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I'm aware of some related questions on this site, e.g., this question, but my following query is somewhat distinct. I also know I only need to prove the relation $[\mathcal{L}_X, \iota_Y] = \iota_{[X,Y]}$ for a function $f$ and an exact 1-form $ df$, since the interior product is an antiderivation, as was done in that post. However, for my own sake I wanted to show the equality for an arbitrary $p$-form $\alpha = a_I dx^I$ (with multi-index $I$) with a brute-force application of Cartan's formula $\mathcal{L}_X = \iota_X \circ d + d \circ \iota_X $, and after expanding all my terms, ended up with the following: $$ [\mathcal{L}_X, \iota_Y] \alpha = a_I \Big(\iota_X \circ d \circ \iota_Y - \iota_Y \circ d \circ \iota_X + d \circ \iota_X \circ \iota_Y \Big) dx^I $$ The problem term I'm finding is mainly $$ a_I d \circ \iota_X \circ \iota_Y dx^I = a_{i_1 \dots i_p} d \circ \iota_X \circ \iota_Y dx^{i_1} \wedge\dots dx^{i_p}. $$ Since this term is trivially zero if $\alpha$ is a 1-form, I'm tempted to believe it should be zero for any $p$-form. However, I am having trouble seeing how to show this. Is this true generally? Or does it cancel non-trivial contributions from the other terms when $\alpha$ is not a zero or one form?

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  • $\begingroup$ Did you include the derivatives on $a_I$? $\endgroup$
    – Keshav
    Jan 6, 2021 at 1:57
  • $\begingroup$ @Keshav yes, I first expanded the LHS as $ \iota_X d \iota_Y + d \iota_X \iota_Y - \iota_Y \iota_X d - \iota_Y d \iota_X $ and then just applied each operation successively. All derivatives of the $a_I$ cancelled. I also expected this since the RHS does not contain any derivatives. $\endgroup$
    – user488914
    Jan 6, 2021 at 5:27

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