We previously computed \begin{align} d^2\rr &= d\left(\ee_j\,\sigma^j \right) \nonumber\\ &= d\ee_j\wedge\sigma^j + \ee_j\,d\sigma^j \nonumber\\ &= \ee_i \left(\omega^i{}_j\wedge\sigma^j + d\sigma^i \right) \end{align} The vector components of this equation are called the first structure equation, which takes the form \begin{equation} \Theta^i = \omega^i{}_j\wedge\sigma^j + d\sigma^i \end{equation} where the $\Theta^i$ are the torsion 2-forms. Since we are assuming that \begin{equation} d^2\rr = 0 \end{equation} the torsion vanishes and the structure equation reduces to \begin{equation} 0 = d\sigma^i + \omega^i{}_j\wedge\sigma^j \end{equation}

We can do a similar computation starting with $d\ee_j$ instead of $d\rr$. We have \begin{align} d^2 \ee_j &= d\left(\ee_k \,\omega^k{}_j \right) \nonumber\\ &= d\ee_k\wedge\omega^k{}_j + \ee_k\,d\omega^k{}_j \nonumber\\ &= \ee_i \left( \omega^i{}_k\wedge\omega^k{}_j + d\omega^i{}_j \right) \end{align} The components of this equation are the second structure equation, which takes the form \begin{equation} \Omega^i{}_j = d\omega^i{}_j + \omega^i{}_k\wedge\omega^k{}_j \end{equation} where the $\Omega^i{}_j$ are the curvature 2-forms. We can not, in general, assume that the curvature 2-forms vanish. However, the curvature 2-forms $\Omega^i{}_j$ are not all independent, as we now show.

Consider first an orthonormal basis, in which metric compatibility takes the form \begin{equation} \omega_{ij} + \omega_{ji} = 0 \end{equation} which clearly implies that \begin{equation} d\omega_{ij} + d\omega_{ji} = 0 \label{dcompat} \end{equation} What about the second term in the curvature 2-forms? We have \begin{equation} \omega_{ik}\wedge\omega^k{}_j = -\omega^k{}_j\wedge\omega_{ik} = \omega^k{}_j\wedge\omega_{ki} = \omega_{kj}\wedge\omega^k{}_i = - \omega_{jk}\wedge\omega^k{}_i \end{equation} Putting these computations together, we have shown that, in an orthonormal basis, we have \begin{equation} \Omega_{ij} = -\Omega_{ji} \label{Omegasym} \end{equation} since \begin{equation} \Omega_{ij} = d\omega_{ij} + \omega_{ik}\wedge\omega^k{}_j \label{ww} \end{equation} which is antisymmetric in $i$ and $j$.

The argument above used properties of an orthonormal basis, but the result is true in any basis. To see this, we repeat the above computation in a general basis. Metric compatibility now takes the form \begin{equation} \omega_{ij} + \omega_{ji} = dg_{ij} \end{equation} which still leads to (\ref{dcompat}), since $d^2g_{ij}=0$. What about the second term in (\ref{ww})? We must carefully lower the index, yielding 1) \begin{equation} \Omega_{ij} = g_{ik}\Omega^k{}_j = g_{ik}d\omega^k{}_j + g_{ik}\omega^k{}_l\wedge\omega^l{}_j = g_{ik}d\omega^k{}_j + \omega_{ik}\wedge\omega^k{}_j \end{equation} Using the product rule on the first term leads to \begin{align} \Omega_{ij} &= d(g_{ik}\omega^k{}_j) - dg_{ik}\wedge\omega^k{}_j + \omega_{ik}\wedge\omega^k{}_j \nonumber\\ &= d\omega_{ij} + (\omega_{ik}-dg_{ik})\wedge\omega^k{}_j = d\omega_{ij} - \omega_{ki}\wedge\omega^k{}_j \end{align} which agrees with (\ref{ww}). Each of these terms is clearly antisymmetric in $i$ and $j$, thus establishing (\ref{Omegasym}) in the general case.

Thus, of the $n^2$ possible curvature 2-forms (with lowered indices, that is, $\Omega_{ij}$) in $n$-dimensions, $n$ are identically zero, and the remaining $n^2-n$ curvature 2-forms are equal and opposite in pairs. Thus, there are only ${n\choose2}=\frac12n(n-1)$ independent curvature 2-forms.