# Christoffel symbol on $T^*M$

I tried to prove the form of the Christoffel symbol on the contangent space given in the book "Elements of Noncommutative Geometry". The Christoffel symbols $$\Gamma^k_{ij}$$ of the Levi-Civita connection are functions in $$C^\infty(U)$$ are defined by $$\nabla^g\partial_j=:\Gamma^k_{ij} dx^i\otimes\partial_k$$

Considering the Levi-Civita connection in the cotangent bundle, we have the following relation $$\nabla^g \alpha(X)= d\alpha(X)-\alpha(\nabla^gX)$$, where $$\alpha\in\Omega^1(M),X\in\mathfrak{X}(M)$$. Using it, one should obtain $$\nabla^g dx^k= -\Gamma^k_{ij} dx^i\otimes dx^j.$$

I used $$\nabla^g f=\nabla^g\alpha(X)=X\otimes d\alpha -\nabla^g X \otimes \alpha$$. The idea was to consider this equation for basis elements, so that I can insert $$\nabla^g\partial_j\otimes dx^l=\Gamma^k_{ij} dx^i\otimes\partial_k\otimes dx^l$$. But then I have problems to continue and get rid of this $$\partial_k$$.

Leibniz rule is, for $$X$$ and $$Y$$ vector fields and $$\alpha$$ a $$1$$-form, $$\nabla_X(\alpha(Y)) = (\nabla_X\alpha)(Y) + \alpha(\nabla_XY).$$ Applying to $$\alpha = dx^k$$, $$X= \partial_i$$ and $$Y=\partial_j$$, and since $$dx^k(\partial_j) = \delta^k_j$$ is constant, one gets $$0 = \nabla_{\partial_i}(dx^k(\partial_j)) = (\nabla_{\partial_i}dx^k)(\partial_j) + dx^k(\Gamma^{\ell}_{ij}\partial_{\ell}),$$ that is, $$\nabla_{\partial_i}(dx^k) = -\Gamma^{\ell}_{ij}dx^k(\partial_{\ell}) = - \Gamma^k_{ij}.$$ Hence, $$\nabla (dx^k) = (\nabla_{\partial_i}(dx^k))(\partial_j) dx^i\otimes dx^j = -\Gamma^k_{ij} dx^i\otimes dx^j.$$