Let there be a Lie-group $G$ and its Lie-algebra $g$. Then the Cartan Maurer form is an 1-form $\omega: T\_gG \rightarrow T\_eG$ for which holds:


$$ (L^\ast\_g)\omega = \omega$$


In Shlomo Sternberg's book "Curvature in Mathematics and Physics" it is said on p. 155:


$$d\omega (X,Y) = X\omega(Y) - Y\omega(X)- \omega([X,Y])$$


But as $X\omega(Y)=0$ and $Y\omega(Y)=0$ the first two terms will disappear.


Why is this valid ? It is even said in the last sentence of the corresponding section of Sternberg's book p.155 that for left-invariant vector fields $\omega(Y) = Y$ and $\omega(X) =X$. The explanation provided by Sternberg is that resulting element of the Lie-algebra is constant, but apparently this element varies if the vector field $X$ plugged into $\omega$ varies.


I could even conclude:


$$X\omega(Y) - Y\omega(X) = XY -YX =[X,Y]\neq 0$$.


Why is this wrong ? I am a physicist and I only have very basic knowledge on differential geometry, so I would appreciate an answer which can be easily understood. Thanks.



Behaviour of the Cartan Maurer form
One should distinguish between a vector $X\in T\_eG$ and the associated left-invariant vector field $\underline{X}$ on $G$. With this notation, the Maurer-Cartan form is the $T\_eG$-valued differential form on $G$ defined by $\omega(\underline X)=X$ for any $X\in T\_eG$. So $\omega(\underline X)$ is not a vector field, it is a function on $G$, valued in the vector space $T\_eG$. As its value doesn't depend on the point $g\in G$, the derivative of this function is $0$.


