Let $(M,g)$ be a Riemannian manifold, and let $(\varphi,U)$ be normal coordinates in $p\in M$. For every $v\in T_p M$, denote $\gamma_v :I_v \to M$ the maximal geodesic with initial point $p$ and initial velocity $v$. Since $U$ is a normal neighborhood of $p$, we have that $\gamma_v ^{-1} (U):=J_v $ is an open interval containing $0$. Now, in normal coordinates, for every $t \in J_v $ we have $\gamma_v (t) \equiv t(v^1 ,…,v^n )$, where $v^i$ are the components of $v$ with respect to the ortonormal basis of $T_p M$ which we used (together with the exp map) to define $\varphi$. So $\gamma _v $ must satisfy the geodesic equation $\ddot \gamma^k _v (t) + \dot \gamma^i _v (t) \dot \gamma^j _v (t) \Gamma^k _{ij} (\gamma_v (t))=0$ for every $t \in J_v$, and using the local expression of $\gamma _v $ and the symmetry of the Levi-Civita connection, we obtain $\Gamma _{ij} ^k (\gamma_v (t))=0$ for every $t\in J_v$. Since for every $q\in U$ there exists a $v\in T_p M$ and a $t\in J_v $ such that $\gamma_v (t)=q$, we have that $\Gamma ^k _{ij} \equiv 0$ in $U$.
The previous reasoning must have something wrong, becouse I know that not every Riemannian manifold is locally flat, but I can't find the mistake. Can you help me?
Best Answer
Let $q\neq p$ be such that $\gamma_v(t)=q$ for some $t$. In normal coordinates (using $\gamma_v (t) = t(v^1 ,...,v^n )$): $$\Gamma^i_{jk}v^jv^j=0$$ But, here, $v$ is fixed, so it can't be concluded that $\Gamma^i_{jk}=0$. But for $p$, it can be applied to every $\gamma_v(t)$ (every geodesic is such that $\gamma_{v'}(0)=p$), i.e., $\forall v$. Then take $v_i=\delta_{ij}$, using the equation above, $$\Gamma^i_{jj}|_p=0 $$ Now set $v_i=(\delta_{ij}+\delta_{ik})$, it follows (using $\Gamma^i_{jj}|_p=0 $) $$\Gamma^i_{(jk)}|_p=0 $$ The symmetric part (assuming $\nabla$ is a general affine connection). So we conclude the symmetric part at $p$ is zero, only at $p$.