Just as the components of the connection 1-forms are the Christoffel symbols, related by \begin{equation} \omega^i{}_j = \Gamma^i{}_{jk}\,\sigma^k \end{equation} the components of curvature also have names. 1) We write \begin{equation} \Omega^i_{}j = \frac12 R^i{}_{jkl}\,\sigma^k\wedge\sigma^l \label{Riem} \end{equation} where the $R^i{}_{jkl}$ are the components of the Riemann curvature tensor. However, Equation (\ref{Riem}) alone only determines the combinations $R^i{}_{jkl}-R^i{}_{jlk}$, so we assume by convention that \begin{equation} R^i{}_{jlk}=-R^i{}_{jkl} \end{equation} The Ricci curvature tensor is defined as the trace of the Riemann tensor, defined by \begin{equation} \Omega^k{}_i(\ee_k) = R^m{}_{imj}\,\sigma^j = R_{ij}\,\sigma^j \end{equation} where the last expression defines the Ricci components $R_{ij}$. Equivalently, we have \begin{equation} g(\Omega^k{}_i,\sigma_k\wedge\sigma_j) = R_{ij} \end{equation}

Finally, the Ricci curvature scalar is the trace of the Ricci tensor, defined by \begin{equation} R = R_{ij}\,g^{ij} = g(\sigma^i,R_{ij}\,\sigma^j) \end{equation}

One way to visualize a 4-index object such as $R^m{}_{inj}$ is to note that if $i$ and $j$ are fixed, then this expression corresponds to a $4\times4$ matrix of coefficients. Thus, for each pair $(i,j)$, we have such a $4\times4$ matrix. But $i$, $j$ themselves denote the components of a $4\times4$ matrix, so $R^m{}_{inj}$ can be thought of as a $4\times4$ matrix (with components labeled by $(i,j)$) of $4\times4$ matrices (with components labeled by $(m,n)$).

If we take the trace of each of the “inner” $4\times4$ matrices, we replace each one with a number, namely its trace. The result? An ordinary $4\times4$ matrix, with components $R_{ij}$ — and trace $R$.