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.
1 Answers
$ \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 $$ \star{\star B} = (-1)^{(p+q)(k-1)+q}B = (-1)^{nk-p}B. $$
- 6,543
- 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.
- 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).
– Filippo Jul 11 '22 at 14:45