As discussed in the previous sections, matter in general relativity is described by a vector-valued 1-form \begin{equation} \Tvec = T^i\,\ee_i \end{equation} where \begin{equation} T^i = T^i{}_j\,\sigma^j \end{equation} are the energy-momentum 1-forms. The components $T^i{}_j$ are (also) components of the energy-momentum tensor. We assume that the energy-momentum tensor is symmetric, that is, that \begin{equation} T_{ji} = T_{ij} \end{equation} which can also be expressed as the requirement that \begin{equation} g(\Tvec,v)\cdot d\rr = \Tvec\cdot\vv \end{equation} for any vector $\vv$. We further assume that matter is conserved, by which we mean that the energy-momentum tensor is divergence free, that is, that \begin{equation} {*}d{*} \Tvec = \zero \end{equation} Introducing the energy-momentum 3-forms \begin{equation} \tau^i = {*}T^i \end{equation} brings the conservation condition to the form \begin{equation} d(\tau^i\,\ee_i) = 0 \end{equation} Finally, expanding $d\ee_i$ in terms of the connection 1-forms and equating components leads to \begin{equation} d\tau^i + \omega^i{}_j\wedge\tau^j = 0 \label{tau} \end{equation} as the condition for matter to be conserved.

We would like to relate matter to curvature, and therefore ask whether we can construct a “curvature 3-form” that satisfies ($\ref{tau}$). As discussed in Bianchi Identities the curvature 2-forms satisfy the (second) Bianchi identity, namely \begin{equation} d\Omega^i{}_j + \omega^i{}_k\wedge\Omega^k{}_j - \Omega^i{}_k\wedge\omega^k{}_j = 0 \end{equation} The simplest 3-form we can construct from the curvature 2-forms is \begin{equation} \alpha^i = \Omega^i{}_j \wedge \sigma^j \end{equation} and we can now compute \begin{align} d\alpha^i &= d\Omega^i{}_j \wedge \sigma^j + \Omega^i{}_j \wedge d\sigma^j \nonumber\\ &= (- \omega^i{}_k\wedge\Omega^k{}_j + \Omega^i{}_k\wedge\omega^k{}_j) \wedge \sigma^j - \Omega^i{}_j \wedge \omega^j{}_n \wedge \sigma^n \end{align} where we have used the first structure equation to differentiate $\sigma^j$. The last two terms cancel, and we are indeed left with ($\ref{tau}$), but for $\alpha^i$ rather than $\tau^i$. It appears that we have found our “curvature 3-form”! However, the (first) Bianchi identity says precisely that $\alpha^i=0$, so this is not in fact what we were looking for.

Nonetheless, we're almost there. What is needed is a slightly different combination of terms of the same form, that is, which look like $\Omega^i{}_j \wedge \sigma^m$. The correct choice turns out to be the Einstein 3-forms \begin{equation} \gamma^i = -\frac12 \Omega_{jk} \wedge {*}(\sigma^i\wedge\sigma^j\wedge\sigma^k) \label{gammadef} \end{equation} which indeed satisfy \begin{equation} d\gamma^i + \omega^i{}_j\wedge\gamma^j = 0 \label{dgamma} \end{equation} To verify (\ref{dgamma}) requires writing out all of the terms on the left-hand side of this expression, using the fact that the indices $i$, $j$, $k$ must be distinct in the definition of $\gamma^i$, (\ref{gammadef}), and showing that the terms cancel in pairs; see §Divergence of the Einstein Tensor for further details.

By analogy with energy-momentum, we can work backward and construct the Einstein 1-forms \begin{equation} G^i = {*}\gamma^i = G^i{}_j \,\sigma^j \end{equation} whose components $G^i{}_j$ define the Einstein tensor, which we can think of as a vector-valued 1-form \begin{equation} \GG = G^i \ee_i = G^i{}_j \,\sigma^j{}_i \end{equation}

The Einstein tensor is closely related to the Ricci tensor, and in fact \begin{equation} G^i{}_j = R^i{}_j - \frac12 \delta^i{}_j \,R \end{equation} where $R=R^m{}_m$ is the Ricci scalar; see §Components of the Einstein Tensor for further details. This equation is more commonly written in covariant form, that is, with all indices down: \begin{equation} G_{ij} = R_{ij} - \frac12 g_{ij} \,R \end{equation} The Einstein tensor is often referred to as the “trace-reversed” Ricci tensor, since \begin{equation} G^m{}_m = R^m{}_m - \frac12 4R = -R^m{}_m \end{equation}