Topological conjugacy between dyadic map and tent map
$begingroup$
For trying to prove that the tent map
$$T(x)=
begin{cases}
2x &text{ if } xin[0,frac{1}{2}]\
2-2x &text{ if } xin[frac{1}{2},1]
end{cases}
$$
is ergodic, I have already shown that the dyadic map (period-doubling map) given by $E(x)=2x$ mod $1$ is ergodic using a Fourier transform of $f$ and comparing coefficients to show that any measurable $f:Xtomathbb{R}$ with $ f circ E = f$ almost everywhere implies $f$ is constant almost everywhere.
Anyways, to show that the tent map is ergodic, I tried topological conjugation with the dyadic map (which is ergodic); I've done the following:
Lemma.
The tent map $T$ is topologically semi-conjugate to the dyadic map $E(x)=2x$ mod 1.
Proof.
Let $E: [0,1] to [0,1]$ be the dyadic map $E(x) = 2x$ mod 1. Let $T$ be the tent map as before.
Let $varphi: [0,1] to [0,1]$ also be the tent map, the same as $T$; i.e. $varphiequiv T$. Since
begin{align*}
varphicirc E(x)=T(2xtext{ mod 1})
&=begin{cases}
2(2xtext{ mod 1}) &text{ if }0leqslant2xtext{ mod } 1leqslantfrac{1}{2}\
2-2(2xtext{ mod 1}) &text{ if }frac{1}{2}leqslant2xtext{ mod } 1leqslant1
end{cases}\
&=
begin{cases}
4x &text{ if }xin[0,frac{1}{4}]cup[frac{1}{2},frac{3}{4}]\
2-4x &text{ if }xin[frac{1}{4},frac{1}{2}]cup[frac{3}{4},1]
end{cases}\
&=T^2(x)=Tcircvarphi(x), %%Do not change, this is best way to write down, I noticed by trial and error
end{align*}
we have that $varphicirc E = Tcircvarphi$; i.e. $T$ is a factor of $E$ (or $E$ is an extension of $T$). $Box$.
I know that this is only semi-conjugation for $varphi$ is not invertible. I think the argument for ergodicty does not go wrong only using semi-conjugation, but I would like to have "full" conjugation. This is the ergodicity argument assuming ergodicity of $E$:
Theorem.
The tent map $T$ is ergodic.
Proof.
Let $A$ be an invariant set in $[0,1]$ for $T$; i.e. $T^{-1}(A)=A$. Since $varphicirc E = Tcircvarphi$ with $varphi, E$ and $T$ as in the lemma above, it follows that
begin{align*}
(varphicirc E)^{-1}&=(Tcircvarphi)^{-1}\ E^{-1}circvarphi^{-1}&=varphi^{-1}circ T^{-1}
end{align*}
which, after plugging in $A$, gives
begin{equation*}
E^{-1}(varphi^{-1}(A))=varphi^{-1}(T^{-1}(A))=varphi^{-1}(A);
end{equation*}
so $varphi^{-1}(A)$ is invariant for $E$. Now since $E$ is ergodic, we have that $varphi^{-1}(A)$ has zero or full Lebesgue measure. As $varphi=T$ (and $varphi^{-1}=T^{-1}$), we have that $varphi^{-1}(A)=T^{-1}(A)=A$
and hence also $A$ has zero or full Lebesgue measure; i.e. $T$ is ergodic.$Box$.
Question 1A: Is the proof of the theorem correct?
Question 1B: Does this last theorem proof that ergodicity is preserved under topological semi-conjugation?
Question 2: How does one prove topological conjugation between the tent map and the dyadic map?
Thanks in advance for time and help!
general-topology measure-theory ergodic-theory
$endgroup$
add a comment |
$begingroup$
For trying to prove that the tent map
$$T(x)=
begin{cases}
2x &text{ if } xin[0,frac{1}{2}]\
2-2x &text{ if } xin[frac{1}{2},1]
end{cases}
$$
is ergodic, I have already shown that the dyadic map (period-doubling map) given by $E(x)=2x$ mod $1$ is ergodic using a Fourier transform of $f$ and comparing coefficients to show that any measurable $f:Xtomathbb{R}$ with $ f circ E = f$ almost everywhere implies $f$ is constant almost everywhere.
Anyways, to show that the tent map is ergodic, I tried topological conjugation with the dyadic map (which is ergodic); I've done the following:
Lemma.
The tent map $T$ is topologically semi-conjugate to the dyadic map $E(x)=2x$ mod 1.
Proof.
Let $E: [0,1] to [0,1]$ be the dyadic map $E(x) = 2x$ mod 1. Let $T$ be the tent map as before.
Let $varphi: [0,1] to [0,1]$ also be the tent map, the same as $T$; i.e. $varphiequiv T$. Since
begin{align*}
varphicirc E(x)=T(2xtext{ mod 1})
&=begin{cases}
2(2xtext{ mod 1}) &text{ if }0leqslant2xtext{ mod } 1leqslantfrac{1}{2}\
2-2(2xtext{ mod 1}) &text{ if }frac{1}{2}leqslant2xtext{ mod } 1leqslant1
end{cases}\
&=
begin{cases}
4x &text{ if }xin[0,frac{1}{4}]cup[frac{1}{2},frac{3}{4}]\
2-4x &text{ if }xin[frac{1}{4},frac{1}{2}]cup[frac{3}{4},1]
end{cases}\
&=T^2(x)=Tcircvarphi(x), %%Do not change, this is best way to write down, I noticed by trial and error
end{align*}
we have that $varphicirc E = Tcircvarphi$; i.e. $T$ is a factor of $E$ (or $E$ is an extension of $T$). $Box$.
I know that this is only semi-conjugation for $varphi$ is not invertible. I think the argument for ergodicty does not go wrong only using semi-conjugation, but I would like to have "full" conjugation. This is the ergodicity argument assuming ergodicity of $E$:
Theorem.
The tent map $T$ is ergodic.
Proof.
Let $A$ be an invariant set in $[0,1]$ for $T$; i.e. $T^{-1}(A)=A$. Since $varphicirc E = Tcircvarphi$ with $varphi, E$ and $T$ as in the lemma above, it follows that
begin{align*}
(varphicirc E)^{-1}&=(Tcircvarphi)^{-1}\ E^{-1}circvarphi^{-1}&=varphi^{-1}circ T^{-1}
end{align*}
which, after plugging in $A$, gives
begin{equation*}
E^{-1}(varphi^{-1}(A))=varphi^{-1}(T^{-1}(A))=varphi^{-1}(A);
end{equation*}
so $varphi^{-1}(A)$ is invariant for $E$. Now since $E$ is ergodic, we have that $varphi^{-1}(A)$ has zero or full Lebesgue measure. As $varphi=T$ (and $varphi^{-1}=T^{-1}$), we have that $varphi^{-1}(A)=T^{-1}(A)=A$
and hence also $A$ has zero or full Lebesgue measure; i.e. $T$ is ergodic.$Box$.
Question 1A: Is the proof of the theorem correct?
Question 1B: Does this last theorem proof that ergodicity is preserved under topological semi-conjugation?
Question 2: How does one prove topological conjugation between the tent map and the dyadic map?
Thanks in advance for time and help!
general-topology measure-theory ergodic-theory
$endgroup$
add a comment |
$begingroup$
For trying to prove that the tent map
$$T(x)=
begin{cases}
2x &text{ if } xin[0,frac{1}{2}]\
2-2x &text{ if } xin[frac{1}{2},1]
end{cases}
$$
is ergodic, I have already shown that the dyadic map (period-doubling map) given by $E(x)=2x$ mod $1$ is ergodic using a Fourier transform of $f$ and comparing coefficients to show that any measurable $f:Xtomathbb{R}$ with $ f circ E = f$ almost everywhere implies $f$ is constant almost everywhere.
Anyways, to show that the tent map is ergodic, I tried topological conjugation with the dyadic map (which is ergodic); I've done the following:
Lemma.
The tent map $T$ is topologically semi-conjugate to the dyadic map $E(x)=2x$ mod 1.
Proof.
Let $E: [0,1] to [0,1]$ be the dyadic map $E(x) = 2x$ mod 1. Let $T$ be the tent map as before.
Let $varphi: [0,1] to [0,1]$ also be the tent map, the same as $T$; i.e. $varphiequiv T$. Since
begin{align*}
varphicirc E(x)=T(2xtext{ mod 1})
&=begin{cases}
2(2xtext{ mod 1}) &text{ if }0leqslant2xtext{ mod } 1leqslantfrac{1}{2}\
2-2(2xtext{ mod 1}) &text{ if }frac{1}{2}leqslant2xtext{ mod } 1leqslant1
end{cases}\
&=
begin{cases}
4x &text{ if }xin[0,frac{1}{4}]cup[frac{1}{2},frac{3}{4}]\
2-4x &text{ if }xin[frac{1}{4},frac{1}{2}]cup[frac{3}{4},1]
end{cases}\
&=T^2(x)=Tcircvarphi(x), %%Do not change, this is best way to write down, I noticed by trial and error
end{align*}
we have that $varphicirc E = Tcircvarphi$; i.e. $T$ is a factor of $E$ (or $E$ is an extension of $T$). $Box$.
I know that this is only semi-conjugation for $varphi$ is not invertible. I think the argument for ergodicty does not go wrong only using semi-conjugation, but I would like to have "full" conjugation. This is the ergodicity argument assuming ergodicity of $E$:
Theorem.
The tent map $T$ is ergodic.
Proof.
Let $A$ be an invariant set in $[0,1]$ for $T$; i.e. $T^{-1}(A)=A$. Since $varphicirc E = Tcircvarphi$ with $varphi, E$ and $T$ as in the lemma above, it follows that
begin{align*}
(varphicirc E)^{-1}&=(Tcircvarphi)^{-1}\ E^{-1}circvarphi^{-1}&=varphi^{-1}circ T^{-1}
end{align*}
which, after plugging in $A$, gives
begin{equation*}
E^{-1}(varphi^{-1}(A))=varphi^{-1}(T^{-1}(A))=varphi^{-1}(A);
end{equation*}
so $varphi^{-1}(A)$ is invariant for $E$. Now since $E$ is ergodic, we have that $varphi^{-1}(A)$ has zero or full Lebesgue measure. As $varphi=T$ (and $varphi^{-1}=T^{-1}$), we have that $varphi^{-1}(A)=T^{-1}(A)=A$
and hence also $A$ has zero or full Lebesgue measure; i.e. $T$ is ergodic.$Box$.
Question 1A: Is the proof of the theorem correct?
Question 1B: Does this last theorem proof that ergodicity is preserved under topological semi-conjugation?
Question 2: How does one prove topological conjugation between the tent map and the dyadic map?
Thanks in advance for time and help!
general-topology measure-theory ergodic-theory
$endgroup$
For trying to prove that the tent map
$$T(x)=
begin{cases}
2x &text{ if } xin[0,frac{1}{2}]\
2-2x &text{ if } xin[frac{1}{2},1]
end{cases}
$$
is ergodic, I have already shown that the dyadic map (period-doubling map) given by $E(x)=2x$ mod $1$ is ergodic using a Fourier transform of $f$ and comparing coefficients to show that any measurable $f:Xtomathbb{R}$ with $ f circ E = f$ almost everywhere implies $f$ is constant almost everywhere.
Anyways, to show that the tent map is ergodic, I tried topological conjugation with the dyadic map (which is ergodic); I've done the following:
Lemma.
The tent map $T$ is topologically semi-conjugate to the dyadic map $E(x)=2x$ mod 1.
Proof.
Let $E: [0,1] to [0,1]$ be the dyadic map $E(x) = 2x$ mod 1. Let $T$ be the tent map as before.
Let $varphi: [0,1] to [0,1]$ also be the tent map, the same as $T$; i.e. $varphiequiv T$. Since
begin{align*}
varphicirc E(x)=T(2xtext{ mod 1})
&=begin{cases}
2(2xtext{ mod 1}) &text{ if }0leqslant2xtext{ mod } 1leqslantfrac{1}{2}\
2-2(2xtext{ mod 1}) &text{ if }frac{1}{2}leqslant2xtext{ mod } 1leqslant1
end{cases}\
&=
begin{cases}
4x &text{ if }xin[0,frac{1}{4}]cup[frac{1}{2},frac{3}{4}]\
2-4x &text{ if }xin[frac{1}{4},frac{1}{2}]cup[frac{3}{4},1]
end{cases}\
&=T^2(x)=Tcircvarphi(x), %%Do not change, this is best way to write down, I noticed by trial and error
end{align*}
we have that $varphicirc E = Tcircvarphi$; i.e. $T$ is a factor of $E$ (or $E$ is an extension of $T$). $Box$.
I know that this is only semi-conjugation for $varphi$ is not invertible. I think the argument for ergodicty does not go wrong only using semi-conjugation, but I would like to have "full" conjugation. This is the ergodicity argument assuming ergodicity of $E$:
Theorem.
The tent map $T$ is ergodic.
Proof.
Let $A$ be an invariant set in $[0,1]$ for $T$; i.e. $T^{-1}(A)=A$. Since $varphicirc E = Tcircvarphi$ with $varphi, E$ and $T$ as in the lemma above, it follows that
begin{align*}
(varphicirc E)^{-1}&=(Tcircvarphi)^{-1}\ E^{-1}circvarphi^{-1}&=varphi^{-1}circ T^{-1}
end{align*}
which, after plugging in $A$, gives
begin{equation*}
E^{-1}(varphi^{-1}(A))=varphi^{-1}(T^{-1}(A))=varphi^{-1}(A);
end{equation*}
so $varphi^{-1}(A)$ is invariant for $E$. Now since $E$ is ergodic, we have that $varphi^{-1}(A)$ has zero or full Lebesgue measure. As $varphi=T$ (and $varphi^{-1}=T^{-1}$), we have that $varphi^{-1}(A)=T^{-1}(A)=A$
and hence also $A$ has zero or full Lebesgue measure; i.e. $T$ is ergodic.$Box$.
Question 1A: Is the proof of the theorem correct?
Question 1B: Does this last theorem proof that ergodicity is preserved under topological semi-conjugation?
Question 2: How does one prove topological conjugation between the tent map and the dyadic map?
Thanks in advance for time and help!
general-topology measure-theory ergodic-theory
general-topology measure-theory ergodic-theory
edited Dec 2 '18 at 19:06
Guus Palmer
asked Nov 30 '18 at 12:16
Guus PalmerGuus Palmer
618319
618319
add a comment |
add a comment |
1 Answer
1
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$begingroup$
1.A. The proof is correct, although in the computation of $varphi circ E(x)$, the cases $1/2leq xleq 3/4$ and $xgeq 3/4$ were forgotten.
1.B. Yes, ergodicity of a system implies ergodicity of its factors. However, it must be pointed out that 'factor' must be understood as factor as for a measure preserving transformation (not topological semi-conjugacy): a system $(Y,nu,S)$ is a factor of $(X,mu,T)$ if there is a map $f:Xto Y$, measurable, such that $f_*mu=nu$, and $fcirc T=S circ f$.
As a matter of fact, in your proof, you did not use the fact that $varphi$ is continuous. However, at the end, you used the specific of the situation ($varphi=T$) to conclude, but could easily have used the fact that $Leb(varphi^{-1}A)=Leb(A)$, i.e. $varphi_*(Leb)=Leb$, instead, and that would have been the general proof of the statement that a factor map is ergodic if the extension is.
- The answer depends on the precise topological model you choose.
model $E_1$ : $E_1:[0,1]to [0,1]$,
model $E_2$ : $E_2:mathbb{R}/mathbb{Z} to mathbb{R}/mathbb{Z}$,
model $T_1$ : $T_1:[0,1]to [0,1]$,
model $T_2$ : $T_2:mathbb{R}/mathbb{Z}to mathbb{R}/mathbb{Z}$.
$E_1$ and $T_2$ cannot be topologically conjugate as a circle is not homeomorphic to a closed interval. Same for $T_1$ and $E_2$. $E_1$ cannot be conjugate to $T_1$, because $T_1$ is continuous but $E_1$ is not. $E_2$ and $T_2$ cannot be conjugate, because the degree of $E_2$ is two, and the degree of $T_2$ is zero.
So unless one considers another topological model for these maps, they do not seem to be topologically conjugated.
$endgroup$
$begingroup$
First of all, thank for considering the proof. 1A: I should have formulated it better, yes, but the mapping just repeats itself after [0,1/2]. It's a subtility well found by you. 1B: why is it not a semi-conjugation since $varphi$ is a surjection? Also, $lambda(varphi^{-1}A)=lambda(A)$ does need $varphi$ to be measure preserving by definition; that may be needed in the proof that a factor map is ergodic.
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:32
$begingroup$
Moreover, on wikipedia it says (in part 1) that the tent map and the dyadic map are topologically conjugate (without proof). How is that different from your statement at 2 that one should use "another topological model"?
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:35
$begingroup$
My answer of 2 may be incomplete: I just wanted to say that one can call "dyadic map" several different map defined on several different spaces (e.g $[0,1]$, $[0,1)$ or $mathbb{R}/mathbb{Z}$ - all of which are measurably isomorphic, but not homeomorphic, what I called a topological model. And for the ones that spring to mind, dyadic map and tent map are not topologically conjugate, but perhaps I missed an obvious model that works...
$endgroup$
– user120527
Nov 30 '18 at 13:58
add a comment |
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$begingroup$
1.A. The proof is correct, although in the computation of $varphi circ E(x)$, the cases $1/2leq xleq 3/4$ and $xgeq 3/4$ were forgotten.
1.B. Yes, ergodicity of a system implies ergodicity of its factors. However, it must be pointed out that 'factor' must be understood as factor as for a measure preserving transformation (not topological semi-conjugacy): a system $(Y,nu,S)$ is a factor of $(X,mu,T)$ if there is a map $f:Xto Y$, measurable, such that $f_*mu=nu$, and $fcirc T=S circ f$.
As a matter of fact, in your proof, you did not use the fact that $varphi$ is continuous. However, at the end, you used the specific of the situation ($varphi=T$) to conclude, but could easily have used the fact that $Leb(varphi^{-1}A)=Leb(A)$, i.e. $varphi_*(Leb)=Leb$, instead, and that would have been the general proof of the statement that a factor map is ergodic if the extension is.
- The answer depends on the precise topological model you choose.
model $E_1$ : $E_1:[0,1]to [0,1]$,
model $E_2$ : $E_2:mathbb{R}/mathbb{Z} to mathbb{R}/mathbb{Z}$,
model $T_1$ : $T_1:[0,1]to [0,1]$,
model $T_2$ : $T_2:mathbb{R}/mathbb{Z}to mathbb{R}/mathbb{Z}$.
$E_1$ and $T_2$ cannot be topologically conjugate as a circle is not homeomorphic to a closed interval. Same for $T_1$ and $E_2$. $E_1$ cannot be conjugate to $T_1$, because $T_1$ is continuous but $E_1$ is not. $E_2$ and $T_2$ cannot be conjugate, because the degree of $E_2$ is two, and the degree of $T_2$ is zero.
So unless one considers another topological model for these maps, they do not seem to be topologically conjugated.
$endgroup$
$begingroup$
First of all, thank for considering the proof. 1A: I should have formulated it better, yes, but the mapping just repeats itself after [0,1/2]. It's a subtility well found by you. 1B: why is it not a semi-conjugation since $varphi$ is a surjection? Also, $lambda(varphi^{-1}A)=lambda(A)$ does need $varphi$ to be measure preserving by definition; that may be needed in the proof that a factor map is ergodic.
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:32
$begingroup$
Moreover, on wikipedia it says (in part 1) that the tent map and the dyadic map are topologically conjugate (without proof). How is that different from your statement at 2 that one should use "another topological model"?
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:35
$begingroup$
My answer of 2 may be incomplete: I just wanted to say that one can call "dyadic map" several different map defined on several different spaces (e.g $[0,1]$, $[0,1)$ or $mathbb{R}/mathbb{Z}$ - all of which are measurably isomorphic, but not homeomorphic, what I called a topological model. And for the ones that spring to mind, dyadic map and tent map are not topologically conjugate, but perhaps I missed an obvious model that works...
$endgroup$
– user120527
Nov 30 '18 at 13:58
add a comment |
$begingroup$
1.A. The proof is correct, although in the computation of $varphi circ E(x)$, the cases $1/2leq xleq 3/4$ and $xgeq 3/4$ were forgotten.
1.B. Yes, ergodicity of a system implies ergodicity of its factors. However, it must be pointed out that 'factor' must be understood as factor as for a measure preserving transformation (not topological semi-conjugacy): a system $(Y,nu,S)$ is a factor of $(X,mu,T)$ if there is a map $f:Xto Y$, measurable, such that $f_*mu=nu$, and $fcirc T=S circ f$.
As a matter of fact, in your proof, you did not use the fact that $varphi$ is continuous. However, at the end, you used the specific of the situation ($varphi=T$) to conclude, but could easily have used the fact that $Leb(varphi^{-1}A)=Leb(A)$, i.e. $varphi_*(Leb)=Leb$, instead, and that would have been the general proof of the statement that a factor map is ergodic if the extension is.
- The answer depends on the precise topological model you choose.
model $E_1$ : $E_1:[0,1]to [0,1]$,
model $E_2$ : $E_2:mathbb{R}/mathbb{Z} to mathbb{R}/mathbb{Z}$,
model $T_1$ : $T_1:[0,1]to [0,1]$,
model $T_2$ : $T_2:mathbb{R}/mathbb{Z}to mathbb{R}/mathbb{Z}$.
$E_1$ and $T_2$ cannot be topologically conjugate as a circle is not homeomorphic to a closed interval. Same for $T_1$ and $E_2$. $E_1$ cannot be conjugate to $T_1$, because $T_1$ is continuous but $E_1$ is not. $E_2$ and $T_2$ cannot be conjugate, because the degree of $E_2$ is two, and the degree of $T_2$ is zero.
So unless one considers another topological model for these maps, they do not seem to be topologically conjugated.
$endgroup$
$begingroup$
First of all, thank for considering the proof. 1A: I should have formulated it better, yes, but the mapping just repeats itself after [0,1/2]. It's a subtility well found by you. 1B: why is it not a semi-conjugation since $varphi$ is a surjection? Also, $lambda(varphi^{-1}A)=lambda(A)$ does need $varphi$ to be measure preserving by definition; that may be needed in the proof that a factor map is ergodic.
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:32
$begingroup$
Moreover, on wikipedia it says (in part 1) that the tent map and the dyadic map are topologically conjugate (without proof). How is that different from your statement at 2 that one should use "another topological model"?
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:35
$begingroup$
My answer of 2 may be incomplete: I just wanted to say that one can call "dyadic map" several different map defined on several different spaces (e.g $[0,1]$, $[0,1)$ or $mathbb{R}/mathbb{Z}$ - all of which are measurably isomorphic, but not homeomorphic, what I called a topological model. And for the ones that spring to mind, dyadic map and tent map are not topologically conjugate, but perhaps I missed an obvious model that works...
$endgroup$
– user120527
Nov 30 '18 at 13:58
add a comment |
$begingroup$
1.A. The proof is correct, although in the computation of $varphi circ E(x)$, the cases $1/2leq xleq 3/4$ and $xgeq 3/4$ were forgotten.
1.B. Yes, ergodicity of a system implies ergodicity of its factors. However, it must be pointed out that 'factor' must be understood as factor as for a measure preserving transformation (not topological semi-conjugacy): a system $(Y,nu,S)$ is a factor of $(X,mu,T)$ if there is a map $f:Xto Y$, measurable, such that $f_*mu=nu$, and $fcirc T=S circ f$.
As a matter of fact, in your proof, you did not use the fact that $varphi$ is continuous. However, at the end, you used the specific of the situation ($varphi=T$) to conclude, but could easily have used the fact that $Leb(varphi^{-1}A)=Leb(A)$, i.e. $varphi_*(Leb)=Leb$, instead, and that would have been the general proof of the statement that a factor map is ergodic if the extension is.
- The answer depends on the precise topological model you choose.
model $E_1$ : $E_1:[0,1]to [0,1]$,
model $E_2$ : $E_2:mathbb{R}/mathbb{Z} to mathbb{R}/mathbb{Z}$,
model $T_1$ : $T_1:[0,1]to [0,1]$,
model $T_2$ : $T_2:mathbb{R}/mathbb{Z}to mathbb{R}/mathbb{Z}$.
$E_1$ and $T_2$ cannot be topologically conjugate as a circle is not homeomorphic to a closed interval. Same for $T_1$ and $E_2$. $E_1$ cannot be conjugate to $T_1$, because $T_1$ is continuous but $E_1$ is not. $E_2$ and $T_2$ cannot be conjugate, because the degree of $E_2$ is two, and the degree of $T_2$ is zero.
So unless one considers another topological model for these maps, they do not seem to be topologically conjugated.
$endgroup$
1.A. The proof is correct, although in the computation of $varphi circ E(x)$, the cases $1/2leq xleq 3/4$ and $xgeq 3/4$ were forgotten.
1.B. Yes, ergodicity of a system implies ergodicity of its factors. However, it must be pointed out that 'factor' must be understood as factor as for a measure preserving transformation (not topological semi-conjugacy): a system $(Y,nu,S)$ is a factor of $(X,mu,T)$ if there is a map $f:Xto Y$, measurable, such that $f_*mu=nu$, and $fcirc T=S circ f$.
As a matter of fact, in your proof, you did not use the fact that $varphi$ is continuous. However, at the end, you used the specific of the situation ($varphi=T$) to conclude, but could easily have used the fact that $Leb(varphi^{-1}A)=Leb(A)$, i.e. $varphi_*(Leb)=Leb$, instead, and that would have been the general proof of the statement that a factor map is ergodic if the extension is.
- The answer depends on the precise topological model you choose.
model $E_1$ : $E_1:[0,1]to [0,1]$,
model $E_2$ : $E_2:mathbb{R}/mathbb{Z} to mathbb{R}/mathbb{Z}$,
model $T_1$ : $T_1:[0,1]to [0,1]$,
model $T_2$ : $T_2:mathbb{R}/mathbb{Z}to mathbb{R}/mathbb{Z}$.
$E_1$ and $T_2$ cannot be topologically conjugate as a circle is not homeomorphic to a closed interval. Same for $T_1$ and $E_2$. $E_1$ cannot be conjugate to $T_1$, because $T_1$ is continuous but $E_1$ is not. $E_2$ and $T_2$ cannot be conjugate, because the degree of $E_2$ is two, and the degree of $T_2$ is zero.
So unless one considers another topological model for these maps, they do not seem to be topologically conjugated.
edited Nov 30 '18 at 13:52
answered Nov 30 '18 at 13:21
user120527user120527
1,611215
1,611215
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First of all, thank for considering the proof. 1A: I should have formulated it better, yes, but the mapping just repeats itself after [0,1/2]. It's a subtility well found by you. 1B: why is it not a semi-conjugation since $varphi$ is a surjection? Also, $lambda(varphi^{-1}A)=lambda(A)$ does need $varphi$ to be measure preserving by definition; that may be needed in the proof that a factor map is ergodic.
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:32
$begingroup$
Moreover, on wikipedia it says (in part 1) that the tent map and the dyadic map are topologically conjugate (without proof). How is that different from your statement at 2 that one should use "another topological model"?
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:35
$begingroup$
My answer of 2 may be incomplete: I just wanted to say that one can call "dyadic map" several different map defined on several different spaces (e.g $[0,1]$, $[0,1)$ or $mathbb{R}/mathbb{Z}$ - all of which are measurably isomorphic, but not homeomorphic, what I called a topological model. And for the ones that spring to mind, dyadic map and tent map are not topologically conjugate, but perhaps I missed an obvious model that works...
$endgroup$
– user120527
Nov 30 '18 at 13:58
add a comment |
$begingroup$
First of all, thank for considering the proof. 1A: I should have formulated it better, yes, but the mapping just repeats itself after [0,1/2]. It's a subtility well found by you. 1B: why is it not a semi-conjugation since $varphi$ is a surjection? Also, $lambda(varphi^{-1}A)=lambda(A)$ does need $varphi$ to be measure preserving by definition; that may be needed in the proof that a factor map is ergodic.
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:32
$begingroup$
Moreover, on wikipedia it says (in part 1) that the tent map and the dyadic map are topologically conjugate (without proof). How is that different from your statement at 2 that one should use "another topological model"?
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:35
$begingroup$
My answer of 2 may be incomplete: I just wanted to say that one can call "dyadic map" several different map defined on several different spaces (e.g $[0,1]$, $[0,1)$ or $mathbb{R}/mathbb{Z}$ - all of which are measurably isomorphic, but not homeomorphic, what I called a topological model. And for the ones that spring to mind, dyadic map and tent map are not topologically conjugate, but perhaps I missed an obvious model that works...
$endgroup$
– user120527
Nov 30 '18 at 13:58
$begingroup$
First of all, thank for considering the proof. 1A: I should have formulated it better, yes, but the mapping just repeats itself after [0,1/2]. It's a subtility well found by you. 1B: why is it not a semi-conjugation since $varphi$ is a surjection? Also, $lambda(varphi^{-1}A)=lambda(A)$ does need $varphi$ to be measure preserving by definition; that may be needed in the proof that a factor map is ergodic.
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:32
$begingroup$
First of all, thank for considering the proof. 1A: I should have formulated it better, yes, but the mapping just repeats itself after [0,1/2]. It's a subtility well found by you. 1B: why is it not a semi-conjugation since $varphi$ is a surjection? Also, $lambda(varphi^{-1}A)=lambda(A)$ does need $varphi$ to be measure preserving by definition; that may be needed in the proof that a factor map is ergodic.
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:32
$begingroup$
Moreover, on wikipedia it says (in part 1) that the tent map and the dyadic map are topologically conjugate (without proof). How is that different from your statement at 2 that one should use "another topological model"?
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:35
$begingroup$
Moreover, on wikipedia it says (in part 1) that the tent map and the dyadic map are topologically conjugate (without proof). How is that different from your statement at 2 that one should use "another topological model"?
$endgroup$
– Guus Palmer
Nov 30 '18 at 13:35
$begingroup$
My answer of 2 may be incomplete: I just wanted to say that one can call "dyadic map" several different map defined on several different spaces (e.g $[0,1]$, $[0,1)$ or $mathbb{R}/mathbb{Z}$ - all of which are measurably isomorphic, but not homeomorphic, what I called a topological model. And for the ones that spring to mind, dyadic map and tent map are not topologically conjugate, but perhaps I missed an obvious model that works...
$endgroup$
– user120527
Nov 30 '18 at 13:58
$begingroup$
My answer of 2 may be incomplete: I just wanted to say that one can call "dyadic map" several different map defined on several different spaces (e.g $[0,1]$, $[0,1)$ or $mathbb{R}/mathbb{Z}$ - all of which are measurably isomorphic, but not homeomorphic, what I called a topological model. And for the ones that spring to mind, dyadic map and tent map are not topologically conjugate, but perhaps I missed an obvious model that works...
$endgroup$
– user120527
Nov 30 '18 at 13:58
add a comment |
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