Normal Coordinates on a Riemannian Manifold

Question

Consider Riemann Normal coordinates on a manifold. Consider a point other than the origin. Given that the metric has vanishing derivatives at this point, is it correct to deduce that the metric is Euclidean at this point? If the deduction is correct, how to prove/ argue this?

Answer 1

If you consider a general point let it be $ x$ in the domain of the normal coordinates, we can't say much. One important relation is given by the Gauss Lemma that can be stated as $$\sum_i x^iu^i = \sum _ix^ig_{i j}(x) u^j \quad \forall u \in \mathbb{R}^n$$

From the Gauss Lemma we have that $$1)\quad \quad x^j = x^ig_{ij}(x).$$ Deriving 1) along $\partial_k$ we obtain $$\partial_k x^j = \delta_k^j = \partial_k x^i g_{i j}(x) + x^i g_{i j,k}(x) = g_{k j}(x) + x^i g_{i j,k}(x)$$ If the derivatives of the metric vanish at $x$, then $g_{ij,k}(x)= 0$ so we obtain that $$\delta_k^j = g_{k j}(x).$$ Notice that also the Christoffel symbols for the Levi-Civita connection vanish at $x$ in this case so you have that for a vector field $X$, $\nabla_{\partial_k} X(x) = \frac {\partial X^i(x)} {\partial x^k} \frac \partial {\partial x^i}|_x$. This can help in computations, also here you can find a proof of an interesting relation The metric tensor at $p$ satisfies $g_{a b, c d} + g_{a d,b c} + g_{a c, d b} = 0$ in a normal coordinate system centered at $p$ .

Answer 2

Clearly: $x_k=x^ig_{ik}$. Consider differentiating on both sides: $\partial_j x_k=(\partial_j x^i)g_{ik}+x^i\partial_j g_{ik}$. Now if the metric derivatives vanish then the last term is zero and we have: $\partial_j x_k=(\partial_j x^i)g_{ik}\Rightarrow\delta_{jk}=\delta_j^ig_{ik}\Rightarrow\delta_{jk}=g_{jk}$.