Hodge star operator

Remarks

Example: in $\R^3$ with the standard orientation, $\star dx = dy \wedge dz$, $\star dy = dz \wedge dx$ and $\star dz = dx \wedge dy$. Check for $dx$: $dx \wedge \star dx = dx \wedge dy \wedge dz = \ip{dx}{dx} \operatorname{vol}$, and $dy \wedge \star dx = 0 = \ip{dy}{dx} \operatorname{vol}$, and the same for $dz$.

Non-example: in $\R^2$ the map with $dx \mapsto -dy$ and $dy \mapsto dx$ is not the Hodge star, since $dx \wedge (-dy) = -\operatorname{vol}$, while $\ip{dx}{dx} \operatorname{vol} = \operatorname{vol}$. The Hodge star of $\R^2$ is $\star dx = dy$, $\star dy = -dx$, a rotation by a right angle.

On orthonormal bases, $\star(e_{i_1} \wedge \dots \wedge e_{i_k}) = \pm e_{j_1} \wedge \dots \wedge e_{j_{n-k}}$, where the $j$'s are the complementary indices and the sign is that of the permutation $(i_1, \dots, i_k, j_1, \dots, j_{n-k})$. It satisfies $\star 1 = \operatorname{vol}$ and $\star\star = (-1)^{k(n-k)}$ on $\Lambda^k V$, and it is an isometry. Reversing the orientation changes the sign of $\star$.