Here the complete problem statement (slightly modified):
The Ricci tensor $R_{ij} \mathrm d x^i \otimes \mathrm d x^j$ of a connexion $D$ on a manifold is the tensor of type (0, 2) obtained by contracting the curvature tensor $R$ as follows: $R_{ij} = R^h_{ihj}$. If $D$ is symmetric use the (algebraic) bianchi identity to show that $R_{ij} - R_{ji} = R^h_{hij}$, so there is only one independent contraction of $R$.
The first part of the question is a straightforward computation, but I don't understand what is meant by independent contraction.
As interchanging the $i$ and $j$ in the contraction does only lead to a transposition of the ricci tensor, the only remaining interesting choices are the three possible positions of the lower $h$, one of which is already the definition of the ricci tensor.
Lets look at the remaining two:
Let $\tilde{R}_{ij} = R^h_{ijh}$ be the components of an alternate ricci tensor. This results in a sign change: $\tilde{R}_{ij} = R^h_{ijh} = -R^h_{ihj} = -R_{ij}$. So information-wise nothing is gained or lost. This contraction is dependent.
Let $\tilde{R}_{ij} = R^h_{hij}$, then I would end up with a skew-symmetric ricci tensor, and $R_{ij} - R_{ji} = 2 R^h_{hij}$ would hold. This seems to be a rather different result. Isn't this independent of the original contraction?
Thanks in advance for any clarification on this.