Hodge star - Duality: Why is $\star{\star\alpha}=\pm\alpha$? 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.
 A: $
\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
$$
  \star{\star B} = (-1)^{(p+q)(k-1)+q}B = (-1)^{nk-p}B.
$$
