Proof that $L_X\omega=\iota_X(\mathrm{d}\omega+\omega\wedge\omega)$ if $\iota_X\omega=0$

Question

Let $\nabla$ be a covariant derivative on a vector bundle $E\to M$ and suppose we are given a trivialization $$E\to M\times V$$ where $V$ is a finite-dimensional vector space. This trivialization defines a decomposition $$\nabla=\mathrm{d}+\omega$$ (both $\mathrm d$ and $\omega$ depend on the chosen trivialization.) Now suppose that $\newcommand{\imult}{\mathbin{\lrcorner}}$ $$\iota_X\omega=0$$ for some vector field $X$ on $M$. In the book Heat Kernels and Dirac Operators (page $27$) it is claimed that this implies $$L_X\omega=\iota_XF$$ where $F=\mathrm{d}\omega+\omega\wedge\omega$ is the curvature of $\nabla$. The following proof is given: $$L_X\omega=[\iota_X,\omega]\omega=\iota_X(\mathrm{d}\omega+\omega\wedge\omega)$$ Unfortunately, this doesn't make sense to me: According to Cartan's formula, we have $$L_X\omega=\iota_X\mathrm{d}\omega+\underbrace{\mathrm{d}\iota_X\omega}_{=0\text{ by assumption}}=\iota_X\mathrm{d}\omega.$$ So I get a different result ($\omega\wedge\omega$ is not necessarely zero, because endomorphisms do not necessarely commute)...

Answer 1

We have

$$ \iota_X (\omega \wedge \omega)=[ \iota_X\omega, \omega] = 0$$

since $\iota_X\omega = 0$. The first equality can be checked directly: write $\omega = A_i dx^i$, where $A_i\in \operatorname{End}(E)$, and $X = X^k \partial_k$, $Y= Y^l\partial_l$,

\begin{split} (\iota_X (\omega\wedge \omega)) (Y)= \omega \wedge \omega (X, Y) &= A_i A_j dx^i \wedge dx^j (X^k \partial_k, Y^l \partial_l) \\ &= A_i A_j (X^i Y^j - X^jY^i) \\ &= \iota_X \omega \omega(Y) - \omega (Y) \iota_X \omega \\ &= [ \iota _X \omega, \omega] (Y) \end{split}