我有以下内容但看起来很糟糕：
\documentclass[11pt, a4paper]{article}
\usepackage{eurosym}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{bm}
\usepackage{amsfonts, graphicx, verbatim, amsmath,amssymb}
\usepackage{color}
\usepackage{lipsum}
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\setlength{\textheight}{22cm}\setlength{\textwidth}{16cm}
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\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}
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\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{corol}[theorem]{Corollary}
\newtheorem{Fact}[theorem]{Fact}
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\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{Definition}[theorem]{Definition}
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\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}}
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\parindent 0pt
\setlength{\parindent}{0pt}
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\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}
我该怎么做才能删除最后一行的间距？
答案1
您不应忽略以下警告
Underfull \hbox (badness 10000) in paragraph at lines 168--169
10000 是最大限度TeX 分配给其输出的不良程度
我可能错过了一些东西，但是，正如你问的......
\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}}
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%Better to use \mathrm than \text so it uses the same font in all contexts
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% \mathcal{K} not {\cal K} %\cal not defined by default since 1993
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% \mathrm{Be} not {\rm BeK} %\cal not defined by default since 1993
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%\newcommand{\Bin}{{\rm Bin}}
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%\newcommand{\Po}{{\rm Pois}}
%\newcommand{\Bi}{{\cal B}}
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\newcommand{\xxi}{\bar{\xi}}
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% use amsmath package (that you have loaded) align environment, not eqnarray
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%\newcommand{\bbb}{\begin{eqnarray}}
%\newcommand{\eee}{\end{eqnarray}}
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% \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}