## The Einstein 3-Forms

Recall that the Einstein 3-forms are given by $$\gamma^i = -\frac12 \sum_{j,k} \Omega_{jk} \wedge {*}(\sigma^i\wedge\sigma^j\wedge\sigma^k)$$ and are related to the components $G_{ij}$ of the Einstein tensor through $$\vec{G} = \sum_{i,j} G^i{}_j\,\sigma^j\,\hat{e}_i = \sum_i {*}\gamma^i\,\hat{e}_i .$$ Expanding the curvature 2-forms $\Omega_{jk}$ in terms of their components, we obtain $$\gamma^i = -\frac14 \sum_{j,k,p,q} R_{jkpq} \,\sigma^p\wedge\sigma^q \wedge {*}(\sigma^i\wedge\sigma^j\wedge\sigma^k) .$$ Thus, $$\sigma^m\wedge\gamma^i = -\frac14 \sum_{j,k,p,q} R_{jkpq} \, g(\sigma^m\wedge\sigma^p\wedge\sigma^q, \sigma^i\wedge\sigma^j\wedge\sigma^k) \, \omega .$$ The right-hand-side vanishes unless $\{m,p,q\}=\{i,j,k\}$, although the order of these distinct elements can be different. Assuming for the moment that $m=i$ (but no sum on $i$), then $\{p,q\}$ must be the same as $\{j,k\}$, and $$\sigma^i\wedge\gamma^i = -\frac12 \sum_{j,k\ne i} R_{jkjk} \, g(\sigma^i,\sigma^i) g(\sigma^j,\sigma^j) g(\sigma^k,\sigma^k) \, \omega ,$$ where the factor of $2$ comes from fixing the order of $p$, $q$, (and with no sum on $i$). If instead $m\ne i$, we obtain nonzero terms only when $j=m$ or $k=m$. Thus, $$\sigma^m\wedge\gamma^i = -\frac12 \sum_k R_{mkki} \, g(\sigma^i,\sigma^i) g(\sigma^m,\sigma^m) g(\sigma^k,\sigma^k) \, \omega -\frac12 \sum_j R_{jmij} \, g(\sigma^i,\sigma^i) g(\sigma^j,\sigma^j) g(\sigma^m,\sigma^m) \, \omega .$$ Using the symmetry of the components of the Riemann tensor, and the fact that $g$ can be used to raise an index, and noting that the last two terms are equal, we can rewrite these expressions as \begin{align} \sigma^i\wedge\gamma^i &= -\frac12 \sum_{j,k\ne i} R^{jk}{}_{jk} \, g(\sigma^i,\sigma^i) \, \omega ,\\ \sigma^m\wedge\gamma^i &= - \sum_k R^{ki}{}_{mk} \, g(\sigma^m,\sigma^m) \, \omega . \end{align} Comparing these expressions with the definition of the inner product in terms of the Hodge dual operator $*$, we see that $$\gamma^i = - \sum_{j,k\ne i} R^{jk}{}_{jk} \,{*}\sigma^i + \sum_{k,m\ne i} R^{ik}{}_{mk} \,{*}\sigma^m .$$ Thus, $$G^i{}_m = \sum_k R^{ik}{}_{mk} = R^i{}_m$$ if $i\ne m$, and $$G^i{}_i = -\sum_{j,k\ne i} R^{jk}{}_{jk} .$$ It is straightforward to write out the last expression, thus verifying that $$G^i{}_j = R^i{}_j - \frac12 \delta^i{}_j \,R .$$