Differential 1-forms and linear independence

Question

We consider the vector space of constant differential $1$-forms of $\mathbb{R}^3$. Then, the $1$-forms $\omega_{1},\omega_{2},\omega_{3}$ are linearly independet if and only if $$\omega_{1}\wedge\omega_{2}\wedge\omega_{3}\neq 0$$ I can see the $"\Leftarrow"$ direction but i'm a bit confused with the other direction. Any advice would be very helpful. Thank you in advance.

Answer 1

We want to show that if the 1-forms are lineraly independent, then the wedge product is not zero. We will use the property that for all $v_1,v_2,v_3\in\mathbb{R}^3$

$$ (\omega_1\wedge \omega_2\wedge\omega_3) (v_1, v_2, v_3) = \det\left((\omega_i(v_j))_{1\leq i,j\leq 3}\right). $$

Suppose the $1$-forms are linearly independent. Since the dual of a finite-dimensional space has the same dimension, $ (\omega_1, \omega_2, \omega_3)$ form a basis of $\Lambda^1\mathbb{R}^3$. Suppose further that $\omega_1\wedge \omega_2\wedge\omega_3 = 0$.

Let $\omega^1, \omega^2, \omega^3 \in\mathbb{R^3}$ be the dual basis elements of $\mathbb{R}^3$ to the basis $ (\omega_1, \omega_2, \omega_3)$, i.e. $\omega_i(\omega^j) = \delta_{ij}$. Then by assumption $$ 0 = (\omega_1\wedge \omega_2\wedge\omega_3)(\omega^1, \omega^2, \omega^3) = \det((\omega_i(\omega^j))_{1\leq i,j\leq 3}).$$ This is a contradiction because the matrix is the identity matrix, therefore has a determinant of $1$. This proves the claim.