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I read the following statement in Methods of Information Geometry by S. Amari, chapter 1 (Screenshot below)

Riemannian connection is flat iff there exists a Euclidean coordinate system

While I am able to intuitively understand the 'if' statement; I cannot quite figure out the 'only if' part.

Here is my approach for the 'if' part:

Let $X,Y,Z$ be vector fields on a manifold $S$. Since Riemannian connection is a metric connection, we have $$Z\langle X, Y\rangle=\left\langle\nabla_{Z} X, Y\right\rangle+\left\langle X, \nabla_{Z} Y\right\rangle$$ If the manifold is flat with respect to the Riemannian connection, we can find an affine coordinate system $\left[\xi^{i}\right]$ such that

$$Z\langle \partial_i, \partial_j\rangle=\left\langle\nabla_{Z} \partial_i, \partial_j\right\rangle+\left\langle \partial_i, \nabla_{Z} \partial_j\right\rangle = 0 + 0 = 0$$

In other words, we have that $\langle \partial_i, \partial_j\rangle$ is constant along any direction. Further, because any affine transformation of the affine coordinate system is also parallel, we can choose $\langle \partial_i, \partial_j\rangle = \delta_{ij}$. Hence, there exists an affine coordinate system which is Euclidean.

My attempt for the 'only if' part:

Given that there is a coordinate system which is Euclidean, we have that $\langle \partial_i, \partial_j\rangle = \delta_{ij}$. We again have, by the metric connection property,

$$\left\langle\nabla_{Z} \partial_i, \partial_j\right\rangle+\left\langle \partial_i, \nabla_{Z} \partial_j\right\rangle = Z\langle \partial_i, \partial_j\rangle = 0$$

This implies that $\nabla_{Z} \partial_i = \nabla_{Z} \partial_j = 0$ as $Z$ is an arbitrary vector field. Further, this implies that the coordinate vectors $(\partial_i,\partial_j)$ are parallel with respect to $\nabla$. Hence, the Riemannian connection $\nabla$ is flat because we are able to find a coordinate system which is parallel.

My doubts:

  1. Is my approach for the 'only if' part correct?
  2. I feel that 'only if' part violates common sense - e.g. on a sphere $S^2$, the unit vectors along the latitudes and longitudes are orthogonal. But the Levi-Civita connection with respect to the induced metric from $\Bbb R^3$ is not flat - which contradicts the statement.

enter image description here

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  • $\begingroup$ For Euclidean coordinates, the local frame $(\partial_1,\dots,\partial_n)$ is orthonormal, but it also commutes (in the sense of Lie brackets). On the sphere (or any curved manifold), some local frames are orthonormal, and some commute, but none do both. $\endgroup$
    – Kajelad
    Aug 26 '20 at 11:35
  • $\begingroup$ @Kajelad Thanks for the additional information about the Euclidean coordinate system. However, the book does not introduce this distinction about Lie brackets. Assuming the only requirement for Euclidean coordinate system is the equation 1.70, I still think that the 'only if' part is wrong - as Sphere has Euclidean coordinate system but still is not flat wrt induced Riemannian metric. Can you please me check if this understanding is correct? $\endgroup$ Aug 26 '20 at 16:23
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    $\begingroup$ What is a Euclidean coordinate system? You're wrong about the sphere. The "only if" part is a result using the Frobenius Theorem (partial differential equations): You have to use curvature = zero to construct the appropriate Euclidean coordinates locally. $\endgroup$ Aug 26 '20 at 17:50
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I think there is a mistake in my premise that the spherical coordinate system is Euclidean. By definition, an Euclidean coordinate system should satisfy $<\partial_i,\partial_j> = \delta_{ij}$. In case of the Spherical coordinates, this does not hold true as $ds^2=d\theta^2+sin^2\theta\,d\phi^2$, which shows that $<\partial_\phi,\partial_\phi>\ne1$.

So, spherical coordinates on a sphere is not a counterexample to the statement that "A Riemannian manifold is flat iff it has an Euclidean coordinate system".

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    $\begingroup$ That is correct. $\endgroup$
    – Deane
    Jan 27 at 5:14

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