Mixing
Homogeneous space and quotient space for spin groups
Clash Royale CLAN TAG#URR8PPP
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1
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We know that
$$
O(n+1)/O(n) simeq SO(n+1)/SO(n) simeq S^n,
$$
based on the result of homogeneous space.
These are in some sense spheres.
If we embed the spin group $Spin(n)$ into $Spin(n+1)$, we may be able to define the quotient space
$$
Spin(n+1)/Spin(n) simeq ?
$$
- Do we have simpler expressions for the above "manifolds" or "quotient space"?
My Trial/Attempt:
We know that
$$
Spin(2)simeq U(1)simeq SO(2) simeq S^1
$$
$$
Spin(3)simeq SU(2)simeq S^3
$$
$$
Spin(4)simeq SU(2) times SU(2)
$$
$$
Spin(5)simeq Sp(2)
$$
$$
Spin(6)simeq SU(4),
$$
while
$$
Spin(3)/Spin(2) simeq S^3/ S^1 simeq S^2
$$
$$
Spin(4)/Spin(3) simeq (SU(2) times SU(2))/ S^3 simeq (S^3 times S^3 )/ S^3 simeq S^3
$$
Could we obtain the generic formulas for $Spin(n+1)/Spin(n) simeq ?
$
manifolds differential-topology lie-groups quotient-spaces homogeneous-spaces
asked Aug 6 at 0:41
wonderich
1,68821226
add a comment |Â
up vote
1
down vote
favorite
We know that
$$
O(n+1)/O(n) simeq SO(n+1)/SO(n) simeq S^n,
$$
based on the result of homogeneous space.
These are in some sense spheres.
If we embed the spin group $Spin(n)$ into $Spin(n+1)$, we may be able to define the quotient space
$$
Spin(n+1)/Spin(n) simeq ?
$$
- Do we have simpler expressions for the above "manifolds" or "quotient space"?
My Trial/Attempt:
We know that
$$
Spin(2)simeq U(1)simeq SO(2) simeq S^1
$$
$$
Spin(3)simeq SU(2)simeq S^3
$$
$$
Spin(4)simeq SU(2) times SU(2)
$$
$$
Spin(5)simeq Sp(2)
$$
$$
Spin(6)simeq SU(4),
$$
while
$$
Spin(3)/Spin(2) simeq S^3/ S^1 simeq S^2
$$
$$
Spin(4)/Spin(3) simeq (SU(2) times SU(2))/ S^3 simeq (S^3 times S^3 )/ S^3 simeq S^3
$$
Could we obtain the generic formulas for $Spin(n+1)/Spin(n) simeq ?
$
manifolds differential-topology lie-groups quotient-spaces homogeneous-spaces
asked Aug 6 at 0:41
wonderich
1,68821226
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We know that
$$
O(n+1)/O(n) simeq SO(n+1)/SO(n) simeq S^n,
$$
based on the result of homogeneous space.
These are in some sense spheres.
If we embed the spin group $Spin(n)$ into $Spin(n+1)$, we may be able to define the quotient space
$$
Spin(n+1)/Spin(n) simeq ?
$$
- Do we have simpler expressions for the above "manifolds" or "quotient space"?
My Trial/Attempt:
We know that
$$
Spin(2)simeq U(1)simeq SO(2) simeq S^1
$$
$$
Spin(3)simeq SU(2)simeq S^3
$$
$$
Spin(4)simeq SU(2) times SU(2)
$$
$$
Spin(5)simeq Sp(2)
$$
$$
Spin(6)simeq SU(4),
$$
while
$$
Spin(3)/Spin(2) simeq S^3/ S^1 simeq S^2
$$
$$
Spin(4)/Spin(3) simeq (SU(2) times SU(2))/ S^3 simeq (S^3 times S^3 )/ S^3 simeq S^3
$$
Could we obtain the generic formulas for $Spin(n+1)/Spin(n) simeq ?
$
manifolds differential-topology lie-groups quotient-spaces homogeneous-spaces
asked Aug 6 at 0:41
wonderich
1,68821226
We know that
$$
O(n+1)/O(n) simeq SO(n+1)/SO(n) simeq S^n,
$$
based on the result of homogeneous space.
These are in some sense spheres.
If we embed the spin group $Spin(n)$ into $Spin(n+1)$, we may be able to define the quotient space
$$
Spin(n+1)/Spin(n) simeq ?
$$
- Do we have simpler expressions for the above "manifolds" or "quotient space"?
My Trial/Attempt:
We know that
$$
Spin(2)simeq U(1)simeq SO(2) simeq S^1
$$
$$
Spin(3)simeq SU(2)simeq S^3
$$
$$
Spin(4)simeq SU(2) times SU(2)
$$
$$
Spin(5)simeq Sp(2)
$$
$$
Spin(6)simeq SU(4),
$$
while
$$
Spin(3)/Spin(2) simeq S^3/ S^1 simeq S^2
$$
$$
Spin(4)/Spin(3) simeq (SU(2) times SU(2))/ S^3 simeq (S^3 times S^3 )/ S^3 simeq S^3
$$
Could we obtain the generic formulas for $Spin(n+1)/Spin(n) simeq ?
$
manifolds differential-topology lie-groups quotient-spaces homogeneous-spaces
asked Aug 6 at 0:41
wonderich
1,68821226
asked Aug 6 at 0:41
wonderich
1,68821226
asked Aug 6 at 0:41
wonderich
1,68821226
asked Aug 6 at 0:41
wonderich
1,68821226
1,68821226
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Recall that $SO(n+1)$ acts smoothly and transitively on $S^n$ with isotropy subgroup $SO(n)$, so $SO(n+1)/SO(n)$ is diffeomorphic to $S^n$. Now note that $operatornameSpin(n+1)$ also acts smoothly and transitively on $S^n$ by first mapping to $SO(n+1)$. The isotropy subgroup is the subgroup of $operatornameSpin(n+1)$ which projects to $SO(n) subset SO(n+1)$, this is precisely $operatornameSpin(n)$. Therefore $operatornameSpin(n+1)/operatornameSpin(n)$ is diffeomorphic to $S^n$.
More generally, suppose $G$ is a compact Lie group which acts smoothly and transitively on a smooth manifold $M$ with isotropy subgroup $H$. If $hatG$ is a compact Lie group and $pi : hatG to G$ is a surjective Lie group homomorphism, then $hatG$ acts smoothly and transitively on $M$ with isotropy subgroup $pi^-1(H)$, so both $G/H$ and $hatG/pi^-1(H)$ are diffeomorphic to $M$.
answered Aug 6 at 15:10
Michael Albanese
61.3k1591290
Thanks, this is really a clean answer. +1.
– wonderich
Aug 6 at 16:10
I accept your answer
– wonderich
Aug 7 at 15:41
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Recall that $SO(n+1)$ acts smoothly and transitively on $S^n$ with isotropy subgroup $SO(n)$, so $SO(n+1)/SO(n)$ is diffeomorphic to $S^n$. Now note that $operatornameSpin(n+1)$ also acts smoothly and transitively on $S^n$ by first mapping to $SO(n+1)$. The isotropy subgroup is the subgroup of $operatornameSpin(n+1)$ which projects to $SO(n) subset SO(n+1)$, this is precisely $operatornameSpin(n)$. Therefore $operatornameSpin(n+1)/operatornameSpin(n)$ is diffeomorphic to $S^n$.
More generally, suppose $G$ is a compact Lie group which acts smoothly and transitively on a smooth manifold $M$ with isotropy subgroup $H$. If $hatG$ is a compact Lie group and $pi : hatG to G$ is a surjective Lie group homomorphism, then $hatG$ acts smoothly and transitively on $M$ with isotropy subgroup $pi^-1(H)$, so both $G/H$ and $hatG/pi^-1(H)$ are diffeomorphic to $M$.
answered Aug 6 at 15:10
Michael Albanese
61.3k1591290
Thanks, this is really a clean answer. +1.
– wonderich
Aug 6 at 16:10
I accept your answer
– wonderich
Aug 7 at 15:41
add a comment |Â
up vote
2
down vote
accepted
Recall that $SO(n+1)$ acts smoothly and transitively on $S^n$ with isotropy subgroup $SO(n)$, so $SO(n+1)/SO(n)$ is diffeomorphic to $S^n$. Now note that $operatornameSpin(n+1)$ also acts smoothly and transitively on $S^n$ by first mapping to $SO(n+1)$. The isotropy subgroup is the subgroup of $operatornameSpin(n+1)$ which projects to $SO(n) subset SO(n+1)$, this is precisely $operatornameSpin(n)$. Therefore $operatornameSpin(n+1)/operatornameSpin(n)$ is diffeomorphic to $S^n$.
More generally, suppose $G$ is a compact Lie group which acts smoothly and transitively on a smooth manifold $M$ with isotropy subgroup $H$. If $hatG$ is a compact Lie group and $pi : hatG to G$ is a surjective Lie group homomorphism, then $hatG$ acts smoothly and transitively on $M$ with isotropy subgroup $pi^-1(H)$, so both $G/H$ and $hatG/pi^-1(H)$ are diffeomorphic to $M$.
answered Aug 6 at 15:10
Michael Albanese
61.3k1591290
Thanks, this is really a clean answer. +1.
– wonderich
Aug 6 at 16:10
I accept your answer
– wonderich
Aug 7 at 15:41
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Recall that $SO(n+1)$ acts smoothly and transitively on $S^n$ with isotropy subgroup $SO(n)$, so $SO(n+1)/SO(n)$ is diffeomorphic to $S^n$. Now note that $operatornameSpin(n+1)$ also acts smoothly and transitively on $S^n$ by first mapping to $SO(n+1)$. The isotropy subgroup is the subgroup of $operatornameSpin(n+1)$ which projects to $SO(n) subset SO(n+1)$, this is precisely $operatornameSpin(n)$. Therefore $operatornameSpin(n+1)/operatornameSpin(n)$ is diffeomorphic to $S^n$.
More generally, suppose $G$ is a compact Lie group which acts smoothly and transitively on a smooth manifold $M$ with isotropy subgroup $H$. If $hatG$ is a compact Lie group and $pi : hatG to G$ is a surjective Lie group homomorphism, then $hatG$ acts smoothly and transitively on $M$ with isotropy subgroup $pi^-1(H)$, so both $G/H$ and $hatG/pi^-1(H)$ are diffeomorphic to $M$.
answered Aug 6 at 15:10
Michael Albanese
61.3k1591290
Recall that $SO(n+1)$ acts smoothly and transitively on $S^n$ with isotropy subgroup $SO(n)$, so $SO(n+1)/SO(n)$ is diffeomorphic to $S^n$. Now note that $operatornameSpin(n+1)$ also acts smoothly and transitively on $S^n$ by first mapping to $SO(n+1)$. The isotropy subgroup is the subgroup of $operatornameSpin(n+1)$ which projects to $SO(n) subset SO(n+1)$, this is precisely $operatornameSpin(n)$. Therefore $operatornameSpin(n+1)/operatornameSpin(n)$ is diffeomorphic to $S^n$.
More generally, suppose $G$ is a compact Lie group which acts smoothly and transitively on a smooth manifold $M$ with isotropy subgroup $H$. If $hatG$ is a compact Lie group and $pi : hatG to G$ is a surjective Lie group homomorphism, then $hatG$ acts smoothly and transitively on $M$ with isotropy subgroup $pi^-1(H)$, so both $G/H$ and $hatG/pi^-1(H)$ are diffeomorphic to $M$.
answered Aug 6 at 15:10
Michael Albanese
61.3k1591290
edited Aug 6 at 15:35
answered Aug 6 at 15:10
Michael Albanese
61.3k1591290
answered Aug 6 at 15:10
Michael Albanese
61.3k1591290
answered Aug 6 at 15:10
Michael Albanese
61.3k1591290
61.3k1591290
Thanks, this is really a clean answer. +1.
– wonderich
Aug 6 at 16:10
I accept your answer
– wonderich
Aug 7 at 15:41
add a comment |Â
Thanks, this is really a clean answer. +1.
– wonderich
Aug 6 at 16:10
I accept your answer
– wonderich
Aug 7 at 15:41
Thanks, this is really a clean answer. +1.
– wonderich
Aug 6 at 16:10
Thanks, this is really a clean answer. +1.
– wonderich
Aug 6 at 16:10
I accept your answer
– wonderich
Aug 7 at 15:41
I accept your answer
– wonderich
Aug 7 at 15:41
add a comment |Â
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