Confusion about quotient of the Lie group $mathbb{S}^1$












1












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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







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  • 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












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














1












1








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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











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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














  • 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










2 Answers
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$begingroup$

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?






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    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$$begin{array}{ccc}S^1/H&longrightarrow&S^1\{z,-z}&mapsto&z^2.end{array}$$






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      2 Answers
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      2 Answers
      2






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      active

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      6












      $begingroup$

      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?






      share|cite|improve this answer









      $endgroup$


















        6












        $begingroup$

        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?






        share|cite|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          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?






          share|cite|improve this answer









          $endgroup$



          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?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 15:32









          Mees de VriesMees de Vries

          17.2k12956




          17.2k12956























              4












              $begingroup$

              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$$begin{array}{ccc}S^1/H&longrightarrow&S^1\{z,-z}&mapsto&z^2.end{array}$$






              share|cite|improve this answer











              $endgroup$


















                4












                $begingroup$

                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$$begin{array}{ccc}S^1/H&longrightarrow&S^1\{z,-z}&mapsto&z^2.end{array}$$






                share|cite|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  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$$begin{array}{ccc}S^1/H&longrightarrow&S^1\{z,-z}&mapsto&z^2.end{array}$$






                  share|cite|improve this answer











                  $endgroup$



                  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$$begin{array}{ccc}S^1/H&longrightarrow&S^1\{z,-z}&mapsto&z^2.end{array}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 10 '18 at 15:56

























                  answered Dec 10 '18 at 15:34









                  José Carlos SantosJosé Carlos Santos

                  159k22126231




                  159k22126231






























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