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!
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What definition of determinant are you using? This is basically the definition of a determinant in some places. – Kenta S May 29 '22 at 08:39
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@KentaS the definition my book gives is for $V$ an $n$-dimensional vector space, $\omega \in \bigwedge^n(V^)$, and $A: V \to V$, $A^\omega = \mbox{det}(A)\omega$. – junglekarmapizza May 29 '22 at 08:43
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So how is $(\omega_1)_p\wedge\dots\wedge (\omega_n)_p$ defined? (And you do not want to assume $n=\dim V$. The formula holds for $n<\dim V$ as well (and that's where it's more interesting). – Ted Shifrin Jun 01 '22 at 23:23
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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.
Mr. Gandalf Sauron
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