Duality of Hodge star.

Question

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 $\sqrt{|\det g|}$.

Answer 1

(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 $\star^2 = (-1)^{p(n-p)+q}$.