Interior product of inner product I need to calculate how the interior product $\iota_{\xi}$ for some vector field $\xi$ acts on the inner product of two forms. That is, if $\alpha$ and $\beta$ are both $p$-forms, then how do I calculate
$$\iota_{\xi}\left(\alpha\wedge\star\beta\right)?$$
In particular, I am mostly interested in the case where $\alpha=\beta$, namely, I want to calculate
$$\iota_{\xi}\left(\alpha\wedge\star\alpha\right),$$
which I have reason to believe is simply $2\iota_{\xi}\alpha\wedge\star\alpha$, but this could be wrong.
My attempt so far has been to use the fact that
$$\iota_{\xi}\left(\alpha\wedge\star\beta\right)=\iota_{\xi}\alpha\wedge\star\beta+(-1)^p\alpha\wedge\iota_{\xi}(\star\beta),$$
but then have no idea how to deal with $\iota_{\xi}(\star\beta)$.
 A: 
Claim: We have $$
i_\xi(\star\omega) = \star(\omega\wedge \xi^\flat),
$$
  where $\xi^\flat$ is the 1-form $g(\xi,-)$.

Proof: Both sides are linear in $\xi,\omega$.  So it suffices to verify it for $\xi=e_1$ and $\omega=e^{i_1\dots i_p}$, $1\leq i_1<i_2<\dots<i_p\leq n$.


*

*If $i_1=1$ then $e^1$ does not appear in $\star\omega$, hence LHS is $0$.  RHS is also $0$ because $\omega\wedge \xi^\flat=e^{1i_2\dots i_p}\wedge e^1=0$.

*If $i_1\neq 1$, then $\star\omega=\varepsilon e^{j_1j_2\dots j_{n-p}}$, $1=j_1<j_2<\dots<j_{n-p}\leq n$ with $(i_1,\dots,i_p,\underbrace{j_1}_{=1},j_2,\dots,j_{n-p})$ a sign $\varepsilon$ permutation of $(1,\dots,n)$.  Then LHS is $\varepsilon e^{j_2\dots j_{n-p}}$.  For the RHS, $e^{i_1\dots i_p}\wedge e^1$ has Hodge star $\varepsilon e^{j_2j_3\dots j_{n-p}}$.  So both sides agree.



You can now applying this to $i_\xi(\alpha\wedge\star\beta)$.
There is a special simplification available with $\alpha=\beta$: $\alpha\wedge\star\alpha=\lvert\alpha\rvert^2\,(\star1)$
so
$$
i_\xi(\alpha\wedge\star\alpha)=\lvert\alpha\rvert^2\,i_\xi(\star1)=\lvert\alpha\rvert^2\,\star(\xi^\flat).
$$
