凯莱在他的原始论文中使用了以下符号 矩阵理论回忆录 1
我尝试重现该符号，如下所示：
\documentclass{article}
\usepackage{amsmath,amssymb,graphicx}
\begin{document}
\scalebox{2.5}{\parbox{0.3\textwidth}{$$
\renewcommand\arraystretch{0}\setlength\arraycolsep{0pt}
(\mathrm{X}, \mathrm{Y}, \mathrm{Z}){=}\lower10pt\hbox{$\begin{array}{r@{}c@{,\mkern10mu}c @{,\mkern10mu}c@{,\mkern10mu}c l}
( &a &b & c & )\\
\lvert& a' &b' & c' &\rvert\\
\lvert& a'' & b'' & c'' & \rvert\\
\end{array}$}\mkern-6mu(x,y,z),
$$}}
\end{document}
正如您所看到的，分隔符需要改进。有什么建议吗？
答案1
我的建议：
\documentclass{article}
\usepackage{amsmath,amssymb,graphicx}
\DeclareMathSymbol{A}{\mathalpha}{operators}{`A}
\DeclareMathSymbol{B}{\mathalpha}{operators}{`B}
\DeclareMathSymbol{C}{\mathalpha}{operators}{`C}
\DeclareMathSymbol{D}{\mathalpha}{operators}{`D}
\DeclareMathSymbol{E}{\mathalpha}{operators}{`E}
\DeclareMathSymbol{F}{\mathalpha}{operators}{`F}
\DeclareMathSymbol{G}{\mathalpha}{operators}{`G}
\DeclareMathSymbol{H}{\mathalpha}{operators}{`H}
\DeclareMathSymbol{I}{\mathalpha}{operators}{`I}
\DeclareMathSymbol{J}{\mathalpha}{operators}{`J}
\DeclareMathSymbol{K}{\mathalpha}{operators}{`K}
\DeclareMathSymbol{L}{\mathalpha}{operators}{`L}
\DeclareMathSymbol{M}{\mathalpha}{operators}{`M}
\DeclareMathSymbol{N}{\mathalpha}{operators}{`N}
\DeclareMathSymbol{O}{\mathalpha}{operators}{`O}
\DeclareMathSymbol{P}{\mathalpha}{operators}{`P}
\DeclareMathSymbol{Q}{\mathalpha}{operators}{`Q}
\DeclareMathSymbol{R}{\mathalpha}{operators}{`R}
\DeclareMathSymbol{S}{\mathalpha}{operators}{`S}
\DeclareMathSymbol{T}{\mathalpha}{operators}{`T}
\DeclareMathSymbol{U}{\mathalpha}{operators}{`U}
\DeclareMathSymbol{V}{\mathalpha}{operators}{`V}
\DeclareMathSymbol{W}{\mathalpha}{operators}{`W}
\DeclareMathSymbol{X}{\mathalpha}{operators}{`X}
\DeclareMathSymbol{Y}{\mathalpha}{operators}{`Y}
\DeclareMathSymbol{Z}{\mathalpha}{operators}{`Z}
\newsavebox{\cayleybox}
\newlength{\cayleyheight}
\newenvironment{cayleymatrix}[1]
{\begin{lrbox}{\cayleybox}
$\begin{array}{@{\enspace}l*{\numexpr#1-1}{@{{,}\enspace}l}@{\enspace}}}
{\end{array}$\end{lrbox}%
\setlength{\cayleyheight}{\dimexpr\ht\cayleybox+\dp\cayleybox}%
\kern-0.8pt
\vtop to \cayleyheight{\hbox{\vphantom{$\between$}}\leaders\vrule\vfill}%
\raisebox{\dimexpr\ht\strutbox-\ht\cayleybox}{\usebox\cayleybox}%
\vtop to \cayleyheight{\hbox{\vphantom{$\between$}}\leaders\vrule\vfill}%
\kern-0.8pt
}
\newcommand{\by}{\kern-0.4pt{\between}\kern-0.4pt}
\thinmuskip=6mu
\medmuskip=0mu
\thickmuskip=0mu
\begin{document}
\[ % page 17
(X,Y,Z)=
(\begin{cayleymatrix}{3}
a & b & c \\
a' & b' & c' \\
a'' & b'' & c''
\end{cayleymatrix}
\by x,y,z)
\]
\[ % page 32
(\begin{cayleymatrix}{2} a & c \\ b & d \end{cayleymatrix} \by
\begin{cayleymatrix}{2} a & b \\ c & d \end{cayleymatrix} \by
\begin{cayleymatrix}{2} a & c \\ b & d \end{cayleymatrix} \by
\begin{cayleymatrix}{2} a & b \\ c & d \end{cayleymatrix} )
=
\bigl(
(\begin{cayleymatrix}{2} a & c \\ b & d \end{cayleymatrix}\by
\begin{cayleymatrix}{2} a & b \\ c & d \end{cayleymatrix})
\bigr)^2
\]
\end{document}
答案2
钛钾Z 提案。类似地Andrew 的回答很好您可以专注于输入矩阵。
\documentclass{article}
\usepackage{tikz}
\makeatletter% https://tex.stackexchange.com/a/85531/121799
\long\def\ifnodedefined#1#2#3{%
\@ifundefined{pgf@sh@ns@#1}{#3}{#2}%
}
\makeatother
\newcounter{CM}
\usetikzlibrary{matrix}
\newcommand{\CayleyMatrix}[2][]{\stepcounter{CM}%
\begin{tikzpicture}[baseline=(mat-\number\value{CM}-1-1.base),inner sep=2pt,#1]
\matrix[matrix of math nodes,ampersand replacement=\&] (mat-\number\value{CM})
{ #2
};
\ifnodedefined{mat-\number\value{CM}-2-1}{%
\draw[semithick] (mat-\number\value{CM}.south west) -- (mat-\number\value{CM}.south west|-mat-\number\value{CM}-1-1.south) to[out=135,in=-135]
(mat-\number\value{CM}.south west|-mat-\number\value{CM}-1-1.north)
(mat-\number\value{CM}.south east) -- (mat-\number\value{CM}.south east|-mat-\number\value{CM}-1-1.south) to[out=45,in=-45]
(mat-\number\value{CM}.south east|-mat-\number\value{CM}-1-1.north) ;}{
\draw[semithick] (mat-\number\value{CM}.south west|-mat-\number\value{CM}-1-1.south) to[out=135,in=-135]
(mat-\number\value{CM}.south west|-mat-\number\value{CM}-1-1.north)
(mat-\number\value{CM}.south east|-mat-\number\value{CM}-1-1.south) to[out=45,in=-45]
(mat-\number\value{CM}.south east|-mat-\number\value{CM}-1-1.north) ;
}
\end{tikzpicture}}
\begin{document}
\[\CayleyMatrix{
X \&Y \& Z \\
}=\CayleyMatrix{
a \&b \& c \\
a' \&b' \& c' \\
a'' \& b'' \& c'' \\
}\!\!\CayleyMatrix{
x \&y \& z \\
}\]
\end{document}
答案3
稍微扩展一下 Bernard 的想法：
...但还没有达到 egreg 的优秀解决方案的标准:)
\documentclass{article}
\usepackage{array}
\newcounter{cayley}
\newcolumntype{L}{>{\stepcounter{cayley}%
\ifnum\value{cayley}=1\raisebox{-0.4ex}{(}\kern0.5ex\else\vline\fi}c@{}}
\newcolumntype{R}{@{}c<{\space\ifnum\value{cayley}=1\kern0.5ex\raisebox{-0.4ex}{)}\else\vline\fi}}
\newenvironment{cayley}[1][3]{\setcounter{cayley}{0}\array{L*{#1}{c}R}}{\endarray}
\begin{document}
\[
\begin{cayley}
&a & b & c &\\
&a' & b' & c' &\\
&a''& b''& c''&\\
\end{cayley}
\qquad
\begin{cayley}[4]
&a & b & c & d &\\
&a' & b' & c' & d' &\\
&a''& b''& c''& d''&\\
\end{cayley}
\]
\end{document}
答案4
以下是一个近似值，包含基本命令：
\documentclass{article}
\usepackage{amsmath,amssymb,graphicx}
\begin{document}
\[ \begin{array}{c@{}ccc@{}c}
(\mkern 4.2mu{} & a &b & c &\mkern 4.2mu)\\[-0.31ex]
\vline & a' &b' & c' & \vline \\
\vline & a'' &b'' & c''& \vline
\end{array} \]%
\end{document}
