We know a hyperbolic toral automorphism is defined to be an automorphism of $\mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n$ induced by integer matrix in $\text{GL}(n,\mathbb{Z})$ which has no eigenvalue of module $1$. One shows that any hyperbolic automorphism $T: \mathbb{T}^n \rightarrow \mathbb{T}^n$ is ergodic w.r.t Lebesgue measure by showing that any $L^2$ function $f:\mathbb{T}^n\to\mathbb{R}$ such that $f\circ T = f$ is constant Lebesgue almost everywhere. Now I need to prove the hyperbolic automorphism has a property stronger than ergodicity:
Exercise: Prove that any hyperbolic automorphism of $\mathbb{T}^n$ is mixing.
(This is Exercise 4.4.3 from p.79 of Brin and Stuck's book Introduction to Dynamical Systems.)
Definition: For $(X, \mathcal{A}, \mu)$ a probability space, a measurable and measure-preserving self-map $T:X\to X$ is defined to be (strongly measure theoretically) mixing, if for any measurable sets $A$ and $B$ we have
$$\lim_{n\to \infty} \mu(T^{-n}(A) \cap B) = \mu(A)\mu(B).$$
I have no idea about the details of image of $A$, hence it is difficult to determine $\mu(T^{-n}(A)\cap B)$. Any suggestion or help is appreciated.
Best Answer
The solution is based on the Fourier series representation in $L^2$ space. In fact, the following important lemma of mixing helps to solve the problem a lot.
Lemma: $T$ is mixing in measure space $(X, \mathcal{L}, \mu)$ iff for any $L^2$ basis of $X$ ${\{\xi_i\}}$ we have $$\lim_N\int_X \xi_i\circ{T^N}\xi_jd\mu = \int_X \xi_i d\mu \int_X \xi_j d\mu$$
Then apply this lemma to the Fourier basis $\{\xi_{m,n}=e^{mx+ny}\}_{m,n \in \mathbb{Z}}$ we WTS $$\lim_N\int_X \xi_{m,n}\circ{T^N}\xi_{k,l}d\mu = \int_X \xi_{m,n} d\mu \int_X \xi_{k,l} d\mu$$
and discuss by two classifications $(m,n) = 0$ and $(m,n) \neq (0,0)$ to get the solution. Note that $\int_X \xi_{m,n}d\mu = 0$ iff $(m,n) \neq 0$ and $\int_X \xi_{m,n}d\mu=1$ iff $(m,n) = 0$. In the case $(m,n) \neq 0$ we can find enough large $N$ such that for any $n > N$ we have $\xi_{m,n}\circ T^n \neq \xi_{k,l}$ for any $(k,l)$, hence the LHS vanishes.
The idea is not belong to me but I think it is easy to learn.