group acts on the contract operator $iota_X_M$

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I am working on equvariant cohomology and get some problem.



Give $M$ as a manifold and the action of a Lie group $G$ acts on $M$, denote the Lie algebra of $G$ by $mathfrakg$. Let $alpha$ be a differential form on $M$ and $X_M$ is the vector field induced by $X in mathfrakg$



My problem is how to prove $g$ $iota_X_M$ $g^-1$ = $iota_gX_Mg^-1$ = $iota_Ad_gX_M$.



Where $Ad_g$ denotes the adjoint action of $g$



My idea is following



($g$ $iota_X_M$ $g^-1$ )($alpha$) = ($L_g$$R_g$$iota_X_M$)($alpha$)



= $iota_X_M$(($L_g$$R_g$)$^*$$alpha$)=(($L_g$$R_g$)$^*$$alpha$)($X_M$)=$alpha$(($L_g$$R_g$)$_*$$X_M$)=($iota_Ad_gX_M$)($alpha$)



I am not familiar with group action. I think it makes sense but I am not sure.



Thanks a lot!




differential-forms group-actions




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    up vote
    0
    down vote

    favorite












    I am working on equvariant cohomology and get some problem.



    Give $M$ as a manifold and the action of a Lie group $G$ acts on $M$, denote the Lie algebra of $G$ by $mathfrakg$. Let $alpha$ be a differential form on $M$ and $X_M$ is the vector field induced by $X in mathfrakg$



    My problem is how to prove $g$ $iota_X_M$ $g^-1$ = $iota_gX_Mg^-1$ = $iota_Ad_gX_M$.



    Where $Ad_g$ denotes the adjoint action of $g$



    My idea is following



    ($g$ $iota_X_M$ $g^-1$ )($alpha$) = ($L_g$$R_g$$iota_X_M$)($alpha$)



    = $iota_X_M$(($L_g$$R_g$)$^*$$alpha$)=(($L_g$$R_g$)$^*$$alpha$)($X_M$)=$alpha$(($L_g$$R_g$)$_*$$X_M$)=($iota_Ad_gX_M$)($alpha$)



    I am not familiar with group action. I think it makes sense but I am not sure.



    Thanks a lot!




    differential-forms group-actions




    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I am working on equvariant cohomology and get some problem.



      Give $M$ as a manifold and the action of a Lie group $G$ acts on $M$, denote the Lie algebra of $G$ by $mathfrakg$. Let $alpha$ be a differential form on $M$ and $X_M$ is the vector field induced by $X in mathfrakg$



      My problem is how to prove $g$ $iota_X_M$ $g^-1$ = $iota_gX_Mg^-1$ = $iota_Ad_gX_M$.



      Where $Ad_g$ denotes the adjoint action of $g$



      My idea is following



      ($g$ $iota_X_M$ $g^-1$ )($alpha$) = ($L_g$$R_g$$iota_X_M$)($alpha$)



      = $iota_X_M$(($L_g$$R_g$)$^*$$alpha$)=(($L_g$$R_g$)$^*$$alpha$)($X_M$)=$alpha$(($L_g$$R_g$)$_*$$X_M$)=($iota_Ad_gX_M$)($alpha$)



      I am not familiar with group action. I think it makes sense but I am not sure.



      Thanks a lot!




      differential-forms group-actions




      share|cite|improve this question














      I am working on equvariant cohomology and get some problem.



      Give $M$ as a manifold and the action of a Lie group $G$ acts on $M$, denote the Lie algebra of $G$ by $mathfrakg$. Let $alpha$ be a differential form on $M$ and $X_M$ is the vector field induced by $X in mathfrakg$



      My problem is how to prove $g$ $iota_X_M$ $g^-1$ = $iota_gX_Mg^-1$ = $iota_Ad_gX_M$.



      Where $Ad_g$ denotes the adjoint action of $g$



      My idea is following



      ($g$ $iota_X_M$ $g^-1$ )($alpha$) = ($L_g$$R_g$$iota_X_M$)($alpha$)



      = $iota_X_M$(($L_g$$R_g$)$^*$$alpha$)=(($L_g$$R_g$)$^*$$alpha$)($X_M$)=$alpha$(($L_g$$R_g$)$_*$$X_M$)=($iota_Ad_gX_M$)($alpha$)



      I am not familiar with group action. I think it makes sense but I am not sure.



      Thanks a lot!





      differential-forms group-actions





      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 19 at 9:02

























      asked Aug 19 at 8:53









      李岳鴻

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