Continuous function $ f：operatorname{SO}(3) to operatorname{SU}(2)$
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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 at 18:58
kot
808
add a comment |
up vote
1
down vote
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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 at 18:58
kot
808
2
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.
– Charlie Frohman
Dec 2 at 19:27
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 at 18:58
kot
808
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 at 18:58
kot
808
asked Dec 2 at 18:58
kot
808
edited Dec 4 at 19:07
asked Dec 2 at 18:58
kot
808
asked Dec 2 at 18:58
kot
808
asked Dec 2 at 18:58
kot
808
808
2
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.
– Charlie Frohman
Dec 2 at 19:27
add a comment |
2
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.
– Charlie Frohman
Dec 2 at 19:27
2
2
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.
– Charlie Frohman
Dec 2 at 19:27
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.
– Charlie Frohman
Dec 2 at 19:27
add a comment |
1 Answer
1
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oldest
votes
up vote
4
down vote
accepted
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 at 19:43
Paul Frost
8,1891528
Thanks @Paul Frost
– kot
Dec 2 at 19:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
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 at 19:43
Paul Frost
8,1891528
Thanks @Paul Frost
– kot
Dec 2 at 19:44
add a comment |
up vote
4
down vote
accepted
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 at 19:43
Paul Frost
8,1891528
Thanks @Paul Frost
– kot
Dec 2 at 19:44
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
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 at 19:43
Paul Frost
8,1891528
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 at 19:43
Paul Frost
8,1891528
answered Dec 2 at 19:43
Paul Frost
8,1891528
answered Dec 2 at 19:43
Paul Frost
8,1891528
answered Dec 2 at 19:43
Paul Frost
8,1891528
8,1891528
Thanks @Paul Frost
– kot
Dec 2 at 19:44
add a comment |
Thanks @Paul Frost
– kot
Dec 2 at 19:44
Thanks @Paul Frost
– kot
Dec 2 at 19:44
Thanks @Paul Frost
– kot
Dec 2 at 19:44
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.
– Charlie Frohman
Dec 2 at 19:27