Given two 1-forms $\phi, \psi \in \bigwedge^1 \mathbb{R}^4$ and the 2-form defined by $\phi \wedge \psi \in \bigwedge^2 \mathbb{R}^4$, then why $(\phi \wedge \psi) \wedge (\phi \wedge \psi) = 0$?
What I've tried is to find for each summand a summand which eliminates with the first one, but it doesn't really work. If I have some vectors $v_1, v_2, v_3, v_4 \in \mathbb{R}^4$ and a permutation $\sigma$ and look at the summand $sgn(\sigma)(\phi \wedge \psi)(v_{\sigma(1)},v_{\sigma(2)})(\phi \wedge \psi)(v_{\sigma(3)},v_{\sigma(4)})$, then if I exchange 1 and 2 (which is a permutation with sign -1), due to the 2-form being alternating I get the same summand (with the same sign so they don't eliminate each other). Also the fact that $\omega \wedge \omega = 0$ for any k-Form $\omega$ with k odd doesn't help here since we're considering a 2-form. I also know that $\omega \wedge \omega = 0$ doesn't hold for every 2-form $\omega$, so the reason should be the basic form of $\phi \wedge \psi$ but that's about it.
Best Answer
The wedge product is antisymmetric (and thus alternating) on one-forms and associative. Hence
$$(\phi \wedge \psi) \wedge (\phi \wedge \psi) = -(\phi \wedge \psi) \wedge (\psi \wedge \phi) = -\phi \wedge (\psi \wedge \psi) \wedge \phi = -\phi \wedge 0 \wedge \phi = 0$$