Since $\sigma^i$ is an (ordinary) differential form, we must have \begin{align} 0 = -d^2\sigma^i &= d\left( \omega^i{}_k\wedge\sigma^k \right) \nonumber\\ &= d\omega^i{}_k\wedge\sigma^k - \omega^i{}_k\wedge d\sigma^k \nonumber\\ &= d\omega^i{}_j\wedge\sigma^j + \omega^i{}_k\wedge\omega^k{}_j\wedge\sigma^j \nonumber\\ &= \left( d\omega^i{}_j + \omega^i{}_k\wedge\omega^k{}_j \right) \wedge\sigma^k \nonumber\\ &= \Omega^i{}_j \wedge \sigma^j \label{bianchiI} \end{align} which is the first Bianchi identity. A similar argument starting with the connection 1-forms yields \begin{align} 0 = d^2\omega^i{}_j &= d\left( \Omega^i{}_j - \omega^i{}_k\wedge\omega^k{}_j \right) \nonumber\\ &= d\Omega^i{}_j -d\omega^i{}_k\wedge\omega^k{}_j + \omega^i{}_k\wedge d\omega^k{}_j \nonumber\\ &= d\Omega^i{}_j -\left(\Omega^i{}_k-\omega^i{}_l\wedge\omega^l{}_k\right)\wedge\omega^k{}_j + \omega^i{}_k\wedge \left(\Omega^k{}_j-\omega^k{}_l\wedge\omega^l{}_j\right) \nonumber\\ &= d\Omega^i{}_j + \omega^i{}_k\wedge\Omega^k{}_j - \Omega^i{}_k\wedge\omega^k{}_j \label{bianchiII} \end{align} since the remaining terms cancel by suitable relabeling of indices; this is the second Bianchi identity, which is often simply called “the” Bianchi identity.

In terms of components, the first Bianchi identity can be written in the form \begin{equation} R^{i}{}_{jkl}\>\sigma^k\wedge\sigma^l\wedge\sigma^j = 0 \end{equation} which implies \begin{equation} R^{i}{}_{jkl} + R^{i}{}_{klj} + R^{i}{}_{ljk} = 0 \end{equation} (since the components $R^{i}{}_{jkl}$ are antisymmetric in their last two indices). The second Bianchi identity involves derivatives of these components, and can be interpreted as the vanishing of the “covariant curl” of the Riemann tensor.

The Bianchi identities imply some further symmetries on the components of the Riemann and Ricci tensors. Lowering an index for convenience, the antisymmetry of 2-forms leads to \begin{equation} R_{ijlk}=-R_{ijkl} \end{equation} by convention, and metric compatibility leads to \begin{equation} R_{jikl}=-R_{ijkl} \end{equation} as shown in §Curvature. We can now show that the components of the Ricci tensor are symmetric, since \begin{equation} R_{ij} = R^m{}_{imj} = -R^m{}_{mji} - R^m{}_{jim} = R^m{}_{jmi} = R_{ji} \end{equation} Finally, we compute \begin{align} R_{ijkl} &= -R_{ijlk} = R_{jilk} \nonumber\\ &= -R_{jlki} - R_{jkil} \nonumber\\ &= R_{ljki} + R_{kjil} \nonumber\\ &= -R_{lkij} - R_{lijk} - R_{kilj} - R_{klji} \nonumber\\ &= 2 R_{klij} + R_{iljk} + R_{iklj} \nonumber\\ &= 2 R_{klij} - R_{ijkl} \end{align} so that \begin{equation} R_{ijkl} = R_{klij} \end{equation}