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My question is motivated by the hypothesis of the Lindenstrauss' proof of arithmetic quantum unique ergodicity, and the answer to my question is certainly known. However, I could not find it in the ergodic theory material that is available to me. So, I would be grateful for any answer or reference.

Here is the question: Given a measure space $(X,\mu)$ and a one-parameter group $H=\{h_t: t\in\mathbb{R}\}$ acting on $X$ (in any nice way that might be necessary). The example I have in mind is $X=\Gamma\backslash SL_2(\mathbb{R})$ for some lattice $\Gamma$ in $SL_2(\mathbb{R})$ and $H$ the diagonal subgroup $diag(e^t, e^{-t})$. Pick an element $h\in H$, $h\not= id$. I want to compare the entropy of the ergodic components of $\mu$ with respect to $H$ to those with respect to $h$. Are the following statements equivalent?

1) For almost all $x\in X$, the entropy of the $H$-ergodic component is positive. 2) For almost all $x\in X$, the entropy of the $h$-ergodic component is positive.

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2 Answers 2

A small precision :

If $\mu$ is an ergodic measure for the action of the one-parameter group $H=(h_t)_{t\in\mathbb{R}}$, then for almost every $t\in \mathbb{ R}$, the single element $h_t$ is ergodic.

Moreover, if $\mu$ is mixing (it is the case for example if $\mu$ is the Liouville measure or any other Gibbs measure, in your favorite case), then for all $t\in\mathbb{R}$, the single element $h_t$ is mixing.

And as said in the previous answer, the entropy of $H$ is the entropy of $h_1$ or the entropy of $h_t$ divided by $|t|$.

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The entropy of a flow $H=(h_t)$ is defined as the entropy of the time 1 transformation $h_1$. For any other $t$ the entropy of $h_t$ is $|t|$ times the entropy of $h_1$ (for example, see the book by Cornfeld-Fomin-Sinai).

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The point in the question is that the measures (the ergodic components of $\mu$) may be different with respect to which one calculated entropy in 1) and 2). An $H$-ergodic measure does not need to be $h_1$-ergodic. –  Roger Weilik Jan 13 '13 at 13:42
    
I see what you mean. It's not a problem, because the entropy of $h_1$-ergodic components is constant along $H$-orbits. –  R W Jan 13 '13 at 14:53

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