Interior product general rule (differential forms)

Question

How is this general form of interior product on forms $$(i_V\omega^{(p)})=\frac{1}{(p-1)!}V^{\mu}\omega_{\mu\mu_1...\mu_{p-1}}dx^{\mu_1}\wedge dx^{\mu_2}\wedge ...\wedge dx^{\mu_{p-1}}$$?

Answer 1

Here is about the simplest example you can take, which is what you were trying to accomplish (I think!). Let $dx$ and $dy$ be the standard $1$-forms on $\mathbb{R}^2$, and let $\partial_x$, $\partial_y$ be the corresponding dual vector fields (i.e. $\partial_x$ is the unit vector field in the $x$ direction). Then

$$i_{\partial_x}(dx \wedge dy)(\partial_x) = dx(\partial_x)dy(\partial_x) - dy(\partial_x)dx(\partial_x) = 0$$

$$i_{\partial_x}(dx \wedge dy)(\partial_y) = dx(\partial_x)dy(\partial_y) - dy(\partial_x)dx(\partial_y) = 1$$

and so $i_{\partial_x}(dx \wedge dy) = dy$.

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Using the formula instead, we have $V^1 = 1$, $V^2 = 0$, $\omega_{1,2} = 1$, and so $$i_V{\omega} = V^1\omega_{1,2}dy + V^2 \omega_{2,1} dx = dy.$$