Affine connection in general relativity
In The GR lecture my teacher deduced the relation between affine connection and the metric tensor according to the following way: He firstly wrote the relationship of two tensors like this (I understand): $$A^{\mu }{}{[P\to Q]}=A^{\mu }{}{[P]}-\Gamma {\nu \sigma }^{\mu }{}{[P]}A^{\nu }{}_{[P]}dx^{\sigma }$$
Then he wrote(I understand): $$g_{\mu \nu (Q)}=dx^{\sigma } g_{\mu \nu ,\sigma (P)}+g_{\mu \nu (P)}$$
If the vector is to have the same length after been transported, we have: $$g_{\mu \nu (Q)}A^{\mu }{}{[P\to Q]}A^{\nu }{}{[P\to Q]}=g_{\mu \nu (P)}A^{\mu }{}{[P]}A^{\nu }{}{[P]}$$
The above three equations give the result which connects the affine connection with metric tensor: $$g_{\mu \nu ,\sigma }-g_{\alpha \nu } \Gamma {\mu \sigma }^{\alpha }-g{\mu \alpha } \Gamma _{\nu \sigma }^{\alpha }=0$$
He had used two conditions:
1. $\Gamma$ is symmetric
2. the length should be same.
My questions are:
1. In fact the derivation of $A^{\mu }{}{[P\to Q]}=A^{\mu }{}{[P]}-\Gamma {\nu \sigma }^{\mu }{}{[P]}A^{\nu }{}{[P]}dx^{\sigma }$ can at the same time give $\Gamma {\mu \nu }^{'\lambda }=\frac{\partial x^{'\lambda }}{\partial x^{\rho }}\frac{\partial x^{\tau }}{\partial x^{'\mu }}\frac{\partial x^{\sigma }}{\partial x^{'\nu }}\Gamma {\tau \sigma }^{\rho }+\frac{\partial x^{'\lambda }}{\partial x^{\rho }}\frac{\partial ^2x^{\rho }}{\partial x^{'\mu }\partial x^{'\nu }}$. Since we can always choose a coordinate such that $\Gamma {\mu \nu }^{'\lambda }=\frac{\partial x^{'\lambda }}{\partial x^{\rho }}\frac{\partial ^2x^{\rho }}{\partial x^{'\mu }\partial x^{'\nu }}$, we can see that $\Gamma$ is already symmetric (since we can exchange the indices), and thus we don't need to add this condition. Is my statement here true?
2. 818
I read from <PERSON>'s Gravitation and Cosmology on P.7 (3.2.4) that he derived $\Gamma {\mu \nu }^{'\lambda }=\frac{\partial x^{'\lambda }}{\partial x^{\rho }}\frac{\partial ^2x^{\rho }}{\partial x^{'\mu }\partial x^{'\nu }}$ simply from geodesics, while for geodesics the length of the vectors need not be the same. Then <PERSON> still derived $g{\mu \nu ,\sigma }-g_{\alpha \nu } \Gamma {\mu \sigma }^{\alpha }-g{\mu \alpha } \Gamma _{\nu \sigma }^{\alpha }=0$ without using the the second condition. But, my teacher did use that one. So what's wrong with the conditions?
3. To explain my last confuse better:
I just want to know the difference between the affine connections defined by my teacher and <PERSON>. It seems that both have the same expression. But my teacher's seemed to need another "same length" condition ,while <PERSON>'s didn't ,to derive the final result of $g_{\mu \nu ,\sigma }-g_{\alpha \nu } \Gamma {\mu \sigma }^{\alpha }-g{\mu \alpha } \Gamma _{\nu \sigma }^{\alpha }=0$.What properties can the $\Gamma$given by my teacher have ( parallel ? the same length? 818