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 $\partial_k$.

Thanks for your help.


1 Answer 1


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, $$ \nabla (dx^k) = (\nabla_{\partial_i}(dx^k))(\partial_j) dx^i\otimes dx^j = -\Gamma^k_{ij} dx^i\otimes dx^j. $$


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