Hodge Dual in Minkowski 2-space

Consider Minkowski 2-space, with line element \begin{equation} ds^2 = dx^2 - dt^2 \end{equation} and (ordered) orthonormal basis $\{dx,dt\}$. The orientation is \begin{equation} \omega = dx\wedge dt \end{equation} and it is straightforward to compute the Hodge dual on a basis. First of all, we have \begin{equation} dx \wedge {*}dx = g(dx,dx)\, dx \wedge dt = dx \wedge dt \end{equation} from which it follows that 1) \begin{equation} {*}dx = dt \end{equation} Similarly, from \begin{equation} dt \wedge {*}dt = g(dt,dt)\, dx \wedge dt = - dx\wedge dt \end{equation} we see that \begin{equation} {*}dt = dx \end{equation} The remaining cases were already worked out in general; for Minkowski 2-space we obtain \begin{align} {*}1 &= dx\wedge dt \\ {*}(dx\wedge dt) &= -1 \end{align}

1) One must also use the fact that $dt\wedge{*}dx=0$ to show that there is no $dx$ term in ${*}dx$. This argument generalizes to other, similar computations, and will not be repeated.

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