Formula for $k$-form

Question

So I have seen the formula $\omega_1 \wedge \cdots \wedge \omega_n(v_1, \cdots, v_n) = $ det($\omega_i(v_j))$ before. However, the book my class is using doesn't give this formula. It states that $dx_i\left(\frac{d}{dx_j}\right)_p = 1$ if $i = j$ and $ = 0$ if $i \neq j$ as well stating that $(\omega_1 \wedge \cdots \wedge \omega_n)_p = (\omega_1)_p \wedge \cdots \wedge (\omega_n)_p$. Is it possible to derive the formula from here? If so, how? I believe I need this for a homework problem, but cannot seem to figure out how to derive it. Thank you!

Answer 1

Let $\omega_{1},\dots\omega_{k}$ be covectors and $v_{1},...,v_{k}\in V$.

Then the formula holds for basis vectors . That is the cases when $\omega_{j}$ is a basis vector of the form $E^{i_{j}}$ . See my answer here for more context.

Now use Multi-linearity of the alternating forms on both sides of the equation to prove .

For more information see John M Lee Introduction to Smooth Manifolds . See the chapter on Differential forms.