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Suppose $T: X \to X$ is a continuous map and $\mu$ a $T$-ergodic probability measure over the Borel sets of $X$. Now, suppose $K \subset \mathrm{Hom}(X)$ is a compact group of measure-preserving homeomorphisms of $X$ commuting with $T$. Consider the factor $Y = K \backslash X = \{Kx | x \in X\}$. Then, defining $\hat{T}(Kx) = T(Kx) = KT(x)$, we have that the canonical projection factors $T$ into $\hat{T}$.

Call $\hat{\mu}$ the $\hat{T}$-invariant probability measure induced on the Borel sets of $Y$. How do I prove that $$h_{\hat{\mu}}(\hat{T}) = 0 \Rightarrow h_\mu(T) = 0?$$

I am trying to make sense of the demonstration of Lemma 7 at Asymptotic Pairs in Positive-Entropy Systems.


Notation: $h_\mu(T)$ is the Kolmogorov-Sinai entropy.

Notice: It is evident that $h_\mu(T) = 0 \Rightarrow h_{\hat{\mu}}(\hat{T}) = 0$.

Notice: The factor $K \backslash X$ is intended to be a group extension.

Edit: Added the fact that elements of $K$ must be measure-preserving.

Edit: Changed $\mu$ from $T$-invariant to $T$-ergodic.

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1 Answer 1

up vote 1 down vote accepted

[I'm assuming G means K]

In case the elements of K are not measure preserving, I doubt that the question is correct. In general, one can easily show that entropy decrease by moving into factors.

Here's an example. Take $T^{1}$ to be Z\R, and the times 2 map. Take a nice measure supported on suitable Cantor set with positive dimension, this in turn implies that the entropy of the times 2 map wrt this measure is positive (by observation of Furstenberg, altough it was known earlier, probably Blinigsley is the right name here, maybe Erdos and maybe some pre-historic british mathematicians). You can also take the usual Lebesgue measure, it is also okay for that matter.

Let K be the (closure of) the subgroup generated by x goes to x+a where a is your favourite irrational number. By Kronceker's lemma, K is identified with $T^{1}$ itself in an obvious manner. Hence your factor space will consist of a point, due to transitivity. The metric entropy of the unit-mass measure is always zero, therefore you can see the decrease of the entropy.

Edit - If you take the Lebesgue measure, the element of K here will be measure preserving also.

Edit 2 - I know Furstenberg's paper, he basically defines there "Kronecker factor" which is basically the factor which controls convergence of his ergodic averages (the multiple recurrence theorem so on), because basically the weak-mixing extensions does not matter for that. But I don't think that this is the correct reference there. Furstenberg's paper does not deal with entropy at all.

Now, I think that in the end we're in the following settings - we are having a skew-product of the system (I love this term more than the categorial approach and the terms Tao uses, Furstenberg and Weiss themselves use the term skew-product), if you want the definition, you can see it in Eli's book about joinings, Ch 3.5, but in general it means that we're having a new system Y=X\times K, and a transformation T'(x,k)=(Tx,f(x)k) where f:X\to K is a continuous map. This what I think is that "group extension theorem" which is found in the reference. Now you need to know which measure there is on Y, it turns out, that it \mu times m, where m is the Haar measure on K, and there's some uniqueness here, you must always choose (in the ergodic case say), an ergodic measure over the factor times the Haar measure, see Eli's book, theorem 3.30.

Moreover, the fact that it is isometric, means that f acts as an isometry over K. Now the theorem goes as follows - the entropy of X is the entropy of Y. Why? informally, the orbits in the K coordinate, won't diverge (it's basically deterministic system), hence the only entropy comes from the X coordinate.

Now a bit more formally, take a nice generating partition for Y, the projection in the X coordinate will also be generating. You can obviously approximate suich a partition by cylinders, and get a partition of the form P\times Q. Computing H_{Y}(P\times Q), divide the sum, so you sum over P and then over Q. In one term you have -\sum \sum \mu(p)m(q)\log(\mu(p)), where the outer sum is over p, the inner over q, so this is the entropy of P over X, by def. In the other term you have -\sum \mu(p) \sum m(q)\log(m(q)), which is basically the entropy of Q over Y. So know you're just dealing with the fact that an isometric system (say over a compact group) have zero entropy, which is also very easy to establish.

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Yes, $G$ means $K$... :-) I will correct it. –  André Caldas Oct 5 '11 at 11:44
    
I do not want to prove the entropy decreases. It is quite the opposite. I want to prove that if the entropy in the quotient is $0$ then it will be $0$ in the original system. –  André Caldas Oct 5 '11 at 11:46
    
In my example, the entropy of the system is positive, but the entropy of the factor is zero! –  Asaf Oct 5 '11 at 11:54
    
@Asaf: Oops... you are right. It is a very good example. I will fix the post. –  André Caldas Oct 5 '11 at 12:01
    
@Andre, actually, if you choose the Lebesgue measure (and not any general times 2 inv. and ergodic measure), you will see that the element of K will be measure-preserving, but still the conclusion is false. Notice also the the times 2 system is weak mixing and disjoint from the Kronceker systems, hence I don't think that you're question can be true in a very general settings. –  Asaf Oct 5 '11 at 12:39

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