Let's turn this problem on its head. We don't yet know how $d$ acts on our basis $\{\ee_i\}$, but let's give this action a name. We must have \begin{equation} d\ee_j = \omega^i{}_j\,\ee_i \end{equation} for some 1-forms $\omega^i{}_j$, which are called connection 1-forms. In fact, any choice of these 1-forms determines an exterior derivative operator satisfying the two conditions in the previous section. As we will see in the next section, further conditions must be imposed in order to determine a preferred connection.

We establish some notation. First of all, let \begin{equation} g_{ij} = \ee_i \cdot \ee_j \end{equation} In an orthonormal basis, the matrix $(g_{ij})$ will of course be diagonal, with diagonal entries $\pm1$. The $g_{ij}$ are the components of an object called the metric tensor, which is closely related to the line element, since \begin{equation} ds^2 = d\rr\cdot d\rr = \left( \sigma^i\,\ee_i \right) \cdot \left( \sigma^j\,\ee_j \right) = g_{ij}\,\sigma^i\,\sigma^j \end{equation}

We then set \begin{equation} \omega_{ij} = \ee_i \cdot d\ee_j \end{equation} so that we also have \begin{equation} \omega_{ij} = g_{ik} \,\omega^k{}_j \end{equation} In Euclidean signature, $(g_{ij})$ is the identity matrix, and there is no difference between “up” and “down” indices, but in other signatures there will be sign differences, so it is important to keep track.