We have seen that, in $\RR^3$, the wedge product of two 1-forms looks very much like the cross product of vectors, except that it is a 2-form. And the wedge product of a 1-form and a 2-form looks very much like the dot product of vectors, except that it is a 3-form. There is clearly some sort of correspondence of the form \begin{align} dx &\longleftrightarrow dy\wedge dz \\ dy &\longleftrightarrow dz\wedge dx \\ dz &\longleftrightarrow dx\wedge dy \\ 1 &\longleftrightarrow dx\wedge dy\wedge dz \end{align} This correspondence is the Hodge dual operation, which we now describe.

What do the above correspondences have in common? For each of the 8 basis forms which appear therein, call it $\alpha$, the corresponding form, call it ${*}\alpha$, satisfies \begin{equation} \alpha \wedge {*}\alpha = \omega \end{equation}

This is almost enough to define the Hodge dual operation, denoted by $*$. Each of the above basis elements $\alpha$ is a unit $p$-form. What happens if we rescale $\alpha$? In order for the Hodge dual to be a linear map, ${*}\alpha$ must be rescaled as well, so we must have \begin{equation} \alpha \wedge {*}\alpha = g(\alpha,\alpha)\,\omega \end{equation} and this is indeed a defining property of the Hodge dual. 1)

We are not quite done. Comparing $(\alpha+\beta)\wedge{*}(\alpha+\beta)$ with $\alpha\wedge{*}\alpha$ and $\beta\wedge{*}\beta$ yields \begin{equation} \alpha\wedge{*}\beta + \beta\wedge{*}\alpha = 2g(\alpha,\beta)\,\omega \end{equation} a technique known as polarization. We would like the left-hand side to be symmetric, 2) so we are finally led to define the Hodge dual operator $*$ by \begin{equation} \alpha \wedge {*}\beta = g(\alpha,\beta)\,\omega \label{hodgedef} \end{equation}

In using this definition, it is convenient to extend the notation so as to include 0-forms. We have already defined $g$ on 0-forms by \begin{equation} g(1,1) = 1 \end{equation} and wedge products involving 0-forms by \begin{equation} f \wedge \alpha = f\,\alpha \end{equation} for any form $\alpha$ and function $f$. It is then straightforward to show that \begin{equation} {*}1 = 1 \wedge {*}1 = g(1,1) \,\omega = \omega \end{equation} in any signature and dimension. Similarly, from \begin{equation} \omega \wedge {*}\omega = g(\omega,\omega) \,\omega = (-1)^s \omega \end{equation} we have \begin{equation} {*}\omega = (-1)^s \label{hodges} \end{equation}

The implicit formula ($\ref{hodgedef}$) above in fact suffices to compute the Hodge dual, and it is of course enough to do so on a basis. Since ($\ref{hodgedef}$) must hold for all forms, each orthonormal basis element of the form $\sigma^{i_1}\wedge … \wedge\sigma^{i_p}$ in $\bigwedge^p$ must be mapped to its “complement” in $\bigwedge^{n-p}$ of the form $\pm\sigma^{i_{p+1}}\wedge … \wedge\sigma^{i_n}$, where $(i_p)$ is a permutation of $(1…n)$; it only remains to determine the signs.