We can now finally write down precise equivalents of the dot and cross products using differential forms. First of all, the definition of the Hodge dual tells us that the Hodge dual, $*$, and the metric, $g$, in fact contain the same information. Interpreting $g$ on 1-forms as the dot product, we have \begin{equation} \alpha \cdot \beta = (-1)^s {*}(\alpha\wedge{*}\beta) \end{equation} where we have used ${*}\omega=(-1)^s$. The dot product is thus defined in any dimension and signature, and reduces to the expected dot product in the positive-definite case. In 3 dimensions, if we identify $\xhat$ with $dx$, etc., we recover the dot product in its usual form.

As for the cross product, recall that $\alpha\wedge\beta$ over $\bigwedge(\RR^3)$ had the form of the cross product, but was a 2-form. We can now use the Hodge dual to turn this into a 1-form, obtaining \begin{equation} \alpha\times\beta = {*}(\alpha\wedge\beta) \end{equation} Note however that this only works in 3 dimensions; otherwise the result is not a 1-form. 1)