The Riemann curvature possesses symmetries like anti-symmetry of the variables or Bianchi identities. The first Bianchi identity seems to imply that:

$$g(R(x,y)z,t)=\sum_i [ g(S_i (x),z)g(S'_i(y),t) +g(S_i(y),t)g(S'_i(x),z)-$$

$$-g(S_i(x),t)g(S'_i(y),z)-g(S_i(y),z)g(S'_i(x),t)]$$

there $g$ is the metric, $R$ is the Riemann curvature and $S_i,S'_i$ are symmetric endomorphisms of the tangent bundle $S_i^*=S_i, (S'_i)^*=S'_i$.

Then, the second Bianchi identity is implied if the $S_i$ are parallel for the Levi-Civita connection $\nabla S_i =\nabla S'_i=0$.

It follows that under conditions the Ricci curvature $Ric$ can be decomposed:

$$Ric= \sum_i [tr(S_i) S'_i +tr(S'_i)S_i- S_i S'_i-S'_i S_i]$$

with $\nabla S_i=\nabla S'_i=0$.

Can such a decomposition of the curvatures always happen when the manifold is Einstein?