I'm reading a paper (https://arxiv.org/pdf/gr-qc/0012037.pdf) and I have a question about equation H3:
$$\star(e_a \rfloor \psi) = (-1)^{p-1}\vartheta_a\wedge\star \psi$$
How can it be possible for the equation to be true when, on the left-hand side of the equation, we contract one of the indices of the component of $\psi$ without contracting the basis $1$-forms, whereas on the right-hand side, the $a$ index remains on the $\vartheta$. Explicitly, in components and in some convention, defining
$$ \psi = \frac{1}{p!}\psi_{a_1\ldots a_p}\vartheta^{a_1}\wedge\ldots\wedge \vartheta^{a_p} $$
$$ \star(e_a\rfloor\psi) = \frac{1}{(p-1)!}\psi_{a\,b_1\ldots b_{p-1}}\frac{1}{(n-(p-1))!}\epsilon^{b_1\ldots b_{p-1}}{}_{c_1\ldots c_{n-p+1}}\vartheta^{c_1}\wedge\ldots\wedge \vartheta^{c_{n-p+1}} $$
Whereas
$$ \vartheta_{a}\wedge\star\psi = \frac{1}{p!}\frac{1}{(n-p)!}\psi_{a_1\ldots a_p}\epsilon^{a_1\ldots a_p}{}_{b_1\ldots b_{n-p}}\vartheta_a\wedge \vartheta^{b_1}\wedge\ldots\wedge\vartheta^{b_{n-p}} $$
How can we reconcile both equations? Is it a matter of combinatorics? Any guidance appreciated! Thanks
Ok, so I had a play around with some formulae, as @TedShifrin suggested, and it turned out fine. I have a possible proof, but it relies on uniqueness of the basis coframe/dual frame. Hopefully, it should be self-explanatory!
$$\star(e_a \rfloor \psi) = (-1)^{p-1}\vartheta_a\wedge\star \psi$$
$$ \psi = \frac{1}{p!}\psi_{a_1\ldots a_p}\vartheta^{a_1}\wedge\ldots\wedge \vartheta^{a_p} $$
$$ \star(e_a\rfloor\psi) = \frac{1}{(p-1)!}\psi_{a\,b_1\ldots b_{p-1}}\frac{1}{(n-(p-1))!}\epsilon^{b_1\ldots b_{p-1}}{}_{c_1\ldots c_{n-p+1}}\vartheta^{c_1}\wedge\ldots\wedge \vartheta^{c_{n-p+1}} $$
And
$$ \vartheta_{a}\wedge\star\psi = \frac{1}{p!}\frac{1}{(n-p)!}\psi_{a_1\ldots a_p}\epsilon^{a_1\ldots a_p}{}_{b_1\ldots b_{n-p}}\vartheta_a\wedge \vartheta^{b_1}\wedge\ldots\wedge\vartheta^{b_{n-p}} $$
If we use the following lemma;
Let $\alpha,\beta\neq 0$ be $p$ forms, $\gamma\neq 0$ a $q$ form, with $p+q \leq n$, where $n$ is the dimension of the Manifold. Then provided
$$ \alpha\wedge\gamma = \beta\wedge\gamma \Longleftrightarrow (\alpha - \beta )\wedge \gamma =0 \Longleftrightarrow \alpha - \beta =0_p \Longleftrightarrow \alpha = \beta $$
We can consider the contractions
$$ \star (e_a \rfloor\psi)\wedge \vartheta^{d_1} \wedge\ldots\wedge \vartheta^{d_{p-1}} $$
And
$$\vartheta_a\wedge\star \psi \wedge \vartheta^{d_1} \wedge\ldots\wedge \vartheta^{d_{p-1}} $$
Upon computation, we find that, using generalised Kronecker delta identities that the result holds true as required. Hope this is useful to anybody!