The definition of the curvature form ( from Wikipedia ) is:
$$ \Omega = d \omega + \frac{1}{2} [ \omega \wedge \omega ] $$
I am puzzled about the rightmost term. According to what I have learned so far, the wedge product of a one-form with itself is always zero. So shouldn't the $ \omega \wedge \omega $ term simply be zero?
My best guess is that my "learnings" don't work on curved manifolds. Since I am a beginner physics student ( with almost zero math knowledge ), I am having a hard time finding the answer in math literature. Can someone give me a low-brow answer to this naive question? Thanks.
Motivation for question: I stumbled upon this term when I was trying to learn about the Chern-Simons theory. There also we have a term like $ A \wedge A \wedge A $.
Repeating comment here:
For a real-valued 1-form, it's true that $\omega\wedge\omega=0$. But the $\omega$ in the curvature form is Lie-algebra-valued, and so this no longer holds in general. See https://wikipedia.org/wiki/Lie_algebra-valued_differential_form