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According to Wikipedia, we have $\star{\star\alpha}=\pm\alpha$ (the sign depends on the dimension of the vector space, the signature of the bilinear form and the degree of $\alpha$). I guess we could prove that by considering an orthonormal basis $e_1,\ldots,e_n$ and prove the equality for all $\alpha$ of the form $e_{i_1}\wedge\cdots\wedge e_{i_k}$, but I was wondering if we can prove this more elegantly by using \begin{equation} \forall\alpha\in\Lambda^k:\bigg[y=\star\alpha\Leftrightarrow\big(\forall x\in\Lambda^k:x\wedge y=\langle x,\alpha\rangle \omega\big)\bigg] \end{equation} where $\omega\in\Lambda^n $ is the volume form. That is, we have to show that \begin{equation} \forall\alpha\in\Lambda^k:\forall x\in\Lambda^{n-k}:x\wedge\alpha=\pm\langle x,\star\alpha\rangle \omega \end{equation} but I don't know how to proceed.

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  • $\begingroup$ I just noticed that the question has already been asked here. I have decided not to delete my question though: 1. I don't remember how I found the other post, but it was a coincidence. That is to say that the other post is rather difficult to find for people with the same question because of the chosen title. 2. The question was asked more than 4 years ago and the OP seems to have lost interest, since the question was never edited to bump up the question (no bounty either). $\endgroup$
    – Filippo
    Commented Jul 11, 2022 at 14:45

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$ \newcommand\grade[1]{\langle#1\rangle} \newcommand\rev\widetilde $

If we use the Clifford algebra associated with the bilinear form, then we can write $$ A\wedge(\star B) = \grade{\rev AB}_0\omega = \rev A\wedge(B\omega) $$ where $\rev A$ is the reversal of $A$ and $\grade{\cdot}_0$ is the projection onto the scalar component. The second equality is a standard identity in geometric algebra but usually in the notation $X{\rfloor}Y\:\omega = X\wedge(Y\omega)$. When $A, B$ are both $k$-vectors, we can move the reversal onto $B$. Also multiplying on the left by $\omega$ gives $$ \omega\:A\wedge(\star B) = \omega\:A\wedge(\rev B\omega), $$$$ \grade{(\omega A)(\star B)}_0 = \grade{(\omega A)(\rev B\omega)}_0. $$ The bilinear form $\grade{(\cdot)(\cdot)}_0$ is an extension of the original bilinear form and is also non-degenerate. Hence $$ \star B = \rev B\omega, $$ and this extends to arbitrary multivectors by linearity. By definition, the Hodge star uses a unital $\omega$ so $\omega^2 = \pm1$. Finally, for any $k$-vector $B$: $$ \star{\star B} = \rev{(\rev B\omega)}\omega = \rev\omega B\omega = (-1)^{n(k-1)}\rev\omega\omega B = \omega^2\rev B. $$ If the bilinear form has the signature $(p,q)$ with $p$ positives and $q$ negatives and $n = p+q$, then the second-to-last expression tells us $$ \star{\star B} = (-1)^{(p+q)(k-1)+q}B = (-1)^{nk-p}B. $$

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