Working with differential forms

  • Thread starter Demon117
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Homework Statement



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


Homework Equations



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



The Attempt at a Solution


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

Answers and Replies

  • #2
165
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Ok, clearly no one else understands this. . . . but it is actually quite straightforward now that I have done it myself.
 

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