\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

Find all the fixed points, 2-cycles and find two 3-cycles for the
hyperbolic toral automorphism defined by the matrix
$A=\left(\begin{array}{cc} 1 & 2\\ 2 & 3 \end{array} \right)$.
\medskip

{\bf Answer}

$A=\left(\begin{array}{cc} 1 & 2\\2 & 3 \end{array} \right)$:
$(A-I)^{-1}=\ds\frac{1}{4} \left(\begin{array}{cc} -2 & 2\\2 & 0
\end{array} \right)$;

fixed points $\left(\begin{array}{c} x\\y
\end{array} \right)=\ds\frac{1}{4} \left(\begin{array}{cc} -2 & 2\\2 & 0
\end{array} \right) \left(\begin{array}{c} k\\l
\end{array} \right)$ mod 1; $k,l \in \bf{Z}$.

Distinct options $k=0,1$ and $l=0,1$ give

$\left(\begin{array}{c} x\\y \end{array}
\right)=\left(\begin{array}{c} 0\\0
\end{array} \right),\ \left(\begin{array}{c}
\frac{1}{2}\\\frac{1}{2} \end{array} \right),
\left(\begin{array}{c} \frac{1}{2}\\0 \end{array} \right),
\left(\begin{array}{c} 0\\\frac{1}{2} \end{array} \right)$.

$A^2=\left(\begin{array}{cc} 5 & 8\\ 8 & 13 \end{array} \right):\
(A^2-I)^{-1}=\ds\frac{1}{16} \left(\begin{array}{cc} -12 & 8\\ 8 &
-4 \end{array} \right)=\ds\frac{1}{4}\left(\begin{array}{cc} -3 &
2\\ 2 & -1 \end{array} \right)$.

The number of fixed points of $f_A^2$ is $|det(A^2-I)|=16$; of
these, 4 are fixed points of $f_A$ which leaves \underline{six}
2-cycles.

Taking $\left(\begin{array}{c} x\\y
\end{array} \right)=\ds\frac{1}{4}\left(\begin{array}{cc} -3 &
2\\ 2 & -1 \end{array} \right)\left(\begin{array}{c} k\\l
\end{array} \right)$ mod 1 with $k,l \in \bf{Z}$ gives factor
$\ds\frac{1}{4}$ times:

\begin{center}
$\left\{\left(\begin{array}{c} 1\\2 \end{array}
\right),\left(\begin{array}{c} 1\\0 \end{array} \right)\right\}$,
$\left\{\left(\begin{array}{c} 2\\3 \end{array}
\right),\left(\begin{array}{c} 0\\1 \end{array} \right)\right\}$,
$\left\{\left(\begin{array}{c} 3\\1 \end{array}
\right),\left(\begin{array}{c} 1\\1 \end{array} \right)\right\}$,

$\left\{\left(\begin{array}{c} 3\\3 \end{array}
\right),\left(\begin{array}{c} 1\\3 \end{array} \right)\right\}$,
$\left\{\left(\begin{array}{c} 2\\1 \end{array}
\right),\left(\begin{array}{c} 0\\3 \end{array} \right)\right\}$,
$\left\{\left(\begin{array}{c} 3\\2 \end{array}
\right),\left(\begin{array}{c} 3\\0 \end{array} \right)\right\}$.
\end{center}

$A^3=\left(\begin{array}{cc} 21 & 34\\ 34 & 55 \end{array}
\right)$
$$(A^3-I)^{-1}=\ds\frac{1}{76} \left(\begin{array}{cc} -54
& 34\\ 34 & -20 \end{array}
\right)=\ds\frac{1}{38}\left(\begin{array}{cc} -27 & 17\\ 17 & -10
\end{array} \right).$$

3-cycles($\times \frac{1}{38}$) $\left\{\left( -\frac{27}{17} \right),
\left(-\frac{7}{3}\right),\left( \frac{1}{5}
\right)\right\}$, $\left\{\left( -\frac{17}{10} \right), \left( -\frac{3}{4}
\right),\left(\frac{5}{6} \right)\right\},
\cdots$


\end{document}

