# Injectivity on the set of conjugacy classes
of some monomorphisms between Artin groups

Eon-Kyung Lee and Sang-Jin Lee  Department of Mathematics, Sejong University,
Seoul, 143-747, Korea<EMAIL_ADDRESS>Department of Mathematics, Konkuk
University, Seoul, 143-701, Korea<EMAIL_ADDRESS>
###### Abstract.

There are well-known monomorphisms between the Artin groups of finite type
$\mathbf{A}_{n}$, $\mathbf{B}_{n}=\mathbf{C}_{n}$ and affine type
$\tilde{\mathbf{A}}_{n-1}$, $\tilde{\mathbf{C}}_{n-1}$. The Artin group
$A(\mathbf{A}_{n})$ is isomorphic to the $(n+1)$-strand braid group $B_{n+1}$,
and the other three Artin groups are isomorphic to some subgroups of
$B_{n+1}$. The inclusions between these subgroups yield monomorphisms
$A(\mathbf{B}_{n})\to A(\mathbf{A}_{n})$, $A(\tilde{\mathbf{A}}_{n-1})\to
A(\mathbf{B}_{n})$ and $A(\tilde{\mathbf{C}}_{n-1})\to A(\mathbf{B}_{n})$.
There are another type of monomorphisms $A(\mathbf{B}_{d})\to
A(\mathbf{A}_{md-1})$, $A(\mathbf{B}_{d})\to A(\mathbf{B}_{md})$ and
$A(\mathbf{B}_{d})\to A(\mathbf{A}_{md})$ which are induced by isomorphisms
between Artin groups of type $\mathbf{B}$ and centralizers of periodic braids.

In this paper, we show that the monomorphisms $A(\mathbf{B}_{d})\to
A(\mathbf{A}_{md-1})$, $A(\mathbf{B}_{d})\to A(\mathbf{B}_{md})$ and
$A(\mathbf{B}_{d})\to A(\mathbf{A}_{md})$ induce injective functions on the
set of conjugacy classes, and that none of the monomorphisms
$A(\mathbf{B}_{n})\to A(\mathbf{A}_{n})$, $A(\tilde{\mathbf{A}}_{n-1})\to
A(\mathbf{B}_{n})$ and $A(\tilde{\mathbf{C}}_{n-1})\to A(\mathbf{B}_{n})$ does
so.

_Key words:_ Artin group, braid group, conjugacy class, Nielsen-Thurston
classification
_MSC:_ 20F36, 20F10

##  1\. Introduction

Let $M$ be a symmetric $n\times n$ matrix with entries
$m_{ij}\in\\{2,3,\ldots,\infty\\}$ for $i\neq j$, and $m_{ii}=1$ for $1\leq
i\leq n$. The _Artin group_ of type $M$ is defined by the presentation

$A(M)=\langle
s_{1},\ldots,s_{n}\mid\underbrace{s_{i}s_{j}s_{i}\cdots}_{m_{ij}}=\underbrace{s_{j}s_{i}s_{j}\cdots}_{m_{ij}}\quad\mbox{for
all $i\neq j$, $m_{ij}\neq\infty$}\rangle.$

The _Coxeter group_ $W(M)$ of type $M$ is the quotient of $A(M)$ by the
relation $s_{i}^{2}=1$. The Artin group $A(M)$ is said to be _of finite type_
if $W(M)$ is finite, and _of affine type_ if $W(M)$ acts as a proper,
cocompact group of isometries on some Euclidean space with the generators
$s_{1},\ldots,s_{n}$ acting as affine reflections. Artin groups are usually
described by _Coxeter graphs_ , whose vertices are numbered $1,\ldots,n$ and
which has an edge labelled $m_{ij}$ between the vertices $i$ and $j$ whenever
$m_{ij}\geq 3$ or $m_{ij}=\infty$. The label 3 is usually suppressed.

There are well-known monomorphisms between the Artin groups of finite type
$\mathbf{A}_{n}$, $\mathbf{B}_{n}=\mathbf{C}_{n}$, and affine type
$\tilde{\mathbf{A}}_{n-1}$, $\tilde{\mathbf{C}}_{n-1}$. The Coxeter graphs
associated to these Artin groups are as follows.

$\mathbf{A}_{n}$ |  |  | $\tilde{\mathbf{A}}_{n-1}$ |
---|---|---|---|---
$\mathbf{B}_{n}$ |  |  | $\tilde{\mathbf{C}}_{n-1}$ |

In this paper, we investigate whether these monomorphisms induce injective
functions on the set of conjugacy classes or not. Before stating our results,
let us review briefly braid groups and monomorphisms between these Artin
groups.

Let $D^{2}=\\{z\in\mathbb{C}\mid|z|\leq n+1\\}$, and let $D_{n}$ be the
$n$-punctured disk $D^{2}\setminus\\{1,\ldots,n\\}$. The braid group $B_{n}$
is defined as the group of self-diffeomorphisms of $D_{n}$ that fix the
boundary pointwise, modulo isotopy relative to the boundary. Equivalently,
$n$-braids can be defined as isotopy classes of collections of pairwise
disjoint $n$ strands $l=l_{1}\cup\cdots\cup l_{n}$ in $D^{2}\times[0,1]$ such
that $l\cap(D^{2}\times\\{t\\})$ consists of $n$ points for each $t\in[0,1]$,
and $l\cap(D^{2}\times\\{0,1\\})=\\{1,\ldots,n\\}\times\\{0,1\\}$. The
admissible isotopies lie in the interior of $D^{2}\times[0,1]$. The $n$-braid
group $B_{n}$ has the well-known Artin presentation [Art25]:

$B_{n}=\left\langle\sigma_{1},\ldots,\sigma_{n-1}\biggm{|}\begin{array}[]{ll}\sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i}&\mbox{if
}|i-j|\geq 2\\\
\sigma_{i}\sigma_{j}\sigma_{i}=\sigma_{j}\sigma_{i}\sigma_{j}&\mbox{if
}|i-j|=1\end{array}\right\rangle.$

######  Definition 1.1.

For $\alpha\in B_{n}$, let $\pi_{\alpha}$ denote its induced permutation on
$\\{1,\ldots,n\\}$. If $\pi_{\alpha}(i)=i$, we say that $\alpha$ is _$i$
-pure_, or the $i$-th strand of $\alpha$ is _pure_. For
$P\subset\\{1,\ldots,n\\}$, we say that $\alpha$ is _$P$ -pure_ if $\alpha$ is
$i$-pure for each $i\in P$.

######  Definition 1.2.

Let $B_{n,1}$ denote the subgroup of $B_{n}$ consisting of 1-pure braids. Let
$\operatorname{lk}:B_{n,1}\to\mathbb{Z}$ be the homomorphism measuring the
linking number of the first strand with the other strands: $B_{n,1}$ is
generated by $\sigma_{1}^{2},\sigma_{2},\ldots,\sigma_{n-1}$, and
$\operatorname{lk}$ is defined by $\operatorname{lk}(\sigma_{1}^{2})=1$ and
$\operatorname{lk}(\sigma_{i})=0$ for $i\geq 2$. A braid $\alpha$ is said to
be _1-unlinked_ if it is 1-pure and $\operatorname{lk}(\alpha)=0$. Let
$\nu:B_{n,1}\to B_{n-1}$ be the homomorphism deleting the first strand.

For example, the braid in Figure 1 is $\\{1,4,5\\}$-pure and 1-unlinked. Note
that $\\{1,\ldots,n\\}$-pure braids are nothing more than pure braids in the
usual sense, and that $\operatorname{lk}$ is a conjugacy invariant of 1-pure
braids because
$\operatorname{lk}(\beta\alpha\beta^{-1})=\operatorname{lk}(\beta)+\operatorname{lk}(\alpha)-\operatorname{lk}(\beta)=\operatorname{lk}(\alpha)$
for $\alpha,\beta\in B_{n,1}$.

Figure 1. This braid is $\\{1,4,5\\}$-pure and 1-unlinked.

From the presentation of $B_{n}$, it is obvious that the Artin group
$A(\mathbf{A}_{n})$ is isomorphic to $B_{n+1}$. The other Artin groups
$A(\mathbf{B}_{n})$, $A(\tilde{\mathbf{A}}_{n-1})$ and
$A(\tilde{\mathbf{C}}_{n-1})$ are known to be isomorphic to subgroups of
$B_{n+1}$ as follows [All02, CC05, BM07].

$\displaystyle A(\mathbf{B}_{n})$ $\displaystyle\simeq$
$\displaystyle\\{\alpha\in B_{n+1}\mid\mbox{$\alpha$ is
1-pure}\\}=B_{n+1,1}\subset B_{n+1};$ $\displaystyle
A(\tilde{\mathbf{A}}_{n-1})$ $\displaystyle\simeq$ $\displaystyle\\{\alpha\in
B_{n+1}\mid\mbox{$\alpha$ is 1-unlinked}\\}\subset B_{n+1,1};$ $\displaystyle
A(\tilde{\mathbf{C}}_{n-1})$ $\displaystyle\simeq$ $\displaystyle\\{\alpha\in
B_{n+1}\mid\mbox{$\alpha$ is $\\{1,n+1\\}$-pure}\\}\subset B_{n+1,1}.$

From now on, we identify $A(\mathbf{A}_{n})$ with $B_{n+1}$, and the Artin
groups $A(\mathbf{B}_{n})$, $A(\tilde{\mathbf{A}}_{n-1})$ and
$A(\tilde{\mathbf{C}}_{n-1})$ with the corresponding subgroups of $B_{n+1}$
induced by the above isomorphisms. Then the inclusions between the subgroups
of $B_{n+1}$ yield monomorphisms $\psi_{1}:A(\mathbf{B}_{n})\to
A(\mathbf{A}_{n})$, $\psi_{2}:A(\tilde{\mathbf{A}}_{n-1})\to
A(\mathbf{B}_{n})$ and $\psi_{3}:A(\tilde{\mathbf{C}}_{n-1})\to
A(\mathbf{B}_{n})$. See Figure 2.

$\textstyle{A(\mathbf{A}_{n})}$$\textstyle{A(\mathbf{B}_{n})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{1}}$$\textstyle{A(\tilde{\mathbf{A}}_{n-1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{2}}$$\textstyle{A(\tilde{\mathbf{C}}_{n-1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{3}}$
---
$\textstyle{A(\mathbf{A}_{md-1})}$$\textstyle{A(\mathbf{A}_{md})}$$\textstyle{A(\mathbf{B}_{md})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu}$$\scriptstyle{\psi_{1}}$$\textstyle{A(\mathbf{B}_{d})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{4}}$$\scriptstyle{\psi_{5}}$$\scriptstyle{\psi_{6}}$

Figure 2. $\psi_{1},\ldots,\psi_{6}$ are monomorphisms between Artin groups.

There are another type of monomorphisms which can be described by isomorphisms
between Artin groups of type $\mathbf{B}$ and centralizers of periodic braids.
The well-known Nielsen-Thurston classification of mapping classes of surfaces
(possibly with punctures) with negative Euler characteristic into periodic,
reducible and pseudo-Anosov ones [Thu88] yields an analogous classification of
braids. An $n$-braid $\alpha$ is said to be _periodic_ if some power of it is
a power of $\Delta^{2}$, where
$\Delta=\sigma_{1}(\sigma_{2}\sigma_{1})\cdots(\sigma_{n-1}\cdots\sigma_{1})$;
_reducible_ if there is an essential curve system in $D_{n}$ whose isotopy
class is invariant under the action of $\alpha$; _pseudo-Anosov_ if there is a
pseudo-Anosov diffeomorphism $f$ defined on the interior of $D_{n}$
representing $\alpha$ modulo $\Delta^{2}$, that is, there exist a pair of
transverse measured foliations $(F^{s},\mu^{s})$ and $(F^{u},\mu^{u})$ and a
real $\lambda>1$, called _dilatation_ , such that
$f(F^{s},\mu^{s})=(F^{s},\lambda^{-1}\mu^{s})$ and
$f(F^{u},\mu^{u})=(F^{u},\lambda\mu^{u})$. Every braid belongs to one of the
following three mutually disjoint classes: periodic; pseudo-Anosov; non-
periodic and reducible. We say that two braids are _of the same Nielsen-
Thurston type_ if both of them belong to the same class.

Let $\delta=\sigma_{n-1}\cdots\sigma_{1}$ and $\epsilon=\delta\sigma_{1}$,
then $\delta^{n}=\Delta^{2}=\epsilon^{n-1}$. If we need to specify the braid
index $n$, we will write $\delta=\delta_{(n)}$, $\epsilon=\epsilon_{(n)}$ and
$\Delta=\Delta_{(n)}$. The braids $\delta$ and $\epsilon$ are induced by rigid
rotations of the punctured disk as in Figure 3 when the punctures are at the
origin or evenly distributed on a round circle centered at the origin. Due to
L.E.J. Brouwer [Bro19], B. de Kérékjártó [Ker19] and S. Eilenberg [Eil34] (see
also [CK94]), it is known that an $n$-braid is periodic if and only if it is
conjugate to a power of $\delta$ or $\epsilon$.

|  |
---|---|---
(a) $\delta_{(n)}=\sigma_{n-1}\sigma_{n-2}\cdots\sigma_{1}\in B_{n}$ |  | (b) $\epsilon_{(n)}=\delta_{(n)}\sigma_{1}\in B_{n}$
Figure 3.  The braids $\delta$ and $\epsilon$ are represented by rigid
rotations of punctured disks.

For a group $G$ and an element $g\in G$, let $Z_{G}(g)$ denote the centralizer
of $g$ in $G$, that is, $Z_{G}(g)=\\{h\in G:gh=hg\\}$. If the group $G$ is
clear from the context, we write simply $Z(g)$. The following theorem is a
consequence of the result of D. Bessis, F. Digne and J. Michel [BDM02] on
irreducible complex reflection groups. See also [GW04].

######  Theorem 1.3 ([BDM02]).

Let $\omega\in B_{n}$ be conjugate to $\delta^{k}$ (resp. $\epsilon^{k}$),
where $k$ is not a multiple of $n$ (resp. $n-1$). Then the centralizer
$Z(\omega)$ is isomorphic to the Artin group $A(\mathbf{B}_{d})$, where
$d=\gcd(k,n)$ (resp. $d=\gcd(k,n-1)$).

Let $m\geq 2$ and $d\geq 1$. The centralizer $Z(\epsilon_{(md+1)}^{d})$
consists of 1-pure braids because the first strand of $\epsilon_{(md+1)}^{d}$
is the only pure strand. By theorem 1.3, we have

$\displaystyle A(\mathbf{B}_{d})$ $\displaystyle\simeq$ $\displaystyle
Z(\epsilon_{(md+1)}^{d})\subset B_{md+1,1}=A(\mathbf{B}_{md})\subset
B_{md+1}=A(\mathbf{A}_{md});$ $\displaystyle A(\mathbf{B}_{d})$
$\displaystyle\simeq$ $\displaystyle Z(\delta_{(md)}^{d})\subset
B_{md}=A(\mathbf{A}_{md-1}).$

Compositions of the above isomorphisms and inclusions yield monomorphisms
$\psi_{4}:A(\mathbf{B}_{d})\to A(\mathbf{B}_{md})$,
$\psi_{5}:A(\mathbf{B}_{d})\to A(\mathbf{A}_{md-1})$ and
$\psi_{6}:A(\mathbf{B}_{d})\to A(\mathbf{A}_{md})$. The monomorphisms
$\psi_{4}$ and $\psi_{5}$ in [BDM02] and [GW04] are such that
$\psi_{4}(\epsilon_{(d+1)})=\epsilon_{(md+1)}$,
$\psi_{5}(\epsilon_{(d+1)})=\delta_{(md)}$ and $\psi_{5}=\nu\circ\psi_{4}$,
where $\nu:B_{md+1,1}\to B_{md}$ is the homomorphism deleting the first strand
(see Definition 1.2). By definition, $\psi_{6}=\psi_{1}\circ\psi_{4}$. See
Figure 2. In [BGG06] J. Birman, V. Gebhardt and J. González-Meneses described
explicitly in terms of standard generators the monomorphism
$\psi_{5}:A(\mathbf{B}_{d})\to A(\mathbf{A}_{md-1})$ when $m=2$.

The following theorem is the main result of this paper.

######  Theorem 1.4.

Let $\omega$ be a periodic $n$-braid, and let $\alpha,\beta\in Z(\omega)$. If
$\alpha$ and $\beta$ are conjugate in $B_{n}$, then they are conjugate in
$Z(\omega)$.

We prove the above theorem, relying on the Nielsen-Thurston classification of
braids. In the case of pseudo-Anosov braids, we obtain a stronger result: if
$\alpha,\beta\in Z(\omega)$ are pseudo-Anosov and conjugate in $B_{n}$, then
any conjugating element from $\alpha$ to $\beta$ belongs to $Z(\omega)$ (see
Proposition 5.1). The case of periodic braids is easy to prove due to the
uniqueness of roots up to conjugacy for Artin groups of type $\mathbf{B}$
[LL07], hence most part of this paper is devoted to the case of reducible
braids.

Using the characterization of $Z(\omega)$ as the Artin group of type
$\mathbf{B}$, we obtain the following corollary which is equivalent to the
above theorem (see Lemma 4.3 for the equivalence).

######  Corollary 1.5.

For $m\geq 2$ and $d\geq 1$, the monomorphisms $\psi_{5}:A(\mathbf{B}_{d})\to
A(\mathbf{A}_{md-1})$ and $\psi_{6}:A(\mathbf{B}_{d})\to A(\mathbf{A}_{md})$
induce injective functions on the set of conjugacy classes.

The above corollary implies that $\psi_{4}$ is also injective on the set of
conjugacy classes because $\psi_{5}=\nu\circ\psi_{4}$ and
$\psi_{6}=\psi_{1}\circ\psi_{4}$.

We also investigate the injectivity for the well-known monomorphisms of
$A(\tilde{\mathbf{A}}_{n-1})$ and $A(\tilde{\mathbf{C}}_{n-1})$ into
$A(\mathbf{B}_{n})$, and a further monomorphism of $A(\mathbf{B}_{n})$ into
$A(\mathbf{A}_{n})$, and have the following.

######  Theorem 1.6.

For $n\geq 3$, none of the monomorphisms $\psi_{1}:A(\mathbf{B}_{n})\to
A(\mathbf{A}_{n})$, $\psi_{2}:A(\tilde{\mathbf{A}}_{n-1})\to
A(\mathbf{B}_{n})$ and $\psi_{3}:A(\tilde{\mathbf{C}}_{n-1})\to
A(\mathbf{B}_{n})$ induces an injective function on the set of conjugacy
classes.

For the proof of the above theorem, we present three examples which show that
the monomorphisms $\psi_{1}$, $\psi_{2}$ and $\psi_{3}$ are not injective on
the set of conjugacy classes. For instance, the first example is a pair of
elements in $B_{n+1,1}$ that are conjugate in $B_{n+1}$, but not conjugate in
$B_{n+1,1}$.

We close this section with a remark on a monomorphism from
$A(\mathbf{A}_{n-1})$ to $A(\mathbf{D}_{n})$. The Coxeter graph of type
$\mathbf{D}_{n}$ is as follows.

$\mathbf{D}_{n}$ |
---|---

Let $\sigma_{1},\ldots,\sigma_{n-1}$ and $t_{1},\ldots,t_{n}$ be the standard
generators of $A(\mathbf{A}_{n-1})$ and $A(\mathbf{D}_{n})$. Let
$\iota:A(\mathbf{A}_{n-1})\to A(\mathbf{D}_{n})$ and $\pi:A(\mathbf{D}_{n})\to
A(\mathbf{A}_{n-1})$ be the homomorphisms defined by
$\iota(\sigma_{i})=t_{i+1}$ for $1\leq i\leq n-1$; $\pi(t_{1})=\sigma_{1}$;
$\pi(t_{i})=\sigma_{i-1}$ for $i\geq 2$. Because $\pi\circ\iota$ is the
identity, $\pi$ is an epimorphism and $\iota$ is a monomorphism, further,
$\iota$ is injective on the set of conjugacy classes. It is known by J. Crisp
and L. Paris that $A(\mathbf{D}_{n})$ is isomorphic to a semidirect product
$F_{n-1}\rtimes A(\mathbf{A}_{n-1})$, where $F_{n-1}$ is a free group of rank
$n-1$ [CP05].

##  2\. Preliminaries

In this section, we review the uniqueness of roots up to conjugacy in some
Artin groups, and some tools useful in handling reducible braids such as
canonical reduction system and standard reduction system.

###  2.1. Periodic braids

The center of $B_{n}$ is the infinite cyclic group generated by $\Delta^{2}$.
An $n$-braid is periodic if and only if it is of finite order in the central
quotient $B_{n}/\langle\Delta^{2}\rangle$. Note that $\delta$ generates a
cyclic group of order $n$ in $B_{n}/\langle\Delta^{2}\rangle$, hence
$\delta^{a}$ and $\delta^{b}$ generate the same group in
$B_{n}/\langle\Delta^{2}\rangle$ if and only if $\gcd(a,n)=\gcd(b,n)$. In
fact, we have the following.

######  Lemma 2.1.

Let $a$ and $b$ be integers. In $B_{n}$, the following hold.

  * (i)

$Z(\delta^{a})=Z(\delta^{b})$ if and only if $\gcd(a,n)=\gcd(b,n)$.

  * (ii)

$Z(\epsilon^{a})=Z(\epsilon^{b})$ if and only if $\gcd(a,n-1)=\gcd(b,n-1)$.

  * (iii)

$Z(\delta^{a})=Z(\delta^{\gcd(a,n)})$ and
$Z(\epsilon^{a})=Z(\epsilon^{\gcd(a,n-1)})$.

In the above, (iii) is a direct consequence of (i) and (ii).

The central quotient $B_{n}/\langle\Delta^{2}\rangle$ is the group of isotopy
classes of self-diffeomorphisms of $D_{n}$ without the condition of fixing
$\partial D_{n}$ pointwise on self-diffeomorphisms and isotopies. If a
diffeomorphism $f:D_{n}\to D_{n}$ does not fix $\partial D_{n}$ pointwise, it
determines a braid modulo $\Delta^{2}$.

###  2.2. Uniqueness of roots up to conjugacy in some Artin groups

It is known that the $k$-th roots of a braid are unique up to conjugacy by J.
González-Meneses [Gon03], and this is generalized to the Artin groups
$A(\mathbf{B}_{n})=A(\mathbf{C}_{n})$, $A(\tilde{\mathbf{A}}_{n})$ and
$A(\tilde{\mathbf{C}}_{n})$ [LL07].

######  Theorem 2.2 (J. González-Meneses [Gon03]).

Let $\alpha$ and $\beta$ be $n$-braids such that $\alpha^{k}=\beta^{k}$ for
some nonzero integer $k$. Then $\alpha$ and $\beta$ are conjugate in $B_{n}$.

######  Theorem 2.3 (E.-K. Lee and S.-J. Lee [LL07]).

Let $G$ denote one of the Artin groups of finite type $\mathbf{A}_{n}$,
$\mathbf{B}_{n}=\mathbf{C}_{n}$ and affine type $\tilde{\mathbf{A}}_{n-1}$,
$\tilde{\mathbf{C}}_{n-1}$. If $\alpha,\beta\in G$ are such that
$\alpha^{k}=\beta^{k}$ for some nonzero integer $k$, then $\alpha$ and $\beta$
are conjugate in $G$.

######  Corollary 2.4.

Let $G$ denote one of the Artin groups of finite type $\mathbf{A}_{n}$,
$\mathbf{B}_{n}=\mathbf{C}_{n}$ and affine type $\tilde{\mathbf{A}}_{n-1}$,
$\tilde{\mathbf{C}}_{n-1}$. Let $\alpha$ and $\beta$ be elements of $G$, and
let $k$ be a nonzero integer. Then, $\alpha$ and $\beta$ are conjugate in $G$
if and only if so are $\alpha^{k}$ and $\beta^{k}$.

###  2.3. Canonical reduction system of reducible braids

Abusing notation, we use the same symbol for a curve and its isotopy class.
Hence, for curves $C_{1}$ and $C_{2}$, $C_{1}=C_{2}$ means that $C_{1}$ and
$C_{2}$ are isotopic, unless stated otherwise explicitly.

A curve system $\mathcal{C}$ in $D_{n}$ means a finite collection of disjoint
simple closed curves in $D_{n}$. It is said to be _essential_ if each
component is homotopic neither to a point nor to a puncture nor to the
boundary. For an $n$-braid $\alpha$ and a curve system $\mathcal{C}$ in
$D_{n}$, let $\alpha*\mathcal{C}$ denote the left action of $\alpha$ on
$\mathcal{C}$. Recall that an $n$-braid $\alpha$ is reducible if
$\alpha*\mathcal{C}=\mathcal{C}$ for some essential curve system $\mathcal{C}$
in $D_{n}$, called a _reduction system_ of $\alpha$.

For a reduction system $\mathcal{C}$ of an $n$-braid $\alpha$, let
$D_{\mathcal{C}}$ be the closure of $D_{n}\setminus N(\mathcal{C})$ in
$D_{n}$, where $N(\mathcal{C})$ is a regular neighborhood of $\mathcal{C}$.
The restriction of $\alpha$ induces a self-diffeomorphism on $D_{\mathcal{C}}$
that is well defined up to isotopy. Due to J. Birman, A. Lubotzky and J.
McCarthy [BLM83] and N. V. Ivanov [Iva92], for any $n$-braid $\alpha$, there
is a unique _canonical reduction system_ $\operatorname{\mathcal{R}}(\alpha)$
with the following properties.

  1. (i)

$\operatorname{\mathcal{R}}(\alpha^{m})=\operatorname{\mathcal{R}}(\alpha)$
for all $m\neq 0$.

  2. (ii)

$\operatorname{\mathcal{R}}(\beta\alpha\beta^{-1})=\beta*\operatorname{\mathcal{R}}(\alpha)$
for all $\beta\in B_{n}$.

  3. (iii)

The restriction of $\alpha$ to each component of
$D_{\operatorname{\mathcal{R}}(\alpha)}$ is either periodic or pseudo-Anosov.
A reduction system with this property is said to be _adequate_.

  4. (iv)

If $\mathcal{C}$ is an adequate reduction system of $\alpha$, then
$\operatorname{\mathcal{R}}(\alpha)\subset\mathcal{C}$.

Note that a braid $\alpha$ is non-periodic and reducible if and only if
$\operatorname{\mathcal{R}}(\alpha)\neq\emptyset$. Let
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ denote the
collection of outermost components of $\operatorname{\mathcal{R}}(\alpha)$.
Then $\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ satisfies the
properties (i) and (ii).

###  2.4. Reducible braids with a standard reduction system

Here we introduce some notions in [LL08] that will be used in computations
involving reducible braids.

######  Definition 2.5.

An essential curve system in $D_{n}$ is said to be _standard_ if each
component is isotopic to a round circle centered at the real axis as in Figure
4 (a), and _unnested_ if none of its components encloses another component as
in Figure 4 (b). Two curve systems $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ in
$D_{n}$ are said to be _of the same type_ if there is a diffeomorphism
$f:D_{n}\to D_{n}$ such that $f(\mathcal{C}_{1})$ is isotopic to
$\mathcal{C}_{2}$.

|  |
---|---|---
(a) a standard curve system |  | (b) the unnested standard curve system $\mathcal{C}_{(1,1,2,1,2,3)}$
Figure 4. Some standard curve systems in $D_{10}$

$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ is unnested for any
non-periodic reducible braid $\alpha$. If $\alpha$ and $\beta$ are conjugate,
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)$ are of the same type.

Recall that an ordered $k$-tuple $\mathbf{n}=(n_{1},\ldots,n_{k})$ is a
$k$-composition of $n$ if $n=n_{1}+\cdots+n_{k}$ and $n_{i}\geq 1$ for each
$i$. Unnested standard curve systems in $D_{n}$ are in one-to-one
correspondence with $k$-compositions of $n$ for $2\leq k\leq n-1$. The
$k$-braid group $B_{k}$ acts on the set of $k$-compositions as

$\alpha*(n_{1},\ldots,n_{k})=(n_{\theta^{-1}(1)},\ldots,n_{\theta^{-1}(k)}),$

where $\alpha\in B_{k}$ and $\theta=\pi_{\alpha}$.

######  Definition 2.6.

For a composition $\mathbf{n}=(n_{1},\ldots,n_{k})$ of $n$, let
$\mathcal{C}_{\mathbf{n}}$ denote the curve system $\cup_{n_{i}\geq 2}C_{i}$
in $D_{n}$, where $C_{i}$ with $n_{i}\geq 2$ is a round circle enclosing the
punctures $\\{m:\sum_{j=1}^{i-1}n_{j}<m\leq\sum_{j=1}^{i}n_{j}\\}$.

For example, Figure 4 (b) shows $\mathcal{C}_{\mathbf{n}}$ for
$\mathbf{n}=(1,1,2,1,2,3)$. For compositions $\mathbf{n}_{1}$ and
$\mathbf{n}_{2}$ of $n$, $\mathcal{C}_{\mathbf{n}_{1}}$ and
$\mathcal{C}_{\mathbf{n}_{2}}$ are of the same type if and only if
$\mathbf{n}_{1}$ and $\mathbf{n}_{2}$ induce the same partition of $n$ such as
$\mathbf{n}_{1}=(2,2,1)$ and $\mathbf{n}_{2}=(2,1,2)$.

######  Definition 2.7.

Let $\mathbf{n}=(n_{1},\cdots,n_{k})$ be a composition of $n$.

  * (i)

Let $\hat{\alpha}=l_{1}\cup\cdots\cup l_{k}$ be a $k$-braid, where the right
endpoint of $l_{i}$ is at the $i$-th position from bottom. We define
$\langle\hat{\alpha}\rangle_{\mathbf{n}}$ as the $n$-braid obtained from
$\hat{\alpha}$ by taking $n_{i}$ parallel copies of $l_{i}$ for each $i$. See
Figure 5 (a).

  * (ii)

Let $\alpha_{i}\in B_{n_{i}}$ for $i=1,\ldots,k$. We define
$(\alpha_{1}\oplus\cdots\oplus\alpha_{k})_{\mathbf{n}}$ as the $n$-braid
$\alpha_{1}^{\prime}\alpha_{2}^{\prime}\cdots\alpha_{k}^{\prime}$, where each
$\alpha_{i}^{\prime}$ is the image of $\alpha_{i}$ under the homomorphism
$B_{n_{i}}\to B_{n}$ defined by
$\sigma_{j}\mapsto\sigma_{(n_{1}+\cdots+n_{i-1})+j}$. See Figure 5 (b).

We will use the notation
$\alpha=\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{1}\oplus\cdots\oplus\alpha_{k})_{\mathbf{n}}$
frequently. See Figure 5 (c). Let $\bigoplus_{i=1}^{m}\alpha_{i}$ denote
$\alpha_{1}\oplus\cdots\oplus\alpha_{m}$, and let $m\alpha$ denote
$\bigoplus_{i=1}^{m}\alpha$. For example,

$\begin{array}[]{rcl}(\alpha_{0}\oplus{\textstyle\bigoplus_{i=1}^{d}}(\alpha_{i,1}\oplus\cdots\oplus\alpha_{i,m}))_{\mathbf{n}}&=&(\alpha_{0}\oplus(\alpha_{1,1}\oplus\cdots\oplus\alpha_{1,m})\oplus\cdots\oplus(\alpha_{d,1}\oplus\cdots\oplus\alpha_{d,m}))_{\mathbf{n}},\\\
(\alpha_{0}\oplus m\alpha_{1}\oplus\cdots\oplus
m\alpha_{r})_{\mathbf{n}}&=&(\alpha_{0}\oplus(\underbrace{\alpha_{1}\oplus\cdots\oplus\alpha_{1}}_{m})\oplus\cdots\oplus(\underbrace{\alpha_{r}\oplus\cdots\oplus\alpha_{r}}_{m}))_{\mathbf{n}}.\end{array}$

|  |  |  |
---|---|---|---|---
(a) $\langle\sigma_{1}^{-1}\sigma_{2}\rangle_{\mathbf{n}}$ |  | (b) $(\sigma_{1}^{3}\oplus\sigma_{1}^{-2}\sigma_{2}^{3}\oplus 1)_{\mathbf{n}}$ |  | (c) $\langle\sigma_{1}^{-1}\sigma_{2}\rangle_{\mathbf{n}}(\sigma_{1}^{3}\oplus\sigma_{1}^{-2}\sigma_{2}^{3}\oplus 1)_{\mathbf{n}}$
Figure 5. $\mathbf{n}=(2,3,1)$

######  Lemma 2.8 ([LL08, Lemmas 3.5 and 3.6]).

Let $\mathbf{n}=(n_{1},\ldots,n_{k})$ be a composition of $n$.

  1. (i)

The expression
$\alpha=\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{1}\oplus\cdots\oplus\alpha_{k})_{\mathbf{n}}$
is unique, that is, if
$\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{1}\oplus\cdots\oplus\alpha_{k})_{\mathbf{n}}=\langle\hat{\beta}\rangle_{\mathbf{n}}(\beta_{1}\oplus\cdots\oplus\beta_{k})_{\mathbf{n}}$,
then $\hat{\alpha}=\hat{\beta}$ and $\alpha_{i}=\beta_{i}$ for $1\leq i\leq
k$.

  2. (ii)

For $\alpha\in B_{n}$, $\alpha*\mathcal{C}_{\mathbf{n}}$ is standard if and
only if $\alpha$ can be expressed as
$\alpha=\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{1}\oplus\cdots\oplus\alpha_{k})_{\mathbf{n}}$.
In this case,
$\alpha*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\hat{\alpha}*\mathbf{n}}$.

  3. (iii)

$\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{1}\oplus\cdots\oplus\alpha_{k})_{\mathbf{n}}=(\alpha_{\theta^{-1}(1)}\oplus\cdots\oplus\alpha_{\theta^{-1}(k)})_{\hat{\alpha}\ast\mathbf{n}}\langle\hat{\alpha}\rangle_{\mathbf{n}}$,
where $\theta=\pi_{\hat{\alpha}}$.

  4. (iv)

$\langle\hat{\alpha}\hat{\beta}\rangle_{\mathbf{n}}=\langle\hat{\alpha}\rangle_{\hat{\beta}*\mathbf{n}}\langle\hat{\beta}\rangle_{\mathbf{n}}$.

  5. (v)

$(\langle\hat{\alpha}\rangle_{\mathbf{n}})^{-1}=\langle\hat{\alpha}^{-1}\rangle_{\hat{\alpha}*\mathbf{n}}$.

  6. (vi)

$(\alpha_{1}\beta_{1}\oplus\cdots\oplus\alpha_{k}\beta_{k})_{\mathbf{n}}=(\alpha_{1}\oplus\cdots\oplus\alpha_{k})_{\mathbf{n}}(\beta_{1}\oplus\cdots\oplus\beta_{k})_{\mathbf{n}}$

  7. (vii)

$(\alpha_{1}\oplus\cdots\oplus\alpha_{k})_{\mathbf{n}}^{-1}=(\alpha_{1}^{-1}\oplus\cdots\oplus\alpha_{k}^{-1})_{\mathbf{n}}$.

  8. (viii)

$\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{1}\oplus\cdots\oplus\alpha_{k})_{\mathbf{n}}=\Delta_{(n)}$
if and only if $\hat{\alpha}=\Delta_{(k)}$ and $\alpha_{i}=\Delta_{(n_{i})}$
for $1\leq i\leq k$.

Lemma 2.8 (iii)–(vii) are similar to the computations in the wreath product
$G\wr B_{k}$, where $k$-braids act on the $k$-cartesian product of $G$ by
permuting coordinates. Let $n=mk$ with $m,k\geq 2$, and let
$\mathbf{n}=(m,\ldots,m)$ be a composition of $n$. Let
$B_{n}(\mathcal{C}_{\mathbf{n}})=\\{\alpha\in
B_{n}\mid\alpha*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}\\}$. Then
Lemma 2.8 shows that $B_{n}(\mathcal{C}_{\mathbf{n}})$ is indeed isomorphic to
$B_{m}\wr B_{k}$.

######  Definition 2.9.

Let $\mathbf{n}=(n_{1},\ldots,n_{k})$ be a composition of $n$, and let
$\alpha\in B_{n}$ be such that $\alpha*\mathcal{C}_{\mathbf{n}}$ is standard.
Then $\alpha$ is uniquely expressed as
$\alpha=\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{1}\oplus\cdots\oplus\alpha_{k})_{\mathbf{n}}$
by Lemma 2.8. We call $\hat{\alpha}$ the _$\mathbf{n}$ -exterior braid_ of
$\alpha$, and denote it by $\operatorname{Ext}_{\mathbf{n}}(\alpha)$. The
braids $\alpha_{1},\ldots,\alpha_{k}$ are called _$\mathbf{n}$ -interior
braids_ of $\alpha$. If $\operatorname{Ext}_{\mathbf{n}}(\alpha)$ is the
identity, $\alpha$ is called an _$\mathbf{n}$ -split braid_.

The lemma below follows from the above lemma and properties of
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\cdot)$.

######  Lemma 2.10.

Let $\alpha$ and $\beta$ be $n$-braids such that
$\alpha*\mathcal{C}_{\mathbf{n}}=\beta*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$
for a composition $\mathbf{n}$ of $n$.

  * (i)

$\operatorname{Ext}_{\mathbf{n}}(\alpha\beta)=\operatorname{Ext}_{\mathbf{n}}(\alpha)\operatorname{Ext}_{\mathbf{n}}(\beta)$
and
$\operatorname{Ext}_{\mathbf{n}}(\alpha^{-1})=\operatorname{Ext}_{\mathbf{n}}(\alpha)^{-1}$.

  * (ii)

If
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)=\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)=\mathcal{C}_{\mathbf{n}}$
and $\gamma$ is an $n$-braid with $\beta=\gamma\alpha\gamma^{-1}$, then
$\gamma*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$ and
$\operatorname{Ext}_{\mathbf{n}}(\beta)=\operatorname{Ext}_{\mathbf{n}}(\gamma)\operatorname{Ext}_{\mathbf{n}}(\alpha)\operatorname{Ext}_{\mathbf{n}}(\gamma)^{-1}$.

##  3\. Proof of Theorem 1.6

In this section, we prove Theorem 1.6 by presenting three examples which show
that none of the monomorphisms $\psi_{1}:A(\mathbf{B}_{n})\to
A(\mathbf{A}_{n})$, $\psi_{2}:A(\tilde{\mathbf{A}}_{n-1})\to
A(\mathbf{B}_{n})$ and $\psi_{3}:A(\tilde{\mathbf{C}}_{n-1})\to
A(\mathbf{B}_{n})$ is injective on the set of conjugacy classes for $n\geq 3$.
Recall that

$\begin{array}[]{l}A(\mathbf{A}_{n})=B_{n+1};\\\
A(\mathbf{B}_{n})=B_{n+1,1};\end{array}\quad\begin{array}[]{l}A(\tilde{\mathbf{A}}_{n-1})=\\{\alpha\in
B_{n+1}\mid\mbox{$\alpha$ is 1-unlinked}\\};\\\
A(\tilde{\mathbf{C}}_{n-1})=\\{\alpha\in B_{n+1}\mid\mbox{$\alpha$ is
$\\{1,n+1\\}$-pure}\\}.\end{array}$

###### Proof of Theorem 1.6.

Consider the following $(n+1)$-braids for $n\geq 3$. See Figure 6.

$\biggl{\\{}\begin{array}[]{l}\alpha_{1}=\sigma_{1}^{2}\sigma_{2}^{4},\\\
\beta_{1}=\sigma_{2}^{2}\sigma_{1}^{4},\end{array}\qquad\biggl{\\{}\begin{array}[]{l}\alpha_{2}=\sigma_{n}^{2}\sigma_{n-1}^{4},\\\
\beta_{2}=\sigma_{n-1}^{2}\sigma_{n}^{4},\end{array}\qquad\mbox{and}\qquad\biggl{\\{}\begin{array}[]{l}\alpha_{3}=\sigma_{n}\cdots\sigma_{2},\\\
\beta_{3}=\sigma_{1}^{-2}\alpha_{3}\sigma_{1}^{2}.\end{array}$

|  |  |  |  |  |
---|---|---|---|---|---|---
(a) $\alpha_{1}=\sigma_{1}^{2}\sigma_{2}^{4}$ |  | (b) $\alpha_{2}=\sigma_{n}^{2}\sigma_{n-1}^{4}$ |  | (c) $\alpha_{3}=\sigma_{n}\cdots\sigma_{2}$ |  | (d) $\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha_{3})$
|  |  |  |  |  |
(e) $\beta_{1}=\sigma_{2}^{2}\sigma_{1}^{4}$ |  | (f) $\beta_{2}=\sigma_{n-1}^{2}\sigma_{n}^{4}$ |  | (g) $\beta_{3}=\sigma_{1}^{-2}\alpha_{3}\sigma_{1}^{2}$ |  | (h) $\chi=(\sigma_{1}\cdots\sigma_{n})(\sigma_{n}\cdots\sigma_{1})$
Figure 6. The examples for Theorem 1.6 when $n=4$

First, let us consider $\alpha_{1}$ and $\beta_{1}$. Because $\alpha_{1}$ and
$\beta_{1}$ are 1-pure, they belong to $B_{n+1,1}$. They are conjugate in
$B_{n+1}$ because
$\beta_{1}=(\sigma_{1}\sigma_{2}\sigma_{1})\alpha_{1}(\sigma_{1}\sigma_{2}\sigma_{1})^{-1}$.
However they are not conjugate in $B_{n+1,1}$ because
$\operatorname{lk}(\alpha_{1})=1$ and $\operatorname{lk}(\beta_{1})=2$. This
shows that $\psi_{1}:A(\mathbf{B}_{n})\to A(\mathbf{A}_{n})$ is not injective
on the set of conjugacy classes.

The second pair $(\alpha_{2},\beta_{2})$ is similar to the first one. Notice
that $n\geq 3$. Both $\alpha_{2}$ and $\beta_{2}$ are $\\{1,n+1\\}$-pure,
hence they belong to $A(\tilde{\mathbf{C}}_{n-1})$. They are conjugate in
$B_{n+1,1}$ because $\sigma_{n}\sigma_{n-1}\sigma_{n}$ is 1-pure and
$\beta_{2}=(\sigma_{n}\sigma_{n-1}\sigma_{n})\alpha_{2}(\sigma_{n}\sigma_{n-1}\sigma_{n})^{-1}$.
However $\alpha_{2}$ cannot be conjugated to $\beta_{2}$ by an
$\\{1,n+1\\}$-pure braid. An easy way to see this is to consider the linking
number, say $\operatorname{lk}^{\prime}$, of $(n+1)$-st strand with the other
strands, which is a well-defined homomorphism from
$A(\tilde{\mathbf{C}}_{n-1})$ to $\mathbf{Z}$. Because
$\operatorname{lk}^{\prime}(\alpha_{2})=1$ and
$\operatorname{lk}^{\prime}(\beta_{2})=2$, the braids $\alpha_{2}$ and
$\beta_{2}$ are not conjugate in $A(\tilde{\mathbf{C}}_{n-1})$. This shows
that $\psi_{3}:A(\tilde{\mathbf{C}}_{n-1})\to A(\mathbf{B}_{n})$ is not
injective on the set of conjugacy classes.

Now, let us consider the last pair $(\alpha_{3},\beta_{3})$. Because both
$\alpha_{3}$ and $\beta_{3}$ are 1-unlinked, they belong to
$A(\tilde{\mathbf{A}}_{n-1})$. They are conjugate in $B_{n+1,1}$ because
$\sigma_{1}^{2}$ is 1-pure and
$\beta_{3}=\sigma_{1}^{-2}\alpha_{3}\sigma_{1}^{2}$.

The claim below shows the structure of $Z_{B_{n+1}}(\alpha_{3})$. It is a
direct consequence of the main result of J. González-Meneses and B. Wiest in
[GW04], however we provide a proof for the reader’s convenience.

_Claim._ $Z_{B_{n+1}}(\alpha_{3})$ is generated by $\alpha_{3}$ and
$\chi=(\sigma_{1}\cdots\sigma_{n})(\sigma_{n}\cdots\sigma_{1})$. In
particular, every element $\zeta$ in $Z_{B_{n+1}}(\alpha_{3})$ is 1-pure with
$\operatorname{lk}(\zeta)\equiv 0\pmod{n}$.

###### Proof of Claim.

Let $\mathbf{n}=(1,n)$ be a composition of $n+1$, and let
$Z(\alpha_{3})=Z_{B_{n+1}}(\alpha_{3})$. Note that
$\alpha_{3}=(1\oplus\delta_{(n)})_{\mathbf{n}}$ and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha_{3})=\mathcal{C}_{\mathbf{n}}$.
See Figure 6 (d). If $\gamma\in Z(\alpha_{3})$, then
$\gamma*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$, hence $\gamma$ is
of the form

$\gamma=\langle\hat{\gamma}\rangle_{\mathbf{n}}(1\oplus\gamma_{1})_{\mathbf{n}}$

for some $\hat{\gamma}\in B_{2}$ and $\gamma_{1}\in B_{n}$. By direct
computation, we can see that $\gamma\in Z(\alpha_{3})$ if and only if
$\hat{\gamma}=\sigma_{1}^{2k}$ for some $k\in\mathbf{Z}$ and $\gamma_{1}$
commutes with $\delta_{(n)}$. Because $Z_{B_{n}}(\delta_{(n)})$ is infinite
cyclic generated by $\delta_{(n)}$ by [BDM02], $Z(\alpha_{3})$ is generated by
$\langle\sigma_{1}^{2}\rangle_{\mathbf{n}}=\chi$ and
$(1\oplus\delta_{(n)})_{\mathbf{n}}=\alpha_{3}$.

Because $\operatorname{lk}(\chi)=n$ and $\operatorname{lk}(\alpha_{3})=0$,
every element in $Z(\alpha_{3})$ has linking number a multiple of $n$. ∎

Assume that $\alpha_{3}$ and $\beta_{3}$ are conjugate in
$A(\tilde{\mathbf{A}}_{n-1})$. Then there exists a 1-unlinked $(n+1)$-braid
$\gamma$ such that
$\alpha_{3}=\gamma^{-1}\beta_{3}\gamma=\gamma^{-1}\sigma_{1}^{-2}\alpha_{3}\sigma_{1}^{2}\gamma$.
Therefore $\sigma_{1}^{2}\gamma\in Z_{B_{n+1}}(\alpha_{3})$. Because
$\operatorname{lk}(\sigma_{1}^{2}\gamma)=\operatorname{lk}(\sigma_{1}^{2})+\operatorname{lk}(\gamma)=1+0=1$,
$\operatorname{lk}(\sigma_{1}^{2}\gamma)$ is not a multiple of $n\geq 3$. It
is a contradiction. This shows that $\psi_{2}:A(\tilde{\mathbf{A}}_{n-1})\to
A(\mathbf{B}_{n})$ is not injective on the set of conjugacy classes. ∎

##  4\. Braids in the centralizer of a periodic braid

In this section, we review isomorphisms between Artin groups of type
$\mathbf{B}$ and centralizers of periodic braids, and make some tools useful
in proving Theorem 1.4.

###  4.1. Discussions

One of the difficulties in proving Theorem 1.4 lies in dealing with reducible
braids. Let $(\alpha,\beta,\omega)$ be a triple of $n$-braids such that
$\omega$ is periodic and both $\alpha$ and $\beta$ are non-periodic reducible
braids in $Z(\omega)$ which are conjugate in $B_{n}$. If
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)$ are standard, it
would be very convenient for computations, but they are not necessarily
standard. Our strategy is to construct another such triple
$(\alpha^{\prime},\beta^{\prime},\omega^{\prime})$ such that both
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha^{\prime})$ and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta^{\prime})$ are
standard. This new triple is related to the original one by conjugations so
that if $\alpha^{\prime}$ and $\beta^{\prime}$ are conjugate in
$Z(\omega^{\prime})$ then $\alpha$ and $\beta$ are conjugate in $Z(\omega)$.
This construction is the main theme of this section.

The following lemma is obvious, hence we omit the proof.

######  Lemma 4.1.

Let $\omega$ be a periodic $n$-braid, and let $\alpha$ and $\beta$ belong to
$Z(\omega)$.

  * (i)

Let $\chi\in B_{n}$. Then $\chi\alpha\chi^{-1}$ and $\chi\beta\chi^{-1}$
belong to $Z(\chi\omega\chi^{-1})$. Furthermore, $\chi\alpha\chi^{-1}$ and
$\chi\beta\chi^{-1}$ are conjugate in $Z(\chi\omega\chi^{-1})$ if and only if
$\alpha$ and $\beta$ are conjugate in $Z(\omega)$.

  * (ii)

Let $\chi_{1},\chi_{2}\in Z(\omega)$. Then $\chi_{1}\alpha\chi_{1}^{-1}$ and
$\chi_{2}\beta\chi_{2}^{-1}$ belong to $Z(\omega)$. Furthermore,
$\chi_{1}\alpha\chi_{1}^{-1}$ and $\chi_{2}\beta\chi_{2}^{-1}$ are conjugate
in $Z(\omega)$ if and only if $\alpha$ and $\beta$ are conjugate in
$Z(\omega)$.

From the above lemma, one can think of two ways in order to make
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)$ standard. First, one
can expect that there exists $\chi\in B_{n}$ such that both
$\chi*\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ and
$\chi*\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)$ are standard. If
it is true, $(\chi\alpha\chi^{-1},\chi\beta\chi^{-1},\chi\omega\chi^{-1})$ can
be taken as a desired triple by Lemma 4.1 (i). Second, one can expect that
there exist $\chi_{1},\chi_{2}\in Z(\omega)$ such that
$\chi_{1}*\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ and
$\chi_{2}*\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)$ are
standard. If it is true,
$(\chi_{1}\alpha\chi_{1}^{-1},\chi_{2}\beta\chi_{2}^{-1},\omega)$ can be taken
as a desired triple by Lemma 4.1 (ii). However the following example shows
that each of these ideas does not work for itself. Our construction will be as
$(\chi\chi_{1}\alpha\chi_{1}^{-1}\chi^{-1},\chi\chi_{2}\beta\chi_{2}^{-1}\chi^{-1},\chi\omega\chi^{-1})$,
where $\chi_{1},\chi_{2}\in Z(\omega)$ and $\chi\in B_{n}$ such that both
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\chi\chi_{1}\alpha\chi_{1}^{-1}\chi^{-1})$
and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\chi\chi_{2}\beta\chi_{2}^{-1}\chi^{-1})$
are standard and $\chi\omega\chi^{-1}$ is a special kind of periodic braid
(see Corollary 4.14).

######  Example 4.2.

Consider the 4-braids

$\omega=\delta^{2}=(\sigma_{3}\sigma_{2}\sigma_{1})^{2},\qquad\alpha=\sigma_{2}\sigma_{1}\sigma_{2}^{-1}\quad\mbox{and}\quad\beta=\sigma_{3}\sigma_{2}\sigma_{3}^{-1}.$

The element $\omega$ is a periodic 4-braid which corresponds to the
$\pi$-rotation of the punctured disk when punctures are distributed as in
Figure 3 (a). $\alpha$ and $\beta$ are reducible braids in $Z(\omega)$ which
are conjugate in $B_{4}$. Let
$C=\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ and
$C^{\prime}=\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)$. They
enclose two punctures as in Figure 7 (a) and (b).

|  |  |  |
---|---|---|---|---
|  |  |  |
(a) $C$ | (b) $C^{\prime}$ | (c) | (d) | (e)
Figure 7. The figures in two rows show the same curves with different location
of punctures.

There are three standard curves in $D_{4}$ enclosing two punctures as in
Figure 7 (c)–(e) and the geometric intersection numbers between two of them
are either 0 or 2. Because the geometric intersection number between $C$ and
$C^{\prime}$ is 4, there is no 4-braid $\chi$ such that $\chi*C$ and
$\chi*C^{\prime}$ are standard.

Assume that there exists $\chi_{1}\in Z(\omega)$ such that $\chi_{1}*C$ is
standard. Because $C$ encloses two punctures, so does $\chi_{1}*C$. Therefore
$\chi_{1}*C$ is one of the three curves in Figure 7 (c)–(e). Moreover,
$\chi_{1}*C$ is invariant under the action of $\omega$ because
$\omega*(\chi_{1}*C)=\chi_{1}*(\omega*C)=\chi_{1}*C$. Because none of the
curves in Figure 7 (c)–(e) is invariant under the action of $\omega$, it is a
contradiction. Similarly, there is no $\chi_{2}\in Z(\omega)$ such that
$\chi_{2}*C^{\prime}$ is standard.

The monomorphisms $\psi_{5}$ and $\psi_{6}$ are compositions of an isomorphism
$\varphi$ and the inclusion $\iota$:

$\textstyle{B_{d+1,1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\scriptstyle{\simeq}$$\textstyle{Z(\omega)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota}$$\textstyle{B_{n},}$

where $n=md$ or $md+1$ with $m\geq 2$; $\omega$ is a periodic $n$-braid
conjugate to $\delta^{k}$ with $d=\gcd(k,n)$ or to $\epsilon^{k}$ with
$d=\gcd(k,n-1)$. Observe the following.

  * (i)

$\iota$ is injective on the set of conjugacy classes if and only if so is
$\iota\circ\varphi$.

  * (ii)

Let $\omega^{\prime}$ be a periodic $n$-braid which is conjugate to $\omega$,
and let $\iota^{\prime}:Z(\omega^{\prime})\to B_{n}$ be the inclusion. Then
$\iota$ and $\iota^{\prime}$ are related by an inner automorphism, say
$I_{\chi}$, of $B_{n}$ and its restriction to $Z(\omega)$ as follows.

$\textstyle{Z(\omega)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\scriptstyle{I_{\chi}}$$\scriptstyle{\iota}$$\textstyle{B_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\scriptstyle{I_{\chi}}$$\textstyle{Z(\omega^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota^{\prime}}$$\textstyle{B_{n}}$

Therefore, $\iota$ is injective on the set of conjugacy classes if and only if
so is $\iota^{\prime}$.

  * (iii)

Consequently, the injectivity of $\iota\circ\varphi$ on the set of conjugacy
classes depends only on the conjugacy class of $\omega$.

Therefore Theorem 1.4 states that $\iota:Z(\omega)\to B_{n}$ is injective on
the set of conjugacy classes for any periodic $n$-braid $\omega$. And
Corollary 1.5 is equivalent to $\iota:Z(\omega)\to B_{n}$ being injective on
the set of conjugacy classes for any noncentral periodic $n$-braid $\omega$.
Because Theorem 1.4 obviously holds for central braids $\omega$, we have the
following.

######  Lemma 4.3.

Theorem 1.4 and Corollary 1.5 are equivalent.

We remark that there are infinitely many isomorphisms from $B_{d+1,1}$ to
$Z(\omega)$, and any two of them are related by an automorphism of
$B_{d+1,1}=A(\mathbf{B}_{d})$. See [CC05] for the classification of
automorphisms of Artin groups of type $\mathbf{A}$, $\mathbf{B}$,
$\tilde{\mathbf{A}}$ and $\tilde{\mathbf{C}}$.

###  4.2. Periodic braids and their centralizers

Let $m$ and $d$ be integers with $m\geq 2$ and $d\geq 1$, and let $n=md+1$.
Here we construct a periodic $n$-braid $\mu_{m,d}$ conjugate to
$\epsilon^{d}$, and an isomorphism $B_{d+1,1}\simeq Z(\mu_{m,d})$ following
the work in [BDM02].

Let $Z_{n}^{(m)}=\\{z_{0}\\}\cup\\{z_{i,j}\mid 1\leq i\leq d,~{}1\leq j\leq
m\\}$, where $z_{0}$ is the origin and $z_{i,j}$ is the complex number with
$|z_{i,j}|=i$ and $\arg(z_{i,j})=2\pi(j-1)/m$. Let

$D_{n}^{(m)}=\\{z\in\mathbb{C}\mid|z|\leq n+1\\}\setminus Z_{n}^{(m)}.$

$\scriptstyle{\eta_{m,d}\qquad\mbox{}}$$\scriptstyle{\phi_{m,d}\mbox{}}$

$D_{d+1}$ $D_{n}^{(m)}$ and $\rho_{m}$ $D_{n}$ and
$\eta_{m,d}\circ\rho_{m}\circ\eta_{m,d}^{-1}$

Figure 8. The punctured spaces $D_{n}^{(m)}$ and $D_{d+1}$, where $n=13$,
$m=3$ and $d=4$

Now we construct a covering map $\phi_{m,d}:D_{n}^{(m)}\to D_{d+1}$ and a
diffeomorphism $\eta_{m,d}:D_{n}^{(m)}\to D_{n}$ as in Figure 8. Let
$\rho_{m}$ be the clockwise $(2\pi/m)$-rotation of $D_{n}^{(m)}$. The
identification space $D_{n}^{(m)}/\langle\rho_{m}\rangle$ is diffeomorphic to
the $(d+1)$-punctured disk $D_{d+1}$. We write the natural projection by

$\phi_{m,d}:D_{n}^{(m)}\to D_{d+1}.$

$\phi_{m,d}$ is a covering map whose group of covering transformations is the
cyclic group generated by $\rho_{m}$. We construct a diffeomorphism

$\eta_{m,d}:D_{n}^{(m)}\to D_{n},$

which fixes $\partial D_{n}^{(m)}=\partial D_{n}$ pointwise, as follows.
First, cut $D_{n}^{(m)}$ and $D_{n}$ by $d$ round circles into one once-
punctured disk and $d$ annuli with $m$ punctures as in Figure 9 (a). Then,
define $\eta_{m,d}$ on each punctured annulus as in Figure 9 (b) such that it
coincides at the intersections of two adjacent annuli, and extend $\eta_{m,d}$
to the once-punctured disk in an obvious way. Such a diffeomorphism
$\eta_{m,d}$ is unique up to isotopy.

$\scriptstyle{\eta_{m,d}}$ | $\scriptstyle{\eta_{m,d}}$
---|---
(a) | (b)
Figure 9. The diffeomorphism $\eta_{m,d}:D_{n}^{(m)}\to D_{n}$

######  Definition 4.4.

Let $m$ and $d$ be integers with $m\geq 2$ and $d\geq 0$. We define an
$(md+1)$-braid $\mu_{m,d}$ recursively as follows: $\mu_{m,0}$ is the unique
braid with one strand; for $d\geq 1$,

$\mu_{m,d}=\mu_{m,d-1}(\sigma_{dm}\sigma_{dm-1}\cdots\sigma_{2}\sigma_{1})(\sigma_{1}\sigma_{2}\cdots\sigma_{(d-1)m+1}),$

where $\mu_{m,d-1}$ is regarded as an $(md+1)$-braid by using the inclusion
$B_{m(d-1)+1}\to B_{md+1}$ defined by $\sigma_{i}\mapsto\sigma_{i}$ for $1\leq
i\leq m(d-1)$. See Figure 10 for $\mu_{m,d}$ with $m=3$ and $1\leq d\leq 4$.

|  |  |
---|---|---|---
$\mu_{3,1}$ | $\mu_{3,2}$ | $\mu_{3,3}$ | $\mu_{3,4}$
Figure 10. The periodic braid $\mu_{3,d}$ for $d=1,2,3,4$

From the construction, we can see the following.

  * (i)

$\eta_{m,d}\circ\rho_{m}\circ\eta_{m,d}^{-1}:D_{n}\to D_{n}$ represents the
braid $\mu_{m,d}$ modulo $\Delta^{2}$.

  * (ii)

An $n$-braid belongs to $Z(\mu_{m,d})$ if and only if it is represented by
$\eta_{m,d}\circ h\circ\eta_{m,d}^{-1}$ for a $\rho_{m}$-equivariant
diffeomorphism $h:D_{n}^{(m)}\to D_{n}^{(m)}$ that fixes the boundary
pointwise.

Because $(\mu_{m,d})^{m}=\Delta_{(n)}^{2}=(\epsilon_{(n)}^{d})^{m}$,
$\mu_{m,d}$ is conjugate to $\epsilon_{(n)}^{d}$, hence $\nu(\mu_{m,d})$ is
conjugate to $\delta_{(n-1)}^{d}$. Recall that the monomorphisms $\psi_{4}$,
$\psi_{5}$ and $\psi_{6}$ are induced by the isomorphisms $B_{d+1,1}\simeq
Z(\epsilon_{(n)}^{d})$ and $B_{d+1,1}\simeq Z(\delta_{(n-1)}^{d})$ in [BDM02,
GW04]. In the same way, we define monomorphisms $\psi_{4}^{\prime}$,
$\psi_{5}^{\prime}$ and $\psi_{6}^{\prime}$ by using isomorphisms
$B_{d+1,1}\simeq Z(\mu_{m,d})$ and $B_{d+1,1}\simeq Z(\nu(\mu_{m,d}))$.

######  Definition 4.5.

Let $m\geq 2$ and $d\geq 1$, and let $n=md+1$. We define an isomorphism

$\varphi:B_{d+1,1}\to Z(\mu_{m,d})$

as follows. Let $\alpha\in B_{d+1,1}$, that is, $\alpha$ is a 1-pure
$(d+1)$-braid. Then $\alpha$ is represented by a diffeomorphism $f:D_{d+1}\to
D_{d+1}$ that fixes the first puncture together with all the boundary points.
Let $\tilde{f}:D_{n}^{(m)}\to D_{n}^{(m)}$ be the lift of $f$ (with respect to
$\phi_{m,d}$) that fixes $\partial D_{n}^{(m)}$ pointwise. Then $\tilde{f}$ is
$\rho_{m}$-equivariant, hence $\eta_{m,d}\circ\tilde{f}\circ\eta_{m,d}^{-1}$
represents a braid in $Z(\mu_{m,d})$. We define $\varphi(\alpha)$ as the braid
represented by $\eta_{m,d}\circ\tilde{f}\circ\eta_{m,d}^{-1}$.

---
$\textstyle{D_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\eta_{m,d}^{-1}}$$\textstyle{D_{n}^{(m)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{m,d}}$$\scriptstyle{\tilde{f}}$$\textstyle{D_{n}^{(m)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{m,d}}$$\scriptstyle{\eta_{m,d}}$$\textstyle{D_{n}}$$\textstyle{D_{d+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{D_{d+1}}$

The monomorphisms $\psi_{4}^{\prime}$ and $\psi_{6}^{\prime}$ are compositions
of $\varphi$ with the inclusions $Z(\mu_{m,d})\subset B_{md+1,1}$ and
$Z(\mu_{m,d})\subset B_{md+1}$, respectively. The monomorphism
$\psi_{5}^{\prime}$ is the composition of $\varphi$ with the homomorphism
$\nu:B_{md+1,1}\to B_{md}$. See Figure 11.

|
---|---
$\textstyle{B_{md}}$$\textstyle{Z(\nu(\mu_{m,d}))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Z(\mu_{m,d})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mbox{}\quad\nu}$$\textstyle{B_{md+1,1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{md+1}}$$\textstyle{B_{d+1,1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\scriptstyle{\simeq}$$\scriptstyle{\psi_{5}^{\prime}}$$\scriptstyle{\simeq}$$\scriptstyle{\psi_{4}^{\prime}}$$\scriptstyle{\psi_{6}^{\prime}}$

Figure 11. Monomorphisms $\psi_{4}^{\prime}$, $\psi_{5}^{\prime}$ and
$\psi_{6}^{\prime}$

By the construction,
$\psi_{4}^{\prime}(\Delta_{(d+1)}^{2})=\psi_{6}^{\prime}(\Delta_{(d+1)}^{2})=\mu_{m,d}$.
The following lemma looks obvious, but we include a sketchy proof for
completeness.

######  Lemma 4.6.

Let $\psi$ denote either $\psi_{4}^{\prime}:B_{d+1,1}\to B_{md+1,1}$,
$\psi_{5}^{\prime}:B_{d+1,1}\to B_{md}$ or $\psi_{6}^{\prime}:B_{d+1,1}\to
B_{md+1}$. The monomorphism $\psi$ preserves the Nielsen-Thurston type, that
is, for $\alpha\in B_{d+1,1}$

  * (i)

$\alpha$ is periodic if and only if $\psi(\alpha)$ is periodic;

  * (ii)

$\alpha$ is pseudo-Anosov if and only if $\psi(\alpha)$ is pseudo-Anosov;

  * (iii)

$\alpha$ is reducible and non-periodic if and only if $\psi(\alpha)$ is
reducible and non-periodic.

###### Proof.

Assume $\psi=\psi_{6}^{\prime}$. The case of $\psi_{4}^{\prime}$ is the same,
and the case of $\psi_{5}^{\prime}$ can be proved similarly. Because the three
classes {periodic braids}, {pseudo-Anosov braids} and {non-periodic reducible
braids} are mutually disjoint, it suffices to prove (i) together with (ii)′
and (iii)′ where

  * (ii)′

If $\alpha$ is pseudo-Anosov, then $\psi(\alpha)$ is pseudo-Anosov;

  * (iii)′

If $\alpha$ is reducible and non-periodic, then $\psi(\alpha)$ is reducible
and non-periodic.

(i)  $\alpha$ is periodic $\Leftrightarrow$ $\alpha^{k}=(\Delta^{2})^{\ell}$
for some $k\neq 0$ and $\ell\in\mathbf{Z}$ $\Leftrightarrow$
$\psi(\alpha)^{k}=\psi(\alpha^{k})=\psi(\Delta^{2\ell})=\mu_{m,d}^{\ell}$ for
some $k\neq 0$ and $\ell\in\mathbf{Z}$ $\Leftrightarrow$ $\psi(\alpha)$ is
periodic

(ii)′ Suppose that $\alpha\in B_{d+1,1}$ is pseudo-Anosov. There is a pseudo-
Anosov diffeomorphism, say $f$, defined on the interior of $D_{d+1}$
representing $\alpha$ modulo $\Delta_{(d+1)}^{2}$. Let $\tilde{f}$ be the lift
of $f$ (with respect to $\phi_{m,d}$) defined on the interior of
$D_{md+1}^{(m)}$. Then $\tilde{f}$ is also a pseudo-Anosov diffeomorphism
because invariant measured foliations of $f$ lift to invariant measured
foliations of $\tilde{f}$ with the same dilatation. Therefore
$\eta_{m,d}\circ\tilde{f}\circ\eta_{m,d}^{-1}$ is a pseudo-Anosov
diffeomorphism. Since $\eta_{m,d}\circ\tilde{f}\circ\eta_{m,d}^{-1}$
represents $\psi(\alpha)$ modulo $\Delta_{(md+1)}^{2}$, $\psi(\alpha)$ is
pseudo-Anosov.

(iii)′ Suppose that $\alpha\in B_{d+1,1}$ is reducible and non-periodic. Then
$\psi(\alpha)$ is non-periodic by (i). Because $\alpha$ is reducible, there
exists a diffeomorphism $f:D_{d+1}\to D_{d+1}$, which represents $\alpha$,
together with an essential curve system $\mathcal{C}$ such that
$f(\mathcal{C})=\mathcal{C}$. Let
$\tilde{\mathcal{C}}=\phi_{m,d}^{-1}(\mathcal{C})$, and let $\tilde{f}$ be a
lift of $f$ with respect to $\phi_{m,d}$. Then $\tilde{\mathcal{C}}$ is an
invariant essential curve system of $\tilde{f}$, hence
$\eta_{m,d}(\tilde{\mathcal{C}})$ is a reduction system of $\psi(\alpha)$. ∎

######  Corollary 4.7.

Let $m\geq 2$ and $d\geq 1$. Let $\omega$ be a 1-pure braid conjugate to
$\epsilon_{(md+1)}^{d}$, and let $\alpha\in Z(\omega)$. Then $\alpha$ and
$\nu(\alpha)$ have the same Nielsen-Thurston type, that is,

  * (i)

$\alpha$ is periodic if and only if $\nu(\alpha)$ is periodic;

  * (ii)

$\alpha$ is pseudo-Anosov if and only if $\nu(\alpha)$ is pseudo-Anosov;

  * (iii)

$\alpha$ is reducible and non-periodic if and only if $\nu(\alpha)$ is
reducible and non-periodic.

###### Proof.

Without loss of generality, we may assume that $\omega=\mu_{m,d}$. Then there
exists $\bar{\alpha}\in B_{d+1,1}$ such that
$\alpha=\psi_{4}^{\prime}(\bar{\alpha})$ and
$\nu(\alpha)=\psi_{5}^{\prime}(\bar{\alpha})$. By Lemma 4.6, $\alpha$ and
$\bar{\alpha}$ (resp. $\nu(\alpha)$ and $\bar{\alpha}$) have the same Nielsen-
Thurston type. Therefore $\alpha$ and $\nu(\alpha)$ have the same Nielsen-
Thurston type. ∎

From Lemma 4.6 and its proof, we can see that the monomorphisms
$\psi_{4}^{\prime}$, $\psi_{5}^{\prime}$ and $\psi_{6}^{\prime}$ send
canonical reduction systems to canonical reduction systems as follows.

######  Corollary 4.8.

Let $m\geq 2$ and $d\geq 1$, and let $\alpha\in B_{d+1,1}$ be reducible and
non-periodic. If $\mathcal{C}$ is the canonical reduction system of $\alpha$,
then $\eta_{m,d}(\phi_{m,d}^{-1}(\mathcal{C}))$ is the canonical reduction
system of both $\psi_{4}^{\prime}(\alpha)$ and $\psi_{6}^{\prime}(\alpha)$,
and the canonical reduction system of $\psi_{5}^{\prime}(\alpha)$ is obtained
from $\eta_{m,d}(\phi_{m,d}^{-1}(\mathcal{C}))$ by forgetting the first
puncture. In particular,
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\psi_{4}^{\prime}(\alpha))=\operatorname{\mathcal{R}}_{\operatorname{ext}}(\psi_{6}^{\prime}(\alpha))=\eta_{m,d}(\phi_{m,d}^{-1}(\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)))$,
and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\psi_{5}^{\prime}(\alpha))$
is obtained from
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\psi_{4}^{\prime}(\alpha))$
by forgetting the first puncture.

###### Proof.

We prove the corollary only for the case of $\psi_{4}^{\prime}$. The other
cases are similar. Recall that the canonical reduction system of a braid is
the smallest adequate reduction system of it.

Let $\mathcal{C}\subset D_{d+1}$ be the canonical reduction system of
$\alpha\in B_{d+1,1}$. Following the construction in Lemma 4.6, let
$\tilde{\mathcal{C}}=\phi_{m,d}^{-1}(\mathcal{C})$ and
$\mathcal{D}=\eta_{m,d}(\tilde{\mathcal{C}})$.

From the proof of Lemma 4.6, $\mathcal{D}$ is a reduction system of
$\psi_{4}^{\prime}(\alpha)$. Because $\mathcal{C}$ is adequate, $\mathcal{D}$
is adequate by Lemma 4.6 (i) and (ii). Hence there is a subset
$\mathcal{D}^{\prime}$ of $\mathcal{D}$ which is the canonical reduction
system of $\psi_{4}^{\prime}(\alpha)$. Let
$\tilde{\mathcal{C}}^{\prime}=\eta_{m,d}^{-1}(\mathcal{D}^{\prime})$ and
$\mathcal{C}^{\prime}=\phi_{m,d}(\tilde{\mathcal{C}}^{\prime})$. Note that
$\tilde{\mathcal{C}}^{\prime}\subset\tilde{\mathcal{C}}$ and
$\mathcal{C}^{\prime}\subset\mathcal{C}$.

---
$\textstyle{D_{n}^{(m)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{m,d}}$$\scriptstyle{\eta_{m,d}}$$\textstyle{D_{n}}$$\textstyle{D_{d+1}}$
---
$\textstyle{\tilde{\mathcal{C}}^{\prime}\subset\tilde{\mathcal{C}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{m,d}}$$\scriptstyle{\eta_{m,d}}$$\textstyle{\mathcal{D}^{\prime}\subset\mathcal{D}}$$\textstyle{\mathcal{C}^{\prime}\subset\mathcal{C}}$

We claim that
$\tilde{\mathcal{C}}^{\prime}=\phi_{m,d}^{-1}(\mathcal{C}^{\prime})$ and that
$\mathcal{C}^{\prime}$ is a reduction system of $\alpha$.

Since $\mathcal{D}^{\prime}$ is the canonical reduction system of
$\psi_{4}^{\prime}(\alpha)$ and since $\psi_{4}^{\prime}(\alpha)$ commutes
with $\mu_{m,d}$,

$\mu_{m,d}*\mathcal{D}^{\prime}=\mu_{m,d}*\operatorname{\mathcal{R}}(\psi_{4}^{\prime}(\alpha))=\operatorname{\mathcal{R}}(\mu_{m,d}\,\psi_{4}^{\prime}(\alpha)\,\mu_{m,d}^{-1})=\operatorname{\mathcal{R}}(\psi_{4}^{\prime}(\alpha))=\mathcal{D}^{\prime}.$

Because the diffeomorphism $\eta_{m,d}\circ\rho_{m}\circ\eta_{m,d}^{-1}$
represents $\mu_{m,d}\bmod\Delta^{2}$, the above formula implies that
$\rho_{m}(\tilde{\mathcal{C}}^{\prime})=\tilde{\mathcal{C}}^{\prime}$, that
is, if $\tilde{\mathcal{C}}^{\prime}$ contains a component of
$\tilde{\mathcal{C}}$ then it contains all the components of
$\tilde{\mathcal{C}}$ in the $\rho_{m}$-orbit of this component. Since the
covering transformation group of $\phi_{m,d}$ is generated by $\rho_{m}$,

$\tilde{\mathcal{C}}^{\prime}=\phi_{m,d}^{-1}(\mathcal{C}^{\prime})\quad\mbox{and
hence}\quad\mathcal{D}^{\prime}=\eta_{m,d}(\phi_{m,d}^{-1}(\mathcal{C}^{\prime})).$

Let $f:D_{d+1}\to D_{d+1}$ be a diffeomorphism representing $\alpha$ such that
$f(\mathcal{C})=\mathcal{C}$. Let $\tilde{f}$ be a lift of $f$ with respect to
$\phi_{m,d}$. Then $\tilde{f}(\tilde{\mathcal{C}})=\tilde{\mathcal{C}}$
because $\tilde{\mathcal{C}}=\phi_{m,d}^{-1}(\mathcal{C})$. The diffeomorphism
$\eta_{m,d}\circ\tilde{f}\circ\eta_{m,d}^{-1}$ represents the braid
$\psi_{4}^{\prime}(\alpha)$. Since
$\mathcal{D}^{\prime}=\eta_{m,d}(\tilde{\mathcal{C}}^{\prime})$ is a reduction
system of $\psi_{4}^{\prime}(\alpha)$, one has
$\tilde{f}(\tilde{\mathcal{C}}^{\prime})=\tilde{\mathcal{C}}^{\prime}$.
Therefore $f(\mathcal{C}^{\prime})=\mathcal{C}^{\prime}$ because
$\tilde{\mathcal{C}}^{\prime}=\phi_{m,d}^{-1}(\mathcal{C}^{\prime})$ and
$\tilde{f}$ is a lift of $f$ with respect to $\phi_{m,d}$. Hence
$\mathcal{C}^{\prime}$ is a reduction system of $\alpha$.

Since $\mathcal{D}^{\prime}$ is adequate, the reduction system
$\mathcal{C}^{\prime}$ is also adequate by Lemma 4.6 (i) and (ii). Because
$\mathcal{C}$ is the canonical reduction system of $\alpha$, it is the
smallest adequate reduction system of $\alpha$, hence
$\mathcal{C}\subset\mathcal{C}^{\prime}$. Because
$\mathcal{C}^{\prime}\subset\mathcal{C}$, this implies
$\mathcal{C}=\mathcal{C}^{\prime}$ and hence
$\mathcal{D}=\mathcal{D}^{\prime}$.

So far, we have shown that if $\mathcal{C}$ is the canonical reduction system
of $\alpha$ then $\eta_{m,d}(\phi_{m,d}^{-1}(\mathcal{C}))$ is the canonical
reduction system of $\psi_{4}^{\prime}(\alpha)$. If
$\mathcal{C}^{\prime\prime}$ is the collection of the outermost components of
$\mathcal{C}$, then $\eta_{m,d}(\phi_{m,d}^{-1}(\mathcal{C}^{\prime\prime}))$
is the collection of the outermost components of
$\eta_{m,d}(\phi_{m,d}^{-1}(\mathcal{C}))$, hence
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\psi_{4}^{\prime}(\alpha))=\eta_{m,d}(\phi_{m,d}^{-1}(\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)))$.
∎

We remark that the above argument actually shows that there is a one-to-one
correspondence

$\\{\mbox{curve systems $\mathcal{C}\subset
D_{d+1}$}\\}\stackrel{{\scriptstyle 1-1}}{{\longleftrightarrow}}\\{\mbox{curve
systems $\mathcal{D}\subset D_{md+1}$ such that
$\mu_{m,d}*\mathcal{D}=\mathcal{D}$}\\}$

sending $\mathcal{C}\subset D_{d+1}$ to
$\eta_{m,d}(\phi_{m,d}^{-1}(\mathcal{C}))$. This correspondence satisfies the
property that $\mathcal{C}$ is a reduction system (resp. an adequate reduction
system, the canonical reduction system) of $\alpha\in B_{d+1}$ if and only if
$\eta_{m,d}(\phi_{m,d}^{-1}(\mathcal{C}))$ is a reduction system (resp. an
adequate reduction system, the canonical reduction system) of
$\psi_{4}^{\prime}(\alpha)$.

For $1\leq i\leq d$ and $1\leq j\leq m$, let $x_{0}$ and $x_{i,j}$ be the
integers such that $x_{0}=1$ and $x_{i,j}=(i-1)m+j+1$. In other words,
$(1,2,\ldots,md+1)=(x_{0},x_{1,1},\ldots,x_{1,m},\ldots,x_{d,1},\ldots,x_{d,m})$.
Note that $\eta_{m,d}(z_{0})=x_{0}$ and $\eta_{m,d}(z_{i,j})=x_{i,j}$ for
$1\leq i\leq d$ and $1\leq j\leq m$. The induced permutation of $\mu_{m,d}$ is

$\pi_{\mu_{m,d}}=(x_{0})(x_{1,m},\ldots,x_{1,2},x_{1,1})\cdots(x_{d,m},\ldots,x_{d,2},x_{d,1}).$

######  Lemma 4.9.

Let $m\geq 2$ and $d\geq 1$, and let $\theta$ be a permutation on
$\\{1,\ldots,md+1\\}$. Then the following conditions are equivalent.

  * (i)

$\theta$ is an induced permutation of an element of $Z(\mu_{m,d})$.

  * (ii)

$\theta$ commutes with the induced permutation of $\mu_{m,d}$

  * (iii)

$\theta$ is such that $\theta(x_{0})=x_{0}$ and
$\theta(x_{i,j})=x_{k_{i},j-\ell_{i}}$ for $1\leq i\leq d$ and $1\leq j\leq
m$, where $0\leq\ell_{i}<m$ and $k_{1},\ldots,k_{d}$ is a rearrangement of
$1,\ldots,d$. (Hereafter we regard the second index $j-\ell_{i}$ of
$(k_{i},j-\ell_{i})$ as being taken modulo $m$.)

###### Proof.

(i) $\Rightarrow$ (ii) $\Leftrightarrow$ (iii) is clear. Let us show that (ii)
$\Rightarrow$ (i). Suppose that (ii) is true.

The centralizer of $\pi_{\mu_{m,d}}$ is generated by two types of permutations
$\theta_{i}$ for $1\leq i\leq d-1$ and $\tau_{i}$ for $1\leq i\leq d$, where

  * (i)

the permutation $\theta_{i}$ interchanges the $i$-th and the $(i+1)$-st cycles
of $\pi_{\mu_{m,d}}$, more precisely, $\theta_{i}(x_{i,k})=x_{i+1,k}$,
$\theta_{i}(x_{i+1,k})=x_{i,k}$ and $\theta_{i}(x_{j,k})=x_{j,k}$ for $1\leq
k\leq m$ and $j\not\in\\{i,i+1\\}$;

  * (ii)

the permutation $\tau_{i}$ shifts the $i$-th cycle of $\pi_{\mu_{m,d}}$ by one
in a way that $\tau_{i}(x_{i,k})=x_{i,k-1}$ and $\tau_{i}(x_{j,k})=x_{j,k}$
for $1\leq k\leq m$ and $j\neq i$.

Notice that $\theta_{i}$ and $\tau_{i}$ are induced permutations of elements
of $Z(\mu_{m,d})$. Let the punctures be located as in Figure 12 (a) so that
$\mu_{m,d}$ corresponds to the clockwise $2\pi/m$-rotation. Figure 12 (b) and
(c) illustrate diffeomorphisms whose induced permutations interchange two
cycles or shift a cycle by one. ∎

|  |  |  |
---|---|---|---|---
(a) |  | (b) |  | (c)
Figure 12. (a) shows the punctured disk $D_{13}^{(3)}$. (b) and (c) illustrate
diffeomorphisms inducing $\theta_{2}$ and $\tau_{3}$, respectively.

###  4.3. Reducible braids in the centralizer of a periodic braid

Let $m$ and $d$ be integers with $m\geq 2$ and $d\geq 0$, and let $n=md+1$.
Let $\mathbf{d}=(d_{0}+1,d_{1},\ldots,d_{r})$ be a composition of $d+1$. We
define a composition $L_{m}(\mathbf{d})$ of $n$ as

$L_{m}(\mathbf{d})=(md_{0}+1,\underbrace{d_{1},\ldots,d_{1}}_{m},\ldots,\underbrace{d_{r},\ldots,d_{r}}_{m}).$

For example, if $\mathbf{d}=(2,2,1)$ then $L_{3}(\mathbf{d})=(4,2,2,2,1,1,1)$,
and if $\mathbf{d}=(3,2,1)$ then $L_{3}(\mathbf{d})=(7,2,2,2,1,1,1)$. Let
$\mathcal{C}_{m,\mathbf{d}}$ denote the curve system
$\phi_{m,d}^{-1}(\mathcal{C}_{\mathbf{d}})$ in $D_{n}^{(m)}$. See Figure 13.

$\scriptstyle{\eta_{m,\mathbf{d}}}$$\scriptstyle{\phi_{m,d}}$

$\mathcal{C}_{\mathbf{d}}$ in $D_{d+1}$ $\mathcal{C}_{m,\mathbf{d}}$ in
$D_{n}^{(m)}$ $\mathcal{C}_{L_{m}(\mathbf{d})}$ in $D_{n}$

Figure 13. The curve system $\mathcal{C}_{\mathbf{d}}$ in $D_{d+1}$ is lifted
to $\mathcal{C}_{m,\mathbf{d}}$ in $D_{n}^{(m)}$, and then mapped to
$\mathcal{C}_{L_{m}(\mathbf{d})}$ in $D_{n}$. Here $n=13$, $m=3$, $d=4$ and
$\mathbf{d}=(2,2,1)$.

We construct a diffeomorphism

$\eta_{m,\mathbf{d}}:D_{n}^{(m)}\to D_{n}$

similarly to $\eta_{m,d}$ such that it fixes the boundary $\partial
D_{n}^{(m)}=\partial D_{n}$ pointwise and sends $\mathcal{C}_{m,\mathbf{d}}$
in $D_{n}^{(m)}$ to $\mathcal{C}_{L_{m}(\mathbf{d})}$ in $D_{n}$. First, we
cut the punctured disk $D_{n}^{(m)}$ (resp. $D_{n}$) by $r$ round circles
disjoint from $\mathcal{C}_{m,\mathbf{d}}$ (resp.
$\mathcal{C}_{L_{m}(\mathbf{d})}$) into one $(md_{0}+1)$-punctured disk and
$r$ annuli such that each annulus contains $m$ punctures or $m$ components of
$\mathcal{C}_{m,\mathbf{d}}$ (resp. $\mathcal{C}_{L_{m}(\mathbf{d})}$) as in
Figure 14 (a). Then, on each annulus, define $\eta_{m,\mathbf{d}}$ as in
Figure 14 (b), and on the $(md_{0}+1)$-punctured disk, define
$\eta_{m,\mathbf{d}}$ as $\eta_{m,d_{0}}$. Such a diffeomorphism
$\eta_{m,\mathbf{d}}$ is unique up to isotopy.

$\scriptstyle{\eta_{m,\mathbf{d}}}$ | $\scriptstyle{\eta_{m,\mathbf{d}}}$
---|---
(a) | (b)
Figure 14. The diffeomorphism $\eta_{m,\mathbf{d}}:D_{n}^{(m)}\to D_{n}$ where
$n=13$, $m=3$, $d=4$ and $\mathbf{d}=(2,2,1)$.

######  Definition 4.10.

Let $n=md+1$ with $m\geq 2$ and $d\geq 1$. Let
$\mathbf{d}=(d_{0}+1,d_{1},\ldots,d_{r})$ be a composition of $d+1$ and
$\mathbf{n}=L_{m}(\mathbf{d})$. We define an $n$-braid $\mu_{m,\mathbf{d}}$ as

$\mu_{m,\mathbf{d}}=\langle\mu_{m,r}\rangle_{\mathbf{n}}(\mu_{m,d_{0}}\oplus(\underbrace{\Delta_{(d_{1})}^{2}\oplus
1\oplus\cdots\oplus
1}_{m})\oplus\cdots\oplus(\underbrace{\Delta_{(d_{r})}^{2}\oplus
1\oplus\cdots\oplus 1}_{m}))_{\mathbf{n}}.$

See Figure 15 (b) for $\mu_{m,\mathbf{d}}$ with $m=3$ and $\mathbf{d}=(2,2,1)$
and compare it with $\mu_{m,r}$ in Figure 15 (a).
$\mu_{m,\mathbf{d}}*\mathcal{C}_{L_{m}(\mathbf{d})}=\mathcal{C}_{L_{m}(\mathbf{d})}$
and $\mu_{m,\mathbf{d}}$ is conjugate to $\epsilon_{(n)}^{d}$. Observe that
$\alpha\in B_{n}$ belongs to $Z(\mu_{m,\mathbf{d}})$ if and only if it is
represented by $\eta_{m,\mathbf{d}}\circ h\circ\eta_{m,\mathbf{d}}^{-1}$ for a
$\rho_{m}$-equivariant diffeomorphism $h:D_{n}^{(m)}\to D_{n}^{(m)}$ which
fixes the boundary pointwise.

|  |  |  |
---|---|---|---|---
(a) $\mu_{3,2}$ |  | (b) $\mu_{3,\mathbf{d}}$ |  | (c) $\mathcal{C}_{L_{3}(\mathbf{d})}$
Figure 15. The braid $\mu_{3,\mathbf{d}}$ with $\mathbf{d}=(2,2,1)$ in (b) is
obtained from the braid $\mu_{3,2}$ in (a) by taking four parallel copies of
the first strand and two parallel copies of the next three strands, and then
by concatenating some interior braids. This braid has the standard reduction
system $\mathcal{C}_{L_{3}(\mathbf{d})}=\mathcal{C}_{(4,2,2,2,1,1,1)}$ as in
(c).

######  Lemma 4.11.

Let $n=md+1$ with $m\geq 2$ and $d\geq 1$. Let
$\mathbf{d}=(d_{0}+1,d_{1},\ldots,d_{r})$ be a composition of $d+1$, and let
$\mathbf{n}=L_{m}(\mathbf{d})$. Let $\alpha$ be an $n$-braid with
$\alpha*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$. Then, $\alpha\in
Z(\mu_{m,\mathbf{d}})$ if and only if $\alpha$ is of the form

$\alpha=\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{0}\oplus{\textstyle\bigoplus_{i=1}^{r}}(\underbrace{\alpha_{i}\Delta_{(d_{i})}^{2}\oplus\cdots\oplus\alpha_{i}\Delta_{(d_{i})}^{2}}_{\ell_{i}}\oplus\underbrace{\alpha_{i}\oplus\cdots\oplus\alpha_{i}}_{m-\ell_{i}}))_{\mathbf{n}}$

such that $\hat{\alpha}\in Z(\mu_{m,r})$, $\alpha_{0}\in Z(\mu_{m,d_{0}})$ and
$d_{k_{i}}=d_{i}$ for $1\leq i\leq r$, where $k_{i}$’s and $\ell_{i}$’s are
integers such that $\pi_{\hat{\alpha}}(x_{i,j})=x_{k_{i},j-\ell_{i}}$ as in
Lemma 4.9.

###### Proof.

Since $\alpha*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$, $\alpha$ is
of the form

$\alpha=\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{0}\oplus\underbrace{\alpha_{1,1}\oplus\cdots\oplus\alpha_{1,m}}_{m}\oplus\cdots\oplus\underbrace{\alpha_{r,1}\oplus\cdots\oplus\alpha_{r,m}}_{m})_{\mathbf{n}}.$

Suppose that $\alpha\in Z(\mu_{m,\mathbf{d}})$. Since
$\alpha\ast\mathcal{C}_{\mathbf{n}}=\mu_{m,\mathbf{d}}\ast\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$,
we have

$\hat{\alpha}\mu_{m,r}=\operatorname{Ext}_{\mathbf{n}}(\alpha\mu_{m,\mathbf{d}})=\operatorname{Ext}_{\mathbf{n}}(\mu_{m,\mathbf{d}}\alpha)=\mu_{m,r}\hat{\alpha},$

hence $\hat{\alpha}\in Z(\mu_{m,r})$. By Lemma 4.9, $\pi_{\hat{\alpha}}$ is of
the form $\pi_{\hat{\alpha}}(x_{0})=x_{0}$ and
$\pi_{\hat{\alpha}}(x_{i,j})=x_{k_{i},j-\ell_{i}}$, where $1\leq k_{i}\leq r$
and $0\leq\ell_{i}<m$. Since
$\pi_{\hat{\alpha}}(x_{i,j})=x_{k_{i},j-\ell_{i}}$ and

$\mathbf{n}=L_{m}(\mathbf{d})=(md_{0}+1,\underbrace{d_{1},\ldots,d_{1}}_{m},\ldots,\underbrace{d_{r},\ldots,d_{r}}_{m}),$

one has

$\hat{\alpha}^{-1}*\mathbf{n}=(md_{0}+1,\underbrace{d_{k_{1}},\ldots,d_{k_{1}}}_{m},\ldots,\underbrace{d_{k_{r}},\ldots,d_{k_{r}}}_{m}).$

Therefore $d_{k_{i}}=d_{i}$ for $1\leq i\leq r$ because
$\hat{\alpha}*\mathbf{n}=\mathbf{n}$.

Now we compute $\alpha\mu_{m,\mathbf{d}}$ and $\mu_{m,\mathbf{d}}\alpha$
straightforwardly:

$\displaystyle\alpha\cdot\mu_{m,\mathbf{d}}$ $\displaystyle=$
$\displaystyle\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{0}\oplus{\textstyle\bigoplus_{i=1}^{r}}(\alpha_{i,1}\oplus\cdots\oplus\alpha_{i,m}))_{\mathbf{n}}$
$\displaystyle\mbox{}\quad\cdot\langle\mu_{m,r}\rangle_{\mathbf{n}}(\mu_{m,d_{0}}\oplus{\textstyle\bigoplus_{i=1}^{r}}(\Delta_{(d_{i})}^{2}\oplus
1\oplus\cdots\oplus 1))_{\mathbf{n}}$ $\displaystyle=$
$\displaystyle\langle\hat{\alpha}\rangle_{\mathbf{n}}\langle\mu_{m,r}\rangle_{\mathbf{n}}\cdot(\alpha_{0}\oplus{\textstyle\bigoplus_{i=1}^{r}}(\alpha_{i,m}\oplus\alpha_{i,1}\oplus\cdots\oplus\alpha_{i,m-1}))_{\mathbf{n}}$
$\displaystyle\mbox{}\quad\cdot(\mu_{m,d_{0}}\oplus{\textstyle\bigoplus_{i=1}^{r}}(\Delta_{(d_{i})}^{2}\oplus
1\oplus\cdots\oplus 1))_{\mathbf{n}}$ $\displaystyle=$
$\displaystyle\langle\hat{\alpha}\mu_{m,r}\rangle_{\mathbf{n}}(\alpha_{0}\mu_{m,d_{0}}\oplus{\textstyle\bigoplus_{i=1}^{r}}(\alpha_{i,m}\Delta_{(d_{i})}^{2}\oplus\alpha_{i,1}\oplus\cdots\oplus\alpha_{i,m-1}))_{\mathbf{n}},$
$\displaystyle\mu_{m,\mathbf{d}}\cdot\alpha$ $\displaystyle=$
$\displaystyle\langle\mu_{m,r}\rangle_{\mathbf{n}}(\mu_{m,d_{0}}\oplus{\textstyle\bigoplus_{i=1}^{r}}(\Delta_{(d_{i})}^{2}\oplus
1\oplus\cdots\oplus 1))_{\mathbf{n}}$
$\displaystyle\mbox{}\qquad\cdot\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{0}\oplus{\textstyle\bigoplus_{i=1}^{r}}(\alpha_{i,1}\oplus\cdots\oplus\alpha_{i,m}))_{\mathbf{n}}$
$\displaystyle=$
$\displaystyle\langle\mu_{m,r}\rangle_{\mathbf{n}}\langle\hat{\alpha}\rangle_{\mathbf{n}}\cdot(\mu_{m,d_{0}}\oplus{\textstyle\bigoplus_{i=1}^{r}}(\underbrace{1\oplus\cdots\oplus
1}_{\ell_{i}}\oplus\Delta_{(d_{i})}^{2}\oplus\underbrace{1\oplus\cdots\oplus
1}_{m-\ell_{i}-1}))_{\mathbf{n}}$
$\displaystyle\mbox{}\qquad\cdot(\alpha_{0}\oplus{\textstyle\bigoplus_{i=1}^{r}}(\alpha_{i,1}\oplus\cdots\oplus\alpha_{i,m}))_{\mathbf{n}}$
$\displaystyle=$
$\displaystyle\langle\mu_{m,r}\hat{\alpha}\rangle_{\mathbf{n}}(\mu_{m,d_{0}}\alpha_{0}\oplus$
$\displaystyle\mbox{}\qquad{\textstyle\bigoplus_{i=1}^{r}}(\alpha_{i,1}\oplus\cdots\oplus\alpha_{i,\ell_{i}}\oplus\Delta_{(d_{i})}^{2}\alpha_{i,\ell_{i}+1}\oplus\alpha_{i,\ell_{i}+2}\oplus\cdots\oplus\alpha_{i,m}))_{\mathbf{n}}.$

Comparing the above two formulae, we have $\alpha_{0}\in Z(\mu_{m,d_{0}})$. In
addition, for $1\leq i\leq r$,

$\displaystyle\alpha_{i,m}\Delta_{(d_{i})}^{2}=\alpha_{i,1}=\alpha_{i,2}=\cdots=\alpha_{i,\ell_{i}}=\Delta_{(d_{i})}^{2}\alpha_{i,\ell_{i}+1},$
$\displaystyle\alpha_{i,\ell_{i}+1}=\alpha_{i,\ell_{i}+2}=\cdots=\alpha_{i,m}.$

Let $\alpha_{i}=\alpha_{i,m}$, then $\alpha$ has the desired form. The other
direction is straightforward. ∎

######  Corollary 4.12.

Let $n=md+1$ with $m\geq 2$ and $d\geq 1$. Let
$\mathbf{d}=(d_{0}+1,d_{1},\ldots,d_{r})$ be a composition of $d+1$, and let
$\mathbf{n}=L_{m}(\mathbf{d})$. Let $\alpha$ be an $n$-braid with
$\alpha*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$, and let the
$\mathbf{n}$-exterior braid of $\alpha$ be pure. Then, $\alpha\in
Z(\mu_{m,\mathbf{d}})$ if and only if $\alpha$ is of the form

$\alpha=\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{0}\oplus
m\alpha_{1}\oplus\cdots\oplus m\alpha_{r})_{\mathbf{n}}$

where $\hat{\alpha}\in Z(\mu_{m,r})$ and $\alpha_{0}\in Z(\mu_{m,d_{0}})$.

###### Proof.

Since $\operatorname{Ext}_{\mathbf{n}}(\alpha)$ is pure, $k_{i}=i$ and
$\ell_{i}=0$ for $i=1,\ldots,r$ in Lemma 4.11. ∎

######  Lemma 4.13.

Let $n=md+1$ with $m\geq 2$ and $d\geq 1$. Let $\mathcal{C}$ be an unnested
curve system in $D_{n}$ such that $\mu_{m,d}*\mathcal{C}=\mathcal{C}$. Then
there exist $\zeta\in Z(\mu_{m,d})$ and a composition
$\mathbf{d}=(d_{0}+1,d_{1},\ldots,d_{r})$ of $d+1$ with $d_{1}\geq\cdots\geq
d_{r}$ such that

$(\chi\zeta)*\mathcal{C}=\mathcal{C}_{L_{m}(\mathbf{d})},$

where $\chi$ is an $n$-braid represented by
$\eta_{m,\mathbf{d}}\circ\eta_{m,d}^{-1}$.

###### Proof.

The construction of $\mathbf{d}$ and $\zeta$ can be viewed in Figure 16. Let
$\mathcal{C}^{\prime}=\eta_{m,d}^{-1}(\mathcal{C})$. It is a curve system in
$D_{n}^{(m)}$ such that $\rho_{m}(\mathcal{C}^{\prime})$ is isotopic to
$\mathcal{C}^{\prime}$.

---
$\mathcal{C}$$\scriptstyle{h}$$\scriptstyle{f}$$\mathcal{C}_{L_{m}(\mathbf{d})}$
$\mathcal{C}^{\prime}$$\scriptstyle{g}$$\scriptstyle{\phi_{m,d}}$$\scriptstyle{\eta_{m,d}}$$\mathcal{C}_{m,\mathbf{d}}$$\scriptstyle{\phi_{m,d}}$$\scriptstyle{\eta_{m,d}}$$\scriptstyle{\eta_{m,\mathbf{d}}}$
$\bar{\mathcal{C}}^{\prime}$$\scriptstyle{\bar{g}}$$\mathcal{C}_{\mathbf{d}}$

Figure 16. This figure shows the constructions in the proof of Lemma 4.13.

_Claim._  _Without loss of generality, we may assume that
$\rho_{m}(\mathcal{C}^{\prime})=\mathcal{C}^{\prime}$._

###### Proof.

Give a complete hyperbolic geometry on the interior of $D_{d+1}$, and lift it
to the interior of $D_{n}^{(m)}$ by using the covering
$\phi_{m,d}:D_{n}^{(m)}\to D_{d+1}$. Then $\rho_{m}$ is an isometry. Let
$\mathcal{C}^{\prime\prime}$ be the unique geodesic curve system isotopic to
$\mathcal{C}^{\prime}$. Because $\rho_{m}(\mathcal{C}^{\prime\prime})$ is
isotopic to $\mathcal{C}^{\prime\prime}$, we have
$\rho_{m}(\mathcal{C}^{\prime\prime})=\mathcal{C}^{\prime\prime}$. ∎

Now we assume that $\rho_{m}(\mathcal{C}^{\prime})=\mathcal{C}^{\prime}$. Let
$\bar{\mathcal{C}}^{\prime}=\phi_{m,d}(\mathcal{C}^{\prime})$. Because
$\bar{\mathcal{C}}^{\prime}$ is an unnested curve system in $D_{d+1}$, there
exists a diffeomorphism $\bar{g}:D_{d+1}\to D_{d+1}$ fixing the first puncture
and the points at $\partial D_{d+1}$ such that
$\bar{g}(\bar{\mathcal{C}}^{\prime})$ is a standard unnested curve system,
hence $\bar{g}(\bar{\mathcal{C}}^{\prime})=\mathcal{C}_{\mathbf{d}}$ for a
composition $\mathbf{d}=(d_{0}+1,d_{1},\ldots,d_{r})$ of $d+1$. Note that
$\bar{g}$ can be chosen so that $d_{1}\geq d_{2}\geq\cdots\geq d_{r}$.

Let $g:D_{n}^{(m)}\to D_{n}^{(m)}$ be the lift of $\bar{g}$ that fixes
$\partial D_{n}^{(m)}$ pointwise. Then $g$ is a $\rho_{m}$-equivariant
diffeomorphism with
$g(\mathcal{C}^{\prime})=\phi_{m,d}^{-1}(\mathcal{C}_{\mathbf{d}})=\mathcal{C}_{m,\mathbf{d}}$.
Let $h=\eta_{m,d}\circ g\circ\eta_{m,d}^{-1}$ and
$f=\eta_{m,\mathbf{d}}\circ\eta_{m,d}^{-1}$. Let $\zeta$ and $\chi$ be
$n$-braids represented by $h$ and $f$, respectively. Because $g$ is
$\rho_{m}$-equivariant, $\zeta\in Z(\mu_{m,d})$. By the construction, $(f\circ
h)(\mathcal{C})=\mathcal{C}_{L_{m}(\mathbf{d})}$, hence
$(\chi\zeta)*\mathcal{C}=\mathcal{C}_{L_{m}(\mathbf{d})}$. ∎

######  Corollary 4.14.

Let $n=md+1$ with $m\geq 2$ and $d\geq 1$. Let $\mathcal{C}_{1}$ and
$\mathcal{C}_{2}$ be unnested curve systems in $D_{n}$ that are of the same
type and invariant under the action of $\mu_{m,d}$. Then there exist $\chi\in
B_{n}$, $\chi_{1},\chi_{2}\in Z(\mu_{m,d})$ and a composition
$\mathbf{d}=(d_{0}+1,d_{1},\ldots,d_{r})$ of $d+1$ such that

$\chi\mu_{m,d}\chi^{-1}=\mu_{m,\mathbf{d}}\quad\mbox{and}\quad(\chi\chi_{1})*\mathcal{C}_{1}=(\chi\chi_{2})*\mathcal{C}_{2}=\mathcal{C}_{L_{m}(\mathbf{d})}.$

###### Proof.

Applying Lemma 4.13 to $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$, we have
$\chi_{1},\chi_{2}\in Z(\mu_{m,d})$ and compositions
$\mathbf{d}=(d_{0}+1,d_{1},\ldots,d_{r})$ and
$\mathbf{d}^{\prime}=(d_{0}^{\prime}+1,d_{1}^{\prime},\ldots,d_{r^{\prime}}^{\prime})$
of $d+1$ with $d_{1}\geq\cdots\geq d_{r}$ and $d_{1}^{\prime}\geq\cdots\geq
d_{r^{\prime}}^{\prime}$ such that

$(\chi\chi_{1})*\mathcal{C}_{1}=\mathcal{C}_{L_{m}(\mathbf{d})}\quad\mbox{and}\quad(\chi^{\prime}\chi_{2})*\mathcal{C}_{2}=\mathcal{C}_{L_{m}(\mathbf{d}^{\prime})},$

where $\chi$ and $\chi^{\prime}$ are represented by
$\eta_{m,\mathbf{d}}\circ\eta_{m,d}^{-1}$ and
$\eta_{m,\mathbf{d}^{\prime}}\circ\eta_{m,d}^{-1}$, respectively. Because
$\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ are of the same type, so are
$\mathcal{C}_{L_{m}(\mathbf{d})}$ and
$\mathcal{C}_{L_{m}(\mathbf{d}^{\prime})}$. This implies
$\mathbf{d}=\mathbf{d}^{\prime}$ because $d_{1}\geq\cdots\geq d_{r}$ and
$d_{1}^{\prime}\geq\cdots\geq d_{r^{\prime}}^{\prime}$. Therefore
$\chi=\chi^{\prime}$.

Notice that the diffeomorphisms
$(\eta_{m,\mathbf{d}}\circ\eta_{m,d}^{-1})\circ(\eta_{m,d}\circ\rho_{m}\circ\eta_{m,d}^{-1})\circ(\eta_{m,\mathbf{d}}\circ\eta_{m,d}^{-1})^{-1}$
and $\eta_{m,\mathbf{d}}\circ\rho_{m}\circ\eta_{m,\mathbf{d}}^{-1}$ represent
$\chi\mu_{m,d}\chi^{-1}$ and $\mu_{m,\mathbf{d}}$, respectively, modulo
$\Delta^{2}$. Since the two diffeomorphisms are the same,
$\chi\mu_{m,d}\chi^{-1}=\mu_{m,\mathbf{d}}\bmod\Delta^{2}$. Therefore

$\chi\mu_{m,d}\chi^{-1}=\mu_{m,\mathbf{d}}$

because $\chi\mu_{m,d}\chi^{-1}$ and $\mu_{m,\mathbf{d}}$ are conjugate. ∎

######  Definition 4.15.

Let $\Lambda$ be a choice function for the set $\cup_{k\geq 1}B_{k}$, hence
$\Lambda(A)\in A$ for $A\subset\cup_{k\geq 1}B_{k}$. For $\alpha\in B_{k}$,
let $[\alpha]$ denote the conjugacy class of $\alpha$ in $B_{k}$. Let $m\geq
2$ and $t\geq 0$. We define

$\lambda=\lambda_{m,t}:\cup_{k\geq 1}B_{k}\to\cup_{k\geq 1}B_{k}$

as $\lambda(\alpha)=\Lambda(Z(\mu_{m,t})\cap[\alpha])$ if $\alpha\in B_{mt+1}$
and $[\alpha]\cap Z(\mu_{m,t})\neq\emptyset$, and
$\lambda(\alpha)=\Lambda([\alpha])$ otherwise. We call $\lambda(\alpha)$ the
_conjugacy representative_ of $\alpha$ with respect to $(m,t)$.

Then $\lambda$ has the following properties.

  * (i)

$\lambda(\alpha)$ is conjugate to $\alpha$. Two braids $\alpha$ and $\beta$
are conjugate if and only if $\lambda(\alpha)=\lambda(\beta)$.

  * (ii)

If $\alpha\in Z(\mu_{m,t})$, then $\lambda(\alpha)\in Z(\mu_{m,t})$.

######  Lemma 4.16.

Let $n=md+1$ with $m\geq 2$ and $d\geq 1$. Let
$\mathbf{d}=(d_{0}+1,d_{1},\ldots,d_{r})$ be a composition of $d+1$, and let
$\mathbf{n}=L_{m}(\mathbf{d})$. Let
$\alpha=\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{0}\oplus
m\alpha_{1}\oplus\cdots\oplus m\alpha_{r})_{\mathbf{n}}$ belong to
$Z(\mu_{m,\mathbf{d}})$ such that $\hat{\alpha}$ is a pure braid. Let
$\lambda=\lambda_{m,d_{0}}$. Suppose that $\alpha_{0}$ and
$\lambda(\alpha_{0})$ are conjugate in $Z(\mu_{m,d_{0}})$. Then there exists
an $\mathbf{n}$-split braid $\gamma$ in $Z(\mu_{m,\mathbf{d}})$ such that

$\gamma\alpha\gamma^{-1}=\langle\hat{\alpha}\rangle_{\mathbf{n}}(\lambda(\alpha_{0})\oplus
m\lambda(\alpha_{1})\oplus\cdots\oplus m\lambda(\alpha_{r}))_{\mathbf{n}}.$

###### Proof.

Since $\alpha\in Z(\mu_{m,\mathbf{d}})$, one has $\alpha_{0}\in
Z(\mu_{m,d_{0}})$ by Lemma 4.11, hence $\lambda(\alpha_{0})\in
Z(\mu_{m,d_{0}})$. By the hypothesis, there exists $\gamma_{0}\in
Z(\mu_{m,d_{0}})$ such that
$\gamma_{0}\alpha_{0}\gamma_{0}^{-1}=\lambda(\alpha_{0})$. For $1\leq i\leq
r$, choose $\gamma_{i}\in B_{d_{i}}$ such that
$\gamma_{i}\alpha_{i}\gamma_{i}^{-1}=\lambda(\alpha_{i})$. Using
$\gamma_{i}$’s, define an $\mathbf{n}$-split braid $\gamma$ as

$\gamma=\langle 1\rangle_{\mathbf{n}}(\gamma_{0}\oplus
m\gamma_{1}\oplus\cdots\oplus m\gamma_{r})_{\mathbf{n}}.$

By Corollary 4.12, $\gamma\in Z(\mu_{m,\mathbf{d}})$. Because the
$\mathbf{n}$-exterior braids of $\alpha$ and $\gamma$ are pure braids, we can
compute the conjugation $\gamma\alpha\gamma^{-1}$ component-wise as follows.

$\displaystyle\gamma\alpha\gamma^{-1}$ $\displaystyle=$ $\displaystyle\langle
1\rangle_{\mathbf{n}}\langle\hat{\alpha}\rangle_{\mathbf{n}}\langle
1\rangle_{\mathbf{n}}^{-1}(\gamma_{0}\alpha_{0}\gamma_{0}^{-1}\oplus
m(\gamma_{1}\alpha_{1}\gamma_{1}^{-1})\oplus\cdots\oplus
m(\gamma_{r}\alpha_{r}\gamma_{r}^{-1}))_{\mathbf{n}}$ $\displaystyle=$
$\displaystyle\langle\hat{\alpha}\rangle_{\mathbf{n}}(\lambda(\alpha_{0})\oplus
m\lambda(\alpha_{1})\oplus\cdots\oplus m\lambda(\alpha_{r}))_{\mathbf{n}}.$

This completes the proof. ∎

##  5\. Proof of Theorem 1.4

In this section we prove Theorem 1.4. Let us explain our strategy.

Let $\omega$ be a periodic $n$-braid, and let $\alpha,\beta\in Z(\omega)$ be
conjugate in $B_{n}$. If $\alpha$ and $\beta$ are periodic, it is easy to
prove the theorem because we may assume that $\alpha$ and $\beta$ are central
by Corollary 2.4. If $\alpha$ and $\beta$ are pseudo-Anosov, then we obtain a
stronger result in Proposition 5.1 that any conjugating element between
$\alpha$ and $\beta$ must be contained in $Z(\omega)$. (This result will be
crucial in dealing with reducible braids.) However a far more laborious task
is required when $\alpha$ and $\beta$ are reducible.

Let us say that a periodic braid is _$\delta$ -type_ (resp.  _$\epsilon$
-type_) if it is conjugate to a power of $\delta$ (resp. a power of
$\epsilon$). The periodic braid $\omega$ is either $\delta$-type or
$\epsilon$-type, and Theorem 1.4 can be proved similarly for the two cases.
One possible strategy for escaping repetition of similar arguments would be
writing an explicit proof for one case and then showing that the other case
follows easily from that. However this strategy hardly works. It seems that
(when $\alpha$ and $\beta$ are reducible) the proof for $\delta$-type $\omega$
must contain most of the whole arguments needed in the proof for
$\epsilon$-type $\omega$, and that the theorem for $\delta$-type $\omega$ is
not a consequence of that for $\epsilon$-type $\omega$.

Proposition 5.2 is a key step in the proof of the theorem when $\alpha$ and
$\beta$ are reducible. The proposition may look rather unnatural at a glance,
but it is written in that form in order to minimize repetition of similar
arguments.

######  Proposition 5.1.

Let $\omega$ be a periodic $n$-braid, and let $\alpha$ and $\beta$ be pseudo-
Anosov braids in $Z(\omega)$. If $\gamma$ is an $n$-braid with
$\beta=\gamma\alpha\gamma^{-1}$, then $\gamma\in Z(\omega)$.

###### Proof.

Let $I_{\omega}:B_{n}\to B_{n}$ be the inner automorphism such that
$I_{\omega}(\chi)=\omega^{-1}\chi\omega$ for $\chi\in B_{n}$. Since
$\alpha,\beta\in Z(\omega)$, we have $I_{\omega}(\alpha)=\alpha$ and
$I_{\omega}(\beta)=\beta$. Therefore

$\gamma\alpha\gamma^{-1}=\beta=I_{\omega}(\beta)=I_{\omega}(\gamma\alpha\gamma^{-1})=I_{\omega}(\gamma)I_{\omega}(\alpha)I_{\omega}(\gamma^{-1})=I_{\omega}(\gamma)\alpha
I_{\omega}(\gamma)^{-1}.$

Let $\gamma_{0}=\gamma^{-1}I_{\omega}(\gamma)$. Then $\gamma_{0}\in
Z(\alpha)$. It is known by J. González-Meneses and B. Wiest [GW04] that the
centralizer of a pseudo-Anosov braid is a free abelian group of rank two.
Since $\alpha$ is pseudo-Anosov and $\gamma_{0},\omega\in Z(\alpha)$,
$\gamma_{0}$ commutes with $\omega$, hence
$I_{\omega}^{k}(\gamma_{0})=\gamma_{0}$ for all $k$. Take a positive integer
$m$ such that $\omega^{m}$ is central. Since $I_{\omega}^{m}(\gamma)=\gamma$,

$\displaystyle\gamma_{0}^{m}$ $\displaystyle=$
$\displaystyle\gamma_{0}\gamma_{0}\cdots\gamma_{0}=\gamma_{0}\cdot
I_{\omega}(\gamma_{0})\cdots I_{\omega}^{m-1}(\gamma_{0})$ $\displaystyle=$
$\displaystyle\gamma^{-1}I_{\omega}(\gamma)\cdot
I_{\omega}(\gamma^{-1})I_{\omega}^{2}(\gamma)\cdots
I_{\omega}^{m-1}(\gamma^{-1})I_{\omega}^{m}(\gamma)$ $\displaystyle=$
$\displaystyle\gamma^{-1}I_{\omega}^{m}(\gamma)=\gamma^{-1}\gamma=1.$

Since braid groups are torsion free (for example see [Deh98]), we have
$\gamma_{0}=1$, hence $\gamma=I_{\omega}(\gamma)$. This means that $\gamma\in
Z(\omega)$. ∎

We remark that, in the above proposition, the pseudo-Anosov condition on
$\alpha$ and $\beta$ is necessary. Let $\omega$ be a non-central periodic
braid, and let $\gamma$ be a braid which does not belong to $Z(\omega)$. Let
$\alpha=\beta=\Delta^{2}$. Because $\Delta^{2}$ is central, we have
$\alpha,\beta\in Z(\omega)$ and $\gamma\alpha\gamma^{-1}=\beta$. However,
$\gamma\not\in Z(\omega)$.

The following proposition is a key step in the proof of Theorem 1.4. Recall
that $\nu:B_{n,1}\to B_{n-1}$ is the homomorphism deleting the first strand.

######  Proposition 5.2.

Let $m\geq 2$ and $d\geq 1$. Let $\omega$ be a 1-pure braid conjugate to
$\epsilon_{(md+1)}^{d}$. Let $\alpha$ and $\beta$ be pure braids in
$Z(\omega)$ such that either $\alpha$ and $\beta$ are conjugate in $B_{md+1}$
or $\nu(\alpha)$ and $\nu(\beta)$ are conjugate in $B_{md}$. Then $\alpha$ and
$\beta$ are conjugate in $Z(\omega)$.

Before proving the above proposition, we prove Theorem 1.4 assuming the
proposition.

###### Proof of Theorem 1.4.

Let $\omega$ be a periodic $n$-braid, and let $\alpha,\beta\in Z(\omega)$ be
conjugate in $B_{n}$. We will show that $\alpha$ and $\beta$ are conjugate in
$Z(\omega)$.

_Claim._ Without loss of generality, we may assume that $\omega$ is non-
central, and both $\alpha$ and $\beta$ are pure braids.

###### Proof of Claim.

If $\omega$ is central, then $Z(\omega)=B_{n}$, hence there is nothing to
prove. Hence we may assume that $\omega$ is non-central.

Assume that Theorem 1.4 holds for pure braids. Choose a positive integer $k$
such that $\alpha^{k}$ and $\beta^{k}$ are pure. By assumption, $\alpha^{k}$
and $\beta^{k}$ are conjugate in $Z(\omega)$. Because $Z(\omega)$ is
isomorphic to $A(\mathbf{B}_{\ell})$ for some $\ell\geq 1$, $\alpha$ and
$\beta$ are conjugate in $Z(\omega)$ by Corollary 2.4. This shows that it
suffices to prove Theorem 1.4 for the case where $\alpha$ and $\beta$ are pure
braids. ∎

Now we assume that $\omega$ is noncentral, and both $\alpha$ and $\beta$ are
pure braids.

By Lemma 4.1 (i), we may assume that $\omega=\delta^{d}$ or
$\omega=\epsilon^{d}$ for some $d$. By Lemma 2.1, we may further assume that
$d$ is a positive integer which is a divisor of $n$ if $\omega=\delta^{d}$ and
a divisor of $n-1$ if $\omega=\epsilon^{d}$. Therefore there is a positive
integer $m$ such that $n=md$ if $\omega=\delta^{d}$ and $n=md+1$ if
$\omega=\epsilon^{d}$. Note that $m\geq 2$ because $\omega$ is central if
$m=1$.

If $\omega=\epsilon^{d}$, then $\alpha$ and $\beta$ are conjugate in
$Z(\omega)$ by Proposition 5.2.

Now, suppose that $\omega=\delta^{d}$. Let $\bar{\alpha}$ and $\bar{\beta}$ be
the elements of $B_{d+1,1}$ such that $\psi_{5}(\bar{\alpha})=\alpha$ and
$\psi_{5}(\bar{\beta})=\beta$. Let $\alpha^{\prime}=\psi_{4}(\bar{\alpha})$
and $\beta^{\prime}=\psi_{4}(\bar{\beta})$. Then $\alpha^{\prime}$ and
$\beta^{\prime}$ are $(n+1)$-braids in $Z(\epsilon_{(n+1)}^{d})$ such that
$\nu(\alpha^{\prime})=\alpha$ and $\nu(\beta^{\prime})=\beta$ are conjugate in
$B_{n}$.

---
$\textstyle{B_{n}\supset
Z(\delta_{(n)}^{d})}$$\textstyle{Z(\epsilon_{(n+1)}^{d})\subset
B_{n+1,1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu}$$\textstyle{B_{d+1,1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{5}}$$\scriptstyle{\psi_{4}}$
---
$\textstyle{\alpha,~{}\beta}$$\textstyle{\alpha^{\prime},~{}\beta^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu}$$\textstyle{\bar{\alpha},~{}\bar{\beta}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{5}}$$\scriptstyle{\psi_{4}}$

Since $\alpha$ and $\beta$ are pure, so are $\alpha^{\prime}$ and
$\beta^{\prime}$. Applying Proposition 5.2 to
$(\alpha^{\prime},\beta^{\prime},\epsilon_{(n+1)}^{d})$, we conclude that
$\alpha^{\prime}$ and $\beta^{\prime}$ are conjugate in
$Z(\epsilon_{(n+1)}^{d})$. This implies that $\bar{\alpha}$ and $\bar{\beta}$
are conjugate in $B_{d+1,1}$, hence $\alpha$ and $\beta$ are conjugate in
$Z(\delta^{d})=Z(\omega)$. ∎

###### Proof of Proposition 5.2.

Let $n=md+1$. Recall that $m\geq 2$ and $d\geq 1$; $\omega$ is a 1-pure braid
conjugate to $\epsilon_{(n)}^{d}$; $\alpha$ and $\beta$ are pure braids in
$Z(\omega)$ such that either $\alpha$ and $\beta$ are conjugate in $B_{n}$ or
$\nu(\alpha)$ and $\nu(\beta)$ are conjugate in $B_{n-1}$. We will show that
$\alpha$ and $\beta$ are conjugate in $Z(\omega)$ by using induction on $d\geq
1$. By Lemma 4.1, we may assume that $\omega=\mu_{m,d}$.

_Claim 1._

  * (i)

$\alpha$ and $\beta$ are conjugate in $Z(\omega)$ if and only if $\nu(\alpha)$
and $\nu(\beta)$ are conjugate in $Z(\nu(\omega))$.

  * (ii)

The braids $\alpha$, $\beta$, $\nu(\alpha)$ and $\nu(\beta)$ have the same
Nielsen-Thurston type.

  * (iii)

The proposition holds if $\alpha$ is periodic or pseudo-Anosov, or if $d=1$.

  * (iv)

If $\alpha$ is non-periodic and reducible, then
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)$ are of the same type.

###### Proof.

Let $\bar{\alpha}$ and $\bar{\beta}$ be the elements of $B_{d+1,1}$ such that
$\alpha=\psi_{4}^{\prime}(\bar{\alpha})$ and
$\beta=\psi_{4}^{\prime}(\bar{\beta})$, hence
$\nu(\alpha)=\psi_{5}^{\prime}(\bar{\alpha})$ and
$\nu(\beta)=\psi_{5}^{\prime}(\bar{\beta})$. We have the following commutative
diagram, where the isomorphisms preserve the Nielsen-Thurston type by Lemma
4.6 and Corollary 4.7.

---
$\textstyle{Z(\nu(\omega))}$$\textstyle{Z(\omega)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu}$$\scriptstyle{\simeq}$$\textstyle{B_{d+1,1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{5}^{\prime}}$$\scriptstyle{\simeq}$$\scriptstyle{\psi_{4}^{\prime}}$$\scriptstyle{\simeq}$
---
$\textstyle{\nu(\alpha),~{}\nu(\beta)}$$\textstyle{\alpha,~{}\beta\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu}$$\textstyle{\bar{\alpha},~{}\bar{\beta}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{5}^{\prime}}$$\scriptstyle{\psi_{4}^{\prime}}$

(i) This follows from the fact that $\nu:Z(\omega)\to Z(\nu(\omega))$ is an
isomorphism.

(ii) Because $\nu$ preserves the Nielsen-Thurston type, $(\alpha,\nu(\alpha))$
and $(\beta,\nu(\beta))$ are pairs of braids with the same Nielsen-Thurston
type. By the hypothesis, either $(\alpha,\beta)$ or $(\nu(\alpha),\nu(\beta))$
is a pair of braids with the same Nielsen-Thurston type. Therefore the braids
$\alpha$, $\beta$, $\nu(\alpha)$ and $\nu(\beta)$ have the same Nielsen-
Thurston type.

(iii) Suppose that $\alpha$ is pseudo-Anosov. Then $\beta$, $\nu(\alpha)$ and
$\nu(\beta)$ are pseudo-Anosov by (ii). If $\alpha$ and $\beta$ are conjugate
in $B_{n}$, they are conjugate in $Z(\omega)$ by Proposition 5.1. If
$\nu(\alpha)$ and $\nu(\beta)$ are conjugate in $B_{n-1}$, they are conjugate
in $Z(\nu(\omega))$ by Proposition 5.1, hence $\alpha$ and $\beta$ are
conjugate in $Z(\omega)$ by (i).

Suppose that $\alpha$ is periodic. Then $\beta$, $\nu(\alpha)$ and
$\nu(\beta)$ are periodic by (ii). Since $\alpha$, $\beta$, $\nu(\alpha)$ and
$\nu(\beta)$ are periodic and pure, they are central. If $\alpha$ and $\beta$
are conjugate in $B_{n}$, then $\alpha=\beta$. If $\nu(\alpha)$ and
$\nu(\beta)$ are conjugate in $B_{n-1}$, then $\nu(\alpha)=\nu(\beta)$, hence
$\alpha=\beta$.

Suppose that $d=1$. Because $\bar{\alpha}\in B_{2,1}=\\{\sigma_{1}^{2k}\mid
k\in\mathbf{Z}\\}=\\{\Delta_{(2)}^{2k}\mid k\in\mathbf{Z}\\}$, $\bar{\alpha}$
is periodic. Hence $\alpha$ is periodic. From the above discussion, $\alpha$
and $\beta$ are conjugate in $Z(\omega)$.

(iv) The braids $\nu(\alpha)$, $\beta$ and $\nu(\beta)$ are non-periodic and
reducible by (ii).

If $\alpha$ and $\beta$ are conjugate in $B_{n}$, then
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)$ are of the same type.

Suppose that $\nu(\alpha)$ and $\nu(\beta)$ are conjugate in $B_{n-1}$. Taking
conjugates of $\bar{\alpha}$ and $\bar{\beta}$ by elements of $B_{d+1,1}$ if
necessary, we may assume that
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\bar{\alpha})$ and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\bar{\beta})$ are standard.
Let $\mathbf{d}_{1}=(d_{0}+1,d_{1},\ldots,d_{r})$ and
$\mathbf{d}_{2}=(e_{0}+1,e_{1},\ldots,e_{s})$ be the compositions of $d+1$
such that
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\bar{\alpha})=\mathcal{C}_{\mathbf{d}_{1}}$
and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\bar{\beta})=\mathcal{C}_{\mathbf{d}_{2}}$.
Let

$\begin{array}[]{*7{l}}\mathbf{n}_{1}&=&(md_{0}+1,\underbrace{d_{1},\ldots,d_{1}}_{m},\ldots,\underbrace{d_{r},\ldots,d_{r}}_{m}),&&\mathbf{n}_{1}^{\prime}&=&(md_{0},\underbrace{d_{1},\ldots,d_{1}}_{m},\ldots,\underbrace{d_{r},\ldots,d_{r}}_{m}),\\\
\mathbf{n}_{2}&=&(me_{0}+1,\underbrace{e_{1},\ldots,e_{1}}_{m},\ldots,\underbrace{e_{s},\ldots,e_{s}}_{m}),&&\mathbf{n}_{2}^{\prime}&=&(me_{0},\underbrace{e_{1},\ldots,e_{1}}_{m},\ldots,\underbrace{e_{s},\ldots,e_{s}}_{m}),\end{array}$

where $md_{0}$ (resp. $me_{0}$) is deleted if $d_{0}=0$ (resp. $e_{0}=0$). Let
us write “$\mathcal{C}_{1}\approx\mathcal{C}_{2}$” if two curve systems
$\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ are of the same type. By Corollary
4.8,

$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)\approx\mathcal{C}_{\mathbf{n}_{1}},\quad\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)\approx\mathcal{C}_{\mathbf{n}_{2}},\quad\operatorname{\mathcal{R}}_{\operatorname{ext}}(\nu(\alpha))\approx\mathcal{C}_{\mathbf{n}_{1}^{\prime}},\quad\operatorname{\mathcal{R}}_{\operatorname{ext}}(\nu(\beta))\approx\mathcal{C}_{\mathbf{n}_{2}^{\prime}}.$

Since $\nu(\alpha)$ and $\nu(\beta)$ are conjugate in $B_{n-1}$, one has
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\nu(\alpha))\approx\operatorname{\mathcal{R}}_{\operatorname{ext}}(\nu(\beta))$,
and hence
$\mathcal{C}_{\mathbf{n}_{1}^{\prime}}\approx\mathcal{C}_{\mathbf{n}_{2}^{\prime}}$.
Thus the following equality holds between multisets consisting of integers:

$\\{md_{0},\underbrace{d_{1},\ldots,d_{1}}_{m},\ldots,\underbrace{d_{r},\ldots,d_{r}}_{m}\\}=\\{me_{0},\underbrace{e_{1},\ldots,e_{1}}_{m},\ldots,\underbrace{e_{s},\ldots,e_{s}}_{m}\\}.$

Since $m\geq 2$, one has $r=s$ and $d_{0}=e_{0}$ and there is an
$r$-permutation $\theta$ such that $d_{i}=e_{\theta(i)}$ for $1\leq i\leq r$.
Therefore $\mathcal{C}_{\mathbf{n}_{1}}$ and $\mathcal{C}_{\mathbf{n}_{2}}$
are of the same type, hence so are
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)$. ∎

By the above claim, the proposition holds if $d=1$, or if $\alpha$ is periodic
or pseudo-Anosov. Thus we may assume that $d\geq 2$ and that $\alpha$ is non-
periodic and reducible.

_Claim 2._ Without loss of generality, we may assume that
$\omega=\mu_{m,\mathbf{d}}$ for a composition
$\mathbf{d}=(d_{0}+1,d_{1},\ldots,d_{r})$ of $d+1$, and that
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)=\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)=\mathcal{C}_{\mathbf{n}}$
where $\mathbf{n}=L_{m}(\mathbf{d})$.

###### Proof of Claim 2.

Recall that we have assumed that $\omega=\mu_{m,d}$. By Claim 1 (iv),
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)$ are of the same type.
They are invariant under the action of $\omega$. By Corollary 4.14, there
exist $\chi\in B_{n}$, $\chi_{1},\chi_{2}\in Z(\omega)$ and a composition
$\mathbf{d}$ of $d+1$ such that $\chi\omega\chi^{-1}=\mu_{m,\mathbf{d}}$ and
$(\chi\chi_{1})*\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)=(\chi\chi_{2})*\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)=\mathcal{C}_{L_{m}(\mathbf{d})}$.
Notice that
$(\chi\chi_{1})*\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)=\operatorname{\mathcal{R}}_{\operatorname{ext}}(\chi\chi_{1}\alpha\chi_{1}^{-1}\chi^{-1})$
and
$(\chi\chi_{2})*\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)=\operatorname{\mathcal{R}}_{\operatorname{ext}}(\chi\chi_{2}\beta\chi_{2}^{-1}\chi^{-1})$.
By Lemma 4.1, it suffices to show that
$\chi\chi_{1}\alpha\chi_{1}^{-1}\chi^{-1}$ and
$\chi\chi_{2}\beta\chi_{2}^{-1}\chi^{-1}$ are conjugate in
$Z(\chi\omega\chi^{-1})=Z(\mu_{m,\mathbf{d}})$. ∎

Let us assume the hypothesis stated in Claim 2. By Corollary 4.12, $\alpha$
and $\beta$ are written as

$\displaystyle\alpha$ $\displaystyle=$
$\displaystyle\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{0}\oplus
m\alpha_{1}\oplus\cdots\oplus m\alpha_{r})_{\mathbf{n}},$ $\displaystyle\beta$
$\displaystyle=$
$\displaystyle\langle\hat{\beta}\rangle_{\mathbf{n}}(\beta_{0}\oplus
m\beta_{1}\oplus\cdots\oplus m\beta_{r})_{\mathbf{n}},$

where $\hat{\alpha},\hat{\beta}\in Z(\mu_{m,r})$ and $\alpha_{0},\beta_{0}\in
Z(\mu_{m,d_{0}})$.

Let $\lambda=\lambda_{m,d_{0}}$. By the induction hypothesis, $\alpha_{0}$ and
$\lambda(\alpha_{0})$ (resp. $\beta_{0}$ and $\lambda(\beta_{0})$) are
conjugate in $Z(\mu_{m,d_{0}})$. By Lemmas 4.16 and 4.1 (ii), we may assume
that $\alpha_{i}$’s and $\beta_{i}$’s are conjugacy representatives, that is,
$\alpha_{i}=\lambda(\alpha_{i})$ and $\beta_{i}=\lambda(\beta_{i})$ for $0\leq
i\leq r$.

The braids $\nu(\alpha)$ and $\nu(\beta)$ are written differently according to
whether $d_{0}\geq 1$ or $d_{0}=0$. Let $\mathbf{n}^{\prime}$ be a
compositions of $n-1$ such that

$\begin{array}[]{lcll}\mathbf{n}^{\prime}&=&(md_{0},\underbrace{d_{1},\ldots,d_{1}}_{m},\ldots,\underbrace{d_{r},\ldots,d_{r}}_{m})&\mbox{if
$d_{0}\geq 1$},\\\
\mathbf{n}^{\prime}&=&(\underbrace{d_{1},\ldots,d_{1}}_{m},\ldots,\underbrace{d_{r},\ldots,d_{r}}_{m})&\mbox{if
$d_{0}=0$}.\end{array}$

By Corollary 4.8,
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\nu(\alpha))$ and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\nu(\beta))$ are obtained
from $\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)$ and
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)$, respectively, by
forgetting the first puncture. Since
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)=\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)=\mathcal{C}_{\mathbf{n}}$,
one has

$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\nu(\alpha))=\operatorname{\mathcal{R}}_{\operatorname{ext}}(\nu(\beta))=\mathcal{C}_{\mathbf{n}^{\prime}}.$

If $d_{0}\geq 1$, then $\nu(\mu_{m,\mathbf{d}})$, $\nu(\alpha)$ and
$\nu(\beta)$ are written as

$\displaystyle\nu(\mu_{m,\mathbf{d}})$ $\displaystyle=$
$\displaystyle\langle\mu_{m,r}\rangle_{\mathbf{n}^{\prime}}(\nu(\mu_{m,d_{0}})\oplus{\textstyle\bigoplus_{i=1}^{r}}(\underbrace{\Delta_{(d_{i})}^{2}\oplus
1\oplus\cdots\oplus 1}_{m}))_{\mathbf{n}^{\prime}},$
$\displaystyle\nu(\alpha)$ $\displaystyle=$
$\displaystyle\langle\hat{\alpha}\rangle_{\mathbf{n}^{\prime}}(\nu(\alpha_{0})\oplus
m\alpha_{1}\oplus\cdots\oplus m\alpha_{r})_{\mathbf{n}^{\prime}},$
$\displaystyle\nu(\beta)$ $\displaystyle=$
$\displaystyle\langle\hat{\beta}\rangle_{\mathbf{n}^{\prime}}(\nu(\beta_{0})\oplus
m\beta_{1}\oplus\cdots\oplus m\beta_{r})_{\mathbf{n}^{\prime}}.$

In this case, $\nu(\alpha_{0}),\nu(\beta_{0})\in Z(\nu(\mu_{m,d_{0}}))$.

If $d_{0}=0$, then $\nu(\mu_{m,\mathbf{d}})$, $\nu(\alpha)$ and $\nu(\beta)$
are written as

$\displaystyle\nu(\mu_{m,\mathbf{d}})$ $\displaystyle=$
$\displaystyle\langle\nu(\mu_{m,r})\rangle_{\mathbf{n}^{\prime}}({\textstyle\bigoplus_{i=1}^{r}}(\underbrace{\Delta_{(d_{i})}^{2}\oplus
1\oplus\cdots\oplus 1}_{m}))_{\mathbf{n}^{\prime}},$
$\displaystyle\nu(\alpha)$ $\displaystyle=$
$\displaystyle\langle\nu(\hat{\alpha})\rangle_{\mathbf{n}^{\prime}}(m\alpha_{1}\oplus\cdots\oplus
m\alpha_{r})_{\mathbf{n}^{\prime}}$ $\displaystyle\nu(\beta)$ $\displaystyle=$
$\displaystyle\langle\nu(\hat{\beta})\rangle_{\mathbf{n}^{\prime}}(m\beta_{1}\oplus\cdots\oplus
m\beta_{r})_{\mathbf{n}^{\prime}}.$

In this case, $\nu(\hat{\alpha}),\nu(\hat{\beta})\in Z(\nu(\mu_{m,r}))$.

_Claim 3._ $\hat{\alpha}$ and $\hat{\beta}$ are conjugate in $B_{mr+1}$.

###### Proof of Claim 3.

If $\alpha$ and $\beta$ are conjugate in $B_{n}$, then the
$\mathbf{n}$-exterior braids $\hat{\alpha}$ and $\hat{\beta}$ are conjugate in
$B_{mr+1}$ by Lemma 2.10.

Suppose that $\nu(\alpha)$ and $\nu(\beta)$ are conjugate in $B_{n-1}$. If
$d_{0}\geq 1$, then $\hat{\alpha}$ and $\hat{\beta}$ are conjugate in
$B_{mr+1}$ by Lemma 2.10 because
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\nu(\alpha))=\operatorname{\mathcal{R}}_{\operatorname{ext}}(\nu(\beta))=\mathcal{C}_{\mathbf{n}^{\prime}}$
and the $\mathbf{n}^{\prime}$-exterior braids of $\nu(\alpha)$ and
$\nu(\beta)$ are $\hat{\alpha}$ and $\hat{\beta}$. Similarly, if $d_{0}=0$,
then $\nu(\hat{\alpha})$ and $\nu(\hat{\beta})$ are conjugate in $B_{mr}$,
hence $\hat{\alpha}$ and $\hat{\beta}$ are conjugate in $Z(\mu_{m,r})$ by the
induction hypothesis. ∎

Since
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)=\mathcal{C}_{\mathbf{n}}$
and $\operatorname{Ext}_{\mathbf{n}}(\alpha)=\hat{\alpha}$, the
$\mathbf{n}$-exterior braid $\hat{\alpha}$ is either periodic or pseudo-
Anosov.

Case 1. $\hat{\alpha}$ is periodic. Let $k=mr+1$. Since $\hat{\alpha}$ and
$\hat{\beta}$ are pure and periodic, they are central. Because they are
conjugate by the above claim, $\hat{\alpha}=\hat{\beta}=\Delta_{(k)}^{2u}$ for
some integer $u$.

_Claim 4._ $\alpha_{0}=\beta_{0}$ and there is an $r$-permutation $\theta$
such that $\beta_{i}=\alpha_{\theta(i)}$ for $1\leq i\leq r$.

###### Proof of Claim 4.

If $\alpha$ and $\beta$ are conjugate in $B_{n}$, then the following equality
holds between multisets consisting of conjugacy classes by J. González-Meneses
[Gon03, Proposition 3.2]:

$\Bigl{\\{}[\alpha_{0}],\underbrace{[\alpha_{1}],\ldots,[\alpha_{1}]}_{m},\ldots,\underbrace{[\alpha_{r}],\ldots,[\alpha_{r}]}_{m}\Bigr{\\}}=\Bigl{\\{}[\beta_{0}],\underbrace{[\beta_{1}],\ldots,[\beta_{1}]}_{m},\ldots,\underbrace{[\beta_{r}],\ldots,[\beta_{r}]}_{m}\Bigr{\\}}.$

Since $m\geq 2$, $[\alpha_{0}]=[\beta_{0}]$ and there is an $r$-permutation
$\theta$ such that $[\beta_{i}]=[\alpha_{\theta(i)}]$ for $1\leq i\leq r$. By
the induction hypothesis, $\alpha_{0}$ and $\beta_{0}$ are conjugate in
$Z(\mu_{m,d_{0}})$.

If $\nu(\alpha)$ and $\nu(\beta)$ are conjugate in $B_{n-1}$ and $d_{0}\geq
1$, then the following equality holds between multisets consisting of
conjugacy classes:

$\Bigl{\\{}[\nu(\alpha_{0})],\underbrace{[\alpha_{1}],\ldots,[\alpha_{1}]}_{m},\ldots,\underbrace{[\alpha_{r}],\ldots,[\alpha_{r}]}_{m}\Bigr{\\}}=\Bigl{\\{}[\nu(\beta_{0})],\underbrace{[\beta_{1}],\ldots,[\beta_{1}]}_{m},\ldots,\underbrace{[\beta_{r}],\ldots,[\beta_{r}]}_{m}\Bigr{\\}}.$

Since $m\geq 2$, $[\nu(\alpha_{0})]=[\nu(\beta_{0})]$ and there is an
$r$-permutation $\theta$ such that $[\beta_{i}]=[\alpha_{\theta(i)}]$ for
$1\leq i\leq r$. By the induction hypothesis, $\alpha_{0}$ and $\beta_{0}$ are
conjugate in $Z(\mu_{m,d_{0}})$.

If $\nu(\alpha)$ and $\nu(\beta)$ are conjugate in $B_{n-1}$ and $d_{0}=0$,
then the following equality holds between multisets consisting of conjugacy
classes:

$\Bigl{\\{}\underbrace{[\alpha_{1}],\ldots,[\alpha_{1}]}_{m},\ldots,\underbrace{[\alpha_{r}],\ldots,[\alpha_{r}]}_{m}\Bigr{\\}}=\Bigl{\\{}\underbrace{[\beta_{1}],\ldots,[\beta_{1}]}_{m},\ldots,\underbrace{[\beta_{r}],\ldots,[\beta_{r}]}_{m}\Bigr{\\}}.$

Then there is an $r$-permutation $\theta$ such that
$[\beta_{i}]=[\alpha_{\theta(i)}]$ for $1\leq i\leq r$. In this case,
$\alpha_{0}$ and $\beta_{0}$ are the unique braid with one strand.

From the above three cases, we can see that $\alpha_{0}$ is conjugate to
$\beta_{0}$ in $Z(\mu_{m,d_{0}})$ and there is an $r$-permutation $\theta$
such that $\beta_{i}$ is conjugate to $\alpha_{\theta(i)}$ for $1\leq i\leq
r$. Because we have assumed that each $\mathbf{n}$-interior braid of $\alpha$
and $\beta$ is the conjugacy representative with respect to $(m,d_{0})$, one
has $\alpha_{0}=\beta_{0}$ and $\beta_{i}=\alpha_{\theta(i)}$ for $1\leq i\leq
r$. ∎

In the above claim, $\theta$ must satisfy $d_{i}=d_{\theta(i)}$ for all $1\leq
i\leq r$. By the claim, we have

$\displaystyle\alpha$ $\displaystyle=$
$\displaystyle\langle\Delta_{(k)}^{2u}\rangle_{\mathbf{n}}(\alpha_{0}\oplus
m\alpha_{1}\oplus\cdots\oplus m\alpha_{r})_{\mathbf{n}},$ $\displaystyle\beta$
$\displaystyle=$
$\displaystyle\langle\Delta_{(k)}^{2u}\rangle_{\mathbf{n}}(\alpha_{0}\oplus
m\alpha_{\theta(1)}\oplus\cdots\oplus m\alpha_{\theta(r)})_{\mathbf{n}}.$

By Lemma 4.9, there exists $\hat{\zeta}\in Z(\mu_{m,r})$ such that the induced
permutation of $\hat{\zeta}$ fixes $x_{0}$ and sends $x_{i,j}$ to
$x_{\theta(i),j}$ for $1\leq i\leq r$ and $1\leq j\leq m$ under the notation
of the lemma. Let $\zeta=\langle\hat{\zeta}\rangle_{\mathbf{n}}$. Then
$\zeta\ast\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$ because
$d_{i}=d_{\theta(i)}$ for all $1\leq i\leq r$, hence $\zeta\in
Z(\mu_{m,\mathbf{d}})$ by Lemma 4.11. On the other hand,

$\displaystyle\zeta\beta\zeta^{-1}$ $\displaystyle=$
$\displaystyle\langle\hat{\zeta}\rangle_{\mathbf{n}}\langle\Delta_{(k)}^{2u}\rangle_{\mathbf{n}}(\alpha_{0}\oplus
m\alpha_{\theta(1)}\oplus\cdots\oplus
m\alpha_{\theta(r)})_{\mathbf{n}}\langle\hat{\zeta}\rangle_{\mathbf{n}}^{-1}$
$\displaystyle=$
$\displaystyle\langle\hat{\zeta}\rangle_{\mathbf{n}}\langle\Delta_{(k)}^{2u}\rangle_{\mathbf{n}}\langle\hat{\zeta}\rangle_{\mathbf{n}}^{-1}(\alpha_{0}\oplus
m\alpha_{1}\oplus\cdots\oplus m\alpha_{r})_{\mathbf{n}}$ $\displaystyle=$
$\displaystyle\langle\Delta_{(k)}^{2u}\rangle_{\mathbf{n}}(\alpha_{0}\oplus
m\alpha_{1}\oplus\cdots\oplus m\alpha_{r})_{\mathbf{n}}=\alpha.$

Therefore $\alpha$ and $\beta$ are conjugate in $Z(\mu_{m,\mathbf{d}})$.

Case 2. $\hat{\alpha}$ is pseudo-Anosov.

_Claim 5._ The braids $\alpha$ and $\beta$ are conjugate in $B_{n}$.

###### Proof of Claim 5.

If $\alpha$ and $\beta$ are conjugate in $B_{n}$, there is nothing to prove.
Therefore, assume that $\nu(\alpha)$ and $\nu(\beta)$ are conjugate in
$B_{n-1}$. Let $\gamma$ be an $(n-1)$-braid with
$\nu(\beta)=\gamma^{-1}\nu(\alpha)\gamma$. Since
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\nu(\alpha))=\operatorname{\mathcal{R}}_{\operatorname{ext}}(\nu(\beta))=\mathcal{C}_{\mathbf{n}^{\prime}}$,
one has
$\gamma*\mathcal{C}_{\mathbf{n}^{\prime}}=\mathcal{C}_{\mathbf{n}^{\prime}}$
and
$\operatorname{Ext}_{\mathbf{n}^{\prime}}(\nu(\beta))=\operatorname{Ext}_{\mathbf{n}^{\prime}}(\gamma)^{-1}\operatorname{Ext}_{\mathbf{n}^{\prime}}(\nu(\alpha))\operatorname{Ext}_{\mathbf{n}^{\prime}}(\gamma)$
by Lemma 2.10.

First, suppose that $d_{0}\geq 1$. Because
$\gamma*\mathcal{C}_{\mathbf{n}^{\prime}}=\mathcal{C}_{\mathbf{n}^{\prime}}$,
$\gamma$ is of the form

$\gamma=\langle\hat{\gamma}\rangle_{\mathbf{n}^{\prime}}(\gamma_{0}\oplus\underbrace{\gamma_{1,1}\oplus\cdots\oplus\gamma_{1,m}}_{m}\oplus\cdots\oplus\underbrace{\gamma_{r,1}\oplus\cdots\oplus\gamma_{r,m}}_{m})_{\mathbf{n}^{\prime}}.$

In this case, $\hat{\beta}=\hat{\gamma}^{-1}\hat{\alpha}\hat{\gamma}$. Because
$\hat{\alpha}$ and $\hat{\beta}$ are pseudo-Anosov braids in $Z(\mu_{m,r})$,
one has $\hat{\gamma}\in Z(\mu_{m,r})$ by Proposition 5.1. In particular,
$\hat{\gamma}$ is 1-pure by Lemma 4.9. Since
$\nu(\beta)=\gamma^{-1}\nu(\alpha)\gamma$ and since $\hat{\alpha}$,
$\hat{\beta}$ and $\hat{\gamma}$ are all 1-pure, we have
$\nu(\beta_{0})=\gamma_{0}^{-1}\nu(\alpha_{0})\gamma_{0}$. By the induction
hypothesis, there exists $\gamma_{0}^{\prime}\in Z(\mu_{m,d_{0}})$ such that
$\beta_{0}=\gamma_{0}^{\prime-1}\alpha_{0}\gamma_{0}^{\prime}$. Let
$\gamma^{\prime}$ be the $n$-braid obtained from $\gamma$ by replacing
$\mathbf{n}^{\prime}$ with $\mathbf{n}$ and $\gamma_{0}$ with
$\gamma_{0}^{\prime}$, that is,

$\gamma^{\prime}=\langle\hat{\gamma}\rangle_{\mathbf{n}}(\gamma_{0}^{\prime}\oplus\underbrace{\gamma_{1,1}\oplus\cdots\oplus\gamma_{1,m}}_{m}\oplus\cdots\oplus\underbrace{\gamma_{r,1}\oplus\cdots\oplus\gamma_{r,m}}_{m})_{\mathbf{n}}.$

We can easily show $\beta=\gamma^{\prime-1}\alpha\gamma^{\prime}$ by using the
fact that $\nu(\beta)=\gamma^{-1}\nu(\alpha)\gamma$, $\hat{\gamma}$ is 1-pure,
and $\beta_{0}=\gamma_{0}^{\prime-1}\alpha_{0}\gamma_{0}^{\prime}$.

Now, suppose that $d_{0}=0$. Because
$\gamma*\mathcal{C}_{\mathbf{n}^{\prime}}=\mathcal{C}_{\mathbf{n}^{\prime}}$,
$\gamma$ is of the form

$\gamma=\langle\hat{\gamma}\rangle_{\mathbf{n}^{\prime}}(\underbrace{\gamma_{1,1}\oplus\cdots\oplus\gamma_{1,m}}_{m}\oplus\cdots\oplus\underbrace{\gamma_{r,1}\oplus\cdots\oplus\gamma_{r,m}}_{m})_{\mathbf{n}^{\prime}}.$

In this case,
$\nu(\hat{\beta})=\hat{\gamma}^{-1}\nu(\hat{\alpha})\hat{\gamma}$. Since
$\hat{\alpha}$ and $\hat{\beta}$ are pseudo-Anosov braids in $Z(\mu_{m,r})$,
$\nu(\hat{\alpha})$ and $\nu(\hat{\beta})$ are pseudo-Anosov braids in
$Z(\nu(\mu_{m,r}))$ by Corollary 4.7. Thus $\hat{\gamma}\in Z(\nu(\mu_{m,r}))$
by Proposition 5.1. Because $\nu:Z(\mu_{m,r})\to Z(\nu(\mu_{m,r}))$ is an
isomorphism, there exists $\hat{\gamma}^{\prime}\in Z(\mu_{m,r})$ with
$\nu(\hat{\gamma}^{\prime})=\hat{\gamma}$. Then
$\hat{\beta}=\hat{\gamma}^{\prime-1}\hat{\alpha}\hat{\gamma}^{\prime}$. In
addition, $\hat{\gamma}^{\prime}$ is 1-pure by Lemma 4.9. Since $d_{0}=0$,
both $\alpha_{0}$ and $\beta_{0}$ are the unique braid with one strand. Let
$\gamma^{\prime}$ be the $n$-braid obtained from $\gamma$ by replacing
$\mathbf{n}^{\prime}$ with $\mathbf{n}$ and $\hat{\gamma}$ with
$\hat{\gamma}^{\prime}$ and by inserting the trivial 1-braid as follows:

$\gamma^{\prime}=\langle\hat{\gamma}^{\prime}\rangle_{\mathbf{n}}(1\oplus\underbrace{\gamma_{1,1}\oplus\cdots\oplus\gamma_{1,m}}_{m}\oplus\cdots\oplus\underbrace{\gamma_{r,1}\oplus\cdots\oplus\gamma_{r,m}}_{m})_{\mathbf{n}}.$

We can easily show $\beta=\gamma^{\prime-1}\alpha\gamma^{\prime}$ by using the
fact that $\nu(\beta)=\gamma^{-1}\nu(\alpha)\gamma$,
$\hat{\beta}=\hat{\gamma}^{\prime-1}\hat{\alpha}\hat{\gamma}^{\prime}$, and
$\hat{\gamma}^{\prime}$ is 1-pure. ∎

Recall that $\alpha$ and $\beta$ are of the form

$\displaystyle\alpha$ $\displaystyle=$
$\displaystyle\langle\hat{\alpha}\rangle_{\mathbf{n}}(\alpha_{0}\oplus
m\alpha_{1}\oplus\cdots\oplus m\alpha_{r})_{\mathbf{n}},$ $\displaystyle\beta$
$\displaystyle=$
$\displaystyle\langle\hat{\beta}\rangle_{\mathbf{n}}(\beta_{0}\oplus
m\beta_{1}\oplus\cdots\oplus m\beta_{r})_{\mathbf{n}},$

where $\hat{\alpha},\hat{\beta}\in Z(\mu_{m,r})$, $\alpha_{0},\beta_{0}\in
Z(\mu_{m,d_{0}})$, $\alpha_{i}=\lambda(\alpha_{i})$ and
$\beta_{i}=\lambda(\beta_{i})$ for $0\leq i\leq r$.

Because $\alpha$ and $\beta$ are conjugate in $B_{n}$ by the above claim,
there is an $n$-braid $\gamma$ such that

$\beta=\gamma\alpha\gamma^{-1}.$

Since
$\operatorname{\mathcal{R}}_{\operatorname{ext}}(\alpha)=\operatorname{\mathcal{R}}_{\operatorname{ext}}(\beta)=\mathcal{C}_{\mathbf{n}}$,
we have $\gamma*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$, hence
$\gamma$ is of the form

$\gamma=\langle\hat{\gamma}\rangle_{\mathbf{n}}(\gamma_{0}\oplus\underbrace{\gamma_{1,1}\oplus\cdots\oplus\gamma_{1,m}}_{m}\oplus\cdots\oplus\underbrace{\gamma_{r,1}\oplus\cdots\oplus\gamma_{r,m}}_{m})_{\mathbf{n}}.$

Since $\hat{\alpha}$ and $\hat{\beta}$ are pseudo-Anosov braids in
$Z(\mu_{m,r})$ and
$\hat{\beta}=\operatorname{Ext}_{\mathbf{n}}(\beta)=\operatorname{Ext}_{\mathbf{n}}(\gamma\alpha\gamma^{-1})=\hat{\gamma}\hat{\alpha}\hat{\gamma}^{-1}$,
we have $\hat{\gamma}\in Z(\mu_{m,r})$ by Proposition 5.1. Let $\theta$ be the
$(mr+1)$-permutation induced by $\hat{\gamma}$. By Lemma 4.9,
$\theta(x_{0})=x_{0}$ and there exist integers $1\leq k_{i}\leq r$ and
$0\leq\ell_{i}<m$ such that $\theta(x_{i,j})=x_{k_{i},j-\ell_{i}}$ for $1\leq
i\leq r$ and $1\leq j\leq m$. Define

$\chi=\langle\hat{\gamma}\rangle_{\mathbf{n}}(1\oplus{\textstyle\bigoplus_{i=1}^{r}}(\underbrace{\Delta_{(d_{i})}^{2}\oplus\cdots\oplus\Delta_{(d_{i})}^{2}}_{\ell_{i}}\oplus\underbrace{1\oplus\cdots\oplus
1}_{m-\ell_{i}}))_{\mathbf{n}}.$

Since
$\chi\ast\mathcal{C}_{\mathbf{n}}=\langle\hat{\gamma}\rangle_{\mathbf{n}}*\mathcal{C}_{\mathbf{n}}=\gamma*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$,
one has $d_{i}=d_{k_{i}}$ for $1\leq i\leq r$. Thus $\chi\in
Z(\mu_{m,\mathbf{d}})$ by Lemma 4.11. Let

$\beta^{\prime}=\chi^{-1}\beta\chi.$

Since $\beta,\chi\in Z(\mu_{m,\mathbf{d}})$ with
$\beta*\mathcal{C}_{\mathbf{n}}=\chi*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$,
one has $\beta^{\prime}\in Z(\mu_{m,\mathbf{d}})$ with
$\beta^{\prime}*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$. Since
$\operatorname{Ext}_{\mathbf{n}}(\beta^{\prime})=\operatorname{Ext}_{\mathbf{n}}(\chi^{-1}\beta\chi)=\hat{\gamma}^{-1}\hat{\beta}\hat{\gamma}=\hat{\alpha}$
is pure, $\beta^{\prime}$ is of the following form by Corollary 4.12:

$\beta^{\prime}=\langle\hat{\alpha}\rangle_{\mathbf{n}}(\beta_{0}^{\prime}\oplus
m\beta_{1}^{\prime}\oplus\cdots\oplus
m\beta_{r}^{\prime})_{\mathbf{n}},\qquad\beta_{0}^{\prime}\in
Z(\mu_{m,d_{0}}).$

Since both $\operatorname{Ext}_{\mathbf{n}}(\beta)$ and
$\operatorname{Ext}_{\mathbf{n}}(\chi)$ are 1-pure, one has
$\beta_{0}^{\prime}=\beta_{0}$, hence
$\beta_{0}^{\prime}=\beta_{0}=\lambda(\beta_{0})=\lambda(\beta_{0}^{\prime})$.
By Lemma 4.16, there exists an $\mathbf{n}$-split braid $\chi^{\prime}$ in
$Z(\mu_{m,\mathbf{d}})$ such that

(1)
$\beta^{\prime\prime}=\chi^{\prime-1}\beta^{\prime}\chi^{\prime}=\langle\hat{\alpha}\rangle_{\mathbf{n}}(\beta_{0}^{\prime\prime}\oplus
m\beta_{1}^{\prime\prime}\oplus\cdots\oplus
m\beta_{r}^{\prime\prime})_{\mathbf{n}},$

where $\beta_{i}^{\prime\prime}$’s are conjugacy representatives, that is,
$\beta_{i}^{\prime\prime}=\lambda(\beta_{i}^{\prime\prime})$ for $0\leq i\leq
r$.

Now we will show that $\beta^{\prime\prime}=\alpha$. Let
$\chi^{\prime\prime}=\gamma^{-1}\chi$. Then
$\chi^{\prime\prime}*\mathcal{C}_{\mathbf{n}}=\mathcal{C}_{\mathbf{n}}$.
Because
$\operatorname{Ext}_{\mathbf{n}}(\chi^{\prime\prime})=\operatorname{Ext}_{\mathbf{n}}(\gamma)^{-1}\operatorname{Ext}_{\mathbf{n}}(\chi)=\hat{\gamma}^{-1}\hat{\gamma}=1$,
$\chi^{\prime\prime}$ is an $\mathbf{n}$-split braid. Note that

$\beta^{\prime\prime}=\chi^{\prime-1}\beta^{\prime}\chi^{\prime}=\chi^{\prime-1}\chi^{-1}\beta\chi\chi^{\prime}=\chi^{\prime-1}\chi^{-1}\gamma\alpha\gamma^{-1}\chi\chi^{\prime}=(\chi^{\prime\prime}\chi^{\prime})^{-1}\alpha(\chi^{\prime\prime}\chi^{\prime}).$

Because both $\chi^{\prime}$ and $\chi^{\prime\prime}$ are $\mathbf{n}$-split
braids, so is $\chi^{\prime\prime}\chi^{\prime}$. Therefore
$\chi^{\prime\prime}\chi^{\prime}$ is of the form

$\chi^{\prime\prime}\chi^{\prime}=\langle
1\rangle_{\mathbf{n}}(\chi_{0}\oplus{\textstyle\bigoplus_{i=1}^{r}}(\chi_{i,1}\oplus\cdots\oplus\chi_{i,m}))_{\mathbf{n}}.$

Because the $\mathbf{n}$-exterior braids of $\alpha$ and
$\chi^{\prime\prime}\chi^{\prime}$ are pure braids, the conjugation
$(\chi^{\prime\prime}\chi^{\prime})^{-1}\alpha(\chi^{\prime\prime}\chi^{\prime})$
can be computed component-wise:

(2)
$(\chi^{\prime\prime}\chi^{\prime})^{-1}\alpha(\chi^{\prime\prime}\chi^{\prime})=\langle\hat{\alpha}\rangle_{\mathbf{n}}(\chi_{0}^{-1}\alpha_{0}\chi_{0}\oplus{\textstyle\bigoplus_{i=1}^{r}}(\chi_{i,1}^{-1}\alpha_{i}\chi_{i,1}\oplus\cdots\oplus\chi_{i,m}^{-1}\alpha_{i}\chi_{i,m}))_{\mathbf{n}}.$

Comparing $(\ref{eq:1})$ and $(\ref{eq:2})$, we can see that $\alpha_{i}$ and
$\beta_{i}^{\prime\prime}$ are conjugate for $0\leq i\leq r$. Because
$\alpha_{i}$ and $\beta_{i}^{\prime\prime}$ are conjugacy representatives with
respect to $(m,d_{0})$, we have $\alpha_{i}=\beta_{i}^{\prime\prime}$ for
$0\leq i\leq r$, hence $\alpha=\beta^{\prime\prime}$. Because

$\alpha=\beta^{\prime\prime}=(\chi\chi^{\prime})^{-1}\beta(\chi\chi^{\prime})$

and $\chi,\chi^{\prime}\in Z(\mu_{m,\mathbf{d}})$, $\alpha$ and $\beta$ are
conjugate in $Z(\mu_{m,\mathbf{d}})$. ∎

We remark that in the above proof we can show by a straightforward computation
that $\beta_{i}^{\prime}=\beta_{k_{i}}$ for $i=1,\ldots,r$. In particular, the
$\mathbf{n}$-interior braids $\beta_{0}^{\prime},\ldots,\beta_{r}^{\prime}$
are conjugacy representatives with respect to $(m,d_{0})$, hence we can take
$\chi^{\prime}=1$ and $\beta^{\prime\prime}=\beta^{\prime}$. This means that
$\alpha=\chi^{-1}\beta\chi$, where $\chi$ depends only on $\mathbf{n}$ and
$\hat{\gamma}$. However we introduced $\chi^{\prime}$ and
$\beta^{\prime\prime}$ because it is much simpler than showing
$\beta_{i}^{\prime}=\beta_{k_{i}}$.

### Acknowledgements

The authors are grateful to the anonymous referee for valuable comments and
suggestions. This work was done partially while the authors were visiting
Department of Mathematical and Computing Sciences, Tokyo Institute of
Technology in February 2008. We thank Sadayoshi Kojima and Eiko Kin for their
hospitality and interest in this work. This work was supported by the Korea
Research Foundation Grant funded by the Korean Government (MOEHRD, Basic
Research Promotion Fund) (KRF-2008-331-C00039).

## References

  * [All02] D. Allcock, _Braid pictures for Artin groups_ , Trans. Amer. Math. Soc. 354 (2002), 3455–3474.
  * [Art25] E. Artin, _Theorie der Zöpfe_ , Hamburg Abh. 4 (1925), 47–72.
  * [BM07] R.W. Bell and D. Margalit, _Injections of Artin groups_ , Comment. Math. Helv. 82 (2007), 725–751.
  * [BDM02] D. Bessis, F. Digne and J. Michel, _Springer theory in braid groups and the Birman-Ko-Lee monoid_ , Pacific J. Math. 205 (2002), 287–309.
  * [BGG06] J.S. Birman, V. Gebhardt and J. González-Meneses, _Conjugacy in Garside Groups III: Periodic braids_ , J. Algebra 316 (2007), 246–276.
  * [BLM83] J.S. Birman, A. Lubotzky and J. McCarthy, _Abelian and solvable subgroups of maping class groups_ , Duke Math. J. 50 (1983), 1107–1120.
  * [Bro19] L.E.J. Brouwer, _Über die periodischen Transformationen der Kugel_ , Math. Ann. 80 (1919), 39–41.
  * [CC05] R. Charney and J. Crisp, _Automorphism groups of some affine and finite type Artin groups_ , Math. Res. Lett. 12 (2005), 321–333.
  * [CK94] A. Constantin and B. Kolev, _The theorem of Kerékjártó on periodic homeomorphisms of the disc and the sphere_ , Enseign. Math. (2) 40 (1994), 193–204.
  * [Cri99] J. Crisp, _Injective maps between Artin groups_ , Geometric group theory down under (Canberra, 1996), 119–137, de Gruyter, Berlin, 1999.
  * [CP05] J. Crisp and L. Paris, _Artin groups of type $B$ and $D$_, Adv. Geo. 5 (2005), 607–636.
  * [Deh98] P. Dehornoy, _Gaussian groups are torsion free_ , J. Algebra 210 (1998), 291–297.
  * [Eil34] S. Eilenberg, _Sur les transformations périodiques de la surface de la sphére_ , Fund. Math. 22 (1934), 28–41.
  * [Gon03] J. González-Meneses, _The $n$th root of a braid is unique up to conjugacy_, Algebr. Geom. Topol. 3 (2003), 1103–1118.
  * [GW04] J. González-Meneses and B. Wiest, _On the structure of the centralizer of a braid_ , Ann. Sci. Ecole Norm. Sup. (4) 37 (2004), 729–757.
  * [Iva92] N.V. Ivanov, _Subgroups of Teichmüller modular group_ , Transl. Math. Monogr. 115, AMS, 1992.
  * [Ker19] B. de Kerékjártó, _Über die periodischen Transformationen der Kreisscheibe und der Kugelfläche_ , Math. Annalen 80 (1919), 3–7.
  * [LL07] E.-K. Lee and S.-J. Lee, _Uniqueness of roots up to conjugacy for some affine and finite type Artin groups_ , arXiv:0711.0091, to appear in Math. Z.
  * [LL08] E.-K. Lee and S.-J. Lee, _A Garside-theoretic approach to the reducibility problem in braid groups_ , J. Algebra 320 (2008), 783–820.
  * [Thu88] W. Thurston, _On the isotopy and geometry of diffeomorphisms of surfaces_ , Bull. Amer. Math. Soc. 19 (1988), 109–140.