Smaller neighborhood around the identity of Lie Group The Next CEO of Stack OverflowAbout connected Lie GroupsIs $G$ a lie group if left multiplication is smooth and multiplication is smooth near $e$?Map $n(g,h) = gh^-1$ is smooth implies $G$ is a Lie Group.(Whitney) Extension Lemma for smooth mapsImmersed subgroup of a Lie group is a Lie group?Fundamental theorem on flows lee's book 2nd edition$G$ a Lie group, $V, S$ submanifolds of $G$ containing $e$, $psi : V times S rightarrow G$; $psi(v,s)=vs$, then $dpsi(X,0)=X$ and $dpsi(0,Y)=Y$Open neighborhood in Lie groupfactoring a neighborhood of identity in a compact connected Lie group with a closed Lie subgroupJohn Lee : Cubical charts and cube in $mathbbR^n$

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Smaller neighborhood around the identity of Lie Group



The Next CEO of Stack OverflowAbout connected Lie GroupsIs $G$ a lie group if left multiplication is smooth and multiplication is smooth near $e$?Map $n(g,h) = gh^-1$ is smooth implies $G$ is a Lie Group.(Whitney) Extension Lemma for smooth mapsImmersed subgroup of a Lie group is a Lie group?Fundamental theorem on flows lee's book 2nd edition$G$ a Lie group, $V, S$ submanifolds of $G$ containing $e$, $psi : V times S rightarrow G$; $psi(v,s)=vs$, then $dpsi(X,0)=X$ and $dpsi(0,Y)=Y$Open neighborhood in Lie groupfactoring a neighborhood of identity in a compact connected Lie group with a closed Lie subgroupJohn Lee : Cubical charts and cube in $mathbbR^n$










2












$begingroup$


This is the problem 7-6 of Lee's Introduction to Smooth Manifolds (2nd edition):




Suppose G is a Lie group and U is any neighborhood of the identity. Show
that there exists a neighborhood V of the identity such that $V subset U$ and $gh^-1 in U$ whenever $g, h in V$.




How do I approach this problem? I tried using the smoothness of $(g,h) mapsto gh^-1$ or the open subgroup generated by $U$, but it didn't get me anywhere.










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    This is the problem 7-6 of Lee's Introduction to Smooth Manifolds (2nd edition):




    Suppose G is a Lie group and U is any neighborhood of the identity. Show
    that there exists a neighborhood V of the identity such that $V subset U$ and $gh^-1 in U$ whenever $g, h in V$.




    How do I approach this problem? I tried using the smoothness of $(g,h) mapsto gh^-1$ or the open subgroup generated by $U$, but it didn't get me anywhere.










    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      This is the problem 7-6 of Lee's Introduction to Smooth Manifolds (2nd edition):




      Suppose G is a Lie group and U is any neighborhood of the identity. Show
      that there exists a neighborhood V of the identity such that $V subset U$ and $gh^-1 in U$ whenever $g, h in V$.




      How do I approach this problem? I tried using the smoothness of $(g,h) mapsto gh^-1$ or the open subgroup generated by $U$, but it didn't get me anywhere.










      share|cite|improve this question











      $endgroup$




      This is the problem 7-6 of Lee's Introduction to Smooth Manifolds (2nd edition):




      Suppose G is a Lie group and U is any neighborhood of the identity. Show
      that there exists a neighborhood V of the identity such that $V subset U$ and $gh^-1 in U$ whenever $g, h in V$.




      How do I approach this problem? I tried using the smoothness of $(g,h) mapsto gh^-1$ or the open subgroup generated by $U$, but it didn't get me anywhere.







      differential-geometry manifolds lie-groups smooth-manifolds






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 19 at 7:51









      Sou

      3,3242923




      3,3242923










      asked Mar 19 at 4:40









      o zo z

      132




      132




















          1 Answer
          1






          active

          oldest

          votes


















          6












          $begingroup$

          You don’t need anything fancy here. Just use continuity and pick $V$ carefully.



          The map $f : G times G to G$ defined as $f(g,h) = gh^-1$ is a smooth map. Suppose $U$ is a neighbourhood of the identity $ein G$. By continuity, $W = f^-1(U)$ is open in $G times G$. Since $(e,e) in W$, there are neighbourhoods $W_1,W_2 subseteq G$ containing $ein G$ such that $W_1times W_2 subseteq W$. Choose $V$ as




          $$V = (W_1 cap W_2) cap U$$




          So $V times V subseteq W=f^-1(U)$ implies $f(V times V) subseteq U$. I.e., $forall g,h in V$, we have $f(g,h)=gh^-1 in U$.






          share|cite|improve this answer











          $endgroup$













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            1 Answer
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            active

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            6












            $begingroup$

            You don’t need anything fancy here. Just use continuity and pick $V$ carefully.



            The map $f : G times G to G$ defined as $f(g,h) = gh^-1$ is a smooth map. Suppose $U$ is a neighbourhood of the identity $ein G$. By continuity, $W = f^-1(U)$ is open in $G times G$. Since $(e,e) in W$, there are neighbourhoods $W_1,W_2 subseteq G$ containing $ein G$ such that $W_1times W_2 subseteq W$. Choose $V$ as




            $$V = (W_1 cap W_2) cap U$$




            So $V times V subseteq W=f^-1(U)$ implies $f(V times V) subseteq U$. I.e., $forall g,h in V$, we have $f(g,h)=gh^-1 in U$.






            share|cite|improve this answer











            $endgroup$

















              6












              $begingroup$

              You don’t need anything fancy here. Just use continuity and pick $V$ carefully.



              The map $f : G times G to G$ defined as $f(g,h) = gh^-1$ is a smooth map. Suppose $U$ is a neighbourhood of the identity $ein G$. By continuity, $W = f^-1(U)$ is open in $G times G$. Since $(e,e) in W$, there are neighbourhoods $W_1,W_2 subseteq G$ containing $ein G$ such that $W_1times W_2 subseteq W$. Choose $V$ as




              $$V = (W_1 cap W_2) cap U$$




              So $V times V subseteq W=f^-1(U)$ implies $f(V times V) subseteq U$. I.e., $forall g,h in V$, we have $f(g,h)=gh^-1 in U$.






              share|cite|improve this answer











              $endgroup$















                6












                6








                6





                $begingroup$

                You don’t need anything fancy here. Just use continuity and pick $V$ carefully.



                The map $f : G times G to G$ defined as $f(g,h) = gh^-1$ is a smooth map. Suppose $U$ is a neighbourhood of the identity $ein G$. By continuity, $W = f^-1(U)$ is open in $G times G$. Since $(e,e) in W$, there are neighbourhoods $W_1,W_2 subseteq G$ containing $ein G$ such that $W_1times W_2 subseteq W$. Choose $V$ as




                $$V = (W_1 cap W_2) cap U$$




                So $V times V subseteq W=f^-1(U)$ implies $f(V times V) subseteq U$. I.e., $forall g,h in V$, we have $f(g,h)=gh^-1 in U$.






                share|cite|improve this answer











                $endgroup$



                You don’t need anything fancy here. Just use continuity and pick $V$ carefully.



                The map $f : G times G to G$ defined as $f(g,h) = gh^-1$ is a smooth map. Suppose $U$ is a neighbourhood of the identity $ein G$. By continuity, $W = f^-1(U)$ is open in $G times G$. Since $(e,e) in W$, there are neighbourhoods $W_1,W_2 subseteq G$ containing $ein G$ such that $W_1times W_2 subseteq W$. Choose $V$ as




                $$V = (W_1 cap W_2) cap U$$




                So $V times V subseteq W=f^-1(U)$ implies $f(V times V) subseteq U$. I.e., $forall g,h in V$, we have $f(g,h)=gh^-1 in U$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 25 at 12:18

























                answered Mar 19 at 7:47









                SouSou

                3,3242923




                3,3242923



























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