Ergodic automorphism is mixing of all orders I'm trying to solve an exercise in "Ergodic Theory with a view towards Number Theory".
The exercise is suppose $T$ is an ergodic automorphism on an compact abelian group, show that it is mixing of all orders.
(The idea is to use the dual basis of characters and ergodicity being equivalent to no finite periods).
I am unable to show the mixing of all orders.
Thus we need to show for character $\chi_i$ that unless they're all trivial, then as $n_i \to \infty$, we can't have $\chi_0 \times \chi_1(T^{n_1}) \times \ldots \times \chi_k(T^{n_1 + \ldots +n_k}) = 1$.
I can't even refute $\chi_0 (T^{2^n}) \chi_1(T^n) = \chi_2$.
Basically I would love to have a repitition of some sort so that I can cancel and induct.
Please give me a hint.
 A: Begin by thinking about the dual automorphism to a hyperbolic toral automorphism. You can project the dual group onto an "expansive" direction. Any relation of the sort you are looking at can be shifted by the automorphism so that all powers are negative except for one large positive one. In the projection, this would mean that a huge number is the sum of very small numbers, a contradiction. The kind of argument works whenever there is some sort of expansiveness. Then you need to show all ergodic group automorphisms do have some sort of expansiveness. 
Much more is true, see my paper Ergodic group automorphisms are exponentially recurrent, Israel J. Math. 41 (1982), 313-320.
What is more interesting is that this fails for commuting group automorphisms (or algebraic ${\Bbb Z}^d$-actions), the classic example due to Ledrappier. Amazingly, it does remain true in the case of commuting toral automorphisms, however the proof uses some very deep number theory about additive relations in fields (see Cor. 27.6 in Dynamical Systems of Algebraic Origin by Klaus Schmidt).
