Working with differential forms

[Demon117]

Homework Statement

Show that
d\omega_{ij}+\sum_{k=1}^{n} \omega_{ik}\wedge\omega_{kj} =0

Homework Equations

Let G be the group of invertible nxn matrices. This is an open set in the vector space
M=Mat(n\times n, R)
and our formalism of differential forms applies there with the coordinate functions now being the entries
X_{ij} of a matrix X.
For every linear function
\lambda : M\rightarrow R,
there is a unique 1-form
\omega_{\lambda}\in G
that has the value\lambda at I and is invariant under all left multiplications
L_{\lambda}:Y\rightarrow YX for X\in G; that is, L_{X}^{*}(\omega_{\lambda})=\omega_{\lambda}. It is given by
\omega_{\lambda}(X):Y|\rightarrow\lambda(D(L_{X^{-1}})_{X}(Y))=\lambda(X^{-1}Y), X\in G, Y\in M.
The basic examples are the Maurer-Cartan forms
\omega_{ij}(X)(Y)=(X^{-1}Y)_{ij}
or
\omega_{ij}(X)=\sum_{k=1}^{n} (X^{-1}_{ik}dX_{kj}).

The Attempt at a Solution

HA HA, I don't know. It has been suggested that I use an equivalent formula, namely
dX_{ij}=\sum_{k=1}^{n} X_{ik}\omega_{kj}.
Then apply d. Now this will hold for all i and j. To obtain a differential d\omega_{rj}, multiply that equation by (X^{-1})_{ri} and sum over i.

This suggestion is great, but I am not even sure I really understand it. Any other suggestions on how to prove this, or maybe some tips about this suggestion?

 

[Demon117]

Ok, clearly no one else understands this. . . . but it is actually quite straightforward now that I have done it myself.