Lie group intuition












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Why do we require that the multiplication map and the inversion map to be smooth on a Lie group? I'm looking for some intuition into that.



Thank you!










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    1












    $begingroup$


    Why do we require that the multiplication map and the inversion map to be smooth on a Lie group? I'm looking for some intuition into that.



    Thank you!










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Why do we require that the multiplication map and the inversion map to be smooth on a Lie group? I'm looking for some intuition into that.



      Thank you!










      share|cite|improve this question









      $endgroup$




      Why do we require that the multiplication map and the inversion map to be smooth on a Lie group? I'm looking for some intuition into that.



      Thank you!







      lie-groups smooth-manifolds






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      asked Jan 23 at 14:28









      mipmip

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          $begingroup$

          I'm not sure if this is a complete answer to your question. But in many cases these assumptions hold automatically.



          Hilbert's fifth problem states that any topological group that is also a manifold is a Lie group (i.e. the multiplication and inversion is automatically smooth).



          We only need to assume that the multiplication and inversion are continuous and this is needed otherwise there is no relation between the topological structure and the algebraic structure of the group.






          share|cite|improve this answer









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            Because we want to be able to do Differential Geometry on groups and that's what's required in order to be able to do that.






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              2 Answers
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              2 Answers
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              2












              $begingroup$

              I'm not sure if this is a complete answer to your question. But in many cases these assumptions hold automatically.



              Hilbert's fifth problem states that any topological group that is also a manifold is a Lie group (i.e. the multiplication and inversion is automatically smooth).



              We only need to assume that the multiplication and inversion are continuous and this is needed otherwise there is no relation between the topological structure and the algebraic structure of the group.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                I'm not sure if this is a complete answer to your question. But in many cases these assumptions hold automatically.



                Hilbert's fifth problem states that any topological group that is also a manifold is a Lie group (i.e. the multiplication and inversion is automatically smooth).



                We only need to assume that the multiplication and inversion are continuous and this is needed otherwise there is no relation between the topological structure and the algebraic structure of the group.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  I'm not sure if this is a complete answer to your question. But in many cases these assumptions hold automatically.



                  Hilbert's fifth problem states that any topological group that is also a manifold is a Lie group (i.e. the multiplication and inversion is automatically smooth).



                  We only need to assume that the multiplication and inversion are continuous and this is needed otherwise there is no relation between the topological structure and the algebraic structure of the group.






                  share|cite|improve this answer









                  $endgroup$



                  I'm not sure if this is a complete answer to your question. But in many cases these assumptions hold automatically.



                  Hilbert's fifth problem states that any topological group that is also a manifold is a Lie group (i.e. the multiplication and inversion is automatically smooth).



                  We only need to assume that the multiplication and inversion are continuous and this is needed otherwise there is no relation between the topological structure and the algebraic structure of the group.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 23 at 14:35









                  YankoYanko

                  7,7751830




                  7,7751830























                      2












                      $begingroup$

                      Because we want to be able to do Differential Geometry on groups and that's what's required in order to be able to do that.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Because we want to be able to do Differential Geometry on groups and that's what's required in order to be able to do that.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Because we want to be able to do Differential Geometry on groups and that's what's required in order to be able to do that.






                          share|cite|improve this answer









                          $endgroup$



                          Because we want to be able to do Differential Geometry on groups and that's what's required in order to be able to do that.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 23 at 14:35









                          José Carlos SantosJosé Carlos Santos

                          167k22132235




                          167k22132235






























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