Standard form of the Riemannian metric in a neighborhood

In Wikipedia it says

In Riemannian geometry, the metric can always be made to take the standard form at any given point, but not always in a neighborhood around that point.

And I am a little confused, because for any $$p\in M$$, where $$(M,g)$$ is a Riemannian manifold, we can find a neighborhood $$U\subset M$$ of $$p$$ and get the natural basis $$\{\partial/\partial x^i\}$$ in $$U$$, and the Gram-Schmidt process gives a local orthonomal basis, under which the metric $$g$$ takes the standard form. Is there anything wrong?

• There is no guarantee that when you move off $p$ to some other $q \in U$, that the metric is still in standard form. – Joppy Nov 9 '18 at 3:44
• If you apply Gram-Schmidt simultaneously at every point in some neighborhood of $p$, you produce a local orthonormal frame $(E_a)$ for which the metric in the dual coframe $(e^a)$ is $\delta_{ab} e^a e^b$, but in general $(E_a)$ will not be the coordinate frame for any coordinate system. In fact, it follows from quickly that this is only possible when $g$ is (locally) flat. – Travis Willse Nov 9 '18 at 4:17

If you apply Gram-Schmidt only at a point $$p$$, then there is no guarantee that it will be orthonormal frame at other $$q$$ in any neighbourhood. But that is a cheap answer, because you have available a parametric version of Gram-Schmidt which is smooth and produce a local orthonormal frame: \begin{align*} V_1(x)&=\frac{\partial}{\partial x^1}&X_1=\frac{1}{\sqrt{g(x)(V_1(x),V_1(x))}}V_1(x)\\ V_2(x)&=\frac{\partial}{\partial x^2}-g^{12}(x)\frac{\partial}{\partial x^1}&X_2=\frac{1}{\sqrt{g(x)(V_2(x),V_2(x))}}V_2(x)\\ &\vdots \end{align*} So suppose you use this and get $$X_i:=A^j_i(x^1,x^2,\dots,x^n)\frac{\partial}{\partial x^j}$$ where $$A^j_i$$ is a pretty much arbitrary matrix of smooth functions of the coordinates $$x^1,\dots,x^n$$ in general -- only that $$A^j_i=0$$ if $$j>i$$ by the nature of Gram-Schmidt, and that the diagonal is nonvanishing (since we want $$X_i$$ a basis). If we can find $$y^i$$ locally such that $$X_i=\frac{\partial}{\partial y^i}$$, then $$[X_i,X_j]=\left[\frac{\partial}{\partial y^i},\frac{\partial}{\partial y^j}\right]=0.$$ Conversely, by Frobenius theorem this is sufficient.
However, this is a very big ask on the $$A$$: $$0=dx^k([X_i,X_j])=dx^k\left[A^{i'}_i\partial_{i'},A^{j'}_j\partial_{j'}\right]=A^{l}_i\partial_{l}A^k_j-A^{l}_j\partial_{l}A^k_i.$$ and the RHS is basically the Christoffel symbol $$\Gamma^k_{ij}$$ of the metric $$g$$. You are asking them to vanish identically on a neighbourhood.
I do not think that you can say that the frame $$\{ \partial/\partial x^{i} \}$$ remains orthonormal throughout the entirety of the neighbourhood $$U$$ of $$p$$, since the Gram-Schmidt process has to be done by specifying some tangent space $$T_{p}M$$. Then it will be orthonormal in $$T_{p}M$$, but not necessarily in $$T_{q}M$$ when $$q \neq p$$.
A neighbourhood $$U$$ centered at $$p$$ that satisfies this property is called a normal neighbourhood at $$p$$, and in J. M. Lee's book Riemannian Manifolds: An Introduction to Curvature, on page 77 he states that
Then Proposition 5.11 (c) of his book staes that "components of the metric at $$p$$ are $$g_{ij} = \delta_{ij}$$", but nothing about the metric away from $$p$$.