Deriving the Christoffel symbol formula from the covariant derivative

[John Creighto]

Can someone explain to me how to go from

$$\Gamma ^k_{ij}=-\bold{e}_j \cdot D_i \bold{e}^k$$

To

$$D_i \bold{e}^k = \Gamma ^k_{ij} \bold{e}^j \ \cdot \ \$$

 

[DavidCantwell]

Try this:
$$\Gamma^k_{ij} = -\textbf{e}_j \cdot D_i\textbf{e}^k$$
$$\Gamma^k_{ij} = -D_i\textbf{e}^k \cdot \textbf{e}_j$$ because $$\textbf{v} \cdot \textbf{u} = \textbf{u} \cdot \textbf{v}$$
$$\Gamma^k_{ij} \textbf{e}^j = -D_i\textbf{e}^k \cdot \textbf{e}_j \textbf{e}^j$$
$$\Gamma^k_{ij} \textbf{e}^j = -D_i\textbf{e}^k \cdot \textbf{1}$$ because $$\textbf{e}_j \textbf{e}^j = \textbf{1}$$
$$\Gamma^k_{ij} \textbf{e}^j = -D_i\textbf{e}^k$$ because $$\textbf{e} \cdot \textbf{1} = \textbf{e}$$
Well, I got close, but I don't know how to drop the minus sign. For the identity tensor, I believe $$\textbf{1} \cdot \textbf{v} = \textbf{v} \cdot \textbf{1}$$ is true (though not true for other second rank tensors)

 

[hanskuo]

I think you are right, Davidcantwell!
$$\Gamma^k_{ij} \textbf{e}^j = -D_i\textbf{e}^k$$
implies
$$\Gamma^k_{ij} = -\textbf{e}_j \cdot D_i\textbf{e}^k$$
The original statement seems not correct.