#  Pointed Hopf algebras with classical Weyl groups (II)

Weicai Wu, Shouchuan Zhang, Zhengtang Tan
Department of Mathematics, Hunan University
Changsha 410082, P.R. China, Emails<EMAIL_ADDRESS>
###### Abstract

We prove that except in several cases Nichols algebras of irreducible Yetter-
Drinfeld modules over classical Weyl groups
$\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}$ are infinite dimensional. We also
prove that except in several cases conjugacy classes of classical Weyl groups
are of type $D$; hence they collapse. We give the relationship between
$\mathfrak{B}({\mathcal{O}}_{\sigma},\rho)$ and ${\mathcal{E}}_{n}$.

2000 Mathematics Subject Classification: 16W30, 16G10

keywords: Rack, Hopf algebra, Weyl group.

##  0 Introduction

This work is a contribution to the classification of finite-dimensional Hopf
algebras over an algebraically closed field of characteristic 0, problem posed
by I. Kaplansky in 1975. N. Andruskiewitsch and H.-J. Schneider classify
finite-dimensional complex Hopf algebras by Lifting method [AS10]. N.
Andruskiewitsch and M. Grana study Nichols algebra of the most important class
of braided vector spaces $(CX,cq)$, where X is a rack and q is a 2-cocycle on
X with values in C in [AG03]. N. Andruskiewitsch, F. Fantino, M. Graña,
L.Vendramin and S. Zhang obtain that Nichols algebra
$\mathfrak{B}({\mathcal{O}}_{\sigma},\rho)$ over symmetry groups have infinite
dimension, except for a small list of examples and remarkable cases
corresponding to ${\mathcal{O}}_{\sigma}$ in [AFGV08, AFZ, AZ07]. Shouchuan
Zhang and Yao-Zhong Zhang show that except in three cases Nichols algebras of
irreducible Yetter-Drinfeld (YD in short) modules over classical Weyl groups
$A\rtimes\mathbb{S}_{n}$ supported by $\mathbb{S}_{n}$ are infinite
dimensional in [ZZ12], but this has not completed for general elements.

This paper follows from [ZZ12] to keep on classifying finite dimensional
complex pointed Hopf algebras with classic Weyl groups.

In this paper we prove that except in several cases Nichols algebras of
irreducible Yetter-Drinfeld modules over classical Weyl groups
$\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}$ are infinite dimensional. We also
prove that except in several cases conjugacy classes of classical Weyl groups
are of type $D$; hence they collapse.

The main results in this paper are summarized in the following statements.

######  Theorem 0.1.

Let $G=\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}$ with $n>4$. Let
$\tau\in\mathbb{S}_{n}$ be of type
$(1^{\lambda_{1}},2^{\lambda_{2}},\dots,n^{\lambda_{n}})$ and
$a\in\mathbb{Z}_{2}^{n}$ with $\sigma=(a,\tau)\in G$ with $\tau\not=1$. If
$\dim\mathfrak{B}(\mathcal{O}_{\sigma}^{G},\rho)<\infty$, then some of the
following hold.

  1. (i)

$(2,3)$; $(2^{3});$

  2. (ii)

$(2^{4});$ $(1,2^{2}),$

  3. (iii)

$(1^{2},2^{2})$, $(1^{n-2},2)$ and $(1^{n-3},3)$ with $a_{i}=a_{j}$ when
$\tau(i)=i$ and $\tau(j)=j$.

Indeed, It follows from Theorem 2.6 and Theorem 3.2.

## Preliminaries and Conventions

Let ${k}$ be the complex field, A quiver $Q=(Q_{0},Q_{1},s,t)$ is an oriented
graph, where $Q_{0}$ and $Q_{1}$ are the sets of vertices and arrows,
respectively; $s$ and $t$ are two maps from $Q_{1}$ to $Q_{0}$. For any arrow
$a\in Q_{1}$, $s(a)$ and $t(a)$ are called its start vertex and end vertex,
respectively, and $a$ is called an arrow from $s(a)$ to $t(a)$. For any $n\geq
0$, an $n$-path or a path of length $n$ in the quiver $Q$ is an ordered
sequence of arrows $p=a_{n}a_{n-1}\cdots a_{1}$ with $t(a_{i})=s(a_{i+1})$ for
all $1\leq i\leq n-1$. Note that a 0-path is exactly a vertex and a 1-path is
exactly an arrow. In this case, we define $s(p)=s(a_{1})$, the start vertex of
$p$, and $t(p)=t(a_{n})$, the end vertex of $p$. For a 0-path $x$, we have
$s(x)=t(x)=x$. Let $Q_{n}$ be the set of $n$-paths. Let ${}^{y}Q_{n}^{x}$
denote the set of all $n$-paths from $x$ to $y$, $x,y\in Q_{0}$. That is,
${}^{y}Q_{n}^{x}=\\{p\in Q_{n}\mid s(p)=x,t(p)=y\\}$.

A quiver $Q$ is finite if $Q_{0}$ and $Q_{1}$ are finite sets. A quiver $Q$ is
locally finite if ${}^{y}Q_{1}^{x}$ is a finite set for any $x,y\in Q_{0}$.

Let ${\mathcal{K}}(G)$ denote the set of conjugacy classes in $G$. A formal
sum $r=\sum_{C\in{\mathcal{K}}(G)}r_{C}C$ of conjugacy classes of $G$ with
cardinal number coefficients is called a ramification (or ramification data )
of $G$, i.e. for any $C\in{\mathcal{K}}(G)$, $r_{C}$ is a cardinal number. In
particular, a formal sum $r=\sum_{C\in{\mathcal{K}}(G)}r_{C}C$ of conjugacy
classes of $G$ with non-negative integer coefficients is a ramification of
$G$.

For any ramification $r$ and $C\in{\mathcal{K}}(G)$, since $r_{C}$ is a
cardinal number, we can choose a set $I_{C}(r)$ such that its cardinal number
is $r_{C}$ without loss of generality. Let
${\mathcal{K}}_{r}(G):=\\{C\in{\mathcal{K}}(G)\mid
r_{C}\not=0\\}=\\{C\in{\mathcal{K}}(G)\mid I_{C}(r)\not=\emptyset\\}$. If
there exists a ramification $r$ of $G$ such that the cardinal number of
${}^{y}Q_{1}^{x}$ is equal to $r_{C}$ for any $x,y\in G$ with $x^{-1}y\in
C\in{\mathcal{K}}(G)$, then $Q$ is called a Hopf quiver with respect to the
ramification data $r$. In this case, there is a bijection from $I_{C}(r)$ to
${}^{y}Q_{1}^{x}$, and hence we write ${\ }^{y}Q_{1}^{x}=\\{a_{y,x}^{(i)}\mid
i\in I_{C}(r)\\}$ for any $x,y\in G$ with $x^{-1}y\in C\in{\mathcal{K}}(G)$.

$(G,r,\overrightarrow{\rho},u)$ is called a ramification system with
irreducible representations (or RSR in short ), if $r$ is a ramification of
$G$; $u$ is a map from ${\mathcal{K}}(G)$ to $G$ with $u(C)\in C$ for any
$C\in{\mathcal{K}}(G)$; $I_{C}(r,u)$ and $J_{C}(i)$ are sets with
$\mid\\!J_{C}(i)\\!\mid$ = ${\rm deg}(\rho_{C}^{(i)})$ and
$I_{C}(r)=\\{(i,j)\mid i\in I_{C}(r,u),j\in J_{C}(i)\\}$ for any
$C\in{\mathcal{K}}_{r}(G)$, $i\in I_{C}(r,u)$;
$\overrightarrow{\rho}=\\{\rho_{C}^{(i)}\\}_{i\in
I_{C}(r,u),C\in{\mathcal{K}}_{r}(G)}\
\in\prod_{C\in{\mathcal{K}}_{r}(G)}(\widehat{{G^{u(C)}}})^{\mid
I_{C}(r,u)\mid}$ with $\rho_{C}^{(i)}\in\widehat{{G^{u(C)}}}$ for any $i\in
I_{C}(r,u),C\in{\mathcal{K}}_{r}(G)$. In this paper we always assume that
$I_{C}(r,u)$ is a finite set for any $C\in{\mathcal{K}}_{r}(G).$ Furthermore,
if $\rho_{C}^{(i)}$ is a one dimensional representation for any
$C\in{\mathcal{K}}_{r}(G)$, then $(G,r,\overrightarrow{\rho},u)$ is called a
ramification system with characters  (or RSC $(G,r,\overrightarrow{\rho},u)$
in short ) (see [ZZC04, Definition 1.8]). In this case, $a_{y,x}^{(i,j)}$ is
written as $a_{y,x}^{(i)}$ in short since $J_{C}(i)$ has only one element.

For ${\rm RSR}(G,r,\overrightarrow{\rho},u)$, let $\chi_{C}^{(i)}$ denote the
character of $\rho_{C}^{(i)}$ for any $i\in I_{C}(r,u)$,
$C\in{\mathcal{K}}_{r}(C)$. If ramification $r=r_{C}C$ and
$I_{C}(r,u)=\\{i\\}$ then we say that ${\rm RSR}(G,r,\overrightarrow{\rho},u)$
is bi-one, written as ${\rm RSR}(G,{\mathcal{O}}_{s},\rho)$ with $s=u(C)$ and
$\rho=\rho_{C}^{(i)}$ in short, since $r$ only has one conjugacy class $C$ and
$\mid\\!I_{C}(r,u)\\!\mid=1$. Quiver Hopf algebras, Nichols algebras and
Yetter-Drinfeld modules, corresponding to a bi-one ${\rm
RSR}(G,r,\overrightarrow{\rho},u)$, are said to be bi-one.

If $(G,r,\overrightarrow{\rho},u)$ is an ${\rm RSR}$, then it is clear that
${\rm RSR}(G,{\mathcal{O}}_{u(C)},\rho_{C}^{(i)})$ is bi-one for any
$C\in{\mathcal{K}}$ and $i\in I_{C}(r,u)$, which is called a bi-one  sub-${\rm
RSR}$ of ${\rm RSR}(G,r,\overrightarrow{\rho},u)$,

For $s\in G$ and $(\rho,V)\in\widehat{G^{s}}$, here is a precise description
of the YD module $M({\mathcal{O}}_{s},\rho)$, introduced in [Gr00, AZ07]. Let
$t_{1}=s$, …, $t_{m}$ be a numeration of ${\mathcal{O}}_{s}$, which is a
conjugacy class containing $s$, and let $g_{i}\in G$ such that $g_{i}\rhd
s:=g_{i}sg_{i}^{-1}=t_{i}$ for all $1\leq i\leq m$. Then
$M({\mathcal{O}}_{s},\rho)=\oplus_{1\leq i\leq m}g_{i}\otimes V$. Let
$g_{i}v:=g_{i}\otimes v\in M({\mathcal{O}}_{s},\rho)$, $1\leq i\leq m$, $v\in
V$. If $v\in V$ and $1\leq i\leq m$, then the action of $h\in G$ and the
coaction are given by

$\displaystyle\delta(g_{i}v)=t_{i}\otimes g_{i}v,\qquad
h\cdot(g_{i}v)=g_{j}(\gamma\cdot v),$ (0.1)

where $hg_{i}=g_{j}\gamma$, for some $1\leq j\leq m$ and $\gamma\in G^{s}$.
The explicit formula for the braiding is then given by

$c(g_{i}v\otimes g_{j}w)=t_{i}\cdot(g_{j}w)\otimes
g_{i}v=g_{j^{\prime}}(\gamma\cdot w)\otimes g_{i}v$ (0.2)

for any $1\leq i,j\leq m$, $v,w\in V$, where $t_{i}g_{j}=g_{j^{\prime}}\gamma$
for unique $j^{\prime}$, $1\leq j^{\prime}\leq m$ and $\gamma\in G^{s}$. Let
$\mathfrak{B}({\mathcal{O}}_{s},\rho)$ denote
$\mathfrak{B}(M({\mathcal{O}}_{s},\rho))$. $M({\mathcal{O}}_{s},\rho)$ is a
simple YD module (see [AZ07, Section 1.2 ]).

Set $sq(x,y):=x\rhd(y\rhd(x\rhd y)).$ $G$ is called collapse if every finite
dimensional pointed Hopf algebra with group $G$ is a group algebra. It follows
from [AFGV10] that $G$ is called collapse if and only if
$\dim\mathfrak{B}(\mathcal{O}_{\sigma}^{G},\rho)=\infty$ for any
$\rho\in\widehat{G^{\sigma}}$. Let
$\mathbb{B}_{n}:=\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}$. Obviously,
$(a,\tau)=a\tau$ for any $(a,\tau)\in\mathbb{B}_{n}.$

Assume that $X$ is a rack. $R\cup S$ is called a subrack decomposition of $X$
if $R$ and $S$ are two disjoint subracks satisfying $y\rhd x\in R$ and $x\rhd
y\in S$, for any $x\in R,y\in S$. $X$ is said to be of type $D$ if there
exists a subrack decomposition $R\cup S$ and two element $r\in R,s\in S$ such
that $sq(r,s)\not=s.$ We say that a finite rack $X$ collapses if for any
finite faithful cocycle $q$ ( associated to any decomposition of $X$ and of
any degree $n$ ) $\dim\mathfrak{B}(X,q)=\infty$ (see [AFGV08, Section 2.4] and
[AG03]).

##  1 Extension

In this section we apply juxtapositions to decide if Nichols algebras
associated to the irreducible Yetter-Drinfeld modules over classic Weyl groups
are finite dimensional or not.

######  Lemma 1.1.

Let $G=\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}$ and $(a,\tau),(b,\mu)\in G$.
Let $(c,\lambda)$ denote $(a,\tau)\rhd((b,\mu)\rhd((a,\tau)\rhd(b,\mu)))$.

(i) Then

$\displaystyle c$ $\displaystyle=$
$\displaystyle(a+\tau\cdot[b+\mu\cdot(a+\tau\cdot b+(\tau\rhd\mu)\cdot
a)+(\mu\rhd(\tau\rhd\mu))\cdot b]$ (1.1) $\displaystyle+$
$\displaystyle(\tau\rhd(\mu\rhd(\tau\rhd\mu)))\cdot a.$

(ii) If $\tau$ and $\mu$ are commutative, then

$\displaystyle c=a+\tau\mu\cdot a+\tau\mu^{2}\cdot a+\mu\cdot a+\tau\cdot
b+\tau^{2}\mu\cdot b+\tau\mu\cdot b.$ (1.2)

(iii) If $\tau$ and $\mu$ are commutative, then $sq(a\tau,b\mu)=b\mu$ if and
only if

$\displaystyle a+\tau\mu\cdot a+\tau\mu^{2}\cdot a+\mu\cdot a=b+\tau\cdot
b+\tau^{2}\mu\cdot b+\tau\mu\cdot b.$ (1.3)

(iv) If $\tau$ and $\mu$ are commutative with $(b,\mu)=\xi\rhd(a,\tau)$ and
$\xi\cdot a=a$, then $c=a+\tau\mu^{2}\cdot a+\mu\cdot a+\tau\cdot
a+\tau^{2}\mu\cdot a.$

(v) If $\tau$ and $\xi$ are commutative with $\tau^{2}=1$ and
$(b,\mu)=\xi\rhd(a,\tau)$ and $\xi\cdot a=a$, then $c=a$.

Proof. It is clear. $\Box$

If lengths of independent sign cycles of $(a,\pi)$ and $(b,\tau)$ are
different each other, then they are called orthogonal each other, written as
$a\pi\bot b\tau.$

For any $a\pi\in\mathbb{B}_{n}$ and $b\tau\in\mathbb{B}_{m}$, define
$a\pi\\#b\tau\in\mathbb{B}_{m+n}$ as follows:
$(a\\#b)_{i}:=\left\\{\begin{array}[]{ll}a_{i}&\hbox{when }i\leq n\\\
b_{i-n}&\hbox{when }i>n\end{array}\right.,$
$(\pi\\#\tau)(i):=\left\\{\begin{array}[]{ll}\pi(i)&\hbox{when }i\leq n\\\
\tau(i-n)+n&\hbox{when }i>n\end{array}\right.$ and
$(a\pi)\\#(b\tau):=(a\\#b,\pi\\#\tau),$ which is called a juxtaposition of
$a\pi$ and $b\tau$. Obviously, $(a\pi)\\#(b\tau)\in\mathbb{B}_{m+n}$. Let
$\overrightarrow{\nu_{n,m}}$ be a map from $\mathbb{B}_{n}$ to
$\mathbb{B}_{m+n}$ by sending $a\pi$ to
$\overrightarrow{\nu_{n,m}}(a\pi):=a\pi\\#1_{\mathbb{B}_{m}}$; let
$\overleftarrow{\nu_{n,m}}$ be a map from $\mathbb{B}_{m}$ to
$\mathbb{B}_{m+n}$ by sending $b\tau$ to
$\overleftarrow{\nu_{n,m}}(b\tau):=1_{\mathbb{B}_{n}}\\#b\tau$.

######  Lemma 1.2.

Assume $a\pi\bot b\tau$ with $a\pi,a^{\prime}\pi^{\prime}\in\mathbb{B}_{n}$
and $b\tau,b^{\prime}\tau^{\prime}\in\mathbb{B}_{m}$. Then

(i) $(a\pi\\#b\tau)(a^{\prime}\pi^{\prime}\\#b^{\prime}\tau^{\prime})=(a\pi
a^{\prime}\pi^{\prime}\\#b\tau b^{\prime}\tau^{\prime})$ .

(ii)
$a\pi\\#b\tau=\overrightarrow{\nu_{n,m}}(a\pi)\overleftarrow{\nu_{n,m}}(b\tau)=\overleftarrow{\nu_{n,m}}(b\tau)\overrightarrow{\nu_{n,m}}(a\pi)$.

(iii)
$\mathbb{B}_{m+n}^{a\pi\\#b\tau}=\mathbb{B}_{n}^{a\pi}\\#\mathbb{B}_{m}^{b\tau}=\overrightarrow{\nu_{n,m}}(\mathbb{B}_{n}^{a\pi})\overleftarrow{\nu_{n,m}}(\mathbb{B}_{m}^{b\tau})$
as directed products.

(iv) For any $\rho\in\widehat{\mathbb{B}_{m+n}^{a\pi\\#b\tau}}$, there exist
$\mu\in\widehat{\mathbb{B}_{n}^{a\pi}}$,
$\lambda\in\widehat{\mathbb{B}_{m}^{b\tau}}$ such that
$\rho=\mu\otimes\lambda$.

(v)
$(a\pi\\#b\tau)\rhd(a^{\prime}\pi^{\prime}\\#b^{\prime}\tau^{\prime})=(a\pi\rhd
a^{\prime}\pi^{\prime})\\#(b\tau\rhd b^{\prime}\tau^{\prime})$

(vi)
$\mathcal{O}_{a\pi\\#b\tau}^{\mathbb{B}_{m+n}}=\mathcal{O}_{a\pi}^{\mathbb{B}_{n}}\\#\mathcal{O}_{b\tau}^{\mathbb{B}_{m}}$.

Proof. (i) , (ii) and (v) are clear.

(iii) By [AZ07, Section 2.2],
$\mathbb{S}_{m+n}^{\pi\\#\tau}=\mathbb{S}_{n}^{\pi}\\#\mathbb{S}_{m}^{\tau}.$
Obviously,
$\mathbb{B}_{n}^{a\pi}\\#\mathbb{B}_{m}^{b\tau}\subseteq\mathbb{B}_{m+n}^{a\pi\\#b\tau}$.
For any $c\xi\in\mathbb{B}_{m+n}^{a\pi\\#b\tau},$ then
$\xi\in\mathbb{S}_{m+n}^{\pi\\#\tau}$ and there exist
$\mu\in\mathbb{S}_{n}^{\pi}$ and $\lambda\in\mathbb{S}_{m}^{\tau}$ such that
$\xi=\mu\\#\lambda$. Consequently, $c\xi=d\mu\\#f\lambda.$ Considering
$c\xi(a\pi\\#b\tau)=(a\pi\\#b\tau)c\xi$, we have
$d\mu\in\mathbb{B}_{n}^{a\pi}$ and $f\lambda\in\mathbb{B}_{m}^{b\tau}$. This
completes the proof.

(iv) It follows from (iii).

(vi) By (v),
$\mathcal{O}_{a\pi}^{\mathbb{B}_{n}}\\#\mathcal{O}_{b\tau}^{\mathbb{B}_{m}}\subseteq\mathcal{O}_{a\pi\\#b\tau}^{\mathbb{B}_{m+n}}.$
Consequently, (vi) follows from (iii). $\Box$

Remark: (i), (ii) and (v) above still hold when $a\pi$ and $b\tau$ are not
orthogonal each other.

######  Theorem 1.3.

If $\mathcal{O}_{a\tau}$ is of type $D$, then $\mathcal{O}_{a\tau\\#b\mu}$ is
of type $D$, too.

Proof. Let $X=R\cup S$ be a subrack decomposition of $\mathcal{O}_{a\tau}$ and
of type $D$. It is clear that $X\\#b\mu=R\\#b\mu\cup S\\#b\mu$ is a subrack
decomposition of $\mathcal{O}_{a\tau\\#b\mu}$ and of type $D$.$\Box$

######  Lemma 1.4.

Assume $a\pi\bot b\tau$. Let
$\rho=\mu\otimes\lambda\in\widehat{\mathbb{B}_{m+n}^{a\pi\\#b\tau}}$ ,
$\mu\in\widehat{\mathbb{B}_{n}^{a\pi}}$,
$\lambda\in\widehat{\mathbb{B}_{m}^{b\tau}}$, $a\pi\in\mathbb{B}_{n}$ and
$b\tau\in\mathbb{B}_{m}$ with $q_{a\pi}id=\mu(a\pi)$ and
$q_{b\tau}id=\lambda(b\tau)$. If
$\dim\mathfrak{B}({\mathcal{O}}_{a\pi\\#b\tau}^{\mathbb{B}_{n+m}},\rho)<\infty$,
then

(i) $M({\mathcal{O}}_{a\pi}^{\mathbb{B}_{n}},\mu)$ is isomorphic to a YD
submodule of $M({\mathcal{O}}_{a\pi\\#b\tau}^{\mathbb{B}_{n+m}},\rho)$ over
$\mathbb{B}_{n}$ when $q_{b\tau}=1$; hence
$\dim\mathfrak{B}({\mathcal{O}}_{a\pi}^{\mathbb{B}_{n}},\mu)<\infty$.

(ii) $q_{a\pi}q_{b\tau}=-1$.

(iii) $q_{b\tau}=1$ and $q_{a\pi}=-1$ when ${\rm ord}(b\tau)\leq 2$ and ${\rm
ord}(q_{a\pi})\not=1$.

(iv) $q_{b\tau}=1$ and $q_{a\pi}=-1$ when ${\rm ord}(a\pi)$ and ${\rm
ord}(b\tau)$ are coprime and ${\rm ord}(b\tau)$ is odd.

Proof. (i) We can show this as the proof in [AZ07, Section 2.2].

(ii) and (iii) are clear.

(iv) Obviously, $ordq_{a\pi}\mid ord(a\pi)$ and $ordq_{b\tau}\mid ord(b\tau)$.
Therefore, $ordq_{b\tau}$ is odd. $(ord(q_{a\pi}),ord(q_{b\tau}))=1$ since
$(ord{(a\pi}),ord({b\tau}))=1$. By Part (ii), $ordq_{a\pi}ordq_{b\tau}=2$.
Consequently, $ordq_{a\pi}=2$ and $ordq_{b\tau}=1$. $\Box$

##  2 Rack of $\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}$

In this section we prove that except in several cases conjugacy classes of
classical Weyl groups are of type $D$. Let
$\alpha:=(1,1,\cdots,1)\in\mathbb{Z}_{2}^{n}.$

######  Lemma 2.1.

Let $n$ be odd with $n\geq 5$ and $a\tau\in\mathbb{B}_{n}$ with $\tau=(1\
2\cdots\ n)$. Then $\mathcal{O}_{a\tau}^{\mathbb{B}_{n}}$ is of type $D$.

Proof. (i) Assume that $a\tau$ is a negative cycle and $b=(1,0,\cdots,0)$.
Thus $\alpha\tau$, $a\tau$ and $b\tau$ are conjugate. We assume $a=\alpha$
without lost generality. Obviously, the side of right hand (1.3) $\not=0$ for
$\mu=\tau^{2}$, i.e. $sq(\alpha\tau,b\tau^{2})\not=b\tau^{2}.$ Let
$R:=\mathbb{Z}_{2}^{n}\rtimes\tau\cap\mathcal{O}_{\alpha\tau}^{\mathbb{B}_{n}}$
and
$S:=\mathbb{Z}_{2}^{n}\rtimes\tau^{2}\cap\mathcal{O}_{\alpha\tau}^{\mathbb{B}_{n}}$.
It is clear that $R\cup S$ is a subrack decomposition and of type $D$.

(ii) Assume that $a\tau$ is a positive cycle. Let $b=(1,0,0,1,0)$ when $n=5;$
$b=(1,1,0,\cdots,0)$ when $n>5.$ Thus $0\tau$, $a\tau$ and $b\tau$ are
conjugate. We assume $a=0$ without lost generality. It is clear that the right
side of (1.3) $\not=0$. Consequently, $R\cup S$ is a subrack decomposition and
of type $D$ as Part (i). $\Box$

######  Lemma 2.2.

If $\sigma$ is of type ($3^{2}$), then
$\mathcal{O}_{a\sigma}^{\mathbb{B}_{6}}$ is of type $D.$

Proof. Let $\tau=(1\ 2\ 3)(4\ 5\ 6)$, $\mu=(1\ 2\ 3)^{2}(4\ 5\ 6)$, $\pi=(1\
2\ 3)$ and $\xi=(4\ 5\ 6)$. It is clear that we have

$\displaystyle(a_{6}+a_{5},a_{4}+a_{6},a_{5}+a_{4})=(b_{6}+b_{5},b_{4}+b_{6},b_{5}+b_{4}).$
(2.1)

by (1.3).

(i) If $a=\alpha$ and $b=(1,0,0,1,0,0)$, then (2.1) does not hold.

(ii) If $a=0$ and $b=(0,0,0,1,1,0)$, then (2.1) does not hold.

(iii) If $a=(1,0,0,0,0,0)$ and $b=(1,0,0,1,1,0)$, then (2.1) does not hold.

(iv) If $a=(0,0,0,1,0,0)$ and $b=(0,0,0,0,1,0)$, then (2.1) does not hold.

Let $R:=\mathbb{Z}_{2}^{6}\rtimes\tau$ and $S:=\mathbb{Z}_{2}^{6}\rtimes\mu$.
It is clear that $R\cup S$ is a subrack decomposition of
$\mathcal{O}_{a\sigma}^{\mathbb{B}_{6}}$. Consequently
$\mathcal{O}_{a\sigma}^{\mathbb{B}_{6}}$ is of type $D.$ $\Box$

######  Lemma 2.3.

If $\sigma$ is of type ($2^{2}\ 3^{1}$), then
$\mathcal{O}_{a\sigma}^{\mathbb{B}_{7}}$ is of type $D.$

Proof. Let $\pi=(5\ 6\ 7)$, $\xi=(1\ 2)(3\ 4)$, $\lambda=(1\ 3)(2\ 4)$,
$\tau=\pi\xi$ and $\mu=\pi\lambda$.

If $a=(a_{1},a_{2},a_{3},a_{4},0,0,0)$, let
$b=(b_{1},b_{2},b_{3},b_{4},1,1,0)$; If $a=(a_{1},a_{2},a_{3},a_{4},,1,1,1)$,
let $b=(b_{1},b_{2},b_{3},b_{4},1,0,0)$. The 5th, 6th, 7th component of (1.3)
are
$(a_{5}+a_{7},a_{6}+a_{5},a_{7}+a_{6})=(b_{5}+b_{7},b_{6}+b_{5},b_{7}+b_{6})$.
Consequently, (1.3) does not hold.

Let $R:=\mathbb{Z}_{2}^{7}\rtimes\tau$ and $S:=\mathbb{Z}_{2}^{7}\rtimes\mu$.
It is clear that $R\cup S$ is a subrack decomposition of
$\mathcal{O}_{a\sigma}^{\mathbb{B}_{7}}$. Consequently
$\mathcal{O}_{a\sigma}^{\mathbb{B}_{7}}$ is of type $D.$ $\Box$

######  Lemma 2.4.

(i) Assume that $\tau$ and $\mu$ are conjugate with $sq(\tau,\mu)\not=\mu$ in
$\mathbb{S}_{n}$ and $\tau(n)=\mu(n)=n$. If $a\in\mathbb{Z}_{2}^{n}$ and there
exist $i,j$ such that $a_{i}\not=a_{j}$ with $\tau(i)=i$ and $\tau(j)=j,$ then
$\mathbb{O}_{a\tau}$ is of type $D.$

(ii) Assume that $n>3$ and $a\tau\in\mathbb{B}_{n}$ with type $(1^{n-2},2)$ of
$\tau.$ If there exist $i,j$ such that $\tau(i)=i$ and $\tau(j)=j$ with
$a_{i}\not=a_{j}$, then $\mathcal{O}_{a\tau}$ is of type $D$.

(iii) Assume that $n>4$ and $a\tau\in\mathbb{B}_{n}$ with type $(1^{n-3},3)$
of $\tau.$ If there exist $i,j$ such that $\tau(i)=i$ and $\tau(j)=j$ with
$a_{i}\not=a_{j}$, then $\mathcal{O}_{a\tau}$ is of type $D$.

(iv) Assume that $n=6$ and $a\tau\in\mathbb{B}_{n}$ with type $(1^{2},2^{2})$
of $\tau.$ If there exist $i,j$ such that $\tau(i)=i$ and $\tau(j)=j$ with
$a_{i}\not=a_{j}$, then $\mathcal{O}_{a\tau}$ is of type $D$.

Proof. (i) Let
$R:=\\{d\xi\in\mathcal{O}_{a\tau}^{\mathbb{B}_{n}}\mid\xi\in\mathbb{S}_{n-1};d_{n}=0\\}$
and
$S:=\\{d\xi\in\mathcal{O}_{a\tau}^{\mathbb{B}_{n}}\mid\xi\in\mathbb{S}_{n-1};d_{n}=1\\}$.
Obviously, $R\cup S$ is a subrack decomposition. It is clear that there exists
$b,c\in\mathbb{Z}_{2}^{n}$, $\xi,\lambda\in\mathbb{S}_{n-1}$ such that
$b\xi\in R$ and $c\lambda\in S$. Since $sq(\tau,\mu)\not=\mu$, we have that
$sq(b\tau,c\mu)\not=c\mu$. That implies that $\mathbb{O}_{a\tau}$ is of type
$D.$

(ii) It is clear that $sq(\tau,\mu)\not=\mu$ with $\tau:=(1,2)$ and
$\mu:=(2,3)$. Applying Part (i) we complete the proof.

(iii)It is clear that $sq(\tau,\mu)\not=\mu.$ with $\tau=(1,2,3)$ and
$\mu=(2,4,3)$. Applying Part (i) we complete the proof.

(iv) It is clear that $sq(\tau,\mu)\not=\mu.$ with $\tau=(1\ 2)(3\ 4)$ and
$\mu=(2\ 3)(4\ 5)$. Applying Part (i) we complete the proof. $\Box$

######  Lemma 2.5.

Let $G=\mathbb{Z}_{2}^{n}\rtimes H$ and $\sigma=(a,\tau)\in G$ with
$a\in\mathbb{Z}_{2}^{n},\tau\in H.$ If $\mathcal{O}_{\tau}^{H}$ is of type
$D$, then so is $\mathcal{O}_{(a,\tau)}^{G}$.

Proof. Let $X=S\cup T$ is a subrack decomposition of $\mathcal{O}_{\tau}^{H}$
and there exist $s\in S,t\in T$ such that

$\displaystyle s\rhd(t\rhd(s\rhd t))\not=t.$ (2.2)

Let $h\rhd\tau=s$ and $g\rhd\tau=t$ with $h,g\in H$. It is clear $sq((h\cdot
a,s),(g\cdot a,t))\not=(g\cdot a,t)$ since (2.2); $(h\cdot a,s)=h\rhd(a,\tau)$
and $(g\cdot a,t)=g\rhd(a,\tau)$.

$(<H\cdot a>,X)$ is a subrack, where $<H\cdot a>$ is the subgroup generated by
set $H\cdot a$ of $\mathbb{Z}_{2}^{n}$. In fact, for any $h,g\in H,\xi,\mu\in
X,$ we have that

$\displaystyle(h\cdot a,\xi)\rhd(g\cdot a,\mu)=((h+\xi g+\xi\mu\xi^{-1}h)\cdot
a,\xi\rhd\mu)\in(<H\cdot a>,X).$

Thus $(<H\cdot a>,X)$ is a subrack. Consequently $(<H\cdot
a>,X)\cap\mathcal{O}_{(a,\tau)}^{G}=(<H\cdot
a>,S)\cap\mathcal{O}_{(a,\tau)}^{G}\cup(<H\cdot
a>,T)\cap\mathcal{O}_{(a,\tau)}^{G}$ is a subrack decomposition of
$\mathcal{O}_{(a,\tau)}^{G}$ of type $D$. $\Box$

######  Theorem 2.6.

Let $G=\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}$ with $n>4$. Let
$\tau\in\mathbb{S}_{n}$ be of type
$(1^{\lambda_{1}},2^{\lambda_{2}},\dots,n^{\lambda_{n}})$ and
$a\in\mathbb{Z}_{2}^{n}$ with $\sigma=(a,\tau)\in G$ and $\tau\not=1$. If
$\mathcal{O}_{\sigma}^{G}$ is not of type $D$, or rack $\mathcal{O}_{a\tau}$
does not collapse, then some of the following hold.

  1. (i)

$(2,3)$; $(2^{3});$

  2. (ii)

$(2^{4});$ $(1,2^{2}),$

  3. (iii)

$(1^{2},2^{2})$, $(1^{n-2},2)$ and $(1^{n-3},3)$ with $a_{i}=a_{j}$ when
$\tau(i)=i$ and $\tau(j)=j$.

Proof. It follows from Lemma 2.3, Lemma 2.4, Lemma 2.5, Lemma 2.1, Lemma 2.2
and [AFGV08, Th. 4.1]. $\Box$

##  3 Nichols algebras over $\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}$

In this section we show that except in three cases Nichols algebras of
irreducible YD modules over classical Weyl groups are infinite dimensional.
Let ${\rm supp}M=:\\{g\in G\mid M_{g}\not=0\\}$ for $G$-comodule $M$.

######  Theorem 3.1.

([AFGV08]) If $G$ is a finite group and $\mathcal{O}_{\sigma}^{G}$ is of type
$D$, then dim $\mathfrak{B}(\mathcal{O}_{\sigma}^{G},\rho)=\infty$ for any
$\rho\in\widehat{G^{\sigma}}$.

Proof. It is clear that $\mathcal{O}_{\sigma}^{G}$ is a subrack of ${\rm
supp}M(\mathcal{O}_{\sigma}^{G},\rho)$. If $G$ is a finite group and
$\mathcal{O}_{\sigma}^{G}$ is of type $D$, then (B) in [AFGV08, Rem 2.3] holds
according to the proof of [AFGV08, Th. 3.6]. Consequently, dim
$\mathfrak{B}(\mathcal{O}_{\sigma}^{G},\rho)=\infty$. $\Box$

######  Theorem 3.2.

Let $G=\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}$ with $n>4$. Let
$\tau\in\mathbb{S}_{n}$ be of type
$(1^{\lambda_{1}},2^{\lambda_{2}},\dots,n^{\lambda_{n}})$ and
$a\in\mathbb{Z}_{2}^{n}$ with $\sigma=(a,\tau)\in G$ and $\tau\not=1$. If dim
$\mathfrak{B}(\mathcal{O}_{\sigma}^{G},\rho)<\infty$, then some of the
following hold.

  1. (i)

$(2,3)$; $(2^{3});$

  2. (ii)

$(2^{4});$$(1,2^{2}),$

  3. (iii)

$(1^{2},2^{2}),$ $(1^{n-2},2)$ and $(1^{n-3},3)$ with $a_{i}=a_{j}$ when
$\tau(i)=i$ and $\tau(j)=j$.

Proof. It follows from Theorem 2.6 and [AFGV08, Th. 4.1]. $\Box$

Assume that $\tau\in\mathbb{S}_{m}$ and $a\mu=c\tau\\#d\xi\in\mathbb{B}_{n}$
with $c\tau\in\mathbb{B}_{m}$, $d\xi\in\mathbb{B}_{n-m}$, $c\tau\bot d\xi$,
$d_{1}=d_{2}=\cdots=d_{n-m}$, $\xi=id$, $m<n$. Obviously,
$\rho=\rho_{1}\otimes\rho_{2}\in\widehat{\mathbb{B}_{n}^{a\mu}}=\widehat{\mathbb{B}_{m}^{c\tau}}\times\widehat{\mathbb{B}_{n-m}^{d\xi}}.$
$\rho_{2}=\chi_{2}\otimes\mu_{2}$ with $\chi_{2}\in\widehat{Z_{2}^{n-m}}$,
$\mu_{2}\in\widehat{\mathbb{S}_{n-m}}.$
$\rho_{1}=(\chi_{1}\otimes\mu_{1})\uparrow_{G^{a\mu}_{\chi_{1}}}^{G^{a\mu}}$
with $\chi_{1}\in\widehat{(Z_{2}^{m})^{\tau}}$,
$\mu_{1}\in\widehat{(\mathbb{S}_{m}^{\tau})_{\chi_{1}}}$ when
$c=(1,1,\cdots,1)$ (see [ZZ12, Section 2.5] and [Se]). Case $a=0$ and
$a=(1,1,\cdots,1)$ were studied in paper [ZZ12, Theorem 1.1 and Table 1]. The
other case are listed as follows:

######  Corollary 3.3.

Under notation above assume
$\dim\mathfrak{B}({\mathcal{O}}_{a\mu}^{\mathbb{B}_{n}},\rho)<\infty$.

(i) Then $\rho_{1}(c\tau)=-id$ when $\rho_{2}(d\xi)=id$ and
$\rho_{1}(c\tau)=id$ when $\rho_{2}(d\xi)=-id$.

(ii) Case $\tau=(1\ 2)$ , $c=0$ and $d=(1,1,\cdots,1)$. Then
$\rho_{1}(c\tau)=\pm id$, $\rho_{2}(d\xi)=\mp id$, $\chi_{1}(c)=1$ and
$\mu_{1}(\tau)=\pm 1.$

(iii) Case $\tau=(1\ 2)$ , $c=(1,1)$ and $d=0$. Then $\rho_{1}(c\tau)=-id$,
$\rho_{2}(d\xi)=id$.

(iv) Case $\tau=(1\ 2\ 3)$ , $c=0$ and $d=(1,1,\cdots,1)$. Then
$\rho_{1}(c\tau)=id$, $\rho_{2}(d\xi)=-id$, $\chi_{1}(c)=1$ and
$\mu_{1}(\tau)=1.$

(v) Case $\tau=(1\ 2\ 3)$ , $c=(1,1,1)$ and $d=0$. Then $\rho_{1}(c\tau)=-id$,
$\rho_{2}(d\xi)=id$, $\chi_{1}(c)=-1$ and $\mu_{1}(\tau)=1.$

(vi) Case $\tau=(1\ 2)(3\ 4)$, $c=0$ and $d=(1,1)$. Then $\rho_{1}(c\tau)=id$,
$\rho_{2}(d\xi)=-id$.

(vii) Case $\tau=(1\ 2)(3\ 4)$, $c=(1,0,1,0)$ and $d=(1,1)$. Then
$\rho_{1}(c\tau)=\pm id$, $\rho_{2}(d\xi)=\mp id$.

(viii) Case $\tau=(1\ 2)(3\ 4)$, $c=(1,0,1,0)$ and $d=0$ . Then
$\rho_{1}(c\tau)=-id$, $\rho_{2}(d\xi)=id$.

(ix) Case $\tau=(1\ 2)(3\ 4)$, $c=(1,0,0,0)$ and $d=(1,1)$. Then
$\rho_{1}(c\tau)=\pm id$, $\rho_{2}(d\xi)=\mp id$.

(x) Case $\tau=(1\ 2)(3\ 4)$, $c=(1,0,0,0)$ and $d=(0,0)$ . Then
$\rho_{1}(c\tau)=-id$, $\rho_{2}(d\xi)=id$.

Proof. (iv) If $\rho_{2}(d\xi)=id$, then
$\dim\mathfrak{B}({\mathcal{O}}_{c\tau}^{\mathbb{B}_{3}},\rho_{1})<\infty$ by
Lemma 1.4 , which constracts to [ZZ12, Th. 1.1].

(v) If $\chi_{1}(c)=1$, then there exists a contradiction by [ZZ12, Pro. 2.4,
Th. 1.1].

The others follow from Lemma 1.4 and [ZZ12, Theorem 1.1 ]. $\Box$

Let $\bar{a}$ denote $(a_{1},\cdots,a_{n},0)$ for
$a=(a_{1},a_{2},\cdots,a_{n})\in\mathbb{Z}_{2}^{n}.$

######  Lemma 3.4.

Let $\varphi$ be a map from $\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}$ to
$\mathbb{Z}_{2}^{n+1}\rtimes\mathbb{S}_{n+1}$ sending $(a,\tau)$ to
$(\bar{a},\tau)$ for any $(a,\tau)\in\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}$.
Then $\varphi$ is monomorphic as groups.

Proof.

$\displaystyle\varphi((a,\tau)(b,\mu))$ $\displaystyle=$
$\displaystyle\varphi(a+\tau\cdot b,\tau\mu)$ $\displaystyle=$
$\displaystyle(\bar{a}+\overline{(\tau\cdot b)},\tau\mu)$ $\displaystyle=$
$\displaystyle\varphi((a,\tau))\varphi((b,\mu)),$

for any $(a,\tau),(b,\mu)\in\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}$. $\Box$

######  Lemma 3.5.

If
$\dim\mathfrak{B}({\mathcal{O}}_{(a,\tau)}^{\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n}},\rho)=\infty$
for any
$\rho\in\widehat{(\mathbb{Z}_{2}^{n}\rtimes\mathbb{S}_{n})^{(a,\tau)}}$ , then
$\dim\mathfrak{B}({\mathcal{O}}_{\bar{a}\tau}^{\mathbb{Z}_{2}^{n+1}\rtimes\mathbb{S}_{n+1}},\mu)=\infty$,
for any
$\mu\in\widehat{(\mathbb{Z}_{2}^{n+1}\rtimes\mathbb{S}_{n+1})^{(\bar{a},\tau)}}$.

Proof. It follows from Lemma 3.4 and [ZZ12]. $\Box$

##  4 Relation between bi-one arrow Nichols algebras and
$\mathfrak{B}({\mathcal{O}}_{s},\rho)$

In this section it is shown that bi-one arrow Nichols algebras and
$\mathfrak{B}({\mathcal{O}}_{s},\rho)$ introduced in [Gr00, AZ07, AFZ] are the
same up to isomorphisms.

For any ${\rm RSR}(G,r,\overrightarrow{\rho},u)$, we can construct an arrow
Nichols algebra $\mathfrak{B}(kQ_{1}^{1},ad(G,r,\overrightarrow{\rho},$ $u))$
( see [ZCZ, Pro. 2.4]), written as $\mathfrak{B}(G,r,\overrightarrow{\rho},$
$u)$ in short. Let us recall the precise description of arrow YD module. For
an ${\rm RSR}(G,r,\overrightarrow{\rho},u)$ and a $kG$-Hopf bimodule
$(kQ_{1}^{c},G,r,\overrightarrow{\rho},u)$ with the module operations
$\alpha^{-}$ and $\alpha^{+}$, define a new left $kG$-action on $kQ_{1}$ by

$g\rhd x:=g\cdot x\cdot g^{-1},\ g\in G,x\in kQ_{1},$

where $g\cdot x=\alpha^{-}(g\otimes x)$ and $x\cdot g=\alpha^{+}(x\otimes g)$
for any $g\in G$ and $x\in kQ_{1}$. With this left $kG$-action and the
original left (arrow) $kG$-coaction $\delta^{-}$, $kQ_{1}$ is a Yetter-
Drinfeld $kG$-module. Let $Q_{1}^{1}:=\\{a\in Q_{1}\mid s(a)=1\\}$, the set of
all arrows with starting vertex $1$. It is clear that $kQ_{1}^{1}$ is a
Yetter-Drinfeld $kG$-submodule of $kQ_{1}$, denoted by
$(kQ_{1}^{1},ad(G,r,\overrightarrow{\rho},u))$, called the arrow YD module.

######  Lemma 4.1.

For any $s\in G$ and $\rho\in\widehat{G^{s}}$, there exists a bi-one arrow
Nichols algebra $\mathfrak{B}(G,r,\overrightarrow{\rho},u)$ such that

$\mathfrak{B}({\mathcal{O}}_{s},\rho)\cong\mathfrak{B}(G,r,\overrightarrow{\rho},u)$

as graded braided Hopf algebras in ${}^{kG}_{kG}\\!{\mathcal{Y}D}$.

Proof. Assume that $V$ is the representation space of $\rho$ with
$\rho(g)(v)=g\cdot v$ for any $g\in G,v\in V$. Let $C={\mathcal{O}_{s}}$,
$r=r_{C}C$, $r_{C}={\rm deg}\rho$, $u(C)=s$, $I_{C}(r,u)=\\{1\\}$ and
$(v)\rho_{C}^{(1)}(h)=\rho(h^{-1})(v)$ for any $h\in G$, $v\in V$. We get a
bi-one arrow Nichols algebra $\mathfrak{B}(G,r,\overrightarrow{\rho},u)$.

We now only need to show that
$M({\mathcal{O}}_{s},\rho)\cong(kQ_{1}^{1},ad(G,r,\overrightarrow{\rho},u))$
in ${}^{kG}_{kG}\\!{\mathcal{Y}D}$. We recall the notation in [ZCZ,
Proposition 1.2]. Assume $J_{C}(1)=\\{1,2,\cdots,n\\}$ and $X_{C}^{(1)}=V$
with basis $\\{x_{C}^{(1,j)}\mid j=1,2,\cdots,n\\}$ without loss of
generality. Let $v_{j}$ denote $x_{C}^{(1,j)}$ for convenience. In fact, the
left and right coset decompositions of $G^{s}$ in $G$ are

$\displaystyle G=\bigcup_{i=1}^{m}g_{i}G^{s}\ \ \hbox{and }\ \ G$
$\displaystyle=$ $\displaystyle\bigcup_{i=1}^{m}G^{s}g_{i}^{-1}\ \ ,$ (4.1)

respectively.

Let $\psi$ be a map from $M({\mathcal{O}}_{s},\rho)$ to $(kQ_{1}^{1},{\rm
ad}(G,r,\overrightarrow{\rho},u))$ by sending $g_{i}v_{j}$ to
$a_{t_{i},1}^{(1,j)}$ for any $1\leq i\leq m,1\leq j\leq n$. Since the
dimension is $mn$, $\psi$ is a bijective. See

$\displaystyle\delta^{-}(\psi(g_{i}v_{j}))$ $\displaystyle=$
$\displaystyle\delta^{-}(a_{t_{i},1}^{(1,j)})$ $\displaystyle=$ $\displaystyle
t_{i}\otimes a_{t_{i},1}^{(1,j)}=(id\otimes\psi)\delta^{-}(g_{i}v_{j}).$

Thus $\psi$ is a $kG$-comodule homomorphism. For any $h\in G$, assume
$hg_{i}=g_{i^{\prime}}\gamma$ with $\gamma\in G^{s}$. Thus
$g_{i}^{-1}h^{-1}=\gamma^{-1}g_{i^{\prime}}^{-1}$, i.e.
$\zeta_{i}(h^{-1})=\gamma^{-1}$, where $\zeta_{i}$ was defined in [ZZC04,
(0.3)]. Since $\gamma\cdot x^{(1,j)}\in V$, there exist
$k_{C,h^{-1}}^{(1,j,p)}\in k$, $1\leq p\leq n$, such that $\gamma\cdot
x^{(1,j)}=\sum_{p=1}^{n}k_{C,h^{-1}}^{(1,j,p)}x^{(1,p)}$. Therefore

$\displaystyle x^{(1,j)}\cdot\zeta_{i}(h^{-1})$ $\displaystyle=$
$\displaystyle\gamma\cdot x^{(1,j)}\ \ (\hbox{by definition of
}\rho_{C}^{(1)})$ (4.2) $\displaystyle=$
$\displaystyle\sum_{p=1}^{n}k_{C,h^{-1}}^{(1,j,p)}x^{(1,p)}.$

See

$\displaystyle\psi(h\cdot g_{i}v_{j})$ $\displaystyle=$
$\displaystyle\psi(g_{i^{\prime}}(\gamma v_{j}))$ $\displaystyle=$
$\displaystyle\psi(g_{i^{\prime}}(\sum_{p=1}^{n}k_{C,h^{-1}}^{(1,j,p)}v_{p}))$
$\displaystyle=$
$\displaystyle\sum_{p=1}^{n}k_{C,h^{-1}}^{(1,j,p)}a_{t_{i^{\prime}},1}^{(1,p)}$

and

$\displaystyle h\rhd(\psi(g_{i}v_{j}))$ $\displaystyle=$ $\displaystyle
h\rhd(a_{t_{i},1}^{(1,j)})$ $\displaystyle=$ $\displaystyle
a_{ht_{i},h}^{(1,j)}\cdot h^{-1}$ $\displaystyle=$
$\displaystyle\sum_{p=1}^{n}k_{C,h^{-1}}^{(1,j,p)}a_{t_{i^{\prime}},1}^{(1,p)}\
\ (\hbox{by \cite[cite]{[\@@bibref{}{ZCZ08}{}{}, Pro.1.2]} and
}(\ref{e1.11})).$

Therefore $\psi$ is a $kG$-module homomorphism. $\Box$

Therefore $\mathfrak{B}({\mathcal{O}}_{s},\rho)$ is viewed as
$\mathfrak{B}(G,r,\overrightarrow{\rho},u)$ sometimes.

######  Remark 4.2.

The representation $\rho$ in $\mathfrak{B}({\mathcal{O}}_{s},\rho)$ introduced
in [Gr00, AZ07] and $\rho_{C}^{(i)}$ in RSR are different. $\rho(g)$ acts on
its representation space from the left and $\rho_{C}^{(i)}(g)$ acts on its
representation space from the right.

Otherwise, when $\rho=\chi$ is a one dimensional representation, then
$(kQ_{1}^{1},ad(G,r,\overrightarrow{\rho},u))$ is PM (see [ZZC04, Def. 1.1]).
Thus the formulae are available in [ZZC04, Lemma 1.9]. That is, $g\cdot
a_{t}=a_{gt_{i},g}$, $a_{t_{i}}\cdot g=\chi(\zeta_{i}(g))a_{t_{i}g,g}$.

##  5 Transposition

In this section we consider Nichols algebra
$\mathfrak{B}({\mathcal{O}}_{\sigma},\rho)$ of transposition $\sigma$ over
symmetry groups, where $\rho=sgn\otimes sgn$ or $\rho=\epsilon\otimes sgn.$
$\mathfrak{B}({\mathcal{O}}_{\sigma},\rho)$ is finite dimensional when $n<5$
according to [MS, FK97, AZ07]. However, it is open whether
$\mathfrak{B}({\mathcal{O}}_{\sigma},\rho)$ is finite dimensional when $n>4.$
We give the relation between $\mathfrak{B}({\mathcal{O}}_{\sigma},\rho)$ and
${\mathcal{E}}_{n}$ defined in [FK97, Def. 2.1].

Let $\sigma=(12)\in S_{n}:=G$, $\mathcal{O}_{\sigma}=\\{(ij)|1\leq i,j\leq
n\\}$, $G^{\sigma}=\\{g\in G\mid g\sigma=\sigma g\\}$.
$G=\bigcup\limits_{1\leq i<j\leq n}G^{\sigma}g_{ij}$.

Let $g_{kj}:=\left\\{\begin{array}[]{lll}id\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
k=1,j=2\\\ (2j)\ \ \ \ \ \ \ \ \ \ \ \ \ \ k=1,j>2\\\ (1j)\ \ \ \ \ \ \ \ \ \
\ \ \ \ k=2,j>2\\\ (1k)(2j)\ \ \ \ \ \ \ \ \ k>2,j>k\\\ \end{array}\right.$
and $t_{ij}=(ij).$

######  Lemma 5.1.

In The following equations in $\mathbb{S}_{m}$ hold.

$(12)id=id(12)$

$(12)(2j)=(2j1)=(1j)(12)$

$(12)(1j)=(1j2)=(2j)(12)$

$(12)(1k)(2j)=(1k2)(2j)=(1k)(2j)(kj)$

$(1j)id=(1j)id$

$(1j)(2j)=(j21)=(2j)(12)$

$(1j)(2j_{1})=(1j)(2j_{1})id\ \ j<j_{1}$

$(1j)(2j_{1})=(1j_{1})(2j)(jj_{1})(12)\ \ j>j_{1}$

$(1j)(1j)=id$

$(1j)(1j_{1})=(1j_{1}j)=(1j_{1})(jj_{1})$

$(1j)(1k)(2j)=(1kj)(2j)=(2k)(12)(kj)$

$(1j)(1k)(2j_{1})=(2j_{1})id\ \ \ j=k$

$(1j)(1k)(2j_{1})=(1k)(kj)(2j_{1})=(1k)(2j_{1})(kj)\ \ j\neq j_{1}$

$(2j)id=(2j)id$

$(2j)(2j)=id$

$(2j)(2j_{1})=(2j_{1}j)=(2j_{1})(jj_{1})$

$(2j)(1j)=(j12)=(1j)(12)$

$(2j)(1j_{1})=(1j)(2j_{1})(jj_{1})(12)\ \ j<j_{1}$

$(2j)(1j_{1})=(1j_{1})(2j)id\ \ j>j_{1}$

$(2j)(1k)(2j)=(1k)=(1k)id$

$(2j)(1k)(2j_{1})=(1k)(12)(2j_{1})=(1k)(1j_{1})(12)=(1j_{1})(12)(kj_{1})\ \
j=k$

$(2j)(1k)(2j_{1})=(2j)(2j_{1})(1k)=(2j_{1})(j_{1}j)(1k)=(2j_{1})(1k)(j_{1}j)\
\ j\neq j_{1},j\neq k$

$(kj)id=id(kj)$

$(kj)(2j)=(2k)(kj)$

$(kj)(2k)=(2j)(kj)$

$(kj)(2j_{1})=(2j_{1})(kj)\ \ k\neq j_{1},j\neq j_{1}$

$(kj)(1j)=(1k)(kj)$

$(kj)(1k)=(1j)(kj)$

$(kj)(1j_{1})=(1j_{1})(kj)\ \ k\neq j_{1},j\neq j_{1}$

$(kj)(1k)(2j)=(1k)(2j)(12)$

$(kj)(1k_{1})(2j_{1})=(1k)(2j_{1})(kj)\ \ k_{1}=j$

$(kj)(1k_{1})(2j_{1})=(1k_{1})(2k)(kj)\ \ k_{1}<k,j_{1}=j$

$(kj)(1k_{1})(2j_{1})=(1k)(2k_{1})(12)(kjk_{1})\ \ k_{1}>k,j_{1}=j$

$(kj)(1k_{1})(2j_{1})=(1j)(2j_{1})(kj)\ \ j_{1}>j,k_{1}=k$

$(kj)(1k_{1})(2j_{1})=(1j_{1})(2j)(12)(jkj_{1})\ \ j_{1}<j,k_{1}=k$

$(kj)(1k_{1})(2j_{1})=(1k_{1})(2j)(kj)\ \ k_{1}\neq j,j_{1}=k$

$(kj)(1k_{1})(2j_{1})=(1k_{1})(2j_{1})(kj)\ \ k_{1}\neq k,k_{1}\neq
j,j_{1}\neq j,j_{1}\neq k$.

Remark: By Lemma above, we can obtain $\zeta_{st}(t_{ij})\in G^{\sigma}$ such
that $g_{st}t_{ij}=\zeta_{st}(t_{ij})g_{s^{\prime}t^{\prime}}$ for any $1\leq
i,j,s,t\leq n.$

Let $a_{ij}$ denote the arrow $a_{t_{ij},1}$ from $1$ to $t_{ij}$. By Lemma
4.1, $\\{a_{ij}\mid i\not=j,1\leq i,j\leq n\\}$ generates algebra in its co-
path Hopf algebra is isomorphic to Nichols algebra
$\mathfrak{B}({\mathcal{O}}_{\sigma},\rho).$

######  Lemma 5.2.

In $\mathfrak{B}({\mathcal{O}}_{(12)},\chi)$ with $\chi=sgn\otimes sgn$ or
$\chi=\epsilon\otimes sgn$,

(i) If $i,j$ and $k$ are different each other, then there exists
$\alpha_{i,j,k}$, $\beta_{i,j,k}\in\\{1,-1\\}$ such that

$\displaystyle
a_{ij}a_{jk}+\alpha_{ijk}a_{jk}a_{ki}+\beta_{ijk}a_{ki}a_{ij}=0.$ (5.1)

(ii)

The left hand side of (5.1) $\displaystyle=$
$\displaystyle(\chi(\zeta_{ij}(t_{jk}))a_{t_{ij}t_{jk},t_{jk}}a_{t_{jk},1}+a_{t_{ij}t_{jk},t_{ij}}a_{t_{ij},1})$
(5.2) $\displaystyle+$
$\displaystyle\alpha_{ijk}(\chi(\zeta_{jk}(t_{ik}))a_{t_{jk}t_{ik},t_{ik}}a_{t_{ik},1}+a_{t_{jk}t_{ik},t_{jk}}a_{t_{jk},1})$
$\displaystyle+$
$\displaystyle\beta_{ijk}(\chi(\zeta_{ik}(t_{ij}))a_{t_{ik}t_{ij},t_{ij}}a_{t_{ij},1}+a_{t_{ik}t_{ij},t_{ik}}a_{t_{ik},1})$

(iii) If
$\chi(\zeta_{ij}(t_{jk}))\chi(\zeta_{jk}(t_{ik}))\chi(\zeta_{ik}(t_{ij}))=-1,$
then Part (i) holds.

(iv) If $i,j$ and $k$ are different each other, then Part (i) holds if and
only if
$\chi(\zeta_{ij}(t_{jk}))\chi(\zeta_{jk}(t_{ik}))\chi(\zeta_{ik}(t_{ij}))=-1.$

(v) If $i,j,k$ and $l$ are different each other, then there exist
$\lambda_{ijkl}\in\\{1,-1\\}$ such that
$a_{ij}a_{kl}=\lambda_{ijkl}a_{kl}a_{ij}.$

Proof. Let $\chi^{\prime}:=sgn\otimes sgn$ and
$\chi^{\prime\prime}:=\epsilon\otimes sgn.$ It is clear that
$M({\mathcal{O}}_{(12)},\chi)$ is a PM $\mathbb{S}_{n}$\- YD module (see
[ZZC04, Def. 1.1]) and $g\cdot a_{ij}=a_{gt_{ij},g}$, $a_{t_{ij}}\cdot
g=\chi(\zeta_{ij}(g))a_{t_{ij}g,g}$ (see [ZZC04, Lemma 1.9]). By [CR02],

$\displaystyle
a_{{ij}}a_{{kl}}=\chi(\zeta_{ij}({t_{kl}}))a_{{t_{ij}}{t_{kl}},{t_{kl}}}a_{{t_{kl}},1}+a_{{t_{ij}}{t_{kl}},{t_{ij}}}a_{{t_{ij}},1}.$
(5.3)

(ii) It follows from (5.3).

(iii) and (iv) follow from Part (ii).

(i) Let $a,b$ and $c$ denote $\zeta_{ij}(t_{jk})$, $\zeta_{jk}(t_{ik})$ and
$\zeta_{ik}(t_{ij})$ in the following Table 1 in short, respectively.
case $a$ $b$ $c$ $\chi^{\prime}(a)$ $\chi^{\prime}(b)$ $\chi^{\prime}(c)$
$\chi^{\prime\prime}(a)$ $\chi^{\prime\prime}(b)$ $\chi^{\prime\prime}(c)$
$2<i<j<k$ $(kj)$ $(12)(ijk)$ $(ij)$ $-1$ $-1$ $-1$ $1$ $-1$ $1$ $i=1,j=2<k$
$(1)$ $(1)$ $(12)$ $1$ $1$ $-1$ $1$ $1$ $-1$ $i=1,2<j<k$ $(kj)$ $(12)(kj)$
$(1)$ $-1$ $1$ $1$ $1$ $-1$ $1$ $i=2<j<k$ $(kj)$ $(1)$ $(12)(jk)$ $-1$ $1$ $1$
$1$ $1$ $-1$ $2<i<k<j$ $(kj)$ $(ik)$ $(12)(ijk)$ $-1$ $-1$ $-1$ $1$ $1$ $-1$
$i=1,k=2<j$ $(1)$ $(12)$ $(1)$ $1$ $-1$ $1$ $1$ $-1$ $1$ $i=1,2<k<j$ $(kj)$
$(1)$ $(12)(jk)$ $-1$ $1$ $1$ $1$ $1$ $-1$ $i=2<k<j$ $(kj)$ $(kj)(12)$ $(1)$
$-1$ $1$ $1$ $1$ $-1$ $1$

$\hbox{Table }1$

By Table 1,
$\chi(\zeta_{ij}(t_{jk}))\chi(\zeta_{jk}(t_{ik}))\chi(\zeta_{ik}(t_{ij}))=-1$.
Consequently, Part (i) holds by Part (iii).

(v) By (5.3), $a_{ij}a_{kl}=\lambda_{ijkl}a_{kl}a_{ij}$ if and only if
$(\chi(\zeta_{ij}(t_{kl}))-\lambda_{ijkl})=(1-\chi(\zeta_{kl}(t_{ij}))\lambda_{ijkl})=0$.
Since $(ij)(kl)=(kl)(ij)$, we have $g_{ij}(kl)=g_{ij}(kl)g_{ij}g_{ij}$ with
$g_{ij}(kl)g_{ij}\in\mathbb{S}_{n}^{(12)}$ and
$\zeta_{ij}((kl))=g_{ij}(kl)g_{ij}.$ See
$\zeta_{ij}(t_{kl})=g_{ij}t_{kl}g_{ij}=$
$\left\\{\begin{array}[]{lll l}(kl)&\hbox{ if }k,l>2\\\ (kl)&\hbox{ if
}i=1,j=2\\\ =(2j)(kl)(2j)=(lj)\hbox{ or }(kj)\hbox{ or }(kl)&\hbox{ if
}i=1,j>2,k=2\hbox{ or }i=1,j>2,l=2\\\ &\hbox{ or }i=1,j>2,k\not=2,l\not=2\\\
=(1j)(kl)(1j)=(lj)\hbox{ or }(kj)\hbox{ or }(kl)&\hbox{ if }i=2,j>2,k=1\hbox{
or }i=1,j>2,l=1\\\ &\hbox{ or }i=1,j>2,k\not=1,l\not=1\\\ \hbox{ is a
transposition of }\\\ \hbox{ two greater numbers than }$2$&\hbox{ if
}i,j>2\end{array}\right..$ Consequently, $\zeta_{ij}(t_{kl})$ is a
transposition of two greater numbers than $2$ and
$\chi^{\prime}(\zeta_{ij}(t_{kl}))=-1$ and
$\chi^{\prime\prime}(\zeta_{ij}(t_{kl}))=1$. Similarly,
$\chi^{\prime}(\zeta_{kl}(t_{ij}))=-1$ and
$\chi^{\prime\prime}(\zeta_{kl}(t_{ij}))=1$. Consequently, it is enough to set
$\lambda_{ijkl}=:\chi(\zeta_{ij}(t_{kl}))$. $\Box$

######  Definition 5.3.

(See [FK97, Def. 2.1]) algebra ${\mathcal{E}}_{n}$ is generated by
$\\{x_{ij}\mid 1\leq i<j\leq n\\}$ with definition relations:

(i) $x_{ij}^{2}=0$ for $i<j.$

(ii) $x_{ij}x_{jk}=x_{jk}x_{ik}+x_{ik}x_{ij}$ and
$x_{jk}x_{ij}=x_{ik}x_{jk}+x_{ij}x_{ik},$ for $i<j<k.$

(iii) $x_{ij}x_{kl}+x_{kl}x_{ij}=0$ for any distinct $i,j,k,l$ and $l,$
$i<j,k<l.$

Equivalently, algebra ${\mathcal{E}}_{n}$ is generated by $\\{x_{ij}\mid
i\not=j,1\leq i,j\leq n\\}$ with definition relations:

(i) $x_{ij}^{2}=0$, $x_{ij}=-x_{ji}$, for $1\leq i,j\leq n.$

(ii) $x_{ij}x_{jk}+x_{jk}x_{ki}+x_{ki}x_{ij}=0,$ for $1\leq i,j,k\leq n.$

(iii) $x_{ij}x_{kl}=x_{kl}x_{ij}$ for any distinct $i,j,k$ and $l.$

By [FK97, Th. 7.1], a subring of $\mathcal{E}_{n}$ is isomorphic to the
cohomology ring of the flag manifold.

By [AFGV08] and [HS08], most of Nichols algebras over $\mathbb{S}_{n}$ are
infinite dimensional when $n>5$. Consequently, we have

######  Conjecture 5.4.

Let $\alpha_{ijk},\beta_{ijk},\gamma_{ij},\lambda_{ijkl}\in\\{1,-1\\}$. Assume
that algebra $A(\alpha,\beta,\gamma,\lambda)$ is generated by $\\{x_{ij}\mid
i\not=j,1\leq i,j\leq n\\}$ with definition relations:

(i) $x_{ij}^{2}=0$, $x_{ij}=\gamma_{ij}x_{ji}$, for $1\leq i,j\leq n.$

(ii) $x_{ij}x_{jk}+\alpha_{ijk}x_{jk}x_{ki}+\beta_{ijk}x_{ki}x_{ij}=0,$ for
$1\leq i,j,k\leq n.$

(iii) $x_{ij}x_{kl}=\lambda_{ijkl}x_{kl}x_{ij}$ for any distinct $i,j,k$ and
$l.$

Then $A(\alpha,\beta,\gamma,\lambda)$ is infinite dimensional when $n>4$.
Furthermore, $\mathfrak{B}({\mathcal{O}}_{(12)},\rho)$ is infinite dimensional
with $\rho=sgn\otimes sgn$ or $\rho=\epsilon\otimes sgn$ when $n>6$.

Obviously, $\mathcal{E}_{n}=A(1,1,-1,1).$

##  6 Appendix

In this section we give another proof of Lemma 2.5.

######  Lemma 6.1.

(i) If $\varphi$ is epimorphic from group $G$ onto group $\bar{G}$, then
$\varphi\mid_{\mathcal{O}_{a}^{G}}$ is epimorphic from $\mathcal{O}_{a}^{G}$
onto $\mathcal{O}_{\varphi(a)}^{\bar{G}}$ as racks.

(ii) If $\pi$ is a map from $\mathbb{Z}_{2}^{n}\rtimes H$ to $H$ by sending
$a\tau$ to $\tau$ for any $a\tau\in A\rtimes H,$ then $\pi$ is epimorphic as
groups.

Proof. (i) For any $\bar{x},\bar{y}\in\mathcal{O}_{\varphi(a)}^{\bar{G}}$ with
$x,y\in G$ and $\bar{x}=\varphi(x),\bar{y}=\varphi(y)$, It is clear
$\varphi(x)\rhd\varphi(y)=\varphi(x\rhd y)$. i.e. $\varphi$ is a homomorphism
of racks. Since $\bar{x}\in\mathcal{O}_{\varphi(a)}^{\bar{G}}$, there exists
$h\in G$ such that $\varphi(h)\rhd\varphi(a)=\bar{x}$. Consequently,
$\bar{x}\in\varphi(\mathcal{O}_{a}^{G})$.

(ii) It is clear. $\Box$

######  Lemma 6.2.

(i) If $\varphi$ is epimorphic from $X$ to $Y$ as racks and $Y$ is of type
$D$, then $X$ is of type $D$.

(ii) Let $G:=\mathbb{Z}_{2}^{n}\rtimes H$ and $\sigma=(a,\tau)\in G$ with
$a\in\mathbb{Z}_{2}^{n},\tau\in H.$ If $\mathcal{O}_{\tau}^{H}$ is of type
$D$, then so is $\mathcal{O}_{(a,\tau)}^{G}$.

Proof. (i) Let $Y=R\cup S$ be a decomposition of subracks with type $D$ and
let $r\in R$, $s\in S$ such that $sq(r,s)\not=s.$ Set
$\bar{R}:=\varphi^{-1}(R)$, $\bar{S}:=\varphi^{-1}(S)$;
$\varphi(r^{\prime})=r$ and $\varphi(s^{\prime})=s$ with
$r^{\prime}\in\bar{R}$ and $s^{\prime}\in\bar{S}$. It is clear that
$\bar{R}\cup\bar{S}$ be a decomposition of subracks and
$sq(r^{\prime},s^{\prime})\not=s^{\prime}.$

(ii) It follows from Part (i) and Lemma 6.1. $\Box$

Remark: Lemma 6.2 (i) appeared in [AFGV10, Section 2.4], but they had not
proof.

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