When comparing the cross product in $\RR^3$ with the wedge product on $\bigwedge^1(\RR^3)$, it was natural to associate $\zhat$ with both $dz\in\bigwedge^1$ and $dx\wedge dy\in\bigwedge^2$. Since $|\zhat|=1$, we expect not only \begin{equation} g(dz,dz) = 1 \end{equation} which we already know, but also something like \begin{equation} g(dx\wedge dy,dx\wedge dy) = 1 \label{g2} \end{equation} which we have yet to define. Consider for simplicity $\bigwedge^2(\RR^2)$, which only has one independent element. Suppose we define a linear operator, also called $g$, on $\bigwedge^2$ by ($\ref{g2}$). What properties does it have? Well, \begin{align} g(\alpha\wedge\beta,dx\wedge dy) &= g\bigl((\alpha_x\,dx+\alpha_y\,dy)\wedge(\beta_x\,dx+\beta_y\,dy), dx\wedge dy)\bigr)\nonumber\\ &= g\bigl((\alpha_x\beta_y-\alpha_y\beta_x)\,dx\wedge dy,dx\wedge dy\bigr) \nonumber\\ &= (\alpha_x\beta_y-\alpha_y\beta_x) g(dx\wedge dy,dx\wedge dy)\nonumber\\ &= \alpha_x\beta_y-\alpha_y\beta_x \nonumber\\ &= \left| \begin{matrix} g(\alpha,dx)& g(\alpha,dy)\cr g(\beta,dx)& g(\beta,dy)\cr \end{matrix} \right| \end{align} It seems reasonable to conjecture that \begin{equation} g(\alpha\wedge\beta,\gamma\wedge\delta) = \left| \begin{matrix} g(\alpha,\gamma)& g(\alpha,\delta)\cr g(\beta,\gamma)& g(\beta,\delta)\cr \end{matrix} \right| \label{g2def} \end{equation} So let's start again, and take ($\ref{g2def}$) as the definition of the metric $g$ acting on 2-forms. A similar $p\times p$ determinant can be used to define $g$ on $\bigwedge^p$. It is straightforward to check that in an orthonormal basis $\{\sigma^i\}$, not only is \begin{equation} g(\sigma^i,\sigma^j) = \pm \delta^{ij} \end{equation} but also \begin{equation} g(\sigma^I,\sigma^J) = \pm \delta^{IJ} \end{equation} so that our canonical basis $\{\sigma^I\}$ of $\bigwedge^p$ is orthonormal as well — precisely what we wanted.

What about the special case $\bigwedge^0$, the space of scalars? We chose the constant function $1$ as the basis of this one-dimensional space, and would like this basis to be normalized. Thus, we set \begin{equation} g(1,1) = 1 \label{gone} \end{equation} so that $g(f_1,f_2)=f_1f_2$ for any functions $f_1$ and $f_2$.