text:
\section{Introduction}
The analysis of textual documents is an important task in the legal domain~\cite{ashley2017artificial}. A plethora of different use cases exist, that heavily rely on analysis and inspection of documents. The importance is not restricted to legal research tasks, such as the analysis of statutory texts, judgments, or contracts, but also includes reviewing the content of text documents with regard to facts and evidences~\cite{ashley2017artificial}. More and more pieces of circumstantial evidence are discovered using technology, especially when it comes up to inspect huge document corpora. The field of e-Discovery, forensics, legal reasoning and argument mining, and information extraction (IE) is well-studied and established in the field of legal informatics \cite{ashley2010emerging}. Especially the usage during due diligence is highly attractive as it helps to save valuable resources, such as money and time.

Technology assists human experts during complex discovery tasks to find the "needle in the haystack". Methods and software tools have been used successfully to detect crime such as organized manipulation, e.g., analysis of a very large collection of e-mails during the VW scandal of manipulated software, or to automatically unveil discrimination in the internet. The latter is especially relevant for social media platforms but also for e-participation initiatives. The past has shown, that anonymity within the internet attracts users for inappropriate, unconstitutional and illegal comments. For example, extremist statements, threats, insults, and so forth. Online platforms are getting more responsible in charge of the content that users create and share with others. The responsibility to delete comments on request but also to proactively delete illegal comments is more and more in the charge of the platform providers, which struggle at this complex and tedious task. This leads to the need of a highly accurate software to perform this task automatically.

This paper contributes to the detection and classification of statements and comments with regard to their sentimental content. The sentiment analysis is restricted to inspect text with regard to illegal content, such as discrimination. The approach described in the paper extends the existing state-of-the-art in the field and uses a multi-task learning architecture based on deep learning (DL). The result is relevant for every social media platform that wants to improve compliance by proactively detecting problematic comments and statements.

The remainder of the paper is structured as follows: Section 2 describes related work and similar approaches in the domain of sentiment analysis to detect illegal statements. It also introduces some recent work on utilizing DL in this matter. Section 3 introduces the multi-label system by describing the dataset that is used, the algorithms, and the topology of the architecture. The experimental setup is described in Section 4, while the results are discussed in Section 5. Based on the results, the limitations and potential arising research directions are described in Section 6.

\section{Related Work}
The computer-assisted analysis of legal documents is highly relevant and has attracted researchers for quite some time. Sentiment analysis, also called opinion mining, is an important field of research, not only in the legal domain \cite{conrad2007opinion}. Pang et al. \cite{pang2008opinion} provide a comprehensive overview of different approaches to sentiment analysis across various domains. This paper deals with sentiment analysis in comments from various online sources in order to detect illegal comments. Hutto \& Gilbert \cite{gilbert2014vader} came up with a rule based system, called VADER, to analyze social media text. A deep dive on supervised sentiment analysis based on various multilingual web texts was made by Boiy \& Moens \cite{boiy2009machine}. They used hand-crafted features to train three different ML classifiers (support vector machines (SVM), multinomial naive bayes, and maximum entropy classifier) on the given task. Even though there are several existing works on this topic, among others \cite{hong2016malicious, Paltoglou:2012:TMD:2337542.2337551, Paltoglou2014, sood2012automatic}, hardly any attempt has been made in the legal domain by utilizing DL.

\begin{comment}
However, research concerning sentiment analysis based on a multi-label setting leveraging DL has been already performed. Liu \cite{liu2015multi} proposed a deep convolution neural network approach for multi-label Text classification sentiment analysis which were tested on two microblog datasets.
\end{comment}
\section{Illegal Content Detection System}

In this section, we are covering the datasets which we used, and a detail description of the Multi-Task Multi-Embedding algorithm that we have used.

\subsection{Datasets}

\textbf{Multi-Label Illegal Comments:} The main dataset was provided by Jigsaw which is a team within Google through one of Kaggle competitions \cite{JigsawDataset2018}. Each sample has a text comment which is labeled with one or more labels among six labels: Toxic, Severe Toxic, Obscene, Threat, Insult and Identity Hate.  The dataset is divided into a training set and a test set. The number of samples of the training dataset and test dataset is 159571 and 63978 respectively. Figure~\ref{fig:multiLabelOccurrence} shows the number of multilabel occurrences on the training dataset.

\begin{figure}[htb!]
\centering
\includegraphics[width=\linewidth]{figures/dataSetMultiLabelOccurance}
\caption{Number of multi-label occurrence on the training dataset}
\label{fig:multiLabelOccurrence}
\end{figure}

The biggest challenge of this dataset is imbalance class distribution, where the number of true classes is between 0.3\% and 10\%. In these cases most machine learning algorithms tend to give very high accuracy because it is easier for the algorithms to just produce one value either 1 or 0 to minimize the loss function. However, in this case, it crucially important to predict the toxicity of the comments and not just predict every comment as clean comment. This is a major problem in the legal domain where the number of positive or negative samples for many problems are very low. In Figure~\ref{fig:classOccurrence}, the number of comments occurrence of each class on the training dataset is presented. For example, the number of threat comments is about 0.3\% which make it difficult for the algorithm to predict it correctly.

\begin{figure}[htb!]
\centering
\includegraphics[width=\linewidth]{figures/dataSetClassOccurance}
\caption{Number of comment occurrence on the training dataset for each class }
\label{fig:classOccurrence}
\end{figure}

The dataset contains text from various languages. The top 30 unicode scripts occurrence on the training dataset is presented in Table~\ref{table:languageOccurrence}. This is another problem since most of the modern machine learning algorithm uses specific language word representations like word embedding. Which means the model will just set all the other languages words to zero. However, since the number of Non-English comments was less than 0.5\% and as shown on the word cloud almost all of the major illegal words were in English, we only focused on English comments.

\begin{table}[htb!]
\centering
\caption{Top 30 unicode scripts occurrence on the train dataset}
\label{table:languageOccurrence}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\textbf{Language} & \textbf{Occurrence} & \textbf{Language} & \textbf{Occurrence} & \textbf{Language} & \textbf{Occurrence} \\ \hline
Latin             & 159564              & Bengali           & 25                  & Kannada           & 4                   \\ \hline
Greek             & 456                 & Runic             & 22                  & Lao               & 4                   \\ \hline
Han               & 344                 & Hangul            & 21                  & Gujarati          & 3                   \\ \hline
Cyrillic          & 272                 & Ethiopic          & 20                  & Telugu            & 3                   \\ \hline
Arabic            & 78                  & Georgian          & 17                  & Tibetan           & 3                   \\ \hline
Hiragana          & 66                  & Tamil             & 9                   & Oriya             & 2                   \\ \hline
Katakana          & 55                  & Thai              & 8                   & Sinhala           & 2                   \\ \hline
Devanagari        & 52                  & Gurmukhi          & 7                   & Bopomofo          & 1                   \\ \hline
Inherited         & 50                  & Armenian          & 6                   & Malayalam         & 1                   \\ \hline
Hebrew            & 48                  & Khmer             & 5                   & Syriac            & 1                   \\ \hline
\end{tabular}
\end{table}

A word cloud for each class is presented in Figure~\ref{fig:wordCloud} for the top 1000 word occurrence. Since the illegal comments were collected from wiki page comments, we can notice that the top keywords on clean comments are "Wikipedia, page and article". A second example, on "Identity Hate" class, the top keywords target specific race or specific religion like "Nigger or Jew". A third example, on "Threat" class, the top keywords were life threatening words like "Die or Kill". In Figure \ref{fig:wordCloudCombined}, we can see a word cloud of over all top 1000 word occurrence, and clearly none of the most occurred words on the six labels appear. This makes the machine learning algorithms difficult to predict them.

\begin{figure}[htb!]
\begin{subfigure}{0.5\textwidth}
  \centering
  \includegraphics[width=\linewidth]{figures/wordCloudClean}
  \caption{Clean}
  \label{fig:wordCloudClean}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\linewidth]{figures/wordCloudToxic}
  \caption{Toxic}
  \label{fig:wordCloudToxic}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\linewidth]{figures/wordCloudSevereToxic}
  \caption{Severe Toxic}
  \label{fig:wordCloudSevereToxic}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\linewidth]{figures/wordCloudObsence}
  \caption{Obsence}
  \label{fig:wordCloudObsence}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\linewidth]{figures/wordCloudThreat}
  \caption{Threat}
  \label{fig:wordCloudThreat}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\linewidth]{figures/wordCloudInsult}
  \caption{Insult}
  \label{fig:wordCloudInsult}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\linewidth]{figures/wordCloudIdentityHate}
  \caption{Identity Hate}
  \label{fig:wordCloudIdentityHate}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\linewidth]{figures/wordCloudCombined}
  \caption{All Combined}
  \label{fig:wordCloudCombined}
\end{subfigure}
\caption{Single gram word cloud of the top 1000 word occurrence for each comments class}
\label{fig:wordCloud}
\end{figure}

\hfill\break
\textbf{Toxic, Attack and Aggression Comments:} Additionally, three datasets were used to analyze if it is beneficial for the main dataset to join training it with these three datasets using Multi-task training. These specific datasets were chosen because generally in multi-task training, the closer the tasks the better the results. All of them were also provided by Jigsaw team. The number of samples of the toxic, attack and aggression datasets are 159686, 115864, 115864 samples. These three datasets also suffer from the imbalance class distribution problem.

\subsection{Multi-Task Multi-Embedding Model}
Figure~\ref{fig:modelArchitecture} shows the Model Architecture, where the model receives four different inputs from four different datasets (Main Multi-Label Illegal, Toxic, Attack and Aggression) and produce four different outputs for each task, while the whole layers are hard shared among all tasks~\cite{ruder2017overview}. Simply, hard sharing means all the internal learning parameters are shared among all tasks. The model can be divided into the following layers:

\subsubsection{Word Representations Layer}
Each input goes through three different word embedding layers (FastTexts~\cite{joulin2016bag} , Glove~\cite{pennington2014glove} and our Glove). The Fast Text is 2 million word vectors trained on Common Crawl with dimension 300, while Glove is 2.2 million word vectors trained on Common Crawl with dimension 300. Furthermore, we trained a custom Glove model with comments from English Wikipedia talk pages~\cite{wulczyn2017ex} which produced 0.5 million word vectors with dimension 300.
Each word embedding model is followed by 20\% Spatial 1D Dropout, which simply tries to hide some words representations' during training to reduce over fitting.
Two ideas are behind using different word representation, first is to capture different semantic and syntactic features of words, second is to cover as many words as possible by training them on different unstructured text sources.

\subsubsection{Recurrent Neural Network (RNN) Layers}
Each word embedding layer is followed by two parallel Bidirectional RNN layers. The first is a GRU and second one is LSTM layer. Both layers has 128 neurons and is followed by 20\% dropout. This creates a total of 3 GRU layers and 3 LSTM layers. The main benefit of using these RNN layers is to allow the model to capture the relationship between the sequence of words.

\subsubsection{Convolutional Neural Network (CNN) Layers}
Afterwards, a separate CNN layer takes the output of each separate RNN layer, which creates a total of 6 CNN layers. The hyper parameters of each CNN are as follows: the number of filters is 64, the kernel size is 2, and the activation function was RELU. Each one is followed by 20\% dropout.
Afterward The output of the CNN layers is concatenated into 2 outputs. The first and second concatenation concatenates the output of the 3 CNN layers coming after the LSTM layers and the output of the 3 CNN layers coming after the GRU layers respectively.

\subsubsection{Pooling Layers}
The output of each concatenation of the CNN layers goes through separate Max Pooling and Average Pooling. Afterwards, the output of the 2 Max Pooling and the 2 Average Pooling layers are concatenated together.

\subsubsection{Dense Layers}
Finally, separate 4 dense layers take the output of the concatenated pooling layers to produce the final outputs. The first dense layer produces 6 outputs for each label of the main illegal comment dataset (Toxic, Severe Toxic, Obscene, Threat, Insult and Identity Hate), while the second, third and fourth dense layers produce only 1 output for each class on each complementary datasets (Toxic, Attack and Aggression).

\begin{figure}[htb!]
\centering
\includegraphics[width=\linewidth]{figures/model}
\caption{Multi-Task Multi-Embedding Deep Learning Architecture}
\label{fig:modelArchitecture}
\end{figure}

\section{Experimental Settings}

\subsection{Training Details}
Several single task algorithms were tested against the multi-task multi-embedding algorthim. All the single task algorithms were trained only on the main dataset, while the multi-task multi-embedding algorithm were trained on at least two data sets together.
Some standard classic approach algorithms were tested like Linear Regression (LR) and Decision Trees (DT). Other single task deep learning algorithms which were tested were reported by top participants in the Kaggle competition including: BI-GRU, CNN and BI-RNN-CNN models. The BI-GRU models is 2 layers of bi-directional GRU with 128 units. The CNN models consist of 4 parallel CNN layers "300 neurons and  kernel sizes are 2,3,4 and 5 respectively" followed by global max pooling, then concatenated together and followed by a final dense layer with 36 neurons. The BI-RNN-CNN models consist of 2 parallel GRU and LSTM "128 neurons" layers, followed separate CNN layers "64 neurons", followed by max and average and max pooling and then concatenated together.

Generally, all the models were trained until they converged using early stop to prevent over-fitting. For neural network based models a mini-batch stochastic gradient decent was used, with a binary cross-entropy loss function. Furthermore, the Adam optimizer was used to optimize the loss function on all of the neural network models with learning rate $ 1e^{-3} $. L1-norm or L2-norm regularization were not used, but dropout was used as discussed on the previous section.
For our Multi-Task model at each training step we trained the model for the same batch size of each problem sequentially.

All Models were trained using one computer with one Nvidia Titan XP GPU with 12 GB Ram, 20 CPU cores and 128 Memory Ram.

\subsection{Metrics}
As stated in the previous subsection the main dataset is skewed, which means the accuracy can't be a reliable measurement. That is why we choose the precision, recall and F1 score to give us a more reliable measurement. The equations for precision, recall and F1 score is shown in equation ~\ref{equ:precisionEqu}, ~\ref{equ:RecallEqu} and ~\ref{equ:F1Equ} respectively.

\begin{equation}
Precision = \frac{\#True Positives}{\#True Positives + \#False Positives}
\label{equ:precisionEqu}
\end{equation}
\begin{equation}
Recall = \frac{\#True Positives}{\#True Positives + \#False Negatives}
\label{equ:RecallEqu}
\end{equation}
\begin{equation}
F_{1} = \frac{Precision \times Recall}{Precision + Recall}
\label{equ:F1Equ}
\end{equation}

\section{Results and Discussion}

Table~\ref{table:MLResults} shows the result of the tested algorithms regarding precision, recall and F1 score matrices. All of the algorithms were able to reach an average 99\% F1 score for all the negative labels "Clean Comment" however, the problems arise with the positive labels "illegal comments" where the average F1 score ranges between 38\% and 59\%.
The Linear regression algorithm using TF-IDF performed the worst with an average positive F1 score 38\%. However, the decision tree using TF-IDF, which is another classical ML algorithm performed well compared to the deep learning models with F1 score 54\%, despite the fact it doesn't use any word embedding layer which transfer a lot of information for all the tokens, after being trained in big unstructured text datasets. The reason is that Decision Trees (DT) algorithms don't affect too much with imbalanced class data compared to other machine learning approaches. Both the BI-GRU and CNN models using fastText performed lower than the DT with an average F1 score 53\% and 52\%. The best model among all the single task approaches was the BI-RNN-CNN model using fastText with an average F1 score 58\%. However, by changing the embedding layer with Glove and our pre-trained model the BI-RNN-CNN F1 score decreased with average F1 score 54\% and 52\%.

The multi-task multi-embedding algorithm when used by combining the main data set with either toxic, attack or aggression didn't out-perform the best single task model with average F1 score 55\%, 56\% and 56\%. However, when the 4 data sets were combined it out-performed the best single task model with an average F1 score 59\%.

First observation, the balance of samples per label affect dramatically the performance of all algorithms. All models were able to perfectly predict a clean comment which was not the case with illegal comments, because the number of clean samples is almost about 97\% of the data set.

Second observation, multi-task algorithms with more data sets generalize much better than when only few data sets are chosen. Despite the fact that the model couldn't outperform all the F1 scores of every label, but it did generalize better on the overall average F1 score.

Third observation, multi-task algorithms which were fed with the 4 datasets had much higher true positive Rate (Recall), especially on the labels with very few samples (severe toxic, threat and identity hate). Which in this case makes it a better choice since the algorithm tends to be sensitive to classify the illegal comments.

\begin{table}[htp!]
\centering
\caption{Precision, Recall and F1 score of base line machine learning models versus our proposed Multi-Task Multi-Embedding Model}
\label{table:MLResults}
\resizebox{\columnwidth}{!}{%

\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\multicolumn{3}{|l|}{\textbf{Archticture}}                                                   & \multicolumn{7}{c|}{\textbf{Single Task}}                                                                  & \multicolumn{4}{c|}{\textbf{Multi-Task}}             \\ \hline
\multicolumn{3}{|l|}{\textbf{Model}}                                                         & \textbf{LR} & \textbf{DT} & \textbf{BI-GRU} & \textbf{CNN} & \multicolumn{3}{c|}{\textbf{BI-RNN-CNN}}      & \multicolumn{4}{c|}{\textbf{Multi-Emb RNN-CNN}}             \\ \hline
\multirow{4}{*}{\textbf{Datasets}}       & \multicolumn{2}{l|}{\textbf{Multi-Label Illegal}} & X           & X           & X               & X            & X             & X             & X             & X             & X    & X             & X             \\ \cline{2-14}
                                         & \multicolumn{2}{l|}{\textbf{Toxic}}               &             &             &                 &              &               &               &               & X             &      &               & X             \\ \cline{2-14}
                                         & \multicolumn{2}{l|}{\textbf{Attack}}              &             &             &                 &              &               &               &               &               & X    &               & X             \\ \cline{2-14}
                                         & \multicolumn{2}{l|}{\textbf{Agression}}           &             &             &                 &              &               &               &               &               &      & X             & X             \\ \hline
\multirow{4}{*}{\textbf{Word Embedding}} & \multicolumn{2}{l|}{\textbf{TF-IDF}}               & X           & X           &                 &              &               &               &               &               &      &               &               \\ \cline{2-14}
                                         & \multicolumn{2}{l|}{\textbf{FastText}}            &             &             & X               & X            & X             &               &               &X               & X    & X             & X             \\ \cline{2-14}
                                         & \multicolumn{2}{l|}{\textbf{Glove}}               &             &             &                 &              &               & X             &               &X               & X    & X             & X             \\ \cline{2-14}
                                         & \multicolumn{2}{l|}{\textbf{Our}}                 &             &             &                 &              &               &               & X             &X               & X    & X             & X             \\ \hline
\multirow{6}{*}{\textbf{Toxic}}          & \multirow{2}{*}{\textbf{P}}      & \textbf{0}     & 0,95        & 0,98        & 0,98            & 0,99         & 0,98          & 0,98          & 0,98          & 0,98          & 0,98 & 0,98          & 0,98          \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,78        & 0,63        & 0,6             & 0,55         & 0,64          & 0,62          & 0,64          & 0,61          & 0,6  & 0,6           & 0,64          \\ \cline{2-14}
                                         & \multirow{2}{*}{\textbf{R}}      & \textbf{0}     & 0,98        & 0,95        & 0,94            & 0,93         & 0,95          & 0,95          & 0,95          & 0,94          & 0,94 & 0,94          & 0,95          \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,56        & 0,77        & 0,85            & 0,87         & 0,81          & 0,81          & 0,78          & 0,83          & 0,82 & 0,83          & 0,8           \\ \cline{2-14}
                                         & \multirow{2}{*}{\textbf{F1}}     & \textbf{0}     & 0,97        & 0,96        & 0,96            & 0,95         & 0,97          & 0,96          & 0,97          & 0,96          & 0,96 & 0,96          & 0,96          \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,65        & 0,7         & 0,7             & 0,67         & \textbf{0,71} & 0,7           & 0,7           & \textbf{0,71} & 0,7  & 0,7           & \textbf{0,71} \\ \hline
\multirow{6}{*}{\textbf{Severe toxic}}   & \multirow{2}{*}{\textbf{P}}      & \textbf{0}     & 1           & 1           & 1               & 1            & 1             & 1             & 1             & 1             & 1    & 1             & 1             \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,37        & 0,35        & 0,45            & 0,39         & 0,51          & 0,43          & 0,45          & 0,45          & 0,45 & 0,35          & 0,35          \\ \cline{2-14}
                                         & \multirow{2}{*}{\textbf{R}}      & \textbf{0}     & 1           & 1           & 1               & 1            & 1             & 1             & 1             & 1             & 1    & 1             & 1             \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,2         & 0,25        & 0,27            & 0,31         & 0,21          & 0,36          & 0,34          & 0,2           & 0,22 & 0,39          & 0,41          \\ \cline{2-14}
                                         & \multirow{2}{*}{\textbf{F1}}     & \textbf{0}     & 1           & 1           & 1               & 1            & 1             & 1             & 1             & 1             & 1    & 1             & 1             \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,26        & 0,29        & 0,34            & 0,35         & 0,3           & \textbf{0,39} & \textbf{0,39} & 0,28          & 0,29 & 0,37          & 0,38          \\ \hline
\multirow{6}{*}{\textbf{Obscene}}        & \multirow{2}{*}{\textbf{P}}      & \textbf{0}     & 0,97        & 0,98        & 0,98            & 0,99         & 0,99          & 0,98          & 0,98          & 0,98          & 0,98 & 0,98          & 0,98          \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,88        & 0,68        & 0,66            & 0,59         & 0,64          & 0,73          & 0,73          & 0,69          & 0,68 & 0,69          & 0,7           \\ \cline{2-14}
                                         & \multirow{2}{*}{\textbf{R}}      & \textbf{0}     & 1           & 0,98        & 0,98            & 0,97         & 0,97          & 0,98          & 0,98          & 0,98          & 0,98 & 0,98          & 0,98          \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,5         & 0,72        & 0,75            & 0,8          & 0,78          & 0,68          & 0,68          & 0,73          & 0,73 & 0,73          & 0,71          \\ \cline{2-14}
                                         & \multirow{2}{*}{\textbf{F1}}     & \textbf{0}     & 0,98        & 0,98        & 0,98            & 0,98         & 0,98          & 0,98          & 0,98          & 0,98          & 0,98 & 0,98          & 0,98          \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,64        & 0,7         & 0,7             & 0,68         & 0,7           & \textbf{0,71} & 0,7           & \textbf{0,71} & 0,7  & \textbf{0,71} & \textbf{0,71} \\ \hline
\multirow{6}{*}{\textbf{Threat}}         & \multirow{2}{*}{\textbf{P}}      & \textbf{0}     & 1           & 1           & 1               & 1            & 1             & 1             & 1             & 1             & 1    & 1             & 1             \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 1           & 0,65        & 0,49            & 0,31         & 0,58          & 0,44          & 0,51          & 0,38          & 0,35 & 0,36          & 0,4           \\ \cline{2-14}
                                         & \multirow{2}{*}{\textbf{R}}      & \textbf{0}     & 1           & 1           & 1               & 1            & 1             & 1             & 1             & 1             & 1    & 1             & 1             \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0           & 0,28        & 0,12            & 0,14         & 0,47          & 0,28          & 0,13          & 0,51          & 0,59 & 0,52          & 0,6           \\ \cline{2-14}
                                         & \multirow{2}{*}{\textbf{F1}}     & \textbf{0}     & 1           & 1           & 1               & 1            & 1             & 1             & 1             & 1             & 1    & 1             & 1             \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,01        & 0,39        & 0,2             & 0,2          & \textbf{0,52} & 0,34          & 0,2           & 0,43          & 0,44 & 0,42          & 0,48          \\ \hline
\multirow{6}{*}{\textbf{Insult}}         & \multirow{2}{*}{\textbf{P}}      & \textbf{0}     & 0,97        & 0,98        & 0,98            & 0,98         & 0,98          & 0,98          & 0,98          & 0,98          & 0,97 & 0,97          & 0,98          \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,83        & 0,74        & 0,72            & 0,6          & 0,73          & 0,68          & 0,69          & 0,76          & 0,8  & 0,76          & 0,77          \\ \cline{2-14}
                                         & \multirow{2}{*}{\textbf{R}}      & \textbf{0}     & 1           & 0,99        & 0,99            & 0,97         & 0,99          & 0,98          & 0,98          & 0,99          & 0,99 & 0,99          & 0,99          \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,4         & 0,58        & 0,65            & 0,73         & 0,64          & 0,68          & 0,65          & 0,57          & 0,54 & 0,54          & 0,56          \\ \cline{2-14}
                                         & \multirow{2}{*}{\textbf{F1}}     & \textbf{0}     & 0,98        & 0,98        & 0,98            & 0,98         & 0,98          & 0,98          & 0,98          & 0,98          & 0,98 & 0,98          & 0,98          \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,54        & 0,65        & \textbf{0,68}   & 0,66         & \textbf{0,68} & \textbf{0,68} & 0,67          & 0,65          & 0,64 & 0,63          & 0,65          \\ \hline
\multirow{6}{*}{\textbf{Identity hate}}  & \multirow{2}{*}{\textbf{P}}      & \textbf{0}     & 0,99        & 0,99        & 0,99            & 0,99         & 1             & 0,99          & 0,99          & 0,99          & 0,99 & 0,99          & 1             \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,73        & 0,78        & 0,72            & 0,59         & 0,59          & 0,73          & 0,8           & 0,61          & 0,6  & 0,54          & 0,54          \\ \cline{2-14}
                                         & \multirow{2}{*}{\textbf{R}}      & \textbf{0}     & 1           & 1           & 1               & 1            & 1             & 1             & 1             & 1             & 1    & 1             & 0,99          \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,09        & 0,39        & 0,48            & 0,51         & 0,58          & 0,32          & 0,3           & 0,47          & 0,55 & 0,48          & 0,64          \\ \cline{2-14}
                                         & \multirow{2}{*}{\textbf{F1}}     & \textbf{0}     & 0,99        & 1           & 1               & 1            & 1             & 1             & 1             & 1             & 1    & 0,99          & 0,99          \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,16        & 0,52        & 0,57            & 0,54         & \textbf{0,59} & 0,44          & 0,43          & 0,53          & 0,57 & 0,51          & 0,58          \\ \hline
\multirow{2}{*}{\textbf{Total Average}}          & \multirow{2}{*}{\textbf{F1}}     & \textbf{0}     & 0,99        & 0,99        & 0,99            & 0,99         & 0,99          & 0,99          & 0,99          & 0,99          & 0,99 & 0,99          & 0,99          \\ \cline{3-14}
                                         &                                  & \textbf{1}     & 0,38        & 0,54        & 0,53            & 0,52         & 0,58          & 0,54          & 0,52          & 0,55          & 0,56 & 0,56          & \textbf{0,59} \\ \hline
\end{tabular}

}

\end{table}

\section{Conclusion}

Multi-task deep learning models are a promising new technology that could pave the way for more deep learning applications in the legal domain by transferring knowledge from large datasets to small datasets. We investigate the transfer learning capabilities of a particular multi-task architecture on a classification task on the publicly available toxic-comments Kaggle challenge dataset.
We explore the toxic-comments dataset in depth and identify the imbalanced classes as a major challenge.Our multi-task approach does not significantly improve upon single-task models in terms of F1-score. However, it improves the recall score significantly for labels with a low number of samples, which is extremely important to this use-case. A key result is that we can observe a significant improvement in F1-score when adding additional, related datasets to train the multi-task model.

These promising results encourage us to further investigate the transfer learning capabilities of multi-task models. In particular, it is unclear, what deep neural network architectures are most suitable for applications in the legal domain. It will be necessary to explore different combinations of tasks, datasets, architectures and hyper parameter selections to better understand the transfer learning capabilities of transfer learning especially with multi-task models.
\section{Acknowledgements}
We gratefully acknowledge the support of NVIDIA Corporation with the donation of the Titan XP Pascal GPU used for this research.


text:
\section*{Abstract}
\begin{quote} We continue our study \cite{Ou4} of $f$-biharmonic\; maps and
$f$-biharmonic\; \\submanifolds  by exploring the applications of $f$-biharmonic\; maps and the relationships among biharmonicity, $f$-biharmonicity and conformality of maps between Riemannian manifolds. We are able to characterize harmonic maps and minimal submanifolds by using the concept of $f$-biharmonic\; maps and prove that the set of all $f$-biharmonic\; maps from $2$-dimensional domain is invariant under the conformal change of the metric on the domain. We  give an improved equation for  $f$-biharmonic\;  hypersurfaces and use it to prove some rigidity theorems about $f$-biharmonic\; hypersurfaces in  nonpositively curved manifolds, and to give some classifications of $f$-biharmonic\; hypersurfaces in Einstein spaces and in space forms. Finally, we also use the improved  $f$-biharmonic\; hypersurface equation to obtain an improved equation and some classifications of biharmonic conformal immersions of surfaces into a $3$-manifold.
{\footnotesize  }
\end{quote}

\section{Introduction}
 {\em Biharmonic maps} and  {\em $f$-biharmonic\; maps}  are maps $\phi: (M,g)\longrightarrow (N,h)$ between Riemannian manifolds which are the critical points of
\begin{eqnarray}\notag
{\rm the\; bienergy:}\;\;\;\;E_{2}(\phi)=\frac{1}{2}\int_\Omega |\tau(\phi)|^2v_g,\;\;\;{\rm and}\\\notag
{\rm the\;} f{\rm-bienergy:}\;\;\;E_{2,f}(\phi)=\frac{1}{2}\int_\Omega f\,|\tau(\phi)|^2v_g,
\end{eqnarray}
respectively, where $\Omega$ is a compact domain of $ M$ and $\tau(\phi)$ is the tension field of the map $\phi$.
The Euler-Lagrange equations of these functionals give the biharmonic map equation (see \cite{Ji1})
\begin{equation}\label{BTF}
\tau_{2}(\phi):={\rm
Trace}_{g}(\nabla^{\phi}\nabla^{\phi}-\nabla^{\phi}_{\nabla^{M}})\tau(\phi)
- {\rm Trace}_{g} R^{N}({\rm d}\phi, \tau(\phi)){\rm d}\phi =0,
\end{equation}
where $R^{N}$ denotes the curvature operator of $(N, h)$ defined by
$$R^{N}(X,Y)Z=
[\nabla^{N}_{X},\nabla^{N}_{Y}]Z-\nabla^{N}_{[X,Y]}Z,$$

and the $f$-biharmonic map equation  ( see\cite{Lu} and also \cite{Ou4})
\begin{equation}\label{fbhe}
\tau_{2,f} (\phi)\equiv f\tau_2(\phi)+(\Delta f)\tau(\phi)+2\nabla^{\phi}_{{\rm grad}\, f}\tau(\phi)=0,
\end{equation}
where  $\tau_2( \phi)$ is the  bitension field of $\phi$ defined in (\ref{BTF}).\\

 As the images of biharmonic isometric immersions, biharmonic submanifolds have become a popular topic of research  with many progresses since the beginning of this century (see a recent survey \cite{Ou5} and the references therein for more details). In a recent paper \cite{Ou4}, we introduced and studied $f$-biharmonic submanifolds as those submanifolds whose isometric immersions $\phi: M^m\longrightarrow (N^n, h)$ are $f$-biharmonic maps for some positive function $f$ defined on the submanifolds.  $f$-Biharmonic submanifolds are generalizations of of biharmonic submanifolds and of minimal submanifolds since (i) $f$-biharmonic\; submanifolds with $f={\rm constant}$ are precisely biharmonic submanifolds, and (ii) any minimal submanifold is $f$-biharmonic for any positive function defined on the submanifold. As it is seen in \cite{Ou4}  and Example 1 of Section \ref{Sect3} of this paper, there are many examples of $f$-biharmonic submanifolds which are neither biharmonic nor minimal. \\

An interesting link among biharmonicity, $f$-biharmonicity and conformality is the following theorem which, although holds only for two-dimensional domain, has played an important role in the study of  biharmonic maps from  two-dimensional domains (see \cite{Ou4} and \cite{WOY}) and will be used to study  biharmonic conformal immersions in the last section of this paper.\\

{\bf Theorem A.}(\cite{Ou4} A map $\phi: (M^2, g)\longrightarrow (N^n, h)$ is an $f$-biharmonic map if and only if $\phi:
(M^2, f^{-1}g)\longrightarrow (N^n,h)$ is a biharmonic map. In particular, if a surface (i.e., an isometric immersion)  $\phi: (M^2, g=\phi^*h)\longrightarrow (N^n, h)$ is an $f$-biharmonic surface if and only if the conformal immersion$\phi:(M^2, f^{-1}g)\longrightarrow (N^n,h)$ is a biharmonic map.\\\

Regarding the topological and the geometric obstacles to the existence of general $f$-biharmonic\; maps, it was proved in \cite{Ou4} that an $f$-biharmonic\; map from a compact manifold into a non-positively curved manifold with constant $f$-bienergy density is a harmonic map, and that any $f$-biharmonic\; map from a compact manifold into Euclidean space is a constant map. \\

In this paper, we continue our work  on $f$-biharmonic\; maps and $f$-biharmonic\; \\submanifolds  by further exploring the applications of $f$-biharmonic\; maps and the relationships among harmonicity, biharmonicity, $f$-biharmonicity and conformality of maps between Riemannian manifolds. We prove that a map between Riemannian manifold is a harmonic map if and only if it is an $f$-biharmonic\; map for any positive function $f$, and that a submanifold of a Riemannian manifold is minimal if and only if it is $f$-biharmonic\; for any positive  function $f$ defined on the submanifold. We show that the set of all $f$-biharmonic\; maps from $2$-dimensional domain is invariant under the conformal change of the metric on the domain. We prove some rigidity theorems about $f$-biharmonic\; hypersurfaces in  nonpositively curved manifolds and give some classifications of $f$-biharmonic\; hypersurfaces in Einstein spaces and in space forms by using an improved form of  $f$-biharmonic\;  \\hypersurface equation. We also use the improved  $f$-biharmonic\; hypersurface equation to obtain an improved  equation and some classifications of biharmonic conformal immersions of surfaces into a $3$-manifold.

\section{Harmonicity, biharmonicity, $f$-biharmonicity and conformality}

It is well known that harmonicity of maps from a $2$-dimensional domain is invariant under conformal transformations of the domain. It is also known that biharmonic maps do not have this kind of invariance regardless of domain dimensions. However,  as it is shown by our next theorem, the set
\begin{eqnarray}\notag
\mathcal{H}_{2, f}(M^2, N^n)=&&\{(\phi, f)|\; \phi: (M^2, g)\longrightarrow (N^n,h),\; f: (M^2, g)\longrightarrow (0,\infty),\\ \notag && \phi \;{\rm is\;an\;f}-{\rm biharmonic\;map}\}
\end{eqnarray}
 of all $f$-biharmonic\; maps is invariant under conformal transformation of the domain.
 \begin{theorem}
 The set $\mathcal{H}_{2, f}(M^2, N^n)$ of all $f$-biharmonic\; maps from a surface into a manifold is invariant under conformal transformation of the domain, more precisely, if $ \phi: (M^2, g)\longrightarrow (N^n,h)$ is $f$-biharmonic\; for a function $f=\alpha$, then $ \phi: (M^2, {\bar g}=\lambda^2 g)\longrightarrow (N^n,h)$ also an $f$-biharmonic\; map for $f=\alpha \lambda^{2}$.
\end{theorem}
\begin{proof}
Let $ \phi: (M^2, g)\longrightarrow (N^n,h)$ be an $f$-biharmonic\; map for a function $f=\alpha$, then by Theorem A, the map $ \phi: (M^2, \alpha^{-1}g)\longrightarrow (N^n,h)$ is a biharmonic map. Noting that $(M^2, \alpha^{-1}g)=(M^2, (\alpha^{-1}\lambda^{-2})\lambda^2g)=(M^2, (\alpha \lambda^{2})^{-1}{\bar g})$, we conclude that the map $ \phi: (M^2, (\alpha \lambda^{2})^{-1}{\bar g})\longrightarrow (N^n,h)$ is a biharmonic map. Using Theorem A again we conclude that the map $ \phi: (M^2, {\bar g}=\lambda^{2}g)\longrightarrow (N^n,h)$ is an $f$-biharmonic\; map with $f=\alpha \lambda^{2}$. Thus, we obtain the theorem.
\end{proof}

It is clear from the definition of  $f$-biharmonic\; maps that a harmonic map $\phi:(M^m, g)\longrightarrow (N^n,h)$ is an $f$-biharmonic map for any function $f:(M^m, g)\longrightarrow (0, \infty)$.  Our next theorem shows that this property can actually be used to characterize harmonic maps between Riemannian manifolds.
\begin{theorem}
A map $\phi:(M^m, g)\longrightarrow (N^n,h)$ between Riemannian manifold is a harmonic map if and only if it is an $f$-biharmonic map for any function $f:(M^m, g)\longrightarrow (0, \infty)$.
\end{theorem}
 \begin{proof}
 We only need to prove the ``if" part of the statement, i.e.,  if a map $\phi:(M^m, g)\longrightarrow (N^n,h)$ is an $f$-biharmonic map for any function $f:(M^m, g)\longrightarrow (0, \infty)$, then it is a harmonic map. To prove this is equivalent to show that if the equation
  \begin{equation}\label{fbihe}
 f\tau_2(\phi)+(\Delta f)\tau(\phi)+2\nabla^{\phi}_{{\rm grad}\, f}\tau(\phi)=0
 \end{equation}
 holds for all positive functions $f$ on $(M^m, g)$, then we have $\tau(\phi)\equiv 0$.

In fact, take a positive constant function $f=C$, then equation (\ref{fbihe}) implies that the map $\phi$ is  a biharmonic map, i.e., $\tau_2(\phi)\equiv 0$.
 It follows that if a map $\phi:(M^m, g)\longrightarrow (N^n,h)$ is an $f$-biharmonic maps for any function $f$, then we have
  \begin{equation}\label{A2}
(\Delta f)\tau(\phi)+2\nabla^{\phi}_{{\rm grad}\, f}\tau(\phi)=0 \;\;\forall\;f:(M^m, g)\longrightarrow (0, \infty).
 \end{equation}
 If there exists a point $p\in M$ such that $\tau(\phi)(p) \ne 0$, then there is an open neighborhood $U$ of $p$ so that $\tau(\phi)(q) \ne 0,\; \forall\; q\in U$. By taking dot product of both sides of (\ref{A2}) with $\tau(\phi)$,  we have, on $U$,
 \begin{equation}\label{A3}
 (\Delta f)|\tau(\phi)|^2+({\rm grad}\, f)|\tau(\phi)|^2=0 \;\;\forall\;f:(U, g|_U)\longrightarrow (0, \infty).
 \end{equation}
It is well known that  there exists a local harmonic coordinate system in a neighborhood of any point of a Riemannian manifold. By this well known fact and a translation in Euclidean spaces, we can choose a neighborhood $V\subset U$ of $p$ with harmonic coordinate $\{x^i\}$ satisfying $x^i>0$ for all $i=1, 2, \cdots, m$. By choosing positive harmonic coordinate functions $f=x^i$ and substituting it into (\ref{A3})  we have
\begin{equation}
g^{ik}\frac{\partial}{\partial x^k} |\tau(\phi)|^2=0, \;\; \forall\; i=1, 2, \cdots, m.
\end{equation}

It follows that $|\tau(\phi)|^2={\rm constant}$, and hence Equation (\ref{A3}) reduces to
  \begin{equation}\label{A4}
 (\Delta f)|\tau(\phi)|^2=0 \;\;\forall\;f:(V, V|g)\longrightarrow (0, \infty).
 \end{equation}
 Now let $f=(x^1)^2$, where $x^1$ is the first coordinate function, and substitute it into Equation (\ref{A4}) to have $ (\Delta f)|\tau(\phi)|^2=2g^{11}|\tau(\phi)|^2=0$ on $V$. Since the Riemannian metric is positive definite, we conclude that $|\tau(\phi)|^2=0$ on $V$, which contradicts the assumption that $\tau(\phi)\ne 0$ on $U$. The contradiction shows that we must have $\tau(\phi)\equiv 0$ on $M$,  i.e., the map $\phi$ is a harmonic map. This completes the proof of the theorem.
\end{proof}

As an immediate consequence, we have the following characterization of minimal submanifolds using the concept of $f$-biharmonic\; submanifolds.
\begin{corollary}\label{23}
A submanifold $M^m\hookrightarrow (N^n, h)$ is minimal if and only if it is $f$-biharmonic\; for any positive function defined on the submanifold.
\end{corollary}

To close this section, we take a look at maps which are $f$-biharmonic\; for two different functions, especially those maps which are both biharmonic (i.e., $f$-biharmonic\; for $f=$constant) and $f$-biharmonic\; for $f\ne constant $.\\

It is shown in \cite{Ou4} that conformal immersions $\phi: \mbox{${\mathbb R}$}^4\setminus\{0\}\longrightarrow \mbox{${\mathbb R}$}^4$ given by the inversion of the sphere $S^3$, $\phi(x)=\frac{x}{|x|^2}$ is a biharmonic map and also an $f$-biharmonic\; map with $f(x)=|x|^4$. Our next theorem shows that this situation cannot happen for pseudo-umbilical isometric immersions.
\begin{theorem}\label{MT1}
If a pseudo-umbilical proper biharmonic submanifold $\phi: M^m\longrightarrow (N^n,h)$ is $f$-biharmonic\;, then $f$ is constant.
\end{theorem}
\begin{proof}
$\phi: M^m\longrightarrow (N^n,h)$ be a pseudo-umbilical submanifold, then, by definition, its shape operator with respect to the mean curvature vector field $\eta$ is given by
\begin{equation}
A_{\eta}(X)=\langle \eta, \eta\rangle X
\end{equation}
for  any vector field  $X$ tangent to the submanifold. It is well known that the tension field of the submanifold is given by $\tau(\phi)=m\eta$. If the submanifold is both biharmonic and $f$-biharmonic\;, then we have both $\tau_2(\phi)=0$ and $\tau_{2,f} (\phi)=0$ identically. Substituting these into Equation (\ref{fbhe}) we obtain
\begin{eqnarray}\label{GD1}
0 &=& (\Delta f)\tau(\phi)+2\nabla^{N}_{{\rm grad}\, f}\tau(\phi)\\\notag
&=& m [(\Delta f)\eta+2\nabla^{N}_{{\rm grad}\, f}\eta]\\\notag
&=& m [(\Delta f)\eta+2\nabla^{\perp}_{{\rm grad}\, f}\eta-2A_{\eta}(\rm{grad}\,f)]\\\notag
&=& m [(\Delta f)\eta+2\nabla^{\perp}_{{\rm grad}\, f}\eta-2\langle \eta, \eta\rangle\,\rm{grad}\,f],
\end{eqnarray}
where the third equality was obtained by using the Weingarten formula for submanifolds and the last equality was obtained by the assumption that the submanifold is pseudo-umbilical. Comparing the normal and the tangent components of the both sides of (\ref{GD1}) we have $\rm{grad}\,f=0$ since the submanifold is non-minimal, and hence $f=\rm constant$. This completes the proof of the theorem.
\end{proof}
\begin{corollary}
 Biharmonic hypersurface $\phi: S^m(\frac{1}{\sqrt{2}})\hookrightarrow S^{m+1},\; \phi(x)=(\frac{1}{\sqrt{2}}, x)$ is an $f$-biharmonic\; hypersurface if and only if $f$ is constant. In particular, there is no nonconstant function $f$ on $S^2(\frac{1}{\sqrt{2}})$ so that the conformal immersion $\phi: (S^2(\frac{1}{\sqrt{2}}), f^{-1}g_0)\longrightarrow (S^3, h_0),\; \phi(x)=(\frac{1}{\sqrt{2}}, x)$ becomes a biharmonic map, where $g_0$ and $h_0$ are the metrics on $S^2(\frac{1}{\sqrt{2}})$ and $S^3$ induced from their ambient Euclidean spaces respectively.
\end{corollary}
\begin{proof}
The first statement of the corollary follows from Theorem \ref{MT1} and the fact that the hypersurface $\phi: S^m(\frac{1}{\sqrt{2}})\hookrightarrow S^{m+1},\; \phi(x)=(\frac{1}{\sqrt{2}}, x)$ is a totally umbilical biharmonic hypersurface of $S^{m+1}$ (See \cite{CMO1}). The second statement of the corollary follows from the first statement and Theorem A.
\end{proof}
\section{ The equations  of $f$-biharmonic\; hypersurfaces and some classification results }\label{Sect3}
In this section, we will give an improved equation for $f$-biharmonic\; hypersurfaces in a general Riemannian manifold and use it to prove some rigidity theorems and  some classifications of $f$-biharmonic\; hypersurfaces in Einstein spaces, including space forms.
\begin{theorem}\label{FBH}
A hypersurface $\phi:M^{m}\longrightarrow (N^{m+1}, h)$  with the mean curvature vector field $\eta=H\xi$ is  $f$-biharmonic if and only if:
\begin{equation}\label{fbh2}
\begin{cases}
\Delta (f\,H)-(f\,H )[|A|^{2}-{\rm
Ric}^N(\xi,\xi)]=0,\\
A\,({\rm grad}(f\,H)) +(f\,H)[\frac{m}{2} {\rm grad}\, H
-({\rm Ric}^N\,(\xi))^{\top}]=0,
\end{cases}
\end{equation}
where ${\rm Ric}^N : T_qN\longrightarrow T_qN$ denotes the Ricci
operator of the ambient space defined by $\langle {\rm Ric}^N\, (Z),
W\rangle={\rm Ric}^N (Z, W)$, $A$ is the shape operator of the
hypersurface with respect to the unit normal vector $\xi$, and $\Delta, \rm grad$ are the Laplace and the gradient operator of the hypersurface respectively.
\end{theorem}
\begin{proof}
It was proved in \cite{Ou4} that an isometric immersion $\phi:M^{m}\longrightarrow N^{m+1}$
 with mean curvature vector $\eta=H\xi$ is $f$-biharmonic if and only if:
\begin{equation}\label{fbh}
\begin{cases}
\Delta H-H |A|^{2}+H{\rm
Ric}^N(\xi,\xi)+H(\Delta f)/f+2({\rm grad} \ln f) H=0,\\
A\,({\rm grad}\,H) +\frac{m}{2}H {\rm grad}\, H
-\, H \,({\rm Ric}^N\,(\xi))^{\top}+H A\,({\rm grad}\,\ln f)=0,
\end{cases}
\end{equation}
where $\xi$, $A$, and $H$ are the unit normal vector field, the
shape operator, and the mean curvature function  of
the hypersurface $\varphi(M)\subset (N^{m+1}, h)$ respectively, and the
operators $\Delta,\; {\rm grad}$ and $|,|$ are taken with respect to
the induced metric $g=\varphi^{*}h=\lambda^{2}{\bar g}$ on the
hypersurface.\\

Multiplying $f$ to both sides of each equation in the system (\ref{fbh}) we can rewrite the resulting system as (\ref{fbh2}).
\end{proof}

As a straightforward consequence, we have to following equations for  $f$-biharmonic\; \\hypersurfaces in Einstein spaces and space forms, which improves the equations given in \cite {PA}, and \cite{Ou4} respectively.
\begin{corollary}
A hypersurface $\phi:M^{m}\longrightarrow (N^{m+1}, h)$ in an Einstein space with ${\rm Ric}^N=\lambda h$  is  $f$-biharmonic if and only if it mean curvature function $H$ solves the following PDEs
\begin{equation}\label{E1}
\begin{cases}
\Delta (f\,H)-(f\,H )[|A|^{2}-\lambda]=0,\\
A\,({\rm grad}(f\,H)) +\frac{m}{2}(f\,H) {\rm grad}\, H
=0.
\end{cases}
\end{equation}
In particular, a hypersurface $\phi:M^{m}\longrightarrow N^{m+1}(C)$ in
a space form of constant sectional curvature $C$ is $f$-biharmonic if and only if its mean curvature function $H$ is a solution of
\begin{equation}\label{fbh3}
\begin{cases}
\Delta (f\,H)-(f\,H) [|A|^{2}-mC]=0,\\
A\,({\rm grad}(f\,H)) +\frac{m}{2}(f\,H){\rm grad}\, H
=0.
\end{cases}
\end{equation}
\end{corollary}

As the first application of the improved $f$-biharmonic\; hypersurface equation, we will prove the following theorem, which gives a rigidity result of $f$-biharmonic\; \\isometric immersions under some constraints of Ricci curvatures of the ambient space. Let us recall the following lemma, which will be used in proving our next theorem.\\

{\bf Lemma A.} (\cite{NU}) Assume that $(M, g)$ is a complete non-compact Riemannian manifold, and $\alpha$ is a non-negative smooth function on $M$. Then, every smooth $L^2$ function $u$ on $M$ satisfying the Schr$\ddot{\rm o}$dinger type equation
\begin{equation}\label{nu}
\Delta_g u = \alpha u
\end{equation}
on $M$ is  a constant.\\
\begin{theorem}\label{Ric}
A complete hypersurface $\phi:M^{m}\longrightarrow (N^{m+1}, h)$ of a manifold $N$  with  ${\rm Ric}^N(\xi,\xi)\le |A|^2$ and the mean curvature vector field $\eta=H\xi$ satisfying $\int_M (fH)^2dv_g<\infty$ is $f$-biharmonic if and only if it is minimal, or
\begin{equation}\label{HP2}
\begin{cases}
{\rm Ric}^N(\xi,\xi)=|A|^{2},\\
({\rm Ric}^N\,(\xi))^{\top}=\frac{m}{2} {\rm grad}\, H,\;\;{\rm or}\;\; H=0.
\end{cases}
\end{equation}
In particular, a complete $f$-biharmonic\; hypersurface $\phi:M^{m}\longrightarrow (N^{m+1}, h)$ of a manifold $N$  of nonpositive Ricci curvature  and the mean curvature function satisfying $\int_M (fH)^2dv_g<\infty$ is minimal.
\end{theorem}
\begin{proof}
A straightforward computation yields
\begin{eqnarray}\label{GD3}
\Delta (fH)^2= 2fH\Delta (fH)+2 |\nabla (fH)|^2.
\end{eqnarray}

If the biharmonic hypersurface $\phi:M^{m}\longrightarrow (N^{m+1}, h)$ is compact, then, by Theorem \ref{FBH} we have Equation (\ref{fbh2}). Substituting the first equation of (\ref{fbh2}) into (\ref{GD3}) we have
\begin{eqnarray}\label{GD30}
\Delta (fH)^2= 2(fH)^2[|A|^2-{\rm Ric}^N(\xi,\xi)]+2 |\nabla (fH)|^2\ge 0
\end{eqnarray}
by the assumption that ${\rm Ric}^N(\xi,\xi)\le |A|^2$. It follows that the function $fH$ is a subharmonic function on a compact manifold, and hence it is constant.\\

If the biharmonic hypersurface $\phi:M^{m}\longrightarrow (N^{m+1}, h)$ is non-compact but complete, then the first equation of (\ref{fbh2}) implies that the function $fH$ is a solution of (\ref{nu}) with $\alpha=|A|^2-{\rm Ric}^N(\xi,\xi)$ being nonnegative. Using Lemma A we conclude that $fH$ is constant. So, in either case, the function $fH=C$, a constant. Substituting this into (\ref{fbh2}) we obtain the first statement of the theorem.\\

To prove the second statement, notice that $fH={\rm constant}$ and $f$ being positive imply that  we either have (i) $H\equiv 0$ in which case the hypersurface is minimal, or (ii) $H$ is never zero. If case (ii) happens, then we use $fH={\rm constant}\ne 0$ and the first equation $f$-biharmonic\; hypersurface equation to conclude that $0\le |A|^2={\rm Ric}^N(\xi,\xi)\le 0$. This implies that $A=0$ and hence the hypersurface is totally geodesic. It follows that $H\equiv 0$, which contradicts the assumption that $H$ is never zero. The contradiction shows that we only have case (i), which gives the second statement.
\end{proof}
From the proof of Theorem \ref{Ric} we have the following
\begin{corollary}
(i) A compact hypersurface $\phi:M^{m}\longrightarrow S^{m+1}$ with squared norm of the second fundamental form $|A|^2\ge m$ is $f$-biharmonic if and only if  it is biharmonic with  constant mean curvature and $|A|^2=m$.
(ii)  a complete hypersurface $\phi:M^{m}\longrightarrow S^{m+1}$ of the sphere $S^{m+1}$  with the mean curvature vector field $\eta=H\xi$ satisfying $ |A|^2\ge m$ and $\int_M (fH)^2dv_g<\infty$ is $f$-biharmonic if and only if the hypersurface is a biharmonic hypersurface of constant mean curvature and $|A|^2=m$.
\end{corollary}
As the second application of the improved $f$-biharmonic\; hypersurface equation, we give some classifications of $f$-biharmonic\; hypersurfaces in Einstein spaces and space forms.
\begin{theorem}
A compact  hypersurface of an Einstein space with  nonpositive scalar curvature is $f$-biharmonic if and only if it is a minimal hypersurface.
\end{theorem}
\begin{proof}
First of all, by Corollary \ref{23},  a minimal hypersurface $\phi:M^{m}\longrightarrow (N^{m+1}, h)$  is  $f$-biharmonic for any positive function $f$ on $M$. Conversely, let $\phi:M^{m}\longrightarrow (N^{m+1}, h)$ be a compact $f$-biharmonic hypersurface with the mean curvature vector field $\eta=H\xi$ on an Einstein manifold. Then, one can easily compute that
\begin{eqnarray}\label{NP}
\Delta((fH)^2)&=& 2(fH)\Delta(fH)+2|{\rm grad}(fH)|^2.
\end{eqnarray}
Using the first  equation of  (\ref{E1}) with $\lambda= \frac{{\rm R}_N}{m+1}$ and the assumption  on the scalar curvature ${\rm R}_N$ we have
\begin{eqnarray}\label{NP1}
\Delta((fH)^2)= 2(fH)^2[|A|^{2}-\frac{{\rm R}_N}{m+1}]+2|{\rm grad}(fH)|^2\ge 0.
\end{eqnarray}
It follows that $(fH)^2$ is a subharmonic function on a compact manifold and hence it is a constant. This implies that $fH=C$, a constant, from which and the fact the $f$ is a positive function we conclude that either (i) $H\equiv 0$, in which case the hypersurface in minimal, or (ii) $H(p)\ne 0$ for any $p\in M^m$. Now if case (ii) happens, then we $fH=C\ne 0$ and the second equation of the $f$-biharmonic equation of Einstein space to conclude that  $H$ is a nonzero constant. Using this and the first equation of (\ref{E1}) again we have $ 0\le |A|^2=\frac{{\rm R}_N}{m+1}\le 0$ and hence $A=0$. This means the hypersurface is actually totally geodesic and hence $H\equiv 0$, a contradiction to the assumption that $H(p)\ne 0$ for any $p\in M^m$. The contradiction shows that case (ii) cannot happen. Thus, we obtain the theorem.
\end{proof}

\begin{theorem}\label{EA}
A totally umbilical hypersurface $\phi:M^{m}\longrightarrow (N^{m+1}, h)$ in an Einstein space with ${\rm Ric}^N=\lambda h$  is  $f$-biharmonic if and only if it is totally geodesic or a biharmonic hypersurface with constant mean curvature $H=\pm \sqrt{\lambda/m}$.
\end{theorem}
\begin{proof}
 If the hypersurface is totally umbilical, then we have $A(X)=HX$ for any vector tangent to the hypersurface and $|A|^2=mH^2$. Substituting these into (\ref{E1}) we have
\begin{equation}\label{E2}
\begin{cases}
\Delta (f\,H)-(f\,H )[mH^{2}-\lambda]=0,\\
H{\rm grad}(f\,H) +\frac{m}{2} (f\,H){\rm grad}\, H
=0.
\end{cases}
\end{equation}
On the other hand, it was proved in \cite{Ko} (see also Lemma 2.1 in \cite{JMS}) that a totally umbilical hypersurface in an Einstein space has constant mean curvature, i.e., $H={\rm constant}$.
If $H= 0$, then it is easily seen that the totally umbilical minimal hypersurface is actually totally geodesic. If $H={\rm constant}\ne 0$, then Equation of (\ref{E2}) is solved by $f={\rm constant}$ and

$mH^{2}-\lambda=0$. This implies that the totally umbilical hypersurface is a biharmonic hypersurafce with constant mean curvature $H=\pm \sqrt{\lambda/m}$. Summarizing the results in the two cases discussed above we obtain the theorem.
\end{proof}

As a consequence of Theorem \ref{EA}, we have
\begin{corollary}
A totally umbilical $f$-biharmonic hypersurface in Euclidean space $\mathbb{R}^{m+1}$ or hyperbolic space $H^{m+1}$ is a totally geodesic hypersurface, any totally umbilical $f$-biharmonic hypersurface in sphere $S^{m+1}$ is, up to isometries,  a part of the great sphere $S^m \subset S^{m+1}$ or a part of the small sphere  $S^m(\frac{1}{\sqrt{2}})$.
\end{corollary}
\begin{proof}
This follows from the well-known facts that space forms $\mbox{${\mathbb R}$}^{m+1},\; H^{m+1}, S^{m+1}$ of constant sectional curvature $0, -1, 1$ are Einstein spaces with $\lambda= 0, -m, m$, and that totally umbilical biharmonic hypersurfaces in Euclidean spaces and hyperbolic spaces are totally geodesic, and the only totally umbilical biharmonic hypersurfaces in $S^{m+1}$ is a part of the  great sphere $S^m \subset S^{m+1}$ or a part of the small  $S^m(\frac{1}{\sqrt{2}})$ (\cite{CMO1}).
\end{proof}

It is well known that biharmonic hypersurfaces of Euclidean space $\mathbb{R}^{m+1}$ with at most two distinct principal curvature are minimal ones. The following example shows that this  is no longer true for $f$-biharmonic hypersurfaces.
\begin{example}
Let $\phi:D=\{ (\theta,x)\in (0,2\pi)\times \mbox{${\mathbb R}$} ^{m-1}\}$ and  $\phi:D\longrightarrow (\mbox{${\mathbb R}$}^{m+1}, h_0)$ with $\phi(\theta,x_1, \cdots, x_{m-1})=(R\cos\,\frac{\theta}{R}, R\sin\,\frac{\theta}{R}, x_1, \cdots, x_{m-1})$ be the isometric immersion into Euclidean space. One can check that it is a hypersurface  with two distinct principal curvatures, which is also an $f$-biharmonic hypersurface for any positive function $f$ from the family $f=C_1e^{x_1/R}+C_2e^{-x_1/R}$, where $C_1, C_2$ are constants.
\end{example}

In fact, we can take
\begin{eqnarray}\notag
 && e_1={\rm d} \phi(\frac{\partial}{\partial \theta})=-\sin \frac{\theta}{R}\frac{\partial}{\partial y^1}+ \cos
\frac{\theta}{R}\frac{\partial}{\partial y^2},\\\notag
&&  e_i= {\rm d} \phi(\frac{\partial}{\partial x^{i-1}})=\frac{\partial}{\partial y^{1+i}},\; i= 2, \cdots, m,\\\notag
&&  \xi= \cos \frac{\theta}{R}\frac{\partial}{\partial y^1}+ \sin
\frac{\theta}{R}\frac{\partial}{\partial y^2}
\end{eqnarray}
as an orthonormal frame adapted to the hypersurface. Then a straightforward computation gives

\begin{eqnarray}\label{EJS}
\begin{cases}
Ae_1=-\frac{1}{R}e_1,\;\;Ae_i=0,\; \forall \; i=2, 3, \cdots, m.\\
H=\frac{1}{m}\sum_{i=1}^m\langle  Ae_i,e_i\rangle=-\frac{1}{mR}\ne 0\\
|A|^2=\sum_{i=1}^m|Ae_i|^2=\frac{1}{R^2},
\end{cases}
\end{eqnarray}
which show indeed that the hypersurface has two distinct principal curvatures and constant mean curvature.
Let $f: (0,2\pi)\times \mbox{${\mathbb R}$} ^{m-1}\longrightarrow (0, \infty)$ be a function that does not depend on $\theta$, i.e., $f=f(x_1, \cdots, x_{m-1})$. Substituting this and (\ref{EJS}) into the $f$-biharmonic\; hypersurface equation (\ref{fbh3}) with $C=0$ we have
\begin{equation}\label{TZ}
\Delta^{\mbox{${\mathbb R}$}^{m-1}} f=\frac{1}{R^2}\,f.
\end{equation}
It follows that the cylinder $S^1\times \mbox{${\mathbb R}$}^{m-1}\hookrightarrow \mbox{${\mathbb R}$}^2\times \mbox{${\mathbb R}$}^{m-1}$ is a an $f$-biharmonic\; hypersurface for any positive eigenfunction which solves equation (\ref{TZ}), in particular, any positive function from the family
$f=C_1e^{x_1/R}+C_2e^{-x_1/R}$, where $C_1, C_2$ are constants, is a solution of (\ref{TZ}).

\section{Biharmonic conformal immersions of surfaces into $3$-manifolds}

It is well known (See, e.g., \cite{Ta}) that a conformal immersion $\varphi : (M^2 ,{\bar g}) \longrightarrow (N^n ,h)$  with  $\varphi^{*}h=\lambda^2{\bar g}$ is harmonic if and only if the surface $\varphi(M^2) \hookrightarrow (N^n,h)$ is a minimal surface, i.e., the isometric immersion $\varphi(M^2) \hookrightarrow (N^n,h)$ is harmonic.  The following corollary, which follows from Theorem A, can be viewed as a generalization of this well-known result.
\begin{corollary}\label{Co41}
A conformal immersion $\varphi : (M^2 ,{\bar g}) \longrightarrow (N^n ,h)$  with  $\varphi^{*}h=\lambda^2{\bar g}$ is biharmonic if and only if the surface (i.e., the isometric immersion) $\varphi(M^2) \hookrightarrow (N^n,h)$ is an $f$-biharmonic\; surface with $f=\lambda^2$.
\end{corollary}

It was proved in \cite{Ou1} that a  hypersurface, i.e., an isometric immersion $M^m\hookrightarrow (N^{m+1}, h)$ with mean curvature $H$ and the shape operator $A$ is biharmonic if and only if
\begin{equation}\label{BHS}
\begin{cases}
\Delta H-H |A|^{2}+H{\rm
Ric}^N(\xi,\xi)=0,\\
 2A\,({\rm grad}\,H) +\frac{m}{2} {\rm grad}\, H^2
-2\, H \,({\rm Ric}^N\,(\xi))^{\top}=0,
\end{cases}
\end{equation}
where ${\rm Ric}^N : T_qN\longrightarrow T_qN$ denotes the Ricci
operator of the ambient space defined by $\langle {\rm Ric}^N\, (Z),
W\rangle={\rm Ric}^N (Z, W)$,  $\xi$ is the unit normal vector field, and $\Delta$ and ${\rm grad}$ denote the Laplace and the gradient operators defined by the induced metric on the hypersurface.\\

Later in \cite{Ou3}, it was proved that a conformal immersion $\varphi:(M^2, {\bar g})\longrightarrow (N^3, h)$ with $\varphi^{*}h=\lambda^2{\bar g}$ is biharmonic if and only if
\begin{equation}\label{BCI3}
\begin{cases}
\Delta H-H[ |A|^{2}-{\rm
Ric}^N(\xi,\xi)-(\Delta \lambda^2)/\lambda^2]+4g({\rm grad} \ln \lambda), {\rm grad} H)=0,\\
A\,({\rm grad}\,H) +H [{\rm grad}\, H
-({\rm Ric}^N\,(\xi))^{\top}]+2 A\,({\rm grad}\,\ln \lambda)]=0,
\end{cases}
\end{equation}
where $\xi$, $A$, and $H$ are the unit normal vector field, the
shape operator, and the mean curvature function  of
the surface $\varphi(M)\subset (N^{3}, h)$ respectively, and the
operators $\Delta,\; {\rm grad}$ and $|,|$ are taken with respect to
the induced metric $g=\varphi^{*}h=\lambda^2{\bar g}$ on the surface.\\

Our next theorem gives an improved equation for biharmonic conformal immersions of surfaces into a $3$-manifold by writing it in a  form similar to biharmonic surface equation (\ref{BHS}),  which  turns out to be very useful in proving some classification and/or  existence results on biharmonic conformal surfaces.\\

\begin{theorem}\label{NEW}
A conformal immersion
\begin{equation}\label{MAP1}
\varphi : (M^{2},{\bar g}) \longrightarrow
(N^3,h)
\end{equation}
 into a $3$-dimensional manifold with
$\varphi^{*}h=\lambda^2{\bar g}$ is biharmonic if and only if the surface $\varphi(M^2)\hookrightarrow (N^3, h)$ is an $f$-biharmonic surface with $f=\lambda^2$, which is characterized by the equation
\begin{equation}\label{M03}
\begin{cases}
\Delta (\lambda^2 H )-(\lambda^2H)[|A|^2-{\rm Ric}^N(\xi,\xi)]=0,\\A({\rm grad} (\lambda^2 H))+ (\lambda^2 H) [{\rm grad}
H- \,({\rm Ric}^N\,(\xi))^{\top}]=0.
\end{cases}
\end{equation}
where $\xi$, $A$, and $H$ are the unit normal vector field, the
shape operator, and the mean curvature function  of
the surface $\varphi(M)\subset (N^3, h)$ respectively, and the
operators $\Delta,\; {\rm grad}$ and $|,|$ are taken with respect to
the induced metric $g=\varphi^{*}h=f{\bar g}$ on the
surface.
\end{theorem}
\begin{proof}
One can easily see that a map $\varphi : (M^{2},{\bar g}) \longrightarrow
(N^3,h)$ is a conformal immersion with conformal factor $\lambda$, i.e., $\varphi^{*}h=\lambda^2{\bar g}$, if and only if the map
\begin{equation} \label{IDI}
\varphi : (M^{2}, g=\lambda^2{\bar g}) \longrightarrow
(N^3,h)
\end{equation}
is an isometric immersion. By Corollary \ref{Co41}, the conformal immersion $\varphi : (M^{2}, {\bar g}) \longrightarrow
(N^3,h)$ is  biharmonic if and only if the surface $\varphi : (M^{2}, \lambda^2{\bar g}) \longrightarrow
(N^3,h)$ is an $f$-biharmonic\; surface. Applying Equation (\ref{fbh2}) with $m=2$ and $f=\lambda^2$ we obtain the biharmonic conformal equation (\ref{M03}) and hence the theorem.
\end{proof}
As an immediate consequence of Theorem \ref{NEW}, we have the following
\begin{corollary}\label{C2}
A conformal immersion $\varphi : (M^{2}, {\bar
g}) \longrightarrow (N^3(C),
h_0)$ into $3$-dimensional space of constant sectional curvature $C$
with $\varphi^{*}h_{0}=\lambda^2{\bar
g}$ is biharmonic if and only if
\begin{equation}\label{C}
\begin{cases}
\Delta (\lambda^2 H )-(\lambda^2H)(|A|^2-2C)=0,\\A({\rm grad} (\lambda^2 H))+ (\lambda^2 H) {\rm grad}
H=0.\\
\end{cases}
\end{equation}
where  $A$ and $H$ are the shape operator and the mean curvature function of the surface
respectively, and the operators $\Delta,\; {\rm grad}$ and $|,|$ are
taken with respect to the induced metric
$g=\varphi^{*}h=\lambda^2{\bar g}$ on the surface.
\end{corollary}

\begin{theorem}\label{MT2}
Let $\varphi : (M^{2},{\bar g}) \longrightarrow (\mbox{${\mathbb R}$}^3, h_0)$ be a
 biharmonic conformal immersion from a complete surface into
$3$-dimensional Euclidean space with
$\varphi^{*}h_0=\lambda^2{\bar g}$. If the mean curvature $H$ of the surface $\varphi : (M^{2}, g=\lambda^2{\bar g}) \longrightarrow (\mbox{${\mathbb R}$}^3, h_0)$ satisfies $\int_M\,\lambda^4H^2 \,dv_{g}<\infty$.  Then the biharmonic conformal immersion $\varphi$ is a minimal immersion.
\end{theorem}

\begin{proof}
Note that $\varphi : (M^{2},{\bar g}) \longrightarrow (\mbox{${\mathbb R}$}^3, h_0)$ is a biharmonic conformal immersion implies that the surface $ (M^{2}, g=\lambda^2{\bar g}) $ can be biharmonically conformally immersed into Euclidean space $ (\mbox{${\mathbb R}$}^3, h_0)$. By Corollary \ref{C2}, we have
\begin{equation}\label{M310}
\begin{cases}
\Delta (\lambda^2 H )=(\lambda^2 H) |A|^2,\\A({\rm grad} (\lambda^2  H))+ (\lambda^2 H) {\rm grad}
H=0.\\
\end{cases}
\end{equation}
where  $A$, and $H$ are  the shape operator, and the mean curvature
function of the surface  $\varphi(M)\subset (\mbox{${\mathbb R}$}^3, h_0)$
respectively, and the operators $\Delta,\; {\rm grad}$ and $|,|$ are
taken with respect to the induced metric $g=\varphi^{*}h=\lambda^2 {\bar g}$ on $M$.\\

From  the assumption $\int_M\,\lambda^4 H^2 \,dv_{g}<\infty$ and the first equation of (\ref{M310}) we see that the function $\lambda^2 H\in L^2(M)$ is a solution of the Schr$\ddot{\rm o}$dinger type equation
\begin{equation}
\Delta u = \alpha u
\end{equation}
for a fixed nonnegative function $\alpha=|A|^2$ on a complete manifold $(M^2, g)$.
By Lemma A (Section 3), we conclude that $\lambda^2H=C$,  a constant. Substituting this into the second equation of (\ref{M310}) we have either $H=0$ which means the conformal immersion is minimal, or $H=C_1$ a constant. If $H=C_1\ne 0$, then $\lambda^2H=C$ implies that $\lambda^2=C/H=C/C_1$ is a constant.  In this case, the conformal immersion is a homothetic immersion. By a well-known result of Chen and Jiang, any biharmonic homothetic immersion of a surface into $\mbox{${\mathbb R}$}^3$ is minimal. Therefore, in either case, we conclude that the conformal biharmonic immersion is a minimal homothetic immersion.
\end{proof}

Motivated by the beautiful theory of minimal surfaces as conformal harmonic immersions and the attempt to understand the relationship among biharmonicity, conformality and $f$-biharmonicity of maps from surfaces, we studied biharmonic conformal immersions of surfaces in \cite{Ou2}, and \cite{Ou3}. A Fundamental question is: what are the geometric and/or topological obstacles for a surface to admit a biharmonic conformal immersion into a ``nice space" like a space form?  Recall (See \cite{Ou3}) that a surface in a Riemannian manifold $(N^{3}, h)$ defined by an isometric immersion $\varphi: (M^{2}, g) \longrightarrow (N^{3},h)$ is said to admit a {\em biharmonic conformal immersion} into $(N^{3}, h)$, if there exists  a function $f :M^2\longrightarrow \mbox{${\mathbb R}$}^{+}$ such that the conformal immersion $\varphi: (M^{2}, {\bar g}=f^{-1}g) \longrightarrow (N^{3},h)$ with conformal factor $f$ is a biharmonic map. In such a case, we also say that the surface $\varphi: (M^{2}, g) \longrightarrow (N^{3},h)$  can be {\em biharmonically conformally immersed} into $(N^{3}, h)$. Later in studying $f$-biharmonic maps and $f$-biharmonic\; submanifolds in \cite{Ou4}, Theorem A was proved. It follows from Theorem A that a surface  that $\varphi: (M^{2}, g) \longrightarrow (N^{3},h)$ that admits a  biharmonic conformal immersion into $(N^{3}, h)$ if and only if the surface is an $f$-biharmonic\; surface.\\

Examples of surfaces that can be biharmonically conformally immersed into a $3$-manifold include the following

\begin{itemize}
\item For any biharmonic conformal immersion  $\varphi: (M^{2}, {\bar g}) \longrightarrow (N^{3}, h)$ with $\varphi^{*}h=\lambda^2{\bar g}$, its associated surface  $\varphi: (M^{2}, g=\varphi^{*}h) \longrightarrow (N^{3}, h)$ admits a biharmonic conformal immersion into $(N^{3}, h)$ as we see that there exists $f=\lambda^2$ such that $f^{-1}(\varphi^{*}h)= {\bar g}$. For examples of biharmonic conformal immersions of surfaces, see \cite{Ou2} and \cite{Ou3}.
\item Any minimal surface (i.e., harmonic isometric immersion) $\varphi:M^2\longrightarrow (N^3,h)$ admits a biharmonic conformal immersion into $(N^3,h)$. This is because, by Corollary \ref{23}, a minimal surface $\varphi:M^2\longrightarrow (N^3,h)$ is $f$-biharmonic\; for any positive function $f$ defined on the surface.
\end{itemize}

Using the improved equation for  biharmonic conformal immersions of surfaces we can prove the following
\begin{theorem}
If a compact surface $M^2\hookrightarrow S^3$ with the squared norm of the second fundamental form $|A|^2\ge 2$ can be biharmonically conformally immersed into $S^3$, then  $M^2$ is minimal, or up to a homothetic change of the metric, $M^2= S^2(\frac{1}{2})$.
\end{theorem}

\begin{proof}
Suppose the surface is given by an isometric immersion $\phi:M^2\longrightarrow (S^3,h)$, where $h$ denotes the standard metric on $S^3$ with constant sectional curvature $1$. By definition, if the surface admits a biharmonic conformal immersion into $S^3$, then there exists a function $\lambda: M\longrightarrow (0, \infty)$ such that the conformal immersion $\phi:(M^2, {\bar g}=\lambda^2 g)\longrightarrow (S^3,h)$ is biharmonic. So,  by (\ref{C}), we have
\begin{equation}\label{S3}
\begin{cases}
\Delta (\lambda^2 H )-(\lambda^2H)(|A|^2-2)=0,\\A({\rm grad} (\lambda^2 H))+ (\lambda^2 H) {\rm grad}
H=0.\\
\end{cases}
\end{equation}
We compute
\begin{eqnarray}\notag
\Delta ((\lambda^2 H )^2)&=&2(\lambda^2 H )\Delta (\lambda^2 H )+2|{\rm grad}  (\lambda^2 H )^2)|^2\\\label{La}
&=&2(\lambda^2 H )^2(|A|^2-2)+2|{\rm grad}  (\lambda^2 H )^2)|^2.
\end{eqnarray}
where in obtaining the second equality we have used the first equation of (\ref{S3}).  From Equation (\ref{La})  and the assumption that $|A|^2\ge 2$ we conclude that $\Delta ((\lambda^2 H )^2)\ge 0$, i.e.,  the function $\lambda^2 H$ is a subharmonic function on the compact manifold $M$. It follows that $\lambda^2 H$ is a constant function. Using this and the second equation of (\ref{S3}) we see that $H$ must be a constant. If $H=0$, then the surface $\phi:M^2\longrightarrow (S^3,h)$ is minimal. Otherwise, if $H=C\ne 0$,  then $\lambda$ is also a constant and hence, up to a homothety, the compact surface $\phi:M^2\longrightarrow (S^3,h)$ is biharmonic. It follows from a well-known result (\cite{CMO1}) of biharmonic surfaces in $S^3$ that $M^2= S^2(\frac{1}{2})$. Thus, we obtain the theorem.
\end{proof}

redpajama_set_name: RedPajamaArXiv