text:
\begin{document}
\title{\huge Successive Optimization Tomlinson-Harashima Precoding Strategies for Physical-Layer Security in Wireless Networks \vspace{-0.5em}}

\author{\IEEEauthorblockN{Xiaotao Lu}
\IEEEauthorblockA{Department of Electronis\\
University of York\\
Email: xtl503@york.ac.uk \vspace{-1.15em}} \and
\IEEEauthorblockN{Rodrigo de Lamare}
\IEEEauthorblockA{CETUC/PUC-Rio\\
Department of Electronis, University of York\\
Email: rodrigo.delamare@york.ac.uk \vspace{-1.75em}} \and
\IEEEauthorblockN{Keke Zu}
\IEEEauthorblockA{Ericsson Research, Sweden\\
Email: zukeke@gmail.com  } \vspace{-1.75em}}

 \maketitle
\begin{abstract}
\boldmath
In this paper, we propose successive optimization non-linear
precoding strategies for physical-layer security in wireless
networks.  We also investigate different precoding techniques for
multi-user MIMO systems under various conditions of channel state
information (CSI) between the access point and the users and the
eavesdroppers. A non-linear precoding technique based on Successive
Optimization Tomlinson-Harashima Precoding (SO-THP) and Simplified
Generalized Matrix Inversion (S-GMI) technique is proposed along
with a strategy for injecting artificial noise prior to
transmission. Simulation results show that the proposed SO-THP+S-GMI
precoding technique outperforms existing non-linear and linear
precoding algorithms in terms of
BER and secrecy rate performances.\\

\end{abstract}
\begin{IEEEkeywords}
Physical-layer security techniques, precoding, transmit processing.
\end{IEEEkeywords}

\section{Introduction}

Data security in wireless systems has been dominated in the past by
encryption methods like Data Encryption Standard (DES) and Advanced
Encryption Standard (AES) which have played very important roles in
the security aspects of data transmission. However, these encryption
algorithms suffer from high complexity and latency. This has
motivated the investigation of novel secrecy techniques which can be
implemented in the physical layer and which are under consideration
to provide secure wireless transmission.

In the 1970s, Wyner posed the Alice-Bob-Eve problem and proposed the
wire-tap transmission system in \cite{wyner}. The wire-tap
transmission system gives a basic description on how to achieve the
secrecy in the physical layer with better Gaussian channels between
users than the Gaussian wire-tap channel. Later on another study
reported in \cite{csiszar} showed that secrecy transmission is
possible even when the eavesdropper has a better channel to the
transmitter in a statistical sense than the receiver with
confidential messages. The secrecy capacity \cite{hellman} for
different kinds of channels, such as the Gaussian wire-tap channel,
the MIMO wire-tap channels have been studied in \cite{hellman} and
\cite{oggier}. In some later works \cite{goel}, \cite{mukherjee}, it
has been found that by adding artificial noise to the system the
secrecy of the transmission can be further enhanced.

\subsection{Related Work}

In recent years, precoding techniques have been widely applied to
multi-user MIMO systems. Precoding techniques rely on pre-processing
the transmitted data with the help of channel state information
(CSI). Different kinds of linear precoding techniques such as Zero-forcing (ZF) precoding, Minimum
Mean Square Error (MMSE) precoding and Block Diagonalization (BD)
precoding are introduced in \cite{stankovic}-\cite{geraci},
respectively. Some non-linear precoding techniques like
Tomlinson-Harashima precoding (THP) \cite{payaro}, Vector
Perturbation (VP) precoding \cite{sebdani} have also been reported.
The precoding techniques can also contribute to the performance of relay systems \cite{cai}
These precoding techniques have been proven effective in improving
the sum-rate and the bit-error rate (BER) performances. However,
precoding strategies require CSI at the transmit side, whereas in
reality  CSI is available to the users of interest but is rarely
available to the eavesdroppers. In most situations, a designer only
has access to imperfect CSI for the users of interest. Dealing with
imperfect CSI and the absence of CSI of the eavesdropper is a major
challenge. Moreover, a technique that is particularly effective for
improving the secrecy rate of physical-layer security is the
introduction of artificial noise at the transmitter \cite{goel}.

\subsection{Contributions}

In this pape we propose a novel non-linear technique based on
Successive Optimization Tomlinson-Harashima Precoding (SO-THP) and
Simplified Generalized Matrix Inversion (S-GMI) techniques along
with a strategy of injecting artificial noise prior to transmission.
We also study precoding techniques for multi-user MIMO systems under
various conditions of CSI between the access point and the users and
the eavesdroppers. Simulation results show that the proposed
SO-THP+S-GMI precoding technique outperforms existing non-linear and
linear precoding algorithms in terms of physical layer secrecy rate
performances.

The remainder of this paper is organized as follows. In Section II,
the system model and the performance metrics are given. The proposed
SO-THP+S-GMI precoding algorithm is introduced in Section III.
In Section IV, we discuss the simulation results.
The conclusions are given in Section V.

Notation: Bold uppercase letters ${\boldsymbol A}\in
{\mathbf{C}}^{M\times N}$ denote matrices with size ${M\times N}$
and bold lowercase letters ${\boldsymbol a}\in {\mathbf{C}}^{M\times
1}$ denote column vectors with length $M$. Conjugate, transpose, and
conjugate transpose are represented by $(\cdot)^\ast$, $(\cdot)^T$
and $(\cdot)^H$ respectively; $\boldsymbol I_{M}$ is the identity
matrix of size $M\times M$; $\rm diag \{\boldsymbol a\}$ denotes a
diagonal matrix with the elements of the vector $\boldsymbol a$
along its diagonal; $\mathcal{CN}(0,\sigma_{n}^{2})$ represents
$i.i.d$ entries with zero mean and $\sigma_{n}^{2}$ variance.

\section{System Model and Performance Metrics}

In this section we introduce the system model considered for the
transmission as well as the performance metrics that are used in the
evaluation of the precoders.

\subsection{System Model}
\begin{center}
\begin{figure}[h]
\vspace{-20pt}
\includegraphics[scale=0.55]{fig1.eps}
\caption{System model of a MU-MIMO system with T users and K eavesdroppers}
\label{fig:sys}
\end{figure}
\vspace{-25pt}
\end{center}
Here we consider an uncoded multi-user multiple input multiple
output (MU-MIMO) broadcast channel. At the Acess Point (AP) a
transmitter with $N_{t}$ antennas is used to transmit $T$ users'
data steams. Each user's data streams are received with $N_{i}$
receive antennas. Along with the T users, $K$ eavesdroppers each
with $N_{k}$ receive antennas also receive data from the
transmitter. In this system we assume the eavesdroppers do not jam
the transmission and each user's or eavesdropper's channel is a
flat-fading MIMO channel. The quantity ${\boldsymbol H}_{i}\in
{\mathbf{C}}^{N_{i}\times N_{t}}$ and ${\boldsymbol H}_{k}\in
{\mathbf{C}}^{N_{k}\times N_{t}}$ corresponds to each user's or the
eavesdropper's channel matrix. The Gaussian noise with $i.i.d$
entries is distributed as $\mathcal{CN}(0,\sigma_{B}^{2})$ and
$\mathcal{CN}(0,\sigma_{E}^{2})$.

In the scenarios without artificial noise, each user's or
eavesdropper's received data are described by
\begin{equation}
{\boldsymbol y}_{r}=\beta_{r}^{-1}({\boldsymbol H}_{r}{\boldsymbol P}_{r} {\boldsymbol s}_{r} + {\boldsymbol H}_{r}\sum \limits_{j=1,j\neq r}^R
{\boldsymbol P}_{j} {\boldsymbol s}_{j}+ {\boldsymbol n}_{r}),\label{eqn:yr}
\end{equation} in (\ref{eqn:yr})
$\beta_{r}=\sqrt{\dfrac{E_{r}}{||\boldsymbol P_{r}||}}$ is used to
guarantee that the transmit power after precoding is the same as the
original transmit power $E_{r}$ of user $r$
 and ${\boldsymbol P}_{r}\in {\mathbf{C}}^{N_{t}\times N_{r}}$ is
the precoding matrix. We use the vector ${\boldsymbol x}_{r}\in
{\mathbf{C}}^{N_{t} \times 1}$ to represent the signal after
precoding. Without artificial noise ${\boldsymbol x}_{r}$ is given
by
\begin{equation}
{\boldsymbol x}_{r}={\boldsymbol P}_{r} {\boldsymbol s}_{r},
\label{eqn:xr1}
\end{equation}
where ${\boldsymbol s}_{r}\in
{\mathbf{C}}^{N_{r}\times 1}$ is the transmit symbols.

If the artificial noise is applied before the transmission,
following the description in \cite{mukherjee}, the signal after
precoding is given by
\begin{equation}
{\boldsymbol x}_{r}={\boldsymbol P}_{r}{\boldsymbol s}_{r}+
{\boldsymbol P}_{r}' {\boldsymbol s}_{r}',
\label{eqn:xr2}
\end{equation}
where ${\boldsymbol P}_{r}\in {\mathbf{C}}^{N_{t}\times N_{r}}$,
${\boldsymbol P}_{r}'\in {\mathbf{C}}^{N_{t}\times m}$ are the
precoding matrices. The vector ${\boldsymbol s}_{r}\in
{\mathbf{C}}^{N_{r}\times 1}$ represents the information to be
transmitted. The vector ${\boldsymbol s}_{r}'\in
{\mathbf{C}}^{m\times 1}$ is the jamming signal. Suppose $0<\rho<1$
is the fraction of the power devoted to the information signal and
${{\boldsymbol P}_{r}'}^H {{\boldsymbol P}_{r}'}=\boldsymbol I$. The
artificial noise then should satisfy
\begin{equation}
E\{{{\boldsymbol s}_{r}'} {{\boldsymbol s}_{r}'}^{H}\}=\boldsymbol Q_{s}',
\label{eqn:eqs}
\end{equation}
\begin{equation}
Tr({\boldsymbol Q}_{s}')=(1-\rho)E_{s},
\label{eqn:trqs}
\end{equation}
In (\ref{eqn:eqs}) and (\ref{eqn:trqs}) ${\boldsymbol Q}_{s}' \in
{\mathbf{C}}^{m\times m}$ is the covariance matrix associated with
the jamming signal $s_{r}'$. The covariance matrix of the transmit
signal ${\boldsymbol Q}_{s} \in {\mathbf{C}}^{N_{r}\times N_{r}}$
satisfies
\begin{equation}
Tr({\boldsymbol Q}_{s})=\rho E_{s},
\end{equation}
and proper artificial noise power ratio $\rho$  will result in an
increase of the secrecy rate. Based on the construction of the
artificial noise, there will be a different effect on the secrecy
rate performance. In \cite{lin}, it shows a generalized  artificial
noise with more flexible covariance matrix to improve the secrecy
rate. Besides the construction introduced in \cite{mukherjee}, the
study in \cite{chorti} reports a system with a helping interferer to
improve the achievable secrecy rate.

\subsection{Performance Metrics}

From \cite{wyner}-\cite{oggier}, we have the definition of the
secrecy capacity as the maximum transmission rate Alice can achieve
during the transmission without any information detected by the
eavesdropper. From \cite{oggier} without artificial noise, we have
the secrecy capacity
\begin{equation}
\begin{split}
C_{sec}&=\max_{{\boldsymbol Q}_{s}\geq 0 , \rm Tr({\boldsymbol Q}_{s})\leq E_{s}}{\textit{I}(X_{s};Y_{i})-\textit{I}(X_{s};Y_{k})}\\
             &=\max_{{\boldsymbol Q}_{s}\geq 0, \rm Tr({\boldsymbol Q}_{s})\leq E_{s}}\log(\det({\boldsymbol I}+{\boldsymbol H}_{ba} {\boldsymbol Q}_{s} {\boldsymbol H}_{ba}^H/\sigma_{B}^{2}))\\
             & \quad -\log(\det({\boldsymbol I}+ {\boldsymbol H}_{ea}{\boldsymbol Q}_{s} {\boldsymbol H}_{ea}^H/\sigma_{E}^{2})),
\end{split}
\label{eqn:Cs}
\end{equation}
In (\ref{eqn:Cs}) $\textit{I}(X_{s};Y_{i})$ and
$\textit{I}(X_{s};Y_{k})$ represent the mutual information between
the transmitter to the user as well as the eavesdropper.
$X_{s},Y_{i},Y_{k}$ are the counterparts of different signal
symbols. ${\boldsymbol Q}_{s}$ is the covariance matrix associated
with the signal after precoding ${\boldsymbol x}_{r}$. The secrecy
capacity achieved in this situation with prior knowledge of the
users' channel ${\boldsymbol H}_{ba}$ as well as the eavesdroppers'
channel ${\boldsymbol H}_{ea}$ can be used as a benchmark for
further study. In reality the channel between the transmitter and
the eavesdroppers is not perfectly known. This situation known as
the imperfect CSI is discussed in \cite{goel}. However, for the
imperfect CSI  ${\boldsymbol H}_{ea}$ we suppose the distribution of
the eavesdroppers' channel is known to the transmitter and users.

Another performance metric is the BER estimated by the different
links. Ideally, we would like the users to experience a reliable
communication and the eavesdropper to have a very high BER
(virtually no reliability when communicating). In this paper,
different precoding techniques are supposed to suppress the
interferences between different users thus leading to a better BER
performance.

Thirdly, the algorithm computational complexity should also take
into consideration. With the increasing of the number of users, the
high computational complexity algorithm will suffer from a long time
delay. Low computational complexity will contribute to reduce the
latency of the system.

\section{Proposed  Precoding Algorithms}

In this section, we detail the proposed precoding algorithms.

\subsection{SO-THP Precoding}

According to \cite{shenouda}, we assume the modified date symbol vector $\boldsymbol u_{r}=\boldsymbol s_{r}+\boldsymbol z_{r}$, where $\boldsymbol u_{r} \in {\mathbf{C}}^{N_{r}\times 1}$ and $\boldsymbol z_{r} \in {\mathbf{C}}^{N_{r}\times 1}$.
With the feedback matrix we can then obtain the output of the modulo operation $\boldsymbol v_{r}$
\begin{equation}
\boldsymbol v_{r}=(\boldsymbol I + \boldsymbol B)_{r}^{-1} \boldsymbol u_{r}
\label{eqn:vr}
\end{equation}
where $\boldsymbol v_{r} \in {\mathbf{C}}^{N_{r}\times 1}$. When we have the feedforward matrix $\boldsymbol F_{r} \in {\mathbf{C}}^{N_{t}\times N_{r}}$,
the transmit signal can be obtained as $\boldsymbol x_{r}=\boldsymbol F_{r} \boldsymbol v_{r}$. where $\boldsymbol x_{r} \in {\mathbf{C}}^{N_{t}\times 1}$
The received signal $\boldsymbol y_{r-pro} \in {\mathbf{C}}^{N_{t}\times 1}$ then can be written in the following form:
\begin{equation}
\boldsymbol y_{r-pro}=\boldsymbol H_{r} \boldsymbol x_{r}+\boldsymbol n_{r}\\
=\boldsymbol H_{r} \boldsymbol F_{r}(\boldsymbol I + \boldsymbol B)_{r}^{-1} \boldsymbol u_{r}+\boldsymbol n_{r}
\label{eqn:yrpro}
\end{equation}
In \cite{haardt}, the SO-THP algorithm is illustrated as a
combination of a successive optimization technique and
Tomlinson-Harashima precoding. The reordered feedforward and
feedback matrices are expressed as
\begin{equation}
{\boldsymbol F}= \left( {\boldsymbol F}_{1}  \cdots  {\boldsymbol F}_{R}  \right)
\label{eqn:fmatrix}
\end{equation}
\begin{equation}
{\boldsymbol B}={\rm lower~triangular}\left({\boldsymbol D}{\boldsymbol H}{\boldsymbol F}\bullet {\rm diag}\left([{\boldsymbol D}{\boldsymbol H}{\boldsymbol F}]_{ii}^{-1}\right)\right)
\label{eqn:bmatrix}
\end{equation}
where the complex unitary matrix ${\boldsymbol D}=\begin{pmatrix}
 {\boldsymbol D}_{1} &         & &\\
         & \ddots & &  \\
         &         & {\boldsymbol D}_{T} &\\ \end{pmatrix} $.

\subsection{S-GMI Precoding}

The concept of generalized matrix inversion precoding has been
introduced in \cite{hakjea}, where two approaches called Generalized
Zero-Forcing Channel Inversion (GZI) and Generalized MMSE Channel
Inversion (GMI) are illustrated. The GMI scheme uses the QR
decomposition to decompose the MMSE channel inversion
$\bar{{\boldsymbol H}_{r}}$ as expressed by
\begin{equation}
\bar{{\boldsymbol H}_{r}}=({{\boldsymbol H}_{r}^H}{\boldsymbol H}_{r}+\alpha \boldsymbol I)^{-1}{\boldsymbol H}_{r}^H,
\label{eqn:parh}
\end{equation}
\begin{equation}
\bar{{\boldsymbol H}_{r}}=\bar{{\boldsymbol Q}_{r}}\bar{{\boldsymbol R}_{r}} \qquad  for \qquad r=1,\cdots,R,
\label{eqn:parhqr}
\end{equation}
where $\bar {{\boldsymbol H}_{r}}\in {\mathbf{C}}^{N_{t}\times
N_{r}}$, $\bar{{\boldsymbol Q}_{r}}\in {\mathbf{C}}^{N_{t}\times
N_{t}}$, $\bar{{\boldsymbol R}_{r}}\in {\mathbf{C}}^{N_{t}\times
N_{r}}$
and GMI takes the noise into account. It overcomes the noise
enhancement in the BD algorithm. Meanwhile extra interference may be
introduced in the following steps. To solve this problem, a transmit
combining matrix ${\boldsymbol T}_{r}$ is applied to
$\bar{{\boldsymbol Q}_{r}}$. Under the total transmit power
constraint we use the minimum total MSE criterion to achieve the
transmit combining matrix ${\boldsymbol T}_{r}$. In \cite{hakjea}
the transmit combining matrix ${\boldsymbol T}_{r}$ is given by
${\boldsymbol T}_{r}=\beta \bar{{\boldsymbol T}_{r}}$ as described
by
\begin{equation}
\bar{{\boldsymbol T}_{r}}=\left(\bar{{\boldsymbol Q}_{r}}^H \sum\limits_{j=1}^R {\boldsymbol H}_{j}^H {\boldsymbol H}_{j}\bar{{\boldsymbol Q}_{r}}+\alpha \boldsymbol I  \right)^{-1} \bar{{\boldsymbol Q}_{r}}^H {\boldsymbol H}_{r}^H {\boldsymbol H}_{r}\bar{{\boldsymbol Q}_{r}},
\label{eqn:Tr}
\end{equation}
where ${{\boldsymbol T}_{r}}$, $ \bar{{\boldsymbol T}_{r}}\in {\mathbf{C}}^{N_{t}\times N_{t}}$.
Once we have $\bar{{\boldsymbol Q}_{r}}$ and ${\boldsymbol T}_{r}$, the following approach is
\begin{equation}
{{\boldsymbol H}_{r}}\bar{{\boldsymbol Q}_{r}}{\boldsymbol T}_{r}={\bar{{\boldsymbol U}_{r}}}{\bar{{\boldsymbol \Sigma}_{r}}}{\bar{{\boldsymbol V}_{r}}}^H,
\label{eqn:Trsvd}
\end{equation}
In \cite{keke1}, it has been shown that the transmit combining
matrix $\boldsymbol T_{r}$ is not necessary. Therefore, a simplified
GMI (S-GMI) is developed in \cite{keke1} as an improvement of the
original RBD precoding in \cite{stankovic}. This is known as S-GMI.
Therefore, the decomposition  in (\ref{eqn:Trsvd}), the precoder and
the receive filter are
\begin{equation}
{{\boldsymbol H}_{r}}\bar{{\boldsymbol Q}_{r}}=\tilde{{\boldsymbol U}_{r}}\tilde{{\boldsymbol \Sigma}_{r}}\tilde{{{\boldsymbol V}_{r}}}^H,
\label{eqn:sgmisvd}
\end{equation}
\begin{equation}
{{\boldsymbol P}_{S-GMI}}=[\bar{{\boldsymbol Q}_{1}}\tilde{{\boldsymbol V}_{1}} \quad \bar{{\boldsymbol Q}_{2}}\tilde{{\boldsymbol V}_{2}}\quad \cdots \quad \bar{{\boldsymbol Q}_{R}}\tilde{{\boldsymbol V}_{R}}],
\label{eqn:psgmi}
\end{equation}
\begin{equation}
{{\boldsymbol M}_{S-GMI}}={\rm diag}\{\tilde{{\boldsymbol U}_{1}}^H \quad \tilde{{\boldsymbol U}_{2}}^H \quad \cdots \quad \tilde{{\boldsymbol U}_{R}}^H\},
\label{eqn:msgmi}
\end{equation}
where ${{\boldsymbol P}_{S-GMI}}\in {\mathbf{C}}^{N_{t}\times
N_{t}}$, ${{\boldsymbol M}_{S-GMI}}\in {\mathbf{C}}^{N_{t}\times
N_{t}}$. In the SO-THP+S-GMI algorithm, the feedforward matrix
(\ref{eqn:fmatrix}) is obtained by reordering the S-GMI precoding
matrix (\ref{eqn:psgmi}).

\section{Simulation Results}

A system with $N_{t}=4$ transmit antennas and $T=2$ users as well as
$K=2$ eavesdroppers is considered. All the transmitted symbols are
modulated with QPSK. In the simulations, we assume that the channel
for each user is uncorrelated and is generated following a complex
Gaussian model with zero mean and variance equal to one.

\subsection{Computational complexity}

In assessing the computational complexity, the number of  FLOPS
(Floating-Point Operations Per Second) is used as a measurement of
the cost of an algorithm. In \cite{golub}, the FLOPS for real QR,
SVD and complex QR algorithms are given. According to \cite{keke2}
for an $m\times n$ complex matrix $\boldsymbol A$ ,the FLOPS of the
SVD of the complex matrix $\boldsymbol A$ are equivalent to its
$2m\times 2n$ extended real matrix. The number of FLOPS of the
multiplication, the QR decomposition, the matrix inversion as well
as SVD with different matrices obtained have been reported in
\cite{golub}.

\begin{center}
\begin{figure}[h]
\vspace{-5pt}
\includegraphics[scale=0.5]{fig2.eps}
\caption{Computational complexity in FLOPs for MU-MIMO systems}
\label{fig:com}
\end{figure}
\vspace{-25pt}
\end{center}

In Fig. \ref{fig:com} it is shown that although the BD precoding
algorithm has the lowest complexity, its BER and secrecy rate
performances are not satisfactory. Among the SO-THP type algorithms
the SO-THP+S-GMI algorithm shows an advantage over the standard
SO-THP algorithm in terms of complexity.

\begin{center}
\begin{figure}[h]
\vspace{-5pt}
\includegraphics[scale=0.5]{fig3.eps}
\caption{BER performance with precoding techniques in $4 \times 4 \times 4$ MU-MIMO broadcast channel, $m=0.5$}
\label{fig:ber1}
\end{figure}
\vspace{-20pt}
\end{center}

\subsection{Perfect Channel State Information}

Fig. \ref{fig:ber1} shows the BER performance for different
precoding techniques. The proposed SO-THP+S-GMI precoding requires
less than 5 dB than the original SO-THP precoding in terms of SNR to
achieve $10^{-2}$ bit error rate. According to Fig. \ref{fig:444P},
the secrecy rate will level out according to the ratio between the
main channel and the wire-tap channel $m$. From Fig. \ref{fig:444P},
we can also see that among all the investigated precoding
algorithms, the proposed SO-THP+S-GMI precoding can achieve a
certain secrecy rate with lower SNR.

\begin{center}
\begin{figure}[h]
\vspace{-5pt}
\includegraphics[scale=0.5]{fig4.eps}
\caption{Secrecy rate performance with precoding techniques in $4 \times 4 \times 4$ MU-MIMO broadcast channel}
\label{fig:444P}
\end{figure}
\vspace{-20pt}
\end{center}

\begin{center}
\begin{figure}[h]
\vspace{-5pt}
\includegraphics[scale=0.5]{fig5.eps}
\caption{Secrecy rate with precoding techniques $4 \times 4 \times 4$ MU-MIMO broadcast channel with imperfect CSI}
\label{fig:444IP}
\end{figure}
\end{center}

\subsection{Imperfect Channel State Information}

In Fig. \ref{fig:444IP}, the secrecy rate performance is evaluated
in the imperfect CSI scenario, which shows a degradation at low SNR.
Furthermore, the secrecy rate requires a very high SNR to converge
to a high secrecy rate. The proposed SO-THP+S-GMI precoder has the
best secrecy rate performance among all the studied precoding
techniques.

\begin{center}
\begin{figure}[h]
\vspace{-5pt}
\includegraphics[scale=0.5]{fig6.eps}
\caption{Secrecy rate with precoding techniques $4 \times 4 \times
4$ MU-MIMO broadcast channel with imperfect CSI adding artificial
noise, $m=0.5$} \label{fig:444AIP}
\end{figure}
\vspace{-20pt}
\end{center}

\begin{center}
\begin{figure}[h]
\vspace{-5pt}
\includegraphics[scale=0.5]{fig7.eps}
\caption{Secrecy rate with precoding techniques $4 \times 4 \times 4$ MU-MIMO broadcast channel with imperfect CSI adding artificial noise, $m=2$}
\label{fig:444m2AIP}
\end{figure}
\vspace{-20pt}
\end{center}

\subsection{Imperfect Channel State Information With Artificial Noise}

In Fig. \ref{fig:444AIP} and Fig. \ref{fig:444m2AIP}, we assume that
artificial noise is generated with extra free antennas. Although the
artificial noise is added, the total transmit power $E_{s}$ will
remain the same. In the simulation $0.4$ of the transmit power
$E_{s}$ is used to generate the artificial noise. In Figure
\ref{fig:444AIP}, with higher SNR, we can have better secrecy rate
performance. Even when the eavesdropper has better statistical
channel, the secrecy rate keeps almost the same according to Figure
\ref{fig:444m2AIP}.

\section{Conclusion}

Precoding techniques used in MU-MIMO broadcast systems can achieve
good BER and secrecy-rate performances. The proposed SO-THP+S-GMI
algorithm requires the lowest computational complexity among
non-linear techniques and outperforms most linear and non-linear
techniques in BER performance. Moreover, SO-THP+S-GMI can achieve a
high secrecy rate, which means the security of the transmission with
SO-THP+S-GMI is superior to existing techniques. When artificial
noise is added, the performance is further improved.


text:
\section{Introduction} \label{sec:introduction}

The discontinuous Galerkin (DG) finite element method has been originally proposed by Reed and Hill \cite{reed}. Later, a solid theoretical framework has been
established by Cockburn and Shu in a well-known series of papers \cite{cbs0,cbs1,cbs2,cbs3,cbs4} for the application of discontinuous Galerkin schemes to nonlinear
hyperbolic systems of conservation laws. A very important property of DG schemes is that they satisfy a local cell entropy inequality for any polynomial degree
$N$ used for the approximation of the discrete solution. As a consequence, this guarantees \textit{nonlinear} stability in $L_2$ norm for arbitrary high order
of accuracy, see the proof by Jiang and Shu \cite{jiangshu} for the scalar case and its subsequent extensions to systems \cite{BarthCharrier,HouLiu}. This means
that the DG scheme is by nature very robust and clearly appropriate for the solution of nonlinear hyperbolic conservation laws. However, being a linear scheme
in the sense of Godunov \cite{godunov}, even the DG method needs some sort of nonlinear limiting to avoid the Gibbs phenomenon at shock waves or other
discontinuities. There is a vast literature on the topic of limiters for DG schemes and a non-exhaustive review on this topic will be presented later in section
\ref{sec:apriori_limiter} of this paper. The key idea of many DG limiters is the following: first an unlimited solution is computed with the DG scheme, then an
indicator detects so-called \textit{troubled cells}, i.e. those zones of the domain which may need limiting, see \cite{Qiu_2005} for a detailed comparison.
For troubled cells, the degrees of freedom of the discrete solution are then modified by some sort of nonlinear \textit{reconstruction} technique, based on the
discrete solution in the troubled cell and its neighbors.

Concerning the time discretization, mostly explicit TVD Runge-Kutta schemes are used, which lead to the so-called Runge-Kutta DG schemes. A review on DG schemes
can be found in \cite{CBS-book,CBS-convection-dominated}. However, explicit DG schemes suffer from a very severe time step restriction where the maximum admissible
Courant number typically scales as approximately $1/(2N+1)$, if $N$ denotes the polynomial degree of the approximation of the DG scheme.
Alternative high order accurate explicit time discretizations for DG schemes have been explored in \cite{QiuDumbserShu} and \cite{taube_jsc,dumbser_jsc}, which, however,
have led to an even more restrictive CFL condition.
While the DG method is mostly used only for spatial discretization, it has been introduced as a uniform discretization of space and time in the global space-time
DG scheme of Van der Vegt et al.  \cite{spacetimedg1,spacetimedg2,KlaijVanDerVegt}, which leads to an implicit method of theoretically arbitrary high order of
accuracy in space and time and which is unconditionally stable. The strategy presented in \cite{spacetimedg1,spacetimedg2}, however, requires the solution of a
global nonlinear algebraic system at each time step. In order to reduce the complexity of globally implicit space-time DG schemes, in
\cite{DumbserEnauxToro,HidalgoDumbser,Dumbser2008} a \textit{local} space-time DG approach has been suggested, which leads only to an element-local implicit method.
However, also in this case the final DG scheme is explicit and thus has to satisfy the typical stability condition of explicit DG schemes.

In the finite volume context, recently a new concept has been proposed, namely the Multi-dimensional Optimal Order Detection (MOOD) approach,
which is an \textit{a posteriori} approach to the problem of limiting. The key idea of this paradigm is to run a spatially \textit{unlimited} high-order finite
volume scheme in order to produce a so-called \textit{candidate solution}. Then the validity of this candidate solution is tested against a set of
predefined \textit{admissibility criteria}. Some cells are marked as 'acceptable' and are therefore valid. Some others may be locally marked as 'problematic'
or 'troubled', if they do not pass the detection process. These cells and their neighbors are consequently \textit{locally recomputed} using polynomial
reconstructions of a lower degree. Thus, after decrementing the polynomial degree and locally recomputing the solution, a new candidate solution is obtained.
That solution is again tested for validity and the decrementing procedure re-applies, if necessary.

Such order decrementing can occur several times within one time step for the same cell, but it will always halt after a finite number of steps:
either the cell is valid for a polynomial degree greater than $0$, or the degree zero is reached. In the worst case,
a cell is updated with a robust and stable first order accurate Godunov-type finite volume scheme, which is supposed to produce always
valid (monotone and positivity-preserving) solutions under CFL condition. This \textit{a posteriori} check and order decrementing loop is called the
'MOOD loop'. We refer the reader to \cite{CDL1,CDL2,CDL3,ADER_MOOD_14} for more details.

The link between the MOOD concept developed in the finite volume framework and the typical strategy adopted in a classical DG limiter seems obvious. In both
approaches first a candidate solution is computed using an \textit{unlimited} scheme. Then, troubled cells are detected based on some criteria and the discrete
solution is corrected. However, there are important differences: the typical DG limiters \textit{postprocess} the candidate solution using a nonlinear
\textit{reconstruction} technique. They furthermore use only \textit{one} time level (the current one) for the detection and postprocessing step. In contrast,
the MOOD approach first of all uses \textit{two} time levels (the old one and the current one) for the detection of troubled cells. Second, the MOOD approach
\textit{recomputes} the solution using a \textit{different numerical scheme}, which is supposed to be more robust at shock waves. The fact of recomputing the
solution and of looking at two different time levels  for the detection of troubled cells makes it in the notation of the authors of \cite{CDL1,CDL2,CDL3,ADER_MOOD_14}
an \aposteriori approach.

The key innovation of the present paper is now the use of the aforementioned \aposteriori MOOD paradigm as a limiter for high order DG schemes. Since simple order
decrementing as in the MOOD approach for finite volume schemes would obviously destroy the natural subcell resolution capability of DG, a more sophisticated strategy is
needed here. We therefore suggest to recompute the solution of troubled cells on a finer \textit{subgrid} inside each cell, using a more robust but still very accurate
one-step ADER-WENO finite volume scheme \cite{titarevtoro,Balsara2013934,AMR3DCL}. The data can be scattered from the main grid to the subgrid and gathered back
via appropriate subcell projection and subcell reconstruction operators. Such operators are in principle well-known from high order finite volume schemes and spectral
finite volume schemes \cite{spectralfv2d,spectralfv3d,spectralfv.dg}. The choice of the subgrid size is very important. In this paper we suggest to choose the subgrid
size so that the local CFL number on the subgrid is as large as possible, hence the maximum admissible time step size of the finite volume scheme on the subgrid matches
the maximum admissible time step of the DG scheme on the main grid. This leads to $N_s=2N+1$ subcells per space dimension for a DG scheme using a piecewise polynomial
approximation of degree $N$. For alternative subcell methods used as limiter for the DG method see \cite{CasoniHuerta2,Sonntag}. However, none of these
uses our \aposteriori detection concept, nor do they recompute the solution in troubled cells via a better than second order accurate subcell finite volume method.
Furthermore, the method proposed in \cite{Sonntag} does not use the optimal subgrid size that allows to get the maximum admissible CFL number on the subgrid.

The rest of this paper is organized as follows.
The one-step discontinuous Galerkin scheme used in this paper (ADER-DG) is presented in section \ref{sec:framework}.
In section \ref{sec:apriori_limiter} we give a thorough but still non-exhaustive overview of existing \apriori limiters for the
DG method, such as slope/moment reduction, artificial viscosity or WENO-like techniques. Then in section \ref{sec:aposteriori_limiter} we
present all details of our new subcell-based \aposteriori approach. Next, section \ref{sec:numerics} gathers all numerical results for a
large set of different test cases in order to assess the validity and robustness of our new \aposteriori subgrid limiter. Smooth and non-smooth
test cases are simulated. Finally, conclusions and perspectives are drawn in section \ref{sec:conclusion}.

config: arxiv


TITLE: A solid strategy to prove: $A \subset B\Leftrightarrow A=A\cap B \Leftrightarrow B= A\cup B $
QUESTION [3 upvotes]: To begin i tried to prove that the first statement implies second, but took a while to type that (now let's prove the other 5 implications, which would fullfill the circle....), I think there must be a short way to do this. Any ideas?

REPLY [1 votes]: Statement $1$ is $x\in A\implies x\in B$
Statement $2$ is $x\in A= x\in A\land x\in B$
Statement $3$ is $x\in B = x\in A\lor x\in B$

$1\to2:$ $x\in A\implies x\in B \therefore x\in A\implies x\in A \land x\in B$
$2\to1:$ $x\in A=x\in A \land x\in B\therefore x\in B$
$1\to3:$ $x\in A\implies x\in B \therefore x\in A \lor x\in B \implies x\in B \lor x\in B \implies x\in B$
$3\to1:$ $x\in A \lor x\in B=x\in B\therefore x\in A\implies x\in A \lor x\in B\implies x\in B$


TITLE: What is an outline of a proof?
QUESTION [3 upvotes]: How does one build an outline of a proof? What should it consist of? Should it proof the result without working out calculations or not? Thank you!

REPLY [3 votes]: I'll give you an example from topology that might help - even if you don't know topology, the distinction between the proof styles should be clear.
Proposition: Let $S$ be a closed subset of a complete metric space $(E,d)$. Then the metric space $(S,d)$ is complete.
Proof Outline: Cauchy sequences in $(S,d)$ converge in $(E,d)$ by completeness, and since $(S,d)$ is closed, convergent sequences of points in $(S,d)$ converge in $(S,d)$, so any Cauchy sequence of points in $(S,d)$ must converge in $(S,d)$.
Proof: Let $(a_n)$ be a Cauchy sequence in $(S,d)$. Then each $a_n\in E$ since $S\subseteq E$, so we may treat $(a_n)$ as a sequence in $(E,d)$. By completeness of $(E,d)$, $a_n\to a$ for some point $a\in E$. Since $S$ is closed, $S$ contains all of its limit points, implying that any convergent sequence of points of $S$ must converge to a point of $S$. This shows that $a\in S$, and so we see that $a_n\to a\in S$. As $(a_n)$ was arbitrary, we see that Cauchy sequences in $(S,d)$ converge in $(S,d)$, which is what we wanted to show.
The main difference here is the level of detail in the proofs. In the outline, we left out most of the details that are intuitively clear, providing the main idea so that a reader could fill in the details for themselves. In the actual proof, we go through the trouble of providing the more subtle details to make the argument more rigorous - ideally, a reader of a more complete proof should not be left wondering about any gaps in logic.
(There is another type of proof called a formal proof, in which everything is derived from first principles using mathematical logic. This type of proof is entirely rigorous but almost always very lengthy, so we typically sacrifice some rigor in favor of clarity.)
As you learn more about a topic, your proofs typically begin to approach proof outlines, since things that may not have seemed obvious before become intuitive and clear. When you are first learning it is best to go through the detailed proof to make sure that you understand everything as well as you think you do, and only once you have mastered a subject do you allow yourself to omit obvious details that should be clear to someone who understands the subject on the same level as you.


TITLE: Are these the limit points of the set S? Finding Limit Points of a Set
QUESTION [0 upvotes]: If we have the subset $S$ of the metric space $(R^2,d_{{\infty}})$  such that $$S=\{(\frac{1}{n}+\frac{1}{m},0): n,m \in \mathbb{N}\}$$,
I have found out that the limit points of $S$, contained in the set $S'$ are: $$S'={ \left\lbrace(0,0) \right\rbrace}\cup{(\frac{1}{m},0):m\in\mathbb{N}}$$
I want to know if that is correct.
To find these limit points,
a) I fixed m, and then I got a sequence $(\frac{1}{n}+\frac{1}{m},0)=(b^{(n)})_{n\in\mathbb{N}}$ which converges to a point of the form $(\frac{1}{m},0)$
b) Next I shew (0,0) is also a limit point, since if we take n=m, we get a sequence $(c^{(n)})_{n\in\mathbb{N}}:(c^{(n)})_{n\in\mathbb{N}}=(\frac{2}{n},0)$, which converges to $(0,0)$

REPLY [1 votes]: It is clear that the set of limit points of $S$ in $\Bbb{R}^2$ has the form $L \times \{ 0 \}$, where $L$ is the set of limit points of $A=\{ \frac{1}{m}+\frac{1}{n} : m,n \in \Bbb{N} \}$ in $\Bbb{R}$. This has been determined in Find a set of limit point of $A=\{ \frac{1}{m}+\frac{1}{n} : m,n \in \Bbb{N} \}$ in $\Bbb{R}$ by using the following way.. It confirms that the set of limit points of $S$ is the set $S'$ in your question.


TITLE: Prove the irreducibility of the polynomial
QUESTION [1 upvotes]: Let $p$ be a prime and set $g = X^2 + X - 1\in \mathbb{F}_p[X]$. Prove the following:
$$\text{$g$ is irreducible in $\mathbb{F}_p[X]$ iff $g$ is irreducible in $\mathbb{F}_{p^3}[X]$}$$
One side is really easy to prove, namely
$\Leftarrow$ We know that $\mathbb{F}_p \subset \mathbb{F}_{p^3}$. As $g$ is irreducible in $\mathbb{F}_{p^3}[X]$ and $\deg(g) = 2$, then $g$ has no zeros in $\mathbb{F}_{p^3}$. Therefore, it has no zeros in $\mathbb{F}_p$, thus it is irreducible in $\mathbb{F}_p$.
Any suggestions about the other implication are welcome.

REPLY [2 votes]: We will assume that $g$ is irreducible in $\mathbb{F}_p[X]$. On the contrary, suppose that $g$ is not irreducible in $\mathbb{F}_{p^3}[X],$ i.e., suppose there exists a root $\alpha$ of $g$ in $\mathbb{F}_{p^3}.$ By definition, $[\mathbb{F}_p(\alpha):\mathbb{F}_p]$ is the degree of the minimal polynomial of $\alpha$ over $\mathbb{F}_p$. As $g$ is irreducible, we conclude that the minimal polynomial is $g$ and $[\mathbb{F}_p(\alpha):\mathbb{F}_p] = \deg(g) = 2$. Also, recall that $[\mathbb{F}_{p^3}:\mathbb{F}_p] = 3$. As $\mathbb{F}_p \subset \mathbb{F}_p(\alpha) \subset \mathbb{F}_{p^3},$ we have that $$\begin{align*} [\mathbb{F}_{p^3}:\mathbb{F}_p] &= [\mathbb{F}_{p^3}:\mathbb{F}_p(\alpha)]\cdot[\mathbb{F}_p(\alpha): \mathbb{F}_p] \text{ so that} \\ 3 &= [\mathbb{F}_{p^3}:\mathbb{F}_p(\alpha)]\cdot 2. \end{align*}$$
But this is a contradiction: $3$ is not divisible by $2$. Hence, $g$ is irreducible in $\mathbb{F}_{p^3}$.


TITLE: Prove $\begin{pmatrix} a&b\\ 2a&2b\\ \end{pmatrix} \begin{pmatrix} x\\y\\ \end{pmatrix}=\begin{pmatrix} c\\2c \\ \end{pmatrix}$
QUESTION [1 upvotes]: Prove that
$$
\begin{pmatrix} a & b \\ 2a & 2b \\ \end{pmatrix}
\begin{pmatrix} x \\ y \\ \end{pmatrix}
=
\begin{pmatrix} c \\ 2c \\ \end{pmatrix}
$$
has solution
$$
\begin{pmatrix} x \\ y \\ \end{pmatrix}
=
\begin{pmatrix} \frac{c}{a} \\ 0 \\ \end{pmatrix}
+t
\begin{pmatrix} -b \\ a \\ \end{pmatrix}
$$
and $t$ is real number.
If I plug some numbers, all $y$ variables are vanish and the matrix has infinitely many solutions, but I have no idea how the solution could be into like that. Also I haven't reached Vector chapter.
Any help is appreciated!

REPLY [0 votes]: The fast approach, as I see it is (in case $a \neq 0$):
you have $ax+by = c$ , just plug in $y=0+ta$, and you get
$$ax + bta = c$$
$$x= \frac ca - bt$$.
alternatively, in case you don't have the solution in advance, use same approach and you will get: $$ x = \frac ca - \frac ba y$$, so the general solution will be $$(x,y) = (\frac ca - \frac bat , t)$$
$$ (x,y) = (\frac ca, 0) + (-\frac ba,1)t$$ where $t$ is free. if you plug $t=a$ you will end up with the same result.


TITLE: What is the motivation behind the definition of a distribution on an algebraic group
QUESTION [2 upvotes]: Definition. Let $M$ be a smooth manifold. Then a distribution on $M$ of rank $k$ is defined to be a (usually smooth) rank-$k$ subbundle of the tangent bundle $TM$.[1]
Definition. Let $X$ be an affine scheme over $k$, and let $x$ be a $k$-valued point. Write $I_x$ for the set of functions $f \in \mathcal{O}(X)$ such that $f(x) = 0$. A distribution on $X$ with support in $x$ of order $\leq n$ is a linear functional $\mu : \mathcal{O}(X) \to k$ such that $\mu(I_x^{n+1}) = 0$.[2]
I can only assume that the second definition is based on the first, but I do not see where the definition comes from. Why is the definition of a distribution on an affine scheme the way it is?
[1]: Lee, John M., Introduction to smooth manifolds, Graduate Texts in Mathematics. 218. New York, NY: Springer. xvii, 628 p. (2002). ZBL1030.53001.
[2]: Jantzen, Jens Carsten, Representations of algebraic groups, Pure and Applied Mathematics, Vol. 131. Boston etc.: Academic Press, Inc. (Harcourt Brace Jovanovich, Publishers). XIII, 443 p.; {$} 59.50 (1987). ZBL0654.20039.

REPLY [4 votes]: This is a big old confusion regarding the two different notions of distributions floating around in mathematics. For one, there are (geometrical) distributions as subbundles of tangent bundles, as in your first definition. On the other hand, there is the notion of (analytical) distributions as functionals on function spaces, for example the Dirac delta functionals and its derivatives, so
$$\delta_p : C^\infty(\mathbb{R}) \to \mathbb{R}, f \mapsto f(p)$$
or
$$\delta_p' : C^\infty(\mathbb{R}) \to \mathbb{R}, f \mapsto f'(p)$$
for $p \in \mathbb{R}$. This is what your second definition generalizes, to affine schemes specifically. The order in your definition then mimics the idea of the order of a differential operator, but in particular, this has nothing to do with the rank of the geometrical distributions.
As far as I know, there is no direct relation in any way. It would be pretty helpful if people added the adjectives "geometrical" or "analytical", but I'm sure someone in the math community will take issue with that, so for now, whenever you come across the word "distribution", be very careful :)


TITLE: Can we Take the Derivative of a Function over a Non-Convex Set?
QUESTION [1 upvotes]: Recently, I found out why (most) Integer Programming and Mixed Integer Programming optimization problems are usually considered as Non-Convex (Why is the Set of Integers Non-Convex?). This is because even though the objective function in such problems (e.g. Travelling Salesman Problem https://en.wikipedia.org/wiki/Travelling_salesman_problem) is a Convex Function - since the set of constraints that this objective function has to be optimized over is Non-Convex (e.g. the set of Integers by definition is a Non-Convex set), the entire problem becomes a Non-Convex problem.
This being said, I have the following question:
1) In the case of the Travelling Salesman Problem, if we look at the "objective function" in this problem:

Is this function sum(sum(c_ij*x_ij)) differentiable when subject to its constraints? Can we take its derivative?
My inclination is "no" - this function is non-differentiable, seeing how I have never seen the Travelling Salesman Problem being solved using an optimization algorithm that requires evaluating a derivative (e.g. gradient descent), and instead always solved using optimization algorithms that do not require evaluating the derivative (e.g. Branch and Bound, Evolutionary and Metaheuristic Algorithms).
2) Consider the "Decision Tree" statistical/machine learning model (e.g. CART https://en.wikipedia.org/wiki/Decision_tree_learning) that is often used for supervised classification and regression tasks.
Given a dataset, a Decision Tree (e.g. CART) is formed by "splitting" variables into smaller subsets (i.e. "nodes") such that "purity" increases in each subsequent subset; "purity" is often measured through some sort of "Information Gain" that is based on measures such as "Gini Index" or "Entropy". Thus, Decision Trees can be interpreted as an optimization problem where "Information Gain" has to be optimized. I have heard that since Decision Trees often have different variable types (e.g. continuous and categorial), searching for the optimal variable splits that optimize "Information Gain" is a Mixed Integer Optimization Problem having an enormous Combinatorial Search Space. For the interest of creating a decent Decision Tree in a reasonable amount of time, "Information Gain" is optimized using a "Greedy Search Algorithm," and as a result, the final Decision Tree (i.e. the answer to this Mixed Integer Optimization Problem) is almost certainly unlikely to be the optimal Decision Tree (as there is very high probability that a better Decision Tree likely exists in this large Discrete Combinatorial Search Space, but finding this Decision Tree would take too much time):

The above equation is the general function for "Information Gain" , again I ask - when subject to its constraints, is the above function differentiable?
Again, my inclination is "no" - this function is not differentiable, because I have never seen any references that show this optimization problem being solved using an optimization algorithm that requires evaluating the derivative (e.g. Gradient Descent), and significant research is being done on improving optimization algorithms for this problem, and these optimization algorithms do not involve derivatives (e.g. Branch and Bound https://arxiv.org/pdf/1904.12847.pdf , Evolutionary Algorithms https://cran.r-project.org/web/packages/evtree/vignettes/evtree.pdf)
Thus, my rationale as to why these above functions are non differentiable are unfortunately non-mathematical reasons, but rather anecdotal reasons - and this of course is not an acceptable way to present arguments in math.
But in general, is it relatively straightforward to establish that:

Are all functions defined over Non-Convex sets are by definition non-differentiable?

Could we technically still use "Gradient Descent" on problems like Travelling Salesman and Decision Trees (e.g. "Relax" the problem, use Gradient Descent, check to see if the solution falls within the feasible region, Explore for solution in neighborhood, Repeat, etc.), with the only problem being that it would be an extremely inefficient compared to Gradient-Free Optimization Methods (e.g. Metaheuristics/Evolutionary Algorithms, Branch and Bound)? Or is it by definition, we are mathematically unable to use Gradient Descent on problems like Travelling Salesman and Decision Trees?

Thanks!
References:

Are all mixed integer nonlinear problems non-convex? If not, is my problem non-convex? And what solvers can we use?

REPLY [2 votes]: First of all, it is not because you have not seen any references that show this optimization problem being solved using an optimization algorithm that requires evaluating the derivative, that the function is nondifferentiable.
It is important to define the domain of the function. If you view the function as a function of real variables on $\mathbb{R}^{n^2}$, yes it is differentiable. If the domain is $\{0,1\}^{n^2}$, it is also trivially differentiable but with the discrete definition for the derivative.
It has nothing to do with the fact that the set is nonconvex. It has to do with the fact that the set is discrete (or continuous). The reason why there is no derivative involved in the traveling salesman problem is that it is not necessary to use it to solve it. Same comment for the decision tree.


TITLE: Primes of form $a^2 + 24b^2$
QUESTION [10 upvotes]: For a prime number $p \neq 2$, $3$, is it necessarily the case the prime number can be written in the form $a^2 + 24b^2$ if and only if $p \equiv 1 \text{ mod }24$?
I think this has to be true based upon the facts that $73 = 7^2 + 24(1^2)$, $241 = 5^2 + 24(3^2)$.

REPLY [11 votes]: If $p = a^2+24b^2$, then since $a^2\equiv 1\pmod{24}$ for all $a$ prime to $2,3$, it follows that $p\equiv 1\pmod{24}$.
Now suppose that $p\equiv 1 \pmod {24}$. Let $K=\mathbb Q(\sqrt{-6})$. The ring $R:=\mathbb Z[\sqrt{-24}]=\mathbb Z[2\sqrt{-6}]$ is a subring of $\mathcal O_K = \mathbb Z[\sqrt{-6}]$. The norm of an element $a+b\sqrt{-24}\in R$ is $a^2+24b^2$. Therefore, we would like to find out which rational primes are norms of elements of $R$.
Suppose that $p$ can be written as $p=a^2+6c^2$. Working modulo $24$, we find
$$1\equiv p \equiv1+6c^2\pmod {24}$$
so $c$ must be even. Hence,

A prime $p\equiv 1\pmod{24}$ can be written as $p=a^2+24b^2$ if and only if it is a norm of an element of $\mathcal O_K$.

Equivalently,

A prime $p\equiv 1\pmod{24}$ can be written as $p=a^2+24b^2$ if and only if $p$ splits completely in $\mathcal O_K$ and the primes above $p$ in $\mathcal O_K$ are principal.

Here, things become a bit tricky because $\mathcal O_K$ is not a PID. To solve this completely (at least using this generalisable method), we need to use class field theory. We will use the following theorem:

Theorem: Let $H$ be the Hilbert class field (i.e. the maximal everywhere unramified extension) of $K$. Then a prime $\mathfrak p\subset \mathcal O_K$ splits completely in $H$ if and only if it is principal.

In other words, we can find out if a prime is principal by seeing whether it splits in $H$. Our problem reduces to

A prime $p\equiv 1\pmod{24}$ can be written as $p=a^2+24b^2$ if and only it splits completely in the Hilbert class field of $K$.

One can check that the Hilbert class field of $K$ is $\mathbb Q(\sqrt2,\sqrt{-3})$. A prime $p$ will split completely in this field if and only if it splits completely in $\mathbb Q(\sqrt2)$ and $\mathbb Q(\sqrt{-3})$. By the Kummer-Dedekind theorem, for $p\ne 2,3$, this occurs if and only if $X^2-2$ and $X^2+3$ both split mod $p$ - i.e. if and only if
$$\left(\frac2p\right)=\left(\frac{-3}p\right) = 1.$$
By quadratic reciprocity, this is equivalent to
$$\left(\frac 2p\right)=\left(\frac p3\right) = 1,$$
and this holds since $p\equiv 1\pmod 8$ and $p\equiv 1 \pmod 3$.


TITLE: Weierstrass product form
QUESTION [3 upvotes]: How to show the Weierstrass product form of the entire function $f(z)= \sinh z$
This question seem so interesting. I would like to write my some ideas, but I dont want to direct incorrectly. Please help me to learn correctly and explicitly. I asked to question in fact in order to learn the topic precisely.

REPLY [5 votes]: Hint: consider the functional relation $$\sinh(z)=-i \sin(iz). $$
EDIT: Here is a more detailed explanation:
I will derive the Weierstrass product for $\sin(z)$, and using the identity above we will get the product for $\sinh (z)$ as well.
The zeros of the entire function $\sin(\pi z)$ are the integers $\mathbb Z$, and the smallest integer $h$ which makes the series $$ \sum_{n \in \mathbb Z \setminus \{0 \}} \frac{1}{n^{h+1}}$$ converge is $h=1$. This number is known as the genus of the entire function $\sin(\pi z)$, and it is needed for applying the Weierstrass factorization theorem effectively.
It follows from the Weierstrass factorization theorem that $$\sin(\pi z)=z e^{g(z)} \prod_{n \in \mathbb Z^*} \left(1-\frac{z}{n} \right) e^{z/n} $$
where $\mathbb Z^*=\mathbb Z \setminus\{0 \}$ and $g(z)$ is an entire function to be determined. Taking the logarithmic derivative of both sides (except at the integers) we find $$ \pi \cot (\pi z)=\frac{1}{z}+g'(z)+ \sum_{n \in \mathbb Z^*} \frac{1}{z-n}+\frac{1}{n} .$$ Consider now the development of $\pi \cot(\pi z)$ into partial fractions (this is another exercise by itself):
$$\pi \cot(\pi z) = \frac{1}{z}+\sum_{n \in \mathbb Z^*} \frac{1}{z-n}+\frac{1}{n}$$ from this it follows that $g'(z) \equiv 0$, thus $g(z)$ is a constant.
Taking the limit of $\frac{\sin(\pi z)}{z}$ as $z \to 0$ we find that $e^{g(z)} \equiv \pi$, thus we have found $$\sin(\pi z)= \pi z \prod_{n \in \mathbb Z^*} \left( 1-\frac{z}{n} \right) e^{z/n} $$
combining the factors $\left(1 - \frac{z}{ \pm n} \right) e^{ \frac{z}{ \pm n}}$ gives the equivalent form $$\sin(\pi z)= \pi z \prod_{n=1}^\infty \left(1-\frac{z^2}{n^2}\right). $$
Now replacing $z \mapsto \frac{z}{\pi}$ gives $$\sin(z)=z \prod_{n=1}^\infty \left(1-\frac{z^2}{n^2 \pi^2} \right) .$$
Thus $$\sinh(z)=-i \sin(iz)= -i \cdot i z \prod_{n=1}^\infty \left(1+\frac{z^2}{\pi^2 n^2} \right)=z \prod_{n=1}^\infty \left(1+\frac{z^2}{\pi^2 n^2} \right) .$$
If you are interested in any more details, you may consult Ahlfors' complex analysis text.
I hope everything is clear now, if not I can try to elaborate further.


TITLE: Why considering the complementary event is the only solution in this case?
QUESTION [0 upvotes]: dear all I am reviewing a probability problem but I am a bit unsure the solution. Here is the problem:
males has 50% of being infected by diseaseX
females has 20% probability of being infected by diseaseX
the probability of a baby born from one infected parent to be infected is 30%;
what is the probability of a random baby be infected?
I am ok with the calculations for the following combinations:
P(baby infected|mother NOT infected, father NOT infected)= 0*(1-0.2)*(1-0.5)=0
P(baby infected|mother infected, father NOT infected)= 0.3*0.2*(1-0.5)=0.03;
P(baby infected|mother NOT infected, father infected)= 0.3*(1-0.2) * 0.5=0.12
But when it comes to the probability of the baby being infected given both the parents are infected I am a bit confused, my first approach would be:
P(baby infected|mother infected, father infected)= 0.2*0.5*0.3^2=0.009
But from the solution I have on the slides I see this last one combination should be calculated as 1- the probability of the baby “escaping” the infection from both the parents:
P(baby infected|mother infected, father infected)= 0.2*0.5*(1-((1-0.3)*(1-0.3)))=0.05
It makes sense but what I am missing is the theoretical reason of why this is the only solution in this last case. I think it has to do with independence of events but I can’t really figure it out.
 Any simple explanation would be highly appreciated!!
Thanks!!

REPLY [1 votes]: I think this comes down to the wording and interpretation of the setup. Perhaps a better wording would be "the probability of an infected parent passing the infection to their child is 30%." Then in order for a child of two infected parents to become infected, at least one of their two parents must pass the infection on to their child. The complement of this event is the event that both parents fail to pass on the infection, which happens with probability $(1-0.3)(1-0.3)=0.49$. Hence the probability that at least one parent passes on the infection is $1-0.49=0.51$.
If you instead calculate $0.3^2=0.09$, that is the probability that both parents pass on the infection. Certainly this would lead to an infected child, but that is not the only way for the child to be infected. It could also happen that the first parent passes on the infection and the second doesn't, which happens with probability $(0.3)(1-0.3)=0.21$. Or it could happen that the first parent fails to pass on the infection while the second parent does pass it on, which happens with probability $(1-0.3)(0.7)=0.21$. Notice if we add up the probabilities of these three cases we get $0.09+0.21+0.21=0.51$, which is the correct probability.
One final note is that the probabilities you've calculated are not conditional probabilities but rather probabilities of the intersection of the events. For instance,
$$P(\text{baby infected} | \text{mother infected, father not infected}) =0.3$$ while
\begin{align*}
&P(\text{baby infected and mother infected and father not infected})\\
&=P(\text{baby infected} | \text{mother infected,father not infected})\cdot P(\text{mother infected, father not infected})\\
&=(0.3)\cdot(0.2)\cdot(0.5).
\end{align*}


text:
\begin{document}

\title{knot modules and ribbon 2-knots}

\author{Jonathan A. Hillman }
\address{School of Mathematics and Statistics\\
     University of Sydney, NSW 2006\\
      Australia }

\email{jonh@maths.usyd.edu.au}

\begin{abstract}
We show that every $\mathbb{Z}$-torsion free knot module
is realized by a ribbon 2-knot with group $\pi$ of geometric dimension $\leq2$,
and give some partial results on the characterization of the knot modules of fibred ribbon 2-knots.
\end{abstract}

\keywords{deficiency, fibred, geometric dimension, 2-knot, ribbon}

\subjclass{57M45}

\maketitle

\section{Introduction}

The question underlying this note is whether knot groups
with deficiency 1 are the groups of ribbon 2-knots.
Every group with deficiency 1 and weight 1 is the group of a 2-knot.
It is the group of a ribbon 2-knot if and only if
it has a Wirtinger presentation of deficiency 1 \cite{Ya69}
(see also \cite [Section 1.7]{Hi1});
equivalently, if there is a deficiency-1 presentation such that the
balanced presentation of the trivial group obtained by killing a meridian
is AC equivalent to the empty presentation \cite{Yo82}.
There is a parallel criterion in the fibred case.
A knot group is the group of a fibred homotopy ribbon 2-knot if and only if
it has deficiency 1 and its commutator subgroup is finitely generated \cite[Theorem 17.9]{Hi2}.
If the Andrews-Curtis Conjecture holds then every group $\pi$
with  deficiency 1 and weight 1 is a ribbon 2-knot group,
while if the Whitehead Asphericity Conjecture holds then
every such group has geometric dimension $\leq2$.

We shall show that every $\mathbb{Z}$-torsion free knot module
is realized by a ribbon 2-knot with group $\pi$ such that $g.d.\pi\leq2$.
After first posting this paper on the arXiv, we learned that
(apart from the consideration of geometric dimension)
Theorem 1 below is a special case of Theorem 3.2 of \cite{Ka08},
which characterizes the homology of infinite cyclic covers of ribbon links of surfaces in $S^4$.
However, our motivation for considering this issue is different,
and we hope that the present simple argument may still be of interest.

Whether knot modules which are finitely generated as abelian groups
can be realized by {\it fibred} ribbon 2-knots seems more difficult,
and we have only partial results on this.
Knot modules which are direct sums of cyclic modules
and which are torsion-free and finitely generated as groups
are realized by fibred ribbon knots.
(This is surely known, but we include a proof as Theorem 2.)
Beyond this, every knot module which is torsion-free and finitely generated as a group
may be realized either by a fibred 2-knot in a homology 4-sphere (Lemma 3)
or by a ribbon 2-knot whose group is an ascending HNN extension
(Lemma 4).

The final section records seven questions on the issues considered here;
deficiency, geometric dimension, knot modules and (fibred) ribbon knots.

\newpage
\section{terminology}
{\it
A Wirtinger presentation\/} for a group $G$ is one in which each relator
expresses one of the generators as a conjugate of itself or another.
Thus it has the form
\[
\langle{x_1,\dots,x_m}\mid{x_j=w_{ij}x_iw_{ij}^{-1}~\forall(i,j)\in{S}}\rangle,
\]
where each $w_{ij}$ is a word in the generators, indexed by a subset $S$ of $\{1,\dots,m\}^2$.
After introducing equal numbers of new generators and relators, we may assume that the
words $w_{ij}$ have length 1. We may then arrange that they have positive exponent.
The resulting presentation is determined by a {\it labeled oriented graph} $(\Gamma,\lambda)$
with vertices $V$, edges $E\subseteq{V^2}$, origin and terminus functions
$o,t:E\to{V}$, and a third (label) function $\lambda:E\to V$,
and may be given by
\[
\langle{V}\mid{t(e)=\lambda(e)o(e)\lambda(e)^{-1)}~\forall{e}\in{E}}\rangle.
\]
If the group $G$ has abelianization $G/G'\cong\mathbb{Z}$ and the presentation has deficiency 1
then $\Gamma$ is a tree, and so we have a {\it labelled oriented tree}, or LOT \cite{Ho85}.

\section{realizing knot modules by ribbon 2-knots}

Let $\Lambda=\mathbb{Z}[t,t^{-1}]$, and let  $\varepsilon:\Lambda\to\mathbb{Z}$
be the canonical augmentation homomorphism, determined by $\varepsilon(t)=1$.
A {\it knot module\/} is a finitely generated $\Lambda$-module $A$ such that
$\mathbb{Z}\otimes_\Lambda{A}=0$.
We shall say that $A$ can be realized by an $n$-knot $K$
if $A\cong\pi'/\pi''$, where $\pi=\pi{K}$ is the knot group.
If $\pi$ has deficiency 1 then $\pi'/\pi''$  is $\mathbb{Z}$-torsion free.
A $\mathbb{Z}$-torsion free knot module has finite rank as an abelian group
and a square presentation matrix as a $\Lambda$-module \cite{Le77}.

\begin{theorem}
Every $\mathbb{Z}$-torsion free knot module is realized by a ribbon $2$-knot
with group $\pi$ such that $g.d.\pi\leq2$.
\end{theorem}

\begin{proof}
Let $A$ be a $\mathbb{Z}$-torsion free knot module.
Then $A$ has a presentation matrix of the form $\alpha=tM+(I_r-M)$,
where $M$ is an integer $r\times{r}$ matrix such that $\det(M)\not=0$ and
$\det(M-I_r)\not=0$ \cite{Tr74}.
Let $\pi$ be the group with presentation
\[
\langle{t,~x_1,\dots,x_r, y_1,\dots,y_r}
\mid{y_i=\Pi_jx_i^{m_{ij}}},
~ty_it^{-1}y_i^{-1}=x_i^{-1},
~\forall1\leq{i}\leq{r}
\rangle.
\]
(The order of the terms in the products $y_i=\Pi_jx_i^{m_{ij}}$
is not important.)
Then $\pi'$ is the normal closure of $\{x_1,\dots,x_r\}$.
Let $Y$ and $Z$ be the subgroups of $F(r)=F(x_1,\dots,x_r)$
generated by $\{y_1,\dots,y_r\}$ and $\{x_1^{-1}y_1,\dots,x_r^{-1}y_r\} $, respectively.
Since each has image of rank $r$ in $F(r)/F(r)'=\mathbb{Z}^r$,
and since  finitely generated free groups are Hopfian,
they are freely generated by these bases.
Hence $\pi$ is an HNN extension with base $F(r)$ and associated subgroups
$Y$ and $Z$, and so there is a $K(\pi,1)$ which is a finite 2-complex.
Moreover, it easy to see that $\pi'/\pi''\cong{A}$.

Now let $s_i=y_ity_i^{-1}$, for $1\leq{i}\leq{r}$.
Then $x_i=s_it^{-1}$, and so we can write each $y_j$ in terms of the $s_i$s and $t$.
Hence $\pi$ also has the deficiency-1 Wirtinger presentation
\[
\langle{t,~s_1,\dots,s_r}
\mid{s_i=y_ity_i^{-1}
~\forall1\leq{i}\leq{r}}
\rangle,
\]
and so $\pi$ is the group of a ribbon 2-knot.
\end{proof}

A similar argument may be used to show that the groups constructed
in \cite[Theorem 11.1]{Le77} have Wirtinger presentations of deficiency 1.

If the knot module is cyclic then the construction of Theorem 1 gives a 1-relator group.
If $\pi$ is any 1-relator knot group then the relator is not a proper power,
and so $g.d.\pi\leq2$ \cite{Ly50}.
An argument related to that of the theorem shows that any such group
has a deficiency-1 Wirtinger presentation.
However it need not have a 1-relator Wirtinger presentation.
Yoshikawa has shown that if a knot group has a 1-relator Wirtinger presentation
then it is an HNN extension with free base and associated subgroups.
He showed also that if $p,q,r,s\not =\pm1$ and $ps-qr=\pm1$ then the groups with presentation
$\langle{a,b,t}\mid{a^p=b^q},~ta^rt^{-1}=b^s\rangle$ have no such free base \cite{Yo88}.
If $r=p+1$ these groups have 1-relator presentations,
but do not have 1-relator Wirtinger presentations.
(For example, taking $p=2$, $q=r=3$ and $s=4$ gives the presentation
$\langle{b,t}\mid{(b^{-3}t^{-1}b^4t)^2=b^3}\rangle$.
Let  $s_i=b^{-i}tb^i$, for $i\in\mathbb{Z}$.
Then we may write the relation as $b=s_5^{-1}s_1s_4^{-1}t$.
Using this relation to eliminate the generator $b$,
we obtain a deficency-1 Wirtinger presentation
with 4 generators.)

If $g.d.\pi=2$ then $c.d.\pi=2$ and $\mathrm{def}(\pi)=1$.
If $\pi$ has deficiency 1 then $g.d.\pi=c.d.\pi$ \cite[Theorem 2.8]{Hi2}.
If $\mathcal{P}$ is a deficiency-1 presentation of a group $\pi$ of weight 1
then adjoining a 2-cell to the  associated 2-complex $C(\mathcal{P})$
gives a contractible 2-complex.
Hence if the (finite) Whitehead Conjecture is true then
$C(\mathcal{P})$ is aspherical and so $g.d.\pi\leq2$.
If the Eilenberg-Ganea conjecture holds, then $c.d.\pi\leq2$
implies $g.d.\pi\leq2$.
It is known that these conjectures cannot both hold in general.
(However, the known counterexamples are not finitely presentable.)

If $c.d.\pi=2$ and $\mathcal{P}$ is any finite presentation for $\pi$
then $\pi_2(C(\mathcal{P}))$ is a finitely generated projective $\mathbb{Z}[\pi]$-module.
If, moreover, $\pi$ is of type $FF$ then $\pi_2(C(\mathcal{P}))$ is
stably free, by a Schanuel's Lemma argument.
After adjoining  2-cells along constant maps we may assume that
$\pi_2(C(\mathcal{P}))$ is a free module.
Adjoining 3-cells along a basis for this module then gives
a finite 3-dimensional $K(\pi,1)$-complex.

\section{finitely generated modules and fibred 2-knots}

It is not known whether  every $\mathbb{Z}$-torsion free knot module $A$ which
is finitely generated as an abelian group is realized
by a high dimensional knot group with finitely generated commutator subgroup.
In this section we shall consider the sharper question,
whether such a knot module is realized by  a fibred ribbon 2-knot.

For knot groups $\pi$ with $\pi'$ finitely generated the conditions
``$\pi'$ {\it is a free group}", ``$g.d.\pi\leq2$" and ``$\mathrm{def}(\pi)= 1$"
are equivalent, while for any knot group , ``$g.d.\pi\leq2$"
if and only if  ``$c.d.\pi\leq2$ {\it and} $\mathrm{def}(\pi)=1$".
(See \cite[Chapter 2]{Hi2} for the less obvious implications.)
Knot groups with $\pi'$ free are the groups of fibred knots with fibre
a once-punctured connected sum of copies of $S^2\times{S^1}$.
A fibred 2-knot with group $\pi$ is homotopy ribbon if and only if $\pi'$ is free \cite{Co83}.
A 2-knot $K$ is $s$-concordant to a fibred homotopy ribbon knot if and only if
$\mathrm{def}(\pi)=1$ and $\pi'$ is finitely generated \cite[Theorem 17.9]{Hi2}.
The next result is largely known, but is included for completeness.

\begin{theorem}
Let $A$ be a knot module which is a direct sum of cyclic $\Lambda$-modules
and is free as an abelian group.
Then $A$ is realized by a fibred ribbon $2$-knot.
\end{theorem}

\begin{proof}
Suppose first that $A$ is cyclic.
We may assume that
$A\cong\Lambda/(\alpha)$,
where $\alpha=\Sigma{a_k}t^k$ is a polynomial
of degree $r$ in $\mathbb{Z}[t]$ with nonzero constant term
and $\varepsilon(\alpha)=1$.
Then $|a_0|=|a_r|=1$,
since $A$ is finitely generated as an abelian group.
Let $\beta=(\alpha-1)/(t-1)$.
Then $\beta=\Sigma_{i=0}^{i=r-1}b_it^i\in\mathbb{Z}[t]$,
and $|b_0|=|b_{r-1}|=1$.
Let $\pi$ be the group with presentation
\[
\langle{t,x}\mid{x=wtw^{-1}t^{-1}}\rangle,
\]
where
$w=\Pi_{i=0}^{i=r-1}{t^i}x^{b_i}t^{-i}$.
Then $\pi$ is the normal closure of $t$, and so has weight 1,
while $\pi'/\pi''\cong\Lambda/(\alpha)$, since $\alpha=(t-1)\beta+1$.
The commutator subgroup is the normal closure of $x$.
We see easily that (since $b_{r-1}=\pm1$) we can solve for
$t^rxt^{-r}$ in terms of the finite set $\{x,txt^{-1},\dots,t^{r-1}xt^{1-r}\}$.
Similary (since $b_0=\pm1$) we may solve for $t^{-1}xt$ in terms of the same set.
It follows that $\pi'$ is finitely generated.

Let $u=xt$ and rewrite $w$ as a word $W$ in $\{t,u\}$.
Then $\pi$ also has the 1-relator Wirtinger presentation
\[
\langle{t,u}\mid{u=WtW^{-1}}\rangle.
\]
Such presentations are realized by ribbon knots of 1-fusion \cite{Ya69}.
(See also \cite[Theorem 7.17]{Hi1}.)
Ribbon knots of 1-fusion with finitely generated commutator subgroup
are fibred \cite{Yo81}.
(See also \cite[Appendix]{AS88}.)

Taking sums of knots shows that if $A$ is a nontrivial direct sum of cyclic $\Lambda$-modules
then $A$ is realized by a fibred ribbon knot.
\end{proof}

In general, it seems difficult to realize a knot module $A$ by a group of deficiency 1
which has both weight 1 and also finitely generated commutator subgroup.
Taken separately, these tasks are easy.
We do not need the weight-1 condition to find a fibred knot in
an homology 4-sphere which realizes $A$.

\begin{lemma}
Let $A$ be a knot module which is free of rank $r$ as an abelian group.
Then $A$ is realized by a fibred $2$-knot in an homology $4$-sphere.
\end{lemma}

\begin{proof}
The action of $t$ on $A$  determines a matrix $T\in{GL(r,\mathbb{Z})}$
such that $T-I_r$ is also invertible.
Since the canonical map from $Aut(F(r))$ to $GL(r,\mathbb{Z})$ is onto,
there is an automorphism $\tau$ of $F(r)$ which abelianizes to $T$.
Let $G=F(r)\rtimes_\tau\mathbb{Z}$.
Then $G/G'\cong\mathbb{Z}$, $G'\cong{F(r)}$ and $G'/G''\cong{A}$.

The automorphisms $\tau$ can be realized by a based self-diffeomorphism
$f_\tau$ of $N=\#^r(S^3\times{S^1})$, and so $G\cong\pi_1(M(f_\tau))$,
where $M(f_\tau)$ is the mapping torus of $f_\tau$.
Surgery on the section of $M(f_\tau)$ determined by the basepoint of $N$
gives a homology 4-sphere $\Sigma$,
and the cocore of the surgery is a fibred 2-knot in $\Sigma$ with group $G$.
\end{proof}

If we try a similar approach to constructing a knot in $S^4$,
we loose some control of the commutator subgroup.
However we can improve slightly on Theorem 1.

\begin{lemma}
Let $A$ be a knot module which is free of rank $r$ as an abelian group.
Then $A$ is realized by a ribbon $2$-knot whose group
is an ascending HNN extension with base $F(r)$.
\end{lemma}

\begin{proof}
Since $A$ is a knot module, $t-1$ acts invertibly,
and so there is a matrix $M\in{GL(r,\mathbb{Z})}$
such that $(t-1)X=MX$ for all $X\in{A}$.
Let $\mu$ be an automorphism of $F(r)$ which induces $M$ on $F(r)/F(r)'=\mathbb{Z}^r$.
Then the group $\pi$ with presentation
\[
\langle
{t,~x_1,\dots,x_r}\mid{tx_it^{-1}=x_i\mu(x_i)}~\forall1\leq{i}\leq{r}
\rangle
\]
is an ascending HNN extension with base $F(r)$,
and $\pi'/\pi''\cong{A}$.
As in Theorem 1, it has a deficiency-1 Wirtinger presentation,
and so is the group of a ribbon 2-knot.
\end{proof}

We may find such an automorphism $\mu$ by
using row reduction to express $M$ as a product of elementary
matrices, and then lifting each factor in the obvious way.

The endomorphism $\theta$ determined by $\theta(x_i)=x_i\mu(x_i)$
induces an isomorphism on abelianiztion.
Since $H_2(F(r);\mathbb{Z})=0$, it also induces automorphisms of all
the quotients $F(r)/F(r)_{[n]}$ of the lower central series \cite{St65}.
Can we choose $\mu$ so that $\theta$ is also an automorphism?

\section{some questions}

It is possible that every 2-knot with group of deficiency 1 may be a ribbon 2-knot.
If so, this would relate the somewhat opaque notion of deficiency
to an immediately understandable geometric condition.
The following questions reflect our lack of a fuller understanding of the connections
between these notions.

Three of these questions are closely related to the Andrews-Curtis Conjecture,
through \cite{Yo82}.
The algebraic version of this conjecture is that every balanced presentation
$\mathcal{P}=\langle{x_1,\dots,x_m}\mid{w_1,\dots,w_m}\rangle$
of the trivial group may be reduced to the empty presentation by a limited class of moves
(AC {\it equivalence}).
We may replace relators by their inverses, conjugates and products with another relator,
and we may introduce or remove a generator $y$ and a relator $yz$ with $z$ a word in the
other generators.
In \cite{Yo82} it is shown that if $\mathcal{P}_o$ is a deficiency-1 presentation of a knot group $G$ and
$\mu$ is a normal generator for $G$ then $G$ has a deficiency-1 Wirtinger presentation
in which $\mu$ is one of the generators if and only if $\mathcal{P}=\mathcal{P}_o\cup\{\mu=1\}$
is AC equivalent to the empty presentation.

\begin{enumerate}

\item{}If  $A$ is a knot module which is free of rank $r$ as an abelian group,
is it realized by a 2-knot with group $\pi$ such that $\pi'$ is free?

\item{}Do deficiency-1 knot groups have deficiency-1 Wirtinger presentations?

\item{}Is every knot group $\pi$ with $\pi'$ free the group of a ribbon 2-knot?

\item{}If $t\in\pi$ normally generates a knot group $\pi$ with a deficiency-1 Wirtinger presentation,
does $\pi$ have such a presentation for which $t$ is the image of a generator?

\item{}If $K$ is a ribbon 2-knot with group $\pi$ such that $\pi'$ is free is $K$ fibred?

\item{}Do deficiency-1 knot  groups have geometric dimension $\leq2$?

\item{}If $\pi$ is a high dimensional knot group such that $c.d.\pi=2$ is $g.d.\pi=2$?

\end{enumerate}

Question (1) may be reformulated as follows.
Let $P\in{GL(r,\mathbb{Z})}$ be a matrix such that $\det(P-I_r)=\pm1$.
Is there an automorphism $\tau$ of $F(r)$ with abelianization $P$
and such that $\langle{x_1,\dots,x_r}\mid{x_i=\tau(x_i)~\forall1\leq{i}\leq{r}}\rangle$
is a presentation of the trivial group?
The results of \S4 above are partial answers to this question.

Questions (2-4) would all have positive answers if the Andrews-Curtis Conjecture is true.
However our questions consider an a priori narrower class of presentations,
and so some may have positive answers even if
the Andrews-Curtis Conjecture fails.
For instance,  (3) may be reformulated as follows.
Suppose that $\theta$ is an automorphism of $F(r)$ such that
$\langle{x_1,\dots,x_r}\mid{x_i=\theta(x_i)~\forall1\leq{i}\leq{r}}\rangle$
is a presentation of the trivial group.
Is this presentation AC equivalent to the empty presentation?

If $\pi$ has a  Wirtinger presentation $\mathcal{P}$ of deficiency 1 and
$g$ is a generator then $\mathcal{P}\cup\{g=1\}$ is AC equivalent to the empty presentation.
Question (4) may be phrased as asking ``is this also so for $\mathcal{P}\cup\{t=1\}$"?
This question reflects the fact that knot groups may have infinitely many essentially distinct
normal generators \cite{Su85}.
Even if (2) holds it may not be true that 2-knots with deficiency-1 groups are ribbon knots.

Resolution of  (5) may depend on progress in 4-dimensional topology,
although it could also be viewed as a question about the geometric information encoded in a LOT.
(See \cite{Ho85}.)

Question (6) is part of the Whitehead Conjecture,
while (7) is more of a wish for simplicity than an expectation.

\newpage


TITLE: Initial step in proving that a partial derivative of a tensor is not a tensor
QUESTION [2 upvotes]: Exercise 91 in Pavel Grinfeld's book Introduction to Tensor Analysis and the Calculus of Moving Surfaces reads

Similarly, for a contravariant tensor $T^i,$ derive the transformation rule for $∂ T^i ∕∂ Z^j,$ and show that it is not a tensor.

The solution (here) seems to proceed by contradiction from
$$S^{ij}\overset{?}{=}∂ T^i ∕∂ Z^j$$
I don't understand this. The contravariant tensor with one single live index is really a vector (and $T^i$ would be the components of the vector), and the operation on the RHS is the partial derivative of its components with respect to the different coordinates.
This is not a tensor, because it depends on the specific coordinates, which may vary from point to point, and this change in basis vectors is not accounted for. There are a number of posts on this (example).
The question is - under the assumption that this was actually true, as in Euclidean space - why would it render a twice contravariant tensor, i.e. $S^{ij},$ as opposed to simply another contravariant tensor of order one, $S^i$?

As an aside, the notation in the book is confusing. He seems to use $Z_i$ or $Z^i$ for covariant and contravariant components, and $\bf Z_i$ and $\bf Z^i$ for covariant and contravariant basis vectors, but I believe the usual notation for basis vectors is $\partial_i$ (alternatively $\frac{\partial}{\partial_i}$) or simply ${\bf e}_i$) for covariant and $\rm dx^i$ (or ${\bf e}^i$) for contravariant. Is this right? The idea of the "Z" I guess is to get rid of any distraction from the indices...

REPLY [0 votes]: The gradient of a vector field is a tensor.
In Cartesian coordinates where there is no change in basis vectors from point to point this is clear:
$$\begin{align}
\nabla \vec v &= \frac{\partial}{\partial Z^1}\vec v\; \vec e_1 + \frac{\partial}{\partial Z^2}\vec v\; \vec e_2 + \frac{\partial}{\partial Z^3}\vec v\; \vec e_3 \\
&= \frac{\partial}{\partial Z^1}\left(v_1 \; \vec e_1 + v_2 \; \vec e_2 + v_3 \; \vec e_3 \right)\; \vec e_1 +\\
&+ \frac{\partial}{\partial Z^2}\left(v_1 \; \vec e_1 + v_2 \; \vec e_2 + v_3 \; \vec e_3 \right)\; \vec e_2 +\\
&+ \frac{\partial}{\partial Z^3}\left(v_1 \; \vec e_1 + v_2 \; \vec e_2 + v_3 \; \vec e_3 \right)\; \vec e_3 \\
&= \frac{\partial v_1}{\partial Z^1} \vec e_1\vec e_1 + \frac{\partial v_2}{\partial Z^1} \vec e_1\vec e_2 +\frac{\partial v_3}{\partial Z^1} \vec e_1\vec e_3 +\\
&+\frac{\partial v_1}{\partial Z^2} \vec e_2\vec e_1 + \frac{\partial v_2}{\partial Z^2} \vec e_2\vec e_2 +\frac{\partial v_3}{\partial Z^3} \vec e_2\vec e_3 +\\
&+\frac{\partial v_1}{\partial Z^3} \vec e_3\vec e_1 + \frac{\partial v_2}{\partial Z^3} \vec e_3\vec e_2 +\frac{\partial v_3}{\partial Z^3} \vec e_3\vec e_3 \\
&=\frac{\partial v_k}{\partial Z^j} \vec e_j\vec e_k\\
&=\frac{\partial v_k}{\partial Z^j} \vec e_j\otimes\vec e_k\\
&=T^{jk}
\end{align}$$
Therefore it makes sense to start off the prove that, besides in Cartesian coordinates, the partial derivatives of a tenson $T^i$ are not a tensor by assuming they do, and seeking a contradiction.
Found here.


TITLE: Sequence of Functions Uniformally converging to $0$ method of proof
QUESTION [0 upvotes]: I am starting to learn about point-wise and uniform convergence. I know uniform is stronger because it only depends on $\epsilon$ while point-wise depends on $x$. I was wondering if my proof is adequate as well as any further comments on the subject is appreciated.
$\{\frac{x}{1+nx}\}$ on $(0,1)$
proof
$| f_n(x) - f_n | = | \frac{x}{1+nx} - 0| < \frac{x}{nx} = \frac{1}{n} \rightarrow 0$
Thus $f_n(x)$ is uniformly continuous to $0$ on $(0,1)$.

REPLY [1 votes]: $$|f_n(x) - 0| = |\frac{x}{1+nx}| < \frac{1}{n}$$
Tells you that $f_n$ converges to $0$ uniformly on $(0,1)$.
If $1+nx \neq 0$, that is $x \neq -\frac{1}{n}$, then the uniform convergence still holds. So actually $f_n \rightarrow 0$ uniformly on $\mathbb R_{\geq 0}$ is also true.