Sign of the permutation when I show that $\star{\star w}= (-1)^{n(n-k)} w$ for the Hodge operator

Question

Let $w=\sum_{I} a_{I}dx_{I}$ be a $k$-form in $\Bbb R ^n$. Let us consider the Hodge operator in a combinatorial form, i.e. as an $(n-k)$ form such that

$$\star(dx_{i_{1}} \wedge \dotsb \wedge dx_{i_{k}})=(-1)^{\sigma} (dx_{j_{1}} \wedge \dotsb \wedge dx_{j_{n-k}})$$

from where $(i_{1}, \dots, i_{k}, j_{1}, \dots, j_{n-k})$ it is a family of the permutation group, i.e. a permutation of the vector of $n$ entries and $\sigma=0$ if the permutation is even and $\sigma=1$ if the permutation is odd. The idea is to prove the property only with that definition.

Note that: $$ \begin{align} \star{\star w}&=\star{\star \left(\sum_{I} a_{I}dx_{I} \right)} \\ &= \star\left(\sum_{I} a_{I} \ast(dx_{i_{1}} \wedge \dotsb \wedge dx_{i_{k}}) \right) \\ &=\star \left(\sum_{I} a_{I} \ast(dx_{i_{1}} \wedge \dotsb \wedge dx_{i_{k}}) \right) \\ &=\star \left(\sum_{I} a_{I} (-1)^{\sigma_{I}} (dx_{j_{1}} \dotsb \wedge dx_{j_{n-k}}) \right) \\ &=\star \left(\sum_{I} a_{I} (-1)^{\sigma_{I}} (dx_{j_{1}} \dotsb \wedge dx_{j_{n-k}} \wedge \dotsb \wedge dx_{i_{k}}) \right) \\ &=\star \left(\sum_{I} a_{I} (-1)^{\sigma_{I}} (-1)^{k(n-k)} (dx_{i_{1}} \dotsb \wedge dx_{i_{k}} \wedge \dotsb \wedge dx_{j_{n-k}}) \right) \quad \text{(wedge antisymmetry)}\\ &=(-1)^{k(n-k)} \sum_{I} a_{I} (-1)^{\sigma_{I}} (-1)^{\sigma_{I}} (dx_{j_{1}} \dotsb \wedge dx_{j_{n-k}} \wedge \dotsb \wedge dx_{i_{k}}). \end{align} $$ From here I think that the result can be deduced, but there is something that does not convince in this proof, note that the permutation is anchored to the counter of $I$, with which, as I can guarantee that when applying once the star comes out $(-1)^{\sigma_{I}}$ it seems to be too convenient, but I can't find the reasoning behind it, any suggestions?

Note: What I want to answer is why when I apply twice the Hodge star, the product $(-1)^{\sigma_{I}} (-1)^{\sigma_{I}}=1$ comes out, because it could happen that when I apply once the star I get the permutation fixed to $I$, i.e. $(\sigma_{I})$ but what guarantees me that when I apply the other star I don't have something like $(-1)^{\sigma_{I}} (-1)^{\sigma_{J}}$?

Answer 1

$\newcommand\dvol{d{\operatorname{vol}}}$If $I \subset \{1, \dotsc, n\}$, write $I^c$ for the complement. Then ${\star dx_I} = \pm dx_{I^c}$, where the sign is such that $dx_I \wedge {\star dx_I} = \dvol_n$. This is the same as the sign of the permutation $\sigma_I$ which first enumerates $I$ then $I^c$.

In particular, $\star{\star dx_I} = \pm dx_I$, where the sign is such that ${\star dx_I} \wedge \star{\star dx_I} = \dvol_n$. By graded commutativity of the wedge product, the LHS is $(-1)^{k(n-k)} \star{\star dx_I} \wedge {\star dx_I}$. But note that $\star{\star dx_I} = \pm dx_I$ and that by definition $dx_I \wedge {\star dx_I} = \dvol_n$. Comparing signs, we see $$ \star{\star dx_I} = (-1)^{k(n-k)} dx_I. $$ This argument can be recast in terms of the sign of the permutations, as you desire. If $I = \{i_1, \dotsc, i_k\}$ and $I^c = \{j_1, \dotsc, j_{n-k}\}$, then $$\sigma_I(m) = \begin{cases} i_m & m \le k \\ j_{m-k} & m > k.\end{cases}$$ If we write $$\tau_{k,n}(m) = \begin{cases} n-k+m & m \le k \\ m - k & m > k \end{cases}$$ then it is a simple symbol-pushing exercise to verify $\sigma_I = \sigma_{I^c} \tau_{k,n}$. Because $\tau_{k,n}$ is a product of $k(n-k)$ transpositions, the desired result follows.