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?
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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 $I=\{i_1,\ldots ,i_p\}\subset [1,n]$).
