Continuous function $ f：operatorname{SO}(3) to operatorname{SU}(2)$
$begingroup$
Given the usual surjective homomorphism $ Φ:operatorname{SU}(2)to operatorname{SO}(3)$ that maps a quaternion to a rotation matrix,
Does there exist a continuous function $f:operatorname{SO}(3)to operatorname{SU}(2)$
such that $ Phi circ f = operatorname{Id}$ on $operatorname{SO}(3)$?
If yes, what would be this map be?
Thank you
general-topology lie-groups smooth-manifolds
asked Dec 2 '18 at 18:58
kotkot
808
$endgroup$
add a comment |
$begingroup$
Given the usual surjective homomorphism $ Φ:operatorname{SU}(2)to operatorname{SO}(3)$ that maps a quaternion to a rotation matrix,
Does there exist a continuous function $f:operatorname{SO}(3)to operatorname{SU}(2)$
such that $ Phi circ f = operatorname{Id}$ on $operatorname{SO}(3)$?
If yes, what would be this map be?
Thank you
general-topology lie-groups smooth-manifolds
asked Dec 2 '18 at 18:58
kotkot
808
$endgroup$
2
$begingroup$
No. It’s like trying to define a continuous square root in the complex plane. An element of SO(3) has an axis and an angle of rotation. The corresponding element of SU(2) has one half the angle.
$endgroup$
– Charlie Frohman
Dec 2 '18 at 19:27
add a comment |
$begingroup$
Given the usual surjective homomorphism $ Φ:operatorname{SU}(2)to operatorname{SO}(3)$ that maps a quaternion to a rotation matrix,
Does there exist a continuous function $f:operatorname{SO}(3)to operatorname{SU}(2)$
such that $ Phi circ f = operatorname{Id}$ on $operatorname{SO}(3)$?
If yes, what would be this map be?
Thank you
general-topology lie-groups smooth-manifolds
asked Dec 2 '18 at 18:58
kotkot
808
$endgroup$
Given the usual surjective homomorphism $ Φ:operatorname{SU}(2)to operatorname{SO}(3)$ that maps a quaternion to a rotation matrix,
Does there exist a continuous function $f:operatorname{SO}(3)to operatorname{SU}(2)$
such that $ Phi circ f = operatorname{Id}$ on $operatorname{SO}(3)$?
If yes, what would be this map be?
Thank you
general-topology lie-groups smooth-manifolds
general-topology lie-groups smooth-manifolds
asked Dec 2 '18 at 18:58
kotkot
808
asked Dec 2 '18 at 18:58
kotkot
808
edited Dec 4 '18 at 19:07
kot
asked Dec 2 '18 at 18:58
kotkot
808
asked Dec 2 '18 at 18:58
kotkot
808
asked Dec 2 '18 at 18:58
kotkot
808
808
2
$begingroup$
No. It’s like trying to define a continuous square root in the complex plane. An element of SO(3) has an axis and an angle of rotation. The corresponding element of SU(2) has one half the angle.
$endgroup$
– Charlie Frohman
Dec 2 '18 at 19:27
add a comment |
2
$begingroup$
No. It’s like trying to define a continuous square root in the complex plane. An element of SO(3) has an axis and an angle of rotation. The corresponding element of SU(2) has one half the angle.
$endgroup$
– Charlie Frohman
Dec 2 '18 at 19:27
2
2
$begingroup$
No. It’s like trying to define a continuous square root in the complex plane. An element of SO(3) has an axis and an angle of rotation. The corresponding element of SU(2) has one half the angle.
$endgroup$
– Charlie Frohman
Dec 2 '18 at 19:27
$begingroup$
No. It’s like trying to define a continuous square root in the complex plane. An element of SO(3) has an axis and an angle of rotation. The corresponding element of SU(2) has one half the angle.
$endgroup$
– Charlie Frohman
Dec 2 '18 at 19:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No. A map $f$ as in your question would induce an embedding of fundamental groups $f_* :pi_1(text{SO(3)}) to pi_1(text{SU}(2))$.
If you look at Fundamental group of $SO(3)$, you will see that $pi_1(text{SU}(2)) = 0$, while $pi_1(text{SO(3)}) = mathbb{Z}_2$. Therefore $f$ cannot exist.
answered Dec 2 '18 at 19:43
Paul FrostPaul Frost
9,9153932
$endgroup$
$begingroup$
Thanks @Paul Frost
$endgroup$
– kot
Dec 2 '18 at 19:44
add a comment |
$begingroup$
Here is another proof. It is known that $Phi$ is a covering projection with two sheets. The map $f$ would be a section of $Phi$, but this would imply that $Phi$ is a homeomorphism. See If a covering map has a section, is it a $1$-fold cover?.
answered 2 days ago
Paul FrostPaul Frost
9,9153932
$endgroup$
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%2f3023045%2fcontinuous-function-f-operatornameso3-to-operatornamesu2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. A map $f$ as in your question would induce an embedding of fundamental groups $f_* :pi_1(text{SO(3)}) to pi_1(text{SU}(2))$.
If you look at Fundamental group of $SO(3)$, you will see that $pi_1(text{SU}(2)) = 0$, while $pi_1(text{SO(3)}) = mathbb{Z}_2$. Therefore $f$ cannot exist.
answered Dec 2 '18 at 19:43
Paul FrostPaul Frost
9,9153932
$endgroup$
$begingroup$
Thanks @Paul Frost
$endgroup$
– kot
Dec 2 '18 at 19:44
add a comment |
$begingroup$
No. A map $f$ as in your question would induce an embedding of fundamental groups $f_* :pi_1(text{SO(3)}) to pi_1(text{SU}(2))$.
If you look at Fundamental group of $SO(3)$, you will see that $pi_1(text{SU}(2)) = 0$, while $pi_1(text{SO(3)}) = mathbb{Z}_2$. Therefore $f$ cannot exist.
answered Dec 2 '18 at 19:43
Paul FrostPaul Frost
9,9153932
$endgroup$
$begingroup$
Thanks @Paul Frost
$endgroup$
– kot
Dec 2 '18 at 19:44
add a comment |
$begingroup$
No. A map $f$ as in your question would induce an embedding of fundamental groups $f_* :pi_1(text{SO(3)}) to pi_1(text{SU}(2))$.
If you look at Fundamental group of $SO(3)$, you will see that $pi_1(text{SU}(2)) = 0$, while $pi_1(text{SO(3)}) = mathbb{Z}_2$. Therefore $f$ cannot exist.
answered Dec 2 '18 at 19:43
Paul FrostPaul Frost
9,9153932
$endgroup$
No. A map $f$ as in your question would induce an embedding of fundamental groups $f_* :pi_1(text{SO(3)}) to pi_1(text{SU}(2))$.
If you look at Fundamental group of $SO(3)$, you will see that $pi_1(text{SU}(2)) = 0$, while $pi_1(text{SO(3)}) = mathbb{Z}_2$. Therefore $f$ cannot exist.
answered Dec 2 '18 at 19:43
Paul FrostPaul Frost
9,9153932
answered Dec 2 '18 at 19:43
Paul FrostPaul Frost
9,9153932
answered Dec 2 '18 at 19:43
Paul FrostPaul Frost
9,9153932
answered Dec 2 '18 at 19:43
Paul FrostPaul Frost
9,9153932
9,9153932
$begingroup$
Thanks @Paul Frost
$endgroup$
– kot
Dec 2 '18 at 19:44
add a comment |
$begingroup$
Thanks @Paul Frost
$endgroup$
– kot
Dec 2 '18 at 19:44
$begingroup$
Thanks @Paul Frost
$endgroup$
– kot
Dec 2 '18 at 19:44
$begingroup$
Thanks @Paul Frost
$endgroup$
– kot
Dec 2 '18 at 19:44
add a comment |
$begingroup$
Here is another proof. It is known that $Phi$ is a covering projection with two sheets. The map $f$ would be a section of $Phi$, but this would imply that $Phi$ is a homeomorphism. See If a covering map has a section, is it a $1$-fold cover?.
answered 2 days ago
Paul FrostPaul Frost
9,9153932
$endgroup$
add a comment |
$begingroup$
Here is another proof. It is known that $Phi$ is a covering projection with two sheets. The map $f$ would be a section of $Phi$, but this would imply that $Phi$ is a homeomorphism. See If a covering map has a section, is it a $1$-fold cover?.
answered 2 days ago
Paul FrostPaul Frost
9,9153932
$endgroup$
add a comment |
$begingroup$
Here is another proof. It is known that $Phi$ is a covering projection with two sheets. The map $f$ would be a section of $Phi$, but this would imply that $Phi$ is a homeomorphism. See If a covering map has a section, is it a $1$-fold cover?.
answered 2 days ago
Paul FrostPaul Frost
9,9153932
$endgroup$
Here is another proof. It is known that $Phi$ is a covering projection with two sheets. The map $f$ would be a section of $Phi$, but this would imply that $Phi$ is a homeomorphism. See If a covering map has a section, is it a $1$-fold cover?.
answered 2 days ago
Paul FrostPaul Frost
9,9153932
answered 2 days ago
Paul FrostPaul Frost
9,9153932
answered 2 days ago
Paul FrostPaul Frost
9,9153932
answered 2 days ago
Paul FrostPaul Frost
9,9153932
9,9153932
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.
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%2f3023045%2fcontinuous-function-f-operatornameso3-to-operatornamesu2%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
2
$begingroup$
No. It’s like trying to define a continuous square root in the complex plane. An element of SO(3) has an axis and an angle of rotation. The corresponding element of SU(2) has one half the angle.
$endgroup$
– Charlie Frohman
Dec 2 '18 at 19:27