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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?

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  • $\begingroup$ There is no guarantee that when you move off $p$ to some other $q \in U$, that the metric is still in standard form. $\endgroup$ – Joppy Nov 9 '18 at 3:44
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    $\begingroup$ 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. $\endgroup$ – Travis Willse Nov 9 '18 at 4:17
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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.

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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

Given p ∈ M and a normal neighborhood U of p, there is a one-to-one correspondence between normal coordinate charts and orthonormal bases at p.

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 $p$.

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