Mixing
Definition of ergodic map
$begingroup$
I ask the similar question before. About definition of Ergodic theorem.
Now just sincerely ask another fundamental problem about the definition of ergodic map.
The following definition is what I read in my textbook:
Let $(X,mathcal{A},mu)$ be a probability space. Let $T: Xrightarrow X$ is $mu$-invariant ($mu$-preserving). Then $T$ is ergodic if for every $Ein mathcal{A}$ with $T^{-1}(E) = E$ if and only if $mu(E)=0$ or $1$.
In this definition, it requires $T^{-1}(E) = E$. However, in the following paper, it seems to use $T(E) = E$ as the definition.
https://www.tandfonline.com/doi/abs/10.1080/10236190601045788 (p.1154, Def 2.16, please also see "$A$ is invariant, i.e. $T(A)=A$ on the top of that page.")
My questions are:
- What is the difference between them? or is the definition in the paper not correct?
- For the definition by using $T^{-1}(E) = E$, does it imply $E$ is not an attractor? since the only initial condition such that $T(x_0) in E$ is $x_0 in E$ instead of $x_0in Xsetminus E$.
So if we want to discuss $E$ being an attractor, we have to use $T(E) = E$ since this could imply the possibility that $T(x)in E$ for $xin Xsetminus E$. Do I make sense?
Thanks in advance.
measure-theory operator-theory ergodic-theory
asked Jan 21 at 9:38
sleeve chensleeve chen
3,14042053
$endgroup$
add a comment |
$begingroup$
I ask the similar question before. About definition of Ergodic theorem.
Now just sincerely ask another fundamental problem about the definition of ergodic map.
The following definition is what I read in my textbook:
Let $(X,mathcal{A},mu)$ be a probability space. Let $T: Xrightarrow X$ is $mu$-invariant ($mu$-preserving). Then $T$ is ergodic if for every $Ein mathcal{A}$ with $T^{-1}(E) = E$ if and only if $mu(E)=0$ or $1$.
In this definition, it requires $T^{-1}(E) = E$. However, in the following paper, it seems to use $T(E) = E$ as the definition.
https://www.tandfonline.com/doi/abs/10.1080/10236190601045788 (p.1154, Def 2.16, please also see "$A$ is invariant, i.e. $T(A)=A$ on the top of that page.")
My questions are:
- What is the difference between them? or is the definition in the paper not correct?
- For the definition by using $T^{-1}(E) = E$, does it imply $E$ is not an attractor? since the only initial condition such that $T(x_0) in E$ is $x_0 in E$ instead of $x_0in Xsetminus E$.
So if we want to discuss $E$ being an attractor, we have to use $T(E) = E$ since this could imply the possibility that $T(x)in E$ for $xin Xsetminus E$. Do I make sense?
Thanks in advance.
measure-theory operator-theory ergodic-theory
asked Jan 21 at 9:38
sleeve chensleeve chen
3,14042053
$endgroup$
1
$begingroup$
If $T$ is invertible then these definition are equivalent.
$endgroup$
– Yanko
Jan 27 at 11:51
add a comment |
$begingroup$
I ask the similar question before. About definition of Ergodic theorem.
Now just sincerely ask another fundamental problem about the definition of ergodic map.
The following definition is what I read in my textbook:
Let $(X,mathcal{A},mu)$ be a probability space. Let $T: Xrightarrow X$ is $mu$-invariant ($mu$-preserving). Then $T$ is ergodic if for every $Ein mathcal{A}$ with $T^{-1}(E) = E$ if and only if $mu(E)=0$ or $1$.
In this definition, it requires $T^{-1}(E) = E$. However, in the following paper, it seems to use $T(E) = E$ as the definition.
https://www.tandfonline.com/doi/abs/10.1080/10236190601045788 (p.1154, Def 2.16, please also see "$A$ is invariant, i.e. $T(A)=A$ on the top of that page.")
My questions are:
- What is the difference between them? or is the definition in the paper not correct?
- For the definition by using $T^{-1}(E) = E$, does it imply $E$ is not an attractor? since the only initial condition such that $T(x_0) in E$ is $x_0 in E$ instead of $x_0in Xsetminus E$.
So if we want to discuss $E$ being an attractor, we have to use $T(E) = E$ since this could imply the possibility that $T(x)in E$ for $xin Xsetminus E$. Do I make sense?
Thanks in advance.
measure-theory operator-theory ergodic-theory
asked Jan 21 at 9:38
sleeve chensleeve chen
3,14042053
$endgroup$
I ask the similar question before. About definition of Ergodic theorem.
Now just sincerely ask another fundamental problem about the definition of ergodic map.
The following definition is what I read in my textbook:
Let $(X,mathcal{A},mu)$ be a probability space. Let $T: Xrightarrow X$ is $mu$-invariant ($mu$-preserving). Then $T$ is ergodic if for every $Ein mathcal{A}$ with $T^{-1}(E) = E$ if and only if $mu(E)=0$ or $1$.
In this definition, it requires $T^{-1}(E) = E$. However, in the following paper, it seems to use $T(E) = E$ as the definition.
https://www.tandfonline.com/doi/abs/10.1080/10236190601045788 (p.1154, Def 2.16, please also see "$A$ is invariant, i.e. $T(A)=A$ on the top of that page.")
My questions are:
- What is the difference between them? or is the definition in the paper not correct?
- For the definition by using $T^{-1}(E) = E$, does it imply $E$ is not an attractor? since the only initial condition such that $T(x_0) in E$ is $x_0 in E$ instead of $x_0in Xsetminus E$.
So if we want to discuss $E$ being an attractor, we have to use $T(E) = E$ since this could imply the possibility that $T(x)in E$ for $xin Xsetminus E$. Do I make sense?
Thanks in advance.
measure-theory operator-theory ergodic-theory
measure-theory operator-theory ergodic-theory
asked Jan 21 at 9:38
sleeve chensleeve chen
3,14042053
asked Jan 21 at 9:38
sleeve chensleeve chen
3,14042053
asked Jan 21 at 9:38
sleeve chensleeve chen
3,14042053
asked Jan 21 at 9:38
sleeve chensleeve chen
3,14042053
asked Jan 21 at 9:38
sleeve chensleeve chen
3,14042053
3,14042053
1
$begingroup$
If $T$ is invertible then these definition are equivalent.
$endgroup$
– Yanko
Jan 27 at 11:51
add a comment |
1
$begingroup$
If $T$ is invertible then these definition are equivalent.
$endgroup$
– Yanko
Jan 27 at 11:51
1
1
$begingroup$
If $T$ is invertible then these definition are equivalent.
$endgroup$
– Yanko
Jan 27 at 11:51
$begingroup$
If $T$ is invertible then these definition are equivalent.
$endgroup$
– Yanko
Jan 27 at 11:51
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081674%2fdefinition-of-ergodic-map%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081674%2fdefinition-of-ergodic-map%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
If $T$ is invertible then these definition are equivalent.
$endgroup$
– Yanko
Jan 27 at 11:51