Covariant derivative of a symmetric tensor

Question

Assume that a symmetric $(0,2)$ satisfies
$$\nabla_iT_{jk}+\nabla_jT_{ik}+\nabla_kT_{ji}=0$$
where $T=T_i^i$ is constant and $\nabla_jT_{ik}\ne 0$.
What are the values of the constants $a,b,c$ such that $$a\nabla_iT_{jk}+b\nabla_jT_{ik}+c\nabla_kT_{ji}=0$$

Is there any difference if the tensor $T$ is the Ricci tensor.

I think $(a,b,c)$ where $a=b=c$ are the only solution set to it however I failed to prove it nor to find the solution.

Thanks in advance.

Answer 1

When $T_{ijk}:=\nabla_{i}T_{jk}$, then $$ T_{ijk} +T_{jki}+ T_{kij} =0 \ (A)$$

$$ aT_{ijk} +b T_{jki}+ cT_{kij} =0 \ (B)$$

When $a=b=0$, then $T_{ijk}=0$.

When $a=0,\ b,\ c\neq 0$, then $T_{ijk}=C\cdot T_{jki},\ C\neq 0$. Then $$T_{ijk}\{1+C+C^2\}=0$$ from $(A)$ so that $T_{ijk}=0$.

When $a,\ b,\ c\neq 0$, then $(A),\ (B)$ implies that $(1-b/a)T_{jki} + (1-c/a)T_{kij}=0$. Hence $a=b=c$ or $T_{ijk}=0$.

Answer 2

I am afraid not. Take $T_{jk}=g_{jk}$, where $g_{jk}$ denotes the metric tensor with which the connection $\nabla$ is compatible. In this case, $a$, $b$, and $c$ could be arbitrarily valued, and $g_{jk}$ also differs from the Ricci tensor.