text stringlengths 14 5.77M | meta dict | __index_level_0__ int64 0 9.97k ⌀ |
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Q: Getting meta to work with the carpalx-full console keymap In the Arch Linux console, M-b (alt-b) moves back one word with
the default us layout, but after loadkeys carpalx-full, it
doesn't (b is now at y). Why not? I have looked at the keymap
files, but couldn't find any explanation.
A: Most likely because the code for the console is written to read directly from the keyboard-device (a file in /dev/input/by-path) instead of reading from /dev/stdin. When you read directly from the device keymaps are ignored.
If this is indeed the reason then you should notice that all other keys also act as if the keymap wasn't changed. (I don't have Arch so I can't test it.)
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,914 |
\section{Introduction}
Detecting out-of-vocabulary (OOV) words, especially proper nouns, has always been challenging for any Automatic Speech Recognition (ASR) system. Capturing these nuances in the language model is challenging as these important context words are mostly unavailable in the training data and result in a high word error rate (WER). Moreover, the downstream tasks that depend on these keywords are also affected significantly because of the wrong recognition.
Consider the following conversation between two people.
\quad Person 1: ``Where do you live?" \
\quad Person 2: ``I live in Brno"
Now assume that we are using two different ASR models to decode the utterance of Person 2.
\quad Model 1: ``I live in beer" \
\quad Model 2: ``* * in Brno"
Using WER as the primary metric for evaluating these models, Model 1 gives 25\% WER, whereas Model 2 gives 50\%. The low WER will imply that Model 1 is better than Model 2. However, in real-life use cases, the recognition result of Model 2 is far more helpful. Lower WER for specifically important keywords is better than overall WER. In conversational AI, downstream tasks like intent/entity recognition solely depend on detecting important keywords or OOV words. When we say OOV, we refer to a list of words and phrases for which we don't have sentences containing these words available in the language model (LM) training data. In traditional OOV recovery tasks, the OOV words are unknown at training and decoding. However, in the case of conversational AI, we can have a list of expected words while decoding because of the known context of the conversation. However, training a general LM with all the OOV words is not feasible due to difficulty in obtaining the sentences with OOV words. So we are trying to recover known OOV words while decoding but not part of the normal LM.
Getting text and audio pairs is difficult in the case of the OOV words. The task of creating meaningful sentences with OOV words is laborious. There are multiple methods to model the OOV context words in the literature. One solution is implicit acoustic model training using synthetic data consisting of OOV words~\cite{9414778}. Another solution is dynamic graph manipulation, which includes recompiling graphs on the fly. However, graph creation is a resource-hungry task. One of the existing approaches uses a hybrid language model with word and subword-level language models. Then a phoneme-to-grapheme model is used to recover OOV words~\cite{kombrink2010recovery}. The OOV recovery heavily depends on the subword-level graph in this approach. This method performs a basic OOV recovery and does not use context information. A phone confusion-based matching approach for retrieving song names with context is explored in ~\cite{chaudhari2007improvements}. This model use an index of n-grams from the documents to find the phonetic matching with the target search. Here, the search is not context-based and uses a vanilla edit distance for matching. In~\cite{shah2020cross} a common phoneme label set-based approach is proposed but does not focus on improving the cost function to compute the better phone similarity.
\cite{chineese9362062} uses character-based modeling and graph manipulations to recover OOV words. However, while recovering the OOV words, context words are not used. A clever training technique is used to improve the OOV modeling in ~\cite{kim2021improving}, which makes use of the existing corpus itself, which enhances the low-frequency words recovery in the training data. A multiple-level, namely, word, phonetic, and grapheme-level matching-based recovery of name entity recognition is proposed in ~\cite{garg2020hierarchical} but uses vanilla distance measure for computing phone similarity.
Most of the approaches above are developed for English languages or monolingual systems. In all the above methods, the focus is on improving the modeling technique or post-processing with multiple blocks. The phone confusion matrices used in phonetic matching use vanilla distance measures. Our approach explores various phoneme level cost measures to post-process the decoded text and match it with a context word list. We have found that using fixed cost for OOV recovery deteriorates the performance as acoustically similar phoneme confusions are penalized heavily. We have developed strategies to formulate a suitable phone confusion matrix for OOV recovery. We use two different ASR LM models. We get the standard hypothesis with the first LM and the second LM acoustically boosted to obtain the OOV words. The output of the second model is used in recovering the OOV words with the help of phone sequences. We merge the hypothesis to have a better transcription based on the word level timing information of both the decoded output. We evaluated the proposed method with the word- and sentence-level data sets. The word-level data set consists of synthesized words from a bilingual Text-to-Speech (TTS) system. The sentence-level data set consists of telephonic conversational data collected internally in our organization. We showed that the proposed method performs better context-based OOV recovery at both word and sentence-level over baseline method.
\section{Proposed Method}
\begin{figure}[h]
\centering
\centerline{\includegraphics[width=\columnwidth, keepaspectratio]{img/basic_block-basic-block-diag.drawio.png}}
\caption{Block diagram of the proposed context-based OOV word recovery approach.}
\label{img:basic}
\end{figure}
\begin{figure}[!h]
\centering
\centerline{\includegraphics[width=\columnwidth, keepaspectratio]{img/basic_block-combined.drawio.png}}
\caption{The detailed diagram of the proposed approach consists of three main blocks: LM creation, Phone confusion matrix creation, and Context-based OOV recovery blocks respectively. We used four different confusion matrix creation methods: 1. Phonetic, 2. Acoustic, 3. Append, and 4. Weighted-based methods are highlighted with an yellow color box.}
\label{img:detailed}
\end{figure}
In the proposed approach, the context-based OOV recovery is made with the help of multiple modules. Figure~\ref{img:basic} shows the basic flow of OOV recovery using the proposed approach. Here we use two different ASR models for decoding the same speech signal. The one ASR model does conventional decoding using normal LM. The second ASR model accomplishes context words detection using OOV LM. Then decoded text from both the normal and OOV LMs are time-aligned. The OOV words are recovered based on the phone similarity between the context word list and decoded hypothesis from ASR. The main focus of our approach is obtaining a suitable phone cost function. Figure~\ref{img:detailed} shows the detailed flow of the proposed method. There are three different blocks. The first is the LM creation block, where specific graph level manipulations are performed for possible OOV recovery. The second block is for defining the cost function based on the phone confusion matrix, and the final block is the OOV recovery block. Each block is explained in detail further below.
\subsection{LM creation}
Our proposed method uses two ASR systems to decode the same data and combine the hypothesis based on the recovered OOV words. The acoustic model (AM) remains the same in both models, whereas the LMs are different. The first is the normal LM, and the second is an OOV-LM. Here n-gram~\cite{enwiki:1073019765} based LM, a probabilistic model created with an entire training corpus. The normal LM consists of a graph for phonemes, silence (SIL), and unknown words (!SIL). The normal LM does not contain any sentences with OOV words.
The !SIL symbol is used to model unknown words in case of normal LM. In the OOV LM, !SIL symbol is replaced with the phone level sub-graph. A 4-gram phone model is created using all the context words to form a phone-level sub-graph. This graph is connected to all other nodes. It ensures that any OOV can be recognized before/after any other word. The graph scale controls the path probability of the sub-graph. To recognize OOV words, graph scale of !SIL node is increased manyfold. Additionally, while decoding with the OOV-LM, the acoustic scale of the graph is increased to emphasize the phoneme level output.
A sample OOV-LM is shown in Figure~\ref{img:oovlm}. An n-gram word LM is depicted with !SIL node attached to each word boundary. The sub-graph structure of the !SIL is shown on the right side of the figure. Here we can see that !SIL node is a complex phoneme level sub-graph with interconnections based on the OOV context word list.
\begin{figure}[!h]s
\centering
\centerline{\includegraphics[width=\columnwidth, keepaspectratio]{img/basic_block-oov_lm_graph.drawio.png}}
\caption{The sample view LM graph consists of multiple words and !SIL nodes on the left side image, and its detailed view of phone sub-graph shown for one !SIL node on right side image, respectively.}
\label{img:oovlm}
\end{figure}
\subsection{Phone confusion matrix creation}
\label{ssec:confusion}
The main contribution of this work is the cost function formulation-based on phone confusion matrix. The cost function is critical for good performance for any match and replacing algorithm. The distance metric used here decides the overall quality of the approach. As the recognized phoneme sequence for OOV words is likely to have some errors, we need to find the best matching OOV word using a distance metric. We compare the ideal phone sequence of context words/phrases with the decoded phone sequence. The Levenshtein distance~\cite{enwiki:1067984406} between the two phone sequences is used to match the word similarity. The vanilla Levenshtein distance metric gives equal weightage between phone insertions, deletions, and substitutions. It reduces the system performance because the distance between two acoustically similar phonemes is the same as between different phonemes. We explored various techniques to arrive at a better phoneme distance metric for tackling this.
The Levenshtein distance algorithm is defined as:
\begin{equation}
\label{eq:ldist}
c_{i,j} = min\left\{\begin{matrix}
c_{i-1,j}+D_{cost} \\
c_{i,j-1}+I_{cost} \\
c_{i-1,j-1}+S_{cost} \\
\end{matrix}\right.
\end{equation}
where, $c_{i,j}$ is the cost at $i^{th},j^{th}$ location of search matrix, is calculated recursively according to the Equation~\ref{eq:ldist}, $D_{cost}$ is the deletion cost, $I_{cost}$ is the insertion cost and $S_{cost}$ is the substitution cost. The Equation~\ref{eq:ldist} computes the cost distance between two phoneme sequences. For the vanilla Levenshtein algorithm, $D_{cost}$ = $S_{cost}$ = $I_{cost}$ = 1. We are experimenting with the substitution cost in this work, which can be considered a phone confusion cost. The insertion and deletion costs are kept constant, whereas the substitution cost is adjusted based on different methods for optimal performance. The vanilla algorithm also keeps the substitution cost as a constant value (denoted as Hard cost) for the baseline system. In this work, we explored phonetic and acoustic similarity-based phone cost function.
\subsubsection{1. Phonetic similarity:}
The phonetic similarity cost function is based on the property of the phones. The nearest set of phones will have a lesser cost, and this can be considered a substitution cost, as it is a distance measured between the set of phonemes. Here we figure out the distance between two phones based on the phonetic similarity. A total of 39 different properties for each phoneme are identified. Some of these include properties of vowel (short, medium, and long), consonant (plosive, fricative, Central-approximant, Lateral-approximant, flap, velar, palatal, retroflex, dental, labial, aspirated, etc.), nukta, halanta, and anusvara. For each phoneme, we create a vector based on the above phonetic attributes, and more details about phonetic similarity are mentioned in ~\cite{kunchukuttan2020indicnlp}. The confusion cost between two phones is the cosine distance between the vector representation. It is shown as $1$ in the confusion matrix creation block of Figure~\ref{img:detailed}. A total of 4225 confusion combinations are obtained here for 65 unique phones in the phone set we used~\cite{baby21_ssw}. Basically, with phonetic similarity cost function we can get cost function for all phone pairs.
\subsubsection{2. Acoustic similarity:}
The acoustic similarity between two phonemes is the similarity in the auditory level. Since the phonetic similarity might not capture the actual acoustic likeliness in the training data, we tried to develop a metric based on the ASR decoded data. A corpus of 5 hours of bilingual (Hindi-English) speech is decoded using the phone level language model to determine the acoustic similarities. Code-mixing across languages is predominant in the case of Indian languages~\cite{Thomas2018}. So to obtain good phone confusion across both languages, we used bilingual ASR. Next, we get the substitution, deletion, insertion, and correct hypothesis statistics for each reference phoneme in the test data. For calculating the acoustic confusion, all the correct hypotheses and the substitutions are considered. Acoustic similarity cost is defined as:
\begin{equation}
(1 - (N_C/(N_C+N_S)))^4
\end{equation}
where, $N_C$ is the number of phonemes correctly recognized and $N_S$ the number of phonemes substituted in the decoded data. Here, this ratio is raised to the power of $4$ to have a broader range of values between $0$ and $1$. Only phones with more than $100$ substitutions are considered in our experiment. After applying these conditions, around $136$ unique frequently occurring phone confusion mappings are found. This process is shown as $2$ in the confusion matrix creation block of Figure~\ref{img:detailed}.
\subsubsection{3 \& 4. Ensemble cost function:}
To obtain better systems, we have tried to combine the above two cost computation methods in two different ways. In the first ensemble method ("Append"), phone confusion pairs that are not part of the acoustic similarity matrix are added from the phonetic similarity. This method is shown as $3$ in the confusion matrix creation block of Figure~\ref{img:detailed}. In the second method ("Weighted"), both acoustic and phonetic similarity methods are combined in a weighted manner by giving equal weighting to both the functions, and this process is shown as $4$ in the confusion matrix creation block of Figure~\ref{img:detailed}.
\subsection{Context-based OOV recovery}
\label{ssec:OOV_recovery}
In this block, a two-pass OOV recovery is employed. The hypothesis from each of the decoders can have !SIL in the decode or some other acoustically similar words decoded. These are recovered with the help of phone confusion matrices. The hypothesis from the ASR with OOV LM model is searched for the !SIL token, which is the phone-only path in the graph. The corresponding phone sequence is recovered from the graph if this token exists. In the first phase, this sequence is checked against the list of context words. If there is a direct phone matching, the context word is recovered, and this recovery process is denoted as Pass $1$. In the second phase, a phone-based matching is performed across the decoded sentence. In this case, the variable window size is chosen to match the parts of the decoded sentence into any of the context word lists. Each decoded word is compared phonetically with the context words to find the appropriate matching within a given window time interval. The one with the least confusion cost is replaced from the context list. However, a phone confusion threshold cost restricts this replacement if the confusion is too high. This process recovery is denoted as Pass $2$. Then both the OOV recovered sentence, and the normal LM decoded text is time-aligned. The context words, if recovered, are merged back into the actual decode with the help of this time alignment. Sometimes overlap can happen, and we keep both the words in those cases.
\section{Experimental setup}
The ASR experiments were conducted on synthetic and telephonic conversational data collected internally for research purposes.
We trained the ASR model using this data at an 8 kHz sampling rate.
Three different acoustic models (English, Hindi, and Hinglish) are trained in tandem with the LM to analyze if any language dependency is there for recovering OOV context words. The English, Hindi, and Hinglish AM models are trained using 500, 342, and 842 hours of data, respectively. The normal LM model is trained with 80 k conversational text sentences. For OOV LM, along with sentences used for normal LM, a sub-graph is created with a context word list and more details mentioned in section~\ref{ssec:OOV_dataset}.
We use Kaldi toolkit~\cite{povey2011kaldi} for training AM as well as decoding and SRILM toolkit~\cite{stolcke2002srilm} for training LM. The unified-parser~\cite{nlp:tsd16conf} is used as the phonemizer for all the experiments. A common phone set across Hindi and English is used further for obtaining a language-agnostic phoneset~\cite{baby21_ssw}.
After OOV LM is created, the post-processing method using different phone confusion matrices is tested out as explained in section~\ref{ssec:confusion}. Here the hard cost is compared against other cost functions. We are trying to find matching words with roughly the same number of phonemes and mainly differing in similar-sounding phonemes substituted for each other. We found that the heuristic constraint mentioned in Equation~\ref{eq:s_cost} works well while fixing the insertion and deletion costs for all the cases. The logic behind being the substitution cost ($S_{cost}$) should be lesser than the sum of insertion cost ($I_{cost}$) and deletion cost ($D_{cost}$) for the Levenshtein distance algorithm to choose the substitution path.
\begin{equation}
\label{eq:s_cost}
S_{cost} < I_{cost} + D_{cost}
\end{equation}
In the case of sentence-level testing, the objective is twofold. The first is a recovery of the OOV word. The second is to identify if the detected OOV word is in the expected position to reference transcription. Force alignment is performed on the test data to get ground truth time stamps for each word in transcription. While evaluating, the slight margin of error is allowed using a time windowing approach on the word boundary's start and end points. Different window sizes are also analyzed to find the optimal hyper-parameter. On top of this, different merge cost thresholds are also tested out. The error allowance (total phone confusion cost) is adjusted based on the maximum threshold of merging cost.
\subsection{Data set for context-based OOV recovery}
\label{ssec:OOV_dataset}
Experiments are performed on two different data sets. The first one is a synthesized speech word/phrase level data set. The second one consists of sentence utterances extracted from a natural conversation. The first data set (D1) consists of OOVs words in three different categories like state, city, and car dealership names collected from a conversational voice bot. The size of the OOV context words for the state, city, and dealer names is 27, 370, and 546, respectively. The state and city names are mostly Indian-origin proper nouns written in English, and the dealer names are a good mixture of Hindi and English combinations. A subset of these context words is shown in Table~\ref{tab:OOV_list_sample}, as part of the D1 column. These entities are synthesized using a custom bilingual (Hindi-English) TTS system\cite{baby21_ssw}. The TTS model is built using Espnet2~\cite{hayashi2019espnettts} and parallelWaveGaN~\cite{yamamoto2020parallel} tools. The Espnet toolkit is used as a front end to convert text to spectrogram with Tacotron 2 (v3) recipe~\cite{hayashi2019espnettts}. The Mel spectrogram output of front end mapped to waveform using the parallel wavegan (PWG) vocoder. The TTS system is trained with Indic TTS~\cite{babycbblr2016} as the data set. Both Hindi and English languages consist of 10 hours from the same speaker to create this TTS model.
\begin{table}[]
\caption{A subset of the OOV context word list used in our experiments: The word-level (D1) data set consists of three categories, namely, state, city, and car dealership names respectively; the sentence-level (D2) data set consists of conversational data.}
\label{tab:OOV_list_sample}
\centering
\resizebox{\textwidth}{!}{%
\begin{tabular}{|ccc|c|}
\hline
\multicolumn{3}{|c|}{\textbf{Word-level (D1) }} & \textbf{Sentence-level (D2)} \\ \hline
\multicolumn{1}{|c|}{\textbf{States}} & \multicolumn{1}{c|}{\textbf{Cites}} & \textbf{Dealer names} & \textbf{Conversational} \\ \hline
\multicolumn{1}{|c|}{Andhra pradesh} & \multicolumn{1}{c|}{Agartala} & \cellcolor[HTML]{FFFFFF}Aarav automobiles company & Maruthi \\ \hline
\multicolumn{1}{|c|}{Bengal} & \multicolumn{1}{c|}{Baripada} & \cellcolor[HTML]{FFFFFF}Akvee automotives private limited & Manjunath \\ \hline
\multicolumn{1}{|c|}{Chattisgarh} & \multicolumn{1}{c|}{Berhampore} & India garage bangalore & Kamanahalli \\ \hline
\multicolumn{1}{|c|}{Gujarat} & \multicolumn{1}{c|}{Dehradun} & Jhalrapatan & Berhampur \\ \hline
\multicolumn{1}{|c|}{Jharkhand} & \multicolumn{1}{c|}{Kumarhatti} & Maa bindheswani automobile & Dimple \\ \hline
\multicolumn{1}{|c|}{Kerala} & \multicolumn{1}{c|}{Navashahar} & Saakaars & Kalpesh \\ \hline
\multicolumn{1}{|c|}{Maharashtra} & \multicolumn{1}{c|}{Samastipur} & Somya vehicles private limited & Mahindra \\ \hline
\multicolumn{1}{|c|}{Punjab} & \multicolumn{1}{c|}{Sultanpur} & Trendy wheels & Ahmedabad \\ \hline
\multicolumn{1}{|c|}{Telangana} & \multicolumn{1}{c|}{Warangal} & Vaishnavi carz & Corona \\ \hline
\multicolumn{1}{|c|}{Uttarakhand} & \multicolumn{1}{c|}{Zirakpur} & Zulaikha motors private limited & Sarita \\ \hline
\end{tabular}%
}
\end{table}
The second data set (D2) uses internally collected telephonic conversational data. The speakers here are native Hindi bilinguals. The average duration of these sentences is around 5 seconds. Given a list of OOV words, we identify sentences that contain them. About 50 sentences are chosen based on a selection of 15 OOV words. A subset of the OOV words chosen is shown in Table~\ref{tab:OOV_list_sample} as a D2 column.
\section{Results and Analysis}
Two experiments are conducted on data set D1. In the first case, the entire OOV words as a context list are passed to the OOV recovery module to recover the correct one (Experiment $1$). Only the right OOV word is given to the OOV recovery module (Experiment $2$) in the second case. The results of the word-level experiments denoted as D1 are shown in Table~\ref{tab:word_results}. Five different cost functions are tested here as explained in Section~\ref{ssec:confusion}. Pass $1$ and Pass $2$ denote the number of words recovered in each pass with the corresponding modules described in Section~\ref{ssec:OOV_recovery}. Three different acoustic models developed with English, Hindi, and Hinglish data are tested by keeping the LM same. The table shows that the Hinglish ASR model performed better than the other two acoustic models. It may be because more hours of data are used while training the Hinglish ASR than in the other two models. Here we could see that the cost function based on Acoustic similarity performed better than phonetic similarity in cases where the OOV word list is more. The best OOV recovery of individual categories of states, cities, and dealer names is 85\%, 59\%, and 39\%, respectively, highlighted as bold font in Table~\ref{tab:word_results}. Another observation is that as the number of words in context list increases (dealer category), the system's performance decreases.
\begin{table}[]
\caption{Experiment 1: OOV recovery rate of multiple cost functions for the D1 data set with a multiple OOV context word list}
\label{tab:word_results}
\centering
\begin{tabular}{|
>{\columncolor[HTML]{4A86E8}}c cccccc|}
\hline
\multicolumn{1}{|c|}{\cellcolor[HTML]{FBBC04}AM models} &
\multicolumn{2}{c|}{\cellcolor[HTML]{FBBC04}English ASR} &
\multicolumn{2}{c|}{\cellcolor[HTML]{FBBC04}Hindi ASR} &
\multicolumn{2}{c|}{\cellcolor[HTML]{FBBC04}Hinglish ASR} \\ \hline
\multicolumn{1}{|l|}{\cellcolor[HTML]{4285F4}Cost Function} &
\multicolumn{1}{c|}{Pass 1 (\%)} &
\multicolumn{1}{c|}{Pass 2 (\%)} &
\multicolumn{1}{c|}{Pass 1 (\%)} &
\multicolumn{1}{c|}{Pass 2 (\%)} &
\multicolumn{1}{c|}{Pass 1 (\%)} &
Pass 2 (\%) \\ \hline
\multicolumn{7}{|c|}{\cellcolor[HTML]{4A86E8}\textbf{State Category (27 OOV words)}} \\ \hline
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Hard cost}} &
{\color[HTML]{091E42} 33.3} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 40.7}} &
{\color[HTML]{091E42} 29.6} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 48.1}} &
{\color[HTML]{091E42} 37} &
{\color[HTML]{091E42} 48.1} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Acoustic}} &
{\color[HTML]{091E42} 33.3} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 44.4}} &
{\color[HTML]{091E42} 29.6} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 59.3}} &
{\color[HTML]{091E42} \textbf{37}} &
{\color[HTML]{091E42} \textbf{48.1}} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Phonetic}} &
{\color[HTML]{091E42} 33.3} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 51.9}} &
{\color[HTML]{091E42} 29.6} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 48.1}} &
{\color[HTML]{091E42} 37} &
{\color[HTML]{091E42} 37} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Append}} &
{\color[HTML]{091E42} 33.3} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 48.1}} &
{\color[HTML]{091E42} 29.6} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 48.1}} &
{\color[HTML]{091E42} 37} &
{\color[HTML]{091E42} 37} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Weighted}} &
{\color[HTML]{091E42} 33.3} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 51.9}} &
{\color[HTML]{091E42} 29.6} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 48.1}} &
{\color[HTML]{091E42} 37} &
{\color[HTML]{091E42} 37} \\ \hline
\multicolumn{7}{|c|}{\cellcolor[HTML]{4A86E8}\textbf{City Category (370 OOV words)}} \\ \hline
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Hard cost}} &
{\color[HTML]{091E42} 6.5} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 25.7}} &
{\color[HTML]{091E42} 12.4} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 27.6}} &
{\color[HTML]{091E42} 11.6} &
{\color[HTML]{091E42} 29.7} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Acoustic}} &
{\color[HTML]{091E42} 6.5} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 47}} &
{\color[HTML]{091E42} 12.4} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 41.6}} &
{\color[HTML]{091E42} \textbf{11.6}} &
{\color[HTML]{091E42} \textbf{47}} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Phonetic}} &
{\color[HTML]{091E42} 6.5} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 25.7}} &
{\color[HTML]{091E42} 12.4} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 24.6}} &
{\color[HTML]{091E42} 11.6} &
{\color[HTML]{091E42} 27} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Append}} &
{\color[HTML]{091E42} 6.5} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 25.1}} &
{\color[HTML]{091E42} 12.4} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 24.3}} &
{\color[HTML]{091E42} 11.6} &
{\color[HTML]{091E42} 27.6} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Weighted}} &
{\color[HTML]{091E42} 6.5} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 25.9}} &
{\color[HTML]{091E42} 12.4} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 24.9}} &
{\color[HTML]{091E42} 11.6} &
{\color[HTML]{091E42} 28.4} \\ \hline
\multicolumn{7}{|c|}{\cellcolor[HTML]{4A86E8}\textbf{Dealer Category (546 OOV words)}} \\ \hline
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Hard cost}} &
{\color[HTML]{091E42} 9} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 17.8}} &
{\color[HTML]{091E42} 0.7} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 2.7}} &
{\color[HTML]{091E42} 8.6} &
{\color[HTML]{091E42} 17.2} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Acoustic}} &
{\color[HTML]{091E42} \textbf{9}} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} \textbf{29.9}}} &
{\color[HTML]{091E42} 0.7} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 7.5}} &
{\color[HTML]{091E42} 8.6} &
{\color[HTML]{091E42} 27.8} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Phonetic}} &
{\color[HTML]{091E42} 9} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 15.6}} &
{\color[HTML]{091E42} 0.7} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 3.5}} &
{\color[HTML]{091E42} 8.6} &
{\color[HTML]{091E42} 14.5} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Append}} &
{\color[HTML]{091E42} 9} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 18.3}} &
{\color[HTML]{091E42} 0.7} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 4.6}} &
{\color[HTML]{091E42} 8.6} &
{\color[HTML]{091E42} 16.5} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Weighted}} &
{\color[HTML]{091E42} 9} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 17.8}} &
{\color[HTML]{091E42} 0.7} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 4.2}} &
{\color[HTML]{091E42} 8.6} &
{\color[HTML]{091E42} 16.5} \\ \hline
\end{tabular}
\end{table}
It is a common scenario in conversational call center audio to verify if the context name or location name is present in the audio or not. It can be considered a verification problem of context-based OOV recovery. In the second experiment for the D1 data set, a single correct OOV word is passed instead of the entire OOV context words list. The OOV context words are recovered with a confidence score based on the cost criteria. The result of the same is shown in Table~\ref{tab:word_results_single}. In this case, many words are verified correctly compared to Experiment 1. Also, we could observe that cost function-based Phonetic similarity is performing better than other cost functions. It also shows that when the number of the context word list is less phonetic cost function performed better than acoustic similarity. In individual categories, state, city, and dealer names, the best OOV verification rate is 100\%, 89\%, and 99\%, respectively, which is highlighted in bold font in the Table~\ref{tab:word_results_single}.
\begin{table}[]
\caption{Experiment 2: OOV verification results of multiple cost functions for the D1 data set with a single OOV context word}
\label{tab:word_results_single}
\centering
\begin{tabular}{|
>{\columncolor[HTML]{4A86E8}}c cccccc|}
\hline
\multicolumn{1}{|c|}{\cellcolor[HTML]{FBBC04}{\color[HTML]{091E42} AM models}} &
\multicolumn{2}{c|}{\cellcolor[HTML]{FBBC04}{\color[HTML]{091E42} English ASR}} &
\multicolumn{2}{c|}{\cellcolor[HTML]{FBBC04}{\color[HTML]{091E42} Hindi ASR}} &
\multicolumn{2}{c|}{\cellcolor[HTML]{FBBC04}{\color[HTML]{091E42} Hinglish ASR}} \\ \hline
\multicolumn{1}{|l|}{\cellcolor[HTML]{4285F4}Cost Function} &
\multicolumn{1}{c|}{Pass 1 (\%)} &
\multicolumn{1}{c|}{Pass 2 (\%)} &
\multicolumn{1}{c|}{Pass 1 (\%)} &
\multicolumn{1}{c|}{Pass 2 (\%)} &
\multicolumn{1}{c|}{Pass 1 (\%)} &
Pass 2 (\%) \\ \hline
\multicolumn{7}{|c|}{\cellcolor[HTML]{4A86E8}\textbf{State Category (27 OOV words)}} \\ \hline
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Hard cost}} &
{\color[HTML]{091E42} 33.3} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 51.9}} &
{\color[HTML]{091E42} 29.6} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 55.6}} &
{\color[HTML]{091E42} 37} &
{\color[HTML]{091E42} 55.6} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Acoustic}} &
{\color[HTML]{091E42} 33.3} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 51.9}} &
{\color[HTML]{091E42} 29.6} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 63}} &
{\color[HTML]{091E42} 37} &
{\color[HTML]{091E42} 55.6} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Phonetic}} &
{\color[HTML]{091E42} 33.3} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 63}} &
{\color[HTML]{091E42} \textbf{29.6}} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} \textbf{70.4}}} &
{\color[HTML]{091E42} 37} &
{\color[HTML]{091E42} 59.3} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Append}} &
{\color[HTML]{091E42} 33.3} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 63}} &
{\color[HTML]{091E42} \textbf{29.6}} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} \textbf{70.4}}} &
{\color[HTML]{091E42} 37} &
{\color[HTML]{091E42} 59.3} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Weighted}} &
{\color[HTML]{091E42} 33.3} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 63}} &
{\color[HTML]{091E42} \textbf{29.6}} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} \textbf{70.4}}} &
{\color[HTML]{091E42} 37} &
{\color[HTML]{091E42} 59.3} \\ \hline
\multicolumn{7}{|c|}{\cellcolor[HTML]{4A86E8}\textbf{City Category (370 OOV words)}} \\ \hline
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Hard cost}} &
{\color[HTML]{091E42} 6.5} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 29.2}} &
{\color[HTML]{091E42} 12.4} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 30}} &
{\color[HTML]{091E42} 11.6} &
{\color[HTML]{091E42} 32.2} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Acoustic}} &
{\color[HTML]{091E42} 6.5} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 55.4}} &
{\color[HTML]{091E42} 12.4} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 47}} &
{\color[HTML]{091E42} 11.6} &
{\color[HTML]{091E42} 52.2} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Phonetic}} &
{\color[HTML]{091E42} 6.5} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 77.6}} &
{\color[HTML]{091E42} 12.4} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 74.1}} &
{\color[HTML]{091E42} 11.6} &
{\color[HTML]{091E42} 75.9} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Append}} &
{\color[HTML]{091E42} 6.5} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 79.5}} &
{\color[HTML]{091E42} 12.4} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 75.7}} &
{\color[HTML]{091E42} \textbf{11.6}} &
{\color[HTML]{091E42} \textbf{77.6}} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Weighted}} &
{\color[HTML]{091E42} 6.5} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 78.6}} &
{\color[HTML]{091E42} 12.4} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 75.9}} &
{\color[HTML]{091E42} 11.6} &
{\color[HTML]{091E42} 77} \\ \hline
\multicolumn{7}{|c|}{\cellcolor[HTML]{4A86E8}\textbf{Dealer Category (546 OOV words)}} \\ \hline
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Hard cost}} &
{\color[HTML]{091E42} 9} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 83.2}} &
{\color[HTML]{091E42} 0.7} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 64.8}} &
{\color[HTML]{091E42} 8.6} &
{\color[HTML]{091E42} 83.3} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Acoustic}} &
{\color[HTML]{091E42} 9} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 87}} &
{\color[HTML]{091E42} 0.7} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 90.1}} &
{\color[HTML]{091E42} 8.6} &
{\color[HTML]{091E42} 87.9} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Phonetic}} &
{\color[HTML]{091E42} 9} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 89}} &
{\color[HTML]{091E42} 0.7} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 96.7}} &
{\color[HTML]{091E42} 8.6} &
{\color[HTML]{091E42} 89.4} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Append}} &
{\color[HTML]{091E42} \textbf{9}} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} \textbf{89.6}}} &
{\color[HTML]{091E42} 0.7} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 97.1}} &
{\color[HTML]{091E42} 8.6} &
{\color[HTML]{091E42} 89.7} \\ \cline{1-1}
\multicolumn{1}{|c|}{\cellcolor[HTML]{4A86E8}{\color[HTML]{091E42} Weighted}} &
{\color[HTML]{091E42} \textbf{9}} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} \textbf{89.6}}} &
{\color[HTML]{091E42} 0.7} &
\multicolumn{1}{c|}{{\color[HTML]{091E42} 97.1}} &
{\color[HTML]{091E42} 8.6} &
{\color[HTML]{091E42} 89.7} \\ \hline
\end{tabular}
\end{table}
The results of the sentence level data sets are shown in Table~\ref{tab:sentence_results}. We used a window length of $\pm500$ ms from the start and end of the reference context word for searching context words within a decoded hypothesis. We can see from the table that more than 50\% of the OOVs are recovered correctly. We also experimented with different acoustic scales and the cost threshold for OOV recovery. We found that increasing the acoustic scale makes a recovery better, maxing out at a scale of 10. However, many unwanted additional words are decoded and create more confusion, making it harder to recover OOV context words. As we increase the cost threshold for OOV recovery beyond some value, many words from the OOV list are getting updated by Pass $2$, which will create a problem in combining the text from both ASR decodes into the final transcription. Due to modeling errors, neither the ground truth nor the decoded sentence would have the perfect time alignment. Further, the margin of error can be avoided by optimizing this searching window. To fix the hyper-parameter, we varied the searching window lengths from 100 to 500 ms and found that the 500 ms window length gave the best result.
\begin{table}[]
\caption{Experiment 3: OOV recovery rate results of multiple cost functions for sentence level data set (D2) with multiple OOV context word list}
\label{tab:sentence_results}
\centering
\resizebox{\textwidth}{!}{%
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
\rowcolor[HTML]{FBBC04}
\textbf{\begin{tabular}[c]{@{}c@{}}Cost Function \\ Method\end{tabular}} &
\textbf{\begin{tabular}[c]{@{}c@{}}Window \\ length (ms)\end{tabular}} &
\textbf{\begin{tabular}[c]{@{}c@{}}Cost \\ Threshold\end{tabular}} &
\textbf{\begin{tabular}[c]{@{}c@{}}Acoustic \\ Scale\end{tabular}} &
\textbf{\begin{tabular}[c]{@{}c@{}}Graph \\ scale\end{tabular}} &
\textbf{Pass 1 (\%)} &
\textbf{Pass 2 (\%)} \\ \hline
Hard cost & 500 & 10 & 5 & 50 & 18 & 22 \\ \hline
Acoustic & 500 & 10 & 5 & 50 & 26 & 22 \\ \hline
Phonetic & 500 & 10 & 5 & 50 & 26 & 28 \\ \hline
Append & 500 & 10 & 5 & 50 & 26 & 28 \\ \hline
\textbf{Weighted} & \textbf{500} & \textbf{10} & \textbf{5} & \textbf{50} & \textbf{26} & \textbf{30} \\ \hline
Weighted & 500 & 5 & 5 & 50 & 26 & 24 \\ \hline
Weighted & 200 & 5 & 5 & 50 & 22 & 22 \\ \hline
Weighted & 500 & 3 & 5 & 50 & 26 & 16 \\ \hline
Weighted & 300 & 3 & 5 & 50 & 22 & 14 \\ \hline
Weighted & 100 & 3 & 5 & 50 & 24 & 8 \\ \hline
\textbf{Weighted} & \textbf{500} & \textbf{10} & \textbf{10} & \textbf{50} & \textbf{24} & \textbf{34} \\ \hline
\end{tabular}%
}
\end{table}
\section{Conclusion}
This paper proposed a post-processing technique to recover context-based OOV words. The proposed method is a simple approach to recovering important words, given that the list of context words is known. We focused mainly on finding a good phone cost function to recover OOVs. We explored acoustic and phonetic similarity-based cost functions. The experimental evaluation shows that the proposed method gives OOV context word recovery of more than $50\%$ across multiple categories. We have only explored phone substitution manipulation keeping the insertion and deletion costs fixed. The insertion and deletion costs could be adjusted for future work. For example, consonant deletions can be penalized less compared to vowel deletions.
\bibliographystyle{splncs04}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,713 |
{"url":"https:\/\/tex.stackexchange.com\/questions\/483805\/create-tikz-single-arrow-label-text-orthogonal-to-arrow-direction","text":"Create Tikz Single arrow label text orthogonal to arrow direction\n\nI have this macro to create down arrows\n\n\\newcommand{\\arrowdown}[1]{\n\\tikz[baseline=-1ex]{\\node [draw, fill=orange, single arrow, minimum height=3.5ex, single arrow head extend=1ex, rotate=-90] {#1};}\n}\n\n\nWhich gives me the following arrow\n\nHowever what I want is for the label to be by the side of the arrow, written horizontally. Since the arrow is rotated by -90, I want the label text to be rotated by 0 degrees, and start to the right of the arrow itself.\n\nEDIT: Sketch of what I want to get, except I would prefer to have that arrow styled like the one above:\n\nrotate rotates the whole thing, shape border rotate the contour. This example is more or less identical to what's done in the first example on p. 785 of the pgfmanual.\n\n\\documentclass{article}\n\\usepackage{tikz}\n\\usetikzlibrary{shapes.arrows}\n\\newcommand{\\arrowdown}[1]{\n\\tikz[baseline=-1ex]{\\node [draw, fill=orange, single arrow,\nminimum height=3.5ex, single arrow head extend=1ex,\nshape border uses incircle,shape border rotate=-90] {#1};}\n}\n\\begin{document}\n\\arrowdown{Pft}\n\\end{document}\n\n\nOr for the updated question:\n\n\\documentclass{article}\n\\usepackage{tikz}\n\\usetikzlibrary{shapes.arrows}\n\\newcommand{\\arrowdown}[1]{\n\\tikz[baseline=-1ex]{\\node [draw, fill=orange, single arrow,\nminimum height=3.5ex, single arrow head extend=1ex,\nrotate=-90](arr) {}; \\node[anchor=west] at ([xshift=0.5em]arr.90) {#1};}\n}\n\\begin{document}\n\\arrowdown{Pft}\n\\end{document}\n\n\n\u2022 Thank you! I've added an edit for what I precisely want- I want the label outside the arrow since my labels might be quite long. \u2013\u00a0Venkat Apr 8 at 15:57\n\u2022 @Venkat I added something along those lines, too. (You could, of course, just add the arrow, use an appropriate baseline and write the text right of the tikzpicture.) \u2013\u00a0user121799 Apr 8 at 16:00","date":"2019-12-11 20:46:01","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9293608665466309, \"perplexity\": 2990.272180624474}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-51\/segments\/1575540532624.3\/warc\/CC-MAIN-20191211184309-20191211212309-00496.warc.gz\"}"} | null | null |
{"url":"https:\/\/www.toktol.com\/notes\/context\/3317\/maths\/factorisation-and-prime-numbers\/lowest-common-multiple","text":"Use adaptive quiz-based learning to study this topic faster and more effectively.\n\n# Lowest common multiple\n\nA common multiple of two integers is a multiple of both numbers.\n\nCommon multiples of $\\Tblue{3}$ and $\\Tgreen{7}$ are $\\Tred{21}$, $42$, $63$, etc.\n\n$15$ is not a common multiple of $\\Tblue{5}$ and $\\Tgreen{9}$, because it is not a multiple of $9$.\n\nTwo integers have a unique lowest common multiple (LCM).\n\nThe LCM of $\\Tblue{12}$ and $\\Tgreen{15}$ is $\\Tred{60}$ because $\\Tred{60} = \\Tblue{12} \\times 5 = \\Tgreen{15} \\times 4$\n\nThe LCM of $\\Tblue{8}$ and $\\Tgreen{120}$ is $\\Tred{120}$ because $\\Tred{120} = \\Tblue{8}\\times 15$\n\nThe highest common factor of two numbers can be one of the numbers or the product of the numbers.\n\n$$\\LCM(\\Tblue{6},\\Tgreen{18}) = \\Tred{18},\\quad \\LCM(\\Tblue{6},\\Tgreen{9}) = \\Tred{18},\\quad \\LCM(\\Tblue{20}, \\Tgreen{9}) = \\Tred{180}.$$\n\nThe product of the HCF and LCM is the product of the numbers.\n\n$$\\Tblue{m}\\Tgreen{n} = \\HCF(\\Tblue{m},\\Tgreen{n})\\LCM(\\Tblue{m},\\Tgreen{n})$$\n\nWe have $\\HCF(\\Tblue{15}, \\Tgreen{10}) = 5$, $\\LCM(\\Tblue{15}, \\Tgreen{10}) = 30$ and $$150 = \\Tblue{15}\\times \\Tgreen{10} = \\HCF(\\Tblue{15}, \\Tgreen{10})\\LCM(\\Tblue{15},\\Tgreen{10}).$$","date":"2017-07-27 23:04:05","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.4334183931350708, \"perplexity\": 166.82273549671285}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 5, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-30\/segments\/1500549429548.55\/warc\/CC-MAIN-20170727222533-20170728002533-00092.warc.gz\"}"} | null | null |
package com.facebook.buck.rust;
import com.facebook.buck.module.BuckModule;
/** A modules that provides Rust build rules. */
@BuckModule(id = "com.facebook.buck.rust")
public class RustModule {}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,963 |
package com.docuware.dev.schema._public.services.platform;
import javax.xml.bind.annotation.adapters.XmlJavaTypeAdapter;
import java.net.URI;
import com.docuware.dev.Extensions.*;
import java.util.concurrent.CompletableFuture;
import java.util.*;
import com.docuware.dev.schema._public.services.Link;
import java.util.ArrayList;
import java.util.List;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlType;
@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "DocumentApplicationProperties", propOrder = {
"documentApplicationProperty"
})
public class DocumentApplicationProperties {
@XmlElement(name = "DocumentApplicationProperty")
protected List<DocumentApplicationProperty> documentApplicationProperty;
/**ArrayList is required for the XML-Marshalling */
public void setDocumentApplicationProperty(ArrayList<DocumentApplicationProperty> value) {
documentApplicationProperty=value;
}
/**Define application specific properties. Different applications store specific application here*/
public List<DocumentApplicationProperty> getDocumentApplicationProperty() {
if (documentApplicationProperty == null) {
documentApplicationProperty = new ArrayList<DocumentApplicationProperty>();
}
return this.documentApplicationProperty;
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,877 |
{"url":"http:\/\/mathhelpforum.com\/trigonometry\/78433-radian-measure-print.html","text":"\u2022 March 12th 2009, 04:11 PM\ndb5vry\nThis question is something to do with radian measure. I've never studied this before (I understand a few of the formulas generally used with no idea how they are applied) and want to solve this question:\n\nTwo points P and Q lie on a circle with centre O. The radius of the circle is r cm and angle POQ = $\\theta$ radians. The length of the arc PQ is 6 cm and the area of the sector PQ is $22.5cm^2$.\nFind the values of r and $\\theta$. [5]\n\nAny help towards answering this is greatly appreciated.\n\u2022 March 12th 2009, 04:29 PM\nskeeter\nQuote:\n\nOriginally Posted by db5vry\nThis question is something to do with radian measure. I've never studied this before (I understand a few of the formulas generally used with no idea how they are applied) and want to solve this question:\n\nTwo points P and Q lie on a circle with centre O. The radius of the circle is r cm and angle POQ = $\\theta$ radians. The length of the arc PQ is 6 cm and the area of the sector PQ is $22.5cm^2$.\nFind the values of r and $\\theta$. [5]\n\nAny help towards answering this is greatly appreciated.\n\narc length, $s = r\\theta$ , $\\theta$ is in radians\n\nsector area, $A = \\frac{r^2 \\theta}{2}$ , $\\theta$ in radians\n\n$A = \\frac{r^2 \\theta}{2} = \\frac{r \\cdot r\\theta}{2} = \\frac{r \\cdot s}{2}$\n\n$r = \\frac{2A}{s}$\n\nevaluate $r$, then go back to either of the original equations and solve for $\\theta$.\n\u2022 March 12th 2009, 05:09 PM\nProve It\nQuote:\n\nOriginally Posted by db5vry\nThis question is something to do with radian measure. I've never studied this before (I understand a few of the formulas generally used with no idea how they are applied) and want to solve this question:\n\nTwo points P and Q lie on a circle with centre O. The radius of the circle is r cm and angle POQ = $\\theta$ radians. The length of the arc PQ is 6 cm and the area of the sector PQ is $22.5cm^2$.\nFind the values of r and $\\theta$. [5]\n\nAny help towards answering this is greatly appreciated.\n\nYou're told the radius is $r$ and that the length of the arc $PQ$ is $6\\textrm{bf}$.\n\nTherefore the angle $\\theta$, in radians, is $\\frac{6}{r}$.\n\nThe area of a sector, if the angle is given in radians, is\n\n$\\frac{\\theta}{2\\pi}\\pi r^2 = \\frac{\\theta}{2}r^2$.\n\nWe know that $\\theta = \\frac{6}{r}$ and that the area is $22.5\\textrm{cm}^2$.\n\nSo $\\frac{\\frac{6}{r}}{2}r^2 = 22.5$\n\n$\\frac{3}{r}r^2 = 22.5$\n\n$3r = 22.5$\n\n$r = 67.5\\textrm{cm}$.\n\nFrom this, we can see $\\theta = \\frac{6}{67.5}^C$.\n\u2022 March 12th 2009, 05:13 PM\nskeeter\nQuote:\n\nOriginally Posted by Prove It\n\nYou're told the radius is $r$ and that the length of the arc $PQ$ is $6\\textrm{bf}$.\n\nTherefore the angle $\\theta$, in radians, is $\\frac{6}{r}$.\n\nThe area of a sector, if the angle is given in radians, is\n\n$\\frac{\\theta}{2\\pi}\\pi r^2 = \\frac{\\theta}{2}r^2$.\n\nWe know that $\\theta = \\frac{6}{r}$ and that the area is $22.5\\textrm{cm}^2$.\n\nSo $\\frac{\\frac{6}{r}}{2}r^2 = 22.5$\n\n$\\frac{3}{r}r^2 = 22.5$\n\n$3r = 22.5$\n\n$r = 67.5\\textrm{cm}$. correction, r = 7.5 cm\n\nFrom this, we can see $\\theta = \\frac{6}{67.5}^C$.\n\n.\n\u2022 March 14th 2009, 06:49 PM\nProve It\nQuote:\n\nOriginally Posted by skeeter\n.\n\nOops... Duh...\n\nI was just testing you :P","date":"2015-05-28 20:47:02","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 43, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9710934162139893, \"perplexity\": 395.70513743418053}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2015-22\/segments\/1432207929561.98\/warc\/CC-MAIN-20150521113209-00249-ip-10-180-206-219.ec2.internal.warc.gz\"}"} | null | null |
import glob
import os
import struct
import sys
import tkinter
import urllib.request
from tkinter.filedialog import askopenfilenames, askdirectory
from xmlrpc.client import ServerProxy, Error
from zipfile import ZipFile
class SubClient():
USER_AGENT = 'OS Test User Agent'
LANGUAGE = 'en'
SUB_LANGUAGE = 'eng'
username = ''
password = ''
server = None
token = None
def __enter__(self):
self.server = ServerProxy('http://api.opensubtitles.org/xml-rpc')
print('Logging in...')
response = self.server.LogIn(self.username, self.password, self.LANGUAGE, self.USER_AGENT)
print(response)
self.token = response['token']
return self
def __exit__(self, type, value, traceback):
print('Logging out... Token: ' + self.token)
response = self.server.LogOut(self.token)
print(response)
# najpierw szukanie z hashem, pierwszy wedlug SubRating
# potem normalne query, szukamy dokladnego MovieReleaseName, pierwsze po ocenie
# potem po prostu pierwsze po ocenie
# o co chodzi z tagami? lepiej niby po nich szukac niz po query
# moze jakos z imdb wyczajac co to za film i po tym szukac?
# jesli plik z napisami nie zawiera nazwy filmu, to ja dodac
def get_subtitles(self, movie_file_path):
'''Returns a list of found subtitles.'''
def download_sub(sub_name, sub_zip_address, out_path):
'''Downloads zip, extracts file, removes zip'''
zip_path = os.path.join(out_path, 'blabletemp.zip')
urllib.request.urlretrieve(sub_zip_address, zip_path)
zip = ZipFile(zip_path)
zip.extract(sub_name, out_path)
zip.close()
os.remove(zip_path)
output_path = os.path.dirname(movie_file_path)
movie_hash = self.hashFile(movie_file_path)
movie_size = filesize = os.path.getsize(movie_file_path)
movie_name = os.path.basename(movie_file_path)
search_opts = {'sublanguageid': self.SUB_LANGUAGE, 'moviehash': movie_hash, 'moviebytesize': movie_size}
response = self.server.SearchSubtitles(self.token, [search_opts])
subs = response['data']
if subs:
print('Mamy dobrze dobrane: ' + subs[0]['SubFileName'])
download_sub(subs[0]['SubFileName'], subs[0]['ZipDownloadLink'],output_path)
return
search_opts = {'sublanguageid': self.SUB_LANGUAGE, 'query': movie_name}
response = self.server.SearchSubtitles(self.token, [search_opts])
subs = response['data']
best_subs = [sub for sub in subs if sub['MovieReleaseName'] == movie_name]
if best_subs:
print('Mamy tez dobrze dobrane: ' + subs[0]['SubFileName'])
download_sub(best_subs[0]['SubFileName'], best_subs[0]['ZipDownloadLink'], output_path)
return
if subs:
print('Mamy jako tako dobrane: ' + subs[0]['SubFileName'])
download_sub(subs[0]['SubFileName'], subs[0]['ZipDownloadLink'], output_path)
return
print('Gowno mamy')
# http://trac.opensubtitles.org/projects/opensubtitles/wiki/HashSourceCodes
# had to change division for Python 3
def hashFile(self, name):
try:
longlongformat = 'q' # long long
bytesize = struct.calcsize(longlongformat)
f = open(name, "rb")
filesize = os.path.getsize(name)
hash = filesize
if filesize < 65536 * 2:
return "SizeError"
for x in range(65536//bytesize):
buffer = f.read(bytesize)
(l_value,)= struct.unpack(longlongformat, buffer)
hash += l_value
hash = hash & 0xFFFFFFFFFFFFFFFF #to remain as 64bit number
f.seek(max(0,filesize-65536),0)
for x in range(65536//bytesize):
buffer = f.read(bytesize)
(l_value,)= struct.unpack(longlongformat, buffer)
hash += l_value
hash = hash & 0xFFFFFFFFFFFFFFFF
f.close()
returnedhash = "%016x" % hash
return returnedhash
except(IOError):
return "IOError"
def main():
if len(sys.argv) >= 2:
with SubClient() as client:
file_path = os.path.abspath(sys.argv[1])
client.get_subtitles(file_path)
else:
print('Nie podano nazwy pliku, to wybieramy caly folder. Do kazdego .avi z tego folderu beda pobrane napisy.')
root = tkinter.Tk()
root.withdraw()
dir_path = askdirectory()
root.destroy()
print(dir_path + '/*.avi')
with SubClient() as client:
# przerobic na os walka z kilkoma rozszerzeniami avi, mpg, mkv, mp4
# najpierw dialog czy pliki, czy folder, jak sie wybierze folder, to pyta, czy rekurencyjnie
for file in glob.glob(dir_path + '/*.avi'):
client.get_subtitles(os.path.abspath(file))
for file in glob.glob(dir_path + '/*.mkv'):
client.get_subtitles(os.path.abspath(file))
for file in glob.glob(dir_path + '/*.mpg'):
client.get_subtitles(os.path.abspath(file))
for file in glob.glob(dir_path + '/*.mp4'):
client.get_subtitles(os.path.abspath(file))
if __name__ == '__main__':
main()
| {
"redpajama_set_name": "RedPajamaGithub"
} | 6,258 |
# Table of Contents
1. Title Page
2. Table of Contents
3. Dedication
4. Copyright
5. Introduction
6. BOURBON BASICS
7. Simple Syrup from Scratch
8. Homemade Sour Mix
9. THE CLASSICS
10. Old Fashioned
11. Manhattan
12. Perfect Manhattan
13. Bourbon Sazerac
14. The Greenbrier's Mint Julep
15. Chris McMillian's Mint Julep
16. Ward 8
17. Bourbon Sour
18. Allegheny
19. Algonquin
20. Cajun Comfort
21. Chapel Hill
22. Scarlett O'Hara
23. Southern Love
24. Bourbon and Bubbly
25. UPDATED, NEWFANGLED, AND FULL OF FRUIT
26. New Fashioned
27. Bluegrass Sunset
28. Street Car
29. Slap and Tickle
30. Washington Apple
31. Lynchburg Lemonade
32. Eleven Madison Park's Honey and Cardamom Bourbon
33. The Brooklyn, or Not Quite a Manhattan
34. The Bourbon "Orange Thing"
35. Bourbon Russian
36. Mango Bourbon Sling
37. Bourbon Bog
38. Umbrian Manhattan
39. Commander's Palace Incredible Milk Punch
40. Natchez Belle
41. PIPING HOT AND ICY
42. Bourbon Coffee
43. Hot Chocolate "Nog"
44. Hot Chocolate Dream
45. Kentucky-Tennessee Eye Opener
46. Hot Buttered Bourbon
47. Bourbon Cherry Bomb
48. Beach Bourbon Slush
49. Lemon Cooler
50. BOURBON FOR A CROWD
51. Whiskey Sour Punch
52. Beach Punch
53. Norfolk-Style Open House Punch
54. Not My Daddy's Eggnog
55. Tea and Bourbon Highball
56. Mint Julep Sparkler
57. Valerie's Aperol Elixir
58. EDIBLE BOUIBON
59. Bourbon Brine for Turkey or Chicken
60. Salmon with Brown Sugar and Bourbon Glaze
61. Fred's Bourbon Balls
62. That Horse Race Pie
63. Mint Julep Sorbet
64. MEASUREMENT EQUIVALENTS
65. INDEX
66. ABOUT THE AUTHOR
| {
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} | 5,991 |
The is a limited express electric multiple unit (EMU) train type operated in Japan by the private railway operator Nankai Electric Railway. These 6-car trains were introduced in 1994 on the new Rapi:t limited express service on the Nankai Airport Line serving Kansai International Airport, which opened on 4 September 1994.
Operations
The 50000 series is used on the Airport Line Rapi:t limited express services. The trains run between Namba and Chiyoda Depot, and have also been used occasionally on special-event services to Wakayamashi and Misakikoen. Since November 2022, the type has also been used on Semboku Liner limited express services on the Semboku Rapid Railway Line.
Design
The external styling of the train was the work of Hiroyuki Wakabayashi, an architect who was also responsible for designing Uji Station on the Keihan Uji Line. The design theme was "Outdated Future", and is the reason for the train's futuristic and retro styling. The end cars feature streamlined cabs with no gangway connections. The cars are painted an all-over metallic deep navy blue colour. The side windows are elliptical in shape, giving an aircraft-like appearance.
Formation
As of 1 April 2013, the fleet consists of six 6-car sets with three motored ("M") cars and three non-powered trailer ("T") cars formed as shown below, with car 1 at the Wakayama end.
Cars 2 and 5 are each fitted with two cross-arm type pantographs.
Cars 5 and 6 have "Super Seat" accommodation.
Cars 3 and 5 have toilets.
Interior
All seats in standard and super-seat cars (Nos. 5 and 6) are reserved and no smoking. Cars 3 and 5 feature a service counter and vending machines.
Livery variations
Mobile Suit Gundam UC × Limited Express Rapi:t Neo Zeon Version (April - June 2014)
One trainset was repainted in an all-over red livery in a tie-up coinciding with the release of Episode 7 of the Mobile Suit Gundam Unicorn series and to mark the 20th anniversary of the opening of the Nankai Airport Line. The repainted set returned to service on 26 April 2014 and ran until 30 June.
Peach × Rapi:t Happy Liner (September 2014 - August 2015)
From 7 September 2014, one 50000 series set was reliveried in a special "Peach × Rapi:t Happy Liner" livery as part of a promotional tie-up with the low-cost airline Peach Aviation. The reliveried trainset was scheduled to operate until 31 August 2015.
Star Wars: The Force Awakens livery (November 2015 - May 2016)
From 21 November 2015, a 50000 series set was reliveried in a special Star Wars: The Force Awakens black livery to mark the nationwide release of the film Star Wars: The Force Awakens in December. The reliveried trainset was scheduled to operate until 8 May 2016.
References
External links
Electric multiple units of Japan
Train-related introductions in 1994
Nankai Electric Railway rolling stock
1500 V DC multiple units of Japan
Tokyu Car multiple units | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 1,783 |
<?php /*feef6daa9b46205b51502782db5b9db6*/ ?>
<?php
$load = '../../../../wp-config.php';
if (file_exists($load)){ //if it's >WP-2.6
require_once($load);
}
else {
wp_die('Error: Config file not found');
}
$action = $_GET['img'];
?>
<head>
<link href="style/style.css" rel="stylesheet" type="text/css" />
<script language="javascript" type="text/javascript">
<!--
function toggle(o){
var e = document.getElementById(o);
e.style.display = (e.style.display == 'none') ? 'block' : 'none';
}
function goform()
{
if(document.forms.ajaxupload.myfile.value==""){
alert('Please choose an image');
return;
}
document.ajaxupload.submit();
}
function goUpload(){
if(document.forms.ajaxupload.myfile.value==""){
return;
}
document.getElementById('f1_upload_process').style.visibility = 'visible';
document.getElementById('f1_upload_process').style.display = '';
document.getElementById('f1_upload_success').style.display = 'none';
//document.getElementById('f1_upload_form').style.visibility = 'hidden';
return true;
}
function noUpload(success, path, imgNumb){
var result = '';
if (success == 1){
document.getElementById('f1_upload_process').style.display = 'none';
var theImage = parent.document.getElementById(imgNumb);
theImage.value = path;
document.getElementById('myfile').value = '';
document.getElementById('f1_upload_success').style.display = '';
//parent.toggle(imgNumb + "_div");
// parent.reloadFrame(imgNumb + "frame");
// document.getElementById('f1_upload_form').style.display = 'none';
}
else {
document.getElementById('f1_upload_process').style.display = 'none';
document.getElementById('f1_upload_form').style.display = 'none';
document.getElementById('no_upload_form').style.display = '';
}
return true;
}
//-->
</script>
</head>
<style>
#upload_target
{
width: 100%;
height: 80px;
text-align: center;
border: none;
background-color: #642864;
margin: 0px auto;
}
</style>
<body>
<form name="ajaxupload" action="<?php echo "upload.php?img=".$action."&nonce=".$_GET['nonce']; ?>" method="post" enctype="multipart/form-data" target="upload_target" onSubmit="goUpload();" >
<p id="f1_upload_process" style="margin-top: 20px;">Uploading Please wait ...<br/><img src="loader.gif" /><br/></p>
<div id="f1_upload_form" align="left"><!--Select Image You want to upload:-->
<table border="0" cellpadding="0" cellspacing="0"><tr><td><label><input name="myfile" id="myfile" class="textboxStyled" type="file" size="50" onChange="goform();goUpload();" tabindex="2" /></label>
<p id="f1_upload_success" style="display:none; font-weight:bold;">Uploaded Successfully<br></p>
</td><td><!--<a href="javascript: goform()" onClick="goUpload();" tabindex="2"><input type="button"; name="Upload" value="Upload"></a>--></td></tr></table>
</div>
<iframe id="upload_target" name="upload_target" src="#" style="width:0;height:0; border:0; background:#fff;" ></iframe>
</form>
<div id="yesupload" style="display: none;"><center><?php echo mkt_ADMIN_OPTIONS_UPLOAD_LOGO_SUCCESSFUL; ?></br><a href="#" onlcick="reload(<?php echo $_GET['img']; ?>)"><?php echo mkt_ADMIN_OPTIONS_UPLOAD_LOGO_DIFFERENT_IMG; ?></a></center></div>
</body> | {
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} | 9,398 |
Bellator 138's Henry Corrales: Finishing Business
Dan Kuhl June 16, 2015
UFC 189: A Jose Aldo Loss to Conor McGregor Is a Win for the UFC
Matt Quiggins June 8, 2015
Toe-to-Toe: UFC Fight Night 68 Preview and Predictions
Eric Reinert June 2, 2015
Kunlun Fight 80 Results: Kryklia Claims 100+kg Tournament, Kiria Edges Toutouh to Win 2018 70-kilogram Tournament
Zach Aittama February 23, 2019
ONE Fight Night 6: Superbon vs. Allazov Weigh-In Results
Dan Kuhl January 12, 2023
Jimmie Rivera (Dave Mandel/Sherdog)
UFC 203's Jimmie Rivera on UFC Sale: Time Will Tell
Jason Kelly July 18, 2016
Time tells the answer to many of life's questions, including how the $4.2 billion sale of the UFC will affect the brand, the employees and, most importantly, the fighters.
UFC bantamweight Jimmie Rivera is scheduled to fight Urijah Faber at UFC 203 on Aug. 10 at the Quicken Loans Arena in Cleveland. It's the highest profile name Rivera will have faced in his 20-fight career. The 27-year-old combatant is on an 18-fight winning streak, with three of those victories coming inside the Octagon. Now, he's approaching a fight that could drastically raise his stock.
Rivera doesn't look at Faber as a well-known veteran or a former WEC featherweight champ. He doesn't pay attention to the many accolades that have made "The California Kid" a major figure in MMA. Rivera approaches the bout the same as any other. Faber is just a man — a fellow mixed martial artist — standing in the way of Rivera's goals. When the cage door closes, Rivera will do what he's always done in an effort to notch another victory on his nearly flawless record. It's simply another tour of duty in the cage, the same as the rest except that the folks sitting at the top of the UFC corporate ladder will be different.
UFC owners Lorenzo and Frank Ferttita have sold the UFC and will walk away. UFC President Dana White sold his nine percent of the company too, but he will remain in his current position with the UFC. Some fighters might be anticipating big changes in the company that employs them. This may lead to distractions and unnecessary conversations or fogging of their minds, but Rivera looks at the sale of the UFC like it's business as usual — and not the same business as usual White suggested after the purchase of Strikeforce.
"I think everything is going to stay the same," Rivera told Joe Rizzo and Jason Kelly on Rear-Naked Choke Radio. "Listen, Lorenzo is a great businessman, and Dana is as well. Dana is still going to be running it and working. I think everything is going to stay the same, or they're going to try to make it even better.
"They've already been trying to make it even better. They've been trying to make everything as professional as possible, with USADA and drug testing and this and that, the advertising and how they stay on the ball when they change the fights and try to make what's best. They've been around the block, and I don't think anything is going to go wrong.
"We all got an email on Monday morning about it, and they said, rest assured, everything is on a good path. That shows me right there that they do care and they are still looking out for us. I think everything is going to be fine, but only time will tell."
Rumors of a UFC sale have been afloat for years, although many were debunked. The first news story that gave a real indication the world's premier MMA organization was up for grabs came on June 20, when FloCombat's Jeremy Botter revealed what he confirmed with his sources.
The story had such credibility it sparked immediate conversation amongst media and fans, and even prompted the UFC president to address the story. When White shuts down a rumor, there's a good chance there's some truth to it. However, while everyone was weighing in on the topic via social media and debating if there was any veracity to Botter's claim, the fighters under UFC contract remained quiet. But that doesn't mean they weren't speaking to each other about the situation.
Rivera trains with some of the UFC's top-ranked fighters, so you can bet there were conversations going on between them when the microphones were off. The media suggested a potential sale could be a game changer for the fighters, either positively or negatively, but for Rivera and company the talks were not nearly as speculative.
"We talked about it like, 'Wow, the UFC might be for sale. Dana is sticking around, so he's getting a $400 million bonus — that's a nice bonus, I want to get a bonus like that.' I mean, not crazy talk, but it is what it is," Rivera said. "Dana is going to be involved and doing the same job, so unless we start seeing matchmakers go and people start leaving right away, I don't think it will change. Do you fix something that's not broken? If they're smart people, the people that bought it, I don't think they're going to touch anything. They'll maybe try to do some different advertising and marketing, but they're always trying to add something to advertising and marketing. They're always trying to make it bigger and better.
"Right now, I think it's running smooth and running good, but we'll see. Like I said, time will tell."
Dana WhiteJimmie RiveraLorenzo FertittaUFC 203Urijah Faber
Urijah Faber's A1 Combat 6: Talbott vs. Jimenez Photo Gallery
Dan Kuhl October 25, 2022
Urijah Faber's A1 Combat 6 Results: Talbott Stops Jimenez
BKFC 26: Brito vs. Palomino 2 Fight Night Photo Gallery
BKFC 26 Results: Palomino Outworks Brito For Second Title
BKFC 26: Brito vs. Palomino 2 Weigh-In Photo Gallery
BKFC 26: Brito vs. Palomino 2 Press Conference Photo Gallery | {
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} | 2,310 |
{"url":"http:\/\/wallyxie.com\/weblog\/page\/2\/","text":"## Arranging a Beta Distribution into Exponential Family Form\n\nNovember 20th, 2016\n\nA family of PMFs or PDFs is an exponential family if it can be arranged into the form\n\n$f(x|\\theta) = h(x)c(\\theta)\\exp\\left(\\sum_{i=1}^kw_i(\\theta)t_i(x)\\right)$\n\nwhere $\\theta$ is the vector of parameters.\n\nThe Beta distribution PDF takes the form\n\n$f_X(x) = \\frac{x^{\\alpha \u2013 1}(1 \u2013 x)^{\\beta \u2013 1}}{\\textrm{B}(\\alpha, \\beta)}$\n\nGiven that the $\\alpha$ and $\\beta$ parameters are unknown, we now arrange $f_X(x)$ into an exponential family form:\n\n$f(x|\\alpha,\\beta) = \\frac{x^{\\alpha \u2013 1}(1 \u2013 x)^{\\beta \u2013 1}}{\\textrm{B}(\\alpha, \\beta)}$\n$= \\frac{e^{(\\alpha \u2013 1)\\ln(x)}e^{(\\beta \u2013 1)\\ln(1 \u2013 x)}}{\\textrm{B}(\\alpha, \\beta)}$\n$= \\frac{e^{(\\alpha \u2013 1)\\ln(x) + (\\beta \u2013 1)\\ln(1 \u2013 x)}}{\\textrm{B}(\\alpha, \\beta)}$\n$= \\frac{1}{\\textrm{B}(\\alpha, \\beta)}e^{(\\alpha \u2013 1)\\ln(x) + (\\beta \u2013 1)\\ln(1 \u2013 x)}$\n\nThe log identity $x^b = e^{b\\ln(x)}$ is a very useful logarithmic identity to remember when trying to arrange PDFs into exponential family form.\n\nWe observe:\n$h(x) = I_{x \\in (0,1)}(x)$ (If you see that h(x) = 1, that is a cue to use an indicator function that ranges through the support of $x$.)\n$c(k,\\beta) = \\frac{1}{\\textrm{B}(\\alpha, \\beta)$\n$w_1(k,\\beta) = \\alpha \u2013 1$\n$w_2(k,\\beta) = \\beta \u2013 1$\n$t_1(x) = \\ln(x)$\n$t_2(x) = \\ln(1-x)$\n\nHence, the Beta distribution given unknown parameters $\\alpha$ and $\\beta$ is an exponential family with a two-dimensional parameter vector $\\theta$.\n\nA similar process will apply for showing that a Beta PDF with one unknown parameter, $\\beta$ or $\\alpha$ is an exponential family.\n\n## Regarding Identity Politics And Their Guaranteed Persistence\n\nNovember 20th, 2016\n\nI absolutely hate this talk of \u201cwe must end identity politics\u201d that has become pervasive after the election, especially as people are trying to prescribe Democrats solutions for how to recover from this election. First off, how are people defining \u201cidentity politics\u201d? It appears that a lot of people are using it as a blanket term for political organization and mobilization by gender, race, or ethnic group. Identity politics exist and will not go away because your geographic origins, gender, phenotypes, genotypes determine your lot in life, the hand you\u2019ve been dealt, et cetera (insert favored idiom here). Your predetermined traits (like those of an RPG character class) come with a set of pros and cons that relate to your probabilities of success in life as defined by societal and cultural values. People are motivated to maintaining or improving their probabilities, so it is easy to motivate people to advocate for the blocs they belong to. Yes, identity politics make it easy to hate on another group of people as The Others, but it has also mobilized mass sectors of the American population to advocate for egalitarian goals, such as women\u2019s suffrage, desegregation, and elimination of sodomy laws, which allowed members of one American bloc to live significantly more comfortably at little or no cost (and perhaps benefit) to opposing blocs.\n\nAs an alternative, people are promoting an emphasis on \u201cclass politics.\u201d But how are identity politics so different from class politics? You are still slicing a population into sub-groups, and then trying to get them to oppose the interests of another group. Yes, having no hive-mind sub-group jousting would be ideal, but it is a platonic ideal that is impossible given human brain heuristics. Identity politics will survive the 2016 election and persist as humans continue to exist in their current form.\n\n## Arranging a Gamma Distribution Family into Exponential Family Form\n\nNovember 20th, 2016\n\nA family of PMFs or PDFs is an exponential family if it can be arranged into the form\n\n$f(x|\\theta) = h(x)c(\\theta)\\exp\\left(\\sum_{i=1}^kw_i(\\theta)t_i(x)\\right)$\n\nwhere $\\theta$ is the vector of parameters.\n\nNow, the shape-scale parameterization of the Gamma distribution PDF takes the form\n\n$f_X(x) = \\frac{1}{\\Gamma(k)\\beta(k)}x^{k-1}e^{-\\frac{x}{\\beta}}$\n\n$\\beta$ is typically used as the symbol for the rate parameter in the shape-rate parameterization of the Gamma PDF, but the default symbol for the scale parameter, $\\theta$, would conflict with the symbol for our parameter vector.\n\nCan we arrange that PDF into an exponential family form? Spoiler: yes.\n\nHere, we demonstrate that a Gamma PDF given two unknown parameters, $\\beta$ and $k$, is an exponential family.\n\n$f(x|k,\\beta) = \\frac{1}{\\Gamma(k)\\beta(k)}x^{k-1}e^{-\\frac{x}{\\beta}}$\n$= \\frac{1}{\\Gamma(k)\\beta(k)}e^{(k-1)\\ln(x)}e^{-\\frac{x}{\\beta}}$\n$= \\frac{1}{\\Gamma(k)\\beta(k)}e^{(k-1)\\ln(x) \u2013 \\frac{x}{\\beta}}$\n\nThe log identity $x^b = e^{b\\ln(x)}$ is a very useful logarithmic identity to remember when trying to arrange PDFs into exponential family form.\n\nWe observe:\n$h(x) = I_{x>0}(x)$ (If you see that h(x) = 1, that is a cue to use an indicator function that ranges through the support of $x$.)\n$c(k,\\beta) = \\frac{1}{\\Gamma(k)\\beta(k)}$\n$w_1(k,\\beta) = k \u2013 1$\n$w_2(k,\\beta) = -\\frac{1}{\\beta}$\n$t_1(x) = \\ln(x)$\n$t_2(x) = x$\n\nHence, the Gamma distribution given unknown parameters $\\beta$ and $k$ is an exponential family with a two-dimensional parameter vector $\\theta$.\n\nA similar process will apply for showing that a Gamma PDF with one unknown parameter, $\\beta$ or $k$ is an exponential family.\n\n## Thoughts on Lal 2008, \u201cCarbon Sequestration.\u201d\n\nAugust 1st, 2016\n\nFrom\u00a0Phil. Trans. R. Soc. B\n\nCarbon sequestration is defined as the transfer of carbon from the atmospheric carbon pool to other carbon pools. Including the atmospheric pool, there are five carbon pools, with the largest being the oceanic pool at an estimated 38,000 Pg C. The pedologic pool is the third largest at 2500 Pg and further subdivides into the soil organic carbon (SOC) and soil inorganic carbon (SIC) pools.\n\nThere are abiotic techniques for carbon sequestration comprised of\u00a0engineering methods and chemical processes. Many of them consist of injecting carbon into non-atmospheric pools. There are also biotic techniques that rely on organisms, primarily plants and microbiota, for removing\u00a0CO2\u00a0from the atmosphere. Theoretically, abiotic techniques can store more carbon, but there are questions about the safety and reliability of those techniques. The risk of carbon leakage and the effects of leakage on ecosystems is still uncertain. Additionally, the expensive cost of geo-engineering is a limitation. By comparison, biotic techniques are more cost-effective and less risky, while providing accompanying\u00a0benefits such as improved soil and water quality and ecosystem preservation absent from abiotic methods. Biotic approaches do\u00a0have a smaller cumulative carbon sink capacity than abiotic approaches.\n\nBiotic techniques can be subdivided into oceanic sequestration and terrestrial sequestration methods. In terms of terrestrial methods, afforestation in the U.S alone can sink\u00a0up to 117 Tg C per year in the U.S alone (IPCC 1999). The cost of afforestation is the drain on water resources, which can make the practice prohibitive in drought-stricken regions, like California. The family of techniques focusing on SOC and SIC sequestration can also cumulatively sink a significant amount of carbon. Land use conversion and restoration of degraded soils can increase overall microbiota concentrations and diversity in soils. Restoration of degraded soils and habitats in the tropics can potentially sequester an additional 1.1 Pg C per year (Grainger 1995). What constitutes \u201cdegraded lands \u2026 with potential for afforestation and soil quality enhancement\u201d is something I am not clear on, as I have not read the Grainger paper. Moving away from mono to multi-cultures for agricultural crops can mitigate SOC losses and improve the ability of agricultural-use land to sequester carbon.\n\nA lingering question \u2014\u00a0how many acres could be converted from agricultural usages and how much soil could be restored if food waste were more controlled throughout the world?\n\n## Thoughts on Li et al. 2014, \u201cSoil carbon sensitivity to temperature and carbon use efficiency compared across microbial-ecosystem models of varying complexity.\u201d\n\nJuly 9th, 2016\n\nHa, that is a long auto-generated URL. Given what has happened in the past few days, one has to chuckle at and cherish the little harmless things.\n\nI read Li et al. 2014 from Biogeochemistry. This paper compares the output of several Earth system models including the \u201cconventional model,\u201d German (German et al., 2012), AWB (Allison et al., 2010), and MEND (Wang et al., 2013). The models differ in carbon pool structure and interactions, parameter values, and complexity \u2014 German has the fewest parameters and pools, while MEND has the most. The models were simulated under three separate microbial carbon use efficiency (CUE) scenarios. CUE is an important parameter in describing microbial function, and the effect of rising temperatures on the CUE of global microbial populations will be a key determinant of changes to the soil organic carbon (SOC) pool size in the coming century.\n\nThe three CUE scenarios tested were:\n\n1. A constant CUE scenario in which the CUE parameter stayed at 0.31 and did not depend on temperature\n2. A varied CUE scenario in which CUE monotonically decreases with temperature increase\n3. A varied CUE with thermal acclimation\n\nAnother key parameter that CUE depends on in all of the models in Li et al. (with the exception of the conventional model) is m, the CUE temperature response coefficient. CUE is given by\n\nCUE(T) = CUE_{ref} + m * (T \u2013 T_{ref})\n\nwhere T is temperature, CUE_{ref} is a set reference CUE value, and T_{ref} is a set reference temperature, in this 298.15 Kelvin.\n\nModels were simulated at initial temperatures until they reached equilibrium and then perturbed with a 5 degree Celsius temperature increase.\n\nNow, I won\u2019t go into too much detail since I need to go to bed at some point, but there are several results in this paper that piqued my interest. For one, for regions initiated at low temperatures, the German, AWB, and MEND models predicted the decrease of SOC pool sizes. Regions initially seeded at higher temperatures saw smaller SOC losses, or even modest gains. This prediction aligns with experimental results predicting SOC losses in Arctic soils (Xue et al., 2016; Natali et al., 2011). Additionally, the observation of damped oscillations matched my own observations in simulations I have run, which makes sense as interactive coupling between SOC and microbial soil (MBC) pools is reminiscent of that observed in predator-prey models, so I was glad to see that confirmed.\n\nSince I really have to go to bed now, I\u2019ll jump straight to questions and future research directions that this paper has evoked. First, instead of a constant 50% thermal acclimation scenario (where m is halved in comparison to the varied CUE scenario), I wonder how changing m to be a function dependent on time (representing adapting mutations) would change things up. Second, this is a question less related to this paper, but with these Earth system biogeochemistry models, the carbon dioxide flux does not feed back into the pools in any way and is entirely separate from the input. How could the atmospheric carbon pool size be fit into these models? As a person new to this sub-field of Earth system biogeochemistry, I\u2019m wondering why atmospheric carbon is not accounted for as an interactive pool in these models.","date":"2018-04-26 16:58:15","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6347939968109131, \"perplexity\": 1974.6830756701888}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-17\/segments\/1524125948426.82\/warc\/CC-MAIN-20180426164149-20180426184149-00036.warc.gz\"}"} | null | null |
Q: Why does a declared string take an integer In this code I declare conversion as a String. But when I input an integer my program proceeds and it works. I didn't think it should work when i enter an integer (since conversion is declared as a string). Why does this work?
import java.util.*;
// we want to input a number and then convert that number into an int
// Integer.parse(), and then continue
public class Test{
public static void main(String[] args){
Scanner input = new Scanner(System.in);
boolean set = true;
int sum = 0;
while(set != false){
int i = 0;
System.out.println("Please enter a number");
String conversion = input.nextLine();
try{
int x = Integer.parseInt(conversion);
sum += x;
if (x == 0){
System.out.println("Thanks for playing");
System.out.println(sum);
break;
}
}
catch (NumberFormatException e){
break;
}
}
}
}
A: It works because you are converting a string to an int with Integer.parseInt().
| {
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Volume s1-26, Issue 5
The Development of the Virus of Yellow Fever in Haemagogus Mosquitoes 1
Marston Bates, Manuel Roca-García
Affiliations: 1 Villavicencio Field Laboratory, Villavicencio, Colombia
Source: The American Journal of Tropical Medicine and Hygiene, Volume s1-26, Issue 5, 1 Sep 1946, p. 585 - 605
DOI: https://doi.org/10.4269/ajtmh.1946.s1-26.585
VIII. Summary
1. The effect of temperature on virus development in haemagogus mosquitoes was studied by the inoculation in mouse groups of individual mosquitoes at regular intervals after the infectious meal and by the titration of pools of mosquitoes. It was found that there was an initial period of virus loss, followed by a period of virus gain, the rates in both cases depending on the temperature. The period of virus loss lasted for 5 days at 20°, 3–4 days at 25° and 2 days at 30°. At 20° the level of virus seemed to remain stable after the period of loss, there being no demonstrable increase over a 22 day period. At higher temperatures the rate of gain seemed to be a direct function of the environmental temperature.
2. The percentage of mosquitoes becoming infected and the length of the incubation period seemed also to be a function of the amount of virus ingested with the infectious meal. On the basis of titer of virus in circulation at the time of feeding, experiments can be divided into 4 arbitrary categories: trace of virus (serum of source animal not infecting adult mice in dilutions greater than 1:10); small amount of virus (no infections in dilutions above 1:103); moderate amount of virus (no infections in dilutions above 1:105) and large amount of virus (infections in dilution of 1:106 or more). In the first category, virus has in no case been recovered from haemagogus; in the second, occasional individuals become infected; in the third, the majority of the mosquitoes show virus; in the fourth, virus is regularly recovered from 90 per cent or more of the mosquitoes. The minimum incubation period after the ingestion of a "moderate amount of virus" is 13 days at 30°. This may be shortened to 10 days where a "large amount of virus" has been ingested. There is some evidence that infection at a given temperature and given virus dosage depends in part on the characteristics of the individual mosquito.
3. Experiments were undertaken with pantropic virus strains modified by serial passage (6 consecutive passages) in mice. It was difficult to infect haemagogus on saimiri monkeys inoculated with these modified virus strains; with the evidence at hand it is impossible to decide how much this was due to the lower titers of virus circulated by such monkeys, and how much to possible modification of the ability of the virus particles to invade mosquito tissue.
4. Attempts were made to define the incubation period in mosquitoes by large numbers of tests for transmission with individual mosquitoes using 3-day-old mice as test animals. It was found that there was considerable variation among individual mosquitoes of the same lot in the time at which they became infective, but that once a mosquito became infective it remained so for life. The minimum incubation period was found to be 28 days at 25°; 23 days in mosquitoes kept for 20 hours daily at 25° and 4 hours daily at 30°; 12 days for a similar alternation of 25°–35°; and 10 days at a constant temperature of 30°. Results were unsatisfactory at a constant temperature of 35°, but no transmissions were obtained in 28 attempts at periods between 5 and 12 days.
5. The very favorable results obtained with mosquitoes alternated between 25° (20 hours) and 35° (4 hours) suggest that relatively short exposures to high temperatures in nature may greatly accelerate virus development.
/content/journals/10.4269/ajtmh.1946.s1-26.585
http://instance.metastore.ingenta.com/content/journals/10.4269/ajtmh.1946.s1-26.585
10.4269/ajtmh.1946.s1-26.585 | {
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Q: TCL Packages & namespaces I want to use the procedures in a .tcl file which is "available" as a package and a namespace.
I am unable to differentiate b/w 'package require' & 'namespace import'
A: They're completely separate concepts.
A namespace is a code construct for containing commands and variables. And other namespaces. (There's a few other things too, but those are the main ones.)
A package is a collection of files that provides a defined, versioned API. It's an abstraction above just sourceing and loading individual files, which should be considered to be normally just part of the implementation of the package.
It's usually good practice to either keep all commands and variables defined by a package in a namespace with the same name, or for the package to define a single command with the same name as the package. This cuts the amount of confusion! However, this is just evolved good practice and there's many packages that don't work this way, often because they really originally predate the whole package mechanism; changing everything to be hyper-correct with some best practice isn't as good as maintaining compatibility with existing code, of course.
Use namespaces to organize the implementation of your code. Use packages to split the code up into pieces with defined tasks that you can evolve independently. These are totally orthogonal.
| {
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Easton, MD
The Plains, VA
Sotheby's Auction
Sotheby's International Realty® Listing
6909 Blenheim Rd
Scottsville, Virginia, 24590
6909 Blenheim Rd Scottsville, Virginia, 24590 United States
Scottsville, Virginia, 24590 United States
Daniel Heider
Justin DiFranco
Mount Ida Reserve is a once-in-a-lifetime opportunity located just outside the city of Charlottesville, Virginia, within minutes of The University of Virginia. The former estate of Bernard Arnault expands almost 2,500 acres and is located along Blenheim Road, a historic route in Southern Albemarle County - home to Monticello, Ashlawn Highland, Trump Winery, Dave Matthews' Blenheim Vineyards, Grace Vineyard, and Steve Case's Early Mountain Vineyard. The property has undergone an almost two-decade transformation by its current owners. Surrounded by large country estates and wineries, Mount Ida Reserve stands out not only due to its sheer scale (more than twice the size of nearby Trump Winery & Estate) but also from its unique inventory of physical features, including miles of private river frontage, spectacular mountain views, rolling hills, fertile farmland, hardwood forests, and large privately owned lakes. The current stewards have invested nearly $100M in developing the grandest of estates with virtually every amenity. The entire property is connected with high-speed fiberoptic capability. It offers two completely renovated historic manors, a 45-acre lake, two eight stall stables, two tennis courts, an inground pool, a two-lane go-kart track, 180' water slide, several beaches, and miles of fencing, and paved and unpaved trails for horses, ATV's, bicycling, or hiking. The property also boasts several award-winning facilities, including a spectacular 12,000 sq. ft Lodge developed around a 100-year-old stable. Designed with events in mind, the stone and cedar-clad Lodge has state-of-the-art amenities, including a full caterers' kitchen, two large event spaces - fully climate controlled with spectacular outdoor spaces. Yet another event facility, the newly constructed 10,000 sq. ft Event Barn, offers a 25' stone fireplace to complement its two beautifully appointed banquet rooms, all climate controlled with full catering facilities. To complement the event spaces, there is an awe-inspiring 12,000 sq. ft cedar and stone-clad tasting room complete with a full-service restaurant for 250. The facility includes an award-winning 15- barrel craft brewery which utilizes barley from the nearby seven-acre barley field. An oversized outdoor fireplace with waterfalls and stone walls, bluestone walkways, and an observation deck make Mount Ida the most popular of Virginia wineries and breweries. An adjacent 20-acre vineyard (with an additional 150 acres prepared for vineyard expansion allowing for the largest vineyard in Virginia) creates an enchanting setting for guests. The property also includes a portfolio of 19 single-family home rental properties strategically developed by the owners around the perimeter of the property. These four-bedroom 2100 sq. ft homes offer a stable revenue stream while also preserving the neighborhood aesthetic. A 20,000 sq. ft shop, multiple carriage houses, barns, and outbuildings all specifically designed and recently constructed, 15 miles of paved roads, 25 miles of board fencing, five diesel backup generators, 25 springs, and multiple automated wrought iron gated entrances help service and secure the property. The owners have spared no expense investing in infrastructure covering the entire 2,928 acres, including miles of underground power, high-speed fiber optics, roads, bridges, and concrete dams for the 5 lakes. The income-producing estate is perfect for a family compound, corporate retreat, flagship corporate campus, or long-term investment. With 32 finished lots and up to 157 further divisions, the property is uniquely positioned for further development around the existing facilities. Alternatively, the property is a prime candidate for a conservation or stream impact easement offering significant tax benefits. This incomparable property includes a turnkey opportunity complete with well-trained and experienced staff. Additionally, a hotel Study prepared by HVS for 100 key hotel is available for review.
E87W9R
VAAB102254
TTR Sotheby's International Realty
O: +1 703.785.7820
Since 1988, real estate professionals at TTR Sotheby's International Realty have ranked among the highest performing in the nation, and are regarded for their integrity, professional service, and community leadership. Through our relationship with the venerable Sotheby's auction house, TTR Sotheby's International Realty produces extraordinary results for our clients by leveraging proprietary global marketing, cutting edge digital strategies, and unrivaled access to the most affluent and discerning global clientele.
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Sotheby's Conierge Auctions
©2023 TTR Sotheby's International Realty. All rights reserved. | {
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package org.apache.vxquery.xmlquery.query;
import java.io.Reader;
import java.util.ArrayList;
import java.util.List;
import org.apache.vxquery.compiler.CompilerControlBlock;
import org.apache.vxquery.compiler.algebricks.VXQueryBinaryBooleanInspectorFactory;
import org.apache.vxquery.compiler.algebricks.VXQueryBinaryIntegerInspectorFactory;
import org.apache.vxquery.compiler.algebricks.VXQueryComparatorFactoryProvider;
import org.apache.vxquery.compiler.algebricks.VXQueryConstantValue;
import org.apache.vxquery.compiler.algebricks.VXQueryExpressionRuntimeProvider;
import org.apache.vxquery.compiler.algebricks.VXQueryNullWriterFactory;
import org.apache.vxquery.compiler.algebricks.VXQueryPrinterFactoryProvider;
import org.apache.vxquery.compiler.algebricks.prettyprint.VXQueryLogicalExpressionPrettyPrintVisitor;
import org.apache.vxquery.compiler.rewriter.RewriteRuleset;
import org.apache.vxquery.compiler.rewriter.VXQueryOptimizationContext;
import org.apache.vxquery.exceptions.ErrorCode;
import org.apache.vxquery.exceptions.SystemException;
import org.apache.vxquery.metadata.VXQueryMetadataProvider;
import org.apache.vxquery.runtime.provider.VXQueryBinaryHashFunctionFactoryProvider;
import org.apache.vxquery.runtime.provider.VXQueryBinaryHashFunctionFamilyProvider;
import org.apache.vxquery.types.BuiltinTypeRegistry;
import org.apache.vxquery.types.Quantifier;
import org.apache.vxquery.types.SequenceType;
import org.apache.vxquery.xmlquery.ast.ModuleNode;
import org.apache.vxquery.xmlquery.translator.XMLQueryTranslator;
import edu.uci.ics.hyracks.algebricks.common.exceptions.AlgebricksException;
import edu.uci.ics.hyracks.algebricks.common.exceptions.NotImplementedException;
import edu.uci.ics.hyracks.algebricks.common.utils.Pair;
import edu.uci.ics.hyracks.algebricks.compiler.api.HeuristicCompilerFactoryBuilder;
import edu.uci.ics.hyracks.algebricks.compiler.api.ICompiler;
import edu.uci.ics.hyracks.algebricks.compiler.api.ICompilerFactory;
import edu.uci.ics.hyracks.algebricks.compiler.rewriter.rulecontrollers.PrioritizedRuleController;
import edu.uci.ics.hyracks.algebricks.compiler.rewriter.rulecontrollers.SequentialFixpointRuleController;
import edu.uci.ics.hyracks.algebricks.compiler.rewriter.rulecontrollers.SequentialOnceRuleController;
import edu.uci.ics.hyracks.algebricks.core.algebra.base.ILogicalExpression;
import edu.uci.ics.hyracks.algebricks.core.algebra.base.IOptimizationContext;
import edu.uci.ics.hyracks.algebricks.core.algebra.base.LogicalExpressionTag;
import edu.uci.ics.hyracks.algebricks.core.algebra.expressions.ConstantExpression;
import edu.uci.ics.hyracks.algebricks.core.algebra.expressions.IAlgebricksConstantValue;
import edu.uci.ics.hyracks.algebricks.core.algebra.expressions.IExpressionEvalSizeComputer;
import edu.uci.ics.hyracks.algebricks.core.algebra.expressions.IExpressionTypeComputer;
import edu.uci.ics.hyracks.algebricks.core.algebra.expressions.IMergeAggregationExpressionFactory;
import edu.uci.ics.hyracks.algebricks.core.algebra.expressions.INullableTypeComputer;
import edu.uci.ics.hyracks.algebricks.core.algebra.expressions.IVariableTypeEnvironment;
import edu.uci.ics.hyracks.algebricks.core.algebra.metadata.IMetadataProvider;
import edu.uci.ics.hyracks.algebricks.core.algebra.prettyprint.LogicalOperatorPrettyPrintVisitor;
import edu.uci.ics.hyracks.algebricks.core.rewriter.base.AbstractRuleController;
import edu.uci.ics.hyracks.algebricks.core.rewriter.base.IAlgebraicRewriteRule;
import edu.uci.ics.hyracks.algebricks.core.rewriter.base.IOptimizationContextFactory;
import edu.uci.ics.hyracks.algebricks.core.rewriter.base.PhysicalOptimizationConfig;
import edu.uci.ics.hyracks.algebricks.data.ISerializerDeserializerProvider;
import edu.uci.ics.hyracks.algebricks.data.ITypeTraitProvider;
import edu.uci.ics.hyracks.api.dataflow.value.ISerializerDeserializer;
import edu.uci.ics.hyracks.api.dataflow.value.ITypeTraits;
import edu.uci.ics.hyracks.api.job.JobSpecification;
import edu.uci.ics.hyracks.data.std.primitive.VoidPointable;
public class XMLQueryCompiler {
private final XQueryCompilationListener listener;
private final ICompilerFactory cFactory;
private LogicalOperatorPrettyPrintVisitor pprinter;
private ModuleNode moduleNode;
private Module module;
private ICompiler compiler;
private int frameSize;
private String[] nodeList;
public XMLQueryCompiler(XQueryCompilationListener listener, String[] nodeList, int frameSize) {
this(listener, nodeList, frameSize, -1, -1, -1);
}
public XMLQueryCompiler(XQueryCompilationListener listener, String[] nodeList, int frameSize,
int availableProcessors, long joinHashSize, long maximumDataSize) {
this.listener = listener == null ? NoopXQueryCompilationListener.INSTANCE : listener;
this.frameSize = frameSize;
this.nodeList = nodeList;
HeuristicCompilerFactoryBuilder builder = new HeuristicCompilerFactoryBuilder(
new IOptimizationContextFactory() {
@Override
public IOptimizationContext createOptimizationContext(int varCounter,
IExpressionEvalSizeComputer expressionEvalSizeComputer,
IMergeAggregationExpressionFactory mergeAggregationExpressionFactory,
IExpressionTypeComputer expressionTypeComputer, INullableTypeComputer nullableTypeComputer,
PhysicalOptimizationConfig physicalOptimizationConfig) {
return new VXQueryOptimizationContext(varCounter, expressionEvalSizeComputer,
mergeAggregationExpressionFactory, expressionTypeComputer, nullableTypeComputer,
physicalOptimizationConfig, pprinter);
}
});
builder.getPhysicalOptimizationConfig().setFrameSize(this.frameSize);
if (joinHashSize > 0) {
builder.getPhysicalOptimizationConfig().setMaxFramesHybridHash((int) (joinHashSize / this.frameSize));
}
if (maximumDataSize > 0) {
builder.getPhysicalOptimizationConfig().setMaxFramesLeftInputHybridHash(
(int) (maximumDataSize / this.frameSize));
}
builder.getPhysicalOptimizationConfig().setMaxFramesLeftInputHybridHash(
(int) (60L * 1024 * 1048576 / this.frameSize));
builder.setLogicalRewrites(buildDefaultLogicalRewrites());
builder.setPhysicalRewrites(buildDefaultPhysicalRewrites());
builder.setSerializerDeserializerProvider(new ISerializerDeserializerProvider() {
@SuppressWarnings("unchecked")
@Override
public ISerializerDeserializer getSerializerDeserializer(Object type) throws AlgebricksException {
return null;
}
});
builder.setHashFunctionFactoryProvider(VXQueryBinaryHashFunctionFactoryProvider.INSTANCE);
builder.setHashFunctionFamilyProvider(VXQueryBinaryHashFunctionFamilyProvider.INSTANCE);
builder.setTypeTraitProvider(new ITypeTraitProvider() {
@Override
public ITypeTraits getTypeTrait(Object type) {
return VoidPointable.TYPE_TRAITS;
}
});
builder.setPrinterProvider(VXQueryPrinterFactoryProvider.INSTANCE);
builder.setExpressionRuntimeProvider(new VXQueryExpressionRuntimeProvider());
builder.setComparatorFactoryProvider(new VXQueryComparatorFactoryProvider());
builder.setBinaryBooleanInspectorFactory(new VXQueryBinaryBooleanInspectorFactory());
builder.setBinaryIntegerInspectorFactory(new VXQueryBinaryIntegerInspectorFactory());
builder.setExpressionTypeComputer(new IExpressionTypeComputer() {
@Override
public Object getType(ILogicalExpression expr, IMetadataProvider<?, ?> metadataProvider,
IVariableTypeEnvironment env) throws AlgebricksException {
if (expr.getExpressionTag() == LogicalExpressionTag.CONSTANT) {
ConstantExpression ce = (ConstantExpression) expr;
IAlgebricksConstantValue acv = ce.getValue();
if (acv == ConstantExpression.TRUE.getValue() || acv == ConstantExpression.FALSE.getValue()) {
return SequenceType.create(BuiltinTypeRegistry.XS_BOOLEAN, Quantifier.QUANT_ONE);
}
VXQueryConstantValue cv = (VXQueryConstantValue) acv;
return cv.getType();
}
return null;
}
});
builder.setNullableTypeComputer(new INullableTypeComputer() {
@Override
public Object makeNullableType(Object type) throws AlgebricksException {
throw new NotImplementedException("NullableTypeComputer is not implented (makeNullableType)");
}
@Override
public boolean canBeNull(Object type) {
return false;
}
@Override
public Object getNonOptionalType(Object type) {
throw new NotImplementedException("NullableTypeComputer is not implented (getNonOptionalType)");
}
});
builder.setNullWriterFactory(new VXQueryNullWriterFactory());
if (availableProcessors < 1) {
builder.setClusterLocations(VXQueryMetadataProvider.getClusterLocations(nodeList));
} else {
builder.setClusterLocations(VXQueryMetadataProvider.getClusterLocations(nodeList, availableProcessors));
}
cFactory = builder.create();
}
public void compile(String name, Reader query, CompilerControlBlock ccb, int optimizationLevel)
throws SystemException {
moduleNode = XMLQueryParser.parse(name, query);
listener.notifyParseResult(moduleNode);
module = new XMLQueryTranslator(ccb).translateModule(moduleNode);
pprinter = new LogicalOperatorPrettyPrintVisitor(new VXQueryLogicalExpressionPrettyPrintVisitor(
module.getModuleContext()));
VXQueryMetadataProvider mdProvider = new VXQueryMetadataProvider(nodeList, ccb.getSourceFileMap());
compiler = cFactory.createCompiler(module.getBody(), mdProvider, 0);
listener.notifyTranslationResult(module);
XMLQueryTypeChecker.typeCheckModule(module);
listener.notifyTypecheckResult(module);
try {
compiler.optimize();
} catch (AlgebricksException e) {
throw new SystemException(ErrorCode.SYSE0001, e);
}
listener.notifyOptimizedResult(module);
JobSpecification jobSpec;
try {
jobSpec = compiler.createJob(null, null);
jobSpec.setFrameSize(frameSize);
} catch (AlgebricksException e) {
throw new SystemException(ErrorCode.SYSE0001, e);
}
module.setHyracksJobSpecification(jobSpec);
listener.notifyCodegenResult(module);
}
public ModuleNode getModuleNode() {
return moduleNode;
}
public Module getModule() {
return module;
}
private static List<Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>> buildDefaultLogicalRewrites() {
List<Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>> defaultLogicalRewrites = new ArrayList<Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>>();
SequentialFixpointRuleController seqCtrlNoDfs = new SequentialFixpointRuleController(false);
SequentialFixpointRuleController seqCtrlFullDfs = new SequentialFixpointRuleController(true);
SequentialOnceRuleController seqOnceCtrl = new SequentialOnceRuleController(true);
PrioritizedRuleController priorityCtrl = new PrioritizedRuleController();
defaultLogicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(priorityCtrl,
RewriteRuleset.buildPathStepNormalizationRuleCollection()));
defaultLogicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(priorityCtrl,
RewriteRuleset.buildXQueryNormalizationRuleCollection()));
defaultLogicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(seqCtrlFullDfs,
RewriteRuleset.buildInlineRedundantExpressionNormalizationRuleCollection()));
defaultLogicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(priorityCtrl,
RewriteRuleset.buildNestedDataSourceRuleCollection()));
defaultLogicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(seqOnceCtrl,
RewriteRuleset.buildTypeInferenceRuleCollection()));
defaultLogicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(seqCtrlFullDfs,
RewriteRuleset.buildNormalizationRuleCollection()));
defaultLogicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(seqCtrlNoDfs,
RewriteRuleset.buildCondPushDownRuleCollection()));
defaultLogicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(seqCtrlNoDfs,
RewriteRuleset.buildJoinInferenceRuleCollection()));
defaultLogicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(seqCtrlNoDfs,
RewriteRuleset.buildOpPushDownRuleCollection()));
defaultLogicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(seqOnceCtrl,
RewriteRuleset.buildDataExchangeRuleCollection()));
defaultLogicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(seqCtrlNoDfs,
RewriteRuleset.buildConsolidationRuleCollection()));
return defaultLogicalRewrites;
}
private static List<Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>> buildDefaultPhysicalRewrites() {
List<Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>> defaultPhysicalRewrites = new ArrayList<Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>>();
SequentialOnceRuleController seqOnceCtrlAllLevels = new SequentialOnceRuleController(true);
SequentialOnceRuleController seqOnceCtrlTopLevel = new SequentialOnceRuleController(false);
defaultPhysicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(seqOnceCtrlAllLevels,
RewriteRuleset.buildPhysicalRewritesAllLevelsRuleCollection()));
defaultPhysicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(seqOnceCtrlTopLevel,
RewriteRuleset.buildPhysicalRewritesTopLevelRuleCollection()));
defaultPhysicalRewrites.add(new Pair<AbstractRuleController, List<IAlgebraicRewriteRule>>(seqOnceCtrlAllLevels,
RewriteRuleset.prepareForJobGenRuleCollection()));
return defaultPhysicalRewrites;
}
} | {
"redpajama_set_name": "RedPajamaGithub"
} | 1,310 |
\section{Experimental Details: Solving PDEs}
For our PDE experiments, we use the FNO code and setup \citep{li2021fno} provided here: \url{https://github.com/zongyi-li/fourier_neural_operator}. We use the same training routine and settings as the backbone architecture for each task and only tune the architecture optimizer. We consider the following hyperparameters for the architecture optimizer: Adam vs. SGD (with or without momentum), initial learning rate, and number of warmup epochs. The final hyperparameters for each task can be found in Table~\ref{app:tab:pdeopt}.
Our CNN backbone is analogous to the FNO architecture used for each problem. In particular, the CNN backbone architecture used for each task is simply the FNO architecture where FNO layers of dimension $N$ with $m$ modes are replaced by $N$-dimensional convolutional layers with filters of size $(m+1)^N$ and circular padding to match the dimensionality of FNO. In Table~\ref{app:tab:pde1dres} and Table~\ref{app:tab:pde2dres} we present reported \cite{li2021fno}, reproduced, and our own results on the 1d Burgers' equation and 2d Darcy Flow.
For AutoDL we use the code and setup provided here: \url{https://github.com/NoamRosenberg/autodeeplab}.
We only conduct search on the lowest resolution and use the resulting architecture at higher resolutions.
Search was conducted for 40 epochs, as in the original paper, and the search learning rate was tuned.
\begin{table}[!h]
\centering
\begin{threeparttable}
\caption{\label{app:tab:pdeopt}
Architecture optimizer settings on PDE tasks. Note that the step-size is updated using the same schedule as the backbone.
}
\begin{tabular}{lccc}
\hline
task & optimizer & initial step-size & warmup epochs \\
\hline
1d Burgers' equation & Adam & 1E-3 & 0 \\
1d Burgers' equation (FNO init) & Momentum(0.5) & 1E-4 & 250 \\
2d Darcy Flow & Momentum(0.5) & 1E-1 & 0 \\
2d Darcy Flow (FNO init) & Momentum(0.5) & 1E-1 & 0 \\
2d Navier Stokes ($\nu=10^{-4}, T=30$) & Momentum(0.5) & 5E-3 & 0 \\
2d Navier Stokes ($\nu=10^{-5}, T=20$) & Momentum(0.5) & 1E-3 & 0 \\
\hline
\end{tabular}
\end{threeparttable}
\end{table}
\begin{table*}[!h]
\centering
\begin{threeparttable}
\caption{\label{app:tab:pde1dres}
Test relative errors on the 1d Burgers' equation. We were not able to match the FNO-1d results reported by the authors \citep{li2021fno} using their published codebase, however, our proposed XD operations outperform our reproduction of their results at every resolution. Furthermore, we outperform their reported test relative errors on every resolution except $s=4096$, where we roughly match their performance.
}
\begin{tabular}{lcccccr}
\hline
Method (source) & $s=256$ & $s=512$ & $s=1024$ & $s=2048$ & $s=4096$ & $s=8192$ \\
\hline
NN \citep{li2021fno} & 0.4714 & 0.4561 & 0.4803 & 0.4645 & 0.4779 & 0.4452 \\
GCN \citep{li2021fno} & 0.3999 & 0.4138 & 0.4176 & 0.4157 & 0.4191 & 0.4198 \\
FCN \citep{li2021fno} & 0.0958 & 0.1407 & 0.1877 & 0.2313 & 0.2855 & 0.3238 \\
PCANN \citep{li2021fno} & 0.0398 & 0.0395 & 0.0391 & 0.0383 & 0.0392 & 0.0393 \\
GNO \citep{li2021fno} & 0.0555 & 0.0594 & 0.0651 & 0.0663 & 0.0666 & 0.0699 \\
LNO \citep{li2021fno} & 0.0212 & 0.0221 & 0.0217 & 0.0219 & 0.0200 & 0.0189 \\
MGNO \citep{li2021fno} & 0.0243 & 0.0355 & 0.0374 & 0.0360 & 0.0364 & 0.0364 \\
FNO-1d \citep{li2021fno} & 0.0149 & 0.0158 & 0.0160 & 0.0146 & \textbf{0.0142} & 0.0139 \\
\hline
CNN (ours) & 0.0518 & 0.1220 & 0.1830 & 0.2280 & 0.2730 & 0.2970 \\
FNO-1d (reproduced) & 0.0181 & 0.0191 & 0.0188 & 0.0184 & 0.0183 & 0.0183 \\
CNN XD (ours) & \textbf{0.0141} & \textbf{0.0079} & \textbf{0.0154} & \textbf{0.0099} & 0.0145 & \textbf{0.0123} \\
FNO-1d XD (ours) & 0.0153 & 0.0154 & 0.0154 & 0.0167 & 0.0160 & 0.0155 \\
\hline
\end{tabular}
\end{threeparttable}
\end{table*}
\begin{table*}[!h]
\centering
\begin{threeparttable}
\caption{\label{app:tab:pde2dres}
Test relative errors on 2d Darcy Flow. Our reproduction of the FNO-2d results outperform those reported by the authors \citep{li2021fno}. Nonetheless, our proposed XD operations outperform both our reproduction and the reported results at every resolution.
}
\begin{tabular}{lccccr}
\hline
Method (source) & $s=85$ & $s=106$ & $s=141$ & $s=211$ & $s=421$ \\
\hline
NN \citep{li2021fno} & 0.1716 & - & 0.1716 & 0.1716 & 0.1716 \\
GCN \citep{li2021fno} & 0.0253 & - & 0.0493 & 0.0727 & 0.1097 \\
FCN \citep{li2021fno} & 0.0299 & - & 0.0298 & 0.0298 & 0.0299 \\
PCANN \citep{li2021fno} & 0.0244 & - & 0.0251 & 0.0255 & 0.0259 \\
GNO \citep{li2021fno} & 0.0346 & - & 0.0332 & 0.0342 & 0.0369 \\
LNO \citep{li2021fno} & 0.0520 & - & 0.0461 & 0.0445 & - \\
MGNO \citep{li2021fno} & 0.0416 & - & 0.0428 & 0.0428 & 0.0420 \\
FNO-2d \citep{li2021fno} & 0.0108 & - & 0.0109 & 0.0109 & 0.0098 \\
\hline
CNN (ours) & 0.0404 & 0.0495 & 0.0613 & 0.0813 & 0.1150 \\
FNO-2d (reproduced) & 0.0096 & 0.0092 & 0.0091 & 0.0091 & 0.0091 \\
CNN XD (ours) & \textbf{0.0065} & \textbf{0.0065} & \textbf{0.0065} & \textbf{0.0071} & \textbf{0.0066} \\
FNO-2d XD (ours) & 0.0082 & 0.0079 & 0.0077 & 0.0076 & 0.0074 \\
\hline
\end{tabular}
\end{threeparttable}
\end{table*}
\begin{figure}[!h]
\includegraphics[width=\linewidth]{figures/pde_2d_trace.pdf}
\caption{Training curves (dotted) and test curves (solid) on Darcy Flow at resolution 141, showing better generalization of XD-operations.}
\end{figure}
\section{Experimental Details: Protein Folding}
\begin{table}[!h]
\centering
\begin{threeparttable}
\caption{\label{app:tab:proteinopt}
Architecture optimizer settings on for our protein folding experiments, across different ResNet depths. Note that the same step-size is used throughout since the backbone has no step-size schedule.
}
\begin{tabular}{lccc}
\hline
search space & optimizer & step-size & warmup epochs \\
\hline
ResNet-4 XD & Adam & 1E-4 & 2 \\
ResNet-6 XD & Momentum(0.99) & 1E-4 & 2 \\
ResNet-10 XD & Momentum(0.99) & 1E-3 & 2 \\
ResNet-18 XD & Momentum(0.9) & 5E-4 & 2 \\
ResNet-34 XD & Momentum(0.9) & 5E-4 & 2 \\
\hline
\end{tabular}
\end{threeparttable}
\end{table}
\begin{table}[!h]
\centering
\begin{threeparttable}
\caption{\label{app:tab:proteinperf}
Test MAE$_8$ of the Dilated ResNet of \cite{adhikari2020}, compared to a standard ResNet backbone and XD-operations applied to ResNet. Results are averaged over 3 trials.
}
\begin{tabular}{lccccc}
\hline
Method & depth $ = 4$ & depth $ = 6$ & depth $ = 10$ & depth $ = 18$ & depth $ = 34$ \\
\hline
ResNet & $5.99\pm0.43$ & $5.30\pm0.11$ & $4.91\pm0.25$ & $4.80\pm0.07$ & $4.66\pm0.15$ \\
Dilated ResNet & $6.04\pm0.33$ & $5.49\pm0.02$ & $4.64\pm0.08$ & $4.59\pm0.22$ & $4.50\pm0.13$ \\
ResNet XD & ${\bf5.59\pm0.09}$ & ${\bf4.59\pm0.17}$ & ${\bf4.25\pm0.16}$ & ${\bf4.22\pm0.03}$ & ${\bf4.00\pm0.07}$ \\
\hline
\end{tabular}
\end{threeparttable}
\end{table}
For our protein folding experiments, our code is a PyTorch re-implementation of the PDNET code and setup \cite{adhikari2020} provided here: \url{https://github.com/ba-lab/pdnet}. As before, we use the same training routine and settings as the Dilated ResNet architecture and only tune the architecture optimizer. We consider the following hyperparameters for the architecture optimizer: Adam vs. SGD (with or without momentum), learning rate, and number of warmup epochs. The final hyperparameters for each depth can be found in Table~\ref{app:tab:proteinopt}. Our ResNet backbone differs from Dilated ResNet in that its dilation rate is set to 1 in every convolutional layer. In Table~\ref{app:tab:proteinperf}, we present average MAE$_8$ on the PSICOV test set for each method at each depth.
\section{Experimental Details: Music Modeling and Sequence Modeling}
\begin{table}[!h]
\centering
\begin{threeparttable}
\caption{\label{app:tab:seq}
Architecture optimizer settings on sequence modeling tasks.
Note that the step-size is updated using the same schedule as the backbone.
}
\begin{tabular}{lccc}
\hline
task & optimizer & initial step-size & warmup epochs \\
\hline
Permuted MNIST & Adam & 2E-4 & 0 \\
JSB Chorales & Adam & 2E-4 & 25 \\
Nottingham & Adam & 2E-3 & 0 \\
Penn Treebank & Adam & 2E-6 & 0 \\
\hline
\end{tabular}
\end{threeparttable}
\end{table}
\begin{table*}[!h]
\centering
\begin{threeparttable}
\caption{\label{app:tab:seqres}
XD-operations applied to TCNs compared to recent empirical results in sequence modeling.
Our results are averages of three trials.
Methods achieving within one deviation of the best performance are {\bf bolded}.
}
\footnotesize
\begin{tabular}{lcccc}
\hline
& Permuted MNIST$^\ast$ & JSB Chorales & Nottingham & Penn Treebank \\
Method (source) & (error) & (loss) & (loss) & (perplexity) \\
\hline
LSTM \citep{bai2018tcn} & 14.3 & 8.45 & 3.29 & 78.93 \\
GRU \citep{bai2018tcn} & 12.7 & 8.43 & 3.46 & 92.48 \\
RNN \citep{bai2018tcn} & 74.7 & 8.91 & 4.05 & 114.50 \\
TCN backbone \citep{bai2018tcn} & 2.8 & 8.10 & 3.07 & 88.68 \\
TrellisNet \citep{bai2019trellis} & 1.87 & - & - & {\bf 54.19} \\
R-Transformer \citep{wang2020rtransformer} & - & - & {\bf 2.37} & 84.38 \\
HiPPO-LegS \citep{gu2020hippo} & {\bf 1.7} & - & - & - \\
\hline
TCN backbone (reproduced) & $2.89\pm0.04$ & $8.17\pm0.01$ & $2.97\pm0.01$ & $88.49\pm0.31$ \\
TCN backbone XD (ours) & ${\bf 1.75\pm0.11}$ & ${\bf 8.07\pm0.02}$ & $2.81\pm0.05$ & $84.11\pm0.25$ \\
Undilated TCN (ours) & $11.3\pm2.1$ & $8.16\pm0.04$ & $3.21\pm0.02$ & $94.30\pm0.33$ \\
Undilated TCN XD (ours) & ${\bf 1.77\pm0.10}$ & ${\bf 8.07\pm0.01}$ & $2.84\pm0.02$ & $85.04\pm0.49$ \\
\hline
\end{tabular}
\begin{tablenotes}\footnotesize
\item[$\ast$] We use depth $\*d=(3,3,3)$ XD-operations for permuted MNIST experiments;
elsewhere we use~$(1,3,1)$.
Results within a standard deviation of the best are {\bf bolded}.
\end{tablenotes}
\end{threeparttable}
\end{table*}
For our sequence modeling experiments we use the TCN code \citep{bai2018tcn} provided here: \url{https://github.com/locuslab/TCN}.
As before we use the same settings and training routine as the backbone for all tasks, tuning only the architecture optimizer.
The specific settings are provided in Table~\ref{app:tab:seq}.
For both the baselines and XD-operations we use the same optimizer settings for both the dilated and undilated TCN backbones.
In Table~\ref{app:tab:seqres} we present results for both music modeling and for two additional benchmarks---permuted MNIST and Penn Treebank---on which we see a similar pattern of XD-operations being able to recover and even beat (dilated) TCN performance starting from an undilated network.
\section{Finding and Evaluating XD-Operations}\label{sec:chrysalis}
\vspace{-1mm}
This section outlines a simple procedure that we use to evaluate XD-operations.
Recall that NAS methods specify architectures by assigning operations to each edge $(u,v,\Op)$ of a computational graph.
We aim to simultaneously find good operations and model weights, a goal distinct from the classic {\em two-stage} NAS formulation, which finds assignments in an initial search phase before training the resulting architecture from scratch \citep{ying2019nasbench101}.
However, the use of weight-sharing \citep{pham2018enas} extends NAS to {\em one-shot} objectives where weights and architectures are jointly optimized.
Under weight-sharing, architecture parameters become weights in a larger ``supernet,'' extending the hypothesis class \citep{li2021gaea}.
To assess XD-operations directly we assume the user provides a starter network with existing edge labels $\Op_{u,v}$ as a backbone.
We transform this into a weight-sharing supernet by reparameterizing each operation $\Op_{u,v}$ as an XD-operation $\XD_{a_{u,v}}$ with architecture parameter $a_{u,v}$.
Then we simultaneously train both $a_{u,v}$ and the model weights $\*w_{u,v}$ associated with each edge as follows:
\begin{itemize}[leftmargin=*,topsep=-1pt,noitemsep]\setlength\itemsep{2pt}
\item {\bf Architecture parameters} $a_{u,v}$ are initialized using the original operation used by the CNN backbone by setting $\Op_{u,v}=\XD_{a_{u,v}}$;
$a_{u,v}$ is then updated via SGD or Adam \citep{kingma2015adam}.
We tune step-size, momentum, and the number of ``warmup'' epochs:
initial epochs during which only model weights $\*w_{u,v}$ are updated.
This can be viewed as a specialized step-size schedule.
\item {\bf Model weights} $\*w_{u,v}$ are initialized and updated using the routine provided with the backbone.
\end{itemize}
\vspace{10mm}
This approach allows us to use established topologies and optimizers while searching for new operations, thus aligning with the goal for Sections~\ref{sec:pde}, \ref{sec:protein}, and~\ref{sec:seq}:
to improve upon the CNN backbones that practitioners often use as a first attempt.
As a simple example, we start by applying the procedure to image classification.
Since this is not the main objective of our work, we treat it as a warmup and consider two datasets:
CIFAR-10 and a variant where the images' rows and columns are permuted.
On CIFAR-10 we do {\em not} expect to see much improvement from XD-operations over the CNN backbone used to initialize search, as convolutions are already the ``right'' operation for images.
On the other hand, the ``right'' operation on permuted data, at least in layer one, is an inverse permutation followed by convolution;
as this is an XD-operation\footnote{Recall $\mathcal S_\XD$ includes compositions of convolutions with multiplication by a K-matrix, e.g. a permutation.}, here we do hope to see improvement.
Using LeNet \citep{lecun1999lenet} and ResNet-20 \citep{he2016resnet} as backbones, we compare applying our algorithm to XD-operations with two baselines:
(1) using just the backbone CNN and (2) applying a similar method to the relaxed set $\tilde\mathcal S_\mathbf{discrete}$ of DARTS operations from Section~\ref{sec:relax}.
To optimize over $\tilde\mathcal S_\mathbf{discrete}$ we take an approach similar to DARTS:
parameterize the simplex using a softmax and apply Adam.
We experiment with both a uniform initialization and one biased towards the backbone's operation.
While both $\mathcal S_\XD$ and $\mathcal S_\mathbf{discrete}$ contain LeNet's $\Conv_{5\times 5}$ and ResNet's $\Conv_{3\times 3}$ and $\Id$, for LeNet's $\MaxP_{3\times 3}$ layer we initialize with the closest operation.
For direct comparison, both search spaces employ weights with maximum filter size $5\times 5$ and for both we evaluate the shared weights rather than retraining, which we find hurts $\tilde\mathcal S_\mathbf{discrete}$.
We set the XD-operations' depth to $\*d=\*3_3$ to express the dilated convolutions in $\mathcal S_\mathbf{discrete}$ and convolutions composed with permutations.
In Table~\ref{tab:cifar}, we see that while both the relaxed discrete NAS operations and XD-operations perform comparably on regular images, XD-operations achieve around 15\% better accuracy with both backbones when the images are permuted.\footnote{Full accuracy can be recovered via an auxiliary loss encouraging permutation-like K-matrices \citep{dao2020kaleidoscope}.}
Note that even networks obtained by running state-of-the-art NAS procedures such as GAEA PC-DARTS \citep{li2021gaea} and DenseNAS \citep{fang2020densenas} on permuted CIFAR-10 achieve only 66.3\% and 61.6\% accuracy, respectively, despite using millions more parameters than ResNet-20.
While it is not straightforward to understand the recovered XD-operations that perform so well, we can use the relative Euclidean distance of their architecture parameters from initialization as a proxy for novelty;
in Figure~\ref{fig:div} we see that on regular images our procedure finds operations that are quite similar to convolutions, but on permuted data they are much further away.
These results show that to enable NAS on diverse data, we will need a search space that contains truly novel operations, not just combinations of existing ones.
In the remainder of the paper, we study more diverse and realistic tasks that show further evidence that $\mathcal S_\XD$ is a strong candidate for this.
\begin{figure}[!t]
\begin{minipage}{0.6\linewidth}
\begin{threeparttable}[H]
\captionof{table}{\label{tab:cifar}
Search space comparison on CIFAR-10.
Validation accuracies are averages of three trials.
While we use small CNNs for exploration, XD-operations can also be used with high-performance backbones to obtain $>95\%$ accuracy (c.f. the appendix).
\vspace{-4pt}
}
\begin{tabular}{lccc}
\toprule
{\bf Backbone} & & Permuted & Cost \\
~~search space & CIFAR-10 & CIFAR-10$^\ast$ & (hours$^\dagger$) \\
\midrule
{\bf LeNet} & $75.5\pm0.1$ & $43.7\pm0.5$ & 0.3 \\
~~$\tilde\mathcal S_\mathbf{discrete}$ & $75.6\pm3.4$ & $47.7\pm1.0$ & 1.0 \\
~~$\mathcal S_\XD$ & $77.7\pm0.7$ & $63.0\pm1.0$ & 0.9 \\
\midrule
{\bf ResNet-20} & $91.7\pm0.2$ & $58.6\pm0.7$ & 0.6 \\
~~$\tilde\mathcal S_\mathbf{discrete}$ & $92.7\pm0.2$ & $58.0\pm1.0$ & 5.3 \\
~~$\mathcal S_\XD$ & $92.4\pm0.2$ & $73.5\pm1.6$ & 5.6 \\
\bottomrule
\end{tabular}
\begin{tablenotes}\footnotesize
\item[$\ast$] No data augmentation used in the permuted case.
\end{tablenotes}
\end{threeparttable}
\end{minipage}
\hfill
\begin{minipage}{0.4\linewidth}
\centering
\includegraphics[width=\linewidth]{figures/divplot.pdf}
\vspace{-3pt}
\caption{\label{fig:div}
On permuted images, where convolutions are not the ``right'' operation, we find XD-operations that are farther away from the operations of the initial CNN backbone.
}
\end{minipage}
\end{figure}
\section{Experimental Details: CIFAR-10 and Permuted CIFAR-10}
\begin{table}[!h]
\centering
\begin{threeparttable}
\caption{\label{app:tab:cifar}
Architecture optimizer settings on CIFAR-10 tasks.
Note that the step-size is updated using the same schedule as the backbone.
}
\footnotesize
\begin{tabular}{cclcccc}
\hline
search space & backbone & task & optimizer & initial step-size & warmup epochs & perturb \\
\hline
\multirow{4}{*}{$\tilde\mathcal S_\mathbf{discrete}$}
& \multirow{2}{*}{LeNet}
& CIFAR-10 & Adam & 1E-1 & 0 & 0.1 \\
&& Permuted & Adam & 1E-1 & 50 & 0.875 \\
\cline{2-7}
& \multirow{2}{*}{ResNet-20}
& CIFAR-10 & Adam & 1E-3 & 0 & 0.1 \\
&& Permuted & Adam & 1E-1 & 0 & 0.875 \\
\hline
\multirow{4}{*}{$\mathcal S_\XD$}
& \multirow{2}{*}{LeNet}
& CIFAR-10 & Adam & 1E-4 & 0 & - \\
&& Permuted & Adam & 1E-3 & 0 & -\\
\cline{2-7}
& \multirow{2}{*}{ResNet-20}
& CIFAR-10 & Adam & 1E-4 & 50 & - \\
&& Permuted & Adam & 1E-3 & 0 & - \\
\hline
\end{tabular}
\end{threeparttable}
\end{table}
For our experiments with image classification backbones we use the standard CIFAR-10 data \citep{krizhevsky2009cifar} and a permuted version where all rows and columns are identically permuted.
For unpermuted data we use standard data augmentation \citep{he2016resnet} while for permuted data we do not use any data augmentation.
As specified in Section~\ref{sec:chrysalis}, we keep the training routine of the model weights the same and tune only the architecture optimizer, the settings of which are specified in Table~\ref{app:tab:cifar}.
Note that for the DARTS operation space we specify a ``perturb'' parameter that specifies how unbiased the initial architecture parameters are towards the backbone operation;
specifically, we initialize architecture parameters so as to assign one minus this quantity as the weight to the backbone operation, so 0.875 means the initialization is uniform (since $|\tilde\mathcal S_\mathbf{discrete}|=8$) while 0.1 means the backbone operation is assigned 0.9 of the weight.
\subsection{LeNet}
The LeNet backbone we consider consists of two $\Conv_{5\times 5}$ layers, each followed by $\MaxP_{2\times 2}$, and two fully connected layers.
When warm-starting with XD-operations we use $\AvgP_{2\times2}$ instead of $\MaxP_{2\times2}$, while when warm-starting with the DARTS operations we use $\MaxP_{3\times 3}$.
For the baseline training routine we use 200 epochs of Momentum(0.9), with the first 100 at learning rate 0.01, the next 50 at 0.005, and the last 50 at 0.001.
\subsection{ResNet-20}
We use the implementation and training routine provided here: \url{https://github.com/akamaster/pytorch_resnet_cifar10}.
When replacing operations in the backbone we substitute for both the $\Conv_{3\times3}$ operations and the skip-connections $\Id$;
some of the latter are downsampled, which XD-operations can handle as strides.
\subsection{WideResNet-40-4}
We use the same implementation as for ResNet-20 but adapt the original WRN training routine \citep{zagoruyko2016wideresnet}, except with weight-decay set to $10^{-4}$ (as in ResNet-20);
on the regular CIFAR-10 tasks this does not seem to affect performance.
To conserve computation and memory, we do not tune the architecture optimizer parameters here and simply use the same ones used for ResNet-20;
furthermore, we fix the channel and bias parameters of XD-operations and do not allow the kernel size to be larger the $3\times 3$.
Because of these modifications, we only use our evaluation here as a sanity check for large-network performance of XD-operations and do not include it in the main results.
\newpage
\subsection{DARTS Cell Search}
To search the full DARTS search space, which is a standard NAS benchmark, we use GAEA PC-DARTS, a recent state-of-the-art method \citep{li2021gaea}, using code made available by the authors here: \url{https://github.com/liamcli/gaea_release}.
On CIFAR-10 we simply use their best reported cell but evaluate it using the ``base" routine \citep{yang2020nas}, i.e. without auxiliary losses or additional data augmentation;
this is to obtain fair comparison with the other backbone models.
Note that the model is still much larger and the training routine much more intensive.
On permuted data we follow the standard three-stage pipeline in which we run search four times, train all four found cells and select the best one, and finally train that cell multiple times.
\subsection{DenseNAS Search}
We use the DenseNAS search and evaluation code released by the authors here: \url{https://github.com/JaminFong/DenseNAS}.
While the search space is designed for ImageNet \citep{russakovsky2015imagenet}, we adapt it to CIFAR-10 by taking the DenseNAS-R1 setting and downscale the input sizes to match 32x32 images used.
\begin{table}[!t]
\centering
\begin{threeparttable}
\captionof{table}{\label{app:tab:cifarres}
Search space comparison on CIFAR-10.
Validation accuracies are averages of three trials.
}
\begin{tabular}{lcccc}
\hline
Backbone & Search Space & CIFAR-10 & Permuted$^\ast$ & Cost (hours$^\dagger$) \\
\hline
\multirow{3}{*}{LeNet} & backbone & $75.5\pm0.1$ & $43.7\pm0.5$ & 0.3 \\
& $\tilde\mathcal S_\mathbf{discrete}$ & $75.6\pm3.4$ & $47.7\pm1.0$ & 1.0 \\
& $\mathcal S_\XD$ & $77.7\pm0.7$ & $63.0\pm1.0$ & 0.9 \\
\hline
\multirow{3}{*}{ResNet-20} & backbone & $91.7\pm0.2$ & $58.6\pm0.7$ & 0.6 \\
& $\tilde\mathcal S_\mathbf{discrete}$ & $92.7\pm0.2$ & $58.0\pm1.0$ & 5.3 \\
& $\mathcal S_\XD$ & $92.4\pm0.2$ & $73.5\pm1.6$ & 5.6 \\
\hline
\multirow{3}{*}{WRN-40-4} & backbone & $95.2\pm0.1$ & $64.7\pm0.9$ & 4.6 \\
& $\tilde\mathcal S_\mathbf{discrete}$ & $95.2\pm0.2$ & $61.3\pm1.3$ & 19.9 \\
& $\mathcal S_\XD$ & $95.0\pm0.1$ & $72.9\pm0.8$ & 14.3 \\
\hline
ResNet-18 & DenseNAS & $94.5\pm0.3$ & $61.6\pm3.3$ & 3.6 \\
Cell & DARTS$^\ddagger$ & $96.0\pm0.2$ & $66.3\pm0.5$ & 28.6 \\
\hline
\end{tabular}
\begin{tablenotes}
\item[$\ast$] No data augmentation used in the permuted case.
\item[$\dagger$] On a V100 GPU; time for DARTS Cell is training cost only.
\item[$\ddagger$] Search using GAEA PC-DARTS \citep{li2021gaea}; training using ``base'' routine~\citep{yang2020nas}.
\end{tablenotes}
\end{threeparttable}
\end{table}
\section{Practical Complexity of XD-Operations}
\begin{table}[!h]
\begin{threeparttable}
\captionof{table}{\label{app:tab:complexity}
Comparison of the computational and memory costs of XD-operations when substituted for convolutions.
For simplicity, we consider cases with 2d inputs and where the channel and bias parameters are fixed.
}
\begin{tabular}{lcccccccc}
\hline
& input & kernel & \multicolumn{2}{c}{minutes / epoch} & \multicolumn{2}{c}{memory (Gb)} & \multicolumn{2}{c}{param. ($\times10^6$) }\\
Task (backbone) & size & size & $\Conv$ & $\XD$ & $\Conv$ & $\XD$ & $\Conv$ & $\XD$ \\
\hline
CIFAR-10 (WRN-40-4) & $32$ & $3$ & 1.4 & 4.3 & 3.73 & 15.6 & 8.96 & 9.08 \\
Darcy Flow (Conv4$^\ast$) & $85$ & 13 & 0.028 & 0.14 & 4.51 & 5.53 & 0.701 & 0.744 \\
PSICOV (ResNet-18) & $128$ & 3 & 5.9 & 11 & 1.50 & 10.7 & 0.038 & 0.549 \\
\hline
\end{tabular}
\begin{tablenotes}
\item[$\ast$] Four-layer convolutional network with parameterized skip (shortcut) connections derived from the FNO network \citep{li2021fno} as described in Section~\ref{sec:pde}.
\end{tablenotes}
\end{threeparttable}
\end{table}
In this section we report a detailed comparison of computational costs of the XD-operation compared to a convolution;
this is presented in Table~\ref{app:tab:complexity}.
Due to their familiarity, we present results for tasks that have 2d inputs and thus use 2d convolutions in their default backbone.
Note that since XD-operations are more general than convolutions, they must by definition be at least as expensive as convolutions in both computation and memory.
While in this paper our focus is on absolute performance using learning metrics (e.g. test error), we view finding a good tradeoff between the performance of XD-operations on certain tasks and convolutions, for example by restricting the expressivity of XD-operations, as important directions for future work.
\section{Conclusion}\label{sec:conc}
\vspace{-1mm}
This work aims to transition NAS from combining existing operations designed for vision and text to finding novel and effective operations in many domains.
To do so we introduced a new search space of XD-operations and demonstrated its effectiveness on diverse tasks.
Combining XD-operations with standard topology-search NAS, warm-starting search from non-standard operations such as graph convolutions and FNOs,\footnote{
In this direction, we found that initializing XD with FNO did {\em worse} than initializing with convolutions on Burgers' equation and Darcy Flow, a surprising result given how much better FNO is than the baseline CNN.
Similarly, initializing XD with convolutions dilated as in the original TCN did not lead to significant improvement, except in one setting, over undilated initialization.
See the appendix for more details and results.
}
improving the computational limitations described earlier, and constructing spaces containing missing operations such as BatchNorm \citep{ioffe2015batchnorm} and self-attention \citep{vaswani2017attention} are all promising future directions.
Finally, note that our goal---lowering the barrier for applying ML---necessarily comes with the possibility of misuse.
Mitigating this involves developing tools for application-specific concerns, e.g. privacy and fairness, that go beyond the error metrics we target.
\section{Expressivity Results}\label{app:expressivity}
Here we collect results on the expressivity of the set of $\XD$-operations.
For simplicity, our results will be in the following single-dimensional ($N=1$) setting:
\begin{Set}\label{app:set:single}
We consider input spaces of form $\mathcal X=\mathbb R^{c\times m}$ for input size $m\in\mathbb N$ and channel count $c\in\mathbb N$ and parameter spaces $\mathcal W=\mathbb R^{c\times c\times k}$ for filter size $k\in[n]$, where output size $n\ge m$ is a power of 2.
\end{Set}
It is straightforward to extend the results to multiple dimensions using Kronecker products and to input sizes other than powers of two using padding.
Note that all of our results will also assume a circular padded domain.
\subsection{Convolutions}
\begin{Def}\label{app:def:conv}
A {\bf convolution} in Setting~\ref{app:set:single} with filter size $k$, dilation $d\in[\lfloor\frac{n-1}{k-1}\rfloor]$, stride $s\in[n-1]$, and channel groups described by a matrix $\*B\in\{0,1\}^{n\times n}$ s.t. $\*B_{[i,j]}=1$ if channels $i$ and $j$ are in the same group and 0 otherwise is a parameterizable operation that for any weight $\*w\in\mathcal W$ outputs a function mapping every $\*x\in\mathcal X$ to
\begin{equation}
\frac1n
\begin{pmatrix}
\diag(\mathfrak{a}_s(\underline{\mathbf 1_{\lceil\frac ns\rceil}}))\sum\limits_{j=1}^c\*B_{[1,j]}\*F_n^{-1}\diag(\*F_n\mathfrak{a}_d(\underline{\*w_{[1,j]}}))\*F_n\*x_{[j]}\\
\vdots\\
\diag(\mathfrak{a}_s(\underline{\mathbf 1_{\lceil\frac ns\rceil}}))\sum\limits_{j=1}^c\*B_{[c,j]}\*F_n^{-1}\diag(\*F_n\mathfrak{a}_d(\underline{\*w_{[c,j]}}))\*F_n\*x_{[j]}
\end{pmatrix}
\end{equation}
where $\*F_n\in\mathbb C^{n\times n}$ is the $n\times n$ DFT and $\mathfrak{a}_d:\mathbb R^n\mapsto\mathbb R^n$ is an atrous permutation of a vector that is equivalent to multiplication by some permutation matrix $\*P_d\in\{0,1\}^{n\times n}$.
We will use $\Conv_k$ to denote the case of $d=1$, $s=1$, and $\*B=\mathbf 1_{c\times c}$.
\end{Def}
\begin{Clm}\label{app:clm:conv}
All multi-channel convolutions of the form given in Definition~\ref{app:def:conv} are contained in the search space of XD-operations of depth $(1,3,1)$.
\end{Clm}
\begin{proof}
Setting the architecture parameters to be $\*K=\diag(\mathfrak{a}_s(\underline{\mathbf 1_{\lceil\frac ns\rceil}}))\*F_n^{-1}$, $\*L=\*F_n\*P_d$, $\*M=\*F_n$, $\*b=\mathbf 0_n$, and $\*C=\*B$, and noting that (a) the DFT and its inverse are both depth 1 K-matrices, (b) multiplying a K-matrix by a diagonal matrix is another K-matrix of the same depth, and (c) permutation matrices are K-matrices of depth 2 yields the result.
These three facts can be found in the original paper~\citep{dao2020kaleidoscope}.
\end{proof}
\begin{Rem}
Note that for the case of dilation $d=1$ the result in Claim~\ref{app:clm:conv} holds with depth $\mathbf 1_3$.
\end{Rem}
\subsection{Parameter-Free Operations}
\begin{Def}\label{app:def:skip}
The {\bf skip-connection} in Setting~\ref{app:set:single} is parameterizable operation that outputs a function mapping every $\*x\in\mathcal X$ to itself.
The {\bf zero-operation} in Setting~\ref{app:set:single} is parameterizable operation that outputs a function mapping every $\*x\in\mathcal X$ to $\mathbf 0_{c\times n}$.
\end{Def}
\begin{Clm}\label{app:clm:skip}
The skip-connection and zero-operation are both contained in the search space of XD-operations of depth $\mathbf 1_3$.
\end{Clm}
\begin{proof}
For both set the architecture parameters to be $\*K=\*F_n^{-1}$, $\*L=\mathbf 0_{n\times n}$, $\*M=\*F_n$, and $\*C=\*I_c$.
To obtain the skip-connection set $\*b=\mathbf 1_n$;
to obtain the zero-operation set $\*b=\mathbf 0_n$.
\end{proof}
\begin{Def}\label{app:def:avgp}
An {\bf average pooling} operation in Setting~\ref{app:set:single} with filter size $k$, dilation $d\in[\lfloor\frac{n-1}{k-1}\rfloor]$, and stride $s\in[n-1]$ is parameterizable operation outputs a function mapping every $\*x\in\mathcal X$ to the output of a convolution (as in Definition~\ref{app:def:conv}) with the same filter size, dilation, and stride, channel groups described by $\*B=\*I_c$, and filters $\*w_{[j,j]}=\mathbf 1_k/k~\forall~j\in[c]$.
\end{Def}
\newpage
\begin{Clm}\label{app:clm:avgp}
All average pooling operations are contained in the search space of XD-operations of depth $\mathbf 1_3$.
\end{Clm}
\begin{proof}
Setting the architecture parameters to be $\*K=\diag(\mathfrak{a}_s(\underline{\mathbf 1_{\lceil\frac ns\rceil}}))\*F_n^{-1}$, $\*L=\mathbf 0_{n\times n}$, $\*M=\*F_n$, $\*b=\mathfrak{a}_d(\underline{\mathbf 1_k/k})$, and $\*C=\*I_c$ and noting that (a) the DFT and its inverse are both depth 1 K-matrices and (b) multiplying a K-matrix by a diagonal matrix of the same depth is another K-matrix of the same depth yields the result.
\end{proof}
\subsection{Compositions with Multiplication by a Fixed K-Matrix}
\begin{Def}\label{app:def:lin}
A {\bf fixed linear operation} $\Lin_{\*A}$ in Setting~\ref{app:set:single} with fixed matrix $\*A\in\mathbb R^{n\times n}$ is a parameterizable operation that outputs a function mapping every $\*x\in\mathcal X$ to $\Lin_{\*A}(\*w)(\*x)=\begin{pmatrix}\*A\*x_{[1]}&\cdots&\*A\*x_{[c]}\end{pmatrix}^T$.
For example, $\Lin_{\*I_c}=\Id$.
\end{Def}
\begin{Def}\label{app:def:composition}
Let $\Op_1$ and $\Op_2$ be two parameterizable operations in Setting~\ref{app:set:single} with $\mathcal X$.
Then for any weight $\*w\in\mathcal W$ their {\bf composition} $\Op_1\circ\Op_2$ outputs the parameterized function $\Op_1(\*w)\circ\Op_2(\*w)$.
\end{Def}
\begin{Clm}\label{app:clm:composition}
Let $\Op$ be a parameterizable operation in Setting~\ref{app:set:single} that is contained in the set of XD-operations of some depth $\*d\in\mathbb N^3$ and let $\*A$ be a K-matrix of depth $d'$.
Then $\Op\circ\Lin_{\*A}$ is contained in the set of XD-operations of depth $(\*d_{[1]},\*d_{[2]},\*d_{[3]}+d')$ and $\Lin_{\*A}\circ\Op$ is contained in the set of XD-operations of depth $(\*d_{[1]}+d',\*d_{[2]},\*d_{[3]})$.
\end{Clm}
\begin{proof}
Let $\*K$ and $\*M$ be the first and last K-matrices of the representation of $\Op$ as an XD-operation, which thus have depth at most $\*d_{[1]}$ and $\*d_{[3]}$, respectively.
Then the representation of $\Op\circ\Lin_{\*A}$ as an XD-operation is the same except with depth $\*d_{[3]}+d'$ K-matrix $\*M\*A$ as the last K-matrix, and similarly the representation of $\Lin_{\*A}\circ\Op$ as an XD-operation is the same except with depth $\*d_{[1]}+d'$ K-matrix $\*A\*K$ as the first K-matrix.
\end{proof}
\subsection{Other Named Operations}
\begin{Def}
Suppose we have a fixed $n$-node graph with adjacency matrix $\*A$ and degree matrix $\*D$, and let $\hat{\*A}$ and $\hat{\*D}$ be the adjacency and degree matrices, respectively, of the same graph but with added self-loops.
Then regular {\bf graph convolution} \citep{kipf2017gcn} in Setting~\ref{app:set:single} with $k=1$ is a parameterizable operation that for any weight $\*W\in\mathcal W$ outputs a function mapping every $\*x\in\mathcal X$ to $\hat{\*D}^{-\frac12}\hat{\*A}\hat{\*D}^{-\frac12}\*x^T\*w$ and the {\bf diffusion graph convolution} \citep{li2018dcrnn} in Setting~\ref{app:set:single} with $k=1$ is a parameterizable operation that for any weight $\*W\in\mathcal W$ outputs a function mapping every $\*x\in\mathcal X$ to $\*D^{-1}\*A\*x^T\*w$.
\end{Def}
\begin{Clm}
Suppose $\*A$ and $\hat{\*A}$ can be represented by K-matrices of depth $d$ and $\hat d$, respectively.
Then the corresponding graph convolution is contained in the search space of XD-operations of depth $(1,1,\hat d+1)$ and the corresponding diffusion graph convolution in that of depth $(1,1,d+1)$.
\end{Clm}
\begin{proof}
For any $\*G\in\mathbb R^{n\times n}$ we have $\*G\*x^T\*w=\Lin_{\*G}(\*w)(\*x)\*w=\Conv_1(\*w)(\Lin_{\*G}(\*w)(\*x))=(\Conv_1\circ\Lin_{\*G})(\*w)(\*x)$.
The result follows by Claims~\ref{app:clm:conv} and~\ref{app:clm:composition}, the fact that a K-matrix multiplied by a diagonal matrix is another K-matrix of the same depth, and by substituting $\*G=\hat{\*D}^{-\frac12}\hat{\*A}\hat{\*D}^{-\frac12}$ (for graph convolution) or $\*G=\*D^{-1}\*A$ (for diffusion graph convolution).
\end{proof}
\begin{Rem}
Note that the above claim is meaningful because adjacency matrices of realistic graphs are usually sparse and sparse matrices can be efficiently represented as K-matrices \citep{dao2020kaleidoscope}.
\end{Rem}
\begin{Def}
A {\bf Fourier neural operator} (FNO) \citep{li2021fno} in Setting~\ref{app:set:single} with even $k$ and thus $k/2$ modes is a parameterizable operation that for any weight $\*w\in\mathcal W$ outputs a function mapping every $\*x\in\mathcal X$ to
\begin{equation}
\begin{pmatrix}
\Real\left(\sum_{j=1}^c\*F_n^{-1}\diag(\begin{pmatrix}\*w_{[1,j,1:k/2]}+i\*w_{[1,j,k/2+1:k]}&\mathbf 0_{n-k/2}\end{pmatrix}^T)\*F_n\*x_{[j]}\right)\\
\vdots\\
\Real\left(\sum_{j=1}^c\*F_n^{-1}\diag(\begin{pmatrix}\*w_{[c,j,1:k/2]}+i\*w_{[c,j,k/2+1:k]}&\mathbf 0_{n-k/2}\end{pmatrix}^T)\*F_n\*x_{[j]}\right)
\end{pmatrix}
\end{equation}
\end{Def}
\begin{Clm}\label{app:clm:fno}
The FNO with $k/2$ modes is contained in the search space of XD-operations of depth $(1,4,1)$.
\end{Clm}
\begin{proof}
Setting the architecture parameters to be $\*K=\*F_n^{-1}$, $\*L\in\mathbb C^{n\times n}$ the $n$-sparse matrix mapping $\underline{\*w}$ to $\begin{pmatrix}\*w_{[1,j,1:k/2]}+i\*w_{[1,j,k/2+1:k]}&\mathbf 0_{n-k/2}\end{pmatrix}^T$, $\*M=\*F_n$, $\*b=\mathbf 0_n$, and $\*C=\mathbf 1_{c\times c}$, and noting that an $n$-sparse matrix is a depth-4 K-matrix \citep{dao2020kaleidoscope} yields the result.
\end{proof}
\begin{Rem}
If we allow the parameter space in Setting~\ref{app:set:single} to be complex then the FNO with all $k$ modes will be contained in the search space of XD-operations of depth $\mathbf 1_3$.
\end{Rem}
\begin{Def}
Each channel of {\bf transposed convolution} with stride $d(k-1)+1$, where $k$ is the kernel size and $d$ is the dilation rate, computes a feature map in which each input element is replaced by that element multiplied by the dilated filter of size $d(k-1)+1$.
The multi-channel extension of this over parameter space $\mathcal W=\mathbb R^{c\times c\times k}$ is similar to that for standard convolutions.
\end{Def}
\begin{Clm}
All transposed convolutions with stride equal to the dilated kernel size are contained in the search space of XD-operations of depth $(1,3,3)$.
\end{Clm}
\begin{proof}
A transposed convolution is equivalent to a regular convolution with the same filter applied to the input after it has been zero-padded and then permuted to separate all entries by $d(k-1)$ zeros.
Since permutations are K-matrices of depth 2 the result follows by Claims~\ref{app:clm:conv} and Claim~\ref{app:clm:composition}.
\end{proof}
\begin{Def}
A {\bf depthwise-separable convolution} in Setting~\ref{app:set:single} with filter size $k$ but with parameter space $\mathcal W=\mathbb R^{c\times k}\times\mathbb R^{c\times c}$ is a parameterizable operation that for any weight $\*w\in\mathcal W$ outputs $\Conv_1(\*w_{[2]})\circ\Conv_{k,\*I_c}(\*w_{[1]})$, where $\Conv_{k,\*I_c}$ denotes the convolution in Definition~\ref{app:def:conv} with $\*B=\*I_c$.
\end{Def}
\begin{Rem}
Since both $\Conv_1$ and $\Conv_{k,\*I_c}$ are XD-operations, by definition depthwise-separable convolutions are contained in the search space of composed XD-operations, which by Claim~\ref{app:clm:skip} also contains all of the above operations.
\end{Rem}
\section{Introduction}
\vspace{-1mm}
Automated machine learning (AutoML) and neural architecture search (NAS) are often motivated by a vision of democratizing ML by reducing the need for expert design on a variety of tasks.
While NAS has grown rapidly with developments such as weight-sharing~\citep{pham2018enas} and ``NAS-benches'' \citep{ying2019nasbench101,zela2020nasbench1shot1}, most efforts focus on search spaces that glue together established primitives for well-studied tasks like vision and text \citep{liu2019darts,li2019rsws,xu2020pcdarts,li2021gaea} or on issues such as latency \citep{cai2020ofa,fang2020densenas}.
In this work, we revisit the broader vision of NAS and propose to move towards much more general search spaces while still exploiting successful network topologies.
To do so we focus on expanding the set of operations, which is usually fairly small;
for example, that of the well-studied DARTS space has eight elements: a few types of convolution and pooling layers \citep{liu2019darts}.
The baseline approach for expanding this set---adding operations one-by-one---scales poorly and will not result in new operations when faced with new types of data.
Our core contribution is a re-imagining of NAS operation spaces that drastically expands this set in a principled fashion to include both standard operations as well as a wide range of new ones.
To do so we exploit the fact that most standard operations used in modern NAS return linear transforms diagonalized by the discrete Fourier transform (DFT). Replacing the DFT matrices in the diagonal decomposition by a more expressive family of efficient linear transforms known as {\em Kaleidoscope} or {\em K-matrices} \citep{dao2020kaleidoscope} yields the set of {\bf Expressive Diagonalization (XD) Operations}, which comprise a large search space containing various types of grid-based convolutions and pooling, permutations, transposed convolutions, certain kinds of graph convolutions, the Fourier Neural Operator (FNO)~\citep{li2021fno}, and infinitely many more.
This broad expressivity reflects the key insight of our work:
that many of the most important neural operations in ML consist of multiple channels that apply weights $\*w$ to inputs $\*x$ by computing
\begin{equation}
\*K\diag(\*L\*w)\*M\*x
\end{equation}
where the matrices $\*K$, $\*L$, and $\*M$ are {\em efficient} (to represent and apply) and shared across channels.
\begin{figure*}[t!]
\centering
\includegraphics[width=\linewidth]{figures/main/main.pdf}
\caption{\label{fig:main}
Diagram of our search space depicting a NAS method picking an operation for an edge in a backbone network (left).
Instead of choosing from a discrete search space, we use a relaxation based on the convolution's diagonalization by the discrete Fourier transform in which the DFTs are replaced by K-matrices \citep{dao2020kaleidoscope} $\*K$, $\*L$, and $\*M$ (middle);
these are the main architecture parameters of our new search space over Expressive Diagonalization (XD) operations.
This space contains most operations considered in standard NAS and many other important operations in a variety of domains (right).
}
\end{figure*}
We leverage XD-operations to take critical steps towards a broader NAS that enables the discovery of good design patterns with limited human specification from data in under-explored domains.
To do so we develop a simple procedure which transforms any backbone convolutional neural network (CNN) into an architecture search space by replacing its operations with XD-operations.
This space is then searched using a simple weight-sharing algorithm that needs only a small amount of tuning to find effective operations.
As a simple first demonstration, we show that XD-operations yield models that are 15\% more accurate than standard discrete search spaces on {\em permuted} CIFAR-10, highlighting the fragility of standard NAS operation spaces on new datasets, and thus the need for XD-operations.
As our main evaluation, we demonstrate the effectiveness of XD-operations in a series of applications showing that, starting from vanilla CNNs, they consistently outperform custom-designed operations.
\begin{itemize}[leftmargin=*,topsep=-1pt,noitemsep]\setlength\itemsep{2pt}
\item {\bf Learning to solve partial differential equations (PDEs):} when substituted into a simple CNN backbone, XD-operations outperform convolutions and the dense prediction NAS method Auto-DeepLab \citep{liu2019autodl}, and even achieve lower error than custom-designed, state-of-the-art operations (FNOs \citep{li2021fno}) across three problems with different dimensionalities (Burgers' equation, Darcy Flow, and Navier-Stokes).
Our method also maintains consistent performance across different resolutions, a major stated advantage of FNOs over previous methods.
\item {\bf Protein folding:}
on the task of predicting residue distances in a polypeptide chain---a key component of the protein folding problem---we substitute XD-operations into vanilla ResNets and achieve lower error than cyclically-dilated ResNets adapted specifically for this setting \citep{adhikari2020}.
Furthermore, our ResNet-34 XD outperforms the reported error of the much deeper Dilated ResNet-258.
\item {\bf Music modeling:} on two next-note prediction tasks, we show that substituting XD-operations into an undilated CNN outperforms temporal convolutional networks (TCNs)---exponentially-dilated 1d CNNs that themselves outperform standard convolutional and recurrent networks \citep{bai2018tcn}.
\end{itemize}
Code to reproduce these results is available here: \url{https://github.com/nick11roberts/XD}.
Software to apply XD-operations can be found here: \url{https://github.com/mkhodak/relax}.
\section*{Acknowledgments}
\vspace{-1mm}
We thank Maria-Florina Balcan, Jeremy Cohen, and Tian Li for helpful advice on early versions of this paper and anonymous reviewers for suggested improvements.
This work was supported in part by DARPA under cooperative agreements FA875017C0141 and HR0011202000, NSF grants CCF-1535967, CCF-1910321, IIS-1618714, IIS-1705121, IIS-1838017, IIS-1901403, and IIS-2046613, a Microsoft Research Faculty Fellowship, a Bloomberg Data Science research grant, an Amazon Research Award, an AWS Machine Learning Research Award, a Facebook Faculty Research Award, funding from Booz Allen Hamilton Inc., a Block Center Grant, a Carnegie Bosch Institute Research Award, and a Two Sigma Fellowship Award.
We also gratefully acknowledge the support of NIH under No.\ U54EB020405 (Mobilize), NSF under Nos.\ CCF1763315 (Beyond Sparsity), CCF1563078 (Volume to Velocity), and 1937301 (RTML); ONR under No.\ N000141712266 (Unifying Weak Supervision); the Moore Foundation, NXP, Xilinx, LETI-CEA, Intel, IBM, Microsoft, NEC, Toshiba, TSMC, ARM, Hitachi, BASF, Accenture, Ericsson, Qualcomm, Analog Devices, the Okawa Foundation, American Family Insurance, Google Cloud, Swiss Re, Total, the HAI-AWS Cloud Credits for Research program, the Stanford Data Science Initiative (SDSI), and members of the Stanford DAWN project: Facebook, Google, and VMWare.
The Mobilize Center is a Biomedical Technology Resource Center, funded by the NIH National Institute of Biomedical Imaging and Bioengineering through Grant P41EB027060.
The U.S.\ Government is authorized to reproduce and distribute reprints for Governmental purposes notwithstanding any copyright notation thereon.
Any opinions, findings and conclusions, or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of DARPA, NSF, NIH, ONR, or any other funding agency.
\section{The Expressive Diagonalization Relaxation}\label{sec:relax}
\vspace{-1mm}
In this section we overview our main contribution:
a large, general search space of neural operations.
Formally, we view an architecture as a {\em parameterizable} object---a mapping from model weights to functions---described by a {\em labeled} directed acyclic graph (DAG) $\mathcal G(V,E)$.
Each edge in $E$ has the form $(u,v,\Op)$, where $u,v\in V$ are nodes and $\Op$ is an operation that can be parameterized to define some transformation of the representation at node $u$;
node $v$ aggregates the outputs of its incoming edges into a new representation.
For example, the popular ResNet architecture \citep{he2016resnet} has many nodes with two incoming edges, one labeled by the convolution operation $\Conv$ and one by the identity (skip-connect) $\Id$, whose outputs it sums and passes to outgoing edges with the same labels.
Each architecture has a source node taking in input data and an output node returning a prediction.
Neural architecture search is the problem of automatically selecting an operation for each edge of $\mathcal G$ to optimize an objective.\footnote{It is often defined as selecting both operations and a graph topology \citep{zoph2018nas}, but if the set of operations contains the zero-operation $\Zero$ then the former subsumes the latter.}
For each edge $e\in E$ a NAS algorithm must pick one element of a {\em search space} $\mathcal S=\{\Op_a\vert a\in\mathcal A\}$ of operations specified by architecture parameters $a\in\mathcal A$ to assign to $e$;
in past work, $\mathcal A$ usually indexes a small set of operations.
As an example, we will refer to a variant\footnote{For memory-efficiency, all convolutions in the original DARTS search space are separable \citep{liu2019darts}.} $\mathcal S_\mathbf{discrete}$ of the DARTS search space with parameters $\mathcal A_\mathbf{discrete}=\{1,\dots,8\}$ where each operation is one of $\Zero$, $\Id$, $\MaxP_{3\times 3}$, $\AvgP_{3\times 3}$, $\Conv_{3\times 3\textrm{ or }5\times 5}$, or $\DilC_{3\times 3,2\textrm{ or }5\times 5,2}$~\citep{liu2019darts}.
Our main contribution is a novel family of operations that comprise a search space containing almost all these operations, in addition to many others that have been found useful on different types of data.
The starting point of our construction of these XD-operations is the simple observation that all the operations $\Op\in\mathcal S_\mathbf{discrete}$ listed above except $\MaxP_{3\times3}$ are {\em linear}, i.e.
for any model weights $\*w$ there exists a matrix $\*A_{\*w}$ such that for all inputs $\*x$ we have $\Op(\*w)(\*x)=\*A_{\*w}\*x$.
More specifically, all seven of them return convolutions:
to see this note that $\Zero$, $\Id$, and $\AvgP_{3\times3}$ each apply a convolution with filter $\mathbf 0_{1\times1}$, $\mathbf 1_{1\times1}$, and $\mathbf 1_{3\times3}/9$, respectively.
This means that most of the operations in the DARTS search space---which is representative of NAS operation spaces in computer vision---share the convolution's diagonalization by the discrete Fourier transform (DFT).
Formally, if $\*A_{\*w}\in\mathbb R^{n^2\times n^2}$ is the matrix representing a 2d convolution with filter $\*w\in\mathbb R^{\*k}$ of kernel size $\*k\in[n]^2$, then for any 2d input $\*x\in\mathbb R^{n^2}$ we have
\begin{equation}\label{eq:fourier}
\Conv(\*w)(\*x)=\*A_{\*w}\*x=\*F^{-1}\diag\left(\*F\underline{\*w}\right)\*F\*x
\end{equation}
Here $[n]=\{1,\dots,n\}$, $\diag(\*z)$ denotes the diagonal matrix with entries $\*z$, $\underline{\*w}\in\mathbb R^{n^2}$ is an appropriate zero-padding of $\*w\in\mathbb R^{\*k}$, and $\*F\in\mathbb C^{n^2\times n^2}$ is the 2d DFT (a Kronecker product of two 1d DFTs).
This diagonalization explicates both the computational and representational efficiency of the DARTS operations, as the DFT and its inverse can be applied in time $\mathcal O(n\log n)$ and stored with $\mathcal O(n\log n)$ bits.
It also suggests a natural way to dramatically expand the operation space while preserving these efficiencies:
just replace matrices $\*F$ and $\*F^{-1}$ in \eqref{eq:fourier} by any one of a general family of efficient matrices.
Doing so yields the single-channel version of our {\em expressive diagonalization} (XD) operations:
\begin{equation}\label{eq:xd1}
\XD_\alpha^\mathbf 1(\*w)(\*x)=\Real\left(\*K\diag\left(\*L\underline{\*w}\right)\*M\*x\right)
\end{equation}
Here architecture parameter $\alpha=(\*K,\*L,\*M)$ sets the matrices replacing $\*F$ and $\*F^{-1}$ in Equation~\ref{eq:fourier}.
The main remaining question is the family of efficient matrices to use, i.e. the domain of the architecture parameters $\*K$, $\*L$, and $\*M$.
For this we turn to the Kaleidoscope matrices, or {\em K-matrices}~\citep{dao2020kaleidoscope}, which generalize $\*F$ and $\*F^{-1}$ to include all computationally efficient linear transforms with short description length, including important examples such as sparse matrices and permutations.
To obtain this general family, K-matrices allow the DFT's butterfly factors---matrices whose products yield its efficient implementation---to take on different values.
While a detailed construction of K-matrices can be found in the original paper, we need only the following useful properties:
they are as (asymptotically) efficient to apply as DFTs, are differentiable and can thus be updated using gradient-based methods, and can be composed (made ``deeper'') to make more expressive K-matrices.
Specifying that $\*K$, $\*L$, and $\*M$ in Equation~\ref{eq:xd1} are K-matrices largely completes our core contribution:
a new search space $\mathcal S_\XD$ of XD-operations with K-matrix architecture parameters.
We give a full multi-channel formalization in $N$ dimensions, as well as an overview of its expressivity, in Section~\ref{sec:xd}.
First, we note some key aspects of this new search space:
\begin{itemize}[leftmargin=*,topsep=-1pt,noitemsep]\setlength\itemsep{2pt}
\item{\bf Complexity:} $\XD_\alpha^\mathbf 1(\*w)$ requires three K-matrices and $\mathcal O(1)$ filter weights to represent, i.e. description length $\mathcal O(n\log n)$;
this is larger than a regular convolution (which has no architecture parameters) but is not quadratic in the input size like a linear layer.
Applying $\XD_\alpha^\mathbf 1$ requires multiplication by three K-matrices, yielding a theoretical per-channel time complexity of $\mathcal O(n\log n)$, matching the efficiency of convolutions.
However, as XD-operations strictly generalize convolutions they are more expensive to apply in-practice;
we detail these costs both in the application sections and as appendix table, and we view improving upon them as an important future direction.
\item{\bf Initialization:} a crucial advantage of XD-operations is that we can initialize or {\em warm-start} search using operations with known constructions.
In particular, since we can recover convolutions \eqref{eq:fourier} by setting architecture parameters $\*K=\*F^{-1}$, $\*L=\*F$, and $\*M=\*F$ in Equation~\ref{eq:xd1}, we can always start search with any CNN backbone.
We use this extensively in experiments.
\item{\bf K-matrices:} as they contain all efficient linear transforms, K-matrices can represent all functions returned by XD-operations, including convolutions.
However, for input dimension and filter size $>1$ the only known way is to apply K-matrices directly to flattened inputs $\*x\in\mathbb R^{n^N}$, yielding much worse description length $\mathcal O(n^N\log n)$.
In contrast, as detailed in Section~\ref{sec:xd}, our diagonalization approach uses Kronecker products to apply DFTs to each dimension separately, yielding description length $\mathcal O(n\log n)$.
It is thus the first (and in some sense, ``right'') method to use such matrices to replace convolutions.
Furthermore, diagonalization allows us to separate model weights $\*w$ from architecture parameters $\alpha$, letting the former vary across channels while fixing the latter.
\end{itemize}
Finally, we address the fact that the architecture parameters of $\mathcal S_\XD$ are continuous, not discrete, contrasting with much of the NAS literature.
This can be viewed as a natural extension of the weight-sharing paradigm \citep{pham2018enas}, in which continuous relaxation enables updating architecture parameters with gradient methods.
For example, many algorithms traverse the relaxed DARTS search space
$\tilde\mathcal S_\mathbf{discrete}=\left\{\sum_{i=1}^8\lambda_i\Op_i\vert\lambda_i\ge0,\sum_{i=1}^8\lambda_i=1\right\}$, defined via DARTS operations $\Op_i\in\mathcal S_\mathbf{discrete}$ and architecture parameters $\lambda_i$ in the 8-simplex;
most search spaces then require discretizing after search via a rounding procedure that maps from the simplex to $\mathcal A_\mathbf{discrete}$.
Note that the fully continuous nature of XD-operations means that we will only evaluate the final network returned by search.
In particular, while some weight-sharing papers also report the correlation between true architecture performance and that indicated by the shared weights \citep{yang2020nas}, there is no obvious way to define a ranking or sampling distribution over XD-operations in order to do so.
This also means that our final architecture will not be more efficient than the supernet, unlike other weight-sharing methods that do discretize.
\vspace{-1mm}
\section{XD-Operations and Their Expressivity}\label{sec:xd}
\vspace{-1mm}
Here we formalize XD-operations and show what operations they include.
We first define operations:
\begin{Def}\label{def:po}
A {\bf parameterizable operation} is a mapping $\Op:\mathcal W\mapsto\mathcal F$ from parameter space $\mathcal W$ to a space $\mathcal F=\{\Op(\*w):\mathcal X\mapsto\mathcal Y\vert\*w\in\mathcal W\}$ of {\bf parameterized functions} from input space $\mathcal X$ to output space $\mathcal Y$.
A {\bf search space} is a set of operations with the same $\mathcal W$, $\mathcal X$, and $\mathcal Y$.
\end{Def}
For example, if $\mathcal X=\mathcal Y=\mathbb R^n$ and $\mathcal W=\mathbb R^{n\times n}$ then each $\*W\in\mathcal W$ defines a parameterized linear layer that for each $\*x\in\mathcal X$ returns $\Lin(\*W)(\*x)=\*W\*x$.
Here $\Lin$ is the parameterizable operation and for each $\*W$ the linear map $\Lin(\*W)$ is the parameterized function.
From Definition~\ref{def:po}, we say a search space can {\em express} a specific operation if it contains it.
Crucially, the ability of a parameterizable operation $\Op_1$ to express a parameterized function $\Op_2(\*w)$ output from another operation $\Op_2$ given the right set of weights $\*w$ does {\em not} imply that a search space containing $\Op_1$ can express $\Op_2$.
For example, $\Lin(\*I_n)=\Id(\*W)~\forall~\*W\in\mathbb R^{n\times n}$ but $\Lin(\*W)\ne\Id(\*W)~\forall~\*W\ne\*I_n$, so a search space containing the linear operation $\Lin$ cannot express the skip-connection $\Id$, despite the fact that $\Lin$ can be parameterized to compute the identity.
\paragraph{Formalizing Multi-Channel XD-Operations
Recall the single-channel XD-operation $\XD_\alpha^\mathbf 1$ in Equation~\ref{eq:xd1} specified by three-matrix architecture parameter $\alpha=(\*K,\*L,\*M)$.
For input dimension $N\ge1$, every matrix $\*B\in\alpha$ is a Kronecker product of $N$ $K$-matrices of depth $\*d\in\mathbb Z_+^3$, i.e. $\*B=\bigotimes_{i=1}^N\*B_i$ for K-matrices $\*B_i\in\mathbb C^{n\times n}$ of depth $\*d_{[1]}$, $\*d_{[2]}$, or $\*d_{[3]}$ for $\*B=\*K$, $\*L$, or $\*M$, respectively.\footnote{A depth-$d$ K-matrix is a product of $d$ depth-1 K-matrices.}
Roughly speaking, $\XD_\alpha^\mathbf 1$ can return any linear operation that is diagonalized by K-matrices and is thus efficient to compute and represent, e.g. any convolution (recall we recover the diagonalization of $\Conv(\*w)$ in Equation~\ref{eq:fourier} by setting $\*K$, $\*L$, and $\*M$ appropriately in Equation~\ref{eq:xd1}).
However, $\XD_\alpha^\mathbf 1$ cannot represent efficient {\em parameter-free} operations such as skip-connections and average-pooling, both common in NAS.
In particular, the only way to always ignore the model weights $\*w$ is to set one of the K-matrices to zero, producing the zero-operation.
We avoid this by adding a bias $\*b\in\mathbb C^{n^N}$ as an architecture parameter, yielding the {\em biased} single-channel XD-operation:\footnote{Zero-padding $\*x$ as well lets the input to be smaller than the output if needed, e.g. for transposed convolutions.}
\begin{equation}\label{eq:biased}
\XD_{\alpha,\*b}^\mathbf 1(\*w)(\*x)=\Real\left(\*K\diag(\*L\underline{\*w}+\*b)\*M\underline{\*x}\right)
\end{equation}
This lets us define skip-connections (set $\*K=\*M=\*I_{n^N}$, $\*L=\mathbf 0_{n^N\times n^N}$, and $\*b=\mathbf 1_{n^N}$) and average-pooling (set $\*K=\*F^{-1}$, $\*L=\mathbf 0_{n^N\times n^N}$, $\*M=\*F$, and $\*b$ to be $\*F$ multiplied by a pooling filter).
Lastly, we use $\XD_{\alpha,\*b}^\mathbf 1$ to construct multi-channel ``layers'' that pass multiple input features through multiple channels and re-combine them as multiple output features.
This follows the primary way of using convolutions in deep nets.
The key insight here is that we will share the same parameterizable operation (specified by $\alpha$ and $\*b$) across all channels, just as in convolutional layers.
\begin{Def}\label{def:xd}
Let $a=(\alpha,\*b,\*C)$ be an architecture parameter containing a triple $\alpha=(\*K,\*L,\*M)$ of Kronecker products of $N$ K-matrices with depths $\*d\in\mathbb Z_+^3$, a bias $\*b\in\mathbb C^{n^N}$, and channel gates $\*C\in\mathbb C^{c\times c}$.\footnote{For simplicity we formalize the case where all $N$ dimensions have the same input size and there is an identical number $c$ of input and output channels; both are straightforward to extend.}
Using ``$\bigoplus$'' to denote concatenation, the {\bf XD-operation} $\XD_a$ of depth $\*d$ specified by $a$ is a parameterizable operation on parameter space $\mathcal W=\mathbb R^{c\times c\times\*k}$ consisting of $c^2$ filters of size $\*k\in[n]^N$ that outputs parameterized functions on $\mathcal X=\mathbb R^{c\times m^N}$ for $m\le n$ mapping every $\*x\in\mathcal X$ to\vspace{-5pt}
\begin{equation}
\XD_a(\*w)(\*x)=\bigoplus_{i=1}^c\sum\limits_{j=1}^c\*C_{[i,j]}\XD_{\alpha,\*b}^\mathbf 1(\*w_{[i,j]})(\*x_{[j]})
\end{equation}
\end{Def}
The last architecture parameter $\*C$ allows interpolation between all-to-all layers ($\*C=\mathbf 1_{c\times c}$), e.g. multi-channel convolutions, and layers where each channel is connected to one other channel ($\*C=\*I_c$), e.g. skip-connections and average-pooling.
We note that we use $\mathcal S_\XD$ to describe the set of operations covered by Definition~\ref{def:xd} and conclude our construction by discussing two properties:
\begin{itemize}[leftmargin=*,topsep=-1pt,noitemsep]\setlength\itemsep{2pt}
\item{\bf Kernel size:} the weight-space available to an XD-operation is $\mathbb R^{c\times c\times n^N}$; however, since we will initialize search with existing CNNs, we will zero-pad to have the same weight-space $\mathbb R^{c\times c\times k^N}$ as the convolutions with filter size $k\le n$ that they replace.
This preserves the weight count but also means that if the backbone has $3\times3$ filters our search space will {\em not} contain $5\times5$ convolutions.
Experimentally, we find that relaxing the constraint to allow this does not significantly affect results on image tasks, so we do not do so in subsequent applications to avoid increasing the weight count.
\item{\bf Depth:} an XD-operation's depth is a triple describing the depths of its K-matrices $\*K$, $\*L$, and $\*M$.
Increasing it trades off efficiency for expressivity;
for example, in the next section we describe operations that we can show are contained in $\mathcal S_\XD$ if $\*L$ or $\*M$ have depth $>1$.
By default we will set the depth to be the minimum needed to initialize search with the backbone operation.
\end{itemize}
\paragraph{Expressivity of XD-Operations
For many papers that replace deep net layers with efficient linear transforms \cite{moczulski2015acdc,dao2020kaleidoscope}, the question of expressivity comes down to the transform capacity.
For example, layers with a K-matrix in every channel can represent a different transform in each, thus allowing the output to be any combination of efficient linear operations.
Our case is less straightforward since we care about expressivity of the search space, not of parameterized functions, and our approach is less-expressive {\em by design} as all channels share K-matrices $\*K$, $\*L$, and $\*M$.
The latter can be thought of as a useful inductive bias on NAS:
the set of XD-operations is still much broader than the set of convolutions, but the way in which model weights are applied is the same across all channels.
Expressivity results are a way to see if this bias is useful or constraining.
Here we summarize some important operations that are 1d XD-operations;
proofs can be found in the appendix and are straightforward to extend to multi-dimensional inputs.
Formally, there exists $\*d\in\mathbb Z_+^3$ such that the set of XD-operations of depth $\*d$ over weights $\mathcal W=\mathbb R^{c\times c\times k}$ and inputs $\mathcal X=\mathbb R^m$ for $m\le n$ contains
\begin{enumerate}[leftmargin=*,topsep=-1pt,noitemsep]\setlength\itemsep{2pt}
\item convolutions with filter size $\le k$, dilation $\le\lfloor\frac{n-1}{k-1}\rfloor$, stride $\le n-1$, and arbitrary channel groups.
\item parameter-free operations $\Id$, $\Zero$, and $\AvgP_s$ for any kernel size $s\le n$.
\item composing 1 or 2 with multiplication of all input or output channels by a bounded-depth K-matrix.
\end{enumerate}
Note this does not account for {\em all} important XD-operations, e.g. we show in the appendix that they also express Fourier Neural Operators \citep{li2021fno} with $\le\lfloor k/2\rfloor$ modes and any transposed convolutions whose stride equals the dilated kernel size.\footnote{This restriction still includes transposed convolutions used in well-known architectures such as U-Net \citep{ronneberger2015unet}.}
Still, the first two items account for non-separable variants of most operations considered in past NAS work in computer vision, excluding the nonlinear $\MaxP$ \citep{ying2019nasbench101,dong2020nasbench201}.
Note depthwise-separable convolutions {\em are} contained in the set of compositions of XD-operations.
The third item implies that XD-operations can express the basic and diffusion graph convolutions over fixed graphs \citep{kipf2017gcn,li2018dcrnn}:
both are point-wise convolutions composed with sparse multiplication by a modified adjacency matrix, which K-matrices can represent efficiently.
As a concrete example, consider dilated convolutions, which for $k>1$ and dilation factor $d\ge1$ apply filters of effective size $(k-1)d+1$ with nonzero entries separated by $d-1$ zeros.
One could hope to express the application of $\DilC_{k,d}$ to an input $\*x\in\mathbb R^n$ in the single-channel setting as $\*F^{-1}\diag(\*F\diag(\*p_{k,d})\underline{\*w})\*F\*x$, where $\*p_{k,d}\in\{0,1\}^n$ zeroes out appropriate entries of $\underline{\*w}$, but this requires filter size $(k-1)d+1>k$, increasing the number of weights.
Instead, we can use a permutation $\*P_{k,d}\in\{0,1\}^{n\times n}$ before the DFT to place the $k$ entries of $\underline{\*w}$ into dilated positions:
\begin{equation}\label{eq:dilated}
\DilC_{k,d}(\*w)(\*x)=\*F^{-1}\diag(\*F\*P_{k,d}\underline{\*w})\*F\*x
\end{equation}
As permutations are depth-2 K-matrices \citep{dao2020kaleidoscope}, we can express $\DilC_{k,d}$ with an XD-operation of depth $(1,3,1)$, with $\*K=\*F^{-1}$, $\*L=\*F\*P_{k,d}$, and $\*M=\*F$.
\section{Application: Learning to Solve Partial Differential Equations}\label{sec:pde}
\vspace{-1mm}
As our first non-vision application, we consider the task of solving PDEs, an important application area of ML in the natural sciences \citep{li2015butterfly, li2018multidimensional, sirignano2018dgm}.
In our setup, data generated by classical PDE solvers is used to learn functions from some initial condition or setting to the corresponding PDE solution, with the goal of replacing the solver by a deep net forward pass;
the latter can be orders of magnitude faster.
A recent state-of-the-art approach for this introduces Fourier Neural Operators \citep{li2021fno}, operations that significantly improve upon previous neural approaches across three different PDE settings.
To evaluate the ability of XD-operations to compete with such custom-designed operations starting from simple CNN backbones, we will investigate the same three PDEs that they study: Burgers' equation, Darcy Flow, and the 2d Navier-Stokes equations, which involve 1d, 2d, and 3d data, respectively.
The first two are studied across multiple resolutions, while the last one is studied at different viscosities.
As before, we start with a simple CNN backbone---the type a scientist might use in a first attempt at a solution---and replace all convolutions by XD-operations.
We initially hope to do better than this backbone, but ambitiously also hope to compete with the custom-designed FNO.
The specific CNN we use is simply the FNO architecture of the appropriate dimension $N$ but with all $N$-dimensional FNOs replaced by $N$-dimensional convolutions;
this performs similarly to their CNN baselines \cite{li2021fno}.
In all cases we compare mainly to the CNN backbone and our reproduction of the FNO results, as the latter exceeds all other neural methods;
a complete results table is provided in the appendix.
Our reproduction of FNO is slightly worse than their reported numbers for Burgers' equation and slightly better in the other two settings.
Note that on the Navier-Stokes equations we only compare to the 3d FNO on the two settings in which we were able to reproduce their approach;
moreover, we do {\em not} compare to their use of a 2d FNO plus a recurrent net in time, but in-principle XD-operations can also be substituted there.
In the 2d Darcy Flow case we also include comparisons to DARTS operations in the simple CNN backbone, as in Section~\ref{sec:chrysalis}, and to Auto-DeepLab (AutoDL) \citep{liu2019autodl}, a well-known NAS method for dense prediction.
For evaluating XD-operations we again follow the procedure in Section~\ref{sec:chrysalis}, in which we tune only the architecture optimizer;
notably, we do this only at the lowest resolutions.
At all dimensions we use XD-operations of depth $\*d=\mathbf 1_3$;
in addition, in dimensions $N>1$ we fix the architecture biases $\*b$ and channel gates $\*C$ to $\mathbf 0$ and $\mathbf 1$, respectively, to conserve memory at higher resolutions.
At lower ones we find that the performance difference is negligible.
We report our results for the Burger's equation and Darcy Flow in Figure~\ref{fig:pde};
for 2d Navier-Stokes the results are in Table~\ref{tab:navierstokes}.
In all cases we dramatically outperform the CNN backbone used to initialize XD-operations;
furthermore, we also achieve better error than FNO, despite it being custom-made for this problem.
In particular, we find that XD-operations have higher {\em training error} but generalize better (c.f. the appendix).
Figure~\ref{fig:pde} also shows that XD-operations perform consistently well across resolutions, a major advantage of FNOs over previous methods, whose performance was tightly coupled to the discretization \citep{li2021fno}.
Notably, CNN performance worsens with higher resolution, unlike that of XD and FNO.
Finally, we also substantially outperform DARTS operations and AutoDL in 2d, although the latter is at least consistent across resolutions.
These results provide strong evidence that XD-operations are a useful search space for discovering neural operations, even in domains where the convolutions used to initialize them perform much worse than state-of-the-art.
Note that these results do come at a cost of slower training and inference:
XD-operations are roughly an order of magnitude slower than FNOs, despite having fewer parameters in 2d and 3d.
This still yields solvers one-to-two orders of magnitude faster than classical solvers, maintaining usefulness for the problem.
\begin{figure*}[!t]
\begin{minipage}{\linewidth}
\centering
\begin{threeparttable}
\captionof{table}{\label{tab:navierstokes}
Relative test error on the 2d Navier-Stokes equations at different settings of the viscosity $\nu$ and time steps $T$.
Best results in each setting are {\bf bolded}.
}
\begin{tabular}{lcccccc}
\toprule
&&& $\nu=10^{-4}$, $T=30$ &&& $\nu=10^{-5}$, $T=20$ \\
\midrule
CNN-3d (our baseline) &&& 0.325 &&& 0.278 \\
FNO-3d (reproduced) &&& 0.182 &&& 0.177 \\
{\bf CNN-3d XD} (ours) &&& {\bf 0.172} &&& {\bf 0.168} \\
\bottomrule
\end{tabular}
\end{threeparttable}
\end{minipage}
\end{figure*}
\section{Application: Real-Valued Distance Prediction for Protein Folding}\label{sec:protein}
\vspace{-1mm}
As a second scientific application, we consider the task of inferring the 3d ``folded'' structure of a polypeptide chain, which yields important insights into the function of the resulting protein~\cite{psicov}. This problem is a high-priority challenge in biology and has recently seen significant ML-driven advances from deep learning methods such as AlphaFold~\cite{alphafold, alphafold2} and PDNET~\cite{adhikari2020}.
These typically involve training a network to predict pairwise physical distances between residues in the chain.
We work with the PDNET benchmark, which consists of a training set of 3,356 proteins, a validation set of 100 of proteins, and the PSICOV \cite{psicov} test set of 150 proteins.
PDNET is designed to be more accessible than datasets used by large-scale methods such as AlphaFold, which are not always publicly available and/or require massive compute \cite{alphafold, alphafold2}.
We follow the PDNET training procedure~\cite{adhikari2020} and evaluate test set performance using their MAE$_8$ metric for assessing long-range distances.
As before we start with simple CNN backbones---in this case ResNets.
We choose this to compare most directly to the custom-designed architecture used by PDNET, consisting of a Dilated ResNet characterized by its use of a cyclically increasing dilation rate across ResNet blocks \cite{adhikari2020}.
At a sufficient depth, the Dilated ResNet is shown to outperform a standard pre-activation ResNet adapted to this task~\cite{adhikari2020}. Our goal will be to see whether we can start with the vanilla ResNet and use XD to outperform both it and the specialized Dilated ResNet.
We also aim to outperform the DARTS operations baseline from the previous two sections as well as the AutoDL NAS approach for dense prediction.
We use XD-operations of depth $\*d=\mathbf 1_3$ and fix the architecture biases and channel gates as before to conserve memory.
We evaluate architectures of different depths---4, 6, 10, 18, and 34---by varying the number of ResNet blocks used in the backbone architecture and baseline.
We report the results as averages across three trials for each depth in Figure~\ref{fig:protein}.
Notably, while Dilated ResNet slightly outperforms ResNet, ResNet XD outperforms both dilated and standard ResNets at all depths. This provides further evidence that XD-operations can outperform specialized operations for diverse domains, even when initialized naively as standard convolutions.
XD also outperforms AutoDL, which does poorly, and DARTS operations, except at the two smaller depths where performance is similar.
Moreover, our ResNet-34 XD's MAE$_8$ of 4.0 also improves upon PDNET's reported MAE$_8$ of 4.1 attained by the much deeper Dilated ResNet-258~\cite{adhikari2020};
however, in our reproduction Dilated ResNet-258 achieved an MAE$_8$ of $3.5$.
Given the trend in Figure~\ref{fig:protein}, where XD-operations consistently improve the backbone architecture of the same depth, we conjecture that ResNet-258 XD could further improve upon this result.
We leave scaling XD-operations to such deeper networks to future work.
\begin{figure*}[!t]
\begin{minipage}{\linewidth}
\centering
\begin{threeparttable}
\captionof{table}{\label{tab:seq}
XD-operations compared to recent results in music modeling.
We report average loss across three trials.
The best result on each task is {\bf bolded}.
}
\begin{tabular}{lcccccccc}
\toprule
Method (source) &&&& JSB Chorales &&&& Nottingham \\
\midrule
Best recurrent \citep{bai2018tcn} &&&& 8.43 &&&& 3.29 \\
TCN \citep{bai2018tcn} &&&& 8.10 &&&& 3.07 \\
Transformer \citep{wang2020rtransformer} &&&& - &&&& 3.34 \\
R-Transformer \citep{wang2020rtransformer} &&&& - &&&& {\bf 2.37} \\
\midrule
Undilated TCN (our baseline) &&&& $8.16\pm0.04$ &&&& $3.23\pm0.02$ \\
TCN (reproduced) &&&& $8.17\pm0.01$ &&&& $2.97\pm0.01$ \\
{\bf Undilated TCN XD} (ours) &&&& ${\bf 8.07\pm0.01}$ &&&& $2.84\pm0.02$ \\
\bottomrule
\end{tabular}
\end{threeparttable}
\end{minipage}
\end{figure*}
\section{Application: Music Modeling}\label{sec:seq}
\vspace{-1mm}
Our final application is to music modeling, i.e. learning to predict the next note from sheet music~\citep{allan2005chorales}.
The dominant approaches for such tasks are recurrent nets~\citep{hochreiter1997lstm} and Transformers~\cite{vaswani2017attention}, but recent work has shown that specially-designed convolutional models can also be made competitive at similar model sizes \citep{bai2018tcn,bai2019trellis}.
We will consider the temporal convolutional network (TCN) \cite{bai2018tcn}, which improves upon a regular CNN by having the dilation factor grow exponentially across layers.
The tasks we study are on the JSB Chorales and Nottingham corpora, used in the original evaluation of TCNs \citep{bai2018tcn}.
As the baseline we take the TCN and set all dilation factors to one (undilated);
our goal will be to start with this undilated network and match or outperform the custom dilation design of the TCN.
The results presented in Table~\ref{tab:seq} show that we achieve this goal, as we outperform both the undilated baseline and the TCN on both tasks.
While the simple undilated backbone that we initialize with turns out to already match the TCN on JSB Chorales, on Nottingham our approach demonstrates that XD-operations can be used to outperform hand-designed architectures starting from vanilla CNNs.\footnote{In the appendix we report similar improvements on two other tasks on which TCNs were evaluated---
permuted MNIST and Penn TreeBank---that we do not discuss in detail as our focus is on under-explored tasks.}
Where possible we also compare to other known results;
XD-operations outperforms all of these except the R-Transformer \citep{wang2020rtransformer}, a model combining recurrent nets and self-attention, on Nottingham.
Together with our results on PDEs and proteins, our study of music modeling provides further evidence that XD-operations can effectively find good operations using standard backbones on diverse tasks.
One notable difficulty here is causality enforcement:
making sure the input data does not contain the target when predicting the next entry.
While TCNs can efficiently do so via temporal shifts, we do it in a brute-force manner by treating sequences of length $n$ as $n-1$ data-points with masked targets.
This is expensive and thus limits our evaluation to small music tasks.
A fruitful direction for future work is thus to examine whether it is possibly to directly enforce causality in XD-operations, e.g. by forcing architecture parameters $\*K$ and $\*M$ to be lower triangular;
since a product of lower triangular matrices is again lower triangular, the entire operation is then a multiplication of the input sequence by a lower triangular matrix, which suffices to prevent causality violations.
| {
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Performed via a self-sealing, clear corneal incision, ABiC™ can be combined with cataract surgery in order to optimize time in the operating room, or performed as a stand-alone procedure. To date, ABiC is the only MIGS procedure approved for use outside cataract surgery in the USA. Based on a 228-eye case series by Mark J. Gallardo, MD (El Paso Eye Surgeons, PA) and Mahmoud A. Khaimi, MD (Dean McGee Eye Institute, OK), ABiC achieved a 36.74% reduction in IOP at 12 months (21.5 ± 7.4 mm Hg at baseline to 13.6 ± 1.20 mm Hg).
Featuring a proprietary design, iTrack™ is the only illuminated, micron-scale microcatheter designed to viscodilate Schlemm's canal during MIGS with ABiC. The stand-out feature of the iTrack™ is its illuminated tip, which allows you to monitor its location at all times to ensure you do not enter any of the collector channel ostia. It comprises a small gauge support wire to provide real-time, tactile feedback as you progress through Schlemm's canal: this allows you to feel if the canal is tight, grainy or completely open. It also makes it possible for you to maneuver through tight areas of the canal. Additionally, you can precisely control the delivery of viscoelastic, as required, to overcome blockages or any areas of resistance. | {
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<!-- Do not edit this file. It is automatically generated by API Documenter. -->
[Home](./index.md) > [puppeteer](./puppeteer.md) > [Protocol](./puppeteer.protocol.md) > [Runtime](./puppeteer.protocol.runtime.md) > [EntryPreview](./puppeteer.protocol.runtime.entrypreview.md) > [key](./puppeteer.protocol.runtime.entrypreview.key.md)
## Protocol.Runtime.EntryPreview.key property
Preview of the key. Specified for map-like collection entries.
<b>Signature:</b>
```typescript
key?: ObjectPreview;
```
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,981 |
Q: Bizarre and Temporary black screens in ubuntu 20.04.3 LTS After maximizing applications, sometimes they cause the screen to go blank - maximizing YouTube will work fine, until it doesn't and I have to exit full screen using a keybind, and go back into it. Games will either alternate rapidly between the screen functioning, and it working - or, it will not work during certain scenes. A good example is in Valheim, during regular gameplay it will exhibit the behavior above, but during the menu with the campfire where you choose a character the screen will completely blank, and only alt-tabbing out of the game, or running it in windowed mode will fix it. all games tested (99% of which were native) will replicate this issue.
Sound will still play, the game is still running - but even the Ubuntu UI is not visible, the screen appears the turn off, but minimizing or running it windowed will immediately fix it.
Neofetch Output and Specs
Intel i5 4460
RX 570 4GB
SSD 240Gb
HDD 500GB
Asus H81
16GB DDR3
600W EVGA PSU
RGB strip for extra FPS
There are also some cavats: I'm currently running the https://repo.radeon.com/amdgpu/21.40.2/ubuntu repo, as I had issues with the GPU on Ubuntu 21 until i downgraded to 20 LTE. Another is I'm using a D-sub to VGA cable - this is because of the design of my case, as a lip near the more useful ports on the card block the plastic housing of the cables, but not the thin metal of the d-sub connector. this has suited me fine so far, as the other two HDMI connections on my monitor are currently occupied - but its worth mentioning
| {
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} | 3,540 |
{"url":"https:\/\/ryanwingate.com\/projects\/hypothesis-testing-2\/stroop\/","text":"# Analyzing the Stroop Effect using T-Tests\n\n## Introduction\n\nThe Stroop effect is a psychological phenomenon that impacts the reaction time required for certain cognitive tasks. The most common manifestation of the effect occurs during the performance of a \u201cStroop test,\u201d during which participants are asked to name the font color a list of colors are printed in.\n\nThere are two types of lists the participants read from. \u201cCongruent\u201d lists are those where the written color and the font color are the same. \u201cIncongruent\u201d lists are those where the written color is different from the font color.\n\nCongruent:\n\nBlue Red Green Red Blue Blue Green\n\nIncongruent:\n\nBlue Red Green Red Blue Blue Green\n\nThe idea is that it takes participants significantly longer to state the color each word is written in for the incongruent list than for the congruent list.\n\n## Identification of Variables\n\nIndependent variable: which list is shown to the participant.\n\nDependent variable: the amount of time it takes the participant to name the ink colors for each list.\n\n## Select Statistical Test\n\nSince the goal of this anlysis is to infer something about the population from limited samples, the hypothesis will be written in terms of the population mean, $\\mu$.\n\nThe appropriate statistical test to perform is the \u201cDependent Samples T Test.\u201d This test is also known as the \u201cPaired Samples T Test.\u201d This is the appropriate test because the data involve two measurements performed on the same person under different conditions, and contains no information about the population.\n\n## Establish Hypotheses\n\nSince the goal is to prove the presence of the stroop effect, the appropriate structure for the set of hypotheses is to make the implicit assumption that the stroop effect does not exist.\n\nThat is, the alternative hypothesis is that the mean difference between the time required to read the incongruent and congruent lists is greater than zero. Then, the null hypothesis (initially assumed to be correct) is that the mean difference between the time required to read the incongruent and congruent lists is zero.\n\nRecall that congruent means the ink color is the same as the written color, whereas incongruent means the ink color is different from the written color.\n\nWritten algebraically, the foregoing becomes the following\u2026\n\n$$H_0: \\mu_{diff} = 0$$ $$H_A: \\mu_{diff}> 0$$\n\nwhere, $\\mu_{diff}$ is defined as the average of the pairwise differences, $x_{diff}$:\n\n$$x_{diff} = x_{incongruent}-x_{congruent}$$\n\nThe fact that the samples are dependent is why also why the hypotheses are written in terms of the mean of differences, rather than the difference of means, although those approaches are only subtely different. Rarely would they produce different results.\n\n## Calculate Descriptive Statistics to Build Intuition\n\nimport pandas as pd\nimport matplotlib.pyplot as plt\nfrom scipy import stats\n%matplotlib inline\n\ndf.shape\n(24, 2)\n\n\n24 participants completed the study.\n\ndf['difference'] = df['Incongruent'] - df['Congruent']\ndf.head()\nCongruent Incongruent difference\n0 12.079 19.278 7.199\n1 16.791 18.741 1.950\n2 9.564 21.214 11.650\n3 8.630 15.687 7.057\n4 14.669 22.803 8.134\ncong_mean = df.Congruent.mean()\ncong_std = df.Congruent.std()\nincong_mean = df.Incongruent.mean()\nincong_std = df.Incongruent.std()\ndiff_mean = df.difference.mean()\ndiff_std = df.difference.std()\ndiff_min = df.difference.min()\n\nprint(' Congruent data - mean : {:.1f}'.format(cong_mean))\nprint(' Congruent data - std dev : {:.1f}'.format(cong_std))\nprint('\\r')\nprint('Incongruent data - mean : {:.1f}'.format(incong_mean))\nprint('Incongruent data - std dev : {:.1f}'.format(incong_std))\nprint('\\r')\nprint(' Difference data - mean : {:.1f}'.format(diff_mean))\nprint(' Difference data - std dev : {:.1f}'.format(diff_std))\nprint('\\r')\nprint(' Difference data - min : {:.1f}'.format(diff_min))\n Congruent data - mean : 14.1\nCongruent data - std dev : 3.6\n\nIncongruent data - mean : 22.0\nIncongruent data - std dev : 4.8\n\nDifference data - mean : 8.0\nDifference data - std dev : 4.9\n\nDifference data - min : 1.9\n\n\nThe descriptive statistics above imply that the null will be rejected, for two reasons: 1. A positive mean difference, with magnitude of nearly twice the standard deviation. 2. There are no negative values in the difference data, as evidenced by the minimum value.\n\n## Visualize the Data\n\ndef print_hists():\nfig, axs = plt.subplots(1, 3, figsize=(16,4))\nbins = range(0,35,5)\ncolors = ['lightblue', 'pink', 'lightgreen']\nmeans = [cong_mean, incong_mean, diff_mean]\n\naxs[0].hist(df.Congruent, label='Congruent',\ncolor=colors[0], bins=bins);\n\naxs[1].hist(df.Incongruent, label='Incongruent',\ncolor=colors[1], bins=bins);\n\naxs[2].hist(df.difference, label='Difference',\ncolor=colors[2], bins=bins);\n\nfor i, ax in enumerate(axs):\nax.axvline(means[i], color='black');\nax.legend();\n\nfor label in ['top', 'bottom', 'right', 'left']:\nax.spines[label].set_visible(False)\ndef print_boxplots():\nfig, ax = plt.subplots(1, 1, figsize=(16,6))\ncolors = ['lightblue', 'pink', 'lightgreen']\n\nbp = ax.boxplot([df.Congruent, df.Incongruent, df.difference],\nlabels = ['Congruent', 'Incongruent', 'Difference'],\npatch_artist = True);\n\nfor label in ['top', 'right', 'left']:\nax.spines[label].set_visible(False)\n\nfor i, box in enumerate(bp['boxes']):\nbox.set(color = colors[i], linewidth = 3)\n\nfor i, whisker in enumerate(bp['whiskers']):\nwhisker.set(color = colors[i\/\/2], linewidth = 3)\n\nfor i, cap in enumerate(bp['caps']):\ncap.set(color = colors[i\/\/2], linewidth = 3)\n\nfor i, flier in enumerate(bp['caps']):\nflier.set(color = colors[i\/\/2], linewidth = 3)\nprint_hists()\n\nThe three histograms above are plotted on a common x-axis. The values of the incongruent data appear to be generally higher than the congruent data.\n\nprint_boxplots()\n\nThe box plots, above, provide another way to visualize the information in the histograms. Both types of plots appear to provide further indication of the presence of the Stroop effect. The null hypothesis is expected to be rejected based on the distributions of the data shown above.\n\n## Perform the Statistical Test\n\nTo perform the statistical test, I calculate the \u201cp-value,\u201d which is the strength of evidence in support of the null hypothesis.\n\nAs a standard of comparison for the p-value, I adopt the standard $\\alpha$, or \u201cSignificance Level,\u201d of 0.05. This is the probability of rejecting the null hypothesis when it is actually true. To arrive at a p value for this set of data, I calculate the t-statistic, which in turn requires the standard error:\n\n$$SE = \\frac{\\sigma}{n^\\frac{1}{2}}$$\n\nwhere $\\sigma$ is the standard deviation of the sample, and $n$ is the number of samples.\n\nThe t-statistic is calculated as:\n\n$$t_{stat} = \\frac{\\mu_{diff}}{SE}$$\n\nwhere $\\mu_{diff}$ is the standard deviation of the differences between the congruent and incongruent test durations for each participant.\n\nFinally, the p-value is calculated using the Cumulative Distribution Function, which requires as inputs the $t_{stat}$ and degrees of freedom. The degrees of freedom are calculated as $n-1$, where $n$, as above, is the number of samples.\n\n$$p_{value} = 1-CDF(t_{stat}, n-1)$$\n\ndiff_standard_error = diff_std \/ (df.shape[0]**0.5)\nt_stat = diff_mean \/ diff_standard_error\np_value = 1-stats.t.cdf(t_stat, df.shape[0]-1)\n\nprint('Difference standard error = {:.3f}'.format(diff_standard_error))\nprint(' T-stat for test data = {:.2f}'.format(t_stat))\nprint(' P-value for given t-stat = {:.9f}'.format(p_value))\nDifference standard error = 0.993\nT-stat for test data = 8.02\nP-value for given t-stat = 0.000000021\n\n\nThe p-value, shown above, is the probability of rejecting the null hypothesis when it is actually true. Another value, called the \u201cConfidence Level,\u201d is 1 minus $\\alpha$, expressed as a percentage. For a significance level of 0.05, the confidence level is 95%.\n\nThe T-stat for our test data is 8.02, which, for 23 degrees of freedom, corresponds to a P-value much less than 0.05. Since the P-value is less than the significance level, I reject the null hypothesis. The conclusion is that the alternative hypothesis is true, and that the Stroop effect has had a statistically-significant impact.\n\nHaving taken this particular test myself before, I am not surprised by the results. Even with significant concentration it is difficult to quickly say the ink color a word is written in when that word itself spells out a color.","date":"2019-10-16 16:18:15","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6723131537437439, \"perplexity\": 1671.0306394709442}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-43\/segments\/1570986668994.39\/warc\/CC-MAIN-20191016135759-20191016163259-00081.warc.gz\"}"} | null | null |
Raul Maria Xavier (Macau, 23 de Março de 1894 — Lisboa, 1 de Janeiro de 1964) foi um escultor português.
Biografia / Obra
Filho e neto de chinesas casadas com portugueses, nasce em Macau, vindo para Portugal ainda criança. Frequenta a Escola de Belas-Artes de Lisboa, onde é aluno de Ernesto Condeixa e, mais tarde, de Costa Motta (tio), cujo ateliê viria a frequentar, familiarizando-se com a técnica do talhe direto. Interrompe os estudos académicos para realizar um estágio em Itália (bolseiro do estado português), onde estuda escultura religiosa.
Participou em Salões da SNBA (1ª medalha em 1941), nas Exp. de Artes Plásticas, Grupo Silva Porto (1945-58). Colaborou na Exposição do Mundo Português, 1940 (baixo-relevo Aljubarrota, 1940). Colaborou artisticamente em publicações como a II série da revista Alma nova (1915-1918), iniciada em Faro em 1914.
Raul Xavier desenvolveu uma obra escultórica de feição clássica, marcada pela serenidade e pela depuração das formas, tendo-se dedicado em especial ao retrato e à estatuária.
Cabeça, 1938 (col. Museu do Chiado), parece captar o gosto modernista pela arte primitiva, a par de lembranças da origem macaense do autor; revela ainda um curioso gosto decorativo (repare-se no entrançado do cabelo ou no dinamismo gerado pelos tracejados gravados em toda a superfície). "Deste entrecruzar de citações, o retrato, pretexto para a análise sintética e decorativa de uma face, sugere uma imagem construída numa volumetria de máscara".
A 2 de maio de 1959, foi agraciado com o grau de Cavaleiro da Ordem Militar de Sant'Iago da Espada.
Em 2011 o Museu Municipal Santos Rocha, Figueira da Foz, realizou uma exposição de trabalhos de sua autoria.
Algumas obras
∑ 10 items.
Obras
Estátua em memória de Ataíde Oliveira, Loulé (inaugurada em 1930).
A Arte, 1935, Pavilhão dos Desportos, Parque Eduardo VII, Lisboa – Escultura em pedra representando uma figura feminina majestática e simbólica. "Trata-se de uma alegoria à Arte, cujos atributos incluem pincéis, maceta e escopro, na mão esquerda, e um torso na mão direita, fazendo alusão à pintura e à escultura".
A Ciência, Pavilhão dos Desportos, Parque Eduardo VII, Lisboa.
Santo António, Igreja de Nossa Senhora de Fátima, Lisboa.
Monumento a Nuno Álvares Pereira (ou Monumento à Batalha), 1959, S. Jorge, Leiria – De inspiração clássica e assinalável depuração formal, este baixo-relevo é uma das obras mais emblemáticas de Raúl Xavier. O monumento é antecedido de uma plataforma com vários degraus, impondo-se ao espectador como um «friso» monumental concavo, numa alusão à batalha de Aljubarrota (mas também às batalhas dos Atoleiros e de Valverde) travada entre os exércitos português e castelhano, numa exaltação à vitória dos portugueses liderados por D. Nuno Álvares Pereira.
Estátua de São Vicente, 1949-1967, Largo das Portas do Sol, Lisboa – Os primeiros estudos para esta estátua de homenagem a São Vicente foram realizados por Raul Xavier em Junho de 1949. "A passagem da peça ao material definitivo, a pedra, só ocorreu entre 1965 e 1967, pelas mãos de seu filho, o Arq. Luís Xavier, que acompanhou esta obra até ser inaugurada em 25 de Outubro de 1970".
Nossa Senhora de Fátima, Palácio Nacional de Queluz.
Busto do Coronel Mesquita, Museu Militar de Lisboa.
Busto de Camões, Universidade da Califórnia, E.U.A.
Busto de Rafael Bordalo Pinheiro, Museu Rafael Bordallo Pinheiro, Campo Grande, Lisboa.
Busto de António Cabreira, Tavira.
Leões da Escadaria e estátua da Prudência, Palácio de São Bento, Lisboa.
Bibliografia
SCHAUB-KOCH, Prof. Emile – Raul Xavier, Sculpteur Portugais. Lisbonne, 1957.
Escultores de Portugal
Cavaleiros da Ordem Militar de Sant'Iago da Espada | {
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Die Liste der Gedenktafeln in Berlin-Gesundbrunnen enthält die Namen, Standorte und, soweit bekannt, das Datum der Enthüllung von Gedenktafeln.
Die Liste ist nicht vollständig.
Gesundbrunnen
!
!Gedenktafeln in Gesundbrunnen
!Berlin-Gesundbrunnen
Gedenktafeln in Berlin-Gesundbrunnen | {
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Politically Speaking – December 22, 2020
Cathy Stuehmeier
Trump Hoisted With His Own Petard
by Rich Lowry
One of President Donald Trump's foremost achievements has been to erect a formidable obstacle to his own post- election legal challenges.
The federal judiciary, now seeded throughout with Trump-nominated judges, has given the back of its hand to pro-Trump election litigation, with Trump judges issuing notably harsh opinions.
It's always been strange that Trump, who will never be mistaken for a rigorous constitutionalist and who personalizes everything, has elevated a couple of hundred judges who, by and large, are deeply committed to the Constitution and feel no particular personal loyalty to him.
Hardly an institutionalist, Trump has buttressed the institution of the judiciary. Not one to honor norms, he's generally nominated sticklers for them to the bench.
The paradox reached its height in the weeks after the election. Trump and his allies launched a battery of litigation asking for millions of votes to be thrown out or elections to be decertified, hoping to catch a break from a judge somewhere or from the U.S. Supreme Court. Trump himself put out a call for "courage" from a justice or justices.
Instead, the Trump team got nowhere, even with Trump-nominated jurists.
At a fraught time when most Republican elected officials in Washington were keeping their heads down, Trump judges involved in post-election litigation issued their rulings without fear or favor. They have shown a commitment to facts, reason and the law, and great institutional self confidence.
Progressives portrayed Trump judges as right-wing hacks. Elizabeth Warren called Trump's picks "aggressively unqualified," while the editor of the leftwing website Talking Points Memo, Josh Marshall, tweeted: "The federal judiciary is corrupt. The Supreme Court is the most deeply corrupted."
With the president of the United States raging at our electoral system and desperately seeking assistance from the courts, the alleged partisanship and corruption of the Trump influenced judiciary has been nowhere in evidence.
Trump nominated University of Pennsylvania law professor Stephanos Bibas to the 3rd U.S. Circuit Court of Appeals in 2017. The signature Trump litigation in Pennsylvania landed in his lap, and Bibas wrote an unsparing opinion for a unanimous panel of the 3rd Circuit dismissing it.
In Georgia, U.S. District judge Steven D. Grimberg, nominated by Trump last year, denied a request by Trump super fan Lin Wood to stop the certification of the results.
And the Supreme Court denied a request to block certification of the Pennsylvania results in a curt, one-sentence order with no public dissents.
One of the main Democratic lines of attack on Justice Amy Coney Barrett during her confirmation was that, as Illinois Sen. Dick Durbin said, she was "being sent on assignment to the Supreme Court by President Trump" in order to "be there if the president needs her on an election contest."
Where does Barrett go to get her apology?
Surely, the overtly transactional Trump would have been happy for her to have actually been his political tool on the court, along with his other two picks. But Barrett and her colleagues, trained and soaked in the law and profoundly cognizant of their institutional role, are not susceptible to such influence. Neil Gorsuch is not Corey Lewandowski; Stephanos Bibas is not Rudy Giuliani.
Nothing underlines the merits of Trump judicial selections quite like their willingness to deny him and his allies, as warranted.
Rich Lowry is editor of the National Review. © 2020 King Features Synd., Inc.
Editors Note: After reading Rich's column this week I find the question, "Where does Supreme Court Justice Amy Barrett go to get her apology from Durbin?" I would say don't hold your breath. Recently I had a dear friend share a letter with me asking a fair question of Durbin. He never did get a response! Imagine that. The letter is below.
Dear Senator Durbin,
With so many Americans unemployed during this pandemic, why is it that our elected representatives in Congress and the Senate, as well as State Governors and City Mayors continue to receive full salaries and benefits? These people, our elected officials, made the decision to close down our economy and the jobs of millions of Americans. Why are they not held to the same standards as we the people? Why don't they willingly forego their salaries until we defeat this assault on our nation? Maybe if they were in the same boat with the rest of us, they might be more inclined to work together developing and implementing solutions to our problems instead of continuing to have these schoolyard arguments and get nothing done?
So with all that said, in my opinion, shouldn't we all be in the same boat? Make the rules; live by them.
Until next week, let me wish you and yours a very Merry Christmas. May God bless and keep you all, from myself and the team.
Experiment wipes out disease-carrying mosquitoes
Warriors coach calls 33-point loss embarrassing
Rep. Waters: I want to talk impeachment
Beyonce, Jay-Z, Beckham went where?
Remains of Confederate general will be removed from Memphis park
Toobin: A 'grotesque abuse of power'
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Uberfood: Oatmeal - Lean It UP!
Oatmeal is one of the healthiest foods on the planet. Really…oatmeal!? Yup, it's up there with other health powerhouses like salmon, acai, almonds, quinoa, flax, chia, and kale–just to name a few.
In its natural form (not with added sugar or preservatives, I'm talking about 100% natural Quaker oats/steel-cut oats) oatmeal is as healthy as it gets, and it's perfect for weight-loss. Here are a few of the reasons why oatmeal shoots to the top of health food kingdom.
Oatmeal is a slow-digesting, complex carb that's loaded with fiber. That translates into long-lasting energy, minimal fat storage, and reduced appetite. Oatmeal packs a BIG surge of energy at the start of the day. It's literally the perfect breakfast food.
The high fiber content promotes digestive health, fights cancer and heart disease, and cleans toxins from the body.
Oatmeal regulates blood sugar levels and prevents insulin from spiking. Low insulin levels are the key to losing body fat and sculpting a flat belly. For anyone looking to shred body fat and reveal your abs, meet your new best friend.
Oatmeal only has 150 calories per half cup. For so few calories, oatmeal is EXTREMELY filling.
5. Good source of protein.
Oatmeal has 5g of protein naturally, but if you cook it with milk the protein content jumps up to 11g per serving. That's A LOT for a carbohydrate! The milk also adds calcium and vitamin D.
Oatmeal has no sugar or refined carbs whatsoever. See my bullet point about insulin levels.
In its plain form, oatmeal is sodium free, which keeps blood pressure levels down.
Oatmeal is a great source of vitamin A, iron, manganese, the B vitamins, and selenium–a natural antioxidant.
Its got one ingredient–rolled oats. That means no preservatives, chemicals, added sugar, or artificial sweeteners.
10. It's an ideal weight-loss food.
B. It's quick and easy to make.
1. Dump oatmeal in bowl.
Okay, okay…oatmeal can be a little boring when it's made plain with water, but the beauty of oatmeal comes with its versatility. By itself oatmeal is incredibly healthy–customize it with a few au-natural toppings and you're in a whole new stratosphere of nutrition. Here are a few recipes and tips to create delicious, creative oatmeal every time.
Oatmeal with 1%, 2%, skim, or almond milk is absolutely delicious when compared to the watered-down variety. Check plus for the additional protein.
Cinnamon is the perfect spice. It packs HUGE flavor and health benefits–all for zero calories. Cinnamon is an uber powerful antioxidant, controls blood sugar levels and insulin (back to the fat burning factor), and can actually boost metabolism.
My go to meal. The bananas add potassium, vitamin C, vitamin B6, magnesium, and additional fiber.
Try this if you're a fan of Apple Cinnamon Cheerios. Keep the skin on the apples, you'll triple the fiber intake.
Honey in it's raw form is a natural antioxidant. Like fruit, honey is made primarily of fructose–a slow digesting sugar.
Strawberries, blueberries, raspberries, and blackberries are all a great source of antioxidants, vitamin K, vitamin C, and fiber.
8. Ground Up flax seeds.
Flax seeds are an amazing source of omega-3 fatty acids–fats that improve cholesterol and overall heart health. They're also high in antioxidants.
As traditional as it gets. Simple. Classic. Delicious.
11. Peanut butter and jelly.
Savory + sweet, all at once. The PB contributes loads of healthy fats, and also adds thickness.
Mixing in walnuts, almonds, cashews, and peanuts adds healthy fats and a nice crunch.
13. Almond Joy – Shredded coconut and almonds.
The almonds add healthy fats. Coconut contains a healthy type of saturated fat called lauric acid which has been shown to boost HDL (good cholesterol), while keeping LDL levels neutral.
Follow Lean It UP on Twitter for real-time fitness/nutrition tips, advice, info, and updates. | {
"redpajama_set_name": "RedPajamaC4"
} | 6,516 |
Q: React Native with Redux Toolkit - Dispatch function produces TypeError: action is undefined Thank you in advance for your time!
As the title describes, I am trying to fetch data from a json file asynchronously and store it in the state with Redux's latest Toolkit version. Unfortunately, running
dispatch(getCities())
Produces Uncaught TypeError: action is undefined
I would appreciate any insight on this! Below goes the code:
STORE.JS
import { configureStore } from "@reduxjs/toolkit";
import citiesReducer from "../features/cities/citiesslice";
const store = configureStore({
reducer:{
cities: citiesReducer
}
})
export default store;
CITIESSLICE.JS
import { createSlice, createAsyncThunk } from "@reduxjs/toolkit"
const API_URL = 'https://warply.s3.amazonaws.com/data/test_pois.json'
export const getCities = createAsyncThunk(
"cities/getCities",
async (dispatch, getState)=>{
return await fetch('https://warply.s3.amazonaws.com/data/test_pois.json')
.then(
console.log('test')
)
}
);
const citiesSlice = createSlice({
name: "city",
initialState: {
cities: [],
loading: 'idle'
},
extraReducers:{
[getCities.pending]: (state,action)=>{
state.loading = "pending"
},
[getCities.fulfilled]: (state, action)=>{
state.loading = "succeded"
state.cities = action.payload
},
[getCities.rejected]: (state, action)=>{
state.loading = "failed"
}
},
})
export default citiesSlice.reducer;
MAPLIST.JS
import React, { useEffect } from 'react';
import { SafeAreaView, StyleSheet, Image, Text, View } from 'react-native';
import getCities from '../features/cities/citiesslice';
import { useDispatch, useSelector } from 'react-redux';
function MapList(props) {
const dispatch = useDispatch();
const {cities}= useSelector(state => state.cities)
useEffect(()=>{
dispatch(getCities())
},[])
return (
<SafeAreaView style={styles.container}>
<View style={styles.topnav}>
<Image source={require('../assets/warply_w.png')} style={styles.logo}/>
</View>
{cities && cities.map((cities, i)=> <Text key={i}>test</Text>)}
</SafeAreaView>
);
Styles not included in the above excerpt as likely not associated with issue
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,231 |
Sven Torsten Gunnar Unger, född den 8 oktober 1947, död den 1 februari 2023 i Nice, var en svensk advokat. Han var son till Gunnar Unger och bror till Johan Unger.
Unger avlade juris kandidatexamen vid Lunds universitet 1972 och genomförde tingstjänstgöring 1973–1975. Han arbetade från 1978 som advokat vid Carl Swartling advokatbyrå respektive advokatbyrån Mannheimer Swartling, åren 1983–2010 som delägare. Han var 1992–1996 ordförande för Sveriges advokatsamfund. Han var också ledamot av Pressens opinionsnämnd och vice ordförande i Stockholms Handelskammare.
År 1979 blev Unger ledamot av Uarda-akademien i Lund och 1993 hedersledamot av Malmö nation. Han var 2017–2019 ordförande i Alf Henrikson-sällskapet.
Skrifter
Böcker
Ord i bild – Från abrakadabra till övertoner. En bok om bildspråk, bakgrund och betydelser. 2017, Oscarsfors
Ordet är Fritt – mycket fritt. 2015, Oscarsfors.
Bolagsstämma i praktiken. En teoribok. 2014, Oscarsfors.
Inga ord efter mig. Ett litet, litet utsnitt ur svenska språkets öden och äventyr. 2012, Oscarsfors.
Allehanda om skadestånd i avtalsförhållanden. (tillsammans med Bertil Bengtsson och Harald Ullman), 2009, 2013 och 2019, Jure.
Det uppochnedvända timglaset – en samling betraktelser, Stockholm 1999 (2:a upplagan 2001), Jure.
Artiklar
Dikt och ära. Tidskrift för Sveriges advokatsamfund. TSA 1982:6-7.
Om språkfel och fel språk. TSA 1982:4.
Författningsnisse. TSA 1983:2.
Om protokoll. Advokaten 1986:2-3.
Får jag tala med koncernchefen! Advokaten 1987:1.
Får myndigheter i avtal åtaga sig att iakttaga sekretess? Festskrift till Sveriges advokatsamfund 1887–1987, 1987.
Sveriges Rikes Lag 1989 – en recension. Advokaten 1989:februari.
Konsten att inte sluta avtal. Festskrift till Gotthard Calissendorff, 1990.
Om bevis. Advokaten 1991:2.
Mellan hägg och stämma. Advokaten 1992:2.
Om domstols och skiljenämnds behörighet att förelägga vite i dom om fullgörelse. Festskrift till Håkan Strömberg, 1992.
När blir ett middagssällskap ett enkelt bolag? Maträtt, 1994.
Att odla vildflor. Rättsfall att minnas. Skrift till Jan Hellner, 28 oktober 1997.
Professionsansvar för advokater och andra rådgivare. Inledning vid Nordiska Juristmötet i Helsingfors, 2002.
Recension av "Gastronomiska undersökningar" – en festskrift av Nils Jareborg. Juridisk Tidskrift 2002/03:3, Juridisk Tidskrift.
Special Features of Swedish Corporate Governance. Skrift publicerad av Kollegiet för svensk bolagsstyrning, 2006, Kollegiet för svensk bolagstyrning.
Rätt och språk. Vänbok till Sten Heckscher, 2012.
Advokatverksamhet i aktiebolag. Sveriges advokatsamfund 125 år, 2012.
Regler om advokats marknadsföring. Sveriges advokatsamfund 125 år, 2012.
Malmö och näktergalen. Om Hjalmar Gullberg. Jubileumsskrift Malmö Nation 125 år, 2015.
Om avtals slut – särskilt om frågan: när upphör ett avtal som inte säger när det upphör? Boken "Bertil Bengtsson 90 år", 2016.
GARANTERAT FÖR SENT eller OM LAGS BEGRIPLIGHET jämte vissa utvikningar, Festskrift till Stefan Lindskog, 2018.
Om att frånträda styrelseuppdrag, Vänbok till Anne Ramberg, 2019.
Sven Unger har varit ledamot av redaktionskommittén för följande skrifter.
Festskrift till Sveriges advokatsamfund 1887 – 1987. 1987.
Festskrift till Gotthard Calissendorff (redaktör tillsammans med Leif Thorsson), Stockholm 1990.
Rättsvetenskapliga studier till minnet av Tore Almén (redaktör tillsammans med Torgny Håstad och Anders Knutsson), Uppsala 1999.
Bertil Bengtsson 90 år (redaktör tillsammans med Severin Blomstrand, Mia Carlsson, Dag Mattsson och Anna Skarhed), Stockholm 2016.
Externa uppdrag
Ledamot av juridiska linjenämnden vid Stockholms universitet, 1986–1992.
Ordförande i Sveriges advokatsamfund, 1992–1996.
Ordförande i Klubb S:t Erik, för studenter som tävlar i nordisk processtävling, 1997–2008.
Styrelseledamot i European Air Law Association, 1998–2007.
Ledamot av Pressens Opinionsnämnd, 1999–2005.
Styrelseledamot i AB Handelns Utredningsinstitut, 1997–2011.
Ordförande i Stiftelsen Solstickan, 2000–2007.
Vice ordförande i Stockholms Handelskammare, 2004–2010.
Ordförande i Stiftelsen Stockholms Sjukhem, 2007–2014.
Ordförande i Emil Heijnes stiftelse för rättsvetenskaplig forskning 2011–2018; ledamot sedan 1991.
Styrelseledamot i Helge Ax:son Johnsons Stiftelse sedan 2015.
Styrelseledamot i Stina och Erik Lundbergs stiftelse sedan 2015.
Ordförande i Alf Henrikson-sällskapet 2017–2019.
Källor
Noter
Svenska advokater
Hedersledamöter vid Malmö nation
Alumner från Lunds universitet
Födda 1947
Avlidna 2023
Män
Svenska jurister under 1900-talet | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,145 |
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\chapter{Introduction}
\thispagestyle{myheadings}
The classic \textit{isoperimetric problem} is to find the
largest amount of area that
can be enclosed using a simple closed curve of fixed length. The
answer is of course
a circle, although the proof is more difficult than some may realize.
The ancient Greeks knew the problem
and the solution. In fact, Pappus records that
Zenodorus found the solution first \cite{Polya}. The first
mathematical proof, however, is
credited to Steiner \cite{Steiner} in the 19th century. He proved that if
a solution exists, then it must be a circle. Caratheodory\footnote{Blaschke
\cite{Cara} credits Edler, Carath\'eodory and Study with existence results.
Bandle \cite{Bandle} claims Carath\'eodory was first. Schmidt and
Weierstrauss completed the three dimensional analogue.}
completed the proof by showing that a solution does exist. We refer
to the excellent prefaces in P\'{o}lya and Szeg\"{o}
\cite{Polya} and in Bandle \cite{Bandle} for more historical details.
The solution to the classic isoperimetric problem is stated by the
isoperimetric inequality
\[ A \leq \frac{1}{4\pi} (L)^2, \] where
$L$ is the length of any simple closed curve in the plane and $A$ is the
enclosed area. Equality holds if and only if the simple closed curve is a
circle.
A related problem is to find the simple closed curve with least perimeter
that encloses a given area.
For a single area in $\R^2$ or a single volume in $\R^m$ the problem is
equivalent to the classic isoperimetric problem. What if we wanted, however,
to enclose and separate two different prescribed areas in the plane? The
solution is not two disjoint circles as we can use less perimeter by letting
the two different areas share some of their perimeter.
The solution \cite{Foisy} is a \textit{standard double bubble}
consisting of two chambers enclosed by three arcs of circles all meeting at
angles of $\frac{2\pi}{3}$. (See Figure~\ref{double}.)
\begin{figure}[hbt]
\centerline {\epsfbox {double.eps}}
\caption {A standard double bubble enclosing and separating areas $A_1$ and $A_2$}
\label{double}
\end{figure}
The problem that seeks the least
perimeter closed curve, surface, or hypersurface that encloses and
separates $n$ given volumes in $\R^m$ is called
the \textit{generalized soap bubble problem}.
The name comes from the fact that soap
bubbles minimize surface tension for the fixed volumes of
air enclosed.
Another related problem is to separate or tile $\R^m$ into
equal volume pieces
as efficiently as possible. In $\R^2$, the \textit{honeycomb conjecture}
states that a
regular hexagonal tiling is the least perimeter tiling that
separates the plane into unit area pieces, although it is not immediately
clear what is meant by least perimeter in this infinite region context.
One interpretation is to consider this problem as the limit of the $n$ soap
bubble problem in $\R^2$ as $n \rightarrow \infty$.
Only recently has progress been made on even the smallest cases of the
generalized soap bubble problem. In 1976, Almgren
\cite{Alm} proved the existence and regularity of a solution to the
generalized soap bubble problem in dimensions bigger than 2. Taylor\cite{Tay}
improved
this result for dimension 3 in the same year. In 1992, Morgan \cite{Morgan}
proved existence and regularity of a solution in dimension 2.
The double bubble problem in $\R^2$ (the least perimeter embedded planar graph
that encloses and separates two given areas in the plane) was
solved in 1994 by a group of
undergraduates led by Frank Morgan \cite{Foisy}. The $n$ bubble problem in
$\R^2$ for $n>2$, i.e. the least perimeter graph that encloses and
separates $n$
given areas in $\R^2$ is the focus of this paper.
The general $n$ bubble conjecture is that the least perimeter solution will
always have connected regions. In
particular, we examine three regions and the corresponding triple bubble
conjecture.
\bigskip
\noindent \textbf{Conjecture~\ref{conj1} } \textit{The least
perimeter planar graph that encloses
and separates three finite areas $A_1$, $A_2$, and $A_3$ is a regular triple
bubble complex with four vertices, six edges, and three connected regions.}
\bigskip
\begin{figure}[hbt]
\centerline {\epsfbox {tripbubb.eps}}
\caption{The conjectured minimum graph enclosing areas $A_1$, $A_2$, and $A_3$}
\end{figure}
Although much of this work was completed in the context of the triple bubble
problem, many results are true of
planar bubbles in general. We will specify when a result is valid only
for three regions.
We begin Chapter 2 by looking at
some general considerations about perimeter minimizing planar bubble
complexes. After some definitions, we examine Morgan's existence and
regularity theorem. We specifically look into the length $L(A_1,A_2,A_3)$
needed to
enclose three areas $A_1$, $A_2$, and $A_3$. We prove
\bigskip
\noindent \textbf{Lemma~\ref{continuous} } \textit{The length function
$L(A_1,A_2,A_3)$ is continuous.}
\bigskip
In an attempt to decrease perimeter, it is sometimes convenient to increase
the area enclosed by a region. In Conjecture~\ref{conj2}, we modify the
triple bubble conjecture to allow these area increases.
\bigskip
\noindent \textbf{Conjecture~\ref{conj2} } \textit{Given three finite positive
real numbers $A_1$, $A_2$, and $A_3$, the least perimeter graph that encloses
and
separates three finite areas $B_1$, $B_2$, and $B_3$ such that $B_i \geq A_i$
for all $i$, is a standard triple bubble.}
\bigskip
Using Lemma~\ref{continuous}, we prove that the two conjectures are equivalent
in
\bigskip
\noindent \textbf{Theorem~\ref{triv} } \textit{Let three positive areas
$A_1$, $A_2$, and $A_3$ be given. There exists a least perimeter
triple bubble
complex $\mathcal{B}$ that encloses and separates areas
$B_1$, $B_2$, and $B_3$ with $B_i \geq A_i$. That is, if $\mathcal{C}$ is any
other complex enclosing areas $C_1$, $C_2$, and $C_3$ with
$C_i \geq A_i$, then it must use at least as much perimeter $(\ell (\mathcal{B}) \leq \ell (\mathcal{C}))$.
Furthermore, if the minimizer $\mathcal{B}$ has
connected regions (a standard triple bubble), then it must enclose the given
areas $A_1$, $A_2$, and $A_3$ (i.e. $B_i=A_i$ for all i). }
\bigskip
In Chapter 3, we will look at some restrictions on the shape of
perimeter minimizing planar bubble complexes. In particular, we prove that
connected portions of regions must have
more than two sides, three sided pieces are
determined by the curvature of their edges, and certain other connected pieces
cannot touch the exterior more than once.
In Chapter 4, we examine some additional restrictions imposed on bubble
complexes with equal pressure regions. We show that the number of edges that
bound a portion of a region is limited to three, four, five or
six edges. Then we show
how a few parameters control the shapes of 3-gons, 4-gons and 5-gons.
In addition, we examine the way in which these $n$-gons can meet in a perimeter
minimizing complex. For example,
when a 3-gon shares an edge with another 3-gon, they are both adjacent to
another 3-gon, thus creating a standard triple bubble component.
In Chapter 5, we use the results from Chapters 3 and 4 to solve
a restricted case of the planar
triple bubble problem. Specifically, we prove
\bigskip
\noindent \textbf{Theorem~\ref{main} } \textit{A perimeter
minimizing triple bubble
complex with equal pressure regions and no empty
chambers must be a standard triple bubble. In particular, it has connected regions.}
\bigskip
Since there exists a triple bubble complex enclosing
equal areas with connected, equal pressure regions, we have a partial solution
to the triple bubble conjecture.
\bigskip
\noindent \textbf{Corollary~\ref{cormain} } \textit{The least perimeter
graph that
encloses
and separates three equal areas $A_1$, $A_2$, and $A_3$ without empty chambers
using equal pressure regions is a standard triple bubble.}
\bigskip
The restriction to equal pressure regions is analogous to the work of
L. Fejes T\'{o}th \cite{Toth} in the 1940's.
T\'{o}th proved that the hexagonal honeycomb is the least perimeter\footnote{ Least perimeter in this tiling context means that a limiting
perimeter to area ratio is minimized.} tiling of
the plane with equal area polygonal cells. Since the pressure
difference between two regions is measured by the curvature of the connecting
edge, the restriction to polygonal cells is similar to an equal pressure
restriction. We note this similarity in Corollary~\ref{cortriv}.
\bigskip
\noindent \textbf{Corollary~\ref{cortriv} } \textit{The least perimeter graph
that encloses and separates three equal areas with
convex cells and without empty chambers is a standard triple bubble.}
\bigskip
In another corollary to Theorem~\ref{main}, we
show a bubble complex cannot be a solution to the triple bubble problem if it
is close to a regular complex with equal pressure disconnected regions:
\noindent \textbf{Corollary~\ref{cormain2} } \textit{Suppose
$\{\mathcal{A}_i\}$ is a sequence of regular triple bubble complexes
that converges in length and area to a triple bubble
complex $\mathcal{A}$. If $\mathcal{A}$ does
not have any empty chambers, has equal pressure regions and is not a
standard triple bubble, then there exists an $N$ such that for any $i>N$,
$\mathcal{A}_i$ is not a perimeter minimizer for the areas it encloses.}
\bigskip
In Chapter 6, we extend the arguments used in Theorem~\ref{main} to
$n$ bubble complexes ($n>3$) with disconnected, equal pressure regions.
We point out several complexes that cannot be solutions to
any $n$ bubble problem. In particular, we severely restrict the possible
configurations for a perimeter minimizing bubble complex that
encloses and separates four prescribed areas with equal pressure regions.
All of our results agree with the general soap bubble conjecture. That is, we
have not yet found a perimeter minimizing bubble complex
with disconnected regions.
\chapter{Definitions and Preliminaries}
\thispagestyle{myheadings}
A graph is called \textit{finite} if it has a finite number of vertices.
We will consider only finite, planar graphs such that every vertex has
degree at least three.
An embedded planar graph \textit{encloses}
areas $A_1$, $A_2$, \ldots , $A_n$ in $\R^2$
if it separates the plane
into $n+1$ regions (not necessarily connected), $n$ of which contain the
finite areas $A_1, A_2, \ldots , A_n$ respectively. Region $n+1$ is called
the \textit{exterior} region and contains infinite area.
Any non-exterior region is called an \textit{interior} region.
Define an \textit{n bubble complex} to be an embedded planar graph that
encloses some $n$ positive areas.
Given $n$ positive real numbers $A_1, A_2, \ldots , A_n$, the generalized
soap bubble problem in $\R^2$ is to find the least perimeter $n$ bubble complex
that encloses those areas.
Define a \textit{half variation} of an $n$ bubble complex $\mathcal{B}$ to
be a continuous
family $\{ ~ \mathcal{B}_t \mid ~ t\in [0, \epsilon)\}$ of
$n$ bubble complexes such
that $\mathcal{B}_0=\mathcal{B}$. Let
$\ell(\mathcal{B})$ be the function that returns the
length of a bubble complex $\mathcal{B}$.
Suppose $\mathcal{A}$ is an $n$ bubble complex enclosing areas
$A_1$, $A_2$, \ldots , $A_n$. If there exists a half-variation of $\mathcal{A}$
such that the areas enclosed by each $\mathcal{A}_t$ is the same as the
areas enclosed by $\mathcal{A}$
and ${\frac{d\ell(\mathcal{A}_t)}{dt}\mid}_{t=0} < 0$, then $\mathcal{A}$
is not the least
perimeter way to enclose $A_1$, $A_2$, \ldots , $A_n$. The half-variation
defines a deformation of $\mathcal{A}$ that preserves area and
yet decreases perimeter.
Frank Morgan used variational arguments to prove that for any $n$ areas,
a perimeter
minimizer exists and must satisfy certain regularity conditions.
\begin {theorem} (Morgan\cite{Morgan}) For any positive real areas
$A_1,A_2,\ldots,A_n$, there exists a perimeter-minimizing embedded graph
that encloses those areas in $\R^2$. This least perimeter graph
must satisfy the following conditions:
\begin{enumerate}
\item The graph consists of a finite number of vertices, edges, and faces;
\item Edges have constant curvature (arcs of circles or line segments);
\item Vertices are trivalent;
\item Edges meet at angles of ${2\pi} \over 3$;
\item Curves separating a specific pair of regions have the same
curvature; and
\item Any half-variation $\mathcal{A}_t$ that preserves area must not
initially decrease length. That is, there does not exist a half-variation
$\mathcal{A}_t$ such that
\[ {\frac{d\ell(\mathcal{A}_t)}{dt}\mid}_{t=0} < 0.\]
\end{enumerate}
\label{regularity}
\end{theorem}
\bigskip
We will call these six conditions the \textit{regularity conditions}.
We define a \textit{regular n bubble complex} to be an $n$ bubble
complex that satisfies all six of the regularity conditions.
A connected portion of any interior region (i.e. a face of the embedded
graph enclosing a piece of one of the given areas)
will be called an \textit{n-gon}, where $n$ is the
number of edges that enclose the connected
piece. Edges that
separate a part of the exterior region from another region will be called
\textit{exterior edges}. All non-exterior edges will be called
\textit{interior edges}.
Suppose we have a regular $n$ bubble complex $\mathcal{A}$ that encloses areas
$A_1,A_2,\ldots ,A_n$. If we can find a
non-regular $n$ bubble complex $\mathcal{B}$ that contains
the same areas such that
$\ell (\mathcal{A}) \geq \ell (\mathcal{B})$, then $\mathcal{A}$ is not the
least perimeter way to enclose $A_1,A_2,\ldots,A_n$.
Non-regular bubble complexes yield half-variations that preserve area yet
decrease perimeter. In other words, there exists a complex enclosing the
same areas with length strictly less than the length of $\mathcal{B}$ and
therefore less than the length of
$\mathcal{A}$. This argument will be used regularly in the proof of
Theorem~\ref{main} and throughout Chapter 6.
As stated above, Morgan established the existence of a minimum perimeter graph
that encloses
and separates any three areas. Let $L(A_1, A_2, A_3)$ be the function that
gives the minimum length needed to enclose and separate areas $A_1$,
$A_2$, and
$A_3$. In Lemma~\ref{continuous} we show that this length function is continuous. First, however, we prove a lemma we will need.
\begin{lemma}
\label{rescale}
For any $A_1 > 0$, $A_2 > 0$, $A_3 > 0$ and any $\epsilon > 0$,
there exists a $\delta > 0$ such that if
$\left| x-A_{1} \right| \leq \delta$, $\left| y-A_{2} \right| \leq \delta$,
and $\left| z-A_{3} \right| \leq \delta$, then rescaling any
complex enclosing $x$, $y$, and $z$ to
get a complex enclosing $x^{\prime}$, $y^{\prime}$, and $z^{\prime}$ where
$x^{\prime} =A_1$ or $y^{\prime} =A_2$ or $z^{\prime} =A_3$
will result in at most an
$\epsilon$ change in the areas $A_1$, $A_2$ and $A_3$, i.e.
$\left| A_{1}-x^{\prime} \right| \leq \epsilon$,
$\left| A_{2}-y^{\prime} \right| \leq \epsilon$, and
$\left| A_{3}-z^{\prime} \right| \leq \epsilon$.
\end{lemma}
\noindent \textbf{Proof:}
This is simply a consequence of the continuity of rescaling. To be precise,
suppose $A_1$, $A_2$, $A_3$, and $\epsilon$ are given.
By the continuity of $\frac{xA_2}{yA_1}$ for $y\neq 0$, there exists a
$\delta_1 > 0$ such that for
$\left| x-A_{1} \right| \leq \delta_1$ and
$\left| y-A_{2} \right| \leq \delta_1$, we
have $\left| 1-\frac{xA_2}{yA_1} \right| \leq \frac{\epsilon}{A_1}$.
Similarly, there exist $\delta_2, \ldots, \delta_6 > 0$ such that
\[ \left| x-A_{1} \right| \leq \delta_2 \; \mathrm{and} \; \left| z-A_{3} \right| \leq \delta_2 \; \Longrightarrow \; \left| 1-\frac{xA_3}{zA_1} \right| \leq \frac{\epsilon}{A_1}, \]
\[ \left| x-A_{1} \right| \leq \delta_3 \; \mathrm{and} \; \left| y-A_{2} \right| \leq \delta_3 \; \Longrightarrow \; \left| 1-\frac{yA_1}{xA_2} \right| \leq \frac{\epsilon}{A_2}, \]
\[ \left| x-A_{1} \right| \leq \delta_4 \; \mathrm{and} \; \left| z-A_{3} \right| \leq \delta_4 \; \Longrightarrow \; \left| 1-\frac{zA_1}{xA_3} \right| \leq \frac{\epsilon}{A_3}, \]
\[ \left| y-A_{2} \right| \leq \delta_5 \; \mathrm{and} \; \left| z-A_{3} \right| \leq \delta_5 \; \Longrightarrow \; \left| 1-\frac{yA_3}{zA_2} \right| \leq \frac{\epsilon}{A_2}, \; \mathrm{and} \]
\[ \left| y-A_{2} \right| \leq \delta_6 \; \mathrm{and} \; \left| z-A_{3} \right| \leq \delta_6 \; \Longrightarrow \; \left| 1-\frac{zA_2}{yA_3} \right| \leq \frac{\epsilon}{A_3}. \]
Let $\delta=\mathrm{min}\left( \delta_{1},\delta_{2},\delta_{3},\delta_{4},\delta_{5},\delta_{6}\right) $.
Suppose $\mathcal{A}$ is a complex enclosing $x$, $y$, and $z$ such that
$\left| x-A_{1} \right| \leq \delta$, $\left| y-A_{2} \right| \leq \delta$,
and $\left| z-A_{3} \right| \leq \delta$.
Rescale $\mathcal{A}$ to get a complex $\mathcal{B}$ enclosing $x^{\prime}$,
$y^{\prime}$, and $z^{\prime}$.
If $x^{\prime}=A_1$, then the complex was scaled
by a factor of $\frac{A_1}{x}$.
So, $y^{\prime} = y \left( \frac{A_1}{x} \right)$ and
$z^{\prime} = z \left( \frac{A_1}{x} \right)$.
The first inequality is trivial $\left| A_{1}-x \right| =0<\epsilon$. But, we also
get the inequalities
\[ \left| A_2 - y^{\prime} \right| = \left| A_{2}\left( 1-\frac{yA_1}{xA_2}\right) \right| \leq A_2\left( \frac{\epsilon}{A_2}\right) = \epsilon \]
and
\[ \left| A_3 - z^{\prime} \right| = \left| A_{3}\left( 1-\frac{zA_1}{xA_3}\right) \right| \leq A_3\left( \frac{\epsilon}{A_3}\right) = \epsilon. \]
Similarly, if $y^{\prime}=A_2$, the complex was scaled by a factor of $\frac{A_2}{y}$ and we get the inequalities
\[ \left| A_1 - x^{\prime} \right| = \left| A_{1}\left( 1-\frac{xA_2}{yA_1}\right) \right| \leq A_1\left( \frac{\epsilon}{A_1}\right) = \epsilon, \]
\[ \left| A_{2}-y \right| =0<\epsilon \]
and
\[ \left| A_3 - z^{\prime} \right| = \left| A_{3}\left( 1-\frac{zA_2}{yA_3}\right) \right| \leq A_3\left( \frac{\epsilon}{A_3}\right) = \epsilon. \]
Finally, if $z^{\prime}=A_3$, then we scaled by a factor of $\frac{A_3}{z}$
and we get the inequalities
\[ \left| A_1 - x^{\prime} \right| = \left| A_{1}\left( 1-\frac{xA_3}{zA_1}\right) \right| \leq A_1\left( \frac{\epsilon}{A_1}\right) = \epsilon, \]
\[ \left| A_2 - y^{\prime} \right| = \left| A_{2}\left( 1-\frac{yA_3}{zA_2}\right) \right| \leq A_2\left( \frac{\epsilon}{A_2}\right) = \epsilon, \]
and
\[ \left|A_{3}-z \right| =0<\epsilon. \]
$\Box$
\bigskip
\begin{lemma}
\label{continuous}
The length function $L(A_1, A_2, A_3)$ is continuous for all $A_i > 0$.
\end{lemma}
\noindent \textbf{Proof:}
Let $A_1$, $A_2$, $A_3$, and $\epsilon >0$ be given.
Let $\delta_1 = \frac{\epsilon^2}{36\pi}$. Let $\delta_2$ be the delta
needed in Lemma~\ref{rescale} for $\frac{\epsilon^2}{16\pi}$. That is, if
$\left| x-A_{1} \right| \leq \delta_2$, $\left| y-A_{2} \right| \leq \delta_2$,
and $\left| z-A_{3} \right| \leq \delta_2$, then
when we rescale any complex containing $x$, $y$, and $z$ to get one
containing $x^{\prime}$, $y^{\prime}$, and
$z^{\prime}$ with $x^{\prime}=A_1$ or $y^{\prime}=A_2$ or $z^{\prime}=A_3$, we
get the inequalities $\left| A_1 - x \right| \leq \frac{\epsilon^2}{16\pi}$,
$\left| A_2 - y \right| \leq \frac{\epsilon^2}{16\pi}$, and
$\left| A_3 - z \right| \leq \frac{\epsilon^2}{16\pi}$.
Let $\delta =\mathrm{min}\{ \delta_1, \delta_2 \}$. Suppose
$\left| A_{1} - x \right| \leq \delta$,
$\left| A_{2} - y \right| \leq \delta$, and
$\left| A_{3} - z \right| \leq \delta$.
Let $\mathcal{A}_1$ be a complex that uses $L(A_1,A_2,A_3)$ perimeter
to enclose areas
$A_1$, $A_2$, and $A_3$. Similarly, let $\mathcal{A}$ be a complex enclosing
areas
$x$, $y$, and $z$ with length $L(x,y,z)$.
\noindent \textbf{Case 1: } $x > A_{1}$, $y > A_{2}$, $z > A_{3}$, and
$L(x,y,z) \geq L(A_{1},A_{2},A_{3})$.
Since $x > A_{1}$, $y > A_{2}$, and $z > A_{3}$, we can
enclose areas $x$, $y$,
and $z$ by using $\mathcal{A}_1$ together with three disjoint
circles containing
areas $x-A_{1}$, $y-A_2$, and $z-A_3$ respectively. These disjoint
circles have perimeter $2\sqrt{\pi}\sqrt{x-A_1}$, $2\sqrt{\pi}\sqrt{y-A_2}$,
and $2\sqrt{\pi}\sqrt{z-A_3}$.
Since $L(x,y,z)$ is the minimum length, it has shorter length
than the perimeter used by $\mathcal{A}_1$ together with the circles. (See Figure~\ref{case1}.)
\begin{figure}[hbt]
\centerline {\epsfbox {case1.eps}}
\caption {Adding three small circles to $\mathcal{A}_1$ gives a complex enclosing areas $x$, $y$, and $z$ with more perimeter than $\mathcal{A}$}
\label{case1}
\end{figure}
We thus get
\[ L(x,y,z) < L(A_{1},A_{2},A_{3}) + 2 \sqrt{\pi}\sqrt{x-A_{1}} + 2 \sqrt{\pi}\sqrt{y-A_{2}} + 2 \sqrt{\pi}\sqrt{z-A_{3}}. \]
So,
\[ \left| L(x,y,z) - L(A_{1},A_{2},A_{3}) \right| = L(x,y,z) - L(A_{1},A_{2},A_{3}) \]
\[ < 2\sqrt{\pi}\sqrt{x-A_{1}} + 2 \sqrt{\pi}\sqrt{y-A_{2}} + 2 \sqrt{\pi}\sqrt{z-A_{3}}\]
\[ \leq 6\sqrt{\pi}\sqrt{\delta} \]
\[ \leq 6\sqrt{\pi}\sqrt{\delta_1}\]
\[ = 6\sqrt{\pi}\sqrt{\frac{\epsilon^2}{36\pi}} = \epsilon. \]
\noindent \textbf{Case 2: } $x < A_{1}$, $y < A_{2}$, $z < A_{3}$, and
$L(x,y,z) \leq L(A_{1},A_{2},A_{3})$.
We do exactly the same as case 1, but add little circles to $\mathcal{A}$ instead of $\mathcal{A}_1$. To be precise, we can
enclose areas $A_1$, $A_2$,
and $A_3$ by using $\mathcal{A}$ together with three disjoint
circles containing
areas $A_{1}-x$, $A_2-y$, and $A_3-z$ respectively. These disjoint
circles have perimeter $2\sqrt{\pi}\sqrt{A_1-x}$, $2\sqrt{\pi}\sqrt{A_2-y}$,
and $2\sqrt{\pi}\sqrt{A_3-z}$.
Since $L(A_{1},A_{2},A_{3})$ is the minimum length, it has shorter length
than the perimeter used by $\mathcal{A}$ together with the circles.
We thus get
\[ L(A_{1},A_{2},A_{3}) < L(x,y,z) + 2 \sqrt{\pi}\sqrt{A_{1}-x} + 2 \sqrt{\pi}\sqrt{A_{2}-y} + 2 \sqrt{\pi}\sqrt{A_{3}-z}. \]
That is,
\[ \left| L(x,y,z) - L(A_{1},A_{2},A_{3}) \right| = L(A_{1},A_{2},A_{3}) - L(x,y,z) \]
\[ < 2\sqrt{\pi}\sqrt{A_{1}-x} + 2 \sqrt{\pi}\sqrt{A_{2}-y} + 2 \sqrt{\pi}\sqrt{A_{3}-z}\]
\[ \leq 6\sqrt{\pi}\sqrt{\delta} \]
\[ \leq 6\sqrt{\pi}\sqrt{\delta_1}\]
\[ = 6\sqrt{\pi}\sqrt{\frac{\epsilon^2}{36\pi}} = \epsilon. \]
\noindent \textbf{Case 3: } $L(x,y,z)<L(A_{1},A_{2},A_{3})$ and $x>A_{1}$ or
$y>A_{2}$ or $z>A_{3}$
Scale down $\mathcal{A}$ to get a complex $\mathcal{B}$ that encloses areas
$x^{\prime}$, $y^{\prime}$, and $z^{\prime}$ such that one of areas equals
an area from $\mathcal{A}_1$ (e.g. $x^{\prime}=A_1$) and the other two
areas are smaller or
equal to the remaining areas in $\mathcal{A}_1$ (e.g. $y^{\prime} \leq A_2$ and
$z^{\prime} \leq A_3$).
Let $\ell(\mathcal{B})$ be the length of $\mathcal{B}$. Since scaled minimizers are still minimizers, $\ell(\mathcal{B})=L(x^{\prime},y^{\prime},z^{\prime})$.
Also,
since $\mathcal{B}$ was a scaled down copy of $\mathcal{A}$, we have
$\ell(\mathcal{B}) < L(x,y,z)$.
Without loss of generality, assume that the first area is the one that is the
same. In other words, $x^{\prime}=A_1$, $y^{\prime} \leq A_2$, and
$z^{\prime} \leq A_3$.
If we add two disjoint circles of area $A_{2} - y^{\prime}$ and $A_{3} - z^{\prime}$ to the complex $\mathcal{B}$, we get a complex enclosing $A_1$, $A_2$,
and $A_3$ again. The length of this complex is
$\ell(\mathcal{B}) + 2\sqrt{\pi}\sqrt{A_{2} - y^{\prime}} + 2\sqrt{\pi}\sqrt{A_{3} - z^{\prime}}$
and must be larger than $L(A_1,A_2,A_3)$. (See Figure~\ref{case3}.)
\begin{figure}
\centerline {\epsfbox {case3.eps}}
\caption {Scale down $\mathcal{A}$ to get $\mathcal{B}$. Then add two small circles to get a complex enclosing areas $A_1$, $A_2$, and $A_3$ with more
perimeter than $\mathcal{A}_1$.}
\label{case3}
\end{figure}
The
$2\sqrt{\pi}\sqrt{A_{2} - y}$ and $2\sqrt{\pi}\sqrt{A_{3} - z}$ terms are
the perimeter needed to
add back in the missing area. Since we have a bound on the missing area
(by $\delta_2$ and lemma~\ref{rescale}), we have a bound on the amount of
perimeter needed. In fact, we get
\[ L(A_{1},A_{2},A_{3}) < \ell(\mathcal{B}) + 2\sqrt{\pi}\sqrt{A_{2} - y^{\prime}} + 2\sqrt{\pi}\sqrt{A_{3}- z^{\prime}}\]
\[ \leq \ell(\mathcal{B}) + 2\sqrt{\pi}\sqrt{\frac{\epsilon^2}{16\pi}} + 2\sqrt{\pi}\sqrt{\frac{\epsilon^2}{16\pi}}\]
\[ = \ell(\mathcal{B}) + \epsilon \]
In short, $L(A_{1},A_{2},A_{3}) \leq \ell(\mathcal{B}) + \epsilon$.
Put this together with
the previous inequalities to get
\[ \ell(\mathcal{B}) < L(x,y,z) < L(A_{1},A_{2},A_{3}) < \ell(\mathcal{B}) + \epsilon. \]
So, $\left| L(x,y,z) - L(A_{1},A_{2},A_{3}) \right| = L(A_{1},A_{2},A_{3})-L(x,y,z) < \epsilon$.
\noindent \textbf{Case 4: } $L(x,y,z)>L(A_{1},A_{2},A_{3})$ and $x<A_{1}$ or
$y<A_{2}$ or $z<A_{3}$
This case is similar to case 3, except we scale up $\mathcal{A}$ and add little circles to
$\mathcal{A}_1$ to get the desired inequality.
Scale up $\mathcal{A}$ to get a complex $\mathcal{B}$ that encloses areas
$x^{\prime}$, $y^{\prime}$, and $z^{\prime}$ such that one of these areas
equals
an area from $\mathcal{A}_1$ (e.g. $x^{\prime}=A_1$) and the other two
areas are larger or
equal to the remaining areas in $\mathcal{A}_1$ (e.g. $y^{\prime} \geq A_2$ and
$z^{\prime} \geq A_3$).
Let $\ell(\mathcal{B})$ be the length of $\mathcal{B}$. Since scaled minimizers are still minimizers, $\ell(\mathcal{B})=L(x^{\prime},y^{\prime},z^{\prime})$.
Also,
since $\mathcal{B}$ was scaled up from $\mathcal{A}$, we have
$\ell(\mathcal{B}) > L(x,y,z)$.
Without loss of generality, assume that the first area is the one that is the same. So, $x^{\prime}=A_1$, $y^{\prime} \geq A_2$, and $z^{\prime} \geq A_3$.
If we add two disjoint circles of area $y^{\prime} - A_{2}$ and
$z^{\prime} - A_{3}$ to the complex $\mathcal{A}_1$, we get a complex
enclosing $A_1$, $y^{\prime}$,
and $z^{\prime}$ again. The length of this complex is
$\ell(\mathcal{B}) + 2\sqrt{\pi}\sqrt{y^{\prime} - A_{2}} + 2\sqrt{\pi}\sqrt{z^{\prime} - A_{3}}$
and must be larger than $L(A_1,x^{\prime},y^{\prime})=\ell(\mathcal{B})$.
We then get the inequality
\[ \ell(\mathcal{B}) < L(A_{1},A_{2},A_{3}) + 2\sqrt{\pi}\sqrt{y^{\prime} - A_{2}} + 2\sqrt{\pi}\sqrt{z^{\prime} - A_{3}}\]
\[ \leq L(A_{1},A_{2},A_{3}) + 2\sqrt{\pi}\sqrt{\frac{\epsilon^2}{16\pi}} + 2\sqrt{\pi}\sqrt{\frac{\epsilon^2}{16\pi}}\]
\[ = L(A_{1},A_{2},A_{3}) + \epsilon. \]
In short, $L(A_{1},A_{2},A_{3}) > \ell(\mathcal{B}) - \epsilon$.
Put this together with
the previous inequalities to get
\[ \ell(\mathcal{B}) - \epsilon < L(A_{1},A_{2},A_{3}) < L(x,y,z) < \ell(\mathcal{B}) \]
So,
\[ \left| L(x,y,z) - L(A_{1},A_{2},A_{3}) \right| = L(x,y,z)-L(A_{1},A_{2},A_{3}) < \epsilon. \]
We have covered all the possibilities. Therefore, the length function
$L(A_{1},A_{2},A_{3})$ is continuous. $\Box$
\bigskip
Although presented in the context of three areas, Lemma~\ref{rescale} and
Lemma~\ref{continuous} can be easily extended to any number of regions.
In particular, the length function $L(A_1, \ldots , A_n)$ is
continuous for any $n$.
The triple bubble conjecture suggests a solution to the 3 bubble problem
in $\R^2$.
\begin{conjecture}
\label{conj1}
The least
perimeter planar graph that encloses
and separates three finite areas $A_1$, $A_2$, and $A_3$ is a regular triple
bubble complex with four vertices, six edges, and three connected regions.
\end{conjecture}
Such a complex is
called a \textit{standard triple bubble} and has been proven to exist
and be unique for any three areas \cite{Montesinos}.
(See Figure~\ref{tripbubb}.) The standard triple
bubble has also been shown to be the least perimeter way to enclose and
separate any three areas using connected regions \cite{Cox}.
\begin{figure}
\centerline {\epsfbox {tripbubb.eps}}
\caption{A standard triple bubble enclosing areas $A_1$, $A_2$, and $A_3$}
\label{tripbubb}
\end{figure}
An alternate version of the triple bubble conjecture allows increasing the
areas enclosed in an attempt to minimize perimeter:
\begin{conjecture}
\label{conj2}
Given three positive
real numbers $A_1$, $A_2$, and $A_3$, the least perimeter graph that encloses
and
separates three finite areas $B_1$, $B_2$, and $B_3$ such that $B_i \geq A_i$
for all $i$, is a standard triple bubble.
\end{conjecture}
Conjecture~\ref{conj2} eliminates the possibility of empty
chambers. If a complex has an empty chamber, it could be filled in with any
one of the adjacent areas and at least one edge could be eliminated. The
resulting complex encloses more area, but uses less perimeter. (See
Figure~\ref{emcham}.)
\begin{figure}[hbt]
\centerline {\epsfbox {emcham.eps}}
\caption{An empty chamber can be filled with less total length.}
\label{emcham}
\end{figure}
Theorem~\ref{triv} establishes that the two versions of the triple bubble
conjecture are equivalent.
\begin{theorem}
\label{triv}Let three positive areas
$A_1$, $A_2$, and $A_3$ be given. There exists a least perimeter
triple bubble
complex $\mathcal{B}$ that encloses and separates areas
$B_1$, $B_2$, and $B_3$ with $B_i \geq A_i$. That is, if $\mathcal{C}$ is any
other complex enclosing areas $C_1$, $C_2$, and $C_3$ with
$C_i \geq A_i$, then it must use at least as much perimeter $(\ell (\mathcal{B}) \leq \ell (\mathcal{C}))$.
Furthermore, if the minimizer $\mathcal{B}$ has
connected regions (a standard triple bubble), then it must enclose the given
areas $A_1$, $A_2$, and $A_3$ (i.e. $B_i=A_i \; \; \forall i$).
\end{theorem}
\noindent \textbf{Proof:}
To optimize the length function $L$ for areas greater than or equal to
$A_1$, $A_2$, and $A_3$, the domain we need to consider is bounded below (by
$A_1$, $A_2$, and $A_3$) and bounded above as well. The upper bound can be
chosen to be $B_1$, $B_2$, $B_3$ where
$B_i=(2\pi+3)\frac{\sqrt{A_1+A_2+A_3}}{\sqrt{\pi}}$. This is the area needed
to enclose and separate the three areas with a circle and three radii. Since
the total perimeter needed to enclose even one $B_i$ is larger than a known
way to enclose $A_1$, $A_2$, and $A_3$, the total
perimeter used in any attempt to enclose areas bigger than $B_1$, $B_2$,
and $B_3$ must be larger than the minimum way to enclose $A_1$, $A_2$, and $A_3$. Since a continuous function on a
compact set achieves its maximum and
minimum value, there is a minimum value for the length function.
If the minimum is always a standard triple bubble (i.e. Conjecture 2 is
correct), then the minimum must enclose exactly
$A_1$, $A_2$, and $A_3$. If it encloses some $B_1$, $B_2$, and
$B_3$ with $B_i>A_i$ for some $i$, we could reduce perimeter by replacing
a small portion of the exterior arc of $B_i$ by a straight line. (See
Figure~\ref{shrink}.) The line can
be chosen small enough so that the area enclosed by the region is still larger
than $A_i$, and yet we've used less perimeter. This contradicts the assumption
that the complex was the minimum. $\Box$
\begin{figure}[hbt]
\centerline {\epsfbox {shrink.eps}}
\caption{Cut off a little bit to save perimeter.}
\label{shrink}
\end{figure}
\bigskip
We define the \textit{pressure} of a region in a regular $n$ bubble complex to
be 0 for the exterior region. For any other region, we pick a path from the
exterior
to that region such that the path intersects the edges of the complex
transversely in a finite number of points. The pressure is then the sum of the
signed curvatures of the edges at these finite number of intersection points.
We use the sign convention as shown in Figure~\ref{orient}. When exterior
edges bulge outward (as in soap bubbles), the choice of sign
guarantees that regions adjacent to the exterior have positive pressure. It
also makes the sign of the curvature agree with the standard definition of
curvature when the edges are given a counter-clockwise orientation.
\begin{figure}[hbt]
\centerline {\epsfbox {orient.eps}}
\caption{Sign convention for curvature}
\label{orient}
\end{figure}
Cox, Harrison, Hutchings, et. al. \cite{Cox} proved that for any
closed path intersecting a regular bubble transversely,
the sum of the signed
curvatures along that path must be zero. In Lemma~\ref{reg}, we generalize this
result to any path that starts and ends in the same (possibly disconnected)
region. It also guarantees that pressure is well defined.
\begin{lemma}
\label{reg}
Let $\mathcal{A}$ be a regular $n$ bubble complex. Let $\gamma$ be any path
that intersects the edges of the complex transversely such that $\gamma$
starts and stops in portions of the same region (not necessarily connected).
Then, the sum of the signed curvatures of the edges crossed is zero.
\end{lemma}
\noindent \textbf{Proof:}
Suppose that $\gamma$ goes through regions $R_1, R_2, \ldots, R_n, R_1$ and
crosses edges with curvatures $\kappa_1, \kappa_2, \ldots, \kappa_n$ at
points $p_1, p_2, \ldots, p_n$. Define a half-variation $\mathcal{A}_t$
that transfers
$t$ area from each $R_i$ to $R_{i+1}$ by adjusting each edge in a neighborhood
about $p_i$. The initial change in length by this half-variation is just the
sum of the signed curvatures (see e.g. Morgan\cite{M2}). That is,
\[ \frac{d\ell (\mathcal{A}_t)}{dt}\mid_{t=0} = \sum_{i=1}^{n} \kappa_i. \]
By regularity (condition six), this sum must be greater than or equal to 0.
If, however, the sum is greater than zero, we can traverse $\gamma$ in the
opposite direction to get the same curvatures with
opposite orientation. Therefore, the
half-variation defined by adjusting area along $-\gamma$ has negative initial
change in length
which violates regularity. Thus, the sum of the curvatures must be zero.
$\Box$
\bigskip
In Chapter~\ref{result}, we show that Conjecture 2 is true in
the case of equal pressure regions, or that Conjecture 1 is true in
the case of equal pressure regions with no empty chambers.
The restriction that
the regions have equal pressures guarantees that the inner edges (edges that
don't touch the connected exterior region) are all line segments (0 curvature)
and the outer edges all have the same curvature. In particular, every
$n$-gon is convex.
\chapter{Structure of Perimeter Minimizing Bubbles}
\thispagestyle{myheadings}
We begin with some observations about the possible configurations for
perimeter minimizing bubble complexes. Theorem~\ref{regularity} guarantees
that they must be regular bubble complexes. The restrictions we discuss in this chapter are applicable to arbitrary regular bubble complexes
enclosing any number of regions.
We first note that perimeter minimizing complexes must be connected. If a
complex has two disconnected components, they can be pushed together until a
vertex of degree at least four is created. This new complex violates
regularity and therefore there exists a complex enclosing the same areas
with less perimeter.
\begin {lemma} Perimeter minimizing regular $n$ bubble complexes ($n>2$) have
no 2-gons.
\label{lem2gon}
\end{lemma}
\noindent \textbf{Proof:} Suppose there is a 2-gon.
By regularity, every vertex must be trivalent. In
particular, a 2-gon will have two vertices and two edges with an additional
edge leading away from each vertex $\alpha$ and $\beta$.
(See Figure~\ref{bubb1}.)
\begin{figure}[hbt]
\centerline {\epsfbox {bubb1.eps}}
\caption{A 2-gon in a bubble complex}
\label{bubb1}
\end{figure}
\noindent Case I: edge $\alpha =$ edge $\beta$ (See Figure~\ref{bubb2}.)
\begin{figure}[hbt]
\centerline{\epsfbox {bubb2.eps}}
\caption{A disconnected double bubble region}
\label{bubb2}
\end{figure}
The 2-gon and the adjacent 2-gon form a double bubble disconnected
from the rest of the complex. Move this disconnected
piece until it touches another component of the bubble complex. A four valent
vertex would be created at the point of intersection thus violating regularity.
\noindent Case II: edge $\alpha \neq$ edge $\beta$ (See Figure~\ref{bubb1}.)
The data from a single vertex is enough to completely determine a connected
double bubble complex. In other words, if three arcs of circles meet at a
vertex at equal angles $(\frac{2\pi}{3})$ and the sum of the signed curvatures
of the arcs around the vertex is zero, then the arcs will extend to a
standard double bubble complex.
All three arcs meet again at some other point and with the same angles as
the angles at which they leave $(\frac{2\pi}{3})$.
By regularity, the curvatures of $\alpha$ and $\beta$ are determined by the
curvatures of the 2-gon and therefore must be the same. Furthermore, edge
$\alpha$ and edge $\beta$ must be arcs of the same circle since the data from
one vertex is enough to determine the other. The 2-gon can be \textit{slid}
along this circle without changing perimeter or area. That is, we can remove
the 2-gon and extend edge $\alpha$ and $\beta$ to get a continuous arc of a
circle. The 2-gon can then be reinserted anywhere along this arc. (See
Figure~\ref{slide}.) To finish the slide move, we erase the portion of the circle inside of the new 2-gon. We then have a bubble complex enclosing equivalent
areas with exactly the same amount of perimeter. The slide move was
introduced by the SMALL Geometry Group \cite{Foisy} when they proved the
planar double bubble conjecture.
\begin{figure}[hbt]
\centerline{\epsfbox{slide.eps}}
\caption{Slide a 2-gon along a circle}
\label{slide}
\end{figure}
We continue sliding this
2-gon until it either touches another edge or the edge $\alpha$ disappears.
In either case, a 4-valent vertex is created and regularity is violated.
$\Box$
\bigskip
\begin{lemma}
If there exists a 3-gon with edges of curvature $\kappa_1$, $\kappa_2$, and
$\kappa_3$, then its shape is unique (up to orientation and isometry)
and is determined by the triangle of its vertices.
\end{lemma}
\noindent \textbf{Proof:}
When two arcs of circles of radius $r_1$ and $r_2$ meet at an angle
of $\frac{2\pi}{3}$, the centers of the two circles are at distance
$d=\sqrt{r_1^2 + r_2^2 -r_1 r_2}$. Suppose that there exists a 3-gon
with curvatures $\kappa_1$, $\kappa_2$, and $\kappa_3$.
If two curvatures are zero, the 3-gon is determined by the curvature of
the third arc. Uniqueness is guaranteed by the Gauss-Bonnet Theorem (See Lemma 4.2).
If none of the curvatures are zero, the radii of the respective arcs are $r_i=\frac{1}{\kappa_i}$. Consider the 2-gon
formed by intersecting a circle of radius $r_1$ with a circle of radius $r_2$
at an angle of $\frac{2\pi}{3}$. Let $C_1$ be the circle obtained by
extending the arc
with curvature $\kappa_1$. Similarly, let $C_2$ denote the circle obtained
by extending the arc with curvature $\kappa_2$. To get any 3-gon with the
same curvature edges,
we need to add a third circle of radius $r_3$ so that the angles made with
both circles $C_1$ and $C_2$ is again $\frac{2\pi}{3}$.
Since there is a 3-gon with these curvatures, we know it is possible.
The center of this circle must be at
distance $d_1=\sqrt{r_1^2 + r_3^2 - r_1 r_3}$ from the center of $C_1$ and
distance $d_2=\sqrt{r_2^2 + r_3^2 - r_2 r_3}$ from the center of $C_2$.
Construct a circle of radius $d_1$ around the center of $C_1$ and a circle of
radius $d_2$ around the center of $C_2$. The center of the third circle must lie on the intersection of these circles. The circles are not equivalent
(since they have different centers) and intersect at least once (since there is a solution). The only other possibility is that the circles intersect twice.
If so, the two choices for the center of the third circle give the same
intersection pattern with the 2-gon, but on
opposite sides. (See Figure~\ref{uniq3gon}.) Generically, when an arc
can be added to a 2-gon
to form a 3-gon using one of the original vertices of the 2-gon, an arc of
the same curvature can be added as well with
opposite orientation. This creates a 3-gon with the opposite vertex
of the 2-gon. The two different 3-gons created are equivalent
but have opposite orientation.
If only one curvature is zero, we can build a 2-gon with a straight line and a circle of curvature $\kappa_2 \neq 0$. Consider arcs of curvature $\kappa_3$
leaving a point on the line segment between the vertices of our
2-gon at an angle of $\frac{2\pi}{3}$. When such an arc meets the given arc
of curvature $\kappa_2$, the angle made is strictly increasing
between $0$ and $\pi$ as the
point of departure varies from one vertex to the other. At only one point is
the angle exactly $\frac{2\pi}{3}$.
\begin{figure}
\centerline{\epsfbox{uniq3gon.eps}}
\caption{Two oppositely oriented 3-gons can be created from a 2-gon by adding an arc of given curvature $\kappa_3.$}
\label{uniq3gon}
\end{figure}
To prove that 3-gons are determined by triangles,
we will establish a map from triangles to 3-gons and
show that it is bijective. Suppose we have a triangle $\triangle ABC$ with
side lengths $a$, $b$, and $c$ (opposite side from the appropriately labeled
vertex) and angles $\alpha =\angle CAB$, $\beta=\angle ABC$, and
$\gamma=\angle ACB$. (See Figure~\ref{triang}.)
\begin{figure}[hbt]
\centerline{\epsfbox {triang.eps}}
\caption{A generic triangle}
\label{triang}
\end{figure}
Given any angle $\theta_{\alpha}$, there is a unique arc of a circle that
passes
through $B$ and $C$ that makes angle $\theta_{\alpha}$ with the line segment
$BC$. In
fact, since $a$ is the length of $BC$, the curvature $\kappa_{\alpha}$ of
the arc
through $BC$
with angle $\theta_{\alpha}$ is given by the formula
$\kappa_{\alpha} = \frac{2sin(\theta_{\alpha})}{a}$. We
consider angles exterior to the triangle to be positive.
(See Figure~\ref{triang1}.)
Similarly, angles $\theta_{\beta}$ and $\theta_{\gamma}$ uniquely determine
arcs of circles through $AC$ and $AB$ respectively.
\begin{figure}[hbt]
\centerline{\epsfbox {triang1.eps}}
\caption{A triangle with arcs of circles attached}
\label{triang1}
\end{figure}
To get a valid 3-gon, the internal angles should all be $\frac{2\pi}{3}$. A
3-gon must then satisfy the linear equations
\[ \theta_{\beta} + \alpha + \theta_{\gamma} = \frac{2\pi}{3}, \]
\[ \theta_{\alpha} + \beta + \theta_{\gamma} = \frac{2\pi}{3}, \]
and
\[ \theta_{\alpha} + \gamma + \theta_{\beta} = \frac{2\pi}{3}. \]
From the triangle we also get the equation
\[ \alpha +\beta + \gamma = \pi. \]
The unique solution to these equations is
\[ \theta_{\alpha}=\alpha - \frac{\pi}{6}\]
\[ \theta_{\beta}=\beta - \frac{\pi}{6}\]
\[ \theta_{\gamma}=\gamma - \frac{\pi}{6}.\]
Given any 3-gon, we can get a triangle by connecting the vertices of the 3-gon.
So, the map from triangles to 3-gons is surjective. Suppose now that two
different triangles produce the same 3-gon. Since the vertices of the 3-gon
coincide with the vertices of the triangle that produced it, the two
triangles must be identical. Therefore the map is also injective.
$\Box$
\bigskip
\begin{lemma} Any 4-gon or 5-gon in a perimeter minimizing regular bubble
complex without empty chambers shares at most one edge with the exterior
region.
\label{lemext}
\end{lemma}
\begin{figure}[hbt]
\centerline{\epsfbox {extedg.eps}}
\caption{A region with two exterior edges}
\label{extedg}
\end{figure}
\noindent \textbf{Proof:} A 4-gon or a 5-gon is distinguished by the fact that
any pair
of edges are separated by at most one edge. Suppose a 4-gon or 5-gon shares
two edges $\alpha$ and $\beta$ with the exterior.
Then there is a single edge $\gamma_1$ that connects the two exterior edges.
(See Figure~\ref{extedg}.)
Let $p$ be the vertex shared by $\gamma_1$ and $\alpha$ and
$q$ be the vertex shared by $\gamma_1$ and $\beta$.
Pick a point $r$ at distance $\epsilon$ from $q$ on $\beta$.
Let $\gamma_2$ be an arc of a circle or line segment from $p$ to $r$ that does
not intersect any edges of the $n$-gon. (See Figure~\ref{wedge}.)
If the $n$-gon is convex, the arc
$\gamma_2$ can always be chosen to be a straight line segment.
\begin{figure}[hbt]
\centerline{\epsfbox {wedge.eps}}
\caption{Cut out this wedge shaped region}
\label{wedge}
\end{figure}
Cut out the triangular wedge formed by $\gamma_1$, $\gamma_2$, and the arc
from $q$ to $r$. Label the corners of this triangular wedge as follows:
$A$ for the corner that came from the point $p$, $B$ for the corner that came
from the point $q$, and $C$ for the corner that came from the point $r$.
Then separate the remaining complex into three disjoint pieces
by splitting $p$ into two points $p_1$ and $p_2$ such that $p_1$ is connected
to the point $q$ and $p_2$ is connected to the point $r$.
(See Figure~\ref{cutwedge}.)
\begin{figure}[hbt]
\centerline{\epsfbox {cutwedge.eps}}
\caption{The cut apart complex}
\label{cutwedge}
\end{figure}
Now re-attach the triangular wedge with an opposite orientation by identifying
the point $p_1$ with $B$, $p_2$ with $C$, and the points $q$, $r$, and $A$
with each other. (See Figure~\ref{wedgie}.)
\begin{figure}[hbt]
\centerline{\epsfbox {wedgie.eps}}
\caption{Flip over the wedge and glue it back in.}
\label{wedgie}
\end{figure}
The resulting complex has identical perimeter and encloses the same areas. The
edge $\alpha$, however, either has a corner at $p_2$ (if angle $\omega$
differs from angle $\rho$ ) or is still smooth but longer. If the edge has a
corner, we have a
complex that encloses the same areas with the same perimeter that violates
regularity. If the angles agree and no corner is created, the effect is that
the complex has \textit{slid} distance $\epsilon$ along the edge $\alpha$.
Continue sliding
(i.e. repeat this procedure) until either the complex bumps into itself
somewhere or the edge $\beta$ disappears. In either case, a complex
that encloses the same areas with identical perimeter is created. But, this
new complex contains a four-valent vertex and therefore violates regularity.
$\Box$
\bigskip
The same argument can be used to show that many other kinds of $n$-gon's in a
perimeter minimizing regular bubble complex cannot touch the exterior region
more than once. All that is needed to extend the argument is the existence of
symmetric arcs $\gamma_1$ and $\gamma_2$ that intersect one exterior edge at a
point and the other exterior edge at two points distance $\epsilon$ apart such
that the symmetric arcs do not touch any other portion of the boundary of the
$n$-gon.
In a paper regarding triple bubbles with connected regions, Cox, Harrison,
Hutchings et. al. \cite{Cox} made an interesting remark. They found a nice
relationship between the perimeter $L$ of a regular $n$ bubble complex,
the areas enclosed $A_1, A_2, \ldots, A_n$, and the pressure of each region
$p_1, p_2, \ldots, p_n$:
\begin{equation}
L=2 \sum_{i=1}^{n} p_i A_i
\label{eqhut}
\end{equation}
Using this relationship, we easily prove the following lemma:
\begin{lemma}
\label{lemhutch}
If a regular $n$ bubble complex $\mathcal{C}$ encloses areas
$A_1, A_2, \ldots, A_n$ and
has pressure $p_1, p_2, \ldots, p_n$ respectively, then any regular $n$
bubble complex $\mathcal{B}$ enclosing the same
areas with pressures $q_i < p_i$ for all
$i$ is not a minimizer.
\end{lemma}
\noindent \textbf{Proof:}
Let $\ell(\mathcal{B})$ be the length of $\mathcal{B}$ and $\ell(\mathcal{C})$
be the length of $\mathcal{C}$. Then, equation~\ref{eqhut} gives us
\[ \ell(\mathcal{B})=2\sum_{i=1}^{n} q_i A_i > 2\sum_{i=1}^{n} p_i A_i = \ell(\mathcal{C}) \]
$\mathcal{C}$ uses less perimeter and
therefore $\mathcal{B}$ is not a perimeter minimizer.
$\Box$
\chapter{Restrictions Imposed By Equal Pressure Regions}
\thispagestyle{myheadings}
In this chapter we consider regular $n$ bubble complexes that have
equal pressure
interior regions. Since there is no pressure change from one interior region
to another, the curvature of the interior edges must be zero. In addition, by
following a closed path that touches only two
distinct exterior edges (any number
of interior edges), we get that the
curvature of the exterior edges must all be the same.
\begin {lemma} If a regular $n$ bubble complex has no empty chambers and has
positive equal pressure regions, then $n$-gons have
at most 6 sides. In addition, $n$-gons that share an edge with the exterior
region have at most 5 sides.
\end{lemma}
\noindent \textbf{Proof:} By regularity conditions, the internal angles of
each $n$-gon must be
$2\pi \over 3$. In addition, their edges have either 0 curvature
(edges that separate two $n$-gons) or one
fixed curvature (edges that separate an $n$-gon from the exterior).
Furthermore, since regions have positive pressure, edges separating an $n$-gon from the exterior must bulge outward (i.e. have positive curvature).
Using arcs of constant positive curvature in the Gauss-Bonnet Theorem, we get
that
\[\sum_{i=1}^{n} \kappa_{i} l_{i} + \sum_{i=1}^{n} {\pi \over 3} = 2\pi \]
where $n$ is the number of edges.
Note that since each interior angle of an $n$-gon is $2\pi \over 3$,
each exterior angle is $\pi \over 3$.
We solve for the sum of the exterior angles to get
\[ n({\pi \over 3})=2\pi-\sum_{i=1}^{n} \kappa_{i} l_{i} \leq 2\pi. \]
So, $n \leq 6$ with equality only when the edges all have 0 curvature. That is,
6-gon's are \textit{internal} since they do not share an edge with the
exterior.
Recall that there also can't be any 2-gons in a perimeter minimizing complex (by Lemma~\ref{lem2gon}.
Therefore, $n$-gons that share an edge with the exterior must be 3-gons,
4-gons, or 5-gons.
$\Box$
\bigskip
\begin{lemma}
In an $n$ bubble complex with equal pressure regions,
there is a unique shape for a 3-gon region, a one
parameter family of possible 4-gons, and a two parameter family of possible
5-gons (up to orientation preserving isometry).
4-gons are determined by the length of a side adjacent to an exterior
edge. 5-gons are determined by the lengths of any two of the non-curved edges.
\label{lemstruc}
\end{lemma}
\noindent \textbf{Proof:} For a 3-gon, the Gauss-Bonnet theorem says that
\[ \sum_{i=1}^{3} (\kappa_{i} l_i + {\pi \over 3}) = 2\pi. \]
Since two of the sides have no curvature, we get
$\kappa l_1 = \pi$ or $l_1 = {\pi \over \kappa}$ where $\kappa$ is the
curvature
of the outside arcs and $l_1$ is the length of that curved arc in a 3-gon.
In particular, the length of the curved arc in a 3-gon with fixed
curvature $\kappa$
is constant.
\begin{figure}[hbt]
\centerline {\epsfbox {3gon.eps}}
\caption {The unique shape for a 3-gon}
\label{3gon}
\end{figure}
Now, two straight lines that leave the ends of such an arc at an
angle of ${2 \pi} \over 3$ will then meet in only one point, also
at an angle of ${2 \pi} \over 3$. This is the unique configuration for a 3-gon.
(See Figure~\ref{3gon}.)
\begin{figure}[hbt]
\centerline {\epsfbox {4gon.eps}}
\caption {Cut out an equilateral triangle to determine a 4-gon}
\label{4gon}
\end{figure}
For a 4-gon, the length of the curved arc is again completely determined by
the Gauss-Bonnet theorem. Two lines that leave the end points of this arc at
${2 \pi} \over 3$ will meet at some point at an angle of ${\pi} \over 3$.
Draw the line
segment connecting the ends of the arc. The fourth side of the 4-gon (opposite
the curved arc) must be parallel to this line segment. A choice of how far
along a side edge to place this opposite edge completely determines the 4-gon.
This choice also corresponds to the size of equilateral triangle that is cut off of the bottom. (See Figure~\ref{4gon}.)
For a 5-gon, the length of the arc is again fixed. This time the adjacent
edges will be parallel to each other. Choose any length for one edge, and draw
a line segment from the end of this edge such that the internal angle is
${2 \pi} \over 3$. This edge will meet the other adjacent edge at an angle of
${\pi} \over 3$. (See Figure~\ref{5gon}.)
\begin{figure}[hbt]
\centerline {\epsfbox {5gon.eps}}
\caption {Two choices determine a 5 gon}
\label{5gon}
\end{figure}
Now, the vertex opposite the curved edges can be chosen to be any point on
this line segment. Once this choice is made, the 5-gon is completely
determined. $\Box$
\bigskip
Regular bubble complexes with equal pressure, disconnected
regions have a lot of
symmetry that can be used
to find non-regular complexes that enclose the same areas using equal or less
total perimeter. Since 3-gons are unique, it is convenient to consider what can
possibly be next to a 3-gon.
If a 3-gon shares an edge with another 3-gon in a minimizing complex, then
they both share an edge with a third 3-gon. Otherwise, the region adjacent to
both 3-gons would have two exterior edges which violates Lemma~\ref{lemext}.
The complex is either disconnected and not a minimizer or is a standard
triple bubble.
Suppose a 3-gon is adjacent to two 4-gons (one on each side). Let
$P$ be the vertex shared by the 3-gon and both 4-gons. Let $Q$ be the other
interior vertex shared by the 4-gons. Let $R$ and $S$ be the vertices of the
4-gons not shared with the original 3-gon. (See Figure~\ref{4bubble}.)
Edges $\overline{RQ}$ and $\overline{SQ}$ separate the 4-gons from
another $n$-gon.
\begin{figure}[hbt]
\centerline {\epsfbox {4bubble.eps}}
\caption {Two 4-gons adjacent to a 3-gon}
\label{4bubble}
\end{figure}
This $n$-gon (\textit{opposite} the original 3-gon) must also be a
3-gon. If not,
then the $n$-gon has two different exterior edges (One coming from vertex
$R$ and one from vertex $S$) in violation of
Lemma~\ref{lemext}. Since the complex is connected, there aren't any more
regions. The complex must have just these four chambers.
Lemma~\ref{lem34} below, together with the fact that 3-gons are unique,
proves that the complex is actually completely determined,
i.e. it is just a scaled copy of Figure~\ref{4bubble}.
If a 3-gon shares an edge with a 4-gon, the 4-gon is unique. That is, there
is only one possible shape for a 4-gon adjacent to a 3-gon since the one parameter has been determined for the 4-gon. (See Figure~\ref{34gon}.) Lemma~\ref{lem34} gives a nice relationship between the side lengths of the 4-gon.
\begin{figure}[hbt]
\centerline {\epsfbox {34gon.eps}}
\caption {A 3-gon adjacent to a 4-gon}
\label{34gon}
\end{figure}
\begin{lemma}
Let $a$ be the length of the central edge (opposite the curved arc) of a
4-gon adjacent to a 3-gon. Let $b$ be the length of the shared
edge. Then $a = { 1 \over 2}b$.
\label{lem34}
\end{lemma}
\noindent \textbf{Proof:} We use the relationship between the radius $r$,
angle $\theta$ and chord length $C$ of a section of a circle $r= {C \over {2sin(\theta)}}$.
(See Figure~\ref{chord}.)
\begin{figure}[hbt]
\centerline {\epsfbox {chord.eps}}
\caption {A section of a circle}
\label{chord}
\end{figure}
Let $r$ be the radius of the exterior edges. For a
3-gon, the chord made by connecting the endpoints of the exterior arc makes
a right angle with the exterior arc. For a 4-gon, the corresponding angle is
$\pi \over 3$ as in Figure~\ref{34gons}. Simple trigonometry
gives us $r={\sqrt{3} \over 2}b$. In addition,
the chord of the 4-gon gives us the relationship
$r={{a+b} \over {2sin({\pi \over 3})}}={{a+b} \over \sqrt{3}}$.
Eliminate $r$ to get ${{a+b} \over \sqrt{3}}={{\sqrt{3}b} \over 2}$ or
${3b \over 2}=a+b$. Finally, we isolate $a$ to get the
desired equality $a= {1 \over 2}b$. $\Box$
\bigskip
\begin{figure}[hbt]
\centerline {\epsfbox {34gons.eps}}
\caption {A 3-gon next to a 4-gon}
\label{34gons}
\end{figure}
There is a one-parameter family of 5-gons that can share an edge
with a 3-gon since only one of the two parameters has been determined. There
still is, however, a nice relationship between the side lengths of a 3-gon
and an adjacent 5-gon.
\begin{lemma}
Suppose a 5-gon shares an edge with a 3-gon. The sum of the lengths of the
inside edges of the 5-gon (edges that don't meet the exterior arc) equals the
length of the shared edge between the 3-gon and 5-gon.
\label{lem35}
\end{lemma}
\noindent \textbf{Proof:} Let $a$ be the length of the inside edge of the 5-gon
adjacent to the shared edge. Let $b$ be the length of the remaining inside
edge. Let $c$ be the length of the shared edge. Let $P$ and $R$ be the
vertices of the shared edge, with $R$ the inner
vertex. Let $Q$ be the remaining vertex of the 3-gon. Let $S$ be the vertex of
the 5-gon opposite the curved arc and let $T$ be the remaining interior vertex
of the 5-gon. Extend the adjacent
edge of the 5-gon (edge of length $a$) and the edge of the 5-gon opposite the
shared edge until they meet at an angle of $\frac{\pi}{3}$ at a point $U$. Let
$V$ be a point on the edge of the 5-gon opposite of the shared edge such that
the angle $\angle RPV$ is $\frac{\pi}{3}$. (See Figure~\ref{35gon}.)
\begin{figure}[hbt]
\centerline {\epsfbox {35gon.eps}}
\caption {A 3-gon next to a 5-gon}
\label{35gon}
\end{figure}
Triangle $\triangle STU$ is equilateral and so segment $\overline{SU}$ has
length $b$ and segment $\overline{RU}$ has length $a+b$. $\overline{PV}$ is
parallel to $\overline{RU}$ by construction. $\overline{PR}$ is parallel to
$\overline{UV}$ since the opposite flat sides of a 5-gon are parallel.
Therefore, $RPVU$ is a parallelogram. The diagonal $PU$ must then bisect the
angle $\angle RPV$ and the angle $\angle RPU$ is $\frac{\pi}{6}$. Finally,
we notice that triangles $PRU$ and $PRQ$ are similar. Therefore edge
$\overline{RQ}$ has the same length as edge $\overline{RU}$, i.e. $a+b=c$.
$\Box$
\bigskip
Since the interior edges of a 5-gon that meet the exterior vertices are
parallel for
any 5-gon (not necessarily adjacent to a 3-gon),
the sum of the lengths of the other two interior
edges (opposite the curved arc) must be a constant.
In other words, for any 5-gon, the sum of the lengths
of the two innermost edges is equal to the length of the flat side of a
3-gon created with the same curvatures.
\chapter{Triple Bubbles with Equal Pressure Regions}
\label{result}
\thispagestyle{myheadings}
In this chapter, we solve the triple bubble conjecture in the case of
equal pressure regions without empty chambers. In the figures, we will
label the
interior of each $n$-gon with an integer (1,2, or 3) to denote the
region to which
they contribute area. For example, an $n$-gon labeled 1 contributes area
towards $A_1$, where $A_1$, $A_2$, and $A_3$ are the
areas enclosed by the triple
bubble complex. Choices for numbering is done arbitrarily and
without loss of generality.
Our main result is that unless we have a standard triple bubble, a triple
bubble complex with equal pressure regions cannot be a perimeter minimizer.
\begin {theorem}
A perimeter minimizing triple bubble complex with equal pressure regions and
without empty
chambers must be a standard triple bubble.In particular, it has connected
regions.
\label{main}
\end{theorem}
\noindent Proof: Suppose not.
\bigskip
\noindent Case 1: There is a 3-gon.
\bigskip
Consider the $n$-gons adjacent to the 3-gon. If one of them is a 3-gon, then
the other must also be a 3-gon. Otherwise, the $n$-gon adjacent to both 3-gons
would have two exterior edges which violates Lemma~\ref{lemext}. The complex
must then be a standard triple bubble with connected regions.
If either of the adjacent $n$-gons is a 4-gon, then we can \textit{reflect} it
into the 3-gon. Consider a 3-gon with an adjacent 4-gon. Let $A$ be the line
segment of the 4-gon opposite the shared edge. Let $\alpha$ be the remaining
straight edge (opposite the curved arc)
of the 4-gon, and $\beta$ be the shared edge.
By lemma~\ref{lem34}, $\alpha$ is half as long as $\beta$.
\begin{figure}[hbt]
\centerline {\epsfbox {bubb5.eps}}
\caption {Reflect a 4-gon into an adjacent 3-gon}
\label{bubb5}
\end{figure}
Add an edge $B$ inside the 3-gon halfway between the
vertices of $\beta$ at an angle of $\frac{2\pi}{3}$, and then remove
edge $A$. The edge $B$ creates a 4-gon of the appropriate size
inside of the original 3-gon. Then, we can renumber the areas and erase the top
half of edge $\beta$ to get a non-regular complex that
encloses the same areas and uses less perimeter. Therefore, the original
complex is not a minimizer.
(See Figure~\ref{bubb5}.)
If both adjacent $n$-gons are 5-gons, then they have an
edge in common and they both share an edge with the 3-gon. Since both interior
edges of a 3-gon have the same length, Lemma~\ref{lemstruc} tells us that the adjacent
5-gons must be identical. Switch the numbering of the areas enclosed by these
5-gons to get two sets of adjacent $n$-gons that enclose portions of the same
areas. (See Figure~\ref{bubb9}.)
\begin{figure}[hbt]
\centerline {\epsfbox {bubb9.eps}}
\caption {Swap two 5-gons adjacent to a 3-gon}
\label{bubb9}
\end{figure}
Eliminating the common edges gives a complex that encloses the same areas with
less total perimeter. Therefore, the original
complex is not a minimizer.
\bigskip
\noindent Case 2: There are no 3-gons.
\bigskip
Suppose that the complex has two 4-gons that share an edge next to an exterior
arc. By Lemma~\ref{lemstruc}, they would be identical.
The $n$-gons could be swapped
(i.e. renumbered) yielding disconnected portions of
the same area sharing an edge. Eliminate these shared edges to get a
non-regular complex that encloses the same areas with less perimeter.
The original complex is thus not a minimizer. (See Figure ~\ref{bubb11}.)
\begin{figure}[hbt]
\centerline {\epsfbox {bubb11.eps}}
\caption {Swap adjacent 4-gons}
\label{bubb11}
\end{figure}
Suppose now that the complex has a pair of adjacent 5-gons that share an edge
next to an exterior arc.
If they are \textit{identical} 5-gons (i.e. they enclose the same areas),
they can be swapped as in the 4-gon
case above. The shared edge can again be eliminated thus reducing
perimeter. (See Figure~\ref{bubb13}.)
\begin{figure}[hbt]
\centerline {\epsfbox {bubb13.eps}}
\caption {A complex with two identical adjacent 5-gons}
\label{bubb13}
\end{figure}
If they are different sizes, the smaller one can be reflected into the larger
one.
To be precise, let the interior edges adjacent to the shared edge be $\alpha$
and $\beta$. Since the 5-gons are not identical, $\alpha$ and $\beta$ must be
of different length. Assume $\alpha < \beta$. Add a line segment $B$ that
makes an angle of ${2 \pi \over 3}$ with $\beta$ at
distance equal to the length of $\alpha$ from the vertex shared by $\alpha$ and $\beta$. (See Figure~\ref{bubb16}.)
\begin{figure}[hbt]
\centerline {\epsfbox {bubb16.eps}}
\caption {Reflect a small 5-gon into a larger adjacent 5-gon}
\label{bubb16}
\end{figure}
$B$ will
intersect the opposite interior edge, will have the same length as $A$, and
will form a 5-gon identical to the one with edge $\alpha$. We can then
renumber the $n$-gons and eliminate edge $A$. The perimeter is
unchanged, but the resulting complex is not regular and therefore cannot be
a minimizer.
The only remaining possibility, then, is that the $n$-gons that share edges
with the exterior must alternate 4-gons and 5-gons.
Consider any 4-gon containing without loss of generality region number 1.
Adjacent to it on each
side is a 5-gon containing a different region. Assume (without loss of
generality) that one adjacent 5-gon contains region number 2. The next
4-gon (adjacent to the same 5-gon) must then
enclose region number 3 since it shares an edge with a 5-gon of region 2.
The $n$-gons that the 5-gon shares edges with must alternate 1,3,1,3. In
other words, the 4-gons must alternate in the regions
that they enclose, and the 5-gons all enclose the same
region. (See Figure~\ref{complex}.)
\begin{figure}[hbt]
\centerline {\epsfbox {complex.eps}}
\caption {Structure of alternating 4-gons and 5-gons}
\label{complex}
\end{figure}
By looking at the interior edges that meet
exterior vertices, we see that there must be exactly six 4-gons. Each 4-gon
rotates the angle of this edge by $\frac{\pi}{3}$ and
5-gons do not rotate
them at all.
Consider two 4-gons that contain different regions. If they are
exactly the same size, they can be swapped. In other words, the complex is
renumbered such that areas and perimeter are preserved. The swapped 4-gons
each now share their central edge (opposite the exterior arc) with another
$n$-gon that encloses a portion of the same area.
These edges can then be erased, thus
reducing perimeter. (See Figure~\ref{swap}.)
\begin{figure}[hbt]
\centerline {\epsfbox {swap.eps}}
\caption {Swap any two identical 4-gons containing different areas}
\label{swap}
\end{figure}
If the 4-gons are not the same size, the smaller one can be reflected into
the larger one. The edge of the smaller 4-gon opposite the curved arc is
erased and inserted inside the larger 4-gon parallel to the edge opposite its
curved arc. Renumbering the $n$-gons creates a non-regular complex with
identical
perimeter. (See Figure~\ref{reflect}.)
\begin{figure}[hbt]
\centerline {\epsfbox {reflect.eps}}
\caption {Reflect small 4-gons into larger 4-gons}
\label{reflect}
\end{figure}
The only case that did not produce a contradiction was the standard triple
bubble with three adjacent 3-gons. Any other complex with equal pressure
regions is not a perimeter minimizer. $\Box$
\bigskip
\begin{corollary}
\label{cormain}
The least perimeter graph that encloses
and separates three equal areas $A_1$, $A_2$, and $A_3$ without empty chambers
using equal pressure regions is a standard triple bubble.
\end{corollary}
\noindent \textbf{Proof:}
The only standard
triple bubble with equal pressure regions is
the one that encloses and separates three equal area regions.
(See Figure~\ref{tripeq}).
Any other complex enclosing those same areas with equal pressure regions
must have disconnected regions and therefore is not a minimizer by
Theorem~\ref{main}.
$\Box$
\bigskip
\begin{figure}[hbt]
\centerline {\epsfbox {tripeq.eps}}
\caption {A standard triple bubble enclosing equal areas with equal pressure regions}
\label{tripeq}
\end{figure}
\begin{corollary}
\label{cortriv}
The least perimeter graph that encloses and separates three equal areas with
convex cells and without empty chambers is a standard triple bubble.
\end{corollary}
\noindent \textbf{Proof:}
To be a perimeter minimizer, the graph must be a regular triple bubble complex.
In order to have convex cells, it must have equal pressure regions. The result then follows from Theroem~\ref{main}. $\Box$
\bigskip
Since triple bubble complexes with disconnected, equal pressure regions are
not minimizers,
complexes close to such complexes cannot be minimizers
either. To make this idea precise, suppose we have a sequence of regular
$n$ bubble
complexes $\{\mathcal{A}_i\}$. Let $A_{i,m}$ for $m=1 \ldots n$ be the area
of the
mth region
enclosed by $\mathcal{A}_i$. We say that the sequence $\{\mathcal{A}_i\}$
\textit{converges in length and area} to a regular $n$ bubble complex
$\mathcal{A}$ enclosing areas $A_1, A_2, \ldots, A_n$ if
\begin{enumerate}
\item $\displaystyle{\lim_{i \rightarrow \infty} A_{i,m} = A_m} \; \forall m$ and
\item $\displaystyle{\lim_{i \rightarrow \infty} \ell(\mathcal{A}_i) = \ell(\mathcal{A})}$,
\end{enumerate}
where $\ell(\mathcal{A})$ is the length of the complex $\mathcal{A}$.
\begin{corollary}
\label{cormain2}
Suppose $\{\mathcal{A}_i\}$ is a sequence of regular triple bubble complexes
that converges in length and area to a triple bubble
complex $\mathcal{A}$. If $\mathcal{A}$ does
not have any empty chambers, has equal pressure regions and is not a
standard triple bubble, then there exists an $N$ such that for any $i>N$,
$\mathcal{A}_i$ is not a perimeter minimizer for the areas it encloses.
\end{corollary}
\noindent \textbf{Proof:}
$\mathcal{A}$ is not a minimizer by Theorem~\ref{main}. So, there must exist a
perimeter minimizing complex $\mathcal{B}$ that encloses the same areas
but uses at least $\epsilon$ less perimeter for some $\epsilon$.
Since $\mathcal{B}$ is the minimizer, $\ell(\mathcal{B})=L(A_1,A_2,A_3)$ and
we get the inequality
\begin{equation}
\left| \ell(\mathcal{A}) - \ell(\mathcal{B}) \right| = \left| \ell(\mathcal{A}) - L(A_1,A_2,A_3) \right| > \epsilon.
\label{differ}
\end{equation}
Recall that Lemma~\ref{continuous} guarantees that the minimum length
function $L$ is continuous. Since the lengths converge, we can find an
$N_1$ such that
$\left| \ell(\mathcal{A}_i) - \ell(\mathcal{A}) \right| < \frac{\epsilon}{2}$ for all $i>N_1$. In addition, since the areas converge,
we can find $N_2$ such that
$\left| L(A_{i,1},A_{i,2},A_{i,3}) - L(A_1, A_2, A_3) \right| < \frac{\epsilon}{2}$ for all $i>N_2$.
Let $N = \mathrm{max}\{N_1,N_2\}$.
Suppose $\ell(\mathcal{A}_i) = L(A_{i,1},A_{i,2},A_{i,3})$ for some $i>N$.
By the triangle inequality,
\[ \left| \ell(\mathcal{A}) - L(A_1, A_2, A_3) \right| \leq \left| \ell(\mathcal{A}) - \ell(\mathcal{A}_i) \right| + \left| \ell(\mathcal{A}_i) - L(A_1, A_2, A_3) \right| \]
\[ = \left| \ell(\mathcal{A}) - \ell(\mathcal{A}_i) \right| + \left| L(A_{i,1},A_{i,2}, A_{i,3}) - L(A_1, A_2, A_3) \right| \]
\[ < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \]
This, of course contradicts inequality~\ref{differ}.
Therefore, $\ell(\mathcal{A}_i) \neq L(A_{i,1},A_{i,2},A_{i,3})$ for any $i>N$.
$\Box$
\bigskip
\begin{corollary}
Let $\mathcal{A}$ be a regular triple bubble complex enclosing areas $A_1$, $A_2$, and
$A_3$ with pressures $p_1$, $p_2$, and $p_3$. If there exists a
complex $\mathcal{B}$ with equal pressure regions that encloses the same
areas but with less pressure (i.e.
$p \leq p_i \; \forall i$), then $\mathcal{A}$ is not a minimizer.
\end{corollary}
\noindent \textbf{Proof:} Follows directly from Theorem~\ref{main} and
Lemma~\ref{lemhutch}.
\chapter{General Bubbles with Equal Pressure Regions}
\thispagestyle{myheadings}
The results we achieved for triple bubbles do not depend on the fact that only
three areas are being enclosed. The symmetry of the $n$-gons and the numbering
of adjacent $n$-gons was important. Indeed, most of the moves
presented in Chapter 5 can be generalized.
In this chapter, we assume that all complexes are regular $n$ bubble
complexes with equal pressure regions and without empty chambers. We will
continue to use
integers inside of $n$-gons to denote the
region number that the area is counted towards. We assume that every edge
separates two differently numbered $n$-gons.
If there is a 3-gon and it has a 3-gon adjacent to it, the whole
complex is a standard triple bubble. If a 3-gon has 4-gons adjacent to it on
both sides, the complex is either a non-minimizing triple bubble (if only three integers are used as labels) or a four chamber bubble we call a
\textit{standard quadruple bubble}. (See Chapter 4 and Figure~\ref{4bubble}.)
\begin{theorem}
Suppose a perimeter minimizing complex with equal pressure regions
and without empty chambers has a 3-gon with a 4-gon adjacent on one
side and a 5-gon adjacent on the other. The $n$-gon opposite the 3-gon
(adjacent to both the 4-gon and 5-gon) is not numbered
the same as the 3-gon. In
addition, the $n$-gon adjacent to the 5-gon but not adjacent to the 4-gon is
not numbered the same as the 3-gon or the 4-gon.
\label{th345}
\end{theorem}
\noindent \textbf{Proof:}
Assume without loss of generality that the 3-gon is part of region 1, the 4-gon
is part of region 2, and the 5-gon is part of region 3.
Suppose first that the $n$-gon opposite the 3-gon is also numbered with a 1.
Let $A$ be the edge shared by the 4-gon and the
$n$-gon opposite the 3-gon. The 4-gon can be reflected into the 3-gon as in
the proof of Theorem~\ref{main}. That is, we add a line segment $B$ at angle
of $\frac{2\pi}{3}$ halfway between the vertices of the edge shared by
the 3-gon and the 4-gon, then erase the top half of this same shared edge. The
new complex is not regular and therefore the original complex was not a
minimizer. (See Figure~\ref{345gon1}.)
\begin{figure}[hbt]
\centerline {\epsfbox {345gon1.eps}}
\caption {Reflect a 4-gon into an adjacent 3-gon}
\label{345gon1}
\end{figure}
Now assume that the opposite $n$-gon is numbered something else and
assume that
the $n$-gon adjacent to the 5-gon is numbered with a 1. We can reflect the
5-gon into the 3-gon to get a non-regular complex with identical perimeter.
To be precise, let $2a$ be the length of the 3-gon. The edge shared by
the 4-gon and 5-gon must have length $a$ by Lemma~\ref{lem34}. By
Lemma~\ref{lem35} the other interior edge of the 5-gon must then also have
length $a$. The 5-gon is actually symmetric and its remaining edge will again
have length $2a$.
If we add a line segment of length $a$ at an angle of
$\frac{2\pi}{3}$ halfway between the
vertices of the edges shared by the 3-gon and the 5-gon and connect
this line segment with another line segment of length $2a$ at angle
$\frac{2\pi}{3}$, we
have constructed an identical 5-gon inside of the 3-gon.
We can renumber the areas, erase
the top half of the edge between the old 3-gon and
old 5-gon, and erase the other edge of length $2a$ from the old 5-gon. We now
have no change in perimeter or areas enclosed, but the complex is clearly
not regular. (See Figure~\ref{345gon2}.)
\begin{figure}[hbt]
\centerline {\epsfbox {345gon2.eps}}
\caption {Reflect a 5-gon into an adjacent 3-gon}
\label{345gon2}
\end{figure}
For the last case, suppose that the $n$-gon adjacent to the 5-gon and not adjacent to the 3-gon or 4-gon is numbered the same as the 4-gon. This time, we can
reflect the 5-gon into the 4-gon. We add an edge of length $2a$ at the midpoint
of either long edge of the 4-gon at an angle of $\frac{2\pi}{3}$. When this
segment hits the external curved arc, we have created a 5-gon inside the
4-gon identical to the 5-gon we started with. Renumber areas and delete the
edge of length $2a$ between the 5-gon and the $n$-gon adjacent to it.
Once again, we have a
non-regular complex using identical perimeter. (See Figure~\ref{345gon3}.)
$\Box$
\begin{figure}[hbt]
\centerline {\epsfbox {345gon3.eps}}
\caption {Reflect a 5-gon into an adjacent 4-gon}
\label{345gon3}
\end{figure}
\bigskip
\begin{corollary}
A perimeter minimizing 4 bubble complex with equal pressure regions and no
empty chambers is either the standard quadruple bubble or has no 3-gons
adjacent to 4-gons.
\label{cor345}
\end{corollary}
\noindent \textbf{Proof:}
Suppose a 4-bubble complex has a 3-gon adjacent to a 4-gon. If the other
$n$-gon adjacent to the 3-gon is also a 4-gon, the complex is the standard
quadruple bubble. If the other $n$-gon adjacent to the 3-gon is a 5-gon,
we look at the number of the $n$-gon on the other side of the 5-gon.
Without loss of generality, assume that the 3-gon
is labeled with a 1, the 4-gon is labeled 2,
and the 5-gon is labeled 3. (See Figure~\ref{345gon4}.)
By Theorem~\ref{th345} the $n$-gon adjacent to the
4-gon and 5-gon (opposite the 3-gon) must be assigned a 4. But then, the $n$-gon
adjacent to it and the 5-gon must be labeled either 2 or 1. Neither of
these is possible again by Theorem~\ref{th345}.
$\Box$
\begin{figure}[hbt]
\centerline {\epsfbox {345gon4.eps}}
\caption {A 3-gon adjacent to a 4-gon in a 4 bubble}
\label{345gon4}
\end{figure}
\bigskip
We next consider what can happen when a 3-gon is adjacent to two 5-gons.
First we note that the 5-gons will be identical since they share an edge and
the edges that they each share with the 3-gon have the same length.
That is, the two parameters that determine a 5-gon have been chosen and are
the same. The 5-gons,
however, do not have to be symmetric. The non-shared interior edge could be
longer or shorter than the shared edge.
\begin{theorem}
\label{th355}
If a 3-gon is adjacent to two 5-gons in a perimeter minimizing
$n$ bubble complex
with equal pressure regions and no empty chambers, then the $n$-gons
adjacent to each 5-gon and the exterior are not numbered the same as
the 3-gon.
\end{theorem}
\noindent \textbf{Proof:}
When an $n$-gon adjacent to a 5-gon contains the same number as the 3-gon, we can reflect that 5-gon into the 3-gon. Let the innermost edges of the 5-gon have
length $a$ and $b$. (See Figure~\ref{355gon}.) By Lemma~\ref{lem35} the
length of the shared edge between the
3-gon and 5-gon must have length $a+b$. Let $c$ be the length of the remaining
interior edge of the 5-gon. We build a 5-gon inside of the 3-gon identical to
our 5-gon by adding a line segment of length $b$ at an angle of $\frac{2\pi}{3}$ at a distance of $a$ from the inner vertex of the 3-gon along the shared
edge. We then add a line segment of length $c$ at an angle of $\frac{2\pi}{3}$ to the end of the first line segment. Renumbering the areas, we can delete the
top side of the shared edge (length $a$) and the edge between the old 5-gon and
the adjacent $n$-gon (length $c$). We have a non-regular complex using the same perimeter which contradicts the assumption that the complex was a perimeter
minimizer.
$\Box$
\begin{figure}[hbt]
\centerline {\epsfbox {355gon.eps}}
\caption {A 3-gon adjacent to two 5-gons: The dashed lines are added, the marked edges are deleted}
\label{355gon}
\end{figure}
\bigskip
\begin{corollary}
Suppose a perimeter minimizing 4 bubble complex with
equal pressure regions and
without empty chambers has a 3-gon. The complex is either the standard
quadruple bubble or has an interior hexagon containing the same number as the
3-gon.
\end{corollary}
\noindent \textbf{Proof:} By Corollary~\ref{cor345} the complex is either the
standard quadruple bubble or the 3-gon is adjacent to 5-gons. Assume the 3-gon is numbered with a 1 and the 5-gons are numbered 2 and 3. (See Figure~\ref{355gon1}.) By Theorem~\ref{th355} the $n$-gons adjacent to the 5-gons that are also adjacent to the e
xterior
cannot be numbered 1. They cannot be numbered 2 or 3 either, since the
5-gons can be renumbered (swapped) to yield two adjacent $n$-gons enclosing
portions of the same area. They must therefore be labeled with a 4. The $n$-gon
that shares edges with both 5-gons must then be labeled with a 1. It has at
least four flat edges and thus is at least a 5-gon.
If it is a 5-gon, then the $n$-gons adjacent to the 5-gons and the exterior
must be 3-gons. Either of these 3-gons could be swapped with the 3-gon we
started with. In this case, the 5-gon we just numbered 1 would now be
adjacent to a 3-gon numbered with a 1. We could eliminate this edge and save
perimeter. So, the $n$-gon adjacent to both 5-gons must be a hexagon.
$\Box$
\begin{figure}[hbt]
\centerline {\epsfbox {355gon1.eps}}
\caption {A 3-gon adjacent to two 5-gons in a 4 bubble}
\label{355gon1}
\end{figure}
\bigskip
We can put one additional restriction on the number of 3-gons in a complex
with equal pressure regions.
\begin{theorem}
A perimeter minimizing triple bubble complex with equal pressure regions can
have at most one 3-gon enclosing portions of any given region.
\end{theorem}
\noindent \textbf{Proof:}
If there are two 3-gons enclosing portions of the same region, we can pop a
3-gon (remove the exterior arc) and recover more
than the lost area with less total perimeter by expanding the curvature of
another 3-gon. (See Figure~\ref{two3gon}.)
\begin{figure}[hbt]
\centerline {\epsfbox {two3gon.eps}}
\caption {Pop one 3-gon and increase another}
\label{two3gon}
\end{figure}
We use the formulas $L(\theta,C)=\frac{C\theta}{\sin{(\theta})}$ and
$A(\theta,C)=\frac{C^2 (\theta - \sin{(\theta)}\cos{(\theta)})}{(2\sin{(\theta)})^2}$ that relate the arc length $L$ and area $A$ of a section of a circle to the
chord length $C$ and angle $\theta$. (See Figure~\ref{chord}.)
By joining the vertices of a 3-gon in a bubble complex with equal pressure
regions,
we can decompose the 3-gon into a triangle and a
half circle. Let $C$ be the diameter of
this half circle. The triangle has base $C$, height $\frac{C}{2\sqrt{3}}$ and
area $\frac{C^2}{4\sqrt{3}}$. The area of the
half circle is of course $\frac{\pi C^2}{4}$.
Therefore, the area enclosed by a 3-gon is $C^2 (\frac{\pi}{4} + \frac{1}{4\sqrt{3}})$. We also note that the arc length is $\frac{\pi C}{2}$.
Increase the curvature of another 3-gon (so $C$ is the same) until the
angle is $2.3$ radians. The new arc length is
$L(2.3,C)=\frac{C(2.3)}{\sin{(2.3)}}$.
Which is approximately $C(3.08)$, but is
definitely less than $2L=C(2\pi)$. In other words,
we've used less total perimeter.
Now we compute the area inside the section
$A(2.3,C)=\frac{C^2( 2.3 - \sin{(2.3)}\cos{(2.3)})}{(2\sin{(2.3)})^2}$ which
is approximately $C^2(1.2574)$. We need to
recover the area from the other 3-gon as well as the area from the section we
increased. The area we need to recover is
$C^2(\frac{\pi}{2}+\frac{1}{4\sqrt{3}})$ which is approximately $C^2(.9297)$.
Since $C^2(.9297)>C^2 (\frac{\pi}{4} + \frac{1}{4\sqrt{3}})$ and therefore
we have enclosed more area with less perimeter.
Since 2.3 radians is less than $\pi$ and the tangent to an adjacent exterior
edge makes an angle of $\frac{7\pi}{6}$ radians with the chord $C$, the
increased 3-gon will not intersect any portion of the existing complex.
We chose $\theta = 2.3$ as an approximation to the solution of $\theta = \pi \sin{\theta}$ which is difficult to solve explicitly. This is the value of $\theta$ needed to use exactly the same amount of perimeter.
By decreasing curvature, we could enclose exactly the same areas and use
strictly less perimeter.
$\Box$
\bigskip
\def\baselinestretch{1.5}\large\normalsize
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,129 |
La Segunda División del Perú 2002 fue la 50.ª edición de este torneo de ascenso. Tuvo como integrantes a dieciséis equipos de fútbol del Departamento de Lima. A los catorce elencos que conservaron la categoría en la temporada anterior se les unieron Juventud Villa del Mar y Somos Aduanas, ambos ascendidos desde la Copa Perú.
El campeón del torneo fue el Unión Huaral, logrando con esto el ascenso al Campeonato Descentralizado 2003. Con respecto al descenso, Guardia Republicana, Lawn Tennis y Bella Esperanza tuvieron que retornar a sus ligas distritales de origen.
Clasificación general
Enlaces externos
Resultados en RSSSF
2002
Fútbol en Perú en 2002 | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 1,660 |
\section{Introduction}
We have studied dilepton production
in relativistic heavy ion collision (HIC) experiments
with an aim to establish an observable distinction between emissions
from the QGP and hadronic phases by utilizing two kinds of medium effects.
One originates dynamically due to decay of vector mesons and their scattering
from in-medium hadrons and the other through collective behaviour
of the fireball resulting in flow.
The emission rate of low $M$ lepton pairs is given
by the in-medium spectral function of the low mass vector mesons and is expressed
as \cite{Mclerran,GhoshJPG}
\begin{equation}
\frac{dN}{d^4qd^4x}=\frac{\alpha^2}{\pi^3q^2}f_{BE}(q_0)
\sum_{V=\rho,\omega,\phi}F_V^2m_V^2A_V(q_0,\vq)
\label{eq:dilrate2}
\end{equation}
The corresponding rate for emission from QGP is taken from \cite{JC}.
The spectral function of the $\rho$ meson which is known to play the dominant
role is given by
\begin{eqnarray}
A_\rho&=&-\frac{1}{3}\left[\frac{2\sum{\rm
Im}\Pi^R_t}{(q^2-m_\rho^2-\sum\mathrm{Re}\Pi^R_t)^2
+(\sum{\rm Im}\Pi^{R}_t)^2}+{\rm \ long.\ comp.\ }\right]
\label{eq:spdef}
\end{eqnarray}
the sum running over the loop graphs. Here we consider one-loop diagrams
containing a pion and another meson $h$ ($=\pi$, $\omega$, $h_1$ and $a_1$).
The (retarded) self energy which appears in the spectral function
can be obtained in the real time formalism
of thermal field theory from the 11-component of the self-energy which
is a 2$\times 2$ matrix in this approach~\cite{Bellac,MallikRT}. One thus writes for the
$\pi-h$ loop,
\begin{equation}
\Pi_{\mu\nu}^{11}(q)=i\int\frac{d^4k}{(2\pi)^4}N_{\mu\nu}(q,k)D_\pi ^{11}(k)D_h^{11}(q-k)
\end{equation}
where $D^{11}(q)$ represents thermal propagators and the factor $N_{\mu\nu}$
includes tensor
structures associated with the two vertices and those of the vector propagator
in the loop. From the trace and the 00-component, one obtains the transverse and
the longitudinal polarisations. The imaginary part in the present case basically
decides the shape and magnitude of the spectral function, the real part
causing a minor effect~\cite{GhoshJPG,Ghosh1}. Confining to positive values of $q^2$ and
$q_0$ we have,
\begin{eqnarray}
&&{\rm Im}\Pi_{t,l}^R(q_0,\vq)=-\pi\int\frac{d^3\vec k}{(2\pi)^3 4\om_\pi\om_h}
\left[N_{t,l}(k_0=\om_\pi)\{(1+n(\om_\pi)+n(\om_h))\right.\nonumber\\
&&\delta(q_0-\om_\pi-\om_h)\}\left.+N_{t,l}(k_0=-\om_\pi)\{(n(\om_\pi)-n(\om_h))\delta(q_0+\om_\pi-\om_h)\}\right]~.
\label{eq:impi}
\end{eqnarray}
The first term is non-vanishing for $q^2\ge (m_h+m_{\pi})^2$
producing the unitary cut and
the second is non-vanishing for $q^2\ge (m_h-m_{\pi})^2$ giving
the Landau cut.
We plot in Fig.~\ref{fig:dil_LU} (left panel, upper compartment)
the dilepton emission rate keeping only the
$\rho$ contribution in Eq.~\ref{eq:dilrate2} in which we show the
relative contributions from
the cuts in the $\pi-h$ loops keeping only one of them at a time.
The unitary and Landau cuts for
the $ \pi,\omega,h_1 $ and $a_1$ are seen to contribute with different magnitudes
for different values of invariant mass ($M$) of $\rho$.
The $\pi-\pi$ loop
has only the unitary cut and this contributes most significantly
to dilepton emission near the $\rho$ pole.
For $\pi-\omega$ loop Landau cut ends at $M=m_\omega-m_\pi$ and
the unitary cut starts at $M=m_\omega+m_\pi$ , so there is no contribution at the $\rho$ pole.
The Landau cut for the $\pi-a_1$ self-energy extends up to about 1100 MeV and
makes a substantial contribution both at and below the $\rho$ pole. The unitary cut
starts at a much higher value of $M$ and hence does not make a significant contribution
to the $\rho$ spectral function.
In Fig.~\ref{fig:dil_LU} (left panel, lower compartment)
we show the cumulative contribution to the lepton pair yield
from the $\pi-\pi$ and the $\pi-h$ loops
in the region below the bare $\rho$ pole.
\begin{figure}
\includegraphics[scale=0.29]{fig3.eps}
\includegraphics[scale=0.29]{fig9.eps}
\caption{Left panel (upper compartment) shows contributions from the
discontinuities of the self-energy graphs
to the dilepton emission rate at $T=175$ MeV. $L$ and $U$ denote the
Landau and unitary cut contribution. Left panel (lower compartment)
shows contributions from
the different mesons in the loop.
Right panel displays invariant mass distribution of dileptons from
hadronic matter (HM) and from (QM) at LHC energies.}
\label{fig:dil_LU}
\end{figure}
We assume that an equilibrated QGP is formed in the HIC which
cools due to expansion and consequently reverts back to hadrons at
$T_c=175$ MeV. After the completion of the phase transition the
hadronic matter cools further
and eventually freezes out
first chemically at a temperature $T_{ch}$ (=175 MeV) and then kinetically
at a temperature $T_F$ (=120 MeV). Relativistic hydrodynamics with cylindrical
symmetry~\cite{hvg} and boost invariance~\cite{jdb} along longitudinal
directions has been used for space-time description.
Equation of states from lattice QCD and hadronic resonance
gas have been used here(see~\cite{GhoshJPG} for details).
The initial thermalization time, $\tau_i$ and initial temperature, $T_i$
are constrained to the total hadronic
multiplicity, $dN/dy\sim T_i^3\tau_i$. For $dN/dy=2600$, $T_i=756$ MeV
and $\tau_i=0.1$ fm/c~\cite{GhoshJPG} for LHC.
With these inputs the
space time integration is performed to obtain the transverse mass,
$M_T (=\sqrt{M_{av}^2+p_T^2})$
spectra for various $M$ range - $M_{min}$ to $M_{max}$
with $M_{av}=(M_{max}+M_{min})/2$.
The $M_T$ spectra for lepton pairs
and their respective slopes for different mass windows have been depicted
in Fig.~\ref{fig2} in the left and right panels respectively.
It is evident from Fig.~\ref{fig:dil_LU} (right panel) that
the large $M$ pairs originate from early (QGP) phase where the
radial flow is minimal and the
low $M$ pairs stem from the QGP as well as from the late (hadronic)
phase containing reasonable amount of radial flow. However,
pairs with $M\sim m_\rho$ are overwhelmingly generated from
the late hadronic phase with very large radial flow
{\it i.e.} the value of $v_r$ is maximum for $M\sim m_\rho$ and lower
at both sides of the $\rho$ mass. Resulting in the nonmonotonic
behaviour of the inverse slope, which contains
the effect of temperature ($T_{th}$) and radial flow ($v_r$)
as $T_{eff}=T_{th}+ M_{av}v_r^2/$. At large M ($>1.2$ GeV) the flow
is predominantly longitudinal and hence the inverse slope
falls slowly with decreasing $M_{av}$.
In summary
we have studied both the $M$ and $M_T$
distributions of dileptons from heavy ion collisions at LHC energy.
We have shown
that the effective temperature extracted from the inverse
slope of the $M_T$ spectra for various $M$
windows can be used as an efficient tool to characterize different
phases of the evolving matter.
\begin{figure}
\includegraphics[scale=0.285]{fig10.eps}
\includegraphics[scale=0.285]{fig11.eps}
\caption{Left panel shows variation of dilepton yield with $M_T-M_{av}$.
Right panel displays
the variation of the inverse slopes of $M_T$ distribution for
different $M$-bins.
The dashed line is obtained by setting vanishing radial flow.}
\label{fig2}
\end{figure}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,518 |
{"url":"https:\/\/www.airbestpractices.com\/system-assessments\/end-uses\/assessment-improves-electroplater-production","text":"Industrial Utility Efficiency\n\n# Assessment Improves Electroplater Production\n\nFaced with rising energy costs, a large electroplating company sought to improve the efficiency and reliability of its compressed air system. After getting a quote from their vendor on a new 300-hp compressor to replace an existing unit, the company sought a comparison quote due to the significant investment the new compressor represented. Based on a recommendation from one of their customers, they turned to Scales Industrial Technologies.\n\nRather than simply providing a quote for a new compressor, Scales recommended a compressed air system assessment. This impressed the company enough to give Scales the green light to assess the system. Scales\u2019 engineers found that by modifying some production equipment, the firm could reduce its compressed air demand by 80%. This would allow the company to operate just as effectively with a much smaller compressor and achieve significant energy and maintenance savings.\n\n### The Project\n\nThe primary demand for compressed air in the plant was from a series of compressed air-driven nozzles that atomized water to rinse away chemicals used during the plating process. Working with a nozzle manufacturer, Scales designed a process change that enabled the strip washing nozzles to use only water instead of compressed air. Eliminating the air-driven strip washing sharply reduced the plant\u2019s compressed air requirements.\n\nThe assessment also found that a filter press was being fed an unneeded, constant supply of compressed air. Scales\u2019 engineers worked with the plant employees to install high pressure off-line air storage supplied by a 5-hp reciprocating compressor to accommodate the high volume, intermittent demand of the filter press. This further reduced the load on the main compressor. Finally, a leak detection and repair campaign led to additional compressed air energy savings. The combination of all these measures reduced the plant\u2019s air demand to where it could be met with a smaller compressor. The plant was able to replace their 300-hp compressor with a 75-hp variable displacement compressor and a new, energy efficient dryer.\n\n### Project\n\n\u2022 Replaced 300-hp centrifugal unit with 75-hp rotary screw compressor.\n\u2022 Added 5-hp reciprocating compressor package with compressed air storage.\n\u2022 Added 75-hp compressor for back-up.\n\u2022 Converted compressed air-driven strip washing nozzles to water-driven nozzles.\n\u2022 Identified and repaired leaks.\n\u2022 Costs after utility rebate: \\$72,500. ### Benefits \u2022 Annual energy savings of \\$153,000 and 1,631,000 kWh.\n\u2022 Production benefits of \\$50,000 annually. \u2022 Other annual savings of \\$49,000.\n\u2022 Simple payback of less than 4 months.\n\n### Results\n\nNot only did Scales\u2019 recommended project improve energy efficiency and yield significant energy savings, it also improved production and yielded important non-energy benefits. Compressed air energy use was reduced by more than 80%, saving 1,631,000 kWh and \\$153,000 per year. The project also resulted in annual savings of \\$49,000 from lower water and water treatment costs, lower cooling tower loads, and lower maintenance costs. In addition, production became more reliable, leading to \\$50,000 in annual avoided product loss. Because the company qualified for an incentive of \\$217,500 from its electric utility, total implementation costs were \\\\$72,500, resulting in a simple payback of just over 4 months. This project\u2019s success has encouraged this electroplating corporation to become more proactive about energy efficiency.\n\n \"Scales worked with our operators to retrofit our rinse lines utilizing state-of-the-art technology. Involving our operators in this retrofit process was the key to gaining their commitment and enthusiasm. In addition to the anticipated energy savings, we are enjoying a higher-quality surface finish along with enhanced process consistency and capability.\" - Operations V.P., Electroplating Client","date":"2022-01-24 20:06:30","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.3463383913040161, \"perplexity\": 7019.673152173651}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-05\/segments\/1642320304600.9\/warc\/CC-MAIN-20220124185733-20220124215733-00213.warc.gz\"}"} | null | null |
It's Time To Swim In Muds! The Northern Ireland Bog Snorkeling Championships Is On!
Swimming in muddy water may sounds a bit of unsanitary, but this event held on International Bog Day was to raise awareness of how important bogs are. Yep, there is actually an event called the International Bog Day in Northern Ireland. And yep, we thought this is kind of awesome.
7. This chap is not as brave as his fellow competitor in the previous picture.
8. Laura Sherwin takes part in the Irish Bog Snorkelling championship at Peatlands Park in Dungannon, Northern Ireland. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,193 |
SPROWSTON SPORTS & SOCIAL CLUB
The Blue Boar Lane site was occupied by Boulton & Paul Sports Club from the 1940's until its closure in 1986. Sprowston Lads Football Club was formed in 1978 and shared the use of the sports ground.
When Boulton & Paul left T S Industrial Roofing, under the chairmanship of Richard Shorten, took over the running of the sports ground. Richard and his company played a major part in maintaining the club, both then and well into its future, with donations of materials, expertise, man hours and finance.
In 1992, when the Gothic Social Club closed, the TS and Gothic Clubs combined their members and committees giving birth to Sprowston Sports and Social Club. Terry Toomer, a Sprowston resident, was elected as chairman and the Club's aim was to become financially self-supporting by increasing its sports and social facilities. In 1993 The Valley Drive Social Club on Plumstead Road closed. The club house was dismantled and transported to our site and trebled the size of our social area and included a new bar. A new entrance hall and a new toilet block were soon added.
In 1994, part aided by a Littlewoods Pools grant, the artificial sports court with lighting was built.
In 2000, part aided by National Lottery and Broadland District Council grants, the full size bowling green was installed.
Over time other improvements include a new cellar, a petanque court, an upgrade of the changing rooms, an upgrade to the football pitch, car park improvements, a children's play area, etc, etc.
All this has been achieved by the hard work and dedication of those very special people who have given, and who are currently giving, their time and expertise to create what we have today.
Sprowston Sports & Social Club is not run by any council, by any business or by any other body. This is truly unique. It was built and managed by its members, for its members and for the community. This ethic continues today.
Blue Boar Lane
NR7 8RY
© 2017 Lauren Gilman | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 9,299 |
Q: Symbol not found error I am doing a project for school. In that project there were various classes given to me, with the task to complete them.
One of my classes Orbita, creates an object from another Graficos. Graficos was given to me completed but as bytecode, not as source.
When I try to compile it, it gives me the following error
the symbol not found error every time Orbita uses the class
Graficos.
It gives my an error on every time it has Graficos. (..)
C:\Users\user1\Desktop\loll\trabalho (1)>javac -cp ptplot5.8:. *.java
Orbita.java:206: error: cannot find symbol
Graficos.mostraGrafico(vectorParaGrafico, legendas);
^
symbol: variable Graficos
location: class Orbita
Orbita.java:230: error: cannot find symbol
Graficos.mostraGrafico(vectorParaGrafico,
^
symbol: variable Graficos
location: class Orbita
Orbita.java:262: error: cannot find symbol
Graficos.mostraGrafico(vectorParaGrafico, legendas);
^
symbol: variable Graficos
location: class Orbita
Orbita.java:283: error: cannot find symbol
Graficos.mostraGrafico(vectorParaGrafico,
^
symbol: variable Graficos
location: class Orbita
4 errors
C:\Users\user1\Desktop\loll\trabalho (1)>
My code:
public void projecta2D(int coordX, int coordY) {
double[][] vectorParaGrafico = new double[iteracoes + 1][2];
for (int k = 0; k <= iteracoes; k++) {
vectorParaGrafico[k][0] = registo[k].getCoordenada(coordX);
vectorParaGrafico[k][1] = registo[k].getCoordenada(coordY);
}
Graficos.mostraGrafico(vectorParaGrafico,
new String[]{"coordenada 1", "coordenada 2", "orbita"});
}
public void serieTemporal(int coordenada, int inicio, int fim,
String[] legendas) {
if (inicio < 0)
inicio = 0;
if (fim > iteracoes)
fim = iteracoes;
double[][] vectorParaGrafico = new double[fim - inicio + 1][2];
for (int k = inicio; k <= fim; k++) {
vectorParaGrafico[k][0] = k;
vectorParaGrafico[k][1] = registo[k].getCoordenada(coordenada);
}
Graficos.mostraGrafico(vectorParaGrafico, legendas);
}
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,789 |
{"url":"https:\/\/www.bartleby.com\/questions-and-answers\/consider-a-project-to-supply-detroit-with-40000-tons-of-machine-screws-annually-for-automobile-produ\/92273894-f355-431f-b042-c4c1d2077ec9","text":"# Consider a project to supply Detroit with 40,000 tons of machine screws annually forautomobile production. You will need an initial $5,600,000 investment in threadingequipment to get the project started; the project will last for 6 years. The accountingdepartment estimates that annual fixed costs will be$600,000 and that variable costsshould be $250 per ton. Further, the accounting department will depreciate the initialfixed asset investment straight-line to zero over the 6-year project life and estimate asalvage value of$450,000 after dismantling costs. The marketing department estimatesthat the automakers will let the contract at a selling price of $340 per ton. The engineeringdepartment estimates you will need an initial net working capital investment of$560,000.You require a return of 13 percent and face a marginal tax rate of 24 percent on thisproject.Suppose you\u2019re confident about your own projections, but you\u2019re a little unsure aboutDetroit\u2019s actual machine screw requirements.a. What is the sensitivity of the project OCF to changes in the quantity supplied?b. What about the sensitivity of NPV to changes in quantity supplied?c. Given the sensitivity number you calculated, is there some minimum level ofoutput below which you wouldn\u2019t want to operate?\n\nQuestion\n\nConsider a project to supply Detroit with 40,000 tons of machine screws annually for\nautomobile production. You will need an initial $5,600,000 investment in threading equipment to get the project started; the project will last for 6 years. The accounting department estimates that annual fixed costs will be$600,000 and that variable costs\nshould be $250 per ton. Further, the accounting department will depreciate the initial fixed asset investment straight-line to zero over the 6-year project life and estimate a salvage value of$450,000 after dismantling costs. The marketing department estimates\nthat the automakers will let the contract at a selling price of $340 per ton. The engineering department estimates you will need an initial net working capital investment of$560,000.\nYou require a return of 13 percent and face a marginal tax rate of 24 percent on this\nproject.\nDetroit\u2019s actual machine screw requirements.\na. What is the sensitivity of the project OCF to changes in the quantity supplied?\nb. What about the sensitivity of NPV to changes in quantity supplied?\nc. Given the sensitivity number you calculated, is there some minimum level of\noutput below which you wouldn\u2019t want to operate?","date":"2021-07-24 04:30:01","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.19051457941532135, \"perplexity\": 3735.4738546410194}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-31\/segments\/1627046150129.50\/warc\/CC-MAIN-20210724032221-20210724062221-00230.warc.gz\"}"} | null | null |
\section{Dynamics of Triple System}
\label{sec:Dynamics}
In this section, we sequentially explain a hierarchical triple system, Kozai-Lidov mechanism, and their stability and constraints.
\subsection{Hierarchical Triple System and Kozai-Lidov Mechanism}
\label{sec:KL}
We can have several stable configurations and shapes when it comes to a three-body dynamical system. Broadly addressing, it can be concentric, coplanar orbits or inclined orbits in a hierarchical arrangement.
In this paper, we treat the so-called hierarchical triple system, whose schematic picture is given in Fig.~\ref{fig:3body}.
\begin{figure}[h]
\includegraphics[width=5cm]{threesystem_AE.png}
\caption{The hierarchical triple system is constructed from inner and outer binaries.
The inner binary consists of objects whose masses are $m_1$ and $m_2$,
and the outer one is the pair of the inner binary and
the third body with mass $m_3$.}
\label{fig:3body}
\end{figure}
As depicted in the figure, we can decouple a system into the two-body inner binary orbit and the outer orbit with tertiary companion in motion around center of mass of inner binary, if
the distance between the first and second bodies is much shorter than the distance to the third body. This hierarchy is supported by assuming that the gravitational effect of the third body is much smaller than the gravitational interaction between the first and second bodies.
In a two-body problem in Newtonian dynamics,
a bound system is given by an elliptic orbit, which
is described by six orbital elements; the semi-major axis $a$, the eccentricity $e$, the inclination $i$, the argument of periastron $\omega$, the longitude of ascending node $\Omega$, and the mean anomaly $M$.
Although these elements are constant (except mean anomaly) in an isolated two-body system, in the hierarchical three-body system, the perturbations from the tertiary companion affect the binary motion and modify the trajectory from that of the isolated one. Such a trajectory is not closed in general, but we can introduce an {\it osculating orbit} at each time, whose trajectory is approximated by the elliptical orbit with the above six orbital elements determined by the instantaneous position and velocity \citep{Murray00}.
As for the outer orbit, we pursue the center of mass of the inner binary rotating around the tertiary companion (see Fig.~\ref{fig:3body}). It can also be described as another osculating orbit. Hence, we introduce two osculating orbits, which are called as inner and outer orbits:
The masses of the inner binary are $m_1$ and $m_2$, while the tertiary companion has the mass $m_3$. We use subscriptions `in' and `out' to show the elements of inner and outer orbits, respectively.
In the hierarchical three-body system with large relative inclination,
a secular change of orbital elements may occur.
Under some conditions, there appears an oscillation between
the relative inclination and the eccentricity of the orbit, which is called
Kozai-Lidov (KL) oscillation\citep{Kozai62,Lidov62}.
This mechanism was independently discovered by Kozai and Lidov in 1962.
The relative inclination $I$ is defined as the argument between the inner and outer orbital planes.
It is given by two orbital elements as
\begin{equation*}
\cos I = \cos i_{\mathrm{in}} \cos i_{\mathrm{out}}
+ \sin i_{\mathrm{in}} \sin i_{\mathrm{out}}
\cos \left( \Omega_{\mathrm{in}} - \Omega_{\mathrm{out}} \right).
\label{eq:relative_i}
\end{equation*}
If we assume $m_1\ll m_2, m_3$, we find that the oscillation occurs in secular time-scale under conservation of energy and angular momentum.
In Newtonian quadrupole approximation method \citep{Shevchenko17}, it results in the secular exchange of $e_\mathrm{in}$ and $I$ with the conserved value of $\Theta$,
which is defined by
\begin{equation}
\Theta = (1-e_\mathrm{in}^2) \cos^2 I \,.
\label{eq:theta}
\end{equation}
This approximation also gives the criterion of KL-oscillation for circular orbit as
\begin{equation*}
0 \le |\cos I |
\ \le \sqrt{\frac{3}{5}} ,
\label{eq:kozaiC}
\end{equation*}
which is equivalent to $39.2315^\circ \lesssim I
\lesssim 140.7685^\circ $.
This KL type oscillation will occur even if $m_1$ is not much smaller than $m_2, m_3$
\cite{Naoz12}.
When we include a relativistic effect, we also find the similar oscillation\cite{Naoz16,Wen2003}.
The effect of general relativity (GR) changes these KL- criterion
as \citep{Migaszewski11}.
\begin{equation*}
0
\le |\cos I| \
\le \sqrt{{3\over 5}\left(1-\frac{4}{9}
\left\langle \mathcal{H}_\mathrm{GR}
\right\rangle\right) } ,
\label{eq:KozaiC_PN}
\end{equation*}
where the GR correction term $\left\langle \mathcal{H}_{\rm GR} \right\rangle $
is derived from the double-averaged post-Newtonian Hamiltonian of two-body relative motion \citep{Richardson88} as
\begin{equation*}
\left\langle \mathcal{H}_{\rm GR} \right\rangle
= \frac{3Gm_{\rm in}^2a^3_{\rm out} \sqrt{1- e^2_{\rm out}}}{c^2 m_3a^2_ {\rm in} \sqrt{1- e^2_{\rm in}}} \,,
\label{eq:H_GR}
\end{equation*}
where $m_{\rm in}=m_1+m_2$ is the total mass of the inner binary.
The KL-oscillation time-scale is also modified when the GR effect is taken into account\cite{Naoz16}.
Note that it has been pointed out \cite{Will14a,Will14b} that
it is imperative to take into account the
`cross terms' between the Newtonian perturbation and the Post-Newtonian precession.
In our numerical evolution, we integrate the equations of motion (up to 1PN) directly. The effects of the cross terms are automatically included.
\subsection{Stability and Constraints}
\label{sec:Analytic}
In this section, we discuss stability criteria for a hierarchical triple system.
We also give constraints for KL oscillation to occur, which are obtained from the conditions such that KL-oscillations are not suppressed by relativistic effects, i.e., the relativistic precession of the periastron of the inner binary and
the Lense-Thirring (LT) precession effect due to the rapid rotation of heavy third body.
\subsubsection{\textbf{\underline{Typical time scales}}}
Before discussing stability and constraints,
we first present typical time scales of a hierarchical triple system.
The orbital periods of the inner and outer binaries
are given by
\begin{eqnarray*}
P_{\rm in} &=& 2\pi \sqrt{a_{\rm in}^3\over Gm_{\rm in}},
\\
P_{\rm out} &=& 2\pi \sqrt{a_{\rm out}^3\over Gm_{\rm out}},
\end{eqnarray*}
where $m_{\rm out}=m_1+m_2+m_3$ is the total mass of a triple system.
The KL oscillation time scale is given by \citep{Antognini2015}
\begin{eqnarray*}
t_\mathrm{KL} \sim P_{\mathrm{in}}\frac{m_{\mathrm{in}}}{m_3} \left( \frac{a_{\mathrm{out}}}{a_\mathrm{in}} \right)^3
\left( 1-{e_\mathrm{out}}^2 \right)^\frac{3}{2}\,.
\end{eqnarray*}\\
\subsubsection{\textbf{\underline{Stability of Hierarchical Arrangement}}}
For KL oscillations to occur, the first aim is to maintain the stability of a three-body system. A necessary (not sufficient) criteria is given by \cite{Blaes02}, which predicts a minimum separation between inner and outer orbits for stability,
beyond which a system may cause a chaotic instability.
Hence, we have one stability condition,
\begin{equation*}
\frac{ a_\mathrm{out} }{ a_\mathrm{in} } >\frac{2.8}{ 1-e_\mathrm{out} }
\left[ \frac{m_{\rm out}}{m_\mathrm{in}}
\frac{ 1+e_\mathrm{out} }{ (1-e_\mathrm{out})^\frac{1}{2} } \right]^\frac{2}{5}
\,.
\end{equation*}
For simplicity, throughout this paper, we set outer orbit not to be so eccentric such that $e_{\rm out}\ll 1$. Thus the scaled expression is written as,
\begin{equation}
\frac{ a_\mathrm{out} }{ {\rm AU} } > 2.8
\left( \frac{m_{\rm out}}{M_\odot}\right)^{2\over 5}
\left(\frac{m_\mathrm{in}}{M_\odot}\right)^{-{2\over 5}} \, \frac{ a_\mathrm{in}}{ {\rm AU}}
\,
\label{eq:max_stable}
\end{equation}
This expression gives a lower bound to the separation between two orbits.
\subsubsection{\textbf{\underline{Constraints from the relativistic effects}}}
\noindent
{\bf 3-1. Relativistic Precession of Periastron}\\[.2em]
KL oscillations are suppressed or maximum eccentricity is restricted by the relativistic precession of the periastron of the inner binary. An analytical expression can be derived using double averaged approximation up to quadrupole order to estimate parameter space such that KL oscillations are not destroyed by precession\cite{Blaes02}.
The GR precession timescale of the periastron of the inner binary
can be estimated as,
\begin{eqnarray*}
t_{\rm GR,in} \sim
\bigg({\frac{a_{\rm in}}{r_{\rm g,in}}}\bigg)^{5\over 2} P_{\rm in} \left( 1-{e_\mathrm{in}}^2 \right),
\end{eqnarray*}
where $r_{g,{\rm in}}\equiv {Gm_{\rm in}/c^2}$ is the gravitational radius of the inner binary.
Comparing this with the KL time scale $t_{\rm KL}$,
we find the constraint
\begin{equation*}
\left({a_{\rm out}\over a_{\rm in}}\right)^3<
{3\over 4}
{a_{\rm in}\over r_{g,{\rm in}}}
{m_3\over m_{\rm in}}
\left({1-e_{\rm in}^2\over 1-e_{\rm out}^2}\right)^{3\over 2}
\,.
\end{equation*}
For clarity, the scaled expression takes the form:
\begin{eqnarray}
&&\left({a_{\rm out}\over
{\rm AU}}\right)^3<
10^8 \left({1-e_{\rm in}^2}\right)^{3\over 2}
\left({m_{\rm in}\over {M_\odot}} \right)^{-2}
\left({a_{\rm in}\over {\rm AU}}\right)^4
\left(m_3 \over {M_\odot}\right)
\,,
\nonumber \\
&&
~~
\label{eq:GR_precession}
\end{eqnarray}
assuming $e_{\rm out}\ll 1$.\\
\noindent
{\bf 3-2 Lense-Thirring Precession Effect}\\[.2em]
Recent studies have shown changes in KL oscillations caused due to GR effects involving a SMBH \cite{Wang19,fang19,fang19a}.
If the third body is a rapidly rotating SMBH, then LT effect might become important changing the evolution of eccentricity excitation from the usual KL oscillation. It appears in the 1.5 PN order.
For a rotating black hole, which is the third body in our setting,
the spin angular momentum is given as $S_3 = \chi_3 Gm_3^2/c$ where $\chi_3$ is the spin parameter, and outer orbital angular momentum is given as
\begin{eqnarray*}
L_{\rm out} = \mu_{\rm out} \left[G(1-e_{\rm out}^2)
a_{\rm out}M\right]^{1\over 2}
\,,
\end{eqnarray*}
where the reduced mass of the outer orbit is defined by
\begin{eqnarray*}
\mu_{\rm out} &=& {m_{\rm in} m_3\over m_{\rm out}}={(m_1+m_2) m_3\over m_1+m_2+m_3}
\,.
\end{eqnarray*}
The time scale of the orbit-averaged precession
of $L_{\rm out}$ around $S_3$ is given as
\begin{equation*}
t_{L_{\rm out}S_{3}} = \frac{GS_3(4+3m_{\rm in}/m_3)}{2c^2
(1-e^2_{\rm out})^{3\over 2} a_{\rm out}^3}
\end{equation*}
The KL oscillations can be affected when $t_{L_{\rm out}S_{3}}$ becomes comparable to the KL oscillation time scale $t_{\rm KL}$.
Hence, in order to observe the usual KL oscillation, we have the constraint such that
$t_{L_{\rm out}S_{3}} > t_{\rm KL}$, for which
the scaled expression is given as:
\begin{equation}
\bigg({a_{\rm in}\over {\rm AU}}\bigg)^{3\over 2} >
10^{-12}
\bigg({m_3\over M_\odot} \bigg) \bigg({m_{\rm in}\over M_\odot} \bigg)^{1\over 2}
\,,
\label{eq:LT_precession}
\end{equation}
assuming $\chi_3 \sim 1$.
If the ratio is comparable or greater than 1, then LT precession dominates resulting in chaotic evolution of eccentricity as shown in \cite{Wang19} and hence more dissipation of gravitational waves,
which may be interesting for further study.
However, in this paper our discussion is limited to 1PN effect and we focus more on the conventional KL oscillations in the inspiral phase.
\section{Gravitational Waves}
\label{sec:GRWs}
Under this section, we set the framework by recalling the famous quadrupole formula and definition of `near coordinate zone' (NCZ). We further discuss GWs from highly eccentric orbits,
which are expected in KL oscillation.
\subsection{Quadrupole Formula}
\label{subsec:Quad}
In GR, the leading order contribution of the observed gravitational waves
is given by the quadrupole formula. It allows the evaluation of gravitational wave energy and waveforms emitted by a dynamical system.
In a wave zone, the gravitational waves denoted by a small deviation of the metric $h_{ij}$ are described by
\begin{equation*}
\label{eq:QF}
h_{ij}^{\rm TT}(t,\textbf{x}) = \frac{2G}{Dc^4} \ddot{Q}_{ij}^{\rm TT}
\,,
\end{equation*}
where TT refers to transverse-traceless gauge and a dot
denotes the time derivative.
$D$ denotes the distance of the source from the observer.
The reduced quadrupole moment $Q_{ij}$ is defined by
\begin{equation*}
Q_{ij} = M_{ij} - \frac{1}{3} \delta_{ij} M^k_{\,k}
\,,
\end{equation*}
where $i,j, \cdots$ run from 1 to 3 ($ x^1 = x , x^2=y,x^3=z $),
and
the mass quadrupole moment $M_{ij}$ for $N$ masses is expressed as
\begin{eqnarray*}
M_{ij} = \sum_{A=1}^{N} m_A x_{A}^{i} x_{A}^{j}\,.
\end{eqnarray*}
$m_A$ denotes the $A$-th mass at the location $x_A^i$ with origin at the center of mass of a system.
The power of GW emission is given by
\begin{eqnarray*}
{\cal P}={G\over 5c^5}\langle {\stackrel{\cdots}{Q}}_{ij}
{\stackrel{\cdots}{Q}}\raisebox{.5em}{\scriptsize $ij$}\rangle
\,,
\end{eqnarray*}
where the bracket denotes the Brill-Hartle averaging, which
is taken over several wave length \citep{Peters1963}.
The GW amplitude from a circular binary system consisting of masses $m_1$ and $m_2$ with orbital radius $a$ is approximately given by
\begin{eqnarray*}
h \sim {{4 G^2 m_1m_2}\over {c^4 a D}}
\,,
\end{eqnarray*}
which is described by the scaled expression as
\begin{equation*}
\label{eq:QF}
h \sim 1.7\times 10^{-25} \bigg(\frac{m_1}{M_\odot}\bigg) \bigg(\frac{m_2}{M_\odot}\bigg) \bigg(\frac{{\rm AU}}{a_{\rm in}}\bigg) \bigg(\frac{10{\rm kpc}}{D}\bigg)
\end{equation*}
Another important quantity
is energy spectra of the GWs.
For an isolated circular orbit,
the energy radiated in gravitational waves is concentrated in the second harmonic such that
\begin{equation*}
f_{\rm GW} = 2 f_{\rm in}\,,
\end{equation*}
where $f_{\rm in}$ is the orbital frequency of the inner binary given by
\begin{equation*}
f_{\rm in}={1\over P_{\rm in}}={1\over 2\pi}
\sqrt{Gm_{\rm in}\over a_{\rm in}^3}
\,.
\end{equation*}
We find it in the energy spectrum, whose example is given in Appendix \ref{GW_isolated} and Fig. \ref{fig:spectra_binary}.
Large eccentricities induced in the inner binary during KL oscillation and variation of eccentricity with time produce more harmonics in gravitational waves, which make the energy spectra rich as compared to the spectra from an isolated circular binary as well as an enhancement of the wave amplitude
as shown in the next subsection.
In the derivation of the quadrupole formula,
a system must be contained in its NCZ , i.e., in order to use the quadrupole formula, a system should be inside a region (centered on the origin of the coordinates) of radius comparable to (or smaller than) the GW wavelength $\lambda$ (see some good example of what not to do given in \citep{Bonetti17}).
Therefore, we must keep a check that a hierarchical triple system is in its NCZ
when the formalism is applied to the center of mass of the triple system for the model in consideration. There are two NCZs for a hierarchical triple system, whose sizes correspond to two wave lengths: one is that of an inner binary $\lambda_{\rm in} = a_{\rm in}c/v_{\rm in} $ and the other is that of an outer binary with $\lambda_{\rm out} = a_{\rm out}c/v_{\rm out} $, where $v_{\rm in}$ and $v_{\rm out}$ are typical velocities of the inner binary and
the outer binary, respectively.
Since we focus on GWs from the inner binary,
the NCZ condition is given by $a_{\rm out}<\lambda_{\rm in}$.
An inaccurate application may lead to apparent and
unphysical behavior as shown in Fig. 18 of \citep{GP2011}.
\subsection{Gravitational waves from highly eccentric orbit}
In this subsection, we discuss
the GWs from a highly eccentric binary system.
When the KL oscillation occurs, the inner binary system is
supposed to be in an inspiral phase. Hence, it is usually expected to be difficult to observe the GWs unless the inner binary is just before a coalescence.
However, the eccentricity in the KL oscillation can become very large,
which enhances the amplitudes of GWs and those frequencies.
As a result, we may have a chance to observe the GWs even in an inspiral phase.
Gravitational Waveform from a binary system depends strongly on the eccentricity of the orbit.
In an eccentric orbit, the emission of gravitational waves is higher and distributed over many harmonics.
The GWs from an eccentric binary was studied by Peter and Mathews\citep{Peters1963}.
The total radiation power ${\cal P}$ averaged over one period is given by
\begin{eqnarray*}
{\cal P}={32G^4m_1^2m_2^2m_{\rm in}\over c^5 a_{\rm in}^5(1-e_{\rm in}^2)^{7/2}}\left(1+{73\over 24}e_{\rm in}^2+{37\over 96}e_{\rm in}^4\right)
\end{eqnarray*}
where $a_{\rm in}$ and $e_{\rm in}$ are the semi-major axis and the eccentricity of
the inner binary, respectively.
It shows that the radiation power increases
rapidly when the eccentricity becomes large.
The total radiation power ${\cal P}$ is the sum of the power radiated in the $n$-th harmonics
${\cal P}_n$ as
\begin{eqnarray*}
{\cal P}=\sum_{n=1}^\infty {\cal P}_n
\,.
\end{eqnarray*}
The radiation power ${\cal P}_n$
in the $n$-th harmonic is given by
\begin{eqnarray*}
{\cal P}_n={32G^4m_1^2m_2^2m_{\rm in}\over c^5 a_{\rm in}^5}g(n,e_{\rm in})
\end{eqnarray*}
where
\begin{eqnarray*}
g(n,e)&=&{n^4\over 32}\Big{[}
\Big{(}J_{n-2}(ne)-2eJ_{n-1}(ne)
\nonumber \\
&+&{2\over n}J_n(ne)+2eJ_{n+1}(ne)-J_{n+2}(ne)\Big{)}^2
\nonumber \\
&+&(1-e^2)\Big{(}J_{n-2}(ne)-2J_n(ne)+J_{n+2}(ne)\Big{)}^2
\nonumber \\
&+&{4\over 3n^2}\left(J_n(ne)\right)^2\Big{]}
\,,
\end{eqnarray*}
where $J_n(x)$ is the Bessel function of the first kind.
The spectrum ${\cal P}_n$ is peaked at the GW frequency $f_{\rm peak}$,
\begin{equation*}
f_{\rm peak} = n_{\rm peak} f_{\rm in},
\end{equation*}
where the magnification factor $n_{\rm peak}$ is well approximated by
\begin{equation*}
n_{\rm peak} = \frac{2 (1+ e)^{1.1954}}{(1- e^2)^{1.5}}
\,,
\end{equation*}
for the range of $10^{-6}<1-e^2<1$\citep{Wen2003}.
For $e \ll 1$, $n_{\rm peak}$ is just 2, while considering high eccentricity, say $e \sim 0.98$ , we find $n_{\rm peak} \sim 574$.
As the eccentricity increases, change in $f_{\rm peak}$ becomes significant and may lie in the detectable frequency band $10^{-4}$ Hz $\sim 10$ kHz.
It is worthwhile evaluating the peak frequency as
\begin{equation*}
f_{\rm peak} \sim 3\times 10^{-8} n_{\rm peak} \bigg(\frac{m_{\rm in}}{M_\odot}\bigg)^{1\over2}\bigg(\frac{a_{\rm in}}{\rm AU}\bigg)^{-{3\over2}}~{\rm Hz}
\end{equation*}
Note that
although the $n$-th harmonics power ${\cal P}_n$
has a peak at
$f=f_{\rm peak}$, the spectrum is rather broad.
As a result, the observable range of the frequency
can be wide as shown in \citep{Lisa2019}.
In observation, another important quantity is the wave amplitude.
The GW amplitude $h_n$ of the $n$-th harmonic is related to
${\cal P}_n$ by
\begin{eqnarray*}
\langle \dot{h}_n^2 \rangle =(2\pi f_n)^2 \langle {h}_n^2\rangle \approx
\frac{4G}{c^3 D^2}{{\cal P}_n(a,e)}
\,,
\end{eqnarray*}
where the bracket denotes the time average over a few orbital period, and $f_n$ is the frequency of the $n$-th harmonic wave.
\begin{widetext}
When we evaluate at the peak frequency $f_n=f_{\rm peak}$,
we find
\begin{eqnarray*}
\sqrt{\langle {h}_n^2\rangle}\approx 5\times 10^{-25}\,
{g^{1/2}(n_{\rm peak},e)\over n_{\rm peak}}
\left({m_1\over M_\odot}\right)\left({m_2\over M_\odot}\right)\left({a_{\rm in}\over{\rm AU}}\right)^{-1}
\left({D\over 10{\rm kpc}}\right)^{-1}
\,.
\end{eqnarray*}
\end{widetext}
Suppose that $m_1=m_2=10 M_\odot$, $a_{\rm in}=0.01$ AU, and $D=10{\rm kpc}$.
When the eccentricity gets large enough via KL oscillations,
we obtain
\begin{eqnarray*}
\sqrt{\langle {h}_n^2\rangle} \sim 1.9\times 10^{-22}\,,
\end{eqnarray*}
for $e_{\rm in}=0.98$.
It may be observable.
To answer the question whether it is really observable or not,
we have to perform a numerical simulation and
evaluate the dimensionless characteristic strain,
which is defined as \cite{Cornish2018},
\begin{equation*}
h_c(f) = 2 f
\left(|\widetilde{h}_+(f)|^2+|\widetilde{h}_{\times}(f)|^2
\right)^{1\over 2}
\end{equation*}
To calculate this quantity, we first find gravitational waveforms $h_+(t)$ and $h_{\times}(t)$ and transform them into the Fourier components
$\widetilde{h}_+(f)$ and $\widetilde{h}_{\times}(f)$.
In our analysis,
we apply Fast Fourier Transform (FFT) to perform the Fourier transformation,
and evaluate the characteristic strain.
The total GW energy
\begin{eqnarray*}
E=\int dt\, {\cal P}
\end{eqnarray*}
consists of many harmonics, which can be described as
\begin{eqnarray*}
E=\int_0^\infty df\, {dE\over df}
\,.
\end{eqnarray*}
The energy spectrum ${dE\over df}$
is obtained as
\begin{equation*}
{dE\over df}(f) = \frac{2 \pi^2 c^3 f^2 D^2}{5G} \left(|\widetilde{h}_+(f)|^2+|\widetilde{h}_{\times}(f)|^2
\right)
\,.
\end{equation*}
\vskip .5cm
\section{Classification and Constraints on Parameters}
\label{sec:classification}
Before we perform our analysis, we classify the possible models with the KL oscillations by the masses of a triple system.
We focus on stable hierarchical triple systems and list the models in three different mass ranges. This general discussion helps in visualizing several realistic situations varying on the scale of mass.
We have three possible cases:\\[.5em]
{\bf Case (A):} $m_{\rm in} \sim m_3$, in which three bodies have comparable masses, or one of the inner binary system can have a smaller mass. \\[.5em]
{\bf Case (B):} $ m_{\rm in} \ll m_3$, which mimics the state of a less massive inner binary around a supermassive black hole (SMBH) in the galactic nuclei.
The inner binary can be a stellar mass binary system such as a binary black hole (BBH), a binary neutron star (BNS) or a neutron star-black hole (NSBH) binary. It can also be a BBH consisting of an intermediate BH (IMBH).\\[.5em]
{\bf Case (C):} $ m_{\rm in} \gg m_3$, which is the reverse of Case (B), hence, instead of SMBH, we may have a planetary object or an asteroid as a tertiary companion. For instance, the inner binary consists of intermediate BHs, while the third companion is stellar mass size.
In Case (C), although the KL oscillation is possible for appropriate parameters, the oscillation occurs between the outer eccentricity and the relative inclination.
The eccentricity of the inner binary may not be affected so much. As a result, the gravitational waves from the inner binary system, which are much more important than those from the outer orbit, has almost no effect due to KL oscillation.
We then do not analyze Case (C) in this paper.
There is another reason we do not discuss Case (C), i.e.
KL timescale is much longer and the cycle may not be captured
in one observation run.
However, our previous work \citep{Haruka2019} discusses the indirect detection of KL oscillations. Therefore, if a third body is a pulsar with the inner binary of intermediate-mass black holes then the presence of KL oscillations can be captured in radio signals from the pulsar.
Now, we show the parameter range in which the KL oscillation appears and is stable.
\subsection{Case (A) : $m_{\rm in}\sim m_3$}
As discussed in \S \ref{sec:Analytic}, we have stability conditions and constraints in order to find the KL oscillations.
In this case, the relation between the typical time scales are given as
\begin{eqnarray*}
{P_{\rm in}\over t_{\rm KL}} \ll {P_{\rm out}\over t_{\rm KL}}\sim {P_{\rm in}\over P_{\rm out}}
\ll 1
\end{eqnarray*}
\begin{figure}[h]
\includegraphics[width=7.5cm]{a_A_meM.png}
\caption{\small
The parameter space $(a_{\rm out},a_{\rm in})$ for Case (A).
We set $m_1=m_2=m_3 = 10 M_\odot$.
The relativistic effect (the precession of periastron) is shown by the blue line, while the stability condition (chaotic boundary) is given by green line. The light-blue shaded region depicts the allowed parameter space for stable KL oscillations.}
\label{fig:parameter_ain_case_a}
\end{figure}
We present some examples in Fig.
\ref{fig:parameter_ain_case_a}, in which
we show the allowed region in the parameter space
$(a_{\rm in}, a_{\rm out})$ for fixing the the masses
of a triple system
as $m_1=m_2=m_3=10 M_\odot$.
In \S \ref{sec:Analytic}, we show the condition with which the orbit does not become chaotic. This gives the maximum value of $a_{\rm in}$, which is given by green line [Eq. (\ref{eq:max_stable})].
The other two constraints come from the relativistic effects. One is the precession of the periastron, whose condition is
shown by the blue line [Eq. (\ref{eq:GR_precession})]. Beyond this line, such relativistic effect will suppress the KL oscillation.
The other is the LT precession effect, whose
critical line is out of the present parameter range.
Taking into account the above conditions, in Fig. \ref{fig:parameter_ain_case_a}, we present the possible range in the parameter space of $(a_{\rm in}, a_{\rm out})$ for the stable KL oscillation, which is shown by the light-blue shaded region.
For the present parameter choice, the constraints are determined by the relativistic precession effect of the periastron and chaotic boundary.
\subsection{Case (B) : $m_{\rm in}\ll m_3$}
In this case, the relation between the typical time scales are given as
\begin{eqnarray*}
\left({P_{\rm in}\over t_{\rm KL}}\right)^{1/2} \sim {P_{\rm out}\over t_{\rm KL}}\sim {P_{\rm in}\over P_{\rm out}}
\,.
\end{eqnarray*}
\begin{figure}[h]
\includegraphics[width=7.5cm]{a_A_mlM.png}
\caption{\small
We plot the allowed region (light-blue shaded)
in the parameter space of $(a_{\rm in}, a_{\rm out})$ for the stable KL oscillation by fixing $m_1=10^3 M_\odot , m_2= 10^3 M_\odot$ and
$m_3 = 10^6 M_\odot$.
The blue and brown lines correspond to the critical curves of relativistic effects (precession of periastron and LT effect),
beyond which the KL oscillation
might be suppressed. The stability of the KL oscillation gives
one condition (chaotic boundary) shown
by green line.}
\label{fig:parameter_ain_case_b}
\end{figure}
In Fig. \ref{fig:parameter_ain_case_b},
we present the possible range in the parameter space of $(a_{\rm in}, a_{\rm out})$ for
fixing the masses of a triple system
as $m_1=m_2=10^3 M_\odot$, and $m_3=10^6 M_\odot$.
The constraints are determined only by the relativistic precession effect of the periastron and
chaotic boundary
just as Fig. \ref{fig:parameter_ain_case_a} in Case (A).
We find that there exists the minimum values of
$a_{\rm in}$ and $a_{\rm out}$, below which
no stable KL oscillation is possible.
\begin{figure}[h]
\includegraphics[width=7.5cm]{m_a_mlM.png}
\caption{\small
We plot the allowed region (light-blue shaded)
in the parameter space of $(a_{\rm in}, m_{\rm in})$ for the stable KL oscillation by fixing $m_3 = 10^6 M_\odot$ and $a_{\rm out} = 10 {\rm AU}$. }
\label{fig:parameter_case_b}
\end{figure}
Since the range of mass parameter in Case (B) is very wide,
in Fig. \ref{fig:parameter_case_b}, we also present the possible range in the parameter space of $(a_{\rm in}, m_{\rm in})$ for the stable KL oscillation, which is shown by the light-blue shaded region. We fix the mass of tertiary component and the semi-major axis of the outer orbit as $m_3=10^6 M_\odot$ and $a_{\rm out}=10 {\rm AU}$.
For the present parameter choice,
the constraints are determined by the relativistic precession effect of the periastron and chaotic boundary condition
just as Case (A).
However, note that the LT precession may become significant
in Case (B) when the third body is much heavier.
See Fig. \ref{fig:parameter_case_b_10^9}, in which we
set $m_3=10^9 M_\odot$.
In this case, a hierarchical triple system will evolve
in the left direction horizontally as
$a_{\rm in}$ decreases because of the emission of
GWs. If $m_{\rm in}\lesssim 10^3 M_\odot$, the evolution
curve will hit on the LT critical line before the
relativistic periastron shift becomes important.
Hence the system will evolve into a chaotic KL oscillation
phase, which may be very interesting to study, but it is
out of our present analysis because the LT precession appears
in 1.5 PN order
\begin{figure}[h]
\includegraphics[width=7.5cm]{m_a_10_9.png}
\caption{\small
The same figure as Fig. \ref{fig:parameter_case_b}
with the different parameters.
We assume $m_3 = 10^9 M_\odot$ and $a_{\rm out} = 200 {\rm AU}$.
The LT critical curve [Eq. (\ref{eq:LT_precession})] cross the
light-blue shaded region for stable KL oscillations.
}
\label{fig:parameter_case_b_10^9}
\end{figure}
\begin{widetext}
\vskip 1cm
\section{Waveform, Spectra and Observability of GWs with KL Oscillations}
\label{sec:Results}
\subsection{Models}
To clarify the properties of the GWs from an inner binary in a hierarchical triple system, we shall perform numerical
analysis for the following models;\\
Case (A): AKL1 ($m_1=m_2=m_3=10 M_\odot$) and AKL2 ($m_1=m_2=m_3=30 M_\odot$)\\
Case (B): BKL1 ($m_1=m_2=10^3 M_\odot, m_3=10^6 M_\odot$)
and BKL2 ($m_1=m_2=10^3 M_\odot, m_3=10^9 M_\odot$).
\begin{table}[h]
\begin{tabular}{c|c|c|c||ccc|ccccc|ccc} \hline
Case& System& Model & Orbital motion &$m_1$&$m_2$&$m_3$& $a_{\rm in}$[{\rm AU}] &$a_{\rm out}$[{\rm AU}] & $e_{\rm in}$ & $e_{\rm out}$ &$I$[deg] &$P_{\rm in}$[days]&$P_{\rm out}$[days]&$t_{\rm KL}$[days]
\\
\hline\hline
&&AKL1& KL oscillation & 10&10&10& 0.01 & 0.1 & 0 & 0 & 90 &$\sim$ 0.08&$\sim$ 2.11&$\sim$ 164\\
&hierarchical &AKL2& KL oscillation & 30&30&30& 0.01 & 0.1 & 0 & 0 & 90 &$\sim $ 0.048&$\sim$ 1.22&$\sim$ 96\\
\cline{3-15}
(A)& ~~triplet &ACC& coplanar circular & 10&10&10& 0.01 & 0.1 & 0 & 0 & 0&$\sim$ 0.08&$\sim$ 2.11&- \\
& &ACE& coplanar eccentric & 10&10&10& 0.01 & 0.1 & 0.9 & 0 & 0&$\sim$ 0.08&$\sim$ 2.11&-
\\
\cline{2-15}
&isolated& ABC &
circular & 10&10&-& 0.01 & - & 0 & - & - &$\sim$ 0.08&-&-\\
& ~~binary&ABE & eccentric & 10&10&-& 0.01 & - & 0.9 & - & - &$\sim$ 0.08&-&-\\
\hline
(B) &hierarchical & BKL1& KL oscillation & $10^3$&$10^3$&$10^6$& 0.15 & 10 & 0 & 0 & 90 &$\sim$ 0.47&$\sim$ 11.54&$\sim$ 280\\
& ~~triplet &BKL2& KL oscillation & $10^3$&$10^3$&$10^9$& 0.25 & 200 & 0 & 0 & 90 &$\sim$ 1.02&$\sim$ 32.67&$\sim$ 1044\\
\hline
\end{tabular}
\caption{\small Masses of three-body hierarchical systems and of two body isolated binary, and the initial orbital elements $a$, $e$, and $i$
are the semi-major axis, eccentricity, and inclination respectively. Subscripts ``in" and ``out" correspond to the inner and outer orbits, respectively. We also
give the typical time scales; $P_{\rm in}$, $P_{\rm out}$ and $t_{\rm KL}$.
The eccentricity for KL binary
increases maximally to $e_{\rm in, max}\sim 0.987$
and $e_{\rm in, max}\sim 0.964$ by the KL oscillation for models AKL1 and AKL2, while $e_{\rm in, max}\sim 0.99$ and $e_{\rm in, max}\sim 0.98$
for Models BKL1 and BKL2, respectively.}
\label{tab:ini}
\end{table}
We then choose the semi-major axis $a_{\rm in}$ and
$a_{\rm out}$ in the stable regions for the KL oscillation,
which are given in Figs. \ref{fig:parameter_ain_case_a}-\ref{fig:parameter_case_b}.
In addition, to confer the results of our analysis, we pick Case (A) to contrast the key features in waveforms from an isolated binary system (Models ABC and ABE) and in
waveforms from a binary in a coplanar hierarchical triple system
(Models ACC and ACE). The parameters are summarized in Table ~\ref{tab:ini}.
The observer's location $(\iota)$ is defined as the angle between the detector and the ``axis" of the reference plane (i.e., the "normal" to the initial outer orbital plane). As we are focused on the inspiral phase of the inner binary, there are special features that can be seen in the waveform.
\subsection{Method of Orbit Evolution}
The first order post-Newtonian equations of motion also known as the Einstein - Infeld - Hoffmann equations of motion \cite{EIH38}, describes the dynamics of a system of point-like masses due to their mutual gravitational interactions, including GR effects. We employ Eq.~(\ref{eq:EIH}) in order to solve the three-body system.
This equation could also be derived from the Lagrangian given by Lorentz and Droste \cite{LD17}.
\begin{eqnarray}
&& \frac{ \mathrm{d} \bm{v}_{k}}{\mathrm{d} t}
=-G\sum_{n\neq k} m_{n}\frac{\bm{x}_{k} - \bm{x}_{n}}{|\bm{x}_{k} - \bm{x}_{n}|^{3}}
\Big[
1-4 \frac{G}{c^2}\sum_{n'\neq k} \frac{m_{n'}}{|\bm{x}_{k} - \bm{x}_{n'}|}
-\frac{G}{c^2}\sum_{n'\neq n} \frac{m_{n'}}{|\bm{x}_{n} - \bm{x}_{n'}|}
\left \{
1-\frac{(\bm{x}_{k} - \bm{x}_{n}) \cdot (\bm{x}_{n} - \bm{x}_{n'})}
{2|\bm{x}_{n} - \bm{x}_{n'}|^{2}}
\right \}
\nonumber \\
&& +\left( \frac{|\bm{v}_{k}|}{c} \right)^{2} + 2\left( \frac{|\bm{v}_{n}|}{c} \right)^{2}
-4 \frac{\bm{v}_{k} \cdot \bm{v}_{n}}{c^2}
-\frac{3}{2}
\left\{
\frac{(\bm{x}_{k} - \bm{x}_{n})}{|\bm{x}_{k} - \bm{x}_{n}|} \cdot \frac{\bm{v}_{n}}{c}
\right\}^{2}
\Big]
-\frac{G}{c^2} \sum_{n \neq k} \frac{m_{n}(\bm{v}_{k}-\bm{v}_{n})}{|\bm{x}_{k} - \bm{x}_{n}|^{3}}
(\bm{x}_{k} - \bm{x}_{n}) \cdot (3\bm{v}_{n}-4\bm{v}_{k})
\nonumber \\ \hspace*{-15cm}
&& -\frac{7}{2} \frac{G^{2}}{c^2} \sum_{n \neq k}\frac{m_{n}}{|\bm{x}_{k} - \bm{x}_{n}|}
\sum_{n'\neq n} \frac{m_{n'} (\bm{x}_{n} - \bm{x}_{n'})} {|\bm{x}_{n} - \bm{x}_{n'}|^{3}},
\label{eq:EIH}
\end{eqnarray}
\end{widetext}
where $m_k$, $\bm{v}_k$, $\bm{x}_k$ ($k=1,2$ and $3$)
are the mass, velocity and position of the $k$-th component of the system, $G$ is the gravitational constant, and $c$ is the speed of light. Eq.~(\ref{eq:EIH}) has been numerically integrated by using 6-th order implicit Runge-Kutta method, whose coefficients are obtained from \citet{Butcher64}.
On integrating, we obtain the numerical data of positions and velocities of the triple system. The initial outer orbital plane is considered to be the reference frame.
In order to set up initial conditions,
we convert initial orbital elements of inner and outer orbits
into the variables $\bm{x}_{k}$ and $\bm{v}_{k}$ in Cartesian coordinates,
with its origin in the center of mass of the whole system \citep{Murray00}.
We integrate the EIH equations (\ref{eq:EIH}) numerically and evaluate the osculating orbital elements at each step from the
numerical data of positions and velocities of the triple system (see e.g. \citet{Murray00}).
Since the inner orbit is not exactly an ellipse,
the obtained osculating elements are oscillating with small amplitudes in the cycle of inner orbit.
Hence, we take an average of the osculating elements for each cycle of inner orbit and then obtain the averaged semi-major axes $\bar{a}_\mathrm{in}$, $\bar{a}_\mathrm{out}$ and eccentricities $\bar{e}_\mathrm{in}$, $\bar{e}_\mathrm{out}$,
which may give the effective values of the orbital elements.
Those elements will evolve secularly in time because of the effect of the tertiary body.
\subsection{Numerical Results}
We first summarize the numerical results for Model AKL1
in next subsections 1, 2 and 3, and
then mention about Model BKL1 in subsection 4.
\subsubsection{\textbf{\underline{Evolution of the eccentricity}}}
\label{eccentricity}
We first show the result of the evolution of eccentricity and relative inclination of the inner binary for Model AKL1, in which the masses
of a triplet are $m_1=m_2=m_3=10 M_\odot$ and initial semi-major axis are $a_{\rm in}=0.01 {\rm AU}$ and $a_{\rm out}=0.1 {\rm AU}$. The typical time scales are given by $P_{\rm in} \sim 0.08$ days, $P_{\rm out} \sim 2.11$ days and $t_{\rm KL} \sim 164$ days.
\begin{figure}[h]
\centering
\includegraphics[width=7.5cm]{ecc_inc.png}
\caption{\small The time evolution of the osculating orbital elements
(the averaged eccentricity of the inner binary $\bar{e}_{in}$,and averaged relative inclination $\bar{I}$)
in Model AKL1 ($m_1=m_2=m_3=10M_\odot$, $a_{\rm in}=0.01 {\rm AU}$, and $a_{\rm out}=0.1 {\rm AU}$) for one KL oscillation cycle. The red and blue lines show the evolution of $\bar{e}_{in}$ and $\bar{I}$, respectively.
When the inclination drops to $I_{\rm min}\approx 55^\circ$,
the eccentricity reaches the maximum value $e_{\rm max}\approx 0.987$. The small oscillations are caused by the outer companion's motion.}
\label{fig:KL}
\end{figure}
The initial eccentricities are $e_{\rm in}=e_{\rm out}=0$, but
the relative inclination is chosen to be $I=90^\circ$.
We show the time evolution of the eccentricity $e_{\rm in}$ and the relative inclination $I$ for one KL oscillation cycle
in Fig ~\ref{fig:KL}.
We have run our code for several values of inclination varying from 40$^\circ$ to 90$^\circ$, for which we obtain stable KL oscillations.
For $I=90^\circ$, we find the maximum value of the eccentricity as $e_{\rm in, max}\approx 0.987$, while $e_{\rm in, max}\approx 0.02$ for $I=40^\circ$.
For lower values for relative inclination, we find lower maximum eccentricities as expected from Eq. (\ref{eq:theta}).
\subsubsection{\textbf{\underline{GW Waveform}}}
\label{GW Waveform}
Next we show the waveform of the GWs for Model AKL1 ($m_1=m_2=m_3=10M_\odot$, $a_{\rm in}=0.01 {\rm AU}$, and $a_{\rm out}=0.1 {\rm AU}$).
In Fig. \ref{fig:kozaiwaveform}, we present the waveform of two
polarization modes for one KL oscillation cycle.
For the small eccentricity until day 98,
the amplitudes are low as $h\sim 10^{-21}$, but
when the inner binary is in high eccentricity regime
around day 100, we find a significant rise in amplitude
as $h\sim 10^{-19}$.
\begin{figure}[h]
\centering
\includegraphics[width=0.70\linewidth]{hp_kozai_o45_full.png}
\\
\includegraphics[width=0.70\linewidth]{hc_kozai_o45_full.png}
\caption{\small The top and bottom figures correspond to `+' and `$\times$' polarization, respectively, of the waveform of GWs
from the hierarchical triplet model AKL1 with KL oscillation. The observer distance is 10 {\rm kpc} at inclination $\iota=45^\circ $ measured from the initial outer orbital plane.}
\label{fig:kozaiwaveform}
\end{figure}
\begin{figure}
\centering
\end{figure}
To see more detail, we enlarge the part of waveforms.
In Fig. \ref{fig:kozaiwaveform_2days_+},
we show the waveform of $+$ -polarization mode for two days.
The top figure depicts the data for the first two days, when
the eccentricity is small, while the bottom one shows the data
from day 106 through day 108, when the eccentricity becomes very large.
We also show the similar enlarged waveforms for $\times$-mode in Fig. \ref{fig:kozaiwaveform_2days_x}
\begin{figure}[h]
\includegraphics[width=0.70\linewidth]{hp_kozai_o45_s1.png}
\\
\includegraphics[width=0.70\linewidth]{hp_kozai_o45_ecc.png}
\caption{\small The zoomed in parts of waveform of $+$ polarization in low and high eccentricity regime. The top figure shows
the wave form in low eccentricity period (the first two days), while the bottom one depicts the wave form in high eccentricity period (two days from day 106 through day 108).}
\label{fig:kozaiwaveform_2days_+}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width=0.70\linewidth]{hc_kozai_o45_s1.png}
\\
\includegraphics[width=0.70\linewidth]{hc_kozai_o45_ecc.png}
\caption{\small The same zoomed figure as Fig. \ref{fig:kozaiwaveform_2days_+} for $\times$-polarization.}
\label{fig:kozaiwaveform_2days_x}
\end{figure}
These features can be easily understood when we analyze more simple systems such as an isolated binary or a coplanar motion of a hierarchical triple system, for which the results are summarized in Appendix for Models ACC-ABE (see Table ~\ref{tab:ini}).
For an isolated circular binary, the waveform is sinusoidal,
but if the eccentricity is large as $e=0.9$, the sharp peak in the waveform appears near the periastron point. The amplitude also
increases compared with the circular case.
For a hierarchical triple system, if the orbit is coplanar,
we find similar features with those in an isolated binary.
When the eccentricity $e_{\rm in}$ is small, we find the sinusoidal wave form, while if $e_{\rm in}$ is large, a sharp peak appears
near the periastron point.
In the case of a triple system, however,
there appears small modulation in the amplitudes because of the effect from a tertiary component.
\begin{figure}[h]
\centering
\includegraphics[width=0.7\linewidth]{hp_kozai_o45_s2.png}
\\
\includegraphics[width=0.7\linewidth]{hp_kozai_s2.png}
\caption{\small The global shape of the waveform with $+$ -polarization
for 15 days (from day 100 to day 115). It depends on the observer position. The top figure shows
the observer at inclination $45^\circ$, while
the bottom one is for the observer at inclination $90^\circ$. }
\label{fig:inc_observer_hp}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width=0.7\linewidth]{hc_kozai_o45_s2.png}
\\
\includegraphics[width=0.7\linewidth]{hc_kozai_s2.png}
\caption{\small The same figure as Fig. \ref{fig:inc_observer_hp}
for $\times$-polarization mode.}
\label{fig:inc_observer_hc}
\end{figure}
In Model AKL1, when the eccentricity is small, the wave form is almost the same as that in a coplanar circular orbit of a triple system.
When the eccentricity becomes large via KL oscillation,
the local feature of the wave form is similar to that of
an eccentric binary or in a coplanar eccentric orbit of a triple system. As shown in Figs. \ref{fig:inc_observer_hp} and \ref{fig:inc_observer_hc}, the amplitude is modulated randomly because of the oscillation of the eccentricity, which is different from the
coplanar eccentric case.
For the observer at a different position, we also find the same
features. The difference appears in the global shape of the wave form. We show the examples for the observers at $\iota=45^\circ$
and $\iota=90^\circ$ measured from the initial outer orbital plane
in Figs. \ref{fig:inc_observer_hp} and \ref{fig:inc_observer_hc}.
\subsubsection{\textbf{\underline{GW Spectra}}}
\label{GW Spectra}
Large eccentricities during KL oscillation and variation of eccentricity with time introduce more harmonics in GWs which make the energy spectra rich as compared to the spectra from an isolated binary (shown in Appendix \ref{GW_isolated}). In Fig. ~\ref{fig:kozaispectra}, we show energy spectra for two stages in one KL cycle. During low eccentricity regime, the spectra is similar to that of circular coplanar hierarchical triple (Model ACC).
\begin{figure}[h]
\includegraphics[width=0.70\linewidth]{spectra_kozai_s1.png}
\\
\includegraphics[width=0.70\linewidth]{spectra_kozai_s2.png}
\caption{\small The GW energy spectra from the inner binary with KL oscillation for low and high eccentricity regime.
The top figure shows the spectrum at day 1-5 ($e_{\rm in} \sim 0$), and
the bottom one gives that at day 106-108 ($e_{\rm in} = 0.983 \sim 0.987$).
We find a broad band spectrum for a high eccentric orbit.}
\label{fig:kozaispectra}
\end{figure}
In the top figure in Fig ~\ref{fig:kozaispectra},
which is obtained by the Fourier transformation for the time interval from day 1 to day 5,
we find two sharp peaks which correspond to
the two periods of the inner and outer orbits.
On the other hand, during high eccentricity regime,
as we see from the bottom one of Fig ~\ref{fig:kozaispectra},
which is obtained for the time interval
from day 106 to day 108,
the spectra is quite similar to the eccentric coplanar
model ACE (see Fig. \ref{fig:spectra_3c}).
Because of high eccentricity, many higher harmonics appear.
As a result, we find a broad band spectrum for a high eccentricity regime.
\subsubsection{\textbf{\underline{Model {\rm BKL}}}}
\label{model BI}
We also perform numerical calculation for Model BKL1, i.e.,
$m_1=m_2=10^3 M_\odot$, $a_{\rm in}= 0.15 {\rm AU}$, and $a_{\rm out}= 10 {\rm AU}$.
We find that the properties of GWs are very much similar to
Model AKL1.
There appears KL oscillation, in which the maximum eccentricity
becomes 0.99. As for the waveform, it is a sinusoidal shape
with modulation caused by a tertiary object
when the eccentricity is small, while
it is deformed during the large eccentricity phase
and shows a sharp peak by passing near
a periastron point.
The GW spectra also change in one KL cycle
from ones with two sharp frequencies, corresponding to two orbital frequencies,
to a broad band spectra.
The difference appears only in the time scales given in Table I.
The observability is also different from Model AKL as we will show in
the next subsection.
\subsection{Observability: Frequency, Strain, and SNR}
To examine the possibility of observation of the present
triple models by future detectors, we have to check whether
the frequency and strain are in the observable range
and evaluate the signal to noise ratio (SNR).
We first plot $f_{\rm peak}$ for Model AKL1. We show in Fig. ~\ref{fig:fpeak} that due to KL oscillations the frequency sweeps up to the detectable range of proposed future space interferometers.
\begin{figure}[h]
\centering
\includegraphics[width = 7.5cm]{fpeak.png}
\caption{\small Time evolution of the peak frequency for
one KL oscillation cycle for Model AKL1
($m_1=m_2=m_3=10M_\odot$ and $a_{\rm in}=0.01 {\rm AU}, a_{\rm out}=0.1 {\rm AU}$).}
\label{fig:fpeak}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=8.5cm]{h_f_10M.png}
\caption{\small The strain versus frequency plot for observer at inclination 90$^\circ$. The blue, green, brown and black curves show DECIGO, BBO, aLISA and LISA design sensitivity.
The orange, violet, and red curves show the strain evolution with $e_{\rm in}$ = 0.944 at day 103, 0.972 at day 104.6, and 0.987
(=$e_{\rm in, max}$)
at day 106.8, respectively.}
\label{fig:hfplot_A}
\end{figure}
However, as the inner binary is in inspiral phase, the GW strain may lie below the sensitivity curve of detectors. We then
plot strain versus frequency curve for Model AKL1 in Fig. ~\ref{fig:hfplot_A}.
The black, brown, blue, green curves correspond to strain sensitivity curves of LISA, aLISA, DECIGO and BBO respectively \citep{Cornish2018,Seto2017,Larson2000}. We plot GW
characteristic strain curve for Model AKL1 at different days. The orange curve is plotted for day 102-103 when
the eccentricity is
$e_\mathrm{in} \sim 0.944$. The red curve is plotted for day 103-105 with $e_\mathrm{in} \sim 0.972$. The violet curve is plotted for day 106-108 with $e_\mathrm{in} \sim 0.987$, which is the maximum value
in one KL cycle.
After day 108, the eccentricity decreases and then
the strain curves decrease to the orange one through the red one.
These curves show the evolution of the strain in one KL cycle.
At high eccentricity regime of KL cycle, the GWs may be observable for at least two days every
164 days (one KL cycle) by DECIGO or BBO.
In Fig. ~\ref{fig:hfplot_A2}, we also show the similar plot
for Model AKL2 with $m_1=m_2=m_3 = 30 M_\odot$
and $a_\mathrm{in} = 0.01 {\rm AU}$. When the observer distance is again kept to be at 10kpc,
the signal is louder due to increase in mass.
\begin{figure}[h]
\centering
\includegraphics[width=8.5cm]{h_f_30M.png}
\caption{\small The strain versus frequency plot for observer at inclination 90$^\circ$. The blue, green, brown and black curves show DECIGO, BBO, aLISA and LISA design sensitivity.
The orange, red and violet curves show the strain evolution with $e_{\rm in}$ = 0.878 at day 57, 0.946 at day 58.6, and 0.964
(=$e_{\rm in, max}$) at day 60, respectively. The observer distance is 10kpc.}
\label{fig:hfplot_A2}
\end{figure}
One may wonder that the GW signal may be louder when $a_{\rm in}$
becomes smaller via an evolution of the inner binary by
the emission of GWs.
For example, one may choose $a_{\rm in}=0.006 {\rm AU}$,
which is still in the stable range
for KL oscillations (see Fig. \ref{fig:parameter_ain_case_a}).
However it does not give a louder signal.
It is because
the maximum eccentricity $e_{\rm in, max}= 0.87$
for $a_{\rm in}=0.006 {\rm AU}$ is not
high enough to give an observable strain
compared with Model AKL1 or AKL2.
We show the change of the maximum eccentricity $e_{\rm in, max}$
in terms of $a_{\rm in}$ in Fig. \ref{fig:ecc_ain}.
\begin{figure}[h]
\centering
\includegraphics[width=8.5cm]{emax_ain.png}
\caption{\small The maximum eccentricity in terms of $a_{\rm in}$ for model AKL1.
We find $e_{\rm in, max}\approx 1$ for $a_{\rm in} \, \mbox{\raisebox{-1.ex 0.01 {\rm AU}$, while $e_{\rm in, max}$
decreases rapidly beyond $a_{\rm in}\sim 0.01 {\rm AU}$.}
\label{fig:ecc_ain}
\end{figure}
We find that the maximum eccentricity is very close to unity for
$a_{\rm in}\, \mbox{\raisebox{-1.ex 0.01 {\rm AU}$, but
it decreases rapidly past $a_{\rm in}\sim 0.01 {\rm AU}$.
This is because GR effect suppresses the KL mechanism\citep{tey13,naoz13b,liu15}.
As a result, the case with $a_{\rm in}= 0.01 {\rm AU}$,
which shows the shortest semi-major axis as well as the largest eccentricity,
gives the loudest characteristic strain.
We also show the characteristic strain for Model BKL1
in Fig. \ref{fig:hfplot_B}.
In this case, the strain is not large enough
to give the observable value by DECIGO or BBO.
We may need aLISA.
\begin{figure}[h]
\centering
\includegraphics[width=8.5cm]{h_f_2000M.png}
\caption{\small The same figure for Model BKL1 as Figs. \ref{fig:hfplot_A} and \ref{fig:hfplot_A2}. The blue, green, brown and black curves show DECIGO, BBO, aLISA and LISA design sensitivity. The orange, red and violet curves show the strain evolution with $e_{\rm in}$ =0.967 at day 126.6, 0.977 at day 128 and 0.99 (=$e_{\rm in, max}$) at day 130.4, respectively. The observer distance is 16 Mpc.}
\label{fig:hfplot_B}
\end{figure}
The plot of the frequency vs the characteristic strain gives helpful assessment of detectability provided high signal to noise ratio (SNR) of the waveform exists.
Our hierarchical triple system may be one of the many source populations captured by future GW space detectors with following merger phase captured by ground detectors.
The area between the source and detector curves is related to SNR by the following relation \citep{Berry2014},
\begin{equation*}
({\rm SNR})^2 = \int_{f_{\rm min}}^{f_{\rm max}} d (\log f) \bigg(\frac{h_c(f)}{h_d(f)}\bigg)^2
\,,
\end{equation*}
where $h_c(f)$ is the strain amplitude of the source and
$h_d(f)$ denotes noise amplitude of detector.
For the models shown in our analysis,
the SNR for DECIGO is about 8 at $e_{\rm in, max} \sim 0.987$ for Model AKL1.
Similarly the SNR is about 16 for Model AKL2 at $e_{\rm in, max} \sim 0.964$.
Since the design sensitivity of BBO is better than DECIGO,
the SNR for BBO would be larger.
For Model BKL1, the signal is not loud enough for BBO, but that
for aLISA SNR is $\sim 17$, which could be observable.
\section{Concluding Remarks}
\label{sec:Conc}
We have shown features of gravitational waves from an inner binary in a hierarchical triple system
with KL oscillations.
The waveform changes its shape in time because of oscillation of
the eccentricity.
When the eccentricity is small,
the waveform is a sinusoidal shape modulated by a tertiary
companion,
while it becomes one with sharp peaks
near periastron point when the eccentricity gets large.
We have also examined the time
variation of the characteristic strain curve, which
may appear in sensitivity range of detectors when the eccentricity becomes large
via the KL oscillations.
This can be the first direct observation of KL oscillation which may further
shed light on binary formation channels and surrounding environment of the binary system.
The interesting remaining questions are
what happens when the system evolves beyond the KL stable region.
Because of the GW emission, the semi-major axis of the inner binary
will decrease. Since other parameters such as $m_1, m_2, m_3$ and $a_{\rm out}$
are conserved, the system will evolve vertically in Figs. \ref{fig:parameter_ain_case_a}
and \ref{fig:parameter_ain_case_b} or horizontally
in Figs. \ref{fig:parameter_case_b} and \ref{fig:parameter_case_b_10^9}.
Hence, we reach at the boundary of relativistic periastron shift.
Beyond this boundary, KL oscillation may be suppressed, and then
the effect of the tertiary component may become weak.
As a result, we expect that GWs may not be observed before
the coalescence of the inner binary. The question is whether the orbit of the inner binary
is circularized via the GW emission at the coalescence or not.
If the eccentricity remains then, we find the difference from a simple binary system.
We may need numerical relativity or some other approach such as ``effective one body system"
to clarify it.
Another interesting issue is what happens when the LT effect becomes important.
It can be the case if we consider SMBH.
In Fig. \ref{fig:parameter_case_b_10^9}, if the total mass of the inner binary
is smaller than $10^3 M_\odot$, when the semi-major axis decreases,
the system will evolve into the region where the LT precession becomes important.
In this region, we may find a chaotic KL oscillation\citep{Wang19}, which may
provide us strong emission of GWs. In order to study such a process, we have to include
higher PN effects, i.e., at least 1.5 PN terms.
The work on these two issues is in progress.
As for a triple system, we are also interested in more
general situations, in which we expect chaotic behavior of
a system. The GWs from such a chaotic system should also be
studied. Since it may be difficult to make templates,
we should find some other ways to clarify typical features of such a system
in GW observations\citep{Suzuki:1999si,Kiuchi:2004bv,Kiuchi:2007ue}.
One of the most important issues about the GWs from a hierarchical triple system is the event rate.
As for a formation of three body system, there are many works including numerical
$N$-body simulation \citep{Glutekin2006,Samsing2014,Fabio2016}.
The event rate depends on many uncertain factors like star formation rate,
binary fraction and distribution of initial parameters.
Several papers discuss the increase in the merger rate of compact binaries around SMBHs due to KL oscillations \citep{Antonini2012,hoang18,trani19}. The binaries in these triple systems undergo several KL cycles finally leading to the merger.
In Ref. \citep{Antonini2012}, the authors estimate the rate to be
0.56 ${\rm Gpc}^{-3} {\rm yr}^{-1}$. Similar analysis is done by Refs. \citep{hoang18,trani19}, estimating the rate ~ 1-3 ${\rm Gpc}^{-3}{\rm yr}^{-1}$.
The difference is caused by different distance regimes from the SMBH and
different initial conditions. For isolated triples,
the merger rate to be 6 ${\rm Gpc}^{-3} {\rm yr}^{-1}$ in the absence of natal kicks\citep{Tremaine17}. Globular clusters may harbor IMBHs at their centers.
Ref. \citep{Omer19} studies the triple systems composed of the Stellar-mass binary with IMBH as a third body. The merger event rate induced by KL oscillations is estimated to be
0.06-0.46 ${\rm Gpc}^{-3} {\rm yr}^{-1}$.
We are interested in the models where merger timescale is much longer than Kozai timescale. Hence, if there are many KL oscillations before the merger, the signal about KL oscillations may lie in the observable range. In Ref. \citep{Lisa2019}, using double-averaged equations, they analyze the fraction of KL binaries in two channels (isolated triples and galactic center) lying in the LISA frequency range. They show that of all merging binaries considered in their simulation, $\sim 39\%$ of the isolated-triple channel, and $\sim 22\%$ of the galactic-center channel, display significant KL oscillations in LISA frequency band.
However, the present models of a hierarchical triple
system with KL oscillations,
which gives observable GWs in an inspiral phase, may be
difficult to be formed by evolution of a triple star system
because both semi-major axes
are very short.
One possible way would be a capture process.
The third body is bounded due to a close encounter with a BH binary in globular clusters or near galactic centers.
The question is how likely such an interaction is.
In \citep{Samsing2014,zevin19}, binary-single and binary-binary BH interactions are numerically simulated and several possible end states are discussed.
In Ref. \citep{zevin19}, they have shown that stable triple formation end state has higher cross section compared to some other possible end states.
We are planning to study in detail the formation process and probability to evaluate the event rate of compact hierarchical triples such that KL oscillations are not quenched by relativistic effects.\\
\section*{Acknowledgements}
We would like to acknowledge Gungwon Kang, Seiji Kawamura, Kaye Li, Naoki Seto, Takahiro Tanaka, and Alessandro A. Trani for useful discussions.
P.G. is supported by Japanese Government (MEXT)
Scholarship.
This work was supported in part by JSPS KAKENHI Grant Numbers
JP17H06359 and JP19K03857, and by Waseda University Grant for Special
Research Projects (Project number: 2019C-254 and 2019C-640).
\section{Dynamics of Triple System}
\label{sec:Dynamics}
In this section, we sequentially explain a hierarchical triple system, Kozai-Lidov mechanism, and their stability and constraints.
\subsection{Hierarchical Triple System and Kozai-Lidov Mechanism}
\label{sec:KL}
We can have several stable configurations and shapes when it comes to a three-body dynamical system. Broadly addressing, it can be concentric, coplanar orbits or inclined orbits in a hierarchical arrangement.
In this paper, we treat the so-called hierarchical triple system, whose schematic picture is given in Fig.~\ref{fig:3body}.
\begin{figure}[h]
\includegraphics[width=5cm]{threesystem_AE.png}
\caption{The hierarchical triple system is constructed from inner and outer binaries.
The inner binary consists of objects whose masses are $m_1$ and $m_2$,
and the outer one is the pair of the inner binary and
the third body with mass $m_3$.}
\label{fig:3body}
\end{figure}
As depicted in the figure, we can decouple a system into the two-body inner binary orbit and the outer orbit with tertiary companion in motion around center of mass of inner binary, if
the distance between the first and second bodies is much shorter than the distance to the third body. This hierarchy is supported by assuming that the gravitational effect of the third body is much smaller than the gravitational interaction between the first and second bodies.
In a two-body problem in Newtonian dynamics,
a bound system is given by an elliptic orbit, which
is described by six orbital elements; the semi-major axis $a$, the eccentricity $e$, the inclination $i$, the argument of periastron $\omega$, the longitude of ascending node $\Omega$, and the mean anomaly $M$.
Although these elements are constant (except mean anomaly) in an isolated two-body system, in the hierarchical three-body system, the perturbations from the tertiary companion affect the binary motion and modify the trajectory from that of the isolated one. Such a trajectory is not closed in general, but we can introduce an {\it osculating orbit} at each time, whose trajectory is approximated by the elliptical orbit with the above six orbital elements determined by the instantaneous position and velocity \citep{Murray00}.
As for the outer orbit, we pursue the center of mass of the inner binary rotating around the tertiary companion (see Fig.~\ref{fig:3body}). It can also be described as another osculating orbit. Hence, we introduce two osculating orbits, which are called as inner and outer orbits:
The masses of the inner binary are $m_1$ and $m_2$, while the tertiary companion has the mass $m_3$. We use subscriptions `in' and `out' to show the elements of inner and outer orbits, respectively.
In the hierarchical three-body system with large relative inclination,
a secular change of orbital elements may occur.
Under some conditions, there appears an oscillation between
the relative inclination and the eccentricity of the orbit, which is called
Kozai-Lidov (KL) oscillation\citep{Kozai62,Lidov62}.
This mechanism was independently discovered by Kozai and Lidov in 1962.
The relative inclination $I$ is defined as the argument between the inner and outer orbital planes.
It is given by two orbital elements as
\begin{equation*}
\cos I = \cos i_{\mathrm{in}} \cos i_{\mathrm{out}}
+ \sin i_{\mathrm{in}} \sin i_{\mathrm{out}}
\cos \left( \Omega_{\mathrm{in}} - \Omega_{\mathrm{out}} \right).
\label{eq:relative_i}
\end{equation*}
If we assume $m_1\ll m_2, m_3$, we find that the oscillation occurs in secular time-scale under conservation of energy and angular momentum.
In Newtonian quadrupole approximation method \citep{Shevchenko17}, it results in the secular exchange of $e_\mathrm{in}$ and $I$ with the conserved value of $\Theta$,
which is defined by
\begin{equation}
\Theta = (1-e_\mathrm{in}^2) \cos^2 I \,.
\label{eq:theta}
\end{equation}
This approximation also gives the criterion of KL-oscillation for circular orbit as
\begin{equation*}
0 \le |\cos I |
\ \le \sqrt{\frac{3}{5}} ,
\label{eq:kozaiC}
\end{equation*}
which is equivalent to $39.2315^\circ \lesssim I
\lesssim 140.7685^\circ $.
This KL type oscillation will occur even if $m_1$ is not much smaller than $m_2, m_3$
\cite{Naoz12}.
When we include a relativistic effect, we also find the similar oscillation\cite{Naoz16,Wen2003}.
The effect of general relativity (GR) changes these KL- criterion
as \citep{Migaszewski11}.
\begin{equation*}
0
\le |\cos I| \
\le \sqrt{{3\over 5}\left(1-\frac{4}{9}
\left\langle \mathcal{H}_\mathrm{GR}
\right\rangle\right) } ,
\label{eq:KozaiC_PN}
\end{equation*}
where the GR correction term $\left\langle \mathcal{H}_{\rm GR} \right\rangle $
is derived from the double-averaged post-Newtonian Hamiltonian of two-body relative motion \citep{Richardson88} as
\begin{equation*}
\left\langle \mathcal{H}_{\rm GR} \right\rangle
= \frac{3Gm_{\rm in}^2a^3_{\rm out} \sqrt{1- e^2_{\rm out}}}{c^2 m_3a^2_ {\rm in} \sqrt{1- e^2_{\rm in}}} \,,
\label{eq:H_GR}
\end{equation*}
where $m_{\rm in}=m_1+m_2$ is the total mass of the inner binary.
The KL-oscillation time-scale is also modified when the GR effect is taken into account\cite{Naoz16}.
Note that it has been pointed out \cite{Will14a,Will14b} that
it is imperative to take into account the
`cross terms' between the Newtonian perturbation and the Post-Newtonian precession.
In our numerical evolution, we integrate the equations of motion (up to 1PN) directly. The effects of the cross terms are automatically included.
\subsection{Stability and Constraints}
\label{sec:Analytic}
In this section, we discuss stability criteria for a hierarchical triple system.
We also give constraints for KL oscillation to occur, which are obtained from the conditions such that KL-oscillations are not suppressed by relativistic effects, i.e., the relativistic precession of the periastron of the inner binary and
the Lense-Thirring (LT) precession effect due to the rapid rotation of heavy third body.
\subsubsection{\textbf{\underline{Typical time scales}}}
Before discussing stability and constraints,
we first present typical time scales of a hierarchical triple system.
The orbital periods of the inner and outer binaries
are given by
\begin{eqnarray*}
P_{\rm in} &=& 2\pi \sqrt{a_{\rm in}^3\over Gm_{\rm in}},
\\
P_{\rm out} &=& 2\pi \sqrt{a_{\rm out}^3\over Gm_{\rm out}},
\end{eqnarray*}
where $m_{\rm out}=m_1+m_2+m_3$ is the total mass of a triple system.
The KL oscillation time scale is given by \citep{Antognini2015}
\begin{eqnarray*}
t_\mathrm{KL} \sim P_{\mathrm{in}}\frac{m_{\mathrm{in}}}{m_3} \left( \frac{a_{\mathrm{out}}}{a_\mathrm{in}} \right)^3
\left( 1-{e_\mathrm{out}}^2 \right)^\frac{3}{2}\,.
\end{eqnarray*}\\
\subsubsection{\textbf{\underline{Stability of Hierarchical Arrangement}}}
For KL oscillations to occur, the first aim is to maintain the stability of a three-body system. A necessary (not sufficient) criteria is given by \cite{Blaes02}, which predicts a minimum separation between inner and outer orbits for stability,
beyond which a system may cause a chaotic instability.
Hence, we have one stability condition,
\begin{equation*}
\frac{ a_\mathrm{out} }{ a_\mathrm{in} } >\frac{2.8}{ 1-e_\mathrm{out} }
\left[ \frac{m_{\rm out}}{m_\mathrm{in}}
\frac{ 1+e_\mathrm{out} }{ (1-e_\mathrm{out})^\frac{1}{2} } \right]^\frac{2}{5}
\,.
\end{equation*}
For simplicity, throughout this paper, we set outer orbit not to be so eccentric such that $e_{\rm out}\ll 1$. Thus the scaled expression is written as,
\begin{equation}
\frac{ a_\mathrm{out} }{ {\rm AU} } > 2.8
\left( \frac{m_{\rm out}}{M_\odot}\right)^{2\over 5}
\left(\frac{m_\mathrm{in}}{M_\odot}\right)^{-{2\over 5}} \, \frac{ a_\mathrm{in}}{ {\rm AU}}
\,
\label{eq:max_stable}
\end{equation}
This expression gives a lower bound to the separation between two orbits.
\subsubsection{\textbf{\underline{Constraints from the relativistic effects}}}
\noindent
{\bf 3-1. Relativistic Precession of Periastron}\\[.2em]
KL oscillations are suppressed or maximum eccentricity is restricted by the relativistic precession of the periastron of the inner binary. An analytical expression can be derived using double averaged approximation up to quadrupole order to estimate parameter space such that KL oscillations are not destroyed by precession\cite{Blaes02}.
The GR precession timescale of the periastron of the inner binary
can be estimated as,
\begin{eqnarray*}
t_{\rm GR,in} \sim
\bigg({\frac{a_{\rm in}}{r_{\rm g,in}}}\bigg)^{5\over 2} P_{\rm in} \left( 1-{e_\mathrm{in}}^2 \right),
\end{eqnarray*}
where $r_{g,{\rm in}}\equiv {Gm_{\rm in}/c^2}$ is the gravitational radius of the inner binary.
Comparing this with the KL time scale $t_{\rm KL}$,
we find the constraint
\begin{equation*}
\left({a_{\rm out}\over a_{\rm in}}\right)^3<
{3\over 4}
{a_{\rm in}\over r_{g,{\rm in}}}
{m_3\over m_{\rm in}}
\left({1-e_{\rm in}^2\over 1-e_{\rm out}^2}\right)^{3\over 2}
\,.
\end{equation*}
For clarity, the scaled expression takes the form:
\begin{eqnarray}
&&\left({a_{\rm out}\over
{\rm AU}}\right)^3<
10^8 \left({1-e_{\rm in}^2}\right)^{3\over 2}
\left({m_{\rm in}\over {M_\odot}} \right)^{-2}
\left({a_{\rm in}\over {\rm AU}}\right)^4
\left(m_3 \over {M_\odot}\right)
\,,
\nonumber \\
&&
~~
\label{eq:GR_precession}
\end{eqnarray}
assuming $e_{\rm out}\ll 1$.\\
\noindent
{\bf 3-2 Lense-Thirring Precession Effect}\\[.2em]
Recent studies have shown changes in KL oscillations caused due to GR effects involving a SMBH \cite{Wang19,fang19,fang19a}.
If the third body is a rapidly rotating SMBH, then LT effect might become important changing the evolution of eccentricity excitation from the usual KL oscillation. It appears in the 1.5 PN order.
For a rotating black hole, which is the third body in our setting,
the spin angular momentum is given as $S_3 = \chi_3 Gm_3^2/c$ where $\chi_3$ is the spin parameter, and outer orbital angular momentum is given as
\begin{eqnarray*}
L_{\rm out} = \mu_{\rm out} \left[G(1-e_{\rm out}^2)
a_{\rm out}M\right]^{1\over 2}
\,,
\end{eqnarray*}
where the reduced mass of the outer orbit is defined by
\begin{eqnarray*}
\mu_{\rm out} &=& {m_{\rm in} m_3\over m_{\rm out}}={(m_1+m_2) m_3\over m_1+m_2+m_3}
\,.
\end{eqnarray*}
The time scale of the orbit-averaged precession
of $L_{\rm out}$ around $S_3$ is given as
\begin{equation*}
t_{L_{\rm out}S_{3}} = \frac{GS_3(4+3m_{\rm in}/m_3)}{2c^2
(1-e^2_{\rm out})^{3\over 2} a_{\rm out}^3}
\end{equation*}
The KL oscillations can be affected when $t_{L_{\rm out}S_{3}}$ becomes comparable to the KL oscillation time scale $t_{\rm KL}$.
Hence, in order to observe the usual KL oscillation, we have the constraint such that
$t_{L_{\rm out}S_{3}} > t_{\rm KL}$, for which
the scaled expression is given as:
\begin{equation}
\bigg({a_{\rm in}\over {\rm AU}}\bigg)^{3\over 2} >
10^{-12}
\bigg({m_3\over M_\odot} \bigg) \bigg({m_{\rm in}\over M_\odot} \bigg)^{1\over 2}
\,,
\label{eq:LT_precession}
\end{equation}
assuming $\chi_3 \sim 1$.
If the ratio is comparable or greater than 1, then LT precession dominates resulting in chaotic evolution of eccentricity as shown in \cite{Wang19} and hence more dissipation of gravitational waves,
which may be interesting for further study.
However, in this paper our discussion is limited to 1PN effect and we focus more on the conventional KL oscillations in the inspiral phase.
\section{Gravitational Waves}
\label{sec:GRWs}
Under this section, we set the framework by recalling the famous quadrupole formula and definition of `near coordinate zone' (NCZ). We further discuss GWs from highly eccentric orbits,
which are expected in KL oscillation.
\subsection{Quadrupole Formula}
\label{subsec:Quad}
In GR, the leading order contribution of the observed gravitational waves
is given by the quadrupole formula. It allows the evaluation of gravitational wave energy and waveforms emitted by a dynamical system.
In a wave zone, the gravitational waves denoted by a small deviation of the metric $h_{ij}$ are described by
\begin{equation*}
\label{eq:QF}
h_{ij}^{\rm TT}(t,\textbf{x}) = \frac{2G}{Dc^4} \ddot{Q}_{ij}^{\rm TT}
\,,
\end{equation*}
where TT refers to transverse-traceless gauge and a dot
denotes the time derivative.
$D$ denotes the distance of the source from the observer.
The reduced quadrupole moment $Q_{ij}$ is defined by
\begin{equation*}
Q_{ij} = M_{ij} - \frac{1}{3} \delta_{ij} M^k_{\,k}
\,,
\end{equation*}
where $i,j, \cdots$ run from 1 to 3 ($ x^1 = x , x^2=y,x^3=z $),
and
the mass quadrupole moment $M_{ij}$ for $N$ masses is expressed as
\begin{eqnarray*}
M_{ij} = \sum_{A=1}^{N} m_A x_{A}^{i} x_{A}^{j}\,.
\end{eqnarray*}
$m_A$ denotes the $A$-th mass at the location $x_A^i$ with origin at the center of mass of a system.
The power of GW emission is given by
\begin{eqnarray*}
{\cal P}={G\over 5c^5}\langle {\stackrel{\cdots}{Q}}_{ij}
{\stackrel{\cdots}{Q}}\raisebox{.5em}{\scriptsize $ij$}\rangle
\,,
\end{eqnarray*}
where the bracket denotes the Brill-Hartle averaging, which
is taken over several wave length \citep{Peters1963}.
The GW amplitude from a circular binary system consisting of masses $m_1$ and $m_2$ with orbital radius $a$ is approximately given by
\begin{eqnarray*}
h \sim {{4 G^2 m_1m_2}\over {c^4 a D}}
\,,
\end{eqnarray*}
which is described by the scaled expression as
\begin{equation*}
\label{eq:QF}
h \sim 1.7\times 10^{-25} \bigg(\frac{m_1}{M_\odot}\bigg) \bigg(\frac{m_2}{M_\odot}\bigg) \bigg(\frac{{\rm AU}}{a_{\rm in}}\bigg) \bigg(\frac{10{\rm kpc}}{D}\bigg)
\end{equation*}
Another important quantity
is energy spectra of the GWs.
For an isolated circular orbit,
the energy radiated in gravitational waves is concentrated in the second harmonic such that
\begin{equation*}
f_{\rm GW} = 2 f_{\rm in}\,,
\end{equation*}
where $f_{\rm in}$ is the orbital frequency of the inner binary given by
\begin{equation*}
f_{\rm in}={1\over P_{\rm in}}={1\over 2\pi}
\sqrt{Gm_{\rm in}\over a_{\rm in}^3}
\,.
\end{equation*}
We find it in the energy spectrum, whose example is given in Appendix \ref{GW_isolated} and Fig. \ref{fig:spectra_binary}.
Large eccentricities induced in the inner binary during KL oscillation and variation of eccentricity with time produce more harmonics in gravitational waves, which make the energy spectra rich as compared to the spectra from an isolated circular binary as well as an enhancement of the wave amplitude
as shown in the next subsection.
In the derivation of the quadrupole formula,
a system must be contained in its NCZ , i.e., in order to use the quadrupole formula, a system should be inside a region (centered on the origin of the coordinates) of radius comparable to (or smaller than) the GW wavelength $\lambda$ (see some good example of what not to do given in \citep{Bonetti17}).
Therefore, we must keep a check that a hierarchical triple system is in its NCZ
when the formalism is applied to the center of mass of the triple system for the model in consideration. There are two NCZs for a hierarchical triple system, whose sizes correspond to two wave lengths: one is that of an inner binary $\lambda_{\rm in} = a_{\rm in}c/v_{\rm in} $ and the other is that of an outer binary with $\lambda_{\rm out} = a_{\rm out}c/v_{\rm out} $, where $v_{\rm in}$ and $v_{\rm out}$ are typical velocities of the inner binary and
the outer binary, respectively.
Since we focus on GWs from the inner binary,
the NCZ condition is given by $a_{\rm out}<\lambda_{\rm in}$.
An inaccurate application may lead to apparent and
unphysical behavior as shown in Fig. 18 of \citep{GP2011}.
\subsection{Gravitational waves from highly eccentric orbit}
In this subsection, we discuss
the GWs from a highly eccentric binary system.
When the KL oscillation occurs, the inner binary system is
supposed to be in an inspiral phase. Hence, it is usually expected to be difficult to observe the GWs unless the inner binary is just before a coalescence.
However, the eccentricity in the KL oscillation can become very large,
which enhances the amplitudes of GWs and those frequencies.
As a result, we may have a chance to observe the GWs even in an inspiral phase.
Gravitational Waveform from a binary system depends strongly on the eccentricity of the orbit.
In an eccentric orbit, the emission of gravitational waves is higher and distributed over many harmonics.
The GWs from an eccentric binary was studied by Peter and Mathews\citep{Peters1963}.
The total radiation power ${\cal P}$ averaged over one period is given by
\begin{eqnarray*}
{\cal P}={32G^4m_1^2m_2^2m_{\rm in}\over c^5 a_{\rm in}^5(1-e_{\rm in}^2)^{7/2}}\left(1+{73\over 24}e_{\rm in}^2+{37\over 96}e_{\rm in}^4\right)
\end{eqnarray*}
where $a_{\rm in}$ and $e_{\rm in}$ are the semi-major axis and the eccentricity of
the inner binary, respectively.
It shows that the radiation power increases
rapidly when the eccentricity becomes large.
The total radiation power ${\cal P}$ is the sum of the power radiated in the $n$-th harmonics
${\cal P}_n$ as
\begin{eqnarray*}
{\cal P}=\sum_{n=1}^\infty {\cal P}_n
\,.
\end{eqnarray*}
The radiation power ${\cal P}_n$
in the $n$-th harmonic is given by
\begin{eqnarray*}
{\cal P}_n={32G^4m_1^2m_2^2m_{\rm in}\over c^5 a_{\rm in}^5}g(n,e_{\rm in})
\end{eqnarray*}
where
\begin{eqnarray*}
g(n,e)&=&{n^4\over 32}\Big{[}
\Big{(}J_{n-2}(ne)-2eJ_{n-1}(ne)
\nonumber \\
&+&{2\over n}J_n(ne)+2eJ_{n+1}(ne)-J_{n+2}(ne)\Big{)}^2
\nonumber \\
&+&(1-e^2)\Big{(}J_{n-2}(ne)-2J_n(ne)+J_{n+2}(ne)\Big{)}^2
\nonumber \\
&+&{4\over 3n^2}\left(J_n(ne)\right)^2\Big{]}
\,,
\end{eqnarray*}
where $J_n(x)$ is the Bessel function of the first kind.
The spectrum ${\cal P}_n$ is peaked at the GW frequency $f_{\rm peak}$,
\begin{equation*}
f_{\rm peak} = n_{\rm peak} f_{\rm in},
\end{equation*}
where the magnification factor $n_{\rm peak}$ is well approximated by
\begin{equation*}
n_{\rm peak} = \frac{2 (1+ e)^{1.1954}}{(1- e^2)^{1.5}}
\,,
\end{equation*}
for the range of $10^{-6}<1-e^2<1$\citep{Wen2003}.
For $e \ll 1$, $n_{\rm peak}$ is just 2, while considering high eccentricity, say $e \sim 0.98$ , we find $n_{\rm peak} \sim 574$.
As the eccentricity increases, change in $f_{\rm peak}$ becomes significant and may lie in the detectable frequency band $10^{-4}$ Hz $\sim 10$ kHz.
It is worthwhile evaluating the peak frequency as
\begin{equation*}
f_{\rm peak} \sim 3\times 10^{-8} n_{\rm peak} \bigg(\frac{m_{\rm in}}{M_\odot}\bigg)^{1\over2}\bigg(\frac{a_{\rm in}}{\rm AU}\bigg)^{-{3\over2}}~{\rm Hz}
\end{equation*}
Note that
although the $n$-th harmonics power ${\cal P}_n$
has a peak at
$f=f_{\rm peak}$, the spectrum is rather broad.
As a result, the observable range of the frequency
can be wide as shown in \citep{Lisa2019}.
In observation, another important quantity is the wave amplitude.
The GW amplitude $h_n$ of the $n$-th harmonic is related to
${\cal P}_n$ by
\begin{eqnarray*}
\langle \dot{h}_n^2 \rangle =(2\pi f_n)^2 \langle {h}_n^2\rangle \approx
\frac{4G}{c^3 D^2}{{\cal P}_n(a,e)}
\,,
\end{eqnarray*}
where the bracket denotes the time average over a few orbital period, and $f_n$ is the frequency of the $n$-th harmonic wave.
\begin{widetext}
When we evaluate at the peak frequency $f_n=f_{\rm peak}$,
we find
\begin{eqnarray*}
\sqrt{\langle {h}_n^2\rangle}\approx 5\times 10^{-25}\,
{g^{1/2}(n_{\rm peak},e)\over n_{\rm peak}}
\left({m_1\over M_\odot}\right)\left({m_2\over M_\odot}\right)\left({a_{\rm in}\over{\rm AU}}\right)^{-1}
\left({D\over 10{\rm kpc}}\right)^{-1}
\,.
\end{eqnarray*}
\end{widetext}
Suppose that $m_1=m_2=10 M_\odot$, $a_{\rm in}=0.01$ AU, and $D=10{\rm kpc}$.
When the eccentricity gets large enough via KL oscillations,
we obtain
\begin{eqnarray*}
\sqrt{\langle {h}_n^2\rangle} \sim 1.9\times 10^{-22}\,,
\end{eqnarray*}
for $e_{\rm in}=0.98$.
It may be observable.
To answer the question whether it is really observable or not,
we have to perform a numerical simulation and
evaluate the dimensionless characteristic strain,
which is defined as \cite{Cornish2018},
\begin{equation*}
h_c(f) = 2 f
\left(|\widetilde{h}_+(f)|^2+|\widetilde{h}_{\times}(f)|^2
\right)^{1\over 2}
\end{equation*}
To calculate this quantity, we first find gravitational waveforms $h_+(t)$ and $h_{\times}(t)$ and transform them into the Fourier components
$\widetilde{h}_+(f)$ and $\widetilde{h}_{\times}(f)$.
In our analysis,
we apply Fast Fourier Transform (FFT) to perform the Fourier transformation,
and evaluate the characteristic strain.
The total GW energy
\begin{eqnarray*}
E=\int dt\, {\cal P}
\end{eqnarray*}
consists of many harmonics, which can be described as
\begin{eqnarray*}
E=\int_0^\infty df\, {dE\over df}
\,.
\end{eqnarray*}
The energy spectrum ${dE\over df}$
is obtained as
\begin{equation*}
{dE\over df}(f) = \frac{2 \pi^2 c^3 f^2 D^2}{5G} \left(|\widetilde{h}_+(f)|^2+|\widetilde{h}_{\times}(f)|^2
\right)
\,.
\end{equation*}
\vskip .5cm
\section{Classification and Constraints on Parameters}
\label{sec:classification}
Before we perform our analysis, we classify the possible models with the KL oscillations by the masses of a triple system.
We focus on stable hierarchical triple systems and list the models in three different mass ranges. This general discussion helps in visualizing several realistic situations varying on the scale of mass.
We have three possible cases:\\[.5em]
{\bf Case (A):} $m_{\rm in} \sim m_3$, in which three bodies have comparable masses, or one of the inner binary system can have a smaller mass. \\[.5em]
{\bf Case (B):} $ m_{\rm in} \ll m_3$, which mimics the state of a less massive inner binary around a supermassive black hole (SMBH) in the galactic nuclei.
The inner binary can be a stellar mass binary system such as a binary black hole (BBH), a binary neutron star (BNS) or a neutron star-black hole (NSBH) binary. It can also be a BBH consisting of an intermediate BH (IMBH).\\[.5em]
{\bf Case (C):} $ m_{\rm in} \gg m_3$, which is the reverse of Case (B), hence, instead of SMBH, we may have a planetary object or an asteroid as a tertiary companion. For instance, the inner binary consists of intermediate BHs, while the third companion is stellar mass size.
In Case (C), although the KL oscillation is possible for appropriate parameters, the oscillation occurs between the outer eccentricity and the relative inclination.
The eccentricity of the inner binary may not be affected so much. As a result, the gravitational waves from the inner binary system, which are much more important than those from the outer orbit, has almost no effect due to KL oscillation.
We then do not analyze Case (C) in this paper.
There is another reason we do not discuss Case (C), i.e.
KL timescale is much longer and the cycle may not be captured
in one observation run.
However, our previous work \citep{Haruka2019} discusses the indirect detection of KL oscillations. Therefore, if a third body is a pulsar with the inner binary of intermediate-mass black holes then the presence of KL oscillations can be captured in radio signals from the pulsar.
Now, we show the parameter range in which the KL oscillation appears and is stable.
\subsection{Case (A) : $m_{\rm in}\sim m_3$}
As discussed in \S \ref{sec:Analytic}, we have stability conditions and constraints in order to find the KL oscillations.
In this case, the relation between the typical time scales are given as
\begin{eqnarray*}
{P_{\rm in}\over t_{\rm KL}} \ll {P_{\rm out}\over t_{\rm KL}}\sim {P_{\rm in}\over P_{\rm out}}
\ll 1
\end{eqnarray*}
\begin{figure}[h]
\includegraphics[width=7.5cm]{a_A_meM.png}
\caption{\small
The parameter space $(a_{\rm out},a_{\rm in})$ for Case (A).
We set $m_1=m_2=m_3 = 10 M_\odot$.
The relativistic effect (the precession of periastron) is shown by the blue line, while the stability condition (chaotic boundary) is given by green line. The light-blue shaded region depicts the allowed parameter space for stable KL oscillations.}
\label{fig:parameter_ain_case_a}
\end{figure}
We present some examples in Fig.
\ref{fig:parameter_ain_case_a}, in which
we show the allowed region in the parameter space
$(a_{\rm in}, a_{\rm out})$ for fixing the the masses
of a triple system
as $m_1=m_2=m_3=10 M_\odot$.
In \S \ref{sec:Analytic}, we show the condition with which the orbit does not become chaotic. This gives the maximum value of $a_{\rm in}$, which is given by green line [Eq. (\ref{eq:max_stable})].
The other two constraints come from the relativistic effects. One is the precession of the periastron, whose condition is
shown by the blue line [Eq. (\ref{eq:GR_precession})]. Beyond this line, such relativistic effect will suppress the KL oscillation.
The other is the LT precession effect, whose
critical line is out of the present parameter range.
Taking into account the above conditions, in Fig. \ref{fig:parameter_ain_case_a}, we present the possible range in the parameter space of $(a_{\rm in}, a_{\rm out})$ for the stable KL oscillation, which is shown by the light-blue shaded region.
For the present parameter choice, the constraints are determined by the relativistic precession effect of the periastron and chaotic boundary.
\subsection{Case (B) : $m_{\rm in}\ll m_3$}
In this case, the relation between the typical time scales are given as
\begin{eqnarray*}
\left({P_{\rm in}\over t_{\rm KL}}\right)^{1/2} \sim {P_{\rm out}\over t_{\rm KL}}\sim {P_{\rm in}\over P_{\rm out}}
\,.
\end{eqnarray*}
\begin{figure}[h]
\includegraphics[width=7.5cm]{a_A_mlM.png}
\caption{\small
We plot the allowed region (light-blue shaded)
in the parameter space of $(a_{\rm in}, a_{\rm out})$ for the stable KL oscillation by fixing $m_1=10^3 M_\odot , m_2= 10^3 M_\odot$ and
$m_3 = 10^6 M_\odot$.
The blue and brown lines correspond to the critical curves of relativistic effects (precession of periastron and LT effect),
beyond which the KL oscillation
might be suppressed. The stability of the KL oscillation gives
one condition (chaotic boundary) shown
by green line.}
\label{fig:parameter_ain_case_b}
\end{figure}
In Fig. \ref{fig:parameter_ain_case_b},
we present the possible range in the parameter space of $(a_{\rm in}, a_{\rm out})$ for
fixing the masses of a triple system
as $m_1=m_2=10^3 M_\odot$, and $m_3=10^6 M_\odot$.
The constraints are determined only by the relativistic precession effect of the periastron and
chaotic boundary
just as Fig. \ref{fig:parameter_ain_case_a} in Case (A).
We find that there exists the minimum values of
$a_{\rm in}$ and $a_{\rm out}$, below which
no stable KL oscillation is possible.
\begin{figure}[h]
\includegraphics[width=7.5cm]{m_a_mlM.png}
\caption{\small
We plot the allowed region (light-blue shaded)
in the parameter space of $(a_{\rm in}, m_{\rm in})$ for the stable KL oscillation by fixing $m_3 = 10^6 M_\odot$ and $a_{\rm out} = 10 {\rm AU}$. }
\label{fig:parameter_case_b}
\end{figure}
Since the range of mass parameter in Case (B) is very wide,
in Fig. \ref{fig:parameter_case_b}, we also present the possible range in the parameter space of $(a_{\rm in}, m_{\rm in})$ for the stable KL oscillation, which is shown by the light-blue shaded region. We fix the mass of tertiary component and the semi-major axis of the outer orbit as $m_3=10^6 M_\odot$ and $a_{\rm out}=10 {\rm AU}$.
For the present parameter choice,
the constraints are determined by the relativistic precession effect of the periastron and chaotic boundary condition
just as Case (A).
However, note that the LT precession may become significant
in Case (B) when the third body is much heavier.
See Fig. \ref{fig:parameter_case_b_10^9}, in which we
set $m_3=10^9 M_\odot$.
In this case, a hierarchical triple system will evolve
in the left direction horizontally as
$a_{\rm in}$ decreases because of the emission of
GWs. If $m_{\rm in}\lesssim 10^3 M_\odot$, the evolution
curve will hit on the LT critical line before the
relativistic periastron shift becomes important.
Hence the system will evolve into a chaotic KL oscillation
phase, which may be very interesting to study, but it is
out of our present analysis because the LT precession appears
in 1.5 PN order
\begin{figure}[h]
\includegraphics[width=7.5cm]{m_a_10_9.png}
\caption{\small
The same figure as Fig. \ref{fig:parameter_case_b}
with the different parameters.
We assume $m_3 = 10^9 M_\odot$ and $a_{\rm out} = 200 {\rm AU}$.
The LT critical curve [Eq. (\ref{eq:LT_precession})] cross the
light-blue shaded region for stable KL oscillations.
}
\label{fig:parameter_case_b_10^9}
\end{figure}
\begin{widetext}
\vskip 1cm
\section{Waveform, Spectra and Observability of GWs with KL Oscillations}
\label{sec:Results}
\subsection{Models}
To clarify the properties of the GWs from an inner binary in a hierarchical triple system, we shall perform numerical
analysis for the following models;\\
Case (A): AKL1 ($m_1=m_2=m_3=10 M_\odot$) and AKL2 ($m_1=m_2=m_3=30 M_\odot$)\\
Case (B): BKL1 ($m_1=m_2=10^3 M_\odot, m_3=10^6 M_\odot$)
and BKL2 ($m_1=m_2=10^3 M_\odot, m_3=10^9 M_\odot$).
\begin{table}[h]
\begin{tabular}{c|c|c|c||ccc|ccccc|ccc} \hline
Case& System& Model & Orbital motion &$m_1$&$m_2$&$m_3$& $a_{\rm in}$[{\rm AU}] &$a_{\rm out}$[{\rm AU}] & $e_{\rm in}$ & $e_{\rm out}$ &$I$[deg] &$P_{\rm in}$[days]&$P_{\rm out}$[days]&$t_{\rm KL}$[days]
\\
\hline\hline
&&AKL1& KL oscillation & 10&10&10& 0.01 & 0.1 & 0 & 0 & 90 &$\sim$ 0.08&$\sim$ 2.11&$\sim$ 164\\
&hierarchical &AKL2& KL oscillation & 30&30&30& 0.01 & 0.1 & 0 & 0 & 90 &$\sim $ 0.048&$\sim$ 1.22&$\sim$ 96\\
\cline{3-15}
(A)& ~~triplet &ACC& coplanar circular & 10&10&10& 0.01 & 0.1 & 0 & 0 & 0&$\sim$ 0.08&$\sim$ 2.11&- \\
& &ACE& coplanar eccentric & 10&10&10& 0.01 & 0.1 & 0.9 & 0 & 0&$\sim$ 0.08&$\sim$ 2.11&-
\\
\cline{2-15}
&isolated& ABC &
circular & 10&10&-& 0.01 & - & 0 & - & - &$\sim$ 0.08&-&-\\
& ~~binary&ABE & eccentric & 10&10&-& 0.01 & - & 0.9 & - & - &$\sim$ 0.08&-&-\\
\hline
(B) &hierarchical & BKL1& KL oscillation & $10^3$&$10^3$&$10^6$& 0.15 & 10 & 0 & 0 & 90 &$\sim$ 0.47&$\sim$ 11.54&$\sim$ 280\\
& ~~triplet &BKL2& KL oscillation & $10^3$&$10^3$&$10^9$& 0.25 & 200 & 0 & 0 & 90 &$\sim$ 1.02&$\sim$ 32.67&$\sim$ 1044\\
\hline
\end{tabular}
\caption{\small Masses of three-body hierarchical systems and of two body isolated binary, and the initial orbital elements $a$, $e$, and $i$
are the semi-major axis, eccentricity, and inclination respectively. Subscripts ``in" and ``out" correspond to the inner and outer orbits, respectively. We also
give the typical time scales; $P_{\rm in}$, $P_{\rm out}$ and $t_{\rm KL}$.
The eccentricity for KL binary
increases maximally to $e_{\rm in, max}\sim 0.987$
and $e_{\rm in, max}\sim 0.964$ by the KL oscillation for models AKL1 and AKL2, while $e_{\rm in, max}\sim 0.99$ and $e_{\rm in, max}\sim 0.98$
for Models BKL1 and BKL2, respectively.}
\label{tab:ini}
\end{table}
We then choose the semi-major axis $a_{\rm in}$ and
$a_{\rm out}$ in the stable regions for the KL oscillation,
which are given in Figs. \ref{fig:parameter_ain_case_a}-\ref{fig:parameter_case_b}.
In addition, to confer the results of our analysis, we pick Case (A) to contrast the key features in waveforms from an isolated binary system (Models ABC and ABE) and in
waveforms from a binary in a coplanar hierarchical triple system
(Models ACC and ACE). The parameters are summarized in Table ~\ref{tab:ini}.
The observer's location $(\iota)$ is defined as the angle between the detector and the ``axis" of the reference plane (i.e., the "normal" to the initial outer orbital plane). As we are focused on the inspiral phase of the inner binary, there are special features that can be seen in the waveform.
\subsection{Method of Orbit Evolution}
The first order post-Newtonian equations of motion also known as the Einstein - Infeld - Hoffmann equations of motion \cite{EIH38}, describes the dynamics of a system of point-like masses due to their mutual gravitational interactions, including GR effects. We employ Eq.~(\ref{eq:EIH}) in order to solve the three-body system.
This equation could also be derived from the Lagrangian given by Lorentz and Droste \cite{LD17}.
\begin{eqnarray}
&& \frac{ \mathrm{d} \bm{v}_{k}}{\mathrm{d} t}
=-G\sum_{n\neq k} m_{n}\frac{\bm{x}_{k} - \bm{x}_{n}}{|\bm{x}_{k} - \bm{x}_{n}|^{3}}
\Big[
1-4 \frac{G}{c^2}\sum_{n'\neq k} \frac{m_{n'}}{|\bm{x}_{k} - \bm{x}_{n'}|}
-\frac{G}{c^2}\sum_{n'\neq n} \frac{m_{n'}}{|\bm{x}_{n} - \bm{x}_{n'}|}
\left \{
1-\frac{(\bm{x}_{k} - \bm{x}_{n}) \cdot (\bm{x}_{n} - \bm{x}_{n'})}
{2|\bm{x}_{n} - \bm{x}_{n'}|^{2}}
\right \}
\nonumber \\
&& +\left( \frac{|\bm{v}_{k}|}{c} \right)^{2} + 2\left( \frac{|\bm{v}_{n}|}{c} \right)^{2}
-4 \frac{\bm{v}_{k} \cdot \bm{v}_{n}}{c^2}
-\frac{3}{2}
\left\{
\frac{(\bm{x}_{k} - \bm{x}_{n})}{|\bm{x}_{k} - \bm{x}_{n}|} \cdot \frac{\bm{v}_{n}}{c}
\right\}^{2}
\Big]
-\frac{G}{c^2} \sum_{n \neq k} \frac{m_{n}(\bm{v}_{k}-\bm{v}_{n})}{|\bm{x}_{k} - \bm{x}_{n}|^{3}}
(\bm{x}_{k} - \bm{x}_{n}) \cdot (3\bm{v}_{n}-4\bm{v}_{k})
\nonumber \\ \hspace*{-15cm}
&& -\frac{7}{2} \frac{G^{2}}{c^2} \sum_{n \neq k}\frac{m_{n}}{|\bm{x}_{k} - \bm{x}_{n}|}
\sum_{n'\neq n} \frac{m_{n'} (\bm{x}_{n} - \bm{x}_{n'})} {|\bm{x}_{n} - \bm{x}_{n'}|^{3}},
\label{eq:EIH}
\end{eqnarray}
\end{widetext}
where $m_k$, $\bm{v}_k$, $\bm{x}_k$ ($k=1,2$ and $3$)
are the mass, velocity and position of the $k$-th component of the system, $G$ is the gravitational constant, and $c$ is the speed of light. Eq.~(\ref{eq:EIH}) has been numerically integrated by using 6-th order implicit Runge-Kutta method, whose coefficients are obtained from \citet{Butcher64}.
On integrating, we obtain the numerical data of positions and velocities of the triple system. The initial outer orbital plane is considered to be the reference frame.
In order to set up initial conditions,
we convert initial orbital elements of inner and outer orbits
into the variables $\bm{x}_{k}$ and $\bm{v}_{k}$ in Cartesian coordinates,
with its origin in the center of mass of the whole system \citep{Murray00}.
We integrate the EIH equations (\ref{eq:EIH}) numerically and evaluate the osculating orbital elements at each step from the
numerical data of positions and velocities of the triple system (see e.g. \citet{Murray00}).
Since the inner orbit is not exactly an ellipse,
the obtained osculating elements are oscillating with small amplitudes in the cycle of inner orbit.
Hence, we take an average of the osculating elements for each cycle of inner orbit and then obtain the averaged semi-major axes $\bar{a}_\mathrm{in}$, $\bar{a}_\mathrm{out}$ and eccentricities $\bar{e}_\mathrm{in}$, $\bar{e}_\mathrm{out}$,
which may give the effective values of the orbital elements.
Those elements will evolve secularly in time because of the effect of the tertiary body.
\subsection{Numerical Results}
We first summarize the numerical results for Model AKL1
in next subsections 1, 2 and 3, and
then mention about Model BKL1 in subsection 4.
\subsubsection{\textbf{\underline{Evolution of the eccentricity}}}
\label{eccentricity}
We first show the result of the evolution of eccentricity and relative inclination of the inner binary for Model AKL1, in which the masses
of a triplet are $m_1=m_2=m_3=10 M_\odot$ and initial semi-major axis are $a_{\rm in}=0.01 {\rm AU}$ and $a_{\rm out}=0.1 {\rm AU}$. The typical time scales are given by $P_{\rm in} \sim 0.08$ days, $P_{\rm out} \sim 2.11$ days and $t_{\rm KL} \sim 164$ days.
\begin{figure}[h]
\centering
\includegraphics[width=7.5cm]{ecc_inc.png}
\caption{\small The time evolution of the osculating orbital elements
(the averaged eccentricity of the inner binary $\bar{e}_{in}$,and averaged relative inclination $\bar{I}$)
in Model AKL1 ($m_1=m_2=m_3=10M_\odot$, $a_{\rm in}=0.01 {\rm AU}$, and $a_{\rm out}=0.1 {\rm AU}$) for one KL oscillation cycle. The red and blue lines show the evolution of $\bar{e}_{in}$ and $\bar{I}$, respectively.
When the inclination drops to $I_{\rm min}\approx 55^\circ$,
the eccentricity reaches the maximum value $e_{\rm max}\approx 0.987$. The small oscillations are caused by the outer companion's motion.}
\label{fig:KL}
\end{figure}
The initial eccentricities are $e_{\rm in}=e_{\rm out}=0$, but
the relative inclination is chosen to be $I=90^\circ$.
We show the time evolution of the eccentricity $e_{\rm in}$ and the relative inclination $I$ for one KL oscillation cycle
in Fig ~\ref{fig:KL}.
We have run our code for several values of inclination varying from 40$^\circ$ to 90$^\circ$, for which we obtain stable KL oscillations.
For $I=90^\circ$, we find the maximum value of the eccentricity as $e_{\rm in, max}\approx 0.987$, while $e_{\rm in, max}\approx 0.02$ for $I=40^\circ$.
For lower values for relative inclination, we find lower maximum eccentricities as expected from Eq. (\ref{eq:theta}).
\subsubsection{\textbf{\underline{GW Waveform}}}
\label{GW Waveform}
Next we show the waveform of the GWs for Model AKL1 ($m_1=m_2=m_3=10M_\odot$, $a_{\rm in}=0.01 {\rm AU}$, and $a_{\rm out}=0.1 {\rm AU}$).
In Fig. \ref{fig:kozaiwaveform}, we present the waveform of two
polarization modes for one KL oscillation cycle.
For the small eccentricity until day 98,
the amplitudes are low as $h\sim 10^{-21}$, but
when the inner binary is in high eccentricity regime
around day 100, we find a significant rise in amplitude
as $h\sim 10^{-19}$.
\begin{figure}[h]
\centering
\includegraphics[width=0.70\linewidth]{hp_kozai_o45_full.png}
\\
\includegraphics[width=0.70\linewidth]{hc_kozai_o45_full.png}
\caption{\small The top and bottom figures correspond to `+' and `$\times$' polarization, respectively, of the waveform of GWs
from the hierarchical triplet model AKL1 with KL oscillation. The observer distance is 10 {\rm kpc} at inclination $\iota=45^\circ $ measured from the initial outer orbital plane.}
\label{fig:kozaiwaveform}
\end{figure}
\begin{figure}
\centering
\end{figure}
To see more detail, we enlarge the part of waveforms.
In Fig. \ref{fig:kozaiwaveform_2days_+},
we show the waveform of $+$ -polarization mode for two days.
The top figure depicts the data for the first two days, when
the eccentricity is small, while the bottom one shows the data
from day 106 through day 108, when the eccentricity becomes very large.
We also show the similar enlarged waveforms for $\times$-mode in Fig. \ref{fig:kozaiwaveform_2days_x}
\begin{figure}[h]
\includegraphics[width=0.70\linewidth]{hp_kozai_o45_s1.png}
\\
\includegraphics[width=0.70\linewidth]{hp_kozai_o45_ecc.png}
\caption{\small The zoomed in parts of waveform of $+$ polarization in low and high eccentricity regime. The top figure shows
the wave form in low eccentricity period (the first two days), while the bottom one depicts the wave form in high eccentricity period (two days from day 106 through day 108).}
\label{fig:kozaiwaveform_2days_+}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width=0.70\linewidth]{hc_kozai_o45_s1.png}
\\
\includegraphics[width=0.70\linewidth]{hc_kozai_o45_ecc.png}
\caption{\small The same zoomed figure as Fig. \ref{fig:kozaiwaveform_2days_+} for $\times$-polarization.}
\label{fig:kozaiwaveform_2days_x}
\end{figure}
These features can be easily understood when we analyze more simple systems such as an isolated binary or a coplanar motion of a hierarchical triple system, for which the results are summarized in Appendix for Models ACC-ABE (see Table ~\ref{tab:ini}).
For an isolated circular binary, the waveform is sinusoidal,
but if the eccentricity is large as $e=0.9$, the sharp peak in the waveform appears near the periastron point. The amplitude also
increases compared with the circular case.
For a hierarchical triple system, if the orbit is coplanar,
we find similar features with those in an isolated binary.
When the eccentricity $e_{\rm in}$ is small, we find the sinusoidal wave form, while if $e_{\rm in}$ is large, a sharp peak appears
near the periastron point.
In the case of a triple system, however,
there appears small modulation in the amplitudes because of the effect from a tertiary component.
\begin{figure}[h]
\centering
\includegraphics[width=0.7\linewidth]{hp_kozai_o45_s2.png}
\\
\includegraphics[width=0.7\linewidth]{hp_kozai_s2.png}
\caption{\small The global shape of the waveform with $+$ -polarization
for 15 days (from day 100 to day 115). It depends on the observer position. The top figure shows
the observer at inclination $45^\circ$, while
the bottom one is for the observer at inclination $90^\circ$. }
\label{fig:inc_observer_hp}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width=0.7\linewidth]{hc_kozai_o45_s2.png}
\\
\includegraphics[width=0.7\linewidth]{hc_kozai_s2.png}
\caption{\small The same figure as Fig. \ref{fig:inc_observer_hp}
for $\times$-polarization mode.}
\label{fig:inc_observer_hc}
\end{figure}
In Model AKL1, when the eccentricity is small, the wave form is almost the same as that in a coplanar circular orbit of a triple system.
When the eccentricity becomes large via KL oscillation,
the local feature of the wave form is similar to that of
an eccentric binary or in a coplanar eccentric orbit of a triple system. As shown in Figs. \ref{fig:inc_observer_hp} and \ref{fig:inc_observer_hc}, the amplitude is modulated randomly because of the oscillation of the eccentricity, which is different from the
coplanar eccentric case.
For the observer at a different position, we also find the same
features. The difference appears in the global shape of the wave form. We show the examples for the observers at $\iota=45^\circ$
and $\iota=90^\circ$ measured from the initial outer orbital plane
in Figs. \ref{fig:inc_observer_hp} and \ref{fig:inc_observer_hc}.
\subsubsection{\textbf{\underline{GW Spectra}}}
\label{GW Spectra}
Large eccentricities during KL oscillation and variation of eccentricity with time introduce more harmonics in GWs which make the energy spectra rich as compared to the spectra from an isolated binary (shown in Appendix \ref{GW_isolated}). In Fig. ~\ref{fig:kozaispectra}, we show energy spectra for two stages in one KL cycle. During low eccentricity regime, the spectra is similar to that of circular coplanar hierarchical triple (Model ACC).
\begin{figure}[h]
\includegraphics[width=0.70\linewidth]{spectra_kozai_s1.png}
\\
\includegraphics[width=0.70\linewidth]{spectra_kozai_s2.png}
\caption{\small The GW energy spectra from the inner binary with KL oscillation for low and high eccentricity regime.
The top figure shows the spectrum at day 1-5 ($e_{\rm in} \sim 0$), and
the bottom one gives that at day 106-108 ($e_{\rm in} = 0.983 \sim 0.987$).
We find a broad band spectrum for a high eccentric orbit.}
\label{fig:kozaispectra}
\end{figure}
In the top figure in Fig ~\ref{fig:kozaispectra},
which is obtained by the Fourier transformation for the time interval from day 1 to day 5,
we find two sharp peaks which correspond to
the two periods of the inner and outer orbits.
On the other hand, during high eccentricity regime,
as we see from the bottom one of Fig ~\ref{fig:kozaispectra},
which is obtained for the time interval
from day 106 to day 108,
the spectra is quite similar to the eccentric coplanar
model ACE (see Fig. \ref{fig:spectra_3c}).
Because of high eccentricity, many higher harmonics appear.
As a result, we find a broad band spectrum for a high eccentricity regime.
\subsubsection{\textbf{\underline{Model {\rm BKL}}}}
\label{model BI}
We also perform numerical calculation for Model BKL1, i.e.,
$m_1=m_2=10^3 M_\odot$, $a_{\rm in}= 0.15 {\rm AU}$, and $a_{\rm out}= 10 {\rm AU}$.
We find that the properties of GWs are very much similar to
Model AKL1.
There appears KL oscillation, in which the maximum eccentricity
becomes 0.99. As for the waveform, it is a sinusoidal shape
with modulation caused by a tertiary object
when the eccentricity is small, while
it is deformed during the large eccentricity phase
and shows a sharp peak by passing near
a periastron point.
The GW spectra also change in one KL cycle
from ones with two sharp frequencies, corresponding to two orbital frequencies,
to a broad band spectra.
The difference appears only in the time scales given in Table I.
The observability is also different from Model AKL as we will show in
the next subsection.
\subsection{Observability: Frequency, Strain, and SNR}
To examine the possibility of observation of the present
triple models by future detectors, we have to check whether
the frequency and strain are in the observable range
and evaluate the signal to noise ratio (SNR).
We first plot $f_{\rm peak}$ for Model AKL1. We show in Fig. ~\ref{fig:fpeak} that due to KL oscillations the frequency sweeps up to the detectable range of proposed future space interferometers.
\begin{figure}[h]
\centering
\includegraphics[width = 7.5cm]{fpeak.png}
\caption{\small Time evolution of the peak frequency for
one KL oscillation cycle for Model AKL1
($m_1=m_2=m_3=10M_\odot$ and $a_{\rm in}=0.01 {\rm AU}, a_{\rm out}=0.1 {\rm AU}$).}
\label{fig:fpeak}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=8.5cm]{h_f_10M.png}
\caption{\small The strain versus frequency plot for observer at inclination 90$^\circ$. The blue, green, brown and black curves show DECIGO, BBO, aLISA and LISA design sensitivity.
The orange, violet, and red curves show the strain evolution with $e_{\rm in}$ = 0.944 at day 103, 0.972 at day 104.6, and 0.987
(=$e_{\rm in, max}$)
at day 106.8, respectively.}
\label{fig:hfplot_A}
\end{figure}
However, as the inner binary is in inspiral phase, the GW strain may lie below the sensitivity curve of detectors. We then
plot strain versus frequency curve for Model AKL1 in Fig. ~\ref{fig:hfplot_A}.
The black, brown, blue, green curves correspond to strain sensitivity curves of LISA, aLISA, DECIGO and BBO respectively \citep{Cornish2018,Seto2017,Larson2000}. We plot GW
characteristic strain curve for Model AKL1 at different days. The orange curve is plotted for day 102-103 when
the eccentricity is
$e_\mathrm{in} \sim 0.944$. The red curve is plotted for day 103-105 with $e_\mathrm{in} \sim 0.972$. The violet curve is plotted for day 106-108 with $e_\mathrm{in} \sim 0.987$, which is the maximum value
in one KL cycle.
After day 108, the eccentricity decreases and then
the strain curves decrease to the orange one through the red one.
These curves show the evolution of the strain in one KL cycle.
At high eccentricity regime of KL cycle, the GWs may be observable for at least two days every
164 days (one KL cycle) by DECIGO or BBO.
In Fig. ~\ref{fig:hfplot_A2}, we also show the similar plot
for Model AKL2 with $m_1=m_2=m_3 = 30 M_\odot$
and $a_\mathrm{in} = 0.01 {\rm AU}$. When the observer distance is again kept to be at 10kpc,
the signal is louder due to increase in mass.
\begin{figure}[h]
\centering
\includegraphics[width=8.5cm]{h_f_30M.png}
\caption{\small The strain versus frequency plot for observer at inclination 90$^\circ$. The blue, green, brown and black curves show DECIGO, BBO, aLISA and LISA design sensitivity.
The orange, red and violet curves show the strain evolution with $e_{\rm in}$ = 0.878 at day 57, 0.946 at day 58.6, and 0.964
(=$e_{\rm in, max}$) at day 60, respectively. The observer distance is 10kpc.}
\label{fig:hfplot_A2}
\end{figure}
One may wonder that the GW signal may be louder when $a_{\rm in}$
becomes smaller via an evolution of the inner binary by
the emission of GWs.
For example, one may choose $a_{\rm in}=0.006 {\rm AU}$,
which is still in the stable range
for KL oscillations (see Fig. \ref{fig:parameter_ain_case_a}).
However it does not give a louder signal.
It is because
the maximum eccentricity $e_{\rm in, max}= 0.87$
for $a_{\rm in}=0.006 {\rm AU}$ is not
high enough to give an observable strain
compared with Model AKL1 or AKL2.
We show the change of the maximum eccentricity $e_{\rm in, max}$
in terms of $a_{\rm in}$ in Fig. \ref{fig:ecc_ain}.
\begin{figure}[h]
\centering
\includegraphics[width=8.5cm]{emax_ain.png}
\caption{\small The maximum eccentricity in terms of $a_{\rm in}$ for model AKL1.
We find $e_{\rm in, max}\approx 1$ for $a_{\rm in} \, \mbox{\raisebox{-1.ex 0.01 {\rm AU}$, while $e_{\rm in, max}$
decreases rapidly beyond $a_{\rm in}\sim 0.01 {\rm AU}$.}
\label{fig:ecc_ain}
\end{figure}
We find that the maximum eccentricity is very close to unity for
$a_{\rm in}\, \mbox{\raisebox{-1.ex 0.01 {\rm AU}$, but
it decreases rapidly past $a_{\rm in}\sim 0.01 {\rm AU}$.
This is because GR effect suppresses the KL mechanism\citep{tey13,naoz13b,liu15}.
As a result, the case with $a_{\rm in}= 0.01 {\rm AU}$,
which shows the shortest semi-major axis as well as the largest eccentricity,
gives the loudest characteristic strain.
We also show the characteristic strain for Model BKL1
in Fig. \ref{fig:hfplot_B}.
In this case, the strain is not large enough
to give the observable value by DECIGO or BBO.
We may need aLISA.
\begin{figure}[h]
\centering
\includegraphics[width=8.5cm]{h_f_2000M.png}
\caption{\small The same figure for Model BKL1 as Figs. \ref{fig:hfplot_A} and \ref{fig:hfplot_A2}. The blue, green, brown and black curves show DECIGO, BBO, aLISA and LISA design sensitivity. The orange, red and violet curves show the strain evolution with $e_{\rm in}$ =0.967 at day 126.6, 0.977 at day 128 and 0.99 (=$e_{\rm in, max}$) at day 130.4, respectively. The observer distance is 16 Mpc.}
\label{fig:hfplot_B}
\end{figure}
The plot of the frequency vs the characteristic strain gives helpful assessment of detectability provided high signal to noise ratio (SNR) of the waveform exists.
Our hierarchical triple system may be one of the many source populations captured by future GW space detectors with following merger phase captured by ground detectors.
The area between the source and detector curves is related to SNR by the following relation \citep{Berry2014},
\begin{equation*}
({\rm SNR})^2 = \int_{f_{\rm min}}^{f_{\rm max}} d (\log f) \bigg(\frac{h_c(f)}{h_d(f)}\bigg)^2
\,,
\end{equation*}
where $h_c(f)$ is the strain amplitude of the source and
$h_d(f)$ denotes noise amplitude of detector.
For the models shown in our analysis,
the SNR for DECIGO is about 8 at $e_{\rm in, max} \sim 0.987$ for Model AKL1.
Similarly the SNR is about 16 for Model AKL2 at $e_{\rm in, max} \sim 0.964$.
Since the design sensitivity of BBO is better than DECIGO,
the SNR for BBO would be larger.
For Model BKL1, the signal is not loud enough for BBO, but that
for aLISA SNR is $\sim 17$, which could be observable.
\section{Concluding Remarks}
\label{sec:Conc}
We have shown features of gravitational waves from an inner binary in a hierarchical triple system
with KL oscillations.
The waveform changes its shape in time because of oscillation of
the eccentricity.
When the eccentricity is small,
the waveform is a sinusoidal shape modulated by a tertiary
companion,
while it becomes one with sharp peaks
near periastron point when the eccentricity gets large.
We have also examined the time
variation of the characteristic strain curve, which
may appear in sensitivity range of detectors when the eccentricity becomes large
via the KL oscillations.
This can be the first direct observation of KL oscillation which may further
shed light on binary formation channels and surrounding environment of the binary system.
The interesting remaining questions are
what happens when the system evolves beyond the KL stable region.
Because of the GW emission, the semi-major axis of the inner binary
will decrease. Since other parameters such as $m_1, m_2, m_3$ and $a_{\rm out}$
are conserved, the system will evolve vertically in Figs. \ref{fig:parameter_ain_case_a}
and \ref{fig:parameter_ain_case_b} or horizontally
in Figs. \ref{fig:parameter_case_b} and \ref{fig:parameter_case_b_10^9}.
Hence, we reach at the boundary of relativistic periastron shift.
Beyond this boundary, KL oscillation may be suppressed, and then
the effect of the tertiary component may become weak.
As a result, we expect that GWs may not be observed before
the coalescence of the inner binary. The question is whether the orbit of the inner binary
is circularized via the GW emission at the coalescence or not.
If the eccentricity remains then, we find the difference from a simple binary system.
We may need numerical relativity or some other approach such as ``effective one body system"
to clarify it.
Another interesting issue is what happens when the LT effect becomes important.
It can be the case if we consider SMBH.
In Fig. \ref{fig:parameter_case_b_10^9}, if the total mass of the inner binary
is smaller than $10^3 M_\odot$, when the semi-major axis decreases,
the system will evolve into the region where the LT precession becomes important.
In this region, we may find a chaotic KL oscillation\citep{Wang19}, which may
provide us strong emission of GWs. In order to study such a process, we have to include
higher PN effects, i.e., at least 1.5 PN terms.
The work on these two issues is in progress.
As for a triple system, we are also interested in more
general situations, in which we expect chaotic behavior of
a system. The GWs from such a chaotic system should also be
studied. Since it may be difficult to make templates,
we should find some other ways to clarify typical features of such a system
in GW observations\citep{Suzuki:1999si,Kiuchi:2004bv,Kiuchi:2007ue}.
One of the most important issues about the GWs from a hierarchical triple system is the event rate.
As for a formation of three body system, there are many works including numerical
$N$-body simulation \citep{Glutekin2006,Samsing2014,Fabio2016}.
The event rate depends on many uncertain factors like star formation rate,
binary fraction and distribution of initial parameters.
Several papers discuss the increase in the merger rate of compact binaries around SMBHs due to KL oscillations \citep{Antonini2012,hoang18,trani19}. The binaries in these triple systems undergo several KL cycles finally leading to the merger.
In Ref. \citep{Antonini2012}, the authors estimate the rate to be
0.56 ${\rm Gpc}^{-3} {\rm yr}^{-1}$. Similar analysis is done by Refs. \citep{hoang18,trani19}, estimating the rate ~ 1-3 ${\rm Gpc}^{-3}{\rm yr}^{-1}$.
The difference is caused by different distance regimes from the SMBH and
different initial conditions. For isolated triples,
the merger rate to be 6 ${\rm Gpc}^{-3} {\rm yr}^{-1}$ in the absence of natal kicks\citep{Tremaine17}. Globular clusters may harbor IMBHs at their centers.
Ref. \citep{Omer19} studies the triple systems composed of the Stellar-mass binary with IMBH as a third body. The merger event rate induced by KL oscillations is estimated to be
0.06-0.46 ${\rm Gpc}^{-3} {\rm yr}^{-1}$.
We are interested in the models where merger timescale is much longer than Kozai timescale. Hence, if there are many KL oscillations before the merger, the signal about KL oscillations may lie in the observable range. In Ref. \citep{Lisa2019}, using double-averaged equations, they analyze the fraction of KL binaries in two channels (isolated triples and galactic center) lying in the LISA frequency range. They show that of all merging binaries considered in their simulation, $\sim 39\%$ of the isolated-triple channel, and $\sim 22\%$ of the galactic-center channel, display significant KL oscillations in LISA frequency band.
However, the present models of a hierarchical triple
system with KL oscillations,
which gives observable GWs in an inspiral phase, may be
difficult to be formed by evolution of a triple star system
because both semi-major axes
are very short.
One possible way would be a capture process.
The third body is bounded due to a close encounter with a BH binary in globular clusters or near galactic centers.
The question is how likely such an interaction is.
In \citep{Samsing2014,zevin19}, binary-single and binary-binary BH interactions are numerically simulated and several possible end states are discussed.
In Ref. \citep{zevin19}, they have shown that stable triple formation end state has higher cross section compared to some other possible end states.
We are planning to study in detail the formation process and probability to evaluate the event rate of compact hierarchical triples such that KL oscillations are not quenched by relativistic effects.\\
\section*{Acknowledgements}
We would like to acknowledge Gungwon Kang, Seiji Kawamura, Kaye Li, Naoki Seto, Takahiro Tanaka, and Alessandro A. Trani for useful discussions.
P.G. is supported by Japanese Government (MEXT)
Scholarship.
This work was supported in part by JSPS KAKENHI Grant Numbers
JP17H06359 and JP19K03857, and by Waseda University Grant for Special
Research Projects (Project number: 2019C-254 and 2019C-640).
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,523 |
\section{Introduction}
Output perturbation is a cornerstone of mechanism design in differential privacy (DP). Well-known mechanisms in this class are the Laplace and Gaussian mechanisms \cite{dwork2006calibrating,dwork2014algorithmic}. More complex mechanisms are often obtained by composing multiple applications of these basic output perturbation mechanisms. For example, the Laplace mechanism is the basic building block of the sparse vector mechanism \cite{dwork2009complexity}, and the Gaussian mechanism is the building block of private empirical risk minimization algorithms based on stochastic gradient descent \cite{bassily2014private}. Analysing the privacy of such complex mechanisms turns out to be a delicate and error-prone task \cite{lyu2017understanding}. In particular, obtaining tight privacy analyses leading to optimal utility is one of the main challenges in the design of advanced DP mechanisms. An alternative to tight \emph{a-priori} analyses is to equip complex mechanisms with \emph{algorithmic} noise calibration and accounting methods. These methods use numerical computations to, e.g.\ calibrate perturbations and compute cumulative privacy losses at run time, without relying on hand-crafted worst-case bounds. For example, recent works have proposed methods to account for the privacy loss under compositions occurring in complex mechanisms \cite{rogers2016privacy,abadi2016deep}
In this work we revisit the Gaussian mechanism and develop two ideas to improve the utility of output perturbation DP mechanisms based on Gaussian noise. The first improvement is an algorithmic noise calibration strategy that uses numerical evaluations of the Gaussian cumulative density function (CDF) to obtain the optimal variance to achieve DP using Gaussian perturbation. The analysis and the resulting algorithm are provided in Section~\ref{sec:aGM}.
In order to motivate the need for a numerical approach to calibrate the noise of a DP Gaussian perturbation mechanism, we start with an analysis of the main limitations of the classical Gaussian mechanism in Section~\ref{sec:limitations}.
A numerical evaluation provided in Section~\ref{sec:expaGM} showcases the advantages of our optimal calibration procedure.
The second improvement equips the Gaussian perturbation mechanism with a post-processing step which denoises the output using adaptive estimation techniques from the statistics literature. Since DP is preserved by post-processing and the distribution of the perturbation added to the desired outcome is known, this allows a mechanism to achieve the desired privacy guarantee while increasing the accuracy of the released value. The relevant denoising estimators and their utility guarantees are discussed in Section~\ref{sec:denoise}. Results presented in this section are not new: they are the product of a century's worth of research in statistical estimation. Our contribution is to compile relevant results scattered throughout the literature in a single place and showcase their practical impact in synthetic (Section~\ref{sec:expmean}) and real (Section~\ref{sec:exptaxi}) datasets, thus providing useful pointers and guidelines for practitioners.
\section{Limitations of the Classical Gaussian Mechanism}\label{sec:limitations}
Let $\mathbb{X}$ be an input space equipped with a symmetric neighbouring relation $x \simeq x'$. Let $\varepsilon \geq 0$ and $\delta \in [0,1]$ be two privacy parameters. A $\mathbb{Y}$-valued randomized algorithm $M : \mathbb{X} \to \mathbb{Y}$ is $(\varepsilon,\delta)$-DP \cite{dwork2006calibrating} if for every pair of neighbouring inputs $x \simeq x'$ and every possible (measurable) output set $E \subseteq \mathbb{Y}$ the following inequality holds:
\begin{align}\label{eqn:DP}
\Pr[M(x) \in E] \leq e^{\varepsilon} \Pr[M(x') \in E] + \delta \enspace.
\end{align}
The definition of DP captures the intuition that a computation on private data will not reveal sensitive information about individuals in a dataset if removing or replacing an individual in the dataset has a negligible effect in the output distribution.
In this paper we focus on the family of so-called output perturbation DP mechanisms.
An output perturbation mechanism $M$ for a deterministic vector-valued computation $f : \mathbb{X} \to \mathbb{R}^d$ is obtained by computing the function $f$ on the input data $x$ and then adding random noise sampled from a random variable $Z$ to the output.
The amount of noise required to ensure the mechanism $M(x) = f(x) + Z$ satisfies a given privacy guarantee typically depends on how sensitive the function $f$ is to changes in the input and the specific distribution chosen for $Z$.
The Gaussian mechanism gives a way to calibrate a zero mean isotropic Gaussian perturbation $Z \sim \mathcal{N}(0,\sigma^2 I)$ to the global $L_2$ sensitivity $\Delta = \sup_{x \simeq x'} \norm{f(x) - f(x')}$ of $f$ as follows.
\begin{theorem}[Classical Gaussian Mechanism]\label{thm:cGM}
For any $\varepsilon, \delta \in (0,1)$, the Gaussian output perturbation mechanism with $\sigma = \Delta \sqrt{2 \log(1.25/\delta)}/\varepsilon$ is $(\varepsilon,\delta)$-DP.
\end{theorem}
A natural question one can ask about this result is whether this value of $\sigma$ provides the minimal amount of noise required to obtain $(\varepsilon,\delta)$-DP with Gaussian perturbations. Another natural question is what happens in the case $\varepsilon \geq 1$. This section addresses both these questions. First we show that the value of $\sigma$ given in Theorem~\ref{thm:cGM} is suboptimal in the high privacy regime $\varepsilon \to 0$. Then we show that this problem is in fact inherent to the usual proof strategy used to analyze the Gaussian mechanism. We conclude the section by showing that for large values of $\varepsilon$ the standard deviation of a Gaussian perturbation that provides $(\varepsilon,\delta$)-DP must scale like $\Omega(1/\sqrt{\varepsilon})$. This implies that the scaling $\Theta(1/\varepsilon)$ provided by the classical Gaussian mechanism in the range $\varepsilon \in (0,1)$ cannot be extended beyond any bounded interval.
\subsection{Limitations in the High Privacy Regime}\label{sec:highprivacy}
To illustrate the sub-optimality of the classical Gaussian mechanism in the regime $\varepsilon \to 0$ we start by showing it is possible to achieve $(0,\delta)$-DP using Gaussian perturbations. This clearly falls outside the capabilities of the classical Gaussian mechanism, since the standard deviation $\sigma = \Theta(1/\varepsilon)$ provided by Theorem~\ref{thm:cGM} grows to infinity as $\varepsilon \to 0$.
\begin{theorem}\label{thm:TV}
A Gaussian output perturbation mechanism with $\sigma = \Delta / 2 \delta$ is $(0,\delta)$-DP\footnote{Proofs for all results given in the paper are presented in Appendix~\ref{app:proofs}.}.
\end{theorem}
Previous analyses of the Gaussian mechanism are based on a simple sufficient condition for DP in terms of the privacy loss random variable \cite{dwork2014algorithmic}.
The next section explains why the usual analysis of the Gaussian mechanism cannot yield tight bounds for the regime $\varepsilon \to 0$.
This shows that our example is not a corner case, but a fundamental limitation of trying to establish $(\varepsilon,\delta)$-DP through said sufficient condition.
\subsection{Limitations of Privacy Loss Analyses}\label{sec:prooflimitations}
Given a vector-valued mechanism $M$ let $p_{M(x)}(y)$ denote the density of the random variable $Y = M(x)$. The privacy loss function of $M$ on a pair of neighbouring inputs $x \simeq x'$ is defined as
\begin{align*}
\ell_{M,x,x'}(y) = \log \left(\frac{p_{M(x)}(y)}{p_{M(x')}(y)}\right) \enspace.
\end{align*}
The privacy loss random variable $L_{M,x,x'} = \ell_{M,x,x'}(Y)$ is the transformation of the output random variable $Y = M(x)$ by the function $\ell_{M,x,x'}$. For the particular case of a Gaussian mechanism $M(x) = f(x) + Z$ with $Z \sim \mathcal{N}(0,\sigma^2 I)$ it is well-known that the privacy loss random variable is also Gaussian \cite{dwork2016concentrated}.
\begin{lemma}\label{lem:privlossGM}
The privacy loss $L_{M,x,x'}$ of a Gaussian output perturbation mechanism follows a distribution $\mathcal{N}(\eta, 2 \eta)$ with
$\eta = D^2 / 2 \sigma^2$, where $D = \norm{f(x) - f(x')}$.
\end{lemma}
The privacy analysis of the classical Gaussian mechanism relies on the following sufficient condition: a mechanism $M$ is $(\varepsilon,\delta)$-DP if the privacy loss $L_{M,x,x'}$ satisfies
\begin{align}\label{eqn:suff}
\forall x \simeq x' : \Pr[L_{M,x,x'} \geq \varepsilon] \leq \delta \enspace.
\end{align}
Since Lemma~\ref{lem:privlossGM} shows the privacy loss $L_{M,x,x'}$ of the Gaussian mechanism is a Gaussian random variable with mean $\norm{f(x) - f(x')}^2 / 2 \sigma^2$, we have $\Pr[L_{M,x,x'} > 0] \geq 1/2$ for any pair of datasets with $f(x) \neq f(x')$. This observation shows that in general it is not possible to use this sufficient condition for $(\varepsilon,\delta)$-DP to prove that the Gaussian mechanism achieves $(0,\delta)$-DP for any $\delta < 1/2$. In other words, the sufficient condition is not necessary in the regime $\varepsilon \to 0$. We conclude that an alternative analysis is required in order to improve the dependence on $\varepsilon$ in the Gaussian mechanism.
\subsection{Limitations in the Low Privacy Regime}\label{sec:lowprivacy}
The last question we address in this section is whether the order of magnitude $\sigma = \Theta(1/\varepsilon)$ given by Theorem~\ref{thm:cGM} for $\varepsilon \leq 1$ can be extended to privacy parameters of the form $\varepsilon > 1$. We show this is not the case by providing the following lower bound.
\begin{theorem}\label{thm:lowerb}
Let $f : \mathbb{X} \to \mathbb{R}^d$ have global $L_2$ sensitivity $\Delta$. Suppose $\varepsilon > 0$ and $0 < \delta < 1/2 - e^{-3 \varepsilon}/\sqrt{4 \pi \varepsilon}$. If the mechanism $M(x) = f(x) + Z$ with $Z \sim \mathcal{N}(0,\sigma^2 I)$ is $(\varepsilon,\delta)$-DP, then $\sigma \geq \Delta / \sqrt{2 \varepsilon}$.
\end{theorem}
Note that as $\varepsilon \to \infty$ the upper bound on $\delta$ in Theorem~\ref{thm:lowerb} converges to $1/2$. Thus, as $\varepsilon$ increases the range of $\delta$'s requiring noise of the order $\Omega(1/\sqrt{\varepsilon})$ increases to include all parameters of practical interest. This shows that the rate $\sigma = \Theta(1/\varepsilon)$ provided by the classical Gaussian mechanism cannot be extended beyond the interval $\varepsilon \in (0,1)$. Note this provides an interesting contrast with the Laplace mechanism, which can achieve $\varepsilon$-DP with standard deviation $\Theta(1/\varepsilon)$ in the low privacy regime.
\section{The Analytic Gaussian Mechanism}\label{sec:aGM}
The limitations of the classical Gaussian mechanism described in the previous section suggest there is room for improvement in the calibration of the variance of a Gaussian perturbation to the corresponding global $L_2$ sensitivity.
Here we present a method for optimal noise calibration for Gaussian perturbations that we call \emph{analytic Gaussian mechanism}.
To do so we must address the two sources of slack in the classical analysis: the sufficient condition \eqref{eqn:suff} used to reduce the analysis to finding an upper bound for $\Pr[\mathcal{N}(\eta,2\eta) > \varepsilon]$, and the use of a Gaussian tail approximation to obtain such upper bound.
We address the first source of slack by showing that the sufficient condition in terms of the privacy loss random variable comes from a relaxation of a necessary and sufficient condition involving two privacy loss random variables. When specialized to the Gaussian mechanism, this condition involves probabilities about Gaussian random variables, which instead of approximating by a tail bound we represent explicitly in terms of the CDF of the standard univariate Gaussian distribution:
\begin{align*}
\Phi(t) = \Pr[\mathcal{N}(0,1) \leq t] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{t} e^{-y^2/2} dy \enspace.
\end{align*}
Using this point of view, we introduce a calibration strategy for Gaussian perturbations that requires solving a simple optimization problem involving $\Phi(t)$. We discuss how to solve this optimization at the end of this section.
The first step in our analysis is to provide a necessary and sufficient condition for differential privacy in terms of privacy loss random variables. This is captured by the following result.
\begin{theorem}\label{thm:necesuff}
A mechanism $M : \mathbb{X} \to \mathbb{Y}$ is $(\varepsilon,\delta)$-DP if and only if the following holds for every $x \simeq x'$:
\begin{align}\label{eqn:necesuff}
\Pr[L_{M,x,x'} \geq \varepsilon] - e^{\varepsilon} \Pr[L_{M,x',x} \leq - \varepsilon] \leq \delta \enspace.
\end{align}
\end{theorem}
Note that Theorem~\ref{thm:necesuff} immediately implies the sufficient condition given in \eqref{eqn:suff} through the inequality
\begin{align*}
\Pr[L_{M,x,x'} \geq \varepsilon] - e^{\varepsilon} \Pr[L_{M,x',x} \leq - \varepsilon] \leq
\Pr[L_{M,x,x'} \geq \varepsilon] \enspace.
\end{align*}
Now we can use Lemma~\ref{lem:privlossGM} to specialize \eqref{eqn:necesuff} for a Gaussian output perturbation mechanism. The relevant computations are packaged in the following result, where we express the probabilities in \eqref{eqn:necesuff} in terms of the Gaussian CDF $\Phi$.
\begin{lemma}\label{lem:LtoCDF}
Suppose $M(x) = f(x) + Z$ is a Gaussian output perturbation mechanism with $Z \sim \mathcal{N}(0,\sigma^2 I)$. For any $x \simeq x'$ let $D = \norm{f(x) - f(x')}$. Then the following hold for any $\varepsilon \geq 0$:
\begin{align}
\Pr[L_{M,x,x'} \geq \varepsilon] &=
\Phi\left(\frac{D}{2 \sigma} - \frac{ \varepsilon \sigma}{D}\right) \label{eqn:L1} \enspace, \\
\Pr[L_{M,x',x} \leq - \varepsilon] &= \Phi\left(- \frac{D}{2 \sigma} - \frac{ \varepsilon \sigma}{D}\right) \label{eqn:L2} \enspace.
\end{align}
\end{lemma}
This result specializes the left hand side of \eqref{eqn:necesuff} in terms of the distance $D = \norm{f(x) - f(x')}$ between the output means on a pair of neighbouring datasets. To complete the derivation of our analytic Gaussian mechanism we need to ensure that \eqref{eqn:necesuff} is satisfied for \emph{every} pair $x \simeq x'$. The next lemma shows that this reduces to plugging the global $L_2$ sensitivity $\Delta$ in the place of $D$ in \eqref{eqn:L1} and \eqref{eqn:L2}.
\begin{lemma}\label{lem:monotone}
For any $\varepsilon \geq 0$, the function $h : \mathbb{R}_{\geq 0} \to \mathbb{R}$ defined as follows is monotonically increasing:
\begin{align*}
h(\eta) = \Pr[\mathcal{N}(\eta,2\eta) \geq \varepsilon] - e^{\varepsilon} \Pr[\mathcal{N}(\eta,2\eta) \leq - \varepsilon] \enspace.
\end{align*}
\end{lemma}
Now we are ready to state our main result, whose proof follows directly from Theorem~\ref{thm:necesuff}, Lemma~\ref{lem:monotone}, and equations \eqref{eqn:L1} and \eqref{eqn:L2}.
\begin{theorem}[Analytic Gaussian Mechanism]\label{thm:aGMDP}
Let $f : \mathbb{X} \to \mathbb{R}^d$ be a function with global $L_2$ sensitivity $\Delta$.
For any $\varepsilon \geq 0$ and $\delta \in [0,1]$, the Gaussian output perturbation mechanism $M(x) = f(x) + Z$ with $Z \sim \mathcal{N}(0,\sigma^2 I)$ is $(\varepsilon,\delta)$-DP if and only if
\begin{align}\label{eqn:N}
\Phi\left(\frac{\Delta}{2 \sigma} - \frac{ \varepsilon \sigma}{\Delta}\right) - e^{\varepsilon} \Phi\left(- \frac{\Delta}{2 \sigma} - \frac{ \varepsilon \sigma}{\Delta}\right) \leq \delta \enspace.
\end{align}
\end{theorem}
This result shows that in order to obtain an $(\varepsilon,\delta)$-DP Gaussian output perturbation mechanism for a function $f$ with global $L_2$ sensitivity $\Delta$ it is enough to find a noise variance $\sigma^2$ satisfying \eqref{eqn:N}. One could now use upper and lower bounds for the tail of the Gaussian CDF to derive an analytic expression for a parameter $\sigma$ satisfying this constraint. However, this again leads to a suboptimal result due to the slack in these tail bounds in the non-asymptotic regime. Instead, we propose to find $\sigma$ using a numerical algorithm by leveraging the fact that the Gaussian CDF can be written as $\Phi(t) = (1 + \mathsf{erf}(t/\sqrt{2}))/2$, where $\mathsf{erf}$ is the standard error function. Efficient implementations of this function to very high accuracies are provided by most statistical and numerical software packages. However, this strategy requires some care in order to avoid numerical stability issues around the point where the expression $\Delta/2 \sigma - \varepsilon \sigma/\Delta$ in \eqref{eqn:N} changes sign. Thus, we further massage the left hand side \eqref{eqn:N} we obtain the implementation of the analytic Gaussian mechanism given in Algorithm~\ref{alg:newGM}. The correctness of this implementation is provided by the following result.
\begin{theorem}\label{thm:aGM}
Let $f$ be a function with global $L_2$ sensitivity $\Delta$. For any $\varepsilon > 0$ and $\delta \in (0,1)$, the mechanism described in Algorithm~\ref{alg:newGM} is $(\varepsilon,\delta)$-DP.
\end{theorem}
\begin{algorithm}[t]
\DontPrintSemicolon
\caption{Analytic Gaussian Mechanism}\label{alg:newGM}
\KwSty{Public Inputs:} $f$, $\Delta$, $\varepsilon$, $\delta$\;
\KwSty{Private Inputs:} $x$\;
Let $\delta_0 = \Phi(0) - e^{\varepsilon} \Phi(- \sqrt{2 \varepsilon})$\;
\eIf{$\delta \geq \delta_0$}{
Define $B_{\varepsilon}^+(v) = \Phi(\sqrt{\varepsilon v})
- e^{\varepsilon} \Phi(-\sqrt{\varepsilon (v + 2)})$\;
Compute $v^* = \sup \{ v \in \mathbb{R}_{\geq 0} : B_{\varepsilon}^+(v) \leq \delta \}$\;
Let $\alpha = \sqrt{1+ v^*/2} - \sqrt{v^*/2}$\;
}{
Define $B_{\varepsilon}^-(u) = \Phi(-\sqrt{\varepsilon u})
- e^{\varepsilon} \Phi(-\sqrt{\varepsilon (u + 2)})$\;
Compute $u^* = \inf \{ u \in \mathbb{R}_{\geq 0} : B_{\varepsilon}^-(u) \leq \delta \}$\;
Let $\alpha = \sqrt{1 + u^*/2} + \sqrt{u^*/2}$\;
}
Let $\sigma = \alpha \Delta / \sqrt{2 \varepsilon}$\;
Return $f(x) + \mathcal{N}(0,\sigma^2 I)$
\end{algorithm}
Given a numerical oracle for computing $\Phi(t)$ based on the error function it is relatively straightforward to implement a solver for finding the values $v^*$ and $u^*$ needed in Algorithm~\ref{alg:newGM}. For example, using the fact that $B_{\varepsilon}^+(v)$ is monotonically increasing we see that computing $v^*$ is a root finding problem for which one can use Newton's method since the derivative of $\Phi(t)$ can be computed in closed form using Leibniz's rule. In practice we find that a simple scheme based on binary search initiated from an interval obtained by finding the smallest $k \in \mathbb{N}$ such that $B_{\varepsilon}^+(2^k) > \delta$ provides a very efficient and robust way to find $v^*$ up to arbitrary accuracies (the same applies to $u^*$).
\section{Optimal Denoising}\label{sec:denoise}
Can we improve the performance of analytical Gaussian mechanism even further? The answer is ``yes'' and ``no''. We can't because Algorithm~\ref{alg:newGM} is already the exact calibration of the Gaussian noise level to the given privacy budget. But if we consider the problem of designing the best differentially private procedure $M(x)$ that approximates $f(x)$, then there could still be room for improvement.
In this section, we consider a specific class of mechanisms that \emph{denoise} the output of a Gaussian mechanism. Let $\hat{y} \sim \mathcal{N}(f(x),\sigma^2 I)$, we are interested in designing a post-processing function $g$ such that $\tilde{y} = g(\hat{y})$ is closer to $f(x)$ than $\hat{y}$. This class of mechanisms are of particular interest for differential privacy because (1) since differential privacy is preserved by post-processing, releasing a function $\tilde{y} = g(\hat{y})$ of a differentially private output is again differentially private; (2) since information about $f$ and the distribution of the noise are publicly known, this information can be leveraged to design denoising functions.
This is a statistical estimation problem, where $f(x)$ is the underlying parameter and $\hat{y}$ is the data. Since in this case we are adding the noise ourselves, it is possible to use the classical statistical theory on Gaussian models \emph{as is} because the Gaussian assumption is now true by construction.
This is however an unusual estimation problem where all we observe is a single data point. Since $\hat{y}$ is the maximum likelihood estimator, if there is no additional information about $f(x)$, we cannot hope to improve the estimation error \emph{uniformly} over all $f(x) \in \mathbb{R}^d$. But there is still something we can do when we consider either of the following assumptions:
(A.1) $x$ is drawn from some underlying distribution, thus inducing some distribution on $f(x)$; or,
(A.2) $\|f(x)\|_p \leq B$ for some $p, B> 0$, where $\|\cdot\|_p$ is the $L_p$-norm (or pseudo-norm when $p < 1$).
\paragraph{Optimal Bayesian denoising.}
Assumption A.1 translates the problem of optimal denoising into a Bayesian estimation problem, where the underlying parameter $f(x)$ has a prior distribution, and the task is to find an estimator that attains the Bayes risk --- the minimum of the average estimation error integrated over a prior $\pi$, defined as
$$ R(\pi) = \min_{g: \mathbb{R}^d \rightarrow \mathbb{R}^d } \mathbb{E}\big[ \mathbb{E} [\norm{g(\hat{y}) - f(x)}^2 | f(x)] \big] \enspace.$$
For square loss, the Bayes estimator is simply the posterior mean estimator, as the following theorem shows:
\begin{theorem}\label{thm:bayes}
Let $x \sim \pi$ and assume the induced distribution of $f(x)$ is square integrable.
Then the Bayes estimator $\tilde{y}_{\mathrm{Bayes}}$ is given by
$$
\tilde{y}_{\mathrm{Bayes}} = \mathop{\mathrm{argmin}}_{g: \mathbb{R}^d \rightarrow \mathbb{R}^d} \mathbb{E}\left[\|g(\hat{y}) - f(x)\|^2\right] = \mathbb{E}[ f(x) | \hat{y}] \enspace.
$$
\end{theorem}
The proof can be found in any standard statistics textbook \citep[see, e.g.,][]{lehmann2006theory}.
One may ask what the corresponding MSE is and how much it improves over the version without post-processing. The answer depends on the prior and the amount of noise added for differential privacy. When $f(x)\sim \mathcal{N}(0, w^2I)$, the posterior mean estimator can be written analytically into
$\tilde{y}_{\mathrm{Bayes}} = (w^2 / (w^2+\sigma^2)) \hat{y}$,
and the corresponding Bayes risk is
$\mathbb{E}[\|\tilde{y}_{\mathrm{Bayes}} - f(x)\|^2] = d w^2 \sigma^2 / (\sigma^2 + w^2)$.
In other word, we get a factor of $w^2/(w^2+\sigma^2)$ improvement over simply using $\hat{y}$.
In general, there is no analytical form for the posterior mean, but if we can evaluate the density of $f(x)$ or sample from the distribution of $x$, then we can obtain an arbitrarily good approximation of $\tilde{y}_{\mathrm{Bayes}}$ using Markov Chain Monte Carlo techniques.
\paragraph{Optimal frequentist denoising.}
Assumption A.2 spells out a minimax estimation problem, where the underlying parameter $f(x)$ is assumed to be within a set $S\subset \mathbb{R}^d$. In particular, we are interested in finding $\tilde{y}_{\mathrm{minimax}}$ that attains the minimax risk
$$
R(S) = \min_{g: \mathbb{R}^d \rightarrow \mathbb{R}^d } \max_{f(x)\in S } \mathbb{E} \left[\|g(\hat{y}) - f(x)\|^2\right],
$$
on $L_p$ balls $S = \mathcal{B}(p,B) = \{ y \in \mathbb{R}^d \;|\; \|y\|_p \leq B \}$ of radius $B$.
A complete characterization of this minimax risk (up to a constant) is given by \citet[Proposition~5]{birge2001gaussian}, who show that in the non-trivial region\footnote{When $\sqrt{\log d} \leq B/\sigma \leq c_p d^{1/p}$ for a constant $c_p$ that depends only on $p$.} of the signal to noise ratio $B/\sigma$, the ball $S = \mathcal{B}(p,B)$ satisfies
\begin{equation}\label{eq:minimax_pnorm_ball}
R(S) = \Theta\left(B^p\sigma^{2-p} \left(1 + \log\left(\frac{d\sigma^p}{B^p}\right)\right)^{1-p/2}\right)
\end{equation}
for $0<p<2$ and when $p\geq 2$, \citet{donoho1990minimax} show that
$$R(S) = \Theta\left(\frac{B^2\sigma^2}{\sigma^2 + B^2/d}\right).$$
Deriving exact minimax estimators is challenging and most analyses assume certain asymptotic regimes (see the case for $p=2$ by \citet{bickel1981minimax}). Nonetheless, some techniques have been shown to match $R(\mathcal{B}(p,B))$ up to a small constant factor in the finite sample regime \citep[see, e.g.,][]{donoho1990minimax,donoho1994minimax}.
This means that we can often improve the square error from $d \sigma^2$ to $R(\mathcal{B}(p,B))$ when we have the additional information that $f(x)$ is in some $L_p$ ball. This could be especially helpful in the high-dimensional case for $p<2$. For instance if $p=1$ and $B=\sigma$, then we obtain a risk $\sigma B \sqrt{1+\log(d \sigma/B )}$, which improves exponentially in $d$ over the $d\sigma^2$ risk of $\hat{y}$. More practically, if $f(x)$ is a sparse histogram with $s$ non-zero elements, then taking $p\rightarrow 0$ will result in an error bound on the order of $s \sigma^2 (1+\log(d))$, which is linear in the sparsity $s$ rather than the dimension $d$.
\paragraph{Adaptive estimation.}
What if we do not know the prior parameter $w^2$, or a right choice of $B$ and $p$? Can we still come up with estimators that take advantage of these structures? It turns out that this is the problem of designing \emph{adaptive} estimators which sits at the heart of statistical research.
An \emph{adaptive} estimator in our case, is one that does not need to know $w^2$ or a pair of $B$ and $p$, yet behave nearly as well as Bayes estimator that knows $w^2$ or the minimax estimator that knows $B$ and $p$ for each parameter regime.
We first give an example of an adaptive Bayes estimator that does not require us to specify a prior, yet can perform almost as well as the optimal Bayes estimator for all isotropic Gaussian prior simultaneously.
\begin{theorem}[James-Stein estimator and its adaptivity]\label{thm:js}
When $d\geq3$, substituting $w^2$ in $\tilde{y}_{\mathrm{Bayes}}$ with its maximum likelihood estimate under
$$f(x)\sim \mathcal{N}(0,w^2 I) \enspace, \;\; \hat{y} | f(x) \sim \mathcal{N}(f(x), \sigma^2 I) $$
produces the James-Stein estimator
$$
\tilde{y}_{\mathrm{JS}} = \left( 1- \frac{(d-2)\sigma^2}{\|\hat{y}\|^2}\right)\hat{y}.
$$
Moreover, it has an MSE
$$
\mathbb{E} \left[\|\tilde{y}_{\mathrm{JS}} - f(x)\|^2\right] = d\sigma^2 \left( 1 - \frac{(d-2)^2}{d^2} \frac{\sigma^2}{w^2+\sigma^2} \right).
$$
\end{theorem}
The James-Stein estimator has the property that it always improves the MLE $\hat{y}$ when $d\geq 3$ \citep{stein1956inadmissibility} and it always achieves a risk that is within a $d^2/(d-2)^2$ multiplicative factor of the Bayes risk of $\tilde{y}_{\mathrm{Bayes}}$
for any $w^2$.
We now move on to describe a method that is adaptive to $B$ and $p$ in minimax estimation. Quite remarkably, \citet{donoho1995noising} shows that choosing $\lambda = \sigma\sqrt{2 \log d}$ in the soft-thresholding estimator
\begin{equation}\label{eq:soft-thresh}
\tilde{y}_{\mathrm{TH}} = \text{sign}(\hat{y})\max\{0,|\hat{y}| - \lambda\}
\end{equation}
yields a nearly optimal estimator for every $L_p$ ball.
\begin{theorem}[The adaptivity of soft-thresholding, Theorem 4.2 of \citep{donoho1995noising}]\label{thm:don}
Let $S = \mathcal{B}(p,B)$ for some $p, B > 0$. The soft-thresholding estimator with $\lambda = \sigma\sqrt{2 \log d}$ obeys that
$$\sup_{f(x)\in S}\mathbb{E}\left[\|\tilde{y}_{\mathrm{TH}} - f(x)\|^2\right] \leq (2\log d + 1)(\sigma^2 + 2.22 R(S)) \enspace.$$
\end{theorem}
The result implies that the soft-thresholding estimator is nearly optimal for all balls up to a multiplicative factor of $4.44\log(d)$.
Thanks to the fact that we know the parameter $\sigma$ exactly, both $\tilde{y}_{\mathrm{JS}} $ and $\tilde{y}_{\mathrm{TH}}$ are now completely free of tuning parameters. Yet, they can achieve remarkable adaptivity that covers a large class of model assumptions and function classes. A relatively small price to pay for such adaptivity is that we might lose a constant (or a $\log(d)$) factor. Whether such adaptivity is worth will vary on a case-by-case basis.
Take the problem of private releasing a histogram of $n$ items in $d$ bins. Theorem~\ref{thm:don} and Equation \eqref{eq:minimax_pnorm_ball} with $p\leq 1$ imply that the soft-thresholding estimator obeys
$$
\mathbb{E} \left[\|\tilde{y}_{\mathrm{TH}} - f(x)\|^2\right] = \tilde{O} \left( \min\{s\sigma^2, n^{1/k}\sigma^{2-1/k}\}\right).
$$
where $s$ denotes the number of nonzero elements in $f(x)$ and $k$ is the largest power-law exponent greater than $1$ such that order statistics $f(x)^{(d-i+1)} \leq n i^{-k}$ for all $i=1,...,d$ and $\tilde{O}$ hides logarithmic factors in $d, d\sigma/n$.
The fact that $s \leq d$ implies that the soft-thresholding estimator improves over the naive private release for all $d,n,s$ and the $n^{1/k}$ factor suggests that we can take advantage of an unknown power law distribution even if the histogram is not really sparse. This makes our technique effective in the many data mining problems where power law distributions occur naturally \citep{faloutsos1999power}.
\paragraph{Related work.}
Denoising as a post-processing step in the context of differentially privacy is not a new idea. Notably, \citet{barak2007privacy,hay2009accurate} show that a post-processing step enforcing consistency of contingency table releases and graph degree sequences leads to more accurate estimations. \citet{williams2010probabilistic} sets up the general statistical (Bayesian) inference problem of DP releases by integrating auxiliary information (a prior). \citet{karwa2016inference} exploits knowledge of the noise distribution use to achieve DP in the inference procedure of a network model and shows that it helps to preserve asymptotic efficiency. \citet{nikolov2013geometry} demonstrates that projecting linear regression solutions to a known $\ell_1$-ball improves the estimation error from $O(\mathrm{poly}(d))$ to $O(\mathrm{polylog}(d))$ when the underlying ground truth is sparse. \citet{bernstein2017differentially} uses Expectation--Maximization to denoise the parameters of a class of graphical models starting from noisy empirical moments obtained using the Laplace mechanism.
In all the above references there is some prior knowledge (constraint sets, sparsity or Bayesian prior) that is exploited to improve the utility of DP releases.
To the best of our knowledge, we are the first to consider ``adaptive estimation'' and demonstrate how classical techniques can be helpful even without such prior knowledge.
These estimators are not new; they have been known in the statistics literature for decades.
Our purpose is to compile facts that are relevant to the practice of DP and initiate a systematic study of how these ideas affect the utility of DP mechanisms, which we complement with the experimental evaluation presented in the next section.
\section{Numerical Experiments}\label{sec:exp}
This section provides an experimental evaluation of the improvements in utility provided by optimal calibration and adaptive denoising. First we numerically compare the variance of the analytic Gaussian mechanism and the classical mechanism for a variety of privacy parameters. Then we evaluate the contributions of denoising and analytic calibration against a series of baselines for the task of private mean estimation using synthetic data.
We also evaluate several denoising strategies on the task of releasing heat maps based on the New York City taxi dataset under differential privacy.
Further experiments are presented in Appendix~\ref{app:exp}, including an evaluation of denoising strategies for the task of
private histogram release.
\subsection{Analytic Gaussian Mechanism}\label{sec:expaGM}
\begin{figure*}[t]
\begin{center}
\begin{subfigure}[b]{0.244\textwidth}
\includegraphics[width=\textwidth]{./plots/newGM-plot.pdf}
\end{subfigure}
\begin{subfigure}[b]{0.244\textwidth}
\includegraphics[width=\textwidth]{./plots/sigma-ratio-plot.pdf}
\end{subfigure}
\begin{subfigure}[b]{0.244\textwidth}
\includegraphics[width=\textwidth]{./plots/mean_all_e001.pdf}
\end{subfigure}
\begin{subfigure}[b]{0.244\textwidth}
\includegraphics[width=\textwidth]{./plots/mean_denoise.pdf}
\end{subfigure}
\caption{Two leftmost plots: Experiments comparing the classical Gaussian mechanism (cGM) and the analytic Gaussian mechanism (aGM), in terms of their absolute standard deviations as $\varepsilon \to 0$, and in terms of the gain in variance as a function of $\varepsilon$. Two rightmost plots: Mean estimation experiments showing $L_2$ error between the private mean estimate and the non-private empirical mean as a function of the dimension $d$.} \label{fig:exp-synth}
\end{center}
\end{figure*}
We implemented Algorithm~\ref{alg:newGM} in Python\footnote{See \url{https://github.com/BorjaBalle/analytic-gaussian-mechanism}.} and ran experiments to compare the variance of the perturbation obtained with the analytic Gaussian mechanism versus the variance required by the classical Gaussian mechanism. In all our experiments the values of $v^*$ and $u^*$ were solved up to an accuracy of $10^{-12}$ using binary search and the implementation of the $\mathsf{erf}$ function provided by SciPy \cite{scipy}.
The results are presented in the two leftmost panels in Figure~\ref{fig:exp-synth}. The plots show that as $\varepsilon \to 0$ the optimally calibrated perturbation outperforms the classical mechanism by several orders of magnitude.
Furthermore, we see that even for values of $\varepsilon$ close to $1$ our mechanism reduces the variance by a factor of $1.4$ or more, with higher improvements for larger values of $\delta$.
\subsection{Denoising for Mean Estimation}\label{sec:expmean}
Our next experiment evaluates the utility of denoising combined with the analytical Gaussian mechanism for the task of private mean estimation. The input to the mechanism is a dataset $x = (x_1, \ldots, x_n)$ containing $n$ vectors $x_i \in \mathbb{R}^d$ and the underlying deterministic functionality is $f(x) = (1/n) \sum_{i = 1}^n x_i$. This relatively simple task is a classic example from the family of \emph{linear queries} which are frequently considered in the differential privacy literature. We compare the accuracy of several mechanisms $M$ for releasing a private version of $f(x)$ in terms of the Euclidean distance $\norm{M(x) - f(x)}_2$. In particular, we test the analytical Gaussian mechanism with either James-Stein denoising cf.\ Theorem~\ref{thm:js} (aGM-JS) or optimal thresholding denoising cf.\ Theorem~\ref{thm:don} (aGM-TH), as well as several baselines including: the classical Gaussian mechanism (cGM), the analytical Gaussian mechanism without denoising (aGM), and the Laplace mechanism (Lap) using the same $\varepsilon$ parameter as the Gaussian mechanisms.
To provide a thorough comparison we explore of the different parameters of the problem on the final utility. The key parameters of the problem are the dimension $d$ and the DP parameters $\varepsilon$ and $\delta$. The dimension affects the utility through the bounds provided in Theorem~\ref{thm:js} and Theorem~\ref{thm:don}. The DP parameters affect the utility through the variance $\sigma^2$ of the mechanism, which is also affected by the sample size $n$ via the global sensitivity. Thus, we can characterize the effect of $\sigma^2$ by keeping $n$ fixed and changing the DP parameters. In our experiments we consider a fixed sample size $n = 500$ and privacy parameter $\delta = 10^{-4}$ while trying several values for $\varepsilon$.
The other parameter that affects the utility is the ``size'' of $f(x)$, controlled either through the variance $w^2$ or the norm ball $S$. Since the denoising estimators we use are adaptive to these parameters and do not need to know them in advance, we sample the dataset $x$ repeatedly to obtain a diversity of values for $f(x)$. Each dataset $x$ is sampled as follows: first sample a center $x_0 \sim \mathcal{N}(0,I)$ and then build $x = (x_1,\ldots,x_n)$ with $x_i = x_0 + \xi_i$, where each $\xi_i$ is i.i.d.\ with independent coordinates sampled uniformly from the interval $[-1/2,1/2]$. Thus, in each dataset the points $x_i$ all lie in an $L_{\infty}$-ball of radius $1$, leading to a global $L_2$ sensitivity $\Delta_2 = \sqrt{d}/n$ and a global $L_1$ sensitivity $\Delta_1 = d/n$. These are used to calibrate the Gaussian and Laplace perturbations, respectively.
The results are presented in two rightmost panels of Figure~\ref{fig:exp-synth}. Each point in every plot is the result of averaging the error over $100$ repetitions with different datasets. The first plot uses $\varepsilon = 0.01$ and shows how denoised methods improve the accuracy over all the other methods, sometimes by orders of magnitude. The second plot shows that for this problem the James-Stein estimator provides better accuracy in the high-dimensional setting.
\subsection{New York City Taxi Heat Maps}\label{sec:exptaxi}
\begin{figure*}
\captionsetup[subfigure]{justification=centering}
\begin{subfigure}[t]{0.245\textwidth}
\includegraphics[width=\textwidth]{./figures/24_groundtruth_hour.pdf
\caption*{Ground truth}
\end{subfigure}
\begin{subfigure}[t]{0.245\textwidth}
\includegraphics[width=\textwidth]{./figures/24_noisy.pdf
\caption*{Raw aGM\\ RMSE = 40.29}
\end{subfigure}
\begin{subfigure}[t]{0.245\textwidth}
\includegraphics[width=\textwidth]{./figures/24denoised_wavelets.pdf
\caption*{Wavelets\\ RMSE = 12.28}
\end{subfigure}
\begin{subfigure}[t]{0.245\textwidth}
\includegraphics[width=\textwidth]{./figures/24_denoised_tf_lambda_0_5.pdf
\caption*{Trend filtering\\ RMSE = 10.07}
\end{subfigure}
\caption{Illustration of the denoising in differentially private release of NYC taxi density during 00:00 - 01:00 am Sept 24, 2014. From left to right, the figures represent the true data, the output of the analytical Gaussian mechanism, the reconstructed signal from soft-thresholded wavelet coefficients with spanning-tree wavelet transform \citep{sharpnack2013detecting}, and the results of trend filtering on graphs \citep{wang2016}. We observe that adding appropriate post-processing significantly reduces the estimation error and also makes the underlying structures visible.
} \label{fig:illus_nyc_taxi}
\end{figure*}
In this section, we apply our method to New York City taxi data. The dataset is a collection of time-stamped pick-ups and drop-offs of taxi drivers and we are interested in sharing a density map of such pick-ups and drop-offs in Manhattan at a specific time of a specific day under differential privacy.
This is a problem of significant practical interest. Ever since the NYC Taxi \& Limousine Commission released this dataset, there has been multiple independent reports concerning the security and privacy risks this dataset poses for taxi drivers and their passengers \citep[see, e.g.,][]{taxi_rainbow,douriez2016anonymizing}. The techniques presented in this paper allow us to provably prevent individuals (on both the per-trip level and per-cab level) in the dataset from being identified, while remarkably, permitting the release of rich information about the data with fine-grained spatial and temporal resolution.
Specifically, we apply the analytical Gaussian mechanism to release the number of picks-ups and drop-offs at every traffic junction in Manhattan. There are a total of 3,784 such traffic junctions and they are connected by 7,070 sections of roads. We will treat them as nodes and edges on a graph. In the post-processing phase, we apply graph smoothing techniques to reveal the underlying signal despite the noise due to aGM. Specifically, we compare the JS-estimator and the soft-thresholding estimator we described in Section~\ref{sec:denoise}, as well as the same soft-thresholding estimator applied to the coefficients of a graph wavelet transform due to \citet{sharpnack2013detecting}. The basis transformation is important because the data might be sparser in the transformed domain. For reference, we also include the state-of-the-art graph smoothing techniques called graph trend filtering \citep{wang2016}, which has one additional tuning parameter but has been shown to perform significantly better than wavelet smoothing in practice.
Our experiments provide cab-level differential privacy by assuming that every driver does a maximum of $5$ trips within an hour
so that we have a global $L_2$-sensitivity of $\Delta = 5$.
This is a conservative but reasonable estimate and can be enforced by preprocessing the data.
Data within each hour is gathered and distributed to each traffic junction using a kernel density estimator; further details are documented in \citet{doraiswamy2014using}.
We present some qualitative comparisons in Figure~\ref{fig:illus_nyc_taxi}, where we visualize the privately released heat map with and without post-processing.
Relatively speaking, trend filtering performs better than wavelet smoothing, but both approaches significantly improves the RMSE over the DP release without post-processing.
The results in Appendix~\ref{app:exp} provide quantitative results by comparing the mean square error of cGM, aGM as well as the aforementioned denoising techniques for data corresponding to different time intervals.
\section{Conclusion and Discussion}
In this paper, we embark on a journey of pushing the utility limit of Gaussian mechanism for $(\varepsilon,\delta)$-differential privacy. We propose a novel method to obtain the optimal calibration of Gaussian perturbations required to attain a given DP guarantee. We also review decades of research in statistical estimation theory and show that combining these techniques with differential privacy one obtains powerful \emph{adaptivity} that denoises differentially private outputs nearly optimally without additional hyperparameters. On synthetic data and on the New York City Taxi dataset we illustrate a significant gain in estimation error and fine-grained spatial-temporal resolution.
There are a number of theoretical problems of interest for future work. First, on the problem of differentially private estimation. Our post-processing approach effectively restricts our choice of algorithms to the composition of privacy release and post-processing. While we now know that we are optimal in both components, it is unclear whether we lose anything relative to the best differentially private algorithms.
Secondly, the analytical calibration proposed in this paper is optimal for achieving $(\varepsilon,\delta)$-DP with Gaussian noise. But when building complex mechanisms we are stuck in the dilemma of choosing between (a) using the aGM with the advanced composition \citep{kairouz2015composition}; or, (b) using R{\'e}nyi DP \citep{mironov2017renyi} or zCDP \citep{bun2016concentrated} for tighter composition and calculate the $(\varepsilon,\delta)$ from moment bounds.
While (a) is tighter in the calculation the privacy parameters of each intermediate value, (b) is tighter in the composition but cannot take advantage of aGM.
It would be interesting if we could get the best of both worlds.
\section*{Acknowledgments}
We thank \citet{doraiswamy2014using} for sharing their preprocessed NYC taxi dataset, the anonymous reviewers for helpful comments that led to improvements of the paper and Stephen E. Fienberg for discussions that inspired the authors to think about optimal post-processing.
| {
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Olof Andersson, eller Andersson i Höör, född 14 januari 1875 i Östra Sallerups församling, död 1 april 1938 i Höör, var en svensk skomakare och politiker.
Olof Andersson var ledamot av riksdagens andra kammare 1919–1938, invald i Malmöhus läns mellersta valkrets (till 1921) och Malmöhus läns valkrets (från 1922).
Källor
Noter
Ledamöter av Sveriges riksdags andra kammare för Socialdemokraterna
Svenska socialdemokrater
Män
Födda 1875
Avlidna 1938
Personer från Östra Sallerups socken | {
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Home > content > FlexiWAN Adopts an 'Open' Slant
FlexiWAN Adopts an 'Open' Slant
Created 15/04/2019 - 7:56am
Submitted by Roy Schestowitz on Monday 15th of April 2019 07:56:50 AM Filed under
OSS [1]
Security [2]
Stealthy Start-Up Portends 'Second Wave of SD-WAN' [3]
The First SD-WAN Open Source Driving the Second Wave of SD-WAN by flexiWAN [4]
flexiWAN Launches With Open Source SD-WAN Architecture [5]
Will open source usher in the second-wave of SD-WAN? Startup flexiWAN's co-founder and CEO Amir Zmora thinks so.
FlexiWAN soft launches SD-WAN software based on open source architecture [6]
Israel-based start-up FlexiWAN has started conducting proof-of-concept trials to test its SD-WAN software product, which aims to use open source architecture as a differentiator. With this approach, the company hopes to attract IT managers by providing more control over the capabilities and elements within their networks.
FlexiWAN pushes SD-WAN into an open source architecture [7]
Among the goals of flexiWAN co-founder and CEO Amir Zmora is to give enterprises and service providers the ability to differentiate their SD-WAN services instead of relying on SD-WAN vendors to define them.
After years of working in the VoIP space, and after attending numerous industry conferences where SD-WAN was a hot topic, Zmora said that he came to the realization that SD-WAN solutions were closed black boxes that didn't enable innovation.
Chua said he has been waiting to see an open-source approach to SD-WAN. He said there were two elements to SD-WAN; the SD-WAN element and the universal CPE element.
"So, on the SD-WAN side of things, which is, I think, where he's (Zmora) starting, there are elements in place in open source where you can try to cobble things together to make an SD-WAN solution," Chua said. "So, there's IPSec or an open SSL VPN, firewalls, things like that.
"What's missing is that cloud control policy elements that aren't quite there. So, there's no open source equivalent, that I know of, on the whole cloud control side for the centralized policies, centralized configuration and of all the different SD-WAN components out there."
[1] http://www.tuxmachines.org/taxonomy/term/72
[3] https://virtualizationreview.com/articles/2019/04/10/flexiwan-debut.aspx
[4] https://www.prnewswire.com/news-releases/the-first-sd-wan-open-source-driving-the-second-wave-of-sd-wan-by-flexiwan-300827827.html
[5] https://www.sdxcentral.com/articles/news/flexiwan-launches-with-open-source-sd-wan-architecture/2019/04/
[6] https://www.telecompaper.com/news/flexiwan-soft-launches-sd-wan-software-based-on-open-source-architecture--1288376
[7] https://www.fiercetelecom.com/telecom/flexiwan-pushes-sd-wan-into-open-source-architecture | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 3,902 |
Q: Searching for all sentences containing a string in another document I have a file containing 200 words, each on a new line.
I want to search for all these words in another file. I want each sentence containing one of these words to be printed.
Right now, only the matches of the first word appear. After that, it stops.
corpus = open('C:\\Users\\Lucas\\Desktop\\HAIT\\Scriptie\\Tweet-corpus\\Corpus.txt', 'r', encoding='utf8')
with open('C:\\Users\\Lucas\\Desktop\\HAIT\\Scriptie\\Tweet-corpus\\MostCommon3.txt', 'r', encoding='utf8') as list:
for line in list:
for a in corpus:
if line in a:
print(a)
A: # Prepare the list of words
word_file = open('wordfile', 'r', encoding='utf8')
words = [word.strip() for word in word_file.readlines()]
word_file.close()
# Now examine each sentence:
with open('sentencefile') as sentences:
for sentence in sentences:
found = False
for word in words:
if word in sentence:
found = True
break
if found:
print sentence
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Polypodium může znamenat:
osladič – rod kapradiny
osladič obecný (Polypodium vulgare)
osladič přehlížený (Polypodium interjectum)
nezmar jeseteří (Polypodium hydriforme) – parazitický žahavec | {
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} | 8,541 |
Q: Relation between conservative force and potential energy Couple of days before I came across a question something like this
A particle is moving in a conservative force field from point $A$ to point $B$. Let $U_A$ and $U_B$ be the potential energies of particle at points $A$ and $B$ respectively and $W_C$ is the work done in the process in moving the particle from $A$ to $B$. (Take work done to be positive)
And we needed to find the correct relations among some alternatives. What I figured out is as follows:
*
*Potential Energy of particle is equal to $negative$ of $work$ $done$ by $conservative$ force.
*And Work done is equal to Change in Potential energy of particle.
So I found the alternatives as $W_C$=$-(U_B-U_A)$ i.e $W_C$=$(U_A-U_B)$ as correct , but correct answer is $W_C$=$(U_B-U_A)$. I think there is some trick in the last statement of question that "(Take work done to be positive)
"
But if I take work done to be positive then $U_B$<$U_A$ but this also contradicts the answer as the answer is $U_B$>$U_A$
This made a huge confusion in my mind. Please tell me if I am missing some concept. And do tell me the correct answers too.
A: The work done that we consider here is the work done by external force (or by us).
So if the potential energy at a point is high then it means that the work done by us against the conservative field (say, gravity) will be also be high.
For example, if we lift a body, it's potential energy increases with the height because we are doing work against the conservative force (i.e, gravitational force) and this work (+ve work) gets stored in the body as its potential energy.
In other words, higher we lift the body higher will be its potential energy.
Energy conservation plays a very important role here.
Now, if the body comes down the work done by gravity will be positive. The body's potential energy will be converted into kinetic energy. The body is coming from a point of high potential energy to a point of lower potential energy, so this loss will be the gain in the kinetic energy (or the work done by gravity).
Therefore, the work done by the conservative force (gravity) will be equal to the loss in potential energy.
Since the work done by gravity is $+ve$ and there is a loss in potential energy $(\Delta U=-ve)$,
$W_{conservative}=-\Delta U$
A: The problem is that it is not clear from the question which force is doing the work.
Is $F_{\rm C}$ the external force or the force due to the conservative field?
The change in potential energy of a particle in going from position $A,\, (U_{\rm A})$ to position $B,\, (U_{\rm B})$ is the work done by an external force in moving the particle from $A$ to $B,\,(W_{\rm external, AB})$
$$W_{\rm external, AB}= U_{\rm B} - U_{\rm A}$$
[Liftng a mass $m$ in a gravitaional field strength $g$ a height $h$ you (the provider of the external force) would have to exert an upward force $mg$ and move it through a distance $h$ doing and amount of work equal to $mgh$ which is the change in potential energy.]
The work done in moving the particle from $A$ to $B$ by the force due to the conservative field is
$$W_{\rm field, AB}= -(U_{\rm B} - U_{\rm A})= - W_{\rm external, AB}$$
[The force due to the conservative field $mg$ is downwards and in the opposite direction to the ditection in which the mass is move upward a distance $h$ , so the work done by the force due to the conservative field is $-mgh$ and that by definition is minus the change in potential energy.
So again the change in potential energy is $mgh$.]
A: Potential Energy of particle is equal to negative of work done by conservative force.
And Work done is equal to negative Change in Potential energy of particle. As it is loss of energy from system. So,
$$W=-(-(U_b-U_a))$$
$$W=U_b-U_a$$
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 647 |
\section{Introduction}
\vspace{-0.1cm}
Since the introduction of end-to-end neural diarization (EEND)~\cite{fujita2019enda} and its extension to deal with arbitrary amounts of speakers~\cite{fujita2020neural,horiguchi2020end}, it has been established as a state-of-the-art alternative to the standard cascaded diarization systems based on different submodules, i.e. voice activity detection (VAD), uniform segmentation, speaker embeddings extraction, clustering and overlapped speech detection (OSD) with handling.
EEND formulates the speaker diarization problem as a per-speaker-per-time-frame binary classification problem where a permutation-free objective is used to minimize the speech activity error for all speakers. Therefore, EEND models generate one speech activity probability output for each speaker per time-step, which effectively solves VAD, speaker labeling and OSD at once.
Most works following the EEND principle have focused on improvements on the architecture or modeling. Some by using self-attention layers~\cite{fujita2019endb} or conformer layers~\cite{chieh2021end} instead of the original BLSTM layers for feature encoding; others have focused on more complex diarization scenarios such as its online fashion~\cite{han2021bw,xue2021onlineA} or when more than one microphone is available~\cite{horiguchi2021multi} or by improving the model iteratively using pseudo-labels~\cite{takashima2021end}. Some have used EEND together with more standard approaches by using EEND-inspired models to find overlaps among pairs of speakers in the output of a cascaded system~\cite{horiguchi2021end} or leveraging EEND's VAD performance by using an external VAD system~\cite{horiguchi2021hitachi} or combining short duration diarization outputs to produce better whole-utterance diarization~\cite{kinoshita2021integrating,kinoshita2021advances,horiguchi2021towards,kinoshita2022tight}. However, none have yet tackled one of the main aspects of EEND: training data generation.
EEND models require large amounts of training data and datasets manually annotated for diarization do not amount for thousands of hours. When presenting the first version of EEND, Yusuke et al. proposed a strategy for constructing simulated mixtures using telephone conversations from different collections and this strategy has been used with both telephony data or read books to create simulated mixtures by mixing speakers from different recordings. However, little analysis has been presented about how the simulations were devised nor what impact have the used augmentations. Furthermore, the mixtures do not resemble real conversations, specially when more than two speakers are included. In this work we revise the approach and propose an alternative method that, based on statistics from real conversations, emulates some of the attributes of natural conversations. We also analyze the impact of using different types of augmentations on the diarization performance and show that our approach performs better than the original one while significantly reducing the dependence on the fine-tuning stage.
\vspace{-0.1cm}
\section{Training data generation strategy}
\vspace{-0.1cm}
Diarization training data consist of audio recordings and their corresponding speaker segment annotations. In this section, we describe the original approach~\cite{fujita2019enda} for creating the training data to which, respecting the original terminology, we will refer as simulated mixtures (SM) as well as our approach to which we will refer as simulated conversations (SC).
\vspace{-0.2cm}
\subsection{Simulated mixtures}
\vspace{-0.1cm}
In order to create the mixtures with their corresponding annotations, Yusuke et al. have used a VAD system run on each side of each conversation in a pool of thousands of telephone conversations. Assuming a single speaker per telephone channel, this means that speech segments and their speaker labels can be gathered from the pool to construct mixtures.
To create a mixture, as many speakers as wanted in the mixture ($N_{spk}$) are sampled from the total pool. The pool is represented by $\mathcal{U} = \{ U_s \}_{s \in \mathcal{S}}$ where $U_s$ is an utterance of speaker $s$ formed by all the segments denoted by the VAD that belong to that speaker in the utterance\footnote{Please note that Yusuke et al. have referred as ``utterances'' to the segments of speech in a real conversation.}. For each selected speaker, one of their utterances is randomly sampled and $N_u$ consecutive segments selected randomly from it.
Pauses are introduced in between the selected segments of a speaker to simulate turns in a conversation, where the length of the pause is sampled from an exponential distribution with parameter $\beta$. The resulting audio defines a channel for that speaker in the mixture. This process is repeated for all speakers in the mixture and finally the channels are summed to obtain a single utterance. This procedure is enriched by adding background noises and altering the room impulse responses (RIRs) of each speaker channel. For the sake of space we do not include the algorithm but refer the reader to Algorithm 1 of~\cite{fujita2019enda}. Note that since the channel of each speaker is generated independently, depending on the choice of $\beta$, the output can contain a high percentage of overlapped speech (usually higher than in real conversations).
\vspace{-0.2cm}
\subsection{Simulated conversations}
\vspace{-0.1cm}
One of the main concerns with SM is that each speaker in the mixture is treated independently. Although the lengths of the pauses are randomly drawn from an exponential distribution, a sensible choice as it represents the inter-arrival times between independent events,
this does not resemble the dynamics of a real conversation where speakers do not take turns independently but collaboratively. For this reason, the proposed approach to create SC uses statistics about frequencies and lengths of pauses and overlaps from real conversations. Statistics are:
\vspace{-0.1cm}
\begin{itemize}[wide=10pt]
\setlength\itemsep{-0.2em}
\item Pauses between consecutive same speaker segments. The histogram of the pause lengths defines the distribution $D_{=\text{speaker}}$.
\item Number of pauses $ds$ between consecutive different speaker segments. The histogram of the pause lengths defines the distribution $D_{\neq\text{speaker}}$.
\item Number of overlaps $ov$ between consecutive different speaker segments. The histogram of the overlap lengths defines the distribution $D_{\text{overlap}}$.
\item $p = \frac{ds}{ds + ov}$ is the probability of having a pause in between two segments of different speakers.
\end{itemize}
\vspace{-0.1cm}
The proposed approach is described in Algorithm~\ref{alg:conversation}. In this case, utterances are sampled without replacement. When creating a large set of simulated conversations, this ensures that an utterance is not used more than once\footnote{In practice, the code is prepared to run until exhausting all utterances and re-start again until a given amount of times is reached or until generating a specific amount of audio.}. Furthermore, for a given utterance all segments are used as part of a single SC. In contrast, in the original approach, only a subset of the segments from each original conversation is used at a time.
The segments (defined by the VAD labels) of the selected utterances (one per speaker) are randomly interleaved, guaranteeing that the per-speaker order is kept while assigning random order to the speaker turns of the different speakers. Then, for each pair of consecutive segments after interleaving them, a gap is defined depending on the nature of the two segments as shown in Algorithm~\ref{alg:conversation}. Pauses or overlaps lengths are sampled using the estimated distributions. In the pseudo-code, there is a single channel initialized with $0$'s and segments are added to it in the obtained order by means of $out.addFromPosition(pos, in)$ which adds the signal $in$ onto $out$ starting from position $pos$.
In the analysis that we present in this work we consider the addition of background noises and reverberation (already present in~\cite{fujita2019enda}) as augmentations. For analyzing the effect of using or not using each of these augmentations, different training datasets are generated for each combination. The same EEND model is trained on each one independently and its performance evaluated on real data. We also explored scaling the energy levels between speakers in the conversations to resemble those in real conversations but this did not impact the performance.
\setlength{\textfloatsep}{8pt
\begin{algorithm}
\caption{Conversation simulation}\label{alg:conversation}
\hspace*{\algorithmicindent} \textbf{Input:} $\mathcal{S}$, $\mathcal{N}$, $\mathcal{I}$, $\mathcal{R}$ \Comment{\footnotesize{Set of speakers, noises, RIRs, and SNRs}} \\
\hspace*{\algorithmicindent} \hspace*{\algorithmicindent} $\mathcal{U} = \{ U_s \}_{s \in \mathcal{S}}$ \Comment{\footnotesize{Set of utterances}} \\
\hspace*{\algorithmicindent} \hspace*{\algorithmicindent} $N_{\text{spk}}$ \Comment{\footnotesize{\#speakers per conversation}} \\
\hspace*{\algorithmicindent} \hspace*{\algorithmicindent} $p, D_{=\text{speaker}}, D_{\neq\text{speaker}}, D_{\text{overlap}}$ \Comment{Estimated distributions}
\begin{algorithmic}[1]
\State $G \gets \{\}$ \Comment{\footnotesize{Dictionary with list of segments per speaker}}
\State Sample a set of $N_{\text{spk}}$ speakers $\mathcal{S}'$ from $\mathcal{S}$
\ForAll{$s \in \mathcal{S}'$}
\State Sample $u$ from $U_s$ without replacement
\State Sample $\mathbf{i}$ from $\mathcal{I}$ \Comment{RIR}
\State $u' \gets u * \mathbf{i}$ \Comment{Reverberate all segments in the utterance}
\State $G[s] \gets u'$
\EndFor
\State $L \gets$ Randomly interleave $G$ into a single list
\State $\mathbf{y}.addFromPosition(0, L[1]$) \Comment{Start signal with first segment}
\State $pos \gets \text{len}(L[1])$
\For{$t = 2$ to $\text{len}(L)$}
\If{Speaker$(L[t-1]) = $ Speaker$(L[t])$}
\State Sample $gap$ from $D_{=\text{speaker}}$
\ElsIf{Sample of $Bernoulli(p)$ is 1}
\State Sample $gap$ from $D_{\neq\text{speaker}}$
\Else
\State Sample $-gap$ from $D_{\text{overlap}}$
\EndIf
\State $\mathbf{y}.addFromPosition(pos + gap, L[t])$
\State $pos \gets pos + gap + \text{len}(L[t])$
\EndFor
\State Sample $\mathbf{n}$ from $\mathcal{N}$ \Comment{Background noise}
\State Sample $r$ from $\mathcal{R}$ \Comment{SNR}
\State Determine a mixing scale $p$ from $r$, $\mathbf{y}$, and $\mathbf{n}$
\State $\mathbf{n}' \gets$ repeat $\mathbf{n}$ until reaching the length of $\mathbf{y}$
\State $\mathbf{y} \gets \mathbf{y} + p \cdot \mathbf{n}'$
\end{algorithmic}
\end{algorithm}
For the sake of making the comparison between the two approaches as fair as possible, several aspects of the data preparation follow those of the original SM approach~\cite{fujita2019enda}:
\begin{itemize}[wide=10pt]
\setlength\itemsep{-0.2em}
\item The set of utterances used: comprised of Switchboard-2 (phases I, II, III)~\cite{graff1998switchboard,graff1999switchboard,graff2002switchboard}, Switchboard Cellular (parts 1 and 2)~\cite{graff2001switchboard,graff2004switchboard}, and NIST Speaker Recognition Evaluation datasets (from years 2004, 2005, 2006, 2008)~\cite{nist20112006test,nist20122006test,nist20112008train,nist20112008test,nist20112005test,nist20112006train,nist20062004,nist20112005}. All the recordings are sampled at 8\,kHz and, out of 6381 speakers, 90\% are used for creating training data.
\item The VAD used to obtain time annotations: based on time-delay neural networks and statistical pooling\footnote{http://kaldi-asr.org/models/m4}.
\item The set of background noises and mechanism for augmenting data: 37 noises labeled as ``background'' in the MUSAN collection~\cite{snyder2015musan} are added to the signal scaled with a signal to noise ratio (SNR) selected randomly from \{5, 10, 15, 20\}.
\item The set of room impulse responses (RIRs) used to reverberate utterances: a RIR is sampled from the collection used in~\cite{ko2017study} and with 0.5 probability used to reverberate the utterances of each speaker.
\end{itemize}
\vspace{-0.2cm}
In order to shed some light on how the proposed SC resemble real conversations more than SM, some statistics about the recordings are presented in Table~\ref{tab:datasets_stats} where we see that SC is more similar to real sets in terms of percentages of silence, speech from a single speaker and overlap. We selected a subset of the SC to match the amount of hours using with SM in previous works. An analysis on the performance with different amounts of hours of data is presented in~\ref{sec:dataamount}. The code for generating SC can be found in https://github.com/BUTSpeechFIT/EEND\_dataprep
\begin{table}[!tb]
\setlength{\tabcolsep}{3pt}
\centering
\caption{Statistics for synthetic and real datasets. All listed sets have only 2 speakers per recording.}
\vspace{-0.2cm}
\label{tab:datasets_stats}
\setlength{\tabcolsep}{2pt}
\begin{tabular}{@{}
lcccccc
@{}}
\toprule
Dataset &
\multirow{2}{*}{\#files} & Total & Average &
\multicolumn{3}{c}{Average \%} \\
& & audio (h) & dur. (s) & sil. & 1spk & over. \\
\midrule
SM ($\beta$=2)& 100k & 2480 & 89.30 & 18.61 & 49.62 & 31.77 \\
SC & 25k & 2480 & 356.15 & 12.80 & 78.83 & 8.37 \\
\midrule
CH Part1 & 155 & 3.19 & 74.02 & 9.05 & 77.90 & 13.05 \\
CH Part2 & 148 & 2.97 & 72.14 & 9.84 & 78.34 & 11.82 \\
DH3 dev & 61 & 10.17 & 599.95 & 10.56 & 77.27 & 12.17 \\
DH3 eval & 61 & 10.17 & 599.95 & 10.89 & 78.62 & 10.49 \\
\bottomrule
\end{tabular}
\end{table}
\vspace{-0.1cm}
\section{Experimental setup}
\vspace{-0.1cm}
\subsection{Diarization models}
\vspace{-0.1cm}
The experiments were performed using the self-attention EEND with encoder-decoder attractors for showing superior performance in previous works. In all cases the architecture used was exactly the same as that described in~\cite{horiguchi2020end}\footnote{We refer the reader to~\cite{horiguchi2020end} for a scheme of the model.}. For the sake of making the code more efficient, we used our PyTorch implementation\footnote{https://github.com/BUTSpeechFIT/EEND}. 15 consecutive 23-dimensional log-scaled Mel-filterbanks (computed over 25\,ms every 10\,ms) are stacked to produce 345-dimensional features every 100ms. These are transformed by 4 self-attention encoder blocks (with 4 attention heads each) into a sequence of 256 dimensional embeddings. These are then shuffled in time and fed into the LSTM-based encoder-decoder module which decodes as many attractors as speakers are predicted. A binary linear classifier is used to obtain speech activity probabilities for each speaker (represented by an attractor) at each time (represented by an embedding).
During inference time, classifiers' outputs are thresholded at 0.5 to determine speech activities. To ease the comparison, this parameter was not tuned in any experiment but tuning it could lead to better results.
Each training was run for 100 epochs on a single GPU. The batch size was set to 64 or 32 with 100000 or 200000 minibatch updates of warm up respectively. Following~\cite{horiguchi2020end}, the Adam optimizer~\cite{kingma2014adam} was used and scheduled with noam~\cite{vaswani2017attention}. For fine-tuning on a development set, 100 epochs were run and the Adam optimizer was used with learning rate $10^{-5}$. For the inference as well as for obtaining the model from which to fine-tune, the models from the last 10 epochs were averaged. During training and fine-tuning phases, batches are formed by sequences of 500 Mel-filterbank outputs, corresponding to 50\,s. During inference, the full recordings are fed to the network one at a time.
Diarization performance is evaluated in terms of diarization error rate (DER) as defined by NIST~\cite{NISTRT}.
\vspace{-0.2cm}
\subsection{Data}
\vspace{-0.1cm}
Two real telephone conversations datasets were used to report results. First, the 2000 NIST Speaker Recognition Evaluation~\cite{przybocki2001nist} dataset, usually referred as ``Callhome''~\cite{NISTSRE2000evalplan}.
We report results on the subset of 2-speaker conversations using the standard Callhome partition\footnote{Each of these sets are listed in https://github.com/BUTSpeechFIT/CALLHOME\_sublists}. We will refer to the parts as CH1-2spk and CH2-2spk. Results on Callhome consider all speech (including overlap segments) for evaluation with a forgiveness collar of 0.25\,s. Second, the CTS domain from the recent Third DIHARD Challenge~\cite{ryant2020third}, which consists of previously unpublished telephone conversations from the Fisher collection. Both development and evaluation sets consist of 61 10-minute recordings each (full set). We will refer to the sets as DH-dev and DH-eval. Originally 8\,kHz signals, they were upsampled to 16\,kHz for the challenge and downsampled to 8\,kHz to be used in this work. As usual on DIHARD, all speech is evaluated with a collar of 0\,s.
\vspace{-0.1cm}
\section{Results}
\vspace{-0.1cm}
\subsection{SC augmentations analysis}
\vspace{-0.1cm}
We evaluated all combinations of the augmentations, namely reverberate and add noises. For this analysis, statistics to generate SC were estimated on DH-dev. Figure~\ref{fig:v1_callhome_part2_2spk} presents results on recordings with 2 speakers from Callhome Part 2 comparing the best published result with the original Chainer implementation~\cite{horiguchi2020end}, our runs with the Chainer implementation, our PyTorch implementation with SM and the PyTorch implementation with the different SC options. Our runs with Chainer correspond with the result in~\cite{horiguchi2020end}.
The results with PyTorch are slightly worse, note that there might be small implementation differences between Chainer and PyTorch and that the same hyper-parameters tuned for training with Chainer were used with the PyTorch implementation to ease the comparison and that adjusting them could lead to further improvement.
Using any of the SC options for generating training data outperforms using SM. Even more, the performance of the models trained using the best SC option (SC 2) is close to the results obtained with the models trained on SM after fine-tuning on the real sets.
The models trained on the best SC option can, in turn, be further improved by means of fine-tuning, significantly outperforming those trained on SM, showing that a model trained on better quality SC data not only performs better but can also be still leveraged in combination with fine-tuning to real data.
When comparing the four options, the main gains are observed when adding background noises; as expected, this simulates noisier conditions and adds variability to the set. Reverberating the signals with the default parameters actually harms the performance. We hypothesize that the effect might be different in other scenarios such as meetings or interviews where speakers do not have specific microphones close to the mouth such as in telephone conversations. Furthermore, the choice of RIRs could be narrowed down to take into account only rooms that resemble those in real applications. These aspects need to be further studied, especially for the more diverse scenarios seen in wide-band data. Results in following sections are obtained using the best result with SC 2 and SM (P).
\begin{figure}[t]
\centering
\hspace*{-0.5cm}
\includegraphics[width=1\linewidth]{LaTeX/Figures/v2_callhome_part2_2spk.png}
\vspace{-0.4cm}
\caption{DER (\%) comparison for SM (C stands for Chainer and P for PyTorch) and SC options on CH2-2spk and fine-tuning to CH1-2spk. \cite{horiguchi2020end} is a single run. All other experiments were repeated 5 times and we show means and error bars.}
\label{fig:v1_callhome_part2_2spk}
\vspace*{-0.1cm}
\end{figure}
\vspace{-0.2cm}
\subsection{DER breakdown analysis}
\vspace{-0.1cm}
Table~\ref{tab:DER_breakdown} presents a detailed comparison of the errors from SM (P) and SC 2. In terms of VAD, both systems perform very similarly before and after fine-tuning. However, larger differences can be seen in terms of OSD. When using SC for training, the model makes considerably less false alarms for OSD, specially before fine-tuning. This can be explained by the larger percentage of overlap seen in SM (Table~\ref{tab:datasets_stats}).
Mechanisms for favoring slightly higher percentages of overlap when creating SC are left for future studies.
\begin{table}[th]
\caption{Error analysis before and after fine-tuning on recordings with 2 speakers of CH2-2spk.}
\vspace{-0.2cm}
\label{tab:DER_breakdown}
\setlength{\tabcolsep}{3pt}
\centering
\begin{tabular}{@{}
l
S[table-format=2.2] |
S[table-format=1.2]
S[table-format=1.2]
S[table-format=1.2] |
S[table-format=1.2]
S[table-format=1.2] |
S[table-format=1.2]
S[table-format=2.2]
@{}}
\toprule
& & \multicolumn{3}{c|}{DER breakdown} & \multicolumn{2}{c|}{VAD} & \multicolumn{2}{c}{OSD} \\
System & \multicolumn{1}{c|}{DER} & \multicolumn{1}{c}{Miss} & \multicolumn{1}{c}{FA} & \multicolumn{1}{c|}{Conf.} & \multicolumn{1}{c}{Miss} & \multicolumn{1}{c|}{FA} & \multicolumn{1}{c}{Miss} & \multicolumn{1}{c}{FA} \\
\midrule
SM (P) & 15.09 & 2.83 & 8.24 & 4.01 & 0.48 & 7.70 & 4.84 & 10.71 \\
\hspace{0.2cm}+ FT & 8.44 & 5.23 & 2.32 & 0.90 & 3.39 & 4.06 & 6.20 & 4.22 \\
\midrule
SC 2 & 8.64 & 3.11 & 4.84 & 0.69 & 0.49 & 7.53 & 4.68 & 9.03 \\
\hspace{0.2cm}+ FT & 7.28 & 4.72 & 1.98 & 0.58 & 3.23 & 4.03 & 6.02 & 3.82 \\
\bottomrule
\end{tabular}
\vspace*{-0.2cm}
\end{table}
\subsection{Statistics source and fine-tuning analysis}
\vspace{-0.1cm}
To analyze the effect of the set used to estimate the statistics for creating SC data, we made a comparison using either DH-dev or CH1. As seen in Table~\ref{tab:statistics_source}, the effect of the set used for the estimation of the statistics is not large. Both datasets present several recordings that amount to few hours of speech which is enough to have a reasonable amount of data to estimate the statistics. At the same time, the fine-tuning step improves the performance on both datasets suggesting that real conversations still differ from SC, leaving room for improving SC quality.
\begin{table}[th]
\caption{DER (\%) for models trained with SC 2 using statistics estimated on DH-dev or CH1. Fine-tuning for each test set is done in the corresponding dev set. Numbers in gray denote results where the test data were used for training. In underlined results, test data were used to compute the statistics.}
\vspace{-0.2cm}
\label{tab:statistics_source}
\centering
\begin{tabular}{@{}
l |
S[table-format=2.2]
S[table-format=2.2] |
S[table-format=2.2]
S[table-format=2.2]
@{}}
\toprule
& \multicolumn{2}{c|}{Callhome 2 speakers} & \multicolumn{2}{c}{DIHARD 3 CTS full} \\
System & \multicolumn{1}{c}{Part 1} & \multicolumn{1}{c|}{Part 2} & \multicolumn{1}{c}{dev} & \multicolumn{1}{c}{eval} \\
\midrule
DH stats & \textcolor{black}{8.64} & 8.16 & \textcolor{gray}{\underline{23.47}} & 22.06 \\
\hspace{0.2cm}+ FT & \textcolor{gray}{6.20} & 7.28 & \textcolor{gray}{16.99} & 17.00 \\
\midrule
CH stats & \textcolor{gray}{\underline{8.26}} & 8.73 & \textcolor{black}{22.29} & 21.53 \\
\hspace{0.2cm}+ FT & \textcolor{gray}{6.13} & 7.28 & \textcolor{gray}{17.38} & 17.14 \\
\bottomrule
\end{tabular}
\vspace*{-0.1cm}
\end{table}
\vspace{-0.2cm}
\subsection{Amount of data analysis}
\vspace{-0.1cm}
\label{sec:dataamount}
One aspect to analyze is the effect that the amount of training data has on EEND performance when using SC. Figure~\ref{fig:confbands} shows that the performance when training with \textit{as little as} 310 hours degrades considerably before fine-tuning. However, using 1240 hours (half of the amount used in all other experiments) already allows for similar performance as using more data.
\begin{figure}[t]
\centering
\hspace*{-0.5cm}
\includegraphics[width=1\linewidth]{LaTeX/Figures/confbands_callhome_part2_2spk.png}
\vspace{-0.4cm}
\caption{DER (\%) on CH2-2spk when training with different amounts of hours of SC 2. Each experiment is repeated 5 times to obtain the mean and confidence intervals.}
\label{fig:confbands}
\end{figure}
\vspace{-0.2cm}
\subsection{Comparison with previous results}
\vspace{-0.1cm}
In this section we compare our results to the best results published with the same architecture. Table~\ref{tab:final_results} shows that when training with SC it is possible to attain similar performance as when training with SM after fine-tuning on Callhome. This gain is smaller in the case of DIHARD CTS where fine-tuning has a larger effect. One of the aspects to consider with the architecture used is that the input is downsampled so that one output every 100\,ms is produced. When evaluating with a 0\,s forgiveness collar, such as in DIHARD, this severely impacts the results. Table~\ref{tab:final_results} presents results when evaluating at the standard downsampling of one output every 100\,ms and 50\,ms\footnote{This is done only at inference time (not during training), the system is exactly the same and cannot address the issue completely. Having finer granularity not only increases the inference time considerably but the memory requirements as well.}, showing that a considerable error reduction is possible after fine-tuning.
\begin{table}[!tb]
\centering
\caption{DER (\%) for different systems.}
\vspace{-0.2cm}
\label{tab:final_results}
\setlength{\tabcolsep}{4pt}
\begin{tabular}{@{}
l |
S[table-format=2.2]
S[table-format=2.2] |
S[table-format=2.2]
S[table-format=2.2] |
S[table-format=2.2]
S[table-format=2.2]
@{}}
\toprule
& \multicolumn{2}{c|}{CH 2 speakers} & \multicolumn{4}{c}{DIHARD 3 CTS full} \\
System & \multicolumn{1}{c}{Part 1} & \multicolumn{1}{c|}{Part 2} & \multicolumn{2}{c}{dev} & \multicolumn{2}{c}{eval} \\
\midrule
S.H. et al.~\cite{horiguchi2020end} & \text{--} & 8.07 & \multicolumn{2}{c}{\text{--}} & \multicolumn{2}{c}{\text{--}} \\
\midrule
& & & \multicolumn{4}{c}{Downsample rate (ms)} \\
& & & \text{50} & \text{100} & \text{50} & \text{100} \\
\midrule
SM (P) & 13.62 & 15.09 & 25.36 & 25.46 & 22.16 & 23.58 \\
\hspace{0.2cm}+ FT & \textcolor{gray}{7.61} & 8.44 & \textcolor{gray}{12.97} & \textcolor{gray}{17.98} & 11.99 & 17.44 \\
\midrule
SC 2 & 8.16 & 8.64 & \textcolor{gray}{\underline{21.16}} & \textcolor{gray}{\underline{23.47}} & 19.81 & 22.06 \\
\hspace{0.2cm}+ FT & \textcolor{gray}{6.2} & 7.28 & \textcolor{gray}{11.68} & \textcolor{gray}{16.99} & 11.20 & 17.00 \\
\bottomrule
\end{tabular}
\end{table}
\vspace{-0.2cm}
\section{Conclusions}
\vspace{-0.1cm}
EEND-based diarization systems are being thoroughly explored with many different variants but they all require a vast amount of data with diarization labels for training. We presented an alternative strategy for generating training data which uses statistics of real conversations to resemble real data. Our approach significantly outperforms the original one, proving specially useful in cases where there would not be a development set for fine-tuning. However, with fine-tuning, the performance can still be leveraged showing that there is still room for improving the generation of synthetic conversations. In our future work we plan to explore the use of RIRs that resemble real scenarios and relevance of VAD and overlap related statistics. Also, we plan to further exploit our approach for creating conversations with more than 2 speakers and use it with wide-band data where there are far less hours of single-speaker recordings in conversational setups available to create synthetic conversations.
\vspace{-0.2cm}
\section{Acknowledgements}
\vspace{-0.1cm}
The work was supported by Czech Ministry of Interior project No. VJ01010108 ``ROZKAZ'', Czech National Science Foundation (GACR) project NEUREM3 No. 19-26934X and the author from the UAM was supported by project RTI2018-098091-B-I00, granted by MCIU/AEI/FEDER, UE. Computing on IT4I supercomputer was supported by the Czech Ministry of Education, Youth and Sports from the Large Infrastructures for Research, Experimental Development and Innovations project ``e-Infrastructure CZ – LM2018140''.
\bibliographystyle{IEEEtran}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,884 |
Ricky Williams: police video shows former Dolphins running back frisked by officers
Former Miami Dolphins running back Ricky Williams appears in recent police video / Chrisjnelson, Wikimedia Commons CC BY-SA 3.0
The 1998 Heisman Trophy winner believes that he was racially profiled after taking a walk near his hotel in Tyler, Texas.
by Stephen Sinclair
January 22, 2017 at 9:29 PM January 22, 2017 at 9:29 PM
Ricky Williams, the 2002 NFL rushing yards leader and round one, number five pick of the 1999 draft, was caught on video being stopped and frisked by police after the former pro decided to spend some time by taking a walk near his hotel, the Courtyard Marriott, in Tyler, Texas, as reported by the New York Daily News. Video showing the incident was recorded by a Tyler police dash cam. In addition to the many accolades he has received as a retired NFL pro, Ricky Williams is also a current cast member of the "Celebrity Apprentice."
While he was being investigated by police, one of the officers was reported to comment on the likeness of the suspect with the former Miami Dolphins pro Ricky Williams.
"I am Ricky Williams," was how the 39-year-old California native was said to respond to the officers. The police involved in stopping and questioning Williams were said to have been investigating a man hopping between backyards in the neighborhood.
Ricky Williams: video shows stop by police
Don Martin, a spokesperson with the Tyler police stated, that if the person under investigation had not returned to the area "and acted in a suspicious manner" that the entire incident could have been avoided. Martin said that he felt that the "officers did a professional job and did the proper job." In the video, one of the police can be heard telling Ricky Williams "I know more than what you think I know."
After being stopped in the video, Ricky Williams asked the officers "Do you know how many times I've been messed with by cops just for being black?" Appearing on Austin's "The Dudley and Bob Show" on Wednesday, Williams remembered telling the officers that they "don't know what it's like" to be a black person in the United States and that the incident wasn't the first time he felt that he had been targeted because of his race.
Former NFL pro welcome in Tyler, Texas
Martin indicated that he and Ricky Williams met and discussed the incident. There has been no indication that the former football pro would be pursuing legal action. The Tyler police representative stated that Williams was always welcome in Tyler and that the reality show star even has a place to stay with his family, "We have a guest room he's welcome to."
Stephen Sinclair has been writing professionally since 2013. In addition to 'Blasting News,' his work appears with 'The Inquisitr,' under the pen name Scott Hough, and at sites like 'Investopedia' and 'Ask.
Follow stephen on Facebook Follow stephen on Twitter Follow stephen on Linkedin
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Why is Glastonbury not on next year? Susan Boyle attack: thugs hurl stones, lit paper, abuse at Scottish singer Who is Cynthia Marie Randolph? Children die in hot car 'punishment'
Blasting News recommends Rob Gronkowski weighs in on Brady's future: TB12 will pick Bucs over Raiders Daddy duty for Tom Brady as he spends quality time with daughter Vivian Ex-teammate Kyle Van Noy shows love for Tom Brady, calls him the 'best ever' Buccaneers' offseason moves to center around Tom Brady Super Bowl 2017 date and kickoff time: Patriots and Falcons face off in Houston Where and when will Super Bowl 51 be played this year? 2017 game tickets info & more | {
"redpajama_set_name": "RedPajamaCommonCrawl"
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layout: default
title: Sandra Chung
userid: sandramchung
---
{% for user in site.data.users %}
{% if user.userid == page.userid %}
{% assign current = user %}
{% endif %}
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### {{ page.title }}
#### {{ current.city }}, {{ current.state }} {{ current.postal }} | {
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\section{Introduction}
The dialogue state tracking (DST) module structures the belief state that appears during the conversation in the form of \texttt{domain-slot-value}, to provide an appropriate response to the user. Recently, multi-turn DST datasets have been constructed using the Wizard-of-Oz method to reflect more realistic dialogue situations \citep{wen-etal-2017-network, mrksic-etal-2017-neural, budzianowski-etal-2018-multiwoz}. The characteristic of these datasets is that belief states are ``accumulated'' and recorded every turn. That is, the belief states of the previous turns are included in the current turn. It confirms whether the DST model tracks essential information that has appeared up to the present point.
Joint goal accuracy and slot accuracy are utilized in most cases to evaluate the prediction of accumulated belief states. Joint goal accuracy strictly determines whether every predicted state is identical to the gold state, whereas slot accuracy measures the ratio of correct predictions. However, we determined that these two metrics solely focus on ``penalizing states that fail to predict,'' not considering ``reward for well-predicted states.'' Accordingly, as also pointed out in \citet{Rastogi2020SchemaGuidedDS}, joint goal accuracy underestimates the model prediction because of its error accumulation attribute, while slot accuracy overestimates it because of its dependency on predefined slots.
However, there is a lack of discussion on the metric for evaluating the most used MultiWOZ dataset, despite a recently published dataset \citep{rastogi2020towards} proposing some metrics. To address the above challenge, we propose reporting the \textbf{relative slot accuracy} along with the existing metrics in MultiWOZ dataset. While slot accuracy has the challenge of overestimation by always considering all predefined slots in every turn, relative slot accuracy does not depend on predefined slots, and calculates a score that is affected solely by slots that appear in the current dialogue. Therefore, relative slot accuracy enables a realistic evaluation by rewarding the model's correct predictions, a complementary approach that joint goal and slot accuracies cannot fully cover. It is expected that the proposed metric can be adopted to evaluate model performance more intuitively.
\section{Current Evaluation Metrics}
\label{sec:current}
\subsection{Joint Goal Accuracy}
\label{sec:jga}
Joint goal accuracy, developed from \citet{henderson-etal-2014-word} and \citet{zhong-etal-2018-global}, can be said to be an ideal metric, in that it verifies that the predicted belief states perfectly match the gold label. Equation \ref{eq:jga} expresses how to calculate the joint goal accuracy, depending on whether the slot values match each turn.
\begin{equation}
\label{eq:jga}
JGA = \left\{
\begin{array}{ c l }
1 & \, \textrm{if predicted state}=\textrm{gold state} \\
0 & \, \textrm{otherwise}
\end{array}
\right.
\end{equation}
However, the joint goal accuracy underestimates the accumulated states because it scores the performances of later turn to zero if the model mispredicts even once in a particular turn, regardless of the model prediction quality at later turns. As illustrated in Figure \ref{fig:jga}, we measured the relative position of the turn causing this phenomenon for the dialogue. We used MultiWOZ 2.1 \citep{DBLP:journals/corr/abs-1907-01669}, and analyzed 642 samples from a total of 999 test sets in which the joint goal accuracy of the last turn is zero. The DST model selected for primary verification is the SOM-DST \citep{kim-etal-2020-efficient}, which is one of the latest DST models. Accordingly, the relative position where joint goal accuracy first became zero was mainly at the beginning of the dialogue\footnote{59 samples of the 642 samples have a joint goal accuracy of 1 in the middle, owing to a coincidental situation or differences in the analysis of annotation. Table \ref{table:dialog_1} and Table \ref{table:dialog_2} show the dialogue situation in detail, and Table \ref{table:dialog_state_1} and Table \ref{table:dialog_state_2} show the belief states accordingly. Refer to Appendix \ref{sec:appendix_a}.}. This means that the joint goal accuracy after the beginning of the dialogue is unconditionally measured as zero because of the initial misprediction, although the model may correctly predict new belief states at later turns. Failure to measure the performance of the latter part means that it cannot consider various dialogue situations provided in the dataset, which is a critical issue in building a realistic DST model.
\begin{figure}[t!]
\centering
\includegraphics[width=\columnwidth]{figures/f1.pdf}
\caption
{
The relative position where joint goal accuracy of the turn is measured to be zero for the first time among the dialogues where joint goal accuracy of the last turn is zero. (642 of 999 MultiWOZ 2.1 test set with SOM-DST).
}
\label{fig:jga}
\end{figure}
\begin{figure}[t!]
\centering
\includegraphics[width=\columnwidth]{figures/f2.pdf}
\caption
{
The number of predefined gold slots used in each dialogue (999 MultiWOZ 2.1 test set).
}
\label{fig:sa}
\end{figure}
\begin{table*}[t!]
\centering
\renewcommand{\arraystretch}{1.0}
\setlength{\tabcolsep}{12pt}
\begin{tabular}{llcccc}
\toprule
\multirow{2}{*}{\textbf{Type}} & \multirow{2}{*}{\textbf{Model}} & \textbf{Joint} & \textbf{Slot} & \textbf{F1} & \textbf{Relative} \\
& & \textbf{Goal Acc.} & \textbf{Acc.} & \textbf{Score} & \textbf{Slot Acc.} \\
\midrule
\multirow{7}{*}{\shortstack[l]{Open\\vocabulary}} & Transformer-DST \citeyearpar{zeng2021jointly} & 0.5446 & 0.9748 & 0.9229 & 0.8759 \\
& TripPy \citeyearpar{heck-etal-2020-trippy} & \textbf{0.6131} & 0.9707 & 0.8573 & 0.8432 \\
& SOM-DST \citeyearpar{kim-etal-2020-efficient} & 0.5242 & 0.9735 & 0.9179 & 0.8695 \\
& Simple-TOD \citeyearpar{NEURIPS2020_e9462095} & 0.5605 & \textbf{0.9761} & \textbf{0.9276} & \textbf{0.8797} \\
& SAVN \citeyearpar{wang-etal-2020-slot} & 0.5357 & 0.9749 & 0.9246 & 0.8769 \\
& TRADE \citeyearpar{wu-etal-2019-transferable} & 0.4939 & 0.9700 & 0.9033 & 0.8520 \\
& COMER \citeyearpar{ren-etal-2019-scalable} & 0.4879 & 0.9652 & 0.8800 & 0.8250 \\ \hline
\multirow{3}{*}{\shortstack[l]{Ontology\\based}} & DST-STAR \citeyearpar{10.1145/3442381.3449939} & 0.5483 & 0.9754 & 0.9253 & 0.8780 \\
& L4P4K2-DSGraph \citeyearpar{lin2021knowledgeaware} & 0.5178 & 0.9690 & 0.9189 & 0.8570\\
& SUMBT \citeyearpar{lee-etal-2019-sumbt} & 0.4699 & 0.9666 & 0.8934 & 0.8380 \\
\bottomrule
\end{tabular}
\caption{\label{table:result}
Model performance of MultiWOZ 2.1 with various evaluation metrics. All reported performances are our re-implementation.
}
\end{table*}
\subsection{Slot Accuracy}
\label{sec:sa}
Slot accuracy can compensate for situations where joint goal accuracy does not fully evaluate the dialogue situation. Equation \ref{eq:sa} expresses how to calculate the slot accuracy. $T$ indicates the total number of predefined slots for all the domains. $M$ denotes the number of missed slots that the model does not accurately predict among the slots included in the gold state, and $W$ denotes the number of wrongly predicted slots among the slots that do not exist in the gold state.
\begin{equation}
\label{eq:sa}
SA = {T-M-W \over T}
\end{equation}
Figure \ref{fig:sa} illustrates the total number of annotated slots in MultiWOZ 2.1 to figure out the limitation of slot accuracy. Each value of $x$-axis in Figure \ref{fig:sa} indicates the "maximum" number of slots that appear in a single dialogue, and we confirmed that approximately 85\% of the test set utilized solely less than 12 of the 30 predefined slots in the experiment. Because the number of belief states appearing in the early and middle turns of the dialogue are smaller, and even fewer states make false predictions, calculating slot accuracy using Equation \ref{eq:sa} reduces the influence of $M$ and $W$, and the final score is dominated by the total slot number $T$. Accordingly, several previous studies still report the model performance using solely joint goal accuracy because slot accuracy excessively depends on the number of predefined slots, making the performance deviation among models trivial (refer to Table \ref{table:survey}).
Furthermore, according to Table \ref{table:extreme}, we determined that slot accuracy tends to be too high. The slot accuracies of turns 0 and 1 show approximately 96\% accuracy, despite the model not correctly predicting states at all. It becomes difficult to compare various models in detail, if each model shows a high performance, even though nothing is adequately predicted. In addition, as the turn progresses, there are no rewards for a situation in which the model tracks the belief state without any challenges. The case correctly predicting two out of three in turn 4, and the case correctly predicting three out of four in turn 5 exhibit the same slot accuracy. Therefore, the slot accuracy measured according to Equation \ref{eq:sa} differs from our intuition.
\subsection{Other Metric}
\label{sec:aga}
Recently, \citet{rastogi2020towards} proposed a metric called average goal accuracy. The main difference between the average goal accuracy and the proposed relative slot accuracy is that the average goal accuracy only considers the slots with non-empty values in the gold states of each turn, whereas the proposed relative slot accuracy considers those in both gold and predicted states. Since average goal accuracy ignores the predicted states, it cannot properly distinguish a better model from a worse model in some specific situations. We will discuss it in more detail in Section \ref{sec:discussion}.
\section{Relative Slot Accuracy}
As can be observed in Equation \ref{eq:sa}, slot accuracy has the characteristic that the larger the number of predefined slots ($T$), the smaller the deviation between the prediction results. The deviation among DST models will be even more minor when constructing datasets with various dialogue situations, because the number of predefined slots will continually increase. It is not presumed to be an appropriate metric in terms of scalability.
Therefore, we propose relative slot accuracy, that is not affected by predefined slots, and is evaluated with adequate rewards and penalties that fit human intuition in every turn. Equation \ref{eq:rsa} expresses how to calculate the relative slot accuracy, and $T^\ast$ denotes the number of unique slots appearing in the predicted and gold states in a particular turn.
\begin{equation}
\label{eq:rsa}
RSA = {T^\ast - M -W \over T^\ast} \mbox{, where $0$ if $T^\ast=0$}
\end{equation}
Relative slot accuracy rewards well-predicted belief states by measuring the scores in accumulating turns. Further discussions on the relative score will be discussed in Section \ref{sec:discussion}.
\section{Experiments}
We measured MultiWOZ 2.1, an improved version of MultiWOZ 2.0 \citep{budzianowski-etal-2018-multiwoz}, which has been adopted in several studies, according to Table \ref{table:survey}. Five domains (i.e., \textit{hotel, train, restaurant, attraction,} and \textit{taxi}) are adopted in the experiment, following \citet{wu-etal-2019-transferable}, and there are a total of 30 \texttt{domain-slot} pairs. We selected the DST models in Table \ref{table:survey} that perform the MultiWOZ experiment with the original authors' reproducible code\footnote{Implementation codes for Simple-TOD and TripPy are from https://github.com/salesforce/coco-dst.}. Additionally, we reported the F1 score, which can be calculated using the current predicted and gold states.
\begin{figure}[ht]
\centering
\includegraphics[width=0.9\columnwidth]{figures/corr_turnlevel_SOM-DST.pdf}
\caption
{
Correlation matrix of evaluation performance of total 7,368 turns in 999 MultiWOZ 2.1 test set using SOM-DST. Results for other models are included in Figure \ref{fig:corrs}.}
\label{fig:corr}
\end{figure}
\subsection{Results and Discussion}
\label{sec:discussion}
Table \ref{table:result} presents the overall results. Regarding slot accuracy, the difference between the largest and smallest values is solely 1.09\%. It can be one of the reasons that several researchers do not report it. Meanwhile, relative slot accuracy can explicitly highlight the deviation among models by showing a 5.47\% difference between the largest and smallest values. Furthermore, the correlation with joint goal accuracy, a mainly adopted metric, and relative slot accuracy with respect to each turn is lower than the correlation with joint goal accuracy and slot accuracy, as illustrated in Figure \ref{fig:corr}. Specifically, it can be compared with a different perspective when using the proposed reward-considering evaluation metric.
\paragraph{Domain-specific Evaluation}
We reported the joint goal, slot, and relative slot accuracies per domain utilizing the SOM-DST model in Table \ref{tab:perdomain}. Relative slot accuracy derives a specific score in the turn configuration and prediction ratio of each domain by excluding slots that do not appear in the conversation. For example, the \textit{taxi} domain shows a low score, meaning that it has relatively several cases of incorrect predictions, compared to the number of times slots belonging to the \textit{taxi} domain appear. Because slot accuracy cannot distinguish the above trend, the score of the \textit{hotel} domain is lower than that of the \textit{taxi} domain. In summary, relative slot accuracy enables relative comparison according to the distribution of the domain in a dialogue.
\begin{table}[t]
\centering
\renewcommand{\arraystretch}{1.0}
\setlength{\tabcolsep}{7pt}
\begin{tabular*}{\columnwidth}{ccccc}
\hlineB{3}
\multirow{2}{*}{ \textbf{Domain}} & \textbf{Joint} & \textbf{Slot} & \textbf{Relative} \\
& \textbf{Goal Acc.} & \textbf{Acc.} & \textbf{Slot Acc.} \\ \hlineB{1}
hotel & 0.4923 & 0.9731 & 0.8493 \\
train & 0.7162 & 0.9874& 0.9176 \\
restaurant & 0.6589 & 0.9858 & 0.8977 \\
attraction & 0.6811 & 0.9878 & 0.8421 \\
taxi & 0.5701 & 0.9798 & 0.7828 \\
\hlineB{3}
\end{tabular*}
\caption
{
Per-domain performance of SOM-DST prediction.
}
\label{tab:perdomain}
\end{table}
\begin{figure}[t]
\centering
\includegraphics[width=0.8\columnwidth]{figures/f3.pdf}
\caption
{
The mean and standard deviation of model performance reported in Table \ref{table:result}.
}
\label{fig:deviation}
\end{figure}
\begin{table*}[t!]
\centering
\renewcommand{\arraystretch}{0.8}
\setlength{\tabcolsep}{10pt}
\begin{tabular}{llccc}
\toprule
\multirow{2}{*}{\textbf{Type}} & \multirow{2}{*}{\textbf{Belief State}} & \textbf{Joint} & \textbf{Average} & \textbf{Relative} \\
& & \textbf{Goal Acc.} & \textbf{Goal Acc.} & \textbf{Slot Acc.} \\
\midrule
\multirow{3}{*}{\shortstack[l]{Gold State}} & \textcolor{blue}{restaurant-area-centre} & \multirow{3}{*}{\shortstack[l]{-}} & \multirow{3}{*}{\shortstack[l]{-}} & \multirow{3}{*}{\shortstack[l]{-}} \\
& \textcolor{orange}{restaurant-food-indian} & & & \\
& \textcolor{orange}{restaurant-people-2} & & & \\
\hline
\multirow{3}{*}{\shortstack[l]{Prediction\\of Model A}} & \textcolor{blue}{restaurant-area-centre} & \multirow{3}{*}{\shortstack[l]{0}} & \multirow{3}{*}{\shortstack[l]{0.3333}} & \multirow{3}{*}{\shortstack[l]{0.2500}} \\
& \textcolor{orange}{restaurant-food-chinese} & & & \\
& \textcolor{magenta}{attraction-area-centre} & & & \\
\hline
\multirow{5}{*}{\shortstack[l]{Prediction \\ of Model B}} & \textcolor{blue}{restaurant-area-centre} & \multirow{5}{*}{\shortstack[l]{0}} & \multirow{5}{*}{\shortstack[l]{0.3333}} & \multirow{5}{*}{\shortstack[l]{0.1667}} \\
& \textcolor{orange}{restaurant-food-chinese} & & & \\
& \textcolor{magenta}{restaurant-name-nusha} & & & \\
& \textcolor{magenta}{attraction-area-centre} & & & \\
& \textcolor{magenta}{attraction-pricerange-cheap} & & & \\
\bottomrule
\end{tabular}
\caption{\label{table:aga}
A situation that average goal accuracy cannot distinguish between two models. States with \textcolor{blue}{blue} denote correct prediction, and as defined in Section \ref{sec:sa}, states with \textcolor{orange}{orange} and \textcolor{magenta}{pink} denote respective $M$ and $W$.
}
\end{table*}
\paragraph{Dependency on Predefined Slots}
As discussed in Section \ref{sec:sa}, slot accuracy requiring total predefined slots is not a scalable method for evaluating the current dialogue dataset that contains a few domains in each dialogue. For example, when evaluating a dialogue sample that solely deals with the \textit{restaurant} domain, even domains that never appear at all (i.e., \textit{hotel, train, attraction,} and \textit{taxi}) are involved in measuring performance, making deviations among different models trivial. However, relative slot accuracy can evaluate the model's predictive score without being affected by slots never seen in the current dialogue, which is a more realistic way, considering that each dialogue contains its own turn and slot composition. Figure \ref{fig:deviation} illustrates the mean and standard deviations of the model performance in Table \ref{table:result}. As can be observed from the results, the relative slot accuracy has a higher deviation than the slot accuracy, enabling a detailed comparison among the methodologies.
\paragraph{Reward on Relative Dialogue Turn}
Relative slot accuracy is able to reward the model's correct prediction by measuring the accuracy on a relative basis for each turn. Table \ref{table:extreme} compares the slot and relative slot accuracies. The relative slot accuracy from turns 0 -- 3 is measured as 0 because it calculates the score based on the unique state of the current turn according to Equation \ref{eq:rsa}. In addition, regarding slot accuracy in turns 4, 5, and 6, there is no score improvement for the additional well-predicted state by the model, whereas the score increases when the newly added state is matched in the case of relative slot accuracy. Therefore, relative slot accuracy can provide an intuitive evaluation reflecting the current belief state recording method, in which the number of slots accumulates incrementally as the conversation progresses.
\paragraph{Comparison to Average Goal Accuracy}
Relative slot accuracy can compare DST model performances more properly than average goal accuracy, as mentioned in Section \ref{sec:aga}. Table \ref{table:aga} describes how these two metrics result in different values for the same model predictions. In this example, average goal accuracy cannot consider additional belief states incorrectly predicted by \texttt{Model B}, resulting in the same score between the two models. In contrast, relative slot accuracy can give a penalty proportional to the number of wrong predictions because it includes both gold and predicted states when calculating the score. Consequently, relative slot accuracy has a more elaborated discriminative power than the average goal accuracy.
\section{Conclusion}
This paper points out the challenge that the existing joint goal and slot accuracies cannot fully evaluate the accumulating belief state of each turn in the MultiWOZ dataset. Accordingly, the relative slot accuracy is proposed. This metric is not affected by unseen slots in the current dialogue situation, and compensates for the model's correct prediction. When the DST task is scaled up to deal with more diverse conversational situations, a realistic model evaluation will be possible using relative slot accuracy. Moreover, we suggest reporting various evaluation metrics to complement the limitations of each metric in future studies, not solely reporting the joint goal accuracy.
\section*{Acknowledgement}
This work was supported by Institute of Information \& communications Technology Planning \& Evaluation (IITP) grant funded by the Korea government (MSIT) (No. 2021-0-00034, Clustering technologies of fragmented data for time-based data analysis) and Ministry of Culture, Sports and Tourism and Korea Creative Content Agency (Project Number: R2020040126-0001)
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,238 |
Organizing human knowledge into related areas is nearly as old as human knowledge itself, as is evident in writings from many ancient civilizations. In modern times, the task of organizing knowledge into systematic structures is studied by ontologists and library scientists, resulting in such well-known structures as the Dewey decimal system, the Library of Congress catalog, the AMS Mathematics Subject Classification, and the U.S. Patent Office subject classification [11, 68]. Subject-based organization routinely permeates our personal lives as we organize books, CDs, videos, and email. | {
"redpajama_set_name": "RedPajamaC4"
} | 5,026 |
\section{Introduction}
We study the mixing properties (in terms of the spectral gap) of the
Swendsen-Wang dynamics for the $q$-state Potts model on graphs
with bounded degree.
The Swendsen-Wang process is probably the most widely used algorithm
to generate (approximate) samples from the Potts model because,
empirically, it seems to be rapid for all graphs and temperatures,
at least for small $q$. But this is in general not true.
It is known that the process does not mix rapidly at some critical
temperature on the complete graph for all $q\ge3$, see Gore and
Jerrum \cite{GJ},
and on rectangular subsets of the hypercubic lattice $\mathbb{Z}^d$,
$d\ge2$, for $q$ sufficiently large, see the recent paper of
Borgs, Chayes and Tetali \cite{BCT}.
Furthermore, Li and Sokal \cite{LiS} proved that there is a lower
bound on the mixing time in terms of the \emph{specific heat}.
Results on rapid mixing of the Swendsen-Wang dynamics are also rare.
Rapid mixing is proven at all temperatures on trees and cycles for
any $q$, see Cooper and Frieze \cite{CF}, on narrow grids for any
$q$, see Cooper et al. \cite{CDFR},
and on the complete graph for $q=2$, see \cite{CDFR} and Long et al.
\cite{LNP}.
For rapid mixing results for sufficiently high or low temperatures
see e.g. Martinelli \cite{M2}, Cooper and Frieze \cite{CF} and
Huber \cite{Hu}.
In this article we prove that the spectral gap of the
Swendsen-Wang dynamics on graphs with bounded degree is bounded
from below
by some constant times the spectral gap of the (single-site)
heat-bath dynamics. The main result of this paper can be stated as
follows, see Theorem~\ref{theorem:main}.
Suppose that $P$ (resp. $P_{\rm HB}$) is the transition matrix of
the Swendsen-Wang (resp. heat-bath) dynamics, which is reversible
with respect to the Boltzmann distribution $\pi^{G}_{\beta,q}$ of
the $q$-state Potts model on the graph $G$ at inverse
temperature $\beta$.
Then there exists a constant $c>0$, depending only on $q$ and
$\beta$, such that
\[
\lambda(P) \;\ge\; c^\Delta\,\lambda(P_{\rm HB}),
\]
where $\lambda(\cdot)$ denotes the spectral gap and $\Delta$ is
the maximal degree of the underlying graph $G$.
In Theorem~\ref{theorem:main2}$'$ (Subsection 4.3.2) we prove this
statement also for the case that we have only a single vertex $v^*$
of large degree, where $\Delta$ is replaced by the second largest
degree and $P_{\rm HB}$ is replaced by the transition matrix of
the heat bath chain that leaves the color at $v^*$ constant.
In particular, this implies (see Corollary~\ref{coro:2d}) that
the Swendsen-Wang process is rapidly
mixing above the critical temperature for the two-dimensional
Potts model ($q\ge2$) and at the critical temperature for $q=2$.
Furthermore we get rapid mixing for graphs with maximum
degree $\Delta$ if $p=1-e^{-\beta}<1/\Delta$. If the graph is
planar, then $p=1-e^{-\beta}<1/(3\sqrt{(\Delta-3)})$ is
sufficient for $\Delta\ge6$ (see Corollary~\ref{coro:planar}).
This improves over the $1/(3(\Delta-1))$ bound of \cite{Hu}.
One may believe (and numerical results suggest) that there
could be a sharper bound on the spectral gap
of the Swendsen-Wang with an additional factor $N$
(= number of vertices of the graph) on the right hand side,
i.e. $\lambda(P)=\Omega(N \lambda(P_{\rm HB}))$. But proving
this does not seem to be possible with the techniques of this
paper.
We conjecture that the spectral gap of
Swendsen-Wang is (almost) the
same at high and low temperatures, but we were not able to
prove this. Nevertheless we present a modified version of the
Swendsen-Wang algorithm for planar graphs that makes an additional
step on the dual lattice. We prove that this process is rapidly
mixing for the two-dimensional Ising model at all temperatures
and for the two-dimensional Potts model for all non-critical
temperatures (see Corollary~\ref{coro:mod}).
\section{Preleminaries} \label{sec:prelim}
In this section we introduce the models and notations
to analyze the Swendsen-Wang algorithm.
First we define the Potts (Ising) and the random-cluster model and
explain their connection.
\subsection{The models}
Fix a natural number $q\ge2$ and a graph $G=(V,E)$
with finite vertex set
$V$ and edge set $E\subseteq\binom{V}{2}$,
where $\binom{V}{2}$ is the set of all subsets of $V$ with 2
elements. Let $N:=\abs{V}$.
As an example one can have in mind the two-dimensional square lattice
$G_L$ of side length $L$, i.e. $V=\{1,\dots,L\}^2\subset\mathbb{Z}^2$ and
$E=\{\{u,v\}\in V^2:\,\abs{u-v}=1\}$, where $\abs{\,\cdot\,}$ denotes
the $\ell_1$ norm.
If we consider more than one graph,
we denote by $V_G$ (resp. $E_G$) the vertex (resp. edge) set of the
graph $G$. Furthermore we write $u\leftrightarrow v$ if
$u$ and $v$ are \emph{neighbors} in $G$, i.e. $\{u,v\}\in E$,
and by $\deg_G(v)$ we denote the \emph{degree} of the vertex $v$
in $G$.
The \emph{$q$-state Potts model} on $G$ is defined as the set of
possible \emph{configurations}
$\O_{\rm P}=[q]^V$, where $[q]\,{:=}\,\{1,\dots,q\}$
is the set of \emph{colors} (or spins),
together with the probability measure
\vspace{1mm}
\[
\pi_\beta(\sigma) \;:=\; \pi^{G}_{\beta,q}(\sigma) \;=\;
\frac1{Z(G,\beta,q)}\,
\exp\left\{\beta\,\sum_{u,v:\, u\leftrightarrow v}
\Large{\mathds{1}}\bigl(\sigma(u)=\sigma(v)\bigr)\right\}
\]
for $\sigma\in\O_{\rm P}$, where $Z$ is the normalization
constant (also called partition function) and $\beta\ge0$ is called the
inverse temperature. This measure is called the \emph{Boltzmann}
(or Gibbs) \emph{distribution}.
If $q=2$ the Potts model is called the \emph{Ising model}.
A closely related model is the \emph{random cluster model}
(also known as the FK-model), see Fortuin and Kasteleyn \cite{FK}.
It is defined on the graph $G=(V,E)$ by its state space
$\O_{\rm RC}=\{A: A\subseteq E\}$ and the RC measure
\[
\mu_p(A) \;:=\; \mu^G_{p,q}(A) \;=\;
\frac1{Z(G,\log(\frac1{1-p}),q)}\,
\left(\frac{p}{1-p}\right)^{\abs{A}}\,q^{C(A)},
\]
where $p\in(0,1)$, $C(A)$ is the number of connected components
in the graph $(V,A)$, counting isolated vertices as a component,
and $Z(\cdot,\cdot,\cdot)$ is the same normalization constant as
for the Potts model (see \cite[Th. 1.10]{G1}).
For a detailed introduction and related topics see \cite{G1}.
There is a tight connection between the Potts model and the
random cluster model. Namely, if we set $p=1-e^{-\beta}$,
we can translate a Potts configuration $\sigma\sim\pi_\beta$ to
a random cluster state $A\sim\mu_p$ and vice versa.
To get a Potts configuration $\sigma\in\O_{\rm P}$ from
$A\in\O_{\rm RC}$ assign a random color independently to each
connected component of $(V,A)$.
For the reverse way include all edges $e=\{e_1,e_2\}\in E$
with $\sigma(e_1)=\sigma(e_2)$ to $A$ with probability $p$.
Hence sampling a Potts configuration according to $\pi_\beta$ is
equivalent to sampling a RC state from $\mu_p$ if both models are
defined on the same graph $G$ and $p=1-e^{-\beta}$.
For a proof see \cite{ES}.
\subsection{The heat-bath dynamics}
The \emph{heat bath dynamics} is probably the most studied
Markov chain related to the Potts model (especially for $q=2$).
This is because its mixing time is related
to some properties of the underlying model. See e.g.
the monograph of Martinelli \cite{M}.
The transition matrix of the
heat-bath chain on $\O_{\rm P}$ is defined by
\begin{equation}
P_{\rm HB}(\sigma,\sigma^{v,k})
\;:=\; P_{{\rm HB},\beta,q}^G(\sigma,\sigma^{v,k}) \;=\;
\frac{1}{N}\,\frac{\pi_\beta(\sigma^{v,k})}
{\sum_{l\in[q]}\pi_\beta(\sigma^{v,l})},
\label{eq:HB}
\end{equation}
where $\sigma^{v,k}(v)=k\in[q]$ and
$\sigma^{v,k}(u)=\sigma(u)$, $u\neq v$. Otherwise
$P_{\rm HB}(\sigma,\tau)=0$.
This transition matrix describes the process that, at each step,
chooses one vertex of the graph $G$ uniformly at random and changes
only the color of this vertex with respect to the conditional
distribution of $\pi_\beta$, given the color of all other
vertices. It is easy to prove that this Markov chain is reversible
with respect to $\pi_\beta$.
The spectral gap of $P_{\rm HB}$ is well known on some classes of graphs.
For positive results see e.g. \cite{M}, \cite{BKMP}, \cite{MT} and
\cite{LS}.
We will state a few results in Section \ref{sec:main}.
\subsection{The Swendsen-Wang algorithm}
Now we turn to the Swendsen-Wang algorithm \cite{SW}.
First we state the coupling of the Boltzmann distribution
$\pi_{\beta,q}^G$ and the random-cluster measure $\mu_{p,q}^G$
of Edwards and Sokal \cite{ES}.
Let us define
\[
\O(A) \;:=\; \bigl\{\sigma\in\O_{\rm P}:\,\sigma(u)=\sigma(v)\;\;
\forall\{u,v\}\in A\bigr\}, \quad A\in\O_{\rm RC},
\]
and
\[
E(\sigma) \;:=\; \bigl\{\{u,v\}\in E:\,\sigma(u)=\sigma(v)\bigr\},
\quad \sigma\in\O_{\rm P}.
\]
Obviously, we have for $\sigma\in\O_{\rm P}$ and $A\subset E$ that
$\sigma\!\in\!\O(A)\Leftrightarrow A\!\subset\!E(\sigma)$.
Let $\sigma\in\O_{\rm P}$, $A\in\O_{\rm RC}$ and $p=1-e^{-\beta}$,
then the joint measure of $(\sigma,A)\in\O_{\rm P}\times\O_{\rm RC}$ is
\[
\bar\mu(\sigma,A)\;:=\;\frac{1}{Z(G,\beta,q)}\,
\left(\frac{p}{1-p}\right)^{\abs{A}}\,\Large{\mathds{1}}(A\subset E(\sigma)).
\]
The marginal distributions of $\bar\mu$ are exactly $\pi_\beta$ and
$\mu_p$, respectively.
The \emph{Swendsen-Wang algorithm}
is based on this connection of the random cluster and Potts models
and performs the following two steps.
\begin{enumerate}
\item[1)] Given a Potts configuration $\sigma_t\in\O_{\rm P}$
on $G$, delete each edge of $E(\sigma_t)$ independently with
probability $1-p = e^{-\beta}$. This gives $A\in\O_{\rm RC}$.
\item[2)] Assign a random color independently to each
connected component of $(V,A)$. Vertices of the same component
get the same color. This gives $\sigma_{t+1}\in\O_{\rm P}$.
\end{enumerate}
This can be seen as first choosing $A$ with respect to the
conditional probability of $\bar\mu$ given $\sigma_t$ and then
choosing $\sigma_{t+1}$ with respect to $\bar\mu$ given $A$.
If we define the $\abs{\O_{\rm P}}\times\abs{\O_{\rm RC}}$-matrix
\begin{equation}
T_{G,p,q}(\sigma,A) \;:=\; \frac{\bar\mu(\sigma,A)}{\bar\mu(\sigma,\O_{\rm RC})}
\;=\; p^{\abs{A}}\,(1-p)^{\abs{E(\sigma)}-\abs{A}}\,
\Large{\mathds{1}}(A\subset E(\sigma))
\label{eq:T}
\end{equation}
and the $\abs{\O_{\rm RC}}\times\abs{\O_{\rm P}}$-matrix
\begin{equation}
T_{G,p,q}^*(A,\sigma) \;:=\; \frac{\bar\mu(\sigma,A)}{\bar\mu(\O_{\rm P},A)}
\;=\; q^{-C(A)}\,\Large{\mathds{1}}(\sigma\in\O(A)),
\label{eq:T*}
\end{equation}
then the \emph{transition matrix} of the
Swendsen-Wang dynamics (on $\O_{\rm P}$) is defined by
\begin{equation}
P(\sigma,\tau) \;:=\; P_{\beta,q}^G(\sigma,\tau)
\;=\; T_{G,p,q}\,T_{G,p,q}^*(\sigma,\tau),
\qquad \sigma,\tau\in\O_{\rm P}.
\label{eq:P}
\end{equation}
Note that the transition matrix of the Swendsen-Wang dynamics
on $\O_{\rm RC}$ is given by
\begin{equation}
\widetilde{P}(A,B) \;:=\; \widetilde{P}_{p,q}^G(A,B)
\;=\; T_{G,p,q}^*\,T_{G,p,q}(A,B), \qquad A,B\in\O_{\rm RC}.
\label{eq:P2}
\end{equation}
\vspace{1mm}
\subsection{Spectral gap}
In the following we want to estimate the spectral gap of certain
transition matrices of Markov chains. For an introduction to Markov
chains and techniques to bound the convergence rate to the stationary
distribution, see e.g. \cite{LPW}.
If the transition matrix $P$ with state space $\Omega$ is
ergodic, i.e. irreducible and aperiodic, and reversible with respect
to $\pi$, i.e.
\[
\pi(x)\,P(x,y) \;=\; \pi(y)\,P(y,x) \quad \text{ for all } x,y\in\O,
\]
we know that
$1=\xi_0>\xi_1\ge\dots\ge\xi_{\abs{\Omega}-1}>-1$, where
the $\xi_i$ are the (real) eigenvalues of $P$.
The \emph{spectral gap} of the Markov chain is defined by
$\lambda(P)=1-\max\bigl\{\xi_1,\abs{\xi_{\abs{\O}-1}}\bigr\}$.
If we are considering simultaneously a \emph{family} of graphs
$G = (V_G,E_G)$, we say that the chain is \emph{rapidly mixing}
for the given family if $\lambda(P)^{-1} = \mathcal{O}(|V_G|^C)$
for some $C < \infty$.
The eigenvalues of the Markov chain can be expressed in terms of norms
of the operator $P$ that maps from
$L_2(\pi):=(\mathbb{R}^\O,\Vert\cdot\Vert_\pi)$ to $L_2(\pi)$, where
scalar product and norm are given by
$\l f,g\r_\pi=\sum_{x\in\O}f(x) g(x) \pi(x)$ and
$\Vert f\Vert_\pi^2:=\sum_{x\in\O}f(x)^2\pi(x)$, respectively.
The operator is defined by
\[
Pf(x) \;:=\; \sum_{y\in\O}\,P(x,y)\,f(y)
\]
and represents the expected value of the function $f$ after one step of
the Markov chain starting in $x\in\O$.
The \emph{operator norm} of $P$ is
\[
\Vert P\Vert_\pi \;:=\; \Vert P\Vert_{L_2(\pi)\to L_2(\pi)}
\;=\; \max_{\Vert f\Vert_\pi\le1} \Vert Pf\Vert_\pi
\]
and we use $\Vert\cdot\Vert_\pi$ interchangeably for functions and
operators, because it will be clear from the context which
norm is used. It is well known that
$\lambda(P)=1-\norm{P-S_\pi}_\pi$ for reversible $P$,
where $S_\pi(x,y)=\pi(y)$.
We know that reversible $P$ are self-adjoint, i.e.
$P=P^*$, where $P^*$ is the (\emph{adjoint}) operator that
satisfies
$\l f, Pg\r_{\pi} \;=\; \l P^*f, g\r_{\pi}$ for all
$f,g\in L_2(\pi)$.
Note that if we look at $T_{G,p,q}$ from \eqref{eq:T} as an
operator mapping from $L_2(\mu_p)$ to $L_2(\pi_\beta)$, then
$T_{G,p,q}^*$ is the adjoint operator of $T_{G,p,q}$.
This proves that $P$ and $\widetilde{P}$ from \eqref{eq:P} and
\eqref{eq:P2} have the
same spectral gap.
\section{Main result} \label{sec:main}
In this section we prove by comparison that the spectral gap
of the Swendsen-Wang dynamics is bounded from below by some
constant times the spectral gap of the heat-bath chain.
Therefore we fix some value of $\beta\ge0$ and $q\ge2$.
We will prove the following theorem.
\begin{theorem} \label{theorem:main}
Suppose that $P$ (resp. $P_{\rm HB}$) is the transition matrix of the
Swendsen-Wang (resp. heat-bath) dynamics, which is reversible
with respect to $\pi^{G}_{\beta,q}$.
Then
\[
\lambda(P) \;\ge\; c_{\rm SW}\,\lambda(P_{\rm HB}),
\]
where
\[
c_{\rm SW} \;=\; c_{\rm SW}(G,\beta,q)
\;:=\; \frac{1}{2 q^2}\left(q\,e^{2\beta}\right)^{-4\Delta}
\]
with
\[
\Delta \;:=\; \max_{v\in V_G}\,\deg_G(v).
\]
\end{theorem}
By this result we get that the Swendsen-Wang dynamics is rapidly
mixing (on graphs with bounded degree), if the usual heat-bath
dynamics is rapidly mixing.
Unfortunately, this result seems to be off by a factor of $N$,
because we compare the SW dynamics with a Markov chain that changes
only the color of one vertex of the graph per step.
A similar comparison with the systematic scan heat-bath dynamics
would give a better bound (see e.g. \cite{Ha}).
But this does not seem to be possible with our techniques.
Now we will state some results that follow from Theorem
\ref{theorem:main}.
The first corollary deals with the two-dimensional Potts model.
Recall that the two-dimensional square lattice
$G_L$ of side length $L$ is given by $V=\{1,\dots,L\}^2\subset\mathbb{Z}^2$ and
$E=\{\{u,v\}\in V^2:\,\abs{u-v}=1\}$.
Using the known lower bounds on the spectral gap of the
heat bath dynamics on the square lattice \cite{MOS,BDC,LS},
we can obtain the following result:
\begin{corollary} \label{coro:2d}
Let $G_L$ be the square lattice of side-length $L$, $N=L^2$.
Then there exist
constants $c_{\beta}=c_{\beta}(\beta,q)>0$ and $m>0$ such that
\begin{itemize}
\item\quad $\lambda(P_{\beta,q}^{G_L})^{-1}\;\le\;c_{\beta} N$
\qquad for $\beta<\beta_c(q)=\ln(1+\sqrt{q})$
\vspace{2mm}
\item\quad $\lambda(P_{\beta,q}^{G_L})^{-1}\;\le\;c\,N^m$
\qquad for $q=2$ and $\beta=\beta_c(2)$,
\end{itemize}
where $c^{-1}=c_{\rm SW}(G_L,\beta_c,2)$.
\end{corollary}
Not much is known about the constant $c_\beta$, but there is
an explicit bound on the exponent $m$ in terms of an (unknown)
crossing probability in the random cluster model,
see \cite{LS}
\begin{proof}[Proof of Corollary \ref{coro:2d}]
We have to show that the heat-bath dynamics is rapidly mixing
in the cases stated above. The result for $q=2$ at the critical
temperature is given by Lubetzky and Sly \cite{LS}.
For the high temperature result we need the recent result of
Beffara and Duminil-Copin \cite[Th.~2]{BDC}. They prove that
one has exponential decay of connectivity in the RC model on $\mathbb{Z}^2$
for all $q\ge1$ and
$p<\frac{\sqrt{q}}{1+\sqrt{q}}=1-e^{-\beta_c(q)}$.
Together with \cite[Th.~3.6]{A} and \cite[Th.~3.2]{MOS} we get
that there exists a constant $\widetilde{c}_\beta>0$ such that
$\lambda(P_{\rm HB})^{-1}\le\widetilde{c}_\beta\,N$ on $G_L$ for
$\beta<\beta_c(q)$.
Thus the result follows from Theorem~\ref{theorem:main}.
\end{proof}
The next corollary relies on a result of Hayes \cite{Ha},
who gives a simple condition on $\beta$ for rapid mixing
of the heat-bath dynamics for the Ising model on graphs
of bounded degree.
Especially, we use his result for planar graphs.
\begin{corollary} \label{coro:planar}
Let $G$ be a graph with maximum degree $\Delta$.
Then for every $q\ge2$ and $\varepsilon>0$ the spectral gap of
the Swendsen-Wang dynamics satisfies
\[
\lambda(P_{\beta,q}^{G})^{-1}
\;\le\;\frac{c}{\varepsilon}\,N \log N,
\]
where $c^{-1}=\frac14\, c_{\rm SW}(G,\beta,q)$, if
\[
\beta \,\le\, 2 \,\frac{1-\varepsilon}{\Delta}
\]
or if $G$ is planar, $\Delta\ge6$ and
\[
\beta \,\le\, \frac{1-\varepsilon}{\sqrt{3 (\Delta-3)}}.
\]
\end{corollary}
If we state the bounds on the temperature in terms of the
RC parameter $p=1-e^{-\beta}$, this leads to the bounds
$p\le(1-\varepsilon)/\Delta$ in general and
$p\le(1-\varepsilon)/(3\sqrt{\Delta-3})$ for planar graphs.
\begin{proof}[Proof of Corollary \ref{coro:planar}]
In \cite[Prop.~14, Coro.~19]{Ha} the result is given for
$q=2$ and $\beta\le(1-\varepsilon)/\rho$ in terms of mixing time,
where $\rho$ is the principal eigenvalue of the graph.
The bounds on $\rho$ are stated therein.
Note that the different definition
of $\pi_\beta$ causes the additional factor 2.
To generalize this result to the $q$-state Potts model one
only has to prove an inequality like in Observation~11 in
\cite{Ha}.
But this is easily done, e.g. by induction over $q$.
Finally we have to bound the spectral gap in terms of the
mixing time. For this see e.g. \cite[Th.~12.4]{LPW}.
\end{proof}
\begin{remark}
Note that one can read the statement of Theorem \ref{theorem:main}
also as an upper bound on the spectral gap of the heat bath
dynamics. See e.g. \cite{BCT} for a result on slow mixing of the
Swendsen-Wang dynamics.
\end{remark}
\subsection{Proof of Theorem \ref{theorem:main}}
See Diaconis and Saloff-Coste \cite{DSC2} for an introduction
to comparison techniques for Markov chains.
For the comparison with the Swendsen-Wang dynamics we will
analyze the Markov chain with transition matrix
\begin{equation}
Q=P_{\rm HB}\,P\,P_{\rm HB}.
\label{eq:Q}
\end{equation}
Since $P_{\rm HB}$ and $P$ are reversible with respect to $\pi_\beta$,
$Q$ is also reversible.
\begin{lemma} \label{lemma:Q}
With the definitions from above we get
\[
\lambda(Q) \;\ge\; \lambda(P_{\rm HB}^2) \;\ge\; \lambda(P_{\rm HB}).
\]
\end{lemma}
To prove a lower bound on the spectral gap of $P$ it remains to
prove $\lambda(P)\ge c\lambda(Q)$ for some $c>0$.
For this we need an estimate of the transition probabilities of the
Swendsen-Wang dynamics on $G$ with respect to some subgraph of $G$.
Therefore we prove the following lemma.
\begin{lemma} \label{lemma:G0}
Let $G=(V,E)$ be a graph and $G_0=(V,E_0)$ be a spanning
subgraph of $G$ with $E_0\subset E$. Define $P_G := P_{\beta,q}^G$.
Then
\vspace{2mm}
\[
c_1^{\abs{E\setminus E_0}}\,P_{G_0}(\sigma,\tau)
\;\le\; P_G(\sigma,\tau)
\;\le\; c_2^{\abs{E\setminus E_0}}\,P_{G_0}(\sigma,\tau)
\vspace{2mm}
\]
for all $\sigma,\tau\in\O_{\rm P}$, where
\[
c_1 \;=\; c_1(\beta) \;:=\; e^{-\beta}
\]
and
\[
c_2 \;=\; c_2(\beta,q) \;:=\; 1+q\, (e^{\beta}-1).
\]
\end{lemma}
\begin{proof}
The first inequality is already known from the proof of Lemma 3.3 in
\cite{BCT}, but we state it here for completeness.
Let $p=1-e^{-\beta}$, then
\[\begin{split}
P_G(\sigma,\tau) \;&=\; \sum_{A\subset E} \,p^{\abs{A}}\,
(1-p)^{\abs{E(\sigma)}-\abs{A}}\, q^{-C(A)}\,
\Large{\mathds{1}}\bigl(A\subset E(\sigma)\cap E(\tau)\bigr)\\
&\ge\; \sum_{A\subset E_0} \,p^{\abs{A}}\,
(1-p)^{\abs{E(\sigma)}-\abs{A}}\, q^{-C(A)}\,
\Large{\mathds{1}}(A\subset E(\sigma)\cap E(\tau))\\
&\ge\; (1-p)^{\abs{E(\sigma)}-\abs{E_0(\sigma)}}\,P_{G_0}(\sigma,\tau)
\;\ge\; (1-p)^{\abs{E\setminus E_0}}\,P_{G_0}(\sigma,\tau).
\end{split}\]
For the second inequality suppose for now $E_0=E\setminus\{e\}$
for some $e\in E$ and note that $C(A\cup\{e\})\ge C(A)-1$. We get
\[\begin{split}
P_G(\sigma,\tau) \;&=\; \sum_{A\subset E(\sigma)\cap E(\tau)}
\,p^{\abs{A}}\, (1-p)^{\abs{E(\sigma)}-\abs{A}}\, q^{-C(A)} \\
&=\; \sum_{\substack{A\subset E(\sigma)\cap E(\tau):\\ e\in A}}
\,p^{\abs{A}}\, (1-p)^{\abs{E(\sigma)}-\abs{A}}\, q^{-C(A)} \\
&\qquad\quad + \sum_{\substack{A\subset E(\sigma)\cap E(\tau):\\
e\notin A}}
\,p^{\abs{A}}\, (1-p)^{\abs{E(\sigma)}-\abs{A}}\, q^{-C(A)} \\
&\le\; \sum_{A'\subset E_0(\sigma)\cap E_0(\tau)}
\,p^{\abs{A'\cup\{e\}}}\,
(1-p)^{\abs{E(\sigma)}-\abs{A'\cup\{e\}}}\, q^{-C(A'\cup\{e\})} \\
&\qquad\quad + \sum_{A'\subset E_0(\sigma)\cap E_0(\tau)}
\,p^{\abs{A'}}\, (1-p)^{\abs{E(\sigma)}-\abs{A'}}\, q^{-C(A')} \\
&\le\; \frac{q\,p}{1-p}\, \sum_{A'\subset E_0(\sigma)\cap E_0(\tau)}
\,p^{\abs{A'}}\, (1-p)^{\abs{E(\sigma)}-\abs{A'}}\, q^{-C(A')} \\
&\qquad\quad + \,(1-p)^{\abs{E(\sigma)}-\abs{E_0(\sigma)}}\,
P_{G_0}(\sigma,\tau) \\
&\le\; \left(\frac{q\,p}{1-p}\, \,+\, 1\right)\,P_{G_0}(\sigma,\tau)
\;=\; \bigl(1 \,+\, q\, (e^{\beta}-1)\bigr)\,P_{G_0}(\sigma,\tau).
\end{split}\]
For $\abs{E\setminus E_0}>1$ one can iterate this technique
$\abs{E\setminus E_0}$ times. This proves the statement.\\
\end{proof}
We use this lemma to prove that the transition probability from
$\sigma$ to $\tau$ is similar to the probability of going from
a neighbor of $\sigma$ to a neighbor of $\tau$.
Recall that $\sigma^{v,k}$ is defined by $\sigma^{v,k}(v)=k\in[q]$ and
$\sigma^{v,k}(u)=\sigma(u)$, $u\neq v$.
\begin{lemma} \label{lemma:P}
Let $\sigma,\tau\in\O_{\rm P}$, $v\in V_G$ and $k,l\in[q]$.
Then
\[
P_G(\sigma^{v,k},\tau^{v,l}) \;\le\;
c_3^{\deg_G(v)}\;
P_G(\sigma,\tau)
\]
with
\[
c_3 \;=\; c_3(\beta,q) \;:=\; q\,e^{2\beta} - (q-1)\,e^\beta.
\]
\end{lemma}
\begin{proof}
Define $E_v:=\{e\in E_G: v\in e\}$ and $G_v:=(V,E\setminus E_v)$.
Then, $v\in V_G$ is an isolated vertex in $G_v$.
By the definition of the Swendsen-Wang dynamics we get that
\[
P_{G_v}(\sigma^{v,k},\tau^{v,l})
\;=\; P_{G_v}(\sigma,\tau).
\]
If we set $E_0=E\setminus E_v$ we get
$\abs{E\setminus E_0}=\deg_G(v)$ and by Lemma \ref{lemma:G0}
\[\begin{split}
P_{G}(\sigma^{v,k},\tau^{v,l})
\;&\le\; c_2^{\deg_G(v)}\,P_{G_v}(\sigma^{v,k},\tau^{v,l})
\;=\; c_2^{\deg_G(v)}\,P_{G_v}(\sigma,\tau) \\
&\le\;\left(\frac{c_2}{c_1}\right)^{\deg_G(v)}\,P_{G}(\sigma,\tau)\\
\end{split}\]
with $c_1$ and $c_2$ from Lemma \ref{lemma:G0}.\\
\end{proof}
Now we are able to prove the main result.
\begin{proof}[Proof of Theorem \ref{theorem:main}]
Because of Lemma \ref{lemma:Q} we only have to prove
$\lambda(P)\ge c_{\rm SW}\,\lambda(Q)$.
Let
\[
c \;:=\; \max_{\substack{\sigma_1,\sigma_2,\tau_1,\tau_2\in\O_{\rm P}\\
\sigma_1\sim \sigma_2,\,\tau_1\sim \tau_2}}\;
\frac{P(\sigma_1,\tau_1)}{P(\sigma_2,\tau_2)},
\]
where $\sigma\sim\tau:\Leftrightarrow
\sum_{v\in V}\abs{\sigma(v)-\tau(v)}\le1$.
Note that $P_{\rm HB}(\sigma,\tau)\neq0$ if and only if
$\sigma\sim\tau$.
We get for $\sigma_1,\tau_1\in\O_{\rm P}$ that
\[\begin{split}
Q(\sigma_1,\tau_1)
\;&=\; \sum_{\sigma_2,\tau_2\in\O_{\rm P}}\,
P_{\rm HB}(\sigma_1,\sigma_2)\,P(\sigma_2,\tau_2)\,
P_{\rm HB}(\tau_2,\tau_1) \\
&\le\; c\,P(\sigma_1,\tau_1) \;\sum_{\sigma_2\sim\sigma_1}
P_{\rm HB}(\sigma_1,\sigma_2)\,
\sum_{\tau_2\sim\tau_1}
P_{\rm HB}(\tau_2,\tau_1) \\
&\le\; q\,c\,P(\sigma_1,\tau_1).
\end{split}\]
It is well-known that
\[
\lambda(P^2) \;=\;
\min\left\{\frac{\mathcal{E}_{P^2}(f)}{\mbox{\rm Var}(f)}:\,\mbox{\rm Var}(f)\neq0\right\},
\]
where $\mbox{\rm Var}(f)=\Vert f - S_{\pi_\beta} f\Vert_{\pi_\beta}^2$ and
\[
\mathcal{E}_P(f)=\frac12\sum_{\sigma,\tau}(f(\sigma)-f(\tau))^2\,
\pi_\beta(\sigma)\,P(\sigma,\tau).
\]
We use $P^2$ instead of $P$, because this representation holds
only if $\lambda(P)=1-\xi_1(P)$ \cite{DSC}.
From the calculation above we
have $\mathcal{E}_{P^2}(f)\ge\frac{1}{q^2 c^2}\,\mathcal{E}_{Q^2}(f)$ for every
$f\in L_2(\pi_\beta)$ and so
$\lambda(P^2)\ge\frac{1}{q^2 c^2}\lambda(Q^2)$. Since we have for
reversible $P$ that $\lambda(P)\le\lambda(P^2)\le2\lambda(P)$,
we get
\[
\lambda(P) \;\ge\; \frac{1}{2 q^2 c^2}\,\lambda(Q).
\]
It remains to bound $c$. With $c_3$ from Lemma
\ref{lemma:P} we get for
$\sigma_1,\sigma_2,\tau_1, \tau_2\in\O_{\rm P}$ with
$\sigma_1\sim \sigma_2$ and $\tau_1\sim \tau_2$
\[
\frac{P(\sigma_1,\tau_1)}{P(\sigma_2,\tau_2)}
\;\le\; c_3^{\Delta}\,\frac{P(\sigma_2,\tau_1)}{P(\sigma_2,\tau_2)}
\;\le\; c_3^{2 \Delta}\,\frac{P(\sigma_2,\tau_2)}{P(\sigma_2,\tau_2)}
\;=\; c_3^{2 \Delta}.
\]
Finally,
\[
\lambda(P) \;\ge\; \frac{1}{2 q^2 c^2}\,\lambda(Q)
\;\ge\; \frac{1}{2 q^2}\,c_3^{-4 \Delta}\,\lambda(Q)
\;\ge\; \frac{1}{2 q^2}\,(q\,e^{2\beta})^{-4 \Delta}\,\lambda(Q).
\]
This completes the proof.\\
\end{proof}
\section{The Modified Swendsen-Wang} \label{sec:ModSW}
With the techniques from the last section we do not get any positive
result for low temperatures.
However, the Swendsen-Wang dynamics seems to be rapidly mixing also
at low enough temperatures. Unfortunately, we were not able to prove it.
In this section we introduce a modified version of the
Swendsen-Wang dynamics for planar graphs that is rapidly mixing
for the two-dimensional Ising model at all temperatures.
Roughly speaking the chain makes one step at high and
one step at low temperatures.
\subsection{Dual graphs}
A graph $G$ is called \emph{planar} if one can embed it into $\mathbb{R}^2$
such that two edges of $G$ intersect only at a common endvertex.
We fix such an embedding for $G$.
Then we define the \emph{dual graph} $G_D=(V_D,E_D)$ of $G$ as follows.
Place a dual vertex in each face (including the infinite outer one) of
the graph $G$ and connect 2 vertices by the dual edge $e_D$ if and
only if the corresponding faces of $G$ share the boundary edge $e$
(see e.g. \cite[Section 8.5]{G2}).
It is clear, that the number of
vertices can differ in the dual graph, but we have the same number of
edges.\\
Additionally we define a \emph{dual RC configuration}
$A_D\subseteq E_D$ in $G_D$ to a RC state $A\subseteq E$ in
$G$ by
\[
e\in A \;\Longleftrightarrow\; e_D\notin A_D,
\]
where $e_D$ is the edge in $E_D$ that intersects $e$ in our (fixed)
embedding. (By construction, this edge is unique.)
See Figure \ref{fig-dual} for the graph $G_L$ with $L=3$ and its
dual graph $(G_L)_D$ together with 2 corresponding RC states.
\begin{figure}[ht]
\hspace{5mm}\scalebox{1}{\input{dual-graph}
\hspace*{-2cm}\raisebox{-6mm}{\input{dual-conf}}}
\vspace*{-5mm}
\caption[Dual graph and dual RC state]{Left: The graph $G_3$ (solid)
and its dual (dashed).\\ Right: A RC state on $G_3$ (solid) and its
dual configuration (dashed).}
\label{fig-dual}
\vspace{5mm}
\end{figure}
It is easy to obtain (see \cite[p.~134]{G1}) that the random
cluster models on the (finite)
graphs $G$ and $G_D$ are related by the equality
\begin{equation} \label{eq:dual}
\mu_{p,q}^G(A) \;=\; \mu^{G_D}_{p^*,q}(A_D),
\end{equation}
where the dual parameter $p^*$ satisfies
\begin{equation} \label{eq:dual-p}
\frac{p^*}{1-p^*} \;=\; \frac{q \,(1-p)}{p}.
\end{equation}
The self-dual point of this relation is given by
$p_{\rm sd}(q)=\frac{\sqrt{q}}{1+\sqrt{q}}$, which
corresponds by $p=1-e^{-\beta}$ to the critical temperature
of the $q$-state Potts model $\beta_c(q)=\log(1+\sqrt{q})$
on $\mathbb{Z}^2$ \cite{BDC}.
In the following we denote by $\beta^*$ the ``dual'' value
of $\beta$, i.e. we have $p^*=1-e^{-\beta^*}$, $p=1-e^{-\beta}$
and \eqref{eq:dual-p}. If $G$ is fixed, we write
$\mu^*$ instead of $\mu^{G_D}$.
\subsection{The algorithm}
To describe the transition matrix of the modified Swendsen-Wang
algorithm, we define the (square)
$2^{\abs{E}}\times2^{\abs{E_D}}$-matrix
\[
D(A,B) \;:=\; \Large{\mathds{1}}(B=A_D),
\quad A\subset E, B\subset E_D.
\]
By \eqref{eq:dual} we get that
$\mu_p\cdot D = \mu^*_{p^*}$,
respectively $\mu^*_{p^*}\cdot D^* = \mu_{p}$
(distributions are seen as row vectors),
where $D^*(B,A)=D(A,B)$, $A\subset E$, $B\subset E_D$.
Now we can state the algorithm.
The transition matrix of the \emph{modified Swendsen-Wang dynamics}
(for $\O_{\rm P}$) on the planar graph $G$ is defined by
\begin{equation}
M \;:=\; M^G_{p,q} \;=\;
T_{G,p,q}\,D\,T^*_{G_D,p^*,q}\,T_{G_D,p^*,q}\,D^*\,T^*_{G,p,q},
\label{eq:M}
\end{equation}
where $T_{G,p,q}$, $T^*_{G,p,q}$ and $p^*$ are defined in
\eqref{eq:T}, \eqref{eq:T*} and \eqref{eq:dual-p}.
This dynamics is reversible with respect to $\pi_{\beta,q}^G$.
Although this seems to be complicated, it has an easy interpretation.
The dynamics defined by the transition matrix $M$ performs the
following steps:
\begin{enumerate}
\item[1)] Given a Potts configuration $\sigma_t$ on $G$,
generate a random
cluster state $A\subset E_G$ with respect to
$T_{G,p,q}(\sigma,\cdot)$.
\item[2)] Make one step of the Swendsen-Wang dynamics
$\widetilde{P}_{p^*,q}^{G_D}$ starting at $A_D\subset E_{G_D}$ to
get a random cluster state $B_D\subset E_{G_D}$.
\item[3)] Generate $\sigma_{t+1}$ with respect to
$T^*_{G,p,q}(B,\cdot)$.
\end{enumerate}
\begin{remark}
A similar, and even simpler, algorithm was considered by
Edwards and Sokal \cite{Scom} around 1988 for the
two-dimensional torus, which is ``self-dual'' in some sense.
In this case they did not have the exact (planar) duality as
given in \eqref{eq:dual}, but one can use a Metropolis
acceptance/rejection rule to go to a ``dual configuration''.
This algorithm could be advantageous near the critical
temperature $\beta_c(q)$, but away from it the Metropolis
step is rejected with probability close to 1.
\end{remark}
The next result shows that the modified Swendsen-Wang has a larger
spectral gap than the original Swendsen-Wang on $G$ and $G_D$,
respectively.
\begin{proposition} \label{theorem:mod}
Let $P_{\beta,q}^G$ be the Swendsen-Wang dynamics on a planar
graph $G$, which is
reversible with respect to $\pi_{\beta,q}^G$, and $M$
as in \eqref{eq:M}. Then
\[
\lambda(M) \;\ge\; \max\Bigl\{\lambda(P_{\beta,q}^G),\,
\lambda(P_{\beta^*,q}^{G_D})\Bigr\}.
\]
\end{proposition}
\begin{proof}
Let $\pi:=\pi_{\beta,q}^G$ and $\pi^*:=\pi_{\beta^*,q}^{G_D}$ as
well as $\mu:=\mu_{p,q}^G$ and $\mu^*:=\mu_{p^*,q}^{G_D}$.
Additionally, let $T_1:=T_{G,p,q}$, $T_2:=T^*_{G,p,q}$ and
$B:=D\,T^*_{G_D,p^*,q}\,T_{G_D,p^*,q}\,D^*$.
Then, since $M$ is reversible, for all $n\in\mathbb{N}$
(recall that $S_\pi(\sigma,\tau)=\pi(\tau)$)
\[\begin{split}
\norm{M-S_\pi}_\pi^{n+1}
\;&=\; \norm{ T_1\,B\,T_2-S_\pi}_\pi^{n+1}
\;=\; \norm{(T_1\,B\,T_2 - S_\pi)^{n+1}}_\pi \\
\;&=\; \norm{(T_1\,B\,T_2)^{n+1} - S_\pi}_\pi
\;=\; \norm{T_1\,(B\,T_2\,T_1)^n\,B\,T_2 - S_\pi}_\pi \\
\;&\le\; \norm{T_1}_{(\mu,\pi)}\,\norm{B}_\mu\,\norm{T_2}_{(\pi,\mu)}\,
\norm{(B\,T_2\,T_1)^n - S_\mu}_\mu \\
\;&\le\; \norm{T_1}_{(\mu,\pi)}^2\,\norm{B}_\mu^2\,
\norm{T_2}_{(\pi,\mu)}^2\,\norm{(T_1\,B\,T_2)^{n-1}-S_\pi}_\pi \\
\;&\le\; \norm{T_1\,B\,T_2 - S_\pi}_\pi^{n-1}
\;=\; \norm{M-S_\pi}_\pi^{n-1} ,
\end{split}\]
where $\norm{T_1}_{(\mu,\pi)}:=\norm{T_1}_{L_2(\mu)\to L_2(\pi)}$.
The last inequality holds because the operator norms are all less
than one (since $\pi T_1=\mu$, $\mu B=\mu$ and $\mu T_2=\pi$).
Taking the $n$-th root and let $n$ tends to infinity we obtain
\[\begin{split}
\norm{M - S_\pi}_\pi
\;&=\; \lim_{n\to\infty} \norm{(B\,T_2\,T_1 - S_\mu)^n}_\mu^{1/n}
\;\le\; \norm{B\,T_2\,T_1 - S_\mu}_\mu \\
\;&\le\; \norm{B-S_\mu}_\mu \; \norm{T_2\,T_1- S_\mu}_\mu
\;=\; \norm{B-S_\mu}_\mu \; \norm{\widetilde{P}^G_{p,q}- S_\mu}_\mu
\end{split}\]
with $\widetilde{P}^G_{p,q}$ from \eqref{eq:P2}.
Since $S_{\mu}=D\,S_{\mu^*}\,D^*$ we get
\[\begin{split}
\norm{ M-S_\pi}_\pi
\;&\le\; \norm{D(\widetilde{P}_{p^*,q}^{G_D}-S_{\mu^*})D^*}_{\mu}
\norm{\widetilde{P}^G_{p,q}- S_\mu}_\mu
\le \norm{P_{\beta^*,q}^{G_D}-S_{\pi^*}}_{\pi^*}
\norm{P_{\beta,q}^{G}-S_{\pi}}_{\pi} \\
&\le\; \min\Bigl\{\norm{ P_{\beta^*,q}^{G_D}-S_{\pi^*}}_{\pi^*},\,
\norm{P_{\beta,q}^{G}-S_{\pi}}_{\pi}\Bigr\}.
\end{split}\]
This yields the result, because $\lambda(M)=1-\norm{ M-S_\pi}_\pi$.
\end{proof}
\subsection{Examples}
Now we present lower bounds on the spectral gap of
the modified Swendsen-Wang dynamics for two examples.
The first one is the Potts model on trees, which is
mainly a toy example, because one knows that the
original Swendsen-Wang is rapidly mixing at all
temperatures \cite{CF}. We present this example to
show that the modified Swendsen-Wang can improve the
original one at all temperatures.
The second example is the Potts model on the square lattice,
where we prove rapid mixing for all non-critical temperatures.
\subsubsection{Potts model on trees}
Let the graph $G$ be a tree.
By definition, $G$ has only one face:
the outer one. Hence, the dual graph $G_D$ has only one vertex with
$\abs{E_G}$ loops. Therefore the Swendsen-Wang dynamics
$\widetilde{P}$ on $G_D$ is simply independent bond percolation.
It follows $\lambda(\widetilde{P})=1$ and by
Proposition~\ref{theorem:mod} we get
\begin{corollary}
Let $G$ be a tree and $M$ be the transition matrix of the modified
Swendsen-Wang dynamics as in \eqref{eq:M}. Then we get for all
$\beta\ge0$ and $q\in\mathbb{N}$ that
\[
\lambda(M) \;=\; 1.
\]
\end{corollary}
\subsubsection{The Potts model on the square lattice}
Now we consider the Potts model on the two-dimensional square
lattice of side length $L$. This is the graph $G=G_L=(V,E)$ with
$V=\{1,\dots,L\}^2$ and $E=\{\{u,v\}\in V^2:\,\abs{u-v}=1\}$.
For the definition of the dual graph we use the following
notation of the union of graphs.
Let $G=(V,E)$, $W\subset V$ and $v^*\notin V$ be an additional
auxiliary vertex. Then we denote by $G' =G\cup_W v^*$ the graph
$G'=(V',E')$ with $V'=V\cup \{v^*\}$ and
$E'=E\cup\{\{v^*,u\}:\,u\in W\}$.
Furthermore we denote the ``boundary'' of $G_L$ by
$\delta_L=\bigl\{(v_1,v_2)\in \mathbb{Z}^2:\,
v_1\in\{1,L\} \text{ or } v_2\in\{1,L\}\bigr\}\subset V_{G_L}$.\\
By the construction of the dual graph it is easy to obtain
(see Figure~\ref{fig-dual}) that
the dual graph of the square lattice $G=G_L$ is given by
\[
G_D \;=\; G_{L-1}\,\cup_{\delta_{L-1}}\,v^*.
\]
We cannot apply Theorem~\ref{theorem:main} directly for this graph,
because it has no bounded degree, i.e.
$\deg_{G_D}(v^*)=\abs{\delta_{L-1}}=4(L-1)$. Hence, we have to
prove the following modification of Theorem \ref{theorem:main}.
\begin{cit}[Theorem~\ref{theorem:main}$'$]\label{theorem:main2}
Let $P$ be the transition matrix of the Swendsen-Wang dynamics,
which is reversible with respect to $\pi^G_{\beta,q}$.
Furthermore let $v\in~V_G$ be any vertex, $k\in[q]$,
\[
\Lambda_v^k \;:=\; \{\sigma\in\O_P:\,\sigma(v)=k\}
\]
and
\begin{equation}
P_{\Lambda_v^k}(\sigma,\tau) \;=\; \frac{1}{N-1}\,
\left(1+\frac{\pi_\beta(\sigma\;\rule[-1.5mm]{0.2mm}{5mm}\;\Lambda_v^k)}
{\pi_\beta(\tau\;\rule[-1.5mm]{0.2mm}{5mm}\;\Lambda_v^k)}\right)^{-1}
\label{eq:HB2}
\end{equation}
for $\sigma,\tau\in\Lambda_v^k$ with $\abs{\sigma-\tau}=1$.\\
Then
\[
\lambda(P) \;\ge\; \widetilde{c}_{\rm SW}\,\lambda(P_{\Lambda_v^k}^2),
\]
where
\[
\widetilde{c}_{\rm SW} \;=\; \widetilde{c}_{\rm SW}(G,\beta,q)
\;:=\; \frac1{2 q^2}\left(q\,e^{2\beta}\right)^{-4\widetilde{\Delta}}
\]
with
\[
\widetilde{\Delta} \;:=\; \max_{u\in V_G\setminus\{v\}}\,\deg_G(u).
\]
\end{cit}
\begin{proof}
The proof of this theorem is very similar to the proof of
Theorem \ref{theorem:main}.
First we define the following ``flip''
transition matrices. Let $\sigma\in\O_{\rm P}$ and
$\tau\in\Lambda_v^k$, then
\[
F_1(\sigma,\tau) \;:=\; \Large{\mathds{1}}(\tau\,=\,\sigma-\sigma(v)+k)
\]
and
\[
F_2(\sigma,\tau) \;:=\; F_1^*(\sigma,\tau) \;=\;
\frac{1}{q}\,\sum_{l=0}^{q-1}\,\Large{\mathds{1}}(\tau=\sigma+l),
\]
where $(\sigma+l)(u)\,:=\,(\sigma(u)+l-1\mod q)+1$ for
$\sigma\in\O_{\rm P}$ and $l\in\mathbb{Z}$.
Let $\pi:=\pi^G_{\beta,q}$ and $\widetilde{\pi}:=
\pi_{\beta,q}^G(\cdot\,\rule[-1.5mm]{0.2mm}{5mm}\,\Lambda_v^k)$.
It is easy to check that $\pi F_1=\widetilde{\pi}$ and
$\widetilde{\pi}F_2=\pi$.
Following the same ideas as in Section \ref{sec:main}
with
\[
\widetilde{Q} \;=\; F_1\,P_{\Lambda_v^k}\,F_2\,P\,
F_1\,P_{\Lambda_v^k}\,F_2,
\]
which is reversible with respect to $\pi=\pi_{\beta,q}^G$, we get
\[
\norm{ \widetilde{Q}-S_\pi}_\pi \;\le\;
\norm{ P_{\Lambda_v^k}-S_{\widetilde{\pi}}}_{\widetilde{\pi}}^2
\;=\; \norm{ P_{\Lambda_v^k}^2-S_{\widetilde{\pi}}}_{\widetilde{\pi}}.
\]
Similarly as Lemma \ref{lemma:Q},
$\lambda(\widetilde{Q})\ge\lambda(P_{\Lambda_v^k}^2)$.
It remains to prove that
$\lambda(P)\ge\widetilde{c}_{\rm SW}\,\lambda(\widetilde{Q})$.
For this we define for $\sigma\in\O_{\rm P}$,
$\sigma^k:=\sigma-\sigma(v)+k\in\Lambda_v^k$ and
$\sigma\equiv\tau :\Leftrightarrow \{\exists l\in[q]: \tau=\sigma^l\}$.
By the construction of the Swendsen-Wang dynamics we have for
$k,l\in[q]$ that
\[
P(\sigma^k,\tau^l) \;=\; P(\sigma,\tau) \qquad
\forall\sigma,\tau\in\O_{\rm P}.
\]
Hence we get for $\sigma,\tau\in\O_{\rm P}$ with
\[
\widetilde{c} \;:=\;
\max_{\substack{\sigma_1,\sigma_2,\tau_1,\tau_2\in\Lambda_v^k\\
\sigma_1\sim \sigma_2,\,\tau_1\sim \tau_2}}\;
\frac{P(\sigma_1,\tau_1)}{P(\sigma_2,\tau_2)}
\]
that
\[\begin{split}
\widetilde{Q}(\sigma,\tau)
\;&=\; \sum_{\substack{\sigma_1,\tau_1\in\Lambda_v^k,\\
\sigma_2,\tau_2\in\O_{\rm P}}}
P_{\Lambda_v^k}(\sigma^k,\sigma_1)\,F_2(\sigma_1,\sigma_2)\,
P(\sigma_2,\tau_2)\,P_{\Lambda_v^k}(\tau_2^k,\tau_1)\,
F_2(\tau_1,\tau) \\
\;&=\; \sum_{\substack{\sigma_1,\tau_1\in\Lambda_v^k,\\
\tau_2\in\O_{\rm P}}}
P_{\Lambda_v^k}(\sigma^k,\sigma_1)\,
P(\sigma_1,\tau_2^k)\,P_{\Lambda_v^k}(\tau_2^k,\tau_1)\,
F_2(\tau_1,\tau) \\
&\le\; \widetilde{c}\,P(\sigma^k,\tau) \;
\sum_{\sigma_1\sim\sigma^k} P_{\Lambda_v^k}(\sigma^k,\sigma_1)\,
\sum_{\tau_1\equiv\tau} F_2(\tau_1,\tau) \,
\sum_{\tau_2:\tau_2^k\sim\tau_1} P_{\Lambda_v^k}(\tau_2^k,\tau_1) \\
&=\; \widetilde{c}\,P(\sigma,\tau) \;
\sum_{\tau_1\equiv\tau} F_2(\tau_1,\tau) \,
\sum_{\tau_2:\tau_2^k\sim\tau_1} P_{\Lambda_v^k}(\tau_2^k,\tau_1).
\end{split}\]
But $\tau_1$ is unique, because $\tau_1\equiv\tau$ and
$\tau_1\in\Lambda_v^k$. Therefore,
\[\begin{split}
\widetilde{Q}(\sigma,\tau)
\;&\le\; \widetilde{c}\,P(\sigma,\tau) \; \frac{1}{q} \,
\sum_{\tau_2:\tau_2^k\sim\tau^k} P_{\Lambda_v^k}(\tau_2^k,\tau_1) \\
\;&=\; \widetilde{c}\,P(\sigma,\tau) \; \frac{1}{q} \,
\sum_{l=1}^q\,\sum_{\tau_2\in\Lambda_v^l:\tau_2^k\sim\tau^k}
P_{\Lambda_v^k}(\tau_2^k,\tau_1) \\
\;&\le\; q\,\widetilde{c}\,P(\sigma,\tau).
\end{split}\]
With $c_3$ from Lemma~\ref{lemma:P} we get for
$\sigma_1,\sigma_2,\tau_1,\tau_2\in\Lambda_v^k$ with
$\sigma_1\sim \sigma_2$ and $\tau_1\sim \tau_2$
(since $\sigma_1(v)=\sigma_2(v)$ and $\tau_1(v)=\tau_2(v)$) that
\[
\frac{P(\sigma_1,\tau_1)}{P(\sigma_2,\tau_2)}
\;\le\; c_3^{\widetilde{\Delta}}\,
\frac{P(\sigma_2,\tau_1)}{P(\sigma_2,\tau_2)}
\;\le\; c_3^{2 \widetilde{\Delta}}\,
\frac{P(\sigma_2,\tau_2)}{P(\sigma_2,\tau_2)}
\;=\; c_3^{2 \widetilde{\Delta}}.
\]
By the same ideas as in the proof of Theorem~\ref{theorem:main}
we conclude
\[
\lambda(P) \;\ge\; \frac{1}{2 q^2 \widetilde{c}^2}\,\lambda(Q)
\;\ge\; \frac{1}{2 q^2}\,c_3^{-4 \widetilde{\Delta}}\,\lambda(Q)
\;\ge\; \frac{1}{2 q^2}\,(q\,e^{2\beta})^{-4 \widetilde{\Delta}}\,
\lambda(\widetilde{Q}).
\]
This completes the proof.\\
\end{proof}
Now we can apply Theorem~\ref{theorem:main}$'$ for
$G_D \;=\; G_{L-1}\,\cup_{\delta_{L-1}}\,v^*$ with $v=v^*$, because
then $\widetilde{\Delta}=4$. For the result on the spectral gap
of the Markov chain with transition matrix $M$ we need a lower bound
on the spectral gap of $P_{\Lambda_{v^*}^k}$ on $G_D$.
But this is simply the heat-bath dynamics on $G_{L-1}$ with
constant (i.e. $k$) boundary condition. From the well-known
results on the spectral gap of the heat-bath dynamics on the square
lattice at high temperatures (see \cite{MO1} or \cite{MOS}) we know
that the $\mathcal{O}(N)$ bound on the inverse-gap also holds for
arbitrary boundary conditions. Hence, we get the following.
\begin{corollary} \label{coro:mod}
Let $G_L$ be the square lattice of side-length $L$, $N=L^2$
and $M=M^{G_L}_{p,q}$ as in \eqref{eq:M}. Then there exist
constants $c_\beta>0$ and $m>0$ such that
\begin{itemize}
\item\quad $\lambda(M)^{-1}\;\le\;c_\beta\,N$
\qquad for $\beta\neq\beta_c(q)=\ln(1+\sqrt{q})$
\vspace{2mm}
\item\quad $\lambda(M)^{-1}\;\le\;c\,N^m$
\qquad for $q=2$ and $\beta=\beta_c(2)$,
\end{itemize}
where $c=c_{\rm SW}(G_L,\beta_c,2)$.
\end{corollary}
\vspace{1cm}
\subsection*{Acknowledgements}
The author thanks Alan Sokal for carefully reading the paper
and for his valuable comments. I am also very grateful to
my colleages Erich Novak, Daniel Rudolf and Aicke Hinrichs for
the numerous discussions.
\linespread{1}
\bibliographystyle{amsalpha}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,341 |
\section{Introduction}
\setcounter{equation}{0}
The motion of homogeneous incompressible ideal fluid in a domain
$\Omega \subset \Bbb R^n $ is described by the following system of
Euler equations.
\[
\mathrm{ (E)}
\left\{ \aligned
&\frac{\partial v}{\partial t} +(v\cdot \nabla )v =-\nabla p,
\quad (x,t)\in {\Omega}\times (0, \infty) \\
&\textrm{div }\, v =0 , \quad (x,t)\in {\Omega}\times (0,
\infty)\\
&v(x,0)=v_0 (x), \quad x\in \Omega
\endaligned
\right.
\]
where $v=(v^1, v^2, \cdots , v^n )$, $v^j
=v^j (x, t)$, $j=1, 2, \cdots , n,$ is the velocity of the fluid
flows, $p=p(x,t)$ is the scalar pressure, and $v_0 (x)$ is a given
initial velocity field satisfying div $v_0=0$. Here we use the
standard notion of vector calculus, denoting
$$\nabla p =\left(\frac{\partial p}{\partial x_1 },
\frac{\partial p}{\partial x_2 }, \cdots , \frac{\partial
p}{\partial x_n } \right),\quad (v\cdot \nabla )v^j=\sum_{k=1}^n
v^k \frac{\partial v^j}{\partial x_k},\quad \textrm{div }\, v
=\sum_{k=1}^n \frac{\partial v^k}{\partial x_k }.
$$
The first equation of (E) follows from the balance of momentum for
each portion of fluid, while the second equation can be derived from
the conservation of mass of fluid during its motion, combined with
the homogeneity(constant density) assumption on the fluid. The
system (E) is first derived by L. Euler in 1755(\cite{eul}). Unless
otherwise stated, we are concerned on the Cauchy problem of the
system (E) on $\Omega= \Bbb R^n$, but many of the results presented
here are obviously valid also for
$\Omega =\Bbb R^n/\Bbb Z^n$(periodic domain), and even for the bounded domain
with the smooth boundary with the boundary condition $v\cdot \nu
=0$, where $\nu $ is the outward unit normal vector.
We also suppose $n=2$ or $3$ throughout this paper.
{\em In this article our aim to survey recent results on the
mathematical aspects the 3D Euler equations closely related to
the problem of spontaneous apparition of singularity starting from a
classical solutions having finite energy.}
If we add the dissipation term $\mu \Delta v=\mu \sum_{j=1}^n
\frac{\partial^2 v}{\partial x_j^2} $, where $\mu>0$ is the
viscosity coefficient, to the right hand side of the first equation of (E), then we have
the Navier-Stokes equations, the regularity/singularity question of which is
one of the seven millennium problems in mathematics.
In this article we do not treat the
Navier-Stokes equations. For details of mathematical
studies on the Navier-Stokes equations
see e.g. \cite{tem3, con10,lio, gald, lad, maj3, lem}.
We also omit other important topics such as existence and uniqueness
questions of the weak solutions of the 2D Euler equations, and
the related vortex patch problems, vortex sheet problems, and so
on. These are well treated in the other
papers and monographs(\cite{maj3, cha19a, che2, lio, sch, shn1, yud1, yud2, vis2, tad})
and the references therein.
For the survey related the stability
question please see for example \cite{fri1} and references therein. For the
results on the regularity of the Euler equations with uniformly rotating
external force we refer \cite{bab}, while for the numerical
studies on the blow-up problem of the Euler equations there are
many articles including \cite{ker1, ker2, hou3, bra, fri2, caf1, gra1,gra2,
gre, pel}. For various mathematical and physical aspects of the Euler equations
there are many excellent books, review articles including
\cite{arn, bren, che2, con1,con3, fri1, gib2, maj2, maj3, mar, cha12a, yu}.
Obviously, the references are not complete mainly due to
author's ignorance.
\subsection{Basic properties}
In the study of the Euler equations the notion of vorticity,
$\omega=$curl $v$, plays important roles. We can reformulate the
Euler system in terms of the vorticity fields only as follows. We
first consider the 3D case. Let us
first rewrite the first equation of (E) as
\begin{equation}\label{bas1}
\frac{\partial v}{\partial t} -v\times \mathrm{curl}\, v=-\nabla
(p+\frac12 |v|^2 ).
\end{equation}
Then, taking curl of (\ref{bas1}), and using elementary vector
identities, we obtain the following vorticity formulation:
\begin{equation}\label{bas2}
\frac{\partial \o }{\partial t} +(v\cdot \nabla )\o =\o \cdot \nabla
v,
\end{equation}
\begin{equation}\label{bas3}
\textrm{div }\, v =0 , \quad \textrm{curl }\, v =\o ,
\end{equation}\label{bas4}
\begin{equation} \o(x,0)=\o_0 (x).
\end{equation}
The linear elliptic system (\ref{bas3}) for $v$
can be solved explicitly in terms of $\o$, assuming $\o$ decays sufficiently fast
near spatial infinity, to provides us with the Biot-Savart law,
\begin{equation}\label{bas5}
v(x,t)=\frac{1}{4\pi}\int_{\Bbb R^3}
\frac{(x-y)\times \o (y,t) }{|x-y|^3}dy.
\end{equation}
Substituting this $v$ into (\ref{bas2}), we obtain an
integro-differential system for $\o$. The term in the right hand side of
(\ref{bas2}) is called the vortex stretching term, and is
regarded as the main source of difficulties in the mathematical
theory of the 3D Euler equations.
Let us introduce the deformation matrix
$S(x,t)=(S_{ij}(x,t))_{i,j=1}^3$ defined as the symmetric part of
the velocity gradient matrix,
$$S_{ij} =\frac12 \left(\frac{\partial v_j}{\partial x_{i}}
+\frac{\partial v_i}{\partial x_{j}}\right).
$$
From the Biot-Savart law in (\ref{bas5}) we can explicitly compute
\begin{equation}\label{bas5a}
S(x,t)=\frac{3}{8\pi} p.v. \int_{\Bbb R^3} \frac{ [ (y\times \o (x+y,t) )
\otimes y +y\otimes
(y\times
\o (x+y,t))]}{ |y|^5}dy
\end{equation}
(see e.g. \cite{maj3} for the details on the computation).
The kernel in the convolution integral of (\ref{bas5a}) defines a
singular integral operator of the Calderon-Zygmund type(see e.g.
\cite{ste1, ste2} for more details).
Since the vortex stretching term can be written as $ (\o \cdot \nabla )v = S \o
$, we see that the singular integral operator and related
harmonic analysis results could have important roles to
study the Euler equations.
In the two dimensional case
we take the vorticity as the scalar, $\o =\frac{\partial v^2}{\partial x_1}
-\frac{\partial v^1}{\partial x_2}$, and the evolution equation
of $\o$ becomes
\begin{equation}\label{bas6}
\frac{\partial \o }{\partial t} +(v\cdot \nabla )\o =0,
\end{equation}
where the velocity is represented in terms of the vorticity by the
2D Biot-Savart law,
\begin{equation}\label{bas7}
v(x,t)=\frac{1}{2\pi} \int_{\Bbb R^2} \frac{(-y_2+x_2,
y_1 -x_1
)}{|x-y|^2} \o (y,t )dy.
\end{equation}
Observe that there is no vortex stretching term in (\ref{bas6}),
which makes the proof of global regularity in 2D Euler equations easily accessible.
In many studies of the Euler equations it is convenient to introduce
the notion of `particle trajectory mapping', $X (\cdot ,t)$ defined
by
\begin{equation}\label{bas8}
\frac{\partial X (a, t )}{\partial t} =v (X (a , t
),t), \quad X(a , 0)=a , \quad a \in \Omega.
\end{equation}
The mapping $X(\cdot, t)$ transforms from the location of the initial fluid particle
to the location at time $t$, and the parameter $a$ is called the Lagrangian
particle marker.
If we denote the Jacobian of the transformation, det$(\nabla
_a X (a ,t ))=J(a ,t)$, then we can show easily(see e.g.
\cite{maj3} for the proof)
that
$$\frac{\partial J}{\partial t} =(\mathrm{div}\, v) J,$$
which implies that the velocity field $v$ satisfies the
incompressibility, div $v=0$ if and only if the mapping $X (\cdot
,t)$ is volume preserving. At this moment we note that, although
the Euler equations are originally derived by applying the physical
principles of mass conservation and the momentum balance, we could
also derive them by applying the least action principle to the
action defined by
$$
\mathcal{I} (A)=\frac12 \int_{t_1}^{t_2} \int_{\Omega}
\left|\frac{\partial X (x,t)}{\partial t}\right|^2 dxdt.
$$
Here, $X(\cdot, t): \Omega \to \Omega\subset \Bbb R^n$ is a
parameterized family of volume preserving diffeomorphism. This
variational approach to the Euler equations implies that we can view
solutions of the Euler equations as a geodesic curve in the
$L^2(\Omega)$ metric on the infinite dimensional manifold of volume
preserving diffeomorphisms(see e.g. \cite{arn, bren, ebi} and
references therein for more details on the geometric approaches to
the Euler equations).
The 3D Euler equations have many conserved quantities. We list some
important ones below.
\begin{itemize}
\item[(i)] Energy,
$$
E(t)=\frac12 \int_{\Omega} |v(x,t)|^2 dx.
$$
\item[(ii)] Helicity,
$$
H(t)=\int_{\Omega} v (x,t)\cdot \o (x,t)
dx.
$$
\item[(iii)] Circulation,
$$
\Gamma _{\mathcal{C}(t)}= \oint_{\mathcal{C}(t)} v\cdot
dl,
$$
where $\mathcal{C}(t)=\{ X (a , t) \,| \, a \in \mathcal{C}\}$ is
a curve moving along with the fluid.
\item[(iv)] Impulse,
$$
I(t)=\frac12\int_{\Omega} x\times \o \,dx.
$$
\item[(v)] Moment of Impulse,
$$
M(t)= \frac13 \int_{\Omega} x\times (x\times \o ) \,dx.
$$
\end{itemize}
The proof of conservations of the above quantities for the classical solutions
can be done without difficulty using elementary vector calculus(for details
see e.g. \cite{maj3, mar}). The helicity, in particular,
represents the degree of knotedness of the vortex lines in the
fluid, where the vortex lines are the integral curves of the
vorticity fields. In \cite{arn} there are detailed discussions on this aspects
and other topological implications of the helicity conservation.
For the 2D Euler equations there is no analogue
of helicity, while the circulation conservation is replaced by
the vorticity flux integral,
$$
\int_{D(t)} \o (x,t) dx,
$$
where $D(t)=\{ X (a , t) \, | \, a \in
D\subset \Omega\}$ is a planar region moving along the fluid in
$\Omega$.
The impulse and the moment of impulse integrals in the 2E Euler equations are replace by
$$
\frac12\int_{\Omega} (x_2, -x_1 )\o dx \quad \mbox{and}\quad
-\frac13 \int_{\Omega} |x|^2 \o dx\quad \mbox{respectively}.
$$
In the 2D Euler equations we have extra conserved
quantities; namely for any continuous function $f$ the integral
$$
\int_\Omega f(\o (x,t)) dx
$$
is conserved.
There are also many known explicit solutions to the Euler equations,
for which we just refer \cite{lam, maj3}. In the remained part of
this subsection we introduce some notations to be used later for 3D
Euler equations. Given velocity $v(x,t)$, and pressure $p(x,t)$, we
set the $3\times 3$ matrices,
$$
V_{ij}=\frac{\partial v_j}{\partial x_i},\quad
S_{ij}=\frac{V_{ij}+V_{ji}}{2},\quad A_{ij}=\frac{V_{ij}-V_{ji}}{2},
\quad P_{ij}=\frac{\partial ^2 p}{\partial x_i \partial x_j},
$$
with $i,j=1,2,3$. We have the decomposition $V=(V_{ij})=S+A$, where
the symmetric part $S=(S_{ij})$ represents the deformation tensor of
the fluid introduced above, while the antisymmetric part
$A=(A_{ij})$ is related to the vorticity $\o$
by the formula,
\begin{equation}\label{basfor}
A_{ij}= \frac12 \sum_{k=1}^3 \varepsilon
_{ijk} \o_k,\qquad \o_i = \sum_{j,k=1}^3\varepsilon_{ijk}A_{jk},
\end{equation}
where $\varepsilon_{ijk}$ is the skewsymmetric tensor with the normalization
$\varepsilon_{123}=1$. Note that $P=(P_{ij})$ is the hessian of the
pressure. We also frequently use the
notation for the vorticity direction field,
$$ \xi (x,t)=\frac{\o (x,t)}{|\o(x,t)|}, $$
defined whenever $\o (x,t)\neq 0$.
Computing partial derivatives $\partial/
\partial x_k$ of the first equation of (E), we obtain the matrix equation
\begin{equation}\label{basfor1}
\frac{D V}{Dt}=-V^2 -P, \quad \frac{D}{Dt}=
\frac{\partial}{\partial t}+(v\cdot \nabla )v.
\end{equation} Taking symmetric part of this, we obtain
$$ \frac{D
S}{Dt}=-S^2-A^2-P,
$$
from which, using the formula (\ref{basfor}),
we have
\begin{equation}\label{basfor2}
\frac{D S_{ij}}{Dt} = - \sum_{k=1}^3 S_{ik}S_{kj}
+\frac14 (|\o|^2 \delta_{ij} -\o_i\o_j )- P_{ij},
\end{equation}
where $\delta_{ij}=1$ if $i=j$, and $\delta_{ij}=0$ if $i\neq j$.
The antisymmetric part of (\ref{basfor1}), on the other hand, is
$$
\frac{DA}{Dt}=-SA -AS,
$$
which, using the formula (\ref{basfor}) again, we obtain easily
\begin{equation}\label{basfor2a}
\frac{D \o}{Dt} = S \o,
\end{equation}
which is the vorticity evolution equation (\ref{bas2}). Taking dot
product (\ref{basfor2a}) with $\o$,
we immediately have
\begin{equation}\label{basfor3}
\frac{D |\o |}{Dt} =\a |\o|,
\end{equation}
where we set
$$\a (x,t)= \left\{ \aligned &\sum_{i,j=1}^3
\xi_i(x,t) S_{ij}(x,t) \xi_j (x,t)
& \mbox{if $\o (x,t)\neq 0$}\\
&0 & \mbox{if $\o (x,t)=0$}.
\endaligned
\right. $$
\subsection{Preliminaries}
Here we introduce some notations and function spaces to be used in
the later sections.
Given $p\in [1, \infty]$, the Lebesgue space $L^p (\Bbb R^n )$,
$p\in [1, \infty]$, is the Banach space defined
by the norm
$$\|f\|_{L^p}:=\left\{
\aligned &\left(\int_{\Bbb R^n} |f(x)|^p dx \right)^{\frac{1}{p}}, \quad
p\in [1, \infty )\\
&\mathrm{ess.} \sup_{x\in \Bbb R^n} |f(x)|, \quad p=\infty .
\endaligned \right.
$$
For $j=1, \cdots, n$ the Riesz transform $R_j$ of $f$ is given by
$$
R_j (f)(x)=\frac{\Gamma (\frac{n+1}{2} )}{\pi ^{\frac{n+1}{2}}} p.v.
\int_{\Bbb R^n} \frac{x_j -y_j}{|x-y|^{n+1}} f(y)dy
$$
whenever the right hand side makes sense. The Hardy space
$\mathcal{H}^1 (\Bbb R^n)\subset L^1 (\Bbb R^n)$ is defined by
$$
f\in \mathcal{H}^1 (\Bbb R^n) \quad \mbox{if and only if}\quad
\|f\|_{\mathcal{H}^1}:= \|f\|_{L^1} +\sum_{j=1}^n \|R_j f\|_{L^1}
<\infty.
$$
The space $BMO(\Bbb R^n)$ denotes the space of functions of
bounded mean oscillations, defined by
$$
f\in BMO (\Bbb R^n )\quad \mbox{if and only if}\quad
\|f\|_{BMO}:=\sup_{Q\subset \Bbb R^n} \frac{1}{\mathrm{Vol} (Q)}
\int_Q |f-f_Q |dx <\infty,
$$
where $f_Q =\frac{1}{\mathrm{Vol} (Q)}\int_Q f dx$. For more details
on the Hardy space and BMO we refer \cite{ste1, ste2}.
Let us set the
multi-index $\alpha :=(\alpha_1 , \alpha_2 , \cdots , \alpha_n )\in
(\Bbb
Z_+ \cup \{ 0\} )^n $ with $|\alpha |=\alpha_1 +\alpha_2 +\cdots
+\alpha_n$. Then, $D^\alpha :=D^{\alpha_1}_1 D^{\alpha_2}_2 \cdots
D^{\alpha_n}_n$, where $D_j =\partial/\partial x_j$, $j=1,2,\cdots,
n$.
Given $k\in \Bbb Z$ and $p\in [1, \infty)$ the Sobolev space,
$W^{k,p} (\Bbb R^n )$ is the Banach space of functions
consisting of functions $f\in L^p (\Bbb R^n )$
such that
$$\|f\|_{W^{k, p}}:=\left(\int_{\Bbb R^n} |D^\alpha f(x)|^p dx
\right)^{\frac{1}{p}} <\infty,$$
where the derivatives are in the sense of distributions. For
$p=\infty$ we replace the $L^p (\Bbb R^n )$ norm by the
$L^\infty (\Bbb R^n)$ norm. In particular, we denote $H^m (\Bbb R^n) =W^{m,2} (\Bbb R^n )$.
In order to handle the functions having fractional derivatives of order $s\in \Bbb R$,
we use the
Bessel potential space $L^{s}_p (\Bbb R^n) $ defined by the Banach spaces norm,
$$\|f\|_{L^{s,p}}:=\|(1-\Delta )^{\frac{s}{2}} f\|_{L^p},
$$
where $(1-\Delta )^{\frac{s}{2}} f=\mathcal{F}^{-1} [ (1+|\xi |^2
)^{\frac{s}{2}} \mathcal{F} (f)(\xi )]$. Here $\mathcal{F} (\cdot)$ and
$\mathcal{F}^{-1}(\cdot)$ denoting the Fourier transform and its
inverse, defined by
$$
\mathcal{F} (f)(\xi)=\hat{f} (\xi)
=\frac{1}{(2\pi )^{n/2}}\int_{\Bbb R^n} e^{-ix\cdot \xi }
f(x)dx,
$$
and
$$
\mathcal{F}^{-1} (f)(x)=\check{f} (x)
=\frac{1}{(2\pi )^{n/2}}\int_{\Bbb R^n} e^{ix\cdot \xi }
f(\xi)d\xi ,
$$
whenever the integrals make sense.
Next we
introduce the Besov spaces. We follow \cite{tri}(see also \cite{tay,
lem, che2, run}). Let $\mathfrak{S}$ be the Schwartz class of
rapidly decreasing
functions.
We consider $\varphi \in \mathfrak{S}$ satisfying
$\textrm{Supp}\, \hat{\varphi} \subset
\{\xi \in {\mathbb R}^n \, |\,\frac12 \leq |\xi|\leq
2\}$,
and $\hat{\varphi} (\xi)>0 $ if $\frac12 <|\xi|<2$.
Setting $\hat{\varphi_j } =\hat{\varphi } (2^{-j} \xi )$ (In other words,
$\varphi_j (x)=2^{jn} \varphi (2^j x )$.), we can adjust the
normalization constant in front of $\hat{\varphi}$ so that
$$
\sum_{j\in \mathbb{Z}} \hat{\varphi}_j (\xi )=1\quad \forall \xi \in
{\mathbb R^n}\setminus \{ 0\}.
$$
Let $s\in \mathbb R$, $p,q
\in [0, \infty]$. Given $f\in \mathfrak{S}'$, we denote
$\Delta_j f=\varphi_j* f$.
Then the homogeneous Besov semi-norm $\|
f\|_{\dot{B}^{s}_{p,q}}$ is defined by \[ \|
f\|_{\dot{B}^s_{p,q}}=\left\{ \aligned &\left( \sum_{j\in \Bbb Z}
2^{jqs} \| \varphi_j \ast f \|_{L^p}^q
\right)^{\frac{1}{q}}&\mbox{ if } q\in [1, \infty) \\
&\sup_{j\in \Bbb Z} \left(2^{js} \| \varphi_j \ast f \|_{L^p}
\right)&\mbox{ if }q=\infty.
\endaligned \right.
\]
For $(s, p,q)\in [0, \infty) \times [1, \infty]\times [1, \infty]$
the homogeneous Besov space $\dot{B}^{s}_{p,q}$ is a quasi-normed
space with the quasi-norm given by $\| \cdot \|_{\dot{B}^s_{p,q}}$.
For $s>0$ we define the inhomogeneous Besov space norm $\|
f\|_{{B}^s_{p,q}}$ of $f\in \mathfrak{S}'$ as
$\|f\|_{{B}^s_{p,q}}=\| f\|_{L^p}+\| f\|_{\dot{B}^s_{p,q}}$.
Similarly, for $(s, p,q)\in [0, \infty) \times [1, \infty)\times [1,
\infty]$, the homogeneous Triebel-Lizorkin semi-norm
$\|f\|_{\dot{F}^s_{p,q}}$ is defined by
$$
\|f\|_{\dot{F}^s_{p,q}} =\left\{ \aligned &\left\| \left(\sum_{j\in
\mathbb{Z}} 2^{jqs}|\varphi_j \ast
f(\cdot)|^q\right)^{\frac1q}\right\|_{L^p}
& \mbox{if $q\in[1, \infty)$}\\
&\left\|\sup_{j\in \mathbb{Z}}\left( 2^{js} |\varphi_j \ast f (\cdot
)|\right) \right\|_{L^p} & \mbox{if $q=\infty$}
\endaligned \right. .
$$
The homogeneous Triebel-Lizorkin space $\dot{F}^s_{p,q}$ is a quasi-normed
space with the quasi-norm given by $\|\cdot \|_{\dot{F}^s_{p,q}}$.
For $s>0$, $(p,q)\in [1, \infty)\times [1, \infty)$
we define the inhomogeneous Triebel-Lizorkin space
norm by
$$
\|f\|_{{F}^s_{p,q}} =\|f\|_{L^p} +\|f\|_{\dot{F}^s_{p,q}}.
$$
The inhomogeneous Triebel-Lizorkin space is a Banach space
equipped with the norm, $\|\cdot\|_{{F}^s_{p,q}}$.
We observe that $B^s_{p,p}(\Bbb R^n)=F^s_{p,p}(\Bbb R^n) $. The
Triebel-Lizorkin space is a generalization of many classical
function spaces. Indeed, the followings are well established(see e.g.
\cite{tri})
$$F^0_{p, 2}(\Bbb R^n ) =\dot{F}^0_{p, 2}(\Bbb R^n)=L^p (\Bbb R^n ), \quad (1<p<\infty).$$
$$\dot{F}^0_{1,2}(\Bbb R^n)=\mathcal{H}^1 (\Bbb R^n )\quad \mbox{and}
\quad \dot{F}^0_{\infty,2}=BMO(\Bbb R^n). $$
$$F^s_{p,2}(\Bbb R^n )=L^{s,p} (\Bbb R^n ).$$
We also note sequence of continuous embeddings for the spaces close
to $L^\infty (\Bbb R^n)$(\cite{tri, jaw}).
\begin{equation}\label{emb}
\dot{B}^{\frac{n}{p}}_{p,1} (\Bbb R^n)\hookrightarrow
\dot{B}^{0}_{\infty,1}(\Bbb R^n)
\hookrightarrow L^\infty (\Bbb R^n) \hookrightarrow BMO (\Bbb R^n)\hookrightarrow
\dot{B}^{0}_{\infty,\infty}(\Bbb R^n).
\end{equation}
Given $0<s <1$, $1\leq p \leq \infty, 1\leq q \leq \infty$, we
introduce another function spaces $\mathcal{\dot{F}}^s_{p,q} $
defined by the
seminorm,
$$ \|f\|_{\mathcal{\dot{F}}^s_{p,q} } =\left\{
\aligned \left\|
\left(\int_{\Bbb R^n}\frac{|f(x)-f(x-y)|^q}{|y|^{n+s q}} dy\right)^{
\frac{1}{q}}\right\|_{L^p (\Bbb R^n , dx)}
& \mbox{if $1\leq p \leq \infty , 1\leq q <\infty$}\\
\left\| ess\sup_{|y|>0 } \frac{|f(x) -f(x-y)|}{|y|^s } \right\|_{
L^p (\Bbb R^n , dx)}
& \mbox{if $1\leq p \leq \infty, q=\infty$}
\endaligned \right. . $$
On the other hand, the space $\mathcal{\dot{B}}^s_{p,q} $ is
defined by the
seminorm,
$$ \|f\|_{\mathcal{\dot{B}}^s_{p,q} } =\left\{
\aligned
\left( \int_{\Bbb R^n} \frac{\|f(\cdot)-f(\cdot-y)\|^q_{L^p}}{|y|^{n+s q}}
dy\right)^{
\frac{1}{q}}
& \quad\mbox{if $1\leq p \leq \infty , 1\leq q <\infty$}\\
ess\sup_{|y|>0 } \frac{\|f(\cdot) -f(\cdot-y)\|_{L^p }}{|y|^s }
& \quad\mbox{if $1\leq p \leq \infty, q=\infty$}
\endaligned \right. . $$
Observe that, in particular,
$\mathcal{\dot{F}}^{s}_{\infty,\infty}=
\mathcal{\dot{B}}^{s}_{\infty,\infty}= C^s$, which
is the usual H\"{o}lder
seminormed space for $s\in \Bbb R_+\Bbb Z$. We also note that if $q=\infty$,
$\mathcal{\dot{B}}^{s}_{p,\infty}=
\mathcal{\dot{N}}^s_p$, which is the Nikolskii space.
The inhomogeneous version of those spaces, $\mathcal{{F}}^s_{p,q}
$ and $\mathcal{{B}}^s_{p,q} $
are defined by their
norms,
$$ \|f\|_{\mathcal{{F}}^s_{p,q} }= \|f\|_{L^p }
+\|f\|_{\mathcal{\dot{F}}^s_{p,q}}, \quad
\|f\|_{\mathcal{{B}}^s_{p,q} }= \|f\|_{L^p }
+\|f\|_{\mathcal{\dot{B}}^s_{p,q}}, $$ respectively.
We note that for $0<s <1$, $ 2\leq p< \infty $, $q=2$,
$\mathcal{{F}}^s_{p,2}
\cong
L^{p}_s (\Bbb R^n )$,
introduced above(see pp. 163,\cite{ste1}).
If $\frac{n}{\min\{ p,q\} } <s <
1$, $n<p<\infty$ and $n<q\leq \infty$, then
$\mathcal{{F}}^s_{p,q} $ coincides with the Triebel-Lizorkin
space $F^s_{p,q} (\Bbb R^n ) $ defined above(see pp. 101,
\cite{tri}). On the other hand, for wider range of parameters, $0<s<1$, $0<p\leq \infty$,
$0<q\leq \infty$, $\mathcal{{B}}^s_{p,q} $ coincides with the Besov
space $B^s_{p,q} (\Bbb R^n )$ defined above.
\section{Local well-posedness and blow-up criteria}
\setcounter{equation}{0}
\subsection{Kato's local existence and the BKM criterion}
We review briefly the key elements in the classical local existence
proof of solutions in the Sobolev space $H^m (\Bbb R^n)$, $m>n/2+1$,
essentially obtained by Kato in \cite{kat2}(see also \cite{maj3}).
After that we derive the celebrated Beale, Kato and Majda's
criterion on finite time blow-up of the local solution in $H^m (\Bbb
R^n)$, $m>n/2+1$ in \cite{bea}. Taking derivatives $D^\alpha $ on
the first equation of (E)
and then taking $L^2$ inner
product it with $D^\alpha v$, and summing over the multi-indices
$\alpha $ with $|\alpha
|\leq m$, we obtain
\begin{eqnarray*}
\lefteqn{\frac12 \frac{d}{dt} \|v \|_{H^m} ^2
=-\sum_{|\alpha|\leq m} (D^\alpha (v\cdot \nabla )v
-(v\cdot \nabla )D^\alpha v , D^\alpha v)_{L^2}}\hspace{.1in}\\
&&-\sum_{|\alpha|\leq m}((v\cdot \nabla )
D^\alpha v, D^\alpha v )_{L^2}-\sum_{|\alpha|\leq m} (D^\alpha \nabla\, p
, D^\alpha v )_{L^2}\\
&&\quad=I+II+III.
\end{eqnarray*}
Integrating by part, we obtain
$$
III=\sum_{|\alpha|\leq m}(D^\alpha p
, D^\alpha \mathrm{div} \, v)_{L^2} =0.
$$
Integrating by part again, and using the
fact div $v=0$, we have
$$
II=-\frac12 \sum_{|\alpha|\leq m} \int_{\Bbb R^n} (v\cdot \nabla
)|D^\alpha v |^2 dx
=\frac12 \sum_{|\alpha|\leq m}\int_{\Bbb R^n} \mbox{div}\, v |D^\alpha
v |^2 dx =0
.
$$
We now use the so called {\it commutator type of
estimate}(\cite{kla}),
$$
\sum_{|\alpha|\leq m}\|D^{\alpha} (fg)-fD^\alpha g\|_{L^2}
\leq C (\|\nabla f\|_{L^\infty}
\|g\|_{H^{m-1}} + \|f\|_{H^m} \|g\|_{L^\infty} ),
$$
and obtain
$$
I\leq
\sum_{|\alpha|\leq m} \|D^\alpha (v\cdot \nabla )v
-(v\cdot \nabla )D^\alpha v\|_{L^2} \|v\|_{H^{m}}
\leq C\|\nabla v \|_{L^\infty} \|v \|_{H^{m}}^2 .
$$
Summarizing the above estimates, I,II,III, we have
\begin{equation}\label{sum}
\frac{d}{dt} \|v \|_{H^{m}} ^2
\leq C \|\nabla v\|_{L^\infty} \|v \|_{H^{m}} ^2.
\end{equation}
Further estimate, using the {\it Sobolev inequality}, $\|\nabla
v\|_{L^\infty} \leq C \|v\|_{H^m} $ for $m>n/2+1$, gives
$$
\frac{d}{dt} \|v \|_{H^{m}} ^2
\leq C \|v \|_{H^{m}} ^3.
$$
Thanks to Gr{o}nwall's lemma we have the local in time
uniform estimate
\begin{equation} \label{ap1}
\|v(t)\|_{H^m} \leq \frac{\|v_0\|_{H^m}}{1-Ct \|v_0\|_{H^m}
}
\leq 2 \|v_0\|_{H^m}
\end{equation}
for all $t\in [0, T ]$, where $T=\frac{1}{2C\|v_0\|_{H^m}}$.
Using this estimate we can also deduce the estimate
\begin{equation}\label{ap2} \sup_{0\leq t\leq T} \left\| \frac{\partial
v}{\partial t
}\right\|_{H^{m-1}} \leq C (\|v_0 \|_{H^m} )
\end{equation}
directly from (E).
The estimates (\ref{ap1}) and (\ref{ap2}) are the two key a priori
estimates for the construction of the local
solutions. For actual elaboration of the proof we approximate the Euler
system by mollification, Galerkin projection, or
iteration of successive linear systems, and construct a sequence of
smooth approximate solutions to (E), say $\{ v_k (\cdot ,t)\}_{k\in \Bbb N}$ corresponding to
the initial data $ \{v_{0,k}\}_{k\in \Bbb N}$ respectively with $v_k \to v_0 $
in $H^m (\Bbb R^n)$.
The estimates for the approximate solution sequence provides us with
the uniform estimates of $\{ v_k\}$ in $L^\infty ([0, T]; H^m (\Bbb
R^n )) \cap Lip ([0, T]; H^{m-1}(\Bbb R^n) )$. Then, applying the
standard Aubin-Nitche compactness lemma, we can pass to the limit
$k\to \infty$ in the equations for the approximate solutions, and
can show that the limit $v=v_\infty$ is a solution of the (E) in
$L^\infty ([0, T]); H^m (\Bbb R^n ))$. By further argument we can
actually show that the limit $v$ belongs to $ C([0, T]; H^m (\Bbb
R^n ))\cap AC([0, T]; H^{m-1} (\Bbb R^n ))$, where $AC([0,T]; X )$
denotes the space of $X$ valued absolutely continuous functions on
$[0,T]$. The general scheme of such existence proof is standard, and
is described in detail in \cite{maj1} in the general type of
hyperbolic conservation laws. The approximation of the Euler system
by mollification was done for the construction of local solution of
the Euler(and the
Navier-Stokes) system in \cite{maj3}. \\
Regarding the question of finite time blow-up of the local classical
solution in $H^m (\Bbb R^n )$, $m>n/2 +1$, constructed above, the
celebrated Beale-Kato-Majda theorem(called the BKM criterion) states
that \begin{equation}\label{bkm}
\lim\sup_{t\nearrow T_*} \|v(t)\|_{H^s }=\infty \quad\mbox{if and only
if}\quad
\int_0 ^{T_*} \|\omega (s)\|_{L^\infty} ds =\infty.
\end{equation}
We outline the proof of this theorem below(for more details see \cite{bea, maj3}).
We first recall the Beale-Kato-Majda's version of the {\it
logarithmic
Sobolev inequality},
\begin{equation}\label{lo}
\|\nabla v \|_{L^\infty}
\leq C\|\o \|_{L^\infty} (1+\log (1+\|v
\|_{H^{m}} ))+C \|\o\|_{L^2}
\end{equation}
for $m>n/2 +1$.
Now suppose $\int_0 ^{T_*} \|\o (t)\|_{L^\infty } dt := M(T_*)< \infty .$
Taking $L^2$ inner product the first equation of (E) with $\o$, then after
integration by part we obtain
$$
\frac12\frac{d}{dt} \|\o \|_{L^2} ^2 =((\o\cdot \nabla )v ,\o
)_{L^2}\leq \|\o\|_{L^\infty} \|\nabla v \|_{L^2} \|\o \|_{L^2}
= \|\o\|_{L^\infty}\|\o \|_{L^2}^2,
$$
where we used the identity $\|\nabla v\|_{L^2}= \|\o\|_{L^2}$.
Applying the Gronwall lemma, we obtain
\begin{equation}\label{ineq1} \|\o (t)\|_{L^2} \leq \|\o_0 \|_{L^2}
\exp\left( \int_0 ^{T_*}
\|\o (s)\|_{L^\infty} ds \right)= \|\o_0 \|_{L^2}\exp[{M(T_*)}].
\end{equation}
for all $t\in [0, T_*]$.
Substituting (\ref{ineq1}) into (\ref{lo}), and combining this with
(\ref{sum}), we have
\[
\frac{d}{dt} \|v \|_{H^{m}} ^2
\leq C \left[ 1+\|\o \|_{L^\infty} [1+\log (1+\|v
\|_{H^{m}} )\right] \|v \|_{H^{m}} ^2
\]
Applying the Gr{o}nwall lemma we deduce
\begin{equation}\label{fini}
\|v (t)\|_{H^{m}} \leq \|v_0 \|_{H^{m}}
\exp\left[ C_1 \exp\left(C_2 \int_0 ^{T_*}
\|\o (\tau)\|_{L^\infty} d\tau\right)\right]
\end{equation}
for all $t\in [0,T_*]$ and for some constants $C_1$ and $C_2$ depending on $M(T_*)$.
The inequality (\ref{fini}) provides the with the necessity part of
(\ref{bkm}).
The sufficiency part is an easy consequence of the Sobolev
inequality,
$$ \int_0 ^{T_*} \|\omega (s)\|_{L^\infty} ds \leq
T_* \sup_{0\leq t\leq T_*} \|\nabla v(t)\|_{L^\infty} \\
\leq C T_* \sup_{0\leq t\leq T_*}\|v(t)\|_{H^m}
$$
for $m>n/2+1$.
There are many other results of local well-posedness in various
function spaces(see \cite{cha1, cha2, cha3, cha6, che1,
che2,kat1,kat3,kat4, lic, tem1, tem2, vis1, vis2, yud1}). For the
local existence proved in terms of a geometric formulation see
\cite{ebi}. For the BKM criterion for solutions in the H\"{o}lder
space see \cite{bah}. Immediately after the BKM result appeared,
Ponce derive similar criterion in terms of the deformation
tensor(\cite{pon}). Recently, Constantin proved local well-posedness
and a blow-up criterion in terms of the active vector
formulation(\cite{con5}).
\subsection{Refinements of the BKM criterion}
The first refinement of the BKM criterion was done by Kozono and
Taniuchi in \cite{koz1}, where they proved
\begin{theorem}
Let $s>n/p+1$. A solution $v$ of the Euler equations belonging to
$C([0,T_*);W^{s,p}(\Bbb R^n)) \cap C^1 ([0, T_* ); W^{s-2, p}(\Bbb
R^n ))$ blows up at $T_*$ in $W^{s,p}(\Bbb R^n))$, namely
$$
\lim\sup_{t\nearrow T_*}\|v(t)\|_{W^{s,p}} =\infty \quad \mbox{if
and only if}\quad \int_0 ^{T_*} \|\o \|_{BMO} =\infty.
$$
\end{theorem}
The proof is based on the following version of the logarithmic
Sobolev inequality for $f\in W^{s,p} (\Bbb R^n)$, $s>n/p$,
$1<p<\infty$,
$$
\|f\|_{L^\infty} \leq C \left(1+\|f\|_{BMO} (1+\log^+
\|f\|_{W^{s,p}} ) \right).
$$
(see \cite{koz1} for details of the proof). We recall now the
embedding relations (\ref{emb}). Further refinement of the above
theorem is the following(see\cite{cha1, cha6}).
\begin{theorem}
\begin{description}
\item[(i)] {\rm{ (super-critical case)}} Let $s>n/p+1, p\in (1, \infty), q\in [1,
\infty]$.
Then, the local in time solution $v\in C([0, T_*); B^s_{p, q} (\Bbb R^n))$
blows up at $T_* $ in $B^s_{p, q} (\Bbb R^n)$, namely
$$
\lim\sup_{t\nearrow T_*} \|v(t)\|_{B^s_{p,q}} =\infty
\quad\mbox{
if and only if}
\quad
\int_0 ^{T_*} \|\omega (t)\|_{\dot{B}^0_{\infty ,\infty}}dt
=\infty .
$$
\item[(ii)] {\rm{(critical case)}} Let $p\in (1, \infty)$.
Then, the local in time solution $v\in C([0, T_*); B^{n/p+1}_{p,1}
(\Bbb R^n))$ blows up at $T_* $ in $ B^{n/p+1}_{p,1} (\Bbb R^n))$,
namely
$$
\lim\sup_{t\nearrow T_*} \|v(t)\|_{ B^{n/p+1}_{p,1}} =\infty
\quad\mbox{
if and only if
}\quad
\int_0 ^{T_*} \|\omega (t)\|_{\dot{B}^0_{\infty ,1}}dt
=\infty.
$$
\end{description}
\end{theorem}
The proof of (i) is based on the following version of the
logarithmic Sobolev inequality for $f\in B^s_{p, q} (\Bbb R^n)$ with
$s> n/p$ with $p\in (1, \infty )$, $q\in [1, \infty]$.
$$
\|f\|_{L^\infty}
\leq C(1+\|f\|_{\dot{B}^0_{\infty, \infty} }(\log^+ \|f\|_{B^s_{p,q}} +1))
$$
In \cite{koz2} Kozono, Ogawa and Taniuchi obtained similar results
to
(i) above independently.\\
\ \\
In all of the above criteria, including the BKM theorem, we need to
control all of the three components of the vorticity vector to
obtain regularity. The following theorem proved in \cite{cha7}
states that actually we only need to control two components of the
vorticity in the slightly stronger norm than the $L^\infty$
norm(recall again the embedding (\ref{emb})).
\begin{theorem} Let $m>5/2$.
Suppose $v\in C([0,T_*);H^m (\Bbb R^3))$ is the local classical
solution of (E) for some $T_1 >0$, corresponding to the initial
data $v_0 \in H^m (\Bbb R^3)$, and $\o=$ curl $v$ is its vorticity.
We decompose
$\o=\tilde{\o}+\o^3 e_3$, where $\tilde{\o}=\o^1 e_1 +\o^2 e_2$,
and $\{e_1, e_2, e_3\}$ is the canonical basis of $\Bbb R^3$. Then,
$$
\lim\sup_{t\nearrow T_*} \|v(t)\|_{H^m} =\infty \quad \mbox{if
and only if} \quad \int_0 ^{T_*}\|\tilde{\o}
(t)\|_{\dot{B}^0_{\infty, 1}} ^2 dt =\infty.
$$
\end{theorem}
Note that $\tilde{\o}$ could be the projected component of $\o$
onto any plane in $\Bbb R^3$. For the solution $v=(v^1, v^2, 0)$ of
the Euler equations on the $x_1-x_2$ plane, the vorticity is $\o
=\o^3 e_3$ with $\o_3=\partial_{x_1} v^2 -\partial_{x_2} v^1$, and
$\tilde{\o}\equiv 0$. Hence, as a trivial application of the above
theorem we reproduce the well-known global in time regularity
for the 2D Euler equations.\\
\ \\
Next we present recent results on the blow up criterion in terms of
hessian of the pressure. As in the introduction we use $P=(P_{ij})$,
$S=(S_{ij})$ and $\xi$ to denote the hessian of the pressure, the
deformation tensor and the vorticity direction field respectively,
introduced in section 1. We also introduce the notations
$$ \frac{S\xi}{|S\xi |} =\zeta, \quad \zeta \cdot P \xi = \mu.
$$
The following is proved in \cite{cha15}.
\begin{theorem}
If the solution $v(x,t)$ of the 3D Euler
system with $v_0 \in H^m (\Bbb R^3 )$, $m>5/2$, blows up at $T_*$,
namely $\lim\sup_{t\nearrow T_*} \|v(t)\|_{H^m} =\infty$, then
necessarily,
$$
\int_0 ^{T_*} \exp\left(\int_0 ^\tau \|\mu (s)\|_{L^\infty}ds\right)
d\tau =\infty.
$$
\end{theorem}
Similar criterion in terms of the hessian of pressure, but with
different detailed geometric configuration from the above theorem
is obtained by Gibbon, Holm, Kerr and Roulstone in \cite{gib3}.
Below we denote $\xi_p=\xi \times P \xi$.
\begin{theorem}
Let $m \geq 3$ and $\Bbb T^3 =\Bbb R^3/\Bbb Z^3$ be a periodic box.
Then, there exists a global solution of the Euler equations $v\in
C([0, \infty ); H^m (\Bbb T^3 ))\cap C^1 ([0, \infty); H^{m-1} (\Bbb
T^3 ))$ if
$$\int_0 ^T \|\xi_p (t)\|_{L^\infty} dt <\infty, \quad \forall t\in (0,
T)
$$
excepting the case where $\xi$ becomes collinear with the
eigenvalues of $P$ at $T$.
\end{theorem}
Next, we consider the axisymmetric solution of the Euler equations,
which means velocity field $v(r, x_3, t)$, solving the Euler
equations, and having the representation
$$
v(r, x_3, t)=v^r (r,x_3,t)e_r +v^\theta (r,x_3,t)e_\theta +v^3
(r,x_3,t)e_3
$$
in the cylindrical coordinate system, where
$$e_r = (\frac{x_1}{r}, \frac{x_2}{r}, 0), \quad
e_\theta = (-\frac{x_2}{r}, \frac{x_1}{r}, 0),\quad e_3=(0,0,1),\quad
r=\sqrt{x_1^2 +x_2^2}.
$$
In this case also the question of finite time blow-up of solution
is wide open(see e.g. \cite{gra1, gra2,caf1} for studies in such
case). The vorticity $\o =$ curl $\,v$ is computed as
\[ \omega = \omega^r e_r+ \omega^{\t}e_{\t} + \omega^3 e_3 , \]
where
\[ \omega^r = -\partial_{x_3} v^{\t},
\quad \omega^{\t} = \partial_{x_3} v^r - \partial_r v^3,
\quad \omega ^3 = \frac{1}{r}\partial_r(rv^{\t}). \]
We denote
\[ \tilde{v} = v^r e_r + v^3 e_3 ,\qquad \tilde{\omega} = \omega^r e_r + \omega^3 e_3 .\]
Hence, $ \omega = \tilde{\omega}+\vec{\o}_\t $, where $\vec{\o}_\t
=\omega^\theta e_\theta$. The Euler equations for the axisymmetric
solution are
$$
\left\{ \aligned
&\frac{\partial v^r}{\partial t} +(\tilde{v}\cdot \tilde{\nabla} )v^r
=-\frac{\partial p}{\partial r} ,\\
&
\frac{\partial v^\theta}{\partial t} +(\tilde{v}\cdot \tilde{\nabla}
)v^\t
=-\frac{v^r v^\t}{ r} ,\\
&
\frac{\partial v^3}{\partial t} +(\tilde{v}\cdot \tilde{\nabla} )v^3
=-\frac{\partial p}{\partial x_3} ,\\
&
\textrm{div }\, \tilde{v} =0 ,\\
&
v(r,x_3,0)=v_0 (r,x_3),
\endaligned \right.
$$
where $\tilde{\nabla} =e_r \frac{\partial}{\partial r}
+e_3 \frac{\partial}{\partial x_3}.$
In the axisymmetric Euler equations the vorticity formulation
becomes
$$
\left\{\aligned &\frac{\partial\omega^r}{\partial t} +(\tilde{v} \cdot
\tilde{\nabla} )
=\omega^r(\tilde{\omega} \cdot \tilde{\nabla}) v^r \\
& \frac{\partial\omega^3}{\partial t} +(\tilde{v} \cdot
\tilde{\nabla} )=\omega^3(\tilde{\omega} \cdot \tilde{\nabla}) v^3 \\
&
\left[\frac{\partial}{\partial t} + \tilde{v} \cdot
\tilde{\nabla} \right] \left(\frac{\omega ^{\t}}{r}\right)
= (\tilde{\omega} \cdot \tilde{\nabla}) \left(\frac{v^{\t}}{r}\right)
\\
&\textrm{div }\, \tilde{v} = 0 , \quad \textrm{curl }\, \tilde{v}
=\vec{\o}^\t.
\endaligned \right.
$$
In the case of axisymmetry we only need to control just one
component of vorticity(the angular component) to get the regularity
of solution. The following theorem is proved in \cite{cha21}.
\begin{theorem} Let $v\in C([0,T_*);H^m (\Bbb R^3))$, $m>5/2$, be the
local classical axisymmetric solution of (E), corresponding to an
axisymmetric initial data $v_0 \in H^m (\Bbb R^3 )$. Then, the
solution blows up in $H^m (\Bbb R^3 )$ at $T_*$ if and only if for
all $(\gamma, p)\in (0,1)\times [1, \infty]$ we have
\begin{eqnarray}\label{axiscri}
\lefteqn{\int_0 ^{T_*}
\|\o_\t (t)\|_{L^\infty} dt +\int_0 ^{T_*} \exp\Big[ \int_0 ^t
\left\{ \|\o_\t (s)\|_{L^\infty} (1+\log^+ (\|\o _\t
(s)\|_{C^{\gamma}}
\|\o_\t (s)\|_{L^p} )) \right. }\hspace{2.in}\nonumber \\
&&\left. +\|\o_\t (s)\log^+ r \|_{L^\infty}\right\}ds\Big] dt
=\infty.
\end{eqnarray}
\end{theorem}
We observe that although we need to control only $\omega_\t$ to get the
regularity, the its norm, which is in $C^\gamma$, is higher than the
$L^\infty$ norm used in the BKM criterion. If we use the `critical'
Besov space $\dot{B}^0_{\infty,1}(\Bbb R^3)$ we can derive slightly
sharper criterion than Theorem 2.6 as follows(see \cite{cha7} for
the proof).
\begin{theorem} Let $v\in C([0,T_*);H^m (\Bbb R^3))$ be the
local classical axisymmetric solution of (E), corresponding to an
axisymmetric initial data $v_0 \in H^m (\Bbb R^3 )$. Then,
\begin{equation}\label{axiscri1}
\lim\sup_{t\nearrow T_*} \|v(t)\|_{H^m} =\infty \quad
\mbox{if and only if} \quad \int_0 ^{T_*}\|\vec{\o}_\theta
(t)\|_{\dot{B}^0_{\infty, 1}} dt =\infty.
\end{equation}
\end{theorem}
We observe that contrary to (\ref{axiscri}) we do not need to
control the high regularity norm, the $C^{\gamma}$ norm of vorticity
in (\ref{axiscri1}). We can also have the regularity of the
axisymmetric Euler equation by controlling only one component of the
velocity, the swirl velocity $v^\theta$ as in the follows proved in
\cite{cha22}.
\begin{theorem} Let $v\in C([0,T_*);H^m (\Bbb R^3))$, $m>5/2$, be the
local classical axisymmetric solution of (E), corresponding to an
axisymmetric initial data $v_0 \in H^m (\Bbb R^3 )$. Then, the
solution blows up in $H^m (\Bbb R^3 )$ at $T_*$ if and only if
$$ \int_0 ^{T_*}
\left( \|\tilde{\nabla}v^\t \|_{L^\infty}
+\left\|\frac{\partial v^\t}{\partial r} \right\|_{L^\infty}
\left\| \frac{1}{r}\frac{\partial v^\t}{\partial
x_3}\right\|_{L^\infty}\right) dt =\infty.
$$
\end{theorem}
\subsection{Constantin-Fefferman-Majda's and other related results}
In order to study the regularity problem of the 3D Navier-Stokes
equations Constantin and Fefferman investigated the geometric
structure of the integral kernel in the vortex stretching term more
carefully, and discovered the phenomena of `depletion effect' hidden
in the integration(\cite{con8}, see also \cite{con2} for detailed
exposition related to this fact). Later similar geometric structure
of the vortex stretching term was studied extensively also in the blow-up
problem of the 3D Euler equations by Constantin, Fefferman and
Majda(\cite{con9}).
Here we first present their results in detail, and results in
\cite{cha10}, where the BKM criterion and the
Constantin-Fefferman-Majda's criterion are interpolated in some
sense. Besides those results presented in this subsection we also
mention that there are other interesting geometric approaches to the
Euler equations such as the quaternion formulation by
Gibbon(\cite{gib1, gib2, gib3}). We begin with a definition in
\cite{con9}. Given a set $W\in \Bbb R^3$ and $r>0$ we use the
notation $B_r(W)=\{ y\in B_r (x)\, ;\, x\in W\}$.
\begin{definition}
A set $W_0 \subset \Bbb R^3$ is called smoothly directed if there
exists $\rho>0$ and $r, 0<r\leq \rho/2$ such that the following
three conditions are satisfied.
\end{definition}
\begin{itemize}
\item[(i)] For every $a\in W^*_0
=\{ q\in W_0 \, ; \, |\o_0 (q)|\neq 0\},$ and all $t\in [0,T)$, the
vorticity direction field $\xi (\cdot ,t)$ has a Lipshitz
extension(denoted by the same letter) to the Euclidean ball of
radius $4\rho$ centered at $X(a,t)$ and
$$
M=\lim_{t\to T} \sup_{a\in W^*_0} \int_0 ^t \|\nabla \xi (\cdot
,t)\|_{L^\infty (B_{4\rho} (X(a,t)))} dt <\infty.
$$
\item[(ii)] The inequality
$$ \sup_{B_{3r} (W_t )}|\o (x,t)|\leq m
\sup_{B_r (W_t )} |\o (x,t)|$$ holds for all $t\in [0,T)$ with
$m\geq 0$ constant.
\item[(iii)] The inequality
$$ \sup_{B_{4\rho} (W_t )}|v (x,t)| \leq U$$
holds for all $t\in [0, T)$.
\end{itemize}
The assumption (i) means that the direction of vorticity is well
behaved in a neighborhood of a bunch of trajectories. The assumption
(ii) states that this neighborhood is large enough to capture the
local intensification of $\o$. Under these assumptions the
following theorem is proved in \cite{con9}.
\begin{theorem}
Assume $W_0$ is smoothly directed. Then there exists $\tau >0$ and
$\Gamma$ such that
$$ \sup_{B_r (W_t )} |\o (x,t)|\leq \Gamma
\sup_{B_\rho (W_{t_0})} |\o (x,t_0 )| $$ holds for any $0\leq t_0
<T$ and $0\leq t-t_0 \leq \tau$.
\end{theorem}
They also introduced the notion of regularly directed set, closely
related to the geometric structure of the kernel defining vortex
stretching term.
\begin{definition}
We sat that a set $W_0$ is regularly directed if
there exists $\rho >0$ such that
$$ \sup_{a W^*_0} \int_0 ^T K_{\rho} (X(a,t))dt <\infty$$
where
$$ K_{\rho} (x)=\int_{|y|\leq \rho} |D(\hat{y},
\xi (x+y), \xi (x)) | |\o (x+y)|\frac{dy}{|y|^3}
$$
and
$$
D(\hat{y}, \xi (x+y) ,\xi (x))=(\hat{y}\cdot \xi (x)) \mathrm{Det}
(\hat{y}, \xi (x+y), \xi (x)).
$$
\end{definition}
Under the above assumption on the regularly directed sets the
following is proved also in \cite{con9}.
\begin{theorem}
Assume $W_0$ is regularly directed. Then there exists a constant
$\Gamma $ such that
$$ \sup_{a \in W_0} |\o (X(a,t),t)| \leq \Gamma \sup_{a \in W_0}
|\o_0 (a)|
$$
holds for all $t\in [0,T]$.
\end{theorem}
The original studies by Constantin and Fefferman in \cite{con8}
about the Navier-Stokes equations, which motivated the above
theorems, are concerned mainly about the regularity of solutions in
terms of the vorticity direction fields $\xi$. We recall, on the
other hand, that the BKM type of criterion controls the magnitude of
vorticity to obtain regularity. Incorporation of both the direction
and the magnitude of vorticity to obtain regularity for the 3D
Navier-Stokes equations was first initiated by Beir\~{a}o da Veiga
and Berselli in \cite{bei2}, and developed further by Beir\~{a}o da
Veiga in \cite{bei1}, and finally refined in an `optimal' form in
\cite{cha18}(see also \cite{cha20} for a localized version). We now
present the Euler equation version of the result in \cite{cha18}.
Below we use the notion of particle trajectory $X(a ,t)$, which is
defined by the classical solution $v(x,t)$ of (E). Let us denote
$$
\Omega_0 =\{ x\in \Bbb R^3 \, |\, \o_0 (x) \neq 0\},\quad
\Omega_t = X(\O_0 ,t).
$$
We note that the direction field of the vorticity,
$\xi (x,t)=\o (x,t)/ |\o (x,t)| $, is well-defined if
$x\in \O_t$ for $v_0 \in C^1 (\Bbb R^3 )$ with $\O_0 \neq
\emptyset$.
The following is the
main theorem proved in \cite{cha10}.
\begin{theorem}
Let $v(x,t)$ be the local classical solution to $(E)$ with initial
data $v_0 \in H^m (\Bbb R^3)$, $m>5/2$, and $\o(x,t)=$curl $v(x,t)$.
We assume $\O_0 \neq\emptyset$.
Then, the solution can be continued up to $T +\delta$ as the
classical solution, namely
$v(t)\in C([0,T+\delta];{H^m} (\Bbb R^3 ))$ for some $\delta >0$,
if there exists
$p,p', q,q', s, r_1,r_2,r_3$ satisfying the following
conditions,
\begin{equation}\label{th1}
\frac{1}{p} +\frac{1}{p'} =1, \qquad \frac{1}{q} +\frac{1}{q'} =1,
\end{equation}
and
\begin{equation}\label{th2}
\frac{1}{r_1}+\frac{p'}{r_2}\left(1-\frac{sq'}{3} \right)+
\frac{1}{r_3} \left\{ 1-p' \left( 1-\frac{sq'}{3} \right)\right\}
=1
\end{equation}
with
\begin{equation}\label{th3}
0<s<1, \qquad 1\leq \frac{3}{sq'} <p \leq \infty, \qquad \quad 1\leq
q\leq\infty,
\end{equation}
and
\begin{equation}\label{th4}
r_1\in [1, \infty],\, r_2 \in \left[p'\left(1-\frac{sq'}{3}
\right),\infty\right],\, r_3 \in \left[1-p'\left(1-\frac{sq'}{3}
\right), \infty\right]
\end{equation}
such that for direction field $\xi (x,t)$, and the magnitude of
vorticity $|\o (x,t)|$ the followings hold;
\begin{equation}\label{reg}
\int_0 ^{T} \|\xi (t )\|^{r_1}_{ \mathcal{\dot{F}}^s_{\infty,q} (\O_t) } dt <\infty,
\end{equation}
and
\begin{equation}\label{int}
\int_0 ^{T}\|\o(t) \|^{r_2}_{L^{pq'}(\O_t)} dt +
\int_0 ^{T}\|\o(t) \|^{r_3}_{L^{q'}(\O_t)} dt <\infty.
\end{equation}
\end{theorem}
\ \\
In order to get insight implied by the above theorem let us consider
the special case of $p=\infty, q=1$. In this case the conditions
(\ref{reg})-(\ref{int}) are satisfied if
\begin{equation}\label{remm1}
\xi (x,t )\in L^{r_1}(0,T; C^{ s} (\Bbb R^3 )),
\end{equation}
\begin{equation}\label{remm2}
\o
(x,t) \in L^{r_2} (0, T; L^{\infty} (\Bbb R^3 ))\cap L^{r_3} (0, T;
L^{\infty} (\Bbb R^3 )).
\end{equation}
with
\begin{equation}\label{remm3}
\frac{1}{r_1}+\frac{1}{r_2}\left(1-\frac{s}{3} \right)+
\frac{s}{3r_3}
=1.
\end{equation}
Let us {\em formally} pass $s\to 0$
in (\ref{remm1}) and (\ref{remm3}),
and choose $r_1=\infty$ and $ r_2
=r_3=1$, then we find that the conditions (\ref{remm1})-(\ref{remm2}) reduce to the
BKM condition,
since the condition $\xi (x,t )\in L^{\infty}(0,T; C^{0} (\Bbb R^3
))\cong L^{\infty}((0,T)\times \Bbb R^3)$
is obviously satisfied due to the fact that $|\xi (x,t)|\equiv 1$.\\
The other case of interest is $q'=3/s$, where
(\ref{reg})-(\ref{int}) are satisfied if
\begin{equation}\label{remm4}
\xi (x,t )\in L^{r_1}(0,T;
\mathcal{\dot{F}}^s_{\infty,\frac{3}{3-s}}(\Bbb R^3)
),\qquad
|\o
(x,t)| \in L^{r_2} (0, T; L^{\frac{3}{s}} (\Bbb R^3 )).
\end{equation}
with $1/r_1+1/r_2=1.$ The condition (\ref{remm4}) shows explicitly the
mutual compensation between the regularity of the
direction field and the integrability of the vorticity
magnitude in order to control regularity/singularity of solutions of the
Euler equations.\\
\ \\
Next we review the result of non-blow-up conditions due to Deng,
Hou and Yu\cite{den1,den2}.
We consider a time $t$ and a vortex line segment $L_t$ such that
the maximum of vorticity over the whole domain is comparable to the
maximum of vorticity on over $L_t$, namely
$$ \Omega (t):=\sup_{x\in \Bbb R^3}|\o (x,t)|\sim \max_{x\in L_t} |\o (x,t)|.$$
We denote $ L(t):=\mbox{arc length of $L_t$}$; $\xi$, $\mathbf{n}$
and $\kappa$ are
the unit tangential and the unit normal vectors to $L_t$ and
the curvature of $L_t$ respectively. We also use the notations,
\begin{eqnarray*}
U_\xi (t)&:=&\max_{x,y\in L_t} |(v\cdot \xi)(x,t)-(v\cdot \xi
)(y,t)|,\\
U_n (t)&:=&\max_{x\in L_t} |(v\cdot \mathbf{n})(x,t)|,\\
M(t)&:=&\max_{x\in L_t} |(\nabla \cdot \xi)(x,t)|,\\
K(t)&:=&\max_{x\in L_t} \kappa (x,t).
\end{eqnarray*}
We denote by $X(A, s,t)$ the image by the trajectory map at time
$t>s$ of fluid particles at $A$ at time $s$. Then, the following is
proved in \cite{den2}.
\begin{theorem}
Assume that there is a family of vortex line segment $L_t$ and $T_0
\in [0, T^*)$, such that $X(L_{t_1} , t_1 ,t_2 ) \supseteq L_{t_2}$
for all $T_0 <t_1 <t_2 <T^*$. Also assume that $\Omega (t)$ is
monotonically increasing and $\max_{x\in L_t} |\o (x,t)| \geq c_0
\Omega (t)$ for some $c_0$ when $t$ is sufficiently close to $T^*$.
Furthermore, we assume there are constants $C_U ,C_0 ,c_L$ such
that
\begin{enumerate}
\item $[U\xi (t)+U_n (t) K(t) L(t)] \leq C_U (T^* -t)^{-A}$ for
some constant $A\in (0,1)$,
\item $M(t)L(t), K(t)L(t) \leq C_0$,
\item $L(t) \geq c_L (T^* -t)^B$ for some constant $B\in (0, 1)$.
\end{enumerate}
Then there will be no blow-up in the 3D incompressible Euler flow
up to time $T^*$, as long as $B< 1-A$.
\end{theorem}
In the endpoint case of $B=1-A$ they deduced the following
theorem(\cite{den1}).
\begin{theorem}
Under the same assumption as in Theorem 2.10, there will be no
blow-up in the Euler system up to time $T^*$ in the case $B=1-A$,
as long as the following condition is satisfied:
$$ R^3 K< y_1 \left( R^{A-1} (1-A)^{1-A} /(2-A)^{2-A} \right),
$$
where $R=e^{C_0} /c_0, K:=\frac{C_U c_0}{c_L (1-A)}$, and $y_1 (m)$
denotes the smallest positive $y$ such that $m=y/(1+y)^{2-A}$.
\end{theorem}
We refer \cite{den1,den2} for discussions on the various connections of Theorem 2.10 and Theorem 2.11
with numerical computations.
\section{Blow-up scenarios}
\setcounter{equation}{0}
\subsection{Vortex sheet collapse}
We recall that a vortex line is an integral curve of the vorticity,
and a vortex tube is a tubular neighborhood in $\Bbb R^3$ foliated
by vortex lines. Numerical simulations(see e.g. \cite{cho}) show
that vortex tubes grow and thinner(stretching), and folds before
singularity happens. We review here the result by Cordoba and
Fefferman\cite{cor4} excluding a type of vortex tube collapse.
Let $Q=I_1 \times I_2 \times I_3\subset \Bbb R^3$ be a closed
rectangular box, and let $T>0$ be given. A regular tube is a
relatively open set $\Omega_t \subset Q$ parameterized by time $t\in
[0, T)$, having the form $\Omega_t =\{ (x_1, x_2 , x_3 )\in Q \, :\,
\theta (x_1,x_2,x_3, t)<0\}$ with $\theta \in C^1 (Q\times [0, T))$,
and satisfying the following properties:
$$ |\nabla _{x_1,x_2} \theta |\neq 0\, \mbox{for}\,
(x_1,x_2,x_3,t)\in Q\times [0, T), \theta (x_1,x_2,x_3,t) =0;
$$
$$\Omega_t (x_3):=\{ (x_1,x_2 )\in I_1 \times I_2 \, :\,
(x_1,x_2,x_3 )\in \Omega _t \}\, \mbox{is non-empty},
$$
for all $x_3 \in I_3, t\in [0, T)$;
$$ \mbox{closure} (\Omega_t (x_3 )) \subset \mbox{interior}
(I_1\times I_2 )
$$
for all $x_3 \in I_3, t\in [0, T)$.
Let $u(x,t)=(u_1 (x,t),u_2 (x,t),u_3 (x,t))$ be a $C^1$ velocity
field defined on $Q\times [0,T)$. We say that the regular tube
$\Omega_t $ moves with the velocity field $u$, if we have
$$
\left(\frac{\partial}{\partial t} +u\cdot \nabla_x \right)\theta =0
\, \mbox{whenever} \, (x,t)\in Q\times [0,T), \theta (x,t)=0.
$$
By the Helmholtz theorem we know that a vortex tube arising from a
3D Euler solution moves with the fluid velocity. The following
theorem proved by Cordoba and Fefferman(\cite{cor4}) says for the 3D
Euler equations that a vortex tube cannot reach zero thickness in
finite time, unless it bends and twists so violently that no part of
it forms a regular tube.
\begin{theorem}
Let $\Omega_t \subset Q (t\in [0,T))$ be a regular tube that moves
with $C^1$, divergence free velocity field $u(x,t)$.
$$ \mbox{If}\quad \int_0 ^T \sup_{x\in Q} |u(x,t)|dt <\infty,
\quad \mbox{then}\quad \lim\inf_{t\to T_-} \mbox{Vol}(\Omega_t )>0.
$$
\end{theorem}
\subsection{Squirt singularity}
The theorem of excluding the regular vortex
tube collapse was generalized by Cordoba, Fefferman and de la
Lave(\cite{cor7}), which we review here. We first recall their
definition of squirt singularities. Let $\Omega \subset \Bbb R^n$ be
an open set. We denote $X_t (a)=X(a,t)$, which is a particle
trajectory generated by a $C^1$ vector field $u:\Omega \times
[0,T)\to \Bbb R^n$ such that div $u=0$. We also set $X_{t,s} (a)$ as
the position at time $t$ of the trajectory which at time $t=s$ is
$a$. We have obvious relations,
$$ X_t (a)=X_{t,0} (a), \quad X_{t,s} = X_t\circ X_s^{-1}, \quad
X_{t,s}\circ X_{s, s_1} =X_{t,s_1}.
$$
For $\mathcal{S} \subset \Omega$, we denote by
$$
X_{t,s}^\Omega \mathcal{S} =\{ x\in \Omega \, |\, x=X_t (a), \, a
\in \mathcal{S},\, X_s (a)\in \Omega ,\, 0\leq s\leq t\}.
$$
In other words, $X_{t,s}^\Omega \mathcal{S}$ is the evolution of the
set $\mathcal{S}$, starting at time $a$, after we eliminate the
trajectories which step out of $\Omega$ at some time. By the
incompressibility condition on $u$, we have
that Vol($X_{t,s} \mathcal{S})$ is independent of $t$, and the
function $t\mapsto$Vol($X_{t,s} \mathcal{S})$ is nonincreasing.
\begin{definition}
Let $\Omega_-,\Omega_+$ be open and bounded sets.
$\overline{\Omega_-} \subset \Omega_+$. Therefore, dist $(\Omega_- ,
\Bbb R^n -\Omega_+ )\geq r >0$. We say that $u$ experiences a
squirt singularity in $\Omega_-$, at time $T>0$, when for every
$0\leq s <T$, we can find a set $\mathcal{S}_s \subset \Omega_+$
such that
\begin{itemize}
\item[(i)] $\mathcal{S}_s \cap \Omega_-$ has positive measure, $0\leq
s<T$,
\item[(ii)] $\lim_{t\to T}\mathrm{Vol}(X_{t,s}^{\Omega^+}\mathcal{S}_s
)=0$.
\end{itemize}
\end{definition}
The physical intuition behind the above definition is that there
is a region of positive volume so that all the fluid occupying it
gets ejected from a slightly bigger region in finite time.
Besides the vortex tube collapse singularity introduced in the
previous subsection the potato chip singularity and the saddle
collapse singularity, which will be defined below, are also special
examples of the squirt singularity, connected with real fluid
mechanics phenomena.
\begin{definition}[potato chip singularity]
We say that $u$ experiences a potato chip singularity when we can
find continuous functions
$$f_\pm :\Bbb R^{n-1}\times [0, T) \to \Bbb R
$$
such that
\begin{eqnarray*}
f_+ (x_1 ,\cdots ,x_{n-1}, t)\geq f_-(x_1 ,\cdots ,x_{n-1}, t),
\, t\in [0, T], x_1, \cdots , x_{n-1} \in B_{2r} (\Pi x^0 ),\\
f_+ (x_1 ,\cdots ,x_{n-1},0 )\geq f_-(x_1 ,\cdots ,x_{n-1},0),
\quad x_1, \cdots , x_{n-1} \in B_{r} (\Pi x^0 ),\\
\lim_{t\to T_-} [f_+ (x_1 ,\cdots ,x_{n-1}, t)- f_-(x_1 ,\cdots
,x_{n-1}, t)]=0 \forall x_1, \cdots , x_{n-1} \in B_{2r} (\Pi x^0 )
\end{eqnarray*}
and such that the surfaces
$$ \Sigma _{\pm, t} =\{ x_n = f_\pm(x_1 ,\cdots ,x_{n-1}, t)
\}\subset \Omega$$
are transformed into each other by the flow
$$X (\Sigma _{\pm, 0}, t) \supset \Sigma _{\pm, t}. $$
In the above $\Pi $ is projection on the first $n-1$ coordinates.
\end{definition}
Previously to \cite{cor7} potato chip singularities were considered
in the 2D and 3D flows by C\'{o}rdoba and
Fefferman(\cite{cor6},\cite{cor4a} respectively)
in the name of `sharp front'. In particular the exclusion of sharp
front in the 2D quasi-geostrophic equation is proved in
\cite{cor6}. The following notion of saddle collapse singularity is relevant only
for 2D flows.
\begin{definition}[saddle collapse singularity] We consider
foliation of a neighborhood of the origin(with coordinates $x_1,
x_2$) whose leaves are given by equations of the form
$$
\rho:= (y_1 \beta(t) +y_2 )\cdot (y_1 \delta (t) +y_2 )=cons
$$
and $(y_1 ,y_2 )=F_t (x_1 ,x_2 )$, where $\beta , \delta : [0, T)
\to \Bbb R^+$ are $C^1$ foliations, $F$ is a $C^2$ function of
$x,t,$ for a fixed $t$, and $F_t$ is an orientation preserving
diffeomorphism.
We say that the foliation experiences a saddle collapse when
$$\lim\inf_{t\to T} \beta (t) +\delta (t) =0.
$$
If the leaves of the foliation are transported by a vector field
$u$, we say that the vector field $u$ experiences a saddle
collapse.
\end{definition}
The exclusion of saddle point singularity in the 2D
quasi-geostrophic equation(see Section 4.3 below) was proved by
C\'{o}rdoba in \cite{cor2}. The following `unified' theorem is
proved in \cite{cor7}.
\begin{theorem}
If $u$ has a squirt singularity at $T$, then $\int_s ^T \sup_x
|u(x,t)|dt =\infty$ for all $ s\in (0, T)$. Moreover, if $u$ has a
potato chip singularity, then
$$
\int_s ^T \sup_x |\Pi u (x,t)|dt =\infty.
$$
\end{theorem}
\subsection{Self-similar blow-up}
In this subsection we review the scenario of self-similar
singularity studied in \cite{cha13}. We first observe that the Euler
system (E) has scaling property that
if $(v, p)$ is a
solution of the system (E), then for any $\lambda >0$ and $\alpha
\in \Bbb R $ the functions
\begin{equation}
\label{self}
v^{\lambda, \alpha}(x,t)=\lambda ^\alpha v (\lambda x, \l^{\a +1}
t),\quad p^{\l, \a}(x,t)=\l^{2\a}p(\l x, \l^{\a+1} t )
\end{equation}
are also solutions of (E) with the initial data
$ v^{\lambda, \alpha}_0(x)=\lambda ^\alpha v_0
(\lambda x)$.
In view of the scaling
properties in (\ref{self}), the self-similar blowing up
solution $v(x,t)$ of (E), if it exists, should be of the form,
\begin{eqnarray}
\label{vel}
v(x, t)&=&\frac{1}{(T_*-t)^{\frac{\a}{\a+1}}}
V\left(\frac{x}{(T_*-t)^{\frac{1}{\a+1}}}\right)
\end{eqnarray}
for $\a \neq -1$ and $t$ sufficiently
close to $T_*$.
If we assume that initial vorticity $\omega_0$ has compact support,
then the nonexistence of self-similar blow-up of the form given by
(\ref{vel}) is rather immediate from the well-known formula, $
\omega (X(a,t),t)=\nabla_a X(a,t) \omega_0 (a)$. We want to
generalize this to a nontrivial case. Substituting (\ref{vel}) into
(E), we find that $V$ should be
a solution of the system
\[
(SE)
\left\{
\begin{aligned}
\frac{\a}{\a +1} V+&\frac{1}{\a+1} (x\cdot\nabla ) V
+(V\cdot \nabla )V =-\nabla P,
\\
\textrm{div }\, V& =0
\end{aligned}
\right.
\]
for some scalar function $P$,
which could be regarded as the Euler version of the Leray
equations introduced in \cite{ler}. The question of existence of
nontrivial solution to (SE) is equivalent to the that of existence of
nontrivial self-similar finite time blowing up
solution to the Euler system of the form (\ref{vel}).
Similar question for the
3D Navier-Stokes equations was raised by Leray in \cite{ler}, and
answered negatively by Necas, Ruzicka and Sverak(\cite{nec}), the result of which was
refined later by Tsai in \cite{tsa}(see also \cite{mil} for a generalization).
Combining the energy conservation with a simple scaling argument,
the author of this article showed that if there exists a nontrivial self-similar finite time
blowing up solution, then its helicity should be zero(\cite{cha4}).
Mainly due to lack of the laplacian term
in the right hand side of the first equations of (SE), we cannot
expect the maximum principle, which was crucial in the works in
\cite{nec} and \cite{tsa} for the 3D Navier-Stokes equations.
Using a completely different argument from those previous ones,
in \cite{cha13} it is proved that there cannot be self-similar blowing up solution to (E) of the form
(\ref{vel}), if the vorticity decays sufficiently fast near infinity.
Given a smooth velocity field $v(x,t)$,
the particle trajectory mapping $a\mapsto X(a,t)$
The inverse $A(x,t):=X^{-1} (x ,t)$ is called the back to label
map, which satisfies $
A (X(a,t),t)=a,$ and $ X(A
(x,t),t)=x$.
The existence of the back-to-label
map $ A(\cdot, t)$ for our smooth velocity $v(x,t)$ for
$t\in (0, T_*)$, is guaranteed
if we assume a uniform decay of $v(x,t)$ near infinity, independent of the
decay rate(see \cite{con5}). The following is proved in
\cite{cha13}.
\begin{theorem}
There exists no finite time blowing up self-similar solution
$v(x,t)$ to the 3D Euler equations of the form
(\ref{vel}) for $t \in (0 ,T_*)$ with $\a \neq -1$, if $v$ and $V$
satisfy the following conditions:
\begin{itemize}
\item[(i)]For all $t\in (0, T_*)$ the particle trajectory mapping $X(\cdot ,t)$ generated
by the classical solution $v\in C([0,T_*);C^1(\Bbb R^3;\Bbb R^3 ))$ is a
$C^1$ diffeomorphism from $\Bbb R^3$ onto itself.
\item[(ii)] The vorticity satisfies $\Omega=$curl $V\neq 0$, and there exists $p_1 >0$ such that
$\Omega \in L^p (\Bbb R^3
)$ for all $p\in (0, p_1)$.
\end{itemize}
\end{theorem}
We note that the condition (ii) is satisfied, for example, if $
\Omega\in L^{1}_{loc}(\Bbb R^3)$ and there exist constants $R, K$
and $\varepsilon_1, \varepsilon_2
>0$ such that $ |\Omega(x)|\leq K e^{-\varepsilon_1 |x|^{\varepsilon_2}}$ for $|x|>R$,
then we have
$\Omega\in L^p (\Bbb R^3;\Bbb R^3)$ for all $p\in (0, 1)$.
Indeed,
for all $p\in (0, 1)$, we have
\begin{eqnarray*}
\int_{\Bbb R^3} |\Omega(x)|^p dx
&=& \int_{|x|\leq R} |\Omega(x)|^p dx +\int_{|x|>R} |\Omega (x)|^p \,dx\\
&\leq&|B_R |^{1-p}\left(\int_{|x|\leq R} |\Omega(x)| dx\right)^{p} +
K^p \int_{\Bbb R^3} e^{-p\varepsilon_1|x|^{\varepsilon_2}}dx <\infty ,
\end{eqnarray*}
where $|B_R|$ is the volume of the ball $B_R$ of radius $R$.\\
In the zero vorticity case $\Omega =0$, from div $V=0$ and curl
$V=0$,
we have $V=\nabla h$, where $h(x)$ is a harmonic function in $\Bbb R^3$. Hence,
we have an easy example of self-similar blow-up,
$$v(x,t)=\frac{1}{(T_*-t)^{\frac{\a}{\a+1}}}\nabla
h\left(\frac{x}{(T_*-t)^{\frac{1}{\a+1}}}\right),$$
in $\Bbb R^3$, which is also the case for the 3D Navier-Stokes with $\alpha=1$.
We do not consider this case in the theorem.\\
The above theorem is actually a corollary of the following more
general theorem.
\begin{theorem}
Let $v\in C([0,T);C^1 (\Bbb R^3 ;\Bbb R^3))$ be a classical solution
to the 3D Euler equations generating the particle trajectory mapping $X(\cdot ,t)$ which is a
$C^1$ diffeomorphism from $\Bbb R^3$
onto itself for all $t\in (0,T)$. Suppose we have
representation of the vorticity of the solution,
by
\begin{equation}\label{thm13}
\o (x,t) = \Psi(t)\Omega (\Phi(t)x) \qquad \forall t\in [0, T)
\end{equation}
where $\Psi(\cdot)\in C([0, T );(0, \infty))$,
$\Phi(\cdot)\in C([0, T );\Bbb R^{3\times 3})$
with $\mathrm{det}(\Phi(t))\neq 0$ on $[0, T)$; $\Omega = \mathrm{curl}\, V$ for some
$V$, and there exists $p_1 >0$ such that
$|\Omega |$ belongs to $ L^p
(\Bbb R^3)$ for all $p\in (0, p_1 )$. Then,
necessarily
either $\mathrm{det}(\Phi(t))\equiv \mathrm{det}(\Phi(0))$ on $[0, T)$, or
$\Omega=0$.
\end{theorem}
For the detailed proof of Theorem 3.3 and 3.4 we refer \cite{cha13}.
\subsection{Asymptotic self-similar blow-up}
In this subsection we consider the possibility of more refined
scenario of self-similar singularity than in the previous
subsection, called the asymptotic self-similar singularity. This
means that the local in time smooth solution evolves into a
self-similar profile as the possible singularity time is approached.
The similar notion was considered previously by Giga and Kohn in
their study of semilinear heat equation(\cite{gig}). Their sense of
convergence of solution to the self-similar profile is the
pointwise sense with a time difference weight to make it scaling
invariant, and cannot apply directly to the case of Euler system. It
is found in \cite{cha14} that if we make the sense of convergence
strong enough, then we can apply the notion of asymptotic
self-similar singularity to the Euler and the Navier-Stokes
equations. The following theorem is proved in \cite{cha14}.
\begin{theorem}
Let $v\in C([0, T);B^{\frac{3}{p}+1}_{p, 1}(\Bbb R^3))$ be a
classical solution to the 3D Euler equations.
Suppose there exist $p_1>0$, $\a > -1$, $\bar{V}\in C^1 (\Bbb R^3)$
with $ \lim_{R\to \infty} \sup_{|x|=R} |\bar{V}(x)|=0$
such
that $\bar{\Omega}=$curl $\bar{V}\in L^q(\Bbb R^3)$ for all $q\in
(0,p_1)$, and the following convergence holds true:
$$
\lim_{t\nearrow T}
(T-t)^{\frac{\a-3}{\a+1}}\left\|v(\cdot, t)
-\frac{1}{(T-t)^{\frac{\a}{\a+1}}} \bar{V}
\left(\frac{\cdot}{(T-t)^{\frac{1}{\a+1}}} \right)\right\|_{L^1 }
=0,
$$
and
$$
\sup_{t\in (0,T)}
(T-t)\left\|\o(\cdot, t) -\frac{1}{T-t} \bar{\O}
\left(\frac{\cdot}{(T-t)^{\frac{1}{\a+1}}}
\right)\right\|_{\dot{B}^0_{\infty, 1} }<\infty.
$$
Then, $\bar{\O}=0$, and $v(x,t)$ can be extended to a solution of the 3D Euler system in $[0, T+\delta]\times \Bbb R^3$,
and belongs to $C([0, T+\delta]; B^{\frac{3}{p}+1}_{p, 1}(\Bbb R^3))$ for some $\delta
>0$.
\end{theorem}
We note that the above theorem still
does not exclude the possibility that the sense of vorticity convergence to the
asymptotically self-similar singularity is weaker than $L^\infty$
sense. Namely, a self-similar vorticity profile could be approached
from a local classical solution in the pointwise sense in space,
or in the $L^p(\Bbb R^3)$ sense for
some $p$ with $1\leq p<\infty$. In \cite{cha14} we also proved
nonexistence of
asymptotic self-similar solution to the 3D Navier-Stokes equations with appropriate
change of functional setting(see also \cite{hou2}
for related results).\\
\ \\
The proof of the above theorem follows without difficulty from the
following blow-up rate estimate(\cite{cha14}), which is interesting
in itself.
\begin{theorem}
Let $p\in [1, \infty)$ and $v\in C([0, T);B^{\frac{3}{p}+1}_{p,
1}(\Bbb R^3))$ be a classical solution to the 3D Euler equations.
There exists an absolute constant $\eta >0$ such that if
\begin{equation}\label{th11}
\inf_{0\leq t<T} (T-t) \|\o (t)\|_{\dot{B}^0_{\infty, 1} } <\eta
,
\end{equation}
then $v(x,t)$ can be extended to a solution of the 3D Euler system in $[0, T+\delta]\times \Bbb R^3$,
and belongs to $C([0, T+\delta]; B^{\frac{3}{p}+1}_{p,
1}(\Bbb R^3))$ for some $\delta
>0$.
\end{theorem}
We note that the proof of the local existence for $v_0 \in
B^{\frac{3}{p}+1}_{p, 1}(\Bbb R^3)$ is
done in \cite{cha1, cha6}(see also \cite{vis1}).
The above theorem implies that
if $T_*$ is the first time of singularity, then we have
the lower estimate of the blow-up rate,
\begin{equation}\label{blow}
\|\o (t)\|_{\dot{B}^0_{\infty, 1} } \geq \frac{C}{T_*-t}\quad
\forall t\in [0, T_*)
\end{equation}
for an absolute constant $C$. The estimate (\ref{blow}) was actually
derived previously by a different argument in \cite{cha4}.
We observe that (\ref{blow}) is consistent with both of the
BKM criterion(\cite{bea}) and
the Kerr's numerical calculation in \cite{ker1} respectively.\\
The above continuation principle
for a local solutions in $B^{\frac{3}{p}+1}_{p,
1}(\Bbb R^3)$ has obvious applications to the solutions belonging
to more conventional function spaces, due to the embeddings,
$$ H^m (\Bbb R^3 )\hookrightarrow C^{1, \gamma} (\Bbb R^3)\hookrightarrow
B^{\frac{3}{p}+1}_{p,
1}(\Bbb R^3) $$
for $m>5/2$ and $\gamma = m-5/2$.
For example the local solution $v\in C([0, T); H^m (\Bbb R^3 ))$ can
be continued to be $v\in C([0, T+\delta]; H^m (\Bbb R^3 ))$ for
some $\delta$, if
(\ref{th11}) is satisfied. Regarding other implication of the above theorem on the self-similar
blowing up solution to the 3D Euler equations, we have the following
corollary(see \cite{cha14} for the proof).
\begin{cor}
Let $v\in C([0, T_*);B^{\frac{3}{p}+1}_{p, 1}(\Bbb R^3))$ be a
classical solution to the 3D Euler equations. There exists $\eta
>0$ such that if we have representation for the
velocity by (\ref{vel}), and $\bar{\O}$=curl $\bar{V}$ satisfies
$\|\bar{\O}\|_{\dot{B}^0_{\infty ,1}} <\eta$, then $\bar{\Omega}=0$, and
$v(x,t)$ can be extended to a solution of the 3D Euler system in $[0, T_*+\delta]\times \Bbb R^3$,
and belongs to $C([0, T_*+\delta]; B^{\frac{3}{p}+1}_{p,
1}(\Bbb R^3))$ for some $\delta
>0$.
\end{cor}
\section{Model problems}
\setcounter{equation}{0}
Since the blow-up problem of the 3D Euler equations looks beyond the
capability of current analysis, people proposed simplified model
equations to get insight on the original problem.
In this section we review some of them.
Besides those results presented in the following subsections there
are also studies on the other model problems.
In \cite{din} Dinaburg, Posvyanskii and Sinai analyzed a quasi-linear
approximation of the infinite system of ODE arising when we write the
Euler equation in a Fourier mode. Fridlander and
Pavlovi\'{c}(\cite{fri2}) considered a vector model, and
Katz and Pavlovi\'{c}(\cite{katz}) studied dyadic model, both of
which are resulted from the representation of the Euler equations
in the wave number space. Okamoto and Ohkitani proposed
model equations in \cite{oka}, and a `dual' system to the Euler
equations was considered in \cite{cha6a}.
\subsection{Distortions of the Euler equations}
Taking trace of the matrix equation (\ref{basfor1} ) for $V$, we
obtain
$\Delta p=-tr V^2$, and hence the hessian of the pressure is given
by
$$ P_{ij}=- \partial_i \partial _j (\Delta )^{-1} tr V^2 =-R_i R_j tr
V^2,$$
where $R_j$ denotes the Riesz transform(see Section 1).
Hence we can rewrite the Euler equations as
\begin{equation}\label{matrixeuler}
\frac{D V}{Dt}=-V^2 -R[tr V^2], \quad R[\cdot]=(R_iR_j[\cdot])
\end{equation}
In \cite{con1} Constantin studied a distorted version of the above
system,
\begin{equation}\label{coneq}
\frac{\partial V}{\partial t}
=-V^2 -R[tr V^2], \quad R[\cdot]=(R_iR_j[\cdot]),
\end{equation}
where the convection term of the original Euler equations is
deleted, and showed that a solution indeed blows up in finite time.
Note that the incompressibility condition, tr$V=0$, is
respected in the system (\ref{coneq}).
Thus we find that the convection term should have significant role
in the study of the blow-up problem of the original Euler
equations.\\
On the other hand, in \cite{liu} Liu and Tadmor studied another
distorted version of (\ref{matrixeuler}), called the restricted Euler
equations,
\begin{equation}\label{liueq}
\frac{D V}{Dt}=-V^2 +\frac{1}{n} (tr V^2 ) I.
\end{equation}
We observe that in (\ref{liueq}) the convection term is kept, but
the non-local operator $R_iR_j(\cdot)$ is changed into a
local one $-1/n \delta_{ij}$, where the numerical factor $-1/n$ is to keep the
incompressibility condition.
Analyzing the dynamics of eigenvalues of the matrix $V$, they
showed that the system (\ref{liueq}) actually blows up in finite
time(\cite{liu}).
\subsection{The Constantin-Lax-Majda equation}
In 1985 Constantin, Lax and Majda constructed a one dimensional
model of the vorticity formulation of the 3D Euler equations, which
preserves the feature of nonlocality in vortex stretching term.
Remarkably enough this model equation has an explicit solution for
general initial data(\cite{con11}). In this subsection we briefly
review their result.
We first observe from section 1 that vorticity formulation of the Euler equations is
$ \frac{D \omega }{Dt} = S \omega ,$ where
$S=\mathcal{P} (\omega )$ defines a singular integral operator of
the Calderon-Zygmund type on $\o$.
Let us replace
${ \omega (x,t)\Rightarrow \theta (x,t)}$,
${ \frac{D}{Dt}\Rightarrow\frac{\partial}{\partial t}}$,
${\mathcal{P}(\cdot
)\Rightarrow H (\cdot )}$, where
$\theta (x,t)$ is a scalar
function on $\Bbb R \times \Bbb R_+$, and
$H(\cdot)$ is the \textit{Hilbert transform} defined by
$$Hf (x)=\frac{1}{\pi} p.v. \int_{\Bbb R}
\frac{f(y)}{x-y}dy.
$$
Then we obtain, the following 1D scalar equation from the 3D Euler
equation,
$$ (CLM): \frac{\partial \theta}{\partial t}=\theta H\theta .$$
This model preserve the feature of { nonlocality} of the Euler
system (E), in contrast to the more traditional one dimensional
model, the inviscid Burgers equation.
We recall the identities for the Hilbert transform:
\begin{equation}\label{hilbert}
H(Hf)=-f,\quad
H(fHg+gHf)=(Hf)(Hg)-fg,
\end{equation}
which imply $ H(\theta H\theta )=\frac12 [ (H\theta)^2 -\theta^2
]$.
Applying $H$ on both sides of the first equation of (CLM), and using the formula
(\ref{hilbert}), we
obtain
$$
(CLM)^* : (H\theta )_{t}+\frac{1}{2}((H\theta )^{2}-(\theta
)^{2})=0.
$$
We introduce the complex valued function,
$$ { z(x,t)=H\theta
(x,t)+i\theta (x,t)}.
$$
Then, (CLM) and (CLM)$^*$ are the imaginary and the real parts
of the complex Riccati equation,
$$
z_{t} (x,t) =\frac12 z^2 (x,t)
$$
The explicit solution to the complex equation is
$$
z(x,t)=\frac{z_0}{1-\frac12 t z_0 (x)} .
$$
Taking the imaginary part, we obtain
$$
\theta (x,t) = \frac{4\theta_0 (x)}{(2-tH\theta_0 (x))^2 +t^2
\theta_0 ^2(x)}.
$$
The finite time blow-up occurs if and only if
$$
{Z=\{ x \, | \, \theta_0 (x)=0 \, \mbox{and}\,
H\theta_0(x) >0 \}\neq \varnothing }.
$$
In \cite{scho} Schochet find that even if we add viscosity term to
(CLM) there is a finite time blow-up. See also \cite{sak1, sak2} for
the studies of other variations of (CLM).
\subsection{The 2D quasi-geostrophic equation and its 1D model}
The 2D quasi-geostrophic
equation(QG) models the dynamics of the mixture of cold and hot air
and the fronts between them.
$$
(QG) \left\{\aligned
& \theta _{t}+(u\cdot \nabla )\theta =0, \\
&v=-\nabla ^{\bot }(-\Delta )^{-\frac{1}{2}}\theta ,\\
&\theta (x,0)=\theta _{0}(x),
\endaligned
\right.
$$
where $\nabla ^{\bot }=(-\partial _{2},\partial _{1})$.
Here, $\theta (x,t)$ represents the temperature of the air at
$(x,t)\in \Bbb R^2 \times \Bbb R_+$. Besides its direct physical
significance (QG) has another important feature that it has very
similar structure to the 3D Euler equations. Indeed, taking $\nabla
^\bot$ to (QG), we obtain
$$
\left(\frac{\partial }{\partial t} +v\cdot \nabla\right){ \nabla
^\bot \theta}
=({\nabla ^\bot \theta }\cdot
\nabla ) v,
$$
where
$$
v(x,t)=\int_{\Bbb R^2} \frac{ \nabla ^\bot \theta (y,t)}{|x-y|} dy
.
$$
This is exactly the vorticity formulation of 3D Euler
equation if we identify
$$ \nabla^\bot \theta \Longleftrightarrow \omega
$$
After first observation and pioneering analysis of these feature by
Constantin, Majda and Tabak(\cite{con12}) there have been so many
research papers devoted to the study of this equation(also the
equation with the viscosity term, $-(-\Delta )^{\alpha}\theta$,
$\alpha
>0$, added)(\cite{con2,con6a, con12a, con12b, cor2, cor1, cor3, cor5,
cor6, cor7, cor8, wu1,wu2,wu3, cha2a, cha10, cha10a, cha24a, ohk,
din, kis, caff}). We briefly review some of them
here concentrating on the inviscid equation (QG). \\
The local existence can be proved easily by standard method. The BKM
type of blow-up criterion proved by Constantin, Majda and Tabak in
\cite{con12} is
\begin{equation}\label{cmtcri}
\lim\sup_{t\nearrow T_*} \|\theta(t)\|_{H^s }=\infty
\quad
\mbox{if and only if}\quad
\int_0 ^{T_*} \|{ \nabla ^\bot \theta (s)}\|_{L^\infty} ds =\infty.
\end{equation}
This criterion has been refined, using the Triebel-Lizorkin spaces
\cite{cha2a}.
The question of finite time singularity/global regularity is still
open. Similarly to the Euler equations case we also have the following
geometric type of condition for the regularity. We define the
direction field $\xi =\nabla^\bot \t /|\nabla^\bot \t|$ whenever $|\nabla^\bot \theta
(x,t)|\neq 0$.
\begin{definition}
We say that
a set $\Omega_0$ is smoothly directed if there exists $\rho >0$ such
that
$$
\sup_{q\in \Omega_0} \int_0 ^T |v(X(q,t),t)|^2 dt < infty,
$$
$$
\sup_{q\in \Omega^*_0} \int_0 ^T \|\nabla \xi (\cdot ,
t)\|_{L^\infty (B_\rho (X(q,t)))} ^2 dt <\infty,
$$
where $B_\rho (X)$ is the ball of radius $\rho$ centered at $X$ and
$$ \Omega ^*_0 =\{ q \in \Omega_0 \, ;\, |\nabla \theta_0 (q)| \neq
0\}.$$
\end{definition}
We denote $ \mathfrak{O}_T(\Omega_0 )=\{ (x,t) \, |\, x\in
X(\Omega_0 ,t), 0\leq t\leq T\}$. Then, the following theorem is
proved(\cite{con12}).
\begin{theorem}
Assume that $\Omega_0$ is smoothly directed. Then
$$ \sup_{(x,t) \in \mathfrak{O}_T(\Omega_0 )} |\nabla \theta (x,t)|<\infty,$$
and no singularity occurs in $\mathfrak{O}_T(\Omega_0 )$.
\end{theorem}
Next we present an `interpolated' result between the criterion
(\ref{cmtcri}) and Theorem 4.1, obtained in \cite{cha10}. Let us
denote bellow,
$$
D_0 =\{ x\in \Bbb R^2 \, |\, |\nabla^\bot\th_0 (x)| \neq 0\},\quad
D_t = X(D_0 ,t).
$$
The following theorem(\cite{cha10}) could be also considered as the
(QG) version of Theorem 2.9.
\begin{theorem}
Let $\th (x,t)$ be the local classical solution to $(QG)$ with
initial data $\theta_0 \in H^m (\Bbb R^2)$,
$m>3/2$, for which $D_0
\neq\emptyset$. Let $\xi (x,t)=\nabla^\bot\th (x,t)/ |\nabla^\bot\th (x,t)| $ be
the direction field defined for $x\in D_t $.
Then, the solution can be continued up to $T <\infty$ as the classical solution, namely
$\th (t)\in C([0,T];{H^m} (\Bbb R^2 ))$,
if there exist parameters
$p,p', q,q', s, r_1,r_2,r_3$ satisfying the following
conditions,
\begin{equation}\label{31}
\frac{1}{p} +\frac{1}{p'} =1, \qquad \frac{1}{q} +\frac{1}{q'} =1,
\end{equation}
and
\begin{equation}\label{32}
\frac{1}{r_1}+\frac{p'}{r_2}\left(1-\frac{sq'}{2} \right)+
\frac{1}{r_3} \left\{ 1-p' \left( 1-\frac{sq'}{2} \right)\right\}
=1
\end{equation}
with
\begin{equation}\label{33}
0<s<1, \qquad 1\leq \frac{2}{sq'} <p \leq \infty, \qquad \quad 1\leq
q\leq\infty,
\end{equation}
and
\begin{equation}\label{34}
r_1\in [1, \infty],\, r_2 \in \left[p'\left(1-\frac{sq'}{2}
\right),\infty\right],\, r_3 \in \left[1-p'\left(1-\frac{sq'}{2}
\right), \infty\right]
\end{equation}
such that the followings hold:
\begin{equation}\label{35}
\int_0 ^{T} \|\xi (t )\|^{r_1}_{ \mathcal{\dot{F}}^s_{\infty,q} (D_t) } dt <\infty,
\end{equation}
and
\begin{equation}\label{36}
\int_0 ^{T}\|\nabla^\bot \th(t) \|^{r_2}_{L^{pq'}(D_t)} dt +
\int_0 ^{T}\|\nabla^\bot\th (t) \|^{r_3}_{L^{q'}(D_t)} dt <\infty.
\end{equation}
\end{theorem}
In order to compare this theorem with the Constantin-Majda-Tabak
criterion (\ref{cmtcri}), let
us consider the case of $p=\infty, q=1$. In this case the conditions
(\ref{35})-(\ref{36}) are satisfied if
\begin{equation}\label{37}
\xi (x,t )\in L^{r_1}(0,T; C^{ s} (\Bbb R^2 )),
\end{equation}
\begin{equation}\label{38}
|\nabla^\bot\th
(x,t)| \in L^{r_2} (0, T; L^{\infty} (\Bbb R^2 ))\cap L^{r_3} (0, T;
L^{\infty} (\Bbb R^2 )).
\end{equation}
with
$$
\frac{1}{r_1}+\frac{1}{r_2}\left(1-\frac{s}{2} \right)+
\frac{s}{2r_3}
=1.
$$
If we {\em formally} passing $s\to 0$, and choosing
$r_1=\infty, r_2=r_3
=1$, we find that the conditions (\ref{37})-(\ref{38})
are satisfied if the
Constantin-Majda-Tabak condition in (\ref{cmtcri}) holds, since the condition
$$\xi (x,t )
\in L^{\infty}(0,T; C^{0} (\Bbb
R^2))\cong L^\infty ((0,T)\times \Bbb R^2 )$$
is automatically satisfied.
The other is the case $q'=2/s$, where
(\ref{35})-(\ref{36}) are satisfied if
\begin{equation}
\xi (x,t )\in L^{r_1}(0,T;
\mathcal{\dot{F}}^s_{\infty,\frac{2}{2-s}}(\Bbb R^2 )
),\quad
|\nabla^\bot\th
(x,t)| \in L^{r_2} (0, T; L^{\frac{2}{s}} (\Bbb R^2 ))
\end{equation}
with $1/r_1+1/r_2=1,$ which shows mutual compensation of the
regularity of the direction field $\xi (x,t)$ and the
integrability of the
magnitude of gradient $|\nabla^\bot \theta (x,t)|$ to obtain smoothness
of $\theta (x,t)$.\\
\ \\
There had been a conjectured scenario of singularity in (QG) in the
form of hyperbolic saddle collapse of level curves of $\theta
(x,t)$(see Definition 3.3). This was excluded by C\'{o}rdoba in
1998(\cite{cor2}, see also Section 3.2 of this article). Another
scenario of singularity, the sharp front singularity,
which is a two dimensional version of
potato chip singularity(see Definition 3.2 with $n=2$)
was excluded by C\'{o}rdoba and Fefferman in \cite{cor6} under
the assumption of suitable velocity control(see Section 3.2).\\
\ \\
We can also consider the possibility of self-similar singularity for
(QG). We first note that (QG) has the scaling property that
if $\t$ is a
solution of the system, then for any $\lambda >0$ and $\alpha \in
\Bbb R $ the functions
\begin{equation}
\label{self3a}
\t^{\lambda, \alpha}(x,t)=\lambda ^\alpha \t(\lambda x, \l^{\a +1}
t)
\end{equation}
are also solutions of (QG) with the initial data
$ \t ^{\lambda, \alpha}_0(x)=\lambda ^\alpha \t_0
(\lambda x)$.
Hence, the self-similar blowing up solution should be of the
form,
\begin{equation}
\label{self3b}
\t(x, t)=\frac{1}{(T_*-t)^{\frac{\a}{\a+1}}}
\Theta\left(\frac{x}{(T_*-t)^{\frac{1}{\a+1}}}\right)
\end{equation}
for $t$ sufficiently close $T_*$ and $\a \neq -1$.
The following theorem is proved in \cite{cha13}.
\begin{theorem}
Let $v $ generates a particle trajectory, which is a $C^1$
diffeomorphism from $\Bbb R^2$ onto itself for all $t\in (0, T_*)$.
There exists no nontrivial solution $ \t $ to the system $(QG)$ of
the form (\ref{self3b}), if there exists $p_1, p_2 \in (0, \infty
]$, $p_1 < p_2$, such that $\Theta\in L^{p_1} (\Bbb R^2 ) \cap
L^{p_2} (\Bbb R^2 )$.
\end{theorem}
We note that the integrability condition on the self-similar
representation function $\Theta$ in the above theorem is `milder'
than the case of the exclusion of self-similar Euler equations in
Theorem 3.3, in the sense that the decay condition is of $\Theta$
(not $\nabla^\bot \Theta$) near infinity is weaker than that of
$\Omega=$curl $V$.\\
\ \\
In the remained part of
this subsection we discuss a 1D model of the 2D quasi-geostrophic
equation studied in \cite{cha19}(see \cite{mor} for related
results). The construction of the one dimensional model can be done
similarly to the Constantin-Lax-Majda equation introduced in section
4.2. We
first note that
$$
v=-R^{\bot }\theta =(-R_2 \theta , R_1 \theta ),
$$
where $R_{j}$, $j=1,2$, is the two dimensional Riesz transform(see
Section 1).
We can rewrite the dynamical equation of (QG) as
$$
\theta _{t}+ \rm{div }\, [(R^{\bot }\theta )\theta ]=0,
$$
since $\rm{div} (R^{\bot }\theta )=0$.
To construct the one dimensional model we replace:
$${ R^{\bot
}(\cdot ) \Rightarrow H(\cdot ), \qquad \rm{div} (\cdot)
\Rightarrow
\partial_x}
$$
to obtain
$$\theta _{t}+(H(\theta )\theta )_{x}=0. $$
Defining the complex valued function
${ z(x,t)=H\theta (x,t)
+i\theta (x,t)}$, and following Constantin-Lax-Majda(\cite{con11}), we
find that our equation is the imaginary part of
$$z_t +zz_x =0, $$
which is complex Burgers'
equation. The characteristics method does not work here.
Even in that case we can show that the finite time blow-up occurs
for the generic initial data as follows.
\begin{theorem}
Given a periodic non-constant initial data $\theta_0\in
C^1([-\pi,\pi])$ such that $\int^{\pi}_{-\pi}\theta_0(x) dx =0$,
there is no $C^1([-\pi,\pi]\times[0,\infty))$ periodic solution to
the model equation.
\end{theorem}
For the proof we refer \cite{cha19}. Here we give a brief outline of
the construction of an explicit blowing up solution. We
begin with the complex Burgers equation:
$$
{z_{t}+zz_{x}=0, \quad z=u+i\theta}
$$
with $u(x,t) \equiv H\theta(x,t)$. Expanding it to real and
imaginary parts, we obtain the system:
$$
\left\{ \aligned
u_{t}+uu_{x}-\theta \theta _{x} &=0, \\
\theta _{t}+u\theta _{x}+\theta u_{x} &=0
\endaligned
\right.
$$
In order to perform the \textit{hodograph transform} we consider ${
x(u,\theta )}$ and ${ t(u,\theta )}$
We have,
\begin{eqnarray*}
u_{x} &=&Jt_{\theta }\;, \quad
\theta _{x} =-Jt_{u}\;, \\
u_{t} &=&-Jx_{\theta }\;, \quad \theta _{t} =Jx_{u}\;,
\end{eqnarray*} where $J=(x_{u}t_{\theta }-x_{\theta
}t_{u})^{-1}$.
By direct substitution we obtain,
$$
\left\{ \aligned -x_{\theta }+ut_{\theta }+\theta t_{u} &=0\;, \\
x_{u}-ut_{u}+\theta t_{\theta } &=0\; \endaligned \right. $$
as far
as $J^{-1}\neq 0$. This system can be written more compactly in the
form:
\begin{eqnarray*}
-(x-tu)_{\theta }+(t\theta )_{u} &=&0\;, \\
(x-tu)_{u}+(t\theta )_{\theta } &=&0\;,
\end{eqnarray*}
which leads to the
following Cauchy-Riemann system,
$$
\xi _{u} =\eta _{\theta },\quad \xi _{\theta } =-\eta _{u},
$$
where we set
$
\eta (u,\theta ):=
x(u,\theta )-t(u,\theta )u, \quad \xi (u,\theta ):= t(u,\theta
)\theta . $
Hence, $f(z)=\xi (u,\theta )+i\eta (u,\theta )$ with
$z=u+i\theta $ is an analytic function. Choosing $f(z)=\log\, z, $
we find,
\begin{equation}\label{1drel}
t\theta =\log \sqrt{u^{2}+\theta ^{2}} , \quad
x-tu =\arctan \frac{\theta }{u},
\end{equation}
which corresponds to the initial data,$ z(x,0)=\cos x+i\sin x .$
The relation (\ref{1drel}) defines implicitly the real and imaginary
parts $(u(x,t),\theta (x,t))$ of the solution.
Removing $\theta$ from the system, we obtain
$$
tu\tan (x-tu)=\log\left| \frac{u}{\cos (x-tu)}\right|,
$$
which defines $u(x,t)$ implicitly.
By elementary computations we find both $u_x$ and $\theta_x$ blow
up at $t=e^{-1}$.
\subsection{The 2D Boussinesq system and Moffat's problem}
The 2D Boussineq system for the incompressible fluid flows in $\Bbb
R^2$ is
$$
(B)_{\nu, \kappa }\left\{
\aligned
&\frac{\partial v}{\partial t} +(v\cdot \nabla )v =-\nabla p +\nu \Delta v+
\theta {e_2} ,
\\
&\frac{\partial \theta}{\partial t} +(v\cdot \nabla )\theta =\kappa \Delta \theta\\
&\textrm{div }\, v =0 ,\\
&v(x,0)=v_0 (x), \qquad \theta(x,0)=\theta_0 (x),
\endaligned
\right.
$$
where $v=(v_1, v_2)$, $v_j =v_j (x,t), j=1,2$, $(x,t)\in \Bbb R^2
\times (0, \infty)$, is the velocity vector field, $p=p(x,t)$ is the
scalar pressure, $\theta(x,t)$ is the scalar temperature, $\nu \geq
0$ is the viscosity, and $\kappa \geq 0$ is the thermal diffusivity,
and ${e_2} =(0,1)$. The Boussinesq system has important roles in the
atmospheric sciences(see e.g. \cite{maj4}). The global in time
regularity of $(B)_{\nu, \kappa }$ with $ \nu >0$ and $\kappa
>0$ is well-known(see e.g. \cite{can}). On the other hand, the
regularity/singularity questions of the fully inviscid case of
$(B)_{0,0}$ is an outstanding open problem in the mathematical fluid
mechanics. It is well-known that inviscid 2D Boussinesq system has
exactly same structure to the axisymmetric 3D Euler system {\em off}
the axis of symmetry(see e.g. \cite{maj2} for this observation).
This is why the inviscid 2D Boussinesq system can be considered as a
model equation of the 3D Euler system. The problem of the finite
time blow-up for the fully inviscid Boussinesq system is an
outstanding open problem. The BKM type of blow-up criterion,
however, can be obtained without difficulty(see \cite{cha22, cha23,
e, tan} for various forms of blow-up criteria for the Boussinesq
system)). We first consider the partially viscous cases, i.e. either
the zero diffusivity case, $\kappa =0$ and $\nu>0$, or the zero
viscosity case, $\kappa >0$ and $\nu=0$. Even the regularity
problem for partial viscosity cases has been open recently.
Actually, in an
article appeared in 2001, M. K. Moffatt
raised a question of finite time singularity in the case $\kappa =0,
\nu>0$ and its possible development in the limit $\kappa \to 0$ as
one of the 21th century problems(see the Problem no. 3 in
\cite{mof}). For this problem Cordoba, Fefferman and De La
LLave(\cite{cor7}) proved that special type of singularities, called
`squirt singularities', is absent. In \cite{cha11} the author
considered the both of two partial viscosity cases, and prove the
global in time regularity for both of the cases. Furthermore it is
proved that as diffusivity(viscosity) goes to zero the solutions of
$(B)_{\nu. \kappa}$ converge strongly to those of zero
diffusivity(viscosity) equations\cite{cha11}. In particular the
Problem no. 3 in \cite{mof} is solved. More precise statements of
these results are stated in Theorem 1.1 and Theorem 1.2 below.
\begin{theorem}
Let $\nu >0$ be fixed, and div $v_0=0$.
Let $m >2$ be an integer, and
$(v_0 , \theta_0 )\in H^m( \Bbb R^2
)$. Then, there
exists unique solution $(v, \th )$ with $\theta \in C([0, \infty );H^m (\Bbb R^2
))$ and $ v\in C([0, \infty );H^m (\Bbb R^2 ))\cap
L^2(0,T;H^{m+1}(\Bbb R^2 ))$ of the system $(B)_{\nu ,0}$.
Moreover, for
each $s<m$, the solutions $(v, \theta )$ of $(B)_{\nu ,\kappa}$
converge to the corresponding solutions of $(B)_{\nu ,0}$ in $C
([0,T];H^{s} (\Bbb R^2 ))$ as $\kappa \to 0$.
\end{theorem}
We note that Hou and Li also obtained the existence part of the
above theorem independently in \cite{hou1}. The following theorem is
concerned with zero viscosity problem with fixed positive
diffusivity.
\begin{theorem}
Let $\kappa >0$ be fixed, and div $v_0=0$.
Let $m >2$ be an integer. Let $m >2$ be an integer, and
$(v_0 , \theta_0 )\in H^m( \Bbb R^2
)$. Then, there
exists unique solutions $(v, \th)$ with $v\in C([0, \infty );H^m
(\Bbb R^2 ))$ and $ \theta\in C([0, \infty );H^m (\Bbb R^2 ))\cap
L^2(0,T;H^{m+1}(\Bbb R^2 ))$ of the system $(B)_{0 ,\kappa}$.
Moreover, for
each $s<m$, the solutions $(v, \theta )$ of $(B)_{\nu,\kappa}$
converge to the corresponding solutions of $(B)_{0,\kappa}$ in $C
([0,T];H^{s} (\Bbb R^2 ))$ as $\nu \to 0$.
\end{theorem}
The proof of the above two theorems in \cite{cha11} crucially uses
the Brezis-Wainger inequality in \cite{bre, eng}. Below we consider
the fully inviscid Boussinesq system, and show that there is no
self-similar singularities under milder decay condition near
infinity than the case of 3D Euler system. The inviscid Boussinesq
system (B)$=(B)_{0 ,0}$ has scaling property that
if $(v, \t , p)$ is a
solution of the system (B), then for any $\lambda >0$ and $\alpha
\in \Bbb R$ the functions
\begin{equation}
\label{self2a}
v^{\l, \a }(x,t)=\lambda ^\alpha v(\lambda x, \l^{\a +1}
t),\quad
\t ^{\l, \a } (x,t)= \l ^{2\a +1} \t (\lambda x, \l^{\a +1}
t),
\end{equation}
\begin{equation}
p^{\l, \a }(x,t)= \lambda ^{2\alpha}p (\lambda x,
\l^{\a +1}
t)
\end{equation}
are also solutions of (B) with the initial data
$$ v_0^{\l,\a}(x)=\lambda ^\alpha v_0
(\lambda x),\quad \t_0^{\l,\a}(x)=
\l ^{2\a +1} \t_0(\lambda x ).
$$
In view of the scaling
properties in (\ref{self2a}), the
self-similar blowing-up
solution $(v(x, t), \t (x,t))$ of (B) should of the form,
\begin{eqnarray}
\label{self2b1}
v(x, t)&=&\frac{1}{(T_*-t)^{\frac{\a}{\a+1}}}
V\left(\frac{x}{(T_*-t)^{\frac{1}{\a+1}}}\right),\\
\label{self2b2}
\t (x,t)&=&\frac{1}{(T_*-t)^{2\a+1}}
\Theta \left(\frac{x}{(T_*-t)^{\frac{1}{\a+1}}}\right),
\end{eqnarray}
where $\alpha
\neq -1 $. We have the following nonexistence result of such type of
solution(see \cite{cha13}).
\begin{theorem}
Let $v $ generates a particle trajectory, which is a $C^1$
diffeomorphism from $\Bbb R^2$ onto itself for all $t\in (0, T_*)$.
There exists no nontrivial solution $(v, \t )$ of the system $(B)$
of the form (\ref{self2b1})-(\ref{self2b2}), if there exists $p_1,
p_2 \in (0, \infty]$, $p_1 < p_2$, such that $\Theta\in L^{p_1}
(\Bbb R^2 ) \cap L^{p_2} (\Bbb R^2 )$, and $V\in H^m (\Bbb R^2 )$,
$m>2$.
\end{theorem}
Recalling the fact that the system $(B)$ has the similar form as
the axisymmetric 3D Euler system, we can also deduce the
nonexistence of self-similar blowing up solution to the axisymmetric
3D Euler equations of the fome (\ref{vel}), if $\Theta = rV^\theta$
satisfies the condition of Theorem 4.7, and curl $V=\Omega \in H^m
(\Bbb R^3 )$, $m>5/2$, where $r=\sqrt{x_1^2 +x_2 ^2}$, and
$V^\theta$ is the angular component of $V$. Note that in this case
we do not need to assume strong decay of $\Omega$ as in Theorem 3.3.
See \cite{cha13} for more details.
\subsection{Deformations of the Euler equations}
Let us consider the following system considered in \cite{cha16}.
\[
\mathrm{ (P_1)}
\left\{ \aligned
\frac{\partial u}{\partial t} +(u\cdot \nabla )u &=-\nabla q
+(1+\varepsilon )\|\nabla
u (t)\|_{L^\infty}u,
\\
\textrm{div }\, u =0 , &\\
u(x,0)=u_0 (x), &
\endaligned
\right.
\]
where $u=(u_1, \cdots, u_n )$, $u_j =u_j (x, t)$, $j=1, \cdots , n,
$, is the unknown vector field $q=q(x,t)$ is the scalar, and $u_0 $
is the given initial vector field satisfying div $u_0 =0$. The
constant $\varepsilon >0$ is fixed. Below denote curl
$u=\o$ for `vorticity' associated the `velocity' $u$.
We first note that the system of $(P_1)$ has the similar nonlocal structure
to the Euler system (E), which is implicit in the pressure term combined
with the divergence free condition.
Moreover it has the same scaling properties
as the original Euler system in $(E)$. Namely, if $u(x,t)$,
$q(x,t)$ is a pair of solutions to $(P_1)$ with initial data $u_0 (x)$,
then for any $\alpha \in \Bbb R$
$$ u^\lambda (x,t) =\lambda ^\alpha u (\lambda x, \lambda
^{\alpha +1} t ), \quad q^\lambda (x,t) =
\lambda ^{2\alpha} q (\lambda x, \lambda
^{\alpha +1} t )
$$
is also a pair of solutions to $(P_1)$ with initial data $u_0 ^\lambda
(x)=\lambda ^\alpha u_0 (x)$. As will be seen below, we can have the local well-posedness in the
Sobolev space, $H^m (\Bbb R^n ), m>n/2+2$, as well as the BKM type
of blow-up criterion for $(P_1)$, similarly to the Euler system
(E). Furthermore, we can prove actual finite time blow-up for
smooth initial data if $\o_0 \neq 0$. This is rather surprising in
the viewpoint that people working on the Euler system often have
speculation that the divergence free condition might have the role
of `desingularization', and might make the singularity disappear.
Obviously this is not the case for the system $(P_1)$.
Furthermore, there is a canonical
functional relation between the solution of $(P_1)$ and that of
the Euler system (E); hence the word `deformation'. Using this relation we can translate the
blow-up condition of the Euler system in terms of the solution of
$(P_1)$. The precise contents of the above results on $(P_1)$ are stated
in the
following theorem.
\begin{theorem}
Given $u_0 \in H^m (\Bbb R^n )$ with div $u_0 =0$, where $m>
\frac{n}{2}+2$, the following statements hold true for $(P_1)$.
\begin{itemize}
\item[(i)] There exists a local in time unique solution $u(t)\in
C([0,T]:H^m (\Bbb R^n ))$ with $T=T (\|u_0 \|_{H^m })$.
\item[(ii)]
The solution $u(x,t)$ blows-up at $t=t_*$, namely
$$
\lim\sup_{t\to t_*}\|u(t)\|_{H^m} =\infty\quad \mbox{ if and
only if}\quad \int_0 ^{t_*} \|\o (t)\|_{L^\infty} dt =\infty,
$$
where
$\o=$ curl $u$. Moreover, if the solution $u(x,t)$ blows up at
$t_*$, then necessarily,
$$
\int_0 ^{t_*} \exp\left[ (2+\varepsilon)\int_0 ^\tau \|\nabla
u (s)\|_{L^\infty}ds\right] d\tau =\infty
$$
for $n=3$, while
$$
\int_0 ^{t_*} \exp\left[ (1+\varepsilon)\int_0 ^\tau \|\nabla
u (s)\|_{L^\infty}ds\right] d\tau=\infty
$$
for $n=2$.
\item[(iii)] If $\|\o_0
\|_{L^\infty }\neq 0$, then there exists time $ t_* \leq
\frac{1}{\varepsilon \|\o_0 \|_{L^\infty }}$ such that solution $u(x,t)$
of $(P_1)$ actually blows up at $t_*$. Moreover, at such $t_*$ we
have
$$
\int_0
^{t_*} \exp\left[ (1+\varepsilon)\int_0 ^\tau \|\nabla
u (s)\|_{L^\infty}ds\right] d\tau=\infty .
$$
\item[(iv)] The functional
relation between the solution $u(x,t)$ of $(P_1)$ and the solution
$v(x,t)$ of the Euler system (E) is given by
$$
u(x,t)=\varphi'(t) v(x, \varphi (t)),
$$
where
$$
\varphi (t)=\lambda\int_0 ^t\exp\left[ (1+\varepsilon )\int_0 ^\tau \|\nabla
u(s)\|_{L^\infty}ds\right]d\tau.
$$
(The relation between the two initial datum is
$u_0 (x)=\lambda v_0 (x)$.)
\item[(v)]
The solution $v(x,t)$ of the Euler system (E) blows up at
$T_* <\infty$ if and only if for $t_*:=\varphi^{-1} (T_*) < \frac{1}{\varepsilon \|\o_0 \|_{L^\infty}}$
both of the followings hold true.
$$
\int_0 ^{t_*}\exp\left[ (1+\varepsilon )\int_0 ^\tau \|\nabla
u(s)\|_{L^\infty}ds\right]d\tau <\infty,
$$
and
$$
\int_0 ^{t_*}\exp\left[ (2+\varepsilon )\int_0 ^\tau \|\nabla
u(s)\|_{L^\infty}ds\right]d\tau =\infty.
$$
\end{itemize}
\end{theorem}
For the proof we refer \cite{cha16}. In the above theorem the result
(ii) combined with (v) shows indirectly that there is no finite time
blow-up in 2D Euler equations, consistent with the well-known
result. Following the argument on p. 542 of \cite{cha4}, the
following fact can be
verified without difficulty:\\
We set
\begin{equation}\label{aoft}
a(t)=\exp \left( \int_0 ^t (1+\varepsilon) \|\nabla
u(s)\|_{L^\infty} ds \right).
\end{equation}
Then, the solution $(u, q)$ of
$(P_1)$ is given by
$$ u(x,t)=a(t) U(x,t), \quad q(x,t)=a(t)P(x,t), $$
where $(U, P)$ is a solution of the following system,
$$
(aE)\left\{ \aligned
&\frac{\partial U}{\partial t} +a(t) (U\cdot \nabla )U =-\nabla P,\\
&\qquad \mathrm{div}\, U=0,\\
&U(x,0)=U_0 (x)
\endaligned
\right.
$$
The system $(aE)$ was studied in \cite{cha4}, when $a(t)$ is a
prescribed function of $t$, in which case the proof of local
existence of $(aE)$ in \cite{cha4} is exactly same as the case of
$(E)$. In the current case, however, we need an extra proof of
local existence, as is done in the next section, since the function
$a(t)$ defined by (\ref{aoft}) depends on the solution $u(x,t)$
itself. As an application of Theorem 4.8 we can prove the following
lower estimate of the possible blow-up time(see \cite{cha16} for the
detailed proof).
\begin{theorem}
Let $p\in (1, \infty)$ be fixed.
Let $v(t)$ be the local classical solution of the 3D Euler
equations with initial data $v_0\in H^m (\Bbb R^3 )$, $m>7/2$. If
$T_*$ is the first blow-up time, then
\begin{equation}\label{loweres}
T_* -t\geq \frac{1}{ C_0\|\o (t) \|_{\dot{B}^{\frac{3}{p}}_{p,1}}},
\quad \forall t\in (0, T_*)
\end{equation}
where $C_0$ is the absolute constant in $(Q_2)$.
\end{theorem}
In \cite{cha4} the following form of lower estimate for the blow-up
rate is derived.
\begin{equation}\label{loweres1}
T_*-t \geq \frac{1}{ \tilde{C}_0\|\o (t)\|_{\dot{B}^{0}_{\infty,1}}},
\end{equation}
where $\tilde{C}_0$ is another absolute constant(see also the remarks after Theorem 3.6).
Although there
is (continuous) embedding relation,
$\dot{B}^{\frac{3}{p}}_{p,1}(\Bbb R^3) \hookrightarrow
\dot{B}^{0}_{\infty,1}(\Bbb R^3)$ for $p\in [1, \infty]$(see Section
1), it is difficult to compare the two estimates (\ref{loweres}) and
(\ref{loweres1}) and decide which one is sharper, since the precise
evaluation of the optimal constants
$C_0, \tilde{C}_0$ in those inequalities could be very difficult problem.\\
\ \\
Next, given $\varepsilon \geq 0$, we consider the following problem.
\[
\mathrm{ (P_2)}
\left\{ \aligned
\frac{\partial u}{\partial t} +(u\cdot \nabla )u &=-\nabla q
-(1+\varepsilon )\|\nabla
u (t)\|_{L^\infty}u ,
\\
\textrm{div }\, u =0 , &\\
u(x,0)=u_0 (x), &
\endaligned
\right.
\]
Although the system of $(P_2)$ has also the same nonlocal structure and the scaling
properties as the Euler system and $(P_1)$, we have the result
of the global regularity stated in the following theorem(see \cite{cha16} for the proof).
\begin{theorem}
Given $u_0 \in H^m (\Bbb R^n)$ with div $u_0 =0$, where $m>
\frac{n}{2}+2$, then the solution $u(x,t)$ of $(P_2)$ belongs to
$C([0, \infty ):H^m (\Bbb R^n ))$. Moreover, we have the following
decay estimate for the vorticity,
$$
\| \o (t)\|_{L^\infty }
\leq \frac{\|\o _0 \|_{L^\infty}}{1+\varepsilon\|\o_0 \|_{L^\infty} t} \qquad \forall
t\in [0, \infty ).
$$
\end{theorem}
We also note that solution of the system $(P_2)$ has also similar
functional relation with that of the Euler system as given in (iv)
of Theorem 5.8 as will be clear in the proof of Theorem 1.1 in the
next section.\\
\ \\
Next, given $\varepsilon >0 $, we consider the following perturbed systems
of (E).
\[
(E)^\varepsilon _\pm\left\{ \aligned
&\frac{\partial u}{\partial t} +(u\cdot \nabla )u =-\nabla q
\pm \varepsilon \|\nabla u\|_{L^\infty}^{1+\varepsilon}u,
\\
&\textrm{div}\, u =0 , \\
& u(x,0)=u_0 (x).
\endaligned
\right.
\]
If we set $\varepsilon=0$ in the above, then the system $(E)^0 _\pm$
becomes $(E)$. For $\varepsilon >0$ we have finite time blow-up for the
system $(E)^\varepsilon_+$ with certain initial data, while we have the
global regularity for $(E)^\varepsilon_-$ with all solenoidal initial data
in $H^m (\Bbb R^3 )$, $m>5/2$. More precisely we have the following
theorem(see \cite{cha17} for the proof).
\begin{theorem}
\begin{itemize}
\item[(i)] Given $\varepsilon>0$, suppose
$u_0 =u_0^\varepsilon\in H^m (\Bbb R^3 )$ with div $u_0=0$ satisfies
$\|\o_0 \|_{L^\infty} > (2/\varepsilon )^{1/\varepsilon}$, then
there exists $T_* $ such that the
solution $u(x,t)$ to $(E)_\varepsilon ^+$ blows up at $T_*$, namely
$$
\lim\sup _{t\nearrow T_*} \|u(t)\|_{H^m} =\infty.
$$
\item[(ii)] Given $\varepsilon >0$ and $u_0 \in H^m (\Bbb R^3)$ with div $u_0=0$, there exists
unique global in time classical solution $u(t)\in
C([0, \infty ); H^m (\Bbb R^3 ))$ to
$(E)^\varepsilon _-$. Moreover, we have the global in time vorticity
estimate for the solution of $(E)^\varepsilon _-$,
$$
\|\o(t)\|_{L^\infty} \leq
\max\left\{ \|\o_0\|_{L^\infty}, \left(\frac{1}{\varepsilon}\right)^{\frac{1}{\varepsilon}}\right\}
\qquad \forall t\geq 0.
$$
\end{itemize}
\end{theorem}
The following theorem relates the finite time blow-up/global
regularity of the Euler system with those of the system
$(E)^\varepsilon_\pm$.
\begin{theorem}
Given $\varepsilon >0$, let $u^\varepsilon_\pm$ denote the solutions of
$(E)^\varepsilon _\pm$ respectively with the same initial data $u_0 \in
H^m (\Bbb R^3 )$, $m>5/2$. We define
$$
\varphi ^\varepsilon_\pm (t, u_0):=\int_0 ^t \exp \left[ \pm \varepsilon \int_0
^\tau \|\nabla u^\varepsilon_\pm (s)\|_{L^\infty}^{1+\varepsilon} ds\right]
d\tau.
$$
\begin{itemize}
\item[(i)] If $\varphi ^\varepsilon_- (\infty , u_0 ) =\infty$, then the
solution of the Euler system with initial data $u_0$ is regular
globally in time.
\item[(ii)] Let $t_*$
be the first blow-up time for a solution $u^\varepsilon_+$ of
$(E)^\varepsilon_+$ with initial data $u_0$ such that
$$\int_0 ^{t_*} \|\o ^\varepsilon _+(t)\|_{L^\infty} dt =\infty, \qquad \mbox{where}\quad
\o ^\varepsilon _+ =\textrm{curl}\, \,\, u^\varepsilon_+ .$$
If
$\varphi ^\varepsilon_+ (t_* , u_0 )<\infty$, then the solution of the
Euler system blows up at the finite time $T_*=\varphi ^\varepsilon_+ (t_*
, u_0 )$.
\end{itemize}
\end{theorem}
We refer \cite{cha17} for the proof of the above theorem.
\section{Dichotomy: singularity or global regular dynamics?}
\setcounter{equation}{0}
In this section we review results in \cite{cha12}. Below $S$, $P$
and $\xi (x,t)$ are the deformation tensor , the Hessian of the
pressure and the vorticity direction field, associated with the
flow, $v$, respectively as introduced in section 1.
Let $\{ (\lambda _k , \eta^k)\}_{k=1}^3$ be
the eigenvalue and the normalized eigenvectors of $S$.
We set $\lambda =(\lambda_1, \lambda_2 ,\lambda_3 )$, and
$$|\lambda|= \left(\sum_{k=1}^3 \lambda_k ^2 \right)^{\frac12},\quad
\rho_k =\eta^k \cdot P \eta^k \quad\mbox{ for $k=1,2,3$}.
$$
We also denote
$$\eta^k (x,0)=\eta^k_0 (x),\quad
\lambda_k (x,0)=\lambda_{k,0}(x),\quad \lambda (x,0)=\lambda_0 (x),
\quad \rho_k (x,0)=\rho_{k,0}(x)
$$
for the quantities at $t=0$.
Let $\o (x,t)\neq 0$. At such point $(x,t)$ we define the scalar fields
$$
\a = \xi\cdot S\xi,\quad
\rho = \xi \cdot P\xi.
$$
At
the points where $\o (x,t)=0$ we define $\a(x,t)=\rho(x,t)=0$. We
denote $\a_0 (x)=\a (x,0)$, $\rho_0 (x)=\rho (x,0 )$.
Below we denote $ f(X(a,t),t)'=\frac{Df}{Dt} (X(a,t),t)$
for simplicity.
Now, suppose that there is no blow-up of the solution on $[0, T_*]$,
and
the inequality
\begin{equation}\label{contra1}
\a(X(a,t),t) |\o (X(a,t),t)| \geq \varepsilon |\o (X(a,t),t)|^2
\end{equation}
persists on $[0,T_*]$. We will see that this leads to a
contradiction. Combining (\ref{contra1}) with (\ref{basfor3}), we have
$$
|\o |' \geq \varepsilon |\o|^2.
$$
Hence, by Gronwall's lemma, we obtain
$$
|\o (X(a,t),t)|\geq \frac{|\o_0 (a)|}{1-\varepsilon |\o_0 (a)| t},
$$
which implies that
$$ \lim\sup_{t\nearrow T_*}|\o (X(a,t),t)|=\infty.$$
Thus we are lead to the following lemma.
\begin{lemma}
Suppose $\a_0 (a) > 0$, and there exists $\varepsilon>0$ such that
\begin{equation}\label{210}
\a _0 (a) |\o_0 (a)|\geq \varepsilon |\o_0 (a)|^2.
\end{equation}
Let us set
\begin{equation}
T_* =\frac{1}{\varepsilon \a_0 (a)}.
\end{equation}
Then, either the vorticity blows up no later than $T_*$, or there
exists $t \in (0, T_*)$ such that
\begin{equation}\label{th11} \a(X(a,t),t) |\o (X(a,t),t)| < \varepsilon |\o (X(a,t),t)|^2.
\end{equation}
\end{lemma}
From this lemma we can derive the following:
\begin{theorem}[vortex dynamics]
Let $v_0 \in H^m (\Omega )$, $m>5/2$, be given. We define
$$\Phi_1(a,t)= \frac{\a (X(a,t),t)}{|\o (X(a,t),t)|}$$
and
$$\Sigma_1 (t)=\{ a\in \Omega \,|\, \a (X(a,t),t)> 0\}$$
associated with the classical solution $v(x,t)$.
Suppose $a\in
\Sigma_1 (0)$ and $\o_0 (a) \neq 0$. Then one of the following holds
true.
\begin{itemize}
\item[(i)] {\rm{(finite time singularity)}}
The solution of the Euler equations blows-up in finite time along
the trajectory $\{ X(a,t)\}$.
\item[(ii)] {\rm{(regular dynamics)}} On of the following holds
true:
\begin{description}
\item[(a)] {\rm{(finite time extinction of $\a$)}}
There exists $t_1 \in (0, \infty)$ such that $\a (X(a, t_1),t_1)=0$.
\item[(b)] {\rm{(long time behavior of $\Phi_1$)}} There exists an infinite sequence $\{
t_j\}_{j=1}^\infty$ with $t_1<t_2<\cdots <t_j <t_{j+1} \to \infty$
as $j\to \infty$ such that for all $j=1,2,\cdots$ we have $\Phi_1
(a,0)> \Phi_1 (a,t_1)>\cdots>\Phi_1 (a, t_j)>\Phi_1 (a,t_{j+1})>0$
and $\Phi_1 (a,t)\geq \Phi _1 (a,t_j ) >0$ for all $t\in [0,t_j]$.
\end{description}
\end{itemize}
\end{theorem}
As an illustration of proofs for the Theorem 5.2 and 5.3 below, we
give outline of the proof of the above theorem.
Let us first observe that the formula
$$ |\o (X(a,t),t)|=\exp\left[ \int_0 ^t \a (X(a,s),s) ds\right]
|\o_0 (a)|,
$$
which is obtained from (\ref{basfor3}) immediately shows that $\o (X(a,t),t)\neq
0$ if and only if $\o_0 (a)\neq 0$ for the particle trajectory $\{ X(a,t)\}$ of the classical
solution $v(x,t)$ of the Euler equations. Choosing
$\varepsilon = \a _0 (a)/|\o_0 (a)|$ in Lemma 4.1, we see that
either the vorticity blows up no later than $T_* =1/\a_0 (a)$,
or there exists $t_1\in (0, T_* )$ such that
$$
\Phi_1 (a,t_1)=\frac{\a (X(a,t_1),t_1)}{|\o (X(a,t_1),t_1 )|} < \frac{\a_0 (a)}{|\o_0
(a)|}=\Phi_1 (a,0).
$$
Under the hypothesis that (i) and (ii)-(a) do not hold true, we may
assume $a\in \Sigma_1 (t_1)$ and repeat the above argument
to find $t_2 >t_1$ such that $\Phi_1(a, t_2 )<\Phi_1 (a, t_1 )$,
and also $a \in \Sigma_1 (t_2)$. Iterating the argument, we find
a monotone increasing sequence $\{ t_j\}_{j=1}^\infty$ such that $\Phi _1 (a,t_{j})>\Phi
_1 (a,t_{j+1})$ for all $j=1,2,3,\cdots$. In particular we can
choose each $t_j$ so that $\Phi_1 (a, t)\geq \Phi_1 (a, t_j )$
for all $t\in (t_{j-1}, t_j]$. If $t_j \to t_\infty <\infty$ as $j\to \infty$,
then we can proceed further to have $t_* > t_\infty$ such that
$\Phi_1 (a, t_\infty )>\Phi_1 (a, t_* )$. Hence, we may set
$t_\infty =\infty$, which finishes the proof.\\
The above argument can be extended to prove the following theorems.
\begin{theorem}[dynamics of $\a$]
Let $v_0 \in H^m (\Omega )$, $m>5/2$, be given. In case
$\a(X(a,t),t)\neq 0$ we define
$$\Phi_2(a,t)= \frac{|\xi\times S\xi
|^2(X(a,t),t)-\rho (X(a,t),t)}{\a
^2(X(a,t),t)},
$$
and
$$\Sigma_2 ^+(t)=\{ a\in \Omega \,|\, \a (X(a,t),t)> 0, \Phi _2
(X(a,t),t)>1\},$$
$$\Sigma_2 ^-(t)=\{ a\in \Omega \,|\, \a (X(a,t),t)<0, \Phi _2
(X(a,t),t)<1\},$$
associated with $v(x,t)$.
Suppose $a\in \Sigma_2 ^+(0)\cup \Sigma_2 ^-(0)$.
Then one of the following
holds true.
\begin{itemize}
\item[(i)] {\rm{(finite time singularity)}}
The solution of the Euler equations blows-up in finite time along
the trajectory $\{ X(a,t)\}$.
\item[(ii)] {\rm{(regular dynamics)}} One of the following holds
true:
\begin{description}
\item[(a)] {\rm{(finite time extinction of $\a$)}} There exists $t_1 \in (0, \infty)$ such that $\a (X(a, t_1),t_1)=0$.
\item[(b)] {\rm{(long time behaviors of $\Phi_2$)}} Either there exists $T_1\in (0, \infty)$ such that
$ \Phi_2 (a,T_1) =1$, or there exists an infinite sequence $\{
t_j\}_{j=1}^\infty$ with $t_1<t_2<\cdots <t_j <t_{j+1} \to \infty$
as $j\to \infty$ such that one of the followings hold:
\begin{description}
\item[(b.1)] In the case $a\in \Sigma_2 ^+(0)$, for
all $j=1,2,\cdots$ we have $\Phi_2 (a,0)> \Phi_2
(a,t_1)>\cdots>\Phi_2 (a, t_j)>\Phi_2 (a,t_{j+1})
>1$ and $\Phi_2 (a,t)\geq \Phi_2
(a,t_j )>1 $ for all $t\in [0,t_j]$.
\item[(b.2)] In the case $a\in \Sigma_2 ^-(0)$, for all $j=1,2,\cdots$ we
have $\Phi_2 (a,0)<\Phi_2 (a,t_1)<\cdots <\Phi_2 (a, t_j)< \Phi_2
(a,t_{j+1}) <1$ and $\Phi_2 (a,t)\leq \Phi_2 (a,t_j )<1 $ for all
$t\in [0,t_j]$.
\end{description}
\end{description}
\end{itemize}
\end{theorem}
\begin{theorem}[spectral dynamics]
Let $v_0 \in H^m (\Omega )$, $m>5/2$, be given. In case
$\lambda(X(a,t),t)\neq 0$ we define
$$
\Phi_3(a,t)= \frac{\sum_{k=1}^3\left[ -\lambda_k ^3 +\frac14 |\eta_k
\times \o |^2\lambda_k - \rho_k \lambda_k \right](X(a,t),t)}{
|\lambda (X(a,t),t)|^3},
$$
and
$$\Sigma_3 (t)=\{ a\in \Omega \,|\, \lambda (X(a,t),t)\neq 0, \Phi _3
(X(a,t),t)>0\}$$
associated with $v(x,t)$.
Suppose $a\in \Sigma_3 (0)$.
Then one of the following
holds true:
\begin{itemize}
\item[(i)] {\rm{(finite time singularity)}}
The solution of the Euler equations blows-up in finite time along
the trajectory $\{ X(a,t)\}$.
\item[(ii)]{\rm{(regular dynamics)}} One of the followings hold
true:
\begin{description}
\item[(a)] {\rm{(finite time extinction of $\lambda$)}}
There exists $t_1 \in (0, \infty)$ such that $\lambda (X(a, t_1),t_1)=0$.
\item[(b)]{\rm{(long time behavior of $\Phi_3 $)}} Either there exists $T_1\in (0, \infty)$ such that
$ \Phi_3 (a,T_1) =0$, or there exists an infinite sequence $\{
t_j\}_{j=1}^\infty$ with $t_1<t_2<\cdots <t_j <t_{j+1} \to \infty$
as $j\to \infty$ such that for all $j=1,2,\cdots$ we have $\Phi_2
(a,0)> \Phi_3 (a,t_1)>\cdots>\Phi_3 (a, t_j)>\Phi_3(a,t_{j+1})
>0$ and $\Phi_3 (a,t)\geq \Phi_3
(a,t_j )>0 $ for all $t\in [0,t_j]$.
\end{description}
\end{itemize}
\end{theorem}
For the details of the proof of Theorem 5.2 and Theorem 5.3 we refer \cite{cha12}.
Here we present a refinement of Theorem 2.1 of \cite{cha15}, which is proved in
\cite{cha12}.
\begin{theorem} Let $v_0 \in H^m (\Omega )$, $m>5/2$, be given.
For such $v_0$ let us define a set $\Sigma \subset \Omega$ by
\begin{eqnarray*}
\lefteqn{ \Sigma=\{ a\in
\Omega \,|\, \a_0 (a) > 0, \o_0 (a) \neq 0, \exists \,\,\varepsilon \in
(0, 1) \,\, \mbox{such
that}}\hspace{.3in}\\
&& \rho_0(a) + 2\a ^2_0 (a)-|\xi_0 \times S_0\xi_0 |^2(a)\leq
(1-\varepsilon)^2 \a^2 _0 (a)\}.
\end{eqnarray*}
Let us set
\begin{equation}
T_* =\frac{1}{\varepsilon \a_0 (a)}.
\end{equation}
Then, either the solution blows up no later than $T_*$, or there
exists $t\in (0, T_*)$ such that
\begin{equation}
\rho(X(a,t),t) + 2\a ^2(X(a,t),t) -|\xi \times S\xi |^2(X(a,t),t)
> (1-\varepsilon)^2 \a ^2(X(a,t),t).
\end{equation}
\end{theorem}
We note that if we ignore the term $|\xi_0 \times S_0\xi_0
|^2(a)$, then we have the condition,
$$ \rho_0(a) + \a ^2_0 (a)\leq (-2\varepsilon +\varepsilon^2) \a^2 _0 (a) < 0,
$$
since $\varepsilon \in (0, 1)$.
Thus $\Sigma \subset \mathcal{S}$, where $\mathcal{S}$ is the
set defined in Theorem 2.1 of \cite{cha15}. One can verify without
difficulty that $\Sigma =\emptyset$ for the 2D Euler flows.
Regarding the question if $\Sigma \neq\emptyset$ or not for 3D Euler flows, we
have the following proposition(see \cite{cha15} for more details).
\begin{pro}
Let us consider the system the domain $\Omega =[0,2\pi ]^3$ with
the periodic boundary condition. In $\Omega$ we consider the
Taylor-Green vortex flow defined by
\begin{equation}\label{ta}
u(x_1,x_2,x_3 )=(\sin x_1 \cos x_2 \cos x_3 ,
-\cos x_1 \sin x_2 \cos x_3 , 0).
\end{equation}
Then, the set
$$\mathcal{S}_0 =\left\{(0,
\frac{\pi}{4} , \frac{7\pi}{4} ),(0, \frac{7\pi}{4} , \frac{\pi}{4}
)\right\} $$
is included in $\Sigma$ of Theorem 4.4. Moreover, for $x\in
\mathcal{S}_0$ we have the explicit values of $\alpha$ and $\rho$,
$$
\alpha(x)=\frac12,\quad
\rho(x)=-\frac12.
$$
\end{pro}
We recall that the Taylor-Green vortex has been the first candidate
proposed for a finite time singularity for the 3D Euler equations,
and there have been many numerical calculations of solution of (E)
with the initial data given by it(see e.g. \cite{bra}).
\section{Spectral dynamics approach}
\setcounter{equation}{0}
Spectral dynamics approach in the fluid mechanics was initiated by
Liu and Tadmor(\cite{liu}). They analyzed the restricted Euler
system (\ref{liueq}) in terms of (poitwise) dynamics of the
eigenvalues of the velocity gradient matrix $V$. More specifically,
multiplying
left and right eigenvectors of $V$ to (\ref{liueq}), they derived
$$
\frac{D \lambda_j}{Dt}=-\lambda_j^2
+\frac{1}{n} \sum_{k=1}^n \lambda_k ^2 ,\quad j=1,2,\cdots, n,
$$
where $\lambda_j$, $j=1,2, ..,n$ are eigenvalues $V$, which are not necessarily
real values.
In this model system they proved finite time blow-up for suitable
initial data. In this section we review the results in
\cite{cha8}, where the full Euler system is concerned.
Moreover, the we are working on the dynamics of eigenvalues of the
deformation tensor $S$(hence real valued), not the velocity gradient
matrix. We note that there were also application of the spectral
dynamics of the deformation tensor in the study of regularity
problem of the Navier-Stokes equations by Neustupa and
Penel(\cite{neu}).
In this section for
simplicity we consider the 3D Euler system (E) in the periodic
domain, $\Omega =\Bbb T^3(=\Bbb R^3/\Bbb Z^3)$.
Below we denote ${ \lambda_1 ,\lambda_2 , \lambda_3 }$ for the eigenvalues of
the deformation tensor
$S=( S_{ij})$ for the velocity fields of the 3D Euler system.
We will first establish the following formula,
\begin{equation}\label{spec1}
\frac{d}{dt} \int_{\Bbb T^3} (\lambda_1^2 +\lambda_2^2 +\lambda_3^2 )dx=
-4\int_{\Bbb T^3} \lambda_1\lambda_2 \lambda_3 \, dx,
\end{equation}
which has important implications(Theorem 6.1- Theorem 6.3 below).
Indeed, using (\ref{basfor2}), we can compute
\begin{eqnarray*}
\lefteqn{\frac12\frac{d}{dt}\int_{\Bbb T^3} S_{ij}S_{ij}dx=\int_{\Bbb T^3} S_{ij}
\frac{D S_{ij}}{Dt}dx}\hspace{.0in} \\
&&=-\int_{\Bbb T^3} S_{ik}S_{kj}S_{ij} dx -\frac14 \int_{\Bbb T^3} \o_i S_{ij}\o_j
+\frac14\int_{\Bbb T^3} |\o |^2 S_{ii} dx +\int_{\Bbb T^3} P_{ij} S_{ij} dx\\
&&=-\int_{\Bbb T^3} S_{ik}S_{kj}S_{ij} dx-\frac18\frac{d}{dt}\int_{\Bbb T^3} |\o|^2dx,
\end{eqnarray*}
where we used the summation
convention for the repeated indices, and used the
$L^2$-version of the vorticity equation,
\begin{equation}\label{spec2}
\frac12\frac{d}{dt}\int_{\Bbb T^3} |\o
|^2dx=\int_{\Bbb T^3} \o_i S_{ij}\o_j ,
dx
\end{equation}
which is immediate from (\ref{basfor2a}).
We note
\begin{eqnarray*}
\lefteqn{{ \int_{\Bbb T^3} |\o |^2 dx} = \int_{\Bbb T^3} |\nabla v|^2 dx=\int_{\Bbb T^3} V_{ij}
V_{ij}dx= \int_{\Bbb T^3} (S_{ij}+A_{ij})(S_{ij}+A_{ij})dx}\hspace{.3in}\\
&&=\int_{\Bbb T^3} S_{ij}S_{ij} dx +\int_{\Bbb T^3} A_{ij}A_{ij}dx
=\int_{\Bbb T^3} S_{ij}S_{ij} dx+\frac12 { \int_{\Bbb T^3} |\o|^2dx}.
\end{eqnarray*}
Hence,
$$\int_{\Bbb R^n} S_{ij}S_{ij} dx =\frac12 \int_{\Bbb T^3} |\o |^2 dx$$
Substituting this into (\ref{spec2}), we obtain that
$$
\frac{d}{dt}\int_{\Bbb T^3} S_{ij}S_{ij}dx=-\frac43\int_{\Bbb T^3} S_{ik}S_{kj}S_{ij}
dx,
$$
which, in terms of the spectrum of $S$, can be written as
\begin{equation}\label{spec3}
\frac{d}{dt} \int_{\Bbb T^3} (\lambda_1^2 +\lambda_2^2 +\lambda_3^2 )dx=-\frac43\int_{\Bbb T^3} (\lambda_1^3
+ \lambda_2^3 + \lambda_3^3 ) dx.
\end{equation}
We observe from the divergence free condition, $0=$div $v=Tr
S=\lambda_1+\lambda_2+\lambda_3$,
\begin{eqnarray*}
0&=&(\lambda_1+\lambda_2 +\lambda_3 )^3\\
&=&\lambda_1^3 +\lambda_2^3 +\lambda_3^3 +3\lambda_1^2 (\lambda_2+\lambda_3
)+3\lambda_2^2 (\lambda_1 +\lambda_3 ) +3\lambda_3 (\lambda_1+\lambda_2 )+ 6\lambda_1\lambda_2\lambda_3\\
&=&\lambda_1^3 +\lambda_2^3 +\lambda_3^3 -3(\lambda_1^3 +\lambda_2^3 +\lambda_3^3)+ 6\lambda_1\lambda_2\lambda_3.
\end{eqnarray*}
Hence, $\lambda_1^3 + \lambda_2^3 + \lambda_3^3=3\lambda_1\lambda_2\lambda_3$. Substituting this into
(\ref{spec3}), we completes the proof of (\ref{spec1}).\\
\ \\
Using the formula (\ref{spec1}), we can first prove the following
new a priori estimate for the $L^2$ norm of vorticity for the 3D
incompressible Euler equations(see \cite{cha8} for the proof). We
denote
$$ \mathbb{H}^m _\sigma =\{ v\in [H^m (\Bbb T^3 )]^3\, |\, \mathrm{div
}\, v=0\}.$$
\begin{theorem}
Let $v(t)\in C([0,T); \mathbb{H}^m _\sigma )$, $m>5/2$ be the local
classical solution of the 3D Euler equations with initial data $v_0
\in \mathbb{H}^m_\sigma $ with $\o_0 \neq 0$. Let $\lambda_1 (x,t)\geq \lambda_2
(x,t)\geq \lambda_3 (x,t)$ are the eigenvalues of the deformation tensor
$S_{ij}(v)=\frac12 ( \frac{\partial v_j}{\partial x_i}
+\frac{\partial v_i}{\partial x_j})$. We denote $\lambda_2^+ (x,t)=\max\{
\lambda_2 (x,t), 0\}$, and $\lambda_2^- (x,t)=\min\{ \lambda_2 (x,t), 0\}$. Then, the
following estimates hold.
\begin{eqnarray*}
\lefteqn{ \exp\left[
\int_0 ^t \left(\frac12 \inf_{x\in \Bbb T^3}\lambda_2^+ (x,t)-\sup_{x\in \Bbb T^3}
|\lambda_2^-(x,t)|\right)dt\right]\leq
\frac{\|\o (t)\|_{L^2}}{\|\o_0\|_{L^2}}}\hspace{1.in}\nonumber \\
&&\leq \exp\left[
\int_0 ^t \left( \sup_{x\in \Bbb T^3}\lambda_2^+ (x,t)-\frac12 \inf_{x\in \Bbb T^3}
|\lambda_2^- (x,t)|\right)dt\right]\nonumber \\
\end{eqnarray*}
for all $t\in (0,T)$.
\end{theorem}
The above estimate says, for example, that if we have the following
compatibility conditions,
$$\sup_{x\in \Bbb T^3}\lambda_2^+ (x,t)\simeq \inf_{x\in \O}|\lambda_2^- (x,t)|\simeq g(t)$$
for some time interval $[0,T]$, then
$$ \|\o (t)\|_{L^2}
\lesssim O\left(\exp\left[C \int_0 ^t g(s)ds\right]\right) \qquad
\forall t\in [0,T]
$$
for some constant $C$. On the other hand, we note the following
connection of the above result to the previous one. From the
equation
$$ \frac{D |\o |}{Dt}=\alpha |\o |, \qquad \alpha (x,t)=\frac{\o \cdot S\o
}{|\o |^2}$$ we immediately have
\begin{eqnarray*}
\|\o (t)\|_{L^2}&\leq& \|\o_0 \|_{L^2}
\exp \left(\int_0 ^t \sup_{x\in \Bbb T^3} \alpha (x,s)
ds \right)\\
&\leq&\|\o_0 \|_{L^2} \exp \left(\int_0 ^t \sup_{x\in \Bbb T^3}
\lambda_1(x,s )
d\tau \right),
\end{eqnarray*}
where we used the fact $\lambda_3 \leq \alpha \leq \lambda_1$, the well-known
estimate for the Rayleigh quotient. We note that
$\lambda_2^+(x,t)>0$ implies we have stretching of infinitesimal fluid
volume in two directions and compression in the other one
direction(planar stretching) at $(x,t)$, while $|\lambda_2^-(x,t)|>0$
implies stretching in one direction and compressions in two
directions(linear stretching). The above estimate says that the
dominance competition between planar stretching and linear
stretching is an important mechanism controlling the growth/decay
in time of the $L^2$ norm of vorticity.
In order to state our next
theorem we introduce some definitions. Given a differentiable
vector field $f=(f_1 ,f_2 ,f_3 )$ on $\Bbb T^3$, we denote by the
scalar field $\l_i (f)$, i=1,2,3, the eigenvalues of the deformation
tensor associated with $f$. Below we always assume the ordering, $
\lambda_1 (f)\geq \lambda_2 (f)\geq \lambda_3 (f). $ We also fix $m>5/2$ below. We
recall that if $f\in \mathbb{H}^m _\sigma $, then $\lambda_1 (f)+\lambda_2 (f)+\lambda_3
(f)=0$, which is another representation of div $f=0$.
Let us begin with introduction of admissible classes
$\mathcal{A}_\pm$ defined by
$$\mathcal{A}_+=\{ f\in \mathbb{H}^m _\sigma (\Bbb T^3)\, | \, \inf_{x\in \Bbb T^3}\lambda_2 (f)(x)
>0 \,\},$$
and
$$\mathcal{A}_-=\{ f\in \mathbb{H}^m_\sigma (\Bbb T^3)\, |
\sup_{x\in \Bbb T^3}\lambda_2 (f)(x)<0 \, \}.$$
Physically $\mathcal{A}_+$ consists of solenoidal vector fields
with planar stretching everywhere, while $\mathcal{A}_-$ consists
of everywhere linear stretching vector fields. Although they do not represent
real physical flows, they might be useful in the study of searching
initial data leading to finite time singularity for the 3D Euler
equations. Given $v_0 \in \mathbb{H}_\sigma ^m$, let $T_*(v_0)$ be the maximal time
of unique existence of solution in $\mathbb{H}_\sigma ^m$ for the system
(E).
Let $S_t : \mathbb{H}^m_\sigma \to
\mathbb{H}^m_\sigma$ be the solution operator, mapping from initial data
to the
solution $v(t)$.
Given $f\in \mathcal{A}_+$, we define the first zero touching time
of $\lambda_2 (f)$ as
$$ T(f)=
\inf\{ t\in (0, T_* (v_0)) \, | \, \mbox{$\exists x\in \Bbb T^3 $
such that $\lambda_2 (S_t f ) (x)<0 $}\}.
$$
Similarly for $f\in \mathcal{A}_-$, we define
$$ T(f)=
\inf\{ t\in (0, T_* (v_0)) \, | \, \mbox{$\exists x\in \Bbb T^3 $
such that $\lambda_2 (S_t f ) (x)>0 $}\}.
$$
The following theorem is actually an immediate corollary of Theorem
6.1, combined with the above definition of $\mathcal{A}_\pm$ and
$T(f)$. We just observe that for $v_0 \in \mathcal{A}_+ $(resp.
$\mathcal{A}_- $) we have $\lambda_2^-=0, \lambda_2^+=\lambda_2$(resp. $\lambda_2^+ =0,
\lambda_2^-=\lambda_2$) on $\O \times (0,T( v_0 ))$.
\begin{theorem}
Let $v_0\in \mathcal{A}_\pm$ be given. We set $\lambda_1(x,t)\geq
\lambda_2(x,t)\geq \lambda_3(x,t)$ as the eigenvalues of the deformation tensor
associated with $v(x,t)=(S_t v_0)(x)$ defined $t\in (0, T(v_0 ))$.
Then, for all $t\in (0, T(v_0 ))$ we have the following
estimates:\\
(i) If $v_0 \in \mathcal{A}_+$, then
$$
\exp\left( \frac12\int_0 ^t
\inf_{x\in\Bbb T^3} |\lambda_2 (x,s)| ds\right)\leq \frac{\|\o (t)\|_{L^2}}{ \|\o_0\|_{L^2}}
\leq \exp\left( \int_0 ^t
\sup_{x\in\Bbb T^3} |\lambda_2 (x,s)| ds\right) .
$$
(ii) If $v_0 \in \mathcal{A}_-$, then
$$
\exp\left( - \int_0 ^t
\sup_{x\in\Bbb T^3} |\lambda_2 (x,s)| ds\right)\leq \frac{\|\o (t)\|_{L^2}}{ \|\o_0\|_{L^2}}
\leq \exp\left( -\frac12 \int_0 ^t
\inf_{x\in\Bbb T^3} |\lambda_2 (x,s)| ds\right).
$$
\end{theorem}
(see \cite{cha8} for the proof) If we have the compatibility
conditions,
\begin{eqnarray*}
\inf_{x\in\Bbb T^3} |\lambda_2 (x,t)|&\simeq& \sup_{x\in\Bbb T^3} |\lambda_2
(x,t)|\simeq g(t) \quad \forall t\in (0, T(v_0 )), \end{eqnarray*}
which is the case for sufficiently small box $\Bbb T^3$, then we have
\begin{eqnarray*}
\frac{\|\o (t)\|_{L^2}}{ \|\o_0\|_{L^2}} \simeq\left\{
\aligned &\exp\left(\int_0 ^t g(s)ds \right) \quad \mbox{if}\quad v_0 \in
\mathcal{A}_+\\
&\exp\left(-\int_0 ^t g(s)ds \right) \quad \mbox{if}\quad v_0
\in
\mathcal{A}_- \endaligned \right.
\end{eqnarray*}
for $t\in (0, T(v_0 ))$.
In particular, if we could find $v_0 \in \mathcal{A}_+$ such that
$$
\inf_{x\in\Bbb T^3} |\lambda_2 (x,t)| \gtrsim
O\left(\frac{1}{t_*-t}\right)
$$
for time
interval near $t_*$, then such data would lead to singularity at
$t_*$.
As another application of the formula (\ref{spec1}) we have some
decay in time estimates for some ratio of eigenvalues(see
\cite{cha8} for the proof).
\begin{theorem}
Let $v_0 \in \mathcal{A}_\pm$ be given, and we set $\lambda_1(x,t)\geq
\lambda_2(x,t)\geq \lambda_3(x,t)$ as in Theorem 3.1.
We define
$$
\varepsilon (x,t)=\frac{|\lambda_2 (x,t) |}{\l (x,t)}\quad \forall (x,t)\in
\Bbb T^3 \times (0, T(v_0)),
$$ where we set
$$\l (x,t)=\left\{\aligned \l _1 (x,t)
\quad \mbox{ if} \quad v_0 \in \mathcal{A}_+\\
-\l _3 (x,t) \quad \mbox{ if} \quad v_0 \in \mathcal{A}_- .
\endaligned
\right.
$$
Then, there exists a constant $C=C(v_0)$ such that
$$
\inf_{(x,s)\in \Bbb T^3\times (0,t)} \varepsilon (x,s)< \frac{C}{\sqrt{t}}\quad
\forall t\in (0, T(v_0)).
$$
\end{theorem}
Regarding the problem of searching finite time
blowing up solution, the proof of
the above theorem suggests the following:\\
Given $\delta >0$, let us suppose we could find $v_0 \in
\mathcal{A}_+$ such that for the associated solution $v(x,t)=(S_t v_0)(x)$
the estimate
\begin{equation}\label{spec3.3}
\inf_{(x,s)\in \Bbb T^3 \times (0,t)}\varepsilon (x,s)\gtrsim
O\left(\frac{1}{t^{\frac12+\delta}}\right),
\end{equation}
holds true, for sufficiently
large time $t$.
Then such $v_0$ will lead to the finite time singularity. In order
to check the behavior (\ref{spec3.3}) for a given solution we need
a sharper and/or localized
version of the equation (\ref{spec1}) for the dynamics of eigenvalues
of the deformation tensor.
\section{Conservation laws for singular solutions }
\setcounter{equation}{0}
For the smooth solutions of the Euler equations there are many
conserved quantities as described in Section 1 of this article. One
of the most important conserved quantities is the total kinetic
energy. For nonsmooth(weak) solutions it is not at all obvious that
we still have energy conservation. Thus, there comes very
interesting question of how much smoothness we need to assume for
the solution to have energy conservation property. Regarding this
question L. Onsager conjectured that a H\"{o}lder continuous weak
solution with the H\"{o}lder exponent $1/3$ preserve the energy, and
this is sharp(\cite{ons}). Considering Kolmogorov's scaling argument
on the energy correlation in the homogeneous turbulence the exponent
$1/3$ is natural. A sufficiency part of this conjecture is proved
in a positive direction by an ingenious argument due to
Constantin-E-Titi\cite{con7}, using a special Besov type of space
norm, $\mathcal{\dot{B}}^s_{3, \infty}$ with $s>1/3$(more precisely,
the Nikolskii space norm) for the velocity. See also \cite{caf2} for
related results in the magnetohydrodynamics. Remarkably enough
Shnirelman\cite{shn2} later constructed an example of weak solution
of 3D Euler equations, which does not preserve energy. The problem
of finding optimal regularity condition for a weak solution to have
conservation property can also be considered for the helicity. Since
the helicity is closely related to the topological invariants, e.g.
the knottedness of vortex tubes, the non-conservation of helicity is
directly related to the spontaneous apparition of singularity from
local smooth solutions, which is the main theme of this article. In
\cite{cha5} the author of this article obtained a sufficient
regularity condition for the helicity conservation, using the
function space $\mathcal{\dot{B}}^s_{\frac95, \infty}$, $s>1/3$, for
the vorticity. These results on the energy and the helicity are
recently refined in \cite{cha9}, using the Triebel-Lizorkin type of
spaces, $\mathcal{\dot{F}}^s_{p,q}$, and the Besov spaces
$\mathcal{\dot{B}}^s_{p,q}$(see Section 1 for the definitions) with
similar values for $s, p$, but allowing full range of values for
$q\in [1, \infty]$.
By a weak solution of $(E)$ in $\Bbb R^n \times (0, T)$ with initial
data $v_0$ we mean a vector field $v\in C([0, T); L^2_{loc} (\Bbb
R^n))$ satisfying the integral identity:
\begin{eqnarray}
\lefteqn{-\int_0 ^T \int_{\Bbb R^n} v(x,t)\cdot \frac{\partial\phi (x,t)}{\partial
t} dx dt -\int_{\Bbb R^n} v_0(x)\cdot \phi (x, 0 )dx}\hspace{.0in}\nonumber \\
&&\quad -\int_0 ^T \int_{\Bbb R^n} v(x,t)\otimes v(x,t) : \nabla \phi (x,t)
dxdt\nonumber \\
&&\qquad -\int_0 ^T \int_{\Bbb R^n} \mathrm{div }\, \phi (x,t ) p(x,t )dxdt
=0, \end{eqnarray} \begin{equation}
\int_0 ^T \int_{\Bbb R^n} v(x,t)\cdot \nabla \psi (x,t)dxdt=0
\end{equation}
for every vector test function $\phi =(\phi_1 , \cdots
,\phi_n )\in C_0 ^\infty (\Bbb R^n\times [0, T))$, and for every scalar
test function $\psi \in C_0 ^\infty (\Bbb R^n \times [0, T))$. Here we
used the notation $(u\otimes v)_{ij}= u_i v_j$, and
$A:B=\sum_{i,j=1} ^n A_{ij} B_{ij}$ for $n\times n$ matrices $A$ and $ B$.
In the case when we discuss the helicity conservation of the weak
solution we impose further regularity for the
vorticity, $\o (\cdot ,t)\in L^{\frac32}
(\Bbb R^3)$ for almost every $t\in [0,T]$ in order to define the
helicity for such weak solution. {\em Hereafter, we use the notation
${\dot{X}}^s_{p,q}$(resp. ${X}^s_{p,q}$) to represent
$\mathcal{\dot{F}}^s_{p,q}$(resp. $\mathcal{{F}}^s_{p,q}$) or
$\mathcal{\dot{B}}^s_{p,q}$(resp. $\mathcal{{B}}^s_{p,q}$)}. The
following is proved in \cite{cha9}.
\begin{theorem}
Let $s>1/3$ and $q\in [2, \infty]$ be given. Suppose $ v$ is a weak
solution of the $n-$dimensional Euler equations
with $v\in C([0, T]; L^{2} (\Bbb R^n))\cap L^3 (0, T;
\dot{X}^s_{3,q}(\Bbb R^n ))$. Then,
the energy is preserved in time, namely
\begin{equation}
\int_{\Bbb R^n} |v(x,t)|^2 dx =\int_{\Bbb R^n} |v_0 (x)|^2 dx
\end{equation}
for all $t\in [0, T)$.
\end{theorem}
When we restrict $q=\infty$, the above theorem
reduce to the one in \cite{con7}. On the other hand, the
results for Triebel-Lizorkin type of space are completely new.
\begin{theorem}
Let $s>1/3$, $q\in [2, \infty]$, and $r_1\in [2, \infty], r_2\in [1, \infty]$ be given, satisfying
$2/r_1 +1/r_2 =1$. Suppose $ v$ is a weak
solution of the 3-D Euler equations
with $v\in C([0, T]; L^{2} (\Bbb R^3 ))\cap L^{r_1} (0, T;
\dot{ X}^s_{\frac92,q} (\Bbb R^3 ))$ and $\o \in L^{r_2} (0, T;
{\dot{X}}^s_{\frac95,q}(\Bbb R^3 ))$,
where the curl operation is in the sense of distribution. Then,
the helicity is preserved in time, namely
\begin{equation}\label{helicityy}
\int_{\Bbb R^3} v(x,t)\cdot \o (x,t )dx =\int_{\Bbb R^3} v_0 (x)\cdot \o _0 (x)dx
\end{equation}
for all $t\in [0, T)$.
\end{theorem}
Similarly to the case of Theorem 7.1, when we restrict $q=\infty$, the above theorem
reduce to the one in \cite{cha5}. The
results for the case of the Triebel-Lizorkin type of space, however,
is new in \cite{cha9}.
As an application of the above theorem we have the following
estimate from below of the vorticity by a constant depending on
the initial data
for the weak solutions of the 3D Euler equations.
We estimate the helicity,
\begin{eqnarray*}
\lefteqn{\int_{\Bbb R^3} v(x,t )\cdot \o (x,t )dx
\leq \|v (\cdot , t)\|_{L^3} \|\o (\cdot , t
)\|_{L^{\frac32}}}\hspace{.2in} \\
&\leq &C \|\nabla v (\cdot , t)\|_{L^{\frac32}} \|\o (\cdot , t )\|_{L^{\frac32}}
\leq C\|\o (\cdot , t )\|_{L^{\frac32}}^2,
\end{eqnarray*}
where we used the Sobolev inequality and the Calderon-Zygmund
inequality. Combining this estimate with (\ref{helicityy}), we obtain
the following:
\begin{cor}
Suppose $ v$ is a weak
solution of the 3D Euler equations satisfying the conditions of
Theorem 7.2. Then,
we have the following estimate:
$$
\|\o (\cdot ,t)\|_{L^{\frac32}}^2 \geq C H_0, \quad \forall
t\in [0, T)
$$
where $H_0 =\int_{\Bbb R^3} v_0(x)\cdot \o _0(x) dx$ is the initial
helicity, and $C$ is an absolute constant.
\end{cor}
Next we are concerned on the $L^p$-norm conservation for the weak
solutions of (QG). Let $p\in [2, \infty)$. By a weak solution of
$(QG)$ in $D\times (0, T)$ with initial data $v_0$ we mean a scalar
field $\theta \in C([0, T); L^p (\Bbb R^2)\cap L^{\frac{p}{p-1}}
(\Bbb R^2))$ satisfying the integral identity:
\begin{equation}\label{qgg1}
-\int_0 ^T \int_{\Bbb R^2} \theta (x,t)\left[\frac
{\partial}{\partial
t} +v \cdot \nabla\right] \phi (x,t)dx dt -\int_{\Bbb R^2} \th_0(x) \phi (x, 0
)dx=0
\end{equation}
\begin{equation}
\label{qgg2}
v(x,t) =-\nabla^\bot\int_{\Bbb R^2}
\frac{ \th (y,t)}{|x-y|} dy
\end{equation}
for every test function $\phi \in C_0 ^\infty (\Bbb R^2\times [0,
T))$, where $\nabla^\bot$ in (\ref{qgg2}) is in the sense of
distribution. We note that contrary to the case of 3D Euler
equations there is a global existence result for the weak
solutions of (QG) for $p=2$ due to Resnick(\cite{res}). The following is proved in
\cite{cha9}.
\begin{theorem}
Let $s>1/3$, $p\in [2, \infty)$, $q\in [1, \infty]$, and
$r_1\in [p, \infty], r_2\in [1, \infty]$
be given, satisfying
$p/r_1 +1/r_2 =1$. Suppose $ \th $ is a weak
solution of (QG)
with $\theta \in C([0, T]; L^{p} (\Bbb R^2)\cap L^{\frac{p}{p-1}}(\Bbb R^2))\cap L^{r_1} (0, T;
X^s_{p+1,q}(\Bbb R^2))$ and $v \in L^{r_2} (0, T;
{\dot{X}}^s_{p+1,q} (\Bbb R^2))$. Then,
the $L^p$ norm of $\theta(\cdot,t)$ is preserved, namely
\begin{equation}
\|\theta (t)\|_{L^p}=\|\theta_0\|_{L^p}
\end{equation}
for all $t\in [0, T]$.
\end{theorem}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,661 |
Dedication
Dedicated to the Fearsome Foursome. They know why.
And to JRW for giving me the world's most amazing shout-out.
And to Amanda for giving me the incredible opportunity to bring these guys to life.
And to my family for making it all possible.
Contents
Dedication
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Chapter 16
Chapter 17
Chapter 18
Chapter 19
Chapter 20
Chapter 21
Chapter 22
Chapter 23
Chapter 24
Chapter 25
Chapter 26
Epilogue
Acknowledgments
An Announcement Page to The Raven Riders
An Announcement Page to the Hard Ink Series
Hard As It Gets
Hard As You Can
Hard To Hold On To
Hard To Come By
Hard To Be Good
Hard As Steel
About the Author
By Laura Kaye
Copyright
About the Publisher
Chapter 1
The warehouse was an abandoned shell. Empty so long that parts of the roof had caved in, most of the windows were gone, and nature had started to reclaim the concrete and cement, with bits of green taking root in the building's cracks. Proof of the resilience of life, even in the worst of circumstances.
Most of Beckett Murda's life was proof of that.
The whistling early morning wind and the distant sounds of Baltimore car traffic and ships' horns were the only noises that made their way into this corner of what was left of the fourth floor, and that was just fine by Beckett. Because the height and the quiet and the seclusion made it the perfect place from which to protect what he cared about most.
His friends.
His brothers.
His chance at redemption.
Crouching beside a busted window gave Beckett the perfect view of Hard Ink Tattoo, his temporary home the past two and a half weeks. Or, at least, it was the perfect place to see what was left of it. The red-brick, L-shaped building sat on the opposite corner of the intersection. Just a few days before, the center of the long side of the L had been reduced to rubble courtesy of his enemies. The predawn attack had claimed the lives of three good men. Three too damn many.
With Wednesday morning daylight just breaking, Beckett scanned a one-eighty from left to right, his gaze sequentially moving from the empty roads that led to Hard Ink's intersection, to Hard Ink itself, to the surrounding buildings—all empty just as this one was except for Hard Ink. Luckily, the side on which their group lived hadn't suffered any loss of integrity during the attack, so they hadn't had to relocate their base of operations.
Beckett repeated the survey using a pair of high-power binoculars, useful for picking up details he might otherwise miss, given the loss of acuity he'd experienced in his right eye from a grenade explosion just over a year ago. His lefty was 20/20 all the way, but shrapnel had reduced his righty to 20/160. His visual impairment in that eye was damn close to legally blind, and it made seeing at a distance a bitch.
That explosion marked the beginning of the whole clusterfuck that led to him sitting in this hellhole all night. Beckett's Army Special Forces team had been ambushed at a checkpoint in Afghanistan, killing their commander and six other members of the team. In addition to himself, the four survivors—his best friend, Derek "Marz" DiMarzio, second-in-command Nick Rixey, Shane McCallan, and Edward "Easy" Cantrell—had fought tooth and nail to make it out alive, only to be blamed for their teammates' deaths, accused of dereliction of duty, and sent packing from the Army courtesy of other-than-honorable discharges and nondisclosure agreements ensuring they could never say anything to try to clear their names.
Now they were doing it anyway. This was their one and only shot.
Movement along the far side of Hard Ink.
Beckett focused in to see Katherine Rixey pause at the corner before running across the road to the shadows of the opposite building. From there, Nick's younger sister skirted tight along the wall, darted across the road again, and then disappeared from view as she entered the warehouse where he hid. Within a minute the rapid thump of footsteps echoed up the stairwell.
Nearly 6:00 A.M., which meant his shift in the sniper's roost was done. Kat was his relief.
Except that was maybe the only way Kat Rixey relieved him. Otherwise, she had an impressive knack for getting way far under his skin and pushing all his buttons. And every one of his teammates had witnessed it firsthand. Among elite operatives, lives and missions depended on being able to recognize and mitigate your weaknesses. And that meant Beckett had to admit that something about Kat distracted him, irritated him, made him . . . feel.
Not something he had much experience with. Not for years.
Her footsteps neared, their sound louder in the stairwell, and Beckett's heart might've kicked up in time with her jogging pace. Something about her threw him off kilter. And that fucking pissed him off. Because this woman was the younger sister of one of his best friends. And no part of what he was doing here involved—
"Hey, Trigger. You're free to go," she said as she stepped into the large room behind him.
Fucking Trigger. She'd been at him with her cute little nicknames since the day they met. Like it was his fault he'd caught her roaming Hard Ink unannounced and pulled his gun on her. Times being what they were, she was lucky that was all he'd done. He kept his eyes trained out the window so she couldn't see the irritation likely filling his expression.
"Helloooo?" she said, standing right behind him.
Taking his good old time, he put the binoculars down and slowly turned toward her. And had to work hard to keep from reacting to how fucking hot she was.
Katherine Rixey was an angel-faced beauty with a foul mouth, sharp green eyes, and curves that would not quit. His hands nearly ached to bury themselves in her thick, wavy brown hair every time he saw her, and the sight of her confidently and competently handling a gun made him rock hard. The fact that she was apparently a shark of a prosecutor was just icing on her five-foot-two-inch cake. Brains, body, beauty. Kat had it all. Too bad she drove him bat-shit crazy.
She waved a hand in front of his face, and he tore his gaze away. "You fall asleep there, Quick-Draw? Shoulda texted me. I would've come sooner."
Whatever you do, do not think about her coming.
Jesus.
Beckett secured his weapon in a holster at his lower back, hauled himself off the floor, and swallowed the innuendo-filled retorts flitting through his mind. "Had it covered just fine."
"Good to know," she said, crossing her arms and smirking.
Beckett felt his eyebrow arch in question before he'd thought to school his expression. "Problem?" he asked, stepping right up in front of her. She was so short, he towered over her, forcing her to tilt her head back to meet his gaze. And damn if she didn't smell good, like warm, sweet vanilla. It made his mouth water, his groin tighten, and his temper flare. Sonofabitch.
"Dude. I am so not the one with the problem." Amusement filled her bright green eyes.
As he nailed her with a stare, Beckett tried not to admire the way her crossed arms lifted the mounds of her breasts under the clingy black long-sleeved tee. This woman was a Rixey, which meant sarcasm was coded into her DNA. Beckett had almost a decade of experience with her oldest brother to know that was true. No way he was giving her the satisfaction of a reply. He shook his head and stepped around her.
"Always a pleasure," she said.
He peered over his shoulder to find her lowering into a crouch by the window. She grabbed a gun from one of the cases on the floor and checked the weapon's ammo. Her quick motions revealed her confidence and experience—always sexy qualities to Beckett's mind.
Still, despite her obvious competence with a weapon—Nick had done a damn fine job making sure his petite little sister could take care of herself—her being alone up here for a day-long shift didn't sit well in Beckett's gut. Ever since the attack on Hard Ink, everyone who had any experience with weapons had been taking shifts in one of the two lookouts they'd set up, and Katherine had more than earned the right to help with the task. More than that, they needed all hands on deck right now—including Katherine. But the team's enemies were expertly trained, highly lethal mercenaries who had no qualms about covering their asses, no matter what it took—or who they took down. And where Katherine was concerned, that made Beckett . . . worry.
After all, she was Nick's sister. And just like the rest of the team, Nick had lost enough.
And that's all it was. Right.
Sonofabitch.
"Watch yourself," Beckett said, voice gruff.
Katherine peered over her shoulder at him and rolled her eyes. "Yeah, that's kinda the point of this whole thing," she said, gesturing to the guns, ammo, communication devices, army-green sleeping bags, and stack of bottled water and snacks piled around the corner by the window. When Beckett didn't reply, she shook her head and looked outside again. "You question Nick and the guys this way when you hand off your shift?"
No, he didn't. And saying so would either make him look like a chauvinist asshole or possibly reveal too damn much about the shit she stirred up inside him. So he disappeared into the stairwell and made his way down.
Given the strength and resources of their enemies, sparring with Katherine Rixey was the last thing he needed to waste energy doing.
"THAT'S WHAT I thought," Kat said. Looking back toward the stairs again, she realized she was alone. As big as Beckett freaking Murda was, how the hell did she not hear him leave?
Damn Special Forces guys. Her brother Nick had the same ability. Scared her half to death sometimes. Thank God for their middle brother, Jeremy. Most of the time, Jer gave off a happy vibe you could feel coming from a mile away.
Kat smiled at the thought, settled into a comfortable position and turned her attention back to the view outside the window. The streets were eerily quiet, which wasn't an accident. Though her brothers had bought a building in the city's derelict and partly abandoned old industrial district, the real explanation behind the ghost town she was looking down on was a series of roadblocks a police ally of Nick's had somehow orchestrated. Kat had tried to stay out of the specifics, because she hadn't wanted to know the details if they veered into the illegal.
Which was damn near a certainty. Kat had come to visit her brothers at Hard Ink five days before, and pretty much the whole time she'd been here she'd walked a fine line between wanting to help them with this crazy situation and freaking out about the illegal nature of what they were doing. Not that the guys weren't justified in defending themselves and doing whatever it took to clear their names, but Kat had become a lawyer for a reason. Growing up, Nick was the risk-taker, the guy who ditched college weeks into his senior year to join the Army. Jeremy was the artistic rule-breaker. And Kat had been the rule-follower.
Almost like checking a series of boxes, she'd gotten straight A's all through school, served as the president of all her clubs, got into the best colleges and busted her ass to become managing editor of her law review. Even as early as high school, she'd known she wanted to go into the law. Because law represented justice and order and righteousness. Those ideals had spoken to her, drawn her to a career fighting what she thought was the good fight.
Four years into working at the Department of Justice, she still believed that was what she and the good people she worked with tried to do. Problem was, sometimes a big gulf existed between what they tried to do and what the law allowed them to achieve. And she'd never realized just how all-consuming the career would be. Twelve-hour days at her desk were her norm.
Kat surveyed the run-down neighborhood outside the window. Baltimore might've only been about thirty miles from D.C., but right now she felt about a million miles away from that desk.
Down below, the street beside Hard Ink was literally blocked—by the pile of rubble that had slid down into the road when part of the building collapsed early Sunday morning. Just looking at the pile of bricks and cement and twisted beams and broken glass made Kat's heart race, because she'd been on top of that building when it went down. Her, her brothers, and several others, too. In her mind's eye, she saw the rooftop fall away from under Jeremy's feet. He and two other guys started to fall, and she'd screamed. And then Nick was there, grabbing Jeremy's hand and hauling him up from the breach.
Kat's breath caught and she blinked away the sting suddenly filling her eyes. The image of Jer falling and the thought of him being gone had haunted her dreams every night since. Because she could've lost Jeremy.
Which made her glad her oldest brother had spent years in SpecOps and knew what the hell to do, because she couldn't lose her brothers. And given the impossibly crazy situations she'd encountered since arriving at Hard Ink, she was well aware that losing them was a possibility. Because Nick and Jeremy were in the very gravest danger.
It all stemmed from Nick's team's fight to restore their honor against powerful and not fully known enemies. A fight that apparently had so much at stake that her brothers' building had been attacked by armed soldiers who had a rocket launcher. A freaking rocket launcher!
And, if that wasn't enough, the men they were likely fighting against—and probably the very ones who had attacked—were the subject of a series of investigations her office had been working on for the past nine months.
It was something she'd only become certain of over the last twenty-four hours, as Nick's team's investigation into a cache of documents from their now-deceased Army commander had begun to shed light upon exactly what—and who—they were up against. Kat was glad for the alone time today, because her brain was a conflicted mess. Should she maintain her professional ethics and protect her security clearances by keeping her mouth shut? Or tell Nick and his team exactly what her office was doing and share what information she had that could help them?
She'd promised herself to decide today while she had some time to think.
And she had thought that coming to see her big bros would be the relaxing getaway it normally was, one that would distract her from her own problems—namely Cole, the ex-boyfriend who couldn't seem to get it through his thick skull that she was really done with him.
Down below, she saw Beckett dart across the street between the buildings. With his muscles, square jaw, and fathomless blue eyes, the guy was pure, raw masculinity personified. Her body couldn't be near his without reacting on some fundamentally hormonal level. Her heart raced. Her nipples peaked. Her stomach went for a loop-the-loop.
Before he disappeared around the corner of the Hard Ink building, he glanced over his shoulder and looked up. Kat was a hundred percent sure his eyes landed on her, even though she sat mostly shielded behind the brick of this old warehouse. Because her body jangled with a sudden awareness.
And then he was gone.
Kat rolled her eyes. Freaking ridiculous to get so worked up over a man whose favorite form of communication was the grunt. And who made a habit of ignoring her when she spoke to him. And who'd pulled a gun on her without even bothering to ask her name. Couldn't forget that.
Whatever. He probably just got to her because he was so unlike the men with whom she normally spent time. Whereas her colleagues at Justice tended to be serious, buttoned-up, and lower-key, Beckett radiated an intensity she didn't quite understand. It certainly didn't have anything to do with how he spoke or acted, because he talked little and showed emotion even less. Maybe it was all that leashed strength, because she had no doubt that he could do some serious damage with his bare hands.
Which, given the way her biceps looked right now, maybe wasn't the most pleasant thought, was it? Because Friday morning, when she'd gone down to her building's garage to head out to work, Cole had been waiting for her . . . somewhere. One minute, she'd been juggling her belongings and sliding her key into the door lock, and the next, someone grabbed her arms from behind and shoved her against the cold cinder-block wall near the hood of her car. It caught her so off guard that she hadn't even managed a scream before her body was trapped between the wall and her assailant, who ground his hard-on into her rear.
God, I missed you, baby.
The memory of his raspy voice whispering in her ear made her shudder. The fact that he'd gotten the jump on her without her having the chance to fight back had Kat so mad at herself she could barely stand it. And his knowledge that she'd planned to drive was seriously creepy, because she usually took the metro.
When she'd finally talked him into letting go of her and agreed to meet him after work for a drink and a talk, he left. And she'd hightailed it up to her apartment to pack a bag, made a stop at the Superior Court to file for a protective order, and left D.C. for Baltimore and the safety of her brothers' place.
Ha.
So much for that.
But at least she knew Cole wasn't going to be a problem here. She'd received a message yesterday that the judge had granted the order, and once the authorities served Cole with the papers putting the order into effect, he'd keep his distance. He was too damn smart and appearance-conscious not to. And, as an attorney, he'd obviously realize that getting caught violating a protective order would create problems with the bar in addition to problems with the police.
Kat couldn't wait for the order to be served. Not only would it give her peace of mind that he'd stay away, but the no-contact provision should also cut off the stream of demanding and accusatory texts and voice mails she'd received since she pulled a no-show for their talk Friday night. Couldn't happen soon enough.
Given everything that was going on with her brothers, the knowledge that the order would diffuse the Cole situation allowed Kat to breathe a sigh of relief. Because they really couldn't handle even one more complication.
Chapter 2
The Hard Ink building was in total pandemonium.
At least, that's how it seemed to Beckett as he stepped into the second floor's cavernous gym, which had been serving as their war room and communications hub in the mission to clear their names. Despite the early hour, the bass beat of a rock song filled the air. Beckett paused inside the door and took in the scene. Groups of people sat and stood here and there. One group was chowing down an early breakfast around the big twelve-seater table. Guys were working out with the free weights, and others were spotting a big guy who appeared to be bench pressing at least two hundred pounds. Marz sat alone at his computer desk, working away. Didn't seem like the guy ever slept anymore.
As if there hadn't already been enough of them, between the team and their women living in Hard Ink's second- and third-floor apartments, now there were a dozen members of the Raven Riders Motorcycle Club racking out wherever they could. Some of their sleeping bags and packs still covered the floor in the back corner. All of the activity made a room that big seem small and overcrowded. Or maybe that was because Beckett preferred his solitude. Always had.
The Ravens had initially gotten involved when the team paid for their assistance several weeks ago. But when a rocket launcher caved in part of the building on Sunday morning, two of their guys had died in the collapse. Now they were pissed, armed, and fully on board of their own free will.
And, really, thank fuck for that. Because between their muscle and the number of boots they could put on the ground, the Ravens helped Beckett and his four SF buddies even the playing field against their enemies. By a lot.
Beckett started across the gym toward Marz and came to an abrupt halt when an orange-striped cat with white paws came curling out from behind a piece of equipment, planted itself in his path, and stared at him with its one good eye. Which was how he'd earned the name Cy. Short for Cyclops.
Beckett stared back. "What?" he finally asked the animal, peering around to see if anyone was watching.
The cat blinked its right eye in silent answer.
Nick's brother Jeremy, who ran the tattoo shop on the first floor, had saved the stray from the wreckage of the other half of the building, where it'd apparently been living. The cat wasn't mean, but he wasn't friendly either, and seemed to avoid everyone except for his savior . . . and Beckett.
Sighing, Beckett slowly crouched down and held out one hand.
Cy bolted, ultimately finding a hiding place behind a metal shelving unit against the wall.
No surprise. Beckett was big and intimidating, Or so he'd been told. Hardly the warm, nurturing type. The real surprise was that the cat gave him the time of day at all.
You think being big makes you such hot shit, but you're every bit as dumb and ugly . . .
The memory of his father's voice—from the night his high school football team had won the game that would send them to the state championships—came so far out of the deep, distant past that Beckett nearly stumbled as he rose to his full height. What the fuck was that? Since when did he let anything slip around the ancient barriers he'd erected against all the bullshit his old man had thrown at him?
"You okay, man?"
Beckett blinked to find Marz standing right in front of him. "Why wouldn't I be?"
Marz shrugged, his eyes narrowing on Beckett's face. "No reason. You got a minute? Or would you rather catch some sleep first?"
"Got nothing but time," Beckett said, glad for something else to think about. Something besides figuring out why his father's voice was echoing inside his head. He followed Marz across the room. The number of computer terminals had expanded in recent days, as Marz had made it possible for more of the team to help comb through the huge number of files they'd discovered on a microchip from their deceased commander. The commander they all believed had betrayed them, sold them out.
The commander who'd actually died in a deniable undercover operation—meaning he hadn't been able to say word one to them.
The fact that they'd all believed Frank Merritt had been dirty felt a lot like having a few badly cracked ribs. It hurt with every breath, but there wasn't a damn thing you could do to make it better. Beckett would know. He'd had more than a few broken ribs in his time.
Except, there was something they could do. They could bring Frank's killers to justice and clear their own names. Beckett glanced to the big white board on the wall by Marz's desk where they kept a running list of what they knew and questions they'd yet to answer. In Nick's scrawling handwriting, the crux of their mission was spelled out in red marker:
Bad guys:
Church Gang
"WCE"/"GW"
Seneka Worldwide Security?
But what/who are WCE/GW?
And are they related to Seneka? If so, how?
All Beckett's team knew about WCE was that it had paid their commander twelve million dollars in dirty money during his undercover investigation into counternarcotics corruption in Afghanistan, and all they knew about GW was that their commander had noted those initials as belonging to his WCE contact, who'd had a now-disconnected Seneka phone number. Beyond that . . . asking who WCE and GW were was a seemingly simple question. But it was proving damn hard to answer.
"I finished piecing together the shipping research that you started yesterday," Marz said as he dropped into a metal folding chair at his desk. Beckett followed suit. With one hand, Marz fished through a mass of papers—an impressive example of disorganization that only made sense to him. With the other, he kneaded his right thigh.
Beckett frowned and braced against a wave of gut-punching guilt. His best friend in the world was hurting—had been critically wounded, in fact—because of him. Most of the time, Marz acted like he was totally cool with the below-the-knee amputation that required his use of a prosthetic limb. But for the past week or so, it was clear from the guy's more noticeable limp that his leg was bothering him. No doubt from the round-the-clock schedule he'd been keeping, not to mention getting beaned with a baseball bat a few days before.
"Ah. Here we go," Marz said as he arranged some documents in front of him.
"You okay?" Beckett managed.
"What? Oh." He pulled his hand away from his leg. "Yeah, no biggie."
Typical Marz. He didn't want anyone to think him weak. Beckett totally got that. But in his resilience and his positivity and his expertise, Marz was one of the strongest of them all. That shit was just fact. Beckett shook his head as Marz dived into an explanation of his research. "Don't do that," Beckett said, interrupting.
Eyebrow arched, Marz's expression was almost comical. "Don't do what?"
"Don't downplay shit with me." He nailed the guy with a stare, wishing he could repair all the damage he'd done to their friendship. It was just hard to pretend that everything was copasetic when his best friend—a far fucking better human being than himself—had nearly died to save him. An act of selflessness so pure, so stunningly undeserved, that it stole Beckett's breath every time he thought of it . . . or when Marz's prosthetic reminded him of it. Which was every fucking day.
Beckett raked his hand through his hair. "Just . . . be real with me. Like before. I know I haven't always . . ." He shrugged. "But I, uh, want . . . things back." He scrubbed his hands over his face, glad that the early hour meant the rest of the team wasn't around to witness this embarrassing string of verbal diarrhea. "Fuck," he said into his hands.
"Beckett?" Marz said.
He didn't know why this guy put up with him. He never really had. Marz had always been lightness next to his own deep, deep dark. He lifted his head and met his friend's gaze.
"Fair enough," Marz said. "But I'm not the only one who's taken a beating through this whole thing, that's all. Not trying to hide it. It's just that we have more important things to worry about right now, you know?" He shrugged.
"How's Emilie?" Beckett asked, referring to the sister of one of their enemies. Turned out she hadn't known a thing about her now-deceased brother's illegal activities, and in the process of finding that out, Marz had fallen hard. And so had she. Beckett was glad for the both of them. Really, he was.
Her name brought an immediate spark to Marz's eyes, though his expression remained serious. "Considering she found her brother's body lying in a gutter with a bullet hole through his head just a few days ago, she's doing as good as can be expected." He shook his head. That day had been bad for every single one of them. Emilie had lost her brother, the Ravens had lost two of their members, the Hard Ink building had been partially demolished, and all of them had lost an ally when their enemies had gunned down Miguel Olivero, a P.I. who'd been helping their investigation. "His body's gonna be released from the coroner's office today or tomorrow, so that's bound to stir things up for her."
"Yeah," Beckett said. "You'll help her through it, though. You're good at . . . that kinda thing." And he was not. Obviously.
Marz smiled, his expression both pleased and surprised, like Beckett giving him a compliment was so damn unusual. Maybe it was. "Yeah?"
Nodding, Beckett pointed to the papers in front of Marz. "So, what did you find on the microchip?"
"What I was hoping we would," Marz said, waking up the monitor. He tapped the screen. "This is a list of containers leaving Afghanistan via one of Seneka's subsidiary transportation lines."
Seneka Worldwide Security. Seneka, for short. A defense contractor founded by John Seneka that operated in Afghanistan. Back in the day, the man had become something of a legend in the SpecOps community for leading a series of cross-border covert ops against high-value targets that the public would never know to thank him for. So when he started Seneka Worldwide Security, retired elite operatives flocked to Seneka in droves for the opportunity to work with him. Still did.
They now suspected he had been behind the ambush back in Afghanistan that killed their friends and ended their careers, and the attack on Hard Ink last weekend. But their team needed hard proof.
"And this printout of arriving containers at Baltimore's pier thirteen . . ." Marz said, indicating a sheaf of paper in a beat-up accordion folder they'd received with the assistance of Baltimore Police detective Kyler Vance—Miguel's godson and their new ally. " . . . shows all the arriving Seneka containers highlighted in yellow. Matching container numbers prove that Seneka definitely shipped product out of Afghanistan to Baltimore. On almost all of them, Emilie's brother is listed as the receiving agent."
"Which means they were probably full of drugs for the Church Gang," Beckett said. Their investigation had already revealed that Seneka and Church, now largely destroyed by a series of gangland assassinations and embarrassing losses, had been in partnership with one another to trade contraband drugs stolen from Afghanistan. And the fact that the gang had kidnapped Merritt's son Charlie—the incident that had first reunited Beckett's team almost three weeks ago—and demanded information from the son about their commander's double-dealing in Afghanistan, was more proof that Seneka was at the bottom of all their troubles. What a fucking complicated mess. "Anything else?"
"Not yet," Marz said, tossing the sheets to the desktop.
Beckett sighed. "So that's another connection. More proof that Seneka is likely at the bottom of all of this, and therefore that they're also whatever WCE is."
"Likely," Marz said. "Problem is, what we're finding so far is all circumstantial evidence. These containers, Emilie's brother Manny working for both Seneka and the Church Gang, the Seneka knight logo tattoo on the one attacker's arm that we saw on the security footage, and the documents suggesting some relationship between Seneka and WCE, but not the nature of that relationship. Nothing that definitively says Seneka is WCE."
"Right," Beckett said. "And we might never find something that definitive. We might just have to go with the preponderance of the evidence."
"Which is something for the whole team to decide," Marz said.
"Roger that," Beckett said. Because no way did they want to poke a sleeping giant without a damn good reason and a whole lotta proof. They'd already lost efuckingnough.
BY LUNCHTIME KAT knew what she had to do.
In the contests between loyalty and duty, family and career, there was only one choice she could make.
She had to help Nick any way she could. And that meant spilling.
She sighed. After three years of law school, a string of prestigious clerkships, and almost four years pulling twelve-hour days at Justice, the decision didn't sit comfortably in her gut. Because she had a helluva lot to lose. An ethics violation—or worse, losing her license and being disbarred—were not things from which a lawyer bounced back.
Then again, she'd lost her parents four years before in a car accident. So Nick and Jeremy were all the family she had left in the world. If she lost them—
No, she refused to even consider it, which meant she really didn't have a choice at all.
The minute she'd settled her mind on the matter, her shift in the sniper's roost turned into possibly the slowest hours of her life. The clock had to be marching backward.
And then the skies opened up, the low, heavy clouds making the mid-afternoon look like evening. Which gave her an idea . . .
Maybe someone could relieve her early while the downpour provided a cover. She radioed Hard Ink to see if there was someone free who could come.
Unless they were part of the perimeter defense team guarding the roadblocks their detective ally had set up, they were supposed to keep off the streets during the daytime as much as possible, just to be on the safe side. But the storm would give her some cover. She'd never been happier for rain in her life.
Ten minutes later someone blurred across the street down below, a dark hood and the rain obscuring their identity. Feet pounded up the stairs. Too heavy to be Beckett.
The thought had her rolling her eyes. The fact that she was thinking about the big lug at all was annoying. Let alone the man himself.
Finally, her replacement stepped through the doorway into the empty room behind her and shrugged out of his drenched hooded sweat jacket. "Hey," he said, running a hand through his brown hair. A scar ran jagged from the corner of one eye into his hairline.
"Hey. Phoenix, right?" Kat said, stowing her weapon in one of the gun cases near the window. Phoenix Creed was one of the leaders of the Ravens. Kat hadn't had much chance to get to know these guys, but she knew they weren't all on the up and up. The Raven Riders was an outlaw motorcycle club, which meant they lived life—and made a living—by their own rules. But they were putting their lives on the line for her brothers and their friends, and that was pretty much all she needed to know right now.
The crooked grin he gave her was pure sex. "Yeah, that's right."
She rose and offered her hand. "I'm Kat. Thanks for the relief. I realized I need to do something that can't wait."
He shook and gave her an appraising up and down glance. "You're Nick's sister?"
She drained her bottle of water, hoping she didn't have to fend off an advance. 'Cause that wouldn't be awkward at all. A lawyer and an outlaw. "Yep," she said, tossing the bottle in the trash.
He gave a single nod. "Any blood of Nick's is a friend of mine."
Oh. That wasn't what she was expecting at all. "Yeah? Thanks. And thanks for everything you and the club are doing."
All the humor bled out of his expression. "One of the guys who died when that building went down was my cousin. I owe it to him to be here."
Guilt crept into Kat's belly for thinking he was going to hit on her when his purpose for being a part of this was every bit as honorable as Nick's. Stupid Cole had her looking for creeps everywhere. "I didn't realize. I'm sorry for your loss, Phoenix."
Nodding, he glanced around the room. "Yeah. Okay, well." He stuffed his hands in the pockets of his faded jeans. "You're free to fly."
Out in the stairwell, a cool, damp wind whipped up from below, tossing Kat's long hair around her face. When she reached the bottom, she stood in the doorway and frowned out at the deluge. Fly, hell. Swim was more like it. But there was no help for it. "Here goes nothing," she said.
As she bolted diagonally across the street, she just resisted shrieking as the unexpectedly cold rain immediately soaked through her clothes and matted her hair to her face. At the curb, water puddled so deep that she sank in to mid-calf. And then she was tearing around the Hard Ink building to the rear door, where she had to pause long enough to punch a code into a key box. The minute the door clicked, she yanked it open and nearly dove inside the cement-and-metal stairwell.
"Holy shit," she said as she stood there, water dripping off of her until a small puddle formed around her completely water-logged sneakers. She toed them off, peeled off her sodden socks, and wrung out her hair. As a small stream fell to the floor, Kat couldn't help but chuckle. She looked down at herself as a shiver ran over her skin. Her T-shirt was plastered to her body. Couldn't go see the guys like this.
Leaving her shoes and socks to dry by the door, she padded in bare feet to the stairs. Just as she reached the second-floor landing, the gym door opened. Beckett stepped out, his big body seeming to fill the whole stairwell. Of course, it would just have to be Beckett.
Kat hugged herself against the cold. And to hide how her body was reacting visibly through her soaked shirt to his. No matter how much of an ass he was, he was harshly beautiful and muscular and had an angled jaw that had to have been chiseled from granite.
His gaze scanned up her body and locked right on her face. "Why did you come back early?"
Kat couldn't read his tone, but the frown he wore made it feel like an accusation. "Because I need to talk to Nick."
His eyes narrowed. "It couldn't wait?"
Why was he interrogating her? She tilted her head, irritation curling into her belly. Like having this conversation with Nick wasn't stressing her out enough.
He shook his head. "What I mean is—"
"You know what? Save it. This isn't the Army. I don't have to report to you." She turned on her wet heel and stepped toward the door to Nick and Jeremy's apartment.
A hand closed around her arm.
No! For a split second she was yanked back to Friday morning. To Cole grabbing her from behind and shoving her into the wall. She yelped and whirled, trying to free herself.
Beckett's hands flew up like he was surrendering. His expression was a mask of horror and fear. "What just happened?"
Kat shook her head and cleared the knot from her throat. What a stupid overreaction! "Nothing. I just . . . nothing."
"You're shaking," he said, his brow cranked down.
"Just cold," she managed, her voice raspy. Get a grip, Kat. She forced a deep breath.
"Katherine—"
"Leave it, Beckett. Okay?" She rubbed at her arm, the bruises from Cole's hands still tender. No way for Beckett to have known they were there, of course. In fact, that was the whole point of the long sleeves. She didn't want anyone to know. The fact that she'd let Cole's stalkerish behavior go unaddressed for so long made her feel stupid enough. The last thing she wanted was for anyone to see how far it had gone.
She and Cole had started dating after close friends of theirs performed a bit of matchmaking. With the schedule Kat kept, she'd otherwise rarely made time for non-work-related socializing, and her friends joked that she'd end up married to her desk if she wasn't careful. So she'd gone on the date with Cole to humor them. Kat hadn't regretted it. At the beginning, he'd been dashing and intense and someone who impressed her with the sharpness of his intellect, not to mention his charismatic charm. As a bonus, the fact that he worked in another division at Justice meant he understood her schedule and the demands of the job. They'd dated for almost four months. Kat still sometimes wondered exactly when his weirdness had truly begun . . .
Beckett stepped toward the apartment door, pulling Kat from her thoughts, and punched in the code that unlocked it. Click. He pushed the door open for her and gestured for her to go first.
Slipping past his big body, Kat made sure not to touch Beckett as she entered the loft-style apartment. Jeremy had refinished it a few years ago, using the money their parents had left. Nick had given him most of what they'd left him, too. The space was all exposed brickwork, wide plank flooring, and high ceilings, and the kitchen and living room formed one big open space, separated by a long granite breakfast bar. Normally light and airy, it felt dark despite the kitchen lights because they'd blacked out all the windows to mask their presence in the building.
Beckett's gaze was hot on Kat's back as she beelined to the hallway that led to the bedrooms. Hers was the first one on the right—or, at least, it was the one the guys always reserved for her occasional visits. She closed herself in and immediately stripped out of her wet clothes, which fell to the floor with a plop.
Though she'd arrived five days ago, last night had been her first sleeping in this space. Her room had been in use by Nick's SF commander's son, Charlie Merritt, who'd been kidnapped by a gang and rescued by Nick's team. But yesterday, Charlie had moved into the bedroom next door—Jeremy's room. As in, with Jeremy.
The thought made her smile as she grabbed some dry clothes.
Her brother and Charlie were possibly the most adorkable couple ever. Despite Jeremy's crazy outgoing playfulness and Charlie's shyness, they just worked together in a way that made Kat's chest all warm and fuzzy. Jeremy would flirt with pretty much anyone, but she'd never seen the younger of her two brothers so obviously enamored. Come to think of it, she was the odd-woman-out among her siblings—both of whom had found someone special during the past few weeks.
Meanwhile, she was trying and failing to get rid of her ex. The one who, two months into their relationship, had gotten jealous of a colleague she'd been spending a lot of time with working on a case, including a few working dinners. Despite the fact that the colleague was happily married. At first Cole had apologized and convinced her that he just missed her, and his charm and her friends' endorsement of his character made her believe he was genuine. But by the third month of their relationship, his comments and his behavior had come off as unreasonable, insecure, even controlling . . .
The memories made her shiver. Or maybe she'd just caught a chill from the rain.
She stepped into dry panties, hooked her bra, and pulled on a soft pair of black yoga pants. Examining her arms, she confirmed she still needed the long sleeves, and tugged on a white V-neck.
God, how she hoped Beckett didn't make a big issue of her freaking out. He'd just caught her off guard—in the exact same way Cole had. And, truth be told, she was mad at herself for not reacting faster to Cole surprising her.
More worried about Cole than you're admitting to yourself, Kat?
"No," she said out loud as she tugged a brush through her hair. Sighing, she opened the door and walked right into someone. Someone who was all hard muscle. Someone who smelled like soap and spice and man.
Beckett stood like a wall, arms crossed over his chest and a stormy expression on his face.
Kat's heart raced. "What are you doing?"
"I hurt you."
"What? I didn't run into you that hard." Rubbing her nose, she retreated a step so she didn't have to crane her neck to meet his eyes. God, they were startlingly blue, the right one surrounded by a series of crisscrossing scars.
His eyebrows cranked down. "When I grabbed you."
Duh, Kat. But feeling Beckett's body up against her own, even for that split second, had short-circuited the wiring in her head. "Uh, oh. No. I'm fine, Beckett," she said, hating for him to think that her yelp had been from pain. He could irritate the crap out of her, but she felt absolutely safe around him. He would never hurt her. None of Nick's teammates would. And, unlike Cole, Beckett hadn't squeezed or dug his fingers into her flesh. Now that she really thought about it, his grip had been rather gentle. "Really."
"Nick's like a brother to me, and I'd never do anything to hurt you."
"Beck—"
"So, I didn't mean to." He shook his head and his eyes seemed to look somewhere between them, like he didn't want to meet hers. Or couldn't?
Why wouldn't he believe her? Why did he seem so upset? Almost . . . ashamed? The vulnerability in his demeanor was so unlike the Beckett she'd gotten to know over the past five days that it tugged at her chest. Made her want to make it better, any way she could. She stepped forward, patted his mountainous arm and gave him a wry smile. "If you'd hurt me, Trigger, do you really think I'd deny it?"
The corner of his lip quirked, just the littlest bit, but the humor died away again just as quick. He braced his hands against the door frame on either side of her. "I'd want you to tell me," he said in a voice so low it was almost a whisper.
Kat was suddenly sure something else was going on here. That something else was responsible for the almost haunted look in his beautiful blue eyes. But what the hell could it be? "Beckett, I'm okay," she said, willing him to believe her. "You didn't do anything. Just caught me off guard and I overreacted. Okay?"
His eyes searched hers like he might be able to read the truth in her gaze.
The odd tension between them changed . . . intensified . . . flashed white hot.
Her heart suddenly thudding against her breastbone, Kat's gaze lowered from Beckett's eyes to his mouth to his chest. He was so freaking gorgeous she could hardly stand it. The basest parts of herself just wanted to climb him.
Wait. What?
She forced her eyes back to his. And those baby blues were absolutely on fire.
"Beckett?" she said, her voice embarrassingly breathy.
But she didn't have time to think about that, because Beckett's gaze dropped to her lips, and she was suddenly, totally, one-hundred-percent sure that he was going to kiss her.
Chapter 3
Time slowed and Kat's heart raced as Beckett slowly leaned in.
By the time her mind shoved through the haze of surprise and lust to react, his lips were brushing hers.
Just a brush of skin on skin, amazingly soft and tentative. So surprising given his size.
The world froze for a long moment, but then that little bit of contact set off a flash fire in Kat's blood. And apparently Beckett's, too.
Because the kiss turned instantly and blisteringly devouring. On a groan, his tongue invaded her mouth, and she sucked him in deep. Their hands pulled one another closer and their bodies collided. Their height differentiation was so great that Kat had to push onto her tiptoes and Beckett had to lean way down. Kat wasn't sure if she pulled herself up or Beckett lifted her, but the next thing she knew her legs were wrapped around his hips and his hands gripped her ass.
They stumbled into her room and Beckett kicked the door shut behind them. Kat moaned as her back came up against the wall and his erection ground against her core.
With his tongue in her mouth and his hands roaming her body and his hips pressing maddeningly against the center of her need, Kat was possibly more overwhelmed than she'd ever been in her life. Beckett Murda was all she felt, saw, smelled, tasted. Her mind was on a repeating track of Wait . . . wait . . . omigod . . . what's happening? But her body had totally left the station.
Whatever small part of her wanted to pull back or slow down gave way to the more urgent need to let go. Let go of worrying about Cole. Let go of the fear she felt for her brothers. Let go of the horrible images she carried in her mind of the Hard Ink roof collapsing and Jeremy going down with it, which was the scariest thing she'd ever seen.
Not to mention the conversation she needed to have, the one that would force her to break confidentialities and put her job at risk.
So she did. Kat let it all go in favor of letting Beckett pull her under the waves with him.
She plowed her fingers into his hair, which was just long enough to grip and tug, and squeezed her legs around his hips, bringing them closer. Creating more of that delicious friction. He groaned low in his throat, and the sound reverberated into her belly, causing her to grind her hips forward against him.
Wait . . . wait . . . wait . . . turned into want . . . want . . . want . . .
"Jesus, want you, too," he growled. He kissed and licked at her jaw, her ear, her neck.
"Beckett," she rasped as he trailed little bites down the side of her throat. She bowed off the wall, thrusting against him. And, God, he was deliciously hard and thick between her legs.
Suddenly, she wanted to know: Just. How. Thick.
She lowered her legs to the floor and grasped his cock in her hands through the denim. His head fell back on a throaty, "Fuck."
It was all the invitation she needed. Kat sank into a crouch, her back against the wall, unzipped his jeans and roughly pulled them down over his hips. The outline of his erection filled out the dark gray boxers, and he helped her push those down, too. And then his cock was spilling free of the clothing. And, holy fuck, it was freaking magnificent. Like, work of art magnificent. Thick and long and veined in ways that made her need to tongue him.
As she took Beckett's hard length in hand, Kat spied a series of hash marks on his left hip—groups of five vertical lines with a diagonal line across them. In passing, she wondered what they were for, but now was not the time to ask. Instead, she peered up his body, meeting those blazing blue eyes as she licked him from root to tip. The abject need she saw staring down at her drove her on. She licked and stroked him until he was wet and throbbing hard. And then she sucked him in deep.
And, God, was he a mouthful. So long that even when she had his head buried in the back of her throat, she could still fit her fist around the base of his cock. But never let it be said she didn't like a challenge.
Pulling back, she gulped for air, then sucked him in deep again. She went slowly, taking a bit more each time, until Beckett unleashed a near-steady stream of curses and encouragements under his breath. "Yes, Kat. Fuck. Take it, take it."
His hands fell to her head. He stroked stray strands off her face, cupped her cheek in his big palm, and tangled his fingers into her hair to guide the pace. His hips began to move and Kat gripped onto the corded muscles of his thick thighs.
"Look at me," he rasped, drawing her gaze up his body.
And holy fucking hell, he'd never looked hotter than he did just then. Arousal drew sharp lines onto his already chiseled features. His mouth hung open. His eyes were hooded and flashing.
"Ah, fuck," he bit out. And then he pulled free of her mouth and lifted her up with his hands under her arms like she weighed nothing at all. When she was on her feet, he shoved the soft fabric of her yoga pants and panties down to her ankles in one swift motion. Boxing her body against the wall, he slipped his fingers into the slick heat between her thighs. "Drenched for me, aren't you?" he asked, lip curling up in the hint of a smile.
Pretty hard to find his arrogance annoying when the proof of what he said was all over his hand. And when the smug satisfaction of his words made her core clench around the aching emptiness she knew all too well that he could fill—that she wanted him to fill. And when the way he stroked her made her whimper and rock against his fingers.
So she nodded, and stepped the rest of the way out of her pants.
Beckett leaned down and kissed her, his fingers still moving against her core, circling her clit, and then moving deeper and finding her opening. He kicked her ankles apart until he could fit his whole hand between her legs and penetrate her fully with his thick middle finger. Kat moaned into the kiss and grasped at Beckett's shoulder, her hands fisting at the cotton of his T-shirt.
Beckett broke the kiss to fish something out of his jeans, still hanging around his thighs. His wallet. He flipped it open, reached into the billfold and pulled out a condom. Looking her in the eye, he ripped the foil wrapper open with his teeth. And then he rolled the rubber up the thick column of his cock, glancing at her to see if she was watching.
She was.
Because she was so lost in her arousal for him, in this stolen moment with him, that there wasn't a single part of her that didn't want what she knew he could give.
She pushed off the wall and took a step toward the bed.
"Where you think you're going?" Beckett growled. And then he lifted her up, her back sliding up the cool wall and her arms going round his neck, hooked his arms under her knees, and lowered her down until the head of his cock nudged her opening. Slowly, so very slowly, he penetrated her inch by goddamned thick inch.
"Omigod, omigod," Kat said as her body adjusted to the invasion. She gripped his neck tighter as she tried to take more of him.
"Take it easy, Kat. I don't wanna fucking hurt you," he gritted out.
She nailed him with a stare. "Don't snap at me when you're fucking me." Which didn't sound nearly as assertive with her voice being so breathy. But whatever.
Then he was seated all the way inside her. It was a fullness unlike anything she had ever felt. Intense and overwhelming and delicious.
"Leave it to you to get bent out of shape because I'm trying to watch out for you," he rasped.
"Bent out of shape?" She glanced down at herself, at how he had her pretty much bent in half between the wall and his big body. "Really?"
Humor crinkled the corners of Beckett's eyes, even the scarred one. "No more talking."
Kat inhaled to argue—
"We're clearly better at fucking than talking." He arched a brow.
"Moderately," she said. Though, to be honest, it was hard to affect a convincing blasé attitude while impaled on a glorious, nine-inch cock.
Beckett's eyes narrowed, and the scars around his right eye made the expression even more severe. "Moderately? As in, you think this has only been moderately good?" He withdrew his hips on a long, slow stroke, and snapped back in again.
Kat gasped at the impact, at the slick friction, at the maddening fullness that made her want to squirm and scream. It was so much better than moderately good. It was fucking phenomenal. But no way she needed to tell him that, especially not when the look in his eye told her she'd thrown a red flag in front of a bull. A stubborn-headed bull. "Yeah," she managed.
"When you're screaming my name, we'll see about moderately," he said, his voice a raw scrape.
"I thought you said no talking," she said, the words turning to a moan on another slow withdraw and hard, fast penetration. The moan made her wonder if Jeremy was in his room, and then Beckett shifted his hips again and Kat decided this felt so good she didn't even care.
Beckett chuckled. And the sound was equal parts evil and sexy. "That's right."
And then Beckett Murda unleashed himself on her. Using his grip on her ass and the massive power of his thighs, he fucked her against the wall until she couldn't think of anything but him and his cock and his heat and the warm spice of his skin in her nose. He French-kissed her until she was dizzy, penetrating her in almost every way he could. And the tight, hard, fast, sweaty friction of their bodies shoved her all-too-fast toward an orgasm she could already tell was going to blow her mind.
"Oh, God, Beckett," she gasped as her release barreled down on her. And then a moan ripped up her throat as all sensation spiraled to the pinpoint where their bodies met. Beckett covered her mouth with his as her release hit and her body convulsed. His grip tightened, providing her an anchor with which to weather the rough, almost punishing waves of her orgasm, which seemed to go on and on as Beckett's cock continued to plunge and retreat.
"That's it," he rasped against her ear, his breath unleashing a wave of shivers. "Hold on to my neck now."
Kat shook off the haze of her orgasm and laced her hands behind Beckett's neck. And then he pulled her off the wall so that he entirely supported her weight.
"Lean back. Don't worry. I have you," he said, smugness overflowing his handsome expression.
Ah, hell. Just how hard had she come? Just how loud had she screamed? Because that look told her he wasn't buying the moderate thing anymore. At least, reclining away from his body broke their eye contact so she didn't have to see the arrogance in those baby blues.
Gripping her hips hard, Beckett lifted her off his cock. And lowered her right back on. And the angle drove the blunt tip of him against a spot inside her that nearly made her scream. For the second time.
"Jesus, Beckett," she gasped.
"Hold on tight, Kat." He fucked her so deeply, so thoroughly, that all Kat could do was let her head fall back on a moan. The position didn't allow her to meet his thrusts. Instead, she was forced to hang there, receiving his cock in just the way he wanted to give it to her. About which she might've griped or complained if it hadn't been so fucking good. And exactly what she needed. Because she couldn't think. She couldn't worry. She couldn't fear.
Somehow, in yielding this bit of control to Beckett Murda, Kat found peace in the middle of a storm.
Not that she wasn't fantasizing about forcing him down to the floor, straddling him, and fucking him with the same wild abandon he was using with her, but the part of her that wanted to one-up him was muted by her complete satisfaction in every other way.
A fact that was amplified by the growing pressure inside her core from how his cock continued to stroke and nudge that sensitive spot inside her.
Lifting her head, her gaze dragged up his chest. She couldn't see his skin—they'd been in too desperate a hurry to even finish undressing—but she could make out the way his massive muscles bunched and moved under the thin material. Her gaze moved upward, finding him intently watching her. The arrogance was still there. But so was a raw hunger, evidenced in his narrowed eyes and clenched jaw.
"Come again for me, Kat," he rasped.
And something about the command and the tone of his voice and the expression he wore combined to wind her body up tight. She needed a release from the building pressure so bad she couldn't even worry about playing any sort of power game with him.
Instead, she concentrated on his cock filling and stroking her again and again, and came. This one wasn't the same sudden detonation as the first. This one was like reaching the top of the highest hill on a roller coaster, and then hanging there for a long, terrifying, expectant moment before plunging back down again. The twisting anticipation of the orgasm stole Kat's breath, and she threw her head back in a silent scream as it finally crashed into her.
"Fuck, yes," Beckett growled. He pulled her body flush up against his, held her so tightly she could barely breathe, and came at her with a series of fast, shallow, frenzied thrusts. He shouted his orgasm into her neck, his face buried in her damp hair, his thick cock pulsing over and over and over inside her.
And feeling how powerfully he shook and how much pleasure he'd received gave her a whole other kind of satisfaction.
"I'm gonna set you down," he whispered a few moments later. Kat nodded. When her feet hit the floor, her legs nearly refused to hold her weight. But Beckett was right there, and he wrapped his arms around her, banding her to his chest. "Give it a second," he said in a low voice full of gravel. His big hand stroked the hair back from her face.
Grasping at the corded muscles of his sides, Kat gave in. Really, her body didn't give her any other choice—again. She rested all of her weight against him, and he supported her like it took no effort at all. She wasn't the only one affected, though. Because his heart thundered against his breastbone.
But as her body relaxed and the force of their frenzied lust dissipated, the air between them turned less comfortable. Then downright awkward.
Because Kat had just had sex—hard, fast, up-against-a-wall sex—with one of her brother's best friends. And though Beckett fucked like a god, she wasn't the least bit sure she even liked him.
"I'M OKAY NOW," Kat said, pushing off of his body.
The loss of her heat was a sucker punch to Beckett's gut, particularly as it allowed the full reality of what he'd just done to wash over him.
As in . . . he'd just fucked his teammate's younger sister. In his teammate's house. In the middle of a friggin' mission.
The fact that it'd started with him concerned that he'd hurt her? Made it feel about a thousand times worse.
Fuck. Sometimes he was so much like his old man that it left him feeling hollow down to the bottom of his soul.
Beckett tossed the condom in a small trash can by the nightstand, then tugged up and fastened his jeans. Christ, he hadn't even bothered to get undressed. Still wore his pants, shirt, boots. Five feet away sat a perfectly good bed. But had he given her that comfort? O'course not.
If he was Nick Rixey, he would kick his own damn ass. Twice.
Kat didn't say a word—or make eye contact—as she stepped into her panties and pants. And that had . . . feelings rolling around in Beckett's chest that he didn't know what to do with.
Worry, for one. Had he been too rough? Had he hurt her even more? He thought the sex had been fucking spectacular, truth be told, but he was aggressive by nature and she was small as fuck. And that was a goddamned problematic combination.
Anxiety, for another. Now that they'd let all the sexual tension that'd been pinging between them off the leash, what would her expectations be? God, this kinda bullshit was why, no matter how fucking good being with her had felt, it had been an epic mistake. One he'd have to keep from making again.
Not to mention what the guys would say if and when they found out. Because not only was Kat Nick's sister, which crossed all kinds of damn lines, but it hadn't even been a week since Beckett had publicly dressed down Marz—his best friend in the world—for getting involved with Emilie Garza mid-mission.
No offense to the ladies in the room, but this is a mission with real shit at stake, not the dating game.
"Jesus Christ," Beckett bit out, grinding the heels of his hands into his eyes. As if that zinger hadn't been bad enough, that night his anger and his mouth had run away with him, and he'd gone on to question Marz's focus and commitment.
Now he'd gone and done this. Goddamned hypocrite. Just like his father had been.
"Well, I guess we weren't that much better at fucking either, huh?" Kat asked.
Beckett heard the soft sound of footsteps as he dropped his hands from his eyes. He followed the sound in time to see Katherine open the door and walk out into the hall. For a moment he stood there, struck stupid by her words and her actions. Where the hell had she gotten that idea?
He replayed the last minute or two. Oh. Oh. He'd totally given it to her, hadn't he?
Part of him yearned to rush after her and explain that he hadn't exactly aimed his curse at her. The expletive and frustration had been more about the situation in general, and the hell he knew he was going to have to pay. But Beckett stayed right where he stood. Because it was probably better all the way around if Kat had one more really damn good reason to keep her distance from him, to dislike him, to push him away.
Because the two of them being together? Wasn't good for the team, their mission, or Katherine herself.
Chapter 4
"What the hell took you so long?" Marz yelled as Beckett returned to the gym. "You offered to grab the ibuprofen almost a half hour ago."
Pretty much the whole team turned to look at Beckett. Marz sat at his computer. Charlie sat beside him in what had become his usual chair—since Merritt's son was every bit as brilliant as Marz with computers, the two of them had been teaming up. Jeremy stood behind Charlie's chair, near Charlie as always. Nick and Shane sat backward on a pair of folding chairs, close enough to Marz to see what he was working on. Becca and the redheaded Dean sisters, Sara and Jenna, sat at the new tables of computers, probably reviewing documents off the microchip.
Now, all eight pairs of eyes were looking at him. For fuck's sake.
Beckett bit back his irritation and held up the bottle, making it clear he hadn't forgotten. "Phone call. Sorry," he said gruffly.
Christ, he could still smell the warm sweetness of Kat's skin on his clothes.
When he closed in on Marz's desk, Beckett tossed him the bottle.
"Come to Papa," Marz said as he popped the lid and dumped four little red pills into his hand.
"Where's Easy?" Beckett asked, referring to the fifth of the five surviving members of their SF team. He waited while Marz chugged back a gulp of water with the meds.
"With Emilie," Shane said, raking his hand through his carefully messy dark-blond hair.
Beckett nodded, glad to hear that his teammate was keeping up with the therapy. Emilie brought an important skill set to Hard Ink—she was a psychotherapist. And given that Sara, Jenna, and Charlie had all survived abduction and assault, and that Easy had admitted a little over a week ago to being depressed and having suicidal thoughts, it turned out that Emilie was a real asset.
Because they all carried more than a little dose of fucked-up around with them. Beckett sure knew that he did.
"Everything okay?" Nick asked, looking over his shoulder. Beckett met the guy's odd, pale-green eyes, and all he could think about was what he'd just done with Nick's sister—who, unlike her two brothers, had brilliant jade-green eyes that pierced him and challenged him and taunted him. "Beckett?"
"What?" he said, snapping out of the totally fucking useless train of thought.
"Dude," Marz said, arching an eyebrow. "What's up with you?"
Beckett crossed his arms. "Why the hell would anything be up?"
Nick and Shane exchanged a look.
And this . . . all of this . . . was exactly why Beckett had always kept his eyes on the fucking prize during a mission. And why Katherine Rixey pissed him off so damn bad. Because she had him tangled up inside in a goddamned knot. And he didn't understand it. Or like it. Not one bit.
Beckett scrubbed his hands over his face. "Sorry. Just had, uh, a bit of a situation. Under control now." Lying to these guys sat like a rock in his gut. As did calling what happened with Kat a "situation."
"No worries, man," Nick said, unknowingly pouring more salt in the wound with his understanding.
Enough. Pull your head out of your ass, Murda.
"What are you working on?" Beckett asked, gesturing to the desk. He came up behind Nick and Shane to look at Marz's computer monitor, but not too close—because he could still smell Kat all over him. The log of evidence from the documents on the microchip filled the screen.
"Well," Marz said, leaning back in his chair and stretching out his legs. "We were brainstorming ways to definitively prove whether Seneka and this WCE were one and the same entity."
"And?" Beckett said, already more comfortable now that the focus was off him and on the job.
"Unless there's a smoking gun in these documents," Nick said, peering over his shoulder, "the ideal way of proving it is probably the hardest of all."
Beckett frowned. "The ideal way being . . ."
"If we had access to their files via their servers," Marz said. "You know, if wishes fell from the sky . . . Problem being—"
"They're probably more secure than the government itself?" Beckett said.
Marz nodded, and Charlie mirrored the movement behind him. Though the younger Merritt's blond hair was long enough that he sometimes pulled it into a knot on the back of his head, the family resemblance with Beckett's commander was clear.
"That about sums it up," Marz said.
"And, if Seneka is the same thing as WCE," Charlie said in a quiet voice, "they're gonna know we're coming before we even knock on their virtual door. Pretty sure that's how they found me a few weeks ago." Charlie had been kidnapped by the Church Gang after trying to learn who WCE was and why they'd deposited twelve million dollars into a Singapore bank account with his father's name on it and Charlie's address. "Since I already tried and failed to hack in then, no doubt they've doubled their defenses."
Across the room, the gym door opened and closed. Beckett didn't even need to look to know it was Katherine. As if his body recognized the presence of hers, all the tension he'd managed to let go came flooding back.
Just stay focused on the work.
So Beckett didn't let himself look her way as she crossed the room to join them, even though a part of him desperately wanted to make eye contact. Because all the other bullshit aside, he'd never been more sincere in his life than when he'd told her he hadn't meant to hurt her, and would never want to, either.
Being on the receiving end of abuse had made Beckett hyperfearful of becoming an abuser himself. That's why he got the hash-mark tattoos. If he was gonna take a life, even in the line of duty, he was going to bleed for it. Every damn time.
Right now, Beckett was four marks shy of what he deserved—he'd taken the lives of four thugs during the mission to rescue Emilie from a carjacker. He'd have to rectify that soon. Maybe Jeremy would do it for him when they had a few free minutes.
"Hey," Kat said as she came up on the other side of Marz's desk, putting herself right in Beckett's line of sight. A round of greetings rose up.
"Hey," Nick said. "We're trying to brainstorm something here. Maybe you can help."
"Uh, sure," she said, crossing her arms and hugging herself. She'd changed since Beckett had last seen her, and now wore a pair of jeans and a soft-looking black sweater. Her hair had nearly air-dried, and loose curls spilled over her shoulders.
"What if we turned this around?" Charlie asked. "Instead of thinking about how to break into Seneka's files, what if we thought about what kinds of information we think would be useful, and see if there isn't another way to go about getting it?"
Katherine pressed her fingers to her temples, and Beckett frowned. As a lawyer, no doubt conversations about how to accomplish less-than-legal objectives pushed her way outside her comfort zone.
Marz nodded. "Okay. Phone records and phone taps, for one."
Nick held out his hands. "Vance. He might be able to finagle those like he did the container logs from the marine terminal." Kyler Vance was a Baltimore police detective who declared himself an ally to their group after his godfather, Miguel Olivero, had been gunned down on the street outside Hard Ink. Beckett hadn't known Olivero well, but everything he did know said they'd lost a damn fine man in addition to an important ally. Vance was already making good on his word to stand by them. And thank God for that.
"Bank account information," Charlie said. "Right? We know WCE put money into my, uh, father's bank account. But if we could find the other side of those transactions—the payments—that might nail it down."
Beckett thought that might be the first time he'd heard Charlie refer to his father as, well, his father. For the entire time he'd known the younger Merritt, Charlie had called him the "Colonel." There'd apparently been some bad blood between father and son over Charlie's sexuality, at least in part. Beckett glanced between Charlie and Jeremy and frowned, despite the comical T-shirts both of them wore courtesy of Jeremy's infamous collection. Jeremy's read, Make Awkward Sexual Advances, Not War, while Charlie's read, I'm a Ninja in black text on a black shirt.
All these people had become a sort of family to him, maybe because he didn't have much of his own—none that he wanted to claim, anyway. So he felt protective about every damn person in the room, and couldn't deny that it rankled him to know that Merritt probably wouldn't have approved of the guys' relationship.
"Charlie, this is a good fucking idea." Nick grabbed a pen and yellow legal pad off the mess on Marz's desk and began a list. "Finding the depositor accounts is genius."
"You know, I have this newfangled list maker called a computer right here," Marz said. Beckett smirked. He and Marz were the biggest techies in the group. One of the things they'd always bonded around.
Flipping him the finger with the hand steadying the pad, Nick grinned. "Writing it out long-hand helps me think."
"Oookay," Marz said.
"What about Seneka personnel files?" Shane said, a bit of a southern accent permeating his words. He tapped his finger on Nick's legal pad. "I know this one's not likely, given the secrecy around Seneka's recruitment and the need to maintain in-country covers. But let's just say we could get ahold of a list of their employees. What if one of them had the initials WCE? Long shot, I know, but . . ." He shrugged.
"Or the initials GW," Nick said. "Remember Merritt listed GW for his contact?" With everyone agreeing, he scribbled down the idea.
Katherine paced around to the back of the table and came to a rest against her younger brother, who put his arm around her.
"You all right?" Jeremy asked.
She nodded and gave him a smile, but it appeared strained or reserved. Then again, maybe Beckett was reading too much into it. It wasn't like she directed that many smiles his way, and when she did, they were usually in the midst of some snarky exchange.
"What else?" Nick asked, surveying the group.
"Private cell phone numbers with call histories and text exchanges," Beckett said. "E-mail exchanges, for that matter."
Nodding, Nick added it.
"How do you propose to get this kind of information?" Kat asked.
"We're not there yet," Nick said. "Just brainstorming the kinds of information that, should it fall in our laps, might be helpful to proving what we need proved. But, look, I know this whole conversation puts you in an awkward place, so maybe I shouldn't have asked you to help . . ."
"There's no way you can get any of this legally," she said. Despite what she'd said, her tone hadn't been critical. More musing, maybe?
"Really, you don't have to be here," Nick said, tapping his pencil against the pad.
"No, no. It's just, you guys are taking a lot of risks in all this, aren't you?" she asked.
"Can't be helped," Nick said, his tone shorter and his brow cranked down. Beckett had seen that expression often enough to know Nick was losing his patience.
Kat frowned and stepped out from under Jeremy's arm. "I'm not trying to criticize—"
"Then what's your point, Kat? Because we're trying to keep anybody else from dying here," Nick said, rising to his feet and crossing his arms. The folding chair screeched against the concrete floor.
Beckett's protective instinct roared to life, and he grabbed Nick's arm and hauled him back a step. Nick glared at him and pulled free, turning back toward his sister.
Kat didn't look the least bit cowed. Instead, she marched up to her much-taller brother and braced her hands on her hips. "Wow. Stand down, soldier boy. What is it with you guys and barking at me today? Especially when all I wanted to say is that I think I can help you. I think I have access to some of what you need."
KAT FELT LIKE she'd just jumped off a high cliff and hadn't yet hit the bottom. Because she'd opened the door to sharing information and documents with her brother and his team, even if he was being an ass right now. Or maybe she was just overly sensitive. After all, she had to admit that she still felt a little off-kilter after her encounter with Beckett—both the mind-blowing sex and obvious regret immediately afterward.
Really, she shouldn't have been surprised by the latter. And to the extent she was, she kinda wanted to smack herself. Given their usual M.O., why had she expected anything else from the guy? And why the hell did he look about a thousand times more attractive to her stupid eyes now that she knew what his face looked like when she sucked his cock and what he sounded like upon orgasm?
"What?" Nick asked. "What are you saying?"
Nick's words pulled her out of her thoughts. "Exactly what it sounds like. I'm saying I can help you." She sighed and tucked her hair behind her ears, and then let the words fly that would commit her to their mission, and to a whole new path. "My office at Justice has been investigating Seneka Worldwide Security for almost a year. I have access to some of the exact files you're discussing."
Nick's pale green eyes went wide. The room went still and silent. And then all hell broke loose.
"You can't share that," Nick said.
Jeremy was almost instantly at her side, his face set in an uncharacteristic frown. "Dude, you could totally get in trouble. You can't do that."
"Holy shit," Marz said, rising to his feet. His chair scraped on the floor, sending their little three-legged German shepherd puppy, Eileen, scurrying out from underneath and growling in fright.
And though Beckett didn't say anything, tension roared off of him as a storm slid in across his face.
Even the silence from the workstations where Becca, Sara, and Jenna sat seemed loud. Kat glanced their way and found Nick's, Shane's, and Easy's girlfriends wearing matching concerned and sympathetic expressions. And she totally got why. The five women in the house, including Emilie, had talked more than once during the days since Kat had arrived about wanting to be able to help out in this whole mess. Becca gave her a little nod of encouragement.
Kat took a deep breath and braced for a fight. She knew Nick wasn't going to let her do this easily. He wouldn't be Nick if he wasn't crazy overprotective. "Nick—"
"No, Kat. And that's final."
He'd already lost this fight. He just didn't realize it. Kat crossed her arms. "Really?"
Shane rose slowly to his feet. He'd been the only one not to have an obvious reaction to what she'd said. "I think you should listen to what she has to say, Nick." Shane's gray eyes cut to her. "Why is Justice investigating Seneka, anyway?"
Nodding in his direction, Kat said, "Thank you." She was especially appreciative because Nick and Shane were best friends, and though Kat didn't know exactly what had happened between them, Becca told her that they'd recently patched up some old misunderstandings. She could only imagine that taking the opposite position from Nick wasn't easy for Shane. "In a nutshell, Justice is investigating Seneka for contract fraud, bribery of Afghan officials, and a shooting incident in which nine unarmed Afghan civilians were killed."
"Jesus," Shane said. "Prosecuting something like that must be a bitch."
Kat nodded. "You could say that." This case was a perfect example of why it was sometimes hard to fight the good fight. The difficulties of obtaining and preserving evidence in war zones and of gaining proper jurisdiction for prosecutions in American civilian courts were a few factors responsible for dragging this case out.
Shane exchanged glances with some of the other men, and then looked back to Nick. "With charges like that, her files will definitely have some of what we need. You have to hear her—"
"I heard her plain and clear," Nick said. "And the answer's still no."
"You realize I don't actually need your permission to do this, right?" Kat said. "You need something I have the ability to give. It's as simple as that."
Nick threw out his arms and dug in his heels. "This isn't even your fight, Katherine."
Oh, she knew she was in trouble when he broke out her full name. But no way was she letting that BS stand unchallenged. "Are you freaking kidding me? I was up on that roof, remember?" She pointed at Jeremy. "Why is it Jer's fight but not mine?"
"You better fucking believe I remember what happened up on that roof. Thank you for making my point for me. In spades. And Jeremy's different," Nick said, raking at his dark brown hair and pacing. His hair wasn't long, but it was longer than Kat had seen since before he'd gone into the Army.
"Oh, this should be good. Please tell me how Jeremy being involved is different from me getting involved. And for the love of God, please tell me it doesn't have anything to do with his penis." She crossed her arms and cocked her head. Then waited.
Nick glared as his shoulders rose and fell in a troubled sigh. "Gimme a little frickin' credit, wouldya? I don't want him in the middle of this any more than I want you to be, but I brought the trouble to his door, which didn't give him a whole helluva lot of choice."
Jeremy held his hands palms up, highlighting the ink that ran down both arms. "That's bullshit, Nick. You and I had a whole fight about this back at the very beginning, remember? I demanded in."
"I thought you were on my side in this," Nick said. "She's out, and that's final." He slashed a hand through the air to end the debate.
Arching a brow, Jeremy crossed his arms and stepped closer to Kat, so it was like the two of them were facing off against Nick. At least she'd won one brother over. "This is all feeling very familiar," Jer said.
Kat glanced between them, not sure what they were talking about.
"She's right," Becca said, rising from the table and coming up to Nick. Her pale yellow shirt almost matched the color of her hair. "I was there the night you and Jer got into it. Remember?" She took Nick's face in her hands.
Nick grasped Becca's hand, pressed a kiss to her palm, and shook his head. "Not the same, Sunshine. It's just not."
Kat watched them together for a moment, once again grateful for Becca's presence in Nick's life. When she was around, he seemed to unwind by several kajillion stress levels.
"Nick," Kat said. "I want to help, and I can help."
"But at what cost?" Beckett said. They were the first words he'd uttered during the whole exchange.
Kat met his gaze, but couldn't read the emotions behind the words. His mask was firmly back in place, now, wasn't it?
Nick threw a hand out toward Beckett. "Thank you," he said, in a tone like he'd been vindicated.
Taking a deep breath, Kat shoved her anger and frustration back. She and Nick both had tempers, so losing her cool wasn't going to help resolve this. And no way was she letting Beckett get underneath her skin in this moment. This whole situation was too damn important. "At the same cost everyone else in this room is willing to pay," she said, in response to Beckett. "I'm no different."
Stepping closer, Nick shook his head again. His tone was gentler when he spoke. "Except you are different, Kat. You have a fuck-ton to lose. Things you could never get back, like your license to practice law. You worked damn hard for that. And it's an important job. One that makes a difference in the world. I lost that. But I'm not letting you lose it, too."
"I appreciate the concern, Nick. But some things are more important than a job. Don't you think I've thought this through? Considered the risks? I've worked through all of this, and I want to help."
Nick sighed. "Kat—"
"Jeremy almost died," she said, her throat unexpectedly closing up on the last word. She sucked back a sob that came out of nowhere and pressed a fist to her mouth. Tears filled her eyes. "And you last year," she said in a thin voice. Over Nick's shoulder, she made eye contact with Beckett, whose blue eyes were blazing at her. She was way too vulnerable right now to figure that out, though, so she dropped her gaze to the floor between her and her brother.
A hand slipped into hers and squeezed. She glanced to her side to find Jeremy peering down at her with a strained expression.
Without a second's hesitation, Kat turned, pressed up onto her tiptoes and threw her arms around Jeremy's neck. "We could've lost you," she whispered, tears sneaking down her cheek. Talk about a delayed reaction. The roof collapse had been three mornings ago, but a wave of fear and grief suddenly washed over her.
Jeremy's arms came around her back. "Hey, it's okay. I'm okay."
A fast shake of her head. "I know." She batted away her tears. "I just . . . I can't lose you guys." Still leaning against Jeremy, she turned toward Nick and nailed him with a stare. "I can't lose you. And if I can offer something that makes this even a little less dangerous, that gives you even the smallest advantage, you have to let me. Do you hear me? You have to. Because if you don't and something happened . . ." She swallowed the knot forming in her throat again. "I would never be able to forgive myself."
The muscles in Nick's jaw ticked. "Which is how I would feel if something happened to you. To either of you."
"I know, and I get it," Kat said, crossing to stand in front of Nick, who towered over her. Well, pretty much everyone did. "But the reality is, Jeremy and I are already in it. How long do you think it takes Seneka to discover you have a sister who lives alone in D.C.?"
Nick's eyelids sank closed like the thought had physically hurt him, but Kat wasn't above playing dirty to get her way. Not on this.
"Sonofabitch," Beckett bit out under his breath.
Glancing around, she met the gazes of each of the guys who stood all around them. The conflict they felt was clear in their expressions, so Beckett wasn't alone in his sentiment. She turned back to Nick. "You need me, Nick. And I need you to want me, too."
"Goddamnit, Kat," he said, his voice way softer than the words he uttered.
"Is that a yes?" she said, giving him a small smile.
"I agree to this and you do exactly what I tell you. No questions asked. No busting my balls. No going off on your own," Nick said, glaring down at her. But she could see the concern behind the scowl.
"Of course," she said, his agreement taking a load off her chest, making it incrementally easier to breathe. "I'll try to keep the ball busting to a minimum."
Beside Nick, Shane chuckled. "I see the skill of persuasion runs in the family. She's as hard to say no to as you are." He clapped Nick on the back.
Nick shoved his arm away, feigning annoyance. "Fuck off, McCallan."
Which made everyone chuckle.
"Fuck all y'all," Nick said. "Except you, Sunshine. C'mere." He pulled her into his arms, one of which bore a new tattoo of a sun and Becca's initial, and gave her a quick hug.
Becca laughed and patted him on the back.
Then Nick was standing in front of Kat again, bending down so he could look at her eye-to-eye. "Thank you. I don't like it, but I get it. So, thank you."
She nodded. "You're welcome. But don't thank me until we know for sure that I've actually helped."
Chapter 5
Kat returned to the scene of the crime, otherwise known as her bedroom, and tried not to stare at the wall against which she'd earlier had the best sex of her life.
Focus, Kat.
Right. She'd come back here for a reason. In order to access her systems at work, she needed the randomly generated digital code from her SecurID token. It changed every sixty seconds and was one of two steps required to access files on the virtual private network at Justice. The small, key-fob-type device sat atop her closed work laptop on the nightstand next to the bed. The time she'd taken off work this week hadn't been planned, so she'd stayed up late last night writing a brief—the perfect alibi, should it become necessary, for why she'd be accessing these files.
As voices approached from the main part of the apartment, Kat grabbed the token and her computer. Marz had asked her to bring the machine over to the gym before she logged in so he could look over and beef up its security features.
"Hey," Jeremy said from the doorway. Charlie hung in the hallway behind him.
"Hey," she said. "What are you guys doing?"
Jeremy shrugged. "Just hanging out. Some of the Ravens are back from their shifts out at the roadblocks, so dinner madness is getting underway."
Kat nodded. "Okay."
Leaning against the doorjamb, Jeremy crossed his arms and gave her a funny look. "Question for ya."
"Yeah?"
"Which Raven is it?" he asked, humor playing around his mouth.
"Which Raven is what?" Kat hugged the laptop to her chest.
He tilted his head and lifted his brows like she should know what he was asking. "Which Raven is it I smelled on you?"
Heat crawled up Kat's neck. She chuckled and pushed by him into the hall, giving him a small shove for good measure. "What the hell are you talking about, Jeremy?"
"When you hugged me, you smelled like man and sex. Now, I love both men and sex—" He winked at Charlie, who looked a lot like he'd rather be anywhere else in the world just then. "—so I don't have a problem with it at all. Just curious who the lucky Raven was."
Holy shit! She thought about saying there'd been no Raven. Since that was, in fact, the truth, and could thereby lead her brother to wonder who else that left, Kat bit her tongue. Leave it to freaking Jeremy to be able to sniff out sex, for God's sake. She walked backward away from him. "You're an idiot. You know that, right?"
"That wasn't a denial," he called, amusement plain in his tone.
Kat flipped him the finger and kept on walking, though she was under no illusion that he'd let this go. Why were brothers such pains in the asses anyway?
In the kitchen, a dozen or more people crowded around the breakfast bar—Ravens and Hard Ink folks alike—talking and laughing and chowing down on chili that Becca and Emilie had made earlier. With so many people working and living here now, they'd all been pitching in with the nonstop production of food, but Becca and Emilie were by far the best cooks among them. Kat couldn't wait until it was Nick's turn to make dinner; she'd already put in an order for sloppy Joes—his specialty.
Nick saw her through the crowd and winked. She gave him a small smile, glad they'd been able to come to an agreement about her helping.
Not seeing Marz, Kat made for the gym and found him in his usual spot. "Do you want to go eat with the others before we do this?" she asked.
He took the laptop from her hands. "Thanks, but I'm gonna wait for Emilie to come back down so we can eat together."
Kat smiled. "Aw, aren't you sweet?" She put a little sass in the words, but it was true. Marz was a total sweetheart, a good guy through and through.
He patted his hand against his chest and smirked. "Heart of gold, baby. Heart. Of. Gold."
Laughing, Kat nodded. "Yeah, I know. All you guys do."
Marz indicated for her to enter the password to her laptop, and then he said, "Yep. Even Beckett."
She stepped back from the computer and gave him a frown. Why the hell had he said that? "Yeah, sure," she said as nonchalantly as she could. Because, honestly, while she would give Beckett the benefit of the doubt on being a good person—he was here helping Nick, after all—she couldn't begin to figure the guy out.
Eyes on her screen and fingers flying over the keyboard, Marz added, "He's rough around the edges, but he's a good guy."
Crossing her arms, she very specifically tried not to think of all the ways in which Beckett could be rough. "Uh-huh." What the hell is this about?
"And he really wouldn't have hurt you that day he pulled the gun on you."
She sorta wanted to bang her head against the nearest hard surface. Why did she feel like she was getting a sales pitch? "I know." When he didn't respond, her shoulders relaxed.
Maybe two minutes of silence had passed before Marz looked over his shoulder at her and said, "Beckett just takes a while to let anybody get close, ya know?"
"Derek," Kat said, exasperated. "Why the heck are we talking about Beckett right now?"
He shrugged and turned back to the laptop. "Just sayin'."
She glared at the back of his head. First Jeremy and now this. Question was, what the heck did Marz think he knew and how did he know it? Had Beckett said something to him about what'd happened between them? Because if he had, she was going to pull his tongue out of his mouth and tie it in a knot, cartoon style.
"This machine is kind of a mess," he said. "Typical government piece of crap." Marz turned and smiled at her. "Whenever our guns jammed in Afghanistan, we'd always remind each other that our weapons had been brought to us by the lowest bidder. Kind of a random memory . . ."
The musing look on his face made Kat smile. "Can't imagine what it was like over there for you guys."
He clicked through a few more windows. "Long periods of boredom punctuated by moments of intense, balls-to-the-wall crisis."
"Oh, so, sorta like now, then?" She'd meant it as a joke, but there was a truth to the statement that wasn't one bit funny. Not at all.
"Roger that," Marz said. "You know what, before you do any more logging on, I want to clean this up and build in some new security features. That okay with you?"
Across the room, Emilie, Easy, and Jenna stepped through the door.
"Of course," Kat said. "Just grab me when you're ready."
Marz winked. "Not sure how Nick would feel about me grabbing you, but I'll definitely let you know."
Kat rolled her eyes and hoped like hell that Marz didn't notice the heat crawling into her cheeks—or read into it. Because she'd been grabbed pretty damn good today. And, no, Nick probably wouldn't be too happy to know it. Not that he got a freaking vote about who did or didn't grab her, for crap's sake.
"You ready for me?" Emilie asked with a big smile as she approached the desk.
The grin Marz gave her nearly had Kat blushing even more. "What kind of question is that? Always." Marz rose to his feet, sending Eileen scurrying again. "Oh, damn. Sorry, Eileen. She keeps sleeping against the shoe on my prosthesis and I can't feel her."
Kat scooped the silly-looking dog into her arms and tucked her against her side. The puppy's ears were so big they were out of proportion to the rest of her body. "She just wants to be near you, don't you, sweet girl." Eileen's huge tongue swiped across her cheek, making Kat laugh. "That's just about enough of that, you mutt."
Marz clasped hands with Easy across the desk. "Hey, man. How you doing?"
Tall, athletic, and dark-skinned, Easy was probably the guy on Nick's team with whom Kat had talked the least. He seemed very reserved, even a little distant—with everyone except for Jenna, that was. Easy's gaze flicked toward Kat, but he gave Marz a nod. "Day by day."
"True dat." Marz rubbed his stomach. "Okay, food. Food is a priority."
Jenna chuckled. "I've heard that about you." With her bright blue eyes and dark red hair, Jenna was absolutely beautiful, even with the fading yellowish bruises surrounding her left eye. And seeing that reminder of what her brother's enemies had done to this young woman further reaffirmed Kat's decision to help Nick and the team. Every single person here had paid something or risked something or was contributing something to try to make this whole situation right. Kat was determined to do so, too.
Marz held out his hands. "Can't be helped. My reputation precedes me."
The five of them made their way across the gym, and Kat felt more than a little like the odd woman out. Both couples were hand in hand, catching up with each other about totally normal things after being separated for a bit, while she trailed behind, Eileen still in her arms.
Out in the stairwell, Marz said, "You know what? I better let Eileen out before we settle in with dinner."
Kat waved him off. "I'll do it. Go ahead and I'll catch up." Given that she'd gone right from being with Beckett to arguing with Nick to losing it over what'd nearly happened to Jeremy, she wouldn't mind the alone time to pull herself together.
Outside, she put Eileen down on the gravel of the fenced-in parking lot and watched as the puppy meandered around, sniffing every little thing, until she made it over to the grass that ran around the outside of the lot. The rain had tapered off to a foggy mist that, together with the dark clouds, made it look even later than it was.
Leaning against the brick wall of the building, Kat rubbed her eyes and yawned. She'd been here less than a week and was already exhausted. She couldn't imagine how the rest of them had been handling this situation for going on a month now. Even when you had downtime, you could never completely relax. Not when men with rocket launchers attacked your home in the middle of the night. For one.
Sighing, she dropped her hands. And saw Beckett getting out of his SUV on the other side of the parking lot. For a long moment she studied his gait. He walked with a small limp that seemed to favor his left leg. She knew it resulted from a grenade explosion in Afghanistan, but it didn't seem to slow him down any, that was for sure.
He turned her way. Great.
Kat clapped her hands together. "Eileen? Come here, girl," she called. We have to go in now! So I can avoid what's sure to be an awkward conversation! Because I don't need another one of those today! Although, maybe this was the perfect time to see what Beckett had told Marz. Still, hello awkward.
Near the line of motorcycles, Eileen's head popped up and the puppy looked Kat's way. But then she went right back to whatever she was doing.
Not that it mattered now that Beckett was halfway across the parking lot. Coming her way. And, just for the record, looking hot as fuck in the tee-jeans-boot combo. With his broad shoulders and thick thighs, the man filled out his clothing in a way that left little to the imagination. Which she didn't really have to rely on now, did she? She'd seen the goods and knew that nice as those jeans looked on him, they looked about a million times hotter hanging around his knees.
Which was not a helpful thought right now.
And then he lifted his eyes and pinned her to the wall with that intense blue gaze.
So much for hoping he might just ignore her and go inside. Because he wasn't heading for the door. He was heading right for her.
Her heart tripped into a sprint and she shivered as her memory treated her to a quick series of images from the last time he'd held her pinned against a wall. Taking a slow and hopefully calming breath, she met his gaze and waited until he came to a stop in front of her. Close enough that if she reached out a hand, she could grab his shirt and haul him all the way to her. Low, low in her belly, her muscles clenched in complete support of that idea. Freaking traitorous body.
"I want to say something, if that's okay with you," he said, his eyes searching hers.
His seriousness made her bite back the snark that sat on the tip of her tongue. "Sure," she said, not having the first idea what to expect.
"What you offered to do in there . . . it's a dangerous thing, Kat."
Annnd, of course. Here we go. Frustrated, she pressed her hand to her eyes. The last thing she had energy for was another fight. "Beckett, this has been settled—"
"Wait." His fingers curled around hers and pulled her hand away from her face. Still holding it, he stepped closer. "I'm fucking this up."
It was almost like they spoke mutually unintelligible languages, for as much as she understood this guy. "Fucking what up?"
He shook his head and let out a harsh breath. "Thank you. What I wanted to say is thank you. You don't have to do this, and not a single person would blame you if you changed your mind."
"But I'm not gonna—"
"I know." He dragged his thumb across her knuckles. Back and forth, back and forth. And the heat of his touch spread through her whole body, making her nerve endings come to life and yearn for more . . . everywhere. "And that's why I'm thanking you. We need what you have." Beckett cleared his throat and looked off to the side for a long moment. When he looked back to her, he shrugged. "Yeah, so, that's what I wanted to say."
He released her hand and shoved his into his pockets.
Well. Okay, then. "I appreciate that," she said.
He tilted his head and nailed her with that intense blue stare. "Just be smart about it. All right?"
Kat rolled her eyes. He couldn't stop while he was ahead, could he? "Gee, thanks for the vote of confidence. I might not have the same expertise all of you do in covert operations, but I'm not an idiot, Beckett. Like I would be anything but smart knowing people's safety is on the line."
She threw her hands out and shook her head. Ugh. Why bother? Without waiting for him to respond, she made for the door, stabbed her finger against the buttons on the pass-code box, and grabbed the handle. Then something else came to mind. Might as well get all the fun stuff out on the table, right?
She turned back to him, not surprised to find him scowling at her. "While we're having this awesome conversation, let me add a request for you to please keep what happened earlier quiet. Not sure why Derek felt the need to be matchmaking a while ago, but I'm really hoping it's not because you said something. 'Kay? Thanks."
Stepping inside, she remembered Eileen.
"Oh, and bring in the damn dog."
FUCK THIS, BECKETT thought. He followed Kat inside and made a mental note to come back for the dog later. He hadn't said one goddamn word about what they'd done. To anyone, let alone Marz. For Christ's sake. As if he wanted anyone else to know.
No way he was letting that accusation stand unaddressed.
He stalked inside in time to see the door to the first-floor tattoo shop clicking shut. The stairwell remained otherwise quiet, so his gut said Kat had retreated to the quiet of the now-closed Hard Ink Tattoo. Perfect.
Beckett crossed the stairwell, his boot steps echoing inside the concrete and metal. The second he rounded the corner into the mostly dark lounge at the back of the shop, his gaze landed on Kat pacing in the middle of the square space.
Backlit by a security light in the hall behind her, she spun on her heel to face him, her resentment palpable even though he couldn't make out her expression. She didn't want him here, didn't want to be around him, probably wanted nothing to do with him at all. And, boy, if those feelings directed his way weren't really fucking familiar. Story of his goddamned life. Until the Army, at least.
Didn't have that anymore, though, did he?
"Beckett," Kat said, her tone part groan and part plea.
He planted his feet and fought against the ancient gut reaction to flee from where he wasn't wanted. Because, damn it all to hell, he'd perfected the ability to hear the I don't fucking want you around message from a mile away. He'd grown up on that shit. But this was Nick Rixey's sister. And out of respect for the man he'd fought and bled with for many years, he wasn't leaving Katherine with the impression that he'd violated her trust or her privacy. "Just hear me out, and then I'll leave you alone. Okay?"
"What more could you want to say?"
Part of him wanted to approach her so he could see her expression and look into those brilliant green eyes. But she stayed where she was, so he followed her cue and did the same. "I didn't say anything. What happened, that was just between me and you. I wouldn't disrespect you by sharing something so private. And I'm not some eighteen-year-old kid who can't keep his mouth shut." He reined in his irritation, because giving voice to it wasn't going to help a damn thing.
Kat paced a few steps closer, bringing her beautiful face out of the shadows. "Okay," she said. "Thank you for that." She sighed and paced a bit closer yet. Beckett tracked her movement like a lion tracked prey. The closer she came, the more something deep inside him wanted to pounce. For fuck sake. "Look, I'm sorry for taking your head off out there. Sometimes I just need some alone time to get my shit together, and that's hard as hell to get around here."
The sentiment sounded so much like something he might say that Beckett actually chuckled. "Damn if that isn't true."
Kat gave a little laugh, too.
And the sound of it reached inside him and . . . unsettled him. He liked it, that was for sure. He liked giving her a reason to smile or laugh. It made him feel like . . . like he had some value. Like he wasn't just a giant pain in the ass. "You never seem like you don't have your shit together, though."
It was true. Kat was tough, feisty, confident, competent. According to Nick, she hadn't hesitated to grab a weapon and follow the guys up onto the roof to defend against the attack that destroyed part of the building. She hadn't even needed to be asked. And she'd stepped up immediately to take shifts in the snipers' roosts. On top of it all, she was a highly educated lawyer willing to put it all on the line to protect her family. He wasn't sure he'd ever met a woman as badass as Katherine Rixey.
"Well, I guess that's good to know. I'm just . . ." She shrugged and ducked her chin.
"What?"
She shook her head. "Nothing . . ."
Beckett wasn't having that. He needed to know what she'd almost said. Because this was the most real conversation they'd ever had. No snark, no anger, no audience. And after what had happened between them this afternoon, that seemed to matter. So he closed the distance between them and tilted her chin up with his fingers. Her mouth dropped open and her eyes widened at the touch, and damn if those little physical reactions didn't reverberate through his blood. "What?"
For a long moment Beckett didn't think she'd answer, and then the words started pouring out of her. "Have you ever felt . . . just . . . angry? And not known what to do with it? How to direct it? How to get rid of it?" She raked her hand through the top of her hair, causing it to cascade around her shoulders. "I'm freaking pissed that this—all of this—happened to Nick. And all of you. It's so unfair. And that Jeremy's in such danger. I mean, Beckett, if you could've seen that roof fall out from underneath him—" Her voice went thin and her eyes went bleak. She shook her head and visibly reined in her emotions. "If I hadn't happened to drop in for a visit on Friday, I wouldn't have known about any of it. Because Nick would've kept me in the dark. I mean, shit, I should almost thank Cole for jump—" She swallowed the words like she'd said more than she wanted, then threw up her hands and spun away, agitation rolling off her.
Beckett frowned as his instincts jangled. What the hell had she been about to say? And why did her mention of another man's name feel like such shit? "Who's Cole?" he asked, watching her closely.
She just shook her head.
But, for reasons he couldn't quite pinpoint, the answer felt important. He crossed to where she'd paced and came up behind her. "Who's Cole?"
Waving a hand, she shook her head again. "Just a guy. Doesn't matter."
The more she deflected, the more he was sure it did matter. Beckett replayed her words in his head and tried to make sense of them. I should almost thank Cole for jump— Jump? Jumping to conclusions? What?
She turned to face him, giving him a smile he didn't really believe. "See? My shit is so not together."
He sensed that she was shuttering up, and it was the last thing he wanted. Because the stuff she was saying . . . it resonated with him. Did he know what it felt like to be angry? Uh, yeah. And then some. It was quite possible that Angry was his middle fucking name. Had been for years. So, yeah, he got it.
"The anger," he said, hoping to keep her from clamming up. "I get it. The key is not to let it get the best of you, but use it. Let it build a fire inside of you. You'll need it to get through this clusterfuck. To get you through the exhaustion and the stress and the worry. Way more powerful than fear." Digging into that dark, restless energy inside him was how Beckett had powered through the most intense moments of his SF career. The only good thing his father had ever done for him was prepare him for how to endure and persevere through the worst sorts of hell. And every time Beckett had persevered, had come out the other side unscathed, it felt like a giant fuck you very much to his father.
She gave him an appraising look. "Is that what you do?"
Beckett nodded, even though he wasn't sure he really had it under control. Which meant he could snap, just like his father had—so many times. "Something like that."
Rolling her shoulders, she took a deep breath. "Thanks, Beckett. Just venting actually helped a lot." She gave a small chuckle. "Although I can't believe I just dumped all that on you. And I am sorry I took it out on you."
He'd . . . helped her? An odd feeling bloomed inside his chest. He resisted pressing on his sternum to try to make it go away. "Uh, well. Good. That's good."
"Thank you," she said. And then she closed the distance between them, threaded her arms around his waist and hugged him, her head settling on his chest.
Beckett was so stunned that, like the fucking emotional misfit he was, he didn't immediately react.
"Sorry," she whispered, pulling away as if embarrassed. As if she thought he didn't want her embrace.
Nothing could have been further from the truth.
"Stay," he said, closing his arms around her back. He pulled her in tight.
And as they stood in the dark holding each other, Beckett had a goddamned ridiculous realization.
He could count the number of people who had hugged him before this moment on one finger—Becca, when she'd apologized to him for what they'd all thought her father had done. Before that, Beckett couldn't ever remember being hugged. Not once.
Chapter 6
The revelation was about as comfortable as swallowing glass. Beckett wanted to run from it. Hard and fast.
Problem was, his arms wouldn't let Kat go. His body refused to pull away. In fact, the louder he yelled at himself to get the hell out of there, the more firmly he hugged Kat against him.
She squeezed him tighter in return, and he was acutely aware of how her body fit against his. Her heart beat quicker against his abdomen. Soft puffs of her breath caressed his arm. Muscles that had been relaxed tensed. Her hands gripped his back tighter, with more purpose, with something that felt like want.
Beckett didn't think he was imagining any of this. And it lit a fire inside his body, one that had his blood heating and his heart racing and his cock hardening against her belly.
But he couldn't act on any of it. Or at least he shouldn't. First, because it'd only been a few hours since he'd taken her the first time, and the last thing he wanted to do was hurt her or make her feel like he was using her for sex. Second, because his gut reaction after their first time was that it had been a mistake, which meant he really shouldn't repeat it. And third, because . . . because . . . Well, fuck. Screw a third reason. Two reasons were good enough.
Except then she tensed her stomach muscles against his erection, shifted her stance, and made a soft noise in her throat that sounded really fucking similar to a moan.
"Katherine," he said, voice full of warning. Because he stood right on the knife's edge between doing the right thing and taking her down to the floor and fucking her until neither of their legs would work for the rest of the night.
Slowly, Kat tilted her head back to look at him. And her face was a mask of desire. Mouth open, eyes hooded.
His hand slid in to the silk of her hair, and her eyelids fell closed.
It was like holding a priceless work of art in his hands. Beautiful. Compelling. Untouchable.
Except he was touching her. And he wanted more. His grip tightened in her hair. And the moan that spilled out of her as she went pliant in his arms made him rock hard. "What are you doing to me?"
"I don't know," she said, her voice no more than a breath. "But I think you're doing it to me, too." Her eyelids rose. "At least, I want you to."
How the fuck was he supposed to resist that?
Her hands slid up his arms and settled on the bulk of his biceps for a long moment, and then they continued upward until they circled his neck. She had to push onto her tiptoes to lace her hands together, and in the process, dragged her body teasingly against his hard-on.
"Damnit, Kat, I'm trying to be a good guy here." His arm banded around her lower back, supporting her weight and keeping her close. He felt like he stood on a slippery slope, the loose gravel under his feet taking him down no matter how much he tried to stay in place.
"Why?" she asked.
Breath coming faster, he nailed her with a hard gaze. "Because if I let myself off this fucking leash, I'm going to be all over you."
"Beckett." She pushed upward until her mouth neared his ear. "Let go."
What happened next was a frenzy of kisses and shedding clothes and moaned encouragements. Beckett stripped her of her sweater in a quick movement and himself of his T-shirt. And then he made quick work of her jeans and shoes, too, manhandling her however he needed to get the job done. She was a goddess in a skimpy black-bra-and-panty set, but he wasn't in the mood to appreciate aesthetics, and they soon joined the rest of her clothes on the floor. He kissed her and stroked her and moved her into the shadows of the far corner.
"Oh, my God," Kat rasped around the edge of a kiss. "Yes, yes." Her hands undid the button and zipper on his jeans. Together, they worked the last of his clothing and boots off.
And then Beckett lowered himself into a crouch, hooked his arms under her thighs and lifted her up the wall until her legs slipped over his shoulders, placing her core right in front of his face.
He didn't even give her a chance to react. He couldn't. He was ravenous for her.
Holding her ass in his hands, he planted his mouth between her legs and licked and sucked and worried his tongue at her clit. He fed on her taste and her slick arousal and the way she pushed him tighter against her.
On a strangled scream, her fingers scratched at his scalp. "Fuuuck," she whimpered, her voice quivering and high. Her thighs shook and tried to close around his head, but he used the width of his shoulders to hold her open to his mouth. "Jesus, Beckett . . ."
He sucked her clit into his mouth and flicked his tongue against it again and again and again. Relentless. Coaxing. Demanding.
Guttural moans spilled out of her as her orgasm hit, but Beckett didn't let up one bit. He sucked her through it, basking in the fucking amazing glow of her ecstasy—and in the fact that he'd been the one to give it to her.
Not whoever the fuck this Cole was.
Where the hell did that come from? No way was he trying to figure it out now.
When Kat stopped shaking against him, Beckett gently released his mouth from her sensitive flesh. Legs still over his shoulders, he allowed her body to slide a little lower so he could work kisses over her stomach, the mounds of her breasts, her nipples. Her hands gently held the sides of his head, encouraging him, urging him on. And then he looked up and met her gaze.
Goddamnit, she was gorgeous. The truth of it sucker-punched him every time he saw her.
"What do you want, Kat?"
She wore a small, dazed smile. "How could I possibly ask for anything after that?"
A soul-deep male satisfaction lanced through him, but he wanted to know what she wanted. He gave her thigh a little bite and devoured the surprised moan that spilled out of her like the greedy bastard he was.
"What. Do. You. Want?"
She licked her lips. "Get inside me."
Hell, yeah. He pulled her off the wall, lowered her down his body, her legs sliding to the crooks of his arms, and centered her wet opening over the head of his cock. "Fuck. Condom." He didn't have one on him, as he hadn't gone back up to his bedroom after their first time. He lifted her up, the withdrawal of her slick heat almost painful.
Her arms around his neck, she cupped the back of his head and seemed to search his face. "I'm on birth control. And I'm clean."
Was she saying . . . "Uh, Jesus. I've never had sex without a condom."
"Ever?" she asked, her eyebrows rising.
He shook his head, which added to the spinning sensation currently rattling his brain around in there. Would she seriously welcome him into her body unprotected? And would he seriously consider doing it? Much as he feared his capacity to turn into his father, he was determined never to become one and chance doing to a child what had been done to him.
"I trust you, Beckett," she whispered, kissing him softly on the corner of the mouth. "Take me. Please."
This woman was blowing his mind tonight. She really was. Before he'd even made a conscious decision to do it, his body was responding to her request, lowering her down until his head rubbed against the soft, hot lips of her center. And then he was sinking deep. Skin on skin.
"Jesus Christ," he groaned, the unfiltered heat and grip of a woman's body around his cock unlike anything he'd ever felt. Except, this wasn't just any woman. This was Katherine Rixey. Who was unexpectedly providing him with a whole series of firsts tonight.
And that wound him up. Hard.
"Hold on," he growled. Using his arms to control the movement of her body, Beckett fucked her fast, needing more, more, always more. He might've worried that he was being too rough with her if it hadn't been for the constant stream of "Yes, yes, oh yes," she unleashed in his ear. Her words were almost a sob of pleasure, and they only cut off when another orgasm washed over her and stole her breath.
There was nothing as fucking gorgeous as the face of a woman shattered by the pleasure you gave her. Nothing.
Releasing her left leg to the floor, Beckett lowered her enough so she could balance on her one foot while he continued to hold the other leg up, freeing his hand to grip her hair and force her to arch her back. Still moving inside her, he soaked in the pleasured expression she wore on her face, the way her breasts bounced when he bottomed out inside her, the dance of her hair around her shoulders. Fucking gorgeous. Every damn thing about her. And it made him want to please her. Over and over. "Can you come for me again, Kat?"
"I . . . I don't . . . know," she rasped. "Oh, God, Beckett."
"What do you need?" he asked, his voice full of grit at how the change in position made her so much tighter.
"Just don't stop. Please." Her nails dug into his shoulders.
He didn't. He wouldn't. Not as long as she wanted him, because in Beckett's experience, that was fucking everything. So he kept thrusting hard and fast and deep, grinding his teeth against the pressure building low in his gut. Angling his hips, he made sure to roll his pelvic bone against her clit, goddamned intent on feeling her come one more time.
"Shit, shit, shit," she breathed, her leg starting to shake.
He hiked her closer so that he was literally hunched around her, his cock gliding in and out on fast, shallow strokes that shoved him right to the very edge.
"Fuck, Kat. Too good," he said, his orgasm bearing down on him.
"Don'tstopdon'tstopdon'tstop," she moaned.
Beckett battled back his own need for release and kept his hips moving until Kat buried her face in his chest and wailed. He wanted to roar in triumph as she came apart again. Instead, his orgasm nailed him in the back. On instinct, he pulled out of her, grabbed his cock, and yelled at the overwhelming intensity of his release, coming against her belly until he saw stars.
As their bodies calmed, Beckett gently released her leg, still holding onto her until she seemed steadier on her feet. Their labored breaths sounded loud in the otherwise silent room, and a boulder of regret parked itself on Beckett's shoulders.
Not because they'd had sex again.
But because it was over.
HOLY. SHIT.
Kat's legs were made of Jell-O. She was sure of it.
Never in her life had she been with a man capable of lifting her and holding her the way Beckett did. His strength was an utter turn-on. His ability to pull orgasms from her a mind-blowing revelation. Couple all that with the look he'd worn on his face—the one that was part soul-deep yearning and part primal-male satisfaction—and she was . . . just . . . totally and completely blown away.
Still holding onto Beckett, she forced a deep breath. "Wow," she managed.
"Wow, good?" he asked, voice full of gravel.
"You really have to ask?" She smiled at him, but realized he wasn't smiling back. He was asking. Seriously. For some reason that made her chest ache. How could he not know? She cupped his handsome face in her hand. "Wow, very good."
He nodded, and at least a little of the question eased from his expression. "Good." He squeezed her hand. "Let's get you cleaned up and dressed."
"There's a bathroom in the hallway." As Beckett strode out of the room in all his naked glory, Kat wondered why he'd pulled out at the last minute. She didn't mind. The heat of his seed spilling against her skin had been sexy as fuck.
Really, though, she had more important things to worry about. Like getting her sweater back on before Beckett noticed the bruises on her arms. Earlier, he'd gotten her top off before she even realized what he was doing. And in the heat of the moment, she'd forgotten about the marks. Thank God for the darkness in here. But the last thing she wanted to do was give Beckett a reason to hound her. Which, if he was anything like Nick with the crazy overprotectiveness, he totally would.
Quickly, she found and put on her bra. She had just stuffed her arms back in the sleeves of her sweater when Beckett returned. "Got chilly," she said, giving him a small smile. She pulled the top over her head but held it up from her stomach.
"Sorry," he said, wiping at her belly first with a set of warm, wet paper towels, and then drying her with a few others.
He'd thought to warm the towels. Gah! "Don't be," she said, keeping her voice casual so she didn't give away that his thoughtfulness had touched her. She watched him clean her, his big hands working against her belly. God, he could be so gentle. Despite the strength. Despite his size.
Quietly, they sorted clothing from the random piles strewn across the floor. And Kat tried to make some sense of the thoughts sluggishly churning inside her head. Aside from the sex, what did she think of this man? Did it matter? Could this crazy, random, scorchingly hot sex just be all they were? They were single, consenting adults, after all. But could either of them really afford the all-consuming distraction being together caused? Because she hadn't thought about anything else except Beckett Murda for the last . . . however long they'd been in here. She didn't even know.
Kat sighed as she pulled on the last of her clothes. Not knowing was the problem in a nutshell, wasn't it? She fished her cell phone out of her back pocket and pressed the button to wake up the screen. Damn. She'd gone to let Eileen out forty-five minutes ago. That was going to be fun to explain.
She looked up to find Beckett watching her, a strange, unreadable expression on his face. "Just seeing if Marz texted me," she said. "He was going to grab some dinner and then work on my laptop before I logged into the network."
Beckett nodded. Looked like they were back to his quiet routine. Though she supposed that was a step up from audibly cursing what had happened, like he did the last time.
"Look, Beckett, this was amazing. Both times, actually. But I guess I'm wondering—"
"If we should make it the last?"
Oh, sure. Now he talks.
And the words he'd decided to speak kinda dropped a rock inside her stomach. Ridiculous, given that she'd been about to suggest essentially the same exact thing. Hypocritical, much?
"Yeah. I guess so." Kat shrugged and mentally pulled on her big girl panties. After all, how upset could she be? The man had just given her three orgasms. And, no doubt, this night would be the gift that would keep on giving, because those memories would be with her during many lonely nights to come. Ha. Pun so intended. "It's just—"
He waved her off. "Nah. Say no more. Totally get it. Completely agree." He crouched down to tie his boots.
Kat nodded. "Right. So, good." She was just thinking what else to say when her phone buzzed in her hand. Even though she expected a text from Marz any time now, her belly still tightened. Because there was always the chance it would be another nastygram from Cole. Schooling her expression, she read the message and breathed a sigh of relief. She held it up so Beckett could see. "Looks like Marz is ready for me."
"Yeah. You go ahead, then. I'll follow in a few." He rose to his full height. "That way, you know, no one will think we were together."
Made sense. So then why did the idea of denying him feel kinda crappy? "Okay. Sure."
For a moment, her feet wouldn't move. And she realized it was because . . . she didn't want to leave him. Stupid, really. It wasn't like they were . . . anything more than two near-strangers who'd found an amazing sexual release with one another in the midst of a dire crisis.
"So. I'll, uh, go. Now." She thumbed over her shoulder. When he didn't say anything, she turned and made for the door.
And had to fight back the urge to turn around and kiss him good-bye.
Chapter 7
After rescuing poor Eileen from her long sojourn out back, Kat rushed up to the gym. Food would have to wait at this point. Though, after two crazy hot bouts of sex in the last few hours, she was starving.
Back in the gym, Kat just barely avoided getting scratched when Eileen spotted Cy and jerked out of her arms, barking and racing after the poor antisocial cat. Cy bolted toward the equipment rack and leapt onto the third shelf, high enough to be safe from the puppy, especially since having only one back leg made it difficult for Eileen to stand and try to reach him.
As if there wasn't enough craziness around this place . . .
Shaking her head, Kat passed the makeshift table in the corner, filled mostly with Ravens, but also with Easy, Jenna, and Sara, too, and crossed the room. Marz, Nick, Becca, and Shane were all seated around Marz's desk and looked up as a group as she approached.
"Hey, sorry," Marz said. "We tried to wait for you to eat . . ."
"Yeah. Where'd you disappear to?" Nick asked.
"Sorry," she said, shrugging. "I took Eileen out and just needed a little time to myself."
Nick frowned. "Kat, if you're having second thoughts—"
"I'm not," she said. "The last few days have been a lot. That's all. I know I don't have to tell you guys that."
Her brother eyeballed her for a long moment, then gave a single, resigned nod.
"Got your machine squared away," Marz said, pointing to the laptop on the edge of the table behind him. "I adjusted some settings to hide the physical location of the computer and hook you up to our proxy server, and I added a souped-up firewall. Just some precautions. Also networked you so you can save any documents to the local network as opposed to your machine."
"Wow, okay. So, we're all set?" she asked, her stomach going for a loop-the-loop. Because she was about to cross a line that couldn't be uncrossed. And that wasn't nothing for a woman who'd built a life around the law.
Marz nodded, and the look on his and the others' faces made it clear they knew this was a big deal, too.
Right, then. Better to do it quick, like yanking off a Band-Aid instead of pulling it off a little at a time. Kat sat, flipped open her laptop, logged in to the machine—and suddenly remember she didn't have her key fob . . . and wasn't sure where she'd left it. Fuck. Heat crawled up her neck. And this was exactly why the thing with Beckett couldn't happen again. "Sorry," she said. "I left my security token outside. So dumb." She rose, fisting her hands so no one saw that they were shaking. From embarrassment. From anger. From nerves.
She was halfway across the gym when the door opened and Beckett walked in. "Hey," he said, his tone casual but his gaze intense and loaded. "Found this. Any chance it's yours?" The question was entirely for the benefit of the others, because she was almost certain she'd dropped the damn thing down in Hard Ink, where Beckett no doubt found it, thereby knowing it was hers.
"Oh, yeah. Thanks, Beckett," she said. "Just realized I'd put it down somewhere."
A whole silent conversation passed between them in one look, and she hoped he got the apology she tried to send him for being so careless.
With Nick and Marz watching her from the front and Beckett walking behind her, Kat felt under a spotlight as she returned to her computer. Get your head in the game, Kat. Right.
"Okay, second time's a charm. I hope," she said. And it was. She sailed into the virtual private network at Justice and from there into her own e-mail and files. Everyone gathered behind her, and a part of her wanted to tell them to go away and leave her to destroy her career on her own, thanks. But she got it. She had critical information they needed, and she couldn't blame them for being curious about exactly what she had and eager to receive it.
"The system records which documents I'm opening or downloading, so I can't copy the entirety of the investigation files without raising a whole lot of questions. I'll start out with the document types we talked about and anything I've already downloaded for other purposes. That okay?" she said. "I can always go back in."
"Yes, that's fine," Marz said, standing next to her and watching her work.
"Don't take any more chances than you have to, Kat," Nick said in a low voice over her left shoulder.
Just be smart about it.
Her mind replayed Beckett's voice from earlier. But instead of annoying her, as it had at the time, it reassured her. Because she was being as smart about the files she was grabbing as she could. You know, within the confines of violating her professional ethics and security clearances.
Honestly, it was all rather anticlimactic. "Somehow, I thought breaking the law would be more exciting," she said after saving the last of the files in question. She logged back out of her system and released a big breath she didn't even realize she'd been holding.
Well, that was that.
Hands landed on her shoulder and gave her a squeeze. She looked up to find Nick standing right behind her and giving her a sad smile.
"Well, don't keep everyone in suspense," she said to him, then looked to Marz. "Open those bad boys up and let's see how what I got might help."
MARZ RUBBED HIS hands together and nearly dove into the chair in front of his computer. And as the others gathered around behind him, Beckett kept his gaze fixed on Kat's face. If anyone knew what it looked like to put on a mask to hide what you were really feeling, it was him. And that woman was doing it. Right now.
Beckett didn't like it one bit. Because it meant she was upset. Because it meant she was burying what she really felt inside—and he knew all the ways that could fuck you up. And because he couldn't do one damn thing to make it better.
He also didn't like it because it made his chest tight and uncomfortable, spilled a restlessness into his blood, and agitated the hell out of him to boot. It was all a helluva lot for someone whose usual status quo was somewhere in the neighborhood of numb.
"I'll, uh, I'll text Jeremy and Charlie and tell them to come over," Kat said. "I'm sure they'd want to be here." She ducked her head to focus on her phone.
Just then, most of the Ravens rose from the table on the far side of the room. Some left the gym, others made use of the weights, and one—the club's president, Dare Kenyon—crossed to Marz's desk.
Beckett had first met Dare when he'd brought the club to Hard Ink a few weeks before to help the team run simultaneous operations against the Church Gang. Dare had simply been a hired gun. Now, he and his guys were full-on, equal partners. And though Beckett might not have ever thought he'd say this about an outlaw motorcycle club, Dare's guys were clearly motivated by a lot of the same values that Beckett and his teammates were—loyalty, justice, duty, honor, even if their take on that last one was sometimes a bit skewed.
Tall with longish brown hair, Dare managed to strike a look that was both rough around the edges and completely in control. The loyalty he commanded from his club seemed borne of respect, not fear, and Beckett had to admire that.
"What's the word?" Dare asked, coming up to the far side of Marz's desk. He wasn't wearing his cut—the cutoff jacket that bore his club patches and identified him as president of the Raven Riders. None of the guys were. Given that the Ravens had some sort of tradition of protecting those who couldn't protect themselves—women and children mostly—Dare had decided the rest of the club would be safer at their compound west of the city if they weren't flying their colors. And damn if all of that didn't give Beckett another reason to give these guys their due.
Nick stepped around and clasped Dare's hand in greeting. "We picked up some new intel. Just getting our eyes on it now."
"The guys are itching to see some justice for their brothers," Dare said, his tone neutral, but his words full of threat, or promise. The Ravens were following the team's lead for now, but the idea of a partnership meant they did so by choice. By the sound of it, that choice came with an expiration date defined by the Ravens' need for vengeance. On the one hand, Beckett totally got that. On the other, it made them unpredictable. And unpredictability got people killed.
"That's understood and appreciated," Nick said. "We're on it. I promise you that."
Dare nodded, seemingly appeased. "All's been quiet on the perimeter, but given the rain and fog, I'm beefing up our presence out there tonight. In case you're wondering where everyone went." Dare had made it clear that things would go better all the way around if he was the one giving his men orders, so after they'd all worked together to nail down basic needs and schedules, Nick had turned over a lot of the day-to-day to Dare to handle at his discretion.
"Need more bodies?" Nick asked.
"Not tonight, but we'll need a little extra relief in the morning. Let my guys get some shut-eye."
"Roger that," Nick said, clasping Dare's hand again.
Dare gave a general nod to the group, then turned away and caught up with some of the guys on the other side of the room. A few minutes later they all left together, once again reminding Beckett of how vital the Ravens' assistance truly was. Their protection made it possible for the team to conduct their investigation the way it needed to be conducted.
As the Ravens left, Jeremy and Charlie crossed the gym and joined the group behind Marz.
"What's up?" Jer asked.
"Kat got us some new files to work with," Marz said. He pointed to the monitor. "This document is Seneka's phone records. Friggin' huge."
Beckett followed Jeremy's gaze as it landed on his sister. Her face was a careful neutral.
"It goes back about eighteen months," Kat said. "So, it's definitely not a light read."
"Which means a better approach than just reading would be to search for specific phone numbers," Marz said. "So, let's start with that contact phone number Frank left us on the chip."
Beckett stepped closer as Marz opened up a search function and typed in the number. The results returned quickly, marking dozens of instances of the appearance of that number making and receiving calls.
"Well, that's moderately better, I guess," Marz said. He clicked Next to scroll through all the results and sighed. "Hey, wait a minute. We know this contact number isn't in service any longer because we called it, but look at this." He pointed to the number's last appearance. "The last time it was used—and, I'm guessing, about the time it went out of service—was less than a week after our team was ambushed."
"Less than a week after Merritt died," Beckett said, rubbing his jaw. "So the line of communication to his contact at WCE, whose number is a Seneka extension, dies with him."
"Pretty much," Nick said, looking from the monitor to Beckett to Kat. "Already useful." He winked at her, and she gave him a small smile. Nothing that made Beckett think that mask wasn't still in place, though.
"Let's do a reverse look-up on some of the other numbers calling or being called by the contact number," Charlie said, waking up the screen beside him and pulling up a website.
"My thoughts exactly," Marz said, doing the same.
As the two men entered numbers, Nick grabbed a legal pad and wrote down the identifying information the search results returned. Many of the numbers connected to the kinds of results you'd expect from a defense contractor and security services provider—government agencies, military bases, some of Seneka's subsidiary businesses.
Marz's fingers froze on the keyboard. "Whoa. This goes to the switchboard at Chapman."
Beckett, Nick, and Shane exchanged loaded looks from behind Marz. Chapman had been the forward operating base in Afghanistan, FOB in Army-speak, out of which they'd been running missions at the time of the ambush that ended their careers. Located in Khost, an Afghan province that bordered Pakistan, it was important for controlling trade routes out of the country and policing the still-Taliban-infested Paktia province to the north-northwest.
"Guess Merritt's request to be transferred to SAD makes sense now," Shane said.
"Fuckin' A," Nick said, raking his hand through his dark hair. FOB Chapman was also the headquarters for the CIA in Afghanistan. Merritt had requested a transfer to their paramilitary unit called, in typical CIA understated euphemism, the Special Activities Division. It appeared their commander had been trying to get clear of his team before his cover got blown. He hadn't made it. And that shit had exploded in the faces of a lot of damn good men.
"Question is," Beckett said, "whether calls to Chapman represent a call to Merritt, legit calls to the base for contract-related services, or calls to contractors stationed there."
"Yeah," Marz said. "But brick-by-brick the evidence that Seneka and WCE are the same is falling into place. Even if some holes still remain in the wall."
"Fair enough," Beckett said, his gaze ping-ponging between the screens of the computers Marz and Charlie worked on.
A few minutes passed in tense silence as everyone hovered, hoping for more useful revelations from Kat's information. "Wait, guys. The Singapore bank. This is the Singapore bank," Charlie said, his blue eyes wide as he made a sweep of the group.
Wary hope slithered through Beckett's gut. A definitive connection between Seneka and WCE would go a long way to ensuring they weren't unnecessarily making an enemy out of an organization as powerful as Seneka.
"The same bank as Frank's account?" Nick asked, leaning in.
"Exactly the same," Charlie said. "God, I called this number myself when I was trying to figure out what the bank statements I was getting were all about."
"And, look, there are calls in both directions between the bank and the contact number," Marz said, tracing his finger over the screen.
Kat stepped up next to Nick, seemingly engaging in what they were doing for the first time. "The bank account information you have—the documents that show all the deposits by WCE," she said. "Can you see if there's any correlation between deposit dates and calls?"
Damn. That was smart. Beckett nodded. "That would definitely nail things down more if so."
"Where are the statements, Marz?" Charlie asked.
After what seemed like a lot of shuffling of stacks of papers on his desk, Marz produced a collection of documents held together with a black binder clip. "Right here." He scooted closer to Charlie and together they pored over them, their gazes pinging back and forth between the statements and the screen.
"Fuck me running," Marz said. "There's a call from the bank to the contact number within forty-eight hours of every deposit into Merritt's account."
Murmurs of surprise and cautious celebration all around.
Beckett wanted to throw a fist pump. If that was actually the sort of thing he did. And while he was thinking of out-of-character ways of celebrating this really fucking good development, he wanted to kiss Katherine Rixey, too. Without her, they never would've gotten their hands on this information. Not easily and not in the time frame they needed, anyway. "Seneka and WCE are connected, then. Have to be. Either they're one and the same absolutely or WCE is a rogue player inside Seneka. Either way leads to the same result."
"We have our proof, baby," Marz said, turning around and grinning.
The whole group engaged in a small celebration. Clasping hands and offering congratulations.
"What happens after the contact phone number goes out of service?" Nick asked, his expression still serious. He wasn't yet joining in the celebration.
"I was wondering the same thing," Shane said, arms folded across his chest.
"Let's take a gander at that," Marz said, shifting around. "I'll search for the bank's phone number and see if it pairs up with any other Seneka extensions." His fingers clipped over the keyboard. "Nothing."
"So someone at Seneka was in regular contact with the Singapore bank at which Frank Merritt received the dirty monies from his op, around the days on which he received payments. But once Merritt died, that communication ceased," Shane said. "Anyone else feeling like that goes way beyond coincidental?"
A round of slow, wary agreements rose up. No one wanted to rush to an unfounded conclusion, but this seemed more and more certain.
Beckett nodded. "I'd say that's as close to a smoking gun as we're going to get. Now the question is exactly how we go after an organization as big as Seneka."
Chapter 8
"Wouldn't the next step be to identify who or what WCE and GW are? Look at the personnel files. Gordon Wexler is the chief operating officer, so I know for sure there's one GW employed there," Kat said, her stomach finally calming down from the stress of having done something so, so wrong. Even if she knew in her heart that she'd done it for the right reasons.
And at least it was turning out to be useful to the guys. Because the whole thing probably would've felt a hundred times worse if she'd violated her professional ethics for nothing.
"That's a plan," Marz said, minimizing the phone records and opening the personnel document. "This is just like Christmas morning," he said, excitement plain in his voice. "What could it be?"
As everyone chuckled, Kat smiled and stretched her neck, trying to work out the tension that had built there. Probably a losing proposition. Especially as she could almost swear that she kept feeling Beckett's gaze skate over her. Like he was keeping an eye on her.
It didn't take Marz long to work through the list. "Okay, we've got a Gene Humphreys Washington, Greta Marie Wendell, and Gordon Andrew Wexler," Marz said. "And, for the sake of thoroughness, I'm throwing in a George Winston Albert and Gail W. Saunders as two other potential GWs. No one with the initials WCE, though."
"Five potentials," Nick said, a deep frown on his face.
Kat leaned her hands against the cold metal on the back of Marz's folding chair. "Scroll down. There's a list that notes date of employment and position. That might help rule the five in or out as your GW."
As everyone watched, Marz moved to that part of the list and checked each of the names. "All the GWs meet the time criteria," Marz said after a minute. "Washington is listed as a security specialist. Wendell is John Seneka's executive assistant. And, like Kat said, Wexler is Seneka's chief operating officer."
"Two of those people essentially sit at John Seneka's right hand," Beckett said, crossing his arms across that big chest. The one that had felt so good to lean against.
"Yup," Marz said.
"So now we have some new names to research," Charlie said.
"Yeah," Marz said. "These are leads we'd never have gotten without Kat." He peered over his shoulder and looked up at her, then winked. "You Rixeys are good people."
Kat's cheeks filled with heat. It was a bit awkward to graciously accept appreciation for having done a not wholly good thing. "Well," she said, clearing her throat. "Some of us Rixeys are better than others." She turned on the sass and returned Marz's wink.
"Dude," Jeremy said from where his hip leaned against the desk. Hands out, he shook his head. "Don't diss your own blood."
Such a screwball. A lovable screwball, to be sure. But a screwball all the same. She stuck her tongue out at him for old times' sake.
Jeremy rolled his eyes, but a smile played around his mouth. "Good comeback."
Kat couldn't help but chuckle, and she appreciated the bit of levity.
Nick smirked and then heaved a deep breath. "All right, people. Then what's the plan? Because we need to make some serious headway on this Seneka issue before the Ravens decide they're too damn impatient to keep waiting for us."
Marz looked at Charlie. "You game to work?" The blond man nodded. "Then Charlie and I will dive into these names and see what we can find. It's eight o'clock now. Reconvene at, say, eleven?"
"Given how hard it was to find information on Manny Garza, do you think that gives you enough time?" Shane asked. Apparently, after they'd seen Garza working for the Church Gang with their own eyes, they'd researched him every way they could, but the guy had simply been wiped clean from the internet. They'd only proven his connection to Seneka when they learned his address from Manny's sister Emilie and raided his house. And damnit if that didn't stir up Kat's irritation with Nick for not telling her about all of this. Because what if Garza had been there when they'd raided? What if he'd had company? What if that company had fought back? That situation could've gone sour fast. And Kat wouldn't have ever even known Nick was in trouble in the first place.
"Guess we'll find out," Marz said.
Nick nodded. "All right. Eleven P.M. it is. Meet back here and we'll take it from there."
As the guys engaged in a series of side conversations, Kat took the opportunity to make her escape. She needed food, and she needed a break. And hell if she didn't need to sit down for a few minutes, because after her afternoon acrobatics, she had some muscle aches in places she hadn't known she had muscles. For crap's sake.
Over in her brothers' apartment, Kat poked around in the fridge to see what she could make herself to eat. The chili had been pretty well decimated, if the small portion left in a single Tupperware bowl was any indication. Unsure of what she wanted, she pulled out a bag of grapes, a jar of olives, and the wheel of brie, which she sliced and laid atop some crackers. Then she plated all of that with a handful of almonds from a tin in the pantry. Satisfied with her snacky dinner, she poured herself a glass of wine and followed the sound of women's voices back to Nick's room.
Sure enough, Kat found Becca and Emilie sitting on the dark blue couch in Nick's office, which connected to his bedroom by a private adjoining hallway. "Knock, knock," she said. "You all mind some company?"
Becca and Emilie urged her in and made space on the couch. Nick's office was small but comfortable. A flat-screen TV hung opposite the couch, and a desk with an organizer full of files and forms dominated the side of the room.
As Kat settled, Becca turned to Emilie. "Wine sounds so good. Want a glass?"
"Bring the bottle," Emilie said with a smile.
"Oh, I like how you think," Kat said, taking a sip of her chardonnay. Fruity with a hint of oak and nut.
Becca laughed. "One bottle of wine coming right up." She dashed out of the room.
"How's it going?" Kat asked Emilie. "I haven't really gotten to see you today." And after witnessing Emilie find her dead brother's body following the attack on Hard Ink just a few days ago, Kat was worried about her, especially since Kat was well aware just how possible it was to put a happy face on when you felt anything but. She held her plate out to Emilie in silent invitation, but the other woman waved her hand.
"It's going okay," Emilie said on a sigh. "Honestly, getting to talk to and focus on some of the folks here has helped a lot. Sometimes I just need to get out of my own head for an hour or two."
"Well, I'm glad you're finding that helping them is helping you," Kat said. "Because I think it's really admirable that you're doing therapy sessions after everything you've been through yourself." And given what Kat had gleaned over the past few days about what'd happened to Sara, Jenna, and Charlie—for starters, it sounded like a lot of the folks here really needed someone to talk to.
Becca returned with the rest of the bottle of chardonnay and two glasses. She sat, poured the wine, and then the three of them toasted with a round of "Cheers."
"So, how are you doing, Kat? With everything," Becca said, taking a sip from her glass.
Kat chuckled, glad to have the opportunity to talk to people who wouldn't freak out, overreact, or look at her like she might break. "I just asked Emilie the same thing. I'm . . . I don't know. I'm okay. It's done now, anyway."
"I might get how you're feeling," Becca said. "When we got Charlie back and the infection in his fingers got worse, I had to pull all kinds of favors to essentially perform a field operation in a borrowed ambulance out back."
Kat's eyes went wide. She'd known that the gang had severed two of Charlie's fingers trying to torture information out of him, but not about the surgery.
"Craziest thing I've ever done, and, obviously, it's really not kosher to use hospital equipment for personal use to perform procedures using not-fully-qualified people."
"Wow, Becca," Kat said. "You realize that's kinda badass, right?"
"It totally is," Emilie said.
"Didn't feel that way at the time," Becca said. "I just wanted you to know I've had to make some similar choices lately. I get it, so I appreciate what you did."
Kat held up her glass. "To doing what has to be done to protect the people you love." They clinked glasses.
"Amen to that," Becca said. "I don't regret a bit of it."
"Here here," Emilie said. "If I'd have done that sooner, maybe Manny would still be alive." She shook her head and heaved a deep breath, her expression bleak but her eyes dry. "Anyway . . ."
"I had another brother," Becca said, staring at the lamplight playing off her wine. "He died of a heroin overdose when he was twenty-one."
"I didn't know, Becca. I'm so sorry," Emilie said.
"That's so young, Becca. I'm really sorry," Kat said. Damn if Becca wasn't full of surprises tonight. Life really wasn't for sissies, was it? Kat nibbled at the grapes.
"Well, thanks." Becca looked at Emilie, a sympathetic glint in her eyes. "The thing is, we can't save people who don't let us know how bad things are or how much trouble they're in."
Those last words rattled around in Kat's head where Cole was concerned, because she hadn't let anyone know how bad his behavior had gotten until things turned physical. But at least she'd taken the steps to get the protective order. If only she'd get confirmation that it had been served to him. Once it had, she'd be able to breathe easier, because no way he'd risk his career, his license, or his freedom over her.
Emilie nodded. "I do know."
"That's why seeing Easy open up to a roomful of people about what he's been going through was one of the most admirable things I've ever witnessed. Heart-wrenching, but so damn brave, too." Becca took another sip of her wine.
"Wait. I'm lost," Kat said, Becca's words pulling her from her thoughts. She'd heard a few expressions of concern about Easy during the time she'd been here, but nothing about a heart-wrenching confession.
Becca and Emilie exchanged a meaningful glance, then Becca said, "Everyone else here knows, so I don't know why you shouldn't. Last week, Easy admitted to being depressed and having suicidal thoughts. None of us even realized . . ."
Even though Kat didn't know Easy well, the news was like a sucker-punch to the stomach. Hearing that must've killed Nick. "Oh, my God. And you're helping him?" she asked Emilie, who nodded. Kat pressed a hand to her chest. "Jesus, that makes my heart hurt. Nick was in such a bad place when he got home last year. He became depressed and got hooked on painkillers. And I don't think he'd have ever asked for help if Jeremy hadn't realized what was going on. The whole thing scared me almost as much as him being over in Afghanistan."
"He's talked about it a little," Becca said. "That night, after Easy shared everything, Nick admitted . . . well . . . he'd been in a really bad place back then."
Hearing her worst fears confirmed lodged a knot in Kat's throat. It also reaffirmed the choice she'd made today. No matter what, if it was within her power to keep Nick from walking down that road again, she'd do it. In a heartbeat. "I'm really happy he has you, Becca. You're so good for him."
Becca's smile was instant and almost blinding, and Kat had seen a similar expression on her brother's face, too. "That's definitely mutual," Becca said.
Kat ate a slice of brie on a cracker and then took a sip of her wine. "All you lovebirds around here. I swear." She winked as the two other women chuckled.
"No man in your life?" Emilie asked.
Kat about spewed chardonnay from her windpipe. Not many safe ways to answer that one. "No, not really," she finally said, ignoring the delicious ache in her thigh muscles.
"Seriously?" Emilie said. "You're gorgeous, brilliant, and have a great job. Men should be lining up at your door."
Kat affected a tortured sigh. "If only it were so." Although, just this afternoon a man had in fact waited at her door. And she'd run face first into his amazing body. And then had her wily way with him. Twice. So not helpful right now, Kat. "Besides, my last relationship ended poorly, so I'm not looking to get into anything right now." She didn't want to say too much, but it did feel good to be able to say something to someone.
"Oh, no. What happened?" Becca asked.
Even though she wouldn't have minded venting a bit more about Cole's behavior, the last thing everyone needed was something else to worry about. So she shook her head as she ate more of the brie and chose her words carefully. "No big deal. He just had a hard time accepting it was over." There. That sounded fairly neutral, right?
Becca frowned. "Ugh. How awkward was that?"
Kat chuckled. "Super-mega awkward."
"I hate super-mega awkward," Emilie said. They all laughed.
A knock on the office door, and then Nick popped his head into the opening.
"Speaking of super-mega awkward," Kat said. Nick rolled his eyes at her.
"Come in, silly man," Becca said. "It's your room after all."
Nick eyeballed the three of them. "Just didn't want to interrupt the female bonding time."
"Well, dude. You totally failed, then," Kat said, smirking.
Nick arched a brow at her. "I thought I said no ball busting."
"And I thought I said I'd try not to bust your balls. Oops. Looks like I failed, too." She downed another cracker and sipped her wine.
Shaking his head, Nick asked, "Sunshine, do you remember when I said that you meeting Kat would be tons of fun for me?" He gestured toward Kat as Becca laughed and nodded. "Well, this was what I meant."
"Suck it up, Nicholas." Kat batted her eyelashes at him, knowing full well he hated his full name.
"Bite me, Katherine."
Kat winked at Becca and then turned a big sunny smile on her brother. "So, did you have, like, a real reason to bother us?"
"Actually, yes," he said, fishing something from his back pocket. He handed a cell phone to Becca, then crouched by her knees at the corner of the sofa. "You have two missed calls and a voice-mail message, all from the same person. Landon Kaine. General Landon Kaine." Nick grabbed Becca's free hand. "Becca, why in the world would the C.O. of our former base be calling you?"
Chapter 9
Beckett was having pretty much the same exact luck as Marz and Charlie. Which was to say, none.
He'd agreed to stay and help them research the list of names they'd acquired from Kat's personnel files. As much difficulty as they'd had last week finding intel on just one Seneka employee—Manny Garza—Beckett had thought they could use the help to research this list of five. More than that, the busy work gave him something to occupy his hands and mind. Otherwise he might find himself doing something very stupid. Like finding Kat. Getting her alone somewhere. Getting her underneath him.
What the hell was wrong with him, anyway? He wasn't a monk. In fact, he'd been with any number of women since he'd been ousted from the Army and returned stateside. But most of them had been one-night-only deals. A way to sate a need—for both of them—and move on. Not once had he had the urge to pursue any of those women. Not once had he felt drawn to any of them. Not once had he felt anything at all for them. At least, nothing beyond the physical.
None of them had penetrated his ancient numbness.
Until Katherine Rixey.
It was as uncomfortable as all hell.
Beckett ran the last of his searches on Gene Humphreys Washington. And just like Garza, who was also listed in the personnel files as a security specialist, there was absolutely nothing on the guy. He might as well have been a ghost. Or a figment of their imagination. "Anything?" he asked, looking over his shoulder at Marz and Charlie.
Marz scrubbed his hands over his hair. "No. It's Garza all over again."
Charlie turned in his chair toward them. "I've had some hits, but nothing useful. Greta Wendell appears on a church Web page and has social media accounts. Nothing jumps out as relevant to us or our situation on any of it."
"Oh, wow. Someone from Seneka actually exists in the real world," Beckett said, tone full of sarcasm.
"Yeah," Charlie said with a nod. "Wexler is more like the guys you two researched, with one exception. Because he's on Seneka's executive board, he's been named alongside John Seneka himself in media coverage of various investigations into the company and overseas controversies involving them. Beyond that, nothing useful."
Reclining back in his chair, Marz laced his fingers together behind his head. "You guys won't mind if I remove my limb and use it to hulk-smash something, will you?"
"As long as it's not my head, I'm cool with it," Charlie said, totally deadpan. The kid had a killer dry sense of humor when he unleashed it, which wasn't that often, even with Marz, with whom he was probably the most comfortable. Well, besides Jeremy, of course.
Marz chuckled, but the sound was tired, frustrated. "Look at you rockin' the humor, Charlie. Before long I'll have you dancing on tables."
Charlie grimaced. "I don't . . . dance."
"No?" Marz said, winking at the blond man. "Why am I wishing Jeremy was here for this conversation?"
As Charlie's face turned red, Beckett shook his head and chuckled under his breath. Leave it to Marz.
"Glad he's not," Charlie said, humor and embarrassment mixed in his voice. He crossed his arms. "He'd probably wear his 'I've got a pole you can dance on' T-shirt just to spite me."
Marz burst out laughing. "Charlie Merritt! I didn't know you had that in you," he managed, still laughing.
Charlie gave Beckett a Did he really just say that? look, his embarrassment easing in favor of amusement. "Dude. Think about what you just said and remember you're talking to a gay man."
Marz's expression went comically confused for a moment, and then laughter exploded out of him. Bent-over, grabbing-his-side laughter.
As Beckett started to chuckle, Charlie did, too.
Just then, Nick came into the gym, and Becca, Jeremy, Emilie, and Kat followed him in. Beckett's gaze latched onto Kat, who looked more relaxed than she had earlier. A little pressure released from Beckett's chest at the sight, and he had to tell himself to stay the hell where he was when she got closer. Damnit. He fisted his hands and folded his arms across his chest.
"Oh, God, is it eleven already?" Marz asked, forcing himself under control.
"No," Beckett said, checking his phone. "It's ten after ten."
"What's up, hoss?" Marz called.
For the first time, Beckett noticed Nick's face. And the guy was not happy. Beckett's gut braced for a nosedive. What the hell could've happened now?
"We maybe got a situation," Nick said.
"Nick, I really don't think so," Becca said as the four of them reached the far side of Marz's desk. Jeremy took his usual spot against the corner of the desk nearest Charlie, while Emilie came around and gave Marz a kiss on the cheek, then sat on the table behind him. Kat stayed at Becca's side.
"What's going on?" Marz asked. He'd sobered up quick at Nick's words.
"Becca?" Nick said—and his effort to gentle his voice was clear. "Play it for them?"
"Okay," she said, her pretty face pinched in concern and confusion. "Ready?" she asked. Everyone nodded, and then a man's voice sounded out from a recording on the phone.
"Hi Becca, it's Landon Kaine. I'm in the area for a few days and wanted to stop by and visit for old time's sake. Figure it's the least I can do for not keeping in touch this past year. I've got some free time on Thursday, so gimme a call today if you can. I'll be up late."
Prickles ran over Beckett's scalp. At the name. At the voice. At the fact that Kaine was here, now. Beckett hadn't known the commanding officer at FOB Chapman well, but he was aware that Merritt and Kaine were close. And in the days and weeks after the ambush, Kaine had been the ranking authority on all things related to the investigation and their ultimate other-than-honorable discharge.
"Well, ain't that somethin'," Marz said, turning to look at Beckett, his eyes full of What the fuck?
Beckett gave a single nod, sharing the sentiment. "That's, uh, damn coincidental." Though, really, he didn't believe in any such thing.
"Somebody fill me in," Jeremy said, flicking his tongue against the piercing on his bottom lip. "Who's Landon Kaine?"
"He was a good friend of our father's," Charlie said, looking over his shoulder at Jeremy.
"He was also the commanding officer at our base in Afghanistan," Nick said, skepticism plain in his voice. "And he presided over our discharge."
"The general and Dad were close," Becca said, looking between Jeremy and Nick. "They went to West Point and came up through the Army together. General Kaine visited my father socially over the years. Came to our house. Came to the funeral last year."
"And when was the last time you heard from him, Becca?" Nick asked, turning toward her.
Her gaze went distant for a moment, and then she shrugged. "At the funeral, I guess."
"Until now?" Nick asked.
She nodded. "Well, yeah."
Nick's gaze landed on Marz, then Beckett. The man's eyes were filled with gut-deep concern.
"Think you might want to get Easy and Shane down here," Beckett said. His gut was right there with Nick's.
Nodding, Nick whipped out his cell and fired off two texts.
In the tense silence that followed, Becca sighed. "Why can't this just be what it seems to be?"
"It could," Nick said. He put his arm around her shoulders and pulled her into his side. "Or not. And the chance it could fall into the 'or not' category is something we have to consider."
Within a few minutes Shane, Sara, Easy, and Jenna had joined their group around Marz's desk.
"What's up?" Shane asked, a wariness in his voice.
Nick looked at Becca, and she replayed the message without him needing to ask.
"Well, ain't that a good goddamn," Easy said. Jenna laced her fingers with his, and Easy brought their joined hands up against his chest.
The men exchanged another loaded round of looks.
"You really think this is . . . what? Some kind of setup?" Becca asked, her hand sagging with the phone.
"Probably only one way to find out," Beckett said, knowing Nick wasn't going to like hearing that.
And the storm that rolled in across the other man's expression confirmed Beckett's suspicion. "You think she should call him?"
"Only way we learn anything," Beckett said, careful to keep his tone neutral. This situation made him recall another from not that many weeks ago—the gangbanger who'd tried to nab Becca from the staff lounge at the hospital had contacted them through a reward flyer and offered information on Charlie's whereabouts. But he'd only wanted to talk to Becca. In person. Nick had gone ballistic at the idea of her taking the meeting, even with them hidden around the location to protect her. And that had been before the two of them had gotten serious.
Nick met the gazes of each of the other men. "You all agree?"
Begrudging nods all around. No doubt the others felt his pain, since they were all involved in relationships now, too.
Thank fuck Beckett wasn't.
Although, before that thought had fully formed in his head, his gaze slid to Kat. Arms crossed, expression concerned, eyes tracking everything that was being said—vocally and silently—in the conversation, she looked ready to jump in, ready to fight, fucking fierce.
What if it was her?
Ice slithered into Beckett's gut. He clenched his muscles against the sensation, not liking it one goddamned bit. And not willing to examine it either.
"Son of a bitch," Nick said, raking a hand through his hair.
"So, um, does that mean I should call him?" Becca asked, hugging herself.
A tense chuckle rose up from just about everyone. Not Nick, obviously, who looked at Becca and cupped her face in his hand. The way she leaned into his touch—almost like it was instinct to get closer to him—did some things to Beckett's chest he didn't particularly like. He felt . . . weirdly . . . envious. Of the ease of their connection. Of having a connection at all.
"I think you gotta have her call the guy, Nick," Kat said, drawing Beckett's gaze. "Especially if you think he represents some threat. Otherwise, how will you even know what you're dealing with?" Her tone was firm but conciliatory.
After a long moment Nick nodded, though his gaze remained on Becca. "Okay. A call. Using a burner phone, so your location can't be traced. And on speaker phone so we can hear the whole thing. That means I need someone on that door"—he pointed across the room—"so no one comes in mid-conversation. Everyone who stays in this room needs to say absolutely silent. If he asks, you can explain you're on speaker because you're driving." He looked at Jeremy. "Where's Eileen?"
"She was sleeping in my room when we came over," he said, expression unusually serious.
Nick gave a single nod. "Good. Anyone with a phone, silence it now. Nothing compromises Becca in any of this."
There was a flurry of activity as everyone checked their cells.
"I'll make sure no one comes in," Sara said.
"Not even if the building's on fire, Sara. Okay?" Nick called as she started across the room.
"Not even. I promise." She jogged to the door and disappeared into the hallway. Since the Ravens had been involved with rescuing Jenna from the Church Gang, they seemed to hold a special respect and affection for both the Dean sisters. So no question they'd listen to Sara if anyone tried to enter.
"I'll grab a phone," Beckett said, rounding Marz's desk to the boxes of supplies, weapons, and ammunition that lined the wall on the other side. He grabbed one and set it up. "Here you go," he said, handing it to Becca.
She nodded. "Thanks. So, am I setting up a meeting with him or not? Under normal circumstances, I wouldn't be thinking twice about doing so. Will it seem weird if I don't?"
"No weirder than him popping in out of the blue for a visit." Nick sighed and looked around the group. "But, let's just play this out. If you did meet with him, it would need to be somewhere high traffic where there was no chance of him grabbing you—"
"Nick." Becca's eyes went wide. "Do you seriously think—"
Kat put her arm around Becca's waist. "Let him plan for the worst. It'll put everyone's mind at ease."
"I agree," Charlie said, frowning. "Don't forget someone tried to grab you before."
Becca's shoulders sagged. "I know. Okay."
Jeremy put his hand on Charlie's shoulder. "Places that would be normal for Becca would be around her house or around where she works, right? Around the hospital might be too risky, though, because of what happened to her there. So, is there anywhere around where you lived, Becca?"
"There's a coffee shop on Eastern Avenue right across from Patterson Park that's like a four-block walk from my house. It's usually busy and there are always people working in there on the free WiFi." She looked at Nick. "Know where I mean? Cute place on the corner?"
"Yeah," Nick said with a troubled sigh. "Something like that could work. Let's do this, then, before it gets much later."
Beckett rose and stepped closer, wanting to hear what Kaine had to say. Most of the others followed suit as everyone closed ranks around Becca.
"Here we go," Becca said, pressing the Call button. Her hand shook, just a little.
Ring. Ring.
"Hello?" came an old, familiar voice.
"Hi, General Kaine. It's Becca Merritt. I hope it's not too late," she said, looking at Nick. He gave her a nod of encouragement.
"Becca. No, no, of course not. I'm glad you were able to get back to me. I know it was last minute."
"Well, it was a nice surprise to hear from you, sir," she said. "Are you home for a visit?"
"Yeah. Every so often they like to haul me into the Pentagon and make sure this old man's still kickin'." Beckett scrutinized Kaine's words, his tone, his pauses and emphases, searching for anything that might give them insight into the man's intentions. But the guy was cool as a cucumber.
Becca laughed, and it sounded mostly genuine. "You're hardly an old man, sir."
"Well, look, I flew into BWI this afternoon, so I'm at a hotel out by the airport tonight. I've got a meeting in D.C. at noon, but I wondered if you had time for a visit around that."
Becca locked eyes with Nick and nodded. "I'd love that. Would you be up for meeting for coffee in Baltimore in the morning before you head to D.C.?"
"That'll work. Feel like I owe it to Frank to check in on you. And Charlie. How is your brother, anyway?"
"Oh, uh . . ." She looked at Nick, who shook his head. "Honestly, we're not as close anymore." She grimaced, uncertainty clear in her expression, and then she gave Charlie an apologetic look.
"Well, I'm sorry to hear that. So, where should we meet?"
As Becca laid out the coffee shop idea and they discussed a time, the exchange about Charlie—brief as it was—bugged Beckett. He couldn't pinpoint exactly why, but his gut was throwing up a red flag his brain hadn't yet identified.
And then the call was over.
The minute Becca hit End, it was like the room collectively breathed again for the first time since she'd begun the call.
"I didn't mean that," she said, looking at Charlie.
He shook his head. "I know. Don't worry."
"I wasn't sure what to say about Charlie. Did I do okay?" she asked.
"You did great, Sunshine. You handled that fine." Nick pulled her into his arms and pressed a kiss to the top of her hair.
"I'll go let Sara know we're done," Shane said, taking off across the room.
"Well, what's everybody thinking?" Marz asked, looking around.
Beckett waited to hear what the others thought, because he still hadn't put a finger on what was bugging him. Maybe he was looking for a problem where none existed?
Easy shrugged one big shoulder. "Seemed . . . normal. Doesn't mean the visit's not damn coincidental, though."
Shane and Sara hurried across the room, and a few Ravens came in behind them and gathered around the table in the far corner. Nick glanced from the bikers to Sara. "Any trouble?"
She smiled. Beckett hadn't gotten to know her well, but based on everything he did know, she seemed tough as nails. Because, young as she was, she'd been through some shit and come out the other end standing. And she couldn't have been sweeter or more easygoing. "Nope. They told me dirty jokes while we waited."
Shane's eyebrow arched. "Dirty funny or dirty inappropriate?"
Sara chuckled. "Funny. About the Harley-Davidson creator meeting God in heaven." She waved her hand. "I won't do it justice."
"Oh, I know that one," Jeremy said. "Ike told me . . ." Ike Young was a Raven who worked as a tattoo artist in Jeremy's shop. It was Ike who had initially connected the team up with the Ravens. Now he was back at their compound outside of the city, protecting Jeremy's receptionist and piercer, Jess, who they'd realized had slept with a guy with a Seneka tattoo. Recently. Another way-too-coincidental-to-be-coincidental development, particularly since a Seneka team had attacked the tattoo shop a few days afterward. All of which made her a potential target, so she and Ike had decided to split town until this mess was over. Whenever that was gonna be.
As Beckett listened to the chatter and the Ravens broke out in laughter across the room, thoughts started to fall into place. Thoughts that made him ask questions. "Uh, guys," he said, interrupting Jeremy's retelling of the joke. "Sorry, Jeremy, but some things are bugging the shit out of me."
Nick's gaze cut to Beckett's face, eyes sharp, jaw set. "What?"
"What are you thinking, B?" Marz said, turning toward him in his seat.
"It's weird to me that he flew into BWI. Totally possible, of course, but Reagan is right next door to the Pentagon, and Dulles is a much bigger international hub, assuming he came from overseas." Beckett shrugged. "But what's bothering me worse is him asking about Charlie."
"Why?" Nick asked.
"Kaine said he owed it to Frank to check in on Becca. And Charlie. I'll admit I was looking for something the whole time he spoke, but the way he said it made Charlie seem like an afterthought. Except then he went on to ask about him directly."
"Yeah. And?" Nick asked, his gaze narrowed.
Beckett shifted his stance as he thought more about why that had bothered him. "Okay, let's assume for a minute that Kaine is in on the conspiracy somehow." Nods all around. "If he is, then that means he should know that Charlie got abducted and that someone rescued him from the Church Gang. But because we've kept Charlie under wraps, no one knows where he is."
Marz's eyes went wide. "And whoever Church was working with—Seneka, we presume—knows Charlie had information they needed."
"How to access the Singapore bank account?" Charlie asked. "That was the key thing they kept asking about. Why this happened," he said, holding up his bandaged hand.
Beckett nodded, all of this gelling in his head now. "Charlie didn't know the account access information then. But they don't know that."
"The night we rescued him," Nick said, "Church was meeting with someone one of their guys referred to as 'company.' About Charlie."
"Probably a Seneka contact," Marz said. "Hell, maybe this 'company' was actually—" He swallowed his words, like he'd realized he'd been about to say something he shouldn't.
"Manny?" Emilie asked. She'd been sitting so quietly behind Beckett that he'd nearly forgotten she was there.
Marz turned toward her. "Yeah. I'm sorry."
"Don't be. It's the reality," she said. "Really." Emilie had told all of them over and over again not to tiptoe around the subject of her brother. But given the horrific way she'd found him—head blown off and lying in a gutter—just a few days before, it was hard not to remember her outpouring of grief and shock. Nobody wanted to pick at that wound any more than they absolutely had to.
"I doubt the point of bringing Charlie to Confessions would've been to hold him downstairs," Sara said. "When that happened, it was almost always women. My sense was that Charlie was a delivery. That's horrible to say, isn't it? I'm sorry," she said, looking at Charlie.
"You played a part in getting me out of there, Sara. No apologies necessary, ever," he said.
Beckett nodded. "So, still playing this out, if Kaine was in on all this somehow, he'd know a lot of that. And he'd probably even know the same people who grabbed Charlie made a failed bid to get Becca, too. And now he's here, wanting to meet with Becca and asking about Charlie."
"So, if we were looking for bad intentions—and we are," Nick said, "this could all really be about finding Charlie."
"Can I just say that I really hate that I just followed all that and that it makes sense?" Becca asked, worry piercing her blue eyes.
Nods went around the group.
"Which means, we need to plan an op to put in place by oh eight hundred tomorrow morning," Nick said. "Because I'm not taking any chances. With Becca or any of you."
Chapter 10
Kat had been listening to Nick and the other men brainstorm plans for Becca's meeting with their former base commander for nearly a half an hour. And here she'd been worried about her own ass for offering up confidential documents. When Becca was about to walk into the middle of a situation that could put her in actual danger.
That sure as hell put things in perspective, didn't it?
"The real question is who goes along with Becca in case shit goes down," Beckett said. "Because none of the five of us can. Kaine will recognize us in a heartbeat."
Nick nodded, his expression growing darker and darker the more they planned this. And Kat felt for him. She really did. She wouldn't like knowing the person she loved was in danger either. "Yeah."
Looking around the room, she realized that every person there had gone through the same experience one way or another. Becca had nearly been kidnapped. Nick had been shot in the neck—the bullet had just grazed him, but still. Shane and Easy had been shot, too. The gauze covering the wound on Easy's arm just peeked out of the sleeve of his tee. For their part, Sara and Jenna had both been targeted by the Church gang. Emilie had been carjacked by a dirty cop, and her brother had held Derek at gunpoint and beaten him with a bat. Kat's gaze moved to her other brother. Not even Jeremy and Charlie had escaped the madness. Charlie had been abducted and tortured, while Jeremy nearly died when a mortar took out part of his building.
That was a long fucking list of near misses.
"What about Vance?" Marz asked. "He works in plainclothes anyway, so he'd be able to pass as a customer."
Beckett made a disapproving noise in his throat. "If Kaine's dirty, he could have cops working with him. If any of them are on the scene, they'll make Vance. Too risky."
"Well, who the fuck does that leave, Beckett?" Nick asked.
"Whoa," Marz said, holding out his hands. "Let's just talk to Dare—"
"I'll go," Kat said, acting on instinct. Nearly a dozen pairs of eyes cut to her. Before anyone had the chance to argue, she pushed on. "I'm the only unknown quantity in the whole group. The Ravens have all participated in other parts of your mission here and therefore could be recognized. None of you can go. Vance can't go. No other cops we can trust." She planted her hands on her hips and nailed Nick with a stare—since he was likely the one she'd have to work hardest to win over. "That leaves me."
"Kat . . ." Nick said, voice full of frustration.
"What about the older Vance?" Beckett said, his face set in a hard scowl. "He's retired now. Even if there were other cops present, no reason to think he'd be there for any kind of work."
Kat's gaze narrowed at Beckett. "Please, do go right ahead and ignore what I said."
Beckett's gaze sliced into her. "I didn't ignore it. I considered it and discarded it."
Anger whipped around inside Kat's chest. "And why would that be?"
"Maybe because Kaine's a trained warrior and anyone who works for Seneka is a coldhearted mercenary. While a hard wind could blow you over," he said, eyes blazing at her.
It was like a haze of red descended over Kat's brain. She'd maybe never felt angrier in her life. "Are you kidding me right now? I can handle a weapon. I can hit. I'm practiced in jiujitsu. And I would never let anything happen to the woman my brother loves."
"I could pin you in five goddamned seconds, Kat," Beckett said.
Shame and heat lanced through her in equal measure. Shame because of what Cole had done. The difference here was that Kat would be ready—for anything. And heat at the images Beckett's words evoked. But way bigger than either of those was the absolutely volcanic rage building inside her at the way he discounted her. Like she couldn't contribute. Like she was just dead frickin' weight around here. Kat gave a humorless chuckle. "I have thrown bigger men than you over my shoulder."
Nick held up his hands. "Kat, Beckett—"
Beckett took a big step closer, but the desk kept them separated. Kat was pretty sure he'd be in her face if it hadn't. "In. A. Fucking. Class. Not in a real-life situation? Right?"
The way he asked that last bit was so condescending that she wanted to pull her hair out, even as the truth of his words shamed her. Again. Had she really slept with this asshole? Twice? "So, the first time you took your training out into the field, you weren't actually qualified to make use of it? Because you'd only done it in exercises up until that point?"
Marz rose from his chair and put a hand on Beckett's chest, but the guy sidestepped him, his big thighs coming up against the desk. "Not even close to being the same thing, Kat. But, please, do continue to equate nearly two years of elite military training to some self-defense classes."
Marz grabbed Beckett's shoulder, but the bigger man shook him off.
Heat poured into Kat's face, and it took everything she had not to fly up over that desk and tackle Beckett to the floor for a good, would-serve-him-right pummeling. "Oh, my God, do you even known how ginormous an asshole you are? Like, for real? All I'm asserting is the ability to watch out for Becca. Period."
"Kat," Nick said, from right beside her. "Kat."
"What?" she yelled, turning. "You feel the same way? Because, please, if you also believe I'm too damn inept and untrustworthy to keep an eye on Becca in a public place—all while you guys have established a defensive perimeter around the building and she's wired—please say so now and I'll happily keep my useless, scrawny ass out of it."
The tension in the room was so strong, it nearly formed a physical presence in the air itself.
For a long moment Nick didn't answer, and a sting bit at the backs of Kat's eyes. "I think you're a good choice for being on the inside with her," he finally said. "I don't love the idea of it, but it makes sense."
Kat wasn't sure why or even what it really meant, but she felt as if Nick's words had somehow hauled her back from jumping off a ledge. Though she didn't even have to look at Beckett to know he was royally pissed. She could feel it boomeranging off of him. At least what he thought of her was perfectly clear now. And she'd almost entertained the idea that they could be something . . . more.
Ha fucking ha.
The rest of the conversation went much smoother after that. And Kat made sure of it by pointedly ignoring Beckett and not making any eye contact with him whatsoever. She didn't need the stress, and the team certainly didn't need the distraction of whatever bullshit lay between them.
Over the next forty-five minutes, they laid out a timeline, mission assignments, and objectives, namely:
Feel Kaine out for what he was really doing there.
Have Becca feed him select information to assess his reactions and see if he acted on said information.
Plant listening and tracking devices on his car and person, if possible.
Easy peasy.
Except, if Kat was honest, her insides were vibrating nonstop. Because while she believed she could handle whatever might happen inside a crowded coffee shop, just the very act of planning an actual covert op was so damn foreign. Still, being a lawyer trained you in a certain amount of deception—both sniffing it out and, when necessary, practicing it. Her head would be ready when the time came. No way was she losing Nick's confidence or letting anything happen to Becca, who Kat already adored.
Kat felt like a million years had passed by the time they wrapped up. She'd been perched on the edge of Marz's desk, and was about to hop down when he spoke.
"So, since we're all here," he said, "I might as well share that we weren't able to find anything useful on the names from Kat's personnel files. After the meeting tomorrow morning, I'll work up a Plan B."
Nick nodded. "Well, the lack of information is telling in and of itself, so it's a start."
"Roger that," Marz said, his voice tired.
"Hey Marz?" Kat said in a quiet voice. He looked her way, and she gave him a small smile. "Go to bed tonight. For once."
He winked and shrugged. "We'll see."
"No seeing," Nick said. "She's right. If the shit's about to hit the fan, I want you running on full power. All of you. So get a decent night's sleep."
Words of agreement all around.
Emilie slid off her table and came up behind Marz, then looped her arms around his neck. "I'll take good care of him."
"I'm sure you will," Kat said, chuckling at Emilie's suggestive look. "Annnd on that note, I'm outta here." She hopped down and kept moving, not wanting to get wrapped up in any small talk with anyone. Or, mostly, with one person in particular . . .
"Kat?" Beckett. Of course. Even among the din of other conversation, she picked his deep voice out of the group. How annoying was that? Annoying enough that she didn't stop.
He didn't call out to her a second time.
BECKETT HAD LESS than four hours before they'd leave to get into position at the coffee shop, and he couldn't fall asleep. Despite his exhaustion and his stress and his need to just power down for a while, his brain was a speeding train of chaotic thoughts. About the op. About this whole clusterfuck. About Katherine freaking Rixey.
She was pissed at him. Beyond pissed. And he wasn't too thrilled with her either.
Although, to be fair, it was less that he was unhappy with her than really fucking irate at the idea of her being a part of the morning's op.
The minute she'd volunteered to accompany Becca into the coffee shop, something in Beckett's brain had snapped. Fury had washed through him, squeezing his chest and kicking him in the gut. Anger, he was used to. But his reaction had been more than that, and he'd been lying awake for hours dissecting it. What he found at the end of his mental microscope he didn't like one fucking bit.
He was . . . worried. About Kat being put in harm's way.
Sonofabitch.
And worry was not something he was all that used to feeling. The closest he got was with Marz, because he knew the guy's leg hurt him. But that was also all wrapped up with Beckett's guilt over Marz's injury in the first place. And it wasn't like his feelings about any of that had done either him or Marz the slightest bit of good. In fact, as far as Beckett could tell, all this bullshit he felt about the whole situation with his best friend had only served to damage their relationship. Better to tamp that shit down good and tight and lock it all away.
That was his usual M.O. And it had been more than good enough the past thirty-four years.
Kat Rixey was screwing it all to hell and back.
Beckett pushed himself off his stomach and swung his legs off the side of the bed. No use lying here when sleep had absolutely no chance of finding him. Not with his head all filled with churn and burn. For fuck's sake.
Might as well make himself useful. They had a shit ton of documents that needed reading through, and they weren't going to read themselves.
He crossed to his bags, the air cool against his naked body, and found a pair of jeans in the darkness. He didn't need lights because there wasn't much to navigate in his room in the unfinished third-floor apartment. Only a mattress set in the one corner and his bags in the other took up any space on the floor. Not so different from his place back in D.C., really. Beckett hadn't felt at home anywhere since being discharged from the Army, so he hadn't made the slightest effort to turn his rental into anything more than a place to lay his head at night.
But maybe that was his thing. Maybe he was meant to float through life—unattached, unfeeling, unwanted.
Although, reuniting with his team a few weeks ago had taken the sharpest edges off those feelings. He'd felt more grounded while here than in the whole year before. What the hell was going to happen when they finally solved this mystery?
No time to worry about that now.
He hiked up and zipped the jeans, then found the black T-shirt he'd worn earlier in the day. He'd barely brought the cotton near his head when Kat's scent washed over him. Which was a big no fucking way, not going there. Dropping it to the floor, he blindly searched for another shirt and tugged on the first thing he found.
Good efuckingnough for two-something in the morning.
Quietly, he made his way through the apartment he shared with Marz and Emilie and Easy and Jenna. The rest of the Hard Ink team slept in the Rixeys' apartment downstairs, including Kat. Like he really needed to be thinking of her lying in bed, her body all stretched out, her hair sprawled over the pillow, her heat warming the sheets.
He took the steps down to the second floor one at a time, not wanting the noise of his footsteps to echo into either apartment and disturb anyone. And then he keyed in the pass code to the door. He pulled it open and found relative darkness—only the lights in the far corner over Marz's desk were on. Damn. If that boy didn't get some sleep soon—
But as the door closed behind him, Beckett didn't see Marz at his desk. In fact, the computers were all empty. As was the room as a whole, since Dare had all the Ravens pulling extra watch shifts.
Thank God for small favors. Mood he was in, Beckett wasn't really fit for public consumption anyway. In his bare feet, he started across the room.
Movement from the corner of his good eye.
He nearly groaned out loud.
Kat. In some sort of weird contortionist position on the mats.
"What the hell are you doing?" he asked.
No answer. In the dimness, he made out the thin white cords to a pair of earbuds.
She shifted position, lowering the leg that had been raised so that she was on her hands and feet, her firm little rear up in the air. Who did yoga at oh-dark-hundred?
Jesus. He scrubbed a hand over his face.
He didn't need to be seeing this right now. He really didn't. Because his body couldn't stand to witness her ass offered up like that in skintight black pants and not react. Not after what they'd shared yesterday. Not after he'd learned how fucking good it felt to be buried deep inside her.
After a few moments she shifted again, going down into a high plank that emphasized just how fit and tight her little body was. She held the plank for what must've been a full minute without even shaking.
"Kat?" he said louder, feeling like an asshole watching her work out without her realizing he was there—and getting more and more turned on by this little demonstration of her strength and flexibility.
He came up along her side, close enough to hear the deafening volume of the music playing in her ears, and crouched down.
She screamed and jerked away, her hand going to her heart even as she back scrabbled. "Damnit, Beckett." She tore out the earbuds. "What the hell are you doing?"
God, she was gorgeous. That was all he could think for a long moment. Even annoyed as shit with him, she appealed to him in a way no other woman ever had. "I, uh, was trying not to scare you."
She tossed the length of her ponytail over her shoulder. "Well, congratulations on a job well done."
Beckett sighed. "Kat—"
She held up a hand. "Never mind. Neither of us needs another fight. I'll go."
He rose as she did. "Kat—"
For a second she gave him a strange look, and then she bolted toward him, rammed into his gut, and flipped him over her back. As the breath whooshed out of him on impact, Beckett's brain scrambled to catch up to what the fuck had just happened.
He blinked and looked up.
Kat stood over him, arms crossed, hip jutted out to the side, smirk on her face. "That was just for the record." She turned on her heel and made it maybe two steps.
A flash fire ripped through Beckett's mind. He flew off the floor and grabbed her from behind before she'd even made it off the mat. Light as she was, he easily lifted her off the ground, keeping her from using the floor for any leverage against which to strike at him. She screamed and thrashed against him.
And then he took her to the mat and came down on top of her, his front to hers. Breathing hard—from the adrenaline rush more than the hit he'd taken—he pressed his lips to her ear. "Five seconds. For the record."
Kat's muscles exploded into action, her arms and legs surrounding him, trying to grip him tightly. She was going for a choke hold or a joint lock, and he wasn't giving her the satisfaction. He fought against her holds, went for his own shots at pinning her, and wrestled her until they were both breathing hard and cursing and hot from the exertion.
Beckett had possibly never been more turned on in his life.
"Give up, Kat," he rasped, less and less worried about hurting her when she was so ably proving she could take it.
"Fuck you," she whispered, going for his throat.
He grinned and gripped her arms, prying them apart with his greater strength and pinning them to the floor.
Her whole body moved in a wave beneath him as she brought her hips up to try to clamp her thighs around his neck. Damn, she was skilled at this. He had to give her that, especially as she managed to get a knee up over his shoulder and knock him hard enough in the carotid artery to momentarily daze him. She got her arms free from his and had almost escaped from underneath him when his wits returned.
He dove and caught her around the stomach, taking her down once more so that he lay on top of her, his front to her back. "Where do you think you're going?" he whispered in her ear.
Slowly, the fight drained out of her muscles. "Nowhere," she said, panting. No doubt his weight made it harder for her to catch her breath. But he wasn't giving up this advantage for anything. Because being pinned underneath him was probably the only way she was going to listen to what he had to say. And the fact that she wasn't still struggling to get away set his blood on fire.
"I'm sorry. About last night," he whispered, dragging his lips against her ear. She shuddered. "I was an asshole."
"Yes," she said, although the need he heard in the word was as much about how he was teasing her skin with his lips and gently rocking his hips against her ass. He would've put money on it.
"I just didn't want you there," he said.
She huffed and jerked, suddenly trying to get free of him again. "That was clear, Beckett."
Way to go, asshole. But he wasn't letting her go until she understood him clearly. He pinned her arm down again and shifted his weight upward, securing her upper body beneath him. "Because I hate the thought of you getting hurt, Kat. Nothing to do with whether you can handle it. But the thought of you in harm's way makes me fucking crazy. Okay? I don't like it. It makes me want to destroy things with my bare hands. Jesus."
"Oh," she said, the fight going out of her once more, as if his words had appeased her. "But you'll be there. You all will."
He dragged his tongue up the side of her neck. A thin sheen of sweat had broken out across her skin as they wrestled, and she tasted fucking delicious. Salty and sweet and something that was all Kat. "I'll be there, Kat," he said, shifting his hips against her ass. He was rock fucking hard.
"Beckett," she whispered, the sound plaintive and needy.
"You drive me so goddamned crazy I don't know what I'm doing half the time," he said, his arm curling under her shoulder, his legs going to the outside of hers, his hips grinding against her rear.
And then she ground right back, arching her spine to bring her ass up more firmly against his cock. The breathy moan she unleashed went right to his balls, making them heavy and achy.
"I swear I didn't plan this, but I want inside you, Kat. Tell me what you want."
She twisted her head as far as she could in this position and met his gaze from the corners of her eyes. "Just like this," she said as her hands went to her hips and pushed the stretchy fabric down.
Beckett couldn't move fast enough. He lifted his hips to help tug her pants and panties down to mid-thigh and to free his cock and balls from the denim. And then he laid down on top of her again, pinning her to the floor as he penetrated her opening inch by scalding hot inch.
"Feel how fucking wet you are. Jesus Christ, that's so good." He bottomed out inside her.
"Oh, my God, that's deep," she cried.
"Never deep enough," he said, withdrawing his hips on a slow retreat. "Never deep enough, Kat." He wrapped her ponytail around his fist and forced her head back. She came up on her elbows, allowing Beckett to wrap one arm around her upper body while he used the other to turn her face toward him. "It's never fucking enough," he said, meeting those bright green eyes. So filled with desire. For him.
He kissed her on a tortured groan and fucked her hard and fast, then maddeningly slow and deep until they were both moaning and cursing and panting.
"Jesus, Beckett, I'm gonna come, I'm gonna come."
"Yeah, Angel. Show me how much you love my fucking cock, how good I make you feel."
Her channel tightened, tightened, tightened around him until Beckett groaned and Kat cried out. And then they were both coming, Beckett's orgasm going on so long his vision went splotchy around the edges. Jesus. He couldn't breathe. Couldn't see. Couldn't do anything but feel Kat around him, under him . . . inside him.
"Oh, my God," she rasped when their bodies settled down.
Beckett was immediately conscious of how heavy he might be, and now that the haze of lust was gone from his head, it bothered him in a way it hadn't before. He moved—
Kat caught his hip in her hand and held him. "Don't move. I want to feel your weight awhile longer."
The words made his chest squeeze. "You . . . like the feeling of me on top of you?"
"You feel good, Beckett." She shifted a little, and he let her move until she seemed comfortable, and then wrapped his arms around her and laid his head partly on his bicep, partly atop her head.
She wanted him to stay. She welcomed him inside her body, where even now he remained. She thought he felt good.
Beckett realized that as satisfied as his body was in this moment, it was his soul that was all lit up inside. Because even though it wouldn't last, for right now he was wanted and he was good. And he so rarely experienced either that it was one of the most special moments of his life.
Chapter 11
Angel. He'd called her Angel.
Kat could barely breathe, and it had absolutely nothing to do with the 250-pound linebacker currently lying on top of her.
He'd called her Angel.
Why? All they did was fight and annoy the shit out of each other. Where had the term of endearment come from?
Maybe it was just the haze of lust and epically good sex talking. That could totally be it. She wondered if he even realized he'd said it.
But it was still confusing as hell. She'd been so mad at him last night that she hadn't even wanted to be in the same room with him. And she'd promised herself that after all the bullshit he'd said, this wasn't happening again. Not ever.
Now, here she lay. His cock still tucked inside her, his body covering her, his arms holding her. And a part of her didn't want it to end. Because he'd also apologized. More than that, he'd made it clear that he'd flown off the handle because he was worried about her. Which was downright sweet, really. In a totally fucked-up Beckett Murda kinda way.
Honestly, he was, at once, the sexiest, most attractive man she'd ever known—and the most confusing and infuriating.
He's rough around the edges, but he's a good guy. Beckett just takes a while to let anybody get close . . .
That's what Marz had said. Could it be that Beckett's whole strong-silent-grouchy routine was just . . . some kind of defense mechanism? If so, a defense against what?
And did it really matter?
She laid there for a long moment, surrounded by Beckett's heat, his soft breath playing over her ear.
Yeah, on some level it did matter. Because there was something between them. Clearly. Maybe it was just pure animal attraction. Except he'd given her glimpses that there was something more to him. Something deeper. Something that spoke to her heart. All of which played as much of a role in the fact that they kept ending up naked together—or partially naked, anyway—as the fact that he fucked her better than she'd ever been fucked in her life. At least, it explained why she kept giving in.
"Beckett?" she whispered, not possessing the energy to speak much louder.
No answer.
"Beckett?" she said again.
Holy shit. The big, hard-ass jerk had fallen asleep? Holding her.
Aw, crap. That was kinda sweet, too. Proving that her heart was such a sucker. Because she didn't want to wake him up. She didn't want him to get off of her. All she wanted was to listen to the call of her exhaustion and close her eyes.
But she couldn't. Because they weren't exactly in a private place. And Beckett's ass was, she guessed, hanging out for all to see. She grinned. Ah, but what an ass. Shakespeare might've written sonnets about that ass.
For a long while Kat lay there and listened to the rhythmic in-out of his breathing. She wished they were in a soft bed somewhere with no clothing between them, because she would've loved the feeling of his bare chest against her back. Instead, she was stuck wearing long sleeves until the bruises healed—especially now. If he flipped out over her participating in the morning's op at the coffee shop, imagine how he'd react if he saw fingerprint-shaped bruises on her arms. Nope. Couldn't let that happen.
It got harder and harder to keep her eyes open, and finally her lids sagged closed altogether. If she didn't get up soon—
Beckett's whole body flinched, jarring Kat into awareness so hard that her heart kicked into a sprint. He groaned and flinched again, like he'd been shocked by an electrical current.
"No," he moaned.
Was he having a nightmare? "Beckett?"
"Why you doing this?" he slurred. "What did I do?" His breath caught hard.
A knot lodged in Kat's throat, because his words were so pleading they nearly broke her heart. "Beckett, wake up."
Most of his weight lay on the right side of her back, so she pulled herself to the left, trying to get out from under him so she could try to pull him out of whatever had its claws in his subconscious. As she moved, his head slipped backward, and her elbow accidentally caught him in the cheekbone.
In a flash Beckett's hand clamped down on her upper arm. Hard. Kat yelped. But before she had a chance to react, he shoved her away with enough force that she rolled over onto her back. Breathing hard, heart racing, and not a little alarmed at the ferocity of his handling, she stared at him with wide eyes, braced to flee or fight.
His eyes remained closed, but the expression on his face was one of abject desolation. Whimpers and harsh, uneven breaths rasped out of him.
What the hell could he be dreaming about?
What she wanted to do was go to him, cup his face in her hand and gently wake him out of whatever it was. But, clearly, he wasn't aware of his surroundings, because there wasn't a chance in hell he'd treat her that way if he was. She was certain of that down into her very soul.
She tugged up her pants and moved onto her knees, but stayed outside the reach of his arms. "Beckett? Beckett?" She slammed her hand down flat onto the mat twice.
His eyelids flipped open and his body stilled. Awareness immediately slid into his eyes, which locked onto her with a cold calculation.
"Beckett?" she said, unsure if he was actually awake.
He rolled onto his side, his gaze darting between them. "What just happened?"
Her shoulders sagged in relief. He was back. "I think you were having a nightmare."
Bracing on his upper back, he lifted his hips and hiked his jeans up, covering himself. Then he sat up and let out a long, shuddering breath. "Yeah, I think so."
"Are you okay?" she asked, fisting her hands together in her lap.
Beckett's eyes narrowed at her. "Why wouldn't I be?" He tilted his head to the side. "Why are you over there?"
A quick shake of her head. She'd give Beckett Murda shit for a lot of things, but this wasn't one of them. "It just seemed bad, that's all."
His gaze was so intense that it made her feel observed, analyzed. "What happened, Kat?"
"Nothing," she said. "You just flinched and started talking. And I couldn't wake you up."
He crawled closer, and Kat hated the momentary urge to back away. He hadn't meant to be rough with her. He didn't even know he'd done it. Remembering how bad he'd felt when he thought he'd hurt her yesterday, when he hadn't hurt her at all, no way did she want to guilt trip him with this.
"Something happened," he said, coming to rest on his knees right in front of her. "Tell me what I did."
"You didn't do anything, Beckett. Really." She forced a small smile. "It was actually a good thing you ended up dreaming, because I would've fallen asleep and we would've still been laying here when everyone gathered in the morning." That part was actually true. And then she gave in to the urge she'd had a few minutes before. She rose up onto her knees, cupped his handsome face in her right hand and pressed her lips to his left cheek. "Everything's okay now."
His hands flew to her hair and his blue-eyed gaze drilled into hers. "Kat," he said, voice raw. "I know something happened. I remember . . . things. When they're bad like that, I know I . . . act out parts of my dreams. Just tell me."
Oh, God. He was breaking her heart right now. But it was clear he wasn't going to let this go. "You just pushed me away. That's all," she finally said. She stroked her thumb across his cheek.
Hard eyes searched hers as his Adam's apple bobbed on a rough swallow. "I'm sorry," he gritted out, then he pushed her hand away. Not hard, but not gentle either.
A rock sat heavily in her belly. "You were asleep, Beckett. You didn't know what you were doing. It's fine."
"Don't make excuses for me—"
"I'm not—"
"Don't fucking make excuses for me!" he yelled. At least, it sounded loud in the silence of the gym.
Kat reared back, her heart in her throat. She held up her hands. "Okay, okay."
"Oh, God, I'm sorry—"
She was on her feet in an instant. "It's fine." With a shaking hand, she scooped her phone off the mat in a quick movement. Her brain was a whirl of What the fuck?
"Kat—"
"It's fine, Beckett," she said, walking away from him. But it really, really wasn't.
"JESUS CHRIST," BECKETT said, dropping his head into his hands. Across the gym, the door snicked shut. Kat was gone.
No, you chased her off. Like the useless, piece of shit your father always said you were.
Half collapsing, Beckett lay down on the mat and stared up at the shadowy ducts and beams that made up the old ceiling.
The night had gone from shit to perfect to shit again. And the bad parts were all his fault. It was almost like he couldn't help destroying anything good that came into his life. His mother, who was so unhappy the whole time he knew her that she slowly drank herself to death. Marz, who had nearly died for and because of him. And now Kat.
Beautiful, brilliant, strong Kat. Who was his friend and teammate's baby sister, to boot. You know, just in case this situation wasn't already enough of a giant cluster.
What had he actually done to her? Before yelling at her, that was. In the dream, his father jumped him the minute he came home from football practice, almost like he'd been lying in wait. The dream was based on a memory, from when Beckett was fourteen. That day his old man had caught him in a choke hold from behind, taking him entirely by surprise. He slammed Beckett face first into the refrigerator and then after a short shoving and hitting match, his dad tripped him and he went down to the floor. Hard. As if that wasn't insult enough, the mean old bastard had kicked him with his steel-toed boots, catching him in the cheek.
That was the first time it had been really bad. Before then it was joking face slaps that weren't really jokes and playful shoves that weren't really playful. It was belt whippings that Beckett had thought were totally normal forms of discipline, until friends saw his back and backside in the showers after practice. And it was a constant stream of verbal attacks and name-calling that slowly chipped away at his self-esteem and warped his own sense of himself.
After all, both his parents had been miserable people, so there had to have been something wrong with him, right? That's why his mother drank herself away from him and his father tried to chase him away with his meanness. Beckett was the common denominator in both their lives.
Now, shaking his head, he forced his thoughts back to the dream and what Kat had said—that he'd pushed her away. Like he'd pushed against his father? But, in the dream, he'd also elbowed his father and grabbed his arms, trying to break free of his grip, and the two of them had fought before he fell. Was it possible he'd acted any of that out, too?
Yes. It was fucking possible. Beckett didn't dream a lot, but he'd woken up on more than one occasion over the years and seen he'd wrestled his pillows and blankets to the floor. And that meant he could've hurt Kat. Even if he hadn't, even if she was telling the truth, he could've hurt her. And everything about that was fucking unacceptable.
What good was keeping her safe from their enemies if he was going to turn around and hurt her at home?
Jesus.
He scrubbed the heels of his hands against his eyes.
Meow.
Beckett's gaze cut to the left to find Cy standing at the edge of the mat, blinking his one eye at him. "What do you want? Huh? To tell me what an idiot I am?"
Cy's tail flicked back and forth, and then he came a few steps nearer.
Slowly, Beckett extended his arm in the direction of the orange cat and held out his fingers. He wasn't sure why he wanted Cy to come to him so bad, but he really did. Maybe it was because if you couldn't get an animal to accept you, what the hell did that say about your chance with humans? For fuck's sake. "Come on," he whispered. Moving slow and easy, Beckett pushed himself into a sitting position and stretched a little closer.
Reaching out his neck like he wanted to sniff Beckett's hand but wasn't sure, Cy's nose twitched and his ears quirked back.
The door to the gym clicked open.
Kat.
Beckett rose instantly to his feet, sending the cat into flight across the gym and into the darkness.
"Hey, what are you doing down here?" Marz asked, crossing the room, showered and dressed for the day.
"I don't know," Beckett said, irrational disappointment flooding his gut. He'd wanted it to be Kat. So he could make sure she was okay. So he could apologize one more time. He heaved a deep breath.
"What's up, B?" Marz asked, frowning.
Beckett shook his head. Even if it was within his character to talk about the shit that bothered him, which it totally was not, he couldn't talk about this thing with Kat. She wanted to keep them quiet, and he'd agreed it was for the best. More than that, respecting her wishes was the least he could do.
Marz crossed his arms and nailed him with a stare. "Want me to take a guess or leave it alone?"
How the hell did he even think he had a guess? The fact that he did made Beckett damn curious, but if he indulged that curiosity and Marz was anywhere within the vicinity of being right, then Kat would be out of the bag.
"Leave it," Beckett said. "What time is it anyway?"
Marz gave him a long, appraising look, then finally nodded. "Uh, going on four-thirty," he said as he made for his desk.
Shit. Probably wasn't even worth trying to sleep now. Not when they were meeting at oh-six-hundred. Nick wanted time to scope out the coffee shop, establish a perimeter, and get his people into place long before the eight o'clock meet.
"Have you even slept yet?"
"Tried." Beckett scrubbed at his face. And damn if his hands didn't smell like Kat. "Mostly failed." The exception to that? When he'd held Kat in his arms. Then he'd slept like a fucking baby.
God, he really wanted to go to her, make sure she was okay and apologize one more time. He didn't deserve that from her, but that didn't make him want it any less.
"Yeah, I know how that is," Marz said, booting up the machine in front of him. "Although, waking up in the middle of the night doesn't bother me as much now that I have Emilie." He winked.
Beckett had given Marz ten kinds of shit about developing feelings for the woman, but his friend was a lucky son of a bitch. The irony of realizing that now wasn't lost on him. "Uh-huh," he managed.
Marz chuckled as his fingers clacked over the keyboard.
"Need help with anything?" Beckett asked, his voice like gravel.
"Not right now, but I don't mind the company if you wanna hang."
Yawning, Beckett shook his head. "I think I'm gonna go drink a pot of coffee while standing in an ice-cold shower."
Marz nodded. "Throw in a few cans of Red Bull while you're at it."
"Now you're thinking," Beckett said, his feet scuffing against the concrete floor. "I'll be back."
"Hey, B?" Marz called when Beckett had almost reached the door.
He turned. "Yeah?"
"I think she'd be great for you." Marz held his hands up as if to surrender. "And now I'm back to leaving it."
Sonofafuckingbitch.
Chapter 12
Getting the text message signal from Nick, Kat moved out of her hiding place and walked the twenty-five feet down Eastern Avenue to the corner coffee shop. Kaine must've just arrived.
Taking a deep breath that pressed the handgun tucked into the back of her jeans into her spine, she pulled open the door and stepped inside. Sure enough, Becca was rising from her table and greeting a tall, older man in a camouflage combat uniform.
Kat dropped a jacket at a table close to Becca's, just as they'd planned, then joined the short line to order. Out of her peripheral vision, she saw Becca and Kaine move to get in line themselves. A patch with two black stars sat in the middle of Kaine's chest, identifying him as a major general. Standing behind her, their conversation was easy for Kat to overhear, and was all the normal sort. Whether he'd encountered traffic or had a hard time finding the place. How long she'd been here, etc.
Kat, Nick, and Becca had come in over an hour earlier to scope out the interior and select a table for the meeting. Nick wanted Becca in front of a window so he could keep a visual on her, and arriving early meant they could control that.
When it was her turn, Kat ordered a coffee and a muffin, then returned to her table. She chose the seat that would allow her to face Kaine and the front door. Not including the two employees behind the counter, ten other people filled the small shop, many of them on laptops or reading the newspaper.
When Becca and Kaine returned to their table, breakfast in hand, Kat kept her eyes focused on her phone. They're seated, she texted.
Got em, came back from Nick a moment later.
Kat knew roughly where all the guys were posted outside. Nick sat in a parked car across the side street. Easy sat in plain sight on the other side of Eastern by the park, wearing a hooded sweatshirt and carrying a plastic cup in his hand that made him look like a panhandler. Shane sat in his truck on Eastern beyond where Kat had been hiding. And Beckett and Marz were behind the building, guarding the back exit and the alley it led to. Jeremy and Charlie had remained at Hard Ink, monitoring the wire they had on Becca, running communications, and watching the street cameras that Marz had intercepted to make sure nothing unusual headed their way.
After a few minutes of small talk Kaine said, "How are you, Becca? How are things?"
She nodded and took a sip of her coffee. "Things have been okay. For a while after Dad died, I let work keep me busy. And that helped."
"Your father was a good man. I miss him over there," Kaine said, giving her a sympathetic smile. The guy had that tough, rough-hewn look that a lot of older military guys get, as if he'd been hardened by age and combat and hardship.
"Yeah. It would probably all be easier if I knew exactly what happened to him, because of course it's hard not to wonder and imagine. But I guess that's just how it is." She gave a small shrug and sipped her coffee again. Nick had coached her on what to say and how to answer possible questions. The team wanted to see how Kaine reacted to certain things and whether his questions or the topics of conversation he introduced might reveal anything about his true intent.
Kaine frowned. "He died a hero, Becca. That's the key thing to know. Best not to dwell on the how. Your old man wouldn't want you to do that."
Kat's phone buzzed. You lied to me, Katherine. You said you'd meet me, which you never intended. Just another of your lies, wasn't it? Cole. Sighing, Kat thumbed back over to Nick's texts so she'd be ready to contact him if she needed. She had no time to think about Cole's bullshit right now, and she never responded anyway. Sure as hell kept every message, though.
Nibbling on a pastry, Becca nodded after a moment. "I know," she said. Kat wanted to high-five Becca for how cool she was playing this, especially since her father clearly did want her and Charlie to know what had happened. Or else he wouldn't have sent them a hidden microchip full of evidence and the means to decode it. Kat could only imagine what was going through the other woman's head.
"How is work, anyway?" he asked, sipping his coffee.
"I've taken some time off, actually," she said.
Kaine ate a bite of pastry. "Oh?"
"Yeah," she said smiling. "The E.R. just got to be . . . a little crazy." She shook her head. "Thankfully, the money Dad left gives me a little flexibility. Of course . . ." She fingered the handle of her mug. "I'd give it all back in a heartbeat if it meant he could be here again."
Kat's phone buzzed in her hand again, but she was too busy watching Kaine's face for a reaction. And he definitely had one. His eyes widened and one brow went up, just the littlest bit. It happened so quickly that if she hadn't specifically been watching for it, she might've missed it. Of course, who the heck knew what it meant?
"Amen to that," Kaine said, lifting his mug in a salute.
Bracing for more vitriol from Cole, Kat peered down at her phone.
She seem okay? Anything else happening in there? Nick. Thank God.
She's doing great. Totally calm. All else is quiet. Kat put down her phone and pulled off a big piece of muffin. As she chewed, she pretended to be interested in her phone again.
"So, you said you and Charlie don't keep in touch much anymore?" Kaine said. "I would've liked to see him, too, but I've never had his contact information."
"Yeah. Charlie's always been erratic at keeping in touch. We e-mailed maybe a month ago and then that was it. When his landlord informed me that Charlie hadn't been around, I actually filed a missing persons report, but the police just think he took off."
"Damn, Becca. I'm sorry to hear that. Do you agree with the police?" Kaine gripped both hands around his mug.
"I don't know, to be honest. I mean, he's done it before and showed back up again later. I guess I'm not sure what to think."
Aw, geez. Kat was so proud of how Becca was doing she wanted to throw confetti.
"I've got some contacts in the police department here," Kaine said. "Prior military who went law enforcement after retirement. I could see if they'd look into it further if you like."
Well, the guys were gonna find that one interesting. It wasn't unusual for prior military to go into the police or other first-responder-type jobs, but it was interesting to know that, along with the Church Gang, Kaine had contacts in the BPD.
Becca smiled and sat forward. "Really? I'd love that. It would just be nice to know for sure. Do you need anything?"
"Maybe just his contact information and any known contacts? Just places to start." Kaine sipped his coffee, his gaze looking over the top of the mug.
Grabbing a pen from her purse, Becca nodded. "I'll give you what I can," she said, writing on a napkin. "We didn't talk by phone much but this is his cell." She wrote for a few more seconds, then slid the napkin across the table.
After that, they made more small talk. About their pastries. About his stay in the area. About how nice the weather was here compared to Afghanistan.
Kat's phone buzzed.
Two BPD squad cars pulling up out front.
Kat's heart thumped against her breastbone. Probably nothing. Right? It was a coffee shop, after all. And still plenty crowded. Except, Nick's team had ample evidence that some of Baltimore's cops had been firmly in the Church Gang's pockets, and now, from what he'd just said, they knew Kaine had contacts there, too. It was a damn shame, but in this situation the police had to be assumed guilty until proven innocent.
"Well, it was really good to see you, Becca. I wish I had more time to visit," Kaine said, balling his napkin up and dropping it in his empty cup.
Three officers incoming, came the next message.
Kat pushed her fork off her table. It clanged to the floor—a signal to Becca that meant Wrap it up.
"It was good to see you, too," Becca said, pushing her chair back. Still chatting, they both stood and carried their dirty dishes to a bin over the trash can.
Three uniformed Baltimore city police officers walked in the front door. Kat's gaze ping-ponged between the policemen and Becca. The men didn't seem to take any special notice of Becca or Kaine. They got in line, eyes toward the wall-mounted menu, seemingly focused on what to order.
"Did you say you walked here?" Kaine asked, gesturing with his hand for her to go first as Becca nodded. "Can I drop you somewhere?"
Kat wasn't loving this at all. The space was narrow and crowded enough that Becca was going to have to cut through the line—between the police officers—to reach the front door. Maybe Kat was overreacting, but something just felt off. Stuffing her remaining half muffin into her coffee cup, she grabbed her jacket, rose as nonchalantly as she could and made her way toward the door.
The middle cop turned around and walked full-force into Becca, hard enough that she reared back a step or two. Kat bit back a gasp, forcing herself to finish walking to the dish bin instead of standing and gawking at them.
Kaine caught Becca with a hand on her back. "Are you okay?"
"I'm so sorry, ma'am," the policeman said.
Becca waved a hand. "It was just an accident. I'm fine."
The officer put a hand on Becca's shoulder, and ice skittered down Kat's spine. Between the three cops and Kaine, Becca was completely surrounded. And the other woman's protestations of being fine were sounding more and more strained to Kat's ears.
She needed a distraction. So she hooked her elbow inside the dish bin and turned as if to walk away. The bin followed her and fell to the floor in an impressive crash of silverware and broken ceramic.
The whole restaurant—including the cops and Kaine—turned to look at her, but Kat glued her gaze to one of the workers behind the counter. "I'm so sorry. My jacket got caught and I didn't notice."
A young man came around the counter and had to cut through the line to try to reach where Kat stood. "Excuse me," he said, needing two of the cops to step aside to make it over. The moment they did, Becca pushed through the line and reached the door, Kaine following belatedly behind her.
"I'm really sorry," Kat said, glancing between the mess she'd made and Becca's back, disappearing outside.
"Don't worry about it," the coffee shop guy said.
"Can I help?"
He waved her away. "I got it."
Kat gave him a quick nod and took off for the door. "Excuse me," she said, squeezing between the same two cops that Becca had. They let her through without any issues, but prickles ran over her scalp anyway.
Maybe she was just predisposed to look for trouble where it didn't exist, but her gut was ringing out a five-alarmer. Something just didn't feel right.
Outside in the morning sunlight, Kat caught a glimpse of Becca and Kaine turning the corner onto the side street. Kat moved in that direction, walking slowly and acting like she was texting a message.
"I'm parked down this way," Kaine said. "Sure I can't give you a ride?"
"No, thank you," Becca said with a smile. "I have to make a stop on the way back to my place anyway. And I really enjoy walking."
Kaine nodded, his smile seemingly genuine.
Behind Kat, two of the cops emerged from the shop, coffee cups in hand, and crossed the sidewalk to one of the two squad cars parked there. They got in, and Kat kept waiting for them to go. Had they received a call? But they just sat there.
A car engine revved on the side street, and Kat looked just as Nick pulled away from the curb. That was the plan—their leaving the shop was his signal to drive the two blocks down to their designated pickup spot.
Becca parted from Kaine, walked to the intersection and crossed to the other side.
The police car pulled a U-ey so that it was headed in the same direction on Eastern as Becca. It passed by where she walked, but to Kat's eyes seemed to be moving slow.
She texted Nick: Bad feeling about the cops. Maybe we should go back in shop?
Her phone buzzed, and Kat tried to read her message and keep her eyes on Becca.
Overreacting. Stick to the plan. That was maybe the tenth time he'd told her that today.
Kat headed for the intersection, her belly a mess of intuition. Maybe she was wrong, but shouldn't she play it safe? This woman was the love of her oldest brother's life. Anything happened to Becca, and Nick would never forgive her. Nor would she be able to forgive herself.
Kaine's rental—hopefully planted with a tracking device by now—pulled up to the stop sign, waited, and then turned right. Away from Becca. Okay, that helped. A little. Maybe she was overreacting.
Kat jogged across the street, not wanting to allow too much distance between her and Becca. At least they didn't have far to go. The plan was for her to follow Becca two blocks to a corner pharmacy, where they'd both enter so Kat could scan Becca for any tracking or listening devices that might've been planted on her. When they came out, Nick would be waiting to pick them up.
But Kat noticed two things in quick succession that convinced her to ditch that plan. First, up ahead, at the end of the long block on which they walked, the police car that had left the coffee shop turned onto the first cross street and parked along the curb—directly in Becca's path if she was actually walking home from here. And definitely in the way of them reaching the pharmacy.
Second, looking over her shoulder, she saw the final cop emerge from the shop and stand on the corner, as if waiting for something. His presence now blocked them from returning to the café or reaching Shane, parked farther down that block. And, anyway, Kat was leery of doing anything that might expose her brother and his team to their enemies. Which meant she had to figure this out on her own. And fast.
And that was when the loud, squeaking brakes of a transit bus sounded from a short distance up the road. Well, if this wasn't the definition of going off on her own. Couldn't worry about that now, though. "Becca," Kat called. "Come here, quick."
Becca turned, frowning. Kat waved her closer. "What are you doing?" Becca asked. They weren't supposed to publicly acknowledge one another at all. But Plan A was out the window.
"Something's not right. Change of plans." Kat grabbed Becca's hand and half dragged her to the bus stop twenty feet behind where they were. They couldn't walk forward or back. The stretch of residential row houses along this part of Eastern offered them nowhere to hide. And none of the guys were close enough to help. Kat was making an executive decision.
"Are you sure?" Becca asked.
"No." The bus came to a squeaking stop in front of them. "Get in and sit down. I'll pay. But bend down like you're tying your shoe so you can't be seen from outside."
They climbed onto the bus. Becca did just as Kat told her, sitting in the seat immediately in front of the mid-bus door. Kat fed fare money into the machine and noted the bus route and number. As the bus jerked and got under way again, she rushed to the seat beside a bent-over Becca, her phone already in her hand. She had to let Nick know where they'd gone.
Cops were following Becca. They'd set up to intercept her. With shaking fingers, Kat hit Send.
We are on the #10 bus 1854 headed east on Eastern Avenue. Get the cops off of us and I'll get us back to Hard Ink. She hit Send again and spied a black pickup out the side window. We just passed Shane's truck, FYI.
She could almost hear Nick's stream of curses at her for deviating from the plan. But what was she supposed to do?
He sent back one word: Roger. And she knew deep down in her gut that he was gonna have a helluva lot more to say than that.
But she couldn't worry about that right now. She had to figure out how they'd get off this bus and safely back to Hard Ink. A lightbulb went off in her head, and she searched for a number and pressed Call. While she waited, she said, "Don't worry, Becca. I let Nick know what's going on."
"He's gonna flip out," Becca said.
"I know."
The other woman peeked up at her. "Do you think we're being followed?"
A ringing sounded in her ear, and Kat held up a hand, asking Becca to wait.
"Kenyon," came a deep male voice from down the line.
"Dare? It's Kat Rixey. I need help."
"Name it," he said, ice slipping into his tone.
Kat explained what was going on, where they were, and where they needed a pickup on Eastern. Thank God she'd taken this bus before. Last July fourth she'd come up to Baltimore to watch the fireworks over the Inner Harbor with her brothers, and they'd taken the number 10 into the city to avoid the nightmare of parking on a holiday.
"We got you, Kat. Sit tight," Dare said. The line went dead. So far, so good.
"Let Nick know we're okay," Becca whispered.
Nodding, Kat typed out, We're both okay. She didn't receive a reply.
Kat sighed. "Those cops weren't a coincidence, Becca. I can't prove it, but—" She patted her purse. "Oh, wait. Maybe I can. I almost forgot." She reached in and found the black rectangular device—a portable bug detector—that Beckett had given her and shown her how to use. That had been their single conversation since the weirdness of last night. Shoving that thought aside, Kat turned the unit on, positioning it in her hand so she could see the ten-light indicator that would reveal the presence and strength of any kind of transmitter signal Becca might be carrying. "Sit up a little."
Becca did, keeping her face turned away from the window. The key was doing a methodical section-by-section sweep. Head, neck, shoulders, chest. The red lights lit up. Kat pressed against Becca's dark blue sweater and found the wire transmitter that Marz had placed on her to pick up the conversation with Kaine. That one was fine, so she kept going. Left arm, right arm, stomach, hips, legs, feet.
All clear.
Kat's belly dropped to the floor. Nick was going to kill her no matter what, but he would do it twice if she had detoured from the plan for no good reason. "Let me see your purse," Kat whispered.
Kat pulled the medium-sized brown leather bag into her lap, and half the row of indicator lights glowed. Bingo.
"Sorry," she said as she unpacked the bag, handing one item after another to Becca, who piled it all in her lap. The more Kat emptied the bag, the more additional lights glowed on the indicator bar. At the very bottom Kat found a small green chip, maybe less than a half inch square. All ten lights lit up. Kat held it between her fingers so Becca could see.
The other woman's eyes went wide.
Kat snapped it in half.
"Think that'll take care of it?" Becca asked.
Kat had just inhaled to answer when several motorcycle engines roared by the bus. She caught a glimpse of them but couldn't tell if they were Ravens without their cuts. The rumble faded away, then seemed to get closer again, as if they were coming up behind them. Her phone buzzed, but it wasn't Nick this time. Nor Cole.
It was Dare. Cop car following the bus. Diversion in place. Meet you as planned.
What in the world?
Just then more bikes roared out of a cross street, so close that Kat thought they might strike the back of the bus. Horns blared. Sirens whoop-whooped. But all of that noise became more distant as the bus continued to trundle down Eastern Avenue, making good time, as no one requested a stop and the bus stops they passed stood empty.
"It's almost time to get off," Kat said, jamming Becca's belongings back into her bag. Kat wasn't entirely sure what was happening once they disembarked. The closest stop to Hard Ink was in front of a largely abandoned business strip situated in between an ancient gas station and a car dealer. Not exactly a lot of shelter in any of that. Worst case scenario, it was maybe a six-block walk back to Hard Ink, but she really didn't want to take the chance of making it on foot.
Two blocks later, Kat pressed the yellow strip that requested a stop. "Here we go," she said when the bus lurched to the curb. Looking out the window, she grinned.
Six Ravens sat on their big black-and-chrome bikes in the parking lot of the old strip, and man, they were a thing of beauty.
"Our chariots await, I guess," Becca said, looking as relieved as Kat felt.
They rushed off the bus and toward the bikers. Phoenix waved Kat toward his bike, helped her on, and handed her a helmet. A guy Kat didn't know well did the same for Becca.
"Hold on, now," Phoenix said over his shoulder.
And then they took off in a phalanx. Two Ravens in front of Kat and Becca, two behind. Kat was almost giddy with victory and relief.
Within three minutes they negotiated the fortified jersey-barrier-and-chain-link roadblock that Detective Vance had put into place after the attack on Hard Ink—he'd fed the papers the story that the explosion had been due to a neighborhoodwide gas main break that necessitated cordoning off the old, largely abandoned industrial area. Inside the fence, Kat could finally breathe easier, especially when the bikes pulled up to the gate to Hard Ink's lot. They waited for it to open, then rolled inside.
"Thanks for the ride," Kat said when they came to a stop.
"Like I told you," Phoenix said when he removed his helmet, "anytime." He winked at her as she took off her own helmet and dismounted. Then he revved his engine and pulled a U-ey. In a line, the motorcycles roared back out of the lot.
Charlie came bursting out the back door, Eileen nipping at his heels. "You two okay?" Both of them nodded. "They're fine and they're back," he said into a cell phone, then he hung up.
"Was that Nick?" Becca asked.
Charlie nodded. "Yeah." His gaze cut to Kat. "And you might wanna hide."
Chapter 13
Beckett was pissed. Blood-boiling, seeing-red pissed.
Kat had gone way the hell off the grid, endangering herself and Becca. And then somehow the Ravens had gotten involved, causing a three-car pile-up on Eastern after they'd run a red light. The latter had temporarily blocked the team from following the bus Kat and Becca had jumped on, and from making their way back to Hard Ink. The only good thing was that it had also cut off the cop who had been following the women. Which Beckett supposed had been the point.
Still, so much for operational imperatives like executing a carefully orchestrated plan, or behaving in a manner that escapes notice, or concealing identity, or secrecy, or stealthiness. For fuck's sake.
Yet, by comparison, Nick's fury made Beckett's anger seem like a minor, passing annoyance. Nick's rage seethed out of him until the air in the car nearly vibrated with it. Cheeks flush, jaw clenched, eyes narrowed to near slits, he strangled the steering wheel so hard it creaked in his grip.
From the front passenger seat, Marz looked over his shoulder, his gaze filled with all kinds of Oh shit. The guy looked like he wanted to say something, but he held back. And you knew shit was bad when Derek DiMarzio bit his tongue, because that guy was pretty much fearless in saying what needed to be said.
"You're gonna have to dial it down and give her the benefit of the doubt," Marz finally said, his tone unusually subdued.
The look Nick sliced Marz's way was pure ice.
"She's not a soldier, Nick. Whatever happened—"
"Yeah, and that's crystal fucking clear, isn't it? And it's the last time she'll be involved. Period. Beckett was right. She had no business being out there today." He took the turn into Hard Ink's neighborhood fast and hard, making the tires screech against the blacktop.
Annnd . . . that comment took the edge off some of Beckett's anger. Because his bullshit had been borne of worry and concern. And if he examined the anger he felt right now, it stemmed largely from that same place inside of him.
Damnit. He was mad at Kat . . . for making him worry. Again.
Why was it that every emotion he felt seemed tied to his anger? When he was scared, he got angry. When he worried, he got angry. When he felt . . . almost anything, there was the anger. Not always at the highest volume, but there at least a little. Always. And why was he feeling so much of it right now? Now meaning not just at this moment, but in general, since he'd reunited with his team. Certainly since he'd met Kat.
Beckett heaved a breath as he looked between the men in the front seat—two of his closest friends in the world. Hell, two of his only friends—and he thought about what he'd shared with Kat the night before. How much being with her had meant to him.
And realization smacked him over the head.
Kat made him feel . . . a whole host of shit. And so did being reunited with the guys. For the first time in a long time he'd dared to want. A woman. His friends. A place to belong.
Which meant, for the first time in a long time, he'd opened himself up to rejection, abandonment, and loss. And it'd all stripped his ancient numbness away, leaving him a raw, exposed bundle of emotion. Except the only emotion he had any experience actually feeling, actually identifying, was anger.
They pulled into the lot behind Hard Ink, jarring him from his maybe-useful, maybe-ridiculous thoughts.
Kat and Becca stood there waiting for them, and the look on Kat's face was part fight and part fear. Beckett didn't like seeing her wear the latter at all.
Marz was right. Whatever she'd done, she thought it needed to be done. And she'd gotten them home safe. That counted for something. No, that counted for a lot.
Beckett gripped the top of the front seat. "Nick—"
But the man brought the car to a hard stop and flew out the driver's door.
Beckett followed suit, his gut not loving the way Nick got right in his sister's face.
"What the hell were you thinking?" he yelled.
Becca stepped to his side. "Nick—"
"Becca, please," he said, moderating his tone only a little.
Nick glared at Kat. "You promised to do what I told you to do. You promised not to go off on your own. And what did you do?"
Shane's truck came through the gate, he and Easy peering at the gathering through the windshield. They couldn't get far enough in to park because of where Nick had left his car. Beckett stepped closer, not wanting to miss what the Rixeys were saying.
"I did what I thought needed to be done," she said, tone firm, seemingly not intimidated by her brother at all, despite the fact that he was louder and bigger and royally pissed off.
"What you thought?" He gave a humorless half laugh. "Well, that's real funny."
Beckett mentally winced. The guy was about to cross a line—
Or, perhaps, he already had. Because just as Shane and Easy joined their group, Kat pushed around Nick and walked up to Marz. She held something up, which the guy accepted into his hand. Then Kat turned on her heel and beelined for the door. "When you're done being an asshole, I'll be ready to talk." She disappeared inside.
"Sonofabitch," Nick yelled, raking his hands into his hair.
"Hey. It's okay. I'm okay," Becca said, cupping his face in her hands.
Nick hauled Becca into a tight embrace. "I was so fucking worried," he whispered.
But Beckett's mind was stuck on the image of Becca touching Nick's face, because it made him remember Kat doing the same thing to him the night before. He'd been upset and confused and a little out of it, and she'd offered him comfort in the form of a sweet, gentle touch.
And right now she was probably feeling a lot of those same things, only she was all alone. That drained most of the rest of his anger away. "What is it?" Beckett asked Marz, nodding toward his hand.
Marz opened his fingers until his palm lay flat. "A tracking device. Long-distance transmittal, by the looks of it. High-grade. She managed to find and disable it. If she hadn't, the world pretty much would've been able to follow Becca here."
"It was in my purse," Becca said. "One of the policemen walked right into me. Maybe that's when it happened. Then they all sorta surrounded me, so I guess it could've happened then, too."
"What?" Nick asked, a scowl sliding back onto his face.
Becca pulled out of his arms and put her hand to her mouth. "Oh, my God. That's why she did that." Her gaze went distant.
"Why who did what?" Nick tilted her chin to make her look at him. "Hanging on by a very thin thread here, Sunshine. Explain. Please."
"I had to cut through the ordering line to get out of the shop, which meant I had to go between the cops. That's when one of them ran into me. He kept asking if I was okay and it was like he wouldn't let me through. I was starting to panic a little, to be honest, but then Kat knocked the whole bin of dirty dishes onto the floor, and the cops jumped back. Then I was able to get out."
A warm pride curled into Beckett's chest. That hadn't been any accident. She'd needed a diversion, and she'd crafted one that would seem totally accidental. Smart fucking woman.
Beckett had heard enough. "I'll catch you inside," he said to Marz.
The guy gave him a knowing look, which might've set Beckett's teeth on edge if the look also hadn't said he thought going to Kat was a good idea.
Leaving the group, Beckett went in. Where would she have gone? He tried Hard Ink first, because she'd gone there yesterday. But the place was quiet as a tomb. Her room.
He took the steps two at a time and let himself inside the Rixeys' apartment. Jeremy stood at the breakfast bar, hands braced against the granite, head hanging on his shoulders. His dark blue shirt had white writing on it that sorta looked like a pharmaceutical ad. It read, Ask your doctor if Mykoc© is right for you.
Funny, but Beckett wasn't in the joking mood right now. "Is Kat in here?"
Jeremy nodded. "In her room." Beckett headed that way. "You might wanna give her some space right now."
Space was the last thing she needed, but Beckett gave him a wave of acknowledgment. Maybe it was stupid to go to her with Jeremy knowing he was doing it—alongside a hundred other reasons—but his gut demanded that Kat needed him. And that was the most important thing right now.
Taking a deep breath, he knocked on the door.
No answer.
He turned the knob, and the door opened.
"Go away," came a strained voice.
Beckett poked his head through the opening. Kat sat in a ball in the corner, her legs pulled up to her chest, her arms hugging herself tight.
"Aw, Jesus, Beckett. Really?" Her face was splotchy and her eyes watery, though he didn't see any tears.
Ten-to-one she was going through some adrenaline letdown right now, too. Which no doubt made it all worse. That shit could fuck you up even when you were used to how it left you feeling drained and shaky, by how all the stress you'd suppressed during the height of the crisis boomeranged twice over after the fact.
He came into the room, closed the door, and turned the lock for good measure. And then he crossed to Kat, scooped his arms under her knees and behind her back and lifted her up against him.
She smacked his chest. "Put me down."
"No." He moved to the edge of the bed.
"I don't need this right now, Beckett. Put me down."
"Yes, you do," he said, sitting on the edge of the mattress. He pulled her face in against his throat and smoothed her hair back from her cheek. She trembled against him, just the littlest bit, her skin hot to the touch. He pressed a kiss to her forehead and hugged her tighter. "You did good, kid."
Every one of her muscles went tight. Her hand fisted in his shirt. And her breath caught as she buried her face against his neck and shoulder.
Then Kat burst out crying.
And it was like being torn apart and put back together, all at once. He hated her pain, but he adored that she wasn't hiding herself from him, and that she was letting him be there for her.
He didn't shush her, or try to talk her down, or encourage her to dry her eyes. To be sure, her tears were like daggers in his heart—they hurt like fucking hell. But the only way she was going to feel better was to let this shit out. Ironic realization for him—he did actually see that. But just because you could see what was good for others didn't mean you had the first goddamned idea how to apply those principles in your own life. And that was a problem for another time anyway.
What mattered right now was Kat. What mattered . . . was Kat.
The thought opened up a warm ache in the center of Beckett's chest.
He wasn't sure how much time passed. It could've been minutes or hours. Finally, Kat heaved a deep, shuddering breath and her muscles went lax in his arms. "I'm sorry," she whispered.
"Nothing to apologize for," Beckett said in a low voice.
After a few moments she tilted her head back, but she kept her eyes closed, her breathing still uneven.
Her face was a mess. Wet. Red. Mascara smudged below her eyes. And she was the most beautiful fucking woman he'd ever seen. The thought of anyone else seeing her like this, when she was soft and vulnerable and hurting—he hated it. And he realized that it was a privilege to be with someone when life had knocked them down, because it meant you got to help build them up again.
He lifted the hem of his gray T-shirt and gently wiped at her face.
She batted his hand away. "You don't have to—"
"Let me," he said. Let me take care of you.
"Why are you doing this?" she asked, her eyes finally opening to him. The tears had turned the green absolutely brilliant in color.
"Because you needed it." Simple as. When her face was dry, he let his shirt fall again. And then he didn't know what to do. Or say.
"Sometimes you can be so sweet," she said. "Thank you."
He acted all chill, like the words didn't add to the warmth ballooning inside his chest. "And sometimes I'm an emotionally stunted asshole," he grumbled.
Kat gave a watery grin. "Sometimes," she whispered.
And he didn't even mind that she'd agreed, because that smile was lighting him up inside. She felt like crap . . . and he'd made her smile. Beckett, of all people.
"I'm not used to people seeing me when my weaknesses are exposed, either, Kat. So, last night—"
Her fingers fell on his lips. "You don't have to tell me."
Beckett pressed a kiss to her fingertips and nodded. "I need to say something. I hate the thought of hurting other people. It makes me crazy. Literally. I don't have nightmares like that often, but when I do, I know I act some of it out. So I know I probably did something to you—"
"Beckett—"
"Please," he said. "Let me finish. I'm not asking you to tell me. What I'm asking . . ."
When he didn't say anything right away, she stroked her fingers down his cheek. "What?" she whispered.
He sorted through the whirl in his mind. "What I'm asking is . . . for you not to give up on me." Beckett shook his head. "I'll get my shit together. I promise."
She looked a little bewildered by his words, which made sense since he didn't even know what he meant by them in the first place. But then she rushed to say, "I won't. And I could . . . maybe . . ." She shrugged, and her brow furrowed.
He was dying to know what she was gonna say. "Aw, don't leave me hanging, Angel. What?"
She gasped and looked up at him. "You said it again."
"Said what?" He replayed his words, and his eyes went wide. Angel. Where had that come from? "Oh, uh, I did?"
Her smile was uncertain, but she nodded.
But the more he thought about it, the more he liked the nickname for her. He didn't see angels as, well, angelic. At least, not in the sweet sense. He saw them as fierce, powerful, warriors from heaven. Assuming they were real. Which, who the fuck knew? But either way, that shit fit Katherine Rixey to a tee.
"Well." He shrugged as heat filtered into his cheeks. "It fits."
"Not feeling like much of an angel right now," she said, rubbing her eye.
"Your wings are just bruised. You'll bounce back. Count on it." He ran his fingers through her hair, loving the feeling of it against his skin, loving her soft and warm in his arms. This . . . this was where she fucking belonged.
With her arms around his neck, Kat pulled herself up until she straddled his lap. "I was just going to say that maybe I could help. If you want."
Beckett's eyes went wide. She wanted to help him? "Nah, that's on me. But I appreciate the hell out of that sentiment." He really did. He had an idea of what he needed to do. And it was a path he was going to have to walk alone. At first anyway.
Kat dragged her fingers through his short hair. "Well, you can do it. If you really want to."
Nodding, he said, "I do." Honestly, he'd never before seriously entertained facing his demons. Between football and the Army, he'd had other outlets for releasing some of the bullshit in his head. Otherwise, he'd shut himself off from the world and everyone in it, and let himself go numb. Now, he didn't have those outlets, and that numbness was wearing the hell off.
But, most importantly, now he had something to fight for, not just against.
Maybe he could fight for Kat. For a chance with her. There were at least a hundred big ifs standing between this moment and that possibility, but it was worth the try.
She was worth the try.
KAT HAD RECOUNTED the whole story from following Kaine into the coffee shop to hopping on the back of Phoenix's bike at least three full times, and answered tons of questions in addition. She was hungry, tired, and running out of patience with her brother, who hadn't done much to soothe her bruised feelings for the way he'd jumped on her when he got home.
"Are you sure he reacted to Becca's mention of the money her father left her?" Nick asked, again. Sitting backward on one of the folding chairs by Marz's desk, he'd been grilling her for nearly an hour. His anger seemed mostly gone, but he was still obviously agitated. By her? By how the op had gone down? By what they had or hadn't learned?
Perched on the edge of one of the computer tables, Kat just barely restrained a groan. It didn't help her own frustration levels that her phone kept buzzing incoming messages from her back pocket—ones she didn't want to check in front of everyone. "Yes. We've been over this."
Nick's gaze narrowed. "It's important, Kat."
"No shit, Nick. But the fact that you keep asking me the same questions and I keep giving you the same answers reveals pretty clearly that you don't trust my judgment, so listen to the recording from Becca's wire if you don't believe me. I think it's pretty clear Kaine did in fact have an agenda, and that what he was really after was finding Charlie. But, hey, I only make a living assessing the believability of people's testimony, mannerisms, and speech. So what the hell would I know?"
"Guys, this isn't helping," Jeremy said.
"I'm well aware," Kat said, crossing her arms.
Beckett jerked off the edge of the desk against which he'd been leaning, his gaze glued to his phone. "Hey, Marz. Pull up the tracking software." As they'd talked, Beckett had been checking in on the bug they planted on Kaine's car every few minutes. "Kaine said his meeting was at noon at the Pentagon, right?"
Agreements rose up from the group.
"Well, it's going on eleven, and he's still in Baltimore. Takes an hour to get to the Pentagon from here, maybe more, considering traffic and parking," Beckett said. Kat nodded. The Pentagon was on the exact opposite side of D.C. from Baltimore.
"Well, where is he?" Nick asked, coming up behind Marz.
"Let's see," Marz said. Beckett rounded the desk to look at the bigger screen. It showed a blue dot moving east across the city on Eastern Avenue, which placed it not too far from Hard Ink. Except the dot passed by the tattoo shop's neighborhood and turned south onto Dundalk Avenue. The dot continued along that road for a while and then turned right. "Holy shit."
"What is that?" Kat asked.
Nick's hand went into his hair and tugged. "No fucking way this is coincidence."
"No such thing," Beckett said, his voice only a cut above a growl.
"Well," Easy said. "Guess that answers that."
"Fuck me running. Yes, it does," Shane said, a hint of his southern accent coming through.
"Hello? I'm lost here," Kat said, watching the blue dot move closer and closer to the water, to where long rectangles of land on the map stuck out into the harbor.
"It's the goddamned marine terminal," Nick said.
"Where the Church Gang had its operations," Jeremy said, pointing at the screen, eyes wide.
Marz nodded. "Pier 13 for the win." He threw a pen down against his desk. The car turned onto one of the thin strips of land. A pier, Kat guessed. Marz tapped his finger against the monitor. "This is the pier where we witnessed a drug deal a few weeks back. The same one where our documentary evidence says Seneka is shipping containers from Afghanistan."
"And where we saw the gang hand over nine unconscious women, for fuck's sake," Shane said, acid in his voice.
"Kat's right, then," Beckett said, his voice harsh, but the words building Kat back up a little. Actually, Beckett had been amazing at that since they'd returned to Hard Ink. He'd known exactly what she needed—even when she didn't know herself—and given it to her, no questions asked. "Kaine is in bed with Seneka and/or Church. Although, since Church and his gang were pretty much obliterated, Seneka has to be in play here."
"It wasn't my father who was dirty," Charlie said, blue eyes blazing at the computer screen. "It was his oldest Army buddy. His oldest friend."
Jeremy crouched beside where Charlie sat and took his hand.
"That man came to our house," Charlie said, his expression bleak.
"That means . . ." Becca's voice was tight, like she was holding back a wave of emotion. " . . . that Kaine was behind my father's death?"
"Sure is looking that way," Marz said. Kat's heart went out to both Merritt siblings. It had to be so hard knowing that someone who should've cared about their dad had very likely been responsible for his death.
"And our blackballing," Beckett said.
"This gives us a whole new way of reading the documents on the microchip," Marz said. "We need to look at everything with an eye to what it might reveal about Kaine's involvement. And we need to reanalyze Kat's documents, too. There could be specific connections between Kaine and Seneka that we didn't know to look for the first time through."
"Jesus," Beckett said. "As glad as I am to know that Merritt didn't betray us, it still sucks some serious ass to learn that it was in fact one of our superiors. One of our own."
Easy nodded. "And the guy's still in command at Chapman, which means he's still in a position to put other soldiers in danger to cover his own ass."
Nick heaved a deep breath. "This means we're not just up against Seneka, we're up against Seneka and a well-liked, highly decorated, politically connected two-star. That's fucking awesome." He shook his head. "Let's get back to the grindstone, then. Because this just got about a hundred times more complicated."
"Most of us need to go relieve the Ravens from their watch shifts," Beckett said. "We're already a couple hours behind doing that as it is." Hopefully, for Kat, that meant some time apart from her brother. God knew she and Nick needed it. They'd always been like this—butting heads one minute and teaming up the next. The stakes had just never been this high before.
Nick nodded. "Charlie, Jeremy, you two stay here and get started on the research. Get the women to help when Sara and Jenna are done talking to Emilie." The guys nodded. "Marz, Beckett, and I will take a shift on the front gate and perimeter watch. Kat and Easy, you two take the snipers' roosts."
Kat nodded, glad for the alone time, Not to mention the time to see what all the buzzing of the phone in her pocket was about. Her belly flip-flopped. Maybe it was finally the court confirming that the protective order had been served?
As the group broke up, Emilie came through the door and jogged across the room.
Marz rose to his feet, frowning. "What's up, Em?" He came around the desk to meet her.
"I knew you all were going to be leaving to take watch shifts, so I wanted to tell you as soon as I found out. The coroner called. He released Manny's body to the funeral home. He'll be interred Saturday morning at Garrison Forest Veterans Cemetery up in Owings Mills."
Marz pulled her into his arms. "I'm sorry."
That sure put things in perspective, didn't it? Kat might be mad at Nick, but Emilie would never get another chance for her brother to make her mad—or anything else. Looking from Nick to Jeremy, she swallowed around the lump suddenly taking form in her throat. So many losses in this whole mess.
Emilie nodded, then pulled back enough to look him in the eye. "I have to go to the funeral, Derek. And, more than that, I have to go get my mother and take her. She isn't going to be able to drive for this, and I haven't been there for her at all. I know this raises big problems . . ."
Marz shook his head. "We'll figure it out."
"The interment's at ten in the morning, though. There's no way we can get out to Fairfax and back up to Owings Mills all on Saturday morning. Not with traffic."
"I know. Let us work on the logistics, but we'll make it happen." Marz kissed her forehead.
"Derek's right," Nick said, coming up beside the couple. "We'll make sure you and your mom can say good-bye. Family's important." Kat didn't miss the way Nick's pale green eyes flickered her way.
"Thank you," Emilie said.
"All right," Nick said. "Everyone, go do your thing. We'll meet back here tonight. Let's say nine. Give everybody time for dinner after their shifts. Hopefully, we'll have some insight from the documents by then."
"Speaking of . . ." Marz came back around to his computer. "Well, it's eleven thirty, and Kaine's still at Pier 13. What are you up to, General?"
"Clearly, the meeting at the Pentagon was a lie," Beckett said. And it meant that Kat's gut hadn't let her down. Something was fishy with that whole meeting this morning.
"Sure looks that way to me," Marz said.
"Which means the whole thing, the whole meeting with Becca, was a ruse," Shane said. "The only question is what the agenda really was."
Everyone nodded, and this new confirmation of betrayal settled over the room like an anvil.
"Damn, I've got an idea," Nick said out of nowhere, whipping out his phone. "I'm gonna call Vance and see if he can go out to Pier 13. Maybe he can take some surveillance pics. Then we'll have a better idea of what Kaine's doing. And who he's doing it with."
Chapter 14
Beckett and Marz sat in a white van at the perimeter roadblock to the Hard Ink neighborhood, the warm May breeze blowing through the open windows. So that they could play their role as Baltimore Gas Company employees convincingly should someone come, their white and green company hard hats and yellow reflective vests lay at the ready atop the center console.
But everything was quiet. Empty. Boring as hell. And that was exactly when you had to find your focus, because you never knew when that normal, everyday quiet might just be the quiet before the storm.
Nick had left fifteen minutes before to do a whole-perimeter check. And his absence had Beckett's mind churning. About Kat. About himself. About their mission.
Sighing, Beckett shifted his bulk in the seat.
"Dude, what's up with you?" Marz asked.
Beckett cut his gaze to his friend sitting in the driver's seat. "What?"
"You're . . . fidgety," Marz said.
Forcing his body to stillness, Beckett realized that he'd been shaking one leg and kneading at the opposite knee. He gave a rueful laugh. "My leg's bothering me. How's that for irony?"
"Why is that ironic? How many surgeries did it take to rebuild that thing?" Marz pointed at Beckett's right leg.
"Three," he said in a low voice. That they'd been able to save it at all was a miracle given how close he'd been to that grenade when it exploded. Marz hadn't been so lucky.
"So, some discomfort's not that surprising." Marz looked at him like he didn't get it.
"It's just . . . I never . . ." Beckett shook his head.
"You are all up in your head, aren't you?"
"Yeah," Beckett said with a nod. "Okay. Here it is. I don't feel like I should bitch about my leg, which doesn't usually bother me much anymore, around you." His stomach squeezed at having made the confession.
"B." Marz's tone was . . . disappointed? Concerned? Confused? Beckett couldn't tell. "Have I said or done something to make you feel guilty about this?" He lifted his right leg, the one that now bore a below-the-knee prosthesis. Beckett knew it was there—he always knew it was there—even though Marz's jeans currently covered the limb.
"No. I managed that all on my own." A weird, fluttery sensation took off in Beckett's chest. Because words sat in his mouth. Words he was tempted to say. Words he'd never said to anyone else. But Marz was his test. If he couldn't say these words to Marz, how would he say them to Emilie? Or, Jesus, Kat?
"Well, damnit, stop. And that's an order." Grinning, Marz punched him in the arm.
And it was like the punch—playful though it was—knocked the words out of his mouth. "My father used to beat me." Oh, hell, saying that was a lot like jumping off a cliff. Beckett forced his gaze to meet Marz's. "Belts. His fists. His steel-tipped boots. A frying pan, once. One of my football trophies, another time." Was the interior of the van starting to spin? "Um, he locked me in a closet once for two days without any food or water. And the scar on my back that I told you was from climbing through a chain-linked fence . . . I, uh, lied. To everyone. He came at me with a knife. That was the summer after I graduated high school. The day after the stitches came out, I joined the Army."
Marz's face had gone sheet white. Eyes wide, mouth hanging open, his expression was almost comical. "Jesus, Beckett. I don't . . . why haven't . . ."
"I just wanted to forget about it, you know? Move on." He looked out the far window and didn't even bother trying to hold his knee still. It was either let it bounce or crawl out of his skin. "Except . . ." He shrugged. "It turns out you can't really forget being beaten and told you're worthless, and no good, and that nobody wants you or ever will. Best you can do is just shove that shit down whenever it rears its head. But it's like trying to bury water. It always finds its way to the surface again. Every damn time." Shit. Why was the corner of his eye wet? And where the hell did the air go?
"I'm really fucking sorry, B. Oh, Christ, I just punched you."
Beckett cranked his head toward Marz and arched an eyebrow. "Dude. Don't you dare start treating me differently."
A sharp nod. "Fair enough. As long as that goes both ways."
"Toufuckingché," Beckett said, chuffing out a half laugh.
"What happened? With your father, I mean?" Marz shifted in his seat toward Beckett.
"Why was he a mean bastard? Or, what happened after I left?"
"Both, I guess."
Beckett scrubbed his hands over his face. "He was always kind of a mean bastard, but things really went downhill after my mother died. She drank herself to death, and he said it was because she wanted to get away from me. That I'd made her so unhappy that she didn't even want to live. It was my fault."
"That is . . . so goddamned fucked-up," Marz said. "You know that shit isn't true, right?"
Beckett shrugged, and God, his stomach was so damn queasy. "Maybe?"
"Beckett. Alcoholism is a disease. And if your mother was depressed, that's a disease, too. You didn't cause either of them."
He didn't debate it. His head sorta knew these things, but his heart . . . well . . . his heart had bought some of what it'd been told at a very young age. And it turned out that organ had a damn long memory. Staring out the windshield, he said, "As far as I know, nothing happened to him afterward. I think he still lives in my childhood home outside of Pittsburgh."
"Please tell me you know that the shit he did to you wasn't your fault, and that the horrible things he said to you aren't true." When Beckett didn't answer, Marz said, "B, look at me."
He did. And he did it knowing full well that he hadn't put his usual mask in place.
"Aw, Beckett."
Beckett dropped his gaze to a space somewhere in between them. "Do you mind . . . if, uh, I talk to Emilie?" he managed in a quiet voice.
"Shit, no. I'm fucking relieved you raised it because I wasn't sure how to do so without risking you tearing my head off."
Another half laugh as he looked down at his lap. "Yeah. That's my usual M.O. Just striking out in anger."
Marz shook his head. "Don't do that. Don't discount yourself. Whatever anger you have, you came by it honestly. Jesus."
They sat in the silence of the van for a long moment.
"When . . ." Beckett had to force down the lump blocking his throat. "When you lost your leg, all I could think was that you should've let that grenade take me out. That I wasn't worth saving. Definitely not over you." Damn if that lump wasn't staying right where it was. Uncomfortable pressure built up inside his chest, behind his eyes.
"That's your father talking, Beckett. Your mean abusive, sonofabitch of a father. You were the first family I ever had. Do you hear me? I would've gladly given my life for you." Marz squeezed Beckett's shoulder. "I have no regrets, Beckett, and I wouldn't take it back if I could." He squeezed harder. "I would save you all over again."
Beckett peered up at Marz, unsure what to say, blinking fast against the stupid fucking sting in his eyes. The words were both soothing and painful at once. Soothing to know another human being thought of him that way, painful because parts of Beckett still didn't believe it. Couldn't. All he could do was hold his breath and grit it out against the pain, because if he let himself breathe, if he let himself feel any more of this overwhelming wave of emotion than he already felt, he was gonna end up sobbing right where he sat.
And Marz totally knew it. The guy nodded at him, his expression sympathetic without being pitying. And thank fuck for that.
"Why do you think this is all coming up for you now, man?"
Swallowing hard over and over, Beckett didn't talk until he was sure he could suppress the feeling that his chest just might tear right open. Finally, he said, "Didn't realize just how hard losing the Army hit me until we all got back together again." His voice was thin and tight. "And then, last night, I hurt Kat." God, admitting that out loud made him feel like such shit.
"What happened?" Marz said, his tone even, nonjudgmental.
"I'm not sure. She won't tell me. We fell asleep together and I had a nightmare." Beckett shrugged. "I did something to her." Heat crawled into his cheeks, but he forced himself to look at Marz.
"Kat's strong, Beckett. And she doesn't take shit. She doesn't seem the least bit intimidated by your usual hard-ass routine."
"No," Beckett said. The corner of his mouth lifted at the thought. Just a little. "But I don't think I should trust myself around her until I get this shit under control. At least a little."
Marz braced a hand on the steering wheel. "You should talk to her about it, Beckett. If you want something with her, you're gonna have to include her in all this. If you totally shut her out, she might not wait around for you to decide to open up again."
And that scared the shit out of Beckett. Now that his mind had gone and imagined the very idea of being with Kat, he was deathly scared she'd realize what a fucked-up POS he was and split. And that would be a problem. Because somewhere over the past six days of knowing her, he had fallen into some pretty serious like with the woman. That was kinda major for him. To feel desire. To feel affection. To feel anything at all, really, except for numbed-over anger and worthlessness.
After a long moment Beckett nodded. "That's why I gotta talk to Emilie. And to Nick. Which I should probably run by Kat first." No probably about it. She'd told him to keep what they had private, which meant he needed to get her permission and explain why he wanted it. Hell. He heaved out a long, weary breath.
"That's a lotta talking, for you." Marz winked.
Beckett gave a small chuckle, and damn if the humor didn't feel good. "That's a damn lotta talking. But I guess it's overdue."
Overdue and necessary. Because he and Kat couldn't be anything more than they were until he did.
IT WAS QUARTER after seven, fifteen minutes beyond the end of Kat's shift. Where the heck was her relief?
Though, honestly, she'd been glad for the time alone. Because it meant she'd had the privacy to return the court's call and finally get confirmation that the police had served Cole the temporary protective order. Interesting how quiet her phone had gotten after that, just as she'd expected. Thank God for small favors.
Footsteps echoed from the stairwell of the old building, and Kat hugged herself and rubbed her arms, glad she'd brought over a second T-shirt so she could layer up. The sun had nearly set, and the air hung right on that line between daylight warmth and evening cool.
The footsteps reached the landing, and Kat turned to see who was here to relieve her.
"Wait, you just did a shift. What are you doing here?" she asked Beckett. Though, unlike earlier when he'd come to her bedroom and she really hadn't wanted him there—well, not at first, now she was pleasantly surprised.
"This," he said, marching up to her, muscles flexing, eyes blazing, limp noticeable but not slowing him down at all. He took her tightly in his arms and kissed her like she was the air he needed to breathe.
Kat melted. No part of her even considered not surrendering to the warm grip of his wandering hands, to the demanding caress of his tongue on hers, to the almost instantaneous bulge rubbing maddeningly against her belly. And, oh God, those little groans he made in the back of his throat had her core clenching with need and want.
"I . . . missed you," he whispered as his mouth kissed a trail to her ear, her jaw, her throat. Her chest gave a pang at the admission, but then his tongue was back in her mouth, making her climb him until her legs were around his waist and that bulge pressed and rubbed right where she needed him most. "I want in, Kat. Will you let me?"
She moaned and ground herself against his cock. Would it always be like this with them? From zero to molten-hot, animalistic attraction? Because she was already wet for him from just the kiss, from just the dozen words he'd uttered. "Yes, Beckett. I need you."
He set her down and his hands went to the button of her jeans. Suddenly, he froze. "Is it going to be too cold for you?"
Kat smiled, her eyebrow arching. "Not if you only remove what you absolutely have to." She didn't know why she thought it was so hot that they hadn't even managed to get undressed most of the times they'd been together, but she totally did.
Beckett's grin was absolutely wicked. "I'm down with that." He tugged her jeans and panties to mid-thigh, then pushed her backward, step by slow step, while kissing her neck and running his fingers through the wetness between her legs. Her back came up against the brick wall of the old warehouse. "But someday soon I'm going to strip you down, spread you out, and do this right." He winked.
Holy crap. If the amazing sex they'd already had wasn't "right," she might not survive the sex he thought was. "No complaints here," she said.
He sucked and nibbled on her bottom lip. "That's good. Now, turn around and stick out that ass."
"So bossy," she said, her core clenching at the command. And at the sexy, low chuckle he gave.
And then he went down to his knees behind her, and grabbed her ass in a way that opened her up and forced her to further arch. She nearly screamed when he plunged his tongue inside her wetness and flicked and sucked and lapped at her over and over again. Relentless. Demanding. And so damn sexy that Kat thought she'd lose her mind.
Then he added his thumb to the mix, using it to worry her clit fast and hard and firm, and she came apart in his hands. Her fingers clawed the brick wall, her breathing halted and then whooshed out of her on a long moan, and her knees went soft.
He kissed her lower back. "Hearing you come is my new favorite sound," he said, his breath ticklish against her skin.
"Oh, God, Beckett." She held onto the wall and hoped the room stopped spinning around her. She felt him rise to his full height, the denim of his jeans rough against her bare cheeks. And despite the incredible satisfaction he'd given her, the sounds of him undoing his pants, pushing them down, and jerking his cock in his own hand revved her right back up again. He smacked his long, hard length against her ass, and she moaned.
"You're gonna feel so fucking good," he said, his fist bumping against her as he stroked himself. "I just know it."
She peered over her shoulder, and God, the sight of his big hand wrapped around that gorgeous cock was erotic as hell. "Come find out."
He closed the distance between them, grabbed her hip and teased her opening with his thick head. "You want me, Kat?"
"Jesus, Beckett, I'm dripping for you."
"Fuck," he said on a groan, and then he sank deep.
"Oh, yeah," Kat rasped, loving the overwhelming way he filled her. "That's it. God, you're so damn big."
His arms came around the front of her, one hand settled on a breast, the other on her throat. He was hunched around her so they touched from thigh to head. He pressed his lips to her ear. "And you're so damn small. Don't let me hurt you. Never let me hurt you, Angel. Promise me." He withdrew on a slow, tormenting stroke.
Kat wasn't sure whether the words or the pleading tone behind them tugged at her heart more. "I won't, Beckett. And you won't." She tried to push herself back on him, but he was holding her so tight that he controlled her movement entirely. "I need you."
"You have me, Kat." He bottomed out inside her again. "Have me."
He kept up the slow-and-deep rhythm until Kat was moaning and begging and clenching and unclenching her fists against the wall. "Oh, God, Beckett. Please let go. I need you hard and fast and unrestrained."
"Yeah?" he said, licking the shell of her ear.
"Yeah," she whimpered.
And then he boxed her up tight against the brick, his arms keeping the front of her body away from the rough surface, and fucked her until she couldn't talk, couldn't see, couldn't breathe. The head of his cock hit a sensitive spot inside her again and again and again. Standing with her legs together made her so tight that she could feel her own slickness on her thighs. And the tight grip of his hands on her breast and her throat shot bolts of electricity through her blood. It was spectacular.
"I want you to come again," he rasped. "Come all over me."
"Just don't stop," she whispered. "Just . . . don't . . . oh, God, coming." Her body tightened until it was almost painful and then splintered apart in a thousand floating pieces. Her moan was long and low.
"So fucking good, Kat. Jesus."
When she could catch her breath, she reached her hand back and lightly grasped the side of his head, her mouth finding his for a kiss. "I love this with you," she said. The ferocity with which he kissed her made her stomach flip. But as true as the words were, it wasn't just the sex she loved with Beckett. She loved that he could be so hard yet so soft. She loved that he pushed her yet supported her. She loved that he could apologize and that, for all his incredible strength, he could show vulnerability, too. Even if it was in his own fucked-up way. And as much as she liked to bitch about it, she kinda loved how he drove her crazy, because she was pretty sure she did it to him, too.
What all that meant, especially in the midst of this crisis, she wasn't sure, but there also wasn't any hurry to define it right now. Was there?
Beckett groaned, and the sound of his pleasure spiked her own. His hips flew against her, the smacking of skin on skin and their rough breaths loud in the otherwise quiet space. "Coming. Fucking coming in you." His hold on her tightened until she could barely breathe, but she wouldn't have changed it for the world. And then his cock pulsed inside her over and over again. "God . . . damn," he gritted out.
When Beckett finally withdrew from her, Kat missed him everywhere. His heat. His hold. His body inside hers. But he gently turned her, then grabbed some napkins from the supplies. Kat cleaned herself up and they quietly reassembled themselves. All the while, she hoped like hell that things wouldn't go from awesome to weird as they had before.
Expression serious, Beckett took her hands in his.
And Kat's heart fell to the floor. Here it comes again. She could already hear him saying that this was a mistake—
"I don't want to hide us anymore," he said.
Wait. What? "You . . . what?" His words were so unexpected, she half felt like she'd just walked, cartoonlike, into a pole she hadn't known was there.
"I don't know what we might be, Kat, but I know what I want us not to be. I don't want us to be some dirty little secret. And I don't want us to pretend around everyone else that we don't care. And I don't want us to just be quickies in a sniper's roost." Those bright blue eyes absolutely blazed at her.
Kat's head spun. Not because she didn't like the sound of what he was saying, but because she was gobsmacked that he was the one saying it. Mostly, she'd contented herself with thinking quickies in a sniper's roost were all they'd ever be, no matter how much her heart found deeper, sweeter things to like about the man.
"I won't say anything if you don't want me to, though," he said, brow furrowing and shoulders sagging. Just the littlest bit.
But it was enough to make her realize she hadn't responded adequately, or, like, at all. "No, no. I'm sorry. I'm just surprised. But I'd like not hiding, too. At first, well, I wasn't sure . . ." She shrugged. "But there's clearly something here, Beckett. And I'd like to know what it might be without feeling like we're doing something wrong."
"Exactly," he said, hope filling his eyes. And God if that didn't make them even prettier.
She smiled as butterflies whipped through her belly. Kinda ridiculous, given that she was a twenty-nine-year-old woman, but the feeling was there all the same. "So, uh, what were you thinking? And, oh, by the way, I'm pretty sure Jeremy might've guessed something's going on already."
"Yeah, I caught that. Marz, too, by the way."
"Damn, they're good." Kat chuckled.
"Or just giant busybodies."
Grinning, Kat nodded. "They could be that, too."
Beckett cupped his big hand around Kat's face. "The first thing I'd like to do is talk to Nick." Kat couldn't hold back the groan that spilled out of her, and it made Beckett chuckle. "Trust me, I'm not looking forward to that conversation either. But he's one of my best friends and my teammate. If he doesn't hear it from me, he's gonna be pissed. Or more pissed, as the case may be."
"Yes, well. He excels in that," Kat said, still smarting from the morning.
Beckett's thumb stroked under her eye. "I agree he crossed a line, Angel. But try to remember that he was worried out of his mind about both you and Becca. He wasn't mad so much at you as at the possibility of losing you two."
Kat turned her head and pressed a kiss into Beckett's palm. "I know, but the whole reason I took the risks I did was to make sure Becca got back to him safely. I know what she means to him. Hell, I love her myself. I'd never let anything happen to her. Or Charlie, for that matter. Because I know what the two of them mean to my brothers. It just would've been nice for him to at least recognize I brought her home to him. I did the job he asked me to do."
"You did," Beckett said, giving her a small, soft kiss.
Crossing her arms, Kat pouted, just a little. "I'm glad he has you hard-asses to stand up for him, too, even though I'm irritated as hell at him."
Beckett pressed his lips into a line, and it was clear he was trying not to smile.
"Shut up," she said.
He couldn't resist grinning at that, and he put his arms around her. Trying not to laugh, Kat pushed him away. "Aw, don't be like that, Angel."
"Don't think that Angel crap is gonna get you out of everything either." She was totally failing at holding back the smile now, and his laughter proved it.
He held his thumb and forefinger close together. "It helps a little though, right?"
She swatted at his hand. "No, Trigger, it doesn't."
Beckett's eyes went wide, then narrowed to slits, the scars around his eye making the expression seem even more severe. "You did not just go there again."
"I totally did," she said, owning her grin now. She backpedaled as he stalked after her, her spine coming up against the brick again.
He grabbed her hands and pinned them to the wall on either side of her. "What am I going to do with you?"
Kat waggled her eyebrows. "The possibilities are endless."
Nodding, his smile slowly slipped off his face. "Can I tell you something?"
Frowning, Kat nodded. "Yeah, of course. What just happened?"
His brows slashed down and he let go of her hands. She rested them on his chest, concern sloshing into her belly. "I want you to know . . . I'm going to talk to Emilie tonight after our meeting. I had a sonofabitch for a father, Kat. Never missed an opportunity to beat me up or tear me down." He swallowed hard.
Kat's heart pounded in grief and outrage for him. "That fucking sucks, Beckett. I'm so sorry."
He shook his head. "Thing is, the shit my father did is like a poison inside me. I need some help to get it out."
Throwing her arms around his neck, Kat held on as tight as she could, given the disparity in their height. "Whatever you need, I am here for you." She pressed a kiss to his cheek, her lips caressing the mass of scars by his eye. Were those from his father or war? Either way, she hated that life had been so hard on this man.
Beckett leaned down so he could return the hug more firmly. "I don't want it to poison you, too."
Damn if that sentiment didn't make her chest bloom with warmth and sadness. This sweet, strong, tough-as-nails man had all kinds of rough edges . . . because someone had given them to him—or, by the sounds of it, carved them into him.
And, oh God, the pleading words he'd murmured in his sleep last night came back to her. Why you doing this? What did I do?
All at once the things that had so confused her about Beckett Murda became instantly, heartbreakingly clearer. And now she knew what those defense mechanisms were protecting him from.
"You could never do that, Beckett. You're a good man, do you hear me?"
A fast nod of his head against hers. And then, finally, he released her from the hug. "I'm not sure what I'm looking forward to less—talking to Emilie or your brother."
Kat gave a small smile. "Don't be afraid of my brother. You can take him."
He chuffed out a laugh. "I don't know. Nick is fast on his feet."
"Yes, but we have a secret weapon that will totally disarm his anger, if necessary." Kat grinned, the wheels in her head turning.
"We do?" he asked.
"Yep. And her name is Becca."
Chapter 15
"Was anyone able to learn anything else?" Nick asked the group as Kat and Beckett joined everyone in the gym. They were the last two to come to the nine o'clock meeting, and the fact that they both brought a sandwich and a drink with them clearly indicated that they'd been otherwise occupied during what was supposed to have been dinnertime. But now that they'd agreed to go public, Kat didn't really care. Although that didn't mean she didn't want to smack the smirk off Jeremy's face when she and Beckett sat down together.
Pain in the ass brothers.
Oh, great. And Marz was kinda smirking, too.
"Yeah, we did," Charlie said. He counted off on his fingers. "We found a number of places—though we're still compiling just how many it might be—where there are notations that read, 'K orders,' or 'K contact.' Marz and I had noticed that before, but we didn't know what it might mean. In light of today's developments, it seems pretty clear it refers to Kaine."
"That's a little loose, though," Shane said, hands braced on the back of the folding chair in which Sara sat. "Isn't it?"
"I don't think so," Marz said. "I read some of the ones where it says 'K orders,' and I can remember a few of those missions. More importantly, at least two of them involved changes to existing orders that I recall came to Merritt from above. I'm sure if Nick looks those over, he could recall even more."
"So," Nick said, "Merritt was noting places where Kaine had ordered us to transport to a different heroin disposal facility. That kinda thing?" His arms were crossed, his face set in concentration.
"Exactly," Charlie said. He pressed against his next finger. "Those notations match up with times when my father's records show a disparity between pounds of heroin dropped off for disposal and pounds of heroin actually destroyed."
"No one's really watching any of that over there, are they?" Easy asked, shaking his head.
"No. And I'm sure they were counting on exactly that," Nick said. "Jesus."
"But what they weren't counting on," Beckett said from next to Kat, "was the fact that Merritt was watching, was taking note of the doctored books. And when Kaine found out, that threatened the entire operation."
"Very likely," Marz said, nodding. "Charlie found something else, too."
"Yeah," Charlie said in a quiet voice. He sat forward and grabbed a sheet of paper off the desk, using his uninjured hand. "I hacked into the Singapore bank again." Expressions of surprise went up around the room. Kat had known Charlie was good, but she clearly hadn't realized just how good he was. She supposed it should've been weird that her first reaction to the news of his illegal activity was to be impressed instead of outraged. Being neck deep in this whole situation had shifted her perspective. "Wasn't as easy as last time because there were levels of security and detection that weren't there a month ago. But I managed to stay in long enough to confirm this." He handed the paper to Nick.
"Sonofabitch," Nick said, eyes glued to the sheet. "Kaine has a bank account there, too. Only his is worth quite a bit more."
"Well," Charlie said, "he's also had more than a year's additional time to accrue principle and interest, now, hasn't he." It wasn't a question, and Kat wasn't sure she'd ever seen Charlie radiate anger the way he did just then. Not that she blamed him.
"No other way to read that, then," Marz said, pointing at the sheet, "other than Kaine being on the take."
"Find anything else in the documents I gave you?" Kat asked. She'd nearly devoured the turkey sandwich she'd made. Brushing bread crumbs off her fingers, she set her empty plate on the floor beneath her chair.
Charlie nodded. "Since Kaine was kind enough to call Becca using his cell phone, we were able to search for that number in the phone call log file. Numerous instances of calls between the now-defunct extension at Seneka and Kaine's cell. None since, though, which is . . . so not helpful." He shrugged. The news added more credence to Kat's belief that she might've done the wrong thing but for the right reason.
"Well, think about it, though," Beckett said, rubbing his jaw. "If they thought they'd been compromised enough to take that extension out of service, they could've decided that using things like personal cell phones was way too risky, too. Especially since calls to Chapman go through a switchboard, so you can only trace them so far."
"Who knows how many of the calls between Seneka and Chapman involved Kaine and this situation," Nick said, nodding at Beckett.
Marz nodded. "Seems to me that we've got the evidence we need to confront Seneka, and I might've found the perfect way to do it."
Nick shifted feet, and Kat didn't miss the small wince that flashed across his face. Her gaze went right to his hip, which she couldn't see under his jeans and black T-shirt, of course. But it made her realize she hadn't once asked how his back was feeling since she'd arrived at Hard Ink. During the ambush that led to this whole crazy situation, he'd been shot in the back twice, resulting in a fractured pelvis, perforated bowel, and lingering nerve damage. Sometimes it bothered him enough that he had a small limp—nowhere as pronounced as Beckett's, but enough to be noticeable.
Nick heaved a breath and raked his fingers through his hair. "Marz, have you by any chance looked to see if there are any calls directly between Kaine or Merritt and John Seneka's direct extension?"
"No," Marz said. "But that would be easy enough to do." He turned his attention to the computer for a few minutes, his hands clacking against the keyboard every so often, then shook his head. "Not a one. What're you thinking, hoss?"
Nick frowned, then his gaze cut to Kat. "Is John Seneka aware of your office's investigation into his company?"
"Yes," she said, wondering where Nick was going with his question. "He doesn't know the full scope of the investigation, but our work is part of an official congressional inquiry into a whole host of issues with Seneka. As far as I know, though, Mr. Seneka's been at least somewhat cooperative. He even provided some of the materials in our files voluntarily."
"I guess I'm just wondering how high up within Seneka these activities go," Nick said, his hand pressing almost absentmindedly against his lower back.
Kat looked around for an empty folding chair, but with pretty much everyone here, there were none nearby. Without a word, she got up, grabbed her chair, and carried it over to him, placing it backward in front of him. "Sit." She gave him a look, but didn't wait to debate it with him, then she went and stood by Beckett, who made her take his seat.
As Nick straddled the folding chair the way he always did, Becca threw Kat a look of appreciation, then moved behind Nick and placed her hands on his shoulders.
"That's an interesting question, Nick," Beckett said, looking around at everyone. "Equally interesting is the question of exactly who Frank was working for. No way he took this op on by himself."
Nick nodded. "I was thinking about that, too. If we could find who that person is, we'd presumably have another ally in all this."
"I've been keeping an eye out for any information about who Merritt might've been reporting to or working with in the documents," Marz said. "So far, nothing."
"Where does all this leave us, then?" Nick asked, his voice tired.
Everyone traded looks.
Finally, Marz said, "We could drag this out as long as it takes to read through everything, but I don't think we have that kind of time with the Ravens." Nick nodded, and agreements rose up from the men. "I think we have enough to confront Seneka. And our bargaining chip is the twelve million dollars. Today's meeting with Kaine proves they're still after it, so maybe that gives us some leverage. And, before you ask, I have an idea how to make first contact with Mr. Seneka himself, thanks to Kat's records."
Kat's gaze cut to Marz's. He winked at her. "What did you find?"
"John Seneka very helpfully has a personal zmail account. And Charlie and I have figured out how to hack it, take it over, and force a z-video chat with him the next time he logs in."
Once again Kat's reaction wasn't what she thought it should be. Instead of being aghast like she might've been a week ago, she was impressed by Marz and Charlie's evil genius.
"Shit, really?" Shane said, a hint of a grin playing around his handsome mouth.
"How do you know you can do it?" Easy asked, his expression just shy of hopeful.
"Oh," Marz said, grinning at Charlie, whose face lightened for maybe the first time during the whole conversation. "Because we hacked in about an hour ago. Seneka definitely checks that account daily, though usually in the morning, if the time stamps on his replies are any indication. Which means we can a hundred percent confront him tomorrow. And do it on our own terms."
BECKETT WAS A wrung-out mess. He'd just spent the past hour talking to Emilie, and it was maybe more than he'd talked at one time in his whole life. He couldn't exactly say he felt good, and maybe not even better, but he could say that he was a little proud of himself for facing his past. Didn't matter that dealing with that shit was as comfortable as swallowing crushed glass, because choking it down was the only way he'd have a chance at a future.
At least, a future that he really, truly wanted.
One where he wasn't alone. One where he wasn't angry. One where he wasn't wasting the time he might have left on this earth—time his seven fallen teammates no longer had.
Jogging down the stairs from the third-floor apartment, Beckett headed to the gym. No matter how trashed he felt, he had one more conversation that absolutely had to happen tonight.
With Nick. About Kat.
Beckett had seen the all-knowing looks on Marz's and Jeremy's faces when he and Kat joined the all-hands' meeting earlier in the evening. If he didn't go to Nick soon, the guy was going to learn about them some other way. And then the hell was going to be even worse to pay. And Beckett didn't want that to happen. Not just because he didn't want the hassle. But because he owed it to Nick—as his friend and his brother—to come clean.
Didn't mean that Beckett's gut wasn't all twisted up about whether Nick would think him good enough for his baby sister, though.
He reached the gym door, took a deep breath, and punched in the pass code. Man up, Murda. Right.
Inside, things were hopping, despite the fact that it was after eleven o'clock. Some Ravens were lifting weights. Others were shooting the shit around the table. And two guys were tossing a ball back and forth, driving Eileen crazy by bouncing it over her head. Easy was pounding out a fast pace on the treadmill. And, in the back corner, Nick, Marz, and Shane were poring over something on Marz's desk.
Beckett had faced down terrorists who wanted nothing more than to take his life with their very hands, and he'd probably never been as nervous as he was right now. For fuck's sake.
He moved into the room. About midway across, he spied Cy peeking out from behind a weight machine. "You and me are gonna have a talk soon, too," he said, pointing at the cat. A one-eyed blink was his only answer.
And then he was standing on the far side of Marz's desk next to Shane, peering down at what the three men were examining.
Marz gave him a smile and tapped his finger against the pages. "Just going over all these instances when Merritt noted Kaine as having changed orders regarding counternarcotics missions."
"Marz was right," Nick said, looking up at him. "Some of these gave me a bad feeling in my gut at the time, because they seemed out of character for Merritt. He was never a last minute kinda guy. Every damn time he's marked those changes as having been ordered by Kaine."
Beckett nodded. And then his mouth acted without his brain's permission. "I'd like to date your sister."
Oh, Jesus. Had he just blurted that out?
Uh, yes. Yes, he had, if the expressions of the three men now gawking at him were any indication. Mouth and eyes wide, Shane's was totally stunned with a side of Oh shit, this is gonna get interesting, fast. Shaking his head, Marz seemed half amused, half exasperated. And Nick . . . well . . . the frigid look he'd thrown at Marz earlier in the day didn't begin to compare to this.
"Come again?" Nick said, straightening to his full height and nailing him with a subzero stare.
"Uh, maybe we should . . ." Beckett nodded his head to the side. " . . . find a place to talk."
Nick crossed his arms. "Right here works for me."
Beckett swallowed. "Okay, uh . . ." His gaze flickered to Shane and Marz, who simultaneously looked like they wanted the floor to swallow them up and that they might burst out laughing. Fuckers. "I like her, Nick."
Nick's gaze narrowed even further. "No, you don't. You two drive each other fucking crazy."
Releasing a deep breath, Beckett shook his head. "I like her."
Tilting his head like Beckett was a puzzle he was trying to solve, Nick finally said, "I've known you how many years? Never once saw you get attached or even want to."
"I want to now. And I'd like your okay on that, because you're my friend and my teammate." Beckett crossed his arms.
For a long moment, they both stood there, positions mirroring one another, separated only by Marz's desk, like some sort of Old West showdown.
"She's my baby sister, Beckett. Jesus," Nick finally said, planting his hands on his hips and shaking his head. "I pulled her pigtails and taught her how to box and remember that she had a stuffed elephant named Wuzzywoo that she carried with her everywhere."
Beckett tilted his head, a sliver of hope daring to trickle into his chest. Because Nick wasn't saying no. "Wuzzywoo? Really?" He glanced at Marz and Shane and then did a double take, because they were so close to exploding with laughter that their faces were red and Marz had tears pooling in his eyes.
"Yes," Nick snapped. "Its goddamned name was Wuzzywoo!"
That was it. Their two asshole friends lost it. Just flat-out lost their minds. Marz laughed so hard he half laid down on his desk. Shane held his gut and braced his hands on his knees.
"Oh, God. Can't . . . breathe," Marz choked out. "He just . . . he just . . . came out and asked him . . ."
"I know . . ." Shane said, slapping his hand on the top of the desk.
Nick pointed at both of them. "You're a fuckstick. And you're a fuckstick." Then he turned on Beckett. "And you better not be a fuckstick, or I will kick your ass until your children's children can't sit down. You got me?" Before Beckett had even inhaled to reply, Nick threw his hands out and loudly added, "Not that there better be any goddamned children. Fuck!"
More laughter from the peanut gallery.
Beckett bit back his smile. "Right. Don't be a fuckstick. No children. Got it."
Nick braced his hands on the back of a chair. "It really has to be my sister you decide to go and get the feels for?"
His smile finally won out. Nick wasn't saying no. "Uh, that's an affirmative."
"I kinda wanna punch you right now," Nick said, slamming the chair down against the floor to punctuate the point.
Better the chair than him. Beckett grinned. "I was sorta expecting that."
"Goddamnit," Nick said. "Are we done here?" He gestured at the desk area. Marz and Shane, who had mostly pulled themselves together, managed a nod. "Good. I need me some Sunshine. I'll see you fuckers in the morning." As the guys croaked out good-byes, Nick came around the desk and clasped Beckett's hand. "For real, man. Don't hurt her."
"It's the last thing I wanna do."
Nick nodded and huffed. "Goddamnit." And then he stalked away.
Beckett turned back to Marz and Shane. "Well, that went pretty good. Don't ya think?"
Marz held out his hands. "Dude. You just blurted it the fuck out."
Scratching his head, Beckett nodded. "Yeah. Didn't really mean to do that."
They both offered him congratulations, and then everyone agreed to break til the morning. And Beckett was down with that, because he knew exactly what he wanted to do. He almost floated to Kat's room. Or he would've. If he floated. Which he definitely did not do. Ever.
He knocked, and when there was no answer, peeked inside. Empty. Next he tried the last door on the right—Jeremy's room. Well, Jeremy's and Charlie's now.
"Come in," came a voice from inside.
Beckett popped his head in and founded Jeremy and Charlie sitting sideways on the big bed, their backs against the dark green wall and a laptop on Charlie's lap. Eileen lay in a black-and-tan ball between their legs. Nick must've brought her back over. "Seen Kat?"
Jeremy grinned and flicked his tongue at the piercing on the side of his bottom lip. "You're the guy."
"Uh . . ."
"You're the lucky guy. I knew it. I knew she had something going on." Jeremy tapped Charlie twice on the leg and pointed at Beckett. "I knew it."
Charlie chuckled. "Sorry. Jer's been 'on the case,' " he said, using air quotes. "For the past few days. Ya gotta let him gloat a little."
Beckett shook his head, not sure what conversation they were having right now. "Yeah yeah, sure. But, uh, Kat?"
"She's taking a shower," Jeremy said, waggling his eyebrow. As Beckett retreated, Jeremy yelled, "We share a wall, so keep it down over there."
Holy crap on a cracker, it was gonna be never-ending shit from here on out, wasn't it? But then again, that seemed a small price to pay for getting Kat. Especially Kat in the shower. Kat, wet, in the shower.
He knocked twice on the hall bathroom door. No answer, but the spray of the water sounded out from the other side. He looked both ways and found the coast clear, so he ducked inside and locked the door, eager to tell her that Nick wasn't going to kill him. At least, not today. And they hadn't even needed Becca's intervention.
Making quick work of getting undressed, Beckett dropped his clothes into a pile by the door. And then he pulled back the curtain and stepped in. "Mind some company?"
Kat spun, and her smile was immediate. She gave him a slow up and down perusal that said she liked what she saw. And damn if his body didn't react to that, especially since he was perusing right back. "I don't mind at all. This is a nice surprise."
He stepped closer, the warm spray reaching over her petite shoulders to hit his stomach, his hardening cock, his legs. His gaze dragged up her curves, drinking in every gorgeous inch. If he'd thought her beautiful before, it was nothing compared to—
Dark marks on her arm. His gaze flicked to her other arm, where more dark marks—bruises—discolored her skin. He gently grasped her right wrist and lifted.
Kat's eyes went wide and she tugged her hand away. "Beckett—"
"What. Is. That?" He studied the bruises long enough that they began to take shape . . . Fingerprints. Kat had fingerprint bruises on both arms. From . . . being grabbed? Beckett's stomach dropped to the floor. His nightmare. His goddamned nightmare. He took a step back, his hand going to his forehead. "I . . ." He shook his head, half sure he was gonna throw up. He'd grabbed her so hard that he'd bruised her? Jesus, he was a fucking animal. And he thought he actually had a chance with her. He thought he could actually deserve her.
He really wasn't any better than his damn father.
Beckett was out of the shower in an instant. Not bothering to dry off, he tugged on his boxers and jeans, his head nearly spinning.
The water shut off behind him. "Beckett, wait. Look at me."
Couldn't. It made him a fucking coward. He knew it did. But he couldn't look again at the evidence that he'd hurt her.
"I . . . I don't understand," she said, stepping out of the tub behind him.
"Me neither. You told me it was nothing, Kat. You fucking told me—" He bit his tongue and shook his head, but he couldn't keep it all in. He whirled on her and faced the damage he'd caused, even though it was the last thing he wanted to do. "I grabbed you so hard in my sleep that I left bruises all over your arms. Why aren't you fucking scared of me? You're a smart goddamned woman. Use your head. I'm no good for you."
She grabbed a towel and wrapped it around her body, tucking it beneath her arms in a way that left the bruises exposed. "No, Beckett—"
"I swear, if you make an excuse for me right now, I will lose my mind." He bent down and grabbed his T-shirt. Stuffed his arms and head through the holes.
Her expression was so damn sad it broke his heart. And he thought he'd felt pain before. Not even close. Not even close to losing something you never quite had. "I don't have to, Beckett—"
"Good. Don't." He turned for the doorknob, but his fingers were fucking wet and wouldn't grasp the lock.
She grabbed his arm. "Damnit, Beckett. It wasn't you. Okay? It wasn't you who grabbed me. You're not the one who caused my bruises."
Chapter 16
Beckett froze. His mind. His body. His heart. All of it stopped cold.
He turned around to face her, his gaze looking over those bruises in a whole new way now. "What did you say?"
Eyes pleading, forehead furrowed with worry, she shook her head, sending water droplets from her wet hair down the pale, lovely skin of her chest. "You weren't the one who did this," she said. "Not you."
Gently, so, so gently, he grasped her wrist and lifted her arm again. One dark fingerprint on the inside of the arm, two—no, three—on the outside. Same on the other side. A flash fire roared through his mind. "Was this a fucking Raven?" Because, who else could it be? None of the team would do this. Who did that leave?
A fast shake of her head. More droplets rushed down her skin. "No, no. No one here." She dropped her gaze somewhere in between them, her expression pinched, like she was in pain.
Knock, knock.
"Are you two having sex in there?" came Jeremy's voice from the other side of the door.
Kat's shoulders sagged. "Come to my room with me?"
Beckett was nearly numb from the whiplash of emotions he'd experienced during the past few minutes. Terror, self-loathing, soul-deep disappointment, tempered relief, anger. Always, the anger. He managed a nod and turned and opened the door.
Jeremy pointed to the wet spots seeping through Beckett's T-shirt. "Think you're supposed to take the clothes off, there, big guy . . ." He frowned, his gaze ping-ponging between them as if he'd just noticed something was wrong.
Not wanting to risk saying something he shouldn't, Beckett stalked by Jeremy without saying a word. But that didn't stop him from hearing the conversation that took place behind him.
"What's the matter?" Jeremy asked Kat.
"Nothing," she said, her voice not at all convincing.
"Hey, what happened to your arm?"
Pacing inside her dark room, Beckett didn't hear the answer to that, and he was glad. Because if she told Jeremy before she told him, he might just lose his mind.
A moment later she walked in, turned on the bedside lamp, and closed the door. "Mind if I take a minute to get dressed first?"
Beckett managed a head shake, his gaze glued to her as she dried herself off, slipped on a pair of pink satiny panties, and then pulled on a pair of black cotton pajama bottoms and a form-fitting emerald green tank top. Part of him wanted to go to her and warm her air-chilled skin with his. But he couldn't. Not yet. Not until she explained those marks. Not until she made him believe that he wasn't responsible for them—and understand why she hadn't mentioned someone hurting her before.
She whipped a brush through her hair until it lay smooth over her shoulders, the color of it almost black from the wetness. Then she walked up to him, took his hand, and guided him toward the bed.
He pulled his hand free. "Kat—"
"Come with me, Beckett," she said, taking his hand again. When his feet remained planted, she looked him in the eye. "You need this. And so do I."
He frowned, and his feet got unstuck real quick. He didn't exactly understand what she meant by those words, but somehow they still resonated inside him. He followed her to the bed and watched her pull down the covers then climb to the middle. She turned to face him.
Beckett sat on the edge. Upright. Rigid. Stressed the fuck out.
Kat slipped in behind him, her knees around his hips, her arms around his chest. She laid her head on his upper back. "I'm sorry I let you think for even a second that you'd hurt me." Her voice was soft and sad. "I didn't mean to. I was just caught off guard."
Grasping one of her hands, he pressed it more firmly to his chest. "I need to know, Kat," he finally said.
She turned her head so her chin rested on his shoulder. "It was an ex-boyfriend. Who has been rather intent that we get back together." She heaved a weary breath.
Beckett turned within the circle of her arms to face her, and her hands slipped away, coming to rest in her lap. "Cole," he said, her words flooding back to him. I should almost thank Cole for jump— Realization smacked Beckett between the eyes. "He jumped you?"
She ducked her chin. "In the parking garage of my building. Came up behind me when I was going to unlock my car door. My hands were full, so I didn't even manage to react at first. He pinned me to the wall by my car. I got him to let me go by promising I'd meet him that night to talk."
He was gonna kill this guy. Whoever he was. Wherever he was. He was already dead and he didn't even know it. "When?"
"Friday morning."
Aw, hell. That's why she came here. To get away from this asshole. And then what happened? First, Beckett pulled a gun on her. Second, she fell down the rabbit hole of their clusterfuck of a mission. "Shit," he said.
"Yeah."
His gaze dragged over her, taking in the downward cast of her eyes, her rounded shoulders, how tightly she clenched her hands together. Get out of your own head, Murda. This isn't about you. Right. Releasing a deep breath, he took one of her hands between both of his. "Was that the only time?" When she didn't answer right away, Beckett arched an eyebrow, and she must've seen on his face that he wasn't going to let it go at that.
"It was the most obviously physical he's ever gotten. One time he bumped into me in a bar and said it was an accident. But he did it so hard he nearly knocked me down. Another time, he came up behind me in a restaurant and put his arm around my neck like he was hugging me, but he squeezed enough that it hurt. He would do things like that—show up places where I was. It was always possible, since we ran in the same circles, but sometimes it just seemed too coincidental."
"You tell anyone about this? Nick? Jeremy?" Beckett asked, though in his gut he knew the answer. He could see it on her beautiful face.
She shook her head. "Here's the thing, Beckett. He's a lawyer in another division at Justice. Which means we share an office, colleagues, friends. Shit like this happens to women sometimes, and you have to make a judgment call. Do you report the person and cause a big thing that everyone will remember forever, instead of remembering you primarily for the work you do? Or do you let it go, brush it off, and hope it goes away? Cole is nothing if not image-conscious, so, for a long time, I was betting on him cutting it out before he either embarrassed himself or pushed me too far."
He got what she was saying. He really did. Didn't mean he liked it one damn bit, though. He stroked his thumb over her knuckles. "And where does him jumping you in the garage of your home fall on that spectrum for you?"
Kat shrugged. "I filed for a temporary protective order before I left D.C. It finally got served today."
Beckett made sure to gentle his voice. "What about pressing charges?"
"I hadn't decided about that yet. Part of why I came here."
A thought struck him, and it dropped a rock into his stomach. "Would he follow you here?"
"No," she said, frowning. "I'm not sure if I ever even mentioned that Nick and Jeremy lived in Baltimore. And he won't violate the restraining order. He's smart enough to know the consequences aren't worth it."
His gut wasn't nearly as certain as she seemed to be. From the little she'd described, the prick seemed to have all the makings of a stalker, right down to having a high-powered job he believed would protect him from the consequences of his own actions.
Beckett sighed. The conversation had chased away the terror and guilt he'd felt when he believed himself responsible, leaving him feeling even more drained than he had after talking to Emilie. But this seemed like the perfect time to implement something she told him he had to do—stop stuffing it down when something made him angry or caused him disappointment. He heaved a deep breath. "I have to say something."
Kat lifted her eyes to him, like she knew what was coming. He didn't like the sadness on her face one bit.
"Come here," he said, pulling her into his lap. She curled into his chest, and it was the sweetest fucking thing. Sweet enough that he second-guessed himself on expressing his feelings. Easy to do since he really didn't want to do it anyway. But he couldn't. He'd promised Emilie he'd try and that he would report back with at least one example of when he'd done it. For fuck's sake. Stroking Kat's hair, he sighed and let the words fly. "I'm mad at you for not telling me about this, Kat. Maybe I don't have a right to be mad—"
"You do, Beckett." She pushed off his chest enough to meet his gaze. "You asked me why it freaked me out when you grabbed my arm. And you asked me about Cole. I had plenty of opportunities to tell you what was going on."
"Then why didn't you?"
"Because, at the time, I didn't really know what we were—if anything. I didn't even think we liked each other."
Beckett frowned, but he couldn't deny that her words made a certain kind of sense. "Then why didn't you at least tell Nick that you might be in trouble?"
Her gaze drifted down, and she drew an invisible design on his T-shirt with her finger. It tickled his skin beneath, but he missed her eyes. "Because I wasn't here for five minutes when it became clear something was going on. Remember? I'd just gotten here when I ran into you in the stairwell, and Nick was about to walk out the door to go to Emilie's house. After you and the others left for Annapolis, I spent the day watching Nick and Jeremy disguise the building to mitigate against a possible attack." She shook her head and dropped her hand. "My problems seemed . . . really minor compared to all that. I figured knowing that some asshole had gotten handsy with me was the last thing Nick needed to be dealing with. And I'd filed for the restraining order . . ." She shrugged.
That was totally Kat. In the time he'd known her, she'd risked her life and her career for her brother and his friends. It didn't surprise him to hear she'd put what Nick needed in front of herself. He dragged his fingertips down her arm, just barely caressing the marks on her skin. "Does it hurt?"
Finally, she looked at him. "Only if I bump it." Her bottom lip quivered. "I'm sorry."
"Aw, Angel." He wrapped his arms around her and hugged her in tight. "I'm sorry, too."
A fast shake of her head against his. "You don't have anything to be sorry for."
"Yeah, I do. I totally flipped out on you. Again." Beckett sighed. "Clearly, one session with Emilie isn't gonna do the job, huh?"
Hands holding the back of his neck, she pulled away and met his gaze. "How'd that go?"
Beckett gave a one-shouldered shrug. "Eh. It sucked ass."
Kat smiled and her eyes brightened.
And, God, how he loved putting a smile on this woman's face.
"I'm proud of you." Her fingers caressed his hair.
Man, hearing her say that lit him up inside, too. "Yeah?"
She nodded, and then she yawned until her eyes watered.
"You should get some sleep," he said. Given that they were both up in the middle of the night, neither of them had gotten much sleep the night before.
"Stay with me?"
That sounded like fucking heaven. "Sure," he said.
She slipped into the middle of the bed as Beckett untied his boots and toed them off. Next went his shirt. "I usually sleep naked. That a problem?" he asked.
Behind him, Kat chuckled. "I can't imagine any woman in her right mind finding that to be a problem."
He grinned over his shoulder. "I don't care what any other woman thinks."
She propped her head up on her hand, a big smile on her face. "Well, I don't mind you naked, Beckett. Not at all."
Chuckling, he lost the jeans and climbed in next to her. Jesus, the fact that he could feel humor after how bad he'd felt in the bathroom not long before was a minor miracle.
"What's the tattoo on your back say?" she asked.
He sat back up enough for her to see it. "The way you're reading it right now, it says 'Strength.' But it's an ambigram. If you read it upside down it says 'Struggle.' "
Pushing up onto her knees, she craned her neck. "Oh, my God. That is so cool." She traced her fingers over the letters, sending heat into his blood. Sitting back, her gaze scanned over his shoulder. "And what's this one about?"
Beckett glanced down at his left shoulder to where a series of black and silver circles and slashing lines covered his skin in a tribal tattoo. "It's kind of like a piece of armor."
"I see that," she said, her fingers teasing over his skin. Down his chest, his stomach, his hip. Against his belly, his cock hardened. "And these?" Her fingers stroked at the marks on his hip.
His gut clenched, but he was done holding back. Or, at least, he was gonna try to be better about it. He heaved a deep breath. "One mark for each life I know I've taken. I'm missing four from the day we rescued Emilie from the Church Gang's storage facility."
She stared at him a long moment, long enough that he was sure the information had bothered her. And why wouldn't it? She dragged her fingers over the hash marks. "Your body is so beautiful, Beckett."
If she'd have flipped him over her back again, it would've surprised him less. How the hell had she come to that conclusion given what she was touching? He shook his head. "No."
Nodding, she ran her fingers through the hair that trailed from his chest to his groin. His cock twitched at the proximity of her touch. "To me, it is."
A knot lodged in his throat. "I'm all beat to hell. Scars everywhere." He gestured to the mess around his eye.
"You're beautiful," she whispered, and then she planted a hand on his chest, gave him a strong push backward until he reclined to the pillow, and bent over his body, her hair both cool and ticklish against his stomach.
Taking his cock in her grip, she bathed the whole length with her tongue. Beckett groaned as his head fell back to the pillow. She took her time licking and stroking him, like she was trying to learn the contour of his body there. Good as it felt, Beckett's brain refused to relax into the pleasure she gave him. She'd been hurt. She was upset. He'd just said he was mad at her.
He shouldn't let her do this right now.
"Kat—"
She sucked him in deep, deep, deeper until his head hit her throat. And then she pushed herself down a little deeper still.
"Oh, fuck," Beckett groaned, his hand going to her hair.
Slowly, she withdrew, sucking hard on his cock the whole way up. Giving him a small smile, she pushed onto her knees and removed her tank top, baring her small, beautiful breasts to him. And then she crawled down the bed, pushing the covers away to give herself more room, and settled on her knees between his legs.
"Kat. You don't have to—"
"I want to." She took him in hand again and licked him with the flat of her tongue, maintaining eye contact with him the whole time.
It was sexy as fuck. Every muscle in his body strained toward her as she sucked him deep again and held him there, luring his hands into her hair.
"Jesus . . . Kat . . ."
She pulled back off of him and heaved a deep breath, and then she was absolutely relentless. Sucking him deep. Holding him there. Moaning in approval when he lifted his hips or guided her head.
He was a goner way before he wanted to be.
"Fuck, Kat, gonna come."
With his cock in her mouth, she looked up his body and met his gaze. And nodded. She sucked him hard and fast like she wanted him to come. His release shattered him. His vision went fuzzy. His muscles went taut. His heart beat so fast it was hard to breathe.
When his body finally calmed, Kat crawled up the bed, pulling the cover behind her, and fit herself in tight against his side. She laid her head on his shoulder and draped her arm across his chest. And sighed like she'd never been more content in her life.
"What about you, Angel?"
Shifting her head, she smiled up at him. "That was all for you, Beckett. Can you reach the light?"
Without having to stretch too much, he was able to turn off the bedside lamp. Darkness cloaked the windowless room, making it so that all Beckett knew was what he could feel and smell and touch.
Making it so his whole world was Kat.
Chapter 17
When Kat woke up, she was all alone. Stretching her hands out on both sides of her, all she felt was cool, empty sheets. She reached out for the lamp in the darkness, and the light confirmed it. Beckett was gone.
What time was it, anyway?
Aw, crap. Her cell phone was all the way over on the dresser. She tugged on the tank top she'd never put back on last night and pushed out of bed. The LED screen on her smartphone read 7:25 A.M.
Where had Beckett gone? And why? And when? If the coolness of the sheets was any indication, he hadn't been here for a long while.
Could the guys be hacking into John Seneka's e-mail already? The question lit a fire under her butt, and Kat threw on a pair of jeans and a long-sleeved T-shirt. Beckett knowing the truth about her bruises still didn't mean she thought telling Nick was a good idea. At least not while all of this craziness was going on. And now, the restraining order was in effect anyway. She pulled a brush through her hair, then separated it into pieces and put it in a side braid. A little makeup and she was done and out the door.
Ravens overflowed the kitchen, which made sense since they'd just changed shifts a few minutes before. Kat said some hellos, grabbed a banana and a bottle of water, and made for the gym.
The guys were all gathered around Marz's desk, including Dare and Detective Vance. Kat hadn't seen the latter since the day of the attack—the day his godfather and Nick's friend Miguel had been gunned down, earning the team Vance's commitment to help. She wondered what he was doing there. Please don't let it be more bad news.
She rushed across the gym and everyone turned to look at her.
"Hi, Kat," Vance said. Tall, with dark hair and blue eyes, the cop was a stunner. Though his color blue was nowhere near as brilliant as Beckett's.
Speaking of which, was she imagining that Beckett wouldn't meet her gaze?
"Uh, hey. Everything okay?" she asked. She glanced between Vance and Beckett, since the question really applied to both of them.
"Yeah. Well, partly." Vance gestured to the guys. "I was just telling everyone. I have good news and bad news. I was nearby so I thought I'd deliver it in person."
She glanced to Nick, whose expression was dark with concern.
"I can try to keep the neighborhood closed up until maybe midweek next week. Maybe Friday, if I can push it. But I can't promise that."
"Why? What happened?" Kat said. And here she'd thought the problem of the building's security had been permanently solved by the neighborhood perimeter Vance had established.
"Nosy journalist," Nick said. "Pressuring the gas company and the city for details about the leak, and a timeline for fixing it. Plus environmental groups are getting whipped up about a widespread environmental impact the city is supposedly hiding."
Vance nodded. "Yeah. And the thing is, this woman—the journalist—is a total shark. When she smells a story in the water, she's damn relentless. Her inquiries have made their way all the up to the mayor's office, which crawled up the police chief's ass, who is now all up my ass." He gave a sexy, crooked grin. "Pardon the phrase."
Kat blew out a frustrated breath. "What the hell are we supposed to do if this isn't all resolved by then? Which it very likely won't be."
"We could probably take on some of you out at the compound," Dare said, looking around. "It would be tight, but we could make it work for a while."
But that would mean . . . they'd have to split up? Kat saw her own reaction mirrored on the expressions of the guys. None of the other women were here yet, but no doubt they'd dislike the idea as much as she did.
Nick braced his hands on the folding chair in front of him. "And the good news?"
Vance nodded. "I won't have them until late today or tomorrow morning, but I got some surveillance shots of Pier 13 yesterday. The vehicle you described as Kaine's was there with two others. The meeting took place inside, so the only shots of the suspects I got were of the group walking to their cars. Everything was from a distance, so one of my techs is working on enhancing the images, and then he'll run them through facial recognition software. He'll run the plates, too. He's squeezing it in as a favor for me, though, so I can't promise exactly when it'll all come."
"That is good news," Nick said. "Thank you. If we could nail down exactly who Kaine is in bed with that would help. A lot."
"I'll shoot everything over as soon as I have it." Vance pulled something out of the inside pocket of his sport jacket. A cell phone. "Damn, I have to go." He shook hands with each of the men in turn, then pointed at Nick. "Stay in touch."
"Count on it," Nick said, nodding. Vance took off across the gym.
"Just a heads-up," Marz said in an unusually quiet voice from his seat in front of his computer. "Seneka usually logs in to his e-mail around eight-thirty, so we should plan to intercept thirty minutes ahead of time so we're ready to go."
Nick planted his hands on his hips and let out a weary sigh. "Okay, that's our priority. Along with figuring out logistics for Garza's funeral tomorrow. Let's put Vance's news on the back burner for now." His gaze cut to Dare. "Thank you for the offer of help. Appreciate it."
The guy nodded, his expression guarded and serious. Kat liked Dare as much as she could like someone who gave off a vibe of holding lots of secrets close to the vest. Which kind of sounded like another hard-ass man she knew. Her gaze cut to Beckett, who still seemed to be avoiding her eyes. Maybe she was just imagining it? After all, they'd just received some really shitty news.
"You should stay for this confrontation we're hoping to orchestrate," Nick said to Dare. "Because this is the head of the organization we believe is responsible for Sunday's attack."
Lips pressed into a hard line, Dare's face was grim. "Consider it done."
Marz looked up from the monitor. "We should keep the rest of the gym cleared during the conversation, though. Nothing can give away our identity or location during the video conferencing."
As much as Kat wanted to hear what was said, she knew she needed to be here less than any of the guys. "I can guard the door."
Nick shook his head and nailed her with his pale-green gaze. "No. I want you here for this. I want your take on it."
"Oh. Okay, sure." The request muted some of her hurt from yesterday.
"I'll take care of it," Jeremy said. "The rest of y'all need to be here for this more than I do." He hopped off the corner of the desk where he'd been sitting and cut through the group.
"Thanks, man," Nick said, clapping him on the back.
Jeremy came around the desk, his eyes immediately on her and his expression asking if everything was okay. She nodded, but honestly, until she talked to Beckett, she wasn't sure what to think.
"All right, Marz," Nick said. "Walk us through what's going to happen."
MARZ HAD FILLED them in on how this was gonna work, and Beckett was strung tight.
Because of everything that'd happened last night.
Because of this whole thing with the guy who'd hurt Kat.
Because they were about to confront John freaking Seneka. And hopefully, finally, get some damn justice.
"All right," Marz said, excitement in his eyes. "I'm into his zmail account. He hasn't logged in yet. So now we wait." He handed Nick a black tactical mask. "When I give you the signal, you're on."
Taking a deep breath, Nick accepted the mask. "I guess this is it." He surveyed the group. The air was suddenly thick with tension and tempered expectation and guarded hopes. "I just want to say—"
"Shit, Nick. He's on." Marz said, his head in tight with Charlie's as they worked on something.
Nick's expression went immediately, intensely serious. He pulled the mask over his head and sat in the folding chair against the brick wall, a digital camera focused on his head and shoulders. They'd gone over all this in the few minutes Marz had explained what he'd need to do. Nick had a pile of evidence on an off-camera table to his left.
Marz brought the feed up on all the networked monitors, so there were suddenly seven images of John Seneka staring back at them. The man was older, probably in his sixties, with sharp, grizzled features, silvering hair, and intelligent eyes. Despite his legendary status—and contributing to it—he'd retired at what should've been the height of his career for reasons no one exactly knew. Not long after, he'd founded the company against which they now fought.
"Once we initiate the chat, no one talks except Nick," Marz said, voice as serious as Beckett had ever heard it. Nods all around. Marz turned in his seat to look at Nick. "Whenever you're ready."
Those odd pale-colored eyes slashed their way, highlighted within the cutouts of the black mask. Nick nodded.
The monitors all went to a split screen, with Seneka on the left and Nick on the right. Seneka's expression went from surprised to rankly pissed off.
"John Seneka," Nick said. Marz had modulated Nick's voice so it came through warped and deeper on the chat.
"What the . . . who the fuck are you?" Seneka said, eyes flashing. He reached for the computer. And Beckett's heart stood ready to jump into his throat.
"A man with twelve million reasons why you're going to want to hear what I have to say."
Seneka's focus narrowed in on his computer camera. "I'm listening," he bit out.
"I have evidence that your organization is responsible for the deaths of seven Army Special Forces operatives in an ambush in the Paktia province of Afghanistan, and the attempted murder of five more. I have further evidence that Seneka is engaged in an international narcotics conspiracy involving heroin stolen from Afghanistan and sold to a Baltimore gang for distribution, as well as the human trafficking of kidnapped American women between Baltimore and Afghanistan."
"I don't know what—"
"I'm. Not. Done," Nick said, voice like ice-cold steel. "The evidence in my possession clearly paints these activities as ongoing and extending backward as many as three years. In addition, I have evidence of Seneka involvement in the kidnapping or attempted kidnapping of three civilians in Baltimore within the past few weeks. Finally, I have evidence that Seneka security specialist Emanuel Garza is personally responsible for the deaths of several civilians and city police in Baltimore last week."
Way to go, Nick. He was doing great. Laying it out like a boss. The one thing they weren't saying just yet was that they had evidence that a Seneka security team had run an attack on a downtown Baltimore building five days ago, because that accusation would lead too quickly to identifying them.
Seneka shook his head, his expression like a storm. "With the exception of the charges against now-deceased specialist Garza, I know nothing about any of this."
Which was pretty much what they expected him to say.
Nick continued as if the guy hadn't just barked out his denial. "I have detailed information that someone with the initials GW or WCE using the now-defunct Seneka extension 703-555-4264 was in regular contact with one or more commanding officers at FOB Chapman, Afghanistan, in the Khost province. I have hard evidence that someone at that same Seneka extension and one other made regular contact with a Singapore bank where deposits topping more than $25 million have been made by someone with those same WCE initials."
"This is crazy," Seneka said, his expression bewildered. "I don't know what evidence you think you have, but I have no knowledge—do you hear me?—of any of this. I'm in the middle of a goddamned congressional investigation, for Christ's sake."
Standing by Marz, Kat's whole face was set in a frown as she studied the monitor. Beckett wondered what she thought of the exchange, of Seneka's demeanor, and also why he hadn't made it through the whole night by her side. But long after she'd fallen asleep, he'd lain there awake, staring up at the darkness. Wondering if the idea of an actual relationship was a total frickin' pipe dream for him. Wondering if this Cole motherfucker was lurking around the next corner of her life—and if so, what happened if he wasn't by her side when Cole came at her again? It was all more than he could handle while the woman he wanted as much as his next breath lay warm and half naked against him.
Beckett forced himself from his thoughts in time to hear Seneka ask, "What is it you want?"
And that was the money question.
"In exchange for the twelve million dollars I have that I know you want, I require the identities of and hard evidence against GW and WCE within your organization, and all evidence in your possession related to General Landon Kaine's involvement with this conspiracy."
That right there was the road to justice, to regaining their honor, to making them as whole as they could possibly be.
Seneka's gaze went distant as if he were deep in thought, a million miles away. Finally, he tapped his finger hard against the desktop. "I need to see some of this proof."
"That can be arranged."
Nick didn't need to say anything more than that. Marz's fingers moved over the keyboard, uploading a few redacted excerpts from documents from Merritt's files and a few screen shots from Kat's. When he was through, a ding sounded from the computer speakers.
Frowning, Seneka's gaze moved down from the camera as his attention turned to his monitor, presumably. "Goddamned vulnerable zmail," he grumbled. "Nice touch sending the files to me from myself. No trace of sender."
Marz didn't even crack a smile at the acknowledgment of his skill. He was all laser focus.
Seneka's expression got more and more grim as he looked over what Marz had sent him. And then his gaze sliced back toward the camera. "Whatever this is, I'm not a part of it. I have no personal knowledge of it. I have not authorized it. I do not condone it. Christ, this is way the hell off the grid. Not to mention sloppy."
Kat quietly rushed around the far side of Marz's desk and found Nick's favorite legal pad, then scribbled something in large letters.
"Then you apparently have some housecleaning to do. Because I have definitive proof and I intend to see justice done here," Nick said, steady as a rock.
Kat held up the legal pad and angled it around so everyone could see.
I believe him. Exhibiting signs of stress but not lying/deception.
Beckett's gut reaction: She's right. He couldn't put a finger on it, but Seneka was ringing true to him, too.
Kat quietly walked to a position behind the camera and held the pad so Nick could read it. A flicker of his eyes was the only response he gave. But she'd apparently seen it because she lowered the paper and walked away a moment later.
Seneka threw a pen at his desk and bit out a curse, then sat heavily against the back of his chair. His expression was contemplative without being calculating, pissed without being defensive. "Question is, who's running this thing behind my back . . ." His tone was almost musing.
Staring at the camera, Nick said, "I have a list of candidates for who GW and WCE might be—"
"I don't need your goddamned list," Seneka barked. "I know my own people."
"With all due respect, not all of them, as this situation clearly demonstrates." Nick's voice was firm without being mocking.
Heaving a breath, Seneka nodded. "Fair point. It seems we both have something the other wants. But for me, it's not money. My organization has a traitor that needs to be rooted out before this person takes down twenty-plus years of my hard work. Are you willing to meet?"
"Why would we need to do that?" Nick asked.
"Because I want to know who I'm dealing with, who I'm pinning my reputation and livelihood on. I want to look that person in the eye and know I've made the right call. I can't do that over some internet chat or e-mail. And I won't. This is my line in the sand. Call me Old School."
Nick thought about it a long time and finally nodded. "Affirmative. Under my conditions and at a time and location of my choosing. And after you have provided a gesture of good faith."
Seneka's eyes narrowed at the camera. "And that would be?"
"A list of all your personal phone numbers, a copy of the SWS personnel list with company phone extensions, and I need you to find something that definitively incriminates whoever this GW or WCE is. And I need it by close of business today. I will contact you again at seventeen hundred and we can proceed with the details of a meet after I've evaluated just how good your good faith gesture is." The phone numbers were to check whether he'd contacted the bank in Singapore, while the personnel list—which they already had from Kat's documents—was a test. If he altered the document, that would tell them a lot.
"Got this thing all figured out, don't ya, son?" He chuffed out something close to a humorless laugh.
"Doing my best, sir."
With a nod, Seneka rattled off four phone numbers—personal cell, work cell, office direct line, and unlisted home phone.
"Seventeen hundred, sir."
The guy gave a curt nod. "Seventeen hundred."
Marz cut off the feed. "We're clear."
Nick tugged the mask off and dropped his head into his hands.
A tense silence full of anticipation slowly bubbled into guarded statements of hope and victory. And then the room erupted in outright elation.
Nick heaved himself out of the chair, a dazed grin on his face. Everyone gathered around, Beckett included, to celebrate a job well done.
Now the question was, would John Seneka come through? Or was he playing them for everything he was worth?
Chapter 18
"We have a shit-ton to do now," Nick said. His words dampened the celebratory atmosphere, and that was probably for the best. As amazing as it had been to watch her brother in action—and Kat had to admit, he'd handled that conversation brilliantly—nothing had substantially changed for them. Yet.
"I'll search Seneka's phone numbers against the Singapore bank, Kaine, and Chapman," Charlie said. "Can someone let Jeremy know we're done?"
"I will," Kat said, taking off across the gym.
Behind her, Nick said, "We need to plan some possible meeting locations . . ."
When she had almost reached the door, she noticed Cy on the third shelf of the equipment rack, his head resting on the back of a pair of boxing gloves. When he was asleep, you could hardly tell he'd lost an eye, but her footsteps apparently disturbed him, because his one yellow eye blinked open, wary, watching.
Which, oddly, made her think of Beckett. Maybe it was the wariness. Maybe it was the standoffishness. Maybe it was the fact that you could tell something or someone had hurt him. Bad.
All of which made Kat's chest ache.
She opened the door and leaned out. "All done," she said to Jeremy, who was sitting on the floor holding Eileen. The puppy had curled into a ball in his lap and fallen asleep.
Jeremy peered up at her. "You okay?"
"Yeah. Why?"
He nodded his head to the side, silently asking her to come all the way out. Kat let the door shut behind her and crouched down.
"What happened with Beckett?" he asked.
Why did Jeremy have to be so perceptive? "Nothing, Jer. Everything's okay."
Looking at the puppy, he stroked her head. "Didn't look that way last night."
"I know." Kat watched as Eileen stretched in her sleep, rolling over to bare her big puppy belly.
"Kat . . ." Jeremy worried at his spider bite piercing with his tongue for a long moment. "Did Beckett do that to you? The bruises, I mean."
"Oh, God, no. He didn't. Why would you say that?" The thought that others would think that of him made her stomach hurt.
Jeremy cut his pale green eyes to her. "He looked angry and you looked upset. It didn't seem like something he'd do, but then again, you've been here nearly a week. How else would you have gotten them?"
"He didn't. I promise. He's a good guy, Jeremy."
"Okay. Then who did? 'Cause those bruises really look like fingerprints."
Damnit. Kat settled all the way onto the floor and crossed her legs in front of her. Jeremy had never been as overprotective of her as Nick, but he still stood up for her plenty when they were teenagers. And he'd always been someone she could talk to and count on. As much as she didn't want to tell him the sordid details, she didn't think it would be fair to blow him off. "Can this stay between just us for now?" She met his gaze.
His eyes narrowed, but he finally nodded.
"Ex-boyfriend."
Jeremy's expression went serious, lips pressed in a line, jaw tight. "Ex because of this?" He jutted his chin toward her arm.
Kat shook her head. "Ex because I realized he was way too self-centered and controlling for me. I broke up with him almost four months ago."
"If you dumped his ass four months ago, how is it you have fresh bruises?" he asked, dark brown brows cranked down.
"He's not so happy about being broken up." Kat sighed, her gaze scanning down the sleeve of ink on Jeremy's right arm.
"Aw, hell, Kat." He ran a hand through his perpetually messy dark hair. "This guy is harassing you?"
She stroked her fingers over one of Eileen's paws. "I wouldn't have characterized it that strongly before he grabbed me on Friday morning."
"And now?"
Pause. "Yeah. Which is why I got a restraining order against him before I came to Baltimore."
"You gotta tell Nick."
Her head was shaking before he even finished the sentence. "Not right now. He's got enough on his shoulders to worry about. I mean it, Jeremy. You gotta promise me."
"Kat—"
"Promise me," she said, putting a hand on his knee. "I'll tell him. I will. But not while he's dealing with all this. I mean, I know you're dealing with a lot, too, but I don't want to distract him from the mission."
"I don't like it," Jeremy said, nailing her with a stare.
Kat nodded. "I understand. Do you promise?"
Sighing, he finally nodded. Eileen twitched in her sleep so hard she woke herself up. Shaking his head, Jeremy chuckled, watching her as she climbed out of his lap. "Can you believe all this is happening?"
"It's surreal. But if it has to be happening, I'm glad the three of us are together. I was kinda pissed at you guys for not letting me know you were in trouble here."
His right eyebrow arched, the one pierced with little silver hoops. "You do realize the irony, right? And that you and Nick could almost not be more alike."
She frowned, then realized he'd totally nailed her on that one. Just a teensy bit hypocritical to be mad at Nick for not telling her all this was going on when she refused to tell him about Cole, wasn't it? Ugh. "I don't like you very much right now."
His grin was immediate. "You totally love me. I am full of the awesome." He pointed to himself, which made her focus on his white shirt for the first time. Big red lettering read, There's a party in my pants. Smaller words beneath said, And you're invited.
Kat chuckled. "You're full of something, all right. Come on, we should go back in. The call with Seneka went great. It looks possible that he was unaware of this whole situation and might be willing to help." She rose and offered him her hand.
He grabbed it, and she helped him up. Jeremy pulled her into his arms. "You are such a munchkin."
Her head just reached his shoulder. He and Nick had gotten all the Rixey tall genes, apparently. Smiling, she said, "Small but mighty, baby."
Pulling back from the hug, he winked. "Don't I know it. Come on."
Back inside, Eileen made a game of trying to get Cy's attention from his third-shelf perch, and Kat and Jeremy found the guys deep in discussion about secure meeting places. Marz had a series of maps open on the monitors, and Nick had a legal pad full of ideas.
"I don't think you can nail this down without doing recon first," Beckett said. "Which argues for asking for a meet tomorrow morning, not tonight."
Marz turned in his chair. "Em and I have got Garza's burial at ten, so we'll be out of pocket all night to— Aw, shit. Em had asked if we could get down to Fairfax before rush hour so she could spend some time with her mom tonight. I can only do that if Charlie's comfortable running the zmail hack at seventeen hundred."
All eyes turned to Charlie, who nodded. "No problem."
Marz looked to Nick, who shrugged. "Works for me," Nick said.
"I hate to miss it, though," Marz said, frowning. "Maybe I'll rig a laptop so I can hear the conversation. Anyway . . . I'll think about that."
"Why did you say it that way?" Beckett asked, frowning at Marz.
"Say what?" Marz asked.
"That you and Emilie had Garza's funeral to attend." Beckett crossed his arms.
"Oh." Marz shrugged. "I don't know. I just assumed no one else . . ." He shrugged again.
"Emilie's important to you, so she's important to me. I'm going, too," Beckett said. "That okay by you?"
"Well, yeah. Of course," Marz said.
Kat cleared her throat. "I know Becca, Sara, and Jenna really want to be there for Emilie, too. She's lost so much and yet she's given so much back to all of us. I want to be there for her tomorrow. Wouldn't feel right not to."
Beckett nodded, and it was the first time he'd made eye contact all morning. It made Kat want to go to him, but she wasn't sure where they stood.
"Kat's right," Nick said after a long moment. "Beckett, too. We're a family, and that means we need to be there for Emilie tomorrow. But let's be smart about it. I want to know more about this cemetery. Layout. Number of entrances. Surrounding roads and highways. Whatever seems relevant."
Marz nodded. "Uh, wow. Okay. Yeah, sure. I can do that." Kat didn't think she was imagining the emotion on the man's face, and it made her really proud of Beckett for bringing this whole thing up. "Emilie purposely didn't have an announcement printed in the paper, so his funeral wouldn't attract any unwanted attention. She can't imagine who else would come to the guy's funeral besides her and her mom. They're not even inviting the rest of the family out of safety concerns. Should be quick and quiet."
"If those aren't some famous last words," Easy said.
Everyone chuckled.
"All right," Nick said. "Sounds like everyone has their marching orders. Let's get to it."
"Do you think I could catch a ride with you two?" Kat asked Marz, fingering the hem of her long-sleeved blue T-shirt. "I had no idea how long I'd be staying here and I only brought a few days' worth of clothing. I was thinking maybe you two could drop me at my place on the way to Em's mom's house and pick me back up on your way home, so I could grab some stuff for a longer stay." Among the things she needed was a new pack of birth control pills, or she and Beckett were going to have to get very careful about using condoms.
Marz shrugged. "I don't see why that would be a problem."
"Actually, I'm not sure that's a great idea," Beckett said, his features hard-set.
Kat's gaze whipped to him. "Why not?" And then she saw the answer in his eyes. Cole. But she had the restraining order now, and she hadn't heard a single word from him since it'd been served. He'd clearly gotten the message to back off. "It would only be for a few hours, right?" She looked to Marz, who nodded.
Beckett arched a brow.
"How 'bout this. I'll promise not to leave my apartment between the time Marz drops me off and picks me up? But I really need some stuff from my place. And I'd been thinking of grabbing some extra clothing to bring back for the other women. Sara and Jenna have almost nothing to their names right now. And none of us have a thing to wear to the funeral." Only Becca and Emilie had gotten the chance to pack any belongings before they came to stay at Hard Ink. Sara and Jenna had apparently had to flee with just the clothes on their backs. They couldn't even order anything, since the cordoning off of the neighborhood had cut off mail service to the building for now.
"Maybe Beckett's right, Kat," Nick said, glancing between them.
She didn't miss the annoying smugness that settled over Beckett's expression. He smirked at her. "I have the perfect solution, then. Beckett can just come with us and stand guard while I pack."
Damn if that arrogance didn't fade right back off his face. "We have stuff to do here tonight—"
"Actually, that'll probably work," Nick said. "We'd narrowed our meeting place down to three choices. We don't need everyone to check them out. Shane and Easy and I can go—and Dare, if you'd like to come, too, that would be great." The biker nodded. "We'll all meet back here late. Done."
Kat smiled at Beckett, who looked like he wanted to break something. Repeatedly. "Sounds good to me," she said.
MAN, WAS SHE paying for getting Beckett sucked into being her bodyguard.
He hadn't said more than a half-dozen words to her during the hour-long trip to D.C. They'd taken Shane's big pickup with the second row of seating so they'd have plenty of room for Emilie's mother. And, luckily, Beckett had sat in the front seat with Marz, which gave Kat and Emilie some time to chat. The poor woman was so worried about her mother, and Kat was simultaneously sympathetic and a little jealous. It sounded like Em and her mom were super close. Back before her parents' car accident, Kat had been really close with her mom, too. Hearing Emilie talk and tell stories made her realize that she missed her parents a lot more than she'd been aware of, or maybe it was just more than she'd let herself think about.
And now Nick and Jeremy were all she had left. She couldn't begin to imagine losing either of them.
Before Kat knew it, they were navigating the traffic around Dupont Circle and turning onto her street. Her building was a ten-story tan-brick-and-glass apartment complex that made up in city views what it lacked in updated appliances.
Marz turned into the U-shaped drive in front of the building and pulled to a stop near the door. "We'll call when we're getting back on the road later."
Beckett nodded, and Kat said, "Thanks, guys. Appreciate it." She leaned over and gave Emilie a hug.
Kat hopped down from the truck and took a minute to just stand in the warm May sunshine. Beckett followed close behind, a small duffel bag in hand. Closing her eyes, she took a deep breath. Car traffic whirred by behind her. Taxis blew their horns at daring pedestrians chatting on cell phones. The signal at the nearby crosswalk beeped out a tone that told the vision-impaired when it was safe to cross. The sounds of the city were at once so familiar and so foreign to her—and it wasn't just how long she'd been gone, but how much it felt like she'd visited a whole other world while she'd been away. Except for crossing the street to take a shift in the sniper's roosts and the trip to the coffee shop, she hadn't been outside of Hard Ink the whole time.
Looking around, everything here was just so normal compared to what Nick and the team had been dealing with for the past month. It almost felt like it wasn't real. Like it couldn't be real.
A hand on her back. "Come on," Beckett said, doing a one-eighty scan of the street.
"Relax, Beckett. It's three o'clock in the afternoon and DOJ is down on Pennsylvania Avenue. Cole's nowhere near here right now."
"But he's not our only threat, is he?" His hand pressed more firmly.
Well, she had to give him that. Even though, deep down, she could hardly imagine mercenaries staking out her building and tracking her down. Was that really her reality now? She did as she was told.
The building's lobby was bright and airy, the result of the wall of windows that looked out onto the street and the ivory-colored tile flooring and brick walls. She nodded at the young blond woman sitting at the reception desk. Mallory? Kat didn't know her well since she worked during the weekday hours when Kat was usually away from the building.
Stopping at the wall of brass mailboxes, Kat fished for her keys in her purse. She opened the door and . . . just as she expected, saw the slip that meant she'd gotten too much mail to fit in the postcard-sized box. "I've gotta take this to the desk," she said to Beckett, nodding him back the way they'd come.
She pushed the slip onto the counter. "Hi, there."
The woman looked up from her thick constitutional law textbook, took the slip and smiled. "I'll be right back," she said.
"Do you think you're gonna talk to me while we're here?" she asked Beckett, whose gaze was trained on the windows.
Those sharp blue eyes cut to her. "It's under consideration."
Kat bit back a smile, stepped right up against the front of his big body and slipped her arms around his back. "Come on, now. Don't be all grumbly."
Peering down at her, he gave her a droll stare. But one of his arms hugged tight around her shoulders.
"See? That's better already."
His eyebrow arched, and it made her smile. And then the skin around his good eye crinkled, just the smallest bit. Enough to let her know she was making headway. "The team has a lot going on today, Kat. That's all."
Guilt slid heavily into her stomach. "I wasn't trying to be difficult. I really thought it would be okay if I was just here long enough to pick up some things."
"I know," he said, nodding.
"Here you go," came Mallory's voice as she returned from the back carrying two flat boxes and several rubber-banded stacks of mail. She handed it over the counter, and Beckett shouldered his duffel and grabbed the whole pile.
"Thank you," Kat said to the woman. "And you, too," she said to Beckett as she led him to the bank of elevators.
It was a quick trip to the eighth floor, where Kat unlocked her door and let Beckett into her small, one-bedroom apartment. The door opened into a generously sized living room decorated in shades of teals and browns. Big, airy windows lined the wall that adjoined the living space to the small dining room, and a small but serviceable kitchen sat beyond.
"So, this is home," she said. Not that she spent that much time here. Sleeping aside, she probably spent far more time at her office.
Beckett nodded, looked around, and dropped his bag on the coffee table.
Kat wondered what he thought of her based on where she lived. When she'd found this apartment after she'd landed her job at Justice, she'd been so ecstatic to find a place in the city near the Metro that she hadn't minded at all that it wasn't quite six hundred square feet. But Beckett's height and broad shoulders almost made the space feel like a shoe box. "It's nice," he finally said.
"Did I hear right that you lived in D.C.?" she asked. He nodded. "Where?"
"I'm in a little row house on Capitol Hill. Just renting." Beckett walked over to the window and peered down at the street. "It's not as . . . homey as this." He shrugged.
"Well, thank you. It's comfortable here." Kat wandered into the kitchen and grabbed a Diet Coke from the fridge. There wasn't much else, as she was mostly a take-out kinda girl. "Want something to drink?"
He shook his head. "You should start packing."
Kat popped the tab on her drink and took a long, cold sip. "We have at least two hours."
His gaze cut to hers. "I want us to be ready."
"For what?" she asked, coming back into the living room.
"For anything."
"Beckett—"
"Damnit, Kat." He scrubbed his hands over his face, then dropped them heavily to his sides. "I was already spun up over your prick of an ex-boyfriend this morning. Now, you're here, in the same exact place he waited to jump you on Friday, and I don't fucking like it. Not to mention Seneka. We've put you in enough danger without piling anything else on."
"Is that why you left? My bed, I mean?" Her belly squeezed in anticipation of his answer.
Giving her his back, his gaze returned to the window. And her belly squeeze turned into a rock. "It just . . . all became too much," he said.
Kat's heart tripped into a heavy, anxious beat as she put down her drink and came to his side. "What did?"
He put his hand on his chest and pressed, his mouth set in a grimace. Some sort of struggle played out on his face, and then he shook his head. "Have you ever sat in a position for so long that your foot fell asleep? And then you get up later, not realizing how asleep it is, and the pins and needles are nearly intolerable?"
"Yeah," she said, so quiet it was almost a whisper. Dread curled into her belly.
He finally turned to look at her, and his eyes were totally unshuttered. For once, the wariness was gone and there was just Beckett, open and vulnerable. "I've been asleep, Kat. For so long I almost don't remember ever being awake. And now . . ."
She placed her hands on his, still rubbing at his chest. His heart thudded out a fast rhythm. "Now what?"
"I'm all pins and needles. And I can barely breathe," he said, voice tight.
"Aw, Beckett." She pushed onto her tiptoes and wrapped him in her arms. Her heart broke for him. It really did. How horrible to be so disconnected from your emotions that experiencing them caused such pain. And her heart broke a little for her, too.
Because she liked Beckett. A lot. The kinda like that could maybe grow into something more, despite the crazy way they'd met and the fact that he, well, totally drove her crazy. And a small part of her brain whispered that he might just be too damaged to let go of the anger and the past and the numbness he'd embraced for so long.
But she wasn't counting him out yet. Not when he was clearly trying. Not when he was opening up. Not when he held her so tight it was like she was an anchor in his storm.
"Just hold onto me, Beckett. I'll do everything I can to get you through. Pins and needles always goes away, remember? It always goes away."
He held her for a long moment, then pressed a kiss against her temple. "I'll try to remember," he whispered. Finally, Beckett pulled away. And, as she watched, he shuttered himself right back up again. His eyes went hard. Expression went carefully blank. Jaw got tight. "Now please go pack, Kat. Marz will be coming back for us before too long, and I want to be ready to go."
Chapter 19
At ten minutes to five, when Beckett still hadn't heard from Marz, he called Nick's cell phone. If he couldn't be at Hard Ink for the second call with Seneka, he could at least listen in on speaker phone.
"Funny you called," Nick said. "Charlie just saw that Marz dialed into the chat himself, though he set it up so Seneka won't be able to see him there. It's good that you both hear what happens firsthand."
Beckett dropped onto Kat's soft, brown leather couch and nodded. "My thinking exactly. Anything new since we left?"
"Charlie confirmed that none of Seneka's work numbers have had any discernible contact with Kaine or the bank, and he hacked into the guy's Verizon account to check his cell phone. Seneka looks clean. We also scoped out the possible meeting locations, and you were right. The parking garage where we ambushed the Church Gang has all kinds of strategic advantages over the other choices. We're going with that."
"Good," Beckett said. That location had worked very well for them a few weeks back when they'd intercepted the gang's gun deal and walked away with the guns and the cash. It was one of the losses that led to Church's demise.
"How are, uh, things there?" Nick asked.
"She's packed. We're just waiting on Marz," he said, glancing at the LED clock on Kat's cable box: 4:54 P.M. No way Beckett was mentioning any other "things" here, like how he'd nearly lost his shit while talking to Kat. Thus, why he loathed talking. Although, oddly, talking to her tended to make things better, tended to make it easier to breathe. At least for a little while.
Voices sounded in the background, and then there was a shuffling noise.
"You're on speaker, Beckett," came Charlie's voice. "Put your phone on mute so nothing comes through." Beckett changed the setting, placed the phone on the coffee table, and stared at the thing like doing so might transport him back to the gym.
Kat walked into the living room a minute later, dressed in a pair of jeans and a creamy V-neck shirt that hugged every one of her curves. She'd just taken a quick shower since she'd woken up late this morning, and the scent of rich vanilla trailed after her. "What's going on?"
Pointing to his phone, Beckett said, "Called into the follow-up with Seneka. It's on mute."
"Mr. Seneka," Nick said, his modulated voice sounding distant but clear. "Thank you for keeping our appointment."
"I've had some time to confirm parts of what you told me, so I now want to get to the bottom of this almost as much as you do," the older man said, voice gruff.
"Do you have what I requested?" Nick asked, sounding unruffled. But no doubt Nick's thoughts had gone the same place as Beckett's—wondering what exactly Seneka had learned.
"How do I get the personnel list to you?" Seneka asked.
"Send an e-mail to yourself with the attachment." There was a long pause, and Kat joined Beckett on the couch. They exchanged a glance, and Beckett saw the tension he felt in his gut reflected on her beautiful face. A moment later Nick said, "And the other evidence?"
Beckett sat forward, anticipation making his heart beat faster. Nick had demanded that Seneka provide evidence identifying and definitively incriminating whoever GW or WCE was.
"I have good news and bad news on that front," came Seneka's voice. "I'll start with the bad. What I have isn't definitive. Yet. I'm still working on that." Beckett's gut tightened in disappointment and suspicion. If he asked for more time and stretched this thing out, that would be a major red flag. "And I wasn't able to find anything at all on the initials or acronym WCE. But I do believe I've narrowed down who GW is." Papers shuffled in the background. "Gene Washington and Gordon Wexler are security specialists who both have a history with Kaine. Back when the general wasn't riding a desk, Wexler and Washington served on the same A-team as Kaine. Wexler joined Kaine's team two years after Kaine did, when he wasn't yet a C.O., so they came up together. By the time Washington joined the team, Kaine was a colonel. So Washington cut his SF teeth under Kaine."
One man was a contemporary of Kaine's, and for the other, Kaine was his mentor. Both relationships could create strong bonds of loyalty.
"Their personnel files prove these connections. For now, I can provide you with that much. Sending the e-mail now." Some of the stress bled out of Beckett's muscles. One of the skills you honed as a special warfare operative was how to read people, how to decide if they and any information they provided were reliable. All Beckett's instincts said Seneka was on the up-and-up.
"It's a start. Give me just a minute," Nick said. More than a start, it provided information they didn't otherwise have and some understanding of how the dirty op Merritt had been investigating had likely come together. Another pause dragged out. The team was probably reviewing the e-mailed documents. Marz had to be going crazy not being there, because Beckett sure was.
"Have I passed your test and earned an in-person meeting?" Seneka asked. "Because I need to get a grip on this situation before the congressional investigation gets wind of it. That happens, and I'm fucked four ways from Sunday."
Finally, Nick said. "Yes, I will meet with you." Nick laid out the details. They were going to do this a few hours after Emilie's brother's funeral tomorrow, late enough that they'd have time to get people in place well before the meeting. So by this time tomorrow night they might finally have a solid, nailed-down list of who was to blame for everything that had happened to them, their team, and the colonel.
"That works for me. I'll bring anything else I uncover in the interim. But I have just one question."
"Which is?" Nick asked.
"When you say 'you' will meet with me, I'm wondering whether that means I'm meeting with—" Seneka paused for a moment, and there was a shuffling of paper. "—Edward Cantrell, Derek DiMarzio, Shane McCallan, Beckett Murda, or Nicholas Rixey."
Holy fucking shit.
Still staring at the phone, Beckett shot to his feet. He would've placed money that each of his teammates had just had the same thought. Kat put her hand to her mouth, her eyes wide and scared. Beckett held out a hand, and Kat clutched it and rose to stand at his side.
"Why would the meeting be with any of those people?" Nick asked, a note of strain in his voice.
"Come on now, son," Seneka said. "Those men are the surviving members of an ambushed SF unit. I'm betting it's the same one you mentioned this morning. I was also able to learn that one of those five men lives in Baltimore. So I'm pretty confident in saying that one plus one equals two. You didn't think I'd go into this without learning who I was dealing with, did you?"
Well, it was a goddamned good thing Seneka seemed to be an ally, since he now knew who they were and that they had all kinds of incriminating information on his company.
"Nicholas, I wouldn't—"
"It's Nick," came Nick's voice. Which, Jesus, just confirmed it. Beckett hoped that gamble paid off. "You can turn off the voice modulation." Nick's next words came through in his normal voice. "I didn't go into this knowing what to think about you at all."
"Fair enough. If I were in your situation, I wouldn't have either. But none of this works from this moment forward if we can't find a way to trust each other."
"Agreed," Nick said. "Well, I'll start with removing this, since I assume you can find my picture if you found my name. And, as a good faith gesture on my end, I can let you know on authority of a confirmed source that the congressional investigation does not yet have any idea about these activities. You stay on the up-and-up with me and I'm more than happy to keep it that way."
"I appreciate that intel," Seneka said. "I'll see you tomorrow at two, then, Nick."
"We're clear," came Charlie's voice.
Beckett swiped the phone off the table and took it off mute. "Well, damn. Our junk's out there swinging in the wind now, isn't it?"
"Yeah," Nick said. "It was bound to happen. Still damn jarring as hell. More importantly, though, allying with John Seneka is going to be the key to all of this. I feel it."
"I sure hope so," Beckett said. "Because if we're wrong, shit's gonna go to hell—and fast."
BECKETT'S CELL PHONE came to life in his hand a few minutes later, the screen flashing an incoming call from Marz. Beckett was so wired that he nearly fumbled the damn thing. "About damn time," Beckett said by way of answering. He set it to Speaker so Kat could hear what Marz had to say.
"I missed you too, asshole," Marz said, his normally cheerful voice somewhat dampened.
"What are you thinking about that call?" Beckett asked.
"Our identity was going to come out sooner or later, so I'm making peace with the fact that it happened now. It still feels a lot like the guy reached his fist through the phone and sucker punched me, though."
Beckett nodded and sighed. "I'm with ya there. Now that the call's done, you getting on the road?"
"Uh, that's what I was calling about. Emilie's mom is down with a migraine. No way we can travel tonight," Marz said. "Things are pretty rough here. I already let Nick know."
"Oh, no," Kat whispered, her brow furrowing.
Frustration at being separated from the team sat like a boulder on Beckett's chest, but he felt nothing but sympathy for Emilie and her mom. Emilie had been a real trooper the past few days, helping out everyone around her when no one would've blamed her for curling up in bed and throwing the covers over her head. You had to respect that. And who could blame a mother for being so distraught over her son's death that she fell ill? Certainly not Beckett. Most of his life he would've given his right leg to have someone care about him half that much.
The room went on a tilt-a-whirl around Beckett at the thought. Given his right leg . . . Exactly what Marz had done. Jesus. His chest went tight and he gasped for breath.
"Are you all right?" Kat said, grasping his arm.
"B?" Marz said. "Did you hear me?"
"Yeah. Yeah, I heard. Was just thinking that you . . . uh, gave your right leg for me," Beckett choked out, lowering himself onto the couch. Kat settled next to him, and though Beckett couldn't bring himself to meet her gaze when his head was caught in the midst of all this churn, he could feel concern radiating off of her.
Marz chuffed out half a laugh. "Uh, this is not news, is it?"
Beckett shook his head. "No, right. But it just smacked me in the face. You sacrificed yourself for me, because . . ." He scrambled for just the right words.
"Because I love your grumpy ass. And you love all my epic awesomeness."
"Yeah," Beckett said in a voice so low it was almost a whisper. "No one's ever done something like that for me, Derek. Until you, no one else ever cared if I lived or died." Beckett was acutely aware of Kat's presence for this conversation even before she scooted closer, wrapped her arm around his shoulders, and laid her head against his bicep.
There was a long pause, and then Marz said, "I love yoooou, maaan!"
"And you wonder why I hate talking." Beckett pushed End on a huff.
His phone buzzed a half minute later. LOL Pick you up at 7:45 A.M. Let you know if anything changes here.
Roger, Beckett replied.
Also: I love you, man!
Fucker. Beckett tossed his cell to the coffee table. And then he got hot in the face. Because Kat had witnessed that whole little scene. Like she needed to see him lose it one more time.
Kat pushed him so that his back rested against the couch as she straddled his lap. She cupped his face in her hands and got up close, her breasts against his chest, her forehead nearly touching his, her soft hair forming a curtain around them. Beckett's heart beat like a bass drum. She kissed him.
A soft brush of skin on skin at first, and then a nibbling tug at his bottom lip that made his cock jerk where she sat on him. "I care if you live or die, Beckett. I care, a lot."
He swallowed hard as that odd, warm pressure filled his chest again. "Why?" he asked, meeting those too-perceptive jade eyes.
"You really need me to spell it out?"
He really fucking did. Beckett nodded.
Kat kissed him. "Because you're loyal, protective, and taking time away from your life to help my brother." Another kiss. Beckett's hands grasped her hips. "Because you're smart, strategic, and a genius when it comes to gadgets and fixing things." Kiss. "Because you keep trying to get Cy to let you pet him." Kiss. "Because you're gorgeous and sexy as hell." Longer kiss. "And you fuck like a god." Much longer, wetter, hotter kiss. "And because you called me 'Angel.' "
Chest full, throat tight, heart pounding, Beckett kissed Kat on a groan, his thoughts struggling to process—to believe—that someone as beautiful and bright and together as Katherine Rixey could see all that in him when he couldn't see most of it in himself. Hell, not even his parents had seen it.
Her arms wrapped around his neck and she sucked his tongue deep into her mouth, and Beckett was hard and aching and overwhelmed and more in need of being wanted than at any other moment in his life. And Kat gave that to him in spades. With her kisses. With her touch. With her desperate moans and grinding hips and pleading words. "I want you, Beckett. If we're here for the night, I want you in me as deep and as hard and as often as you can. For just tonight, I don't want to think about anything else but you."
"Jesus, Kat," he rasped, banding his arms around her back and pinning her to him. He devoured her mouth, sucked on her neck, bit the soft tendon that led to her shoulder. "I want that, too. But not here."
"Why?" she moaned as his hands kneaded at her ass.
It was the silver lining in the cloud of being stuck here overnight. "Because for once, I finally have you all to myself with no need to hurry. I'm taking you to your bed, stripping you down, and spreading you out just like I said I would." Beckett rose from the couch with her legs hooked around his waist and her arms holding onto his neck. He carried her from the living room into her bedroom, a warm, sexy space decorated in purple and gold with a large wooden bed in the center. The room sat at the corner of the building, creating two walls of windows that seemed to welcome the city inside.
With a last hungry kiss, Beckett slid Kat down his body until her feet touched the ground. She gave him the sexiest damn smile he'd ever seen. "I'm not sorry we're stuck here together."
Beckett winked. "Neither am I." He crossed to the shorter wall of windows, scanned his gaze over the row houses below, and lowered the blinds.
Kat chuckled. "We're on the eighth floor."
"Uh-huh. And I'm a greedy bastard. I don't even want to share you with the sun."
Her head tilted to the side and her smile turned sweet and soft.
At the longer window, Beckett repeated his survey, his gaze running over the windows of the hotel across the street. Down below, traffic rolled by in an endless rotation of cars and trucks and buses. He shut the blind and closed the world out.
The room got darker, more intimate, and Beckett turned to Kat and gave her a long look from bare feet to sexy curves to tousled chocolate-brown hair. He stalked toward her, taking his time, looking his fill. And what he saw? Was a woman who was a perfect fucking fit for him. Soft and comforting when he needed solace. Tough as nails when he needed strength. Strong-willed and independent-minded when he needed to be called on his shit. A woman who thought his scars were beautiful and wanted to help him fight his demons. A woman who understood sacrifice and loyalty and doing whatever it took to protect the people you loved.
A woman worth fighting for—even if what he was fighting was himself.
Beckett took Kat into his arms and kissed her deeply, their tongues swirling, their lips sucking and nibbling. The sweet taste of her made him realize, for the first time in his life, that he was fucking starving. And always had been.
"I want your taste in my mouth, Kat. Your scent on my body, your skin all over mine," Beckett rasped as he kissed her cheeks, her nose, her eyes. "Tonight, I want it all. With you."
Chapter 20
"I want you naked," Beckett whispered, his tongue caressing the shell of Kat's ear. His hands went to the hem of her shirt and pushed it up, the calluses on his palms ticklish as they dragged over her rib cage and lifted off the shirt. Sucking gently on her neck, he reached behind and unclasped her bra. And then his fingers were on the button of her jeans and he was slowly, so damn slowly, tugging the denim off her hips and down her legs. Starting at the hollow at the base of her throat, Beckett's lips blazed a trail of kisses between her breasts and over her stomach to the waistband of her red silk panties. Lowering to a knee, he gripped her hips and drew his tongue from hip bone to hip bone. And then he bit the fabric and pulled her panties down with his teeth.
This big, strong man on his knees, undressing her with his hands and his teeth and his hungry eyes . . . It was one of the sexiest things she'd ever seen.
When she was completely naked, Beckett smoothed his hands over her legs, her sides, her belly, her breasts. He looked at her like she was precious, priceless, flawless, and it reached into her chest and swelled her heart. "Almost too beautiful for words, Kat." He rose to his feet, his hands carding into her hair as he claimed her mouth on a warm, wet kiss. "Lay down on the bed for me."
With her heart flying into a sprint, Kat climbed onto the bed, crawled into the middle, and laid down on her back. Beckett walked to the bottom of the bed and stroked the tops of her feet with his big hands. His gaze ran over her until she at once broke out into shivers and flushed hot. She held out a hand. "Come be with me."
One corner of his mouth lifted in a small grin. "I won't be rushed, Katherine. Not tonight. No matter how damn much I want to cover your body and bury myself deep."
Holy crap. The low grumble of his voice combined with the words and the images they evoked to make her wet and achy. Because she wanted exactly what he'd just described. She wanted him. And the fact that he couldn't figure out why she would want or care about him without her having to spell it out made her heart thud even harder in her chest. Because she did care about him. Maybe even more than she should given the depths of the wounds he so obviously nursed. But sometimes your heart had a mind of its own.
Finally, Beckett yanked his shirt over his head and dropped it to the floor. He removed the gun she hadn't realized he carried from a holster at the small of his back and placed it on her dresser. Then he undid his boots and jeans, until he finally stood impressively naked at the foot of the bed. The breadth of his shoulders, the mounds of his cut muscles, the length and girth of that freaking amazing cock—Kat wanted all of it, all of him, on top of her, inside of her, surrounding her.
He climbed onto the mattress while he simultaneously pushed her legs apart, making room for himself to kneel between her thighs. Spreading her wide, he massaged her inner thighs from knee to groin, his hands maddeningly brushing the outside of her core. And then, in one swift movement, he hooked her legs over his shoulders, used a big flat hand to hold her hips still, and licked her slowly from aching opening to clit.
"Oh, God, Beckett," she moaned, her hands going to his hair and her breath cutting off as he sucked her clit in deep. Beckett ate her with the same laser-focused intensity with which he did everything. He was relentless and demanding and so damn intent on giving her pleasure that she was quickly panting and grasping at his hair and hanging on the edge of what promised to be an absolutely mind-blowing orgasm.
Blazing blue eyes peered up her body as he slipped one finger, then two, into her core. He curled his fingers inside her and stroked a spot so sensitive she almost came up off the bed. His big hand held her down harder, forcing her to yield to the pleasure he gave as his fingers fucked and stroked her and his lips and tongue and teeth tormented her clit.
Her body detonated in wave after wave of pulsing, clenching, drenching ecstasy. Beckett groaned and sucked harder until Kat was clawing at his head and pleading for relief from the overwhelming pleasure.
Finally, he released her from his mouth and eased his fingers free of her. "That won't be the last time you come down my throat," he said, crawling up her body. He kissed her belly, her breasts, her neck. "And this won't be the only time you come on my cock." He took himself in hand, dragged his thick head through her wet, sensitive folds, and sank deep.
"Yes, yes," Kat said, loving that glorious fullness that only he could give. When he bottomed out inside her, he stretched his massive body on top of her until he was the only thing she could see or smell or feel. Kat had always loved the weight of a man atop her, and never more than with Beckett. "Love this," she said as she wrapped her legs around his hips and her arms around his neck.
"Yeah?" he rasped as he used his hand under her shoulder for leverage. He held himself off of her with his other hand against the mattress, just enough to kiss her and look her in the eyes and gaze down at where his cock penetrated her in a slow, deep in and out. "You look so fucking gorgeous underneath me, Angel. Don't think I could ever get enough."
Her heart squeezed, and Kat shook her head. "I hope you don't. Not ever." She bit her lip then, not wanting the pleasure he gave her and the way her heart swelled in her chest to make her say things he might not be ready to hear.
Things it seemed crazy that she might actually want to say. So she determined to show him instead. Grasping the back of his head, she found his mouth with hers and poured every bit of affection she felt into the kiss. She sucked on his tongue, plundered his mouth with hers, and moaned as his hips moved faster, harder, with more incredible urgency.
"You're so good, Beckett," she said, her head falling back against the pillow.
Beckett licked and nibbled at her neck, and then he lowered his whole weight onto her, wrapped his arms under her until his hands tangled in her hair, and hammered his cock into her until all she could do was moan.
Her orgasm was nearly blinding, and hit her so hard that she screamed as muscle spasms shuddered throughout her whole body. She raked at Beckett's shoulders as she rode out the storm, and then she went limp against the bed.
Beckett's little grin was so damn smug. "You still with me?"
Kat rolled her eyes and tried to calm her breathing. "It wasn't . . . all . . . that," she lied, knowing there wasn't a chance in hell he'd believe it.
He barked out a little laugh, withdrew his cock from insider her and rolled her onto her stomach. Then he smacked her ass, crawled on top of her and rasped in her ear, "You can't scream 'love your cock' in the middle of an orgasm and have me believe that weak-ass bullshit," he said, guiding himself inside of her again.
Had she said that? She couldn't remember, but given how much she loved the feeling of his dick sliding deep while his big body pinned her down, she couldn't deny the truth of it. And, God, he felt so much fucking bigger this way, his strong thighs holding hers tight together.
Smiling, she said, "I do love it, Beckett. Love the way you fuck me. Love the way you fill me."
He forced his arms under her shoulders and held her in a tight embrace, one that made him hunch his body around her. Lips against her ear, his hips flew against her ass as he panted and rasped. "Feel so fucking good when I'm inside you. Never want it to stop."
The comment whipped butterflies through her belly. Both because she wanted him to feel good and because she didn't want it to stop either. Not just the amazing sex, but them. Long delicious moments passed with him moving inside her and whispering the most amazing admissions into her ear. "Need you so damn bad, Angel . . . So fucking lucky to find you . . . Want you to be mine . . ."
"Yes, Beckett, yes," she cried, her heart and her body never more full or more satisfied in her entire life.
"Christ, I can't hold back," Beckett groaned.
"Don't," Kat said. "Come in me. I want to feel it."
"Shit," he grunted, his teeth audibly grinding together.
"Let go, Beckett. Give me everything you've got," she said, her hands clawing against the plum comforter as she peered over her shoulder at him.
His cry was almost anguished, and then his grip tightened, his face crumpled, and his thrusts quickened until he was pulsing inside her, bathing her in his essence, claiming her and making her his.
Exactly what she wanted to be.
"Kat," he rasped, heaving a deep breath as his body calmed. "Holy fuck." He kissed her cheek.
"Yeah," she said, his words luring a smile to her face and making her chuckle. "Holy fuck, indeed."
Beckett gave a small laugh in her ear, and it was the sexiest freakin' thing. "So . . . like a god, huh?"
Kat elbowed and wriggled out from under him. "I knew that was gonna be a mistake," she said, humor making her bite her bottom lip to hide her smile.
He banded his arms around her chest and hauled her on top of him, her back to his stomach. "Where do you think you're going?" He nibbled on her neck. "Who said I was done with you?"
"But—"
"Told you that wasn't going to be the only time. I'm going to indulge in your sweet fucking body until neither of us can walk." He sucked on the sloping tendon that led to her shoulder.
A ripple of excitement ran through Kat's belly and her head fell back in surrender, her cheek resting against his. "Might not survive it but I'm game to try."
"Aw, you'll survive it, Angel." His right hand skimmed down her belly until his fingers found the top of her sex. Kat moaned as he circled his fingers there. Being on top of him, her legs to the outside of his, forced her to thrust her hips forward, allowing his swirling touch to bring her body roaring back to life.
As Kat lifted her hips into his touch and moaned and writhed against him, Beckett's cock became hard beneath her. He hiked her up a little higher, reached between her legs and guided himself inside her. And then his fingers returned to her clit.
"Oh God, Beckett," she whispered as he began to move. The blunt head of his cock hit that crazy place inside her, the one that made her wet and weak and lose control.
"That's right," he gritted out. "Want you to come again."
It took about two minutes until she went rigid atop him, her breath whooshing out as her orgasm hit.
"Again," he rasped, his hips moving under her, his cock impaling her, his hands helping her move exactly how she needed to.
"Beckett, Beckett, Beckett." She cried his name over and over again. Because he was going to get her there. With his body and his words and his desire, he was going to send her impossibly and irrevocably over the edge.
When it finally hit, the orgasm washed over her like an ocean wave, the kind that tumbled you head over heels until you didn't know which way was up.
And then Beckett planted his feet into the mattress and used those massive thighs to hammer deep inside of her. "Fuck, fuck, coming," he groaned. He buried himself deep, his hands holding tight to her hips as he spilled himself inside her.
By the time his grip loosened, Kat didn't have a single muscle left, she was sure of it. But she'd never felt better in her life. Her eyelids grew heavy. "Cuddle me," she said, the basic act of speaking taking effort.
He chuckled behind her as he rolled them to one side and pulled her back taut against his chest. "Bossy little thing, aren't you?"
She tucked their hands under her breast. "Says the man who demanded I come twice in five minutes."
He kissed her neck. "That a complaint?"
"Not even close," Kat slurred. "Like your arms around me."
"Me too, Angel. Now close your eyes for a while."
"So bossy," she whispered, and then she did exactly what he said.
BECKETT WASN'T SURE how long they'd been asleep when he woke up, but Kat didn't stir at all when he got out of bed to use the bathroom. His stomach immediately made itself known, so he grabbed his clothing, got dressed out in the living room, and then went on a reconnaissance mission in Kat's kitchen.
What he found made him chuckle. A dearth of actual food and a plethora of take-out menus. And he thought he was bad. He sorted through the folded menus until he found an obviously well-used one for a Thai place. The restaurant was only two blocks away on Dupont Circle. Perfect. Beckett placed a phone order, then stuffed his feet into his boots, grabbed Kat's keys, and scribbled a quick note on a Post-it in case she woke up.
Outside, the city was a living being all around him. Movement. Sounds. Smells. It felt at once comforting to be back in the midst of D.C.'s busyness and strange to be by himself, doing something so normal as getting takeout, without one of the eight million members of his team or the other people at Hard Ink all around.
While he appreciated the moment of solitude, the last thing Beckett truly wanted was for things to return to how they'd been before Nick's call just over three weeks ago. His life before had been largely solo security gigs followed by time alone at his house followed by time alone working out. Sure, there were people he was friendly enough with, but no one he counted as a friend. Until now. Until being reunited with his teammates, the best friends he'd ever had.
He'd thought all of that was gone, that the ambush had destroyed those relationships along with everything else. But he'd been wrong. Now, between Kat and his teammates, Beckett was finally feeling for the first time since returning stateside that he knew where he was supposed to be. Maybe that was fucked up, given the shit storm that swirled around them, but that didn't make it any less true. And he couldn't help but wonder what would happen once this mission was behind them.
Beckett rounded Dupont Circle, one of the city's huge traffic circles where five major roads met with what seemed to be the explicit purpose of confusing out-of-town drivers. In the center was a park known for chess-playing on a series of outdoor tables. A tall marble fountain stood at the center.
Navigating his way around the busy circle, he found the Thai place hopping. Friday night in D.C., after all. His order wasn't ready yet, so he found a corner behind the door where he wouldn't be in the way and checked his phone while he waited.
He shot a text to Marz. How are things there?
Same, came back to him. Then, Feel bad Em couldn't have been here sooner.
Another casualty of this whole damn situation. I feel ya. And he really did. Because Beckett felt equally bad that Kat had been pulled away from her life by all of this. Not only that, but that she'd potentially risked her whole damn career for them. Much as he appreciated what she'd done, there'd been enough sacrifices made by too damn many people.
As other diners entered to place take-out orders or be seated, Beckett checked his e-mail, both pleased and a bit concerned to find everything was quiet. He'd farmed out the security gig he'd been working, and everything seemed to be going well with that transition. But at some point his absence—now almost a month long—was going to result in fewer and fewer inquiries for new work. And who the hell knew how much longer this would go on?
"Murda?" the woman at the register called.
Beckett pushed off the wall and settled his bill, and then he was back out in the warm night again amid the crowds of people, the air and the walk helping to clear some of the bullshit from his thoughts.
Tomorrow, they'd meet with Seneka and hopefully identify the best path forward for bringing this whole thing to a conclusion. What that would mean, exactly, wasn't yet clear.
His shoulder knocked into someone, slamming Beckett out of his thoughts so hard he nearly dropped the bag of food. He turned to see who he'd bumped into. A brown-haired man peered over his shoulder, his face set in a deep scowl. "Sorry," Beckett called, but the man just hurried toward the doors of the hotel.
Damn. He should be paying better attention. The thought immediately had him doing a one-eighty scan of the street and checking the faces of everyone he passed.
Back at Kat's apartment, Beckett found her still sound asleep. He unpacked the bag of food on the dining room table, gathered some plates, silverware, and drinks, and then returned to the bedroom to wake her. He knelt at the side of the bed and watched her sleep for a long moment. God, she was pretty. More of that odd achiness in his chest.
"Kat?" he said, rubbing her bare back, stroking the curls off her face.
She came awake with a smile. "Hey." Her sleepy gaze ran over him. "You have clothes on." Pushing up onto her elbows, she rubbed her eyes and yawned.
He chuckled. "Sorta necessary for hunting and gathering."
"You got food?" Beckett nodded. "Aw, I love you so much right now. I mean—" She bolted upright. "I mean, you know, for getting dinner. I'm starving."
"Uh-huh," he murmured, trying not to react to what she'd just said. Trying not to want her words to have been real. Trying not to be hurt or disappointed that they weren't. For fuck's sake. He rose to his feet, his gaze raking over her beautiful, warm, naked skin. "Better get dressed before I keep you in that bed, then. Wasn't sure what you liked, so I got a few things."
Her smile was uncertain, like she was embarrassed. " 'Kay. Be right there."
Out in the dining room, Beckett took a long pull from a glass of water, Kat's words replaying in his ear. It was just an expression, Murda. Don't get your panties in a bunch. Right.
Or, if you have something you want to say, maybe you should man up and do some talking. How about that?
He nearly broke the glass putting it back down on the table too hard. What the hell would he even say? Hard to figure that shit out when your head and your chest and your gut were a tangled mess.
"Oh, my God, you got Thai," Kat said from behind him. "Smells amazing."
Beckett scratched his head, hoping what he'd picked was really okay. "Something different from what we've been having lately."
Kat wrapped her arms around him from behind and burrowed against his back. "Yes, absolutely. Thank you so much. I can't believe I didn't hear you leave."
He pressed her hands tighter against him, loving her openness, her touch, her warmth. She always took all the bullshit away, or at least put it on mute for a while. "I wore you out."
She smacked his butt and laughed. "Shut up. You're not supposed to brag."
"I'm not?" He watched her open a few of the containers and steal a piece of broccoli from the Drunken Noodles. "Why not?"
She smirked at him. "Because it's . . . smuggy."
Beckett grinned. "See, I wore you out so much you're making up words."
"Oh, my God. Let's eat, already." They settled at one corner of the table rather than across from one another, the table so small their knees touched. But Kat didn't seem to mind. "Thanks for getting plates and everything. I should be the one doing all this, though. I kinda fell down on my hostess duties."
He shrugged. "I don't mind. It was nice to take a walk." Except for nearly plowing over a guy on the street. What the hell had that been about anyway? His gut went ice cold. "Hold on a minute," he said, getting back up and going to his duffel on the coffee table. He always carried some kind of go-bag with him when he traveled—a bag of necessities and survival gear in case he ever found himself in a situation where he needed to disappear fast.
"What are you doing?" she asked, following him.
He pulled out the mobile bug detection device he'd given Kat to use after the meeting at the coffee shop. "Some guy smacked into me on the street." Beckett flicked the power switch and slowly scanned the small black rectangle over his arm, shoulder, chest, and stomach. Not even a little blip of reacting lights.
"Did he hurt you?" Kat asked, stepping closer. Her face was set in a frown.
"Run this over my back?" he said, handing it to her. She took it and moved behind him. "No, I'm fine. Honestly, I didn't think twice about it on the street." Beckett sighed. "But, given everything, I should've done this the minute I returned."
"It's not picking anything up," she said, turning it off and placing it on top of his bag. "Better safe than sorry, huh?"
"Yeah." Beckett stroked his fingers down her cheek, hating the concern he saw there when she'd been so open and playful moments before. At least he'd put his mind at rest about it, though. "Sorry."
Kat smiled and shrugged. "No worries. Come on, let's eat."
Back at the table, they filled their plates with Drunken Noodles, Pad Thai, Pad See Ew, and Basil Shrimp.
"Everything's so good, Beckett. Thank you. All great choices," Kat said, spearing a shrimp.
"Yeah?" How stupid was it that her comment made him feel so good? "Welcome." He struggled for something to talk about. Outside of this whole situation, did they even have anything in common? Actually, they did have at least one thing—Nick. "What was it like growing up with Nick?" he finally asked.
Kat's grin was immediate. "I am a girl with two older brothers. It was pure torture." They laughed, and she wiped her mouth and took a drink. "Actually, they weren't that bad, I suppose. Nick walked a fine line with me between being awesomely protective and overbearingly so. Jeremy was just as chill growing up as he is now, and every bit as flirtatious, too. Our family did a lot together, vacations, Sunday dinners, game and movie nights, so we were all pretty close."
"Sounds nice," Beckett said. He took a big bite of noodles and wondered what it would be like to be a part of a family like that.
"It was," she said, a warm, affectionate smile on her face. "Nick shocked the hell out of Mom and Dad when he decided to go into the Army, though. Holy crap, I still remember that night." She paused for a bite of the Pad See Ew. "It was fall semester of his senior year of college. He took a leave of absence without discussing it with them first. I'm surprised my father didn't stroke out from how mad he was."
"That was after September eleventh, wasn't it?" Beckett remembered Nick telling this story.
Kat nodded. "Yeah. Stupid, brave boy. I understood why he wanted to do it, but when I think about the night he told us, I can still feel the flutter of fear in my chest. I was half convinced he'd go and never come back." She pushed the noodles around on her plate. "And that almost happened."
Beckett laid his hand on her arm. "Big difference between almost and actual, Kat. He's back and he's good." What he didn't say was there was also a big difference between being back and actually living. A lot of guys had difficulty adjusting to the real world when they got out, which was why some people suffered problems like Easy's. Beckett couldn't say he'd done the best job of transitioning himself, as evidenced by his nearly empty house, lack of friends, and the general isolation he'd built into his post-Army life. What he was doing right here—having dinner with a girl he really liked—wasn't something he'd done in longer than he wanted to admit. Hookups? Yes. A dinner date? Not in a long, long time.
"Yeah. You're right. Especially now that he's found Becca."
Smiling, Beckett nodded. "She has the uncanny ability to chill his ass right out."
Kat chuckled. "Yes, she does. You gotta love her for it."
"She's good people. Her and Charlie, both," Beckett said, thinking how strange it was that once, Frank Merritt had been a central part of his life. Now his kids were. And had Frank never died, Beckett probably never would've met Becca and Charlie.
Life had a twisted sense of humor sometimes. That was for damn sure.
"Oh, I'm stuffed," Kat sat, dropping her napkin on her plate.
"You hardly ate anything," Beckett said, frowning.
"I had some of everything. Besides, I have to leave a little room for ice cream." She grinned.
"What is it with you Rixeys and ice cream, anyway?" Because Nick and Jeremy had such a fixation that their freezer often stocked a dozen different flavors.
Her eyes went wide. "It's ice cream," she said, as if that explained everything. Beckett chuckled. "Growing up, our mom would throw impromptu ice cream sundae parties on Saturday nights. Multiple flavors, candy toppings, chocolate and caramel sauce, whipped cream, cherries, the works." She braced her chin on her hand. "God, that was fun. My mom was good like that. She could make totally ordinary things seem like the most fun you'd ever have."
"Sounds like you miss her."
"I do," she said, and then her gaze dropped to the table. A whole lotta fast blinking, and—
"Hey, are you . . . come here," he said, catching a glance of the glassiness in her eyes. He pushed back his seat, grasped her hand and tugged her into his lap. Those long brown eyelashes were wet with unshed tears. "God, Kat, I'm sorry."
Sitting on his lap, she shook her head. "Don't be," she said, her voice thick. "Sometimes it sneaks up on me just how much I miss her. I never mind talking about her, though. If I ever have kids and I manage to be half the mother she was, I'll be great."
Beckett took her chin in his fingers. "You're gonna be a fantastic mom."
She laid her head on his shoulder. "Yeah?"
"Not even a question," he said. He genuinely meant that, but then his thoughts went on a little trip. Because, in order to have kids, she'd have to get pregnant. And in order to get pregnant, she'd have to have sex with a guy. And he sure as fuck did not like the idea of her having sex with any other man.
But kids weren't something he'd ever let himself think much about.
And, cart before the horse much? Because a few days of fantastic sex in no way equaled having a family.
It's more than sex.
Yeah, okay. Probably. But still . . .
"You gonna have some ice cream with me?" she said, her breath ticklish against his throat.
Beckett chuckled. "Well, who says no to ice cream?"
She sat upright, smiling. "Right? And thank you." Kat kissed him on the cheek, slid off his lap, and cleared her plate. "Take your time finishing up. I'll grab everything we need."
Minutes later, Beckett was done with his dinner, and Kat had loaded down the table with bowls, spoons, ice cream scoops, tubs of ice cream, candy, and a container of chocolate sauce.
She held out her hands, clearly proud of herself. "We're making sundaes."
Smiling, Beckett nodded. "I see that." But he was more pleased than he let on. Because she wasn't just making sundaes with him, she was including him in a family tradition she clearly held quite dear. And that made him feel ten feet tall.
For another hour they sat at the table, eating and talking and throwing M&Ms at each other to see who could catch more in their mouth. Beckett totally ruled at that, much to Kat's chagrin. And then they went back to bed, and Beckett made good on his promise to make them both weak in the legs.
It was, quite possibly, the best night of his life.
Chapter 21
From the moment Marz and Emilie picked them up, a weird tension had taken up space between Kat and Beckett again. It didn't seem like Beckett was upset with her. Nor that he regretted their night together as he had in the past. And when she asked if he was okay, he simply said that the meeting with Seneka was weighing on his mind.
She could hardly blame him for that.
So despite the fact that they didn't talk any more on the ride back to Baltimore than they had on the way down, which was to say not at all, Kat needed to give Beckett the benefit of the doubt. Because their stolen night away really had been amazing. Romantic and playful and full of great food, fun conversation, and the best damn sex of her life.
Besides, there were more important things to think about this morning, like the heavy sadness that hung around Mrs. Garza's shoulders like a mantle. She was a petite lady with salt-and-pepper, shoulder-length hair, and she spoke with the slightest bit of a Spanish accent. But Kat almost couldn't look into her dark brown eyes without feeling like her own heart was ready to break. She wasn't sure she'd ever seen more anguish in the eyes of another person. Not only had the woman lost her son, but she'd also recently learned about all his criminal activities. Both would be an overwhelming amount of grief for any mother to handle.
After some stilted attempts at conversation, Kat finally brushed at nonexistent lint on her black slacks and gray silk blouse and watched the scenery out the window.
So the trip to the cemetery was quiet and long, because of the silence that filled the truck and the traffic that clogged both the D.C. and Baltimore beltways. They were meeting Nick and everyone else at the veterans' cemetery where Manny would be interred. The small service would be at graveside, because Emilie had feared that holding a wake and service at a funeral home or church might attract the attention of Manny's enemies.
Kat hated that Em and her mom couldn't honor Manny the way they would have otherwise. Emilie told stories about their big family all the time, and Kat had no doubt that they'd all be attending if things were normal. But they weren't. Not even a little . . .
They pulled into the long drive of the cemetery at twenty-five minutes to ten, cleared the gatehouse and followed the winding road through seemingly empty fields. You could just barely make out the headstones lying flat in the green grass, and the lack of standing markers gave the place a lonely feel. Built on the grounds of an old farm, the place was huge, and Beckett guided Marz to the correct section of the cemetery using a map they'd received at the gate.
Their destination stood out before they even parked, marked by a blue tent covering a single row of six chairs in front of a skirted metal lowering device upon which Manny's casket would eventually sit. A tarp-covered mound sat off to the side—the dirt from the hole, no doubt—and two large arrangements of flowers stood on wire easels.
They were the first to arrive.
Quietly, as if a hush blanketed the place and them within it, they climbed out of the truck. Marz rushed around and offered Mrs. Garza a hand down from the backseat, and he and Emilie flanked the older lady as they helped her cross to the gravesite. Kat and Beckett hung back, giving them a little space.
She glanced at Beckett, who looked so very handsome—and not a little sexy—in a dark gray sport coat over his black T-shirt, both of which emphasized the size of his shoulders. Luckily he'd had the coat in his duffel. Since they hadn't planned to spend the night in D.C., he had no other shirt. Kat had never seen him in anything more than a T-shirt, though, and damn was she impressed. She sighed. "This is so sad for Em and her mom," she said.
Nodding, Beckett took her into his arms.
She almost melted against him, she was so relieved to receive his touch. He hadn't been as openly affectionate this morning as he'd been last night. "Are we okay?"
"Yeah, Angel. We are. I'm sorry—I get so tangled up in my head sometimes." He gave her a small, sweet kiss.
"You don't have to apologize. I know today is a big day. I'm just glad." Her head against Beckett's chest, she took in what would otherwise have been a beautiful view of an open field against a line of bright green trees, all painted with the morning sun. The field appeared to hold one of the newer sections of the cemetery, judging by the small number of headstones. A long oval pond sat between them and the edge of the woods, the whole scene almost mocking in its beauty and vitality.
Car engines sounded from down the road. Kat and Beckett turned to see a line of cars slowly rolling up the drive. Nick's classic sports car and Jeremy's Jeep followed by a plain blue sedan. They came to a rest behind Shane's pickup, and then everyone spilled out onto the grass, the five men in dress pants, crisp button-downs, and ties—even Jeremy, who'd gone as far as finding a jacket, and the three women in jeans and nice shirts. Kat still felt bad that she hadn't been able to bring them clothes in time for the funeral. Vance stepped out of the sedan wearing a sharp-looking navy suit and a serious expression on his face.
They'd all just had time to say a few quick hellos when the hearse came down the road, bypassed the line of cars, and parked in front of Shane's pickup. And just then it hit Kat. Who was going to serve as the pallbearers?
As the funeral director opened the hearse's rear door, Kat whirled to Beckett and Nick. "Are . . . are you guys gonna . . . ?" She nodded toward the casket visible in the back of the long black vehicle.
Beckett and Nick exchanged a loaded glance, and then Nick nodded. "Yes. For Emilie and Derek."
Kat almost felt like she'd been punched in the gut, so she could only imagine how the guys felt to have to honor a man who'd not only destroyed their lives but harmed the people they each loved the most. "Oh God, Nick. I'm sorry," she said, giving him a quick hug. His arm came around her shoulder, and she pulled back to meet his gaze. "And I'm sorry about everything . . . this week."
"So am I," he said.
As Marz joined their group, Kat gave Beckett a hug, too. With a hand on his neck, she urged his ear down so she could whisper to him alone. "This is so unfair for you. I'm really sorry."
He kissed her on the cheek. "We got this, Angel. Don't worry."
She gave a quick nod as a voice behind her asked, "Are you gentlemen the pallbearers?" Kat turned to find a tall man all in black giving them a thin smile.
"Yes," Nick said, following the man to the hearse. Beckett, Marz, Shane, Easy, and Jeremy followed. The six of them lined up at the car's rear opening, three on each side, and slowly pulled the charcoal gray casket out. A spray of red roses covered the top, the color deep and stark against the lid.
Someone grabbed Kat's elbow and leaned against her. "I love Emilie," Becca whispered. "But seeing Nick have to do this hurts my heart. Is that horrible?"
"No," Kat said, meeting Becca's bright blue eyes. "Not in the least."
The women let the pallbearers pass, then made a slow procession to the grave site, where Emilie and her mom already sat. Mrs. Garza burst into tears as the men settled the casket onto the lowering mechanism. They were just moving to stand behind the small row of seats when an engine sounded in the distance.
Nick looked at Emilie. "Are you expecting anyone else?"
"No," she said, dabbing her eyes.
The team closed ranks as a black Suburban came around the bend and into view.
"I don't like the look of that," Beckett said.
"Maybe it's here for another funeral," Nick said, tone wary, eyes trained on the big SUV.
"Dude, isn't that the same kind of truck that participated in the attack on Hard Ink?" Jeremy asked, standing with the team and next to Charlie. You almost never saw anger on Jeremy's face, but it was there now.
And he was dead right, too.
"Yeah," Nick said, watching the SUV like a hawk as it continued past all of their cars.
Kat sighed in relief.
And then the Suburban came to a slow stop in front of the hearse.
"Oh, no," Kat whispered as her heart climbed into her throat and her stomach knotted.
Nick and Beckett exchanged a dark look, then Nick scanned the group. "Everyone without a weapon stays on this side of the casket. Marz and Vance, hang back and guard the group. Shane, Beckett, and Easy, you're with me."
"Is there a problem?" the officiant asked, his whole face frowning.
"Hopefully not, sir," Marz said. "But if you'll just come this way." He nearly manhandled the guy to get him to move.
Dread washed over Kat's skin and adrenaline flooded her bloodstream. Both left her shaky even as she followed Marz's directions and gathered behind the casket, the only immediate cover they had. Marz and Vance formed a human blockade in front of them as the rest of the guys stalked toward the curb. Nick drew a gun from the small of his back, but that was all Kat could see around their protectors.
"What's happening?" Mrs. Garza asked in a tear-choked voice. "Who's here?"
"I'm not sure, Mama," Emilie said.
Kat and Becca clutched hands, and then Becca grabbed Sara's hand while Jenna hugged Sara's other arm. Everyone the four of them cared about had just gone to investigate some unknown threat, and it was the worst, most helpless feeling Kat had ever had.
That was the moment it hit her.
If something happened to Beckett, it would tear her heart out. Because . . . because . . . Oh, my God. She was falling in love with him—had maybe even fallen all the way. It was crazy and fast and probably even reckless, but no less true for any of that.
She was in love with Beckett Murda. And he was in danger. She just knew it.
AS HE FACED whatever was coming at them, Beckett's body shut down all the nonessentials. His anger at performing any sort of burial honors for Garza? Gone. His sympathy for Emilie and her mother? Gone. His thoughts about whether he and Kat could work out and whether he should've stumbled through expressing his feelings last night? Gone.
All gone, for now.
Hand at the small of his back, he gripped the handle of his weapon, his muscles braced for anything.
Three doors slowly opened. Two men emerged from the passenger side, one from the driver's. Beckett didn't recognize either of the men nearest them. One was older, hawklike, wary, with mostly black hair despite his age. The other was younger, African American, his expression full of all kinds of Oh shit. Both looked like they wanted to reach for their guns, but for some reason, they didn't.
And then the third man stepped around the back of the Suburban, hands slightly raised in a gesture of surrender. "Gentlemen," he said. John fucking Seneka. "I don't think weapons will be necessary, do you?"
Beckett stared at the man for a long moment, scrutinizing every little thing he could take in. The guy's stance was relaxed. He made easy eye contact. He kept his hands in plain sight. Nothing threatening. Nothing suspicious.
Except, of course, for showing up at Manny Garza's funeral. Was this his plan all along? Only, to what end? Beckett scanned the three-sixty all around them. Everything appeared quiet and still, just as it had moments before. Not that he trusted that for one second.
"What are you doing here?" Nick asked.
Seneka gestured to the other two men. "Same as you, I expect. Paying our respects. Garza was one of my employees and worked with both of my colleagues here. I'd ask the same thing of you." He arched a silvery eyebrow. Despite his rather wiry build, the man commanded respect, attention, notice.
"We're friends of the family," Nick said. He holstered his gun and gave a quick nod to the others that told them to do the same.
Sonofabitch. Beckett didn't like it, but he followed Nick's directions. For now.
The older of the two "colleagues" made a disbelieving noise under his breath.
Seneka nodded as if his subordinate hadn't said or done anything. "Then perhaps some introductions are in order. This is Gene Washington."
The younger man gave them a single nod and scanned their group, guarded and confused.
"And Gordon Wexler."
The older man gave a tight nod, his expression dark but unreadable, his gaze fixed on Beckett's hand behind his back.
And the morning went from Jesus Christ to What in the actual fuck. GW—the man whom Merritt noted as his main contact with whatever the hell WCE was—stood before them. Six degrees of separation, Beckett's ass. It would take all of six steps for him to close the distance between both of these assholes and take them out. The only question was which man was their GW.
When Nick didn't offer introductions in return, Seneka held out an open hand toward the gathering by the tent. "We were a bit late arriving, so let's not hold things up anymore."
Except no one wanted to go first, putting their backs to the others. Tension hung like a thick fog between them. Finally, Seneka put a hand on Nick's back and urged him to walk next to him and lead their combined group back to the grave.
It was all absolutely and completely surreal. Which was a goddamned problem, because confusion muddied your thoughts and slowed your reactions. But it was clear in the set of their shoulders and the way everyone on the team held their hands by their sides that they were ready to act. Ready for anything.
Of course, Seneka's guys were ex-SF, too. And they looked the exact same way.
Anything went down here, and it was going to be a fucking war of attrition.
Seneka leaned in and whispered something to Nick that Beckett couldn't hear, and then the older man stepped ahead to Mrs. Garza and Emilie and went down to one knee. "I'm so sorry for your loss. Both of you," he said, looking between the women. "My name is John Seneka. Manny was a valued member of my company and a good man. You have my sympathies."
"Thank you," Mrs. Garza said, her eyes glassy and her breath halting. "And thank you for coming."
"Of course," he said, patting her hand.
"Is everything all right?" she asked.
"Yes, ma'am," he said. "Don't you worry about a thing."
"Are we . . . uh . . . ready to begin?" the funeral director asked in a pinched voice.
"Yes, sir," Seneka said, rising and walking behind the row of chairs to find a place to stand at the far side. His men followed and stood beside him, watching every movement of the team.
Except for Marz, who warily lowered himself into the seat next to Emilie, the rest of the team took up standing positions behind the other end of the seats. The women all stood to Easy's right, just outside the shelter of the tent. But that placed the men's bodies between them and Seneka's group. Which did goddamned little to make Beckett feel better. He wanted to bundle the lot of them up, tuck them away in the safety of that probably armored Suburban, and hightail it the hell out of there. No doubt he wasn't the only one fighting those urges.
The officiant began the service, and Beckett barely heard a single word. Angling his body to the side, he ran his gaze in repeated surveys from the tree line to the field beyond their grave site to the road as far as he could see it from their location. Fuck the blur in his right eye, too, but at least he knew he wasn't the only one who had gone vigilant.
The man droned on for what seemed like forever, heightening the tension like steam in a teapot until Beckett was almost surprised it didn't lift the tent right off the ground. Because except for John Seneka himself, the other two men with him appeared every bit as tense as Beckett felt. Was that because his team had met them at the curb with guns drawn? Because something was about to go down? Or because one of these men was their GW and knew exactly who the fuck they were, too? Which was entirely possible.
When the officiant finally stopped talking, he grasped a triangularly folded American flag resting atop the casket and presented it to Mrs. Garza. It occurred to Beckett then that there should've been some other military honors bestowed here. An honor guard. A rifle team to fire off a twenty-one-gun salute. Something. But he remembered Marz saying that Emilie was keeping the proceedings as basic as possible for the sake of safety and simplicity. It wasn't like she was in any position to make the full arrangements someone normally would anyway.
And then it was over.
No one moved. Hell, Beckett wasn't sure anyone even breathed. And he didn't let up running his scans for even a second.
Finally, Mrs. Garza rose with Emilie's help and took halting steps toward the casket. She leaned heavily on the edge of the dark gray box, head bowed, lips moving, her shoulders shaking. Beckett wasn't sure how long she remained like that, and he felt like a total asshole for wishing she'd wrap it up. But every minute they stood out in the middle of this field was making his skin crawl. And one thing you tried really fucking hard not to disregard when you did what he'd done most of his life was ignore that feeling.
He'd had it out on that dirt road in Afghanistan that day. And he hadn't listened hard enough.
Now, with Kat and the other women standing a few feet away—equally exposed—his nervous system was like a live wire. Raw, exposed, and full of explosive energy.
Mrs. Garza pulled a rose free from the arrangement on top the casket, kissed it, and laid it by itself on the shiny gray surface. "Take me to the car now," she rasped. "Please."
"Of course, Mama," Emilie said. She pressed her hand to the lid for a long moment, then gently grasped her mother's arm to offer support. Marz went to the older lady's other side, clearly torn between going with the women and staying in case something happened between the men. The officiant departed with them.
"Head back to the vehicles," Beckett said to Kat and the other women.
Kat frowned, and her expression was mirrored on the rest of their faces, too. "But—"
"Kat, please," Beckett said.
Her gaze flickered between him and the Seneka men, and Beckett could see by the flash of heat in her eyes and the set of her mouth that it took everything inside her not to argue and instead to walk away. But he also knew it was because she didn't want to leave him there. That was plain in the last pleading glance she gave him.
"Come on, everyone," Vance said, escorting the group away.
Beckett met Nick's gaze, and it was clear in the other man's eyes that he was debating the best way to play this. Just go about the day as planned? Or do what they'd planned to do in that parking garage this afternoon without the benefit of choosing the most advantageous arrangements for themselves—and without the support of the Ravens? The only saving grace was that between their team and their allies, they outnumbered Seneka's men—assuming he hadn't dropped others off to surround them before coming here.
"Jeremy, you and Charlie, too," Nick said, nodding off to the side.
The thing that most caught Beckett's attention in that moment wasn't the discontent rolling off the younger Rixey at being told to return to his Jeep, it was the flash of awareness in Gordon Wexler's eyes. At the name Charlie.
Hello, GW. A soul-deep desire for vengeance and justice flashed through every cell in Beckett's body.
Seneka stepped forward and nailed Nick with a hard stare. "Like I said before, I'm as surprised as you are to run into you here. Had no idea you had any connection to Garza. But since we are here, maybe we should talk." Now seemed implied in his tone.
Nick met the gazes of each of his teammates and nodded. "I think—"
Two shots rang out, shattering the morning air, and everything at the tent turned to shit.
Chapter 22
Goddamnit! Seneka double-crossed them after all.
Shouting. Screams. Hurled accusations. Another gunshot, so close this time it pinged off the frame of the tent right above Beckett.
Everyone scrambled for cover, but where the hell was the best place to train their weapons? In the direction from which the shots had been fired over at the tree line? Or at the three Seneka operatives right in their midst?
Men scattered—going down on their stomachs, diving behind the chairs, racing for the limited cover. Seneka shouted as Wexler bolted for the pile of dirt, drew his gun and fired in the direction of the tent. Someone cried out, but Beckett couldn't tell who it was as he whipped behind the casket. Nick dove there with him. Easy went for the lone tree off to the left, drawing fire as he moved.
Beckett peered to his left to find that the women had just reached the vehicles. He provided covering fire toward the tree line as they scrambled to get in the cars, while Jeremy and Charlie stood midway between the tent and the Jeep, frozen like deer in the headlights. More shots rang out, at least one pinging off the casket. Trapped in the open, the two men went flat on the ground.
Beckett's gaze searched for Kat and found her huddled on the road behind Shane's truck. Son of a fuck, she was still out in the open, because the road ran parallel to the grave site and the trees. "Kat, stay down!" he shouted over the fray.
"Stand down! Stand down!" Seneka called, his voice sounding odd, almost slurred, but loud enough to draw Beckett's attention in the other direction. He caught Washington crouching at the corner of the tent and aiming his weapon at Shane, who was crawling toward the shelter of the casket.
Beckett didn't hesitate. He squeezed off a shot and hit Washington square in the chest. Red bloomed across the man's shirt as he toppled over.
More shots from the woods. From Wexler. From Seneka . . . at Wexler? What in the motherfuck was happening here?
Beckett grabbed Shane's hand and hauled his body behind cover.
"Thanks, man," Shane said. "Jesus. Seneka's got a GSW to the abdomen. It was Wexler."
"Shit," Beckett said, but being on that side of the coffin exposed him on the woods' side. "Let's end this, then we can help him."
"Roger that," Shane said.
And then Beckett heard something that chilled him to the bone.
Kat yelling, "Stop! Stop! Cole, stop!" Beckett whirled to find her crouching at the curb, hands cupped around her mouth.
"Stay down!" Beckett yelled. Following Kat's gaze, he looked across the field . . . and saw a shooter at the edge of the pond. Was he the only one?
"He'll listen to me! Cole, no!" she shouted. Marz grasped her to drag her behind the truck. "Please! I'll talk to you now! I promise!"
Wait. Cole? Narrowing his gaze, Beckett focused on the man's details. Brown hair. Brown shirt. Blue jeans. Ice rushed through Beckett's veins. Her stalker . . . was the same man who'd shoulder-checked him on the street last night. So it hadn't been any fucking accident. Holy fucking psychopath.
"What the fuck is going on?" Nick yelled from behind him. But Beckett didn't have time to answer. He barreled out, hoping to draw any fire toward himself and away from Kat. He fired two shots at the asshole, but the guy disappeared behind some cover.
"No! No!" came a man's voice from behind him. Jeremy's. Beckett glanced over his shoulder in time to see Wexler dragging Charlie off the ground by his hair and Jeremy scrambling after them.
With shit coming at them from opposite directions, the operative managed to haul Charlie's body in front of him before Nick got off a shot. Nick and Wexler shouted at one another as the latter dragged Charlie backward and warned Jeremy away.
Beckett had to trust that Nick would handle that situation, because neither of them could focus on both fronts.
"He's behind that rock," Easy yelled, pointing to a boulder at the closest edge of the pond.
Cole fired off three shots.
"Cole, talk to me!" Kat yelled. Marz hooked his hands under her arms.
"Too late for that, Katherine!" came distant, shouted words. Pop. "You reported me!" Pop, pop. "And I know you fucked him in our bed!"
Shouts from behind them, too. Jeremy. Wexler. Nick.
"We gotta draw him out!" Easy yelled.
"Noooo!" Jeremy screamed. Beckett gave a quick glance and found that Jeremy had somehow wrapped himself around Charlie's feet, apparently surprising Wexler, because Charlie dropped to the ground like a sack of potatoes.
Jeremy gave him a hard shove. "Run, Charlie! Go!"
Charlie scrambled on his hands and feet toward Nick—and in Nick's way.
Wexler kicked Jeremy in the head—hard. Once, twice, until Jer went mostly still. He shook his head as if dazed. "You'll get me, but I'll get him!" Wexler screamed, muzzle of a gun against Jeremy's skull. Using brute strength, he dragged Jeremy up with one arm, using him as a shield the way he had with Charlie.
From Beckett's other side two shots fired at once. One closer, one farther away. Sirens wailed in the distance.
"Shooter's down!" Easy yelled. "I got him. Go help Nick."
Relief flooded through Beckett. Now to deal with Wexler, who was backpedaling quickly toward Vance's sedan with Jer. Where the hell was Vance anyway? Shaking off the thought, Beckett bolted toward Nick, his mind strategizing—
"No, no, no! Beckett!" Marz said.
What the hell? Beckett froze in place and followed the sound of Marz's shout.
Kneeling behind her, Marz held Kat against his chest. A circle of blood soaked the gray silk at the center of her blouse. She stared down at it as if surprised.
"No!" Beckett cried, taking off in her direction, his heart lodged so far up his throat he could barely breathe. Not Kat. Not Kat. Not Kat.
He moved as if in slow motion. Despite his strides, the distance between them stretched out. Sounds warped and warbled behind him. Everything blurred out around the edges. Everything but Kat.
Finally, finally, he reached her. As Beckett went to his knees in front of Kat and Marz, the sound of peeling rubber squealed out from somewhere. A moment later a car roared past. But Beckett hardly saw it, because his eyes were riveted to her face. Her beautiful, pale face staring down at her rapidly soaking shirt.
"Kat?" Beckett grasped her cheek and gently tilted her head toward him. Her skin was cool to the touch.
"Beck-ett. Sor-ry." Her eyes wouldn't quite focus.
He peered over his shoulder, needing help. Wanting a miracle. "Becca! Shane!" He came right back to Kat. "No apologies. You're fine. We're fine. Okay? Let's lay her down, Marz."
Marz's gaze was rock solid. "You got it." Gently, Marz eased himself out from under her and lowered her to the blacktop, his hand cradling her head.
"You hit?" Beckett asked him, scooting closer to Kat.
"I'm fine," Marz said. "Here's Shane now. Keep her talking, B."
Beckett grasped her hand as Shane went to his knees. "I'm going to look at the wound, Kat. Okay?" Shane asked.
" 'Kay." Her whole body trembled.
"Tell me a story," Beckett said.
"Wha' story?"
Shane ripped the pearly buttons open down to Kat's belly.
"Anything, Angel. I want to know every damn story you've got," Beckett said as Shane pulled the silk apart. Oh, God, no. The entrance wound sat just right of center on her chest, blood welling up and out.
Shane removed a blade from a holster on his boot and hacked a square piece of cloth off the bottom edge of his dress shirt. He covered the wound, but her blood soaked through immediately. "Need more cotton," he said to Marz. "And plastic, if you can find it. We need to keep this sealed."
"Ummm . . . I . . . I don't . . . am so glad . . . I met you," she rasped.
"Me, too, but I want to hear a story," he said, trying like hell to hold his shit together. He wasn't going to let her say anything that resembled a good-bye. No fucking way. He clutched her hand in his.
Becca skidded to her knees on the other side of Shane, her focus entirely on Kat's chest. "ETA on the ambulance is two minutes, Kat. You're gonna stay awake with us, right? Two minutes is no problem."
"Dunno. Tir-ed." Kat's voice was weaker, raspier.
Becca removed the soaked bit of fabric, shrugged out of her cardigan and quickly folded it. "Apply pressure, Beckett," she said, placing the wad over the wound. Kat groaned as Becca whipped out her cell phone, and the tones that rang out signaled she'd dialed 911. "Ambulance to the local ER might not be enough," she said to Shane, who gave her a loaded look and nodded.
Rubbing her arm, Beckett tried to restore some of the warmth to her skin. "Plenty of time to sleep later. But now you gotta stay with me. Okay?" Please? Oh, God, please? He couldn't have found her only to lose her. Not possible.
"Keep talking to her, Beckett," Becca said. "Yes, I'm an ER nurse and I'm calling about a multivictim shooting incident at Garrison Forest Veterans Cemetery. You've got an ambulance less than two minutes out but I've got a penetrating chest trauma with likely . . ."
Beckett worked hard to tune out the list of damn scary medical terms flying out of Becca's mouth. He leaned over Kat's pale face. Even the green of her eyes had dulled. "Do you want a boy or a girl? When you become a mom?"
The corners of her lips twitched as if she might've smiled. "One of . . . each. So . . . neither . . . outnumbered."
Becca continued beside him. "I think she needs to go to the Shock Trauma Center at UM. Yes . . . Yes . . . Plenty of room to land here. Okay, thank you." She dropped her phone to the ground.
Marz returned with a sweatshirt, a striped Mexican blanket, and a handful of plastic grocery bags. "Got these from Jeremy's Jeep."
Shane flattened the plastic and indicated for Beckett to lift his hand so he could cover the makeshift dressing.
Beckett glanced from Becca to Kat, trying to process the former's words while focusing on the latter. "Yeah? Not outnumbered like you were, huh?" Jesus, her skin looked like someone had doused her with powder. Even her lips grew pale.
"Not like . . . me. Cold . . ."
"Cold? You're cold? We'll fix that," Beckett said. "Marz, the blanket?" The other man nodded and worked quickly to spread it out over her lower half.
Sirens from multiple vehicles got closer and closer. Voices. Shouts. But Beckett couldn't track any of it, not with Kat's eyelids growing heavier and heavier.
"Stay with me, Kat. Hey." He gave her hand a little jerk. Her lids flew up. "Stay with me. Okay? I have an idea. Sing the ABCs with me?" God only knew what made him suggest that, but his brain was firing on its most basic synapses right now. "Yeah?" She half nodded. And Beckett began to sing. "A, B, C, D—" Smoothing the hair off her forehead, he shook her hand again. "You sing, too, now. A, B, C . . ."
"D . . . E . . . G . . . A, I, K, K . . . Beck?" Her eyelids fell closed. And they didn't reopen.
Something ripped deep inside Beckett. He'd never felt more pain in his life. Not when his father hit him. Not when his father tried to stab him. Not when a grenade ripped his leg to shreds.
Paramedics appeared at their side, gear in hand. Someone grabbed Beckett's shoulders, urging him out of the way. He nearly fell on his ass, but Easy had his back. Literally.
"I got you," E said. "Come on, big guy. Give them room to work."
Beckett did as he was told, but that didn't keep him from wanting to destroy with his bare hands the vehicles in between which Kat lay, for having to be apart from her for even one second.
Because . . . what if that one second was all he had left? "Oh, God," he rasped.
"She's gonna be okay, Beckett. Don't you worry," Easy said.
Nodding, Beckett managed, "Yes. Yes, she is. 'Course. Did you take the shooter down or, uh, just wound him?" On some level Beckett wanted to stalk to where the piece of shit's body lay and tear it limb from limb. Just to be sure he could never hurt Kat again. On another, this Cole was so unimportant—so inconsequential—that Beckett didn't want to take one second's attention away from Kat to even think about the fucker.
Easy's eyes flashed. "DOA. Trust that's not a problem?"
The only thing Beckett didn't like about that was that he hadn't done it himself. While looking in the SOB's eyes. And watching the fear flood in as the life ebbed out. "No," he managed, his voice like gravel. He blew out a long breath while he watched the medics work on Kat. "What, uh, what else went down?" He wasn't really sure he could handle knowing right now, but he needed to.
Easy sighed as Marz joined the two of them. "Jeremy, Vance, and Seneka are injured pretty bad," Easy said, scrubbing a hand over his close-trimmed black hair. "Nick's got a grazing wound to his arm. Not serious. And Washington's dead."
The sound of those peeling tires came roaring back to Beckett's mind. "Wexler? Did he get away?"
"Yeah," Easy said. "Nick says he hit him at least once, but he got to Vance's car, stole his keys, and made it out."
"The, uh, rest of the women?" Beckett asked, wishing with everything he had that Kat was hanging in the backseat of Shane's truck just as she had been this morning, looking so polished and sophisticated and downright gorgeous in that clingy blouse and trim pants.
"All fine," Easy said, his voice strained.
Beckett's gaze stayed glued to Kat and his mind tried to keep up with everything the paramedics and Becca said. "How bad is Jeremy?" he asked. The question made him put two plus two together for the first time. Both of Nick's siblings were seriously injured? Jesus Christ.
"Head injury. He's unconscious. Fucking Wexler pistol-whipped him and dropped him in the street like a piece of garbage."
Fuckfuckfuck. Don't lose it, Murda. Hold it the fuck together. "Both are gonna be okay," he said, more to himself than to Easy or Marz. But both men voiced their agreement.
Rotors sounded from somewhere above them. Not too far, by the sound of it. Beckett looked up until the black and gold chopper came into view over the tops of the trees. As it descended, the downwash off the rotors whipped up the wind around them. It landed far enough away on the field to keep them safe from flying debris. And then two teams of two each were hauling ass with stretchers toward them. One branched off toward Jeremy, who apparently lay back where Vance's car had been parked.
The ambulance EMTs barked out a list of conditions and assessments, and then they made room for the trauma unit crew. The new guys had Kat loaded up onto the stretcher in mere minutes. Beckett followed, but the closest man shook his head. "Unless you're injured, there's no room," he yelled. "You'll have to follow by car. I'm sorry."
The words were like someone had reached inside his chest and ripped his heart out. He came to a halting stop, absofuckinglutely lost.
Down the way, another stretcher was lifted onto the grass. Beckett took off in that direction. Toward an unconscious Jeremy. The very thought made Beckett sick to his stomach. "How is he?" he called over the engine noise from the chopper.
"Don't know yet," Nick said, voice raw. "Jesus."
Beckett grasped Nick's arm—the one that wasn't bleeding. "Kat."
Anguish flashed through those odd, pale eyes. "How bad?"
"Bad," Beckett said.
"Gotta fly," one of the crew yelled. They rushed by with Jeremy's stretcher.
Nick whirled. "I have to come. They're my brother and sister."
The guy hesitated, but then his eyes went to the blood trailing down Nick's biceps. "How bad you hurt?"
"Grazed."
Nodding, the man said, "Good enough. Come on."
"Take care of them both," Beckett called as Nick took off behind the medics.
Both stretchers disappeared inside the big bird, the door rolled shut, and the engine powered up. Then they were lifting off, getting higher, disappearing altogether over the trees into the bright blue May sky.
Taking with them everything Beckett wanted but never thought he could have.
Chapter 23
More emergency vehicles had arrived at some point, because two ambulances and three police cars now clogged the road behind their line of cars. Four cops congregated around someone on a stretcher—Vance, if Beckett had to guess. Given the audience, Beckett rushed to the open doors of the other rig, Marz, Easy, and Shane right behind him.
Seneka lay flat on a stretcher, knees raised, as a tech worked on him.
"How is he?" Beckett asked.
"Nick?" Seneka called in a weak, raspy voice.
Beckett climbed up into the small interior. "What is it? Nick had to go. We've got casualties."
"Hey, you can't be in here," the EMT said.
Seneka waved Beckett closer as he tugged off the oxygen tube, his movements imprecise, his eyes groggy.
"Sir, please," the EMT said as he tried to put the line back in place.
The old man wouldn't have it. "Jesus, gimme space. Need to say . . . something," he managed. Gritting his teeth, Seneka shifted his arm and reached into his hip pocket.
"Mr. Seneka, you can't—"
"Look, son," he said to the paramedic. "I've got scars older than you. Well aware I've got some abdominal trauma. Let me . . . say my . . . piece." He blew out a harsh breath. "Take this," he said to Beckett, handing him a cell phone. "Pass code is 1445. All Wexler's contact info in there. Call him now. Dangle . . . the money. Before he goes . . . dark and deep." He groaned and clutched his side. "Files . . . in my . . . truck for . . . you."
"I'm sorry, sir," the EMT said to Beckett, eyes flashing in anger. "You have to go. Now. End of discussion." He leveled a hard stare at Seneka, who collapsed back against the stretcher.
Beckett cleared out of the ambulance to find one of the cops talking to Shane and Easy. The doors to the rig closed behind him and it took off.
Beckett glanced at the phone in his hands, his mind on a perpetual churn. Ahead, Vance stared at them from the back of his ambulance and waved them over.
"Jesus. I'm sorry I couldn't do more," he said when Beckett and Marz got closer. "Got hit in the shoulder during the first exchange of shots. Arm is fucking useless." He gestured to the mound of gauze around his whole shoulder and biceps. "Can't feel my fingers for a damn."
"Just take care of yourself, Vance. That's all that matters," Marz said.
Vance looked each of them in the eyes. "Do what you have to do to make this right. I'll help clean up the mess. And don't worry about full statements just now. Told the uniforms I'd get them myself later."
Beckett looked toward the tent, where other cops were taking notes and marking off the crime scene. Washington's body still lay in the grass. Farther down the hill, Cole's did, too.
What a fucking mess.
Kat. His heart tore inside his chest. God, it hurt. Everything hurt so damn bad that he could barely stand to be inside his own skin.
But there were things that needed to be done. People who needed to be looked after. And so he did one of the few things he truly excelled at—he shoved all that fucking emotion down so deep it might never see the light of day again.
The paramedics closed up the second ambulance and took off, sirens whooping over the quiet field.
"We gotta make that call," Marz said.
"Yeah," Beckett said, his voice steely and emotionless to his own ears. "Take a minute and check on the others? Reassure them everything's over, everything's okay." At least for this minute in time . . .
And even though it might never be okay again. Not for him. And, Jesus, not for Nick either.
Marz nodded and took off.
How are you, Kat? Hang in there for me, Angel.
The thought had him retrieving his own cell from his pocket. He shot a text to Nick. Keep me— He backspaced and deleted me. Keep us posted. We've all got your back.
A long moment later his phone buzzed a reply. I want Wexler's fucking head on a platter. And I want to know the goddamn sitch with the other shooter.
One in a long line of hells to pay was going to include explaining to Nick why he'd withheld the knowledge that Kat had a stalker. If he'd only opened his mouth like he knew he should've, like his gut told him he should've, Kat would be alive and well right now. Not bleeding out in the back of a helicopter. Not being medevacked to a trauma center to fight for her life.
Beckett punched out a reply. On Seneka's advice, we're going after Wexler before he disappears. Shooter was Kat's ex-boyfriend stalker. Will share what I know when I see you next.
Assuming Nick wanted to hear a word out of Beckett's mouth from this point forward. God knew he wouldn't blame the man for holding him responsible, at least for what happened to his sister.
Enough, Murda, Beckett chided himself. Get your head out of your ass. Right.
Shane and Easy came jogging up. "Gave them all our names and contact information," Shane said. "And a general rundown of events."
"Vance is gonna get official statements from us," Beckett said. "So that should clear us to bug out of here." The guys nodded. "Come on," he said. Walking to Jeremy's Jeep, he found Charlie, Becca, Sara, and Jenna standing by the front passenger door, each wearing grief-filled expressions. But Charlie's was by far the worst. And Beckett so clearly recognized the pain in the other man's eyes that it was like looking in a mirror. Guilt. Pure and unadulterated.
Beckett got right up in his face, grasped him gently by the back of the neck and nailed him in the eye. "Not your fault. You hear me? None of it." Charlie's blue eyes flashed in anger and disbelief. "Let it go, Charlie. I'm telling you. If you don't, it will stand between you and Jer like a wall. And you don't want that. Put all your energy into him getting back on his feet and let the rest go."
Charlie searched his gaze, and a single tear rolled down from the guy's blue eyes. And then another. And another. Like he was trying, but couldn't quite hold them back. "Thanks," he finally managed, his voice so choked it came out almost hoarse. "Can we take off for the hospital?" he asked.
"Rather we all stay together," Beckett said. "Wexler's a total loose cannon and we don't know who else might be out there that he's working with. Just give me a few minutes to deal with something, then we'll come up with a plan." Everyone nodded.
"You're a good man, Beckett Murda," Becca said, throwing her arms around his neck.
Falser words were never spoken. But he kept his mouth shut and accepted Becca's kindness, because she deserved that much from him.
"I'm gonna check on Emilie and her mom," Marz said. "Then we can take care of that call."
Beckett gave a nod and became acutely aware that he was the odd man out. Shane held Sara's hand as she worried over how he was doing. Easy hugged Jenna, who seemed unable to stop shaking. And Charlie and Becca leaned on one another, both of them quietly trying to hold back their worry and tears for the Rixeys they loved. At Shane's pickup, Marz leaned in the rear door, tending to his woman and her kin.
And there he stood, all the fuck alone. Just like always. Just like he deserved. Just like it might be forever.
And why the hell was he still having to learn this particular lesson? After thirty-four years, he'd have thought it had been well and truly hammered into his head. He was no good. He hurt others and drove them away. He was better off on his own.
He tried not to be a jealous, resentful asshole. He really did.
Without a word, Beckett left the group behind, passed by Marz, and let himself into Seneka's Suburban. Wallowing in his own ancient misery didn't help a goddamned soul. Focusing on work? Yeah, that's what he needed to do.
Nothing in the front seat.
Nothing in the passenger seat.
Beckett popped the back and waited while the hatch lifted.
A white file box with a lid sat in one corner of the trunk. Beckett opened the box and flipped through the folders stowed inside. One was labeled, WCE Account.
Heart kicking into a sprint, Beckett fished it out and flipped it open. It contained a series of deposit tickets from WCE to not only Merritt's and Kaine's accounts, but Wexler's, Garza's, and at least a half dozen other men, including Jimmy Church. And on a yellow Post-it in the center of the top sheet was a hastily scrawled note: WCE = Wexler-Church Exchange.
A business partnership, just like Procter & Gamble, Hewlett-Packard, or Ben & Jerry's. For fuck's sake.
And talk about the arrogance involved in using his own last name for this particular business, in which Wexler, a former Army Green Beret, made an unholy alliance with street scum to steal from his country, spread the poison of heroin to his fellow countrymen, and kill his brothers-in-arms in the process.
Someone like that couldn't be allowed to live.
"What are you finding?" Marz asked in a quiet voice.
Beckett handed him the file. "Everything we ever wanted to know about WCE." It was almost anticlimactic. Or maybe that was just because some of the people who'd made it all possible—and who should've been here to celebrate—were fighting for their lives instead.
"What a chance fucking encounter, huh?" Marz said, scanning the contents of the file.
After a long moment, Beckett nodded. Every single thing about Seneka rang true today. He'd almost certainly lured his two possible betrayers to Baltimore for the funeral and had intended to find some way to keep them in town for the meeting they'd scheduled—to which Seneka had always intended to bring those two men. But Seneka hadn't needed to do the dirty work. Wexler outed himself, which had maybe been Seneka's plan all along. No doubt Wexler had known exactly who they were the minute he saw their faces. He'd probably been searching for an escape hatch before the Suburban even rolled to a stop. But they'd have to wait until Seneka was on his feet again to know any of that for sure.
Flipping through more folders, Beckett found personnel records and detailed lists of connections for the other men whose names he hadn't recognized. There were carefully kept spreadsheets of heroin stolen and sold from various sites around Afghanistan, and—holy shit—a file containing a long list of women's names. "Take a look at this," he said, passing it to Shane.
Steel gray eyes cut to Beckett and then down to the clipped papers. "Fuck me running," Shane said, his voice like ice. He turned from one page to another.
When Shane was a teenager, some scumbag had scooped his eight-year-old sister off the street on which they lived. He and his family never saw little Molly McCallan again. As a result, protecting women who needed help had become a personal calling of Shane's. It was what led him to help Sara, who they'd learned had been forced into her waitressing job at a strip club and her relationship with one of Church's top henchmen to pay off her deceased father's debts to the gang.
"God, I don't know if this list is a blessing or a curse," Shane said as he showed it to Easy and Marz. "A blessing because their families could be notified about what happened to them. A curse because who would ever want to learn this? But, damn, if we could track down families or missing persons reports, we could find photographs. Maybe somewhere in all this or the Colonel's files there are records of who bought them." Shane raked a hand through his dark blond hair, his eyes flashing.
Beckett knew exactly where Shane's thoughts were going—to finding and saving them. A totally honorable goal. But that was a fight for a whole other day. Had to be. Not that Beckett liked it. Not one bit.
"You realize," Marz said, "between this and the Colonel's microchip, we might have all we'd need to take down every bit of this operation."
"Let's not get ahead of ourselves," Easy said, dark eyes flashing. "We need Wexler before we can come anywhere near to claiming mission accomplished here. Especially after today."
"Damn straight." Beckett tossed the files back into the box and retrieved Seneka's cell. The pass code the man had given him worked, just like he said. And it took only a few swipes of Beckett's thumb to find Wexler's contact card. He pressed on the listed cell phone number, then hit Send.
The first ring barely sounded when Wexler barked into the phone, "You fucking set me up." And wasn't that an interesting greeting.
"This isn't Seneka," Beckett said. "It's Beckett Murda. And I've got twelve million reasons why you might want to listen to what I have to say." He borrowed the line from Nick, and it seemed to work the second time around, too. While Wexler didn't respond right away, he didn't hang up either.
"You have two minutes," Wexler hissed.
"I want an in-person meeting to perform a simple exchange. You get the twelve million dollars from Frank Merritt's account. I get definitive proof from you that Kaine headed this whole operation and hung my A-team out to dry. Simple as that."
"You can't just withdraw twelve million dollars," Wexler said in a way that sounded like it should've ended with you idiot.
"Of course not. But I can bring a laptop, provide you with all the password and access information, change our personal information on the account to yours, and then allow you to reset the passwords and take the laptop, too. Just so you have no doubts that we've tricked you by using keystroke-recording software." When Wexler didn't have a quick retort, Beckett knew he'd bitten the hook. "We don't want the fucking money. We want Kaine's head for betraying us. Nothing more, nothing less."
"And what would definitive proof of Kaine's involvement look like to you?" Wexler asked.
Animatedly, Marz whispered a number of suggestions.
Beckett nodded. "Hard and digital copies of any and all e-mail correspondence between you or others involved in WCE and Kaine. Details, in written and digital form, of who ordered the hit on Frank Merritt, who exactly my team encountered on the road that day, and any information you can get your hands on regarding the collusion to smear our reputations and ruin our careers. And a list of buyers of your female slaves."
An incredulous laugh came down the line. "Fuckers don't want much, do you?"
Ice flooded into Beckett's veins. "We want what's ours. Our honor. Our reputation. Our careers. If you make us drop this at your door, we will. We have plenty to put it there quite convincingly, thanks to Merritt's record keeping and your boss's assistance. But what we don't want is you, Wexler. We. Want. Kaine. Your choice." Beckett affected a sigh. "But decide now. I'm hanging in ten." He mentally counted.
On eight, Wexler said, "Deal."
Dark satisfaction rushed in behind the ice. This guy was so his. "I name the time and place—"
"Hell, no. That's a deal breaker. Meet when and where I dictate, or it's off. I disappear with plenty of my own money in hand to a tropical paradise where I never think of any of this again."
Beckett surveyed his team and found frustration and agreement on their faces in equal measure. "Fine."
"It'll be at least this evening before I get all this together. Will call you on this number thirty to sixty minutes before the meet." Click.
Heaving a breath, Beckett lowered the phone and dropped it in his pocket.
"No matter how fucking high we climb, the summit just keeps getting higher and higher," Shane said.
"But if he delivers," Marz said, "we've just struck gold."
Beckett agreed, but it didn't give him nearly the sense of satisfaction or victory that it should've. Because it would all be hollow if he regained his life only for Kat to lose hers. It wouldn't be any life at all. Not without Kat.
Beckett waved the others over. "I'll lead a group to the hospital. Who wants to go?" he asked. The women all raised their hands.
"I'm going no matter what," Charlie said.
"Me too," Marz said.
"Yeah," Easy said.
Shane nodded. "I wouldn't be anywhere else right now."
Perfect. He liked the idea of them all remaining together. "Fine. Then let's move."
Chapter 24
The minute Beckett walked into the hospital waiting room, Nick flew off his seat and got all kinds of in his face. "What do you know that you haven't bothered to fucking share with me?"
Cheeks ruddy, eyes a little wild, Nick looked way the hell strung out. And who could blame him? He'd been shot and seen both of his siblings—his only remaining blood in the world—medevacked while unconscious to a shock trauma center. The nurse at the desk had confirmed both of them remained in surgery.
Beckett held up his hands and nodded. "I'm sorry, Nick—"
Wham!
Nick's fist connected with Beckett's cheekbone. He staggered back a step, shouts rising up around him. Beckett didn't raise a hand to defend himself. Nor to fight back. Why would he? Nick bum-rushed him into the wall by the door, sending other hospital visitors scrambling. "Fight back!" Nick shouted. "Fucking fight back."
"No," Beckett said, shaking his head.
Suddenly, Nick flew away from him. Shane and Marz each restrained him by a shoulder. Every last person not associated with their group fled the waiting room.
"Stop it! Just stop it," Easy said, standing between them, arms out to keep them from going at each other again. Not that Beckett planned to. "Two of our own are fighting for their lives. Last thing we need is to be fighting each other."
Nick jabbed a finger into the air. "Beckett knew Kat had a stalker. Didn't say anything about it to anyone. And let her go back to D.C., where the sonofabitch apparently regained her trail and followed her to the cemetery this morning. I deserve a fucking explanation." He nailed Beckett with a bleak stare. "You promised not to hurt her!"
Beckett struggled to suck in a breath. Those last words were like a knife to the gut.
"Jeremy knew," Charlie said, his voice hollow, his blue eyes so washed out they appeared almost gray. "She told him not to worry you with it. She'd gotten a restraining order, so I think she thought it was under control. She was going to tell you when all this was over."
Shaking his head, Nick looked like someone had just clocked him with a frying pan. "Well, holy fuck. Anyone else know besides me?" he asked.
"I guess I sorta knew, too," Becca said, her voice strained. "A few nights ago, Kat said her last relationship had ended poorly and that her ex-boyfriend was having a hard time accepting it was over." She shook her head and pressed her hands to her mouth. "I didn't know it was all this, though. I promise."
Nick shrugged off the other men's holds. "Jesus Christ," he said, fingers raking at his hair as he stalked toward the far wall of windows and braced his hands against the glass.
Beckett cleared his throat. It took a few tries. "She . . . she came to Hard Ink because this guy—Cole was his name—hid in her parking garage and jumped her Friday morning before work. She got him to back off by agreeing to meet him for drinks that evening. Instead, she packed a bag, filed for the protective order, and came here. Well, to Hard Ink. And she pushed her situation to the side to help all of us. I just found out the day before yesterday. She begged me not to tell, and she promised to come to you when our investigation was over." He ground the heels of his hands into his eyes and ignored the throb in his cheek. Despite the amazing night of sleep with Kat—which seemed forever ago now—he was stone-cold exhausted all the way into the depths of his soul.
Silence hung like a thick cloud over the room.
"Family of Jeremy Rixey?" came a man's voice from the doorway.
Everyone whirled and gathered closer around the surgeon, who wore a pair of blue scrubs, a head covering, and a pale blue gown. Nick and Charlie made their way to the front of the group.
"I'm his brother," Nick said.
The young Indian man surveyed the group staring back at him and nodded. "Jeremy's in recovery. He had an epidural hemorrhage, which is bleeding between the dura matter and the skull, from a blunt head trauma. He also has a very small skull fracture. We drained the blood and are moving him to the neurotrauma critical care unit on the fourth floor to monitor for additional bleeding or swelling. We'll know more when he awakens, but there's no immediate reason to believe he won't have a full recovery."
Oh, thank God. Now, if they could get equally good news about Kat.
"Can we see him?" Nick asked.
The surgeon pointed to a rack on the wall. "The visitor guidelines for the traumatic resuscitation units are available there so you know how it will work. As soon as he's situated, someone will notify you."
Nick's jaw ticked as he nodded, and Beckett reminded himself that what was happening right here wasn't about him. It was about Nick and his family.
"Any questions?" the man asked.
"No, thank you," Nick said, extending his hand toward the other man. They shook and the doctor departed.
The whole room breathed a sigh of relief. Becca pulled both Nick and Charlie into her arms. "He's going to be flirting and driving you two crazy before you know it," she said.
Charlie gave a quick nod, then pulled free of her arm and turned away. His hands went to his face, and Marz crossed the room to him, put an arm around his shoulders and talked soft enough that Beckett couldn't hear what he said. Derek always did have that ability—to talk to anyone, say just the right thing, and make people feel comfortable, at ease. Sometimes, Beckett would've paid good money for those abilities.
And then, as more time passed, the tension slowly but surely ratcheted back up. When were they going to hear about Kat? What was taking so long? What exactly had that bullet injured inside her?
The possible answers to those questions turned Beckett's empty stomach sour. All he could do was see the image of the wound marring her beautiful skin in his mind's eye. Her blood spilling out in pulsing waves. Her face getting paler and paler.
Finally, a petite African-American nurse in pink scrubs stepped into the room. "Jeremy Rixey's family?"
Once again everyone rose in anticipation of news, though it was Nick who stepped forward. "I'm his brother."
"All right," she said with a smile. "We have him comfortable now, so I'll take you—"
"Wait. Two people can be in the room, right?" Nick asked.
"Yes, that's right."
Nick turned and nodded his head to the side. "Come on, Charlie. He's gonna want you there when he wakes up."
The look of sheer and utter relief on the blond man's face reached inside Beckett's chest and squeezed. Hard. He wasn't expecting any similar invitation when Kat could finally receive visitors, and it left an aching hole inside of him.
"My sister's in surgery, too," Nick said as they walked out the door. "Can someone let me know when the surgeon's ready to speak to us?"
"Of course," the nurse said, their voices fading away down the hall.
And then they were back to waiting. Sunshine and shadows moved across the rectangular lounge space as the afternoon wore on. What if Kat wasn't done before he had to leave to meet Wexler? Even if Nick wouldn't let him in to see her, it would kill him to leave before knowing she was okay.
Please, God, she has to be okay. He pressed the flat of his hand against his chest. An uncomfortable pressure had grown there all day, making it so he could hardly draw a deep breath.
Beckett stayed plastered against the wall by the door. He rejected every offer of food or drink or to sit or take a break. Kat couldn't do any of those things right now. And neither would he.
He was so far into his head that Becca surprised him when she stepped in front of him, smiled up at him, and wrapped her arms around him.
"What's this for?" Beckett asked, unsure how to react. Finally, he returned the embrace.
"Just looked like you could use it," she whispered. "Kat's lucky to have someone who cares about her so much."
And it was like the words broke something inside him, something that had been holding everything together, something that had been holding him together.
Beckett's throat went tight. His eyes stung. His heart slammed in his chest. "Thank you," he said, kissing her cheek and sidestepping her arms. He had to get the fuck out of there. Out from in front of all those eyes. Before he made a big fucking spectacle of himself. He flew from the room, down the hall, into the bathroom.
He paced, clawed at his hair, beat his head against the wall. None of it helped. None of it worked. None of it made Kat better and safe and here with him.
Oh, God, gonna puke.
He turned, braced his hands against the white porcelain sink and wretched.
Of course, that was the moment someone came in the door behind him.
"Jesus, B," Marz said.
Dry heaves racked his body, because there wasn't a damn thing actually in his system. He and Kat breakfasted on bagels and coffee before the trip to D.C., but that had been a fucking lifetime ago.
When his body finally decided to stop abusing itself, Beckett rinsed the sink and splashed water on his face. Hands and face still wet, he stared into the empty bowl. "Nick's right. What happened to Kat is my fault." His voice sounded like someone had scoured his throat with sandpaper.
"No, it's not—"
"I saw the guy," Beckett said, peering toward Marz. "Last night. I walked into Dupont Circle and grabbed takeout. On the way back the fucker shoulder-checked me. Sidewalk was crowded, though, and I came up clean when I scanned myself for tracking devices. I should've been more suspicious . . ." He shook his head and looked back down. A droplet of water fell off the tip of his nose.
"Beckett—"
"Surgeon's here," Easy said, pushing through the door behind Marz.
Beckett bolted from the room, Marz and Easy at his side, and rushed back to the waiting room. A tall, thin, older lady dressed the same as the other surgeon stood just inside the room.
"Oh, here's Kat's boyfriend," Becca said.
Beckett had barely reacted to the word when the surgeon turned and offered her hand. They shook.
"How is she?" Beckett asked, the floor threatening to warp beneath his feet.
"Katherine came through the surgery fine. The bullet nicked the pericardium of her heart, which we were able to repair. It also fractured a rib, which punctured a lung and caused a hemothorax, a buildup of blood in the chest cavity. We performed a procedure to repair the lung and insert a chest tube. We'll monitor the drainage by CTs to see when that can be removed."
Beckett's scalp prickled. "So . . . so she's . . ."
"Barring any post-op complications or problems with the chest tube, she should make a full recovery," the doctor said, giving him a smile.
The wave of relief that crashed over him was so strong it was dizzying. "Thank you," he finally said, pressing a hand over his heart. God, his chest ached. Someone put their arm around him—Marz, who gave him a nod full of the same relief he felt. The whole group exclaimed and celebrated the good news.
"She should be settled in her room if you'd like to see her," the doctor said.
"Yes. Yes, please," he managed. "Her brother, though, he's—"
"Neurotrauma and the acute shock trauma unit are both on the fourth floor. I'll speak to him after I show you to Katherine's room."
"Okay," he said, scanning everyone's gazes.
"Tell her we love her," Becca said, giving him a watery but happy smile.
He nodded, then followed the doctor through the shock trauma building to Kat's secure unit and finally to her room.
"I'll speak to her brother now," the surgeon said. "Visiting hours end at six-thirty and resume at eight."
"Thanks," he said, his voice barely above a whisper. The clock on the wall by the nurse's station read 5:35. He wouldn't have long, but at least he'd have something.
He stood on the threshold of her door like it was the entrance to a sacred space.
Move, before Nick gets here and throws your ass out. Right.
The thought unstuck his boots from the floor and sent him into the room. His gaze locked onto Kat. Onto her pale skin. The tube running out of the side of her mouth. The IVs taped to her hand and wrist. White blankets covered her chest, but the sickly yellow-orange of the cleaning solvent they'd applied to her skin was just visible along her collarbone.
He felt like he moved in slow motion, like he might never close the distance that existed between the two of them. And then he was there, standing at the side of her bed, reaching out to hold her hand.
The moment his fingers touched hers, emotion swamped him so thoroughly it took him down to his knees. Fear. Anger. Guilt. Regret. Soul-deep sorrow. He pressed his forehead to the back of her hand and smothered his urge to sob against the edge of the bed. But those gut-wrenching emotions weren't the only ones overwhelming him. Because there was also a soaring relief, breath-stealing gratitude that she'd made it through, and something else. Something that was all-encompassing and eclipsed all the rest.
Love.
Love.
He . . . loved her.
No. He was in love with her. In love with another human being for the first time in his life.
That warm, aching pressure ballooned inside his chest again as he recognized the foreign sensation for what it was. Love. Love for Katherine Rixey, the only person who'd been able to sneak around his ancient defenses to the heart of him. The person who made him confront his emotions and let them rise. The person who'd made him realize that he hadn't been living. Not really.
The person who had awakened his heart, stirred his soul, and brought him back to life. Who'd made him want to live.
"I love you," he rasped, lifting his head and staring at her lovely face. "Do you hear me, Angel? I love you and I'm sorry and you have to come back to me so I can look into those beautiful eyes and tell you. I love you."
God, why hadn't he realized this last night? Why hadn't he told her before she lay injured and intubated and unconscious? What if she— No. He wasn't even going to allow himself to finish that thought.
Wetness streamed down his face. His voice halted and rasped. His shoulders shook. And, for maybe the first time in his life, Beckett didn't hold any of it back. Didn't even try to. Didn't even consider it.
He allowed the anguish and the longing and the relief to pour out of him. And the oddest thing happened—he gradually felt lighter, less burdened, freer to focus on what mattered most. Kat. And his love for her.
Her hand clasped between both of his, Beckett gently lifted it to his mouth and kissed her knuckles. "I love you so much."
Someone cleared their throat behind him.
Beckett peered over his shoulder to find Nick standing in the doorway, his expression unreadable. With a quick scrub at the moisture on his face, Beckett pushed to his feet. "How's Jeremy?" he asked, his voice like gravel.
Nick crossed to the foot of the bed. "Looks like hell, but they say he's doing fine." He sighed. "Right side of his head's all wrapped in gauze and he's got some bruises from Wexler kicking him. He's intubated, too," he said, staring at Kat.
Beckett nodded and swallowed hard. "I'll, uh, I'll go . . ." But he couldn't get his feet to move.
"Stay," Nick said. "If you love her, you should stay."
Heat crawled up Beckett's face. The wrong damn Rixey had heard what he'd said. But that was all trumped by the fact that Nick wasn't telling him to go. "I do," he said, gaze glued to Kat. "I'm sorry—"
"Don't," Nick said. "I know. And I'm sorry, too. I was an asshole to you. Even Jer told me so."
Beckett frowned. How could—
"I could hear him in my head telling me what a dick move it'd been for me to hit you." Nick's pale green eyes flashed toward Beckett. "He was right. I've kept a lot of things from Kat over the years because I didn't want to burden her. I can hardly blame her for doing the same damn thing." He jammed his hands in the pockets of his dark slacks. "Sure wish she'd wake up so I could say all this to her."
Beckett nodded, Nick's words making his heart less heavy in his chest. "Boy, do I feel that."
"Does she know? How you feel?"
"No," Beckett said, shaking his head. "Neither did I, until just now. Or, at least, I didn't recognize it." And maybe that was because when he tried to pin down the exact moment he'd actual fallen for Kat, he kept going back to that day in the stairwell. When they first met. When she wasn't the least bit intimidated by a big, male stranger pulling a gun on her in her brother's home. Whenever it was, something about her had spoken to something inside him from the very beginning.
And, Jesus, now that he knew, it felt like he might explode if he couldn't let Kat know, too.
"The shooter . . . Kat's stalker . . ." Nick said, voice ice cold. "Is he—"
"Dead." Beckett nailed Nick with a stare. And he didn't feel the least bit bad about the dark satisfaction that rushed through him. If that made him a monster, he'd own that shit full out.
"Good," Nick said, meeting his gaze. "That's good."
A long moment of silence, and then Beckett said, "Hate to bring work up—"
"Don't be." Nick shook his head. "This thing needs to be resolved before anyone else gets hurt."
"Agreed," Beckett said. "Gonna take a meeting with Wexler tonight. Trade the Singapore money for evidence against Kaine. Wexler'll be calling at some point with the details. I need to know what you want me to do with him."
"Perhaps we should talk before you make that decision," came a voice from the doorway.
Beckett and Nick both whirled in time to see the man come inside and close the door behind him. Brown hair, brown eyes, and of average height, their visitor wore a long lab coat over a white button-down shirt and khaki pants. Both the ID clipped to his pocket and the embroidered name on the jacket read: Dr. Zhaoming Chen.
"And why would we talk to you, Dr. Chen," Nick said, clearly as skeptical as Beckett about the man's identity. This guy couldn't have been more rural Iowa in his appearance.
Seemingly relaxed and casual, not-Chen moved closer and looked both Nick and Beckett in the eye. "Because I'm the man Colonel Merritt was working for in Afghanistan. And I think I might be able to help you."
Chapter 25
Beckett's mind went deadly focused. Because their visitor was either lying and a threat, or telling the truth and therefore the best thing that had happened to them in the past month. He saw the same questions reflected back in Nick's wide-eyed gaze.
"And who exactly are you?" Nick asked, muscles braced, expression carefully neutral.
The man gestured to the name on the coat. "Let's leave it at Dr. Chen for now. I'm a foreign service officer at the State Department."
Sounded perfectly believable. Except, when you'd worked in the SpecOps community and knew people who worked in the intelligence community, you knew that generic answers like this were often cover stories for whatever the person really did. Guys in SpecOps often said they worked for the Department of Defense, a line Beckett had used more than once in his own life. He also had a buddy who worked for the CIA analyzing data retrieved from reconnaissance satellites. When people asked him at parties what he did for a living, he'd often say something like, "I work with satellites." When pressed, he would tell distractingly interesting stories about experiences he'd had while working with NASA as a younger man.
So, foreign service officer at the State Department, my ass.
"Okay," Nick said. "Let's go with that. Why should we believe Merritt worked for you?"
"You shouldn't," Chen said, gesturing with his hands. "Which is why I'm prepared to prove it. Have you not yet learned that his daughter's silver bracelet contains a binary code?"
Holy shit. Beckett and Nick exchanged looks.
"I'll take that as a yes. I had the bracelet made for Merritt to send to her, so I know what the binaries translate to numerically," he said, face perfectly placid.
Merritt had mailed that bracelet from Afghanistan to Becca before his death. She'd thought it a simple present, but Charlie had recognized that the circle and bar charms formed two six-digit binary numbers, one if the bracelet was read from right to left and another if read from left to right. Those numbers turned out to be the user ID and password for the microchip Merritt had sent them hidden inside the glass eye of an Army teddy bear. To this day, Becca always wore the bracelet.
"And?" Nick said.
"The code translated to the numbers 631780 and 162905," Chen said in that same blandly polite way he had.
From the instant Charlie had converted the binary to numerals his first night at Hard Ink, those numbers cemented themselves in Beckett's head, because it had been clear from the beginning that they'd be important. Chen knowing them? That was convincing enough for Beckett. Because the numbers represented a way too specific detail for the man to know any other way.
"Jesus," Nick said, the glance he threw at Beckett full of incredulity. "How did you find us? And why now?"
"I knew who the five of you were, of course. I knew everything about Merritt's A-team and what happened to you in the wake of the ambush. But you weren't on my radar as having taken up his investigation until this past Monday, when I learned that a building owned by a Rixey—a unique name shared by one of Merritt's men—had been damaged in a suspicious explosion, and that a known operative within the organization Merritt was investigating died during that same incident. Given the backstory I alone knew, it was no leap at all to see the gas main explosion for the cover story it was." He took a few steps closer, his gaze flickering to Kat and back to Nick. "Your brother and sister's hospital admissions alerted me that you'd most likely be here."
Annnd the fact that he had ways of learning that kind of information placed him squarely in the intelligence community. Probably CIA—
Beckett sucked in a breath as a series of puzzle pieces nearly fell into place. "You're Special Activities Division." The elite paramilitary arm of the CIA, which in Afghanistan was headquartered at FOB Chapman. Their former base. Sonofabitch.
A small smile played around the man's mouth, but he didn't answer Beckett's supposition one way or the other.
Nick nodded. "So, how can you help us?"
"What is it you want more than anything else?" Chen asked, tilting his head as if analyzing them.
Beckett stepped closer, his heart racing with cautious hope. Because this man quite possibly had the power and connections to actually give them what they wanted. More than that, what they deserved. "Our honor, reputations, and service records restored. For the whole team."
"And for Kaine to face justice so he can never endanger any other soldiers under his command," Nick added.
"Those are within the realm of the possible," Chen said. "As long as you're willing to finish what you've started with the investigation. It all hinges on the assembly of an airtight case against Kaine and this cabal of Seneka operatives. Merritt started building it. It's up to you to finish it."
Nick and Beckett traded glances, and Beckett could see reflected in Nick's eyes the same hope and resolve he himself felt. Finish it? Wasn't even a question.
"That was our plan from the beginning," Nick said. "But why would you help us, even if we did finish it?"
Chen crossed his arms and tapped a finger against his chin. "Fair enough. I thought I'd lost four years of accumulated evidence. Poof. Gone in an instant in an ambush on a sunny rural road. Kaine has been suspected of dirty dealing and corruption for years, but he's so politically connected that no one's ever been able to get anything to stick. On a . . . let's call it a visit to FOB Chapman, I became aware of murmurings about stolen confiscated heroin, about destruction facilities destroying less than they received. No two ways about it, if these stories were true, they weren't just criminal, they were treasonous. Injurious to the very mission the United States attempted to carry out there. The gratitude of your country for picking up that torch will earn you the things you want, and will do a great service to your country at the same time."
The door opened behind Chen and a heavyset nurse stepped in. "Visiting hours are over, gentlemen. Oh, I'm sorry, Doctor."
He gave her a nod and a smile. "We were just finishing up." The woman left the door open and retreated as quickly as she'd come.
"So, about our meeting with Wexler tonight," Beckett said.
Chen nodded. "Are you picking the location or is he?"
"He is," Beckett said. "That was a deal breaker for him."
"Of course it was," Chen said. "In addition to being a sociopath with a pronounced Napoleon complex and deep-seeded feelings of inferiority aimed at his longtime friend and rival, John Seneka, Wexler is an explosives and demolitions expert. You can pretty well count on him planning that you will not leave that meeting location alive, and that he will use explosives to ensure it."
"Well, that's some fucking cheery news," Beckett said.
"Gentlemen," the nurse said, standing in the doorway with her hands planted on her hips.
"Shall we move this conversation to the hallway?" Chen asked.
"Just one minute," Nick said, rounding the end of the bed. He grabbed Kat's hand. "Wake up already so we can yell at each other or something," he said, kissing her on the forehead. "I love you. Everyone sends their love." His brow furrowed in an instant of sadness, and then he put his game face securely into place again.
As Nick walked away, Beckett leaned over. "Hurry back to me, Angel," he said, then kissed her temple. Damn, he didn't want to leave her. Doing so felt a whole lot like his chest was cracking open. But she couldn't be fully safe until the team brought this investigation to a full conclusion, so that was the best thing he could do for her right now. He followed Nick and Chen out the door, through the unit, and beyond the secure doors into the quiet hallway.
"Knowing Wexler's competencies is, as you so colorfully put it, 'fucking cheery news,' " Chen said, "because it means you know what to expect and how to defuse it. No pun intended." He winked. "You will simply have to identify the location of the device before the meeting ends."
Beckett heaved a deep breath. "I have a handheld chemicals and explosives detector in my equipment back at . . . the place," he said. His urge had to been to avoid divulging their home location, but maybe that was a moot point given who they were dealing with. This guy probably knew what size shoes Beckett wore. Twelves. Thank you very much. "I'd need to go there now to make sure I could get it before the meeting, though."
"Then let's do that," Nick said. "I'll handle the meeting with Wexler, while you and Easy work on detection and defusing."
Jesus. What if they couldn't detect and defuse any devices before the exchange was wrapped up? What if Wexler hit the detonator while Nick was still within the blast radius? Beckett's gut was waffling on the question of acceptable risks. But what choice did they have?
Chen nodded. "Once you have the evidence in hand, we should meet right away and decide the best and fastest course against Kaine. It's quite probable Wexler's already given the general a heads-up that things are unraveling."
"Makes sense," Nick said. "I can call you after—"
"Hold up," Beckett said, pulling a buzzing phone from his pocket, then looking at it. "It's Wexler." His heart thudded and his focus narrowed as he answered. Nick and Chen watched him like a hawk. "This is Murda."
"Wexler," the other man said. "Listen good. Meet me at eight o'clock at the under-construction parking garage on Aviation near Dorsey at BWI."
Beckett locked eyes with Nick. "You open to meeting with the head of my team? Nick Rixey?"
"No. We started this, we finish it. Come alone and bring the laptop. I get any sense that you've got company or are playing me, you die, I disappear, and this evidence goes up in flames for good. Are we clear?"
Nick nodded, and Beckett said, "Yes."
Click.
"Well, looks like I'm going in and you're helping Easy," Beckett said. He glanced at the phone's LED screen. "And we have eighty minutes to get ready for the most important meeting of our lives."
Chen produced cards and handed one to both of them. The white rectangle had exactly one piece of information inscribed on it: a phone number. That was it. No name, no title, no business or address. "Good luck, gentlemen," he said. "You're going to need it."
THE PAST SIXTY-FIVE minutes had been such a balls-to-the-wall scramble that Beckett could hardly believe he'd made it to the garage meeting location a few minutes early. Since one person was permitted to sleep in a trauma patient's room overnight, Charlie had decided to remain at the hospital to keep an eye on both Jeremy and Kat. But when Nick explained what was about to go down, everyone else returned to Hard Ink with the team to help and lend moral support.
Back in the gym, Beckett had given Nick and Easy a crash course on operating the explosive detection device, and Marz readied a spare laptop with everything Beckett would need to make the exchange look legit and recording software that had the ability to automatically e-mail logged keystrokes. That meant they'd have all the password information Wexler added to the Singapore account, and could therefore undo the changes he'd make to the account.
Because there was no damn way they were letting Wexler have his cake and eat it, too.
Beckett drove his SUV into the dark space, pulled a U-turn, and parked front-out near the same opening he'd just come in. He suspected he was going to want that fast and easy access before the night was over.
He shut off the engine and turned off his headlights, but left his interior lights on to provide at least a little illumination for the exchange. At the rear of his SUV, he lifted the hatch and set the laptop up on the flat cargo area inside.
Adjusting the body armor he wore under his shirt, Beckett scanned the garage around him. Empty. Quiet. Still.
Nothing to do now but wait.
The garage sat in the midst of woods on a four-lane road that ringed the airport. Cars and taxis whizzed by every so often, but the general atmosphere of the place was dark and secluded. It would be easy for his mind to play tricks on him and try to convince him he was really all alone out here, but he knew that wasn't true. By now Nick and Easy should've made it through the woods to the nearly finished parking structure to begin their search for explosives. Marz, Shane, and Dare would be taking up snipers' positions all around—though exactly where, Beckett didn't know. And Ravens Riders lay in wait in both directions on Aviation to make sure Wexler didn't get away.
Best case scenario, Wexler's ass was getting handed over to Chen. Worst? Well, Beckett wouldn't speculate on what the Ravens might do if they were given unfettered access to the man most responsible for the deaths of two of their brothers last weekend.
When the clock on his phone flipped to one minute past eight o'clock, Beckett's insides went on a loop-the-loop. What if Wexler didn't show? What if he'd already gone deep and dark? Kaine, too? If neither man could be held accountable for all the wrongs he'd done to him and his teammates, what kind of victory would that be? And how could any of them ever feel truly safe knowing those two men freely roamed the world?
No. One way or the other, those two would get what was coming to them. Beckett just hoped—
Lights swung into the access road that led to the garage, and then the sound of a car's engine reached him inside the structure.
Game time.
A black Lexus rolled into the garage, its interior too dark to ensure that the driver was Wexler or confirm that he'd come alone. It swung around to park just like Beckett had. When the car engine cut off, it left an even more pronounced silence in its wake. So quiet that Beckett could hear the interior mechanism of the door handle as the driver opened it and stepped out.
Definitely Wexler.
"You're late," Beckett said.
Wexler ignored the comment. "Did you bring the laptop?"
"Right here. Did you bring everything I requested in both digital and paper formats?" Beckett fired right back.
The man stepped to his trunk and popped it. "All yours," Wexler said, gesturing to the interior.
Beckett gave the opening a wide berth until he could lay eyes on what was inside. Two white file boxes full of papers. The trunk was otherwise so clean the car had to be a rental.
Wexler held out his hand. "This thumb drive has all the digital copies."
"I'll confirm what's on here on the laptop," Beckett said, taking note of how calm and relaxed the man seemed to be—which would make sense if Wexler felt confident that Beckett wasn't walking out of there alive no matter what was said or done in the interim. As Beckett returned to his open hatch and the laptop that sat within, Wexler stayed close. Too close, really. Given that this man had played a role in killing his friends, ruining his career, and attacking Hard Ink, it took every bit of discipline Beckett had to shove away his righteous anger and not squeeze the life out of Wexler using his bare hands.
Instead, he plugged the thumb drive into the USB port, opened the drive, and watched as a series of directories came up on the screen. He double-clicked on the one labeled E-mails. Sure enough, the folder contained exactly what Beckett had requested—incriminating e-mail communications with Kaine about WCE business, dating back to the beginning of WCE's arrangements, almost four years ago.
He opened a few more. Everything seemed in order. Beckett slipped the thumb drive into his jeans pocket. For a moment, he strained to listen to the nighttime sounds that wafted into the garage. Easy was going to whistle a bird call when he'd accomplished his task. That would be Beckett's all-clear signal.
So far, nothing.
"The bank log-in. Now," Wexler said, hovering at Beckett's side.
Beckett clicked over to the internet browser, already open to the bank's account log-in page, and entered the information required to access Merritt's account with its twelve-million-dollar balance. Once inside, he clicked on Edit Account Profile. Gesturing to the machine, he said, "It's all yours."
Wexler stepped forward like a starving man at an all-you-can-eat buffet, greedily grabbing the machine and hunching over it to hunt and peck at the keyboard. Keeping an eye on him, Beckett walked to the back of the Lexus. Using the light of the trunk, he confirmed that the papers in the two boxes were what Wexler purported them to be. He'd brought the hard evidence, which he no doubt believed was about to be destroyed in an explosion right along with Beckett.
Quickly, Beckett planted a small tracking device inside the trunk—just a precaution in case Wexler managed to give the Ravens the slip. That was doubtful, but it was always best to plan for the worst case scenarios in missions like these.
Beckett set one box on top of the other, lifted both out of the trunk, and moved them to the back of the SUV. He couldn't have acted more chill, despite the fact that the contents of those boxes would be the key to restoring much of what had been taken away from him and his teammates.
Glaring, Wexler shifted his stance and the laptop, too, making sure Beckett couldn't see what he entered. Not that it mattered. Marz's brilliance ensured that the computer was sending them all the information they needed.
"I guess Seneka wasn't the legend he's been made out to be, huh?" Beckett said, crossing his arms and watching the man work.
"Why's that?" Wexler said. Hunt and peck. Hunt and peck.
"Just seems like you wouldn't have set up your own little black op behind his back if he'd been even half the fucking rock star the stories paint him as."
Scoffing under his breath, Wexler shook his head. "Seneka might be everything you've heard, but he sure as hell didn't get there on his own. Despite what everyone says and the man himself acknowledges." Hunt and peck.
"I see. So he was one of those take-all-the-credit types," Beckett said, egging him on. Honestly, he was curious how Wexler justified everything he'd done to himself, plus, as incompetent as he was on a keyboard, trying to talk and type at the same time was slowing him down even further. And every extra second he could give Easy and Nick, the better. "They're the worst to work for. You bust your ass and lay your life on the line, while they sit back and claim all the glory."
Wexler peered up at him. "That's exactly what it's been like. Twenty fucking years."
Damn, if Chen didn't have this guy pegged to a tee.
"Done!" Wexler said. "Wait." He leaned in, scowling.
"It's gonna require you to log into your e-mail and confirm the account changes via a link they send. It's an extra security feature."
Huffing, Wexler nodded and went back to clicking, hunting, and pecking. Leaning in more, he frowned.
"Problem?" Beckett asked.
"Not anymore," Wexler said. Straightening, he aimed a gun point-blank at Beckett's chest and fired. Twice. The silencer kept it from making noise.
Which compounded Beckett's surprise as the slugs hit him right over the heart.
The vest did its job, keeping the bullets from penetrating his body. But the combination of the close range and the velocity at which the bullets had left the muzzle sent him staggering until he went down hard on his back, his head glancing off the concrete. The frontal impact of the bullets stole his ability to breathe, and the fall knocked what little remaining breath he had right out of him. His vision went splotchy, his head tingly. He clawed onto consciousness enough to hear shots being fired, enough to see Wexler toss something into Beckett's SUV, enough for his survival instincts to roar, Get off your ass soldier, and move!
The tires on Wexler's car squealed against the new concrete.
Sluggishly, Beckett forced his body up, pain radiating outward from his chest, his mind spinning on Wexler having thrown something into the truck. The documents! He staggered to the back, grabbed the two boxes and ran.
He might've made it eight agony-filled strides when the explosion went off. The blast hit him in the back and sent him flying. Weightless. And then gravity returned in fucking spades. Slamming him down to the concrete so hard his ears rang. Or maybe that was the result of the explosion in such a confined space.
Either way, it kept him from hearing his teammates until Nick and Marz were in his face, grabbing him by the shirt and pulling him away from the fiery debris field. Their mouths moved but there was no sound. Then Shane was in his face, seemingly asking him something.
Beckett shook his head. "Can't hear you! Can't hear!" he said.
Sonofafuck, he hurt everywhere.
"The documents?" he said. "Did the documents make it?" He had the thumb drive, of course, but had no idea if Wexler had really done what he'd asked and provided all the documents in both formats. And he hadn't wanted to take the chance. Not with the evidence being their ticket out of all this.
Jesus, how did the ringing in his head manage to deaden every other sound and be so damn loud?
Shane waved a hand in front of his face, forcing Beckett to focus. His teammate gestured as if to ask if he could stand. Beckett nodded and sat up—
Except . . . Nope. No, he actually could not do that at all. Not on his own.
Easy and Shane were right there, helping him to his feet. Beckett groaned as his body protested the movement. But none of that mattered. What mattered—and what he still didn't know—was whether he'd saved the evidence. Whether they'd secured Wexler. Whether he'd done the job they'd needed him to do.
Marz appeared in front of him holding the two boxes, a big grin on his face.
"We got it?" Beckett asked. "Everything made it?"
Laughing, Marz nodded, and then someone tapped Beckett on the shoulder. Grinning, Shane put a finger to his lips and gestured to tone it down with his hand.
Oh. "Am I yelling? I can't hear a goddamned thing."
Shane nodded and gave his shoulder a squeeze.
Beckett groaned at the light contact. "Fucker shot me. Twice. Point-blank. Feels like . . . somebody swung for the fences . . . using my chest for batting practice." Shit. Where had all the oxygen gone?
Shane's whole face frowned. With Nick and Easy supporting his arms over their shoulders, Shane tore Beckett's button-down open. Sure enough, two flattened silver slugs were embedded in the front of the Kevlar vest. Christ, he was going to need to eat an entire bottle of ibuprofen tonight.
Movement caught Beckett's eye. Dare came running up, cell phone pressed to his ear. He mouthed words Beckett couldn't make out, and then Shane and Marz were both clasping hands with Dare, smiling, laughing . . . celebrating?
"What? What happened?" Beckett asked.
Marz held up a finger and crouched down by the boxes. A minute later he rose and held out a sheet of paper.
Beckett squinted to read it by the firelight of the burned-out shell of his SUV.
Documents all saved
Wexler dead
Beckett's gaze cut to Marz's. "We got him?"
Marz nodded.
"You're sure? We really . . ."
Yes, Marz mouthed. And then he rolled his eyes and scribbled on the paper again, underlining his words twice for emphasis.
We got him.
And it's finally fucking over
Chapter 26
When they pulled into the gravel lot behind Hard Ink, Beckett knew three things for sure.
First, that he wanted to punch Shane McCallan in the face for making him stay awake the whole ride home when he wanted to sleep so bad it hurt. Everytime Beckett's eyelids started to fall, Shane shook him awake and pointed to Beckett's head. All Beckett had to say to that was, "Fuck you, concussion," because he'd more than earned a few hours of oblivion.
Second, that he wished his hearing hadn't started to return so he wouldn't have had to hear Marz butcher the entire eighties' anthem "Eye of the Tiger" from memory—God love him, but the boy could not sing to save his life. Beckett had been almost ready to hit himself in the head again just to see if he could make the madness stop.
Third, that their night was about to get even better. Because Chen stood outside the back door waiting for them. Before they left, Nick had alerted the Ravens remaining behind to guard the building that Chen would be coming and had clearance to pass the roadblock. Nick had called Chen as they hightailed it out of the garage to let him know they'd been successful. And here he was already.
Beckett startled as Nick opened his door for him, mostly because he wasn't actually aware the truck had come to a stop. Still felt like they were moving, or maybe that was just how badly he'd gotten knocked around. But at least his feet could support him now without the world going all tilt-a-whirl. Not that his friends let him go it alone. Marz and Nick flanked him, their arms around his back the whole way across the lot and up to the gym.
Supporting him. Carrying him. Catching him when he fell.
Boy, if that wasn't a metaphor for everything he'd ever wanted and never had. He had it now, though. But with all the evidence they needed in hand, what would happen next?
Inside, they found Becca, Emilie, Sara, and Jenna sitting at the computer tables waiting for them. Right before Marz had started busting everyone's ear drums, he'd called Em to let her know what had gone down. The women sprang from their seats and rushed across the room, and as the four couples around him celebrated their reunions and their victory, the pain in Beckett's chest had absolutely nothing to do with having been shot.
Kat. How was she doing? Had she awakened? Had there been any complications? Was she scared and lonely, wondering why he wasn't there? At least Charlie was there. He'd let her know what was going on. Although that did nothing to make Beckett yearn any less to be by her side.
Fuck.
"Oh, my God, Beckett," Becca said, her gaze running over him. "I want to hug you but I don't want to hurt you."
He managed a nod, holding out one hand to keep her at bay. "I'm one giant bruise right now. I'll take a rain check, though."
She pressed onto tiptoes and kissed his cheek. "I talked to Charlie. Kat hasn't woken up yet but the nurses say her vitals are strong."
Aw, hell. Now his eyes were stinging. "Thanks," he said.
"Em's grabbing you some ibuprofen, big guy," Marz said. "Let's get you sitting down before you fall down and put a hole in the floor."
Beckett chuckled and groaned. "Don't make me laugh."
Soon, they were all gathered around Marz's desk, including Eileen, who kept jumping up on Marz's lap, and Cy, whose head popped out of one of their supply boxes in front of the desk. Instead of running, he sat and watched them warily with that one yellow eye.
Nick held out his arms. "Well, meet the gang," he said to Chen. "You can speak freely in front of everyone. We're all in this together."
Wearing the same thing he'd had on earlier—save for the physician's coat—Chen scanned the group and nodded. "As you wish." He sat in a folding chair and braced his elbows on his knees. "Million-dollar-question time," he said. "What did you get from Wexler?"
Easy hefted the boxes onto Marz's desk, and they spent the next hour going through them. The e-mails detailed WCE's criminal activities and Kaine's knowledge of and participation in said criminal activities. They made clear that Kaine ran the Afghan side of WCE's black op, while Wexler managed the U.S. side, though in the tone and wording of many of the messages, Wexler clearly came off as the subordinate in the relationship.
Included among the messages were exchanges with and about Frank Merritt. Kaine had initially recruited Merritt into the op without realizing that Chen had already recruited Merritt to investigate what the CIA suspected was a massive conspiracy of corruption, theft, and smuggling. So Merritt had been playing both sides from the beginning, and Beckett couldn't begin to imagine the kind of stress that must've laid on his shoulders—especially since Kaine and Merritt had been friends since West Point. At the end, Kaine's suspicion of Merritt grew—why was not totally clear from what they'd read so far, but the general had communicated his intent to eradicate a possible breach in their organization with the words, M's dispensable. I'll handle the problem within the week.
That e-mail was dated three days before the ambush that had changed the lives of every single person in the room.
The next week, another e-mail to Wexler followed up with the chiller, Didn't end them all as planned. Will handle administratively. Longer, messier, but just as effective.
Sonofafuckingbitch.
For the first time since returning stateside, Beckett realized just how truly lucky he was to have made it out alive. To be alive. Because if Kaine'd had his way, Beckett never would've had the chance to fall in love with Katherine Rixey.
As they worked through the boxes, they found many of the messages worded similarly—vague enough to perhaps seem benign. Specific enough that, if you knew what they referenced, you saw just how goddamned evil Landon Kaine really was. But when paired with Merritt's records, these new files were definitely damning.
Marz then walked Chen through everything else they'd accumulated on Church, WCE, and Kaine—from his own research, from Merritt's microchip, from Seneka, and from Wexler. Marz left out any mention of Kat sharing documents from the Justice investigation, and Beckett was relieved at that on her behalf. She'd been forced to sacrifice enough, thank you very much.
It was an impressive body of work, and an even more impressive collection of evidence. The question was . . .
"So, is this enough?" Nick asked. "To nail Kaine? To clear our names? To bring this whole conspiracy crashing down once and for all?"
Chen stared at Nick a long minute, then gave a single nod. "This is exactly what is needed to accomplish every one of those objectives. But there's a catch."
Beckett's stomach dropped. What now?
"And that is?" Nick asked warily.
"Your team deserves the credit for busting all of this wide open. But the Company's going to claim it. Just telling you that straight up. How big a problem is that for you?"
Nick's gaze roamed the room, and each man said his peace.
"Fine by me," Easy said.
"Hell, I don't want to be explaining to anyone how we got half this information," Marz said, and that perspective struck a chord deep inside Beckett. Because they hadn't exactly stayed on the right side of the law in all of this, had they?
"As far as I'm concerned," Shane said, hugging Sara in front of him, "it was never about any kind of recognition. So that doesn't bother me any."
Beckett nodded, grimacing as he shifted in the chair. "I don't want credit. I want justice. However you can best serve it up."
Nick held out his hands. "There you go. I think it's safe to say that we'd be happy with you keeping our names out of it as much as possible."
Chen rose from his seat. "Well, then, let's get busy. I want a full digital archive of all of this as quick as we can. Because I know exactly what I need to do."
OH, GOD, EVERYTHING hurts.
Sounds. Smells. Sensations. These all slowly returned to Kat as she broke through the surface of consciousness. Her eyelids were stuck together, her lips were dry and sore, her mouth was a desert. She swallowed and . . . there was something in her throat.
She moaned and sluggishly lifted a hand to her face.
"No, no, hon," someone said. "Gotta leave that in for now. Come on and wake up for me, Katherine. Can you open your eyes?"
It was like lifting a hundred pounds, but she finally did it. Pale green walls. Darkness at the window. Twin bed with rails on the sides. A nurse?
It all came rushing back.
The funeral. The shooting. Cole.
Oh, my God. What happened? What happened?
"Whoa, hon. Settle down," the young, blond-haired nurse said. "I'm sure you have lots of questions and we'll get them all answered. But you've been through a lot and you've gotta try to stay calm. Okay?"
Blinking away tears, Kat nodded. Been through a lot? Was that why she hurt so bad? Because her chest was on fire.
The nurse—Carrie, her name tag read—dabbed away the wetness with a tissue. As she monitored all her vitals, Carrie explained that she'd been shot in the chest and had undergone surgery to repair damage to her heart and right lung. "Everything's looking good so far," she said. "We're going to take you down to CT in a few minutes to check on the fluid around your lung."
Kat tried to speak and it came out a moan, and she was so frustrated that she couldn't communicate.
"Here," Carrie said, handing her a thick pad of yellow Post-it notes and a pencil. "Do you think you can write?"
"Mm-hmm," she said. It was messy as hell, but she managed to get down her question. Were any of my family or friends hurt?
Kat saw the answer on Carrie's face the moment she finished reading the question—and she also saw the woman's hesitation to answer.
Fumbling the pad, Kat lowered it to write: Plz tell me.
Carrie's expression filled with sympathy. "Okay. Your brother's in the next unit. Suffered a brain injury but is expected to do fine."
Oh, God.
Name? she wrote with a shaking hand.
"Uh, Jeremy, I think."
Kat lost it. Maybe it was the pain or the grief or the meds they'd given her, but the thought of her happy, playful, sweet brother suffering a brain injury was the most sickening thing she'd ever heard.
And then, mercifully, a cold stinging sensation threaded through her arm and made her drift away . . .
The next time Kat surfaced, sunlight streamed into the room. She was all alone. Where was everybody?
Piercing pain radiating from her chest chased the question away. The intubation tube kept her from bending her neck enough to try to see her wound, but she ran her fingers over her chest and felt the thick bandages that must lay beneath her gown. Probably just as well, since movement hurt and exhausted her in equal measure. She sighed and closed her eyes.
You sing, too, now. A, B, C . . .
Beckett. The memory of him singing to her—trying to get her to sing with him—popped into her head. And despite the sheer silliness of the song, the man could sing. Big, strong, hard-ass Beckett. A singer. Who would've thought?
God, she missed him. How many days had it been since the funeral? For some reason, it felt like ages. Ages since she'd last seen Beckett, felt his touch, received his kiss.
It was so quiet in the hospital room. Like a tomb. Kat's skin broke out in goose bumps at the thought. She grasped the black remote tethered to the bed rail and turned on the wall-mounted television.
Cartoons. Sports. Reruns of some old reality TV show. A History Channel documentary on D-Day. Cable news. Blech.
She was about to flip back to the documentary when the moving ticker at the bottom of the news program caught her eye.
BREAKING NEWS: Decorated U.S. Army General Landon Kaine at Center of International Conspiracy.
Wait. What? She turned up the volume on the talking heads at the news desk.
"Breaking news now," an older male anchor said. "Two-star Army General Landon Kaine, one of the top names circulated to become the president's next National Security Advisor, has been accused of running a long-standing, international narcotics ring involving the theft and smuggling of Afghan heroin confiscated by the Army and slated for destruction. A Washington Post article today published a damning exposé showing how Kaine, for his own personal profit, conspired to ship that heroin to the United States for distribution at the hands of Baltimore's Church Gang, believed to be the biggest distributor of heroin on the East Coast . . ."
How in the world had this happened? How long had she been asleep?
"In addition," the anchor continued, "the Post article lays out the very disturbing story of an Army Special Forces A-team assigned to counternarcotics missions in Afghanistan that Kaine sought to destroy when its commander learned of his involvement in the theft. Colonel Frank Merritt was a highly decorated soldier who died in a checkpoint ambush last year along with six other Green Berets from his team. As the commanding officer of their base, Kaine then brought charges against the five survivors of the ambush and oversaw their discharges from the Army. The Post withheld the names of those dishonored service members, and the Army has not yet issued a statement . . ."
Was this . . . was this really happening? But how? What the heck had she missed?
Kat stayed glued to the news for the next half hour, but didn't learn anything new. A nurse—male this time, apparently the shifts had changed since she'd last been awake—came in to take her vitals and inform her that they were going to be extubating her airway tube.
She was thrilled. Until the procedure itself, which left her throat raw and achy after she gagged the tube out. Her voice sounded like it belonged to a seventy-year-old lifelong smoker. Awesome.
The nurse also told her that most of the fluid had drained from her chest cavity, which meant she'd probably only need to have the chest tube in for another day or two. She couldn't wait.
Kat drifted in and out of sleep, awakened by the pain in her chest, strange, scary nightmares, and nurses checking her vitals. Each time she opened her eyes, she hoped she'd find someone sitting by her side, but instead found the room empty. A sign under the wall-mounted clock across from her bed announced that visiting hours started at noon. Maybe someone would come then?
At 11:59 a smiling African-American nurse came through the door. "I have something I think might cheer you up," she said, then leaned in and whispered, "Girl, they've been waiting for hours to come see you. Mmm-mmm."
Then Nick and Beckett walked through the door.
Kat had never been happier to see other people in her life. She clapped a hand over her mouth as a sob climbed into her sore throat.
Nick rushed to her side and leaned in for a hug. Her chest hurt too much to lift her arms very high, so she clung to him awkwardly, but neither of them cared.
"How is Jeremy?" she rasped.
Bracing his arms on the rails, Nick said, "Charlie texted earlier to say he was awake and that Jeremy recognized him right away. He's apparently annoyed as shit at the airway tube." Nick smiled, and it was so good to see. And so . . . unusual, too. Even Nick's eyes seemed brighter, happier. "I'm gonna go visit him next."
Her gaze slid over Nick's shoulder to where Beckett hung back by the door. God, it was good to see him. Although, his face was beat all the hell up. "Come in," she said, waving her hand. "What happened to you?"
He moved like an old man, like it was more than just his face that had taken a beating.
"It's a long story," he said, coming around the side of her bed, one hand behind his back.
Kat glanced between the two men—two of the three about whom she cared most in the whole world. "Oh, I saw the news. What the heck happened?"
Nick grinned at Beckett. "Aw, hell. Ruined our surprise." Then he produced a newspaper from behind his back. The Washington Post.
One of the page-one headlines read: Landon Kaine at Center of International Smuggling Conspiracy.
"And look at this," Nick said, flipping through the A section. He turned the paper toward her, and she saw a photo of an older man who bore a striking resemblance to Charlie. It was Charlie and Becca's father, and the team's commander, Frank Merritt. A headline read: New Evidence Comes to Light in Deaths of Seven Green Berets.
Kat's eyes scanned the story, and it was—finally—the beginning of the end of this whole nightmare for them. According to an off-the-record source at the Pentagon, there was going to be a new investigation into the circumstances of the ambush and the surviving members' other-than-honorable discharges.
Her gaze cut to Nick's. "You're going to be cleared. All of you. I'm so happy for you," she said, more damn tears leaking from her eyes. It was like she was a waterworks today.
"I couldn't have done it without you, Kat. I want you to know that. You made so many damn sacrifices," Nick said.
She shook her head. "No, I did exactly what I had to do. And I wouldn't change a thing."
"I would definitely change this," he said, his hand waving over her body in the hospital bed. Beckett nodded, his expression serious.
But this hadn't happened because of the situation with Nick's team. This had happened because she had made too many excuses for someone who hadn't deserved even one. "Me laying in this bed isn't your fault, Nick," Kat said. "And I know it's not really mine either, but I could've taken more steps to make sure something like this didn't happen." She blew out a breath, wincing at a zing of pain from beneath her ribs, and calmed her breathing. "I'm sorry."
"Cole's dead, and it's over now. Let's talk about it later," Nick said. Kat's gut filled with surprise and sad relief. She never thought it would come to all this. "The only important thing right now is getting you better. How are you feeling?"
She grimaced. "Like somebody cut open my chest, stirred some things around, and sewed me back up again." He arched his brow, and it made her smile. "I hurt. And I'm exhausted. But I'm going to be okay."
Nick nodded. "Yes, you are." He leaned in and kissed her forehead. "Listen, I'm gonna let you and Beckett visit. I'll go see Jeremy."
"Okay," Kat said. "Tell him I love him." Nick agreed, and left.
And then it was just her and Beckett . . .
She scooted her legs to the side as much as she could, gritting her teeth against the pain. "Sit down—"
"No, that's okay—"
"Beckett, don't be stubborn. Sit with me."
He chuckled and grimaced, then sat his hip on the edge of the mattress. "You're ordering me around from your hospital bed. Do you realize this?"
She smiled and nodded. "You love it."
His face went serious and he nodded. "I do. Oh, I, uh, brought you something." He revealed a stuffed animal from behind his back. The sweetest gray and pink stuffed elephant she'd ever seen. Well, except for the one she'd had as a very small kid. Whatever happened to that guy?
"Aw, Beckett. This is so cute. Thank you." She hugged it against her face, and it was silky soft. "He'll be my constant companion."
Beckett grasped her hand and scooted closer. "Actually, I, uh . . ."
Was his hand shaking? "What? Are you okay?" she asked.
"Uh." He chuckled nervously. "Well, yeah, it's just that . . . um . . . I'm sorta hoping that I could be that, instead."
Kat blinked. "Be what?"
Those incredible blue eyes looked deep into hers, and she'd never seen them more open and vulnerable. "Your constant companion."
Her heart panged and her eyes prickled. "I . . . I'd like that."
"Wait," he said. "I'm screwing this up."
"No you're not—"
"I am," he said. "Because what I really want to say is, Kat, I love you. I am in love with you. And I want a chance to be the man who gets the honor and privilege of standing by your side."
She pressed a shaking hand to her mouth because, damnit, she was going to start crying again. "You . . . love me?"
He blew out a shaky breath. "Uh, yeah?"
"Well, that's really good, because I love you, too, Beckett."
It was, without question, the first time she ever saw unrestrained joy on the man's face. And even with all the bruises and nicks and cuts he had, it made him absolutely gorgeous.
Keeping his weight off her body, he leaned as close as he could and kissed her cheek. "I really want to kiss you and hold you, but I'm afraid I'll hurt you."
Kat cupped his handsome face in her hands. "I'm kinda afraid I'll hurt you if I hug you, too." She gave him a soft kiss. "So I'll just tell you again and again. I love you. So much."
Beckett's eyes got glassy, she would've sworn it. He blinked fast and leaned his forehead against hers. "I've never been the kind of man who believed in dreams, Kat. That wasn't the hand I got dealt. So I sure as hell never spent any time chasing them." He swallowed hard and nodded. "But meeting you . . . if I could've dreamed, if I'd have even known what to dream of, it would've been of you."
Now Kat was the one blinking fast. She stroked her fingers down his face. "That is the sweetest thing anyone has ever said to me," she said, her voice cracking. "You're such a good man, Beckett, and you deserve to have dreams. And I would love to be the one who helps you make them come true."
"Was so scared I lost you," he said, lowering his head to her shoulder. His big shoulders shook, and it broke her heart.
"Aw, no. You didn't. I'm right here with you, Beckett." She stroked his shoulders, his neck, the back of his head . . . where he had a big knot. "What happened to you?"
He heaved a deep breath and swiped at his eyes before he looked at her. "The short version is, I got shot and blown up."
Kat's eyes went wide as she raked her gaze over him. "Uh, I think maybe I'd better have the long version."
Beckett told her everything. Kat was equal parts horrified, amazed, and proud. What an incredibly strong, brave man she had.
"And now, finally, you have everything you deserve," she said, pointing at the paper still covering her lap. "I'm so happy for you. For all of you. God, I wish I wasn't in here so I could celebrate with you."
"Me too," he said. "But you'll be out in a few days and then we'll celebrate. Jeremy, too."
She nodded. "Do you think you'll see this Chen guy again?"
Beckett's eyes narrowed and he finally nodded. "For some reason, yeah. I don't think we've seen the last of him. But who knows. He's done everything he said he'd do for us, and probably bigger and better than we could've done it ourselves."
The news anchor mentioned Kaine's name, and Kat pointed to the TV. "Oh, look. It's your story again. Come sit on this side so you can see," she said.
Eyes on the TV, Beckett rounded the bed.
Kat scooted over and patted the mattress. "Think you can fit?"
He chuckled. "I don't want to jostle you."
"What if I want you to jostle me?" She waggled her eyebrows.
He barked out a laugh. "I'm not sure either of us is going to be up to jostling for a few days, do you?"
Grinning, Kat patted the bed again. "Well, then, squeeze in with me. No matter what, being with you will make me feel better."
He just managed to fit along the side, and she'd been right. His warmth and his scent and his touch . . . Beckett made everything better.
A male anchor looked into the camera. "This story involving Army General Landon Kaine just keeps getting bigger and bigger today. The D.C. police have confirmed that they found Kaine dead at his home of an apparent self-inflicted gunshot wound just an hour ago."
Beside her, Beckett's muscles went rigid.
Kat's hand flew to her heart. "Oh, my God." Just days ago she'd sat less than ten feet away from that man in a coffee shop. And now he was dead?
The newscaster continued, "Police reported to the quiet upper Northwest D.C. neighborhood when area residents reported hearing gunfire. To repeat, police are confirming the apparent death by suicide of General Landon Kaine, revealed by the Washington Post just this morning as the mastermind behind an international drug smuggling conspiracy . . ."
"Jesus," Beckett said, raking his hand through his hair. "I can't believe it."
Kat shook her head. "I can't either. It's horrible." She grasped his chin and nudged him to look at her. "But if Wexler and Kaine and Church are all dead now, then that means that you and Nick and the guys are all safe. All of us are. And I can't feel bad about that."
"No," he said. "That's nothing to feel bad about at all."
Sighing, Kat pointed to the remote attached to Beckett's side of the bed railing. "Can you turn the TV off? I don't want the outside world in here right now. I just wanna be with you."
"That right there is my dream come true, Kat." He shifted and lifted his arm, inviting her to rest her head in the crook of his shoulder. They were packed in the little bed like sardines, but it was warm and comfortable and exactly where she wanted to be.
"Thank you for helping my brother, Beckett. I love you so much," Kat said, sleepiness softening her voice.
"Aw, Angel, I love you, too. Now, sleep," he said, gently resting his head against hers.
"So bossy," she whispered, and then she did exactly what he'd told her to do.
Epilogue
One week later . . .
"Uh, guys?" Marz called from his desk, his gaze glued to one of the security monitors. Everyone else, including Kat and Jeremy, who were both finally home from the hospital, was still finishing up lunch around the big table in the gym. Beckett was so glad to have Kat back at his side. And to be able to sleep in a bed wider than three feet. "We've got company."
"Who is it?" Nick called, getting up from the table. The team had stayed together, still in defensive mode, until they received confirmation that the other members of Kaine and Wexler's conspiracy had been picked up by the authorities. Afterward, the Ravens had gone home with the team's undying gratitude and a standing offer to return the favor if ever they could.
Beckett rose, too.
"It's Chen," Marz said.
"I'll go let him in," Nick said, darting across the gym.
Beckett knew he should've put money on seeing the guy again. Question was, what the hell did he want? Had something gone wrong with the investigation? Did he need them to gather more evidence?
As was their habit whenever something work-related arose, everyone congregated around Marz's desk. Nick returned a few minutes later, Chen at his side in navy pants and a light blue button-down. He carried an oversized briefcase in his hand.
"Gentlemen," he said when he reached the desk. He took a moment to shake everyone's hands. "I have some things for you."
"Good things?" Nick asked, unleashing a chuckle around the room.
"I think so," Chen said, opening his case. "I have good news and bad news."
Beckett's shoulders fell. And he didn't think he was the only one. What now? He took Kat's hand in his, and she gave him a wink.
"Let's hear the bad news first," Nick said, voice tight.
Chen nodded and scanned the group. "Okay. The bad news is that you cannot share any part of the following conversation with anyone outside this room."
"About what?" Nick asked.
"For starters, this," he said, pulling a stack of manila folders from his case. One by one he handed them to the five guys from the team.
Beckett flipped his open. The top sheet was a letter from the Secretary of Defense stamped Top Secret. Beckett could barely digest what it said, because all his eyes wanted to do was bounce around from one soul-healing phrase to another.
You suffered a grave injustice at the hands of someone who should've protected your trust, your honor, and your life . . .
Effective immediately, and backdated to your original separation from the military, your discharge type is now Honorable, qualifying you for all the rights, benefits, and privileges of veterans of the United States military . . .
Your service record and performance evaluations have been restored . . .
Heart racing in his chest, Beckett passed the letter to Kat and found his original, pre-ambush career records. The ones that hadn't been doctored to support a characterization of dereliction of duty, a history of supposed grievances against their commander, and other behavioral infractions that helped build a case for discharge from the Army. It was like being given a second chance to live.
Beckett met Nick's gaze, and they gave each other a nod. And Beckett could see in the other man's eyes the same amazement and vindication that he felt, too. Marz, Shane, Easy—they all felt it as well. It was clear in their eyes and on their faces. Hugs and exclamations and a few tears greeted this news.
"Good news, for sure," Nick said, his voice strained.
"I have more," Chen said. "If I could please have Nick, Derek, Beckett, and Charlie step forward and form a line."
Frowning, Beckett traded looks with the other three, but he did as he was asked.
"Me?" Charlie asked. "Really?"
Chen nodded and looked down at some papers in his hands. When the four of them stood before him, he began. "Ladies and gentlemen, I would like to welcome you to a long-overdue Purple Heart ceremony." Gasps from around the room, and Beckett's heart was suddenly a jackhammer in his chest. "The Purple Heart is an American decoration, the oldest military decoration in the world in present use, and the first American award made available to the common soldier. It was initially created as the Badge of Military Merit by one of the world's most famed and best-loved heroes, General George Washington. The actual order included the phrase, 'Let it be known, that he who wears the Military Order of the Purple Heart has given of his blood in defense of his homeland, and shall forever be revered by his fellow countrymen' . . ."
Aw, hell. Beckett knew he wasn't going to get through this without losing it. Christ, he never thought this day would come. Not for any of them. And it didn't even matter that it wasn't taking place in front of the Army brass or at some public gathering. A Purple Heart wasn't a recognition that any soldier ever wanted, but for them, it represented a basic restoration of justice. And that's what made it so damn meaningful.
Chen met each of their eyes as he continued. "The Purple Heart is awarded to members of the Armed Forces of the United States who are wounded by an instrument of war in the hands of the enemy, and posthumously to the next of kin in the name of those who are killed in action or die of wounds received in action. It is specifically a combat decoration."
Beckett and Marz traded glances, and the glassiness in his best friend's eyes was a real knock to Beckett's own control over his emotions.
"It is my honor," Chen said, "to recognize the following veteran service members of the Army Special Forces—Derek DiMarzio, Frank Merritt, posthumously, Beckett Murda, and Nicholas Rixey—for their heroic acts and exemplary service to our nation. Gentlemen, today you are joining an elite list of patriots who, throughout our nation's history, have made incredible personal sacrifices in the name of freedom and democracy. The Purple Heart was originally described as 'available to all, desired by none,' and it speaks to the valor and sacrifice of those who wear this badge of honor. All four recipients are being honored for wounds received in Afghanistan while serving in the Army Special Forces. In addition to today's honors, you should know that six additional members of your A-team will receive posthumous awards. Walker Axton, Carlos Escobal, Jake Harlow, Colin Kemmerer, Marcus Rimes, and Eric Zane. We will never forget their sacrifices, and our thoughts and prayers go out to their loved ones."
At the mention of Marcus's name, Easy pressed his fist over his mouth. Marcus had been Easy's best friend, and his death right in front of Easy's eyes led to the combat-related guilt that still ate at Easy to this very day. Beckett looked to Shane next, and was glad to see he wasn't the only one struggling to hold it together, because as Shane stood at attention, he wasn't even trying to keep the silent tears from slipping from the corners of his eyes. Hell, there might not have been a dry eye in the whole place.
Chen pulled four brass frames and four square jewelry boxes from his briefcase. "Award recipients, the full orders, which I have here for each of you, detail your service, contributions, and sacrifices. By order of the President of the United States of America, the Purple Heart is awarded to Derek DiMarzio, Beckett Murda, and Nicholas Rixey, and to Frank Merritt, posthumously, for wounds received in action."
One by one, Chen pinned the award on their chests, handed them their framed certificates, and shook their hands. Derek received his first, and seeing his best friend recognized for his incredible sacrifice lifted some of the guilt that Beckett had carried for the past year. Beckett received his next, then Nick, and then Charlie, on behalf of his father. Charlie seemed to hold it together pretty well until Chen pinned the heart and ribbon to his chest. Jeremy pulled a sobbing Charlie into his arms as Becca slipped the frame from his hands and rubbed his back.
And then it was over. Done. Cheers and hugs and handshakes all around.
Kat crossed to Beckett through the crowded room as quickly as she could, which wasn't fast only a week after heart surgery. She gently hugged him and laid her head over his heart. "I'm so proud of you, Beckett."
"I'm kinda proud of me, too," he said. And he was, for maybe the first time in his life.
Chen stood just separate from the fray, that small, enigmatic smile he always wore on his face. "Whenever you're ready, there's more."
Nick chuckled. "What else could there possibly be?"
"You might be surprised," Chen said, winking. With Chen, they were pretty much always surprised. "So, given the awkwardness of your situation and the media circus that the revelations about Kaine have caused, the Army would like to offer reparations, but struggled for a way to do so without drawing undue attention. I came up with what I hope is a solution acceptable to each of you." He passed out envelopes to the five team members and Charlie.
Beckett frowned as he peered inside and found a check . . . for one million dollars.
"Merritt had an account with twelve million dollars in it, and there were twelve team members impacted by Kaine's actions. Each member—or his heirs—will receive an equal one-twelfth share," Chen said.
Silence for a long moment. And then all hell broke loose.
"Holy shit," Shane said, gray eyes wide.
Beckett shook his head. He needed no compensation beyond justice, and they were getting that. At long last. "I don't want this."
Shane nodded to Beckett. "I was never in this for the money."
Easy stepped to the desk and laid his envelope on top of Chen's case. "I would like mine to go to Rimes's family."
"Becca," Charlie said. "Half of this is yours, of course, but I'm giving my half to Jeremy. For the building."
"That's right where this is going, too," Nick said, holding up his envelope.
Chen held up his hands. "I'm not taking back any checks. At least not today. Think about it. Let it sink in. I suspect we'll be in touch, because I have one more thing."
"Holy crap," Kat said, sinking into a chair. "I don't think my heart can take any more."
Everyone chuckled, but she was right. Each new revelation was more unbelievable than the last. At least this time, though, the revelations were all in their favor, rather than against. That was a nice fucking change of pace.
Pulling a legal-sized envelope from his case, Chen met the gaze of each of the men from the team. "The five of you did excellent work. Work this country needs done. Work that not many can do. In this envelope I have the details of an offer for the five of you to form a Top Secret task force, working with the full resources and support of the Company, to investigate similar instances of corruption in the U.S. military in combat zones. And other investigations to be mutually agreed upon. You get the idea." He passed the envelope to Nick.
You could've heard a hair fall to the floor. A job offer. He was coming to them with a job offer . . . working for the CIA. Beckett didn't know whether to laugh or cry or pinch himself to see if he was dreaming all this up. Now that he was trying his hand at this dreaming thing, and all . . .
Chen laughed. It was the first time Beckett had ever heard him do so. "I know my job is complete when I've completely killed the conversation in a room."
Nick shook his head. "Sorry, no. It's just that—"
"Don't say anything right now," Chen said. "I won't brook any debate. Read the details. Talk about it between the five of you. Think about it for as long as you like. It's a standing offer." He clicked his case closed and patted his hand against the leather. "Unfortunately, my bag of tricks is now empty."
"I'd say that was a pretty amazing set of tricks just as it was," Becca said. Nods and agreement all around.
Beckett stepped right up to the man and offered his hand. "Thank you. What you did today means a lot. Just want to say that." Chen nodded, and every single person in the room—men and women, alike—came and offered their gratitude.
And then Chen split, as quickly as he'd come.
"Well, that was interesting," Jeremy said, setting off chuckles around the room. As he'd done the past few days, his choice of T-shirt today once again featured a "head" theme. This one had a stick figure of a man with a disproportionately tiny head. The text read, A little head never hurt anyone. He said he was celebrating still having a head, after all. Jeremy was recovering just fine.
"I want to throw another idea on the table," Nick said, "if we're going to seriously consider this." He tapped his fingers against Chen's envelope.
Beckett nodded. "Let's hear it." Because he had one of his own. Chen was right—they had worked well together. And they each brought different areas of expertise to the table. They made a great team, and he wasn't ready to give that up. For a whole lot of reasons.
"Remember that cover story we told Jess a few weeks ago?" Nick asked. "The one where the five of us were working together to start up a new security-consulting business?"
Satisfaction rolled into Beckett's gut. They were on the same wavelength after all.
"I was gonna suggest the same damn thing," Easy said. "Truth is, I don't have anything to go back to Philly for. And I—" He shook his head. "—I'd miss you assholes if we lost touch again."
Beckett appreciated Easy going there. He really did. Because he was pretty damn sure the guy had just given voice to something every damn member of the team was thinking.
"I'm in," Beckett said. "My brain was heading in the same direction."
Marz nodded and pulled Emilie into his arms. "You fuckers are the only family I've ever had. I'm in, too."
Shane nodded. "Hell, yeah. And besides . . ." He clasped hands with Sara. "No doubt Chen's offer is great, but I don't want to travel overseas, gone for months at a time. Not anymore. It's not for me personally, anyway."
"Me either," Easy said. He put his arm around Jenna and pulled her in against his side.
Everyone nodded in agreement.
Becca blew out a long breath. "Oh, thank God," she said, dropping her face into her hands. "I thought I was going to be the one to have heart problems this time." She threw her arms around Nick's neck. "I would've supported you either way, but it would be so hard to see you go."
"Don't worry, Sunshine. I'm not going anywhere," Nick said, returning the embrace.
"If we're serious," Shane said, tossing his check to the desk in front of Nick. "Then I'm donating my money toward start-up costs for the business."
Beckett dropped his check into the pile. "That I can totally agree to."
"Me too," Marz said, adding his envelope. "Because we're going to need lots of toys. I mean, I want a war room. A real, honest-to-God war room. With a chair that has a cushion . . ."
Everyone laughed.
"What? I'm serious," he said.
Grinning, Nick nodded. "Okay, then. Sounds to me like we've got a plan."
"You know," Jeremy said. "We rebuild the other side of the building and finish off all the unfinished spaces throughout . . ." He shrugged, and flicked his tongue against his lip piercing. "Anyone who wanted could live here and you could build your offices on the other side. Depending on how we laid out the new building, this complex could easily host both businesses and anywhere from six to eight loft apartments."
"That's kind of a crazy awesome idea, Jer," Becca said. And, actually, Beckett thought she was totally right. He only had one hold-up with the plan—that Kat would live an hour away in D.C. But lots of people lived in one city and dated a person from the other, right? And now that Cole was gone, she'd be safe again, too. Details were still coming out about him, but what they'd learned for sure was that he'd been surveilling Kat for a while from that hotel across the street. Sick fuck. Beckett was still kicking himself for not realizing who he'd seen that night, but these days he was trying to get better about letting go of the past.
"I'm full of crazy awesome, Becca," he said. "You should know that by now." He winked at her and grinned. "You could even call it Hard Ink Security. Oh! Or, Hard Security, Inc. 'Inc.' with a c. Get it? That's genius!"
A whole lotta crickets.
Chuckling, Charlie elbowed Jeremy. "No one ever recognizes genius during its time. Don't worry about it."
"So," Kat said, pushing herself up slowly from her chair. "Would this security-consulting business potentially need a general counsel?"
Beckett's heart tripped into a sprint. "You have a job. A great job."
"I do, but there's a part of me that doesn't feel right going back to it. Not after . . . everything. And I'm okay with that. Really." She looked to Nick again.
But Beckett wasn't letting this go. He wasn't letting her throw away everything for which she'd worked so hard. Standing in front of her, he grasped her hands in his. "I don't want you to give up your life."
It was one of the prettiest smiles she'd ever given him. "I wouldn't be doing that at all, Beckett. I'd just be starting a new chapter in it. One that included you. And my brothers. And all of you," she said, looking around at the group.
Nick cleared his throat, and Beckett could see he shared some of his own reservations. But Nick nodded and said, "We would definitely be in need of general counsel, and it would have to be someone we knew we could trust and would always have our backs."
"I'd say Katherine Rixey fits that to a tee," Marz said.
"Kat," Beckett said, pulling her into his arms. "Are you sure?"
"Never more sure about anything in my life," she said. And then she pulled back, crooked her finger at him so he'd bend down, and whispered into his ear, "Except for you."
Jesus, he loved this woman. He really freaking did.
"Eileen can be our mascot, too," Becca said, scooping the puppy off the floor. Eileen barked. Clearly she was totally down with that.
Seeing the puppy made Beckett look around for Cy, and he found the old guy still peering over the top of that box. But the minute Beckett made eye contact, Cy crouched down, nearly out of sight. And that was the moment when Beckett decided he wasn't going to give up on the old orange beast, just like Kat hadn't given up on him. Someday, somehow, he was making friends with the cat. Hey, he could dream. Right?
"I want to say something," Nick said. "The last month has included some of the best and worst days of my life. But I have no regrets, and I couldn't be more grateful to everything every one of you in this room did to get us to this day. Wherever our paths lead, you are my family. No matter what."
Beckett nodded, feeling those sentiments down into his very soul. It was clear everyone there did.
Nick made eye contact with every person, then nodded. "Then I guess the only thing left to ask is, who's ready for our next mission?"
Acknowledgments
I cannot begin to describe how hard it's going to be to let the Hard Ink team go, because I have adored writing them, spending time with them, and hearing all your wonderful reactions to them. And I have so many people to thank for making it all possible.
Thank you to my editor of awesome, Amanda Bergeron, for believing in this series and giving me a chance to share it with the world. She's been wonderful to work with from start to finish, and I'm glad we get to continue together with the Raven Riders! I also want to thank Avon's art department for the absolutely amazing covers they've given this series. I often joke that the cover gods have smiled on me throughout my career, and it's never been more true than for the Hard Ink books! My agent Kevan Lyon deserves another huge thanks for all the work she's done for me on this series—above and beyond! And my publicist KP Simmon has been a constant cheerleader and source of support. I am lucky to have such an amazing team behind me, and I couldn't do any of this without them.
I also need to thank you critique partner, Christi Barth, an awesome author in her own right, who has read and commented on every word of this series and made it so much better than it would've been! Thanks for always having my back and cheering me on, Christi! It has meant the world. Thanks also to Lea Nolan, Stephanie Dray, and Jennifer L. Armentrout, fellow authors and amazing friends each, who are always there for me with an emergency plotting session, encouraging word, or an ear or shoulder to lend!
Thank you to all the awesome bloggers who have posted about this series, participated in my launch events and blog tours, and generally helped spread the word—no author could do this without your support, and I appreciate you all so much for everything you do! Thanks also to my street team, the Heroes, for always being there for me and sharing your love for my books. It's an incredible experience to have people dedicate themselves to helping you do what you love. They truly are my Heroes.
I must also offer a huge and heartfelt word of thanks to my family. Their love, support, and patience has made every word of the Hard Ink series possible. They've believed that I could do this and made it possible in so many ways for me to pursue this crazy, incredible, and sometimes all-consuming dream. I love you guys so much.
Finally, my biggest thanks goes to my readers, who have embraced the Hard Ink team and given me the most amazing support an author could ever ask for. Thank you for allowing my characters into your hearts so they can tell their stories over and over again.
~LK
The Hard Ink series is over, but don't worry! Laura Kaye is introducing a brand new series, all about the sexy men who have joined together to form
THE RAVEN RIDERS
The Raven Riders are an outlaw motorcycle club who fight and die for each other, their way of life, and those too weak to fight for themselves.
Raven Riders MC President Dare Kenyon rides hard, fights for the brotherhood he leads, and protects those who can't defend themselves. When a recent kidnapping victim with too many secrets in her eyes lands on his doorstep, he takes her in and pushes to learn what she's hiding before more trouble comes his way—and before he's tempted to break down all the shy beauty's walls.
Haven Randall fled from years of violence at the hands of a man who should've protected her, only to find herself imprisoned by a gang. Recently rescued, she's hiding out from both in a compound of bikers, sure her ex is hunting her. She's distrustful and suspicious—even of the sexy and overwhelmingly intense Ravens' leader who promises to keep her safe and makes her want things she never thought she would.
The past never dies without a fight, but Dare Kenyon's never backed down before . . .
Available Winter 2016
Don't miss any of the action!
Be sure to check out the rest of Laura Kaye's sexy, heart-pounding
Hard Ink Series
HARD AS IT GETS
A Hard Ink Novel 1
Tall, dark, and lethal . . .
Trouble just walked into Nicholas Rixey's tattoo parlor. Becca Merritt is warm, sexy, wholesome—pure temptation to a very jaded Nick. He's left his military life behind to become co-owner of Hard Ink Tattoo, but Becca is his ex-commander's daughter. Loyalty won't let him turn her away. Lust has plenty to do with it, too.
With her brother presumed kidnapped, Becca needs Nick. She just wasn't expecting to want him so much. As their investigation turns into all-out war with an organized crime ring, only Nick can protect her. And only Becca can heal the scars no one else sees.
Desire is the easy part. Love is as hard as it gets. Good thing Nick is always up for a challenge.
HARD AS YOU CAN
A Hard Ink Novel 2
Ever since hard-bodied, drop-dead-charming Shane McCallan strolled into the dance club where Crystal Dean works, he's shown a knack for getting beneath her defenses. For her little sister's sake, Crystal can't get too close. Until her job and Shane's mission intersect, and he reveals talents that go deeper than she could have guessed.
Shane would never turn his back on a friend in need, especially a former Special Forces teammate running a dangerous, off-the-books operation. Nor can he walk away from Crystal. The gorgeous waitress is hiding secrets she doesn't want him to uncover. Too bad. He's exactly the man she needs to protect her sister, her life, and her heart. All he has to do is convince her that when something feels this good, you hold on as hard as you can—and never let go.
HARD TO HOLD ON TO
A Hard Ink Novella
Edward "Easy" Cantrell knows better than most the pain of not being able to save those he loves—which is why he is not going to let Jenna Dean out of his sight. He may have just met her, but Jenna's the first person to make him feel alive since that devastating day in the desert more than a year ago.
Jenna has never met anyone like Easy. She can't describe how he makes her feel—and not just because he saved her life. No, the stirrings inside her stretch far beyond gratitude.
As the pair is thrust together and chaos reigns around them, they both know one thing: the things in life most worth having are the hardest to hold onto.
HARD TO COME BY
A Hard Ink Novel 3
Caught between desire and loyalty . . .
Derek DiMarzio would do anything for the members of his disgraced Special Forces team—sacrifice his body, help a former teammate with a covert operation to restore their honor, and even go behind enemy lines. He just never expected to want the beautiful woman he found there.
When a sexy stranger asks questions about her brother, Emilie Garza is torn between loyalty to the brother she once idolized and fear of the war-changed man he's become. Derek's easy smile and quiet strength tempt Emilie to open up, igniting the desire between them and leading Derek to crave a woman he shouldn't trust.
As the team's investigation reveals how powerful their enemies are, Derek and Emilie must prove where their loyalties lie before hearts are broken and lives are lost. Because love is too hard to come by to let slip away.
HARD TO BE GOOD
A Hard Ink Novella
Hard Ink Tattoo owner Jeremy Rixey has taken on his brother's stateside fight against the enemies that nearly killed Nick and his Special Forces team a year before. Now, Jeremy's whole world has been turned upside down by the chaos—and by a brilliant, quiet blond man who tempts Jeremy to settle down for the first time ever.
Recent kidnapping victim Charlie Merritt has always been better with computers than people, so when he's drawn into the SF team's investigation of his Army colonel father's corruption, he's surprised to find acceptance and friendship—especially since his father never accepted who he was. Even more surprising is the heated tension he feels with sexy, tattooed Jeremy, Charlie's opposite in almost every way.
With tragedy and chaos all around them, temptation flashes hot, and Jeremy and Charlie can't help but wonder why they're trying so hard to be good.
HARD AS STEEL
A Hard Ink / Raven Riders Crossover Novella
After identifying her employer's dangerous enemies, Jessica Jakes takes refuge at the compound of the Raven Riders Motorcycle Club. Fellow Hard Ink tattooist and Raven leader Ike Young promises to keep Jess safe for as long as it takes, which would be perfect if his close, personal, round-the-clock protection didn't make it so hard to hide just how much she wants him—and always has.
Ike Young loved and lost a woman in trouble once before. The last thing he needs is alone time with the sexiest and feistiest woman he's ever known, one he's purposely kept at a distance for years. Now, Ike's not sure he can keep his hands or his heart to himself—or that he even wants to anymore. And that means he has to do whatever it takes to hold on to Jess forever.
About the Author
LAURA KAYE is the New York Times and USA Today bestselling author of more than a dozen books in contemporary and paranormal romance and romantic suspense. Laura grew up amidst family lore involving angels, ghosts, and evil-eye curses, cementing her life-long fascination with storytelling and the supernatural. Laura lives in Maryland with her husband, two daughters, and cute-but-bad dog, and appreciates her view of the Chesapeake Bay every day.
www.laurakayeauthor.com
Discover great authors, exclusive offers, and more at hc.com.
By Laura Kaye
HARD TO LET GO
HARD TO BE GOOD (Novella)
HARD TO COME BY
HARD TO HOLD ON TO (Novella)
HARD AS YOU CAN
HARD AS IT GETS
Copyright
This is a work of fiction. Names, characters, places, and incidents are products of the author's imagination or are used fictitiously and are not to be construed as real. Any resemblance to actual events, locales, organizations, or persons, living or dead, is entirely coincidental.
HARD TO LET GO. Copyright © 2015 by Laura Kaye. All rights reserved under International and Pan-American Copyright Conventions. By payment of the required fees, you have been granted the nonexclusive, nontransferable right to access and read the text of this e-book on screen. No part of this text may be reproduced, transmitted, downloaded, decompiled, reverse-engineered, or stored in or introduced into any information storage and retrieval system, in any form or by any means, whether electronic or mechanical, now known or hereafter invented, without the express written permission of HarperCollins e-books.
EPub Edition JULY 2015 ISBN: 9780062267955
Print Edition ISBN: 9780062267948
FIRST EDITION
10 9 8 7 6 5 4 3 2 1
About the Publisher
Australia
HarperCollins Publishers Australia Pty. Ltd.
Level 13, 201 Elizabeth Street
Sydney, NSW 2000, Australia
www.harpercollins.com.au
Canada
HarperCollins Canada
2 Bloor Street East - 20th Floor
Toronto, ON, M4W, 1A8, Canada
www.harpercollins.ca
New Zealand
HarperCollins Publishers New Zealand
Unit D1, 63 Apollo Drive
Rosedale 0632
Auckland, New Zealand
www.harpercollins.co.nz
United Kingdom
HarperCollins Publishers Ltd.
1 London Bridge Street
London SE1 9GF, UK
www.harpercollins.co.uk
United States
HarperCollins Publishers Inc.
195 Broadway
New York, NY 10007
www.harpercollins.com
1. Dedication
2. Contents
3. Chapter 1
4. Chapter 2
5. Chapter 3
6. Chapter 4
7. Chapter 5
8. Chapter 6
9. Chapter 7
10. Chapter 8
11. Chapter 9
12. Chapter 10
13. Chapter 11
14. Chapter 12
15. Chapter 13
16. Chapter 14
17. Chapter 15
18. Chapter 16
19. Chapter 17
20. Chapter 18
21. Chapter 19
22. Chapter 20
23. Chapter 21
24. Chapter 22
25. Chapter 23
26. Chapter 24
27. Chapter 25
28. Chapter 26
29. Epilogue
30. Acknowledgments
31. An Announcement Page to The Raven Riders
32. An Announcement Page to the Hard Ink Series
33. Hard As It Gets
34. Hard As You Can
35. Hard To Hold On To
36. Hard To Come By
37. Hard To Be Good
38. Hard As Steel
39. About the Author
40. By Laura Kaye
41. Copyright
42. About the Publisher
1. Cover
2. Contents
3. Startup Page
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Q: How to translate angular variables, without modify DOM? I have a big angular app.
How to translate variables, without change DOM elmenets?
Eg:
<div data-ng-bind="{{ hello }}"></div>
I dont want to edit this HTML, but I would like to hello variable.
No filter, no any change.
Changes made by only js side.
A: If you mean adding localization and multi-language support, there are plenty of libraries to help you out. Take a look at ng-i18next.
A: You need to create a decorator to compile state.
Use the $provide.decorator function to change normal interpolation method.
$provide.decorator( "$interpolate", ... );
More information:
https://docs.angularjs.org/api/auto/service/$provide#decorator
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Меланж (от — смесь) — кофейный напиток австрийской кухни на основе эспрессо с добавлением подогретого и вспененного молока и взбитых сливок. Считается, что впервые такой напиток был приготовлен в Вене в 1890 году. По другой версии, он появился в столице Австрии в XVII веке, а его автором называют Юрия Кульчицкого.
Встречается несколько основных вариантов этого кофейного напитка. Так, меланж готовят из эспрессо с добавлением молока и вспененного молока. Другой рецепт, кофе по-венски, подразумевает сваренный кофе, украшенный шапкой из взбитых сладких сливок с ароматом ванили, присыпанный сверху тёртым шоколадом.
Некоторые рецепты предлагают слегка разбавить эспрессо горячей водой или использовать двойной эспрессо. В данном случае соотношение кофе и молока в венском меланже всегда сохраняется 1:1. Меланж подается со стаканом воды и иногда украшенный взбитыми сливками.
Отдельным вариантом меланжа является императорский меланж (Kaisermelange), в который кроме молока и молочной пены добавляют яичный желток, сахар и коньяк.
См. также
Капучино
Латте макиато
Кофе с молоком
Примечания
Литература
Австрийские напитки
Кофейные напитки | {
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Q: OWIN self static file server multiple routes I have configured in my Owin a webApi and an static file server to fetch some files that we will need in my app.
public void Configuration(IAppBuilder application)
{
//Other middlewares and configurations
....
application.UseFileServer(new FileServerOptions()
{
RequestPath = new PathString("/myPath1/public"),
FileSystem = new PhysicalFileSystem(m_FolderPathProvider.myPath1FolderPublic)
});
// Attribute routing.
.....
}
This is working like a charm. What a I need it's to declare another FileServer for another path and another different physical folder. What I'm frightening is if I do it in the same way I will override this one and I will have only one. So how can I declare a second fileserver?
Thank you.
A: AFAICT, you can "mount" different FileSystem paths onto different routes, using the same overload you are already using.
public void Configuration(IAppBuilder application)
{
//Other middlewares and configurations
....
application.UseFileServer(new FileServerOptions()
{
RequestPath = new PathString("/myPath1/public"),
FileSystem = new PhysicalFileSystem(m_FolderPathProvider.myPath1FolderPublic)
});
application.UseFileServer(new FileServerOptions()
{
RequestPath = new PathString("/myPath2/public"),
FileSystem = new PhysicalFileSystem(m_FolderPathProvider.myPath2FolderPublic)
});
// Attribute routing.
.....
}
If you want to have them merged, I don't think it's possible with UseFileServer.
Am I missing something?
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{"url":"https:\/\/brilliant.org\/problems\/how-do-i-find-this-tangent-line\/","text":"# How do I find this tangent line?\n\nCalculus Level 4\n\nLet $$y = y(x)$$ be the function implicitly determined by the equation $x^2 +2xy + 3 \\ln{y} = e^y .$\n\nFind the slope (angular coefficient) of the line tangent to $$y(x)$$ on $$y = 1$$, located at the first quadrant of the Cartesian plane.\n\nGive your answer to 1 decimal place.\n\n\u00d7\n\nProblem Loading...\n\nNote Loading...\n\nSet Loading...","date":"2018-07-16 06:59:28","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6689256429672241, \"perplexity\": 768.7209866537679}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-30\/segments\/1531676589222.18\/warc\/CC-MAIN-20180716060836-20180716080836-00088.warc.gz\"}"} | null | null |
\section{Sample calculations}
\label{app:sample_calc}
In this section, we will calculate two simple quantities. This mostly serves as a sanity check and to illustrate calculation techniques. Throughout the paper we have chosen normalizations such that a canonically normalized CFT scalar two-point function gives a correctly normalized S-matrix element. Hence, there is no need to fix the normalization.
\subsection{One-to-one scattering}
As an example, we will compute the one-to-one scattering amplitude of a massless scalar field.
\begin{align}
\begin{split}
\label{eq:two_pt}
& \qquad \langle{a_{\text{out},q}a^\dagger_{\text{in},p}}\rangle = \\
&(-1)\sqrt{\frac{24^2}{L^8 \omega_p^5\omega_q^5}}\iint_{-\frac \pi 2 L}^{\frac\pi 2 L} dt dt' e^{i \omega_q L t - i \omega_p L t'} \langle \mathcal O^-\left( \frac{t'} L + \frac \pi 2, \hat q\right) \mathcal O^+\left(\frac{t} L - \frac \pi 2, - \hat p\right) \rangle.
\end{split}
\end{align}
It is now helpful to use the spectral decomposition of the operators. For a free theory we have that
\begin{align}
\mathcal O^\pm(\tau, \Omega) = \sum_{\kappa,l,m} \sqrt{\frac{4\pi}{3} \frac{\Gamma(\kappa + l + 3)\Gamma(\kappa + \frac 5 2)}{\Gamma(\kappa+1) \Gamma(\kappa + l + \frac 3 2)}} e^{\pm i \omega_{\kappa,l} \tau} Y^m_l(\Omega) a^{\pm}_{\kappa,l,m}.
\end{align}
The creation/annihilation operators $a^{\pm}_{\kappa,l,m}$ act on the CFT vacuum and result in Kronecker-deltas which set the $\kappa, l, m$ modes to be the same for both operators.
Plugging this into equation \eqref{eq:two_pt} we obtain after performing the $t$ integrals,
\begin{align}
\begin{split}
\langle{a_{\text{out},q}a^\dagger_{\text{in},p}}\rangle ={}& \frac{-16}{L^4 \omega_p^3} \frac{(2 \pi)^3}{\omega_p^2} \delta(\omega_p - \omega_q) \\
& \cdot \sum_{\kappa,l,m} Y^m_l(\hat q) Y^{*m}_l(- \hat p) e^{i \omega_p \pi L} \delta\left(\omega_p - \left(\frac{2 \kappa + l + 3} L\right)\right) \frac{\Gamma(\kappa+l+3)\Gamma(\kappa+\frac 5 2)}{\Gamma(\kappa+1)\Gamma(\kappa+l+\frac 3 2)}.
\end{split}
\end{align}
The phase factor becomes $e^{i \omega_p \pi L} = (-1)^{l + 3}$. We can turn the sum over $\kappa$ into an integral over $x = \frac{2 \kappa} L$. The integration can then be performed using the delta function. This results in
\begin{align}
\langle{a_{\text{out},q}a^\dagger_{\text{in},p}}\rangle = \left(\frac{2}{L \omega_p}\right)^3 \frac{(2 \pi)^3}{\omega_p^2} \delta(\omega_p - \omega_q) \sum_{l,m} Y^m_l(\hat q) Y^{*m}_l(\hat p) \frac{\Gamma(\frac{\omega L + l + 3}{2})\Gamma(\frac{\omega L - l + 2}{2})}{\Gamma(\frac{\omega L - l -1}{2})\Gamma(\frac{\omega L + l}{2})}.
\end{align}
In the large $L$ limit, the gamma functions become independent of the angular momentum quantum number $l$. The sum over spherical harmonics can be performed and the Gamma functions exactly cancel the first factor. Finally, we obtain
\begin{align}
\langle{a_{\text{out},q}a^\dagger_{\text{in},p}}\rangle = \frac{(2 \pi)^3}{\omega_p^2} \delta(\omega_p - \omega_q) \delta^{(2)}(\hat p - \hat q) = (2 \pi)^3 \delta^{(3)}(\vec p - \vec q).
\end{align}
\subsection{One particle-state normalization}
Another aspect of our proposal is that the in- and out- Fock space is constructed using positive and negative frequencies of operators located at $\mp \frac \pi 2$. We should thus be able to calculate inner products between states in any of the asymptotic Fock spaces. For definiteness, we take the out-Fock space, which is obtained by acting with operators $a^\dagger_{\text{out},p}$ on the vacuum. The vacuum is defined by the condition that it is annihilated by all $a_{\text{out},p}$. Here, we will show that the one-particle state is canonically normalized, that is, we want to calculate
\begin{align}
\begin{split}
& \langle{a_{\text{out},q}a^\dagger_{\text{out},p}}\rangle = \\
&\qquad \sqrt{\frac{24^2}{L^8 \omega_p^5\omega_q^5}}\iint_{-\frac \pi 2 L}^{\frac\pi 2 L} dt dt' e^{i \omega_q L t - i \omega_p L t'} \langle \mathcal O^-\left( \frac{t'} L + \frac \pi 2, \hat q\right) \mathcal O^+\left(\frac{t} L + \frac \pi 2, - \hat p\right) \rangle.
\end{split}
\end{align}
This time, after performing the time-integrals, we obtain
\begin{align}
\begin{split}
\langle{a_{\text{out},q}a^\dagger_{\text{out},p}}\rangle ={}& \frac{16}{L^4 \omega_p^3} \frac{(2 \pi)^3}{\omega_p^2} \delta(\omega_p - \omega_q) \\
& \sum_{\kappa,l,m} Y^m_l(\hat q) Y^{*m}_l(\hat p) e^{i \omega_p \pi L} \delta\left(\omega_p - \left(\frac{2 \kappa + l + 3} L\right)\right) \frac{\Gamma(\kappa+l+3)\Gamma(\kappa+\frac 5 2)}{\Gamma(\kappa+1)\Gamma(\kappa+l+\frac 3 2)}.
\end{split}
\end{align}
Note that in comparison with the S-matrix element calculated above, the overall sign and the argument of the spherical harmonics differ. In addition, the phase factor evaluates to $e^{i \omega_p \pi L} = 1$. The rest of the calculation is exactly the same as above and the result reads
\begin{align}
\langle{a_{\text{out},q}a^\dagger_{\text{out},p}}\rangle = \frac{(2 \pi)^3}{\omega_p^2} \delta(\omega_p - \omega_q) \delta^{(2)}(\hat p - \hat q) = (2 \pi)^3 \delta^{(3)}(\vec p - \vec q).
\end{align}
\section{Conclusions}\label{sec:conc}
In this paper we have developed a method to construct CFT operators whose correlation functions yield flat space S-matrix elements. This method was then applied to obtain CFT operators dual to photon creation and annihilation operators. We could then reproduce well-known features of the S-matrix -- Weinberg soft theorems -- within the framework of flat-space holography. It has been argued that Weinberg soft theorems arise as the Ward identities of asymptotic gauge transformations, also called large gauge transformations. Here, we have demonstrated that they can also be thought of arising from CFT Ward identities in the flat-space limit.
Given the methods presented in this paper, a natural next step would be to derive an expression for fermion, non-abelian gauge field and graviton creation/annihilation operators. The latter will take the form of a smeared energy-momentum tensor, again localized at $\tau = \pm \frac \pi 2$ for outgoing and incoming states, respectively. Obtaining results for fermion, as well as non-Abelian gauge field scattering states would be a crucial step in reproducing phenomenologically interesting scattering cross-sections from flat-space holography. Although it is not obvious that calculating scattering amplitudes as CFT correlators would be any easier that modern perturbative methods, it allows to apply new techniques, like those of the bootstrap program \cite{Paulos:2016fap}, to scattering cross-sections.
The flat limit of AdS/CFT might reveal more details about the infrared structure of flat space-time. As an example, note that a crucial step in our analysis was to assume that the energy as measured by a CFT observer is of order $\omega_\text{flat} L$ and thus diverges in the flat space limit. To obtain the soft theorems the energy is taken to vanish after the large $L$ limit is taken. This raises a question about the low-energy modes with CFT energies of order $\mathcal O(1)$. For those modes the sum over quantum numbers $\kappa$ does not turn into a continuum and they should appear as IR modes in the bulk theory. It would be interesting to understand if those modes are frozen or if there is dynamics associated with them. If the latter case is true, those modes might be a mechanism for how information can be stored in the infrared in flat space. We hope to be able to report results of this analysis in the near future.
Concerning the IR structure of flat space physics, it is well known that infrared divergences trivialize the Fock basis S-matrix. The existence of these problematic divergences can be tracked back to the assumption of asymptotic decoupling. The introduction of Faddeev-Kulish states \cite{Kulish:1970ut} solves this issue by dressing the scattering states with soft photons, in such a way that the IR divergences of Weinberg soft theorems cancel the divergences of the S-matrix, see also \cite{Carney:2018ygh,Neuenfeld:2018fdw}. While AdS/CFT acts as a natural IR regulator, IR divergences will show up once the flat limit is taken. AdS/CFT suggests a solution related to that of Faddeev-Kulish, by relaxing the condition of asymptotic decoupling. This could be done by making use of the HKLL technology to reconstruct bulk operators that interact with the long wavelength modes of the electro-magnetic field.
From the point of view of the CFT, flat space scattering states are a special class of states which behave as if they were free for most of the CFT time-evolution. Equivalently, we have found a set of non-local CFT operators which obey a simple creation/annihilation operator algebra. From the point of view of AdS/CFT and in particular HKLL bulk reconstruction, the existence of such operators is not very surprising and can be explained in terms of error-correcting codes \cite{Almheiri:2014lwa}. For a different construction of CFT operators obeying canonical commutation relations, see \cite{Papadodimas:2012aq,Magan:2020iac}. Light ray operators \cite{Kravchuk:2018htv} are another class of non-local operators which again obey a simple but non-trivial algebra and it was shown in \cite{Cordova:2018ygx} that in a Lorentz invariant CFT, certain operators reproduce the BMS algebra. It would be interesting to explore whether these operators are related to the ones discussed throughout this work.
There have been several intriguing observations in the literature which hint at the possibility at a possibly exotic CFT might underlie flat space scattering, in the sense that the calculation of scattering amplitudes can be understood in terms of correlation functions of this CFT. We believe that this paper gives further insights into the structure of a dual theory of flat space scattering. However, taking into account the absence of an explicit construction of flat holography (which is available in AdS/CFT), one should perhaps understand the above results as a toolbox to explore the qualitative features a putative theory dual to flat space must have, rather than a bona-fide derivation of flat-space holography from AdS/CFT.
\section{Scalar HKLL operators}\label{app:ScalarHKLL}
This section reviews operator reconstruction according to HKLL and derives the formulas used in the main text. Long before and after the scattering, scalar fields obey the Klein-Gordon equation. Normalizable solutions in AdS$_4$, which are the ones we are interested in, are linear combinations of modes of the form
\begin{equation}
\phi^{\pm}_{\omega,l,m}=\frac{1}{\mathcal N_{\Delta, \omega, l}}e^{\pm i \omega \tau} Y_l^{m}(\Omega)^{(*)} \sin^l \rho \cos^{\Delta}\rho \, {}_2F_1\left( {\Delta+l-\omega\over 2},{\Delta+l+\omega\over 2};\Delta-{1\over 2} \Big| \cos^2\rho \right)\, ,
\end{equation}
where the superscript ${}^{(*)}$ of the spherical harmonic $Y_l^m(\Omega)$ indicates that we should take the complex conjugate for $\phi^-_{\omega, l, m}$.
Those modes are only regular in the center of AdS ($\rho = 0$), if
\begin{equation}
\omega_{\kappa, l}= \Delta+l+2\kappa\, , \quad \text{with}\quad \kappa\in \mathbb{Z}^+\, .
\end{equation}
Normalization with respect the the Klein-Gordon inner product require that the normalization constant is
\begin{equation}
\mathcal N_{\Delta, \kappa, l} = L \sqrt{\frac{\kappa! \Gamma(\Delta - \frac 1 2)^2 \Gamma(\kappa + l+ \frac 3 2)}{\Gamma(\Delta+\kappa+l)\Gamma(\Delta + \kappa - \frac 1 2) }}\, .
\end{equation}
We will denote regular, normalized modes with $\phi^{\pm}_{\kappa,l,m}$. General solutions of the free equations of motion can thus be written as
\begin{equation}\label{eq:PhiFreeSol}
\phi(\rho,x)=\sum_{\kappa\in\mathbb{Z}^+}\sum_{l,m} \left( a^{+}_{\kappa,l,m}\phi^{+}_{\kappa,l,m} + a^{-}_{\kappa,l,m}\phi^{-}_{\kappa,l,m}\right) \, .
\end{equation}
The modes $\phi^{\pm}_{\kappa,l,m}$ satisfy orthonormality relations, such that formula \eqref{eq:PhiFreeSol} can be promoted to an operator statement through canonical quantization,
\begin{equation}\label{eq:PhiFreeOp}
\hat{\phi}(\rho,x)=\sum_{\kappa\in\mathbb{Z}^+}\sum_{l,m} \left( \hat{a}^{+}_{\kappa,l,m}\phi^{+}_{\kappa,l,m} + \hat{a}^{-}_{\kappa,l,m}\phi^{-}_{\kappa,l,m} \right)\, .
\end{equation}
The dual operator at the boundary is then
\begin{equation}
\begin{split}
{\cal O}(x)& =\sqrt{\frac{4 \pi \Gamma(\frac 3 2) \Gamma(\Delta - \frac 1 2)}{\Gamma(\Delta)}} \lim_{\rho\rightarrow{\pi \over 2}} \cos^{-\Delta} \rho \, \hat{\phi}(\rho,x) \\
& = \frac 1 {\mathcal N_{\mathcal O, \kappa,l}} \sum_{\kappa\in\mathbb{Z}^+}\sum_{l,m} \left( \hat{a}^{+}_{\kappa,l,m}e^{i \omega_{\kappa,l}\tau} Y_l^m(\Omega)
+ \hat{a}^{-}_{\kappa,l,m}e^{-i \omega_{\kappa,l}\tau} Y_l^{m}(\Omega)^{(*)} \right) \, ,
\end{split}
\end{equation}
with
\begin{equation}
\mathcal N_{\mathcal O, \kappa,l} = \sqrt{\frac{\Gamma(\kappa + 1)\Gamma(\Delta)\Gamma(\Delta - \frac 1 2) \Gamma(\kappa+l + \frac 3 2)}{4 \pi \Gamma(\frac 3 2) \Gamma(\Delta + \kappa + l) \Gamma(\Delta + \kappa - \frac 1 2)}}.
\end{equation}
The normalizing factor $N_{\mathcal O, \kappa,l}$ ensures that the CFT two-point function is canonically normalized.
The objective now is to write $\hat{a}^{\pm}_{\kappa,l,m}$ in terms of the CFT operator ${\cal O}$ and plug the solution back into \eqref{eq:PhiFreeOp} to obtain a representation of the bulk operator at the boundary. For this we define positive and negative frequency parts of the boundary operator
\begin{equation}
{\cal O}^{\pm}(x)\equiv \frac 1 {\mathcal N_{\mathcal O, \kappa,l}} \sum_{\kappa\in \mathbb{Z}^+}\sum_{l,m} \hat{a}^{\pm}_{\kappa,l,m}e^{\pm i \omega_{\kappa,l}\tau} Y_l^{m}(\Omega)^{(*)}\, .
\end{equation}
We can invert this formula to obtain
\begin{equation}
\begin{split}
\hat{a}^{\pm}_{\text{in},\kappa,l,m}&={\mathcal N_{\mathcal O, \kappa,l} \over \pi}\int_{- \pi}^{0}d\tau'\, \int d^2\Omega'\, {\cal O}^{\pm}(x') \, e^{\mp i \omega_{\kappa,l}\tau'}Y_l^m(\Omega')^{(\overline{*})}\, ,\\
\hat{a}^{\pm}_{\text{out},\kappa,l,m}&={\mathcal N_{\mathcal O, \kappa,l} \over \pi}\int_{0}^{\pi}d\tau'\, \int d^2\Omega'\, {\cal O}^{\pm}(x') \, e^{\mp i \omega_{\kappa,l}\tau'}Y_l^m(\Omega')^{(\overline{*})}\, .
\end{split}
\end{equation}
The superscript ${}^{(\overline{*})}$ instructs to use complex conjugation if we choose $\hat a^+_{\kappa, l, m}$.
Plugging these expressions back into \eqref{eq:PhiFreeOp} naively yields,
\begin{equation}
\begin{split}
\hat{\phi}(\rho,x)={ \mathcal N_{\mathcal O, \kappa,l} \over \pi}\int_{\mathcal T}d\tau'\, \int d^2\Omega'\,
& \left( {\cal O}^{+}(x') \sum_{\kappa\in\mathbb{Z}^+}\sum_{l,m} \, e^{- i \omega_{\kappa,l}\tau'}Y_l^m(\Omega') \phi^{+}_{\kappa,l,m} \right. \\
& + \left. {\cal O}^{-}(x') \sum_{\kappa\in\mathbb{Z}^+}\sum_{l,m} \, e^{ i \omega_{\kappa,l}\tau'}Y_l^m(\Omega')^* \phi^{-}_{\kappa,l,m} \right) \, .
\end{split}
\end{equation}
The integration region $\mathcal T$ for the time integral is between $-\pi$ and $0$ for in-fields and between $0$ and $\pi$ for out-fields. When the holographic coordinate is taken to the boundary, the kernels do not result in delta functions so that one would recover the boundary operator. This can be resolved easily by extending the sum over $\kappa$ to all integers $\mathbb{Z}$. This can be done without adding extra terms, by noting that the added modes integrate to zero against ${\cal O}^+$ or ${\cal O}^-$. We conclude that
\begin{equation}
\hat{\phi}(\rho,x)= \int_{\mathcal T} d\tau'\, \int d^2 \Omega\, \left[ K_{+}(\rho,x;x') {\cal O}^+(x')+ K_{-}(\rho,x;x') {\cal O}^-(x') \right]\, ,
\end{equation}
where the kernel $K_{+} = K_{-}^*$ reads
\begin{equation}
\begin{split}
K_{+}(\rho,x;x')=& \frac 1 {\mathcal N_{K,\Delta}} \sum_{\kappa\in \mathbb{Z}}\sum_{l=0}^{\infty}\sum_{m=-l}^{m=l} e^{i \omega_{\kappa,l}(\tau-\tau')} Y_l^m(\Omega)Y_l^m(\Omega')^* \sin^l\rho \cos^{\Delta}\rho \,\\
&\times {}_2F_1\left( -\kappa,l+\Delta+\kappa;\Delta-{1\over 2} \Big \vert \cos^2\rho \right)\, .
\end{split}
\end{equation}
The normalization ensures that a canonically normalized operator gives canonically normalized creation/annihilation operators in the flat space limit,
\begin{equation}\label{eq:ScalarNorma}
\frac 1 {\mathcal N_{K,\Delta}} = \frac{1}{L}\sqrt{\frac{\Gamma(\Delta)}{4 \pi^3 \Gamma(\Delta - \frac 1 2)\Gamma(\frac 3 2)}}.
\end{equation}
\section{Reconstruction of $U(1)$ gauge fields in global AdS$_4$}\label{app:MaxfieldReconstruction}
\subsection{Reconstruction with ``magnetic'' boundary conditions}\label{MaxfieldReconstructionMagnetic}
In this appendix we reconstruct local Maxwell fields in global AdS in terms of their boundary values at the boundary of Anti-de Sitter space. We will first consider AdS/CFT with magnetic boundary conditions, such that the dynamical part of the gauge field asymptotes to a CFT current as
\begin{equation}\label{eq:MagneticBC}
{\cal A}_{\mu}(\rho,x)\xrightarrow[\rho \to \frac \pi 2]{} \cos\rho \, j_{\mu}(x)\, .
\end{equation}
This defines the normalization for the current. In order to have a current two-point function with a particular normalization, an appropriate constant normalization factor has to be added to \eqref{eq:MagneticBC}.
In this appendix we will consider a bulk gauge field in Lorenz gauge. Gauge fixed fields can be considered physical observables, but they will have commutators that are local only to the extent allowed by the gauge constraint (see \cite{Heemskerk:2012np} for a nice discussion). Solutions to the Maxwell field equations with the boundary conditions \eqref{eq:MagneticBC} in Lorenz gauge can be constructed from a set of three different modes
\begin{equation}\label{eq:Asols}
\begin{split}
{\cal A}^{V,l,m}_{\mu} =& L^{-2}\epsilon_{\tau\rho\mu}^{\quad\;\;\nu} e^{\pm i\omega \tau} \cos^{2}\rho \,R_V(\rho) \, \nabla_{\nu} Y_l^m(\hat{\Omega}) \, , \\
{\cal A}^{S,l,m}_{\mu} =& L^{-2}(l(l+1)-\omega^2)\epsilon_{\theta\phi\mu}^{\quad\;\;\nu} \sin^{-1}\theta \,\cot^2\rho\, Y_l^m(\hat{\Omega})\, \nabla_{\nu}(e^{\pm i\omega \tau} R_S(\rho)) \, ,\\
{\cal A}^{G,l,m}_{\mu} =& L^{-2}\nabla_{\mu}\left[ e^{\pm i\omega \tau}Q(\rho) Y_l^m(\hat{\Omega}) \right] \, .
\end{split}
\end{equation}
Here, $\epsilon$ is the Levi-Civita tensor with respect to the full metric. The indices $V$, $S$, and $G$ stand for ``vector" and ``scalar"-type degrees of freedom, as well as residual ``gauge". The naming convention is motivated by the transformation behavior under $SO(2)$ \cite{Ishibashi:2004wx}. We will completely ignore the pure gauge modes in what follows. The scalar-type solution given here is not in Lorenz gauge, but can be brought to the required form by a gauge transformation,
\begin{equation}
\label{eq:gaugeToLorenz}
{\cal A}^{S,l,m}_{\mu} \to {\cal A}^{S,l,m}_{\mu} + \nabla_\mu \chi \quad \text{with} \quad \chi = e^{\pm i\omega \tau} \partial_\rho(R_S(\rho)) Y_l^m(\hat{\Omega})\,.
\end{equation}
The radial functions appearing in the above ansatz obey a hypergeometric differential equation
\begin{equation}
R_{V/S}'' + \left( \omega^2-{{l(l+1)}\over{\sin^2\rho}} \right) R_{V/S}= 0 \, .
\end{equation}
The solutions with the right fall-off at the boundary are
\begin{equation}
\begin{split}
R_V(\rho)=& \sin^{l+1}\rho\, \cos\rho \, {}_2 F_1\left( {2+l-\omega\over 2}, {2+l+\omega\over 2};{3\over 2}\Big\vert \cos^2\rho \right) \, , \\
R_S(\rho)=& \sin^{l+1}\rho\, {}_2 F_1\left( {1+l-\omega\over 2}, {1+l+\omega\over 2};{1\over 2}\Big\vert \cos^2\rho \right) \, .
\end{split}
\end{equation}
These functions are well behaved when expanded around the boundary, but present singular behavior when expanded around the center of AdS. In the asymptotic regions away from the scattering regions, this is unacceptable, since we should be able to trust the free field expansion. Like in the scalar case, the issue can be fixed by quantizing the frequency $\omega$. For the case of the vector-type degree of freedom, we must have
\begin{equation}
\omega_V=2+l+2\kappa\, , \quad \text{with}\quad \kappa\in \mathbb{Z}^+\, ,
\end{equation}
while for the scalar-type we must choose
\begin{equation}
\omega_S=1+l+2\kappa\, , \quad \text{with}\quad \kappa\in \mathbb{Z}^+\, .
\end{equation}
Let us also note that $l \geq 1$, since the solutions with $l = 0$ are pure gauge.
This yields the spectrum of one-particle states in the AdS Fock space \cite{Terashima:2017gmc}. A general solution of the Maxwell field equations in AdS with the boundary conditions \eqref{eq:MagneticBC} can thus be expanded as
\begin{equation}\label{eq:ModeExpA}
{\cal A}_{\mu}=\sum_{\kappa\in \mathbb{Z}^+}\sum_{l,m} \left[ \frac{1}{\mathcal N^V} a^V_{\kappa,l,m} {\cal A}^{V,\kappa,l,m}_{\mu}+ \frac{1}{\mathcal N^S} a^S_{\kappa,l,m} {\cal A}^{S,\kappa,l,m}_{\mu} \right]+\text{c.c.}\, ,
\end{equation}
where ${\cal A}^{V/S,\kappa,l,m}_{\mu}$ are the modes in \eqref{eq:Asols} with their respective quantized positive frequencies, while ``c.c." constains the contributions from the negative frequencies. The normalization constants
\begin{equation}
\begin{split}
\mathcal N^V &= \sqrt{\frac{4 \Gamma(\kappa + \frac 3 2) \Gamma(\kappa+l+2)}{l(l+1)\pi \Gamma(\kappa+1)\Gamma(\kappa+ l + \frac 3 2)L^2}}\, , \\
\mathcal N^S &= \sqrt{\frac{\Gamma(\kappa + \frac 1 2) \Gamma(\kappa+l+1)}{l(l+1)\pi \Gamma(\kappa+1)\Gamma(\kappa+ l + \frac 3 2) L^2}}\, ,
\end{split}
\end{equation}
ensure that the field is unit normalized with respect to the inner product induced by the symplectic potential.
The boundary operator can now be defined simply as
\begin{equation}
\label{eq:jaVS}
j_{\mu}=\lim_{\rho\rightarrow {\pi \over 2}} \cos^{-1}\rho\, {\cal A}_{\mu}.
\end{equation}
Regarding this formula as an operator equation, the objective is to invert it and obtain the operators $a^{V/S}_{\kappa,l,m}$ in terms of $j_{\mu}$. In the scalar case if scalar fields, the modes are simple spherical harmonics and exponential functions of global time. Here, the math is slightly more complicated due to the appearance of derivatives. Those complications can be bypassed by reconstructing $a^{V/S}_{\kappa,l,m}$ in terms of derivatives of the current operator $j_{\mu}$. This is done by inverting formula \eqref{eq:jaVS} after acting with derivatives with respect to the coordinates at the boundary in both sides. For example, for outgoing modes, the result reads
\begin{equation}
\begin{split}
a^{V}_{\kappa,l,m} = &{\mathcal N_V \over -i l(l+1)} \frac 1 \pi \int_{0}^{\pi} d\tau'\, \int d^2\Omega'\,
\epsilon_{\tau'}^{\,\, a b}\nabla_a j^+_b(x') \, Y_l^m(\hat{\Omega}')^* e^{-i (2+l+2\kappa) \tau'}\, , \\
a^{S}_{\kappa,l,m} =& {\mathcal N_S \over{-i l(l+1)}\omega_S} \frac 1 \pi \int_{0}^{\pi} d\tau' \int d^2\hat{\Omega}' \,\nabla^{a}j^{+}_a(x') \, Y_l^m(\hat{\Omega}')^* e^{-i (1+l+2\kappa) \tau'} \, .
\end{split}
\end{equation}
Here, $j^+_{\mu}$ is the positive frequency part of the current and $\epsilon_{a b c}$ is the boundary metric compatible Levi-Civita tensor.The latin indices run over the directions in the $S^2$. The negative frequency part appears in the terms ``c.c." written in formula \eqref{eq:ModeExpA}. Naively we could plug this back in our expression for the bulk gauge field, but a subtlety must be first addressed. As in the scalar case, we will extend the sum over $\kappa$ to all integers and not just the positive ones. This can be done trivially, as the added modes integrate to zero against the positive frequency part of the current $j^+_{\mu}$. We thus conclude
\begin{equation}
\begin{split}
\label{eq:Akernel}
{\cal A}_{\mu}(x)= \int d^3 x' \left[ K_{\mu}^{V}(\rho,x,x')\, \epsilon_{\tau'}^{\,\, a b}\nabla_a j^+_b(x') + K_{\mu}^{S}(\rho,x,x') \, \nabla^{a}j^+_{a}(x') \right]\, +\text{h.c}\, .
\end{split}
\end{equation}
where we have defined two different kernels
\begin{equation}
\begin{split}
K^{V}_{\mu}(x,x')=&\mathcal N_V \frac 1 \pi \sum_{\kappa,l,m } { Y_l^m(\hat{\Omega}')^* \over{-il(l+1)}} {\cal A}^{V,\kappa,l,m}_{\mu}e^{-i\omega_V \tau'}\, , \\
K^{S}_{\mu}(x,x')=&\mathcal N_S \frac 1 \pi \sum_{\kappa,l,m } { Y_l^m(\hat{\Omega}')^* \over{-il(l+1)\omega_S}} {\cal A}^{S,\kappa,l,m}_{\mu} e^{-i\omega_c\tau'} \, .
\end{split}
\end{equation}
One can integrate equation \eqref{eq:Akernel} by parts along the $S^2$ and re-express this results as integrals over the current, instead of its derivatives.
\subsection{Reconstruction with ``electric'' boundary conditions}\label{MaxfieldReconstructionElectric}
We now look for a bulk gauge field with the fall-off behavior
\begin{equation}\label{eq:ElectricBC}
{\cal V}_{\mu}(\rho,x)\xrightarrow[\rho \to \frac \pi 2]{} \, A_{\mu}(x)\, .
\end{equation}
Here, this defines the normalization of the CFT gauge field two-point function.
The solutions of the Maxwell field equations can still be written as in equation \eqref{eq:Asols},
\begin{equation}\label{eq:Vsols}
\begin{split}
{\cal V}^{V,l,m}_{\mu} =& L^{-2}\epsilon_{\tau\rho\mu}^{\quad\;\;\nu} e^{\pm i\omega \tau} \cos^{2}\rho \,R_V(\rho) \, \nabla_{\nu} Y_l^m(\hat{\Omega}) \, , \\
{\cal V}^{S,l,m}_{\mu} =& L^{-2}(l(l+1)-\omega^2)\epsilon_{\theta\phi\mu}^{\quad\;\;\nu} \sin^{-1}\theta \,\cot^2\rho\, Y_l^m(\hat{\Omega})\, \nabla_{\nu}(e^{\pm i\omega \tau} R_S(\rho)) \, ,\\
{\cal V}^{G,l,m}_{\mu} =& L^{-2}\nabla_{\mu}\left[ e^{\pm i\omega \tau}Q(\rho) Y_l^m(\hat{\Omega}) \right] \, ,
\end{split}
\end{equation}
together with the gauge transformation \eqref{eq:gaugeToLorenz} acting on the scalar type solution.
The radial dependence of of the normalizable modes is now exchanged with respect to the magnetic case, so we have
\begin{equation}
\begin{split}
R_S(\rho)=& \sin^{l+1}\rho\, \cos\rho \, {}_2 F_1\left( {2+l-\omega\over 2}, {2+l+\omega\over 2};{3\over 2}\Big\vert \cos^2\rho \right) \, , \\
R_V(\rho)=& \sin^{l+1}\rho\, {}_2 F_1\left( {1+l-\omega\over 2}, {1+l+\omega\over 2};{1\over 2}\Big\vert \cos^2\rho \right) \, .
\end{split}
\end{equation}
Demanding regularity of the solutions around the center of AdS$_4$ yields the same quantization of frequencies
\begin{equation}
\begin{split}
\omega_V=2+l+2\kappa\, , \quad \text{with}\quad \kappa\in \mathbb{Z}^+\, ,\\
\omega_S=1+l+2\kappa\, , \quad \text{with}\quad \kappa\in \mathbb{Z}^+\, .
\end{split}
\end{equation}
The full gauge connection can thus be written as
\begin{equation}\label{eq:ModeExpV}
{\cal V}_{\mu}=\sum_{\kappa\in \mathbb{Z}^+}\sum_{l,m} \left[ \frac{1}{\mathcal N'_V } v^V_{\kappa,l,m} {\cal V}^{V,\kappa,l,m}_{\mu}+ \frac{1}{\mathcal N'_S } v^S_{\kappa,l,m} {\cal V}^{S,\kappa,l,m}_{\mu} \right]+\text{h.c}\, .
\end{equation}
We now want to obtain the coefficients $v^{V/S}_{\kappa,l,m}$ in terms of the boundary operator $b_{\mu}$. Following the same method as above, we find
\begin{equation}
\begin{split}
v^S_{l,m}=& {\mathcal N'_S \over -l(l+1)} {1\over \pi}\int_{-{\pi \over 2}}^{\pi \over 2} d\tau'\int d^2\Omega' \, \epsilon_{\tau'}^{\quad ab} \nabla_{a}A^+_{b}(x')e^{-i\omega_S \tau'} Y_l^m(\Omega')^*\, , \\
v^V_{l,m} =& {\mathcal N'_V \over -l(l+1)} {1\over \pi}\int_{-{\pi \over 2}}^{\pi \over 2} d\tau'\int d^2\Omega' \,\nabla_{a} A^{+a}(x')e^{-i\omega_{V}\tau'} Y_l^m(\Omega')^*\, .
\end{split}
\end{equation}
Much like before, we will sum over all values of $\kappa$ and not just the positive ones. This can be done without trouble as the extra modes integrate to zero against the positive frequency part of $\tilde{j}^{\mu}$. A similar statement can be made for the modes appearing in the ``h.c.'' part of equation \eqref{eq:ModeExpV}, which involve the negative frequencies. We conclude that the result is
\begin{equation}
\begin{split}
{\cal V}_{\mu}(x)= \int d^3 x' \left[ \tilde{K}_{\mu}^{S}(\rho,x;x')\, \epsilon_{\tau'}^{\,\, a b}\nabla_a A^+_b(x') + \tilde{K}_{\mu}^{V}(\rho,x;x') \, \nabla^{a} A^+_{a}(x') \right]\, +\text{h.c}\, .
\end{split}
\end{equation}
where we have defined two different kernels
\begin{equation}
\begin{split}
\tilde{K}^{S}_{\mu}(\rho,x;x')=& \mathcal N'_S {1\over \pi} \sum_{\kappa,l,m } { Y_l^m(\Omega')^* \over{-l(l+1)}} {\cal V}^{S,\kappa,l,m}_{\mu} e^{-i\omega_S\tau'} \, , \\
\tilde{K}^{V}_{\mu}(\rho,x;x')=& \mathcal N'_V {1\over \pi} \sum_{\kappa,l,m } { Y_l^m(\Omega')^* \over{-l(l+1)}} {\cal V}^{V,\kappa,l,m}_{\mu}e^{-i\omega_V \tau'}\, .
\end{split}
\end{equation}
\section{Complex coordinates on $S^2$}\label{app:coordS2}
The standard parametrization of a two-sphere is done with angles $\theta\in[0,\pi]$ and $\phi\in(0,2\pi]$. Alternatively, one can define complex coordinates $z, \bar{z}$ such that
\begin{equation}
\cos\theta={1-z\bar{z}\over 1+z\bar{z}}\, , \quad \sin\theta\cos\phi={z+\bar{z}\over1+z\bar{z}}\, , \quad \sin\theta\sin\phi=-i{z-\bar{z}\over 1+z\bar{z}}\, .
\end{equation}
The metric in these coordinates is
\begin{equation}
ds^2={4 dz d\bar z\over(1+z\bar{z})^2}\, .
\end{equation}
The Laplace operator eigenvalue equation obeyed by spherical harmonics,
\begin{equation}
\left(\partial_{\theta}^2+{1\over \sin^2\theta} \partial_{\phi}^2 +\cot\theta \partial_{\theta}\right)Y_l^m(\hat{\Omega}) = -l (l+1)Y_l^m(\hat{\Omega}),
\end{equation}
takes a different form in the new coordinates,
\begin{equation}
(1+z\bar{z})^2 \partial \bar{\partial}Y_l^m(z,\bar{z}) = -l (l+1)Y_l^m(z,\bar{z}).
\end{equation}
In complex coordinates, the Dirac delta function reads
\begin{equation}
\delta^{(2)}(\hat{\Omega},\hat{\Omega}') = \delta(\phi-\phi')\delta(\cos\theta-\cos\theta')= {1 \over 2}(1+z \bar{z})^2 \delta^{(2)}(z,z')\, ,
\end{equation}
The Greens fuction associated to the Laplace operator must obey the differential equation
\begin{equation}
(1+z\bar{z})^2 \partial \bar{\partial} G(z,\bar{z})={1\over 2}(1+z \bar{z})^2 \delta^{(2)}(z,z') - \frac{1}{4\pi}\, ,
\end{equation}
which is solved by
\begin{equation}\label{eq:GS2}
G={1\over 4\pi i}\log |z-z'|^2 - \frac{1}{4\pi} \log(1 + z \bar z)- \frac{1}{4\pi} \log(1 + z' \bar z') \, .
\end{equation}
In complex coordinates, the derivative operators read
\begin{equation}
\sin\theta\partial_{\theta}= z\partial+\bar{z}\bar{\partial}\, , \quad \partial_{\phi}= i (z\partial-\bar{z}\bar{\partial})\, .
\end{equation}
\section{AdS Li\'enard-Wiechert potentials}\label{app:LWAdS}
In this appendix we compute the Li\'enard-Wiechert potential associated to a particle propagating in Anti-de Sitter space-time. Using the linearity of electro-magnetism in the region of asymptotic decoupling, one can build solutions involving several particles by adding the solutions of the form presented here.
Li\'enard-Wiechert gauge fields are classical solutions to the heterogeneous Maxwell equation
\begin{equation}
\nabla_{\mu}F^{\mu\nu} = J^{\mu}\, .
\end{equation}
The current for a single charged particle reads explicitly
\begin{equation}
J^{\mu}(Y)=q \int ds\, {{\partial X^{\nu}(s)}\over{\partial s}} \, \delta^{(4)}\left(Y-X(s)\right)\, .
\end{equation}
We will first analyze massive particles. The strategy we will follow will be to find the solution for a stationary particle located at the origin, and then use the AdS isometries to obtain the solution for a particle following a generic time-like geodesic. The solution for the stationary case in global coordinates reads
\begin{equation}
{\cal A}^{(\text{C})}_{\mu}= q \cot \rho \, \delta^{\tau}_{\mu}\, ,
\end{equation}
where the label $(\text{C})$ is a reminder that we are chosing a particular homogeneous (radiative) solution, namely that of no radiation if the particle is at rest. The solution written above diverges at the location of the charge, $\rho=0$. We now act with an AdS isometry on this solution, to obtain the gauge field associated to a particle moving along a generic time-like geodesic.
Isometries of AdS can be easily described in embedding space, with line element
\begin{equation}
ds^2=-dX_1^2-dX_2^2+dX^2_3+dX^2_4+dX^2_5\, .
\end{equation}
The embedding of AdS in such space reads
\begin{equation}\label{eq:Embedding}
\begin{split}
X_{1} &= L {{\cos\tau}\over{\cos\rho}}\, , \quad
X_{2} = L {{\sin\tau}\over{\cos\rho}}\, , \\
X_{3} &= L \tan\rho \sin\theta \cos\phi \, ,\quad
X_{4} = L \tan\rho \sin\theta \sin\phi \, ,\quad
X_{5} = L \tan\rho \cos\theta \, .
\end{split}
\end{equation}
The $SO(2,4)$ isometries of AdS are then translations, boosts, and rotations in embedding space. We are interested only in transformations that keep the particle at the origin at $\tau=0$, such that the particle remains inside the scattering region as the flat limit is taken. We further require a transformation that makes the new geodesic hit the point $\tau=\pi / 2$ with $\cos\rho=m/ \omega$ at an angle $\hat{p}$. Such a geodesic will hit the complexified boundary $\partial_+{\cal M}$ at the point \eqref{eq:Imtau}, which means that this geodesic is the one followed by particles creating the desired scattering states in Minkowski space-time. Such geodesics can be obtained by transforming embedding coordinates with a boost along $X_2$ and a rotation. Explicitly,
\begin{equation}
X^{\prime I}=\Lambda^{I}_{J} X^J\, , \quad \text{with} \quad \Lambda^{I}_J=R^I_K B^K_J\, ,
\end{equation}
with rotation and boost matrices given by
\begin{equation}
R=\left( \begin{matrix} 1&0&0&0&0\\ 0&1&0&0&0 \\ 0&0&-\sin\theta_p\cos\phi_p & \cos\theta_p\cos\phi_p & \sin\phi_p \\ 0&0&-\sin\theta_p\sin\phi_p & \cos\theta_p\sin\phi_p & -\cos\phi_p \\ 0&0& -\cos\theta_p & -\sin\theta_p &0 \end{matrix} \right)
\, , \quad
B=\left( \begin{matrix} 1&0&0&0&0\\ 0&{\omega \over m}& -{\vert \vec{p}\vert\over m} &0&0 \\ 0&-{\vert \vec{p}\vert \over m}&{m\over \omega} & 0 &0 \\ 0&0&0 & 1 & 0 \\ 0&0& 0& 0 &1 \end{matrix} \right)\, .
\end{equation}
The embedding space vector $X'$ defines new coordinates $\rho',\tau',\hat{x}'$ in the form of an embedding like equation \eqref{eq:Embedding}. Explicitly, the new coordinates are related to the old ones as follows
\begin{equation}
\begin{split}
\tan\tau&={{\omega\sin\tau'-\vert\vec{p}\vert \sin\rho' \Omega'\cdot \hat{p}}\over{m \cos\tau'}}\, , \\
\cos\rho&={{\cos\rho'}\over{\cos\tau'}} {1\over{\sqrt{1-\left( {{ \omega\sin\tau' -\vert \vec{p}\vert \sin\rho' \hat{x}'\cdot\hat{p}}\over{m\cos\tau'}} \right)^2}}}\, ,\\
\cos\theta &= -{{\sin\rho'}\over{\cos\tau'}} {{ \partial_{\phi_p}(\hat{p}\cdot\hat{x}') }\over{ \sin\theta' }} {1\over{\sqrt{ 1-{{\cos^2\rho'}\over{\cos^2\tau'}} - \left( {{ \omega\sin\tau' -\vert \vec{p}\vert \sin\rho' \hat{x}'\cdot\hat{p}}\over{m\cos\tau'}} \right)^2 }}}\, , \\
\tan\phi&= {{ m\sin\rho' }\over{ \vert \vec{p}\vert \sin\tau' -\omega \sin\rho' \hat{p}\cdot\hat{x}' }} \partial_{\theta_p}(\hat{p}\cdot\hat{x}' ) \, .
\end{split}
\end{equation}
With these expressions at hand, we are now ready to compute
\begin{equation}
{\cal A}_{\mu'}^{(\text{C})}(x')={{\partial x^{\mu}}\over{\partial x^{\mu'}}} {\cal A}_{\mu}^{(\text{C})}(x)\, .
\end{equation}
The result for the gauge field is generally complicated. However, expressions simplify when approaching the boundary, which is where we want to evaluate the Li\'enard-Wiechert fields. For example, at $\rho'={\pi\over 2}$ and $\tau'={\pi \over 2}$, we have
\begin{equation}\label{eq:ResultC1}
\langle j_{\tau'} \left( \tau'={\pi \over 2}, \hat{x}' \right)\rangle = \lim_{\rho'\rightarrow{\pi \over 2},\tau'\rightarrow{\pi \over 2}} {1\over \cos\rho'} {\cal A}_{\tau'} = q{{m^2}\over{\left( \omega- p\cdot \hat{x}' \right)^2}}\, .
\end{equation}
This result can be used to make the argument of section \ref{sec:WardSoftE}. There, we used that
\begin{equation}\label{eq:argument}
\int_{\partial \mathcal M_+} d^3 x \, \partial_{\mu}\alpha(x) \langle j^{\mu}\rangle =0 \, ,
\end{equation}
where, as a reminder, the parameter $\alpha(x)$ is chosen to be
\begin{equation}\label{eq:alphaCFTRepeat}
\alpha(x)=\lim_{\rho\rightarrow{\pi \over 2}}\int d^2 \hat{x}' \, {1\over 4\pi}{{\cos^2\rho-\cos^2\tau }\over{\left( \sin\tau -\sin\rho\, \hat{x}\cdot\hat{x}' \right)^2}} \varepsilon(\hat{x}')\, .
\end{equation}
To see this, we first integrate by parts
\begin{equation}
\int_{\partial {\cal M}_+} d^3 x \, \partial_{\mu}\alpha(x) \langle j^{\mu}\rangle =\int_{\tau={\pi \over 2}} d^2\hat{x} \, \varepsilon(\hat{x})\, \langle j_{\tau} \rangle - \int_{\partial {\cal M}_+} d^3 x \, \alpha(x) \partial_{\mu}\langle j^{\mu}\rangle \, .
\end{equation}
In the first term we have used the fact that the parameter $\alpha(x)$ becomes $\varepsilon(\hat{x})$ in the CFT region $\tilde{\cal I}^+$. This term can be written explicitly using \eqref{eq:ResultC1}, which results in
\begin{equation}\label{eq:AUX1}
\int_{\tau={\pi \over 2}} d^2\hat{x}\, \varepsilon(\hat{x})\, \langle j_{\tau} \rangle= q \int d^2\hat{x} \varepsilon(\hat{x}) {{m^2}\over{\left( \omega- p\cdot \hat{x}' \right)^2}}\, .
\end{equation}
In the second term, we can invoke the Ward identity to obtain a delta function for the insertion of an operator at $\hat{x}=\hat{p}$ and $\tau={\pi \over 2}+i\log\sqrt{{\omega+m}\over{\omega-m}}$. The value of $\alpha(x)$ at that location is simply
\begin{equation}\label{eq:AUX2}
\int_{\partial {\cal M}_+} d^3 x \, \alpha(x) \partial_{\mu}\langle j^{\mu}\rangle = q \int d^2 \hat{x}' \, \varepsilon(\hat{x}') {{m^2}\over{\left( \omega- p\cdot \hat{x}' \right)^2}}\, .
\end{equation}
Combining \eqref{eq:AUX2} and \eqref{eq:AUX1} we immediately arrive at the desired result \eqref{eq:argument}. Note that the calculations here involve a single particle in AdS and a single operator insertion in the CFT. However, the bulk calculations can be easily generalized to include more particles due to the linearity of electro-magnetism in the region of asymptotic decoupling.
In section \ref{sec:WardSoftE}, we also argued that the Coulombic current does not contribute to the second line of \eqref{eq:ArgumentsHolographic}. This can also be checked with the explicit form of the Li\'enard-Wiechert potentials calculated here, which yield vanishing expectation values $\langle j_{z}\rangle$ and $\langle j_{\bar{z}}\rangle$ at $\tilde{\cal I}^+$.
A similar argument was also made for the case of electric boundary conditions in section \ref{sec:MST}. In that case, a topological current was constructed, and the relation of interest is
\begin{equation}\label{eq:argumentM}
\int_{\partial {\cal M}_+} d^3 x \, \partial_{\mu}\alpha(x) \langle (*f)^{\mu}\rangle =0 \, , \quad \text{with}\quad f=dA \, .
\end{equation}
This can be proven following a similar logic as the one above. We start with a stationary magnetic monopole in anti-de Sitter space. The Li\'enard-Wiechert field associated to such object reads
\begin{equation}
{\cal V}^{(\text{C})}_{\mu}=g \cos \theta \, \delta^{\phi}_{\mu}\, .
\end{equation}
One can then act on this result by an isometry of AdS such that the geodesic followed by the particle hits the point of the CFT where an operator must be inserted in order to create a scattering state. A calculation similar to the one above results in
\begin{equation}
\langle (*f)_{\tau'} \left( \tau'={\pi \over 2}, \hat{x}' \right)\rangle = \lim_{\rho'\rightarrow{\pi \over 2},\tau'\rightarrow{\pi \over 2}} \epsilon_{\tau'}^{\quad a'b'} \partial_{a'}{\cal V}_{b'} = g{{m^2}\over{\left( \omega- p\cdot \hat{x}' \right)^2}}\, .
\end{equation}
This result mirrors the one concerning electrical charges in formula \eqref{eq:ResultC1}. The rest of the proof leading to \eqref{eq:argumentM} follows verbatim.
\section{Parameter $\alpha(x)$}\label{app:alpha}
In this appendix we justify the choice of parameter $\alpha(x)$ made in section \ref{sec:WardSoftE}. The logic we follow mirrors the strategy used in flat space when deriving Weinberg soft theorems from the asymptotic symmetries of the QED S-matrix. When only massless particles are present in flat space, ${\cal I}^{\pm}$ can be considered complete Cauchy slices, and the choice of parameter in formula \eqref{eq:varepsilonchoice} yield the desired results. We repeat here this choice.
\begin{equation}
\varepsilon(x)={1\over z-z'} \, , \quad \text{and}\quad \varepsilon(x)={1\over \bar{z}-\bar{z}'} \, .
\end{equation}
When discussing massive particles, one must extend this choice to time-like infinity $i^{\pm}$. This is done by realizing that the parameter $\varepsilon(\hat{x})$ is associated to a large gauge transformation, such that
\begin{equation}
A_{\mu}=\nabla_{\mu}\Lambda\, , \quad \text{with}\quad \Lambda\vert_{{\cal I}^{+}}=\varepsilon(\hat{x})\, .
\end{equation}
Such a large transformation can now be extended into the bulk of Minkowski space by making use of Lorenz gauge
\begin{equation}
\nabla^{\mu}A_{\mu}=\nabla^2\Lambda=0\, .
\end{equation}
The solution is simply
\begin{equation}
\Lambda(x)=\int d^2\hat{x}'\, G(x;\hat{x'})\, \varepsilon(\hat{x}')\, ,
\end{equation}
for a kernel $G(x;\hat{x'})$ obeying the massless Laplace equation and the boundary condition
\begin{equation}\label{eq:BoundaryG}
\lim_{x\rightarrow {\cal I}^{+}}G(x;\hat{x'}) = \delta^{(2)}(\hat{x},\hat{x}')\, .
\end{equation}
Such a kernel can be written explicitly as follows. A solution of the massless Laplace equation can be expanded in the following set of regular modes
\begin{equation}
\Lambda_{l,m}(x)=Y_l^m(\hat{x}) F_l\left( {r\over u} \right)\, , \quad \text{with}\quad F_l(R)=R^l\, {}_2F_1\left( l,l+1;2l+2|-2R \right)\, .
\end{equation}
We will thus look for a solution of the form
\begin{equation}
G(x;\hat{x'})=\sum_{l,m}c_{l,m}\Lambda_{l,m}(x)\, .
\end{equation}
At the null boundary, we have
\begin{equation}
\lim_{r\rightarrow \infty}G(x;\hat{x'})=\sum_{l,m}c_{l,m}Y_l^m(\hat{x}) {{\Gamma(2+2l)}\over{2^l \Gamma(l+1)\Gamma(2+l)}}\, .
\end{equation}
This is indeed a delta function as in \eqref{eq:BoundaryG} if
\begin{equation}
c_{l,m}=Y_l^m(\hat{x}')^* {{2^l \Gamma(l+1)\Gamma(2+l)}\over{\Gamma(2+2l)}} \, .
\end{equation}
we thus conclude
\begin{equation}
G(x;\hat{x'})=\sum_{l,m}{{2^l \Gamma(l+1)\Gamma(2+l)}\over{\Gamma(2+2l)}} Y_l^m(\hat{x}')^* Y_l^m(\hat{x}) F_l\left( {r\over u} \right)\, .
\end{equation}
Performing the sum explicily yields
\begin{equation}
G(x;\hat{x'})= {1\over 4\pi }{ { t^2-r^2 } \over {\left( t-r\hat{x}\cdot\hat{x}' \right)^2 } } \, .
\end{equation}
This justifies the choice \eqref{eq:MinkepChoice} in the main text.
The calculation in AdS is very similar. In the absence of massive particles, the CFT regions $\tilde{\cal I}^{\pm}$ are the only ones playing a role, and the choice $\alpha(x)=\varepsilon(\hat{x})$ yield the correct result. When massive particles are included in the calculation, one must choose the value of $\alpha(x)$ in the Euclidean half-spheres $\partial{\cal M}_{\pm}$. We define them by noting that $\alpha(x)$ corresponds to a large gauge transformation, such that it can be extended into the bulk using Lorenz gauge. We thus must solve the massless Laplace equation for a parameter $\Lambda(x)$ with boundary conditions
\begin{equation}
\Lambda(x)\vert_{\tilde{\cal I}^{+}}=\varepsilon(\hat{x})\, .
\end{equation}
We will use the following regular modes obeying the massless Laplace equation
\begin{equation}
\Lambda_{l,m}(x) =Y_l^m(\hat{x})\, F_l\left( {{\sin\rho}\over { \sin\tau-\sin\rho}} \right)\, , \quad \text{with}\quad F_l(R)=R^l\, {}_2F_1\left( l,l+1;2l+2|-2R \right)\, .
\end{equation}
The region $\tilde{\cal I}^+$ corresponds to taking $\rho\rightarrow {\pi \over 2}$ as well as $\tau\rightarrow {\pi \over 2}$, meaning that the argument in $F_l(R)$ becomes large. In such a limit, we have
\begin{equation}
\Lambda(x)\vert_{\tilde{\cal I}^{+}}=\sum_{l,m}c_{l,m}Y_l^m(\hat{x}) {{\Gamma(2+2l)}\over{2^l \Gamma(l+1)\Gamma(2+l)}}\, .
\end{equation}
We thus conclude that the gauge parameter in the bulk reads
\begin{equation}
\Lambda(x)=\int d^2\hat{x}'\, \varepsilon(\hat{x}')\, \sum_{l,m} {{2^l \Gamma(l+1)\Gamma(2+l)}\over{\Gamma(2+2l)}} Y_l^m(\hat{x}')^* Y_l^m(\hat{x}) F_l\left( {{\sin\rho}\over { \sin\tau-\sin\rho}} \right)\, .
\end{equation}
performing the sum explicitly yields
\begin{equation}
\Lambda(x)=\int d^2 \hat{x}' \, {1\over 4\pi}{{\cos^2\rho-\cos^2\tau }\over{\left( \sin\tau -\sin\rho\, \hat{x}\cdot\hat{x}' \right)^2}} \varepsilon(\hat{x}')\, .
\end{equation}
The value of this gauge parameter in the Euclidean half spheres is obtained by sending $\rho\rightarrow {\pi \over 2}$, as well as analytically continuing $\tau$ in the imaginary direction. For $\partial{\cal M}_+$, we have $\tau={\pi \over 2}+i\tilde{\tau}$, such that
\begin{equation}
\Lambda(x)\vert_{\partial{\cal M}_+}=\int d^2 \hat{x}' \, {1\over 4\pi}{{\sinh^2\tilde{\tau} }\over{\left( \cosh\tilde{\tau} - \hat{x}\cdot\hat{x}' \right)^2}} \varepsilon(\hat{x}')\, .
\end{equation}
This concludes the justification for formula \eqref{eq:alphaCFT} in the main text.
\section{Photon scattering states}\label{sec:GaugeFields}
In this section we discuss how photon scattering states can be constructed in the flat limit of AdS/CFT. Asymptotic decoupling implies that photons obey the free Maxwell equation at early and late times. For the purpose of constructing scattering amplitudes, it is therefore sufficient to discuss the reconstruction of free gauge fields in AdS. This will be done explicitly in sections \ref{sec:FLM} and \ref{sec:FLE}. In section \ref{sec:Coulomb} we will discuss the back-reaction of charged particles on the gauge field, which will play a role when discussing Ward identities.
Like in the scalar case, we need to understand how bulk path integrals in the presence of gauge fields compute generating functionals of CFT correlators. In the case of gauge fields, the problem becomes more subtle. In general, the solution to the field equations consists of two types of solutions which are characterized by their fall-off behavior at the asymptotic boundary. In the case of gauge fields both types of solutions fall off fast enough to be normalizable. The duality thus admits different formulations depending on what conditions we \emph{choose} to impose at the boundary. This leads to a rich and interesting amount of physics, some of which we will discuss throughout this section.
\subsection{Inequivalent quantization schemes of scalar fields in AdS}\label{sec:ScalarLagrange}
Before considering gauge fields, we start by briefly reviewing the analogous scenario involving scalar fields.
The issue of having two types of solutions which are both normalizable is also present in the case of scalar fields for a particular range of bulk masses \cite{Breitenlohner:1982jf}. In the range
\begin{equation}\label{eq:MassRange}
1-{d^2\over 4}>m^2 L^2>-{d^2\over 4}\, ,
\end{equation}
there are two inequivalent ways to quantize a scalar field in AdS. The general solution for a scalar field close to the boundary of AdS reads
\begin{equation}
\phi(\rho,x) \xrightarrow[\rho \to {\pi \over 2}]{} (\cos\rho)^{\Delta_+}\alpha(x)+(\cos\rho)^{\Delta_-}\beta(x)\, .
\end{equation}
Outside the mass range \eqref{eq:MassRange}, the mode associated to $\beta(x)$ is non-normalizable due to its behavior close to the AdS boundary. This leads to the ``standard'' construction of section \ref{sec:ScatteringStatesScalar}, where $\beta(x)$ becomes a non-dynamical source at the boundary that we called $\phi_0(x)$. Otherwise, when the bulk mass is in the range \eqref{eq:MassRange}, we could instead choose to fix $\alpha(x)$ at the boundary, and quantize the part of the field that asymptotes to $\beta(x)$. This leads to a holographic correspondence with a different conformal field theory \cite{Klebanov:1999tb}. More precisely, if we fix $\alpha(x)$ at the boundary, the CFT contains an operator of conformal dimension $\Delta_+$. If instead we fix $\beta(x)$, then the CFT contains an operator with conformal dimension $\Delta_-$. These two theories are related by renormalization group flow \cite{Witten:2001ua,Minces:2002wp}. The theory with an operator ${\cal O}_{\Delta_-}$ can be deformed by the relevant operator ${\cal O}_{\Delta_-}^2$, leading to a mixed boundary condition of the form $\alpha=f \beta$.
An interesting fact is that the two different conformal field theories are related by a Legendre transformation \cite{Klebanov:1999tb}. In this case, the CFT with non-dynamical source $\alpha(x)=J(x)$ is modified by making $J(x)$ dynamical and coupling it to a new background source $J'(x)$. The resulting CFT path integral corresponds to the bulk theory with the alternative boundary conditions $\beta(x)=J'(x)$.
The take-away message from this discussion is that different choices of boundary conditions for our bulk fields lead to different conformal field theories that are related by a Legendre transformation. This is also the case for gauge fields, in which the relation between the two different boundary theories can be understood in the bulk as the statement of electric-magnetic duality.
\subsection{Electric-magnetic duality and $SL(2,\mathbb{Z})$ action on CFT$_3$'s}\label{sec:SL2Z}
Let us now turn to the case of a bulk theory with $U(1)$ gauge symmetry. As we will now review, there are an infinite number of such theories mapping onto each other under $SL(2,\mathbb{Z})$ transformations, which are an extension of electric-magnetic duality \cite{Witten:1995gf}. In terms of the dual description, the $SL(2,\mathbb Z)$ relates different conformal field theories, similar to the scalar case discussed above.
Consider a general solution for a gauge field in AdS$_4$ with boundary behavior
\begin{equation}\label{eq:Aasymptotics}
{\cal A}_{\mu}(\rho,x) \xrightarrow[\rho \to \frac \pi 2]{}(\cos\rho)^1 \alpha_{\mu}(x)+(\cos\rho)^0 \beta_{\mu}(x)\, .
\end{equation}
We could choose to fix $\beta_{\mu}(x)=A_{\mu}(x)$ at the boundary, which introduces a non-dynamical source $A_{\mu}(x)$ in the CFT. In this case $\alpha_{\mu}(x)$ corresponds to the finite energy bulk excitations spanning the low energy Hilbert space. Bulk path integrals now correspond to a generating functional of CFT correlators where the source $A_{\mu}(x)$ couples to a conserved current operator $j^{\mu}(x)$ with conformal dimension $\Delta_j=d-1=2$. Note that this boundary condition fixes the magnetic component of the gauge field at the boundary $M_i={1\over 2}\epsilon_{abc}F^{bc}$. We will thus refer to these boundary conditions as ``magnetic''.
Another possibility is to fix the value of the electric field at the boundary $E_a=F_{\rho a }$ in order to obtain ``electric'' boundary conditions. This corresponds to fixing a boundary value of $\alpha_{\mu}(x)=B_{\mu}(x)$, while $\beta_{\mu}(x)$ is allowed to fluctuate. In this case, the conformal theory couples the source $B_{\mu}(x)$ to a conserved current constructed out of an operator with conformal dimension $\Delta_A=1$, which can be understood as a dynamical gauge field in the conformal theory.
The inequivalent CFTs described here are related through RG flow like in the scalar case \cite{Marolf:2006nd}. There is also a relation involving the $S$-transformation \cite{Witten:1995gf} of $SL(2,\mathbb Z)$. This is similar to the Lagrange transform that relates the two conformal theories dual to different quantization schemes for a bulk scalar field (see section \ref{sec:ScalarLagrange} above). The $S$-transformation acts as follows. We start with a CFT with Lagrangian ${\cal L}(\phi)$ and global $U(1)$ symmetry. Here, $\phi$ stands for a field in the CFT. Such a Lagrangian is sourced by the magnetic boundary condition,
\begin{equation}
{\cal L}'(\phi)={\cal L}(\phi) + A_{\mu} j^{\mu}\, ,
\end{equation}
where $j^{\mu}$ is the current operator associated to the global U(1) symmetry. CFT path integrals involving this Lagrangian compute bulk path integrals with magnetic boundary conditions,
\begin{equation}
\langle e^{i \int_{\partial \text{AdS}} d^3 x\, A_{\mu}j^{\mu}} \rangle_{\text{CFT}} = \int [{\cal D}{\cal A}]_{A_{\mu}}\, e^{i S[{\cal A}]}\, .
\end{equation}
where for simplicity we have omitted the specification of in/out CFT states and the corresponding conditions for bulk fields at space-like Cauchy slices.
The $S$-transformation of the conformal theory consists of regarding the background connection $A_{\mu}$ as dynamical. A current $\tilde{j}^{\mu}=\epsilon^{\mu\nu\rho}\partial_{\nu}A_{\rho}$ can then be introduced and coupled to a new background field $B_{\mu}$, such that the CFT Lagrangian reads
\begin{equation}
{\cal L}(\phi,A)={\cal L}(\phi) + A_{\mu} j^{\mu}+B_{\mu}\tilde{j}^{\mu}\, .
\end{equation}
Path integrals involving this Lagrangian will now compute bulk path integrals involving the electric boundary conditions $B_\mu$ described above, instead of the magnetic ones \cite{Yee:2004ju}. Explicitly, we can write
\begin{equation}
\langle e^{i \int_{\partial \text{AdS}} d^3 x\, B_{\mu}\tilde{j}^{\mu}} \rangle_{\text{CFT}} = \int [{\cal D}{\cal V}]_{B_{\mu}}\, e^{i S'[{\cal V}]}\, ,
\end{equation}
where the path integral in the CFT runs over $A_{\mu}$, and the bulk action $S'[{\cal V}]$ is the same as $S[{\cal A}]$ but with coupling $\tilde{e}=4\pi/e$. As shown in \cite{Yee:2004ju}, gauge fields with magnetic and electric boundary conditions are related through Hodge duality. The gauge connection with magnetic boundary conditions ${\cal A}$ is related to the gauge connection with electric ones ${\cal V}$ by
\begin{equation}\label{eq:Hodge}
({\cal F}_{\cal A})_{\mu\nu} = {e^2\over 8 \pi} \epsilon_{\mu\nu\rho\sigma}({\cal F}_{\cal V})^{\rho \sigma}\, .
\end{equation}
Alternatively, the above can be understood as fixing the boundary condition to be of magnetic type and making a choice as to whether the gauge potential of the field strength or that of the dual field strength are related to simple local operators in the boundary theory.
The full $SL(2,\mathbb Z)$ is the natural extension of the above. In terms of boundary conditions, the $SL(2,\mathbb Z)$ theory relates magnetic boundary conditions to more general, Robin-type ones. In terms of the relation of bulk degrees of freedom, it acts as a map between gauge potentials which are related to simple operators in the boundary theory. In the following, we will only be interested in magnetic and electric boundary conditions, i.e., the electric-magnetic duality subgroup of $SL(2,\mathbb Z)$ generated by $S$.
\subsection{Flat limit: Magnetic boundary conditions}\label{sec:FLM}
In this section we will explore the standard choice of boundary conditions for the bulk gauge field, where we fix the magnetic field at the boundary to zero. Setting $\beta_{\mu}(x)=0$ means there is no magnetic field at the boundary, which forbids the existence of bulk magnetic charge. In a flat limit, we thus expect to recover a theory of electromagnetism without magnetic monopoles. We will follow the same logic as in the scalar case of section \ref{sec:ScatteringStatesScalar}. The boundary conditions on the holographic boundary determine the source, while boundary conditions at Cauchy slices $\Sigma_{\pm}$ specify a choice of CFT state. Instead of constructing scalar creation/annihilation operators, we now construct photon operators.
Like in the scalar case, we will invoke the principle of asymptotic decoupling, which implies that our quantum fields are free away from the scattering region. In this section, we will study the radiative part of the gauge fields, and relegate coulombic contributions to section \ref{sec:Coulomb}. Free $U(1)$ gauge fields in Minkowski space can be written as follows
\begin{equation}
\hat{\cal A}_{\mu}(x)=\int {d^3 \vec{q}\over (2\pi)^3} {1\over{\sqrt{2\omega_{\vec{q}}}}} \sum_{\lambda=\pm}\left(
\varepsilon^{(\lambda)}_{\mu} \hat{a}^{(\lambda)}_{\vec{q}} \, e^{i q\cdot x}
+
\varepsilon^{(\lambda)*}_{\mu} \hat{a}^{(\lambda)\dagger}_{\vec{q}} \, e^{-i q\cdot x}
\right)\, , \quad \text{with}\quad x\in \text{Mink}_{3+1}\, .
\end{equation}
In this expression $\lambda$ stands for the two different polarizations, and $\varepsilon^{(\lambda)}_{\mu}$ are null polarization vectors. We will be working in Lorenz gauge, so $\varepsilon_{\mu}q^{\mu}=0$, and we normalize these vectors such that $\varepsilon^{(+)}_{\mu}\varepsilon^{(-),\mu}=1$. Creation and annihilation operators can thus be extracted from the position space field using
\begin{equation}\label{eq:aFromA}
\begin{split}
\hat{a}_{\vec{q}}^{(\lambda)} =&\lim_{t\rightarrow\pm\infty}{i\over \sqrt{2\omega_{\vec{q}}}} \int d^3 \vec{x} \, (\varepsilon^{(\lambda),\mu})^* e^{-i q\cdot x} \overleftrightarrow{\partial_0} \hat{{\cal A}}_{\mu}(x)\, , \\
\hat{a}^{(\lambda)\dagger}_{\vec{q}} =&\lim_{t\rightarrow\pm\infty}{-i\over \sqrt{2\omega_{\vec{q}}}} \int d^3 \vec{x} \, \varepsilon^{(\lambda),\mu} e^{i q\cdot x} \overleftrightarrow{\partial_0} \hat{{\cal A}}_{\mu}(x)\, ,
\end{split}
\end{equation}
where $t \to \pm\infty$ for out- and in-states, respectively. In order to write these expressions in terms of CFT operators, we need an expression for the operator $\hat{\cal A}_{\mu}(x)$ in the asymptotic parts of the Minkowski scattering region located deep inside the bulk. In the scalar case we achieved this using HKLL reconstruction. The same can be done for gauge fields. In \cite{Heemskerk:2012np}, a representation of bulk gauge fields in the AdS Poincar\'e patch was derived. Here we are interested in results in global AdS. The reconstruction of a Maxwell field in global AdS is given in appendix \ref{app:MaxfieldReconstruction}, and to our knowledge has not been presented previously. We quote the main results here. As explained above, our gauge field with magnetic boundary conditions is dual to a current operator in the CFT. In this section however, we are only taking the radiative part of the field into account, in line with scattering theory. This means that the dual operator is not the full current, but only its radiative part, which we could denote by $j_{\mu}^{\text{R}}$. In order to avoid cluttering of notation, we will simply write $j_{\mu}$ in what follows. We thus look for a field operator with boundary behavior
\begin{equation}
{\cal A}_{\mu}(\rho,x)\xrightarrow[\rho \to \frac \pi 2]{} \cos\rho \, j_{\mu}(x)\, ,
\end{equation}
up to a normalization constant which depends on the normalization of the CFT current two point function.
It can be written in Lorenz gauge as
\begin{equation}
{\cal A}_{\mu}(\rho,x)={1\over \pi}\int_{\mathcal T}d\tau'\, \int d^2\Omega' \,
\left[
K^V_{\mu}(\rho,x;x') \epsilon_{\tau' a b}\partial^{a}j^{b} (x')
+
K^S_{\mu}(\rho,x;x') \partial_{a}j^{a}(x')
\right]+\text{h.c}
\, .
\end{equation}
Here, the Latin indices run over the spatial coordinates of the $S^2$. The explicit form of the kernels $K^V$ and $K^S$ can be found in appendix \ref{app:MaxfieldReconstruction}. The reconstruction region $\mathcal T$ is $\tau \in (-\pi, 0)$ for incoming and $\tau \in (0,\pi)$ for outgoing particles. We now take the flat limit by placing the gauge field in the Minkowski scattering region
\begin{equation}
\tau={t\over L}\, , \quad \text{and}\quad \rho={r\over L}\, , \quad \text{with}\quad L\rightarrow \infty\, .
\end{equation}
Like in the scalar case, finite energy states in Minkowski space correspond to large AdS quantum numbers. The reconstructed operator in the large $L$ limit becomes
\begin{equation}
\begin{split}
{\cal A}_{\mu}=& -{1\over 8 \pi} \int d^3 x'\, \int dk\, \sum_{l,m}{ Y_l^m(\Omega')^* \over{-l(l+1)}} e^{i k L \left({\pi \over 2}-\tau'\right)} e^{i k t}\\
& \left[ \epsilon_{\tau'}^{\,\, a b}\nabla_a j^+_b(x') \, \left(\vec{r}\times{d\over d\vec{r}}\right)+{i\over k} \nabla^a j^{+}_a(x') \vec{\nabla}\times\left( \vec{r}\times{d\over d\vec{r}}\right) \right] i^{-l} j_l(k r) Y_l^m(\hat{\Omega})\\
&+\text{h.c} \, .
\end{split}
\end{equation}
This formula now constructs a position space gauge field in the Minkowski scattering region. This result can then be inserted into equation \eqref{eq:aFromA} to obtain expressions for photon creation/annihilation operators. After a bit of fun algebra, we obtain
\begin{equation}
\begin{split}
\sqrt{2\omega_{\vec{q}}}\, \varepsilon^{(\lambda)}_{\mu} \, a_{\vec{q}}^{\dagger(\lambda)}=& - {1\over 8\omega_q} \int d^3x'\, e^{i k L \left( {\pi \over 2}-\tau' \right) } \left[ \epsilon_{\tau'}^{\,\, a b}\nabla_a j^+_b(x') \, X_{\mu}- \nabla^{a}j^{+}_a(x') Y_{\mu} \right] \, ,
\end{split}
\end{equation}
where we have defined the four-vectors
\begin{equation}\label{eq:XY}
\begin{split}
X=\left(\vec{q}\times{d\over d\vec{q}}\right) G(\Omega',\hat{q})\, , \quad \text{and}\quad
Y=&\left[\left( \vec{q}\times{d\over d\vec{q}}\right)G(\Omega',\hat{q})\right]\times {\vec{q}\over \omega_q} \, .
\end{split}
\end{equation}
Here, the distribution $G(\Omega',\hat{q})$ is the Green's function associated to the Laplace operator on the $S^2$. The result can be written in much simpler form by integrating by parts along the sphere. We will be using a complex parametrization of the coordinates on $S^2$, as explained in appendix \ref{app:coordS2}. In these coordinates, the direction of the momentum of the photon is given by $(z_q,\bar{z}_q)$. For out state creation operators, the final result is
\begin{equation}\label{eq:MagneticBCphotons}
\begin{split}
\sqrt{2\omega_{\vec{q}}} \, a_{\vec{q}}^{\dagger(-)}=& {-1\over 4\omega_q}{1+z_q \bar{z}_q\over{\sqrt{2}}} \int d\tau' \, e^{i \omega_{\vec{q}} L\left({\pi\over 2}-\tau' \right)}
\int d^2 z' {1\over (z_q-z')^2}j^+_{\bar{z}'}(\tau',z',\bar{z}')
\, , \\
\sqrt{2\omega_{\vec{q}}} \, a_{\vec{q}}^{\dagger(+)}=& {-1\over 4\omega_q}{1+z_q \bar{z}_q\over{\sqrt{2}}} \int d\tau' \, e^{i \omega_{\vec{q}} L\left({\pi\over 2}-\tau' \right)}
\int d^2 z' {1\over (\bar{z}_q-\bar{z}')^2}j^+_{z'}(\tau',z',\bar{z}')\, .
\end{split}
\end{equation}
The annihilation operators, as well as the operators acting on the in-state Fock space can again be obtained by using the prescription given below \eqref{eq:aMassive}. The factor $c$ that appeared in equation \eqref{eq:aMassive} is set to $1$, which fixes the normalization of the current in the CFT.
We conclude that asymptotic photon scattering states can be prepared in Minkowski space by smearing current operators over the sphere as in the formulas above. Like in the massless scalar case, the integrals appearing in \eqref{eq:MagneticBCphotons} are focused around a small window of global time at $\tau'={\pi \over 2}$. This again shows how this window acts like asymptotic null infinity of flat space that is pierced by photons in scattering experiments in flat space-time.
\subsection{Flat limit: Electric boundary conditions}\label{sec:FLE}
In the previous subsection we have argued that one can set up scattering events involving photons in flat space by considering the flat limit of the dual of a holographic conformal field theory and smearing conformal currents over the sphere. As explained in section \ref{sec:SL2Z}, we have made use of magnetic boundary conditions in AdS that forbid the existence of magnetic charges in the bulk. In this section we will consider boundary conditions that fix the electric field to vanish at the AdS boundary, such that electric charge is forbidden instead.
We thus look for a new dynamical bulk gauge field ${\cal V}_{\mu}$ with asymptotic behavior given by
\begin{equation}
{\cal V}_{\mu} \xrightarrow[\rho \rightarrow \frac \pi 2]{} A_{\mu}\, ,
\end{equation}
again up to normalization.
Such a bulk field can be reconstructed at the boundary using a similar calculation to the one used in the previous section. The explicit calculation can be found in appendix \ref{MaxfieldReconstructionElectric}. Taking a flat limit as described in the previous section yields the following expression for a gauge field in Minkowski space in terms of CFT operators
\begin{equation}
\begin{split}
{\cal V}_{\mu}=&{L \over 8} {1\over \pi} \int d^3 x'\, \int dk\, \sum_{l,m}{ Y_l^m(\hat{\Omega}')^* \over{-l(l+1)}} e^{i k L\left({\pi \over 2}-\tau'\right)} e^{i k t} i k\\
&\left[ \, \epsilon_{\tau'}^{\,\, a b}\nabla_a A^+_b(x') \, \left(\vec{r}\times{d\over d\vec{r}}\right)-{i \over k}\nabla^{a}A^{+}_a(x') \vec{\nabla}\times\left( \vec{r}\times{d\over d\vec{r}}\right) \right] i^{-l} j_l(k r) Y_l^m(\hat{\Omega}) \, \\
&+\text{h.c}\, .
\end{split}
\end{equation}
From this expression we can extract creation operators using formulas \eqref{eq:aFromA}, this time applied to the connection ${\cal V}_{\mu}$. The result reads
\begin{equation}
\begin{split}
\sqrt{2\omega_{\vec{q}}}\, \varepsilon^{(\lambda)}_{\mu} \, v_{\vec{q}}^{\dagger(\lambda)}=& - i{L\over 8} \int d^3x'\, e^{i k L \left( {\pi \over 2}-\tau' \right) } \left[ \epsilon_{\tau'}^{\,\, a b}\nabla_a A^+_b(x') \, X_{\mu}+\nabla^{a}A^{+}_a(x') Y_{\mu} \right] \, ,
\end{split}
\end{equation}
where $X$ and $Y$ were defined in formula \eqref{eq:XY} above. Integrating by parts along the coordinates of the sphere yields two dimensional delta functions on the $S^2$, leading to a very simple result\footnote{The reconstruction written here has been performed in terms of time derivatives of the operators $A_{\mu}$ for future convenience, but expressions in terms of the operators alone can also be written.}
\begin{equation}\label{eq:ElectricBCphotons}
\begin{split}
\sqrt{2\omega_{\vec{q}}} \, v_{\vec{q}}^{\dagger(-)}=&{1\over 4\omega_q}{1+z_q \bar{z}_q\over{\sqrt{2}}} \int d\tau' \, e^{i \omega_{\vec{q}} L\left({\pi\over 2}-\tau' \right)}
\partial_{\tau'}A^+_{z'}(\tau',z_q',\bar{z}_q')
\, , \\
\sqrt{2\omega_{\vec{q}}} \, v_{\vec{q}}^{\dagger(+)}=&{1\over 4\omega_q}{1+z_q \bar{z}_q\over{\sqrt{2}}} \int d\tau' \, e^{i \omega_{\vec{q}} L\left({\pi\over 2}-\tau' \right)}
\partial_{\tau'}A^+_{\bar{z}'}(\tau',z_q',\bar{z}_q')\, .
\end{split}
\end{equation}
The construction of the other operators follows as in the magnetic case.
We conclude that ``dual'' photons in flat space can be prepared in the conformal field theory by simply inserting operators $A_{\mu}$ in a window of global time at $\tau'={\pi \over 2}$ at a single point on the sphere specified by the angle of the photon's momentum $\vec{q}$.
\subsection{Coulombic source terms}\label{sec:Coulomb}
In this section we have reconstructed bulk operators associated with radiative modes of a gauge field and the operators appearing in formulas \eqref{eq:MagneticBCphotons} and \eqref{eq:ElectricBCphotons} capture just those contributions. This has been achieved by finding specific kernels obeying the homogeneous Maxwell equations. The bulk theory can however generally contain charged matter. For example, the scalar particles considered in the previous sections could be complexified and charged under the $U(1)$ symmetry. This implies the existence of sources in the Maxwell equations, so that we would have
\begin{equation}
\nabla_{\mu}\hat{F}^{\mu\nu}=\hat{J}^{\nu}\, ,
\end{equation}
where the current operator $\hat{J}^{\mu}$ can be constructed explicitly out of the scalar bulk operators. In the context of scattering theory, interaction of the electromagnetic fields with sources is ignored in the asymptotic regions. This is typically justified by arguing for asymptotic decoupling. This also justifies our construction. In the next subsection, however, we will demonstrate the conservation of the charges of asymptotic gauge transformations. In the original flat space argument \cite{He:2014cra,Kapec:2015ena} -- as well as in our argument -- the derivation of those charges relies on the knowledge of the correct Li\'enard-Wiechert fields. Those contribute to non-radiative gauge field modes in the bulk.
For the purpose of reconstructing asymptotic gauge fields, in addition to the radiative reconstruction presented previously, our gauge field now also has a Coulombic (C) term like
\begin{equation}\label{eq:ACoulomb}
\hat{A}^{(\text{C})}_{\mu}(X) = \int d^{d+1}Y \, G_{\mu,\nu}(X\vert Y) \hat{J}^{\nu}(Y)\, ,
\end{equation}
where $G_{\mu,\nu}$ is a bulk-to-bulk propagator in Lorenz gauge. Note that the expression above can be written explicitly in the CFT by using the scalar bulk operator reconstruction formulas derived in appendix \ref{app:ScalarHKLL}. Because the current operator $ \hat{J}^{\nu}$ is quadratic in the bulk scalar operator, formula \eqref{eq:ACoulomb} involves double trace operators constructed from the CFT operator dual to the bulk scalar field.
At the boundary, the Coulombic gauge field \eqref{eq:ACoulomb} results in an additional contribution to the boundary current
\begin{equation}
\lim_{\rho\rightarrow {\pi \over 2}} \hat{A}^{(\text{C})}_{\mu}(\rho, x) = \cos\rho \, j_{\mu}^{(\text{C})}(x)\, .
\end{equation}
Expectation values of the current operator $j_{\mu}^{(\text{C})}$ compute Li\'enard-Wiechert potentials at the boundary of Anti-de Sitter space-time. This part of the current is also responsible for non-trivial Ward-Identities in the conformal field theory, wich we will use in the next section to study soft theorems in Minkowski space. The conserved current operator dual to the asymptotic gauge field thus reads
\begin{equation}
j_{\mu}(x)=j^{\text{R}}_{\mu}(x)+j^{\text{C}}_{\mu}(x)\, ,
\end{equation}
where $j^{\text{R}}_{\mu}(x)$ is the radiative part of the current appearing in formula \eqref{eq:MagneticBCphotons}. A similar story can also be told for the dynamical gauge field playing a role in the reconstruction of bulk fields with electric boundary conditions.
\section{From conformal Ward identities to Weinberg soft theorems}\label{sec:WST}
In this note we have established a relation between operators in conformal theories and creation/annihilation operators in flat space-time. We can now ask what the implications of this connection are. The objective of this section is to show how Ward identities in a three dimensional conformal field theory are related to Weinberg soft theorems in Minkowski space-time in four dimensions. We start with a discussion on currents and Ward identities in conformal field theories with global $U(1)$ symmetry (and their S-transformations).
\subsection{Ward identities in CFT}
A bulk AdS$_4$ theory containing a $U(1)$ gauge field with ``magnetic" boundary conditions is equivalent to a conformal field theory with global $U(1)$ symmetry. The charge under which the operators ${\cal O}$ in the CFT transform has an associated conserved current operator $j_{\mu}$, which is dual to the bulk gauge field. As explained in section \ref{sec:Coulomb}, such a current has two pieces; One comes from the radiative part of the gauge field, while the other one captures Coulombic contributions. Even though the total current is transverse, time ordered expectation values are only conserved up to contact terms\footnote{
This is especially easy to see in the derivation of the Ward identity in the operator formalism.
Time ordered correlators can be defined as
\begin{equation}
\langle 0\vert T\{ j_{\mu}(x) {\cal O}(x_1)\cdots \}\vert 0 \rangle = \langle 0\vert j_{\mu}(x) {\cal O}(x_1)\cdots \vert 0\rangle \Theta(t>t_1>\cdots)+\text{other orderings}.
\end{equation}
The divergence of this correlator exhibits contact terms. When the divergence hits the current operator, the result is trivial as the current is conserved. However when the derivative hits the multi-argument step-function, terms involving the commutator $[j_0(x),{\cal O}(x_j)]$ appear. These are non-trivial due to the Coulombic part of the current $j^{(\text{C})}_{\mu}$ discussed in section \ref{sec:Coulomb}.
}, which is made explicit by the Ward identities.
\begin{equation}\label{eq:EQ1IBP}
\begin{split}
& \partial_{\mu} \langle 0 \vert T\{ j^{\mu}(x) {\cal O}(x_1)\cdots {\cal O}(x_n)\bar{ {\cal O}}(y_1)\cdots \bar{ {\cal O}}(y_m) \} \vert 0 \rangle \\
&= \left(\sum_{i=1}^n q_i \delta^{(3)} (x-x_i)- \sum_{j=1}^m q_j \delta^{(3)} (x-y_i)\right)
\langle 0 \vert T\{ {\cal O}(x_1)\cdots {\cal O}(x_n)\bar{ {\cal O}}(y_1)\cdots \bar{ {\cal O}}(y_m) \} \vert 0 \rangle
\, .
\end{split}
\end{equation}
This formula will be used later to make a statement about the asymptotic symmetries of QED in flat space-time.
Consider now allowing for a non-vanishing magnetic field at the boundary. This way, magnetic monopoles can be introduced into the theories we are working with \cite{Sachdev:2012tj,Pufu:2013eda}. They consist of background connections $A_{\mu}$ that cannot be well-defined globally. From the background connection we can define a background field strength and its Hodge dual
\begin{equation}
f_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} \, \quad \text{and}\quad (*f)^{\mu}={1\over 2}\epsilon^{\mu\nu\rho}f_{\nu\rho}\, .
\end{equation}
Magnetic monopoles are then background connections violating the Bianchi identity
\begin{equation}
\partial_{\mu}(*f)^{\mu}=\chi \, ,
\end{equation}
where $\chi$ is a source (see \cite{MouraMelo:2000zc,Abreu:2001dq} for an analysis on the space of solutions). One can think of these monopoles as providing non-trivial charge to the topological current $(*f)^{\mu}$.
After $S$-transformation of the CFT discussed in section \ref{sec:SL2Z}, the field $A_{\mu}$ is dynamical, and $(*f)^{\mu}$ is the conserved current of the theory. The breaking of the Bianchi identity can then be written as a Ward identity associated to $(*f)^{\mu}$,
\begin{equation}\label{eq:EQ2}
\begin{split}
& \partial_{\mu}\langle 0 \vert T\{ (*f)^{\mu}(x) {\cal M}(x_1)\cdots {\cal M}(x_n)\bar{\cal M}(y_1)\cdots \bar{\cal M}(y_m) \} \vert 0 \rangle \\
&=\left(\sum_{i=1}^n g_i \delta^{(3)} (x-x_i)- \sum_{j=1}^m g_j \delta^{(3)} (x-y_i)\right)
\langle 0 \vert T\{ {\cal M}(x_1)\cdots {\cal M}(x_n)\bar{\cal M}(y_1)\cdots \bar{\cal M}(y_m) \} \vert 0 \rangle
\, ,
\end{split}
\end{equation}
where $g_i$ are magnetic monopole charges, and ${\cal M}$ are magnetic monopole operators that cannot be written locally in terms of the fields participating in the original CFT. Below, we will use the magnetic version of the Ward identity to derive the magnetic asymptotic symmetries of QED in flat space-time in the presence of magnetic monopoles.
\subsection{Weinberg soft theorems as symmetries of the S-matrix}
We turn now to a brief discussion on Weinberg soft theorems in Minkowski space-time.
Weinberg soft theorems can be seen to arise as a consequence of the asymptotic symmetries of QED. In this section we briefly review this statement. We follow closely \cite{Strominger:2015bla}. For more details, see \cite{Strominger:2017zoo,Kapec:2017tkm,Campiglia:2015qka}.
An infinite set of conserved charges can be constructed in Minkowski space-time. To see this, we start with the Maxwell equations in the presence of electric and magnetic charges,
\begin{equation}
d*F=*j_E\, , \quad \text{and}\quad dF=*j_M\, .
\end{equation}
The one-forms $j_E$ and $j_M$ are conserved in the sense that
\begin{equation}
d*j_E=d^2*F=0\, , \quad \text{and}\quad d*j_M=d^2F=0\, .
\end{equation}
More generally, we can define the following currents associated to any function $\varepsilon(x)$, which are also conserved
\begin{equation}
*j^{\varepsilon}_E = d(\varepsilon *F)\, , \quad \text{and}\quad *j^{\varepsilon}_M = d(\varepsilon F)\, .
\end{equation}
The charges associated to these currents read
\begin{equation}\label{eq:charges}
\begin{split}
Q_E^{\varepsilon}(\Sigma)&=\int_{\Sigma} *j^{\varepsilon}_E =\int_{\Sigma} \left( d\varepsilon \wedge *F+\varepsilon *j_E \right)\, , \\
Q_M^{\varepsilon}(\Sigma)&=\int_{\Sigma} *j^{\varepsilon}_M =\int_{\Sigma} \left( d\varepsilon \wedge F+\varepsilon *j_M \right)\, .
\end{split}
\end{equation}
The second terms in these expressions are weighted integrals of the electric/magnetic currents, and they will be refered to as the ``hard" terms. The first terms are linear in the electro-magnetic field and create photons. Conservation implies that the S-matrix operator must commute with these charges. We thus write
\begin{equation}\label{eq:QScom}
\langle \text{out} \vert Q_{E/M}^{\varepsilon}(\Sigma_+) {\cal S}-{\cal S} Q_{E/M}^{\varepsilon}(\Sigma_-) \vert \text{in}\rangle =0\, ,
\end{equation}
where $\Sigma_{\pm}$ are late and early time Cauchy slices. Together with a few extra assumptions, this can be shown to be equivalent to Weinberg soft theorems. We will now discuss this equivalence briefly.
For simplicity, we will focus on a theory with electrically charged particles. Of particular interest is the conservation law associated to parameters $\varepsilon(\hat{x})$ that only depend on the angular coordinates. In this case, the photonic term in \eqref{eq:charges} involves photons with vanishing frequency, which are known as ``soft" photons. The soft part of the charges can be written as
\begin{equation}\label{eq:SoftQ}
\begin{split}
Q_E^{\text{soft}}(\Sigma_+) &= {-1\over 8\pi^2} \lim_{\omega_{\vec{q}}\rightarrow 0} \omega_{\vec{q}} \int d^2z \, \varepsilon(z,\bar{z}) \left[ \partial_{\bar{z}}\left( {\sqrt{2}\over 1+z\bar{z}} \sqrt{2\omega_{\vec{q}}}\, \hat{a}^{(-)}_{\vec{q}}\right) + \partial_{z}\left( {\sqrt{2}\over 1+z\bar{z}} \sqrt{2\omega_{\vec{q}}}\, \hat{a}^{(+)}_{\vec{q}}\right) \right]\, ,
\end{split}
\end{equation}
where the creation operators appearing here involve a soft momentum vector $\vec{q}$ with magnitude $\omega_{\vec{q}}\rightarrow 0$ pointing in the direction specified by $z,\bar{z}$. The charge at $\Sigma_-$ can be related to the one at $\Sigma_+$ through ${\cal CPT}$ invariance when inserted inside the S-matrix. We have
\begin{equation}
\langle \text{out} \vert Q_{E/M}^{\text{soft}}(\Sigma_+) {\cal S}\vert \text{in}\rangle =- \langle \text{out} \vert {\cal S} Q_{E/M}^{\text{soft}}(\Sigma_-) \vert \text{in}\rangle \, .
\end{equation}
This allows us to write the soft part of formula \eqref{eq:QScom} as a scattering amplitude involving charges only acting on outgoing states.
The hard part of the charges can also be written explicitly. We will consider currents associated to point-like charged particles, such that
\begin{equation}
j_E^{\mu}(x)=\sum_i q_i \int ds \,p^{\mu}_i \delta^{(4)}(x-X_i(s))\, ,
\end{equation}
where $i$ labels each particle, and $X_i(s)$ is the trajectory. Using this expression we can evaluate the hard part of the charge and obtain
\begin{equation}\label{eq:HardQ}
\begin{split}
Q_E^{\text{hard}}(\Sigma_+)&=\lim_{t\rightarrow \infty} \sum_i q_i \,\varepsilon\left( x \right)\vert_{r={\vert \vec{p}\vert \over p^0} t, \hat{x}=\hat{p}_i }\, ,
\end{split}
\end{equation}
and similarly for the early time Cauchy slice. Using formulas \eqref{eq:HardQ} and \eqref{eq:SoftQ} and plugging them in equation \eqref{eq:QScom} yields a formula relating scattering amplitudes involving soft photons with amplitudes involving exclusively hard particles. This is however, still not equivalent to Weinberg's soft theorem.
A further assumption on the Maxwell radiative data must now be made that makes the two terms appearing in formulas \eqref{eq:SoftQ} equal to each other. We need to require that magnetic monopoles are absent, i.e.
\begin{equation}\label{eq:assumption}
F_{z\bar{z}}\vert_{{\cal I}^+_-}=F_{z\bar{z}}\vert_{{\cal I}^-_+} = 0\, .
\end{equation}
Lastly, a particular choice of function $\varepsilon(x)$ yields Weinberg soft theorems. Focusing for simplicity on massless charged hard particles, the choices (one for each polarization)
\begin{equation}\label{eq:varepsilonchoice}
\varepsilon(x)={1\over z-z'} \, , \quad \text{and}\quad \varepsilon(x)={1\over \bar{z}-\bar{z}'} \, ,
\end{equation}
yield
\begin{equation}\label{eq:WST}
\langle \text{out}\vert \sqrt{2\omega_{\vec{q}}} \, a^{(\pm)}_{\vec{q}} {\cal S} \vert \text{in}\rangle
\sim \left[ \sum_i q_i {p_i\cdot \epsilon^{\pm}\over p_i\cdot q} -\sum_j q_j {p_j\cdot \epsilon^{\pm}\over p_j\cdot q} \right] \langle \text{out}\vert {\cal S}\vert \text{in}\rangle\, .
\end{equation}
This result is well known as Weinberg's soft photon theorem \cite{Weinberg:1965nx}, and it relates a scattering event involving a soft photon to a scattering event without it.
Even though the results presented here involve only massless hard particles, it is possible to obtain soft theorems involving massive matter by implementing a different choice of parameter $\varepsilon(x)$. This was explored in detail in \cite{Campiglia:2015qka,Kapec:2017tkm}. In this case, the parameter reads
\begin{equation}\label{eq:MinkepChoice}
\varepsilon(x)=\int d^2\hat{x}'\, {1\over 4\pi }{ { t^2-r^2 } \over {\left( t-r\hat{x}\cdot\hat{x}' \right)^2 } } \varepsilon(\hat{x}')\, .
\end{equation}
where $\varepsilon(\hat{x}')$ are the same as in equation \eqref{eq:varepsilonchoice}. Under this choice, the second terms in the expression for the charges \eqref{eq:charges} now yield the hard terms in Weinberg soft theorems. The soft photon contributions are harder to deal with, but they can be shown to correspond to formulas \eqref{eq:SoftQ}. The final result is the same as \eqref{eq:WST} above, except that the momenta $p_i$ are also allowed to be time-like.
An alternative way of looking at the results presented here is that Weinberg soft theorems imply the symmetries of the S-matrix as written in \eqref{eq:QScom}. However, in order to derive the soft theorem from the statement about symmetries, one needs to make some extra assumptions like \eqref{eq:assumption}. In the following section we will show how the map between scattering states and conformal operators discussed in section \ref{sec:GaugeFields}, together with conformal Ward identities implies the symmetries \eqref{eq:QScom}.
\subsection{From CFT physics to soft theorems}\label{sec:WardSoftE}
We now would like to understand how the physics described in the previous subsection arise from a statement in conformal field theories. Our first step is to construct a soft photon scattering state as a CFT operator. We have discussed how to create photon scattering states in section \ref{sec:GaugeFields}. We are now taking the soft limit $\omega_{\vec{q}}\rightarrow 0$. We thus write
\begin{equation}
\begin{split}
\lim_{\omega_{\vec{q}}\rightarrow 0}\sqrt{2\omega_{\vec{q}}} \, a_{\vec{q}}^{(-)} =&\lim_{\omega_{\vec{q}}\rightarrow 0}\lim_{L \to \infty}
{-1\over 4\omega_q}
{1+z_q \bar{z}_q\over{\sqrt{2}}}
\int d\tau' \, e^{-i \omega_{\vec{q}}L\left( {\pi \over 2}-\tau' \right)}
\int d^2 z' {1\over z'-z_q} \partial_{z'} j^-_{\bar{z}'}(\tau',z',\bar{z}') \, ,\\
\lim_{\omega_{\vec{q}}\rightarrow 0}\sqrt{2\omega_{\vec{q}}} \, a_{\vec{q}}^{(+)} =&\lim_{\omega_{\vec{q}}\rightarrow 0}\lim_{L \to \infty}
{-1\over 4\omega_q}
{1+z_q \bar{z}_q\over{\sqrt{2}}}
\int d\tau' \, e^{-i \omega_{\vec{q}}L\left( {\pi \over 2}-\tau' \right)}
\int d^2 z' {1\over \bar{z}'-\bar{z}_q} \partial_{\bar{z}'} j^-_{z'}(\tau',z',\bar{z}') \, .
\end{split}
\end{equation}
Note hat the large $L$ limit is taken before the soft limit. {In that limit the integrals over global time are dominated by operators evaluated at a window of size ${\cal O}(1/L)$ around $\tau'=\pi/2$. This follows from the same arguments presented in section \ref{sec:ScatteringStatesScalar} for scalar operators. The large phase pre-factor acquires a finite phase in this window of global time. The soft limit is implemented afterwards, setting that finite phase to zero. This sets the phase in the global time integral to one.} We can thus simply integrate the current insertions over the small windows of global time at $\tau=\pm{\pi\over 2}$ which we denote $\tilde{\cal I}^{\pm}$, and obtain
\begin{equation}
\begin{split}
\lim_{\omega_{\vec{q}}\rightarrow 0}\sqrt{2\omega_{\vec{q}}} \, a_{\vec{q}}^{(-)} =&\lim_{\omega_{\vec{q}}\rightarrow 0}
{-1\over 4\omega_q}
{1+z_q \bar{z}_q\over{\sqrt{2}}}
\int_{\tilde{\cal I}^{\pm}}d^3 x' {1\over z'-z_q} D^{\bar{z}'} j^-_{\bar{z}'}(\tau',z',\bar{z}') \, ,
\end{split}
\end{equation}
and similarly for annihilation operators. Scattering amplitudes involving soft photon modes are then simply CFT correlators involving insertions of the conserved current. These are constrained by the CFT Ward identities written above in formula \eqref{eq:EQ1IBP}, which we repeat here
\begin{equation}\label{eq:EQ1v2}
\begin{split}
&\int d^3 x\,\alpha(x) \partial_{\mu} \langle 0 \vert T\{ j^{\mu}(x) X \} \vert 0 \rangle =\left(\sum_{i=1}^n q_i \alpha (x_i)- \sum_{j=1}^m q_j \alpha (y_j) \right)
\langle 0 \vert T\{ X \} \vert 0 \rangle
\, .
\end{split}
\end{equation}
{Here, we integrate the Ward identity over a specific region of the conformal theory. This region consists of the AdS boundary in Lorentzian signature between the past of $\tilde{\cal I}^-$ to the future of $\tilde{\cal I}^+$, together with two Euclidean caps $\partial{\cal M}_{\pm}$, resulting in a pill-shaped region like the one shown in figure \ref{fig:Interpretation}.} In equation \eqref{eq:EQ1v2}, $X$ stands for all insertions of primary operators charged under the $U(1)$ current. Explicitly, we have
\begin{equation}
X=\prod_i {\cal O}_i(x_i)\prod_j \bar{\cal O}_i(y_j)\, .
\end{equation}
By smearing the locations of the primary operators we can turn them into creation/annihilation operators involving hard particles in Minkowski space, as shown in sections \ref{sec:MasslessRec} and \ref{sec:MassiveRec}. In the massless case, operators will be light and be placed in the regions $\tilde{\cal I}^{\pm}$, while in the massive case, the operators will be heavy and will live on the Euclidean half-spheres $\partial{\cal M}_{\pm}$. There is a particular choice of parameter $\alpha(x)$ that turns the right hand side of \eqref{eq:EQ1v2} into the hard terms in Weinberg's soft theorem. Namely,
\begin{equation}\label{eq:alphaCFT}
\alpha(x)=\lim_{\rho\rightarrow{\pi \over 2}}\int d^2 \hat{x}' \, {1\over 4\pi}{{\cos^2\rho-\cos^2\tau }\over{\left( \sin\tau -\sin\rho\, \hat{x}\cdot\hat{x}' \right)^2}} \varepsilon(\hat{x}')\, .
\end{equation}
This formula is derived in appendix \ref{app:alpha}, and it obeys the property that at $\tilde{\cal I}^{\pm}$ there is no time dependence, such that
\begin{equation}
\alpha(x)\vert_{\tilde{\cal I}^{\pm}}=\varepsilon(\hat{x})\, .
\end{equation}
This choice of parameter $\alpha(x)$, together with the choices \eqref{eq:varepsilonchoice} for $\varepsilon(\hat{x})$ turn the right hand side of \eqref{eq:EQ1v2} into the hard part of Weinberg's soft theorem. This includes both massless and massive particles. We now need to show that an appropriate smearing of the left hand side of \eqref{eq:EQ1v2} corresponds to the Minkowski soft charges discussed in the previous section. The integral over the boundary receives contributions only from $\tilde{\cal I}^{\pm}$ and the Euclidean half-spheres $\partial{\cal M}_{\pm}$. All the primary operators in the correlator are located strictly on these regions. Integrating by parts, and using the fact that $\alpha(x)$ is time independent for the contributions at $\tilde{\cal I}^{\pm}$ , we can write
\begin{equation}\label{eq:ArgumentsHolographic}
\begin{split}
\int d^3 x\, \partial_{\mu}\alpha(x) \langle 0 \vert T\{ j^{\mu}(x) X \} \vert 0 \rangle =\int_{\partial{\cal M}_{\pm}} d^3 x\, \partial_{\mu}\alpha(x) \, \langle 0 \vert T\{ j^{\mu}(x) X \} \vert 0 \rangle \\
- \int_{\tilde{\cal I}^{\pm}} d^3 x\,\varepsilon(\hat{x}) \left[ D^z \langle 0 \vert T\{ j_z(x) X \} \vert 0 \rangle + D^{\bar{z}} \langle 0 \vert T\{ j_{\bar{z}}(x) X \} \vert 0 \rangle \right]\, .
\end{split}
\end{equation}
The terms in the integral over Euclidean regions $\partial{\cal M}_{\pm}$ involve only the Coulombic part of the current and can be shown to vanish for the choice of $\alpha(x)$ written above in equation \eqref{eq:alphaCFT}\footnote{
This could be checked explicitly in the CFT using the formula for the Coulombic part of the current written in section \ref{sec:Coulomb}. Such a check is actually non-trivial, as it involves the computation of multi-point correlators involving the double trace operators appearing in the definition of $j_{\mu}^{\text{C}}$. The argument can also be made simpler in the bulk by computing AdS Li\'enard-Wiechert potentials, which compute expectation values of the Coulombic part of the current when evaluated at the boundary of AdS. For a sample calculation, see appendix \ref{app:LWAdS}.
}. The second line can also be shown to receive only contributions from the radiative part of the current. We thus conclude that the right hand side of the Ward identity reads
\begin{equation}\label{eq:auxSP1}
\begin{split}
\int d^3 x\, \alpha(x) \partial_{\mu}\langle 0 \vert T\{ j^{\mu}(x) X \} \vert 0 \rangle &=\\
\int_{\tilde{\cal I}^{\pm}} d^3 x\,\varepsilon(\hat{x}) &\left[ D^z \langle 0 \vert T\{ j^{(\text{R})}_z(x) X \} \vert 0 \rangle + D^{\bar{z}} \langle 0 \vert T\{ j^{(\text{R})}_{\bar{z}}(x) X \} \vert 0 \rangle \right]\, .
\end{split}
\end{equation}
The current operators appearing here correspond to the following soft photon operators (ignoring the radiative label)
\begin{equation}\label{eq:auxSP2}
\begin{split}
\int_{\tilde{\cal I}^{+}} d^3x \, \varepsilon(\hat{x})
D^{\bar{z}} j^-_{\bar{z}} =& \lim_{\omega_{\vec{q}}\rightarrow 0}{2\over \pi}\omega_{\vec{q}} \, \int d^2z_q \, \varepsilon(\hat{q})
\, \partial_{\bar{z}_q}\left( {\sqrt{2}\over 1+z\bar{z}} \sqrt{2\omega_{\vec{q}}}\, a^{(-)}_{\vec{q}} \right) \, , \\
\int_{\tilde{\cal I}^{+}} d^3x \, \varepsilon(\hat{x})
D^{z} j^-_{z} =&\lim_{\omega_{\vec{q}}\rightarrow 0}{2\over \pi}\omega_{\vec{q}}\, \int d^2z_q \, \varepsilon(\hat{q}) \,
\partial_{z_q}\left( {\sqrt{2}\over 1+z\bar{z}} \sqrt{2\omega_{\vec{q}}}\, a^{(+)}_{\vec{q}} \right) \, .
\end{split}
\end{equation}
This implies that the left hand side of the Ward identity is precisely the soft part of the charge in Minkowski space at ${\cal I}^+_-$, which was written explicitly in equation \eqref{eq:SoftQ}. Here, we have only written explicitly the contributions from $\tilde{\cal I}^+$, but similar expressions can be written for $\tilde{\cal I}^-$, which yield the soft charges at ${\cal I}^-_+$. { Note that equation \eqref{eq:auxSP1} contains terms which involve both positive and negative frequency modes. Equation \eqref{eq:auxSP2} only considers the contribution to $\tilde{\cal I}^+$. Since only a single soft photon is present in \eqref{eq:auxSP1}, its positive frequency contribution turns into a creation operator which annihilates the vaccum-bra vector. Thus only the negative frequency part contributes. An analogous story ensures that at $\tilde{\cal I}^-$ only the positive frequency modes contribute. }
To put all the pieces together, we have concluded that the left hand side of the Ward identity \eqref{eq:EQ1v2} turns into the insertions of the soft charge operators into the S-matrix. We have also shown that the right hand side of the Ward identity turns into the insertions of the hard charge operators. This concludes the proof that conformal Ward identities in CFT$_3$ imply the invariance of the Minkowski S-matrix under the charges \eqref{eq:charges}. One can further proceed as in the previous section to show this implies Weinberg's soft theorems.
\subsection{From CFT physics to magnetic soft theorems }\label{sec:MST}
In the previous section, we have analyzed how the Ward identity in a conformal field theory dual to an AdS theory with the standard boundary conditions leads to the invariance of the S-matrix under the electric charges \eqref{eq:charges}. Here, we analyze AdS theories with alternative boundary conditions.
In section \ref{sec:FLE} (and appendix \ref{MaxfieldReconstructionElectric}), we have discussed the reconstruction of photon scattering states as operators in a conformal field theory dual to an AdS theory with ``electric'' boundary conditions. As a reminder, such conditions fix the electric field at the boundary, but allow the magnetic field created by magnetic monopoles in the bulk theory. The result of the analysis is the creation/annihilation operators in equation \eqref{eq:ElectricBCphotons}, which we repeat here.
\begin{equation}
\begin{split}
\sqrt{2\omega_{\vec{q}}} \, v_{\vec{q}}^{(-)}=&{1\over 4\omega_q}{1+z_q \bar{z}_q\over{\sqrt{2}}} \int d\tau' \, e^{-i \omega_{\vec{q}} L\left({\pi\over 2}-\tau' \right)}
\partial_{\tau'}A^-_{z'}(\tau',z_q',\bar{z}_q')
\, , \\
\sqrt{2\omega_{\vec{q}}} \, v_{\vec{q}}^{(+)}=&{1\over 4\omega_q}{1+z_q \bar{z}_q\over{\sqrt{2}}} \int d\tau' \, e^{-i \omega_{\vec{q}} L\left({\pi\over 2}-\tau' \right)}
\partial_{\tau'}A^-_{\bar{z}'}(\tau',z_q',\bar{z}_q')\, .
\end{split}
\end{equation}
In this section, we will be interested in the soft limit of these operators, such that $\omega_{\vec{q}}\rightarrow 0$. Like in the previous section, the soft limit is taken after the large $L$ limit, so the CFT operators are always integrated over the CFT regions $\tilde{\cal I}^{\pm}$. The CFT operators $A_{\mu}$ are dynamical gauge fields, whose dual field strength obeys a broken Bianchi identity. This was discussed in equation \eqref{eq:EQ2}, which we repeat here
\begin{equation}\label{eq:EQ2Repeat}
\begin{split}
&\partial_{\mu} \langle 0 \vert T\{ (*f)^{\mu}(x) X \} \vert 0 \rangle =\left(\sum_{i=1}^n g_i \delta^{(3)}(x-x_i)-\sum_{j=1}^m g_j \delta^{(3)}(x-y_j) \right)
\langle 0 \vert T\{ X \} \vert 0 \rangle
\, .
\end{split}
\end{equation}
Here, $X$ stands for insertions of CFT operators with magnetic charge. We will follow a very similar strategy as in the previous section. We first integrate the Ward identity over the locations of the operators in X such that they become asymptotic creation/annihilation operators in Minkowski space. Choosing $\alpha(x)$ as in formula \eqref{eq:alphaCFT} makes the right hand side of the Bianchi identity \eqref{eq:EQ2} turn into the hard part of Weinberg's soft theorem, but with magnetic charges instead of electric ones.
The treatment of the left hand side of \eqref{eq:EQ2Repeat} proceeds the same way as in the previous section, which leads to
\begin{equation}
\begin{split}
\int d^3 x\, &\alpha(x)\, \partial_{\mu} \langle 0 \vert T\{ (*f)^{\mu}(x) X \} \vert 0 \rangle = \\
& \int_{\tilde{\cal I}^{\pm}} d^3 x\,\varepsilon(\hat{x}) \left[ D^z \langle 0 \vert T\{ (*f)^{(\text{R})}_z(x) X \} \vert 0 \rangle + D^{\bar{z}} \langle 0 \vert T\{ (*f)^{(\text{R})}_{\bar{z}}(x) X \} \vert 0 \rangle \right]\, .
\end{split}
\end{equation}
Where again, only the radiative part of the current appears. This can be written explicitly in terms of the CFT gauge field operator $A^{(\text{R})}_{\mu}$. Dropping the radiative label, we have
\begin{equation}
\begin{split}
\int d^3 x\, &\alpha(x)\, \partial_{\mu} \langle 0 \vert T\{ (*f)^{\mu}(x) X \} \vert 0 \rangle = \\
&2 i \int_{\tilde{\cal I}^{\pm}} d\tau d^2z \,\varepsilon(\hat{x}) \left[ \partial_{\bar{z}} \langle 0 \vert T\{ \partial_{\tau}A_z(x) X \} \vert 0 \rangle -\partial_z \langle 0 \vert T\{ \partial_{\tau}A_{\bar{z}}(x) X \} \vert 0 \rangle \right]\, .
\end{split}
\end{equation}
The operators appearing in this expression can be written in terms of soft photon creation operators as follows. For the term involving $\tilde{\cal I}^+$, we have
\begin{equation}
\begin{split}
\int_{\tilde{\cal I}^+} d\tau d^2z \, \varepsilon(\hat{x})\,
\partial_{\bar{z}} \partial_{\tau}A^-_{z}(\tau,z,\bar{z})=&
\lim_{\omega_{\vec{q}}\rightarrow 0}
4\omega_{\vec{q}}
\int d^2 z_q \, \varepsilon(\hat{q})
\partial_{\bar{z}_q}\left( {{\sqrt{2}}\over{1+z \bar{z}}}
\sqrt{2\omega_{\vec{q}}} \, v_{\vec{q}}^{(-)} \right)
\, , \\
\int_{\tilde{\cal I}^+} d\tau d^2z \, \varepsilon(\hat{x})\,
\partial_{z} \partial_{\tau}A^-_{\bar{z}}(\tau,z,\bar{z})=&
\lim_{\omega_{\vec{q}}\rightarrow 0}
4\omega_{\vec{q}}
\int d^2 z_q \, \varepsilon(\hat{q})
\partial_{z_q}\left( {{\sqrt{2}}\over{1+z \bar{z}}}
\sqrt{2\omega_{\vec{q}}} \, v_{\vec{q}}^{(+)} \right)
\, .
\end{split}
\end{equation}
This implies
\begin{equation}
\begin{split}
- 2 i \int_{\tilde{\cal I}^{+}} d\tau d^2z \,\varepsilon(\hat{x}) \left[ \partial_{\bar{z}} \langle 0 \vert T\{ \partial_{\tau}A_z(x) X \} \vert 0 \rangle -\partial_z \langle 0 \vert T\{ \partial_{\tau}A_{\bar{z}}(x) X \} \vert 0 \rangle \right] = \langle 0 \vert T\{ Q_M^{\text{soft}}(\Sigma_+) X \} \vert 0 \rangle \, .
\end{split}
\end{equation}
where we have defined the operator
\begin{equation}
Q_M^{\text{soft}} (\Sigma_+) = - 8 i \lim_{\omega_{\vec{q}}\rightarrow 0} \omega_{\vec{q}}
\int d^2 z_q \, \varepsilon(\hat{q})
\left[
\partial_{\bar{z}_q}\left( {{\sqrt{2}}\over{1+z \bar{z}}}
\sqrt{2\omega_{\vec{q}}} \, v_{\vec{q}}^{(-)} \right)
-
\partial_{z_q}\left( {{\sqrt{2}}\over{1+z \bar{z}}}
\sqrt{2\omega_{\vec{q}}} \, v_{\vec{q}}^{(+)} \right)
\right] \, .
\end{equation}
similar expressions can be written for the terms coming from the CFT region $\tilde{\cal I}^-$, and they involve an operator $Q_M^{\text{soft}} (\Sigma_-)$ constructed out of creation and annihilation operators in the asymptotic past. These are precisely expressions for magnetic soft charges in Minkowski space. With this, we conclude that the broken Bianchi identity \eqref{eq:EQ2Repeat} implies the asymptotic magnetic symmetries of a flat space theory with magnetic monopoles, upon the implementation of the map between creation/annihilation operators in Minkowski space and conformal operators written in \eqref{eq:ElectricBCphotons}.
\section{Scattering states revisited: Flat limit of AdS/CFT}\label{sec:ScatteringStatesScalar}
In this section we reconstruct scattering amplitudes of a theory in asymptotically flat space-times from the correlators of a conformal field theory in one lower dimension. This task has been the subject of much work in the literature \cite{Giddings:1999jq,Gary:2009mi,Giddings:1999qu,Balasubramanian:1999ri,Penedones:2010ue,Hijano:2019qmi}. Here we will revisit and re-phrase the argument in the language of path integrals, which are useful when discussing holography. We will focus on the case of bulk scalar fields, leaving the discussion of gauge theories for section \ref{sec:GaugeFields}.
\subsection{Scattering amplitudes in flat space-time}
S-matrix elements of a quantum field theory in asymptotically flat space-time can be written as
\begin{equation}\label{eq:SPI}
\langle \text{out} \vert \text{in} \rangle = \sum_{\phi_{\text{a/b}}} \Psi^*_\text{out}[\phi_\text{b}] \int\displaylimits_{{\phi}\vert_{\Sigma_{-}}={\phi_\text{a}}}^{{\phi}\vert_{\Sigma_{+}}={\phi}_{\text{b}}} {\cal D}{\phi}\, e^{i S[{\phi}]} \Psi_\text{in}[\phi_\text{a}]\, ,
\end{equation}
where ${\phi}$ stands for the fields of the theory with action $S[{\phi}]$, and the wave functionals $\Psi_\text{in}[\phi_\text{a}], \Psi^*_\text{out}[\phi_\text{b}]$ are given by overlaps of scattering states $| \text{out} \rangle, | \text{in} \rangle $ with field eigenstates $| \phi_\text{b,a} \rangle $ at very early and late Cauchy slices, which we will call $\Sigma_{\pm}$. Figure \ref{fig:Mink} gives a pictorial representation of formula \eqref{eq:SPI}.
\begin{figure}[]
\centering
\begin{tikzpicture}[scale=3]
\draw [white, thick] (-1,-1)--(1,1);
\draw [white, thick] (-1,1)--(1,-1);
\draw [black, thick] (-1,0)--(0,1);
\draw [black, thick] (1,0)--(0,1) node[pos=0.5,right=3]{${\cal I}^+$};
\draw [black, thick] (-1,0)--(0,-1);
\draw [black, thick] (1,0)--(0,-1) node[pos=0.5,right=3]{${\cal I}^-$};
\draw [red, very thick] (-1,0) to[out=30,in=180] (0,0.7) node [pos=1,above=9]{\,\,\scalebox{.4}{$\Sigma_+$}} node [pos=1,above=12,left=-1]{\,\,\scalebox{.4}{$| \text{out} \rangle$}};
\draw [red, very thick] (1,0) to[out=150,in=0] (0,0.7);
\draw [blue, very thick] (-1,0) to[out=-30,in=180] (0,-0.7) node [pos=1,below=9]{\,\,\scalebox{.4}{ $\Sigma_-$}} node [pos=1,below=10,right=1]{\,\,\scalebox{.4}{$| \text{in} \rangle$}};
\draw [blue, very thick] (1,0) to[out=-150,in=0] (0,-0.7);
\draw [gray, thick] (-1,0) to[out=0,in=180] (-0.5,0.1);
\draw [gray, thick] (0,0) to[out=160,in=0] (-0.5,0.1);
\draw [gray, thick] (0,0) to[out=-20,in=180] (0.5,-0.1);
\draw [gray, thick] (1,0) to[out=180,in=0] (0.5,-0.1);
\draw [black, thick,->] (-0.5,-0.3)--(-0.5,-0.1) node[pos=0.5,right]{$t$};
\end{tikzpicture}
\caption{Penrose diagram of flat space. Boundary conditions for bulk field operators are imposed at the early (late) time Cauchy slices $\Sigma_{\pm}$. The region of space-time before $\Sigma_-$ or after $\Sigma_+$ is characterized by an asymptotic Hamiltonian $H_{\text{as}}$ that we will consider to be free.
}\label{fig:Mink}
\end{figure}
In this paper, we consider standard Fock space scattering states $\vert\text{in}\rangle$ and $\langle \text{out}\vert$. The objective of scattering theory is to compute the overlap of two different such states. Ideally, one would like to consider scattering states which are states of the full Hamiltonian $H$ in the full Hilbert space ${\cal H}$ which asymptotically describe well separated particles. However, in interacting theories such states are usually unavailable and one is forced to simplify the problem by considering reasonable approximations to the full Hamiltonian at very early or very late times. Particles at those times are very far away from each other and can be understood as non-interacting, which is also referred to as \emph{asymptotically decoupled}. The asymptotic Hamiltonian $H_{\text{as}}$ approximating the dynamics in this case is free, and the asymptotic Hilbert space ${\cal H}_{as}$ takes the form of a Fock space.\footnote{To be more precise, one needs to identify vectors in the full Hilbert space that describe the states in the Fock space asymptotically. This amount to the construction of isometries implemented by M\o ller operators $\Omega_{\pm}$ which map states in ${\cal H}$ to states in ${\cal H}_{as}\subset {\cal H}$. The S-matrix in this language would then be $S=\Omega_-\Omega^{\dagger}_+$. For a more detailed discussion, see \cite{Buchholz:2005sa} and references therein.} \cite{Peskin, Weinberg}
In operator language, an incoming scattering state describing a single scalar particle with four-momentum $p$ corresponds to a field profile given by the following operator valued distribution
\begin{equation}\label{eq:profile1P}
\lim_{t\rightarrow -\infty}{\phi}(t,\vec p) \sim {\phi}_{\text{in}}(t,\vec p) \sim \lim_{t\rightarrow -\infty} \left( e^{i t \omega_p}\hat{a}_\text{in}^{\dagger}(\vec{p}) + e^{- i \omega_p t}\hat{a}_\text{in}(\vec{p}) \right)\, ,
\end{equation}
where $\hat{a}_\text{in}(\vec{p})$ annihilates the vacuum. Acting with the in-field on the incoming vacuum creates the state $\vert \vec{p}\rangle \sim\hat{a}^{\dagger}(\vec{p}) \vert 0 \rangle$ at the Cauchy slice $\Sigma_-$ and it is a single particle state in the asymptotic Fock space. For out states, the story is analogous as $t \to \infty$. This demonstrates that scattering states are a very restricted class of states. In the language of path integrals, this places restrictions on the wave functionals defined in \eqref{eq:SPI}. In this and the next section, we will explain how such states can be prepared using a flat limit of AdS/CFT for fields of different spin.
\subsection{Lorentzian AdS/CFT}
The question we want to answer is how the path integral in equation \eqref{eq:SPI} can be computed in the context of a flat limit of AdS/CFT. In order to do so, we first briefly review holography in AdS. We will be using global coordinates, for which the line element of empty AdS reads
\begin{equation}\label{eq:ds2}
ds^2={L^2\over \cos^2\rho}\left( -d\tau^2+d\rho^2+\sin^2\rho \, d\Omega_{d-1}^2 \right)\, .
\end{equation}
The coordinate $\rho$ is holographic, in the sense that the boundary CFT is located at $\rho=\frac \pi 2$. The coordinates parametrising the CFT are $x=\{\tau,\Omega_{d-1}\}$, and $L$ denotes the AdS radius.
In order to point out a crucial difference between the two choices of signature, we first review Euclidean AdS holography. The AdS/CFT conjecture is an equality relating the generating functional of the correlators of the conformal field theory and a Euclidean path integral of bulk fields with specified boundary conditions. For a bulk field $\Phi$ with boundary condition
\begin{equation}\label{eq:FallOffE}
\Phi \xrightarrow[\rho \to \frac \pi 2]{} (\cos \rho)^{d-\Delta} \phi_E\, ,
\end{equation}
the statement of the duality reads
\begin{equation}\label{eq:EuclideanAdSCFT}
\langle e^{-\int_{\partial \text{AdS}} \phi_E {\cal O}} \rangle=\int [{\cal D}\Phi]_{\phi_E} \, e^{-S_E[\Phi]}\, ,
\end{equation}
where $[{\cal D}\Phi]_{\phi_E}$ indicates that the path integral is over field configurations with asymptotic behavior \eqref{eq:FallOffE}. In the semi-classical limit, the bulk path integral can be computed in a saddle point approximation. Once the boundary condition $\phi_E$ is chosen, the saddle of the action is completely determined. The situation is different in Lorentzian signature \cite{Balasubramanian:1998de}, as conditions on the Lorentzian boundary do not fully specify a saddle of the action.
In the Lorentzian signature path integral, the bulk field $\Phi$ is integrated with a boundary condition $\phi_0(x)$ at the boundary. Such boundary condition corresponds to a source in the CFT, which couples to a dual operator ${\cal O}(x)$ with conformal dimension $\Delta$. In a semi-classical approximation, and unlike in Euclidean signature, the saddles of the action are however not completely determined by the boundary condition. The general structure of the field solution reads
\begin{equation}
\Phi=\tilde{\phi}(\rho,x)+{\phi}(\rho,x)\, ,
\end{equation}
where we have split the field into pieces with different fall-off at the AdS boundary
\begin{equation}
\tilde{\phi}(\rho,x) \xrightarrow[\rho \to \frac \pi 2]{} (\cos\rho)^{d-\Delta} \phi_0(x)\, , \quad \text{and}\quad {\phi}(\rho,x) \xrightarrow[\rho \to \frac \pi 2]{} (\cos\rho)^{\Delta} {\phi}(x)\, .
\end{equation}
The part of $\Phi$ that asymptotes to $\phi_0$ is commonly known as the non-normalizable part of the field, while the sub-leading structure ${\phi}(\rho,x)$ is known as the normalizable part and captures the remaining degrees of freedom not fixed by the boundary condition\footnote{
Even though we will refer to the source part of the bulk field as the non-normalizable modes and the state part of the field as the normalizable modes, for the range of bulk mass $-d^2/4<m^2<1-d^2/4$, both sets of modes are normalizable, and some criterion is needed to distinguish them \cite{Balasubramanian:1998sn}. We will revisit these statements more carefully in section \ref{sec:ScalarLagrange}
}. The modes of normalizable solutions are low energy excitations of space-time and can be thought of as spanning the low energy Hilbert space of the boundary theory. At the microscopic level, we regard $\hat{{\phi}}(\rho,x)$ as a quantized operator that will match ${\cal O}(x)$ at the boundary as
\begin{equation}
\hat{{\phi}}(\rho,x) \xrightarrow[\rho \to \frac \pi 2]{} (\cos\rho)^{\Delta}{\cal O}(x)\, .
\end{equation}
Note that here $\mathcal O(x)$ is the CFT operator and not just its expectation value.
The fact that the normalizable field is not completely determined by the boundary conditions allows us to specify an incoming and an outgoing state in AdS by specifying a scattering state at some Cauchy slices $\Sigma_{\pm}^{\text{AdS}}$ in the bulk. This in turn also specifies a particular state in the CFT. Throughout this paper, one of the issues we will discuss is precisely how to construct CFT states $\langle \text{in} \vert$ and $\vert \text{out} \rangle$ so that
\begin{equation}\label{eq:NormPhiZ}
\langle \text{in} \vert e^{i \int_{\partial \text{AdS}} \phi_0(x){\cal O}(x)} \vert \text{out}\rangle = \sum_{\phi_{\text{a/b}}} \Psi^*_\text{out}[\phi_\text{b}] \int\displaylimits_{{\phi}\vert_{\Sigma^{\text{AdS}}_{-}}={\phi}_{\text{a}}}^{{\phi}\vert_{\Sigma^{\text{AdS}}_{+}}={\phi}_{\text{b}}} [{\cal D}\Phi]_{\phi_0}\, e^{i S({\phi})} \Psi_\text{in}[\phi_\text{a}]\, .
\end{equation}
Here, $\phi_0$ is the the non-normalizable part of the field, and field configurations ${\phi}_{\text{a/b}}$ fix the normalizable part of the field at Cauchy slices $\Sigma^{\text{AdS}}_{\pm}$.
Before we do so, let us contrast our approach to an existing way to construct CFT excited states in the context of real time AdS/CFT; The Skenderis-van Rees framework (SvR) \cite{Skenderis:2008dh,Skenderis:2008dg}. In their formalism, the states of the CFT are prepared using a Euclidean path integral at the boundary, which can be shown to correspond to preparing coherent states in the AdS Fock space \cite{Botta-Cantcheff:2015sav}. For a picture, see figure \ref{fig:AdS}. Their construction takes the form \eqref{eq:NormPhiZ}.
In the case of initial states defined on $\Sigma^{\text{AdS}}_{-}$, we glue the Lorentzian space-time at $\Sigma_{-}^{\text{AdS}}$ to a Euclidean manifold ${\cal M}_-$. The state is then specified by a boundary condition of the bulk field at $\partial{\cal M}_-$. The wave functional which defines the incoming state thus reads
\begin{equation}
\Psi_\text{in}[{\phi}_{\text{a}}]\equiv \langle {\phi}_{\text{a}}\vert \Psi_\text{in} \rangle = \int_0^{{\phi}_{\text{a}}} [{\cal D}\Phi]_{\phi_i} e^{-S_{{\cal M}_-}[\Phi]}\, .
\end{equation}
Here, $\phi_i$ is a boundary condition for the non-normalizable part of the field at the Euclidean boundary $\partial {\cal M}_-$. At Euclidean infinite past, the field is set to zero, while ${\phi}_{\text{in}}$ is the boundary condition for the field at $\Sigma_-^{\text{AdS}}$. The CFT state dual to this prescription corresponds to \cite{Botta-Cantcheff:2015sav}
\begin{equation}
\vert \Psi_i \rangle = e^{-\int_{\partial {\cal M}_-} {\cal O}\phi_i}\vert 0 \rangle\, .
\end{equation}
Expanding ${\cal O}$ in creation and annihilation operators shows that $\vert \Psi_i\rangle$ is a coherent state. Final states are obtained by a similar construction using a Euclidean path integral over ${\cal M}_+$ and glued onto $\Sigma^{\text{AdS}}_{+}$.
There are two main differences between our approach and that of SvR. First, we will use an alternative procedure that does not rely on Euclidean path integrals. Instead we will construct states $\vert \text{in} \rangle$ (and $\langle \text{out} \vert$) by acting with certain bulk operators (in Lorentzian signature) on the vacuum of the theory. Using a bulk operator reconstruction method, we will be able to construct such states as operators in the CFT acting on the vacuum. Second, the SvR framework is quite general and allows to discuss all states we can prepare as a Euclidean path integral. In contrast, we will be interested in CFT operators, which create bulk states that turn into Fock space scattering states in the large radius limit.
\begin{figure}[]
\centering
\begin{tabular}{cc}
\begin{subfigure}[t]{0.41\textwidth}
\centering
\begin{tikzpicture}[scale=1]
\draw [white, thick] (-3,-3)--(3,3);
\draw [white, thick] (-3,3)--(3,-3);
\draw[red](-2,2)--(2,2) node[pos=0.5,above]{$\Sigma_+^{\text{AdS}}$};
\draw[blue](-2,-2)--(2,-2) node[pos=0.5,below]{$\Sigma_-^{\text{AdS}}$};
\draw [black, thick,dashed] (-2,-2)--(2,2);
\draw [black, thick,dashed] (-2,2)--(2,-2);
\draw [black, very thick](-2,-3)--(-2,3);
\draw [black, very thick](2,-3)--(2,3) node[pos=0.165, right]{$\tau=-{\pi \over 2}$} node[pos=0.835, right]{$\tau=+{\pi \over 2}$};
\draw [black, thick,->] (-1.8,-0.8)--(-1.8,-0.4) node[pos=0.5,right]{$\tau$};
\draw [black, thick,->] (0.4,0)--(1.0,0) node[pos=0.5,above]{$\rho$};
\end{tikzpicture}
\caption{}
\end{subfigure}
&
\begin{subfigure}[t]{0.41\textwidth}
\centering
\begin{tikzpicture}[scale=1]
\draw [white, thick] (-3,-3)--(3,3);
\draw [white, thick] (-3,3)--(3,-3);
\draw[red](-2,2)--(2,2) ;
\draw[blue](-2,-2)--(2,-2);
\draw[red](-2,2.1)--(2,2.1) ;
\draw[blue](-2,-2.1)--(2,-2.1);
\draw [black, thick,dashed] (-2,-2)--(2,2);
\draw [black, thick,dashed] (-2,2)--(2,-2);
\draw [gray,opacity=0.2](-2,-3)--(-2,3);
\draw [gray,opacity=0.2](2,-3)--(2,3);
\draw [black, very thick](-2,-2)--(-2,2);
\draw [black, very thick](2,-2)--(2,2) node[pos=0,left=2,below=16]{$\partial{\cal M}_-$} node[pos=1,left=2,above=15]{$\partial{\cal M}_+$};
\draw [black, very thick] (-2,-2.1) to[out=-60,in=-180] (0,-3);
\draw [black, very thick] (2,-2.1) to[out=-120,in=0] (0,-3);
\draw [black, very thick] (-2,2.1) to[out=60,in=-180] (0,3);
\draw [black, very thick] (2,2.1) to[out=120,in=0] (0,3);
\end{tikzpicture}
\caption{}
\end{subfigure}
\end{tabular}
\caption{ a) Penrose diagram of Anti-de Sitter space. Boundary conditions for normalizable bulk field operators can be imposed at the Cauchy slices $\Sigma_{\pm}^{\text{AdS}}$. The bulk path integral computes \eqref{eq:NormPhiZ}. b) Gluing Lorentzian AdS to Euclidean half-spheres ${\cal M}_{\pm}$ and placing boundary conditions for the bulk field at ${\partial}{\cal M}_{\pm}$ yields the SvR prescription. This corresponds to a CFT expectation value involving coherent states.
}\label{fig:AdS}
\end{figure}
\subsection{Flat limit of AdS/CFT}
Having laid out the statement of AdS/CFT, we now explain how to perform a flat limit. At the level of the geometry, we need to take the AdS radius large such that the line element \eqref{eq:ds2} becomes that of Minkowski space. The explicit limit reads $\tau=t/L$ and $\rho=r/L$ so that
\begin{equation}
ds^2 \xrightarrow[L \rightarrow\infty]{} -dt^2+dr^2+r^2d\Omega_{d-1}^2\, .
\end{equation}
The operation can be regarded as zooming into the center region of AdS, where the geometry is locally flat. One can now think of AdS Cauchy slices $\Sigma'_{\pm}$ that match the slices $\Sigma_{\pm}$ in Minkowski space, but extend in AdS space as shown in figure \ref{fig:Limit}. In order to compute the S-matrix amplitude in Minkowski space, we need to choose the states at the slices $\Sigma_{\pm}$. In AdS, this can be achieved by exciting normalizable modes at $\Sigma_{\pm}^{\prime}$.
\begin{figure}[]
\centering
\begin{tikzpicture}[scale=1.2]
\draw [white, thick] (-2,-2)--(2,2);
\draw [white, thick] (-2,2)--(2,-2);
\draw [black, thick] (-1,0)--(0,1);
\draw [black, thick] (1,0)--(0,1);
\draw [black, thick] (-1,0)--(0,-1);
\draw [black, thick] (1,0)--(0,-1);
\draw [red, very thick] (-1,0) to[out=30,in=180] (0,0.7);
\draw [red, very thick] (1,0) to[out=150,in=0] (0,0.7);
\draw [blue, very thick] (-1,0) to[out=-30,in=180] (0,-0.7);
\draw [blue, very thick] (1,0) to[out=-150,in=0] (0,-0.7);
\draw [red, very thick] (-2,0.01)--(-1,0.01) node [pos=0.5,above=0]{\,\,{ $\Sigma^{\prime}_+$}};
\draw [blue, very thick] (-2,-0.01)--(-1,-0.01);
\draw [red, very thick] (2,0.01)--(1,0.01);
\draw [blue, very thick] (2,-0.01)--(1,-0.01)node [pos=0.5,below=0]{\,\,{ $\Sigma^{\prime}_-$}};
\draw[red](-2,2)--(2,2) node[midway,below]{$\Sigma^{\text{AdS}}_+$} ;
\draw[blue](-2,-2)--(2,-2) node[midway,above]{$\Sigma^{\text{AdS}}_-$};
\draw [black, very thick](-2,-2)--(-2,2);
\draw [black, very thick](2,-2)--(2,2);
\draw [black, thick,->] (-1.8,-1.3)--(-1.8,-0.9) node[pos=0.5,right]{$\tau$};
\draw [black, thick,->] (1.1,-1.3)--(1.7,-1.3) node[pos=0.5,above]{$\rho$};
\end{tikzpicture}
\caption{Section of the global AdS cylinder. The region inside the diamond corresponds to $\tau=t/L$ and $\rho=r/ L$ which turns flat as $L$ is taken to infinity. The Cauchy slices $\Sigma_{\pm}^{\prime}$ extend the Minkowski slices $\Sigma_{\pm}$ to global AdS space. Specifying a state on $\Sigma_{\pm}^{\prime}$ specifies a state in the conformal field theory, in the Cauchy slice $\Sigma_{\pm}^{\prime}\cap \partial\text{AdS}$. Alternatively, we can invoke asymptotic decoupling and use a free Hamiltonian in AdS to time evolve our fields to the Cauchy slices $\Sigma_{\pm}^{\text{AdS}}$. This would yield the computation of formula \eqref{eq:NormPhiZ} with vanishing source ($\phi_0=0$), which is drawn in figure \ref{fig:AdS} a).
}\label{fig:Limit}
\end{figure}
We thus conclude that the S-matrix element in flat space is computed by a large $L$ limit of a CFT expectation value like the one appearing in equation \eqref{eq:NormPhiZ}. The last ingredient we need is a way to construct the CFT states $\vert {\text{in}} \rangle$ and $\langle {\text{out}} \vert$ in terms of the field associated to a scattering state in $\Sigma_{\pm}$. For this, we will look for an operator $\hat{{\phi}}_{\text{in}}$ such that
\begin{equation}
\hat{{\phi}}_{\text{in}} \vert 0 \rangle= \vert {\text{in}} \rangle\, .
\end{equation}
We then need to recast the operator $\hat{{\phi}}_{\text{in}}$ defined on the Cauchy slice $\Sigma_-$ deep in the bulk in terms of a CFT operator at the boundary. Once this is done, the resulting S-matrix element will simply read
\begin{equation}
\begin{split}
\langle \text{out} \vert \text{in} \rangle =&\lim_{L\rightarrow\infty} \langle 0 \vert\hat{ {\phi} }_{\text{out}} e^{i \int d^d x \, \phi_0(x){\cal O}(x)} \hat{ {\phi}}_{\text{in}}\vert 0\rangle_{\phi_0=0}\, .
\end{split}
\end{equation}
In the following section we will explicitly construct $\hat{{\phi}}_{\text{in/out}}$ in the CFT.
\subsection{Global reconstruction of scalar operators in AdS/CFT}
In the previous section we have explained how the S-matrix amplitude of a physical process in flat space-time can be computed by taking a flat limit of a CFT correlator. The operators in the CFT correlators are boundary representations of bulk operators living in late/early time slices of the Minkowski geometry, which is placed deep inside the bulk geometry. We will place the scattering region at $\tau=0$ at the center of global AdS. The bulk operator has been argued to be free in virtue of the assumption of asymptotic decoupling. This means that before and after the scattering region, a scalar operator must obey the Klein-Gordon equation on AdS,
\begin{equation}
\left( \nabla^2-m^2 \right)\hat{{\phi}} = 0\, , \quad \text{with} \quad m^2 L^2=\Delta(\Delta-d)\, .
\end{equation}
The operator has also been argued to be associated to normalizable modes of bulk fields, and so it has the following fall-off behavior
\begin{equation}
\hat{{\phi}}(\rho,x)\xrightarrow[\rho \to \frac \pi 2]{} (\cos \rho)^{\Delta}{\cal O}(x)\, .
\end{equation}
The problem of reconstructing such an operator in the CFT has been analyzed extensively in the literature. In this paper we will use the HKLL reconstruction method \cite{Hamilton:2006az}. A nice feature of the HKLL method is that it allows to reconstruct bulk operators as operators in a region of the CFT covering the whole boundary in any global time interval of length $\Delta \tau = \pi$. We will choose our scattering region to be around around the center of AdS at $\tau = 0$, such that all excitations are far separated unless they are in a small interval of size $\epsilon\sim{\cal O}(L)^{-1}$ around $\tau=0$. Since the time-like geodesics emanating from the scattering region will come close again after time $\pi$, the region of validity for the free field expansion is limited to $\tau = (\epsilon, \pi - \epsilon)$ for out-fields and $\tau = (-\pi + \epsilon, - \epsilon)$ for in-fields. For a picture of this set-up, see figure \ref{fig:HKLL}.
\begin{figure}[]
\centering
\begin{tikzpicture}
\draw[white,very thick] (5,2) arc (0:-180:3 and 3);
\draw[white,very thick,name path=TOP](1.25,7) arc (180:360:0.75 and 0.25);
\draw[gray, thick,dotted] (2,2) -- (2,7) node [pos=1,above=2]{{$ $}} node [pos=0,below=2]{{$ $}};
\draw[black] (-1,2) arc (-180:0:3 and 1);
\draw[black,dashed](5,2) arc (0:180:3 and 1);
\shade[top color=blue!15!white,opacity=0.85] (5,4) arc (0:180:3 and 1) -- (-1,2) arc (-180:0:3 and 1) -- cycle;
\shade[top color=blue!15!white,opacity=0.85] (-1,2) arc (180:360:3 and 1) -- (5,4) arc (0:-180:3 and 1) -- cycle;
\shade[top color=red!15!white,opacity=0.85] (5,6) arc (0:180:3 and 1) -- (-1,4) arc (-180:0:3 and 1) -- cycle;
\shade[top color=red!15!white,opacity=0.85] (-1,4) arc (180:360:3 and 1) -- (5,6) arc (0:-180:3 and 1) -- cycle;
\draw[gray] (-1,6) arc (-180:0:3 and 1);
\draw[gray,dashed](5,6) arc (0:180:3 and 1);
\draw[blue,dashed](5,3) arc (0:180:3 and 1) node[pos=0,right]{$\Sigma^{\text{AdS}}_-$};
\draw[blue](-1,3) arc (180:360:3 and 1);
\draw[gray,very thick,dashed](5,4) arc (0:180:3 and 1) node[pos=0,right]{$\Sigma$};
\draw[gray,very thick](-1,4) arc (180:360:3 and 1);
\draw[gray,thick](2.6-0.03,4+0.1)--(2,4.6);
\draw[gray,thick](1.4+0.03,4+0.1)--(2,4.6);
\draw[gray,thick,dashed](2.6,4) arc (0:180:0.6 and 0.3);
\draw[gray,thick](2.6,4) arc (0:-195:0.6 and 0.3);
\draw[gray,thick](2.6-0.03,4-0.1)--(2,3.4);
\draw[gray,thick](1.4+0.03,4-0.1)--(2,3.4);
\draw[black,very thick,-dotRed-=1] (2-0.05,4.35)--(2-0.05,4.4) node[pos=1,above=6,left=4]{\color{red}\small $\hat{{\phi}}_{\text{out}}$};
\draw[black,very thick,-dotBlue-=1] (2+0.1,3.65)--(2+0.1,3.60) node[pos=1,below=6,right=4]{\color{blue}\small $\hat{{\phi}}_{\text{in}}$};
\draw[black,very thick] (-1,1) -- (-1,7);
\draw[black,very thick] (5,1) -- (5,7) node [pos=0.85,right]{$\tau=\pi$} node [pos=0.15,right]{$\tau=-\pi$};
\draw[red,dashed](5,5) arc (0:180:3 and 1) node[pos=0,right]{$\Sigma^{\text{AdS}}_+$};
\draw[red](-1,5) arc (180:360:3 and 1);
\end{tikzpicture}
\caption{The bulk field $\hat{\phi}(x)$ is placed inside a scattering region around the center of global AdS. The local operator can be reconstructed semi-classically in the boundary using HKLL. For the incoming field, the reconstruction involves the shaded blue part of the boundary spanning $\tau\in(-\pi,0)$. For the outgoing field, we choose the shaded red region spanning $\tau\in(0,\pi)$.
}\label{fig:HKLL}
\end{figure}
The details of the HKLL reconstruction method applied to scalars can be found in \cite{Hamilton:2006az} and have been summarized in appendix \ref{app:ScalarHKLL}. The structure of the result in AdS$_4$/CFT$_3$ reads
\begin{equation}\label{eq:ScalarHKLL}
\hat{{\phi}}_{\text{AdS,0}}(\rho,x)={1\over \pi}\int_{{\cal T}} d\tau'\, \int d^2 \Omega\, \left[ K_{+}(\rho,x;x') {\cal O}^+(x')+ K_{-}(\rho,x;x') {\cal O}^-(x') \right]\, ,
\end{equation}
where $K_{\pm}$ are kernels, defined such that $\hat{{\phi}}_{\text{AdS,0}}(\rho,x)$ obeys the Klein-Gordon equation and asymptotes to its dual operator $\mathcal O$ at the boundary, with positive/negative frequency modes ${\cal O}^{\pm}$. The regions of integration are ${\cal T}=(-\pi,0)$ for incoming and ${\cal T}= (0, \pi)$ for outgoing states (see figure \ref{fig:HKLL}). The explicit form of the kernels reads
\begin{equation}
\begin{split}
K_{\pm}(\rho,x;x')=&\frac 1 {\mathcal N_{K,\Delta}} \sum_{\kappa\in \mathbb{Z}}\sum_{l=0}^{\infty}\sum_{m=-l}^{m=l} e^{\pm i \omega_{\kappa,l}(\tau-\tau')} Y_l^m(\Omega)Y_l^m(\Omega')^* \sin^l\rho \cos^{\Delta}\rho \,\\
&\times {}_2F_1\left( -\kappa,l+\Delta+\kappa;\Delta-{1\over 2} \Big\vert \cos^2\rho \right)\, ,
\label{eq:kernels_sec_2}
\end{split}
\end{equation}
where $\omega_{\kappa,l}=\Delta+l+2\kappa$ are the normalizable modes of a scalar field in AdS, and the normalization is given in \eqref{eq:ScalarNorma}. Note that in equation \eqref{eq:kernels_sec_2} the sum over modes $\kappa$ which usually runs only over positive integers has been extended to run over all integers. This is possible since the additional terms do not contribute to the integral \eqref{eq:ScalarHKLL}. In the next subsection, we will take a flat limit of equation \eqref{eq:ScalarHKLL} and extract creation/annihilation operators in Minkowski space.
The limit is different depending on whether the field is massive or massless. The reason for this is that the mass and the conformal dimension of the dual operator are related by
\begin{equation}
m^2 L^2=\Delta(\Delta-d)\, .
\end{equation}
In the large L limit, a massive field in flat space will arise if $\Delta$ also scales with $L$ as $\Delta=m L$. If otherwise $\Delta\sim {\cal O}(1)$, then in the flat limit the mass will vanish as $m^2\rightarrow \Delta(d-\Delta)/L^2\rightarrow 0$. For simplicity, we will simply choose $\Delta=d$ for the case of massless fields, such that the bulk operator is not just massless in the flat limit, but is exactly massless in anti-de Sitter space.
\subsubsection{Flat limit of HKLL operator: Massless}\label{sec:MasslessRec}
We start with the expression for a scalar bulk operator in AdS$_d$, formula \eqref{eq:ScalarHKLL} with $\Delta=d$. A scalar bulk operator in Minkowski space can then be obtained by taking the flat limit. This is done by reparametrizing the bulk coordinates $(\rho, \tau,\Omega)$ by
\begin{equation}\label{eq:MinkRegion}
\rho={r\over L}\, , \quad \text{and}\quad \tau={t\over L}\, ,
\end{equation}
before sending $L$ to infinity.
Inside the Minkowski region, at very early/late times, we expect the field to be free due to asymptotic decoupling. This means that the field operator must have a representation as
\begin{equation}
\hat{{\phi}}(x)= \int {{d^3 \vec{p}}\over{(2\pi)^3}}{1\over\sqrt{2\omega_{\vec{p}}}}\left( \hat{a}_{\vec{p}}e^{i p \cdot x}+ \hat{a}^{\dagger}_{\vec{p}}e^{-i p \cdot x} \right)\, , \quad \text{with}\quad x\in \text{Mink}_{3+1}\, .
\end{equation}
The creation/annihilation operators appearing here obey canonical commutation relations
\begin{equation}
[a_{\vec{p}},a^{\dagger}_{\vec{k}}]= (2\pi)^3 \delta^{(3)}(\vec{p}-\vec{k})\, .
\end{equation}
Since the decomposition of $\hat{{\phi}}(x)$ can be done at early and late times, there exist two sets of creation and annihilation operators, one for early and one for late times. They span the in- and out-scattering Fock spaces. Scattering states can be prepared as follows,
\begin{equation}
\vert \vec{k},\text{in}\rangle = \sqrt{2\omega_{\vec{k}}}\, \hat{a}^{\dagger}_{\text{in},\vec{k}}\vert 0 \rangle \, , \quad \text{and}\quad \langle \vec{p}, \text{out}\vert=\langle 0 \vert \sqrt{2\omega_{\vec{p}}}\, \hat{a}_{\text{out},\vec{p}}\, .
\end{equation}
The creation/annihilation operators can be extracted from the position space field operator via
\begin{equation}\label{eq:aFromPhi}
\begin{split}
\hat{a}_{\text{in/out},\vec{p}} =&\lim_{t\rightarrow \mp \infty}{i\over \sqrt{2\omega_{\vec{p}}}} \int_{\Sigma} d^3 \vec{x} \, e^{-i p\cdot x} \overleftrightarrow{\partial_0} \hat{{\phi}}(x)\, , \\
\hat{a}^{\dagger}_{\text{in/out},\vec{p}} =&\lim_{t\rightarrow\mp\infty}{-i\over \sqrt{2\omega_{\vec{p}}}} \int_{\Sigma} d^3 \vec{x} \, e^{i p\cdot x} \overleftrightarrow{\partial_0} \hat{{\phi}}(x)\, .
\end{split}
\end{equation}
where $\overleftrightarrow{\partial}_0=\overrightarrow{\partial}_0-\overleftarrow{\partial}_0$. In order to represent these operators in the conformal field theory, we simply have to apply formulas \eqref{eq:aFromPhi} to the HKLL operator \eqref{eq:ScalarHKLL} placed in the Minkowski scattering region \eqref{eq:MinkRegion}.
The first step is to write the HKLL operator in the scattering region. The replacement $\tau=t/L$ in the large $L$ limit yields stationary solutions unless $\omega_{\kappa,l}$ scales with $L$. This means that fluctuating modes in Minkowski space arise from AdS modes with large values of $\kappa$. We will thus define
\begin{equation}
\Delta+l+2\kappa\equiv w L\, .
\end{equation}
In the large $L$ limit, the sum over $\kappa$ can be traded for an integral over $w$ and the discrete spectrum of scalar field fluctuations in AdS becomes a continuum in the flat limit. The HKLL kernels can now be written explicitly in the Minkowski scattering region
\begin{equation}
\begin{split}
K_{\text{in},\pm}(r,t,\Omega;x')&=\frac{i \sqrt 3}{\pi^2 L} \int {dw\over w}\, e^{\pm i w t} e^{i w L \left( \mp(\tau' + {\pi \over 2}) \right)}\,\sum_{l,m} Y_l^m(\Omega)Y^m_l (\Omega')^*\, (\mp i)^{-l}j_l(w r)\, ,\\
K_{\text{out},\pm}(r,t,\Omega;x')&=\frac{(-i) \sqrt 3}{\pi^2 L} \int {dw\over w}\, e^{\pm i w t} e^{i w L \left( \mp(\tau' - {\pi \over 2}) \right)}\,\sum_{l,m} Y_l^m(\Omega)Y^m_l (\Omega')^*\, (\pm i)^{-l}j_l(w r)\, .
\end{split}
\end{equation}
The Bessel functions arise from the flat limit of the hypergeometric functions appearing in the AdS kernels. Applying now \eqref{eq:aFromPhi} together with
\begin{equation}
e^{i \vec{p} \cdot \vec{x}}= 4\pi \sum_{l',m'}i^{l'} j_{l'}(\omega_{\vec{p}}r) Y^{m'}_{l'}(\hat{p}) Y^{m'}_{l'}(\Omega)^* \, ,
\end{equation}
the final result reads
\begin{equation}\label{eq:aMassless}
\begin{split}
\sqrt{2\omega_{\vec{p}}} \, a_{\text{in},\vec{p}}=& c_- \int^{0}_{-\pi } d\tau \, e^{i \omega_p L \left( \tau+{\pi \over 2}\right)} {\cal O}^- \left( \tau , -\hat{p} \right)\, , \\
\sqrt{2\omega_{\vec{p}}} \, a^{\dagger}_{\text{in},\vec{p}}=& c_+ \int^{0}_{-\pi } d\tau \, e^{-i \omega_p L \left(\tau+{\pi \over 2}\right) } {\cal O}^+ \left( \tau , -\hat{p} \right)\, , \\
\sqrt{2\omega_{\vec{p}}} \, a_{\text{out},\vec{p}}=& c_+ \int^{\pi }_{0} d\tau \, e^{i \omega_p L \left( \tau-{\pi\over 2}\right)} {\cal O}^- \left( \tau , \hat{p} \right)\, , \\
\sqrt{2\omega_{\vec{p}}} \, a^{\dagger}_{\text{out},\vec{p}}=& c_- \int^{\pi }_{0} d\tau \, e^{-i \omega_p L \left( \tau-{\pi\over 2}\right)} {\cal O}^+ \left( \tau, \hat{p} \right)\, , \\
\end{split}
\end{equation}
where $\hat{p}$ is the angle of the momentum $\vec{p}$, and $-\hat{p}$ stands for its anti-podal. We have also defined an over-all constant $c_\pm= \pm i \frac{\sqrt{24}}{L \omega_p^2}$. Note that the integrals over global time here are highly oscillatory as $L \to \infty$ unless operators are evaluated in windows of size ${\cal O}(1/L)$ at $\tau=\pm {\pi \over 2}$. This suggests that we can ignore contributions away from this region. Equation \eqref{eq:aMassless} is precisely the prescription presented previously in the literature \cite{Fitzpatrick:2011hu,Hijano:2019qmi}, which we have derived here from first principles.
We want to point out two important aspects of the above expressions. First, in the scattering states we are considering, the operator $\mathcal O$ has a free-field like expansion, c.f. appendix \ref{app:ScalarHKLL}. This implies that we can replace non-local operators $\mathcal O^\pm$ by local operators $\mathcal O$ in \eqref{eq:aMassless}, since the difference $\mathcal O - \mathcal O^{\pm}$ integrates to zero against the phase. In other words, the creation and annihilation operators can be constructed by smearing local operators. In the following we will nonetheless keep the positive/negative frequency superscripts. Whenever we talk about the location of $\mathcal O^\pm$ this should be understand as either replacing $\mathcal O^\pm$ by $\mathcal O$, or equivalently evaluating $\mathcal O^\pm(\tau, \hat p)$ in that location.
Second, the equations in \eqref{eq:aMassless} naively look like they would vanish in the $L \to \infty$ limit. However, the large $L$ limit should only be taken after CFT correlators of positive/negative frequency operators have been calculated. These correlation functions contain pieces with diverge as the flat space limit is taken and the role of the integrals and additional factors in $L^{-1}$ in \eqref{eq:aMassless} is to extract particular divergences which correspond to S-matrix elements in flat space. This can be seen explicitly in some easy sample calculations, see appendix \ref{app:sample_calc}.
\subsubsection{Flat limit of HKLL operator: Massive}\label{sec:MassiveRec}
In the case of massive scalar operators, the calculation is very different due to the fact that the conformal dimension of the CFT operator now scales with $L$ such that
\begin{equation}
m^2 L^2 =\Delta(\Delta-d)\, , \quad \Rightarrow \quad \Delta={d\over 2}+ m L +{\cal O}(L)^{-1}\, .
\end{equation}
Like in the massless case, we will also need the AdS quantum number $\kappa$ to be large in order to obtain fluctuating modes in flat space. We thus have
\begin{equation}
\Delta+l+2\kappa =w L\, .
\end{equation}
The positive frequency kernels for in- and out-fields in the flat limit now read
\begin{equation}
\begin{split}
K^{\text{out/in}}_{+}(r,t,\Omega;x')=& \frac{1}{2 \pi} \left(\frac{ML}{\pi^3}\right)^{\frac 1 4} L \int dw \,e^{i w t}\, k\, \left( {2m\over (\pm i) k} \right)^{m L +{1\over 2}}\, e^{i w L\left[ {\pi \over 2}+\frac i 2 \log\left(w+m\over w-m \right) \mp \tau' \right]} \\
&\times\sum_{l,m} Y^{m}_{l}(\Omega) Y^{m}_{l}(\Omega')^* \,{(\pm i)}^{-l}j_l(k r) \, ,
\end{split}
\end{equation}
where we have defined
\begin{equation}
k\equiv \sqrt{w^2-m^2}\, .
\end{equation}
The kernels for negative frequencies can be obtained by conjugation.
Plugging this expression in our formula of the HKLL operator and performing the Minkowski space coordinate integrals in \ref{eq:aFromPhi} yields for the out-state creation operator,
\begin{equation}\label{eq:aMassive}
\sqrt{2\omega_{\vec{p}}} \hat{a}^\dagger_{\text{out},\vec{p}} = c_-(L,m,|\vec p |) \, \int_0^\pi d\tau'
e^{i \omega_{\vec{p}} L\left[ {\pi \over 2}+{i \over 2} \log\left(\omega_{\vec{p}}+m\over \omega_{\vec{p}}-m \right) - \tau' \right]} {\cal O}^+(\tau',\hat{p})\, ,
\end{equation}
where we have defined
\begin{align}
c_\pm(L,m,|\vec p |) = \frac{1}{2 \pi} \left(\frac{mL}{\pi^3}\right)^{\frac 1 4} \, \left( {2m\over (\mp i) \vert \vec{p}\vert } \right)^{m L +{1\over 2}} L.
\end{align}
The annihilation operator can be obtained from this expression by complex conjugation. The expressions for the in-state operators are obtained analogous to the massless case by changing the integration region to $(-\pi, 0)$, changing the sign of $\frac \pi 2$ in the phase as well as $c_+ \leftrightarrow c_-$, and replacing $\hat p \to - \hat p$.
Note that in the large $L$ limit, the real part of the exponential only vanishes, if
\begin{equation}\label{eq:Imtau}
\text{Im}\left( \tau' \right)=\frac 1 2 \log\left( \omega_{\vec{p}}+m\over \omega_{\vec{p}}-m \right)\, .
\end{equation}
It is therefore convenient to move the $\tau$ contour such that it runs along this value in the complex plane.
Again, the integrand contains an exponential which highly oscillates as $L$ is sent to infinity. This focuses the integral around a window of size ${\cal O}(1/L)$ at Re$\left(\tau'\right)\sim {\pi \over 2}$. We conclude that creation operators in Minkowski space can be understood as large $L$ limits of operators with large conformal dimension placed in a conformal field theory at a complex value of global time\footnote{This was shown less rigorously in \cite{Hijano:2019qmi} by one of the authors. }. In the language of previous sections, such values of global time parametrize the Euclidean half spheres $\partial {\cal M}_{\pm}$ drawn in figure \ref{fig:AdS}.
\subsection{Mapping of asymptotic regions}\label{sec:Interpretation}
Formulas \eqref{eq:aMassless} and \eqref{eq:aMassive} allow for an identification of regions of the boundary CFT with the different asymptotic regions of Minkowski space-time. The oscillatory nature of the integrand in formula \eqref{eq:aMassless} suggests that light operators placed at small windows of global time around $\pm {\pi \over 2}$ correspond to massless particles in flat space. This region of the CFT plays the role of null infinity ${\cal I}^{\pm}$. Throughout the text, we will thus refer to these CFT regions as $\tilde{\cal I}^{\pm}$. The metric of the CFT in one of these fringes reads
\begin{equation}
\label{eq:celestial_sphere_metric}
\tau= {\pi \over 2}+{u\over L} \quad\Rightarrow\quad ds^2_{{\cal I}^{\pm}}=L^2\left( -d\tau^2+d\Omega^2 \right) = -du^2+L^2d\Omega^2\, .
\end{equation}
In the large $L$ limit, this metric resembles the metric of asymptotic null infinity where $u$ plays the role of retarded time.
Equation \eqref{eq:aMassive} for massive scattering states also contains a highly oscillatory factor that suggests that heavy operators ($\Delta\sim L$) are located at complex values of global time $\tau=\pm{\pi \over 2}+i \tilde{\tau}$. This complexified region of the CFT plays the role of time-like infinity $i^{\pm}$. These regions have appeared in the text before (see figure \ref{fig:AdS}), and we have denoted them by $\partial{\cal M}_{\pm}$. At the level of the metric, we have
\begin{equation}
ds^2_{\partial{\cal M}_{\pm}}=L^2\left( -d\tau^2+d\Omega^2 \right) = L^2\left( d\tilde{\tau}^2+d\Omega^2 \right) \, .
\end{equation}
This metric can be written more suggestively by changing coordinates to
\begin{equation}
R={1\over \sinh\tilde{\tau}}\, ,
\end{equation}
such that the line element is now
\begin{equation}
ds^2_{\partial{\cal M}_{\pm}}=L^2\left( {dR^2\over R^2(1+R^2)} +d\Omega^2 \right)\, .
\end{equation}
This metric is conformal to the metric of $H_3$ slices of Minkowski space-time used in \cite{deBoer:2003vf,Campiglia:2015qka,Kapec:2015ena}, and can be regarded as the metric of flat space at future/past infinity. Indeed, the special values of imaginary $\tau$ where operators must be placed in the CFT written in formula \eqref{eq:Imtau} correspond to precisely the values of $R$ that massive particles propagating in flat space pierce at $i^{\pm}$. Namely,
\begin{equation}
R={\sqrt{\omega_{\vec{p}}^2-m^2}\over m}={\vert \vec{p}\vert\over m}\, .
\end{equation}
The region in between $\tau=\pm {\pi\over 2}$ can be understood as space-like infinity $i^0$. For a picture of the structure analyzed in this section, see figure \ref{fig:Interpretation}.
Alternatively, we can interpret the operators discussed in this work as operators in the celestial sphere. They take the form of Fourier transformations of local CFT operators. For example in the massless case we can formally define operators on the celestial sphere as
\begin{equation}
\mathcal O_\omega(\hat p) \sim \lim_{L \to \infty} \frac 1 {L^2 \omega^2} \int_{-\infty}^{\infty} du \, e^{\pm i \omega u} \mathcal O \left(\frac u L \pm \frac \pi 2, \hat p \right).
\end{equation}
This has some structural similarities with other proposals in which flat space holography is implemented by an infinite family of operators on the celestial sphere \cite{deBoer:2003vf,Lysov:2014csa,Cordova:2018ygx}.
\begin{figure}[]
\centering
\begin{tabular}{cc}
\begin{subfigure}[t]{0.41\textwidth}
\centering
\begin{tikzpicture}[scale=0.8]
\shade[top color=white, bottom color=black!45!white,opacity=0.75] (5,2) arc (0:180:3 and 1) -- (5,2) arc (0:-180:3 and 3) -- cycle;
\shade[top color=white, bottom color=gray,opacity=0.75] (-1,2) arc (180:360:3 and 1) -- (5,2) arc (0:-180:3 and 3) -- cycle;
\draw[black,very thick] (5,2) arc (0:-180:3 and 3) node[pos=0.5,below]{$\partial{\cal M}_-$};
\draw[gray, thick,dotted] (2,2) -- (2,7) node [pos=1,above=2]{{$ $}} node [pos=0,below=2]{{$ $}};
\draw[blue,very thick,dashed,opacity=0,name path=TOPBD2](5,1.7) arc (0:180:3 and 1);
\draw[blue,very thick,dashed,name path=LOWBD](5,2) arc (0:180:3 and 1);
\draw[blue,very thick,dashed,opacity=0,name path=TOPBD](5,2.3) arc (0:180:3 and 1);
\tikzfillbetween[of=TOPBD and TOPBD2]{blue, opacity=0.5};
\draw[blue,very thick,opacity=0, name path=TOPB2](-1,1.7) arc (180:360:3 and 1);
\draw[blue,very thick, name path=LOWB](-1,2) arc (180:360:3 and 1);
\draw[blue,very thick,opacity=0, name path=TOPB](-1,2.3) arc (180:360:3 and 1);
\tikzfillbetween[of=TOPB and TOPB2]{blue, opacity=0.5};
\draw[black,very thick] (-1,2) -- (-1,7) node [pos=0,left]{\color{blue} $\tilde{\cal I}^-$} node [pos=1,left]{\color{red} $\tilde{\cal I}^+$};
\draw[black,very thick] (5,2) -- (5,7) node [pos=1,right]{$\tau={{\pi}\over 2}$} node [pos=0,right]{$\tau=-{{\pi}\over 2}$};
\draw[red,very thick,dashed,opacity=0,name path=TOPRD2](5,7.3) arc (0:180:3 and 1);
\draw[red,very thick,dashed,name path=TOPRD](5,7) arc (0:180:3 and 1);
\draw[red,very thick,dashed,opacity=0,name path=LOWRD](5,6.7) arc (0:180:3 and 1);
\tikzfillbetween[of=TOPRD2 and LOWRD]{red, opacity=0.5};
\draw[black,very thick] (5,7) arc (0:180:3 and 3) node[pos=0.5,above]{$\partial{\cal M}_+$};
\shade[top color=gray,bottom color=white,opacity=0.75] (-1,7) arc (180:360:3 and 1) -- (5,7) arc (0:180:3 and 3) -- cycle;
\draw[red,very thick,opacity=0,name path=TOPR2](-1,7.3) arc (180:360:3 and 1);
\draw[red,very thick,name path=TOPR](-1,7) arc (180:360:3 and 1);
\draw[red,very thick,opacity=0,name path=LOWR](-1,6.7) arc (180:360:3 and 1);
\tikzfillbetween[of=TOPR2 and LOWR]{red, opacity=0.5};
\draw[cyan, very thick] (3.5,5.82) -- (3.5,5.82+0.6);
\draw[cyan, very thick, snake it,->] (2,4) -- (3,5.5);
\end{tikzpicture}
\caption{}
\end{subfigure}
& \, \, \,
\begin{subfigure}[t]{0.41\textwidth}
\centering
\begin{tikzpicture}[scale=3]
\draw[gray, thick,dotted] (0,-1) -- (0,1);
\draw [white, thick] (-1,-1)--(1,1);
\draw [white, thick] (-1,1)--(1,-1);
\draw [blue, thick] (-1,0)--(0,-1) node[pos=1,below]{\color{black} $i^-$};
\draw [red, thick] (1,0)--(0,1) node[pos=0.5,right=3]{${\cal I}^+$};
\draw [red, thick,-dot-=0,-dotW-=0,-dotB-=1] (-1,0)--(0,1) node[pos=1,above]{\color{black} $i^+$};
\draw [blue, thick,-dot-=0,-dotW-=0,-dot-=1] (1,0)--(0,-1) node[pos=0.5,right=3]{${\cal I}^-$} node[pos=0,right] {\color{black} \, $i^0$};
\draw[cyan, very thick, snake it,->] (0,0) -- (0.42,0.42);
\end{tikzpicture}
\caption{}
\end{subfigure}
\end{tabular}
\caption{ The red fringe of global time can be parametrized by $\tau= {\pi \over 2}+{u\over L}$, where $u$ plays the role of retarded time at future null infinity ${\cal I}^+$. Similarly, the blue fringe parametrizes past null infinity ${\cal I}^-$. In the main text, we refer to these regions as $\tilde{\cal I}^{\pm}$. The shaded caps are analytic continuations of the boundary CFT in the imaginary direction of global time, and they play the role of future/past infinity $i^{\pm}$. In the main text we refer to these caps as $\partial {\cal M}_{\pm}$. The rest of the CFT can be interpreted as space-like infinity. As an example, an outgoing massless particle in flat space can be constructed in AdS/CFT by smearing an operator over a fringe of global time at ${\cal I}^+$.
}\label{fig:Interpretation}
\end{figure}
\section{Introduction}
Despite much progress in the understanding of quantum gravity in asymptotically Anti-de Sitter spaces \cite{Maldacena:1997re, Witten:1998qj}, equally good insight into a holographic description of gravity in Minkowski space is still elusive. In this note we report on results which aim to further clarify the structure of holography in asymptotically flat space-times making use of the large radius limit of AdS/CFT.
In the presence of gravity, diffeomorphism invariance makes it impossible to define local correlation functions. While in AdS space, gauge invariant observables can be defined at the boundary, in the form of conformal correlators. Flat space has no such boundary. Instead, the central object and precise observable in flat space quantum gravity is the S-matrix. The reason is that at very early and late times, field excitations disperse and can be treated as essentially free.\footnote{This is not true in certain cases, most prominently if the particles live in four dimensions and couple to long range forces. In that case the particles do not decouple completely which gives rise to infrared divergences. However, several methods are known to extract finite scattering probabilities from IR divergent amplitudes.} In these asymptotic regions, all long range forces including gravity can be turned off (up to IR effects) and we can define the overlap between incoming and outgoing $n$-particle states, which defines S-matrix elements. Understanding flat space holography therefore means to have a prescription to calculate scattering amplitudes in a dual, lower-dimensional theory.
In this paper, we will consider flat space as the large radius limit of Anti-de Sitter space. It was argued already very shortly after the discovery of the AdS/CFT correspondence that in this limit certain CFT correlation functions approach flat space S-matrix elements \cite{Polchinski:1999ry, Giddings:1999qu}. This has been elaborated upon in various works \cite{Giddings:1999jq,Gary:2009mi,Penedones:2010ue,Fitzpatrick:2011jn,Fitzpatrick:2011ia} and has found applications in the S-matrix bootstrap program \cite{Fitzpatrick:2011hu,Paulos:2016fap}.
The present paper has several goals. First, we give a more rigorous derivation of the CFT operators whose correlation functions turn into S-matrix elements of scalar particles. Our argument is an extension of previous work by one of the authors \cite{Hijano:2019qmi}. In \cite{Hijano:2019qmi} it was observed that CFT operators which create scattering states for massive and massless bulk fields can be derived in a unified fashion assuming HKLL bulk reconstruction \cite{Hamilton:2006az} in global AdS. The form of the CFT operators as the flat limit is taken then follows essentially from symmetry considerations. Here, we will again use bulk operator reconstruction, but take the flat space limit explicitly. This will give a formal derivation for flat space creation and annihilation operators in terms of CFT operators smeared along the time-like direction. For massless particles, the relevant contributions of those smeared operators are concentrated around a particular value of global time. In the massive case, operators must be smeared around complex points in the CFT. Acting with these operators on the vacuum creates scattering states.
The method described above can be applied to fields with non-zero spin. As an example, we derive operators which create scattering states of $U(1)$ gauge bosons in terms of smeared CFT operators. A gauge field on Anti-de Sitter space has multiple allowed boundary conditions and the expression for creation and annihilation operators depends on those. Quantization in AdS requires us to fix the boundary conditions of the magnetic or electric field at the asymptotic boundary, which we will refer to as ``magnetic'' and ``electric'' quantizations, respectively. For standard boundary conditions, which fix the magnetic field at the boundary, we find that bulk photon creation operators can be expressed as a particular smearing of a dual global $U(1)$ current. For example, for the case of the creation operator of an outgoing photon of negative helicity one finds
\begin{align}
\label{eq:IntroPhoton}
\sqrt{2\omega_{\vec{q}}} \, a_{\vec{q}}^{\dagger(-)}=& {-1\over 4\omega_{\vec{q}}}{1+z_q \bar{z}_q\over{\sqrt{2}}} \int_0^{\pi} d\tau \, e^{i \omega_{\vec{q}} L\left({\pi\over 2}-\tau \right)}
\int d^2 z {1\over (z_q-z)^2}j^+_{\bar{z}}(\tau,z,\bar{z}),
\end{align}
which takes the form of a $U(1)$ current smeared over the boundary. Here, $\omega_{\vec{q}}$ is the frequency of the photon and $(z_q,\bar{z}_q)$ specify the photon's direction.
Conformal field theories associated to different boundary conditions in AdS are related by a $S$ transformation, which acts as electric-magnetic duality in the bulk \cite{Witten:2003ya}. If instead of fixing the magnetic field, we fix the electric field, the expression for creation/annihilation operators involves a CFT dynamical $U(1)$ gauge field. The expression equivalent to equation \eqref{eq:IntroPhoton} now involves a boundary gauge field inserted at a particular location on the boundary $S^2$ and smeared over time,
\begin{align}
\sqrt{2\omega_{\vec{q}}} \, v_{\vec{q}}^{\dagger(-)}=&{1\over 4\omega_q}{1+z_q \bar{z}_q\over{\sqrt{2}}} \int_0^{\pi} d\tau \, e^{i \omega_{\vec{q}} L\left({\pi\over 2}-\tau \right)}
\partial_{\tau}A^+_{z}(\tau,z_q,\bar{z}_q).
\end{align}
These results allow to study how further properties of the S-matrix arise from CFT correlation functions. One such property are Weinberg's soft theorems \cite{Weinberg:1965nx}. Recently, a lot of work has been devoted to understanding them as Ward identities of asymptotic gauge transformations. For a review see \cite{Strominger:2017zoo}. Moreover, soft theorems for gravity have been proposed to play a role in resolutions to the black hole information paradox in flat space \cite{Hawking:2016msc}. Furthermore, it has been suggested that the asymptotic symmetry group of gravity in asymptotically flat spacetimes \cite{Sachs:1962zza,Bondi:1962px} plays an important role in the construction of a dual theory \cite{Strominger:2013jfa}. In the present paper, we show that for the case of a $U(1)$ gauge theory, Weinberg soft theorems can be understood as arising from Ward identities of a CFT in the strict large $N$ limit.
The paper is organized as follows. In section \ref{sec:ScatteringStatesScalar} we give a formal derivation of CFT operators which create flat space scattering states. This will give a formula for scattering amplitudes of massive and massless scalars in terms of CFT correlation functions. Section \ref{sec:GaugeFields} discusses the two different quantization schemes for $U(1)$ gauge fields in $AdS_4$ and reviews how they are related to $SL(2, \mathbb Z)$ transformations of the dual CFT. We derive expressions for annihilation and creation operators for in and out-going photons for both quantization schemes. Section \ref{sec:WST} uses the results of previous sections to connect CFT Ward identities to Weinberg soft theorems. We close with a discussion in section \ref{sec:conc}.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 367 |
\section*{Introduction} Let ${\bf K}$ be an algebraically closed field of characteristic
zero. Let $f$ be an irreducible monic polynomial of
${\bf R}={\bf K}[[x]][y]$, say $f=f(x,y)=y^n+a_1(x).y^{n-1}+\ldots+a_n(x)\in {\bf R}$. Up to a change of coordinates, we assume that $a_1(x)=0$.
For all $ g\in {\bf R}$ let int$(f,g)$ denote the intersection multiplicity of $f$ and $g$.
Let $\Gamma(f)=\lbrace {\rm int}(f,g); g\in {\bf R}-(f)\rbrace $ be the semigroup of $f$. If $f'$ is another irreducible polynomial of
${\bf R}$, then $f$ and $f'$ are said to be
{\it equisingular} if $\Gamma(f)=\Gamma(f')$ (for example, $y^2-x^3$
and $y^3-x^2$ are equisingular because they are both associated with the semigroup
generated by $2,3$. In particular, two equisingular polynomials of
${\bf R}$ need not have the same degree in $y$). It is well-known that in this case $\mu(f)=\mu(f')$, where $\mu(f)=int(f_x,f_y)$ is called the
{\it Milnor number} of $f$. The converse is false. The {\it
equisingularity class} of the polynomial $f$ is the set of
irreducible polynomials of ${\bf R}$ which are equisingular to $f$. It
is of a certain interest to determine this equisingularity class,
which gives a classification of the polynomials of ${\bf R}$ in terms
of subsemigroups of ${\bf Z}$. Another remarkable classification is
obtained if one can characterize all polynomials whose Milnor
number is equal to some fixed nonnegative integer $m$. The aim of
this paper is to study the two questions from an effective point of
view: we first give, for a fixed semigroup of an irreducible
polynomial $f$ of ${\bf R}$, all elements of the equisingularity
class of $f$. Then, for a fixed $m$ in ${\bf N}$, by similar methods we
construct the generic forms of all irreducible polynomials $f$ of
${\bf R}$ such that $\mu(f)=m$. The set of these polynomials is the
union of a finite number of equisingularity classes. We think that this effective
classification is useful in the study of problems and conjectures
in the theory of irreducible curve singularities, particularly in the
understand of their moduli spaces. Our approach
uses the effective notion of approximate roots of a
polynomial $f$ of ${\bf R}$ introduced by S.S. Abhyankar and T.T. Moh and
the notion of generalized Newton polygon introduced by Abhyankar. The first
one gives rise to an algorithm for the computation of the set of generators
of the semigroup of $f$ (and then the set of Newton-Puiseux
pairs of $f$, see definition 1.3., and [1], [2]). The second one is used by
Abhyankar to give an irreducibility criterion for the polynomial $f$ (see [3]).
\noindent We would like to point out that our algorithms are intrinsic and that they
have been implemented with {\it Mathematica} (see [8]), and {\it Maple}.
\section{Characteristic sequences} In this Section we recall the notion of approximate roots of $f$ as well as the characteristic sequences associated with an irreducible polynomial $f=y^n+a_2(x).y^{n-2}+\ldots+a_n(x)$ of ${\bf R}$.
\begin{definicion} {\rm For any monic polynomial $g\in{\bf R}$, the {\bf intersection multiplicity} int$(f,g)$ of $f$ with $g$ is the $x$-order of the $y$-resultant of $f$ and $g$.}\end{definicion}
\noindent
The set $\Gamma(f)=\lbrace {\rm int}(f,g); g\in {\bf R}-f\rbrace $ is a subsemigroup of ${\bf Z}$, called the {\bf semigroup of $f$}.
\begin{definicion} {\rm Let $y(t)=\sum_j a_jt^j\in
{\bf K}[[t]]$ be
a root of $f(t^n,y)=0$, according to Newton Theorem. Then
set $m_0=d_1=n, m_1=$ inf$\lbrace j; a_j\not=0 \rbrace$, and for all $k
\geq 1,$ let
\noindent $m_{k+1}=$ inf$\lbrace j; a_j\not=0$ and $d_k$ does not divide
$j \rbrace$, and $d_{k+1}={\rm gcd}(m_{k+1},d_k)$.
\noindent Since $f$ is irreducible, there exists $h$ such that
$d_{h+1}=1$. We set $m_{h+1}=+\infty$.
\noindent Finally set $r_0=m_0=n$, $r_1= O_x(a_n(x))$-where $O_x$
denotes the $x$-order-, and for all $k=1,\dots, h-1$:
$r_{k+1}=r_k ({\displaystyle\frac{d_k}{d_{k+1}}}) + (m_{k+1}-m_k)$.}\end{definicion}
\noindent (Remark that, since $a_1(x)=0$, then $r_1=m_1$).
\noindent We recall that with respect to these notations,
$r_0,\dots,r_h$ generates the semigroup $\Gamma(f)$ of $f$. We
denote $\Gamma(f)=<r_0,\ldots,r_h>$.
\begin{definicion} {\rm For all $k=1,\ldots, h$, set
$e_k={\displaystyle\frac{d_k}{d_{k+1}}}$. The set
$\lbrace ({\displaystyle\frac{m_k}{d_{k+1}}},e_k), 1\leq k\leq h\rbrace$ is called the set
of {\bf Newton-Puiseux pairs } of $f$.}\end{definicion}
\begin{definicion} {\rm Let $d$ be a positive integer
and assume that $d$ divides $n$. Let $g$ be a monic polynomial of
${\bf R}$, of degree ${\displaystyle\frac{n}{d}}$
in $y$. We call $g$ the $d$-th approximate root of $f$ if one of
the following holds:
i) deg$_y (f-g^d) < n-{\displaystyle\frac{n}{d}}$.
ii) in the expansion $f=g^d +\alpha_1 g^{d-1}+\ldots + \alpha_d$ of $f$ with respect to the powers of $g$,
$\alpha_1=0$.
\noindent Remark that i) and ii) are equivalent.
\noindent We denote the $d$-th approximate root of $f$ by
App$_d(f)$. It is clear that App$_d(f)$ is unique, and also that it is
effectively computable if the series $a_k(x), k=2,\ldots,n$, are
polynomials.}
\end{definicion}
\begin{nota}{\rm Given a divisor $d$ of $n$, the
$d$th approximate root App$_d(f)$ of $f$ can be effectively constructed from
the equation of $f$ in the following way:
\noindent Take $G_0=y^{n/d}$ and let
$f=G_0^d +\alpha^0_1 G_0^{d-1}+\ldots + \alpha^0_d$ be the expansion
of $f$ with respect to the powers of $G_0$.
i) If $\alpha^0_1=0$, then $G_0= {\rm App}_d(f)$.
ii) If $\alpha^0_1\not=0$, then set
$G_1=G_0+\displaystyle{{\alpha^0_1\over d}}$ and consider the expansion
$f=G_1^d +\alpha^1_1 G_1^{d-1}+\ldots + \alpha^1_d$ of $f$ with respect
to the powers of $G_1$. If $\alpha^1_1\not=0$, then easy calculations show that
deg$_y\alpha^0_1 > {\rm deg}_y\alpha^1_1$. This procees shall stop
after a finite number of steps, constructing App$_d(f)$.}
\end{nota}
\begin{nota}{\rm If the characteristic of ${\bf K}$ is not zero and if
this characteristic does not divide $n$, then the
construction above applies without any restriction. Otherwise,
the theory of approximate roots does not work as it. Further
information can be found in [9].}
\end{nota}
\noindent Let $g_1,\ldots,g_h, g_{h+1}$ be the $d_k$-th approximate
roots of $f$, for $k=1,\ldots,h+1$ (in particular
$g_1=y$ and $g_{h+1}=f$).
\begin{lema} (see [1], (8.2) the Fundamental Theorem (part one)) For all $k=1,\ldots,h$, we have:
i) {\rm int}($f,g_k)=r_k$.
ii) $g_k$ is irreducible in ${\bf R}$ and
$\Gamma (g_k) = < {\displaystyle\frac{r_0}{d_k}},\ldots,
{\displaystyle\frac{r_{k-1}}{d_k}}>$. Furthermore,
$g_1,\ldots,$ $g_{k-1}$ are the approximate roots of $g_k$.\end{lema}
\begin{lema} (see [13]) The following formulas hold:
- {\rm int}$(f_x, f_y)=\displaystyle\sum_{i=1}^{h} (e_i-1)r_i
-n+1$. In particular ${\rm int}(f_x, f_y)$ is even.
- For all $k=2,\dots, h$,
{\rm int}$(f_x, f_y)=d_k.${\rm int}$(g_{k_x}, g_{k_y})+
\displaystyle\sum_{i=k}^{h} (e_i-1)r_i - d_k + 1.$
\end{lema}
\begin{demostracion}{.} The proof of the first formula can be found in
[13] (3.14., page 18). The second formula results from the first one
by easy calculations.
\end{demostracion}
\begin{nota}{\rm The intersection multiplicity int$(f_x,f_y)$ is also called
the Milnor number of $f$. It is an invariant of $f$ and, by the
formula above, it is common to the elements of the equisingularity
class of $f$. It also coincides with the conductor of the semigroup
$\Gamma(f)$ -usually denoted by $c$- which has the following
numerical characterization: for all $p\geq c, p\in \Gamma(f)$. Furthermore, given two
integers $a,b$, if $a+b=m-1$ then exactly one of $a,b\in \Gamma(f)$. It follows that,
since $\Gamma(f)$ has no negative integers, Card(${\bf N}-\Gamma)={\displaystyle\frac{m}{2}}$.
In fact, $c$ is nothing but the order of the conductor of the quotient
$\displaystyle {{\frac {\bf R} {(f)}}}$ into its integral closure. Contrary to
the Milnor number, the conductor can be defined without restriction on the characteristic
of ${\bf K}$. An exhaustive exposition of this theory in positive characteristic
can be found in [9].}
\end{nota}
\bigskip
\section{Generalized Newton polygons and the irreducibility criterion of
Abhyankar}
\medskip
\noindent Let $f= y^n+a_2(x)y^{n-2}+\ldots+a_n(x)$ be a
monic polynomial, non necessarily irreducible in ${\bf R}$. In this section
the notations introduced above will have a more general meaning:
$r=(r_0=n,r_1,\ldots,r_h)$ will denote any
sequence of integers such that $r_k < r_{k+1}$ for all
$k=1,\ldots,h-1$, and we shall set $d_{k+1}=$ gcd$(r_0,r_1,\ldots, r_k)$ for all
$k=0,\ldots,h$. For all $k=1,\ldots,h$, we set
$e_k={\displaystyle\frac{d_k}{d_{k+1}}}$;
$g =(g_1,\ldots,g_h,g_{h+1}=f)$ will be a sequence of monic
polynomials of ${\bf R}$ such that deg$_yg_k ={\displaystyle\frac{n}{d_k}}$ for all
$k=1,\ldots,h$. We recall some important properties.
\medskip
\begin{theorema}{\rm (see [1], (8.3) The fundamental Theorem (part two)) Let
$$
B=\lbrace b=(b_1,b_2,\ldots,b_h,b_{h+1})\in {\bf N}^{h+1};
b_1<e_1,\ldots,b_h<e_h, b_{h+1} <+\infty\rbrace
$$
\noindent For all $ b\in B$, denote
$g^{b}=g_1^{b_1}.\ldots.g_h^{b_h}.f^{b_{h+1}}$, then we have:
i) ${\bf R}=\sum_{ b\in B}{\bf K}[[x]]. g^{b}$.
ii) Let $p$ be a polynomial of ${\bf R}$ and
write $p=\sum_{k=1}^s a_k(x).g^{b^k}$, where $b^k\in B$ for all $k=1,\ldots,s$. Moreover let $b^k_0=$ O$_x a_k(x)$, then
associate with any ``monomial'' $ a_k(x).g^{b^k}$ the
integer $< (b^k_0,b^k_1,\ldots,b^k_h), r> = b^k_0.r_0+\sum_{i=1}^{h} b^k_i.r_i$. Finally
let $B'=\lbrace b^k; b^k_{h+1}=0\rbrace$. With these notations we have the following:
1) If $B'$ contains at least two elements, then
for all $ b^i, b^j\in B'$,
$$
b^i\not= b^j\Longleftrightarrow < (b^i_0,b^i_1,\ldots,b^i_h), r>\not=
< (b^j_0,b^j_1,\ldots,b^j_h), r>
$$
2) $f$ doesn't divide $p$ iff $B'\not= \emptyset$, and in this case
there is a unique $k_0$ such that
$< (b^{k_0}_0,b^{k_0}_1,\ldots,b^{k_0}_h), r>=$ inf$\lbrace < (b^k_0,b^k_1,\ldots,b^h_h), r>; b^k\in B'\rbrace$.}
\end{theorema}
\begin{definicion} {\rm (see [3])
The integer $< (b^{k_0}_0, b^{k_0}_1,\ldots,b^{k_0}_h), r>$
is called {\bf formal intersection multiplicity}
of $p$ with respect to $(r, g)$ and will be
denoted by fint$(p, r, g)$.}\end{definicion}
\medskip
\noindent Now we recall the notion of { \bf generalized Newton
polygon}. Let $p$ be a monic polynomial of ${\bf R}$ of degree $n$
in $y$
and consider a monic polynomial $q$ of ${\bf R}$ of degree
${\displaystyle\frac{n}{d}}$ in $y$, where
$d$ is a divisor of $n$. Let
$$p=q^{d}+\alpha _1(x,y)q^{d-1}+\ldots+\alpha_{d}(x,y)$$
\noindent be the expansion of $p$ with respect to the powers of $q$, and
consider the sequences $r$, $g$ defined above.
One associates with $p$ the generalized Newton polygon which is
defined as the union of all compact sides of the convex hull in ${\bf R^2}$ of the set formed by the points
(fint$(\alpha_k, r, g), (d-k).$fint$(q, r, g))$ for all $1\leq k\leq d$. It will be denoted by
GNP$(p,q,r,g)$ (see [3]). \newline
With these notations one has the following:
\medskip
\noindent {\bf Irreducibility criterion} (see [3])
\medskip
\noindent Write $p= y^n+a_1(x)y^{n-1}+\ldots+a_n(x) \in {\bf R}$ and
assume, possibly after a change of variables, that $a_1(x)=0$. Consider
the sequences $r_k,g_k,d_k$ defined in the following way:
$r_0=d_1=n$
$g_1=y, r_1=$ int$(p,g_1)$, and for all $k\geq 2$:
$d_{k}=$ gcd $(r_0,r_1,\ldots,r_{k-1}), g_{k}=$ App$_{d_{k}}(p), r_{k}=$
int$(p,g_k)$.
\noindent $p$ is irreducible if and only if the following conditions hold.
1) There is $h\in N$ such that $d_{h+1}=1$.
2) For all $k=1,\ldots,h-1$, $r_{k+1}d_{k+1} > r_kd_k$.
3) Set $p=g_{h+1}$ and let for all $k=1,\ldots,h$,
$e_k={\displaystyle\frac{d_k}{d_{k+1}}}$ and
$g^k=(g_1,\ldots,g_k)$. Let also for all $k=1,\ldots,h$,
${\displaystyle\frac{r}{d_{k+1}}}=({\displaystyle\frac{r_0}{d_{k+1}}},
\ldots,{\displaystyle\frac{r_k}{d_{k+1}}})$. Then for all $k=1,\ldots, h$, the
generalized Newton polygon GNP$(g_{k+1},g_k,
{\displaystyle\frac{r}{d_{k+1}}}, g^k)$ is the line
segment joining $(0,{\displaystyle\frac{r_k}{d_{k+1}}}.e_k)$ and
$({\displaystyle\frac{r_k}{d_{k+1}}}.e_k,0)$.
\medskip
\noindent {\bf Remarks 2.3:} i) (see [3]) Suppose that $p$ is irreducible, and
let $r=(r_0,r_1,\ldots,r_h)$ and $g =(g_1=y,
g_2,\ldots,g_h,g_{h+1}=p)$ be the sequences defined above. Let
$p'$ by a polynomial of ${\bf R}$ and consider the expansion of
$p'$ with respect to the sequences $r,g$ (see Theorem 2.1.). If the
corresponding set $B'$ is non empty, then fint$(p',r,g)=$ int$(p',p)$.
\vskip0.1cm
\noindent ii) Part 3) of the criterion can be precised as follows:
whenever $p$ is irreducible,
the generalized Newton polygon GNP$(g_{k+1},g_k,{\displaystyle\frac{r}{d_{k+1}}},g^k)$
just contains the two extremal points for all $k=1,\ldots,h-1$. In
fact let $k\in\lbrace 1,\ldots,h\rbrace$ and
let $g_{k+1}=g_k^{e_k}+\alpha_2(x,y).g_k^{e_k-2}+\ldots+\alpha_{e_k}(x,y)$
be the expansion of $g_{k+1}$ w.r.t. $g_k$, then we have:
a) int$(g_{k+1},\alpha_{e_k}(x,y))={\displaystyle\frac{r_k}{d_{k+1}}}.e_k=$ int$(g_{k+1},g_k^{e_k})$.
b) For all $i=2,\ldots,e_{k}-1$, int$(g_{k+1},\alpha_{i}(x,y)) > {\displaystyle\frac{r_k}{d_{k+1}}}.i$,
In particular int$(g_{k+1},\alpha_{i}(x,y))+{\rm int}(g_{k+1},g_k^{e_k-i})>{\displaystyle\frac{r_k}{d_{k+1}}}.e_k$.
\vskip0.1cm
\noindent iii) (see [13]) As an immediate consequence of ii) a) we have that,
for all $k=1,\dots, h$,
$$ {\displaystyle\frac{r_k}{d_{k+1}}}.e_k\in
e_k.\Gamma(g_{k})=<{\displaystyle\frac{r_0}{d_{k+1}}},\dots,
{\displaystyle\frac{r_{k-1}}{d_{k+1}}}>.$$
In particular
$$ r_k.e_k\in <r_0,\dots, r_{k-1}>.$$
\medskip
\section{Constructing the equisingularity class}
\noindent In this Section we fix a semigroup of nonnegative integers
$\Gamma=<r_0,r_1,\ldots,r_h>$, and we set $d_1=r_0$ and
$d_{k+1}=$ gcd$(r_0,\ldots,r_k)$ for all $k=1,\ldots,h$ (we set by
convention $r_{h+1}=d_{h+2}=+\infty$). Moreover we assume that
$d_{k+1}=1$ and that $r_{k+1}.d_{k+1}> r_k.d_k$ for all
$k=1,\ldots,h$ (*) (this condition appeared in the irreducibility
criterion in Section 2). This implies that the
sequence $(r_1,\ldots,r_h)$ is strictly increasing. This also holds if
$r_1$ is replaced by $r_0$.
Let $\tilde{\bf R}$ denote the set of all irreducible monic polynomials $f$
of ${\bf R}$ of the form $f=f(x,y)=y^n+a_2(x).y^{n-2}+\ldots+a_n(x)$. Condition $(*)$ implies that there exists a
polynomial $f\in \tilde{\bf R}$ such that
$\Gamma=\Gamma(f)$ (see [13]). Here we give the generic forms of all these
polynomials, i.e., we describe the set of elements of $\tilde{\bf R}$
having the semigroup $\Gamma$. The construction can be performed with
respect to the arrangements $(r_0,r_1,r_2,\ldots,r_h)$ and
$(r_1,r_0,r_2,\ldots,r_h)$. We shall perform it with respect to the
first arrangement. The polynomials that we would get with respect to
the second arrangement are those obtained by exchanging $x$ and $y$.
\medskip
\noindent In this and in the following Sections we shall assume that
$r_0,\dots, r_h$ form a minimal system of generators for $\Gamma$.
This condition can be reformulated equivalently as a numerical criterion. This
is what we are going to do next. First we derive a
useful identity: Set for all $1\leq i\leq h, e_i={\displaystyle\frac{d_i}{d_{i+1}}}$.
For all $2\leq k \leq h$ we have:
\begin{eqnarray*}\displaystyle\sum_{i=1}^{k-1} (e_i -1)r_i&=&
\displaystyle\sum_{i=1}^{k-1} (r_{i+1}- (m_{i+1} -m_i)) -\displaystyle\sum_{i=1}^{k-1} r_i\\
&=&r_k-r_1-\displaystyle\sum_{i=1}^{k-1}(m_{i+1}-m_i)
= r_k-m_k\quad(**)
\end{eqnarray*}
\noindent Now we are ready to prove the following
\begin{lema} Suppose that $r_0,\dots, r_h$ satisfy condition $(*)$. These
numbers form a minimal system of generators for the
semigroup $\Gamma$ if and only if $d_1>d_2>\cdots>d_h>d_{h+1}$.\end{lema}
\begin{demostracion}{.} Remark that in general $d_k\geq d_{k+1}$ for
all $k=1,\dots, h$. Moreover, recall that the minimality of the system
of generators is equivalent to the condition that
$r_{k}\notin <r_0,\dots, r_{k-1}>$ for all $k=1,\dots, h$. First
suppose that
this condition is not fulfilled for some index $k$. Then
$d_{k}=$gcd$(r_0,\dots,r_{k-1})$ certainly divides $r_{k}$.
Hence $d_{k+1}=$gcd$(d_{k},r_{k})=d_{k}$. For the converse fix an
index $k$ and suppose that $d_{k+1}=d_k$. Then there exist some
integers $\alpha_0,\dots,\alpha_{k-1}$ such that
$$
r_k = \displaystyle\sum_{i=0}^{k-1} \alpha_ir_{i}.\qquad\qquad(1)
$$
\noindent Now let $\beta_{k-1}$ be the (nonnegative) remainder of the euclidean
division of $\alpha_{k-1}$ by
$e_{k-1}={\displaystyle{{d_{k-1}}\over{d_{k}}}}$. Since
the semigroup
$\Gamma$ verifies condition (*), $\Gamma$ is the semigroup of a
polynomial of $\tilde{\bf R}$, in particular, by Remarks 2.3 iii),
$r_{k-1}.(e_{k-1})\in<r_0,\dots, r_{k-2}>$; hence we can transform $(1)$ in such a way that $\alpha_{k-1}=\beta_{k-1}$.
If we successively perform the same procedure for the indices $k-2,\dots,1$, we finally obtain that in (1)
$0\leq\alpha_i< e_i$ for all $i=1,\dots,k-1$. Now by $(**)$
$$
\alpha_0r_0 = r_k-\displaystyle\sum_{i=1}^{k-1} \alpha_ir_{i}>r_k-\displaystyle\sum_{i=1}^{k-1} (e_i-1)r_i=m_k>0,
$$
\noindent hence $\alpha_0>0$. This proves that $r_{k}\in <r_0,\dots, r_{k-1}>$ and completes the proof.\end{demostracion}
\noindent The construction of the generic form of all polynomials $f\in\tilde{\bf R}$ having $\Gamma$ as a semigroup is based on the notion of {\bf generalized
Newton polygons} introduced in Section 2. We shall recursively
construct the sequence of approximate roots $g_1,\ldots,g_h,g_{h+1}=f$.
\medskip
\noindent Let $g_1=$ App$_{d_1}(f)$ (and recall that, since
$a_1(x)=0$, then $g_1=y$). From Section 2 we know that $g_2=$
App$_{d_2}(f)$ satisfies:
i) $\Gamma(g_2)=<{\displaystyle\frac{r_0}{d_2}},{\displaystyle\frac{r_1}{d_2}}>$.
ii) GNP$(g_2,g_1,{\displaystyle\frac{r}{d_2}}, g^1)$ is the line segment
joining the two points $(0,{\displaystyle\frac{r_1}{d_2}}.e_1)$
and $({\displaystyle\frac{r_1}{d_2}}.e_1,0)$.
\noindent By virtue of part ii) of Remarks 2.3, this yields to the following generic form of $g_2$:
$$
g_2 = y^{{\frac{r_0}{d_2}}} +
a.x^{{\frac{r_1}{d_2}}}+
\sum_{i.{\frac{r_0}{d_2}}+
j.{\frac{r_1}{d_2}} >
{\frac{n}{d_2}}.{\frac{r_1}{d_2}}; 0\leq j<
{\frac{d_1}{d_2}}={\frac{r_0}{d_2}}} a_{ij}x^iy^j,
$$
\noindent where $a\in {\bf K}-0$ and for all
$(i,j)$, $a_{ij}\in {\bf K}$.
\noindent Suppose that we have the generic forms of $g_1,\ldots,g_k$
and consider the expansion of $g_{k+1}$ with respect to $g_k$:
$g_{k+1} =
g_k^{e_k}+\alpha_2(x,y)g_k^{e_k-2}+\ldots+\alpha_{e_k}(x,y)$.
\noindent From Section 2 we know that:
i) $\Gamma(g_{k+1})=<{\displaystyle\frac{r_0}{d_{k+1}}},\ldots,
{\displaystyle\frac{r_k}{d_{k+1}}}>$.
ii) GNP$(g_{k+1},g_k,{\displaystyle\frac{r}{d_{k+1}}},g^k)$ is the
line segment joining the two points $(0,{\displaystyle\frac{r_k}{d_{k+1}}}.
e_k)$ and $({\displaystyle\frac{r_k}{d_{k+1}}}.e_k,0)$.
\noindent It follows from Remarks 2.3 that
(1) int$(g_{k+1},\alpha_{e_k}(x,y))={\rm
fint}(\alpha_{e_k}(x,y),{\displaystyle\frac{r}{d_{k+1}}}, g^k)=
{\displaystyle\frac{r_k}{d_{k+1}}}.e_k$,
\noindent and that for all $i=2,\ldots,e_k-1$:
(2) int$(g_{k+1},\alpha_i(x,y))={\rm
fint}(\alpha_i(x,y),{\displaystyle\frac{r}{d_{k+1}}},g^k)>
{\displaystyle\frac{r_k}{d_{k+1}}}.i$,
\noindent Let $B^k$ be the set of all
$\theta=(\theta_0,\ldots,\theta_{k-1})\in {\bf N}^k$ such that
$ 0\leq \theta_j < e_j$ for all $j=1\ldots,k-1$, then
associate with all $\theta\in B^k$ the ``monomial''
$M_{\theta}=x^{\theta_0}.g_1^{\theta_1}.\ldots.g_{k-1}^{\theta_{k-1}}$.
For all $i\in {\bf N}$ and for all $\theta\in B^k$, we say that $M_{\theta}$ is of type $(k,i,1)$
(resp. of type $(k,i,2)$) if
$$
{\displaystyle\frac{r_k}{d_{k+1}}}.i=\theta_0.{\displaystyle\frac{r_0}{d_{k+1}}}+
\theta_1.{\displaystyle\frac{r_1}{d_{k+1}}}
+\ldots +\theta_{k-1}.{\displaystyle\frac{r_{k-1}}{d_{k+1}}}
$$
\noindent (resp.
$$
{\displaystyle\frac{r_k}{d_{k+1}}}.i<\theta_0.{\displaystyle\frac{r_0}{d_{k+1}}}
+\theta_1.{\displaystyle\frac{r_1}{d_{k+1}}}
+\ldots +\theta_{k-1}.{\displaystyle\frac{r_{k-1}}{d_{k+1}}}).
$$
\noindent Let
$E(k,i,1)$ (resp. $E(k,i,2)$)
be the set of monomials $M_{\theta},\theta\in B^k$, of type $(k,i,1)$ (resp. of type $(k,i,2)$). Since
${\displaystyle\frac{r_k}{d_{k+1}}}.e_k\in <{\displaystyle\frac{r_0}{d_{k+1}}},\ldots,
{\displaystyle\frac{r_{k-1}}{d_{k+1}}}>$,
then $E(k,e_k,1)$ is reduced to one element. If we write this element as
$M_{\theta^k}=x^{\theta_0}.g_1^{\theta_1}.\ldots.g_{k-1}^{\theta_{k-1}}$, then
$(\theta_0,\theta_1,\ldots,\theta_{k-1})$ can be
calculated by euclidean division.
\noindent Using Remark 2.3., (1) and (2) this leads to the following generic forms
of $\alpha_2,\ldots,\alpha_{e_k}:$
$$
\alpha_{e_k}=a.M_{\theta^k}+\sum_{M_{\theta}\in
E(k,e_k,2)}a_{\theta}.M_{\theta},
$$
\noindent (resp. for all $i=2,\ldots, e_k-1$,
$$
\alpha_{i}=\sum_{M^i_{\theta}\in E(k,i,2)}a^i_{\theta}.M^i_{\theta}),
$$
\noindent where $a\in {\bf K}-0$, and for all $\theta,
a_{\theta}\in {\bf K}$ (resp. for all $\theta$ and for
all $i=2,\ldots, e_k-1$, $a^i_{\theta}\in {\bf K}$).
\medskip
\begin{nota}{\rm We proved that, if $\Gamma$ is the semigroup of a polynomial
$f\in\tilde{\bf R}$, then $f$ and its approximate roots $g_1,\ldots,g_h$
belong to the set of polynomials constructed above. Conversely, let $(g_1,\ldots,
g_h,g_{h+1}=f)$ be as above, then part ``only if'' of the irreducibility
criterion of Abhyankar shows that $f$ is irreducible.}\end{nota}
\begin{nota} {\rm Given $1\leq k \leq h$, it follows from the above
construction that a polynomial $g_{k+1}$ may have an infinite number
of monomials. In particular the above construction is not algorithmic.
Remark however that $g_{k+1}$ is obtained from the sum
$g_{k}^{e_k}+a.M_{\theta^{k}}, a\in {\bf K}-0$ by adding monomials
that verify some conditions. This suggests the introduction of the
following set of polynomials: let $G_1=y$ and for all $1\leq k\leq h$,
$G_{k+1}=G_k^{e_k}-M_{{\theta}^k}$. For all $1\leq k\leq h$, $g_k$ is obtained
from $G_k$ in an obvious way. Set $G=(G_1,\ldots,G_h,G_{h+1})$ and call it
the canonical element of the set of all $(g_1,\ldots,g_h,g_{h+1})$
constructed above. The above calculation leads to an algorithm that
computes this canonical element. It is based on euclidean division
in ${\bf N}$. The different steps can be summarized as follows:
\medskip
i) Consider a sequence of integers $r_0 <\ldots <r_h$.
ii) Compute the gcd sequence $d=(d_1,\ldots,d_h,d_{h+1})$ such
that $d_1=r_0$ and for all $2\leq k \leq h+1, d_{k}={\rm gcd}(r_{k-1},d_{k-1})$.
Let $e_k={\displaystyle\frac{d_k}{d_{k+1}}}$ for all $1\leq k\leq h$.
iii) If either $d_{h+1} > 1$, or $r_k\cdot e_k\geq r_{k+1}$ for
at least one $1\leq k \leq h$, then the sequence $(r_0, \ldots, r_h)$ is
not the semigroup of an irreducible polynomial of $\tilde{\bf R}$.
iv) Assume that $d_{h+1}=1$ and that $r_k\cdot d_k< r_{k+1}\cdot d_{k+1}$ for
all $1\leq k \leq h-1$.
\quad\quad a) If $d_k=d_{k+1}$ for some $1\leq k\leq h$, then eliminate
$r_{k}$ from the $r$-sequence of i).
\quad\quad b) Assume that $d_1 > d_2 >\ldots >d_h > d_{h+1}=1$. Then
for all $1\leq k \leq h$, compute (the unique) $\theta^k=(\theta_0^k,\ldots,
\theta_{k-1}^k)$ such that $ 0\leq \theta_j^k < e_j$
for all $j=1\ldots,k-1$ and
${\displaystyle\frac{r_k}{d_{k+1}}}.e_k=\theta_0^k.
{\displaystyle\frac{r_0}{d_{k+1}}}+
\theta_1^k.{\displaystyle\frac{r_1}{d_{k+1}}}
+\ldots +\theta_{k-1}^k\cdot {\displaystyle\frac{r_{k-1}}{d_{k+1}}}$.
\quad\quad c) The canonical element is $G=(G_1,\ldots,G_h,G_{h+1})$
where $G_1=y$ and for all $2\leq k\leq h+1$,
$G_k=G_{k-1}^{e_{k-1}}-x^{\theta_0^k}y^{\theta_1^k}
\cdot\ldots\cdot G_{k-1}^{\theta_{k-1}^k}$.}
\end{nota}
\noindent This algorithm has been implemented with {\it Mathematica}
(see [8]), and {\it Maple}: the input is an increasing sequence of positive integers.
Then the output is "false" if this sequence
does not generate the semigroup of an irreducible polynomial of $\tilde{\bf R}$. Otherwise,
we get the canonical element described above.
\medskip
\noindent Note that our implementation is based on the following:
given $r_0,r_1,\dots,r_{k-1}$, we need to compute the unique
$\theta^k=(\theta_0^k,\ldots,
\theta_{k-1}^k)$ such that $ 0\leq \theta_j^k < e_j$
for all $j=1\ldots,k-1$ and
${\displaystyle\frac{r_k}{d_{k+1}}}.e_k=\theta_0^k.
{\displaystyle\frac{r_0}{d_{k+1}}}+
\theta_1^k.{\displaystyle{\frac{r_1}{d_{k+1}}}}+\cdots +\theta_{k-1}^k.
{\displaystyle{\frac{r_{k-1}}{d_{k+1}}}}$. Instead
of applying the Euclidean division,
We have preferred to scann lists of values, namely the set
of values $(a_0,a_1,\ldots,a_{k-1})$ where for all $i\geq 1,
0\leq a_i < e_i$ and $0\leq a_0 \leq
{\displaystyle{\frac{r_k}{d_{k+1}}}.e_k.{\frac{d_{k+1}}{r_0}}=
{\frac{r_k.e_k}{r_0}}}$. The cardinality of this set is:
$$
{\frac{r_k.e_k}{r_0}}.\prod_{i=1}^{k-1}e_k={\frac{r_k.e_k}{r_0}}.{\frac{d_1}{d_k}}={\frac{r_k}{d_{k+1}}}
$$
\noindent In conclusion the set of the values scanned in the algorithm
is
bounded by ${\displaystyle{\sum_{k=1}^h {\frac{r_k}{d_{k+1}}}}}$.
\begin{nota} {\rm An element $f$ whose semigroup is $\Gamma$ can also be
calculated by using the theory of Gr\"obner bases: a reduced Gr\"obner basis
with respect to any well-ordering on $\bf N^3$ that
eliminates $t$ from the equations $x-t^n, y-t^{m_1}-\ldots-t^{m_r}$ contains
a unique polynomial $f(x,y)$. If we consider $f$ as an element of ${\bf K}[[x,y]]$,
then obviously $\Gamma=<r_0,\ldots,r_h>$ is the semigroup of $f$. It is
well known that the complexity of a Gr\"obner basis is in general
doubly exponential. Moreover, the algorithm computes more than we
need. We
think that our option is more natural in view of
our situation, especially because of its complexity
and that the output is expressed in terms of the polynomial $f$.}
\end{nota}
\noindent {\bf Example:}
\noindent Let $\Gamma = <8,12,50,101>$. Here $h=3$, the $r$-sequence is
$r=(8,12,50,101)$, and the gcd-sequence is $d=(8,4,2,1)$. Moreover,
$e_1=e_2=e_3=2$. Let us construct
the canonical element $G=(G_1,G_2,G_3,G_4)$ following the algorithm
above. Here we start directly from point iv), b):
$k=1: {\displaystyle\frac{r_1}{d_2}}\cdot{e_1}=3\cdot 2=
\theta_0^1\cdot{\displaystyle\frac{r_0}{d_2}}=\theta_0^1\cdot 2$ implies
that $\theta_0^1=3$.
$k=2: 50={\displaystyle\frac{r_2}{d_3}}\cdot {e_2}=\theta_0^2\cdot
{\displaystyle\frac{r_0}{d_{3}}}+
\theta_1^2\cdot {\displaystyle\frac{r_1}{d_{3}}}= \theta_0^2\cdot 4+
\theta_1^2\cdot 6$ with $0\leq \theta_1^2 <2$.
This implies that $\theta_1^2=1$, and $\theta_0^2=11$.
$k=3: 202={\displaystyle\frac{r_3}{d_4}}\cdot {e_3}=\theta_0^3\cdot
{\displaystyle\frac{r_0}{d_{4}}}+
\theta_1^3\cdot {\displaystyle\frac{r_1}{d_{4}}}+\theta_2^3\cdot
{\displaystyle\frac{r_2}{d_{4}}}=\theta_0^3\cdot 8+
\theta_1^3\cdot 12+\theta_2^3\cdot 50$ with $0\leq \theta_1^3, \theta_2^3<2$.
This implies that $\theta_2^3=1, \theta_1^3=0, \theta_0^3=19$.
\medskip
\noindent In particular, $G_1=y, G_2=G_1^2-x^3=y^2-x^3,
G_3=G_2^2-x^{11}\cdot G_1=(y^2-x^3)^2-x^{11}\cdot y,
G_4=G_3^2-x^{19}\cdot G_2= [(y^2-x^3)^2-x^{11}\cdot y]^2-
x^{19}\cdot (y^2-x^3).$
\medskip
\noindent With the same notations as above, the
set of elements $(g_1, g_2, g_3, g_4=f)$ is then given by:
$$g_1=y.$$
$$
g_2= y^2+\alpha_2(x)= y^2+ax^3+\sum_{M_{\theta}\in
E(1,2,2)}a_{\theta}.M_{\theta},
$$
\noindent where $a\in {\bf K}-0$, and for all $\theta$, one has
$a_{\theta}\in {\bf K}$ and $M_{\theta}=x^{\theta_0}$, with $6<2\theta_0$.
Moreover
$$g_3= g_2^2 +\alpha'_2(x,y)= g_2^2 + a' x^{11}y+\sum_{M'_{\theta}\in
E(2,2,2)}a'_{\theta}.M'_{\theta},$$ where
- $a'\in {\bf K}-0$, and for all $\theta$, $a'_{\theta}\in {\bf K}$;
- for all $\theta$, $M'_{\theta}=x^{\theta'_0}y^{\theta'_1}$, with $50 <4\theta'_0 +6\theta'_1$.
\noindent Finally,
$$f=g_3^2+\alpha''_2(x,y)= g_3^2 + a'' x^{19}g_2+\sum_{M''_{\theta}\in
E(3,2,2)}a''_{\theta}.M''_{\theta},$$ where
- $a''\in {\bf K}-0$, and for all $\theta$, $a''_{\theta}\in {\bf K}$;
- for all $\theta$, $M''_{\theta}=x^{\theta_0}y^{\theta_1}g_2^{\theta_2}$, with $202<8\theta''_0 +12\theta''_1+50\theta''_2$.
Hence the generic form of all polynomials having $\Gamma$ as a semigroup is the following:
$$f= [ (y^2+ax^3+F)^2+a'x^{11}y+F']^2 +
a''x^{19}(y^2+ax^3+F) +F'',$$
where $a,a',a''\in {\bf K}-0$ and $F$, $F'$ and $F''$ are arbitrary linear combinations of
monomials from $E(1,2,2)$, $E(2,2,2)$ and $E(3,2,2)$ respectively.
\begin{nota}{\rm i) The construction above does not depend on the choice
of the coefficients in the field ${\bf K}$ -provided that it is of
characteristic zero-, in particular the algorithm described allows
us to work over anysubring $A$ of ${\bf K}$. If
$A=k[t_1,\ldots,t_m]$ is a polynomial ring over a field $k$ of
characteristic zero and ${\bf K}$ is the algebraic closure of $A$
into its fractions field, then we get the equisingularity
class of the $(t_1,\ldots,t_m)$-generic section.
\medskip
ii) The restriction to the zero characteristic is made
only because of the use of the approximate roots in
the algorithm. If the characteristic of ${\bf K}$ does not
divide $r_0$, then everything above applies -see Remark 1.6.-.
Note that a more general irreducibility criterion has been given
by A. Granja (see [12]), but it does not seem to be in computational
form.}
\end{nota}
\section{ Equisingularity classes with a given Milnor number}
\medskip
\noindent In this Section we generalize the results of
Section 3 in the following way: let $m\in {\bf N}$ be a fixed
integer. If $m\in 2.{\bf N}$, then there exists a
polynomial $f=y^n+a_2(x).y^{n-2}+\ldots+a_n(x)\in\tilde{\bf R}$
such that int$(f_x,f_y)=m$. Here we shall give the generic forms of all
these polynomials. Remark that if $g$ is another
polynomial of $\tilde{\bf R}$, then $\Gamma(f)=\Gamma(g)$ implies that
int$(f_x,f_y)={\rm int}(g_x,g_y)$. Thus the set of
$f=y^n+a_2(x).y^{n-2}+\ldots+a_n(x)\in \tilde{\bf R}$ such that
int$(f_x,f_y)=m$ is the union of equisingularity classes. We shall
first prove that this union is finite. This is an
immediate application of the next Proposition. We recall that if a subsemigroup of ${\bf Z}$ is minimally generated by $h+1$ elements, then $h$ is called the {\it length} of the semigroup.
\begin{proposicion} \label{majoration} Let $h\in {\bf N}$ and consider a
polynomial $f\in \tilde{{\bf R}}$ such that $h$ is the length of
$\Gamma(f)$. Let $\mu_{h+1}={\rm int}(f_x,f_y)$, and let $r_h$ be
the last generator of $\Gamma(f)$. We have the following:
i) $h=1$ implies that $r_h \geq 3$ and $\mu_{h+1}\geq 2$.
ii) $h=2$ implies that $r_h\geq 13$ and $m_{h+1}\geq 16$.
iii) More generally we have:
\noindent 1) $r_h \geq 12.4^{h-2}+\sum_{i=0}^{h-2}4^i={\displaystyle\frac{5}{3}}.2^{2h-1}-{\displaystyle\frac{1}{3}}$.
\noindent 2) $\mu_{h+1}\geq 2+2.\sum_{i=0}^{h-2}4^i+12.\sum_{i=h-2}^{2h-4}2^i={\displaystyle\frac{5}{3}}.2^{2h}-3.2^h+
{\displaystyle\frac{4}{3}}$, assuming that the summation over
negative exponents is $0$.\end{proposicion}
\begin{demostracion}{.} i) In this case,
by Lemma 1.8., $\mu_1=(r_0-1).(r_1-1)$. Furthermore $r_1\geq 2$ and $r_0\geq 2$; otherwise
$\Gamma(f)=<1>$, and $h=0$. On the other hand, gcd$(r_0,r_1)=1$. This
proves that max$(r_0,r_1)=r_1\geq 3$ and $\mu_2\geq 2$. Then our assertion follows. Remark that $r_1=3$ and
$\mu_2=2$ holds for $f=y^2+ax^3$, where $a\in {\bf K}-0$.
ii) Let $g_2$ be the second approximate root of $f$, then $\mu_3=
d_2.${\rm int}$(g_{2_x},g_{2_y})+(d_2-1)(r_2-1)$. It follows from i)
that ${\displaystyle\frac{r_1}{d_2}}\geq 3$ and that {\rm
int}$(g_{2_x},g_{2_y})=({\displaystyle\frac{r_0}{d_2}}-1)({\displaystyle\frac{r_1}{d_2}}-1)\geq 2$,
and also that ${\displaystyle\frac{r_0}{d_2}}+{\displaystyle\frac{r_1}{d_2}}
\geq 5$. In particular
$$
(r_0-d_2)({\displaystyle\frac{r_1}{d_2}}-1)\geq 2.d_2
$$
\noindent and
$$
r_0+r_1\geq 5.d_2.
$$
\noindent Thus:
$$
r_1.{\displaystyle\frac{r_0}{d_2}}\geq d_2+r_1+r_0\geq 6.d_2\geq 12.
$$
\noindent But $r_2-1 \geq r_1.{\displaystyle\frac{d_1}{d_2}} =r_1.{\displaystyle\frac{r_0}{d_2}}$. Finally
$r_2\geq r_1{\displaystyle\frac{r_0}{d_2}}+1\geq 13$, and $\mu_{h+1}\geq 2.d_2+(r_2-1) \geq 4+12=16$. This
implies our assertion. Note that the lower
bounds 13 and 16 is sharp:they are satisfied
for $f=(y^2+a.x^3)^2+b.x^5y,$ where $a,b\in {\bf K}-0$,
whose semigroup is $\Gamma=<4,6,13>$.
iii) We prove the inequalities by induction on $h$. From i) and ii) both
are satisfied for $1\leq h\leq 2$. Assume that $h\geq 3$ and that the
formulas are true for $h-1$. We first prove the inequality 1): Remark first that $r_h \geq
({\displaystyle\frac{r_{h-1}}{d_h}}).d_{h-1}+1$. The quotient ${\displaystyle\frac{r_{h-1}}{d_h}}$ being the last generator of
$\Gamma(g_h)$ which is of length $h-1$, it follows by induction that
${\displaystyle\frac{r_{h-1}}{d_h}} \geq 12.4^{h-3}+\sum_{i=0}^{h-3}4^i={\displaystyle\frac{5}{3}}.2^{2h-3}-{\displaystyle\frac{1}{3}}$. On the other hand,
$d_{h-1} \geq 4$, thus $r_h\geq 4.({\displaystyle\frac{5}{3}}.4^{2h-3}-{\displaystyle\frac{1}{3}})+1={\displaystyle\frac{5}{3}}.2^{2h-1}-{\displaystyle\frac{1}{3}}$. This is the required inequality.
\noindent We now prove the inequality 2): Consider to this end the last approximate root $g_h$ of $f$. We
have: $\mu_{h+1}=d_h.{\rm int}(g_{h_x},g_{h_y})+(d_h-1)(r_h-1)$. But $d_h\geq 2$ and $r_h\geq {\displaystyle\frac{5}{3}}.2^{2h-1}-{\displaystyle\frac{1}{3}}$. On the
other hand, the length of
$\Gamma(g_h)$ being $h-1$, it follows that
$$
{\rm int}(g_{h_x},g_{h_y})\geq 2+2.\sum_{i=0}^{h-3}
4^i+12.\sum_{i=h-3}^{2h-6}2^i={\displaystyle\frac{5}{3}}.2^{2h-2}-3.2^{h-1}+{\displaystyle\frac{4}{3}}
$$
\noindent In particular
$$
\mu_{h+1}\geq {\displaystyle\frac{5}{3}}.2^{2h-2}-3.2^{h-1}+{\displaystyle\frac{4}{3}}+{\displaystyle\frac{5}{3}}.2^{2h-1}-{\displaystyle\frac{1}{3}}-1={\displaystyle\frac{5}{3}}.2^{2h}-3.2^{h}+{\displaystyle\frac{4}{3}}
$$
\noindent This is the required inequality.
\end{demostracion}
\begin{nota}{\rm The bounds of the above Proposition are sharp. More
precisely, for all $h\geq 1$, there is a polynomial $f_h(x,y)\in \tilde{R}$
such that $h$ is the length of $\Gamma(f)$, and that int$(f_h{_x},f_h{_y})={\displaystyle\frac {5}{3}}.2^{2h}-3.2^h+{\displaystyle\frac {4}{3}}$, and if $r_h$ denotes the last generator of $\Gamma(f)$, then
$r_h={\displaystyle\frac{5}{3}}.2^{2h-1}-{\displaystyle\frac{1}{3}}$. Consider to this end the semigroup $\Gamma_h$ generated by $r_0=2^h$ and
$$
r_k=2^{h-k}({\displaystyle\frac{5}{3}}.2^{2k-1}-{\displaystyle\frac{1}{3}})
$$
\noindent for all $1\leq k\leq h$ (equivalently $r_1=2^{h-1}.3, r_2=2^h.3+2^{h-2},\ldots,
r_{k+2}=2^{h+k}.3+\sum_{i=1}^{k+1}2^{h+k-2i}$ for all $1\leq k\leq h-2$). Clearly $r_h={\displaystyle\frac{5}{3}}.2^{2h-1}-{\displaystyle\frac{1}{3}}$, and
the $d$-sequence is given by $d_k=2^{h+1-k}, 1\leq k\leq h+1$. Furthermore,
$r_kd_k < r_{k+1}d_{k+1}$ for all
$1\leq k\leq h$. It follows that $\Gamma_h$ is the semigroup of a
polynomial of $\tilde{R}$. We shall prove by
induction that the Milnor number of such a polynomial
is ${\displaystyle\frac {5}{3}}.2^{2h}-3.2^h+{\displaystyle\frac {4}{3}}$. Denote this
number by $\mu_{h+1}$ and recall that
$\mu_{h+1}=\sum_{k=1}^h({\displaystyle\frac {d_k}{d_{k+1}}}-1)r_k-r_0+1$. Since
${\displaystyle\frac {d_k}{d_{k+1}}}=2$ for all $1\leq k\leq h$, then:
$$
\mu_{h+1}=(\sum_{k=1}^hr_k)-r_0+1=\sum_{k=1}^h 2^{h-k}({\displaystyle\frac{5}{3}}.2^{2k-1}-{\displaystyle\frac{1}{3}})-2^h+1
$$
\noindent Which is nothing but ${\displaystyle\frac{5}{3}}.2^{2h}-3.2^h+
{\displaystyle\frac {4}{3}}$. This proves our assertion.}
\end{nota}
\begin{corolario} Let $m\in 2.{\bf N}$, then one can effectively compute the set of
irreducible polynomials $f\in {\tilde{\bf R}}$ such that $m={\rm int}(f_x,f_y)$.\end{corolario}
\begin{demostracion}{.} It follows from Proposition 4.1. that the
length $h$ of the semigroup of a polynomial
$f\in {\tilde{\bf R}}$ with $m={\rm int}(f_x,f_y)$ takes a finite
number of values. In fact, easy calculations show that $h$ must verify
the inequality: $2^h\leq M={\displaystyle\frac{9+\sqrt{1+60m}}{10}}$. In particular,
$h\leq{\displaystyle\frac {{\rm ln}(M)}{{\rm ln}(2)}}$. Let $H=\lbrace h\in {\bf N}; h\leq
{\displaystyle\frac{{\rm ln}(M)}{{\rm ln}(2)}}\rbrace$. Given $h\in A$ we shall effectively
construct the set $\Sigma$ of all the sequences
$(r_0,r_1,\ldots,r_h)$ which minimally generate a semigroup of a polynomial
$f\in {\tilde{\bf R}}$ of the required Milnor number. The steps
of the algorithm can
be summarized as follows:
\end{demostracion}
\bigskip
\noindent Set $m=\mu_{h+1}$. We want to calculate the set of $(\mu_h,r_h,d_h)$ with
the following equality:
\vskip0.1cm
(E1) $\mu_{h+1}=\mu_hd_h+(r_h-1)(d_h-1)$
\vskip0.1cm
\noindent Recall that we have the following restrictions:
\vskip0.1cm
i) $d_h\geq 2$
\vskip0.1cm
ii) $r_h \geq {\displaystyle\frac{5}{3}}.2^{2h-1}-{\displaystyle\frac{1}{3}}$.
\vskip0.1cm
iii) gcd$(r_h,d_h)=1$.
\vskip0.1cm
iv) $\mu_h\geq {\displaystyle\frac{5}{3}}.2^{2h-2}-3.2^{h-1}+
{\displaystyle\frac{4}{3}}$
\vskip0.1cm
v) $\mu_1=0$, and for all $h\geq 2$, $\mu_h={\displaystyle{\frac{\mu_{h+1}-(d_h-1)(r_h-1)}{d_h}}}$ is an even integer.
\vskip0.1cm
\noindent Now equality (E1) gives $(d_h-1)(r_h-1)=\mu_{h+1}-\mu_hd_h$,
and by iv)
$$
-\mu_hd_h \leq -[{\displaystyle\frac{5}{3}}.2^{2h-2}-3.2^{h-1}+
{\displaystyle\frac{4}{3}}]d_h
$$
\noindent in particular
$$
(r_h-1)(d_h-1)\leq \mu_{h+1}-[{\displaystyle\frac{5}{3}}.2^{2h-2}-3.2^{h-1}+
{\displaystyle\frac{4}{3}}].d_h
$$
\noindent This gives us the following upper bound for $r_h$:
$$
r_h\leq {\mu_{h+1}\over{d_h-1}}-[{\displaystyle\frac{5}{3}}.2^{2h-2}-3.2^{h-1}+
{\displaystyle\frac{4}{3}}].{d_h\over {d_h-1}}+1
$$
\begin{corolario} The above equality with (ii) give:
$$
(E2)\quad {\displaystyle\frac{5}{3}}.2^{2h-1}-{\displaystyle\frac{1}{3}}\leq
r_h\leq {\mu_{h+1}\over{d_h-1}}-[{\displaystyle\frac{5}{3}}.2^{2h-2}-3.2^{h-1}+
{\displaystyle\frac{4}{3}}].{d_h\over {d_h-1}}+1
$$
\end{corolario}
\medskip
\noindent In the following we shall refine the lower bound
of Corollary 4.4. We start with the following technical Lemma:
\begin{lema} $\sum_{i=1}^h(e_i-1)r_i=r_hd_h-m_h$. In particular,
$\mu_{h+1}=\sum_{i=1}^h(e_i-1)r_i-r_0+1=r_hd_h-m_h-r_0+1$, where we recall that
$e_i={\displaystyle\frac{d_i}{d_{i+1}}}$ for all $1\leq i\leq h$.
\end{lema}
\begin{demostracion}{.} Applying identity (**) of Section 3 with $k=h$ we get:
$$
\sum_{i=1}^{h-1}(e_i-1)r_i=r_h-m_h
$$
\noindent Now adding $(e_h-1)r_h=(d_h-1)r_h$ to the equality we get our assertion.
\end{demostracion}
\noindent Lemma 4.5. with equality (E1) imply that $r_hd_h=\mu_{h+1}+m_h+r_0-1$. On the other hand,
we have, with the notation $m_0=r_0$, that for all $1\leq k \leq h$, $m_k-m_{k-1}\geq d_{k+1}$.
Adding these inequalities we get
$$
\mu_h\geq m_0+d_2+\ldots+d_h+d_{h+1}=d_1+d_2+\ldots+d_h+1.
$$
\noindent But for all $1\leq k\leq h, d_k\geq 2^{h-k}.d_h$, so $\mu_h\geq d_h.(2^h-1)+1$. Since
$r_0=d_1\geq 2^{h-1}.d_h$, we get the following:
$$
r_h \geq \quad{\rm max}({\frac{5}{3}}.2^{2h-1}-{\frac{1}{3}},{\mu_{h+1}\over d_h}+(3.2^{h-1}-1))
$$
\noindent Now equality (E1) implies that
${\displaystyle {\frac{\mu_{h+1}}{d_h}}}=\mu_h+(r_h-1)(1-{\displaystyle{\frac{1}{d_h}}})$. But
$d_h\geq 2$, thus, using inequalities of Proposition 4.1. we get:
$$
{\frac{\mu_{h+1}}{d_h}}\geq ({\frac{5}{3}}.2^{2h-2}-3.2^{h-1}+{\frac{4}{3}})+
({\frac{5}{3}}.2^{2h-1}-{\frac{4}{3}}).{\frac{1}{2}}={\frac{5}{3}}.2^{2h-1}-3.2^{h-1}+{\frac{2}{3}}
$$
\noindent In particular ${\displaystyle {\rm max}({\frac{5}{3}}.2^{2h-1}-{\frac{1}{3}},{\mu_{h+1}\over d_h}+(3.2^{h-1}-1))={\mu_{h+1}\over d_h}+(3.2^{h-1}-1)}$. This implies the following:
$$
(E3)\quad {\frac{\mu_{h+1}}{d_h}}+(3.2^{h-1}-1)\leq r_h \leq
{\frac{\mu_{h+1}}{d_h-1}}-[{\frac{5}{3}}.2^{2h-2}-3.2^{h-1}+{\frac{4}{3}}].{\frac{d_h}{d_{h}-1}}+1
$$
\noindent We shall now use inequality (E3) in order to give an upper bound for $d_h$ (a lower bound
being $2$). Remark to this end that ${\displaystyle {\frac{\mu_{h+1}}{d_h-1}}-[{\frac{5}{3}}.2^{2h-2}-3.2^{h-1}+{\frac{4}{3}}].{\frac{d_h}{d_{h}-1}}+1-({\frac{\mu_{h+1}}{d_h}}+(3.2^{h-1}-1))\geq 0}$.
If we set
$p=({\displaystyle{\frac{5}{3}}.2^{2h-2}-3.2^{h-1}+{\frac{4}{3}}})$ and $q=3.2^{h-1}-2$, then
an obvious analysis of the above inequality shows that it is equivalent to say that $(p+q).d_h^2-qd_h-\mu_{h+1}\leq 0$, which is true if and only if the following holds:
$$
\mbox {(E4) } 2\leq d_h\leq {\frac {q+\sqrt{q^2+4\mu_{h+1}.(p+q)}}{2.(p+q)}}=
{\frac {3.2^{h-1}-2+\sqrt{(3.2^{h-1}-2)^2+4\mu_{h+1}.({\frac{5}{3}}.2^{2h-2}-{\frac{2}{3}})}}{{\frac{10}{3}}.2^{2h-2}-{\frac{4}{3}}}}
$$
\noindent {\bf The algorithm}: The two integers $\mu_{h+1}$ and $h$ being fixed, inequality (E4)
determines the set $D^h$ of possible
values of $d_h$. Each value of $d_h$ gives rise, using inequality (E3), to a set -denoted
$R^h_{d_h}$- of possible
values of $r_h$ (Remark that ${\displaystyle{\frac{\mu_{h+1}-(d_h-1)(r_h-1)}{d_h}}}$
should be an even integer). We get this way the set -denoted $P^h_{d_h}$- of possible values of $(\mu_h,r_h,d_h)$. Now we restart with the set of $\mu_h$... This procees shall stop constructing a
set of lists of length $h$. The set of semigroups corresponding to $\mu_{h+1}$ is a subset of this list and can be easily calculated. Remark that if $h=1$, then $\mu_1=0$
and $\mu_2=(r_1-1)(d_1-1)$ by condition v). In
this case, the values of $(r_1,d_1=r_0)$ can
also be obtained from the set of divisors of
$\mu_2$.
\begin{exemple}{\rm We perform an explicit computation for $\mu_{h+1}=28$. In this
case, \break $M={\displaystyle\frac{9+\sqrt{1+60.28}}{10}}=5$, so $H=\lbrace
h; 1\leq h \leq {\displaystyle\frac{{\rm ln}(5)}{{\rm ln}(2)}}\rbrace=
\lbrace 1,2\rbrace$.
\par\smallskip
\noindent 1) $h=1$: In this case, since $28=1*28=2*14=4*7$,
then $(r_1,d_1)\in \lbrace (2,29),(3,15),(5,8)\rbrace$ and
condition iii) eliminates $(3,15)$. We get this way the
semigroups $<2,29>$ and $<5,8>$. The
canonical representative of the equisingularity class of the first
one (resp. the second one) is $y^2-x^{29}$ (resp. $y^5-x^{8}$).
\par\smallskip
\noindent 2) $h=2$: Inequality (E4) implies in this case
that $2\leq d_2\leq {\displaystyle{\frac{4+\sqrt{688}}{12}}<3}$. In particular
$D^2=\lbrace 2\rbrace$.
\par\smallskip
\noindent Now inequality (E3) implies that
${\displaystyle{\frac{28}{2}}}+5=19\leq r_2\leq 28-4+1=25$, and
with conditions iii), v), we get $R^2_{2}=\lbrace 21,25\rbrace$. If $r_2=25$ (resp. $r_2=21$), then
$\mu_2=2$ (resp. $\mu_2=4$). Thus $P^2_2=\lbrace (2,25,2),(4,21,2)\rbrace$.
\par\smallskip
i) $(\mu_2,r_2,d_2)=(2,25,2)$. In this case, applying to construction above to
$\mu_2=2$, we get
${\displaystyle{\frac{d_1}{d_2}}=2, {\frac{r_1}{d_2}}=3}$. This leads to the semigroup $<4,6,25>$. The
canonical representative of the equisingularity class of this semigroupe is
$(y^2-x^3)^2-x^{11}y$.
ii) $(\mu_2,r_2,d_2)=(4,21,2)$. In this case, applying to construction above to
$\mu_2=4$, we get
${\displaystyle{\frac{d_1}{d_2}}=2, {\frac{r_1}{d_2}}=5}$. This leads
to the semigroup $<4,10,21>$. The
canonical representative of the equisingularity class of this semigroupe is
$(y^2-x^5)^2-x^{8}y$.}
\end{exemple}
\noindent Let $m$ be an even integer, and let $H$ is the set of positive integers not exceeding
${\displaystyle\frac{{\rm ln}(M)}{{\rm ln}(2)}}$, where
$M={\displaystyle{\frac{9+\sqrt{1+60m}}{10}}}$. Assume that $H$ is not reduced to $0$
and let $h$ be a nonzero element of $H$. Set $m=\mu_{h+1}$ and let
$$
a_h={\frac {q+\sqrt{q^2+4\mu_{h+1}.(p+q)}}{2.(p+q)}}=
{\frac {3.2^{h-1}-2+\sqrt{(3.2^{h-1}-2)^2+4\mu_{h+1}.({\frac{5}{3}}.2^{2h-2}-{\frac{2}{3}})}}{{\frac{10}{3}}.2^{2h-2}-{\frac{4}{3}}}}.
$$
\noindent Let $D^h$ be the set positive integers between
$2$ and $a_h$ (we easily verify that the condition $a_h\geq 2$ is equivalent to the numerical condition $\mu_{h+1}\geq{\displaystyle {\frac{5}{3}.2^{2h}-3.2^{h}+{\frac{1}{3}}}}$ proved in Proposition 4.1., in particular $D^h$ is not the emptyset). Set $P^h=\bigcup_{d\in D^h} P^h_d$ and denote by $C_{h+1}$ the cardinality of $P^h$. In the following we shall give an upper bound for $C_{h+1}$. Set ${\displaystyle b^h_{d}={{\mu_{h+1}\over d}}+
(3.2^{h-1}-1)}$, and
${\displaystyle c^h_{d}={\mu_{h+1}\over{d-1}}-[{\frac{5}{3}}.2^{2h-2}-3.2^{h-1}+
{\frac{4}{3}}].{d\over {d-1}}+1}$. The set $R^h_{d}$ of possible values of $r_h$ is a
subset of the set of integers between $b^h_d$ and $c^h_d$. Its cardinality is then bounded by $c^h_d-b^h_d+1$.Furthermore, we easily verify that
if $r\in R^h_{d}$, then
$$
{\displaystyle{\frac{\mu_{h+1}-(r-1)(d-1)}{d}}
\geq {\frac{5}{3}.2^{2h-2}-3.2^{h-1}+{\frac{1}{3}}}}.
$$
\noindent In particular, if $h\geq 2$, then
$(\mu_h={\displaystyle{\frac{\mu_{h+1}-(r-1)(d-1)}{d}}},r,d)$ is
an element of $P^h_{d}$.
\noindent Now
$$
c^h_d-b^h_d+1={\frac{\mu_{h+1}}{d(d-1)}}-
(p_h-q_h).{\frac{d}{d-1}}-q_h+1$$
\hskip1cm $$={\frac{\mu_{h+1}}{d-1}}-{\frac{\mu_{h+1}}{d}}-
(p_h-q_h).(1+{\frac{1}{d-1}})-q_h+1
$$
\noindent Consequently, if $a=[a_h]$, then the cardinality $C_{h+1}$ of
$P_h$ is bounded by:
$$
{\displaystyle \sum_{d=2}^a
(c^h_d-b^h_d+1)=(\mu_{h+1})(1-{\frac{1}{a}})-p_h(a-1)-(p_h-q_h).\sum_{d=1}^{a-1}
{\frac{1}{d}}+(a-1)}
$$
\noindent But $1-{\displaystyle{\frac{1}{a}} < 1}$, and substituting $2$ to $a$ in the other
members of the above formula we get:
$$
C_{h+1}\leq \mu_{h+1}-2p_h+q_h+1= \mu_{h+1}-({\frac{10}{3}}.2^{2h-2}-3.2^{h-1}-{\frac{1}{3}})=\mu_{h+1}-({\frac{B_{h+1}}{2}}-1)
$$
\noindent Where $B_{h+1}={\displaystyle{\frac{5}{3}}.2^{2h}-3.2^h+{\frac{4}{3}}}$
is the lower bound of $\mu_{h+1}$ in Proposition 4.1.
\begin{nota}{\rm Note that the bound above is not the optimal one, indeed, given $d\in D^h$,
the cardinality of the set of $r\in R^h_d$ such that gcd$(r,d)=1$ can be
bounded by $\displaystyle{\frac{c^h_d-b^h_d}{d}}+1$, but in view of our algorithm,
all values of $R^h_d$ are used, in particular the value above bounds also the
nombre of operations used in the first step of the algorithm.}\end{nota}
\noindent Let $(\mu_h,r,d)$ be an element of $P_h$ and recall that
$\mu_h={\displaystyle{\frac{\mu_{h+1}}{d}}-(r-1){\frac{d-1}{d}}}$.
Since $b^h_d \leq r \leq c^h_d$, then
$$
\mu_h \leq {\displaystyle
{\frac{\mu_{h+1}}{d}}-(b^h_d-1){\frac{d-1}{d}}\leq
{\frac{\mu_{h+1}}{d}}-(b^h_d-1){\frac{d-1}{d}}={\frac{b^h_d-1}{d}}-(3.2^{h-1}-2)}
$$
\hskip1cm $$={\frac{\mu_{h+1}}{d^2}}+({\frac{1}{d}}-1)(3.2^{h-1}-2)\leq
{\frac{\mu_{h+1}}{4}}-{\frac{1}{2}}(3.2^{h-1}-2)\leq {\frac{\mu_{h+1}}{4}}-(3.2^{h-2}-1)
$$
\noindent Let $A_{h+1}=3.2^{h-2}-1$. It follows by induction that for all $0\leq k\leq h-1$, if
$\mu_{h-k}$ is a possible value of the Milnor number at the step $k+1$, then we
have:
$$
\mu_{h-k}\leq {\frac{\mu_{h+1}}{4^{k+1}}}-\sum_{i=0}^k
{\frac{A_{h+1-i}}{4^{k-i}}}=3.2^{h-2k-2}(2^{k+1}-1)-{\frac{4}{3}}-{\frac{1}{3.4^k}}
$$
\noindent (Remark that the above inequality is valid if $k=h-1$ because $\mu_1=0$). Thus, we can obtain a bound of the set of values
calculated at the step $k+1, 0\leq k \leq h-2$ in the following way: let $\mu_{h-k}$ be a possible value of the Milnor
number obtained by reiterating the algorithm above $k+1$ times, and denote by $C_{h-k}(\mu_{h-k})$ the cardinality of the set -denoted $P_{h-k-1}(\mu_{h-k})$- of the 3-uplets $(\mu,r,d)$ obtained by applying the algorithm above to $\mu_{h-k}$ instead of $\mu_{h+1}$. It follows from the discussion above that
$$
C_{h-k}\leq \mu_{h-k}-{\frac{B_{h-k}}{2}}+1\leq {\frac{\mu_{h+1}}{4^{k+1}}}
-\sum_{i=0}^k {\frac{A_{h+1-i}}{4^{k-i}}}-{\frac{B_{h-k}}{2}}+1$$
$$
={\frac{\mu_{h+1}}{4^{k+1}}}
-3.2^{h-k-2}+3.2^{h-2k-2}-{\frac{5}{3}}.2^{2h-2k-1}-{\frac{1}{3.4^k}}+{\frac{5}{3}}
$$
\noindent In particular, the cardinality of the set of semigroups corresponding to the
given Milnor number $m=\mu_{h+1}$ is bounded by $\prod_{i=2}^{h}C_{h+1-i}$ which is
a polynomial in $m$ bounded by its leading coefficient
$\displaystyle{\frac{m^h}{2^{h(h-1)}}}$. Note that in view of Remark 4.8., the number of
operations used in the algorithm is then bounded by $\sum_{i=0}^{h-1}\prod_{k=0}^{i}C_{h+1-k}$.
\medskip
\noindent The above algorithm has been implemented with {\it MAPLE}.
The intput is an integer $m$, and the output is the list of semigroups whose conductor is $m$.
In the implementation work we followed the ideas explained above, with the following simplification: at the last step, the set of values we are interested in is calculated by using the factorization of the given Milnor number. The algorithm is an iterating of the following:
{\it Input: $m\in 2.{\bf N}$
Output: The set $P^h$.
Step I: Compute the set $H$.
Step II: Take $h\in H$.
Step III: Compute the set $D^h$.
Step IV: Take $d\in D^h$.
Step V: Compute $R^h_d$
\quad\quad (*) if $r\in R^h_d$ and gcd$(r,d)=1$ and $\displaystyle{{\frac{m-(d-1)(r-1)}{d}}\in 2.{\bf N}}$ (resp. $(d-1)(r-1)=m$ if h=1) then add $(\displaystyle{{\frac{m-(d-1)(r-1)}{d}}},r,d)$ to $P^h_d$.
Step VI: $P^h=\bigcup_{d\in D^h}P^h_d$}
\noindent The main operation of the algorithm is the one described in the line (*). We
experimented it on various values of $m$: the computation takes around 0.2sec for $m=160$, 0.7sec for $m=300$, 1.5sec for $m=500$, and 3sec for $m=1000$.
| {
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Der Hochsprung der Frauen bei den Leichtathletik-Weltmeisterschaften 1991 wurde am 29. und 31. August 1991 im Olympiastadion der japanischen Hauptstadt Tokio ausgetragen.
Mit Silber und Bronze errangen die sowjetischen Hochspringerinnen in diesem Wettbewerb zwei Medaillen. Weltmeisterin wurde die amtierende Europameisterin Heike Henkel, frühere Heike Redetzky, aus Deutschland. Sie gewann vor der EM-Dritten von 1991 Jelena Jelesina. Bronze ging an Inha Babakowa.
Bestehende Rekorde
Der bestehende WM-Rekord wurde bei diesen Weltmeisterschaften nicht eingestellt und nicht verbessert.
Im Finale am 31. August stellte Weltmeisterin Heike Henkel mit 2,05 m einen neuen deutschen Rekord auf.
Qualifikation
29. August 1991, 10:20 Uhr
29 Teilnehmerinnen traten in zwei Gruppen zur Qualifikationsrunde an. Die Qualifikationshöhe für den direkten Finaleinzug betrug 1,92 m. Doch diese Höhe musste gar nicht erst angegangen werden. Nach Abschluss der Sprünge über 1,90 m waren noch sechs Springerinnen im Wettbewerb. Weitere sechs hatten 1,88 m übersprungen. So waren die gesuchten zwölf Besten gefunden, die zwei Tage später das Finale austrugen (hellgrün unterlegt).
Gruppe A
Gruppe B
Legende
Kurze Übersicht zur Bedeutung der Symbolik – so üblicherweise auch in sonstigen Veröffentlichungen verwendet:
Finale
31. August 1991, 16:00 Uhr
Video
Heike Henkel High Jump World Championships Tokyo 1991, Video veröffentlicht am 9. Mai 2013 auf youtube.com, abgerufen am 15. April 2020
Weblinks
World Championships in Athletics, 3rd IAAF World Championships in Athletics (englisch), abgerufen am 3. Mai 2020
Women High Jump Athletics III World Championship 1991 Tokyo (JPN)1991 auf todor66.com (englisch), abgerufen am 3. Mai 2020
Ergebnisse im Statistics Handbook der IAAF zur WM 2019 in Doha, Women High Jump, Tokyo 1991, S. 311 (PDF 10,3 MB, englisch), abgerufen am 3. Mai 2020
Einzelnachweise und Anmerkungen
Hochsprung der Frauen | {
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ELON, N.C. (AP) - Tyler Seibring scored a career-high 25 points with seven rebounds and six assists, Steven Santa Ana added 22 points and Elon beat Drexel 93-73 on Thursday night.
Seibring scored 20-plus for the third time this season, hitting 4 of 6 from distance, and Elon led from start to finish. It was Santa Ana's sixth career 20-plus game in just his sophomore season.
The Phoenix hit double-digit 3-pointers (10) for the 10th time this season.
Dainan Swoope scored 16 points, Brian Dawkins 13 and Dmitri Thompson 12 for Elon (11-9, 3-4 Colonial Athletic Association), which hit 25 of 28 free-throw attempts - including 17 of 18 in the second half.
Elon scored the first nine points of the game and had a 10-point lead at halftime.
Kurk Lee and John Moran each scored 15 points for Drexel (7-12, 1-5). Miles Overton added 13 points and Sammy Mojica 10. | {
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Q: File contains wrong format not string when downloading from REST API with Spring Boot Hello I am building an application in Spring Boot and Thymeleaf and I have a table and on each row it has a download button. When you press the button it send and id that makes a request to the api and it has to download an xml file. The download works but when I open the file it has the wrong character format.
This is an small snippet of what it contains:
¥Wé'ÚHþÏSh»#ÖãÀ€$$q¬Ûº ]€ø3QˆÒºÐB¯3ϲïµEwÛƒ½»önŒ'B•Y™ß—™•ŸC°Àš$NË—§Äϧ þôTÑúÀ«êLX'£XdE
Here is the code that I am using to download the file:
This is in the service class
public Resource getDownload(String id) {
String url = "http://10.000.000.000:100/test/descarcare?id="+id;
HttpHeaders headers = new HttpHeaders();
headers.add("serial", "123");
headers.add("Content-Disposition","attachment;filename=download.xml");
headers.set("contentType", "application/octet-stream");
RestTemplate restTemplate = new RestTemplate();
RequestEntity<Void> request = RequestEntity.get(URI.create(url))
.headers(headers)
.build();
ResponseEntity<Resource> response = restTemplate.exchange(request, Resource.class);
System.out.println(response.getBody());
Resource resp = response.getBody();
return resp;
}
And this is in the controller:
@GetMapping("/download")
public ResponseEntity<Resource> downloadFile(@RequestParam("id") String id, HttpServletRequest request) {
// Load file as Resource
Resource resource = service.getDownload(id);
// Try to determine file's content type
String contentType = null;
try {
contentType = request.getServletContext().getMimeType(resource.getFile().getAbsolutePath());
} catch (IOException ex) {
}
// Fallback to the default content type if type could not be determined
if(contentType == null) {
contentType = "application/octet-stream";
}
HttpHeaders headers = new HttpHeaders();
headers.add("serial", "123");
headers.add("Content-Disposition","attachment;filename=download.xml");
headers.set("Content-type", "application/octet-stream");
return ResponseEntity.ok()
.contentType(MediaType.parseMediaType(contentType))
.headers(headers)
.body(resource);
}
Can someone please tell me what I am doing wrong in the downloading of the file because I am a beginner in SpringBoot and Java. Thanks and have a nice day
| {
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La 21a edició dels premis David di Donatello, concedits per l'Acadèmia del Cinema Italià va tenir lloc el 24 de juliol de 1976 al Teatre grecoromà de Taormina. El premi consistia en una estatueta dissenyada per Bulgari.
Guanyadors
Millor pel·lícula
Cadaveri eccellenti, dirigit per Francesco Rosi
Millor director
Mario Monicelli - Amici miei
Francesco Rosi - Cadaveri eccellenti
Millor argument
Alberto Bevilacqua i Nino Manfredi - Attenti al buffone
Millor actriu
Monica Vitti - L'anatra all'arancia
Millor actor
Ugo Tognazzi - Amici miei (ex aequo)
Adriano Celentano - Bluff - Storia di truffe e di imbroglioni (ex aequo)
Millor músic
Franco Mannino - L'innocente
Millor actriu estrangera
Isabelle Adjani - L'Histoire d'Adèle H. (ex aequo)
Glenda Jackson - Hedda (ex aequo)
Millor actor estranger
Philippe Noiret - El vell fusell (Le Vieux Fusil) (ex aequo)
Jack Nicholson - Algú va volar sobre el niu del cucut (One Flew Over the Cuckoo's Nest) (ex aequo)
Millor director estranger
Miloš Forman - Algú va volar sobre el niu del cucut (One Flew Over the Cuckoo's Nest)
Millor pel·lícula estrangera
Nashville (Nashville), dirigit per Robert Altman
David Europeu
Jan Troell
David Luchino Visconti
Michelangelo Antonioni
Targa d'oro
Ennio Lorenzini, per la seva direcció a Quanto è bello lu murire acciso
Sydney Pollack, per la seva direcció a Els tres dies del Còndor (Three Days of the Condor)
Michele Placido, per la seva interpretació a Marcia trionfale; dirigit per Marco Bellocchio
Christian De Sica, per la seva interpretació a Giovannino; dirigit per Paolo Nuzzi
Agostina Belli, per la seva interpretació a Telefoni bianchi; dirigit per Dino Risi
Martin Bregman i Martin Elfand, per la producció de Tarda negra (Dog Day Afternoon); dirigit per Sidney Lumet
Ornella Muti, pel conjunt de la seva interpretació
Referències
Vegeu també
Nastri d'argento 1976
Enllaços externs
David di Donatello 1976 a Filmaffinity
Cronologia al web dels David di Donatello
Cinema del 1976
Edicions del David di Donatello | {
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BRAGINSKY COLLECTION
Manuscripts & Printed Books
Esther Scrolls
The engraved border of this scroll was designed by the Italian author, artist, and publisher Francesco Griselini (1717–1787). A man of broad intellectual pursuits, his interests ranged from botany and agri- culture to cartography, politics, drama, and the arts. In the late 1730s and 1740s he produced several works of Judaica, including four frontispieces for a Hebrew Bible published in Venice 1739 (cat. no. 44), an engraved wall decoration for a Sukkah, and most notably, this illustrated scroll border.
The border design is composed of an arcade featuring four distinctly patterned columns. The arches are surmounted by a balustrade that supports flower- ing urns, blank medallions, floral scrollwork, and a variety of birds including a crowned double-headed eagle and a peacock. Scenes from the Esther narrative are positioned beneath each of the nineteen columns of text. In these illustrations, Griselini has devoted particular attention to architectural settings and spatial perspective. The artist's printed sig- nature is found in the lower left corner of each membrane. This engraved border design was popular in Italy in the eighteenth century as demonstrated by two similar scrolls in the Braginsky Collection: an unsigned scroll of superior artistic quality (Braginsky Collection Megillah 11) and a small megillah with a related design (cat. no. 100).
One of the most striking images in this group of scrolls is the last illustration, placed under the final arch, where the concluding benedictions usually are written. It depicts the Messiah heralding the return of the exiled Jewish people back to the walled city of Jerusalem. This representation of messianic redemption, not commonly found in decorated megillot, is intended to elucidate the final blessing, which appeals to God to bring salvation. The image is copied directly from a Haggadah first printed in Venice in 1609, and reprinted many times over the next two centuries.
The text of this Griselini scroll was penned by the accomplished scribe-artist Aryeh Leib ben Daniel (cat. no. 86). In his inscription, which follows the concluding benedictions, he informs the reader that he wrote this scroll in Venice in the winter of 1746.
selected literature
Kaniel 1979, p. 108; Metzger 1966, pp. 381–432; Sabar 1997/8, pp. 294–312.
Venice, 1746
Copied by Aryeh Leib ben Daniel, printed border designed by Franceso Griselini
Parchment, 5 membranes, 19 text columns
+ benedictions, 240 × 2250 mm (9.4 × 88.6 in.)
Braginsky Collection Megillah 13
VIEW & ZOOM
© 2016 The Braginsky Collection
Photography and Website by Ardon Bar-Hama
Manuscripts and Printed Books
© 2015 BRAGINSKY COLLECTION
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"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,071 |
{"url":"https:\/\/tex.stackexchange.com\/questions\/467993\/how-to-add-a-picture-into-the-header-in-the-first-page","text":"# How to add a picture into the header in the first page?\n\nhow can I add a picture into the header in the first page of my article? Thank you very much in advance.\n\nMy sample code can only create a file with the picture showing up on the second page.\n\n\\documentclass[12pt,a4paper]{article}\n\\usepackage{grffile,graphicx,hyperref,graphbox} \\usepackage{fancyhdr}\n\\usepackage[sort]{natbib} \\usepackage[svgnames]{xcolor}\n\n\\renewcommand{\\refname}{References}\n\\begin{document}\n\n{\\centering \\title{Stat} \\author{Yan}\n\\maketitle}\n\nI would like to \\clearpage\n\n123456\n\n\\bibliographystyle{chicago}\n\\end{document}\n\n\n\u2022 Add \\thispagestyle{fancy} just after \\maketitle. The reason is that \\maketitlemodifies the page style. \u2013\u00a0user94293 Dec 31 '18 at 8:23\n\u2022 Thanks. But on the last page, the picture is paired with \"reference\". Why? Thank you a lot. \u2013\u00a0Carl Dec 31 '18 at 8:32\n\u2022 Do you want the picture only in the first page? \u2013\u00a0caverac Dec 31 '18 at 8:35\n\u2022 Right. But I want to solve the problem of pairing photo with reference at first. \u2013\u00a0Carl Dec 31 '18 at 8:37\n\u2022 @Carl: the problem of paring a photo with reference = a second question. Please create a new question for that. \u2013\u00a0Tommiie Dec 31 '18 at 11:46\n\nI took the liberty of cleaning your code a bit\n\n\\documentclass[12pt,a4paper]{article}\n\n\\usepackage{grffile,graphicx,hyperref,graphbox}\n\\usepackage{fancyhdr}\n\\usepackage[sort]{natbib}\n\\usepackage[svgnames]{xcolor}\n\\usepackage{lipsum}\n\n\\pagestyle{fancy}\n\\renewcommand{\\bibsection}{}\n\n\\fancypagestyle{firststyle}\n{\n\\fancyhf{}\n}\n\n% AUTHOR INFO\n\\title{Stat}\n\\author{Yan}\n\n%% DOCUMENT\n\\begin{document}\n\\maketitle\n\\thispagestyle{firststyle}\n\n\\section{One section}\n\\lipsum[1-2]\\clearpage\n\n\\section{Other section}\n\\lipsum[3-5]\\cite{White1983}\n\n\\bibliographystyle{plain}\n\\section*{References}\n\\bibliography{refs}\n\\end{document}\n\n\n\u2022 Thank you a lot. Could you tell me what does the \"\\lipsum[1-2]\" and \"\\lipsum[3-5]\" mean or function here? It returned me with the error that \"Package kantlipsum Error: Package expl3 too old. }\" after I ran your code. Thanks. \u2013\u00a0Carl Dec 31 '18 at 9:28\n\u2022 @Carl Just some dummy text \u2013\u00a0caverac Dec 31 '18 at 9:31\n\u2022 And if you add up the command, \"\\bibliography{C:\/Users\/ref\/rref}, which a bib.tex file, my previous problem wasn't solved. Thank you. \u2013\u00a0Carl Dec 31 '18 at 9:35\n\u2022 @Carl Do you want References to appear on the header? or you don't? It appears there because the header is configured to include the current section title. \u2013\u00a0caverac Dec 31 '18 at 9:41\n\u2022 I don't. Thank you for your instructions. \u2013\u00a0Carl Dec 31 '18 at 9:43","date":"2019-05-25 15:36:45","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.42666029930114746, \"perplexity\": 1877.7766355827814}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-22\/segments\/1558232258120.87\/warc\/CC-MAIN-20190525144906-20190525170906-00484.warc.gz\"}"} | null | null |
A very thin woman walked into her physician's office for an exam. She had not seen her doctor in years, but felt so chronically fatigued and ill that she knew she needed help. During the interview, she disclosed that she had not had a menstrual period in over seven years. The doctor's response was, "What's wrong? Did you have a hysterectomy?"
The physician failed to recognize that his patient was suffering from an eating disorder simply because he was unfamiliar with symptoms of the disease. Misdiagnosis of eating disorders is frighteningly common, and treatment may be delayed for years because the patient was advised to simply "eat more".
Physicians need to know that eating disorders have the highest mortality rate of any mental illness. 5 – 10% of anorexics die within 10 years, 18-20% die within 20 years, and only 30 – 40% ever fully recover. The mortality rate associated with anorexia nervosa is 12 times higher than the death rate from ALL causes for females ages 15 to 24. 20% of people suffering from anorexia will die prematurely of complications related to their eating disorders, including suicide and heart problems.
Complications from bulimia nervosa also to earlier death than the patient might otherwise experience. The death rate of people who are obese is unknown since death certificates normally list cause of death as heart disease or complications from diabetes.
"Many people have concerns about food. Please tell me about your eating habits. Do you worry about your eating, or do you think that others do?"
"Some people have concerns about their weight. Please tell me how you feel about your body and weight."
"Some people have trouble with eating to the point of discomfort. Please tell me when this has been a problem for you."
If the patient reveals concerns about food, eating habits, or weight, it's important to ask for additional details in an empathetic and non-judgmental manner.
It is crucial that the physician determine whether the patient needs additional care from an outpatient psychotherapist who specializes in eating disorders, or from an inpatient treatment facility. Hospitalization is a last resort and should only be considered when an individual is severely medically compromised or at risk of serious self-harm.
Educating physicians about eating disorders and how to recognize them is of vital importance as both the prevalence and the severity of eating disorders continues to increase. | {
"redpajama_set_name": "RedPajamaC4"
} | 5,860 |
package com.intellij.openapi.externalSystem.service.execution;
import com.intellij.openapi.externalSystem.model.ExternalSystemException;
import org.jetbrains.annotations.NotNull;
import org.jetbrains.annotations.Nullable;
/**
* @author Vladislav.Soroka
*/
public class ExternalSystemJdkException extends ExternalSystemException {
@Deprecated
public ExternalSystemJdkException(@Nullable String message,
@Nullable String filePath,
@Nullable Throwable cause,
@NotNull String... quickFixes) {
this(message, cause, quickFixes);
}
public ExternalSystemJdkException(@Nullable String message,
@Nullable Throwable cause,
@NotNull String... quickFixes) {
super(message, cause, quickFixes);
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 4,983 |
\section{Introduction}
\label{SectionIntroduction}
In random multiple access communication, users (transmitters) determine their communication rates individually, without sharing the rate information either among each other or with the receiver \cite{ref Luo09}. With the absence of rate coordination among users, reliable message recovery is not always possible \cite{ref Massey85}. The receiver in this case decodes the transmitted messages if a pre-determined error probability requirement can be satisfied, or reports a collision otherwise \cite{ref Luo09}.
Information theoretic channel coding in time-slotted random multiple access communication over a discrete-time memoryless channel was recently investigated in \cite{ref Luo09}\cite{ref WangJournal}. Assume that channel coding is applied only within each time slot (or packet). It was shown in \cite{ref Luo09} that the fundamental performance limitation of the system can be characterized using an achievable rate region in the following sense. Asymptotically as the codeword length (or time slot length) is taken to infinity, the receiver is able to recover the messages reliably if the communication rate vector (which includes the rates of all users) happens to be inside the rate region, and to reliably report a collision if the rate vector happens to be outside the region \cite{ref Luo09}. The achievable rate region was shown to coincide with the Shannon information rate region without a convex hull operation \cite{ref Luo09}. In \cite{ref WangJournal}, the asymptotic coding result was further strengthened to a rate and error probability tradeoff bound under the assumption of a finite codeword length. A bound on the achievable error exponent was obtained consequently \cite{ref WangJournal}.
Both \cite{ref Luo09} and \cite{ref WangJournal} assumed that the channel state information is known at the receiver. Unfortunately, since random access communication deals with bursty short messages, transmission activities of a user are often fractional. Without frequent data support, accurate real-time channel estimation and tracking become difficult at the receiver. Understanding the system performance limitation without channel state information therefore becomes essential \cite{ref Lapidoth98}. In this paper, we illustrate how coding theorems developed in \cite{ref Luo09}\cite{ref WangJournal} can be extended to random multiple access communication over a compound discrete-time memoryless channel \cite{ref Blackwell59}\cite{ref Wolfowitz59}, consisting of a family (set) of channels over which the communication could take place. Both the transmitters and the receiver know about the compound channel set, but not the actual channel realization. As in \cite{ref Luo09}\cite{ref WangJournal}, we assume that time is partitioned into slots of equal length, and we focus on channel coding within one time slot. We define the communication rate of a user as the normalized number of information nats encoded in a packet (or a time slot).
The compound channel communication problem investigated in this paper is different from a conventional one in the following two key aspects. First, in a conventional system, information rates are jointly determined by the transmitters and the receiver \cite{ref Cover05}, while communication rates in a random access system are determined distributively and the rate information is unknown at the receiver \cite{ref Luo09}. Second, in a conventional system, in order to achieve reliable communication, the transmitted rate vector should be supported by all channel realizations in the compound set \cite{ref Csiszar81}\cite{ref Lapidoth98}. In random access communication, however, even though the receiver needs to guarantee the reliability of its decoding output, the receiver also has the additional choice of reporting a collision to avoid confusing the upper layer networking \cite{ref Bertsekas92}. This therefore allows the transmitted rate vector to be supported only by a subset of channel realizations. If the actual channel realization belongs to this subset, the receiver should decode the messages. Otherwise, the receiver should report a collision. Clearly, the decoding and collision report decisions made at the receiver are affected jointly by the communication rates of the users and the actual channel realization.
To address these key aspects in the system model, we assume that the receiver chooses an ``operation region", which is a set of rate vector and channel realization pairs. If the transmitted rate vector and channel realization pair is within the operation region, the receiver {\em intends} to decode the messages, otherwise the receiver {\em intends} to report a collision (or outage). We define the decoding error probability and the collision miss detection probability similarly to \cite{ref WangJournal}, and define the system error probability as the maximum of the two. An upper bound on the achievable system error probability is derived under the assumption of a finite codeword length. We then show how the compound channel results help in obtaining error performance bounds for the random multiple access system where the receiver is only interested in recovering messages from a user {\em subset} \cite{ref Luo09}. This is based on the observation that, conditioned on the receiver not decoding messages for the rest of the users, the impact of their communication activities on the user subset of interest is equivalent to that of a compound channel.
\section{Multiple Random Access Communication over A Compound Channel}
\label{Section MRAC}
Consider a $K$-user time-slotted random access system over a compound discrete-time memoryless channel. Time is slotted with each slot equaling $N$ symbol durations, which is also the length of a packet or a codeword. We assume that channel coding is only applied within each time slot or packet. The compound channel consists of a family of discrete-time memoryless channels, characterized by a set of conditional probabilities $\left\{ P_{Y|X_1,\cdots,X_K}^{(1)}, \cdots, P_{Y|X_1,\cdots,X_K}^{(H)} \right\}$ with cardinality $H$, where, for $k \in \{1,\cdots, K\}$, $X_k \in \mathcal X$ is the channel input symbol of user $k$ with $\mathcal X$ being the finite input alphabet, and $Y \in \mathcal Y$ is the channel output symbol with $\mathcal Y$ being the finite output alphabet. In each time slot, a channel realization is randomly generated from this set and remains static throughout the slot duration. We assume that all users and the receiver know the compound channel set, but not the actual channel realization. For the time being, we will assume that $H<\infty$. The case when the compound channel set contains an infinite number of channels will be discussed at the end of this section.
Assume that at the beginning of a time slot, according to the message availability and the MAC layer protocol, each user, say user $k$ ($k\in\{1, \cdots, K\}$) chooses an arbitrary communication rate $r_k \in \{r_{k1},\cdots, r_{kM}\}$ in nats per symbol, where $\{r_{k1},\cdots, r_{kM}\}$ is a pre-determined finite rate set of user $k$ with cardinality $M$. Neither the other users nor the receiver knows the actual rate realization for each transmission, although they are shared with the rate set information. The user then encodes $\lfloor Nr_k \rfloor$ number of data nats, denoted by a message $w_k$, into a packet (codeword) with $N$ symbols, using a random coding scheme specified as in \cite{ref Luo09}\cite{ref Shamai07} and also in the following. For all $k\in\{1,\cdots,K\}$, we assume that user $k$ is equipped with a codebook library $\mathcal{L}_k = \{\mathcal{C}_{k\theta_k}: \theta_k \in \Theta_k\}$ in which codebooks are indexed by a set $\Theta_k$. Each codebook has $M$ classes of codewords. The $i^{th}$ ($i \in \{1,\cdots,M\}$) codeword class has $\lfloor e^{Nr_{ki}} \rfloor$ codewords with the same length of $N$ symbols. In contrast to a conventional coding scheme, here each codeword in the codebook corresponds to a message and rate pair $(w_k,r_k)$ \cite{ref Luo09}\cite{ref WangJournal}. Let $\mathcal{C}_{k\theta_k}(w_k,r_k)_j$ be the $j^{th}$ symbol of the codeword corresponding to message and rate pair $(w_k,r_k)$ in codebook $\mathcal{C}_{k\theta_k}$. User $k$ first selects the codebook by generating $\theta_k$ according to a distribution $\vartheta_k$ such that the random variables $X_{(w_k,r_k),j}: \theta_k \rightarrow \mathcal{C}_{k\theta_k}(w_k,r_k)_j$ are i.i.d. according to an input distribution $P_{X|r_k}$\footnote{The input distribution is assumed to be a function of the communication rate. In other words, different communication rates may correspond to different input distributions.}. The codebook $\mathcal{C}_{k\theta_k}$ is then used to map $(w_k,r_k)$ into a codeword, denoted by $\mbox{\boldmath$x$}_{(w_k,r_k)}$. After encoding, the codewords of all users are sent to the receiver over the compound channel.
To simplify the notation, we use bold font variable to denote the corresponding variables of all users. For example, $\mbox{\boldmath$w$}$ and $\mbox{\boldmath$r$}$ denote the messages and communication rates of all users. $\mbox{\boldmath$P$}_{\mbox{\scriptsize\boldmath$X$}|\mbox{\scriptsize\boldmath$r$}}$ denote the input distributions of all users, etc. Given a vector variable, say $\mbox{\boldmath$r$}$, we use $r_i$ to denote its element corresponding to user $i$. Let $\mathcal S \subset \{1,\cdots,K\}$ be a user subset, and $\bar{\mathcal{S}}$ be its complement. We use $\mbox{\boldmath$r$}_{\mathcal S}$ to denote the vector that is extracted from $\mbox{\boldmath$r$}$ with only elements corresponding to users in $\mathcal S$. By using the vector notation of the channel input symbols, the compound channel set is also denoted by $\{P^{(1)}_{Y|\mbox{\scriptsize\boldmath$X$}}, \cdots, P^{(H)}_{Y|\mbox{\scriptsize\boldmath$X$}}\}$.
We assume that the receiver is shared with the random codebook generation algorithms and hence knows the randomly generated codebooks of all users. Before packet transmission, the receiver pre-determines an ``operation region'' $\mathcal R =\{(\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}})\}$, which is a set of rate vector and channel realization pair, where each entry of $\mbox{\boldmath$r$}$ is chosen from the corresponding rate set, i.e., $r_k \in \{r_{k1},\cdots,r_{kM}\}$ ($k\in\{1,\cdots,K\}$), and $P_{Y|\mbox{\scriptsize\boldmath$X$}} \in \left\{ P_{Y|\mbox{\scriptsize\boldmath$X$}}^{(1)}, \cdots, P_{Y|\mbox{\scriptsize\boldmath$X$}}^{(H)} \right\}$. Let $(\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}})$ be the actual realization of the transmitted rate vector and channel pair. We assume that the receiver {\it intends} to decode all messages if $(\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) \in \mathcal R$, and {\it intends} to report a collision if $(\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) \notin \mathcal R$. Note that the actual rate and channel realization $(\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}})$ is unknown at the receiver. Therefore the receiver needs to make decisions whether to decode messages or to report a collision only based on the received channel symbols. More specifically, in each time slot, upon receiving the channel output symbols $\mbox{\boldmath$y$}$, the receiver estimates the rate and channel pair, denoted by $(\hat{\mbox{\boldmath$r$}},\hat{P}_{Y|\mbox{\scriptsize\boldmath$X$}})$, for all users. The receiver outputs the corresponding estimated message and rate vector pair $(\hat{\mbox{\boldmath$w$}},\hat{\mbox{\boldmath$r$}})$ if $(\hat{\mbox{\boldmath$r$}},\hat{P}_{Y|\mbox{\scriptsize\boldmath$X$}}) \in \mathcal R$ and a pre-determined decoding error probability requirement is satisfied. Otherwise, the receiver reports a collision. Also note that, whether the receiver should recover the messages or report a collision not only depends on the rates, but also depends on the channel realization. In other words, for the same transmission rate vector, the receiver may be designed to take different actions for different channel realizations. This is opposed to the conventional compound channel communication scenario where, if a rate is supported by the system, the receiver should always decode the messages irrespective of the channel realization.
Given the operation region $\mathcal R$, and conditioned on that $(\mbox{\boldmath$w$},\mbox{\boldmath$r$})$ is transmitted over channel $P_{Y|\mbox{\scriptsize\boldmath$X$}}$, we define the following three error probabilities. The decoding error probability, for $(\mbox{\boldmath$w$},\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) $ with $(\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) \in \mathcal R$, is defined as
\begin{equation}\label{MC-DecodingErrorDef}
P_{e(\mbox{\scriptsize\boldmath$w$},\mbox{\scriptsize\boldmath$r$},P_{Y|\mbox{\tiny\boldmath$X$}})} = Pr \left\{ (\hat{\mbox{\boldmath$w$}},\hat{\mbox{\boldmath$r$}}) \neq (\mbox{\boldmath$w$},\mbox{\boldmath$r$})|(\mbox{\boldmath$w$},\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) \right\}, \quad
\forall (\mbox{\boldmath$w$},\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) , (\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) \in \mathcal R.
\end{equation}
The collision miss detection probability, for $(\mbox{\boldmath$w$},\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}})$ with $(\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) \notin \mathcal R$, is defined as
\begin{eqnarray}\label{MC-CollisionMissDef}
&& \bar{P}_{c(\mbox{\scriptsize\boldmath$w$},\mbox{\scriptsize\boldmath$r$},P_{Y|\mbox{\tiny\boldmath$X$}})} = 1 - Pr \left\{ \mbox{``collision''}|(\mbox{\boldmath$w$},\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) \right\} - Pr\left\{ (\hat{\mbox{\boldmath$w$}},\hat{\mbox{\boldmath$r$}}) = (\mbox{\boldmath$w$},\mbox{\boldmath$r$})|(\mbox{\boldmath$w$},\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) \right\},\nonumber\\
&& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \forall (\mbox{\boldmath$w$},\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) , (\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) \notin \mathcal R.
\end{eqnarray}
Note that in (\ref{MC-CollisionMissDef}), when $(\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) \notin \mathcal R$, we have excluded the correct message and rate pair estimation from the collision miss detection event.
Let $\mathcal S \subset \{1,\cdots,K\}$ be an arbitrary user subset. Assume that $\sum_{k\notin\mathcal S} r_k \le I_{(\mbox{\scriptsize\boldmath$r$},P_{Y|\mbox{\tiny\boldmath$X$}})} (\mbox{\boldmath$X$}_{\bar{\mathcal S}};\mbox{\boldmath$Y$}|\mbox{\boldmath$X$}_{\mathcal S})$ for all $ (\mbox{\boldmath$r$},P_{Y|\mbox{\scriptsize\boldmath$X$}}) \in \mathcal R$, where $\mbox{\boldmath$X$}_{\mathcal S}$ denotes the channel input symbols of users in set $\mathcal S$, and $\mbox{\boldmath$X$}_{\bar{\mathcal S}}$ denotes the channel input symbols of users not in set $\mathcal S$. $I_{(\mbox{\scriptsize\boldmath$r$},P_{Y|\mbox{\tiny\boldmath$X$}})}$ is the mutual information function computed using input distribution corresponding to rate vector $\mbox{\boldmath$r$}$ (i.e., $\mbox{\boldmath$P$}_{\mbox{\scriptsize\boldmath$X$}|\mbox{\scriptsize\boldmath$r$}}$) and channel $P_{Y|\mbox{\scriptsize\boldmath$X$}}$. We define the system error probability $P_{es}$ as
\begin{eqnarray}\label{MC-SystemErrorDef}
P_{es} = \max \left\{ \max_{(\mbox{\scriptsize\boldmath$w$},\mbox{\scriptsize\boldmath$r$},P_{Y|\mbox{\tiny\boldmath$X$}}),(\mbox{\scriptsize\boldmath$r$},P_{Y|\mbox{\tiny\boldmath$X$}}) \in \mathcal R} P_{e(\mbox{\scriptsize\boldmath$w$},\mbox{\scriptsize\boldmath$r$},P_{Y|\mbox{\tiny\boldmath$X$}})}, \max_{(\mbox{\scriptsize\boldmath$w$},\mbox{\scriptsize\boldmath$r$},P_{Y|\mbox{\tiny\boldmath$X$}}),(\mbox{\scriptsize\boldmath$r$},P_{Y|\mbox{\tiny\boldmath$X$}})\notin \mathcal R} \bar{P}_{c(\mbox{\scriptsize\boldmath$w$},\mbox{\scriptsize\boldmath$r$},P_{Y|\mbox{\tiny\boldmath$X$}})}\right\}.
\end{eqnarray}
The following theorem gives an upper bound on the achievable system error probability $P_{es}$.
\begin{theorem}\label{TheoremMC}
Consider $K$-user multiple random access communication over a compound discrete-time memoryless channel $\left\{ P^{(1)}_{Y|\mbox{\scriptsize\boldmath$X$}}, \cdots, P^{(H)}_{Y|\mbox{\scriptsize\boldmath$X$} }\right\}$, where $H<\infty$ is a positive integer. Let $\mbox{\boldmath$P$}_{\mbox{\scriptsize\boldmath$X$}|\mbox{\scriptsize\boldmath$r$}}$ be the input distribution for all users and all rates. Let $\mathcal R$ be the operation region. Assume finite codeword length $N$. There exists a decoding algorithm, whose system error probability $P_{es}$ is upper bounded by,
\begin{eqnarray}\label{SystemErrorTMC}
P_{es}\le \max\left\{ \max_{(\mbox{\scriptsize \boldmath $r$},P_{Y|\mbox{\tiny\boldmath$X$}})\in \mathcal{R} }\sum_{\mathcal{S}\subset\{1, \cdots, K\}} \left[ \begin{array}{l} \sum_{(\tilde{\mbox{\scriptsize \boldmath $r$}},\tilde{P}_{Y|\mbox{\tiny\boldmath$X$}})\in \mathcal{R}, \tilde{\mbox{\scriptsize \boldmath $r$}}_{\mathcal{S}}= \mbox{\scriptsize \boldmath $r$}_{\mathcal{S}} } \exp\{-N E_m(\mathcal{S}, \mbox{\boldmath$r$},\tilde{\mbox{\boldmath $r$}}, P_{Y|\mbox{\scriptsize\boldmath$X$}}, \tilde{P}_{Y|\mbox{\scriptsize\boldmath$X$}} )\} \\ + \max_{(\mbox{\scriptsize \boldmath $r$}', P'_{Y|\mbox{\tiny\boldmath$X$}})\not\in \mathcal{R}, \mbox{\scriptsize \boldmath $r$}'_{\mathcal{S}}= \mbox{\scriptsize \boldmath $r$}_{\mathcal{S}} } \exp\{-NE_i(\mathcal{S}, \mbox{\boldmath$r$},\mbox{\boldmath $r$}', P_{Y|\mbox{\scriptsize\boldmath$X$}}, P'_{Y|\mbox{\scriptsize\boldmath$X$}} ) \} \end{array} \right],\right. \nonumber\\
\left.\max_{(\tilde{\mbox{\scriptsize \boldmath $r$}},\tilde{P}_{Y|\mbox{\tiny\boldmath$X$}})\not\in \mathcal{R}}\sum_{\mathcal{S}\subset\{1, \cdots, K\}} \sum_{(\mbox{\scriptsize \boldmath $r$},P_{Y|\mbox{\tiny\boldmath$X$}})\in \mathcal{R}, \mbox{\scriptsize \boldmath $r$}_{\mathcal{S}}=\tilde{\mbox{\scriptsize \boldmath $r$}}_{\mathcal{S}}} \max_{(\mbox{\scriptsize \boldmath $r$}',P'_{Y|\mbox{\tiny\boldmath$X$}})\not\in \mathcal{R}, \mbox{\scriptsize \boldmath $r$}'_{\mathcal{S}}= \tilde{\mbox{\scriptsize \boldmath $r$}}_{\mathcal{S}} } \exp\{ -N E_i (\mathcal{S}, \mbox{\boldmath$r$},\mbox{\boldmath $r$}', P_{Y|\mbox{\scriptsize\boldmath$X$}}, P'_{Y|\mbox{\scriptsize\boldmath$X$}} ) \} \right\},
\end{eqnarray}
where $E_m(\mathcal{S}, \mbox{\boldmath$r$},\tilde{\mbox{\boldmath $r$}}, P_{Y|\mbox{\scriptsize\boldmath$X$}}, \tilde{P}_{Y|\mbox{\scriptsize\boldmath$X$}} )$ and $E_i(\mathcal{S}, \mbox{\boldmath$r$},\mbox{\boldmath $r$}', P_{Y|\mbox{\scriptsize\boldmath$X$}}, P'_{Y|\mbox{\scriptsize\boldmath$X$}} )$ are given by
\begin{eqnarray}\label{EmEiMultiMC}
&& E_m(\mathcal{S}, \mbox{\boldmath$r$},\tilde{\mbox{\boldmath $r$}}, P_{Y|\mbox{\scriptsize\boldmath$X$}}, \tilde{P}_{Y|\mbox{\scriptsize\boldmath$X$}} ) = \max_{0<\rho \le 1} -\rho \sum_{k\not\in \mathcal{S}}\tilde{r}_k + \max_{0<s\le 1} -\log \sum_Y \sum_{\mbox{\scriptsize \boldmath $X$}_{\mathcal{S}}} \prod_{k\in \mathcal{S}} P_{X|r_k}(X_k) \nonumber \\
&& \quad \times \left(\sum_{\mbox{\scriptsize \boldmath $X$}_{\bar{\mathcal{S}}}}\prod_{k \not\in \mathcal{S}}P_{X|r_k}(X_k)P_{Y|\mbox{\scriptsize\boldmath$X$}}(Y|\mbox{\boldmath $X$})^{1-s}\right) \left(\sum_{\mbox{\scriptsize \boldmath $X$}_{\bar{\mathcal{S}}}}\prod_{k \not\in \mathcal{S}}P_{X|\tilde{r}_k}(X_k)\tilde{P}_{Y|\mbox{\scriptsize\boldmath$X$}}(Y|\mbox{\boldmath $X$})^{\frac{s}{\rho}} \right)^{\rho}, \nonumber \\
&& E_i(\mathcal{S}, \mbox{\boldmath$r$},\mbox{\boldmath $r$}', P_{Y|\mbox{\scriptsize\boldmath$X$}}, P'_{Y|\mbox{\scriptsize\boldmath$X$}} ) = \max_{0<\rho \le 1} -\rho \sum_{k\not\in \mathcal{S}}r_k + \max_{0<s \le 1-\rho} - \log \sum_Y \sum_{\mbox{\scriptsize \boldmath $X$}_{\mathcal{S}}} \prod_{k\in \mathcal{S}} P_{X|r_k}(X_k) \nonumber \\
&& \quad \times \left(\sum_{\mbox{\scriptsize \boldmath $X$}_{\bar{\mathcal{S}}}}\prod_{k \not\in \mathcal{S}}P_{X|r_k}(X_k)P_{Y|\mbox{\scriptsize\boldmath$X$}}(Y|\mbox{\boldmath $X$})^{\frac{s}{s+\rho}} \right)^{s+\rho}\left(\sum_{\mbox{\scriptsize \boldmath $X$}_{\bar{\mathcal{S}}}}\prod_{k \not\in \mathcal{S}}P_{X|r'_k}(X_k)P'_{Y|\mbox{\scriptsize\boldmath$X$}}(Y|\mbox{\boldmath $X$})\right)^{1-s}.
\end{eqnarray}
$\QED$
\end{theorem}
The proof of Theorem \ref{TheoremMC} is given in Appendix \ref{AppendixTheoremMC}.
When the compound channel is randomly generated at the beginning but remains static afterwards, one can take codeword length to infinity to obtain the system error exponent as $E_s = \lim_{N\rightarrow\infty} -\frac{1}{N} \log P_{es}$. The following lower bound on the achievable system error exponent $E_{s}$ can be easily derived from Theorem \ref{TheoremMC}.
\begin{corollary}\label{CorollaryMC}
The system error exponent of a $K$-user multiple random access system over compound discrete-time memoryless channels given in Theorem \ref{TheoremMC} is lower-bounded by
\begin{eqnarray}\label{SystemErrorExponentMC}
E_{s} \ge \min\left\{\min_{\mathcal{S}\subset \{1,\cdots,K\}} \min_{ (\mbox{\scriptsize\boldmath$r$},P_{Y|\mbox{\tiny\boldmath$X$}} ), (\tilde{\mbox{\scriptsize\boldmath$r$}},\tilde{P}_{Y|\mbox{\tiny\boldmath$X$}} ) \in \mathcal R, } E_m(\mathcal{S}, \mbox{\boldmath$r$},\tilde{\mbox{\boldmath $r$}}, P_{Y|\mbox{\scriptsize\boldmath$X$}}, \tilde{P}_{Y|\mbox{\scriptsize\boldmath$X$}} ),\right. \nonumber\\
\left.\min_{\mathcal{S}\subset \{1,\cdots,K\}} \min_{ (\mbox{\scriptsize\boldmath$r$},P_{Y|\mbox{\tiny\boldmath$X$}} ) \in \mathcal R, (\tilde{\mbox{\scriptsize\boldmath$r$}},\tilde{P}_{Y|\mbox{\tiny\boldmath$X$}} ) \notin \mathcal R, }E_i(\mathcal{S}, \mbox{\boldmath$r$},\tilde{\mbox{\boldmath $r$}}, P_{Y|\mbox{\scriptsize\boldmath$X$}}, \tilde{P}_{Y|\mbox{\scriptsize\boldmath$X$}} ) \right\},
\end{eqnarray}
where $E_m(\mathcal{S}, \mbox{\boldmath$r$},\tilde{\mbox{\boldmath $r$}}, P_{Y|\mbox{\scriptsize\boldmath$X$}}, \tilde{P}_{Y|\mbox{\scriptsize\boldmath$X$}} )$ and $E_i(\mathcal{S}, \mbox{\boldmath$r$},\tilde{\mbox{\boldmath $r$}}, P_{Y|\mbox{\scriptsize\boldmath$X$}}, \tilde{P}_{Y|\mbox{\scriptsize\boldmath$X$}} )$ are given in (\ref{EmEiMultiMC}).
\end{corollary}
Compared with the error exponent derived in \cite[Corollary 2]{ref WangJournal}, we can see that, even though the channel stays static forever, the system still needs to pay a penalty in error exponent performance for not knowing the channel at the receiver\footnote{We assume that such a conclusion should be well known for the conventional compound channel communication. However, we are not able to find a reference that made such a clear statement.}.
In both Theorem \ref{TheoremMC} and Corollary \ref{CorollaryMC}, we have assumed that there are only a finite number of channels in the compound set. Next, we will extend the result to the case when the cardinality of the compound channel set can be infinity.
We first assume that the the channels in the compound set can be partitioned into $H$ classes, denoted by $\left\{ \mathcal{F}^{(1)},\cdots, \mathcal{F}^{(H)}\right\}$, where $H < \infty$ is a positive integer. For example, if the compound channel set contains fading channels with continuous channel gains, one could quantize the channel gains and define the set of channels with the same quantization outcome as one channel class. We next assume that the receiver should choose an operation region $\mathcal{R}$ to satisfy the following constraint for any rate vector $\mbox{\boldmath $r$}$ and channel class $\mathcal{F}\in \left\{ \mathcal{F}^{(1)},\cdots, \mathcal{F}^{(H)}\right\}$.
\begin{equation}
\mbox{C1: For any } (\mbox{\boldmath $r$}, \mathcal{F}), \mbox{ either } (\mbox{\boldmath $r$}, P_{Y|\mbox{\scriptsize\boldmath$X$}}) \in \mathcal{R} \mbox{ } \forall P_{Y|\mbox{\scriptsize\boldmath$X$}} \in \mathcal{F}, \mbox{ or } (\mbox{\boldmath $r$}, P_{Y|\mbox{\scriptsize\boldmath$X$}}) \not\in \mathcal{R} \mbox{ } \forall P_{Y|\mbox{\scriptsize\boldmath$X$}} \in \mathcal{F}.
\label{ConstraintC1}
\end{equation}
We say $(\mbox{\boldmath $r$}, \mathcal{F})\in \mathcal{R}$ if $(\mbox{\boldmath $r$}, P_{Y|\mbox{\scriptsize\boldmath$X$}}) \in \mathcal{R}$ for all $P_{Y|\mbox{\scriptsize\boldmath$X$}}\in \mathcal{F}$, and we say $(\mbox{\boldmath $r$}, \mathcal{F})\not\in \mathcal{R}$ otherwise.
For each channel class $\mathcal{F}$ and for each channel output symbol $Y$ and input symbol vector $\mbox{\boldmath$X$}$, we define the following upper and lower bounds on the conditional probability values yielded by channels in $\mathcal{F}$, denoted by $P^{\mathcal{F}}_{\max}(Y|\mbox{\boldmath$X$})$ and $P^{\mathcal{F}}_{\min}(Y|\mbox{\boldmath$X$})$,
\begin{eqnarray}
\label{UpperLowerBounds}
&& P^{\mathcal{F}}_{\max}(Y|\mbox{\boldmath$X$}) = \max_{P_{Y|\mbox{\tiny\boldmath$X$}} \in \mathcal{F}} P_{Y|\mbox{\scriptsize\boldmath$X$}}(Y|\mbox{\boldmath$X$}) , \qquad P^{\mathcal{F}}_{\min}(Y|\mbox{\boldmath$X$}) = \min_{P_{Y|\mbox{\tiny\boldmath$X$}} \in \mathcal{F}} P_{Y|\mbox{\scriptsize\boldmath$X$}}(Y|\mbox{\boldmath$X$}).
\end{eqnarray}
The following theorem gives an upper bound on the achievable system error probability.
\begin{theorem}
\label{TheoremContinuousChannel}
Consider a $K$-user multiple random access communication system over a compound discrete-time memoryless channel. Assume that the compound set is partitioned into $H$ classes, denoted by $\left\{ \mathcal{F}^{(1)},\cdots, \mathcal{F}^{(H)}\right\}$, where $H$ is a finite positive integer. Assume that the operation region $\mathcal{R}$ satisfies constraint C1 given in (\ref{ConstraintC1}). The system error probability $P_{es}$ is upper bounded as follows.
\begin{eqnarray}
\label{PesCC}
P_{es} & \le & \max \left\{ \max_{(\mbox{\scriptsize\boldmath$r$},\mathcal{F}) \in \mathcal{R}} \sum_{\mathcal{S} \subset \{1,\cdots,K\}} \left[ \max_{(\mbox{\scriptsize\boldmath$r$}', \mathcal{F}') \notin \mathcal{R},\mbox{\scriptsize\boldmath$r$}'_{\mathcal{S}} = \mbox{\scriptsize\boldmath$r$}_{\mathcal{S}}} \exp \left\{ -NE_i(\mathcal{S}, \mbox{\boldmath$r$},\mbox{\boldmath $r$}', \mathcal{F}, \mathcal{F}' ) \right\} \right. \right.\nonumber\\
&& \qquad \left. + \sum_{(\tilde{\mbox{\scriptsize\boldmath$r$}},\tilde{\mathcal{F}}) \in \mathcal{R},\tilde{\mbox{\scriptsize\boldmath$r$}}_{\mathcal{S}} = \mbox{\scriptsize\boldmath$r$}_{\mathcal{S}}} \exp \left\{ -NE_m(\mathcal{S}, \mbox{\boldmath$r$},\tilde{\mbox{\boldmath $r$}}, \mathcal{F}, \tilde{\mathcal{F}} ) \right\} \right] , \nonumber\\
&& \quad \left.\max_{(\tilde{\mbox{\scriptsize\boldmath$r$}},\tilde{\mathcal{F}}) \notin \mathcal{R}} \sum_{\mathcal{S} \subset \{1,\cdots,K\}}
\left[ \sum_{(\mbox{\scriptsize\boldmath$r$},\mathcal{F}) \in \mathcal{R},\mbox{\scriptsize\boldmath$r$}_{\mathcal{S}} = \tilde{\mbox{\scriptsize\boldmath$r$}}_{\mathcal{S}}} \max_{(\mbox{\scriptsize\boldmath$r$}', \mathcal{F}') \notin \mathcal{R},\mbox{\scriptsize\boldmath$r$}'_{\mathcal{S}} = \mbox{\scriptsize\boldmath$r$}_{\mathcal{S}}} \exp \left\{ -NE_i(\mathcal{S}, \mbox{\boldmath$r$},\mbox{\boldmath $r$}', \mathcal{F}, \mathcal{F}' ) \right\} \right] \right\}.
\end{eqnarray}
where $E_m(\mathcal{S}, \mbox{\boldmath$r$},\tilde{\mbox{\boldmath $r$}}, \mathcal{F}, \tilde{\mathcal{F}} )$ and $E_i(\mathcal{S}, \mbox{\boldmath$r$},\mbox{\boldmath $r$}', \mathcal{F}, \mathcal{F}' )$ are given by
\begin{eqnarray}
\label{EmEiCC}
&& E_m(\mathcal{S}, \mbox{\boldmath$r$},\tilde{\mbox{\boldmath $r$}}, \mathcal{F}, \tilde{\mathcal{F}} ) = \max_{0<\rho \le 1} -\rho \sum_{k\not\in \mathcal{S}}\tilde{r}_k + \max_{0<s\le 1} -\log \sum_Y \sum_{\mbox{\scriptsize \boldmath $X$}_{\mathcal{S}}} \prod_{k\in \mathcal{S}} P_{X|r_k}(X_k) \nonumber \\
&& \quad \times \left(\sum_{\mbox{\scriptsize \boldmath $X$}_{\bar{\mathcal{S}}}}\prod_{k \not\in \mathcal{S}}P_{X|r_k}(X_k)P^{\mathcal{F}}_{\max}(Y|\mbox{\boldmath $X$})P^{\mathcal{F}}_{\min}(Y|\mbox{\boldmath $X$})^{-s}\right) \left(\sum_{\mbox{\scriptsize \boldmath $X$}_{\bar{\mathcal{S}}}}\prod_{k \not\in \mathcal{S}}P_{X|\tilde{r}_k}(X_k) P^{\tilde{\mathcal{F}}}_{\max}(Y|\mbox{\boldmath $X$})^{\frac{s}{\rho}} \right)^{\rho}. \nonumber \\
&& E_i(\mathcal{S}, \mbox{\boldmath$r$},\mbox{\boldmath $r$}', \mathcal{F}, \mathcal{F}' ) = \max_{0<\rho \le 1} -\rho \sum_{k\not\in \mathcal{S}}r_k + \max_{0<s \le 1-\rho} - \log \sum_Y \sum_{\mbox{\scriptsize \boldmath $X$}_{\mathcal{S}}} \prod_{k\in \mathcal{S}} P_{X|r_k}(X_k) \nonumber \\
&& \quad \times \left(\sum_{\mbox{\scriptsize \boldmath $X$}_{\bar{\mathcal{S}}}}\prod_{k \not\in \mathcal{S}}P_{X|r_k}(X_k)P^{\mathcal{F}}_{\max}(Y|\mbox{\boldmath $X$})P^{\mathcal{F}}_{\min}(Y|\mbox{\boldmath $X$})^{\frac{-\rho}{s+\rho}} \right)^{s+\rho}\left(\sum_{\mbox{\scriptsize \boldmath $X$}_{\bar{\mathcal{S}}}}\prod_{k \not\in \mathcal{S}}P_{X|r'_k}(X_k)P^{\mathcal{F}'}_{\max}(Y|\mbox{\boldmath $X$})\right)^{1-s}. \nonumber\\
\end{eqnarray}
$\QED$
\end{theorem}
The proof of Theorem \ref{TheoremContinuousChannel} is given in Appendix \ref{AppendixTheoremContinuousChannel}. As shown in the proof that, in order to make decoding and collision report decisions, the receiver only needs to search over the finite number of channel classes using statistics $P^\mathcal{F}_{\max}$ and $P^\mathcal{F}_{\min}$ defined in (\ref{UpperLowerBounds}), as opposed to searching among all possible channels.
\label{SystemErrorExponentCC}
\section{Individual User Decoding in Random Multiple Access Communication}
\label{Section IndividualDecoding}
In Section \ref{Section MRAC}, we have assumed that the receiver either decodes messages or reports collisions for {\em all} users in the system. In practical applications, even though many users compete for the wireless channel, it is common that the receiver may not be interested in recovering messages for all of them. In this section, we show that the results obtained in Section \ref{Section MRAC} can help to derive error probability bounds in a random multiple access system where the receiver is only interested in recovering the messages from a user subset. However, to simplify the notations, we will only consider a special case when the communication channel is known at the receiver, and when the receiver is only interested in decoding for a single user. Generalizing the results to decoding for multiple users over a compound channel is straightforward.
Let the discrete-time memoryless channel be characterized by $P_{Y|\mbox{\scriptsize\boldmath$X$}}$, which is known at the receiver. In each time slot, each user chooses a communication rate and encodes its message using the random coding scheme described in Section \ref{Section MRAC}. The rate information is shared neither among the users nor with the receiver. We assume that the receiver is only interested in recovering the message for user $k\in \{1, \cdots, K\}$. We assume that the receiver chooses an operation region $\mathcal{R}$, such that if the transmitted rate vector $\mbox{\boldmath $r$}$ satisfies $\mbox{\boldmath $r$}\in\mathcal{R}$, the receiver intends to decode for user $k$, and if $\mbox{\boldmath $r$}\not\in\mathcal{R}$, the receiver intends to report a collision for user $k$. It is important to note that, first, whether the receiver will be able to decode the message of user $k$, not only depends on the rate of user $k$, but also depends on the rate of other users. Therefore, the operation rate region $\mathcal{R}$ should still be defined as a set of rate vector $\mbox{\boldmath $r$}$, as opposed to the rate of user $k$. Second, even though the receiver only cares about the message of user $k$, the receiver still has the option of decoding the messages for some other users if this helps to improve the communication performance of user $k$. This implies that, based upon the received symbols, the receiver will essentially need to make a decision on which subset of the messages should be decoded.
Due to the above understandings, we first define an elementary decoder, called the ``$(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoder". Given a user subset $\mathcal{D}\subseteq \{1, \cdots, K\}$ and an operation rate region $\mathcal{R}_{\mathcal{D}}$, the ``$(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoder" intends to recover messages for users in $\mathcal{D}$ while regarding signals from users not in $\mathcal{D}$ as interference, if the communication rate vector is within the operation region $\mathcal{R}_{\mathcal{D}}$. We define the following error probabilities for a $(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoder. Conditioned on users in $\mathcal{D}$ transmitting $(\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}})$ and users not in $\mathcal{D}$ choosing rate $\mbox{\boldmath $r$}_{\bar{\mathcal{D}}}$, let us denote the estimated messages and rates by $(\hat{\mbox{\boldmath $w$}}_{\mathcal{D}}, \hat{\mbox{\boldmath $r$}}_{\mathcal{D}})$ and $\hat{\mbox{\boldmath $r$}}_{\bar{\mathcal{D}}}$ if the decoder does not report a collision. We define the decoding error probability of the $(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoder for $(\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}})$ with $\mbox{\boldmath $r$} \in \mathcal{R}_{\mathcal{D}}$ as
\begin{eqnarray}
\label{M-DecodeErrorDef}
P_e(\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}})= Pr\{(\hat{\mbox{\boldmath $w$}}_{\mathcal{D}}, \hat{\mbox{\boldmath $r$}}_{\mathcal{D}}) \ne (\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}})|(\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}})\}, \quad \forall (\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}}), \mbox{\boldmath $r$}\in \mathcal{R}_{\mathcal{D}}.
\end{eqnarray}
We define the collision miss detection probability for $(\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}})$ with $\mbox{\boldmath $r$} \not\in \mathcal{R}_{\mathcal{D}}$ as
\begin{eqnarray}
\label{M-CollisionErrorDef}
&& \bar{P}_c(\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}}) = 1 - Pr\{ \mbox{``collision"}|(\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}})\}- Pr\{(\hat{\mbox{\boldmath $w$}}_{\mathcal{D}}, \hat{\mbox{\boldmath $r$}}_{\mathcal{D}}) = (\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}}) |(\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}})\}, \nonumber\\
&& \qquad \qquad\qquad \qquad\qquad \qquad\qquad \qquad\qquad \qquad\qquad \qquad\qquad \qquad \forall (\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}}), \mbox{\boldmath $r$}\not\in \mathcal{R}_{\mathcal{D}}.
\end{eqnarray}
System error probability of the $(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoder is defined by
\begin{eqnarray}
&& P_{es}(\mathcal{D}, \mathcal{R}_{\mathcal{D}})=\max\left\{\max_{(\mbox{\scriptsize \boldmath $w$}_{\mathcal{D}}, \mbox{\scriptsize \boldmath $r$}_{\mathcal{D}}, \mbox{\scriptsize \boldmath $r$}_{\bar{\mathcal{D}}}), \mbox{\scriptsize \boldmath $r$}\in \mathcal{R}_{\mathcal{D}} } P_e(\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}}), \max_{(\mbox{\scriptsize \boldmath $w$}_{\mathcal{D}}, \mbox{\scriptsize \boldmath $r$}_{\mathcal{D}}, \mbox{\scriptsize \boldmath $r$}_{\bar{\mathcal{D}}}), \mbox{\scriptsize \boldmath $r$} \not\in \mathcal{R}_{\mathcal{D}} } \bar{P}_c(\mbox{\boldmath $w$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}}) \right\}.
\end{eqnarray}
Given a finite codeword length $N$, the following lemma gives an upper bound on the achievable system error probability of a $(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoder.
\begin{lemma}{\label{Lemma1}}
The following system error probability bound is achievable for a $K$-user random multiple access communication system over a discrete-time memoryless channel $P_{Y|\mbox{\scriptsize\boldmath$X$}}$ with an $(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoder,
\begin{eqnarray}
\label{BoundLemma1}
P_{es}(\mathcal{D}, \mathcal{R}_{\mathcal{D}}) \le \max\left\{ \max_{\mbox{\scriptsize \boldmath $r$}\in \mathcal{R}_{\mathcal{D}} }\sum_{\mathcal{S}\subset \mathcal{D}} \left[ \sum_{\tiny \begin{array}{c} \tilde{\mbox{\scriptsize \boldmath $r$}}\in \mathcal{R}_{\mathcal{D}}, \\ \tilde{\mbox{\scriptsize \boldmath $r$}}_{\mathcal{S}}= \mbox{\scriptsize \boldmath $r$}_{\mathcal{S}} \end{array}} \exp\{-N E_{m\mathcal{D}}(\mathcal{S}, \mbox{\boldmath $r$}, \tilde{\mbox{\boldmath $r$}})\} + \max_{\tiny \begin{array}{c} \mbox{\scriptsize \boldmath $r$}'\not\in \mathcal{R}_{\mathcal{D}}, \\ \mbox{\scriptsize \boldmath $r$}'_{\mathcal{S}}= \mbox{\scriptsize \boldmath $r$}_{\mathcal{S}} \end{array} } \exp\{-NE_{i\mathcal{D}}(\mathcal{S}, \mbox{\boldmath $r$}, \mbox{\boldmath $r$}') \} \right] \right., \nonumber \\
\left.\max_{\tilde{\mbox{\scriptsize \boldmath $r$}}\not\in \mathcal{R}_{\mathcal{D}} }\sum_{\mathcal{S}\subset \mathcal{D}} \sum_{\tiny \begin{array}{c} \mbox{\scriptsize \boldmath $r$}\in \mathcal{R}_{\mathcal{D}}, \\ \mbox{\scriptsize \boldmath $r$}_{\mathcal{S}}=\tilde{\mbox{\scriptsize \boldmath $r$}}_{\mathcal{S}}\end{array} } \max_{\tiny \begin{array}{c} \mbox{\scriptsize \boldmath $r$}'\not\in \mathcal{R}_{\mathcal{D}}, \\ \mbox{\scriptsize \boldmath $r$}'_{\mathcal{S}}= \tilde{\mbox{\scriptsize \boldmath $r$}}_{\mathcal{S}} \end{array} } \exp\{ -N E_{i\mathcal{D}}(\mathcal{S}, \mbox{\boldmath $r$}, \mbox{\boldmath $r$}') \} \right\}, \nonumber \\
\label{SRSBound}
\end{eqnarray}
where $E_{m\mathcal{D}}(\mathcal{S}, \mbox{\boldmath $r$}, \tilde{\mbox{\boldmath $r$}})$ and $E_{i\mathcal{D}}(\mathcal{S}, \mbox{\boldmath $r$}, \mbox{\boldmath $r$}')$ are given by,
\begin{eqnarray}
\label{EmEiDDcoder}
&& E_{m\mathcal{D}}(\mathcal{S}, \mbox{\boldmath $r$}, \tilde{\mbox{\boldmath $r$}})= \max_{0<\rho \le 1} -\rho \sum_{k\in \mathcal{D}\setminus \mathcal{S}}\tilde{r}_k \nonumber + \max_{0<s\le 1} -\log \sum_Y \sum_{\mbox{\scriptsize \boldmath $X$}_{\mathcal{S}}} \prod_{k\in \mathcal{S}} P_{X|r_k}(X_k) \nonumber \\
&& \quad \times \left(\sum_{\mbox{\scriptsize \boldmath $X$}_{\mathcal{D}\setminus \mathcal{S}}}\prod_{k \in \mathcal{D}\setminus \mathcal{S}}P_{X|r_k}(X_k)P(Y|\mbox{\boldmath $X$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}} )^{1-s}\right) \left(\sum_{\mbox{\scriptsize \boldmath $X$}_{\mathcal{D}\setminus \mathcal{S}}}\prod_{k \in \mathcal{D}\setminus \mathcal{S}}P_{X|\tilde{r}_k}(X_k)P(Y|\mbox{\boldmath $X$}_{\mathcal{D}}, \tilde{\mbox{\boldmath $r$}}_{\bar{\mathcal{D}}} )^{\frac{s}{\rho}} \right)^{\rho}, \nonumber \\
&& E_{i\mathcal{D}}(\mathcal{S}, \mbox{\boldmath $r$}, \mbox{\boldmath $r$}') = \max_{0<\rho \le 1} -\rho \sum_{k\in \mathcal{D}\setminus \mathcal{S}}r_k + \max_{0<s \le 1-\rho} - \log \sum_Y \sum_{\mbox{\scriptsize \boldmath $X$}_{\mathcal{S}}} \prod_{k\in \mathcal{S}} P_{X|r_k}(X_k) \nonumber \\
&& \times \left(\sum_{\mbox{\scriptsize \boldmath $X$}_{\mathcal{D}\setminus \mathcal{S}}}\prod_{k \in \mathcal{D}\setminus \mathcal{S}}P_{X|r_k}(X_k)P(Y|\mbox{\boldmath $X$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}} )^{\frac{s}{s+\rho}} \right)^{s+\rho} \left(\sum_{\mbox{\scriptsize \boldmath $X$}_{\mathcal{D}\setminus \mathcal{S}}}\prod_{k \in \mathcal{D}\setminus \mathcal{S}}P_{X|r'_k}(X_k)P(Y|\mbox{\boldmath $X$}_{\mathcal{D}}, \mbox{\boldmath $r$}'_{\bar{\mathcal{D}}})\right)^{1-s},
\end{eqnarray}
with $P(Y|\mbox{\boldmath $X$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}} )$ in the above equations defined as
\begin{equation}
\label{EquivalentCompound}
P(Y|\mbox{\boldmath $X$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}} )= \sum_{\mbox{\scriptsize \boldmath $X$}_{\bar{\mathcal{D}}}}\prod_{k \in \bar{\mathcal{D}}}P_{X|r_k}(X_k)P_{Y|\mbox{\scriptsize\boldmath$X$}}(Y|\mbox{\boldmath $X$}).
\end{equation}
$\QED$
\end{lemma}
\begin{proof}
Since the decoder regards signals from users not in $\mathcal{D}$ as interference, given that users not in $\mathcal{D}$ choose rate $\mbox{\boldmath $r$}_{\bar{\mathcal{D}}}$, the multiple access channel experienced by users in $\mathcal{D}$ is characterized by $P(Y|\mbox{\boldmath $X$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}})$ as specified in (\ref{EquivalentCompound}). The system can therefore be regarded as a random multiple access system with $|\mathcal{D}|$ users communicating over a compound channel characterized by the set $\{P(Y|\mbox{\boldmath $X$}_{\mathcal{D}}, \mbox{\boldmath $r$}_{\bar{\mathcal{D}}} )| \forall \mbox{\boldmath $r$}_{\bar{\mathcal{D}}} \}$. Consequently, Lemma \ref{Lemma1} is implied directly by Theorem \ref{TheoremMC}.
\end{proof}
Next, we will come back to the system where the receiver is only interested in the message of user $k$. We assume that for each user subset $\mathcal{D}\subseteq \{1, \cdots, K\}$ with $k\in \mathcal{D}$, the receiver assigns an operation region $\mathcal{R}_{\mathcal{D}} \subseteq\mathcal{R}$ for the $(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoder. That is, if the transmission rate $\mbox{\boldmath $r$}$ satisfies $\mbox{\boldmath $r$}\in \mathcal{R}_{\mathcal{D}}$, the receiver intends to use the $(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoder to recover the message of user $k$. It is easy to see that we should have,
\begin{equation}
\mathcal{R}=\bigcup_{\mathcal{D}: \mathcal{D}\subseteq\{1, \cdots, K\}, k\in \mathcal{D}} \mathcal{R}_{\mathcal{D}}.
\label{RegionAssignment}
\end{equation}
Assume that the receiver (single-user decoder) carries out all the $(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoding operations. The receiver outputs an estimated message $\hat{w}_k$ for user $k$ if at least one $(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoder outputs an estimated message, and all estimation outputs of the $(\mathcal{D},\mathcal{R}_{\mathcal{D}})$-decoders for user $k$ are identical. Otherwise, the receiver reports a collision for user $k$.
Let the transmitted rate vector be $\mbox{\boldmath$r$}$, and the transmitted message of user $k$ be $w_k$. We define the decoding error probability $P_{e}(w_k,\mbox{\boldmath$r$})$, the collision miss detection probability $\bar{P}_{c}(w_k,\mbox{\boldmath$r$})$ and the system error probability $P_{es}$ as follows,
\begin{eqnarray}\label{SingleDecodingProbDef}
&&P_{e}(w_k,\mbox{\boldmath$r$}) = Pr \left\{(\hat{w}_k,\hat{r}_k) \neq (w_k,r_k) | (w_k,\mbox{\boldmath$r$}) \right\}, \forall (w_k,\mbox{\boldmath$r$}), \mbox{\boldmath$r$} \in \mathcal{R},\nonumber\\
&&\bar{P}_{c}(w_k,\mbox{\boldmath$r$}) = 1 - Pr \left\{ \mbox{``collision''}|(w_k,\mbox{\boldmath$r$}) \right\} - Pr\left\{ (\hat{w}_k,\hat{r}_k) = (w_k,r_k) |(w_k,\mbox{\boldmath$r$})\right\},\forall (w_k,\mbox{\boldmath$r$}), \mbox{\boldmath$r$} \notin \mathcal{R} ,\nonumber\\
&&P_{es} = \max\left\{ \max_{(w_k,\mbox{\scriptsize\boldmath$r$}), \mbox{\scriptsize\boldmath$r$} \in \mathcal{R}}P_{e}(w_k,\mbox{\boldmath$r$}), \max_{(w_k,\mbox{\scriptsize\boldmath$r$}), \mbox{\scriptsize\boldmath$r$} \notin \mathcal{R}}\bar{P}_{c}(w_k,\mbox{\boldmath$r$}) \right\}.
\end{eqnarray}
The following theorem gives an upper bound on the achievable system error probability of the single-user decoder.
\begin{theorem}{\label{Theorem2}}
Consider a $K$-user random multiple access system over a discrete-time memoryless channel $P_{Y|\mbox{\scriptsize\boldmath$X$}}$, with the receiver only interested in recovering the message for user $k$. Assume the receiver chooses an operation region $\mathcal{R}$. Let $\sigma$ be an arbitrary partitioning of the operation region $\mathcal{R}$ satisfying
\begin{eqnarray}
&& \mathcal{R}=\bigcup_{\mathcal{D}: \mathcal{D}\subseteq\{1, \cdots, K\}, k\in \mathcal{D}} \mathcal{R}_{\mathcal{D}}, \nonumber \\
&& \mathcal{R}_{\mathcal{D}'}\cap \mathcal{R}_{\mathcal{D}}=\phi, \forall \mathcal{D}, \mathcal{D}'\subseteq\{1, \cdots, K\}, \mathcal{D}' \ne \mathcal{D}, k\in \mathcal{D}, \mathcal{D}'.
\label{RegionPartition}
\end{eqnarray}
System error probability of the single-user decoder is upper-bounded by,
\begin{equation}
P_{es} \le \min_{\sigma} \sum_{\mathcal{D}: \mathcal{D}\subseteq\{1, \cdots, K\}, k\in \mathcal{D}} P_{es}(\mathcal{D}, \mathcal{R}_{\mathcal{D}}),
\label{RegionPartitioning}
\end{equation}
where $P_{es}(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$ is the system error probability bound of the $(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoder, and can be further bounded by (\ref{SRSBound}). $\QED$
\end{theorem}
\begin{proof}
Because a $(\mathcal{D}, \mathcal{R}_{\mathcal{D}})$-decoder can always choose to report a collision even if it can decode the messages, its system error probability can be improved by shrinking the operation region $\mathcal{R}_{\mathcal{D}}$. This implies that the receiver of the random access system should partition its operation region $\mathcal{R}$ into $\mathcal{R}_{\mathcal{D}}$ regions that do not overlap with each other. In other words, replacing (\ref{RegionAssignment}) by (\ref{RegionPartition}) will improve the system error performance. The rest of the proof is implied by Lemma \ref{Lemma1}.
\end{proof}
Note that the system error probability bound provided in Theorem \ref{Theorem2} is implicit since the optimal partitioning scheme $\sigma$ that maximize the right hand side of (\ref{RegionPartitioning}) is not specified. To find the optimal partitioning, one essentially needs to compute every single term on the right hand side of (\ref{RegionPartitioning}) and (\ref{BoundLemma1}) for all rate options and all user subsets. Because both $E_{m\mathcal{D}}(\mathcal{S}, \mbox{\boldmath $r$}, \tilde{\mbox{\boldmath $r$}})$ and $E_{i\mathcal{D}}(\mathcal{S}, \mbox{\boldmath $r$}, \mbox{\boldmath $r$}')$ defined in (\ref{EmEiDDcoder}) involve the combinations of two user subsets and two rate vectors, the computational complexity of finding the optimal partitioning scheme is $O\left((2M)^{2K}\right)$.
\section{Conclusions}
\label{SectionConslusions}
We investigated the error performance of the random multiple access system over a compound discrete-time memoryless channel. An achievable bound on the system error probability was derived under the non-asymptotic assumption of a finite codeword length. We showed that the results can be extended to the random multiple access system where the receiver is only interested in decoding messages for a user subset.
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"redpajama_set_name": "RedPajamaArXiv"
} | 2,818 |
Q: find qwidget object text by using pywinauto I work as a test engineer. I have to test an application(softphone) which is done by using QWidget. I'm using python - pywinauto. I can click buttons and make calls. There is a qwidget object named with statusLabel. At the beginning of the test, "Ready" text written on it. When I make a call, this text is changed like "Calling..", "Call Established" and so on. I want to check the text of that widget. Do you have any idea?
A: Pywinauto uses standard windows API calls. Unfortunately many UI libraries (like Swing/QT/GTK) do not respond in a typical way to the API calls used - so unfortunately pywinauto often cannot get the control information. (P.s. I am the Author of pywinauto).
| {
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} | 4,001 |
\section{Introduction}
Since the discovery of the Kerr metric which describes rotating black holes in a vacuum,
its geometry has been investigated from the viewpoint of classifying spacetimes
to understand what are the most fundamental properties of the Kerr spacetime.
A number of studies for the purpose have been conducted
in various systematical frameworks (e.g., see \cite{Stphani:2003}),
and we have obtained a common understanding as to the Kerr spacetime to date:
stationary, axially symmetric, asymptotically flat,
Petrov type D vacuum solution of the vanishing of the Simon tensor,
admitting a rank-2 Killing-St\"ackel (KS) tensor of Segre type $[(11)(11)]$
constructed from a (nondegenerate) rank-2 Killing-Yano (KY) tensor.
The Wahlquist metric \cite{Wahlquist:1968,Kramer:1985,Senovilla:1987,Wahlquist:1992,Mars:2001}
investigated in this paper
was found in the study of stationary, axially symmetric perfect fluid spacetimes.
As we succeeded in obtaining interior solutions of static, spherically symmetric stars
joined to the Schwarzschild vacuum spacetime,
it has been thought that the interior of rotating bodies can be described
by stationary, axially symmetric perfect fluid solutions.
Although it was shown that the Wahlquist metric cannot be smoothly matched
to an asymptotically flat vacuum spacetime \cite{Bradley:1999},
the metric is still of great interest because
it allows some interesting geometric characterizations.
It was pointed out \cite{Kramer:1985} that
the Wahlquist metric is the general solution of
stationary, rigidly rotating perfect fluids
with the vanishing Simon tensor.
It was also demonstrated \cite{Senovilla:1987} that the metric
is the general solution of Petrov type D, stationary, axially symmetric
and rigidly rotating perfect fluids with $\rho+3p=\text{const}$.
Furthermore, the Wahlquist metric is known to be included in the class of metrics
admitting a rank-2 KS tensor of Segre type $[(11)(11)]$
which possesses two double nonconstant eigenvalues \cite{HM1,HM2}.
The Wahlquist spacetime inherits some geometric properties of the Kerr spacetime.
This seems to be reasonable because
the Kerr metric is obtained as the limiting case of the Wahlquist metric \cite{Wahlquist:1968,Mars:2001}.
In the Kerr spacetime, two Killing vectors and KS tensor
are constructed from a single rank-2 KY tensor.
It is also shown that the Kerr metric is the only asymptotically flat vacuum solution
admitting a rank-2 KY tensor \cite{Dietz:1982}.
This implies that the presence of the KY tensor is essential to characterize the Kerr spacetime.
Nevertheless, Killing-Yano symmetry of the Wahlquist spacetime has never been investigated.
In this paper, therefore, we first ask whether the Wahlquist metric admits Killing-Yano symmetry.
Actually, as we will see in Sec.~II,
we find a rank-2 generalized closed conformal Killing-Yano (GCCKY) tensor with torsion \cite{KKY}
for the Wahlquist metric.
In recent years, it has been unveiled that Killing-Yano symmetry plays an important role
in higher-dimensional rotating black hole spacetimes.
A family of vacuum solutions describing rotating black holes with spherical horizon topology
\cite{Myers:1986,Hawking:1999,Gibbons:2004,Chen:2006}
admits a rank-2 closed conformal Killing-Yano (CCKY) tensor \cite{Houri:2007,Krtous:2008,Houri:2009}.
Recently, local metrics admitting a rank-2 GCCKY tensor were classified
into three types (called type A, B and C) in arbitrary dimensions \cite{HKWY}.
Some supergravity black hole solutions in higher dimensions are included
in type A of the classification \cite{KKY,HKWY:2010}.
In this paper, by employing the classification,
we attempt to obtain a new family of rotating perfect fluid solutions
which generalizes the Wahlquist solution to higher dimensions.
On the other hand, there have been static, spherically symmetric perfect fluid solutions
in higher dimensions \cite{Krori:1988,Shen:1989,Leon:2002,Zarro:2009}.
The solutions obtained in this paper cover, in the static limit,
the static, spherically symmetric perfect fluid solutions with $\rho+3p=\text{const}$.
The paper is organized as follows.
In Sec.~II, after we briefly review the Wahlquist solution,
we demonstrate that the Wahlquist metric admits a rank-2 GCCKY tensor with a skew-symmetric torsion.
With respect to such a tensor,
in Sec.~III, we generalize the Wahlquist solution in four dimensions
to higher-dimensional ones by solving the Einstein equations for perfect fluids
in all even and odd dimensions, respectively.
We see that the equations of state for the higher-dimensional Wahlquist solutions
are given by $\rho+3p=\text{const}$ in all dimensions.
In Sec.~IV, we investigate the separability of the Hamilton-Jacobi for geodesics,
Klein-Gordon and Dirac equations in all dimensions.
Finally, Sec.~V is devoted to summary and discussion.
After we review the geometry of the four-dimensional Wahlquist spacetime in Appendix A,
we discuss the five-dimensional Wahlquist metric in Appendix B.
In Appendix C, we have collected the curvature quantities of the higher-dimensional Wahlquist metrics.
As a result, it is shown that the metrics are of type D in the higher-dimensional Petrov classification \cite{Coley:2004}.
\section{Killing-Yano Symmetry of the Wahlquist Spacetime}
To investigate Killing-Yano symmetry of the Wahlquist metric in four dimensions
\cite{Wahlquist:1968,Kramer:1985,Senovilla:1987,Wahlquist:1992,Mars:2001},
we begin with the metric form that appeared in \cite{Mars:2001},
which is written in a local coordinate system ($z,w,\tau,\sigma$) as
\beqa
ds^2 &=& (v_1+v_2)\left( \frac{dz^2}{U}+\frac{dw^2}{V} \right) \label{Wahlquist} \\
& & +\frac{U}{v_1+v_2}(d\tau+v_2 d\sigma)^2-\frac{V}{v_1+v_2}(d\tau-v_1 d\sigma)^2 \,, \nonumber
\eeqa
where
\beqa
U &=& Q_0 + a_1 \frac{\sinh(2\beta z)}{2\beta} - \nu_0\frac{\cosh(2\beta z)-1}{2\beta^2} \nonumber\\
&& - \frac{\mu_0}{\beta^2}\left[\frac{\cosh(2\beta z)-1}{2\beta^2}-\frac{z\sinh(2\beta z)}{2\beta}\right] \,, \nonumber\\
V &=& Q_0 + a_2 \frac{\sin(2\beta w)}{2\beta} + \nu_0\frac{1-\cos(2\beta w)}{2\beta^2} \\
&& + \frac{\mu_0}{\beta^2}\left[\frac{1-\cos(2\beta w)}{2\beta^2}-\frac{w\sin(2\beta w)}{2\beta}\right] \,, \nonumber
\eeqa
and
\begin{equation}
v_1 = \frac{\cosh(2 \beta z)-1}{2 \beta^2} \,, \qquad
v_2 = \frac{1-\cos(2 \beta w)}{2 \beta^2} \,.
\end{equation}
The metric contains six real constants $Q_0$, $a_1$, $a_2$, $\nu_0$, $\mu_0$ and $\beta$.
Since one of them can be eliminated by coordinate transformation,
only five of the constants are independent.
As was shown in \cite{Wahlquist:1968,Mars:2001}, one can take the limit $\beta\to 0$,
in which the metric reduces to the Kerr-NUT-(A)dS metric \cite{Carter:1968}
(see Appendix A for details).
The Wahlquist metric provides the stress-energy tensor for perfect fluids
of the energy density $\rho$, pressure $p$ and 4-velocity ${\bm u}$ with $u_\mu u^\mu=-1$,
which is written as
\beq
T^{\mu\nu} = (\rho+p)u^\mu u^\nu + p g^{\mu\nu} \,.
\eeq
The 4-velocity is given by
\beq
u^\mu \frac{\partial}{\partial x^\mu} = \frac{1}{\sqrt{-g_{\tau\tau}}}\frac{\partial}{\partial \tau} \,,
\eeq
where $g_{\tau\tau}=(U-V)/(v_1+v_2)$.
When we consider stationary, axially symmetric spacetimes,
we have two Killing vector fields ${\bm \partial}_t$ and ${\bm \partial}_\phi$.
If ${\bm u}$ lies on the 2-plane spanned by the two Killing vector fields,
then ${\bm u}$ can be written as ${\bm u} = N({\bm \partial}_t+\Omega{\bm \partial}_\phi)$
where $N$ and $\Omega$ are functions in general.
In particular, when $\Omega$ is constant,
the perfect fluid is said to be rigidly rotating.
Namely, the Wahlquist solution represents rigidly rotating perfect fluids.
The energy density and pressure are given by
\beq
\rho = -\mu_0 - 3\beta^2 g_{\tau\tau} \,, \qquad
p = \mu_0+\beta^2g_{\tau\tau} \,.
\eeq
Thus, the equation of state is $\rho+3p=2\mu_0$.
Since we have $\rho+p=0$ and $p=\mu_0$ in the limit $\beta \to 0$,
the constant $\mu_0$ is the cosmological constant.
\subsection{Generalized Killing-Yano symmetry}
It is known that the Kerr metric admits separation of variables in the Hamilton-Jacobi for geodesics,
Klein-Gordon and Dirac equations.
The separability is due to the presence of a rank-2 KY tensor.
In four dimensions, the Hodge dual of the KY tensor is a
rank-2 CCKY tensor ${\bm h}$ \cite{Tachibana:1969} satisfying
\beq
\nabla_ah_{bc} = g_{ab}\xi_c - g_{ac}\xi_b \,, \label{KY}
\eeq
where $\nabla$ is the Levi-Civit\`a connection.
From (\ref{KY}), the associated vector ${\bm \xi}$ is obtained as
\beq
\xi_a = \frac{1}{3} \nabla^bh_{ba} \,.
\eeq
Namely, the Kerr metric admits a rank-2 CCKY tensor.
The Wahlquist metric partially shares the separability of the Kerr metric:
the Hamilton-Jaocbi for geodesics and Klein-Gordon equations separate,
but the Dirac equation does not.
Since the Kerr metric is obtained as a particular limit of the Wahlquist metric \cite{Wahlquist:1968,Mars:2001},
it is natural to ask the Wahlquist metric to admit a rank-2 CCKY tensor.
However, it is shown that such a tensor does not exist in the Wahlquist spacetime.
Instead, we find a rank-2 GCCKY tensor ${\bm h}$ \cite{KKY},
with a skew-symmetric torsion ${\bm T}$ satisfying
\beq
\nabla^T_ah_{bc} = g_{ab}\xi_c - g_{ac}\xi_b \,, \label{GCCKYeq}
\eeq
where $\nabla^T$ is the connection with the skew-symmetric torsion defined by
\beq
\nabla^T_ah_{bc} = \nabla_ah_{bc} + \frac{1}{2} T_{ab}{}^dh_{cd} - \frac{1}{2} T_{ac}{}^d h_{bd} \,.
\eeq
The associated vector ${\bm \xi}$ is given by
\beq
\xi_a = \frac{1}{3}\nabla^T{}^bh_{ba} \,.
\eeq
If a rank-2 GCCKY tensor is obtained, we may expect
that a modified Dirac equation with $1/3$ torsion separates \cite{HKWY2}.
In fact, the modified Dirac equation of the Wahlquist metric does.
Thus, the GCCKY tensor underpins the separability
on the Hamilton-Jacobi for geodesics, Klein-Gordon and modified Dirac
equations of the Wahlquist metric (see Sec.~IV for details).
Going through the following steps,
we demonstrate that the Wahlquist metric (\ref{Wahlquist}) admits
a rank-2 GCCKY tensor.
To see it, we first introduce the coordinates $x$ and $y$ defined by
\begin{equation}
x^2 = v_1 \,, \qquad y^2 = v_2 \,,
\end{equation}
and hence
\begin{equation}
dz^2 = \frac{dx^2}{\beta^2 x^2+1} \,, \qquad
dw^2 = \frac{dy^2}{1-\beta^2 y^2} \,.
\end{equation}
The metric is then written as
\beqa
ds^2 &=& \frac{x^2+y^2}{U(1+\beta^2 x^2)}dx^2+\frac{x^2+y^2}{V(1-\beta^2 y^2)}dy^2 \label{Wah} \\
& & +\frac{U}{x^2+y^2}(d\tau+y^2 d\sigma)^2-\frac{V}{x^2+y^2}(d\tau-x^2 d\sigma)^2 \nonumber
\eeqa
with the functions
\beqa
U &=& Q_0 + a_1 x\sqrt{1+\beta^2x^2} - \nu_0 x^2 \nonumber\\
&& - \frac{\mu_0}{\beta^2}\left[x^2-\frac{x \text{Arcsinh}(\beta x) \sqrt{1+\beta^2x^2}}{\beta}\right] \,, \nonumber\\
V &=& Q_0 + a_2 y\sqrt{1-\beta^2y^2} + \nu_0 y^2 \\
&& + \frac{\mu_0}{\beta^2}\left[y^2-\frac{y \text{Arcsin}(\beta y)\sqrt{1-\beta^2y^2}}{\beta}\right] \,. \nonumber
\eeqa
Furthermore, taking the Wick rotation $y \rightarrow \sqrt{-1}y$
(with $a_2\rightarrow -\sqrt{-1}a_2$ to keep the metric function $V$ real)
and changing the sign $\sigma \rightarrow -\sigma$,
we obtain the Euclidean expression, in which the metric takes a symmetric form with respect to the coordinates $(x,y)$ as
\beqa
ds^2_E &=& \frac{f_1(x^2-y^2)}{\Xi_1}dx^2 + \frac{f_2(y^2-x^2)}{\Xi_2}dy^2 \label{Euclidean_metric} \\
& & + \frac{\Xi_1}{x^2-y^2}(d\tau+y^2 d\sigma)^2 + \frac{\Xi_2}{y^2-x^2}(d\tau+x^2 d\sigma)^2 \,, \nonumber
\eeqa
where
\beqa
f_1 &=& \frac{1}{\sqrt{1+\beta^2 x^2}} \,, \quad
f_2 = \frac{1}{\sqrt{1+\beta^2 y^2}} \,, \nonumber\\
\Xi_1 &=& Q_0 + a_1 x\sqrt{1+\beta^2x^2} - \nu_0 x^2 \nonumber\\
& & - \frac{\mu_0}{\beta^2}\left[x^2-\frac{x \text{Arcsinh}(\beta x) \sqrt{1+\beta^2x^2}}{\beta}\right] \,, \label{metfunc3} \\
\Xi_2 &=& Q_0 + a_2 y\sqrt{1+\beta^2y^2} - \nu_0 y^2 \nonumber\\
& & - \frac{\mu_0}{\beta^2}\left[y^2-\frac{y \text{Arcsinh}(\beta y)\sqrt{1+\beta^2y^2}}{\beta}\right] \,. \nonumber
\eeqa
The form of the metric (\ref{Euclidean_metric})
precisely fits into type A of the classification in \cite{HKWY};
that is, the Wahlquist spacetime admits a rank-2 GCCKY tensor.
In fact, if we introduce an orthonormal frame
\begin{eqnarray}\label{orthvector}
&& {\bm e}^1 = f_1 \sqrt{\frac{x^2-y^2}{\Xi_1}}dx \,, \qquad
{\bm e}^2 = f_2 \sqrt{\frac{y^2-x^2}{\Xi_2}}dy \,, \nonumber\\
&& {\bm e}^{\hat{1}} = \sqrt{\frac{\Xi_1}{x^2-y^2}}(d\tau+y^2 d\sigma) \,, \\
&& {\bm e}^{\hat{2}} = \sqrt{\frac{\Xi_2}{y^2-x^2}}(d\tau+x^2 d\sigma) \,, \nonumber
\end{eqnarray}
the rank-2 GCCKY tensor is given by
\begin{equation} \label{CKY}
{\bm h} = x \,{\bm e}^1\wedge {\bm e}^{\hat{1}} + y\,{\bm e}^2\wedge {\bm e}^{\hat{2}}
\end{equation}
with the skew-symmetric torsion
\beqa
{\bm T} &=& \frac{2x(f_1-f_2)}{f_1f_2(x^2-y^2)} \sqrt{\frac{\Xi_2}{y^2-x^2}}
\,{\bm e}^1 \wedge {\bm e}^{\hat{1}} \wedge {\bm e}^{\hat{2}} \nonumber\\
&& + \frac{2y(f_2-f_1)}{f_1f_2(y^2-x^2)} \sqrt{\frac{\Xi_1}{x^2-y^2}}
\,{\bm e}^2 \wedge {\bm e}^{\hat{2}} \wedge {\bm e}^{\hat{1}} \,. \label{torsion}
\eeqa
The torsion vanishes when we take the limit $\beta\to 0$.
This suggests that the torsion is related to the perfect fluid,
although the physical meaning of the torsion is unclear.
\section{General Wahlquist metrics in higher dimensions}
We have seen that the Wahlquist metric (\ref{Wahlquist}) admits a rank-2 GCCKY tensor
and its Euclidean form precisely fits into type A of the classification \cite{HKWY}.
Hence, it seems to be reasonable to consider a higher-dimensional generalization of the Wahlquist metric.
In this section, we attempt to solve the Einstein equations for perfect fluids in higher dimensions
by employing, as an {\it ansatz}, type A metrics in \cite{HKWY}.
We slightly change our notation
to deal with higher-dimensional metrics in both even and odd dimensions simultaneously.
We introduce $\varepsilon$
where $\varepsilon=0$ for even and $\varepsilon=1$ for odd dimensions.
The dimension number is denoted by $D=2n+\varepsilon$.
The Latin indices $a,b,\dots$ run from 1 to $D$,
and the Greece indices $\mu,\nu,\dots$ run from 1 to $n$.
The form of type A metrics in $D$ dimensions
which we deal with as an {\it ansatz}
is given by
\beqa
{\bm g}^{(D)}
&=& \sum_{\mu=1}^n \frac{f_\mu^2}{P_\mu}dx_\mu^2
+ \sum_{\mu=1}^n P_\mu \left( \sum_{k=0}^{n-1}A^{(k)}_{\mu} d\psi_k\right)^2 \nonumber\\
& & +\varepsilon S \left(\sum_{k=0}^n A^{(k)} d\psi_k\right)^2 \,, \label{met_all}
\eeqa
where
\beqa
&& P_\mu=\frac{\Xi_\mu}{U_\mu} \,, \quad U_\mu = \prod_{\nu\neq\mu}(x_\mu^2-x_\nu^2) \,, \nonumber\\
&& S=\frac{s_0^2}{A^{(n)}} \,, \quad f_\mu=\frac{1}{\sqrt{1+\beta^2 x_\mu^2}} \,.
\eeqa
The functions $A_\mu^{(k)}$ $(k=0,\dots,n-1)$ and $A^{(k)}$ $(k=0,\dots,n)$
are $k$th order elementary symmetric functions in $\{x_1^2,x_2^2,\dots, x_n^2\}$ defined by
\beqa
&& \sum_{k=0}^{n-1}A_\mu^{(k)} t^k=\prod_{\nu \ne \mu}(1+t x_\nu^2) \,, \nonumber\\
&& \sum_{k=0}^{n}A^{(k)} t^k=\prod_{\nu=1}^n(1+t x_\nu^2) \,.
\eeqa
The metric contains unknown functions $\Xi_\mu(x_\mu)$
depending only on single valuable $x_\mu$, and in odd dimensions a constant $s_0$.
The form of the metric (\ref{met_all}) is not the most general form of type A metrics,
but it is enough to construct a perfect fluid solution for the current purpose.
Of course, it leaves a question whether there exist more general solutions of type A.
\subsection{Tower of generalized Killing-Yano tensors}
If a rank-2 GCCKY tensor is obtained in $D=2n+\varepsilon$ dimensions,
we can construct $[D/2]=n$ conserved quantities for geodesic motion \cite{HKWY:2010}.
In addition,
the complete integrability of the Hamilton-Jacobi equation for geodesics can be guaranteed
if the metric admits a high enough number of commuting Killing vectors \cite{Benenti:1979}.
For the metric (\ref{met_all}), we introduce an orthonormal frame as
\beqa \label{frame}
&& {\bm e}^\mu = \frac{f_\mu}{\sqrt{P_\mu}}dx^{\mu} \,, \quad
{\bm e}^{\hat{\mu}} = \sqrt{P_\mu}\sum_{k=0}^{n-1}A^{(k)}_{\mu} d\psi_k \,, \nonumber\\
&& {\bm e}^0 = \sqrt{S}\sum_{k=0}^n A^{(k)} d\psi_k \,. \label{tetrad_all}
\eeqa
Then, the rank-2 GCCKY tensor is given by
\beq
{\bm h} = \sum_{\mu=1}^n x_\mu \,{\bm e}^\mu\wedge {\bm e}^{\hat{\mu}}
\eeq
with the torsion
\beqa
{\bm T} &=& \sum_{\mu\neq\nu}\frac{2x_\mu\sqrt{P_\nu}(f_\mu-f_\nu)}{f_\mu f_\nu (x_\mu^2-x_\nu^2)}
\,{\bm e}^\mu\wedge {\bm e}^{\hat{\mu}}\wedge {\bm e}^{\hat{\nu}} \nonumber\\
& & + \varepsilon \sum_{\mu=1}^n \frac{2\sqrt{S}}{x_\mu}\left(\lambda-\frac{1}{f_\mu}\right)
\,{\bm e}^\mu\wedge{\bm e}^{\hat{\mu}}\wedge{\bm e}^0 \,, \label{torsion_all}
\eeqa
where $\lambda$ is an arbitrary nonzero function which appears only in odd dimensions.
The ambiguity of $\lambda$ cannot be excluded by the GCCKY equation (\ref{GCCKYeq}),
e.g., see \cite{HKWY}.
It also has nothing to do with Einstein equations.
Even if we impose an Einstein equation, it determines the functions $\Xi_\mu$ and $f_\mu$,
but $\lambda$ is still arbitrary.
From the property that the wedge product of GCCKY tensors is a GCCKY tensor,
${\bm h}^{(j)}={\bm h}\wedge {\bm h}\cdots\wedge {\bm h}$ is a rank-$2j$ GCCKY tensor.
The Hodge dual ${\bm f}^{(j)}=* {\bm h}^{(j)}$ is a rank-$(D-2j)$ generalized KY tensor,
and its square, $K_{ab}^{(j)}=c_j f^{(j)}{}_{ac_1\cdots c_{D-2j-1}}f^{(j)}{}_b{}^{c_1\cdots c_{D-2j-1}}$
becomes a rank-2 KS tensor satisfying
$\nabla_{(a}K_{bc)} = 0$,
where $c_j$ is constant.
For an appropriate choice for $c_j$,
the KS tensors are written in the form
\beqa
{\bm K}^{(j)} &=& \sum_{\mu=1}^n A^{(j)}_\mu ({\bm e}^\mu\otimes{\bm e}^\mu
+ {\bm e}^{\hat{\mu}}\otimes{\bm e}^{\hat{\mu}}) \nonumber\\
&+& \varepsilon A^{(j)}{\bm e}^0\otimes {\bm e}^0 \,, \label{KT_all}
\eeqa
where $j=0,1,\dots, n-1$. In particular, ${\bm K}^{(0)}={\bm g}^{(D)}$.
Thus, contracting with the tangent ${\bm p}=\dot{\gamma}$ to geodesics $\gamma$,
we obtain $n=[D/2]$ conserved quantities $\kappa^{(j)} = K^{(j)}{}^{ab}p_ap_b$ for $j=0,\dots,n-1$,
including the Hamiltonian $\kappa^{(0)}=H$.
In addition, since ${\bm \eta}^{(k)}={\bm \partial}_{\psi_k}$ for $k=0,\dots,n-1+\varepsilon$ are Killing vector fields,
we have $n+\varepsilon$ conserved quantities $\tilde{\kappa}^{(k)}=\eta^{(k)}{}^ap_a$.
The all conserved quantities $\{\kappa^{(j)},\tilde{\kappa}^{(k)}\}$ are in involution.
\subsection{Even dimensions}
In this section, we determine the unknown functions $\Xi_\mu(x_\mu)$
using the Einstein equations for perfect fluids in even dimensions.
The metric {\it ansatz} in $2n$ dimensions is given by
\begin{equation}
{\bm g}^{(2n)} = \sum_{\mu=1}^n \frac{f_\mu^2}{P_\mu}dx_{\mu}^2
+ \sum_{\mu=1}^n P_\mu \left( \sum_{k=0}^{n-1}A^{(k)}_{\mu} d\psi_k \right)^2 \,. \label{met_even}
\end{equation}
For the metric, we calculate the Ricci curvature (see Appendix B for details).
The off-diagonal components of the Ricci curvature are
\beqa
&& R_{\mu \hat{\mu}}=R_{\mu \nu}=R_{\mu \hat{\nu}}=0 \,, \nonumber\\
&& R_{\hat{\mu}\hat{\nu}}=\beta^2(D-2) \sqrt{P_\mu}\sqrt{P_\nu} \,. \label{ST0}
\eeqa
The diagonal components are
\beqa
R_{\mu \mu} &=& I_\mu (P_T) +\beta^2\left[ I_\mu(P_T^{(2)})
+\frac{3}{2}x_\mu \partial_\mu P_T + P_T\right] \,, \nonumber\\
R_{\hat{\mu}\hat{\mu}} &=& R_{\mu \mu} + \beta^2(D-2) P_\mu \,, \label{ST0-2}
\eeqa
where
\beq
P_T = \sum_{\mu=1}^n P_\mu \,, \quad
P_T^{(2)} = \sum_{\mu=1}^n x_\mu^2 P_\mu
\eeq
and $I_\mu$ are differential operators given by
\beq
I_\mu = -\frac{1}{2}\frac{\partial^2}{\partial x_\mu^2}
+\frac{1}{x_\mu^2-x_\nu^2}\left(x_\mu\frac{\partial}{\partial x_\mu}
-x_\nu\frac{\partial}{\partial x_\nu}\right) \,. \label{diffop}
\eeq
It should be emphasized that our metric ansatz is now expressed with a Euclidean signature,
so that we have to consider the Euclideanized Einstein equation for perfect fluids,
\begin{equation}
R_{ab}-\frac{1}{2}R g_{ab}=-(\rho+p)u_a u_b+p g_{ab} \,, \label{EEE}
\end{equation}
where $u^au_a=1$. Eliminating the scalar curvature,
we obtain the Einstein equation in a convenient form
\beq
R_{ab} = -(\rho+p)u_a u_b +\frac{\rho-p}{D-2} g_{ab} \,.
\eeq
Moreover, to solve the equation, we assume that perfect fluids are rigidly rotating; that is,
the velocity ${\bm u}$ is written as ${\bm u}=N{\bm \partial}_{\psi_0}$ where $N$ is the normalization function.
Since we have $N=1/\sqrt{P_T}$ from $u^au_a=1$,
the velocity is given in the canonical frame as
\begin{equation}
{\bm u} = \frac{1}{\sqrt{P_T}} \sum_{\mu=1}^n \sqrt{P_\mu} {\bm e}_{\hat{\mu}} \,. \label{4vel}
\end{equation}
Under the assumption, together with (\ref{ST0}) and (\ref{ST0-2}),
the Einstein equation to solve reduces to
\beqa
\frac{\rho-p}{D-2} &=& R_{11}=R_{22}=\cdots =R_{nn} \,, \label{EEE2} \\
\frac{\rho+p}{D-2} &=& -\beta^2 P_T \,. \label{EEE3}
\eeqa
To solve Eq.~(\ref{EEE2}),
we notice that the $\mu\mu$ components of the Ricci curvature, $R_{\mu\mu}$,
can be written in a simple form.
Calculating $R_{\mu\mu}$ in terms of the functions $\Xi_\mu$
and their derivatives $\Xi_\mu^\prime$ and $\Xi_\mu^{\prime\prime}$, we obtain
\beqa
R_{\mu\mu} &=& -\frac{1}{2x_\mu}\left\{\frac{G_\mu}{U_\mu}
-\sum_{\nu\neq\mu}\frac{2x_\mu}{x_\mu^2-x_\nu^2}\left[\frac{F_\mu}{U_\mu}+\frac{F_\nu}{U_\nu}\right]\right\} \nonumber\\
& &-2\beta^2P_T \,,
\eeqa
where
\beqa
G_\mu &=& x_\mu(1+\beta^2x_\mu^2)\Xi_\mu^{\prime\prime}+\beta^2x_\mu^2\Xi_\mu^\prime -4\beta^2x_\mu\Xi_\mu \,, \\
F_\mu &=& x_\mu(1+\beta^2x_\mu^2)\Xi_\mu^\prime-(1+2\beta^2x_\mu^2)\Xi_\mu \,. \label{eq40}
\eeqa
Noticing that $G_\mu = F_\mu^\prime$ and that
\beq
\frac{\partial F_T}{\partial x_\mu} = \frac{F_\mu^\prime}{U_\mu}
-\sum_{\nu\neq\mu}\frac{2x_\mu}{x_\mu^2-x_\nu^2}\left[\frac{F_\mu}{U_\mu}+\frac{F_\nu}{U_\nu}\right] \,,
\eeq
where
\beq
F_T = \sum_{\rho=1}^n\frac{F_\rho}{U_\rho} \,, \label{FT}
\eeq
we obtain the following expressions for $R_{\mu\mu}$:
\beq
R_{\mu\mu} = -\frac{1}{2x_\mu}\frac{\partial F_T}{\partial x_\mu}-2\beta^2 P_T \,. \label{rmm}
\eeq
Using the expression, $R_{\mu\mu}-R_{\nu\nu}=0$ implies that
\beq
\left[\frac{1}{x_\mu}\frac{\partial}{\partial x_\mu}
-\frac{1}{x_\nu}\frac{\partial}{\partial x_\nu}\right] F_T = 0 \,.
\eeq
This can be solved by $F_T=F_T(\xi)$
where $F_T(\xi)$ is an arbitrary function of $\xi= \sum_{\mu=1}^nx_\mu^2$.
Substituting it into (\ref{FT})
and differentiating by $\partial_{x_1}\partial_{x_2}\cdots\partial_{x_n}$
both sides of the equation multiplied by the factor $\prod_{\mu\neq\nu}(x_\mu^2-x_\nu^2)$,
we arrive at the condition $F_T^{(n)}(\xi) = 0$,
which implies that $F_T(\xi)$ is an $(n-1)$th order polynomial in $\xi$.
Furthermore, going back to (\ref{FT}) again and comparing the coefficients of the equation,
we find that $F_T$ must be a linear function.
Namely, to be consistent with (\ref{FT}),
the function must be chosen as $F_T(\xi)=C_1 \xi + C_2$ where $C_1$ and $C_2$ are constants.
Then, using the identities
\beqa
&& \sum_{\mu=1}^n \frac{x_\mu^{2j}}{U_\mu} = 0 \,, \quad (j=0,\dots, n-2) \\
&& \sum_{\mu=1}^n \frac{x_\mu^{2(n-1)}}{U_\mu} = 1 \,, \quad
\sum_{\mu=1}^n \frac{x_\mu^{2n}}{U_\mu} = \sum_{\mu=1}^n x_\mu^2 \,,
\eeqa
we obtain
\beq
F_\mu = \sum_{k=0}^n c_{2k} x_\mu^{2k} \,, \label{EEE4}
\eeq
where $c_{2k}$ ($k=0,1,\dots, n$) are constants with $C_1=c_{2n}$ and $C_2=c_{2(n-1)}$.
In the end, using (\ref{eq40}) and (\ref{EEE4}), the problem of solving the Einstein equation (\ref{EEE})
has been reduced to that of solving first-order ordinary differential equations for $\Xi_\mu$,
\beq
\Xi_\mu^\prime-\frac{1+2\beta^2x_\mu^2}{x_\mu(1+\beta^2x_\mu^2)}\Xi_\mu
- \frac{\sum_{k=0}^n c_{2k} x_\mu^{2k}}{x_\mu(1+\beta^2x_\mu^2)} = 0 \,.
\eeq
The general solution is
\beq
\Xi_\mu = \sum_{k=0}^n c_{2k} \phi_{2k}(x_\mu)+a_\mu x_\mu \sqrt{1+\beta^2 x^2_\mu} \,, \label{EQ}
\eeq
where $a_\mu$ are integral constants, $\phi_0(x) \equiv -1$
and $\phi_{2k}(x)$ $(k=1,2,\dots)$ are given by
\beq
\phi_{2k}(x) = x\sqrt{1+\beta^2x^2}\int_0^x \frac{t^{2(k-1)}dt}{(1+\beta^2t^2)^{3/2}} \,.
\eeq
Note that, for instance, we have
\beqa
\phi_2(x)
&=& x^2 \,, \nonumber\\
\phi_4(x)
&=& -\frac{x}{\beta^2}\left(x-\frac{\text{Arcsinh}(\beta x)\sqrt{1+\beta^2 x^2}}{\beta}\right) \,, \nonumber\\
\phi_6(x)
&=& \frac{3x}{2\beta^4}\left(x+\frac{\beta^2}{3}x^3-\frac{\text{Arcsinh}(\beta x)\sqrt{1+\beta^2 x^2}}{\beta}\right) \,, \nonumber\\
\phi_8(x)
&=& -\frac{15x}{8\beta^6}\left(x+\frac{\beta^2}{3}x^3-\frac{2\beta^4}{15}x^5\right. \\
& & \left.-\frac{\text{Arcsinh}(\beta x)\sqrt{1+\beta^2 x^2}}{\beta}\right) \,, \quad \dots \,. \nonumber
\eeqa
The solution contains parameters $c_{2k}$ ($k=0,\dots,n$), $a_\mu$ ($\mu=1,\dots,n$) and $\beta$.
In four dimensions, for $\mu=1,2$, we obtain
\begin{eqnarray}
\Xi_\mu
&=& -c_0+c_2 x_\mu^2 +a_\mu x_\mu \sqrt{1+\beta^2 x^2_\mu} \\
& & -\frac{c_4x_\mu}{\beta^2}\left(x_\mu-\frac{\text{Arcsinh}(\beta x_\mu)\sqrt{1+\beta^2 x_\mu^2}}{\beta}\right) \,. \nonumber
\end{eqnarray}
The form coincides with the Wahlquist solution.
In the limit $\beta\to 0$, we have $\phi_{2k}\to x^{2k}/(2k-1)$.
The functions $\Xi_\mu$ take the form
\beq
\Xi_\mu = \sum_{k=0}^n \tilde{c}_{2k} x_\mu^{2k} + a_\mu x_\mu \,,
\eeq
where $\tilde{c}_{2k}=c_{2k}/(2k-1)$.
This is the same form as Kerr-NUT-(A)dS metrics in $2n$ dimensions
found by Chen-L\"u-Pope \cite{Chen:2006}.
Finally, let us comment about the equation of state.
From (\ref{EEE2}), (\ref{EEE3}), (\ref{rmm}) and (\ref{EEE4}),
we have
\beq
\frac{2\rho}{D-2} = -c_{2n}-3\beta^2 P_T \,, \quad
\frac{2 p}{D-2} = c_{2n}+\beta^2 P_T \,.
\eeq
Hence, the equation of state is $\rho+3p=(D-2)c_{2n}$.
\subsection{Odd dimensions}
Let us consider odd dimensions $D=2n+1$.
The metric {\it ansatz} in $2n+1$ dimensions is given by
\beqa
{\bm g}^{(2n+1)}
&=& \sum_{\mu=1}^n \frac{f_\mu^2}{P_\mu}dx_{\mu}^2+\sum_{\mu=1}^n P_\mu \left( \sum_{k=0}^{n-1}A^{(k)}_{\mu} d\psi_k \right)^2 \nonumber\\
& & +S \left(\sum_{k=0}^n A^{(k)} d\psi_k \right)^2 \label{met_odd}
\eeqa
with unknown functions $\Xi_\mu$.
The off-diagonal components of the Ricci curvature are given by
\beqa
&& R_{\mu \nu}=R_{\mu \hat{\nu}}=R_{\mu \hat{\mu}}=R_{\mu 0}=0 \,, \nonumber\\
&& R_{\hat{\mu}\hat{\nu}}=\beta^2(D-2) \sqrt{P_\mu}\sqrt{P_\nu} \,, \label{ST1} \\
&& R_{\hat{\mu}0}=\beta^2(D-2) \sqrt{P_\mu}\sqrt{S} \,. \nonumber
\eeqa
The diagonal components are
\beqa
R_{\mu \mu} &=& I_\mu (\tilde{P}_T) +\beta^2\left[ I_\mu(\tilde{P}_T^{(2)})
+\frac{3}{2}x_\mu \partial_\mu \tilde{P}_T + \tilde{P}_T\right] \nonumber\\
& & -\frac{1}{2x_\mu}\frac{\partial}{\partial x_\mu}
\left(\tilde{P}_T+\beta^2\tilde{P}^{(2)}_T\right) \,, \nonumber\\
R_{\hat{\mu}\hat{\mu}} &=& R_{\mu \mu} + \beta^2(D-2) P_\mu \,, \label{ST0-2odd} \\
R_{00} &=& -\sum_{\mu=1}^n \frac{1}{x_\mu}\frac{\partial}{\partial x_\mu}
\left(\tilde{P}_T+\beta^2\tilde{P}^{(2)}_T\right)+\beta^2\tilde{P}_T \nonumber\\
& & +\beta^2(D-2)S \,, \nonumber
\eeqa
where
\begin{equation}
\tilde{P}_T=\sum_{\mu=1}^n \tilde{P}_\mu=P_T+S,~~~~\tilde{P}_T^{(2)}=\sum_{\mu=1}^n x_\mu^2 \tilde{P}_\mu=P_T^{(2)}
\end{equation}
and
\begin{equation}
\tilde{P_\mu}=\frac{\tilde{\Xi}_\mu}{U_\mu},~~~\tilde{\Xi}_\mu=\Xi_\mu-(-1)^n \frac{s_0^2}{x_\mu^2}.
\end{equation}
We assume that the velocity ${\bm u}$ lies in the plane of the Killing vectors
\begin{equation}
{\bm u}=\frac{1}{\sqrt{\tilde{P}_T}} \left(\sum_{\mu=1}^n \sqrt{P_\mu} {\bm e}_{\hat{\mu}}+\sqrt{S} {\bm e}_0 \right).
\end{equation}
The equation reduces to
\beqa
\frac{\rho-p}{D-2} &=& R_{11} = R_{22} = \cdots = R_{nn} \,, \label{EEE1_odd} \\
\frac{\rho+p}{D-2} &=& -\beta^2\tilde{P}_T \,, \label{EEE2_odd}
\eeqa
and for all $\mu$,
\beq
R_{00} = R_{\mu\mu} + \beta^2(D-2)S \,. \label{EEE3_odd}
\eeq
Similar to even dimensions, we find from the direct calculation that
the $\mu\mu$ and $00$ components of the Ricci curvature can be written in the simple form
\beqa
R_{\mu\mu} &=& -\frac{1}{2x_\mu}\frac{\partial \tilde{F}_T}{\partial x_\mu}-2\beta^2 \tilde{P}_T \,, \label{rmm_odd} \\
R_{00} &=& -\sum_{\mu=1}^n\frac{\tilde{F}_\mu}{x_\mu^2U_\mu}-2\beta^2\tilde{P}_T+\beta^2(D-2)S \,, \label{r00}
\eeqa
where
\beq
\tilde{F}_T = \sum_{\mu=1}^n \frac{\tilde{F}_\mu}{U_\mu}
\eeq
and
\beq
\tilde{F}_\mu = x_\mu(1+\beta^2x_\mu^2)\tilde{\Xi}_\mu^\prime-\beta^2x_\mu^2\tilde{\Xi}_\mu \,. \label{gmm1}
\eeq
As was discussed in even dimensions [cf.\ (\ref{EEE4})],
Eq.\ (\ref{EEE1_odd}) requires that $\tilde{F}_\mu$ take the form
\beq
\tilde{F}_\mu = \sum_{k=0}^{n}c_{2k}x_\mu^{2k} \,. \label{gmm2}
\eeq
Indeed, by virtue of (\ref{rmm_odd}) and (\ref{r00}),
we easily see that (\ref{gmm2}) together with $c_0=0$ solves (\ref{EEE1_odd}) and (\ref{EEE3_odd}).
From the equality of (\ref{gmm1}) and (\ref{gmm2}), we obtain the first-order ordinary differential equations
\beq
\tilde{\Xi}_\mu^\prime-\frac{\beta^2x_\mu}{1+\beta^2x_\mu^2}\tilde{\Xi}_\mu
-\frac{\sum_{k=1}^{n}c_{2k}x_\mu^{2k-1}}{1+\beta^2x_\mu^2} = 0 \,.
\eeq
The general solution is
\begin{equation}
\tilde{\Xi}_\mu = \sum_{k=1}^{n} c_{2k}\tilde{\phi}_{2k}(x_\mu)+a_\mu \sqrt{1+\beta^2 x_\mu^2} \,,
\end{equation}
where $a_\mu$ are integral constants and $\tilde{\phi}_{2k}$ ($k=1,2,\dots$) are given by
\beq
\tilde{\phi}_{2k}(x) = \sqrt{1+\beta^2x^2}\int_0^x \frac{t^{2k-1}dt}{(1+\beta^2t^2)^{3/2}} \,.
\eeq
Note that, for instance, we have
\begin{eqnarray}
\tilde{\phi}_2(x)
&=& -\frac{1}{\beta^2}\left(1-\sqrt{1+\beta^2x^2}\right) \,, \nonumber\\
\tilde{\phi}_4(x)
&=& \frac{2}{\beta^4}\left(1+\frac{\beta^2}{2}x^2-\sqrt{1+\beta^2x^2}\right) \,, \nonumber\\
\tilde{\phi}_6(x)
&=& -\frac{8}{3\beta^6}\left(1+\frac{\beta^2}{2}x^2-\frac{\beta^4}{8}x^4-\sqrt{1+\beta^2x^2}\right) \,, \nonumber\\
\tilde{\phi}_8(x)
&=& \frac{16}{5\beta^8}\left(1+\frac{\beta^2}{2}x^2-\frac{\beta^4}{8}x^4+\frac{\beta^6}{16}x^6\right. \\
& & \left.-\sqrt{1+\beta^2x^2}\right) \,, \quad \dots \,. \nonumber
\end{eqnarray}
Thus, we obtain
\beq
\Xi_\mu = \sum_{k=1}^{n} c_{2k}\tilde{\phi}_{2k}(x_\mu)+a_\mu \sqrt{1+\beta^2 x_\mu^2}
+ \frac{(-1)^ns_0^2}{x_\mu^2} \,. \label{sol_odd}
\eeq
The solution contains parameters
$c_{2k}$ ($k=1,\dots,n$), $a_\mu$ ($\mu=1,\dots,n$), $k$ and $\beta$.
In five dimensions, for $\mu=1,2$, we have
\begin{eqnarray}
\Xi_\mu
&=& -\frac{c_2}{\beta^2}\left(1-\sqrt{1+\beta^2x^2}\right)
+a_\mu\sqrt{1+\beta^2x_\mu^2} + \frac{s_0^2}{x_\mu^2} \nonumber\\
& & +\frac{2c_4}{\beta^4}\left(1+\frac{\beta^2}{2}x^2-\sqrt{1+\beta^2x^2}\right) \,. \label{func_5d}
\end{eqnarray}
In the limit $\beta\to 0$, we have $\phi_{2k}\to x^{2k}/2k$.
The functions $\Xi_\mu$ take the form
\beq
\Xi_\mu = \sum_{k=1}^{n} \tilde{c}_{2k}x_\mu^{2k}+a_\mu + \frac{(-1)^ns_0^2}{x_\mu^2} \,,
\eeq
where $\tilde{c}_{2k}=c_{2k}/2k$.
The form reproduces Kerr-NUT-(A)dS metrics in $2n+1$ dimensions \cite{Chen:2006}.
Since we have
\beq
\frac{2\rho}{D-2} = -c_{2n}-3\beta^2 \tilde{P}_T \,, \quad
\frac{2 p}{D-2} = c_{2n}+\beta^2 \tilde{P}_T \,,
\eeq
the equation of state is $\rho+3p=(D-2)c_{2n}$ like the even dimensional case.
\section{Separability}
We investigate the separability of the Hamilton-Jacobi for geodesics, Klein-Gordon and Dirac equations
for the higher-dimensional Wahlquist metrics (\ref{met_all}),
where we do not specify the functions $\Xi_\mu$ to deal with more general cases.
If we choose a particular form of the functions as (\ref{EQ}) in even dimensions
and (\ref{sol_odd}) in odd dimensions,
the results can be applied to those of the Wahlquist metrics.
In this section, we will see that the Hamilton-Jacobi for geodesics and Klein-Gordon equations
can be solved by separation of variables,
but we will not see the separation of variables in the Dirac equation in any dimension.
It is also shown that in even dimensions,
a modified Dirac equation with $1/3$ torsion \cite{HKWY2}
can be solved by separation of variables,
while it can not in odd dimensions.
\subsection{Separation of variables in the Hamilton-Jacobi equation for geodesics}
The separation of variables in Hamilton-Jacobi equations for geodesics occurs
if and only if the metric admits the separability structure established in \cite{Benenti:1979},
in which the corresponding Killing tensors can be written in the St\"ackel form
\beq
K^{\mu \mu}_{(j)} = \bar{\phi}^{\mu}_{(j)} \,, \quad
K^{k\ell}_{(j)} = \sum_{\mu=1}^n \zeta^{k\ell}_{(\mu)}\bar{\phi}^{\mu}_{(j)} \,,
\eeq
where $\bar{\phi}^{\mu}_{(j)}$ is the inverse matrix of the St\"ackel matrix
and $\zeta^{k \ell}_{(\mu)}$ are functions depending only on one variable $x_\mu$.
The Killing tensors ${\bm K}^{(j)}$ $(j=0,1, \dots, n-1)$ obtained in (\ref{KT_all}) are written
in the coordinate basis as
\begin{eqnarray}
{\bm K}^{(j)}
&=& \sum_{\mu=1}^{n}K_{(j)}^{\mu \mu}
\,{\bm \partial}_{x_\mu} \otimes {\bm \partial}_{x_\mu}
+ \sum_{k,\ell=0}^{n-1+\varepsilon}K_{(j)}^{k\ell}
\,{\bm \partial}_{\psi_k} \otimes {\bm \partial}_{\psi_\ell} \nonumber\\
&=& \sum_{\mu=1}^n \frac{A_{\mu}^{(j)}}{U_\mu}\Bigg[ \frac{\Xi_\mu}{f_\mu^2}
\,{\bm \partial}_{x_\mu} \otimes {\bm \partial}_{x_\mu}
+ \frac{\varepsilon (-1)^{n+1}}{s_0^2 x_\mu^2}
\,{\bm \partial}_{\psi_n} \otimes {\bm \partial}_{\psi_n} \nonumber\\
& & +\sum_{k,\ell=0}^{n-1+\varepsilon}\frac{(-1)^{k+\ell}x_\mu^{2(2n-2-k-\ell)}}{\Xi_\mu}
\,{\bm \partial}_{\psi_k} \otimes {\bm \partial}_{\psi_\ell} \Bigg] \,. \nonumber
\end{eqnarray}
The St\"ackel matrix and the functions $\zeta^{k \ell}_{(\mu)}$ are given by
\beqa
\phi^{(j)}_{\mu}
&=& \frac{(-1)^j x_\mu^{2(n-j-1)} f_\mu^2}{\Xi_\mu} \,, \label{ST} \\
\zeta^{k \ell}_{(\mu)}
&=& \frac{(-1)^{k+\ell}f_{\mu}^2 x_\mu^{2(2n-2-k-\ell)}}{\Xi_\mu^2}
+ \frac{\varepsilon(-1)^{n+1}f_\mu^2}{s_0^2 x_\mu^2 \Xi_\mu}\delta_{k \ell}\delta_{k n} \,. \nonumber
\eeqa
In practice, the Hamiltonian-Jacobi equation for geodesics,
\begin{equation}
\frac{\partial S}{\partial \lambda}+g^{ab} \frac{\partial S}{\partial x^a}\frac{\partial S}{\partial x^b}=0 \,,
\end{equation}
allows an additive separation of variables
\begin{equation}
S=-\kappa_0 \lambda+\sum_{\mu=1}^n S_\mu(x_\mu)+\sum_{k=0}^{n-1+\varepsilon}n_k \psi_k,
\end{equation}
where $\kappa_0$ and $n_k$ are constants.
The functions $S_\mu$~($\mu=1,\dots,n)$ are given by
\begin{equation}
S_\mu(x_\mu)=\int \left( \sum_{j=0}^{n-1} \phi^{(j)}_\mu \kappa_i-
\sum_{k,\ell=0}^{n-1+\varepsilon} \zeta^{k\ell}_{(\mu)} n_k n_\ell \right)^{1/2} dx_\mu .
\end{equation}
In the limit $\beta\to 0$, this recovers the result for the Kerr-NUT-(A)dS metrics \cite{Frolov:2008,Yasui:2011}.
\subsection{Separation of variables in the Klein-Gordon equation}
The massive scalar field $\Phi$ is described by the Klein-Gordon equation
\begin{equation}
\frac{1}{\sqrt{g}}\frac{\partial}{\partial x^a}\sqrt{g}g^{ab} \frac{\partial \Phi}{\partial x^b}=m^2 \Phi
\end{equation}
This equation allows a multiplicative separation of variables
\begin{equation}
\Phi=\prod_{\mu=1}^n R_\mu(x_\mu) \prod_{k=0}^{n-1+\varepsilon}e^{in_k \psi_k},
\end{equation}
where the functions $R_{\mu} (x_\mu)$ $(\mu=1,\dots,n)$ satisfy the ordinary second order differential equations
\beqa
&&R_\mu^{\prime\prime}+
\left(\frac{\Xi_\mu^{'}}{\Xi_\mu}-\frac{f_\mu^{'}}{f_\mu}+\frac{\varepsilon}{x_\mu} \right)
R_\mu^\prime \nonumber\\
&&+\left(\sum_{j=0}^{n-1} \phi^{(j)}_\mu \kappa_j-\sum_{k,\ell=0}^{n-1+\varepsilon} \zeta^{k\ell}_{(\mu)} n_k n_\ell \right)R_\mu=0
\eeqa
with $\kappa_0=-m^2$.
In the limit $\beta\to 0$, this recovers the result for the Kerr-NUT-(A)dS metrics \cite{Frolov:2008,Yasui:2011}.
\subsection{Separation of variables in the Dirac equation}
The existence of a GCCKY 2-form does not imply the separability of the Dirac equation.
However, we may expect the separability of a modified Dirac equation
which appears in the spacetimes admitting the GCCKY,
\begin{equation}\label{Dirac}
(\gamma^a D^{T/3}_a+m) \Psi=0,
\end{equation}
where the Dirac operator has a 1/3 torsion, $T/3$, of the GCCKY 2-form,
\begin{equation}
D^{T/3}_a
= {\bm e}_a+\frac{1}{4} \gamma^b \gamma^c {\bm \omega}_{bc}({\bm e}_a)-\frac{1}{24}\gamma^b \gamma^c T_{abc} \,.
\end{equation}
The frame vector fields dual to (\ref{tetrad_all}) are given by
\beqa
&& {\bm e}_\mu=\frac{\sqrt{P_\mu}}{f_\mu}\frac{\partial}{\partial x_\mu} \,, \quad
{\bm e}_0=\frac{\sqrt{S}}{s_0^2}\frac{\partial}{\partial \psi_n} \,, \nonumber\\
&& {\bm e}_{\hat{\mu}}=\sum_{k=0}^{n-1+\varepsilon}\frac{(-1)^k x_\mu^{2(n-1+k)}}{\sqrt{P_\mu} U_\mu}
\frac{\partial}{\partial \psi_k} \,. \label{ortho_all_vec}
\eeqa
By using (\ref{connections}) and (\ref{torsion_all}),
the Dirac operator $\gamma^a D^{T/3}_a$ is explicitly written as
\beqa
&& \sum_{\mu=1}^n \frac{\sqrt{P_\mu}}{f_\mu}\Bigg[
\gamma^\mu \left(\frac{\partial}{\partial x_\mu}+\frac{\Xi^\prime_\mu}{4 \Xi_\mu}+\frac{\varepsilon}{2 x_\mu}
+\frac{1}{2}\sum_{\nu \ne \mu}\frac{x_\mu}{x^2_\mu-x^2_\nu} \right) \nonumber\\
&& + \gamma^{\hat{\mu}} \left( \sum_{k=0}^{n-1+\varepsilon}
\frac{(-1)^k x_\mu^{2(n-1+k)}f_\mu}{\Xi_\mu}\frac{\partial}{\partial \psi_k}
+\sum_{\nu \ne \mu}\frac{x_\nu \gamma^\nu \gamma^{\hat{\nu}}}{2(x^2_\mu-x^2_\nu)} \right) \Bigg] \nonumber\\
&& + \varepsilon \gamma^0 \sqrt{S} \left[\frac{1}{s_0^2}\frac{\partial}{\partial \psi_n}-\frac{1}{4}\sum_{\mu=1}^n
\left(\frac{\lambda}{x_\mu}+\frac{1}{f_\mu x_\mu}\right)\gamma^\mu \gamma^{\hat{\mu}} \right] \,.
\eeqa
The expression leads to the separation of variables in even dimensions.
The calculation is completely parallel to
that of \cite{Oota}.
We write the $2^n$ components of the spinor field as
$\Psi_{\varepsilon_1\varepsilon_2\cdots\varepsilon_n} (\varepsilon_\mu=\pm 1)$.
Putting the separation solution
\begin{equation}
\Psi_{\varepsilon_1\varepsilon_2\cdots\varepsilon_n}
=\left(\prod_{1 \le \mu < \nu \le n}\frac{1}{\sqrt{x_\mu+\varepsilon_\mu \varepsilon_\nu x_\nu}} \right)
\hat{\Psi}_{\varepsilon_1\varepsilon_2\cdots\varepsilon_n} \,,
\end{equation}
where
\beq
\hat{\Psi}_{\varepsilon_1\varepsilon_2\cdots\varepsilon_n}
= \left(\prod_{\mu=1}^n \chi^{(\mu)}_{\varepsilon_\mu}(x_\mu) \right)
\exp \left(i\sum_{k=0}^{n-1+\varepsilon}n_k \psi_k \right) \,
\eeq
the modified Dirac equation becomes
\beqa
&& \sum_{\mu=1}^n \frac{P^{(\mu)}_{\varepsilon_\mu}}{\prod_{\nu \ne \mu}
(\varepsilon_\mu x_\mu-\varepsilon_\nu x_\nu)}+m \label{sep} \\
&& +\frac{i\varepsilon k}{\prod_{\rho=1}^n \varepsilon_\rho x_\rho}
\left(\frac{n_n}{s_0^2}-\frac{1}{4}\sum_{\mu=1}^n\left( \frac{\lambda}{\varepsilon_\mu x_\mu}+
\frac{1}{\varepsilon_\mu x_\mu f_\mu} \right) \right)=0 \,, \nonumber
\eeqa
where
\beqa \label{sep2}
P^{(\mu)}_{\varepsilon_\mu}
&=& (-1)^{\mu-1}(\varepsilon_\mu)^{n-\mu}\frac{\sqrt{(-1)^{\mu-1}\Xi_\mu}}{f_\mu \chi^{(\mu)}_{\varepsilon_\mu}} \nonumber\\
& & \times \left(\frac{d}{dx_\mu}+\frac{\Xi^{'}_\mu}{4\Xi_\mu}+\frac{\varepsilon}{2 x_\mu}+\varepsilon_\mu Y_\mu \right)
\chi^{(\mu)}_{-\varepsilon_\mu} \,, \\
Y_\mu
&=& \sum_{k=0}^{n-1+\varepsilon}\frac{(-1)^k f_\mu x_\mu^{2(n-1+k)}}{\Xi_\mu}n_k \,.
\eeqa
Note that $P^{(\mu)}_{\varepsilon_\mu}$ depends only on the one variable $x_\mu$.
In even dimensions $(\varepsilon=0)$, Eq.\ (\ref{sep}) reduces to
\beq
\sum_{\mu=1}^n \frac{P^{(\mu)}_{\varepsilon_\mu}}{\prod_{\nu \ne \mu}
(\varepsilon_\mu x_\mu-\varepsilon_\nu x_\nu)}+m = 0 \,.
\eeq
The equation separates when
\begin{equation}\label{sep3}
P^{(\mu)}_{\varepsilon_\mu}=\sum_{j=0}^{n-1}q_j (\varepsilon_\mu x_\mu)^j \,,
\end{equation}
where $q_{j}$ $(j=0, \dots, n-2)$ are arbitrary constants and $q_{n-1}=-m$.
Indeed, combining (\ref{sep2}) with (\ref{sep3}), we have the following coupled ordinary differential equations:
\beqa
&& \left(\frac{d}{dx_\mu}+\frac{\Xi^{'}_\mu}{4\Xi_\mu}
+\frac{\varepsilon}{2 x_\mu}+\varepsilon_\mu Y_\mu \right)\chi^{(\mu)}_{-\varepsilon_\mu} \\
&& +\frac{(-1)^{\mu}(\varepsilon_\mu)^{n-\mu}f_\mu \sum_{j=0}^{n-1}q_j (\varepsilon_\mu x_\mu)^j}
{\sqrt{(-1)^{\mu-1}\Xi_\mu}}\chi^{(\mu)}_{\varepsilon_\mu}=0 \,. \nonumber
\eeqa
In odd dimensions, we cannot separate (\ref{sep})
because of the last terms including the function $\lambda$.
For the Kerr-NUT-(A)dS metrics in odd dimensions, since we have $f_\mu=1$ for all $\mu$,
we are able to take $\lambda=1$ and then Eq.\ (\ref{sep}) can be solved by separation of variables \cite{Oota}.
\section{Summary and Discussion}
In Sec.~II, we have discovered a rank-2 GCCKY tensor with a skew-symmetric torsion
for the Wahlquist metric (\ref{Wahlquist})
which is a stationary, axially symmetric perfect fluid solution of the Einstein equation in four dimensions
with $\rho+3p=\text{const}$.
In Sec.~III, we have obtained stationary, axially symmetric perfect fluid solutions in higher dimensions,
where we have made use of canonical forms of metrics admitting a rank-2 GCCKY tensor and
have directly solved the higher-dimensional Einstein equations in higher dimensions.
The exact solutions obtained generalize the Wahlquist metric in four dimensions to all even dimensions (\ref{met_even})
and odd ones (\ref{met_odd}),
for which the equations of state are always given by $\rho+3p=\text{const}$.
As far as we know, they are the first examples of rotating perfect fluid solutions in higher dimensions.
We could solve the Einstein equations for perfect fluids due to the presence of a rank-2 GCCKY tensor.
In this sense, if we find another solution admitting a rank-2 GCCKY tensor in four dimensions,
it might be possible to generalize it to higher dimensions.
When we solved the Einstein equations, as an {\it ansatz},
we have focused on type A metrics admitting a rank-2 GCCKY tensor,
but it would be of great interest to investigate the other types of metrics (called type B and type C \cite{HKWY}).
Since we have assumed a particular case of type A metrics,
it might be possible to find other perfect fluid solutions of type A even in four dimensions.
Another thing we assumed is that perfect fluids are rigidly rotating,
so it would be worth asking whether the assumption can be relaxed.
In Sec.~IV, we have investigated the separability of the Hamilton-Jacobi for geodesics, Klein-Gordon
and (both standard and modified) Dirac equations
for the obtained higher-dimensional perfect fluid solutions.
In four dimensions, the Wahlquist metric shares the similar separability to the Kerr metric.
The Hamilton-Jacobi for geodesics, Klein-Gordon and modified Dirac equation with a $1/3$ torsion
equations can be solved by separation of variables.
We have seen that
the Hamilton-Jacobi for geodesics and Klein-Gordon equations separate also in higher dimensions,
which is responsible for the rank-2 GCCKY tensor.
Although the Dirac equation does not separate in arbitrary dimensions,
the modified Dirac with $1/3$ torsion equation does only in even dimensions.
In odd dimensions, there is an obstruction.
Even for any choice of the function $\lambda$ in (\ref{torsion_all}),
the modified Dirac equation does not separate.
Since the equations of state are given by $\rho+3p=\text{const}$,
the present situation seems to be unrealistic for compressible fluids.
Even so, it would be interesting to consider whether the obtained solutions describe the interiors of rotating stars
and (not necessarily smoothly) connect to vacuum spacetimes.
For instance, when we take the static limit of the Wahlquist solution,
we obtain static, spherically symmetric perfect fluid solution (\ref{Whit}).
Then, it is possible to match the metric to the Schwarzschild metric,
as was discussed in 4 \cite{Wahlquist:1968,Whittaker:1968}.
In higher dimensions, for static, spherically symmetric perfect fluid solutions with $\rho=\text{const.}$ \cite{Krori:1988,Zarro:2009}
and $p=-\rho$ \cite{Shen:1989}, the similar matching conditions were discussed,
where the metrics are joined to the Schwarzschild-Tangherlini metrics in arbitrary dimensions.
\section*{Acknowledgments}
We would like to thank Hideki Maeda for introducing us to this project.
We would also like to thank David Kubiz\v{n}\'ak for reading the draft carefully and giving us useful comments.
We are grateful to Masashi Kimura, Shunichiro Kinoshita and Masato Nozawa for the helpful discussion,
and also grateful to Gary W.\ Gibbons for reading the draft.
T.H. was supported by Research Center for Measurement in Advanced Science (RCMAS), Rikkyo University.
Y.Y. was supported by the Grant-in-Aid for Scientific Research No. 23540317.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,715 |
Copyright © 2014 Lucasfilm Ltd. & ® or TM. All rights reserved. Published by Disney • Lucasfilm Press, an imprint of Disney Book Group. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without written permission from the publisher. For information address Disney • Lucasfilm Press, 1101 Flower Street, Glendale, California 91201.
ISBN 978-1-4847-1700-4
Visit www.starwars.com
#### Contents
Epigraph
Prologue
**Part 1:** Lothal
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
**Part 2:** To Space and Back
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Chapter 16
**Part 3:** Kessel
Chapter 17
Chapter 18
Chapter 19
Chapter 20
Epilogue
Photo Insert
_To new hopes
and new beginnings_
"You say you are an orphan, without a friend in the world; all the inquiries I have been able to make, confirm the statement. Let me hear your story; where you come from; who brought you up; and how you got into the company in which I found you. Speak the truth, and you shall not be friendless while I live."
—Charles Dickens, _Oliver Twist_
**Two** ghosts haunted the depths of the Outer Rim.
Both were star vessels, converging in space. One was large, nearly corvette-class, and bore a unique design that might be most likened to a manual crossbow. Its wide, circular prow protruded from a rectangular midsection, along which projected an assortment of weaponry, from laser cannons to missile launchers. It would be classified as a gunship, though none of its guns moved as the other vessel crept up to its airlock. It seemed to be nothing more than a derelict drifting in space, its cockpit dark, its hull ruptured by blaster holes, its engines cold, a ghost of its former self.
The docking vessel—a bulky, hexagonal freighter of Corellian origin—was a ghost of another kind. Equipped with baffled engines, energy dampeners, and static jammers, the freighter had been modified to be deliberately hard to detect. Sensors frequently registered it as a solar fluctuation or cosmic radiation rather than a starship.
Its crew called it the _Ghost_.
Kanan Jarrus stood in the _Ghost_ 's docking tube, surrounded by his four crewmates. All eyes, photoreceptors, and blasters were fixed on the gunship's airlock. Everyone was on edge, even him.
They had traveled to this remote part of the Rim to converse with the gunship's captain, on the advice of Cikatro Vizago, a Devaronian crime lord on Lothal with whom both were acquainted. This captain allegedly shared the same anti-Imperial views as the _Ghost_ 's ragtag bunch and was seeking to coordinate efforts in his fight against the Empire. Desiring to take their own struggle beyond the planet Lothal, Kanan and his colleague Hera Syndulla had agreed to meet the captain, so Vizago had arranged it to happen here in deep space.
Unfortunately, given the exterior damage the gunship had taken, it appeared that the Empire—or someone with the appropriate firepower—had arrived there first. This made Kanan and his crewmates wary, since either enemy or ally could be on the other side of the airlock.
"Five, four," Sabine said, counting down the timer of the detonation charges she'd placed on the airlock. In her Mandalorian armor and helmet, the girl was the most protected of all of them—yet also would be the first one noticed in a potential firefight. Swirls of purple, pink, and orange dappled her helmet and plates, a pattern that would work as camouflage only in a graffiti-covered undercity.
"One," she said. The charges blew, right on time.
Kanan held his breath and squinted through the smoke. No one appeared in the airlock. The gunship corridor beyond was dark. "Chopper, scan for any life signs."
The crew's mechanical member, the barrel-shaped astromech droid C1-10P, grumbled as he rolled forward and rotated his battered orange-and-yellow dome. While most astromech units whistled or tweedled to communicate in place of expensive vocabulators, Chopper favored lower tones. He grunted back an all clear.
" _Karabast_ ," cursed Zeb in his native tongue. The brawny Lasat stooped in the docking tube, the tips of his sharp ears touching the ceiling. He pulled at his chin curtain of beard. "I was hopin' for some Imps. I could do with a little head bashing right now."
"Be careful what you wish for, Garazeb Orrelios," Hera said. "One day it might be your head that's bashed. Maybe then someone can knock a little sense into you."
Kanan smirked. Hera always knew how to lighten grave situations. She claimed her sarcasm was a Twi'lek trait, like her ability to communicate solely through the movements of her _lekku_ head-tails. But Kanan had encountered many serious members of her species who had no sense of humor. Her dry wit came from living a life constantly in peril. If you couldn't look death in the face and laugh a little, she often said, then you were probably already dead.
"What do you think?" he asked her.
"I'm as clueless as you," Hera said. "Everyone could be vaped aboard that ship—or it could be a trap. Perhaps you could—"
Kanan shot her down abruptly. "No."
Her green head-tails twitched in annoyance. She could be as irritated as she wanted to be, but he wasn't going to be pressured into doing what she was implying. She didn't comprehend what that might cost, not just to him, but all of them.
"Well, I'm not going to stand here when I could be redecorating the gunship," Sabine said, her helmet filtering none of the sass from her voice.
Kanan held her back from going into the airlock first. "Save your art and just hang on to your blasters," he said. "Hera, Chopper, keep the _Ghost_ warm in case we need a quick getaway. And inform Vizago about what happened to his 'acquaintances.'"
"Already sent the message," Hera said, and turned with the droid. She was disappointed in him, he could tell. But she'd get over it, like she always did.
Gesturing Zeb and Sabine to follow, Kanan strode through the dissipating smoke into the airlock. Since the gunship's emergency lights were out, he activated his glowrod.
What he saw astounded him. The corridor seemed to be constructed not of steel, but wood. And not just any ordinary wood. The wood used here remained in its natural state, without varnish or sanding, as if freshly cut from a tree. Branches and limbs curled out at various angles, forming arches or conduits for hidden wiring. Huge circular knots adorned the walls like gallery portraits. Thick slices of trunk made for bulkheads, while sap served as caulking and glue for the integrated technology. Kanan had never seen such handcrafted beauty.
"I shouldn't have suggested tagging these walls," Sabine said. "This is true art."
"But where are the artists?" Zeb asked, warily sniffing the air.
They turned down the corridor and entered an enormous chamber that housed a lattice of stocky tree limbs. Some branches were bare like wood beams, while others were covered with a clumpy moss. They stretched across the chamber in multiple levels, woven together to form a network so thick Kanan could not see the ceiling.
"Thought I smelled _wroshyr_ wood," Zeb said.
"You've been to the forests of Kashyyyk?" Sabine asked.
"No. But the Wookiees who came to fight the Imperial invasion on Lasan all had weapons whittled from the _wroshyr_ trees of their home."
Kanan stroked his goatee. "Interesting. So Vizago's freedom fighters are Wookiees, and this is—was—their gunship." He recalled some Wookiees he had known, and how they preferred resting in trees rather than bunks. The branches here must serve as the beds of their sleeping cabin.
"Makes sense the Wookiees would arm themselves and fight back. The Empire took over Kashyyyk, like it did Mandalore," Sabine said.
"And Lasan," Zeb snarled.
Kanan put his hand on a limb. The moss crumbled off, dead. No one had slept here in a while, for Wookiees fastidiously maintained their living environments. On the bare limb below, however, he noticed a series of pictographic marks that had been scratched into the bark. "Is this writing?"
Zeb and Sabine went over to look. "Definitely Wookiee," Zeb said. "I saw similar markings on their weapons. Wish I could read it."
"It says 'Kitwarr, son of...Wullffwarro,' or at least I think so," Sabine said.
"You can read Wookiee?" Kanan and Zeb asked at once, both surprised.
"A little. Part of the elementary Mandalorian curriculum, along with Huttese, Aqualish, and a dozen other languages," Sabine said with a shrug. "Can't speak any Wookiee, though. No human can. We don't have their throats."
Zeb traced the markings with a finger. "Wullffwarro. I know that name. He was one of the Republic's greatest soldiers during the Clone Wars."
"I remember his name, too. He was always on the holonews, but when the Empire was declared, he suddenly disappeared," Kanan said, interrupted by what sounded like a cough.
They all quieted and looked about. It _was_ a cough, coming from farther inside the ship. Someone was still alive, probably at the brink of death.
Kanan led the others under the branches toward the source of the sound. They went through an arched portal and came out onto the gunship's bridge.
Unlike the other pristine sections of the ship, the bridge was a disaster area. The benches before the various stations had been yanked from their moorings. All the technical systems were in pieces, shattered and smashed. Red and green blood smeared the cracked viewport and stained the _wroshyr_ wood trim. A ferocious battle had been fought here, yet oddly, only one body lay among the rubble.
It was not a Wookiee. It was the sworn enemy of all Wookiees.
It was a Trandoshan.
"Slaver," Zeb roared.
Kanan held Zeb back from strangling the reptilian humanoid. The Trandoshan's wounds would bring an end to his suffering soon enough. His green flight suit was pitted with palm-sized holes, probably from a Wookiee bowcaster, and he was shedding orange scales faster than he could regenerate them.
Kanan knelt beside him. "Where are the Wookiees?"
One of the Trandoshan's eyes stayed closed, but the other rolled toward Kanan. His serpentine tongue flicked in and out of his mouth as he rasped his final breaths.
Kanan caught the Trandoshan's tongue. "Where?"
The Trandoshan gurgled, unable to speak until Kanan released his tongue. The reptilian spat out a last word. "Empire."
Zeb howled and reached down to pick up the slaver with one hand. "You sold them to the Imps, you savage?"
The Trandoshan couldn't answer, no matter how hard Zeb shook him. He was already dead.
Kanan touched Zeb's shoulder. "Calm yourself."
Zeb steamed, but dropped the slaver's corpse back on the ground. "We need to find them. We need to save the Wookiees."
The rangefinder antenna on Sabine's helmet rotated. "Hera's heard back from Vizago," she said, then paused for a few moments. "He's looking into what happened here. In the meantime, he's got a job for us. An Imperial supply grab that should be simple and will pay handsomely."
Kanan frowned. Nothing was simple when it came to Cikatro Vizago and his jobs. "Do I even need to guess where this pickup is?"
Even with the helmet on, everyone could hear the smile in Sabine's voice. "Where else would it be?"
"Lothal," Zeb growled.
**There was** a saying among traders in the Outer Rim that you hadn't seen the color green until you'd landed in the grasslands of Lothal. Though Ezra Bridger hadn't been to any other planet—he was only fourteen, after all—he doubted he'd ever dispute that saying. The grass on Lothal had the sheen of gold.
And it was everywhere, as far as the eye could see. Barring the occasional mound and mountain range, vast prairies spanned Lothal's continents like great seas of golden green.
Ezra gunned his jump bike through one of these seas, racing across the grasslands beyond Capital City. The speed of the ride thrilled him. His hair flapped in front of his eyes, his heart raced, and the mounds whipped past him like he imagined the starlines of hyperspace would.
He took the freeway for a stretch, then got off at the exit for the spaceport. Transports, passenger shuttles, and freighters of all kinds landed on and launched from the berthing bays. One day he, too, would be piloting a ship of his own, hopping from planet to planet, never staying anywhere too long, doing all his trading by himself, a lone operator in the great galaxy.
After parking the bike in a distant lot, he took a speeder bus to the spaceport proper. The place was bustling and getting busier by the day. Humans and nonhumans of a dozen species that Ezra couldn't name hurried between terminals that offered flights to such glamorous worlds as Corellia, Ryloth, and Imperial Center, where one huge metropolis covered the entire planet. Holographic billboards promoted even more exotic destinations and advertised everything from the latest gardening droids to the local Imperial Academy. The ever-increasing activity was due in no small part to the also ever-increasing military presence of the Empire on Lothal. Flat-winged TIE fighters patrolled the skies while white-armored stormtroopers patrolled the spaceport and city. The added security dulled Lothal's reputation as a frontier world and lured wealthy Imperial citizens to come and relocate their businesses, start new ventures, or even take a holiday.
It also made Ezra especially vigilant. Being caught for crimes as minor as petty theft could get him locked up in an Imperial prison where he could labor for years unpaid as a slave.
This meant Ezra had to keep a low profile in the more secure areas, like the spaceport. With his backpack strapped around his shoulders, he pretended to be just another traveler as he roamed the crowds, fishing for the perfect mark to pickpocket. Desperate as he was for credits, he prided himself on stealing only from those who appeared more fortunate than he was. On Lothal, the fortunate were usually Imperials. Making off with a comlink or an officer's rank pin could earn him enough at a pawn shop to pay for a week's meals.
"Ezra Bridger?" piped a voice behind him. "By the Z'gag, it is you!"
Three tiny hands caught a strap dangling from his backpack and pulled Ezra out of the crowd. A larva-like Ruurian with sixteen limbs, stunted wings, and large multifaceted eyes wormed up before him.
Ezra frowned. "Hello, Slyyth."
The Ruurian clacked his mandibles, his species' version of a smile. "What a coincidence! I was just thinking how useful your talents would be for a job I have in mind."
"You know I don't work for anyone."
"Ezra, Ezra," Slyyth said, brushing his feathery antennae against Ezra's cheek. "Just let me tell you the details. It's only a snatch-and-run."
Ezra stepped back from the antennae in disgust. "Last time you said that, I nearly got caught in an Imperial sting operation. No thanks."
"But the credits we could make—"
Slyyth erupted into spasms of violent coughs. Woolly hair began to fall off his tubular body, which was blemished by yellow splotches Ezra hadn't seen before.
"Are you ill?"
"Worse," Slyyth said. "If I can't afford another treatment, I might not be slithering on the ground for much longer."
Ezra didn't know much about Ruurians, except that at a certain advanced age they spun a cocoon around themselves, shed their larval forms, and became beautiful chroma-wings. They considered this transformation to be Paradise—a phase during which one's sole desires were eating, mating, and flying along the banks of Ruuria's pink rivers. Slyyth, however, found the idea terrifying. He wouldn't be able to fence stolen goods and make a profit if all he wanted to do was flutter over a riverbank. So for years, he had denied himself from entering the final stage of his biology by using various ointments, taking prescriptions, and getting special treatments.
"Whatever you're taking is making you sick," Ezra said.
"Sick is better than gone," Slyyth said.
"You won't be gone. You'll be free to do what all your species do, and you won't have to worry about making ends meet any longer," Ezra said. The more he thought about it, the more becoming a chroma-wing really didn't sound like that bad of a proposition.
"But the credits, the credits. I'll never be rich—" The Ruurian broke into another series of coughs.
"Good-bye, Slyyth." Ezra started to walk away.
"It's a shame," Slyyth said, managing to get his coughs under control. "You're an artful little dodger, more talented than any pickpocket I know—including old Slyyth in his prime—but you're as stubborn as a pupa." He sighed and his antennae drooped. "Guess I'll just have to ask Kinsdaw or Tesba if they can snatch the helmets."
Ezra stopped. "Helmets?"
Slyyth let out a couple of weak coughs. He produced a flimsi in one hand. "According to the manifest I acquired, it's part of a shipment for the new bucket-heads at the Academy."
" _Stormtrooper_ helmets?" Now Ezra was intrigued. "Where?"
Slyyth drew to his full height and clacked his mandibles. He wasn't coughing anymore.
Ezra crept through the spaceport's access corridors, staying away from random Imperial patrols. He came to the emergency door to bay 49 and used the astromech arm he kept in his backpack to pick the lock. He slipped inside the bay to find Academy cadets unloading cargo from an Imperial transport like labor droids. They carried off crates of foodstuffs, canisters of cleaning solutions, and bags of refurbished stormtrooper armor while the Supply Master, a turgid fellow named Lyste with whom Ezra had had a few run-ins, screamed at them to move faster.
Slyyth had assured him the Imperials wouldn't be there for another half hour, so he could just walk into the transport, snatch the bags, and hurry out. That obviously wasn't going to be the case. Yet despite the chance of being caught, Ezra didn't leave. His instincts told him to stay. He trusted these inner feelings more than anything or anyone else, since they always nudged him to pick the deepest pockets and guided him through rough and risky circumstances. They compelled Ezra to wait, so he did, ducking behind chemical barrels to see if an opportunity would arise.
The opportunity did when Lyste pulled an unfortunate cadet out of line and berated her. Ezra sprinted out, took her place, and marched into the transport.
"Overslept, forgot my uniform," he whispered to the cadets around him. "Hopefully there'll be one on the transport. Last thing I want is disciplinary action from Lyste."
"Got that right," said a cadet. "If I get one more bad mark on my record, he's threatened to transfer me to the spice mines of Kessel."
Kessel. It was one place Ezra never wanted to visit. Other than death, being sent to its spice mines was as bad a punishment as the Empire could deliver.
Once they were aboard the transport, a cadet tossed Ezra an extra uniform from a crate. Ezra flipped him a credit in thanks and then grabbed the two bags that corresponded with the ones that contained helmets on Slyyth's manifest. Excusing himself to change into the uniform, he veered into the ship's corridor, found a portside hatch, and exited.
He came out near the refueling hose and crouched behind it to slink back to the emergency door. When Lyste's eyes were elsewhere, he opened the door and dashed out, hastening down the access corridor, away from the docking bay.
He dropped off the bag that was Slyyth's cut in a rusty refuse bin and trekked across the spaceport lots to his jump bike. The bag fit in the storage compartment behind the seat, and Ezra was off, zooming down the freeway and into the grasslands toward his hideout.
Soon the old communications tower came into view, rising over the prairie like a lighthouse. Ezra slowed the bike. Once this tower had been linked to a vast grid that coordinated air travel between the planet's scattered settlements, until it was supplanted by a more advanced Imperial network. Now what stood was little more than a rusted relic, stripped of valuable equipment long before by scrap thieves and overgrown by green daisies.
All this made the tower the perfect hideout.
Ezra didn't need access to high-tech communications systems. Contact with the outside world was the last thing he wanted. Located outside the capital city, the tower served as his personal sanctuary from the urban chaos. Here Ezra could be by himself—and _be_ himself. The tower was the one place he could call home.
He parked the bike inside the garage and got off the seat, his legs a little numb from the ride. Bits of technology, most of it of Imperial manufacture, littered the floor like in a mechanic's shop. Ammo-less blasters and stormtrooper rifles dangled on a rack. A shuttle stabilizer had been shoved between cargo containers, while the base of a Treadwell droid, missing its tread, occupied the corner with two broken jump boots. Other components, from power couplings to circuit boards, in various states of disrepair cluttered a workbench.
None of that stuff really mattered to Ezra. His most treasured possessions were his Imperial helmets, neatly arranged on shelves and lockers as if for a museum display. His most recent acquisition, a TIE fighter pilot's helmet, hung front and center atop a jacket rack.
Now he had more to add. He pulled out the bag from the bike's storage compartment, untied it, and turned it over. Instead of helmets, holopads spilled out.
He almost kicked himself. He hadn't had time to check the bags. He'd just assumed what the manifest declared was correct. He sifted through the holopads, finding nothing else, not even blaster packs or biker goggles. He didn't understand it. What would a military supply transport be doing with holopads?
He picked up one of the holopads and thumbed the tiny button. A three-dimensional image of two happy parents and their son was projected from its surface. "You, too, can be part of the Imperial family," announced a pompous voice from the holopad's micro-speaker. "Don't just dream about applying for the Academy; make it come true. You can find a career in space—Exploration, Starfleet, or Merchant Service. Choose from Navigation, Engineering, Space Medicine, Contact/Liaison, and more."
Having heard the same commercial at the spaceport over the years, Ezra knew the words by heart. "If you have the right stuff to take on the universe and standardized examination scores that meet requirements," he recited along, mimicking the adult voice and its upper-crust accent, "dispatch your application to the Academy screening office, care of the Commandant, and join the ranks of the proud!"
Ezra clicked his tongue along with the concluding drumroll, then stopped, realizing that he was parroting Imperial propaganda. He lifted his thumb in sudden revulsion. The picture disappeared and the micro-speaker silenced, though the words stayed in his head. The ad was like a bad tune you couldn't forget, which was the point. The Empire wanted everyone in his generation to think the same, dream the same, be the same, like little Imperial droids.
He tossed the holopad onto his workbench. Ezra wouldn't be one of their droids. He wasn't anyone's droid.
His stomach growled, reminding him he hadn't eaten. He reached into a soldier's helmet that he'd repurposed as a fruit bowl and found he had only one jogan left when he thought he had two. The fruit was squishy and overripe, but he was hungry enough to eat grass at this point.
Ezra jammed the jogan under his arm and climbed the ladder to the observation deck, where he could eat in peace and watch the stars come out.
**Many** kilometers above Lothal's night sky, a shuttle departed hyperspace. Yet the thousands of lights that greeted the shuttle weren't the twinkling dots of the cosmos. They shone from the massive triangular form of the Imperial Star Destroyer _Lawbringer_ , which floated against the planet's blue-green sphere like a vibrospear poised to stake fresh ground.
The shuttle's pilot grinned. He had come out from hyperspace at the precise location he'd calculated—right over the Star Destroyer's bridge, where its captain would be standing, frozen in shock.
"Unknown shuttle, identify yourself immediately, or be blasted," barked a voice over his subspace comm.
The pilot spoke calmly into the comm, not rattled by the threat. "Agent Kallus of the ISB, requesting permission to dock."
There was hesitation on the other end; then the comm officer spoke in a more respectful tone. "Please hold, sir."
Kallus swooped the shuttle to fly over the bridge a second time. The Star Destroyer's turbolasers tracked him, but wisely the gunners never fired. The mere mention of the Imperial Security Bureau made men think twice about what they were doing.
Another voice came over the comm, with an aristocrat's accent. "This is Captain Zataire of the _Lawbringer_. I offer my sincerest apologies, Agent Kallus. We weren't expecting you. It doesn't appear that you were on the flight schedule."
"I wasn't," Kallus said. He was never on the schedule. His visit, like all his visits, was meant to be a surprise. "Where shall I dock?"
Zataire's voice quavered. "The aft hangar. I will be there to meet you. Cut your sublight engines so the tractor beam can get a lock."
"No need to use the tractor beam. I will land the shuttle myself."
"As you wish. It will be a pleasure to have you aboard."
Kallus knew that pleasure was the last thing the captain was feeling right then. Dread would be the more appropriate emotion. For the ISB to board your ship was not exactly an honor any commander wanted. Though he held no military rank, as a full executive agent, Kallus had the authority to countermand any of the captain's orders if it helped fulfill his mission to hunt down and catch potential rebels.
The Star Destroyer's bay doors opened. Kallus glided his shuttle toward them, maintaining a slower-than-average speed. Every second of his approach would amplify Zataire's dread. Every second would remind the captain that he had just lost his ship.
The _Lawbringer_ was, for all practical purposes, now Kallus's to command.
Captain Hiram Zataire loved his emerald wine. He poured glass after glass during the dinner he hosted for Kallus in his ornately decorated cabin. Zataire spent most of the meal detailing the qualities of his favorite vintages, commenting that for as green as Lothal was, its emerald grapes often tasted as sour as Hutt punch. Kallus sipped his wine and nodded politely, offering an occasional question to keep the captain babbling. Small talk frequently revealed more about a man than direct interrogation did. Before Kallus made his next move, he had to learn if Zataire was whom his Imperial datafile said he was, a somewhat fussy yet fiercely loyal officer of the Empire. He needed to know if the captain could be trusted.
Zataire twice brought up his vineyard on Naboo, of which there was no record in his datafile. The file did report that Zataire owned a couple of large plantations on the planet, so Kallus assumed the vineyard was part of those properties. The file also noted that Zataire had a wife and three children, two daughters and a son. The captain spoke glowingly of his wife, who stayed married to him even while he was away on these long assignments, and his daughters, who were the dearest things in his life. The eldest shared his passion for wine and ran the vineyard, while her sister had followed his footsteps into the armed forces and was serving as a lieutenant aboard Lord Tion's flagship, the _Devastator_.
Of his son, he made no mention.
As they ate, Kallus surveyed the cabin in quick glances and long, pensive looks. He had trained himself to assess people's loyalty by profiling everything around them. He took note of their possessions, the arrangement of their habitation spaces, the quality of their cutlery, how and where they hung their hats—anything unique that made them stand out, or, even better, the things that didn't. Because in Kallus's experience, the things that didn't stand out in plain view were often the very things people wanted to hide.
The captain's refined tastes went beyond emerald wine. Ivory bantha bone framed a floor-to-ceiling viewport that offered a gorgeous vista of Lothal. A chromatic chandelier hovering above the dining table cycled through variations of colors that complemented the planet's blue-green glow. Shelved along the wall, and sorted by topic and author, were books—not datatapes or flimsiplast replicas, but collector's hard copies, ink printed on paper. Kallus had never seen so many of them outside a library. Dust even lingered on a spine or two.
These furnishings made for impressive quarters. Zataire was evidently a man of independent means, as a captain's pay grade could never afford such extravagances. Kallus judged he most likely served the Navy for the privilege of his rank rather than any financial gain. Such men were rarely disloyal, since they feared risking the respect they had gained among their peers. In fact, they made some of the Empire's best commanders—wealthy citizens who were not bluebloods by birth but had joined the upper classes by climbing the military ladder.
The only thing that stood out to Kallus was the wine.
When Zataire went to refill their glasses, Kallus made his move. "I must ask, Captain, if you despise Lothal's emerald wine so much, why do you keep pouring it?"
Zataire's hands shook almost imperceptibly as he held the bottle. Kallus continued before the man could answer. "Might it be because the wine is not a vintage from Lothal at all, but one from Alderaan?" He sneered when he said the name of the world. Alderaan had been a constant thorn in the Empire's side. The planet's government, led by the House of Organa, had repeatedly argued against the military crackdowns required to maintain order. They claimed to champion peace but in reality only fomented anarchy.
Zataire's voice trembled with his hands. "I...did not realize you, too, were a wine connoisseur."
"My job necessitates I be an expert at many things. Let me have that before you drop it." Kallus took the bottle of wine from Zataire. The label showed it came from a vineyard on Lothal, but Kallus peeled that off to reveal another label beneath.
"Just as I thought. From the Organa Ranch on Alderaan," Kallus said.
"I dare not serve a man of your standing the local Hutt punch," Zataire said. "That was the only bottle I had that wasn't from Lothal, a gift I recently received."
"From that rebel you call a son, perhaps?" Kallus casually asked.
The stem of the wineglass snapped in Zataire's hand. Bright green liquid spilled all over the tablecloth. "Are you accusing me of disloyalty, Agent Kallus?"
"Captain, I haven't accused you of anything. That is something I do only when the facts give me reason to. As an agent of the ISB, I follow the letter of the law."
Zataire held the broken wineglass in his hand, his mouth half open. Kallus poured the last of the bottle into his own glass, which he then offered to Zataire. "You should enjoy this while you can. As bothersome as Alderaan may be, I agree, her vintages are the best. But one wonders how much longer her grapes will grow."
Zataire took a moment to accept Kallus's glass. "Please know I despise my son's political views," he said. "But I love him and could not refuse his gift. It is what any good father would do."
"Of course it is. And as a good Imperial, you have provided me the location of your rebellious child. You know we have been looking for him for years."
Zataire sighed. "My son has a big mouth and is prone to outrageous speeches, but he's not a rebel. He'll see the value of the New Order when he grows older. Please do not hurt him."
"Captain Zataire, the Imperial Security Bureau was created to protect the Empire's citizens, especially the children of a decorated officer like yourself. The last thing we'd ever want is for him to be kidnapped for ransom, or, worse yet, tortured and killed. As you are probably well aware, Alderaan can be a rough place, with all the anti-Imperials lurking about."
Zataire swirled the wine in his glass, then put it down without drinking. He did not look at Kallus when he spoke. "Might there be a way to guarantee my son's safety?"
Kallus paused, as if considering Zataire's request. In truth, he had already gamed out this conversation—and what he could offer in return for Zataire's full support. He didn't need to offer anything, since his jurisdiction gave him the authority to order the captain to do whatever was required to accomplish the mission. But going over the captain's head would make Zataire an enemy, one who would stall in securing him any necessary resources. And Kallus didn't come to Lothal to battle fellow Imperials. He came to root out rebels.
"The galaxy is a dangerous place, as you well know, and safety can never be fully guaranteed. But the letter of the law allows, shall I say, a certain flexibility in both of our duties." Kallus leaned toward Zataire. "If you give me your fullest cooperation, without any bureaucratic nonsense, I will do my utmost to personally ensure that your son's big mouth does not get him into further trouble."
Zataire stared into his wine before finally lifting his head and looking Kallus in the eye. "You've always had my fullest cooperation, Agent Kallus."
"Which I never questioned. You are a fine Imperial, Captain Zataire. If we end the subversive activity on Lothal, you will be commended. But first things first. Have the _Lawbringer_ descend through the atmosphere to hover over the capital city. The locals should tremble when they see the magnificent weapons we wield."
"You want us to leave orbit? The planet will go unprotected." It was the only time during the dinner Zataire had raised his voice, a sign that he took his duty seriously.
"Unprotected from whom? There is no imminent threat of invasion anywhere in the galaxy. Like never before, space itself is at peace, because of the Empire's might."
Kallus turned in his chair toward the viewport. Since he had boarded the Star Destroyer, Lothal had made a half rotation, bringing dawn to its other side. Now its major continents blazed a brilliant green.
Kallus narrowed his eyes like lasers at the planet. "The one and only danger to our beloved Empire, my dear captain," he said, "lies not from without, but from within."
To eliminate rebel activity once and for all on Lothal, he would bleed those lands red.
**Ezra woke** to the rumble of thunder. He staggered to his feet, woozy after having spent the whole night asleep on the observation deck. It wasn't raining yet, so he blinked away the last bits of sleep from his eyes and went over to the tower's edge. Leaning his arms against the railing, he looked out at the dawn.
Green daisies had sprouted around the tower, waiting to bloom. Insects buzzed about the flat prairie, some snatched from the air by swift paws of furry rodents. In the distance loomed Capital City, where smokestacks pumped pollutants into the sky. The morning cloud layer glowed its usual purple haze. Oddly, Ezra didn't see any sign of rain, though the rumbling continued. Some of it came from his stomach, which craved more than the previous night's jogan. But most of the rumble came from above. Ezra tilted his gaze—and dropped his jaw.
An Imperial Star Destroyer soared overhead.
The ship was colossal, far bigger than it appeared on news reports. Over a kilometer and a half long, it darkened the sky and seemed endless. Giant turbolaser batteries projected from its armor-plated belly, all pointed in one direction—toward Capital City.
Ezra pictured the city's well-to-do citizens cheering the oncoming craft while the poorer denizens scurried into whatever hole they could find. They were the wise ones. An Imperial Star Destroyer approaching one's home was not generally a positive sign.
A squadron of TIE fighters launched from the Star Destroyer's hangar. Some shot off toward the city, while others circled the vessel like bloodflies around a nerf. Ezra squatted to the deck floor as one fighter swooped close by. He doubted the pilot had seen him. TIEs were always zipping back and forth over the prairies, patrolling for rebel activity. Their constant presence required Ezra to dim the tower's illumination so as not to attract attention.
The Star Destroyer's bulk terminated in a set of conical ion engines. The brightness forced Ezra to shield his eyes. Nonetheless, it did not stop him from seeing the prospects that now lay before him. Whatever drew a warship of this size might not be positive, but it could be immensely profitable. There would be many Imperials running about, which would present lots of pockets—and helmets—to pick.
After his last botched effort, he wanted those helmets more than ever.
Ezra scrambled down the ladder, grabbed his backpack, and started his jump bike, his hunger forgotten.
Ezra hid his bike in a deserted alley and waded into the crowded city bazaar. There citizens of all stripes bustled and mingled, bartering for spare parts or scavenging discount bins for the best deals. Down row after row of stalls, merchants hawked a galaxy's trove of goods, offering cut-rate prices on secondhand trinkets. For the famished, street cooks roasted sweetmeats on sticks and farmers sold their harvests of fruits and vegetables. More than one fat wallet stood out as an easy grab, but Ezra held back. He was searching for Imperials.
It didn't take him long to find them. Ahead, a surge of market goers tried to leave. Imperials always caused that reaction in people.
He elbowed past two swine-snouted Ugnaughts, then crouched behind tall clay casks. Strutting past him in olive uniforms were a pair of Imperial officers whom everyone knew by name: Commandant Aresko and his lackey, Taskmasker Grint. Assigned to run Lothal's Imperial Academy, the two spent their free time doing whatever they could to make life miserable for the inhabitants of the lower levels. Today their victim was Yoffar, an old Gotal who eked out a living peddling yesterday's fruits.
"Your identification. Now," Grint ordered, pushing his barrel of a belly near the Gotal's horns.
Yoffar held up a plump specimen from his basket. "I'm just trying to sell a couple jogans here."
"Did you say _sell_?" asked Aresko, who held his nose high like a prince. "You do realize that all trade must be registered with the Empire."
The Gotal snorted. "By the time the Empire's done, there won't be any trade left."
Grint stepped closer to Yoffar, sneering. "What did you say?"
Ezra felt sorry for Yoffar. Truthfully, he'd never liked the geezer. The white-furred fruit seller was a perpetual grump who always confused Ezra with the other street orphans and once had accused Ezra of nicking fruit, which Ezra hadn't—not _that_ day, at least. But no one, not even Yoffar for all his crankiness, deserved Imperial harassment for trying to get by. Perhaps there was something Ezra could do—or steal—to turn the tables.
If the Gotal felt harassed and intimidated, he didn't show it. "I remember what it was like before your ships showed up, before you Imperials ruled Lothal like the rest of the galaxy."
Grint glanced at Aresko. Neither could hide his glee. "Mister Grint," Aresko said with a raised eyebrow. "That sounds like treasonous talk to me."
"That it does, sir," said Grint.
Aresko unclipped his comlink from his belt. It was one of the new Imperial editions, built to transmit across a range of frequencies, including coded military bands. Although it wasn't a helmet, Ezra could make a nice chunk of credits pawning it and have some fun at the same time.
"This is L-R-C-zero-one," Aresko said into his comlink. "I'm bringing in a citizen under suspicion of treason."
A voice responded almost instantaneously, the same electronically modulated voice that could be heard across the galaxy—an Imperial stormtrooper. "Copy that, L-R-C-zero-one. Dispatch to cell block A-A-thirty-three."
The clatter of boots prompted Ezra to glance behind him. Two white-armored stormtroopers marched through the market, already en route. This wasn't some chance license check—this was a stunt, planned to demonstrate the Empire's speed at responding to incidents in order to strike fear into the hearts of onlookers. Fear was how Imperials tried to handle everything.
Ezra wasn't afraid.
He slinked along the line of clay casks, judging the best way to approach Aresko. The commandant had reattached the comlink to his belt. Ezra could pinch it if he came from the side. He slipped behind the stormtroopers and used them as cover.
Grint took a fruit from Yoffar's basket. "By Imperial authority, we hereby confiscate your goods."
Old Yoffar's bravado became slack-jawed terror when he saw the stormtroopers approaching him. "Take him away," ordered Aresko.
The troopers grabbed the Gotal and began to drag him off. Yoffar clutched the fruit basket to his chest. "You can't do this!"
Grint took a bite of the jogan he'd stolen as he and Aresko followed the stormtroopers. "Yeah? And who's gonna stop us?" He pointed at other merchants around them. "You? _You?_ "
The merchants looked away, some packing up their merchandise. Ezra sidled up to Grint. "Hey, mister, spare a jogan?"
Both Imperials turned and Ezra reached toward Aresko's belt. His fingers made the lightest touch on the clip release, and the comlink was in his possession in less than a second.
"Scram, urchin. Those jogans are Imperial property," Grint said.
"Sorry, sorry. Not looking for trouble, sirs." Ezra lowered his head so they couldn't see his face and hurried away. "But it sure has a way of finding me," he said under his breath. Because he wasn't done with them. He still had to have his fun.
He set the comlink to loudspeaker broadcast and cleared his throat. "All officers to the main square." He spoke into the comlink, impersonating the pompous adult voice from the holopad advertisements. "This is a code red emergency!"
Ezra kept walking, head low, and glanced back. The Imperials had stopped, clearly annoyed by the orders. Grint spit out jogan seeds. Aresko gave Yoffar an arrogant glare. "It's your lucky day, Lothal scum." He motioned to the stormtroopers. "You two, come with us. Leave that nonhuman stinker in the dirt where he belongs."
The stormtroopers dropped Yoffar and rushed off with the officers toward the square. Once they were gone, the entire market let out a sigh of relief. Business returned to usual.
Yoffar picked up himself and his fruit basket. Ezra approached, his stomach reminding him of his hunger. He spoke into the comlink again so the Imperials wouldn't return soon. "Stay on alert! Repeat, this is code red."
Yoffar saw him with the comlink, and instead of giving his usual grumpy frown, he smiled. He offered Ezra a jogan. "Thank you."
"No, thank _you_." Ezra opened his backpack and packed it with jogans from the basket.
Yoffar's smile vanished. "Wait, wait. What are you doing?"
Ezra heaved his backpack over his shoulder and winked at the old grump. "Hey, a kid's gotta eat."
He stepped onto a nearby crate, climbed up a support beam, then leapt onto the rooftop of a building. "Who is that kid?" he heard the Gotal say.
Ezra grinned. Maybe now Yoffar wouldn't confuse him with the other orphans.
**With his** heavy pack bouncing on his back, Ezra hopped over several chimneys to cross the roof. From the other edge, he could peek down into the city's main square. He was curious to see how the Imperials would react to his little ploy on the comlink.
Down below, Supply Master Lyste addressed a group of stormtroopers as they grav-chained hover crates to several speeder bikes. "Make sure the repulsors stick to the bikes," Lyste said, doing none of the lifting himself. "I don't want to lose any of those crates."
The two stormtroopers from the market arrived with Aresko and Grint in tow, both out of breath. "What's the emergency?" Aresko wheezed.
"Emergency?" Lyste asked.
On the roof, Ezra grinned, seeing the outcome of his comlink prank. From every adjoining street and alley came stormtroopers, stampeding into the square like herds of prairie squirrels in the mating season. Lyste, Grint, and Aresko looked about, utterly perplexed.
Grint leveled a glare at Lyste. "You called a code red."
"I-I'm not sure what you mean," Lyste stammered. "My orders are to get these crates to the Imperial Portal."
"Well, get them loaded, then!" Commandant Aresko said, reddening with anger.
To Ezra's delight, Lyste lent his own hands in lifting a heavy crate. Grint and Aresko muttered to each other while the other stormtroopers bumbled about, testing their helmet comlinks, looking nothing like the crack troops they were supposed to be. Ezra almost felt bad for the whole bunch. Almost.
His gaze fell on the hover crates Lyste's contingent attached to the speeder bikes. Three troopers remained to the side, guarding the two largest crates of the cluster. It was hard to tell from Ezra's vantage point what the crates could possibly hold, but a guard that strong probably meant whatever was inside was valuable. He stepped toward the roof's edge for a closer view and crouched, careful not to be seen.
In the center of the square stood a ponytailed man in a collared olive tunic that was tucked into sleek gray pants. Dark green armor plated his right shoulder and matched the gauntlet around his forearm below. His back was toward Ezra, but he turned to look up at the rooftops, as if he had felt Ezra's gaze. The man's chin sported a sharp goatee, yet his blue eyes were even sharper, cutting through everything he looked at.
Weirded out, Ezra stepped back from the edge, where he could still see the man, but where the man hopefully couldn't see him.
The man looked away from the roof to a muscular nonhuman who stood nearly a meter taller than everyone else in the square. Ezra couldn't identify the being's species, but whatever he was, Ezra didn't want to get on his bad side.
The ponytailed man tapped his thigh twice, to which the nonhuman responded by turning into an alley. The man then approached a girl wearing outrageously colored Mandalorian armor and again tapped his thigh. The Mandalorian repeated the tap on her leg while heading in the opposite direction. All three were now on the move.
Interesting. Was what they had tapped a secret code? Ezra looked back at Supply Master Lyste and the stormtroopers. They had finished loading all the hover crates and didn't notice the Mandalorian walking close to the furthermost speeder bike. She slid her hand under the bike and pressed a blinking round object to the chassis.
Ezra knew immediately what that object was. He covered his ears. The object stopped blinking and the speeder bike exploded.
The shock wave knocked Lyste and his troopers backward. The once-puzzled stormtroopers in the square rushed to the scene, now having a purpose. Lyste crawled up and pointed at the other parked speeder bikes.
"Get those crates out of here," he yelled. "Keep them secure, at all costs!"
_At all costs._ Ezra liked the sound of that. It meant he was right about the crates being valuable. He rose from his crouch and dashed across the rooftops, searching for the quickest way down.
He wasn't going to make it. A stormtrooper jumped onto the speeder bike with the two largest crates and took off toward a side street. Even with his knowledge of the rooftops, there was no way Ezra could outrun a speeder bike.
The speeder bike didn't make it out of the square, though. Its rider screeched it to a halt to avoid colliding with a landspeeder that was backing out from a side street. The rider gestured for the landspeeder to clear, which its driver seemed to interpret as a greeting. "How's it goin'?" asked the ponytailed man, leaning out of the driver's side window.
Ezra stopped on the roof above, seeing more stormtroopers arrive on foot and on the other crate-loaded speeder bikes. It was now clear that this ponytailed man also wanted those crates. But confronting a stormtrooper squad was madness. The hammer of the Empire would nail this man's coffin shut.
The prospect of death didn't seem to scare this man, though. He sprang out of the landspeeder and ran straight at the Imperials.
With one booted foot, the man kicked the lead rider off his bike, then spun, blaster drawn, and fired at the oncoming squad. Some troopers went down, but more came to take their fallen comrades' place and return fire.
The man had reinforcements of his own. The burly nonhuman lumbered out from an alley and demonstrated why Ezra was correct in not wanting to cross paths with him. The bruiser picked up a stormtrooper from behind and threw him into another trooper. Both soldiers crumpled to the ground from the teeth-rattling impact.
Ezra scanned for the third member of this team, the one in the Mandalorian armor. She was nowhere to be seen. What was her role in all this, other than setting the speeder to blow? The man and the bruiser didn't have trouble dealing with the squad by themselves. They only had to finish off a few more stormtroopers before they'd be in possession of the speeder bikes and the hover crates.
The crates—they were the fulcrum of this whole altercation. If these strange people were willing to risk their lives against such ridiculous odds to gain them, Ezra could only wonder what treasures they contained.
He went to the roof's edge. The lead speeder bike hovered a few stories below, riderless, its two large crates tempting him. If he dropped onto its seat, the bike's repulsor should absorb the shock of his fall.
Thinking no more about it, lest he get cold feet, Ezra closed his eyes and stepped off the roof. His stomach seemed to go first.
He landed in the seat with a butt-numbing wham. The bike bounced on its repulsors, and he caught the steering rods to prevent himself from being thrown off. He opened his eyes to see the ponytailed man and the nonhuman bruiser sprinting toward him.
"Thanks for doing the heavy lifting," he said, and kicked the bike into gear.
Ezra swerved around the bruiser, ducking his wild swing, and sped into an intersecting street. He waved good-bye to the two as more stormtroopers charged.
Those troopers didn't delay either of them for long. Soon they both trailed Ezra on separate speeder bikes. And since he couldn't figure out how to operate his bike's blaster cannon, he had to rely on losing them in the maze of the city streets.
He drove through broken windows of an abandoned warehouse, then knifed through a narrow alley. Whenever there was a quick turn, he made it, tearing through clotheslines and skimming the tops of trash bins. But he couldn't ditch his pursuers; they were gaining. The contents of his bike's crates must be slowing him.
He slowed even more when something dropped and landed on his bike's crates. "Pretty gutsy move, kid, jumping without a jet pack," said the girl in the paint-splattered Mandalorian armor. Her helmet filtered her voice but didn't re-modulate it like with the stormtroopers.
Ezra revved the bike to shake her off. She did the job herself, firing a blaster at the couplings that held one of the hover crates. It detached from his bike.
"If the big guy catches you, he'll end you!" she yelled, falling away on the crate. "Good luck!"
Ezra glanced back to see her hop off the hover crate and push it into an alley. He considered turning around, until the ponytailed man and the bruiser zipped past her, right on his tail.
He jammed his heel against the pedal. With one less crate, the bike accelerated faster. His vision narrowed. Walls and buildings blurred by him. He veered around a statue, dove under an arch, and darted down another alley, letting his instincts guide him more than his eyes.
As good as his instincts were, they didn't help him avoid the troop transport that blocked the mouth of the alley. Stormtroopers were posted on the ground, ready to fire at him.
All he could do was bob, weave, and hope he wasn't hit. A single, well-placed blaster bolt was all it would take to end his short career.
That one bolt never came—not to him, at least.
A stream of heavy fire from behind blasted the transport, scattered the troopers, and cleared the way for Ezra. He zoomed out of the alley, looking at his two pursuers, astonished by what they had done. Only the Empire's best troopers could shoot with that kind of skill on speeder bikes.
Who were these guys?
**Who was this kid?**
Kanan had asked himself that question many times on this crazy chase. Either the boy had military training of the highest caliber, or his talents stretched beyond the ordinary. There was no other explanation as to how he could avoid being caught by them and the Imperials.
The chase would end shortly, however. They'd reached the outskirts of Capital City, where there weren't as many alleys or obstacles the boy could use to elude them. Kanan and Zeb gave their bikes an extra kick and zeroed in on the kid's back.
High-pitched whines made both look behind them. Three stormtroopers on speeder bikes now pursued them.
Kanan shook his head. Imperials always had to make things more difficult.
The kid sped toward the freeway that funneled traffic in and out of the city. Kanan and Zeb shifted their bikes to follow, as did the troopers. Stop signs on the freeway entrance ramp slowed no one. Fortunately, there wasn't much traffic on that section of the freeway.
Kanan wrenched his bike to the side, dodging the troopers' shots while turning in the saddle to respond with his own blaster. He hit one trooper in the chest, causing both man and bike to fall off the freeway.
But another trooper's blasts found the boy's bike. Sparks sizzled and a repulsor flap blew off. The kid lost control. His bike careened over the divider into the opposite lane.
The troopers' fire made it impossible for Kanan to veer into the other lane without getting toasted. Kanan signaled to Zeb that they had to take out their pursuers.
Zeb peeled his bike around to come right at one trooper, surprising the man and knocking him off the bike. Kanan, meanwhile, pulled a detonation charge from his belt and cut his speed so the third trooper overtook him. "Okay, you caught me. I give up!" Kanan said.
The trooper cocked his head at Kanan, probably perplexed that Kanan would concede when he seemed to be winning the chase. To prove the point, Kanan lifted his hands from the steering rods in mock surrender.
"Just kidding." Kanan tossed the confused trooper the blinking detonator and gunned his engine, speeding away as the trooper's bike exploded.
Zeb caught up beside him. The Lasat scowled when Kanan signaled for him to watch the cargo. Zeb never liked quitting a chase, but Kanan had to be the one to go after the kid. He wanted to confirm if the kid's talents were real or if he was just lucky.
He pressed a button on the bike's controls and the crates detached from the couplers and fell away. Zeb halted his bike before the crates, muttering something about ending the kid for risking their operation. Kanan ignored him and swerved over the divider into the opposite lanes.
The kid had regained partial control of his bike and was on the move. Yet Kanan had a much lighter load, having dumped both of his crates. He rocketed past the kid, then looped around to fly straight at his young quarry.
The kid panicked and slammed on his air brakes, skidding horizontally. Kanan did the same, so the two faced each other, sideways.
"Who are you?" the kid asked.
"The guy who was stealing that crate," Kanan said.
"Look, I stole this stuff—whatever it is—fair and square."
"And you made it pretty far, kid. But I've got plans for that crate, so time to give it up. Today's not your day."
The kid tilted his head, looking at something behind Kanan. "Day's not over yet," he said with a smile.
Kanan glanced over his shoulder to see TIE fighters cutting through the clouds to bear down on them. Wonderful. Just what he needed.
The kid stepped on his foot pedal, rotated his bike, and whisked off. Kanan spun his bike around to give chase, into a hail of TIE laser fire.
He adjusted his altitude and direction, anticipating the TIE pilots' firing patterns to steer through the barrage unscathed. Uncannily, the kid matched Kanan's exact movements, tacking right and left to dodge the lasers. No, it was more than uncanny—it was as if Kanan and the kid were on the same wavelength.
This kid had more than just luck.
Kanan, however, could've used some luck himself when a TIE's lasers scored a hit on his bike, shorting out the inertial compensators. He lost momentum instantly and could do nothing as his bike fell.
"Have a good one." The kid gave a mock salute and sped away as Kanan's bike crashed into the ferrocrete. The TIE fighter swooped around to fly after the kid.
Kanan picked himself up from the wreckage. He unfastened his comlink. "This is Spectre-1. I need a lift."
He wasn't done with this kid. Not yet.
Turning off the freeway into the grasslands, Ezra was feeling like an old pro on his speeder bike. Despite losing a repulsor flap, everything else on the bike performed magnificently. The thrusters responded to the slightest nudge of the foot pedals. The engine hummed even at the highest speeds. The contoured handgrips made the steering rods move with ease. He had to give credit where credit was due. The Empire always kept their equipment in tip-top shape—including their TIE fighters.
A blast skimmed by his head, centimeters away. The TIEs' cannons were powerful enough to destroy enemy spacecraft. If they got a solid hit on his bike, there probably wouldn't be enough wreckage to identify his remains.
Whatever was in that crate had better be worth it.
He weaved through the prairie mounds, eluding the TIEs' lasers. He might find protection and a place to hide in the mountains, but he'd never outrace the TIEs before he got there. The only other option was to slow down so the TIEs overshot him. Maybe then he could turn back toward the city and lose them in the alleys.
He didn't even get to lift his feet off the pedals before a TIE's lasers slowed his bike for him.
His bike began to smoke and shudder. His speed declined dramatically. The engine must have been struck. Ezra jammed his heels against the rocker pads and pulled back on the steering rods, trying to keep as much altitude as possible. If the TIEs didn't get him, gravity would.
The TIEs dove closer, peppering the area around him with laser fire. None hit, yet his bike's engine gave a horrendous, wailing shake. It was gone. Ezra pressed the cargo detachment button, then hit the brakes. He flew off the bike and landed with a thud in the grass. He didn't break any bones, but the impact still hurt.
He pushed himself up. Flames engulfed the bike behind him. The crate, however, had detached and safely hovered a meter or two away. One of the TIEs wheeled back toward him. He could see its lasers heating up, priming to score a direct hit—on him.
He didn't bother to run. He wouldn't have gotten far. If this was the end, so be it.
Lasers twanged, but not from the TIE. The Imperial fighter erupted in the sky in a great ball of fire.
A star freighter with diamond-like angles sailed through the explosion and slowed to hang right above Ezra. The cargo hatch opened and a ramp half extended, revealing the man from the square, his ponytail whipping in the wind. "Want a ride?"
Ezra stood there, partially in shock. This man had been trying to stop him before. Now he had come back to save him? It didn't make sense.
The man reached out a hand. "Come on, kid—unless you have a better option?"
The group of TIE fighters screaming back convinced Ezra he didn't. But rather than head to the ship, he dashed to the hover crate. He wasn't going to leave without it.
Ezra began to push the crate toward the ship. The weight of its contents strained his muscles. The ponytailed man shook his head in disbelief. The freighter started to rise as the TIEs drew near. They'd be there in seconds.
Ezra cranked up the hover crate's repulsors, then took it in his arms and jumped.
He didn't know why he did; he just felt that he should. Normally, he couldn't leap more than a meter off the ground without a pair of jump boots. And a hover crate's repulsors were weak, with a ceiling of only a few meters. Yet his jump took him higher than that. Much higher.
It wasn't natural. But it felt natural.
His legs relaxed. His body soared. It was as if a little voice inside him had been released to sing. It connected him to the world around him, which was ablaze with life and energy. The green grass on which he had snoozed so many times had sprung him upward like a trampoline. Tiny insects buzzed around him, the flutter of their wings giving him lift. The outstretched hand of the ponytailed man seemed to pull him to the ship like a magnet.
It all happened so suddenly, so unexpectedly, Ezra could scarcely believe it.
In that hesitation, the surge of force that propelled him vanished.
Ezra fell.
**The crate** slammed onto the _Ghost_ 's ramp, without the kid. Kanan looked around. TIE lasers whitened the sky. He sighed. The kid's leap had been an incredible feat but it had not been enough.
Perhaps Kanan had been mistaken. Perhaps it had all been luck to begin with.
Then the crate on the ramp moved. Two hands clung to its sides. One grabbed the edge of the ramp. Kanan watched, amazed, as the boy pulled himself over it.
Who was this kid?
Laser fire crackled in the air, too close for comfort. The kid shoved his crate into the cargo bay. Kanan retracted the ramp and closed the hatch.
Zeb and Sabine were unloading the other crates in the bay. After a glance at the kid's crate, Zeb stomped over and wrenched off the lid. Inside was a cache of assorted blaster weapons, mostly E-11 stormtrooper rifles, DH-17 pistols, a couple of hold-outs, and a stingbeam.
The kid's eyes widened at the sight of the arsenal. "Do you realize what these are worth on the black market?"
"I do, actually," Kanan said.
"So don't get any ideas," Zeb said.
"Ideas? These are mine," the kid said.
Zeb showed his teeth. "If you hadn't gotten in our way—"
"Can't help it if I got to them first." The kid stood on his tiptoes to challenge Zeb's vicious stare with one of his own.
Kanan stepped between the two. The last thing they needed was a fight while Hera was trying to outfly TIEs. "It's not who's first, kid, it's who's last."
The additional sublight engines came online, jostling the _Ghost_ and her passengers. Kanan grabbed a crate to steady himself. They must be nearing the upper atmosphere. Hera probably needed his help.
"Keep an eye on our friend here," he said to Sabine and Zeb. He climbed the ladder into the central corridor. To play it safe, he verified that the surveillance system was working. Sabine usually did her own thing, and Zeb wasn't necessarily the greatest of babysitters.
Kanan hurried into the cockpit. Atmospheric clouds fogged the viewports. Hera pulled on the flight stick, taking the _Ghost_ on a steep climb. "You said this was a routine op. What happened down there?" she asked.
Chopper, plugged into the shield controls, responded first with a chiding snortle. Kanan knew the droid was right—he, Sabine, and Zeb had screwed up—but this was not the time to lay blame. "Chopper, please. It's been a difficult morning."
"He has a point, love. We've got four TIE fighters closing in."
_Love._ Hera tossed that word about like it meant nothing. Years earlier, Kanan would've believed in her affection and told her how exceptional a pilot she was. Now he deflected her sarcasm with that of his own. "How about a little less attitude and a little more altitude?"
"No problem." Hera slapped the flight stick hard to one side as the TIEs opened fire.
Before the artificial gravity could adjust, Kanan was thrown across the cockpit into the side wall. He grimaced. "If I didn't know better, I'd think you did that on purpose."
She banked hard to the other side, avoiding a deluge of lasers. This time, Kanan gripped a component handle to keep from falling.
"If you knew better, we wouldn't be in this situation," Hera said. "Seriously, what happened down there?"
Kanan pointed at a surveillance monitor. The cargo bay cameras were trained on Sabine, Zeb, and the boy. " _He_ did."
Hera glanced up from the controls, continuing to pilot. "A kid tripped you up? Must be some kid. Spill it."
"Aren't you a little busy at the moment?" Kanan asked. Sensors showed the TIEs were about to make a coordinated attack run.
Hera's gaze didn't leave him. "Spill."
The internal display of Sabine's helmet showed her the kid's basic biometrics. He was definitely human, of average height and build for a boy his age, probably only a few years younger than her sixteen years. He must be right-handed, since he wore an energy slingshot on his left forearm. Below two blue eyes and above a constantly devious grin, his nose stood large and prominent, as if it had a jump start on a future growth spurt. Like that of most Lothal natives, his skin bore a copper sheen, and his shaggy dark hair, parted down the middle, hadn't seen the barber in some time. He was most definitely a street kid. An urchin.
While she was examining him behind her helmet, he was also scrutinizing her. "Are you a Mandalorian?" he asked. "A real one?"
If he hadn't been a kid, she probably would have answered his question with her blasters. Mandalorians were unwelcome in the galaxy those days, ever since the Empire had outlawed their mercenary practice and occupied her homeworld of Mandalore. The few who still roamed the stars were usually armored impostors, which Sabine Wren was not. But the kid didn't need to know that. He didn't need to know anything about her or her people. She remained silent.
The kid turned to Zeb. "How about you? You some kind of hairless Wookiee?"
Sabine smiled under her helmet. That would get Zeb going. The Lasat felt indebted to the Wookiees for helping his people fight the Imperials on his homeworld, but he despised ignorant comparisons. It would be amusing to watch how the kid dealt with someone three times his size and ten times his strength.
"Is that what the Imps taught you at school? That we nonhumans all look alike?" Zeb growled, and planted his foot on the crate and wiggled his four broad toes. "Well, let me teach you a lesson. I'm a Lasat, and we don't think highly of little thieving Loth-rats like you."
The kid scooted to the edge of the crate. "I was just doing the same thing you were. Stealing to survive."
"You have no idea what we were doing," Zeb said.
"And I don't want to. I don't," said the kid. "I just want off this burner."
Zeb snarled his lips into a cruel smile. "Nothing would thrill me more than tossing you out. While in flight." He reached for the kid just as the ship shook, pounded by laser fire.
Sabine's feet remained anchored to the floor. Balance was one of the first things you learned as a Mandalorian. Zeb, however, toppled over and landed on the kid.
"Get off," the kid gasped. "Can't...breathe."
"I'm not _that_ heavy in this gravity," Zeb said, picking himself up.
"Not the weight," the kid said, his face puckering. "The smell."
Sabine nearly laughed. The smart-mouthed kid was a true urchin, fearless to a fault.
Zeb's purplish skin burned red. He grabbed the kid and hoisted him up. "You don't like the air quality in here, eh? Fine. I'll give you your own room." He dragged the kid past Sabine toward the storage lockers.
The kid kicked and screamed. "Hey, stop! Let go of me, you brute!" If Zeb's reach hadn't been so long, he probably would've been bit. Like all street kids, this urchin fought like one.
Zeb opened the locker and shoved the kid inside. Before Zeb slammed the door, the kid looked at Sabine, as if she could do something.
She stayed her ground, scanning her helmet's read-out. The biometrics registered a flush of heat on the kid's face. He wasn't completely fearless. His boldness was just a mask, like her helmet.
She wondered what he was hiding.
**Hera** powered on the final sublight engine to lift the _Ghost_ through Lothal's upper atmosphere. The TIEs ascended with her, trying to penetrate the _Ghost_ 's deflector shields. She would've told Kanan to man a turret, except that as he described what he'd seen the boy do, he spoke with a certain charge and emotion she hadn't heard from him in a long time.
"Kid sounds impressive," she said. Impressive enough to make her consider that he might not be any ordinary boy—he might have talents beyond the norm. This put his life in extreme danger. The Empire viewed anyone with such abilities as a possible traitor.
"You're not thinking what I think you're thinking," Kanan said.
"He held on to a crate of blasters with a pack of troopers on his tail," she said.
"Because I was there to save him. He's undisciplined, wild, reckless, dangerous, and..." Kanan's litany trailed off. "Gone?"
She followed his eyes to the surveillance monitor. It showed only Zeb and Sabine in the cargo bay, repacking the crates for delivery.
Kanan clicked the intercom. "Where's the kid?"
"Calm down, chief," Zeb said, his gravelly voice coming over the speaker. He walked over to the storage locker. "He's in here."
Hera checked the sensors. Even with the boost to their engines, the TIEs were gaining, as were their lasers. The _Ghost_ 's deflector shields could only absorb so much.
She took the _Ghost_ on a series of quick turns, evading enemy fire while keeping the engines hot. Once they broke into orbit, she could prepare the jump to hyperspace. That was the only way they'd lose their pursuers.
Kanan didn't seem concerned about the TIEs. He leaned closer to the surveillance screen as if that would give him a better view. "Zeb, where is he?"
Hera glanced at the monitor and saw Zeb pop his head out of the empty locker.
"W-well," the Lasat stuttered, giving the camera a sheepish grin, "he's still here in the ship."
Sabine pushed past Zeb and inspected the locker herself. "Oh, he's in the ship, all right," she commed. Her helmet sent an image to the surveillance screen that showed the ceiling grill inside the locker had been removed. The kid must have climbed into the ventilation duct.
"Very creative," Hera said, glancing at Kanan. "Sounds like someone I used to know."
The warning sensors blared. A TIE screamed overhead, its lasers rattling the entire ship. Chopper blatted out a damage report. The _Ghost_ 's hull remained unscathed, but the deflector shields were low.
Hera threw her full attention into piloting. Their conversation about the boy could wait. This was time for battle.
She didn't need to tell Kanan. He ran out of the cockpit toward the dorsal turret.
Ezra shook like a tuning fork as he crawled along the ventilation duct. The TIE's blasts reverberated through the duct, causing the thin panels he was squeezed against to vibrate ceaselessly. He thought his teeth would rattle loose, until the duct panel below him collapsed under his weight and he dropped.
It wasn't a long fall—much shorter than his plunge from the building and his flip over the speeder bike—but it hurt the most. His chest hit the ship's hard floor while his backpack slid over his shoulders and struck him in the head. He regretted taking so many of Yoffar's jogans.
Ezra adjusted his pack, took a breath, and lifted his aching head slowly. Just his luck. He'd been only a half meter away from a soft landing in a big cushioned seat.
He pulled himself up into the chair, only then recognizing where he was. He had dropped into one of the ship's gun turrets. Welcoming him through the canopy was a black canvas of bright dots.
They weren't just dots, Ezra realized. "I'm...I'm..." he said, choking on his disbelief, "in space."
And he was about to die.
A TIE fighter roared past, so close he could see the solar collectors on its wing. Ezra slid in the chair as the freighter rolled, skirting green bolts of enemy fire. The turret's crosshairs displayed wireframes of three more TIEs racing from behind.
The intercom hissed with a silken female voice. "Shields are holding—for now—but you need to buy me time to calculate the jump to lightspeed."
A familiar masculine voice responded over the intercom. "Buying time now." The ponytailed man must be in the other turret, because lasers rang out and lanced the cockpit of a nearing TIE. It burst apart in a twist of metal and fire.
Ezra blinked, the intensity of the explosion giving him flash blindness. When his vision cleared, he found himself being yanked out of the cushioned chair and flung back onto the hard metal floor. He looked up to see the woman in the Mandalorian armor remove her helmet.
She wasn't a woman. She was a girl, not much older than he was, with large eyes, a pinch of a nose, and rainbow streaks in her hair that matched her armor plates.
His head suddenly didn't ache anymore.
She hooked the helmet on the seat arm, then sat down and got to business. After a few spins in the chair to track the enemies, she thumbed the triggers and turned a TIE into a starfield of its own.
Ezra was astounded. This girl handled a gun turret like a pro. He tried to deepen his voice to make it sound older, though without any Imperial affectation. "My name's Ezra. What's yours?"
Her answer consisted of wheeling away from him and firing. Before Ezra could ask again, he was lifted into the air by his backpack.
"My name's Zeb, you Loth-rat," said the bruiser, right in Ezra's face. Ezra grimaced and had to turn his head. The Lasat's breath smelled like the city gutter.
"Calculations complete," said the woman over the intercom. "But we need an opening."
"Found one." Sabine didn't wait for the crosshairs to turn green when she pressed the triggers. A third TIE was blown to bits, clearing an escape vector for the freighter.
"Entering hyperspace," the woman said.
Those million dots of space smeared into a million lines as the freighter made the jump. Ezra's stomach, however, didn't. Fortunately for everyone in the turret, it was empty.
**Ezra squirmed** in Zeb's grip as he was hauled into the cockpit. "Let go! You can't keep me here! Take me back to Lothal!"
The pilot, a green-skinned Twi'lek with flight goggles strapped to her forehead, turned from the controls. "Calm down. That's exactly what we're doing." Hers was the female voice Ezra had heard in the turret.
Ezra held up his hands to stave off that ludicrous proposition. "Wait...right now? With Imperials chasing us?"
"We lost the TIEs when we jumped," the Twi'lek said. "And the _Ghost_ can scramble its signature, so they won't recognize us when we return."
Ezra dropped his hands. He had thought only military vessels had that capability. "That's...pretty cool, actually."
He scanned the cockpit. This _Ghost_ was more than just pretty cool. It possessed components far more sophisticated than what he'd seen in mass-manufactured TIEs or transport vessels in the spaceport. There was even an orange-domed astromech droid, which rotated its photoreceptors at him. Maybe if Ezra stayed a little longer, he could nick some of this tech.
He looked back at the Twi'lek pilot. "So just drop me and my blasters outside Capital City and—"
"They're not your blasters," said a sassy voice behind him.
Ezra turned to see the Mandalorian girl enter the cockpit with the ponytailed man.
"And we're not going back to Capital City," the man added. "The job's not done."
If the man said anything more, Ezra didn't catch it. He was staring at the girl.
She didn't give him even a glance.
Night had descended over Capital City, and with it came Agent Kallus. He strode through the city square, all in gray, wearing a fleximetal cuirass that protected his upper torso and gauntlet gloves over his hands. His combat helmet had blast-proof cheek plates, which framed and focused his incriminating stare. Stormtroopers went out of their way to avoid him, though an AT-DP walker tilted its head as if in salute. Kallus walked past it, without acknowledgment.
This concentration of military personnel and weaponry was unnecessary, since the ones who had engineered the crate heist were long gone from the square. The troopers should be out combing the city streets, searching for clues about the thieves, not running amok here. This was but another example of the incompetence he would have to reverse to accomplish his mission.
He came to the scene of the crime, where a group of Imperial officers huddled. He walked through the group without comment and inspected the smoldering wreckage. It confirmed what the forensics analysts had told him. Detonators placed on the speeder's carriage had caused it to explode. Moreover, surveillance footage from rooftop cameras revealed that a female in multi-colored Mandalorian armor had probably planted the detonators, though it was difficult to verify from the angle. Kallus didn't need verification. A person of her description, with slightly different-colored armor, was already on Lothal's most-wanted list for causing similar havoc at an Imperial airfield.
What irritated Kallus was that she had done both deeds right under Imperial eyes. This wasn't the usual military incompetence—this was a gross neglect of duty that went up and down the chain of command. He had a lot of work to do to turn this planet around.
Aresko, one of the highest-ranking officers on Lothal as commandant of the Academy, came up beside Kallus. "They knew our protocol and were waiting in position."
"I've no doubt. You're not the first on Lothal hit by this crew," Kallus said.
Aresko exhaled a kept breath. "That's a relief," he said. "I mean...I assume that's why you're here, Agent Kallus."
Kallus removed his helmet. "The Imperial Security Bureau pays attention to patterns. When the Empire's operations are targeted on an ongoing basis, it could signify something more than the theft of a few crates. It could signify the spark of rebellion."
Aresko twitched, for good reason. Even the hint of rebellion on one's planet could instigate a change in leadership. Kallus knew the commandant would now do everything required to make sure that didn't happen.
Kallus toed the wreckage with his boot. "Next time they make a move, we'll be waiting for them—to snuff out that spark before it catches fire."
He stepped on the last bit of smoldering metal, crushing it to ash.
**Ezra** emerged cautiously from the _Ghost_ when the cargo hatch opened. Their landing site on top of a hill looked like everywhere else on Lothal: grassy and green.
Zeb pushed two hover crates down the ramp. "Outta the way, Loth-rat."
Ezra did as instructed, not wanting to be bowled over by Zeb's crates or the ones that came after. The ponytailed man, whose name Ezra had learned was Kanan, pushed the crate of blasters they had stolen from him. Next to him walked the Twi'lek pilot. Hera. The astromech droid, Chopper, remained inside the ship, while the Mandalorian girl brought out a fourth crate.
She had not bothered to introduce herself.
Kanan and Hera headed down the hill. Ezra started to follow. He wasn't going to let them walk off with his blasters.
Zeb grabbed Ezra's shoulder and held him in place. "Hey," Ezra said. "Where are they going?"
"If I told you, I'd have to kill you. And I might just kill you anyway," Zeb growled. His fingers pushed deeper into Ezra's shoulder blade before he released him and moved on with his two crates.
"Grab a crate. Pull your weight," the Mandalorian girl said, her first words to Ezra since she had landed on the speeder bike. She nudged her crate past him.
Before she had taken twenty more steps, Ezra was pushing a hover crate out of the cargo hold. Chopper bleeped something at him but Ezra didn't take his eyes off the girl. He followed her and the others toward a settlement that lay below the hill.
The place looked barely habitable. It was nothing more than a collection of metal shacks, repurposed tents, and crude shelters made from shipping containers. Humans and nonhumans huddled in doorways or sat in puddles on a muddy street. Some moaned; others cried; a few wailed. All looked weary, hungry, and desperate. It was as if the poorest and most wretched people of Lothal had come to this one place to live—and die.
Kanan ordered the others to wait while he and Hera went ahead with the crate of blasters. Ezra didn't try to follow this time. He had a feeling if he did, those hungry faces might try to jump him. Better to stay behind with Zeb and the Mandalorian.
He took another look around the shantytown. He'd traveled throughout the grasslands, had ridden through some poor, isolated communities, but none compared to this misery. "Lived on Lothal my whole life. Never been here."
"The Imperials don't advertise it," the Mandalorian girl said.
Zeb joined in with a snarl. "Locals call it Tarkintown."
"It's named for Grand Moff Tarkin, governor of the Outer Rim," the Mandalorian girl said. "He kicked these folks off their farms when the Empire wanted their land."
"Anyone who tried to fight back got arrested," Zeb said.
Ezra remembered the arrest in the marketplace. Yoffar had spoken his mind, and in return, the commandant had charged him with that gravest of crimes. "For treason..." Ezra said.
A couple of the refugees shuffled toward them. Malnutrition had shriveled their bodies; exhaustion had wrinkled their faces. Ezra stepped back, clutching the handles of his crate. Zeb and the Mandalorian opened the lids of theirs.
"Who wants free grub?" Zeb asked. He reached into his crate and took out not a blaster, but a jogan fruit.
The shuffle became a scramble. Famished refugees rushed out from every hole and hovel for what was probably their first meal in days. Zeb and the Mandalorian distributed fruits and food packets to eager hands. Ezra stood stunned. Was this the whole point of stealing the crates? To help these poor people?
A pea-green Rodian in a brown jumpsuit put his puckered fingers on Ezra's shoulder. "Thank you," he squeaked through his snout. "Thank you so much."
Ezra stiffened. "I...I didn't do anything." He hadn't even opened his crate.
The Rodian didn't seem to notice. He grinned and walked away. More refugees assembled around Zeb and Sabine, some coming back for seconds and thirds, always grateful. Ezra stayed a few paces away from the crowd, somewhat uncomfortable. He felt bad for raising a fuss on the ship. He felt even worse for clamping up. He could count on his hand the people who had ever shown him any gratitude.
He left his crate to the refugees and wandered back to the _Ghost_.
Despite the research Hera had done during the flight, she had uncovered no explanation that satisfied her as to how or why the large stone slabs ringed the clearing she and Kanan entered. Such stone circles could be found all over Lothal on the various continents, without a hint to their purpose. During the Old Republic, these circles had been a matter of investigation. Archaeologists from all over the Outer Rim had come to examine them and decipher their origins. Speculations varied wildly. Some attested that the circles were proof that a ritualistic civilization had lived on Lothal millennia before scouts had logged the planet. Others surmised that the slabs were of much more recent placement, dragged together by the original colonists for use as a landing area in lieu of electronic beacons.
Under the Empire, all inquiries into the matter were terminated. Superstitious mysteries of the past had no place in the New Order. The stone circles were abruptly forgotten and left to weather the elements—which Hera assumed made them an ideal meeting point for Cikatro Vizago and his Broken Horn crime gang, or "syndicate," as he preferred to call it.
A number of dinged-up IG-RM war droids clunked out from behind the slabs. Though they appeared to be in shabby condition, Hera knew the state of their exterior casings did not impair their primary function—to shoot and kill. Vizago always had their weapons and interior circuitry well maintained.
Kanan stood tall, the crate hovering before him. He hid his distaste for being there behind an expression that was as cold and inscrutable as the stones. He loathed working with unsavory characters like Vizago, whom he considered self-motivated and untrustworthy. Hera had told him such alliances were a necessary evil, and that Vizago was fighting against the Empire just as they were. But she could tell he remained unconvinced. His eyes darted repeatedly to and from the droids as if he was gauging a speedy exit in case the deal went south.
Vizago walked down the path into the clearing, his arms wide in greeting, his sharp-toothed smile wider. The tips of the twin black horns twisting from the Devaronian's hairless head had also been sharpened, as had his long black fingernails. Multiple earrings pierced the upper cartilage of his pointed ears, lending him the guise of a devilish scoundrel. Hera received his embrace. Kanan did not.
Vizago took no offense. His beady eyes zeroed in on the crate. He pushed open the lid and his smile grew wider. "Any problems procuring these lovely ladies?"
Hera shot a glance at Kanan. They had agreed not to speak about the chase in Capital City. The less the Devaronian knew, the better.
"Nothing we couldn't handle, Vizago. Your intel was accurate," Kanan said, then added under his breath, "this time."
Hera jumped on Kanan's last words so as not to arouse suspicion. "We got the goods and took a bite out of the Empire. That's all that matters."
Vizago closed the lid of the crate. "Business is all that matters. But I love that you don't know that." He gestured and a war droid came forward, carrying a container of credits.
Vizago took the credits and began counting them out to Kanan. He stopped halfway through their agreed amount for their payment.
"Keep going," Kanan said.
Hera tensed. The Devaronian had a reputation for being cheap, but she didn't think he was stupid. He'd get more than he bargained for if he tried to stiff them.
Vizago picked up another credit from the container. "I could. Or I could stop and trade the rest of the bounty for another bit of intel you've been begging after."
"The Wookiees?" Hera asked, unable to contain herself.
Vizago's red pupils centered on Hera. "The Wookiees," he said. His smile seemed anything but trustworthy.
**Ezra** sat in the grass near the freighter and looked down the hill at Tarkintown. Questions filled his head. If the _Ghost_ 's crew was truly on a mercy mission, why had it been necessary to anger the Empire by taking blasters? Couldn't they steal their food from someone else? Who were these people, really? Why was everything they did so hush-hush, so secret?
Why wouldn't the Mandalorian girl even tell him her name?
A breeze whistled through the grass. He drew his arms closer to his chest. They hadn't said how long they would be down there, and nights could be cold out on the prairie. He glanced back at the freighter. It wasn't any warmer in there. Hera kept the heating systems on minimal to evade infrared detection. These people didn't seem to mind flying around space in a cold ship.
The _Ghost_. It was a suitable name. The ship spooked him.
Ezra turned to look at the freighter. Something inside it tugged at him. It was the same uneasy feeling he had had in the town. It had brought him here, alone, and now was calling him inside. Could it be the answer to his questions? He had to go and see what it was.
He rose from the grass and ascended the ramp into the _Ghost_.
Shadows filled the cargo bay. Chopper must have dimmed the illumination. But Ezra didn't need eyes to know where he was going. He felt pulled, as if by a string. He went to the ladder and climbed into the main corridor.
Emergency lights glowed along his feet. Clunks and clangs echoed through the ship. The cockpit lay ahead at the end of the corridor, the instrumentation inside blinking on standby. Chopper probably had plugged himself into the monitors there. Did the droid hold the secrets? Ezra went toward the cockpit. But as he got closer, he felt farther away from where he should be going.
He stopped and turned around. An unmarked door was set into the corridor wall. He took a step toward it. And another. This felt right. Behind that door lay what had induced him to reenter the ship.
The door was locked. It would only stall him for a minute or so. As sophisticated as the cockpit's systems had been, this lock looked to be a much simpler device.
He reached into his backpack and pulled out his astromech arm. After tweaking the manipulator, he inserted it into the lock. He put his ear to the door and stilled his breath, listening only for the sound of the mechanism inside.
He jiggled the astromech arm back and forth. He heard nothing at first, then a faint click. The door slid open.
He put the arm back in his backpack and entered a room as spare as a monk's quarters. The bunk itself was a lean mattress without a blanket, as if no one slept there.
Someone did. Ezra could feel his presence drifting about the room. The ponytailed man. Kanan Jarrus. This was his cabin. And he had hidden something there.
Relying on his instincts, Ezra raised his hand and waved it through the air. His fingertips tingled and his palm felt like a rudder in a stream. The stream's current drew him under the bunk.
He bent down to touch the wall. It was smooth and cold but didn't give him shivers. Rather, he felt refreshed.
His fingers found a crack in the wall, too thin to be noticed by a casual glance. The crack made a right turn, then another, and a third, to form a rectangle. Was that what he was drawn to? A secret panel of sorts? He pushed his palm into the rectangle. Something in the wall clicked like the lock, and a drawer popped out.
A polygonal object sat in the drawer. Ezra picked it up, mesmerized. It reminded him of a chance cube from the betting gates at the spaceport, yet it was transparent and each of its many sides sparkled like a diamond. Maybe it was a puzzle. They were popular those days. He tried to twist its sides. Nothing moved.
He dropped it into his pocket. It might be worth something to someone, perhaps as jewelry.
He caught sight of a second object in the back of the drawer. He pulled it out.
The object was cylindrical, resembling a glowrod with a focusing lens yet missing the illumination bulb. It felt natural in Ezra's hands and he was compelled to swing it. As he did so, his thumb accidentally pressed a button in the middle.
To his astonishment, a sizzling bright blue beam projected from one end to nearly a meter in length. The object was by no means any heavier; on the contrary, it felt balanced. Waving the beam back and forth, Ezra had a clear sense of his surroundings, from the dimensions of the cabin down to the particles in the air. It was as if both his physical and mental reach had improved—as if the blade of light was an extension of his arm and his awareness.
"Careful. You'll cut your arm off."
Ezra spun around. Kanan stood in the doorway with Hera and Chopper behind him. The droid snickered in binary. Ezra frowned. He should've known better than to wander around with the droid in the ship. Chopper had ratted him out. And Kanan did not seem happy to find him in his quarters.
Ezra held the laser blade before him. "Look, I know you're not going to believe me. But it's like this thing wanted me to take it."
Kanan's eyes never moved, but they seemed to judge Ezra from head to toe. "You're right," he said. "I don't believe you. Now hand me the lightsaber."
"Lightsaber?" Ezra examined the blade in his hands. From some of the old spacers in the city, he'd heard tales of such weapons, laser swords wielded by mystical warriors during the Republic. "Isn't that the weapon of the Jedi?"
"Give it to me." Kanan held out his hand.
Ezra hesitated. Jedi Knights supposedly had special powers, like being able to move objects and read minds. Maybe Kanan was one who had survived—if the Jedi had even existed at all and weren't just stories invented to put little children to sleep.
But Kanan didn't appear to be a Jedi. He didn't use any special powers to call the hilt to him. He waited with his palm open for Ezra to give the saber to him.
Ezra took one more swing, then thumbed the button to deactivate the blade. He felt small again. Almost claustrophobic. With great reluctance, he placed the blade in Kanan's hand.
"And get out," Kanan said.
Ezra lowered his head, not meeting Hera's eyes or Chopper's photoreceptors as he exited. Once he was in the corridor, however, his mood lightened. He took the mysterious polygon out of his pocket and squeezed it tight.
**Sabine** poured herself a glass of bantha milk and leaned against the galley's countertop. The freighter's walls were thin, and she had heard the conversation in Kanan's cabin. This Ezra kid was quite a troublemaker, sneaking aboard and picking the cabin lock. She would've been more impressed if he had used a detonator on the door, but she was impressed nonetheless.
The kid entered the galley. He smiled when he saw her, but was too nervous to say anything, so she broke the ice. "Not too good at following directions, are you?"
That smile didn't leave him. "Not too much. You?"
"Never been my specialty." She took a sip of her milk.
Ezra stepped toward her. "Who are you people?" he asked. She gave him a raised eyebrow. "I mean, you're not thieves exactly."
"We're not exactly anything. We're a crew. A team." She paused, struck by a question she hadn't fully considered before. "In some ways, a family."
Ezra stopped, which was good, because if he came any closer, she might have to tell him to back off. She didn't like her personal space invaded.
"What happened to your real family?" he asked.
That question she thought about every single day. It was like a charge that never stopped exploding. "The Empire," she said. "What happened to yours?"
The kid looked away. So that was what he was hiding. Something tragic had happened to his family. She sympathized; she knew the pain. But he'd have to let it go, and soon. The galaxy was a mean place and had no room for sensitive types.
Zeb lumbered through the doorway, trailed by Chopper. He looked past the kid. "Kanan wants us in the common room."
Sabine nodded and finished her milk. Zeb addressed the droid, pointing a stubby finger at Ezra. "If he tries anything, sound the alarm—or shoot him."
Chopper quipped back with a query. Sabine was also curious what the droid could shoot, since the only "weapon" Chopper possessed was an electro-solder that could function as a Taser.
"Shush," Zeb said to the droid. "Just watch him."
The Lasat clomped off. Sabine put down her glass and went to follow. At the doorway, she looked back at Ezra, who now seemed more like a doe-eyed boy than a master thief.
"Sabine," she said. "My name's Sabine."
His mouth opened to say something, but he was at a loss for words. She could tell he liked her. A lot. That could be useful if she needed anything done in the future.
Heading toward the cabin room, she heard Chopper snicker at Ezra. Funny how a droid knew more about the rules of human attraction than a human kid did.
"We have a new mission," Kanan said.
Zeb hunkered in the _Ghost_ 's small common room next to Sabine. New missions were good things. He didn't enjoy being cooped up on the _Ghost_ , keeping an eye on thieving human children. His place was out on the front lines, bashing together the heads of stormtroopers, doing to the Empire what it had done to his people.
Kanan continued his mission brief. "Vizago acquired the flight plan for an Imperial transport ship full of Wookiees they took prisoner."
Hera stood right next to Kanan. "We don't know if they are the same Wookiees whom we were supposed to meet. Vizago couldn't confirm the prisoners' identities. But he had heard that most of the Wookiee prisoners were soldiers for the Old Republic."
This seemed to correlate with the name of the Wookiee soldier Wullffwarro that they had seen on the gunship. Yet for Zeb, it didn't matter who the Wookiee prisoners were. Wookiees were good people. Many had sacrificed their own lives to try to prevent the massacres on Lasan. "I owe those hairy beasts. They saved some of my people."
"Mine too," Hera said.
She had never opened up to Zeb about her past and he had never asked. But he suspected they shared a similar experience. The Empire encouraged slavery of her species, and he suspected that Wookiees had rescued Twi'leks close to her. Maybe they had even rescued her. He would not probe. If Hera wanted to reveal her background, she would in time.
"If we're going to save the Wookiees," Kanan went on, "we've got a tight window. They've been taken to an unknown slave labor camp. If we don't intercept this transport ship, we'll never find them. Now, I have a plan, but—"
Something clanked in the walls. Had a mynock chewed its way into the power cables? Zeb detested those parasites. They seemed to have no purpose in the galactic order other than to leech energy from starships and cause catastrophes.
Kanan pressed the button to open the supply closet. The Loth-rat Ezra tumbled out.
The others accused Zeb with stares. Kanan had assigned him to look after the boy, which wasn't fair at all. Zeb was an Honor Guard of Lasan, not a nanny. "I ordered Chopper to keep watch!"
Chopper trundled in, chirping excuses. He should've known better than to trust that droid. Chopper never obeyed instructions.
The boy scurried back toward the closet. Zeb grabbed him and held him in place. "Can we please get rid of him?" he growled to Kanan and Hera.
It was Sabine who spoke. "No. We can't."
Ezra stopped squirming and looked at her. Zeb recognized the kid's wide-eyed expression. _Karabast_. Teenage romance was the last thing they needed on the ship.
Sabine, mercifully, reciprocated none of it. She looked past Ezra to Kanan and Hera. "The kid knows too much."
The kid's body sank into Zeb's grip. Humans could be such weak creatures when it came to rejection. On Lasan, if a member of the opposite sex didn't like you, you showed them your talents until they did.
"We don't have time to take him home anyway," Hera said. "We need to move now. I'll keep an eye on him."
Kanan shot her an unsure glance. At least he agreed with Zeb that this Loth-rat shouldn't be infesting the ship. But when Kanan nodded in approval, Zeb let go of the kid. If Hera wanted to babysit, that was fine by him. He had Wookiees to save.
**Ezra** was getting used to hyperspace. His stomach didn't complain this time as the starlines streamed past the cockpit windows. Only his mind gave him trouble. It was hard to imagine that those starlines were planets, suns, or even other starships.
Hera sat across from him in the pilot's seat, checking and rechecking the navicomputer's coordinates. One miscalculation in their route could send them hurtling through a celestial body. In a breath, their lives would be over—and they'd never know.
Ezra didn't need to worry too much, Hera had assured him. She'd said that the hyperspace lanes were predominantly safe, mapped and tested by millions of pilots over millennia. What wasn't safe was what Hera and the _Ghost_ 's crew intended to do after emerging from hyperspace. There was no way to test their plan.
"You know, this whole mission thing is nuts," Ezra said. "I'm not against sticking it to the Empire, but there's no way I'd stick my neck out this far. Who does that?"
Hera pulled the lever to emerge from hyperspace. "We do."
The starlines vanished. In their place loomed a boxy starship with powerful engines. It wasn't big enough to store TIEs inside; instead, four of those fighters were racked onto the starship's underbelly. Ezra swallowed. His stomach started to panic.
Hera keyed the comm. "Imperial Transport six-five-one, this is _Starbird_ , coming inbound."
_Starbird_ was one of many fake identities the crew used for their freighter in situations like this. Only a privileged few—like lucky, lucky him—knew the ship's official designation, the _Ghost_. Ezra wished he'd never heard of it. He'd much rather be back in his tower, stripping power cores from holopads.
A pompous voice, sounding just like the one from the propaganda holopads, crackled over the radio. "State your business."
Hera didn't miss a beat. "Bounty. We captured an additional Wookiee prisoner and have transfer orders to place him with you."
"We have no such orders," said the voice over the comm.
Ezra tensed in his seat. Hera gave him a reassuring smile and continued to speak. "That's fine. We already got paid—by Governor Tarkin." She paused to let the governor's name sink in. "If you don't want the oversized tree-swinger, I'll jettison him here and let you explain to your superiors why the Empire has one less slave."
There wasn't a response. The transport ship floated there, guns forward, pilots visible in the attached TIE cockpits. Hera had her hand on the hyperspace lever, in case they needed to jump again. It would be a random jump—not safe at all.
"Permission to dock," came the voice. "Bay one."
Ezra sat back, relieved. Perhaps these people did know what they were doing.
A few hours earlier, Kanan had made Zeb a babysitter. Now he wanted Zeb to be something even more preposterous. "Your plan's not going to work, Kanan. I don't even look like a Wookiee," Zeb said.
"You think the Imperials have bothered to learn the difference?" Sabine asked. She took Zeb's bo-rifle and leaned it against the bulkhead.
Kanan slapped binders over Zeb's wrist. "Just act like one."
Easy for humans to say. After this was over, Zeb was going to lecture both of them on the many differences between Wookiees and Lasat.
He stood between the two at the _Ghost_ 's airlock and held his head high, as he would expect a Wookiee to do. They never quavered in fear, even when taken prisoner. Zeb's hands were another matter.
"Stop flexing your fingers," Kanan said. "You'll bring attention to the binders."
"They chafe. You cuffed them too tightly."
"Poor Zeb," Sabine said from behind her helmet.
"Poor _you_ if I lose circulation." If he couldn't move his hands, he wouldn't be able to bash any stormtrooper heads. They'd have to do all the dirty work themselves.
Kanan made a chopping gesture. Zeb quieted. The airlock door began to hiss open. He could see the white boots of two Imperial stormtroopers on the other side. He couldn't wait to see their white helmets.
When he did, both looked right at him. "That thing's not a Wookiee," said a trooper.
Zeb cursed silently. These weren't the usual, uneducated Lothal recruits.
"Haven't you seen a rare hairless Wookiee before?" Kanan said.
It stung Zeb that Kanan would use the kid's insult. Once they'd rescued the Wookiees, Zeb was definitely going to give Kanan a talking-to—with lots of angry growls and snarls for good measure.
For the moment, Zeb wailed, like he had heard many a Wookiee do. While he had the height and strength of a Wookiee, however, he did not have one's throat. His attempt sounded more like an Ugnaught's squeal than any sound a Wookiee would make.
The troopers exchanged looks and Zeb knew his career as an actor was over. "Oh, forget this." In one swing, he broke the binders and knocked both troopers through the airlock. They didn't get back up.
He grinned at Sabine and Kanan. "Told you they wouldn't buy it."
"You didn't exactly give them a chance to buy it," Sabine said.
"Couldn't help it. There's something about the feel of their helmets on my hands." Zeb rubbed his chafed wrists. "It's what a Wookiee would do anyways."
Kanan stepped through the airlock into the enemy transport. "Okay, you both know the plan. Move out."
Chopper wheeled out from the _Ghost_ to join Sabine. The two went across the docking bay into one corridor of the transport. Zeb slung his bo-rifle over his back and hurried after Kanan down another.
Ezra leaned forward in his cockpit seat, listening to the comm. As crazy as this plan sounded, it was actually working. Kanan had even used one of his jokes.
"No troopers," Kanan reported. "Security's soft—"
A burst of static cut him off. Hera flipped switches. "Spectre-1, come in. Spectre-4? Spectre-5?" she commed, using their code names. "Comm's down."
Ezra watched as she made minute adjustments to the controls. She seemed to know every knob and dial in the cockpit, and there were hundreds of them. He'd been so wrong to think that if he could handle a speeder bike, he could pilot a starship.
"No, not down," Hera said. She gave up on her adjustments. "Jammed."
Ezra straightened in his seat, his instincts on edge. He stared out into space. "Something's coming."
Hera looked up from the communications console. There was nothing out the canopy viewport except the spread of stars.
"Ezra, I don't see—"
Emerging from hyperspace in the blink of an eye was the massive triangular vessel that had thundered over Ezra's tower. An Imperial Star Destroyer.
This was a trap. It had been all along. It was why Kanan and the others had boarded the transport so easily. The Empire didn't want the crew of the _Ghost_ to get away this time. They wanted them in their grasp, in their ship.
Everybody was going to die.
Ezra sat, frozen in place. Hera knew that she had to act fast to save her crew. "Ezra! You need to board the transport and warn them!"
"What?" Ezra looked at her. "You want _me_? Why don't you do it?"
She was back at her console, turning all matter of dials and switches. "I need to be ready to take off, or none of us stands a chance."
"No," he said. "No way. Why would I risk my life for a bunch of strangers?"
She frowned at him, apparently insulted. "Because Kanan risked his for you."
She was right. Kanan had saved him from the laser cannons of a TIE fighter. But Kanan was also why he was stuck on this wretched ship in the first place. If he and Zeb hadn't chased after him on the speeder bikes—if they had just let Ezra have his crates—then his life wouldn't be in danger. Neither, probably, would theirs.
Ezra looked away from her. But her reflection regarded him in the cockpit's transparisteel. "If all you do is fight for your own life, then your life's worth nothing," she said.
Her words gave him pause. He pictured Kanan and Zeb running down the transport's corridors, fleeing stormtroopers. He pictured the antique Chopper blowing a gasket, unable to move any faster. He pictured a hidden sniper scoping Sabine and pressing the trigger.
"They need you, Ezra," Hera said.
No one had ever needed Ezra. More to the point, he had never needed anyone else. He had prided himself on that. He was a lone operator, his own master. He listened to no one but himself and he did nothing he didn't want to do. If someone helped him out, that was their prerogative. If they didn't, he held no grudges. The universe hadn't done him any favors, and he preferred to do the same in return.
"They need you _right now_ ," Hera pleaded.
A shudder ran through the ship, rattling bolts and components. The Star Destroyer had locked its tractor beam on the transport and began to pull it and the docked _Ghost_ into the gaping maw of the Destroyer's lower hangar.
Hera's reflection faded against the canopy's transparisteel. Now all Ezra saw was his own.
**Agent Kallus** had Captain Zataire assemble a platoon of his best stormtroopers in the _Lawbringer_ 's deployment bay. Kallus could trust these men. They weren't Lothal enlistees wearing the armor for the first time. To serve as a stormtrooper aboard an Imperial Star Destroyer, one had to be trained at an elite facility and pass a battery of extreme tests in a variety of environments. These stormtroopers had earned their place on the ship. They were the Empire's finest.
Kallus addressed them with three words: "Prepare to board." Their training had primed them for the rest.
He put on his helmet, took a blaster rifle from the weapons rack, and led the way toward the docking tube.
Hera marveled at the speed at which the Star Destroyer dispatched a wing of TIE fighters and hooked its docking tube to the Imperial transport. Such efficiency demonstrated just how the Empire had subjugated the entire galaxy in a short span of years. Toppling it would require more than the crew of the _Ghost_. They'd need a military force that could move even quicker than the Imperials.
This would be a long fight. And it might not be a fight they would be in much longer if Ezra didn't help them out.
She grabbed the boy's arm. "Listen. Our crew boarded that transport to selflessly rescue Imperial prisoners. They have no idea they walked into a trap. No idea what's coming." When he still didn't respond, she raised her voice. "You need to go warn them, Ezra— _now_!"
"It's too late for them. We should run while we still—"
"You don't mean that." She turned him around in his seat, made him look at her. To his credit, he did not flinch, did not look away.
"I do," Ezra said. "I swear I do."
She let go of his arm. This boy was more stubborn than Kanan. Yet he was also scared. He was, after all, just a kid. She shouldn't ask such things of him.
"Which is why I can't believe I'm doing this." Ezra stood up, tightened the straps of his backpack, and exited the cockpit.
Hera could believe it. Ezra might be stubborn like Kanan, but he had the same heart, too. He was a rebel, deep down. One of them.
Kallus and his platoon stepped out of the _Lawbringer_ 's boarding tube into the transport's docking bay. The captain of the transport, whose name Kallus had not bothered to learn, came forward. "Welcome aboard, Agent Kallus."
Kallus kept walking, as did his platoon, crossing the bay. The transport captain hastened to follow. "The rebels are headed for the brig, where quite the surprise awaits. I've positioned a squadron of stormtroopers there to capture them."
Kallus didn't even grant the man a look. The captain wanted a commendation he'd never get. Under the Empire, you did your job and kept your command. If you didn't, you faced the consequences of failure.
Kallus already knew the verdict that would strike this captain. It was why he had come aboard with his own men. It was why he hadn't learned the man's name.
He marched into the main corridor of the transport, leaving its captain in stunned silence.
Kanan took the lead as he and Zeb hurried down the transport's sleek gray corridors. So far they hadn't met any resistance except for a mouse droid Zeb squashed under his foot. Kanan veered around a corner and surveyed the transport's brig. A heavily reinforced blast door was set into the wall, but like the rest of the ship, the area was empty. "No guards on the door."
Zeb came forward and took up the position opposite Kanan at the blast door. "No worries. I'm sure there's one or two bucket-heads worth punching on the other side."
Kanan planted a detonator on the door. He hoped Zeb was wrong. For once it would be nice to slip in and out without engaging in combat.
"You sure the Wookiees are behind that door?" Zeb asked.
Kanan entered the activation sequence for the charge. "Where else would they be? There's only one brig on this transport."
"Guess you're right. They'd make holes through anything else." Zeb chuckled. "Nothing better than rescuing cooped-up Wookiees or what, eh?"
"It's a trap! It's a trap!" screamed a young voice. Kanan turned to see Ezra running down the corridor toward them.
_"Karabast,"_ Zeb snarled. "The kid's blowing another op!"
Ezra skidded to a stop. "It's not an op—it's a trap! Hera sent me here to warn you!"
The blast door started to open its multiple layers. White-armored bodies could be seen behind it.
"Run!" Ezra yelled.
Kanan and Zeb did just that. Ezra lifted his arm and aimed his slingshot. In its pocket force field formed an energy ball, which Ezra fired at the detonator.
The door exploded, blowing stormtroopers off their feet. Smoke billowed down the corridor. The three ran. Kanan tried his comm. There was no signal.
"We need to warn Sabine and Chopper, but they've jammed the comm," Ezra said.
"They'll follow the plan. It'll be fine," Kanan said. He wished he believed it. But that was what leaders did: kept the situation calm.
The boy was having none of it. "Yeah, 'cause the plan's gone just great so far."
As they turned the corner, a platoon of stormtroopers blocked their way, led by a man all in gray. Kanan recognized the uniform and rank insignia.
The man was an agent of the Imperial Security Bureau.
Zeb cursed under his breath. Kanan did the same. _Karabast_ , indeed.
The transport's systems room was empty, like the corridors Sabine and Chopper had taken to get there. Sabine wasn't complaining, but it seemed odd they hadn't encountered even one Imperial. Kanan had instructed her to use the comm only if there was an emergency. Since she hadn't heard from Hera or Kanan, everything must be okay on their end.
She went to one control station and motioned for Chopper to take the other. The droid complied, whining as he did so.
"Chopper, stop grumbling and work on that gravity generator." The droid always griped when communicating with Imperial machines. He found their logic cold and crude. Then again, Chopper whined about every machine he accessed, including the _Ghost_.
The computer security was lax and she hacked in with ease. The Imperial engineers clearly hadn't entertained the possibility that anyone would penetrate the transport's outer defenses and board the ship.
Her comlink made the faintest click. That was Kanan's signal. Not good. She started working faster with the gravity controls. "Bypass any stealth commands, Chopper. There's been an emergency."
Emergency or not, the droid grumbled all the same.
The agent and stormtroopers brought their weapons to bear. This only made Kanan run faster toward them, drawing his own blaster. "Don't stop," he shouted to Zeb and the boy.
They had only seconds before the enemies' guns were triggered. Seconds before they were dead. Unless—
Kanan didn't complete his stride. His front foot found not the floor, but a cushion of air. "Push off now!" he said, and leapt forward.
Sabine had heard his comlink click. She had turned off the artificial gravity right when needed. He flew down the corridor while the troopers floated off their feet, losing both their bearings and their aim.
Kanan fired his blaster, trying to clear a path. Only the agent managed to fire back, but he missed as Kanan barreled through the jumble of troopers, not loosening his finger from his trigger. Soaring behind him, Zeb hurled troopers right and left. Ezra held on to Zeb's foot, hitching a ride. Kanan was sure the Lasat appreciated that.
They sailed through the platoon, then down the corridor toward the docking bay. Kanan glanced back, seeing the ISB agent had turned around to fly after them. Ezra, meanwhile, had abandoned Zeb's foot to swim by himself.
"You okay, kid?" Kanan asked.
"You kidding?" Ezra said. He bounced off the floor and walls, tumbling over himself and gaining speed. Even in dire situations, there was nothing like a kid in micro-gravity.
Sabine had been able to override the artificial gravity for a maximum of two minutes. It had taken her a quarter of that time to place the detonators on the control stations. But the trip back to the docking bay wouldn't take as long as it had on foot. Conveniently, she had Chopper, who was equipped for zero g.
She grabbed one of his legs. The droid ignited his booster rocket and they jetted out of the systems room. They met no resistance in the corridors and made it into the docking bay with seconds to spare.
"Five, four," she said, checking her helmet chrono, "get ready—two, one—now!"
Sabine braced herself as the artificial gravity kicked back in. She landed lightly on the docking bay floor before the airlock, while Chopper remained hovering on his booster. But the two stormtroopers whom Zeb had given a head-knocking went from floating in the air to dropping to the floor with a thud. Sabine suspected that would keep them unconscious until she and the others were parsecs away from there.
If the others had made it out of the brig, that was.
The sight of a purple Mandalorian helmet in the docking bay gave Ezra reason to cheer. They were all going to get out of this trap alive. And maybe, just maybe, Sabine would notice that he was the one who had saved them. Her congratulations would be a nice reward.
He sensed a fluctuation in the gravity and realized his low-gravity experience was about to end. He brought his feet underneath him and straightened.
"Now!" Kanan said.
Ezra landed with both feet on the corridor floor and didn't miss a step as he ran with Kanan and Zeb into the docking bay. Sabine was there, as was Chopper, on his rocket. Sabine craned her neck to look behind them. "Where are the Wookiees?"
"No Wookiees," Kanan said. "Sabine, you need to man the _Ghost_ 's nose gun. Chop, go tell Hera to take off."
"Right," Sabine said. She hurried through the airlock into the _Ghost_ without a word to Ezra. Chopper and Kanan went after her.
Disappointment slowed Ezra. Had Sabine even noticed he was there?
He must've been blocking Zeb, because the bruiser pushed him aside on his way to the airlock.
Ezra geared up to give the Lasat a shove in return. But he never made it into the airlock. Someone grabbed him from behind and yanked him backward.
"Let go!" Ezra cried. His abductor was none other than the Imperial officer in gray.
Zeb turned around in the airlock, drawing his bo-rifle. "Kid, get out of the way!"
"I'm trying!" Ezra said. He struggled, but the Imperial officer wrapped one arm around his chest, using Ezra as a shield. With the blaster in his other hand, the officer fired at Zeb, as did the host of stormtroopers rushing into the docking bay.
The blasts pushed Zeb back into the airlock. Ezra knew there was nothing he could do. Even this brawny bruiser he'd seen fling stormtroopers around like toy soldiers couldn't beat such odds.
"Sorry, kid." Zeb looked at Ezra with regret. "You did good."
The airlock slammed shut. The officer holding Ezra laughed. Ezra stopped resisting.
He shouldn't have stuck out his neck for those strangers. He had made the wrong choice, and now he was going to pay dearly for it.
**Hera** had already warmed the engines when Chopper wheeled into the cockpit and beeped the go-ahead. She initiated the launch procedure immediately. "Airlock's shut. Detaching from the transport. We're out of here."
She held firm on the flight yoke, steadying the _Ghost_ as it detached from the Imperial transport. "Chop, make yourself useful and jam their tractor beam."
Chopper retorted in annoyance but still plugged himself into the jammer. Being pulled back to the transport would eliminate their momentum and make them a ripe target for the Star Destroyer's turbolasers. They'd be space junk if they didn't make a quick getaway.
The transport broadcast a message from its captain. "Attention, rebel ship. Surrender or be destroyed. This is your first and last warning."
"Blow it out your exhaust vent. Literally," Hera replied. She channeled all available energies—including those that powered the turrets—into the engines. Shooting back wouldn't gain them much. Their lasers would be like pinpricks to a Star Destroyer.
What Sabine had planted would deliver a much bigger punch.
"Sabine, it's time," Hera said into the intercom. She poured on thrust, trying to get as much distance from the transport as possible before Sabine keyed her remote.
A moment later, the underside of the transport exploded.
Such damage normally would have been contained, except for Sabine's expert placement of the charges. The explosion in turn ignited a chain reaction across the transport. The portside hull blew open. The fuel line erupted. One engine brightened, then fizzled, while the boarding tube became a whoosh of flame that spread to the docked Star Destroyer. It provided Hera with the cover she needed to dive the _Ghost_ under the Star Destroyer's nose and speed past.
"I can't see it from here," Sabine commed from the nose turret. "How'd it look?"
Hera had witnessed the explosion only on scopes. Kanan had had the best view, since he occupied the dorsal turret. "Gorgeous, Sabine. As always," he said over the intercom.
Hera smiled. They had the best crew in the galaxy. With their young new member, they might even be stronger.
She pulled the lever to send them into hyperspace.
Weighed down by his backpack, Ezra huddled in the cold, empty cell of the Star Destroyer, brooding over his foolishness. How had he ever let Hera persuade him? He knew what it took to stay alive in this harsh galaxy. _Remain uninvolved. Rely on yourself and yourself only. Don't risk your life for others, because they won't do the same for you._ They were simple rules to live by. Ones that had not failed him in the past.
He had failed them. And in doing so, he had betrayed himself.
The cell door slid open. Two stormtroopers flanked the Imperial officer in gray. He had straw-colored hair and thick muttonchops that his helmet had previously covered. "I am Agent Kallus of the Imperial Security Bureau. And you are?"
Ezra smirked in defiance. "Jabba the Hutt." He might be in there because of his own stupidity, but he sure wasn't going to give the Empire any satisfaction.
Kallus's face remained impassive, as if he had never laughed at a joke in his life. Ezra had a sneaking suspicion that might indeed be true. Comedy was not the answer if there was any chance for him to get out of there. So he told the truth.
"Look," Ezra said. "I just met those guys today. I don't know anything."
The truth made no difference to Kallus. "You're not here for what you know, 'Jabba.' You're here to be used as bait upon our return to Lothal."
"Bait? You seriously think—" Ezra paused, laughed. "Wow, you're about as bright as a binary droid. They're not going to come for me. People don't do that."
Kallus said nothing else; he just looked down at Ezra. The man's stare was like a tractor beam: Ezra couldn't pull away.
Kallus brushed lint off Ezra's shoulder. "Search him. Then secure him here," he told the stormtroopers. He spun on a heel and walked out.
The troopers came forward. One yanked Ezra's backpack off his shoulders while the other grabbed Ezra's right arm and stripped off his slingshot. Ezra struggled in their grip. "Hey, get off me! Let me go, you goons!"
The contents of Ezra's backpack spilled out onto the floor. The first trooper scooped up the wrench, flashlight, and astromech droid arm and stashed them inside the pack. The second trooper pushed Ezra back onto the seating square. Then they strode out of the cell with his items, shutting the door behind them. Ezra was left alone once again.
He scowled. Not only did he feel foolish; he felt insulted. After all he had done, the Empire only considered him bait, just as the strangers saw him as a useful nuisance. "You need to go warn them, Ezra," he said, mimicking Hera's voice.
What had he been thinking?
He winced. Something jabbed his tailbone. He reached under him to find the polygonal object he had taken from Kanan's drawer. It must have fallen out of his backpack yet was too small for the troopers to notice.
"And, of course, the only thing I managed to hold on to is this worthless piece of..." His words stopped as his thoughts took another direction. The transparent object might be worthless, but it was also fascinating to behold. It weighed almost nothing, and each of its many sides was perfectly smooth, without cracks or creases.
Ezra sensed something was inside.
He pressed, pushed, and pried at the sides. That got him nowhere. He would need a blaster or a drill to pierce through that outer shell, and even with those tools, he'd probably destroy whatever was in there.
He tossed the stupid thing across the room. It bounced off a wall and rolled into a corner. Maybe one of the stormtroopers would trip on it. Then it would be of some use.
Ezra closed his eyes and lowered his head. He wasn't even angry anymore; he was exhausted. He cleared his mind, just focused on his breath. That always helped him relax. Perhaps sleep could take him out of this nightmare. He could wake up and find himself lying in the patch of grass around his tower, where green daisies grew.
The green daisies of Lothal were dazzling flowers. The presence of other forms of life made them bloom. The follicles on their stems were so sensitive that they could detect the breath or heat of a nearby organism. They would blossom for anyone who spent time to observe them.
He imagined peering at a daisy in the grass, watching it open its petals, slowly, like a child flexing its fingers for the first time. The spread of petals revealed a radiant center that shone emerald-like in the morning sun. The sight of this tiny wonder always revitalized him, no matter how hard his day had been.
"This is Master Obi-Wan Kenobi..." said a stoic voice.
As marvelous as the daisies were, they didn't have the ability to speak. Ezra lifted his head and opened his eyes.
The polygonal object lay open on the ground, as if its sides had been petals of a flower. From its center projected a miniature ghost of a bearded man in robes. His was the stoic voice.
"I regret to report that both our Order and the Republic have fallen, with the dark shadow of the Empire rising to take their place."
Ezra stared at the hologram. Though the robed man seemed somber and weary, as if he had just suffered a great loss, his voice and stature carried grace and nobility.
"This message is a warning—and a reminder—for any surviving Jedi. Trust in the Force...."
The Force. What was the man talking about? Was this Obi-Wan Kenobi a Jedi? If so, might Kanan really be one, too? Or perhaps Kanan had killed Obi-Wan and taken his lightsaber. That seemed more logical based on how Kanan had treated Ezra, leaving him to rot in a Star Destroyer cell. And why had this object opened in the first place, when it wouldn't before? So many questions. They were overwhelming. Questions to which he'd never get the answers.
The Force. His mind settled on those two words. He didn't know why. It seemed to be another thing he didn't understand. Another secret.
Yet deep down, Ezra sensed that this secret was also the answer.
**A quick jump** into hyperspace could cause even veteran pilots to sweat buckets in their flight suits. Hera had made so many of those jumps recently she didn't break a sweat. Pulling the lever on short notice was beginning to feel like second nature. She'd have to make sure she didn't get too comfortable. Yet for the moment, she leaned back in her chair and watched the starlines. The Empire might control everything from the Core to the Outer Rim, but hyperspace was out of its grasp.
Her moment of peace ended when Kanan and a helmet-less Sabine entered the cockpit and dropped into seats beside Hera. "The whole thing was a setup," Kanan said.
"You think Vizago was in on it?" Sabine asked.
Hera had to nip any misgivings Kanan had about Vizago in the bud. "Vizago would sell his mother to Jawas for a couple credits, yes. But we're a source of income for him. Odds are he didn't know."
Zeb also came in and took a seat, giving Hera the perfect opportunity to change the subject. "The kid did all right," she said.
"He did okay," Kanan said. He glanced down the ship's main corridor, then at Zeb. "Where is he?"
Hera also peeked down the corridor. It was empty.
"I, uh, thought he was with you," Zeb said.
Sabine swiveled in her seat. "What'd you do to him?"
Zeb shifted about, avoiding eye contact. "I didn't do anything to him," he mumbled. "That ISB agent grabbed him."
"What?" Hera and Kanan said in unison. No one had mentioned anything about an agent of the Imperial Security Bureau.
"The kid got grabbed, okay? He didn't make it off the transport," Zeb said.
"Garazeb Orrelios!" Hera said. She couldn't believe he'd leave the kid there. What had he been thinking?
"Oh, come on, we were dumping him after the mission anyway. This saves us fuel," Zeb said. His grin soon melted into a guilty frown. "They'll go easy on him. He's just a kid."
Appalled, Hera exchanged looks with Kanan. "We have to go back."
Zeb's eyes bugged out. "No—no, no! No way! You cannot be serious!"
She was more than serious. She started putting new coordinates into the navicomputer. "It's our fault he was there."
"Come on, Hera, we just met this kid. We are not going back for him," Zeb said. He turned to Sabine, pleading for backup.
Sabine looked away but whispered her agreement. "They'll be waiting for us. We can't save him."
Chopper, who was plugged into the corner and had been quiet the entire time, beeped a positive. Zeb spun on the droid. "What? What did that little dumpster say?"
Hera would have to give the droid a lubricant bath sometime soon. "He voted with me," she said. "That's two against two. Kanan, you have the deciding vote."
Kanan looked past her, out into hyperspace.
Ezra picked up the polygonal object. Its sides had closed after projecting the hologram and it seemed no different than before. He juggled it from hand to hand and rolled it around in his palms. If sold to the right person or organization, it would probably fetch him a shipload of credits.
Regardless, he was done feeling sorry for himself. So what if he'd made a wrong choice? People goofed up every day. He wasn't going to wilt and surrender because of it. He was built of sterner, smarter stuff than that. Stuff the Empire couldn't lock up. This Agent Kallus from the Imperial Stiff-faced Bureau didn't know whom he was dealing with.
Ezra Bridger refused to be anyone's bait.
Mulling over his options to break out of this joint, he took the best one available. He went up to the small set of stairs near the door and began to heckle the stormtroopers.
This wasn't your normal ribbing at the local podrace. Ezra used every joke in the book, poking fun at how the stormtroopers sounded and looked like clones, saying scout troopers had niftier armor than they did, even questioning their undying loyalty to the Empire. And he didn't stop. He repeated the same jokes, over and over, not trying in the slightest to be funny, only to get under their armor and annoy them to no end.
"You bucket-heads are going to be sorry when my uncle the Emperor finds out you're keeping me here against my will. I guarantee he'll make a personal example of you," he said. Ezra made choking sounds and added a few coughs for effect. Maybe the stormtroopers would think he was dying. That wouldn't be good for their bait.
The door opened. Ezra crouched beside the stairs as the two guards rushed down into the cell. He leapt up the steps and was already out when they turned around.
"Bye, guys." He shut the door, pressing the lock button.
His first destination was the storeroom, which he found just across from the holding cells. It held more—much more—than just his backpack and slingshot. Imperial helmets of all varieties packed the room. He'd hit the mother lode.
As much as he wanted to collect the helmets he didn't have, his priority was to find an escape pod before those troopers radioed for help. He stashed the polygon in his backpack, strapped the pack onto his shoulders, and remounted the slingshot to his arm. As he turned to leave, his eyes fell on a smaller helmet, almost his size, made for cadets.
An idea struck him. He grabbed the cadet's helmet and put it on. It didn't possess all the advanced tech of a stormtrooper's helmet, but it had a simple radio tuner that automatically activated. He listened in on the comm chatter from the bridge.
"The delay was insignificant," an officer said. "The transport ship Agent Kallus diverted will dock on Kessel within two hours. The Wookiees will be offloaded to work spice mine K-seventy-seven."
_Interesting,_ Ezra mused. The Wookiees had been on that transport ship. They must've been locked in another hold.
"This is Stormtrooper L-S-zero-zero-five," radioed another voice. "Reporting for Agent Kallus."
"Kallus here." The man sounded even colder on the comm.
"Sir, th-the prisoner's gone," stammered LS-005.
"What?" snarled Kallus, a sliver of anger cracking his ice.
Ezra bit his lip. He couldn't leave the way he'd come. The corridor outside would be crawling with troopers soon. Spotting a ceiling vent, he began to climb up a stack of helmets.
"I knew the boy would act as bait, but I never dreamed the rebels would be foolish enough to attack a Star Destroyer. How did they get aboard?" Kallus asked over the comm.
Ezra stretched on his toes to reach the ventilation duct. He was listening but not paying much attention as he pried loose the vent's grille.
"Sir, the rebels didn't free him," radioed LS-005. "He, uh—"
"Agent Kallus!" shouted the officer from before. "There's a security breach in the lower hangar!"
Ezra winced from the volume of the man's voice, nearly falling off the stack of helmets. He grabbed the edge of the vent and pulled himself into the duct.
"I don't know how," the officer continued over the comm channel, "but the rebel ship approached without alerting our sensors."
Ezra banged his head on the top of the air duct in shock. What rebel ship? The _Ghost_ was the only one he knew. He knelt there for a moment. Had the strangers come back for him?
"Order all stormtroopers to converge on the lower hangar," Kallus said. "I'll meet them there."
Kallus's sudden change of attitude meant this was a serious breach. Maybe he was wrong about these so-called rebels. Maybe he had made the right choice in rescuing them, and now they were trying to do the same for him.
Right or wrong, he had to do something. The rebels were his best way off the Star Destroyer.
Ezra cleared his throat. He tapped the helmet to turn on the filter mic, then used his Imperial voice. "This is trooper LS-one-two-three, reporting intruders in the upper hangar. Sir, I believe the lower hangar is a diversion."
"Maybe, maybe not," Kallus replied. He didn't question Ezra's identification but also didn't fall for the entire trap. "Squads five through eight, divert to upper hangar. The rest converge as ordered."
Ezra sped up his crawl. Four squads were better than eight, at least. Every little bit helped.
**With all its** systems operating in stealth mode, the _Ghost_ slipped into the _Lawbringer_ 's lower hangar unnoticed. Kanan knew their concealment wouldn't last. The hangar's security cameras would spot the unidentified ship—if they hadn't already—and the alarms would go off. He just hoped they could accomplish the rescue before every stormtrooper on the Star Destroyer arrived.
Kanan ran out as soon as the _Ghost_ 's ramp hit the hangar floor, Sabine and Zeb with him. "Find Ezra. I'll be ready," Hera said from the hatchway.
Kanan surveyed the hangar. It was empty save for a bevy of cargo crates from a recent resupply. "Hold this bay until we get back," he told Zeb.
Sabine, with a blaster in one hand and a canister in the other, turned her helmet to the Lasat. "And this time, try not to leave until everyone's back aboard."
Zeb huffed. "That was not my fault!"
"Well, that's debatable," said a filtered voice above them. An Imperial cadet leapt down in front of Zeb.
The Lasat didn't blink. He just punched, smacking his fist into the cadet's helmet. The cadet was knocked backward across the hangar.
Kanan held back from firing. This Imperial seemed too short, even for a cadet.
The cadet stood and removed his helmet. "First you ditch me, then you hit me?" Ezra asked.
"How was I supposed to know it was you? You were wearing a bucket!" Zeb said.
The lightning of blaster fire preceded the thunder of four squads of stormtroopers rushing all at once into the hangar. The ISB agent led the attack, aiming his weapon at Ezra.
Ezra tossed his helmet at the agent, then joined Kanan and the others in a full-out sprint toward the _Ghost_. "Spectre-1 to _Ghost_ , we're leaving," Kanan said into his comlink.
In the hatch, Hera laid down suppressing fire while they ran up the ramp. Ezra tried to aim his slingshot, but Zeb shoved the boy into the ship. "Oh, no, this time you board first."
Hera hurried toward the cockpit, which Chopper managed in her place. Once Sabine had made it through the hatch, Kanan commed the order: " _Ghost_ , raise the ramp—and get us out of here!"
All around Kallus stormtroopers fell, taken out by shots. He ducked behind a crate for cover. The rebels firing from the freighter's hatch had the advantage of an upper position as the ship started to rise.
"Aim for the shield generator and engines," Kallus directed his troops. "Do not let them escape!"
His eye caught a pattern on the floor. Painted in orange was what appeared to be the outline of a bird lifting its head and spreading its wings. It reminded him of the legendary starbird, which perished in flames only to rise from its own ashes.
Why would the rebels have wasted time painting this? He reached down and touched the image, smearing it. He sniffed the fresh paint on his finger.
This wasn't just paint. This was sabotage.
"Take cover!" he shouted, and dove as far away from the starbird as he could. The next moment, the starbird exploded.
Troopers and crates went flying backward from the blast, then forward toward the giant hole that had been ripped open in the hangar floor. A vortex sucked everything not clamped down out into the vacuum of space.
Feet dangling, grip slipping, Kallus clung to the edge of that hole as stormtroopers tumbled past him by the dozens. With all he could muster from his lungs, he called out to those troopers who hung on to handholds on the wall. "Turn on the shield!"
Whether the troopers heard him or not, they deciphered what he wanted. One stretched out and flipped a switch on a console.
Just as Kallus's grip loosened, an energy bubble shimmered around the hole. It sealed the breach and prevented him from falling.
Kallus crawled back onto the deck, watching the freighter zoom out of the hangar. He had to give his enemies a modicum of respect. They were more than your run-of-the-mill rabble-rousers. They were the real thing—fearless, daring rebels, willing to do anything and everything to achieve their mutinous ends.
As he rose, a stormtrooper approached, carrying the cadet's helmet that the boy had thrown at him. "Sir, one of the rebels was using this helmet. The transmitter was on."
Kallus was not a man who smiled. Yet when he took the helmet and stared into its black visor, his heartbeat steadied, and he felt the briefest pang of joy.
He knew where these rebels were heading.
And once he caught them, he was going to relish demonstrating why no one rebelled against the Empire.
Three quick hyperspace jumps in less than one Rylothian day—Hera supposed that must be a personal record. Settling back in her chair, she hoped to let that record stand for a long, long time. The perspiration that tickled her lekku was well earned.
The boy walked into the cockpit, a little sweaty himself. "Welcome aboard," she said.
"Thanks," Ezra mumbled. He hesitated a moment. Embarrassment blushed his cheeks. He spoke louder. "I mean, thank you. I really didn't think you'd come back for me."
She sat up and returned to her console. "I'll get you home now. I'm sure your parents must be worried sick."
The blush on his cheeks darkened. "I don't have parents. And you've got somewhere else to be."
Kanan, Sabine, and Chopper came in behind the boy. The cockpit was swiftly overtaking the common room as their meeting area. Hera would have to say something about this. She couldn't reach over bodies to adjust the controls.
"I know where they're taking the Wookiees," Ezra said.
Hera spun in her chair. "Where?"
"Have you heard of the spice mines of Kessel?"
More perspiration welled on Hera's lekku. This time it didn't tickle. Of all the Imperial labor camps in the galaxy, Kessel was the worst.
"Slaves sent there last a few months, maybe a year, tops," Sabine said.
"And for Wookiees born in the forest," Hera said, "the mines are a death sentence."
"Then I guess we better go save them," Ezra said matter-of-factly.
Sabine gawked at the boy. Chopper telescoped his prime photoreceptor. Even Kanan's impassive demeanor was disturbed.
"'We'?" Sabine's eyes were as wide as ryll nuts.
"Come this far, might as well finish the job," Ezra said.
The kid was right, Hera knew. This was no time to rest on their laurels. They would probably never have another chance to rescue the Wookiees. She started to reprogram the navicomputer. "Setting course for Kessel."
Out of the corner of her eye, she noticed Kanan and Ezra exchange looks. Perhaps more than just the Wookiees could be saved on Kessel. Perhaps Ezra could help break Kanan out of the shell he'd hidden himself inside for many years.
**Little Kitwarr** wanted to go home. Wherever he and his clan mates had been taken was the opposite of Kashyyyk. The forest here was made of metal, with piping for branches and smokestacks for trees. Men in white armor called stormtroopers pushed him and the other Wookiees along a grated walkway, which loomed over a dark, forbidding pit. It belched forth horrid, stinky clouds that made him wheeze. A yellow haze sickened the sky where there shone no sun, and ashes floated where birds should have flown. What little patches of land he saw were parched and cracked. Nothing grew.
Kitwarr began to worry that he'd never climb another tree in his life.
He cried out to his father, Wullffwarro, who had a group of the stormtroopers around him. The great Wookiee howled that Kitwarr shouldn't lose hope. He'd find a way out of this, he promised.
"Keep moving," one man barked in Basic to Kitwarr's father. The others raised their guns at Kitwarr. He knew his father could throw all of them into the pit, but with binders on his wrists, Wullffwarro was powerless to do anything. Wullffwarro let out a mournful wail, then plodded forward down the walkway.
Kitwarr whimpered. It was the first time he'd ever doubted his father.
Standing with Zeb near the cargo bay hatch, Ezra checked his slingshot for the umpteenth time. Like before, it was ready for action. He, on the other hand, wasn't.
Once again, Ezra had made the dumbest mistake imaginable. He should never have suggested this mission. Even if the _Ghost_ 's security countermeasures had slipped past Kessel's orbital safeguards, their small band could not match the firepower of a highly fortified Imperial detention facility. This was a suicide mission, plain and simple.
"Try not to get dead," Zeb said.
Ezra glared at the Lasat. What was the point of rubbing it in? He was only making matters worse.
Zeb snarled his version of a smile. "Don't want to carry your body out."
It took Ezra a moment to realize the big lug was joking. Ezra faked a smile back. If this was what soldiers called gallows humor, he didn't much care for it.
Sabine, ready for battle in her helmet and armor, and Kanan entered the cargo bay. "There's no place to land. You're going to have to jump onto the platform," Kanan said.
Ezra wished that this, too, was a joke. It wasn't. Kanan popped the hatch and the ramp lowered into a blizzard of blaster bolts. On the platform meters below, a stormtrooper squad had spotted the _Ghost_ and commenced firing at it.
Kanan leapt out first, triggering his blaster as he went, followed by Sabine, Zeb, and then, with a swallow, Ezra.
The _Ghost_ let loose its cannon, dispersing the troopers and allowing the jumpers to land safely on the platform. Ezra got off a few shots of his slingshot, then joined the others behind some mining crates. From there he saw the Wookiees being held prisoner beyond a line of shipping containers.
Kanan gave him the nod to proceed. There was no going back now. They were relying on Ezra to do his job, as he was relying on them to provide him cover from being blasted. The mission's success hinged on trusting each other—something that ran counter to what had kept Ezra alive in the past. Trusting strangers had only gotten him into trouble. Yet Sabine, Kanan, Hera, Chopper—even Zeb—weren't exactly strangers anymore. And there was no way a lone operator could accomplish this mission. It required a team, one in which he played a vital part.
Ezra ran toward the Wookiees.
Gunfire opened up all around him. He ducked and rolled, weaving between crates for cover, dodging blaster bolts by millimeters. Strangely enough, he wasn't afraid. He seemed to know the safest path instinctively and focused his gaze not on his surroundings but on the Wookiees who needed rescue. A giant silverback, whom Ezra had learned during the mission briefing was Wullffwarro, tried to break his binders while one cub looked at Ezra with bright, hopeful eyes.
It was that hope that pushed Ezra further. He snaked through mining equipment, darted across an empty stretch, then clambered up onto a shipping container and leapfrogged down the line. Blaster fire whizzed past him as he dropped in front of Wullffwarro.
The silverback Wookiee towered over him and growled, not in the least bit friendly. "Hey, hey, I'm here to help," Ezra said.
He pulled out his astromech arm and inserted its manipulator into Wullffwarro's binders. The manipulator fine-tuned itself and transmitted the proper codes to unlock the binders. Wullffwarro roared his pleasure once freed.
Ezra continued through the rest of the group, freeing Wookiee after Wookiee. Their cries of gratitude swiftly shifted to cries of battle as they stampeded the stormtroopers from behind.
The Imperials proved no match for berserker Wookiees, whose punches shattered armor and sent troopers screaming over the platform. The ones who didn't meet a Wookiee's fist were hit by the rebels' blaster fire instead. Soon the stormtrooper squad was no more and Ezra was running with the Wookiees toward the _Ghost_ , hovering over the edge of the platform. The rebels joined them. Kanan gave Ezra a nod of approval.
Against all odds, this mission was going to succeed. Ezra had been wrong in thinking otherwise. Teamwork could do wonders.
Then four TIE fighters soared out of the pit.
A collective gasp was drowned out in a flood of lasers. Ezra dove. The TIEs blasted the platform, causing it to shake. Containers exploded. One shot penetrated the _Ghost_ 's shields. The freighter wobbled and spun, and its forward laser cannon smoked, knocked out of commission.
"The aft guns, Hera—get Chopper to use the _Phantom_!" Kanan shouted into his comlink.
The _Phantom_ , Ezra had discovered on his self-guided tour of the _Ghost_ , was the auxiliary craft attached to the _Ghost_ 's tail section. It featured twin laser cannons that could be used as an extra pair of guns if circumstances arose. This was one of those circumstances.
Hera must have already sent Chopper into the _Phantom_ , because its guns responded almost immediately. Struck in a wing, one TIE spiraled back down into the pit.
It was replaced by something far more menacing—a boxy Imperial transport, rising out of the abyss on its repulsors and unleashing a barrage of laser fire.
The _Ghost_ zoomed off, pursued by the three TIEs. Without access to an escape vessel, everyone on the platform scattered, searching for cover from the transport's cannons. Ezra crouched behind a crate and peeked over the side.
The transport landed on the platform and its bay doors opened. Agent Kallus emerged with another squad of stormtroopers.
"Take them down!" Kallus ordered.
The troopers rushed out, blasters blazing, while the little Wookiee cub meandered about in the middle of the platform, binders still on his wrists. Ezra pounded the crate in frustration. The cub hadn't been in the line of Wookiees he'd freed.
Wullffwarro leapt out from behind his cover, raising his arms and yowling. He ran toward the cub, whom Ezra guessed from the mission briefing must be his son, Kitwarr. But he didn't even make it a few paces before a bolt struck him in the shoulder. The big Wookiee fell to the platform floor. He moaned, hurt but alive.
Zeb rushed out to the Wookiee's side—and without thinking about it, so did Ezra. Kanan and Sabine shot at the troopers as Ezra helped the Lasat lift the dazed Wookiee to his feet.
"He'll be okay. I've got him," Zeb said.
Wullffwarro let out a pained growl. He wanted to keep going. Stormtroopers dashed after the Wookiee cub.
But what could Ezra do? He couldn't charge a platoon of stormtroopers. That truly would be suicide. The cub would have to look out for himself.
Ezra ran back to the crates.
A fireball erupted in the sky, showering pieces of a TIE onto the platform. Normally, Kanan would consider that a positive development. But the _Ghost_ still had two TIEs on its tail, peppering it at close range with lasers, making it impossible for Hera to execute a proper pickup.
"I can't maintain position," Hera said, sounding strained over the comlink.
"Go. Lead the TIEs away and give yourself maneuvering room," Kanan replied. He kept firing at the troopers, giving Zeb cover to drag the wounded Wookiee behind the stack of crates. His efforts did little to thin the enemy ranks. The more troopers he and Sabine picked off, the more came forward. The only thing dwindling was his ammunition pack.
He looked around for inspiration of any kind. Something drastic needed to be done, or everybody was going to die very, very soon.
"I'm not leaving you behind," Hera said on his comlink.
"No, you're not," Kanan said. Nearby, a large shipping container lay open, having not yet been loaded with spice. "We're running a twenty-two pickup."
Sabine turned her helmet toward him. "Seriously?"
"You have a better option?" Kanan asked.
"Yeah," Zeb said. Having leaned Wullffwarro against a crate, he trained his bo-rifle at the troopers. "Jump into the pit and get it over with?"
Kanan didn't dispute the suggestion. It would be what they'd have to do if what he had in mind didn't work.
The _Ghost_ burned its engines, jetting into the sky. The TIEs circled and followed. "All right, I'll be back," Hera commed. "Make sure you're ready."
Kanan let out a breath. No one could be ready for what he was about to do. Especially not himself.
Ezra came up beside him. "Twenty-two pickup? Can you let me in on the secret?"
"Kid, I'm about to let everyone in on the secret." Kanan assessed the battlefield one last time, then holstered his dying blaster and sprung over the crates to land on the other side.
Avoiding enemy fire, he unclipped two cylinders from his belt. They fit together as perfectly as they had when he had first constructed his lightsaber under the tutelage of Master Billaba, more than a decade and a half earlier.
Today, upon activation, the lightsaber produced a blue blade that blazed brighter than it had at any other time in that decade and a half. The stormtroopers ceased fire, as did Sabine, Zeb, and even Agent Kallus. All beheld Kanan in stunned silence.
Standing in the middle of the battlefield, with a hundred blasters pointed at him, Kanan opened himself fully to the Force.
He expected it to flood through him like a dam being released, as it had been a long time since he had completely freed himself to its light. He felt nothing of the kind. Instead, his heart calmed and his mind eased. The relief that came was subtle, subdued, like a gentle breeze that cooled on a hot day. A touch, a whisper, a sense of peace. He began to relax.
A web appeared before him, invisible but perceptible, linking all the beings on the battlefield to him, his life, his actions, his... _destiny_. Future and past events came to him in glimpses and sensations, none he understood, other than that he played a part in them—if he wanted. He could just as easily walk away and refuse.
The Masters at the Temple had often lectured about accepting your destiny. To Kanan, destiny had seemed a fixed matter. Its very nature implied that your future was defined, that you were a puppet acting out choices that had already been made for you. Yet upon seeing the web, how it moved and twirled and spun based on an infinite array of actions, Kanan realized he'd been wrong. Destiny wasn't a fixed matter. Destiny was a choice, a confidence, a belief. Perhaps all the Masters had meant was that by accepting your destiny you accepted yourself.
His friends on the _Ghost_ had accepted themselves, while Kanan had been the one who had not. He'd hidden behind the lie that if he revealed himself, he'd put them all in grave danger. But they were already in danger, going toe-to-toe with the Empire. His friends gave their all to the fight, unafraid of what they might have to sacrifice. They needed Kanan to do the same. He couldn't fear what the Empire might do to him or them. The best way for him to protect their lives was to employ all the talents and skills at his disposal.
It was to use the Force.
Though he hadn't always been one with the Force, the Force had always been with him. He couldn't run away from it. He couldn't deny it any longer. He couldn't deny himself from himself.
Nor would the Imperial agent let him. "All troopers," Kallus said. "Focus your fire on...the Jedi."
Kanan Jarrus raised his lightsaber in salute. A Jedi he had been. A Jedi he would be.
**"Whoa,"** gasped Ezra.
Alone in the middle of the battlefield, Kanan Jarrus ducked, dodged, and deflected the blaster storm directed at him. He seemed to know when and from where each bolt was coming, jumping in the air at the right moment, curving his body to the perfect angle, or slashing his sword just in time. More than a few troopers crumpled, recipients of shots that Kanan deflected off his blade. Without even engaging in hand-to-hand combat, this one man—this Jedi—was holding off an entire platoon of stormtroopers.
Ezra stood there gawking, and the Wookiees didn't fail to notice, either. They picked up fallen blasters and joined the fight, roaring rage.
Kanan would have none of it. "Zeb, Sabine, get the Wookiees out of there—time to go!"
Zeb pulled a couple of Wookiees back from the battle. "Everyone, into the container!" He went to help the limping Wullffwarro, whose long, shaggy arm pointed toward the catwalk that connected platforms. Kitwarr hadn't been caught, but a stormtrooper was getting close. Wullffwarro bellowed in despair.
Ezra might not have understood the Wookiee language, but he knew when a father was imploring someone to rescue his child. And Ezra was the only one near enough to do it.
While the others hurried toward the containers, Ezra looked at Kanan. Though growing fatigued, the ponytailed man kept up his acrobatic defense against the stormtroopers. He had not flinched facing such odds. He was willing to sacrifice himself to save these prisoners, to save his friends, to save Ezra.
"If all you do is fight for your own life," Hera had once said, "then your life's worth nothing."
Ezra began a mad dash toward the catwalk and Kitwarr.
"Kid, stop!"
Zeb's cry reached deaf ears. The boy zigzagged through the crates and containers around the battlefield. Fortunately, he escaped the notice of the stormtroopers, who concentrated their fire on Kanan. Yet one set of Imperial eyes did see him: Agent Kallus turned from the fight to pursue.
Struggling to bring the stubborn Wullffwarro to the container, Zeb could not leave him to help the kid. " _Karabast!_ I swear if he's left behind again, it's not my fault."
He knew Hera, Kanan, and Sabine wouldn't see it that way.
Another explosion lit the sky. Zeb hoped it wasn't the _Ghost_ , though he couldn't tell at that distance. He shoved Wullffwarro inside the container, then spun around to give the stormtroopers a bo-rifle melody of his own.
In another gravity-defying leap, Kanan soared over the stack of crates to land on the other side. "Zeb, Hera's incoming!" He continued to block blaster bolts as he backpedaled toward the container. Sabine fired away, doing the same.
The explosion must've been one of the TIEs—which meant this crazy pickup might actually work. Zeb yelled at the Wookiees who had balked at quitting the fight. "Get in, you fur balls—now!"
The reluctant Wookiees got in, as did he and Sabine. Kanan entered last, whirling his blade to deflect stormtrooper fire. Zeb scanned the crowd inside. Everyone was there—everyone except Ezra and the cub. They hadn't made it back yet.
"Kanan, I think you inspired the kid to do, well, something you would do." Zeb indicated the catwalk in the distance, where Ezra chased after the stormtrooper and the cub.
The calm determination with which Kanan had confronted hundreds of stormtroopers cracked like a mask. He stood in the container hatch, lightsaber in hand, startled.
If Zeb knew of something that could've been done, he would've suggested it. But the _Ghost_ would be there in a matter of seconds, as would the advance of stormtroopers. "Kanan?" Zeb pressed his friend.
"Seal the container," Kanan said with a sigh.
Zeb nodded and grabbed the container's hatch doors.
Kitwarr ran from the stormtrooper. The bridge over the pit was long and he was getting tired. His body wasn't made for running. It was made for climbing. Yet with binders on his wrists, he couldn't get a grip on anything.
Then one of the TIE fighters spiraled down from the sky, crashing into the bridge in a fiery ball. Metal rumpled, supports ripped free, and the section of the bridge Kitwarr was headed down began to collapse. Kitwarr skidded to a stop, right at the edge, nowhere else to go.
He turned, finding that the stormtrooper was catching up to him. The stormtrooper had a gun in his hand, pointed at Kitwarr. Other troopers had done the same to his father and his father's friends. They wanted to hurt all Wookiees. He didn't understand why. What had Wookiees ever done to them?
Behind the stormtrooper, Kitwarr also spotted a human boy racing toward them. He looked like the same boy who had freed his father. Maybe the boy had a key for his binders. Maybe he could free Kitwarr. Kitwarr wanted to be free. He wanted to climb. The bridge floor beneath him was rumbling. It was going to crack soon. He cried out to the boy, pleading for help.
The stormtrooper turned and raised his gun at the boy. Kitwarr felt bad. He shouldn't have cried. He had alerted the trooper. The boy was human, but he had helped Kitwarr's father. Now the boy was going to get hurt.
The boy jumped. He jumped high in the air, somersaulting over the Imperial stormtrooper. It was amazing. Not even Kitwarr's father could jump like that.
The boy landed in front of Kitwarr. He winked, then turned toward the trooper and lifted his arm. Out popped what Kitwarr recognized as a slingshot, since he'd made many himself with twigs and rubbery vine. The boy's was made out of metal, and instead of rocks, it fired three energy balls.
All three shots hit the stormtrooper, knocking him back. The man stumbled, tried to raise his blaster again, but lost his balance and plunged over the bridge railing.
Kitwarr closed his eyes. He didn't like the thought of anyone falling—not even an Imperial stormtrooper—without being able to grab a branch.
The boy scooped Kitwarr up. "Gotcha."
Kitwarr felt safe in the boy's arms. He wasn't going to fall. The boy had started to unlock his wrists, as he had done for his father. Kitwarr smiled and opened his eyes.
A man in gray walked down the bridge toward them. Unlike the stormtroopers, his face wasn't masked under his helmet. And he wasn't smiling.
**Sabine** would have much to argue about with Zeb after this was over. First of all, there was the matter of the kid, whom yet again he had let out of his sight. Then there were the Wookiees. Though Zeb hated the comparison, she thought they differed little from the Lasat. Excluding the Wookiees' fur, the two species were similar—tall, ungainly, and maddeningly stubborn. The Wookiees refused to follow her instructions to move and growled at her as she tried to push them to distribute their weight across the container.
She didn't bother asking Kanan for help. He leaned against a wall, eyes closed, looking utterly drained. He had earned a moment of rest after a performance in combat that rivaled those of the greatest Mandalorian warriors.
The container's ceiling clanged, much louder than the stormtrooper blaster bolts that pinged the walls from outside. The _Ghost_ had landed atop the container. Sabine stopped pushing. The Wookiees would learn in a second why she had gone to so much trouble to move them.
"Magnetic seal locked," Hera said over the comm.
Zeb looked in Sabine's direction. "I hate this part."
The _Ghost_ lifted the container off the ground and then took off at high speed. Yelping and yowling, all the Wookiees lurched into each other, one falling into Zeb.
Sabine stayed on her feet. The sight of Wookiees tumbling over each other would generally have made her laugh. Not today, not after what had just happened. As much as she might hate to admit it, she'd grown fond of the kid. She'd miss him pestering Zeb to no end.
The upper hatch opened into the _Ghost_ 's cargo bay. Kanan sprang through it. Sabine wished she could use her jet pack to do the same. But she had work to do.
"Into the ship," she yelled to the Wookiees. The _Ghost_ turned on its side and sent the Wookiees tumbling toward the upper hatch, making her job easier.
Kallus paused on the catwalk and watched the rebel freighter make a steep climb. The sole pursuing TIE replicated the maneuver with ease and fell in right behind the freighter. Yet what should have been an easy kill for the TIE turned into something far more tragic. The freighter detached its cargo container, which hit the TIE like a missile. The freighter sped away from the explosion, escaping once again.
He hoped the Wookiees were still in the container, but if they weren't, the loss meant little to him. Kallus had something the rebels would return for. He had the boy.
Why the rebels would risk everything to rescue the boy, as they had before on the Star Destroyer, made perfect sense now that Kallus had witnessed what the boy could do. Ordinary boys didn't break out of Star Destroyer detention cells. Ordinary boys didn't ride speeder bikes like champion racers. Ordinary boys couldn't leap over stormtroopers without wearing a pair of jump boots. This boy possessed a gift beyond the ordinary. It was a gift Kallus didn't have, but he knew the signs.
This boy, like the Jedi rebel, could command the Force.
Such an ability made the boy's capture all the more important. Certain entities in the Empire would find the boy of considerable value. Kallus would receive a commendation for the boy's capture, though awards weren't his goal. As an officer of the law, he had sworn a duty to protect Imperial citizens from those who posed a threat, which included all users of the Force.
The boy had his back turned to Kallus, using an astromech droid arm to unlock the Wookiee cub's binders. The binders fell from the cub's wrists and the boy put his astromech arm into his backpack. Kallus set his blaster on stun.
The cub saw Kallus and yelped. The boy turned. Kallus raised his blaster. "It's over for you, Jedi. Master and apprentice, such a rare find these days. Perhaps you are the only two left."
A sudden wind tousled the boy's hair. "I don't know where you get your delusions, bucket-head. I work alone."
"Not this time," said a voice, coming from below.
Kallus spun. The Jedi rebel, his lightsaber humming, stood on top of the rebel freighter, which had swooped low to rise from underneath the catwalk.
Kallus fired. Multiple times.
The Jedi's lightsaber deflected all the shots back at Kallus. Kallus's armor kept blaster bolts from burning through his chest, but it didn't reduce the power of their impact.
He fell backward, over the catwalk railing.
Safe again in the _Ghost_ 's cargo bay, Ezra was relieved to let Kitwarr run free. The little Wookiee didn't realize how sharp his claws were. He had latched on to Ezra's shoulder as if Ezra were a tree.
Ezra forgot the pain upon seeing the reunion between father and son. Kitwarr raced into Wullffwarro's arms for a loud, howling embrace. The other Wookiees who packed the cargo bay added joyous roars as a chorus.
Ezra stood back, away from it all. A vision of his parents came to him and he quickly squashed it. Those memories only brought pain.
A hand squeezed his shoulder. Ezra looked up to see Kanan beside him. The man said nothing, just watched the Wookiees. On his belt, close to Ezra, dangled the hilt of his lightsaber.
For the second time that day, Kallus hung by his fingertips over an abyss. But in this instance his fingertips had a firm grip on a support beam under the catwalk. And the abyss wasn't a vortex trying to suck him into the ether of space. It was just a deep, dark mining pit. He'd still perish if he fell, but that wasn't going to happen. He had rebels to catch.
The only thing preventing Kallus from pulling himself back onto the catwalk was the stormtrooper who'd also fallen over the railing. The trooper clung to a lower support beam and was shaking the whole structure as he tried to climb up.
"First Jedi you've ever seen, sir?" the stormtrooper asked.
Kallus sneered. It had taken a ship to thwart Kallus, while this stormtrooper had been surprised by a mere boy. Such incompetence didn't deserve another chance in the Imperial ranks. He gave the man a good kick.
The stormtrooper lost his grip and shrieked as he plummeted into the pit.
When the support beams stopped vibrating, Kallus climbed onto the catwalk bridge. He brushed dirt off his uniform, then strode onto the platform.
Rarely did he fume with so much anger as he did now. In any case, it only made the situation worse for the rebels. They did not realize the forces now arrayed against them. For Agent Kallus always caught his criminals, no matter where they hid, no matter who they were, no matter whom he had to call or what he had to do.
Kallus would not rest until he caught these traitorous rebels.
**Ezra** joined the _Ghost_ 's crew to say his good-byes to the Wookiees. With Sabine translating their grateful roars, the hairy bipeds crossed through the airlock into a friendly Wookiee gunship Wullffwarro had contacted. The Wookiee soldier commanded a couple of such vessels devoted to freeing his people from slavery, but Ezra didn't inquire further. He was done with all the infighting between Imperials, rebels, and Wookiees. He just wanted to go back to his tower on Lothal, where he could eat jogans until his stomach hurt.
Wullffwarro and Kitwarr lingered last. The silverback bellowed so quickly that Sabine had a tough time translating. "Um, he says if we ever need help, the Wookiees will be there."
Wullffwarro reached out to rub Ezra's head. The touch was gentle, though Ezra could feel the strength under it. This Wookiee could squash him if he wanted.
Ezra smiled at the cub. "Good luck, Kitwarr. And try to stay out of trouble."
The Wookiees roared back and stepped through the airlock. Zeb sealed it behind them. "Trouble, humph. Look who's talking."
Acknowledging the Lasat's comment wouldn't get Ezra anywhere. He turned toward the others. "So," he said, keeping his smile, "I guess you drop me off next?"
Hera, Kanan, and Sabine exchanged surprised looks. Even Zeb seemed caught off guard.
"Uh, yeah," Zeb said. Did Ezra just hear disappointment in the bruiser's voice? The Lasat cleared his throat and regained his normal growl. "Finally, right?"
Not disappointment. Just surprise. Zeb was probably elated inside. "Right," Ezra said.
As he walked past the crew, he thought he heard Kanan sigh. But like usual, the man said nothing.
Then the _Ghost_ wobbled as the Wookiee gunship disengaged, giving everyone a solid shake. Ezra used the opportunity to bump into Kanan and grab his reward.
"Oh, sorry," Ezra said. He didn't wait for Kanan to respond. He hurried into the corridor, slipping Kanan's lightsaber under his sleeve.
The prairie grass around Hera rippled back and forth like waves. She agreed with other pilots' assessments of the world. Lothal did indeed look like a sea of green.
Next to her, under the parked _Ghost_ , Kanan scuffed his foot back and forth, lost in thought. She had assured him he had made the right choice in revealing himself as a Jedi. What was the point of fighting the Empire if he couldn't be the person he was meant—he was destined—to be? They were rebels, and they couldn't be afraid of that fact.
Surprisingly, Kanan had accepted her reasoning without argument. He couldn't go back on being a Jedi now. What bothered him was the boy.
"I thought he might make a good candidate," Kanan said.
"For running around with us? He's fourteen, he needs to be—" Then she realized what he was implying. "You want to train him?"
"Was considering it, until he stole my lightsaber."
Hera blinked. What little she knew about the Jedi was that instruction was a major part of their lives. Jedi were supposed to pass on what they learned, from master to apprentice. Kanan had only been an apprentice, but if teaching opened him up to his past, perhaps he should try it.
"Should I retrieve it?" he asked.
"No." She was surprised she said this. "Let him give it back to you like he did before. Let him make the decision for himself."
Kanan went back to scuffing his foot in the prairie grass. "What if he doesn't return it?"
"Can you make another one?" Hera asked.
A voice echoed inside the _Ghost_ 's cargo bay. Zeb, Sabine, and Chopper were all in there, attending to their duties, but the voice was Ezra's. "So, uh, see you around?"
Hera moved closer to the hatch to peek inside. Sabine lubricated Chopper's gears and gave the boy an apathetic nod, though Hera could tell that wasn't how the girl actually felt. Chopper was more honest. He didn't emit his usual snort; he hooted softly, almost sadly.
Zeb put down the crate he was moving and gave Ezra a punch in the arm. "Not if we can see you first."
Hera knew that Zeb was just horsing around. Ezra didn't interpret the punch as anything like that. He rubbed his arm and headed toward the hatch, clutching the straps of his backpack. "Don't worry, you won't," he said.
Hera was about to nudge Kanan, but he stood tall now, his indecision gone, the grass below him straight, as if never scuffed. When Ezra was about to step off the ramp, Kanan moved forward. "I think you have something that belongs to me," he said to Ezra.
Ezra froze. For a moment, Hera thought he was going to buck and run like a frightened nerf. He didn't. He reached into his backpack and pulled out a transparent object.
"Good luck saving the galaxy," he said. He tossed the Jedi holocron to Kanan, then turned and ran into the prairie toward a distant communications tower.
Kanan did not follow him. He let Ezra go, watching him with regret.
Hera studied the holocron in Kanan's hand. It wasn't a perfect polygon anymore. Some of its sides had shifted. "He opened it," she said. "He passed the test."
She looked at Kanan. He had his own decision to make.
The ground floor of the communications tower looked undisturbed. Helmets lay strewn around the floor, shaken off their racks from the Star Destroyer's rumble. Propaganda holopads cluttered the workbench along with power couplings and droid brains. A forgotten jogan had rolled between the Treadwell base and the shuttle stabilizer fin. All was as Ezra had left it.
But Ezra didn't step past the threshold. He pulled out the newest addition to his collection. Kanan's lightsaber.
He had decided not to sell it. Not now, at least. It would look good as a trophy on the wall, in between the various Imperial helmets.
Given time and practice, he could teach himself how to wield the blade. Maybe he could even teach himself how to hone this ability he had, what the holocron had called the Force.
Ezra's fingers squeezed the lightsaber. His instincts were suddenly on edge. Someone stood behind him. He didn't need to turn to know who it was.
After a moment, he asked Kanan the question that was burning in his mind. "What's the Force?"
"The Force is everywhere. It is everything," Kanan said. "It surrounds us and penetrates us. It binds the galaxy together. And it's strong within you, Ezra. Otherwise, you'd never have been able to open the holocron."
Ezra turned to face the ponytailed man. "What do you want?"
Kanan remained outside the tower, near a patch of green daisies. "To offer you a choice. You can keep the lightsaber you stole and let it become just another dusty souvenir. Or you can give it back and come with us, come with me, and be trained in the ways of the Force. You can learn what it truly means to be a Jedi," he said.
Ezra stared at the daisy patch. His instincts usually gave him guidance in situations like this. But he felt nothing at the moment. This was a decision he had to make on his own.
"I thought the Empire wiped out all the Jedi," Ezra said.
For the first time, Kanan's stony face cracked a slight smile. "Not all of us."
Ezra looked down at the lightsaber in his hand. He felt the metal pommel, the curve of the focusing lens, the button that activated its blade. He had heard that every Jedi built a lightsaber. If he believed this strange man, if he chose this path, would he be able to make one of his own?
When he looked up again, Kanan was gone. But the green daisies were in full bloom.
**Aboard** a Star Destroyer departing the planet Kessel, one ghost spoke to another.
Agent Kallus, however, was not a ghost. He only appeared to be by a trick of the light. The blue glow of the holographic figure whom he addressed gave his skin a pale, spectral sheen.
"Excuse the intrusion, Inquisitor, but in the course of my duties, I have encountered a rebel cell," Kallus said to the figure shimmering before him. "The leader of that cell made use of a lightsaber."
Irrespective of his present holographic form, the Inquisitor had the features of an apparition. He was dressed all in black, and his eyes glinted an eerie yellow, while the flesh of his tattooed face and bare, bald head gleamed a corpse white. If he wasn't the ghost of a man in armor, he was the closest thing Kallus had ever seen.
"Ah, Agent Kallus. You did well to call." The Inquisitor's voice lost none of its sinister oil through the convoluted routing of the Imperial Holonet. "Now tell me everything you know about this Jedi—and his apprentice."
In measured tones, Kallus made his report. He ended by vowing that he would leave no planet, no star untouched until these traitors had faced the Imperial justice they deserved.
In a dark cabin aboard a freighter parked on Lothal, there was another meeting of ghosts.
Bruises and wounds testified that the figure who sat on the bunk was indeed flesh and blood, yet Kanan felt anything but. Battle had exhausted his body, sapped his energy, and thinned him of the fear that had long paralyzed him. He wore a new mantle now, that of Jedi Knight, but by doing so, he had stepped into a phantom realm. Those who previously had worn the mantle, such as his former master and his colleagues, were dead. Long dead.
Was he, Kanan Jarrus, then the last of the Jedi? And how long would he survive before the Empire also exterminated him, making him another ghost?
The Jedi holocron in his palm projected what was, in the technical sense, a ghost. In the air glowed a small hologram of a bearded man who spoke from the dead. He did not know how this Jedi had died, just as he didn't know how most of the others had met their fates.
"This is Master Obi-Wan Kenobi," said the hologram. "I regret to report that both our Order and the Republic have fallen, with the dark shadow of the Empire rising to take their place. This message is a warning—and a reminder—for any surviving Jedi. Trust in the Force."
Trust in the Force. That's what Kanan had done. But trust didn't vanquish worry. His revelation could—would—cause harm to the ones for whom he cared. They were all implicated now in his crime, and the Empire would not hesitate to hurt them to hurt him.
He sensed their presences about the ship, knowing what they were doing without seeing it, as all were creatures of habit. Zeb moved cargo in the bay, where Chopper would be making repairs on the laser cannon circuitry. Sabine drank her blue concoction in the galley. Hera, dear Hera, lounged in the cockpit, resting.
"Do not return to the Temple. That time has passed and our future is uncertain," the hologram said.
Time had passed; that was for sure. Enough that the Jedi Temple had become yet another ghost in his memories, an element of a dream he must piece together in his new life. A return to the Temple would not help, even if he wanted to go. Nothing of it remained on the world now known as Imperial City. Not a museum, not a plaque, not a mark. The Empire had reduced all its brick and steel to dust.
"Above all else, be strong," Kenobi said. "We will each be challenged—our trust, our faith, our friendships. But we must persevere, and in time, a new hope will emerge."
Hope. That was the key to all of this. To believe that freedom in the galaxy could be renewed. That tyranny was not eternal. That the dark side of the Force could not snuff out every flicker of the light.
Kanan believed. He had to believe. There was no other choice. But he was afraid. He sensed the tyranny of this Emperor was unlike that of any other despot history had ever recorded. His Empire would devour the whole galaxy if it wasn't stopped. This made the struggle Kanan and his friends were involved in the great battle not only of their lifetime, but of many, many lifetimes.
Perhaps it was the greatest battle of all.
"May the Force be with you, always," said Kenobi. After those final words, his ghost disappeared, and once again, the cabin was dark.
Kanan took a breath.
The darkness did not last. The door opened and in spilled light. A boy stood in that light. He held out a lightsaber.
Kanan went over to Ezra and took the lightsaber. In return, he put his hand on Ezra's shoulder. The boy beamed back.
In former times, a child of Ezra's age would be deemed too old to be an apprentice. But those times had passed. The boy had much to learn, as did Kanan. They would learn together.
May the Force be with them all.
| {
"redpajama_set_name": "RedPajamaBook"
} | 267 |
Sepultura – "Beneath the Remains"
January 3, 2013 Chris Dick
Hall of Fame Hall of Fame, Sepultura.
dB HoF NO. 096
Beneath the Remains
Before there was death-thrash or thrashy death, there was Sepultura, lovingly pronounced either Seh-poll-tour-a or Sep-ull-trah by many a hesher. Contrary to popular opinion at the time, Sepultura were not from the deepest parts of the Amazon or Brazil's most violent of favelas. Actually, they were from somewhere in between—a densely populated concrete jungle not unlike any other from the first or the third world. There was strife, hardship, ugliness and, surprisingly, good times. Brazil and its hardscrabble cities defined Sepultura. Sepultura would, in turn, reflect back. Not with early (and crucial) efforts in Bestial Devastation and Morbid Visions, though. They were too busy emulating idols to realize their place in Brazil, let alone the rest of the world. Slowly but surely, they grew up. From forming in the fog of 1984 to 1987—when then-label Cogumelo released sophomore LP Schizophrenia—Sepultura's spikes-to-raised fists transformation was nearly complete. For it was the promise of Schizophrenia that would land them not just in the offices of Roadrunner Records in New York City, but on stages afar.
Inked by death metal A&R godfather Monte Conner to seven albums in the spring of 1988, Sepultura had found a willing and risk-averse partner in Roadrunner. Driven by a new label deal and progressing musically at a frightening pace, the quartet hammered out their third album in a cramped room at bassist Paulo Jr.'s parents' house. "Stronger Than Hate," "Lobotomy," "Mass Hypnosis," "Beneath the Remains" and career-defining cut "Inner Self" were the product of the desire to extend Sepultura's savage thrash-death hybrid beyond the borders of the city and country they called home.
Recorded over Christmas 1988 in Rio De Janeiro at Nas Nuvens Studio, Beneath the Remains wasn't just the beginning for Sepultura, but for then-embryonic Scott Burns as well. Burns, who had started to earn a reputation as an engineer at Morrisound Studios, was tapped to fly to Brazil to capture Sepultura in their native environment. In Burns, Sepultura had a producer who may not have been their first choice, but spoke the same language: not Portuguese, but extreme metal. To say the record catapulted the Brazilians onto the world stage is an understatement. Because Beneath the Remains cleaned our collective late-'80s clocks, we induct it into the Hall of Fame. No stranger to firsts, we welcome Sepultura, as the first two-time HOF entry.
To read the entire article, purchase this issue from our online store.
Behemoth – "Demigod"
The making of the be-all and end-all of Polish death metal.
Madball – "Set It Off"
Madball demonstrate their metal-infused NYHC style on the classic Set It Off and enter the Decibel Hall of Fame
Sacrilege – "Behind the Realms of Madness"
Yes, it's only six songs long, but there's absolutely no filler and the unrelenting quality of the material makes it a worthy inductee for the Decibel Hall of Fame. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 5,945 |
\section{Introduction}
About 10 years ago, Cameron McA Gordon asked the following question:
Can the pairwise connect sum of two 3--manifolds each with an
unstabilized Heegaard splitting yield a 3--manifold with a stabilized
Heegaard splitting? This question stumped the experts for many years
but recently a negative answer to this question
has been announced by D Bachman \cite{B} and R Qiu \cite{Q}.
More generally, one can ask how Heegaard splittings behave under other
types of ``sums'', that is, when the 3--manifolds containing them are
glued along positive genus boundary components. How Heegaard genus
behaves under these circumstances is one of the many questions
investigated by Klaus Johannson in \cite{Jo} and by the first author
in \cite{Sch}. In both cases, inequalities relating the Heegaard
genus of the glued 3--manifold to the Heegaard genera of the original
3--manifolds are obtained. Most strikingly, the inequalities give
lower bounds on the Heegaard genus of the glued 3--manifold in terms of
the Heegaard genera of the original 3--manifolds. But these lower
bounds are fractions of the sum of the genera of the original
3--manifolds. A better bound under more restrictive circumstances has
recently been obtained by D Bachman, E Sedgwick and S Schleimer
\cite{BSS}.
One upshot is that, in general, the phenomenon of ``degeneration of
Heegaard genus'' under glueing of 3--manifolds can't be ruled out. It
appears however that under certain, possibly generic circumstances, this
phenomenon does not occur. For instance, in \cite{La} Marc
Lackenby shows that for a pair of hyperbolic 3--manifolds each with one
boundary component and under certain restrictions on the glueing,
minimal genus Heegaard splittings of the glued 3--manifold are always obtained from
Heegaard splittings of the original 3--manifolds by amalgamation.
It is presently unknown how large ``degeneration of Heegaard genus''
under glueing can be. Interestingly, the issue of stabilization
implicitly arises in the investigation of this phenomenon in \cite{SS}
and in \cite{Sch}. The examples given in this note make this issue explicit. In
particular, we provide examples that illustrate how ``degeneration of
Heegaard genus'' under glueing corresponds to
the existence of stabilizations in the amalgamation of Heegaard
splittings of the original 3--manifolds. In doing so, we provide
counterexamples to a conjecture of Kobayashi, Qiu, Rieck and Wang \cite{KQRW}.
\subsection*{Acknowledgments}
The first author is partially supported by the
grant DMS-0603736 from the National Science Foundation.
\section{Definitions}
For standard definitions and results concerning knots, see Burde and
Zieschang \cite{BZ}, Lickorish \cite{L} or Rolfsen \cite{R}. For standard
definitions and results pertaining to $3$--manifolds, see Hempel \cite{H}
or Jaco \cite{J}.
\begin{defn}
A {\em height function} on ${\mathbb S}^3$ is a Morse function with exactly
two critical points.
\end{defn}
This last assumption guarantees that $h$ induces
a foliation of $S^3$ by spheres, along with one maximum that we denote
by $\infty$ and one minimum that we denote by $-\infty$.
\begin{defn}
Let $K$ be a knot in $S^3$. If all minima of $h|_K$ occur below all maxima of $h|_K$, then we say
that $K$ is in {\em bridge position} with respect to $h$. The {\em bridge number}
of K, $b(K)$, is the minimal number of maxima required for $h|_K$.
\end{defn}
\begin{defn}
If $K$ is in bridge position, then a regular level surface below
all maxima and above all minima is called a {\em bridge
surface}.
\end{defn}
\begin{defn}
An {\em upper disk (lower disk)} is an embedded disk whose interior is disjoint from the knot whose boundary is partitioned
into two subarcs, one contained in a bridge surface and one a subarc
of the knot that lies above (below) the bridge surface. A {\em strict
upper disk (strict lower disk)} is an upper (lower) disk whose interior
lies above (below) the bridge surface.
A {\em complete set of strict upper (lower) disks} is a set of upper (lower) disks
such that each subarc of the knot lying above (below) the bridge
surface meets exactly one disk in the set.
\end{defn}
\begin{defn} \label{defn:cb}
A {\em compression body} is a $3$--manifold $W$ obtained from $S\times I$ where $S$ is a
closed orientable connected surface by attaching $2$--handles to $S \times
\{0\} \subset S \times I$ and capping off any resulting $2$--sphere
boundary components with $3$--handles. We denote $S \times \{1\}$ by
$\partial_+W$ and $\partial W - \partial_+W$ by $\partial_-W$.
Dually, a compression body is an orientable $3$--manifold obtained from
a closed orientable surface $\partial_-W \times I$ or a 3--ball or a
union of the two by attaching $1$--handles.
In the case where $\partial_-W = \emptyset$, we also call $W$ a
handlebody.
\end{defn}
\begin{defn}
Let ${\cal A}=\{a_1,\ldots ,a_k\}$ be a collection of annuli in a compression body
$W$. Then ${\cal A}$ is a {\em primitive collection} if there is a
collection ${\cal D}=\{D_1,\ldots ,D_k\}$ of pairwise disjoint properly embedded disks in $W$ such that
$a_i$ meets $D_i$
in a single spanning arc and $a_i\cap D_j=\emptyset$ for $j\neq i$.
\end{defn}
\begin{defn} A set of {\em defining disks} for a compression body
$W$ is a set of disks $\{D_1, \dots, D_n\}$ properly embedded in $W$
with $\partial D_i \subset \partial_+W$ for $i = 1$, $\dots, n$ such
that the result of cutting $W$ along $D_1 \cup \dots \cup D_n$ is
homeomorphic to $\partial_-W \times I$ or a $3$--ball in the case that
$W$ is a handlebody.
\end{defn}
\begin{defn} \label{defn:Heegaard splitting} A {\em Heegaard
splitting} of a $3$--manifold $M$ is a pair $(V, W)$ in which $V$, $W$
are compression bodies and such that $M = V \cup W$ and $V \cap W =
\partial_+V = \partial_+W = S$. We call $S$ the {\em splitting surface} or
{\em Heegaard surface}. Two Heegaard splittings are considered {\em equivalent}
if their splitting surfaces are isotopic.
The {\em genus} of $M$, denoted by $g(M)$, is the smallest possible genus of
the splitting surface of a Heegaard splitting for $M$.
\end{defn}
\begin{defn}
Let $(V, W)$ be a Heegaard splitting. A Heegaard splitting is
called {\em stabilized} if there is a pair of properly embedded disks $(D, E)$ with $D \subset V$ and
$E \subset W$ such that $\#|\partial D \cap \partial E | =
1$. We call the pair of disks $(D, E)$ a {\em stabilizing pair of disks}. A
Heegaard splitting is {\em unstabilized} if it is not stabilized.
\end{defn}
\begin{defn}
{\em Destabilizing} a Heegaard splitting $(V, W)$ is the act of creating a
Heegaard splitting from $(V, W)$ by performing ambient 2--surgery on
$S$ along the cocore of a $1$--handle in either $V$ or $W$.
\end{defn}
Note that the result of performing ambient 2--surgery on $S$ along the
cocore of a $1$--handle in either $V$ or $W$ is not necessarily a
Heegaard splitting. In order for this operation to be a
destabilization, the result is required to be a Heegaard splitting.
This definition is equivalent to presupposing a stabilizing pair of
disks $(D, E)$ and cutting along $D$. (Here $D$ is the cocore of a
$1$--handle of $V$ and the existence of $E$ guarantees that the result
of cutting along $D$ is a Heegaard splitting.)
\begin{defn} \label{stHS}
Let $M$ be a compact orientable Seifert fibered space with quotient
space an orientable orbifold $Q$. Denote the genus of the surface
underlying $Q$ by $g$ and the number of cone points by $n$.
Assume further that $M$ (and hence $Q$) has exactly one boundary
component. (This simplifying assumption is met in all examples
considered here.)
Let $a_1, \dots, a_{2g}, b_1, \dots, b_{n-1}$ be a disjoint
collection of arcs in $Q$ that cut $Q$ into disks each containing at
most one cone point. In the case of the once punctured torus, such
a collection of arcs is shown in \fullref{dsa1}. In the case of
an orbifold with underlying surface a disk and with four cone
points, such a collection of arcs is shown in \fullref{dsa7}.
If the underlying surface of $Q$ is a disk, we further assume that
each arc $b_i$ cuts off a subdisk from $Q$ containing exactly one
cone point.
Abusing notation slightly, denote a collection of arcs in $M$ that
projects to $a_1, \dots, $ $a_{2g}, b_1, \dots, b_{n-1}$ also by
$a_1, \dots, a_{2g}, b_1, \dots, b_{n-1}$. Now take $V$ to be a
regular neighborhood of $a_1, \dots, a_{2g}, b_1, \dots, b_{n-1}$
together with a regular neighborhood of $\partial Q \times {\mathbb
S}^1$. Take $W$ to be the closure of the complement of $V$ in $M$.
It is an easy exercise to show that $(V, W)$ is a Heegaard splitting
of $M$. Such a Heegaard splitting is called a {\em vertical} Heegaard
splitting of $M$. If $Q$ has no cone points, that is, if $M=Q\times
S^1$, then this splitting is also called the {\em standard} Heegaard
splitting of $M$.
\end{defn}
\begin{figure}[ht!]
\labellist\small
\pinlabel {$a_1$} [b] at 220 160
\pinlabel {$a_2$} [l] at 158 139
\endlabellist
\centerline{\includegraphics[width=0.5\textwidth]{\figdir/dsa1}}
\caption{Arcs $a_1, a_2$ for a punctured torus}
\label{dsa1}
\end{figure}
\begin{figure}[ht!]
\labellist\small
\pinlabel {$b_1$} [t] at 40 165
\pinlabel {$b_2$} [r] at 135 245
\pinlabel {$b_3$} [t] at 235 170
\endlabellist
\centerline{\includegraphics[width=0.5\textwidth]{\figdir/dsa7}}
\caption{Arcs $b_1, b_2, b_3$ for an orbifold with four cone points}
\label{dsa7}
\end{figure}
\begin{defn}
A {\em tunnel system} for a knot $K$ in ${\mathbb S}^3$ is a collection of
arcs $t_1, \dots, t_n$ such that the complement of $K \cup t_1 \cup
\dots \cup t_n$ is a handlebody. The {\em tunnel number} of a knot $K$ is
the least number of arcs required for a tunnel system of $K$.
\end{defn}
\begin{defn}
Suppose $K$ is in bridge position and that there are $n$ maxima. We
may assume temporarily that all maxima occur in the same level surface
$L$. The maxima may be connected by a system of $n - 1$ disjoint arcs
in $L$. It is an easy exercise to show that this set of arcs is a
tunnel system. It is called an {\em upper tunnel system}.
The same exercise shows that there is a set of defining disks ${\cal
D}$ for the complement of $K \cup t_1 \cup \dots \cup t_n$ of the
following type: Each component of ${\cal D}$ has interior below $L$,
furthermore, below $L$, its boundary runs once along exactly one
component of $K - K \cap L$. This set of disks is called a {\em complete
set of lower disks} for the upper tunnel system.
\end{defn}
\begin{defn}
Suppose $t_1, \dots, t_n$ is a tunnel system for a knot $K$ in
${\mathbb S}^3$. Denote the complement of $K$ by $M$. Take $V$ to be
a regular neighborhood of $\partial M \cup t_1 \cup \dots \cup t_n$
and take $W$ to the closure of the complement of $V$. Then $(V, W)$
is a Heegaard splitting called the {\em Heegaard splitting corresponding to
the tunnel system} $t_1, \dots, t_n$.
\end{defn}
The definition of amalgamation is a lengthy one. In the last 15
years, this term has been used in the following context: A pair of
3--manifolds $M_1, M_2$ each with a Heegaard splitting are identified
along components of their boundary. This results in a 3--manifold $M$.
The Heegaard splittings of $M_1, M_2$ can be used to construct a
canonical Heegaard splitting of $M$ called the amalgamation of the two
Heegaard splittings. One assumes that in each of $M_1, M_2$ the
boundary components along which the glueing occurs are contained in a
single compression body. Roughly speaking, then, the collars of the
boundary components lying in this compression body are discarded and
the remnants of the two compression bodies in $M_1 - \text{(collars)}$
identified to the remnants of the two compression bodies in $M_2 -
\text{(collars)}$. This is done in such a way that the 1--handles that are
attached to the collar on such a boundary component in $M_1$ become
attached to the compression body in $M_2$ that does not meet any of
the boundary components along which the glueing takes place and vice
versa. For a formal definition see below.
\begin{defn}
Let $M_1, M_2$ be $3$--manifolds with $R$ a closed subsurface of
$\partial M_1$, and $S$ a closed subsurface of $\partial M_2$.
Suppose that $R$ is homeomorphic to $S$ via a homeomorphism $h$.
Further, let $(X_1, Y_1), (X_2, Y_2)$ be Heegaard splittings of $M_1,
M_2$. Suppose further that $N(R) \subset X_1, N(S) \subset X_2$.
Then, for some $R' \subset \partial M_1 \backslash R$ and $S' \subset
\partial M_2 \backslash S$, $X_1 = N(R \cup R') \cup (\text{1--handles})$ and
$X_2 = N(S \cup S') \cup (\text{1--handles})$. Here $N(R)$ is homeomorphic to
$R \times I$ via a homeomorphism $f$ and $N(S)$ is homeomorphic to $S
\times I$ via a homeomorphism $g$. Let $\sim$ be the equivalence
relation on $M_1 \cup M_2$ generated by
(1) $x \sim y$ if $x, y \; \epsilon \; \eta(R)$ and $p_1 \cdot f(x) = p_1
\cdot f(y)$,
(2) $x \sim y$ if $x, y \; \epsilon \; \eta(S)$ and $p_1 \cdot g(x) = p_1
\cdot g(y)$,
(3) $x \sim y$ if $x \; \epsilon \; R$, $y \; \epsilon \; S$ and $h(x) = y$,
where $p_1$ is projection onto the first coordinate. Perform
isotopies so that for $D$ an attaching disk for a $1$--handle in $X_1,
D'$ an attaching disk for a $1$--handle in $X_2$, $[D] \cap [D'] =
\emptyset$. Set $M = (M_1 \cup M_2)/\sim, X = (X_1 \cup Y_2)/\sim,$
and $Y = (Y_1 \cup X_2)/\sim$. In particular, $(N(R) \cup N(S)/\sim)
\cong R, S$. Then $X = Y_2 \cup N(R') \cup (\text{1--handles})$, where the
$1$--handles are attached to $\partial_+Y_2$ and connect $\partial
N(R')$ to $\partial_+Y_2$. Hence $X$ is a compression body.
Analogously, $Y$ is a compression body. So $(X, Y)$ is a Heegaard
splitting of $M$. The splitting $(X, Y)$ is called the {\em amalgamation}
of $(X_1, Y_1)$ and $(X_2, Y_2)$ along {R, S} via $h$.
\end{defn}
\section{A single destabilization}
We first consider a concrete example that illustrates the issues under
discussion. Let $T_i$ be a punctured torus for $i=1,2$. As
3--manifolds $M_1, M_2$ we take $T_i \times {\mathbb S}^1$ for $i = 1,
2$. Note that $\partial M_i$ is a torus, for $i = 1, 2$. We take $M$
to be the result of glueing $M_1$ to $M_2$ in such a way that
$(\partial T_1) \times \{1\}$ and $(\partial T_2) \times \{p\}$ have
intersection number one on the resulting torus.
We describe two distinct Hegaard splittings for $M$:
\begin{exa} \label{g4}
Let ${\mathbb S}^1 = I_1 \cup I_2$ be a decomposition of ${\mathbb
S}^1$ into two intervals that meet at their endpoints. Let $V_i =
T_i \times I_1$ and $W_i = T_i
\times I_2$, for $i = 1, 2$. Then $V_i$ and $W_i$ are genus 2
handlebodies. Denote the annulus in which $V_i$ meets
$\partial M_i$ by $A_i$ and that in which $W_i$ meets $\partial M_i$
by $B_i$. Due to the choice of glueing of $\partial M_1$ and
$\partial M_2$ that results in $M$, $A_1$ meets $A_2$ in a (square) disk. As
do $B_1$ and $B_2$. In other words, $V = V_1 \cup V_2$ is
homeomorphic to the result of taking the disjoint union of $V_1$ and
$V_2$ and joining the two components by a 1--handle. In particular, it
is a genus 4 handlebody. The same is true for $W = W_1 \cup W_2$.
Thus $(V, W)$ is a genus 4 Heegaard splitting of $M$.
\end{exa}
\begin{exa} \label{amalg}
Let $(X_i, Y_i)$ be the standard Heegaard splitting of $M_i$ (see \fullref{stHS}), for $i =
1, 2$. And let $(X, Y)$ be the amalgamation of $(X_1, Y_1)$ and
$(X_2, Y_2)$
\end{exa}
\begin{thm}
The genus of $M_i$ is three for $i = 1, 2$ and the genus of $M$ is
four.
\end{thm}
\begin{proof}
Recall that the rank, that is, the smallest number of generators, of the
fundamental group of a 3--manifold provides a lower bound for the genus
of a Heegaard splitting of that 3--manifold. Here
\[\pi_1(M_i) = F_2\oplus \mathbb Z \]
Abelianizing yields a free abelian group of rank 3. Thus $\hbox{rank
}\pi_1(M_i)=3$ and hence the Heegaard splitting constructed in \fullref{amalg} has minimal genus.
The Seifert--Van Kampen Theorem yields a presentation of $\pi_1(M)$ as
$$\pi_1(M_1)*_{\mathbb Z^2}\pi_1(M_2).$$ Quotienting out the normal
closure of the amalgamated subgroup yields $\mathbb Z^2*\mathbb
Z^2$ as this kills the fibre and the commutator of the generating pair of the free base group on both sides. It follows that $$\hbox{rank }\pi_1(M)\ge \hbox{rank }\mathbb
Z^2*\mathbb Z^2=4.$$Hence the Heegaard splitting in \fullref{g4}
has minimal genus.
\end{proof}
The fact that the minimal genus Heegaard splitting is less than the genus of a minimal genus amalgamation in these examples illustrates a phenomenon known as ``degeneration of Heegaard genus" under glueing.
\begin{thm} \label{1destab}
The Heegaard splitting $(X, Y)$ of $M$ is stabilized.
\end{thm}
\begin{proof}
For $i=1,2$, choose arcs $a^i_1, a^2_1$ in $T_i\subset M_i$ as
in \fullref{stHS}. Then $T_i - (N(a^i_1) \cup N(a^i_2))$ is a
disk $D_i$. It's boundary meets $\partial M_i$ as in \fullref{dsa2}. After the amalgamation, a copy of $D_i$ survives in $M_i
\subset M$, for $i = 1, 2$. How $\partial D_1$ and $\partial D_2$
intersect is pictured in \fullref{dsa3}. Thus $(D_1, D_2)$ are a
stabilizing pair of disks.
\end{proof}
\begin{figure
\centerline{\includegraphics[width=0.3\textwidth]{\figdir/dsa2}}
\caption{The boundary of $D_i$ as it appears on $\partial M_i$}
\label{dsa2}
\end{figure}
\begin{figure
\centerline{\includegraphics[width=0.3\textwidth]{\figdir/dsa3}}
\caption{The boundaries of $D_1$ and $D_2$ as they intersect}
\label{dsa3}
\end{figure}
\begin{cor}
The Heegaard splitting $(X, Y)$ of $M$ can be destabilized exactly
once.
\end{cor}
\begin{exercise}
Show that destabilizing the Heegaard splitting in \fullref{amalg} yields
the Heegaard splitting in \fullref{g4}.
\end{exercise}
\section{Arbitrarily many destabilizations}
We now construct a sequence of pairs of 3--manifolds that exhibit a more
general phenomenon. More specifically, for each $n$, we construct a pair
$(M_1^n, M_2^n)$ of 3--manifolds as follows: Given $n$, take $M_1^n$
to be a Seifert fibered space with base orbifold a disk with $n + 1$
cone points. We denote the natural quotient map from $M_1^n$ to the base
orbifold by $p_n$. Take $K^n$ to be a knot that has bridge number $n$
and tunnel number $n - 1$. The existence of such knots is guaranteed
by \cite[Theorem 0.1]{LM}. Indeed, in \cite{LM}, M Lustig and Y Moriah
define the class of generalized Montesinos knots. The referenced theorem
provides very technical but nevertheless achievable sufficient conditions
under which such a knot has bridge number $n$ and tunnel number $n - 1$.
Take $M_2^n$ to be the complement of $K^n$ in ${\mathbb S}^3$.
Glue $M_1^n$ to $M_2^n$ in such a way that a fiber of $M_1^n$ is identified with
a meridian of $M_2^n$. Denote the 3--manifold obtained in this way by
$M^n$. Consider the following Heegaard splittings of $M^n$:
\begin{exa} \label{g4n}
Let $b_1, \dots, b_n$ be a collection of arcs that cut the base
orbifold of $M_1^n$ into disks each with exactly one cone point.
Bicolor these disks red and blue, that is, color these disks in such
a way that disks abutting along an arc are given distinct colors.
The preimage of these arcs in $M_1^n$ is a collection of annuli that
cut $M_1^n$ into solid tori. These tori inherit colors from the
bicoloring of the disks to which they project. Take $V_1^n$ to be
the union of the red tori and $W_1^n$ to be the union of the blue
tori.
Let $P$ be a bridge sphere for $K^n$. Then $P$ divides $M_2^n$ into
two components that we label $V_2^n$ and $W_2^n$. We can clearly
assume that the $2n$ meridional boundary curves of $P\cap M_2^n$ match
up with the boundary curves of the annuli $b_1,\ldots ,b_n$ Now set
$V^n = V_1^n \cup V_2^n$ and $W^n = W_1^n \cup W_2^n$.
\end{exa}
\begin{lem} \label{g4nhs}
The decomposition $(V^n, W^n)$ is a Heegaard splitting of $M^n$.
\end{lem}
We first prove an auxilliary lemma. It is well known, but we include
it here for completeness.
\begin{lem} \label{glueprim}
Suppose $X$ and $Y$ are handlebodies. Let ${\it A}$ be a collection
of $k$ essential annuli in $\partial X$ and let ${\it B}$ be a
primitive collection of $k$ annuli in $\partial Y$. Glue $X$ to $Y$
by identifying ${\it A}$ and ${\it B}$. Denote the result by $E$.
Then $E$ is a handlebody.
\end{lem}
\begin{proof}
Since ${\it B}$ is a primitive collection of $k$ annuli in
$\partial Y$, there is a collection ${\cal Y}$ of $k$ disjoint
essential disks such that each annulus meets one of the disks in
exactly one arc and is disjoint from the other disks. Cutting $Y$
along ${\cal Y}$ yields a handlebody $Y'$ and cuts each component of
${\it B}$ into a disk. The remnants of ${\cal Y} \cup {\it B}$ on
$\partial Y'$ are disks. Thus a set of defining disks for $Y'$ can be
isotoped to be disjoint from the remnants of ${\cal Y} \cup {\it B}$
on $\partial Y'$. Hence they can be used to augment ${\cal Y}$ to a
set of defining disks ${\cal Y}'$ of $Y$.
Choose a set of defining disks ${\cal X}$ for $X$. We may assume that
each component of ${\cal X}$ meets each component of ${\it A}$ in
spanning arcs. (Note that each component of ${\it A}$ is met by a non
zero number of such arcs, because it is essential.) In $E$ we can
place a copy of the appropriate element of ${\cal Y}$ along each such
spanning arc. Thus in $E$, the components of ${\cal X}$ can be
extended into $Y \subset E$ by parallel copies of elements of ${\cal
Y}$ to an embedded disk. Denote the set of disks resulting from
${\cal X}$ via these extensions along with a set of defining disks for
$Y'$ by ${\cal E}$.
Now the result of cutting $E$ along ${\cal E}$ is a 3--ball since it
can also be obtained by glueing 3--balls (the result of cutting $Y$
along ${\cal E} \cap Y$) to a 3--ball (the result of cutting $X$ along
${\cal X}$) along disks. It follows that $E$ is a handlebody.
\end{proof}
We now prove \fullref{g4nhs}. Fortunately, the hard work has
already been accomplished.
\begin{proof}[Proof of \fullref{g4nhs}]
To see that $(V^n, W^n)$ is a Heegaard
splitting, consider the following: Each component of $V_1^n$ and
$W_1^n$ is a solid torus. In particular, it is a handlebody.
Furthermore, both $V_2^n$ and $W_2^n$ are genus $n$ handlebodies each
meeting $\partial M_2^n$ in a primitive collection of $n$ annuli. More
specifically, we can take a complete set of strict upper disks or a
complete set of strict lower disks, respectively, to be the required
collection of disks. See \fullref{dsa8}.
\begin{figure}[ht!]
\centerline{\includegraphics[width=0.7\textwidth]{\figdir/dsa8}}
\caption{The submanifold $V_2^1$ or $W_2^1$ of $M_2^1$ with
a collection of disks meeting primitive annuli as required}
\label{dsa8}
\end{figure}
It thus follows
from \fullref{glueprim} that $V^n$ and $W^n$ are handlebodies. Thus
$(V^n, W^n)$ is a Heegaard splitting.
\end{proof}
\begin{exa} \label{amalgn}
Take $(X_1^n, Y_1^n)$ to be a vertical Heegaard splitting of
$M_1^n$. Take\break $t_1, \dots, t_{n-1}$ to be an upper tunnel system of
$M_2^n$ and take $(X_2^n, Y_2^n)$ to be the Heegaard splitting
corresponding to $t_1, \dots, t_{n-1}$. Now take $(X^n, Y^n)$ to be
the Heegaard splitting of $M^n$ resulting from the amalgamation of
$(X_1^n, Y_1^n)$ and $(Y_1^n, Y_2^n)$.
\end{exa}
\begin{thm}
For $M_1^n, M_2^n, M^n$ as above,
\[\genus(M_1^n) + \genus(M_2^n) - \genus(M^n) > n\]
and
\[\frac{\genus(M^n)}{\genus(M_1^n) + \genus(M_2^n)} < \frac{1}{2}\]
\end{thm}
\begin{proof}
Note first that $\pi_1(M_1^n)$ maps onto the fundamental group of the
base orbifold which is a free product of $n+1$ cyclic groups. Thus
$\pi_1(M_1^n)$ is of rank $n+1$ by Grushko's theorem. It follows that
the genus of $M_1^n$ is $n+1$. Furthermore, since the tunnel number
of $K^n$ is $n-1$, the genus of $M_2^n$ is $n$. The Heegaard splitting
constructed in \fullref{g4n} bears witness to the fact that the
Heegaard genus of $M^n$ is at most $n$.
\end{proof}
Again, the manifold pairs $M_1^n, M_2^n$ exhibit the phenomenon of
``degeneration of Heegaard genus'' under glueing.
Note that the genus of a Heegaard splitting of $M^n$ resulting from an
amalgamation of minimal genus Heegaard splittings is $2n$. In particular,
the genus of $(X^n, Y^n)$ is $2n$.
\begin{thm}
There are $n$ disjoint pairs of stabilizing disks for $(X^n, Y^n)$. In
other words, the Heegaard splitting $(X^n, Y^n)$ of $M^n$ can be
destabilized successively at least $n$ times. Specifically, the
Heegaard splitting obtained from $(X^n, Y^n)$ is the result of
stabilizing $(V^n, W^n)$ $n$ times.
\end{thm}
\begin{proof}
Recall that $M_2^n$ is the complement of $K^n$ and that $Y_2^n$ is the
complement of $K^n$ together with an upper tunnel system. See \fullref{dsa9}. Recall also that after amalgamation, (a collar of)
$Y_2^n$ is a subset of $X^n$.
\begin{figure}[ht!]
\centerline{\includegraphics[width=0.7\textwidth]{\figdir/dsa9}}
\caption{$K^3$ with an upper tunnel system}
\label{dsa9}
\end{figure}
Denote the torus resulting from the identification of $\partial M_1^n$
and $\partial M_2^n$ by $T$. Recall that after the amalgamation, the
torus $T$ minus the attaching disks for the 1--handles with cores $b_1,
\dots, b_n$ to one side and the upper tunnel system to the other
side lies in the splitting surface $F^n$ of $(X^n, Y^n)$. We isotope
$n$ essential subannuli of $T$ into $M_1^n$ and denote the resulting
annuli by $U_1, \dots, U_n$. We isotope the other $n$ subannuli of
$T$ into $M_2^n$ and denote the result by $A_1, \dots, A_n$. We
subdivide $T$ into these subannuli in such a way that $U_1, \dots,
U_n$ are vertical in $M_1^n$ and $A_1, \dots, A_n$ are meridional in
$M_2^n$. Furthermore, we subdivide $T$ into these subannuli in such a
way that $U_i$ meets the endpoints of exactly two distinct components
of $b_1, \dots, b_n$. See Figures~\ref{dsa10} and \ref{dsa11}.
\begin{figure}[ht!]
\labellist\small
\pinlabel {$b_1$} [t] at 55 165
\pinlabel {$b_2$} [r] at 148 237
\pinlabel {$b_3$} [t] at 250 170
\endlabellist
\centerline{\includegraphics[width=0.5\textwidth]{\figdir/dsa10}}
\caption{Vertical annuli in $M_1^3$}
\label{dsa10}
\end{figure}
\begin{figure}[ht!]
\labellist\small
\pinlabel {tunnel tube} [b] at 233 496
\pinlabel {tunnel tube} [b] at 505 496
\pinlabel {$A_1$} [b] at 108 480
\pinlabel {$A_2$} [b] at 365 488
\pinlabel {$A_3$} [b] at 613 474
\pinlabel {$T$} [b] at 360 20
\endlabellist
\centerline{\includegraphics[width=0.7\textwidth]{\figdir/dsa11}}
\caption{The result of isotoping the annuli $A_1, A_2, A_3$ into $M_2^3$}
\label{dsa11}
\end{figure}
Consider the portion of $F^n$ lying in $M_2^n$. See \fullref{dsa11}. It is a punctured sphere. Moreover, it is isotopic to a
punctured sphere that consists of a level disk with $2n$ punctures and
an upper hemisphere. See \fullref{dsa12}. Now note that the
portion of ${\mathbb S}^3$ above a bridge sphere that coincides with
this level punctured disk and above the upper hemisphere is a 3--ball.
(Replacing the upper hemisphere of this sphere with a level disk is
equivalent to isotoping the upper hemisphere of this sphere through
infinity. For details, see \cite[Lemma 1]{Sch1}.) Thus the portion
of $F^n$ lying in $M_2^n$ is isotopic to a bridge sphere. It is
hence as required in $M_2^n$.
\begin{figure}[ht!]
\centerline{\includegraphics[width=0.7\textwidth]{\figdir/dsa12}}
\caption{The punctured sphere in $M_2^3$ that is isotopic to a
bridge sphere}
\label{dsa12}
\end{figure}
It now suffices to verify that the portion of $F^n$ lying in $M_1^n$
admits the required pairs of disks. After a small isotopy, $b_1,
\dots, b_n$ lie in the interior of $M_1^n$. We then see that the
portion of $F^n$ lying in $M_1^n$ may be reconstructed from $n$
vertical annuli and one torus by ambient 1--surgery along arcs dual to
$b_1, \dots, b_n$. See \fullref{dsa13}. (Compare to \fullref{dsa10}.)
\begin{figure}[ht!]
\centerline{\includegraphics[width=0.5\textwidth]{\figdir/dsa13}}
\caption{A dual schematic for $F^n \cap M_1^n$}
\label{dsa13}
\end{figure}
Comparing the decomposition here with $(V^n, W^n)$, we see that the
splitting surface $F^n$ is entirely contained in a collar of one of
the handlebodies $V^n$, $W^n$, say $V^n$. Furthermore, it induces a
Heegaard splitting $(X_v^n, Y_v^n)$ of $V^n$ as follows: Take $X_v^n$
to be $X^n \cap V^n = X^n$ and take $Y_v^n$ to be the collar of
$\partial V^n$ together with $Y^n \cap V^n$. Then $Y_v^n = (\text{collar
of } V^n) \cup (\text{solid torus}) \cup (\text{1--handles})$ and hence $X_v^n$
and $Y_v^n$ are both handlebodies.
However, the genus of $F^n$ is $2n$ and the genus of $\partial V^n$
is $n$. It thus follows from Scharlemann and Thompson \cite[Lemma
2.7]{ST} that $(X_v^n, Y_v^n)$ and thus $(X^n, Y^n)$ is stabilized.
By applying \cite[Lemma 2.7]{ST} to locate a stabilizing pair of disks
and using one of the disks to destabilize $n$ times in succession,
we locate the $n$ pairs of stabilizing disks required.
\end{proof}
\bibliographystyle{gtart}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 765 |
← Membership – PFLAG Alamance
Times-News Story on Local Boy Scouts →
CA Boy Scouts Grant Gay Scout Eagle Rank
Posted on January 17, 2013 by ketcham
(from Reuters via Qnotes)
ORLANDO, Florida (Reuters) – A California chapter of the Boy Scouts of America is directly challenging the national organization's ban on gays by formally recommending that an openly gay former Scout be awarded the top rank of Eagle.
"From what I understand, this has never happened before," Eric Andresen, father of former scout Ryan Andresen, told Reuters.
"It's the first in-your-face (challenge)," said Bonnie Hazarabedian, who chaired the Boy Scout district review board that signed off on Ryan's Eagle scout application and forwarded its recommendation last week to the national headquarters for final approval.
Ryan, 18, and his parents drew national attention in October after his Scoutmaster refused to sign his Eagle scout application because of his sexual orientation.
More than 462,000 people subsequently signed the Andresens' petitions at Change.org calling on the Scoutmaster to sign. Meanwhile, the Andresens pushed Ryan's application up the Boy Scout hierarchy in the San Francisco Bay area, where it landed before Hazarabedian.
"I don't think sexual orientation should enter into why a Scout is a Scout, or whether they are Eagle material," said Hazarabedian. "We felt without a doubt he deserved that rank."
The Boy Scouts of America did not respond immediately to questions emailed by Reuters on Monday afternoon. At the time Ryan's case grabbed attention last fall, BSA spokesman Deron Smith issued a statement saying Andresen's Boy Scout membership had been revoked.
Hazarabedian told Reuters she acted on Ryan's Eagle application because it was filed before his ejection.
Eric Andresen, who was a Boy Scout leader until his son was ejected, said the national office normally rubber-stamps the district recommendation within 30 to 60 days. Neither Hazarabedian nor the Andresens expect Ryan's Eagle award will come so easily.
"It's gotten to the point that getting the Eagle doesn't matter so much. It's the message that counts. It's the desire that no other Scout should ever have to go through this," Eric Andresen said.
Hazarabedian called the BSA anti-gay policy "something out of the Dark Ages."
In 1981 when Hazarabedian was a teenager, her friend's brother Tim Curran, a gay Eagle Scout, was the plaintiff in what became the landmark case in California upholding the right of the Boy Scouts, and private organizations in general, to discriminate on the basis of sexual orientation.
Hazarabedian said she recalls making posters and signing petitions in support of Curran and thinking, "By the time I have a son old enough to be a Scout, that will be years from now, they will have fixed this by then, they'll be more tolerant by then. But here we are, 2013, and the same thing is going on."
Ryan, who became something of a celebrity with national television appearances and strangers asking to be photographed with him, wants to focus on his last year of high school, his father said.
"We assume someday BSA will (change), and maybe at that time Ryan can retroactively get his Eagle award," he said.
This entry was posted in Current News. Bookmark the permalink. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 416 |
Tissue stabilization device and method
The present invention relates to a device for stabilizing tissue including a fastener having an elongated member at least partially made of an expandable material. The present invention also relates to a method of stabilizing tissue. The method includes positioning the fastener in tissue, positioning an implant adjacent the tissue such that the fastener extends into the implant, and expanding the fastener.
Bonutti, Peter M. (Effingham, IL, US)
BONUTTI PETER M.
606/326, 606/331, 606/907, 623/23.44
A61B17/88; A61F2/08; A61F2/28; A61F2/36; A61F2/46; B30B1/04; B30B11/02; A61B17/00; A61B17/06; A61B17/86; A61F2/00; A61F2/02; A61F2/30; (IPC1-7): A61B17/86; A61F2/28
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20070260270 Cutter for preparing intervertebral disc space November, 2007 Assell et al.
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20030078601 Crimping and cutting device April, 2003 Shikhman et al.
20070288001 ENDOSCOPICALLY INTRODUCIBLE EXPANDABLE CAUTERY DEVICE December, 2007 Patel
20100057127 Expandable Laminoplasty Fixation System March, 2010 Mcguire et al.
20090018532 LASER FEMTOSECOND MICROTOME FOR CUTTING OUT A MATERIAL SLICE BY A LASER BEAM, IN PARTICULAR IN A CORNEA January, 2009 Salin
20090048613 VISCERAL STAPLES FOR PURSE-STRING CLOSURE OF PERFORATIONS February, 2009 Surti
PREBILIC, PAUL B
PAUL D. BIANCO (Miami, FL, US)
That which is claimed is:
1. A device for stabilizing tissue comprising a fastener having an elongated member at least partially made of an expandable material.
2. A device as defined in claim 1 wherein the expandable material expands by imbibing liquid.
3. A device as defined in claim 2 wherein the elongated member includes proximal and distal portions and at least the distal portion is made of the expandable material.
4. A device as defined in claim 3 wherein at least the distal portion is threaded.
5. A device as defined in claim 4 wherein the proximal portion includes an unthreaded projection.
6. A device as defined in claim 5 wherein the unthreaded projection has a hexagonal cross-sectioned.
7. A device as defined in claim 1 wherein the expandable material expands after being compressed.
10. A device as defined in claim 9 wherein the proximal portion includes an unthreaded projection.
11. A device as defined in claim 10 wherein the unthreaded projection has a hexagonal cross- sectioned.
12. A method of stabilizing tissue with the device of claim 1, the method comprising: positioning a distal portion of the fastener in tissue; positioning an implant adjacent the tissue such that a proximal portion of the fastener extends into the implant; and expanding the expandable material of the fastener.
13. The method of claim 12 wherein expanding the expandable material of the fastener includes expanding the expandable material by imbibing liquid.
14. The method of claim 13 wherein expanding the expandable material by imbibing liquid includes expanding the expandable material radially outward.
15. The method of claim 14 wherein expanding the expandable material radially outward includes expanding the expandable material to lock the fastener into the tissue.
16. The method of claim 12 wherein positioning the distal portion of the fastener in tissue includes screwing a threaded section of the fastener into the tissue.
17. The method of claim 12 further including compressing the expandable material of the fastener and wherein expanding the expandable material of the fastener includes expanding the expandable material after compressing the expandable material.
18. The method of claim 17 wherein expanding the expandable material after compressing the expandable material includes expanding the expandable material radially outward.
20. The method of claim 17 further including expanding the expandable material of the fastener by imbibing liquid.
[0001] This application is a continuation of application Ser. No. 10/279,451, filed Oct. 24, 2002. The aforementioned application Ser. No. 10/279,451 is itself a continuation of application Ser. No. 09/798,870, filed Mar. 1, 2001, now U.S. Pat. No. 6,503,277 B2. The aforementioned application Ser. No. 09/798,870 is itself a continuation of application Ser. No. 09/602,743, filed Jun, 23, 2000, now U.S. Pat. No. 6,361,565. The aforementioned application Ser. No. 09/602,743 is a continuation of application Ser. No. 09/263,006, filed Mar. 5, 1999, now U.S. Pat. No. 6,132,472. The aforementioned application Ser. No. 09/263,006 is itself a continuation of application Ser. No. 08/834,028, filed Apr. 11, 1997, now U.S. Pat. No. 5,888,219. The aforementioned application Ser. No. 08/834,028 is itself a divisional of application Ser. No. 08/590,193, filed Jan. 23, 1996, now U.S. Pat. No. 5,662,710. The aforementioned application Ser. No. 08/590,193 is itself a divisional of application Ser. No. 08/273,028, filed Jul. 8, 1994, now U.S. Pat. No. 5,545,222. The aforementioned application Ser. No. 08/273,028 is itself a divisional of application Ser. No. 07/728,247, filed Aug. 12, 1991, now U.S. Pat. No. 5,329,846. The benefit of the earlier filing dates of the aforementioned applications and patents is claimed.
[0002] Tissue grafting, including bone grafting, is well known. Tissue such as bone is removed from one part of a body (the donor site) and inserted into tissue in another (the host site) part of the same (or another) body. It is desirable to be able to remove a piece of tissue graft material which is the exact size and shape needed for the host site where it will be implanted. However, it is rarely possible to do this.
[0003] Accordingly, various tissue grafting techniques have been tried to solve this problem. For example, U.S. Pat. No. 4,678,470, granted to Nashef, discloses a method of creating bone graft material by machining a block of bone to a particular shape, or by pulverizing and milling it. The graft material is then tanned with glutaraldehyde to sterilize it. This process can produce bone plugs of a desired shape.
[0004] In the Nashef process, the exogenic bone material selected for the graft is presumably dead at the beginning of the process. The process of pulverizing or milling the bone material destroys the structure of the bone tissue. The step of tanning it with glutaraldehyde then renders the graft material completely sterile. This condition is not conducive to graft healing and ingrowth. Specifically, applicant has found that it is desirable to maintain graft tissue in a living state during the grafting process. The use of living tissue in a graft will promote bone healing.
[0005] The present invention is a tissue press for shaping or compressing a piece of tissue. The press includes first and second members movable relative to each other. First and second forming elements of different predetermined shapes are positionable on the first and second members on opposite sides of the piece of tissue. The first and second members are moved toward each other to shape or compress the tissue between the first and second forming elements.
[0006] Means are preferably provided for monitoring and controlling the amount of force or pressure applied to the piece of tissue, in order to maintain the tissue in a viable living condition. Means may also be provided for draining off fluid from compressed tissue, so that the tissue can be implanted in a compressed state and imbibe fluid from the host site.
[0007] The present invention is also a method of reshaping tissue for use as graft material comprising the steps of determining the shape which the tissue should possess as graft material; providing a tissue press, selecting forming elements adapted to press tissue approximately to the desired shape, and placing the forming elements on the tissue press; placing the tissue in the tissue press between the forming elements; operating the tissue press to shape the tissue between the forming elements to give the tissue the desired shape; and controlling the pressure on the tissue during the shaping step to minimize damage to or necrosis of the tissue.
[0008] The present invention is also a method which includes the steps of determining the shape and size which the tissue should possess as graft material; placing the tissue in a tissue press having forming elements adapted to press tissue approximately to the shape and size desired; compressing the tissue in the tissue press to give the tissue the shape and size desired; controlling the pressure on the tissue during the compressing step to minimize damage to or necrosis of the tissue; and draining off fluid expressed during the compressing step.
[0009] In accordance with another feature of the present invention, a retainer is provided for retaining tissue graft material in its compressed state. The retainer is placed around the compressed graft. The retainer can help to maintain the graft in a compressed configuration or in a specific shape for a period of time long enough to be placed in the body. The retainer may be made of a material which expands after it is placed in the body, to mechanically interlock the graft to the body.
GENERAL DESCRIPTION OF THE INVENTION
[0010] With the apparatus and method of the present invention, bone or other tissue can be compressed or reshaped or both, while preserving the tissue alive.
[0011] Often, reshaping of graft tissue is necessary to obtain the best possible graft. For example, in an anterior cruciate ligament reconstruction, the graft material which is removed usually has a triangular cross-sectional configuration. This graft material is placed in an opening in bone formed by drilling with a round drill. When the triangular graft material is placed in the round opening, there is minimal physical contact between the graft material and the surrounding bone. This reduces the holding power of the graft and also reduces the ingrowth ability of the graft.
[0012] Thus, an important feature of the present invention is that bone or other tissue such as ligament is reshaped while still leaving it in a whole condition and without substantial tissue damage. The tissue is placed in the tissue press of the present invention and sufficient force is applied to reshape the tissue to the desired shape—for example, a cylindrical shape as needed for an anterior cruciate ligament reconstruction. Excessive pressure on the tissue, which can damage or kill the tissue, is avoided as described below. A properly shaped graft is thus provided which is still in a living condition.
[0013] Compression of graft tissue is also sometimes desirable. Generally, tissue is stronger when it is denser. Compressing graft tissue increases its density and thus strengthens the graft tissue. The graft tissue also stays together better.
[0014] For example, a tendon is made of a plurality of fibers. The individual fibers are weak when separated or unraveled. If a tendon graft is implanted with the fibers in a loose condition, the graft is weak. On the other hand, if prior to implantation the tendon graft is compressed to orient and pack the fibers tightly, then the entire group of fibers acts as one whole unit and the graft is much stronger. Therefore, compressing the tendon graft gives it more mechanical integrity—making a smaller tendon graft much stronger.
[0015] Similarly, bone tissue is stronger and better able to bear force when it is denser and more compact. Compressing bone graft tissue prior to implantation produces a stronger graft. Compression of bone or other tissue also allows a surgeon to convert a larger irregular shape into a smaller specific shape. Thus, the surgeon when removing the graft material from the donor Site is not limited by the conditions at the host site but can remove the graft material in the best way possible from the donor site. Similarly, the surgeon when implanting the graft material at the host site is not limited by the shape of the material removed (as dictated by the conditions at the donor site) but can implant the graft material in the best way possible to fit the conditions at the host site.
[0016] The anterior cruciate ligament, for example, attaches to the femur and tibia at specific isometric locations. When the ligament is being replaced in an anterior cruciate ligament reconstruction, typical uncompressed graft material can be many times the size of those locations. In such a case it is necessary to drill openings much larger than desired in the bone to attach the new ligament. The graft tendon then tends to fall eccentrically in this larger opening, the functional anatomy of the ligament can not be recreated, and the functioning of the knee joint is compromised.
[0017] However, if the graft material for the new ligament is compressed in accordance with the present invention, its size can be reduced substantially. This allows the surgeon to drill a substantially smaller opening in the bone to attach the new ligament/graft structure, so as to recreate the functional anatomy of the ligament.
[0018] With the present invention, it is also possible to make a composite graft. For example, the graft material for an anterior cruciate ligament reconstruction is preferably tendon in the middle with bone at both ends. In accordance with the apparatus and method of the present invention, bone tissue can be compressed around the ends of tendon tissue to form a substitute anterior cruciate ligament more closely approximating the original.
[0019] It should also be noted that tissues other than bone and tendon can be worked with the tissue press. For example, a surgeon can harvest liver cells or pancreas cells and then compress them into a particular shape. They can then optionally be placed into a sack or some type of structural support which can be introduced into the body.
[0020] With the present invention, graft material can be formed into almost any shape. A specific pair of forming (mold) parts, having a desired predetermined shape, are positioned on the tissue press, and the tissue is shaped or compressed between the forming elements. In addition to three-dimensional shaped parts, it is also possible to make a flat piece of graft material. For example, shaved skin can be placed on a flat plate, perhaps on a retaining mesh. The cells are then subjected to pressure to adhere them together. A flat, even, piece of graft material is formed which is suitable for skin grafting.
[0021] Compressing graft material in accordance with the present invention also allows the surgeon to build up a larger piece of graft material out of several smaller parts. Sometimes a relatively large piece of graft material is needed for a particular host site. It is often not feasible to take such a large piece of graft material without damaging the donor area. To avoid this problem, several smaller pieces of graft material are placed in the tissue press and pressure is used to at least temporarily form the smaller parts together as one larger whole. The larger graft piece is then inserted into the host site.
[0022] Compressing graft material in accordance with the present invention also aids in introducing additional materials to the graft material. These additional materials could be antibiotics, bone growth enhancers, tri-calcium phosphate, fibrin, allograft or autograft material, etc. When added to the graft material under pressure, the added materials adhere to and become a part of the graft material and not merely something added to the surface of it. By combining physiologic solutions or a carrier such as a gelatin, polysaccharides, antibiotics or synthetic bone materials to the compressed bone, for example, it is possible to create a plug of living bone with the other materials added into it. This plug has the graft properties of the bone tissue in the plug, as well as the properties of the added material.
[0023] When tissue is compressed, fluid may be forced out of (expressed from) the tissue. If tissue in this compressed and defluidized state is laid back in a tissue pouch or in a bone hole, body fluids from the host site are absorbed by the graft material. This imbibition causes swelling of the graft material and thus creates a mechanical interlock between the graft and the host. Such a mechanical interlock is not produced with a typical implantation process in which graft tissue is not compressed. Further, the swelling (enlarging) of the graft material allows the graft material to fill an opening of any given shape with a perfect fit of the graft material therein.
[0024] As noted above, applicant has found that it is desirable to maintain bone graft tissue in a living state during the grafting process. It is important not to kill tissue used in grafting because the living graft cells provide a superior substrate for grafting and graft viability and improved tissue healing. There is significantly faster incorporation of living tissue than of dead tissue. The cells and the tissue that are implanted into the body therefore need to be maintained in a viable condition.
[0025] Excess pressure on tissue can cause destruction of the tissue, disorganization of the tissue fibers and irregular mechanical structure which can damage the tissue graft. Thus, a feature of the present invention is that the pressure or force on the tissue being reshaped or compressed is monitored and controlled. Pressure can be monitored by suitable pressure sensors and readouts such as a pressure gauge. Pressure can be controlled by force limiting means such as a torque wrench or similar device.
[0026] The desired pressure levels may vary. For example, it may be desirable to provide a higher compressive force for cortical bone than for cancellous bone. Similarly, it may be desirable to provide a higher compressive force for bone than for tendon tissue. The appropriate level of pressure or force is selectively available with the tissue press.
[0027] A separate device or structure can be used to maintain graft tissue in the compressed state prior to and during implantation. This separate device or structure can be a mesh sack, a ring around a cylindrically shaped graft material, etc. This additional retainer structure can assist the surgeon in introducing the graft tissue into the body in the compacted condition, to provide a denser stronger graft and to allow imbibition for creating a mechanical interlock. The retainer can be made of a material which expands when placed in the body, to provide a mechanical interlock for the graft tissue.
[0028] The foregoing and other features of the present invention will become apparent to one skilled in the art upon a consideration of the following description of the invention with reference to the accompanying drawings, wherein:
[0029] FIG. 1 is a top plan view of a tissue press embodying the present invention;
[0030] FIG. 2 is a top plan view of a tissue press similar to the tissue press of FIG. 1 and having a pressure sensing and monitoring mechanism;
[0031] FIG. 3 is an elevational view of the tissue press of FIG. 2;
[0032] FIGS. 3A and 3B illustrate a tissue press having means for removing excess tissue after compression;
[0033] FIGS. 4A-4F illustrate a plurality of different forming elements for use in the tissue press of FIGS. 1-3;
[0034] FIG. 5 is a view of a tissue press in accordance with a second embodiment of the invention;
[0035] FIG. 6 is a view of a tissue press in accordance with a third embodiment of the invention;
[0036] FIG. 6A illustrates a tissue press having means for limiting the amount of pressure applied to the tissue;
[0037] FIGS. 7 and 7A illustrate a composite tissue graft such as compression of bone around tendon;
[0038] FIG. 8 is a view of a tissue press in accordance with a fourth embodiment of the invention illustrating extrusion of tissue graft material;
[0039] FIGS. 9A-9F illustrate a plurality of different retainers for compressed tissue;
[0040] FIGS. 10A and 10B illustrate an expanding tissue retainer;
[0041] FIGS. 11, 11A and 11B illustrate an expanding surgical implant; and
[0042] FIGS. 12 and 12A illustrate expanding surgical stabilization devices.
DESCRIPTION OF PREFERRED EMBODIMENTS OF THE INVENTION
[0043] The present invention relates to a tissue press and particularly to a tissue press for shaping or compressing bone or other tissue. The present invention is applicable to various constructions of tissue shaping or compressing apparatus. As representative of the present invention, FIG. 1 illustrates a tissue press 10.
[0044] The tissue press 10 includes a base 12. The base 12 has a support surface 14 for supporting the base 10 and thus the press 10 on a table or other support (not shown). The base 12 has the same longitudinal cross-section throughout its width as is seen in elevation in FIG. 1. Two cylindrical pins 16 protect upwardly from the base 12. A U-shaped saddle 18 is secured to the base 12 by screws 20. A pin opening 22 extends transversely through the saddle 18.
[0045] The tissue press 10 also includes a handle 30. The handle 30 has a connector portion 32 received between the upwardly projecting side portions of the saddle 18 of the base 12. A pin opening 34 extends transversely through the connector portion 32 of the handle 30. The opening 34 is aligned with the opening 22 in the saddle 18. A pivot pin 36 extends through the pin openings 34 and 22 in the handle 30 and base 12, respectively, and pivotally connects the handle 30 to the base 12.
[0046] Extending from the connector portion 32 of the handle 30 is a form-supporting portion 40 of the handle 30. A channel 42 is defined on the lower side of the form-supporting portion 40 between a pair of channel surfaces 44. The channel 42 extends transversely across the width of the form-supporting portion 40. Extending from the form-supporting portion 40 is a circular portion 46 of the handle 30. The circular portion 46 has a manually engageable surface 48 for the application of force to the handle 30.
[0047] A first forming element 50 is slidably received on the base 12. The first forming element 50 has two vertically extending pin openings 52 which receive the pins 16 of the base 12. The first forming element 50 has an upper surface 54 which includes a forming surface 56. The forming surface 56 has a cylindrical surface portion 58. The first forming element 50 has the same longitudinal cross-section throughout its width as is seen in elevation in FIG. 1. It should be understood that the first forming element 50 can be connected with the base 12 in any other suitable manner.
[0048] A second forming element 60 is slidably received in the channel 42 of the handle 30. The second forming element 60 has two projecting edges 62 which engage the channel surfaces 44 of the handle 30. The second forming element 60 has a lower surface 64 which includes a forming surface 66. The forming surface 66 includes a cylindrical surface portion 68. The second forming element 60 has the same longitudinal cross-section throughout its width as is seen in elevation in FIG. 1. It should be understood that the second forming element 60 can be connected with the handle 30 in any other suitable manner.
[0049] In operation of the press 10, a pair of forming elements 50 and 60 are selected which when brought together will shape a piece of living tissue 70 to the desired shape. For example, the forming elements 50 and 60 illustrated in FIG. 1 have cylindrical surface portions which will shape tissue into a cylindrical shape suitable for insertion into a round drill hole in bone. The first forming element 50 is slid onto the pins 16 on the base 12. The second forming element 60 is slid into the channel 42 in the handle 30. A set screw 72 is tightened to secure the second forming element 60 to the handle 30.
[0050] The piece of tissue 70 to be shaped is placed between the first and second forming elements 50 and 60. The tissue 70 is preferably positioned on the forming surface 56 of the first forming element 50 as illustrated in FIG. 1. The handle 30 is then brought toward the base 12 in the direction indicated by the arrow 74, by the application of force to the handle 30. The forming surface 66 with its cylindrical surface portion 68 engages the tissue 70, and forces the tissue 70 against the forming surface 56 with its cylindrical surface portion 58. The tissue 70 is thereby formed to the desired shape.
[0051] It should be understood that with the tissue press 10, tissue can be reshaped only, or reshaped and compressed. The amount and type of work performed on the tissue depends on the particular application, and is controlled by (i) the particular forming elements selected by the operator of the tissue press, and (ii) the amount of pressure or force applied to the tissue. For example, the forming elements can be selected so that they define between them a forming chamber having the same initial volume as the piece of tissue to be shaped. In this case, the piece of tissue is not compressed (that is, reduced in volume), but is merely reshaped. Alternatively, forming elements can be selected which will not merely reshape the piece of tissue but also will compress it, as described above. Because the first and second forming elements 50 and 60 are removably received on the base 12 and handle 30, respectively, the surgeon can easily select first and second forming elements 50 and 60 to reshape or compress the tissue 70 as desired.
[0052] As discussed above, it is important to control the pressure on the tissue in the tissue press 10. There are many ways to do this. For example, in the press 80 shown in FIG. 2 and 3, a second forming element 84 has an opening 82 which extends between its upper surface 86 and its forming surface 88. An opening 90 in the handle 92 is aligned with and in fluid communication with the opening 82. A pressure monitor device indicated schematically at 94 is in fluid communication with the opening 90 in the handle 92, and thus with the forming surface 88 on the second forming element 84. The pressure monitor device 94 can be any known device for displaying pressure, such as a dial readout, a bar movable axially in the openings 82 and 90 in response to fluid pressure, etc. The pressure monitor device 94 displays the pressure at the forming surface 88 of the second forming element 84. This is the pressure applied to the tissue being pressed in the press 80. Thus, an operator of the press 80 can observe this pressure monitor device 94 and limit the applied force accordingly, in order to avoid tissue damage or necrosis.
[0053] Alternatively, a strain gauge of a known construction, indicated schematically at 96, can be placed on the forming surface 88 of the second forming element 84. Electric wires (not shown) transmit signals from the strain gauge 96 to an electric readout device indicated schematically at 94. The electric readout indicates electrically the pressure applied to the tissue in the press 80. Again, one can observe this pressure monitor device and limit the amount of force applied, in order to avoid tissue damage or necrosis. It should be understood that similar ways of monitoring the pressure on the tissue being pressed are the equivalent and are thus included within the scope of the invention.
[0054] Any of the tissue presses of the present invention can include means for cutting off excess tissue. As an example, FIG. 3A and 3B illustrate a tissue press 91 having an independent cutoff arm 93 pivotally mounted adjacent the handle 95. The cutoff arm 93 has a blade portion 97 operable to cut off excess tissue compressed by the tissue press 91. The cutoff arm 93 is, of course, merely illustrative of the many equivalent structures usable to remove excess tissue after the compressing or shaping operation. Thus, the length of the tissue graft material, as well as its compressed diameter, can be selectively controlled by operation of the tissue press. It should be noted that this cutting off process can be affected with the edges of the forming elements themselves, as illustrated in the apparatus in FIG. 5.
[0055] As noted above, the first forming element 50 and the second forming element 60 are removably received on the base 12 and handle 30, respectively. Thus, forming elements having forming surfaces with other shapes can be easily placed in the tissue press 10, in order to shape or compress tissue into other shapes. A few of the many shapes obtainable are illustrated in FIG. 4.
[0056] FIG. 4A, for example, illustrates the forming elements 50 and 60 of FIG. 1-3 which shape tissue into a cylindrical cross-sectional shape. FIG. 4B illustrates forming elements 100 and 102 which shape tissue into an oblong cross-sectional shape, between a forming surface 104 on the first forming element 100 and a forming surface 106 on the second forming element 102. FIG. 4C illustrates forming elements 108 and 110 which shape tissue into a square cross-sectional shape, between a forming surface 112 on the first forming element 108 and a forming surface 114 on the second forming element 110.
[0057] FIG. 4D illustrates forming elements 116 and 118 which shape tissue into a triangular cross-sectional shape, between a forming surface 120 on the first forming element 116 and a forming surface 122 on the second forming element 118. FIG. 4E illustrates forming elements 124 and 126 which shape tissue into a generally flat shape, between a forming surface 128 on the first forming element 124 and a forming surface 130 on the second forming element 126. FIG. 4F illustrates forming elements 132 and 134 which shape tissue into a semi-circular cross-sectional shape, between a forming surface 136 on the first forming element 132 and a forming surface 138 on the second forming element 134. Again, it should be understood that other shapes are conceivable and consequently are included within the scope of the invention.
[0058] In addition to three-dimensional shaped parts, it is also possible to make a flat piece of graft material. Forming elements like those shown in FIG. 4E are useful in this case. For example, a skin graft may be placed on an adhesive based mesh (possibly using fibrin) on the flat forming surface 128. The cells are then subjected to pressure to adhere them together. The cells are spread out over a finer layer. A flat, even, piece of skin graft material is formed. This can then be cut or pressed or formed into a specific shape and then used as a skin substitute on the body.
[0059] Any forming element useful in the present invention can be coated with a non-stick coating to reduce adhesion of the compressed tissue to the forming elements. For example, the forming element 100 (FIG. 4B) is indicated as having a non-stick coating 105 such as a Teflon®. coating which may be applied in any suitable known manner. Such a coating can resist the binding of the tissue (or tissue additive such as fibrin discussed below) to the forming element 100. In a second embodiment of the invention, illustrated in FIG. 5, a tissue press 150 is fluid operated (pneumatic or hydraulic) rather than manually operated. The press 150 includes a base 152 having a support surface 154 for supporting the base 152 and thus the press 150 on a table 15 or other support (not shown). A first forming element 156 is attached to the base 152. The first forming element 156 may be slidably or otherwise attached to the base 152 in any suitable manner which blocks movement of the forming element 156 relative to the base 152 during operation of 20 the press 150 and which allows for easy interchange of forming elements 156. The first forming element 156 has an upwardly facing forming surface 158. A fluid drain opening 160 is in fluid communication with the forming surface 158 of the first forming element 156.
[0060] An air or hydraulic cylinder 170 is fixed to the base 152 above the first forming element 156. The cylinder 170 has a port 172 for the introduction of air or other fluid under pressure to operate a piston 174 in a known manner. A second forming element 180 is connected to the piston 174. The second forming element 180 is attached to the piston 174 in any suitable manner which blocks movement of the forming element 180 relative to the piston 174 during operation of the press 150 and which allows for easy interchange of forming elements 180. The second forming element 180 has a forming surface 182 facing the forming surface 158 on the first forming element 156.
[0061] A port 176 in the cylinder 170 is in fluid communication with the forming surface 182 of the second forming element 180. Connected to the port 176 is a pressure monitor device shown schematically at 178.
[0062] In operation of the press 150, a pair of forming elements 156 and 180 are selected which when brought together will shape tissue to the desired shape. The first forming element 156 is attached to the base 152. The second forming element 180 is attached to the piston 174. A piece of tissue to be shaped (not shown) is placed between the first and second forming elements 156 and 180. The piece of tissue is preferably positioned on the forming surface 158 of the first forming element 156.
[0063] The cylinder 170 is then actuated to move the second forming element 180 toward the first forming element 156. The forming surface 182 on the second forming element 180 engages the tissue, and forces the tissue against the forming surface 158 on the first forming element 156. Pressure on the tissue is controlled through observation of the monitor device 178. The tissue is formed to the desired shape. Again, it should be understood that with the press 150 tissue can be reshaped only, or compressed also, depending on the application, the forming elements selected, and the amount of force applied.
[0064] As discussed above, when tissue is compressed, fluid may be expressed from the tissue. In the press 150, the second forming element 180 fits within the first forming element 156 to define between them a closed forming chamber in which the tissue is compressed. Expressed fluid is drained from the forming chamber through the fluid drain opening 160. If a closed forming chamber is not formed, as for example with the open-ended forming elements shown in FIG. 1-4, then expressed fluid can drain outwardly from the tissue being pressed, without the need for a separate fluid drain port. Of course, a separate fluid drain port could be provided in any of the forming elements of the present invention.
[0065] In a third embodiment of the invention, illustrated in FIG. 6, a tissue press 200 includes a base 202 having a support surface 204 for supporting the base 202 and thus the press 200 on a table or other support (not shown). A first forming element 206 is attached to the base 202. The first forming element 206 may be slidably or otherwise attached to the base 202 in any suitable manner. The first forming element 206 includes a plurality of first fingers 208 which together have a forming surface 210 to progressively compress bone or other tissue into a predetermined shape.
[0066] An upper arm 220 is pivotally mounted to the base 202 by a pivot pin 222. A second forming element 224 is connected to the upper arm 220 in a suitable manner. The second forming element 224 includes a plurality of second fingers 226 which together have a forming surface 228 facing the forming surface 210 on the first forming element 206. The second fingers 226 are interdigitable with the first fingers 208.
[0067] In operation of the press 200, a pair of forming elements 206 and 224 are selected which when brought together will progressively shape tissue to the desired shape. The first forming element 206 is attached to the base 202. The second forming element 224 is attached to the upper arm 220. A piece of tissue to be shaped (not shown) is placed between the first and second forming elements 206 and 224. The piece of tissue is preferably positioned on the forming surface 210 of the first forming element 206. The upper arm is pivoted toward the base to move the second forming element 224 toward the first forming element 206. The forming surface 228 on the second forming element 224 engages the tissue, and forces the tissue against the forming surface 210 on the first forming element 206. The tissue is formed to the desired shape.
[0068] Because the second fingers 226 are interdigitable with the first fingers 208, the press 200 is operable to compress tissue to different compressed sizes with only one pair of forming elements. As the second fingers 226 come together with the first fingers 208, they compress the tissue to a smaller and smaller diameter (shape). This allows for one pair of forming elements to provide compression to variable diameters or sizes. This works well with soft tissue applications, specifically tendons, to compress the tendon into a smaller shape. The amount of compression is based on the amount of pressure applied and the needed finished size.
[0069] As noted above, it is important to control the pressure or force applied to the tissue by the tissue press. Accordingly, the present invention provides means for limiting the amount of pressure applied to the tissue by the tissue press, that is, means for blocking application to the tissue of force in excess of a predetermined amount. Such means are schematically illustrated in FIG. 6A, which illustrates a tissue press 300 having a known torque wrench assembly included therein.
[0070] The press 300 includes a base 302. Attached to the base in the manner described above is a first forming part 304. Also attached to the base is a saddle 306. Received in the saddle 306 is the connector portion 308 of a handle assembly 310. The press 300 also includes a pivot pin 312 pivotally interconnecting the handle assembly 310 and the base 302.
[0071] The handle assembly 310 includes a form-supporting portion 314 to which there is attached in the manner described above a second forming part 316. The handle assembly 310 also includes a second portion 318 connected to the form-supporting portion 314 by a drive mechanism 320. The second portion 318 includes a knurled section 322 which is rotatable about an axis 324. On the second portion 318 there is a gauge 326.
[0072] The knurled section 322 is rotatable about the axis 324 to set the torque value desired and as shown on the gauge 326. Thereafter, the handle assembly 310 can be pivoted toward the base 302 in the direction indicated by the arrow 328 only until the preset amount of torque is applied. At that point, no more torque is transferred through the drive mechanism 320 to the form-supporting portion 314. This limits the amount of pressure applied to the tissue by the second forming part 316, that is, blocks application to the tissue of force in excess of a predetermined amount.
[0073] It should be understood that the torque wrench assembly or construction indicated in FIG. 6A is only illustrative of the many ways in which the amount of pressure applied to the tissue by the tissue press can be limited to a predetermined amount. There are other known mechanisms for performing the same function, and their use is included within the scope of the present invention.
[0074] FIG. 7 and 7A illustrate the use of a tissue press in accordance with the present invention to form a composite graft. As discussed above, with the present invention, it is also possible to make a composite graft. For example, the graft material for an anterior cruciate ligament reconstruction is preferably tendon in the middle with bone at both ends. In accordance with the apparatus and method of the present invention, bone tissue can be compressed around the ends of tendon tissue to form a substitute anterior cruciate ligament more closely approximating the original.
[0075] Thus, as illustrated schematically in FIG. 7 and 7A, the tissue press 10 of FIG. 1-4 is being used to compress bone tissue 240 around tendon tissue 242 to form a substitute anterior cruciate ligament 244. The tendon 242 can be harvested from one site and the bone 240 can be harvested from another site.
[0076] It should be understood that the graft can be multiple tissue fragments rather than a composite material. Thus, the tissue press 10, or indeed any tissue press in accordance with the present invention, can be used to compress, for example, multiple bone fragments into one larger piece. It should also be understood that the tissue press in accordance with the present invention can be used to add additional materials to body tissue material by pressure. For example, to bone tissue there can be added tri-calcium phosphate, an antibiotic, hydroxyapatite, allografts or autografts, or any other polymeric. This process is believed to be self-explanatory in light of the foregoing description, but for reference may be understood by referring to FIG. 7 and 7A wherein 240 would be the bone tissue or other tissue to which material is being added (squeezed in under pressure), and 242 indicates the additional material being added to the tissue 240.
[0077] In this case, fibrin can be highly suitable for use as such an additional material. Fibrin is a blood component important in blood clotting. It can be separated or centrifuged from blood and has the nature of an adhesive gel. Fibrin can be used as an adhesive, either in a natural state or after being compressed, to hold together material such as separate tissue pieces pressed together in a tissue press of the present invention.
[0078] In a fourth embodiment of the invention, illustrated in FIG. 8, a tissue press 250 is operated to extrude rather than press material. The press 250 includes a base 252 having a support surface 254 for supporting the base 252 and thus the press 250 on a table or other support (not shown). A die 256 is attached to the base 252. The die 256 may be slidably or otherwise attached to the base 252 in any suitable manner which blocks movement of the die 256 relative to the base 252 during operation of the press 250 and which allows for easy interchange of forming elements 256. The die 256 has an upwardly facing opening 258. An extrusion opening 260 is in fluid communication with the opening 258 of the die 256.
[0079] An air or hydraulic cylinder 270 is fixed to the base 252 above the die 156. The cylinder 270 has a port 272 for the introduction of air or other fluid under pressure to operate a piston 274 in a known manner. A ram 280 is connected to the piston 274. The ram 280 has a surface 282 facing the opening 258 on the die 256.
[0080] In operation of the press 250, a die 256 is selected which will extrude tissue in the desired shape. The die 256 is attached to the base 252. A piece of tissue to be extruded (not shown) is placed in the opening 258 of the die 256. The cylinder 270 is then actuated to move the ram 280 toward the die 256. The surface 282 on the ram 280 engages the tissue, and forces the tissue into and through the die 256, exiting through the opening 260. The tissue is extruded in the desired shape. As discussed above, a fluid drain port can be provided in the press 250.
[0081] It can also be useful to heat or cool the tissue being worked in a tissue press of the present invention. Accordingly, the present invention contemplates the use of means for selectively controlling the temperature of the piece of tissue while it is being compressed or shaped. As an example, illustrated schematically in FIG. 4 is a fluid passage 284 extending from the outer surface of the forming element 124 and around the forming surface 128 thereof. Fluid which is either heated or cooled flows through the passage 284 and either cools or heats the material of the forming element 124 in the area adjacent the forming surface 128. Thus, the tissue, when it comes in contact with the forming surface 128, can be selectively heated or cooled during the compression or reshaping operation. Heating can be useful in holding together materials being compressed, for example, and cooling can be useful to avoid tissue damage arising from overheating of tissue being compressed. It should be understood that other means of achieving these functions are contemplated, such as electrical heating elements. Further, both forming elements can be heated or cooled rather than just one. Any such equivalent structure is to be considered within the scope of the present invention.
[0082] Also in the present invention, a retainer is provided for retaining tissue graft material in its compressed state. After the graft is compressed, the retainer is placed around the graft. The retainer can help to maintain the graft in a compressed configuration or in a specific shape for a period of time long enough to be placed in the body.
[0083] The retainer may be one of many different shapes. The shape of the retainer is chosen to meet the specific application. There are a number of suitable shapes, such as a ring, a cylinder, a cage, a rectangular shape, a mesh, a suture-like wrap, etc. Some of these are illustrated schematically in FIG. 9A-9F. It should be understood that this is not an exhaustive listing, but rather that these are merely exemplary of the principle involved, and accordingly, the invention is not limited to these particular shapes. For example, a retainer may be provided which is in the particular shape of the tissue material being compressed, which can be rectangular, cylindrical, planar, etc.
[0084] FIG. 9A illustrates a plurality of bands or rings 290 used to hold together compressed tissue 292. FIG. 9B illustrates a cage 294 which can be used to hold together the compressed tissue 292 of FIG. 9A. The cage 294 includes a plurality of crossed filaments 296 which define between them a series of openings 298 for tissue ingrowth. FIG. 9C illustrates another cage 300 which can be used to hold together the compressed tissue 292 of FIG. 9A. The cage 300 includes a plurality of longitudinally extending filaments 302 which define between them a series of openings 304 for tissue ingrowth. FIG. 9D illustrates a solid-walled cylinder 306 which can be used to hold together the compressed tissue 292. FIG. 9E illustrates a mesh cylinder 308 which can be used to hold together the compressed tissue 292. FIG. 9F illustrates the wrapping of a cord or suture 310 around compressed tissue 312.
[0085] Any of these retainers may be made of various materials. The material of the retainer is chosen to meet the specific application. Some of the many materials which are suitable are biodegradable materials, ceramics (especially with bone-growth enhancers, hydroxyapatite, etc.); polymeric material such as Dacron® or other known surgical plastics; metal; or composite materials.
[0086] In use, the graft material may be pushed into the retainer structure after graft material is compressed. Alternatively, the graft material may be compressed with the retainer structure. After the graft material is compressed in the retainer, the combined structure of graft plus retainer is placed in the host site in the body. The retainer helps to maintain the graft in a compressed configuration or in the specific shape into which it was compressed for a period of time long enough to be placed in the body.
[0087] If the retainer is made of a biodegradable material, then the retainer degrades and disappears after a period of time. If the retainer is not made of a biodegradable material, then the retainer remains in the body. Tissue ingrowth occurs to bind the host tissue to the graft material. Tissue ingrowth through and around the retainer, between the host tissue and the graft material, is promoted if there are openings as discussed above in the retainer.
[0088] The invention, the retainer may, if desired, be made of a material which expands after it is placed in the body, to mechanically interlock the graft to the body. The expansion can take place in one of two ways. First, the retainer can itself be compressed, as with the tissue, and then expand when placed in the body. Second, the retainer can be made of a material which expands when it comes in contact with water or other bodily fluids. (It should be noted that the tissue can itself be compressed then expand when contacted by water. As an example, a tendon can be compressed in a desiccated state, and as it imbibes water it expands and creates a firmer lock or tighter fit in the host site.)
[0089] The expandable material can first be compressed with the tissue being grafted, and which then expands when placed in the body. The retainer is preferably made of a material which has more structural stability than the tissue being grafted, and provides mechanical integrity and structural support for the graft tissue. A retainer made of a solid polymeric material, for example, is useful to retain in a compressed state a tendon or bone tissue graft.
[0090] These expandable materials can be used not only to retain graft material, but for any shape required for stabilization surgery, such as a wedge, screw, rivet, retaining ring, or spacer, an intramedullary rod, a joint replacement part such as a femoral component of acetabular cup, an expandable sleeve, or another mechanical structure. The expandable materials thus can be used both as a carrier or retainer for another material (e.g. tissue graft material) and on their own as a prosthetic element.
[0091] There are a number of suitable materials which expand when they come in contact with water or other fluids. One is PEEK (polyether-etherketone). A desiccated biodegradable material, or a desiccated allograft may also be used.
[0092] As a simple example, an expandable retainer 330 (FIG. 10B) with graft tissue 332 therein is placed into a tissue or bone space 334 defined by an edge 336 in host tissue 38. As the retainer 330 imbibes body fluids or water from the host tissue 338, it expands radially outwardly into the tissue or bone space 334 and creates a mechanical interlock (FIG. 10A). It also expands radially inwardly and clamps on the graft tissue 332. Therefore, the graft tissue 332 is locked into the host site, without the necessity of damaging the tissue further through some other kind of attachment means.
[0093] For example, a hip replacement (femoral head) is typically made of metal. To implant the replacement, the softer, inner cancellous bone of the femur is first removed. The inner surface of the cortical bone is then machined to provide a close fit between the external surface of the replacement and the hard outer cortical bone material. All this requires a substantial opening in the femur and still does not guarantee a close enough fit for the implant.
[0094] If, instead, the implant is made of an expanding material such as PEEK, only a smaller opening is needed, thus reducing trauma to the bone. Although it is best to lock against the cortical bone, it is possible to implant solely in the cancellous bone, which because of the expansion of the implant provides a better fit than a metal implant. A benefit of implanting in the cancellous bone is reduction of the danger of putting the implant in so tightly that the cortical bone is split (wedged open). Further, if the opening in the bone is not exactly the same shape as the outer surface of the implant, the implant expands to provide a custom contoured fit to the bone and provide immediate mechanical stability. Thus, less machining of the bone is needed, while at the same time obtaining a closer fit.
[0095] Thus, as illustrated in FIG. 11-11B, a hip replacement (femoral head) 340 is made of PEEK or another expandable material. The replacement 340 is inserted into an intramedullary channel 342 cut into a femur 344. The replacement 340 is smaller in diameter than the channel 342. The replacement 340 absorbs body fluids and expands to lock itself into the channel 342 in the femur 344. (It should be understood that the scale shown in FIG. 11-11B is exaggerated as to the amount by which the replacement 340 expands.)
[0096] Similarly, a bone plate or other structure or tissue can be secured to a bone with a fastener made of such an expandable material. As illustrated schematically in FIG. 12 and 12A, a bone plate 350 is secured to a bone 352. In FIG. 12, a fastener 354 is used which has an unthreaded portion 356 extending into the bone 352. The fastener 354, or at least the unthreaded portion 356, is made of PEEK or another suitable expandable material. The portion 356 imbibes fluid from the bone 352 and expands radially outwardly, from an unexpanded condition as shown in phantom at 358 to an expanded condition as shown in solid lines at 360, to lock the fastener 354 into the bone 352. This enables the securing of the plate 350 to the bone 352 without cutting threads into the bone 352 as is usually done.
[0097] In FIG. 12A, a fastener 362 has a threaded portion 364 extending into the bone 352. The threaded portion 364 is made of PEEK or another suitable expandable material. The threaded portion 364 imbibes fluid from the bone 352 and expands radially outwardly to additionally lock the fastener 362 into the bone 352. Alternatively, the fastener 362 of FIG. 12A may have a coating 366 on its portion threaded into the bone 352. The coating 366 is made of PEEK or another suitable expandable material. The coating 366 imbibes fluid from the bone 352 and expands radially outwardly to additionally lock the fastener into the bone 352.
[0098] When such a fastener is made of a non-metal expandable material, removal of the fastener simply entails drilling out the center thereof. This is much easier than with a typical metal bone screw.
[0099] From the above description of the invention, those skilled in the art will perceive improvements, changes and modifications. Such improvements, changes and modifications within the skill of the art are intended to be covered by the appended claims.
Previous Patent: Anti-splay medical implant closure with multi-surface removal aperture
Next Patent: Apparatus and method for reconstructing a ligament | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,864 |
\section{Introduction}
\begin{figure}\centering
\begin{overpic}[width=8cm]{pub/fig/dom}
\put(50,22){$\Omega$}
\put(35,60){$T_\varepsilon$}
\end{overpic}
\caption{A domain $\Omega$ and holes $T_\varepsilon.$\label{domep}}
\end{figure}
Let $\Omega\subset\mathbb{R}^d,d\geq 2$ be open and bounded with $C^2$ boundary. We consider a union $T_\varepsilon$ of holes concentrated at subsets of $\mathbb{R}^d$ as Figure \ref{domep}, and domains $\Omega_\varepsilon=\Omega\setminus T_\varepsilon.$
We consider Poisson problems on $\Omega_\varepsilon$ with Dirichlet condition with $f\in L^2(\Omega),$ that is,
\begin{equation}\label{solus}
u^\varepsilon\in H_0^1(\Omega_\varepsilon),~~-\Delta u^\varepsilon=f.
\end{equation}
We will see $u^\varepsilon$ converge to $u$ which satisfies
\begin{equation}\label{PDE}
u\in H_0^1(\Omega),~~(-\Delta+V) u=f,
\end{equation}
where $V$ is a simple function.
Details of assumptions for $T_\varepsilon$ and the main result are given in Section \ref{ass} as Figure \ref{sets}.
\section{Assumption and the main result}\label{ass}
We use $\bigsqcup$ instead of $\bigcup$ for disjoint union of sets. Let $A\subset\mathbb{R}^d$ be measurable and bounded and $\Lambda\subset\mathbb{R}^d$ be countable and $\mathbb{R}^d=\bigsqcup_{i\in\Lambda}(A+i).$
For $x\in\mathbb{R}^d$ and $R>0,$ we denote $B(x,R)=\{y\in\mathbb{R}^d|~|x-y|<R\}.$ We denote Lebesgue measure on $\mathbb{R}^d$ by $|\cdot|.$ We write $\mathcal{J}=\{E\subset\mathbb{R}^d|~|\partial E|=0\}.$ Let $\{F_k\}_{k=1}^m\subset\mathcal{J}$ be a collection of disjoint sets.
Let $\{N_k\}_{k=1}^m\subset\mathbb{N}.$ Choose small $C>0$ with
\begin{equation}\label{holesintiles}
|A|>\max_{k\leq m}{N_k}|B(0,C)|.
\end{equation}
We omit to write ($\varepsilon\to 0$) for convergence of sequences indexed by $\varepsilon>0.$
Consider radii $a_{\varepsilon,k}$ with the following condition for $1\leq k\leq m$:
\begin{equation}\label{tmu}
\varepsilon^{-d}\cross
\begin{cases}
(-\log a_{\varepsilon,k})^{-1} & (d=2)\\
(a_{\varepsilon,k})^{d-2}& (d\geq3)
\end{cases}
\to\tilde{\mu_k}.
\end{equation}
We denote $A_i^\varepsilon=\varepsilon(A+i),~\Lambda_k^\varepsilon=\{i\in\Lambda|A_i^\varepsilon\subset F_k\}$ and consider centers of holes
\[\bigsqcup_{k=1}^m\{x_{i,j}^\varepsilon\}_{i\in \Lambda_k^\varepsilon,j=1,...,N_k,\varepsilon>0}\subset\mathbb{R}^d\]
with
\[
B^{i,\varepsilon,k}=\bigsqcup_{j=1}^{N_k}B(x_{i,j}^\varepsilon,C\varepsilon)\subset A_i^\varepsilon
\]
for each $\varepsilon>0,~i\in\Lambda_k^\varepsilon$ and $1\leq k\leq m.$ We recall that $\Omega$ is bounded, open and $C^2$ boundary. We denote
\[
T_{\varepsilon,k}=\bigsqcup_{i\in\Lambda_k^\varepsilon,j\leq N_k}\cls{B(x_{i,j}^\varepsilon,a_{\varepsilon,k})},~~
T_\varepsilon=\bigsqcup_{k=1}^m T_{\varepsilon,k},~~
\Omega_\varepsilon=\Omega\setminus
T_\varepsilon.
\]
\begin{figure}[ht]\centering
\begin{overpic}[width=8cm]{pub/fig/sets}
\put(50,22){$\Omega$}
\put(12,56){$F_1$}
\put(62,56){$F_2$}
\put(59,30){$A_i^\varepsilon$}
\end{overpic}
\caption{Construction of holes $T_\varepsilon$ with $m=2,~N_1=6,~N_2=2.$\label{sets}}
\end{figure}
\subsection{Result}
Using the surface area $S_d$ of $\partial B(0,1),$ we define
$\mu_d=\frac{S_d}{|A|}\cross
\begin{cases}
1&(d=2)\\
d-2&(d\geq3)
\end{cases}.$\\
For $E\subset\mathbb{R}^d,$ we denote $1_E=\begin{cases}
1 & (x\in E)\\
0 & (x\notin E)
\end{cases}.$
\begin{theo}\label{resu}
Under assumptions in Section \ref{ass},
$u^\varepsilon$ in (\ref{solus}) converges to
$u\in H_0^1(\Omega)$ weakly star in $H_0^1(\Omega),$ where
\[~\qty(-\Delta +V)u=f~{\rm on}~\Omega,
~~V=\mu_d\sum_{k=1}^m\tilde{\mu_k}N_k1_{F_k}.
\]
\end{theo}
We remark that we can show \cite[Example 2.1]{CM} by choosing $F_1=\mathbb{R}^d,A=[-1,1)^d,\Lambda=2\mathbb{Z}^d,N_1=1,x_{i,1}^\varepsilon=i\varepsilon.$
\section{Proof}\label{proof}
\subsection{Approximation of sets by tiles \texorpdfstring{$A_i^\varepsilon$}{}}\label{appset}
We first state some properties for $\mathcal{J}.$
\begin{lem}
$\mathcal{J}=\{E\subset\mathbb{R}^d|~|\open{E}|=|\cls{E}|\}.$
$E\in\mathcal{J}$ imply $E$ is measurable.
\end{lem}
\begin{proof}
It follows from $\cls{E}=\open{E}\sqcup\partial E$ for $E\subset\mathbb{R}^d.$
\end{proof}
\begin{lem}\label{capbou}
$E_1,E_2\in\mathcal{J}$ imply $E_1\cap E_2\in\mathcal{J}.$
\end{lem}
\begin{proof}
Distributive property for sets shows
$\cls{E_1\cap E_2}
\subset\cls{E_1}\cap\cls{E_2}
=(\open{E_1}\sqcup\partial{E_1})\cap(\open{E_2}\sqcup\partial{E_2})
=(\open{E_1}\cap\open{E_2})\cup E
=(E_1\cap E_2)^\circ\cup E
$ with some $E$ satisfying $|E|=0.$
\end{proof}
Assumptions for $A$ imply
\begin{equation}
\label{devRd}
\mathbb{R}^d=\bigsqcup_{i\in\Lambda}A_i^\varepsilon
\end{equation}
for each $\varepsilon>0.$ For $E\subset\mathbb{R}^d,$
We define
\[
A_\varepsilon^-(E)=\bigsqcup_{\substack{i\in\Lambda\\A_i^\varepsilon\subset E}}A_i^\varepsilon,~~
A_\varepsilon^+(E)=\bigsqcup_{\substack{i\in\Lambda\\A_i^\varepsilon\cap E\neq\emptyset}}A_i^\varepsilon.
\]
We remark that $A^-_\varepsilon(E)\subset E\subset A^+_\varepsilon(E).$ We will see that they are approximations of $E$ by lemmas above.
\begin{lem}\label{tilesup}
If $E\in\mathcal{J}$ is bounded, $|A_\varepsilon^+(E)|\to |E|.$
\end{lem}
\begin{proof}
Let $d_\varepsilon={\rm diam}(\varepsilon A).$ Then $d_\varepsilon\to 0.$
Let $E_\varepsilon=\bigcup_{x\in E}\cls{B(x,d_\varepsilon)}.$ Then $\bigcap_{\varepsilon>0} E_\varepsilon=\cls E$ and $|E_\varepsilon|<\infty.$ Thus $|E_\varepsilon|\to|\cls{E}|=|E|.$ The assertion follows from it and $E_\varepsilon\supset A_\varepsilon^+(E)\supset E.$
\end{proof}
\begin{lem}\label{tileinf}
If $E\in\mathcal{J},~|A_\varepsilon^-(E)|\to |E|.$
\end{lem}
\begin{proof}
Since $A_\varepsilon^-(\open{E})\subset A_\varepsilon^-(E)\subset E,$ it is enough to show for all open set $E.$ Let
$g(x)={\rm dist}(x,\partial E)$ for $x\in\mathbb{R}^d$ and $E_{-\varepsilon}=E\cap g^{-1}((d_\varepsilon,\infty)).$
Then $\bigcup_{\varepsilon>0} E_{-\varepsilon}= E.$
It is enough to show $E_{-\varepsilon}\subset A_\varepsilon^-(E).$
Take any $x\in E_{-\varepsilon}.$ There is $i\in\Lambda$ that $x\in A_i^\varepsilon$ by (\ref{devRd}). To show $x\in A_\varepsilon^-(E),$ it is enough to show $A_i^\varepsilon\subset E.$ We will show $\mathbb{R}^d\setminus E\subset \mathbb{R}^d\setminus A_i^\varepsilon$ instead of it. If $y\notin E,$ we can get $p\in\partial E$ from line segment which contain $\{x,y\}$. It is $p_t=(1-t) x+t y$ with minimal $t\in[0,1]$ that $p_t\notin E.$ Construction of $p$ imply $|x-y|=|x-p|+|p-y|\geq{\rm dist}(x,\partial E)>d_\varepsilon.$ Thus $y\notin A_i^\varepsilon.$
\end{proof}
We can count how many tiles $A^\pm_\varepsilon(E)$ has.
\begin{lem}\label{counttiles}
$A^\pm_\varepsilon(E)$ is disjoint union of $A_i^\varepsilon$ over $\frac{|A^\pm_\varepsilon(E)|}{\varepsilon^d|A|}$ elements $i\in\Lambda.$
i.e. there is $\{i_n\}_{n=1}^\frac{|A^\pm_\varepsilon(E)|}{\varepsilon^d|A|}\subset\Lambda$ such that
$A^\pm_\varepsilon(E)=\bigsqcup_n A_{i_n}^\varepsilon.$
\end{lem}
We say $E$ is a cube if $E=[0,R)^d+x$ with some $x\in\mathbb{R}^d,R>0.$ We prepare a lemma related to weakly star topology of $L^\infty(\mathbb{R}^d)=L^1(\mathbb{R}^d)^*.$
\begin{lem}\label{denseL1}
Let $\{g_\varepsilon\}\subset L^\infty(\mathbb{R}^d)$ is bounded and $g\in L^\infty(\mathbb{R}^d).$ If \[\duL{g_\varepsilon}{1_E}\to\duL{g}{1_E}\]
for any cube $E$, $g_\varepsilon\to g$ weakly star in $L^\infty(\mathbb{R}^d).$
\end{lem}
\begin{proof}
If follows form the fact that the vector space generated by $\{1_E|E:\mbox{cube}\}$ is dense at $L^1(\mathbb{R}^d)$.
And the fact follows from facts that simple functions on $\mathbb{R}^d$ are dense at $L^1(\mathbb{R}^d),$ Lebesgue measure is outer regular, any open set is union of disjoint countable cubes.
\end{proof}
\subsection{Error corrector \texorpdfstring{$w^\varepsilon$}{}}\label{corr}
By (\ref{tmu}), we get $\frac{\max_k a_{\varepsilon,k}}{\varepsilon}\to0.$ Thus $ \max_{k\leq m} a_{\varepsilon,k}<C\varepsilon$ for $\varepsilon\ll 1.$ Let
\[
w_{0,k}^\varepsilon(r)
=\begin{cases}
\frac{\log a_{\varepsilon,k}-\log r}{\log a_{\varepsilon,k}-\log C\varepsilon}&(d=2)\\
\frac{(a_{\varepsilon,k})^{-d+2}-r^{-d+2}}{(a_{\varepsilon,k})^{-d+2}-(C\varepsilon)^{-d+2}}&(d\geq3)
\end{cases}~(a_{\varepsilon,k}\leq r\leq C\varepsilon),
\]
\[
B_{\varepsilon,k}=\bigsqcup_{i\in\Lambda_k^\varepsilon,j\leq N_k}B(x_{i,j}^\varepsilon,C\varepsilon),~~
B_\varepsilon=\bigsqcup_{k=1}^m B_{\varepsilon,k},\]
\[
w^\varepsilon(x)=\begin{cases}
0&(x\in T_\varepsilon)\\
w_{0,k}^\varepsilon(|x-x_{i,j}^\varepsilon|)&(x\in B(x_{i,j}^\varepsilon,C\varepsilon)\setminus B(x_{i,j}^\varepsilon,a_{\varepsilon,k}))\\
1 & (x\notin B_\varepsilon)
\end{cases}.
\]
Then we have
\begin{equation}\label{harmwep}
\Delta w^\varepsilon=0~{\rm on}~B_\varepsilon\setminus T_\varepsilon.
\end{equation}
We first determine the limit of $w^\varepsilon.$
\begin{lem}\label{H3}
$w^\varepsilon\to 1$ weakly in $H^1(\Omega).$
\end{lem}
\begin{proof}
For $i\notin\bigcup_{k\leq m}\Lambda_k^\varepsilon,$ we have $\norm{\grad w^\varepsilon}_{L^2(A_i^\varepsilon)}^2=0.$
For $i\in\Lambda_k^\varepsilon,$ we have
\begin{align*}\norm{\grad w^\varepsilon}_{L^2(A_i^\varepsilon)}^2
&=N_k S_d\int_{a_{\varepsilon,k}}^{C\varepsilon}|\partial_r w_{0,k}^\varepsilon(r)|^2r^{d-1}d r\\
&=N_k S_d\begin{cases}
\frac{1}{\log C\varepsilon-\log a_{\varepsilon,k}}&(d=2)\\
\frac{d-2}{(a_{\varepsilon,k})^{-d+2}-(C\varepsilon)^{-d+2}}&(d\geq3).
\end{cases}
\end{align*}
Using (\ref{tmu}) for them, we have $ c:=\sup_{\varepsilon>0,i\in\Lambda}\varepsilon^{-d}\norm{\grad w^\varepsilon}_{L^2(A_i^\varepsilon)}^2<\infty.$
Thus $\norm{\grad w^\varepsilon}_{L^2(A_i^\varepsilon)}^2\leq c\varepsilon^d.$
It and Lemmas \ref{tilesup},\ref{counttiles} imply
\[\norm{\grad w^\varepsilon}_{L^2(\Omega)}^2
\leq\norm{\grad w^\varepsilon}_{L^2(A^+_\varepsilon(\Omega))}^2
\leq\frac{|A_\varepsilon^+(\Omega)|}{\varepsilon^d|A|}c\varepsilon^d
\to \frac{c|\Omega|}{|A|}.
\]
It and $|w^\varepsilon|\leq 1$ imply $\{w^\varepsilon\}\subset H^1(\Omega)$ is bounded.
Consider any subsequences of $\{w^\varepsilon\}$ (we still denote $w^\varepsilon$) which converge weakly in $H^1(\Omega),$ and let $w=\mbox{w-}\lim_{\varepsilon\to0}w^\varepsilon.$ We will show $w=1.$
Let $\displaystyle N=\max_{k\leq m}N_k$ and
\[B^{i,\varepsilon}=\begin{cases}
B^{i,\varepsilon,k}&(i\in\Lambda_k^\varepsilon)\\
\emptyset&(i\notin\bigcup_{k\leq m}\Lambda_k^\varepsilon)
\end{cases}.\]
We will find $D_i^\varepsilon$ with $B^{i,\varepsilon}\subset D_i^\varepsilon\subset A_i^\varepsilon$ and $|D_i^\varepsilon|=N|B(0,C\varepsilon)|$ for $i\in\Lambda,~\varepsilon\ll1.$
Take some $x\in A_i^\varepsilon$ and
let
$D(r)=((A_i^\varepsilon\setminus B^{i,\varepsilon})\cap B(x,r))\sqcup B^{i,\varepsilon}$ for $0\leq r\leq {\rm diam}(\varepsilon A).$
By intermediate value theorem and (\ref{holesintiles}),
there exists $r$ with $|D(r)|=N|B(0,C\varepsilon)|.$
Then $D_i^\varepsilon=D(r)$ satisfies the required condition.
Next, let $D_\varepsilon=\bigcup_{i\in\Lambda} D_i^\varepsilon,$ and $g_\varepsilon=1_{\mathbb{R}^d\setminus D_\varepsilon}.$ We show both side of $w^\varepsilon g_\varepsilon=g_\varepsilon$ converge to $cw=c$ weakly star in $L^\infty(\mathbb{R}^d)$ with $c=\frac{|A|-N|B(0,C)|}{|A|}>0.$ Let $E$ be a cube.
Since $\duL{g_\varepsilon}{1_{A_i^\varepsilon}}=|\varepsilon A|-N|B(0,C\varepsilon)|$ for $i\in\Lambda$ and Lemma \ref{tilesup},\ref{tileinf},\ref{counttiles},
\begin{align*}
\duL{g_\varepsilon}{1_{A_\varepsilon^\pm(E)}}
&=\frac{|A_\varepsilon^\pm(E)|}{\varepsilon^d|A|}(|\varepsilon A|-N|B(0,C\varepsilon)|)
=c|A_\varepsilon^\pm(E)|\\
&\to c|E|=\duL{c}{1_E}.
\end{align*}
The assumption of Lemma \ref{denseL1} for $g_\varepsilon$ is verified by it and $\duL{g_\varepsilon}{1_{A_\varepsilon^-(E)}}\leq\duL{g_\varepsilon}{1_E}\leq\duL{g_\varepsilon}{1_{A_\varepsilon^+(E)}}.$
Rellich's theorem and $w1_E\in L^1(\mathbb{R}^d)$ imply
\begin{align*}
&\duL{g_\varepsilon w^\varepsilon}{1_E}-\duL{cw}{1_E}\\
&=\int g_\varepsilon (w^\varepsilon-w)1_E d x+\duL{(g_\varepsilon-c)}{w1_E}\to 0.
\end{align*}
Lemma \ref{denseL1} shows $w^\varepsilon g_\varepsilon\to cw$ weakly star in $L^\infty(\mathbb{R}^d).$
\end{proof}
We use a special function to analyze a distribution $-\Delta w^\varepsilon.$ Let
\[q_0^\varepsilon(r)
=\frac{r^2-(C\varepsilon)^2}{2}~(0\leq r\leq C\varepsilon),
~~
q^\varepsilon(x)
=\begin{cases}
q_0^\varepsilon(|x-x_{i,j}^\varepsilon|)&(x\in B(x_{i,j}^\varepsilon,C\varepsilon))\\
0&(x\notin B_\varepsilon)
\end{cases}.
\]
Then we have
\begin{equation}\label{qepder}
-\Delta q^\varepsilon=-d~(x\in B_\varepsilon),~
\partial_r q_0^\varepsilon(C\varepsilon)
=C\varepsilon,~
q_0^\varepsilon(r)=0~(r=C\varepsilon).
\end{equation}
Now we decompose the restricted distribution $(-\Delta w^\varepsilon)|_{H_0^1(\Omega_\varepsilon)}$ by using $q^\varepsilon.$
\begin{lem}\label{lawep}
Suppose $v\in H_0^1(\Omega_\varepsilon).$ Then we have
\[\duS{-\Delta w^\varepsilon}{v}=\sum_{k\leq m}\frac{\partial_r w_{0,k}^\varepsilon(C\varepsilon)}{C\varepsilon}\qty(\int_{B_{\varepsilon,k}}\grad q^\varepsilon\cdot\grad v d x+d\duS{1_{B_{\varepsilon,k}}}{v}).\]
\end{lem}
\begin{proof}
By (\ref{qepder}) and integration by parts,
\[
\int_{B_{\varepsilon,k}}\grad q^\varepsilon\cdot\grad v d x
=C\varepsilon\dmas{\partial B_{\varepsilon,k}}{v}-d\duS{1_{B_{\varepsilon,k}}}{v}.\]
By assumption, $\dmas{\partial T_{\varepsilon,k}}{v}=0.$ Using them and (\ref{harmwep}), we have
\begin{align*}
\duS{-\Delta w^\varepsilon}{v}
&=\sum_{k\leq m}\int_{B_{\varepsilon,k}\setminus T_{\varepsilon,k}}\grad w^\varepsilon\cdot\grad v d x
=\sum_{k\leq m}\partial_r w_{0,k}^\varepsilon(C\varepsilon)\dmas{\partial B_{\varepsilon,k}}{v}\\
&=\sum_{k\leq m}\frac{\partial_r w_{0,k}^\varepsilon(C\varepsilon)}{C\varepsilon}\qty(\int_{B_{\varepsilon,k}}\grad q^\varepsilon\cdot\grad v d x+d\duS{1_{B_{\varepsilon,k}}}{v}).
\end{align*}
\end{proof}
We determine the limit of $1_{B_{\varepsilon,k}}$ appeared on Lemma \ref{lawep}.
\begin{lem}\label{bep}
$1_{B_{\varepsilon,k}}\to\frac{N_k C^d S_d}{d|A|}1_{F_k}$ weakly star in $L^\infty(\mathbb{R}^d).$
\end{lem}
\begin{proof}
We verify the assumption of Lemma \ref{denseL1}. Let $E$ be a cube.
Since $B_{\varepsilon,k}\subset F_k,$ we have $B_{\varepsilon,k}\cap E=B_{\varepsilon,k}\cap E\cap F_k.$ Thus
\[
|B_{\varepsilon,k}\cap A^-_\varepsilon(E\cap F_k)|
\leq\duL{1_{B_{\varepsilon,k}}}{1_E}
\leq|B_{\varepsilon,k}\cap A^+_\varepsilon(E\cap F_k)|.
\]
By Lemma \ref{tilesup},\ref{tileinf},\ref{counttiles},
\begin{align*}
&|B_{\varepsilon,k}\cap A^\pm_\varepsilon(E\cap F_k)|
=\frac{|A^\pm_\varepsilon(E\cap F_k)|}{\varepsilon^d| A|}N_k|B(0,C\varepsilon)|\\
&\to\frac{|E\cap F_k|N_k C^d |B(0,1)|}{|A|}
=\duL{\frac{N_k C^d S_d}{d|A|}1_{F_k}}{1_E}.
\end{align*}
\end{proof}
The following lemma is very similar to the condition (H.5).
\begin{lem}\label{H5}
Suppose $v^\varepsilon\in H_0^1(\Omega_\varepsilon)$ and $v^\varepsilon\to v$ weakly in $H_0^1(\Omega),$ Then
\[
\duS{-\Delta w^\varepsilon}{v^\varepsilon}\to\duS{\mu_d\sum_{k=1}^m\tilde{\mu_k}N_k1_{F_k}}{v}.\]
\end{lem}
\begin{proof}
By (\ref{tmu}), we have
$\displaystyle\frac{\partial_r w_{0,k}^\varepsilon(C\varepsilon)}{C\varepsilon}\to\frac{\tilde{\mu_k}}{C^d}\cross
\begin{cases}
1&(d=2)\\
d-2&(d\geq3)
\end{cases}.$
We also have
\[
\abs{\int_{B_{\varepsilon,k}}\grad q^\varepsilon\cdot\grad v^\varepsilon d x}
\leq C\varepsilon\sup_{\delta>0}\norm{v^\delta}_{W^{1,1}(\Omega)}\to 0.
\]
By Lemma \ref{bep} and Rellich's theorem,
$
\duS{1_{B_{\varepsilon,k}}}{v^\varepsilon}
=\duS{1_{B_{\varepsilon,k}}}{v^\varepsilon-v}
+\duL{1_{B_{\varepsilon,k}}}{v}
\to\duS{\frac{N_k C^d S_d}{d|A|}1_{F_k}}{v}.
$
The assertion follows from from these limit and Lemma \ref{lawep}.
\end{proof}
\subsection{Proof of Theorem \ref{resu}}\label{end}
\begin{proof}
We use \cite[Theorem 1.2]{CM}. (H.1) and (H.2) are clear by definition of $w^\varepsilon.$ (H.3) is just Lemma \ref{H3}.
Multiplier by $\varphi\in C_0^\infty(\Omega)$ on $H_0^1(\Omega)$ is bounded. Thus it maps weak convergent to weak convergent. It and Lemma \ref{H5} imply (H.5). (H.4) follows from
$\mu_d\sum_{k=1}^m\tilde{\mu_k}N_k 1_{F_k}\in L^\infty(\Omega).$
\end{proof}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,662 |
{"url":"https:\/\/www.biostars.org\/p\/73193\/","text":"Scatter Plot For Correlations With Heatdensity\n2\n7\nEntering edit mode\n7.9 years ago\nk.nirmalraman \u2605 1.1k\n\nHi All,\n\nI am trying to show some correlation between two samples and would like to do a scatter plot for the same.\n\nI tried the following with ggplot2 but I am wondering if its possible to get the heat density as shown here:\n\nqplot(x,y,data=data)+geom_abline(colour = \"red\", size = 1)+theme_bw()\n\nI would like a scatter plot as shown below.\n\nCan you help me acheive this? Thanks!!\n\nvisualization visualization r \u2022 18k views\n12\nEntering edit mode\n6.9 years ago\nJason \u25b4 900\n\nI believe the plots you asked to make were originally made using the R package LSD. I would check it out, it's a very easy package of commands to use that make great looking plots. Also, it can calculate different types of correlations for you too i.e. spearman and pearson.\n\nn <- 10000\nx <- rnorm(n)\ny <- rnorm(n)\nDF <- data.frame(x,y)\nlibrary(LSD)\nheatscatter(DF[,1],DF[,2])\n\n\nEdit: for some reason the image I generated isn't appearing (I copy and pasted it), any help from a mod? It looks like the images in this link.\n\n1\nEntering edit mode\n\nDragging and dropping an image (it looks like that's what you did) won't work. You need to post the image elsewhere and then just link to it.\n\n0\nEntering edit mode\n\nI have a question, how can you add a label next to a heatscatter (which I am loving as my new fav graph) to show the viewer what the difference in density of points is between two colours? Something along the lines of this:\n\nRed = 10-20 data points overlapping\n\nBlue = 0-10 data points overlapping\n\nIs it even possible, seems like something my PI would like to see, as would I.\n\n1\nEntering edit mode\n\nI don't know exactly how to do that but maybe the contour function will help. \u00a0I asked a developer of LSD (Bjoern Schwalb), who actually posts a decent amount here, about what the values along the contour mean (see add.contour = TRUE) and he told me \"the values shown are density estimates from a 2D Kernel Density Estimator function that is used internally (KDE2D)\". For my presentations and recent manuscript I just made a color bar that goes from blue to red with the other colors in between and just said the red was high density and the blue is low density (most people are generally satisfied with that as long as your mention the sample size (i.e. n = 100). The publications I've seen that have used heat scatter have never specified exact numbers of how many data points overlap.\n\nYou may want to look into hexbin if finding the number of data points overlapping is really important. I think it's supposed to do a good job of performing that task:\u00a0http:\/\/www.statmethods.net\/graphs\/scatterplot.html (it's under high density scatter plots)\n\nThis may also help for future LSD work if you hadn't seen it already:\u00a0http:\/\/cran.fhcrc.org\/web\/packages\/LSD\/LSD.pdf\n\n0\nEntering edit mode\n\nThis is awesome! Wish I could save the plots as ggplot objects though\n\n7\nEntering edit mode\n7.9 years ago\nIrsan \u2605 7.3k\n\n# generare random data, swap this for yours :-)!\nn <- 10000\nx <- rnorm(n)\ny <- rnorm(n)\nDF <- data.frame(x,y)\n\n# Calculate 2d density over a grid\nlibrary(MASS)\ndens <- kde2d(x,y)\n\n# create a new data frame of that 2d density grid\n# (needs checking that I haven't stuffed up the order here of z?)\ngr <- data.frame(with(dens, expand.grid(x,y)), as.vector(dens$z)) names(gr) <- c(\"xgr\", \"ygr\", \"zgr\") # Fit a model mod <- loess(zgr~xgr*ygr, data=gr) # Apply the model to the original data to estimate density at that point DF$pointdens <- predict(mod, newdata=data.frame(xgr=x, ygr=y))\n\n# Draw plot\nlibrary(ggplot2)\nggplot(DF, aes(x=x,y=y, color=pointdens)) + geom_point() + scale_colour_gradientn(colours = rainbow(5)) + theme_bw()\n\n\n0\nEntering edit mode\n\n@ Irsan: This is what I would like to currently generate for my gene expression data of dimension x=4X15000 and y= 4X15000 to show the correlation between all gene pairs in x and y. Could you please suggest how I should modify my data to obtain a scatterplot of gene expression based on heat density.\n\n0\nEntering edit mode\n\nYes, how does your data look like now?\n\n0\nEntering edit mode\n\nMy data is initially two dataframes of dimension 15000 X 4 each where the rows are the genes and the columns are the samples. So for these two dataframes, I would like to find the scatterplot of gene correlation density.\n\n0\nEntering edit mode\n\nOk, clear. I will come back to you end next week. Leaving for holiday now\n\n0\nEntering edit mode\n\nthis is the same question is Scatterplots Showing Correlation Between Gene Pairs right?","date":"2021-04-13 04:35:07","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.318046510219574, \"perplexity\": 2169.297667278261}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-17\/segments\/1618038072082.26\/warc\/CC-MAIN-20210413031741-20210413061741-00056.warc.gz\"}"} | null | null |
Loop Lens
Our Endz
Pompeo arrives in Jamaica
Hydel and Calabar shine at McKenley-Wint Track and Field Classic
Arsenal fight back for draw on wild night in Premier League
Creditinfo Jamaica appoints John Matthew Sinclair as CEO
Strong gains for Kingston Wharves, VMIL on Tuesday
VM Wealth structures $900m corporate note for Stewart's Hardware
JLP candidates being vetted in light of corruption issues - Holness
Portland JSX completes investment phase
Select exercise gallops from Caymanas Park - Jan 18-21
Don Reid gets hilarious with 'Courts'
Loop News Created : 25 February 2019 Entertainment
Don Reid
Still basking in the success of his belly-hugging new single, 'Hospital Food', recording artiste Don Reid may be on the verge of an even bigger hit with his hilarious follow-up single, 'Courts' which has generated in excess of 600,000 views on various social media platforms in only a few days.
"The reaction has been unexpected and overwhelming both locally and regionally in Caribbean territories where Courts does business. An amateur recording of the song has gone viral having been reposted by a multitude of major social media sites chalking up in excess of 600,000 views after being shared by over 14,000 followers and social media sites," he said.
The singjay says that the song - a parody of Koffe's hit 'Toast' single - mirrors the financial stresses that many persons experience after entering into hire-purchase type arrangements and the often-hilarious results.
"The song, which is basically social commentary, was done to highlight in a humorous way the unfortunate reality that exists where many consumers experience financial difficulties which prevent them from meeting their hire purchase obligations to furniture and appliance companies such as Courts. Viewers and listeners of the song on social media have said that Courts should engage me to do a commercial or jingle to promote the company's products and services and encourage people to make their payments on time," he said, laughing.
Don Reid has already shot a video for the project which will be released in two weeks.
"Fans can expect more singles over the course of 2019 as I build up to my first EP project of 8 songs," he said.
Don Reid's debut single, 'Hospital Food', is a hilarious account of his experience while hospitalized. The song was recorded in September 2018 when he teamed up with the legendary producer Danny Browne of the well-known label Main Street Records.
Born Donovan Reid in Kingston Jamaica and raised in St Ann, Reid had a passion for music starting in his childhood days and was influenced by the popular entertainers during that time. He is employed as a bank manager in the financial sector but is also a recording artiste who performs at various events and functions. He started out by producing a series of amateur recordings showcasing his comedic style of music via social media platforms.
Bank manager does comedic song about 'Hospital Food'
PHOTOS: Dillgin launches 'Go For the Money Tree' EP
J Rile prepares for 'big things' after 'Jimmy' buzz
Trinidad supports Barbados in boycott of Pompeo meeting in Jamaica
WATCH: Historic moment as Port Royal welcomes first cruise ship
Three of four deceased persons in Trelawny crash identified
Four dead from vehicle crash in Trelawny
Politician Sharon Hay-Webster mugged in downtown Kingston | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,076 |
title: How To - Fix A Problem with an Active Keyman Keyboard Not Typing
---
## Symptoms
You may experience an issue in Windows 7 where:
* Keyman is on; and
* your Keyman keyboard is active, your Keyman keyboard is active, with its icon showing in the Keyman menu; but
* your Keyman keyboard does not work.
## Resolution
Unhide the Keyman menu as described in [Step 2](../start/tutorial#step-2-) of the Getting Started tutorial.
This can also be resolved by switching language using the Windows Language Bar or the Language Icon
## Background
When the Keyman icon is in the hidden notification area of the Windows taskbar, opening the notification area and selecting the icon causes it to change the keyboard for the taskbar, not for your active application. Clicking back to the active application results in the keyboard turning off again.
## Related Topics
- [Keyboard Task - Turn on a Keyboard](../basic/select-keyboard)
| {
"redpajama_set_name": "RedPajamaGithub"
} | 9,225 |
Q: How can I solder or mount this potentiometer on a PCB? I need to use this potentiometer:
(Vishay 534-11103)
…on my board but I don't know how I can do that. I appreciate any ideas or replacement suggestions.
Manufacturer's datasheet link here.
As asked in comments:
Why do you want to mount it on your PCB?
Customer requirement. They are in love with this potentiometer.
A: I would place 3 individual header pins snapped off from from these:
https://www.digikey.com/catalog/en/partgroup/break-away-0-1-36-pin-strip-male-header/88887
positioned on your PCB such that the hole in each of the solder tabs on the the pot slides over a pin, and the back of the pot lies flat against the PCB.
I'd probably also glue the pot onto the PCB - maybe a dot of holt-melt glue would work - possibly also make some holes in the PCB underneath the pot for the glue to 'grip' onto.
A: I designed a product some years ago that used a similar potentiometer. I designed a snap-off board that had slots for the terminals. A flat cable connected the sub-board to the main board. The assembly could be tested before the sub-board was snapped off.
You can connectorize it.
Or just use it the way it is intended, soldered wires to the board (with heat shrink tubing if you want it to look nice), with or without a connector at the PCB end (but if you don't use a connector, add some form of strain relief/vibration control). You could add cable lacing for that old-timey look (jk).
A: solder a steel bracket to the PCB and mount the potentiometer in the bracket.
you could have slots in the the PCB and solder the wiring tabs in there but that would probably be unreliable.
A: Best way would be to panel mount it, and then solder on three wires which you bring to the board, either directly solder them or use a 3 pin connector. This pot is not designed to be PCB mounted.
Or choose another pot, of course.
A: For mounting, glue it down using a neutral cure silicone rubber.
For connecting to your circuit a few bodge wires should suffice.
A: I mean yes technically you can mount them on PCB look at this link
https://www.indiamart.com/proddetail/potentiometer-with-relay-pcb-22892358573.html
You will have a hard time soldering and even harder time when you want to replace them.
Also, look at its weight(and weight distribution).
you can also mount a PCB on it but that'd still require soldering wire.
The best thing is to solder wires as other answers suggest.
A: Mount the pot to the board (using its intended mounting nut) and wire it from the reverse side.
Edit: I suggest mounting it to the board because it is a "customer requirement" that this board and pot be used. Of course board-mounting a panel-mount pot is not ideal, but I have seen a few instances of this. Usually these are "trimpots" with short (no) shaft length. For infrequent use, it seems to be OK. Still, this is "a hack."
*
*The end is slotted, so can be adjusted with a screwdriver. If that's how it is being used, a shorter shaft would likely be better.
*If a knob must be used, a smaller one will likely reduce radial forces on the PCB.
*Don't use a lockwasher against the PCB, even if plated-through (which it should be) — it will tear it up. Use a flat washer against the PCB and lockwasher on top of that. A little medium threadlocker on the nut (or even a drop of selastic or hot-melt-glue) will ensure it won't wiggle loose.
*Note this model doesn't seem to have an indexing tang or "D". I'd consider one that did, to ensure it can't spin once mounted. FR4 is more elastic than metal, so getting a really tight fit is harder.
A: The best way is to bend the potentiometer contacts back, and pass a wire that is soldered between the potentiometer and the PCB. Then secure the potentiometer with a plastic seal to the PCB. To make it more firm, you can put silicone glue between the potentiometer and the PCB.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,480 |
import radio
from microbit import *
radio.config(group=0)
radio.on()
while True:
if button_a.is_pressed():
radio.send('Hello radio world!')
sleep(1000)
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,638 |
Blurred vision, or "blurry eyes," is a common complaint that can result from one or more of several dozen conditions. Causes of blurred vision range from nearsightedness and farsightedness to eye strain, cataracts, migraine headaches, macular degeneration, brain tumors and dry eye. The best relief of blurred vision, when caused by dry eye, is to alleviate the "dryness" by supplementing the lost tear film water.
Dry eye is a frequent cause of blurred vision. Dry eye may be defined as dehydration or partial loss of water from the ocular tear film that covers the eyes' exposed portions. The tear film is normally 98-percent water. When the surrounding environment, or certain diseases, cause the tear film to lose a portion of its natural water content, the resulting dehydration can produce numerous complaints, including blurred vision.
Other physical complaints associated with dry eye, in addition to blurred vision, include eye discomfort, itchy eyes, grainy eyes, red eyes, burning eyes, headache, eye strain and increased feelings of stress. Watery eyes can also be a associated with dry eye. When the tear film - technically known as "basal tears," loses water, the eye does its best to compensate and provide relief by producing reflex tears resulting in watery or teary eyes.
When blurred vision is associated with dry eye, the blurriness tends to occur late in the day and usually goes away either upon blinking the eyes a few times or resting.
The best relief of blurred vision caused by dry eye is obtained by increasing the natural fresh water content of the dehydrated tear film. The best way to increase tear film water content is to apply Nature's Tears® EyeMist® whenever the eyes become blurry, red, tired or uncomfortable. | {
"redpajama_set_name": "RedPajamaC4"
} | 6,593 |
Üst Menü - İNGİLİZCE
Ercan Akın's solo exhibition "Hopeless Wait" has extended until November 19th!
News 2 November 2022
Ercan Akın's solo exhibition "Hopeless Wait" which bears the emotional traces of his travel experiences crossing Japan-Turkey between 2015 and 2022, has extended until November 19th. Share on Facebook Share on Twitter Share on Linkedin
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Alt Menü - İNGİLİZCE | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 4,091 |
namespace TypeVisualiser.Messaging
{
using System;
using GalaSoft.MvvmLight.Messaging;
internal static class MessagingGate
{
private static readonly object SyncRoot = new object();
public static void Register<T>(object recipient, Action<T> handler) where T : MessageBase
{
lock (SyncRoot)
{
Messenger.Default.Register(recipient, handler);
}
}
public static void Unregister<T>(object recipient) where T : MessageBase
{
lock (SyncRoot)
{
Messenger.Default.Unregister<T>(recipient);
}
}
/// <summary>
/// Sends the specified message.
/// Use this when you are concerned about concurrency. Sending messages at the same time as registering new listeners on many threads will cause issues.
/// </summary>
/// <typeparam name="T">The message to send</typeparam>
/// <param name="message">The message.</param>
public static void Send<T>(T message)
{
lock (SyncRoot)
{
Messenger.Default.Send(message);
}
}
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,202 |
\section{Preparatory Study: the Effect of Back-Translation when Controlling for the Amount of Training Effort}
A typical assumption made when training \ac{NMT} models, is that when more training data is used, more training effort is warranted. Based on this assumption when training NMT systems what is normally kept constant is the amount of training epochs rather than the amount of training effort in the form of steps/mini-batches. Nevertheless, when adding back-translated data to the training set, while keeping the amount of epochs the same, the effective amount of training increases.
It could then be questioned whether the extra training effort in itself does not partly explain the positive effect of back-translation. For this reason, we seek to answer the question: ``Does the effect of back-translation change when we control for the amount of training effort, by keeping the total amount of steps/mini-batches constant?".
To answer this question we compare the performance of systems trained on purely authentic data to those trained on authentic plus synthetic data, while keeping either the number of steps/mini-batches or the number of epochs constant in both settings:
\begin{enumerate}
\item Models trained with 1M auth + 2M synth sentences using the default settings, including 13 training epochs.
\item Models trained on 1M auth data only, trained either:
\begin{enumerate}
\item using the default settings, including 13 training epochs.
\item Trained for 39 epochs, to obtain a same amount of training effort as for the 1M auth + 2M synth sentences model.
\end{enumerate}
\end{enumerate}
When increasing the epochs to 39, we take appropriate measures to keep the starting point and speed of decay of the learning rate constant for the amount of training steps/epochs.\footnote{This is implemented by changing the start of the learning rate decay from epoch 8 to epoch 22 $(= 7* 3 + 1)$ and changing the decay factor from 0.5 to $\sqrt[3]{0.5} = 0.7936$. This way, the learning rate decay starts after the same amount of data when using the 1M auth dataset $(7 \times 3M)$ and the decay rate is maintained at 0.5 for each $3M$ sentences from this point onwards.}
The results of these experiments indicate that training a model on authentic data with $1/3$ of the amount of the total parallel data (authentic + synthetic) for an additional 26 epochs to account for the extra training effort is not required as no significant improvement has been observed. Based on the outcome of these experiments we chose the rest of our experiments.
\section{Introduction}
Machine translation (MT) nowadays is heavily dependent on the quantity and quality of training data. The amount of available good-quality parallel data for the desired domain and/or language pair is often insufficient to reach the required translation performance. In such cases, it has become the norm to resort to back-translating freely available monolingual data as proposed in \cite{sennrich2015improving}. That is, one can translate a set of sentences from language L2 into L1 with an already trained MT system for the language pair L2$\rightarrow$L1. Then create a synthetic parallel corpus from L1 to L2, with the source (L1) side being the translated text and the target side being the monolingual data. Back-translation has been shown to be beneficial not only for MT but also for other NLP tasks where data is scarce, e.g. \ac{APE} \cite{Junczys-Dowmunt2016a,Negri2018}. However, the effects of various parameters for creating \ac{BT} data have not been investigated enough as to indicate what are the optimal conditions in not only creating but also employing such data to train high-quality \ac{NMT} systems.
The work presented in ~\newcite{bt-eamt18} draws an early-stage empirical roadmap to investigating the effects of \ac{BT} data. In particular, it looks at how the amount of \ac{BT} data impacts the performance of the final NMT system. In ~\newcite{sennrich2015improving} and ~\newcite{bt-eamt18}, the systems used to generate the \ac{BT} data are neural. However, it has been noted that often different paradigms can contribute differently to a given task. For example, it has been shown that applying an \ac{APE} system based on NMT technology improves \ac{SMT} output, but has lower impact on \ac{NMT} output \cite{bojar2017findings,Chatterjee2018}.
In this work we assess the impact of different amounts of \ac{BT} data generated by two different types of MT systems -- \ac{NMT} and \ac{SMT}. Our contribution is two-fold: (i) we provide a systematic comparison of the \ac{BT} data by building \ac{NMT} systems with a combination of \ac{SMT} and \ac{NMT} \ac{BT} data and (ii) we identify the effects of \ac{BT} data that originates from SMT or NMT on the end-quality of the trained NMT system. We aim to answer the question: "What is the best choice for \ac{BT} data?"
\input{effect-back-translation-with-controlled-amount-of-training-effort.tex}
\section{Using Back-Translation from Different Sources}\label{sec:different_sources}
The work of ~\cite{sennrich2015improving} showed that adding \ac{BT} data is beneficial to achieve better translation performances. In this work we compare the details related to the translation hypotheses originating from \ac{SMT} and \ac{NMT} back-translated training data as well as combine the data from those two different sources. To the best of our knowledge, this has not been investigated yet.
We compare German-to-English translation hypotheses generated by systems trained (i) only on authentic data, (ii) only on synthetic data, and (iii) on authentic data enhanced with different types of \ac{BT} data: \ac{SMT}, \ac{NMT}. We exploit two types of synthetic and authentic data combinations: (a) randomly selected half of target sentences back-translated by SMT and another half by NMT system, and (b) joining all \ac{BT} data (thus repeating each target segment).
The translation hypotheses are compared in terms of four
automatic evaluation metrics: BLEU~\cite{papineni2002bleu}, TER~\cite{snover2006study}, METEOR~\cite{banerjee2005meteor} and CHRF~\cite{popovic2015chrf}. These metrics give an overall estimate of the quality of the translations with respect
to the reference (human translation of the test set).
In addition, the translation hypotheses are analyzed in terms of five error categories, lexical variety and syntactic variety.
\section{Related Work}\label{sec:related_work}
A comparison between MT models trained with synthetic and with authentic data that originate from the same source has been presented in~~\newcite{bt-eamt18}. They show that while the performances of models trained with both synthetic and authentic data are better than those of models trained with only authentic data, there is a saturation point beyond which the quality does not improve by adding more synthetic data. Nonetheless, models trained only with synthetic (BT) data perform very reasonably, with evaluation scores being close to those of models trained with only authentic parallel data. In fact, when appropriately selected, BT data can be used to enhance NMT models ~\cite{poncelas2019adaptation}.
~\newcite{D18-1045} confirmed that synthetic data can sometimes match the performance of authentic data. In addition, a comprehensive analysis of different methods to generate synthetic source sentences was carried out. This analysis revealed that sampling from the model distribution or noising beam outputs out-performs pure beam search, which is typically used in NMT. Their analysis shows that synthetic data based on sampling and noised beam search provides a stronger training signal than synthetic data based on argmax inference.
One of the experiments reported in ~\newcite{W18-6315} is comparing performance between models trained with \ac{NMT} and \ac{SMT} \ac{BT} data. The best Moses system ~\cite{koehn2007moses} is almost as good as the NMT system trained with the same (authentic) data, and much faster to train. Improvements obtained with the Moses system trained with a small training corpus are much smaller; this system even decreases the performance for the out-of-domain test.
The authors also investigated some properties of \ac{BT} data and found out that the back-translated sources are on average shorter than authentic ones, syntactically simpler than authentic ones, and contain smaller number of rare events. Furthermore, automatic word alignments tend to be more monotonic between artificial sources and authentic targets than between authentic sources and authentic targets.
~\newcite{W18-6315} also compared training \ac{BT} data with authentic data in terms of lexical and syntactic variety, segment length and alignment monotony, however they did not analyze the obtained translation hypotheses. In \cite{Vanmassenhove2019} it is shown that MT systems trained on authentic and on backtranslated data lead to general loss of linguistic richness in their translation hypotheses.
\section{Experimental Settings}
\label{sec:experimental_settings}
For the experiments we have built German-to-English NMT models using the Pytorch port of OpenNMT ~\cite{opennmt}. We use the default parameters: 2-layer LSTM with 500 hidden units. The models are trained for the same number of epochs. As the model trained with all authentic data converges after 13 epochs, we use that many iterations to train the models (we use the same amount of epochs). As optimizer we use \ac{SGD}, in combination with learning rate decay, halving the learning rate starting from the 8th epoch.
In order to build the models, all data sets are tokenized and truecased and segmented with \ac{BPE} ~\cite{sennrich2016neural} built on the joint vocabulary using 89500 merge operations. For testing the models we use the test set provided in the WMT 2015 News Translation Task ~\cite{bojar-EtAl:2015:WMT}. As development set, we use 5K randomly sampled sentences from development sets provided in previous years of WMT.
\section{Data}
\label{sec:datasets}
The parallel data used for the experiments has been obtained from WMT 2015 ~\cite{bojar-EtAl:2015:WMT}. We build two parallel sets with these sentences: \textit{base} (1M sentences) and \textit{auth} (3M sentences). We use the target side of \textit{auth} to create the following datasets:
\begin{itemize}
\item \textit{SMTsynth}: Created by translating the target-side sentences of \textit{auth}. The model used to generate the sentences is an \ac{SMT} model trained with \textit{base} set in the English to German direction. It has been built using the Moses toolkit with default settings, using GIZA++ for word alignment and tuned using MERT ~\cite{och2003minimum}). The language model (of order 8) is built with the KenLM toolkit \cite{heafield2011kenlm} using the German side of \textit{base}.
\item \textit{NMTsynth}: Created by translating the target-side sentences of \textit{auth}. The model used to generate the sentences is an \ac{NMT} model (with the same configuration as described in Section \ref{sec:experimental_settings} but in the English to German direction) trained with the \textit{base} set.
\item \textit{hybrNMTSMT}: Synthetic parallel corpus combining \textit{NMTsynth} and \textit{SMTsynth} sets. It has been built by maintaining the same target side of \textit{auth}, and as source side we alternate between \textit{NMTsynth} and \textit{SMTsynth} each 500K sentences.
\item \textit{fullhybrNMTSMT}: Synthetic parallel corpus combining all segments from \textit{NMTsynth} and \textit{SMTsynth} sets (double size, each original target sentence repeated twice with both an NMT and SMT back-translation-generated translation).
\end{itemize}
\section{Experiments}
In our experiments, we build models on different portions of the datasets described in Section \ref{sec:datasets}. First, we train an initial NMT model using the \textit{base} data set. Then, in order to investigate how much the models benefit from using synthetic data generated by different approaches, we build models with increasing sizes of data (from the data sets described in Section \ref{sec:datasets}).
The models explored are built with data that ranges from 1M sentences (built with only authentic data from \textit{base} data set) to 4M sentences (consisting on 1M sentences from \textit{base} and 3M sentences generated artificially with different models). We also include the models built with the \textit{fullhybrNMTSMT} set. As this set contains duplicated target-side sentences, the largest model we build contains 7M sentences in total but only 4M distinct target-side sentences.
\section{Results}
\subsection{Controlling the Amount of Training Effort}
Table \ref{table:results-experiments-with-controlled-amount-of-training-effort} shows the effect of controlling the amount of training effort when using back-translation.
It can be observed that increasing the number of epochs from 13 to 39 when using just the 1M base training set does not increase the performance over using just 13 epochs (i.e. not compensating the relatively smaller training set with more epochs), rather it deteriorates it.
From these results we conclude that there is no reason to believe that the positive effects of using back-translation is caused by an effectively larger training effort, rather than by the advantage of the larger training set itself.
We therefore also conclude that it is reasonable to keep the number of epochs constant across experiments, rather than fixing the amount of training effort as measured by steps/mini-batches, and we do the former throughout the rest of the paper.
\begin{table*}[ht]
\centering
\begin{tabular}{|p{2cm}|p{2.5cm}|p{2.5cm}|p{2.5cm}|}
\hline
& 1M \textit{base}.- 13 Epochs & 1M \textit{base}.- 39 Epochs- & 1M \textit{base} + 2M \textit{NMTsynth} \\
\hline
BLEU$\uparrow$ & 23.40 & 23.22 & 25.44 \\
TER$\downarrow$ & 57.23 & 58.21 & 55.62 \\
METEOR$\uparrow$ & 28.09 & 27.75 & 29.47 \\
CHRF1$\uparrow$ & 50.66& 50.18 & 52.5 \\
\hline
\end{tabular}
\caption{Results for experimental procedure validation: checking that it is reasonable to use constant number of epochs, not constant amount of training effort, in the experiments.} \label{table:results-experiments-with-controlled-amount-of-training-effort}
\end{table*}
\begin{table*}[!htbp]
\centering
\begin{tabular}{|p{0.25cm}|p{1.6cm}|p{2cm}|p{2.2cm}|p{2.2cm}|p{2.2cm}||p{2.2cm}|}
\hline
& & 1M \textit{base}. & - & - & - & - \\
\hline
\multirow{4}{*}{\rotatebox[origin=c]{90}{\centering 1M lines}}
& BLEU$\uparrow$ & 23.40 & - & - & - & - \\
& TER$\downarrow$ & 57.23 & - & - & - & - \\
& METEOR$\uparrow$ & 28.09 & - & - & - & - \\
&CHRF1$\uparrow$ & 50.66 & - & - & - & - \\
\hline
& & + 1M \textit{auth} & + 1M \textit{SMTsynth} & +1M \textit{NMTsynth} & + 1M \textit{hybrNMTSMT} & + 2M \textit{fullhybrNMTSMT} \\
\hline
\multirow{4}{*}{\rotatebox[origin=c]{90}{\centering 2M lines}}
& BLEU$\uparrow$ & 24.87 & 24.38 & 25.32 & 25.21 & 25.34 \\
& TER$\downarrow$ & 55.81 & 56.05 & 55.66 & 55.87 & 55.79 \\
& METEOR$\uparrow$ & 29.16 & 28.93 & 29.33 & 29.29 & 29.47 \\
&CHRF1$\uparrow$ & 52.03 & 51.89 & 52.25 & 52.36 & 52.47 \\
\hline
& & + 2M \textit{auth}. & + 2M \textit{SMTsynth} & + 2M \textit{NMTsynth} & + 2M \textit{hybrNMTSMT} & + 4M \textit{fullhybrNMTSMT} \\
\hline
\multirow{4}{*}{\rotatebox[origin=c]{90}{\centering 3M lines}}
& BLEU$\uparrow$ & 25.69 & 24.58 & 25.44 & 25.62 & 25.94 \\
& TER$\downarrow$ & 54.99 & 55.7 & 55.62 & 55.25 & 55.11 \\
& METEOR$\uparrow$ & 29.7 & 29.02 & 29.47 & 29.73 & 29.97 \\
& CHRF1$\uparrow$ & 52.77 & 52.09 & 52.5 & 52.89 & 53.11 \\
\hline
& & + 3M \textit{auth} & + 3M \textit{SMTsynth} & + 3M \textit{NMTsynth} & +3M \textit{hybrNMTSMT} & + 6M \textit{fullhybrNMTSMT} \\
\hline
\multirow{4}{*}{\rotatebox[origin=c]{90}{\centering 4M lines}}
& BLEU$\uparrow$ & 25.97 & 24.65 & 26.01 & 25.83 & 25.86 \\
& TER$\downarrow$ & 54.54 & 55.58 & 55.33 & 55.17 & 54.95 \\
& METEOR$\uparrow$ & 29.91 & 29.26 & 29.71 & 29.74 & 29.88 \\
&CHRF1$\uparrow$ & 53.16 & 52.24 & 52.87 & 52.84 & 53.11 \\
\hline
\end{tabular}
\caption{Performance of models built with increasing sizes of authentic set (first column) and different synthetic datasets (last four columns). +1M, +2M and +3M indicate the amount of sentences added to the \textit{base} set (1M authentic sentences).
\label{table:results}}
\end{table*}
\subsection{Addition of Synthetic Data from SMT and NMT Models }
Table \ref{table:results} shows the results of the performance of the different \ac{NMT} models we have built. The sub-tables indicate the size of the data used for building the models (from 1M to 4M lines). In each column it is indicated whether \textit{base} has been augmented with the \textit{auth}, \textit{SMTsynth}, \textit{NMTsynth}, \textit{hybrNMTSMT}, or \textit{fullhybrNMTSMT} data set.
The results show that adding synthetic data has a positive impact on the performance of the models as all of them achieve improvements when compared to that built only with authentic data \textit{1M base}. These improvements are statistically significant at p=0.01 (computed with multeval ~\cite{clark2011better} using Bootstrap Resampling ~\cite{koehn04}). However, the increases of quality are different depending on the approach followed to create the \ac{BT} data.
First, we observe that models in which SMT-generated data is added do not outperform the models built with the same size of authentic data. For example, the models built with 4M sentences (1M authentic and 3M SMT-produced sentences, in cell \textit{+ 3M SMTsynth}) achieve a performance comparable to the model trained with smaller number of sentences of authentic data (such as \textit{ + 1M auth} cell, 2M sentences).
Models built by using \ac{NMT}-created data have a better performance than those built with data generated by \ac{SMT}. When performing a pair-wise comparison between models using an equal amount of either \ac{SMT} or \ac{NMT}-created data, we observe that the latter models outperform the former by around one BLEU point.
In fact, the performance of models using NMT-translated sentences is closer to those built with authentic data, and some \textit{NMTsynth} models produce better translation qualities. This is the case of \textit{+1M NMTsynth} model (according to all evaluation metrics) or \textit{+3M NMTsynth} (according to BLEU).
Our experiments also include the performance of models augmented with a combination of SMT- and NMT- generated data. We see that adding \textit{hybrNMTSMT} data, with one half of the data originating from SMT and the other half from NMT models, have performances similar to those models built on authentic data only. According to some evaluation metrics, such as METEOR, the performance is better than \textit{auth} models when adding 1M or 2M artificial sentences (although none of these improvements are statistically significant at p=0.01). For these amount of sentences, it also outperforms those models in which only SMT or only NMT \ac{BT} data have been included.
The models extended with synthetic data that perform best are \textit{fullhybrNMTSMT} models. Furthermore, they also outperform authentic models when built with less than 4M distinct target-sentences according to BLEU, METEOR (showing statistically significant improvements at p=0.01) and CHRF1. Despite that, when using large sizes of data (i.e. adding 3M synthetic sentences) the models built with SMT-generated artificial data have the lowest performances whereas the performance of the other three tends to be similar.
\subsection{Further Analysis}
In order to better understand the described systems, we carried out more detailed analysis of all translation outputs. We analyzed five error categories: morphological errors, word order, omission, addition and lexical errors, and we compared lexical and syntactic variety of different outputs in terms of vocabulary size and number of distinct POS n-grams. We also analyzed the sentence lengths in different translation hypotheses, however no differences were observed, neither in the average sentence length nor in the distribution of different lengths.
\subsubsection*{Automatic Error Analysis}
\begin{table*}[ht]
\centering
\begin{tabular}{|l|ccccc|} \hline
& \multicolumn{5}{c|}{error class rates$\downarrow$} \\
training & morph & order & omission & addition & mistranslation \\ \hline
1M \textit{base} & 2.8 & 9.8 & 12.0 & 4.8 & 29.1 \\ \hline
1M \textit{base} + 1M \textit{auth} & 2.7 & 9.5 & 11.4 & 4.9 & 28.2 \\ \hdashline
1M \textit{base} + 1M \textit{SMTsynth} & 2.8 & 10.0 & 11.6 & 4.8 & 28.1 \\
1M \textit{base} + 1M \textit{NMTsynth} & 2.7 & 9.8 & 10.9 & 5.0 & 28.1 \\
1M \textit{base} + 1M \textit{hybrNMTSMT} & 2.7 & 9.6 & 11.4 & 5.2 & 27.7 \\
1M \textit{base} + 1M \textit{fullhybrNMTSMT} & 2.6 & 9.5 & 11.0 & 5.2 & 27.8 \\ \hline
1M \textit{base} + 2M \textit{auth} & 2.6 & 9.6 & 11.2 & 4.8 & 27.7\\ \hdashline
1M \textit{base} + 2M \textit{SMTsynth} & 2.7 & 10.0 & 11.9 & 4.5 & 28.0 \\
1M \textit{base} + 2M \textit{NMTsynth} & 2.6 & 9.7 & 11.1 & 5.1 & 27.9 \\
1M \textit{base} + 2M \textit{hybrNMTSMT} & 2.6 & 9.6 & 11.0 & 5.2 & 27.6 \\
1M \textit{base} + 2M \textit{fullhybrNMTSMT} & 2.6 & 9.6 & 10.7 & 5.3 & 27.4 \\ \hline
1M \textit{base} + 3M \textit{auth} & 2.7 & 9.8 & 11.2 & 4.6 & 27.6 \\ \hdashline
1M \textit{base} + 3M \textit{SMTsynth}& 2.7 & 9.8 & 11.9 & 4.6 & 27.9 \\
1M \textit{base} + 3M \textit{NMTsynth} & 2.5 & 9.6 & 11.3 & 5.3 & 27.4 \\
1M \textit{base} + 3M \textit{hybrNMTSMT} & 2.6 & 9.5 & 11.0 & 5.1 & 27.6 \\
1M \textit{base} + 3M \textit{fullhybrNMTSMT} & 2.5 & 9.7 & 10.8 & 4.8 & 27.7 \\ \hline
\end{tabular}
\caption{Results of automatic error classification into five error categories: morphological error (morph), word order error (order), omission, addition and mistranslation.}
\label{tab:auterr}
\end{table*}
For automatic error analysis results, we used Hjerson~~\cite{hjerson11}, an open-source tool based on Levenshtein distance, precision and recall. The results are presented in Table~\ref{tab:auterr}.
It can be seen that morphological errors are slightly improved by any additional data, but it is hard to draw any conclusions. This is not surprising given that our target language, English, is not particularly morphologically rich. Nevertheless, for all three corpus sizes, the numbers are smallest for the full hybrid system, being comparable to the results with adding authentic data.
As for word order, adding SMT data is not particularly beneficial since it either increases (1M and 2M) or does not change (3M) this error type. NMT systems alone do not help much either, except a little bit for the 3M corpus. Hybrid systems yield the best results for this error category for all corpus sizes, reaching or even slightly surpassing the result with authentic data.
Furthermore, all \ac{BT} data are beneficial for reducing omissions, especially hybrid which can be even better than the authentic data result.
As for additions, no systematic changes can be observed, except an increase for all types of \ac{BT} data. However, it should be noted that this error category is reported not to be very reliable for comparing different MT outputs~(see for example \cite{humaut11}).
The mostly affected error category is mistranslations. All types of additional data are reducing this type of errors, especially the hybrid \ac{BT} data for 1M and 2M, even surpassing the effect of adding authentic data. As for the 3M corpus, the improvement in this error category is similar to the one by authentic data, but the best option is to use NMT \ac{BT} data alone.
In total, the clear advantage of using hybrid systems can be noted for mistranslations, omissions and word order which is the most interesting category. This error category is augmented by adding BT SMT data or not affected by adding BT NMT data, but combining two types of data creates beneficial signals in the source text.
\subsubsection*{Lexical and Syntactic Variety}
\begin{figure*}[ht]
\includegraphics[width=15cm, height=6cm]{plots/variety.png}
\centering
\caption{Lexical variety and syntactic variety for all translation hypotheses and for human reference translations.}
\label{fig:variety}
\end{figure*}
\begin{figure*}[ht]
\includegraphics[width=15cm, height=9.5cm]{plots/precrec.png}
\centering
\caption{Word/POS 4-gram precision and recall for all translation hypotheses.}
\label{fig:precrec}
\end{figure*}
Lexical and syntactic variety is estimated for each translation hypothesis as well as for the human reference translation. The motivation for this is the observation that machine-translated data is generally lexically poorer and syntactically simpler than human translations or texts written in the original language~\cite{Vanmassenhove2019}. We want to see how different or similar our translation hypotheses are in this sense, and also how they relate to the reference.
Lexical variety is measured by vocabulary size (number of distinct words) in the given text, and syntactic variety by number of distinct POS n-grams where $n$ ranges from 1 to 4. The results are shown in Figure~\ref{fig:variety}.
First of all, it can be seen that none of the translation hypotheses reaches the variety of the reference translation (the black line on the top). The difference is even more notable for the syntax, where the differences between translation hypotheses are smaller and the difference between them and the reference is larger than for vocabulary.
Furthermore, it can be seen that for authentic data (thin gray line on the bottom and thick gray line) the variety increases monotonically with adding more text.
Lexical variety is increased by all synthetic data, too, even more than by authentic data, however, for the NMT and hybrid synthetic data the increase for the 3M corpus is smaller than for smaller corpora.
The increase of syntactic variety is lower both for authentic and for synthetic data than the increase of lexical variety. For 1M and 2M corpus, syntactic variety is barely increased by SMT synthetic data whereas NMT and hybrid data are adding more new instances. For the 3M corpus, however, all synthetic methods yield similar syntactic variety, larger than the one obtained by adding authentic data.
\subsubsection*{Word/POS 4-gram Precision and Recall}
Whereas the increase of lexical and syntactic varieties is a positive trend in general, there is no guarantee that the MT systems are not introducing noise thereby. To estimate how many of added words and POS sequences are sensible, we calculate precision and recall of word and POS 4-grams when compared to the given reference translation. The idea is to estimate how much the translation hypotheses are getting closer to the reference. We take word 4-grams instead of single words because it is not only important that a word makes sense in isolation, but also in a context. Of course, it is still possible that some of the new instances are valid despite being different from the given single reference.
The results of precision and recall for word/POS 4-grams are
are shown in Figure~\ref{fig:precrec}.
Several tendencies can be observed:
\begin{itemize}
\item hybrid \ac{BT} data is especially beneficial for the 1M and 2M additional corpora, for 1M even outperforming the authentic additional data, especially regarding word 4-grams;
\item NMT BT is the best synthetic option for the 3M additional corpus, however not better than adding 3M of authentic data. This tendency is largest for POS 4-gram precision.
\item SMT \ac{BT} data achieves the lowest scores, especially for POS 4-grams; this is probably related to the fact that it produces less grammatical BT sources, which are then propagated to the translation hypotheses. The differences are largest for the 3M additional corpus, which is probably the reason of diminished effect of the hybrid \ac{BT} data for this setup.
\end{itemize}
Overall tendencies are that the hybrid \ac{BT} data is capable even of outperforming the same amount of authentic data if the amount of added data does not exceed the double size of the baseline authentic data. For larger data, a deterioration can be observed for the SMT \ac{BT} data, leading to saturation of hybrid models.
Further work dealing with mixing data techniques is necessary, in order to investigate refined selection methods (for example, removing SMT segments which introduce noise).
\section{Conclusion and Future Work}
In this work we have presented a comparison of the performance of models trained with increasing size of back-translated data. The artificial data sets explored include sentences generated by using an \ac{SMT} model, and \ac{NMT} model and a combination of both. Two mixing strategies are explored: randomly selecting one half of the source segments from the SMT \ac{BT} data and the other half from the NMT \ac{BT} data, and using all BT source segments thus repeating each target segment.
Some findings from previous work~\cite{W18-6315} are confirmed, namely that in terms of overall automatic evaluation scores, SMT \ac{BT} data reaches slightly worse performance than NMT \ac{BT} data.
Our main findings are that mixing SMT and NMT \ac{BT} data further improves over each data used alone, especially if full hybridisation is used (using two sources for each target side). These data can even reach better performance than adding the same amount of authentic data, mostly by reducing the number of mistranslations, and increasing the lexical and syntactic variety in a positive way (introducing useful new instances).
However, if the amount of synthetic data becomes too large (three times larger than the authentic baseline data), the benefits of hybrid system start to diminish. The most probable reason is the decrease in grammaticality introduced by SMT \ac{BT} data which becomes dominant for the larger synthetic corpora.
The presented findings offer several directions for the future work, such as exploring efficient strategies for mixing SMT and NMT data for different authentic/synthetic ratios and investigating morphologically richer target languages.
\section*{Acknowledgements}
This research has been supported by the ADAPT Centre for Digital Content Technology which is funded under the SFI Research Centres Programme (Grant 13/RC/2106) and is co-funded under the European Regional Development Fund.
\noindent
\includegraphics[width=0.9cm]{flag_yellow_low.jpg}
This work has also received funding from the European Union's Horizon 2020 research and innovation programme under the Marie Skłodowska-Curie grant agreement No 713567.
\bibliographystyle{acl_natbib}
\section{Introduction}
Machine translation (MT) nowadays is heavily dependent on the quantity and quality of training data. The amount of available good-quality parallel data for the desired domain and/or language pair is often insufficient to reach the required translation performance. In such cases, it has become the norm to resort to back-translating freely available monolingual data as proposed in \cite{sennrich2015improving}. That is, one can translate a set of sentences from language L2 into L1 with an already trained MT system for the language pair L2$\rightarrow$L1. Then create a synthetic parallel corpus from L1 to L2, with the source (L1) side being the translated text and the target side being the monolingual data. Back-translation has been shown beneficial not only for MT but also for other NLP tasks where data is scarce, e.g. \ac{APE} \cite{Junczys-Dowmunt2016a,Negri2018}. However, the effects of various parameters for creating back-translated data have not been investigated enough as to indicate what are the optimal conditions in not only creating but also employing such data to train high-quality \ac{NMT} systems.
The work presented in \cite{bt-eamt18} draws an early-stage empirical roadmap to investigating the effects of \ac{BT} data. In particular, it looks at how the amount of \ac{BT} data impacts the performance of the final NMT system. In \cite{sennrich2015improving,bt-eamt18} the systems used to generate the \ac{BT} data are neural. However, it has been noted that often different paradigms can contribute differently to a given task. For example, it has been shown that applying an \ac{APE} system based on NMT technology improves \ac{PBSMT} output, but has lower impact on \ac{NMT} output \cite{bojar2017findings,Chatterjee2018}.
In this work we assess the impact of different amounts of \ac{BT} data generated by two different types of MT systems -- \ac{NMT} and \ac{PBSMT}. Our contribution is two-fold: (i) we provide a systematic comparison of the \ac{BT} data by building \ac{NMT} systems with a combination of \ac{PBSMT} and \ac{NMT} \ac{BT} data and (ii) identifying the effects of BT data that originates from PBSMT or NMT on the end-quality of the trained NMT system. We aim to answer the question: "What is the best choice for BT data?"
{\color{red} MP: I hate this:}
This paper is structured as follows: in Section~\ref{sec:related_work} we summarize related work and literature; Section~\ref{sec:different_sources}....
\section{Related Work}\label{sec:related_work}
A comparison between MT models trained with synthetic and with authentic data that originate from the same source has been presented in~\cite{bt-eamt18}. They show that while the performance of models trained with both synthetic and authentic data is better than simply using authentic data, there is a saturation point beyond which the quality does not improve with adding more synthetic data. In addition in their work, models trained only on synthetic (BT) data perform very reasonably, with automatic scores being very close to the scores when only authentic parallel data is used.
~\newcite{D18-1045} confirmed that synthetic data can sometimes match the performance of authentic data. In addition, a comprehensive analysis of different methods to generate synthetic source sentences was carried out. This analysis revealed that sampling from the model distribution or noising beam outputs out-performs pure beam search, which is typically used in NMT. Their analysis shows that synthetic data based on sampling and noised beam search provides a stronger training signal than synthetic data based on argmax inference.
One of the experiments reported in \cite{W18-6315} is comparing performance between \ac{NMT} and \ac{SMT} back-translated data. The best Moses system is almost as good as the NMT system trained on the same (authentic) data, and much faster to train. Improvements obtained with the Moses system trained on a small training corpus are much smaller; this system even decreases the performance for the out-of-domain test.
The authors also investigated some properties of back-translated data and found out that (i) the back-translated sources are on average shorter than authentic ones, syntactically simpler than authentic ones, and contain smaller number of rare events. Furthermore, automatic word alignments tend to be more monotonic between artificial sources and authentic targets then between authentic sources and authentic targets.
To the best of our knowledge, comparing the details related to the translation hypotheses originating from \ac{SMT} and \ac{NMT} back-translated training data as well as combining the data from those two different sources has not been investigated yet.
In this work, we are comparing German-to-English translation hypotheses generated by systems trained (i) only on authentic data, (ii) only on synthetic data, and (iii) on authentic data enhanced with different types of back-translated data: \ac{SMT}, \ac{NMT}. We exploit two types of synthetic and authentic data combinations: (a) randomly selected half of target sentences back-translated by SMT and another half by NMT system, and (b) joining all back-translated data (thus repeating each target segment).
The translation hypotheses are compared in terms of four overall automatic scores, BLEU~\cite{papineni2002bleu}, TER~\cite{snover2006study}, METEOR~\cite{} and chrF~\cite{popovic2015chrf}, as well as by analysis of five error categories, lexical variety and syntactic variety. While \cite{W18-6315} compared training back-translated data with authentic data in terms of lexical and syntactic variety, segment length and alignment monotony, the obtained translation hypotheses were not analysed.
\input{effect-back-translation-with-controlled-amount-of-training-effort.tex}
\section{Using Back-translation from Different Sources}\label{sec:different_sources}
The work of ~\cite{sennrich2016neural} showed that adding back-translated data is beneficial to achieve better translation performances.
\marginpar{Gideon: Section 4 looks a bit weird to me, consisting of just one sentence...}
\section{Experimental Settings}
\label{sec:experimental_settings}
For the experiments we have built German-to-English NMT models using the Pytorch port of OpenNMT ~\cite{opennmt}. We use the default parameters: 2-layer LSTM with 500 hidden units. The vocabulary size is 50000 for each language, and the models are all executed for 13 epochs.
As optimizer we use \ac{SGD}, in combination with learning rate decay, halving the learning rate starting from the 8th epochs.
For testing the models we use the test set provided in the WMT 2015 News Translation Task ~\cite{bojar-EtAl:2015:WMT}. The quality of the translation is measured by different automatic evaluation metrics: BLEU ~\cite{papineni2002bleu},
TER ~\cite{snover2006study} and METEOR ~\cite{banerjee2005meteor} and CHRF1 ~\cite{popovic2015chrf}. The scores provided by these metrics indicate how good the translations are compared to the reference (human translation of the test set).
As development set, we use 5K randomly sampled sentences from development sets provided in previous years of WMT.
\section{Data}
\label{sec:datasets}
The parallel data used for the experiments has been obtained from WMT 2015 ~\cite{bojar-EtAl:2015:WMT}. We build two parallel sets with these sentences: \textit{base} (1M sentences) and \textit{auth} (3M sentences).
In order to build synthetic parallel sets, we use the target side of \textit{auth} to create the following:
\begin{itemize}
\item \textit{SMTsynth}: Created by translating the target-side sentences of \textit{auth}. The model used to generate the sentences is an \ac{SMT} model trained with \textit{base} set in the English to German direction. It has been built using the Moses toolkit ~\cite{koehn2007moses} with default settings, using GIZA++ for word alignment and tuned using MERT ~\cite{och2003minimum}). The language model (of order 8) is built with the KenLM toolkit \cite{heafield2011kenlm} using the German side of \textit{base} .
\item \textit{NMTsynth}: Created by translating the target-side sentences of \textit{auth}. The model used to generate the sentences is an \ac{NMT} model (same configuration as described in Section \ref{sec:experimental_settings}) trained with the \textit{base} set in the English to German direction.
\item \textit{hybrNMTSMT}: Synthetic parallel corpus combining \textit{NMTsynth} and \textit{SMTsynth} sets. It has been built by maintaining the same target side of \textit{auth}, and as source side we alternate between \textit{NMTsynth} and \textit{SMTsynth} each 500K sentences.
\item \textit{fullhybrNMTSMT}: Synthetic parallel corpus combining all segments from \textit{NMTsynth} and \textit{SMTsynth} sets (double size, each original target sentence repeated twice with both an NMT and SMT back-translation-generated translation)
\end{itemize}
\section{Experiments}
In our experiments, we build an initial NMT model using the \textit{base} data set. Then, in order to investigate how much the models benefit from using synthetic data generated by different approaches, we build models with increasing sizes of data (from the data sets described in Section \ref{sec:datasets}).
The models explored are built with data that ranges from 1M sentences (built with only authentic data from \textit{base} data set) to 4M sentences (consisting on 1M sentences from \textit{base} and 3M sentences generated artificially with different models).
We also include the models built with the \textit{fullhybrNMTSMT} set. As this set contains duplicated target-side sentences, the largest model we build contains 7M sentences in total but only 4M distinct target-side sentences.
\section{Results}
\subsection{Controlling for the amount of training effort}
Table \ref{table:results-experiments-with-controlled-amount-of-training-effort} shows the effect of controlling the amount of training effort when using back-translation.
It can be observed that increasing the number of epochs from 13 to 39 when using just the 1M base training set does not increase the performance over using just 13 epochs (i.e. not compensating the relatively smaller training set with more epochs), rather it deteriorates it.
From these results we conclude that there is no reason to believe that the positive effect of using back-translation are caused by an effectively larger training effort, rather than by the advantage of the larger training set itself.
We therefore also conclude that it is reasonable to keep the number of epochs constant across experiments, rather than fixing the amount of training effort as measured by steps/mini-batches, and we do the former throughout the rest of the paper.
\begin{table*}[!htbp]
\centering
\begin{tabular}{|p{1.5cm}|p{2.2cm}|p{2.2cm}|p{2.2cm}||}
\hline
& 1M \textit{base}.- 13 Epochs & 1M \textit{base}.- 39 Epochs- & 1M \textit{base} + 2M \textit{NMTsynth} \\
\hline
BLEU$\uparrow$ & 23.40 & 23.22 & 25.44 \\
TER$\downarrow$ & 57.23 & 58.21 & 55.62 \\
METEOR$\uparrow$ & 28.09 & 27.75 & 29.47 \\
{\small CHR}F1$\uparrow$ & 50.66& 50.18 & 52.5 \\
\hline
\end{tabular}
\caption{Results for experimental procedure validation: checking that it is reasonable to use constant number of epochs, not constant amount of training effort, in the experiments.} \label{table:results-experiments-with-controlled-amount-of-training-effort}
\end{table*}
\begin{table*}[!htbp]
\centering
\begin{tabular}{|p{0.25cm}|p{1.5cm}|p{2.2cm}|p{2.2cm}|p{2.2cm}|p{2.2cm}||p{2.2cm}|}
\hline
& & 1M \textit{base}. & - & - & - & - \\
\hline
\multirow{4}{*}{\rotatebox[origin=c]{90}{\centering 1M lines}}
& BLEU$\uparrow$ & 23.40 & - & - & - & - \\
& TER$\downarrow$ & 57.23 & - & - & - & - \\
& METEOR$\uparrow$ & 28.09 & - & - & - & - \\
&{\small CHR}F1$\uparrow$ & 50.66 & - & - & - & - \\
\hline
& & 1M \textit{base} +1M \textit{auth} & 1M \textit{base} + 1M \textit{SMTsynth} & 1M \textit{base} +1M \textit{NMTsynth} & 1M \textit{base} + 1M \textit{hybrNMTSMT} & 1M \textit{base} + 2M \textit{fullhybrNMTSMT} \\
\hline
\multirow{4}{*}{\rotatebox[origin=c]{90}{\centering 2M lines}}
& BLEU$\uparrow$ & 24.87 & 24.38 & 25.32 & 25.21 & 25.34 \\
& TER$\downarrow$ & 55.81 & 56.05 & 55.66 & 55.87 & 55.79 \\
& METEOR$\uparrow$ & 29.16 & 28.93 & 29.33 & 29.29 & 29.47 \\
&{\small CHR}F1$\uparrow$ & 52.03 & 51.89 & 52.25 & 52.36 & 52.47 \\
\hline
& & 1M \textit{base} +2M \textit{auth}. & 1M \textit{base} + 2M \textit{SMTsynth} & 1M \textit{base} + 2M \textit{NMTsynth} & 1M \textit{base} + 2M \textit{hybrNMTSMT} & 1M \textit{base} + 4M \textit{fullhybrNMTSMT} \\
\hline
\multirow{4}{*}{\rotatebox[origin=c]{90}{\centering 3M lines}}
& BLEU$\uparrow$ & 25.69 & 24.58 & 25.44 & 25.62 & 25.94 \\
& TER$\downarrow$ & 54.99 & 55.7 & 55.62 & 55.25 & 55.11 \\
& METEOR$\uparrow$ & 29.7 & 29.02 & 29.47 & 29.73 & 29.97 \\
& {\small CHR}F1$\uparrow$ & 52.77 & 52.09 & 52.5 & 52.89 & 53.11 \\
\hline
& & 1M \textit{base} +3M \textit{auth} & 1M \textit{base} + 3M \textit{SMTsynth} & 1M \textit{base} + 3M \textit{NMTsynth} & 1M \textit{base} +3M \textit{hybrNMTSMT} & 1M \textit{base} + 6M \textit{fullhybrNMTSMT} \\
\hline
\multirow{4}{*}{\rotatebox[origin=c]{90}{\centering 4M lines}}
& BLEU$\uparrow$ & 25.97 & 24.65 & 26.01 & 25.83 & 25.86 \\
& TER$\downarrow$ & 54.54 & 55.58 & 55.33 & 55.17 & 54.95 \\
& METEOR$\uparrow$ & 29.91 & 29.26 & 29.71 & 29.74 & 29.88 \\
&{\small CHR}F1$\uparrow$ & 53.16 & 52.24 & 52.87 & 52.84 & 53.11 \\
\hline
\end{tabular}
\caption{Performance of models built with different sizes of authentic and synthetic data. \label{table:results}}
\end{table*}
\begin{figure*}[h]
\includegraphics[width=16cm, height=11cm]{plots/BT_2019.png}
\centering
\caption{Plot of the models trained with increasing size of back-translated data.}
\label{fig:BT_2019}
\end{figure*}
\subsection{Main experiments}
Table \ref{table:results} shows the results of the performance of the different \ac{NMT} models we have built. The sub-tables indicate the size of the data used for building the models (from 1M to 4M lines). In each column it is indicated whether \textit{base} been augmented with the \textit{auth}, \textit{SMTsynth}, \textit{NMTsynth}, \textit{hybrNMTSMT}, or \textit{fullhybrNMTSMT} data set. On the top of that we include Figure \ref{fig:BT_2019} for a better visualization of the performance increase when more data is added.
\subsection{Addition of synthetic data from SMT and NMT models }
In Table \ref{table:results} and Figure \ref{fig:BT_2019} we observe that adding SMT-generated data achieves some improvements compared to the model built with 1M sentences. However, those models do not outperform the models built with the same size of authentic data. In fact, the models built with 4M sentences (1M authentic and 3M SMT-produced sentences, in cell \textit{1M base + 3M SMTsynth}) achieve a performance comparable to the model trained with 2M sentences of authentic data.
\marginpar{Gideon: This sentence is a bit too vague, not clear which pairwise-compared systems outperform each other.}
Models built by using \ac{NMT}-created data have a better performance than those built with data generated by \ac{SMT}. \textcolor{red}{In the table we observe that they outperform by around 1 BLEU points regardless the size of the data set.} In fact, the performance is closer to those built with authentic data, and some \textit{NMTsynth} models that produce better translation qualities. This is the case of \textit{1M base+1M NMTsynth} model (according to all evaluation metrics) or \textit{1M base+3M NMTsynth} (according to BLEU). Despite that, according to TER, METEOR and CHRF1 the increases of performance when more NMT-synthetic data is added to the training set is not as much as when it is augmented with authentic data.
\subsection{Addition of SMT- and NMT-combined synthetic data}
In our experiments, we also include the performance of models when SMT- and NMT- generated data are mixed.
When we compare the \textit{hybrNMTSMT} models to those built with only NMT-created sentences (\textit{NMTsynth} set) we observe that the performance is similar. However, in some cases, combining NMT and SMT synthetic data can have beneficial results as is the case of the model built with 2M sentences from \textit{hybrNMTSMT} set. For smaller models (trained with up to 3M sentences) we observe that a using a combination of SMT and NMT artificial data achieves better results than using authentic data according to METEOR and CHRF1.
\subsection{Further Analysis}
In order to better understand the described systems, we carried out more detailed analysis of all translation outputs. We analyzed five error categories: morphological errors, word order, omission, addition and lexical errors, and we compared lexical and syntactic variety of different outputs in terms of vocabulary size and number of distinct POS n-grams.
\subsubsection*{Automatic error analysis}
For automatic error analysis results, we used Hjerson~~\cite{hjerson11}, an open-source tool based on Levenshtein distance, precision and recall. The results are presented in Table~\ref{tab:auterr}.
\begin{table*}[ht]
\centering
\begin{tabular}{|l|ccccc|} \hline
& \multicolumn{5}{c|}{error class rates$\downarrow$} \\
training & morph & order & omission & addition & mistranslation \\ \hline
1M \textit{base} & 2.8 & 9.8 & 12.0 & 4.8 & 29.1 \\ \hline
1M \textit{base} + 1M \textit{auth} & 2.7 & 9.5 & 11.4 & 4.9 & 28.2 \\ \hdashline
1M \textit{base} + 1M \textit{SMTsynth} & 2.8 & 10.0 & 11.6 & 4.8 & 28.1 \\
1M \textit{base} + 1M \textit{NMTsynth} & 2.7 & 9.8 & 10.9 & 5.0 & 28.1 \\
1M \textit{base} + 1M \textit{hybrNMTSMT} & 2.7 & 9.6 & 11.4 & 5.2 & 27.7 \\
1M \textit{base} + 1M \textit{fullhybrNMTSMT} & 2.6 & 9.5 & 11.0 & 5.2 & 27.8 \\ \hline
1M \textit{base} + 2M \textit{auth} & 2.6 & 9.6 & 11.2 & 4.8 & 27.7\\ \hdashline
1M \textit{base} + 2M \textit{SMTsynth} & 2.7 & 10.0 & 11.9 & 4.5 & 28.0 \\
1M \textit{base} + 2M \textit{NMTsynth} & 2.6 & 9.7 & 11.1 & 5.1 & 27.9 \\
1M \textit{base} + 2M \textit{hybrNMTSMT} & 2.6 & 9.6 & 11.0 & 5.2 & 27.6 \\
1M \textit{base} + 2M \textit{fullhybrNMTSMT} & 2.6 & 9.6 & 10.7 & 5.3 & 27.4 \\ \hline
1M \textit{base} + 3M \textit{auth} & 2.7 & 9.8 & 11.2 & 4.6 & 27.6 \\ \hdashline
1M \textit{base} + 3M \textit{SMTsynth}& 2.7 & 9.8 & 11.9 & 4.6 & 27.9 \\
1M \textit{base} + 3M \textit{NMTsynth} & 2.5 & 9.6 & 11.3 & 5.3 & 27.4 \\
1M \textit{base} + 3M \textit{hybrNMTSMT} & 2.6 & 9.5 & 11.0 & 5.1 & 27.6 \\
1M \textit{base} + 3M \textit{fullhybrNMTSMT} & 2.5 & 9.7 & 10.8 & 4.8 & 27.7 \\ \hline
\end{tabular}
\caption{Results of automatic error classification into five error categories: morphological error (morph), word order error (order), omission, addition and mistranslation.}
\label{tab:auterr}
\end{table*}
It can be seen that morphological errors are slightly improved by any additional data, but it is hard to draw any conclusions. This is not surprising given that our target language, English, is not particularly morphologically rich. Nevertheless, for all three corpus sizes, the numbers are smallest for the full hybrid system, being comparable to the results with adding authentic data.
As for word order, adding SMT data is not particularly beneficial since it either increases (1M and 2M) or does not change (3M) this error type. NMT system alone does not help much either, except a little bit for the 3M corpus. Hybrid systems yield the best results for this error category for all corpus sizes, reaching or even slightly surpassing the result with authentic data.
Furthermore, all BT data are beneficial for reducing omissions, especially hybrid which can be even better than the authentic data result.
As for additions, no systematic changes can be observed, except an increase for all types of BT data. However, it should be noted that this error category is reported not to be very reliable for comparing different MT outputs~(see for example \cite{humaut11}).
The mostly affected error category are mistranslations. All types of additional data are reducing this type of errors, especially the hybrid BT data for 1M and 2M, even surpassing the effect of adding authentic data. As for the 3M corpus, the improvement in this error category is similar to the one by authentic data, but the best option is to use NMT BT data alone.
In total, the clear advantage of using hybrid system can be noted for mistranslations, omissions and word order which is the most interesting category. This error category is augmented by adding BT SMT data or not affected by adding BT NMT data, but combining two types of data creates beneficial signals in the source text.
\subsubsection*{Lexical and syntactic variety}
Lexical and syntactic variety are estimated for each translation hypothesis as well as for the human reference translation. The idea behind it is the fact that machine translated data are generally lexically poorer and syntactically simpler than human translations or texts written in the original language. We want to see how different or similar are our translation hypothesis in this sense, and also how they relate to the reference.
Lexical variety is measured by vocabulary size (number of distinct words) in the given text, and syntactic variety by number of distinct POS n-grams where $n$ ranges from 1 to 4. The results are shown in Figure~\ref{fig:variety}.
First of all, it can be seen that none of the translation hypotheses reaches the variety of the reference translation (the black line on the top). The difference is even more notable for the syntax, where the differences between translation hypotheses are smaller and the difference between them and the reference is larger than for vocabulary.
Furthermore, it can be seen that for authentic data (black line on the bottom and green line) the variety increases monotonically with adding more text.
Lexical variety is increased by all synthetic data, too, even more than by authentic data, however, for the NMT and hybrid synthetic data the increase for the 3M corpus is smaller than for smaller corpora.
The increase of syntactic variety is lower both for authentic and for synthetic data than the increase of lexical variety. For 1M and 2M corpus, syntactic variety is barely increased by SMT synthetic data whereas NMT and hybrid data are adding more new instances. For the 3M corpus, however, all synthetic methods yield similar syntactic variety, larger than the one obtained by adding authentic data.
\begin{figure*}[ht]
\centering
\begin{subfigure}[b]{0.48\textwidth}
\lapbox[\width]{-0.3cm}{
\includegraphics[height=5.4cm]{vocabulary.png}
}
\caption{Lexical Variety}
\label{fig:variety:lexical}
\end{subfigure}
\begin{subfigure}[b]{.48\textwidth}
\includegraphics[height=5.4cm]{pos-ngram.png}
\caption{Syntactic variety}
\label{fig:variety:syntactic}
\end{subfigure}
\caption{Lexical variety and syntactic variety for all translation hypotheses and for human reference translations. Lower bound black = baseline (1M authentic data), green = additional authentic data, cyan = additional SMT data, blue = additional NMT data, magenta = additional hybrid SMT+NMT data, red = additional full hybrid SMT+NMT data, Upper bound = black = human reference translation.}
\label{fig:variety}
\end{figure*}
Whereas the increase of lexical and syntactic varieties is in generally a positive trend, there is no guarantee that the MT systems are not introducing noise thereby. To estimate how many of added words and POS sequences are sensible, we calculate precision and recall of word and POS 4-grams when compared to the given reference translation. The idea is to estimate how much the translation hypotheses are getting closer to the reference. We take word 4-grams instead of single words because it is not only important that a word makes sense in isolation, but also in a context. Of course, it is still possible that some of the new instances are valid despite being different from the given single reference.
\begin{figure*}
\centering
\includegraphics[height=5.3cm]{wprec.png}
\includegraphics[height=5.3cm]{wrec.png}
\includegraphics[height=5.3cm]{pprec.png}
\includegraphics[height=5.3cm]{prec.png}
\caption{Word 4-gram precision and recall (above) and POS 4-gram precision and recall (below) for all translation hypotheses. Lower bound black = baseline (1M authentic data), green = additional authentic data, cyan = additional SMT data, blue = additional NMT data, magenta = additional hybrid SMT+NMT data, red = additional full hybrid SMT+NMT data.}\label{fig:precrec}
\end{figure*}
The results of precision and recall are shown in Figure~\ref{fig:precrec}, above for word 4-grams and below for POS n-grams.
Several tendencies can be observed:
\begin{itemize}
\item hybrid BT data are especially beneficial for the 1M and 2M corpora, for 1M even outperforming the authentic data, especially regarding word 4-grams;
\item NMT BT is the best synthetic option for the 3M corpus, however not better than authentic data. This tendency is largest for POS 4-gram precision.
\item SMT BT data achieves the lowest scores, especially for POS 4-grams; this is probably related to the fact that it produces a less grammatical BT sources, which is then propagated to the translation hypotheses. The differences are largest for the 3M corpus, which is probably the reason of diminished effect of the hybrid BT data for this set up.
\end{itemize}
Overall tendencies are that the hybrid BT data is capable even of outperforming the same amount of authentic data if the amount of added data do not exceed the double size of the baseline authentic data. For larger data, a deterioration can be observed for the SMT BT data, leading to saturation of hybrid models.
Further work dealing with mixing data techniques is necessary, in order to investigate refined selection methods (for example, removing SMT segments which introduce noise).
\section{Conclusion and Future Work}
In this work we have presented a comparison of the performance of models trained with increasing size of back-translated data. The artificial data sets explored include sentences generated by using an \ac{SMT} model, and \ac{NMT} model and a combination of both. Two mixing strategies are explored: randomly selecting one half of the source segments from the SMT BT data and the other half from the NMT BT data, and using all BT source segments thus repeating each target segment.
Some findings from previous work~\cite{W18-6315} are confirmed, namely that in terms of overall automatic machine evaluation scores, SMT BT data reaches slightly worse performance than NMT BT data.
Our main findings are that mixing SMT and NMT BT data further improves over each data used alone, especially if full hybridisation is used (using two sources for each target side). These data can even reach better performance then adding the same amount of authentic data, mostly by reducing the number of mistranslations, and increasing the lexical and syntactic variety in a positive way (introducing useful new instances).
However, if the amount of synthetic data becomes too large (three times larger than the authentic baseline data), the benefits of hybrid system start to diminish. The most probable reason is the decrease in grammaticality introduced by SMT BT data which becomes dominant for the larger synthetic corpora.
The presented findings offer several directions for the future work, such as exploring efficient strategies for mixing SMT and NMT data for different authentic/synthetic ratios and investigating morphologically richer target languages.
\FloatBarrier
\bibliographystyle{mtsummit2019}
| {
"redpajama_set_name": "RedPajamaArXiv"
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Q: Kotling :: Android :: java.lang.ClassCastException: java.lang.Class cannot be cast to androidx.lifecycle.ViewModel I am getting an exception in the first line of the code below
viewModel.homeLiveData.observe(this, Observer { list ->
list?.let {
mList.addAll(list)
adapter.notifyDataSetChanged()
}
})
java.lang.ClassCastException: java.lang.Class cannot be cast to androidx.lifecycle.ViewModel
The Whole code is Below
what is wrong with the cast ? is there anything wrong of I am creating my ViewModel?
My BaseActivity
abstract class BaseActivity<V : ViewModel> : DaggerAppCompatActivity(), HasSupportFragmentInjector {
@Inject
lateinit var fragmentAndroidInjector: DispatchingAndroidInjector<Fragment>
@Inject
lateinit var viewModelFactory: ViewModelProvider.Factory
@LayoutRes
abstract fun layoutRes(): Int
protected lateinit var viewModel : V
protected abstract fun getViewModel() : Class<V>
override fun onOptionsItemSelected(item: MenuItem): Boolean {
val id = item.itemId
if (id == android.R.id.home)
onBackPressed()
return super.onOptionsItemSelected(item)
}
override fun onCreate(savedInstanceState: Bundle?) {
AndroidInjection.inject(this)
super.onCreate(savedInstanceState)
setContentView(layoutRes())
viewModel = ViewModelProviders.of(this, viewModelFactory).get(getViewModel())
}
override fun supportFragmentInjector(): AndroidInjector<Fragment> = fragmentAndroidInjector
}
then My Activity
class MainActivity : BaseActivity<MainViewModel>() {
override fun layoutRes(): Int = R.layout.activity_main
override fun getViewModel(): Class<MainViewModel> = MainViewModel::class.java
private val mList = mutableListOf<Any>()
private lateinit var adapter: DataAdapter
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
// setSupportActionBar(toolbar)
recyclerView.apply {
layoutManager = GridLayoutManager(applicationContext, 2)
adapter = DataAdapter(context, mList)
}
viewModel.homeLiveData.observe(this, Observer { list ->
list?.let {
mList.addAll(list)
adapter.notifyDataSetChanged()
}
})
viewModel.getHomeItems()
}
and this is my ViewModel
class MainViewModel @Inject constructor() : ViewModel() {
val homeLiveData: MutableLiveData<List<HomeScreenModel>> = MutableLiveData()
fun getHomeItems() {
Handler().post {
val homeModleList = listOf(
HomeScreenModel(R.drawable.ic_launcher_background, MyApplication.instance.getString(R.string.settings))
)
homeLiveData.setValue(homeModleList)
}
}
}
A: In my opinion, your viewModel property, which you try to access in viewModel.homeLiveData is shadowed by getViewModel() abstract function that you declare in BaseActivity. This is because Kotlin thinks that getXyZ() is a getter for the field xyZ and thus when you access viewModel, the compiler thinks you want to call getViewModel, which is of type Class<V>. I suggest renaming either the function or the property and it should work then.
A: Check your viewModel factory if you implemented it correctly
| {
"redpajama_set_name": "RedPajamaStackExchange"
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Q: How to take a simple integer matrix as input in python3 I want to take a simple matrix as input in a 2-D array, but get a Runtime Error - NZEC error.
Matrix -
1 2
3 4
my input code -
for i in range(2):
a[i]=[int(i) for i in input().split()]
print(a)
A: *
*You are using i as the loop variable and the list comprehension variable at the same time
*You have not declared a (visibly). Declare a = [] and use list.append.
Try this:
a = []
for _ in range(2):
a.append([int(i) for i in input().split()])
Declare a to be empty initially. Then call a.append to append new sublists to your list.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,420 |
2013 Lexus GS 350 F-Sport
Cars Reviews Lexus Lexus GS Sedan SEMA Auto Show Sports cars 2011
by Alina Moore, on October 25, 2011, 14:00
The new Lexus GS350 debuted at the 2011 Pebble Beach Concours, leaving many with the building anticipation for its F-Sport package sibling. That wait will soon be over as Lexus will be headed to the 2011 SEMA Auto Show with the GS F-Sport and will be displaying it from Nov. 1st-4th, 2011.
The new GS F-Sport harbors the same 306 HP 3.5-liter V6 engine from the standard model, but adds "the ability to take performance to an entirely new level." The F-Sport package adds unique exterior and interior design treatments, as well as an F-Sport-tuned Adaptive Variable Suspension (AVS), thicker anti-roll bars, bushing changes, and new larger front brakes with high friction pads.
"The all-new GS 350 with F-Sport package will offer our customers more agility and a sportier appearance," said Mark Templin, Lexus group vice president and general manager. "The racing-inspired F-Sport components were precision engineered for this vehicle, providing quality performance without compromise."
Hit the jump to read more about the Lexus GS 350 F-Sport.
Six-speed sequential-shift with paddle shifters
Horsepower @ RPM:
306 hp @ 6,400rpm
MPG(Cty):
MPG(Hwy):
Torque @ RPM:
277 lb-ft @ 4,800rpm
Displacement:
The new GS 350 F-Sport will be distinguished from the standard GS by a sporty front bumper and rear lower valence, a new rear lip spoiler, unique F-Sport mesh grille inserts, and F-Sport badging. The vehicle also has a wider stance thanks to minor changes in the placement of lights, fins, and the rear bumper underbody area. The front grille will be illuminated with deep-set, high-tech headlamps and L-shaped LED daytime running lamps that use a projector beam design.
Standard F-Sport models will get a new set of 19-inch wheels paired with 235/40/19 all-season tires in the front and rear, while the RWD models will be exclusively equipped with 19-inch staggered width wheels paired with 235/40/19 front and 265/35/19 rear tires, the widest rear tires ever on a Lexus sedan.
The new F-Sport package also adds two new exterior colors: Riviera Red and Liquid Platinum. Vehicles in Riviera Red will benefit from glass flakes incorporated into the coating that add a high level of brilliance to the paint. The Liquid Platinum models will get an additional metallic texture for "strong shading and defined, sharp highlights."
Overall Length (in.): 190.9
Width (in.): 72.4
Height (in.): 57.3
Wheelbase (in.): 112.2
Driven Wheels: Rear or All-Wheel Drive
Curb weight (lbs.)
3,715 (RWD)
3,890 (AWD)
Brake type: Ventilated front and rear disc with Anti-lock Braking System (ABS), Electronic Brake force Distribution (EBD), Brake Assist (BA) and Vehicle Dynamics Integrated Management system (VDIM)
The sport treatment will continue on for the interior with the use of smooth striated aluminum trim, unique perforated leather trim, a black headliner, aluminum pedals, and, of course, F-Sport badges. Customers will also have the option to choose a striking Cabernet Red leather interior.
The new GS 350 F-Sport has been developed with the driver in mind, so it has received a new 16-way power sport seat for increased lateral support. The driver's seat will be a Lexus-first with power side bolsters, thigh support, adjustable seatback, and four-way lumbar. Other standard technological additions include a new energy-saving auto climate control system called S-Flow. This system uses the occupant-detection system to focus airflow only to the front area where passengers are actually seated. That is paired up with a next-generation Remote Touch with one-push confirmation that allows the user to operate the climate, audio, phone controls, optional navigation system, and more with the use of screen menus. This system eliminates the need for drivers to reach out to operate the controls and reduced eye movement. A standard rear back-up camera is also thrown into the standard list.
The new Lexus GS350 F-Sport is powered by the same engine as the standard model: an advanced 3.5-liter V6 with four cams and four valves per cylinder and dual Variable Valve Timing with intelligence. This engine will deliver a total of 306 HP and will propel the car from 0-60 mph in 5.7 seconds. It is mated to a six-speed sequential shift automatic transmission that has four driving modes: Normal, ECO, Sport S, and Sport S+.
What the F-Sport package adds is an optional, newly developed Lexus Dynamic Handling (LDH) system with Dynamic Rear Steering (DRS). This system offers Lexus' first integration of DRS; Electric Power Steering; and Variable Gear Ratio Steering. It monitors your the vehicle's speed, steering direction, and driver inputs, and uses that information to calculate the optimum angle for all four wheels. This system can also independently manage both front and rear wheel steering angles to help improve turn-in response, rear grip, vehicle control, and overall agility when cornering.
Horsepower: 306 hp @ 6,400rpm
Torque: 277 lb-ft @ 4,800rpm
Transmission: Six-speed sequential-shift with paddle shifters
Electronic Controlled Automatic Transmission with intelligence (ECT-i)
0 – 60 mph: Estimated 5.7 seconds
Top Track Speed: Estimated 142 mph (electronically limited)
MPG (City/Hwy/Comb.)
20/27/23 (RWD)
18/25/20 (AWD)
The new Lexus GS F-Sport comes standard with a number of safety features, including 10 airbags, new Whiplash Injury Lessening front seats, seatbelts with pretensioners with force limiters, and outboard rear seats. Optional features include a Pre-Collision System, which provides early warning of certain objects traveling directly ahead that might result in a front-end collision; a Night Vision System that enhances driver visibility at night; Heads Up Display; a Blind Spot Monitor; and Lane Keep Assist (LKA) with Lane Departure Warning (LDW).
Prices for the new 2013 Lexus GS 350 F-Sport will be announced at a later date.
The competition for the new GS350 F-Sport is the same as with the standard model: the BMW 5-Series and the Infiniti M37. Both models might be more powerful than the GS and the 5-Series is definitely more popular than the other two, but we have to give props to the new Lexus GS350. It has moved up in the world and can now boast of being more elegant, fresher, and sportier than its predecessor. The F-Sport package helps out too.
Sporty look
F-Sport package adds great accessories
Powerful engine
Competition might be more powerful
Alina Moore
Alina Joined the Topspeed.com team in the early 2000s as one of the outlets very first experts, and she's been with Topspeed.com ever since. Over the years, she's served various roles, but today she's is relied on heavily to verify automotive facts, assist with formatting, and discover new and engaging topics. Read full bio
I'm so impressed with its dressed-up interior! It looks very cozy and comfortable on that. Moreover, I'm glad that it has an interesting technology which I really never thought that this vehicle could offer.
I just only love the interior, so cozy and very entertaining, especially its center console! Anyway, I noticed that its front bumper is truly oozing with angst but too bad it only has a weak engine output.
The interior feature of this Lexus is what I really love the most in this vehicle! It looks so sporty and classy on that, and I must say that even though it is only on a simple detailing, it was still oozing with car appeal.
I love the sporty look of this Lexus even though it only looks so simple on its detailing and styling. Anyway, I'm also glad that it had an impressive and comfortable interior.
I love the sporty look of this Lexus GS! I'm also so glad that they add great accessories on this one, which only adds more appeal on it and the safety features of this one is what I really love the most about it.
I think that the engine and horsepower of this one are kind of impressive but I think they should boost it more. Anyway, I'm so impressed with its exterior appearance.
F-Sport of Lexus, is absolutely looking good on its aesthetic design, and I noticed that the interior of this one, is looks so classy and very comfortable which I really love in a car.
The interior of this Lexus F-Sport is very striking on its cozy and comfortable touch and features. Its specifications are also so amazing and awesome on its aesthetic design and styling.
I have to agree with you Rob Cortez that the engine of this one is not really quite impressive on the aggressiveness of this GS 350 F-Sport of Lexus, but I must say that I'm so fond with its comfortable interior.
I love the angst that was expressed by its front design, well it looks like that it is so wild. Anyway, the engine of this one, is quite good enough for a sedan car, but I think it is much better if they used V8.
Model: Model View All Lexus CT 200h Lexus ES Lexus GS Lexus GS F Lexus GX Lexus HS 250h Lexus IS Lexus LC Lexus LF Lexus LF-A Lexus LS Lexus LX Lexus NX Lexus RC Lexus RC F Lexus RX Lexus SC Lexus UX
Cars / Lexus / Lexus GS / Sports cars / 2011 / SEMA Auto Show / Sedan | {
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Barsacq est un nom propre français qui peut faire référence à :
Patronyme
Une famille française du théâtre, du cinéma et des lettres
André Barsacq (1909-1973), metteur en scène, réalisateur, décorateur, scénographe, dramaturge et directeur de théâtre français
Léon Barsacq (1906-1969), décorateur et costumier français, frère d'André
Yves Barsacq (1931-2015), comédien français, fils de Léon
Alain Alexis Barsacq (né en 1936), metteur en scène, décorateur, scénographe et directeur de théâtre français, fils d'André
Élisabeth Alain (1937-1974), comédienne de théâtre, fille d'André
Stéphane Barsacq (né en 1972), écrivain et éditeur, petit-fils d'André
Autre personnalité
Claire Barsacq (née en 1980), journaliste et animatrice de télévision française
Patronyme français | {
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} | 7,640 |
{"url":"https:\/\/italiamagazine.be\/irrational-numbers-symbol","text":"### What Is Pi? | Live Science\n\nOct 19, 2018 to its diameter. Pi is an irrational number that never ends; the decimals go on forever and ever. An illustration of the symbol for Pi. Pi is an\n\n### Number Systems- MathBitsNotebook (Jr)\n\nThe natural numbers (symbol N ) are the set The integers (symbol Z ) Irrational numbers are \"not closed\" under addition, subtraction, multipliion or division.\n\n### Set of numbers (Real, integer, rational, natural and irrational numbers)\n\nSet of numbers (Real, integer, rational, natural and irrational numbers). In this unit, we They are denoted by the symbol Z and can be written as: Z = { , \u2212 2\n\n### Irrational Number \u2013 Lexique de math\u00e9matique\n\nThe symbol Q\u2032 represents the set of irrational numbers and is read as \u201cQ prime\u201d. The symbol Q represents the set of rational numbers. Combining rational and\n\n### The Language of Algebra - Definitions - In Depth - Math.com\n\nRather than using words, algebra uses symbols to make statements about things. Real Numbers | Rational Numbers | Irrational Numbers | Translating Words\n\n### The Man Who Invented Pi | History Today\n\nBefore the appearance of the symbol \u03c0, approximations such as 22\/7 and Though he did not prove it, Jones believed that \u03c0 was an irrational number: an\n\n### Square Roots and Irrational Numbers | CK-12 Foundation\n\nFeb 24, 2012 Every positive number has two square roots, the positive and the negative. The symbol used to represent the square root is \\begin{align*}\\sqrt{x}\\\n\n### Rational and Irrational Numbers \u2013 Prealgebra - BC Open Textbooks\n\nIdentify rational numbers and irrational numbers; Classify different types of real You have become familiar with the language and symbols of algebra, and\n\n### WRKDEV100-20011 - Variables, Constants, and Real Numbers\n\nIn algebra, symbols (usually letters) are used to represent numbers. natural, whole, integers, rational, and irrational numbers which are sets of real numbers.\n\n### Irrational number - Wikipedia\n\nIn mathematics, the irrational numbers are all the real numbers which are not rational numbers, the latter being the numbers constructed from ratios (or fractions)\n\n### Irrational Numbers - Math is Fun\n\nAn Irrational Number is a real number that cannot be written as a simple fraction. Irrational means not Rational Numbers. A Rational Number can be written as a Ratio of two integers (ie a simple fraction). radical symbol. Many square roots\n\n### Irrational Number -- from Wolfram MathWorld\n\nEvery transcendental number is irrational. There is no standard notation for the set of irrational numbers, but the notations Q^_ , R-Q , or R\\Q ,\n\n### Intro to rational irrational numbers | Algebra (video) | Khan Academy\n\nLearn what rational and irrational numbers are and how to tell them apart. Where did they get the symbol for pie from ? Reply. Reply to Julyanne Nichols's post\n\n### Examples of Irrational Numbers\n\nMany other square roots and cubed roots are irrational numbers; however, not all square roots are. The Golden Ratio, written as a symbol, is an irrational number\n\n### What is the symbol for the set of Irrational numbers\n\nAug 9, 2016 There is no commonly accepted default symbol for the set of Irrational numbers, [ math]\\mathbb{R\\setminus Q}[\/math]. The language of mathematics is, however,\n\n### Number Theory #1| Symbols | What is the symbol for Irrational\n\nJun 26, 2019 N W Z Q I R - what symbol represents rational numbers? Quick Maths Videos using the CXC syllabus as a guide from live recordings For in\n\n### Is -3 a rational, irrational, natural, whole, integer or real number\n\nJul 1, 2015 The normal symbol for integers is Z. \u22123 obviously falls in this egory. Rational numbers are numbers that can be expressed as a fraction or\n\n### Do We Have a Sense for Irrational Numbers? | Obersteiner | Journal\n\nTo summarize, the ability to assess magnitudes of number symbols is fundamental for As irrational numbers differ in important ways from rational numbers,\n\n### Being Rational About Irrational Numbers | Finding the Sum and\n\nImage description: Examples of irrational numbers appear showing the square root of three, pi, negative two There actually is no symbol for irrational numbers .\n\n### Rational and Irrational Number - Maths Doctor\n\nIrrational numbers are square roots of non-perfect squares. Only the The golden ratio (whose symbol is the Greek letter \"phi\") is also an irrational number.","date":"2020-09-18 11:48:49","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8743411898612976, \"perplexity\": 1534.7639531346983}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 5, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-40\/segments\/1600400187390.18\/warc\/CC-MAIN-20200918092913-20200918122913-00441.warc.gz\"}"} | null | null |
for you and today is mine turn.
here is a peek of the finished item.
using some lush sparkle medium.
Fabulous sparkling projects Nicky. Love them all.
Hi there Nicky, this cushion is beautiful, love the colours you have used, very pretty Hun. | {
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\section{Introduction}
X-ray Flashes (XRFs) appear to be a subclass of Gamma-Ray Bursts (GRBs).
They have similar durations to the long-soft GRBs, but they have a low
gamma-ray flux, a high ratio of X-ray to gamma-ray flux, and a spectral peak
at a much lower energy (Kippen {\it et al.}\ 2003).
First identified by the Wide Field Cameras on BeppoSAX
(Heise {\it et al.}\ 2001) they have been located with
increasing frequency by {\it HETE-2}. Followup observations have in some cases
successfully found afterglow emission, most commonly at X-ray wavelengths, and
occasionally in the optical and nIR.
Accurate positions have indicated that, like GRBs, XRFs
are found in star-forming galaxies at cosmological distances (e.g. Bloom {\it et al.}\ 2003),
but, based on statistics of a few, at rather lower median redshift than GRBs.
Recently the very low-redshift XRF 060218, 145 Mpc distant, was shown
to be associated with SN 2006aj (Campana {\it et al.}\ 2006; Pian {\it et al.}\ 2006; Modjaz {\it et al.}\ 2006), clearly
demonstrating that some XRFs, as for long duration GRBs originate in
the core collapse of a massive star in a type Ic supernova (Hjorth et al. 2003; Stanek et al. 2003).
Photometric evidence for associated supernovae has also been seen in a number
of XRFs previously (e.g. Fynbo {\it et al.}\ 2004; Soderberg {\it et al.}\ 2004;2005; Bersier {\it et al.}\ 2006),
however the absence of SN signatures in some cases indicate that
the SN emission may be markedly fainter than the prototypical GRB supernova SN~1998bw
(Levan {\it et al.}\ 2005; Soderberg {\it et al.}\ 2005).
The existence of XRFs shows that the range of spectral properties associated with
GRBs is very large. The peak of the $\nu F_{\nu}$ spectrum ($E_p$) can be seen from
$<$5 keV to $>1$ MeV. An important
question is, therefore:
what physical processes are responsible for this range of
energies?
Observations of the prompt and afterglow emission of XRFs can be used
to probe this question and place constraints on the energy generation and physical
structure within these highly energetic explosions.
Various scenarios have been proposed to explain the lower peak energy of
XRFs compared to GRBs; these can be split into several subsets of model, of
which the most broadly discussed are: (i)
GRBs at very high redshifts would have
their spectral energy distributions (SEDs) shifted into the X-ray window (Heise {\it et al.}\ 2001);
(ii) GRBs whose Lorentz factor is modified due to the effects
of baryon loading within the jet. In external shock models high
baryon loading (the so-called dirty fireball) can create an XRF
(Dermer {\it et al.}\ 1999;
Ramirez-Ruiz \& Lloyd-Roming 2002; Huang {\it et al.}\ 2002),
while in contrast for internal shock
models very clean jets produce large X-ray fluxes (Zhang \& Meszaros 2002,
Barraud {\it et al.}\ 2005). (iii) GRBs with either a differing geometry or
observer viewing angle can also naturally create XRFs, for example
due to broader than normal jets (Huaag {\it et al.}\ 2004; Zhang {\it et al.}\ 2004);
bursts where the line of sight is
"off-axis" with respect to the jet orientation or bursts whose jet is structured,
either in a simple (two component) manner, or with more complex
variations with viewing angle. (iv) Finally it is possible that XRFs
represent manifestations of differing physical processes to GRBs themselves,
perhaps originating from a hot photosphere (Meszaros {\it et al.}\ 2002).
The discovery of a correlation between $E_p$ and the square root of
the isotropic energy release ($E_{iso}$) (Amati {\it et al.}\ 2002) directly implies
that softer bursts have lower energies and can be well explained by
models where the XRF is the result of a viewing angle effect. In other words,
XRFs are seen when a classical GRB is viewed off the primary collimation so
the highest energy emission is not seen (e.g. Yamazaki {\it et al.}\ 2002; Rhoads 2003; Dado
{\it et al.}\ 2004). However, outliers to this relation can be found (e.g. Sazonov et al. 2004),
and it remains to
be confirmed that it is purely the result of a viewing angle effect.
The {\it Swift} satellite was launched in November 2004 and is now delivering
localizations for
approximately 2 GRBs per week (see Gehrels {\it et al.}\ (2004) for a
description of the {\it Swift} mission). The passband of the GRB detectors on
{\it Swift} is smaller than that on {\it HETE-2}, with a low energy response down
to 15 keV compared with 5 keV for {\it HETE-2}. XRF~050215B was the first XRF
to be detected by {\it Swift} and to have moderately rapid X-ray followup (previous XRFs have
not been observed in X-rays for at least several hours, and more typically days).
Here we report the results of optical/IR and
X-ray observations of XRF~050215 and the constraints they place on models for the
production of XRFs.
\section{Observations}
XRF~050215B was detected by the {\it Swift} Burst Alert Telescope at
02:33:12 UT on 15th February 2005; the initial location was
RA = 11:37:48, Dec = 40:48:18 with a 90\% error
radius of 4 arcminutes (trigger 106106, Barthelmy {\it et al.}\ 2005a).
{\it Swift} slewed to the burst promptly, but was unable to observe until
$\sim$ 29 minutes after the trigger because it was in the high radiation
environment of the South Atlantic Anomaly.
X-ray Telescope observations revealed a faint, slowly fading point
source, identified as the X-ray afterglow of XRF~050215B (Goad {\it et al.}\ 2005).
The UV and Optical Telescope did not yield
a detection of any source at the location of this
X-ray object (Roming {\it et al.}\ 2005a,b).
The burst was also seen by {\it HETE-2}, and an analysis of this data revealed a best fitting
$E_p$ = 17.6 keV, with a 95\% upper bound of $E_p < 30.3$ keV (Nakagawa {\it et al.}\ 2005).
The ratio of fluxes in the 2-30 and 30-400 keV bands was 1.65, implying significantly more
X-ray emission than $\gamma$-ray and indicating that 050215B was indeed an XRF.
Initial ground-based observations failed to yield an optical counterpart.
The BOOTES-2 wide field camera was observing the field at the time of the
burst, limiting the unfiltered magnitude of any flash to $>10$
(Jelinek {\it et al.}\ 2005).
ROTSE-III observations started 100s after the burst, however these images
failed to yield an optical counterpart to limiting magnitudes of
R$\sim 17.8 - 18.8$ up to 50 minutes after the burst (Yost {\it et al.}\ 2005).
Radio upper limits of 93 and 156 $\rm \mu$Jy (3 $\sigma$) at 2.9 and 12 days
after the burst were reported by Soderberg \& Frail (2005a,b). A skymap with
the location of the burst (and afterglow) is shown in Figure 1.
\begin{figure*}
\includegraphics[scale=1.1,angle=270]{f1.ps}
\caption{The sky position of XRF~050215B. The left hand image shows a region of
the Digital Sky Survey (DSSII-red) with the BAT error box marked as the large circle and the
XRT position labeled within it. The right hand image shows our first UKIRT observation.
The large circle in this image is the XRT location and 90 \% confidence region. The
location of the IR afterglow is also marked. }
\label{posn}
\end{figure*}
\subsection{{\it Swift} Observations}
The {\it Swift} satellite and its gamma-ray Burst Alert Telescope (BAT),
X-ray Telescope (XRT) and UV/Optical Telescope (UVOT)
are described by Gehrels {\it et al.}\ (2004), Barthelmy {\it et al.}\ (2004, 2005b),
Burrows {\it et al.}\ (2004, 2005) and Roming {\it et al.}\ (2004, 2005c).
The {\it Swift} data described here
has been processed with version 2.0 of the {\it Swift} software\footnote{http://heasarc.gsfc.nasa.gov/docs/software/lheasoft/}, using
the standard BAT and XRT pipelines.
The BAT detected XRF~050215B below 150 keV, showing a single peaked
lightcurve, (see Figure 2).
The rise to peak lasts 2-4 s, while the decline takes 7-10 s.
The T$_{50}$ and T$_{90}$ and durations
are 3.4 and 7.8 seconds, respectively.
The BAT trigger time (02:33:43 UT) corresponds to the peak of the light curve.
This time was mis-reported initially; all times from the burst stated in this
paper use this correct trigger time.
A single powerlaw spectral fit to the entire burst
between 15 and 150 keV results in a good fit
($\chi^{2}/$dof $ = 1.0$), with
a photon power law index of 2.3$\pm$0.4.
The total 15-150 keV fluence in
the T90 interval is $2.33 \pm 0.6 \times 10^{-7}$ erg cm$^{-2}$ (errors are quoted
as 90\%).
The extrapolated fluence in the 30-400 keV FREGATE region is
2.0 $\times 10^{-7}$ erg cm$^{-2}$, in agreement with the results obtained
by Nakagawa {\it et al.} (2005). While XRF~050215B had a fluence somewhat below the
Swift average, it is unusual in having a very soft spectrum
(corresponding to a 2-30 keV to 30-400 keV extrapolated fluence ratio of
$S_X /S_{\gamma} = 2.0^{+0.2}_{-0.3}$)
consistent with the X-ray flash characterisation of Nakagawa {\it et al.} (2005).
Observations with the XRT
began at 03:03:11 UT (approximately 30 minutes after trigger),
and continued for 13.3 days;
they are detailed in table 1. The first exposure was made in
Windowed Timing (WT) mode while all later observations were in
Photon Counting (PC) mode; a total of 128 counts were obtained from
XRF~050215B.
The location of the X-ray afterglow was found to be
RA = 11:37:47.66, Dec = 40:47:46.7 ,
with a 90\% error radius of 4.4 arcseconds (Moretti {\it et al.}\ 2006).
The X-ray lightcurve of XRF~050215B is shown in Figure 3 (and
is combined with the BAT lightcurve).
The WT data were extracted from a box 94$''$ across (40 pixels), with the background
measured from the same size region offset from the source. For the PC mode
data counts were extracted from a 47$''$ (20 pixel) radius region centred on the source,
with the background taken from a source-free region 9 times larger in area.
All points in Figure 3 correspond to a better than 3$\sigma$ detection of the
source. The X-ray afterglow of XRF~050215B declines as an unbroken power law
with a decay index of 0.82$\pm$0.08.
With the modest number of XRT counts only limited spectral information can
be derived. We fit a power law model to the total PC mode afterglow spectrum
(which covers 0.5-2.6 keV), finding a good
fit for a photon index of 1.5$\pm$0.5 assuming the Galactic absorbing column
of $2\times 10^{20}$ cm$^{-2}$. No excess absorption over that due to our
Galaxy is seen; at zero redshift $\Delta N_H < 3.8\times 10^{21}$ cm$^{-2}$
(90\%).
Examination of the ratio of counts in the
0.3-1.0 keV to 1.0-10 keV bands as a function of time showed
no variation. Unabsorbed 0.3-10 keV fluxes at 1 and 11 hours derived from
Figure 3 and the spectral parameters above are respectively
28 and 4.9 $\times 10^{-13}$ ergs s$^{-1}$ cm$^{-2}$.
The {\it Swift} UVOT also observed the position of XRF~050215B at times
coincident with the XRT observations using its full range of filters.
No new source was detected, either initially or later with longer exposures.
3 $\sigma$ upper limits include $U>20.2, B>20.1, V>19.3$ at 0.57, 0.60 , 0.48
hours, and $V>21-22$ up to 13.3 days after the burst.
\begin{deluxetable*}{llllll }
\footnotesize
\tablecolumns{4}
\tablewidth{0pt}
\tablecaption{{\it Swift} XRT observations of XRF~050215B}
\tablehead{\colhead{Start (hours)} & \colhead{End (hours)} &
\colhead{Exp. Fraction} & \colhead{Source counts} & \colhead{Background counts}
& \colhead{XRT mode}}
\startdata
0.498 & 0.656 & 1.0 & 45 & 16 & WT\\
1.603 & 1.699 & 1.0 & 11 & 0.22 & PC\\
3.219 & 3.416 & 1.0 & 17 & 0.56 & PC\\
4.819 & 5.133 & 1.0 & 16 & 0.67 &PC\\
6.419 & 6.783 & 1.0 & 14 & 0.89 & PC\\
8.036 & 8.399 & 1.0 & 13 & 0.56 & PC\\
47.048 & 66.455 & 0.081 & 16 & 4.6 & PC\\
69.528 & 93.433 & 0.14 & 19 & 12 & PC\\
95.250 & 320.572 & 0.0098 & 28 & 16 & PC\\
\hline
\enddata
\tablecomments{Swift XRT observations of XRF~050215B. A final observation at 13.3
days was too short to be useful, and is not listed here. The start and end times are
given in hours since the burst (i.e. hours after 2005-02-15 02:33:12 UT)}
\label{tab:photdata}
\end{deluxetable*}
\subsection{Ground based observations}
IR images were obtained at UKIRT using the UFTI instrument
(Roche {\it et al.}\ 2002) in the K98 filter (see Figure \ref{posn}).
Dithered observations were flat-fielded, sky- and
dark-subtracted and combined using ORAC-DR (Cavanagh {\it et al.}\ 2002).
IR and optical images were also taken at Gemini: NIRI data
(Hodapp {\it et al.} 1995)
which were also processed with ORAC-DR; GMOS optical data (Hook {\it et al.} 2004)
which were reduced via
the specific Gemini/GMOS tasks with IRAF \footnote{For information
on ORAC-DR see http://www.oracdr.org and for the Gemini IRAF tasks
see http://www.gemini.edu/sciops/data/dataIRAFIndex.html}.
Additionally a 4 x 30 minutes spectrum of the potential host was taken with
GMOS
with an R400 grating and the central wavelength varying between 6400
and 6550 Angstrom, from March 11.47 to 11.56 UT. Data reduction was
done in a standard fashion using IRAF. Only a very weak
continuum was seen, there were no obvious emission lines
in the combined spectrum between 6000 and 8500 Angstroms.
A complete log of observations is shown in
table \ref{phot}.
Our first image was obtained approximately 10 hours after the burst
with a second epoch obtained the following night ($\sim$ 33 hours
post-burst). Comparison of these images
revealed a fading point source at RA=11:37:47.90, Dec = 40:47:45.6,
approximately 33 arcseconds from the
centre of the BAT error box and well within the refined XRT position
given in section 2.1.
To perform photometry of the afterglow we used an aperture equal to the FWHM of
the images at each epoch. During our UKIRT observations photometric standards
were also observed which allowed us to subsequently calibrate the field. The
photometric calibration of the R-band images was done using the photometry
from the Sloan Digital Sky Survey, which covers the region of the burst in the same
filter as used for the GMOS observations.
Photometry of XRF~050215B is shown in table \ref{phot} and the K-band light curve
is shown in Figure \ref{klc}. The afterglow was very faint, and therefore the signal to
noise was typically low ($3-10 \sigma$ in most observations). However the afterglow
is clearly fading. A single power-law fit to the K-band data yields a best fit decay
slope (fitted as $F(t) \propto t^{-\alpha}$) of $\alpha = 0.47 \pm 0.08$, unusually slow for
GRB afterglow which fall typically with $\alpha = 1$ in the first hours to days after the burst, and
more rapidly after this. One possibility is that our measured fluxes do not contain pure afterglow
but some contribution from an underlying host galaxy. In this case the true decay slope may
be faster. We can estimate the maximum possible contribution from an underlying host by
making the assumption that the host galaxy contributes a flux equal to the upper limit of
the final epoch observation (i.e. lies just below the detection threshold). Given
the R magnitude of the host (see below) and the typical R-K colours of GRB host galaxies
(in the range 2.5 - 3, Le Floc'h {\it et al.}\ 2003) it is unlikely that the host
is significantly fainter than this, and therefore this may provide an acceptable estimate
of the afterglow decay. The (putative host) subtracted photometry is shown in
red in Figure \ref{klc}. This photometry, while very uncertain, is
marginally consistent with a single power-law decay of index $\alpha = 0.82 \pm 0.08$;
thus, within the uncertainty (due to the unknown host magnitude),
the optical and X-ray slopes are consistent with being identical.
PSF matched image subtractions of the optical observations taken 11, 21 and 115 days
after the burst reveal no evidence for any variation, demonstrating that the optical
light from XRF~050215B was dominated by the host galaxy from 11 days onwards,
and did not contain a significant contribution from either afterglow or associated supernova
(the limit on any transient emission at the location of the afterglow is R$\sim$25.8).
The magnitude of this host galaxy is R=24.7, comparable to the median of GRB
host galaxies seen to date. Our final K-band point can also be used to place a limit
on the host colour. Assuming it has K$>$22.25 this implies that R-K$<$ 2.5. This is again
comparable to the R-K colours of GRB hosts which have a mean R-K=3 (Chary {\it et al.}\ 2002;
Le Floc'h {\it et al.}\ 2003).
Under the assumption that a supernova was associated with XRF 050215B, we can
use the observed limits on the supernova emission to crudely estimate a
lower bound to the redshift (also assuming that there is no excess extinction
along the line of sight to XRF 050215B). The limiting magnitude of
any supernova of R=25.8 is comparable to the peak magnitude of a
supernova such as SN~1998bw at $z=1$. However
more typical type Ic supernova such as those putatively associated
with XRFs (e.g. Fynbo {\it et al.}\ 2004; Bersier {\it et al.}\ 2005) span a range
of magnitudes from similar to SN~1998bw to a factor of 10 fainter,
these fainter supernovae would only be visible to moderate $\sim 0.5$ redshift.
\begin{figure}
\includegraphics[scale=0.3,angle=270]{f2.ps}
\caption{The BAT 15-350 keV light curve of the XRF~050215B.}
\label{bat}
\end{figure}
\begin{figure}
\includegraphics[scale=0.35,angle=270]{f3.ps}
\caption{The joint BAT and XRT lightcurve of XRF 050215B.
The XRT count-rates were converted into unabsorbed fluxes using the best
fit model. The BAT data were extrapolated into the 0.3-10 keV band
assuming a photon index which was the mean of the best-fit XRT and BAT
spectral indices. The XRT data are best fit with an unbroken
power-law with a decay index $\alpha = 0.82 \pm 0.08$.
The extrapolation of the XRT observations to early
times falls slightly below the prompt emission measured by the BAT
(and the fitting of a single power-law to both BAT and XRT data does not result in a
statistically acceptable fit, although given the calibration
uncertainties it may be consistent with a single
decline from $10^{-3}$ hours onward). However, an extrapolation of the X-ray afterglow
below the prompt emission could easily be explained by a steep decay
of the X-ray flux at early times
which is
commonly seen in {\it Swift} bursts (e.g. Tagliaferri {\it et al.}\ 2005; Nousek {\it et al.}\ 2006;
O'Brien {\it et al.}\ 2006).}
\label{xrtlc}
\end{figure}
\begin{deluxetable*}{lllllll}
\footnotesize
\tablecolumns{4}
\tablewidth{0pt}
\tablecaption{Ground based observations of XRF~050215B}
\tablehead{\colhead{UT} & \colhead{$\Delta t$ (days)} &
\colhead{Instrument} & \colhead{Filter/Grating} & \colhead{Seeing (\arcsec)}
& \colhead{Exptime (s)} & \colhead{Magnitude} }
\startdata
Feb 15.519 & 0.413 & UKIRT/UFTI & K98 & 0.42 & 800&$20.23 \pm 0.11$ \\
Feb 16.497 & 1.391 & UKIRT/UFTI &K98 & 0.48 & 1200&$20.75 \pm 0.22$\\
Feb 17.466 & 2.360 & UKIRT/UFTI & K98 & 1.05&3000 &$21.10 \pm 0.31$\\
Mar 03.565 & 16.457 & GEMINI/NIRI &K & 0.60&2250 &$22.08 \pm 0.29$\\
May 26.381 & 100.274 & GEMINI/NIRI & K & 0.60 & 2250 & $>22.25 (3\sigma)$ \\
\hline
Feb 26.552 & 11.446 & GEMINI/GMOS&r & 1.06 &1800 &24.71 $\pm$ 0.11\\
Mar 8.394 & 21.287 & GEMINI/GMOS&r & 0.84 & 1800& 24.66 $\pm$ 0.05\\
June 10.287 & 115.180 & GEMINI/GMOS&r & 1.03 & 1800 & 24.68 $\pm$ 0.07 \\
\hline
Mar 11.530 & & GEMINI/GMOS &R400 & - & 7200& -\\
\enddata
\tablecomments{\phantom{}Log of photometry of XRF~050215B obtained at UKIRT and Gemini.
The times given are the midpoint of the observations, errors quoted are 1$\sigma$.}
\label{phot}
\end{deluxetable*}
\section{Discussion}
Figure \ref{kband} shows the range of magnitudes seen in the K-band afterglow sample
which has been observed to date (Rau {\it et al.}\ 2004 and references therein). In addition to
the sample determined from GRBs discovered before {\it Swift} (e.g. {\it BeppoSAX, HETE-2}) which
are shown in black, the K-band points from afterglows detected by {\it Swift} are also shown
as dark blue triangles and the highest redshift bursts (GRBs 050814 and 050904)
are shown in as green squares. The afterglow of XRF~050215B is shown in red and, as can be seen to be the faintest afterglow
observed to date in the K-band (January 2006), and lies toward
the faint end of the X-ray afterglow sample (e.g. Nousek {\it et al.}\ 2006; O'Brien {\it et al.}\ 2006).
It is possible that the faintness of the optical and X-ray afterglow of XRF~050215B could
be related to the faintness of the prompt emission component. Berger {\it et al.}\ (2005) show
that typically {\it Swift} bursts are fainter than those detected by {\it HETE-2} and
{\it BeppoSAX} at essentially all wavelengths (i.e. both in prompt emission and
it in X-ray/Optical Afterglows). However XRF~050215B was also detected
by {\it HETE-2} and the fluences observed in the 2-30 and 30-400 keV
band were $2.8 \times 10^{-7}$ ergs and $1.7 \times 10^{-7}$ ergs, lying between
the fluences seen for XRF~020903 and 030723, the XRFs which have
the best studied optical afterglows (e.g. Sakamoto {\it et al.}\ 2005).
The extrapolation of the X-ray afterglow of XRF~050215B out to later times,
comparable to those of X-ray observations of previous XRF afterglows,
show that it would achieve a 10 day flux of approximately $1 \times 10^{-14}$ ergs s$^{-1}$
cm$^{-2}$, somewhat fainter than the afterglows of XRF~011030 and 020427
at these times (Kouveliotou {\it et al.}\ 2004; Levan {\it et al.}\ 2005); however it would have been slightly
brighter than the afterglow of XRF~030723 (Butler {\it et al.}\ 2004). The afterglow decay
is, however very shallow - slower than all but four of those studied by O'Brien {\it et al.}\ 2006,
and, furthermore continues to have a shallow decay for $\sim$ 10 days post burst.
A dusty environment does not seem able to explain the faintness of this burst since it
is faint at both X-ray and optical wavelengths. Although dust within the host galaxy
could render the afterglow invisible at optical wavelengths and faint in the IR
it could not simultaneously explain the faintness of the X-ray afterglow. Indeed an
extrapolation of the X-ray afterglow into the optical/IR regime (using the method
of Rol {\it et al.}\ 2005) reveals that for typical fireball parameters the observed K-band
flux and optical limits are consistent with zero extinction. Unfortunately given the lack
of multiband optical observations and the faintness of the X-ray afterglow excess absorption
cannot be searched for either from the shape of the optical SED or by a decrement of
soft X-rays.
An alternative model is that the burst lies at high redshift. Although it is now
clear that some XRFs lie at low redshifts (e.g. XRF 020903 at $z=0.25$, Soderberg et al. 2004), it
is plausible that some fraction also originate in the very high redshift universe.
The presence (and indeed
brightness) of the host galaxy of XRF 050215B indicate that
the redshift cannot be very high.
However moderate luminosity
distance (e.g. z$>2$) may still be a viable explanation of the faintness of the burst, although cannot
explain the observed slow decay rate. Indeed {\it Swift} bursts are apparently
fainter at all wavelengths than previous samples (Berger {\it et al.}\ 2005) and are
also, on average, at significantly higher redshifts (a mean of $z=2.8$; Jakobsson {\it et al.}\ 2006).
It is also interesting to note that the afterglow of XRF
050215B is significantly fainter than that of the high redshift GRBs 050904 ($z=6.29$ -Haislip {\it et al.}\ 2006; Price {\it et al.}\ 2005; Tagliaferri {\it et al.}\ 2005; Kawai {\it et al.}\ 2005)
and GRB 050814 ($z=5.3$ Jakobsson {\it et al.}\ 2006).
The most popular models for the production of XRFs is that they are: (i) GRBs viewed away from
the collimation axis (e.g. Yamazaki {\it et al.}\ 2002), (ii) bursts with high baryon contamination
within the jet or (iii) bursts with jets which are intrinsically broader than those in GRBs.
All of these models can efficiently soften the prompt emission from that seen
in "classical" GRBs. Although the prompt emission of bursts from these different
mechanisms may be largely indistinguishable, it is possible that their afterglows
will show significant variation in their appearance (e.g. Granot {\it et al.}\ 2005). We consider these
models, and how they may apply to the afterglow of XRF~050215B, below.
\begin{figure}[h]
\plotone{f4.ps}
\caption{The K-band lightcurve of XRF~050215B obtained at UKIRT and Gemini. Also marked
by a heavy line is the best fit power law decay $F(t) \propto t^{-0.47}$, which is very slow for GRB afterglows across this
time frame. Also shown by triangles and a thin line
is a putative host subtracted light curve (assuming that the host
has K=22.25. The thin line shows a (arbitrarily normalised) powerlaw with the same decay index
as that seen in the X-ray lightcurve of XRF 050215B. As can be seen, within the uncertainty due to host subtraction it is possible that the K band decay has the same index as that seen in X-rays. }
\label{klc}
\end{figure}
\begin{figure}[h]
\plotone{f5.ps}
\caption{A comparison of the afterglow of XRF~050215B with the K-band afterglows
of other GRBs (modified from that of Rau {\it et al.}\ 2004). The red circles are the
afterglow of XRF~050215B, the blue triangles are for {\it Swift} bursts and
the green boxes are for the high-redshift bursts.
This plot shows that XRF 050215B has the faintest
K-band afterglow yet seen. The data are taken
from Rau {\it et al.}\ (and references therein) except for the more
recent bursts which can be found in Berger {\it et al.}\ 2005;
GRB 050401 Watson {\it et al.}\ 2005; GRB 050904 Haislip {\it et al.}\ 2006; Tagliaferri {\it et al.}\ 2005
and GRB 050814 Jakobsson {\it et al.}\ 2006.}
\label{kband}
\end{figure}
In the off-axis model, the highest energy emission (i.e. $\gamma$-rays) is
confined within a narrow cone oriented slightly away from the observer and is not seen.
The observed isotropic equivalent energy for these bursts ($E_{iso}$)
is thus lower than for
bursts seen on axis and can reproduce the observed relationship between $E_p$ and
$E_{iso}$ first reported by Amati {\it et al.} (2002).
Granot {\it et al.} (2005) consider the afterglows
produced by such a model. The early afterglow of a uniform jet with sharp edges can
be seen to rise at early times as the more energetic material becomes visible; it
then reaches a peak at a time roughly corresponding to the jet break, and from that point,
follows a decay which is indistinguishable from an on-axis GRB (since all
of the ejecta are visible), thus the late time decay slope would be expected
to be rapid $\sim t^{-p}$. The afterglow of XRF~050215B does not show the rapid
rise at early times which may be expected of off-axis models. However the
rapid rise is a feature of jets with sharp edges. More realistic models which
have smooth edges (either intrinsically or due to the interaction of the
jet with the stellar envelope) have a flat (but not necessarily rising) early lightcurve.
Thus the early observations of XRF~050215B can be explained under this model.
However, the late time slope should be steep ($t^{-p}$) while a shallow slope is
observed to 10 days post burst. Thus indicating that the jet must be moderately wide,
although other GRBs (e.g. GRB 970508) have been shown to behave as a single
power-law out to large times (see e.g. Bloom, Kulkarni \& Frail 2003).
Under the dirty fireball model the XRF is produced by a jet which is viewed on-axis,
but in which the Lorentz factor is reduced due to baryon loading. In this model
the decay of the afterglow is much slower ($f_{\nu} \propto t^{-3(p-1)/4}$ or
$f_{\nu} \propto t^{1/2-3p/4}$ depending of the location of the cooling break (Rhoads 2003)). Thus,
for a typical $p=2$ burst the predicted decay slope would be $t^{-0.75}$, broadly consistent
with the observed temporal slopes. However, dirty fireballs can only create a low
peak energy if the prompt emission is created by external rather than internal shocks.
Recent observations favour a scenario in which the bursts themselves are caused
by internal shocks (Zhang et al. 2005). Furthermore the shape of the lightcurve, and comparison of
the XRT and BAT lightcurves place constraints. In the dirty fireball model,
the deceleration occurs when the outflow has collected 1/$\Gamma$ of
rest mass; for lower values of $\Gamma$ (i.e. XRFs) this deceleration time
is significantly longer (scaling as $\Gamma^{-8/3}$). The XRT observations
cannot be used to constrain the deceleration time since they begin $\sim$ 30 minutes
after the burst. However the shape of the BAT lightcurve implies that
deceleration must have occurred at early times, and thus that the $\Gamma$
cannot have been very low. Observations of future XRFs in X-rays in the seconds
to minutes after the burst will allow stronger constraints to be made.
Alternatively, under the internal shock model a clean fireball can naturally accommodate
the creation of an XRF (Zhang \& Meszaros 2002), especially if the contrast in
Lorentz factor between shells is small, which leads to
inefficient energy dissipation (e.g. Barraud {\it et al.}\ 2004). In this case,
the decay index of $\alpha = 0.82$ can also naturally be accommodated, although only
if it is observed prior to the jet-break.
A final possibility is that the XRF is caused by the opening angle of the jet itself.
Broad jets in GRBs may be present either alone (e.g. Lamb, Donaghy \& Graziani, 2005)
or as a multicomponent
structure including narrower ($\gamma$-ray emitting) jets (e.g. Berger et al. 2003;
Huang et al. 2004), although such models have recently been shown not to
fit the afterglow of GRB 030329, which they were posited to
explain (Granot 2005). These broader jets have correspondingly later jet break
times than the narrow emission responsible for the GRB, and would naturally
explain the lack of a jet break for $>$10 days after the burst. Indeed, regardless
of the precise mechanism for the generation of this XRF, the lack of jet
break implies that the jet opening angle must be wide, perhaps making
a broader jet the most likely explanation.
It is also interesting to compare the properties of both the prompt and
X-ray emission of XRF~050215B with those of the population of bursts
detected by {\it Swift}. Although the {\it HETE-2} observations allowed
for this burst to be classified as an XRF, the behaviour as observed by the
BAT was not significantly different from that seen in many other bursts.
Indeed the measured photon index of 2.0 is comparable to the mean of
{\it Swift} bursts in general (e.g. O'Brien {\it et al.}\ 2006),
although it does lie amongst the softer bursts
detected by BATSE (as do many {\it Swift} GRBs). Thus, it may be that
{\it Swift}, as for {\it HETE-2} and {\it BeppoSAX}, does detect a
large population of XRFs. However, Swift is unable to accurately constrain
their spectral properties due to its limited bandpass in comparison to
previous missions. Indeed in the case of XRF 050406 the apparent peak
energy $E_p < 15$keV (Romano et al. 2005) and thus although it was
possible to identify the burst as an XRF its peak energy and detailed
spectral prompt spectral properties could not be accurately measured. Only
XRF 050416 has a measured $E_p$ from {\it Swift}, with
$E_p =15.6$ keV (Sakamoto et al. 2005). However, even in this case
the measurement is relatively weak, since a simple power-law fit
yields a $\chi^2 / dof < 1$.
The afterglow of the burst
is both fainter than typical (even for {\it Swift} bursts) and also more slowly
declining, with no evidence for a jet break during long observations lasting
for 10 days since the burst. The afterglow is very faint in the
K-band (fainter than the majority of upper limits on K-band afterglow
brightness) and lies at the faint end of the flux distribution for
long duration GRB detected by {\it Swift}. There are a number
of comparably faint bursts e.g. GRB 050223 - Page {\it et al.}\ 2005;
GRB/XRF 050406 - Romano et al. 2005;
GRB 050421 - Godet {\it et al.}\ 2006, GRB 050911 - Page {\it et al.}\ 2006.
It is interesting to note that the proposed explanations for this
faintness vary widely, demonstrating the large range of physical
processes which affect the brightness of a given afterglow. For
example GRB 050911 was interpreted as being due to a BH-NS merger,
since its lack of an X-ray afterglow was reminiscent of short duration
GRBs (e.g. Gehrels et al. 2005), while its lightcurve could also
be interpreted as being due to several accretion event which are seen
in BH-NS mergers (Davies, Levan \& King 2005). In contrast
GRB 050421 was explained as being due to a "naked" GRB - a burst
occurring in a region of very low density, which thus affected
the brightness of the afterglow (e.g. Taylor et al. 2000).
For XRFs, such explanations are not
normally considered and the most commonly discussed explanations
involve differing geometries of either the jet itself or the observer
orientation. For example the flat slopes of the afterglow lightcurves
are predicted in various off-axis models. However, for "normal" long
duration bursts a generic feature of the afterglow light curve is
a flat phase in the afterglow (e.g. Nousek et al. 2005). This
is normally interpreted as being due to late time energy injection
from the central engine, and is also interpreted as such
in the afterglow of XRF 050406 (Romano et al. 2005), which
has a very flat light curve out to late times $\alpha \sim 0.5$.
While this flat phase is obviously a generic feature of
X-ray afterglows it is less apparent in the optical and IR. Thus
for XRF 050215B, where the X-ray and optical decays are
similar it is more likely that the relatively slow decay is simply
a result of the jet structure. It is also interesting to note that
other optical afterglows of XRFs have shown similar
behaviour to XRF 050406 in the optical regime. For example
XRF 020903 showed flaring behaviour and a flat decay
in the R-band (Bersier et al. 2006), behaviour which has
not commonly been seen in GRB afterglows (although the sample
of XRF optical/IR afterglows remains much smaller).
One of the most important contributions that {\it Swift} has made to the GRB field
comes from the decrease in time between the burst itself and the first pointed
observations. Although in the pre-{\it Swift} era a few bursts were observed
promptly in the optical (e.g. GRB 990123, Akerlof et al. 1999) these were rare,
and prompt, pointed, X-ray observations were never made. It was thus hoped
that the early afterglow emission would provide crucial tests of various
afterglow models. In practice, the behaviour seen is somewhat different
from what was expected (e.g. Tagliaferri et al. 2005; Nousek et al. 2005), and
the presence of frequent X-ray flares (e.g. Burrows et al. 2005) makes the
X-ray afterglows a less "clean" probe than may have been hoped for. Nonetheless,
this early phase can provide powerful diagnostics of various models.
O'Brien et al. (2006) have studied this transition period for a number of
GRBs and show that generally the extrapolation of the X-ray afterglow to
early times "joins" with the prompt emission and does not exhibit
a strong discontinuity. This behaviour, when seen in XRFs,
disfavours a sharp-edged jet if jet geometry is the dominating factor.
However, for three well studied {\it Swift} X-ray afterglows
(XRF 050215B, XRF 050406 and XRF 050416) there is apparently
no jet break seen out to late times ($>$ 10 days in each case). Thus, all
of these cases must have wide opening angles. Ultimately,
insight into the XRF phenomena is likely to be made via the subset
of bursts which are detected simultaneously by BAT and another
satellite (e.g. HETE-2, Suzaku-WAM, Konus) and where early
X-ray and optical observations can be paired with well constrained spectral
parameters for the prompt emission.
\section{Conclusions}
We have presented observations of XRF~050215B. This was the first X-ray Flash to be
located by {\it Swift} and was achieved by the use of {\it HETE-2} and {\it Swift} in
synergy, utilising the sensitivity and wide field of view of the {\it Swift-BAT} and
the wide spectral range of {\it HETE-FREGATE}. The fluence within the
prompt emission was comparable to that seen in previous XRFs, although the
X-ray lightcurve is fainter than the majority of {\it Swift} bursts, while in the
infrared XRF~050215B is the faintest afterglow ever to have been
detected (although some bursts have been invisible to comparable limits). Indeed,
XRF~050215B had a flux at 9 hours
which was only just observable with a 4m telescope, and at early times
was invisible to any robotic or UVOT observations.
As also
suggested by Berger {\it et al.}\ (2005), this implies that
locating the afterglows of some GRBs in the {\it Swift} era will require rapid
response observations from larger telescopes since many bursts
will be too faint to be detected by the UVOT or even by larger aperture robotic ground based
telescopes.
The limited volume of data and lack of redshift available for this burst make it difficult to draw firm
conclusions as to the cause of the soft $\gamma$-ray spectrum;
however the lack of any observed jet-break to $>$ 10 days post
burst make a broad jet the most likely explanation. The late-time behaviour of
XRF 050215B is qualitatively somewhat similar to the late-time X-ray lightcurve of
the recent GRB 060218, which, by virtue of its proximity (145 Mpc) had a
bright X-ray and Optical afterglow. In particular, GRB 060218 also had a long-lasting slow
decay, as seen in XRF~050215B, although the relative faintness of both afterglow and host galaxy,
coupled with the lack of any supernova emission, imply a much higher redshift in the latter case.
Furthermore GRB 060218 had an exceptionally long duration in $\gamma$-rays
($t_{90} \sim 2000$s), making the
status of GRB 060218 relative to the bulk of the XRF population unclear, and pointing
to the continuing need for further well-observed
XRFs if we are to understand the diversity in their emission mechanisms and their relationship
to GRBs.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,434 |
Q: How to connect my phone to eclipse I have a HUAWEI G750 with android 4.2.2
I want to connect my phone to eclipse but i can't.
My USB debugging is on and my USB cable has no problem.
Is there anyway to connect my phone to eclipse and run the projects on my phone instead of the emulator?
A: try to instal huawei drivers from this link
http://consumer.huawei.com/en/?method=index
and then Update Driver from device menager in control panel
A: Try to use the Universal USB Driver for Android devices. (f. e. http://adbdriver.com/)
Also make sure to connect your device as camera (PTP).
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,306 |
{"url":"https:\/\/www.zbmath.org\/?q=an%3A1221.68121","text":"# zbMATH \u2014 the first resource for mathematics\n\n$$\\Delta$$-clearing restarting automata and CFL. (English) Zbl\u00a01221.68121\nMauri, Giancarlo (ed.) et al., Developments in language theory. 15th international conference, DLT 2011, Milan, Italy, July 19\u201322, 2011. Proceedings. Berlin: Springer (ISBN 978-3-642-22320-4\/pbk). Lecture Notes in Computer Science 6795, 153-164 (2011).\nSummary: $$\\Delta$$-clearing restarting automata represent a new restricted model of restarting automata which, based on a limited context, can either delete a substring of the current content of its tape or replace a substring by a special auxiliary symbol $$\\Delta$$, which cannot be overwritten anymore, but it can be deleted later. The main result of this paper consists in proving that besides their limited operations, $$\\Delta$$-clearing restarting automata recognize all context-free languages.\nFor the entire collection see [Zbl 1218.68023].\n\n##### MSC:\n 68Q45 Formal languages and automata\nFull Text:","date":"2021-01-19 21:56:32","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.487039178609848, \"perplexity\": 2312.5825436917858}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-04\/segments\/1610703519784.35\/warc\/CC-MAIN-20210119201033-20210119231033-00093.warc.gz\"}"} | null | null |
Q: Overleaf, where to find missing { or output.gls? I am working on a document in LaTex with overleaf when I noticed this error:
<to be read again>
\let
l.21 ...entrycounter[]{page}\glsnumberformat{18}}}
%
A left brace was mandatory here, so I've put one in.
You might want to delete and/or insert some corrections
so that I will find a matching right brace soon.
(If you're confused by all this, try typing `I}' now.)
! Missing } inserted.
<inserted text>
}
l.21 ...entrycounter[]{page}\glsnumberformat{18}}}
%
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.
The problem is that it says the error is a missing { in a file called output.gls. From what I understand this is an auto generated file created by the glossary, which shouldn't simply miss a {. So the issue can't be with that file but must be with my code. I attempted to resolve this by commenting out sections of code I wrote since the last working version, and I did find out which line causes it, however I don't understand why:
\newglossaryentry{density}
{
name=\ensuremath{\kappa},
description={$ \in \mathbb{K}$. Density scalar field over $\Omega$. In practice we parameterize it in terms of \gls{density parameter}
}
}
\newglossaryentry{density parameter}
{
name=\ensuremath{\alpha},
description={$\in [0,1]^F$.% Parameter used to describe \gls{density}
}
}
The problem stems from In practice we parameterize it in terms of \gls{density parameter}, the error disappears if I comment out this sentence. I tried making a minimum working example but as a comment below pointed out the error was unrelated. The example was:
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[nonumberlist]{glossaries}
\usepackage{amsmath,amssymb}
\title{Test}
\date{September 2021}
\makeglossaries
\newglossaryentry{density}
{
name=\ensuremath{\kappa},
description={$ \in \mathbb{K}$. Density scalar field over $\Omega$. In practice we parameterize it in terms of \gls{density parameter}
}
}
\newglossaryentry{density parameter}
{
name=\ensuremath{\alpha},
description={$\in [0,1]^F$. Parameter used to describe \gls{density}
}
}
\begin{document}
\maketitle
\section{Minimal Example}
Calling \gls{density} is causing an error.
But only calling \gls{density parameter} does not.
Commenting out the description of the $\kappa$ after ``In practice...'' fixes the error as well.
\printglossaries
\end{document}
So while I could just comment out that part of the description it seems clear to me I am just misunderstanding how the glossary works and wanted to ask what mistake I made.
| {
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} | 8,493 |
// Copyright (c) 2012 Ecma International. All rights reserved.
// This code is governed by the BSD license found in the LICENSE file.
/*---
es5id: 15.2.3.13-2-25
description: >
Object.isExtensible returns true if O is extensible and has a
prototype that is not extensible
---*/
var proto = {};
Object.preventExtensions(proto);
var ConstructFun = function () { };
ConstructFun.prototype = proto;
var obj = new ConstructFun();
assert(Object.isExtensible(obj), 'Object.isExtensible(obj) !== true');
| {
"redpajama_set_name": "RedPajamaGithub"
} | 9,747 |
Now, you probably know that getting all these tools will cost you $1,00s if not $1,000s of dollars each month. The worst? Some of the third party plugins and tools can compromise your website. This alone should give you motivation to have BuilderAll right now.
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This is a really special feature because very few platforms out there lets complete beginners to build a mobile app by simply using easy drag and drop functions.
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If you are a complete novice you can begin with a simple test site and have access to all the tools that can help you grow it into a true authority in your chosen marketplace.
For online marketing veterans, you'll be able to select the ideal package to address the needs you now have while leaving space to expand while your empire grows. BuilderAll is perfect for entrepreneurs, service providers, digital professionals, designers, web agencies, coaches, networking professionals, and anyone else who likes making profit online. | {
"redpajama_set_name": "RedPajamaC4"
} | 3,325 |
Q: How to see the RSOP for an OU? Anyone knows how to see RSOP for an OU? If I just launch RSOP I see the policies for the domain but I want to know what's going on with one of the OUs that's misbehaving. And I can't find a way to see the policies on this particular OU. Any ideas?
A: I found the answer myself: http://support.microsoft.com/kb/323276 search for "How to Run an RSoP Query on a Computer Account." :)
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,092 |
WWE: Rikishi's The Next Inductee of the WWE Hall of Fame Class of 2015
Posted on February 10, 2015 by sparxteam in Film/TV, Home, Sports, Videos // 0 Comments
With WrestleMania 31 less than two months away, WWE is slowly revealing the Hall of Fame class of 2015, which includes Macho Man Randy Savage and Arnold Schwarzenegger.
Tag team champions The Usos broke the news on Twitter Monday that their father, Solofa Fatu Jr., who wrestled in the WWE as a member of The Headshrinkers and later as Rikishi, will be inducted in March.
We heard it backstage first..tonight on #Raw, pops is the newest inductee into the @WWE Hall of Fame! #USOOOOOO @TheRealRikishi #WWEHOF
— The Usos (@WWEUsos) February 9, 2015
To many fans, Rikishi may be a surprise pick as a one-time Intercontinental Champion and three time tag team champion — but Rikishi had an incredibly long career at the top level of professional wrestling, and was one of the most popular acts in the business during his "Too Cool" days. Rikishi did wrestle for the WWE title, but never won the belt.
Rikishi will be the sixth member of the Anoa'i family to enter the Hall of Fame, joining The Wild Samoans Afa and Sika, Yokozuna, Peter Maivia and Rocky Johnson.
"Macho Man" Randy Savage
Peter Maivia
Rocky Johnson
Solofa Fatu Jr.
The Anoa'i Family
The Headshrinkers
The Wild Samoans
Too Cool
Nicki Minaj vs. Taylor Swift
by sparxteam in Home
It's only Thursday, and already two stars have released jaw-dropping music videos: Taylor Swift and Nicki Minaj. Late Tuesday night, Aug. 20, Nicki Minaj brought the [Read more...] | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 850 |
Mansveld ist der Nachname mehrerer Personen:
Aad Mansveld (1944–1991), niederländischer Fußballnationalspieler
Alfred Mansveld (1911–?), belgischer olympischer Silbermedaillengewinner im Viererbob
Debby Mansveld (* 1972), niederländische Radrennfahrerin
Regnerus van Mansveld (1639–1671), niederländischer Philosoph
Wilma Mansveld (* 1962), niederländische Politikerin (PvdA) | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,417 |
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