Jacobi Identity of Commutator of Vector Fields












0












$begingroup$


I want to proof the Jacobi identity of Lie brackets of vectorfields on smooth manifolds.



Let $ X=sum xi^ifrac{partial}{partial x^i} $, $ Y=sum eta^ifrac{partial}{partial x^i} $ and $ Z=sum zeta^ifrac{partial}{partial x^i} $. We defined $[X,Y]=sum(X(eta^i)-Y(xi^i)) frac{partial}{partial x^i}$.



Show that $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]=0$.




I tried to find antisymmetric parts*, but i faild. My try was:



$[[X,Y],Z]=[sum_i (X(eta^i)-Y(xi^i)frac{partial}{partial x^i}$,Z]=



$sum_j(sum_i(X(eta^i)-Y(xi^i))frac{partial}{partial x^i}(zeta^j))-sum zeta^ifrac{partial}{partial x^i}(X(eta^j)-Y(xi^i))$



I didn't write the other two doublebrackets because it was just too much chaos.
Should i try to go one layer down (eg. with $X(f)(x)=df_x(X(x))$) or am i on the wrong way?



*(like it's done with the matrix commutator)






differential-geometry lie-groups smooth-manifolds vector-fields







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Have you tried simplifying the associator?
    $endgroup$
    – J.G.
    Jan 7 at 19:52










  • $begingroup$
    Sorry, what do you mean by that?
    $endgroup$
    – Timmathstf
    Jan 7 at 20:03






  • 1




    $begingroup$
    @Timmathsf The associator quantifies how much associativity fails, just as the commutator quantifies how much commutativity fails. Ultimately, you used associativity to answer your own question.
    $endgroup$
    – J.G.
    Jan 7 at 20:06










  • $begingroup$
    Ah ok, thank you. I will add that to my vocabulary. :)
    $endgroup$
    – Timmathstf
    Jan 7 at 20:08
















0












$begingroup$


I want to proof the Jacobi identity of Lie brackets of vectorfields on smooth manifolds.



Let $ X=sum xi^ifrac{partial}{partial x^i} $, $ Y=sum eta^ifrac{partial}{partial x^i} $ and $ Z=sum zeta^ifrac{partial}{partial x^i} $. We defined $[X,Y]=sum(X(eta^i)-Y(xi^i)) frac{partial}{partial x^i}$.



Show that $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]=0$.




I tried to find antisymmetric parts*, but i faild. My try was:



$[[X,Y],Z]=[sum_i (X(eta^i)-Y(xi^i)frac{partial}{partial x^i}$,Z]=



$sum_j(sum_i(X(eta^i)-Y(xi^i))frac{partial}{partial x^i}(zeta^j))-sum zeta^ifrac{partial}{partial x^i}(X(eta^j)-Y(xi^i))$



I didn't write the other two doublebrackets because it was just too much chaos.
Should i try to go one layer down (eg. with $X(f)(x)=df_x(X(x))$) or am i on the wrong way?



*(like it's done with the matrix commutator)






differential-geometry lie-groups smooth-manifolds vector-fields







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Have you tried simplifying the associator?
    $endgroup$
    – J.G.
    Jan 7 at 19:52










  • $begingroup$
    Sorry, what do you mean by that?
    $endgroup$
    – Timmathstf
    Jan 7 at 20:03






  • 1




    $begingroup$
    @Timmathsf The associator quantifies how much associativity fails, just as the commutator quantifies how much commutativity fails. Ultimately, you used associativity to answer your own question.
    $endgroup$
    – J.G.
    Jan 7 at 20:06










  • $begingroup$
    Ah ok, thank you. I will add that to my vocabulary. :)
    $endgroup$
    – Timmathstf
    Jan 7 at 20:08














0












0








0





$begingroup$


I want to proof the Jacobi identity of Lie brackets of vectorfields on smooth manifolds.



Let $ X=sum xi^ifrac{partial}{partial x^i} $, $ Y=sum eta^ifrac{partial}{partial x^i} $ and $ Z=sum zeta^ifrac{partial}{partial x^i} $. We defined $[X,Y]=sum(X(eta^i)-Y(xi^i)) frac{partial}{partial x^i}$.



Show that $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]=0$.




I tried to find antisymmetric parts*, but i faild. My try was:



$[[X,Y],Z]=[sum_i (X(eta^i)-Y(xi^i)frac{partial}{partial x^i}$,Z]=



$sum_j(sum_i(X(eta^i)-Y(xi^i))frac{partial}{partial x^i}(zeta^j))-sum zeta^ifrac{partial}{partial x^i}(X(eta^j)-Y(xi^i))$



I didn't write the other two doublebrackets because it was just too much chaos.
Should i try to go one layer down (eg. with $X(f)(x)=df_x(X(x))$) or am i on the wrong way?



*(like it's done with the matrix commutator)






differential-geometry lie-groups smooth-manifolds vector-fields







share|cite|improve this question











$endgroup$




I want to proof the Jacobi identity of Lie brackets of vectorfields on smooth manifolds.



Let $ X=sum xi^ifrac{partial}{partial x^i} $, $ Y=sum eta^ifrac{partial}{partial x^i} $ and $ Z=sum zeta^ifrac{partial}{partial x^i} $. We defined $[X,Y]=sum(X(eta^i)-Y(xi^i)) frac{partial}{partial x^i}$.



Show that $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]=0$.




I tried to find antisymmetric parts*, but i faild. My try was:



$[[X,Y],Z]=[sum_i (X(eta^i)-Y(xi^i)frac{partial}{partial x^i}$,Z]=



$sum_j(sum_i(X(eta^i)-Y(xi^i))frac{partial}{partial x^i}(zeta^j))-sum zeta^ifrac{partial}{partial x^i}(X(eta^j)-Y(xi^i))$



I didn't write the other two doublebrackets because it was just too much chaos.
Should i try to go one layer down (eg. with $X(f)(x)=df_x(X(x))$) or am i on the wrong way?



*(like it's done with the matrix commutator)






differential-geometry lie-groups smooth-manifolds vector-fields




differential-geometry lie-groups smooth-manifolds vector-fields






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 19:21







Timmathstf

















asked Jan 7 at 18:17









TimmathstfTimmathstf

307




307








  • 1




    $begingroup$
    Have you tried simplifying the associator?
    $endgroup$
    – J.G.
    Jan 7 at 19:52










  • $begingroup$
    Sorry, what do you mean by that?
    $endgroup$
    – Timmathstf
    Jan 7 at 20:03






  • 1




    $begingroup$
    @Timmathsf The associator quantifies how much associativity fails, just as the commutator quantifies how much commutativity fails. Ultimately, you used associativity to answer your own question.
    $endgroup$
    – J.G.
    Jan 7 at 20:06










  • $begingroup$
    Ah ok, thank you. I will add that to my vocabulary. :)
    $endgroup$
    – Timmathstf
    Jan 7 at 20:08














  • 1




    $begingroup$
    Have you tried simplifying the associator?
    $endgroup$
    – J.G.
    Jan 7 at 19:52










  • $begingroup$
    Sorry, what do you mean by that?
    $endgroup$
    – Timmathstf
    Jan 7 at 20:03






  • 1




    $begingroup$
    @Timmathsf The associator quantifies how much associativity fails, just as the commutator quantifies how much commutativity fails. Ultimately, you used associativity to answer your own question.
    $endgroup$
    – J.G.
    Jan 7 at 20:06










  • $begingroup$
    Ah ok, thank you. I will add that to my vocabulary. :)
    $endgroup$
    – Timmathstf
    Jan 7 at 20:08








1




1




$begingroup$
Have you tried simplifying the associator?
$endgroup$
– J.G.
Jan 7 at 19:52




$begingroup$
Have you tried simplifying the associator?
$endgroup$
– J.G.
Jan 7 at 19:52












$begingroup$
Sorry, what do you mean by that?
$endgroup$
– Timmathstf
Jan 7 at 20:03




$begingroup$
Sorry, what do you mean by that?
$endgroup$
– Timmathstf
Jan 7 at 20:03




1




1




$begingroup$
@Timmathsf The associator quantifies how much associativity fails, just as the commutator quantifies how much commutativity fails. Ultimately, you used associativity to answer your own question.
$endgroup$
– J.G.
Jan 7 at 20:06




$begingroup$
@Timmathsf The associator quantifies how much associativity fails, just as the commutator quantifies how much commutativity fails. Ultimately, you used associativity to answer your own question.
$endgroup$
– J.G.
Jan 7 at 20:06












$begingroup$
Ah ok, thank you. I will add that to my vocabulary. :)
$endgroup$
– Timmathstf
Jan 7 at 20:08




$begingroup$
Ah ok, thank you. I will add that to my vocabulary. :)
$endgroup$
– Timmathstf
Jan 7 at 20:08










1 Answer
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$begingroup$

Ok, a buddy helped me out. Apperently it is easier to go a slightly different way.



The key in his solution is that $Xequiv 0 Leftrightarrow X(f)=0quadforall fin C^infty(M)$.



Note: $[X,Y](f)=X(Y(f))-Y(X(f))$



Now : $[[X,Y],Z]=X(Y(Z(f)))-Y(X(Z(f)))-Z(X(Y(f)))-Z(Y(X(f)))$



Notation $Z(Y(X(f)))=:ZYX$



And finaly $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]= XYZ-YXZ+ZYX-ZXY+YZX-XZY+YXZ-XYZ+ZXY-ZYX+XZY-YZX=0$






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    $begingroup$

    Ok, a buddy helped me out. Apperently it is easier to go a slightly different way.



    The key in his solution is that $Xequiv 0 Leftrightarrow X(f)=0quadforall fin C^infty(M)$.



    Note: $[X,Y](f)=X(Y(f))-Y(X(f))$



    Now : $[[X,Y],Z]=X(Y(Z(f)))-Y(X(Z(f)))-Z(X(Y(f)))-Z(Y(X(f)))$



    Notation $Z(Y(X(f)))=:ZYX$



    And finaly $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]= XYZ-YXZ+ZYX-ZXY+YZX-XZY+YXZ-XYZ+ZXY-ZYX+XZY-YZX=0$






    share|cite|improve this answer









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      0












      $begingroup$

      Ok, a buddy helped me out. Apperently it is easier to go a slightly different way.



      The key in his solution is that $Xequiv 0 Leftrightarrow X(f)=0quadforall fin C^infty(M)$.



      Note: $[X,Y](f)=X(Y(f))-Y(X(f))$



      Now : $[[X,Y],Z]=X(Y(Z(f)))-Y(X(Z(f)))-Z(X(Y(f)))-Z(Y(X(f)))$



      Notation $Z(Y(X(f)))=:ZYX$



      And finaly $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]= XYZ-YXZ+ZYX-ZXY+YZX-XZY+YXZ-XYZ+ZXY-ZYX+XZY-YZX=0$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Ok, a buddy helped me out. Apperently it is easier to go a slightly different way.



        The key in his solution is that $Xequiv 0 Leftrightarrow X(f)=0quadforall fin C^infty(M)$.



        Note: $[X,Y](f)=X(Y(f))-Y(X(f))$



        Now : $[[X,Y],Z]=X(Y(Z(f)))-Y(X(Z(f)))-Z(X(Y(f)))-Z(Y(X(f)))$



        Notation $Z(Y(X(f)))=:ZYX$



        And finaly $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]= XYZ-YXZ+ZYX-ZXY+YZX-XZY+YXZ-XYZ+ZXY-ZYX+XZY-YZX=0$






        share|cite|improve this answer









        $endgroup$



        Ok, a buddy helped me out. Apperently it is easier to go a slightly different way.



        The key in his solution is that $Xequiv 0 Leftrightarrow X(f)=0quadforall fin C^infty(M)$.



        Note: $[X,Y](f)=X(Y(f))-Y(X(f))$



        Now : $[[X,Y],Z]=X(Y(Z(f)))-Y(X(Z(f)))-Z(X(Y(f)))-Z(Y(X(f)))$



        Notation $Z(Y(X(f)))=:ZYX$



        And finaly $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]= XYZ-YXZ+ZYX-ZXY+YZX-XZY+YXZ-XYZ+ZXY-ZYX+XZY-YZX=0$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 20:02









        TimmathstfTimmathstf

        307




        307






























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