text:
\begin{document}
\title{Hybrid Beamforming Structure for Massive MIMO System: Full-connection v.s. Partial-connection}

\author{\IEEEauthorblockN{Zheda Li, Shengqian Han, \textit{Member, IEEE}, Andreas F. Molisch, \textit{Fellow,~IEEE}}
\thanks{Z. Li is with DOCOMO Innovations, Inc., Palo Alto, California 94304, USA (email: zli@docomoinnovations.com). S. Han is with the School of Electronics and Information Engineering, Beihang University, Beijing 100191, China (e-mail:sqhan@buaa.edu.cn). A. F. Molisch is with the Department of Electrical Engineering, University of Southern California, Los Angeles, California 90089, USA (e-mail: molisch@usc.edu).}}

\maketitle

\begin{abstract}
In this article we compare the performance of two typical hyrbid beamforming structures for multiuser massive MIMO systems, i.e., the full- and partial-connection structures. Under the assumption of small angular spread for mmWave channels, given the analog precoder formed towards users and the zero-forcing digital precoder, we develop an explicit upper bound for the analog-and-digital-precoded channel gain of users, based on which the relationship between the two structures is investigated. The analysis results show that the full-connection structure is not always better than the partial-connection structure, and the regimes suitable for each structure are revealed. Simulations are conducted to validate the analysis results.
\end{abstract}

\begin{IEEEkeywords}
	Hybrid beamforming, full-connection, partial-connection, massive MIMO.
\end{IEEEkeywords}

\section{Introduction}
To explore the large available bandwidth in the millimeter (mm)-wave bands~\cite{FCC2016} for the fifth generation (5G) cellular systems, massive multiple-input multiple-output (MIMO) technology is essential since the high
free-space pathloss at those frequencies necessitates large array gains to obtain sufficient signal-to-noise ratio (SNR)~\cite{Marzetta2010,Molisch2017}. By employing dozens or hundreds of antenna elements at the base station (BS), massive MIMO is able to significantly increase the spectral efficiency and simplify signal processing~\cite{Marzetta2010,Molisch2017,Fredrik2013}. However, implementation of such systems suffers from the prohibitive cost and high energy consumption to build a complete radio frequency (RF) chain for each antenna element, especially at mm-wave frequencies. A promising solution to these problems lies in the concept of hybrid beamforming (HB) transceivers, which uses a combination of analog beamformers in the RF domain, together with a smaller number of RF chains. This concept was first proposed in~\cite{Zhang2005} and~\cite{Sundarshan2006}. While formulated originally for MIMO with arbitrary number of antenna elements, the approach is applicable in particular to massive MIMO, and in that context interest in hybrid transceivers has surged over the past years.

Two typical HB structures at the BS, namely full- and partial-connection, are illustrated in Fig.~\ref{F:structure}, where the downlink transmission is assumed while the structure for uplink transmission is
similar. The difference between the two structures lies in the phase-shifter network for the analog beamformer. For a full-connection HB structure as in Fig.~\ref{F:structure}(a), each analog precoder output can be
a linear combination of all RF signals. Complexity reduction can be achieved when
each RF chain can be connected only to a subset of antenna elements, as in Fig.~\ref{F:structure}(b).

\begin{figure}
	\centering
    \subfigure[]
        {
        \includegraphics[width=0.47\textwidth]{HBF_full.pdf}
        }
    \subfigure[]
        {
        \includegraphics[width=0.47\textwidth]{HBF_partial.pdf}
        }

	\caption{Hybrid beamforming structures with (a) full-connection and (b) partial-connection.}\label{F:structure}

\end{figure}

The majority of existing works on hybrid beamformer design considered the full-connection structure, e.g.,~\cite{Zhang2005,Ayach2014,Alkhateeb2015,Sohrab2016,Bogale2016,Ni2015,Sundarshan2006,Ansuman2013,Ansuman2014,Li2017,Li2018a,Li2018,Ioushua2019}, and some considered the partial-connection structure, e.g.,~\cite{partial-Gao2016,parital-Yu2016,partial-Park2017,Majidzadeh2018,Dsouza2018,Zhang2018,Ioushua2019}.

It is commonly accepted that the partial-connection structure enjoys the reduced cost and complexity compared to the full-connection structure with the penalty of certain data rate loss. This is true when the hybrid beamformers for the two structures are optimized under the identical objective function and constraints, as demonstrated in e.g.,~\cite{partial-Gao2016,parital-Yu2016,partial-Park2017,Majidzadeh2018,Dsouza2018,Zhang2018,Ioushua2019}. The reason is obvious, i.e., the analog beamformer for the full-connection structure has larger optimization freedoms (i.e., more variables to be optimized) than that for the partial-connection structure. In practical applications, however, to reduce the processing complexity and signaling overhead for CSI acquisition, suboptimal analog beamformers are usually applied. For example, in the 5G New Radio systems, the analog precoder is a combination of Discrete Fourier Transform (DFT) vectors selected from a predefined codebook by user equipments (UEs)~\cite{38.214}. In such scenarios, it is unclear whether the partial-connection structure always sacrifices the performance or not compared to the full-connection structure.

In this article, we compare the downlink performance of two HB structures for multiuser massive MIMO systems. By assuming a small angular spread for the mm-Wave channels, which leads to the analog precoder at the BS formed towards UEs, we derive an explicit upper bound for the analog-and-digital-precoded channel gain, where the zero-forcing digital precoder is employed. Based on the result, the gain of the full-connection structure over the partial-connection structure is analyzed. The analysis results show that the full-connection structure is not always better, and the regimes suitable for the two structures are revealed.

\section{System model}

\subsection{Signal Model}
Consider the downlink transmission in a single cell with $K$ single-antenna UEs and a BS equipped with $M$ antennas and $l_{\text{BS}}$ RF chains. The baseband digital precoder $\mathbf{F}_\text{d}$ processes the data streams $\mathbf{x}\sim\mathcal{CN}(\mathbf{0}, \mathbf{I})$ to produce $l_{\text{BS}}$ outputs,
which are upconverted to RF and mapped via an analog precoder $\mathbf{F}_\text{a}$ to $M$ antenna elements
for transmission. We consider both the full-connection and partial-connection HB structures for the BS, as shown in Fig.~\ref{F:structure}.

The received signal at the $\text{UE}_i$ is
\begin{align}
y_i&=\mathbf{h}_{i}^\dag\mathbf{F}_\text{a}\mathbf{F}_\text{d}\sqrt{p}\mathbf{x}+n_i
=\mathbf{h}_{i}^\dag\mathbf{F}_\text{a}\mathbf{f}_{\text{d,}i}\sqrt{p}x_i+\sum_{j\neq i} \mathbf{h}_i^\dag\mathbf{F}_\text{a}\mathbf{f}_{\text{d,}j}\sqrt{p}x_j+n_i,
\end{align}
where $\mathbf{h}_i\in\mathbb{C}^{M\times 1}$ is the downlink channel of $\text{UE}_i,  \forall i$, $\mathbf{F}_{\text{a}}\in\mathbb{C}^{M\times l_{\text{BS}}}$ is the analog precoder, $\mathbf{F}_\text{d}=[\mathbf{f}_{\text{d,}1},...,\mathbf{f}_{\text{d,}K}]\in\mathbb{C}^{l_{\text{BS}}\times K}$ indicates the digital precoder, $\mathbf{f}_{\text{d,}i}$ is the digital precoder for $\text{UE}_i$, $n_i\sim\mathcal{CN}(0,\delta_n^2)$ is the complex Gaussian noise at $\text{UE}_i$, and
$p = \frac{P_t}{\|\mathbf{F}_\text{a}\mathbf{F}_\text{d}\|_F^2}$ scales the transmitted signal to satisfy the transmit power constraint of the BS with $P_t$ denoting the total transmit power of the BS and $\|\cdot\|_F$ denoting the Frobenius norm.

Define the analog-preocded effective channel as $\bar{\mathbf{h}}_i\triangleq \mathbf{F}_\text{a}^\dag\mathbf{h}_i, \forall i$, which has a low dimension and can be estimated with low overhead. Assume that the BS knows $\bar{\mathbf{h}}_i, \forall i$, and employs the zero-forcing digital precoding. Then, the digital precoder to serve $\text{UE}_i$ can be expressed as
\begin{align}
\mathbf{f}_{\text{d,}i}=\frac{\Pi(\tilde{\mathbf{H}}_i)\bar{\mathbf{h}}_i}{\|\Pi(\tilde{\mathbf{H}}_i)\bar{\mathbf{h}}_i\|},
\end{align}
where $\tilde{\mathbf{H}}_i\triangleq [\bar{\mathbf{h}}_1,...,\bar{\mathbf{h}}_{i-1},\bar{\mathbf{h}}_{i+1},...,\bar{\mathbf{h}}_K]$, and $\Pi(\tilde{\mathbf{H}}_i)\triangleq \mathbf{I}_{l_\text{BS}}-\tilde{\mathbf{H}}_i(\tilde{\mathbf{H}}_i^\dag \tilde{\mathbf{H}}_i)^{-1}\tilde{\mathbf{H}}_i^\dag$ indicates the null space of $\tilde{\mathbf{H}}_i$.
Then, the received signal at $\text{UE}_i$ can be represented as
\begin{align}
y_i=&\bar{\mathbf{h}}_i^\dag\mathbf{f}_{\text{d,}i}\sqrt{p}x_i+\sum_{j\neq i}^K \bar{\mathbf{h}}_i^\dag\mathbf{f}_{\text{d,}j}\sqrt{p}x_j+n_i = \bar{\mathbf{h}}_i^\dag\mathbf{f}_{\text{d,}i}\sqrt{p}x_i+n_i.
\end{align}

We define the analog-digital-precoded effective channel gain of $\text{UE}_i$ as
\begin{align} \label{E:effective-gain}
h_i\triangleq &|\bar{\mathbf{h}}_i^\dag\mathbf{f}_{\text{d,}i}|^2 = \frac{|\bar{\mathbf{h}}_i^\dag\Pi(\tilde{\mathbf{H}}_i)\bar{\mathbf{h}}_i|^2}
{\|\Pi(\tilde{\mathbf{H}}_i)\bar{\mathbf{h}}_i\|^2}=
\bar{\mathbf{h}}_i^\dag\Pi(\tilde{\mathbf{H}}_i)\bar{\mathbf{h}}_i=
\bar{\mathbf{h}}_i^\dag\tilde{\mathbf{U}}_i\tilde{\mathbf{U}}_i^\dag\bar{\mathbf{h}}_i
=\|\tilde{\mathbf{U}}_i^\dag\bar{\mathbf{h}}_i\|^2,
\end{align}
where the projection matrix $\Pi(\tilde{\mathbf{H}}_i)$ is decomposed as $\Pi(\tilde{\mathbf{H}}_i)=\tilde{\mathbf{U}}_i\tilde{\mathbf{U}}_i^\dag$, and $\tilde{\mathbf{U}}_i\in\mathbb{C}^{l_\text{BS}\times (l_\text{BS}-K+1)}$ is semi-unitary.

\subsection{Channel Model}
We consider the following propagation channel description:
\begin{align}
\mathbf{h}_i=\frac{1}{\sqrt{\alpha_i}}\sum_{l=1}^{L_i}a_{il}\mathbf{b}_M(\theta_{il}),
\label{Eq.channel_prop}
\end{align}
where $L_i$ is the number of scatterers between the BS and $\text{UE}_i$, $a_{il}\sim\mathcal{CN}(0,\delta_{il}^2)$ indicates the fading of $l$-th MPC to $\text{UE}_i$, $\alpha_i$ indicates the overall large-scale loss, and $\mathbf{b}_M(\theta_{il})$ is the $M$-by-$1$ steering vector with respect to angle of departure (AOD) $\theta_{il}$, $\forall i, l$. We consider a one-ring cluster model, where every subpath has the equal power, i.e., $\delta_{il}^2=\frac{1}{L_i}$, and the AODs of $\text{UE}_i$ follow uniform distribution $\mathcal{U}(\theta_i-\Delta_i,\theta_i+\Delta_i)$ with $\theta_i$ denoting the center angle and $\Delta_i$ denoting the angle spread. For the uniform linear array (ULA), the $m$-th entry of $\mathbf{h}_i$ can be represented as
\begin{align}
\mathbf{h}_i(m)=\frac{1}{\sqrt{\alpha_i}}\sum_{l=1}^{L_i}a_{il}e^{-j2\pi(m-1)d\cos(\theta_i+\tilde{\theta}_{il})},
\label{Eq.channelEntry}
\end{align}
where $\tilde{\theta}_{il}\sim \mathcal{U}(-\Delta_i,\Delta_i)$.

Assuming a small AOD spread for the mmWave channels, i.e., $\Delta_i\approx 0$, we can approximate $\cos(\theta_i+\tilde{\theta}_{il})$ in \eqref{Eq.channelEntry} as
\begin{align}
&\cos(\theta_i+\tilde{\theta}_{il}){=}\cos(\theta_i)\cos(\tilde{\theta}_{il})-\sin(\theta_i)
\sin(\tilde{\theta}_{il}){\approx}\cos\theta_i-\sin\theta_i\tilde{\theta}_{il},
\label{Eq.cosSimplify}
\end{align}
where the approximation holds when $\tilde{\theta}_{il}$ approaches zero, $\forall i,l$. Substitute (\ref{Eq.cosSimplify}) into (\ref{Eq.channelEntry}), we can approximate $\mathbf{h}_i(m)$ by $\tilde{\mathbf{h}}_i(m)$ as
\begin{align}
\tilde{\mathbf{h}}_i(m)=&\frac{1}{\sqrt{\alpha_i}}\sum_{l=1}^{L_i}a_{il}e^{-j2\pi(m-1)d(\cos\theta_i-\sin\theta_i
\tilde{\theta}_{il})},

\label{Eq.channelApprox}
\end{align}
and the steering vector ${\mathbf{b}}_M(\theta_{il})\triangleq [1,e^{-j2\pi d\cos\theta_{il}},...,e^{-j2\pi(M-1)d\cos\theta_{il}}]^T$ can be approximated by $\tilde{\mathbf{b}}_M(\theta_i,\tilde{\theta}_{il})$ as
\begin{align}
    \tilde{\mathbf{b}}_M(\theta_i,\tilde{\theta}_{il})\triangleq [1,e^{-j2\pi d(\cos\theta_i-\sin\theta_i\tilde{\theta}_{il})},...,e^{-j2\pi(M-1)d(\cos\theta_i-\sin\theta_i\tilde{\theta}_{il})}]^T.
\end{align}

\subsection{Analog Precoder}
Under the scenarios with small AOD spread $\Delta_i\approx 0$, we consider that the analog precoder is formed towards the central angle of UEs.
For the full-connection structure, the analog precoder, denoted by $\mathbf{F}_\text{aF}$, consists of the normalized steering vectors of UEs, which is
\begin{align}
&\mathbf{F}_\text{aF}=[\mathbf{f}_{\text{aF,}1},...,\mathbf{f}_{\text{aF,}K}] \ \text{with} \  \mathbf{f}_{\text{aF,}k}=\frac{1}{\sqrt{M}}\mathbf{b}_M(\theta_k),\forall k.\label{E:Analog-full}
\end{align}

For the partial-connection structure, suppose that each UE is served by a subarray as commonly considered in the literature, e.g., \cite{partial-Gao2016,Majidzadeh2018}, where full multiplexing is assumed to fully exploit all RF and antenna elements. In this paper, we will consider both full and partial multiplexing cases with $K\leq l_{\text{BS}}$. If $K < l_{\text{BS}}$, then there will be $l_{\text{BS}}-K$ unused RF chains and $(l_{\text{BS}}-K)M_{\text{P}}$ unused antenna elements, where $M_{\text{P}} = M/l_{\text{BS}}$ is the number of antenna elements per subarray.
The analog precoder, denoted by $\mathbf{F}_\text{aP}$, can be expressed as
\begin{align}
&\mathbf{F}_\text{aP}=\left[\diag{([\mathbf{f}_{\text{aP},k}^T]_{k=1}^K)} \ \mathbf{0}\right]^T  \ \text{with} \  \mathbf{f}_{\text{aP},k}=\frac{1}{\sqrt{M_\text{P}}}\mathbf{b}_{M_\text{P}}(\theta_k), \forall k,\label{E:Analog-part}
\end{align}
where $\diag(\cdot)$ generates the block diagonal matrix. It can found that $\mathbf{F}_\text{aP}$ has $(l_{\text{BS}}-K)M_{\text{P}}$ rows of zeros, corresponding to unused antenna elements.

\section{Performance Analysis for Different HB Structures}
In this section we conduct theoretical analyses for the performance of the hybrid precoders under the full- and partial-connection HB structures. We first derive an upper bound for the sum rate under the zero-forcing digital precoder, then given the analog precoder as the steering vectors toward the central angles of UEs, we compare the performance of the two HB structures.

\subsection{Upper Bound of Average Sum Rate }
With the zero-forcing digital precoder, the effective channel gain $h_i$ defined in~\eqref{E:effective-gain} can be bounded as follows.

\emph{\textbf{Lemma 1:} For a given UE set and analog precoder, the effective channel gain is upper bounded by}
\begin{align}
h_i\leq \|\bar{\mathbf{h}}_i\|^2- \frac{|\bar{\mathbf{h}}_i^\dag\bar{\mathbf{h}}_j|^2}{\|\bar{\mathbf{h}}_j\|^2}, \forall j\neq i.
\end{align}
\begin{IEEEproof}
 The Lemma follows from the fact that the orthogonal projection of a vector to a subspace is not larger than that to a vector involved in the subspace. A detailed proof is given in Appendix~\ref{A:proof-lemma1}.
\end{IEEEproof}

According to Lemma 1, we obtain the following inequality
\begin{align}
h_i\leq \|\bar{\mathbf{h}}_i\|^2-\frac{1}{K-1}\sum_{j\neq i}^K\frac{|\bar{\mathbf{h}}_j^\dag\bar{\mathbf{h}}_i|^2}{\|\bar{\mathbf{h}}_j\|^2}.
\end{align}

Then, we can acquire an upper bound for the achievable rate
\begin{align}
R_\text{sum}=\sum_{k=1}^K\log(1+\frac{h_kp}{\delta_\text{n}^2})\leq \sum_{k=1}^K\log(1+\frac{\|\bar{\mathbf{h}}_k\|^2-\frac{1}{K-1}\sum_{j\neq k}^K\frac{|\bar{\mathbf{h}}_j^\dag\bar{\mathbf{h}}_k|^2}{\|\bar{\mathbf{h}}_j\|^2}}{\delta_\text{n}^2}p)
\end{align}

Taking the expectation at both sides, we can get an upper bound of the average sum rate by Jensen's inequality as
\begin{align}
\mathbb{E}[R_\text{sum}]\leq &\sum_{k=1}^K\mathbb{E}\left[\log\left(1+\frac{\|\bar{\mathbf{h}}_k\|^2-\frac{1}{K-1}\sum_{j\neq k}^K\frac{|\bar{\mathbf{h}}_j^\dag\bar{\mathbf{h}}_k|^2}{\|\bar{\mathbf{h}}_j\|^2}}{\delta_\text{n}^2}p\right)\right]\nonumber\\

\leq & \sum_{k=1}^K \log\left(1+\frac{\mathbb{E}\left[\|\bar{\mathbf{h}}_k\|^2\right]-\frac{1}{K-1}\sum_{j\neq k}^K \mathbb{E}\left[\frac{|\bar{\mathbf{h}}_j^\dag\bar{\mathbf{h}}_k|^2}{\|\bar{\mathbf{h}}_j\|^2} \right]}{\delta_\text{n}^2}p\right)\triangleq \bar{R}_{\text{sum}}^{\text{ub}}.
\label{Eq.sumRateUpper}
\end{align}

Next, we develop analytic expressions of $\mathbb{E}[\|\bar{\mathbf{h}}_k\|^2]$ and $\mathbb{E}[\frac{|\bar{\mathbf{h}}_j^\dag\bar{\mathbf{h}}_k|^2}{\|\bar{\mathbf{h}}_j\|^2}]$ in (\ref{Eq.sumRateUpper}).

With (\ref{Eq.channelApprox}), $\mathbb{E}[|\bar{\mathbf{h}}_j^\dag\bar{\mathbf{h}}_k|^2]$ can be expanded as
\begin{align} \label{E:cross}
\mathbb{E}[|\bar{\mathbf{h}}_j^\dag\bar{\mathbf{h}}_k|^2]
\approx&\frac{1}{\alpha_j\alpha_k}\mathbb{E}[|\sum_{x=1}^{P_j}\sum_{y=1}^{P} a_{jx}^\ast a_{ky}\tilde{\mathbf{b}}_M^\dag(\theta_j,\tilde{\theta}_{jx})\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag\tilde{\mathbf{b}}_M(\theta_k,\tilde{\theta}_{ky})|^2]\nonumber\\
=&\frac{1}{\alpha_j\alpha_k}\sum_{x_1=1}^{P_j}\sum_{y_1=1}^{P}\sum_{x_2=1}^{P_j}\sum_{y_2=1}^{P}\mathbb{E}[ a_{jx_1}^\ast a_{jx_2}]\mathbb{E}[ a_{ky_1} a_{ky_2}^\ast]\cdot\nonumber\\
&\qquad\mathbb{E}[\tilde{\mathbf{b}}_M^\dag(\theta_j,\tilde{\theta}_{jx_1})\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag\tilde{\mathbf{b}}_M(\theta_k,\tilde{\theta}_{ky_1})\tilde{\mathbf{b}}_M^\dag(\theta_k,\tilde{\theta}_{ky_2})\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag\tilde{\mathbf{b}}_M(\theta_j,\tilde{\theta}_{jx_2})]\nonumber\\

=&\frac{1}{\alpha_j\alpha_k}\tr(\mathbf{F}_\text{a}^\dag\mathbb{E}[\tilde{\mathbf{b}}_M(\theta_k,\tilde{\theta}_{ky_1})\tilde{\mathbf{b}}_M^\dag(\theta_k,\tilde{\theta}_{ky_1})]\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag\mathbb{E}[\tilde{\mathbf{b}}_M(\theta_j,\tilde{\theta}_{jx_1})\tilde{\mathbf{b}}_M^\dag(\theta_j,\tilde{\theta}_{jx_1})]\mathbf{F}_\text{a})\nonumber\\
\triangleq&\frac{1}{\alpha_j\alpha_k}\tr(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_j\mathbf{F}_\text{a}),
\end{align}
where we define $\tilde{\mathbf{K}}_j\triangleq \mathbb{E}[\tilde{\mathbf{b}}_M(\theta_j,\tilde{\theta}_{jx_1})\tilde{\mathbf{b}}_M^\dag(\theta_j,\tilde{\theta}_{jx_1})]$ with the $(m,n)$-th entry
\begin{align} \label{E:self}
\tilde{\mathbf{K}}_j(m,n)&=\mathbb{E}[e^{-j2\pi(m-1)d(\cos\theta_j-\sin\theta_j\tilde{\theta}_{jx})}e^{j2\pi(n-1)d(\cos\theta_j-\sin\theta_j\tilde{\theta}_{jx})}]\nonumber\\

&=e^{j2\pi(n-m)d\cos\theta_j}\mathbb{E}[e^{j2\pi(m-n)d\sin\theta_j\tilde{\theta}_{jx}}]\nonumber\\
&=e^{j2\pi(n-m)d\cos\theta_j}\frac{\sin(2\pi d(m-n)\sin\theta_j\Delta_j)}{2\pi d(m-n)\sin\theta_j\Delta_j},
\end{align}
where the final equality is obtained by considering $\tilde{\theta}_{jx}\sim\mathcal{U}(-\Delta_i,\Delta_i)$.

The term $\mathbb{E}[\|\bar{\mathbf{h}}_i\|^2]$ can be derived as
\begin{align}
\mathbb{E}[\|\bar{\mathbf{h}}_i\|^2]&\approx\frac{1}{\alpha_i}\mathbb{E}[\sum_{x=1}^{L_i}\sum_{y=1}^{L_i} a_{ix}^\ast a_{iy}\tilde{\mathbf{b}}_M^\dag(\theta_i,\tilde{\theta}_{ix})\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag\tilde{\mathbf{b}}_M(\theta_i,\tilde{\theta}_{iy})]\nonumber\\
&=\frac{1}{\alpha_i}\mathbb{E}[\tilde{\mathbf{b}}_M^\dag(\theta_i,\tilde{\theta}_{ix})\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag\tilde{\mathbf{b}}_M(\theta_i,\tilde{\theta}_{ix})]

=\frac{1}{\alpha_i}\tr(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_i\mathbf{F}_\text{a}).
\end{align}

With \eqref{E:cross} and \eqref{E:self}, the upper bound given in (\ref{Eq.sumRateUpper}) can be rewritten as
\begin{align}

 \bar{R}_{\text{sum}}^{\text{ub}}& \approx\sum_{k=1}^K \log(1+\frac{\tr(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{a})-\frac{1}{K-1}\sum_{j\neq k}^K\frac{\tr{(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag
\tilde{\mathbf{K}}_j\mathbf{F}_\text{a})}}{\tr(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_j\mathbf{F}_\text{a})}}
{\alpha_k\delta_n^2}p)\nonumber\\
 &\triangleq\sum_{k=1}^K\log(1+\frac{g_kp}{\alpha_k\delta_n^2}),
 \label{Eq.upperBound}
\end{align}
where we define $g_k\triangleq \tr(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{a})-\frac{1}{K-1}\sum_{j\neq k}^K\frac{\tr{(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag
\tilde{\mathbf{K}}_j\mathbf{F}_\text{a})}}{\tr(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_j\mathbf{F}_\text{a})}$, which determines the data rate of UE$_k$.

To compare the performance under two HB structures, we need to obtain an explicit expression of $g_k$ as defined below \eqref{Eq.upperBound}. Under the scenarios with small AOD spread $\Delta_i\approx 0$,

we can obtain from \eqref{E:self} that
\begin{align}
\tilde{\mathbf{K}}_j(m,n)\approx e^{j2\pi(n-m)d\cos\theta_j}, \ \ \tilde{\mathbf{K}}_j\approx \mathbf{b}_M(\theta_j)\mathbf{b}_M^\dag(\theta_j).
\end{align}
Then, we can further simplify (\ref{Eq.upperBound}) by considering that
\begin{align} \label{E:simplify}
\tr(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{a})

=\|\mathbf{F}_\text{a}^\dag\mathbf{b}_M(\theta_k)\|^2,
\ \
\tr{(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_j
\mathbf{F}_\text{a})}
=|\mathbf{b}_M^\dag(\theta_k)\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag\mathbf{b}_M(\theta_j)|^2.
\end{align}

We first consider the full-connection structure by setting $\mathbf{F}_\text{a}=\mathbf{F}_\text{aF}$.
Upon substituting \eqref{E:Analog-full} into \eqref{E:simplify}, we can update $\tr(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{a})$ as
\begin{align}
&\tr(\mathbf{F}_\text{aF}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{aF})=\frac{1}{M}\sum_{j=1}^K\left|\mathbf{b}_M^\dag(\theta_j)\mathbf{b}_M(\theta_k)\right|^2\nonumber\\
&=\frac{1}{M}\sum_{j=1}^K\left|\sum_{n=1}^M e^{j2\pi d(n-1)(\cos\theta_j-\cos\theta_k)}\right|^2
=\frac{1}{M}\sum_{j=1}^K\left|\frac{\sin(\pi dM(\cos\theta_j-\cos\theta_k))}{\sin(\pi d(\cos\theta_j-\cos\theta_k))}\right|^2, \label{E:F-1}
\end{align}
and update $\tr{(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_j
\mathbf{F}_\text{a})}$ as
\begin{align}
&\tr{(\mathbf{F}_\text{aF}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{aF}\mathbf{F}_\text{aF}^\dag
\tilde{\mathbf{K}}_j\mathbf{F}_\text{aF})}=\left|\frac{1}{M}\mathbf{b}_M^\dag(\theta_k)\sum_{i=1}^K\mathbf{b}_M(\theta_i)\mathbf{b}_M^\dag(\theta_i)\mathbf{b}_M(\theta_j)\right|^2\nonumber\\
&=\frac{1}{M^2}\left|\sum_{i=1}^K\sum_{n=1}^Me^{j2\pi d(n-1)(\cos\theta_k-\cos\theta_i)}\sum_{m=1}^Me^{j2\pi d(m-1)(\cos\theta_i-\cos\theta_j)}\right|^2\nonumber\\
&=\frac{1}{M^2}\left|\sum_{i=1}^K{e^{j\pi d(M-1)(\cos\theta_k-\cos\theta_j)}}\frac{\sin(\pi M d(\cos\theta_k-\cos\theta_i))}{\sin(\pi d(\cos\theta_k-\cos\theta_i))}\frac{\sin(\pi M d(\cos\theta_i-\cos\theta_j))}{\sin(\pi d(\cos\theta_i-\cos\theta_j))}\right|^2\nonumber\\
&=\frac{1}{M^2}\left|\sum_{i=1}^K\frac{\sin(\pi M d(\cos\theta_k-\cos\theta_i))}{\sin(\pi d(\cos\theta_k-\cos\theta_i))}\frac{\sin(\pi M d(\cos\theta_i-\cos\theta_j))}{\sin(\pi d(\cos\theta_i-\cos\theta_j))}\right|^2. \label{E:F-2}
\end{align}

With \eqref{E:F-1} and \eqref{E:F-2}, we can obtain the expression of $g_k$ as defined below \eqref{Eq.upperBound} for the full-connected structure, denoted by $g_{\text{F},k}$, as
\begin{align} \label{E:g-full}
&g_{\text{F},k}=\frac{1}{M}\sum_{i=1}^K\left|\frac{\sin(\pi dM(\cos\theta_i-\cos\theta_k))}{\sin(\pi d(\cos\theta_i-\cos\theta_k))}\right|^2-\nonumber\\
&\qquad \qquad \qquad \frac{1}{K-1}\sum_{j\neq k}^K \frac{\left|\sum_{i=1}^K\frac{\sin(\pi M d(\cos\theta_k-\cos\theta_i))}{\sin(\pi d(\cos\theta_k-\cos\theta_i))}\frac{\sin(\pi M d(\cos\theta_i-\cos\theta_j))}{\sin(\pi d(\cos\theta_i-\cos\theta_j))}\right|^2}{M\sum_{i=1}^K\left|\frac{\sin(\pi dM(\cos\theta_i-\cos\theta_j))}{\sin(\pi d(\cos\theta_i-\cos\theta_j))}\right|^2}.
\end{align}

We proceed to consider the partial-connection structure by setting $\mathbf{F}_\text{a}=\mathbf{F}_\text{aP}$. Upon substituting \eqref{E:Analog-part} into \eqref{E:simplify}, we can update $\tr(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{a})$ as
\begin{align}
\tr(\mathbf{F}_{\text{aP}}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{aP})&=\sum_{i=1}^K\left|\frac{1}{\sqrt{M_\text{P}}}e^{-j2\pi d(i-1)M_\text{P}\cos\theta_k}e^{j\pi d (M_\text{P}-1)(\cos\theta_i-\cos\theta_k)}\frac{\sin(\pi d M_\text{P}(\cos\theta_i-\cos\theta_k))}{\sin(\pi d (\cos\theta_i-\cos\theta_k))}\right|^2\nonumber\\
&=\frac{1}{M_\text{P}}\sum_{i=1}^K \left|\frac{\sin(\pi d M_\text{P}(\cos\theta_i-\cos\theta_k))}{\sin(\pi d (\cos\theta_i-\cos\theta_k))}\right|^2, \label{E:P-1}
\end{align}
and update $\tr{(\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{a}\mathbf{F}_\text{a}^\dag\tilde{\mathbf{K}}_j
\mathbf{F}_\text{a})}$ as
\begin{align}
&\tr{(\mathbf{F}_\text{aP}^\dag\tilde{\mathbf{K}}_k\mathbf{F}_\text{aP}\mathbf{F}_\text{aP}^\dag\tilde{\mathbf{K}}_j
\mathbf{F}_\text{aP})}
=\left|\sum_{i=1}^K\frac{1}{\sqrt{M_\text{P}}}e^{j2\pi d(i-1)M_\text{P}\cos\theta_k}\sum_{n=1}^{M_\text{P}}e^{j2\pi d (n-1)(\cos\theta_k-\cos\theta_i)}\cdot\right.\nonumber\\
&\qquad\qquad\qquad\qquad\qquad\qquad\left.\frac{1}{\sqrt{M_\text{P}}}e^{-j2\pi d(i-1)M_\text{P}\cos\theta_j}\sum_{m=1}^{M_\text{P}}e^{j2\pi d (m-1)(\cos\theta_i-\cos\theta_j)}\right|^2\nonumber\\

&=\frac{1}{M_\text{P}^2}\left|\sum_{i=1}^Ke^{j2\pi d(i-1)M_\text{P}(\cos\theta_k-\cos\theta_j)}\frac{\sin(\pi d M_\text{P}(\cos\theta_k-\cos\theta_i))}{\sin(\pi d (\cos\theta_k-\cos\theta_i))}\frac{\sin(\pi d M_\text{P}(\cos\theta_j-\cos\theta_i))}{\sin(\pi d (\cos\theta_j-\cos\theta_i))}\right|^2.
\end{align}

Consequently, we can obtain the expression of $g_k$ for the partial-connected structure, denoted by $g_{\text{P},k}$, as
\begin{align}\label{E:g-part}
&g_{\text{P},k}=\frac{1}{M_\text{P}}\sum_{i=1}^K\left|\frac{\sin(\pi dM_\text{P}(\cos\theta_i-\cos\theta_k))}{\sin(\pi d(\cos\theta_i-\cos\theta_k))}\right|^2-\nonumber\\
&\qquad  \frac{1}{K-1}\sum_{j\neq k}^K \frac{\left|\sum_{i=1}^Ke^{j2\pi d(i-1)M_\text{P}(\cos\theta_k-\cos\theta_j)}\frac{\sin(\pi M_\text{P} d(\cos\theta_k-\cos\theta_i))}{\sin(\pi d(\cos\theta_k-\cos\theta_i))}\frac{\sin(\pi M_\text{P} d(\cos\theta_i-\cos\theta_j))}{\sin(\pi d(\cos\theta_i-\cos\theta_j))}\right|^2}{M_\text{P}\sum_{i=1}^K\left|\frac{\sin(\pi dM_\text{P}(\cos\theta_i-\cos\theta_j))}{\sin(\pi d(\cos\theta_i-\cos\theta_j))}\right|^2}.

\end{align}

\subsection{Structure Comparison in Special Cases}
Based on the obtained expressions of $g_{\text{F},k}$ and $g_{\text{P},k}$, we compare the two HB structures by considering some special cases in this subsection, and will consider the general case in the next subsection.

\subsubsection{\underline{Case 1: $M_\text{P}\rightarrow \infty$, $M\rightarrow \infty$, arbitrary $K$ UEs}}

When the number of antennas is sufficiently large, we can readily obtain from \eqref{E:g-full} and \eqref{E:g-part} that
\begin{align}
  &\lim_{M\rightarrow \infty} g_{\text{F},k}=M, \ \ \lim_{M_\text{P}\rightarrow\infty}g_{\text{P},k}=M_\text{P},  \ \
  \lim_{M_\text{P}\rightarrow\infty}\frac{g_{\text{F},k}}{g_{\text{P},k}}=l_\text{BS}.
\end{align}
The results show that when the antenna array has sufficiently high spatial resolution with a large number of antennas, the full-connection structure outperforms the partial-connection structure by the beamforming gain at the order of $l_\text{BS}$.

\subsubsection{\underline{Case 2: Arbitrary $M_\text{P}$ and $M$, $K=2$ UEs}}

Consider that the BS serves $K=2$ UEs. For an arbitrary UE, say UE$_1$, we can obtain from \eqref{E:g-full} and \eqref{E:g-part} that

\begin{align}
g_{\text{F},1}&=M+\frac{1}{M}Z_M^2(\beta_{12})-\frac{4MZ_M^2(\beta_{12})}{Z_M^2(\beta_{12})+M^2}=\frac{(M^2-Z_M^2(\beta_{12}))^2}{M(M^2+Z_M^2(\beta_{12}))},\nonumber\\
g_{\text{P},1}&=M_\text{P}+\frac{1}{M_\text{P}}Z_{M_\text{P}}^2(\beta_{12})-\frac{4M_\text{P}Z_{M_\text{P}}^2(\beta_{12})\cos^2(\pi dM_\text{P}\beta_{12})}{Z_{M_\text{P}}^2(\beta_{12})+M_\text{P}^2},\nonumber\\
&=\frac{(M_\text{P}^2+Z_{M_\text{P}}^2(\beta_{12}))^2-4M_\text{P}^2Z_{M_\text{P}}^2(\beta_{12})\cos^2(\pi dM_\text{P}\beta_{12})}{M_\text{P}(M_\text{P}^2+Z_{M_\text{P}}^2(\beta_{12}))}.
\end{align}
where we define the function $Z_M(x)\triangleq\frac{\sin(\pi d Mx)}{\sin(\pi d x)}$, and $\beta_{12}=\cos\theta_1-\cos\theta_2$ reflects the proximity of the two UEs.

Consequently, the ratio gap between $g_{\text{F},1}$ and $g_{\text{P},1}$ is
\begin{align} \label{E:ratio-1}
&\frac{g_{\text{F},1}}{g_{\text{P},1}}=\frac{(M^2-Z_M^2(\beta_{12}))^2}{(M_\text{P}^2+Z_{M_\text{P}}^2(\beta_{12}))^2-4M_\text{P}^2Z_{M_\text{P}}^2(\beta_{12})\cos^2(\pi d M_\text{P}\beta_{12})}\frac{M_\text{P}(M_\text{P}^2+Z_{M_\text{P}}^2(\beta_{12}))}{M(M^2+Z_M^2(\beta_{12}))}.
\end{align}

We next consider a toy example to simplify \eqref{E:ratio-1} to gain some insight, where the BS has $M=2$ antennas and $l_{\text{BS}}=2$ RF chains. In this case, each RF chain connects a single antenna in the partial-connection structure, i.e., $M_{\text{P}} = 1$. We can obtain that $Z_{M}(\beta_{12}) = \frac{\sin(2\pi d \beta_{12})}{\sin(\pi d \beta_{12})} = 2\cos(\pi d \beta_{12})$ and $Z_{M_\text{P}}(\beta_{12}) = 1$, with which \eqref{E:ratio-1} can be updated as
\begin{align}
    \frac{g_{\text{F},1}}{g_{\text{P},1}}

    =\frac{1-\cos^2(\pi d \beta_{12})}{ 1+\cos^2(\pi d \beta_{12})}.
\end{align}
It is obvious that $0\leq \frac{g_{\text{F},1}}{g_{\text{P},1}}\leq 1$ for the toy example, which indicates that the full-connection structure is worse than the partial-connection structure. Although the former is able to provide a larger beamforming gain, it also enhances the collinearity between the effective channels of the two UEs, making the overall performance worse.

For general cases, the value of $\frac{g_{\text{F},1}}{g_{\text{P},1}}$ in \eqref{E:ratio-1} depends on $M$, $l_{\text{BS}}$ and $\beta_{12}$. Numerically, we evaluate the value of $\frac{g_{\text{F},1}}{g_{\text{P},1}}$ for different $\kappa=\beta_{12}M$ in Fig. \ref{Fig:ratio_vs_beta}, where $l_{\text{BS}}=K=2$ is given. Herein, the x-axis $\kappa$ can be interpreted as the normalized angle separation between UEs by the angle resolution of the BS array, which is on the order of $\frac{1}{M}$.
We can see that the full-connection structure is not always better than the partial-connection structure as revealed by the toy example. The latter is better for small $\beta_{12}$, i.e., when the central angles to the two UEs, $\theta_1$ and $\theta_2$, are close. The regime of $\beta_{12}$ where the partial-connection structure is better decreases with the increase of $M$. When $M$ is very large, say 32, the ratio approaches a constant $l_\text{BS}$ as shown in Fig.~\ref{Fig:ratio_vs_beta}(d).

\begin{figure}
\centering
\includegraphics[width=0.85\textwidth]{gRatio_vs_kappa.eps}
\caption{Ratio $\frac{g_{\text{F},1}}{g_{\text{P},1}}$ v.s. $\kappa$ for $M=4, 8, 16$, and $32$.}
\label{Fig:ratio_vs_beta}
\end{figure}

\subsection{Structure Comparison in General Cases}
The above special-case analysis implies that the full-connection structure may be interior to the partial-connection structure when the UEs have close central angles $\theta_i, \forall i$. In this subsection we focus on the regime where $\theta_i\approx \theta_j$, $\forall i\neq j$, aimed at investigating the condition of $\theta_i$ and $\theta_j$ when the full-connection structure performs better for a general case with arbitrary numbers of UEs $K\leq l_{\text{BS}}$ and antennas $M$.

Specifically, we assume that $\cos\theta_i-\cos\theta_j=\beta\approx 0$, $\forall i\neq j$. Then, we can simplify the expression of $g_{\text{F},k}$ in \eqref{E:g-full} with some regular manipulations as
\begin{align}
g_{\text{F},k}

&=\frac{[(2K-3)\zeta^2+2M\zeta+M^2](\zeta-M)^2}{M[M^2+(K-1)\zeta^2]}\approx \frac{M(\eta_\text{F}-1)^2[(2K-3)\eta_\text{F}^2+2\eta_\text{F}+1]}{1+(K-1)\eta_\text{F}^2}, \label{E:g-full-approx}
\end{align}
where $\zeta\triangleq \frac{\sin(\pi d \kappa)}{\sin(\pi d \frac{\kappa}{M})}$, $\kappa = \beta M$ is the normalized angle separation between UEs as defined in Fig.~\ref{Fig:ratio_vs_beta}, and the approximation follows from the first-order Taylor approximation of $\sin(\pi d \frac{\kappa}{M})$, which is accurate since $\frac{\kappa}{M} = \beta$ is small as we assumed, i.e., $\zeta\approx M\frac{\sin(\pi d\kappa)}{\pi d\kappa} \triangleq M\eta_\text{F}$. Herein, we define $\eta_\text{F} \triangleq \frac{\sin(\pi d\kappa)}{\pi d\kappa}$, and we can numerically obtain that its value ranges in $[-0.22,1]$.

Similarly, considering the approximation $\frac{\sin(\pi d \beta M_{\text{P}})}{\sin(\pi d \beta)} = \frac{\sin(\pi d \frac{\kappa}{l_{\text{BS}}})}{\sin(\pi d \frac{\kappa}{M})} \approx M\frac{\sin(\pi d \frac{\kappa}{l_{\text{BS}}})}{\pi d \kappa} =  M_{\text{P}}\frac{l_{\text{BS}}\sin(\pi d \frac{\kappa}{l_{\text{BS}}})}{\pi d \kappa} \triangleq M_{\text{P}}\eta_\text{P}$, we can simplify the expression of $g_{\text{P},k}$ in \eqref{E:g-part} with some regular manipulations as
\begin{align}
g_{\text{P},k} \approx M_\text{P}[1+(K-1)\eta_\text{P}^2]-\frac{M_\text{P}\eta_\text{P}^2}{K-1}
\frac{f_k(\frac{\kappa}{l_\text{BS}})}{1+(K-1)\eta_\text{P}^2},
\end{align}
where we define the function $f_k(\frac{\kappa}{l_\text{BS}})\triangleq \sum_{j\neq k}^K |e^{j2\pi d(k-1)\frac{\kappa}{l_\text{BS}}}+e^{j2\pi d(j-1)\frac{\kappa}{l_\text{BS}}}+\eta_\text{P}\sum_{i\neq j,k}e^{j2\pi d(i-1)\frac{\kappa}{l_\text{BS}}}|^2$ and $\eta_\text{P} = \frac{l_{\text{BS}}\sin(\pi d \frac{\kappa}{l_{\text{BS}}})}{\pi d \kappa}$.

Therefore, the ratio between $g_{\text{F},k}$ and $g_{\text{P},k}$ approximates
\begin{align} \label{E:approx-ratio}
\frac{g_{\text{F},k}}{g_{\text{P},k}}\approx\frac{l_\text{BS}(\eta_\text{F}-1)^2[(2K-3)
\eta_\text{F}^2+2\eta_\text{F}+1]}{[1+(K-1)\eta_\text{F}^2][1+(K-1)\eta_\text{P}^2-
\frac{\eta_\text{P}^2}{K-1}\frac{f_k(\frac{\kappa}{l_\text{BS}})}{1+(K-1)\eta_\text{P}^2}]}.
\end{align}

Since we are investigating the smallest separation between UEs to make the full-connection structure better, the interested normalized angle separation $\kappa = \beta M$ is generally small. Then, it is reasonable to assume $\beta M_{\text{P}} = \frac{\kappa}{l_{\text{BS}}}\approx 0$ since the number of RF chains $l_{\text{BS}}$ can be large for massive MIMO systems. With the approximation $\frac{\kappa}{l_{\text{BS}}}\approx 0$, we can obtain $\eta_\text{P}=\frac{l_{\text{BS}}\sin(\pi d \frac{\kappa}{l_{\text{BS}}})}{\pi d \kappa}\approx 1$ and then approximate $f_k(\frac{\kappa}{l_\text{BS}})$ by
\begin{align}
f_k(\frac{\kappa}{l_\text{BS}})&\approx \sum_{j\neq k}^K \left|\sum_{i}e^{j2\pi d(i-1)\frac{\kappa}{l_\text{BS}}}\right|^2
=\sum_{j\neq k}^K \left|\frac{1-e^{j2\pi dK\frac{\kappa}{l_\text{BS}}}}{1-e^{j2\pi d\frac{\kappa}{l_\text{BS}}}}\right|^2\nonumber\\
&=(K-1)\frac{\sin^2(\pi dK\frac{\kappa}{l_\text{BS}})}{\sin^2(\pi d\frac{\kappa}{l_\text{BS}})}
\approx (K-1){l_\text{BS}^2}\frac{\sin^2(\pi dK\frac{\kappa}{l_\text{BS}})}{(\pi d\kappa)^2} , \label{E:approx-f}
\end{align}
where the first approximation follows from $\eta_{\text{P}}\approx1$ and the second approximation comes from the first-order Taylor approximation of $\sin(\cdot)$.

Upon substituting \eqref{E:approx-f} and $\eta_{\text{P}}\approx1$  into \eqref{E:approx-ratio}, we obtain
\begin{align} \label{Eq.simpleRatio}
\frac{g_{\text{F},k}}{g_{\text{P},k}}\approx\frac{l_\text{BS}(\eta_\text{F}-1)^2[(2K-3)
\eta_\text{F}^2+2\eta_\text{F}+1]}{[1+(K-1)\eta_\text{F}^2] [K-\frac{l_\text{BS}^2{\sin^2(\pi dK\frac{\kappa}{l_\text{BS}})}}{K(\pi d\kappa)^2}] }.
\end{align}

For the sanity check of the approximation by (\ref{Eq.simpleRatio}), we compare the real value of $\frac{g_{\text{F},k}}{g_{\text{P},k}}$ obtained by simulations and the approximation by (\ref{Eq.simpleRatio}) in Fig.~\ref{Fig:ratio_compare1}, where $M_\text{P}=16$, $d=\frac{1}{2}$, and $K=l_\text{BS}$. We consider different settings of $l_\text{BS}$ and $\kappa$. It is shown that in the interested regime, i.e., the values of $\kappa$ making $\frac{g_{\text{F},k}}{g_{\text{P},k}}$ close to 1, (\ref{Eq.simpleRatio}) approximates well for different settings of $l_\text{BS}$, and the approximation accuracy improves with $l_\text{BS}$, e.g., more than 8.

\begin{figure}
\centering
\includegraphics[width=0.8\textwidth]{ratio_kappa_K2to10.pdf}
\caption{$\frac{g_{\text{F},k}}{g_{\text{P},k}}$ vs. $\kappa$ for different setting of $l_{\text{BS}}$, $M_{\text{P}} = 16$, $d=\frac{1}{2}$, and $K=l_{\text{BS}}$.}
\label{Fig:ratio_compare1}
\end{figure}

Based on (\ref{Eq.simpleRatio}), we next examine the condition to determine which HB structure performs better. We consider the following two cases according to whether the system is fully multiplexing or not.

\subsubsection{\underline{Full-multiplexing: $K=l_\text{BS}$}}
In the full-multiplexing mode, the number of the served users equals to the number of RF chains, i.e., $K=l_\text{BS}$. Then, (\ref{Eq.simpleRatio}) reduces to
\begin{align}
\frac{g_{\text{F},k}}{g_{\text{P},k}}\approx\frac{(\eta_\text{F}-1)^2[(2l_\text{BS}-3)
\eta_\text{F}^2+2\eta_\text{F}+1]}{[1+(l_\text{BS}-1)\eta_\text{F}^2](1-\eta_\text{F}^2)},
\label{Eq.ratioAsy}
\end{align}
where $\eta_\text{F} = \frac{\sin(\pi d\kappa)}{\pi d\kappa}$, as defined below \eqref{E:g-full-approx}, is used.

Based on (\ref{Eq.ratioAsy}), to achieve $\frac{g_{\text{F},k}}{g_{\text{P},k}}>1$, we need to satisfy the following condition:
\begin{align}
    l_\text{BS}(1-3\eta_\text{F})>4(1-\eta_\text{F}),\ \eta_\text{F}\neq 0,
\end{align}
from which we obtain the following proposition.

\emph{\textbf{Proposition 1:} Given the analog precoder formed by steering vectors towards UEs, the following operating region needs to be satisfied under the full-multiplexing mode, so that the full-connection structure performs better than the partial-connection structure:}
\begin{align} \label{E:condition1}
    \l_\text{BS}>\frac{4(1-\eta_\text{F})}{1-3\eta_\text{F}},\ \eta_\text{F}<\frac{1}{3},\ \eta_\text{F}\neq 0.
\end{align}

\emph{\textbf{Remark 1:}} Proposition 1 indicates that the full-connection structure is always inferior to the partial-connection structure when $\eta_\text{F}>\frac{1}{3}$, corresponding to the normalized angle separation between UEs $\kappa < 1.45$ for the antenna spacing $d=\frac{1}{2}$ wavelength. For $\kappa > 1.45$, in order to ensure the full-connection structure is better, more RF chains are required. For instance, when $\kappa = 1.6$ and $d=\frac{1}{2}$, i.e., $\eta_\text{F} = 0.23$, $\l_\text{BS}$ should be larger than $10$.

To do the sanity check for proposition 1,  we plot the values of $\frac{g_{\text{F},k}}{g_{\text{P},k}}$ for different $\eta_\text{F}$, where both the real value obtained by simulations and the approximation in \eqref{Eq.ratioAsy} are exhibited, $M_\text{P}=16$, $d=\frac{1}{2}$, and $K=l_\text{BS}$. It is shown that $\frac{g_{\text{F},k}}{g_{\text{P},k}}<1$ always holds when $\eta_\text{F} > \frac{1}{3}$. For $\eta_\text{F} < \frac{1}{3}$, $\frac{g_{\text{F},k}}{g_{\text{P},k}}$ may be still less than 1, e.g., when $l_\text{BS}=16$ and $\eta_\text{F}$ is close to $\frac{1}{3}$. Yet, increasing $l_\text{BS}$ can improve $\frac{g_{\text{F},k}}{g_{\text{P},k}}$, for instance, which is always not smaller than $1$ when $l_\text{BS}=128$ for any $\eta_\text{F} < \frac{1}{3}$. The results are consistent with Proposition~1.
In addition, one can find that $\frac{g_{\text{F},k}}{g_{\text{P},k}} = 1$ holds when $\eta_\text{F}=0$ for all $l_\text{BS}$. This can be readily verified from \eqref{Eq.ratioAsy}.

\begin{figure}
\centering
\includegraphics[width=0.8\textwidth]{Ratio_eta_fullMultiplex.pdf}
\caption{$\frac{g_{\text{F},k}}{g_{\text{P},k}}$ vs. $\eta_\text{F}$ for different $l_{\text{BS}}$, $M_{\text{P}} = 16$, and $K=l_{\text{BS}}$.}
\label{Fig:ratioEta_asymptotic_Msub=4}
\end{figure}

\subsubsection{\underline{Arbitrary $K\leq l_\text{BS}$}}
We now consider a general case with the number of scheduled users not larger than the number of RF chains, i.e., $K\leq l_\text{BS}$. For notational simplicity, define $\rho\triangleq \frac{K}{l_\text{BS}}$, where $\frac{1}{l_\text{BS}}\leq \rho\leq 1$. Substitute $\rho$ into (\ref{Eq.simpleRatio}), we acquire that
\begin{align} \label{E:g-K}
&\frac{g_{\text{F},k}}{g_{\text{P},k}}\approx\frac{(\eta_\text{F}-1)^2[(2\rho l_\text{BS}-3)\eta_\text{F}^2+2\eta_\text{F}+1]}{\rho[1+(\rho l_\text{BS}-1)\eta_\text{F}^2](1-\frac{\sin^2(\pi d\rho \kappa)}{(\pi d\rho\kappa)^2})}.
\end{align}

We first examine the value of $\frac{g_{\text{F},k}}{g_{\text{P},k}}$ when $\eta_{\text{F}} = 0$, which requires $\sin(\pi d \kappa)=0$ and $\kappa \neq 0$ according to the definition of $\eta_{\text{F}}$ as defined below \eqref{E:g-full-approx}, i.e., $\kappa = \frac{n}{d}$ should hold for $n\in\mathbb{Z}^+$. In this case, \eqref{E:g-K} reduces to
\begin{align}
\frac{g_{\text{F},k}}{g_{\text{P},k}}\approx\frac{1}{\rho(1-\frac{\sin^2(\pi n \rho)}{(\pi n \rho)^2})}\geq \frac{1}{\rho}\geq 1.
\end{align}

When $\eta_{\text{F}} \neq 0$, i.e., $\kappa \neq \frac{n}{d}, \forall n\in\mathbb{Z}^+$, the condition to ensure $\frac{g_{\text{F},k}}{g_{\text{P},k}}\geq 1$ can be derived from \eqref{E:g-K}~as
\begin{align}
    (2-A)l_\text{BS}\geq \frac{(3-A)\eta_\text{F}^2-2\eta_\text{F}-1+A}{\eta_\text{F}^2\rho},
\label{Eq.boundOriginal}
\end{align}
where $A\triangleq \frac{1-\frac{\sin^2(\pi d\rho \kappa)}{(\pi d\rho \kappa)^2}}{(1-\eta_\text{F})^2}\rho$.

In Appendix~\ref{A:proof1}, we prove that the condition \eqref{Eq.boundOriginal} is infeasible when $A > 2$. When $A < 2$, \eqref{Eq.boundOriginal} can be transformed into the condition for $l_\text{BS}$ as
\begin{align}
    l_\text{BS}\geq \frac{(3-A)\eta_\text{F}^2-2\eta_\text{F}-1+A}{\eta_\text{F}^2\rho(2-A)},\ A<2.
\label{Eq.feasible}
\end{align}
When $A=2$, (\ref{Eq.boundOriginal}) becomes $0\geq \frac{(\eta_\text{F}-1)^2}{\eta_\text{F}^2\rho}$, which is infeasible except for $\eta_{\text{F}}=1$ and thus indicates that the full-connection structure cannot perform better in this case.

The above analysis leads to the following proposition.

\emph{\textbf{Proposition 2:} Given analog precoder formed by steering vectors towards UEs, the following two operating regions need to be satisfied when $K\leq l_\text{BS}$, so that the full-connection structure performs better than the partial-connection structure:}
\begin{align} \label{E:condition2}
    \begin{cases}
          l_\text{BS}\geq \frac{(3-A)\eta_\text{F}^2-2\eta_\text{F}-1+A}{\eta_\text{F}^2\rho(2-A)} \triangleq l_\text{BS}^{\text{th}}, &\ A<2, \kappa \neq \frac{n}{d}, \forall n\in\mathbb{Z}^+, \\
          \text{any}\ l_\text{BS}, & \ \kappa = \frac{n}{d}, \forall n\in\mathbb{Z}^+.
    \end{cases}
\end{align}

\emph{\textbf{Remark 2:}} When $\rho=1$, i.e., $K=l_\text{BS}$, we have $A = \frac{1+\eta_{\text{F}}}{1-\eta_{\text{F}}}$, with which condition \eqref{E:condition2} reduces to condition \eqref{E:condition1} in Proposition~1.

In Fig.~\ref{Fig:ratioEta_non}, we plot the values of $\frac{g_{\text{F},k}}{g_{\text{P},k}}$ for different $A$, where both the real value obtained by simulations and the approximation in \eqref{E:g-K} are exhibited, $M_\text{P}=16$, $d=\frac{1}{2}$, and $l_\text{BS} = 16$. We consider $K = 8$ and $16$, corresponding to $\rho = 0.5$ and $1$, respectively. When $\rho = 1$, we can find that $\frac{g_{\text{F},k}}{g_{\text{P},k}} < 1$ for $A \geq 2$, while $\frac{g_{\text{F},k}}{g_{\text{P},k}}$ is not always larger than 1 when $A < 2$, depending on whether the condition $l_\text{BS}\geq l_\text{BS}^{\text{th}}$ is satisfied or not. This coincides with Proposition 2. When $\rho = 0.5$, we can find that $\frac{g_{\text{F},k}}{g_{\text{P},k}} > 1 $ always holds when $A < 2$, which is consistent with the analysis in Remark 2 for small $\rho$.

\begin{figure}
\centering
\includegraphics[width=0.8\textwidth]{Ratio_eta_nonfullMultiplex.pdf}
\caption{$\frac{g_{\text{F},k}}{g_{\text{P},k}}$ vs. $A$, $M_{\text{P}} = 16$, $l_{\text{BS}} = 16$, and $K=8$ and $16$ (i.e., $\rho = 0.5$ and $1$).}
\label{Fig:ratioEta_non}
\end{figure}

\section{Conclusions}
This article compared the downlink performance of two hybrid beamforming structures, namely full- and partial-connections, where it is assumed that the angular spread for the mmWave channels is small so that the analog precoder is formed towards users. Given the analog precoder and zero-forcing digital precoder, we developped an upper bound for the analog-and-digital precoded channel gain of users, based on which the precoded channel gain ratio of the full-connection structure over the partial-connection structure was analyzed. We find that the full-connection structure does not always achieve a larger precoded channel gain, which depends on both the angular separation between users and the number of RF chains. We revealed the regimes suitable for the two structures, which are validated by simulations.

\appendices
\section{Proof of Lemma 1} \label{A:proof-lemma1}
Recall that $h_i =\|\tilde{\mathbf{U}}_i^\dag\bar{\mathbf{h}}_i\|^2$ and $\tilde{\mathbf{U}}_i$ is semi-unitary that are orthogonal to the subspace of $\tilde{\mathbf{H}}_i$. Let ${\mathbf{U}}_i$ denote a semi-unitary matrix that spans the subspace of $\tilde{\mathbf{H}}_i$, which can be constructed in the form ${\mathbf{U}}_i=\left[\frac{\bar{\mathbf{h}}_j}{\|\bar{\mathbf{h}}_j\|},{\mathbf{U}}_{i,j}\right]$, $j\neq i$, where ${\mathbf{U}}_{i,j}$ is orthogonal to $\frac{\bar{\mathbf{h}}_j}{\|\bar{\mathbf{h}}_j\|}$.
Then, we can obtain that
\begin{align}
  h_i &=\|\tilde{\mathbf{U}}_i^\dag\bar{\mathbf{h}}_i\|^2 = \bar{\mathbf{h}}_i^\dag\tilde{\mathbf{U}}_i\tilde{\mathbf{U}}_i^\dag\bar{\mathbf{h}}_i = \bar{\mathbf{h}}_i^\dag\left(\mathbf{I}-{\mathbf{U}}_i{\mathbf{U}}_i^\dag\right)\bar{\mathbf{h}}_i \nonumber\\
  &=  \|\bar{\mathbf{h}}_i\|^2 - \bar{\mathbf{h}}_i^\dag \left[\frac{\bar{\mathbf{h}}_j}{\|\bar{\mathbf{h}}_j\|},{\mathbf{U}}_{i,j}\right]
  \left[\frac{\bar{\mathbf{h}}_j}{\|\bar{\mathbf{h}}_j\|},{\mathbf{U}}_{i,j}\right]^\dag \bar{\mathbf{h}}_i\nonumber\\
  &\leq \|\bar{\mathbf{h}}_i\|^2 - \frac{|\bar{\mathbf{h}}_j^\dag\bar{\mathbf{h}}_i|^2}{\|\bar{\mathbf{h}}_j\|^2}.
\end{align}

\section{Infeasibility of \eqref{Eq.boundOriginal} for $A > 2$} \label{A:proof1}

When $A>2$, the left-hand side of \eqref{Eq.boundOriginal} is negative. We next prove that the right-hand side of \eqref{Eq.boundOriginal} is non-negative for $A>2$, making \eqref{Eq.boundOriginal} infeasible.

First, when $2<A<3$, the numerator of the right-hand side of \eqref{Eq.boundOriginal}, denoted by $\Omega(\eta_\text{F})=(3-A)\eta_\text{F}^2-2\eta_\text{F}-1+A$, is monotonically decreasing with $\eta_\text{F}$, because its first-order derivation is $2(3-A)\eta_{\text{F}}-2\leq0$ by recalling that $\eta_\text{F}\in[-0.22,1]$. Then, $\Omega(\eta_\text{F})$ satisfies
\begin{align}
\Omega(\eta_\text{F})\geq \Omega(1)=0.
\end{align}

Second, when $A=3$, we have $\Omega(\eta_\text{F})=-2\eta_\text{F}+2\geq0$.

Third, when $A>3$, in $\Omega(\eta_\text{F})$ the term $(3-A)\eta_\text{F}^2$ is lower bounded by $3-A$, the term $-2\eta_\text{F}$ is lower bounded by $-2$, and thus $\Omega(\eta_\text{F})$ is lower bounded by $\Omega(\eta_\text{F})\geq (3-A)-2-1+A=0$.

In summary, $\Omega(\eta_\text{F})$ is non-negative for $A>2$, which completes the proof.

\bibliography{IEEEabrv,MIMO}

\end{document}

config: arxiv


TITLE: Find the limit: $\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$
QUESTION [2 upvotes]: Find the limit:
$$\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$$
I really have no idea what to do here, since I obviously can't plug in $x=0$ nor divide by highest power...Any help is appreciated!

REPLY [3 votes]: $$\lim_{x\to0} \frac{8^x-7^x}{6^x-5^x} = \lim_{x\to0} \frac{8^x-7^x}{6^x-5^x} \cdot \frac{x}{x} = \lim_{x\to0} \frac{8^x-1-(7^x-1)}{6^x-1-(5^x-1)}\cdot\frac{x}{x} = \lim_{x\to0} \frac{8^x-1-(7^x-1)}{x}\cdot\lim_{x\to0}\frac{x}{6^x-1-(5^x-1)} = \bigg(\lim_{x\to0}\frac{8^x-1}{x}-\frac{7^x-1}{x}\bigg)\cdot\lim_{x\to0} \frac{1}{\frac{6^x-1}{x}-\frac{5^x-1}{x}} = (\ln8-\ln7)\cdot\frac{1}{\ln6-\ln5} = \frac{\ln\frac{8}{7}}{\ln\frac{6}{5}}$$


TITLE: Estimate fraction of standard normal PDFand standard normal CDF $\frac{\varphi(x)}{\Phi(x)}$
QUESTION [0 upvotes]: Let $\varphi(x)$ be the PDF of the standard normal distribution and $\Phi(x)$ the respective CDF.
I am looking for estimates of $\frac{\varphi(x)}{\Phi(x)}$ in $x$, i.e., something in the form of $\frac{\varphi(x)}{\Phi(x)}\leq f(x)$ or $\frac{\varphi(x)}{\Phi(x)}\geq g(x)$ with $g,f$ functions of your choice.
Anything help.
To provide you with a larger picture: I am trying to find the sign of $\frac{\varphi(z+b)}{\Phi(z+b)}-z$.

REPLY [1 votes]: Asymptotic expansions for $\Phi$ and the related $\operatorname{erfc}$ are well known: see for instance the handbook entry
and its wikipedia equivalent.
These expansions cover the case of $\Phi(x)$ for large $|x|$.
Feller's textbook (p166 in the 2nd edition) summarizes the simpler form of this:
If $x>0$, then $$ \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \left\{ \frac 1 x - \frac 1 {x^3}\right\} < 1-\Phi(x) <  \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \frac 1 x .$$


TITLE: Are there real-world devices similar to the thrust reversers in this scene in Star Wars?
QUESTION [0 upvotes]: In Star Wars 3 in the initial crash scene where a spaceship enters the atmosphere for an emergency landing the thrust engine is blocked by a part of the ship and absorbs a part of the thrust which suggests that it's used as a brake.
I'd be interested if that could work inside or outside an atmosphere. However I have few ideas and practical experience approaching such a topic.
I'm aware that Star Wars is fictional, but the pictured engine and brake are a feasible design. Maybe it doesn't make sense, but it was so long ago that they didn't know any better. I spent some time to figure out whether the question belongs here, in case it doesn't I'm sorry.

REPLY [2 votes]: The devices pictured in that scene are clearly modelled on thrust reversers, which are real devices used in a wide variety of jet engines.
Here is one of them in action on a regional jet airliner:

Image source: Wikipedia
For more details, see the first link above.


TITLE: Leibniz integral rule?
QUESTION [1 upvotes]: Suppose that $a$ and $b$ are fixed constants and let $f(x) = \int_{x'=a}^{x'=b} g(x,x') \, dx'$. If x' is in the same direction as x, why is it true that
\begin{equation}
\frac{df}{dx} = g(x,x') \Big|_{x'=a}^{x'=b}?
\end{equation}

REPLY [1 votes]: If $g=g(x+x')$, then
$$\begin{align}f'(x) &= \frac{d}{dx}\int_a^b g(x+x') dx'\\\\
&=\int_a^b \frac{dg(x+x') }{dx}dx'\\\\
&=\int_a^b \frac{dg(x+x') }{dx'}dx'\\\\
&=g(x+b)-g(x+a)
\end{align}$$


TITLE: $\max(a,b)=\frac{a+b+|a-b|}{2}$ generalization
QUESTION [15 upvotes]: I am aware of an occasionally handy identity: $$\max(a,b)=\frac{a+b+|a-b|}{2}$$
However, I have found I'm unable to come up with a nice similar form for $\max(a,b,c)$. Of course I could always use the fact that $\max(a,b,c)=\max(\max(a,b),c)$ to write
$$\begin{align}\max(a,b,c)&=\frac{\frac{a+b+|a-b|}{2}+c+\left|\frac{a+b+|a-b|}{2}-c\right|}{2}
\\&=\frac{a+b+2c+|a-b|+|a+b-2c+|a-b||}{4}\end{align}$$
but this lacks elegance and in particular it is not clear to me just from the formula that if I permute the variables I get the same result.
The following doesn't work, but it would be nice if I could write $\max(a,b,c)$ in some form like
$$\frac{a+b+c+|a-b|+|a-c|+|b-c|}{3}$$

Is there a good generalization of this to $n$ variables? That is given $x_1,x_2,\dots,x_n\in\mathbb{R}$, is there a way to write $\max(x_1,x_2,\dots,x_n)$ in a clearly symmetric form using addition, subtraction, division, and the absolute value function?

REPLY [5 votes]: Symmetrizing in the obvious way:
$$\frac{a+b+c}3+\frac{|a-b|+|b-c|+|a-c|}{12}+\\\frac{|a+b-2c+|a-b||+|a+c-2b+|a-c||+|b+c-2a+|b-c||}{12}$$
For four variables, I found:
$$\tfrac14\left(a+b+c+d+|a-b|+|c-d|+|a+b-c-d+|a-b|-|c-d||\right)$$
Of course you don't have to use absolute value signs: $$\lim_{m\to\infty}\sqrt[m]{a^m+b^m+c^m}$$


text:
\begin{document}

\title[Reducing subspaces on $D$]
 {Reducing subspaces of multiplication \\operators on the Dirichlet space}

\author[Shuaibing Luo]{Shuaibing Luo}

\address{
College of Mathematics and Econometrics\\
Hunan University\\
Changsha, Hunan, 410082\\
PR China}

\email{shuailuo2@126.com}

\thanks{This work is supported by the Young Teacher Program of Hunan University.}

\subjclass{47B35, 46E22}

\keywords{Dirichlet space, Reducing subspaces, Finite Blaschke products.}

\date{\today}

\begin{abstract}
In this paper, we study the reducing subspaces for the multiplication operator by a finite Blaschke product $\phi$ on the Dirichlet space $D$. We prove that any two distinct nontrivial minimal reducing subspaces of $M_\phi$ are orthogonal. When the order $n$ of $\phi$ is $2$ or $3$, we show that $M_\phi$ is reducible on $D$ if and only if $\phi$ is equivalent to $z^n$. When the order of $\phi$ is $4$, we determine the reducing subspaces for $M_\phi$, and we see that in this case $M_\phi$ can be reducible on $D$ when $\phi$ is not equivalent to $z^4$. The same phenomenon happens when the order $n$ of $\phi$ is not a prime number. Furthermore, we show that $M_\phi$ is unitarily equivalent to $M_{z^n} (n > 1)$ on $D$ if and only if $\phi = az^n$ for some unimodular constant $a$.
\end{abstract}

\maketitle

\section{Introduction}
Let $\D = \{z \in \C: |z|<1\}$ be the open unit disc in $\C$ and let $dA$ be the normalized Lebesgue area measure on $\D$. The Dirichlet space $D$ consists of all holomorphic functions $f$ on $\D$ such that
\begin{align*}
D(f) = \int_{\D} |f'|^2 dA = \sum\limits_{n = 1 }^\infty n |\hat{f}(n)|^2 < \infty.
\end{align*}
Endow $D$ with the norm
\begin{align}\label{snorod}
\|f\|_D^2 = \|f\|_{H^2}^2 + \int_{\D} |f'|^2 dA = \sum_{n = 0 }^\infty (n+1) |\hat{f}(n)|^2,
\end{align}
where $H^2$ is the Hardy space. It follows that the reproducing kernel of $D$ has the following form
$$K_\lambda(z) = \frac{1}{\overline{\lambda}z}\log\frac{1}{1-\overline{\lambda}z}.$$

The Dirichlet space has received a lot of attention over the years, see e.g. \cite{ARSW10, ARSW11, ARSW112, EKR09, LR, Ro06}. Articles \cite{ARSW112} and \cite{Ro06} are good surveys about the Dirichlet space which have a number of results and some open questions. Articles \cite{EKR09} and \cite{LR} also contain some open questions for the Dirichlet space.

In a Hilbert space $\HH$, a closed subspace $\HM$ is called a reducing subspace of an operator $T$ if $\HM$ is invariant for both $T$ and $T^*$. An operator $T$ is called reducible if $T$ has a nontrivial reducing subspace. And a nontrivial reducing subspace $\HM$ is called minimal for $T$ if the only reducing subspaces contained in $\HM$ are $\HM$ and $0$.

A function $\phi$ is called a multiplier of $D$ if $\phi D \subseteq D$. Let $M(D)$ denote the multipliers of $D$. By the closed graph theorem, every function $\varphi \in M(D)$ defines a bounded linear operator on $D$. Note that each multiplier in $M(D)$ is a bounded holomorphic function on $\D$, hence $M_\phi$ is also bounded on the Hardy space $H^2$ and the Bergman space $L^2_a$, where the Bergman space $L^2_a$ consists of all square integrable analytic functions on $\D$ and it has the reproducing kernel
$$Q_\lambda(z) = \frac{1}{(1-\overline{\lambda}z)^2}.$$

Given $\lambda \in \D$, let $\varphi_\lambda(z) = \frac{\lambda - z}{1 - \overline{\lambda}z}$ be the M\"{o}bius transform. For finitely many points $\lambda_1, \cdots, \lambda_n \in \D$, let $\phi = \varphi_{\lambda_1}\cdots\varphi_{\lambda_n}$ be the finite Blaschke product with zeros $\lambda_1, \cdots, \lambda_n$.

It is known that for each inner function $\phi$ which is not a M\"{o}bius transform, the reducing subspaces of $M_\phi$ are in one-to-one correspondence with the closed subspaces of $H^2 \ominus \phi H^2$ (\cite{Ha61, No67}). On the Bergman space, for a finite Blaschke product $\phi$, there always exists a nontrivial minimal reducing subspace $M_0(\phi)$ of $M_\phi$ on $L^2_a$ (\cite{GSZZ09, SZ03}). Moreover, Douglas et al.\cite{DPW12} showed that for a finite Blaschke product $\phi$, the number of minimal reducing subspaces of $M_\phi$ on $L^2_a$ equals the number of connected components of the Riemann surface $\phi^{-1}\circ \phi$. We refer the readers to the recent monograph \cite{GH15} for more information about the multiplication operator on the Bergman space.

But on the Dirichlet space $D$, the case is quite different. In \cite{SZ02} Stessin and Zhu showed that the multiplication operator $M_{z^n} (n > 1)$ has exactly $2^n - 2$ proper reducing subspaces on $D$. When $\phi$ is a finite Blaschke product other than $z^n$, little is known about the reducing subspaces of $M_\phi$ on $D$. When $\phi = \varphi_{\lambda_1}\varphi_{\lambda_2}$ is a Blaschke product of order $2$, Zhao \cite{Zh09} proved that on the Dirichlet space with the norm $\|\cdot\|_1$, $M_\phi$ is reducible if and only if $\lambda_1 = - \lambda_2$; Chen and Lee \cite{CL14} obtained a similar result on the Dirichlet space with the norm $\|\cdot\|_D$ defined by (\ref{snorod}). When $\phi$ is a Blaschke product of order $n =2$ or $3$, Chen and Xu \cite{CX13} showed that on the Dirichlet space with the norm $\|\cdot\|_2$, $M_\phi$ is reducible if and only if $\phi$ is equivalent to $z^n$ (see the definition below). However, when order $\phi \geq 3$, it remains open when $M_\phi$ is reducible on the Dirichlet space with the norm $\|\cdot\|_1$ or $\|\cdot\|_D$, see the remark on Page 111 \cite{GH15}.

In this paper, we will use a different method to show that for $\phi = \varphi_{\lambda_1}\varphi_{\lambda_2}$, $M_\phi$ is reducible on the Dirichlet space with the norm defined by (\ref{snorod}) if and only if $\lambda_1 = - \lambda_2$. We also have a similar characterization for $\phi$ with three zeros. To state our results in a unified manner, we introduce some notation. We say that two Blaschke products $\phi_1$ and $\phi_2$ are equivalent if there exist $\lambda \in \D, |a| = 1$ such that
$$\phi_2 = a \varphi_\lambda \circ \phi_1.$$
It can be verified that for two equivalent Blaschke products $\phi_1$ and $\phi_2$, $M_{\phi_1}$ and $M_{\phi_2}$ has the same reducing subspaces, see e.g. \cite{Zh09}.

\begin{thm}\label{prodortt}
Let $\phi$ be a finite Blaschke product of order $n =2$ or $3$. Then $M_\phi$ is reducible on $D$ if and only if there exists $\lambda \in \D$ such that $\phi = a \varphi_\lambda(z^n), |a| = 1$, i.e. $\phi$ is equivalent to $z^n$.
\end{thm}

To prove Theorem \ref{prodortt}, we need the following important observation. Let $U: D \rightarrow L^2_a$ be defined by $Uf = (zf)'$, then $U$ is a unitary operator from $D$ onto $L^2_a$.

\begin{thm}\label{rdsbdbs}
Let $\phi$ be a finite Blaschke product. If $\HM$ is a reducing subspace of $M_\phi$ on $D$, then $U\HM = (z\HM)'$ is a reducing subspace of $M_\phi$ on $L^2_a$.
\end{thm}

Let $\phi$ be a finite Blaschke product of order $n$. If there exists $\lambda \in \D$ such that $\phi = \varphi_\lambda(z^n)$, then by \cite{SZ02}, $M_\phi$ has $2^n - 2$ proper reducing subspaces on $D$. When $2 \leq n \leq 3$, by Theorem \ref{prodortt}, $M_\phi$ has nontrivial reducing subspaces on $D$ if and only if $\phi$ is equivalent to $z^n$. Thus it is natural to ask whether this is the case when $n \geq 4$. Surprisingly, the case is different when $n = 4$.

For a finite Blaschke product $\phi$ we say $\phi$ is decomposable if there exist two Blaschke prodcuts $\psi_1$ and $\psi_2$ with orders greater than $1$ such that $\phi(z) = \psi_1(\psi_2(z))$.

\begin{thm}\label{scnfrsd}
Let $\phi$ be a finite Blaschke product of order $n = 4$. Then one of the following holds.\\
(i) If $\phi$ is equivalent to $z^4$, i.e. there are $\lambda \in \D, |a| = 1$ such that $\phi = a \varphi_\lambda(z^4)$, then $M_\phi$ has exact four nontrivial minimal reducing subspaces on $D$.\\
(ii) If $\phi = \psi_1\circ \psi_2$ is decomposable and $\psi_2$ is equivalent to $z^2$, furthermore, if $\phi$ is not equivalent to $z^4$, then $M_\phi$ has exact two nontrivial minimal reducing subspaces on $D$.\\
(iii) If $\phi = \psi_1\circ \psi_2$ is decomposable and $\phi$ is equivalent to $(z\varphi_\gamma)^2$ for some $\gamma \in \D \backslash \{0\}$, then $M_\phi$ has exact two nontrivial minimal reducing subspaces on $D$.\\
(iv) If $\phi = \psi_1\circ \psi_2$ is decomposable, furthermore, if $\psi_2$ is not equivalent to $z^2$ and $\phi$ is not equivalent to $(z\varphi_\gamma)^2$ for any $\gamma \in \D \backslash \{0\}$, then $M_\phi$ is irreducible on $D$.\\
(v) If $\phi$ is not decomposable, then $M_\phi$ is irreducible on $D$.
\end{thm}

We remark here that in the case (ii) above, the two minimal reducing subspaces $M_1$ and $M_2$ satisfy $\dim (\HM_i \ominus \phi \HM_i) = 2, i = 1, 2$, but for the case (iii) above, the two minimal reducing subspaces $M_1$ and $M_2$ satisfy $\dim (\HM_1 \ominus \phi \HM_1) = 1, \dim (\HM_2 \ominus \phi \HM_2) = 3$.

It is known that for a finite Blaschke product $\phi$, $M_\phi$ is unitarily equivalent to $M_{z^n} (n > 1)$ on $L^2_a$ if and only if $\phi = a \varphi_\lambda^n$ for some $\lambda \in \D$ and $|a| = 1$ (\cite{GZ11, SZZ08}). But this is not the case in the Dirichlet space. On the Dirichlet space with the norm $\|\cdot\|_1$ and $\|\cdot\|_2$, Zhao \cite{Zh11}, Chen and Xu \cite{CX13}, respectively, showed that for a finite Blaschke product $\phi$ of order $n > 1$, $M_\phi$ is unitarily equivalent to $M_{z^n}$ on $D$ if and only if $\phi$ is a constant multiple of $z^n$. On the Dirichlet space with the norm $\|\cdot\|_D$, Chen and Lee \cite{CL14} obtained the same result when the order of $\phi$ is $2$, we show that this is actually true when the order of $\phi$ is $n > 1$.
\begin{thm}\label{fbueztcd}
Let $\phi$ be a finite Blaschke product of order $n > 1$. Then $M_\phi$ is unitarily equivalent to $M_{z^n}$ on $D$ if and only if $\phi = az^n, |a| = 1$.
\end{thm}

\section{The case $n = 2$ or $3$}
We first establish some results which are of independent interest. The following lemma is straightforward and it will be used frequently in this paper.
\begin{lem}\label{iiphbd}
Let $p, q$ be polynomials. Then
$$\langle p, q\rangle_D = \langle (zp)', q\rangle_{H^2} = \langle (zp)', (zq)'\rangle_{L^2_a}.$$
\end{lem}

It was shown in \cite{GZ11} that for a finite Blaschke product $\phi$, the set $\{\phi^k: k = 0, 1, \cdots\}$ is an orthogonal set in $L^2_a$ if and only if $\phi = a z^n$ for some unimodular constant $a$ and $n \geq 1$. This result is nontrivial in the Bergman space, but the analogous result in $D$ is not complicated. Let $\T$ be the unit circle and let $P_\lambda(\zeta) = \frac{1-|\lambda|^2}{|\zeta - \lambda|^2}, \lambda \in \D, \zeta \in \T$ be the Poisson kernel.

\begin{thm}\label{bpoiddfb}
Let $\phi$ be a finite Blaschke product. Then $\{\phi^k: k = 0, 1, \cdots\}$ is an orthogonal set in $D$ if and only if $\phi(0) = 0$.
\end{thm}

\begin{proof}
If $\{\phi^k: k = 0, 1, \cdots\}$ is an orthogonal set in $D$, then
$$\phi(0) = \langle \phi, 1\rangle_D = 0.$$
Conversely, if $\phi(0) = 0$, suppose $\phi = \varphi_{\lambda_1}\cdots \varphi_{\lambda_n}$. Let $k, j \in \N, k < j$, then
\begin{align*}
&\langle \phi^k, \phi^j\rangle_D\\
&= \langle(z\phi^k)', \phi^j\rangle_{H^2}\\
& = \langle z k \phi^{k-1}\phi', \phi^j\rangle_{H^2}\\
& = k \langle z\phi', \phi^{j-k+1}\rangle_{H^2}\\
& = k\int_\T \sum_{i = 1}^n P_{\lambda_i}(\zeta) \overline{\phi^{j-k}(\zeta)}\frac{|d\zeta|}{2\pi}\\
& = k\sum_{i=1}^n \overline{\phi^{j-k}(\lambda_i)}\\
& = 0,
\end{align*}
where we have used Lemma \ref{iiphbd} and the fact that $z\phi' \overline{\phi} = \sum\limits_{i = 1}^n P_{\lambda_i}(z)$ on $\T$.
\end{proof}

The following result is proved in \cite{SZ03}, we prove it in a slightly different way. Suppose $\HH$ be a Hilbert space, $A \subseteq \HH$, we denote by $\text{span}A$ the closed linear span of $A$ in $\HH$.
\begin{thm}[\cite{SZ03}]\label{ddrfbib}
Let $\phi$ be a finite Blaschke product of order $n \geq 2$. Then $M_0(\phi) = \text{span}\{\phi'\phi^j: j = 0, 1, \cdots\}$ is a nontrivial minimal reducing subspace of $M_\phi$ on $L^2_a$.
\end{thm}
\begin{proof}
It is clear that $M_0(\phi)$ is $M_\phi$ invariant, we show that $M_0(\phi)$ is $M_\phi^*$ invariant. Suppose $\phi = \varphi_{\lambda_1}\varphi_{\lambda_2}\cdots \varphi_{\lambda_n}, \lambda_i \in \D$ and $\phi(\varphi_{\lambda_1}) = a z \varphi_{\alpha_1}\cdots \varphi_{\alpha_{n-1}}, \alpha_i \in \D, |a| =1$. Let $\psi = a \varphi_{\alpha_1}\cdots \varphi_{\alpha_{n-1}}$. Then $\phi(\varphi_{\lambda_1}) = z \psi$ and for any polynomial $p$, we have
\begin{align*}
&\langle M_\phi^* \phi', p\rangle_{L^2_a}\\
& = \langle \phi', \phi p\rangle_{L^2_a}\\
& =  \langle \phi'(\varphi_{\lambda_1}) \varphi_{\lambda_1}', \phi(\varphi_{\lambda_1}) p(\varphi_{\lambda_1}) \varphi_{\lambda_1}'\rangle_{L^2_a}\\
& = \langle (z\psi)', z\psi p(\varphi_{\lambda_1}) \varphi_{\lambda_1}'\rangle_{L^2_a}\\
& = \langle \psi, z\psi p(\varphi_{\lambda_1}) \varphi_{\lambda_1}'\rangle_{H^2}\\
& = 0,
\end{align*}
therefore $M_\phi^* \phi' = 0$.

For $j \geq 1$, $\forall k \geq 0$, we have
\begin{align*}
&\langle M_\phi^* (\phi^{j+1})', z^k\rangle_{L^2_a} = \langle (\phi^{j+1})', \phi z^k\rangle_{L^2_a}\\
& = \langle [(z\psi)^{j+1}]', z\psi \varphi_{\lambda_1}^k \varphi_{\lambda_1}'\rangle_{L^2_a}\\
& = \langle \psi(z\psi)^j, z\psi \varphi_{\lambda_1}^k \varphi_{\lambda_1}'\rangle_{H^2}\\
& = \langle (z\psi)^j, z \varphi_{\lambda_1}^k \varphi_{\lambda_1}'\rangle_{H^2}\\
& = \int_\T (z\psi)^j \overline{z \varphi_{\lambda_1}^k \varphi_{\lambda_1}'} \frac{|dz|}{2\pi} \quad \text{let}~~ z = \varphi_{\lambda_1}(w)\\
& = \int_\T \phi^j(w) \overline{\varphi_{\lambda_1}(w) w^k \varphi_{\lambda_1}'(\varphi_{\lambda_1}(w))} \frac{1-|\lambda_1|^2}{|w - \lambda_1|^2}\frac{|dw|}{2\pi}\\
& = \int_\T \phi^j(w) w^{k+1}\frac{|dw|}{2\pi}\\
& = \widehat{(\phi^j)}(k+1),
\end{align*}
where $\widehat{(\phi^j)}(k+1)$ denotes the $(k+1)$-th coefficient of $\phi^j$, hence
\begin{align*}
M_\phi^* (\phi^{j+1})' (z)= \sum_{k=0}^\infty \widehat{(\phi^j)}(k+1) (k+1) z^k = (\phi^j)'(z).
\end{align*}
Thus $M_0(\phi)$ is $M_\phi^*$ invariant and $\dim M_0(\phi) \ominus \phi M_0(\phi) = 1$. It is clear that $M_0(\phi)$ is not $L^2_a$ or see the following remark, therefore $M_0(\phi) = \text{span}\{\phi'\phi^j: j = 0, 1, \cdots\}$ is a nontrivial minimal reducing subspace of $M_\phi$ on $L^2_a$.
\end{proof}

\begin{rem}\label{rkodsrobn}
$M_0(\phi)$ is called the distinguished reducing subspace of $M_\phi$ on $L^2_a ($\cite{GSZZ09}$)$, and $M_\phi|M_0(\phi)$ is unitarily equivalent to $M_z$ on $L^2_a$. In fact,
\begin{align*}
\|\phi'\phi^j\|_{L^2_a}^2& = \frac{1}{j+1}\langle (\phi^{j+1})', \phi^j\phi'\rangle_{L^2_a}\\
& = \frac{1}{j+1} \|\phi'\|_{L^2_a}^2\\
& = \frac{n}{j+1},
\end{align*}
therefore $\{\frac{\sqrt{j+1}}{\sqrt{n}}\phi' \phi^j\}$ is an orthonormal basis of $M_0(\phi)$, then $V: M_0(\phi) \rightarrow L^2_a$ defined by $V \frac{\sqrt{j+1}}{\sqrt{n}}\phi' \phi^j = \sqrt{j+1}z^j$ is unitary and satisfies $V^* M_z V = M_\phi$.

If $\phi = z \varphi_{\lambda_1} \cdots \varphi_{\lambda_{n-1}}$ and $\lambda_1, \cdots, \lambda_{n-1}$ are nonzero and distinct, then $M_0(\phi)^\perp = \text{span}\{\frac{\phi^j}{1- \overline{\lambda_i}z}: 1 \leq i \leq n-1, j =0, 1, \cdots\}$ $($\cite{SZ03}$)$.
\end{rem}

The following two theorems will be used in the proof of Theorem \ref{maintm1od2}.
\begin{thm}\label{dgridtvb}
Let $\phi = z\varphi_{\lambda_1}\cdots\varphi_{\lambda_{n-1}}$ be a finite Blaschke product. Then $\HM = \text{span}\{\frac{\phi^{j+1}}{z}: j = 0, 1, \cdots\}$ is a reducing subspace of $M_\phi$ on $D$ if and only if $\lambda_1 = \cdots = \lambda_{n-1} = 0$.
\end{thm}
\begin{proof}
If $\lambda_1 = \cdots = \lambda_{n-1} = 0$, then it is clear that $\HM$ is a reducing subspace of $M_\phi$ on $D$.

Suppose $\HM = \text{span}\{\frac{\phi^{j+1}}{z}: j = 0, 1, \cdots\}$ is a reducing subspace of $M_\phi$ on $D$. Let $\psi = \varphi_{\lambda_1}\cdots\varphi_{\lambda_{n-1}}$. Then for any $k \geq 0$, we have
\begin{align*}
\langle M_\phi^* \frac{\phi}{z}, z^k\rangle_D &= \langle \frac{\phi}{z}, \phi z^k\rangle_D\\
&= \langle\phi', \phi z^k\rangle_{H^2}\\
& = \langle z\phi', \phi z^{k+1}\rangle_{H^2}\\
& = \int_\T [1 + \sum_{i=1}^{n-1} P_{\lambda_i}(\zeta)] \overline{\zeta^{k+1}}\frac{|d\zeta|}{2\pi}\\
& = \sum_{i=1}^{n-1} \overline{\lambda_i^{k+1}},
\end{align*}
thus
$$M_\phi^* \frac{\phi}{z} = \sum_{k=0}^\infty \sum_{i=1}^{n-1} \overline{\lambda_i^{k+1}} \frac{1}{k+1}z^k = \sum_{i=1}^{n-1} \overline{\lambda_i}K_{\lambda_i}.$$
If $\lambda_i \neq 0$, then $K_{\lambda_i} \in \HM^\perp$. Since $M_\phi^* \frac{\phi}{z} \in \HM$, we conclude that $\lambda_i = 0, i = 1, \cdots, n-1$.

\end{proof}

Let $\phi = \varphi_{\lambda_1}\varphi_{\lambda_2}\cdots \varphi_{\lambda_n}, \lambda_i \in \D$. Then for $f,g \in D, k \geq 0$, we have
$$D(\phi^k f, \phi^k g) = \int_\T k \sum_{i=1}^n P_{\lambda_i}(\zeta) f(\zeta) \overline{g(\zeta)} \frac{|d\zeta|}{2\pi} + D(f,g),$$
see \cite{Zh09} or \cite{RS91}.

Recall that $U: D \rightarrow L^2_a$ defined by $Uf = (zf)'$ is unitary.
\begin{thm}\label{viorbdb}
Let $\phi$ be a finite Blaschke product. If $\HM$ is a reducing subspace of $M_\phi$ on $D$, then $U\HM = (z\HM)'$ is a reducing subspace of $M_\phi$ on $L^2_a$.
\end{thm}

\begin{proof}
Let $f \in \HM, g \in \HM^\perp$. Then for any $k, j \in \N$, we have $\langle \phi^k f, \phi^jg\rangle_D = 0$. Thus for any $m >0$,
\begin{align*}
0 &= \langle \phi^{m+k}f, \phi^{m+j}g\rangle_D\\
& = \langle \phi^{m+k}f, \phi^{m+j}g\rangle_{H^2} + D(\phi^{m+k}f, \phi^{m+j}g)\\
& = \langle \phi^{k}f, \phi^{j}g\rangle_{H^2} + \int_\T m \sum_{i=1}^n P_{\lambda_i}(\zeta) \phi^{k}(\zeta)f(\zeta) \overline{\phi^{j}(\zeta)g(\zeta)} \frac{|d\zeta|}{2\pi} + D(\phi^{k}f,\phi^{j}g)\\
& = \langle \phi^{k}f, \phi^{j}g\rangle_{D} + \int_\T m \sum_{i=1}^n P_{\lambda_i}(\zeta) \phi^{k}(\zeta)f(\zeta) \overline{\phi^{j}(\zeta)g(\zeta)} \frac{|d\zeta|}{2\pi}\\
& = \int_\T m \sum_{i=1}^n P_{\lambda_i}(\zeta) \phi^{k}(\zeta)f(\zeta) \overline{\phi^{j}(\zeta)g(\zeta)} \frac{|d\zeta|}{2\pi},
\end{align*}
therefore
$$ 0 = \int_\T \sum_{i=1}^n P_{\lambda_i}(\zeta) \phi^{k}(\zeta)f(\zeta) \overline{\phi^{j}(\zeta)g(\zeta)} \frac{|d\zeta|}{2\pi} = \langle z\phi'\phi^kf, \phi\phi^jg\rangle_{H^2}.$$
This implies for any $l \geq 1$, $\langle z(\phi^l)'f, g\rangle_{H^2} = 0$. Note that
\begin{align*}
\langle z(\phi^l)'f, g\rangle_{H^2} &= \langle (z\phi^lf)' - (zf)'\phi^l, g\rangle_{H^2}\\
&= \langle f\phi^l,g\rangle_D - \langle(zf)'\phi^l, g\rangle_{H^2}\\
& = - \langle(zf)'\phi^l, (zg)'\rangle_{L^2_a},
\end{align*}
hence $\langle(zf)'\phi^l, (zg)'\rangle_{L^2_a} = 0$, and so $U\HM = (z\HM)'$ is $M_\phi$ invariant. Similarly, $U\HM^\perp = (z\HM^\perp)'$ is $M_\phi$ invariant. Therefore $U\HM = (z\HM)'$ is a reducing subspace of $M_\phi$ on $L^2_a$.
\end{proof}

Let $\{M_\phi\}' = \{T \in B(D): M_\phi T = T M_\phi\}$ be the commutant of $M_\phi$. Since there is a one to one correspondence between the reducing subspaces of $M_\phi$ and the projections in $\{M_\phi\}'$, the problem of classifying the reducing subspaces of $M_\phi$ is equivalent to finding the projections in $\{M_\phi\}'$. Let $\HA_\phi = \{M_\phi, M_\phi^*\}'$. Then $\HA_\phi$ is a von Neumann algebra. Zhu \cite{Zh00} conjectured that for a finite Blaschke product $\phi$ of order $n$, there are exactly $n$ distinct minimal reducing subspaces of $M_\phi$ on $L^2_a$. This is equivalent to the statement that $\HA_\phi$ has exactly $n$ minimal projections. Zhu's conjecture does not hold in general, and it is modified as follows: $M_\phi$ has at most $n$ distinct minimal reducing subspaces on $L^2_a$ (\cite{GH15}). This modified conjecture was proved in \cite{DPW12}, they showed that the von Neumann algebra $\HA_\phi$ in $B(L^2_a)$ is commutative of dimension $q$, where $q$ is the number of connected components of the Riemann surface $\phi^{-1}\circ \phi$. We have a similar result in the Dirichlet space.

\begin{thm}\label{cotovnaod}
Let $\phi$ be a finite Blaschke product of order $n$. If $\HM_1$ and $\HM_2$ are two distinct nontrivial minimal reducing subspaces of $M_\phi$ on $D$, then $\HM_1$ and $\HM_2$ are orthogonal.
\end{thm}
\begin{proof}
Let $\HM_1$ and $\HM_2$ be two distinct nontrivial minimal reducing subspaces of $M_\phi$ on $D$. If $\HM_1$ and $\HM_2$ are not orthogonal, then $\HM_1$ is unitarily equivalent to $\HM_2$, i.e. there is a unitary operator $U$ from $\HM_1$ onto $\HM_2$ commuting with $M_\phi$ (\cite[Theorem3.3]{GH11}). Let $\HM = \HM_1 \cap \HM_2$, then $\HM$ is $\{0\}$, otherwise $\HM$ is a nontrivial reducing subspace of $M_\phi$ on $D$, hence $\HM = \HM_1 = \HM_2$ by the minimality of $\HM_1$ and $\HM_2$.

Now for each $0 < a <1$, set $\HM_1 = \{f + a U f: f \in \HM_1\}$. Then $\HM_a$ is a reducing subspace of $M_\phi$ on $D$. Moreover, if $0 < a < b <1$, then $\HM_a \neq \HM_b$. Let $\HN_a = (z\HM_a)'$. Then for each $0 < a <1$, $\HN_a$ is a reducing subspace of $M_\phi$ on $L^2_a$ and $\HN_a \neq \HN_b$ if $0 < a < b <1$. But this contradicts the fact that $\HA_\phi$ in $B(L^2_a)$ is commutative and has dimension $q \leq n$ (\cite{DPW12}). Thus $\HM_1$ and $\HM_2$ are orthogonal.
\end{proof}

It follows from the above theorem that the von Neumann algebra $\HA_\phi$ in $B(D)$ is commutative.

Now we prove Theorem \ref{prodortt} when $\phi$ has order $2$.
\begin{thm}\label{maintm1od2}
Let $\phi = \varphi_\alpha \varphi_\beta$ be a Blaschke product of order $2$. Then $M_\phi$ is reducible on $D$ if and only if $\phi$ is equivalent to $z^2$.
\end{thm}
\begin{proof}
As noted before, we only need to prove the "only if" part. Suppose $M_\phi$ is reducible on $D$. Let $\HM$ be a nontrivial reducing subspace of $M_\phi$ on $D$, then $(z\HM)'$ is a nontrivial reducing subspace of $M_\phi$ on $L^2_a$. Since $M_\phi$ has only two minimal reducing subspaces $M_0(\phi)$ and $M_0(\phi)^\perp$ on $L^2_a$. Without loss of generality, assume that $(z\HM)' = M_0(\phi)$.

Note that $\varphi_{\alpha\beta}(\phi) = az \varphi_\gamma$ for some $|a| =1, \gamma \in \D$. Let $\varphi = \varphi_{\alpha\beta}(\phi)$. Then $M_0(\varphi) = \text{span}\{\varphi'\varphi^j: j \geq 0\} \subseteq M_0(\phi)$, since $M_0(\varphi)$ is reducing for $M_\phi$ and $M_0(\phi)$ is minimal, we have $ M_0(\varphi) = M_0(\phi)$. Therefore $\HM = \text{span}\{\frac{\varphi^{j+1}}{z}: j = 0, 1, \cdots\}$. It follows from Theorem \ref{dgridtvb} that $\gamma = 0$, so $\phi$ is equivalent to $z^2$.
\end{proof}

Before we prove Theorem \ref{prodortt} when $\phi$ has order $3$, we need one more lemma.
\begin{lem}\label{pfnznnrd}
Given $\alpha \in \D \backslash \{0\}$, let $\phi = \varphi_\alpha^3$. Then $M_\phi$ is irreducible on $D$.
\end{lem}
\begin{proof}
Let $U_\alpha: L^2_a \rightarrow L^2_a$ be defined by $U_\alpha f = f(\varphi_\alpha) q_\alpha$, where $q_\alpha(z) = \frac{1-|\alpha|^2}{(1-\overline{\alpha}z)^2}$ is the normalized reproducing kernel for $L^2_a$. Then $U_\alpha$ is a unitary operator and $U_\alpha^* M_\phi U_\alpha = M_{\phi \circ \varphi_\alpha} = M_{z^3}$.

Since $M_{z^3}$ has exact three nontrivial minimal reducing subspaces on $L^2_a$, we obtain that $M_\phi$ has exact three nontrivial minimal reducing subspaces on $L^2_a$ (\cite{GSZZ09}). These reducing subspaces are $M_0(\phi) = U_\alpha (M_0(z^3))$, $M_1 = U_\alpha [\text{span}\{1,z^3,z^6,\cdots\}] = \text{span}\{q_\alpha \phi^j: j \geq 0\}$ and $M_2 = U_\alpha [\text{span}\{z,z^4,z^7,\cdots\}] = \text{span}\{\varphi_\alpha q_\alpha \phi^j: j \geq 0\}$.

Let $\HM$ be a nontrivial reducing subspace of $M_\phi$ on $D$, then $(z\HM)'$ is a nontrivial reducing subspace of $M_\phi$ on $L^2_a$. There are essentially three cases: $(z\HM)' = M_0(\phi), M_1$ or $M_2$. By the same argument as in Theorem \ref{maintm1od2}, we see that $(z\HM)' \neq M_0(\phi)$.

If $(z\HM)' = M_1$, then
$$\HM = \text{span} \{\frac{\varphi_\alpha^{3j+1} - \varphi_\alpha^{3j+1}(0)}{z}: j = 0, 1, \cdots\}.$$
Since $\frac{\varphi_\alpha - \varphi_\alpha(0)}{z} \in \HM$, we have
$$\phi \frac{\varphi_\alpha - \varphi_\alpha(0)}{z} = \frac{\varphi_\alpha^4 - \varphi_\alpha^4(0)}{z} + \varphi_\alpha(0) \frac{\phi(0) - \phi}{z} \in \HM.$$
It follows that $\frac{\phi - \phi(0)}{z} \in \HM$, but $\frac{\phi - \phi(0)}{z} \in \HM^\perp$. This is a contradiction, therefore $(z\HM)' \neq M_1$.

If $(z\HM)' = M_2$, then
$$\HM = \text{span} \{\frac{\varphi_\alpha^{3j+2} - \varphi_\alpha^{3j+2}(0)}{z}: j = 0, 1, \cdots\}.$$
Similarly, we conclude that this is impossible. So $M_\phi$ is irreducible on $D$.
\end{proof}

Let $\phi$ be a finite Blaschke product of order $n$, then by Bochner's theorem (\cite{Wa18, Wa50}), $\phi'$ has exactly $n -1$ zeros in $\D$, i.e. $\phi'$ has $n -1$ critical points in $\D$.

Now we prove Theorem \ref{prodortt} when $\phi$ has order $3$.
\begin{thm}\label{maintm2od3}
Let $\phi = \varphi_{\alpha_1} \varphi_{\alpha_2} \varphi_{\alpha_3}$ be a Blaschke product of order $3$. Then $M_\phi$ is reducible on $D$ if and only if $\phi$ is equivalent to $z^3$.
\end{thm}
\begin{proof}
As noted before, we only need to prove the "only if" part. Suppose $M_\phi$ is reducible on $D$. If $\phi$ has no multiple critical point in $\D$, then by \cite[Theorem 4]{GSZZ09}, $M_\phi$ has only two minimal reducing subspaces $M_0(\phi)$ and $M_0(\phi)^\perp$ on $L^2_a$. By the same argument as in Theorem \ref{maintm1od2}, we conclude that $\phi$ is equivalent to $z^3$.

If $\phi$ has a multiple critical point $c$ in $\D$, let $\lambda = \phi(c)$, then $\varphi_\lambda(\phi) = a \varphi_c^3$ for some $|a| = 1$. It follows from Lemma \ref{pfnznnrd} that $c = 0$, so $\phi$ is equivalent to $z^3$.
\end{proof}

\section{The case $n = 4$}
Let $\phi$ be a finite Blaschke product. Recall that if $\HM$ is a reducing subspace of $M_\phi$ on $D$, then $(z\HM)'$ is a reducing subspace of $M_\phi$ on $L^2_a$. Thus it is natural to ask the following question: if $\HM$ is a nontrivial minimal reducing subspace of $M_\phi$ on $D$, then is it true that $(z\HM)'$ is a nontrivial minimal reducing subspace of $M_\phi$ on $L^2_a$? We will see that this is not always the case.

We break the proof of Theorem \ref{scnfrsd} into a series of lemmas.

First let's look at an observation. Let $\phi$ be a finite Blaschke product of order $4$. Suppose $\phi = \psi_1 \circ \psi_2$ is decomposable, then $M_\phi$ has reducing subspaces $M_0(\phi), M_0(\psi_2) \ominus M_0(\phi)$ and $M_0(\psi_2)^\perp$ on $L^2_a$ (\cite[Theorem 2.1]{SZZ10}). Suppose $M_\phi$ is reducible on $D$ and let $\HM$ be a reducing subspace of $M_\phi$ on $D$. We wonder whether $(z\HM)'$ can be $M_0(\psi_2)$. If this is the case, then $\HM = \text{span} \{\frac{\psi_2^{j+1} - \psi_2^{j+1}(0)}{z}: j \geq 0\}$ and it is $M_\phi$ invariant. Suppose $\psi_2(z) = \varphi_{\alpha_3}(z) \varphi_{\alpha_4}(z), \alpha_3, \alpha_4 \in \D$, by a little calculation, we have
$$M_\phi^* \frac{\psi_2 - \psi_2(0)}{z} = \overline{\psi_1(0)}\frac{\psi_2 - \psi_2(0)}{z} + \overline{\psi'_1(0)}(\overline{\alpha_3}K_{\alpha_3} + \overline{\alpha_4}K_{\alpha_4}).$$
Thus if $\alpha_3 = \alpha_4 = 0$, then $M_\phi^* \frac{\psi_2 - \psi_2(0)}{z} \in \HM$. Similarly, we have $M_\phi^* \frac{\psi_2^{j+1} - \psi_2^{j+1}(0)}{z} \in \HM, j \geq 1$. Therefore if $\psi_2(z) = z^2$, then $\HM = \text{span} \{z, z^3, z^5, \cdots\}$ is a reducing subspace of $M_\phi$ on $D$. More generally, we have the following result.

\begin{lem}\label{rdcfbofd}
Let $\phi$ be a finite Blaschke product of order $4$. If $\phi = \psi_1 \circ \psi_2$ is decomposable and $\psi_2$ is equivalent to $z^2$, furthermore, if $\phi$ is not equivalent to $z^4$, then $M_\phi$ has exact two nontrivial minimal reducing subspaces on $D$.
\end{lem}
\begin{proof}
Suppose $\phi = \psi_1 \circ \psi_2$ is decomposable and $\psi_2$ is equivalent to $z^2$, then there are $\alpha \in \D, |a| = 1$ such that $\psi_2 = a\varphi_\alpha(z^2)$. Thus $\psi_1(\psi_2(z)) = \psi_1 (a\varphi_\alpha(z^2)) = b \widetilde{\psi_1} (z^2)$, where $b \widetilde{\psi_1} = \psi_1 (a\varphi_\alpha)$ is a Blaschke product of order $2$ and $|b| = 1$.

Note that $M_0(\psi_2) = M_0(z^2) = \text{span}\{z, z^3, \cdots\}$ in $L^2_a$. Let $\HM \subseteq D$ be such that $(z\HM)' = M_0(\psi_2)$, then $\HM = \text{span}\{z, z^3, z^5, \cdots\}$ in $D$. It is clear that $\HM$ is $M_\phi$ invariant, also $\HM^\perp = \text{span}\{1, z^2, z^4, \cdots\}$ in $D$ is $M_\phi$ invariant, thus $\HM$ is a nontrivial reducing subspace of $M_\phi$ on $D$.

Next we show that $\HM$ is a minimal reducing subspace. Let $0 \neq \HN \subseteq \HM$ be a reducing subspace of $M_\phi$ on $D$, then $0 \neq (z\HN)' \subseteq (z\HM)'$ is a reducing subspace of $M_\phi$ on $L^2_a$. Since $M_\phi$ has exact three minimal reducing subspaces $M_0(\phi), M_0(\psi_2) \ominus M_0(\phi)$ and $M_0(\psi_2)^\perp$ on $L^2_a$ (\cite[Theorem 2.1]{SZZ10}), it follows that $(z\HN)' = M_0(\phi)$. By the same argument as in Theorem \ref{maintm1od2}, we conclude that $\phi$ is equivalent to $z^4$, this contradicts the assumption that $\phi$ is not equivalent to $z^4$. So $\HM$ is a nontrivial minimal reducing subspace of $M_\phi$ on $D$. It also follows from \cite[Theorem 2.1]{SZZ10} that $\HM^\perp = \text{span}\{1, z^2, z^4, \cdots\}$ in $D$ is a nontrivial minimal reducing subspace of $M_\phi$ on $D$ and $M_\phi$ does not have any other minimal reducing subspaces on $D$. This completes the proof.
\end{proof}

\begin{rem}\label{minrddnb}
As in the above proof we have $\phi(z) = b \widetilde{\psi_1}(z^2)$. Suppose $\widetilde{\psi_1} = \varphi_{\alpha_1^2} \varphi_{\alpha_2^2}, \alpha_1, \alpha_2 \in \D \backslash \{0\}$, then
$$\phi(z) = b\varphi_{\alpha_1^2} (z^2) \varphi_{\alpha_2^2}(z^2) = b \varphi_{\alpha_1}(z) \varphi_{-\alpha_1}(z) \varphi_{\alpha_2}(z) \varphi_{-\alpha_2}(z).$$
If $\alpha_1 \neq \alpha_2$ and $\alpha_1 \neq -\alpha_2$, then $\ker M_\phi^* \cap \HM = \text{span} \{K_{\alpha_1} - K_{-\alpha_1}, K_{\alpha_2} - K_{-\alpha_2}\}$ is two dimensional, and $\ker M_\phi^* \cap \HM^\perp$ is also two dimensional. If $\alpha_1 = \alpha_2$ or $\alpha_1 = -\alpha_2$, then $\ker M_\phi^* \cap \HM = \text{span} \{K_{\alpha_1} - K_{-\alpha_1}, \frac{1}{1-\overline{\alpha_1}z} - \frac{1}{1+\overline{\alpha_1}z}\}$ is two dimensional, and $\ker M_\phi^* \cap \HM^\perp$ is also two dimensional. Thus $M_\phi$ has two nontrivial minimal reducing subspaces $\HM$ and $\HM^\perp$ on $D$, and they satisfy that $\dim (\HM \ominus \phi\HM) = \dim (\HM^\perp \ominus \phi\HM^\perp) = 2$, and $(z\HM)'$ is not a minimal reducing subspaces of $M_\phi$ on $L^2_a$.
\end{rem}

\begin{rem}\label{rdswoninpod}
Suppose $n$ is not a prime number, then $n = pq$ for some $p, q > 1$. Let $\psi_1$ be a Blaschke product of order $p$ and let $\phi = \psi_1(z^q)$. Let
$$\HM_j = \text{span}\{z^j, z^{q+j}, z^{2q+j}, \cdots\}, \quad j =0, 1, \cdots, q-1.$$
Then each $M_j$ is a reducing subspace of $M_\phi$ on $D$. Thus $M_\phi$ is reducible on $D$ when $\phi = \psi_1(z^q), q >1$.
\end{rem}

\begin{lem}\label{nrfdczpess}
Given $\gamma \in \D \backslash \{0\}$, let $\psi = z \varphi_\gamma$ and $\phi = \psi^2$. Then $M_\phi$ has exact two nontrivial minimal reducing subspaces on $D$.
\end{lem}

\begin{proof}
By \cite[Theorem 2.1]{SZZ10}, $M_\phi$ has exact three minimal reducing subspaces on $L^2_a$, $M_0(\phi), M_0(\psi) \ominus M_0(\phi)$ and $M_0(\psi)^\perp$.

Let $\HM \subseteq D$ be such that $(z\HM)' = M_0(\psi) \ominus M_0(\phi)$. Note that
$$M_0(\psi) = \text{span}\{\psi'\psi^j: j\geq 0\},$$
$$M_0(\phi) = \text{span}\{\phi'\phi^j: j\geq 0\} = \text{span}\{\psi'\psi^{2j+1}: j\geq 0\},$$
thus $M_0(\psi) \ominus M_0(\phi) = \text{span} \{\psi'\psi^{2j}: j\geq 0\}$, and so $\HM = \text{span} \{\frac{\psi^{2j+1}}{z}: j \geq 0\}$. It is clear that $\HM$ is $M_\phi$ invariant.

Now we show that $\HM$ is $M_\phi^*$ invariant. For any $k \geq 0$,
\begin{align*}
\langle M_\phi^* \frac{\psi}{z}, z^k\rangle_D& = \langle \frac{\psi}{z}, \phi z^k\rangle_D = \langle \psi', \psi^2 z^k\rangle_{H^2}= \langle z\psi', \psi^2 z^{k+1}\rangle_{H^2}\\
& = \int_\T [1+P_\gamma(\zeta)] \overline{\psi(\zeta)\zeta^{k+1}} \frac{|d\zeta|}{2\pi}\\
&= \overline{\psi(\gamma)\gamma^{k+1}} = 0,
\end{align*}
hence $M_\phi^* \frac{\psi}{z} = 0$.

For any $j \geq 1, k \geq 0$,
\begin{align*}
\langle M_\phi^* \frac{\psi^{2j+1}}{z}, z^k\rangle_D& = \langle \frac{\psi^{2j+1}}{z}, \phi z^k\rangle_D = \langle(2j+1)\psi^{2j} \psi', \psi^2 z^k\rangle_{H^2}\\
&= \langle (2j+1)\psi^{2j-2} \psi', z^{k}\rangle_{H^2}\\
& = \frac{2j+1}{2j-1} \widehat{(\psi^{2j-1})'}(k),
\end{align*}
therefore
$$M_\phi^* \frac{\psi^{2j+1}}{z} = \frac{2j+1}{2j-1} \sum_{k=0}^\infty \widehat{(\psi^{2j-1})'}(k) \frac{1}{k+1} z^k = \frac{2j+1}{2j-1} \frac{\psi^{2j-1}}{z}.$$
It follows that $\HM$ is $M_\phi^*$ invariant and $\dim \HM \ominus \phi\HM =1$. So $\HM$ is a nontrivial minimal reducing subspace of $M_\phi$ on $D$. Then $\HM^\perp$ is also a reducing subspace of $M_\phi$ on $D$, and $(z\HM^\perp)' = M_0(\phi) \oplus M_0(\psi)^\perp$. Since $M_\phi$ is not equivalent to $z^4$, there is no reducing subspace $\HN$ of $M_\phi$ on $D$ such that $(z\HN)' = M_0(\phi)$, we conclude that $\HM^\perp$ is a nontrivial minimal reducing subspace of $M_\phi$ on $D$ and it satisfies $\dim \HM^\perp \ominus \phi\HM^\perp =3$. This finishes the proof.
\end{proof}

The following theorem is of independent interest.
\begin{thm}\label{nripizf}
Given $\alpha \in \D \backslash \{0\}$, let $\phi = \varphi_\alpha^4$. Then $M_\phi$ is irreducible on $D$.
\end{thm}
\begin{proof}
Suppose $M_\phi$ is reducible on $D$ and $\HM$ is a nontrivial reducing subspace of $M_\phi$ on $D$, then $(z\HM)'$ is a reducing subspace of $M_\phi$ on $L^2_a$. Note that $M_\phi$ has exact four minimal reducing subspaces on $L^2_a$, $M_0(\phi), M_1 = \text{span}\{q_\alpha \phi^j: j \geq 0\}, M_2 = \text{span}\{q_\alpha\varphi_\alpha \phi^j: j \geq 0\}$ and $M_3 = \text{span}\{q_\alpha \varphi_\alpha^2 \phi^j: j \geq 0\}$. By the same argument as in Lemma \ref{pfnznnrd}, we see that $(z\HM)' \neq M_i, 0 \leq i \leq 3$.

Suppose $(z\HM)' = M_0(\phi) \oplus M_1 = \text{span}\{q_\alpha \varphi_\alpha^3 \phi^j, q_\alpha \phi^j: j \geq 0\}$ in $L^2_a$, then $\HM = \text{span}\{\frac{\phi^{j+1} - \phi^{j+1}(0)}{z}, \frac{\varphi_\alpha^{4j+1} - \varphi_\alpha^{4j+1}(0)}{z}: j \geq 0\}$. It can be verified that $M_\phi^* \left[\frac{\phi - \phi(0)}{z}\right] = 4 \overline{\alpha} K_\alpha$. Then $(zK_\alpha)' = \frac{1}{1-\overline{\alpha}z} \in M_0(\phi) \oplus M_1$. Recall that $U_\alpha: L^2_a \rightarrow L^2_a, U_\alpha f = f\circ \varphi_\alpha q_\alpha$ is unitary, it follows that
\begin{align*}
U_\alpha \frac{1}{1-\overline{\alpha}z} = \frac{1}{1-\overline{\alpha}z} &\in U_\alpha \left[M_0(\phi) \oplus M_1\right]\\
 &= \text{span} \{1, z^3, z^4, z^7, z^8, \cdots\}.
\end{align*}
This is impossible, thus $(z\HM)' \neq M_0(\phi) \oplus M_1$. Similarly, $(z\HM)' \neq M_0(\phi) \oplus M_2$ and $(z\HM)' \neq M_0(\phi) \oplus M_3$. Since $M_0(\phi), M_1, M_2$ and $M_3$ are the exact four minimal reducing subspaces of $M_\phi$ on $L^2_a$, we conclude that $M_\phi$ is irreducible on $D$.
\end{proof}

\begin{rem}\label{irdnznpc}
Following the same argument as above, we conclude that if $\phi = \varphi_\alpha^n$ for some $n>1, \alpha \in \D \backslash \{0\}$, then $M_\phi$ is irreducible on $D$.
\end{rem}

\begin{lem}\label{nrfdczes}
Let $\phi$ be a finite Blaschke product of order $4$. Suppose $\phi = \psi_1\circ \psi_2$ is decomposable. If $\psi_2$ is not equivalent to $z^2$ and $\phi$ is not equivalent to $(z\varphi_\gamma)^2$ for any $\gamma \in \D \backslash \{0\}$, then $M_\phi$ is irreducible on $D$.
\end{lem}
\begin{proof}
Suppose $\psi_2 = \varphi_{\alpha_3}\varphi_{\alpha_4}, \alpha_3, \alpha_4 \in \D$, then $\alpha_3 \neq - \alpha_4$. Assume that $\varphi_{\alpha_3\alpha_4}(\psi_2) = az\varphi_\gamma$ for some $\gamma \in \D \backslash \{0\}$ and $|a| = 1$, then $\psi_1 \circ \psi_2 = \psi_1 \circ \varphi_{\alpha_3\alpha_4}(az\varphi_\gamma) = \widetilde{\psi_1}(az\varphi_\gamma)$, where $\widetilde{\psi_1} = \psi_1 \circ \varphi_{\alpha_3\alpha_4}$ is a Blaschke product of order $2$.

Suppose $\varphi_\lambda \circ \widetilde{\psi_1} = b z \varphi_\alpha$ for some $\lambda, \alpha \in \D$ and $|b| = 1$, then
\begin{align}\label{erodtbot}
\varphi_\lambda \circ \phi = \varphi_\lambda \circ \widetilde{\psi_1}(az\varphi_\gamma) = baz\varphi_\gamma \cdot \varphi_\alpha(az\varphi_\gamma) := c z\varphi_\gamma \varphi_{\beta_1} \varphi_{\beta_2},
\end{align}
where $\beta_1, \beta_2 \in \D, |c| = 1$. Since $\gamma \neq 0$, we have $\beta_1 \neq -\beta_2$. We mention here that if $\beta_1\beta_2 = 0$ or $\varphi_{\beta_1}(\gamma) = 0$ or $\varphi_{\beta_2}(\gamma) = 0$, then $\alpha = 0$.

Suppose $M_\phi$ is reducible on $D$ and let $\HM$ be a nontrivial reducing subspace of $M_\phi$ on $D$, then $(z\HM)'$ is a nontrivial reducing subspace of $M_\phi$ on $L^2_a$. Since $M_\phi$ has exact three nontrivial minimal reducing subspaces $M_0(\phi), M_0(\psi_2) \ominus M_0(\phi)$ and $M_0(\psi_2)^\perp$ on $L^2_a$ (\cite{SZZ10}), and $(z\HM)' \neq M_0(\phi)$, note that $[(z\HM)']^\perp = (z\HM^\perp)'$, we conclude that there are essentially two cases, either $(z\HM)' = M_0(\psi_2)$ or $(z\HM)' = M_0(\phi) \oplus M_0(\psi_2)^\perp$.

Let $\psi = \varphi_{\alpha_3\alpha_4}(\psi_2) = a z \varphi_\gamma$ and let $\varphi = \varphi_\lambda \circ \phi = (b z \varphi_\alpha)\circ (az\varphi_\gamma) = c z\varphi_\gamma \varphi_{\beta_1} \varphi_{\beta_2}$, then $M_0(\psi_2) = M_0(\psi)$ and $\varphi = c \overline{a} \psi \varphi_{\beta_1} \varphi_{\beta_2}$. If $(z\HM)' = M_0(\psi_2)$, then
$$\HM = \text{span} \{\frac{\psi^{j+1}}{z}: j \geq 0\}.$$

Note that $K_\gamma \in \HM^\perp$, we obtain that $\varphi K_\gamma \in \HM^\perp$, then
\begin{align*}
0 & = \langle \frac{\psi}{z}, \varphi K_\gamma\rangle_D = \langle \psi', \varphi K_\gamma\rangle_{H^2}\\
& = \langle z \psi',  zc \overline{a} \psi \varphi_{\beta_1} \varphi_{\beta_2} K_\gamma\rangle_{H^2}\\
& = \int_\T [1+P_\gamma(\zeta)] \overline{c \overline{a} \zeta \varphi_{\beta_1}(\zeta) \varphi_{\beta_2}(\zeta) K_\gamma(\zeta)} \frac{|d\zeta|}{2\pi}\\
& = \overline{c \overline{a} \gamma \varphi_{\beta_1}(\gamma) \varphi_{\beta_2}(\gamma) K_\gamma(\gamma)},
\end{align*}
therefore $\varphi_{\beta_1}(\gamma) \varphi_{\beta_2}(\gamma) = 0$, it follows that $\alpha = 0$. Hence $\varphi = \varphi_\lambda \circ \phi = (b z^2)\circ (az\varphi_\gamma) = b (az\varphi_\gamma)^2$ which contradicts the assumption that $\phi$ is not equivalent to $(z\varphi_\gamma)^2$ for any $\gamma \in \D \backslash \{0\}$. Thus $(z\HM)' \neq M_0(\psi_2)$.

If $(z\HM)' = M_0(\phi) \oplus M_0(\psi_2)^\perp$. Let $\HM_1 = \text{span}\{\frac{\varphi^{j+1}}{z}: j \geq 0\}$, then $\HM_1 \subseteq \HM$ and $(z\HM_1)' = M_0(\varphi) = M_0(\phi)$. By the calculation in Theorem \ref{dgridtvb}, we have
\begin{align}\label{ieunues}
M_\varphi^* \frac{\varphi}{z} = \overline{\gamma}K_\gamma + \overline{\beta_1}K_{\beta_1} + \overline{\beta_2}K_{\beta_2} \in \HM_1^\perp \cap \HM.
\end{align}
Let $\HM_2 \subseteq D$ be such that $(z\HM_2)' = M_0(\psi_2)^\perp$, then $\HM = \HM_1 \oplus \HM_2$. Note that $M_0(\psi_2)^\perp = \text{span}\{\frac{\psi^m}{1-\overline{\gamma}z}: m \geq 0\}$ (\cite{SZ03}), we have $K_\gamma \in \HM_2$, hence by (\ref{ieunues}), we obtain $\overline{\beta_1}K_{\beta_1} + \overline{\beta_2}K_{\beta_2} \in \HM_2$ and so $g: = \overline{\beta_1} \frac{1}{1 - \overline{\beta_1} z} + \overline{\beta_2} \frac{1}{1 - \overline{\beta_2} z} \in M_0(\psi_2)^\perp$. Then $g \psi \in M_0(\psi_2)^\perp$, it follows that
\begin{align*}
0 = \langle g \psi, \varphi'\rangle_{L^2_a} &= \langle g \psi, (c z\varphi_\gamma \varphi_{\beta_1} \varphi_{\beta_2})'\rangle_{L^2_a}\\
& = \langle a z \varphi_\gamma g, c\varphi_\gamma \varphi_{\beta_1} \varphi_{\beta_2}\rangle_{H^2}\\
& = \langle a z g, c \varphi_{\beta_1} \varphi_{\beta_2}\rangle_{H^2}\\
& = \langle a\overline{c} g,  \frac{\varphi_{\beta_1} \varphi_{\beta_2} - \varphi_{\beta_1}(0) \varphi_{\beta_2}(0)}{z}\rangle_{H^2}\\
& = a\overline{c} (-2\overline{\beta_1}\overline{\beta_2}),
\end{align*}
thus $\beta_1 \beta_2 = 0$. From (\ref{erodtbot}), we conclude that $\alpha = 0$. Therefore $\varphi = \varphi_\lambda \circ \phi = b (az\varphi_\gamma)^2$ which contradicts the assumption that $\phi$ is not equivalent to $(z\varphi_\gamma)^2$ for any $\gamma \in \D \backslash \{0\}$. Hence $(z\HM)' \neq M_0(\phi) \oplus M_0(\psi_2)^\perp$ and so $M_\phi$ is irreducible on $D$.
\end{proof}

If $\phi = \varphi_\lambda^2 (z\varphi_\gamma)$ for some $\lambda \in \D \backslash \{0\}, \gamma \in \D \backslash \{0\}$, then $\phi$ satisfies the assumption in the above lemma, thus $M_\phi$ is irreducible on $D$. We can also verify directly that $M_\phi$ is irreducible on $D$ in this case.

Now we can prove Theorem \ref{scnfrsd}:
\begin{proof}[Proof of Theorem \ref{scnfrsd}]
(i) If $\phi$ is equivalent to $z^4$, then by \cite{SZ02}, we see that the conclusion is true. (ii), (iii) and (iv) follow from Lemmas \ref{rdcfbofd}, \ref{nrfdczpess} and \ref{nrfdczes} respectively.

(v) If $\phi$ is not decomposable, then by \cite[Theorem 2.1]{SZZ10}, $M_\phi$ has exact two nontrivial minimal reducing subspaces on $L^2_a$, $M_0(\phi)$ and $M_0(\phi)^\perp$. And as noted before, we have $M_\phi$ is irreducible on $D$ in this case. The proof is complete.
\end{proof}

\section{The unitary equivalence relation}

\begin{thm}\label{petplznidd}
Given $\lambda \in \D \backslash \{0\}$, let $\phi = \varphi_\lambda(z^n)$ for some $n > 1$. Then $M_\phi$ is not unitarily equivalent to $M_{z^n}$ on $D$.
\end{thm}
\begin{proof}
Let $\omega = e^{\frac{2\pi i}{n}}$ be a primitive $n$-th root of unity and suppose $\alpha \in \D \backslash \{0\}$ satisfies $\alpha^n = \lambda$. Assume that $M_\phi$ is unitarily equivalent to $M_{z^n}$ on $D$. Note that $\phi$ is unitarily equivalent to $\varphi_\alpha \varphi_{\omega\alpha} \cdots \varphi_{\omega^{n-1}\alpha}$, thus $\varphi_\alpha \varphi_{\omega\alpha} \cdots \varphi_{\omega^{n-1}\alpha}$ is unitarily equivalent to $M_{z^n}$. Without loss of generality, suppose $\phi = \varphi_\alpha \varphi_{\omega\alpha} \cdots \varphi_{\omega^{n-1}\alpha}$ and there is a unitary operator $U$ on $D$ such that $U^* M_\phi U = M_{z^n}$.

Note that $M_\phi$ and $M_{z^n}$ has exact $n$ nontrivial minimal reducing subspaces on $D$,
$$M_j = \text{span} \{z^j, z^{n+j}, z^{2n+j}, \cdots\}, \quad 0 \leq j \leq n-1.$$
Let $f = U1$. Then $f$ is in one of the $M_j$'s. By $M_{z^n}^* 1 = 0$, we have $M_\phi^* f =0$, hence $f = c_1 K_\alpha + c_2 K_{\omega\alpha} + \cdots + c_n K_{\omega^{n-1}\alpha}, c_i \in \C$. Observe that $UM_{z^n} 1 = M_\phi U1 = \phi f, M_{z^n}^*(z^n) = n + 1$, therefore
$$(n+1) f = U M_{z^n}^*(z^n) = M_\phi^* U(z^n) = M_\phi^* (\phi f).$$

If $f \in \HM_0$, by $\dim \HM_j \ominus \phi \HM_j = 1, 0 \leq j \leq n-1$ and $\sum_{k=0}^\infty \frac{\overline{\alpha}^{nk}z^{nk}}{nk+1} \in \ker M_\phi^* \cap \HM_0$, we conclude that $f = c \sum_{k=0}^\infty \frac{\overline{\alpha}^{nk}z^{nk}}{nk+1}$ for some $c \in \C$. From
\begin{align*}
&\langle (n+1) f, 1\rangle_D = \langle M_\phi^* (\phi f), 1\rangle_D = \langle \phi f, \phi \rangle_D \\
&= \langle \phi, 1\rangle_D + \int_\T [p_\alpha(\zeta) + P_{\omega\alpha}(\zeta) + \cdots + P_{\omega^{n-1}\alpha}(\zeta)] f(\zeta) \frac{|d\zeta|}{2\pi},
\end{align*}
we obtain that
$$\langle n f, 1\rangle_D = \int_\T [p_\alpha(\zeta) + P_{\omega\alpha}(\zeta) + \cdots + P_{\omega^{n-1}\alpha}(\zeta)] f(\zeta) \frac{|d\zeta|}{2\pi},$$
i.e. $nf(0) = nf(\alpha)$, so $\alpha = 0$. This is a contradiction.

If $f \in \HM_1$, then by the same reasoning as above, $f = c \sum_{k=0}^\infty \frac{\overline{\alpha}^{nk+1}z^{nk+1}}{nk+2}$ for some $c \in \C$. From
\begin{align*}
&\langle (n+1) f, z\rangle_D = \langle M_\phi^* (\phi f), z\rangle_D = \langle \phi f, \phi z\rangle_D \\
&= \langle \phi, z\rangle_D + \int_\T [p_\alpha(\zeta) + P_{\omega\alpha}(\zeta) + \cdots + P_{\omega^{n-1}\alpha}(\zeta)] f(\zeta)\overline{\zeta} \frac{|d\zeta|}{2\pi},
\end{align*}
we get
$$\langle n f, z\rangle_D = \int_\T [p_\alpha(\zeta) + P_{\omega\alpha}(\zeta) + \cdots + P_{\omega^{n-1}\alpha}(\zeta)] f(\zeta)\overline{\zeta}  \frac{|d\zeta|}{2\pi},$$
i.e. $n \langle f, z\rangle_D = n \frac{f}{z}(\alpha)$, so
\begin{align}\label{scludis}
\sum_{k=0}^\infty \frac{|\alpha|^{2nk}}{nk+2} = 1.
\end{align}
Also
\begin{align*}
&\langle (n+1) f, z^{n+1}\rangle_D = \langle M_\phi^* (\phi f), z^{n+1}\rangle_D = \langle \phi f, \phi z^{n+1}\rangle_D \\
&= \langle \phi, z^{n+1}\rangle_D + \int_\T [p_\alpha(\zeta) + P_{\omega\alpha}(\zeta) + \cdots + P_{\omega^{n-1}\alpha}(\zeta)] f(\zeta)\overline{\zeta^{n+1}} \frac{|d\zeta|}{2\pi},
\end{align*}
then
$$n \langle f, z^{n+1}\rangle_D = \int_\T [p_\alpha(\zeta) + P_{\omega\alpha}(\zeta) + \cdots + P_{\omega^{n-1}\alpha}(\zeta)] f(\zeta)\overline{\zeta^{n+1}} \frac{|d\zeta|}{2\pi},$$
by a little calculation, we have $\frac{1}{2}|\alpha|^{2n}= \sum_{k=1}^\infty \frac{|\alpha|^{2nk}}{nk+2}$. By (\ref{scludis}), we conclude that $|\alpha|^{2n} = 1$. This is a contradiction.

If $f \in M_j, 2 \leq j \leq n-1$, similarly, we have $|\alpha|^{2n} = 1$ which is a contradiction.

From the above $n$ cases, we conclude that $M_\phi$ is not unitarily equivalent to $M_{z^n}$ on $D$.
\end{proof}

Now we can prove Theorem \ref{fbueztcd}.
\begin{proof}[Proof of Theorem \ref{fbueztcd}]
We only need to prove the "only if" part. Suppose $M_\phi$ is unitarily equivalent to $M_{z^n}$ on $D$. Since $M_{z^n}$ has exact $n$ nontrivial orthogonal minimal reducing subspaces on $D$ whose direct sum is $D$, we have $M_\phi$ has exact $n$ nontrivial orthogonal minimal reducing subspaces on $D$ whose direct sum is $D$, hence $M_\phi$ has $n$ nontrivial orthogonal reducing subspaces on $L^2_a$ whose direct sum is $L^2_a$. Since the order of $\phi$ is $n$, it follows that each of these nontrivial reducing subspace in $L^2_a$ is minimal. Thus by \cite[Theorem 2.4]{DPW12} or \cite[Lemma 4.2]{GH11} or \cite[Theorem 3.1]{SZZ08}, we have $\phi$ is equivalent to $\varphi_\alpha^n$ for some $\alpha\in \D$. It follows from Remark \ref{irdnznpc} that $\alpha = 0$. Theorefore by Theorem \ref{petplznidd}, we conclude that $\phi = az^n$ for some $|a| = 1$.
\end{proof}