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.

  • 1
    $\begingroup$ The pull back map works in the opposite direction. In fact it maps the Euclidean metric g to the Riemannian metric on the manifold M if we assume F to be a diffeomorphism from an open subset U of M to an open subset of R^n. So be careful with your hypothesis... $\endgroup$
    – user22351
    Jan 5, 2012 at 14:06
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    $\begingroup$ Is (i) implies (ii) part correct? $\endgroup$
    – user20353
    Jan 7, 2012 at 22:00
  • $\begingroup$ In $(i) \Rightarrow (ii)$ you are not using the fact that you have a coordinate frame, and your diffeomorphism $F$ comes from nowhere (you don't have any information about it), so you can't claim that $\tilde{g}$ is a Euclidean metric. $\endgroup$ Jan 13, 2012 at 5:38
  • $\begingroup$ 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 $F$. $\endgroup$ Jan 13, 2012 at 5:42

2 Answers 2


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 $(R^n, g^E)$.


$(i) \Rightarrow (ii)$ 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!


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 $$F^*g(E_i,E_j)= F^*g((F^{-1})_* X_i,(F^{-1})_* X_j)) = \tilde{g} (F_* (F^{-1})_* X_i,F_* (F^{-1})_* X_j)) = \tilde{g}(X_i,X_j)=\delta_{ij}$$

  • $\begingroup$ Thanks. By the way, is my trial correct? At least partly? $\endgroup$
    – user20353
    Dec 25, 2011 at 16:30
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    $\begingroup$ You have to be careful. If you have a local frame $Z_i$ which is a basis for each $T_pM$, then you can apply Gram-Schmidt to obtain an orthonormal frame (you need a bit of additional reasoning to show it's smooth, but that does work out). So the existence of an orthonormal moving frame alone does not imply that $M$ is flat. This means that if your reasoning is correct, you need to use the fact that your $E_i$ are obtained from a coordinate system, which is, at least as far as I can see, not explicit in your reasoning. The basic steps are there, but you should chase through the definitions. $\endgroup$
    – user20266
    Dec 25, 2011 at 16:50
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    $\begingroup$ Extra info: In fact, Gram-Schmidt can always turn any local coordinate frame into a local orthonormal frame. That is, the existence of a metric implies (is equivalent to) the existence of local orthonormal frames. So of course the existence of a frame cannot show that a manifold is flat. The obstruction is that an orthonormal frame cannot, in general, be integrated to coordinate functions. So what you really want to show is that the fields of an orthonormal frame commute iff the metric is flat. $\endgroup$
    – Neal
    Dec 25, 2011 at 17:06
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    $\begingroup$ @Neal he (or she) explicitly assumes that the frame he obtains from the coordinate system is orthonormal ('the coordinate frame'), so I think it's easier than you say. You just have to write down what that means. $\endgroup$
    – user20266
    Dec 25, 2011 at 17:13
  • $\begingroup$ Is (i) implies (ii) part correct? $\endgroup$
    – user20353
    Jan 7, 2012 at 21:58

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