Homogeneous space and nice manifolds

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite
1












We know that
$$
O(n+1)/O(n) simeq SO(n+1)/SO(n) simeq S^n,
$$
based on the result of homogeneous space.



Also
$$
PO(n+1)/PO(n) simeq P^n,
$$
$P^n$ is the projective space.



Do we have similar nice results for the following:
$$
U(n+1)/U(n) simeq?
$$
and
$$
SU(n+1)/SU(n) simeq?
$$
$$
Sp(n+1)/Sp(n) simeq?
$$




differential-geometry lie-groups quotient-spaces homogeneous-spaces




share|cite|improve this question





















  • What is $PO(n)$?
    – John Ma
    Jul 30 at 15:19










  • Projectivized orthogonal group?
    – John Hughes
    Jul 30 at 15:22










  • Have you looked in Husemolller or Steenrod? I think that all this stuff's discussed in the first few chapters of those books, but it's been a while.
    – John Hughes
    Jul 30 at 15:23














up vote
2
down vote

favorite
1












We know that
$$
O(n+1)/O(n) simeq SO(n+1)/SO(n) simeq S^n,
$$
based on the result of homogeneous space.



Also
$$
PO(n+1)/PO(n) simeq P^n,
$$
$P^n$ is the projective space.



Do we have similar nice results for the following:
$$
U(n+1)/U(n) simeq?
$$
and
$$
SU(n+1)/SU(n) simeq?
$$
$$
Sp(n+1)/Sp(n) simeq?
$$




differential-geometry lie-groups quotient-spaces homogeneous-spaces




share|cite|improve this question





















  • What is $PO(n)$?
    – John Ma
    Jul 30 at 15:19










  • Projectivized orthogonal group?
    – John Hughes
    Jul 30 at 15:22










  • Have you looked in Husemolller or Steenrod? I think that all this stuff's discussed in the first few chapters of those books, but it's been a while.
    – John Hughes
    Jul 30 at 15:23












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





We know that
$$
O(n+1)/O(n) simeq SO(n+1)/SO(n) simeq S^n,
$$
based on the result of homogeneous space.



Also
$$
PO(n+1)/PO(n) simeq P^n,
$$
$P^n$ is the projective space.



Do we have similar nice results for the following:
$$
U(n+1)/U(n) simeq?
$$
and
$$
SU(n+1)/SU(n) simeq?
$$
$$
Sp(n+1)/Sp(n) simeq?
$$




differential-geometry lie-groups quotient-spaces homogeneous-spaces




share|cite|improve this question













We know that
$$
O(n+1)/O(n) simeq SO(n+1)/SO(n) simeq S^n,
$$
based on the result of homogeneous space.



Also
$$
PO(n+1)/PO(n) simeq P^n,
$$
$P^n$ is the projective space.



Do we have similar nice results for the following:
$$
U(n+1)/U(n) simeq?
$$
and
$$
SU(n+1)/SU(n) simeq?
$$
$$
Sp(n+1)/Sp(n) simeq?
$$





differential-geometry lie-groups quotient-spaces homogeneous-spaces





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 30 at 15:20









John Ma

37.5k93669




37.5k93669









asked Jul 30 at 15:12









wonderich

1,60011226




1,60011226











  • What is $PO(n)$?
    – John Ma
    Jul 30 at 15:19










  • Projectivized orthogonal group?
    – John Hughes
    Jul 30 at 15:22










  • Have you looked in Husemolller or Steenrod? I think that all this stuff's discussed in the first few chapters of those books, but it's been a while.
    – John Hughes
    Jul 30 at 15:23
















  • What is $PO(n)$?
    – John Ma
    Jul 30 at 15:19










  • Projectivized orthogonal group?
    – John Hughes
    Jul 30 at 15:22










  • Have you looked in Husemolller or Steenrod? I think that all this stuff's discussed in the first few chapters of those books, but it's been a while.
    – John Hughes
    Jul 30 at 15:23















What is $PO(n)$?
– John Ma
Jul 30 at 15:19




What is $PO(n)$?
– John Ma
Jul 30 at 15:19












Projectivized orthogonal group?
– John Hughes
Jul 30 at 15:22




Projectivized orthogonal group?
– John Hughes
Jul 30 at 15:22












Have you looked in Husemolller or Steenrod? I think that all this stuff's discussed in the first few chapters of those books, but it's been a while.
– John Hughes
Jul 30 at 15:23




Have you looked in Husemolller or Steenrod? I think that all this stuff's discussed in the first few chapters of those books, but it's been a while.
– John Hughes
Jul 30 at 15:23










1 Answer
1






active

oldest

votes

















up vote
2
down vote













There are all spheres. The relevant technical lemma is the follow:




Suppose $G$ is a compact Lie group which acts smoothly and transitively on a manifold $M$. Let $pin M$ and set $G_p = gin G: gp = p$. Then $G/G_p$ is canonically diffeomorphic to $M$ via the diffeomorphism $g G_pmapsto gp$.




For $G = U(n+1)$ or $G = SU(n+1)$, the relevant $M$ is the the sphere $S^2n+1subseteq mathbbC^n+1$, where the $G$ action is given by usual matrix multiplication on vectors in $mathbbC^n+1$. For $G = Sp(n+1)$, $M = S^4n+3subseteq mathbbH^n+1$.



One must verify that the $G$ action is transitive, and then compute $G_p$ for a single $p$.



Technically, the notation $U(n+1)/U(n)$, etc, is ambiguous until one specifies an embedding $U(n)rightarrow U(n+1)$. Fr the examples at hand, it is often the case (but not always the case) that there is a unique embedding $G_prightarrow G$, up to conjugacy. Conjugate embeddings always give rise to diffeomorphic manifolds: $G/G_pcong G/gG_p g^-1$.



One the other hand, the two embeddings of $U(1)$ into $U(2)$ given by $zmapsto operatornamediag(z,1)$ or $operatornamediag(z,z)$ give rise to homogeneous spaces which are not even homotopy equivalent. One has an analogous result for $Sp(1)rightarrow Sp(2)$. But these are the only exceptions: e.g. $U(2)$ embeds into $U(3)$ in only way one, up to conjugacy.






share|cite|improve this answer





















  • Thanks +1, how about $mathbbCP^n+1/mathbbCP^n$?
    – wonderich
    Jul 30 at 17:51











  • $mathbbCP^n$ is never a Lie group for any $n > 0$. What do you mean by your notation?
    – Jason DeVito
    Jul 30 at 18:08











  • $mathbbCP^1 simeq S^2$ is a 2-sphere. It is a complex projective space
    – wonderich
    Jul 30 at 18:18











  • typo fixed -thanks.
    – wonderich
    Jul 30 at 18:18






  • 1




    One can consider the topological quotient $mathbbCP^n+1/mathbbCP^n$ for some prescribed embedding $mathbbCP^nrightarrow mathbbCP^n+1$, but that is very different from Lie theoretic kind of quotient.
    – Jason DeVito
    Jul 30 at 18:22










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













There are all spheres. The relevant technical lemma is the follow:




Suppose $G$ is a compact Lie group which acts smoothly and transitively on a manifold $M$. Let $pin M$ and set $G_p = gin G: gp = p$. Then $G/G_p$ is canonically diffeomorphic to $M$ via the diffeomorphism $g G_pmapsto gp$.




For $G = U(n+1)$ or $G = SU(n+1)$, the relevant $M$ is the the sphere $S^2n+1subseteq mathbbC^n+1$, where the $G$ action is given by usual matrix multiplication on vectors in $mathbbC^n+1$. For $G = Sp(n+1)$, $M = S^4n+3subseteq mathbbH^n+1$.



One must verify that the $G$ action is transitive, and then compute $G_p$ for a single $p$.



Technically, the notation $U(n+1)/U(n)$, etc, is ambiguous until one specifies an embedding $U(n)rightarrow U(n+1)$. Fr the examples at hand, it is often the case (but not always the case) that there is a unique embedding $G_prightarrow G$, up to conjugacy. Conjugate embeddings always give rise to diffeomorphic manifolds: $G/G_pcong G/gG_p g^-1$.



One the other hand, the two embeddings of $U(1)$ into $U(2)$ given by $zmapsto operatornamediag(z,1)$ or $operatornamediag(z,z)$ give rise to homogeneous spaces which are not even homotopy equivalent. One has an analogous result for $Sp(1)rightarrow Sp(2)$. But these are the only exceptions: e.g. $U(2)$ embeds into $U(3)$ in only way one, up to conjugacy.






share|cite|improve this answer





















  • Thanks +1, how about $mathbbCP^n+1/mathbbCP^n$?
    – wonderich
    Jul 30 at 17:51











  • $mathbbCP^n$ is never a Lie group for any $n > 0$. What do you mean by your notation?
    – Jason DeVito
    Jul 30 at 18:08











  • $mathbbCP^1 simeq S^2$ is a 2-sphere. It is a complex projective space
    – wonderich
    Jul 30 at 18:18











  • typo fixed -thanks.
    – wonderich
    Jul 30 at 18:18






  • 1




    One can consider the topological quotient $mathbbCP^n+1/mathbbCP^n$ for some prescribed embedding $mathbbCP^nrightarrow mathbbCP^n+1$, but that is very different from Lie theoretic kind of quotient.
    – Jason DeVito
    Jul 30 at 18:22














up vote
2
down vote













There are all spheres. The relevant technical lemma is the follow:




Suppose $G$ is a compact Lie group which acts smoothly and transitively on a manifold $M$. Let $pin M$ and set $G_p = gin G: gp = p$. Then $G/G_p$ is canonically diffeomorphic to $M$ via the diffeomorphism $g G_pmapsto gp$.




For $G = U(n+1)$ or $G = SU(n+1)$, the relevant $M$ is the the sphere $S^2n+1subseteq mathbbC^n+1$, where the $G$ action is given by usual matrix multiplication on vectors in $mathbbC^n+1$. For $G = Sp(n+1)$, $M = S^4n+3subseteq mathbbH^n+1$.



One must verify that the $G$ action is transitive, and then compute $G_p$ for a single $p$.



Technically, the notation $U(n+1)/U(n)$, etc, is ambiguous until one specifies an embedding $U(n)rightarrow U(n+1)$. Fr the examples at hand, it is often the case (but not always the case) that there is a unique embedding $G_prightarrow G$, up to conjugacy. Conjugate embeddings always give rise to diffeomorphic manifolds: $G/G_pcong G/gG_p g^-1$.



One the other hand, the two embeddings of $U(1)$ into $U(2)$ given by $zmapsto operatornamediag(z,1)$ or $operatornamediag(z,z)$ give rise to homogeneous spaces which are not even homotopy equivalent. One has an analogous result for $Sp(1)rightarrow Sp(2)$. But these are the only exceptions: e.g. $U(2)$ embeds into $U(3)$ in only way one, up to conjugacy.






share|cite|improve this answer





















  • Thanks +1, how about $mathbbCP^n+1/mathbbCP^n$?
    – wonderich
    Jul 30 at 17:51











  • $mathbbCP^n$ is never a Lie group for any $n > 0$. What do you mean by your notation?
    – Jason DeVito
    Jul 30 at 18:08











  • $mathbbCP^1 simeq S^2$ is a 2-sphere. It is a complex projective space
    – wonderich
    Jul 30 at 18:18











  • typo fixed -thanks.
    – wonderich
    Jul 30 at 18:18






  • 1




    One can consider the topological quotient $mathbbCP^n+1/mathbbCP^n$ for some prescribed embedding $mathbbCP^nrightarrow mathbbCP^n+1$, but that is very different from Lie theoretic kind of quotient.
    – Jason DeVito
    Jul 30 at 18:22












up vote
2
down vote










up vote
2
down vote









There are all spheres. The relevant technical lemma is the follow:




Suppose $G$ is a compact Lie group which acts smoothly and transitively on a manifold $M$. Let $pin M$ and set $G_p = gin G: gp = p$. Then $G/G_p$ is canonically diffeomorphic to $M$ via the diffeomorphism $g G_pmapsto gp$.




For $G = U(n+1)$ or $G = SU(n+1)$, the relevant $M$ is the the sphere $S^2n+1subseteq mathbbC^n+1$, where the $G$ action is given by usual matrix multiplication on vectors in $mathbbC^n+1$. For $G = Sp(n+1)$, $M = S^4n+3subseteq mathbbH^n+1$.



One must verify that the $G$ action is transitive, and then compute $G_p$ for a single $p$.



Technically, the notation $U(n+1)/U(n)$, etc, is ambiguous until one specifies an embedding $U(n)rightarrow U(n+1)$. Fr the examples at hand, it is often the case (but not always the case) that there is a unique embedding $G_prightarrow G$, up to conjugacy. Conjugate embeddings always give rise to diffeomorphic manifolds: $G/G_pcong G/gG_p g^-1$.



One the other hand, the two embeddings of $U(1)$ into $U(2)$ given by $zmapsto operatornamediag(z,1)$ or $operatornamediag(z,z)$ give rise to homogeneous spaces which are not even homotopy equivalent. One has an analogous result for $Sp(1)rightarrow Sp(2)$. But these are the only exceptions: e.g. $U(2)$ embeds into $U(3)$ in only way one, up to conjugacy.






share|cite|improve this answer













There are all spheres. The relevant technical lemma is the follow:




Suppose $G$ is a compact Lie group which acts smoothly and transitively on a manifold $M$. Let $pin M$ and set $G_p = gin G: gp = p$. Then $G/G_p$ is canonically diffeomorphic to $M$ via the diffeomorphism $g G_pmapsto gp$.




For $G = U(n+1)$ or $G = SU(n+1)$, the relevant $M$ is the the sphere $S^2n+1subseteq mathbbC^n+1$, where the $G$ action is given by usual matrix multiplication on vectors in $mathbbC^n+1$. For $G = Sp(n+1)$, $M = S^4n+3subseteq mathbbH^n+1$.



One must verify that the $G$ action is transitive, and then compute $G_p$ for a single $p$.



Technically, the notation $U(n+1)/U(n)$, etc, is ambiguous until one specifies an embedding $U(n)rightarrow U(n+1)$. Fr the examples at hand, it is often the case (but not always the case) that there is a unique embedding $G_prightarrow G$, up to conjugacy. Conjugate embeddings always give rise to diffeomorphic manifolds: $G/G_pcong G/gG_p g^-1$.



One the other hand, the two embeddings of $U(1)$ into $U(2)$ given by $zmapsto operatornamediag(z,1)$ or $operatornamediag(z,z)$ give rise to homogeneous spaces which are not even homotopy equivalent. One has an analogous result for $Sp(1)rightarrow Sp(2)$. But these are the only exceptions: e.g. $U(2)$ embeds into $U(3)$ in only way one, up to conjugacy.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 30 at 17:13









Jason DeVito

29.5k473129




29.5k473129











  • Thanks +1, how about $mathbbCP^n+1/mathbbCP^n$?
    – wonderich
    Jul 30 at 17:51











  • $mathbbCP^n$ is never a Lie group for any $n > 0$. What do you mean by your notation?
    – Jason DeVito
    Jul 30 at 18:08











  • $mathbbCP^1 simeq S^2$ is a 2-sphere. It is a complex projective space
    – wonderich
    Jul 30 at 18:18











  • typo fixed -thanks.
    – wonderich
    Jul 30 at 18:18






  • 1




    One can consider the topological quotient $mathbbCP^n+1/mathbbCP^n$ for some prescribed embedding $mathbbCP^nrightarrow mathbbCP^n+1$, but that is very different from Lie theoretic kind of quotient.
    – Jason DeVito
    Jul 30 at 18:22
















  • Thanks +1, how about $mathbbCP^n+1/mathbbCP^n$?
    – wonderich
    Jul 30 at 17:51











  • $mathbbCP^n$ is never a Lie group for any $n > 0$. What do you mean by your notation?
    – Jason DeVito
    Jul 30 at 18:08











  • $mathbbCP^1 simeq S^2$ is a 2-sphere. It is a complex projective space
    – wonderich
    Jul 30 at 18:18











  • typo fixed -thanks.
    – wonderich
    Jul 30 at 18:18






  • 1




    One can consider the topological quotient $mathbbCP^n+1/mathbbCP^n$ for some prescribed embedding $mathbbCP^nrightarrow mathbbCP^n+1$, but that is very different from Lie theoretic kind of quotient.
    – Jason DeVito
    Jul 30 at 18:22















Thanks +1, how about $mathbbCP^n+1/mathbbCP^n$?
– wonderich
Jul 30 at 17:51





Thanks +1, how about $mathbbCP^n+1/mathbbCP^n$?
– wonderich
Jul 30 at 17:51













$mathbbCP^n$ is never a Lie group for any $n > 0$. What do you mean by your notation?
– Jason DeVito
Jul 30 at 18:08





$mathbbCP^n$ is never a Lie group for any $n > 0$. What do you mean by your notation?
– Jason DeVito
Jul 30 at 18:08













$mathbbCP^1 simeq S^2$ is a 2-sphere. It is a complex projective space
– wonderich
Jul 30 at 18:18





$mathbbCP^1 simeq S^2$ is a 2-sphere. It is a complex projective space
– wonderich
Jul 30 at 18:18













typo fixed -thanks.
– wonderich
Jul 30 at 18:18




typo fixed -thanks.
– wonderich
Jul 30 at 18:18




1




1




One can consider the topological quotient $mathbbCP^n+1/mathbbCP^n$ for some prescribed embedding $mathbbCP^nrightarrow mathbbCP^n+1$, but that is very different from Lie theoretic kind of quotient.
– Jason DeVito
Jul 30 at 18:22




One can consider the topological quotient $mathbbCP^n+1/mathbbCP^n$ for some prescribed embedding $mathbbCP^nrightarrow mathbbCP^n+1$, but that is very different from Lie theoretic kind of quotient.
– Jason DeVito
Jul 30 at 18:22












 

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