Topological amenability vs amenability of an action
Let $G$ be a discrete group and let $X$ be a compact, Hausdorff space.
Assume that $G$ acts on $X$ by homeomorphisms.
Consider the following two definitions:
- [$C^*$-algebras and finite dimensional approximations- Brown and Ozawa- Definition 4.3.5.]
The action is (topologically) amenable if there exists a net of
continuous maps $m_i: Xto Prob(G)$ s.t. for each $sin G$: $$
lim_{ito infty} (sup_{xin X} |s.m_i^x-m_i^{s.x}|_1)=0$$ where
$s.m_i^x(g)=m_i^x(s^{-1}g)$.
Now, regard $X$ as a $G$-set. Then one can define:
2.[Amenability of Groups and G-Sets, see definition in the introduction, for example]:
The action is amenable if there exists an invariant mean on the power
set of $X$.
I would expect that (1) will imply (2). Namely, that if the action of $G$ on $X$ is topologically amenable then it is amenable (set-theoretically).
However, I know very little about amenable actions and I would appreciate any help.
Thanks.
gr.group-theory ds.dynamical-systems c-star-algebras group-actions amenability
edited Dec 9 '18 at 18:00
Martin Sleziak
2,91032028
asked Dec 9 '18 at 17:28
13829
461
add a comment |
Let $G$ be a discrete group and let $X$ be a compact, Hausdorff space.
Assume that $G$ acts on $X$ by homeomorphisms.
Consider the following two definitions:
- [$C^*$-algebras and finite dimensional approximations- Brown and Ozawa- Definition 4.3.5.]
The action is (topologically) amenable if there exists a net of
continuous maps $m_i: Xto Prob(G)$ s.t. for each $sin G$: $$
lim_{ito infty} (sup_{xin X} |s.m_i^x-m_i^{s.x}|_1)=0$$ where
$s.m_i^x(g)=m_i^x(s^{-1}g)$.
Now, regard $X$ as a $G$-set. Then one can define:
2.[Amenability of Groups and G-Sets, see definition in the introduction, for example]:
The action is amenable if there exists an invariant mean on the power
set of $X$.
I would expect that (1) will imply (2). Namely, that if the action of $G$ on $X$ is topologically amenable then it is amenable (set-theoretically).
However, I know very little about amenable actions and I would appreciate any help.
Thanks.
gr.group-theory ds.dynamical-systems c-star-algebras group-actions amenability
edited Dec 9 '18 at 18:00
Martin Sleziak
2,91032028
asked Dec 9 '18 at 17:28
13829
461
1
I usually call 1 "Zimmer-amenable" and 2 "amenable". Actually the suffix -able is not well-suited to definition 1, and I think of def 1 as "ameaning" (moyennant in French, vs moyennable in sense 2). Introducing a confusing definition is somewhat hopelessly irreversible. Fortunately the definitions are so opposite that the confusion is somewhat limited, in the sense that any interesting statement using one definition is trivially false or tautologic in the other direction.
– YCor
Dec 9 '18 at 20:55
@YCor Did anyone really use "moyennant" in this context?
– R W
Dec 9 '18 at 21:34
@RW I don't think so. But I would if I had to use this property at some point.
– YCor
Dec 9 '18 at 21:46
add a comment |
Let $G$ be a discrete group and let $X$ be a compact, Hausdorff space.
Assume that $G$ acts on $X$ by homeomorphisms.
Consider the following two definitions:
- [$C^*$-algebras and finite dimensional approximations- Brown and Ozawa- Definition 4.3.5.]
The action is (topologically) amenable if there exists a net of
continuous maps $m_i: Xto Prob(G)$ s.t. for each $sin G$: $$
lim_{ito infty} (sup_{xin X} |s.m_i^x-m_i^{s.x}|_1)=0$$ where
$s.m_i^x(g)=m_i^x(s^{-1}g)$.
Now, regard $X$ as a $G$-set. Then one can define:
2.[Amenability of Groups and G-Sets, see definition in the introduction, for example]:
The action is amenable if there exists an invariant mean on the power
set of $X$.
I would expect that (1) will imply (2). Namely, that if the action of $G$ on $X$ is topologically amenable then it is amenable (set-theoretically).
However, I know very little about amenable actions and I would appreciate any help.
Thanks.
gr.group-theory ds.dynamical-systems c-star-algebras group-actions amenability
edited Dec 9 '18 at 18:00
Martin Sleziak
2,91032028
asked Dec 9 '18 at 17:28
13829
461
Let $G$ be a discrete group and let $X$ be a compact, Hausdorff space.
Assume that $G$ acts on $X$ by homeomorphisms.
Consider the following two definitions:
- [$C^*$-algebras and finite dimensional approximations- Brown and Ozawa- Definition 4.3.5.]
The action is (topologically) amenable if there exists a net of
continuous maps $m_i: Xto Prob(G)$ s.t. for each $sin G$: $$
lim_{ito infty} (sup_{xin X} |s.m_i^x-m_i^{s.x}|_1)=0$$ where
$s.m_i^x(g)=m_i^x(s^{-1}g)$.
Now, regard $X$ as a $G$-set. Then one can define:
2.[Amenability of Groups and G-Sets, see definition in the introduction, for example]:
The action is amenable if there exists an invariant mean on the power
set of $X$.
I would expect that (1) will imply (2). Namely, that if the action of $G$ on $X$ is topologically amenable then it is amenable (set-theoretically).
However, I know very little about amenable actions and I would appreciate any help.
Thanks.
gr.group-theory ds.dynamical-systems c-star-algebras group-actions amenability
gr.group-theory ds.dynamical-systems c-star-algebras group-actions amenability
edited Dec 9 '18 at 18:00
Martin Sleziak
2,91032028
asked Dec 9 '18 at 17:28
13829
461
edited Dec 9 '18 at 18:00
Martin Sleziak
2,91032028
asked Dec 9 '18 at 17:28
13829
461
edited Dec 9 '18 at 18:00
Martin Sleziak
2,91032028
edited Dec 9 '18 at 18:00
Martin Sleziak
2,91032028
edited Dec 9 '18 at 18:00
Martin Sleziak
2,91032028
2,91032028
asked Dec 9 '18 at 17:28
13829
461
asked Dec 9 '18 at 17:28
13829
461
asked Dec 9 '18 at 17:28
13829
461
461
1
I usually call 1 "Zimmer-amenable" and 2 "amenable". Actually the suffix -able is not well-suited to definition 1, and I think of def 1 as "ameaning" (moyennant in French, vs moyennable in sense 2). Introducing a confusing definition is somewhat hopelessly irreversible. Fortunately the definitions are so opposite that the confusion is somewhat limited, in the sense that any interesting statement using one definition is trivially false or tautologic in the other direction.
– YCor
Dec 9 '18 at 20:55
@YCor Did anyone really use "moyennant" in this context?
– R W
Dec 9 '18 at 21:34
@RW I don't think so. But I would if I had to use this property at some point.
– YCor
Dec 9 '18 at 21:46
add a comment |
1
I usually call 1 "Zimmer-amenable" and 2 "amenable". Actually the suffix -able is not well-suited to definition 1, and I think of def 1 as "ameaning" (moyennant in French, vs moyennable in sense 2). Introducing a confusing definition is somewhat hopelessly irreversible. Fortunately the definitions are so opposite that the confusion is somewhat limited, in the sense that any interesting statement using one definition is trivially false or tautologic in the other direction.
– YCor
Dec 9 '18 at 20:55
@YCor Did anyone really use "moyennant" in this context?
– R W
Dec 9 '18 at 21:34
@RW I don't think so. But I would if I had to use this property at some point.
– YCor
Dec 9 '18 at 21:46
1
1
I usually call 1 "Zimmer-amenable" and 2 "amenable". Actually the suffix -able is not well-suited to definition 1, and I think of def 1 as "ameaning" (moyennant in French, vs moyennable in sense 2). Introducing a confusing definition is somewhat hopelessly irreversible. Fortunately the definitions are so opposite that the confusion is somewhat limited, in the sense that any interesting statement using one definition is trivially false or tautologic in the other direction.
– YCor
Dec 9 '18 at 20:55
I usually call 1 "Zimmer-amenable" and 2 "amenable". Actually the suffix -able is not well-suited to definition 1, and I think of def 1 as "ameaning" (moyennant in French, vs moyennable in sense 2). Introducing a confusing definition is somewhat hopelessly irreversible. Fortunately the definitions are so opposite that the confusion is somewhat limited, in the sense that any interesting statement using one definition is trivially false or tautologic in the other direction.
– YCor
Dec 9 '18 at 20:55
@YCor Did anyone really use "moyennant" in this context?
– R W
Dec 9 '18 at 21:34
@YCor Did anyone really use "moyennant" in this context?
– R W
Dec 9 '18 at 21:34
@RW I don't think so. But I would if I had to use this property at some point.
– YCor
Dec 9 '18 at 21:46
@RW I don't think so. But I would if I had to use this property at some point.
– YCor
Dec 9 '18 at 21:46
add a comment |
1 Answer
1
active
oldest
votes
The terminology is unfortunately confusing as these properties are in a sense "orthogonal". The definition in terms of existence of an invariant mean on the action space (your definition 2) goes back to von Neumann, whereas your definition 1 goes back to Zimmer who introduced it (in the measure category and in a somewhat different form though) in 1977. Amenability of the group $G$ is equivalent to the amenability in the sense of Definition 1 of its trivial action on the one-point space and to the amenability in the sense of Definition 2 of its action on itself.
For an explicit counterexample to your claim take the boundary action of a free group. It is topologically amenable in the sense of Definition 1, but not amenable in the sense of Definition 2.
A counterexample in the opposite direction is provided just by the trivial action of any non-amenable group on a singleton.
answered Dec 9 '18 at 19:27
R W
10.1k22047
add a comment |
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: "504"
};
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%2fmathoverflow.net%2fquestions%2f317259%2ftopological-amenability-vs-amenability-of-an-action%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The terminology is unfortunately confusing as these properties are in a sense "orthogonal". The definition in terms of existence of an invariant mean on the action space (your definition 2) goes back to von Neumann, whereas your definition 1 goes back to Zimmer who introduced it (in the measure category and in a somewhat different form though) in 1977. Amenability of the group $G$ is equivalent to the amenability in the sense of Definition 1 of its trivial action on the one-point space and to the amenability in the sense of Definition 2 of its action on itself.
For an explicit counterexample to your claim take the boundary action of a free group. It is topologically amenable in the sense of Definition 1, but not amenable in the sense of Definition 2.
A counterexample in the opposite direction is provided just by the trivial action of any non-amenable group on a singleton.
answered Dec 9 '18 at 19:27
R W
10.1k22047
add a comment |
The terminology is unfortunately confusing as these properties are in a sense "orthogonal". The definition in terms of existence of an invariant mean on the action space (your definition 2) goes back to von Neumann, whereas your definition 1 goes back to Zimmer who introduced it (in the measure category and in a somewhat different form though) in 1977. Amenability of the group $G$ is equivalent to the amenability in the sense of Definition 1 of its trivial action on the one-point space and to the amenability in the sense of Definition 2 of its action on itself.
For an explicit counterexample to your claim take the boundary action of a free group. It is topologically amenable in the sense of Definition 1, but not amenable in the sense of Definition 2.
A counterexample in the opposite direction is provided just by the trivial action of any non-amenable group on a singleton.
answered Dec 9 '18 at 19:27
R W
10.1k22047
add a comment |
The terminology is unfortunately confusing as these properties are in a sense "orthogonal". The definition in terms of existence of an invariant mean on the action space (your definition 2) goes back to von Neumann, whereas your definition 1 goes back to Zimmer who introduced it (in the measure category and in a somewhat different form though) in 1977. Amenability of the group $G$ is equivalent to the amenability in the sense of Definition 1 of its trivial action on the one-point space and to the amenability in the sense of Definition 2 of its action on itself.
For an explicit counterexample to your claim take the boundary action of a free group. It is topologically amenable in the sense of Definition 1, but not amenable in the sense of Definition 2.
A counterexample in the opposite direction is provided just by the trivial action of any non-amenable group on a singleton.
answered Dec 9 '18 at 19:27
R W
10.1k22047
The terminology is unfortunately confusing as these properties are in a sense "orthogonal". The definition in terms of existence of an invariant mean on the action space (your definition 2) goes back to von Neumann, whereas your definition 1 goes back to Zimmer who introduced it (in the measure category and in a somewhat different form though) in 1977. Amenability of the group $G$ is equivalent to the amenability in the sense of Definition 1 of its trivial action on the one-point space and to the amenability in the sense of Definition 2 of its action on itself.
For an explicit counterexample to your claim take the boundary action of a free group. It is topologically amenable in the sense of Definition 1, but not amenable in the sense of Definition 2.
A counterexample in the opposite direction is provided just by the trivial action of any non-amenable group on a singleton.
answered Dec 9 '18 at 19:27
R W
10.1k22047
edited Dec 9 '18 at 20:00
answered Dec 9 '18 at 19:27
R W
10.1k22047
answered Dec 9 '18 at 19:27
R W
10.1k22047
answered Dec 9 '18 at 19:27
R W
10.1k22047
10.1k22047
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- 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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2fmathoverflow.net%2fquestions%2f317259%2ftopological-amenability-vs-amenability-of-an-action%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
I usually call 1 "Zimmer-amenable" and 2 "amenable". Actually the suffix -able is not well-suited to definition 1, and I think of def 1 as "ameaning" (moyennant in French, vs moyennable in sense 2). Introducing a confusing definition is somewhat hopelessly irreversible. Fortunately the definitions are so opposite that the confusion is somewhat limited, in the sense that any interesting statement using one definition is trivially false or tautologic in the other direction.
– YCor
Dec 9 '18 at 20:55
@YCor Did anyone really use "moyennant" in this context?
– R W
Dec 9 '18 at 21:34
@RW I don't think so. But I would if I had to use this property at some point.
– YCor
Dec 9 '18 at 21:46