2-forms as wedge product of 1-forms Why can all $2$-forms on $\operatorname{T}_p (\mathbb{R}^3)$ be written as product of $1$-forms? 
What is a counter-example to show this isn't true in a space other that $\operatorname{T}_p (\mathbb{R}^3)$?
 A: Suppose $\omega \in \Lambda^2 T_p^* \mathbb{R}^3$ is a $2$-form at $p$.  There's a map $T_p^* \mathbb{R}^3 \rightarrow \Lambda^3 T_p^* \mathbb{R}^3$ given by $\alpha \mapsto \omega \wedge \alpha$, and by rank-nullity the kernel must have dimension at least $2$.  So we can pick $\alpha_1, \alpha_2$ in the kernel which are linearly independent, and then pick $\alpha_3$ to complete a basis of $T_p^* \mathbb{R}^3$.
We can then write $\omega$ as $a \alpha_2 \wedge \alpha_3 + b \alpha_3 \wedge \alpha_1 + c \alpha_1 \wedge \alpha_2$ for some $a, b, c$.  Since $\omega \wedge \alpha_1 =0$, we get $a=0$.  Similarly $\omega \wedge \alpha_2=0$ implies $b=0$.  Therefore $\omega$ is the wedge of two $1$-forms: $c \alpha_1$ and $\alpha_2$.
In four dimensions the $2$-form $\mathrm{d}x_1 \wedge \mathrm{d}x_2 + \mathrm{d}x_3 \wedge \mathrm{d}x_4$ cannot be written as a wedge of two $1$-forms.  To see this either compute the kernel of the 'wedge with $\omega$' map on $1$-forms and see that it is zero, or note that this $2$-form has non-zero wedge product with itself.  Both of these things are impossible for a wedge of two $1$-forms.
