Duality of Hodge star. Can someone please provide a rigorous proof that:
$$\star(\star A) = (-1)^{p(n-p)}sA$$
where $\star : \Lambda^{n}\to \Lambda^{n-p
}$
and $s$ is the the sign of the determinant of the metric.
I am using the following definition of Hodge operator:
$$\star A_{\mu_1,\ldots , \mu_{n-p}} = \frac{1}{p!} \epsilon^{\nu_1\ldots \nu_{p}}_{\mu_1,\ldots,\mu_{n-p}}A_{\nu_1,\ldots , \nu_p}$$
So my brute force approach is the following:
$$(\star(\star A))_{\mu_1,\ldots , \mu_p} = \ldots = \frac{1}{(n-p)!p!}\epsilon^{\nu_1,\ldots , \nu_{n-p}}_{\mu_1,\ldots , \mu_p}\epsilon^{\alpha_1\ldots \alpha_p}_{\nu_1\ldots \nu_{n-p}}A_{\alpha_1\ldots \alpha_p}$$
How to proceed? I need to use here the definition of the determinant, but I am not sure how?
Thanks in advance.
Edit:
$\epsilon$ is the Levi-Civita completely symmetric tensor. (i.e the Levi-Civita SYmbol multiplied by $\sqrt{|\det g|}$.
 A: (I am actually working with this right now for a paper, here's an excerpt) We define the hodge dual on an $n$ dimensional manifold diffeomorphic to $\mathbb{R}^n$ $\star : \Lambda^k(M) \longrightarrow \Lambda^{n-k}(M)$ in the following manner. Let $(-1)^q =s$ represent the signature of the metric $f$ will be short for $dx$'s, the basis 1-forms
Let $\alpha$ be a $p$-form:
\begin{equation}
\alpha = \frac{1}{p!}\alpha_{a_1\ldots a_p}f^{a_1}\wedge\ldots \wedge f^{a_p}
\end{equation}
We have that the hodge dual acts on the basis $p$-form as
\begin{equation}
\star (f^{a_1}\wedge\ldots\wedge f^{a_p})=\frac{1}{(n-p)!}{\epsilon}_{b_1\ldots b_{n-p}}{}^{a_1\ldots a_p}f^{b_1}\wedge\ldots\wedge f^{b_{n-p}}
\end{equation}
Hence, we have that the $n-p$ form
\begin{equation}
\star\alpha = \frac{1}{p!(n-p)!}{\epsilon}_{b_1\ldots b_{n-p}}{}^{a_1\ldots a_p}\alpha_{a_1\ldots a_p}f^{b_1}\wedge\ldots\wedge f^{b_{n-p}}
\end{equation}
Such that the components of the dual form is
\begin{equation}
(\star\alpha)_{b_1\ldots b_{n-p}} =\frac{1}{p!}{\epsilon}_{b_1\ldots b_{n-p}}{}^{a_1\ldots a_p} \alpha_{a_1\ldots a_p}
\end{equation}
We find then by action twice
\begin{equation}
(\star\star\alpha)_{a_1\ldots a_p}=\frac{1}{p!(n-p)!}{\epsilon}_{a_1\ldots a_p}{}^{b_1\ldots b_{n-p}}{\epsilon}_{b_1\ldots b_{n-p}}{}^{c_1\ldots c_p}\alpha_{c_1\ldots c_p}
\end{equation}
By passing through each index on the second LC tensor, we pick up a $(-1)^{n-p}$ factor giving
\begin{equation}
\begin{aligned}
(\star\star\alpha)_{a_1\ldots a_p}&=\frac{1}{p!(n-p)!}{\epsilon}_{a_1\ldots a_p}{}^{b_1\ldots b_{n-p}}{\epsilon}^{c_1\ldots c_p}{}_{b_1\ldots b_{n-p}} \alpha_{c_1\ldots c_p} \\
&=\frac{(-1)^{p(n-p)}}{p!(n-p)!} \, (n-p)!p! \,(-1)^q \delta^{c_1\ldots c_p}_{a_1\ldots a_p}\alpha_{c_1\ldots c_p}\\
&=(-1)^{p(n-p)+q}\,\delta^{c_1\ldots c_p}_{a_1\ldots a_p}\alpha_{c_1\ldots c_p}\\
&= (-1)^{p(n-p)+q}\,\alpha_{a_1\ldots a_p}
\end{aligned}
\end{equation}
So that $\star^2 = (-1)^{p(n-p)+q}$.
