Let $S$ be a manifold on which there is given a Riemannian metric $g=<,>$ and two affine connections $\nabla$ and $\nabla^*$. If for all vector field $X,Y,Z$ of $S$ $$Z<X,Y>=<\nabla_{Z}X,Y>+<X,\nabla_{Z}^{*}Y>$$ holds then we say that $\nabla$ and $\nabla^*$ are duals of each other with respect to $g$.
Let the curvature tensor of $\nabla$ and $\nabla^*$ be denoted by $R$ and $R^*$ respectively,then we have $R=0\iff R*=0$ on the other hand it is not true for the torsion tensor $T$ and $T^*$.
Can someone explain this that how to prove this.
We have \begin{align*} \langle \nabla_i\nabla_j\partial_k,\partial_l\rangle &=\partial_i\langle\nabla_j\partial_k,\partial_l\rangle-\langle\nabla_j\partial_k,\nabla_i^*\partial_l\rangle \\ &=\partial_i\partial_j\langle\partial_k,\partial_l\rangle-\partial_i\langle\partial_k,\nabla_j^*\partial_l\rangle \\&\quad-\partial_j\langle\partial_k,\nabla_i^*\partial_l\rangle+\langle\partial_k,\nabla_j^*\nabla_i^*\partial_l\rangle \end{align*} Subtracting this from the same equation with $i,j$ swapped we get $$ R_{ijkl}=-R_{ijkl}^*. $$