TITLE: Maurer-Cartan form
QUESTION [8 upvotes]: I suppose given a Lie Group ($G$) and its corresponding Lie Algebra ($\mathfrak{g}$) every element in its dual defines a Maurer-Cartan form on the whole Lie Group? 
Let $\omega \in \mathfrak{g}^*$ be a Maurer-Cartan form and let $X$ and $Y$ be two elements of $g$ then in what sense are $\omega(X)$ and $\omega(Y)$ "constant functions" on $G$ ? (such that we can write $X\omega(Y)=Y\omega(X) =0$)
Assuming the above one can immediately write the Maurer-Cartan equation, $d\omega(X,Y)= -\frac{1}{2}\omega([X,Y])$ 
Thinking of $\omega$ as Lie Algebra valued 1-form on the Lie Group and using the fact from linear algebra that $V^* \otimes W = Hom(V,W)$ one can write them as $\omega = \sum _i \omega_i \otimes B_i$ where $\omega_i$ are ordinary 1-forms on $G$ and $B_i$ are a basis on $\mathfrak{g}$. (should there be some arbitrary coefficients in front of every term in the above sum?) 
Say $c^i_{jk}$ are the structure constants of the Lie Algebra then I do not understand how the Maurer-Cartan equations can be recast as,
$$d\omega_i = -\frac{1}{2}\sum_{j,k} c^i_{jk}  \omega_j \wedge \omega_k$$
which apparently can be further recast as the equation,
$$d\omega = -\frac{1}{2} [\omega,\omega]$$
I would be happy if someone can explain how the above two forms of the Maurer-Cartan equation can be obtained knowing the first form which is more familiar form to me. 
Also finding the structure constants of a Lie Algebra is not so hard for at least the common ones. Knowing that one fully "knows" the Maurer-Cartan Equation.  Now is there any sense in which one can "solve" this to find out the Maurer-Cartan forms? (I would guess a basis might be obtainable)

REPLY [3 votes]: The Maurer-Cartan form for matrix groups is very well explained in Santalo's book about Integral Geometry.