Gamma How to Calculate Gamma: Step by Step

[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!
<br /> \Gamma^k_{ij} \textbf{e}^j = -D_i\textbf{e}^k <br />
implies
<br /> \Gamma^k_{ij} = -\textbf{e}_j \cdot D_i\textbf{e}^k<br />
The original statement seems not correct.