I'm trying to prove that:
$\partial_{c} \vec{e^{a}} = - \Gamma_{bc}^{a} \vec{e^{b}}$
Where $\vec{e^{i}}$ represents one of the vectors from a basis of a dual vector space. To do the proof, I'm doing that:
I know that the affine connection can be defined as:
$\frac{\partial \vec{e_{a}}}{\partial x^{c}} = \Gamma_{ac}^{b} \vec{e_{b}} $
So I start with this and do a dot product with a vector from the dual basis, and with that I have:
$ \vec{e^{d}} \cdot \partial_{c} \vec{e_{a}} = \Gamma_{ac}^{b} \vec{e^{d}} \cdot \vec{e_{b}} = \Gamma_{ac}^{b} \delta_{b}^{d}$
where I used the reciprocal relation $\vec{e^{d}} \cdot \vec{e_{b}} = \delta_{b}^{d}$. And so, the equation turns out to be:
$ \vec{e^{d}} \cdot \partial_{c} \vec{e_{a}} = \Gamma_{ac}^{d}$ or
$\Gamma_{ac}^{b} = \vec{e^{b}} \cdot \partial_{c} \vec{e_{a}}$
as b and d represents just dummie indices. To continue with my proof I differentiated the reciprocal relation with respect to the coordinate $x^{c}$, to find that:
$\partial_{c}(\vec{e^{b}}\cdot\vec{e_{a}}) = (\partial_{c}\vec{e^{b}})\cdot\vec{e_{a}}+\vec{e^{b}}\cdot(\partial_{c}\vec{e_{a}}) = 0 $
Now using that, we can write:
$\Gamma_{ac}^{b} = -(\partial_{c}\vec{e^{b}})\cdot\vec{e_{a}}$
So, how can I go from that to the formula I want to prove? I imagined to do another dot product with a vector from the dual basis but I didn't work out.