Mixing
$S^2=SO(3)/SO(2)$. Does this mean that $S^2 = SU(2)/U(1) $?
$S^2=SO(3)/SO(2)$. Does this mean that $S^2 = SU(2)/U(1) $ since $SO(3) approx SU(2)$ and $SO(2) approx U(1)$? Is there some more generic rule on how to relate
$S^{n-1} = SO(n)/SO(n-1)$ to the corresponding (special) unitary groups? Also, is there a way to write $S^{n-1}$ as a quotient or so of $Sp(n)$ and $Spin(n)$ (e.g. in specific dimensions)?
differential-geometry algebraic-topology lie-groups
edited Mar 26 '15 at 4:31
Qiaochu Yuan
277k32583919
asked Mar 25 '15 at 17:28
MarionMarion
825618
add a comment |
$S^2=SO(3)/SO(2)$. Does this mean that $S^2 = SU(2)/U(1) $ since $SO(3) approx SU(2)$ and $SO(2) approx U(1)$? Is there some more generic rule on how to relate
$S^{n-1} = SO(n)/SO(n-1)$ to the corresponding (special) unitary groups? Also, is there a way to write $S^{n-1}$ as a quotient or so of $Sp(n)$ and $Spin(n)$ (e.g. in specific dimensions)?
differential-geometry algebraic-topology lie-groups
edited Mar 26 '15 at 4:31
Qiaochu Yuan
277k32583919
asked Mar 25 '15 at 17:28
MarionMarion
825618
6
Just a few comments: (i) Some of the "$S^{1}$"s in your post should perhaps be "$S^{2}$"? (ii) $SO(3)$ and $SU(2)$ are not isomorphic as groups (or even homeomorphic as manifolds; $SU(2)$ double covers $SO(3)$). (iii) Whether or not $SU(2)/U(1) approx S^{2}$ depends on how $U(1)$ is embedded. If $U(1)$ is the diagonal subgroup, then "yes"; the coset space is essentially the Hopf fibration. (iv) If memory serves, Einstein Manifolds by Besse addresses your questions about spheres and Spin coset spaces.
– Andrew D. Hwang
Mar 25 '15 at 17:40
1
If you take $mathbb{R}^{n}$ with standard euclidean norm then $SO(n)$ acts on the vectors of unit length (the sphere $S^{n-1}$. So you have to figure out the stabilizer of a point. In particular choosing co-ordinates and embedding $SO(n-1)$ as the first $n times n$ principal diagonal you get that it fixes the vector $(0, ldots, 1) in mathbb{R}^{n}$. So we see the isomorphism. There are some coincidental isomorphims in low dimensions between spin groups and SO(n) which might give you isomorphic quotients but in general this is not true in higher dimensions.
– DBS
Mar 25 '15 at 17:42
add a comment |
$S^2=SO(3)/SO(2)$. Does this mean that $S^2 = SU(2)/U(1) $ since $SO(3) approx SU(2)$ and $SO(2) approx U(1)$? Is there some more generic rule on how to relate
$S^{n-1} = SO(n)/SO(n-1)$ to the corresponding (special) unitary groups? Also, is there a way to write $S^{n-1}$ as a quotient or so of $Sp(n)$ and $Spin(n)$ (e.g. in specific dimensions)?
differential-geometry algebraic-topology lie-groups
edited Mar 26 '15 at 4:31
Qiaochu Yuan
277k32583919
asked Mar 25 '15 at 17:28
MarionMarion
825618
$S^2=SO(3)/SO(2)$. Does this mean that $S^2 = SU(2)/U(1) $ since $SO(3) approx SU(2)$ and $SO(2) approx U(1)$? Is there some more generic rule on how to relate
$S^{n-1} = SO(n)/SO(n-1)$ to the corresponding (special) unitary groups? Also, is there a way to write $S^{n-1}$ as a quotient or so of $Sp(n)$ and $Spin(n)$ (e.g. in specific dimensions)?
differential-geometry algebraic-topology lie-groups
differential-geometry algebraic-topology lie-groups
edited Mar 26 '15 at 4:31
Qiaochu Yuan
277k32583919
asked Mar 25 '15 at 17:28
MarionMarion
825618
edited Mar 26 '15 at 4:31
Qiaochu Yuan
277k32583919
asked Mar 25 '15 at 17:28
MarionMarion
825618
edited Mar 26 '15 at 4:31
Qiaochu Yuan
277k32583919
edited Mar 26 '15 at 4:31
Qiaochu Yuan
277k32583919
edited Mar 26 '15 at 4:31
Qiaochu Yuan
277k32583919
277k32583919
asked Mar 25 '15 at 17:28
MarionMarion
825618
asked Mar 25 '15 at 17:28
MarionMarion
825618
asked Mar 25 '15 at 17:28
MarionMarion
825618
825618
6
Just a few comments: (i) Some of the "$S^{1}$"s in your post should perhaps be "$S^{2}$"? (ii) $SO(3)$ and $SU(2)$ are not isomorphic as groups (or even homeomorphic as manifolds; $SU(2)$ double covers $SO(3)$). (iii) Whether or not $SU(2)/U(1) approx S^{2}$ depends on how $U(1)$ is embedded. If $U(1)$ is the diagonal subgroup, then "yes"; the coset space is essentially the Hopf fibration. (iv) If memory serves, Einstein Manifolds by Besse addresses your questions about spheres and Spin coset spaces.
– Andrew D. Hwang
Mar 25 '15 at 17:40
1
If you take $mathbb{R}^{n}$ with standard euclidean norm then $SO(n)$ acts on the vectors of unit length (the sphere $S^{n-1}$. So you have to figure out the stabilizer of a point. In particular choosing co-ordinates and embedding $SO(n-1)$ as the first $n times n$ principal diagonal you get that it fixes the vector $(0, ldots, 1) in mathbb{R}^{n}$. So we see the isomorphism. There are some coincidental isomorphims in low dimensions between spin groups and SO(n) which might give you isomorphic quotients but in general this is not true in higher dimensions.
– DBS
Mar 25 '15 at 17:42
add a comment |
6
Just a few comments: (i) Some of the "$S^{1}$"s in your post should perhaps be "$S^{2}$"? (ii) $SO(3)$ and $SU(2)$ are not isomorphic as groups (or even homeomorphic as manifolds; $SU(2)$ double covers $SO(3)$). (iii) Whether or not $SU(2)/U(1) approx S^{2}$ depends on how $U(1)$ is embedded. If $U(1)$ is the diagonal subgroup, then "yes"; the coset space is essentially the Hopf fibration. (iv) If memory serves, Einstein Manifolds by Besse addresses your questions about spheres and Spin coset spaces.
– Andrew D. Hwang
Mar 25 '15 at 17:40
1
If you take $mathbb{R}^{n}$ with standard euclidean norm then $SO(n)$ acts on the vectors of unit length (the sphere $S^{n-1}$. So you have to figure out the stabilizer of a point. In particular choosing co-ordinates and embedding $SO(n-1)$ as the first $n times n$ principal diagonal you get that it fixes the vector $(0, ldots, 1) in mathbb{R}^{n}$. So we see the isomorphism. There are some coincidental isomorphims in low dimensions between spin groups and SO(n) which might give you isomorphic quotients but in general this is not true in higher dimensions.
– DBS
Mar 25 '15 at 17:42
6
6
Just a few comments: (i) Some of the "$S^{1}$"s in your post should perhaps be "$S^{2}$"? (ii) $SO(3)$ and $SU(2)$ are not isomorphic as groups (or even homeomorphic as manifolds; $SU(2)$ double covers $SO(3)$). (iii) Whether or not $SU(2)/U(1) approx S^{2}$ depends on how $U(1)$ is embedded. If $U(1)$ is the diagonal subgroup, then "yes"; the coset space is essentially the Hopf fibration. (iv) If memory serves, Einstein Manifolds by Besse addresses your questions about spheres and Spin coset spaces.
– Andrew D. Hwang
Mar 25 '15 at 17:40
Just a few comments: (i) Some of the "$S^{1}$"s in your post should perhaps be "$S^{2}$"? (ii) $SO(3)$ and $SU(2)$ are not isomorphic as groups (or even homeomorphic as manifolds; $SU(2)$ double covers $SO(3)$). (iii) Whether or not $SU(2)/U(1) approx S^{2}$ depends on how $U(1)$ is embedded. If $U(1)$ is the diagonal subgroup, then "yes"; the coset space is essentially the Hopf fibration. (iv) If memory serves, Einstein Manifolds by Besse addresses your questions about spheres and Spin coset spaces.
– Andrew D. Hwang
Mar 25 '15 at 17:40
1
1
If you take $mathbb{R}^{n}$ with standard euclidean norm then $SO(n)$ acts on the vectors of unit length (the sphere $S^{n-1}$. So you have to figure out the stabilizer of a point. In particular choosing co-ordinates and embedding $SO(n-1)$ as the first $n times n$ principal diagonal you get that it fixes the vector $(0, ldots, 1) in mathbb{R}^{n}$. So we see the isomorphism. There are some coincidental isomorphims in low dimensions between spin groups and SO(n) which might give you isomorphic quotients but in general this is not true in higher dimensions.
– DBS
Mar 25 '15 at 17:42
If you take $mathbb{R}^{n}$ with standard euclidean norm then $SO(n)$ acts on the vectors of unit length (the sphere $S^{n-1}$. So you have to figure out the stabilizer of a point. In particular choosing co-ordinates and embedding $SO(n-1)$ as the first $n times n$ principal diagonal you get that it fixes the vector $(0, ldots, 1) in mathbb{R}^{n}$. So we see the isomorphism. There are some coincidental isomorphims in low dimensions between spin groups and SO(n) which might give you isomorphic quotients but in general this is not true in higher dimensions.
– DBS
Mar 25 '15 at 17:42
add a comment |
1 Answer
1
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In general, to identify, say, a smooth manifold $X$ with a quotient $G/H$ of a Lie group $G$ by a closed Lie subgroup $H$, you need to produce a smooth transitive action of $G$ on $X$ with stabilizer $H$. For the classical compact Lie groups there are three standard series of transitive actions on spheres which go like this.
The special orthogonal group $SO(n)$ acts on $mathbb{R}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{n-1}$ with stabilizer $SO(n-1)$, and hence
$$S^{n-1} cong SO(n) / SO(n-1).$$
Furthermore, the natural map $Spin(n) to SO(n)$ means we can also think about this action as an action of $Spin(n)$; then its stabilizer turns out to be $Spin(n-1)$, and so we can also identify
$$S^{n-1} cong Spin(n) / Spin(n-1).$$
When $n = 3$ there is an exceptional isomorphism $Spin(3) cong SU(2)$, and $Spin(2)$ inside $Spin(3)$ can be identified with the diagonal copy of $U(1)$ inside $SU(2)$. Note that the map $Spin(2) to SO(2)$ is multiplication by $2$, as opposed to the standard map $U(1) to SO(2)$ which is an isomorphism.
Similarly, the special unitary group $SU(n)$ acts on $mathbb{C}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{2n-1}$ with stabilizer $SU(n-1)$, and hence
$$S^{2n-1} cong SU(n) / SU(n-1).$$
Finally, the compact symplectic group $Sp(n)$ acts on $mathbb{H}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{4n-1}$ with stabilizer $Sp(n-1)$, and hence
$$S^{4n-1} cong Sp(n) / Sp(n-1).$$
answered Mar 26 '15 at 3:57
Qiaochu YuanQiaochu Yuan
277k32583919
Let me ask another related question. Is $S^2$ considered a Lie manifold? Generally the symmetries of spheres are rotations. Intuitively this would tell me that they are Lie manifolds. I heard today that only $S^0,S^1,S^3$ are Lie manifolds. Why is that?
– Marion
Mar 27 '15 at 14:30
1
@Marion: no. Lie groups of rotations are more complicated than spheres. They act on spheres, but that doesn't mean that the spheres themselves have Lie group structures. It's known that any compact connected Lie group has Euler characteristic $0$ (e.g. by the Poincare-Hopf theorem), and this rules out even-dimensional spheres having a Lie group structure. For the odd-dmensional spheres the argument is more involved.
– Qiaochu Yuan
Mar 28 '15 at 3:11
1
QiaochuYuan thanks a lot. Would you be able to provide a good reference maybe?
– Marion
Mar 28 '15 at 10:23
add a comment |
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In general, to identify, say, a smooth manifold $X$ with a quotient $G/H$ of a Lie group $G$ by a closed Lie subgroup $H$, you need to produce a smooth transitive action of $G$ on $X$ with stabilizer $H$. For the classical compact Lie groups there are three standard series of transitive actions on spheres which go like this.
The special orthogonal group $SO(n)$ acts on $mathbb{R}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{n-1}$ with stabilizer $SO(n-1)$, and hence
$$S^{n-1} cong SO(n) / SO(n-1).$$
Furthermore, the natural map $Spin(n) to SO(n)$ means we can also think about this action as an action of $Spin(n)$; then its stabilizer turns out to be $Spin(n-1)$, and so we can also identify
$$S^{n-1} cong Spin(n) / Spin(n-1).$$
When $n = 3$ there is an exceptional isomorphism $Spin(3) cong SU(2)$, and $Spin(2)$ inside $Spin(3)$ can be identified with the diagonal copy of $U(1)$ inside $SU(2)$. Note that the map $Spin(2) to SO(2)$ is multiplication by $2$, as opposed to the standard map $U(1) to SO(2)$ which is an isomorphism.
Similarly, the special unitary group $SU(n)$ acts on $mathbb{C}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{2n-1}$ with stabilizer $SU(n-1)$, and hence
$$S^{2n-1} cong SU(n) / SU(n-1).$$
Finally, the compact symplectic group $Sp(n)$ acts on $mathbb{H}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{4n-1}$ with stabilizer $Sp(n-1)$, and hence
$$S^{4n-1} cong Sp(n) / Sp(n-1).$$
answered Mar 26 '15 at 3:57
Qiaochu YuanQiaochu Yuan
277k32583919
Let me ask another related question. Is $S^2$ considered a Lie manifold? Generally the symmetries of spheres are rotations. Intuitively this would tell me that they are Lie manifolds. I heard today that only $S^0,S^1,S^3$ are Lie manifolds. Why is that?
– Marion
Mar 27 '15 at 14:30
1
@Marion: no. Lie groups of rotations are more complicated than spheres. They act on spheres, but that doesn't mean that the spheres themselves have Lie group structures. It's known that any compact connected Lie group has Euler characteristic $0$ (e.g. by the Poincare-Hopf theorem), and this rules out even-dimensional spheres having a Lie group structure. For the odd-dmensional spheres the argument is more involved.
– Qiaochu Yuan
Mar 28 '15 at 3:11
1
QiaochuYuan thanks a lot. Would you be able to provide a good reference maybe?
– Marion
Mar 28 '15 at 10:23
add a comment |
In general, to identify, say, a smooth manifold $X$ with a quotient $G/H$ of a Lie group $G$ by a closed Lie subgroup $H$, you need to produce a smooth transitive action of $G$ on $X$ with stabilizer $H$. For the classical compact Lie groups there are three standard series of transitive actions on spheres which go like this.
The special orthogonal group $SO(n)$ acts on $mathbb{R}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{n-1}$ with stabilizer $SO(n-1)$, and hence
$$S^{n-1} cong SO(n) / SO(n-1).$$
Furthermore, the natural map $Spin(n) to SO(n)$ means we can also think about this action as an action of $Spin(n)$; then its stabilizer turns out to be $Spin(n-1)$, and so we can also identify
$$S^{n-1} cong Spin(n) / Spin(n-1).$$
When $n = 3$ there is an exceptional isomorphism $Spin(3) cong SU(2)$, and $Spin(2)$ inside $Spin(3)$ can be identified with the diagonal copy of $U(1)$ inside $SU(2)$. Note that the map $Spin(2) to SO(2)$ is multiplication by $2$, as opposed to the standard map $U(1) to SO(2)$ which is an isomorphism.
Similarly, the special unitary group $SU(n)$ acts on $mathbb{C}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{2n-1}$ with stabilizer $SU(n-1)$, and hence
$$S^{2n-1} cong SU(n) / SU(n-1).$$
Finally, the compact symplectic group $Sp(n)$ acts on $mathbb{H}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{4n-1}$ with stabilizer $Sp(n-1)$, and hence
$$S^{4n-1} cong Sp(n) / Sp(n-1).$$
answered Mar 26 '15 at 3:57
Qiaochu YuanQiaochu Yuan
277k32583919
Let me ask another related question. Is $S^2$ considered a Lie manifold? Generally the symmetries of spheres are rotations. Intuitively this would tell me that they are Lie manifolds. I heard today that only $S^0,S^1,S^3$ are Lie manifolds. Why is that?
– Marion
Mar 27 '15 at 14:30
1
@Marion: no. Lie groups of rotations are more complicated than spheres. They act on spheres, but that doesn't mean that the spheres themselves have Lie group structures. It's known that any compact connected Lie group has Euler characteristic $0$ (e.g. by the Poincare-Hopf theorem), and this rules out even-dimensional spheres having a Lie group structure. For the odd-dmensional spheres the argument is more involved.
– Qiaochu Yuan
Mar 28 '15 at 3:11
1
QiaochuYuan thanks a lot. Would you be able to provide a good reference maybe?
– Marion
Mar 28 '15 at 10:23
add a comment |
In general, to identify, say, a smooth manifold $X$ with a quotient $G/H$ of a Lie group $G$ by a closed Lie subgroup $H$, you need to produce a smooth transitive action of $G$ on $X$ with stabilizer $H$. For the classical compact Lie groups there are three standard series of transitive actions on spheres which go like this.
The special orthogonal group $SO(n)$ acts on $mathbb{R}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{n-1}$ with stabilizer $SO(n-1)$, and hence
$$S^{n-1} cong SO(n) / SO(n-1).$$
Furthermore, the natural map $Spin(n) to SO(n)$ means we can also think about this action as an action of $Spin(n)$; then its stabilizer turns out to be $Spin(n-1)$, and so we can also identify
$$S^{n-1} cong Spin(n) / Spin(n-1).$$
When $n = 3$ there is an exceptional isomorphism $Spin(3) cong SU(2)$, and $Spin(2)$ inside $Spin(3)$ can be identified with the diagonal copy of $U(1)$ inside $SU(2)$. Note that the map $Spin(2) to SO(2)$ is multiplication by $2$, as opposed to the standard map $U(1) to SO(2)$ which is an isomorphism.
Similarly, the special unitary group $SU(n)$ acts on $mathbb{C}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{2n-1}$ with stabilizer $SU(n-1)$, and hence
$$S^{2n-1} cong SU(n) / SU(n-1).$$
Finally, the compact symplectic group $Sp(n)$ acts on $mathbb{H}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{4n-1}$ with stabilizer $Sp(n-1)$, and hence
$$S^{4n-1} cong Sp(n) / Sp(n-1).$$
answered Mar 26 '15 at 3:57
Qiaochu YuanQiaochu Yuan
277k32583919
In general, to identify, say, a smooth manifold $X$ with a quotient $G/H$ of a Lie group $G$ by a closed Lie subgroup $H$, you need to produce a smooth transitive action of $G$ on $X$ with stabilizer $H$. For the classical compact Lie groups there are three standard series of transitive actions on spheres which go like this.
The special orthogonal group $SO(n)$ acts on $mathbb{R}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{n-1}$ with stabilizer $SO(n-1)$, and hence
$$S^{n-1} cong SO(n) / SO(n-1).$$
Furthermore, the natural map $Spin(n) to SO(n)$ means we can also think about this action as an action of $Spin(n)$; then its stabilizer turns out to be $Spin(n-1)$, and so we can also identify
$$S^{n-1} cong Spin(n) / Spin(n-1).$$
When $n = 3$ there is an exceptional isomorphism $Spin(3) cong SU(2)$, and $Spin(2)$ inside $Spin(3)$ can be identified with the diagonal copy of $U(1)$ inside $SU(2)$. Note that the map $Spin(2) to SO(2)$ is multiplication by $2$, as opposed to the standard map $U(1) to SO(2)$ which is an isomorphism.
Similarly, the special unitary group $SU(n)$ acts on $mathbb{C}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{2n-1}$ with stabilizer $SU(n-1)$, and hence
$$S^{2n-1} cong SU(n) / SU(n-1).$$
Finally, the compact symplectic group $Sp(n)$ acts on $mathbb{H}^n$ preserving the standard inner product. This naturally induces a transitive action on the unit sphere $S^{4n-1}$ with stabilizer $Sp(n-1)$, and hence
$$S^{4n-1} cong Sp(n) / Sp(n-1).$$
answered Mar 26 '15 at 3:57
Qiaochu YuanQiaochu Yuan
277k32583919
answered Mar 26 '15 at 3:57
Qiaochu YuanQiaochu Yuan
277k32583919
answered Mar 26 '15 at 3:57
Qiaochu YuanQiaochu Yuan
277k32583919
answered Mar 26 '15 at 3:57
Qiaochu YuanQiaochu Yuan
277k32583919
277k32583919
Let me ask another related question. Is $S^2$ considered a Lie manifold? Generally the symmetries of spheres are rotations. Intuitively this would tell me that they are Lie manifolds. I heard today that only $S^0,S^1,S^3$ are Lie manifolds. Why is that?
– Marion
Mar 27 '15 at 14:30
1
@Marion: no. Lie groups of rotations are more complicated than spheres. They act on spheres, but that doesn't mean that the spheres themselves have Lie group structures. It's known that any compact connected Lie group has Euler characteristic $0$ (e.g. by the Poincare-Hopf theorem), and this rules out even-dimensional spheres having a Lie group structure. For the odd-dmensional spheres the argument is more involved.
– Qiaochu Yuan
Mar 28 '15 at 3:11
1
QiaochuYuan thanks a lot. Would you be able to provide a good reference maybe?
– Marion
Mar 28 '15 at 10:23
add a comment |
Let me ask another related question. Is $S^2$ considered a Lie manifold? Generally the symmetries of spheres are rotations. Intuitively this would tell me that they are Lie manifolds. I heard today that only $S^0,S^1,S^3$ are Lie manifolds. Why is that?
– Marion
Mar 27 '15 at 14:30
1
@Marion: no. Lie groups of rotations are more complicated than spheres. They act on spheres, but that doesn't mean that the spheres themselves have Lie group structures. It's known that any compact connected Lie group has Euler characteristic $0$ (e.g. by the Poincare-Hopf theorem), and this rules out even-dimensional spheres having a Lie group structure. For the odd-dmensional spheres the argument is more involved.
– Qiaochu Yuan
Mar 28 '15 at 3:11
1
QiaochuYuan thanks a lot. Would you be able to provide a good reference maybe?
– Marion
Mar 28 '15 at 10:23
Let me ask another related question. Is $S^2$ considered a Lie manifold? Generally the symmetries of spheres are rotations. Intuitively this would tell me that they are Lie manifolds. I heard today that only $S^0,S^1,S^3$ are Lie manifolds. Why is that?
– Marion
Mar 27 '15 at 14:30
Let me ask another related question. Is $S^2$ considered a Lie manifold? Generally the symmetries of spheres are rotations. Intuitively this would tell me that they are Lie manifolds. I heard today that only $S^0,S^1,S^3$ are Lie manifolds. Why is that?
– Marion
Mar 27 '15 at 14:30
1
1
@Marion: no. Lie groups of rotations are more complicated than spheres. They act on spheres, but that doesn't mean that the spheres themselves have Lie group structures. It's known that any compact connected Lie group has Euler characteristic $0$ (e.g. by the Poincare-Hopf theorem), and this rules out even-dimensional spheres having a Lie group structure. For the odd-dmensional spheres the argument is more involved.
– Qiaochu Yuan
Mar 28 '15 at 3:11
@Marion: no. Lie groups of rotations are more complicated than spheres. They act on spheres, but that doesn't mean that the spheres themselves have Lie group structures. It's known that any compact connected Lie group has Euler characteristic $0$ (e.g. by the Poincare-Hopf theorem), and this rules out even-dimensional spheres having a Lie group structure. For the odd-dmensional spheres the argument is more involved.
– Qiaochu Yuan
Mar 28 '15 at 3:11
1
1
QiaochuYuan thanks a lot. Would you be able to provide a good reference maybe?
– Marion
Mar 28 '15 at 10:23
QiaochuYuan thanks a lot. Would you be able to provide a good reference maybe?
– Marion
Mar 28 '15 at 10:23
add a comment |
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6
Just a few comments: (i) Some of the "$S^{1}$"s in your post should perhaps be "$S^{2}$"? (ii) $SO(3)$ and $SU(2)$ are not isomorphic as groups (or even homeomorphic as manifolds; $SU(2)$ double covers $SO(3)$). (iii) Whether or not $SU(2)/U(1) approx S^{2}$ depends on how $U(1)$ is embedded. If $U(1)$ is the diagonal subgroup, then "yes"; the coset space is essentially the Hopf fibration. (iv) If memory serves, Einstein Manifolds by Besse addresses your questions about spheres and Spin coset spaces.
– Andrew D. Hwang
Mar 25 '15 at 17:40
1
If you take $mathbb{R}^{n}$ with standard euclidean norm then $SO(n)$ acts on the vectors of unit length (the sphere $S^{n-1}$. So you have to figure out the stabilizer of a point. In particular choosing co-ordinates and embedding $SO(n-1)$ as the first $n times n$ principal diagonal you get that it fixes the vector $(0, ldots, 1) in mathbb{R}^{n}$. So we see the isomorphism. There are some coincidental isomorphims in low dimensions between spin groups and SO(n) which might give you isomorphic quotients but in general this is not true in higher dimensions.
– DBS
Mar 25 '15 at 17:42