# Working with differential forms

## 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?

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## Answers and Replies

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