I want to proof the Jacobi identity of Lie brackets of vectorfields on smooth manifolds.
Let $ X=\sum \xi^i\frac{\partial}{\partial x^i} $, $ Y=\sum \eta^i\frac{\partial}{\partial x^i} $ and $ Z=\sum \zeta^i\frac{\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]=
$\sum_j(\sum_i(X(\eta^i)-Y(\xi^i))\frac{\partial}{\partial x^i}(\zeta^j))-\sum \zeta^i\frac{\partial}{\partial x^i}(X(\eta^j)-Y(\xi^i))$
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)
Ok, a buddy helped me out. Apperently it is easier to go a slightly different way.
The key in his solution is that $X\equiv 0 \Leftrightarrow X(f)=0\quad\forall f\in C^\infty(M)$.
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 $[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]= XYZ-YXZ+ZYX-ZXY+YZX-XZY+YXZ-XYZ+ZXY-ZYX+XZY-YZX=0$