Formula for the interior product of a p-form

Question

Let $X: \mathbb{R}^n \to \mathbb{R}^n$ be a vector field and let $$\omega = \sum_{i_1 < \dots < i_p} f_{i_1\dots i_p} dx_{i_1} \wedge \dots \wedge dx_{i_p}$$ be a $p$-form over $\mathbb{R}^n$. I am interested in a formula for the interior product $i_X\omega$.

(Choose $n=3$ and $X(x,y,z) = (z,y,-x)$ for all examples below.)

$p = 1$:

\begin{align} \omega &= \sum_i f_i dx_i &\Rightarrow \quad i_X\omega &= \sum_i f_iX_i\\ \omega &= dx + zdy + zdz &\Rightarrow \quad i_X \omega &= X_x + zX_y + zX_z\\ & & &= z + yz - xz \end{align}

$p = 2$:

\begin{align} \omega &= \sum_{i < j} f_{ij} dx_i \wedge dx_j &\Rightarrow \quad i_X\omega &= \sum_{i < j} f_{ij} \left( X_i dx_j - X_j dx_i \right)\\ \omega &= dx \wedge dy + dx \wedge dz &\Rightarrow \quad i_X \omega &= X_xdy - X_ydx + X_xdz - X_zdx\\ & & &= zdy - ydx + zdz + xdx \\ & & &= \left( x - y \right)dx + zdy + zdz \end{align}

$p = 3$:

\begin{align} \omega &= \sum_{i < j < k} f_{ijk} dx_i \wedge dx_j \wedge dx_k &&\Rightarrow \quad i_X\omega =\ ???\\ \omega &= 2zdx \wedge dy \wedge dz &&\Rightarrow \quad i_X\omega =\ ??? \end{align}

Questions

1. Are my formulas and examples for $p=1$ and $p=2$ correct?
2. What is the formula for $p \geq 3$?

Answer 1

The formulas and examples for $p=1$ and $p=2$ are correct.

$p = 3$:

\begin{align} \omega &= \sum_{i < j < k} f_{ijk} dx_i \wedge dx_j \wedge dx_k\\ \Rightarrow \quad i_X\omega &= \sum_{i < j < k} f_{ijk} \left( X_i dx_j \wedge dx_k - X_j dx_i \wedge dx_k + X_k dx_i \wedge dx_j \right)\\\\ \omega &= 2zdx \wedge dy \wedge dz\\ \Rightarrow \quad i_X\omega &= 2z\left( X_x dy \wedge dz - X_y dx \wedge dz + X_z dx \wedge dy \right)\\ &= 2z\left( zdy \wedge dz - ydx \wedge dz - xdx \wedge dy \right)\\ &= -2xzdx\wedge dy - 2yzdx \wedge dz + 2z^2dy\wedge dz \end{align}

$p \in \mathbb{N}$:

\begin{align} \omega &= \sum_{i_1 < \dots < i_p} f_{i_1\dots i_p} dx_{i_1} \wedge \dots \wedge dx_{i_p} \\ \Rightarrow \quad i_X\omega &= \sum_{i_1 < \dots < i_p} f_{i_1\dots i_p} \sum_{j=1}^p (-1)^{j-1} X_{i_j} dx_{i_1} \wedge \dots \wedge dx_{i_{j-1}} \wedge dx_{i_{j+1}} \wedge \dots \wedge dx_{i_p} \end{align}

Answer 2

For $p = 3$: $$\omega = \sum_{i < j < k} f_{ijk} dx_i \wedge dx_j \wedge dx_k \quad \Rightarrow \quad i_X\omega =\ \sum_{i < j < k} f_{ijk} (X_i dx_j \wedge dx_k - X_j dx_i \wedge dx_k + X_k dx_i \wedge dx_j)$$

Generally, let $\omega$ be a $k$-form $\alpha_1 \wedge \alpha_2 \wedge \dots \wedge \alpha_k$, where $\alpha_i$'s are $1$-forms and $X$ be vector field. Then, $$i_X\omega = \sum_{i=1}^{k} (-1)^{i-1} \alpha_i(X)\cdot\alpha_1\wedge\dots\wedge\alpha_{i-1}\wedge\alpha_{i+1}\wedge\dots\alpha_k$$