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I'm asked to give an example, that the product of two ergodic systems is not ergodic in general. I know that for $X_1=X_2=(S^1,B,m,R_a)$ (the irrational rotation on the unit circle with Lebesgue measure), the product is not ergodic, while the irrational rotation is. But how to show? I don't know, but my guess is that maybe the orbit of any point $x \in S^1\times S^1$ is not dense? (I start at a point $(x,y)$, while iterating the process i get somehow a "ellipse on the torus" or not?). This would imply non-ergodicity. Any ideas?

Davide Giraudo
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crank
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  • You need to find a invariant subset with measure greater than 0 and less 1. At first you may think about the closure of the orbit of $(x,x)$ which is an invariant set but it is of measure 0. You can try to make it larger so that the measure is a bit larger than 0. – Siming Tu Nov 02 '14 at 00:41
  • Okay…but shouldn't it be sufficient to show that the orbit of ever point is not dense? Because if the system would be ergodic, then almost every point would have a dense orbit…the orbit of the point (x,y) is clearly contained in the set ${(tx,ty):t \in R}$ mod 1 which is obviously not dense (its just one "ellipse" on the torus) – crank Nov 02 '14 at 09:08
  • Ah okay,..i take the union of the orbits of an "intervall" on the torus, for example (0,0) to (1/2,1/2), the measure of the union of the orbits is 1/2 and it is invariant under T (since any orbit is invariant) – crank Nov 02 '14 at 09:20
  • The dense orbit is a topological thing, while ergodicity is a measure theoretic thing, they are different. In fact, later in the course, you will see that any compact topological dynamical system has a ergodic measure on it. – Siming Tu Nov 02 '14 at 11:26

1 Answers1

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Let $f(x,y)=x-y$ on the torus $S^1\times S^1$ (I consider $S^1=\mathbb{R}/\mathbb{Z}$). This function is not constant, but it is invariant: it is easy to check that $f(x+\alpha, y+\alpha)=f(x,y)$.

Stéphane Laurent
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