Let $(V,< \cdot, \cdot >)$ be an inner product space over a field ${K}$. As usual, we can extend $< \cdot, \cdot >$ to a mapping on the exterior algebra of $V$ using the usual matrix determinant, and we then get a corresponding Hodge map $\ast$. Now is it true in general that $\ast$ is unitary, that is, does it hold that
$$
< \ast(v), \ast(w) > = <v, w >,
$$
and if so, why?



Is the Hodge Map Unitary?
Yes. A cheap way is to choose an orthonormal basis $(e\_1,\ldots ,e\_n)$ of $V$ (replacing $K$ by its algebraic closure). Then the $e\_I$ for $I\subset [1,n]$ and $\#I=p$ form an orthonormal basis of $\wedge^pV$, and $\ast$ maps $e\_I$ to $\pm e\_{I^c}$ (as usual, I put $e\_I=e\_{i\_1}\wedge\ldots \wedge e\_{i\_p}$ for $I=\{i\_1,\ldots ,i\_p\}\subset [1,n]$).


