Is there a group whose manifold is a fiber bundle with base is $S_1$ and fiber $mathbb{Z_2}?$












2












$begingroup$


Let's consider a fiber bundle with base $S_1$ and fiber $mathbb{Z}_2$. I want this manifold to be topologically non-trivial, the edge of the Möbius strip.



How do I know if is it possible to introduce a group structure on such a manifold? So that the manifold would turn into a principle bundle.










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$endgroup$

















    2












    $begingroup$


    Let's consider a fiber bundle with base $S_1$ and fiber $mathbb{Z}_2$. I want this manifold to be topologically non-trivial, the edge of the Möbius strip.



    How do I know if is it possible to introduce a group structure on such a manifold? So that the manifold would turn into a principle bundle.










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let's consider a fiber bundle with base $S_1$ and fiber $mathbb{Z}_2$. I want this manifold to be topologically non-trivial, the edge of the Möbius strip.



      How do I know if is it possible to introduce a group structure on such a manifold? So that the manifold would turn into a principle bundle.










      share|cite|improve this question









      $endgroup$




      Let's consider a fiber bundle with base $S_1$ and fiber $mathbb{Z}_2$. I want this manifold to be topologically non-trivial, the edge of the Möbius strip.



      How do I know if is it possible to introduce a group structure on such a manifold? So that the manifold would turn into a principle bundle.







      group-theory fiber-bundles principal-bundles






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      asked Dec 5 '18 at 21:23









      mavzolejmavzolej

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          $begingroup$

          This manifold, considered independent of its fiber bundle structure, is just $S^1$ again, so it admits the same group structure as $S^1$. You can think of the resulting bundle as the short exact sequence



          $$1 to mathbb{Z}_2 to S^1 xrightarrow{x mapsto x^2} S^1 to 1.$$






          share|cite|improve this answer









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            1 Answer
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            1 Answer
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            active

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            active

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            $begingroup$

            This manifold, considered independent of its fiber bundle structure, is just $S^1$ again, so it admits the same group structure as $S^1$. You can think of the resulting bundle as the short exact sequence



            $$1 to mathbb{Z}_2 to S^1 xrightarrow{x mapsto x^2} S^1 to 1.$$






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              This manifold, considered independent of its fiber bundle structure, is just $S^1$ again, so it admits the same group structure as $S^1$. You can think of the resulting bundle as the short exact sequence



              $$1 to mathbb{Z}_2 to S^1 xrightarrow{x mapsto x^2} S^1 to 1.$$






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                This manifold, considered independent of its fiber bundle structure, is just $S^1$ again, so it admits the same group structure as $S^1$. You can think of the resulting bundle as the short exact sequence



                $$1 to mathbb{Z}_2 to S^1 xrightarrow{x mapsto x^2} S^1 to 1.$$






                share|cite|improve this answer









                $endgroup$



                This manifold, considered independent of its fiber bundle structure, is just $S^1$ again, so it admits the same group structure as $S^1$. You can think of the resulting bundle as the short exact sequence



                $$1 to mathbb{Z}_2 to S^1 xrightarrow{x mapsto x^2} S^1 to 1.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 5 '18 at 22:29









                Qiaochu YuanQiaochu Yuan

                278k32585922




                278k32585922






























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