We have argued that 1-forms are “like” vectors, and that in particular $dx$ is “like” $\xhat$. Since $\{\xhat,\yhat,\zhat\}$ is an orthonormal basis for vectors in $\RR^3$, we would like to regard $\{dx,dy,dz\}$ as an orthonormal basis for 1-forms.

We therefore introduce an inner product $g$ on $\bigwedge^1(\RR^3)$ by defining the rectangular basis $\{dx,dy,dz\}$ to be orthonormal under this product, that is \begin{align} g(dx,dx) = g(dy,dy) = g(dz,dz) &= 1\\ g(dx,dy) = g(dy,dz) = g(dz,dx) &= 0 \end{align} To be an inner product, $g$ must be linear, that is: \begin{equation} g(f\alpha+\beta,\gamma) = f g(\alpha,\gamma) + g(\beta,\gamma) \label{glin} \end{equation} symmetric, that is \begin{equation} g(\beta,\alpha) = g(\alpha,\beta) \end{equation} and nondegenerate, that is \begin{equation} g(\alpha,\beta) = 0  \forall\beta \Longrightarrow \alpha=0 \label{gnond} \end{equation} and it is easily checked that (\ref{glin})–(\ref{gnond}) follow from the corresponding properties of the dot product. However, we do not require that $g$ be positive definite, which would force $g(\alpha,\alpha)\ge0$, with $g(\alpha,\alpha)=0$ if and only if $\alpha=0$. As we will see, this property does not hold in either special or general relativity.