Proof of theorem 4.1 Fundamentally, throughout this proof, we will be concerned with the evaluation of $\psi _n^{(k)}(1)$, where the superscript means derivative taken w.r.t. $t$, entrywise, and the resulting matrix in $GL_{n-1}({\Bbb Z}[t^{\pm 1}])$ is evaluated at $t=1$.

We prepare the following formulas for $\psi _n(\kappa _{1,l})$, $l=2,\ldots,n-1$.

(5.1)\begin{equation} \psi_n(\kappa_{1,l})= \left[\begin{array}{cccccc} t^l & 0 & 0 & \cdots & 0 & 1-t \\ t^l-t^2 & t & 0 & \cdots & 0 & 1-t \\ t^l-t^3 & 0 & t & & & 1-t \\ \vdots & \vdots & & \ddots & & \vdots \\ t^l-t^{l-1} & 0 & \cdots & 0 & t & 1-t \\ 0 & 0 & \cdots & 0 & 0 & 1 \\ \end{array}\right]\oplus Id_{n-1-l}, \end{equation}

whence

\[ \hat A_l =\psi_n'(\kappa_{1,l})(1)=\left[\begin{array}{cccccc} l & 0 & 0 & \cdots & 0 & -1 \\ l-2 & 1 & 0 & & 0 & -1 \\ l-3 & 0 & 1 & & 0 & -1 \\ \vdots & \vdots & & \ddots & & \vdots \\ 1 & 0 & \cdots & 0 & 1 & -1 \\ 0 & & \cdots & & 0 & 0 \end{array}\right]\oplus 0_{n-1-l}. \]

One can check (5.1) as follows. There are explicit matrices $X_n$ so that the embedding

(5.2)\begin{equation} \iota_1:B_{n-1}\simeq B_{1,n-1}\subset B_n \end{equation}

gives

(5.3)\begin{equation} \psi_{n}\circ \iota_1 =X_n(\psi_{n-1}\oplus 1)X_n^{{-}1}. \end{equation}

The matrices are not difficult to find directly, and will also be explained in [Reference Stoimenow27]. We just give the formula here. With $s=\sqrt {t}$,

\[ \left\langle{k}\right\rangle=s^{k}-s^{{-}k}\,\quad\text{and}\quad a_k=\frac{s^{{-}k}\left\langle{n-1-k}\right\rangle}{\left\langle{n-1}\right\rangle}, \]

we have

(5.4)\begin{equation} X_n=\left[\begin{array}{ccccc} 1 & & \cdots & 0 & a_{n-2} \\ 0 & 1 & \cdots & 0 & a_{n-1} \\ & & \ddots & & \vdots \\ & & & 1 & a_1 \\ & & & & 1 \\ \end{array}\right]. \end{equation}

(The splitting behaviour of $\psi _{n}\circ \iota _1$ is very well known [Reference Jones15].)

With (2.9), one can then calculate $\psi _n(\delta ^2_{[1,n-1]})$ and $\psi _n(\delta ^2_{[1,n-2]})=\psi _{n-1} (\delta ^2_{[1,n-2]}) \oplus 1$. The rest for obtaining (5.1) follows from (2.15).

We have the (non-commutative) matrix Leibniz rule (where prime refers to the derivative in $t$),

(5.5)\begin{equation} (AB)'=AB'+A'B(\ne AB'+BA'). \end{equation}

By differentiating $AA^{-1}=Id$ for $A=\psi _n(\kappa _{1,l})$, and using $A(1)=Id$, we see also

(5.6)\begin{equation} \psi_n'(\kappa_{1,l}^{{-}1})(1)={-}\psi_n'(\kappa_{1,l})(1)={-}\hat A_l. \end{equation}

It will be better in the following to move from $\psi _n$ to the non-reduced Burau representation $\psi ^{\times }_{n}$ of $B_n$ acting on ${\Bbb C}^n$ by

\[ \psi^{{\times}}_{n}(\sigma_i)(t)=Id_{i-1}\oplus \left[\begin{array}{cc} 1-t & 1 \\ t & 0 \end{array}\right] \oplus Id_{n-i-1}. \]

The form (2.8) comes from looking at the action of $\psi ^{\times }_{n}$ on (the linear span of) the set of vectors $\mathbf {e}_i-\mathbf {e}_{i+1}$, with $\mathbf {e}_i$ the standard basis vector. There is an extra dimension coming from the fixvector $\sum \mathbf {e}_i$. Thus

(5.7)\begin{equation} \psi^{{\times}}_n=T(1\oplus \psi_n)T^{{-}1}\quad T=\left[\begin{array}{ccccc} 1 & 1 & 0 & \cdots & 0 \\ 1 & -1 & 1 & & \vdots \\ 1 & 0 & \ddots & \ddots & 0 \\ \vdots & \vdots & \cdots & -1 & 1 \\ 1 & 0 & \cdots & 0 & -1 \\ \end{array}\right]. \end{equation}

This map $(.)^\times$ augments matrix size, however, adding this extra dimension will simplify calculation enormously. For instance, the embedding $\iota _1$ in (5.2) simplifies (5.3) to $\psi ^{\times }_{n+1}\circ \iota _1=\psi ^{\times }_n\oplus 1$, while the one

(5.8)\begin{equation} \iota_2:B_{n-1}\simeq B_{2,n}\subset B_n \end{equation}

gives $\psi ^{\times }_{n+1}\circ \iota _2=1\oplus \psi ^{\times }_n$. Also

(5.9)\begin{equation} \Pi=\Pi(b)=\psi^{{\times}}_n(b)(1) \end{equation}

is just the permutation matrix of $\pi (b)$, with $\Pi _{i,\pi (b)(i)}=1$ and $\Pi _{ij}=0$ otherwise. Since the transition matrix $T$ does not depend on $t$, it is also clear that $(\psi ^{\times }_n)^{(k)}=(\psi _n^{(k)})^\times$.

One can calculate straightforwardly that

(5.10)\begin{equation} A_l=\psi^{{\times}}_n(\kappa_{2,l+1})'(1)=0\oplus (\hat A_l)^\times \ominus 0, \end{equation}

where $(M\oplus N)\ominus N=M$ and (with a matrix having $l-1$ rows and columns with a ‘$-1$’ entry)

\[ (\hat A_l)^\times{=}\left[\begin{array}{ccccc} l-1 & -1 & -1 & \cdots & -1 \\ -1 & 1 & 0 & \cdots & 0 \\ -1 & 0 & \ddots & & 0 \\ \vdots & \vdots & \cdots & \ddots & \vdots \\ -1 & 0 & \cdots & 0 & 1 \\ \end{array}\right]\oplus 0_{n-l}. \]

Letting the commutator be

(5.11)\begin{equation} [A,B]=AB-BA, \end{equation}

we also prepare for $l=2,\ldots,n-2$,

(5.12)\begin{equation} [A_l,A_{n-1}]=0\oplus \left[\begin{array}{c|c|c} 0 & n-1-l & 1-l \\ \hline l-n+1 & & \\ \vdots & 0 & 1 \\ l-n+1 & & \\ \hline l-1 & & \\ \vdots & -1 & 0 \\ l-1 & & \end{array}\right] \end{equation}

with the blocks of (both horizontal and vertical) size $1,l-1,n-1-l$, respectively (and all entries within each block identical).

It will be easier to work with the form (2.24), which can be obtained from (2.1) by conjugating $\beta$ by

(5.13)\begin{equation} \sigma_{1}\sigma_2\,\cdots\, \sigma_{n-1}, \end{equation}

and isolating the two letters $\sigma _1^{\pm 1}$ [so that (4.3) remains the same], while changing $\alpha$ to $\iota _2(\alpha )$ from (5.8). For the rest of the proof, we will use (2.24), but continue writing $\alpha$ for $\tilde \alpha$ and $\beta$ for $\tilde \beta$. Thus, for $\alpha,\beta \in B_{2,n}$, write

(5.14)\begin{equation} b(\alpha,\beta)=\alpha\sigma_{1}\beta\sigma_1^{{-}1}. \end{equation}

Then by (2.24),

(5.15)\begin{equation} b_0=b(\alpha,\beta). \end{equation}

Furthermore, we keep $n$ fixed and often simplify notation $\psi ^{\times }=\psi ^{\times }_n$.

Consider now the combed normal form for $\alpha$. In (2.16), we can assume $\alpha _1=Id$ because $\alpha \in B_{2,n}$. Also, since any right factor of $\alpha$ in $B_{3,n}$ can be moved into $\beta$ [and, being pure, will not affect (4.3)], we can assume w.l.o.g. $\alpha _i=Id$ for $i>2$.

We write thus

(5.16)\begin{equation} \alpha=\prod_{i=1}^k \kappa_{2,j_i}^{\varepsilon_i}. \end{equation}

Using that $\delta _{[2,n]}^2$ commutes with $\alpha$, we will also rewrite (2.24) in the form that, to obtain $b_1$ from $b_0$, one adds a $\kappa _{2,n}^{-1}$ at the beginning and a $\kappa _{2,n}$ at the end of the product in (5.16). Thus, in conformance with (5.14), and as a generalization of (5.15),

(5.17)\begin{equation} b_1=b(\kappa_{2,n}^{{-}1}\alpha\kappa_{2,n},\beta), \quad b_m=b(\kappa_{2,n}^{{-}m}\alpha\kappa_{2,n}^{m},\beta). \end{equation}

This is equivalent to (2.24), and thus also to (2.1), under conjugacy, and we will afford to maintain the same notation $b_m$.

Next, we will start evaluating

(5.18)\begin{equation} (\psi^{{\times}})^{(k)}(b_{1})(1)-(\psi^{{\times}})^{(k)}(b_{0})(1) \end{equation}

by using the product (5.16) and its modification $\kappa _{2,n}^{-1}\kappa _{2,j_1}^{\varepsilon _1}\!\cdots \kappa _{2,j_k}^{\varepsilon _k} \kappa _{2,n}$ in $b_1$ explained below (5.16).

The case $k=0$ is completely trivial, thus let $k=1$. Using (5.5) iteratedly on the form (5.16), one can see that the terms for $(\psi ^{\times })'(b_{1})(1)$ are the same as for $(\psi ^{\times })'(b_{0})(1)$, except those coming from taking the derivative in the factors $\kappa _{2,n}^{\pm 1}$ added. But since $\alpha$ is pure (and $\psi ^{\times }_{n}(\kappa _{2,j})=Id$) and because of (5.6), these two terms cancel as well. The result of (5.18) is thus $0$ for $k=1$.

But so prepared, we examine now $k=2$. This requires somewhat more careful collection of terms, but the procedure is clear. Since we need them often later, let us fix the permutation matrices of the cycle and transposition

(5.19)\begin{equation} P=\Pi((2\ 3\ \cdots\ n))\quad\text{and}\quad \Gamma=\Pi ((1\ 2)). \end{equation}

Only products involving the terms $\kappa _{2,n}^{\pm 1}$ added to $\alpha$ will contribute. We use again $\psi ^{\times }_n(\alpha )(1)=Id$, and it is helpful to note that

\[ (\psi^{{\times}})''(\kappa_{2,n}^{{-}1})(1)+2(\psi^{{\times}})'(\kappa_{2,n}^{{-}1})(1)(\psi^{{\times}})'(\kappa_{2,n})(1) +(\psi^{{\times}})''(\kappa_{2,n})(1)=0, \]

which is obtained like (5.6) taken one derivative further.

The result is given thus. Let in (5.16) for $l=2,\ldots,n-2$

\[ \eta_l = \sum_{j:k_j=l+1} \varepsilon_j \]

be the exponent sum of $\kappa _{2,l+1}$, also written as

(5.20)\begin{equation} \eta_l = lk_{2,l+1}-lk_{2,l+2}, \end{equation}

from (2.18). Then let $A_l$ be as in (5.10), and with (5.11) and (5.12),

(5.21)\begin{equation} \Omega = \sum_{l=2}^{n-2}\eta_l [A_l,A_{n-1}]. \end{equation}

Then we have with (5.19)

(5.22)\begin{equation} (\psi^{{\times}})''(b_{1})(1)-(\psi^{{\times}})''(b_{0})(1)=\Omega \Gamma P\Gamma. \end{equation}

The only conjugacy invariant we can really control from this is the trace, since it is linear:

\[ \text{tr}((\psi^{{\times}}_n)^{(k)}(b_m)(1))= (\text{tr}\psi^{{\times}}_n(b_m))^{(k)}(1). \]

Assume the non-vanishing condition

(5.23)\begin{equation} \text{tr}\left((\psi^{{\times}})''(b_{1})(1)-(\psi^{{\times}})''(b_{0})(1)\right)=\text{tr}(\Omega \Gamma P\Gamma)\ne 0. \end{equation}

Now note that this expression does not change when we modify $b_1$ to $b_{m+1}$ and $b_0$ to $b_m$, since the contribution of $\kappa _{2,n}^{\pm m}$ added in (5.16) does not change (5.21). Then (5.23) will imply that $\text {tr}((\psi ^{\times })''(b_{m}))(1)$ is a linear progression in $m$, so all $\text {tr}(\psi ^{\times }(b_{m}))$ will be distinct, all $b_m$ will be non-conjugate, and the ES will be TNC. We will now regard the opposite of (5.23) via (5.21) as a linear condition on the $\eta _l$.

The goal is to collect enough similar linear conditions, until only the trivial solution $\eta _l=0$ remains. Now obviously, $\text {tr}((\psi ^{\times })''(b_{m})(1))$ will give a scalar, which is utterly insufficient. To help ourselves, we replace $b_m$ by $b_m^p$ and repeat the process. This means that for $p>0$ we consider (3.5) for the series of conjugacy invariants

(5.24)\begin{equation} \upsilon(b)=\upsilon_p(b)=\text{tr}((\psi^{{\times}}_n)''(b^p)(1)). \end{equation}

It is a technical, but with the above explanation straightforward, thought, that

(5.25)\begin{equation} (\psi^{{\times}})''(b_{1}^p)(1)-(\psi^{{\times}})''(b_{0}^p)(1)=\sum_{q=0}^{p-1}\Gamma P^q\Gamma \Omega \Gamma P^{p-q}\Gamma. \end{equation}

The matrices summed are conjugated, thus the generalization of the logical negation of (5.23) becomes

(5.26)\begin{equation} p\cdot \text{tr}(\Omega \Gamma P^p\Gamma)=0, \end{equation}

wherein we remove the unnecessary first factor. Since $\pi (b)=\pi (\beta )$ has order $n-1$, it is also clear that only

(5.27)\begin{equation} 1\le p< n-1 \end{equation}

makes sense. (The case $p=0$ will give nothing.) This is again a linear condition

\[ \sum_{l=2}^{n-2}q_{p,l}\eta_l=0, \]

where

\[ q_{p,l}=\text{tr}([A_l,A_{n-1}]\cdot \Gamma P^p\Gamma). \]

Since $\Gamma P^p\Gamma$ is a permutation matrix from (5.19), and the commutator is given in (5.12), one can directly evaluate $q_{p,l}$. We obtain a homogeneous linear system with $n-2$ equations for (5.27) and $n-3$ variables for $l=2,\ldots,n-2$. The matrix $(q_{p,l})$ of this system (with rows indexed by $p$ and columns by $l$) looks

(5.28)

(When $n$ is odd, there will be a zero row in the middle.) Its kernel elements $(\eta _2,\ldots,\eta _{n-2})^T$ satisfy

(5.29)\begin{equation} \eta_l={-}\eta_{n-l}, \end{equation}

which with (5.20) (and shifting indices down by $1$, to undo the change of form of the exchange move we performed at the beginning of the proof), gives the palindromicity of (4.4). In this situation, (5.24) run out of use.

This palindromicity problem was well-known in [Reference Stoimenow28], with an example showing that then indeed two distinct $b_m$ can be conjugate. Under such circumstance for any conjugacy invariant $\upsilon$ a linear progression (3.5) will be trivial. So we look for a quadratic one.

We will take $k=3$, in the context of (5.18), and replace in (3.5)

(5.30)\begin{equation} \upsilon(b)=\upsilon_p(b)=\text{tr}((\psi^{{\times}}_n)'''(b^p)(1)). \end{equation}

Our goal is, instead of (5.18), to determine

(5.31)\begin{equation} \upsilon_p(b_{m+2})-2\upsilon_p(b_{m+1})+\upsilon_p(b_m) \end{equation}

first for $m=0$, and notice that this expression again does not depend on $m$ (for fixed $p$), and to show that it is not zero (for some $p$), unless all $\eta _l=0$. This will imply that $m\mapsto \upsilon _p(b_m)$ is a non-constant quadratic polynomial in $m$, and thus complete the proof.

Write

(5.32)\begin{equation} \overline \alpha_{m}=\kappa_{2,n}^{{-}m}\alpha\kappa_{2,n}^{m}\,, \quad \overline \beta_{m'}=\kappa_{2,n}^{m'}\beta\kappa_{2,n}^{{-}m'}, \end{equation}

and set using (5.14)

(5.33)\begin{equation} b(m,m')=b\left(\overline \alpha_{m},\ \overline\beta_{m'}\right), \end{equation}

so that with (2.24)

\[ b_{m+m'}\sim b(m,m'). \]

Instead of (5.31), we will evaluate, equivalently up to sign,

(5.34)\begin{equation} \left(\upsilon_p(b(1,1))-\upsilon_p(b(1,0))\right)-\left(\upsilon_p(b(0,1))-\upsilon_p(b(0,0))\right). \end{equation}

The calculation will also make soon clear that this expression (5.34) remains the same if we replace $1$ by $m+1$ and $0$ by $m$ in the first argument of $b(.,.)$.

It will thus be enough to show that $\eqref {eqn83}\ne 0$ for some $p$.

Again apply the Leibniz rule on the product form (5.16). The linear combination (5.34) was designed again so as to see that most terms cancel.

Since $\beta$ is not pure, the above argument for $k=1$ does not fully apply to $\overline \beta _{m'}$; rather one is left to cancel terms of differentiated $\kappa _{2,n}^{\pm 1}$ in $\overline \beta _{m'}$ [occurring in the explicit form (5.32), not inside $\beta$] in a different way.

In the end, when writing $b(m,m')^p$ as $p$ copies of the r.h.s. of (5.33) one after the other, one is left with terms, for $m,m'=0,1$,

(5.35)\begin{align} & (\psi^{{\times}}_n)'''(b(1,1)^p)(1)- (\psi^{{\times}}_n)'''(b(1,0)^p)(1)- (\psi^{{\times}}_n)'''(b(0,1)^p)(1)+ (\psi^{{\times}}_n)'''(b(0,0)^p)(1)\nonumber\\ & \quad =\sum_{i,j=1}^p\left(\begin{array}{c} \mbox{2 derivatives}\\ \mbox{in $i$-th}\ \overline\alpha_{m}\end{array}\right)\cdot \left(\begin{array}{c} \mbox{one derivative}\\ \mbox{in $j$-th}\ \overline\beta_{m'} \end{array}\right).\end{align}

Hereby we understand $\overline \alpha _{m}$ and $\overline \beta _{m'}$ differentiated, using (5.5), in their product forms (5.32), using the representation (5.16) for $\alpha$. The number of derivatives ‘in’ $\overline \alpha _{m}$ is the number of factors differentiated in the subword corresponding to one of the $p$ copies of $\overline \alpha _{m}$.

With a bit of care in collecting terms one sees that, the way the linear combination on the left of (5.35) was designed, the terms with no derivative in $\overline \beta _{m'}$ and at most one derivative in $\overline \alpha _{m}$ cancel out, as well as those with two derivatives in two different copies of $\overline \alpha _{m}$.

The two derivatives in $\overline \alpha _{m}$ are known to give $\Omega$ from (5.21), while those of $\kappa _{2,n}^{-1}$ in the $j$-th $\overline \beta _{1}$ and of $\kappa _{2,n}$ in the $j+1$-st copy (up to cyclic permutations) cancel (in the trace). There remain one derivative of $\kappa _{2,n}^{-1}$ in a $\overline \beta _{1}$ and one of $\kappa _{2,n}$ for each $\Omega$, those not separated by a copy of $P$ from $\Omega$ (but by a copy of $\Gamma$). We set

(5.36)\begin{equation} \Xi_l=[\Gamma[A_l,A_{n-1}]\Gamma,A_{n-1}]= \left[\begin{array}{c|c|c|c} 0 & 0 & n-1-l & 1-l \\ \hline 0 & 0 & l+1-n & l-1 \\ \hline n-l-1 & l+1-n & & \\ \vdots & \vdots & 0 & 0 \\ n-l-1 & l+1-n & & \\ \hline 1-l & l-1 & & \\ \vdots & \vdots & 0 & 0 \\ 1-l & l-1 & & \\ \end{array}\right], \end{equation}

with the blocks of sizes $1,1,l-1,n-1-l$, respectively (in both rows and columns, and all entries within each block identical).

When the $\kappa _{2,n}^{-1}$ in the last copy of $\overline \beta _{1}$ is differentiated, we cycle it to the left, and get, up to this modification (which does not affect the trace),

(5.37)\begin{equation} \text{tr}(5.35)=\text{tr}\left(\sum_{d=0}^{p-1} \Gamma P^d [\Gamma\Omega\Gamma,A_{n-1}] P^{p-d}\Gamma\right). \end{equation}

As $\Omega$ does not depend on $m$, it is also already clear that (5.31) does not either. We will evaluate it for $m=0$ via (5.34).

When taking traces, one can, as passing from (5.25) to (5.26), cycle powers of $\Gamma P\Gamma$ (the remnants, under setting $t=1$, of undifferentiated copies of $\beta$) to one side. We obtain thus

(5.38)\begin{equation} (5.34)= p\cdot \text{tr} \left(\sum_{l=2}^{n-2}\eta_l\Xi_l P^p \right). \end{equation}

For the purpose of testing whether this is zero or not, we can again discard the factor $p$.

Build a matrix $M$ (of $n-2$ rows), whose $n-3$ columns,

\[ \left[\begin{array}{c} \phi(1) \\ \phi(2) \\ \vdots \\ \phi(n-2) \\ \end{array} \right], \]

for $l=2,\ldots,n-2$ and $\phi =\phi _l$, are given by

(5.39)\begin{equation} \phi_l(p)=\text{tr}(\Xi_l P^p)=\left\{\small\displaystyle \begin{array}{lll} 2l-n & p\le l-1 & \\ 2l-2 & p>l-1,\ p\le n-1-l; & l-1\le n-1-l \\ 2l-n & p>n-l-1 \\ \hline 2l-n & p\le n-1-l & \\ 2l+2-2n & p>n-1-l,\ p\le l-1; & l-1>n-1-l \\ 2l-n & p>l-1 \\ \end{array}\right. . \end{equation}

The matrix $M$ is the analogue of (5.28), but is now antisymmetric when rows are reflected. (The dotted diagonals do not meet at the same element, and when $n$ is even, the middle column is zero.)

(5.40)

The resulting equation system from setting (5.38) to $0$ becomes

(5.41)\begin{equation} M\cdot (\eta_2,\dots,\eta_{n-2})^T = 0, \end{equation}

and it does enough to enforce

\[ \eta_l=\eta_{n-l}. \]

With (5.29), this implies that all $\eta _l=0$, and thus completes the proof of theorem 4.1.

Proof of theorem 4.2 We will adapt the proof of theorem 4.1.

Again, it will be easier to work with the form (2.24). Since we can w.l.o.g. conjugate $\alpha,\beta$ in (2.24) by $B_{3,n}$, we can assume (4.5) turning into

(6.1)\begin{equation} C_0=(2\ 3\ \ldots\ n'-1\ n'), \end{equation}

with $n'-2=n-n_0>0$ and $\lambda _0=n'-1$. A few (obvious) modifications from the formulation of the theorem have then to be made below.

We have now $lk_2=0$ [instead of $lk_n=0$ with (2.1)]. Let us first argue that

(6.2)\begin{equation} lk_3=\cdots=lk_{n'}. \end{equation}

For every $n_0$, we described in the proof of theorem 4.1 an invariant $\upsilon _{[n_*]}$ on $B_{n_*}$ with $\upsilon _{[n_*]}(b_m)$ being a quadratic polynomial in $m$, whose quadratic term we proved to be non-trivial to establish SS.

For $b\in B_n$, define a conjugacy invariant

(6.3)\begin{equation} \xi(b)=\sum_{D_{1,2}}\upsilon_{[|D_1|+|D_2|]}(b_{[D_1\cup D_2]}), \end{equation}

where the sum goes over unordered pairs of (distinct) cycles $D_{1,2}$ of $\pi (b)$ whose lengths are

\[ \{|D_1|,|D_2|\}=\{1,n'-1\}. \]

(This is not immediately relevant, but let us specify already here that in selecting collections of cycles, their length condition is meant as a multiset, with repeated elements counted, i.e., it is an equality of tuples modulo permutations of their entries.)

This $\xi$ is a well-defined conjugacy invariant, and in $\xi (b_m)$ the contribution of all pairs $D_{1,2}$ of cycles will vanish, unless $1\in C_{-1}=D_1$ and $2\in C_0=D_2$. This identifies

\[ \xi(b_m)=\upsilon_{[n']}((b_m)_{[D_1\cup D_2]}), \]

and the rest follows from the proof of theorem 4.1. This establishes (6.2). (The non-palindromicity assertion follows similarly.) So we obtain part 1.

For the rest of the proof we deal with part 2 and assume (4.11).

Also assume (6.2), so that

(6.4)\begin{equation} \eta_l=0\ {\rm for}\ l=2,\ldots,n'-2. \end{equation}

We can try to repeat the calculation in the proof of theorem 4.1, with the only difference being that now $P$ in (5.19) is the matrix of the new permutation

\[ \pi(b)=C_0\,C_1\cdots\, C_r. \]

If we try to evaluate (5.24), we can see that we get nothing new. However, for (5.34), we do.

Assuming from (6.4) that $l\ge n'-1=\lambda _0$, we see that the minor of the matrix in (5.36) consisting of rows and columns $2,\ldots,n'$ is comprised within the four central blocks in the form on the right of (5.36). It includes the upper left one among the four blocks, which is a single $0$ but, since $n'\ge 3$, has at least one further row and column. This matrix does not change with $P$.

Then for any $p$ not divisible by $n'-1$, say, $p=1$ (here we need $n'>2$), we get instead of (5.39), with our new $P$,

(6.5)\begin{equation} \phi_l(1)=2(l+1-n)\,\quad l=n'-1,\ldots,n-2. \end{equation}

What remains of (5.41) is the scalar equation

(6.6)\begin{equation} 0 = 2\sum_{l=n'-1}^{n-2} (l+1-n)\eta_l, \end{equation}

which with (5.20) can be rewritten as

(6.7)\begin{equation} (n-n')lk_{n'}=lk_{n'+1}+\dots+ lk_n. \end{equation}

(Note that the contributions from $C_i$, $i>0$ in $P$ give nothing for the trace. This observation will reappear for modification later.)

Also note that the symmetry of the right in $n'< j\le n$ is necessary because of the freedom to conjugate with $B_{n'+1,n}$. If (6.7) fails, which is the same as saying

\[ \sum_{i=1}^r \tau(C_i)\ne 0, \]

we have in (5.30) that $\upsilon (b_m)$ has a non-trivial quadratic term in $m$.

The expression on the right of (6.7) goes over all cycles $C_i$, $0< i\le r$. However, the idea now is to apply this argument to $(b_m)_{[D_1\cup D_2\cup D_3]}$ where $1\in D_1=C_{-1}$, $2\in D_2=C_0$ and $D_3=C_{i_0}$ is a proper single cycle (different from $D_{1,2}$).

The statement of part 2 is then what one can do about it. A problem is that when one tries to create an invariant like (6.3), we need to specify some condition, preserved under conjugacy, to tell our favourite cycle $C_{i_0}$ apart from others. However, this is not a problem if we can ‘confuse’ $C_{i_0}$ only with ‘similar’ cycles that give the same contribution to the $m$-quadratic term of $m\mapsto \upsilon _{[n^*]}(b_m)$. Then this quadratic term only multiplies by some (non-zero) number.

To this vein, fix [using the notation (4.8)]

\[ lk=\frac{lk(C_{{-}1},C_0)}{\lambda_0-1}, \]

and an $i_0$ in (4.11). Modify (6.3) to

(6.8)\begin{equation} \xi(b)=\sum_{D_{1,2,3}}\upsilon_{[|D_1|+|D_2|+|D_3|]} (b_{[D_1\cup D_2\cup D_3]}) , \end{equation}

where the sum goes over ordered triples $(D_1,D_2,D_3)$ of distinct cycles $D_{1,2,3}$ of $\pi (b)$, with the following conditions:

1. (a) $|D_1|=\lambda _0$, $|D_2|=1$, $|D_3|=\lambda _{i_0}$

2. (b) $lk(D_1,D_2)=(|D_1|-1)\cdot lk$ and

   $lk(D_2,D_3)=lk(C_{-1},C_{i_0})$.

This is again a (suitably chosen) conjugacy invariant of $b\in B_n$.

Note again that the contribution of all $(D_1,D_2,D_3)$ in (6.8) to $\xi (b_m)$ is constant (in $m$) unless both $C_{-1}$ and $C_0$ are among the $D_i$. Also, $C_{-1}$ and $C_0$ are distinguished by $\lambda _{-1}=1<\lambda _0$. Because of (4.11), the $m$-quadratic contribution of

(6.9)\begin{equation} D_1\mapsto C_0,\quad D_2\mapsto C_{{-}1},\quad D_3\mapsto C_{i_0} \end{equation}

will be non-zero. Keep in mind that this contribution (as well as the choice of invariant $\upsilon _{[n^*]}$ contributed to) depends only on the ordered quadruple

(6.10)\begin{equation} (\lambda_0,\lambda_{i_0},lk(C_{{-}1},C_0),lk(C_{{-}1},C_{i_0})). \end{equation}

To specify what other matchings are possible, let us say that $C_{i_0'}$ is similar to $C_{i_0}$ for $1\le i_0'\le r$, if $\lambda _{i_0'}=\lambda _{i_0}$ and $lk(C_{-1},C_{i_0'})=lk(C_{-1},C_{i_0})$. The contribution of

(6.11)\begin{equation} D_1\mapsto C_0,\quad D_2\mapsto C_{{-}1},\quad D_3\mapsto \text{cycle}\ C_{i_0'}\ \text{similar to}\ C_{i_0} \end{equation}

should then be clearly equal to the one of (6.9).

Only in few situations do extra matchings occur, and they create no harm either.

If $\lambda _{i_0}=1$ and $\omega :=lk(C_{i_0},C_0)=lk(C_{-1},C_0)$, then there is the additional possibility

\[ D_1\mapsto C_0,\quad D_2\mapsto C_{i_0},\quad D_3\mapsto C_{{-}1}, \]

but its contribution remains the same because (6.10) does not change compared to (6.9); it is the vertical symmetry of the subgraph

of $\Upsilon (b)=\Upsilon (b_m)$.

Similarly if $\lambda _{i_0}=\lambda _0$ and $\omega :=lk(C_{-1},C_{i_0})=lk(C_{-1},C_0)$, one can have

\[ D_1\mapsto C_{i_0},\quad D_2\mapsto C_{{-}1},\quad D_3\mapsto C_0. \]

This corresponds to the vertical reflection of

and does not change (6.10) either. How to proceed with cycles $C_{i_0'}$ similar to $C_{i_0}$ instead of $C_{i_0}$ should be clear.

Observation 7.1 Every non-degenerate ES is an example where for $|m|\to \infty$, the inequality (7.1) on $b_m$ becomes arbitrarily unsharp.

Proof. It follows from Ito's proof of theorem 3.5 that $h(b_m)\to \infty$ when $|m|\to \infty$. On the other hand, one can see using (2.9) and the embedding properties (5.3) that for fixed unit norm $t\in {\Bbb C}$, all entries of $\psi _n(b_m)(t)$ will remain bounded as $|m|\to \infty$. Thus so need to behave its eigenvalues as well. This bound is also universal (i.e. depends on $b$ only) for all such $t$.

Proof. Since $L$ has 2 components, $\pi (b)(1)=1$ and $\pi (b)(n)=n$ will imply $n=2$, which is irrelevant in (2.11). If (3.4) holds, apply theorem 3.4. Thus we are left in the situation of theorem 4.1. If the ES is not SS, then all $lk_j$ are equal, and $\ell =\sum _{j=2}^{n-1} lk_j$ is divisible by $n-2$.

Proof. Because no $n\ge b(L)$ can satisfy $n-2\mid \ell$.

Proof. The case $n=b(L)$ follows from combining theorem 3.1 and corollary 8.1. Let $n>b(L)$. Let $T(p,q)$ for the $(p,q)$-torus link. Since $n-1\ge 3$ from (2.11), which is more than the components of $L$, we have that $L\ne T(n-1,p(n-1))$ is not such a torus link. Then $L$ has infinitely many conjugacy classes of (reducible, and hence exchangeable) $n$-braid representatives by [Reference Stoimenow25].

Proof. Let us first argue that $\beta \in P_n$. Assume $\pi (\beta )$ has a non-trivial cycle $C$. W.l.o.g. conjugate $\beta$ so that $1\in C$. Take some $\alpha \in B_{2,n+1}$ with $\pi (\beta \alpha )$ being a single cycle. Then consider

\[ b_m=\delta_{[2,n]}^{2m}\beta\delta_{[2,n]}^{{-}2m}\alpha =\kappa_{1,n}^{{-}m}\beta\kappa_{1,n}^{m}\alpha. \]

They satisfy (3.4), with $n$ replaced by $n+1$. On the other hand, since $\psi _n(\beta )$ commutes with $\psi _n(\kappa _{1,n})$, all $\psi _{n+1}(b_m)$ are equal, in particular by the remarks in §2.9, so are $\nabla (L_{b_m})$. Under (3.4) we proved theorem 3.4 in [Reference Shinjo and Stoimenow24] (as explained below its formulation in §3) by using some coefficient of $\nabla (L_{b_m})$. (Since in our case, $\hat b_m$ is a knot, no subbraids of $b_m$ were taken.) This gives a contradiction. So $\beta$ is pure.

The assertion $\beta \in \Lambda _n$ now can be read as saying that all linking numbers in $\beta$ are equal. This is most easily proved by assuming the opposite. Assume again w.l.o.g. by conjugacy (in $B_n$) that $lk_{1,j}\ne lk_{1,k}$ and choose the maximal such $j< k$. So by combed normal form,

\[ \beta = \beta_0\cdot \beta'\,,\quad\beta_0=\prod_{i=1}^{k_1}\kappa_{1,j_i}^{\varepsilon_i}, \]

with $\beta '\in P_{2,n}=B_{2,n}\cap P_n$, and the exponent sums

\[ \eta_l=\sum_{\substack{i=1\\ j_i=l}}^{k_1}\varepsilon_i, \]

satisfying $\eta _j\ne 0$. But by looking at the form (5.12), we see that $[A_l,A_n]$ are linearly independent matrices for $1< l< n$, and

(9.3)\begin{equation} (\psi^{{\times}}_n)''(\beta_0\kappa_{1,n})(1)- (\psi^{{\times}}_n)''(\kappa_{1,n}\beta_0)(1)=\sum_{l=2}^{n-1}\eta_l [A_l,A_n]\ne 0. \end{equation}

(Since we assume equality of matrices, and not only conjugacy, we do no longer need to restrict ourselves to their traces.) Thus

\[ \psi_n(\beta_0\kappa_{1,n})\ne \psi_n(\kappa_{1,n}\beta_0), \]

and by noting that $\beta '\in P_{2,n}$ commutes with $\kappa _{1,n}\in P_n$,

\[ \psi_n(\beta\kappa_{1,n})=\psi_n(\beta_0\kappa_{1,n}\beta')\ne\psi_n(\kappa_{1,n}\beta_0\beta')=\psi_n(\kappa_{1,n}\beta). \]

So $\psi _n(\beta )$ does not commute with $\psi _n(\kappa _{1,n})$. This gives the contradiction to finish the proof of (9.2).

Since we proved in [Reference Stoimenow25] that $\psi _n$ is dense in a unitary group for proper $t$ even on some subgroups of $P_n$, it also follows that $\psi _n$ is irreducible on $P_n$, and that thus, for $\beta \in P_n$ as we already argued, $\psi _n(\beta )$ is a scalar matrix.

Proof. If $\beta \in P_n$, then (9.5) follows from the remainder of the proof of theorem 9.1. One can generalize (9.3) to

\[ (\psi^{{\times}}_n)''(\beta_0^p\kappa_{1,n})(1)- (\psi^{{\times}}_n)''(\kappa_{1,n}\beta_0^p)(1)= p \cdot \sum_{l=2}^{n-1}\eta_l [A_l,A_n]\ne 0. \]

[Note that thus we may even allow in (9.4), $p$ to depend on $\gamma$, but this remains equivalent, as $P_n$ is finitely generated.]

Proof. If $\psi _n(b)$ is idempotent, then because of the determinant (2.10), we have $e(b)=0$, and the part of $b$ in the second factor of the decomposition on the right of (9.2) is trivial. This leaves $b\in P_n^c$.

For the second assertion, observe that $b\in \ker (\psi _n)$ readily implies $b\in P_n$ by setting $t=1$ [and keeping (5.7) and (5.9) in mind].

Proof. We return to the explanation in remark 7.3, which now brings something.

Let $C$ be a cycle of $\pi (b)$ with $\lambda =|C|$ and $e(b_{[C]})\ne 0$. Then, because of (2.19), for $l=\lambda$, in $b^l$, we have

(9.7)\begin{equation} lk_{ij}\ne 0\ \text{for some}\ i,j\in C. \end{equation}

If (9.6) fails, then take $l=\mathop {lcm}\mathord {}(\lambda,\lambda ')$ for $\lambda '=|C'|$, showing in $b^l$

(9.8)\begin{equation} lk_{ij}\ne 0\ \text{for some}\ i\in C, j\in C'. \end{equation}

Then there is some $p>0$ so that $b^*=b^{pl}$ is pure, and (9.7) or (9.8) still holds in $b^*$. Now, if $\psi _n(b^*)$ were idempotent, then by corollary 9.4, $b^*\in P_n^c$, contradicting (9.7) or (9.8). Thus $\psi _n(b^*)$ is not idempotent, and neither is $\psi _n(b)$.

If $e(b_{[C]})=0$, then it is easy to see, essentially because even cycles require an odd number of transpositions, that $\lambda$ must be odd.