黎曼流形上的法坐标系：非原点处度量导数为零可推欧氏度量吗？

Great question! Let's break this down step by step, since it touches on some fundamental properties of Riemann normal coordinates (RNCs) and how metric derivatives interact with curvature.

Given a Riemannian manifold $(M,g)$ and a point $p\in M$, Riemann normal coordinates centered at $p$ are defined via the exponential map $\exp_p: T_pM\to M$. In these coordinates:

• At the origin $p$ (where coordinates are $x^i=0$), the metric is Euclidean: $g_{ij}(p) = \delta_{ij}$.
• All Christoffel symbols vanish at $p$: $\Gamma^k_{ij}(p)=0$.
• The metric has a well-known Taylor expansion around $p$ in terms of the Riemann curvature tensor at $p$:
  $$g_{ij}(x) = \delta_{ij} - \frac{1}{3}R_{ikjl}(p)x^k x^l + O(|x|^3)$$
  where $R_{ikjl}$ are the components of the curvature tensor at $p$.

The answer depends entirely on what you mean by "the metric's derivatives are zero":

Case 1: Only first-order partial derivatives are zero at $q$ #

Suppose at $q$ (with coordinates $x^a\neq0$), all first derivatives $\partial_k g_{ij}(q)=0$. From the Taylor expansion above, we can compute the first derivative:
$$\partial_m g_{ij}(x) = -\frac{2}{3}R_{ikjm}(p)x^k + O(|x|^2)$$
Setting this to zero at $q$ gives $R_{ikjm}(p)x^k=0$ for all $i,j,m$. This means the vector $v=x^k\partial_k\in T_pM$ is "curvature-isotropic" at $p$ (i.e., $R(u,v,w,v)=0$ for all $u,w\in T_pM$).

However, this does not force $g_{ij}(q)=\delta_{ij}$. While the quadratic correction term in the metric at $q$ vanishes (since $R_{ikjl}(p)x^k x^l=0$ from the first-derivative condition), higher-order terms (order $|x|^3$ and above) can still be non-zero. We can construct smooth Riemannian metrics where:

• At $p$, the metric is Euclidean with vanishing Christoffel symbols (so we're in valid RNCs).
• At $q\neq p$, all first derivatives of the metric are zero, but higher-order terms in the Taylor expansion make $g_{ij}(q)\neq\delta_{ij}$.

Thus, first-order derivatives being zero at $q$ is not sufficient to conclude $g(q)$ is Euclidean.

Case 2: All derivatives (of every order) are zero at $q$ #

If every partial derivative of $g_{ij}$ (of any order) is zero at $q$, the conclusion splits based on metric regularity:

• If $g$ is real-analytic: The Taylor series of $g$ at $q$ simplifies to $g_{ij}(x)=g_{ij}(q)$ for all $x$ in a neighborhood of $q$ (since all derivative coefficients are zero). Since we're in RNCs centered at $p$, the metric at $p$ is $\delta_{ij}$. A constant metric on a connected manifold (required for normal coordinates to be defined) must be Euclidean everywhere—so $g_{ij}(q)=\delta_{ij}$.
• If $g$ is only smooth (not analytic): We can build counterexamples. For instance, take $M=\mathbb{R}^2$ and define:
  $$g_{ij}(x,y) = \delta_{ij} + f(x,y)(x^2+y2)^3$$
  where $f$ is a smooth function that is zero everywhere except a small neighborhood around $q=(1,0)$, where $f(1,0)=1$ and all derivatives of $f$ at $q$ are zero. At $p=(0,0)$, the metric is Euclidean with vanishing Christoffel symbols. At $q$, all derivatives of $g_{ij}$ are zero, but $g_{11}(q)=2\neq1$.

You can conclude $g(q)$ is Euclidean if:

• You mean all derivatives (of every order) are zero at $q$ and the metric is real-analytic, or
• You add extra constraints (like the manifold being flat, so the curvature tensor is zero everywhere, making the metric Euclidean in RNCs at every point).

内容的提问来源于stack exchange，提问作者Kong