I'm finishing a course project on Lie groups and I'm quite stuck with the following problem.
Consider the trilinear, antisymmetric form $\omega (x,y,z) = \text{trace}(\left[ x,y \right] \cdot z )$ on the space $\text{Mat}_{n \times n}(\mathbb{R})$. (Here "$\cdot$" denotes matrix multiplication, and $\left[ \cdot, \cdot \right]$ denotes the matrix commutator.)
i) Prove that $\omega$ is defines a $3$-form on $G = SO(n)$.
ii) For $G = SO(3)$, prove that $\omega$ is closed, but not exact.
Some background: Part 1 seems like a very strange question to me since, well, any element of $SO(n)$ is an $n\times n$-matrix, so $\omega$ is already a $3$-form on a space that contains $SO(n)$, so what is the problem? My professor gave a hint that it would be enough to prove that $\omega$ is invariant under translations, but I'm not sure how that helps, and I'm not even sure how to prove that.
Given part 1 and given that $\omega$ is invariant under translations, I feel part 2 is quite simple: $\omega$ is a $3$-form on a $3$-dimensional manifold, so $d \omega$ is a $4$-form on the manifold, hence the $0$ form. But $\omega$ cannot be exact since, if $\omega = d \eta$, then \begin{align*} \int_{SO(3)} \omega = \int_{SO(3)} d \eta = \int_{\partial SO(3)} \eta = 0 \quad \quad (*) \end{align*} since $SO(3)$ has no boundary. On the other hand, since $\omega$ is non-zero and translation invariant, $\omega$ must equal $c \cdot dx \wedge dy \wedge dz$ for some constant $c \neq 0$, which contradicts $(*)$.
My question(s): What is there even to prove in part 1? And how can one realize that $\omega$ must be invariant under translations?
Any thoughts and hints are welcome! Thank you.