Find an $\omega \in \text{Alt}^2(\Bbb R^4)$ such that $\omega \wedge \omega \ne 0$.

Question

Find an $\omega \in \text{Alt}^2(\Bbb R^4)$ such that $\omega \wedge \omega \ne 0$.

What is the intuition for the problem? Since $\omega \in \text{Alt}^2(\Bbb R^4)$ we know that $\omega : \Bbb R^4 \to \Bbb R$ is an alternating $4$-linear map. That is $$\omega(x_1,x_2,x_3,x_4) \in \Bbb R$$ and that $$\omega(x_1,x_2,x_3,x_4) = -\omega(x_2,x_1,x_3,x_4)$$ for example.

I've also seen that this $\omega \wedge \omega$ can be tought of as the volume of a parallelotope so is the question asking me to find an $\omega \in \text{Alt}^2(\Bbb R^4)$ such that the "volume" with itself is nonzero?

Was it not true that the wedge product of any $n$-form with itself was always $0$?

Answer 1

You can easily see from the definition of the exterior (wedge) product that it is antisymmetric on 1-forms, i.e. $\omega\wedge\eta = -\eta\wedge\omega \;\,\forall\omega,\eta\in\mathrm{Alt}(\mathbb{R}^4)$, which implies that this product is alternating on 1-forms, i.e. $\omega\wedge\omega = 0 \;\,\forall\omega\in\mathrm{Alt}(\mathbb{R}^4)$.

This result is not valid in general; indeed, taking a $k$-form $\omega\in\mathrm{Alt}^k(\mathbb{R}^4)$ and a $l$-form $\eta\in\mathrm{Alt}^l(\mathbb{R}^4)$, you can prove that $\omega\wedge\eta = (-1)^{kl}\,\eta\wedge\omega$. In consequence, for higher-degree forms, the exterior product is alternating for odd degrees only, i.e. $\omega\wedge\omega = 0 \;\,\forall\omega\in\mathrm{Alt}^{2k+1}(\mathbb{R}^4)$. Note that the product with itself can vanish for an even-degree form, but it is not true in general.

As Didier suggests in the comments, we could find a form $\omega\in\mathrm{Alt}^k(\mathbb{R}^4)$ whose "square" $\omega\wedge\omega$ gives the volume form (determinant) $e_1^* \wedge e_2^* \wedge e_3^* \wedge e_4^* \in\mathrm{Alt}^4(\mathbb{R}^4)$, where the vectors $e_i^*$ constitute the canonical dual basis of $\mathbb{R}^4$. As $\omega\wedge\omega\in\mathrm{Alt}^{2k}(\mathbb{R}^4)$, we must take $k=2$. Moreover, since the exterior product is alternating on 1-forms, as previously said, no dual basis vector should appear twice, that is why we could take $\omega = e_1^* \wedge e_2^* + e_3^* \wedge e_4^*$ for instance. Let's now compute $$ \begin{array}{rcl} \omega\wedge\omega &=& (e_1^* \wedge e_2^* + e_3^* \wedge e_4^*) \wedge (e_1^* \wedge e_2^* + e_3^* \wedge e_4^*) \\ &=& e_1^* \wedge e_2^* \wedge e_3^* \wedge e_4^* + e_3^* \wedge e_4^* \wedge e_1^* \wedge e_2^* \\ &=& 2\, e_1^* \wedge e_2^* \wedge e_3^* \wedge e_4^* \neq 0 \end{array} $$ since $(e_3^* \wedge e_4^*) \wedge (e_1^* \wedge e_2^*) = (-1)^{2\cdot2}\, (e_1^* \wedge e_2^*) \wedge (e_3^* \wedge e_4^*) = e_1^* \wedge e_2^* \wedge e_3^* \wedge e_4^*$.

Answer 2

Some extra info about this problem

1. A two form $\omega \in \wedge^2(V)$ is decomposable if and only if $\omega \wedge \omega = 0$ ( per the Plucker conditions)

2. If $\dim V = 2k$ and $\omega = \sum m_{ij}\cdot v_i \wedge v_j \in \wedge^2(V)$, then $\omega\wedge \omega \wedge \cdots \wedge \omega= k! \cdot Pf(M) \cdot v_1 \wedge \cdots \wedge v_{2k}$, where $Pf(M)$ is the Pffafian of the skew-matrix $M= (m_{ij})$.