# why is the qed symbol a line or two below when using enumerate in the proofs?

    \documentclass[11pt, a4paper]{report}
% \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, amsthm}
\usepackage{color}
% \usepackage{lipsum} % only for demos
\usepackage{array}
\usepackage{setspace}% if you must (for double spacing thesis)
\usepackage{fancyhdr}
\usepackage{enumitem}
\usepackage{tikz}
\usepackage{parskip}

%\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}{-0.1cm}\setlength{\evensidemargin}{-0.1cm}

% 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

\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{\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{\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{\cross}{\mathbin{\tikz [x=1.4ex,y=1.4ex,line width=.075ex] \draw (0,0) -- (1,1) (0,1) -- (1,0);}}%
% \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

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

%\input{tcilatex}

\makeatletter
\renewenvironment{proof}[1][\normalfont\bfseries\proofname]{\par
\pushQED{\qed}%
\normalfont \topsep6\p@\@plus6\p@\relax
\trivlist
\item\relax
{\itshape
}{%
\popQED\endtrivlist\@endpefalse
}

\def\thm@space@setup{%
\thm@preskip=1.25\parskip \thm@postskip=0pt
}
\makeatother

\begin{document}
\newpage
\pagestyle{fancy}
\fancyhf{}
\title{Introduction to Representation theory}
\maketitle
\begin{corollary}
Let $h$ be a linear mapping of $V_1$ into $V_2$. Let
$h^0 = \frac{1}{g} \sum_{t\in G} (\rho_t^2)^{-1} h \rho_t^1$
Then
\begin{enumerate}[label=(\roman*)]
\item If $\rho^1$ and $\rho^2$ are not isomorphic then we have $h^0=0$
\item If $V_1 = V_2$ and $\rho^1 = \rho^2$, $h^0$ is a scalar multiple of the identity, the scalar being $\frac{1}{n} Tr(h)$ where $n = \dim(V_1)$
\end{enumerate}
\end{corollary}

\begin{proof}
\begin{enumerate}[label=(\roman*)]
\item For any $s$, we have
$(\rho_{s}^2)^{-1} h^0 \rho_s^1 = \frac{1}{g} \sum_{t \in G}(\rho_{s}^2)^{-1}(\rho_{t}^2)^{-1} h \rho_t^1 \rho_s^1 = \frac{1}{g} \sum_{t \in G}(\rho_{ts}^2)^{-1}h \rho_{ts}^1 = h^0$
Applying part $(i)$ of Schur's lemma to $f=h^0$, we see that in case $(i)$ that $h^0 = 0$.
\item Now by part $(ii)$ of Schur's lemma we have $h^0 = \lambda I$. Let us take the trace of both sides
$Tr(h^0) = \frac{1}{g} \sum_{t \in G} Tr((\rho_{t}^1)^{-1}h \rho_{t}^1)=Tr(h)$
We also know that $Tr(\lambda I) = n \cdot \lambda$ hence $\lambda = \frac{1}{n} Tr(h)$
\end{enumerate}
\end{proof}
\end{document}

• all I needed to do is add \qedhere. – Maths Jan 3 at 15:36
• as I said in the comment that you post above % it would be better to use amsthm package for the theorem definitions % That defines a more extensive proof environment – David Carlisle Jan 3 at 15:36
• @DavidCarlisle yes I switched to amsthm package – Maths Jan 3 at 15:37
• I would start by deleting all of it and just put things back if you need them. amsthm package defines a proof environment that you have over-ridden, there is no point in having things like % This discards its argument, is that intended? % \U{wibble} is same as \U{zzzzz} \providecommand{\U}[1]{\protect\rule{.1in}{.1in}}  It does nothing useful then you keep my documentation that it does nothing useful, better simply to delete it. – David Carlisle Jan 3 at 15:43
• no that's not possible, it is something you should do, almost none of the environments and macros defined in that preamble are used in this example, so they could be deleted and this example would not change. Only you can say if that would have an effect on your real document. – David Carlisle Jan 3 at 16:01