Decomposing a 2-form into a product of two 1-forms

Question

I'm trying to decompose the 2-form $\omega = dx \wedge dy + 4dx \wedge dz + 3dz \wedge dy$ (in $\mathbb{R}^{3}$) as the product of two 1-forms, but get stuck. Is it posible to do this?

Answer 1

$$\omega = (\mathrm{d}x+3\mathrm{d}z)\wedge (\mathrm{d}y+4\mathrm{d}z)$$

Answer 2

Consider the first two terms. They both begin with $dx$ so we can 'factorise' to obtain $$dx\wedge dy + 4dx\wedge dz = dx\wedge(dy + 4dz).$$ Now, if I can write $dz\wedge dy$ as $\alpha\wedge(dy + 4dz)$ for some one form $\alpha$ I'm done because then my original two-form becomes $(dx + \alpha)\wedge(dy+4dz)$. I want to get $3 dz\wedge dy$, so I will try $\alpha = dz$ (I know that this will give me a $dz\wedge dy$ and the other term from the expansion will disappear as it contains $dz\wedge dz$). So $$dz\wedge(dy+4dz) = dz\wedge dy + 4dz\wedge dz = dz\wedge dy.$$ Now, to obtain $3dz\wedge dy$ instead of $dz\wedge dy$, I just adjust the constant and choose $\alpha = 3dz$. In summary, we have $$(dx + 3dz)\wedge(dy + 4dz) = dx\wedge dy + 4dx\wedge dz + 3dz\wedge dy.$$

---

More generally, if this approach doesn't work, you can always consider $\alpha = \alpha_1dz + \alpha_2dy + \alpha_3dz$ and $\beta = \beta_1dz + \beta_2dy + \beta_3dz$, then try to find $\alpha_1, \alpha_2, \alpha_3, \beta_1, \beta_2, \beta_3$ such that $\alpha\wedge\beta = \omega$ where $\omega$ is your given two form.