# Why does the Christoffel symbols in normal coordinate charts vanish only at point $p$, not vanish identically in an open neighborhood of $p$?

Let $$(M, g)$$ be a Riemannian $$n$$-manifold, and let $$\left(U,\left(x^{i}\right)\right)$$ be any normal coordinate chart centered at $$p \in M$$. Then the Christoffel symbols in these coordinates vanish at $$p$$.

Proof. Let $$\displaystyle v=v^{i} \left.\frac{\partial}{\partial x^{i}}\right|_p \in T_{p} M$$ be a tangent vector. Consider the geodesic $$\gamma_{v}:(-\varepsilon, \varepsilon) \rightarrow M$$. In the normal coordinate $$(U,\phi)$$, the geodesic takes a simple form $$\phi \circ \gamma_{v}(t)=\left(v^{1} t, \cdots, v^{n} t\right) \in \mathbb{R}^{n}$$ since $$\gamma_{v}(t)=\exp(tv)$$, where $$\exp$$ is the exponential map defined by $$v\mapsto\gamma_{v}(1)$$. Therefore, $$\ddot{\gamma}(t)=0, \dot{\gamma}^{k}(t)=v^{k}.$$

Substitue $$\ddot{\gamma}(t)=0, \dot{\gamma}^{k}(t)=v^{k}$$ in the geodesic equation $$\ddot{\gamma}^{k}(t)+\dot{\gamma}^{i}(t) \dot{\gamma}^{j}(t) \Gamma_{i j}^{k}(\gamma(t))=0$$, we have $$v^{i} v^{j} \Gamma_{i j}^{k}(\gamma_v( t ))=0$$. Evaluate at $$t=0$$, we have $$v^{i} v^{j} \Gamma_{i j}^{k}(p)=0$$ for $$k=1, \cdots, n$$. Now for a fixed $$k$$ we view $$\Gamma_{i j}^{k}$$ as a bilinear form. Since $$\left\{v^{i}\right\}$$ can be arbitrary chosen, it follows that $$\Gamma_{i j}^{k}(p)=0$$ for all $$i, j, k$$.

My question is: Which step prevents me from proving $$\Gamma_{i j}^{k}$$ vanishing identically in an open neighborhood of $$p$$? And why? In the proof, we have $$v^{i} v^{j} \Gamma_{i j}^{k}(\gamma_v(t))=0$$, which seems to imply $$\Gamma_{i j}^{k}$$ vanishing identically in an open neighborhood of $$p$$. But after substituting $$t=0$$, we only get $$v^{i} v^{j} \Gamma_{i j}^{k}(p)=0$$.

## 1 Answer

The relation $$\ddot{\gamma}(t)=0, \dot{\gamma}^{k}(t)=v^{k}$$ only holds for $$t = 0$$. If all the Christoffel symbols vanish in a neighborhood of a point, then the curvature tensor is zero in that neighborhood. The manifold is flat around the point, and there is a local isometry sending that neighborhood to the standard euclidean plane. This won't happen for the sphere or the hyperbolic plane. I suggest that you do again your computation in the concrete case of the sphere to understand what is going on. When you are stuck on some abstract problem, consider a simple example.