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I'm quite new to differential forms/interior derivatives so if anyone could help with understanding the intuition behind proving this certain property step by step would be great.

Let $\iota_X$ denote the interior derivative of differential forms with a vector field $X$. Show that the operator: $$L_X:= \iota_X \ \circ d \ + \ d \ \circ \ \iota_X$$

satisfies $$L_X(\omega_1 \ \wedge \ \omega_2)=(L_X \omega_1) \ \wedge \omega_2 \ + \ \omega_1 \ \wedge \ (L_X \omega_2)$$

So far I have gotten to the expression:

$$\iota_X \ \circ \ [(d\omega_1) \ \wedge \ \omega_2 \ + \ (-1)^k\omega_1 \ \wedge \ d(\omega_2)] \ + d \ \circ \ [(\iota_X\omega_1) \ \wedge \ \omega_2 \ + \ (-1)^k\omega_1 \ \wedge \iota_X(\omega_2)]$$

by using the following properties:

$$\text{(1)} \ \ \ d(\omega_1 \wedge \omega_2)=d\omega_1 \wedge \omega_2 +(-1)^k(\omega_1 \wedge d\omega_2)$$

$$\text{(2)} \ \ \ \iota_X(\omega_1 \wedge \omega_2) = \iota_X(\omega_1) \wedge \omega_2 + (-1)^k\omega_1 \wedge \iota_X(\omega_2)$$

Edit: after substituting the properties again for a second time I am left with:

$$\iota_x \ \circ \ d\omega_1 \ \wedge \ \omega_2 \ + \ (-1)^k d\omega_1 \ \wedge \ \iota_x(\omega_2) \ + \ (-1)^k \iota_x \omega_1 \ \wedge d\omega_2 \ + \ (-1)^{2k} \omega_1 \ \wedge \ (\iota_X \ \circ \ d\omega_2) \ + \ d \ \circ \ \iota_X \omega_1 \ \wedge \ \omega_2 \ + \ (-1)^k(\iota_X \omega_1 \ \wedge \ d\omega_2) \ + \ (-1)^k(d\omega_1) \ \wedge \ \iota_X \omega_2 \ + \ (-1)^{2k}\omega_1 \ \wedge \ (d \ \circ \ \iota_x\omega_2)$$

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    $\begingroup$ Have you written down the right-hand side? How do the terms compare? $\endgroup$ Commented Feb 18, 2022 at 19:29
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    $\begingroup$ What do you mean “get rid of”? What you’ve written down so far expands to 8 terms. Do they simplify/combine to give the 4 terms on the right-hand side? Write it all out explicitly. $\endgroup$ Commented Feb 18, 2022 at 19:43
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    $\begingroup$ Your two formulas below show you how each type of term turns into two! $\endgroup$ Commented Feb 18, 2022 at 19:49
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    $\begingroup$ No, you already used property (2) to get this. But you need to compute $d(\dots)$ here, so use property (1). In the first term, you need to use (2) again, right? $\endgroup$ Commented Feb 18, 2022 at 22:32
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    $\begingroup$ Good job, you two $\endgroup$ Commented Feb 19, 2022 at 1:31

1 Answer 1

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Following from above:

\begin{align} L_X(\omega_1 \wedge \omega_2) &= (\iota_X \circ d + d \circ \iota_X)(\omega_1 \wedge \omega_2) = (\iota_X \circ d)(\omega_1 \wedge \omega_2)+ (d \circ \iota_X)(\omega_1 \wedge \omega_2)\\ &= \iota_X \circ [(d\omega_1) \wedge \omega_2 + (-1)^k\omega_1 \wedge d(\omega_2)]+ d\circ [(\iota_X \omega_1) \wedge \omega_2+(-1)^k\omega_1 \wedge\iota_X(\omega_2)]\\ &= \iota_X \circ d\omega_1 \wedge \omega_2 + (-1)^{k+1} d\omega_1 \wedge \iota_X\omega_2 + (-1)^k \iota_X \omega_1 \wedge d\omega_2 + (-1)^{2k} \omega_1 \wedge \iota_X \circ d\omega_2 \\ & \qquad + d\circ\iota_X \omega_1 \wedge \omega_2 + (-1)^{k-1}\iota_X \omega_1 \wedge d\omega_2 + (-1)^k d\omega_1 \wedge \iota_X \omega_2 + (-1)^{2k}\omega_1 \wedge d\circ\iota_X\omega_2\\ &= (\iota_X \circ d\omega_1 \wedge \omega_2 + d\circ \iota_X\omega_1 \wedge \omega_2) + (\omega_1 \wedge \iota_X \circ d\omega_2 + \omega_1 \wedge d\circ \iota_X \omega_2)\\ &= (L_X\omega_1) \wedge \omega_2 + \omega_1 \wedge (L_X \omega_2)\\ \end{align}

Thank you for the help Ted!:)

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