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Let $(M,g)$ be a Riemannian manifold. Then I want to show that these are equivalent:

(i) Each point of $M$ has a smooth coordinate neighborhood in which the coordinate frame is orthonormal.

(ii) $g$ is flat.

Here's what I've tried:

(i)$\Rightarrow$(ii): Let $p$ be a point in $M$. Assume it has a smooth coordinate neighborhood in which the coordinate frame is orthonormal, i.e., a local frame $(E_1,...,E_n)$ on an open set $U$ around $p$ such that $\langle E_i,E_j\rangle_g=\delta_{ij}$. Since $M$ is a smooth manifold, $U$ is diffeomorphic to an open subset of $\mathbb{R}^n$ by the diffeomorphism $F$. So, it suffices to show that $F^\ast g= \tilde{g}$, where $\tilde{g}$ is Euclidean metric. By the definition of Riemannian metric, this is true. (Is this true really?)

(ii)$\Rightarrow$(i): Assume $g$ is flat, i.e., every point $p$ in $M$ has a neighborhood $U$ such that $(U,g|_U)$ is isometric to an open subset of $\mathbb{R}^n$ with the Euclidean metric. So $U$ is diffeomorphic to an open subset of $\mathbb{R}^n$ and $F^\ast g= \tilde{g}$, where $\tilde{g}$ is Euclidean metric. Since ($\partial/\partial x^i$) is an orthonormal frame in $\mathbb{R}^n$, it gives an orthonormal frame in $U$. (I don't know how.)

Thanks in advance.

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    In $(ii) \Rightarrow (i)$ your reason would work if you used the fact that $\partial/\partial x^i$ is the pullback of the standard orthonormal frame in $mathbb{R}^n$ by your isometry $F$.2012-01-13

2 Answers 2

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The key word in the question is the coordinate neighborhood, so a clear definition of it would help.

Definition 1. A coordinate neighborhood of a point $p \in M$ is an open subset $U \subset M$ endowed with a collection of functions $x^i \colon U \rightarrow \mathbb{R}$, where $i=1,\dots,n$, such that the map $ \textbf{x} \colon U \rightarrow \mathbb{R}^n \colon p \mapsto \left(\begin{array}{c} x^{1}\\ \vdots\\ x^{n} \end{array}\right) $ is a diffeomorphism onto its image. Briefly it is denoted by $(U,x^i)$. The functions $x^i$ are called the coordinate functions. The corresponding coordinate vector fields $\partial_i := \frac{\partial}{\partial{x^i}}$ are defined as the partial derivatives w.r.t. to coordinate $x^i$, so that $\partial_i x^j = \delta_i^j$. No Riemannian structure is involved so far.

There is so called the standard frame $(E_i \in \Gamma(T \mathbb{R}^n))$ in $\mathbb{R}^n$ such that $E_i = (0,\dots,1,\dots,0)$ with $1$ in the $i$-th position. The Euclidean metric $g^E \in \Gamma(S^2 T \mathbb{R}^n)$ is defined by $ g^E(E_i,E_j)=\delta_{ij} $

The coordinate frame $(\partial_i)$ is the pullback of the standard frame $(E_i)$ by map $\textbf{x}$, that is $ \frac{\partial}{\partial{x^i}} = \textbf{x}^*E_i $

Now, let $U$ be an open subset of a Riemannian manifold $(M,g)$. A smooth map $ F \colon (U,g|_U) \rightarrow (\mathbb{R}^n, g^E) $ is an isometry onto its image if $F_*g=g^E$ or, equivalently, $g = F^*g^E$. Recall, that for a diffeomorphism $F$ the pull-back is the inverse of the pushforward: $F^* = (F_*)^{-1}$.

As one can see from the question, the OP uses the following

Definition 2. A Riemannian metric $g$ on a smooth manifold $M$ is called locally flat if for any point $p \in M$ there is an open neighborhood $U$ of $p$ such that $U, g|_U$ is isometric to an open subset of $(\mathbb{R}^n, g^E)$. For brevity, the term "flat metric" is often used instead.

Let me restate slightly the fact in the question as the following

Proposition. For an open subset $U$ of a Riemannian manifold $(M.g)$ the following conditions are equivalent.

(i) $U$ is a "coordinate neighborhood" (of any of its points) in which the coordinate frame is orthonormal;

(ii) $(U,g|_U)$ is isometric to an open subset of $(R^n, g^E)$.

Proof.

$(i) \Rightarrow (ii)$ Check that map $\textbf{x} \colon U \rightarrow (R^n, g^E)$ provides the necessary isometry, i.e. $g = \textbf{x}^* g^E$. Indeed, $ g_{ij}=g(\partial_i,\partial_j)=\textbf{x}^* g^E(\partial_i,\partial_j) = g^E(\textbf{x}_* \partial_i, \textbf{x}_* \partial_j) = g^E (E_i, E_j) = \delta_{ij} $ which exactly means that the coordinate frame $(\partial_i)$ is orthonormal.

$(ii) \Rightarrow (i)$ Let $F: (U,g|_U) \rightarrow (\mathbb{R}^n,g^E)$ be an isometry. Define $ x^i (p) := F^i (p) $ i.e. $\mathbf{x} = F$. Now $(U,x^i)$ is a "coordinate neighborhood". QED.

As one can see, this is in fact a tautology: everything is hidden in the definitions!

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It seems you only want to know how to get that frame? Assume $F:U\rightarrow V\subset \mathbb{R}^n$ is a diffeomorphism on all of $U$ with image (precisely) $V$. Let $X_i := \frac{\partial}{\partial x_i}$ (easier to LaTex...)

If you then let $E_i = (F^{-1})_* X_i$ (that is: $E_i(p) := (F^{-1})_{*, F(p)} X_i(F(p))$ for $q= F(p) \in V$ ) then you have by definition of the pullback $F^*g(E_i,E_j)= F^*g((F^{-1})_* X_i,(F^{-1})_* X_j)) = \tilde{g} (F_* (F^{-1})_* X_i,F_* (F^{-1})_* X_j)) = \tilde{g}(X_i,X_j)=\delta_{ij}$

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    @user20353 As already mentioned in the comments, the ingredients are there, but they are put together in a way which I'd say do not prove the statement. Actually you don't need your $F$, you have to work directly with the coordinate system the existence of which you assume (or it's inverse, this depends on the conventions you are working with). This (the coordinate system or it's inverse) already is a local diffeomorphism $M\rightarrow \mathbb{R}^n$. Just use this as isometry.2012-01-08