Diagram displayed as an equation
If you draw the diagram with TikZ, you can just but the tikzpicture
into the equation
environment. I believe that the same thing works for xy.
I guess the real question you are asking is how to draw a commutative diagram. Here is one way to do it with TikZ.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{matrix}
\begin{document}
\begin{equation}
\begin{tikzpicture}[baseline=-0.8ex]
\matrix (m) [
matrix of math nodes,
row sep=2em,
column sep=4em,
text height=1.5ex, text depth=0.25ex
] {
x & x' \\
\hat x & \hat x' \\
};
\path[->]
(m-1-1) edge node[above] {$L$} (m-1-2)
(m-2-1) edge node[left] {$T$} (m-1-1)
(m-2-2) edge node[right] {$T$} (m-1-2)
(m-2-1) edge node[below] {$R$} (m-2-2);
\end{tikzpicture}
\end{equation}
\end{document}
Let's dissect this a bit:
- To draw a picture using TikZ, we use the
tikzpicture
environment. This environment can be used in math mode without problems. - In the picture we first define a matrix with the vertices of the commutative diagram (using
\matrix
). Then we draw the arrows and label them (using\path
). - For the matrix, we give a name (
m
) and several options (the part in square brackets). Then we define the matrix like in the math command of that name (except that it is important to finish the last line with\\
). The options do the following things:matrix of math nodes
tells TikZ to interpret the code in the actual matrix in math mode (without it, TikZ would want to get drawing commands in the matrix).row sep
andcolumn sep
set how much the the rows/columns should be spaced apart.text height
andtext depth
set just that. This is useful for TikZ to figure out where the center of each cell should be (otherwise it would just center them in the apparent center, which is lower forx
than forx'
).
- Normally one uses
\draw
to draw things in TikZ, but in this case\path
works better.- The
->
tells TikZ to draw arrows instead of just normal lines. (m-i-j)
accesses the cells of the matrix (labeled like in mathematics).- A line is inserted with
(start position) edge (end position)
. - We want to add labels to the arrow. For this we use the
node
syntax. It inserts a node (=label) in the middle of the edge (or above, below,... it, depending on the options) with the text in curly brackets in it.
- The
- Every drawing command must be finished with a semicolon
;
. By default TeX puts the image on top of the baseline. We don't want that, because then the equation counter “(1)” wouldn't be centered. We use the
baseline
option to tell TikZ where the baseline of the surrounding text should be with respect to its internal coordinate system. As we didn't specify the position of the matrix, it's center is at (0,0) in TikZ's internal coordinate system. Sobaseline=0
would center the matrix around the baseline of the text/equation. This is better, but still not quite what we want, as we want the center to be on the centerline, i.e. on the height of the middle of the 1. So we move it up by 0.8ex, which corresponds to telling TikZ that the baseline should be at 0.8ex below 0.As egreg points out, for the centering one could also use
gathered
fromamsmath
instead of baseline shifting.
For more information on TikZ, see its manual. A good introduction to producing commutative diagrams with TikZ is ”Commutative Diagrams using TikZ”.
With Xy-pic it's quite easy
\usepackage{amsmath}
\usepackage[all,cmtip]{xypic}
\xyoption{pdf} % for pdflatex
...
\begin{document}
...
\begin{equation}\label{commdiag}
\begin{gathered}
\xymatrix{
\mathbf{x} \ar[r]^{\mathbf{L}} & \mathbf{x}' \\
\hat{\mathbf{x}} \ar[r]_{\mathbf{R}} \ar[u]^{\mathbf{T}} &
\hat{\mathbf{x}}' \ar[u]_{\mathbf{T}}
}
\end{gathered}
\end{equation}
The gathered
environment is used for centering correctly the equation number. If you really need longer horizontal arrows, write something like \xymatrix@C+12pt
instead of \xymatrix
(the +12pt
is the extra length).
I'd like to share with you a commutative diagram made with eplain:
\input eplain
\noindent Such a transformation can be represented by the commuting diagram
$$
\varrowlength=1cm % Vertical arrow length
\harrowlength=3cm % Horizontal arrow length
\commdiag{
x & \mapright^L & x' \cr
\mapup\lft T & & \mapup\rt T \cr
\hat x & \mapright_R & \hat x' \cr
}
\eqno(1.6.4)
$$
It must be pointed out that the concept of conjucation or matrices is of \dots
\bye