Question about duality of connections in a manifold Let $S$ be a manifold on which there is given a Riemannian metric $g=\left\langle\ \,,\ \right\rangle$ and two affine connections $\nabla$ and $\nabla^*$. If for all vector field $X,Y,Z$ of $S$
$$Z\left\langle X,Y\right\rangle=\left\langle\nabla_{Z}X,Y\right\rangle+\left\langle X,\nabla_{Z}^{*}Y\right\rangle$$ holds then we say that $\nabla$ and $\nabla^*$ are duals of each other with respect to $g$.
It is given that if a connection $\nabla'$ has the same torsion as $\nabla^*$ and if $\frac{\nabla+\nabla'}{2}$ is metric connection then $\nabla^*=\nabla'$. Someone, please give hint on how to approach this one. Thanks.
 A: Here is a hint: the fundamental result in Riemannian Geometry is that for every Riemannian manifold $(M,g)$ there is a unique torsion-free metric connection. This is usually shown in textbooks by using the Koszul formula, see the section titled "Invariant formulation" in this wikipedia article.
A slightly less known fact is that for every Riemannian manifold and every vector-valued $2$-form $T$, there is a unique metric connection having $T$ as torsion. This can be seen from the proof of the Koszul formula, not using the identity $\nabla_XY-\nabla_YX=[X,Y]$ but rather $\nabla_XY-\nabla_YX=[X,Y]+T(X,Y)$.
So, let's get to the proof. Consider the two connections $D':=\frac{1}{2}(\nabla+\nabla')$ and $D^*:=\frac{1}{2}(\nabla+\nabla^*)$. As $\nabla$ and $\nabla^*$ are dual, $D^*$ is a metric connection (exercise for you!). Also, our hypothesis is that $D'$ is a metric connection.
Since $\nabla^*$ and $\nabla'$ have the same torsion, also $D^*$ and $D'$ have the same torsion. But they are also both metric, so $D^*=D'$ by the remark in the second paragraph, and this clearly implies $\nabla^*=\nabla'$.
