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The Hodge star $*$ can be defined in various ways. One of them is by its action on an arbitrary wedge product of basis elements $\{e^i\}$ of $V^*$. This definition is the following: $$*(e^{i_1}\wedge...\wedge e^{i_k})=\frac{\sqrt{|\det g_{ij}|}}{(n-k)!}\sum_{\sigma\in S_n}\text{sgn}(\sigma)\tilde g^{i_1i_{\sigma(1)}}...\tilde g^{i_ki_{\sigma(k)}}e^{i_{\sigma(k+1)}}\wedge...\wedge e^{i_{\sigma(n)}},$$ where the $\sqrt{\det g}$ factor arises due to the fact that we have not demanded that the $V$ basis $\{e_i\}$ is $g$-orthonormal. Of course, for a $g$-orthonormal basis $\{e_i\}$ this factor gives $1$. Also, $\tilde g$ denotes the inverse metric used tot define the inner product between forms. The above definition includes some components of $\tilde g$ and, thus, we don't require the basis $\{e^j\}$ of $V^*$ to be either $\tilde g$-orthogonal or $\tilde g$-orthonormal. Property-wise, the metric and inverse metric are just symmetric $(0,2)$- and $(2,0)$-tensors.

These being said, the Hodge dual of a $k$-form $\alpha^{(k)}$ can be worked out to be $$*\alpha^{(k)}=\hspace{-0.4cm}\sum_{1\leq i_{1}<...<i_k\leq n}\hspace{-0.4cm}(\alpha^{(k)})_{i_{1}...i_k}*(e^{i_1}\wedge...\wedge e^{i_k}) =\hspace{-0.4cm}\sum_{1\leq i_1<...<i_k\leq n}\hspace{-0.4cm}(\alpha^{(k)})_{i_1...i_k}\frac{\sqrt{|\det g_{ij}|}}{(n-k)!}\sum_{\sigma\in S_n}\text{sgn}(\sigma)\tilde g^{i_1i_{\sigma(1)}}...\tilde g^{i_ki_{\sigma(k)}}e^{i_{\sigma(k+1)}}\wedge...\wedge e^{i_{\sigma(n)}},$$

while the Hodge dual of the Hodge dual of $\alpha^{(k)}$ is

$$**\alpha^{(k)}=\hspace{-0.4cm}\sum_{1\leq i_1<...<i_k\leq n}\hspace{-0.4cm}(\alpha^{(k)})_{i_1...i_k}\frac{\sqrt{|\det g_{ij}|}}{(n-k)!}\sum_{\sigma\in S_n}\text{sgn}(\sigma)\tilde g^{i_1i_{\sigma(1)}}...\tilde g^{i_ki_{\sigma(k)}}*(e^{i_{\sigma(k+1)}}\wedge...\wedge e^{i_{\sigma(n)}}) =\hspace{-0.4cm}\sum_{1\leq i_1<...<i_k\leq n}\hspace{-0.4cm}(\alpha^{(k)})_{i_1...i_k}\hspace{0.1cm}\frac{|\det g_{ij}|}{k!(n-k)!}\sum_{\sigma\in S_n}\text{sgn}(\sigma)\tilde g^{i_1i_{\sigma(1)}}...\tilde g^{i_ki_{\sigma(k)}} \hspace{0.8cm}\sum_{\lambda\in S_n}\text{sgn}(\lambda)\tilde g^{i_{\sigma(k+1)}i_{\lambda(\sigma(1)))}}...\tilde g^{i_{\sigma(n)}i_{\lambda(\sigma(n-k)))}}e^{i_{\lambda(\sigma(n-k+1)))}}\wedge...\wedge e^{i_{\lambda(\sigma(n)))}}.$$ I know that there are more elegant formulas for the above containing the Levi-Civita tensor, but I just prefer to write everything down with respect to some sums over permutations $\sigma$ containing some $\text{sgn}(\sigma)$. It is essentially and practicaly the very same.

Now, here are my questions:

  1. Is there any mistake in the above formulae?
  2. Is there a way to simplify the above formulae (apart from using the Levi-Civita tensor and imposing any summation convention)? I think that I might be able to simplify them using the determinant formula $$\det A=\sum_{\sigma\in S_n}\text{sgn}(\sigma)\prod_{i=1}^na_{i,\sigma(i)}.$$ I was trying to use the above formula to include the determinant of $\tilde g$, but it is not that trivial (for me). It is also expected that a determinant of $g$ or $\tilde g$ would come into play since there is the well-known identity $$**\alpha^{(k)}=(-1)^{k(n-k)+s}\alpha^{(k)},$$ where $s$ is the metric signature. This signature would naturally appear through a metric determinant.
  3. Does the identity $$**\alpha^{(k)}=(-1)^{k(n-k)+s}\alpha^{(k)}$$ hold irregardless of the choice of a $\tilde g$-orthonormal basis? I mean, if I take the basis to be orthonormal everything simplifies tremendously and the above property seems to hold (as expected). But in the arbitrary basis it seems non trivial to me how one can prove it.

I hope I am not asking really trivial things.

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