What are the orbits for this action of $mathrmSO(2)$ on $S_3$? The Next CEO of Stack OverflowUnderstanding what an action is?Counting the number of orbits for the group actionGroup action on $mathbb R^2$: are my thoughts correct?How to prove this action is discontinuous?Normal subgroup $H$ of $G$ with same orbits of action on $X$.What are the fixed points of this action?Group Orbits of this space under O(3)Meaning of this exercise in group theoryOrbits and stabilizers of a group actionCompute the size of the orbit under a finite group action
Are the names of these months realistic?
Yu-Gi-Oh cards in Python 3
Calculate the Mean mean of two numbers
Is dried pee considered dirt?
What day is it again?
Is there a way to save my career from absolute disaster?
Is it ever safe to open a suspicious HTML file (e.g. email attachment)?
How can the PCs determine if an item is a phylactery?
What is the process for purifying your home if you believe it may have been previously used for pagan worship?
Raspberry pi 3 B with Ubuntu 18.04 server arm64: what chip
What are the unusually-enlarged wing sections on this P-38 Lightning?
Could a dragon use its wings to swim?
If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?
Towers in the ocean; How deep can they be built?
Strange use of "whether ... than ..." in official text
Inductor and Capacitor in Parallel
Man transported from Alternate World into ours by a Neutrino Detector
how one can write a nice vector parser, something that does pgfvecparseA=B-C; D=E x F;
Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?
Purpose of level-shifter with same in and out voltages
How to use ReplaceAll on an expression that contains a rule
What would be the main consequences for a country leaving the WTO?
Is there an equivalent of cd - for cp or mv
Does the Idaho Potato Commission associate potato skins with healthy eating?
What are the orbits for this action of $mathrmSO(2)$ on $S_3$?
The Next CEO of Stack OverflowUnderstanding what an action is?Counting the number of orbits for the group actionGroup action on $mathbb R^2$: are my thoughts correct?How to prove this action is discontinuous?Normal subgroup $H$ of $G$ with same orbits of action on $X$.What are the fixed points of this action?Group Orbits of this space under O(3)Meaning of this exercise in group theoryOrbits and stabilizers of a group actionCompute the size of the orbit under a finite group action
$begingroup$
The group action defined by: $Phi: SO(2) times S_3 rightarrow S_3$ such that: ($A, r$) $rightarrow$ [$ mathbb1 otimes A$]$cdot$[$r$], where A is a standard matrix $2 times 2$ of 2D rotations, $r$ is a point on $S_3$ such that $r_1^2+r_2^2+r_3^2+r_4^2=1$ and “$cdot$“ means a normal multiplication.
My question is: what actually is an orbit of this action?
group-theory group-actions
edited Mar 19 at 22:25
Travis
63.8k769151
asked Mar 19 at 21:58
BartekBartek
83
$endgroup$
add a comment |
$begingroup$
The group action defined by: $Phi: SO(2) times S_3 rightarrow S_3$ such that: ($A, r$) $rightarrow$ [$ mathbb1 otimes A$]$cdot$[$r$], where A is a standard matrix $2 times 2$ of 2D rotations, $r$ is a point on $S_3$ such that $r_1^2+r_2^2+r_3^2+r_4^2=1$ and “$cdot$“ means a normal multiplication.
My question is: what actually is an orbit of this action?
group-theory group-actions
edited Mar 19 at 22:25
Travis
63.8k769151
asked Mar 19 at 21:58
BartekBartek
83
$endgroup$
$begingroup$
You write the action from $SO(3)$, meaning it is a $3times 3$ matrix, a 3D rotation.
$endgroup$
– Wojowu
Mar 19 at 22:20
$begingroup$
@Wojowu I edited it, thank you.
$endgroup$
– Bartek
Mar 19 at 22:22
add a comment |
$begingroup$
The group action defined by: $Phi: SO(2) times S_3 rightarrow S_3$ such that: ($A, r$) $rightarrow$ [$ mathbb1 otimes A$]$cdot$[$r$], where A is a standard matrix $2 times 2$ of 2D rotations, $r$ is a point on $S_3$ such that $r_1^2+r_2^2+r_3^2+r_4^2=1$ and “$cdot$“ means a normal multiplication.
My question is: what actually is an orbit of this action?
group-theory group-actions
edited Mar 19 at 22:25
Travis
63.8k769151
asked Mar 19 at 21:58
BartekBartek
83
$endgroup$
The group action defined by: $Phi: SO(2) times S_3 rightarrow S_3$ such that: ($A, r$) $rightarrow$ [$ mathbb1 otimes A$]$cdot$[$r$], where A is a standard matrix $2 times 2$ of 2D rotations, $r$ is a point on $S_3$ such that $r_1^2+r_2^2+r_3^2+r_4^2=1$ and “$cdot$“ means a normal multiplication.
My question is: what actually is an orbit of this action?
group-theory group-actions
group-theory group-actions
edited Mar 19 at 22:25
Travis
63.8k769151
asked Mar 19 at 21:58
BartekBartek
83
edited Mar 19 at 22:25
Travis
63.8k769151
asked Mar 19 at 21:58
BartekBartek
83
edited Mar 19 at 22:25
Travis
63.8k769151
edited Mar 19 at 22:25
Travis
63.8k769151
edited Mar 19 at 22:25
Travis
63.8k769151
63.8k769151
asked Mar 19 at 21:58
BartekBartek
83
asked Mar 19 at 21:58
BartekBartek
83
asked Mar 19 at 21:58
BartekBartek
83
83
$begingroup$
You write the action from $SO(3)$, meaning it is a $3times 3$ matrix, a 3D rotation.
$endgroup$
– Wojowu
Mar 19 at 22:20
$begingroup$
@Wojowu I edited it, thank you.
$endgroup$
– Bartek
Mar 19 at 22:22
add a comment |
$begingroup$
You write the action from $SO(3)$, meaning it is a $3times 3$ matrix, a 3D rotation.
$endgroup$
– Wojowu
Mar 19 at 22:20
$begingroup$
@Wojowu I edited it, thank you.
$endgroup$
– Bartek
Mar 19 at 22:22
$begingroup$
You write the action from $SO(3)$, meaning it is a $3times 3$ matrix, a 3D rotation.
$endgroup$
– Wojowu
Mar 19 at 22:20
$begingroup$
You write the action from $SO(3)$, meaning it is a $3times 3$ matrix, a 3D rotation.
$endgroup$
– Wojowu
Mar 19 at 22:20
$begingroup$
@Wojowu I edited it, thank you.
$endgroup$
– Bartek
Mar 19 at 22:22
$begingroup$
@Wojowu I edited it, thank you.
$endgroup$
– Bartek
Mar 19 at 22:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint Show that the given action of $SO(2)$ on $S^3$ is free. Thus, all of its orbits are embedded copies of $SO(2)$. Then, show directly that any orbit is contained within a $2$-dimensional subspace of $Bbb R^4$.
Additional hint Hence the orbits are the intersections of $S^3$ with these planes, that is, they comprise a family of great circles in $S^3$. In fact, space of orbits is homemorphic to $S^2$, and the canonical quotient map $S^3 to S^3 / SO(2) cong S^2$ is the classical Hopf fibration. If you're already familiar with quaternions, it is an instructive exercise to identify $SO(2)$ with the space of unit complex numbers and $S^3$ with the space of unit quaternions and rewrite the action in terms of quaternion multiplication.
answered Mar 19 at 22:48
TravisTravis
63.8k769151
$endgroup$
$begingroup$
If I'm interpreting everything correctly, if we identify $SO(2) cong U(1) subset Bbb C$ then this action is given (in terms of quaternionic multiplication) by $z cdot r := z r$.
$endgroup$
– Travis
Mar 20 at 22:20
$begingroup$
Why is space of orbits is homeomorphic to $S_2$?
$endgroup$
– Bartek
Mar 20 at 22:26
$begingroup$
See en.wikipedia.org/wiki/Hopf_fibration#Direct_construction
$endgroup$
– Travis
Mar 20 at 22:32
add a comment |
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%2f3154673%2fwhat-are-the-orbits-for-this-action-of-mathrmso2-on-s-3%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
$begingroup$
Hint Show that the given action of $SO(2)$ on $S^3$ is free. Thus, all of its orbits are embedded copies of $SO(2)$. Then, show directly that any orbit is contained within a $2$-dimensional subspace of $Bbb R^4$.
Additional hint Hence the orbits are the intersections of $S^3$ with these planes, that is, they comprise a family of great circles in $S^3$. In fact, space of orbits is homemorphic to $S^2$, and the canonical quotient map $S^3 to S^3 / SO(2) cong S^2$ is the classical Hopf fibration. If you're already familiar with quaternions, it is an instructive exercise to identify $SO(2)$ with the space of unit complex numbers and $S^3$ with the space of unit quaternions and rewrite the action in terms of quaternion multiplication.
answered Mar 19 at 22:48
TravisTravis
63.8k769151
$endgroup$
$begingroup$
If I'm interpreting everything correctly, if we identify $SO(2) cong U(1) subset Bbb C$ then this action is given (in terms of quaternionic multiplication) by $z cdot r := z r$.
$endgroup$
– Travis
Mar 20 at 22:20
$begingroup$
Why is space of orbits is homeomorphic to $S_2$?
$endgroup$
– Bartek
Mar 20 at 22:26
$begingroup$
See en.wikipedia.org/wiki/Hopf_fibration#Direct_construction
$endgroup$
– Travis
Mar 20 at 22:32
add a comment |
$begingroup$
Hint Show that the given action of $SO(2)$ on $S^3$ is free. Thus, all of its orbits are embedded copies of $SO(2)$. Then, show directly that any orbit is contained within a $2$-dimensional subspace of $Bbb R^4$.
Additional hint Hence the orbits are the intersections of $S^3$ with these planes, that is, they comprise a family of great circles in $S^3$. In fact, space of orbits is homemorphic to $S^2$, and the canonical quotient map $S^3 to S^3 / SO(2) cong S^2$ is the classical Hopf fibration. If you're already familiar with quaternions, it is an instructive exercise to identify $SO(2)$ with the space of unit complex numbers and $S^3$ with the space of unit quaternions and rewrite the action in terms of quaternion multiplication.
answered Mar 19 at 22:48
TravisTravis
63.8k769151
$endgroup$
$begingroup$
If I'm interpreting everything correctly, if we identify $SO(2) cong U(1) subset Bbb C$ then this action is given (in terms of quaternionic multiplication) by $z cdot r := z r$.
$endgroup$
– Travis
Mar 20 at 22:20
$begingroup$
Why is space of orbits is homeomorphic to $S_2$?
$endgroup$
– Bartek
Mar 20 at 22:26
$begingroup$
See en.wikipedia.org/wiki/Hopf_fibration#Direct_construction
$endgroup$
– Travis
Mar 20 at 22:32
add a comment |
$begingroup$
Hint Show that the given action of $SO(2)$ on $S^3$ is free. Thus, all of its orbits are embedded copies of $SO(2)$. Then, show directly that any orbit is contained within a $2$-dimensional subspace of $Bbb R^4$.
Additional hint Hence the orbits are the intersections of $S^3$ with these planes, that is, they comprise a family of great circles in $S^3$. In fact, space of orbits is homemorphic to $S^2$, and the canonical quotient map $S^3 to S^3 / SO(2) cong S^2$ is the classical Hopf fibration. If you're already familiar with quaternions, it is an instructive exercise to identify $SO(2)$ with the space of unit complex numbers and $S^3$ with the space of unit quaternions and rewrite the action in terms of quaternion multiplication.
answered Mar 19 at 22:48
TravisTravis
63.8k769151
$endgroup$
Hint Show that the given action of $SO(2)$ on $S^3$ is free. Thus, all of its orbits are embedded copies of $SO(2)$. Then, show directly that any orbit is contained within a $2$-dimensional subspace of $Bbb R^4$.
Additional hint Hence the orbits are the intersections of $S^3$ with these planes, that is, they comprise a family of great circles in $S^3$. In fact, space of orbits is homemorphic to $S^2$, and the canonical quotient map $S^3 to S^3 / SO(2) cong S^2$ is the classical Hopf fibration. If you're already familiar with quaternions, it is an instructive exercise to identify $SO(2)$ with the space of unit complex numbers and $S^3$ with the space of unit quaternions and rewrite the action in terms of quaternion multiplication.
answered Mar 19 at 22:48
TravisTravis
63.8k769151
answered Mar 19 at 22:48
TravisTravis
63.8k769151
answered Mar 19 at 22:48
TravisTravis
63.8k769151
answered Mar 19 at 22:48
TravisTravis
63.8k769151
63.8k769151
$begingroup$
If I'm interpreting everything correctly, if we identify $SO(2) cong U(1) subset Bbb C$ then this action is given (in terms of quaternionic multiplication) by $z cdot r := z r$.
$endgroup$
– Travis
Mar 20 at 22:20
$begingroup$
Why is space of orbits is homeomorphic to $S_2$?
$endgroup$
– Bartek
Mar 20 at 22:26
$begingroup$
See en.wikipedia.org/wiki/Hopf_fibration#Direct_construction
$endgroup$
– Travis
Mar 20 at 22:32
add a comment |
$begingroup$
If I'm interpreting everything correctly, if we identify $SO(2) cong U(1) subset Bbb C$ then this action is given (in terms of quaternionic multiplication) by $z cdot r := z r$.
$endgroup$
– Travis
Mar 20 at 22:20
$begingroup$
Why is space of orbits is homeomorphic to $S_2$?
$endgroup$
– Bartek
Mar 20 at 22:26
$begingroup$
See en.wikipedia.org/wiki/Hopf_fibration#Direct_construction
$endgroup$
– Travis
Mar 20 at 22:32
$begingroup$
If I'm interpreting everything correctly, if we identify $SO(2) cong U(1) subset Bbb C$ then this action is given (in terms of quaternionic multiplication) by $z cdot r := z r$.
$endgroup$
– Travis
Mar 20 at 22:20
$begingroup$
If I'm interpreting everything correctly, if we identify $SO(2) cong U(1) subset Bbb C$ then this action is given (in terms of quaternionic multiplication) by $z cdot r := z r$.
$endgroup$
– Travis
Mar 20 at 22:20
$begingroup$
Why is space of orbits is homeomorphic to $S_2$?
$endgroup$
– Bartek
Mar 20 at 22:26
$begingroup$
Why is space of orbits is homeomorphic to $S_2$?
$endgroup$
– Bartek
Mar 20 at 22:26
$begingroup$
See en.wikipedia.org/wiki/Hopf_fibration#Direct_construction
$endgroup$
– Travis
Mar 20 at 22:32
$begingroup$
See en.wikipedia.org/wiki/Hopf_fibration#Direct_construction
$endgroup$
– Travis
Mar 20 at 22:32
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.
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%2f3154673%2fwhat-are-the-orbits-for-this-action-of-mathrmso2-on-s-3%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
$begingroup$
You write the action from $SO(3)$, meaning it is a $3times 3$ matrix, a 3D rotation.
$endgroup$
– Wojowu
Mar 19 at 22:20
$begingroup$
@Wojowu I edited it, thank you.
$endgroup$
– Bartek
Mar 19 at 22:22