Confusion about quotient of the Lie group $mathbb{S}^1$
$begingroup$
I have read that given a Lie group $G$ and a closed subgroup $H$ then $G/H$ is a smooth manifold. I cannot explain though the following example: take as $G = mathbb{S}^1$ and as $H ={pm 1}$, $H$ is a closed subgroup but the quotient should look like figure $8$, and thus is not even a topological manifold.
I am surely making some mistake, can you help me in finding it?
differential-geometry differential-topology lie-groups
asked Dec 10 '18 at 15:27
Warlock of Firetop MountainWarlock of Firetop Mountain
2,300915
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add a comment |
$begingroup$
I have read that given a Lie group $G$ and a closed subgroup $H$ then $G/H$ is a smooth manifold. I cannot explain though the following example: take as $G = mathbb{S}^1$ and as $H ={pm 1}$, $H$ is a closed subgroup but the quotient should look like figure $8$, and thus is not even a topological manifold.
I am surely making some mistake, can you help me in finding it?
differential-geometry differential-topology lie-groups
asked Dec 10 '18 at 15:27
Warlock of Firetop MountainWarlock of Firetop Mountain
2,300915
$endgroup$
1
$begingroup$
You're mistaken about what $G/H$ looks like : this is not a quotient where only $H$ is identified to a point, it's a quotient as a group. In particular, the top hemisphere is identified with the bottom hemisphere through $xsim -x$
$endgroup$
– Max
Dec 10 '18 at 15:39
add a comment |
$begingroup$
I have read that given a Lie group $G$ and a closed subgroup $H$ then $G/H$ is a smooth manifold. I cannot explain though the following example: take as $G = mathbb{S}^1$ and as $H ={pm 1}$, $H$ is a closed subgroup but the quotient should look like figure $8$, and thus is not even a topological manifold.
I am surely making some mistake, can you help me in finding it?
differential-geometry differential-topology lie-groups
asked Dec 10 '18 at 15:27
Warlock of Firetop MountainWarlock of Firetop Mountain
2,300915
$endgroup$
I have read that given a Lie group $G$ and a closed subgroup $H$ then $G/H$ is a smooth manifold. I cannot explain though the following example: take as $G = mathbb{S}^1$ and as $H ={pm 1}$, $H$ is a closed subgroup but the quotient should look like figure $8$, and thus is not even a topological manifold.
I am surely making some mistake, can you help me in finding it?
differential-geometry differential-topology lie-groups
differential-geometry differential-topology lie-groups
asked Dec 10 '18 at 15:27
Warlock of Firetop MountainWarlock of Firetop Mountain
2,300915
asked Dec 10 '18 at 15:27
Warlock of Firetop MountainWarlock of Firetop Mountain
2,300915
asked Dec 10 '18 at 15:27
Warlock of Firetop MountainWarlock of Firetop Mountain
2,300915
asked Dec 10 '18 at 15:27
Warlock of Firetop MountainWarlock of Firetop Mountain
2,300915
asked Dec 10 '18 at 15:27
Warlock of Firetop MountainWarlock of Firetop Mountain
2,300915
2,300915
1
$begingroup$
You're mistaken about what $G/H$ looks like : this is not a quotient where only $H$ is identified to a point, it's a quotient as a group. In particular, the top hemisphere is identified with the bottom hemisphere through $xsim -x$
$endgroup$
– Max
Dec 10 '18 at 15:39
add a comment |
1
$begingroup$
You're mistaken about what $G/H$ looks like : this is not a quotient where only $H$ is identified to a point, it's a quotient as a group. In particular, the top hemisphere is identified with the bottom hemisphere through $xsim -x$
$endgroup$
– Max
Dec 10 '18 at 15:39
1
1
$begingroup$
You're mistaken about what $G/H$ looks like : this is not a quotient where only $H$ is identified to a point, it's a quotient as a group. In particular, the top hemisphere is identified with the bottom hemisphere through $xsim -x$
$endgroup$
– Max
Dec 10 '18 at 15:39
$begingroup$
You're mistaken about what $G/H$ looks like : this is not a quotient where only $H$ is identified to a point, it's a quotient as a group. In particular, the top hemisphere is identified with the bottom hemisphere through $xsim -x$
$endgroup$
– Max
Dec 10 '18 at 15:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The quotient $G/H$ here does not identify just the points in $H$. Like an ordinary quotient of groups, it identifies all the points in the cosets $gH$ for $g in G$. In this case, it means that for each $z in G$, the set ${z, -z}$ is identified to a point. Can you figure out what the resulting figure is?
answered Dec 10 '18 at 15:32
Mees de VriesMees de Vries
17.2k12956
$endgroup$
add a comment |
$begingroup$
You are misunderstanding the meaning of quotient. The quotient $S^1/H$ is the set of all sets ${z,-z}$, with $zin S^1$. It turns out to be homeomorphic to $S^1$ itself. Just consider the map$$begin{array}{ccc}S^1/H&longrightarrow&S^1\{z,-z}&mapsto&z^2.end{array}$$
answered Dec 10 '18 at 15:34
José Carlos SantosJosé Carlos Santos
159k22126231
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
The quotient $G/H$ here does not identify just the points in $H$. Like an ordinary quotient of groups, it identifies all the points in the cosets $gH$ for $g in G$. In this case, it means that for each $z in G$, the set ${z, -z}$ is identified to a point. Can you figure out what the resulting figure is?
answered Dec 10 '18 at 15:32
Mees de VriesMees de Vries
17.2k12956
$endgroup$
add a comment |
$begingroup$
The quotient $G/H$ here does not identify just the points in $H$. Like an ordinary quotient of groups, it identifies all the points in the cosets $gH$ for $g in G$. In this case, it means that for each $z in G$, the set ${z, -z}$ is identified to a point. Can you figure out what the resulting figure is?
answered Dec 10 '18 at 15:32
Mees de VriesMees de Vries
17.2k12956
$endgroup$
add a comment |
$begingroup$
The quotient $G/H$ here does not identify just the points in $H$. Like an ordinary quotient of groups, it identifies all the points in the cosets $gH$ for $g in G$. In this case, it means that for each $z in G$, the set ${z, -z}$ is identified to a point. Can you figure out what the resulting figure is?
answered Dec 10 '18 at 15:32
Mees de VriesMees de Vries
17.2k12956
$endgroup$
The quotient $G/H$ here does not identify just the points in $H$. Like an ordinary quotient of groups, it identifies all the points in the cosets $gH$ for $g in G$. In this case, it means that for each $z in G$, the set ${z, -z}$ is identified to a point. Can you figure out what the resulting figure is?
answered Dec 10 '18 at 15:32
Mees de VriesMees de Vries
17.2k12956
answered Dec 10 '18 at 15:32
Mees de VriesMees de Vries
17.2k12956
answered Dec 10 '18 at 15:32
Mees de VriesMees de Vries
17.2k12956
answered Dec 10 '18 at 15:32
Mees de VriesMees de Vries
17.2k12956
17.2k12956
add a comment |
add a comment |
$begingroup$
You are misunderstanding the meaning of quotient. The quotient $S^1/H$ is the set of all sets ${z,-z}$, with $zin S^1$. It turns out to be homeomorphic to $S^1$ itself. Just consider the map$$begin{array}{ccc}S^1/H&longrightarrow&S^1\{z,-z}&mapsto&z^2.end{array}$$
answered Dec 10 '18 at 15:34
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
add a comment |
$begingroup$
You are misunderstanding the meaning of quotient. The quotient $S^1/H$ is the set of all sets ${z,-z}$, with $zin S^1$. It turns out to be homeomorphic to $S^1$ itself. Just consider the map$$begin{array}{ccc}S^1/H&longrightarrow&S^1\{z,-z}&mapsto&z^2.end{array}$$
answered Dec 10 '18 at 15:34
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
add a comment |
$begingroup$
You are misunderstanding the meaning of quotient. The quotient $S^1/H$ is the set of all sets ${z,-z}$, with $zin S^1$. It turns out to be homeomorphic to $S^1$ itself. Just consider the map$$begin{array}{ccc}S^1/H&longrightarrow&S^1\{z,-z}&mapsto&z^2.end{array}$$
answered Dec 10 '18 at 15:34
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
You are misunderstanding the meaning of quotient. The quotient $S^1/H$ is the set of all sets ${z,-z}$, with $zin S^1$. It turns out to be homeomorphic to $S^1$ itself. Just consider the map$$begin{array}{ccc}S^1/H&longrightarrow&S^1\{z,-z}&mapsto&z^2.end{array}$$
answered Dec 10 '18 at 15:34
José Carlos SantosJosé Carlos Santos
159k22126231
edited Dec 10 '18 at 15:56
answered Dec 10 '18 at 15:34
José Carlos SantosJosé Carlos Santos
159k22126231
answered Dec 10 '18 at 15:34
José Carlos SantosJosé Carlos Santos
159k22126231
answered Dec 10 '18 at 15:34
José Carlos SantosJosé Carlos Santos
159k22126231
159k22126231
add a comment |
add a comment |
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$begingroup$
You're mistaken about what $G/H$ looks like : this is not a quotient where only $H$ is identified to a point, it's a quotient as a group. In particular, the top hemisphere is identified with the bottom hemisphere through $xsim -x$
$endgroup$
– Max
Dec 10 '18 at 15:39