On the flat manifold, one can construct a divergence-free symmetric $(2,0)$ tensor in this way.
- Step 1: Define an arbitrary tensor $K^{abcd}$ with Riemann tensor like symmetry $$K^{(ab)cd}=K^{ab(cd)}=0\, , K^{abcd}=K^{cdab}.$$
- Step 2: Define a $(2,0)$ tensor $H^{ab}$ such that $$H^{ab}=\nabla_{c}\nabla_{d}K^{acbd}\ .$$ One easily check $H^{ab}=H^{ba}$ by $$H^{ab}=\nabla_{c}\nabla_{d}K^{acbd}=\nabla_{c}\nabla_{d}K^{bdac}=H^{ba}.$$ And it is divergence-free \begin{align} \nabla_a\nabla_{c}\nabla_{d}K^{acbd}&=-\nabla_a\nabla_{c}\nabla_{d}K^{acbd}+[\nabla_a,\nabla_c]\nabla_{d}K^{acbd}\\ &=-\nabla_a\nabla_{c}\nabla_{d}K^{acbd}\\ \nabla_aH^{ab}&=\nabla_a\nabla_{c}\nabla_{d}K^{acbd}\\ &=0\,. \end{align} from the first line to the second line, we have used the flat condition $[\nabla_a,\nabla_c]=0$.
My question is, is there a similar method to construct a divergence free symmetric $(2,0)$ tensor on a curved manifold with a zero Ricci but allowing a non-zero Riemann tensor? And this construction should satisfy that
dont include the Riemann tensor and any tensor in its family (Ricci, Weyl...)
linear on an arbitrary tensor (any type) and its derivatives. Thus the freedom of adapting this tensor is only to set its anti/symmetry.
The above construction on a flat manifold absolutely satisfies those conditions. But, with the appearance of a curved connection, both symmetric and divergence-free conditions are broken.