Let $\star$ be the Hodge star operator. Prove that
$$\star \star \omega \enspace = \enspace (-1)^{p(n-p)} \cdot \omega$$
for some $\omega \in \Lambda^p$.
What I have so far is the following: Without loss of generality, I assume $\omega = dx^{i_1} \wedge \ldots \wedge dx^{i_p}$. Then I have
\begin{align}
\star \omega \enspace &= \enspace \frac{\sqrt{\det g}}{(n-p)!} \cdot \varepsilon^{i_1 \ldots i_p}{}_{j_{p+1} \ldots j_n} \cdot dx^{j_{p+1}} \wedge \ldots \wedge dx^{j_n} \\
\end{align}
Renaming the indices yields
\begin{align}
\star \omega \enspace &= \enspace \frac{\sqrt{\det g}}{(n-p)!} \cdot \varepsilon^{i_1 \ldots i_p}{}_{k_1 \ldots k_{n-p}} \cdot dx^{k_1} \wedge \ldots \wedge dx^{k_{n-p}} \\
\end{align}
It follows that
\begin{align}
\star \star \omega \enspace &= \enspace \frac{\det g}{(n-p)! \; p!} \cdot \varepsilon^{i_1 \ldots i_p}{}_{k_1 \ldots k_{n-p}} \cdot \varepsilon^{k_1 \ldots k_{n-p}}{}_{\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\
&= \enspace \frac{\det g}{(n-p)! \; p!} \cdot \Big( (-1)^{p(n-p)} \cdot \varepsilon_{k_1 \ldots k_{n-p}}{}^{i_1 \ldots i_p} \Big) \cdot \varepsilon^{k_1 \ldots k_{n-p}}{}_{\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\
&= \enspace \frac{\det g}{(n-p)! \; p!} \cdot (-1)^{p(n-p)} \cdot (n-p)! \cdot \delta^{i_1 \ldots i_p}_{\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\
&= \enspace \frac{\det g}{p!} \cdot (-1)^{p(n-p)} \cdot p! \cdot dx^{i_1} \wedge \ldots \wedge dx^{i_p}\\
&= \enspace \det g \cdot (-1)^{p(n-p)} \cdot \omega
\end{align}
which is nearly what I want, but how do I get rid of that $\det g$? Any suggestions — or errors that I missed?
${}$
P.S.: I know that this question has already been asked on this platform, but I cannot follow all of the steps in these cases, and I would really appreciate knowing what went wrong in my approach.
Best Answer
I will assume you are in the Riemannian setting, where the metric is positive definite and so on.
The notation $\varepsilon^{i_1\dots i_p}{}_{j_{p+1}\dots j_n}$ is kind of weird, because the permutation symbol $\varepsilon_{i_1\dots i_n}$ is "not a tensor" in the sense that you cannot manipulate its indices at will. Instead, they satisfy the identities
$$ g_{i_1j_1}\cdots g_{i_nj_n}\varepsilon^{j_1\dots j_n} = (\det g)\varepsilon_{i_1\dots i_n}, $$ $$ g^{i_1j_1}\cdots g^{i_nj_n}\varepsilon_{j_1\dots j_n} = \frac{1}{\det g}\varepsilon^{i_1\dots i_n}. $$ Even when they satisfy this "weird rule", their product is a tensor $$ \varepsilon_{i_1\dots i_n}\varepsilon^{j_1\dots j_n} = \delta^{j_1\dots j_n}_{i_1\dots i_n}. $$
Since you cannot raise and lower the indices of the $\epsilon$s, I suggest you always write them with either all indices up or all indices down. Then your Hodge dual looks like $$ (\star\omega)_{a_{p+1}\dots a_n} = \frac{\sqrt{\det g}}{(n-p)!} \omega_{c_1\dots c_p}g^{a_1c_1}\cdots g^{a_pc_p} \varepsilon_{a_1\dots a_p a_{p+1}\dots a_n} $$ instead. Then, using the second identity I wrote, you have
\begin{align} \star \star \omega \enspace &= \enspace \frac{\det g}{(n-p)! \; p!} g^{i_1 r_1}\cdots g^{i_pr_p}\cdot \varepsilon_{r_1\dots r_pk_1 \ldots k_{n-p}} \cdot g^{s_1k_1}\cdots g^{s_{n-p}k_{n-p}}\cdot \varepsilon_{s_1 \ldots s_{n-p}\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\ &= \enspace \frac{1}{(n-p)! \; p!} \varepsilon^{i_1\dots i_ps_1 \ldots s_{n-p}} \cdot \varepsilon_{s_1 \ldots s_{n-p}\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\ &= \enspace \frac{1}{(n-p)! \; p!} \delta^{i_1\dots i_ps_1 \ldots s_{n-p}}_{s_1 \ldots s_{n-p}\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\ &= \enspace \frac{1}{(n-p)! \; p!} (-1)^{p(n-p)}\delta^{s_1 \ldots s_{n-p}i_1\dots i_p}_{s_1 \ldots s_{n-p}\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\ &= \enspace \frac{1}{p!}(-1)^{p(n-p)}\delta^{i_1\dots i_p}_{\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\ &= \enspace (-1)^{p(n-p)} \cdot \omega \end{align}
Alternatively, you can do the whole derivation using the tensors $\def\sgn{\mathrm{sgn}\,}$ $$E_{i_1\dots i_n} = \sqrt{\det g}\,\varepsilon_{i_1\dots i_n}$$ $$E^{i_1\dots i_n} = \frac{1}{\sqrt{\det g}}\,\varepsilon^{i_1\dots i_n}$$ instead of the permutation symbols. Then the Hodge star looks like $$ (\star\omega)_{a_{p+1}\dots a_n} = \frac{1}{(n-p)!} \omega_{c_1\dots c_p}g^{a_1c_1}\cdots g^{a_pc_p} E_{a_1\dots a_p a_{p+1}\dots a_n} $$ and you can freely raise and lower the indices of the $E$s using the metric.
Note: in the semi-Riemannian case, all the $\sqrt{\det g}$ get replaced with $\sqrt{|\det g|}$ and one of the Levi-Civita tensors (conventionally the one with upper indices) has to incorporate a factor of $\sgn g$ in order to obtain one from the other raising an lowering indices. In this case the formula you want is $$\star\star=(-1)^{k(n-k)}(\sgn g)$$ and the Levi-Civita tensors satisfy $$E_{j_1\dots j_n}E^{i_1\dots i_n} = (\sgn g)\delta_{j_1\dots j_n}^{i_1\dots i_n}$$