_1^{-1}\rho (\sigma _i))^k = q_1^{-k} \rho (\sigma _i^k)$ belongs to $\overline{G}$ for all i and all $k\ge 0$. By Lemma A.1, it follows by an easy calculation that

$$\begin{aligned} (q_1^{-1}\rho (\sigma _i))^k&= (1-\delta _{1,i})b(1-(-q)^k)e_{i,i-1} + (-q)^ke_{ii}\\&\quad + (1-\delta _{i,n-1})a(1-(-q)^k)e_{i,i+1} + E(i). \end{aligned}$$

Notice that this matrix depends only on $q = -q_2/q_1$ and lies in $H_i$ for any $k\ge 0$. Since $-q$ is not a root of unity, it follows that the intersection $H_i \cap \overline{G}$ is infinite. Thus, $H_i \subset \overline{G}$, since $H_i$ is a closed one-parameter group.

(b) Since $q_1^{n-2}q_2$ is a root of unity, there is a positive integer d such that $(q_1^{n-2}q_2)^d = 1$, hence $q_2^{d}=q_1^{d(2-n)}$. Since $q=-q_2/q_1$, it follows that

$$\begin{aligned} q^{d} = (-1)^d \, q_1^{d(1-n)}. \end{aligned}$$

Now q is not a root of unity, so $q_1$ cannot be a root of unity. By Lemma A.1, if k is a non-negative integer we have

$$\begin{aligned} \rho (\sigma _i^{kd})&= (1-\delta _{i,1})b(q_1^{kd}-q_1^{kd(2-n)})e_{i,i-1} + q_1^{kd(2-n)}e_{ii} \\&\quad + (1-\delta _{i,n-1})a(q_1^{kd}-q_1^{kd(2-n)})e_{i,i+1} + q_1^{kd} E(i). \end{aligned}$$

Since $q_1$ is not a root of unity, the powers $q_1^{kd}$ are distinct for all k and thus the matrices $\rho (\sigma _i^{kd})$ are also distinct for all k. Notice that $\rho (\sigma _i^{kd})$ belongs to $K_i$ for all k. Hence, the cardinality of $K_i \cap \overline{G}$ is infinite. This forces $K_i \subset \overline{G}$, as $K_i$ is a closed one-parameter group.

$\square$

Corollary A.4

Assume that $q_1q_2 \ne 0$ and $q = -q_2/q_1$ is not a root of unity. Let $\overline{G}$ be the Zariski closure of $G = \rho (B_n)$.

• (a) If $q_1^{n-2}q_2$ is not a root of unity, then the Lie algebra ${\text {Lie}}(\overline{G})$ contains the elements

  $$\begin{aligned} u_i = (1-\delta _{i,1}) b e_{i,i-1} + e_{ii} + (1-\delta _{i,n-1}) a e_{i,i+1} \end{aligned}$$

  for $i = 1, \ldots , n-1$.

• (b) If $q_1^{n-2}q_2$ is a root of unity, then ${\text {Lie}}(\overline{G})$ contains the elements

  $$\begin{aligned} v_i&= (1-\delta _{i,1}) b(n-1) e_{i,i-1} + (2-n)e_{ii} \\&\quad + (1-\delta _{i,n-1}) a(n-1) e_{i,i+1} + E(i) \end{aligned}$$

  for $i = 1, \ldots , n-1$.

Proof

For (a), take the derivative $d/dz_i$ at $z_i=1$ of the one-parameter subgroup $H_i$. Similarly, for (b) take the derivative $d/dw_i$ at $w_i=1$ of the one-parameter subgroup $K_i$. $\square$

Proposition A.5

Assume that $q_1q_2 \ne 0$ and q is not a root of unity. Suppose that $n \ge 3$.

• (a) The Lie algebra generated by $u_1, \ldots , u_{n-1}$ equals $\mathfrak {gl}_{n-1}$.

• (b) The Lie algebra generated by $v_1, \ldots , v_{n-1}$ equals $\mathfrak {sl}_{n-1}$.

Proof

(a) Let $\mathfrak {g}$ be the Lie algebra generated by $u_1, \ldots , u_{n-1}$. Since $\mathfrak {g}\subseteq \mathfrak {gl}_{n-1}$, it suffices to show the reverse containment. We will argue that $e_{ij} \in \mathfrak {g}$, for all $i,j = 1, \ldots , n-1$, making use of the standard commutator formula $[e_{ij}, e_{kl}] = \delta _{jk} e_{il} - \delta _{li} e_{kj}$.

Assume that $n \ge 4$. Direct computation shows that

$$\begin{aligned} {[}u_1, [u_1,u_2]] + [u_1,u_2] = 2a(ab - 1) e_{12} - 2a^2 e_{13}. \end{aligned}$$

As $a \ne 0$, it follows that

$$\begin{aligned} A'_2 = \tfrac{1}{2a}\big ( [u_1, [u_1,u_2]] + [u_1,u_2] \big ) = (ab - 1) e_{12} - a e_{13} \end{aligned}$$

is in $\mathfrak {g}$. Then, $A_1 = u_1$ and $A_2 = bu_1 - A'_2 = be_{12}+e_{13}+ae_{14}$ are both in $\mathfrak {g}$. Clearly $A_1, A_2$ are linearly independent.

Next, we recursively compute elements $A_k$ for $k = 2, \ldots , n-1$ by defining

$$\begin{aligned} A_k = \tfrac{1}{a} [A_{k-1},u_k] \quad \text {for all } k = 3, \ldots , n-1. \end{aligned}$$

Then, by direct computation we have

$$\begin{aligned} A_k = {\left\{ \begin{array}{ll} b e_{1,k-1} + e_{1k} + a e_{1,k+1} &{}\quad \text {if } k = 3, \ldots , n-2,\\ b e_{1,n-2} + e_{1,n-1} &{}\quad \text {if } k = n-1. \end{array}\right. } \end{aligned}$$

We claim that the elements $A_1, A_2, \ldots , A_{n-1}$ are linearly independent.

To see this, identify each $A_k$ with its coordinate vector with respect to the basis $\{e_{11}, \ldots , e_{1,n-1}\}$ of the first row of matrix space. Let M be the matrix whose rows are those coordinate vectors in order. Then, M is the $(n-1) \times (n-1)$ tridiagonal banded matrix

$$\begin{aligned} M= \begin{bmatrix} 1&{}\quad a&{}\quad 0&{}\quad \cdots &{}\quad 0\\ b&{}\quad 1&{}\quad a&{}\quad \ddots &{}\quad \vdots \\ 0&{}\quad b&{}\quad 1&{}\quad \ddots &{}\quad 0\\ \vdots &{}\quad \ddots &{}\quad \ddots &{}\quad \ddots &{}\quad a\\ 0&{}\quad \cdots &{}\quad 0&{}\quad b&{}\quad 1 \end{bmatrix} \end{aligned}$$

with a’s on the super-diagonal and b’s on the sub-diagonal. Setting $D_n = \det M$ we see by Eq. (7) that $D_n$ satisfies

$$\begin{aligned} D_n = D_{n-1} - ab D_{n-2} \quad (n\ge 5), \end{aligned}$$

(27)

where $D_3 = 1-ab$ and $D_4 = 1-2ab$. Thus, we obtain the following. $\square$

Lemma A.6

$D_n = \dfrac{[n]_q}{(1+q)^{n-1}}$ for all $n \ge 3$.

Proof

First check directly that $D_3$ and $D_4$ satisfy the given formula. Assume by induction that $D_{n-2}$ and $D_{n-1}$ satisfy the formula. Applying the recursion (27) gives

$$\begin{aligned} D_n = \frac{[n-1]_q}{(1+q)^{n-2}} - \frac{q}{(1+q)^2} \frac{[n-2]_q}{(1+q)^{n-3}} = \frac{(1+q)[n-1]_q - q[n-2]_q}{(1+q)^{n-1}} \end{aligned}$$

and the result follows from the identity $(1+q)[n-1]_q - q[n-2]_q = [n]_q$. $\square$

Lemma A.6 proves the claim, as q is not a root of unity. The claim implies that $\mathbb {C}A_1 + \mathbb {C}A_2 + \cdots + \mathbb {C}A_{n-1} = \sum _{j=1}^{n-1} \mathbb {C}e_{1j}$. Hence, $e_{1j} \in \mathfrak {g}$ for all $j = 1, \ldots , n-1$. Now we finish the proof quite easily, by observing that

$$\begin{aligned} {[}e_{ij}, u_{i+1}] = {\left\{ \begin{array}{ll} -b e_{i+1,j} &{} \quad \text {if } i+1 \ne j, \\ -b e_{jj} + be_{j-1,j-1} + e_{j-1,j} + ae_{j-1,j+1} &{}\quad \text {if } i+1 = j \ne n-1, \\ -b e_{n-1,n-1} + be_{n-2,n-2} + e_{n-2,n-1} &{} \quad \text {if } i+1 = j = n-1. \end{array}\right. } \end{aligned}$$

This implies that $[e_{ij}, u_{i+1}] = -b e_{1+1,j}$ modulo a linear combination of elements of the form $e_{kl}$ where $k < i+1$. Assuming by induction that the $e_{kl} \in \mathfrak {g}$ for all $k < i+1$, it follows from the fact that $b \ne 0$ that $e_{i+1,j} \in \mathfrak {g}$ for all j. This completes the proof of (a) in case $n \ge 4$.

The case $n=3$ must be handled separately, and is simpler. Notice that in this case

$$\begin{aligned} A'_2 = \tfrac{1}{2a}\big ( [u_1,[u_1,u_2]] + [u_1,u_2]\big ) = (ab-1) e_{12} \in \mathfrak {g}. \end{aligned}$$

Since $ab-1 = 0$ if and only if $[3]_q = 1+q+q^2 = 0$, we see that $ab-1 \ne 0$ as q is not a root of unity. Hence, $e_{12} \in \mathfrak {g}$, and $e_{11} = u_1 -a e_{12} \in \mathfrak {g}$. Thus, $[e_{11},u_2] = -b e_{21} \in \mathfrak {g}$, which implies that $e_{21} \in \mathfrak {g}$ and $e_{22} = u_2 - b e_{21} \in \mathfrak {g}$, completing the proof of (a) in the case $n=3$.

(b) Now let $\mathfrak {g}_1=\langle u_1,\ldots u_{n-1}\rangle$ be the Lie algebra in part (a), and set $\mathfrak {g}_2=\langle v_1,\ldots v_{n-1}\rangle$. Notice that

$$\begin{aligned} u_k =\tfrac{1}{n-1}\left( I_{n-1} -v_k\right) \end{aligned}$$

for all k. Thus, $[u_i,u_j]=c[v_i,v_j]$ with $c=1/(n-1)^2$. Hence, $\mathfrak {g}_1'=\mathfrak {g}_2'$, that is, the derived algebras of $\mathfrak {g}_1$, $\mathfrak {g}_2$ coincide. By part (a), $\mathfrak {g}_1=\mathfrak {gl}_{n-1}$, so $\mathfrak {g}_1'=\mathfrak {gl}_{n-1}'=\mathfrak {sl}_{n-1}$. Thus, $\mathfrak {g}_2'=\mathfrak {sl}_{n-1} \subset \mathfrak {g}_2.$ As it is obvious that $\mathfrak {g}_2\subset \mathfrak {sl}_{n-1}$, it follows that $\mathfrak {g}_2 =\mathfrak {sl}_{n-1}$. This completes the proof of Proposition A.5. $\square$

We now have all the tools needed to prove Theorem 5.2.

Proof of Theorem 5.2

It is well known (see, e.g., [16, 20]) that closed subgroups of ${\text {GL}}(\mathbf {F})$ contain the one-parameter subgroups generated by all elements of their Lie algebra; indeed, that can be taken as an equivalent definition of the Lie algebra for such groups. By Proposition A.5, it easily follows that $\overline{G}$ contains ${\text {SL}}_{n-1} \cong {\text {SL}}(\mathbf {F})$, so we are done. (If $n = 2$ there is nothing to prove, as $\dim \mathbf {F}= 1$ in that case.) $\square$

Remark A.7

• (i) It is easy to see that $\overline{G} = {\text {GL}}(\mathbf {F})$ if and only if $q_1^{n-2}q_2$ is not a root of unity. The sufficiency of this condition follows from Proposition A.5(a). For its necessity, observe that if $\xi = q_1^{n-2}q_2$ is a root of unity (of order d, say) then elements of G satisfy the polynomial equation

  $$\begin{aligned} \prod _{p=0}^{d-1} (\det (X_{ij}) - \xi ^p) = 0 \end{aligned}$$

  and not all elements of ${\text {GL}}(\mathbf {F})$ satisfy it.

• (ii) When $q_1^{n-2} q_2$ is a primitive dth root of unity, for $d>1$, we have a strict containment $\overline{G} \supsetneqq {\text {SL}}(\mathbf {F})$, because the generators of G do not have determinant one.