I currently struggeling with the last exercise on my assignment:
Fix $\omega\in\bigwedge^3(\mathbb R^n)^*$. Let $G$ be a Lie group, $\rho\colon G\to GL(n,\mathbb R)$ a Lie group morphism such that $$ \omega(\rho(g)w_1, \rho(g)w_2, \rho(g)w_3) = \omega(w_1,w_2,w_3) $$ for all $g\in G$ and $w_i\in\mathbb R^n$. Now we take the associated Lie algebra morphism $R = (\rho)_{*e} \colon \mathfrak{g}\to Mat(n,\mathbb R)$.
Show that $$ \omega(R(v)w_1,w_2,w_3)+\omega(w_1,R(v)w_2,w_3)+\omega(w_1,w_2,R(v)w_3) = 0 $$ for all $v\in\mathfrak g$ and $w_i\in\mathbb R^n$.
It looks to me like somehow differentiating the first equation and ending up with the second but to apply the exterior derivative of a k-form we need another representation. Another thought was using the relation $$ \omega(\rho(g)w_1, \rho(g)w_2, \rho(g)w_3) = \det(\rho(g))\omega(w_1,w_2,w_3) $$ thus we see that $\det(\rho(g))=1$ by the first equation and we get $R(v)$ has trace zero.
Any help would be very much appreciated. :-)
If $v\in\mathfrak g$ and $t\in\mathbb R$, then$$\omega\bigl(\rho(e^{tv})w_1,\rho(e^{tv})w_2,\rho(e^{tv})w_3\bigr)=\omega(w_1,w_2,w_3).$$Differentiating both sides at $t=0$, you get$$\omega\bigl(R(v)w_1,w_2,w_3\bigr)+\omega\bigl(w_1,R(v)w_2,w_3\bigr)+\omega\bigl(w_1,w_2,R(v)w_3\bigr)=0.$$