Why are hyperbolic toral automorphisms (e.g. Arnold's cat map) ergodic?

Question

Let $\varphi : \mathbb{T}^2 \to \mathbb{T}^2$ be a hyperbolic automorphism of the torus, induced by a linear map $A : \mathbb{R}^2 \to \mathbb{R}^2$ of determinant $\pm 1$ with no eigenvalues of modulus 1. What is an easy way to prove that $\varphi$ is ergodic?

It's true that the stable and unstable manifolds at $(0,0) \in \mathbb{T}^2$ (projections of the eigenspaces of $A$ to the torus) are dense in the torus, which can be used to prove that such maps are topologically mixing. An example is Arnold's cat map.

Arnold and Avez show in Ergodic Problems in Classical Mechanics that Arnold's cat map is ergodic by proving that it has "Lebesgue spectrum", which implies that it is strong mixing, which implies that it is ergodic. Is there a more direct way to prove this?

Answer 1

I suggest using the following characterization of ergodicity:

$\varphi$ is ergodic if and only if every $f\in L^2(\mathbb{T}^2)$ such that $f\circ \varphi = f$ is constant function.

Now let's use this criterion to prove $\varphi$ is ergodic. Suppose $f\in L^2$ with $f\circ \varphi = f$. Decompose $f$ into its Fourier series $$f = \sum_{m,n\in \mathbb{Z}} = \alpha_{(m\,n)}e^{2\pi imx}e^{2\pi iny},$$ with coefficients $\alpha_{(m\,n)}\in \mathbb{C}$. Then, if $\varphi$ is given by the matrix $$A = \left(\begin{matrix} a & b\\c & d\end{matrix}\right),$$ it is easy to compute that $$f\circ \varphi = \sum_{m,n}\alpha_{(m\,n)}e^{2\pi i(ma+nc)x}e^{2\pi i(mb+nd)y}.$$ Since $f$ is invariant, the Fourier series for $f$ and $f\circ \varphi$ must agree, so $\alpha_{(m\,n)} = \alpha_{(ma+nc\,mb+nd)}$ for all $m,n\in \mathbb{Z}$. We can express this more simply as follows. If $v = (m\,\,n)\in \mathbb{Z}^2$, then $\alpha_v = \alpha_{vA}$. By iterating, $\alpha_v = \alpha_{vA^k}$ for each $k\in \mathbb{Z}$.

Suppose that $v\in \mathbb{Z}^2$. Either the sequence $vA^k$ of vectors with integer coordinates is periodic, or else $\|vA^k\|\to \infty$ as $k\to \infty$. Note that the first case cannot happen unless $v = 0$, since if $v = vA^k$ for some $k$, then $A^k$ would have $1$ as an eigenvalue, which contradicts the assumption of hyperbolicity. Thus either $v = 0$, or $\|vA^k\|\to \infty$. Suppose $v\neq 0$. Since $f\in L^2$, the coefficients $\alpha_{(m\,n)}\to 0$ as $\|(m\,\,n)\|\to \infty$, and thus $\alpha_v = \alpha_{vA^k}\to 0$, i.e., $\alpha_v = 0$. We have therefore shown that the only way $\alpha_v$ can be nonzero is if $v = 0$. The Fourier series for $f$ is then $f = \alpha_0$, so $f$ is a constant function.