I read about the dual connection and do not understand some local expressions.
The Levi-Civita connection $\nabla^g$ is a connection on $TM$ with the metric compatibility $g(\nabla^g X,Y)+g(X,\nabla^g Y)=d(g(X,Y))$. Then for the corresponding Levi-Civita connection on $\Omega^1(M)$ it holds that $\nabla^g\alpha(X):=d(\alpha(X))-\alpha(\nabla^gX)$ for $\alpha\in\Omega^1(M)$, $X\in\mathfrak{X}(M)$ resp. $\nabla^g_Z\alpha(X):=Z(\alpha(X))-\alpha(\nabla^g_ZX)$ after contracting with vector field $Z$.
If $U$ is the domain of a chart of $T^*M$, one can choose orthonormal basisi of 1-forms $\theta^1,\dots,\theta^n$. Then the connection can be expressed by $\nabla^g=-\tilde{\Gamma}^\beta_{i\alpha} dx^i\otimes\theta^\alpha$. Then it was mentioned that the metric compatibility can be expressed by
$$g^{-1}(\nabla^g_{\partial_i}\,\theta^\alpha,\theta^\beta)+g^{-1}(\theta^\alpha ,\nabla^g_{\partial_i}\,\theta^\beta)=\partial_i(\delta^{\alpha\beta})=0$$
and hence $\tilde{\Gamma}^\alpha_{i\beta}+\tilde{\Gamma}^\beta_{i\alpha}=0$. I was thinking about the musical isomorphisms, but I couldn't get it right. How can I show this?
Moreover it was said that $\nabla^g=d-\tilde{\Gamma}$ holds over the chart domain $U$. This would mean that $\tilde{\Gamma}\alpha(X)=\alpha(\nabla^g X)$. Why does that hold?
Thanks for your help.