Christoffel symbols vanishing in normal coordinates

Question

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?

Answer 1

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 $p$.

Answer 2

Giving a coordinate system $x$ in which $\Gamma^c _{i j}=0$ at ${x=0}$ and $g_{ij}=\delta^j_i$, where is $\Gamma^c _{i j}$ is the Christoffel symbol of the second kind. We can construct another coordinate system $x'$ in which at $x'=0$ $\Gamma'^c _{i j}=0$ , $\Gamma'^c _{i j,v}+\Gamma'^c _{i v,j}+\Gamma'^c _{j v,i}=0$and $g'_{ij}=\delta^j_i$ or in general $$\sum_{cyc}\Gamma'^c _{i j,klmn..}=0$$ using series expansion.

Let $x=\{x^1,x^2,x^3,x^4\}$ ,$b^{u}_{\alpha \beta k}=\frac{1}{6}\Gamma^{u}_{\alpha \beta k}|_{x=0}$ and using Einstein's summation notation

$$\tag{$1$} x^{'u}=x^{u}+b^{u}_{\alpha \beta , k}x^{\alpha}x^{\beta}x^{k} +C^{u}_{\alpha \beta k m}x^{\alpha}x^{\beta}x^{k}x^{m}+ ...... $$

$$\delta^u_i=\frac {\partial x^{u}}{\partial x'^{i}}|_{x=0}$$

$$\frac {\partial^2 x^{u}}{\partial x'^{i} \partial x'^{j}}|_{x=0}=0$$

$$-2(b^{u}_{i jv} +b^{u}_{i vj}+b^{u}_{jvi})=\frac {\partial^3 x^{u}}{\partial x^{'i} \partial x^{'j} \partial x^{'v}}|_{x=0}$$

$$3\frac {\partial^3 x^{u}}{\partial x^{'i} \partial x^{'j} \partial x^{'v}}|_{x=0}=-(\Gamma^u _{i j,v}+\Gamma^u _{i v,j}+\Gamma^u _{j v,i})$$

Now the Christoffel symbol transforms as:

$$\tag{$2$} \Gamma^{'u}_{\nu\kappa} = {\partial x^{'u} \over \partial x^\alpha} \left [ \Gamma^\alpha_{\beta \gamma}{\partial x^\beta \over \partial x^{'\nu}}{\partial x^\gamma \over \partial x^{'\kappa}} + {\partial ^2 x^\alpha \over \partial x^{'\nu} \partial x^{'\kappa}} \right ]$$

$$ \Gamma^{'u}_{\nu\kappa} = {\partial x^{'u} \over \partial x^\alpha} \left [ \Gamma^\alpha_{\beta \gamma}{\partial x^\beta \over \partial x^{'\nu}}{\partial x^\gamma \over \partial x^{'\kappa}}|_{x=0} + {\partial ^2 x^\alpha \over \partial x^{'\nu} \partial x^{'\kappa}}|_{x=0} \right ]=0$$

since covariant derivative of $g'_{uv}$ is $0$, this implies $\frac {\partial g'_{uv}}{\partial x^{'i}}|_{x=0}=0$

Also $g_{uv}$ transforms as :

$$g_{uv}^{'}=\frac {\partial x^i}{\partial x^{'u}}\frac {\partial x^j}{\partial x^{'v}}g_{ij}$$

$$g_{uv}^{'}=g_{uv}|_{x=0}=\delta^u_v$$

taking partial derivative of equation 2 at $x=0$:

$$\Gamma^{'u}_{\nu\kappa,a} = {\partial x^{'u} \over \partial x^\alpha} \left [ \Gamma^\alpha_{\beta \gamma ,c}{\partial x^\beta \over \partial x^{'\nu}}{\partial x^\gamma \over \partial x^{'\kappa} }\frac{\partial x^c }{\partial x'^a}+ {\partial ^3 x^\alpha \over \partial x^{'\nu} \partial x^{'\kappa} \partial x'^a} \right ]$$

$$\Gamma^{'u}_{\nu\kappa,a} = {\partial x^{'u} \over \partial x^\alpha} \left [ \Gamma^\alpha_{vk,a}+ {\partial ^3 x^\alpha \over \partial x^{'\nu} \partial x^{'\kappa} \partial x'^a} \right ]$$

$$\Gamma^{'u}_{\nu\kappa,a}+\Gamma^{'u}_{\nu a,\kappa}+\Gamma^{'u}_{\kappa a,\nu}={\partial x^{'u} \over \partial x^\alpha} \left [ \Gamma^\alpha_{vk,a}+ \Gamma^{u}_{\nu a,\kappa}+\Gamma^{u}_{\kappa a,\nu}+3{\partial ^3 x^\alpha \over \partial x^{'\nu} \partial x^{'\kappa} \partial x'^a} \right ]=0$$

if you want $$\sum_{cyc}\Gamma'^u_{ij,km}=0$$ we end up with $$C^{u}_{ijkm}=\frac{\Gamma^u_{ij,km}}{4!}=\frac{\Gamma^u_{ij,km}}{24}$$

Therefore equation 1 becomes : $$x^{'u}=x^{u}+\frac{\Gamma^{u}_{\alpha \beta , k}}{3!}x^{\alpha}x^{\beta}x^{k} +\frac{\Gamma^{u}_{\alpha \beta , k m}}{4!}x^{\alpha}x^{\beta}x^{k}x^{m}+ ...... $$