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:
- Is my approach for the 'only if' part correct?
- 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.
Best Answer
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".