Group of orientation preserving homeomorphisms of circle $S^1$ acts transitively on the set of closed...
The closed intervals here mean the arcs including endpoints on the circle. I tried to do it by taking the inverse image of those two closed intervals from $S^1$ to its covering space $mathbb{R}$ and then constructed a homeomorphism of $mathbb{R}$ that takes these closed intervals to each other. But the issue is that while projecting it back to $S^1$, I am unable to ensure that this will result in a homeomorphism.
Any other approach is also welcome.
general-topology differential-geometry geometric-topology
edited Nov 28 '18 at 14:36
amWhy
192k28224439
asked Nov 28 '18 at 13:27
Sanjay Kapoor
264
add a comment |
The closed intervals here mean the arcs including endpoints on the circle. I tried to do it by taking the inverse image of those two closed intervals from $S^1$ to its covering space $mathbb{R}$ and then constructed a homeomorphism of $mathbb{R}$ that takes these closed intervals to each other. But the issue is that while projecting it back to $S^1$, I am unable to ensure that this will result in a homeomorphism.
Any other approach is also welcome.
general-topology differential-geometry geometric-topology
edited Nov 28 '18 at 14:36
amWhy
192k28224439
asked Nov 28 '18 at 13:27
Sanjay Kapoor
264
I assume that the induced action is via taking image: $phicirc I:=phi(I)$? Then this is false, no proper subset of $S^1$ can be homeomorphic to $S^1$. Or are you saying that these intervals have to be proper subsets? What's the definition of interval in $S^1$ cause its confusing to me? As in: proper, infinite, connected and closed subsets of $S^1$?
– freakish
Nov 28 '18 at 14:52
@freakish That is likely what OP means, yes. The space of intervals in $S^1$ should be the same as the space of ordered pairs of distinct points.
– Mike Miller
Nov 28 '18 at 20:18
So first of all you can assume that two such intervals have common end at $v=(1,0)$ and one is contained in another. This is due to applying rotations (which preserve orientation). Then you look at the preimage of the covering map in $[0,1)$ or $(0,1]$ depending on whether $0$ or $1$ is isolated, shrink one interval to another via $[0,1)to[0,1)$ homeomorphism and recreate a homeomorphism $S^1to S^1$ from that by applying the covering map.
– freakish
Nov 28 '18 at 21:04
Yes, you are right. Proper connected and closed sets.
– Sanjay Kapoor
Nov 29 '18 at 11:52
add a comment |
The closed intervals here mean the arcs including endpoints on the circle. I tried to do it by taking the inverse image of those two closed intervals from $S^1$ to its covering space $mathbb{R}$ and then constructed a homeomorphism of $mathbb{R}$ that takes these closed intervals to each other. But the issue is that while projecting it back to $S^1$, I am unable to ensure that this will result in a homeomorphism.
Any other approach is also welcome.
general-topology differential-geometry geometric-topology
edited Nov 28 '18 at 14:36
amWhy
192k28224439
asked Nov 28 '18 at 13:27
Sanjay Kapoor
264
The closed intervals here mean the arcs including endpoints on the circle. I tried to do it by taking the inverse image of those two closed intervals from $S^1$ to its covering space $mathbb{R}$ and then constructed a homeomorphism of $mathbb{R}$ that takes these closed intervals to each other. But the issue is that while projecting it back to $S^1$, I am unable to ensure that this will result in a homeomorphism.
Any other approach is also welcome.
general-topology differential-geometry geometric-topology
general-topology differential-geometry geometric-topology
edited Nov 28 '18 at 14:36
amWhy
192k28224439
asked Nov 28 '18 at 13:27
Sanjay Kapoor
264
edited Nov 28 '18 at 14:36
amWhy
192k28224439
asked Nov 28 '18 at 13:27
Sanjay Kapoor
264
edited Nov 28 '18 at 14:36
amWhy
192k28224439
edited Nov 28 '18 at 14:36
amWhy
192k28224439
edited Nov 28 '18 at 14:36
amWhy
192k28224439
192k28224439
asked Nov 28 '18 at 13:27
Sanjay Kapoor
264
asked Nov 28 '18 at 13:27
Sanjay Kapoor
264
asked Nov 28 '18 at 13:27
Sanjay Kapoor
264
264
I assume that the induced action is via taking image: $phicirc I:=phi(I)$? Then this is false, no proper subset of $S^1$ can be homeomorphic to $S^1$. Or are you saying that these intervals have to be proper subsets? What's the definition of interval in $S^1$ cause its confusing to me? As in: proper, infinite, connected and closed subsets of $S^1$?
– freakish
Nov 28 '18 at 14:52
@freakish That is likely what OP means, yes. The space of intervals in $S^1$ should be the same as the space of ordered pairs of distinct points.
– Mike Miller
Nov 28 '18 at 20:18
So first of all you can assume that two such intervals have common end at $v=(1,0)$ and one is contained in another. This is due to applying rotations (which preserve orientation). Then you look at the preimage of the covering map in $[0,1)$ or $(0,1]$ depending on whether $0$ or $1$ is isolated, shrink one interval to another via $[0,1)to[0,1)$ homeomorphism and recreate a homeomorphism $S^1to S^1$ from that by applying the covering map.
– freakish
Nov 28 '18 at 21:04
Yes, you are right. Proper connected and closed sets.
– Sanjay Kapoor
Nov 29 '18 at 11:52
add a comment |
I assume that the induced action is via taking image: $phicirc I:=phi(I)$? Then this is false, no proper subset of $S^1$ can be homeomorphic to $S^1$. Or are you saying that these intervals have to be proper subsets? What's the definition of interval in $S^1$ cause its confusing to me? As in: proper, infinite, connected and closed subsets of $S^1$?
– freakish
Nov 28 '18 at 14:52
@freakish That is likely what OP means, yes. The space of intervals in $S^1$ should be the same as the space of ordered pairs of distinct points.
– Mike Miller
Nov 28 '18 at 20:18
So first of all you can assume that two such intervals have common end at $v=(1,0)$ and one is contained in another. This is due to applying rotations (which preserve orientation). Then you look at the preimage of the covering map in $[0,1)$ or $(0,1]$ depending on whether $0$ or $1$ is isolated, shrink one interval to another via $[0,1)to[0,1)$ homeomorphism and recreate a homeomorphism $S^1to S^1$ from that by applying the covering map.
– freakish
Nov 28 '18 at 21:04
Yes, you are right. Proper connected and closed sets.
– Sanjay Kapoor
Nov 29 '18 at 11:52
I assume that the induced action is via taking image: $phicirc I:=phi(I)$? Then this is false, no proper subset of $S^1$ can be homeomorphic to $S^1$. Or are you saying that these intervals have to be proper subsets? What's the definition of interval in $S^1$ cause its confusing to me? As in: proper, infinite, connected and closed subsets of $S^1$?
– freakish
Nov 28 '18 at 14:52
I assume that the induced action is via taking image: $phicirc I:=phi(I)$? Then this is false, no proper subset of $S^1$ can be homeomorphic to $S^1$. Or are you saying that these intervals have to be proper subsets? What's the definition of interval in $S^1$ cause its confusing to me? As in: proper, infinite, connected and closed subsets of $S^1$?
– freakish
Nov 28 '18 at 14:52
@freakish That is likely what OP means, yes. The space of intervals in $S^1$ should be the same as the space of ordered pairs of distinct points.
– Mike Miller
Nov 28 '18 at 20:18
@freakish That is likely what OP means, yes. The space of intervals in $S^1$ should be the same as the space of ordered pairs of distinct points.
– Mike Miller
Nov 28 '18 at 20:18
So first of all you can assume that two such intervals have common end at $v=(1,0)$ and one is contained in another. This is due to applying rotations (which preserve orientation). Then you look at the preimage of the covering map in $[0,1)$ or $(0,1]$ depending on whether $0$ or $1$ is isolated, shrink one interval to another via $[0,1)to[0,1)$ homeomorphism and recreate a homeomorphism $S^1to S^1$ from that by applying the covering map.
– freakish
Nov 28 '18 at 21:04
So first of all you can assume that two such intervals have common end at $v=(1,0)$ and one is contained in another. This is due to applying rotations (which preserve orientation). Then you look at the preimage of the covering map in $[0,1)$ or $(0,1]$ depending on whether $0$ or $1$ is isolated, shrink one interval to another via $[0,1)to[0,1)$ homeomorphism and recreate a homeomorphism $S^1to S^1$ from that by applying the covering map.
– freakish
Nov 28 '18 at 21:04
Yes, you are right. Proper connected and closed sets.
– Sanjay Kapoor
Nov 29 '18 at 11:52
Yes, you are right. Proper connected and closed sets.
– Sanjay Kapoor
Nov 29 '18 at 11:52
add a comment |
1 Answer
1
active
oldest
votes
Let $p : mathbb{R} to S^1, p(t) = e^{it}$, be the standard covering. An interval in $S^1$ is the image $I = e([a,b])$ of an interval $[a,b] subset mathbb{R}$ such that $0 < b - a < 2pi$. The restriction $e_I : [a,b] to I$ of $e$ is a homeomorphism. Moreover we have $S^1 = I cup I'$ with $I' = e([b,a+2pi])$. Note that $I cap I' = { e^{ia}, e^{ib} }$.
Consider two intervals $I_k = e([a_k,b_k])$. Define "linear" homeomorphisms $u : [a_1,b_1] to [a_2,b_2], u(t) = a_2 +frac{b_2-a_2}{b_1-a_1}(t - a_1)$ and $u' : [b_1,a_1+2pi] to [b_2,a_2+2pi], u'(t) = b_2 +frac{a_2+2pi-b_2}{a_1+2pi-b_1}(t - b_1)$. These induce orientation preserving homeomorphisms $U : I_1 to I_2$ and $U' : I'_1 to I'_2$. Now $U$ and $U'$ can be pasted together to an orientation preserving homeomorphisms $h : S^1 = I_1 cup I'_1 to I_2 cup I'_2 = S^1$. By construction $h(I_1) = I_2$.
answered Nov 29 '18 at 0:10
Paul Frost
9,1961631
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: "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%2f3017150%2fgroup-of-orientation-preserving-homeomorphisms-of-circle-s1-acts-transitively%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
Let $p : mathbb{R} to S^1, p(t) = e^{it}$, be the standard covering. An interval in $S^1$ is the image $I = e([a,b])$ of an interval $[a,b] subset mathbb{R}$ such that $0 < b - a < 2pi$. The restriction $e_I : [a,b] to I$ of $e$ is a homeomorphism. Moreover we have $S^1 = I cup I'$ with $I' = e([b,a+2pi])$. Note that $I cap I' = { e^{ia}, e^{ib} }$.
Consider two intervals $I_k = e([a_k,b_k])$. Define "linear" homeomorphisms $u : [a_1,b_1] to [a_2,b_2], u(t) = a_2 +frac{b_2-a_2}{b_1-a_1}(t - a_1)$ and $u' : [b_1,a_1+2pi] to [b_2,a_2+2pi], u'(t) = b_2 +frac{a_2+2pi-b_2}{a_1+2pi-b_1}(t - b_1)$. These induce orientation preserving homeomorphisms $U : I_1 to I_2$ and $U' : I'_1 to I'_2$. Now $U$ and $U'$ can be pasted together to an orientation preserving homeomorphisms $h : S^1 = I_1 cup I'_1 to I_2 cup I'_2 = S^1$. By construction $h(I_1) = I_2$.
answered Nov 29 '18 at 0:10
Paul Frost
9,1961631
add a comment |
Let $p : mathbb{R} to S^1, p(t) = e^{it}$, be the standard covering. An interval in $S^1$ is the image $I = e([a,b])$ of an interval $[a,b] subset mathbb{R}$ such that $0 < b - a < 2pi$. The restriction $e_I : [a,b] to I$ of $e$ is a homeomorphism. Moreover we have $S^1 = I cup I'$ with $I' = e([b,a+2pi])$. Note that $I cap I' = { e^{ia}, e^{ib} }$.
Consider two intervals $I_k = e([a_k,b_k])$. Define "linear" homeomorphisms $u : [a_1,b_1] to [a_2,b_2], u(t) = a_2 +frac{b_2-a_2}{b_1-a_1}(t - a_1)$ and $u' : [b_1,a_1+2pi] to [b_2,a_2+2pi], u'(t) = b_2 +frac{a_2+2pi-b_2}{a_1+2pi-b_1}(t - b_1)$. These induce orientation preserving homeomorphisms $U : I_1 to I_2$ and $U' : I'_1 to I'_2$. Now $U$ and $U'$ can be pasted together to an orientation preserving homeomorphisms $h : S^1 = I_1 cup I'_1 to I_2 cup I'_2 = S^1$. By construction $h(I_1) = I_2$.
answered Nov 29 '18 at 0:10
Paul Frost
9,1961631
add a comment |
Let $p : mathbb{R} to S^1, p(t) = e^{it}$, be the standard covering. An interval in $S^1$ is the image $I = e([a,b])$ of an interval $[a,b] subset mathbb{R}$ such that $0 < b - a < 2pi$. The restriction $e_I : [a,b] to I$ of $e$ is a homeomorphism. Moreover we have $S^1 = I cup I'$ with $I' = e([b,a+2pi])$. Note that $I cap I' = { e^{ia}, e^{ib} }$.
Consider two intervals $I_k = e([a_k,b_k])$. Define "linear" homeomorphisms $u : [a_1,b_1] to [a_2,b_2], u(t) = a_2 +frac{b_2-a_2}{b_1-a_1}(t - a_1)$ and $u' : [b_1,a_1+2pi] to [b_2,a_2+2pi], u'(t) = b_2 +frac{a_2+2pi-b_2}{a_1+2pi-b_1}(t - b_1)$. These induce orientation preserving homeomorphisms $U : I_1 to I_2$ and $U' : I'_1 to I'_2$. Now $U$ and $U'$ can be pasted together to an orientation preserving homeomorphisms $h : S^1 = I_1 cup I'_1 to I_2 cup I'_2 = S^1$. By construction $h(I_1) = I_2$.
answered Nov 29 '18 at 0:10
Paul Frost
9,1961631
Let $p : mathbb{R} to S^1, p(t) = e^{it}$, be the standard covering. An interval in $S^1$ is the image $I = e([a,b])$ of an interval $[a,b] subset mathbb{R}$ such that $0 < b - a < 2pi$. The restriction $e_I : [a,b] to I$ of $e$ is a homeomorphism. Moreover we have $S^1 = I cup I'$ with $I' = e([b,a+2pi])$. Note that $I cap I' = { e^{ia}, e^{ib} }$.
Consider two intervals $I_k = e([a_k,b_k])$. Define "linear" homeomorphisms $u : [a_1,b_1] to [a_2,b_2], u(t) = a_2 +frac{b_2-a_2}{b_1-a_1}(t - a_1)$ and $u' : [b_1,a_1+2pi] to [b_2,a_2+2pi], u'(t) = b_2 +frac{a_2+2pi-b_2}{a_1+2pi-b_1}(t - b_1)$. These induce orientation preserving homeomorphisms $U : I_1 to I_2$ and $U' : I'_1 to I'_2$. Now $U$ and $U'$ can be pasted together to an orientation preserving homeomorphisms $h : S^1 = I_1 cup I'_1 to I_2 cup I'_2 = S^1$. By construction $h(I_1) = I_2$.
answered Nov 29 '18 at 0:10
Paul Frost
9,1961631
answered Nov 29 '18 at 0:10
Paul Frost
9,1961631
answered Nov 29 '18 at 0:10
Paul Frost
9,1961631
answered Nov 29 '18 at 0:10
Paul Frost
9,1961631
9,1961631
add a comment |
add a comment |
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.
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%2fmath.stackexchange.com%2fquestions%2f3017150%2fgroup-of-orientation-preserving-homeomorphisms-of-circle-s1-acts-transitively%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
I assume that the induced action is via taking image: $phicirc I:=phi(I)$? Then this is false, no proper subset of $S^1$ can be homeomorphic to $S^1$. Or are you saying that these intervals have to be proper subsets? What's the definition of interval in $S^1$ cause its confusing to me? As in: proper, infinite, connected and closed subsets of $S^1$?
– freakish
Nov 28 '18 at 14:52
@freakish That is likely what OP means, yes. The space of intervals in $S^1$ should be the same as the space of ordered pairs of distinct points.
– Mike Miller
Nov 28 '18 at 20:18
So first of all you can assume that two such intervals have common end at $v=(1,0)$ and one is contained in another. This is due to applying rotations (which preserve orientation). Then you look at the preimage of the covering map in $[0,1)$ or $(0,1]$ depending on whether $0$ or $1$ is isolated, shrink one interval to another via $[0,1)to[0,1)$ homeomorphism and recreate a homeomorphism $S^1to S^1$ from that by applying the covering map.
– freakish
Nov 28 '18 at 21:04
Yes, you are right. Proper connected and closed sets.
– Sanjay Kapoor
Nov 29 '18 at 11:52