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
10k3932
$endgroup$
$begingroup$
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 Jan 12 at 13:09
Paul FrostPaul Frost
10k3932
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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
10k3932
$endgroup$
$begingroup$
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
10k3932
$endgroup$
$begingroup$
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
10k3932
$endgroup$
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
10k3932
answered Dec 2 '18 at 19:43
Paul FrostPaul Frost
10k3932
answered Dec 2 '18 at 19:43
Paul FrostPaul Frost
10k3932
answered Dec 2 '18 at 19:43
Paul FrostPaul Frost
10k3932
10k3932
$begingroup$
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 Jan 12 at 13:09
Paul FrostPaul Frost
10k3932
$endgroup$
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 Jan 12 at 13:09
Paul FrostPaul Frost
10k3932
$endgroup$
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 Jan 12 at 13:09
Paul FrostPaul Frost
10k3932
$endgroup$
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 Jan 12 at 13:09
Paul FrostPaul Frost
10k3932
answered Jan 12 at 13:09
Paul FrostPaul Frost
10k3932
answered Jan 12 at 13:09
Paul FrostPaul Frost
10k3932
answered Jan 12 at 13:09
Paul FrostPaul Frost
10k3932
10k3932
add a comment |
add a comment |
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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