Christoffel symbol / scalar product of co- and contravarient basis vectors

Question

I am currently learning about the Christoffel symbole.

Given is the following equation:

$\Gamma_{nm}^{l} = e^l \cdot \Gamma_{nm}^{k} e_k$

with $e$ being an arbitrary co-/contravariant basis vector in $\mathbb{R}^3$.

My approach to getting this equation would be:

$e^l \cdot \Gamma^{k}_{nm} e_k = e^l \cdot \partial_m e_n = e^l \cdot \Gamma^{l}_{nm} e_l = \Gamma_{nm}^{l}\delta_l^l \stackrel{?}{=} \Gamma_{nm}^{l}$

My question is: Can i just use $e^l \cdot e_l = \delta_l^l = 1$? In my opinion $\delta_l^l=1$ shouldn't be correct because the basis vector is part of $\mathbb{R}^3$ which would suggest:

$\delta_l^l = \sum{}_{l=1}^3 \delta_l^l =3 $

but i can't see another way of getting the result give above.

Answer 1

I am going to summarize what @ghoster and @basics have said in the comments. As Ghoster points out, $$e^l\cdot \Gamma ^k_{nm}e_k = \Gamma ^k_{nm} (e^l \cdot e_k)$$ We know that, by construction, of basis vectors, this term is nonzero only when $k = l$, thus we have $$(e^l \cdot e_k) = \delta^l_k$$ giving us $$\Gamma ^k_{nm} (e^l \cdot e_k) = \Gamma ^k_{nm}\delta^l_k$$ By contraction of indices, we then have $$\Gamma ^k_{nm}\delta^l_k = \Gamma ^l_{nm}$$

I would encourage you to take @basics advice and think carefully about what each entity is (vector, scalar, tensor, etc.) when you do these manipulations. :)