\documentclass[letterpaper,9pt,twoside,printwatermark=false]{pinp}

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\usepackage{longtable}

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\title{Exercise 1 - How Deep is the Ocean? September 17, 2018.}

\author[a]{EPIB607 - Inferential Statistics}

  \affil[a]{Fall 2018, McGill University}

\setcounter{secnumdepth}{5}

% Please give the surname of the lead author for the running footer
\leadauthor{Bhatnagar and Hanley}

% Keywords are not mandatory, but authors are strongly encouraged to provide them. If provided, please include two to five keywords, separated by the pipe symbol, e.g:
 \keywords{  Sampling distribution |  Means |  Proportions |  Standard error |  Standard deviation |  Parameter |  Statistic  }

\begin{abstract}
This in-class exercise will introduce you to sampling distributions for
means and proportions.
\end{abstract}

\dates{This version was compiled on \today}
\doi{\url{https://sahirbhatnagar.com/EPIB607/}}

\pinpfootercontents{Exercise 1: Sampling Distributions for Means and Proportions}

\begin{document}

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\maketitle
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\ifthenelse{\boolean{shortarticle}}{\ifthenelse{\boolean{singlecolumn}}{\abscontentformatted}{\abscontent}}{}

% If your first paragraph (i.e. with the \dropcap) contains a list environment (quote, quotation, theorem, definition, enumerate, itemize...), the line after the list may have some extra indentation. If this is the case, add \parshape=0 to the end of the list environment.

\section{What percentage of the world's surface is covered by
water?}\label{what-percentage-of-the-worlds-surface-is-covered-by-water}

The data provided by the
\href{https://topex.ucsd.edu/cgi-bin/get_srtm30.cgi}{Scripps Institution
of Oceanography} can provide an answer, but some work is required on
your part. James Hanley (JH) has randomly sampled \(n=5\) and \(n=20\)
latitudes and longitudes for every student in the class. This document
contains unique latitudes and longitudes for

\begin{ShadedResult}
\begin{verbatim}
[1] "Bhatnagar, Sahir"
\end{verbatim}
\end{ShadedResult}

and was sent in an email (using the
\href{https://cran.r-project.org/package=gmailr}{gmailr} package) as a
pdf attachment to the following address:

\begin{ShadedResult}
\begin{verbatim}
[1] "sahir.bhatnagar@mcgill.ca"
\end{verbatim}
\end{ShadedResult}

\subsection*{A sample of 5}\label{a-sample-of-5}
\addcontentsline{toc}{subsection}{A sample of 5}

\begin{ShadedResult}
\begin{verbatim}
Latitude.n.5 <- c(8.042,45.329,37.743,-12.237,21.552)
\end{verbatim}
\end{ShadedResult}\begin{ShadedResult}
\begin{verbatim}
Longitude.n.5 <- c(35.241,125.735,-134.383,-23.623,-5.339)
\end{verbatim}
\end{ShadedResult}

\subsection*{A sample of 20}\label{a-sample-of-20}
\addcontentsline{toc}{subsection}{A sample of 20}

\begin{ShadedResult}
\begin{verbatim}
Latitude.n.20 <- c(40.428,59.851,-38.35,-8.862,-26.312,-49.721,-4.02,44.028,-36.308,22.071,
44.765,-18.818,37.185,41.768,-5.255,14.199,61.165,10.874,-27.083,15.792)
\end{verbatim}
\end{ShadedResult}\begin{ShadedResult}
\begin{verbatim}
Longitude.n.20 <- c(57.814,40.026,131.012,-145.304,-130.685,48.011,59.257,111.303,41.137,-61.05,
159.666,-110.195,-87.961,-118.8,99.592,-102.281,159.827,-166.667,-2.171,-6.664)
\end{verbatim}
\end{ShadedResult}

\subsection{Determine the proportion of water from the sample of
5}\label{determine-the-proportion-of-water-from-the-sample-of-5}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Using the sample of 5, manually enter the latitudes and longitudes in
  \href{https://www.google.com/maps}{Google maps} and record if you land
  on water or land. Figure\ref{maps} shows how to enter them.
\end{enumerate}

\begin{figure}[H]
  \begin{center}
    \includegraphics[scale=0.40]{lat-long.png}
  \end{center}
  \caption{Latitude and longitude in Google maps. Latitudes range from (-90,90) and longitudes range from (-180, 180). Latitude is entered first followed by longitude and separated by a comma.}\label{maps}
\end{figure}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{1}
\tightlist
\item
  In \texttt{R}, store your results in a binary vector of length 5. For
  example, if your sample landed on water 3 times out of 5, then you
  would enter the following in \texttt{R} (1 for water and 0 for land):
\end{enumerate}

\begin{ShadedResult}
\begin{verbatim}
landed_in_water <- c(1,1,1,0,0)
\end{verbatim}
\end{ShadedResult}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{2}
\item
  Using \texttt{R}, estimate the percentage of the world's surface
  covered by water from your sample of 5. This can be done by simply
  taking the mean of the binary vector created in Step 2:
  \texttt{mean(landed\_in\_water)}.
\item
  Enter your estimate in this
  \href{https://docs.google.com/spreadsheets/d/1Mnxeq9nQcTdQycZ7S_62fYFiNC5_a3fibsyodzfwO58/edit?usp=sharing}{Google
  sheet} next to your name and in the column titled
  \texttt{PropnW.5.locations}.
\end{enumerate}

\subsection{Determine the proportion of water from the sample of
20}\label{determine-the-proportion-of-water-from-the-sample-of-20}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Repeat the above steps for the sample of 20. \textbf{Before you do},
  take a moment to think about how tedious a process it can be to
  manually enter 20 latitudes and longitudes into Google maps. JH and SB
  hope you can appreciate the parallels between this toy exercise and
  that of collecting data for your research projects, i.e., it becomes
  increasingly difficult (effort and money!) to collect more and more
  samples.
\end{enumerate}

Now that you appreciate the amount of work it takes to estimate the
proportion of water from 20 samples, JH and SB think it is sufficient
for you to use some automatic procedures to complete this task, which
are further described in step 2.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{1}
\tightlist
\item
  Create an \texttt{R} script and copy the following index vector into
  it:
\end{enumerate}

\begin{ShadedResult}
\begin{verbatim}
index.n.20 <- c(2131,2132,2133,2134,2135,2136,2137,2138,2139,2140,
2141,2142,2143,2144,2145,2146,2147,2148,2149,2150)
\end{verbatim}
\end{ShadedResult}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{2}
\tightlist
\item
  Load a function into your \texttt{R} session that JH and SB created to
  automate the process using the following command:
\end{enumerate}

\begin{Shaded}
\begin{Highlighting}[]
\KeywordTok{source}\NormalTok{(}\StringTok{"https://github.com/sahirbhatnagar/EPIB607/raw/master/exercises/water/automate_water_task.R"}\NormalTok{)}
\end{Highlighting}
\end{Shaded}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{3}
\tightlist
\item
  Now that the function \texttt{automate\_water\_task} has been loaded
  into your environment, you can use it to automatically determine which
  of the locations in your sample of 20 are on water. This function
  requires the unique index vector shown in Step 2 above as input and
  returns a binary vector of length 20 (1 for water, 0 for land). You
  can call the function as follows in \texttt{R}:
\end{enumerate}

\begin{Shaded}
\begin{Highlighting}[]
\NormalTok{landed_in_water <-}\StringTok{ }\KeywordTok{automate_water_task}\NormalTok{(}\DataTypeTok{index =}\NormalTok{ index.n}\FloatTok{.20}\NormalTok{)}
\end{Highlighting}
\end{Shaded}

As before, enter your estimate of the proportion of the earth's surface
covered by water in this
\href{https://docs.google.com/spreadsheets/d/1Mnxeq9nQcTdQycZ7S_62fYFiNC5_a3fibsyodzfwO58/edit?usp=sharing}{Google
sheet} next to your name, but in the column titled
\texttt{PropnW.20.locations}.

\section{What is the average depth of the
ocean?}\label{what-is-the-average-depth-of-the-ocean}

We will now turn to estimating the average depth of the ocean. You will
again make use of the \texttt{automate\_water\_task} function.

\subsection{Determine the average depth of the ocean from the sample of
5 and
20}\label{determine-the-average-depth-of-the-ocean-from-the-sample-of-5-and-20}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  In the same \texttt{R} script as before, copy the following index
  vector into it:
\end{enumerate}

\begin{ShadedResult}
\begin{verbatim}
index.n.5 <- c(1,2,3,4,5)
\end{verbatim}
\end{ShadedResult}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{1}
\tightlist
\item
  Use the \texttt{automate\_water\_task} function to get a sample of 5
  depths. Note: some of the returned samples will not correspond to the
  same latitudes and longitudes provided to you earlier. This is because
  we need to restict our sample to locations on water only in order to
  estimate the mean depth of the ocean. Here we show some example code
  and its output. You need to specify the \texttt{type} and
  \texttt{student\_id} argument:
\end{enumerate}

\begin{Shaded}
\begin{Highlighting}[]
\CommentTok{# be sure to provide your own student id}
\NormalTok{depths.n}\FloatTok{.5}\NormalTok{ <-}\StringTok{ }\KeywordTok{automate_water_task}\NormalTok{(}\DataTypeTok{index =}\NormalTok{ index.n}\FloatTok{.5}\NormalTok{, }\DataTypeTok{student_id =} \DecValTok{222333444}\NormalTok{, }\DataTypeTok{type =} \StringTok{"depth"}\NormalTok{)}
\end{Highlighting}
\end{Shaded}

\newpage

The \texttt{alt} column gives the depth in meters:

\begin{longtable}[]{@{}lrrrrr@{}}
\toprule
& X & lon & lat & alt & water\tabularnewline
\midrule
\endhead
3 & 3 & -134.38257 & 37.742717 & 5028 & 1\tabularnewline
4 & 4 & -23.62332 & -12.237161 & 5358 & 1\tabularnewline
17391 & 17391 & -124.01387 & 11.596521 & 4577 & 1\tabularnewline
46573 & 46573 & -167.19004 & -9.532013 & 3850 & 1\tabularnewline
23312 & 23312 & 83.01109 & 7.885076 & 3895 & 1\tabularnewline
\bottomrule
\end{longtable}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{2}
\item
  Repeat step 2 for a sample of 20 using the \texttt{index.n.20} vector
  specified above.
\item
  Calculate an estimate of the mean depth of the ocean from your samples
  of 5 and 20 using the \texttt{mean} function, e.g.,
  \texttt{mean(depths.n.5\$alt)}.
\item
  Enter your estimates of the mean depth of the ocean from your samples
  of 5 and 20 in this
  \href{https://docs.google.com/spreadsheets/d/1Mnxeq9nQcTdQycZ7S_62fYFiNC5_a3fibsyodzfwO58/edit?usp=sharing}{Google
  sheet} next to your name, in the columns titled \texttt{Mean.5.depths}
  and \texttt{Mean.20.depths}, respectively.
\end{enumerate}

\section{Plot the sampling distribution of the proportion and
mean}\label{plot-the-sampling-distribution-of-the-proportion-and-mean}

It is now time to plot the sampling distributions of the proportions and
means. Once everyone has filled in the
\href{https://docs.google.com/spreadsheets/d/1Mnxeq9nQcTdQycZ7S_62fYFiNC5_a3fibsyodzfwO58/edit?usp=sharing}{Google
sheet}, export the sheet as a \texttt{.csv} file by clicking on
\texttt{File\ -\/-\textgreater{}\ Download\ as\ -\/-\textgreater{}\ Comma-separated\ values\ (.csv,\ current\ sheet)}.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Read in the data:
\end{enumerate}

\begin{Shaded}
\begin{Highlighting}[]
\CommentTok{# read in the results from the Google sheet }
\NormalTok{water_results <-}\StringTok{ }\KeywordTok{read.csv}\NormalTok{(}\StringTok{"EPIB607_FALL2018_water_exercise - water.csv"}\NormalTok{, }\DataTypeTok{as.is=}\OtherTok{TRUE}\NormalTok{)}
\NormalTok{water_results <-}\StringTok{ }\NormalTok{water_results[,}\DecValTok{1}\OperatorTok{:}\DecValTok{6}\NormalTok{]}
\NormalTok{water_results <-}\StringTok{ }\NormalTok{water_results[}\KeywordTok{complete.cases}\NormalTok{(water_results), ]}

\CommentTok{# count the number of students who provided a mean and proportion}
\NormalTok{N.r <-}\StringTok{ }\KeywordTok{nrow}\NormalTok{(water_results)}
\end{Highlighting}
\end{Shaded}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{1}
\tightlist
\item
  Plot the students' estimates of the proportion covered by water for
  samples of size 5. You may use the following code or run your own:
\end{enumerate}

\begin{Shaded}
\begin{Highlighting}[]
\KeywordTok{plot}\NormalTok{(}\KeywordTok{table}\NormalTok{(water_results[,}\StringTok{"PropnW.5.locations"}\NormalTok{]), }
     \DataTypeTok{xlim =} \KeywordTok{c}\NormalTok{(}\DecValTok{0}\NormalTok{,}\DecValTok{1}\NormalTok{),}
     \DataTypeTok{xlab =} \StringTok{"Students' Estimates of Proportion Covered by Water"}\NormalTok{,}
     \DataTypeTok{main =} \StringTok{"n = 5"}\NormalTok{, }
     \DataTypeTok{ylim =} \KeywordTok{c}\NormalTok{(}\DecValTok{0}\NormalTok{, N.r}\OperatorTok{/}\FloatTok{1.5}\NormalTok{), }
     \DataTypeTok{ylab =} \StringTok{"Frequency"}\NormalTok{)}
\end{Highlighting}
\end{Shaded}

\begin{enumerate}
\def\labelenumi{\alph{enumi})}
\tightlist
\item
  Comment on this graph. Does this shape look sensible to you?
\end{enumerate}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{2}
\tightlist
\item
  Now plot the students' estimates of the mean depth of the ocean for
  samples of size 5. You may use the following code or run your own:
\end{enumerate}

\begin{Shaded}
\begin{Highlighting}[]
\NormalTok{d.BREAKS <-}\StringTok{ }\KeywordTok{seq}\NormalTok{(}\DecValTok{1000}\NormalTok{,}\DecValTok{6000}\NormalTok{,}\DecValTok{500}\NormalTok{)}
\KeywordTok{hist}\NormalTok{(water_results[,}\StringTok{"Mean.5.depths"}\NormalTok{], }
     \DataTypeTok{xlim =} \KeywordTok{c}\NormalTok{(}\DecValTok{0}\NormalTok{,}\DecValTok{6000}\NormalTok{),}
     \DataTypeTok{ylim =} \KeywordTok{c}\NormalTok{(}\DecValTok{0}\NormalTok{, N.r}\OperatorTok{/}\FloatTok{1.5}\NormalTok{), }
     \DataTypeTok{breaks =}\NormalTok{ d.BREAKS,}
     \DataTypeTok{xlab =} \StringTok{"Students' Estimates of Mean Ocean Depth (m)"}\NormalTok{,}
     \DataTypeTok{main =} \StringTok{"n = 5"}\NormalTok{)}
\end{Highlighting}
\end{Shaded}

\begin{enumerate}
\def\labelenumi{\alph{enumi})}
\tightlist
\item
  Calculate the mean and the standard error of the mean depth for
  samples of size 5
\item
  Comment on this graph (e.g.~range, variability)
\end{enumerate}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{3}
\tightlist
\item
  Repeat Steps 2 and 3 for samples of size 20.
\end{enumerate}

\begin{enumerate}
\def\labelenumi{\alph{enumi})}
\tightlist
\item
  Compare the two graphs for proportions, and the two graphs for means.
  What do you notice? You might find it helpful to overlay the
  distributions on the sample plot. You may use the following code or
  run your own:
\end{enumerate}

\begin{Shaded}
\begin{Highlighting}[]
\KeywordTok{library}\NormalTok{(mosaic)}
\KeywordTok{library}\NormalTok{(tidyr)}

\CommentTok{# first 'melt' the data to get it in plotting form}
\NormalTok{m.melt <-}\StringTok{ }\NormalTok{water_results }\OperatorTok{%>%}\StringTok{ }\NormalTok{tidyr}\OperatorTok{::}\KeywordTok{gather}\NormalTok{(}\DataTypeTok{key =} \StringTok{"type"}\NormalTok{, }\DataTypeTok{value =} \StringTok{"value"}\NormalTok{, }\OperatorTok{-}\NormalTok{X., }\OperatorTok{-}\NormalTok{student)}

\CommentTok{# subset for means}
\NormalTok{m.melt.means <-}\StringTok{ }\KeywordTok{subset}\NormalTok{(m.melt, type }\OperatorTok{%in%}\StringTok{ }\KeywordTok{c}\NormalTok{(}\StringTok{"Mean.20.depths"}\NormalTok{,}\StringTok{"Mean.5.depths"}\NormalTok{))}

\CommentTok{# plot for means}
\KeywordTok{gf_density}\NormalTok{(}\OperatorTok{~}\StringTok{ }\NormalTok{value, }\DataTypeTok{data =}\NormalTok{ m.melt.means, }\DataTypeTok{fill =} \OperatorTok{~}\StringTok{ }\NormalTok{type) }\OperatorTok{+}\StringTok{ }\KeywordTok{theme_bw}\NormalTok{()}

\CommentTok{# subset for proportions}
\NormalTok{m.melt.props <-}\StringTok{ }\KeywordTok{subset}\NormalTok{(m.melt, type }\OperatorTok{%in%}\StringTok{ }\KeywordTok{c}\NormalTok{(}\StringTok{"PropnW.20.locations"}\NormalTok{,}\StringTok{"PropnW.5.locations"}\NormalTok{))}

\CommentTok{# plot for proportions}
\KeywordTok{gf_histogram}\NormalTok{(}\OperatorTok{~}\StringTok{ }\NormalTok{value, }\DataTypeTok{data =}\NormalTok{ m.melt.props, }\DataTypeTok{fill =} \OperatorTok{~}\StringTok{ }\NormalTok{type, }\DataTypeTok{position =} \StringTok{"dodge"}\NormalTok{) }\OperatorTok{+}\StringTok{ }\KeywordTok{theme_bw}\NormalTok{()}
\end{Highlighting}
\end{Shaded}

%\showmatmethods

\bibliography{pinp}
\bibliographystyle{jss}

\end{document}


\chapter{Compute the energy loss}

 The energy loss processes are very similar for \(e+/e-\) , \(\mu+/\mu-\) and
charged hadrons , so it is quite natural to have a common description for them
and we have chosen this way.

\section{Energy loss of electrons/positrons}

 The G4eEnergyLoss class computes the continuous energy loss \linebreak
  of electrons/positrons.
 The continuous energy loss is calculated as
a sum of the contribution of the different processes.
 At present there are two processes
contributing to the continuous energy loss , they are:

            the ionisation process (class G4eIonisation ) and
            the bremsstrahlung process (class G4eBremsstrahlung).

 The class G4eIonisation calculates the contribution due to ionisation and
it simulates the 'discrete' part of the ionisation , the Moller and Bhabha
scattering or delta ray production by e+/e-.

 The class G4eBremsstrahlung computes the energy loss contribution due to
soft bremsstrahlung and it simulates the 'discrete' or hard bremsstrahlung.

 The formulae used to compute the continuous energy loss are the same than
those in GEANT3, but the two programs are not the same.
 The main difference between GEANT3 and GEANT4 is that while in GEANT3 the
continuous energy loss, delta ray production and bremsstrahlung are three
independent processes , in GEANT4 these processes are closely connected,
they form a unified scheme.

 The G4eEnergyLoss class constructs dE/dx and range tables for every material.
 First the dE/dx tables are costructed and filled , the energy loss class
  simply summes up of the contributions computed by the ionisation and
 bremsstrahlung processes. After this step G4eEnergyLoss creates range tables
 of \(e+/e-\) for every material and computes the inverse of these range
 tables as well. All the tables are constructed
 at the beginning of the GEANT4 run , at initialisation time. Later, during
the simulation the energy loss process performs two tasks : it imposes a
limit on the stepsize of the particle and computes the energy loss after
a step travelled by the particle.

 \subsection{Step limit due to the continuous energy loss}

 The continuous energy loss imposes a limit on the stepsize.
 The reason of this is the
energy dependence of the cross sections. It is generally assumed in MC programs that
the cross sections of the particles are approximately constant during a step , i.e.
 the step size should be so small that the relative difference of the cross sections
at the beginning of the step and at the end should be small enough.In principle one
has to use very small steps in order to have an accurate simulation , but the computing
time increases if the stepsize decreases. As a good compromise the stepsize is
limited here by the requirement that the stopping range of the particle can
decrease by not more than 20 \% during the step. This condition works fine for
a particle of kinetic energy \(> 0.5 MeV - 1. MeV\) , but for low energy it
gives very short step sizes.
 To cure this problem a lower limitation on the stepsize is also introduced.
The choice of this lower limit is quite natural , the stepsize can not be
smaller than the {\em cut in range} parameter of the program. The stepsize
limit varies smoothly with decreasing energy from the value given by the
condition \(\Delta range/range=0.20\) to the lowest possible value
 {\em cut in range} .
It is worthwile to mention , that the lower limit used here is much more
natural and physical , than the one is used in GEANT3 at the automatic
calculation of the tracking parameters.

 \subsection{Computation of the energy loss}

 The computation of the {\em mean energy loss} after a given step is done
by using the range and inverse range tables. The mean loss \(\Delta T\)
 can be written as

  \begin{equation}
       \Delta T = T_0 - f_T(r_0-step)
  \end{equation}

 where \(T_0\) is the kinetic energy,
       \(r_0\)    the range           at the beginning of the step \(step\),
        the function \(f_T(r)\) is the inverse of the range table (i.e. it
        gives the kinetic energy of the particle for a range value of r) .

 After the mean energy loss has been calculated the process computes the
 {\em actual} energy loss , i.e. the loss with fluctuation. The fluctuation
 is computed from a fluctuation model , from the model {\em GLANDZ} of
 the GEANT3 code \cite{eloss.geant3} , \cite{eloss.kati} .

 \section{Energy loss of muons}

  The energy loss of muons is computed by the class G4MuEnergyLoss. The
 scheme of the computation is the same as in the case of \(e+/e-\),
 the only difference is that now there are three processes contributing
 to the total continuous energy loss , these are

   the ionisation process (class G4MuIonisation),
   the bremsstrahlung process (class G4MuBremsstrahlung) and
   the direct pair production of muons (class G4MuPairProduction).

  The G4MuIonisation class computes the contribution to the continuous energy
 loss due to ionisation and simulates the corresponding 'discrete' process,
 the knock-on electron or \(\delta\)-ray production by muons.

  The G4MuBremsstrahlung class calculates the continuous loss due to the
 {\em soft} bremsstrahlung and simulates the 'discrete', {\em hard}
 bremsstrahlung.

  The G4MuPairProduction class gives the contribution to the continuous
 energy loss due to {\em sub-cutoff} \(e+/e-\) pairs and performs the
 simulation of the pair production.

 \section{Energy loss of charged hadrons}

 The continuous energy loss of the charged hadrons are calculated by the class
 G4hEnergyLoss . The algorithm is the same than in the case of \(e+/e-\)
 energy loss. Here there is only one process , the ionisation (class G4hIonisation)
 which  contributes to the
 continuous energy loss .

 The G4hIonisation class computes the continuous energy loss and simulates
 the \(\delta\)-ray producion by hadrons.

  In the case of energy loss of the hadrons two dE/dx , range and inverse range
 tables are constructed only , these are the tables for {\em proton} and
 {\em antiproton} . The energy loss for othe charged hadrons are computed
 from these tables at the  {\em scaled kinetic energy} \(T_{scaled}\)

   \begin{equation}
     T_{scaled} = \frac{ M_{proton} T}{ M_{particle}}
   \end{equation}

  where \(T\) is the kinetic energy of the particle, $M_{proton}$ and
  $M_{particle}$ are the masses of the proton and particle, respectively.

  There is an important constraint for this process:
  {\em the cut in range must be the same for all the charged hadrons} .
  This condition is meaningful physically and it makes possible to use
  the proton/antiproton tables only when computing the energy loss.
\section{Status of this document}
 9.10.98  created by L. Urb\'an.

\begin{thebibliography}{99}

\bibitem[GEANT3]{eloss.geant3}
  GEANT3 manual ,CERN Program Library Long Writeup W5013 (October 1994).
\bibitem[La95]{eloss.kati}
  K. Lassila-Perini and L. Urban , NIM A362 (1995),416.

\end{thebibliography}


\documentclass[a4paper]{article}

\input{temp}

\setcounter{section}{-1}

\begin{document}

\title{Introduction to Approximate Groups}

\maketitle

\newpage

\tableofcontents

\newpage

\section{Introduction}

---Lecture 1---

Example classes: 12 Feb, 5 Mar, 2-3pm.\\
Venue will be confirmed later.\\
Examinable material is exactly what is on board (as usual).\\
No plan for printed notes (as usual).

After 3.5 years of tripos, we finally know what a subgroup is: a \emph{subgroup} $H<G$ is a non-empty set closed under products and inverses.\\
We would then expect an \emph{approximate subgroup} to be a subset that is only \emph{approximately} closed under products. We'll make this precise soon. Such sets arise naturally in a number of branches of maths, and as such approximate groups have had broad range of applications. In this course we'll look in detail, for example, at applications to polynomial growth (fundamental in geometric group theory), and touch on construction of expander(?) graphs (important in theoretical CS).\footnote{Lecturer actually spent the time to write all this on board, probably implying a slower-paced course than category theory! But every course is slower than category theory.}

\newpage

\section{Small Doubling}
To start with, we'll look at a preliminary notion of approximate closure called \emph{small doubling}.\\
In this course, $G$ is always a group, arbitrary unless specified otherwise. Given $A,B \subset G$, we write
$$AB=\{ab:a \in A, b \in B\}$$
called the product set,
$$A^n = \underbrace{A \cdot ... \cdot A}_{n\text{ times}}$$
$$A^{-1} = \{a^{-1}:a \in A\}$$
$$A^{-r} = (A^{-1})^n$$

When $G$ is abelian, we often switch to additive notion, for example $A+B, nA, -A,-nA$ in place of the above (\emph{sum sets}).\\
To say $A$ is \emph{closed} is to say $A^2 = A$.\\
If $A$ is finite, one way to say that $A$ is \emph{approximately closed} is to say that $|A^2|$ is \emph{not much bigger} than $|A|$. This is the notion of approximate closure that arises when studying polynomial growth or expansion, for example.

To get a feel for what this should mean, let's look at the possible values of $|A^2|$. Trivially we have $|A| \leq |A^2| \leq |A|^2$, and both bounds are attained. However, although the quadratic upper bound on $|A^2|$ in terms of $|A|$ is extremal in a strict sense, it should not be seen as atypical for the size of $A^2$.\footnote{We will see in sheet 1 that random $A$ have quadratic size for $|A^2|$, in the sense of: if $A$ is a set of size $n$ chosen uniformly at random from $\{1,...,n^{100}\}$ (we'd like to choose from all integers, but there's no uniform measure there), then $\E(|A+A|)$ is \emph{closed to} $\frac{1}{2} |A|^2$. Note that this is the largest we can get, since in an abelian group we have $a+b=b+a$.} Therefore, we can view sets $A$ satisfying
$$|A^2| = o(|A|^2)\ \ \ (1.1)$$
as being \emph{exceptional}, and so condition (1.1) can already be seen as a form of \emph{approximate closure}. In this course, we will concentrate on the strongest form of (1.1), where $|A^2|$ is \emph{linear} in $|A|$, in the sense that
$$|A^2| \leq K|A|\ \ \ (1.2)$$
for some constant $K \geq 1$ fixed a priori.\\
Obviously, such sets are very far from random, and we can expect (1.2) to impose a significant restriction on $A$. The main aim of the course is to work out how significant the restriction is.

\begin{defi} (Small doubling)\\
Given $A \subset G$, the ratio $|A^2| / |A|$ is called the \emph{doubling constant}.\\
If $A$ satisfies (1.2), we'll say that $A$ has \emph{doubling} at most $K$, or simply \emph{small doubling}.\\
\end{defi}

\begin{eg}
$\bullet$ $A$ a finite subgroup;\\
$\bullet$ $|A| \leq K$;\\
$\bullet$ $A \subset \Z, A = \{-n,...,n\}$, $|A+A|\leq 2|A|$.\\

This last example is especially important as it shows that the theory does not just reduce to subgroups and small sets. We'll develop these examples later in the course.
\end{eg}

Our main aim will be to prove theorems along the lines of: if $A$ has small doubling, then $A$ has a certain structure. When $K$ is very small, this is quite easy, as follows.

\begin{thm} (1.1, Freiman\footnote{The proof presented here is by Tao instead of the original version. Lecturer has no way to tell if the original proof is close to this, especially because the original proof is in Russian.})\\
Let $K< \frac{3}{2}$. Suppose $A \subset G$ and $|A^2| \leq K|A|$ (by writing like this, we're obviously assuming $A$ is finite). Then there is a subgroup $H<G$ with $|H| = |A^2|$ ($\leq K|A|$) such that $A \subset aH = Ha \forall a \in A$.\\
(i.e., $A$ is a large portion of a coset of a finite subgroup).
\end{thm}

\begin{rem}
Converse: if $A \subset xH = Hx$ for $x \in G$, $H<G$, $|H| \leq K|A|$, then $|H^2| \leq K|A|$. So this is a complete classsification of sets of very small doubling.
\end{rem}

\begin{lemma} (1.2, identify $H$)\\
If $|A^2| < \frac{3}{2}|A|$, then $H=A^{-1}A$ is a subgroup. Moreover, $A^{-1}A = AA^{-1}$, and $|H| < 2|A|$.
\begin{proof}
Let $a,b \in A$. The hypothesis gives $|aA \cap bA| > \frac{1}{2} |A|$, so there are $> \frac{1}{2}|A|$ pairs of $(x,y) \in A \times A$ s.t. $ax = by$, i.e. $a^{-1}b=xy^{-1}$. This immediately implies $A^{-1}A \subset AA^{-1}$ (as we just needed one such pair $(x,y)$), and replacing $A$ by $A^{-1}$ we get the other inclusion; so $A^{-1}A = AA^{-1}$ as required.

Since $|A \times A| = |A|^2$, it also implies that $|A^{-1}A| \leq \frac{|A|^2}{\frac{1}{2}|A|} = 2|A|$, as claimed. Note also that $A^{-1}A$ is symmetric, so it remains to show that $A^{-1}A$ is closed under products.\\
Let $c,d \in A$. As above, there exist $>\frac{1}{2}|A|$ pairs $(u,v) \in A \times A$ s.t. $c^{-1} d = uv^{-1}$. This means for at least one pair $(x,y)$ and one pair $(u,v)$, we have $y=u$. In particular, $a^{-1}bc^{-1}d = xv^{-1} \in AA^{-1} = A^{-1}A$.
\end{proof}
\end{lemma}

\begin{lemma} (1.3, size bound)\\
If $|A^2| < \frac{3}{2} |A|$, then $A^2 = aHa$ $\forall a \in A$ ($H$ as before). In particlar, $|H| = |A^2|$.
\begin{proof}
First, note that $A \subset aH \cap Ha$ (label this 1.3) by definition of $H$ and $A^{-1}A = AA^{-1}$, so certainly $A^2 \subset aHa$. For the reverse inclusion, let $z \in aHa$. Since $H$ is a subgroup, there exists $|H|$ pairs $(x,y) \in aH \times Ha$ s.t. $z=xy$. Moreover, by (1.3) and the bound $|H| < 2|A|$ from lemma (1.2), more than half of those $x$ and half of those $y$ belong to $A$. In particular, this means that for at least one pair $(x,y)$, both have to belong to $A$. Hence $z=xy \in A^2$ as required.
\end{proof}
\end{lemma}

Proof of theorem (1.1):
\begin{proof}
Given $a \in A$, we have $Aa^{-1} \subset aHa^{-1} \cap H$, so $|aHa^{-1} \cap H| \geq |A| > \frac{1}{2}|H|$ by lemma (1.2). But the only subgroup of $H$ of size $>\frac{1}{2}|H|$ is $H$ itself. Hence $aHa^{-1} = H$, so indeed $A \subset aH = Ha$ by (1.3).
\end{proof}

Classifying the sets of small doubling is much harder than this in general, and uses a much wider range of techniques e.g. group theory, harmonic analysis, geometry of numbers, etc.

\newpage

\section{Covering and Higher Sum and Product Sets}

---Lecture 2---

Today we will introduce two techniques we'll use repeatedly: \emph{covering}, and \emph{bounding higher product sets}. A nice way to do this is by proving the following theorem.

\begin{thm} (2.1, Rusza)\\
    Suppose $A \subseteq \F^r_p$ satisfies $|A+A| \leq K|A|$. Then there is a subgroup $H \subseteq \F_p^r$ with $|H| \leq p^{K^4}K^2 |A|$ s.t. $A \subseteq H$.
\end{thm}

\begin{rem}
    It's not ideal that $|A|/|H|$ depends on $p$. We'll remove this dependency in a few lectures' time.
\end{rem}

We'll start by proving the following weaker version:

\begin{prop} (2.2)\\
    Suppose $A \subseteq \F_p^r$ satisfies $|2A-2A| \leq K|A|$. Then there is a subgroup $H \subseteq \F_p^r$ with $|H| \leq p^K |A-A|$ ($\leq p^K K|A|$) such that $A \subseteq H$.
\end{prop}

We'll prove this using 'covering', encapsulated by the following lemma:

\begin{lemma} (2.3, Rusza's Covering Lemma)\\
    Suppose $A,B \subseteq G$, and $|AB| \leq K|B|$. Then, $\exists X \subseteq A$ with $|X| \leq K$ s.t. $A \subseteq XBB^{-1}$. Indeed, we may take $X \subseteq A$ maximal such that the sets $xB$ ($x \in X$) are disjoint.\\
    The term \emph{covering} refers to the conclusion $A \subseteq XBB^{-1}$, which says \emph{$A$ can be covered by  few left-translates of $BB^{-1}$.}
    \begin{proof}
        First, disjointness of $xB \implies |XB| = |X||B|$. Since $X \subseteq A$, $|XB| \leq |AB| \leq K|B|$, so $|X| \leq K$. Maximality implies $\forall a \in A$, $\exists x \in X$ s.t. $aB \cap xB \neq \phi$; and hence $a \in xBB^{-1}$. So $A \subseteq XBB^{-1}$.
    \end{proof}
\end{lemma}

Now we prove some lemmas for proposition 2.2:

\begin{lemma} (2.4)\\
    Suppose $A \subseteq G$ satisfies $|A^{-1}A^2A^{-1}| \leq K|A|$. Then $\exists X \subseteq A^{-1} A^2$, $|X| \leq K$ such that $A^{-1} A^n \subseteq x^{n-1} A^{-1}A \forall n \in \N$.
    \begin{proof}
        Lemma 2.3 gives (as $|A| = |A^{-1}|$) the existence of an $X \subseteq A^{-1} A^2$, $|X| \leq K$ s.t. $A^{-1}A^2 \subseteq XA^{-1}A$. We then have
        \[
            A^{-1} A^n &= A^{-1} A^{n-1} A\\
            &\subseteq X^{n-2} A^{-1} A^2 \text{ by induction}\\
            &\subseteq X^{n-1} A^{-1} A
        \]
    \end{proof}
\end{lemma}

\begin{proof} (of proposition 2.2)\\
    Note that $\F_p^r$ is abelian, so we can apply Lemma 2.4 which gives the existence of $X$, $|X| \leq K$ s.t. $nA - A \subseteq (n-1)X+A-A \forall n \in \N$. As we are in a finite vector space, this means that $\bra A \ket \subseteq \bra X \ket +A-A$, so $|\bra A\ket| \leq |\bra X \ket| |A-A| \leq p^K K|A|$, as claimed.
\end{proof}

To strenghthen prop 2.2 to theorem 2.1, we use the technique of bounding higher sum/product sets. The key result, at least in the abelian case, is the following:

\begin{thm} (2.5, Pl\"{u}necke-Rusza)\\
    Suppose $A \subseteq G$ (abelian) and $|A+A| \leq K|A|$, then $|mA-nA| \leq K^{m+n} |A|$ $\forall m,n \geq 0$.\\
    This was proven intro to discrete analysis last term. We won't redo the whole proof, but we will reprove some parts of it.
\end{thm}

\begin{proof}
    2.5 gives that $|2A-2A| \leq K^4 |A|$, and $|A-A| \leq K^2|A|$. Then immediately it follows from prop 2.2.
\end{proof}

We'll spend the rest of teh lecture discussing theorem 2.5 and variants of it. We've seen that it is useful, at least in one context. To see more properly why it's useful, let's think about what the genuine closure of subgroups under products and inverse means. One useful feature is that it can be iterated: given $h_1,h_2,... \in H$, a subgroup, this means that $h_1^{\varepsilon_1}...h_m^{\varepsilon_m} \in H$ $\forall \varepsilon_i = \pm 1$ $\forall m \forall h_i \in H$. Then theorem 2.5 allows us to \emph{iterate the approximate closure} of a set of small doubling: $a_1+...+a_m-a_1'-...-a_n'$ may not belong to $A$, but at least belongs to $mA-nA$ which is (a) not too large, and (b) itself a set of small doubling ($|2(mA-nA)| \leq K^{2m-2n} |mA-nA|$). This is an important part of why the theory works so well.\\
It is therefore unfortunate that theorem 2.5 doesn't hold for non-abelian groups.

\begin{eg} (2.6)\\
    Let $x$ generate an infinite cyclic group $\bra x \ket$, and $H$ be a finite group. Set $G=H *\bra x\ket$, the free product (just keep in mind that $x^{-1} Hx \neq H$). Set $A = H \cup \{x\}$, then $A^2 = H \cup xH \cup Hx \cup \{x^2\}$, so $|A^2| \leq 3|A|$. But $A^3$ contains $HxH$, which has size $|H|^2 \sim |A|^2$. So as $|H| \to \infty$, theorem 2.5 cannot hold.
\end{eg}

Nonetheless, if we strenghten small doubling slightly, we can recover a form of theorem 2.5. One way is to replace it with \emph{small tripling}, i.e. $|A^3| \leq K|A|$:

\begin{prop} (2.7)\\
    Suppose $A \subseteq G$, $|A^3| \leq K|A|$. Then $|A^{\varepsilon_1} ... A^{\varepsilon_m}| \leq K^{3(m-2)} |A|$, $\forall \varepsilon_i =\pm 1, \forall m \geq 3$.
\end{prop}

The key ingredient is the following:

\begin{lemma} (2.8, Rusza's Triangle Inequality)\\
    Given $U,V,W \subseteq G$ all finite, $|U| |V^{-1} W| \leq |UV||UW|$.
    \begin{proof}
        We'll define an injection $\varphi:U \times V^{-1} W \to UV \times UW$. First, for $x \in V^{-1} W$, set $V(x) \in V$, and $W(x) \in W$, s.t. $x=V(x)^{-1} W(x)$. Set $\varphi(u,x) = (uV(x),uW(x))$. To see injectivity, first note that $(uV(x))^{-1} (uW(x)) = x$, so $x$ is determined by $\varphi(u,x)$, then $(uV(x))(v(x)^{-1}) = u$, so $u$ is also determined by $\varphi(u,x)$.
    \end{proof}
\end{lemma}

\begin{proof} (of 2.7)\\
    First, we'll do the case $m=3$.\\
    $\bullet$ $|A^3| = |A^{-3}| \leq K|A|$;\\
    $\bullet$ apply lemma 2.8 with $U=W=A$, $V=A^2$. Get $|A||A^{-2}A|\leq |A^3| |A^2| \leq K^2|A|^2$, so $|S^{-2}A| \leq K^2 |A|$;
    $\bullet$ note that $(A^{-2}A)^{-1} = (A^{-1}A^2)$, so $|A^{-1}A^2||A^{-2}A| \leq K^2|A|$;\\
    $\bullet$ replace $A$ by $A{-1}$ to get $|AA^{-2}| = |A^2 A^{-1}| \leq K^2|A|$;\\
    $\bullet$ Finally, apply lemma 2.8 with $U=V=A, W=AA^{-1}$, we get $|A||A^{-1}AA^{-1}| \leq |A^2||A^2 A^{-1}| \leq K^3 |A|^2$. So, $|A^{-1} AA^{-1}| \leq K^3 |A|$. For the last case, swap $A \leftrightarrow A^{-1}$.

    For $m \geq 4$, lemma 2.8 gives $|A||A^{\varepsilon_1} ... A^{\varepsilon_m}| \leq |AA^{-\varepsilon_2} A^{-\varepsilon_1}| |AA^{\varepsilon_3}...A^{\varepsilon_m}| \leq K^3|A| K^{3(m-3)}|A|$ by induction.
\end{proof}

\newpage

\section{Approximate groups}

---Lecture 3---

Last time we saw that assuming small tripling instead of small doubling allowed us to control higher product sets of the form $A^{\varepsilon_1}...A^{\varepsilon_m}$. In this lecture, we'll see another possible strengthening of small doubling. We also saw, in the proofs of theorem 2.1 and proposition 2.2, that the advantage of having a \emph{covering} condition in place of a size bound. This motivates the following definition:

\begin{defi} (Approximate groups)\\
A set $A \subset G$ is called a \emph{$K$-approximate group} (or $K$-approximate subgroup) if $1 \in A$, $A^{-1} = A$, and $\exists X \subset G$ $|X| \leq K$ s.t. $A^2 \subset XA$.
\end{defi}

Note that $A$ need not be finite, although in this course it almost always will be. Also, if $A$ is finite, then $|A^2| \leq K|A|$.\\
The conditions $1 \in A$ and $A^{-1} = A$ are convenient notationally: for example, we can write $A^m$ in stead of $A^{\varepsilon_1}...A^{\varepsilon_m}$, and $1 \in A \implies A \subset A^2 \subset A^3 \subset ...$, which is also convenient at times. It's the condition $A^2 \subset XA$ that is most important.

For approximate groups, bounding higher product sets is easy:

\begin{lemma} (3.1)\\
If $A$ is a finite $K$-approximate group, then $|A^m| \leq K^{m-1} |A|$.
\begin{proof}
If $X$ is as in definition of approximate group, in fact we have $A^m \subset X^{m-1}A$:
\[
A^m &= A^{m-1}A\\
&\subset X^{m-2} A^2 \text{ induction}\\
&\subset X^{m-1}A \text{ definition of X}
\]
\end{proof}
\end{lemma}

Another advantage is that if $\pi: G \to H$ is a homomorphism and $A$ is a $K$-approximate group, then $\pi(A)$ is also trivially a $K$-approximate group (although we'll see that there exists a version of this for small tripling).\\
It turns out that sets of small tripling and approximate groups are essentially equivalent, in the following sense:

\begin{prop} (3.2)\\
Let $A \subset G$ be finite. If $A$ is a $K$-approximate group then $|A^3| \leq K^2 |A|$. Conversely, if $|A^3| \leq K|A|$ then there exists $O(K^{12})$-approximate groups $B$ with $A \subset B$ and $|B| \leq 7K^3 |A|$ (\emph{$A$ is a large proportion of an approximate group}). In fact, we may take $B=(A \cup \{1 \} \cup A^{-1})^2$.
\begin{proof}
First part is just lemma 3.1. For the converse, set $\hat{A} = A \cup \{1\} \cup A^{-1}$, and note that $A^2 = \{1\} \cup A \cup A^{-1} \cup A^2 \cup A^{-1}A \cup AA^{-1} \cup A^{-2}$. Each set in this union has size $\leq K^2|A|$, by prop 2.7, so $|B| \leq 7K^3|A|$, as claimed. Similarly,
\[
A^4 = \bigcup_{\varepsilon_i = \pm 1, 0 \leq m \leq 4} A^{\varepsilon_1}...A^{\varepsilon_m}
\]
and the sets in this union have size $\leq K^6|A|$. It follows that $|\hat{A}^4| \leq O(K^6) |\hat{A}|$. Lemma 2.4 then tells us that $\exists X \subset G$, $|X| \leq O(K^6)$ s.t. $\hat{A}^n \subset X^{n-2} \hat{A}^2$ for every $n \geq 2$.\\
In particular, $|X^2| \leq O(K^{12})$, and $\hat{A}^4 = (\hat{A}^2)^2 \subset X^2 \hat{A}^2$, so $\hat{A}^2$ is an $O(K^{12})$-approximate group, as claimed.
\end{proof}
\end{prop}

This is all well and good, but what if we are faced with a set like that from example 2.6, which only has small doubling (but not tripling)? In that specific example, a large proportion of $A$ was a set of small tripling, namely $H$. Rather helpfully, that turns out to be a general phenomenon:

\begin{thm} (3.3)\\
If $A \subset G$ satisfies $|A^2| \leq K|A|$, then $\exists U \subset A$ with $|U| \geq \frac{1}{K}|A|$ s.t. $|U^m| \leq K^{m-1} |U|$ $\forall m \in \N$.
\end{thm}

So small doubling reduces to small tripling, which reduces to approximate groups. In sheet 1 we'll see a direct reduction from small doubling to approximate groups.

Tao proved a version of theorem 3.3 when he introduced the definition of approximate groups. We'll use instead a lemma of Petridis, which he proved when proving the Pl\"{u}nnecke-Ruzsa inequalities.

\begin{lemma} (3.4, Petridis)\\
Suppose $A,B \subset G$ are finite, let $U \subset A$ be non-empty, chosen to minimise the ratio $|UB| / |U|$, and write $R=|UB|/|U|$. Then for every finite $C \subset G$, we have $|CUB| \leq R|CU|$.
\begin{proof}
It's trivial if $C=\phi$, so we may assume $\exists x \in C$. Defining $C' = C \{x\}$, we may also assume by induction that $|C'UB| \leq R|C'U|$.\\
Set $W = \{u \in U: xu \in C'U\}$. Then $CU=C'U \cup (xU \setminus xW)$ is a disjoint union, so in particular
\[
|CU| = |C'U| + |U|-|W| \ \text{ (3.1)}
\]
We also have $xWB \subset C'UB$ by definition of $W$, so $CUB \subset C'UB \cup (xUB \setminus xWB)$, and hence
\[
|CUB| &\leq |C'UB|+|UB|-|WB|\  \text{ (3.2)}
\]
We have $|C'UB| \leq R|C'U|$ by induction hypothesis, we have $|UB|=R|U|$ by definition oF $R$, and $|WB| \geq R|W|$ by minimality in the definition of $U$. So
\[
|CUB| &\leq R(|C'U| + |U| - |W|) \text{ by (3.2)}\\
&= R|CU| \text{ by (3.1)}
\]
\end{proof}
\end{lemma}

\begin{proof} (of 3.3)\\
Set $U \subset A$ to be non-empty, minimizing $|UA| / |U|$, and write $R=|UA|/|U|$, noting that $R \leq K$ by minimality (since $U$ must have beaten $A$). Also, $U$ non-empty implies $|UA| \geq |A|$, so $|U| \geq \frac{|A|}{K}$, as required. Lemma 2.4 also implies that $|U^mA| \leq K|U^m| \forall m$ (taking $C=U^{m-1}$, and since $U \subset A$, this gives $|U^{m+1}| \leq K|U^m| \forall m$, so $|U^m| \leq K^{m-1} |U|$.
\end{proof}

Non-examinable: the reason $A$ in example 2.6 failed to have small tripling was the existence of $x \in A$ with $AxA$ large. It turns out that this is the only obstruction to small doubling having small tripling.

\begin{thm} (3.5, Tao, Petridis)\\
If $|A^2| \leq K|A|$, and $|AxA| \leq K|A|$ $\forall x \in A$, then $|A^m| \leq K^{O(m)}|A|$ $\forall m \geq 3$.
\end{thm}

Also non-examinable (it's in intro to DA, but lecturer thought it's was showing):

\begin{thm} (2.5')\\
After writing a few words, lecturer decided that there's not enough time, and we have written enough this lecture, so wiped everything out from the board.
\end{thm}

Check example sheets at $tointon.neocities.org$; they'll be up some time this afternoon.

\newpage

\section{Stability of approximate closure under basic operations}

---Lecture 4---

There's a misprint in sheet 1 q7, but it has been corrected last friday.

Two familiar properties of genuine subgroups are that they behave well under quotients and intersections:\\
$\bullet$ if $H<G$, and $\pi:G \to \Gamma$ is a homomorphism, then $\pi(H) < \Gamma$;\\
$\bullet$ if $H_1,H_2 < G$, then $H_1 \cap H_2 < G$.

In this lecture we'll see versions of these properties for approximate groups and sets of small tripling.

It's trivial that if $A \subset G$ is a $K$=approximate group, then $\pi(A)$ is also a $K$-approximate group. The following is the corresponding result for sets of small tripling.

\begin{prop} (4.1, stability of small tripling under homomorphisms)\\
Let $A \subset G$ be finite, symmetric, containing the identity. Suppose $\pi:G \to \Gamma$ is a homomorphism. Then
\[
\frac{|\pi(A)^m|}{|\pi(A)} \leq \frac{|A^{m+2}|}{|A|}
\]
for all $m \in \N$.\\
In particular, if $|A^3| \leq K|A|$, then $|\pi(A)^3| \leq K^9|\pi(A)|$ by proposition 2.7.
\end{prop}

We'll prove this using an argument of Helfgolt. We'll start with a simple observation that we'll use repeatedly in the course.

\begin{lemma} (4.2)\\
Let $H < G$, let $A \subset G$ be finite, and let $x \in G$. Then $|A^{-1}A \cap H| > |A \cap xH|$.
\begin{proof}
We have $(A \cap xH)^{-1} (A \cap xH) \subset A^{-1} A \cap H$.
\end{proof}
\end{lemma}

\begin{rem}
Most of the lemmas and propositions in this lecture will have familiar/trivial analogues for genuine subgroups.
\end{rem}

\begin{lemma} (4.3)\\
Let $H < G$, write $\pi:G \to G/H$ for the quotient map, let $A \subset G$ be finite.\\
Then $|A^{-1}A \cap H| \geq |A|/|\pi(A)|$.\\
Note that $H$ is not assumed normal, so $G/H$ is just the \emph{space} of left cosets of $xH$, not necessarily a group.
\begin{proof}
By pigeonhole principle there exists $x \in G$ s.t. $|A \cap xH| \geq |A|/|\pi(A)|$. Then apply lemma (4.2).
\end{proof}
\end{lemma}

\begin{lemma} (4.4)\\
Let $H < G$, write $\pi:G \to G/H$ for the quotient map, and let $A \subset G$ be finite.\\
Then $|\pi(A^m)||A^n \cap H| \leq |A^{m+n}| \forall m,n \geq 0$.
\begin{proof}
Define $\varphi:\pi(A^m) \to A^m$ by picking arbitrarily for each $x \in \pi(A^m)$, some $\varphi(x)$ s.t. $\pi(\varphi(x)) = x$. Then the cosets of $\varphi(x) H$ for $x \in \pi(A^m)$ are all distinct by definition, so in particular,
\[
|\varphi(\pi(A^m)) (A^n \cap H)| = |\pi(A^m)| |A^n \cap H|
\]
But also, $\varphi(\pi(A^m)) (A^n \cap H) \subset A^{m+n}$.
\end{proof}
\end{lemma}

\begin{proof} (of 4.1)\\
Write $H=\ker \pi$. Lemma (4.4) gives $|\pi(A^m)| \leq \frac{|A^{m+2}|}{|A^2 \cap H|}$. Lemma 4.3 gives $|A^2 \cap H| \geq |A| / |\pi(A)|$ (?). The proposition follows from combining these two inequalities.
\end{proof}

Now we'll look at intersections:

\begin{prop} (4.5, stability of small tripling under intersections with subgroups)\\
Let $A \subset G$ be finite, symmetric, containing 1. Let $H \leq G$. Then
\[
\frac{|A^m \cap H|}{|A^2 \cap H|} \leq \frac{|A^{m+1}|}{|A|}
\]
In particular, by prop 2.7, if $|A^3| \leq K|A|$, then
\[
|(A^m \cap H)^3| \leq K^{9m} |A^m \cap H|
\]
$\forall m \geq 2$.
\begin{rem}
We'll see in example sheet 1 that even if $A$ has small tripling, $A \cap H$ need not. So $m \geq 2$ is really important for this last conclusion.
\end{rem}
\begin{proof}
Lemma 4.4 gives $|A^m \cap H| \leq |A^{m+1}| / |\pi(A)|$, where $\pi:G \to G/H$ is the quotient map as before).\\
Lemma 4.3 gives $|A^2 \cap H| \geq |A| / |\pi(A)|$.\\
The proposition follows from combining these two inequalities.
\end{proof}
\end{prop}

\begin{prop} (4.6, stability of approximate groups under intersections with subgroups)\\
Let $H < G$, let $A \subset G$ be a $K$-approximate group. Then $A^m \cap H$ is covered by $\leq K^{m-1}$ left-translates of $A^2 \cap H$.\\
In particular, $A^m \cap H$ is a $K^{2m-1}$-approximate group (since $A^2 \cap H \subset A^m \cap H$ and $(A^m \cap H)^2 \subset A^{2m} \cap H$).
\begin{proof}
By definition, $\exists X \subset G$ with $|X| = K^{m-1}$ s.t. $A^m \subset XA$. In particular, $A^m \cap H \subset \cup_{x \in X} (xA \cap H)$.\\
For each $xA \cap H$ non-empty, $\exists h = xa \in H$ for some $a \in A$. this means that $xA \cap H \subset h(a^{-1} A \cap H) \subset h(A^2 \cap H)$.\\
Hence each set $xA \cap H$ in this union is contained in a single left translate of $A^2 \cap H$.
\end{proof}
\end{prop}

In intro to discrete analysis last term, you saw that when studying small doubling/tripling, there is a more general notion of homomorphism that comes into play -- the Freiman homomorphism. To motivate this, consider two sets $A=\{-n,...,n\} \subset \Z/p\Z$, and $B = \{-n,...,n\} \subset \Z/q\Z$ for $p,q$ two large primes $\geq 10n$, say. These two sets are intuitively \emph{isomorphic} from perspective of $A+A$, or $B+B$, but there is no way of encoding this with a group homomorphism $\Z/p\Z \to \Z/q\Z$.

\begin{defi} (Freiman homomorphism)\\
Let $m \in \N$, let $A,B$ be subsets of groups. Then a map $\varphi:A \to B$ is a \emph{Freiman $m$-homomorphism} if for $x_1,...,x_m,y_1,...,y_m \in A$ with $x_1...x_m = y_1...y_m$, we have $\varphi(x_1)...\varphi(x_m) = \varphi(y_1)...\varphi(y_m)$.\\
If $1 \in A$ and $\varphi(1) = 1$, then we say that $\varphi$ is \emph{centred}. If $\varphi$ is injective, and its inverse $\varphi(A) \to A$ is also a Freiman $m$-homomorphism, then we say $\varphi:A \to \varphi(A)$ is a \emph{Freiman $m$-isomorphism}.
\end{defi}

\begin{rem}
1) Every map is trivially a $1$-homomorphism, so we only care about $m \geq 2$.\\
2) This notion gets stronger as $m$ increases: we may assume $A \neq \phi$, and then, picking $a \in A$ arbitrarily, if $x_1...x_k = y_1...y_k$ for $k \leq m$, then
\[
x_1...x_k \underbrace{a...a}_{m-k} = y_1...y_k \underbrace{a...a}_{m-k}
\]
3) If $\varphi$ is centred and $a,a^{-1} \in A$, then exercise to check $\varphi(a^{-1}) = \varphi(a)^{-1}$ ($m \geq 2$). When we say $\varphi$ is a Freiman homomorphism, we mean it is a $2$-homomorphism.
\end{rem}

\begin{lemma} (4.7)\\
Suppose $\varphi:A \to \Gamma$ is a Freiman $m$-homomorphism. Then $|\varphi(A)^m| \leq |A^m|$.\\
In particular, if $\varphi$ is injective, then
\[
\frac{|\varphi(A)^m|}{|\varphi(A)|} \leq \frac{|A^m|}{|A|}
\]
and if $\varphi$ is a Freiman $m$-isomorphism, then equality holds.
\end{lemma}
\begin{proof}
Exercise.
\end{proof}

---Lecture 5---

\begin{lemma} (4.8)\\
Let $A \subset G$ be a $K$-approximate group, suppose $\varphi:A^3 \to \Gamma$ is a centred ($1 \in A$ and $\varphi(1) = 1$) Freiman 2-homomorphism. Then $\varphi(A)$ is also a $K$-approximate group.
\begin{proof}
$\exists X \subset G$, $|X| \leq K$ s.t. $A^2 \subset XA$. So given $a_1,a_2 \in A$, $\exists x \in X, a_3 \in A$ s.t. $a_1a_2 = xa_3$. In particular, $x \in A^3$, so $\varphi(x)$ is defined; and $\varphi(a_1)\varphi(a_2) = \varphi(x) \varphi(a_3)$.\\
Hence $\varphi(A)^2 \subset \varphi(X \cap A^3) \varphi(A)$.\\
Also, $\varphi$ is centred, so $\varphi(A)$ is symmetric and contains 1.
\end{proof}
\end{lemma}

\newpage

\section{Coset progressions, Bohr sets and the Freiman-Green-Ruzsa theorem}
Today we will introduce some non-trivial examples of sets of small doubling in abelian groups.

\begin{defi} (coset progression)\\
Let $G$ be abelian, $x_1,...,x_r \in G$, $L_1,...,L_r \in \N$. Then the set
\[
P(x;L) = P(x_1,...,x_r;L_1,...,L_r) = \{l_1x_1+...+l_rx_r : |l_i| \leq L_i  \forall i\}
\]
is called a \emph{progression of rank $r$}. If in addition $H \subset G$ is finite, then $H+P(x;L)$ is called a \emph{coset progression} of rank $r$.
\end{defi}

It is useful to think of $P(x;L)$ as a homomorphic image of a \emph{box} in $\Z^r$. For example, if $G=\Z$ and $r=2$:\\
(diagram)

It's easy to see that such a box $B$ in $\Z^r$ is a $2^r$-approximaate group, e.g. in $r=2$:\\
(diagram)

Hence $P(x;L)$ is also a $2^r$-approximate group, as is $H_P(r;L)$.

Remarkably, these are essentially the only examples:

\begin{thm} (5.1, Frieman ($G=\Z$); Green-Rusza (arbitrary abelian $G$))\\
Suppose $A \subset G$ (abelian) satisfies $|A+A| \leq K|A|$. Then there exists a coset progression $H+P$ of rank $\leq O(K^{O(1)})$ s.t. $A \subset H+P \subset O(K^{O(1)})(A \cup \{0\} \cup -A)$.\\
In particular, theorem 2.5 gives that $|H+P| \leq \exp(O(K^{O(1)})) |A|$. So $A$ is a large proportion of $H+P$.
\end{thm}

A substantial part of this result was proved in intro to discrete analysis, but with a slightly less explicit version of coset progressions:

\begin{defi} (Bohr set)\\
Let $G$ be a finite abelian group. Let $\Gamma = \{\gamma_1,...,\gamma_r\} \subset \hat{G} = Hom(G,\R/\Z)$, and let $\rho \in [0,\frac{1}{2}]$. Then the set
\[
B(\Gamma,\rho) = \{g \in G: || \gamma_i(g)||_{\R/\Z} \leq \rho \forall i \}
\]
is called a Bohr set of rank $r$. Here, given $x \in \R/\Z$ with representative $\hat{x} \in (-\frac{1}{2},\frac{1}{2}]$, we write $||x||_{\R/\Z} = |\hat{x}|$.
\end{defi}

We'll see in sheet 1 that $B(\Gamma,\rho)$ is a $4^r$-approximate group, whereas progressions were homomorphic images of boxes, $B(\Gamma,\rho)$ is the pullback(??) of $[-\rho,\rho]^r$ under $(\gamma_1,...,\gamma_r) = \hat{G}^r$.

It turns out that the notions of coset progression and Bohr set are essentially equivalent. In sheet 2, we'll see that every coset progression is a Freimana image of a Bohr set of the same rank. Moreover, every Freiman image of a Bohr set is a large proportion of some coset progression. We'll see a special case of that shortly.

\begin{prop} (5.2, from intro to DA)\\
Suppose $A \subset G$ (abelian) with $|A+A| \leq K|A|$. Then $\exists B \subset 2A-2A$ a finite abelian group $Z$ with $|Z| \geq |A|$, a set $\Gamma \subset \hat{Z}$ with $|\Gamma| \leq O(K^{O(1)})$, some $\rho \geq \frac{1}{O(K^{O(1)})}$, and a centred Freiman 2-isomorphism $\varphi:B(\Gamma,\rho) \to B$.\\
(\emph{$2A-2A$ contains a large set isomorphic to a Bohr set of bounded rank}).\\
In intro to DA you saw this in the special case of $G$ torsion-free. The general case is harder, but nonetheless conceptually very similar, so we'll assume this result from now on.
\end{prop}

To pass from prop 5.2 to theorem 5.1, we use the following results:

\begin{prop} (5.3)\\
Suppose $G$ is a finite abelian group, $\Gamma \subset \hat{G}$ is of size $r$, $\rho<\frac{1}{10}$\footnote{Lecturer thinks he might also get away with a larger choice.}. Then there exists a coset progression $H+P \subset B(\Gamma,\rho)$ with rank $r$ and $|H+P| \geq (\rho/r)^r |G|$.
\end{prop}

We'll prove prop 5.3 in the next couple of lectures.

\begin{lemma} (5.4)\\
Suppose $H+P$ is a coset progression of rank $r$, and $\varphi:H+P \to G$ (abelian) is a centred Freiman 2-homomorphism. Then $\varphi(H_P)$ is also a coset progression of rank $r$.
\begin{proof}
Exercise: if $H$ is a group and $\varphi:H \subset G$ is a centred Freiman 2-homomorphism, then $\varphi$ is also a group homomorphism. In particular, in this lemma, $\varphi(H)$ is a finite subgroup. Therefore it's sufficient to show that $\varphi(H+P(x;L)) = \varphi(H) + P(\varphi(x_1),...,\varphi(x_r);L_1,...,L_r)$.\\
In fact, we'll show that $\forall h \in H$, $|l_i| \leq L_i$, we have
\[
\varphi(h+l_1x_1+...+l_rx_r) = \varphi(h) + l_1 \varphi(x_1)+...+l_r \varphi(x_r) \ \text{(5.1)}
\]
Since $\varphi$ is centred, $\varphi(-x_i) = -\varphi(x_i)$, so we may assume that $l_i \geq 0 \forall i$. Also, (5.1) is trivial if $l_i = 0 \forall i$, so we may also assume that $\exists l_j > 0$. Then
\[
\varphi(h+l_1x_1+...+l_rx_r) &= \varphi(h+l_1x_1+...+l_rx_2)+\varphi(0)\\
&= \varphi(h+l_1x_1+...+(l_j-1) x_j + ... + l_r x_r) + \varphi(x_j)
\]
So lemma follows by induction on $\sum_i l_i$.
\end{proof}
\end{lemma}

\begin{proof} (of theorem 5.1)\\
By proposition 5.2 and 5.3 and lemma 5.4, $\exists H+P$ coset progression of rank $\leq O(K^{O(1)})$ s.t. $H+P \subset 2A-2A$ and $|H+P| \geq \exp(-O(K^{O(1)}))|A|$.\\
We'll now apply a version of Ruzsa's covering lemma due to Chang. Define recursively sets $S_1,S_2,... \subset A$ s.t. $S_i$ is a maximal subset of size $\leq 2K$ s.t. the translates $x+S_{i-1} + ... + S_1 + H + P$ are all disjoint. If ever $|S_i| < 2K$, we stop. Now suppose we get as far as $S_1,...,S_t$. Then
\[
S_t+...+S_1 + H+P \subset 2A-2A+tA
\]
So prop 2.5 $\implies$ $|S_t+...+S_1+H+P| \leq K^{4+t} |A|$.\\
On the other hand, disjointness of the translates in the definition of $S_i$ means that
\[
|S_t+...+S_1 + H+P| \geq |S_t|...|S_1||H+P| \geq (2K)^{t-1} \exp(-O(K^{O(1)})) |A|
\]
Putting these together, we have $2^{t-1} \leq K^5 \exp(O(K^{O(1)}))$, hence $t \leq O(K^{O(1)})$.\\
In particular, the process terminates, at $S_t$ say.\\
But also, since $S_t$ is therefore maximal among all subsets of $A$ s.t. $x+S_{t-1} + ... +S_1 + H+P$ are disjoint for $x \in S_t$, Ruzsa's covering lemma from lecture 2 implies that
\[
A \subset H+2P +S_1-S_1 + ... + S_{t-1}-S_{t-1} + S_t
\]
Enumerating $\bigcup_i S_i$ as $s_1,...,s_d$, we have $d \leq O(K^{O(1)})$, and
\[
A \subset H + 2P + P(s_1,...,s_d;1,...,1) \subset O(K^{O(1)})(A \cup\{0\} \cup -A)
\]
as claimed (subject to proving 5.3).
\end{proof}

Exercise: see what bounds you get if you apply Ruzsa's covering lemma directly, instead of Chang's argument.

---Lecture 6---

\begin{prop}
    Let $G$ be a finite abelian group. Suppose $\Gamma \subseteq \hat{G}$, $|\Gamma| = r$, and let $\rho < 1/2$. Then $\exists$ coset progression of $H+P \subseteq B(\Gamma,\rho)$ (Bohr set, see previous lecture) of rank $r$ with $|H+P| \geq (\rho/r)^r |H|$.
\end{prop}

To prove this, we'll use a field called geometry of numbers, which is concerned with lattices in $\R^d$.\\
For us, a lattice $\Lambda \subseteq \R^d$ will simply be the additive subgroup (not \emph{subspace}) generated by some basis $x_1,...,x_d$ for $\R^d$. So $\Lambda = \{l_1x_1+...+l_dx_d|l_i \in i\}$. If $\Gamma \subset \Lambda$ is another lattice, then we'll call it a sublattice, and just write $\Gamma \subset \Lambda$. It is an exercise (in sheet 2) to check that if $\Gamma \subset \Lambda$ with basis $y_1,...,y_d$, say, then $\det(y_1,...,y_d)/\det(x_1,...,x_d) = [\Lambda : \Gamma]$. In particular, two different bases of the same lattice have the same determinant. So we can define this to be $\det \Lambda$.

The relevance of lattices to 5.3 is the following:

\newpage

\section{Geometry of Numbers}

\begin{lemma} (6.1)\\
    Let $G,\Gamma$ be as in 5.3, and set $\gamma: G \to \R^d / \Z^d$ by enumerating $\Gamma$ as $\{\gamma_1,..,\gamma_d\}$ and setting $\gamma = (\gamma_1,...,\gamma_d)$. Then $\Lambda = \gamma(G) + \Z^d$ is a lattice with determinant $|\ker \gamma| / |G|$.
    \begin{proof}
        $\Lambda$ is finitely generated as $G$ is finite, and torsion-free as it's in $\R^d$, so isomorphic to $\Z^k$ for some $k$. Also, $\Lambda$ has $\Z^d$ as a finite index subgroup. So $k=d$, $span_\R(\Lambda) = \R^d$. So we can take a generating set for $\Lambda$ of size $d$, which is then a basis for $\R^d$. The determinant follows by above exercise, as $\det(\Z^d)=1$.
    \end{proof}
\end{lemma}

We'll investigate the intersection of $[-\rho,\rho]$ with $\Lambda$. To do this, we introduce some definitions:

\begin{defi}
    A set $A \subseteq \R^d$ is \emph{convex} if $\forall x \in \R^d \setminus A^0$, $\exists$ hyperplane $h_x$ with $x \in h_x$ and $h_x \cap A^0 = \phi$ ($A^0$ is the interior of $A$).
\end{defi}

\begin{defi}
    A set $B \subseteq \R^d$ is a \emph{convex body} if it is bounded, convex and $B^0 \neq \phi$.\\
    It is \emph{symmetric} if $\forall x \in B$, we have $-x \in B$.
\end{defi}

Given a symmetric body $B$ and a lattice $\Lambda$, define the \emph{successive minima} $\lambda_1 \leq ... \leq \lambda_j$ of $B$ w.r.t. $\Lambda$ as
\[
    \lambda_i = \inf \{\lambda>0 \dim span_\R(\lambda \cdot B \cap \Lambda) \geq i\}
\]

We may then inductively define $v_1,...,v_d \in \Lambda$, s.t. $v_1,...,v_i \in \lambda_i \bar{B}$. We'll call this a \emph{directional basis} for $\Lambda$ w.r.t. $B$. Note that this is not unique, and not necessarily a basis for $\Lambda$ in the earlier sense.

\begin{thm} (6.2, Minkowski's second theorem)\\
    Suppose $B$ is a symmetric convex body, $\Lambda$ a lattice in $\R^d$, and $\lambda_1,...,\lambda_d$ are the successive minima. Then $\lambda_1,...,\lambda_d vol(B) \leq 2^d \det(\Lambda)$.
\end{thm}

\begin{lemma} (6.3, Blichfeldt)\\
    Suppose $A \subseteq \R^d$ is a measurable set, $\Lambda$ a lattice, and $\forall a,b \in A$ distinct, $a-b \not\in \Lambda$. Then $vol(A) \leq \det(\Lambda)$.
    \begin{proof}
        Fix a basis $x_1,...,x_d$ for $\Lambda$. Define the \emph{fundamental parallelopiped} $P$ w.r.t. $x_1,...,x_d$ via $P=\{l_1x_1+...+l_dx_d | l_i \in [0,1)\}$. Since $x_1,...,x_d$ is a basis for $\R^d$, $\forall v \in \R^d$, there is a unique $x_v \in \Lambda$ and $p_v \in P$ s.t. $v = x_v + p_v$.\\
        Define a map $\varphi: \R^d \to P$ via $\varphi(v) = p_v$. This cuts $A$ into countably many measurable pieces, and translates these pieces into $P$. It is injective, by hypothesis, hence volume preserving. So $vol(A) = vol(\varphi(A)) \leq vol(P) = \det(\Lambda)$.
    \end{proof}
\end{lemma}

\begin{proof} (of 6.2)\\
    Let $v_1,...,v_d$ be a directional basis for $\Lambda$ w.r.t. $B$. Set $V_i = span_\R (v_1,...,v_i)$ ($v_0 = \{0\})$, and set $\Lambda_i = \Lambda \cap (V_i \setminus V_{i-1})$, then $\Lambda$ is a disjoint union $\bigcup_{i=0}^d \Lambda_i$.

    Claim 1: We have $\lambda_d B^0 \cap (\lambda_d B^0 + \alpha x) = \phi$ whenever $x \in \Lambda$, and $\alpha \geq 2\lambda_d / \lambda_j$.\\
    Proof of claim: given $x \in \Lambda_j$, by definition, $x \not\in \lambda_j B^0$/ So by convexity, there exists hyperplane $h_x$ s.t. $x \in h_x$ and $h_x \cap A_j B^0 = \phi$.\\
    By symmetry, we can take $h_{-x} = -hx$. But note that $-hx = hx - 2x$. This means that $\lambda_j B^0$ is contained in the slice of space $S_x$ between the two parallel hyperplanes $h_x$ and $h_x-2x$. Clearly $S_x \cap (S_x + \alpha_x) = \phi$ $\forall \alpha \geq 2$, so in particular $\lambda_j B^0 \cap (\lambda_j B^0 + \alpha x) = \phi$ for all such $\alpha$ as well. Scaling by $\lambda_i/\lambda_j$, then $\lambda_d B^0 \cap (\lambda_d B^0 + \alpha x) = \phi$ whenever $x \geq 2\lambda_d / \lambda_j$.

    Claim 2: there exist sets $B_1 \subset ... \subset B_d = \lambda_i B^0$ s.t. \\
    (1) $vol(B_i) = (\lambda_i/{\lambda{i+1}})^i vol(B_{i+1}) \forall i$;\\
    (2) $B_i \cap (B_i + \alpha x) = \phi$ whenever $x \in \Lambda_i$ and $\alpha \geq 2 \max\{\lambda_i/\lambda_j,1\}$.
    Proof of claim: Define operations $\gamma_1,...,\gamma_{d-1}$ on suitable subsets of $\R^d$ as follows:  given $L$ bounded and open, define $\gamma_i$ separately on each affine subspace $z+V_i$, with $z \in L$. For each affine subspace, fix a particular $z \in L$, and define $\varphi(z+v) = z+\frac{\lambda_i}{\lambda_{i+1}} v \forall v \in V_i$. Note the following properties:\\
    (i) $vol(\gamma_i(L)) = (\lambda_i/\lambda_{i+1})^i vol(L)$ by Fubini;\\
    (ii) If $L \cap (z+V_i)$ is open and convex $\forall z$, then $\gamma_i(L) \subset L$, because $z \in L$;\\
    (iii) If $L \cap (z+V_i)$ is open and convex, then so is $\gamma_i(L) \cap (z+V_i)$, and indeed so is $\gamma_i(L) \cap (z+V_j \forall j < i)$.\\
    Set $B_d = \lambda_d B^0$, and $B_i = \sigma_i (B_{i+1})$ otherwise. Conclusion (1) is immediate from property (i); conclusion (2) follows from claim 1 when $i=d$. For $i<d$, it follows by induction and repeated application of (ii) and (iii): indeed, (2) for $i \geq j$ follows from (2) f0r $(i+1)$, because $\sigma_i$ scaleds by $\lambda_i/\lambda_{i+1}$ in direction $x$. For $i<j$, it follows from $B_i \subseteq B_{i+1}$.

    Now $vol(B_1) = \lambda_1...\lambda_d vol(B)$, and by (2), $a-b \not\in 2\Lambda$ $\forall a,b \in B_1$, so Blichfeldt gives that $vol(B_1) \leq 2^d \det(\Lambda)$.
\end{proof}

---Lecture 7---

Example class Tuesday 1400-1530 MR14. Work to lecturer at pembroke plodge by midday monday.

Some errata for last lecture (correct directly later):\\
(1)When we had $\sigma_i$ and $B_1 \subset B_2 \subset ...$ last time, we assume $z$ is the centre of mass of $L \cap (z+V_i)$. To be on safe side, in statement of Minkowski's second theorem, let's assume that $B$ is a polytope (?).\\
(2) the assumption $h_x \cap A^0 = \phi$ is not sufficient; we also need $A^0$ is contained in one of the two half spaces into which $h_x$ divides $\R^d$.

\begin{proof} (of 5.3)\\
Write $\gamma = (\gamma_1,...,\gamma_r) \in \hat{G}^r$. Define $\Lambda = \gamma(G) + \Z^r$, which is a lattice of determinant $|\ker \gamma| / |G|$ by lemma 6.1. Let $\lambda_1,...,\lambda_r$ be the successive minima $[-1,1]^r$ w.r.t. $\Lambda$, and $v_1,...,v_r$ a directional basis. Set $L_i = \lfloor\frac{\rho}{r \lambda_i} \rfloor$ for each $i$. Then $P(v_1,...,v_r;L_1,...,L_r) \subset [-\rho,\rho]^r$ (recall LHS is formula for a coset progression -- see start of chapter 5). Pick, for each $i$, $x_i \in G$ s.t. $\gamma(x_i) = v_i$, and set $H=\ker \gamma$, write $P=P(x_1,...,x_r;L_1,...,L_r)$. Then $H+P \subset B(\Gamma,\rho)$.\\
Now we just need to justify the size constraint. We claim that if $l_1,...,l_r$ and $l'_1,...,l'_r$ satisfy $|l_i|,|l'_i| \leq L_i$, and $l_1x_1+...+l_rx_r \in l'_1x_1+...+l'_r x_r+H$ (call this (*)), then in fact $l_i = l'_i \forall i$. Indeed, (*) implies that $(l_1-l'_1)v_1+...+(l_r-l'_r)v_r \in \Z^r \in \Z^r \cap [-2\rho,2\rho]^r$. But $\rho < \frac{1}{2}$, this last intersection is just $\{0\}$. Also $v_i$ are linearly independent, so $l_i=l'_i$ for each $i$.\\
Then
\[
|H+P| &\geq |H|(L_1+1)...(L_r+1)\\
&\geq |H| ( \frac{\rho}{r})^r \frac{1}{\lambda_1...\lambda_r}\\
&\geq |G| (\rho/r)^r
\]
by Minkowski's second theorem.
\end{proof}

\newpage

\section{Progressions in the Heisenberg group}

We know the Heisenberg group is
\[
H(\Z) = \begin{pmatrix}
1 & \Z & \Z\\
0 & 1 & \Z\\
0 & 0 & 1
\end{pmatrix}
\]
where $\Z$ denotes any integer.\\
Set
\[
u_1 = \begin{pmatrix}
1&0&0\\
&1&1\\
&&1
\end{pmatrix}
u_2 = \begin{pmatrix}
1&1&0\\
&1&0\\
&&1
\end{pmatrix}
u_3 = \begin{pmatrix}
1&0&1\\
&1&0\\
&&1
\end{pmatrix}
\]
and note that any element of $H(\Z)$ can be expressed in the form
\[
\begin{pmatrix}
1&n_2&n_3\\
&1&n_1\\
&&1
\end{pmatrix}
= u_1^{n_1}u_2^{n_2}u_3^{n_3}
\]
and we have the following formula for multiplying elements in this form:
\[
(u_1^{n_1}u_2^{n_2}u_3^{n_3})(u_1^{n'_1}u_2^{n'_2}u_3^{n'_3}) = u_1^{n_1+n'_1}u_2^{n_2+n'_2}u_3^{n_3+n'_3+n_1'n_2} \ (7.1)
\]
(just by doing matrix multiplication). There is a more abstract reason for it. To see this, given $x,y \in G$, define the commutator $[x,y] = x^{-1}y^{-1}xy$. In light of the identity $xy = yx[x,y]$, we can view the commutator as being the 'error' or 'cost' incurred when interchanging two elements. For example, the fact that commutators are trivial in abelian group can be viewed as capturing the notion that elements can be interchanged freely in an abelian group. The $n_1'n_2$ term in (7.1) arises because we swap the order of $n_1'n_2$ pairs of elements of $u_1$ and $u_2$.

Now let's see one possible generalisation of progression to non-abelian groups.

\begin{defi} (ordered progression)\\
Given $x_1,...,x_r \in G$, $L_1,...,L_r \geq 0$, we define the \emph{ordered progression of rank $r$}
\[
P_{ord} (x,L) = P_{ord}(x_1,...,x_r;L_1,...,L_r) = \{x_1^{l_1}...x_r^{l_r}; |l_i| \leq L_i \forall i\}
\]
(almost the same as the abelian case, but now we care about the order).\\
Now consider $P=P_{ord}(u_1,u_2;L_1,L_2)$ for $u_1,u_2 \in H(\Z)$ as before, and $L_1,L_2 \geq 0$. We have $(u_1^{l_1} u_2^{l_2}) (u_1^{l'_1}u_2^{l'_2}) = u_1^{l_1+l'_1} u_2^{l_2+l'_2} u_3^{l'_1l_2}$ (7.2), and it is then easy to check that $|P^2|/|P| \to \infty$ as $L_1,L_2 \to \infty$, essentially because we can change $l_1,l_2,l'_1,l'_2$ without changing $l_1+l'_1$ and $l_2+l'_2$ but changing $l'_1l_2$, so in some we have an extra degree of freedom in $P^2$ compared to in $P$.

Coming back to commutators and recalling that $u_3 = [u_2,u_1]$, we see that this corresponds to the freedom to interchange the order of some of the $u_1,u_2$ in $P^2$, as seen in the LHS of (7.2). This is a freedom that the definition of ordered progression explicitly denies us.
\end{defi}

It turns out that if we introduce this freedom to $P$ as well, then this does force $P$ to have small tripling.

\begin{defi} (nonabelian progression)\\
Given $x_1,...,x_r \in G$, $L_1,...,L_r \geq 0$, the \emph{nonabelian progression} $P(x;L) = P(...)$ of rank $r$ is defined to consist of all those elements of $G$ expressible as products of $x_1^{\pm 1},...,x_r^{\pm r}$, in which each $x_i,x_i^{-1}$ appear at most $L_i$ times between them.

It turns out that $P(u_1,u_2;L_1,L_2)$ \emph{does} have small tripling (sheet 2).

A note of caution: nonabelian progressions don't always have small tripling: consider $P(x_1,x_2;L_1,L_2)$ for $x_1,x_2$ generators of a nonabelian free group. In the case of $H(\Z)$, the formula (7.1) is simplified by the fact that $u_3=[u_2,u_1]$ is central in $H(\Z)$. If this were not the case, we'd end up with elements of the form $[[u_2,u_1],u_1]$, for example. This is in fact a specific example of a property called \emph{nilpotence}.

To define this, first define a \emph{normal series} for a group $G$ to be a sequence $G = G_1 > G_2 > ...$ of normal subgroups $G_i \triangleleft G$ and a \emph{central series} to be such a normal series in which each $G_i/G_{i+1}$ is central in $G/G_{i+1}$.
\end{defi}

\begin{defi} (nilpotent group)\\
A group $G$ is nilpotent if $\exists$ finite central series $G=G_1 >...>G_{s+1} = \{1\}$. The smallest $s$ which such series exists is called the \emph{step} or \emph{nilpotency class} of $G$.
\end{defi}

Exercise: $H(\Z)$ is $2$-step nilpotent.

\newpage

\section{Nilpotent groups}

---Lecture 8---

Example class 1400-1530 MR14 today!

Last time, we said $G$ was \emph{nilpotent} if $\exists $ finite central series $G=H_1 > H_2 > ... >H_{s+1} = \{1\}$. We defined the smallest $s$ for which such a series existed the \emph{step} of $G$. Today we'll look into more detail at nilpotent groups.

The reasons we focus on this setting are twofolds: there's a clean generalization of Frieman-Green-Ruzsa Theorem to nilpotent groups, and a deep theorem of Brenillard, Green and Tao essentially reduces the general case to the nilpotent case.

Given $x_1,...,x_k \in G$, we define the simple commutator $[x_1,...,x_k] (= [x_1,...,x_k]_k) $ recursively as follows:\\
$[x_1] = x_1$;\\
$[x,y] = x^{-1}y^{-1}xy$;\\
$[x_1,...,x_k] = [[x_1,...,x_{k-1}],x_k]$.\\
Given subgroups $H,N < G$, define $[H,N] = \bra[h,n]:h \in H,n \in N\ket$, and then given $H_1,...,H_k <G$, define similarly\\
$[H_1] = H_1$,\\
$[H_1,...,H_k] = [[H_1,...,H_{k-1}],H_k]$.

Note that
\[
[H,N] = [N,H] \ (8.1)
\]
since $[h,n] = [n,h]^{-1}$.

\begin{lemma} (8.1)\\
Let $H_1,...,H_k,N \triangleleft G$, let $S_i$ be a generating set for $H_i$ for each $i$. Suppose $[s_1,...,s_k] \in N$ whenever $s_i \in H_i \forall i$. Then $[H_1,...,H_k] < N$.
\begin{proof}
Case $k=1$ is trivial, so consider $k>1$. If $[s_1,...,s_k] \in N \forall s_i \in S_i$, then we have $[[s_1,...,s_{k-1}],s_k] \in N \forall s_i \in S_i$, and hence
\[
[s_1,...,s_{k-1}] \in C_{G/N} (H_k) = \{g \in G: [g,h] \in N \forall h \in H_k\}
\]
The centraliser of a normal subgroup is itself normal, so by induction, we have $[H_1,...,H_{k-1}] \subset C_{G/N} (H_k)$. Hence $[H_1,...,H_k] \subset N$ as claimed.
\end{proof}
\end{lemma}

\begin{defi}
Given a gruop $G$, we define the \emph{lower central series} $G=G_1 >G_2>...$ of $G$ via $G_k = \bra [g_1,...,g_k]: g_i \in G\ket$.
\end{defi}

Note that $G_k > G_{k+1}$ because $[g_1,...,g_{k+1}] = [[g_1,g_2],g_3,...,g_{k+1}]$. Also, since $[g_1,...,g_k]^h = [g_1^h,...,g_k^g]$ (where the notation $\cdot^h$ means conjugating by $h$), each $G_k$ is normal in $G$. The fact that this is a \emph{central series} (i.e. $G_k/G_{k+1}$ is central in $G/G_{k+1}$ $\forall K$ follows from the following result:

\begin{prop} (8.2)\\
We have $G_{k+1} = [G_k,G]$ $\forall k$. In particular, $G_k = [G,...,G]_k$.
\begin{proof}
First, $G_{k+1} < [G_k,G]$ by definition. The fact that $[G_k,G] < G_{k+1}$ follows from lemma 8.1, since $[g_1,...,g_{k-1}]$ generate $G_k$, and $G_1,G_k,G_{k+1}$ are normal.
\end{proof}
\end{prop}

\begin{prop} (8.3)\\
Let $G$ be a group generated by $S$. Then $G_k = \bra [s_1,...,s_k] G_{k+1}: s_i \in S \forall i \ket$.\\
(This basically says \emph{$G_k$ is generated mod $G_{k+1}$ by simple commutators of generators of $G$}.)
\begin{proof}
Note that $[s_1,...,s_k]^g \in [s_1,...,s_k]G_{k+1}$ by definition of $G_{k+1}$.\\
So $\bra [s_1,...,s_k] G_{k+1} : s \in S\ket$ is normal in $G$.\\
Moreover, $[s_1,...,s_k] \in \bra [t_1,...,t_k]G_{k+1}: t_i \in S\ket$ whenever $s_i \in S$ $\forall i$. So lemma 8.1 implies that $[G,...,G]_k \subset \bra[s_1,...,s_k] G_{k+1}: s_i \in S \ket$. Proposition 8.2 gives that $[G,...,G]_k = G_k$, so we have $G_k \subset \bra [s_1,...,s_k] G_{k+1} : s_i \in S \ket$.\\
The reverse inclusion is immediate.
\end{proof}
\end{prop}

\begin{prop} (8.4)\\
We have $[G_i,G_j] \subset G_{i+j} \forall i,j$.\\
For this we'll need the following commutator identity, which we can just check directly by writing it out:
\[
[x,y^{-1},z]^y[y,z^{-1},x]^z[z,x^{-1},y]^x = 1 \ (8.2)
\]
\begin{proof}
The case $j=1$ is proposition 8.2. So we can assume $j>1$, and by induction, that
\[
[G_k,G_{j-1}] \subset G_{k+j-1} \forall k \ (8.3)
\]
Now note that
\[
[G_i,G_j] = [G_i,[G_{j-1},G]] = [[G,G_{j-1}],G_i] \ (8.4)
\]
by proposition 8.2 and formula (8.1) respectively. We also have
\[
[[G_{j-1},G_i],G] = [[G_i,G_{j-1},G] \subset [G_{i+j-1},G] = G_{i+j} \ (8.5)
\]
by formula (8.1), (8.3), and proposition (8.2) respectively, and
\[
[[G_i,G],G_{j-1}] = [G_{i+1},G_{j-1}] = G_{i+j} \ (8.6)
\]
by proposition 8.2 and formula (8.3).\\
Given $x \in G$, $y \in G_{j-1}$ and $z \in G_i$, we therefore have
\[
[x,y,z] &= (([y^{-1},z^{-1},x]^z[z,x^{-1},y^{-1}]^x)^{-1})^y\\
&\subset G_{i+j}
\]
by formula (8.2) and formula (8.5) plus (8.6) respectively. The proposition follows from (8.4) and Lemma (8.1).
\end{proof}
\end{prop}

\begin{defi}
Given a group $G$, the \emph{upper central series} $\{1\} = Z_0(G) < Z_1(G) < Z_2(G)<...$ is defined recursively by setting $Z_{i+1} (G)$ so that $Z_{i+1} (G) / Z_i (G)$ is the centre of $G/Z_i (G)$. Note that each $Z_i(G)$ is normal by induction, since the centre of any group is normal.
\end{defi}

\begin{prop} (8.5)\\
Let $G=H_1 > H_2>...>H_{r+1} = \{1\}$ be a finite central series for $G$ (so $G$ is nilpotent). Then we have $H_i \supset G_i$ $\forall i = 1,...,r+1$, and $H_{r+1-i} \subset Z_i(a)$ $\forall i = 0,...,r$.\\
This justifies the names \emph{upper} and {lower} central series: $H_1>H_2>...>H_r>H_{r+1}$ and $Z_{r-i+1}(G) \supset H_i \supset G_i$ for each $i$.
\end{prop}

\begin{coro}
If $G$ is $s$-step nilpotent, then both the upper and lower central series have length 5.1.
\end{coro}

\begin{proof} (of 8.5)\\
$H_1 \supset G_1$ by definition, so we may assume $i>1$, and then we have
\[
H_i &\supset [H_{i-1},G]\ (\text{central series})\\
&\supset [G_{i-1},G]\ (\text{induction})\\
&=G_i \ (\text{prop 8.2})
\]
We also have $Z_0(G) > H_{r+1}$ by definition, so we may assume $i>0$ and, by induction, that $H_{r+2-i} \subset Z_{i-1}(G)$. But then we have
\[
G/Z_{i-1}(G) = \frac{G/H_{r+2-i}}{Z_{i-1}(G)/H_{r+2-i}}
\]
Because $(H_j)$ is a central series, $H_{r+1-i}/H_{r+2-i}$ is central in $G/H_{r+2-i}$, so its image in $G/Z_{i-1}(G)$ in the above quotient is central. But the centre of $G/Z_{i-1}(G)$ is $Z_i(G)/Z_{i-1}(G)$, so $H_{r+1-i} \subset Z_i(G)$ as required.
\end{proof}

These results say:\\
$\bullet$ $G$ is nilpotent of step $\leq s$ $\iff G_{s+1} = \{1\} \iff Z_s(G) = G$;\\
$\bullet$ If $G=\bra s \ket $, we can verify this just by checking that $[t_1,...,t_{s+1}] = 1$ $\forall t_i \in s$;\\
$\bullet$ If $G$ is nilpotent of step $\leq s$, then any commutator like $[[[g_1,g_2],g_3],[g_t,g_s]]$ with more than $s$ variables is trivial.

\newpage

\section{Torsion-free approximate groups -- an overview}

---Lecture 9---

Recall form lecture 7 that if $x_1,...,x_n \in G$ and $L_1,...,L_r \geq 0$, then the \emph{nonabelian progression} $P(x;L)$ consists of those elements of $G$ expressible as products of $x_i^{\pm 1}$ in which each $x_i$ and $x_i^{-1}$ appear at most $l_i$ times between them.

\begin{defi} (nilprogession)\\
If $\bra x_1,...,x_r\ket$ is $s$-step nilpotent then $P(x;L)$ is called a \emph{nilprogerssion} of rank $r$ and step $s$. In this case, we'll write $P_{nil}(x;L)$ instead of $P(x;L)$.
\end{defi}

\begin{prop} (9.1) (non-examinable)\\
Given $r,s \in \N, \exists \lambda = \lambda_{r,s}$ s.t. if $x_1,...,x_r$ generate an $s$-step nilpotent group and $L_1,...,L_r \geq \lambda_{r,s}$, then $P_{nil}(x;L)$ is an $O_{r,s}(1)$-approximate group (implied constant can depend on $r$ and $s$, but nothing else).\\
We won't have time to prove this, but we'll do a special case on sheet 2, where we'll also see that it's necessary to have $L_i \geq \lambda_{r,s}$.
\end{prop}

As in the abelian setting, it turns out that these are essentially the only examples of finite nilpotent approximate groups, apart from genuine subgroups.

\begin{thm} (9.2)\\
Let $G$ be $s$-step nilpotent, $A \subset G$ a finite $K$-approximate group. Then there exists $H \triangleleft \bra A \ket$ and a nilprogression $P_{nil}$ of rank $\leq K^{O_s(1)}$ s.t.$A \subset HP_{nil} \subset A^{K^{O_s(1)}}$. In particular, $|HP_{nil}| \leq \exp(K^{O_s(1)})|A|$.
\begin{rem}
If $K<2$ then this theorem is trivial. For $K \geq 2$, we have $O(K^{O(1)}) \subseteq K^{O(1)}$, i.e. multiplicative constants can be absorbed into exponents.\\
We won't have time to prove this in full, but in the next few lectures we'll prove some special cases that contain most of the main ideas. We'll start with the case where $G$ is torsion-free, where theorem 9.2 is originally due to Brenillard and Green, although we'll give a different proof.
\end{rem}
\end{thm}

We start with the following weakened version:

\begin{thm} (9.3)\\
Let $G$ be torsion-free $s$-step nilpotent, $A \subset G$ a finite $K$-approximate group. Then there exists ordered progression $P_{ord}$ of rank $\leq K^{O_s(1)}$ s.t. $A \subset P_{ord} \subset A^{K^{O_s(1)}}$.\\
The basic idea is to write $A$ as a product of approximate groups of step $<s$, and then apply induction to reduce to the step-1 case, aka the abelian case, where we can apply the Freiman-Green-Rusza theorem.
\end{thm}

\begin{prop} (9.4)\\
Let $G$ be \emph{torsion-free} $s$-step nilpotent, $A \subset G$ a finite $K$-approximate group. Then $\exists k \leq K^{O(1)}$ and $K^{O(1)}$-approximate subgroups $A_1,...,A_k \subset A^{O(1)}$ s.t. $A \subset A_1...A_k$, and $\bra A_i \ket$ is of step $<s$ $\forall i$.
\end{prop}

\begin{proof} (of 9.3)\\
An easy induction gives $K^{O_s(1)}$-approximate groups $B_1,...,B_m \subset A^{O_s(1)}$ with $m \leq K^{O_s(1)}$. FGR then gives abelian progressions, in particular, ordered progressions $P_1,...,P_m$, each of rank $\leq K^{O_s(1)}$, s.t. $B_i \subset P_i \subset B_i^{K^{O_s(1)}}$, and hence $A \subset P_1...P_m \subset A^{K^{O_s(1)}}$. $P_1...P_m$ is an ordered progression of rank $\leq mK^{O_s(1)} \leq K^{O_s(1)}$, so done.
\end{proof}

Recall that in proving $FGR$, we wanted \emph{$A \subset$ small progression}, but we first proved \emph{$A^c \supset$ large progression}. We then use Chang's covering argument to swap things around to get what we wanted. We'll use a similar result approach here, starting with the following.

\begin{prop} (9.5)\\
Let $G$ be torsion-free nilpotent, $A \subset G$ finite $K$-approximate group. Then $\exists r \leq K^{O(1)}$ and $K^{O(1)}$-approximate groups $A_0,A_1,...,A_r \subset A^{O(1)}$, each generating a group of step $<s$, s.t. $|A_0A_1...A_r| \geq \exp(-K^{O(1)})|A|$.
\end{prop}

Next time, we'll see that passing from prop 9.5 to prop 9.4 is very similar to the Chang covering part of the proof of FGR.

In proving prop 9.5, we actually use the preliminary version of FGR in which $A$ which $A^c$ contains a large progression. As we noted in that proof, combining Props 5.2 and 5.3 and Lemma 5.4 gives the following result.

\begin{thm} (9.6, Green-Rusza)\\
Let $G$ be abelian, and $A \subset G$ be a finite $K$-approximate group. Then $\exists H < G$ and $x_1,...,x_r \in G$ with $r \leq K^{O(1)}$, and $l_1,...,l_r \in \N$ s.t.
\[
HP(x;L) \subset A^4
\]
and
\[
|HP(x;L)| \geq \exp(-K^{O(1)})|A|
\]
\end{thm}

We'll apply this to prop 9.5 via the following result.

\begin{prop} (9.7)\\
Let $G$ be $s$-step nilpotent and $A \subset G$ a finite $K$-approximate group. Write $\pi:G \to G/[G,G]$ for the quotient homomorphism. Noting that $G/[G,G]$ us abelian and that $\pi(A)$ is a $K$-approximate group, let $H < G/[G,G]$ and $x_1,...,x_r \in G/[G,G]$ be as coming from applying theorem 9.6 to $\pi(A)$. Then
\[
\left|(A^{24} \cap \pi^{-1}(H)) \prod_{i =1}^r(A^{24} \cap \pi^{-1}(\bra x_i\ket))\right| \geq \exp(-K^{O(1)})|A|
\]
\end{prop}

We'll prove prop 9.7 next time. For now, let's see how it implies prop 9.5. Prop 4.6 immediately tells us that $A^{24} \cap \pi^{-1}(H)$ and $A^{24} \cap \pi^{-1}(\bra x_i\ket)$ are $K^{O(1)}$-approximate groups.

It turns out that they also generate subgroups of step $<s$, at least when $G$ is torsion-free.

\begin{lemma} (9.8)\\
Let $G$ be $s$-step nilpotent, and write $\pi:G \to G/[G,G]$ as before. Then\\
(i) $\forall x \in G/[G,G]$, $\pi^{-1}(\bra x\ket)$ is of step $<s$;\\
(ii) If $H \leq G/[G,G]$ is a finite subgroup and $G$ is torison-free, then $\pi^{-1}(H)$ is of step $<s$.
\end{lemma}

\begin{lemma} (9.9)\\
Let $G$ be an arbitrary group. Then the simple commutator map $[,...,]:G^k \to G_k$ by $(x_1,...,x_k) \to [x_1,...,x_k]$ is a homomorphism in each variable modulo $G_{k+1}$2. Moreover, $[G,G]$ is contained in the kernel of each of thses homomorphisms.
\end{lemma}

Next time we'll prove that, use it to prove lemma 9.8 (which you may want to do yourselves in the mean time), and

\end{document}