# Prove that Anosov Automorphisms are chaotic

Let me add some detail first.

An Anosov automorphism on $R^2$ is a mapping from the unit square $S$ onto $S$ of the form

$\begin{bmatrix}x \\y\end{bmatrix}\rightarrow\begin{bmatrix}a && b \\c && d \end{bmatrix}\begin{bmatrix}x \\y\end{bmatrix}mod \hspace{1mm}1$

in which (i) $a, b, c, and \hspace{1mm} d$ are integers, (ii) the determinant of the matrix is $\pm$1, and (iii) the eigenvalues of the matrix do not have magnitude $1$.

It is easy to show that Arnold's cat map is an Anosov automorphism, and that it is chaotic.

To define "chaotic" in this context,

A mapping $T$ of $S$ onto itself is said to be chaotic if:

(i) $S$ contains a dense set of periodic points of the mapping $T$

(ii) There is a point in $S$ whose iterates under $T$ are dense in $S$.

That said, it is said that all Anosov automorphisms are chaotic mappings. Based on the definition of chaotic, how can one prove that statement?

Any feedback will be appreciated.

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What about the rationals and irrationals? –  Berci Dec 2 '12 at 1:08
What are you referring to? The two conditions in the definition of chaotic, or using those concepts to prove it? –  Mlagma Dec 2 '12 at 1:10
Just some random thoughts: there is a "larger" eigenvalue, which will come to dominate as you take higher and higher powers of your matrix. This eigenvalue will also be irrational, and so maybe a 2-dimensional density argument will work (since you are really working on the torus)? –  user641 Dec 2 '12 at 1:13
@SteveD That's true that there will be a larger eigenvalue. However, I think the proof will involve more the argument of the rationals and irrationals like Berci said as they really define chaos in this context. –  Mlagma Dec 2 '12 at 1:17
Actually, rational/irrational doesn't seem to have anything to do with it. Let me elaborate on my earlier comment: by conjugating, you can assume your matrix is diagonal (this will change the square to a quadrilateral, but it doesn't matter). You then have a matrix with two irrational diagonal entries. Then a density argument shows the orbit of any point will be dense (in the quotient torus). –  user641 Dec 2 '12 at 2:43

The proof that there is a dense set of periodic points should go something like this. Consider a point with rational components, both with denominator $q$ (I don't insist the rationals be in lowest terms). Then every point in the orbit of that point will also have rational components with denominator $q$. But there are only finitely many points in $S$ with rational components with denominator $q$, so the orbit must visit some point twice. Once it visits some point twice, it must keep visiting the same points over and over, periodically. So this proves every point with rational components is pre-periodic. But $T$ is one-one so pre-periodic points are periodic. As the points with rational components are dense, we're done.