Maurer Cartan $1$-form explanation of notation












0












$begingroup$


This is the construction I am given:





Let $G$ be a Lie group, $mathfrak{g}:=T_eG$ its Lie algebra. We define $1$-form $w_G in Omega^1(G) otimes mathfrak{g}$, i.e. closed $1$-forms with value in $mathfrak{g}$ by the rule:
At each $g in G$,
$$(w_G)_g: X_g mapsto (L_{g^{-1}})_* X_g $$
where $X_g$ is a vector in $T_gG$.






  1. It doesn't seem to me clear that $w_G$ is a smooth form.

  2. It is claimed that when $G=GL_n(Bbb R)$, $w_G = g^{-1}dg$. What does the $dg$ even mean in this case?



My reference: page 103, Definition 7.4.13, Prop 7.4.15



EDIT: I became aware of this post for my second question. Still I am confused, why and how can we regard $g:G rightarrow G$ as the identity map?






differential-geometry lie-groups lie-algebras differential-forms







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$endgroup$

















    0












    $begingroup$


    This is the construction I am given:





    Let $G$ be a Lie group, $mathfrak{g}:=T_eG$ its Lie algebra. We define $1$-form $w_G in Omega^1(G) otimes mathfrak{g}$, i.e. closed $1$-forms with value in $mathfrak{g}$ by the rule:
    At each $g in G$,
    $$(w_G)_g: X_g mapsto (L_{g^{-1}})_* X_g $$
    where $X_g$ is a vector in $T_gG$.






    1. It doesn't seem to me clear that $w_G$ is a smooth form.

    2. It is claimed that when $G=GL_n(Bbb R)$, $w_G = g^{-1}dg$. What does the $dg$ even mean in this case?



    My reference: page 103, Definition 7.4.13, Prop 7.4.15



    EDIT: I became aware of this post for my second question. Still I am confused, why and how can we regard $g:G rightarrow G$ as the identity map?






    differential-geometry lie-groups lie-algebras differential-forms







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      This is the construction I am given:





      Let $G$ be a Lie group, $mathfrak{g}:=T_eG$ its Lie algebra. We define $1$-form $w_G in Omega^1(G) otimes mathfrak{g}$, i.e. closed $1$-forms with value in $mathfrak{g}$ by the rule:
      At each $g in G$,
      $$(w_G)_g: X_g mapsto (L_{g^{-1}})_* X_g $$
      where $X_g$ is a vector in $T_gG$.






      1. It doesn't seem to me clear that $w_G$ is a smooth form.

      2. It is claimed that when $G=GL_n(Bbb R)$, $w_G = g^{-1}dg$. What does the $dg$ even mean in this case?



      My reference: page 103, Definition 7.4.13, Prop 7.4.15



      EDIT: I became aware of this post for my second question. Still I am confused, why and how can we regard $g:G rightarrow G$ as the identity map?






      differential-geometry lie-groups lie-algebras differential-forms







      share|cite|improve this question











      $endgroup$




      This is the construction I am given:





      Let $G$ be a Lie group, $mathfrak{g}:=T_eG$ its Lie algebra. We define $1$-form $w_G in Omega^1(G) otimes mathfrak{g}$, i.e. closed $1$-forms with value in $mathfrak{g}$ by the rule:
      At each $g in G$,
      $$(w_G)_g: X_g mapsto (L_{g^{-1}})_* X_g $$
      where $X_g$ is a vector in $T_gG$.






      1. It doesn't seem to me clear that $w_G$ is a smooth form.

      2. It is claimed that when $G=GL_n(Bbb R)$, $w_G = g^{-1}dg$. What does the $dg$ even mean in this case?



      My reference: page 103, Definition 7.4.13, Prop 7.4.15



      EDIT: I became aware of this post for my second question. Still I am confused, why and how can we regard $g:G rightarrow G$ as the identity map?






      differential-geometry lie-groups lie-algebras differential-forms




      differential-geometry lie-groups lie-algebras differential-forms






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 21 at 18:51







      CL.

















      asked Jan 21 at 17:54









      CL.CL.

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      2,3142925






















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          $begingroup$

          Since $GL_n(Bbb R)$ is an open subset of $Bbb R^{n^2}$, you can think of the inclusion map $gcolon Gto Bbb R^{n^2}$ and its derivative $dg$ as a vector-valued $1$-form. Of course, this derivative of the inclusion map is just the identity map on tangent spaces.



          Notice that if $ain G$ is fixed, and $X_ain T_aG$, then $omega(a)(X_a) = a^{-1}dg(a)(X_a) = L_{a^{-1}*}(X_a)$, as desired.



          Since $g^{-1}$ is a smooth function and the components of $dg$ are smooth, the matrix-valued $1$-form $g^{-1}dg$ is smooth. You can see it, as well, from your original definition. By definition the smooth function $L_g$ induces a smooth bundle map on the tangent bundle, and applying it to a smooth vector field $X$ gives a smooth function.






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            $begingroup$

            Since $GL_n(Bbb R)$ is an open subset of $Bbb R^{n^2}$, you can think of the inclusion map $gcolon Gto Bbb R^{n^2}$ and its derivative $dg$ as a vector-valued $1$-form. Of course, this derivative of the inclusion map is just the identity map on tangent spaces.



            Notice that if $ain G$ is fixed, and $X_ain T_aG$, then $omega(a)(X_a) = a^{-1}dg(a)(X_a) = L_{a^{-1}*}(X_a)$, as desired.



            Since $g^{-1}$ is a smooth function and the components of $dg$ are smooth, the matrix-valued $1$-form $g^{-1}dg$ is smooth. You can see it, as well, from your original definition. By definition the smooth function $L_g$ induces a smooth bundle map on the tangent bundle, and applying it to a smooth vector field $X$ gives a smooth function.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Since $GL_n(Bbb R)$ is an open subset of $Bbb R^{n^2}$, you can think of the inclusion map $gcolon Gto Bbb R^{n^2}$ and its derivative $dg$ as a vector-valued $1$-form. Of course, this derivative of the inclusion map is just the identity map on tangent spaces.



              Notice that if $ain G$ is fixed, and $X_ain T_aG$, then $omega(a)(X_a) = a^{-1}dg(a)(X_a) = L_{a^{-1}*}(X_a)$, as desired.



              Since $g^{-1}$ is a smooth function and the components of $dg$ are smooth, the matrix-valued $1$-form $g^{-1}dg$ is smooth. You can see it, as well, from your original definition. By definition the smooth function $L_g$ induces a smooth bundle map on the tangent bundle, and applying it to a smooth vector field $X$ gives a smooth function.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Since $GL_n(Bbb R)$ is an open subset of $Bbb R^{n^2}$, you can think of the inclusion map $gcolon Gto Bbb R^{n^2}$ and its derivative $dg$ as a vector-valued $1$-form. Of course, this derivative of the inclusion map is just the identity map on tangent spaces.



                Notice that if $ain G$ is fixed, and $X_ain T_aG$, then $omega(a)(X_a) = a^{-1}dg(a)(X_a) = L_{a^{-1}*}(X_a)$, as desired.



                Since $g^{-1}$ is a smooth function and the components of $dg$ are smooth, the matrix-valued $1$-form $g^{-1}dg$ is smooth. You can see it, as well, from your original definition. By definition the smooth function $L_g$ induces a smooth bundle map on the tangent bundle, and applying it to a smooth vector field $X$ gives a smooth function.






                share|cite|improve this answer









                $endgroup$



                Since $GL_n(Bbb R)$ is an open subset of $Bbb R^{n^2}$, you can think of the inclusion map $gcolon Gto Bbb R^{n^2}$ and its derivative $dg$ as a vector-valued $1$-form. Of course, this derivative of the inclusion map is just the identity map on tangent spaces.



                Notice that if $ain G$ is fixed, and $X_ain T_aG$, then $omega(a)(X_a) = a^{-1}dg(a)(X_a) = L_{a^{-1}*}(X_a)$, as desired.



                Since $g^{-1}$ is a smooth function and the components of $dg$ are smooth, the matrix-valued $1$-form $g^{-1}dg$ is smooth. You can see it, as well, from your original definition. By definition the smooth function $L_g$ induces a smooth bundle map on the tangent bundle, and applying it to a smooth vector field $X$ gives a smooth function.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 23 at 21:27









                Ted ShifrinTed Shifrin

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                64.2k44692






























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