\section{Introduction}
Let $\mathbb{B}_{N}$ be the unit ball in $\mathbb{C}_{N}$ and $S_N$ denote the unit sphere. Let $H(\mathbb{B}_{N})$ be
the space of all holomorphic functions on $\mathbb{B}_{N}$. Denote by $S(\mathbb{B}_{N})$ the set of all holomorphic self-maps of $\mathbb{B}_{N}$. The Dirichlet space $\mathcal{D}(\mathbb{B}_{N})$ is defined as
$$
\mathcal{D}(\mathbb{B}_{N})=\{f\in H(\mathbb{B}_{N}):|f(0)|^2+\int_{\mathbb{B}_{N}}|\nabla f(z)|^2d\nu_N(z) < \infty \},
$$
where $d \nu_N(z)$ is the normalized volume measure on $\mathbb{B}_{N}$. The Hardy space $H^2(\mathbb{B}_{N})$  is defined as
$$
H^2(\mathbb{B}_{N})=\{f\in H(\mathbb{B}_{N}):\sup_{0 < r < 1}\int_{S_{N}}|f(r\zeta)|^2d\sigma(\zeta) < \infty \},
$$
where $d\sigma(\zeta)$ is the normalized surface measure on $S_N$.
Recall that for any analytic self-map $\varphi\in S(\mathbb{B}_{N})$ and any analytic function $\psi\in H(\mathbb{B}_{N})$, the weighted composition operator is given by
$$W_{\psi,\varphi}f=\psi\cdot f\circ\varphi.$$
If $\psi\equiv1$, we get the composition operator $C_\varphi$.

In the past five decades, the study of (weighted) composition operator attracted attention of the researchers. It is very interesting to explore how the function theoretic behavior of $\varphi$ affects the properties of $C_\varphi$ on various holomorphic function spaces. For general information about composition operator, we refer the readers to book \cite{CM1} for more details.

An anti-linear operators $C$ on a complex Hibert space $\mathcal{H}$ is called a conjugation if it satisfies the following conditions:
\begin{itemize}
\item[(i)]  involution, i.e. $C^2=I$.
\item[(ii)] isometric, i.e. $ \langle Cx, Cy \rangle = \langle y, x \rangle$ for all $x,y\in\mathcal{H}$.
\end{itemize}
A bounded linear operator $T$ on $\mathcal{H}$ is called complex symmetric if there exists a conjugation $C$ such that $TC=CT^*$ $(CTC=T^*)$. We also say that $T$ is a $C$-symmetric operator.

It is well known that the general study of complex symmetric weighted composition operators on $\mathcal{H}$ are derived from the work of Gracia and Putinar in \cite{GP}, the authors show that every normal operator is complex symmetric. Since then, many significant results about the complex symmetric (weighted) composition operators are obtained. In \cite{GW}, Garica and Wogen got that if an operator $T$ is algebraic of order $2$, then $T$ is complex symmetric, so $C_{\varphi_a}$ is complex symmetric when $\varphi_a$ is an
involutive automorphism. Furthermore, an explicit conjugation  operator $J_a=JW_a$ on $H^2(\mathbb{B}_{N})$  such that $C_{\varphi_a}=J_aC_{\varphi_a}^*J_a$ was given in \cite{N2}. Very recently, in \cite{JK} Jung et al. studied  which combinations of weights $\psi$ and maps of the disk $\varphi$ give rise to complex symmetric weighted composition operators with  respect to classical conjugation $Jf(z)=\overline{f(\overline{z})}$.  In \cite{GZ1}, Gao and Zhou  gave a complete description of complex symmetric composition operators on $H^2(\mathbb{D})$  whose symbols are linear fractional.

Since all conjugation can be considered as a product of a $J$-symmetric unitary operator $U$ and the conjugation $J$. Fatehi in \cite{F} find all unitary weighted composition operators which are $J$-symmetric and consider complex symmetric weighted composition operators with special conjugation $W_{k_a,\varphi_a}J$ on $H^2(\mathbb{D})$. Moreover, a criterion for complex symmetric structure of $W_{\psi,\varphi}$ on $H_\gamma(\mathbb{D})$ (with reproducing kernels $K_w^{\gamma}=(1-\overline{w}z)^{-\gamma}$, where $\gamma\in \mathbb{N}$) was discovered in \cite{LK}. In \cite{ZY}, Yuan and Zhou characterized the adjoint of linear fractional composition operators $C_\varphi$ acting on $\mathcal{D}(\mathbb{B}_{N})$. In \cite{LNN}, the authors showed that no nontrivial normal weighted composition operator exist on the Dirichlet space in the unit disk when $\varphi$ is linear-fractional with fixed point $p\in\mathbb{D}$.

 Motivated by these researches, we attempt to generalize some discussion into more spaces, such as $\mathcal{D}(\mathbb{B}_{N})$,
$H^2(\mathbb{B}_{N})$. The rest of the paper is organized as follows: First, we recall some fundamental definitions and theorems concerning our results. Then we examine the question ``Which weighted composition operator on $\mathcal{D}(\mathbb{B}_{N})$ are complex symmetric?". We prove that $W_{\psi,\varphi}$ is $J$-symmetric ($JC_{Uz}$-symmetric) if and only if $W_{\psi,\varphi}$ is a multiple of the corresponding complex symmetric composition operator $C_\varphi$. In addition, we also show that $C_\varphi$ is  $J$-symmetric if and only if $\varphi(z) = \varphi'(0)z$ for $z\in\mathbb{B}_{N}$ and $\varphi'(0)$ is a symmetric matrix with $||\varphi'(0)||\leq1$. We then provide characterizations of Hermitian  weighted composition operators on $\mathcal{D}(\mathbb{B}_{N})$. Moreover,  we study  when the class of $J$-symmetric weighted composition operator to be  unitary or Hermitian. By providing some sufficient conditions for weighted composition operators to be both unitary and $J$-symmetric, then we get some new examples of complex symmetric weighted composition operators on  $H^2(\mathbb{B}_{N})$. Finally, we discuss the normality of complex symmetric weighted composition operators on $H^2(\mathbb{B}_{N})$.
\section{Preliminaries}

\subsection{Linear fractional map}
\begin{defn} A linear fractional map $\varphi$ of $\mathbb{C}^N$ is a map of the form
$$\varphi(z)=\frac{Az+B}{\langle z,C\rangle+D},$$
where $A = (a_{j,k})$ be an $N \times N$-matrix, $B=(b_j)$, $C =(c_i)$ be $N$-column vectors, $D$ be a complex number, and $\langle . , .\rangle$ indicates the usual Euclidean inner product in  $\mathbb{C}^N$. If $\varphi(\mathbb{B}_{N}) \in \mathbb{B}_{N}$, $\varphi$ is said to be a linear fractional self-map of $\mathbb{B}_{N}$ and signed as $\varphi(z)\in \textrm{LFT} (\mathbb{B}_{N})$.

In this paper, we identify $N \times N$-matrices with linear transformations of $\mathbb{C}_{N}$  via the standard basis of $\mathbb{C}_{N}$.
\end{defn}

\begin{defn}
If $\varphi(z)=\frac{Az+B}{\langle z, C \rangle + D}$ is a linear fractional map, the matrix
$$
 m_{\varphi}=\left(
  \begin{array}{ccc}
  A & B\\
    C^*& D\\
  \end{array}
\right)
$$
will be called a matrix associated with $\varphi$.  If $\varphi(z)\in$ LFT$(\mathbb{B}_{N})$, the adjoint map $\sigma=\sigma_\varphi$ is defined by
$$\sigma(z)=\frac{A^*z-C}{\langle z, -B \rangle + D^*}$$
and the associated matrix of $\sigma$ is
$$
 m_{\sigma}=\left(
  \begin{array}{ccc}
  A^* & -C\\
    -B^*& D^*\\
  \end{array}
\right),
$$
where $A^* = (\overline{a_{ji}})$ denote the conjugate transpose matrix of $A$.
\end{defn}
\begin{thm}{\rm ([\citealp{CM2}, Theorem 4])}\label{Th2.7}
If the matrix
$$
 m_{\varphi}=\left(
  \begin{array}{ccc}
  A & B\\
    C^*& D\\
  \end{array}
\right)
$$
is a multiple of an isometry on the Kre$\breve{{\i}}$n space with
$$
J=\left(
  \begin{array}{ccc}
  I & 0\\
   0& -1\\
  \end{array}
\right)
,$$
then $\varphi(z)=\frac{Az + B}{\langle z, C \rangle + D}$ maps the unit ball $\mathbb{B}_N$ onto itself. Conversely, if $\varphi(z)=\frac{Az+B}{\langle z, C \rangle + D}$ is a linear fractional map of the unit ball onto itself, then $m_\varphi$ is a multiple of an isometry.
\end{thm}

Fix a vector $a\in\mathbb{B}_N$, we denote by $\varphi_a: \mathbb{B}_N\rightarrow\mathbb{B}_N$ the linear fractional map
$$\varphi_a(z)=\frac{a-P_az-s_aQ_az}{1-\langle z, a \rangle},$$
where $s_a=\sqrt{1-|a|^2}$, $P_a$ be the orthogonal projection of $\mathbb{C}^n$ onto the complex line generated by $a$ and $Q_a=I-P_a$, or equivalently,
$$\varphi_a(z)=\frac{a-Tz}{1-\langle z, a \rangle},$$
where $T$ is a self-adjoint map depending on $a$. We use Aut$(\mathbb{B}_N)$ to denote the set of all automorphism of $\mathbb{B}_N$.

\subsection{Spaces and weighted composition operator}
In the Dirichlet space $\mathcal{D}(\mathbb{B}_N)$, evaluation at $w$ in the unit ball is given by $f(w)=\langle f, K_w \rangle$ where
$$K_w(z)=1+\ln\frac{1}{1 - \langle z, w \rangle} \ \ \  \textrm{and} \ \ \  ||K_w||^2= 1 + \ln \frac{1}{{1-|w|^2}}.$$
Let $k_w$ be the normalization of $K_w$, then
$k_w(z)=\frac{(1-\ln(1-|w|^2))^{-\frac{1}{2}}}{1-\ln{1-\langle z, w \rangle}}$.
And in $H^2(\mathbb{B}_{N})$ the kernel for evaluation at $w$ is given by
$$K_w(z)=\frac{1}{(1-\langle z, w \rangle)^N} \ \ \ \textrm{and} \ \ \  ||K_w||^2=(1-|w|^2)^{-N},$$
then $k_w=\frac{(1-|w|^2)^{\frac{N}{2}}}{(1-\langle z, w \rangle)^N}$.

Next we list some fundamental  properties of bounded weighted
composition operators.
\begin{Prop}{\rm ([\citealp{GZ2}, Remark 2.4])}\label{prop2.4}
If $\varphi$ is an automorphism of $\mathbb{B}_N$ and $\psi\in A(\mathbb{B}_N)$ where $A(\mathbb{B}_N)$ denotes the set of functions that holomorphic on $\mathbb{B}_N$ and continuous up to the boundary $S_N$, then
$$W_{\psi,\varphi}=M_\psi C_\varphi.$$
\end{Prop}

\begin{Prop}{\rm ([\citealp{GZ2}, Proposition 2.5])}\label{prop2.5}
 Suppose that $W_{\psi,\varphi}$ is bounded on $\mathcal{D}(\mathbb{B}_N)(H^2(\mathbb{B}_N))$, then we have
$$
W_{\psi,\varphi}^*K_w(z)=\overline{\psi(w)}K_{\varphi(w)}(z)
$$
for all $z,w\in \mathbb{B}_N$. In particular, since $C_\varphi=W_{1,\varphi}$, we get $C_\varphi^*K_w(z)=K_{\varphi(w)}(z)$.
\end{Prop}

\begin{thm}{\rm ([\citealp{CM2}, Theorem 16])}\label{Th2.6}
Suppose  $\varphi(z)=\frac{Az+B}{\langle z, C \rangle + D}$ is a linear fractional map of $\mathbb{B}_N$ into itself for which $C_\varphi$ is a bounded operator on $\mathcal{H}$. Let $\sigma(z)=\frac{A^*z-C}{\langle z,-B \rangle + D^*}$ be the adjoint mapping.
Then $C_\sigma $ is a bounded operator on $\mathcal{H}$, $g(z)=(\langle z, -B\rangle + D^*)^{-r}$ and
$h(z)= (\langle z, C \rangle + D)^r$ are in $H^\infty(\mathbb{B}_N)$, and
$$
C_\varphi^*=T_gC_\sigma T_h^*.
$$
\end{thm}
\subsection{Some others notations}
In this section, we first recall the first partial  derivative reproducing kernel on $\mathcal{D}(\mathbb{B}_N)$.
Let $K_a^{D_1}, K_a^{D_2},\ldots, K_a^{D_n}$ denote the kernels for the first partial derivatives at $a$, that is
$$
\langle f,K_a^{D_j}\rangle=\frac{\partial f}{\partial z_j}(a), \ \ \ \ \ \ \ \   j=1,2,\ldots,n.
$$

It can be shown that
\begin{align}\label{Eq2.1}
K_a^{D_j } = \frac{z_j} {1-\langle z,a \rangle}, \ \ \ \ \ \ \ \ \  j=1,2,\ldots,n.
\end{align}

Continue to the kernels for the first partial derivatives at $a$, we have
\begin{align*}
\langle f , W_{\psi,\varphi}^*K_a^{D_k} \rangle &=\langle W_{\psi,\varphi}f,K_a^{D_k}\rangle\\
&=\frac{\partial\psi(a)}{\partial z_k}f(\varphi(a))+\psi(a)\sum_{j=1}^{n}\frac{\partial f(\varphi(a))}{\partial \varphi_j(z)}\frac{\partial\varphi_j}{\partial z_k}(a)\\
&=\langle f,\overline{\frac{\partial\psi(a)}{\partial z_k}}K_{\varphi(a)}+\overline{\psi(a)}\sum_{j=1}^{n} \overline{\frac{\partial\varphi_j(a)}{\partial z_k}}K_{\varphi(a)}^{D_j} \rangle
\end{align*}
for any $k=1,2,\ldots,n$ and $f\in \mathcal{D}(\mathbb{B}_{N}).$

Thus, we have
\begin{align}\label{Eq2.2}
W_{\psi,\varphi}^*K_a^{D_k} = \overline{\frac{\partial\psi(a)}{\partial z_k}}K_{\varphi(a)}+\overline{\psi(a)}\sum_{j=1}^{n} \overline{\frac{\partial\varphi_j(a)}{\partial z_k}}K_{\varphi(a)}^{D_j}
\end{align}
for any $k=1,2,\ldots,n$ and $f\in \mathcal{D}(\mathbb{B}_{N}).$

In the same way,  we will write
$\langle f, K_a^{D_{i,j}} \rangle = \frac{\partial^2 f}{\partial z_i z_j}$ and
\begin{align}\label{Eq2.3}
 K_a^{D_{i,j}}= \frac{z_i z_j} {(1-\langle z,a \rangle)^2}
\end{align}
for $i,j=1,2,\ldots,n.$
Also, we find that
\begin{align}\nonumber
\langle f , W_{\psi,\varphi}^* K_a^{D_{11}} \rangle \nonumber
&= \langle W_{\psi,\varphi}f , K_a^{D_{11}} \rangle\\ \nonumber
&=\frac{\partial(\frac{\partial \psi(z)} {\partial z_1} f(\varphi(z)) + \psi(z)\sum\limits_{j=1}^{n}\frac{\partial f(\varphi(z))}{\partial z_j}\frac{\partial \varphi_j}{\partial z_1})} {\partial z_1}\Big{|}_{z=a}\\ \nonumber
&=\{\frac{\partial^2 \psi(z)}{\partial z_1 ^2} f(\varphi(z)) + 2 \frac{\partial \psi(z)}{\partial z_1}\sum\limits_{j=1}^{n}\frac{\partial f(\varphi(z))} {\partial z_j} \frac{\partial \varphi_j}{\partial z_1} \\ \nonumber
&+ \psi(z) \frac{\partial (\sum\limits_{j=1}^{n}\frac{\partial f(\varphi(z))} {\partial z_j} \frac{\partial \varphi_j}{\partial z_1}) }{\partial z_1}\}\Big{|}_{z=a}\\ \nonumber
&=\frac{\partial^2 \psi(a)}{\partial z_1 ^2} f(\varphi(a)) + 2 \frac{\partial \psi(a)}{\partial z_1}\sum\limits_{j=1}^{n}\frac{\partial f(\varphi(a))} {\partial z_j} \frac{\partial \varphi_j(a)}{\partial z_1}\\ \nonumber
&+ \psi(a) \sum\limits_{j=1}^{n} \frac{\partial^2 \varphi_j (a)}{ \partial z_1^2} \frac{ \partial f(\varphi(a))}{ \partial z_j}
+ \psi(a)\frac{ \partial \varphi_1(a)}{ \partial z_1} (\sum\limits_{j=1}^{n} \frac{\partial^2 f(\varphi(a))}{  \partial z_1 \partial z_j}\frac{\partial \varphi_j(a)}{\partial z_1}) \\ \nonumber
& +\ldots + \psi(a) \frac { \partial \varphi_n(a)} { \partial z_1} (\sum\limits_{j=1}^{n} \frac{\partial^2 f(\varphi(a))}{ \partial z_n \partial z_j} \frac{\partial \varphi_j(a)}{\partial z_1})\\ \nonumber
&=\langle f, \overline{\frac{\partial^2 \psi(a)}{\partial z_1 ^2}} K_{\varphi(a)}  +  2\overline{\frac{\partial \psi(a)}{\partial z_1}}\sum\limits_{j=1}^{n}\overline{ \frac{\partial \varphi_j(a)}{\partial z_1}} K_{\varphi(a)}^{D_j}\\ \nonumber
&+ \overline{\psi(a)} \sum\limits_{j=1}^{n} \overline{\frac{\partial^2 \varphi_j (a)}{ \partial z_1^2}}  K_{\varphi(a)}^{D_j}
+ \overline{\psi(a)} \overline{ \frac{ \partial \varphi_1(a)} { \partial z_1} }(\sum\limits_{j=1}^{n} \overline{\frac{\partial \varphi_j(a)}{\partial z_1}} K_{\varphi(a)}^{D_{1j}})\\ \nonumber
& +  \ldots  +   \overline{\psi(a)} \overline{ \frac{ \partial \varphi_n(a)} { \partial z_1} }(\sum\limits_{j=1}^{n} \overline{\frac{\partial \varphi_j(a)}{\partial z_1}} K_{\varphi(a)}^{D_{nj}}) \rangle.
\end{align}

Therefore,
\begin{align}
W_{\psi,\varphi}^* K_a^{D_{11}} \nonumber
&=\overline{\frac{\partial^2 \psi(a)}{\partial z_1 ^2}} K_{\varphi(a)}  +  2\overline{\frac{\partial \psi(a)}{\partial z_1}}\sum\limits_{j=1}^{n}\overline{ \frac{\partial \varphi_j(a)}{\partial z_1}} K_{\varphi(a)}^{D_j}\\ \nonumber
& + \overline{\psi(a)} \sum\limits_{j=1}^{n} \overline{\frac{\partial^2 \varphi_j (a)}{ \partial z_1^2}}  K_{\varphi(a)}^{D_j}
+ \overline{\psi(a)} \overline{ \frac{ \partial \varphi_1(a)} { \partial z_1} }(\sum\limits_{j=1}^{n} \overline{\frac{\partial \varphi_j(a)}{\partial z_1}} K_{\varphi(a)}^{D_{1j}}) \\\label{Eq2.4}
& + \ldots   +  \overline{\psi(a)} \overline{ \frac{ \partial \varphi_n(a)} { \partial z_1} }(\sum\limits_{j=1}^{n} \overline{\frac{\partial \varphi_j(a)}{\partial z_1}} K_{\varphi(a)}^{D_{nj}}).
\end{align}
\section{Complex symmetric composition operators on $\mathcal{D}(\mathbb{B}_N)$}

\subsection{Complex symmetric composition operators on $\mathcal{D}$}
Let us start by characterize composition operators on the Dirichlet space in the unit disk which are complex symmetric with respect to the conjugation $Jf(z)=\overline{f(\bar{z})}$. First  we  give the following theorem, which limits the kinds
of maps that can induce complex symmetric composition operators on $\mathcal{D}(\mathbb{D})$.
\begin{thm}
Let $\varphi$ be an analytic self-map of the unit disk. Suppose that $C_\varphi$ is a complex symmetric operator on $\mathcal{D}(\mathbb{D})$, then $\varphi$ has a fixed point in the unit disk.
\end{thm}
\textit{Proof.}
Since $C_\varphi$ is complex symmetric with non-empty point spectrum. [\citealp{N1}, Proposition 3.1] shows that $C_\varphi$ is not hypercyclic. By [\citealp{YR}, Theorem 1.1], we obtain $\varphi$ has a fixed point in  the unit disk.
\qed
\begin{thm}
Let $\varphi$ be an analytic self-map of the unit disk. Then $C_\varphi$ is $J$-symmetric on $\mathcal{D}$ if and only if $C_\varphi$ is normal.
\end{thm}
{\textit{Proof.}}
Since  $C_\varphi$ is $J$-symmetric, it follows from [\citealp{GH}, Proposition 2.4] that $\varphi(z)=az$ for some $|a|\leqslant1$, and so $C_\varphi$ is normal.

For the converse, suppose $C_\varphi$  is normal on $\mathcal{D}$, by [\citealp{CM1}, Theorem 8.2], we conclude that $\varphi(z)=az$ with $|a|\leqslant1$. An easy calculation gives $C_\varphi JK_w(z)=JC_\varphi^*K_w(z)$, for all $z,w \in \mathbb{D}$ and hence $C_\varphi$ is $J$-symmetric.
\qed
\begin{rem}
Indeed, every normal operator is complex symmetric, so it is natural to ask ``if there any complex symmetric but not normal composition operator $C_{\varphi}$ whose symbol is not a constant and also not involution?"
\end{rem}

\subsection{Complex symmetric weighted composition operators on $\mathcal{D}(\mathbb{B}_{N})$}
Following this idea, one is interested in determining whether the $J$-symmetric of composition operator is equivalent to its normality for the dimension greater than 1. Now, we begin with the theorem that gives the sufficient and necessary condition for weighted composition operators $W_{\psi,\varphi}$ to be $J$-symmetric.
\begin{thm}\label{Th3.4}
Let $\varphi$ be an analytic self-map of $\mathbb{B}_N$  and $\psi$ be an analytic function on $\mathbb{B}_{N}$ for which $W_{\psi,\varphi}$ is bounded on $\mathcal{D}(\mathbb{B}_{N})$. Then $W_{\psi,\varphi}$ is complex symmetric with conjugation $J$ if and only if $\psi(z)=c$ and $\varphi(z)=\varphi'(0)z,$ where $c$ is a constant and $\varphi'(0)$ is a symmetric matrix with $||\varphi'(0)||\leq1$.
\end{thm}
\textit{proof}.
 If $W_{\psi,\varphi}$ is complex symmetric with conjugation $J$, then we have
$$W_{\psi,\varphi}JK_w(z)=JW_{\psi,\varphi}^*K_w(z)$$
 for all $z,w \in \mathbb{B}_N$, which implies that
 \begin{equation} \label{Eq3.001}
\psi(z)(1+\ln\frac{1}{1-\langle \varphi(z),\overline{w}\rangle})=\psi(w)(1+\ln\frac{1}{1-\langle z, \overline{\varphi(w)} \rangle}).
\end{equation}

 Putting $w=0$ in Equation (\ref{Eq3.001}), then we have
 \begin{equation} \label{Eq3.002}
 \psi(z)=\psi(0)+\psi(0)\ln\frac{1}{1-\langle z, \overline{\varphi(0)}\rangle}.
\end{equation}

Subsituting the formula for $\psi(z)$ into Equation (\ref{Eq3.001}), we obtain
\begin{align}\nonumber
&(1-\ln(1-\langle z, \overline{\varphi(0)}\rangle))(1-\ln(1-\langle \varphi(z), \overline{w}\rangle))\\ \label{Eq3.003}
&=(1-\ln(1-\langle w, \overline{\varphi(0)}\rangle))(1-\ln(1-\langle z, \overline{\varphi(w)}\rangle)).
\end{align}
Taking partial derivate with respect to $w_1$ on the both sides of Equation (\ref{Eq3.003}), we get
 \begin{align} \nonumber
&(1-\ln(1-\langle z, \overline{\varphi(0)}\rangle))\frac{\varphi_1(z)}{1-\langle\varphi(z),\overline{w}\rangle}\\ \nonumber
&=(1-\ln(1-\langle w, \overline{\varphi(0)}\rangle))\frac{\frac{\partial\varphi_1(w)}{\partial w_1}z_1+\ldots+\frac{\partial\varphi_n(w)}{\partial w_1}z_n}{1-\langle z, \overline{\varphi(w)}\rangle}\\ \nonumber
&+(1-\ln(1-\langle z, \overline{\varphi(w)}\rangle))\frac{\varphi_1(0)}{1-\langle w, \overline{\varphi(0)}\rangle}.
\end{align}
Setting $w=0$ in the above equation, we have
\begin{align} \nonumber
\varphi_1(z)=\varphi_1(0) + \frac{\frac{\partial\varphi_1(0)}{\partial w_1}z_1+\ldots+\frac{\partial\varphi_n(0)}{\partial w_1}z_n}{(1-\langle z, \overline{\varphi(0)}\rangle)(1-\ln(1-\langle z, \overline{\varphi(0)}\rangle))}.
\end{align}
Similarly, we get
$$\varphi_k(z)=\varphi_k(0)+\frac{\frac{\partial\varphi_1(0)}{\partial w_k}z_1+\ldots+\frac{\partial\varphi_n(0)}{\partial w_k}z_n}{(1-\langle z, \overline{\varphi(0)}\rangle)(1-\ln(1-\langle z, \overline{\varphi(0)}\rangle))}$$
for $k=1,2,\ldots,n$. So we have
\begin{align}\label{Eq3.004}
\varphi(z)=\varphi(0)+\frac{\varphi'(0)^Tz}{(1-\langle z, \overline{\varphi(0)}\rangle)(1-\ln(1-\langle z, \overline{\varphi(0)}\rangle))},
\end{align}
here $\varphi'(0)^T$ denote the transpose matrix of $\varphi'(0)$.

Next, we claim  that $\varphi(0)=0$ when $W_{\psi,\varphi}$ is complex symmetric with conjugation $J$. Note that
\begin{equation} \label{Eq3.005}
 W_{\psi,\varphi}J \left(
  \begin{array}{ccc}
  K_a^{D_{11}}\\
  \vdots\\
  K_a^{D_{nn}}
\end{array} \right) = JW_{\psi,\varphi}^*\left(
  \begin{array}{ccc}
  K_a^{D_{11}}\\
  \vdots\\
  K_a^{D_{nn}}
\end{array}\right).
\end{equation}

Putting $a=0$ in Equation (\ref{Eq3.005}), and by Equation (\ref{Eq2.3}), we have
\begin{equation} \label{Eq3.006}
 W_{\psi,\varphi} \left(
  \begin{array}{ccc}
  z_1^2\\
  \vdots\\
  z_n^2
\end{array}\right)=JW_{\psi,\varphi}^*\left(
  \begin{array}{ccc}
  K_0^{D_{11}}\\
 \vdots\\
  K_0^{D_{nn}}
\end{array}\right).
\end{equation}
It follows from  Equation (\ref{Eq3.006})  that

\begin{equation} \label{Eq3.007}
\psi(z) \varphi_1^2(z) = J W_{\psi,\varphi}^* K_0^{D_{11}}(z).
\end{equation}

Since we obtain a precise formula for $\psi(z) \ \textrm{and} \  \varphi(z)$ when $W_{\psi,\varphi}$ is complex symmetric with conjugation $J$, thus by Equation (\ref{Eq3.002}) and (\ref{Eq3.004}) we get
\begin{align} \nonumber
&\psi(z) \varphi_1^2(z)\\ \nonumber
&=\psi(0)\varphi_1^2(0) ( 1 + \ln \frac {1} {1-\langle z, \overline{\varphi(0)} \rangle} ) + 2 \psi(0) \varphi_1(0) \sum\limits_{j=1}^{n} \frac{ \partial \varphi_j(0)} {\partial z_1} \frac{z_j} {1-\langle z, \overline{\varphi(0)} \rangle }+\\ \nonumber
& \psi(0) \frac{ \partial \varphi_1(0)} {\partial z_1} \sum\limits_{j=1}^{n} \frac{ \partial \varphi_j(0)} {\partial z_1} \frac{z_1 z_j} {(1-\langle z, \overline{\varphi(0)} \rangle )^2 (1- \ln (1-\langle z, \overline{\varphi(0)} \rangle)) } + \ldots +\\ \label{Eq3.008}
&\psi(0) \frac{ \partial \varphi_n(0)} {\partial z_1} \sum\limits_{j=1}^{n} \frac{ \partial \varphi_j(0)} {\partial z_1} \frac{z_n z_j} {( 1 - \langle z, \overline{\varphi(0)} \rangle )^2 (1- \ln (1-\langle z, \overline{\varphi(0)} \rangle)) }.
\end{align}

On the other hand, by Equation (\ref{Eq2.4}), we have
\begin{align} \nonumber
&JW_{\psi,\varphi}^* K_0^{D_{11} }(z)\\ \nonumber
&=\frac{\partial^2 \psi(0)}{\partial z_1 ^2} K_{\overline{\varphi(0)}}(z)  +  2\frac{\partial \psi(0)}{\partial z_1}\sum\limits_{j=1}^{n} \frac{\partial \varphi_j(0)}{\partial z_1} K_{\overline{\varphi(0)}}^{D_j}(z)  +  \psi(0) \sum\limits_{j=1}^{n} \frac{\partial^2 \varphi_j (0)}{ \partial z_1^2} K_{\overline{\varphi(0)}}^{D_j} (z) + \\ \label{Eq3.009}
&  \psi(0)  \frac{ \partial \varphi_1(0)} { \partial z_1} (\sum\limits_{j=1}^{n} \frac{\partial \varphi_j(0)}{\partial z_1} K_{\overline{\varphi(0)}}^{D_{1j}}(z) ) +
\ldots   +  \psi(0)  \frac{ \partial \varphi_n(0)} { \partial z_1} (\sum\limits_{j=1}^{n} \frac{\partial \varphi_j(0)}{\partial z_1} K_{\overline{\varphi(0)}}^{D_{nj}}(z) ).
\end{align}

By Equation (\ref{Eq3.002}), we have
\begin{align}\nonumber
\frac{\partial \psi(z)}{\partial z_1} = \psi(0) \frac{\varphi_1(0)}{1-\langle z, \overline{\varphi(0)} \rangle}
\end{align}
and
\begin{align}\nonumber
\frac{\partial^2 \psi(z)}{\partial z_1^2} = \psi(0) \frac{\varphi_1^2(0)}{(1-\langle z, \overline{\varphi(0)} \rangle)^2}.
\end{align}
Thus,
\begin{align} \label{Eq3.0010}
\frac{\partial \psi(0)}{\partial z_1} = \psi(0)\varphi_1(0)
\end{align}
and
\begin{align} \label{Eq3.0011}
\frac{\partial^2 \psi(0)}{\partial z_1^2} = \psi(0)\varphi_1(0)^2.
\end{align}

Then by Equation (\ref{Eq3.004}), a calculation gives
\begin{align}\nonumber
&\frac {\partial^2 \varphi_1(z)} {\partial z_1^2} \\ \nonumber
&= \frac{-(\frac{\partial\varphi_1(0)}{\partial z_1}z_1+\ldots+\frac{\partial\varphi_n(0)}{\partial z_1}z_n ) \frac{\partial^2 F(z) }{ \partial z_1^2}F(z)^2} { {F(z)}^4}-\\ \nonumber
&\frac{\{\frac{\partial\varphi_1(0)}{\partial z_1} F(z)
-{(\frac{\partial\varphi_1(0)}{\partial z_1}z_1 + \ldots + \frac{\partial\varphi_n(0)}{\partial z_1}z_n) \frac{ \partial F(z)}{ \partial z_1}\}2F(z)\frac{\partial F(z)}{ \partial z_1}}} { {F(z)}^4 },
\end{align}
where $F(z) = (1-\langle z, \overline{\varphi(0)}\rangle)(1-\ln(1-\langle z, \overline{\varphi(0)}\rangle))$. Since $\frac{\partial F(0)}{ \partial z_1}=0$, so we have
\begin{align}\label{Eq3.0012}
\frac{\partial^2 \varphi_1(0)}{\partial z_1^2} = 0.
\end{align}

Similarly, we have
\begin{align}\label{Eq3.0013}
\frac{\partial^2 \varphi_j(0)}{\partial z_1^2} = 0
\end{align}
for $j=1,2,\ldots, n.$

Putting all our information together and returning to the Equation (\ref{Eq3.009}), we get
\begin{align} \nonumber
&JW_{\psi,\varphi}^* K_0^{D_{11}}(z)\\ \nonumber
&=\psi(0)\varphi_1(0)^2 (1 + \ln \frac {1} {1-\langle z, \overline{\varphi(0)} \rangle}) + 2 \psi(0) \varphi_1(0) \sum\limits_{j=1}^{n} \frac{ \partial \varphi_j(0)} {\partial z_1} \frac{z_j} {1-\langle z, \overline{\varphi(0)} \rangle }\\ \nonumber
& + \psi(0) \frac{ \partial \varphi_1(0)} {\partial z_1} \sum\limits_{j=1}^{n} \frac{ \partial \varphi_j(0)} {\partial z_1} \frac{z_1 z_j} {(1-\langle z, \overline{\varphi(0)} \rangle )^2  } + \ldots +\\ \label{Eq3.0014}
&\psi(0) \frac{ \partial \varphi_n(0)} {\partial z_1} \sum\limits_{j=1}^{n} \frac{ \partial \varphi_j(0)} {\partial z_1} \frac{z_n z_j} {( 1 - \langle z, \overline{\varphi(0)}\rangle )^2 }.
\end{align}

Combining Equation (\ref{Eq3.008}) and Equation (\ref{Eq3.0014}), we have
\begin{align}  \label{Eq3.0015}
\varphi(0)=0.
\end{align}
Finally, from Equation (\ref{Eq3.002}),  Equation (\ref{Eq3.004}) and  Equation (\ref{Eq3.0015}) we easily deduce that
$$ \psi(z) = \psi(0) = c \ \ \ \ \ \textrm{and} \ \ \ \ \  \varphi(z) = \varphi'(0)z $$
where $c$ is constant and $\varphi'(0)$ is a symmetric matrix with $||\varphi'(0)||\leq1$. Indeed, notice that if $C_{\varphi'(0)z} $ is $J$-symmetric, it is easy to show that $\varphi'(0)=\overline{\varphi'(0)}^*$, that is, $\varphi'(0)$ is a symmetric matrix.

The converse is clear.
\qed
\begin{cor}
Let $\varphi$ be an analytic self-map of $\mathbb{B}_N$, if $C_\varphi$ is bounded on  $\mathcal{D}(\mathbb{B}_{N})$, then $C_\varphi$ is $J$-symmetric on  $\mathcal{D}(\mathbb{B}_{N})$ if and only if $\varphi(z)=\varphi'(0)z$
for $z\in\mathbb{B}_{N}$ and $\varphi'(0)$ is a symmetric matrix  with $||\varphi'(0)||\leq1$.
\end{cor}
 Next we consider the unitary composition operator $C_\varphi$  on $\mathcal{D}(\mathbb{B}_{N})$. Then we can use the unitary composition operator to construct another conjugation operator on  $\mathcal{D}(\mathbb{B}_{N})$.

\begin{Lemma}{\rm ([\citealp{ZY}, Theorem 4.1])}
Let $\varphi$ be an analytic self-map of $\mathbb{B}_{N}$. Then  $C_\varphi$  is unitary on $\mathcal{D}(\mathbb{B}_{N})$  if and only if  $\varphi(z)=Uz$ where $U$ is a unitary matrix.
\end{Lemma}
\begin{Prop} If $U$ is a unitary symmetric matrix, then $JC_{Uz}$ is a conjugation.
\end{Prop}

Using similar proof of Theorem \ref{Th3.4}, we also easily prove the following theorem.

\begin{thm}
Let $\varphi$ be an analytic self-map of $\mathbb{B}_N$  and $\psi$ be an analytic function on $\mathbb{B}_{N}$ for which $W_{\psi,\varphi}$ is bounded on $\mathcal{D}(\mathbb{B}_{N})$. If $W_{\psi,\varphi}$ is complex symmetric with conjugation $JC_{Uz}$ if and only if $\psi(z)=c$ and $\varphi(z)=\varphi'(0)\overline{U}z,$
where $c$ is constant, $\varphi'(0)$ is a symmetric matrix with $||\varphi'(0)||\leq1$ and $\varphi'(0) \overline{U}= \overline{U}\varphi'(0)$.
\end{thm}

\textit{proof}.
Since $W_{\psi,\varphi}$ is complex symmetric with conjugation $JC_{Uz}$, then
$$JC_{Uz}W_{\psi,\varphi}JC_{Uz}=(W_{\psi,\varphi})^*$$
which means that

$$JC_{Uz}W_{\psi,\varphi}=(C_{Uz}W_{\psi,\varphi})^*J.$$

It follows from  Theorem \ref{Th3.4} that
$$\psi(Uz)=\psi(0)$$
and
$$\varphi(Uz)=\varphi'(0)z .$$

Therefore, replace $Uz$ by $z$,  we have
$$\psi(z)=c \ \ \ \ \ \textrm{and} \ \ \ \ \ \ \varphi(z)=\varphi'(0)\overline{U}z,$$
where $c$ is constant, $\varphi'(0)$ is a symmetric matrix with $||\varphi'(0)||\leq1$ and $\varphi'(0)\overline{U} = \overline{U}\varphi'(0)$. In fact, notice that if $C_\varphi$ is $JC_{Uz}$-symmetric, it is  easy to check that  that  $\varphi'(0)\overline{U} = \overline{U}\varphi'(0)$.

The converse direction follows readily from a simple calculation, so we omit the proof.
\qed
\subsection{Hermitian weighted composition operators on $\mathcal{D}(\mathbb{B}_{N})$}
In this section, we will find out the functions $\psi$ and $\varphi$ when $W_{\psi,\varphi}$ are bounded Hermitian weighted composition operators. Not surprisingly, we will prove that no nontrivial Hermitian weighted composition operator exist on  $\mathcal{D}(\mathbb{B}_{N})$.
\begin{thm}
Let $\varphi$ be an analytic self-map of $\mathbb{B}_N$  and $\psi$ be an analytic function on $\mathbb{B}_{N}$ for which $W_{\psi,\varphi}$ is bounded on $\mathcal{D}(\mathbb{B}_{N})$. Then $W_{\psi,\varphi}$ is a Hermitian weighted composition operator on $\mathcal{D}(\mathbb{B}_{N})$ if and only if
 $$\psi(z)=c \ \ \ \textrm{and}\ \ \ \varphi(z)=\varphi'(0)z$$
where $c$ is a real constant and $\varphi'(0)$ is a Hermitian matrix.

\end{thm}
\textit{Proof.}
Since $W_{\psi,\varphi}$ is a bounded Hermitian weighted operator on $\mathcal{D}(\mathbb{B}_{N})$, then we have
$$W_{\psi,\varphi}K_w(z)=W_{\psi,\varphi}^*K_w(z)$$
for all $z,w\in\mathbb{B}_{N}$, that is
$$\psi(z)K_w(\varphi(z))=\overline{\psi(w)}K_{\varphi(w)}(z)$$
for all $z,w\in\mathbb{B}_{N}$. Thus
\begin{equation} \label{Eq3.02}
\psi(z)(1+\ln\frac{1}{1-\langle\varphi(z),w\rangle})=\overline{\psi(w)}(1+\ln\frac{1}{1-\langle z,\varphi(w)\rangle}).
\end{equation}

Letting $w=0$ in Equation (\ref{Eq3.02}) gives
\begin{equation} \label{Eq3.03}
 \psi(z)=\overline{\psi(0)}(1+\ln\frac{1}{1-\langle z,\varphi(0)\rangle})
\end{equation}
for all $z\in\mathbb{B}_{N}$. Putting $z=0$, we get $\psi(0)=\overline{\psi(0)}$, i.e. $\psi(0)$ is a real number.

Substituting the formula for $\psi(z)$ into  Equation (\ref{Eq3.02}), we obtain
\begin{align}\nonumber
&(1-\ln(1-\langle z,\varphi(0)\rangle))(1-\ln(1-\langle\varphi(z),w\rangle)\\\nonumber
&=(1-\ln(1-\langle\varphi(0),w\rangle))(1-\ln(1-\langle z,\varphi(w)\rangle)).
\end{align}
Taking partial derivative with respect to $\overline{w_1}$, we obtain
\begin{align} \nonumber
&(1-\ln(1-\langle z,\varphi(0)\rangle))\frac{\varphi_1(z)}{1-\langle\varphi(z),w\rangle}\\ \nonumber
&=(1-\ln(1-\langle\varphi(0),w\rangle))\frac{\frac{\partial\overline{\varphi_1(w)}}{\partial \overline{w_1}}z_1+\ldots+\frac{\partial\overline{\varphi_n(w)}}{\partial {\overline{w_1}}}z_n}{1-\langle z,\varphi(w)\rangle}\\ \nonumber
&+(1-\ln(1-\langle z,\varphi(w)\rangle))\frac{\varphi_1(0)}{1-\langle\varphi(0),w\rangle}.
\end{align}

Putting $w=0$ in the above equation, we get
$$\varphi_1(z)=\varphi_1(0)+\frac{\frac{\partial\overline{\varphi_1(0)}}{\partial \overline{w_1}}z_1+\ldots+\frac{\partial\overline{\varphi_n(0)}}{\partial \overline{w_1}}z_n}{(1-\langle z,\varphi(0)\rangle)(1-\ln(1-\langle z,\varphi(0)\rangle))}.$$

Similarly, we get
$$\varphi_k(z)=\varphi_k(0)+\frac{\frac{\partial\overline{\varphi_1(0)}}{\partial \overline{w_k}}z_1+\ldots+\frac{\partial\overline{\varphi_n(0)}}{\partial \overline{w_k}}z_n}{(1-\langle z,\varphi(0)\rangle)(1-\ln(1-\langle z,\varphi(0)\rangle))}$$
for $k=1,2,\ldots,n$. Therefore we obtain
\begin{equation}\label{Eq3.04}
\varphi(z)=\varphi(0)+\frac{\varphi'(0)^*z}{(1-\langle z,\varphi(0)\rangle)(1-\ln(1-\langle z,\varphi(0)\rangle))}.
\end{equation}
Taking derivative with respect to $z$, and putting $z=0$, we have
$$\varphi'(0)=\varphi'(0)^*,$$
that is, $\varphi'(0)$ is a Hermitian matrix.

Furthermore,  we obtain $\varphi(0)=0$, the proof is similar to that of Theorem \ref{Th3.4}, so we omit the details. It follows from  Equation (\ref{Eq3.03}) and Equation (\ref{Eq3.04}) that
 $$
 \psi(z)=c \ \ \ \textrm{and} \ \ \ \varphi(z)=\varphi'(0)z
 $$
where $c$ is a  real constant and $\varphi'(0)$ is a Hermitian matrix.

Conversely, if $\psi(z)=c$ and $\varphi(z)=\varphi'(0)z,$ where $c$ is a real constant and $\varphi'(0)$ is a Hermitian matrix. It is easy to see that
$$W_{\psi,\varphi}^*K_w(z)=W_{\psi,\varphi}K_w(z)$$
for all $z,w\in\mathbb{B}_{N}$,  which completes the proof of the theorem.
\qed
\begin{cor}
Let $\varphi$ be an analytic self-map of $\mathbb{B}_N$, if $C_\varphi$ is bounded on  $\mathcal{D}(\mathbb{B}_{N})$, then $C_\varphi$ is Hermitian on  $\mathcal{D}(\mathbb{B}_{N})$ if and only if
$$\varphi(z)=\varphi'(0)z$$
for $z\in\mathbb{B}_{N}$ and some Hermitian matrix $\varphi'(0)$ with $||\varphi'(0)||\leq1$.
\end{cor}
\section{Complex symmetric weighted composition operators on $H^2(\mathbb{B}_{N})$}

  In this section, we begin with some results about complex symmetric weighted composition operators with respect to the conjugation $J$. Then we give some  nontrivial sufficient conditions for weighted composition operators to be unitary and $J$-symmetric. Based upon this, we obtain more  new examples of complex symmetric weighted composition operators on $H^2(\mathbb{B}_{N})$. Finally, we characterize the normality of $C_{Uz}J$-symmetric weighted composition operators.
\subsection{ Unitary and Hermitian complex symmetric weighted composition operators on $H^2(\mathbb{B}_{N})$}
 We first point out when the class of $J$-symmetric weighted composition operator is unitary or Hermitian.
\begin{thm}
Let $\psi(z)=\frac{a_1}{(1-\langle z,\overline{a_{0}}\rangle)^{N}}$, and $\varphi(z)=\frac{a_0-Az}{1-\langle z,\overline{a_{0}}\rangle}$, where $a_{1}\neq0, a_{0}\in{\mathbb{B}}_{N}$ and $A$ is a symmetric matrix such that $\varphi$ is a self-map of ${\mathbb{B}}_{N}$. Then $W_{\psi,\varphi}$ is unitary if and only if
$$a_1=\lambda(1-|a_{0}|^2)^{\frac{N}{2}}, \ \overline{A}A-\overline{a_{0}}a^{T}_{0}=(1-|a_{0}|^2)I_{N} \ \textrm{and }\ A\overline{a_{0}}-{a_{0}}=0,$$
where $a_{0}\in \mathbb{B}_{N}-\{0\}$.

In particular, if $a_0=0$, $W_{\psi,\varphi}$ is unitary if and only if $\psi(z)=\lambda$ for some $\lambda\in\mathbb{C}$ with $|\lambda|=1$ and $\varphi(z)=Az$ where $A$ is a  unitary and symmetric matrix.
\end{thm}
\textit{Proof.}
First recall that the adjoint of $\varphi$ is defined by
 $$ \sigma(z)=\frac{\overline{a_0}-A^{*}z}{1-\langle z,a_{0}\rangle}, \ \  z\in \mathbb{B}_{N}.$$

 Our hypotheses show that $W_{\psi,\varphi}$ is $J$-symmetric (see [\citealp{WY}, Theorem 3.1]), then by [\citealp{L}, Corollary 3.6], we have $W_{\psi,\varphi}$ is a unitary operator if and only if $\psi(z)=\lambda k_{\varphi^{-1}(0)}^{N}(z)$ for some complex number $\lambda$ with $|\lambda|=1$ and $\varphi(z)$ is an automorphism of ${\mathbb{B}}_{N}$. Therefore $\varphi^{-1}(0)=\sigma(0)=\overline{a_{0}}=\overline{\varphi(0)}$, $a_1=\lambda(1-|a_{0}|^2)^{\frac{N}{2}}$. Since $\varphi$ is an analytic self-map of $\mathbb{B}_{N}$ and $\varphi$ is one-to-one (also see [\citealp{WY}, Theorem 3.1]), due to the Theorem \ref{Th2.7}, we have $\varphi(z)\in$ Aut$(\mathbb{B}_{N})$ if and only if
$$
 m_{\varphi}=\left(
  \begin{array}{ccc}
  -A & a_0\\
    -\overline{a_0}^*& 1\\
  \end{array}
\right)
$$
is a non-zero multiple of Kre$\breve{{\i}}$n isometry on Kre$\breve{{\i}}$n space $\mathbb{C}^{N+1}$. So we have
 $$
 |k|^2\left(\begin{array}{cc}
 -A^*&-\overline{a_0}\\
a_0^*&1\\
\end{array} \right)
\left( \begin{array}{cc}
I_N& 0\\
0&-1\\
\end{array} \right)
\left( \begin{array}{cc}
-A & a_0\\
    -\overline{a_0}^*& 1\\
\end{array}
\right)=
\left( \begin{array}{cc}
I_N & 0\\
0& -1\\
\end{array}
\right).
$$

Then we can obtain from direct computations that
\begin{equation} \label{4.1}
|k|^2(A^*A-\overline{a}_0 \overline{a}_0^*)=I_N,
\end{equation}
\begin{equation} \label{4.2}
|k|^2(-A^*a_0+\overline{a}_0)=0,
\end{equation}
\begin{equation} \label{4.3}
|k|^2(a_0^*a_0-1)=-1.
\end{equation}

If $a_0=0$ in Equation (\ref{4.3}), we get
$$\psi(z)=a_1=\lambda$$
for some $\lambda\in\mathbb{C}$ with $|\lambda|=1$ and
 $$A^*A=AA^*=I_N.$$
Thus we have
 $\varphi(z)=Az$, where $A$ is a symmetric unitary matrix.

If $a_0\neq0$ in Equation (\ref{4.3}), we get $|k|^2=\frac{1}{1-|a_0|^2}$, then substituting the expression of $|k|^2$ into Equation (\ref{4.1}) and (\ref{4.2}) we have
$$\overline{A}A-\overline{a}_0a_0^{T}=(1-|a_0|^2)I_N, \ \ \    A\overline{a}_0-a_0=0.$$

Combining these two cases, we have our conclusion.
\qed
\begin{thm}
Let $\psi(z)=\frac{a_1}{(1-\langle z,\overline{a_{0}}\rangle)^{N}}$, and $\varphi(z)=\frac{a_0-Az}{1-\langle z,\overline{a_{0}}\rangle}$, where $a_{1}\neq0, \  a_{0}\in{\mathbb{B}}_{N}$ and $A$ is a symmetric matrix such that $\varphi$ is a self map of ${\mathbb{B}}_{N}$. Then $W_{\psi,\varphi}$ is Hermitian if and only if $a_1$ is a real number, $a_0$ is a real vector, and $A$ is a real matrix.
\end{thm}
\textit{Proof.}
 We know from the [\citealp{WY}, Theorem 3.1] that $W_{\psi,\varphi}$ is complex symmetric with conjugation $J$, and note that $W_{\psi,\varphi}$ is Hermitian (i.e. $W_{\psi,\varphi}=W_{\psi,\varphi}^*$) if and only if
 $$W_{\psi,\varphi} JK_w(z)=JW_{\psi,\varphi} K_w(z)$$
 for any $z,w\in \mathbb{B}_N$, which implies
 \begin{equation} \label{Eq4.4}
\frac{a_1}{(1-\langle z,\overline{a_{0}}\rangle)^{N}(1-\langle \varphi(z),\overline{w}\rangle)^N}=\frac{\overline{a_1}}{(1-\langle\overline{a_{0}},\overline{z}\rangle)^{N}(1-\langle w,\varphi(\overline{z})\rangle)^N}.
\end{equation}
Putting $w=0$ in Equation (\ref{Eq4.4}) we have
$$\frac{a_1}{(1-\langle z,\overline{a_{0}}\rangle)^{N}}=\frac{\overline{a_1}}{(1-\langle\overline{a_{0}},\overline{z}\rangle)^{N}}.$$
Thus, $a_1$ is a real number and $a_0$ is a real vector.

Combining these with Equation (\ref{Eq4.4}) we get
$$\frac{1}{(1-\langle \varphi(z),\overline{w}\rangle)^N}=\frac{1}{(1-\langle w,\varphi(\overline{z})\rangle)^N}.$$
Hence, we have $\varphi(z)=\overline{\varphi(\overline {z})}$, that is
$$\frac{a_0-Az}{1-\langle z,a_0\rangle}=\frac{a_0-\overline{A}z}{1-\langle z,a_0\rangle}.$$
It follows that $A$ is a real matrix, so we complete our proof.
\qed
\subsection{ Some new examples of complex symmetric weighted composition operators on $H^2(\mathbb{B}_{N})$}
In this section, we will give some new examples of complex symmetric weighted composition operators on $H^2(\mathbb{B}_{N})$. For this purpose, we first present some sufficient conditions for weighted composition operators to be  both unitary and $J$-symmetric.

\begin{Prop}\label{Prop4.2}
Let $\Phi$ be an analytic self-map of $\mathbb{B}_{N}$ and $\Psi$ ba an analytic function. If $\Psi(z)=\lambda$,\ $\Phi(z)=Uz$ where $|\lambda|=1$, $U$ is a unitary and symmetric matrix or $\Psi(z)=\mu\frac{(1-|a|^2)^{\frac{N}{2}}}{(1-\langle z,a\rangle)^{N}}$,\ $\Phi(z)=U\frac{a-Tz}{1-\langle z,a\rangle}$, where $\mu \in \mathbb{C}$, $|\mu|=1$, $Ua=\overline{a}$ and $U$
is a symmetric unitary matrix, and $T$ is a self-adjoint map depending on $a$. Then the weighted composition operator $W_{\Psi,\Phi}$ is unitary and $J$-symmetric.
\end{Prop}
\textit{Proof.}
Clearly  if $\Psi(z)=\lambda$ and $\Phi(z)=Uz$, where $|\lambda|=1$, $U$ is a unitary and symmetric matrix, then $W_{\Psi,\Phi}$ is unitary and $J$-symmetric.

Suppose that
$$\Psi(z)=\mu\frac{(1-|a|^2)^{\frac{N}{2}}}{(1-\langle z,a\rangle)^{N}},\  \  \ \ \  \Phi(z)=U\frac{a-Tz}{1-\langle z,a\rangle},$$
where $|\mu|=1$, $Ua=\overline{a}$. From [\citealp{L}, Corollary 3.6], we  see that  $W_{\Psi,\Phi}$ is unitary.
Notice that $Ua=\overline{a}$, so we have
\begin{align} \nonumber
UT&=\sqrt{1-|a|^2}U+(1 - \sqrt{1-|a|^2}) \frac{Ua\overline{a}^T}{|a|^2}\\
&=\sqrt{1-|a|^2}U+(1 - \sqrt{1-|a|^2}) \frac{\overline{a}a^*}{|a|^2}\label{Eq4.05}
\end{align}
Since $U$ is a symmetric unitary matrix,  we can then use Equation (\ref{Eq4.05})  to obtain $UT$ is a symmetric matrix. Then [\citealp{WY}, Theorem 3.1] implies that $W_{\Psi,\Phi}$ is a $J$-symmetric weighted composition operator.
\qed
\begin{pro}
Try to give a sufficient and necessary condition for a weighted composition operator to be both unitary and $J$-symmetric. Note that the case for $H^2(\mathbb{D})$ was solved by Fatehi \cite{F}.
\end{pro}
Before beginning any proofs, we present an example that is simple enough that unitary and $J$-symmetric weighted composition operator $W_{\Psi,\Phi}$ can be carried out concretely.
\begin{ex}
\begin{itemize}
\item[(i)]Let $\Phi$ be an analytic self-map of $\mathbb{B}_{N}$ and $\Psi$ ba an analytic function. If  $\Psi(z)=\mu\frac{(1-|a|^2)^{\frac{N}{2}}}{(1-\langle z,a\rangle)^{N}}$,\ $\Phi(z)=\frac{a-Tz}{1-\langle z,a\rangle}$, where $\mu \in \mathbb{C}$, $|\mu|=1$ and $a\in\mathbb{R}^N \cap \mathbb{B}^N $. Then the weighted composition operator $W_{\Psi,\Phi}$ is unitary and $J$-symmetric.
\item[(ii)]Let $\Phi$ be an analytic self-map of $\mathbb{B}_{N}$ and $\Psi$ ba an analytic function. If  $\Psi(z)=\mu\frac{(1-|a|^2)^{\frac{N}{2}}}{(1-\langle z,a\rangle)^{N}}$,\ $\Phi(z)=U\frac{a-Tz}{1-\langle z,a\rangle}$, where $\mu \in \mathbb{C}$, $|\mu|=1$, $ a=(a_j) $ with $ a_j \neq 0$ for $j=1,2,\ldots,n$ and
    \setcounter{MaxMatrixCols}{20}
\setlength{\arraycolsep}{6pt}
\newcommand*\matrixrightlabel[2][1em]{\text{\makebox[0em][l]{\hspace{#1}$\leftarrow(#2)$}}}

$$
U=\left(
\begin{array}{c}
e^{-2iarga_1} \qquad \qquad \qquad \\
\qquad e^{-2iarga_2} \qquad \qquad \\
\qquad \ddots \qquad \\
\qquad \qquad \qquad e^{-2iarga_n}
\end{array}
\right).
$$
Then the weighted composition operator $W_{\Psi,\Phi}$ is unitary and $J$-symmetric.
\end{itemize}

\setcounter{MaxMatrixCols}{20}
\setlength{\arraycolsep}{6pt}
\newcommand*\matrixrightlabel[2][1em]{\text{\makebox[0em][l]{\hspace{#1}$\leftarrow(#2)$}}}
\end{ex}

 Next, we will use the unitary and $J$-symmetric weighted composition operator in constructing some special conjugations.

\begin{cor}
If $\Psi$ and $\Phi$ satisfy the conditions in Proposition \ref{Prop4.2}, then $W_{\Psi,\Phi}J$ is a conjugation.
\end{cor}
\textit{ Proof. }
The conclusion follows directly from [\citealp{GPP}, Lemma 3.2].
\qed

Next we present a sufficient and necessary condition for weighted composition operators to be $W_{\Psi,\Phi}J$-symmetric, where $\Psi$ and $\Phi$ satisfy the conditions in Proposition \ref{Prop4.2}.
Most of them are direct extensions of results in \cite{F}, so we omit the details.
\begin{thm}\label{Th4.5}
Let $\psi(z)=\frac{a_1}{(1-\langle z,\overline{a_{0}}\rangle)^{N}}$, and $\varphi(z)=\frac{a_0-Az}{1-\langle z,\overline{a_{0}}\rangle}$, where $a_{1}\neq0, a_{0}\in{\mathbb{B}}_{N}$ and $A$ is a symmetric matrix such that $\varphi$ is a self map of ${\mathbb{B}}_{N}$.
\begin{itemize}
\item[(i)]   For  $a\neq0$, the weighted composition operator $W_{\widetilde{\psi},\widetilde{\varphi}}$ is complex symmetric with conjugation $W_{\Psi,\Phi}J$ if and only if $\widetilde{\psi}=\Psi\cdot\psi\circ\Phi$, $\widetilde{\varphi}=\varphi\circ\Phi$.
\item[(ii)]  If $U$ is a unitary symmetric matrix, the weighted composition operator $W_{\widetilde{\psi},\widetilde{\varphi}}$ is complex symmetric with conjugation $C_{Uz}J$ if and only if
$\widetilde{\psi}(z)=\psi(Uz)$ and $\widetilde{\varphi}(z)=\varphi(Uz).$
\end{itemize}

\end{thm}
Due to a result of [\citealp{NST}, Theorem 2.10], Narayan, Sievewright, and Thompson give some examples of linear-fractional, not automorphic maps that induce complex symmetric composition operators on $H^2(\mathbb{D})$. The ideals in \cite{NST} may be adapted to prove when $C_{Az+c}$ is $JW_{\psi_{b},\varphi_{b}}$-symmetric on $H^2(\mathbb{B}_{N})$, where $\psi_{b}(z)= k_b^N(z)=\frac{(1-|b|^2)^{\frac{N}{2}}}{(1-\langle z,b\rangle)^N}$
,\ $\varphi_{b}(z)=\frac{b-Tz}{1-\langle z,b\rangle}$ with $b=(I-A)^{-1}c $, and $T = \sqrt{1-|b|^2}I + (1 - \sqrt{1-|b|^2}) \frac{bb^T}{|b|^2}$.
However, the result can fail when $N > 1$, we do have the following restricted version.
\begin{Lemma}\label{lem2.7}
 Let $\gamma >0$ be given. Suppose $\varphi$ is a linear fractional map of $\mathbb{B}_{N}$  into itself, then $C_\varphi$  is bounded on $H_\gamma$.
\end{Lemma}
\begin{Lemma}
 Let $\sigma(z)=Az+c$ is a linear fractional map of $\mathbb{B}_{N}$ into $\mathbb{B}_{N}$. Then $C_\sigma$ is bounded on $H^2(\mathbb{B}_{N})$ and  we have
 $$C_\sigma^*=T_\psi C_\varphi,$$
 where $\psi(z)=\frac{1}{(1+\langle z,-c\rangle)^N}$, $\varphi(z)=\frac{A^*z}{1+\langle z,-c\rangle}.$
\end{Lemma}
\textit{Proof.}
We know from Lemma \ref{lem2.7} that $C_\sigma$ is bounded on $H^2(\mathbb{B}_{N})$, then by the adjoint formula on $H^2(\mathbb{B}_{N})$ given by Theorem \ref{Th2.6}, we will get this result by a simple computation.
\begin{thm}\label{Th4.8}
Let $A=(a_{i,j})$ be $N\times N$-symmetric matrix, and 1 is not a eigenvalue of $A$, $c=(c_i)$ be $N$-column vectors. Let $b=(I-A)^{-1} c\in \mathbb{B}_{N}\cap \mathbb{R}_{N}$, and $Ab=\lambda b$ for some $\lambda\in\mathbb{C}$, let $\sigma(z)=Az+c$, then $C_{\sigma}$ is $JW_{\psi_{b},\varphi_{b}}$-symmetric.
\end{thm}

\textit{Proof.}
 we need to prove that
 \begin{equation} \label{Eq4.5}
 C_\sigma JW_{\psi_{b},\varphi_{b}}K_w(z)=JW_{\psi_{b},\varphi_{b}}C_\sigma^*K_w(z)
\end{equation}
for all $z, w\in\mathbb{B}_{N} $, where $C_{\sigma}^*=M_\psi C_\varphi=\frac{1}{(1+\langle z,-c\rangle)^N}\frac{A^*z}{1+\langle z,-c\rangle}$, $\psi_{b} (z)=k_b^N(z)=\frac{(1-|b|^2)^{\frac{N}{2}}}{(1-\langle z,b\rangle)^N}$, and $\varphi_{b}(z)=\frac{b-Tz}{1-\langle z,b\rangle}$ with $T$ is a self-adjoint operator depending on $b$.

For the left side of of Equation (\ref{Eq4.5}), we have
\begin{align*}
&C_{Az+c}JW_{\psi_{b},\varphi_{b}}K_w(z)\\
&=C_{Az+c}J\frac{(1-|b|^2)^{\frac{N}{2}}}{(1-\langle z,b\rangle)^N}K_w(\frac{b-Tz}{1-\langle z,b\rangle})\\
&=\frac{(1-|b|^2)^{\frac{N}{2}}}{(1-\langle b,\overline{Az+c}\rangle)^N}\overline{K_w(\frac{b-T(\overline{Az+c})}{1-\langle\overline{Az+c},b\rangle})}\\
&=\frac{(1-|b|^2)^{\frac{N}{2}}}{(1-\langle b,\overline{Az+c}\rangle)^N}\frac{1}{(1-\langle w,\frac{b-T(\overline{Az+c})}{1-\langle\overline{Az+c},\overline{b}\rangle}\rangle)^N}\\ &=\frac{(1-|b|^2)^{\frac{N}{2}}}{(1-\langle b,\overline{Az+c}\rangle-\langle w,b-T\overline{Az}-T\overline{c}\rangle)^N}.
\end{align*}

For the right side of of Equation (\ref{Eq4.5}), we have
\begin{align*}
&JW_{\psi_{b},\varphi_{b}}C_{Az+c}^*K_w(z)\\
&=J\frac{(1-|b|^2)^{\frac{N}{2}}}{(1-\langle z,b\rangle)^N}C_\frac{b-Tz}{1-\langle z,b\rangle} \frac{1}{(1+\langle z,-c\rangle)^N} K_w \Big(\frac{A^*z}{1+\langle z,-c\rangle} \Big) \\
&=\frac{(1-|b|^2)^{\frac{N}{2}}}{(1-\langle b,\overline{z}\rangle)^N} \frac{1}{(1+\langle-c,\frac{b-T\overline{z}}{1-\langle\overline{z},b\rangle}\rangle)^N} \frac{1}{(1-\langle w,\frac{A^*b-A^*T\overline{z}}{1-\langle\overline{z},b\rangle+\langle b-T\overline{z},-c\rangle}\rangle)^N} \\
&=\frac{(1-|b|^2)^{\frac{N}{2}}}{(1-\langle b,\overline{z}\rangle+\langle-c,b\rangle+\langle c,T\overline{z}\rangle-\langle Aw,b\rangle+\langle Aw,T\overline{z}\rangle)^N}.
\end{align*}

To see Equation (\ref{Eq4.5}) holds, by the previous calculations, it is enough to show that
\begin{equation} \label{Eq4.6}
-z^TA^Tb - b^Tw + z^TA^TTw + c^TTw = -z^Tb + z^TTc - b^TAw + z^TTAw.
\end{equation}

Since $T = \sqrt{1-|b|^2}I + (1 - \sqrt{1-|b|^2}) \frac{bb^T}{|b|^2}$ and $Ab = \lambda b$ for some $\lambda \in C$, we have $c = kb$ with $k = 1 - \lambda$ and
\begin{align}
Tc &= \sqrt{1-|b|^2}c + \frac{(1 - \sqrt{1-|b|^2})bb^T}{|b|^2}kb \nonumber \\
&= \sqrt{1-|b|^2}kb + (1 - \sqrt{1-|b|^2})kb  \nonumber \\
&= kb = c .\label{Eq4.7}
\end{align}

By now, if we can verify that $A$ and $T$ are commute, i.e.
\begin{equation} \label{Eq4.8}
AT = TA,
\end{equation}
then by Equations (\ref{Eq4.6}), (\ref{Eq4.7}) and (\ref{Eq4.8}), we get Equation (\ref{Eq4.5}) holds, immediately.

Indeed, we just need to prove $Abb^T = bb^TA$ holds. Since
$$
Abb^T = A \frac{c}{k} \Big(\frac{c}{k}\Big)^T = \Big(\frac{c}{k}- c\Big) \frac{c^T}{k} = \frac{cc^T}{k^2} - \frac{cc^T}{k}
$$
and
$$
bb^TA = \frac{c}{k} \Big(\frac{c}{k}\Big)^TA = \frac{c}{k} \Big(\frac{Ac}{k}\Big)^T = \frac{c}{k} \Big(\frac{c}{k} - c\Big)^T = \frac{cc^T}{k^2} - \frac{cc^T}{k},
$$
where we use the fact that $A$ is a symmetric matrix, so we have $Abb^T = bb^TA$, as desired.
\qed

\subsection{Normality of complex symmetric weighted composition operators on $H^2(\mathbb{B}_{N})$}
We turn next to the problem of identifying the when the class of complex symmetric weighted composition operators mentioned in Theorem \ref{Th4.5} (ii) is normal.
\begin{thm}
Let $\psi(z)=\frac{a_1}{(1-\langle Uz,\overline{a_{0}}\rangle)^{N}}$, and $\varphi(z)=\frac{a_0-AUz}{1-\langle Uz,\overline{a_{0}}\rangle}$, where $a_{1}\neq0, a_{0}\in{\mathbb{B}}_{N}$, $A$ is a symmetric matrix and $U$ is a unitary symmetric matrix such that $\varphi$ is a self map of ${\mathbb{B}}_{N}$.  Then $W_{\psi,\varphi}$ is normal if and only if
$$\overline{UA}AU-\overline{Ua_0}a_0^TU=A\overline{A}-a_0{a_0^*}$$
and
$$\overline{UA}a_0 - \overline{Ua_0} = A\overline{a_0} - a_0.$$
\end{thm}
\textit{Proof.}
 Since $\varphi(z)=\frac{a_0-AUz}{1-\langle Uz,\overline{a_{0}}\rangle}$, the associated matrix of $\varphi$ is
$$
 m_{\varphi}=\left(
  \begin{array}{ccc}
  -AU &a_0\\
   {-a_0^T}U& 1\\
  \end{array}
\right)
$$
and the  adjoint map $\sigma=\sigma_\varphi$ is defined by
 $$\sigma(z)=\frac{(-AU)^*z+\overline{Ua_0}}{1+\langle z,-a_{0}\rangle}$$
 with the associated matrix of $\sigma$ is

 $$
 m_{\sigma}=\left(
  \begin{array}{ccc}
  -(AU)^* &\overline{Ua_0}\\
   {-a_0}^*& 1\\
  \end{array}
\right).
$$

 Therefore we have  $|\varphi(0)|=|a_0|=|Ua_0|=|\sigma(0)|$, $ \psi(z)=a_1K_{\overline{Ua_0}}(z)=a_1K_{\sigma(0)}(z)$, \   by  [\citealp{L}, Proposition 4.6], we see that $W_{\psi,\varphi}$ is normal if and only if $\varphi\circ\sigma=\sigma\circ\varphi$. Since the multiple of $m_{\varphi\circ\sigma}$ give the same linear fractional map, we must have $W_{\psi,\varphi}$ is normal if and only if $m_{\varphi\circ\sigma}=k m_{\sigma\circ\varphi}$ for some $k\neq0$.

 A calculation shows that
 \begin{equation}\label{Eq4.10}
\overline{UA}AU-\overline{Ua_0}a_0^TU=k(A\overline{A}-a_0{a_0^*}),
\end{equation}
\begin{equation}\label{Eq4.11}
-\overline{UA}a_0+\overline{Ua_0}=k(-A\overline{a_0}+a_0),
\end{equation}
\begin{equation}\label{Eq4.12}
-\overline{a_0}^Ta_0+1=k(-a_0^T\overline{a_0}+1).
\end{equation}
Because $|a_0|=|\varphi(0)|\neq 1$, by  Equation (\ref{Eq4.12}), we have $k=1$. Then combining this with  Equation (\ref{Eq4.10}) and (\ref{Eq4.11}), we have $W_{\psi,\varphi}$ is normal if and only if  $\overline{UA}AU-\overline{Ua_0}a_0^TU=A\overline{A}-a_0{a_0^*}$  and $\overline{UA}a_0 - \overline{Ua_0} = A\overline{a_0} - a_0$, as desired.
\qed


\section{Introduction}\label{Introduccion}

The braid group of $n$ strings ${\mathbb B}_n$, is defined by generators
and relations as follows
$$
{\mathbb B}_{n}=<\tau_{1}, \dots,\tau_{n-1}>_{/\sim}
$$

$$
\sim=\{ \tau_k\tau_j=\tau_j\tau_k, \textrm{ if }|k- j|>1;  \ \
        \tau_k\tau_{k+1}\tau_k=\tau_{k+1}\tau_{k}\tau_{k+1} \ \ 1\leq k\leq n-2\ \}
$$

We will consider finite dimensional complex representations of
${\mathbb B}_n$; that is pairs $(\phi,V)$ where
$$
\phi : {\mathbb B}_n\rightarrow \Aut(V)
$$
is a morphism of groups and $V$ is a complex vector space of
finite dimension.

In this paper, we will construct a family of finite dimensional complex representations of ${\mathbb B}_n$ that contains the standard representations. Moreover, we will give necessary conditions for a member of this family to be irreducible. In this way, we can find explicit families of irreducible representations. In particular, we will define a subfamily of irreducible representations $(\phi_m, V_m)$, $1\leq m<n$, where $\dim V_m=\left( \begin{smallmatrix} n\\\noalign{\medskip}m \end{smallmatrix}\right)$ and the corank of $\phi_m$ is equal to $\frac{2(n-2)!}{(m-1)!(n-m-1)!}$.

This family of representations can be useful in the progress of classification of the irreducible representations of ${\mathbb B}_n$. As long as we known, there are only few contributions in this sense, some known results are the following ones. Formanek classified all the irreducible representations of ${\mathbb B}_n$ of dimension lower than $n$ \cite{F}. Sysoeva did it for dimension equal to $n$ \cite{S}. Larsen and Rowell gave some results for unitary representation of ${\mathbb B}_n$ of dimension multiples of $n$. In particular, they prove there are not irreducible representations of dimension $n+1$. Levaillant proved when the Lawrence-Krammer representation is irreducible and when it is reducible \cite{L}.

\section{Construction and Principal Theorems}

In this section, we will construct a family of representations of ${\mathbb B}_n$ that we believe to be new,
and we will obtain a subfamily of irreducible representations.

We choose $n$ non negative integers $z_1, z_2, \dots, z_n$, not necessarily different. Let $X$ be the set of all
the possible $n$-tuples obtained by permutation of the coordinates of the fixed $n$-tuple $(z_1, z_2, \dots, z_n)$. For example, if the $z_i$ are all different, then the cardinality of $X$ is $n!$. Explicitly, if $n=3$,
$$
X=\{(z_1, z_2, z_3), (z_1, z_3, z_2), (z_2, z_1, z_3), (z_2, z_3, z_1), (z_3, z_1, z_2), (z_3, z_2, z_1)\}
$$

Or if $z_1=z_2=1$ and $z_i=0$ for all $i=3, \dots, n$, then the cardinality of $X$ is $\left( \begin{smallmatrix}
n\\\noalign{\medskip}2 \end{smallmatrix}\right)=\frac{n(n-1)}{2}$. Explicitly, for $n=3$
$$
X=\{(1,1,0), (1,0,1), (0,1,1)\}
$$

Let $V$ be a complex vector space with orthonormal basis $\beta=\{v_x : x\in X\}$. Then the dimension of $V$ is
the cardinality of $X$.

We define $\phi:{\mathbb B}_n \rightarrow \Aut (V)$, such that
$$
\phi(\tau_k)(v_x)= q_{x_k, x_{k+1}} v_{\sigma_k(x)}
$$
where $q_{x_k, x_{k+1}}$ is a non-zero complex number that depends on $x=(x_1, \dots, x_n)$, but, it only depends on the places
$k$ and $k+1$ of $x$; and
$$
\sigma_k(x_1, \dots, x_n)=(x_1, \dots,x_{k-1}, x_{k+1}, x_k, x_{k+2}, \dots, x_n)
$$
With this notations, we have the following theorem,

\begin{theorem}
$(\phi, V)$ is a representation of the braid group ${\mathbb B}_n$.
\end{theorem}
\begin{proof}
We need to check that $\phi(\tau_k)$ satisfy the relations of the braid group. We have for $j\neq k-1, k,
k+1$ that
$$
\phi(\tau_k)\phi(\tau_j)(v_x)= \phi(\tau_k)(q_{x_j, x_{j+1}} v_{\sigma_j(x)})=q_{x_j, x_{j+1}} q_{x_k, x_{k+1}}
v_{\sigma_k \sigma_j(x)}
$$
On the other hand
$$
\phi(\tau_j)\phi(\tau_k)(v_x)= \phi(\tau_j)(q_{x_k, x_{k+1}} v_{\sigma_k(x)})=q_{x_k, x_{k+1}} q_{x_j, x_{j+1}}
v_{\sigma_j \sigma_k(x)}
$$
As $\sigma_k \sigma_j(x)=\sigma_j \sigma_k(x)$, if $|j-k|>1$, then $\phi(\tau_k)\phi(\tau_j)=\phi(\tau_k)\phi(\tau_j)$ if $|j-k|>1$.

In the same way, we have
$$
\begin{aligned}
\phi(\tau_k)\phi(\tau_{k+1})\phi(\tau_k)(v_x)&= \phi(\tau_k)\phi(\tau_{k+1})(q_{x_k, x_{k+1}}
v_{\sigma_k(x)})\\
&=\phi(\tau_k)(q_{x_k, x_{k+1}} q_{x_{k}, x_{k+2}} v_{\sigma_{k+1} \sigma_k(x)})\\
&= q_{x_k, x_{k+1}} q_{x_{k}, x_{k+2}} q_{x_{k+1}, x_{k+2}} v_{\sigma_k \sigma_{k+1} \sigma_k(x)}
\end{aligned}
$$
Similarly,
$$
\begin{aligned}
\phi(\tau_{k+1})\phi(\tau_k)\phi(\tau_{k+1})(v_x)&= \phi(\tau_{k+1})\phi(\tau_k)(q_{x_{k+1}, x_{k+2}}
v_{\sigma_{k+1}(x)})\\
&=\phi(\tau_{k+1})(q_{x_{k+1}, x_{k+2}} q_{x_{k}, x_{k+2}} v_{\sigma_k \sigma_{k+1}(x)})\\
&= q_{x_{k+1}, x_{k+2}} q_{x_{k}, x_{k+2}} q_{x_k, x_{k+1}} v_{\sigma_{k+1} \sigma_k \sigma_{k+1}(x)}
\end{aligned}
$$
As $\sigma_k \sigma_{k+1} \sigma_k(x)=\sigma_{k+1} \sigma_k \sigma_{k+1}(x)$, for all $k$ and $x\in X$, then
$\phi(\tau_k)\phi(\tau_{k+1})\phi(\tau_k)=\phi(\tau_{k+1})\phi(\tau_k)\phi(\tau_{k+1})$ for all $k$.
\end{proof}

As $\beta$ is an orthonormal basis, we have that,
$$
<\phi(\tau_k)v_y, v_x> = <q_{y_k, y_{k+1}}v_{\sigma_k(y)}, v_x> = <v_y, \overline{q_{x_{k+1}, x_k}} v_{\sigma_k(x)}>
$$
then,
$$
(\phi(\tau_k))^*(v_x)= \overline{q_{x_{k+1}, x_k}} v_{\sigma_k(x)}
$$
therefore, $\phi(\tau_k)$ is self-adjoint if and only if $q_{x_{k+1}, x_k}=\overline{q_{x_k, x_{k+1}}}$ for all $x\in
X$. In particular, if $x_k=x_{k+1}$ then $q_{x_k, x_{k+1}}$ is a real number. In the same way, $\phi(\tau_k)$ is
unitary if and only if $|q_{x_k, x_{k+1}}|^2=1$ for all $x\in X$.

Now, we will give a subfamily of irreducible representations.

\begin{theorem} \label{irreducible}
If $\phi(\tau_k)$ is a self-adjoint operator for all
$k$, and for any pair $x,y\in X$, there exists $j$, $1\leq j\leq n-1$, such that $|q_{x_j, x_{j+1}}|^2\neq |q_{y_j, y_{j+1}}|^2$,
then $(\phi, V)$ is an irreducible representation of the
braid group ${\mathbb B}_n$.
\end{theorem}
\begin{proof}
Let $W\subset V$ be a non-zero invariant subspace. It is enough to prove that $W$
contains one of the basis vectors
$v_x$. Indeed, given $y\in X$, there exists a permutation $\sigma$ of the
coordinates of $x$, that sends $x$ to $y$. This happens because the elements of $X$ are $n$-tuples obtained by permutation of the coordinates of the fixed $n$-tuple $(z_1, \dots, z_n)$. Suppose that $\sigma=\sigma_{i_1} \dots \sigma_{i_l}$, then
$\tau:=\tau_{i_1} \dots, \tau_{i_l}$ satisfies that $\phi(\tau)(v_x)=\lambda v_y$, for some
non-zero complex number $\lambda$. Then $W$ contains $v_y$ and therefore, $W$ contains the basis
$\beta=\{v_x : x\in X\}$.

As $\phi(\tau_k)$ is a self-adjoint operator, it commutes with
$P_W$, the orthogonal projection over the subspace $W$. Therefore,
$(\phi(\tau_k))^2$ commute with $P_W$. On the other hand, note
that $(\phi(\tau_k))^2(v_x)=|q_{x_k, x_{k+1}}|^2 v_x$, hence,
$(\phi(\tau_k))^2$ is diagonal in the basis $\beta=\{v_x : x\in
X\}$. Then, the matrix of $P_W$ has at least the same blocks than
$(\phi(\tau_k))^2$ for all $k$, $1\leq k \leq n-1$.

If for some $k$, the matrix of $(\phi(\tau_k))^2$ has one block of
size $1\times 1$, then the matrix of $P_W$ has one block of size
$1 \times 1$. In other words, there exists $x\in X$ such that $v_x$ is an
eigenvector. If the eigenvalue associated to $v_x$ is non-zero,
then $v_x\in W$.

It rest to see that the matrix of $(\phi(\tau_k))^2$ has all its blocks
of size $1\times 1$. By hypothesis, for each pair of vectors in the basis $\beta$,  $v_x$ and
$v_y$, there exists $k$, $1\leq k \leq n-1$, such that $|q_{x_k, x_{k+1}}|^2\neq |q_{y_k, y_{k+1}}|^2$.
Fix any order in $X$ and let $x$ and $y$ the first and second element
of $X$. Then there exists $k$ such that $v_x$ and $v_y$
are eigenvectors of $(\phi(\tau_k))^2$ of different eigenvalue. Hence $(\phi(\tau_k))^2$ has the first block of size $1\times 1$. As $(\phi(\tau_j))^2$ commute with $(\phi(\tau_k))^2$ for all $j$, $(\phi(\tau_j))^2$ also has this property.

By induction, suppose that for all $j$ $(\phi(\tau_j))^2$ has its $r-1$ first blocks of size $1\times 1$. Let $x', y'$ the elements $r$ and $r+1$ of $X$, then there exists $k'$ such that $v_{x'}$ and $v_{y'}$ are eigenvectors of $(\phi(\tau_{k'}))^2$ of different eigenvalue. Hence, $(\phi(\tau_{k'}))^2$ has the $r$ block of size $1\times 1$. Therefore $(\phi(\tau_j))^2$ too because it commute with $(\phi(\tau_{k'}))^2$, for all $j$. Then we obtain that all the blocks are of size $1\times 1$.

\end{proof}

Note that if the numbers $q_{x_k, x_{k+1}}$ are all equal and $|X|>1$, then $\phi$ is not irreducible because the
subspace $W$, generated by the vector $v=\sum_{x\in X} v_x$, is an invariant subspace.

\subsection{Examples}
We are going to compute some explicit examples of this family of representations. We will show that the
standard representation (\cite{S}, \cite{TYM}) is a member of this family.

\subsubsection{\it{Standard Representation}}
Let $z_1=1$ and $z_j=0$ for all $j=2, \dots, n$. Then the cardinality of $X$ is $n$ and $\dim V=n$ too. For each
$x\in X$, let $q_{x_k, x_{k+1}}=1 +(t-1) x_{k+1}$, where $t\neq 0,1$ is a complex number. Therefore $\phi:{\mathbb B}_n
\rightarrow \Aut(V)$, given by $\phi(\tau_k)v_x=q_{x_k, x_{k+1}} v_{\sigma_k(x)}$, is equivalent to the standard
representation $\rho$, given by
$$
\rho(\tau_k)= \left(\begin{array}{ccccccc}
          \  1      &\  \   &\  \      &\  \   &\  \   &\  \        &\  \  \\
          \  \      &\  1   &\  \      &\  \   &\  \   &\  \        &\  \  \\
          \  \      &\  \   &\  \ddots &\  \   &\  \   &\  \        &\  \  \\
          \  \      &\  \   &\  \      &\  0   &\  t   &\  \        &\  \  \\
          \  \      &\  \   &\  \      &\  1   &\  0   &\  \        &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  \ddots   &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  \        &\  1
\end{array}\right)
$$ where $t$ is in the place $(k, k+1)$.
In fact, if $\{\beta_j: j=1, \dots, n\}$ is the canonical basis of ${\mathbb C}^n$, and if $x_j$ is the element of $X$
with $1$ in the place $j$ and zero elsewhere, define
$$
\begin{array}{rcl}
 \alpha :{\mathbb C}^n & \rightarrow &V  \\
      \beta_j & \mapsto     &v_{x_j}
\end{array}
$$

Then $\alpha (\rho(\tau_k) (\beta_j))=\phi(\tau_k)(\alpha(\beta_j))$ for all $j=1, \dots, n$. Hence the
representations are equivalent.

\subsubsection{Example}

Let $z_1, \dots, z_n\in \{0, 1\}$, such that $z_1=z_2=\dots =z_m=1$ and $z_{m+1}=\dots =z_n=0$. Then the cardinality
of $X$ is $\left( \begin{smallmatrix} n\\\noalign{\medskip}m \end{smallmatrix}\right)=\frac{n!}{m! (n-m)!}$.
If $V_m$ is the vector space with basis $\beta_m=\{v_x : x\in X\}$, then $\dim V_m=\frac{n!}{m! (n-m)!}$.

For each $x:=(x_1, \dots, x_n)\in X$, let
$$q_{x_k, x_{k+1}}=\left
\{\begin{array}{ll}
    1       &\text{    if   } x_k=x_{k+1}   \\
    t       &\text{    if   } x_k\neq x_{k+1}
\end{array}\right.
$$ where $t$ is a real number, $t\neq 0, 1, -1$.

We define $\phi_m: {\mathbb B}_n\rightarrow \Aut (V_m)$, given by
$$
\phi_m(\tau_k)v_x=q_{x_k, x_{k+1}} v_{\sigma_k(x)}
$$
For example, fixing the lexicographic order in $X$, if $n=5$ and $m=3$, then $\dim V_m=10$, the ordered basis is
$$
\begin{aligned}
\beta:=\{&v_{(0,0,1,1,1)}, v_{(0,1,0,1,1)}, v_{(0,1,1,0,1)},v_{(0,1,1,1,0)}, v_{(1,0,0,1,1)}, \\
     &v_{(1,0,1,0,1)},v_{(1,0,1,1,0)},v_{(1,1,0,0,1)}, v_{(1,1,0,1,0)}, v_{(1,1,1,0,0)}\}
\end{aligned}
$$
and the matrices in this basis are
$$
\phi_3(\tau_1)= \left(\begin{array}{cccccccccc}
          \  1      &\  \   &\  \      &\  \   &\  \   &\  \        &\  \   &\  \   &\  \    &\  \  \\
          \  \      &\  0   &\  \      &\  \   &\  t   &\  \        &\  \   &\  \   &\  \    &\  \  \\
          \ \       &\  0   &\  0      &\  \   &\  0   &\  t        &\  \   &\  \   &\  \    &\  \  \\
          \ \       &\  0   &\  0      &\  0   &\  0   &\  0        &\  t   &\  \   &\  \    &\  \  \\
          \ \       &\  t   &\  0      &\  0   &\  0   &\  0        &\  0   &\  \   &\  \    &\  \  \\
          \ \       &\  \   &\  t      &\  0   &\  \   &\  0        &\  0   &\  \   &\  \    &\  \  \\
          \ \       &\  \   &\  \      &\  t   &\  \   &\  \        &\  0   &\  \   &\  \    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  \        &\  \   &\  1   &\  \    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  \        &\  \   &\  \   &\  1    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  \        &\  \   &\  \   &\  \    &\  1
\end{array}\right)
$$
$$
\phi_3(\tau_2)= \left(\begin{array}{cccccccccc}
          \  0      &\  t   &\  \      &\  \   &\  \   &\  \        &\  \   &\  \   &\  \    &\  \  \\
          \  t      &\  0   &\  \      &\  \   &\  \   &\  \        &\  \   &\  \   &\  \    &\  \  \\
          \  \      &\  \   &\  1      &\  \   &\  \   &\  \        &\  \   &\  \   &\  \    &\  \  \\
          \  \      &\  \   &\  \      &\  1   &\  \   &\  \        &\  \   &\  \   &\  \    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  1   &\  \        &\  \   &\  \   &\  \    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  0        &\  \   &\  t   &\  \    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  0        &\  0   &\  0   &\  t    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  t        &\  0   &\  0   &\  0    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  \        &\  t   &\  \   &\  0    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  \        &\  \   &\  \   &\  \    &\  1
\end{array}\right)
$$
$$
\phi_3(\tau_3)= \left(\begin{array}{cccccccccc}
          \ 1      &\ \   &\ \      &\ \   &\ \   &\ \        &\ \   &\ \   &\ \     &\ \   \\
          \ \      &\ 0   &\ t      &\ \   &\ \   &\  \       &\ \   &\ \   &\ \     &\ \   \\
          \  \     &\ t   &\ 0      &\ \   &\ \   &\ \        &\ \   &\ \   &\  \    &\ \   \\
          \  \     &\ \   &\ \      &1 \   &\ \   &\ \        &\ \   &\ \   &\ \     &\ \   \\
          \  \     &\ \   &\ \      &\ \   &\ 0   &\ t        &\ \   &\ \   &\ \     &\ \   \\
          \  \     &\ \   &\ \      &\ \   &\ t   &\ 0        &\ \   &\ \   &\ \     &\ \   \\
          \  \     &\  \  &\ \      &\ \   &\ \   &\ \        &\ 1   &\ \   &\ \     &\ \    \\
          \  \     &\  \  &\ \      &\ \   &\ \   &\ \        &\ \   &\ 1   &\ \     &\ \    \\
          \  \     &\  \  &\ \      &\ \   &\ \   &\ \        &\ \   &\ \   &\  0    &\  t  \\
          \  \     &\ \   &\ \      &\ \   &\ \   &\ \        &\ \   &\ \   &\  t    &\  0
\end{array}\right)
$$
$$
\phi_3(\tau_4)= \left(\begin{array}{cccccccccc}
          \  1      &\  \   &\  \      &\  \   &\  \   &\  \        &\  \   &\   &\  \    &\  \  \\
          \  \      &\  1   &\  \      &\  \   &\  \   &\  \        &\  \   &\   &\  \    &\  \  \\
          \  \      &\  \   &\  0      &\  t   &\  \   &\  \        &\  \   &\   &\  \    &\  \  \\
          \  \      &\  \   &\  t      &\  0   &\  \   &\  \        &\  \   &\   &\  \    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  1   &\  \        &\  \   &\   &\  \    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  0        &\ \  t   &\   &\  \    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  t        &\ \  0   &\   &\  \    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  \        &\  \   &0   &\  t    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  \        &\  \   &t   &\  0    &\  \  \\
          \  \      &\  \   &\  \      &\  \   &\  \   &\  \        &\  \   &\  \   &\    &\  1
\end{array}\right)
$$

With this notation, we have the following results,

\begin{theorem}
Let $n>2$, then $(\phi_m, V_m)$ is an irreducible representation of ${\mathbb B}_n$, for all $1\leq m<n$.
\end{theorem}
\begin{proof}
We analyze two cases, $n\neq 2m$ and $n=2m$. Suppose that $n\neq 2m$. Let $x\neq y\in X$, then
there exists $j$, $1\leq j\leq n$, such that $x_j\neq y_j$. If $j>1$, we may suppose that
$x_{j-1}=y_{j-1}$,  then $q_{x_{j-1}, x_j}\neq q_{y_{j-1}, y_j}$, therefore $|q_{x_{j-1}, x_j}|^2\neq |q_{y_{j-1}, y_j}|^2$. If $j=1$, and $n\neq 2m$, there exists $l=2, \dots, n$ such that $x_{l-1}\neq y_{l-1}$ and $x_l=y_l$,  then $|q_{x_{l-1}, x_l}|^2\neq |q_{y_{l-1}, y_l}|^2$. Then, by theorem \ref{irreducible}, $\phi_m$ is an irreducible representation.

Note that if $n=2m$, $x_0=(1, \dots,1,0,\dots,0)$ and $y_0=(0, \dots, 0,1,\dots, 1)$ satisfy $x_0\neq y_0$ but $q_{x_{j-1}, x_j}= q_{y_{j-1}, y_j}$ for all $j$. So, we can not use theorem \ref{irreducible}. But in the proof of the theorem, we really use that $x$ and $y$ are consecutive in some order. Considering the lexicographic order, $x_0$ and $y_0$ are not consecutive. In general, for each $x\in X$, there exists $y_x\in X$ such that $q_{x_j, x_{j+1}}= q_{y_j, y_{j+1}}$ for all $j=1, \dots, n-1$. We define $y_x$ changed in $x$ the zeros by ones and the ones by zeros. For example, if $x=(1,0,0,1,0,1)$, then $y_x=(0,1,1,0,1,0)$. However, only $x=(0, 1, \dots, 1, 0, \dots,0)$ satisfies that $y_x$ is consecutive to $x$. Therefore
$P_W$, the projection on the invariant subspace $W$, has its blocks $1\times 1$, except the block $2\times 2$ associated to
$\{v_x, v_{y_x}\}$. If some block $1\times 1$ of $P_W$ is non-zero, then $P_W$ contains some $v_x'$ of the basis $\beta_m$. Hence $W=V_m$.
On the other case, $W\subseteq \{v_x,v_{y_x}\}$. If the equality holds, $v_x\in W$ and $W=V_m$. Suppose that $W$ is generated by
$v=a v_x+ b v_{y_x}$, with $a,b\neq 0$. But $\phi_m(\tau_1) v= t(a v_{\sigma_1(x)} + b v_{\sigma_1(y_x)})$, with $\sigma_1(x)\neq x$, $\sigma_1(y_x)\neq y_x$ and $\sigma_1(x)\neq y_x$ (if $n>2$). Therefore $\phi(\tau_1)v\neq \lambda v$, for all $\lambda \in {\mathbb C}$.
This is a contradiction because $W$ is an invariant subspace.
\end{proof}

The {\it{corank }} of a finite dimensional representation $\phi$ of ${\mathbb B}_n$ is the rank of $(\phi(\tau_k)-1)$.
This number does not depend on $k$ because all the $\tau_k$ are conjugate to each other (see p. 655 of \cite{C}).

\begin{theorem}
If $n>2$ and $1\leq m <n$, then $(\phi_m, V_m)$ is an irreducible representation
of dimension $\left( \begin{smallmatrix} n\\\noalign{\medskip}m \end{smallmatrix}\right)$ and corank $\frac{2(n-2)!}{(m-1)!(n-m-1)!}$.
\end{theorem}
\begin{proof}
By theorem before, $(\phi_m, V_m)$ is an irreducible representation. The dimension of $\phi$ is the cardinality of $X$, then

$$
\dim V_m= \left( \begin{smallmatrix} n\\\noalign{\medskip}m \end{smallmatrix}\right)=\frac{n!}{m! (n-m)!}
$$

 We compute the corank of $\phi_m$. Let $x\in X$ such that $\sigma_k(x)= x$, then $x_k=x_{k+1}$ and $q_{x_k, x_{k+1}}=1$.
Therefore $\phi_m(\tau_k)(v_x) =v_x$. Hence the corank of $\phi_m$ is equal to the cardinality of $Y=\{x\in X : \sigma_k(x)\neq x
\}$. But it is equal to the cardinality of $X$ minus the cardinality of $\{x\in X : x_k=x_{k+1}=0 \text{  or   }
x_k=x_{k+1}=1 \}$. Therefore
$$
\begin{aligned}
cork(\phi_m)=rk(\phi_m(\tau_k) -1)&=\frac{n!}{m!(n-m)!}-\frac{(n-2)!}{m! (n-m-2)!}-\frac{(n-2)!}{(m-2)!(n-m)!}\\
                                  &=\frac{2(n-2)!}{(m-1)!(n-m-1)!}
\end{aligned}
$$
\end{proof}

In the example $n=5$ and $m=3$, we have that $cork(\phi_m)=6$.

Note that if $m=1$, the dimension of $\phi_m$ is $n$ and the corank is $2$. Therefore $\phi_1$ is equivalent to the
 standard representation, because this is the unique irreducible representations of ${\mathbb B}_n$ of dimension $n$ \cite{S}.

\section*{Acknowledgment}
The authors thanks to Aroldo Kaplan for his helpful comments.


\section{\label{sec:level1}First-level heading}
In the 1980s, studying the effect of the volume cap-
ture of relativistic protons into the channeling regime,
Taratin and Vorobiev [4, 5] demonstrated the possibility
of volume reflection, i.e., the coherent small-angle scattering of particles at angle $\theta \; \tilde{<} \; 2\theta_L$ ($\theta_L$
 is the Lindhard critical angle) to the side opposite to the bending of the
crystal. Recent experiments reported in [1–3] confirm
the presence of this effect for 1-, 70-, and 400-GeV proton beams in a Si crystal. The conclusions made in [4,
5] were based primarily on the numerical simulation. In
view of this circumstance, the aim of this work is to
derive analytical expressions for the deflection function
of relativistic particles. At first sight, the perturbation
theory in the potential can be applied at relativistic
energies and weak crystal potential [$U(r)\approx 10-100$]. However, the relativistic generalization of the known classical formula for small-angle scattering in
the central field [6],
\[
\chi = -b\int^{\infty}_{b}\frac{d\phi}{d r} \frac{ d r}{\sqrt{r^2-b^2}},
\]
where $b$  is the impact parameter and  $\phi(r)=\frac{2U(r) E}{p_{\infty}^2 c^2}$, $U(r), E, p_{\infty}$  are the centrally
symmetric continuous potential of bent planes, total
energy, and particle momentum at infinity, respectively,
is inapplicable for the entire range of impact parameters. Indeed, the above formula is the first nonzero term of the expansion of the classical deflection function
\begin{eqnarray}
\chi(b)=\pi-2b\int^{\infty}_{r_{o}}\frac{d r }{r \sqrt{r^2[1-\phi(r)]-b^2}},
\label{deflection_function}
\end{eqnarray}
in the power series in the “effective” interaction potential  $\phi(r)$. The crystal interaction potential  $U(r)$ is the
sum of the potentials of individual bent planes concentrically located in the radial direction with period $d$.  It has no singularities (i.e., is bounded in magnitude) and
is localized in a narrow ring region at distances $R-N d <r< R$
 (where the crystal thickness $N d << R$ and $N$
 is the number of planes). In this region, $U(r)>0$ and $U(r)< 0$ for the positively and negatively charged scattered particles, respectively. Beyond the ring region, it
vanishes rapidly. The perturbation theory in the interaction potential is obviously well applicable if the impact
parameter satisfies the inequality $b < (R-Nd)$.  In this
case, the scattering area localized in the potential range
is far from the turning point  $r_o$  determined from the
relation  $b=r_o \sqrt{1-\phi(r_o)}$  and the root singularity of the
turning point does not contribute to integral (1). In the
general case, it can be shown [7, 8] that the condition of
the convergence of the power series of $\phi$
 is a monotonic
increase in the function $u(r) = r\sqrt{1-\phi(r)}$ (e.i.  $u(r)^{'} > 0$) in the $r$
 regions substantial for integral (1). Such a
monotonicity is achieved if the energy and momentum
of the relativistic particle satisfy the inequality
\begin{eqnarray}
\frac{p_{\infty}^2c^2}{2E}> U(r)+\frac{r}{2} U(r)^{'}.
\label{OrbittingCondition}
\end{eqnarray}
The derivation of this condition is omitted, because it
was given in the Appendix in [8], but the nonrelativistic
case was considered in that work. Taking into account
that the inequality  $U(r)<< r U(r)^{'}$ is satisfied for large
distances  $r\approx R$, relation (2) is transformed to the known
Tsyganov criterion [9] [$R < \frac{p_{\infty}^2c^2}{E U(r)^{'}}$] of the disappearance of channeling in a strongly bent crystal. Thus,  the perturbation theory is obviously applicable only for
strongly bent crystals, where channeling is absent. It is
interesting that the nonrelativistic variant [8] of condition (2) corresponds to the criterion of the absence of
the so-called spiral scattering appearing in small-energy chemical reactions [10].
For this reason, the exact solution of the problem of
the relativistic scattering on the model potential of the
periodic system of rectangular rings
\begin{eqnarray}
U(r)= U_{o} \left\{
\begin{array}{ll}
	1,&  R-i d-a < r < R-i d,  \\
	0,&  R-(i+1)d < r < R-i d-a,\\
	0,&	 r < R-Nd, r > R,
\end{array} \right.
\label{Potential_M}
\end{eqnarray}
where $i=0,1,...,(N-1)$ and $a$ and $d$ is the thickness
of a single plane and the interplanar distance, is considered below. With the use of the representation
\begin{eqnarray}
\frac{\pi}{2}=b\int^{\infty}_{r_{o}}\frac{d r }{r \sqrt{r^2[1-\phi(r_{o})]-b^2}},
\end{eqnarray}
particle deflection function (1) can be rewritten in the
form
\begin{eqnarray}
2b\int^{\infty}_{r_{o}}\frac{d r }{r}(\frac{1}{ \sqrt{r^2[1-\phi(r_o)]-b^2}}-\frac{1}{ \sqrt{r^2[1-\phi(r)]-b^2}}).
\label{deflection_functionM1}
\end{eqnarray}
Let us consider the range of small impact parameters,
 $b < R-Nd$, when particles penetrate into the inner
region of the potential ring and twice intersect each
plane. This geometry is not used in the experiments,
because it requires very thin crystals. However, such an
approach to solve the problem makes it possible to easily calculate the exact deflection function on potential
(3) for all impact parameters and, moreover, to demonstrate an interesting property of the scattering on potentials with an empty core. This property does not seemingly attract particular attention. In this case, the turning point is located in the inner region of the potential
ring, where $\phi(r_o)=0$, and, as easily seen from the equation  $b=r_o \sqrt{1-\phi(r_o)}$
, the shortest distance from the
center is $r_o=b$ (see Figs. 1C and 1D).

\begin{figure}
	\centering
		\includegraphics{ScatteringRing3.eps}
	\caption{ Scattering of the charged particles on the ring potential: (A) reflection,  $\chi>0$, and (B) refraction,  $\chi<0$. The
effect of the empty core for the (C) positively and (D) negatively charged relativistic particles.}
	\label{fig:ScatteringRing}
\end{figure}

The deflection function for such impact parameters
has the form
\begin{eqnarray}
\chi(b)=2b\int^{\infty}_{b}\frac{d r }{r}(\frac{1}{ \sqrt{r^2-b^2}}-\frac{1}{ \sqrt{r^2[1-\phi(r)]-b^2}}).
\label{deflection_functionM2}
\end{eqnarray}
Let us consider an arbitrary positive centrally symmetric potential $\phi(r)\geq 0$. Then, the inequality $\frac{1}{ \sqrt{r^2-b^2}} \leq \frac{1}{ \sqrt{r^2[1-\phi(r)]-b^2}}$
 is always valid in the integrand in Eq. (6). As immediately follows from this inequality,
the scattering angles for impact parameters  $0 < b\; \tilde{<} \; (R-Nd)$ are always negative, $\chi < 0$ for any form of the
positive potential $\phi(r)$;  i.e., an arbitrary positive potential is attractive. On the contrary, an arbitrary negative potential, $\phi(r)\leq 0$, is repulsive. This seeming paradox
is called the empty core effect. It is schematically
shown in Figs. 1C and 1D and in Figs. 2A–2D for the scattering on the rectangular potential and can be
treated in various ways, one of which is as follows. The
integral action of the positive potential deflects a particle passing through it to the left from the initial direction (see Fig. 1A); i.e., the particle is reflected. The
potential in the core is absent and, according to the conservation laws, the absolute value of the momentum
and angular momentum should coincide with the
respective initial values. Therefore, the particle trajectory in the core should touch a circle with the radius
equal to the impact parameter. This is possible only if
the particle intersecting the inner boundary of the
potential is deviated to the right from the initial direction. The total deflection angle is equal to the doubled
angle to the turning point. Therefore, the total rotation
is clockwise; i.e., the particle is attracted to the center.
The opposite situation occurs for the negative potential.

Let us introduce the convenient notation
\begin{eqnarray}
\Phi = 1-\phi_o, \; \; \phi_o=\frac{2U_{o} E}{p_{\infty}^2 c^2},\; \;
\hat{r}=\frac{r}{R}, \; \;\hat{b}=\frac{b}{R}, \; \; \hat{a}=\frac{a}{R},  \nonumber\\
\hat{d}=\frac{d}{R}, \; \; \hat{b}_{i} =\frac{\hat{b}}{1-i \hat{d}},\; \; \hat{b}_{a} =\frac{\hat{b}}{1-\hat{a}}, \; \; \hat{b}_{ia}=\frac{\hat{b}}{1-\hat{a}-i \hat{d}}.
\label{definitions}
\end{eqnarray}
For potential (3), the scattering problem is solved
exactly and deflection function (6) for b < R – Nd is represented in the form of the sum,
\begin{eqnarray}
\chi(b)=2 \alpha(b)=2 \sum_{i=0}^{N-1} \alpha_i(b),
\label{Deflection_General_Crystal}
\end{eqnarray}
of the integrals over the regions filled with the potential:
\begin{eqnarray}
\alpha_i(b)=\hat{b}\int^{1-i\hat{d}}_{1-i\hat{d}-\hat{a}}\frac{d \hat{r} }{\hat{r}}(\frac{1}{ \sqrt{\hat{r}^2-\hat{b}^2}}-\frac{1}{ \sqrt{\Phi \hat{r}^2-\hat{b}^2}}).
\label{Deflection_i_Ring}
\end{eqnarray}
Integral (9) is easily calculated and, which is most
important, has an analytic continuation valid for any
impact parameter:
\begin{eqnarray}
\alpha_{i}(b) =arcsin(\frac{\hat{b}_{i}(\sqrt{1-\hat{b}^2_{i}}-\sqrt{\Phi-\hat{b}^2_{i}})}{{\sqrt{\Phi}}})-\nonumber\\ arcsin(\frac{\hat{b}_{ia}(\sqrt{1-\hat{b}_{ia}^2}-\sqrt{\Phi-\hat{b}_{ia}^2})}{{\sqrt{\Phi}}}).
\label{DeflectionGen_i}
\end{eqnarray}
Note that only the real parts of the deflection function
are meaningful. Therefore, if the impact parameter $\hat{b}$ is
such that any root in Eq. (10) is imaginary, it should be
rejected. In what follows, an approximation that is not
strictly necessary is used. All below formulas can be
derived without this approximation, but it significantly
simplifies the form and use of all of the expressions, is
accepted. For small angles entering into Eq. (10) and
$\Phi \approx 1$ in the denominators in Eq. (10), Eq. (10) can be
represented in the form
\begin{eqnarray}
\alpha_{i}(b) =\hat{b}_{i}(\sqrt{1-\hat{b}^2_{i}}-\sqrt{\Phi-\hat{b}^2_{i}})-\nonumber\\ \hat{b}_{ia}(\sqrt{1-\hat{b}_{ia}^2}-\sqrt{\Phi-\hat{b}_{ia}^2}).
\label{DeflectionGen_i_app}
\end{eqnarray}
Let us consider the deflection function given by
Eq. (10) for one ring ($N = 1$ and $i = 0$) in more detail:
\begin{eqnarray}
\alpha(b) =arcsin(\frac{\hat{b}(\sqrt{1-\hat{b}^2}-\sqrt{\Phi-\hat{b}^2})}{{\sqrt{\Phi}}})-\nonumber\\ arcsin(\frac{\hat{b}_{a}(\sqrt{1-\hat{b}_{a}^2}-\sqrt{\Phi-\hat{b}_{a}^2})}{{\sqrt{\Phi}}}).
\label{DeflectionFunction_Ring_One}
\end{eqnarray}
If the inner ring radius is equal to zero, then $\hat{a}=1$ and
last two radicals in Eq. (12) become minimal. According to the above rule, they should be rejected. The
resulting exact function describes the scattering of the
particles on a rectangular cylindrical disc ($U_o >0 $) or
well ($U_o <0 $) [6] (see Fig. 2).

\begin{figure}
	\centering
		\includegraphics{DeflectionFunctionF.eps}
	\caption{Deflection function $\alpha(\hat{b})$ for positively charged particles from the relativistic particle at (A)  $\hat{a} > \phi_o/2$ and  (B)
  $\hat{a} < \phi_o/2$ and for negatively charged particles at (C)
 $\hat{a} > |\phi_o|/2$, (D)- $\hat{a} < |\phi_o|/2$. The dotted lines correspond to the
scattering on (A and B) a disc ($U_o > 0$) and (C and D) a well
($U_o < 0$). The region of the empty core effect for positively
charged particles $0 < \hat{b} <\sqrt{\Phi}(1-\hat{a})$ is smaller than the
inner radius $(1-\hat{a})$.}
	\label{fig:DeflectionFunctionF}
\end{figure}

Furthermore, let us analyze the range of impact parameters in the ring region
$(1-N\hat{d})\leq \hat{b}\leq 1$, where    $\hat{b}_{i}$ and $\hat{b}_{ia} \cong 1$
 and can be taken
due to the smallness   $N\hat{d}<<1$ and these factors in
Eq. (11) can also be omitted. In this approximation,
Eq. (12) has the form
\begin{eqnarray}
\alpha(b) =(\sqrt{1-\hat{b}^2}-\sqrt{\Phi-\hat{b}^2})-\nonumber\\ (\sqrt{1-\hat{b}_{a}^2}-\sqrt{\Phi-\hat{b}_{a}^2}).
\label{DeflectionFunction_Ring_app}
\end{eqnarray}

For various $\phi_o$ values ($\phi_o$ is the square of the Lindhard
critical angle with the sign of the charge of the scattered
particle), there are two sequences of critical points that
can pass through the impact parameter  $\hat{b}$  when increasing from 0 to 1. For the positive potential $U_o > 0$ ($\Phi < 1$), the critical points form a sequence $\sqrt{\Phi} (1-\hat{a})<  \sqrt{\Phi} < 1$. Thus, the deflection function for positively charged particles on one ring has the form
\begin{equation}
\alpha(\hat{b})_{+}=
\label{DeflectionFunction_Ring_v1}
\end{equation}
\begin{eqnarray}
\left \{
\begin{array}{ll}
(\sqrt{1-\hat{b}^2}-\sqrt{\Phi-\hat{b}^2})-(\sqrt{1-\hat{b}_{a}^2}-\sqrt{\Phi-\hat{b}_{a}^2}),& \\  for \; \; 0 < \hat{b} <\sqrt{\Phi}(1-\hat{a});\nonumber\\
\sqrt{1-\hat{b}^2}-\sqrt{\Phi-\hat{b}^2}, \; \;for \; \;\sqrt{\Phi}(1-\hat{a}) < \hat{b}< \sqrt{\Phi};\\
\sqrt{1-\hat{b}^2},\; \;for\; \;  \sqrt{\Phi} < \hat{b}< 1;\\
0, \; \;for \; \;1 < \hat{b}.
\end{array} \right.
\end{eqnarray}
For negative potential $U_o < 0$ ($\Phi > 1$), the critical points
form another sequence $(1-\hat{a})<(1-\hat{a})\sqrt{\Phi}<1$. It provides the deflection function for the scattering of negatively charged particles on one ring:
\begin{equation}
\alpha(\hat{b})_{-}=
\label{DeflectionFunction_Ring_v2}
\end{equation}
\begin{eqnarray}
\left \{
\begin{array}{ll}
(\sqrt{1-\hat{b}^2}-\sqrt{\Phi-\hat{b}^2})-(\sqrt{1-\hat{b}_{a}^2}-\sqrt{\Phi-\hat{b}_{a}^2}),& \\  for \; \; 0 < \hat{b} <(1-\hat{a});\\
\sqrt{1-\hat{b}^2}-\sqrt{\Phi-\hat{b}^2}+\sqrt{\Phi-\hat{b}_{a}^2},\nonumber\\for \; \;(1-\hat{a}) < \hat{b} <(1-\hat{a})\sqrt{\Phi};\nonumber\\
\sqrt{1-\hat{b}^2}-\sqrt{\Phi-\hat{b}^2},\; \;for\; \; (1-\hat{a})\sqrt{\Phi} < \hat{b}< 1;\\
0, \; \;for \; \;1 < \hat{b}.
\end{array} \right.
\end{eqnarray}
Figures 2A and 2C show deflection functions (14) and
(15), respectively, for $|\phi_o|/2 < \hat{a}$ . If the potential is sufficiently large, $|\phi_o|/2 > \hat{a}$ , the arc of the reflection of the
positively charged particles from the outer wall of a
bent plane extends to the left and can be larger than the
width of the ring (and the distance between rings if the
system of rings is considered). This effect causes the
reflection of relativistic particles in the crystal. For negatively charged particles, when $|\phi_o|/2 > \hat{a}$, the corresponding reflection from the inner wall of the potential
is shifted to the right and, correspondingly, the impact
parameter region for the refraction of negatively
charged particles is narrowed. However, at least the narrow refraction region always exists. These two variants
are shown in Figs. 2B and 2D. The substitution of  $\hat{b}=\sqrt{\Phi}$,  $\hat{b}=\sqrt{\Phi}(1-\hat{a})$   and $\hat{b}=(1-\hat{a})$, $\hat{b}=1$  into
Eqs. (14) and (15), respectively, provides the maximum
(minimum) deflection angles for the positively and negatively charged particles, respectively:
\begin{eqnarray}
\alpha_{max +} =-\alpha_{min -}=\sqrt{|\phi_o|},& &\nonumber\\
\alpha_{min +} =-\alpha_{max -} =\frac{|\phi_o|}{2\sqrt{2\hat{a}}} -\sqrt{|\phi_o|}.
\label{MaxMin}
\end{eqnarray}
The deflection functions for the system of bent planes
forming the crystal are similarly obtained from Eqs. (8)
and (11). In this case, the summation in Eq. (8) should
be performed from $0$ to $k – 1$, where $k$ is the ordinal
number of the radial period containing the turning
point. The sequences of the critical points mentioned
before Eqs. (14) and (15) and reflection functions given
by Eqs. (14) and (15) refer to the $k$-th period. The formulas for the reflection functions are omitted in this short
paper, but the corresponding plots are presented in
Fig. 3.
Two length parameters,  and $\hat{a}$, $\hat{d}$ ($\hat{a}<\hat{d}$), exist in
the periodic system of bent planes. Hence, three different variants of the curves can exist with (i)   $\hat{d} > \hat{a} > \phi_o/2$, (ii)  $\hat{d}> \phi_o/2 > \hat{a}$, and (iii) $\phi_o/2 >\hat{d}>  \hat{a}$. Figure 3
shows the final result for the reflection function in the first and third variants.

\begin{figure}
	\centering
		\includegraphics{DeflectionFunctionS.eps}
	\caption{Deflection function for the ring crystal and positively
charged particles at (A)  $\hat{a} > \phi_o/2$ and (B) $\hat{d} < \phi_o/2$ and for
negatively charged particles at (C)  $\hat{a} > |\phi_o|/2$ and $\hat{d} < |\phi_o|/2$.}
	\label{fig:DeflectionFunctionS}
\end{figure}

As mentioned above, under the condition
\begin{eqnarray}
\phi_o >\frac{2d}{R},
\label{ConditionOfReflection}
\end{eqnarray}
the refraction regions of positively charged particles
disappear in the pattern of the deflection function (see
Fig. 3B). Since this effect is of an applied interest for
controlling the relativistic particle beams, let us calculate the average reflection angle under these conditions.
For a rough estimate, the extreme ring where the deflection function has the simplest form of Eq. (14) is used.
The average deflection angle in the impact parameter
range $\sqrt{\Phi} < \hat{b}< 1$  is determined by the integral
\begin{eqnarray}
\bar{\alpha}_{+} = \frac{1}{1-\sqrt{\Phi}} \int^{1}_{\sqrt{\Phi}}
\sqrt{1-\hat{b}^2} d\hat{b} \cong \frac{2\sqrt{\phi_o}}{3}.
\label{AverageAnglePReflectLast}
\end{eqnarray}
Thus, the average reflection angle is $\chi_{+}=2\bar{\alpha}_{+}=4\sqrt{\phi_o} / 3$, which is equal to $1.33\cdot \theta_L$. A more accurate estimate of the reflection angle can be obtained by averaging over any inner period of the reflection function. Let
us use the second impact parameter period from the
edge, $(1-\hat{d})\sqrt{\Phi} < \hat{b}< \sqrt{\Phi}$
 (see Fig. 3B).   In this case,
it is unnecessary to calculate the total sum in Eq. (8); it
is sufficient to calculate the reflection function at two
extreme rings. This problem is solved by calculating
two integrals
\begin{eqnarray}
I_{1} = \int^{(1-\hat{a})\sqrt{\Phi}}_{(1-\hat{d})\sqrt{\Phi}}
\Bigl((\sqrt{1-\hat{b}^2}-\sqrt{\Phi-\hat{b}^2})-\nonumber\\
(\sqrt{1-\hat{b}^2_{a}}-\sqrt{\Phi-\hat{b}^2_{a}})+\nonumber\\ \sqrt{1-\hat{b}_{1}^2}\Bigl) d\hat{b},
\label{AverageAnglePI1}
\end{eqnarray}
and
\begin{eqnarray}
I_{2} = \int^{\sqrt{\Phi}}_{(1-\hat{a})\sqrt{\Phi}}
\Bigl(\sqrt{1-\hat{b}^2}-\sqrt{\Phi-\hat{b}^2}\Bigl) d\hat{b},
\label{AverageAnglePI2}
\end{eqnarray}
with subsequent expansion in small parameters $\hat{a},\hat{d}$ and  $\phi_o$. As a result, the average deflection angle
$\bar{\alpha}_{+}=(I_1+I_2)/(\hat{d}\sqrt{\Phi})$ is obtained in the form
\begin{eqnarray}
\bar{\alpha}_{+} = \frac{1}{3\hat{d}}\Bigl(\phi_o^{3/2}+ (2\hat{d}+\phi_o)^{3/2}+(2\hat{d}-2\hat{a})^{3/2}-\nonumber\\-2\sqrt{2}\hat{d}^{3/2}-(2\hat{d}-2\hat{a}+\phi_o)^{3/2}-(2\hat{a}-2\hat{d}+\phi_o)^{3/2}\Bigl).
\label{AverageAnglePReflectF}
\end{eqnarray}

The experiment with 1-GeV protons in a $<\!111\!>$ Si crystal with a bending radius of $R=0.33$m provides an average reflection angle of $236 \mp 6.0$ $\mu$rad [2]. In view
of the data  $\phi_o=\theta_{L}^{2}=0.289\cdot10^{-7}$, $a=0.78 \AA$, $d=3.136 \AA$, Eq. (21) provides $\chi_{+}=2\cdot\bar{\alpha}_{+}=318.8$  $\mu$rad.

Note that rough formula (18) provides the value $\chi_{+}=2\cdot\bar{\alpha}_{+}=226.6$ $\mu$rad  that is closer to the measured value. The
experiment with 70-GeV protons in a  $<\!111\!>$ Si crystal
with a bending radius of $R = 1.7$ m provides an average
reflection angle of $39.5\mp2.0$ $\mu$rad.
In view of the data
$\phi_o=\theta_{L}^{2}=0.58\cdot10^{-9}$, $a=0.78 \AA$, $d=3.136 \AA$,
Eq. (21) provides$\chi_{+}=37.3$ $\mu$rad, which is close to the
experimental value. Formula (18) provides a smaller
angle of 32.0 $\mu$rad.

For the latest CERN experiment for
the reflection of 400-GeV protons [3] in a Si crystal oriented in the  $<\!110\!>$ direction ($R = 18.5$ m,  $\phi_o=\theta_{L}^{2}=0.1132\cdot10^{-9}$, $a=0.48 \AA$, $d=1.92 \AA$) and the  $<\!111\!>$ direction ($R = 11.5 $m, $\phi_o=\theta_{L}^{2}=0.1008\cdot10^{-9}$, $a=0.78 \AA$, $d=3.136 \AA$), Eq. (21) provides $\chi_{+}=2\cdot\bar{\alpha}_{+}=19.0$
and $16.0$  $\mu$rad, respectively.  Rough estimate (18) provides values $14.1$ and $13.3$  $\mu$rad, which is very close to
the experimental values $13.9\mp0.2$ and $13.0$  $\mu$rad, for
the  $<\!110\!>$ and  $<\!111\!>$ orientations, respectively.

Comparison of the theoretical estimates and experimental data shows that estimate formula (18) yields a smaller reflection angle over the entire range of
scanned energies than that assumingly more accurate
formula (21). The experimental data reported in [1, 11]
are sufficiently well reproduced by formula (21), but
the results reported in [2, 3] are closer to estimate (18).
This circumstance requires a more detailed analysis of
the experimental conditions and the accuracy of measurements.

I am grateful to Yu.M. Ivanov for the possibility of
a reading preprint of [11].