1

enter image description hereI have the following but it looks awful:

    \documentclass[11pt, a4paper]{article}
\usepackage{eurosym}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{bm}
\usepackage{amsfonts, graphicx, verbatim, amsmath,amssymb}
\usepackage{color}
\usepackage{lipsum}
\usepackage{array}
\usepackage{setspace}

\setcounter{MaxMatrixCols}{10}
%TCIDATA{OutputFilter=Latex.dll}
%TCIDATA{Version=5.50.0.2953}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{LastRevised=Sunday, November 26, 2017 16:01:29}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}

\setlength{\textheight}{22cm}\setlength{\textwidth}{16cm}
\setlength{\topmargin}{-1.5cm}
\setlength{\oddsidemargin}{-0.5cm}\setlength{\evensidemargin}{-0.5cm}
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
\setlength{\textheight}{24cm}\setlength{\textwidth}{16.5cm}
\setlength{\topmargin}{-1.5cm}
\setlength{\oddsidemargin}{0.5cm}\setlength{\evensidemargin}{0.5cm}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{Theorem}[theorem]{Theorem}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{corol}[theorem]{Corollary}
\newtheorem{Fact}[theorem]{Fact}
\newtheorem{Corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{Definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{Lemma}[theorem]{Lemma}
\newtheorem{fact}[theorem]{Fact}
\newtheorem{lma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{prop}[theorem]{Proposition}
\newtheorem{Property}[theorem]{Property}
\newtheorem{property}[theorem]{Property}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{Comment}[theorem]{Comment}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\newcommand{\ve}{\varepsilon}
\newcommand{\cvgpr}{\xrightarrow{\text{\upshape\tiny P}}}
\newcommand{\cvgdist}{\xrightarrow{\mathrm{d}}}
\newcommand{\G}{{\mathcal{G}}}
\newcommand{\Kx}{{\cal K}}
\newcommand{\tod}{\to^{\cal D}}
\newcommand{\ls}{\limsup_{n\to\infty}}
\newcommand{\rE}{\mathbb{E}}
\newcommand{\A}{{\mathcal{A}}}
\newcommand{\rP}{\mathbb{P}}
\newcommand{\p}{{\mathbb{P}}}
\newcommand{\Z}{{\mathbb{Z}}}
\newcommand{\Be}{{\rm Be}}
\newcommand{\re}{\mathrm{e}}
\newcommand{\ep}{\varepsilon}
\newcommand{\Bin}{{\rm Bin}}
\newcommand{\qand}{\quad\mbox{and}\quad}
\newcommand{\quso}{\quad\mbox{so}\quad}
\newcommand{\Nn}{{\bf N}}
\newcommand{\St}{\underline{\rm S}}
\newcommand{\Rt}{\underline{\rm R}}
\newcommand{\It}{\underline{\rm I}}
\newcommand{\one}{{\bf 1}}
\newcommand{\Ups}{{\Upsilon}}
\newcommand{\iu}{{i\mkern1mu}}
\newcommand{\II}{{\mathcal{I}}}
\newcommand{\Var}{{\rm Var}}
\newcommand{\var}{{\rm Var}}
\newcommand{\Cov}{{\rm cov}}
\newcommand{\cov}{{\rm cov}}
\newcommand{\corr}{{\rm corr}}
\newcommand{\lhs}{{\rm lhs}}
\newcommand{\rhs}{{\rm rhs}}
\newcommand{\ra}{\rightarrow}
\newcommand{\I}{{\mathbf 1}}
\newcommand{\R}{{\mathbb R}}
\newcommand{\N}{{\mathbb N}}
\newcommand{\LL}{{\mathbb L}}
\newcommand{\E}{{\mathbb{E}}}
\newcommand{\bin}{{\rm Bin}}
\newcommand{\Pois}{{\rm Pois}}
\newcommand{\Po}{{\rm Pois}}
\newcommand{\Bi}{{\cal B}}
\newcommand{\ri}{\mathrm{i}}
\newcommand{\rd}{\mathrm{d}}
\newcommand{\XXi}{\Xi_{k,m}^{(n)}}
\newcommand{\xxi}{\bar{\xi}}
\newcommand{\qedhere}{{\diamond}}
\newcommand{\eqdef}{\stackrel{\mathrm{def}}{=}}
\newcommand{\eqdist}{\stackrel{\mathrm{D}}{=}}
\newcommand{\braket}[2]{{\langle{#1|#2}\rangle}}
\newcommand{\independent}{\perp}
\newcommand{\bb}{\begin{eqnarray*}}
\newcommand{\ee}{\end{eqnarray*}}
\newcommand{\bbb}{\begin{eqnarray}}
\newcommand{\eee}{\end{eqnarray}}
\newcommand{\F}{{\mathcal{F}}}
\newcommand{\qed}{$\diamond$}
\parindent 0pt
\setlength{\parindent}{0pt}
\newcommand{\forceindent}{\leavevmode{\parindent=3em\indent}}
\usepackage{enumitem}

%\input{tcilatex}


\begin{document}

\section{Introduction}
\bigskip
\smallskip
\subsection{Definitions and prerequisites}
Let us recall some basic knowledge, we begin by giving some definitions. 
\medskip
\begin{definition} 
A group is a set $G$ together with a binary operation $*$ on $G$ satisfying
the following properties:
\begin{enumerate}[label=(G\arabic*),series=group]
\item Closure: $\forall x,y \in G, x * y \in G$.
\item Associativity: $\forall x,y, z \in G, (x * y) * z = x * (y * z)$.
\item Identity: There is an element $e \in G$ such that $e * x = x * e = x$ for all $x \in G$.
\item Inverses: For any $x \in G$ there is an element $y \in G$ such that $x * y = y * x = e$.
\end{enumerate}
\end{definition}

\begin{definition}
A group $G$ is called an abelian group if the following axiom is satisfied:
\begin{enumerate}[label=(G\arabic*),resume=group]
\item Commutativity: $\forall x,y \in G, x * y = y * x$.
\end{enumerate}
\end{definition}
\medskip
Let $V$ be a vector space over the field $\mathbb{C}$ (unless stated otherwise) of complex numbers and let $GL(V)$ be the group of \textit{isomorphisms} of $V$ onto itself. An element $\alpha$ of $GL(V)$ is, by definition, a linear mapping of $V$ into $V$ which has an inverse $\alpha^{-1}$; this inverse is linear. When $V$ has a finite basis $(e_i)$ of $n$ elements, each linear map $\alpha: V \rightarrow V$ is defined by a square matrix $(\alpha_{ij})$ of order $n$. The coefficients $\alpha_{ij}$ are complex numbers; they are obtained by expressing the images $\alpha(e_j)$ in terms of the basis $(e_i)$:
\smallskip
$$\alpha(e_j)=\sum_i \alpha_{ij}e_i$$
\medskip
Saying that $\alpha$ is an isomorphism is identical to saying that the determinant $det(\alpha) = det(\alpha_{ij})$ of $\alpha$ is non zero. The group $GL(V)$ - the general linear group on $V$ is thus the group of \textit{invertible (or non-singular) square matrices of order $n$}.\\
\begin{definition}Suppose now $G$ is a finite group, with identity element $1$ and with composition $(s,t) \rightarrow st$. A linear representation of $G$ in $V$ is a homomorphism $\rho$ from the group $G$ into the group $GL(V)$(i.e. $\rho:G \rightarrow GL(V)$). In other words, we associate with each elements $s \in G$ and element $\rho(s)$ of $GL(V)$ in such a way that we have the equality:
$$\rho(st)=\rho(s)\cdot\rho(t)\;\; for\;\; s,t \in G. $$
Often we will use $\rho_s$ as an alternate to $\rho(s)$. We notice the above formula implies the following:
$$\rho(1)=1, \;\; \rho(s^{-1})=\rho(s)^{-1}.$$
\end{definition}
\medskip\
\\When $\rho$ is given, we simply say, by abuse of language, $V$ is a \textit{representation} of $G$ (or formally $V$ is a \textit{representation space} of $G$). Let us impose the condition that we are only to consider representations of finite degree, that is, in \textit{finite-dimensional} vector spaces; and these will almost always be vector spaces over $\mathbb{C}$. Therefore, to avoid repetition, let us agree to use the term "representation" to mean representation of finite degree over $\mathbb{C}$, unless stated otherwise. This isn't a strict limitation as for almost all applications we are interested in finite number of elements $x_i$ of $V$ hence we can find a a \textit{subrepresentation}.\\

Suppose now that $G$ has finite dimension, and let $n$ be its dimension; we say also that $n$ is the degree of the representation. Let $(e_i)$ be a basis of V, and let $R_s$ be the matrix of $\rho_s$ with respect to this basis. We have:\\
$$det(R_s)\neq 0, \;\;\;\;\;\;\;\; R_{st} = R_s \cdot R_t \;\;\;\; if\; s,t \in G$$\\

Let $r_{ij}(s)$ denote the coefficients of the matrix $R_s$, the second formula becomes:
$$r_{ik}(st)= \sum_j r_{ij}(s)\cdot r_{jk}(t)$$

Conversely, given the invertible matrices $R_s = (r_{ij}(s))$ satisfying the preceding identities, there is a corresponding linear representation of $\rho$ of $G$ in $V$; this is what we mean by giving a representation in matrix form.

\begin{definition}
let $\rho$ and $\rho\,'$ be two representations of the same group $G$ in vector spaces $V$ and $V\,'$. These representations are said to be similar (or isomorphic) if there exits a linear homomorphism $\tau: V \rightarrow V\,'$ which "transforms" $\rho$ to $\rho\,'$, that is which satisfies the identity:\
$$\tau \circ \rho(s) = \rho(s)\,'\circ \tau \;\;\;\;\; \forall s \in G$$ 
\end{definition}

When $\rho$ and $\rho\,'$ are given in matrix form by $R_s$ and $R_s\,'$, this means that that there exists an invertible matrix $T$ such that:
\smallskip
$$T\cdot R_s = R_s\,' \cdot T \;\;\;\;\; \forall s \in G$$

\subsection{Basic examples}



\end{document}

what must I do to remove the spacing in the last line?

Leo
  • 815
  • 1
    while we wait for an example, note that $$, \rm , \cal etc should not appear in any latex document written this century (\rm and \cal have not been defined by default since 1993) – David Carlisle Dec 11 '18 at 12:40
  • 1
    @A.E All this manual messing with skips looks also fishy... -> this is also the sourc of your problem. The '\medskip` is placed incorrectly. – samcarter_is_at_topanswers.xyz Dec 11 '18 at 12:40
  • @A.E The trick is to let latex do the spacing for you and not to manually interefere. Indicate the end of a paragraph by leaving an empty line and latex will insert the correct spaces – samcarter_is_at_topanswers.xyz Dec 11 '18 at 12:44
  • @A.E If you want bigger gaps, adjust the parskip, see e.g. https://tex.stackexchange.com/questions/240751/how-to-change-separation-space-between-paragraphs – samcarter_is_at_topanswers.xyz Dec 11 '18 at 12:46
  • @A.E Sure, see https://tex.stackexchange.com/questions/464310/spacing-randomly-appeared-inbetween-a-sentence?noredirect=1#comment1168266_464310 – samcarter_is_at_topanswers.xyz Dec 11 '18 at 12:47
  • 1
    Start by removing all explicit vertical spaces. – egreg Dec 11 '18 at 13:45
  • really you should remove all the definitions between proof and \begin{document} and remove all the \\ and \...skip commands and \; commands in the document, LaTeX can make beautiful documents but if you instruct it to make documents with arbitrary inconsistent spacing then you will find that it does what you ask. – David Carlisle Dec 11 '18 at 14:04
  • 1
    The spacing in this document is so messed up that you can't get an idea of the normal spacing that tex would use. The spacing tex does use is almost always customisable you should only very rarely need to add space in a document. – David Carlisle Dec 11 '18 at 14:18
  • 1
    @A.E the spacing (and fonts) are just completely wrong in ;\; for\;\; s,t \in G. $$ you have set for in math italic, which is designed to make adjacent letters not look like a word, but like a product of variables, and for the same reason the spaces in the input are ignored. try \text{ for } – David Carlisle Dec 11 '18 at 14:20
  • @DavidCarlisle if it doesn't require too much work, could you please show me a better version? this will help me a lot – Leo Dec 11 '18 at 14:22
  • @A.E Maybe you are interested in having a look at this book: https://tex.stackexchange.com/a/160999/36296 – samcarter_is_at_topanswers.xyz Dec 11 '18 at 14:39

1 Answers1

4

You should not ignore warnings such as

Underfull \hbox (badness 10000) in paragraph at lines 168--169

10000 is the maximum level of badness that TeX assigns to its output

I probably missed some things but, as you asked...

\documentclass[11pt, a4paper]{article}
% \usepackage{eurosym}% you probably don't need this (most fonts have euro)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%% \usepackage{amsmath} % you load this below
%%\usepackage{amsfonts} % you load this below
\usepackage{bm}
\usepackage{amsfonts, graphicx, verbatim, amsmath,amssymb}
\usepackage{color}
% \usepackage{lipsum} % only for demos
\usepackage{array}
\usepackage{setspace}% if you must (for double spacing thesis)

%\setcounter{MaxMatrixCols}{10} 10 is teh default anyway
%TCIDATA{OutputFilter=Latex.dll}
%TCIDATA{Version=5.50.0.2953}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{LastRevised=Sunday, November 26, 2017 16:01:29}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}

% These are sort of OK, but better to use geometry package
% to set a consistent set of page dimensions
\setlength{\textheight}{22cm}\setlength{\textwidth}{16cm}
\setlength{\topmargin}{-1.5cm}
\setlength{\oddsidemargin}{-0.5cm}\setlength{\evensidemargin}{-0.5cm}
\setlength{\textheight}{24cm}\setlength{\textwidth}{16.5cm}
\setlength{\topmargin}{-1.5cm}
\setlength{\oddsidemargin}{0.5cm}\setlength{\evensidemargin}{0.5cm}


% This discards its argument, is that intended?
% \U{wibble}  is same as \U{zzzzz}
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}


\newtheorem{theorem}{Theorem}[section]
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{Theorem}[theorem]{Theorem}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{Fact}[theorem]{Fact}

% Why all these variant corollary forms?
\newtheorem{corollary}[theorem]{Corollary}
%\newtheorem{corol}[theorem]{Corollary}
%\newtheorem{Corollary}[theorem]{Corollary}

\newtheorem{criterion}[theorem]{Criterion}

% why variant definition forms?
\newtheorem{definition}[theorem]{Definition}
%\newtheorem{Definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
% as above
\newtheorem{lemma}[theorem]{Lemma}
%\newtheorem{Lemma}[theorem]{Lemma}
\newtheorem{fact}[theorem]{Fact}
\newtheorem{lma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
% as above
\newtheorem{proposition}[theorem]{Proposition}
%\newtheorem{prop}[theorem]{Proposition}
%as above
\newtheorem{Property}[theorem]{Property}
%\newtheorem{property}[theorem]{Property}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{Comment}[theorem]{Comment}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}

% it would be better to use amsthm package for the theorem definitions
% That defines a more extensive proof environment
\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\newcommand{\ve}{\varepsilon}

%Better to use \mathrm than \text so it uses the same font in all contexts
\newcommand{\cvgpr}{\xrightarrow{\text{\upshape\tiny P}}}
\newcommand{\cvgdist}{\xrightarrow{\mathrm{d}}}
\newcommand{\G}{{\mathcal{G}}}

% \mathcal{K} not {\cal K}  %\cal not defined by default since 1993
%\newcommand{\Kx}{{\cal K}}
%\newcommand{\tod}{\to^{\cal D}}
\newcommand{\ls}{\limsup_{n\to\infty}}
\newcommand{\rE}{\mathbb{E}}
\newcommand{\A}{{\mathcal{A}}}
\newcommand{\rP}{\mathbb{P}}
\newcommand{\p}{{\mathbb{P}}}
\newcommand{\Z}{{\mathbb{Z}}}

% \mathrm{Be} not {\rm BeK}  %\cal not defined by default since 1993
%\newcommand{\Be}{{\rm Be}}
\newcommand{\re}{\mathrm{e}}
\newcommand{\ep}{\varepsilon}
%\newcommand{\Bin}{{\rm Bin}}
\newcommand{\qand}{\quad\mbox{and}\quad}
\newcommand{\quso}{\quad\mbox{so}\quad}
%\newcommand{\Nn}{{\bf N}}
%\newcommand{\St}{\underline{\rm S}}
%\newcommand{\Rt}{\underline{\rm R}}
%\newcommand{\It}{\underline{\rm I}}
%\newcommand{\one}{{\bf 1}}
\newcommand{\Ups}{{\Upsilon}}
\newcommand{\iu}{{i\mkern1mu}}
\newcommand{\II}{{\mathcal{I}}}
%\newcommand{\Var}{{\rm Var}}
%\newcommand{\var}{{\rm Var}}
%\newcommand{\Cov}{{\rm cov}}
%\newcommand{\cov}{{\rm cov}}
%\newcommand{\corr}{{\rm corr}}
%\newcommand{\lhs}{{\rm lhs}}
%\newcommand{\rhs}{{\rm rhs}}
\newcommand{\ra}{\rightarrow}
\newcommand{\I}{{\mathbf 1}}
\newcommand{\R}{{\mathbb R}}
\newcommand{\N}{{\mathbb N}}
\newcommand{\LL}{{\mathbb L}}
\newcommand{\E}{{\mathbb{E}}}
%\newcommand{\bin}{{\rm Bin}}
%\newcommand{\Pois}{{\rm Pois}}
%\newcommand{\Po}{{\rm Pois}}
%\newcommand{\Bi}{{\cal B}}
\newcommand{\ri}{\mathrm{i}}
\newcommand{\rd}{\mathrm{d}}
\newcommand{\XXi}{\Xi_{k,m}^{(n)}}
\newcommand{\xxi}{\bar{\xi}}
\newcommand{\qedhere}{{\diamond}}
\newcommand{\eqdef}{\stackrel{\mathrm{def}}{=}}
\newcommand{\eqdist}{\stackrel{\mathrm{D}}{=}}
\newcommand{\braket}[2]{{\langle{#1|#2}\rangle}}
\newcommand{\independent}{\perp}

% use amsmath package (that you have loaded) align environment, not eqnarray
%\newcommand{\bb}{\begin{eqnarray*}}
%\newcommand{\ee}{\end{eqnarray*}}
%\newcommand{\bbb}{\begin{eqnarray}}
%\newcommand{\eee}{\end{eqnarray}}
\newcommand{\F}{{\mathcal{F}}}
\newcommand{\qed}{$\diamond$}

% \parindent 0pt % this is just low level version of following line
% \setlength{\parindent}{0pt}% 
% use parskip package (if you must) to stop indent and put vertical space betwen paragraphs
% although most documents lok better with traditional typesetting with indentation and no vertical space
\usepackage{parskip}

%\newcommand{\forceindent}{\leavevmode{\parindent=3em\indent}}%eek

\usepackage{enumitem}

%\input{tcilatex}


\begin{document}

\section{Introduction}
%\bigskip
%\smallskip
\subsection{Definitions and prerequisites}
Let us recall some basic knowledge, we begin by giving some definitions. 
%\medskip
\begin{definition} 
A group is a set $G$ together with a binary operation $*$ on $G$ satisfying
the following properties:
\begin{enumerate}[label=(G\arabic*),series=group]
\item Closure: $\forall x,y \in G, x * y \in G$.
\item Associativity: $\forall x,y, z \in G, (x * y) * z = x * (y * z)$.
\item Identity: There is an element $e \in G$ such that $e * x = x * e = x$ for all $x \in G$.
\item Inverses: For any $x \in G$ there is an element $y \in G$ such that $x * y = y * x = e$.
\end{enumerate}
\end{definition}

\begin{definition}
A group $G$ is called an abelian group if the following axiom is satisfied:
\begin{enumerate}[label=(G\arabic*),resume=group]
\item Commutativity: $\forall x,y \in G, x * y = y * x$.
\end{enumerate}
\end{definition}

Let $V$ be a vector space over the field $\mathbb{C}$ (unless stated
otherwise) of complex numbers and let $GL(V)$ be the group of
\textit{isomorphisms} of $V$ onto itself. An element $\alpha$ of
$GL(V)$ is, by definition, a linear mapping of $V$ into $V$ which has
an inverse $\alpha^{-1}$; this inverse is linear. When $V$ has a
finite basis $(e_i)$ of $n$ elements, each linear map
% colon
$\alpha\colon V \rightarrow V$ is defined by a square matrix
$(\alpha_{ij})$ of order $n$. The coefficients $\alpha_{ij}$ are
complex numbers; they are obtained by expressing the images
$\alpha(e_j)$ in terms of the basis $(e_i)$:
% $$ is not latex (and \medskip here will break tex's space insertion
\[
\alpha(e_j)=\sum_i \alpha_{ij}e_i
\]
Saying that $\alpha$ is an isomorphism is identical to saying that the determinant
% never use math italic for words use \det which is the right font, with operator spacing
$\det(\alpha) = \det(\alpha_{ij})$ of $\alpha$ is non zero. The group $GL(V)$
% en-dash not hyphen
 -- 
the general linear group on $V$ is thus the group of \textit{invertible (or non-singular) square matrices of order $n$}.
% almost never use \\ in text (and really never use it at an end of paragraph as here)
\begin{definition}Suppose now
  $G$ is a finite group, with identity element
  $1$ and with composition $(s,t) \rightarrow
  st$. A linear representation of $G$ in $V$ is a homomorphism
  $\rho$ from the group $G$ into the group $GL(V)$
  (i.e. $\rho\colon G  \rightarrow
  GL(V)$). In other words, we associate with each elements $s \in
  G$ and element $\rho(s)$ of
  $GL(V)$ in such a way that we have the equality:
\[\rho(st)=\rho(s)\cdot\rho(t) \text{ for } s,t \in G. \]
Often we will use $\rho_s$ as an alternate to $\rho(s)$.
We notice the above formula implies the following:
\[\rho(1)=1,\quad \rho(s^{-1})=\rho(s)^{-1}.\]
\end{definition}
% eek\medskip\
%\\

When $\rho$ is given, we simply say, by abuse of language,
$V$ is a \textit{representation} of $G$ (or formally
$V$ is a \textit{representation space} of
$G$). Let us impose the condition that we are only to consider
representations of finite degree, that is, in
\textit{finite-dimensional} vector spaces; and these will almost
always be vector spaces over
$\mathbb{C}$. Therefore, to avoid repetition, let us agree to use the
term 
% never use " in tex sourcde `` and ''
``representation'' to mean representation of finite degree over
$\mathbb{C}$, unless stated otherwise. This isn't a strict limitation
as for almost all applications we are interested in finite number of
elements $x_i$ of
$V$ hence we can find a a \textit{subrepresentation}.
%\\

Suppose now that $G$ has finite dimension, and let $n$ be its
dimension; we say also that $n$ is the degree of the
representation. Let $(e_i)$ be a basis of V, and let $R_s$ be the
matrix of $\rho_s$ with respect to this basis. We have:
%\\
\[
\det(R_s)\neq 0, \quad R_{st} = R_s \cdot R_t \quad \text{if }s,t \in G
\]

Let $r_{ij}(s)$ denote the coefficients of the matrix $R_s$, the second formula becomes:
\[r_{ik}(st)= \sum_j r_{ij}(s)\cdot r_{jk}(t)\]

Conversely, given the invertible matrices $R_s = (r_{ij}(s))$
satisfying the preceding identities, there is a corresponding linear
representation of $\rho$ of $G$ in $V$; this is what we mean by giving
a representation in matrix form.

\begin{definition}
  let $\rho$ and $\rho\,'$ be two representations of the same group
  $G$ in vector spaces $V$ and $V\,'$. These representations are said
  to be similar (or isomorphic) if there exits a linear homomorphism
  $\tau: V \rightarrow V\,'$ which ``transforms'' $\rho$ to $\rho\,'$,
  that is which satisfies the identity:% no\
\[\tau \circ \rho(s) = \rho(s)\,'\circ \tau \quad \forall s \in G\]
\end{definition}

When $\rho$ and $\rho\,'$ are given in matrix form by $R_s$ and
$R_s'$, this means that that there exists an invertible matrix $T$
such that:% \smallskip
\[T\cdot R_s = R_s' \cdot T \quad \forall s \in G\]

\subsection{Basic examples}



\end{document}
David Carlisle
  • 757,742