If $Ttimes T$ is ergodic, then so is $Ttimes Ttimes T$
$begingroup$
Given a measure preserving system $(X, mathcal{F}, mu, T)$, show that $T times T$ is ergodic with
respect to $mutimes mu$ if and only if $T times T times T$ is ergodic with respect to $mutimes mutimes mu$.
I have already shown that the reverse implication holds. For "$Rightarrow$", maybe it suffices to show that for all $Atimes Btimes Cin mathcal{F}times mathcal{F}times mathcal{F}$ there is a set of full measure, $M$ say, such that all $(a,b,c)in M$ have an $ninmathbb{N}: (T^{3})^n(a,b,c)in Atimes Btimes C$. It surely suffices to show that a.e. all $(a,b,c)$ will enter arbitary sets in $sigma(mathcal{F}^3)$, but I am not sure whether we can do it for only a generating $pi$-system.
Since $Ttimes T$ is ergodic, then there is $n: (T^2)^nin Atimes B$ and there is $m: (T^2)^m in Btimes C$. But then we can't conclude that $T^3(a,b,c)$ eventually enters $Atimes Btimes C$?
Is there anyone who has a further idea? Thanks in advance.
ergodic-theory
asked Jan 15 at 15:43
Rocco van VreumingenRocco van Vreumingen
928
$endgroup$
|
show 3 more comments
$begingroup$
Given a measure preserving system $(X, mathcal{F}, mu, T)$, show that $T times T$ is ergodic with
respect to $mutimes mu$ if and only if $T times T times T$ is ergodic with respect to $mutimes mutimes mu$.
I have already shown that the reverse implication holds. For "$Rightarrow$", maybe it suffices to show that for all $Atimes Btimes Cin mathcal{F}times mathcal{F}times mathcal{F}$ there is a set of full measure, $M$ say, such that all $(a,b,c)in M$ have an $ninmathbb{N}: (T^{3})^n(a,b,c)in Atimes Btimes C$. It surely suffices to show that a.e. all $(a,b,c)$ will enter arbitary sets in $sigma(mathcal{F}^3)$, but I am not sure whether we can do it for only a generating $pi$-system.
Since $Ttimes T$ is ergodic, then there is $n: (T^2)^nin Atimes B$ and there is $m: (T^2)^m in Btimes C$. But then we can't conclude that $T^3(a,b,c)$ eventually enters $Atimes Btimes C$?
Is there anyone who has a further idea? Thanks in advance.
ergodic-theory
asked Jan 15 at 15:43
Rocco van VreumingenRocco van Vreumingen
928
$endgroup$
1
$begingroup$
The argument I know goes like this: $Ttimes T$ ergodic implies that $T$ is weakly mixing (because the operator on $L^2(X)$ has no eigenvalues) which implies that $Ttimes Ttimes T$ is ergodic. (Have you heard about any of these implications?)
$endgroup$
– Yanko
Jan 15 at 15:45
$begingroup$
@Yanko No, but that is because we have never had any exercise about mixing. I'm gonna look at the topic mixing to check if there is such an argument in one of the books we use.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 15:48
$begingroup$
@Yanko The first implication is an exercise in one of the books, which was never homework. I don't know about the second.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 15:58
$begingroup$
There's a theorem that the following are equivalent for a system $(X,T)$. 1. $X$ is weakly mixing. 2. $(Xtimes X,Ttimes T)$ is ergodic 3. For every ergodic system $(Y,S)$ the system $(Ytimes X,Stimes T$ is ergodic.
$endgroup$
– Yanko
Jan 15 at 15:59
$begingroup$
From 2 you have 1, and then apply 3 with $Y=Xtimes X$. I don't know any other way to do this.
$endgroup$
– Yanko
Jan 15 at 16:00
|
show 3 more comments
$begingroup$
Given a measure preserving system $(X, mathcal{F}, mu, T)$, show that $T times T$ is ergodic with
respect to $mutimes mu$ if and only if $T times T times T$ is ergodic with respect to $mutimes mutimes mu$.
I have already shown that the reverse implication holds. For "$Rightarrow$", maybe it suffices to show that for all $Atimes Btimes Cin mathcal{F}times mathcal{F}times mathcal{F}$ there is a set of full measure, $M$ say, such that all $(a,b,c)in M$ have an $ninmathbb{N}: (T^{3})^n(a,b,c)in Atimes Btimes C$. It surely suffices to show that a.e. all $(a,b,c)$ will enter arbitary sets in $sigma(mathcal{F}^3)$, but I am not sure whether we can do it for only a generating $pi$-system.
Since $Ttimes T$ is ergodic, then there is $n: (T^2)^nin Atimes B$ and there is $m: (T^2)^m in Btimes C$. But then we can't conclude that $T^3(a,b,c)$ eventually enters $Atimes Btimes C$?
Is there anyone who has a further idea? Thanks in advance.
ergodic-theory
asked Jan 15 at 15:43
Rocco van VreumingenRocco van Vreumingen
928
$endgroup$
Given a measure preserving system $(X, mathcal{F}, mu, T)$, show that $T times T$ is ergodic with
respect to $mutimes mu$ if and only if $T times T times T$ is ergodic with respect to $mutimes mutimes mu$.
I have already shown that the reverse implication holds. For "$Rightarrow$", maybe it suffices to show that for all $Atimes Btimes Cin mathcal{F}times mathcal{F}times mathcal{F}$ there is a set of full measure, $M$ say, such that all $(a,b,c)in M$ have an $ninmathbb{N}: (T^{3})^n(a,b,c)in Atimes Btimes C$. It surely suffices to show that a.e. all $(a,b,c)$ will enter arbitary sets in $sigma(mathcal{F}^3)$, but I am not sure whether we can do it for only a generating $pi$-system.
Since $Ttimes T$ is ergodic, then there is $n: (T^2)^nin Atimes B$ and there is $m: (T^2)^m in Btimes C$. But then we can't conclude that $T^3(a,b,c)$ eventually enters $Atimes Btimes C$?
Is there anyone who has a further idea? Thanks in advance.
ergodic-theory
ergodic-theory
asked Jan 15 at 15:43
Rocco van VreumingenRocco van Vreumingen
928
asked Jan 15 at 15:43
Rocco van VreumingenRocco van Vreumingen
928
asked Jan 15 at 15:43
Rocco van VreumingenRocco van Vreumingen
928
asked Jan 15 at 15:43
Rocco van VreumingenRocco van Vreumingen
928
asked Jan 15 at 15:43
Rocco van VreumingenRocco van Vreumingen
928
928
1
$begingroup$
The argument I know goes like this: $Ttimes T$ ergodic implies that $T$ is weakly mixing (because the operator on $L^2(X)$ has no eigenvalues) which implies that $Ttimes Ttimes T$ is ergodic. (Have you heard about any of these implications?)
$endgroup$
– Yanko
Jan 15 at 15:45
$begingroup$
@Yanko No, but that is because we have never had any exercise about mixing. I'm gonna look at the topic mixing to check if there is such an argument in one of the books we use.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 15:48
$begingroup$
@Yanko The first implication is an exercise in one of the books, which was never homework. I don't know about the second.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 15:58
$begingroup$
There's a theorem that the following are equivalent for a system $(X,T)$. 1. $X$ is weakly mixing. 2. $(Xtimes X,Ttimes T)$ is ergodic 3. For every ergodic system $(Y,S)$ the system $(Ytimes X,Stimes T$ is ergodic.
$endgroup$
– Yanko
Jan 15 at 15:59
$begingroup$
From 2 you have 1, and then apply 3 with $Y=Xtimes X$. I don't know any other way to do this.
$endgroup$
– Yanko
Jan 15 at 16:00
|
show 3 more comments
1
$begingroup$
The argument I know goes like this: $Ttimes T$ ergodic implies that $T$ is weakly mixing (because the operator on $L^2(X)$ has no eigenvalues) which implies that $Ttimes Ttimes T$ is ergodic. (Have you heard about any of these implications?)
$endgroup$
– Yanko
Jan 15 at 15:45
$begingroup$
@Yanko No, but that is because we have never had any exercise about mixing. I'm gonna look at the topic mixing to check if there is such an argument in one of the books we use.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 15:48
$begingroup$
@Yanko The first implication is an exercise in one of the books, which was never homework. I don't know about the second.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 15:58
$begingroup$
There's a theorem that the following are equivalent for a system $(X,T)$. 1. $X$ is weakly mixing. 2. $(Xtimes X,Ttimes T)$ is ergodic 3. For every ergodic system $(Y,S)$ the system $(Ytimes X,Stimes T$ is ergodic.
$endgroup$
– Yanko
Jan 15 at 15:59
$begingroup$
From 2 you have 1, and then apply 3 with $Y=Xtimes X$. I don't know any other way to do this.
$endgroup$
– Yanko
Jan 15 at 16:00
1
1
$begingroup$
The argument I know goes like this: $Ttimes T$ ergodic implies that $T$ is weakly mixing (because the operator on $L^2(X)$ has no eigenvalues) which implies that $Ttimes Ttimes T$ is ergodic. (Have you heard about any of these implications?)
$endgroup$
– Yanko
Jan 15 at 15:45
$begingroup$
The argument I know goes like this: $Ttimes T$ ergodic implies that $T$ is weakly mixing (because the operator on $L^2(X)$ has no eigenvalues) which implies that $Ttimes Ttimes T$ is ergodic. (Have you heard about any of these implications?)
$endgroup$
– Yanko
Jan 15 at 15:45
$begingroup$
@Yanko No, but that is because we have never had any exercise about mixing. I'm gonna look at the topic mixing to check if there is such an argument in one of the books we use.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 15:48
$begingroup$
@Yanko No, but that is because we have never had any exercise about mixing. I'm gonna look at the topic mixing to check if there is such an argument in one of the books we use.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 15:48
$begingroup$
@Yanko The first implication is an exercise in one of the books, which was never homework. I don't know about the second.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 15:58
$begingroup$
@Yanko The first implication is an exercise in one of the books, which was never homework. I don't know about the second.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 15:58
$begingroup$
There's a theorem that the following are equivalent for a system $(X,T)$. 1. $X$ is weakly mixing. 2. $(Xtimes X,Ttimes T)$ is ergodic 3. For every ergodic system $(Y,S)$ the system $(Ytimes X,Stimes T$ is ergodic.
$endgroup$
– Yanko
Jan 15 at 15:59
$begingroup$
There's a theorem that the following are equivalent for a system $(X,T)$. 1. $X$ is weakly mixing. 2. $(Xtimes X,Ttimes T)$ is ergodic 3. For every ergodic system $(Y,S)$ the system $(Ytimes X,Stimes T$ is ergodic.
$endgroup$
– Yanko
Jan 15 at 15:59
$begingroup$
From 2 you have 1, and then apply 3 with $Y=Xtimes X$. I don't know any other way to do this.
$endgroup$
– Yanko
Jan 15 at 16:00
$begingroup$
From 2 you have 1, and then apply 3 with $Y=Xtimes X$. I don't know any other way to do this.
$endgroup$
– Yanko
Jan 15 at 16:00
|
show 3 more comments
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1
$begingroup$
The argument I know goes like this: $Ttimes T$ ergodic implies that $T$ is weakly mixing (because the operator on $L^2(X)$ has no eigenvalues) which implies that $Ttimes Ttimes T$ is ergodic. (Have you heard about any of these implications?)
$endgroup$
– Yanko
Jan 15 at 15:45
$begingroup$
@Yanko No, but that is because we have never had any exercise about mixing. I'm gonna look at the topic mixing to check if there is such an argument in one of the books we use.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 15:48
$begingroup$
@Yanko The first implication is an exercise in one of the books, which was never homework. I don't know about the second.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 15:58
$begingroup$
There's a theorem that the following are equivalent for a system $(X,T)$. 1. $X$ is weakly mixing. 2. $(Xtimes X,Ttimes T)$ is ergodic 3. For every ergodic system $(Y,S)$ the system $(Ytimes X,Stimes T$ is ergodic.
$endgroup$
– Yanko
Jan 15 at 15:59
$begingroup$
From 2 you have 1, and then apply 3 with $Y=Xtimes X$. I don't know any other way to do this.
$endgroup$
– Yanko
Jan 15 at 16:00