Covariant derivative and tensor symmetries Suppose we have a tensor field $T^{ab}$ such that $T^{ab} = T^{ba}$ everywhere. Then from the definition of the Riemannian covariant derivative in terms of a map between tensors, why must we then have $(\nabla_c T)^{ab} = (\nabla_c T)^{ba}$? 
 A: Strangely the result follows rather quickly from one of the questions you linked.  Use
$$\nabla_{c} T^{ab} = \partial_{c}T^{ab}+ \Gamma^a{}_{cd} T^{db} + \Gamma^b {}_{dc}T^{ad}.$$
From this we can get $$\nabla_{c} T^{ba} = \partial_{c}T^{ba}+ \Gamma^b{}_{cd} T^{da} + \Gamma^a {}_{dc}T^{bd}.$$
Then subtract to get
$\begin{array}\nabla_{c} T^{ab}-\nabla_{c} T^{ba} &=& \partial_{c}T^{ab}+ \Gamma^a{}_{cd} T^{db} + \Gamma^b {}_{dc}T^{ad} -\partial_{c}T^{ba}- \Gamma^b{}_{cd} T^{da} - \Gamma^a {}_{dc}T^{bd} \\
 &=&\partial_{c}(T^{ab}-T^{ba})+ \Gamma^a{}_{cd} T^{db} - \Gamma^a {}_{dc}T^{bd} + \Gamma^b {}_{dc}T^{ad} - \Gamma^b{}_{cd} T^{da}\\
 &=&\partial_{c}(T^{ab}-T^{ba})+ (\Gamma^a{}_{cd}  - \Gamma^a {}_{dc})T^{bd} + (\Gamma^b {}_{dc}- \Gamma^b{}_{cd}) T^{da} \\
 &=&\partial_{c}(T^{ab}-T^{ba})+ 0T^{bd} + 0 T^{da} \\ 
&=&\partial_{c}(T^{ab}-T^{ba}) =0.  \end{array}$
Now the first line is from the linked posts, the second from linearity of differentiation, the third is from point wise symmetry of $T$ and the next is from the connection being torsion free (symmetry of Christoffel symbols) and the final is from neighborhood wise symmetry of T
