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 1


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.


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