1

I tried to use package paracol because I want to put similar theorems near one another, so reader can compare them. But I put attention that original text grow up from 290 to 320. Looking what happens I find small example

\documentclass{amsart}
\scrollmode
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\usepackage{xcolor}
\usepackage{framed}
\usepackage{paracol}
\begin{document}
\title{Test File}

\begin{abstract}
In this paper I test shaded environment.
\end{abstract}
\maketitle


\newenvironment{Shaded}{%
\def\FrameCommand{\fboxsep3pt \colorbox{shadecolor}}%
 \MakeFramed {\FrameRestore}}%
{\endMakeFramed}


\newenvironment{framedPage}[1][\hsize]
{\MakeFramed{\hsize#1\advance\hsize-\width \FrameRestore}}%
{\endMakeFramed}


\newenvironment{shadedPage}[1][\hsize]
{
\def\FrameCommand{\colorbox{shadecolor}}%
\MakeFramed{ \FrameRestore}}%
{\endMakeFramed}


\setlength{\columnseprule}{0.5pt}


\begin{paracol}{2}
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
Therefore, automorphisms of left


It remains to  prove that
the kernel of inefficiency consists only of identity.
Identity transformation
satisfies to equation
[
v^i=v^ja_i^j
]
Choosing values of coordinates as
$v^i=\delta^i_k$
where we selected $k$ we get
\begin{equation}
\label{identity col}
\delta^i_k=\delta^j_ka^i_j
\end{equation}
From \eqref{identity col} it follows
[
\delta^i_k=a^i_k
]
Since $k$ is arbitrary, we get the conclusion $a=\delta$.
\switchcolumn%
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $a*b$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
Therefore, automorphisms of left 
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
representation.
Therefore, automorphisms of left


It remains to  prove that
the kernel of inefficiency consists only of identity.
Identity transformation
satisfies to equation
[
v_i=v_ja^j_i
]
Choosing values of coordinates as
$v_i=\delta^k_i$
where we selected $k$ we get
\begin{equation}
\label{identity row}
\delta^k_i=\delta^k_ja^j_i
\end{equation}
From \eqref{identity row} it follows
[
\delta^k_i=a^k_i
]
Since $k$ is arbitrary, we get the conclusion $a=\delta$.
\end{paracol}


\end{document}

You can see that on second page there few empty lines.

Aleks Kleyn
  • 158

1 Answers1

2

There are exactly equal if you use the same content for both columns.

z

\documentclass{amsart}
\scrollmode
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\usepackage{xcolor}
\usepackage{framed}
\usepackage{paracol}
\begin{document}
\title{Test File}

\begin{abstract}
In this paper I test shaded environment.
\end{abstract}
\maketitle


\newenvironment{Shaded}{%
\def\FrameCommand{\fboxsep3pt \colorbox{shadecolor}}%
 \MakeFramed {\FrameRestore}}%
{\endMakeFramed}


\newenvironment{framedPage}[1][\hsize]
{\MakeFramed{\hsize#1\advance\hsize-\width \FrameRestore}}%
{\endMakeFramed}


\newenvironment{shadedPage}[1][\hsize]
{
\def\FrameCommand{\colorbox{shadecolor}}%
\MakeFramed{ \FrameRestore}}%
{\endMakeFramed}


\setlength{\columnseprule}{0.5pt}


\begin{paracol}{2}
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear GL-representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left It remains to  prove that
the kernel of inefficiency consists only of identity.
Identity transformation satisfies to equation
[v^i=v^ja_i^j]
Choosing values of coordinates as
$v^i=\delta^i_k$
where we selected $k$ we get
\begin{equation}
\label{identity col}
\delta^i_k=\delta^j_ka^i_j
\end{equation}
From \eqref{identity col} it follows
[\delta^i_k=a^i_k]
Since $k$ is arbitrary, we get the conclusion $a=\delta$.


\switchcolumn%
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear GL-representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left It remains to  prove that
the kernel of inefficiency consists only of identity.
Identity transformation satisfies to equation
[v^i=v^ja_i^j]
Choosing values of coordinates as
$v^i=\delta^i_k$
where we selected $k$ we get
\begin{equation}
\label{identity col}
\delta^i_k=\delta^j_ka^i_j
\end{equation}
From \eqref{identity col} it follows
[\delta^i_k=a^i_k]
Since $k$ is arbitrary, we get the conclusion $a=\delta$.
\end{paracol}

In order not to be confused with the text of two similar theorems one after the other, I suggest defining new commands, for example \firsttheorem and \secondtheorem each one holding the content of its theorem.

Then use

\begin{paracol}{2}
  \firsttheorem 
\switchcolumn%
  \secondtheorem
\end{paracol}

As in this code.

\documentclass{amsart}
\scrollmode
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\usepackage{xcolor}
\usepackage{framed}
\usepackage{paracol}
\begin{document}
\title{Test File}

\begin{abstract}
In this paper I test shaded environment.
\end{abstract}
\maketitle


\newenvironment{Shaded}{%
\def\FrameCommand{\fboxsep3pt \colorbox{shadecolor}}%
\MakeFramed {\FrameRestore}}%
{\endMakeFramed}


\newenvironment{framedPage}[1][\hsize]
{\MakeFramed{\hsize#1\advance\hsize-\width \FrameRestore}}%
{\endMakeFramed}


\newenvironment{shadedPage}[1][\hsize]
{
\def\FrameCommand{\colorbox{shadecolor}}%
\MakeFramed{ \FrameRestore}}%
{\endMakeFramed}


\setlength{\columnseprule}{0.5pt}


\newcommand{\firsttheorem}{% first theorem
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear GL-representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left It remains to  prove that
the kernel of inefficiency consists only of identity.
Identity transformation satisfies to equation
[v^i=v^ja_i^j]
Choosing values of coordinates as
$v^i=\delta^i_k$
where we selected $k$ we get
\begin{equation}
\label{identity col}
\delta^i_k=\delta^j_ka^i_j
\end{equation}
From \eqref{identity col} it follows
[\delta^i_k=a^i_k]
    Since $k$ is arbitrary, we get the conclusion $a=\delta$.

    }


\newcommand{\secondtheorem}{% a similar theorem
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear
GL-representation.
According to the theorem
the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of rows form
a right-side linear GL-representation.
According to the theorem the product of automorphisms $a$ and $b$
has matrix $ab$.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left $A$-vector space of columns form
a right-side linear representation.
Therefore, automorphisms of left It remains to  prove that
the kernel of inefficiency consists only of identity.
Identity transformation satisfies to equation
[v^i=v^ja_i^j]
Choosing values of coordinates as
$v^i=\delta^i_k$
where we selected $k$ we get
\begin{equation}
\label{identity col}
\delta^i_k=\delta^j_ka^i_j
\end{equation}
From \eqref{identity col} it follows
[\delta^i_k=a^i_k]
Since $k$ is arbitrary, we get the conclusion $a=\delta$.

}


\begin{paracol}{2}
\firsttheorem

\switchcolumn%
\secondtheorem
\end{paracol}


\end{document}

UPDATE

The gap in the second page is originated by the combination of paracol and amsart.

The gap disappears using a standard class as article and add \usepackage{amsmath, amsthm} to make the amslatex commands available.

See also Article vs amsart

d

Simon Dispa
  • 39,141