Let $\nabla$ be a linear connection on a manifold $M$.
I want to prove that:
$$\Gamma^k_{ij}(p)+\Gamma^k_{ji}(p)=0,$$
if in $p\in M$ normal local coordinates are defined.
I have a solution which involves the use of a geodesic curve, using the exp map, but I do not understand the connection with the normal coordinates. Could you please give me some hints to resolve this problem?
Thank you!
Let $x_1,x_2,\cdots,x_n$ be normal coordinates centered at $p$. I am going to show that $$\Gamma_{1 2}^k(p) + \Gamma_{2 1}^k(p) = 0$$ for $k=1,\cdots,n$. The general case i.e. replacing $1,2$ by $i,j$ is analogous. Being normal coordinates mean that the curve $ \gamma: t \to (t,t,0,0,\cdots)$ is a geodesic. Then the velocity vector of $\gamma$ at $p$ is parallel w.r.t. itself. Namely $$\nabla_{\gamma'} \gamma' = 0 \, .$$ Now notice that $\gamma' = \frac{\partial}{\partial x_1} + \frac{\partial}{\partial x_2}$. So at $p$ $$\nabla_{\frac{\partial}{\partial x_1} + \frac{\partial}{\partial x_2}} (\frac{\partial}{\partial x_1} + \frac{\partial}{\partial x_2}) = \nabla_{\frac{\partial}{\partial x_1}}(\frac{\partial}{\partial x_1}) + \nabla_{\frac{\partial}{\partial x_1}}(\frac{\partial}{\partial x_2}) + \nabla_{\frac{\partial}{\partial x_2}}(\frac{\partial}{\partial x_1}) + \nabla_{\frac{\partial}{\partial x_2}}(\frac{\partial}{\partial x_2}) = 0 \, \, .$$
Also the curves $(t,0,\cdots,0)$ and $(0,t,0,\cdots,0)$ are geodesics. So you also have at $p$ $$ \nabla_{\frac{\partial}{\partial x_1}}(\frac{\partial}{\partial x_1}) = \nabla_{\frac{\partial}{\partial x_2}}(\frac{\partial}{\partial x_2}) = 0 \, .$$ Then you get at $p$ $$\nabla_{\frac{\partial}{\partial x_1}}(\frac{\partial}{\partial x_2}) + \nabla_{\frac{\partial}{\partial x_2}}(\frac{\partial}{\partial x_1}) = 0$$ Taking components w.r.t. $\frac{\partial}{\partial x_k}$ you get $$\Gamma_{1 2}^k(p) + \Gamma_{2 1}^k(p) = 0 \, \, .$$